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i
^
PRINCIPLES AND PRACTICE
OF
ELECTRICAL ENGINEERING
': • • • • •
•
\^''
'<>>
COPTRIGHT, 1914, 1916, 1917, bt thb
McGrawHill Book Company, Inc.
PRINTED IN THB UNITED STATES OF AMEBIOA
r
I
THK MAPLB PRB99  YORK PA
PREFACE TO SECOND EDITION
I take this opportunity of expressing my thanks to my fellow
teachers for the cordial reception they have given this book.
Especially am I grateful to those who have pointed out errors and
have offered suggestions for the improvement of the work.
The principal changes considered necessary at this time are
the elimination of the left hand rule for the direction of the force
on a conductor in favor of the conception of strained lines of
force, a change in the method of presenting the subject of direct
current drum windings, a few additions to the chapters on alter
nating current circuits and the addition of a considerable number
of problems without answers.
A. G.
Ithaca, N. Y.
July, 1017.
519909
PREFACE TO THE FIRST EDITION
The following work is based on a lecture and laboratory course
given to the senior civil, mechanical, and mining students at
McGill University. It is therefore suited for men who desire
to obtain a broad idea of the principles and practice of electrical
engineering and who have only a limited amount of time to spend
on the subject. For such men it is necessary to emphasize the
fundamental principles, and to develop the subject by elaborating
on these principles rather than by the solution of mathematical
equations, because only in this way can the student be given such
a grip of the subject in the short time available, that he is able
thereafter to make intelligent use of the data contained in the
electrical handbooks, or take up with advantage a further study
of the special treatises on the subject.
The book gives a selfcontained lecture and laboratory course.
The chapters on the control and applications of electrical ma
chinery have been so written that large sections of these chapters
may be set for private reading. In the laboratory course, com
plete references are given to the theory and purpose of each ex
periment, and these references in no case go beyond the text
contained in the body of the work.
The author wishes to acknowledge his indebtedness to Mr.
A. M. S. Boyd and to Mr. R. Kraus for their help and criticism.
A. G.
McGiLL University,
Sept, 1, 1914.
n
CONTENTS
Preface v
Introduction xxi
CHAPTER I
Magnetism and Magnetic Units
Article Page
1. Magnets 1
2. Coulomb's Law 1
3. The Magnetic Field 1
4. lines of Force 2
5. lanes of Force from a Unit Pole 3
CHAPTER II
Electromagnetism
6. Direction of an Electric Current 4
7. Magnetic Field Surrounding a Conductor Carrying Current ... 4
8. Force at the Center of a Circular Loop Carrying Current ... 4
9. Electromagnets 5
10. Force on a Conductor Carrying Current in a Magnetic Field . . 6
11. Moving Coil Ammeters 7
CHAPTER III
Electromagnetic Induction
12. Electromagnetic Induction • . . 9
13. The Direction of the Induced Electromotive Force 9
14. Mutual Induction 11
15. Self Induction 11
CHAPTER IV
Work and Power
16. Transformation of Mechanical into Electrical Energy 13
17. Unit of Work 14
18. Heat Energy and Electrical Energy 15
19. Conversion Factors 15
20. Problems on Work and Power 15
..
Vll
viii CONTENTS
CHAPTER V
Electric Circuits and Resistance
Article Page
21. The Flow of Electricity 18
22. Ammeters and Voltmeters 10
23. Resistance Circuits 10
24. Ohm's Law 20
25. Specific Resistance 20
26. Variation of Resistance with Temperature 21
27. Power Expended in a Resistance 21
28. Insulating Materials 22
20. Dielectric Strength of Insulating Material 22
30. Series and Parallel Circuits 22
31. Voltage Drop in a Transmission lone 24
CHAPTER VI
Rheostats and Resistors
32. Rheostats . . , 25
33. Resistors 26
34. Heater Units 27
35. Castiron Grid Resistance 27
36. Carbon Pile Rheostat 20
37. Liquid Rheostat 20
38. Sire of a Rheostat 31
CHAPTER VII
Magnetic Circuits and Magnetic Properties of Iron
30. Magnetic Field due to a Solenoid 32
40. Permeability 33
41. Reluctance of a Magnetic Circuit 33
42. Magnetization Curves 34
43. Residual Magnetism 35
44. Molecular Theory of Magnetism 36
45. Hysteresis 36
CHAPTER VIII
Solenoids and Electromagnets
46. Pull of Solenoids 37
47. Electric Hammer 38
48. Variation of the Pull of a Solenoid . . 38
40. Circuit Breaker 30
50. Laws of Magnetic Pull 40
51. Solenoids with Long and Short Plungers 41
52. Ironclad Solenoids 41
CONTENTS ix
Article Page
53. Lifting and Holding Magnets 43
54. Saturation of a Magnetic Circuit 46
55. Electromagnetic Brakes and Clutches 46
56. Magnetic Separator 47
CHAPTER IX
ARBiATURE Windings for Directcurrent Machinery
57. Principle of Operation of the Electric Generator : 48
58. Gramme Ring Winding 48
59. Commutator and Brushes 50
60. Multipolar Windings 51
61. Drum Wmdings 52
62. Lamination of the Armature Core 55
CHAPTER X
Construction and Excitation of Directcurrent Machines
63. Multipolar Construction 56
64. Armature Construction 58
65. Commutator 58
66. The Brushes 58
67. Poles and Yoke 58
68. Large Generators 58
69. Excitation 69
CHAPTER XI
Theory op Commutation
70. Commutation 62
71. Theory of Commutation . 63
72. Shifting of the Brushes 63
73. Interpole Machines 65
74. Carbon Brushes 66
CHAPTER XII
■
Armature Reaction
75. The Crossmagnetizing Effect 67
76. The Demagnetizing Effect 68
77. Effect of Armature Reaction on Commutation 68
CHAPTER XIII
Characteristics of Directcurrent Generators
78. Magnetization or Noload Saturation Curve 70
79. Self Excitation 71
X CONTENTS
Article Page
80. Regulation Curve of a Separately Excited or of a Magneto
Generator 71
81. Regulation Curve of a Shunt Generator 72
82. To Maintain the Terminal Voltage Constant 74
83. Compound Generators 74
84. The Regulation Curve of a Series Generator 75
85. Problem on Generator Characteristics 76
CHAPTER XIV
Theory of Operation of Directcurrent Motors
86. Driving Force of a Motor 78
87. Driving and Retarding Forces in Generators and Motors .... 78
88. The Back E.M.F 79
89. Theory of Motor Operation 80
90. Speed and Torque Formulse 82
91. Improvement of Commutation by Shifting of the Brushes. ... 83
92. Armature Reaction in Generators and Motors 83
CHAPTER XV
Characteristics of Directcurrent Motors
93. The Starting Torque 85
94. The Starting Resistance 85
96. Motor Starter 87
96. Novoltage Release 87
97. Load Characteristics 88
98. Effect of Armature Reaction on the Speed 89
99. Variable Speed Operation 89
100. The Starting Torque 90
101. The Starting Resistance 91
102. Load Characteristics 92
103. Speed Adjustment 92
104. The Compound Motor 93
CHAPTER XVI
Losses, Efficiency and Heating
105. Mechanical Losses in Electrical Machinery 95
106. Copper Losses ^ 95
107. Hysteresis Loss 95
108. Eddy Current Loss 96
109. Stray Loss 96
110. The Efficiency of a Machine 97
111. Heating of Electrical Machinery 99
112. Permissible Temperature Rise 99
CONTENTS xi
CHAPTER XVII
Motor Applications
Article Page
113. Limits of Output 100
114. Open, Semienclosed and Totally Enclosed Motors 100
115. Intermittent Ratings 101
116. Effect of Speed on the Cost of a Motor ^ .... 101
117. Choice of Type of Motor 101
118. line Shaft Drive 102
119. Woodworking Machinery 102
120. Reciprocating Pumps 103
121. Traction Motors 103
122. Crane Motors 103
123. Express Passenger Elevators 103
124. Shears and Punch Presses 103
CHAPTER XVIII
Adjustable Speed Operation of Directcurrent Motors
125. Speed Variation of Shunt Motors by Armature Control .... 105
126. Speed Variation of Shunt Motors by Field Control 107
127. Speed Regulation of an Adjustable Speed Shunt Motor. .... 107
128. Electric Drive for Lathes and Boring Mills 108
129. Multiple Voltage Systems 109
130. Ward Leonard System 110
131. Drive for Ventilating Fans Ill
132. Armature Resistance for Speed Reduction 112
133. Motors for Small Desk Fans 112
134. Printing Presses 113
CHAPTER XIX
Handoperated Face Plate Starters and Controllers
135. Knife Switches 114
136. Auxiliary Carbon Contacts 114
137. Blowout Coils 115
138. Horn Gaps 116
139. Fuses 117
140. Circuit Breakers 117
141. Motor Starters 117
142. The Sliding Contact Type of Starter 117
143. Starting Resistance 118
144. Overload Release 118
145. Multiple Switch Starters 119
146. Compound Starters 120
147. Speed Regulators 121
148. Controllers for Series Motors 122
xii CONTENTS
CHAPTER XX
Drum Type Controllers
Article Page
149. Drum Type Controllers 124
150. Novoltage and Overload Release 125
151. Street Car Controller for Series Parallel Control 126
152. Reversing Drum 129
153. Mechanical Features of Drum Controllers 129
CHAPTER XXI
AiTTOMATic Starters and Controllers
154. Automatic Solenoid Starter 130
155. Float Switch Control 130
156. Magnetic Switch Controller 131
157. Multiple Unit Control of Railway Motors 133
158. Automatic Magnetic Switch Starters 133
159. Automatic Starter with Series Switches 135
CHAPTER XXII
Electrolysis and Batteries
160. Electrolysis 139
161. Voltameter 139
162. Electric Battery 140
163. Theory of Battery Operation 140
164. Polarization 141
165. The E.M.F. and Resistance of Cells 141
166. The DanieU Cell 141
167. Calculation of the E.M.F. of a DanieU Cell 142
168. Local Action 143
169. Leclanch6 Cell 143
170. Dry Cells 143
171. Edison Lalande Cell 144
172. Power and Energy of a Battery 144
173. Battery Connections 144
CHAPTER XXIII
Storage Batteries
174. Action of the Lead CeU 146
175. Storage or Secondary Battery 147
176. Sulphation 147
177. Construction of the Plates 148
178. Construction of a Lead Battery 149
179. Voltage of a Lead Battery 161
180. Capacity of a Cell . 163
CONTENTS xiii
Article Page
181. Amperehour Efficiency 153
182. Watthour Efficiency 164
183. Effect of Temperature on the Capacity 155
184. Limit of Discharge 166
185. Treatment of Lead Cells 156
186. Action of the Edison Battery 167
187. Construction of the Plates 158
188; Construction of an Edison Battery 158
189. The Voltage of an Edison Battery 160
190. Characteristics of an Edison Battery 160
CHAPTER XXIV
Opebation of Generators
191. Operation of the Same Shunt Machine as a Generator or as a Motor 162
192. Loading Back Tests 163
193. Parallel Operation 164
194. Shunt Generators in Parallel 164
195. Division of Load among Shunt Generators in Parallel 165
196. Compound Generators in Parallel 166
197. Division of Load among Compound Generators 167
CHAPTER XXV
Operation of Generators and Batteries in Parallel
198. Isolated Lighting Plants 169
199. Lighting Plants for Farm Houses 169
200. Lamp Circuit Regulator 170
201. Small Isolated Power Stations 171
202. Resistance Control 171
203. End CeU Control 172
204. Booster Charge, End Cell Discharge 173
205. Capacity of Battery 175
206. Batteries for Rapidly Fluctuating Loads 176
207. The Differential Booster 176
208. Carbon Pile Regulator 176
209. Floating Batteries 179
CHAPTER XXVI
Car Lighting and Variable Speed Generators
210. Systems of Vehicle Lighting 181
211. Straight Storage for Trains 181
212. Head and End System 181
213. Carbon Pile Lamp Regulator 182
214. The Axle Generator Systems 182
215* Automatic Switch 183
XIV CONTENTS
Article Page
216. Generator Regulator 183
217. Pole Changer 184
218. The Stone Generator 185
219. Lighting Generators for Motor Cars 186
220. Constant Speed Generators 186
221. Bucking Field Coils ^ 186
222. Vibrating Contact Regulator ;87
223. The Rosenberg Generator 188
CHAPTER XXVII
Alternating Voltages and Currents
224. The Simple Alternator 191
226. The Wave Form 193
226. The Oscillograph 193
227. Frequency 194
228. Vibrating Reed Type of Frequency Meter 195
229. Average Value of Current and Voltage 197
230. The Heating Effect of an Alternating Current 197
231. Symbols 198
232. Voltmeters and Ammeters for Alternatingcurrent Circuits . . . 198
CHAPTER XXVIII
Representation of Alternating Currents and Voltages
233 200
234. Electrical Degrees 200
235. Vector Representation of Alternating Voltages and Currents . . 201
236. The Sum of Two Alternating Voltages of the Same Frequency . . 203
CHAPTER XXIX
Inductive Circuits
237. Inductance . * 205
238. Make and Break Spark Ignition 206
239. The Coefficient of Self Induction 206
240. Alternating Currents in Inductive Circuits 207
241. Voltage and Current Relations 208
242. Power in an Inductive Circuit 209
243. Examples of Inductive and Noninductive Circuits 209
244. Voltage, Current and Power in Resistance Circuits 211
245. Resistance and Inductance in Series 212
246. The Power Factor 213
247. The Wattmeter 214
248. Transmission line Regulation and Losses 215
249. Resistance and Inductance in Parallel 216
CONTENTS XV
CHAPTER XXX
Capacity Circuits
Article Page
250. Condensers 218
251. Capacity Circuits with Direct and with Alternating Currents . . 219
252. Phase Relation between Voltage and Current in Capacity Circuits . 220
258. Voltage and Current Relations in Capacity Circuits 221
254. Parallel Plate Condenser 222
255. Power in Capacity Circuits 223
256. The FormulsB Used in Circuit Problems 223
257. Resistance, Inductance and Capacity in Series 224
258. Resistance, Inductance and Capacity in Parallel 226
CHAPTER XXXI
Alternators
259. Alternator Construction 229
260. Twophase Alternator 230
261. Threephase Alternators 231
262. YConnection 233
263. Deltaconnection 233
264. Voltages, Currents and Power in a YConnected Machine . . . 235
265. Voltages, Currents and Power in a Deltaconnected Machine . . 236
266. Connection of a Threephase Load 237
267. Power Measurement in Polyphase Circuits 238
268. Alternator Construction , . . . . 239
269. The Revolving Armature TyP^ of Alternator .... i ... . 240
270. The Inductor Alternator 241
271. Magneto Alternators 242
CHAPTER XXXII
Alternator Characteristics
272. Armature Reaction 244
273. Vector Diagram at Fullload 245
274. Regulation Curves of an Alternator 245
275. Experimental Determination of Alternator Reactance 246
276. Automatic Regulators 249
277. Efficiency 250
278. Rating of Alternators 251
CHAPTER XXXIII
Stnchrgnous Motors and Parallel Operation
279. Principle of Operation of Synchronous Motors 252
280. The Back E.M.F. of a Synchronous Motor 253
281. Mechanical Analogy 254
xvi CONTENTS
Article Page
282. Vector Diagram for a Synchronous Motor 254
283. Maximum Output 255
284. Operation of a Synchronous Motor when Under and Overexcited 256
285. Use of the Synchronous Motor for Power Factor Correction . . 256
286. Synchronizing 258
287. Hunting 258
288. Parallel Operation of Alternators 259
CHAPTER XXXIV
Tranbfobmer Characteristics
289. The Transformer 261
290. Constant Potential Transformer 261
291. Vector Diagram for a Transformer 263
292. Induction Furnace 263
293. Leakage Reactance 264
294. Leakage Reactance in Standard Transformers and in Induction
Furnaces 266
295. The Constantcurrent Transformer 267
296. The Efficiency of a Transformer 268
297. Hysteresis Loss 269
298. Eddy Current Loss 269
299. Iron Losses 269
300. The AUday Efficiency 269
301. Cooling of Transformers 270
CHAPTER XXXV
Transformer Connections
302. Lighting Transformers 273
303. Connections to a Twophase Line 273
304. Connections to a Threephase Line 276
305. Advantages and Disadvantages of the Y and Deltaconnection . 278
306. Types of Transformer 279
307. The Autotransf ormer 279
308. Boosting Transformers and Feeder Regulators 281
CHAPTER XXXVI
Polyphase Induction Motors
309. The Induction Motor 283
310. The Revolving Field 284
311. The Revolving Field of a Threephase Motor 285
312. Multipolar Machmes 287
313. The Starting Torque 287
314. The Wound Rotor Motor 28S
CONTENTS xvii
Article Page
315. Running Conditions 200
316. Vector Diagrams for the Induction Motor 291
317. Adjustable Speed Operation 293
318. Induction Generator 294
319. Self starting Synchronous Motors 294
320. Dampers for S3rnchronous Machines 295
CHAPTER XXXVII
Induction Motor Applications and Control
321. Choice of Type of Motor 296
322. Line Shaft Drfve 297
323. Woodworking Machinery 297
324. Cement Mills 297
326. Motors for Textile Machinery 298
326. Adjustable Speed Motors 298
327. Crane Motors 298
328. Shears and Punch Presses 299
329. Adjustable Speed Service 299
330. Resistance for Adjustable Speed Motors 300
331. Switches for Alternatingcurrent Circuits 300
332. Starting of Squirrelcage Induction Motors 301
333. Starting Compensator 302
334. The Stardelta Method of Starting 304
335. Starter for a Wound Rotor Motor 305
336. Automatic Starters 306
CHAPTER XXXVIII
Singlephase Motors
337. Singlephase Induction Motors 308
338. Splitphase Method of Starting 308
339. Running Torque of a Singlephase Motor 308
340. Singlephase Series Motor 310
341. Armature Reaction 312
342. The Repulsion Motor 313
343. Commutation of Series and Repulsion Motors 314
344. Wagner Singlephase Motor 314
CHAPTER XXXIX
Motorgenerator Sets and Rotary Converters
345. Motorgenerator Set 315
346. The Booster Set 315
347. The Balancer Set 316
348. Threewire Generator 317
349. To Transform from Alternating to Direct Current 318
xviii CONTENTS
Article Page
350. Rotary Converter 318
351. Motorgenerator Sets and Rotary Converters 319
352. Polyphase Rotary Converter 320
353. SpUtpole Rotary Converter 320
354. Frequency Changers 321
CHAPTER XL
Electric Traction
355. Tractive Effort 322
356. Speed Time Curve 323
357. Energy Required by a Car 326
358. Characteristics Desired in Railway Motors 327
359. Motor Construction 329
360. Distribution to the Cars 329
361. Alternating and Directcurrent Traction 329
362. Motor Car Trains 332
363. Electric Locomotives 332
364. Crane and Hoist Motors 332
365. Braking 333
366. Flywheel Motorgenerator Sets for Mine Hoisting 335
367. Safety Devices 337
CHAPTER XLI
Transmission and Distribution
368. Directcurrent Stations 338
369. Alternatingcurrent Stations 339
370. The Voltages Used in Practice 341
371. Comparison between Singlephase and Threephase Transmission. 341
372. Lightning Arresters 343
373. Switches 345
374. Overhead Line Construction 346
375. Underground Construction 347
376. Switchboards 348
377. Instrument Transformers 351
CHAPTER XLII
Electric Lighting
378. The Carbon Incandescent Lamp 352
379. The Tungsten Lamp 352
380. Gasfilled Tungsten Lamp 353
381. The Unit of light 354
382. Arc Lamps 354
383. The Directcurrent Open Arc 355
384. Directcurrent Enclosed Arc 355
CONTENTS XIX
Article Page
3S5. Alternatingcurrent Enclosed Arc 356
386. Hame Arc Lamps 356
387. Luminous Arc Lamp 357
388. Mercury Vapor Converter 357
389. Mercury Vapor Lamp 359
390. Shades and Reflectors 359
391. Efficiency of Illuminants 360
392. Light and Sensation 361
393. Reflection and Color 362
394. Principles, of Illumination 362
395. Quality of the Light 363
396. Glare 363
397. Shadows 363
398. Litensity of Illumination 364
399. Lines of Illumination 364
400. Power Distribution for Lighting 365
CHAPTER XLIII
Laboratory Course
401. Protection of Circuits 368
402. Ammeter Shunts 368
403. Safe Carrying Capacity of Copper Wires 368
404. Control of the Current in a Circuit 369
Exp. 1. Measurement of the Resistance of the Field Coil Circuit . . . 370
Exp. 2. Measurement of the Resistance of the Armature Circuit . . 370
Exp. 3. Speed Adjustment of a Directcurrent Shunt Motor .... 371
Exp. 4. Voltage of a Directcurrent Generator 371
Exp. 5. Regulation of Directcurrent Generators 372
Exp. 6. Brake Tests on Directcurrent Motors 373
Exp. 7. Starting Torque Tests on Directcurrent Motors 373
Exp. 8. Stray Loss and Efficiency of a Directcurrent Motor .... 374
Exp. 9. Heat Run on a Directcurrent Generator 374
Exp. 10. Voltage Regulation of a Threewire System 374
Exp. 11. Fuse Testing 375
Exp. 12. Calibration of a Circuit Breaker 375
Exp. 13. Alternatingcurrent Series Circuit 376
Exp. 14. Predetermination of the Characteristics of an Alternating
Current Circuit 376
Exp. 15. Characteristics of a Constant Potential Transformer .... 377
Exp. 16. Regulation of an Alternator s . 377
Exp. 17. Starting and Running Characteristics of a Synchronous
Motor 378
Exp. 18. Characteristics of a Rotary Converter 379
Exp. 19. Starting and Running Characteristics of a Polyphase Induc
tion Motor 380
Exp. 20. Transformer Connections 381
Index » , »  423
INTRODUCTORY
Before a study of electric circuits and machinery can be
made, it is necessary to define the electric and the magnetic
units and express them in terms of the fundamental units and
derived mechanical units which are given below.
Quantity Practical units Practical units
c.g.s. system ft. lb. sec. system
Length 1 cm. 1 f t = 30.48 cm.
Mass 1 gm. 1 lb = 453.6 gm.
Time 1 sec. 1 sec.
Force 1 dyne 1 poundal = 1/32.2 lb.
1 gm. = 981dynes 1 lb. = 4.448 X 10* dynes
Work or
energy 1 erg = 1 dyne cm. 1 ft. lb. = 1.356 X 10^ ergs
Power 1 erg per sec. 1 h.p. = 550 ft. lb. per sec.
= 746 X 10^ ergs per sec.
In the first few chapters of this work some of the fundamental
principles of electricity and magnetism are briefly discussed.
Parts of these chapters are difficult and are of theoretical im
portance only. These are printed in small type and may be
omitted if the student is willing to consider as experimental
laws what are really laws depending on the definitions of the
electric and the magnetic units and on their interrelations.
PRINCIPLES AND PRACTICE
OF
ELECTRICAL ENGINEERING
CHAPTER I
MAGNETISM AND MAGNETIC UNITS
1. Magnets. — The power of a magnet to attract or repel is
concentrated at certain points called poles. A simple magnet
has two poles which are equal and opposite and the line joining
them points north and south when the magnet is allowed to
swing freely in a horizontal plane. The pole pointing toward
the north is called the north (N) pole, that pointing toward
the south is called the south (S) pole.
Like poles repel one another, unlike poles attract one another.
2. Coulomb's Law states that the force between two magnetic poles is
directly proportional to the strengths of the poles and inversely proportional
to the square of the distance between the poles, thus, in Fig. 1,
m mi
Fig. 1.
mmi
r*
where / is the force between the poles,
r is the distance between the poles,
m and m\ are the strengths of the poles,
ib is a constant which depends on the surrounding mediiun and on the
units chosen.
The c.g.s. unit of pole strength is chosen so as to make A; » 1 when/ is in
dynes, r in cm. and the medium is air, then/ = — — .
r*
A imit pole therefore acts on an equal pole in air, at a distance
of 1 cm. from it, with a force of 1 dyne.
3. The magnetic field is the name given to the space surround
ing a magnet, but is limited in practice to the space within which
the force of the magnet is perceptible. A magnetic pole placed
in a magnetic field is acted on by a force which is proportional
1
2 PRINCIPLES OF ELlSCTRiCAL ENGINEERING [Chap, i
to the strength of the magnetic pole and to the strength or in
tensity of the magnetic field.
The intensity of a magnetic field at any point is taken as the
force in dynes on a unit pole at that point; therefore, a unit field
will act on a unit pole in air with a force of 1 dyne.
The direction of a magnetic field at any point is taken as the
direction of the force on a north pole at that point.
Fig. 2. — Direction of the field of a magnet.
Let NS, Fig. 2, be a magnet of pole strength tn, and n a unit north pole.
The pole N of the magnet repels the unit pole with a force = m/ri* dynes,
represented in magnitude and direction by the line na; the pole S of the mag
net attracts the unit pole with a force » m/rs* dynes, represented in magni
tude and direction by the line n&; the resultant force on the unit pole, which
is a measure of the field intensity, is represented in magnitude and direction
by the line nc.
Fig. 3. — Lines of force surrounding a bar magnet.
4. Lines of Force. — In dealing with magnetic problems it
is found convenient to represent the magnetic field diagram
matically by what are called lines of force. These are con
tinuous lines whose direction at any point in the field is that of
the force on a north pole placed at the given point. The number
of lines crossing 1 sq. cm. placed perpendicularly to this direction
is made proportional to the field intensity at the point and unit
magnetic field is represented by one line per sq. cm.
aht.s] magnetism and magnetic units 3
In Fig. 3 the intensity of the magnetic field is greatest at the
poles and decreases as the distance from the poles increases, so
that the lines of force which represent this field spread out from
the poles as shown. Since a north pole n placed in this field is
repelled by the pole N and attracted by the pole S, the lines of
force, being drawn in the direction of
the force on a north pole placed in the
field, must leave the N pole and enter
the S pole.
The total number of lines of force
leaving or entering a magnetic pole is
called its magnetic flux ^.
The flux density <S> at any point in
a magnetic field is the number of lines
of force crossing unit area placed per
pendicular to the direction of the lines Fia. 4.
of force at that point.
S. Liaes of Force from « Unit Pole. — ^If a unit pole w«rB aiurounded by a
sphere of 1 cm. radius, as in Fig. 4, another unit pole placed on the surface of
this sphere would be acted on witb unit force &ud so the field intensity at
this surface must be imity; there must therefore be one line of force pw
sq. cm. of sphere surfaoe or a total of 4t lines, as the surface area of a sphere
of 1 cm. radius is 4r sq. cm.
Since the number of lines from a unit pole is 4«', therefore the number from
ft pole of strength m is 4rtn.
CHAPTER II
ELECTROMAGNETISM
6. Direction of an Electric Cnrrent^P and Q, Fig. 5, are
conductors carrying current; the current is going down in con
ductor P and coming up in conductor Q. Let the direction
of the current be represented by an arrow; at the end. of con
ductor P one would see the tail of the arrow, represented by a
cross, while at the end of conductor Q the point of the arrow
would be seen, this is represented by a dot.
— Direction of an electric current
7. Magnetic Field Stirrounding a Conductor Canying Current.
— A conductor carrying ciurent is surrounded by a magnetic
field represented by lines of force as shown in Fig. 6. To deter
Ourrent Down Current Up
Fig. 6. — ^Pield surrounding a conductor carrying current.
mine the direction of these lines the following rule is used: If
a corkscrew is screwed into the conductor in the direction of
the current then the head of the corkscrew has to be turned in
the direction of the lines of force.
8. Force at the Centre of a Circular Loop Carrying Current — Fig. 7 shows
a wire, carrying a current i, and bent to form a circular loop of radius r.
The direction of the magnetic field produced is found by the rule in the last
paragraph.
4
Art. 0]
ELECTR0MAQNETI8M
An element a& acts on a unit pole at the centre of the loop with a force /
which is found to be = h — ^ ; and the total force F on this pole due to the
complete loop
,2irrXt
= k r
r
where A; is a constant which depends on the medium and on the imits chosen.
The unit of current is chosen so as to make A; — 1 when F is in dynes, r is
Fig. 7. — Magnetic field produced by a loop carrying current.
2t%
in cm. and the medium is air, then F = — dynes; a unit current is therefore
of such value that, when flowing in a loop of 1 cm. radius, it acts on a unit
pole at the centre of the loop with a force of 2r dynes. This is called the
e.g. s. unit of current; the practical imit, called the ampere, is equal to one
tenth of a c.g.s. imit.
9. Electromagnets. — The loop carrying current, shown in
Fig. 7, acts like a magnet and is called an electromagnet. The
ig^ft
A^«
FiQ. 8. — ^The polarity of an electromagnet.
strength of an electromagnet may be increased by increasing
the cmrent or, as in Fig. 8, by increasing the number of turns.
The direction of the magnetic field may be conveniently found by
another corkscrew law which states that if the head of the cork
screw is turned in the direction of the current then the screw
6
PRINCIPLES OF ELECTRICAL ENGINEERING IChap.ii
itself will move into the magnetic field in the direction of the lines
of force; the direction of the field produced by the righthand
spiral in diagram A is the same as that produced by the left
hand spiral in diagram B; this direction may be reversed by
reversing the current.
10. Force on a Conductor Carrying Current in a Magnetic Field. — In
Fig. 9, the unit pole n is acted on by the current in the loop with a force of
••aS
S,
I
Votceoa Wica
■JS. Dynes
and Act! Up
Force on
Folo
Fio. 9. — ^Force on a conductor carrying current in a magnetic field
2r%/r dynes (page 5) at right angles to the plane of the paper, where i
is the current in c.g.s. units. The loop itself must be reacted on by the unit
pole with an equal force in the opposite direction.
The flux density (B at the wire, in lines per sq. cm.^ due to the unit pole
flux from the unit pole
" surface of a sphere of r cm. radius
4ir 1
r«
4irr»
As shown above, the force acting on the wire in dynes
« —
r
 , X 2rr X i
(BLi.
Since (B ^2
where (B is the flux density at the wire in lines per sq. cm., L is the length of
wire that is in the magnetic field in cm. = 2irr in the case of a circular loop
and i is the current in the wire in c.g.s. units.
Am. 11] ELBCTROMAONETISM 7
When a conductor is carrying current and is in a magnetic
field, as in Fig, 10, it is acted on by a force which is proportional
to the current and to the strength of the field. The direction
of this force may be determined as follows:
Diagram A shows the magnetic field between the poles of the
magnet when there is no current in the conductor.
Fio. 10, — Feree on a conductor cawying current in a magnetic field.
Diagram B shows the magnetic field produced by the current
in the conductor.
Dii^am C shows the resultant distribution of magnetic flux
when the conductor carrying current is placed into the magnetic
field of diagram A, the lines of force tend to straighten and force
the conductor to move in the direction shown.
Fio. 11. — MoTii^ coil ammeter.
11. Moving Coil Ammeters. — The above principle ia applied
in one of the most satisfactory types c^ instrument for the
measurement of direct current.
Such an instrument is shown in I^. 11, ,^;S'is a permanent
horseshoe magnet with pole shoes bored out cylindrical^ and
8 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, ii
£ is a cylindrical soft iron core concentric with the pole faces,
lines of force therefore pass as shown in diagram A and the flux
density in the air gaps is uniform. In this magnetic field a coil
C is placed and is supported on jewelled bearings. The coil
consists of a number of turns of fine insulated wire wound on a
light aluminium frame and the current to be measured is intro
duced to the coil through the spiral springs D, diagram B. Since
the sides of the coil are carrying current and are in a magnetic
field they are acted on by forces which turn the coil through an
angle against the torsion of the springs D and this angle may be
read on a scale over which plays a pointer B attached to the
coil
CHAPTER III
ELECTROMAGNETIC INDUCTION
12. Electromagnetic Induction. — Faraday's experiments
showed that when the magnetic flux threading a coil undergoes a
change, an electromotive force (e.m.f.) is generated or induced in
the coil and that this e.m.f. is pro
portional to the time rate of change
of the flux. If the coil A, Fig. 12,
be moved from position 1 where the
flux threading the coil is lines, to
position 2 where the flu;K threading
the coil is zero, in a time of t seconds,
then the average rate of change of
flux is <l>/t lines per sec.
The c.g.s. unit of e.m.f. is that
generated in a coil of one turn when
the flux threading the coil is chang
ing at the rate of one line per sec.
The practical unit, called the volt,
is equal to 10^ c.g.s. units so that
when the flux threading a coil of one
turn changes at the rate of lines
in t seconds the average e.m.f. in
duced in the coil = y 10""^ volts and
dt
.*^t
.0^
©.
8
Fig. 12. — Generation of elec
tromotive force. Right Hand
Rule: thumb — motion; fore
the e.m.f. at any instant = ^ 10
volts.
That portion of the coil wherein the ftj^Sromotivl'fo^e.'"'
e.m.f. is actually induced is the con
ductor zy which cuts the lines of force, and the quantity d<l>/dt is
the rate at which the lines are cut.
13. The direction of the induced electromotive force may
be determined by Fleming's threefinger righthand rule which
states that if the thumb, the forefinger and the middle finger of
the right hand be placed at right angles to one another so as to
9
10
PRINCIPLES OF ELECTRICAL ENGINEERING IChap. m
represent three coordinates in space; with the thumb pointed
in the direction of motion of the conductor relative to the
magnetic field and the forefinger in the direction of the lines of
force, then the middle finger will point in the direction of the
induced e.m.f.
The direction of the current in ^, as determined by the righthand rule, is
shown in Fig. 13 for the case where the coil is moving downward and the
number of lines of force threading the coil is decreasing. This current sets
up a magnetic flux 0c, the direction of which, found by the corkscrew law
(page 5) is the same as that of the main flux ^ and tends to prevent the flux
threading the coil from decreasing.
Pig. 13. Fig. 14.
Direction of the generated electromotive force.
If now the direction of motion of the coil be reversed so that the number of
lines of force threading the coil is increasing, the current will be reversed, as
shown in Fig. 14, and the magnetic flux 0« will oppose the main flux ^ and
tend to prevent the flux threading the coil from increasing.
The general law for the direction of the induced e.m.f. in a coil,
known as Lenz^a Law, states that the induced e.m.f. tends to
send an electric current in such a direction as to oppose the change
of flux which produces it.
If the coil abed, Fig. 15, is moved from m to n, the flux threading
the coil does not change and the resultant e.m.f. generated in the
coil is zero; the portions ab and cd of the coil are cutting lines
of force but the e.m.fs. generated in these portions are equal
and opposite.
Abt. 15]
ELECTROMAGNETIC INDUCTION
11
14. Mutual Induction. — The flux threading a coil may be
changed without moving the coil. Suppose a constant current is
flowing in the coil A, Fig. 16, this produces a constant flux which
threads coils A andBbut noe.m.f. is generated in coil B since there
Direction q
Lines of Force
Fig. 16.
is no change in the flux. If the current in coil A is increased, the
flux threading coil B will increase and this change of flux will in
duce an e.m.f . in coil B which will cause a current 72 to flow in such
a direction as to oppose the increase in flux. If the current in coil
A is decreased, the flux threading coil B will decrease and this
change of flux will induce in coil B an e.m.f. which will send a cur
rent 72 in such a direction as to oppose the decrease in flux.
Flax ia
IncreMing i {
±
fUUUi.
a a li a ,i
s
«
rrrrrr
□
Flax ia
Decreaimg
A
MUUUi
,, ^, yf \r \f
B
fUUV
u
/kJuk
a
FiQ. 16. — Direction of electromotive force of mutual induction.
15. Self Induction. — When the current in a coil is changed, an
e.m.f. is generated in the coil itself in such a direction as to oppose
the change in the current. In Fig. 17, for example, when the
switch h is closed, the current flowing in the coil does not reach its
final value instantaneously because, as the current increases in
value, the flux 4> threading the coil increases and causes an e.m.f.
to be induced in the coil in such a direction as to oppose the in
12 PRINCIPLES OF ELECTRICAL ENGINEERim IChap. in
crease of the current. This opposing e.m.f., called the e.m.f. of
self induction, exists only while the current is changing.
If, after the current has reached its final value, the switch k is
suddenly opened, the current in the coil tries to decrease suddenly
to zero but, as it decreases, the flux threading the coil decreases
and causes an e.m.f . to be induced in the coil in such a direction
Tim«
FiQ. 18. — Growth of current in a coil.
as to oppose the decrease of the current; this e.m.f. is generally
large enough to maintain the current between the switch contacts
as they are being separated and accounts for much of the flashing
that is seen when a switch is opened in a circuit carrying current.
When the switch is closed, the current increases to its final value
as shown in Fig. 18. As the current i increases, the correspond
ing increase of the fiux <f> threading the coil induces an e.m.f. of
self induction which is proportional to d<l>/dty the rate of change
of the fiux. When the current has ceased to change the e.m.f.
of self induction becomes zero.
CHAPTER IV
WORK AND POWER
16. Transformation of Mechanical into Electrical Energy. —
If the conductor xy, Fig. 19, be moved downward so as to cut at
a constant rate the lines of force passing from NtoS,s, constant
e.m.f. is induced in the conductor and, by adjusting the resistance
R, the current in the circuit may be maintained at the value i
Fig. 19. — Right Hand Rule for generation of e.m.f.: thumb — motion;
forefinger — ^lines of force; middle finger — e.m.f.
in the direction shown; the direction of the current may be de
termined by the righthand rule (page 9).
As this conductor is carrying current in a magnetic field, it is
acted on by a force F the direction of which may be determined
as in diagram C, Fig. 10 (page 7). This force, as shown in Fig.
19, opposes the motion of the conductor and hence mechanical
enei^gy must be expended in moving the conductor.
13
14 PRINCIPLES OF ELECTRICAL ENGINEERING IChap. iv
If (B is the density of the magnetic field in lines per sq. cm.
L is the length in cm. of that part of the conductor which is cutting lines of
force
V is the velocity of the conductor in cm. per sec.
i is the current in the conductor in c.g.s. units, then
€, the e.m.f . generated in the conductor in c.g.s. units,
a the lines of force cut per sec.
= (BL7
Now F, the force acting on the conductor — CBIi dynes (page 6) and
the mechanical power in dyne cm. per sec. required to keep the
conductor moving
= FF
 ((BZa)7
 ((ELF)t
s ei
B (volte X 10") (amperes/10); pages and 5
= volts X amperes X 10^
The mechanical power required to obtain / amperes at a
difference of potential of E volts from an electrical machine which
has an efficiency of 100 per cent.
= EI W ergs per sec.
= EI watts
where the watt, the practical unit of power, is equal to 10^ ergs
per second.
The power developed by large electrical machines is expressed
in kilowatts, where 1 kw. is equal to 1(X)0 watts.
The horsepower = 560 ft. lb. per sec.
= 746 X 10^ ergs per sec.
= 746 watts.
this result gives a connecting link between the electrical and the
mechanical units.
17. Unit of Work. — ^Work is done when a force is moved
through a distance. The c.g.s. unit of work is the erg, which
is the work done in moving a force of 1 dyne through a distance
of 1 cm.
Power is the rate at which work is done and is expressed
either in ergs per sec, in watts (10^ ergs per sec), or in horse
power (746 X 10^ ergs per sec).
When the power, or rate at which work is being done, is 1
watt, or 10^ ergs per sec, then the work done in 1 sec.
is 1 wattsecond or 10^ ergs and is called 1 joule. A more
Art. 20] WORK AND POWER 18
convenient unit for practical work is the kilowatthour (3600 X 10*
joules) and this will gradually replace the horsepowerhour
because it is based on a system of international units.
18. Heat Energy and Electrical Energy.^~The energy required
to raise the temperature of 1 lb. of water by 1® F. is called
the British Thermal Unit (B.T.U.) and is equal to 780 ft. lb.
The energy required to raise 1 gm. of water through 1® C.
is called the granune calorie and is equal to 4.2 X 10^ ergs so
that
1 gm. calorie = 4.2 X 10^ ergs
= 4.2 wattseconds (joules).
19. Conversion Factors* — ^Although the c.g.s. system of units
is the only possible international system, much calculation
work is still carried out in the footpoundsecond system. The
conversion factors given below help to simplify the work of
changing from one system to another.
C.g.s. unit Other units
Length 1 cm. 1 in. = 2.64 cm.
Mass 1 gm. 1 lb. = 453.6 gm.
Time 1 sec.
Force 1 dyne 1 gm. = 981 dynes
1 lb. = 444,800 dynes
= 453.6 gm.
Work or energy 1 erg = 1 dynecm. 1 joule = 1 wattsec.
10^ ergs
1ft. lb. = 1.356 X 10^ ergs
1 kw.hour = 3600 X 10» joules
1 gm. calorie = 4.2 joules
1 lb. calorie = 1900 joules
Power 1 erg. per sec. 1 watt = 10^ ergs per sec.
1 kw. = 1000 watts
1 h.p. = 550 ft. lb. per sec.
= 746 watts
20. Problems on Work and Power.
1. A hoist raises a weight of 2000 lb. through a distance of 300 ft. in a
time of 1 min. Find the work done and the power expended.
If the efficiency of the hoist is 75 per cent, and that of the motor is 90 per
cent, find the horsepower of the motor and also the current taken by the
motor if the voltage is 110.
o. Work done > 2000 X 300
 60Q.000 ft. lb.
16 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.iv
6. Power expended — 600,000 ft. lb. per 60 sec.
= 10,000 ft. lb. per sec.
« 18.2 h.p.
550
18.2 X 746
« 13.6 kw.
1000
c. The power input to the hoist = — £ = 18.1 kw= 24.3 h.p.
0.76
The power input to the motor = — ^ = 20 kw.
d. Since watts — volts X amperes, therefore 20 X 1000 = 110 X amperes
and the current in amperes » 182.
2. An electric iron takes 5 amp. at 110 volts. What does it cost to operate
this iron for 2 hours if the cost of energy is 6 cents per kw.hour.
The rate at which energy is used » 110 X 5 — 550 watts
= 0.55 kw.
The energy used in 2 hours = 0.55 X 2 = 1.1 kw.hour
The cost of this energy = 1.1 X 6 = 6.6 cents
3. A 32candle power, 110volt tungsten lamp requires 40 watts. What is
the current taken by this lamp and what is the cost of energy for 15 lamps
burning for an average time of 4 hours if the cost of energy is 5 cents per
kw.hour?
watts 40 ^ ^^
amperes = — ; — = —  = 0.36 amp.
^ volts 110
power = 40 X 15 = 600 watts
= 0.6 kw.
energy used = 0.6 X 4 = 2.4 kw.hours
cost of energy = 2.4 X 5 = 12 cents.
4. An electric water heater has an efficiency of 80 per cent, and takes 3
amp. at 110 volts. How long will it take to raise 1 pint (1.25 lb.) of
water from 20^ C. to the boiling point and what will this cost when the
rate is 5 cents per kw.hour?
energy required = 1.25 (100  20) = 100 lb. calories
= 100 X 1900 = 190,000 wattsec.
100
energy delivered = 190,000 X rr
oU
= 238,000 wattsec.
= 238 kw.sec.
= 0.066 kw.hours
cost of energy = 0.066 X 5 = 0.33 cents
Now 238,000 wattsec. are supplied at the rate of 110 X 3 » 330 watts
therefore the time during which energy must be applied
238,000
330
12 min.
= 720 sec.
Art. 20] WORK AND POWER 17
5. If a ton (2000 lb.) of ooal heats a house for a month what would it cost
to give exactly the same heating effect electrically if the cost of energy is 3
cents per kw.hour?
With a good heating system 1 lb. of coal burnt on the grate will deliver
8000 B.T.U. or 4450 lb. calories to the house.
The energy required per month therefore — 4450 X 2000 lb. calories
» 8,900,000 lb. calories.
= 8,900,000 X 1900 wattnaec.
8,900,000 X 1900
3600 X 1000
kw.hours
« 4,700 kw.hours
the cost of this energy » 4,700 X 3 := 14,100 cents
= 141 dollars.
The reason for this enormous difference in the cost of heating by the two
methods is that the efficiency of a heating system is about 60 per cent.
while that of an electric generating station is about 6 per cent.; moreover
the cost of the coal required per kw.hour is about 0.5 cents or 1/6 of
the selling price of the electrical energy.
CHAPTER V
ELECTRIC CIRCUITS AND RESISTANCE
21. The flow of dectricity through electric circuits is similar
in many ways to the flow of water through hydraulic circuits.
This may be seen by a coTiparison between the circuits shown
diagrammatically in Fig. 20.
V)
Fam
Turbine
W
•©
Ammeter
B ^
C^YoItmeter
Motor
^en^rafcor
FiQ. 20. — Hydraulic and electric circuits.
To maintain a steady current
of w gm. of water per sec.
through the hydraulic circuit
and to raise the water from h to
a through a diflference of poten
tial of A cm., an amount of power
= wh gm. cm. per sec. must be
put into the circuit by the
pump.
In returning from a to 6
through the external part of the
circuit, the water falls through
a diflference of potential of h cm.
and supplies an amount of
power = wh gm. cm. per sec. to
drive the turbine and to supply
the frictional resistance loss in
the pipes.
To maintain a steady elec
tric current of I coulombs per
sec. (amperes^) through the
electric circuit and to raise the
electricity through a diflference
of potential of E volts, an
amount of power = EI watts
must be put into the circuit by
the electric generator.
In returning from a to 6
through the external part of the
circuit, the electricity falls
through a diflference of poten
tial of E volts and supplies an
amount of power = EI watts
to drive the motor and to
supply the resistance loss in the
connecting wues.
^A current of electricity is expressed in amperes; there is no corresponding
unit for a current of water which must therefore be expressed in gm. per
sec. The quantity of electricity which passes any point in a circuit is ex
pressed in coulombs where 1 coulomb is 1 amp.sec. A larger unit is the
amperehour.
18
Abt. 23] ELECTRIC CIRCUITS AND RESISTANCE
19
The current of water (the
quantity passmg any point per
sec.) is the same at all points
in the circuit since the circuit is
closed.
The electric current (the
quantity passing any point per
sec.) is the same at all points
in the circuit since the circuit is
closed.
22* Ammeters and Voltmeters. — The current in a circuit may
be measured by means of an instrument such as that described
on page 8, connected directly in the circuit as shown at A,
Fig. 20, while the diflference of potential between two points
may be measured by means of a similar instrument connected
directly between the points as shown at B. The essential diflfer
ence between the two instruments is that the ammeter must
carry the total current in the circuit with only a small diflference
of potential across its terminals and must therefore oflfer a
small resistance to the fllow of current through it, the volt
meter on the other hand must divert only a small portion of
the current from the circuit and must therefore oflfer a large
resistance to the fliow of current through it.
Fvin
irator
PiQ. 21. — Hydraulic and electric circuits.
23. Resistance Circuits. — Consider the case represented dia
grammatically in Fig. 21 where there is no turbine in the hydraulic
circuit nor any potor in the electric circuit.
The diflference of potential of The diflference of potential of
h cm. maintained by the pump
is used up in forcing w gm. of
water per sec. against the fric
tional resistance of the pipe, and
h = wr
E volts maintained by the elec
tric generator is used up in
forcing I amperes against the
resistance of the wires, and
E = IR
where r is called the resistance where R is called the resistance
of the pipe circuit. of the electric circuit and is a
constant for a given circuit.
20 PRINCIPLES OF ELECTRICAL ENGINEERING IChap. v
This resistance increases with This resistance increases
the length and decreases with with the length and decreases
the cross section of the pipe, with the cross section of the
wire, or
length
section
24. Ohm's Law. — The above relation E ^ IR \a known as
Ohm's law and the unit of resistance, called the ohm, is chosen
of such a value that a circuit with a resistance of 1 ohm will
allow 1 amp. to flow when the difference of potential between
the ends is 1 volt, therefore
volts = amperes X ohms
If for example the current in the heating coil of a 110volt
electric iron is 5 amp., then the resistance of this coil = 110/5
= 22 ohms.
26. Specific Resistance. — ^As pointed out in art. 23, the re
sistance of a wire is directly proportional to its length and
inversely proportional to its cross section or
where R is the resistance of the wire in ohms
L is the length of the wire
A is the cross section of the wire
k \a B, constant called the specific resistance and depends
on the material and on the units chosen. If centimeter units are
used then the specific resistance is the resistance of a piece of
the material 1 cm. long and 1 sq. cm. in cross section and is
expressed in ohms per cm. cube.
In practice the unit of cross section is generally taken as the
circular mil which is defined as the cross section of a wire 1 mil
(1/1000 in.) in diameter.
Since a wire 1 mil in dia. has a section of 1 cir. mil a wire 1
inch in dia. has a section of 10^ cir. mils and a wire 1 sq. inch in
4
section has a section of  10* cir. mils.
The specific resistance of copper wire * is 1.6 X 10"* ohms per
cm. cube or 9.7 ohms per cir. mil foot at 0® C; the specific
> For values of specific resistance of various materials see Standard Hand
book for Electrical Engineers.
Abt. 271 ELECTRIC CIRCUITS AND RESISTANCE 21
resistance of cast iron is 80 X 10~* ohms per cm. cube or 480
ohms per cir. mil foot, approx., at 0® C.
26. Variation of Resistance with Temperature. — ^The resist
ance of most materials varies with the temperature and
Rt = Ro{l + od)
where Rt is the resistance at (^ C.
Ro is the resistance at 0^ C.
t is the temperature of the material in deg. C.
a is called the temperature coefficient of resistance.
For all pure metals the resistance increases with the tempera
ture and a is approximately equal to 0.004. The resistance of
carbon and of liquid conductors decreases with increase of tem
perature, while the resistance of special alloys such as manganin
remains approximately constant at all operating temperatures.
A ooU has 1000 turns of copper wire with a cross section of 1288 cir. mils
and a length of mean turn of 15 in.
a. Find the resistance of the ooil at 0° C.
h. Find the resistance of the coil at 25° C.
c. Find the current that will flow through the coil at 25° C. and 110 volts.
d. After current has passed through the coil for some time it is found that
its value has dropped to amp., find the average temperature of the coil
under these conditions.
a. The resistance of 1 cir. mil foot — 0.7 ohms at 0° C.
r^ . . ^ ., ., ^o ^ 07 X 1000 X 15 ^ , ^
The resistance of the cou at C, = — — ^ 9.4 ohms.
' 1288 X 12
h. The resistance of the coU at 25° C. » 0.4(1 +0.004 X 25)  10.3 ohms.
^ 110
c. The current = — — = 10.7 amp.
10.3
110
d. The hot resistance of the coil = —— == 12.2 ohms at t^ C.
9
The resistance of the coil also — 9.4 ohms at 0° C.
Therefore the resistance of the coil at t° C. = 9.4(1 + 0.0040
» 12.2 ohms
from which 1 + 0.004/ » 12.2/9.4 » 1.3
and t » 0.3/0.004 = 76° C.
27. Power Expended in a Resistance. — To force a current of
I amperes through a circuit which has a resistance of R ohms, a
voltage E = IRiQ required so that
the power expended in the circuit = EI watts
= (IR)I
= PR watts.
This power is transformed into heat.
22 PRINCIPLES OF ELECTRICAL ENQINEERINQ [Chap, v
In the electric fiat iron and other such heating apparatus this
heat is utilized, the heatii^ element consisting of a coil of high
resistance wire, insulated with heat resisting insulation such as
asbestos, or mica, and embedded in the iron.
28. Insulating materials are materials which offer a very large
resistance to the fiow of electric current and for that reason they
are used to keep the current in its properpath. Inatransmission
line, current is prevented from passing between the wires by
porceWn insulators attached to cross arms as shown in Fig. 22.
When the wires are placed close to one another, as in house wiring,
the^ are covwed throughout their entire length with insulating
material such as paper, rubber or cotton; the electrical resistance
of these materials is greater than 10^' ohms per cm. cube.
Fia. 22. — loaulators for transmission lines.
29. Dielectric Strength of Insulatdng Material. — If a sheet of
insulating material is placed between two terminals and the
voltage between the terminals is gradually raised the material
will finally break down and a hole be burnt throt^h it, the
material is then said to be punctured, and a large current will flow
through the puncture if the voltage is maintained. The property
of an insulating material by virtue of which it resists breakdown
is called its dielectric strength.
30. Series and Parallel Circuits. — If several conductors are
connected in series as shown in Fig. 23, then the current is the
Abt. 31] ELECTRIC CIRCUITS AND RESISTANCE
23
same in each conductor while the total voltage is the sum of the
voltages across the different. parts of the circuit so that
E = El [ E2 "h E9 [ Ei
If several conductors are connected in parallel as shown in Fig.
24, then the voltage across each conductor is the same while the
total current is the sum of the currents in the different paths so
that
I ^h + h + h + Ii
\Ri Rt Rz R%/
It has been found convenient in problems on parallel circuits
to use a quantity called the conductance of a circuit where the
conductance G is the reciprocal of the resistance R and then the
above expression becomes
/ = E{Gi + G2 + G, + G4)
Pig. 23. — Series circuit.
FiQ. 24. — Parallel circuit.
Four coils having resistances of 3, 5, 10 and 12 ohms respectively are
connected in series across 120 volts, find the current in the circuit and the
voltage drop across each coil.
therefore
120 = /(3 + 6 + 10 + 12)
= 30/
7=4 amp .
^1 « 4 X 3 = 12 volts
^1 = 4 X 5 = 20 volts
Ba = 4 X 10 = 40 volts
&4 = 4 X 12 = 48 volts
If these ooils are now connected in parallel across 120 volts, find the current
in each coil and also the total current.
24 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, v
Ji »= 120/3 = 40 amp.
It = 120/5 = 24 amp.
/, = 120/10 = 12 amp.
I A = 120/12 = 10 amp.
total current / » 86 amp.
31. Voltage Drop in a Transmission Line. — When electric
energy is transmitted from one point to another over wires, a
voltage, called the drop in the line, is required to force the cur
rent through the wires. This voltage = IR where R is the total
resistance of the connecting wires and I is the current flowing,
so that if Eg, Fig. 25, is the voltage at the generating station, and
Er is the voltage applied to the load in the receiving station,
then Eg^ Er + IR.
I
Sff
r
I
FiQ. 25.
The power to be delivered at the end of a 2 mile line is 30 kw. If the
receiver voltage is 600, find the size of wire required to limit the voltage drop
in the line to 5 per cent., find also the power loss in the line.
30 X 1000
Current in line » — = 50 amp.
600
^ The voltage drop in the line ^ 5 per cent, of 600 » 30 volts
The resistance of the wire in the line = 30/50 » 0.6 ohms
The resistance of copper » 0.7 ohms per cir. mil foot at 0° C.
= 9.7 (1 + 0.004 X 25)
— 10.6 ohms per cir. mil foot at 26* C.
The resistance of 2 miles of line or 4 miles of wire
10.6X4X5280 ^^ ^
. , 0.6 ohms
cii, mils
From which cir. mils  370,000
The loss in the line = 30 volts X 50 amp.
» 1500 watts
» 5 per cent, of the power delivered.
The reader will find it of interest to repeat this problem with a receiver
voltage of 1200. The size of wire is inversely proportional to the square
of the voltage and this explains why high voltages are used for long distance
power transmission.
CHAPTER VI
RHEOSTATS AHD RESISTORS
32. Rheostats.— A rheostat is an adjustable resistance of
such a form that it can be conveniently used. In the rheostat
ehowa in F^s. 26 and 27, the resistance ab is tapped at eight
Fig. 27. — Sliding contact t3^ of rheostat.
points which are connected to contact studs s over which the
handle H is free to move.
Such a rheostat is used to control the current in a circuit.
26 PRINCIPLES OF ELECTRICAL ENGINEERINa IChap. yi
When the handle is in the position shown in F^. 26, all the re
sistance ab is in the circuit and the current / has its minimum
value. When the handle is in the position H\, current flows
through the path TcdehQ so that the resistance between a and
e has been cut out. As the handle is moved further over, the
resistance in the circuit is further decreased until finally, when
the handle is in the position H^, the resistance is all cut out and
the current in the circuit has its maximum value.
33. RerastoTS. — ^For economy in manufacture, resistances such
as (d>, Fig. 26, are generally built up of standard resistance
Via. Sa Fio. 29. Fio. 31.
Resistance units.
units called resistors, which may be connected in series or in
parallel as desired. Different types of resistance units are shown
in F^8. 28 to 33.
The unit shown in Fig. 28 consists of a length of wire wound
on an iron tube, from which it is insulated by fireproof insulation
such as asbestos.
In Fig. 29 a similar unit is shown which consists of a length of
wire wound in a spiral groove cut on the surface of a tube of
porcelain or some other such material, adjoinii^ turns of the
wire being thereby separated from one another.
Such units are mounted in frames as shown in Fig. 27. They
are always placed vertically so that air can circulate freely
through the tubes and over the surface of the resistance wire
and thereby keep the temperature of the rheostat within
reasonble limits.
For carrying comparatively small currents, round wire ia
Abt. 35]
RHEOSTATS AND RESISTORS
27
siiitable; strip metal is preferred for larger currents, as it gives
a larger surface for a given section. An excellent type of con
struction is shown in Fig. 32 where the resistance unit consists
of a length of resistance strip metal wound on a frame consisting
of an iron plate A insulated at the edges with porcelain sup
pK)rting pieces B. These units may be mounted on iron rods
which pass through the holes C
34. Heater Units. — Fig. 30 shows the external appearance
of a type of resistor which is largely used for electric irons
and other such heating appliances. It is constructed of re
sistance strip wound in the form of a helix and placed in a metal
tube which is lined with mica, the tube is then packed with fire
Forcelain
m^
Iron
i
FiQ. 32. — Resistance unit.
proof cement to insulate adjacent turns from one another, and
the open end of the tube is closed with a cement plug through
which the leading in wires are brought.
Another type of heater unit is shown in Fig. 31 and consists
of a length of resistance wire wound into a helix of small diameter,
which helix is then coiled into a flat spiral and mounted in a
frame with mica between the convolutions. This unit is held
against a layer of quartz grains which are embedded in enamel on
the bottom of the heater.
36. Castiron Grid Resistance. — When large currents have to be
controlled, the necessary cross section to carry the current and the
necessary radiating surface to dissipate the heat are best obtained
by the use of zigzag units of the shape shown in Fig. 33. For
small rheostats, these zigzag pieces may be punched out of sheet
metal, but for larger sizes they are generally of cast iron as
shown in Fig. 34.
The method or assembling these castings is shown in Fig. 35
which is a plan of a rheostat similar to that in Fig. 34. The units
PRINCIPLES OF ELECTRICAL ENGINEERING [Chap. «
A are mounted on iron rods B which are insulated throughout their
entire length by mica or asbestos tubes C. The individual units
Q U U U U O
FiQ. 33. — Zigzag reaiatatice unit.
Fio. 34. — Caatiron grid resistance.
Fio. 36. — Carbon pile rheostat.
are septmited from one another by washers which are either of
metal as at Z) and E or of insulating materials as at F, depending
Art. 37]
RHEOSTATS AND RESISTANCE
29
on the direction in which it is desired to make the current flow.
The four metal washers E act as terminals from which leads
can be taken to the contacts on the control faceplate.
36. Carbon Pile Rheostat. — An entirely different type of rheo
stat is shown in Fig. 36 and consists of a column of graphite discs
A, enclosed in a steel tube B which is lined with fireproof insula
tion such as asbestos. The resistance of such a pile decreases as
the mechanical pressure between the ends increases, because the
contact between adjacent discs improves. In the type of rheo
stat shown in Fig. 36 the pressure is applied by turning the hand
wheel D and is communicated to the carbon pile through the
plungers E. The two units shown may be connected in series or
in parallel as desired, and the resistance of such a rheostat can be
changed gradually through a total range of about 100 to 1.
Pia. 37. — Liquid rheostat.
37. Liquid Rheostats. — Such a rheostat is shown in Fig. 37
and consists of a castiron trough A which contains a solution of
caustic soda or some similar material which does not attack iron,
and an iron plate B which is insulated from the tank as shown at E
and which dips into the liquid. Between the terminals Tiand Tt
therefore there is the resistance of the path through the liquid
between A and £, and the section of this path can be increased or
decreased by lowering or raising the plate B, The resistance may
be finally short circuited by lowering B far enough to allow the
contact H to close, then current can pass direct from Ti to Ts
without passing through the liquid.
Another type of liquid rheostat is shown in Figs. 38 and 39.
30 PRINCIPLES OF ELECTRICAL ENGINEERING (Chap, vi
In this case the plates are fixed but the level of the liquid is
varied. The pump D sends a continuous stream of liquid from
the cooling chamber A through the resistance chamber B, and
the level of the liquid in this latter chamber may be raised
or lowOTed by a weir C.
Fio. 39. — ^Liquid rheogtat
The liquid is cooled in the lower chamber by water which flows
through cooling pipes. This cooling chamber sometimes takes the
form of a concrete tank made large enough to allow the rheostat
to be self coolii^.
Art. 38] RHEOSTATS AND RESISTORS 31
Liquid rheostats are largely used in making load acceptance
tests on generators. Two electrodes in a barrel of water in which
a handful of common salt has been dissolved will dissipate about 5
kw. if the water is stationary. The type of temporary rheostat
most generally used however consists of a bank of castiron grids
mounted in a wooden frame and placed in running water, the grids
will carry about four times as much current under these conditions
as when air cooled.^
38. The size of a rheostat depends principally on the amount
of power which it is required to dissipate. If two rheostats have
to dissipate the same amount of power but one has only half as
much current flowing as the other then, since the loss in the
rheostat = PR watts, the former rheostat must have four times
the resistance of the latter, that is the wire must have half the
section and twice the length, but the weight of wire and the space
occupied by the rheostat will be approximately the same in each
case.
^ For design data on such temporary rheostats see the Standard Handbook
for Electrical Engineers.
CHAPTER VII
MAGNETIC CmCniTS AND MAGNETIC PROPERTIES
OF IRON
39. Magnetic Field due to a Solenoid. — ^A solenoid is a coil of
wire wound in the form of a helix as shown in Fig. 8, page 5.
When an electric current is passed through such a coil it acts as
an electromagnet and the direction of the magnetic field may be
found by the corkscrew law, page 5.
A Bq.Cm.
T Tuna
Fig. 40. — Closed solenoid.
The solenoid in Fig. 40 has T turns wound on a cardboard spool and is
bent to form an annular ring. A current of t c.g.s. units flowing through
these T turns produces a magnetic field of intensity JC which field can
therefore be represented by JC lines of force.
If a unit pole n be moved once round the magnetic circuit through a dis
tance of 2irr centimeters in a time of i seconds then, since the force on this
pole due to the electromagnet is JC dynes, the work done in moving the
pole » X X 2irr ergs. But a unit pole has 4t lines of force, see page 3,
and while this pole is moved once around the magnetic circuit these lines
cut the T turns of the coil in a time of i seconds and generate in the coil an
e.m.f. e, which in c.g.s. units — ^T/ty the number of lines cut per second.
The coil therefore acts as a generator and supplies an amount of power — ei
ergs per second so long as the unit pole is moving, that is for a time of i
seconds. This power must be obtained at the expense of the power ex
pended in keeping the unit pole moving so that
32
Abt. 41] MAGNETIC PROPERTIES OF IRON 33
5C X 2r r » ei< ergs
= {4^T/t)U
= 4irrt
Ti
therefore X, ^ 4ar y where t is m c.g.s. units and L » 2vr
» jx ¥ where / is in amperes.
In a magnetic circuit such as that shown in Fig. 40 the field
intensity JC is given by the fonnula
^"^ 10 L
where / is the current in amperes
T is the number of turns of the solenoid
TI is called the amperetiuns
L is the length of the magnetic circuit in cm. = 2irr in the
above case
dC is the field intensity in the magnetic circuit and is also
the flux density or the number of lines of force per sq.
cm. of solenoid cross section.
The total magnetic flux threading the magnetic circuit is
_ 4x17
"■ 10 L"^
where A is the cross section of the solenoid in sq. cm.
40. Permeability. — If the solenoid is wound on a core of mag
netic material such as iron or steel it is found that for the same
number of exciting ampereturns a much larger magnetic flux is
produced and that
4t ti
(B, the flux density = ^n ~T~ ^ ^^^^^ P®^ ^' ^°^*
4t ti
0, the magnetic flux = ^ jr A ii lines of magnetic flux.
where /i is a quantity called the permeability of the material and
is equal to unity for air and is greater than imity for magnetic
materials such as iron and steel.
41. Reluctance of a Magnetic Circuit. — ^The above general law
for the magnetic circuit may be expressed in slightly different form
namely
34 PRINCIPLES OF ELECTRICAL ENGINEERING IChap.vh
or m,mj, = ^
where m.m.f. called the magnetomotive force, is that which
4t
produces the magnetic flux and = TFy^^ ampereturns
<f> is the number of lines of magnetic flux in the magnetic circuit
61 called the reluctance of the magnetic circuit = j
From its similarity to the law for the electric circuit, namely
e.m.f. = IR, the above law is sometimes called Ohm's law for
the magnetic circuit.
Since the permeability of iron is much greater than that of air,
the reluctance of an iron path is much lower than that of an air
path of the same dimensions.
18
KlO
8
Iliot*


16
3
12^
^titoa
^
V^
teel
1"
_j
^
y
B
J
/
/
•J
/
/
//
f—
/
r
)n
^
<
;s^
y
i 6
^
^
I*
>
A
/^
2
7
10 20 80 40 60 60 70 80 90 100
Ampere Tami per Cm.
Fig. 41. — Magnetization curves.
42. Magnetization Curves. — The magnetic properties of iron
and steel are generally shown by means of magnetization curves
such as those in Fig. 41; the data from which these curves are
plotted is determined in the foUowing way.
Test pieces of iron are made in the form of an annular ring
with a cross section of A sq. cm. and a mean length of magnetic
path of L cm. These rings are then wound uniformly with T
turns of wire as in Fig. 40 and the flux ^ is measured for different
values of the current / by means of special instruments.
Abt.43]
MAGNETIC PROPERTIES OF IRON
35
The value of <I>/A, the flux density, is then plotted against
corresponding values of TI/L, the ampereturns per unit length
of magnetic path, as shown in Fig. 41, to give what is called the
magnetization curve of the material.
The permeabiUty /x = ^ X ^ may then be determined.
When this value is plotted against flux density, as in Fig. 42,
it may be seen that, once a particular density has been reached,
the permeability decreases rapidly with increase of flux density.
Permeability curves are seldom used in practice, it is found to
be more convenient to work with magnetization curves such as
aooo
1800
1600
1100
gl200
£
1 1000
800
000
400
\
\
\
\
%
%
\
\
\
\
•V
1
fe
\
s^
V
"^
•
X
2 4 6 8 10 12 14
Flux Density In Llnei per 8q. Om.
16 18xl0<
Fig. 42. — Permeability curves.
those shown in Fig. 41. An example of the use of such curves
is given on page 44.
43. Residual Magnetism. — If, after a piece of iron has been
magnetized by means of an exciting coil, the exciting current is
reduced to zero, it will be found that the magnetism has not be
come zero but that some of it, called the residual magnetism,
remains. If the iron is soft and annealed, this residual magnetism
will be of n^ligible amount and the last traces of it may be made
to disappear if the iron is subjected to vibration. If hard tool
steel is used the residual magnetic field will be strong and the
residual magnetism can be removed only with difficulty so that
permanent magnets are generally made of this material.
36 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, to
44. Molecular Theory of Magnetism. — ^To account for the
peculiar magnetic behavior of iron, Ewing suggested that mole«
cules of iron are natural magnets each with its own north and
south pole. When the iron does not exhibit magnetic properties
then the molecular magnets are arranged in groups as shown in
diagram A^ Fig. 43, and their magnetic effects neutralize each
other.
If the iron b placed in a strong magnetic field the molecular
magnets will turn and point in the direction of the field as shown
in diagram B, Fig. 43.
If a piece of iron is placed in an exciting coil, a small current in
this coil will turn these molecular magnets which are not strongly
held together and will line them up in the direction of the mag
Direction
of
A not Magnetised B Macnetlied
FiQ. 43. — ^Arrangement 6f the molecules of an iron bar.
netizing force, these magnets will then add their own magnetic
flux to that which the coil would produce if no iron were present.
As the exciting current is increased, more of these magnets are
lined up until, when the point B has been reached on the curve
in Fig. 41 all but the most rigid of the molecular magnets have
been lined up and the magnetic flux can then increase but little
even for a large increase in the excitation.
When the exciting current is reduced to zero and the magnetiz
ing force thereby removed, the molecular magnets reform into
groups but, on account of molecular friction, they do not return
quite to their original position but have a slight permanent dis
placement in the direction in which they have been magnetized
and this accounts for the residual magnetism.
46. Hysteresis. — There is another phenomenon in connection
with the magnetization of iron which can readily be explained
by the molecular magnet theory, namely, that if the magnetism
of a piece of iron is reversed rapidly the iron becomes hot. What
is called hysteresis energy has to be expended in overcoming the
molecular friction of the magnets and this appears in the form of
heat.
CHAPTER VIII
SOLENOIDS AND ELECTROMAGNETS
16. Pull (^ Solenoids. — A solenoid is a conductor wound in
the fonn of a helix. When an electric current is passed round a
solenoid a magnetic field is produced, the direction of which may
be determined by the corkscrew law, page 5. This field may
be represented by lines of force as shown in diagram A, Fig. 44.
FiQ. 44. — Action of a solenoid.
If loi^ bar magnets are placed in the solenoid field as shown in
diagram B, Fig. 44, then the n pole of magnet x will tend to move
in the direction of the lines of force, see page 2, and be pulled
into the solenoid, while the s pole of magnet y will tend to move
in a direction opposite to that of the fines of force so that it also
tends to move into the solenoid.
If the current hi the solenoid is reversed, the magnetic field of
the solenoid will reverse and the magnets x and y will be repelled.
37
38 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, viii
If as in diagram C, fig. 44, soft iron plungers are used instead
of bar magnets, then the lines of force produced by the solenoid
will pass through the plungers and cause magnetic poles to be
induced; north poles will be formed where the lines of force leave
the iron and south poles where they enter, see page 3. The
induced polarity of the plimgers shown in diagram C is the same
as the polarity of the bar magnets in diagram B so that the
plungers are pulled into the solenoid.
If the current in the solenoid is now reversed, the magnetic
field of the solenoid will reverse but, since the induced polarity
of the plungers will also reverse, the direction of the pull on the
plungers will be unchanged.
47, Electric Hammer. — ^The two types of electric hammer
shown diagrammatically in Figs. 45 and 46 illustrate the action
of a solenoid with a magnet plunger and with a soft iron plunger
respectively.
A B
If 1 3
 J J J J J J 1 — J J J J J J 1
1
FiQ. 45. Fig. 46.
Diagrammatic representation of electric hammers.
In Fig. 45, current passed through coil C makes the iron plunger
into a magnet with the polarity b& shown. If now a current / is
passed through coils A and B in the direction indicated by the
arrows, then the plunger p will be attracted by A and repelled
by B and will move toward the left. If this current / is now r^
versed, the plimger will move in the opposite direction, so that,
by continually reversing the current that flows through A and B,
the plunger p may be made to reciprocate.
Another type of hammer is shown in Fig. 46. The soft iron
plunger is pulled into the position shown when coil A is excited,
while if coil B is excited the plunger is pulled into this latter
coil. By alternately exciting the two coils, the plunger may be
made to reciprocate.
48. Variation of the Pull of a Solenoid. — When the plimger is
in the position shown in A, Fig. 47, the reluctance of the magnetic
Abt. 49]
SOLENOIDS AND ELECTROMAGNETS
39
circuit is large since the path of the lines of flux is nearly all
through air, so that the magnetic field and the plunger poles are
both weak. As the plunger moves toward F, the reluctance of
the magnetic circuit decreases because the amount of iron in the
magnetic path is increasing, so that the magnetic field and the
plunger poles become stronger.
With further motion of the plunger in the same direction, the
reluctance of the magnetic circuit continues to decrease and the
A
Fia. 47. — Pull of a solenoid.
strengths of the magnetic field and of the plunger poles to
increase, but the induced south pole of the plunger now begins
to come under the influence of the solenoid field and is repelled
so that, although the north pole is still attracted, the resultant
pull decreases and finally becomes zero when the plunger is in the
position shown in diagram B; the reluctance of the magnetic
circuit has then its minimimi value.
The pull on the plunger varies with its position as shown in
diagram C; over a considerable range the pull is constant.
49. Circuit Breaker. — The variation in the pu,ll of a solenoid is
taken advantage of in the type of circuit breaker shown dia
graromatically in Fig. 48. Such a circuit breaker consists of the
40 PRINCIPLES OF ELECTRICAL ENGINEERING [CHAP.vra
switch C closed against the force of the spring S and held closed
by the latch d. This latch is released by the plungo* p which is
l^ted when the line current passing round the solenoid M reaches
a predetermined value, the spring S then forces the switch opao.
If the plunger p is moved further into the solenoid by means of
the adjusting screw a then the current required to lift this plunger
will be decreased, by this means the circuit breaker can be
adjusted to open with different currents.
Fio. 49. — Electromagnetic motor.
60. Laws of Magnetic PuU. — The law of inverse squares, art.
2, page 1, applies only to ima^naiy point magnets; in prac
tical work the following laws are applied.
The force on a piece of iron in a magnetic field in air tends to
move the iron in such a direction as to reduce the reluctance of
the magnetic circuit.
The magnitude of this force at any point is proportional to the
space rate of change of the magnetic flux as the iron passes the
given point.
An interesting application of this rule is shown diagrammatic
ally in Fig. 49, The lines of force due to the coils A and B pass
thioi^ the magnetic circuit as shown by the arrows and the
pivoted piece of iron p tends to move until the reluctance of the
magnetic path is a minimum, that is, until the air gaps between
n and s have their minimum value and p is pointing in the
direction ah. If the shape of the curved parts from a to 6 is such
that the magnetic flux ^ increases uniformly with the angle turned
Abt. 51]
SOLENOIDS AND ELECTROMAGNETS
41
through by p then the tummg force, being proportional to the
space rate of change of flux, will be constant over the whole
range of motion.
The above principle is frequently used in toy electromagnetic
motors, provision being made for cutting off the current in the
exciting coils when p approaches close to the position ab and for
switching the current on again when this point is passed.
51. Solenoids with Long and with Short Plungers. — When the
plunger is of the same length as the solehoid, the pull becomes zero
when the plunger is m the position shown m diagram B. Fig.
47, the position of minimum reluctance.
2
Distance X
FiQ. 50. — Pull of a solenoid.
When the plimger is longer than the solenoid, as is generally
the case in practice, the reluctance does not becomje a mini
mum until the plunger projects equally from both ends as
shown in diagram B, Fig. 50, so that the range of the solenoid
is increased, as may be seen by a comparison between the curves
in Pig. 47 and Fig. 50.
62. Ironclad Solenoids. — In order to reduce the reluctance of
the return part of the magnetic circuit and at the same time to
protect the windings, the ironclad construction shown in Fig.
5 1 is used. When the plunger is in the position shown in diagram
A, the reluctance of the magnetic circuit is nearly all in the air
path ab. As the plimger moves toward 6, the flux increases and
changes very rapidly toward the end of the stroke so that, while
42 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, viii
the average pull is not much higher than that of the same solenoid
with an air return path, a large pull over a short distance is ob
^Imple Ironclad Type
B
Ouahlon IVpe
Stopped Ironclad Type
FiQ. 51. — ^Types of ironclad electromagnet.
tained at the end of the stroke as shown in curve B, Fig. 62.^
If a hole for the plunger be bored through the iron cover as
4 5 6 7 8
Distance x In Inches
10 U 12
Fig. 52. — Pull of electromagnets.
shown in diagram B, Fig. 51, then there is no sudden jar at the end
of the stroke but rather a cushion effect; the large increase of pull
^Taken from an article by Underbill, Electrical World and Engineer,
Vol. 45, p. 934 (1905.)
Art. 531 SOLENOIDS AND ELECTROMAGNETS
43
at the end of the stroke b lost however, although this is seldom
a disadvantage.
Fig. 53 shows a series of test curves on ironclad magnets of
the cushion type and will give the reader some idea of the magni
tude and range of pull that can be obtained.
Fia. 53. — Pull of cushion type of electromagnet.
S3. Lifting and holding magnets are generally of the horse
shoe or of the annular type shown diagrammatically in F^s. 54
and 55. As the iron to be lifted moves from a tab, Fig. 55, there
ta
I
is little change in the flux threading the coil M and therefore only
a small pull; when the iron approaches close to th' poJes of the
magnet, however, the flux increases rapidly and the pull, beii^
proportional to the space rate of change of flux, becomes large.
44 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, vui
For such magnets the holdmg power may be determined very
closely by Maxwell's formula
pull m dynes =
where (B is the flux density across the contact surface in lines per
sq. cm.
A is the total pole face area in sq. cm.
In the case of the magnet shown in Fig. 56, the scale on the iron to be lifted
is assumed to be 0.05 cm. thick, it is required to determine how the pull varies
with the exciting current, the number of turns bemg 1000.
To solve this problem it is necessary to assume different values for the
total flux in the magnetic circuit then calculate the pull by the use of the
above formula and the excitation by the use of the curves in Fig. 41, page 34.
WM
kUBlMl
"C—J
^
,OMt Iroa
1.0 2.0 3.0
Amperei
Fig. 56. — Full of a horseshoe magnet.
lif the length of the cast steel path ^ '20 cm.
li, the length of each air gap = 0.05 cm.
li, the length of the cast iron path = 12 cm.
Aif the cross section of the cast steel path == 2 X 4 » 8 sq. cm.
Ai, the cross section of each air path == 8 sq. cm.
A s, the cross section of the cast iron path = 4 X 5 = 20 sq. cm.
If 0, the total flux = 80,000 lines
then CBi, the flux density in the cast steel » 10,000 lines per sq. cm.
(Ba, the flux density in the air gaps = 10,000 lines per sq. cm.
(Bs, the flux density in the cast iron — 4,000 Hues per sq. cm.
and the ampere turns per cm. for the cast steel » 7, see Fig. 41.
*u * * *u • CB2XIO ^„
the ampere turns per cm. for the air gaps = — j — , see page 33.
= 8000
SOLENOIDS AND ELBCTROMAONBTS
the ampere tume per cm. Cor the cast iron = 13, see I^lg. 41
and the total ampere turns = 7 X 20. + 8000 X 2 X 0.05 + 13 X 12
= 140 + 800 + 156
Fig, 57. — Annular type of electromagnet.
(10,000)' X 2 X S
the magnetic pull = 
= 64,000,000 dynes
 65,000 gm.
65 kg.
46 PRINCIPLES OF ELECTRICAL ENOINEERINO [Chap, vm
Other values are worked out in the same way, the work generally being
carried out in tabular form as below:
^
Flux density
Ampere turns per
centimeter
Total ampere turns
Am
peres
Pull
Steel
Air
Iron
Steel
Air
Iron
Steel
Air
Iron
Cir
cuit
64,000
80.000
•96,000
112.000
128,000
8.000
10.000
12,000
14.000
16,000
8,000
10,000
12.000
14.000
16,000
3,200
4,000
4.800
6,600
6.400
5
7
11
19
65
6.400
8,000
9.600
11,200
12,800
9
13
19
26
34
100
140
222
380
1,100
640
800
960
1,120
1,280
108
166
228
300
408
848
1,096
1,408
1,800
2,788
0.848
1.096
1.408
1.800
2.788
42 kg.
65 kg.
94 kg.
126 kg.
164 kg.
These results are plotted in Fig. 56.
The possibilities of the annular type of magnet are illustrated
in Fig. 57. A magnet which weighs 2250 lb. will lift skull cracker
balls up to 12,000 lb., billets and slabs up to 20,000 lb. and mis
cellaneous scrap up to 500 lb. The power required to operate the
magnet being 11 amp. at 220 volts or 2.42 kw.
64. Saturation of a Magnetic Circuit. — ^From the figures in the
last problem, under the heading of total ampere turns, it may be
noted that, when the flux densities in the steel and iron are low,
most of the excitation is required for the air path or most of the
reluctance of the magneiiic circuit is in the air gap.
When the densities exceed 15,000 lines per sq. cm. for cast steel
and 6000 for cast iron, the reluctance of the magnetic circuit
increases rapidly and the curve showing the relation between flux
and excitation, called the magnetization curve of the circuit,
bends over rapidly as shown in Fig. 56, the circuit is then said to
be nearly saturated.
55. Electromagnetic Brakes and Clutches. — One type of brake
used on crane motors is shown in Fig. 58. The annular steel
frame A of the electromagnet is fastened to the housing of the
motor and carries the exciting coil E. The sliding disc B is
fastened to the frame A of the magnet by means of a sliding key
F and is free to move axially but cannot rotate.
When the motor is disconnected, the magnet is not excited and
the springs S push the disc B mto the ring C which is keyed to the
motor shaft, the motor is thereby braked and brought rapidly to
rest. When current is applied to start the motor, the coil E is
excited at the same time and the disc B is attracted, releasing the
ring C, so that the motor shaft is then free to rotate. Electro
magnetic clutches are built on the same principle.
SOLENOIDS AND ELECTROMAGNSTS
66. Magnetic Separator. — A useful application of the electro
magnet is shown diagrammatically ia Fig. 59. The magnetic
Fta. 58. — Electromagnetic brake,
pulley consists of an iron shell containii^ an exciting coil C which
produces the magnetic field shown. Any iron particles carried
over this pulley by the conveyer belt are attracted and are
FiQ. 59. — Magnetic aeparator.
therefore carried further round than the nonmagnetic materials
with which they are mixed.
CHAPTER IX
ARMATURE WINDINGS FOR DIRECTCURRENT
MACHINERY
67. Principle of Operation of the Electric Generator. — The
simplest type of electric generator is shown diagrammatically in
Fig. 60. If the conductor ab is moved alternately up and down so
as to cut the lines of force that pass from N to S, an e.m.f . will
be generated or induced in the conductor which will cause an
electric current to flow in the closed circuit abed.
The direction of the current in the conductor ab may be deter
mined by the righthand rule, page 9. The current will reverse
Fig. 60. — Generation of electromotive force.
when the direction of motion of the conductor is reversed, so that
the current will flow first in one direction and then in the other;
such a current is said to be alternating.
68, Gramme Ring Winding. — The following are the stages in
the development of the above machine into one that will give a
direct current, that is a current which flows continuously in
one direction. The poles are bored out as shown in Fig. 61 and
an iron core is placed concentric with the pole faces so as to re
duce the reluctance of the magnetic circuit. The several con
ductors c mounted on this core rotate with it and cut the lines
of force that pass from iV to S so that e.m.fs. are generated in
these conductors the direction of which, determined by the right
hand rule, is shown in Fig. 61 at a particular instant.
The conductors are now connected together as shown in Fig.
62 to form an endless helix. Since the lines of force pass through
48
Art. 60] ABMATVBS WJNDINQS 51
coils of cotton covered wire. The commutator segments being
part of the winding must also be insulated from one another
and from the iron shell which supports them, the construction
used ia shown in Fig, SOi
Fia, 66. — Armature with a Graname ring winding.
60. Multipolar Windings. — It has been found economical in
practice to build machines with more than two poles, see page
56, the poles being arranged in pairs alternately N and S. In
Fig. 68 the winding for a fourpole machine is shown diagrammat
Osmplele WluUiia
Fio. 68.
Fig. 69.
Gramme ring winding for a four pole raachioe.
ically. The direction of the lines of force and the e.m.f. in the
conductors is shown in Fig. 67, from which diagram it may be
seen that no current can Sow in the closed winding because the
voltages in the conductors under the N poles are opposed by
52 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, nt
equal voltages in the conductors under the S poles. A difference
of potential however will be found between a and 6 due to the
conductors cutting lines of force under pole iSi and there is an
equal difTerence of potential between a and d due to the conduct
ors cutting lines of force under pole Nt so that b and d are at the
same potential and may be connected together. For the same
reason the stationary contacts a and c may also be connected
together as shown in Pig. 68. The external circuit to be supplied
with current is connected between the terminals 7"+ and T.
This current will divide when it enters the machine and pass
Pig. 70. Flo. 71.
through the four paths in the winding as shown in Fig. 68 and
also di^rammatically in Fig. 69.
61. Drum Windings. — One obvious objection to the ring wind
ing, as shown in Fig. 70, is that only the outer conductors 1, 3,
5, etc., cut lines of force, the remainder of the winding being in
active. If the coils were stretched and then placed on the core
as in Fig, 71 there would be two active conductors per coil and
the voltages induced therein would act in the same direction
around the coil so that the total voltage of the machine would
be doubled without any great increase in the amount of copper
used. Since the end connections of this winding have to cross
over one another as shown in Fig. 71 it is usual to arrange that
the winding lie on the armature surface in two layers, the even
Art. 611 ARMATURE WINDINGS 63
numbered conductors being placed below the odd numbered
conductors instead of alongside.
Fia. 72 — Lap or multiple winding,
A perspective drawing such as that in Fig. 71 showB only a
small part of the winding and it does not show the paths through
Rq. 73. — Wave or twocircuit winding.
the armature at all clearly. A special type oi drawing shown in
Fig. 72 is obtained by supposing the armature to be placed in a
54 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, tx
press and the commutator end of the machine forced back into
the winding until the whole machine has been squeezed flat onto
a plane, the rear end of the machine expanding and the commu
tator end contracting in the proceSB,
Fio. 75.— Lap winding.
There is no reason why conductor 3 should not be placed under
pole Ni, Fig. 72, instead of under Nj. If this be done and the wind
ing completed as shown in Fig. 73 what is called a wave winding
will result. Fig. 72 represents what is called a lap winding.
Fm. 76. — Wave winding.
The position of the brushes for these two types of winding
may be found as follows. The direction of the e.m.f. in each
conductor is shown by an arrow and brushes are placed at points
where the current tends to leave the winding and at points where
ARMATURE WINDINGS
55
it tends to enter. It will then be found that in the lap winding,
Fig. 72, the number of parallel paths through the winding is the
same aa the number of poles whereas in the wave winding. Fig,
73, there are only two paths in parallel. Complete drum windings
are shown in Figs. 75 and 76.
61a. The E.m.f . Equation. — ^Let ^ be the m^netic flux per pole
Z the number of face conductors
then theaverage voltage perconductor = (lineacutpersec.) X10~*
= <>XpoleaX^X10s
Since the number of conductors in series between a positive and
a negative brush is equal to ^/(paths through winding) therefore
the voltage generated between the terminals is equal to
X^d. 77. — ^dy currents in a. solid armature
By a suitable choice of the constants in this equation the de
signer is able to'wind armatures for different voltages. It may
be noted, however, that although all the coils shown in the diagrams
have only one turn between adjacent commutator segments it is
generally nece^ary to make the coils with severaltuma betweenseg
ment3 in order that the voltages used in practice may be attained.
62. Lamination of the Armature Core. — Fig. 77 shows an
armattu^ core on which may be placed either a ring or a drum
winding. If this core is made of a solid block of iron, then, as
it rotates, e.m.fs. are induced in the surface layers and force
current through the iron in the direction shown, which direction
may be determined by the righthand rule. These currents can
not be collected and utilized but power is required to maintain them.
To keep these eddy currents small, a high resistance is placed
in their path by laminating the core as in Fig. 7S, the laminatiooa
beii^ separated from one another by varnish.
CHAPTER X
CONSTRUCTION AND EXCITATION OF DIRECT
CURRENT MACHINES
63. Multipolar Construction. — ^Fig. 79 shows a two^pole ma
chine and also a sixpole machine built for the same output^ the
machines having the same armature diameter and the same total
number of lines of force crossing the air gaps. The armature core
of the twopole machine must be deep enough to carry half of the
total flux, while in the sixpole machine the total flux divides up
among six paths so that the core need be only onethird of the
Twopole machine. Sixpole machine.
Fig. 79. — Machines with the same output.
depth of that of the twopole machine. For the same reason the
sixpole machine has the smaller cross section of yoke.
By the use of the multipolar construction therefore there is a
considerable saving in material, but this is at the expense of an
increase in the cost of labor because of the increased number of
parts to be machined and handled. The number of poles is
chosen by the designer to give the cheapest machine that will
operate satisfactorily.
56
CONSTRUCTION AND EXCITATION
58 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.x
64. Armature Construction. — Fig. 80 shows the type of con
struction generally adopted for small machines. The armature
core M is built up of sheet steel laminations which are separated
from one another by layers of varnish, see page 55.
The winding shown is of the drum type, see Fig. 75, page 54,
and the armature coils G are carried in slots F from which they
are insulated by paper, cotton and mica. It is found that, even
when embedded in slots, the conductors cut the lines of force
crossing from pole to pole.
The core is divided into sections by spacers P, so that air can
circulate freely through the machine and keep it cool. The core
laminations and the spacers P are clamped between end heads N
which carry coil supports L attached by arms shaped like fans.
The coils are held against these supports by steel band wires W.
65. Commutator. — The commutator is built of segments J,
see page 50, which are of harddrawn copper. These seg
ments are separated from one another by mica strips and are then
clamped between two cones S from which they are separated by
mica, the segments being thereby insulated from one another
and from the frame of the machine. The segments are connected
to the winding through the leads H which, in modem machines,
have air spaces between one another as shown, so that air is
drawn across the commutator and between the leads thereby
keeping the commutator cool.
66. The brushes, see page 50, are attached to the studs X,
which studs are insulated from the supporting arm F, and con
nection is made from these studs to the external circuit.
67. Poles and Yoke. — The armature revolves in the magnetic
field produced by the exciting or field coils A which are wound on
the poles B. These poles must have sufficient cross section to
carry the magnetic flux without the flux density becoming too
high, the same applies to the cast steel yoke C to which the poles
are attached by screws.
68. Large generators are similar to small generators such as
that described above; some changes are generally required in the
mechanical design because of the heavier parts to be supported
and also because of the different kinds of service for which ma
chines have to be built. In the case of the engine type generator
shown in Fig. 81 for example, the armature core is built up of
segments instead of complete rings, while the commutator is
Abt. 69]
CONSTRUCTION AND EXCITATION
59
supported from the armature spider since the shaft is supplied by
the engine builder.
Gommatatoi
Fig. 81. — ^Large directcurrent generator.
Fio. 82. — Separately excited Fig. 83. — Shunt excited
macnine. machine.
69. Excitation. — Permanent magnets are used as field poles
for small machines called magnetos; large machines are supplied
60
PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, x
with electromagnets the excitation of which can readily be
controlled.
When the generator itself supplies this exciting current it is said
to be self excited; when the exciting current is supplied from
some external source the machine is said to be separately excited.
The different connections used are shown in Figs. 82, 83, 84
and 85.
Fig. 82 shows a separately excited machine.
Fig. 83 shows a shunt machine in which the field coils form a
Short Shunt
Or:
Fig 84. — Series excited
machine.
Fig. 86. — Compound excited
machine.
shunt across the armature terminals and have many turns of small
wire carrying a current 1/ = Et/R/f the terminal voltage divided
by the resistance of the field coil circuit. This exciting current
seldom exceeds 5 per cent, of the fullload current supplied to the
external circuit.
Fig. 84 shows a series machine in which the field coils are in
series with the armature and have only about 5 per cent, of the
number of turns that a shunt winding would have, but employ a
larger size of wire because they have to carry the total current of
the machine.
Art. 69] CONSTRUCTION AND EXCITATION 61
Fig. 85 shows a compound machine in which there are both
shunt and series field coils. When the shunt coils are connected
outside of the series coils the machine is said to have a long shunt
connection; when connected inside of the series coils the connec
tion is said to be short shunt.
CHAPTER XI
THEORY OF COMMUTATION
70. Commutation. — As the armature of a directcurrent genera
tor revolves, the du'ection of the current in each conductor
changes while that conductor passes from one pole to that
adjoining. In Fig. 86 for example, the direction of the current
in the coil M is shown at three consecutive instants in diagrams
A, B and C
As the armature moves from A to C the brush changes from seg
Fia. 86. — Diagram showing the reversal of the current in coil M.
ment 1 to segment 2 and the current in the coil M is automatically
reversed. For a short period, as at B, the brush is in contact with
both segments and, during this interval of time, the coil M is short
circuited, but no e.m.f. is generated in the coil since it is not
cutting lines of force, so that no current passes through the short
circuit.
When in position By the coil is said to be in the neutral position
and the line L is called the neutral line.
The operation of reversing the current in an armature coil by
62
Art. 71] THEORY OF COMMUTATION 63
means of the brush and commutator segments is called commuta
tion. Unfortunately the operation is not so simple as described
above, because the coils have self induction and resist a change of
current, and this we shall see causes sparking and gradual deteri
oration of the brushes and commutator. It is therefore neces
sary to make a detailed study of the subject because of its
importance.
To study the variation of the current in the coil being com
mutated, the student should draw the brushes B^ and B^ and
also the poles N and S, Fig. 87, on a piece of heavy paper, and the
armature and commutator on tracing paper. The armature
should then be placed in the magnetic field and the direction
of the current in a particular coil noted as the armature
goes through one revolution. Such a model illustrates the
operation much better than any set of diagrams such as those
in Figs. 88, 89 and 90.
71. Theory of Commutation. — ^Fig. 88 shows part of a machine
with a ring winding having two turns per coil and with the current
in the coil M undergoing commutation. The brush B is made of
copper so that the resistance of the contact between the brush and
the commutator is negligible.
In diagram A, the currents I enter the brush through the com
mutator lead a.
In diagram B, the brush makes contact with two segments and
the* current flowing to the brush through the coils under the S
pole no longer requires to flow round coil M because it has an
easier path through the lead 6, the current in coil M therefore dies
down to zero because, being in the neutral position, the coil M is
not cutting lines of force so that no e.m.f. is generated in it to
maintain the current.
In diagram D, segment 1 of the commutator is about to break
contact with the brush, and the coil M carrying no current is
about to be thrown in series with the coils under the N pole.
At the instant the contact is broken, as shown in diagram E, the
current in coil M tries to increase suddenly from zero to a value
7, but this change of current is opposed by the self induction of
the coil Af , so that the current prefers to pass to the brush across
the air space a;, causing sparking.
72. Shifting of the Brushes. — For sparkless commutation, it
is necessary that the current in the coil M shall be reduced to zero
and then, by some means or other, raised to a value I in the oppo
64 PRINCIPLES OF ELECTRICAL BNOINEERINO IChap.w
Fig. 88. Fio. SO. Fia. 00.
Fia. 88. — With low lesistance brushes on neutral lioe.
Fio. 89. — With low resistance bruehea shifted in the direction lA motioo.
Via. 90. — With high resistance brushes.
Stssea in the process of commutation.
Abt. 73]
THEORY OF COMMUTATION
65
site direction during the time the coil is short circuited by the
brush, so that when the contact at x is broken, there shall be no
sudden change of current in the coil.
This result may be obtained by shifting the brushes forward
in the* direction of motion as shown in Fig. 89 so that, while coil
M is short circuited, it is in a magnetic field and an e.m.f. is gen
erated in it which will produce the required growth of current.
This magnetic field is called the reversing field. The correspond
ing diagrams in Figs. 88 and 89, and particularly diagrams D
and Ey should be carefully compared.
If the current taken from the generator is increased, the
strength of the reversing field must also be increased if commuta
tion is to be sparkless, so that the brushes must be moved nearer
to the pole tips and further from the noload position. The
brush position must therefore be changed with change of load.
73. Inteipole Machines. — It has been shown in the last para
graph that the commutation of a generator is improved if the
Fig. 91. Fia. 92.
Diagrams illustrating the principle of the interpole generator.
brushes are shifted forward in the direction of motion of the
machine so that the short circuited coils are in a reversing mag
netic field, thus in Fig. 91 the brush B+ is moved so as to come
luxder the tip of the N pole and the brush jB_ is moved so as to
come under the tip of the S pole. The same result may be accom
plished by leaving the brushes in the neutral position and
bringing an auxiliary n pole over the brush B^ and an auxiliary «
pole over the brush jB, as shown in Fig. 92. These auxiliary
poles are called interpoles and a machine so equipped is called an
interpole machine.
It is desirable that the strength of the reversing field increase
with the current drawn from the armature and to obtain this
66 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xi
result the interpoles are supplied with series field coils as shown
m Fig. 92.
74. Carbon brushes have a contact resistance which is gener
ally about ten times that of copper brushes. The effect of this
high contact resistance is to improve commutation, as may be
seen by a comparison between Figs. 88 and 90.
In diagram B, Fig. 90, the brush makes contact with segment 2
and some of the current / that was flowing round coil M now
flows directly to the brush through lead 6. Since however the
contact with segment 2 is small in area, the current ib flowing
through this contact is small and some of the current / continues
to flow around coil M.
As the armature rotates and the contact area between the
brush and segment 2 increases, the current 4 increases and that
in coil M decreases until, at the instant shown in diagram C,
the current in this coil has become zero.
In diagram D, the contact area between the brush and segment 1
is small while that between segment 2 and the brush is large so
that the current in lead a is throttled by the high resistance of the
small contact area and current is forced around coil M in the
direction shown. As the contact area between the brush and
segment 1 decreases, the current ia decreases and that in coil M
increases until, at the instant shown in diagram E, when the
contact at a: is broken, the current in coil M has been raised to the
value / by this slide valve action of the high resistance brush.
The contact can then be broken without causing any sudden
change of current in the coil M and therefore without sparking.
In the above theory, the action at the positive brush has been
considered; the action at the negative brush is similar and need
not be considered separately. The theory applies to a drum
winding as well as to a ring winding, the only difference between
the two cases being in the shape of the coil, see Fig. 71.
By the use of carbon brushes it is possible to operate generators
from noload to fullload without shifting of the brushes during
operation, and for that reason carbon brushes have superseded
copper brushes on modern machines.
CHAPTER XII
ARMATURE REACTION
76. The Crossmagnetizing Effect — In Fig. 93, diagram A
shows the distribution of the magnetic flux in a twopole machine
when the field coils are excited and no current is flowing in the
armature winding; the flux density is uniform under the pole face
so that the same number of lines of force cross each square centi
meter of the air gap between the pole face and the armature
surface.
Diagram B shows the distribution of magnetic flux when the
armature is carrying current, the brushes being in the neutral
N
4 ' B C
Mux distribution due Flux distribution due to the
to the field coils armature winding
Fia. 93. — Armature reaction with the brushes in the neutral position.
Resultant flux
distribution
position and the field coils not excited. The ciurent passing
downward in the conductors under the S pole of the machine and
up in those which are under the N pole causes the armature to
become an electromagnet with lines of force which pass through
the armature in a direction determined by the corkscrew
law, see page 6, and which return across the pole faces to com
plete the circuit.
Diagram C shows the resultant distribution of magnetic
flux when, as imder load conditions, the armature is carry
ing current and the field coils are excited; C is obtained by
combining the magnetic fields of A and B. Under pole tips a and
c the magnetic field due to the current in the armature is opposite
67
68 PRINCIPLES OF ELECTRICAL ENGINEERING [Ch4p. hi
in direction to that due to the current in the field coils while under
tips h and d the two magnetic fields are in the same direction.
Since the armature magnetic field ia at right angles to that pro
duced by the field magnets, the effect produced is called the crosa
magnetizing eSect of armature reaction.
76. The Deuugnetiziiig Efitect. — In a direct current generator
the brushes are shifted from the noload neutral in the direction of
motion so as to improve conmiutation, see page 68, the distribu
tion of the magnetic flux when the armature is canyii^ current
and the field coils are not excited will then be as shown in diagram
A, Fig. 94. The armature field is no longer at right angles to that
produced by the field magnets but acts In the direction oz, it
may however be considered as the resultant of two magnetic fields,
one in the direction oy, called the crossmagnetizing component
and the other in the direction ox, called the demagnetizing com
ponent because it is directly opposed to the field produced by the
field magnets. Diagram 6, Fig. 94, shows the armature divided
80 as to produce these two components; the belts of conductors ab
and cd, when carrying current, tend to demagnetize the machine,
while the belts ad and 6c are crossmagnetizii^ in effect.
77. Effect of Armature Reaction on Commutation. — When
interpoles are not supplied, the brushes are shifted forward in the
direction of motion so that commutation takes place in a revering
magnetic field under pole tips a and c, Fig. 93. But it may be
seen from diagram C, Fig. 93, that the effect of armature reaction
is to weaken the magnetic field under these pole tips and so impair
the commutation.
This effect must be minimized by making the air gap clear
ances 5 as large as possible so that there is a large reluctance in the
path of the cross field. Increasing the air gap also increases
Abt. 77] ARMATURE REACTION 69
the reluctance of the main magnetic path and, in order to pro
duce the required main flux, it is then necessary to increase
the number of exciting ampere turns on the poles. The machine
is then said to have a stiff magnetic field because it is not greatly
affected by armature reaction.
CHAPTER Xni
CHARACTERISTICS OF DIRECTCURRENT GENERATORS
78. Magnetization or Noload Saturation Curve.— The voltage
generated in the armature of a directcurrent machine, being
proportional to the rate of cutting lines of force, is proportional to
the speed and to the flux per pole or
^ = a const. X X r.p.m. for a given machine.
The flux per pole, and therefore the voltage, increase with the
excitation, and the curve showing the relation between noload
voltage and excitation, the speed being constant, is called the mag
netization or the noload saturation curve of the machine. Such
O o. Exciting Current If
Fig. 95. — Magnetization curve of a direct current generator.
a curve is shown in Fig. 95. With no excitation there is a voltage
Cr due to' residual magnetism; as the exciting current is increased,
the flux per pole and the voltage increase in the same ratio until^
with an exciting current of oa amperes the magnetic circuit begins
to saturate and the voltage to increase more slowly.
To obtain such a curve experimentally, the generator is driven
at a constant speed with no connected load. The field coils
are separately excited as shown in Fig. 95 and the exciting cur
rent is increased by gradually cutting out the resistance n
70
Art. 80]
DIRECTCURRENT GENERATORS
71
Simultaneous readings of the voltage Eo and the current // are
taken and the results plotted as shown.
79. Self excitation is made possible by virtue of the residual
magnetism in the magnetic circuit of the machine. If for example
a shunt generator, connected as in Fig. 96, is rotating, a small vol
tage Br is generated in the armature even with no exciting ciurent
in the field coils, because the lines of force of residual magnetism
are being cut. This small voltage sends a small current through
the field coils, which increases the magnetic flux and causes the
generated voltage to increase, and this in turn further increases
the excitation, and so the voltage of the machine builds up.
The voltage and the exciting current cannot build up indefi
nitely because, as the exciting current increases, the magnetic
Exciting Current If
Pig. 96 — Magnetization curve of a shunt generator.
circuit becomes more nearly saturated and the voltage increases
by a smaller and smaller amount until finally, when the point A is
reached at which Eo/If = J?/, the voltage and exciting current
can increase no further.
It frequently happens that, when a generator is started up for
the first time, the e.m.f. generated in the armature due to residual
magnetism sends a current through the field coils in such a direc
tion as to oppose the residual flux, and the voltage, instead of
building up, is reduced to zero. In such a case it is necessary
to reverse the connections of the field coils so as to pass current
through them in the opposite direction.
80. Regulation Curve of a Separately Excited or of a Magneto
Generator. — ^This curve, sometimes called the external character
72 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xni
aj
istic, gives the relation between Et, the terminal voltage, and J,
the line current, and is shown in Fig. 97 for the case of a separately
excited machine operating at constant speed and with constant
excitation. The terminal voltage drops as the current taken from
the machine is increased because:
a. The flux per pole is reduced by armature reaction, see page
68; so that Egy the e.m.f. generated by cutting this flux, is also
reduced.
b. The terminal voltage Et is less than the generated voltage
Eg by the armature resistance drop laRa, that part of the gener
ated voltage required to force the armature current through the
resistance of the armature winding and of the brush contacts.
X
T^I^Crop Dae to Armatore Beftctioa
Ltmatnre Beaiitance Drop /^ Rq
If It
Line Carrent Jq
FiQ. 97. — Regulation curve of a separately excited generator.
To obtain such a curve experimentally, the generator is loaded
on a bank of lamps, or some other suitable load that can readily be
adjusted, as shown in Fig. 97. The speed and the exciting cur
rent If are kept constant, while the current taken from the
machine is gradually increased by connecting an increasing num
ber of lamps in parallel across the terminals, that is by providing
more paths through which current can pass. Simultaneous
readings of the voltage Et and of the current la are taken and the
results plotted as in Fig. 97.
The regulation of the above generator is defined as the per cent.
change in voltage when fullload is thrown off the machine, the
speed and the field circuit being unchanged. The regulation
therefore = {Eo  E^/Et.
81. Regulation Curve of a Shunt Generator. — ^This curve is
shown in Fig. 98 for a constant speed shunt excited generator.
Abt. 81]
DIRECTCURRENT GENERATORS
73
The terminal voltage drops as the current taken from the machine
is increased because :
a. The flux per pole is reduced by armature reaction.
6. The armature drop laRa is used up in the machine itself.
c. The exciting current 7/ is equal to Et/R/, where R/ is the
constant resistance of the shunt field circuit, so that as the ter«
minal voltage drops the exciting current decreases and causes the
voltage to drop still further. Because of this third effect the
terminal voltage of a generator with a given load will be lower
when the machine is shunt excited than when separately excited.
4!!tr"Drop Dae to Armatare
K fieaction
'Armatare Beilitance
Drop la Ra
Drop Due to Decreaie
in Excitation
Szternal
Circait
Line Current /; ^m
FiQ. 98. — ^Regulation curve of a shunt generator.
To obtain such a curve experimentally the machine is connected
up as shown. The speed and the resistance of the shunt circuit
are kept constant while the current taken from the machine is
gradually increased, and simultaneous readings are taken of the
voltage E^ and the current Ii, these results are plotted as in
Pig. 98.
As the resistance of the external circuit is decreased, the current
supplied by the machine increases and the terminal voltage
dropNS until point d is reached. A further reduction in the external
resistance allows an increased current to flow for an instant, but
this increase of current reacts by armature reaction and causes
such a large drop in voltage and in exciting current that the arma
ture current cannot be maintained. In the extreme case when
the generator is short circuited, that is, the terminals of the ma
chine are connected through a circuit of negUgible resistance, then
the terminal voltage must be zero and there can be no field excita
74 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xiu
tion, so that the only current that can flow in the short circuit is
that produced by the voltage due to the residual magnetism,
thus c, Fig. 98, is a point on load curve.
If a shunt generator is short circuited, it carries the maximum
current Jm for only a short interval of time before it loses its
voltage and is thereby protected from injury. For the same
reason a shunt generator will not build up if a very low resistance
is connected across its terminals. Because of this self protecting
power, shunt generators are used with advantage for electric fuiv
nace and other work during the process of which the machine is
liable to be short circuited.
82. To maintain the terminal voltage constant, shunt gen
erators are operated with an adjustable resistance, called a field
OT«r Compoood
Generator
Tlat Compoand
Oanaratox
LlneCarrent Jt
Fig. 99. — Regulation curves of compound generators.
rheostat, placed in the field coil circuit. As the load on the
machine increases and the voltage drops, some of this resistance
may be cut out either automatically or by hand so as to increase
the excitation. Automatic regulators for this purpose are used
with alternatingcurrent generators, see page 249, but are sel
dom used with directcurrent generators because the same result
may be obtained more cheaply by the use of compound windings.
83. Compound generators, operated without a r^ulator,
maintain the terminal voltage approximately constant from no
load to fullload, because the line current passes through the series
field coils and causes the total excitation to increase with the
load. By the use of a large number of series turns, the total
excitation may increase so much with the load that the terminal
voltage will rise, as shown in curve B Fig. 99, the machine is then
Abt. 84]
DIRECTCURRENT GENERATORS
75
said to be overcompounded. When the terminal voltage has the
same value at fullload as at noload, the machine is said to be fiat
compounded.
Generators for lighting and power service are generally flat
compound machines wound for 126 or for 260 volts. For railway
service the generators are overcompounded so as to maintain the
trolley voltage at some distance from the power bourse. Street
railway generators are invariably wound for 600 volts at full
load, while for interurban and trunk line work 2400 volts has been
used.
84. The Regulation Curve of a Series Generator. — Curve A,
Fig. 100, shows what the relation between voltage and current
LnRrr<>l
Drop Dne to Armature
Reaction
Armature Beiiitance
Drop
9^^
Oo&nectlon for Oarreul
Coanection tor Oarre B
Line Current Jq
Fig. 100. — ^Regulation curve of a series generator.
in a series generator would be if armature resistance and armature
reaction were negligible; the voltage would increase with the load
current since this is also the exciting current. Curve A is really
the noload saturation curve of the machine and is determined by
separately exciting the field coils, as shown in diagram A, so that
no current fiows in the armature. Curve B shows the actual rela
tion between terminal voltage and load current; the drop of
voltage between curves A and B consists of the portion due to the
reduction in the fiux per pole caused by armature reaction, and
laBa the drop of voltage in the armature winding, brush contacts
and series field coils.
Series generators are never found in modern stations but were
formerly used as constantcurrent generators for the operation
of arc lamps in series. They were operated with automatic
regulators so as to have the line ab, Fig. 100, nearly vertical, and
the current practically constant for all voltages up to Em*
7
76 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xiii
A very simple type of regulator for this purpose is shown diar
grammatically in Fig. 101, where C is a carbon pile rheostat and
B is a solwioid carrying the line current. If the current in the
external circuit increases, the pull of the solenoid B also increases
and the carbon pile is compressed and its resistance thereby
decreased, see page 29, so that it shunts more of the current from
the series field coils. The flux in the machine is therefore reduced,
and the voltage drops until the current in the line reaches the
value for which the pull of the solenoid was adjusted.
Oarbon File
rOta^
Pig. 101. — Automatic regulator for a constantcurrent generator.
86. Problem on Generator Characteristics. — a. A directcurrent shunt
generator was tested with the brushes shifted forward in the direction of
motion. The voltage drop between noload and fidlload was 6 volts, what
are the causes of this drop in voltage?
b. If the brushes had been placed on the neutral position, why would the
voltage drop have been different and what would you expect its value to be.
Why are the brushes not placed on the neutral position in noninterpole
machines?
c. Series field coils were added to the machine and the voltage dropped 20
volts from noload to fullload, what was the cause of thjs excessive drop?
d. After the series field coil circuit has been fixed, the voltage was found
to increase by 6 volts from noload to fullload while flat compounding was
desired, what changes would you suggest should now be made?
a. The voltage drop is due to:
1. The reduction in the flux per pole due to the demagnetizing effect of
armature reaction, since the brushes are shifted forward.
2. The armature resistance drop.
3. The reduction in the exciting current which causes the flux per pole to
decrease and the voltage to drop still further, see page 73.
b. When the brushes are placed in the neutral position, the armature reao
tion has no demagnetizing effect, see page 68, so that the voltage drop will be
less than when the brushes are shifted forward, and will probably not exceed
4 volts.
Aat. 85]
DIRECTCURRENT GENERATORS
77
Series Shant
Machines have the brushes shifted forward in order to improve com
mutation and, unless interpoles are supplied, these machines will generally
spark at the commutator on fullload if the brushes are kept on the noload
neutraL
c. Since the voltage drop when the series
field was added was greater than before,
it is evident that this field has been con
nected backward so as to oppose the
shunt field instead of assist it, the series
connections must therefore be reversed so
that the current passes through the series
coils in the proper direction.
d. Since the compoimding effect of the
series coils is too large, it will be necessary
to reduce the number of series tm*ns or to
reduce the current flowing through these
turns. The latter method is that generally adopted, a shunt being placed
in parallel with the series coils, as shown in Fig. 102, so that, of the total
current /i, only a fixed portion passes through the series field coils.
Fig. 102. — Series shunt to
vary the series excitation.
CHAPTER XIV
THEORY OF OPERATION OF DIRECTCURRENT MOTORS
86. Driving Force of a Motor. — An electric generator and an
electric motor are identical in structure. The generator is used
to transform mechanical energy into electrical energy, while the
same machine operating as a motor can be used to transform elec
trical energy into mechanical energy.
If a voltage is applied at the terminals of the machine in
diagram B, Fig. 103, so as to send current through the armature
conductors in the direction shown, then, since these conductors
are carrying current and are in a magnetic field, they are acted on
by forces all of which act in the same direction around the shaft
and so cause the armature to rotate.
87. Driving and Retarding Forces in Generators and Motors. —
The generator in diagram A, Fig. 103, driven by an engine in the
^.etardmgForce
ptWing force
Fig. 103. — Driving and retarding forces in a generator and in a motor.
direction shown, supplies electric power to a circuit, and current
flows through the armature conductors in the direction indicated
by the crosses and dots. There is alForce exerted on these conduc
tors in as much as they are carrying current in a magnetic field,
which force is opposed to the direction of motion, see page 13,
and the larger the current the greater is this retarding force. To
keep the generator running, the driving force of the engine must
78
Art. 88] OPERATION OF DIRECTCURRENT MOTORS 79
be great enough to overcome this retarding force and also to over
come the friction force of the machine.
The same machme opiating as a motor is shown in diagram B.
A voltage applied at the motor terminals from some external
source forces current through the armature conductors in the di
rection shown and, since these conductors are carrying current in a
magnetic field, they are acted on by forces which cause the arma
ture to rotate in a direction that may be determined by the left
hand rule, page 7. Now the conductors, rotating with the
armature, cut lines of force, and an e.m.f. is generated in the wind
ing in exactly the same way as if the machine was driven by an
engine. This e.m.f. acts in the same direction as in diagram A
since the machines have the same polarity and rotate in the same
direction. This generated e.m.f. is therefore opposed to the
current in the conductors and opposed to the applied e.m.f.,
for which reason it is called the back or counter e.m.f. of the
motor.
In the case of both a generator and a motor, there is a force
acting on the conductors of the armature in as much as they are
carrying current and are in a magnetic field. This is the driving
force in the case of a motor and the retarding force in the case of a
generator. There is also an e.m.f. generated in the armature of
each machine in as much as it is rotating in a magnetic field.
This e.m.f. acts in the direction of the current flow in the case of
a generator but opposes the current flow in the case of a motor.
In order that current may flow through a motor armature, the
applied e.m.f. Ea must be greater than the back e.m.f. Eh and
where Ea is the applied voltage
Eh is the back e.m.f.
IJta is the voltage required to force the armature current
la through the armature resistance Ra and is called the arma
ture resistance drop.
In the above equation it is most important to note that, of the
applied voltage Eaj the part which forces the current through the
resistance of the armature is laRa and seldom exceeds 5 per cent,
of Ea) the remaining part of the applied voltage is required to
overcome Ehi the back generated voltage of the machine.
88. The Back E.M.F. — The existence of the back e.m.f. may
readily be shown by experiment. If for example a motor, con
80 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xiv
nected as shown in Fig. 104, is driving a flywheel, and the switch
S is suddenly opened so as to disconnect the armature from tho
power mains, the flywheel will keep the machine running, but the
ammeter reading will become zero and the voltmeter reading will
drop suddenly from Ea to Eb, the voltage generated by the rotat
ing armature, and will then drop slowly to zero as the motor comes
to rest.
Belt
\j/7oUmeter
Fig. 104. — ^Experimental determination of the back e.m.f .
Example: Let Eat the applied voltage, be 110, la the Armature current,
100 amp., and Ra the armature resistance, 0.04 ohms. The voltmeter
reading will be 110 volts while switch 8 is closed but will drop suddenly to
110 — (100 X 0.04) » 106 volts at the instant the switch is opened, and will
then drop slowly to zero as the motor slows down.
89. Theory of Motor Operation. — The power taken by a motor
from the mains changes automatically to suit the mechanical
load. Consider the case of a motor connected as shown in Fig.
105, the applied voltage, the exciting current //, and the magnetic
flux per pole being constant. If the motor is at standstill and
the switch S is closed, a large current /« = Ea/Ra will flow
through the armature, the back voltage Eb being zero since the
armature conductors are not cutting lines of force. The arma
ture conductors carrying current, being in a magnetic field, are
acted on by forces which overcome the resisting forces of friction
and of the load and cause the motor to rotate. As the motor
increases in speed, the back e.m.f. Eb also increases since it is pro
portional to the rate at which the armature conductors cut lines of
force, and therefore the current /« == (Ea — Eb)/Ra, see page
79, decreases. The motor will stop accelerating when this cur
rent has dropped to such a value that the total force developed is
just sufficient to overcome the retarding force.
If now the load on the motor is increased, the driving force due
to the armature current is not sufficient to overcome the increased
resisting force and the motor must slow down. As the speed
abt.89.] operation of directcurrent motors 81
decreases, howerver, the back e.m.f. Eh also decreases and allows a
larger current to flow through the armature, smce la = {Ea —
Eb)/Ra. The motor finally settles down to such a speed that the
increased current in the armature again produces a driving force
which is just sufficient to overcome the increased retarding force.
If the load on the motor is decreased, the driving force due to
the armature current is more than sufficient to overcome the
decreased resisting force and the motor must accelerate. As it
increases in speed, however, the back e.m.f. Et also increases and
causes the armature current /« to decrease. The motor stops
accelerating and the speed and
armature current remain con
stant when the driving force
due to the current has dropped
to such a value that it is just
sufficient to overcome the de
creased retarding force. The
electrical power taken by the
motor from the mains there
fore changes automatically to
suit the mechanical load on the
motor. The back e.m.f . of the
motor regulates the flow of current in the same way as the gover
nor regulates the flow of steam in a steam engine.
A 110volt directcurrent motor, connected to the mains as shown in Fig. 106
delivers 10 h.p. If the efficiency is 88 per cent., the exciting current
is 2 amp. and the armatiu*e resistance is 0.08 ohms find:
a. The motor input
b. The current taken from the mains
c. The armature current
d. The back e.m.f.
a. the motor output = 10 h.p.
output
Fig. 105.
the motor input —
efficiency
10
= 11.35 h.p.
0.88
= 11.35 X 746 = 8480 watts
h, Ii, the current from the mains ^
watts input
applied voltage
8480
110
= 77 amp.
82 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xnr
c. the armature current » the total current — the exciting current
» 77  2 = 76 amp.
d. the applied voltage » 110
the voltage to overcome the resistance drop — laRa
= 76 X 0.08
= 6 volts
the back e.m.f . Eb = Ea — laRa
= 110  6
= 104 volts.
90. Speed and Torque Fonnulas. — ^The force on a conductor
carrying current in a magnetic field is proportional to the current
and to the strength of the magnetic field, see page 7, so that the
torque developed by a given motor may be expressed by the
equation
r = a const. X 4>X la
where T is the torque in lb. at 1 ft. radius, or in kg. at Im, radius
4> is the flux per pole of the motor
la is the armature current
the constant depends on the construction of the machine and
on the units chosen, but with its actual value we are not
concerned.
When a motor is running, the back e.m.f. is always less than
the applied e.m.f. by laRa the armature resistance drop, see page
79, so that
Eh ^ Ea — laRa
Now Eh is generated in the motor armature because the con
ductors are cutting lines of force, see page 79, so that, in a given
machine, Eh is proportional to the flux per pole and to the speed
or
Eh = k4> r.p.m. where fc is a constant
and Eh = Ea — laRa as shown above
E,
therefore r.p.m. =
k<i>
(Ea — laRa)
= a const. X
«
where r.p.m. is the motor speed in revolutions per minute
Ea is the voltage applied at the motor terminals
laRa is the armature resistance drop.
<f> is the flux per pole of the motor.
Abt.921 operation OF DIRECTCURRENT MOTORS 83
Theae speed and torque fonnuUe will be used in the next
chapter for the determination of the charactwistica of diffOTent
types of motors.
91. Improvement of Commutation by Shifting of the Brushes. —
la a directcurrent generator the brushes are shifted from the
neutral in the direction of motion so that the coil in which the
current is being reversed is in what has been called a reversing
£eld, see page 65; in the case of the generator shown in diagram
A Fig. 106, this reversing field is under the tip of the north pole.
The same machine operating as a motor is shown in diagram B,
the direction of motion and the polarity of the poles being
unchanged, while the direction of the current is reversed in order
to make the motor rotate in the desired direction. If then the
reversing field for the conductor at brush B of the generator
n a generator and in a motor.
is under tip of the JV pole, that for the conductor at the same brush
of the motor must be under the tip of the S pole since the motor is
rumui^ in the same direction as the generator but with a reversed
current. From this result the rule is obtained that in a generator
the brushes should be shifted forward in the direction of motion
whereas in a motor they should be shifted backward against the
duection of motion.
92. Armature Reaction in Generators and Motors. — In Fig.
106, which shows a generator and a motor respectively with the
brushes shifted so as to improve commutation, the distribution of
magnetic flux due to the armature acting alone is as shown by the
lines of force. The armature field acts in the direction oz and
may be considered as the resultant of a crossmagnetizing com
ponent in the direction oy and of a demagnetizing component in
84 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xiv
the direction ox, see page 68, and, in the case of both the genera
tor and the motor, the most important effects of the reaction of
the armature field on that due to the exciting current in the field
coils are, that the demagnetizing effect reduces the flux per pole,
while the crossmagnetizing effect causes the flux density to
decrease under the pole tips toward which the brushes have been
shifted, a condition which tends to cause poor commutation,
see page 68.
CHAPTER XV
CHARACTERISTICS OF DIRECTCURRENT MOTORS
SHUNT WOUND MOTORS
93. The Starting Torque. — The shunt motor is connected
to the power mains as shown diagrammatically in Fig. 107.
The applied voltage Ea and the exciting current 1/ are constant
and are independent of the armature current /«. /
To start such a machine, the field coils are fully excited so
that the magnetic flux has its normal value and then the re
sistance Rt is gradually decreased and current allowed to flow
through the armature. The torque developed increases directly
as the armature current is increased and the motor will start to
rotate when the current has such a value that the torque devel
oped is large enough to overcome the resisting torque of friction
and of the load.
The torque developed, being equal to fc^/o, see page 82,
depends only on the flux per pole and on the armature current
and, since the exciting current and therefore the flux per pole are
constant, fullload current in the machine produces the same
torque at starting as when the motor is running at fullload and
normal speed, or fullload torque is developed with fullload
current; similarly twice fullload torque is developed with twice
fullload current in the armature.
94. The Starting Resistance. — If a motor armature at stand
still were connected directly to the power mains then, since its
resistance is small, a large current would flow through the armature
and burn the windings and the brushes. To limit the starting
current, a starting resistance must be inserted in series with the
armature as shown in Fig. 107 and, if fullload torque is required
at starting, this resistance must limit the current to its normal
fullload value.
As soon as the armature begins to rotate, a back e.m.f. is
generated in it which tends to make the current decrease since
la — rJ" , DN but, to maintain fullload torque until the
85
86
PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xv
motor is up to speed, fullload current must be maintained in
the armature, so that the starting resistance must gradually be
decreased as the motor comes up to speed and the back e.m.f.
increases.
A 10h.p., 110 volt, directcurrent shunt motor has an efficiency of 88
per cent., an exciting current of 2 amp. and an armature resistance of 0.08
ohms find:
a. The starting resistance required for fullload torque
h. The starting current if no starting resistance was used
,, ... output
a. the motor mput = ^ •
= 088 ^ ^^^^ ^•^'
= 11.35 X 746 = 8480 watts
watts input
the current from the mains =
applied voltage
8480 ^^
JlQ = 77 amp.
t I ^A/ww\
Rs
Fia. 107. — Connec
tions of a shunt motor
during starting.
Fia. 108.— Starter for
a shunt motor.
Fig. 109.— Starter
with a novoltage re
lease.
the armature current = the total current — the exciting current
= 77 — 2 = 75 amp.
4.U 4. X 1 4. i. i. *• applied voltage
the total resistance at startmg = f^inoad armature current
= =r = 1.47 ohms.
75
the starting resistance = the total resistance — the armature resistance
= 1.47  0.08
» 1.39 ohms
Art. 95] DIRECTCURRENT MOTORS 87
5. the starting current if no starting resistance is used
— applied voltage
~ armature resistance
"^^^ ^^^ *°^P
= 18.4 times fullload current, which would burn up
the winding.
96. Motor Starter. — ^A starter which may be used to perform
the operations described above is 'shown diagrammatically in
Fig. 108, When the handle A, which is made of metal, is moved
into position Ai, the field coils are fully excited while the arma
ture and the whole starting resistance are put in series across the
power mains. As the handle is gradually moved over to position
A 2, the starting resistance is gradually cut out of the armature
circuit, but the current 7/ in the field coils remains practically
unchanged since the starting resistance R» is small compared
with Rf, the resistance of the field coils; in the above problem, for
EJ 110
example, B, = 1.39 ohms while B/ = y^ = —^ = 55 ohms.
The starting handle must not be moved over too rapidly, or the
starting resistance will be cut out before the speed and therefore
the back e.m.f . have time to increase and limit the current. The
handle however must not be left on one of the intermediate
notches between Ai and A 2 because, in order to keep down the
cost of the starting resistance, it is made small and will not carry
fullload current without injurious heating for more than about
15 sec.
96. Novoltage Release. — Suppose that a motor is running at
normal speed and that the power supply is interrupted due to some
trouble in the power house or in the line, the motor will stop, but
the starting handle will remain in the running position. If
the power supply is now reestablished, the armature will be at
standstill and there will be no starting resistance in series with it
to limit the current. To take care of such a contingency the
staxter is changed by the addition of what is called the novoltage
release. A starter with this attachment is shown diagrammat
ically in Fig. 109. The starting handle is moved from the starting
to the running position against the tension of the spring S and is
held in the running position by the electromagnet M. If the
power supply is now interrupted, the exciting current will
decrease, the magnet M will not be able to hold the handle
88 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xv
against the pull of the spring, and the handle will be pulled back
to the starting position. The magnet M is generally so designed
that it will release the starting handle should the applied voltage
drop below 30 per cent, of its normal value. Such starters are
described more fully in Chapter 19, page 114.
97. Load Characteristics. — ^The characteristic curves of a motor
show how the torque and the speed vary with the armature
current, the applied voltage being constant. These curves may
readily be determined from the formulse:
a
m
«
&
/j^^ Torqae ■ k <^Ia
Armatare Oarreat I^
Fig. 110. — Characteristic curves of a shunt motor.
torque = fc^/a
r.p.m. =«! 7
9
where Ea is the applied voltage
la is the armature current in amperes
Ra is the armature resistance in ohms
laRat the armature resistance drop, seldom exceeds 6 per
cent, of Ea when the motor is carrying fullload
4> is the flux per pole
k and ki are constants
In the case of the shunt motor, see Fig. 110, the applied voltage
Ea and the exciting current 7/ are constant and so also is the flux
per pole, the efifect of armature reaction being neglected, then:
torque = k<t)Ia
= a const. X la
(Ea — laRa)
r.p.m. = ky
<t>
— a const. (Ea — laR^
Art. 98]
DIRECTCURRENT MOTORS
89
The curves corresponding to these equations are shown in
Fig. 110. The fullload speed is less than that at noload by
about 5 per cent, since the back e.m.f. Eh has to drop this
amount in order that fullload current may flow through the
armature.
98. Effect of Armature Reaction on the Speed. — When the
effect of armature reaction is neglected, the speed characteristic
of a shunt motor is as shown in Fig. 110; the drop in speed seldom
exceeds 6 per cent, at fullload. When the brushes are shifted
backward from the neutral so as to improve commutation, arma
ture reaction causes the flux per pole to decre^i^e as the load in
creases, see page 84, so that the speed, being eqi\al to h{Ea —
IaJia)/4> remains approximately constant from noload to full
load, since the decrease in the value of (Ea — laRa) is compen
sated for by the decrease in the value of <f>.
Shunt motors are suited for constantspeed work such as the
driving of line shafts and woodworking machinery.
99. Variable Sfpeed operation can best be investigated by
means of the equation r.p.m. = k{Ea — IaR^/4>i see page 82.
Oe
\^v\Awv\
Fig. 111. — ^Insert resis Fig. 112. — Insert resistance in the
tance in the field circuit to armature circuit to decrease the
increase the speed. speed.
Methods of adjusting the speed of a shunt motor.
To increase the speed, 4> the flux per pole must be reduced by in
serting a resistance in series with the field coils as in Fig. 111.
To decrease the speed below the value which it has when the flux
per pole is a maximum, the voltage Ea applied to the motor
terminals must be decreased by inserting a resistance in
series with the armature as shown in Fig. 112; this resistance must
be able to carry the fullload current without injury so that the
starting resistance must not be used since it is designed for start
ing duty only, see page 87.
While the formula shows that the speed increases when the
90 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xv
flux is decreased, it is advisable to study more fully how this
takes place. If the flux per pole is suddenly decreased, the back
e.m.f. of the motor drops and allows more current to flow in the
armature. The increase in the armature current is greater than
the decrease in the flux, so that the torque developed is greater
than necessary for the load and the motor accelerates. The
following problem illustrates this.
A 10h.p., 110 volt, 900 r.p.m. directcurrent shunt motor has an
armature resistance of 0.08 ohms and takes an armature current of 75
amp. at full load. Find :
a. The torque at fullload
6. The back e.m.f. at fullload
If the flux per pole is suddenly reduced to 80 per cent, of normal find:
c. The back e.m.f. at the instant the flux is changed
d. The armature current at the same instant
e. The torque at the same instant
, ^ horsepower X 33,000
o. the torque =
2rr.p.m.
10 X 33,000
2ir900
= 58.5 lb. at 1 ft. radius
&. the back e.m.f. Eb ^ E a— laRa
= 110  (75 X 0.08) = 104 volts at fuUload
At the instant the flux is reduced
c. the back e.m.f. Eb ^ 104 X 80/100  83.2 volts, since 4> is reduced
d. the armature current = {Ea — Eb)/Ra
= (110  83.2) /0.08
= 335 amp. or 4.46 times fullload current
e. the torque = 58.5 X 80/100 X 335/75, since it is proportional to the
flux and to the armature current
= 209 lb. at 1 ft. radius or 3.6 times fullload torque.
/. At the instant the flux per pole is reduced to 80 per cent, of its normal
value, the armature current increases to 4.46 normal and the torque to 3.6
times normal value. The driving torque being then larger than the retarding
torque of the load, the motor must accelerate.
SERIESWOUND MOTORS
100. The Starting Torque. — ^The series motor is connected to
the power mains as shown diagrammatically in Fig. 113. The
applied voltage Ea is constant while the field excitation increases
with the load.
Art. 101]
DIRECTCURRENT MOTORS
91
The torque developed, being equal to fc^/o, see page 82,
increases directly with ^ the flux per pole and with /« the
armature current. Now ^ increases with /« since that current
is also the exciting current and, if the magnetic circuit of the
^machine is not saturated, ^ is directly proportional to la and the
torque is therefore proportional to /«*. In an actual motor, the
flux per pole does not increase as rapidly as the exciting current,
due to saturation of the magnetic circuit, but varies with /«
as shown in curve 1, Fig. 113. Using this relation between
<t> and lay the relation between torque {k<l>I^ and 7« has been
determined and is plotted in curve 2.
Fullload current in the machine produces the same flux per
a
Pi
M
s
«
p
hi
o
H
M
P
Torque aft 0f a
«
Ea^^ Conitant
Increaiei with la
Armatare Current la
Fig. 113 — Characteristic curves of a series motor.
pole and therefore the same torque at starting as when the motor
is running at fullload and normal speed, or fullload torque is
developed with fullload current. Since the torque is approxi
naately proportional to /a^, twice fullload torque is developed
with approximately V2 times or 1.414 times fullload current.
In the case of the shunt motor, the flux per pole is constant and
the torque is directly proportional to /«, see page 85, so that
tvidce fullload torque requires twice fullload current. For
heavy starting duty, therefore, the series motor is better than the
shunt motor in that it takes less starting current from the line.
101. The Starting Resistance. — As in the case of the shunt
motor, see page 85, a starting resistance must be inserted
in series with the armature so as to limit the starting current.
92 PRINCIPLES OF ELECTRICAL ENGINEERING (Chap, xv
This resistance must be gradually decreased as the motor comes
up to speed.
102. Load Characteristics. — ^The characteristic curves of a
series motor may readily be determined from the fundamental
formulsB:
torque = fc^/a
^iEa — laRa)
1 7
where Ea is the applied voltage
la is the armature current in amperes
Ra is the combined resistance of the armature and
the series field coils
laRa) the armature and series field drop, seldom ex
ceeds 7 per cent, of Ea when the motor is carry
ing fullload
4> is the flux per pole
k and ki are constants
In the case of the series motor, the applied voltage Ea is
constant, while the flux per pole varies with /« as shown in curve
1, Fig. 113. The relation between torque (fc^/a) and arma
ture current is plotted in curve 2, while curve 3 shows the rela
tion between r.p.m. (fci —  — * " ) and the armature current.
It is important to note that, as the load and therefore the
armature current decrease, the flux per pole decreases and the
machine must speed up to give the required back e.m.f. At light
loads the speed becomes dangerously high and for this reason a
series motor should always be geared or direct connected to the
load. If a series motor were belted to the load and the belt broke
or slipped ofif, then the motor would run away and would prob
ably burst.
Series motors are suited for crane work because they develop
a large starting torque, slow down when a heavy weight is being
lifted and speed up with light loads. Crane motors are geared
to the hoisting drum and are always under the control of
the operator.
103. Speed Adjustment. — The speed of a series motor is
proportional to {Ea — IaRa)/<t>i see page 82, so that, for a given
current Jo, the speed may be changed by altering Ea the applied
voltage, or 4> the flux per pole.
Art. 104]
DIRECTCURRENT MOTORS
93
If a resistance R^ is inserted in series with the armature as
shown in Fig. 1 14, then the voltage applied at the motor terminals
is reduced by IJi. and the lower back e.m.f. required is obtained
at a lower speed.
With constant applied voltage and a given amature current
the speed may be increased by decreasing the flux per pole. This
may be done as shown in Fig. 116 by shunting the series field
winding with a resistance so that, of the total current /«, only
part is allowed to pass through the field winding. The flux per
pole may also be reduced by short circuiting part of the field wind
ing as shown in Fig. 116, if the switch S is closed, the current
^
Fia. 114. Fia. 115. Fia. 116.
Fig. 114 — ^Insert resistance in the armature circuit to reduce the speed.
Fio. 115. — Shunt the field coils to reduce the excitation and increase the
speed.
Fio. 116 — Short circuit part of the field winding to reduce the excitation
and increase the speed.
Methods of adjusting the speed of a series motor.
passing through the machine is not changed so long as the load is
kept constant, but the exciting ampere turns are reduced and so
therefore is the flux per pole.
COMPOUND MOTORS
104. The compound motor is a compromise between the shunt
and the series motor and is connected to the power mains as
shown diagrammatically in Fig. 117. The applied voltage £« is
constant and so also is the shunt current //, but the current in
the series field coils increases with the load, so that the flux per
pole increases with the load but not so rapidly as in the series
motor.
If a shunt and a compound motor have duplicate armatures
and the same excitation at fullload, then at this load they will
94 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xv
develop the same torque and run at the same speed since
torque = fc^/a
r.p.m. = tci 7
For loads greater than fullload, the flux per pole of the shunt
motor is unchanged while that of the compound motor is in
creased due to the series field coils, therefore the compound motor
has the greater torque but the lower speed. For loads less than full
load on the other hand, the flux per pole of the compound motor is
less than that of the shunt motor due to the decrease of the
current in the series field coils, so that the torque is less and the
speed is greater than in the shunt machine.
The speed and torque characteristics of a compound motor
Armature Gorrent Iq
Fig. 117. — Characteristic cxirves of compound motors.
are shown in Fig. 117. Unlike the series motor, the compound
motor has a safe maximum speed at noload and so cannot run
away on light loads. The speed of a compound motor may be
decreased below normal by means of a resistance inserted in the
armature circuit, and increased above normal by means of a
resistance in the field coil circuit.
Compound motors are suitable for driving such machines as rock
crushers which may have to be started up full of rock, because
they develop the large starting torque with a smaller current than
the shunt motor, while they drop in speed as the load comes on
and thereby allow a flywheel connected to the shaft to take the
peak of the load.
CHAPTER XVI
LOSSES, EFFICIENCY AND HEATING
106. Mechanical Losses in Electrical Machinery. — In order to
keep the armature of an electrical machine rotating, power is
required to overcome the windage or air friction, the bearing
friction, and the friction of the brushes on the commutator. This
power is not available for useful work and is called the mechan
ical loss in the machine.
In a given machine this loss increases with the speed, but at a
given speed it is practically independent of the load.
106. Copper Losses. — If Ra is the resistance of the armature
circuit, including the armature winding, the brush contacts, and
the series field coils then, to force a current /« through this circuit,
a voltage Ca = laBa is required. This armature circuit drop at
fullload seldom exceeds 6 per cent, of Et, the terminal voltage.
The power expended in overcoming this vol
tage drop is equal to eJa = IJRa watts and,
since this power is not usefully employed, it
is called the copper loss in the armature circuit.
If again, R/ is the resistance of the shunt
field coil circuit, Fig. 118, and 1/ is the shunt
current, then the power expended in exciting
the machine is equal to I/^R/ watts where //, which is equal
to Et/Rf, seldom exceeds 5 per cent, of the current in the ar
mature of the machine.
107. Hysteresis Loss. — Fig. 119 shows an armature which is
rotating in a twopole magnetic field. If we consider a small
block of iron ab then, when it is under the N pole as shown, lines of
force pass through it from a to b; half a revolution later the same
piece of iron is under the S pole and the lines of force then pass
through it from 6 to a so that the magnetism in the uon is re
versed. To continually reverse the molecular magnets of the
iron in the armature an amount of power is required which is called
the hysteresis loss in the machine, see page 36.
The hysteresis loss increases with the number of reversals per
95
96
PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xvi
second, that is with the speed, it also increases with the flux
density.
108, Eddy Current Loss. — If the armature A, Fig. 119, were
made of a solid block of iron then, as it rotated, e.m.fs. would be
induced in the surface layers of the iron and eddy currents would
flow through the solid mass, see page 66. The power required
to maintain these currents is called the
eddy current loss in the armature and
is kept below 3 per cent, of the arma
ture output by lamination of the core,
see page 68.
Since the e.m.fs. generated in the
eddy current circuits depend on the
rate of cutting lines of force, the eddy
current loss will increase with the
speed and with the flux density.
109. Stray Loss. — The total loss in a directcurrent machine
consists of
Windage
Bearing friction
Brush friction
Hysteresis loss
Eddy current loss
la'RJ + If'Rf
Fig. 119. — ^Reversal of
flux in the armature core as
the armature rotates.
Stray loss
Mechanical losses
Iron losses
Copper loss
The term stray loss is used in practice to include the mechanical
and the iron losses. These losses do not vary with the load so
that they have practically the same value at noload as at full
load. The stray loss can readily be measured at noload by
running the machine idle as a motor at normal speed and normal
voltage. The motor armature input EJaj Fig. 120, is then equal
to the mechanical losses, the iron losses and the small noload
armature circuit copper loss, which latter loss may be neglected
as may be seen from the following problem:
If a 50kw., 110volt, shunt generator requires an armature current
of 30 amp. when run as a motor at noload and normal speed and
voltage, find the stray loss, the resistance of the armature circuit being 0.008
ohms.
The armature input at noload = 110 X 30 » 3300 watts
= stray loss + la^Rn
= stray loss + (30« X 0.008 = 7.2)
from which the stray loss = 3300 — 7 = 3293 watts.
Abt. 110] LOSSES, EFFICIENCY AND HEATING
97
In the case of large machines^ the stray loss is generally de
termined by driving the machine at normal voltage and speed
by means of a small shunt motor the losses of which are known.
The machines are connected up as shown in Fig. 121, the generator
is then excited to give normal voltage Eo and the input to the
small motor armature is determined from readings of the voltage
£m and the current Ia» Under these conditions the input to
the generator must be equal to the windage, friction and iron
losses of the machine and this input is also equal to EnJ^
minus the motor losses, which latter losses are known.
Bell
Motor
Generator
Fia. 120. — Machine Fig. 121. — Machine driven by a small motor the
runs idle as a motor. efficiency of which is known.
Measurement of the stray loss in a directcurrent machine.
110. The efficiency of a machine = output/input and may be
calculated from test data as in the following example.
Draw the efficiency curve for a directcurrent flatcompounded generator
rated at 1000 kw., 600 volts, given the following data
stray loss — 30 kw.
Ra — the resistance of the armature winding, brush contacts and
series field coils <= 0.006 ohms
R/ » the resistance of the field coil circuit « 20 ohms
At full>load, the load current
1000X1000
600
1666 amp.
of which //, the shunt field current —
600
2Q = 30 amp.
and la, the armature ciu*rent — 1696 amp.
then the stray loss « 30 kw.
//»JB/ « 30« X 20 « 18 kw.
I Ma  1696« X 0.006 = 17.2 kw.
total loss » 65.2 kw.
generator output *» IQOO kw.
generator input » 1065.2 kw.
fullload efficiency = 94 per cent.
The mechanical losses, the iron losses and the shunt field copper loss are all
independent of the load on the machine and are often classed together as the
98
PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xvi
r.,1
constant loss, the variable loss being Im^Ra the loss in the armature circuit,
iiinn \^ lonn
so that at halfload, the load current = = 833 amp.
600
of which the shunt field current » 30 amp.
and the armature current — 863 amp.
then the stray loss = 30 kw.
/;«JB/ = 18 kw.
/a'JB. = 863« X 0.006 = 4.5 kw.
total loss = 52.5 kw.
generator output — 500 kw.
generator input = 552.5 kw.
half load efficiency » 90.5 per cent.
Other values are worked out in a similar way and the results plotted as in
Fig. 122.
100
y
^
80
1 60
(
11
1 40
1
20
\
'
250 500 750 1000
Kilowatt Output
J250
Fig. 122. — ^Efficiency curve of a 1000 kw., 600 volt directcurrent generator.
Approximate values for the fullload efficiency of standard
generators and motors are:
Kilowatts Fullload efficiency
1
5
25
100
500
1000
80 per cent.
83 " ''
88 '* "
91 '' ''
94 '' "
95 '' ''
The efficiency of a motor may be determined by loading the
motor with a prony brake and measuring the total electrical input
and the corresponding mechanical output. Such a test however
is rarely carried out except in the case of small motors, the effi
11
Art. Ill]
LOSSES, EFFICIENCY AND HEATING
99
ciency of large machines is more readily determined from meas
urements of the losses and is worked up as in the last problem.
111. Heating of Electrical Machineiy. — ^The losses in an
electrical machine are transformed into heat which causes the
temperature of the machine to rise above that of the surrounding
air. The temperature becomes stationary when the rate at which
heat is generated is equal to the rate at which it is dissipated.
The rate at which heat is dissipated depends on the difference
between the temperature of the machine and that of the sur
roimding air. During the brief interval after starting under load
this temperature difference is small, very little heat is dissipated
and the temperature rises rapidly as shown in Fig. 123. As the
temperature increases, more of the heat is dissipated and the
temperature rises more slowly as from 6 to c. If the load is now
taken off the machine, the temperature will drop rapidly at first
so
°40
«•
•
«
§ 10
la
20 40 eo 80 100 120 140 160 180 aoo
Minatei
Fig. 123. — ^Heating curves of electrical machines.
and then more slowly as shown in Fig. 123, the temperature drop
being more rapid when the machine is rotating than when
stationary because of the better ventilation and the better
convection of heat.
112. Permissible Temperature Rise. — Insulating materials lose
their mechanical and dielectric strengths at high temperatures,
for example, cotton becomes brittle at temperatures greater than
95® C. and begins to char at slightly higher temperatures, so that,
when cotton is used to insulate machines, the permissible rise
of temperature is 55** C. above an air temperature of 40** C.^
Methods of insulating have been devised whereby cotton, paper
and other materials that become brittle and char are not used,
fireproof materials such as enamel, asbestos and mica being used
entirely, with such insulation higher temperatures are permissible.
^ For further information see Standardization Rules of the American
Society of Electrical Engineers (West 39th St., New York City).
o
y
^
\
^
'\
/
/
>
y^
■>
/
V5
k
(.
V,
^
CHAPTER XVII
MOTOR APPLICATIONS
113. Limits of Output. — If the load on a motor is increased, the
armature current and the armature copper loss both increase and
the temperature of the machine rises. The maximum load that
can be put on a motor is that with which the temperature of the
machine reaches its safe maximum value; a greater load raises
the temperature to such a value that the insulation of the
machine is permanently injured.
The output of a motor is often limited by commutation. When
interpoles are not supplied, the brushes are shifted from the
neutral so that commutation takes place in a reversing or com
mutating field. Now the effect of armature reaction is to weaken
this reversing field, see page 84, so that as the armature current
increases, the reversing field becomes weaker and the motor
finally begins to spark at the brushes, after which the load can be
increased no further without injury to the commutator.
If the commutation limit of output is reached before the
heating limit, then interpoles may be supplied to improve commu
tation, see page 65, and the motor output may be increased until
the temperature limit is reached.
114. Open, Semienclosed and Totally Enclosed Motors. —
The cooling of a motor depends largely on the circulation of air
through the core and windings, so that the frame should be as
open as possible.
If chips and flying particles are liable to get into the windings,
the openings in the frame should be covered with perforated sheet
metal; the motor is then said to be semienclosed. This screen
throttles the air supply on which the cooling of the machine
largely depends so that, in order to keep down the temperature
rise, the output of a motor has to be lower when semienclosed
than when of the open type.
When a motor has to be totally enclosed, as for openair service,
the output of the machine has to be considerably reduced so as to
keep the temperature down to a safe value thus:
100
abt.1151 motor applications 101
A 10h.p., 220.volt, 600r.p.m. motor with 40^ C. rise on fullload
as an open machine can be used to deliver 9 h.p. when semi
enclosed and about 6 h.p. when totally enclosed at the same
voltage and speed and with the same rise in temperature.
116. Intermittent Ratings. — It was pointed out on page 99 that
it takes a considerable time for a motor to attain its final tempera
ture so that, if a motor has to be operated intermittently for short
periods, its output may be considerably increased.
With suitable windings a particular motor frame was given the
following ratings
10 h.p., 220 volts, 600 r.p.m. continuous duty
17 h.p., 220 volts, 600 r.p.m. for 1 hour
22 h.p., 220 volts, 600 r.p.m. for 1/2 hour
the temperature rise at the end of the specified time being the
same in each case.
116. Effect of Speed on the Cost of a Motor. — For a given horse
power output, a high speed motor is always cheaper than a slow
speed motor thus:
A 10h.p., 220volt, 1200r.p.m. shunt motor weighs 750 lb. and
costs $150.
A 10h.p., 220volt, 600r.p.m. shunt motor weighs 1250 lb. and
costs $250.
A 10h.p., 220volt, 300r.p.m. shunt motor weighs 1800 lb. and
costs $350.
The reason for this is as follows:
If a given motor frame is supplied with two armatures, one of
which A has half as many conductors as B but the conductors
have twice the cross section and can therefore carry twice the
currrent, then, when run at the same voltage, armature A with half
the conductors must run at twice the speed of B to give the same
back e.m.f., but since armature A can carry twice the current
of B it can therefore deliver twice the output. Thus the armature
of a 10 h.p., 600 r.p.m. motor could be rewound to deliver
20 h.p. at 1200 r.p.m., 15 h.p. at 900 r.p.m. or 5 h.p. at 300
r.p.m. and these armatures would all have approximately the
same weight and cost.
117. Choice of Type of Motor. — The characteristic curves of a
shunt, a series, and a compound motor are shown in Fig. 124, the
motors having the same torque and speed at fullload.
The shunt motor takes a current which is proportional to the
102 PRINCIPLES OF ELECTRICAL ENGINEERING [Ghap.xvii
torque required, and operates at practically constant speed at all
loads. It must be noted however that the speed of such a motor
increases slowly as the field coils heat up because their resistance
increases and causes the excitation to decrease. If the motor
has been started cold, the speed may increase 10 per cent, in 3
hours due to this cause.
The series motor is the best for heavy starting duty because, for
any torque greater than the fullload value, it takes a smaller
current from the line than either the shunt or the compound
machine. The speed of the series motor decreases rapidly with
increase of load and becomes dangerously high at light
loads. For this latter reason, a series motor should always be
(geared or direct connected to the load.
The compound motor is a compromise between the shunt and
Full Loftd. Speed
Shunt
Compound
Series
A.rmature Current A.rmature Current
Fig. 124. — Characteristic curves of direct current motors.
the series motor. It is better than the shunt motor for heavy
starting duty but not so good as the series motor. The speed
drops somewhat with the load but the motor runs at a safe maxi
mum speed even at noload.
The service for which each type of motor is suited can best be
illustrated by a discussion of a few typical motor applications.
118. A line shaft should run at practically constant speed at
all loads and so is driven by a shunt motor. The starting torque
required will seldom exceed 1.5 times fullload torque and this
can be obtained with 1.5 times fullload current in the armature,
which is a reasonable starting overload.
119. Woodworking machinery such as planers and circular
saws run at practically constant speed and so are suitably
I '
/ '\
<l
Art. 120] MOTOR APPLICATIONS 103
driven by shunt motors. In the case of heavy planing mills, the
starting torque required is sometimes excessive due to the inertia
of the machine, in which case it may be advisable to use a
compound motor because it requires a smaller starting current
for the same torque.
120. Reciprocating pumps, which have to start up against full
pressure, require a large starting torque, so that although a shunt
motor is often used for such service yet a compound motor would
take less starting current from the line. A series motor would be
suitable so far as starting torque is concerned, but if the suction
pipe were to leak so that the load on the motor became light,
then the motor would run away.
121. Traction Motors. — For traction service, the torque re
quired to start and accelerate a car is much greater than that
required to keep the car moving, so that a series motor is used
since it is the best for heavy starting duty, see page 91. The
subject of traction is discussed more fully in Chapter 40, page
322.
122. Crane Motors. — The characteristics which make the series
motor suitable for traction work also make it suitable for crane
service. The motor is able to develop a large starting torque
without taking an excessive current from the line; it also oper
ates at a slow speed when the load to be lifted is heavy and runs
at a high speed when the load is light.
Both crane and traction motors are geared to the load and
moreover are always under the control of the operator.
123. Express Passenger Elevators. — An express elevator has to
be accelerated rapidly, so that a large starting torque is required.
After the car has moved through about 20 ft., its velocity has
reached about 500 ft. per min., and has to be kept constant at this
value. This result is obtained by the use of a heavily cojnpounded
motor, by means of which a large starting torque is developed
without an excessive current being taken from the line; when
acceleration is complete, the series winding is short circuited, and
the motor operates thereafter as a constant speed shunt machine.
A series motor would not be suitable for such service because,
during the rush hours when the car is heavily loaded the motor
would slow down, whereas at times of light load the car would run
at an excessive speed, unless specially controlled.
124. Shears and Punch Presses. — The load curve of a punch
press is shown in Fig. 125. In order that the peak load may be
104 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xvh
carried by the motor without sparking, a motor of about 15
h.p. would be required, which is much greater than the
average load of 6.5 h.p. To take the peak load off the
motor, a flywheel is generally supplied with the press and, in
order that the flywheel may be effective, the speed of the motor
must drop as the load comes on.
A shunt motor is not suitable for such service as it does not drop
in speed and so does not cause the flywheel to take the load.
A series motor cannot be used because it would run away when
the clutch of the press was released, and would probably cause
the flywheel to burst.
The motor generally used on large presses is compound wound,
which motor drops in speed as the load comes on and thereby
causes the flywheel to give up energy, while the maximum
speed at noload cannot exceed a safe value.
Fig. 125. — Load curve of a punch press.
Fig. 126.
A drooping speed characteristic may be obtained from a shunt
motor by connecting a resistance permanently in series with the
arniature as shown in Fig. 126, the resistance having such a value
that the voltage drop Cr at fullload is about 5 per cent, of E.
When the load on the motor increases, the voltage drop across
the resistance increases, that applied to the motor terminals
decreases, and the speed of the motor drops. When the load on
the motor decreases, the voltage across the motor increases, the
speed rises, and energy is again stored in the flywheel. The
resistance in the power mains supplying the motor may often be
sufficient to produce this effect. The only objection to the
method is that an amount of power — eJa watts is lost in the
control resistance.
CHAPTER XVIII
ADJUSTABLE SPEED OPERATION OF DIRECTCURRENT
MOTORS
For the driving of lathes and other such machine tools, ad
justable speed motors are largely used, it is therefore necessary to
discuss the different methods of speed control before the subject
of machine tool driving can be profitably taken up.
126. Speed Variation of Shunt Motors by Armature Control. —
The speed of a motor may be lowered by decreasing the voltage
77 Amp.
>>
Ja m 75 Amp
>
, Jf^ 2 Amp
4>
ii Constant
fe
"' 'i VVNAWV
Fig. 127. — Resistance inserted in Fig. 128. — Resistance inserted in
the armature circuit causes the the field coil circuit causes the
speed to decrease. speed to increase.
Methods of adjusting the speed of a direct current shunt motor.
applied to the motor terminals. This may be done by connecting a
resistance in the armature circuit as shown in Fig. 127.
The speed is given by the formula r.p.m. = kr^^ — x^ — ^
where laBa seldom exceeds 5 per cent, of Ea, so that, to obtain
half speed, the applied voltage Ea must be reduced to about 50 per
cent, of normal, the other 50 per cent, of the line voltage being
absorbed by the resistance inserted in the circuit; under these
conditions, the loss in the resistance, which is equal to eja, is also
equal to the armature input EJat and the efficiency of the system
is less than 50 per cent. The actual efficiency may be figured out
as in the following problem.
106
106 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xviu
A 10h.p., 110volt, 900r.p.m. shunt motor has an efficiency of 88 per cent.,
an armature resistance of 0.08 ohms and a shunt field current of 2 amp.
If the speed of this motor is reduced to 450 r.p.m. by inserting a resistance in
the armature circuiti the torque of the load being constant, find the motor
output, the armature current, the eicternal resistance and the overall
efficiency.
At normal load:
the motor output = 10 h.p.
the motor input = 10/0.88 =11.35 h.p. = 8480 watts
the total current == 8480/110 = 77 amp.
the shunt current = 2 amp.
the armature current = 75 amp., see page 82
the torque = ^z^ — ^t^qqq = 58.5 lb. at 1 ft. radius, see page 90
the back e.m.f. ^ Ea — laRa
= 110  (75 X 0.08) = 104 volts, see page 82.
At half speed:
The horsepower output = 3^ 000 — *"^^' since the torque
is constant, the output is proportional to the speed and is equal to 5 h.p.
The torque = kiftia, and, since the torque is constant and so also is the
excitation, therefore /«, the armature current, is the same as at full speed and
is equal to 75 amp.
The back e.m.f. Eb is generated in the armature due to the cutting of lines
of force and is equal to a const. X^ X r.p.m., see page 82, and since the flux
is constant, therefore Eb is proportional to the speed and is equal to 0.5 X 104
= 52 volts.
The voltage applied to the motor ^ Eb + laRa
= 52 + (75 X 0.08)
= 58 volts.
The voltage drop across the external resistance = 110 — 58
= 52 volts
the current in this resistance = 75 amp.
the resistance = 52/75 = 0.7 oh^ms.
the loss in the resistance = 52 X 75 = 3900 watts,
the total input = 110 X (75 + 2), see Fig. 127, = 8500 watts,
the motor output = 5 hp. = 5 X 746/1000 = 3.7 kw.
the overall efficiency = 3.7/8.5 = 44 per cent.
Since the armature current and therefore the armature copper
loss have the same value at half speed as at full speed, the torque
being constant, the temperature rise will be the greater at the
slow speed because of the poorer ventilation.
From the above problem it may be seen that, when the speed of
a motor is reduced by armature resistance, the output is decreased
and is directly proportional to the speed while the temperature
Art. 127] ADJUSTABLE SPEED OPERATION 107
rise, even with this reduced output, is greater than normal
because of the poorer ventilation. The overall efficiency also is
exceedingly low, the per cent, loss in the resistance being ap
proximately equal to the per cent, reduction in speed, that is,
being 50 per cent, of the total input at half speed and 75 per cent,
of the total input at quarter speed.
126. Speed Variation of Shunt Motors by Field Control. —
By inserting a resistance in the field coil circuit of a shunt motor,
as in Fig. 128, the excitation and therefore the magnetic flux are
reduced and the motor has to rim at a higher speed in order to
generate the necessary back e.m.f., see page 89.
When interpoles are not supplied, the brushes are shifted from
the neutral position so that commutation takes place in a reversing
field, but if the excitation is decreased, then this reversing field is
decreased and the conmiutation is impaired; furthermore, the
higher the speed, and therefore the more rapidly the current in the
coils is being commutated, the greater is the voltage of self induc
tion opposing the change of current and the greater the tendency
for sparking to take place at the brushes, see page 63. The range
of speed variation is therefore limited by commutation and an
increase in speed of about 70 per cent, is about all that can
generally be obtained by field weakening from a standard motor;
the flux is then reduced to 1/1.7 = 60 per cent, of its normal value.
When a greater speed range is required, it will generally be neces
sary to use an interpole motor.
When the speed is controlled by a field rheostat the efficiency
is not impaired because, as shown in Fig. 128, the control rheostat
carries only the small shimt current and not the large armature
current.
127. Speed Regulation of an Adjustable Speed Shunt Motor. —
When the speed of a motor varies considerably with change of load
the speed regulation is said to be poor. When the speed is prac
tically constant at all loads the speed regulation is said to be good.
Suppose that the speed of a shunt motor has been adjusted by
means of a resistance in the armature circuit, as shown in
Fig. 127, so as to give a definite speed at a definite load, then, when
the load is increased, the armature current /« and the voltage a,
will increase and therefore the voltage Ea will decrease and the
speed of the motor will drop; the speed regulation is therefore poor
when armature resistance control is used.
If, for example, this method of control is used when a forging
9
I
I
I '
108 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xvin
such as that in Fig. 129 is being turned in a lathe, then the
motor will slow down when the cut is deep and will speed up when
the cut is light, so that the speed will be very irregular.
When the speed of a shunt motor is adjusted by means of a
resistance in the field coil circuit, as in Fig. 128, the speed
regulation is good. The speed is given by the formula r.p.m. =
(E —InR )
fci . — — so that since Ea is constant, as also is the flux
once the field circuit resistance is adjusted, therefore the drop in
speed between noload and fullload will seldom exceed 5 per cent.,
since laRa at fullload seldom exceeds
5 per cent, of Ea
128. Electric Drive for Lathes and
Boring Mills. — Such machines require
constant horsepower at all speeds of
the machine spindle so long as the
Fig. 129. amount of metal removed per minute
remains unchanged. The motor must
therefore be large enough to develop the necessary horsepower
at the lowest operating speed without excessive heating.
The maximum speed of a given motor is limited by centrifugal
force or by the speed of the gear, while the minimum speed may
be as low as desired, but when constant horsepower is required at
all speeds the cost of the motor increases as the minimum speed
is decreased as may be seen from the following table:
A. A 10h.p., 220volt, 1200r.p.m. shunt motor weighs 750 lb.
and costs $150.
B. A 10h.p., 220volt, 600/ 1200.r.p.m. shunt motor weighs 1260
lb. and costs $250.
C. A 10h.p., 220volt, 300/ 1200r.p.m. shunt motor weighs 1800
lb. and costs $390.
the latter motor is necessarily an mterpole machine because of the
large speed variation required, see page 107, and costs about 10
percent, more than a constantspeed 300r,p.m. motor of the same
output.
The cheapest drive from the point of view of motor cost is that
obtained by the use of a constantspeed motor such as A in the
above table, with change gears or coned pulleys to give the
necessary speed range. If some of the gears are eliminated, and
a motor such as B in the above table is used, which has a speed
range of 2 to 1, then the cost of the motor is increased 66 per cent.,
Art. 129]
ADJUSTABLE SPEED OPERATION
109
while for a speed range of 4 to 1 by motor control the cost of the
motor is 2.6 times greater than if a constant speed motor with
jchange gears had been used.
The speed range of the motor may be obtained by armature
control, by field control, or by a combination of the two. Arma
ture control is used as little as possible because of the low overall
eflSciency of the method and also because of the poor speed
regulation obtained, see pages 105 and 107.
129. Multiple Voltage Systems. — When a large number of
adjustable speed motors are in operation in a machine shop, the
three wire system shown diagrammatically in Fig. 130 may be
used with advantage. Two generators in the power house are
connected in series and three leads a. 6 and c are taken to each
Yolti
m
Qj
A B
Fig. 130. — Multiple voltage system.
adjustable speed motor, the voltage between a and c being 220
and between a and 6 and also between h and c being 110 volts.
To obtain the lowest speed from a motor operating on this
system, the field coils are connected across the 220volt mains so
as to give the maximum flux while the armature is connected
across either of the 110volt circuits as shown at A, Fig. 130.
The speed may then be gradually increased by inserting resistance
R in the field coil circuit as shown at JS. When the speed has
been doubled in this way, the armature is then connected across
the 220volt mains and all the resistance R is cut out of the field
coil circuit, and the speed may again be gradually increased by
once more reducing the field excitation by means of the resistance
R as shown at (7. By this means a total speed range of 4 to 1 may
be obtained without the magnetic flux being reduced at any time
110 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xvm
to less than half its normal value, and the efficiency is high over
the whole range of speed because no resistance is inserted in the
armature circuit at any time.
Another multiple voltage system is shown diagrammatically
in Fig. 131 and is in use in a considerable number of machine
shops. The voltages available in this case are 90, 160 and 250
so that a total speed range of 4 to 1 may readily be obtained from
a standard motor since the flux is never reduced below about 60
per cent, of its normal value, see page 107. The twovoltage
system is to be preferred however since 110 and 220 are standard
voltages and still more so because the two voltages may readily
be obtained from a single machine called a threewire generator,
which is much cheaper than two single
voltage machines each of half the output.
This threewire generator is described on
page 317.
130. Ward Leonard System. — One
method of obtaining a wide speed range,
without the use of armature resistance,
would be to use a separate generator with
eso
Fig. ISl.Multiplevolt ^^^^ adjustable speed motor and to vary
age system. the excitation of the generator so as to
vary the voltage applied to the motor ter
minals. Such a system, shown diagrammatically in Fig. 132 is
called the Ward Leonard System after its inventor. The outfit
consists of a highspeed motor generator set supplied with each
Motor Oeneretor
Set
Adjnstable Speed
Motor
Fig. 132. — Ward Leonard system.
adjustable speed motor, the Bet consisting of a highspeed gener
ator G direct connected to a motor M.
To obtain slow speeds, the field excitation of G is reduced so
as to reduce the voltage Eg which is applied to the motor JIf i.
As the field excitation of G is increased, the voltage Eg increases
Aet.131] adjustable speed operation 111
and with it the speed of the motor Mi. The motor M\ can
readily be reversed by reversing the excitation of G^, so that the
whole control is handled through a small field circuit rheostat
and the efficiency is comparatively high.
The Ward Leonard system has been used for printing press
work to obtain the very low speeds required when the paper is
being fed into the machine, it has also been used to secure the
delicate speed adjustment necessary for operating gun turrets
in battleships. Because of the ease with which the motor Mi
can be reversed, this system has been used recently for the drive of
large reversing planers, so as to eliminate the crossed belts.
131. Drive for VentUating Fans. — For ventilating purposes it
is necessary to control the volume of air delivered, to suit the
requirements. This is done either by reducing the speed of the
fan or by throttling the orifice.
When a fan is operated with a fixed orifice, the volume of air
delivered is proportional to the speed, the pressure is proportional
to the square of the speed, and the power required is proportional
to the cube of the speed approximately.
When the volume of air is reduced by throttling the discharge,
then the power taken is directly proportional to the volume
delivered, the speed of the fan being constant.
For adjustable speed operation the shunt motor is used and, in
the case of a fan drive, the speed reduction is obtained by the
armature resistance method of control because of its simplicity;
the total loss in the controlling resistance being small because
of the large reduction in the load and therefore in the armature
current as the speed drops. This may be seen from the following
example, which is worked out by the same method as that on page
106:
The following data is taken from page 106:
The output is 10 h.p., 110 volts, 900 r.p.m. and the motor is shimt wound,
the armature resistance = 0.08 ohms
the exciting current — 2 amp.
the fullload armature current » 75 amp.
the back e.m.f. at fullload and full speed » 104 volts.
If the motor is driving a fan and the speed is reduced to 450 r.p.m. by
inserting a resistance in the armature circuit, find the motor output, the
armature current, the external resistance and the overall efficiency.
The horsepower output is proportional to the cube of the speed approxi
mately and is therefore equal to ^, = 1.25 horsepower.
112 PRINCIPLES OF ELECTRICAL ENGINEERING [CHAP.xvni
,, . horsepower output X 33,000 , . . , . i x * n
The torque = s and is therefore equal to fuU
^ Jirr.p.m. ^
1 J X _ 1.25 h.p. OOOr.p.m. ^ « ^. „, .^
load torque X ^^ u Xjtk — — = 0.25 times fullload torque.
lU n.p. 4oUr.p.m. ^
The torque » k<t>Ia and, since the flux <t> is constant, therefore the current is
proportional to the torque, so that the armature current is equal to 0.25 (full
load current) or 18.8 amp.
The back e.m.f. at half speed « 104/2 « 52 volts, see page 106.
The voltage applied to the motor — Eb + la Ra
= 52 + (18.8 X 0.08)
=53.5 volts
The voltage drop across the external resistance » 110 — 53.5
— 56.5 volts,
the current in this resistance — 18.8 amp.
the resistance » 56.5/18.8 — 3 ohms
the loss in the resistance — 56.5 X 18.8 — 1060 watts
the total input = 110 X (18.8 + 2) = 2290 watts
the motor output = 1.25 h.p. = 1.25 X 746/1000 = 0.93 kw.
the overall efficiency « 0.93/2.29 = 41 per cent.
so that, although the overall efficiency is still low, the total loss in the
control resistance is comparatively small.
132. Armature Resistance for Speed Reduction. — ^From the
two problems on pages 106 and 111 it may be seen that the
resistance required to reduce the speed to 50 per cent, of normal
may be 0.7 ohms or 3 ohms, the corresponding losses in these
resistances may be 3.9 kw. or 1.06 kw., and the currents 75 amp.
or 18.8 amp.^ depending entirely on the kind of load.
If a rheostat built originally for constant torque duty is used
with the same motor for fan operation, it will not have sufficient
resistance to reduce the speed to the desired value; if a fan duty
rheostat is used for constant torque service it will have to carry
a larger current than it was designed for and will burn out.
It is therefore very essential when specifying armature rheostats
to specify the type of service for which it will be used.
133. Motors for Small Desk Fans. — These are usually series
motors and are connected directly to the line without a starting
resistance. When the current is switched on, the motor is at
standstill and its back e.m.f. is zero, but the growth of the cur
rent in the machine is opposed by the self induction of the field
and armature windings which are connected in series, while the
inertia of the armature and fan are so small that the machine is
up to speed before the current has time to reach a dangerous
value. The load on a fan increases as the motor speeds up so
that the motor cannot run away.
Abt. 134]
ADJUSTABLE SPEED OPERATION
113
134* Printixig presses must be run very slowly while being
made ready, that is, while the web is being threaded, after which
they must be smoothly accelerated to the desired running
speed.
Large presses are equipped with two shunt motors, a main driv
ing motor which is direct connected to the driving shaft, and an
auxiliary starting motor which is connected to the driving shaft
through a reduction gear and an automatic clutch.
While the press is being made ready, the auxiliary motor alone
is connected to the power mains and drives the press at about
10 per cent, of normal speed. When the press is ready, the main
motor is connected to the power circuit and is gradually accele
rated, and, when the speed of the press has increased sightly above
the make ready speed, the au
tomatic clutch is released due
to centrifugal force and the
small motor is thereby dis
connected mechanically from
the press; it may then be dis
connected from the power
mains.
For small presses, the aux
iliary starting motor is dis
pensed with, the make ready
speed being obtained by in
serting a resistance in the ar
mature circuit. One objec
tion to armature control is that the speed regulation is poor,
see page 107, and, when suflBicient resistance is inserted to ob
tain 10 per cent, of normal speed, the regulation is very poor
and the speed of the press is irregular. To obtain low speeds
which are not irregular, the connection shown in Fig. 133 is used.
If the current h is large compared with /«, then a considerable
change in the value of la will have comparatively little effect on
the total current /i so that the voltage er will not be greatly
affected, the voltage Ea will therefore remain approximately con
stant and the speed regulation will be fairly good. The method
is not economical since the current 1 2 does no useful work but the
larger the value of /2 relative to la the better is the speed regula
tion. This method of control is used only where slow speeds are
required for short intervals.
Fig. 133. — Connection for slow speed
operation of shunt motors.
CHAPTER XIX
HANDOPERATED FACE PLATE STARTERS AND CON
TROLLERS
If a switch is opened in a circuit carrying current, an arc will
be formed and, unless proper precautions are taken, the switch
contacts will be burned.
135. Knife switches such as that shown in Fig. 134 are
seldom used to open a circuit through which current ts flowing;
they are used to isolate a circuit after the current has been reduced
to zero.
The quickbreak switch shown in Fig. 134 has a main contact
a
Fio. 134. — Quick break type of knife switch.
blade A and an auxiliary contact blade B held together by
the spring C. When this switch is opened, the blade B is re
tained by friction until A has been withdrawn, the spring C
then pulls out the blade B so quickly that no appreciable arc is
formed. This quickbreak principle in different forms is largely
used when circuits carrying current have to be opened.
136. Auxiliary Carbon Contacts. — In the switch shown in Fig.
135, the main contact blocks a and b are bridged by an arch c of
leaf copper, and an auxiliary carbon contact d is in parallel with
the contact a. When this switch is opened, the contact a is the
first to be broken, but the circuit is not interrupted since current
can still pass through the contact d. This latter contact is broken
114
Abt. 137] HANDOPERATED FACE PLATE STARTERS
115
when the switch opwis further, and an arc is formed which bums
the carbon tips. Since these tips volatilize without melting,
they remain in fairly good shape and, when badly burned, can
readily be replaced; the carbon contacts have the additional
Fia. 135. — Circuit breaker with auxiliary carbon contacts.
advantage that by their means a comparatively high resistance
is inserted in the circuit and the current is reduced before the
circuit is broken.
137. Blowout Coils. — If the conductor a6, Fig. 136, is carrying
Fia. 136. — Principle of the blowout coil.
current and is in the magnetic field NS, it la acted on by a force
which, according to the lefthand rule, page 7, tends to move it
upward. If at is an arc formed between two contacts x and y
03 they are separated, this arc will be forced upward and will
116 PRINCIPLES OF ELECTRICAL ENGINEERINQ [Chap.xix
lengthen and break. The coil A which produces tbemagnetic
field ia called the magnetic blowout coil.
The application of this principle to a contactor switch is shown
in Fig. 137. The blowout coil A produces the magnetic field
NS, so that if an arc passes between the switch contacts it will be
forced upward and broken on the contact tips. The polarity of
the magnetic field must be such that the arc is blown away from
the switch contacts and not into them.
138. Horn Gaps. — The intense heat of an arc causes convection
currents of air to flow upward so that, if the switch jaws are
shaped as shown in Fig. 138, the arc stream will be blown upward
and will finally break between the arcing tips c, which tips may be
removable.
This effect may be exaggerated by enclosing the contact in an
Abt. 142] HANDOPERATED FACE PLATE STARTERS 117
J
A
V
arc chute of such shape that the gases, expanding suddenly, can
pass out only through the switch contacts. Such arc chutes, as
for example that in diagram A, Fig. 137, when combined with
magnetic blowout coils, are very effective.
139. Fuses are used to protect electric circuits from overloads.
A fuse is a piece of metal of such size and composition that it
will melt and open the circuit when the current flowing becomes
large enough to endanger the circuit.
The melting of a fuse is accompanied by
an arc and by spattering of the fused metal
so that it is generally advisable to mount
the fuse in the center of a fiber tube and
surround it with a fireproof powder to
quench the arc, terminals being supplied as
at a, Fig. 143, so that the fuse may readily
be removed and replaced. A 10 amp. fuse
is expected to carry 12.5 amp. continuously
and 20 amp. for a period not greater than
2 minutes.
140. Circuit Breakers. — Circuits which
are subject to frequent overloads are gen
erally protected by automatic circuit
breakers such as that shown in Fig. 135,
rather than by fuses. The operation of such a circuit breaker has
been described on page 40.
141. Motor Starters. — The requirements of a motor starter
have been discussed on pages 85 and 87. These requirements
have been met in many different ways but it is impossible in the
space available to describe more than a few standard types.
142. The sliding contact type of starter, as used for shunt
motors, is shown in Fig. 109. As the contact arm A is moved from
the starting position Ai to the running position A 2 the following
operations are performed:
1. The field coils are fully excited as soon as contact is
made with the first segment.
2. The starting resistance is gradually cut out as the contact
arm is moved from A 1 to 4.2; during this interval of time the motor
should come up to speed.
3. The contact arm is held in the running position by the no
voltage release magnet M,
To stop the motor, the main switch is opened. This de
FiG. 138. — Remova
ble horn tips for con
tactor switches.
118 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xix
energises the release coil M; the contact arm is then pulled
back to the starting position by the spring S.
143. Starting Resistance. — The resistance used with a starter
must have sufficient current carrying capacity to allow the motor
with which it is used to start up every 4 min. for an hour with
out overheating, the load being such that the motor shall come
up to speed in less than 15 sec. with a starting current not greater
than 1.5 times fulUoad current.
144. Overload Release. — The current in a shunt motor in
creases with the load and may damage the machine if the load
Fig. 139. Fig. 140.
Starting boxes with both a novoltage and an overload release.
becomes too large. To protect the motor, an overload release
should be supplied to cut the machine out of operation as soon
as the load becomes excessive. This overload release may take
the form of fuses or of a circuit breaker, placed in the circuit as
shown in Fig. 143, or it may consist of a special attachment
on the starter as shown in Fig. 139.
In this type of starter the contact arm A performs the functions
of a starting arm while the arm B, used to close the main circuit
at C is connected to A by the spring S. To operate this starter
Akt. 1461 HAND^PBRATED FACE PLATE STARTERS 119
the contact C is closed by the arm B and ia held closed by the
latch L, the arm A is then moved over the contact buttons so as
to cut out the armature resistance, until it makes contact with
the novoltage release magnet M by which it is held againt the
tension of the spring S,
The latch L is released by the plunger p which 1b lifted when
the line current passing round the solenoid reaches a prede
termined value; the arm B then fiies up, opens the main circuit
at C and at the same time deenergizes the magnet M. Before
the motor can be started again the circuit must be closed at C;
the spring 5 at the same time returns the contact arm A to the
starting position thereby inserting the starting resistance in the
armature circuit.
Another type of overload release is shown in Fig. 140. In
this type, the motor current passes round the magnet and,
A Complete starter with reaiatance B Diagrammatic repreaentation
Fio. 141, — Multiple switch type of starter.
when it reaches a predetermined value, the arm p is lifted to
close the contacts H, the novoltage release coil M is thereby
short circuited, the exciting current no longer passes around it,
and a spiral spring in the hub brings the contact arm A back
to the starting position and cuts the motor out of circuit. Such an
overload release is shown on the starter in Fig. 143 and may be
adjusted to open the circuit for any current up to 1.5 times full
load current. This typeof release is cheaper than that in Fig. 139
but it has the disadvantage that it does not protect the motor
during the starting period.
116. Multiple Switch Starters. — The sliding contact typ« of
starter is liable to give trouble due to arcing at the contacts if
used for motors larger than 35 h.p. at 110 volts or 50 h.p. at
120 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
220 to 500 volts; for such service the multiple switch type
shown in Fig. 141 is to be preferred.
With this type of starter, each step of the armature resistance
is cut out by a separate lever and the levers are so interlocked
that they cannot be closed except in the proper order, thus switch
b cannot be closed before a because of the stop 8. Starting with
a, the switches are closed hand over hand, each switch as it is
closed keeping the one to the left of it in contact by means of
the stops s; the last switch e is held closed by the latch /. Such
a starter cannot be left partly closed with part of the resistance in
circuit because none of the switches can stay closed until the last
switch e is latched.
When the circuit is interrupted, the magnet M is deenergized,
the latch / is released, and the switch e opens; the other switches
then open one after the other.
146. Compound Starters. — The speed of a shunt motor may
be adjusted by means of a rheostat in the field circuit (page
89). When this rheostat is incorporated in the starter, as
shown in Fig. 142, the resulting piece of apparatus is called a
compound starter.
The field contact arm A and the armature contact arm B are
mounted on the same hub post; there is a spiral spring in the hub
of B but none in A. To start the motor the two arms, which are
interlocked, are moved over together, their motion being opposed
by the spiral spring, and the armature resistance is gradually cut
out by the arm B which finally makes contact with the novoltage
release magnet M by which it is held. The contact arm A is then
free to move backward over the field contacts thereby weakening
the shunt field and increasing the speed of the motor. During
the operation of starting, the field resistance is short circuited by
the switch s which is kept closed by a spiral spring in the hub h;
this short circuit is removed while the starting arm B moves on to
the last contact c, the switch s being then tripped by the projec
tions g.
When the circuit is interrupted, the magnet M is deenergized,
and the arm B is pulled back by the spiral spring in the hub
and carries the field arm with it. With such a starter it is im
possible to start up except with full field, it is also impossible to
leave the starting arm B in any position intermediate between
the starting and the running positions.
When a motor has to be operated with a weak magnetic field.
AST. 1471 HANDOPERATED FACE PLATE STARTERS
121
for adjustable speed operation, it is generally advisable to con
nect the novoltage release coil M directly across the line so that
its holding power is not weakened when resistance is put in the
field coil circuit.
Fig, 142. — Compound starter.
147. Speed Regulators. — To obtain speeds lower than normal
a resistance must be inserted in series with the armature, see page
89 Any of the starters already described can be used as a
speed regulator if the resistance is able to carry the current of the
machine continuously without overheating and if provision is
made to return the contact arm to the starting position should the
voltage fail.
A sliding contact type of speed regulator is shown in Fig. 143.
Resistance in the armature circuit is used to cut down the
speed below normal, while speeds higher than normal are ob
tained by inserting resistance in the field coil circuit on the steps
between c and d.
This regulator operates in exactly the same way as a starter of
the sliding contact type except that the handle can stay on any
contact and can still be released from that contact by the no
122 PRINCIPLES OF ELECTRICAL BNGINBSRING [Chap, nx
voltage release. Attached to the starting arm is tbe circular
ratchet C which moves with the arm and, engaging in the ratchet
from below, is a pawl which is caused to press against the ratchet
by the novoltage release magnet M and to notch into the ratchet
when the contact arm is on a contact segment. Should the circuit
be interrupted, the magnet M is deenergized, the pawl is released,
and the starting arm is returned to the off position by a spring
in the bub.
Fio. 143. — Sliding contact type of speed regulator.
148. Controllers for Series Motors. — Such motors are largely
used for crane service and the controller used with them must be
arranged to reverse the direction of rotation by reversing the
armature connections and must also give speed regulation by
means of resistance in the armature circuit.
A simple type of controller for this purpose is ehown in Fig. 144.
The path of the current through the controller is shown in
diagram A when the motor is running in one direction and in
diagram B when the direction is reversed.
The controller shown in Fig. 144 is supplied with a blowout
coil A which produces a magnetic field that passes from the front
arms B to the back arms C, which arms are of iron. This magnetic
field passes vertically through the slate front and so is at r^ht
Art. 1481 HANDOPERATED FACE PLATE STARTERS 123
angles to the arc formed when a sliding contact leaves a segment,
it therefore acts to blow out the arc.
A Hoisting B Lowering
Fio. 144. — Face plate controller for a small reversing serieo motor.
The operation of crane and hoist motors is taken up in
greater detail in Chapter XL.
CHAPTER XX
DRUM TYPE CONTROLLERS
149. Dnim ^e controllers are particularly suited for adj ustablo
speed motors which have to be started and stopped frequently,
because the various operations are performed readily by the move
ment of a single handle and take place in their proper order The
controller is entirely enclosed and can readily be made weather
proof, while contact with the live parts is prevented.
A simple type of drum controller is shown in Fig. 146. It
consists of a castiron drum cylinder A, insulated from a central
shaft to which the operating handle B is keyed. To this drum,
the copper contact segments a, 6, c, d and e are attached; these
are in electrical contact with the drum and therefore with one
another. The drum carries also a brush contact m which slides
over stationary field resistance contacts that arc mounted on
the slate Cjthe contact m is not visible, being hidden by the drum.
The armature resistance is connected to the stationary fingers
/, g, k, j and k which are insulated from one another and mounted
on a wooden base.
The action of such a controller may readily be understood
from Fig. 145 which shows the controller drum developed on to
Art. 150]
DRUM TYPE CONTROLLERS
125
a plane; the vertical dotted lines indicate the successive positions
of the contact drum with respect to the row of stationary
fingers.
In position 1, the fingers/ and g make contact with segments
a and b of the drum, and the armature current passes through
the whole armature resistance, while the field coils are fully excited,
the exciting current passing through the contact m. In position
4, the armature resistance is all cut out but the field coils are still
fully excited. In position 5, the brush m makes contact with field
segment 6 and the resistance ri is inserted in the field coil circuit.
With further motion of the drum from position 5 to position 13,
the resistance in the field coil circuit is gradually increased, the
magnetic field is weakened, and the speed of the motor is thereby
increased above normal.
160. Novoltage and Overload Release.— The controller
shown in Fig. 146 is not provided with either a novoltage or an
Fig. 147. — Novoltage and
overload release panel.
Fig. 148. — Connections of novoltage and
overload release panel.
overload release. These are sometimes incorporated in the
controller but are more often supplied separately on a panel such
as that shown in Fig. 147, the equipment consisting of a single
pole magnetic switch A and an overload release coil B. The
connections of this panel are shown diagrammatically in
Fig. 148.
The contact b is kept open by means of a spring. When this
contact is closed, current passes through the control circuit
abcdefgh, from the positive to the negative side of the line, and
126 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xx
excites the electromagnet ilf , which closes the switch A and
allows current to pass to the motor. When the switch A closes,
the contact s also closes and current now passes through the
circuit ascdefgh and excites the magnet M, so that the contact b
can be opened. In this latter circuit is a resistance r, which
reduces the current in the holding coil M after the switch A has
been closed so that this switch can open the more readily should
power go ofif from the line; this magnetic switch A therefore acts
as a novoltage release.
The total current in the machine passes round the overload
coil B and, when this current reaches a predetermined value,
the plunger p is raised to strike the lever q and open the control
circuit at e, thereby deenergizing the magnet M and allowing
the switch A to open.
■»■
2/
<— t
2/
AAA/\AAr
'WWNA—
«— €
A Motors in Parallel B Motors in Series
Fig. 149. — Seriesparallel system of motor controL
To stop the motor, the contact d is opened; this opens the
control circuit and allows the main switch A to open.
The contacts b and d can be embodied in the controller in
such a way that the contact b and therefore the switch A cannot
be closed except when the controller handle is in the off position
and all the armature resistance is inserted in the armature
circuit.
151. Street Car Controller for Series Parallel Control. —
Street cars are equipped with series motors, and the number of
motors per car is a multiple of two. To start these machines, a
resistance is placed in series with the armatures as shown in
diagram A, Fig. 149, and is then gradually cut out as the motors
come up to speed.
The torque required to start and accelerate a car is much
greater than that required to keep the car in motion so that the
current I in the motor is large at starting and, if the motors are
Art. 151] DRUM TYPE CONTROLLERS 127
connected as shown in diagram A, a large current 2/ is taken
from the line. To reduce this current for half of the starting
period, the series parallel method of control is adopted.
Fio. 150.— Street railway controller.
During the first half of the starting period, the motors are con
nected in series, as shown in diagram B, so that the total line ,
current passes through both machines; when the startii^ re
sistance is all cut out, each motor has half of the normal voltage
Series connection Complete diagram of connections Parallel connec
Fia. 151. — Diagram of connectiona of a street railway controller.
applied across its terminals and runs at half speed. To obtain
higher speeds, the motors are now connected in parallel and the
resistance is E^ain inserted in the circuit as shown in diagram A;
128 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xx
this resistance is gradually cut out as the motors come up to speed
and, when it is all cut out, each motor is running on normal line
voltage.
These operations are performed with a drum controller of the
type shown in Fig. 150, which controller is shown developed in
Fig. 151.
f
0>
c:
r
c;
6 CH
I.
r^WWV
The direction of the cur
rent when the controller is in
position 1 is shown in diar
gram A. The current passes
through the whole armature
resistance R and then through
 the motors in series. The
armature resistance is gradu
ally cut out as the controller
FiQ. 152. — Diagram of connections of is moved to position 5, when
reversing ^j^^ motprs are in series across
the line and are running at half speed.
The direction of the current when the controller is in position 8
is shown in diagram B. The current passes through the whole
, armature resistance and then divides up and passes through the
I
Fia. 153. — Section through drum type of controller, showing the blow out
coil.
two motors in parallel. The resistance is gradually cut out as the
controller is moved to position 12, the motors are then in parallel
across the line and are running at full speed.
The complete controller is shown developed in diagram C.
The student should draw the drum on tracing paper and move it
over the stationary contacts and note how the various combina
tions are obtained.
Abt. 153] DRUM TYPE CONTROLLERS 129
162. Reversing Drum.rTo reverse the motors, the armature
connections axe reversed relative to the field connections. This is
done by means of the auxiliary drum B, Fig. 150, which is separate
from the main drum but is so interlocked with it that the motors
cannot be reversed except when the main drum is in the ofif posi
tion. This drum is shown developed in Fig. 152.
When the reversing drum is in position X, current passes
through the armature from a to 6; when in position Y, the current
passes from 6 to a. A double set of such contacts are required
for a pair of motors.
163. Mechanical Features of Drum Controllers. — The con
troller frame is of cast iron and has an asbestos lined removable
cover A, Fig. 150; the back of the frame is pierced with a verti
cal row of holes lined with insulating bushings through which
the necessary leads run.
The contact cylinder D with the copper contacts is supported
by and insulated from a central shaft to which the operating
handle is attached.
The blowout coil is shown at C. Hinged to the case of the blow
out magnet is a steel plate E extending vertically the entire
length of the cylinder and constituting such a magnetic circuit
that a powerful magnetic field is maintained across the finger
contacts as shown in Fig. 153. This steel plate is lined with as
bestos on its inner side and carries moulded refractory insulating
arc barriers F which project between the contact rings.
CHAPTER XXI
AUTOMATIC STARTERS AND CONTROLLERS
The tendency in the operation of electrical machinery is to
make the starters and controllers self governing so that the
machinery cannot be injured by careless or unskilled operators.
154. Automatic Solenoid Starter. — ^Fig. 154 shows a sliding
contact type of starter in which the contact arm is moved by
means of a solenoid. When the main switch K is closed, the
solenoid A is excited and pulls
the contact arm upward
thereby cutting the resistance
R out of the armature circuit;
the rate at which the contact
arm moves may be regulated
by the dash pot D.
At the end of its travel, the
contact arm presses on the
stop 8 and opens the contact
c and thereby inserts the re
sistance r in the solenoid cir
cuit so that the current in
this circuit is reduced and is
not larger than necessary to
hold the arm in the running
position. When the power is
ofif, or when the switch K is
opened, the solenoid is deen
ergized, and the starting arm falls by gravity to the starting
position.
166. Float Switch Control. — With such an automatic starter
it is not necessary to run the power mains to the point from which
the motor has to be started. The main switch K may be mag
netically operated and placed on the same panel as the starter
as shown in Fig. 155. This switch is operated by a control cir
cuit as shown diagrammatically in Fig. 156.
130
Fig. 154. — Automatic solenoid starter.
Abt. Ifitl] AUTOMATIC STARTERS AND CONTROLLERS 131
When the switch s is closed, the magnetic switch K is excited
and closes the main circuit so that power is available at the motor
Fig. 155. — ^Automatic solenoid Btarter with main magnetic switch.
tenninals. The switch s may readily be opened and closed by
means of a float, a pressure gauge or some other device, so as to
maintain the water level or
the air pressure in a tank
within prescribed limits.
When the float F in Fig. 156
falls below a certain point, the
projection e raises the lever I
and W is moved over. When
W passes the vertical position
it drops over and the projection
g snaps the switch s into con
tact and closes the control cir
cuit. When the water level
reaches the upper limit, the
projection / trips the lever W ^"aMriTr*
which opens the switch S and Fig. 156.— Float switch,
causes the motor to stop.
156. Magnetic Switch Controller. — Starters for large motors
are of the multiple switch type, see page 119, and, when the
132 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxi
motors have to be operated from a distance, the switches are
magnetically operated and are closed in their proper order by a
master controller operating on a control circuit. Fig. 157
shows such a starter used to control a shunt motor by means of
resistance in the armature circuit.
The switches A, B,C, D, E and F are magnetic contactor
switches which close when the electromagnets a, b, c, d, e and/
are excited. These magnets are connected across the mains
xy in the proper order by means of a small drum controller M
called a master controller.
As the drum M is turned, the first contacts to be made are x
and 1 and the electromagnet / is excited and the main switch
Fig. 157. — Magnetic switch controller.
F is closed. In the next position of the drum, the contact 2 is
closed and the magnet a is thereby excited which closes the
switch A and connects one terminal of the motor to the line.
On contact 3 being closed, the magnet b is excited which closes
the switch B and connects the motor armature across the line
with all the armature resistance R in series. With further motion
of the drum M, the resistances Ri, R2 and R3 may be cut out
one after the other.
The drum M has to handle only the current in the control
circuits which current is of the order of 1 amp., the drum may
therefore be small in size.
With such a system of control, practically any series of opera
tions can be performed in their proper order and several motors
can be controlled from a single master controller.
Abt. 168] AUTOMATIC STARTERS AND CONTROLLERS 133
167. Multiple Unit Control of Railway Motors. — The pos
sibilities of magnetic switch controllers are well illustrated in
the multiple unit system of car control. An electric train, made
up of a number of cars each with its own motors and magnetic
switch controller, can readily be controlled by a single operator
at the head of the train.
The control circuit carries only a small current so that the leads
are light and flexible; these leads are carried the whole length of
the train. As each contact of the master controller is closed,
the corresponding electromagnets under each car are excited
and the switches closed. If for example, in Fig. 158, the contacts
m
r^
WWuW
r^
IIu
f
Matter
Controller
luu
f
Fig. 158. — Principle of the multiple unit system of control.
a and h of the control circuit are energized by the master controller
then the magnetic switches A on each of four separate cars will
close simultaneously.
168. Automatic Magnetic Switch Starters. — ^With certain
automatic features attached, the magnetic switch starter may be
so constructed that the operator has only to close the main switch,
after which the starting resistance is cut out automatically by the
controller and the motor is brought up to speed without the
starting current exceeding a predetermined value.
The operation of such automatic starters depends on the
current changes in the armature circuit when the motor with
which the starter is used is brought up to speed. In the hand
operated starter in Fig. 159, when the switch A is closed, a current
of about one and a half times fullload current flows through the
armature and the starting resistance in series, and the motor
starts up. As it gains in speed, the back e.m.f. increases and the
current drops and, when this current has reached fullload
value, the switch B is closed and the same current cycle is
134 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxi
again passed through. Fig. 160 shows the current cycle for a
four switch starter such as that in Fig. 159.
In the automatic starter shown diagrammatically in Fig. 161,
when the switch p is closed, the solenoid a is excited and the
contactor switch A is closed thereby connecting one terminal of
the motor to the positive side of the line and at the same time
— T)wn
s
Pia. 159. — Multiple Fio. 160. — Current cycle in the
switch type of starter, armature circuit during starting.
closing the field coil circuit and exciting the machine; the field coil
circuit is not shown in Fig. 161.
The switch A, when open, supports the relay e and when A
closes e is dropped and closes the contacts/ so that the solenoid b
is excited and the switch B is closed thereby connecting the other
Fig. 161 . — Automatic starter with shunt magnetic switches and series relays .
terminal of the motor to the negative side of the line with all the
starting resistance in series. Current then passes through the
armature, Rzy R2, Riy m and q.
When the switch. J5 closes, a current of about one and a half
times fullload current flows through the armature, and the motor
starts up. Now the solenoid g is connected across the resistance
Art. 169] AUTOMATIC STARTERS AND CONTROLLERS 135
R, so that the voltage across this coil is the drop across the resist
ance, and the current which it sends through the coil is large
enough to hold up the plunger g even although the mechanical
support was removed when the switch B closed. As the motor
gains in speed, its back e.m.f. increases and the drop across the
resistance decreases so that the current in coil g decreases and,
when this current has reached a predetermined value, the
plunger of g drops and closes the contacts h, the solenoid c is
thereby excited and the switch C is closed cutting out fli, the first
step of the resistance.
After B closed but before c was excited, the motor current
passed through the solenoid m and held up this plunger, but after
the switch C has closed, this current passes through only the lower
half of coil m. At the instant C closes, the current is large and
although passing through only half of coil m it still holds the relay
open. As the motor accelerates however, the current in the
armature circuit decreases and finally reaches a value with which
the solenoid m is no longer able to support its plunger, which
plunger therefore drops and closes the contacts n, the solenoid d
is thereby excited and the switch D is closed cutting out fl2, the
second step of the resistance.
The switch E operates in exactly the same way as D and cuts
out the last step of the starting resistance. Such a relay switch
as that used in this type of automatic starter is shown in Fig. 137.
169. Automatic Starter with Series Switches. — ^The closing
electromagnets of the starter shown in Fig. 161 are shunt wound
and the starter is said to be of the shunt switch type with series
relays.
Another type of starter is shown in Fig. 163, the switches in this
case carry the line current and are called series switches. They
are so constructed that they will not close when the current
flowing in the exciting coil exceeds a predetermined value. Such
a switch is shown in Fig. 162. The upper end of the iron plunger
E carries a nonmagnetic stem G to which is attached a copper
plate H which makes contact with the brushes k when the
plunger is raised.
When a small current passes in the coil M, the flux in the
magnetic circuit passes as shown in diagram A and the plunger
tends to move upward so as to reduce the reluctance of the
magnetic circuit.
When a large current passes in the coil, the flux in the magnetic
136 PRINCIPLES OF ELECTRICAL EKOWEBRING [Chap.xxi
circuit passes as shown in diagram B; the reduced stem F becomes
highly saturated so that a large part of the total flux passes across
the gap U without entering the stem and the plunger tends to
move upward to reduce the gap d and downward to reduce the
gap It. The larger the current, the lai^r the value of if* relative
to 01 + ^, and the less the tendency for the plunger to move up.
With such a solenoid then, when the current exceeds a pre
determined value, the downward pull plus the weight of the
plunger keeps the plunger from being lifted, but as the current is
decreased, the fiux ^ and the downward pull both decrease
D
Fia. 182. — Series automatic switches.
rapidly as the stem F ceases to be saturated, until finally the
upward pull is able to raise the plunger.
The critical value of current with which the plunger can be
lifted may be adjusted by raising or lowering the iron plug C
so as to change the value of ifn relative to that of *i + ift.
A starter made with three such switches is shown in Fig. 163.
When the switch A is closed, a large current flows through the
armature, the starting resistance and the coil Ci in series and the
motor starts up; the switch j/i however does not close until the
motor has gained in speed and the current has dropped to the
value for which the solenoid Ci was set.
When Ci closes, the first step of the starting resistance is cut
out and the current, which increased considerably when the
switch closed, now passes through Ci and Ci and the two re
maining steps of the resistance. This current decreases as the
Abt. 159] AUTOMATIC STARTERS AND CONTROLLERS 137
motor speeds up and then d closes the contacts gt. The
same current cycle is again passed through after which Ct closes
Fio. 163. — Automatic starter with series switches.
and cuts out the last step of the starting resistance. The con
tact gs is kept closed by means of the small shunt solenoid s.
The contacts g, are the only ones that carry current continu
ously because, when C« closes, current ao longer passes through
138 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxi
Ci and d because it has an easier path through the circuit abc,
the plungers of Ci and d therefore drop.
When the main switch A is opened, the current in the motor
circuit drops, but the switch Cs requires only a small current to
hold up the plunger which therefore does not drop until the
current has become almost zero, so that blowout coils are not
required to protect the contacts gz. A complete starter is
shown in !Fig. 164.
The novoltage and overload release attachments are generally
supplied on a separate panel such as that shown in Fig. 147,
page 125, the main switch A being then of the magnetic switch
type is operated from a pushbutton circuit.
CHAPTER XXII
ELECTROLYSIS AND BATTERIES
160. Electrolysis. — Certain liquids conduct electricity but in
doing so they undergo decomposition. Such liquids are called
electrolytes and include bases, acids and salts, in solution or in
the molten state. The conductors by which the current enters
or leaves the electrolyte are called electrodes; that connected to
the positive line terminal is called the anode and is the one at
which the current enters, the other is called the kathode. The
name electrolysis is given to the whole process.
It would seem that the electrolyte, in addition to containing
complete molecules of the substance in solution, contains also
molecules which are dissociated into ions (atoms carrying positive
or negative charges) and that the metal and hydrogen atoms
carry positive charges while non metals, the hydroxyl group
(OH) and the acid radicals (SO 4, NOs, etc.) carry negative
charges. If then a difference of potential is established between
the electrodes, the positively charged ions will be attracted to
the negative electrode and the negatively charged ions to the
positive electrode, where they give up their charges. When
this occurs, the particle or group ceases to be an ion and dis
plays at once its ordinary chemical properties.
If a direct current is passed through a solution of hydrochloric acid (HCl),
using platinum electrodes, then the + H ions will be attracted to the nega
tive electrode where they will give up their charges and then appear as hydro
gen gas, similarly the — CI will appear at the positive electrode.
If a solution of sulphuric acid (H2SO4) is used, then the +H will be liber
ated at the negative electrode and the — SO 4 at the positive electrode.
The SO4, however, acts on the water of the solution to form sulphuric acid
and oxygen, which latter gas is liberated while the acid goes into the solution.
If the positive electrode had been of copper, then the SO4 would have acted
on the copper to form copper sulphate which would have gone into the
solution.
161. Voltameter. — If a negative electrode of paltinum and a
positive electrode of pure silver are used in a solution of silver
nitrate (Ag NOs)! then silver is deposited on the negative plati
139
11
140 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxn
num electrode and the — NOs acid radical, liberated at the posi
tive electrode, acts on the silver to form more silver nitrate, which
goes into solution and thereby keeps the concentration of the
electroljrte constant.
Faraday's experiments showed that the mass of material
deposited is proportional to the quantity of electricity (current
X time) so that the apparatus described above, called the silver
voltameter, can be used as a measure of quantity of electricity.
This instrument is used as a primary standard.
162, Electric Battery. — An electric battery is a device for
transforming chemical energy into electrical energy and consists
essentially of two dissimilar plates in a solution which acts more
readily on one plate than on the other. A difference of potential
is found between two such plates so that if they are joined by a
wire an electric current will flow in this wire. The magnitude of
this e.m.f. depends only on the material of the plates and on the
electrolyte and, for a given pair of plates, is independent of their
area.
There are innumerable types of battery, but in nearly every
case one plate is of zinc and the other of either carbon or copper.
If a battery is made of copper, zinc and dilute sulphuric acid, it
will be found that, when current flows in a conducting wire con
necting the copper and the zinc plates, the zinc goes into solution
as ZnSOi while hydrogen is given off at the copper plate. The
zinc and the sulphuric acid are therefore used up and electrical
energy is obtained at the expense of the chemical energy which
was contained in these materials.
When the cell becomes exhausted, the zinc plate and the elec
trolyte have to be renewed. When fresh materials are used, the
battery is called a primary battery; when the materials are re
newed by electrolysis in a way that shall be described later, the
battery is called a secondary or storage battery.
163, Theory of Battery Operation. — If we consider a battery
made up of copper, zinc and dilute sulphuric acid, then the essen
tial difference between these metals so far as battery operation is
concerned is that the zinc is the more readily acted on by oxygen
or has the greater chemical attraction for oxygen so that, while
both copper and zinc attract the —O ions in the solution, the
attraction of the zinc is the greater. As both metals combine
with the attracted oxygen they become negatively charged and
soon repel the negative oxygen ions as strongly electrically as
Art. 166] ELECTROLYSIS AND BATTERIES 141
they attract them chemically. When equilibrium is established,
both metals are negatively charged but the negative charge on the
zinc is the greater and its potential is therefore the lower. If the
two metals are now joined by a connecting wire outside of the
solution, electricity flows from the copper to the zinc and the
plates tend to come to the same potential, so that the state of
equilibrium is disturbed; the potential of the zinc rises slightly
above its potential of equilibrium so that it is able to attract
more negative oxygen, while the potential of the copper falls
slightly below its potential of equilibrium so that it now attracts
the positive hydrogen ions, the voltage between the plates is
therefore maintained and so also is the current in the conducting
wire.
The zinc oxide formed is acted on by the sulphuric acid to form
zinc sulphate which goes into the solution and is no longer an
active constituent of the cell.
164. Polarization. — ^The action of this battery weakens after a
few minutes of operation, and this weakening is found to be due
to a layer of hydrogen bubbles which cling to the copper plate
after giving up their charge. The cell is then one which has ac
tive plates of hydrogen and zinc and gives a much lower e.m.f.
than one of copper and zinc. Hydrogen has also a high electrical
resistance so that the current that can be drawn is small. This
defect of the battery is called polarization, and different methods
are used to keep the hydrogen bubbles away from the copper
plate, generally by the use of a depolarizing substance containing
an excess of oxygen,, which substance is placed around the copper
plate.
166. The E.M.F. and Resistance of Cells. — The electromotive
force depends only on the materials of the cell and is independent
of their size, shape or arrangement.
The internal resistance of a cell of given materials is proportional
to the distance between the plates and inversely proportional to
their area and, in order that a cell may have a low internal resist
ance, the plates should have a large surface and should be close
together.
166. The Daniell cell is a commercial type of copper, zinc and
sulphuric acid battery largely used for telegraph work. The
depolarizing substance in this cell is copper sulphate (CUSO4).
In one form of this battery the sulphuric acid with the zinc are
placed in a porous pot and this in turn is placed in a saturated
142 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxn
solution of copper sulphate contained in a glass jar, the copper
plate is immersed in this latter solution. By means of the porous
pot the liquids are kept from mixing but the action of the battery
is not impaired since the ions pass freely through the walls of the
pot.
The hydrogen liberated at the copper electrode acts on the
adjoining copper sulphate to form copper and sulphuric acid; the
copper is deposited on the electrode so that there is no hydrogen
layer formed and therefore no polarization of the cell until all
the copper sulphate in the solution has been exhausted.
The usual size of Daniell cell has a voltage on open circuit of
about 1.1 and an internal resistance of about 2 ohms so the largest
current that can be obtained from such a cell is 0.55 amperes on
short circuit.
167. Calculation of the E.M.F. of a Daniell Cell. — Faraday's
experiments showed that, when electrolysis takes place, the num
ber of gm. of substance separated out per coulomb i3 equal to
atomic weight
"—  — * — ; , or the coulombs required to produce a number
96,540 X valency' ^ ^
of grams equal to the atomic weight is 96,540 X the valency.
Copper (Cu) and zinc (Zn) have a valency of two in compounds
such as CuSOi and ZnS04, because one atom of Cu or Zn replaces
two atoms of H from H2SO4, and, since the atomic weight of Cu
is 63.75 and that of Zn is 65.37 therefore 96,540 X 2 coulombs
will separate out 63.75 gm. of Cu from CuS04and 65.37 gm.
of Zn from ZnS04.
When 63.75 gm. of Cu are formed into CuSOi, 197,500 gm.
calories of energy are liberated and when 65.37 gm. of Zn are
formed into ZnS04, the energy liberated is 248,000 gm. calories.
In a Daniell cell then, when 65 . 37 gm. of Zn have been con
sumed, 96,540 X 2 coulombs have passed and 63.75 gm. of Cu
have been separated out from the CuSOi, the energy given up
by the cell must therefore be
248,000  197,500 = 50,500 gm. calories
= 50,500 X 4.186 watt sec. see page 15
= 211,400 watt sec.
and 96,540 X 2 X voltage of cell = energy obtained
= 211,400 watt sec.
from which the voltage of the cell = rrrrr r =1.1 volts
96,540 X 2
assuming that no energy was lost in the form of heat.
Art. 170] ELECTROLYSIS AND BATTERIES 143
168. Local Action. — In a welldesigned and constructed cell,
action takes place only when the cell is deUvering energy. Com
mercial zinc, however, dissolves in sulphuric acid even when the
external circuit is not closed because the zinc is impure and local
action is set up between the impurities and the zinc and a number
of small internal batteries are formed in which the zinc is con
sumed but no voltage is available at the terminals. To prevent
this action the zinc is amalgamated, that is covered with a layer
of mercury. To amalgamate zinc, clean it with sandpaper, then
immerse in dilute sulphuric acid and while still wet apply mercury
to it with a rag. So far as the action of the battery is concerned
the mercury is inert.
169. Leclanch6 Cell.^ — This battery consists of carbon (C)
and Zinc (Zn) in ammonium chloride (NHiCl). The zinc is
converted into zinc chloride (Zn CI) while ammonia (NHs) and
hydrogen (H) appear at the carbon plate. To prevent polariza
tion, the carbon is surrounded with an oxidizing agent in the form
of manganese dioxide (Nn02), the oxygen of which attacks the
hydrogen to form water. In the usual form the carbon is placed
in a porous pot and is surrounded with granules of manganese
dioxide and of carbon. The action of this depolarizer is slow
so that, when used to supply current for a considerable time, the
cell becomes polarized and runs down; it will recover, however,
if left on open circuit.
The battery gives a voltage of about 1.5 on open circuit and
has an internal resistance of about 1.5 ohms. It is much used for
bell ringing and other intermittent work and requires little atten
tion beyond the addition of water as the solution evaporates, and
the renewal of the zinc as it is used up. The zinc generally sup
plied is not amalgamated, and nonconducting crystals of ZnCl
stick to its surface instead of dropping to the bottom; the opera
tion is improved if amalgamated zinc is used.
170. Dry Cells. — These are largely used when the battery is
subject to motion, as in motor cars and motor boats. There are
innumerable types but nearly all consist of carbon, zinc and
sal ammoniac with other ingredients, in fact they are Leclanch6
cells. The form usually taken consists of an outer cell of zinc
which is one electrode and is lined with blotting paper. The car
bon stick is placed in the center and is surrounded with a mixture
of manganese dioxide and powdered carbon. The blotting paper
and the powdered mixture are saturated with the electrolyte and
144 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxn
a layer of sawdust is placed on the top after which the cell is
sealed with pitch. The outside is then wrapped with paper so
as to insulate the zinc. The current depends on the size of the
celly the usual size^ 2.5 in. dia. and 6 in. high, will give about
15 amperes on short circuit for a few seconds.
Such cells deteriorate in storage, to prevent which one type has
a hollow carbon rod with a stopper; it is shipped dry and so is
chemically inactive until ready for use when it is filled with water
through the hole in the carbon.
These cells run down as the active materials are consumed
but may often be recuperated temporarily by the addition of
sal ammoniac to the blotting paper through a hole made in the
pitch, even water will help in an emergency.
171. Edison Lalande Cell. — ^The active materials in this cell
are copper oxide (CuO) and zinc in a solution of caustic potash
KOH. The oxygen ions combine with the zinc to form zinc
oxide and this combines with the potash to form a soluble salt.
The hydrogen reduces the copper oxide to metallic copper and so
does not polarize the cell. The construction is very rigid. The
positive plate of CuO and the negative Zn plate are separated
by porcelain spacers and rigidly fastened to the top of the con
taining jar which jar is of enameled steel. A layer of mineral oil
is placed on the top of the electrolyte to keep out the air and to
prevent the formation of creeping salts.
The internal resistance of this cell is low so that large currents
can be drawn from it; thus a 600. amperehour battery has an
opencircuit voltage of 0.95 and an internal resistance of about
0.02 ohm; this battery is designed to deliver 7 amp. for 85 hours
although the battery would give over 40 amp. on short circuit.
172. Power and Energy of a Battery — The active materials
in a battery contain a definite quantity of chemical energy
which may be transformed into electrical energy, so that a bat
tery can give a definite number of watthours or a definite num
ber of amperehours at normal voltage.
This energy may be taken as a small current for a long time
or a large current for a short time so that while the total energy
in the battery is fixed by the weight of the active materials the
power of the battery (volts X amp.) may vary over a wide range.
173. Battery Connections. — If Eo is the opencircuit voltage
of a battery, Rj, is the internal resistance and R is the resistance
Abt. 173]
ELECTROLYSIS AND BATTERIES
145
of the external circuit then the current / =
E,
maximum value on short circuit =
E.
Ri
Rb + R
and has a
If n batteries are coimected in series, then the current I =
and an increase in the number of batteries does not
nRb + R
produce any considerable increase in the current unless R is large
compared with Rb.
If n batteries are connected in parallel, then the current I =»
^^Vo and an increase in the number of batteries does not
Rb/n + R
produce any considerable increase in the current unless R is
small compared with Rb.
The internal resistance of a Daniell cell is 2 ohms and the noload voltage
is 1.1 volts. The current when the batteries are connected in series and
in parallel is given in the following table (a) when the external resistance
is 10 ohms and is greater than that of the battery; (b) when the external
resistance is 1 ohm and is less than that of the battery :
Number of cells
Total
resistance
of cells
External
resistance
Amperes
Terminal voltage
Open circuit
With
current
(a)
10 in series
10 in parallel
(b)
10 in series
10 in parallel
2 ohms
20 ohms
0.2 ohm
2 ohms
20 ohms
0.2 ohm
10 ohms
10 ohms
10 ohms
1 ohm
1 ohm
1 ohm
0.092
. 365
0.108
0.365
0.52
0.92
1.1
11.0
1.1
1.1
11.0
1.1
0.92
3.65
1.08
0.365
0.52
0.92
CHAPTER XXIII
STORAGE BATTERIES
174, Action of the Lead Cell. — If a plate of lead peroxide
(Pb02) and one of lead (Pb) are placed in a solution of
sulphuric acid (H2SO4) a battery is formed with the peroxide
plate at the higher potential. The generally accepted theory of
operation is as follows :
The sulphuric acid acts on the lead plate to form lead sulphate,
which material stays on the plate
H2SO4 + Pb = PbS04 + H2
The hydrogen ions have their charge neutralized at the peroxide
plate which they reduce to lead monoxide
H2 + Pb02 = PbO + H2O
The lead monoxide thus formed is acted on by the acid to form
lead sulphate, which material stays on the plate
PbO + H2SO4 = PbS04 + H2O
The final result may therefore be represented by the equation
Pb + Pb02 + 2H2SO4 = 2PbS04 + 2H2O
so that, during discharge, sulphuric acid is taken from the elec
trolyte, water is added to it, and the specific gravity of the
electrolyte is thereby decreased, while both plates are converted
into lead sulphate.
If now current from some external source is forced through
the cell in the opposite direction so as to cause electrolysis,
the action is completely reversed:
The positive H ions of the acid are attracted to the negative
PbS04 and reduce it to lead
H2 + PbS04 = Pb + H2SO4
The negative SO4 ions of the acid are attracted to the positive
146
Art. 176] STORAGE BATTERIES 147
PbSOi and, being liberated there, act on the water of the solution
to form sulphuric acid and oxygen .
2SO4 + 2H2O = 2H2SO4 + O2
This oxygen, with more water from the electrolyte, act on the
sulphate plate to form peroxide of lead and sulphuric acid
O2 + 2H2O + 2PbS04 = 2Pb02 + 2H2SO4
The final result may therefore be represented by the equation
2PbS04 + 2H2O = Pb + PbOa + 2H2SO4
so that during charge the plates are reformed while sulphuric
acid is added to the electrolyte, water is taken from it, and the
specific gravity of the electrolyte is thereby increased.
After the plates have been completely reformed, further charg
ing will cause the hydrogen and oxygen to appear as gases which
bubble up through the electrolyte from the surfaces of the plates,
the cell is then said to be gassing.
176. Storage or Secondary Battery. — An electric battery which
can be reformed by chemical means is called a storage or sec
ondary battery. There is no essential difference between a
primary and a secondary battery. In the former the active
materials themselves are renewed when the cell is exhausted,
whereas the latter is designed to permit of the materials being
brought back to their original state by electrolysis, that this
may be possible, no product formed during discharge must be
lost.
176. Sulphation. — There would appear to be two forms of lead
sulphate, an unstable electrolytic form which is readily reduced
by an electric current, and the lead sulphate formed by chemical
precipitation which latter is a nonconducting substance not
decomposed by an electric current. This latter substance must
not be allowed to accumulate on the plates.
The electrolytic form changes slowly into the insoluble form
and for that reason a lead battery must not be left discharged for
any length of time; insoluble sulphate also tends to form if the
battery is discharged too far.
The formation of this insoluble sulphate is called sulphation
and must be prevented by proper operation of the battery. If
sulphation has commenced on some of the plates and has not
gone too far, the plates may be cleared by overcharging the
148 PRINCIPLES OF ELECTRICAL ENGINEERINQ [CHAP.jcxra
battery for a long time, the hydrogen and oxygen formed when
the cell is overcharged tear off the insoluble sulphate.
177. Construction of the Plates. — When fully charged, the posi
tive plates are chocolate in color and the peroxide is hard while
the negative plates are gray in color and the spongy lead is so soft
that itcan be scraped off with thefinger n^l. During discharge,
these plates are converted into lead sulphate, which is bulky, so
that the plates expand and, unless carefully deagned, are liable
to buckle, especially if the cell is discharged rapidly so that the
Fio. 165.— 'Plants plate, showing cross section.
sulphate is formed rapidly and loosely. The negative plate in
addition must be so designed that the soft material will not be
readily washed off.
There are two ways of forming the active material. By the
Plants process the material is formed electrochemically out of the
lead plate itaeif, the plates being grooved or made in the form of a
grill so as to have a large exposed surface. The plates shown in
Fig. 165 are made out of pure rolled lead passed backward and
forward through grooving rolls which spin the lead into ribs the
pitch of which is made to suit the service. The peroxide and
spoi^y lead, formed electrochemically on the positive and nega
Art. 178] STORAOE BATTERIES 149
tive plates respectively, pack into the grooves and are tightly held
in the narrow spaces.
Pasted plates are made by spreading a paste of the active
material on to a supporting grid of lead hardened with a small
quantity of antimony, this being the only commercially available
material which will resist the action of sulphuric acid and will not
set up local action with either the spongy lead or with the
peroxide. Most of the processes by which these plates are pre
pared are secret, but by such processes plates can be made soft
and porous or hard and dense according to the service for which
they are required.
Plants plates are used largely for stationary batteries; they are
heavier and more costly than pasted plates but are also more
Positive group Negative group.
Fig. 166. — Groups of plates tor a lead battery.
durable and less liable to lose active material by rapid charging
and discharging. For automobile and motor truck service,
pasted plates are generally used because they are lighter than
Plants plates.
178. Construction oi a Lead Battery. — To obtain large capacity
from a battery, a large surface must be exposed to the electrolyte,
and, since the size of a single plate is limited, increased capacity
must be obtained by connecting a number of plates in parallel to
form a group as shown in Fig. 166, there being one more plate in
the negative than in the positive group. Two sets of plates are
then sandwiched together, adjoining plates being separated from
one another by glass rods in the case of large powerhouse cells or
150 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxiu
by wooden and rubber separators in smaller cells. The wooden
separators, see Fig. 167, are specially treated and are grooved
vertically to allow the gases and the electrolyte to circulate freely;
the Sat side is placed against the soft negative plate while between
the po«tive plate and the corrugations on the wood a sheet of
perforated hard rubber is placed, as shown in Fig 167, which helps
to prevent washing out of the active material.
The plate groups are placed in an acidproof tank, generally
of glass, hard rubber, or of wood lined with lead, and are aup
Fio. 167. — Portable type of lead battery.
ported in various ways as shown in Figs. 168 and 169, ample
space being left below the plates for the accumulation of sedi
ment which must not be allowed to short circuit the plates,
To minimize leakage of electricity, the tanks are insulated from
one another. Small cells are generally carried on shallow trays
filled with sand and supported on glass insulators as shown in
Fig. 168. Large leadlined tanks are generally mounted as shown
in Fig. 169 with a double set of insulators between the tank and
the ground.
Abt.1791 storage BATTERIES 151
To prevent loss of electrolyte due to sprayii^ when the
cells are gassii^ freely, glass sheets are placed over the tanks.
For automobile work, hard rubber jara are used. These are
placed in a wooden box and compound is poured around the
jars and flooded over the top so as to hold them securely and
also to seal the battery and thereby prevent loss of electrolyte.
To provide for the escape of the gases generated during over
charge, vents are provided, constructed so as to allow the gases
to escape but prevent the escape of the electrolyte.
1 a glass jar Fio. 169.—
179. Voltage of a Lead Battery. — The terminal voltage of a
battery
= Et + IR while the battery is charging
= El, — IR while the battery is discharging
where Et is the internal generated voltage of the battery
/ is the current in amperes
R is the effective internal resistance in ohms
so that the larger the current the greater the difference between
the charge and discharge voltages.
152 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxhi
The value of Eb, the generated voltage, depends on the strength
of the electrolyte and increases slightly as the acid becomes
stronger, it therefore increases during charge and decreases
during discharge.
The internal resistance of the cell depends on the specific
gravity of the electrolyte and is a minimum when this has a
value of about 1.22, which is about the normal value used in
cells. During discharge the acid becomes weak, particularly in
the pores where the action is taking place, and sulphate is formed
which is a comparatively poor conductor of electricity, so that,
towards the end of discharge, the internal resistance of the cell
increases rapidly. During charge the acid becomes strong,
particularly in the pores, so that toward the end of charge the
internal resistance again rises.
I
a
B
M A
Charged at 32 Amp.
8.0
y
b
^
d
:::^
/^
2.0
—
—
y
^
—
7~^
■^S^
e
«.7
imp.
e
•
29 An
P.
1.0
■

:
t
2 ;
i
H(
b (
inrs
J
1
10
Fig. 170.^ — Charge and discharge curves of the same lead cell at different
discharge rates.
The curves in Fig. 170 show how the voltage varies during
charge and discharge, the charge being continued until the voltage
and the specific gravity have become constant and the active
materials therefore completely formed, while the discharge is
stopped when the terminal voltage has dropped to about 1.8,
the safe minimum value at the normal discharge rate. If the
discharge is carried much further than this it becomes difficult
to clear the plates of sulphate on recharging.
average voltage on discharge Eh — IR .
The ratio
is called the
average voltage on chaf ge Eb + IR
volt efficiency and is lower the larger the current 7, that is, the
higher the rate of charge and discharge. At the normal rates of
* Secondary Cells by Aspinall Parr; Journal of the Inst, of Elect. Eng.,
Vol 36, p. 406; sec. 1905.
Abt. 181 .] STORAGE BA TTERIES 1 53
charge and discharge this efficiency is seldom less than 80 per
cent.
180. Capacity of a Cell. — The current that can be drawn from a
cell depends on the plate surface exposed to the electrolyte, on
the porosity of the plates, and on the rate of discharge. An
excessive current is liable to buckle the plates, see page 148, while
a short circuit will cause such violent action and such a sudden
evolution of gas in the body of the active material as to cause
parts of this material to be ejected from the plates.
The capacity of a battery is given in amperehours at a definite
rate of charge and discharge, generally an 8hour rate, except
in the case of automobile batteries which are generally rated at a
4hour rate as this more nearly corresponds to the actual service
conditions. A battery with a rating of 100 amp.hours will
deliver 12.5 amp. for 8 hours before the voltage drops below
1.8. Theoretically this same battery, after being fully charged,
should give the same total quantity of electricity at all rates of
discharge, that is, it should give 25 amp. for 4 hours or 50
amp. for 2 hours, but it is found that the amperehour capacity
decreases as the discharge rate is increased as may be seen from
the curves in Fig. 170. This particular battery had a capacity of
29 amp. for 10 hours or 290 amp.hours at a 10hour rate,
42.6 amp. for 6 hours or 255 amp.hours at a 6hour rate,
70 amp. for 3 hours or 210 amp.hours at a 3hour rate,
134 amp. for 1 hour or 134 amp.hours at a 1hour rate.
The cause of this loss of capacity at high rates of discharge
is that the active acid in the pores becomes used up in forming
sulphate, so that only water is left in the pores since the acid
is not replenished fast enough by diffusion from the bulk of acid
in the tank, the action is therefore limited to the surface layers
when the rate of discharge is high and does not penetrate into
the active material.
181. Amperehour Efaciency. — ^Test data is given in Fig. 170 on
a particular battery which, after being completely charged, was
able to deliver 134 amp. for 1 hour before the voltage had
dropped to 1.7. To recharge this battery required 32 amp.
for 6 hours or 150 amp.hours before the voltage and the specific
gravity had become constant. The reason for this additional
quantity of electricity is that, toward the end of a charge, hydro
gen and oxygen gases are given off and these require a definite
154 PRINCIPLES OF ELECTRICAL ENGINEERINQ [CHAP.xxm
quantity of electricity for their formation, which electricity is
not available on discharge because the gases have escaped.
If the charged battery is now discharged at 42.5 amp. the
voltage does not drop to li75 until after 6 hours and the output
is 42.5 X 6 or 255 amp.hours. The action being slower has
gone deeper into the plates and more active material has been
turned into sulphate, but, just because of this increased depth of
action, it is not advisable to allow the voltage to drop so far.
To recharge this battery now requires 32 amp. for 9 hours or
288 amp.hours.
The number of amperehours required to charge a battery is
greater than the number taken out during the previous discharge
amperehours output
and the ratio
is called the am
amperehours input to recharge
perehour efficiency of the battery.
From the test curves in Fig. 170 the following results are
determined:
Discharge
Charge
Amp.
Hours
Amp.
hours
Amp.
Hours
Amp.
hours
Am p. hour
efl&ciency,
per cent.
134
70
42.5
29
1
3
6
10
134
210
255
290
32
32
32
32
6.1
7.7
9.0
10.
164
246
288
320
82
85
89
91
From these figures it may be seen that the higher the rate of
discharge the lower the amperehour efficiency, the charging
rate being the same in each case. The reason for this is that when
the discharge rate is high the output is small so that the useful
input of the next charge is small and the portion used in forming
gases is proportionately large.
182. Watthour efficiency. — This quantity
watthours output
watthours input to recharge
amp.hours output X average discharge voltage
"" amp.hours input X average charging voltage
= amp.hour efficiency X volt efficiency
both of which quantities are lower, the higher the rate of charge
and discharge.
With floating batteries, which are used to carry peak Ibads
Art. 184]
STORAGE BATTERIES
155
of short duration and are not charged and discharged completely,
there is little or no gassing and the amperehour efficiency is
almost 100 per cent., while the peak voltages at the end of a
charge and the very low voltages at the end of a discharge are
avoided and the volt efficiency is high. Thus, while a battery
in a central station supplying a lighting load of several hours'
duration will have an average watthour efficiency over a period
of 12 months of 74 per cent., a similar battery in a central station
supplying a traction load, the load on the battery being inter
mittent, will have an average watthour efficiency of 84 per cent,
over a period of 12 months.
183. Effect of Temperature on the Capacity. — In general, the
cooler a battery is kept the longer is its life but the lower the
capacity. Lowering the temperature of the electrolyte increases
its internal resistance and causes an increased drop for a given
current. The curves in Fig. 171 show how large is the drop in
capacity. when the temperature is lowered; the temperature was
2.0
&
o
1
u
1.0
*^
,^ o
r^
40*C.
f^OT"
10 20
30 40
50
eo W
Fig. 171.* — ^Discharge curves of a lead cell with the same current but at differ
ent temperatures.
maintained constant during any one test by means of circulating
water.
The salvation of a battery in cold weather lies in the fact that
it is self warming, the internal resistance and therefore the
I^R loss become greater as the temperature decreases. For
electrictruck work in cold climates it is advisable to lag the
batteries, or at least to place them in a windproof compartment.
The temperature on charge should be limited to about 40° C;
continual operation at higher temperatures tends to reduce
the life of the cell.
184. Limit of Discharge. — ^As pointed out on page 152, a
battery should not be allowed to discharge to a voltage below
about 1.8 because there is then an excess of sulphate formed on
J, L'Bclwrage Electrique, Vol. 29, p. 160; Nov. 2, 1901.
IS
156 PRINCIPLES OF ELECTRICAL ENGINEERING [CHAP.xxra
the plates and a tendency for the plates to buckle and to sulphate
permanently. The amount of charge in a battery is best deter
mined by measurements of the specific gravity of the electrolyte
since this gives a measure of the amount of acid that has gone
to form sulphate on the plates. The specific gravity may
readily be measured by a hydrometer of the type shown in
Fig. 172.
The voltage of a battery is not an accurate index of its condition
because the voltage depends largely on the rate
of discharge. Voltage readings on open circuit
are of no value because this voltage is almost
independent of the amount of charge still in the
battery.
186. Treatment of Lead CeUs. — From a study
of the action of lead cells, the treatment they
should receive may be determined. The manu
facturers' instructions should be followed in every
ease but the following points require special at
tention.
A lai^e part of the wear of plates is due to gas
sing so that, while the beginning of a charge may
be at the 2bour rate, it is advisable to keep
the chatting rate slow toward the end of a
charge; the average charge rate of 8 hours should
be used whenever possible.
Cells should be overcharged about once a
month to get rid of the last traces of sulphate
_ „ and also to even up the cells and make sure
dronieter for that they are all charged up to their full capacity.
ttsting the ape "Poo rapid discharging causes the sulphate to
the electrolyte. form rapidly and tends to cause buckling, this
will seldom cause trouble in a welldesigned bat
tery. Overdischarge however must be avoided, the terminal volt
age not being allowed to drop below 1.8 at the normal 8hour
rate nor below 1.75 at the 4hour discharge rate.
The cell should not be allowed to stand discharged for any
length of time. If a battery has to stand idle for several months it
should be charged monthly because even a small leakage current
will cause enough sulphate to form on the plates to cause trouble
if it turns into the insoluble form. This monthly charge should
be continued until there is no further rise in voltage or of specific
Art. 186] STORAGE BATTERIES 157
gravity and until the cell has been gassing for about 5 hours, one
may then be reasonably sure that no sulphate has been left on the
plates.
Evaporation of the electrolyte should be made good by the
addition of pure water; the acid does not evaporate and, unless
there is excessive spraying due to the gases given off, the quantity
of acid in the cell will not change.
The specific gravity of the acid used depends on the use to
which the cell will be put while the permissible change in the
specific gravity depends on the bulk of acid in the cell and
should be obtained from the maker. When the cell has to stand
inactive for long periods, a weak acid is used to lessen the risk of
sulphation. For automobile work, the quantity of acid in the
cell should be such that the specific gravity shall not change more
than from 1.28 to 1.17 between full charge and full discharge; for
other service a range of from 1.23 to 1.15 is more usual, the gravity
being measured at a temperature of 70*^ F.
Before removing sediment from a cell, the plates should be
fully charged, then taken out and the separators removed. The
plates and tank should then be washed with water and the whole
battery put back into commission before the plates have time to
dry.
The gases formed during overcharge are explosive so that a
naked flame should be kept away from the battery room. The
room also should be well ventilated and the floor and walls should
be of some acidresisting material such as vitrified brick. The
room should be obscured from direct sunlight, which tends to
warp the plates; a heating system should be put in if the battery
has to operate in a cold climate, so as to maintain the capacity
under all conditions.
186. Action of the Edison Batteiy. — If a plate of nickel oxide
(Ni02) and one of iron (Fe) are placed in a solution of caustic
potash (KOH), a battery is formed with the nickel oxide plate
at the higher potential.
If current is drawn, from this battery, the oxide (Ni02) is re
duced to a lower oxide (NiaOs) while the iron is oxidized to form
FeO, and the cell is gradually discharged;
2Ni02 + Fe = NiaOs + FeO
If now current from some external source is forced through
the cell in the opposite direction so as to cause electrolysis, the
158 PRINCIPLES OF ELECTRICAL ENGINEERING [CnAP.xxin
action is completely reversed. The n^ative ions are attracted
to the positive Ni208 and the higher oxide Ni02 is reformed while
the positive H ions are attracted to the n^ative FeO and reduce it
to iron
NijO, + FeO = 2Ni02 + Fe
the result of charge and discharge is a transfer of oxygen from one
plate to the other; the strength of the electrolyte is not changed
so that the quantity required is less than for an equivalent lead
cell.
After the battery has been completely charged, the hydrogen
and oxygen appear as gases which bubble up through the
electrolyte just as in the lead cell.
187. Construction of the Plates. — The positive or nickel plate
shown in Fig. 173 consists of a nickelplated steel grid carrying
perforated steel tubes, one of which is shown in diagram B . These
tubes are heavily nickel plated and are filled with alternate layers
of nickel hydroxide and flaked metallic nickel. The hydroxide is
acted on electrochemically and becomes nickel oxide. This oxide is
such a poor conductor of electricity that the flaked nickel is added
to bring the inner portions of the oxide into metallic contact with
the surface of the tubes and thereby reduce the internal resistance
of the cell.
Each tube has a lapped spiral seam to allow for expansion, and
is reenforced with steel rings to prevent the tube from expand
ing away from and breaking contact with the enclosed active
material.
The negative or iron plate shown in Fig. 173 consists of a nickel
plated steel grid holding a number of rectangular pockets filled
with powdered iron oxide. Each pocket is made of two pieces of
perforated steel ribbon flanged at the side to forma little flat box
which may be filled from the end.
188. Construction of an Edison Battery. — A number of like plates
are connected in parallel to form a group, there being one more
plate in the negative than in the positive group. Two sets of
plates are then sandwiched together as shown in Fig. 174, ad
joining plates being separated from one another by strips of hard
rubber. End insulators A are provided with grooves which
carry the edges of the plates, and thereby act as spacers and at
the same time insulate the plates from the steel tank. The out
side negative plates are insulated from the tank by sheets of hard
Art. 188] STORAGE BATTERIES 159
rubber, while the whole unit rests on the rubber rack B by which
the plates are insulated from the bottom of the tank; this rack is
shallow since very little space is required for sediment in ao
Edison cell.
A. Pocket for negative plate.
B. Tube for the positive plate.
Positive plate N^ative plate.
Fig. 173.— Plates of an Edison Battery.
The tank, which is made of cold rolled steel welded at the joints,
is corrugated for strength as shown in Fig. 175 and is heavily
nickel plated as a protection against rust. The cover is of the
160 PRINCIPLES OF ELECTRICAL BNOINEERIffO [Chap.xxid
same material and is welded to the rest of the tank after the plates
have been put in place. This cover carries two terminab, as
well as a combined gas vent and filling aperture A. When the
cover b is closed, the hemispherical valve a closes the aperture
and prevents the escape of electrolyte, but allows the gases
generated on overcharge to escape as soon as the pressure in
the tank becomes high enough to raise the valve.
The electrolyte used consists of a 21 per cent, solution of
potash in distilled water to which a small amount of lithia is
added. No corrosive fumes are given off from this electrolyte
so that no special care need be taken in mounting the cells.
189. The Voltage of an Edison Battery.— Fig. 176 shows how
the voltage of an Edison battery changes when the battery is
charged and then discharged. The voltage characteristics axe
similar to those of a lead battery.
There is no lower limit to the voltage of an Edison battery
because in it there is nothing eqivalent to sulphation, but dis
charge is not continued below a useful lower limit.
190. Characteristics of an Edison Battery. — These batteries
are rated at a 7hour charging rate and a 5hour discharge rate
Art. 190.]
STORAGE BATTERIES
161
with the same current in each case, the amperehour efficiency
being about 82 per cent, at this rate and the internal heating
not more than permissible. A higher rate of discharge may be
used so long as the internal temperature does not exceed about
45*^ C; continual operation at higher temperatures shortens
the life of the cell. A longer charge rate than 7 hours should
not be used because, with low currents, the iron element is not
completely reduced; this however does not permanently injure
the cell but makes it necessary to overcharge the cell at normal
rate and then discharge it completely to bring it back to normal
condition.
Because of the comparatively high internal resistance of the
•5
1
lu;
xZ
_
^
2.0
ijead
Cell
,^^^
"^
i.0
"^
—
Bdl
sonC
bU
•
8 4 5 6 7 8
Honca
Fig. 176. — Charge and discharge curves of a lead cell and an Edison cell.
Edison battery, the volt efficiency is lower than in the lead cell,
as may readily be seen from Fig. 176, and, since the ampere
hour efficiency is not any higher, the watthour efficiency of the
Edison cell is also lower.
The great advantages of the Edison cell are that it is lighter than
the lead cell and is more robust, it can remain charged or dis
charged for any length of time without injury, and so little sedi
ment is formed that the makers seal it up. Since no acid fumes
are given off, the cell may be placed in the same room as other
machinery without risk of corrosion of that machinery.
The chief disadvantage of the Edison cell, in addition to its
high cost, is that its efficiency is lower than that of the lead cell.
\
CHAPTER XXIV
OPERATION OF GENERATORS
191. Operation of the Same Shunt Machine as a Generator or
as a Motor. — The generator G, Fig. 177, driven in the direction
shown, supplies power to the mains ran. The same machine,
operating as a motor from mains of the same voltage and polarity,
is shown at M\ the direction and the strength of the shunt field
are unchanged, the direction of the armature current is reversed, but
m
i
Generator Motor
Fig. 177. — Operation of the same shunt machine as a generator and as a
motor.
the direction of motion, determined by the lefthand rule, page 7
is the same as in (?. Since the back e.m.f. when the machine is
operating as a motor has to be practically equal to E, the e.m.f.
of the machine when operating as a generator, the machine must
run at the same speed in each case.
In diagram A, Fig. 178, m and n are two mains kept at a
constant voltage E by the generators in a power house, and D
is a single shunt generator of the same voltage running at normal
speed. If the voltage Ed is exactly equal to E and the polarity
162
Abt. 192]
OPERATION OF GENERATORS
163
is as shown, then no current will flow in the lines a and 6 when
the switch S is closed. If the excitation of D is now increased
slightly so that the voltage generated in the machine is greater
than the line voltage E, then current will flow in the direction
of the generated voltage, as shown in diagram B, and the machine,
operating as a generator, will supply power to the circuit mn.
If now the excitation of D is decreased so that the voltage gen
erated in the machine is less than the line voltage £, then current
will flow in the direction of the greater voltage, that is, in a
direction opposite to that of the generated voltage, as in diagram
C, and the machine, operating as a motor in the same direction
as before, will take power from the circuit mn. Thus by merely
varying the excitation of D it may be made to act as a generator
or as a motor.
m
n
D
m
n
ri
Fig. 178. — Operation of the same shunt machine as a generator and as a
motor.
192. Loading Back Tests. — ^Load tests on large electrical ma
chines must be made by some method whereby the power de
veloped by the machine is not dissipated but is made available
for the test, otherwise the powerhouse capacity may not be
large enough to allow many machines to be tested, while the cost
of such tests will be excessive.
If the machine to be tested is a gejierator, it is driven at normal
speed by a motor of the same voltage but of larger capacity, and
both machines are connected to the power house mains as
shown in Fig. 179. The motor M is started up by means of a
starting box in the usual way, and the generator G is excited until
its voltage is equal to £, the switch Si is then closed and a volt
meter V is placed across the switch S2. If the reading of this
voltmeter is twice normal voltage, then the polarity of G must be
164 PRINCIPLES OF ELECTRICAL ENGINEERING Chap, xxiv
reversed, see problem 96, page 395, but if the reading is zero, then
the switch Si may be closed and the generator G thereby con
nected to the mains. There will then be no current in the leads
a and b so that the generator will be running light, and the motor
will be taking from the power house only that power required to
operate the two machines at noload.
If the excitation of G is now increased so as to increase its
generated voltage, then current will flow in the direction shown;
the machine will deliver power to the circuit mn while the motor
M, which drives G, will take from this same circuit an amount of
power equal to the output of G plus the losses in the two machines,
of which the portion required to supply the losses is all that is
taken from the power house since
To Power>hoaie ^
the output of G is sent back into the
power mains.
193. Parallel Operation.— The
load on a power station is generally
distributed among several genera
tors connected in parallel with one
another, so that a breakdown of
one unit will not seriously cripple
Fig. 179.Loading^back test on ^y^^ ^^^^^^^ Parallel operation of
generators has the additional adi.
v^tftgpi Ihat the number of generating units in operation can be
changed with the load, so as to maintain the individual machines
at approximately fullload, at or near which load they operate
with their highest efficiency.
194. Shunt Generators in ParalleL — A and B, Fig. 180, are
shunt generators which feed into the same mains m and n. Sup
pose that A has been carrying all the load and that it has become
necessary to connect generator B to the mains to share the load.
This latter machine is brought up to speed with the switch S open,
its field rheostat is adjusted until Ei is equal to E, and the switch
S is closed. The load on B is then zero. To make the two
machines divide the load, the excitation of B is increased so as to
increase its generated voltage and thereby cause the machine to
deliver current to the mains.
If, due to a momentary increase in speed or for some other reason,
machine A takes more than its proper share of the total load, the
voltage of A drops since it is a shunt generator, see page 73, and
part of this load is automatically thrown on B, the machine with
Abt. 195]
OPERATION OF GENERATORS
165
i
E
T
I
the higher voltage at that instant. Furthermore, if the engine
connected to B fails for an instant, that machine slows down, its
generated voltage drops, and the load is automatically thrown on
A; if this generated voltage drops far enough, then current flows
from the line to operate machine S as a motor at normal speed and
in the same direction as before, see
page 162, but, as soon as the engine
recovers, this machine again takes
its share of the load. The opera
tion of two shunt machines in paral
lel is therefore stable, each machine
refuses to take more than its proper
share of the load and yet helps the
other machine when necessary.
To disconnect machine B, its ex
citation should be reduced until A
is carrying all the load, the switches
S may then be opened.
195. Division of Load among
Shunt Generators in Parallel. — The external characteristics of
the two shunt generators are shown in Fig. 181. When the line
voltage is E, the currents in the machines are la and h and the
line current is la + h' If the current drawn from the mains
^
Fig. 180. — Parallel operation of
shunt generators.
«
o
a
B
b
^
P:=::
•^b
•
1
a
1
1
/a
h
«
>
9
a
«
Armature Carrent
Fig. 181
100
400
EilowattR
Armature Current
Fio. 182 Fig. 183
FiQs. — 181, 182 and 183. — Division of load between two shunt generators in
parallel.
decreases, the voltage E rises and the currents in the two ma
chines are then to and 4 when the line current is ia + 4.
To make machine A take a larger portion of the total load, its
excitation must be raised so as to raise its characteristic as shown
in Fig. 182.
If a 100kw. and a 400kw. machine have the same regulation
166 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxiv
and therefore the same drop in voltage from noload to full
load then, as shown in Fig. 183, the machines will divide the
load according to their respective capacities.
196. Compound Generators in Parallel. — A and B, Fig. 184,
are two compound generators which are operating in parallel.
If, due to a momentary increase in speed, machine A takes
more than its proper share of the total load, the series excita^
tion of A increases, its voltage rises, and it takes still more of
the load, so that the operation is unstable.
To prevent this instability, the points e and /, Fig. 185, are
joined by a connection of large cross section and of negligible
resistance, called an equalizer connection. The series coils P
and Q are thus connected in parallel with one another, between
the equalizer and the negative main n, as is shown more clearly
— +
^ 1
i
1
:
Fig. 184. — Compound generators
in parallel.
H
£qaaliter
b\C
+"•
n
Fig. 185. — Compound generators
in parallel, an equalizer being sup
plied and the machine A being
overloaded.
in Fig. 186, and the total current from the negative main n
always passes through these coils in one direction and divides
up between them inversely as their resistance, independently of
the distribution of the load between the machines. If now, due
to a momentary increase in speed, machine A takes more than
its proper share of the total load, as shown in Fig. 185, and
therefore less is left for machine B, the series excitation of the
two machines is unchanged, since the total load is unchanged,
so that the machines act as shunt generators with a constant
superimposed excitation and the voltage of A decreases and that
of B increases, and part of this load is automatically thrown
on B, the machine with the higher voltage; the operation of the
machines has therefore been made stable by the addition of the
equalizer connection.
Art. 197]
OPERATION OF GENERATORS
167
To connect machine B in parallel with machine A which is
already running, bring the machine up to speed with the switches
a, 6, and c open, close switches b and c in order to excite the series
coils, then adjust the shunt excitation until Eb is equal to E,
and finally close switch a, the machine may then be made to take
its share of the load by increasing its shunt excitation. To
disconnect the machine, its shunt excitation should be reduced
until all the load has been transferred to A, the switches should
then be opened in the reverse order.
For large machines three separate switches are generally used.
For smaller machines the switches b and c are often combined
to form a double pole switch. When the machines are a con
siderable distance from the mains m and n, the equalizer is often
run straight between the machines as shown in Fig. 186.
m
— n
— ff
Fig. 186. — Compound generators in parallel, showing methods of changing
the compoimding.
197. Division of Load among Compound Generators. — When a
single compound generator has too much compounding, a shunt
in parallel with the series field coils will reduce the current in these
coils and so reduce the compounding, page 77.
When one of a number of compound generators in parallel
is found to take more than its share of the load, then its com
pounding must be reduced, this, however, can no longer be ac
complished by placing a shunt in parallel with the series coils of
that machine, for example the shunt S will not only reduce the
current in the series coils Q but will at the same time reduce the
current in the series coils P since the two sets of series coils and
the shunt iS are then all connected in parallel between the negative
main and the equalizer, as shown in diagram X, and the total
line current will divide among them inversely as their resist
168 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxiv
ance. The compounding of both machines will therefore be
reduced.
To reduce the current in the series coils Q without at the same
time reducing that in the coils P, a resistance must be placed in
series with Q as shown in diagram Y.
CHAPTER XXV
OPERATION OF GENERATORS AND BATTERIES IN
PARALLEL
198. Isolated Lighting Plants. — ^The engine and generator
capacity in such plants is generally sufficient for the total
number of lamps connected^ but since the lamps are seldom all
in use at one time, the plant operates at partial load and con
sequently with low efficiency. When a suitable storage battery
is installed, the generator may be operated for a few hours to
charge the battery and may then be shut down, the battery
being left connected to supply the load current.
199. Lighting Plants for Farm Houses. — The equipment for
such plants is shown diagrammatically in Fig. 187, 30volt
tungsten lamps being used since they have stronger filaments
than 110 volt lamps and can therefore be made in smaller sizes,
see page 353.
The voltage of a lead cell varies from about 2.65 volts on full
charge to 2.2 volts at the beginning of discharge and 1.8 volts
at the end of discharge, so that if 16 cells are connected in series
then: the battery voltage on full charge = 16 X 2.65 = 42.5 volts,
at the beginning of discharge = 16 X 2,2 = 35.2 volts,
at the end of discharge = 16 X 1.8 = 28.8 volts.
If the voltage across the lamps is raised much above 35 volts,
then the 30volt lamps will be burnt out, so that the lamp circuit
must be disconnected while the generator is charging the battery.
Specify the generator and battery for a lighting plant with a connected
load of 24 tungsten lamps of 15 watts and 12 candle power each,
watts per lamp = 15
current per lamp = 15/30 = 0.6 amp.
current for 24 lamps = 0.5 X 24 = 12 amp.
battery capacity at normal 8hour rate = 12 X 8 = 96 amp .hours
maximum generator voltage » 16 cells at 2.65 volts » 42.5 volts
charging current at 8hour rate = 12 amp.
generator output « 12 X 42.5 » 510 watts
169
170 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxv
If the average daily load on the battery is 6 lamps for 4 hours then the
amperehours taken each day = (6 X 0.5) X 4 » 12 and a 96amp.hour
battery will supply this load for 96/12 or 8 days without having to be re
charged.
To prevent the generator from being connected in parallel
with the battery except when its voltage is higher than the
battery voltage, an automatic switch S is supplied. To charge
the battery, the lights are disconnected, the generator which
is shunt wound is started up by the engine, and the shunt
rheostat r is gradually cut out to raise the generator voltage
until the value is reached for which the solenoid F was set, the
pull of this solenoid then closes the switch S and connects the
generator in parallel with the battery. The generator then
delivers current to the battery, which current flowing through
the coil H, adds to the pull of F and helps to keep the switch S
closed.
Lamps
Fig. 187. — Connection diagram for a small iso
lated lighting plant.
Fig. 188. — Lamp circuit
regulator.
If, due to a loose field connection or for some other reason, the
generator voltage drops below that of the battery then current
flows back through the coil H in such a direction as to oppose the
pull of F; the switch S is thereby released and the generator dis
connected from the circuit. The switch S therefore acts as a
reverse current circuit breaker.
200. Lamp Circuit Regulator. — One objection to the above
system is that the voltage across the lamps varies with the
battery voltage from 35.2 to 28.8 and the life of the lamps is
shortened due to the high voltage while the lighting is unsatis
factory when the voltage is below 30 volts. This trouble may
be overcome by the addition of aii automatic regulator. A very
simple type of regulator for this purpose is shown diagram
matically in Fig. 188 and consists of a carbon pile resistance
Abt. 202] OPERA TION OF GENERA TORS AND BA TTERIE8 171
R inserted in the lighting circuit, and a shunt solenoid P by which
the pressure on the carbon pile is varied. If the voltage across
the lamp circuit increases, the current in the solenoid P increases
and lifts the lever L, thereby reducing the pressure on the
carbon pile R and increasing its resistance, so that the voltage
drop across the carbon pile increases and the lamp voltage re
mains approximately constant. A more elaborate regulator
of this type is described on page 182,
201. SmaU Isolated Power Stations. — In such stations,
provision must be made for charging the battery and also for
carrying the day load at the same time. This is accomplished
either by resistance control, end cell control or booster control.
202. Resistance Control. — To take the case of a 110 volt plant.
The number of cells in series is 110/1.8 = 60 and, to charge
them in series, would require a maximum voltage of 60 X 2.65 =
r*l'l'l»
Hi'l»r
o
*♦■
il Charging
fl»l'l« ^'t4
Orerload and BeTorse
Gorreut Circuit Breaker
1
i>.
£• Discharging
Fig. 189. — Resistance system of battery control.
160 volts. But since the day load has to be carried by the
generators while the battery is being charged, it is not permis
sible to raise the generator voltage above 110 volts. This
difficulty is overcome by dividing the battery into two halves
for charging purposes and connecting them to the generator
8.8 shown in Fig. 189, the maximum battery voltage during charge
will then be 80 volts and the remaining part of the 110 volts must
be used up in the resistance R, by means of which resistance
the charging current may be regulated.
When the battery is fully charged the cells are reconnected
in series by throwing the switches over into position Si and,
since the battery voltage at the beginning of discharge is
13
172 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxv
60 X 2.2 or 132 volts while the generator voltage is only 110 volts,
the same resistance R must be kept in the battery circuit to
control the discharge.
203. End Cell ControL — ^With this system of control, shown
diagrammatically in Fig. 190, the battery is charged in series from
a generator which has a voltage range up to 160 volts, the charg
ing current being regulated by the generator field rheostat. Two
end cell switches Ci and Cj are required so that the 110 volt load
may be supplied while the battery is being charged.
When the battery is at the end of discharge, the number of
OTerload
and.
Beyenre
Carrent
e.b.
IFWW
J
Oyerli ad
e.lT\
R
t
S
s
W~f
no I
Voltil
110 VoUt
onLoadClrcttit
»— = — I
t— = — I
^= — •
ib==;^«
Ci t
I
L
110 Volts
Load
T
C' Bnd Cell
Switch
166 Volts T
i
110 Volta
A Qhsrsing
BDiaohtkt8tnt[
Fig. 190. — End cell system of battery control.
cells for 110 volts is 110/1.8 or 61 while at the beginning of
discharge 110/2.2 or 50 cells are all that are required, and 11 end
cells must be so arranged that they can be gradually connected
in circuit as the battery discharges. When the end cell switches
are in the position shown in Fig. 190, there are 50 main cells
and 4 end cells on the lighting circuit while all the cells are being
charged in series; these 4 cells on the lighting circuit will
gradually be cut out as the battery becomes more fully charged
and the voltage rises.
When the battery is fully charged, the switches are thrown over
Art. 204] OPERATION OF GENERATORS AND BATTERIES 173
into the position Si and the generator and battery are thereby
connected in parallel across the mains.
In order that the end cell switch will not open the circuit when
passing from one contact to that adjoining nor yet will it short
circuit any one cell, this switch is generally constructed as shown
diagrammatically in Fig. 19(> with a main cdntact a and an aux
iliary contact 5 electrically connected through a resistance r but
otherwise insulated frorii bne another. As the switch is moved
over the contacts, it stands for an instant in the position shown
and bridges one cell, but the resistance r keeps the current that
flows through this cell from being dangerously large.
Since the end cells are gradually put in circuit as discharge
proceeds, they are never so completely discharged as the rest of
the battery, so that, when the battery is being recharged with all
the cells in series, the end cells should be cut out one by one as
they become fully charged and begin to gas freely.
Soottei
TsFl
=1 J=?
^^ k
1
Oenerators
W
I _ 
<y
4. i
110 Volta
in
A Batt«ry ChKTgDxg
I — =
• — =
' — =
I — =
y
Ky
UOVoItft
T
B Battery Diicharging
Fig. 191. — ^Booster charge and end cell discharge system of battery control.
With this system of control there is not the resistance loss that
is present in the resistance control system, but the outfit is more
costly.
204. Booster Charge, End Cell Discharge. — ^For larger self
contained plants, the high voltage for charging is obtained by
connecting an additional generator B, called a booster, so that
its voltage is added to that of the main generators. To charge
the battery in Fig. 191, the switches must be thrown over into
position Si iand then, if the generated voltage is 110 and the maxi
mum voltage on charge is 160, the booster has to supply 60 volts.
The booster is generally driven by a constant speed motor and
174 PRINCIPLES OF ELECTRICAL ENGINEERING (Chap, xiv
is excited from the generator mains, the field rheostat having
sufficient resistance to allow the booster voltage to be reduced to
about 2 volts.
Fio. 192. — Large end cell switch.
When the battery ia fully charged, the double throw switch
is thrown over into the position shown in Fig. 191 so as to connect
the battery directly across the mains, the discharge voltage is
1
i
1
A B
Dll
.CI..T
J>.rn
i
l»
/"■■
lorLo
ti
eoo
f
^
s
/
\
MD
B
w
ys
litlenC
b.r«.
X
"^
.y
It S 6 9 AJS. 12
Fia. 193.— Daily load curve
a small station.
then regulated by the end cell switch C. The booster may be
used for this purpose if the discharge current is not too large^ see
next article.
Art. 205] OPERATION OF GENERATORS AND BATTERIES 175
The end cell switch for such plants, when of large size, gener
ally takes the form shown in Fig. 192, a laminated brush being
moved across a series of contacts by a motordriven operating
screw. This motor may be provided with pushbutton control if
desired.
206. Capacity of Battery. — ^The irregular curve in Fig. 193 is
the load curve on a small power station, and a battery is required
to supply all the current over 1000 amp.
The battery discharge » cross hatched area A
— 1140 amp. hours
the time of discharge » 2.7 hours
the average discharge current » 420 amp.
the maximum discharge current « 800 amp.
When the discharge takes place at the 2.7hour rate, the capacity of the
battery is only 75 per cent, of the normal capacity, see page 153, therefore
the normal capacity of the above battery «= 1140/0.75
s= 1500 amp.hours.
To recharge the battery, about 20 per cent, more amperehours must be
put in than were takeii out on the previous discharge when this discharge
was at the 2.7hour rate, see page 154.
therefore:
the chai'ge required » 1140 X 1.20
«= 1370 amp.hours
SB shaded area B
the battery charges for 8.7 hours'
the maximum charging current » 230 amp.
or is at the 1500/230 = 6.5hour rate
the average charging current » 1370/8.7 = 157 amp.
the booster has to carry a maximum of 230 amp. and must be designed for
a maximum voltage of 50 the output is therefore 50 X 230 » 11.5 kw.
This booster could not be used to help the battery to discharge
because the 800amp. discharge current would burn up a 230
amp. booster, end cells must therefore be supplied.
In working out this problem it has been assuilied that the battery
will never be called on to deliver more than 1140 amp.hours at
the 2.7hour rate.
No attempt has been made to determine whether or not the
above outfit is the most suitable for the particular service, this
is a question which can be answered only by trial of different com
binations of generating and storage equipment, the total operat
ing cost including maintenance and depreciation being figured out
in each case.
176 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxv
206. Batteries for Rapidly Fluctuating Loads. — ^The load on the
power house of plants such as rolling mills fluctuates so rapidly
that handoperated boosters and end cells cannot follow the
fluctuations and automatic control becomes necessary. Only
two of the many systems that have been devised will be described,
the principle in each case being to control the field excitation of
the booster by means of the load current in such a way that,
when the load is heavy the booster will assist the battery to
discharge, and when the load is light the booster will assist the
generators to charge the battery.
207. The Differential Booster in its simplest form is connected
as shown in Fig. 194 and is supplied with a set of shunt coils A
and a set of series coils B which coils are connected so that their
m.m.fs. are in opposition. With normal load on the generator.
Berlet Coilt Shant Coils
Oenerfttor
Fig. 194. — Differential booster system of battery control.
the m.m.f. of the series coils is equal and opposite to that of the
shunt coils and the resultant magnetic field is zero, so that the
booster voltage is zero and the battery neither charges nor
discharges.
If the load on the generator increases, the series excitation of
the booster becomes greater than the shunt excitation and the
booster voltage is then added to the battery voltage and causes
the battery to discharge and carry the larger part of the excess
load. If the generator load now becomes less than normal then
the series excitation decreases and the shunt excitation is now the
greater so that the booster voltage is reversed, opposes the battery
voltage, and thereby helps the generator to charge the battery.
208. Carbon Pile Regulator. — The diagram of connections for
a booster system controlled by a carbon pile regulator is shown in
Fig. 195. The booster B is excited by the coil E which takes
Art. 2081 OPERATION OF GENERATORS AND BATTERIES 177
current from a small exciter, the field excitation of which is con
trolled by a regulator consisting of two carbon piles ri and r^.
These carbon piles are subjected to pressure by the lever L, to
Fia, 195. — C&rbon pile r^:uIator for the automatic control of a battery.
one end of which is attached the plunger of the operatii^ so
lenoid S and to the other end a spring T, the tension of which may
be adjusted to counterbalance the pull of the solenoid when any
Fig. 196. — Carbon pile regulator for a battery.
desired current is flowing through it. The diagram of con
nections of the exciter field coil circuit is shown in diagram A.
The voltage «i tends to send a current ii = ei/{r + rj) through
178 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
the coil F while the voltage e^ tends to send a current it —
ei/(r+ri) through the same coil in the opposite direction.
If the generator load is normal then the pull of the solenoid S
is equal to the pull of the spring T and n is then equal to r^, ti is
equal to i^, diagram A, and no current flows in the coil F, so that
the exciter voltage and the booster voltage are both zero.
If the generator load increases, the pull on ri increases and its
resistance decreases and is now less than r^ so that t'l becomes
greater than it and current flows in the coil F in the direction of
ii and the booster voltage adds to the battery voltage and causes
the battery to discharge and carry the larger part of the excess
load.
eooo
01284 66789
Minute*
Fig. 197. — ^Load curves on the power house of a steel mill.
If the generator load now becomes less that normal, the pull on
ri decreases and its resistance increases and is now greater than
r2 and current now flows in the coil F in the direction of 1*2, the
booster voltage is thereby reversed and adds to the generator
voltage and thereby helps the generators to recharge the battery.
The curves in Fig. 197 show how nearly constant the generator
load may be maintained by such a regulator, while the indicator
cards in Fig. 198 show the effect of the regulator in maintaining
the steam consumption constant, an operating condition which is
favorable to economy in steam consumption.
Art. 2091 OPERATION OF GENERATORS AND BATTERIES 179
The average load on the generator 13 that at which the pull of
the solenoid S is exactly counterbalanced by the tension of the
spring T, for then the booster field excitation is zero; this load
may be adjusted by varying the tension of the spring.
One advantage of the externally controlled booster over the
differential booster is that the former machine is a standard shunt
generator whereas the latter has special series field coils and
heavy cables leading to these coils; some idea of the section of
copper required for these coils and cables may be obtained from
Fig. 196 which shows the section of copper required to carry the
current in a carbon pile regulator.
A — Battery in circuit. B — Battery out of circuit.
Fio. 198. — Indicator cards from the eteam enginea in the power house of a
Btoel mill.
209. Floating Batteries. — If a battery is connected in par
allel with the generator and with the load, aa shown in Fig.
199, then the condition to be fulfilled before the battery
will carry the peak load is that the voltage regulation of the
generator shall be worse than that of the battery, so that, as the
load increases, the generator voltage drops below that of the
battery and the battery has to discharge, while if the load
decreases, the generator voltage rises above that of the battery
and the battery is charged.
Such poor regulation is not desired in a power house, but is
often found at the end of a transmission line. If the length
L of this line is considerable and the section of the wire is small,
then a heavy load on the line will cause the voltage to drop suf
ficiently to allow the battery to discharge, while with a light
load on the line the drop is small and the voltage is then high
enough to recharge the battery.
The exact equivalent of a line with poor regulation is shown
in Fig. 200, which represents diagrammatically a hotel power
plant supplying lamps L and elevator and other motors M.
180 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxv
The generator is flat compounded to maintain the voltage con
stant across the lamps, while a battery is inserted to carry the
peak loads produced by the starting of the elevators. In order
that the battery may operate properly, there must be a con
siderable drop in voltage between the generator and the battery.
This is arranged for by placing a series wound booster in the
Generator T To Load Generator
I
To Motor* M
Booster
To LiihttBf
Circoitt L
Fig. 199. — Battery floating at the Fia. 200. — Series booster system of
end of a line. battery control.
circuit, the booster being driven at constant speed with its
voltage opposing that of the generators, so that if the elevator
motors take a large current the strength of the booster series
field is increased, this causes the booster voltage to increase
and the voltage Ei to drop and thereby allows the battery to
discharge.
■»
CHAPTER XXVI
CAR LIGHTING AND VARIABLE SPEED GENERATORS
210. The essential condition to be satisfied by any system
of electric lighting for vehicles is that the voltage across the
lamps shall be approximately constant for all speeds of the
vehicle from zero up to the maximum value.
For train lighting on steam railroads the three methods at
present in use are:
a. Lighting from storage batteries, called the straight storage
system.
b. Lighting from a constant voltage generator placed on the
locomotive or in the baggage car, called the head and end system.
c. Lighting from generators which are belted to the car axle.
For motorcar lighting, power is supplied by a generator which
is driven by the engine.
211. Straight Storage for Trains. — ^The power for the lighting
load in this case is supplied entirely by storage batteries which
are carried under the cars. At the end of each run the batteries
must be recharged or else replaced by fully charged batteries.
212. Head and End System. — ^The lighting load in this case
is supplied by a 100volt compound wound generator driven at
constant speed by a turbine which takes steam from the loco
motive. The generating unit may be mounted on the locomo
tive or in the baggage car.
In order that the lights on the train may not go out if the
train is parted or if the locomotive is disconnected, storage
batteries must be placed on at least the front and the rear cars,
the general practice being to place a battery on each car. The
system is then suitable for long runs where cars are not parted and
where no interchange of equipment is made with other roads.
The standard battery equipment for such service, if a lead
battery is used, consists of 32 cells in series so that
the battery voltage on full charge = 32 X 2.65 = 85 volts
at the beginning of discharge . = 32 X 2.2 = 70 volts
at the end of discharge == 32 X 1.8 =57 volts
181
182 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxvi
Since 60 volt lamps are used, an automatic regulator must be
placed in the lamp circuit of each car so as to maintain the lamp
voltage approximately constant.
213. Carbon Pile Lamp Regulator. — One type of regulator
which depends on the variable resistance of a carbon pile for its
operation is shown on the middle panelof Fig. 202 and is also
shown diagrammatically in Fig. 201.
If the battery is being charged, then the voltage Eh increases
and the lamp voltage Ei also increases slightly. This increases
the pull on the plunger a which increases the pressure on the
carbon pile r and decreases its resistance, more current therefore
flows in the coil h which is connected in series with r across the
lamp circuit, so that the plunger h is raised and the pressure on the
carbon pile R is decreased and its resistance increased, the
U— Cr*J
Fig. 201. — ^Lamp circuit regulator.
greater part of the increased battery voltage is therefore ab
sorbed by the resistance R and the lamp voltage remains apn
proximately constant. Due to the multiplying effect of the aux
iliary solenoid a and carbon pile r, a slight change in the voltage
across the lamps produces a considerable change in the pull of the
solenoid 6.
214. The Axle Generator Systems. — With these systems, the
lighting equipment of each car is self contained and consists of a
generator driven from the car axle and a storage battery to supply
power when the car is at standstill.
An automatic switch must be provided so that the generator
shall be connected to the battery only when the generator voltage
is higher than the battery voltage and shall be disconnected
when its voltage is lower. The generator also must be so con
trolled that the charging current shall not be excessive and
Aet. 2161 . CAR LIGHTING 183
shall moreover be reduced automatically aa soon as the battery
is fully charged. The combined automatic switch and generator
regulator shown on the bottom panel of Fig. 202 and diagram
matically in Fig. 203 is typical of the
several carbon pile regulating ^sterns
on the market.
216. Automatic Switch. — ^As the car
speeds up, the generator voltage increases
and when it reaches a value which is
greater than the maximimi battery volt
age the pull on the plunger P due to the
current in F becomes large enough to
close the switch S and connect the gen
erator and the battery in parallel. The
generator then delivers current to the
lamps and to the battery which current,
flowing through the coil H, adds to the
pull on the plunger and helps to keep
the switch closed.
If the car now slows down, the speed
finally reaches a value below which the
generator voltage is less than that of
the battery, and current flows back
through the coil H in such a direction
as to oppose the pull of F, so that the
switch S is released and the generator
disconnected from the circuit, the bat
tery being left connected to supply
power to the lamps.
216. Generator Regulator. — As the
speed of the generator increases, the
voltage of the generator and the charging
current in the battery both increase.
This charging current is hmited by the
regulator shown diagrammatically in Fig. 203 which consists of
a carbon pile rheostat R in the field coil circuit and a solraioid
B in the battery circuit. The weight of the plunger of B keeps
the carbon pile compressed, while the upward pull of the sole
noid decreases the pressure and thereby inserts resistance in the
ahunt cml circuit and cuts down the voltage of the machine.
The r^ulator is adjusted so that the plunger is pulled up and
184 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxvi
the pressure on the carbon pile is relieved as soon as the current
reaches the safe maximum charging rate of the battery.
If this were the only provision made for regulating the
generator, then the battery current would be maintained even
after the battery was fully charged, this would cause excessive
passing of the battery and a loss of water from the electrolyte;
the generator voltage would then be 32 X 2.65 = 85 volts with a
32cell battery.
In order to decrease the charging current once the battery is
fully charged, an additional relay is provided which limits the
generator voltage to some value lower than 85 volts. This is
accomplished by the shunt solenoid A which is connected across
the generator terminals as shown in Fig. 203. The weight of
Fio. 203. — Automatic switch, and generator regulator.
the plunger of this solenoid keeps the carbon pile compressed,
but when the generator voltage reaches a value somewhat lower
than 85 volts the plunger is pulled up and the pressure on the
carbon pile is relieved so that the voltage cannot increase further.
217. Pole Changer. — The polarity of the leads a and 6, Fig.
203, must be independent of the direction of motion of the car
in order that the batterycharging current may flow in the
proper direction, and the generator shunt field coils are connected
across ab instead of across the generator terminals in order that
the magnetic field produced may always be in the same direction
as that due to residual magnetism and may therefore build up
properly, see page 71. If then the direction of motion of the
car reverses, the polarity of the generator will also reverse, so that
Art. 218] CAR LIOHTINO 186
provision must be made for reversing the connections between
the brushes and the leads a and b. This may be accomplished
by a double throw switch such as that shown at T, Fig. 203,
which is thrown over by a mechanism on the generator shaft
whenever the direction of rotation of the generator is reversed.
218. The Stone Generator. — ^The voltage and current of this
generator are controlled by the slipping of the driving belt. The
generator is suspended by an adjustable linkL^ Fig. 204, and is
therefore free to swing toward or away from the driving pulley
on the axle. The belt is then adjusted to pull the generator out
of the position in which it would naturally hang and the tension
put on the belt by this means may be so adjusted by the linkL
that the belt will slip when the load exceeds a certain value.
The combined automatic switch and pole changer is at the
commutator end of the machine, see Fig. 205. The contacts B
B\lpplng_
Belt
Fig. 204. — Suspension of a slippingbelt type of generator.
are fixed while the contacts A are carried on a rocker arm C which
is loose on the shaft and is carried around by friction in the direc
tion of rotation until arrested in the position shown by a stop,
the blade A i is then opposite Bi and A 2 opposite B2. If the rota
tion had been in the opposite direction, the arm C would have
been carried around in this direction until Ai was opposite fis
and A 2 opposite Bi when the motion of the arm would have been
arrested by another stop.
When the speed and the voltage of the generator reach such a
value that the generator is able to charge the battery, then the
weights w are thrown out and the arm C with the switch blades
attached is pushed along so as to connect the generator and the
battery in parallel. If the speed decreases to such a value that
the generator is no longer able to charge the battery, then the
spring S pulls the switch blades out of contact and disconnects
the generator.
The various operations take place in the following order. When
186 PRINCIPLES OF ELECTRICAL ENGINEERING IChap. kcw
the car starts up, the rocker arm C moves around to its prop^
poeitioQ, and when the speed becomes high enough, the switches
AB are closed and the generator charges the battery. As the
speed increases, the generator voltage and the charging current
both increase until the load becomes large enough to cause the
belt to slip, the speed of the generator will then increase no
further nor will the chai^ng current in the battery increase.
This system has given satisfaction although its efficiency is not
80 high as that of some of the other systems because of the loss
in the shpping belt, it also has the objectionable feature that the
charging current is not reduced aft^ the battery is fully charged.
Pia. 205. — The Stone train lighting generator.
219. Lighting Generators for Motor Cars. — The equipment for
motorcar lighting is similar to that supplied for train lighting
^cept that a pole changer is not required since the generator,
which is driven by the engine, always rotates in one direction
when the speed is such that the generator is connected in parallel
with the battery.
220. Constant Speed Genorators. — Several machines operate
on the same principle as the Stone train lighting generator, the
slipping belt being replaced by a slipping clutch in order to make
the outfit as compact as possible. This clutch slips when the
generator output reaches a predetermined value.
221. Bucking Field Coils.— Differentially compounded genera
tors have been successfully used for vehicle lighting, the field
windings being connected as shown in Fig. 206 where A is a shunt
winding which is connected across the battery and therefore
An. 222] CAR UGHTING 187
gives apfgorimatdy constant excitation wtule fi is a series wind
ing which acts in (^position to or bucks the shunt winding.
When the speed of the generator reaches such a value that
the geDNattH* is able to charge the battery, the autcmatic switch
closes. As the speed increases furtho, the generator voltage
and the charging current both increase, but this charging cur
rent passes through the coils B
and reduces the recitation of the
machine, so that the charging
current is limited. The charging
current can never exceed the
value at which the amperetums
of winding B is equal to the am
peretums of winding A for then ^luh
the flux in the machine and the p,o. 206.— Generator with bucking
generated voltage would both be field coils,
zero,
222. ^^brating Contact Regulator. — ^A compact and eflicient
regulator for avariable speed generatwis shown diagrammatically
in Fig. 207, the car being at standstill. As the car speeds up,
the generator voltage increases and, when it reaches a predeter
mined value, the pull of the magnet M due to the shunt coil a
closes the main switch S and connects the battery and the generator
Fro, 207. — Vibrating contact regulator for a variable speed generator.
in parallel. The generator then delivers current to the battery
and to the lamps which current, Sowing through the coil b, adds
to the pull of the magnet and helps to keep the switch S closed.
As the speed of the generator increases, the voltage of the
generator and the chargii^ current in the battery both increase.
This current flows through the coil c and, when it becomes equal
188 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxvi
to the safe charging current of the battery, the pull of the magnet
N closes the. contact T and short circuits the generator field coils
thereby reducing the excitation and limiting the voltage of the
machine, the battery is therefore protected from excessive charg
ing currents.
In order to decrease the charging current once the battery is
fully charged, an additional coil d is placed on the magnet N and
is connected across the generator terminals. The voltage across
coil d increases as the battery charges until finally the current in
coil d is able to close the contact T, the voltage can then increase
no further so that the charging current gradually decreases and
is comparatively small when the battery is fully charged.
For motorcar work, a lamp circuit regulator is considered to
be an unwarranted complication since the battery voltage changes
slowly and the lamp voltage may be adjusted by means of a
hand operated rheostat if desired.
223. The Rosenberg Generator. — Diagram A, Fig. 208, shows
a generator with the field coils excited from a battery so as to
produce a flux 0. When the armature revolves in the direction
of the arrow, e.m.fs. are induced in the conductors, the directions
of which are shown by crosses and dots, and the voltage between
the brushes BB is Ei, while that between the brushes 66 is
zero since the e.m.fs. in the conductors from 6i to a are opposed
by those in the conductors from a to 62. If the brushes BB are
joined, then current will JBow through the armature and will
produce an armature cross field 0ia, see page 67.
Consider now the effect of this cross field 0ia acting alone.
Since this field is stationary in space it can be represented by
poles NiS2 as in diagram B, and the lines of force <t>ia are cut
by the armature conductors and e.m.fs. are induced in the
directions shown. The voltage between the brushes 66 is Et
while that between the brushes BB is zero. If the brushes 66
are joined, then current will flow through the armature and will
produce an armature cross field ^2a.
In one form of the actual machine shown in diagram C, the
poles NiSi are excited from the battery while N2S2 have no field
coils but carry the cross flux <t>ia. The brushes BB are short
circuited while the brushes 66 are connected to the load.
As the speed of the generator increases then, referring to dia
gram A, the voltage Ei increases causing the current /i and the
flux 01a to increase; referring now to diagram B, the voltage
CAR LIGHTING
Et iacreaaes with the flux <fna and causes 1% and ^2. to increase.
But 0ja opposes and demagnetizes the machine, and the
greater the speed the greater the demagnetizing effect, so that
It increases by a smaller and smaller amount and over a wide
range of speed remains practically constant. The current It
can never exceed the value with which tpsa is equal to ^ because
then the flux in the machine would be zero.
Such a machine is therefore a constant current generator,
C Complets Machine
Fio. 20S. — Rosenberg train lighting generator.
and by adjusting the exciting current //, the current It can be
limited to the normal charging current of the battery; the charg
ing current however does not decrease as the battery becomes
fully chained, a disadvantage this generator possesses in common
with the slipping belt type of generator.
The operation of this generator is described in detail because
it has several interesting applications. It isusedfortrainhghtiog,
when it is called the Kosenberg generator; a similar machine called
the C. A. y. generator is used for motor cars. The machine has
190 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xrvi
also been used as a constant current generator for the operation of
the arc of large searchlights and also for arc welding, since it is
practically foolproof and can be short circuited across the operat
ing brushes bb without the current exceeding the normal value.
CHAPTER XXVII
ALTERNATING VOLTAGES AND CURRENTS
224. The Simple Alternator.— If the coil abed, Fig. 209, be
rotated between the poles N and S so that the conductors ab
and cd cut lines of force, an alternating e.m.f. will be found be
tween/ and g, the ends of the coil. The direction of the e.m.f. in
each conductor, found by the righthand rule (page 9), is shown
in diagrams A, B, C and Dfor different positions of the conductors
relative to the poles. In diagram A the conductors are not cut
aO
S N
2 BeT.
Fia. 209. — Simple twopole alternator
ting lines of force, and the e.m.f. between / and g is zero. In
diagram B the e.m.fs. in the conductors are in such a direction
as to force current from/ to g through the external circuit, there
fore / is the positive and g the negative terminal of the machine.
In diagram C the e.m.f . between / and g is again zero. In dia
191
192 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
gram D the e.ni.fs. in the conductors are in such a direction as to
force the current from gto f through the external circuit, therefore
g is now the positive and / the negative terminal. The current in
the external circuit connecting / and g therefore alternates; that
is, electricity oscillates backward and forward in the circuit.
If the e.m.f. between / and g is plotted against time then a
curve such as that in Fig. 209 is obtained.
FiQ. 210. — Twopole revolvingfield alternator.
Alternators are generally made with stationary conductors
and a revolving field, as shown diagrammatically in Fig. 210.
The direction of the e.m.f. in each conductor, shown at different
instants in diagrams A, B, C and D, may be found by the right
hand rule;^ it must be noted that in the case of a revolvingfield
machine the thumb is pointed in a direction opposite to that of
the motion of the poles since, according to the rule, it must be
^ Thumb — direction of motion of conductor relative to magnetic field.
Forefinger — direction of lines of force.
Middle finger — direction of e.m.f .
Art. 226] ALTERNATING VOLTAGES AND CURRENTS
193
pointed in the direction of motion of the conductors relative to
the poles.
225. The Wave Form. — If the airgap clearance under the pole
is uniform in thickness then the lines of force crossing the air
gap are spaced as shown in Fig. 211, and the e.m.f. in a conductor,
being proportional to the rate of cutting of lines of force, varies
as in curve A. By shaping the pole face, however, as in Fig. 212,
the flux density in the air gap and therefore the rate of cutting of
lines of force may be so regulated that the e.m.f. in a conductor
shall vary according to a sine law as shown in curve B. The e.m.f.
is then said to be simple harmonic and may be represented by the
formula 6 = ^m sin 6.
Fia, 211. Fig. 212.
FiQS. 211 AND 212. — Wave form of electromotive force
226. The Oscillograph. — The shape or form of the e.m.f. wave
of an alternator may readily be determined by means of an instru
ment called an oscillograph, the essential parts of which are shown
in Fig. 213.
In the narrow gap between the poles NS of a magnet are
stretched two parallel conductors ss formed by bending a strip
of phosphor bronze back on itself over an ivory pulley P. A
spiral spring attached to this pulley serves to keep a uniform
tension on the strips, and a guide piece L limits the length of the
vibrating portion to the part actually in the magnetic field. A
small mirror M bridges across the strips as shown.
If current is passed through the strips sa then one strip will
194 PRINCIPLES OF ELECTRICAL ENGINEERING (Chap, xxvn
advance and the other will recede and the mirror will thereby
be tilted about a vertical axis. If the current is alternating then
the mirror will tilt backward and forward with a frequency equal
to that of the current, and the deflection will be proportional to
the current. (The natural frequency of vibration of the mirror
is at least fifty times the frequency of the current.)
If now a beam of light is directed on the mirror, the reflected
beam will move to and fro in the horizontal plane, its displace
Fig. 213. — The oscillograph.
Fia. 214. — Sixpole alternator.
ment from the zero position x being proportional to the current
flowing, so that if a photographic film / is moved downward at a
constant speed a curve will be traced on it by the beam of light,
which curve will be the wave of the e.m.f. applied at the oscillo
graph terminals.
227. Frequency. — ^In the twopole machine shown in Fig. 209,
the e.m.f. between the terminals passes through a complete cycle
while the machine makes one revolution. In the sixpole machine
Art. 2281 ALTERNATING VOLTAGES AND CURRENTS
195
shown in Fig. 214, the e.m.f. in any conductor a passes through
three cycles, one cycle per pair of poles, while the machine makes
one revolution.
If, in an alternator, p is the number of poles then
the cycles per revolution = ^
U IT U ITh
and the cycles per sec, called the frequency, = o ^ f{0 *
P X r.p.m. io
120
and is represented by the symbol /.
The frequencies generally found in practice in America are
25 and 60 cycles per sec, while in Europe 25 and 50 cycles per
sec are more common.
A 60cycle alternator has 24 poles, at what r.p.m. must it be run?
/ = p X r.p.m/120
therefore 60 = 24 X r.p.m/120
and r.p.m = 300.
The following table gives the relation between poles, speed and
frequency:
Revolutions per minute
Poles
25 cydes
50 cycles
60 cycles
2
1500
3000
3600
4
750
1500
1800
6
500
1000
1200
8
375
750
900
P
3000/p
6000/p
7200/p
It is important to note that an alternator has a definite speed
for a given frequency and cannot be run above or below that
speed without changing the frequency. In a directcurrent
generator, the voltage may be varied by varying the speed, but
in the case of an alternator this cannot be done without at the
same time changing the frequency.
228. Vibrating Reed Type of Frequency Meter. — ^In this type
of instrument a number of steel strips are fastened at one end as
shown in Fig. 216, while the current whose frequency is to be
determined is passed through the coil A . The reeds are attracted
twice in a cycle by the electromagnet B and that reed which has
a natural frequency equal to twice the frequency of the current
196 PRINCIPLES OF ELECTRICAL ENaiNEERINQ IChap. xxvn
Tlbntlu Strlpa
A — Interior construction.
B — Scale when the frequency is LOO cycles. C — External appearance.
FiQ. 216. — Vibrating reed type of frequency meter.
/ i 1
X"
Fig. 216. — Average value of an alternating electromotive foroe.
Art. 230] ALTERNATING VOLTAGES AND CURRENTS
197
will be set in violent vibration. The reeds have their free ends
whitened and appear as white bands when vibrating. The ex
ternal appearance of such an instrument is shown in diagram C.
229. Average Value of Current and Voltage. — ^The average
value of an alternating current or voltage is zero because similar
sets of positive and negative values occur. The term average
is generally applied to the average value during the positive part
of a cycle as indicated in Fig. 216.
2
Eav, the average e.m.f. =  En^
IT
2
lav, the average current =  Im
IT
230. The Heating Effect of an Alternating Current. — If a
direct current / is forced through a resistance of R ohms then the
power transformed into heat = PR watts.
/"f\
\ Ayerage 
I Vftlae of t*
Fia. 217.
If an alternating current i = Im sin 6 is forced through the
same resistance, then the power transformed into heat at any in
^Eav X TT
=  E^cos e
Ew»m e de
= 2E.
therefore
Eav = "~ Em
7C
198 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxvn
stant = i^R watts and the average power transformed into heat
= the average value of i^R, see Fig. 217.
== the average value of (/« sin oy R
= 7m* R (average value of sin* oy
^ 2
= PeffR
where ley/ = Im/'42 is called the effective current.
When an alternating current or voltage is specified it is always
the effective value that is meant unless there is a definite state
ment to the contrary, thus an alternating current of 10 amp. is
one which has the same heating effect as 10 amp. direct cur
rent and has a maximum value of 10 V2 or 14.1 amp. and an
2
average value of  X 14.1 or 9 amp.
231. Symbols. — Hereafter in the text the following symbols
will be used for alternating voltages and currents:
e si^msin^; i =/mSin9; the instantaneous value
Em Im ; the maximum value
2 2
Bav ^  Em lav ^  Im ) the avcrage value
IT IT
E = j^ I = TTT ; the effective value
232. Voltmeters and Ammeters for Alternatingcurrent Cir
cuits. — The moving coil permanent magnet type of instrument
as used for directcurrent circuits was described on page 8.
If this type of meter was connected into an alternatingcurrent
circuit the moving coil would be acted on by forces tending to
turn it first in one direction and then in the other but, due to
its inertia, the coil itself would not move and the reading would
be zero.
* The average value of sin' 6 Xr =1 sin*d de
I
=  i(j8in2^^j
"2
therefore the average value of sin* $ =» 1/2.
r
i (cos 2$  1) d$
Abt. 2321 ALTERNATING VOLTAGES AND CURRENTS 199
In order that a moving coil instrument may be used for the
measurement of alternating currents, the magnetic field and the
current in the moving coil must alternate together. This result
is obtained by replacing the permanent magnet by an electro
magnet as shown in Fig. 218. The current to be measured is
passed through the stationary coils A and through the moving
coil C in series, and the sides of the moving coll are then acted on
by forces which turn the coil against the tension of the spring S.
Fia. 218. — Electrodynamometer type of inatrumeot.
These forces are proportional to the current i in the coil C
and to the flux ^ produced by the current i in the coils A; the
forces are therefore proportional to i* and the average turning
force on the coil C while the current alternates is proportional to
the average value of i* or to the effective current.
Such an instrument may be used to measure both direct and
alternating currents and the reading would be the same for 10
amp. direct current aa for 10 amp. efifective alternating current.
CHAPTER XXVIII
REPRESENTATION OF ALTERNATING CURRENTS AND
VOLTAGES
233. Part of a rotating field alternator is shown diagram
matically in Fig. 219. As the field rotates, stationary conductors
FiQ. 219.
FiQ. 220.
such as a are cut by lines of force and the e.ni.f. in these con
ductors varies as shown in Fig. 220 and goes through one cycle
for every pair of poles that pass.
A B
Fia. 221. — Representation of an alternating voltage.
234. Electrical Degrees. — If the vector op rotate in the coun
terclock direction and the angle d is measured from the xaxis
then om, the projection of op on the z/axis = op sin 6 and its value,
plotted in diagram B, passes through one cycle while changes
through 360 degrees. If now op is drawn to scale equal to
200
Am. 2351 ALTERNATING CURRENTS AND VOLTAGES
201
Em, Fig. 220y then om — Em sin B and therefore represents the
instantaneous e.m.f. 6, the curves in Figs. 220 and 221 will
therefore be alike in every respect if the angle the machine
moves through in generating one cycle of e.m.f. is called 360
electrical degrees; in the machine in Fig. 219 this angle is that
between two consecutive like poles. From this it follows that
a curve such as that in Fig, 221 may be used to represent the
voltage generated by an alternator with any number of poles and
is the curve that would be obtained by an oscillograph.
235. Vector Representation of Alternating Voltages and Cur
rents. — It is generally assumed that these quantities vary accord
Fig. 222. — Representation of an alternating voltage and current.
ing to a sine law and can therefore be represented by sine curves
as shown in Fig. 222, where
f, the current at any instant = /« sin B
e, the voltage at any instant = Em sin
For much of the work on alternatingcurrent circuits and
machines it is more convenient to represent alternating voltages
and currents by the corresponding vectors^ I and E, Fig. 222,
from which vectors the sine curves may be obtained when desired
by plotting the vertical components i and e against the angle d,
the vectors being rotated in the counterclock direction.
If two oscillographs are used as in Fig. 223 one of which, A,
gives the voltage curve while the other, B, gives the current curve,
it will be found that the current and voltage do not necessarily
reach their maximum values at the same instant but that curves
such as those in diagrams A, B and C, Fig. 223, may be obtained,
depending on the kind of load connected to the circuit. The
reasons for the displacement of the current relative to the voltage
are taken up in Chapters 29 and 30 ; it is necessary, however, to take
up at the point the method of representing such curves by vectors.
^ Vectors are generally drawn to represent the effective value of current
and voltage rather than the maximum values.
202 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxvin
If two vectors such as E and /, Fig. 224, be rotated in the
counterclock direction at the same rate, the angle a between them
will remain unchanged and the vertical components e = Em sin 6
and i = Im sin 6, which are shown at three instants a, b and c,
I
B
\
M^
E
r
A Onrrent Laga Yoltace by
by a Degrees
I
B Onneni In Phase
with Voltage
J^E
ly.
Fig. 223.
C Oatxent Leads Voltage
by a JDtegziees
Phase relation between current and voltage.
when plotted against the angle through which the vectors have
moved measured from any base line ox, will give the sine curves
of E and / which curves represent a voltage and a current of the
same frequency, the voltage E reaching its maximum value a
degrees before the current becomes a maximum.
Art. 2361 ALTERNATING CURRENTS AND VOLTAGES
203
When the current and voltage reach their maximum values at
the same instant they are said to be in phase with one another.
When they reach their maximum values at different instants
they are out of phase and the current is said to be leading or lag
Fia. 224. — Representation of a lagging current.
FiQ. 227. — The sum of two alternating voltages of the same frequency.
ging according as it becomes a maximum before or after the vol
tage has reached its maximum value.
236. The Sum of Two Altemati:^ Voltages of the Same
Frequency. — If two directcurrent generators are connected in
204 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxviii
series as in Fig. 225, the resultant voltage E^ is the numerical
sum of El and Ei the voltages of the two machines.
If two alternators are connected in series as in Fig. 226, the
voltage 6s at any instant is the numerical sum of ei and €2 the
voltages of the two machines at that instant. In the particular
case shown, the voltage of the second machine lags, or reaches
its maximum value later than that of the first machine by the
time required for the poles to pass through a degrees, and the
curves representing the voltages of the two machines are ei and
62, Fig. 227. The points on the resultant curve e^ are obtained
by adding together the values of ci and 62 at different instants.
The voltages of the two machines may be represented by
the vectors Ei and E2 drawn to scale with an angle a between
them because, if these two vectors are rotated together in the
counterclock direction with this fixed angle a between them, then
the vertical components ei and 62, when plotted against the
angle turned through by the vectors, will give the curves d
and 62 in their proper phase relation.
The resultant of these two electromotive forces is the vector
^um obtained by the parallelogram law and is Ez which repre
sents the resultant voltage both in magnitude and in phase
relation.
CHAPTER XXIX
INDUCTIVE CIRCUITS
237. Inductance. — It has been shown that whenever there is a
change in the current flowing through a circuit, an e.m.f. of self
induction is induced which opposes the change of the current,
see page 11.
In Fig. 228, when the switch k is closed a current begins to
flow in the coil and as this current increases in value the flux *
Fia. 228. — Growth and deca7 of current in an inductive circuit.
threading the coil also increases. Due to the change in the flux,
an e.m.f. of self induction is induced in the coil which, accord
ing to Lenz's law, page 10, acts in such a direction as to oppose
the increase of the current.
If, after the current has reached its final value, the switch k
is suddenly opened, the current in the coil decreases, the flux
threading the coil also decreases and causes an e.ni.f, of self
induction to be induced in such a direction as to oppose the de
crease of the current. This e.m.f. is generally large enough to
maintain the current between the switch contacts for a short
205
206 PRINCIPLES OF ELECTRICAL ENOINEBRINO [Chap, xxn
interval as they are opened and accounts for much of the flash
ing that is seen when a switch is opened in a circuitcanying cur
rent. The growth and decay of current in auch a drouit is shown
by the curves in Pig. 228.
That property of an electric circuit whereby it opposes a change
in the current flowing is called the self induction or the inductance
of the circuit; the two terms have the same meaning but the
term inductance is generally used in en^neering work.
238. Make and Break Spark IgnMon. — One method of igniting
the gas in a gas engine cylinder is based on the above properties of
the inductive circuit; the essential parts of the mechanism are
shown in Fig, 229.
""TMIJ 4
Section through cylinder.
Fia. 229. — Make and break method of gae ignition.
When the contact at x is closed, current flows in the direction
of the arrow. When the cam c has reached a predetermined
position, the spring s opens the contacts and the current is maia
tained across the gap by the inductance coil L.
The current in the circuit changes as shown in Fig. 228 so that
the contact x must be closed long enough to allow the current to
grow to its full value, but should not be closed too long or the
batteries will run down.
239. The Coefficient of Self Induction.— If in Fig. 228 the
current / is changing, the flux ^ is also changing and a voltage
of self induction is induced in the coil which is equal to
e,i = — n "jtIO"* volts, see page 9'
It is often desirable to express this voltage in terms of the
changing current rather than of the changing flux produced by
the current
■ The minus sign is used because the e.m.f . oppoBee the change of the fiux.
Art. 240] INDUCTIVE CIRCUITS 207
so that we may write
where L is a constant called the coefficient of self induction^
which, from the above relations, has the value
L ==' nrr 10~^ henries.
at
In many cases in engineering practice the magnetic circuit is
not saturated and then the flux 4> is directly proportional to the
current / which produces it and in such cases we may write
L = y10""® in henries
= (flux interlinkages per unit current) X 10"^
240. Alternating Currents in Inductive Circuits. — If an al
ternating current is flowing in an inductive circuit then, since the
current is always changing, there must be an induced e.m.f. of
self induction opposing the change. If the current is represented
Fig. 230. — Voltage and current relations in an inductive circuit.
by curve /, Fig. 230, then between a and 6, during which interval
of time the current is decreasing, the e.m.f. of self induction, to
oppose this decrease, must be positive, while between 6 and c, dur
ing which interval of time the current is increasing, the e.m.f. of
self induction, to oppose this increase, must be negative. At the
instants a, b and c the current is not changing and at these in
stants the e.m.f. of self induction must be zero. The e.m.f. which
satisfies all those conditions is represented by the curve £,»,*
Fig. 230.
In order to force an alternating current / through a circuit, the
applied e.m.f. must be large enough to overcome the e.m.f. of
self induction and also the resistance of the circuit and, in the
^The coefficient of self induction of a circuit, often called its inductance,
is as much a constant of the circuit as is its electrical resistance.
* It is important to note that the generated voltage Eti lags the current
i and therefore the flux by 90 degrees.
208 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
extreme case of an inductive circuit of negligible resistance, the
applied e.m.f. must be equal and opposite to the e.m.f. of self
induction. The applied e.m.f. in this latter case is represented by
the curve Ea, Fig. 230, and it is important to note that, in the
case of an inductive circuit of negligible resistance, the current /
lags the applied voltage Ea by 90 degrees.
241. Voltage and Current Relations.— The current / in Fig. 230
changes from — /m to + /» in the time of half a cycle or in
Kf seconds so that
Eav, the average voltage of self induction between b and c,
= L (average value of ^ « l (T') = 4/L/«
\2//
and Emax, the maximum voltage of self induction
« I X J?a^ = I X 4/L/« = 2 TfLIn.
therefore Ee// = 2 vfLIe//
In direct current circuits, E = 7/2. In inductive circuits of
negligible resistance, E « IX where X, called the inductive
reactance, is expressed in ohms and is numerically equal to
2vfL.
An alternating e jni. of 110 volts sends 2.2 amperes through an inductance
coil of negligible resistance at 60 cycles. Find the reactance at 60 cycles
and find also the coefficient of self induction, and the current at 30 cycles.
X, the reactance = E/I = 110/2.2 = 50 ohms
X
L, the coefficient of self induction » o~~>
50
" 2T60 0.133 henry
/, the current = E/X is inversely proportional to frequency and so has
values of 2.2 amp. at 60 cycles and 4.4 amp. at 30 cycles.
The results of Arts. 240 and 241 may readily be found as follows:
Let the current flowing in the inductive circuit be i = /« sin 6 where the
angle has a value of 2x per cycle or 2ir/ per sec.
The applied voltage required to overcome the voltage of self induction is
, di
e = +L
+ L
dt
d(J« sin 2ir/0
dt
« 2irfLIm cos 2Trft
« En, cos e ^ Em sin {e + 90)
the sine wave of current tjierefore lags 90 degrees behind the sine wave of
voltage which produces it also E = 2ir/L/ » IX,
Art. 243]
INDUCTIVE CIRCUITS
209
The current is inversely proportional to the frequency because, the greater
the frequency the smaller the current required to give the same voltage of
self induction.
242. Power in an Inductive Circuit. — ^The power in a circuit
at any instant in watts is the product of e and i the voltage and
current at that instant. In an inductive circuit of negligible
resistance the current lags the applied voltage by 90 degrees and
the curves representing e and i are shown by light lines in
Fig. 231.
At the instants a and b the voltage is zero so that the power
is zero at these instants; it is also zero at instants g, d and /
Fig. 231. — Voltage, current and power in an inductive circuit.
when the current is zero. Between g and a the voltage and cur
rent are in the same direction so that power is positive or energy
is being put into the circuit, while between a and d the current
and voltage are in opposite directions so that power is negative
or energy is being taken from the circuit; the average power
in the jircuit is zero.
A hypothetical mechanical circuit with somewhat similar
properties is shown in Fig. 232. As the weight W falls, work is
done on the flywheel and the velocity increases until the rope is
all unwound; the flywheel continues to rotate and now raises
the weight, so that work is done by the flywheel until the weight
has been lifted to the original position, the same cycle is then
repeated. During one half cycle the power is positive or energy
is put into the flywheel while during the next half cycle the power
is negative or energy is being taken from the flywheel, so that the
average power is zero.
243. Examples of Inductive and Noninductive Circuits. —
The coeflScient of self induction L = y 10 ~^ so that, to have a
210 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
large inductance, a circuit must be linked by a large flux 4> for
a given current /• The inductance of coil B Fig. 233 is much
greater than that of the duplicate coil A because the flux 4>
has been greatly increased by the addition of the iron core (7.
One form of adjustable inductance is shown in Fig. 234.
If the iron cross piece mn is brought nearer to the poles pq, the
«+■
^
Wood
Imb
i »
A
Fia. 233. — Inductive circuits.
reluctance of the magnetic circuit is decreased, so that a larger
flux 4> is produced by a given current / and the inductance is
thereby increased.
An incandescent lamp filament has an inductance which is
negligible compared with its resistance. The number of turns
n
T P
(
>
r
V
*
J
Fig. 234. — Adjustable inductance.
linked is small, while the flux 4> has to pass through a path con
taining no iron and moreover is produced by an exciting coil
having only two or three turns, so that ^ is small and L =
{n4>/I) 10~^ is negligible. The resistance on the other hand is
high, that of a 16 candlepower carbon lamp being about 200 ohms.
Abt. 244]
INDUCTIVE CIRCUITS
211
While a transmission line has only one turn, its inductance
is not negligible because that one turn is very long and is linked
by a large flux particularly if the wires are spaced far apart
because then, as shown in Fig. 235, there is room for a large
flux to pass between the wire&
Diagram C
Dimgram B
Wires far apart. Wires close together.
Fio. 235. — Flux linking a transmission line.
A simple long loop such as that in Fig. 236 has a negligible
inductance because there is Uttle room for flux to pass between
the wires. Noninductive resistances are made in the form of a
long narrow loop and are then coiled up for convenience as shown
A B
Fia. 236. — Noninductive resistance.
in diagram B. The resistance between a and 6 may be large,
but the inductance is negligible.
244. VoltagCi Current and Power in Resistance Circuits. — If
an alternating voltage is applied to a noninductive circuit of
resistance R then, since there is no e.m.f. of self induction oppos
ing the change of current, the current i at any instant = e/R
212 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxdc
and increases and decreases with the voltage; oris in phase with
the voltage, as shown in Fig. 237.
The power in such a circuit is zero at the instants a, b, c and d
when both voltage and current are zero but is positive at all other
instants, that is energy is put into the circuit but none is taken out
again. The average power
= the average value of ei
= the average value of E^ sin 6 X Im sin 6
= Emim (average value of sin^ 6)
Htm Im
= EI and is also = PR since E ^ IR
Fig. 237. — Voltage c, current % and power e X i in a resistance circuit.
where E and I are efifective values, see page 198. Thus, in a non
inductive circuit, the power is the product of the effective voltage
and the effective current.
246. Resistance and Inductance in Series. — If an alternating
current / is flowing in a circuit with a resistance R and a reactance
X in series, as shown in Fig. 238, then alternating voltages Er =
IR and Ex = IX will be found across the two parts of the circuit.
The applied voltage E is the vector sum of the two components
Er and Ex and may be determined as follows:
A vector / is drawn in any direction.
A vector Er == IR is drawn to scale and in phase with /, see
above.
A vector Ex = IX is drawn to scale in such a direction that I
lags Ex by 90 degrees, see page 207.
Akt. 246]
INDUCTIVE CIRCUITS
213
Then E = th e vector su m of Er and E^
= V jgr^ + EJ^
2
+ x
The current now lags the applied voltage by an angle a, as
shown by the vectors and by the curves in Fig. 238. The power
curve e X t is also shown from which it may be seen that, although
the power is negative during a portion of the cycle yet the average
power is positive.
r — Sjp *4* —Er «
X R
♦*»./
Fia. 238. — Voltage, current and power in a circuit which has resistance and
.inductance in series.
The voltage E is the resultant of two components one of which
Er ^ E cos ai& in phase with / while the other component Ex =
E sin a leads / by 90 degrees. The average power due to the in
phase component = {E cos a) X /, see page 212; that due to the
other component is zero, see page 209, so that the total average
power = EI cos or.
246. The power factor in an alternatingcurrent circuit is
, actual power
denned as the ratio t
apparent power
In any circuit in which the phase angle between the voltage E
and the current / is a degrees then
214 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
the apparent power
but the actual power
therefore the power factor
= EI watts
s= EI cos a watts
EI cos a
EI
= cos a
and can never be greater than unity.
If a resistanoe of 25 ohms and an inductive reactance of 60 ohms at 60
cycles are put in series across 110 volts; find the current, the voltages across
the two parts of the circuit and the power in the circuit at 30, 60 and 120
cycles.
The work is carried out in tabular form as follows:
Frequency  A ohms jr2r/L Va *+Xs 1 I \ Br \ Bs
COS a 1 Watte 
30
60
120
25
25
25
25
50
100
35.4
56.0
103.0
3.10
1.96
1.07
78
49
27
78
98
107
0.71
0.44
0.24
240
96
29
247. The Wattmeter. — Power in alternatingcurrent circuits
may be measured by means of an electrodynamometer type of
Wattmeter
A
A
'JJJJJ
Fig. 239. — Wattmeter connections.
instrument called a wattmeter, constructed as described on page
199, and connected as shown in Fig. 239. The line current / is
passed through the stationary coils il, while the current which
passes through the moving coil C is proportional to the voltage E
and is in phase with it since the inductance of the coil C is negligi
ble compared with the additional resistance r.
Since the moving coil C is carrying current and is in a magnetic
field it is acted on by a force tending to turn it about a vertical
axis and this force is proportional to the current in the coil and to
the strength of the magnetic field. When the instrument is
connected as shown in Fig. 239, the magnetic field is proportional
to the current / while the current in the moving coil is propor
tional to the voltage E, and the average turning force is propor
A«r. 2481 INDUCTIVE CIRCUITS 215
tional to the average value of 6 X i or to EI cos a, the average
power in the circuit.
If in the circuit shown in Fig. 239
^ = 100 volts
/ B 50 amp.
W = 4000 watts, measured by a wattmeter.
I
then the power factor of the circuit = r — =r~
*^ apparent power
4000
100X50
= 0.8
and the phase angle between current and voltage is the angle whose cosine is
0.8 or is 37 degrees.
248. Transmission Line Regulation and Losses. — ^A transmis
sion line has resistance and inductance and may therefore be
\ix
E
■^Twr > A\vw
Et
X R
nwnnnn < vvvw
A B
Fig. 240. — ^Vector diagram for a transmission line.
represented as in Fig. 240. Eg, the voltage at the generating
station, is the vector sum of the terminal voltage Et, the resistance
drop IR and the reactance drop IX \ the phase relation between
these voltages is shown in diagram B,
A vector / is drawn in any direction.
A vector Et is 4rawn to scale equal to the receiver voltage, the
angle a depending on the resistance and inductance of the load
connected to the line;.
A vector Er = IR is drawn to scale in phase with /.
A vector £, = IX is drawn to scale in such a direction that /
lags E9 by 90 degrees.
The vector Eg is the vector sum of Et, Er and E^ and may be
sealed oflf or determined by calculation.
Since there is no power loss in the inductance of the line, the
total loss is in the resistance and is equal to PR watts.
Values of line resistance and line reactance are generally given
in ohms per mile as in the following table:
216 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxdc
Sise of wire
Reeiatanoe
at 20 deg. C.
Inductive
reactance per mile of wire at 60 cycle*
B. and S.
gauge
Cir. mils
Ohms per
mile of wire
24"
48"
72"
96" spacing in
inches
0000
000
00
1
2
4
6
211600
167800
133100
105500
83690
66370
41740
26250
0.258
0.326
0.411
0.518
0.653
0.824
1.309
2.082
0.594
0.608
0.622
0.637
0.650
0.665
0.693
0.721
0.678
0.692
0.706
0.721
0.735
0.749
0.777
0.805
0.728
0.742
0.756
0.770
0.784
0.797
0.826
0.854
0.763
0.776
0.790
0.804
0.819
0.833
0.860
0.889
75 kw. at 2200 volts and 60 cycles has to be delivered at the end of a 5
mile line, the size of wire being No. B. and S. gauge and the spacing 48
in. Find the voltage in the generating station and the power loss in the
line if the load current b lagging and the power factor is 0.8.
watts = El X / X cos a, see page 214,
or 75 X 1000 = 2200 X / X 0.8
and the current » 42.5 amp.
the resistance « 0.518 ohms per mile of wire
B 5.18 ohms for a 5mile line
the reactance » 0.721 ohms per mile of wire at 60 cycles
» 7.21 ohms for a 5mile line
IR = 42.5 X 5.18 = 220 volts
IX = 42.5 X 7.21 = 307 volts
aby Fig. 240 = (2200 X 0.8) + 220  1980
6c, Fig. 240 = (2 200 X 0.6) + 30 7 « 1627
Eg = V{ab)*+{bcy = V1980« + 1627«
» 2560 volts
ab 1980 ^_^
the power factor at the generatmg station ■* ri = 2560 ~ "• « * °
the power put into the line = 2560 X 42.5 X 0.775 « 84.4 kw.
the power delivered «= 75 kw.
the loss in the line = 84.4 — 75 = 9.4 kw.
and this is equal to PR = (42.5)* X 5.18 = 9.4 kw.
249. Resistance and Inductance in Parallel. — If an alternating
voltage E is applied to a circuit which has a resistance R and a
reactance X in parallel, as shown in Fig. 241, then a current
Ir = E/R will flow through the resistance and will be in phase
with E while a current Ix = E/X will flow through the reactance
and will lag E by 90 degrees. The total current drawn from the
source is the vector sum of the two components /, and /, ajid
may be determined as follows:
A vector E is drawn in any direction.
A vector Ir = E/R is drawn to scale and in phase with jB,
see Art. 244.
Art. 2491
INDUCTIVE CIRCUITS
217
A vector /« = E/X is drawn to scale in such a direction that
Ix lags E by 90 degrees, see page 207
Then / = the vector sum of Ir and /»
= VTT+1?
'^W^'
In series circuits the current flowing is common to all parts
of the circuit and so the current vector is taken as the basis
for the vector diagram, in parallel circuits on the other hand the
voltage applied to each branch of the circuit is the same so that
the voltage vector is taken as the basis for the vector diagram.
If a resistance of 25 ohms and an inductive reactance of 50 ohms at 60
cycles are put in parallel across 110 volts, find the current in each part of the
circuit and also the total current at 30, 60 and 120 cycles.
»'♦
I"
I:
Fig. 241. — Vector diagram for a parallel circuit.
The work is carried out in tabular form as follows :
Frequency  fi ohma I X 2ir/L  Jr 1 I, 1 I I cos a Watts 
30
60
120
25
25
25
25
50
100
4.4
4.4
4.4
4.4
2.2
1.1
6.20
4.93
4.56
0.710
0.890
0.965
485
485
485
CHAPTER XXX
CAPACITY CIRCUITS
260. Condensers. — Two conducting bodies separated by insu
lating material form what is known as an electrostatic condenser.
In diagram A, Fig. 242, a and b, two plates of a condenser, are
at the same potential. When the switch k is closed a momentary
current i passes in the direction of the arrows in diagram B and
the condenser is said to be chai^ged; the potential of plate a is
raised to that of the positive line terminal, the potential of plate
b is lowered to that of the negative line terminal and the differ
ence of potential between the plates becomes equal to the line
voltage.
+ — '^ »
i
•:—
t ^
A.  Gondeniw B  Oondenier Chftrglng C Oondenter Dlichargtttg
without Charge and Storing Xlectricltf and glTtng op Klectrlclty
Fig. 242. — Charge and discharge of a condeiiBer.
If the switch k is now opened, the voltage between the plates
remains unchanged and a quantity, or charge, of electricity re
mains stored in the condenser.
To make the condenser give up its charge, the insulated plates
must be connected by a conducting material, such as the wire d
in diagram C, so as to bring them to the same potential. When
this is done, a momentary current passes in the direction shown,
from the positive to the negative plate.
The quantity of electricity stored in a condenser, called the
charge, is equal to the average current flowing into the condenser
multiplied by the time during which it flows or is equal to J idt
where i is the charging current at any instant. In any condenser
this charge is found to be directly proportional to the applied
voltage or
q = Ce
218
2511 CAPACITY CIRCUITS 219
where q is the chai^ in coulombs (amperes X seconds)
e is the applied voltage
C is a constant called the capacity of the condenser
and is expressed in farads.
A condenser of 1 farad capacity will hold a charge of 1 coulomb
if a difference of potential of 1 volt is applied between the plates.
The capacity of a condenser of given dimensions is found to
depend on the insulating material, or dielectric, between the
plates. If a condenser with air as dielectric has a capacity of
F farads then the capacity becomes equal to kF farads when
another dielectric is used. The constant k is called the specific
inductive capacity of the material. Average values of k are
given in the following table for the materials generally used
in commercial condensers.
Material Specific inductive capacity
Air 1
Glass 4 (varies considerably with the
quality of the glass)
Mica 6
Paraflined paper 2
251. Capacity Circuits with Direct and with Alternating
Currents. — A capacity circuit is one which contains a condenser*
V
A B
Fig. 243. — ^How of current in a capacity circuit.
If a constant e.m.f . is applied across the terminals of the circuit
shown in Fig. 243,. then a momentary current wUl flow in the direc
tion shown in diagram A to charge the condenser but current
will not flow continuously since the circuit is broken by the insu
lating material between the plates.
If the applied voltage is now reversed, current will flow in the
direction shown in diagram B until the condenser has given up
its charge, and will continue to flow in this direction until the
condenser is recharged in the opposite direction. If then the
appUed voltage is alternating, a charging current will flow in and
16
220 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
out of the wires x and y with a frequency which is the same as
that of the applied voltage, and the lamps L will light up if
sufficient current flows to heat the filaments.
The greater the capacity of a condenser, the more electricity
it can hold, and the larger the charging current that passes
through the connecting wires. Furthermore, the greater the
frequency of the applied voltage, the shorter the time available
in which to charge the condenser, and therefore the larger the
current that must flow. With a given applied alternating vol
tage, and therefore with a definite alternating charge (since q =
Ce), the alternating current i is proportional to the capacity
and to the frequency or
i = a const. X C X /
the constant will be determined later.
262. Phase Relation between Voltage and Current in Capacity
Circuits. — ^If the voltage applied to the condenser in Fig. 244 be
represented by curve E then the charge, which is proportional to
the voltage, is represented by the curve Q; the condenser is
Smf. if ALSL Charge Ii
Tncreailng ~5 Incceaiiag
Emf. \t Tl Charge li
Decreaiing "15 Decreasing
>^ •
V
Fig. 244. — Phase relation between the voltage and current in a capacity
circuit.
charged alternately in opposite directions, thus between the
instants m and n the plate a is positive while between n and p
the plate a is negative.
At the instants q and r the charge in the condenser is not chang
ing, the currents in the leads x and y must therefore be zero.
Between.m and q the voltage and the charge are increasing and
current flows in the positive direction, from the positive to the
negative terminal as shown in diagram A, until, at the instant
q, the charge is complete and the current has become zero; this
gives the part fq of the current curve, see diagram C.
Between q and n the voltage and the charge are decreasing so
that current must now be flowing out of the condenser or in the
Abt. 253] CAPACITY CIRCUITS 221
negative direction as shown in diagram B; this gives the part
qg of the current curve.
During the next half cycle between n and p the condenser
charges and discharges in the opposite direction, so that the cur
rent curve gh is the same as the curve fg except that the sign is
reversed.
From these curves it may be seen that the current leads the vol
tage by 90 degrees.
In the above discussion of phase relation between current and
voltage in capacity circuits it is assumed that the current has been
flowing for a few seconds. It is obvious that, at the instant the
switch in a circuit is closed, the ciurent in that circuit must be
zero no matter what value the e.m.f . may have, so that the cur
rent waves are generally abnormal for a few cycles after the clos
ing of the switch, but they gradually change and become regular
waves leading the e.m.f . by 90 degrees.
263« Voltage and Current Relations in Capacity Circuits. — ^The
charge in the condenser shown in Fig. 244 changes from zero to
Qm = CJS?« coulombs in the time of onequarter of a cycle, or
in 1/4/ seconds, so that, since charge = average current X time,
therefore Q« = C^» ^ ^"^^ Tf
and lavi the average charging current = ^fCEm amp.
Now the maximum charging current /» » lav X o' ^^ P&S^ 197.
= I >C ^fCEn. = 27r/CB«; therefore /.//. = 2TfCEeff.
In direct current circuits E = IR; in capacity circuits E = IX
where X, called the capacity reactance, is expressed in ohms and
1
is numerically equal to K—fp
The results of Arts. 252 and 253 may readily be found as follows:
It is shown on page 218 that g = Ce = 1 idt, therefore i = C de/dt.
If the voltage applied to the capacity circuit be e = Em sva. where the
angle $ has the value of 2t per cycle or 2«/ per sec. the current flowing
will be equal to
_ djEm sin 2t ft)
" ^ dt
= 2t/CE« cos 2rft
= /« cos d = /« sin {e + 90)
the sine wave of current therefore leads the sine wave of voltage by 90
degrees ftteo E ^ IX where X = 1/2t/C.
222 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxx
An alternating e.m.f. of 110 volts sends 2.2 amp. through a capacity
circuit at 60 cycles. Find the reactance at 60 cycles and find also the
capacity of the condenser.
Xj the reactance « E/I
 110/2.2  50 ohms.
C, the capacity ■» o /•y
" 2ir6Q fiO " ^*^ ^ ^^* farads « 53 microfarads^
If the voltage applied to the above circuit is kept constant at 110, find the
current that will flow through the capacity at 30, 60 and 120 cycles.
Xy the reactance — ^ — fc ^ ii^versely proportional to frequency =50 ohms
at 60 cycles, from last problem
/, the current « E/X is therefore proportional to frequency and so has
values of 1.1 amp. at 30 cycles; 2.2 amp. at 60
cycles; 4.4 amp. at 12Q cycles.
254* Parallel Plate Condenser. — ^As shown in the last problem,
a condenser with a capacity of 0.53 microfarads will take a cur
FiG. 245. — One method of constructing condensers.
rent of 2.2 amp. at 110 volts and 60 cycles. It is desirable to
know the approximate dimensions of such a condenser.
The capacity of a parallel plate condenser is given by the
formula
Cinfarad8=^X9>~nxV
where A is the area of the active surface of one plate in sq. cm.
t is the distance between plates in cm.
k is the specific inductive capacity, see page 219.
A condenser constructed as in Fig. 245 has plates of tin foil which aie 4C
ft. long and 3 in. wide and are separated by paraffined paper 0.0025 in. thick.
Since both sides of each plate are active,
4=2X40X12X3= 2,880 sq. in.
B 18,600 sq. cm.
Art. 256] CAPACITY CIRCUITS 223
t = 0.0025 in. = 0.0063 cm.
ik = 2.0
111 ft Ann
therefore C in farads = 4;^ Xi^Xlon X 5:0663 ^ ^
= 0.53 X 10« farads
» 0.53 microfarads
Such a condenser will go into a tin case 1.75 in. square by 4 in. deep.
266. Power in Capacity Circuits. — The power in a circuit at
any instant is the product of e and i the voltage and the current at
that instant. In a capacity circuit the current leads the applied
voltage by 90 degrees and the curves representing e, i and ei are
shown in Fig. 246. This latter curve is obtained by multiplying
together corresponding values of e and i at different instants;
Fig. 246. — Voltage e, current i and power e X t in a capacity circuit.
at m and n the voltage and therefore the power are zero; the
power is also zero at instants q and r when the current is zero.
Between m and q energy is stored in the condenser while between
q and n the same energy is given up by the condenser, so that the
average value of the energy used is zero and so also is the average
power in the circuit.
266. The Formulae used in Circuit Problems are :
Resistance Circuit:
E =^ IR
current is in phase with voltage, see page 211
power = EI watts.
Circuit with Inductive Reactance :
E ^ IX where X = 27r/L, see page 208
current lags voltage by 90 degrees, see page 207
power is zero.
> 
224 PRINCIPLES OP ELECTRICAL ENGINEERING [Chap.
Circuit witfa Capacity Reactance :
E ^ IX where X = 2~7r' ®^ ^^^® ^^
current leads voltage by 90 degrees, see page 220
power is zero.
yAAAAA^A y^TJWinnr^ r —
E
I
E
•^I
/*
Beititance
Oircait
InductlTo
Circuit
Fig. 247.
Capacity
Circuit
257. Resistance^ Inductance and Capacity in Series. — In the
solution of such a circuit as that shown in Fig. 248, the current
vector has to be taken as a basis for phase relation since it is the
same in all three parts of the circuit. The voltage E is the vec
tor sum of Er, El and Ee and is determined as follows:
A vector / is drawn in any direction.
A vector Er = IR is drawn to scale in phase with 7.
A vector Ei = IXi is drawn to scale in such a direction that I
lags El by 90 degrees.
A vector Ee = IXo is drawn to scale in such a direction that /
leads Ee by 90 degrees.
Then E = the vector sum of E^ Ei and Ec
= y lEr'+jEiEey
= V(/ii!)2+ (iXt  IXc)^
and the current will lead or lag the applied voltage according as
Xe is greater or less than Xj.
When Xe = Xi the capacity and the inductive reactances
exactly neutralize one another and the current has its maximum
value and is equal to E/R. The circuit is then said to be in
resonance.
Art. 257]
CAPACITY CIRCUITS
225
The inductive reactance Xi is directly proportional to the fre
quency and is equal to 2ir/L whereas the capacity reactance Xe is
inversely proportional to the frequency and is equal to l/2ir/C.
If then in Fig. 248, the voltage E across the terminals is kept con
stant and the frequency is increased, Xi will increase and Xc will
decrease until when
Xi = Xc
2irfL = ^
or
and
/ =
2jr/C
1
2irVLC
•— Er
■*^"4*fH
1
Si
EiEc,
E
"""""'
^
'A
y
fc r
S
s
si
Ea
Er
^^^^ A
Frequency
Fig. 248.—
Voltage and current in a circuit with resistance R, inductive
reactance Xe and capacity reactance Xe in series.
the circuit is said to be in resonance and the current has its
maximum value.
The same problem is found in mechanics. An alternating
force applied to a spring will cause the spring to oscillate. As
the frequency of the applied force increases, the amplitude in
creases and reaches its maximum value when the applied fre
quency is the same as the natural frequency of vibration of the
spring; with further increase of the frequency, the amplitude
will decrease. This principle is made use of in the instrument
shown in Fig. 215.
226 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
The frequency / = p=^ is called the frequency of resonance
2tVLC
and is also called the natural frequency of the circuit.
If a resistance of 2 ohms, an inductive reactance of 12 ohms at 60 cycles
and a capacity reactance of 20 ohms at 60 cycles are put in series across 110
volts plot the current I and also the voltages Er, Ei and Ee against fre
quency.
Several points for these curves are determined as follows:
s
M
•
>)
o
w
><
1
«
1
03
1
*4
>
4
H
I
H
1
+
•*
0$
B4
M
••
1
1?
1
Bq
>
1
20.0
2
4.0
60.0
56
56+
1.97
3.94
7.9
118
40.0
2
8.0
30.0
22
22.1
4.98
9.96
40.0
150
60.0
2
12.0
20.0
8
8.24
13.4
26.8
161.0
268
80.0
2
16.0
15.0
1
2.24
49.2
98.4
788.0
738
100.0
2
20.0
12.0
8
8.24
13.4
26.8
268.0
161
77.5
2
15.5
. 15.5
2.0
55.0
110.0
852.0
852
the frequency of resonance can be determined readily by trial and is 77.5
cycles, because then Xi » Xe <** 15.5 ohms.
If the circuit is in resonance and the resistance is low then a
large current will flow, and if in addition the reactances are large
then the voltage drops across these reactances are large and may
have several times the value of the applied voltage, this result is
shown in the above problem; the inductance coil and the con
denser must be insulated to withstand 852 volts and not merely
the applied 110 volts.
In a circuit which contains only resistance, E ^ IR
in a circuit which contains only inductance, E = IXi
in a circuit which contains only capacity, E = IX c
in a circuit which contains all three in series, E == IZ
where Z, called theimpedenceof the circuit = ViZ^ + {Xi — Xe)*
268. Resistance, Inductance and Capacity in Parallel. — In
the solution of such a circuit as that shown in Fig. 249, the voltage
vector has to be taken as a basis for phase relation since it is the
same for all three parts of the circuit. The current / is the vector
sum of Ir, 1 1 and /« and is determined as follows:
A vector E is drawn in any direction.
Art. 258]
CAPACITY CIRCUITS
227
A vector /,
A vector Ii
E/R is drawn to scale iq phase witii E.
E/Xi is drawn to scale and lagging f by 90
A vector /« = E/X^ is drawn to scale and leading £ by 90
Then / = the vector sum of 1^ Ii and /«
= Vv + (/,  i,y
= ^{E/RY +iE/X,E/Xcy
Ict,
t'J
Freqaency
Fig. 249. — Circuit with resistance, inductance and capacity in parallel.
and the current will lead or lag the applied voltage according as
7c is greater or smaller than Ii.
When the circuit is in resonance, Xe=Xi and the current in
the line has its minimum value and is equal to E/R. If the re
actances are then low compared with the resistance R, the cur
rents Ii and Ic may be much larger than the line current I as
shown in the following example.
17
228 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxx
If a resistance of 20 ohms, an inductive reactance of 2.4 ohms at 60 cycles
and a capacity reactance of 4 ohms at 60 cycles are put in parallel across 110
volts, plot the currents in the different parts of the circuit against frequency.
Several points for those curves are determined as follows:
Frequency
R
Xi  2t/L
^' 2wfC
Ir
Ii
/.
Ji J.
I
20.0
20
0.8
12.0
5.5
137
9
128
128 +
40.0
20
1.6
6.0
5.5
69
18
51
61.2
60.0
20
2.4
4.0
5.6
46
28
18
18.8
77.6
20
3.1
3.1
6.5
35
35
5.5
80.0
20
3.2
3.0
5.5
34
37
 3
6.2
100.0
20
4
2.4
5.5
28
46
18
18.8
CHAPTER XXXI
ALTERNATORS
269. Altenutor Constructioii. — The essential parts of a re
volving field type of alternator are shown in Fig. 250. The
Ro. 250. — Bevolvingfield type of altern&tor.
Fia. 251 —Winding diagram.
Fia. 252. — Alternator coil.
stationary part which carries the conductors that are cut by the
revolving field is called the stator; the revolving field system is
called the rotor.
230 PRINCIPLES OF ELECTRICAL ENGINEERING IChap. xxxi
The stator core B ia built up of soft steel laminations and has
slots on the inner periphery in which the stator coils are placed.
One type of coil is shown in Fig. 252 and consists of several turns
of copper wire which are insulated from one another and are then
taped up with cotton and other such insulating material. The
machine shown in Fig. 250 has four of these coils which are con
nected in series so that their voltages add up.
Since a connection diagram such as Fig. 250 shows only one
end of the machine, it is found desirable in practice to show the
coils and connections by means of a developed diagram such as
Fig. 251; this diagram shows what would be obtained if the
winding in Fig. 250 were spht at xy and then flattened out on a
plane; the two diagrams are lettered similarly.
The voltage between the terminals Si and Fi varies as shown
in Fig. 250 and goes through four cycles per revolution.
260. Twophase Alternator. — In order to utilize more of the
stator surface, a duplicate winding B is placed on the stator as
Fig, 253. — Twophaae alternator.
shown in Fig. 253. This machine has twice as many conductors
as that in Fig. 250, but if the coils A and B are connected in
series, it will be found that the voltage of the machine has not
been doubled but has been increased only 41 per cent.
It was pointed out on page 200 that the distance between two
adjacent like poles is 360 electrical degrees, therefore the distance
between similar points on windings A and 5 is 90 electiical
d^rees. If then the voltage generated in the four coils A in
series ia represented by the curve K, Fig. 25i, that generated
in the four coils B in aeries has the same magnitude but lags £„ by
90 degrees and is therefore represented by the curve Ei,; when the
poles are in the position shown, for example, the voltage in the
Art. 261]
ALTERNATORS
231
winding A is a maximum while that in £ is zero, these values are
obtained at the instant c, Fig. 254.
The resultant voltage when the two windings are connected in
series is the vector sum of Ei and E2 and is equal to V2J? = 1.414E
and if / is the maximum safe current the conductors of the
winding can carry then the maximum output is 1.414 EI watts.
It is therefore desirable to use the two windings A and B as if
they belonged to separate alternators and then, by dividing up
the load between them as shown diagrammatically in Fig. 256,
each winding can be made to deliver EI watts or the whole
machine be made to deUver 2EI watts.
Fig. 254. — Voltage curves of a
twophase alternator.
Fig. 255.— Fig. 256.— Diagram
Voltage vector matic representation of
diagram for a a twopha^ alternator,
twophase al
ternator.
Since the voltage of winding B is out of phase with that of
winding A, the machine operating as shown in Fig. 263 gives two
phases of voltage and is called a twophase machine, whereas that
shown in Fig. 250 is a singlephase machine. The former machine
requires four wires for the load while the latter requires only two.
261. Threephase Alternators. — If three similar and independ
ent singlephase stators are mounted beside one another as in
Fig. 257 in such a way that their conductors are cut by the same
revolving field, then three separate singlephase e.m.fs. may be
obtained, one from each winding. If further these stators are
mounted so that S^ S% and Sg, the starts of the windings of the
three phases, are spaced 120 electrical degrees apart, then the
e.m.f . in winding B will reach a maximum 120 degrees after that
in winding A has reached its maximum value, and the e.m.f. in
winding C will lag that in winding B by 120 degrees, as shown in
Fig. 259.
232 PRINCIPLES OF ELECTRICAL ENOINEERINO [Chj
Fia. 267. — Three ainglephase statora.
iiitfimiiiii''^l
Fio. 258. — Threephase alternator.
Fia 259.— Voltage curves ot a three
phase alteniator.
Pio. 260.— Voltage vector
diagram for a threephase al
temator.
Art. 282] ALTERNATORS 233
In practice the three windings are placed on the same core as in
Fig. 258, but it must be noted that in the resulting threephase
machine the windings are independent of one another and supply
distinct and independent e.m.fs. to three distinct and independent
circuits so that the machine is exactly equivalent to three separate
singlephase machines and is therefore called a threephase ma
chine. Part of the winding for such a machineis shown in Fig. 261.
Fig. 261.— Part of the stator of a large threephase alternator.
262, YConnection. — A threephase machine is conveniently
represented by a diagram such as that in Fig. 262, the three vec
tors in Fig. 260 being replaced by three separate and independent
windings; such a machine has six terminals and six leads, two for
each phase.
In order to reduce the number of leads, the three return wires Oi,
bt and Ct may be connected together to form a single wire n. The
current in this wire at any instant is therefore the sum of ii, u apd
it, the currents in the three phases. But it may be seen from
diagram B that, at any instant, the sum of these three currents is
zero; at instant a for example ii is equal and opposite to it h it
while at instant b, ii is equal and opposite to it and I'l is zero, the
wire n therefore carries no current and may be dispensed with.
The resultant connection, shown in Fig. 263, is called the Y
connection and requires only three leads to supply the load, one
lead always acting as the return for the other two.
263. Deltaconnection. — Another method of connecting the
three windings of a threephase machine is shown diagrammat
ically in Fig, 264, the windings being connected in series in the
234 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
Yoltag e Correi of a Three Phase
M Machine
B  Ourrent Oarrei of a Three Phaie
Machine
FiQ 262. — Diagrammatic representation of a threephase machine.
J ?« ^"y/zE
Fio. 263.— YConnection.
"IplVai
j?^E
I/V3/
K^
Si
Fia. 264. — Deltaconnection.
Art. 264]
ALTERNATORS
235
following order S^2,S^zy SiFu The wires making these connec
tions may be shortened and the terminals connected directly to
one another as shown in diagram B, the slope of the vectors being
unchanged. On account of the appearance of this latter diagram,
this threephase connection is called the deltaconnection.
Although the winding has been closed on itself, no current flows
through this closed circuit. The resultant voltage in the closed
circuit is the sum of the voltages in the three phases, but it may be
seen from diagram A, Fig. 262, that, at any instant, ei + 62 + ez
is zero, the voltage in one phase being always equal and opposite
to the sum of the voltages in the other two phases. If, however,
an external circuit is connected between any two leads, then the
voltage across that circuit will be Ey the voltage of one phase,
and current will flow through the circuit.
264. Voltagesi Currents and Power in a TConnected Machine.
— If two coils SiFi and S2F2 are connected as shown in Fig. 265
k 'T> Er^^
I ' I
Fi 8% Fa
Fig. 266.
1^ 4\.._&f — J
^^c£?.
Ef^s/SE
B
Fig. 266. — The vector difference between the two voltages Ei and ^2, each
equal to ^, is ^( = y/sE.
and the voltage E2 lags Ei by 120 degrees then the resultant
voltage Er is the vector sum of Ei and E2 and may be determined
as shown in diagram A.
If, however, the second coil is connected backward as shown in
Fig. 266, then the resultant voltage Et is no longer the vector
sum but is the vector difference and is obtained by reversing the
236 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
vector to be subtracted and then taking the sum as shown in
diagram B, where
Ei = 2jB cos 30
= V3E
= 1.73B
This latter connection is the Y connection, thus in Fig. 263, Fi
and F2 are connected together and the voltage between Si and S%
= 1.73B. The current Ii in the line is the same as the current I
in the winding.
If the current I in each phase of the machine lags the voltage E
of that phase by an angle a as shown in diagram B, Fig. 262, then
the power in each phase = EI cos a watts
the power in the three phases = SEI cos a
= 3(;^j J/ cos a
= ^EJi cos a
since Et = V3^ and Ii = I.
266. Voltages, Currents and Power in a Deltaconnected
Machine. — If two coils SiFi and SjFj are connected as shown in
St F.
Fig. 267.
5i Fi
/i
Lr
Ii
^^
ijVs/
D
Fig. 268. — The vector difference between the two currents /i and Ii, each
equal to /, is /^ = Va/.
Fig. 267, and the current h lags Ji by 120 degrees then the result
ant current Ii is the vector sum of /i and I2 and may be deter
mined as shown in diagram C.
If, however, the second coil is connected backward as shown
Abt. 266]
ALTERNATORS
237
in Fig. 268, then the resultant current Ii is no longer the vector
sum but is the vector difference as shown in diagram D where
1 1 = 2 J cos 30
= V3/
= 1.737
This latter connection is the deltaconnection, thus in Fig. 264,
S\ and F2 are connected together and the current Ii in the line
connected to that point = 1.73 J. The voltage Et is the same as
the voltage E of the winding.
If the current I in each phase of the machine lags the voltage
E of that phase by an angle a then
the power in each phase = EI cos a watts
the power in the three phases = ZEI cos a
= ^Et (~7^) cos a
m
= V3 EJi cos a
since Et = E and li = V3/
With the same current in the line and the same voltage between
lines, the power is the same no matter whether the machine is
connected Y or delta.
266. Connection of a Threephase Load. — ^The load on a three
phase line may be connected Y as in diagram A, Fig. 269, or
CD
10.
.57
0.91
o.n. I
^
0.91
A " y Connected f Delta Connected C Delta Connected
Load Load Load
Fio. 269. — Connection of the load to a threephase circuit.
delta as in diagram B. If the lamps shown are 100 watt lamps
then, when Yconnected, the voltage per lamp is llO/Vs or 63.5
100
volts and the current per lamp is ^o~E = 1.57 amp. When delta
100
connected, 110volt lamps are required and they take y\h ~ ^'^^
238 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxi
amp. The deltaconnection is generally used for the connection
of individual loads across the phases, the distribution circuits
being connected to the power mains as shown in diagram C
If one of the lamps in diagram B burns out, the two remaining
lamps will bum with their normal brilliancy, but if one of the
lamps in diagram A bums out, then the two remaining lamps will
be in series across 110 volts and will bum dimly since they are
then operating at 55 volts instead of at the normal 63.5 volts.
267. Power Measurement in Polyphase Circuits. — ^A poly
phase circuit is one with more than one phase. To measure the
rV=
■Q>
w^
Wi
^A
H^^
3 Phaae Clrcuft
2 Fhaie Clrcolt
Fig. 270. Fia. 271.
FiQB. 270 AND 271. — Wattmeter connections in polyphase circuits.
power in a twophase circuit, each phase must be considered
separately and two wattmeters used as shown in Fig. 270. If
the load is balanced, that is divided equally between the two
phases, then only one set of instruments is required.
If in a balanced twophase circuit
E = 100 volts
/ » 50 amp.
W = 4000watts
then the total power » 4000 X 2 = 8000 watts
the apparent power = (100 X 60) X 2 = 10,000 volt amperes
the power factor = 8000/10,000 =« 0.8
and the current in each phase lags the voltage of that phase by an angle whose
cosine is 0.8 or by 37 degrees.
In a threephase circuit it is usually impossible to reach the
two leads of each phase and the power has to be measured out on
the line where only three leads are available. One line, for ex
ample 6, Fig. 271, is supposed to be the return line for the other
two and two wattmeters are connected as shown to measure the
power going out on these lines, the total power in the threephase
circuit is the sum of the readings obtained from the two meters.
J
Art. 268]
ALTERNATORS
239
If in Fig. 271 E  100 volts
7 — 60 amp.
Wi = 5000 watts
W% = 2500 watts
then the total power = 7500 watts
the apparent power » 1.73 x 100 X 50 » 8650 volt amperes.
the power factor = 7500/8650 = 86.6 per cent,
and the current in each phase lags the voltage of that phase by an angle whose
cosine is 0.866 or by 30 degrees.
268. Alternator Constructioii. — The construction of a revolving
field type of alternator is shown in Fig. 272.
FiQ. 272. — Revolvingfield type of alternator.
The stator core B is built up of sheet steel laminations which
are dovetailed into a castiron yoke A and clamped between two
iron end heads E. These laminations have slots C on their
inner periphery and in these slots are placed the armature con
ductors D which are insulated from the slots and are connected
together to form a winding from which e.m.f . is supplied to an
external circuit. The stator core is divided into blocks by means
of vent segments F and the ducts thereby provided allow air to
circulate freely through the machine and keep it cool.
240 PRINCIPLES OP ELECTRICAL ENGINEERING [Chap.
The rotor or revolving field system consists of a series of N
and S poles carrying exciting coils H and mounted on an iron
field ring. An alternator has to be excited with direct current,
it cannot therefore be self exciting. The exciting current, gener
ally supplied by a small directcurrent generator called an exciter,
is led into the field coils through brushes M which bear on slip
rings insulated from the shaft.
The exciter voltage is independent of that of the alternator and
is generally chosen as 120 volts so that, in the case of high voltage
alternators, the exciting current may be larger than the fullload
current of the machine as in the following case:
A singlephase alternator has an output of 1000 kw. at 13,200 volts and
100 per cent, power factor, find the current at fullload. K the exciter vol
tage is 120 and the excitation loss is 2 per cent, find the output of the exciter
and also the exciting current.
o. Watts = volts X amperes X power factor,
therefore 1000 X 1000 = 13,200 X amperes X 10
and amperes at fullload — 76
6. The exciter output = 2 per cent, of 1000 kw. =» 20 kw.
X 20 X 1000 ,^^
The excitmg current = joq = 167 amp.
269. The revolving armature type of alternator is generally
cheaper than the revolving field type of machine for small outputs
at low voltages. Such a machine is shown diagrammatically in
Fig. 273; the armature is the same as that of a directcurrent
generator except that the commutator is removed and the arma
ture is tapped at two diametrically opposite points m and n which
are connected to slip rings 1 and 2.
The e.m.f. between these slip rings is a maximum when the
armature is in the position shown, and is zero when the armature
has moved through quarter of a revolution from this position
because then the voltages generated in the conductors between m
and c are opposed by the equal voltages in the conductors between c
and n. The e.m.f. again becomes a maximum after the armature
has moved through half of a revolution from the position shown in
Fig. 273 but the polarity of the slip rings is now reversed. The
e.m.f. between the slip rings is therefore alternating and goes
through one cycle per pair of poles passed.
If the armature is tapped at four points as shown in Fig. 274,
the voltage Ei between the slip rings 1 and 2 is a maximum when
the armature is in the position shown, while the voltage Et
between the rings 3 and 4 is zero at the same instant, and E%
Art. 270]
ALTERNATORS
241
lags El by 90 degrees bo that the machine is now a twophase
alternator.
To obtain threephase currents the armature must be tapped
at three points as shown in Fig. 275, it then becomes a three
phase deltaconnected armature. At the instant shown, the
Fio. 273. — Single^hase. Fig. 274. — Twophase. Pia. 275. — Three^baBe.
Profl, 273276.— Revolving annature type of alternator.
Fig. 278. — Inductor alternator.
voltf^ Et is zero while Ei is positive and decreasing and Et is
n^atlve and increasing, this corresponds to instant a on the
voltage curve diagram.
270. The inductor alternator, one type of which is shown dia
grammatically in Fig. 276, has been found suitable for the gener
242 PRINCIPLES OF ELECTRICAL ENQINEERINO (Chap, xxxi
ation of high frequency e.m.f8., because of the simplicity of the
mechanical construction.
The stationary field coil F, when excited, produces a magnetic
flux ^ which causes all the inductors N to have the same polarity.
The coils C are cut by the lines of force as the inductors rotate,
and the generated voltage is a maximum when the inductors are
in the position shown and one side of each coil is cutting lines of
force, the voltage is zero when the poles are in position y and is a
maximum a^jain but in the opposite direction when the poles
ore in the position 2 and the other side of each coil is now cutting
the lines of force. The voltage therefore passes through one
half cycle while the inductors move from a; to ? or 5 cycles are
passed through per revolution, so that a machine with five in
Fio. 277. — MfLgneto Alternator,
ductors is equivalent to a tenpole revolving field machine. Since
only one side of each coil is active at any instant in the case of the
inductor alternator, it is the heavier of the two machines for a
given output and is therefore used only when simplicity of con
struction is essential.
271. Magneto Alternators.— Two types of alternatingcurrent
magnetos used for gasengine ignition are shown in Figa. 277 and
278. In the former machine, the armature coil C is stationary
and the flux threading this coil is varied by the rotating inductor
ab, whereas in the latter machine the coil is wound on the inductor
and rotates with it.
In each case, when the inductor is in the position shown, the
flux ^ passes through the coil C from atob; half a revolution later
Art 271]
ALTERNATORS
243
h is under the N pole and a under the S pole and the flux ^ now
passes from & to a and therefore passes through coil C in the oppo
site direction.
A high peak of e.m.f. is obtained from such machines by shaping
the pole faces of the revolving parts so that the flux threading the
coil C changes as shown in curve a, Fig. 279; the flux changing
very rapidly as the inductor moves from under the poles into the
neutral position. The e.m.f. in the coil C, being proportional to
the rate of change of the flux, has then its maximum value as shown
Fig. 278. — Magneto alter Fio. 279. — Voltage and current
nator. curves in a magneto alternator.
in curve 6, Fig. 279. The current lags the voltage by an angle a
which increases as the reactance of the circuit increases and there
fore increases with the frequency of the alternator or the speed of
the engine.
18
CHAPTER XXXII
ALTERNATOR CHARACTERISTICS
272. Armature Reactance. — Part of the winding of an alter
nator is shown in Fig. 280, the poles being stationary and the
field coils not excited. If an alternating current I is passed
through this winding from an external source then lines of force
will encircle the coils as shown. This magnetic field is alter
nating and induces in the coils an e.m.f . of self induction which
opposes the applied e.m.f. and is equal to JXySee page 208, where
FiQ. 280. — Magnetic flux due to
the armature current.
Fia. 281.— Dia Fia. 282.— Vec
grammatic repre tor diagram for an
sentation of an al alternator,
temator.
I is the current flowing and X is the reactance of the winding due
to its self induction. An alternator may therefore be considered
as a circuit with a resistance B and a reactance X as shown
diagrammatically in Fig. 281. The value of X is generally from
4 to 10 times the value of R.
If now an alternator is operating under normal conditions, fully
excited, generating voltage and supplying current then, of the
total voltage generated, a portion IX is required to overcome the
244
Art. 274]
ALTERNATOR CHARACTERISTICS
245
reactance of the winding and another portion IR to overcome the
resistance; the terminal voltage Et is obtained from the generated
voltage Eo by subtracting IR and IX as vectors,
273. Vector Diagram at Fullload.— If in Fig. 282, Ea is the
voltage generated by an alternator at noload, and a circuit is then
connected across the alternator terminals which takes a current I
from the machine, the armature resistance drop IR is in phase
with the current while the current lags the armature reactance
drop IX by 90 degrees, see page 207, and the terminal voltage
Et is obtained by subtracting IX and IR from Eo as shown in
Fig. 282.
Three cases are shown in Figs. 283, 284, and 285, in which an
alternator has the same terminal voltage Et and delivers the same
Fio. 283.— Lagging Fia. 284.— 100 per Fia. 285.— Leading
current. cent, power factor. current.
Figs. 283285. — ^Effect of the power factor of the load on the regulation of
an alternator.
current 7, but different circuits are used in the three cases so that
the phase angles a are different.
It may be seen from these diagrams that the regulation of an
alternator, namely, the ratio (Eo — Et)/Et, depends largely on
the power factor of the load, and becomes negative if the current
is leading considerably, as shown in Fig. 285 where Et is greater
than Eo'
274. Regulation Curves of an Alternator. — Since the regulation
of an alternator depends on the power factor of the load as well
as on the current, the external characteristics have to be given
with different power factors as shown in Fig. 286. These curves
are generally determined by calculation after the resistance and
reactance of the winding have been measured.
246 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap xxxii
A singlephase alternator with an output of 416 amp. at 2400 volts has a
resistance of 0.2 ohms and a reactance of 2.3 ohms. Find the regulation at
100 per cent, power factor, and also at 80 per cent, power factor with a lag
ging current, the fullload voltage being 2400 volts in each case.
Fullload current » 416 amp.
the resistance drop IR » 416 X 0.2 » 83 volts
the reactance drop IX == 416 X 2.3 = 960 volts.
At 100 per cent, power factor, see Fig. 284
Eo^ = {Et + iRy + {ixy
= 2483* + 960*
and Eo « 2660
the regulation = jp — = 2400 ~ ^^'^ ^^ ^^**
Et
^
80 % Power Factor, Leading
Armature Current
FiQ. 286. — Regulation curves of an alternator.
At 80 per cent, power factor with lagging current, see Fig. 283
Eo^ = a5* + be*
= (Et cos « + IRy + {Et sin a + IXy
= (2400 X 0.8 + 83)* + (2400 X 0.6 + 960)*
and Eo = 3120
31202400 ^^
the regulation = — oiOO ~ P®^ cent.
276. Experimental Determination of Alternator Reactance. —
The noload saturation curve in Fig. 287 is determined in the
same way as for a directcurrent generator, see page 70. The
alternator is then shortcircuited through an ammeter as shown
in diagram B, and run at normal speed, while simultaneous read
ings are taken of the armature current /« and the exciting cur
rent // from which the shortcircuit curve is plotted.
Art. 275]
ALTERNATOR CHARACTERISTICS
247
From these two curves the reactance of the alternator may
readily be determined. With an exciting current oa for example,
the voltage generated by the alternator is ab = 2400 volts at no
load. With the same excitation and with the armature short
circuited, the terminal voltage is zero and the generated voltage
ab is used up in sending a current ac = 1040 amp. through the
resistance and reactance of the winding or
voltage ab = current ac X ^JR^ + X^
2400 = 1040 X V/22 + i2
and Vi?^ + X^ = 2.3 ohms
from which X may be determined if the value of R is known, and
this may readily be measured by passing a direct current 1
through the alternator winding and measuring the voltage drop
E, since R = E/L
A. ' Connection for No Load
Saturation Teit
^
B Connection for Short Circuit Test
Kzcittng Current If
Fia. 287. — Determination of the itupedence of an alternator.
It may be seen from Fig. 287 that the reactance X decreases
as the magnetic circuit of the machine becomes saturated. The
reactance is caused by the flux produced by the armature current,
as shown in Fig. 280. When the field coils are excited so that
the main flux of the machine is passing through the poles, then
a smaller additional armature flux is produced with a given
armature current than when the poles are not excited, and a
smaller armature flux produces a smaller reactance.
1. A threephase Y connected alternator has an output of 240 amp. at
2400 volts. With a certain field excitation the noload voltage between
terminals was 2400 volts and the current in each line on shortcircuit was
600 amp. The resistance of each phase is 0.2 ohms. Find the reactance
D«r phase.
248 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
Ett the terminal voltage at noload « 2400 volts
E, the voltage per phase at noload — 2400/\/3 page 236
= 1390 volts
7i, the line current on shortcircuit = 600 amp.
/, the current per phase on shortcircuit = 600 amp., page 236
1390
Z, the impedence per phase = ^^ = 2.3 ohms
X, the reactance per phase = ^^^.ZL^L^
= \/2.3»  0.2«
= 2.3
Find the regulation of this machine at fullload and 100 per cent, power
factor, the fullload voltage between terminals being 2400 volts.
Fullload current in line « 240 amp.
the fullload current per phase = 240 amp.
the resistance drop IR per phase = 240 X 0.2 = 48 volts
the reactance drop IX per phase = 240 X 2.3 = 550 volts
the fullload voltage between terminals = 2400 volts
the fuUload voltage per phase = 2400/ V3 = 1390 volts
then at 100 per cent, power factor, see Fig. 284
Eo* = (1390 f 48)» f 550*
and Eo = 1540 volts per phase
and the noload voltage between terminals » 1540 X a/S
= 2660 volts
2660 2400
the regulation = 0400 " ^^'^ ^^ *'®°**
2. A threephase deltaconnected alternator has an output of 240 amp. at
2400 volts. With a particular field excitation the noload voltage between
terminals was 2400 volts and the current in each line on shortcircuit was 600
amp. The resistance of each phase was 0.6 ohms. Find the reactance per
phase.
Ely the terminal voltage at noload = 2400 volts
Ef the voltage per phase at noload — 2400 volts, page 237
Iif the line current on shortcircuit = 600 amp.
/, the current per phase on shortcircuit = 600/ V^3, page 237
» 346 amp.
2400
Z, the impedence per phase = "qIa" — ^'^ ohms
X, the reactance per phase = '\/6.9*— 0.6»
= 6.9
Find the regulation of this machine at fullload and 100 per cent, power
factor, the fullload voltage between terminals being 2400 volts.
Fullload current in the line = 240 amp.
the fullload current per phase = 240/\/3, page 237
B 139 amp.
Art. 276]
ALTERNATOR CHARACTERISTICS
249
the resistance drop IR per phase = 139 X 0.6 = 83 volts
the reactance drop IX per phase = 139 X 6.9 = 960 volts
the fullload voltage between terminals = 2400 volts
the fullload voltage per phase == 2400 volts, page 237
at 100 per cent, power factor, see Fig. 284,
Eo^ =
Eo =
the regulation =
(2400 f 83)« f 960*
2660 volts
26602400 ,^^
— 2400 — ~ ^^^ ^
276. Automatic Regulators. — To maintain the voltage of an
alternator constant, the field excitation must be increased as the
armature current increases and as the power factor decreases.
This cannot be done by adding series field coils as in the case of
the directcurrent generator, see page 74, because the line cur
rent is alternating and not suitable for excitation purposes.
Fig. 288. — ^Automatic voltageregulator.
Automatic regulators are used with alternators. The essen
tial parts of such a regulator are shown diagrammatically in Fig.
288. To keep the voltage Et constant, some means must be
provided to close the contact c so as to shortcircuit the resistance
r and thereby increase the exciter voltage and also the exciter
current i, when the alternator voltage is too low, and to open
this contact and insert the resistance r in the exciter field circuit
when the alternator voltage is too high.
The contact c is opened by the electromagnet M on which are
two opposing windings A and B. When the contact a is open, B
alone is excited and opens the contact c against the tension of the
spring d, but when a is closed, the coil A also is excited and neu
tralizes the pull of B and the spring d closes the contact c.
If then the alternator voltage rises, the pull of the solenoid S is
250 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxn
increased and its plunger is raised so as to open the contact a,
the coil A is then deenergized and B opens the contact c and
inserts the resistance r in the exciter field circuit. The alternator
voltage now drops, the pull of the solenoid S decreases, and the
weight of the plunger closes the contact a and thereby excites
coil A which neutralizes the pull of B, the spring d then closes the
contact c and shortcircuits the resistance r and thereby increases
the field excitation.
When additional attachments are added to this regulator to
make it more sensitive, the voltage fiudtuations on which the
operation of the regulator depends cannot be detected on a sensi
tive voltmeter.
277. Efficiency. — The losses in an alternator are the same as in
a directcurrent generator and consist of the stray loss, the arma
ture copper loss, and the field excitation loss. These losses are
determined in the same way as for the directcurrent machine, see
page 97. It must be noted however that the efficiency depends
on the power factor of the load, as may be seen from the follow
ing example.
In a twophase alternator which has an output of 83 amp. per phase at
2400 volts
the stray power loss =« 16 kw.
the exciting current at fullload = 68 amp. at 100 per cent, power factor
= 85 amp. at 80 per cent, power factor
the exciter voltage is 1 10 and no automatic regulator is supplied
the resistance of each phase of the armature winding, measured by
direct current = 0.4 ohms.
Find the efficiency when the power factor of the load is 100 per cent, and
also when 80 per cent.; find also the horsepower of the driving engine.
at 100 per cent, power factor at 80 per cent, power factor
The output = 2 X 2400 X 83 = 2 X 2400 X 83 X 0.8
= 400 kw. = 320 kw.
The stray loss = 16 kw. =* 16 kw.
The excitation loss
= 110 X 68 = 7.5 kw. = 110 X 85 = 9.4 kw.
The armature copper loss
= (83« X 0.4) X 2 =5.5 kw. = 5.5 kw.
The total loss = 29 kw. = 30.9 kw.
The input = 429 kw. = 350.9 kw.
The horsepower of the
driving engine = 575 h.p. = 470 h.p.
400 320
The efficiency = — = 93.4 per cent. = =91.4 per cent.
429 350.9
Art. 278] ALTERNATOR CHARACTERISTICS 251
The lower the power factor, the smaller is the power output, and
at the same time the greater the excitation loss because of the
increase in the exciting current required to maintain the voltage.
278. Rating of Alternators. — An alternator is designed so as to
give normal voltage and normal current without overheating, but
the output in kilowatts will depend entirely on the power factor
of the connected load. It is usual to specify the output at 100
per cent, power factor and then, to emphasize the fact that this
output cannot be obtained from the machine at lower power fac
tors, the unit of output is taken as the kilovolt ampere (kv.a.)
and not as the kilowatt, where (kv.a. X power factor) = kw.
A singlephase alternator can give 100 amp. at 2400 volts. What is the
output of the machine in kv.a. and also in kw. if the power factor of the load
is 80 per cent.
2400 X 100 „,^
^^•* = 1000^=240
kw. = 240 X 0.8 = 192
A threephase alternator can give 100 amp. from each terminal with a vol
tage between terminals of 2400 then
*i. * *• V 1.73 X 2400 X 100 , .._ ._
the output m kv.a. = rncvS ^'^ page 236, = 415
at 80 per cent, power factor the output would be 415 X 0.8 « 332 kw.
CHAPTER XXXIII
SYNCHRONOUS MOTORS AND PARALLEL OPERATION
279. Principle of Operation of Synchronous Motors. — An
alternatingcurrent generator may be made to operate as a motor.
When an alternating e.m.f . is applied to the winding of the single
phase machine shown in Fig. 289; an alternating current flows
through that winding. The conductors a, b, c, and d are then
canying current and are in a magnetic field so that a force acts on
each conductor, while an equal and opposite force acts on the
poles and tends to turn the rotor. The current however is
alternating, so that the force on the rotor is alternating in direc
tion unless the polarity of the poles is changed at the instant the
current reverses.
Fig. 289. — The synchronous motor.
This would be the case if the machine was already running at
such a speed that, during the time of half a cycle or in ^ seconds,
the rotor moves through the distance between two adjacent poles
or through 1/p of a revolution. This speed, called the synchron
ous speed, is therefore equal to
 X 2/ rev. per sec.
P
120/
or rev. per mm.
V
252
Art. 280]
SYNCHRONOUS MOTORS
253
and is the speed at which the machine would have to run as an
alternator in order to generate an e.m.f. of/ cycles per second, see
the formula on page 195. The table on page 195 therefore applies
to synchronous motors, as this type of machine is called, as well
as to alternators. When a synchronous motor is running at
synchronous speed it is said to be in step with the alternators
driving it.
A synchronous motor is not self starting, but will develop a
torque continuously in one direction if running at synchronous
speed. Such a machine is generally brought up to speed by a
small selfstarting motor which is direct connected to the shaft
280. The Back E.m.f . of a Synchronous Motor. — If the motor
M, Fig. 290, is rotating at synchronous speed, then its stator wind
Diagram matic Sepresentation
►
Em ]M
Alternator Synch rouoas Motors
A No Load B Full Load
FiQ. 290. — Alternator driving synchronous motors.
ing is being cut by lines of force and an e.m.f. is generated in the
machine in the same way as if it were driven by an engine. This
e.m.f . Efnf called the back e.m.f. of the motor, opposes the applied
e.m.f. Eg and is of the same frequency, and, in the case where the
two machines are equally excited, then E^ = Eg.
If the motor is running on noload, the load current that it takes
from the line is practically zero, being merely sufficient to over
come the friction of the machine, and in such a case E^ is exactly
equal and opposite to Eg at every instant as shown by the vectors
in Fig. 292, for, under these conditions, there is no resultant e.m.f.
and no flow of current through the machines. The poles of the
254 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxii
two machines will then rotate together with the relative position
shown in diagram A, Fig. 290.
If now the motor is loaded, it will slow down for an instant
and will swing back relative to machine G as shown in diagram B^
Fig. 290. The back e.m.f. En^, although still equal to Eg is now
no longer opposite in phase as at noload, but lags its noload
value by an angle 9 as shown by the vectors in Fig. 292. There
is now a resultant e.m.f. Er which sends a current through the
machines, and the torque developed due to this current keeps
the motor running at synchronous speed but always with such a
lag behind the generator as to allow a current to flow large enough
to develop a driving torque equal to the retarding torque of the
load. The motor therefore automatically takes from the genera
tor a current corresponding to the mechanical load.
AEi
FiQ. 291. — Mechanical anal
ogy to a synchronous motor.
^m Em
No load. Full load.
Fig. 292. — Voltage vector dia
grams for a synchronous motor.
281. Mechanical Analogy. — The transmission of power by
means of an alternator and a synchronous motor is similar in
many ways to the transmission of power by means of a flexible
spring coupling such as that shown in Fig. 291. If the load on
the side M is increased, the spring stretches and M drops back
through a small angle relative to (?, but both continue thereafter
to rotate at normal speed.
282. Vector Diagram for a Synchronous Motor. — In Fig. 293
E, is the e.m.f. generated in the alternator winding.
Em is the back e.m.f. generated in the motor winding.
Er, the resultant e.m.f., sends an alternating current I through
both machines.
Abt. 283]
SYNCHRONOUS MOTORS
255
Rm and X^ are the resistance and reactance of the motor winding
Rg and Xg are the resistance and reactance of the generator
winding.
and since the resistances are generally small compared with the
reactances, see page 244, therefore
T — ^r
Since the circuit is almost entirely inductive, the resistance being
negligible, the current / must lag the voltage Er by 90 degrees.
The power developed by the generator = Egl cos a.
Generator
Motor
Em
Em
Fig. 293. Fig. 294. Fig. 295.— Load greater than
Light load. Heavy load. the maximum output.
Figs. 293295. — Vector diagram for the synchronous motor at various loads.
If the load on the motor is now increased, the motor swings
back relative to the generator by a greater angle 6, as shown in
Fig. 294. The voltage Er and the current I are now larger than
before and so also is Egl cos a the power developed by the
generator and put into the circuit.
283. Maximum Output — An extreme case is shown in Fig. 295
where the load has caused the motor to swing back by a large
angle 6, but the power put into the line, namely Egl cos a, is
256 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxra
now less than in the case represented m Fig. 294; so that, while
the angle 9 increases as the load is increased, the power input
Egl cos a increases only up to a certain point called the break
down point or point of maximum output. When the load exceeds
this value the motor slows down and stops.
The maximum output is generally more than twice the normal
output of the machine as fixed by the heating of the windings.
284. Operation of a Synchronous Motor when Under and
Overexcited. — In diagram B, Fig. 296, the excitation of the
motor M is such that Em = Eg. Two other conditions of opera
tion are represented in diagrams A and C. In the former case
the motor is overexcited and, since the speed cannot change
but must always be synchronous speed, the voltage Em must
CO! AT
' ^/cos cr /^
A B C
FiQ. 296. — Effect of excitation on the power factor of a synchronous motor.
increase with the excitation and must now be greater than Eg.
In the latter case the motor is underexcited and Em must be less
than Eg. The load on the motor however is unchanged so that
Egl cos a is constant.
It is important to note that, in the case of the overexcited
motor, diagram A, the current / leads the generator voltage Eg^
or an overexcited synchronous motor draws a leading current
from the line. Now a condenser always draws a leading current,
see page 220, so that an overexcited synchronous motor acts to a
certain extent like a condenser.
286. Use of the Synchronous Motor for Power Factor Correc
tion. — If the load connected to an alternator has a low power fac
tor^ it is often advisable to arrange that some of the load shall be
Art. 286]
SYNCHRONOUS MOTORS
257
carried by synchronous motors so as to improve the power factor
of the whole system'. '.
If 1000 horsepower of 2200volt singlephase induction motors are
operating at the end of a transmission line, find the current in the line and
also the generator capacity required if the average power factor is 80 per
cent, and the average efficiency is 90 per cent. (The induction motor, see
Chap. 36, takes a lagging current, and its power factor cannot be controlled.)
The output of the motors is 1000 h.p.
the input to the motors is "ttq" = 1115 h.p.
= 830 kw.
830
the generator capacity required = ?r^ = 1040 kv.a.
., . . ^, ,. 1040 X 1000 _
the current m the Ime = t^kt^t^ = 473 amp.
2200
^ Et  2200
%
Fig. 297.
B
If 500 horsepower of the load is driven by a synchronous motor, the power
factor of the whole system may be raised if this motor is overexcited and made
to act as a condenser.
If the power factor of the synchronous motor be made 80 per cent., with
the current leading, then the vector diagram for the load is as shown in
diagram B, Fig. 297.
The induction motor output = 500 h.p.
the induction motor input = xg X Tqqq = 415 kw.
the current for these motors = 2 20Q y Q S ~ ^^^ amp.
The current for the synchronous motor also is 236 amp., but it leads by an
angle whose cosine is 0.8 whereas the current for the induction motor lags by
the same angle
the resultant current in the line »= 2 (236 X 0.8)
= 378 amp.
.u ^ '^ ' A 378X2200
the generator capacity required = Tom —
= 830 kv.a.
258 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxiii
By the use of an overexcited synchronous motor, the power fac
tor of the system is improved, the generator capacity required
for the load is reduced, and so also is the current in the line, so that
overexcited synchronous motors may be used with advantage at
the end of a long line.
286. Synchronizing. — Before a synchronous motor can carry
a load it must be started and brought up to synchronous speed,
and the switch S, Fig. 298, should not be closed until E^ is equal
and opposite to Eg. If this switch were closed when Em and Eg
are acting in the same direction round the closed circuit, then the
resultant voltage Er would be equal to E^ + Eg and would send
a destructive current / through the circuit. When E^ and Eg
are equal and opposite, then Er is zero, and no current / will
flow when the switch S is closed. To test for this condition, a
lamp L, or an indicating device called a synchroscope, is placed
Fia. 298. — Alternator driving Fig. 299. — Alternators in par
a synchronous motor. allel.
across the switch S. When the lamp is brightest, then Er =
Em + Eg'y when darkest, then Er = zero and the switch S may
be closed, after which the load may be put on the motor. The
operation described above is called synchronizing.
287. Hunting. — In the case of an enginedriven alternator, and
particularly if the engine is a gas engine, the angular velocity is not
uniform but consists of a uniform angular velocity with a super
imposed oscillation, the frequency of the generated e.m.f. there
fore is not constant, but rises and falls regularly.
If this e.m.f. is applied to a synchronous motor, the synchron
ous speed of the motor tends to rise and fall regularly with the
frequency, and the motor tends to have a superimposed oscilla
tion similar to that of the alternator. If the natural period of
oscillation of the motor has the same frequency as this forced
oscillation then the effect will be cumulative and the motor will
oscillate considerably.
Art. 288] SYNCHRONOUS MOTORS 269
A similar result would be found with the model shown in Fig.
291. If the torque applied to 6 is not uniform, then G will
oscillate about its position of mean angular velocity and M will
have an oscillating force impressed on it by the spring. If the
moment of inertia of the flywheel M is such that its natural fre
quency of oscillation is the same as the frequency of the impressed
oscillation then G and M will swing backward and forward relative
to one another through a considerable angle.
As the two machines M and G oscillate relative to one another,
the angle 6, Fig. 294, increases and decreases regularly, and the
value of both Er and of the current / vary above and below the
average value required for the load. This surging of current is of
comparatively low frequency and is indicated by an ammeter
placed in the circuit. Due to this surging, the circuit breakers
protecting the machines may be opened although the load is not
greater than normal, while the cumulative swinging of the ma
chines relative to one another, called hunting, may cause the
motor to drop out of step.
To prevent hunting, the impressed oscillations must be elim
inated or the natural frequency of vibration of the motor must
be changed. The methods used in practice to minimize hunt
ing are:
1. Dampen the governor if the impressed oscillations are found
to be caused by a hunting governor.
2. Change the natural period of vibration of the machine by
changing the flywheel; the larger the moment of inertia of the
rotating part of the motor, the longer is its natural period of
vibration.
3. Dampen the oscillations electrically by the use of pole
dampers such as those described on page 295.
288. Parallel Operation of Alternators. — Two alternators
connected to operate in parallel are shown in Fig. 299. If the
voltage of machine B is not exactly equal and opposite to that of
A at every instant then current will flow in the local circuit
between the two machines just as in Fig. 298. To prevent this
the two machines must be synchronized in the same way as an
alternator and a synchronous motor; when the lamp L in Fig. 299
is dark, then Eb is exactly equal and opposite to Ea and the
machines have the same frequency; the switch S may then be
closed. If now the engine of B fails for an instant, then generator
B will tend to slow down and will swing back relative to 4, so that
260 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxiii
curreDt will flow in the local circuit and B will be driven as a syn
chronous motor in the same direction as before and at the same
speed. As soon as the engine of B recovers, the generator will
swing forward again and cariy its share of the load.
If two directcurrent generators are operating in parallel, and
the field excitation of one of the machines is increased, then the
voltage of that machine will be raised and it will take a larger
portion of the load. An increase in load makes the engine and
generator slow down and allows the engine to draw the additional
amount of steam required for the additional load.
If two alternators are operating in parallel, an increase in the
excitation of one machine raises the voltage of that machine and
tends to make it carry more of the load. But the machine can
not slow down and allow the engine to take more steam because it
can run only at synchronous speed. An increase in excitation of
one machine therefore increases its voltage and at the same time
makes the current in that machine lag further behind the voltage
so as to maintain the load EI cos a constant at the value corre
sponding to the steam supply. To change the distribution of
load between two alternators operating in parallel, the governors
of the driving engines must be manipulated so as to change the
distribution of the steam supply.
As the total load on the two alternators increases, they both
slow down, and the engine governors automatically allow the
necessary amount of steam to flow, while the frequency of the
generated e.m.f. decreases slightly. In order that the two
machines may divide the load properly, the engines should have
the same per cent, drop in speed between noload and fullload.
The same applies to alternators driven by water wheels. To
make any one of a number of turbinedriven alternators take a
larger portion of the total load, the governor of that machine
must be manipulated to allow the turbine to take more water.
CHAPTER XXXIV
TRANSFORMER CHARACTERISTICS
Secondary
289, In order that electric energy may be transmitted
economically over long distances, high voltages must be used;
but in order that electric circuits may be safely handled, low vol
tages are necessary for distribution. The__altematingrcurrent
transformer is a piece of apparatus by means j)f_whjijch. electricity
can be received at one voltage and delivered at c^nother voltage
either higher or lower It consists essentially of two coils wound
on an iron core; one coil receives energy and is called the primary
coil, the other delivers energy and is called the secondary coil.
290, Constant Potential Transformer.— In Fig. 300, C is a
closed magnetic circuit on which
are wound two coils having rii and
n2 turns respectively. When an
alternating e.m.f. d is applied to
the coil Til while coil n2 is closed
through a circuit as shown, then a
current ii flows in the primary coil
and produces an alternating mag
netic flux 4> which threads both
coils and generates in them electromotive forces en and €2 which
are proportional to ni and 712 the number of turns.
Now eib is called the back e.m.f. of the primary and, according
to Lenz's law, opposes the change of the flux which produces it
and therefore opposes ei which produces the change of flux; db
is less than ei by the e.m.f. required to send the current ti through
the primary coil, which e.m.f. is small in modern transformers and
seldom exceeds 1 per cent, of ei even at fullload.
The e.m.f. 62 sends a current t2 through the secondary winding
in such a direction as to oppose the change of the flux <t> which
produces it, and therefore to oppose ii; but the magnetizing
effect of ii must always be greater than the demagnetizing effect
of it by the amount necessary to produce the flux in the mag
netic circuit.
261
Fig. 300. — The transformer.
262 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
If now the impedence of the secondary circuit is decreased, the
current i% will increase and thereby reduce the flux 0. But a
reduction in <t> causes the back e.m.f. eu to decrease and thereby
allow a larger current t'l to flow in the primary winding, so that
the primary current always adjusts itself to suit the requirements
of the secondary circuit.
Now the primary resistance is so small that en, and do not
drop more than about 1 per cent, between noload and fullload,
so the statement may be made that 4> is constant at all loads and
therefore the resultant of nit'i and niUy the primary and the
secondary ampereturns, must always be equal to some quantity
niu which produces the constant flux 4>.
The quantity rtiU may readily be found because, if the second
ary circuit be opened, no current can flow in that winding, and
the current in the primary under these conditions has merely to
produce the flux <t>. This current must therefore be zo and is
called the magnetizing current of the transformer.
Since 6i6 and et are produced by the same magnetic flux they
are proportional to the number of turns ni and n2, and since 615 is
practically equal to ei therefore
Ci _ ni
and is called the ratio of transformation.
It has also been shown that the resultant of niii and of n^it
must always be equal to niu where Hy called the magnetizing
current, is comparatively small, so that if to be neglected then
rii ia 61
or _—__
and eii\ = 62^*2
According to the law of conservation of energy, the input to a
transformer must be exactly equal to the output, the losses being
neglected (the efficiency of a 50kilovolt ampere transformer is
98 per cent.) therefore
eiii cos «! = 62^2 cos a2
and Ciii = e^i^
therefore cos a\ = cos a^
so that, neglecting the magnetizing current and the losses, the
power factor of the primary is the dame as that of the secondary.
Art. 292]
TRANSFORMER CHARACTERISTICS
263
291. Vector Diagram for a Transformer. — (a) Noload condi
tions. For a transformer which has a negligible noload loss, the
voltage and current phase relations are shown in Fig. 301.
it> is the magnetic flux threading both coils.
io is the magnetizing current which produces <t>,
€2 and ei6 are the e.m.fs. generated in the coils rtt and rii
respectively, and lag the flux which produces them by 90 degrees,
see the footnote on page 207.
6i, the applied primary e.m.f., is equal and opposite to eib.
(b) Fullload conditions. Let the secondary circuit now be
closed and let its resistance and reactance be such that it lags
62 by a2 degrees, then the voltage and current phase relations are
as shown in Fig. 302.
is the magnetic flux threading both coils and has practically
the same value at fullload as at noload.
»0
»16 9i
FiQ. 301.— No load. Fia. 302.— Full load.
Vector diagrams for a transformer.
to is the component of the primary current required to produce
the flux 0.
ei is the applied primary e.m.f.
62 is the secondary generated e.m.f.
il is the secondary current, whose value and whose phase angle
«2 depend on the constants of the connected circuit.
ill is the component of the primary current required to neu
tralize the demagnetizing effect of the current {2;
ii is the primary current and is the resultant of to and in.
If the current io is small, then ai = at and riiii = n2i2.
292. Induction Furnace. — An electric furnace which operates
as a transformer is shown diagrammatically in Fig. 303 and is
called an induction furnace, the secondary winding in this case
being the charge which is contained in the annular channel A
264 PRINCIPLES OF ELECTRICAL ENQINEERINQ [Chap, xxxiv
and is heated by the secondary current. The amount of energy
put into the secondary can be varied by varying the applied
primary voltage.
Fig. 304 shows diagrammatically an electric welder which
operates on the same principle. The single turn A in this case
is open, and is closed by the two pieces to be welded, which
pieces are held in the clamps B and are forced together under
pressure while the welding current passes across the contact and
heats the ends to be joined.
It might seem that, since the secondary load in this case is a
resistance load being the resistance of the molten metal, the
power factor of the transformer would be high at fullload, and
yet in practice it seldom exceeds 70 per cent. To explain the
Secondary
'WinJiDg
Ss^ Primary
Winding
Fia. 303. — Induction furnace. Fig. 304. — ^Induction welder.
cause of this low power factor it is necessary to take up the subject
of leakage flux in transformers.
293. Leakage Reactance. — ^Fig. 306 shows the actual flux
distribution in a transformer. The ampereturns Uiii produce
a flux 4>iiy called the primary leakage flux, which is proportional
to ii and which threads the coil Ui but does not thread n^.
The ampereturns 712^2 produce a flux 02z, called the secondary
leakage flux, which is proportional to U and which threads the
coil n2 but does not thread ni.
The ampereturns Uiii and n2t2 acting together produce a
magnetic flux which threads both coils Ui and n^ and which is
practically constant in magnitude; the effect of this constant flux
has already been considered.
Now any coil in which a current i produces a flux 4> which is
proportional to the current is said to have self inductioni see
Abt. 293]
TRANSFORMER CHARACTERISTICS
265
Art, 239, page 206, and the voltage to send an alternating cur
rent / through such a coil = IX where X is the reactance of
the coil; the current lags this voltage by 90 degrees, see page
208.
In Fig. 306, the flux 0iz is proportional to the current ti and
its effect is the same as if the coil Ui had a reactance Xi so that,
instead of considering the effect of the flux 0iz, the effect of the
Iia
I
<P
izRz
IE" Et
I
J.
Fig. 305. Fig. 306. Fig. 307.
Fig. 305. — Ideal transformer.
Fig. 306. — Actual flux distribution in a transformer.
Fig. 307. — Transformer showing the resistances and reactances diagram
matically.
equivalent reactance Xi may be considered. In the same way
the leakage flux 02z may be represented by an equivalent react
ance X2. Fig. 307 shows the diagram of an actual transformer
in which the leakage fluxes 4>ii and <i>2i are replaced by the equiva
lent reactances Xi and X2 which, along with the resistances Ri
and R2 of the coils, are placed for convenience outside of the actual
winding.
Fig. 308. — Vector diagram
of an ideal transformer.
Fig. 309. — Vector diagram
of an actual transformer.
Between the terminals ah and cd, the transformer diagram in
Fig. 307 is the same as the ideal diagram in Fig. 305. The vector
diagram for this ideal transformer is shown in Fig. 308 which is
the same as Fig. 302, page 263.
The actual terminal voltage Et is obtained by subtracting from
Et the vectors /2B2 and /2X2, the voltages to overcome the sec
266 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxnv
ondary resistance and reactance respectively, where ItRt is in
phase with 1% and It lags ItXt by 90 degrees.
The applied primary voltage Ea is obtained by adding to Ei
the vectors IiRi and /iXi, the primary resistance and reactance
drops.
In the elementary discussion of transformer operation, the
resistance and reactance drops were neglected and it was found
that ^ = — In Fig. 309 this relation still holds, but the ratio
El Til
— is less than — and so less than — , so that the transformer
Ea El Til
ratio decreases as the load increases or the secondary voltage
drops with increase of load; this drop however is small, being
less than 2 per cent, for a 5 kv.a. transformer at 100 per cent,
power factor, while for leading currents in the secondary circuit
the secondary voltage may rise with increase of load just as in
the case of the alternator, see page 245; the student may satisfy
himself on this point by drawmg the vector diagram.
294. Leakage Reactance in Standard Transformers and in In
duction Furnaces. — In power transformers, the leakage reactances
Primary
Secondary
J
A B
Fio. 310. — ^Leakage flux of transformers.
Xi and X2 are kept small by constructing the transformer so
that 4>u and 4>2i are small. Diagram Aj Fig. 310, shows a trans
former with a primary and a secondary coil on each leg and shows
also the leakage fluxes. It may be seen from this diagram that,
on each leg, <t>ii and i^i act in opposite directions, so that if n^
were interwound with tii then the leakage fluxes would neutralize
and only the main flux be left. This result is approximated in
practice by constructing the transformer as shown in diagram B,
where half of the primary and half of the secondary winding are
placed over one another on each leg of the transformer core, the
leakage fluxes have then to crowd into the space x between the
Abt. 295] TRANSFORMER CHARACTERISTICS 2&7
windings, and the smaller the space x, the smaller the leakage
fluxes and the smaller the leakage reactances.
In the induction furnace, unfortunately, the distance x cannot
be made small, so that the reactances are comparatively large.
The vector diagram for such a furnace is shown in Fig. 311. The
secondary winding is shortcircuited since it consists of a ring of
molten metal, so that Etj the terminal voltage, is zero, and the
secondary generated voltage E2, is therefore made up of the two
components I%Ri, and /2X2, where R2 is the resistance of the ring'
of molten metal and X2 its reactance due to the leakage flux 02z,
Fig. 307. The remainder of this diagram is determined in the
same way as in Fig. 309, and the power factor of the furnace is
the cos ai and seldom exceeds 70 per cent.
Fig. 311. — Vector diagram for an induction furnace.
296. The Constantcurrent Transformer. — For the opera
tion of arc lamps in series, the constantcurrent transformer
shown in Fig. 312 is used. The primary coil is stationary and
receives power at constant potential, while the secondary coil,
which is suspended and is free to move toward or away from
the primary, delivers a constant current to the lighting
circuit.
When the secondary coil is close to the primary, the react
ances of the transformer are small and the secondary voltage
is approximately equal to the primary voltage multiplied by
the ratio of the number of turns. As the distance between
the coils is increased, the leakage flux and the reactances
increase and the secondary voltage drops, even although the
primary voltage remains constant.
The primary and the secondary currents are opposite in
direction and, under thetee conditions, the primary and the
secondary coils repel one another. The counterweight on the
secondary coil is so adjusted that, when the desired current is
268 PRINCIPLES OF ELECTRICAL BNGINEERINC [Chap, xnciv
Sowing in this coil, the force of repulsion keeps the secondary
coil suspended. If then some of the lamps in the circuit are
cut out, the current tends to increase, but any increase in the
current separates the coils and the voltage drops due to the
increased reactance and so Is unable to maintain the current
at the increased value. The power factor of such s transfonum
Fig. 312. — Conetantcurrent transformer.
is high when the coils are close together but decreases as the
distance between the coils is increased.
296. The efficiency of a transfonner
where the losses are:
Iron losses; the hysteresis and eddy current losses in the core.
Copper losses; these are /I'fli and /I'iJj watts respjectively for
Abt. 300]
TRANSFORMER CHARACTERISTICS
269
(S)
the primary and secondary windings. There is no power loss due
to the primary and secondary reactances, see page 209.
297. Hysteresis Loss. — Since the flux in a transformer core
is alternating, power is required to continually reverse the mole
cular magnets of the iron, this power is called the hysteresis loss.
298. Eddy Current Loss. — If the transformer core in Fig. 313
is made of a solid block of iron, then the alternating flux thread
ing this core causes currents to flow as shown at 4, in the same
way as through a shortcircuited secondary winding. Power is
required to maintain these eddy currents
which power is called the eddy current loss.
To keep these eddy currents small, a high
resistance is placed in their path. This is
accomplished by laminating the core, as shown
at B, the laminations being separated from one
another by varnish.
299. Iron Losses. — The hysteresis and the
eddy current losses taken together constitute
what is called the iron loss. Since the flux
per pole is practically constant at all loads,
see page 262, therefore the iron loss caused by
this flux is also constant at all loads. This
loss may readily be determined by operating
the transformer at noload with normal voltage and frequency,
the input under these conditions, measured by a wattmeter, is
equal to the iron loss, the small copper loss due to the noload
current being neglected.
300. The allday efficiency is defined as the ratio of the total
energy used by the customer to the total energy input to the trans
former, during twentyfour hours. This efficiency is of impor
tance in the case of lighting transformers, which are connected
to the mains for twentyfour hours a day but which supply
energy for about five hours a day, under such conditions the all
, ^ . output X h
y e ciency — Q^^p^^ x A + iron loss X 24 + copper loss x h
where h is the number of hours per day during which energy is
taken from the transformer.
In a 60kv.a. 2200 to 220 volt transformer the iron loss is 300 watts, the
primary resistance is 0.5 ohms and the secondary resistance is 0.005 ohms.
Find
a. The efficiency when the load is 50 kw. and the power factor 100 per
cent.
FiQ. 313.— Trans
former core.
270 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxiv
h. The efficiency when the load is 5 kw. and the power factor 100 per
cent.
c. The efficiency when the load is 50 kv.a. and the power factor 80 per
cent.
d. The allday efficiency in the latter case if the load is constant and con
nected for 5 hours a day while the transformer is connected to the line for
24 hours a day.
a. It, the secondary current » 50000/220 » 227 amp.
7i, the primary current — 50000/2200 = 22.7 amp. approximately
Copper loss = 227* X 0.005 f 22 7* X 0.5 = 514 watts.
Iron loss = 300 watts
Total loss = 814 watts
Output " 50,000 watts
Input — 50,814 watts
Efficiency = 98.5 per cent.
6. It = 5000/220 « 22.7 amp. and 7i  5000/2200 = 2.27 amp. approx.
Copper loss = 22.7* X 0.005 + 2.27* X 0.5 = 5.14 watts
Iron loss « 300 watts
Total loss » 305 watts
Output = 5000 watts
Input = 5305 watts
Efficiency = 94.4 per cent.
c. It = 50000/220 = 227 amp. and 7i = 50000/2200 = 22.7 amp. approx.
Copper loss  227* X 0.005 f 22.7* X 0.5 = 514 watts
Iron loss » 300 watts
Total loss » 814 watts
Output = 40,000 watts
Input = 40,814 watts
Efficiency = 98 per cent.
40000 X 5
d. Allday efficiency
40000 X 5 + 300 X 24 + 514 X 5
= 95.5 per cent.
301. Cooling of Transformers. — Transformers become heated
up due to the losses. This heat must be dissipated and the
temperature of the transformer windings kept below the value
at which the insulation begins to deteriorate. Transformers with
an output of less than 1 kv.a. can dissipate their heat by direct
radiation^ but it is usual to place all transformers up to 500 kv.a.
into a steel tank, which is then filled with insulating oil above the
level of the windings. The oil improves the insulation, and con
vection currents are set up in the oil by means of which the heat
is carried from the surface of the transformer to the larger surface
of the tank, from which it is dissipated to the surrounding air.
Such a transformer is said to be self cooled.
The losses in a transformer are proportional to the volume of
Akt. 3011 TRAffSFORMER CHARACTERISTICS 271
the transformer, while the radiating surface is equal to the super
ficial area, so that, as the size of the transformer increases, the
losses increase more rapidly than the radiating surface; special
arrangements must therefore be made for coolii^ transformers of
large output.
For outputs from 50 to 500 kv.a., corrugated tanks such as
that in Fig, 3 14 are used, while for outputs from 500 to 2000 kv.a.,
FiQ. 314.— Corrugated tank. Fig. 315.
more surface must be provided than can be obtained from a cor
rugated tank and the construction shown in Fig, 315 is used; for
outputs greater than 2000 kv.a., Other methods of cooling allow
the use of a smaller and cheaper transformer.
The watercooled type is shown in F^. 316; cold water is cir
culated through the water coil and takes the heat from the hot
upper layers of the oil. About 2 1/2 lb. or a quarter of a gallon
of water is required per minute per kilowatt loss in the transformer.
The airblast type of transformer is shown in Fig. 317. The
272 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxiv
transformer is well supplied with ducts so that the air can reach
the points at which the heat is generated. This type of trans
former is lighter than the oilfilled type but cannot be satis
factorily insulated for voltages greater than 30,000 volts. One
hundred and fifty cubic feet of air is required per minute per
kilowatt loss.
oter Colli
Fig. 316. — Diagram
matic representation of a
watercooled transformer.
A
»•*»«=» g'O^
Air Duct
Fig. 317. — Diagrammatic represen
tation of an airblast transu)rmer.
BeierToir
Transforiner
1
^n
Cooler
Oil Pump
"JJJ^J^JJJ^,!^
'Jij'/j/jfk VSv
Water
Pump
MjtjfjjffiTyA. vM
Fig. 318. — Method of cooling a transformer by circulating the oil.
The circulating oil type of transformer is shown in Fig. 318
and may be used with advantage where only hard water is avail
able for cooling purposes. When such water is used in water
cooled transformers, salts are liable to deposit inside the cooling
coils and throttle the supply of water. With the circulating
oil method of cooling, these salts will deposit on the outside of the
cooling coils.
CHAPTER XXXV
TRANSFORMER CONNECTIONS
302. Lighting Transformers. — These are generally built to
transform from 2200 to 110 volts, but both primary and secondary
windings are divided as shown diagrammatically in Fig. 319, and
these windings may be so connected that a standard transformer
can operate on the hightension side at either 2200 or 1100 volts
and on the lowtension side at either 220 or 110 volts. The differ
ent connections used are shown in Fig. 319.
UOO VoltB to UO Yolta
to 22D YoltB to 220 Volta
3 Wire
toaaOVolti
3 Wire
2200 YoltB to UO YoltB to 220 YoltB
Fio. 3119. — Standard connections for a lighting transformer.
303. Connections to a Twophase Line. — In Fig. 320, diagram
A shows the method of transforming from high voltage twophase
to low voltage twophase, for the operation of motors.
Specify the transformers for a twophase motor which delivers 50 h.p. at
440 volts with an eflSciency of 90 per cent, and a power factor of 88 per cent.^
the line voltage being 2200.
273
274 PRINCIPLES OP ELECTRICAL ENGINEERING ICkap.xxxv
motor output — 50 h.p.
 SO X 0.748  37.2 kw.
motor input = 37.2/0.9 " 41.4 kw.
 41.4/0.88 = 47 kv.a.
therefore two transformers are required each of 23.5 kv.a. output. The
secondary current = 23500/440 — 53 amp., the primary current, neglect
ing transformer losses  23500/2200  10.7 amp.
Diagram B shows the method of comiecting a lowvoltage
singlephase load to a twophase line; the load should be divided
as equally as possible between the two phases.
Diagram C shows the Scott connection used to transform
from twophase to three phase. The two transfonuers x and y
are Wound for the same primary voltage, namely that of the
twophase line, but the secondary voltage of y is only V3/2 or
Fio. 320. — Connection of transformers to a twophase lina.
0.86 times that of x. One end of the secondary of y is connected.
to the middle point of x.
If the secondary voltage of x is £ volta then that of y is V3E/2
volts and, in diagram D, Fig, 320,
the difference of potential between a and b = E volts
the difference of potential between a and c = VCoo)' + (oc)*
= E volts
the difference of potential between b and c = E volts
and the phase relations between these voltages is the same as in
a deltaconnected bank of transformers,
A load which is balanced on the threephase side is also balanced os the two
phase side. Fig. 321 shows the currents and voltages in a Scott ooDnected
Art. 303]
TRANSFORMER CONNECTIONS
276
group which is used to transform from twophase at E volts to threephase
at E volts.
Since the threephase load is balanced, therefore /i,/s and 7s are all equal,
and the kv.a. output on the threephase side = 1.73^/.
Now the magnetizing effect of the primary of a transformer is always
equal to the magnetizing effect of the secondary, so that in transformer y
and
therefore
niy X ly  nty X I
rii,
E
ly^IX \/3/2
In transformer x, the current in oa = 7 and that in oh also » 7, but these
currents are out of phase by 120 degrees and are in opposite directions there
/i/
TT^
a
nsy Tama
woowoww<r<N^ >
niy Tucal'
/•I
>^
E
,?
Fig. 321. — Voltage and current relations in a Scottconnected bank of
transformers.
fore the resultant magnetizing effect is not equal to n2x7 but to nsx7 X V^/2
see diagram B
now
ni£
and Hulx
therefore Ix
E ^
= nul X \/3/2
= 7 X \/3/2 = 7,
and the total twophase kv.a. =» 2^7, = V3B7, the same as on the three
phase side.
If a 50h.p., 440volt, threephase motor is supplied from a 2200 volt,
twophase line, find the current 7, Fig. 321, and also the currents 7« and ly,
if the efficiency of the motor is 90 per cent, and the power factor is 88 per
oent.
276 PRINCIPLES OF SLSCTRICAL ENGINEERING [Chai. xxja
motor output " 50 h.p.
 50 X 0.746  37.2 kw.
motor input  37.2/0.9  41.4 kw.
 41.4/0.88  47 kv.ft.
1.73 X 440 X /^
1000 ^■*
and /  47,000/1.73 X 440  62 amp.
neglecting transformer loaaea, 2 X 2200 X /. = 47 kv.s.,
and h  47,000/2 X 2200  10.7 amp.
301. Connecttoiis to a Threephase Line. — In Fig. 332:
Diagram A shows three transformers connected Y on both
primary and secondary sides. If Ei is the voltage between the
Fia. 322. — Connection of transformers to a threephase Ime.
primary lines and Et is that between the secondary lines then
The primary voltage of each transformer = EifJZ, see page 236.
The secondary voltage of each transformer = Eifyl^.
The transformation ratio of each transformer = Et/Eg.
Diagram B shows the transformers connected delta on both
primary and secondary sides :
The primary voltage of each transformer = Ei, see page 237.
The secondary voltage of each transformer = Ei.
The transformation ratio of each transformer = EJEt.
f)iagram C shows the transformers connected Y on the prim
ary side and delta on the secondary aide:
The primary voltage of each transformer = J?i/V3.
The secondary voltage of each transformer = ffj.
Abt. 304]
TRANSFORMER CONNECTIONS
277
The transformation ratio of each transformer = Ei/ylSEi*
In a delta connection, the voltage of any one phase at any
instant is equal and opposite to the sum of the voltages in the
other two phases, see page 233, so that if, in Fig. 323, one trans
former A of a bank of deltaconnected transformers is discon
nected, the difference of potential between a and b is unchanged,
being maintained by the transformers B and C in series, so that
threephase power can still be obtained from the lines a, b and c.
This connection is called the V or open delta connection and is
shown in diagram D, Fig. 322.
Primary Secondary
Delta connection.
J t 6
b li
Primary Secondary
Opendelta connection.
Fig. 323. — Comparison between the closeddelta and the opendelta
connections.
In the open delta connection in Fig. 323.
The current in each transformer = Ii,
The voltage of each transformer = Et
Eth
kv.a.
The rating of each transformer =  ^^^
But the total load on the threephase line =
l.lSEJi
1000
kv.a..
page 237, so that the rating of each transformer is greater than
half the total load by the ratio 2/1.73 or by 15 per cent.
If 30 kv.a. has to be transmitted, then three deltaconnected transformers
may be used each with a capacity of 10 kv.a.
278 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
If one of these transformers is burnt out, then the power that can be trans
mitted by the remaining two transformers is not 20 kv.a., but
 20 X V3/2  17.3 kv.a.
306. Advantages and Disadvantages of the T and Delta Con
nection. — For a given voltage of 110,000 volts between lines,
each tran^ormer in a Yconnected bank has to withstand
110,000/V3 or 64,000 volts and is therefore less liable to break
down than the transformers in a deltaconnected bank which
have to withstand the full 110,000 volts.
When one transformer in a Yconnected bank breaks down, the
system can no longer operate threephase, whereas, if the trans
formers were connected delta, the faulty transformer could be
u
26 Amp.
15
Amp.' '
15
Amp
no.,
n
65
'Amp.
(jaJUNJuttJ
JL' Connection Dlacram
Fio. 324.
B Vector Diagnm
disconnected and about 58 per cent, of the load could still be
carried by the two remaining transformers connected open delta.
Deltaconnected transformers are used for practically all low
voltage distribution. The Y to delta connection is often used
for high voltage work, the transformers being Y connected on the
highvoltage side.
The load on a 2200yolt threephase line consists of 1800, 1/2 amp., 110
volt lamps on three circuits, and 200 h.p. of tbreephase motors with an
average power factor of 80 per cent, and an average efficiency of 88 per cent,
all on one circuit. Draw the diagram of connections, specify the trans
formers, and find the current in the mains and also the resultant power
factor. The connection diagram is shown in Fig. 324.
Art. 307] TRANSFORMER CONNECTIONS 279
The kv.a. output of each motor transformer
= 1/3 (motor kv.a.)
= 1/3(200 X 0.746 X g^ X ^g)
= 70 kv.a.
4.x. V * ^ *v 4. 3 X 70 X 1000
the une current for the motors = i 7q y 2200 ~ amp.
the kv.a. output of each lighting transformer
600 X 0.5 X 110
1000
= 33 kv.a.
33 000
the primary current in each transformer = o^vq = 15 amp., but these
transformers form a delta connected load on the Ime, see Fig. 269, page 237,
therefore the current in each line « 1.73 X 15 = 26 amp.
the resultant current in the line is the resultant of 26 amp. at 100 per cent
power factor and of 55 amp. at 80 per cent, power factor and is equal tc^
/ = Va6« + be*, diagram B
m
'^ V(0.8 X 55 + 26)« + (0.6 X 55)«
= 77.5 amp
. _, ah 0.8 X 55 + 26
the power factor « — = ===
= 90.5 per cent.
806. Types of Transformer. — The transformer shown in Fig.
325 is said to be of the core type. If the coil on limb B is placed
on A, and the iron of B is split and bent over as shown in Fig. 326,
the resulting transformer is said to be of the shell type.
For threephase transmission there is a considerable saving in
cost and floor space if, instead of a bank of three transformers
each in its own tank, a threephase transformer is used in which
the windings of the three phases are all placed on the same core
as shown in Figs. 327 and 328. The principal objection to the
threephase transformer is that a breakdown in the winding of
one phase puts the whole transformer out of commission, so that
such transformers are used only in large central stations where
there is ample reserve capacity.
307. The Autotransformer. — If a reactance coil is placed across
an alternatingcurrent line as shown in Fig. 329, the voltage E%
obtained by tapping the coil as shown may have any value less
than Ely and, as in the transformer, et = — = t". Since the
Jli2 712 Jl
280 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxv
primary and secondary currents oppose one another, the current
in the section ab is the difference between the currents /i and /j.
Iron Core
Ooil
A
T_
OotI
B
\r.
vr=yv
Fig. 325. — CJore type of trans
former.
/ \
V y
Fig. 326.— Shell type of trans
former.
£
Core I
s&
Primary
unji
I I il Ijl H II Secondary
Fig. 327. — Threephase core
type of transformer.
Core
I
I
IZI
i
I
ca
m
m
m
■nriiii J
Primary
Seccoidary
Fig. 328. — Threephase shell
type of transformer.
J"
t=a Jt
r* = **« Tumi
^^ b
/«/]
4^9
Fig. 329. — Autotransformer.
When the transformation ratio is comparatively small, the
autotransformer, as this reactance coil is called, is cheaper than
Am. 3081 TRANSFORMER CONNECTIONS 281
the equivalent transformer with separate windings. It is much
used for reducing the voltage applied to alternatingcurrent
motors during the starting period, see page 301.
308. Boosting Transfonners and Feeder Regulators. — The
voltage of a line may be raised by a small amount if a standard
transformer is connected as shown in Fig. 330. The voltage of
the line is boosted by the amount of the secondary voltage Ei.
Fig. 330. — Boosting transfoimer.
When the secondary side of the transformer is tapped so that
the boosting effect can be adjusted, the resulting piece of appara
tus is the Stillwell feeder regulator. It is used to raise the feeder
voltage above that of the power house, so as to compensate for
the voltage drop in the feeder and maintain the voltage at the
load.
Another type of feeder regulator is shown in Fig, 331. The
Fig. 331. — Induction regulator.
secondary coil in this case is movable relative to the primary.
When the coils are in the relative position shown in diagram A,
the voltage Ei has its maximum value; when the secondary coil
is moved through 90 degrees relative to the primary, as shown in
diagram B, then the flux ^ which threads the secondary coil is
zero, and no voltage is induced in that coil. Such regulators may
be made automatic if the core is turned by means of a small motor
282 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxv
which is controlled by a solenoid connected across the line to be
regulated. When the line voltage increases, the plunger of the
solenoid is raised and closes the motor circuit, and the motor
then turns the regulator in such a direction as to lower the volt
age. When the line voltage decreases, the plunger of the solenoid
drops and reverses the motor, so that it now turns the regulator
in the opposite direction.
CHAPTER XXXVI
POLYPHASE INDUCTION MOTORS
309. llje induction motor is used for practic^ly all the power
work when only alternating current is available. The essential
Fio. 332. — SquirrelcEtge induction motor.
parts of such a machine are shown in Fig. 332. The stator or
stationary part is exactly the same as that of an alternator, the
Fio. 333. — Squirrelcage rotor.
rotor, however, is entirely different and the type most generally
used, called the squirrelcage type, consists of a cylindrical core
284 PRINCIPLES OF ELECTRICAL ENGINBERINO IChap.xxxv?
which carries a large number of copper bars on its periphery
which bars are all joined together at the ends by copper or braas
end connectors as shown in F^. 333. The action of this machine
will be explained in detail. >
310. The Revolving Field.— Digram P, Fig. 334, shows the
essential parts of a twopole, twophase induction motor. The
stator carries two windings M and N which are spaced 90 elec
trical degrees apart and, in the actual machine, are bent back so
that the rotor may readily be inserted. These windings are
Fia. 334. — Revolving field of a twopole, twophase induction motor.
connected by wires to twophase mains and the currents which
flow at any instant in the coils M and N are given by the curves
in diagram Q; at instant A for example the current in phase 1 =
f /m while that in phase 2 is zero.
The windings of each phase are marked S and F at the terminals
and these letters stand for start and finish respectively; a +
current is one that goes in at S and a — current one that goes in
&tF.
The resultant m^netic field produced by the windings M and
A"" at instants A, B, C and D is shown in diagram B from which it
may be seen that, although the windings are stationary, a revolv
ing field is produced which is of constant strength and which
Art. 311]
POLYPHASE INDUCTION MOTORS
285
goes through one revolution while the current in one phas
through one cycle.
311. The Revolving Field of a Threephase Motor. —
P, Fig. 335, shows the winding for a twopole threephase motor;
M, N and Q, the windings of the three phases, are spaced 120
electrical degrees apart. These windings are connected to the
threephase mains and may be connected either Y or delta, see
page 237, in either case the currents which flow at any instant in
il Im IiVilm Jl.H'm 'lIv,
Jl— Him IiH'm h Im t2HIm
Ji—Hlm li'Im S Ii—m„ la'Hlm
Fia. 335. — Revolving field of a twopole, threephase induction motor.
the coils M, N and Q are given by the curves in diagram R; at
instant A for example the current in phase 1 = +/„ that in phase
2 = ^ and that in phase 3 is also = ^■
The resultant magnetic field produced by the windings at
instants A, B, C and D is shown in diagram S from which it may
. be seen that, just as in the case of the twophase machine, a
revolving field is produced which is of constant strength and
which goes through one revolution while the current in one phase
passes through one cycle.
286 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvi
Fia. 336. — Revolving field of an eightpole, twophase induction motor.
Art. 313]
POLYPHASE INDUCTION MOTORS
287
312. Multipolar Machines. — Fig. 336 shows part of the wind
ing of an eightpole, twophase induction motor and also the re
sultant magnetic field at the instants A and B, Fig. 334. The
field moves through the angle 6, or through half the distance
between two adjacent poles or «" of a revolution, in the time
interval between the instants A and B or in 7> seconds, therefore
the speed of the revolving field
= 2" X 4/ rev. per sec.
120/
= rev per. mm.
This is called the synchronous speedy and is the speed at which
an alternator with the same number of poles must be run in order
to give the same frequency as that applied to the motor, the table
on page 195 therefore applies to induction motors as well as to
alternators.
313. The Starting Torque.— Let the revolving field of a two
pole machine, produced by either a two or a threephase stator,
Fig. 337. — Direction Fig. 338. — Direction Fig. 339. — Direction
of the e.m.f. in the of the currents in the of the currents in the
rotor bars. bars of a low resistance bars of a high resistance
rotor. rotor.
be represented by a revolving north and south pole as shown in
Fig. 337. The field is moving in the direction of the arrow and
therefore cuts the stationary rotor bars and generates in them
e.m.fs. which are shown by crosses and dots. The e.m.f. will be
a maximum in the conductor which is in the strongest field and the
direction of the e.m.f. in each conductor may be found by the right
hand rule, see page 192. Since the rotor winding forms a closed
288 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxyi
circuit, the generated e.m.fs. will cause currents to flow in. the
rotor bars.
The frequency of the e.m.fs. generated in the stationary rotor
bars by the revolving field = i ^/T^ — f s®® P^g® l^^? which
is the same as the frequency of the e.m.f . applied to the stator,
/ X 120
since the syn. speed = , see page 287, so that the fre
quency of the rotor currents is high, and the reactance of
the rotor, which is proportional to the rotor frequency, is
large compared with the rotor resistance, the rotor current in
each bar therefore lags considerably behind the e.m.f. in that bar.
In conductor a for example, the e.m.f. has just reached its maxi
mum value, but the current in that conductor lags the e.m.f. by
an angle 6 which is almost 90 degrees and so does not become a
maximum until the poles have moved into the position shown in
Fig. 338, which is almost 90 degrees from the position shown in
Fig. 337.
Since the conductors in Fig. 338 are carrying current and are in
a magnetic field, they are acted on by forces, the direction of which
may be determined by the lefthand rule, from which it is found
that while the force in the belts be and de tends to make the rotor
follow the revolving field, that on the conductors in the belts cd
and be acts in the opposite direction. The former force is the
larger, so that the rotor tends to follow the revolving field.
The current in the rotor conductors at standstill
rotor voltage at standstill
rotor impedence at standstill
and the rotor impedence is large enough to limit the current to
about five times the fullload value. On account of the large
torque opposing the starting of the rotor, this large starting cur
rent produces an effective starting torque which seldom exceeds
one and a half times the fullload torque of the motor.
314. The Wound Rotor Motor. — The starting torque for a
given current can be increased if that current is brought more in
phase with the e.m.f. This may be seen from Fig. 339 in which
the current lags by such a small angle 6 that there is practi
cally no opposing torque.
The angle of lag in a circuit may be decreased by increasing the
resistance of the circuit without changing the value of the react
Art. 314]
POLYPHASE INDUCTION MOTORS
289
ance. In Fig. 340 for example, when the resistance R is in
creased, the vector diagram changes from A to B; the current /
is decreased and so also is the angle $. If then sufficient resist
nnnpH
A B
Fig. 340. — Effect of resistance on the magnitude and on the phase angle of
current in a series circuit.
ance is put into the rotor bars to reduce the starting current to
the fullload value, the angle of lag of the rotor currents will be so
small that the torque is all effective, and fullload torque is
.Botor Bar
Ci^
iiliiA^^
JEtotor
T/Broah
Slip Bing
Fig. 341. — Squirrelcage rotor with bars that have an adjustable resistance.
Fig. 342.— Wound rotor.
developed with fullload current. By using a lower resistance,
twice fullload torque may be obtained with about twice full
load current.
290 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvi
When a motor is running under load, a large rotor resistance is
undesirable because it causes large copper loss, low efficiency and
excessive heating, so that some method must be devised whereby
resistance can be inserted in the rotor conductors during the start
ing period and cut out during the running period.
This result could be attained by leaving the rotor bars open at
one end and connecting that end to a slip ring on the shaft,
as shown in Fig. 341, between which ring and the stationary
end connector C an adjustable resistance R could be inserted.
This construction would necessitate the use of as many slip
rings as there are rotor bars so that it is modified in practice to
reduce the number of slip rings.
The bars are connected so as to form a winding of three groups,
as shown diagrammatically in Fig. 342. One end of each group
is connected to the end connector Ci, while the free ends are
connected to slip rings and then through adjustable resistances
Ri to the other end connector C, which is now merely a short
piece of wire connecting the resistances together. The resistances
Ri are gradually cut out as the motor comes up to speed, and are
finally shortcircuited. This type of motor is called the wound
rotor type of induction motor and is to be preferred to the squirrel
cage type for heavy starting duty.
316. Running Conditions. — It was shown on page 288 that
the resultant starting torque is in such a direction as to make
the rotor follow up the revolving field. When the motor is not
carrying any load, the rotor will revolve at practically the speed
of the revolving field, that is at synchronous speed; it can never
run at a speed greater than that of the revolving field. If the
motor is then loaded, it will slow down and slip through the
revolving field, the rotor bars will cut lines of force, the e. m. fs.
generated in these bars will cause currents to flow in them and a
torque will be produced. The rotor will slow down until the point
is reached at which the torque developed by the rotor is sufficient
to overcome the retarding torque of the load.
^, .. syn. r.p.m. — r.p.m. of rotor . „ , ,,
The ratio is called the per cent.
syn. r.p.m.
slip, and is represented by the symbol s, its valuejit/iriWoadJs
generally about 4 per cent., that is, when the ^revol ving field
make one revolution, the rotor makes 0.96 of aTrevolufion and
the field moves relative to the rotor bars through 0.04 of a
revolution.
Abt. 316]
POLYPHASE INDUCTION MOTORS
291
When the rotor is at standstill, the rotor frequency is / cycles
per second, see page 288, where / is the frequency of the e.m.f .
applied to the stator, but at fullload the rotor frequency is only
sf cycles per second because then the velocity of the field relative
to the rotor is only s per cent, of the relative velocity at standstill.
As the load on the motor increases, the motor slows down and
the rotor slips more rapidly through the revolving field and causes
the rotor current to increase, but the frequency of the rotor cur
rent increases as well as its numerical value and this causes the
rotor reactance to increase and the current to lag. Now the
torque developed by the rotor tends to increase due to the increase
in the rotor current and to decrease due to the increase in the
£
B
600 800
1200 UOO
Starting ourrent » 160 amp. » 5 times full load current
Starting torque * 1.9 times full load torque
Maximum torque ■ 3.1 times full load torque
Fio. 343. — The relation between torque, current and speed in a 25 h.p.,
440 volt, Sphase, 60 cycle induction motor.
current lag, see page 288. Up to a certain point, called the break
down point or point of maximum torque, the effect of the current
is greater than that of the current lag, beyond that point the
effect of the current lag is the greater, so that, after the break
down point is passed, the torque actually decreases even although
the current is increasing. The relation between speed, torque and
current is shown in Fig. 343 for a 25h.p., 440 volt, threephase,
60cycle, 1200 syn. r.p.m induction motor.
316. Vector Diagrams for the Induction Motor. — a. Noload
conditions. Alternating e.m.fs. jBi, applied to the stator, cause
alternating currents h to flow in the stator windings and pro
duce the revolving field <t>, but no work is done if the noload
losses are neglected, so that the current flowing in each phase
must lag the voltage applied to that phase by 90 degrees, as
31
292 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvi
shown in Fig. 344. This no load current is kept as small as pos
sible by the use of a small air gap clearance between the stator
and rotor; a 60h.p., 900r.p.m. induction motor will have a rotor
diameter of 20 in. (51 cm.) and an air gap clearance of 0.03
in. (0.075 cm.) and, in spite of this small air gap, the magnetizing
current Jo will not be less than 30 per cent, of fullload current.
The revolving field, in addition to cutting the rotor conductors,
cuts also those of the stator and generates an e.m.f . Etb, called the
i k^i
A^i
Fig. 344.— Noload
vector diagram.
100
Fig. 345.— Ideal. Fig. 346.— Actual.
Fullload vector diagrams.
80
£ 60
O
04
o
O
40
20
>'
,_. ■— — '
—'""'
^x^
/
•c
a
o
■J
/
H Vz H X
Fig. 347. — The power factor of an induction motor.
Fall Load
back e.m.f., in each phase of the stator winding. Ev, is less
than El by the e.m.f. required to send the magnetizing current
/o through the impedence of the stator winding, which e.m.f. is
comparatively small. The revolving field is therefore propor
tional to the voltage Ei.
6. Fullload conditions. When the motor is loaded it slows
down and current passes through the rotor windings and develops
Art. 317] POLYPHASE INDUCTION MOTORS 293
the driving torque of the machine. This current, like that in the
secondary of a transformer, tends to demagnetize the machine or
to oppose the flux which produces it, but a reduction in the
flux causes the back e.m.f. Ev, to decrease and allow a larger
current to flow in the primary or stator winding, so that the
stator current always adjusts itself to suit the requirements of
the secondary or rotor.
The reduction in the value of is comparatively small, so that
at fullload, a magnetizing component of current h is still required
to produce the revolving field while a power component /«, in
phase with the applied voltage is required for the load, as shown in
Fig. 345. The power factor of an induction motor, namely, the
cos a, is therefore practically zero at noload and gradually
increases as the load increases, as shown in curve A, Fig. 347.
Due to the reactance of the stator winding, and to that of the
rotor winding which, as pointed out on page 291, increases with
the load, the current lags more than shown in Fig. 345 and the
power factor in an actual machine changes with the load as shown
in curve B.
317. Adjustable Speed Operation. — The squirrel cage motor is
essentially a constant speed machine. For adjustable speed
operation the wound rotor motor must be used. If such a motor
is operating with a load having a constant torque, and a resistance
is inserted in the rotor circuit, the rotor current will decrease and
the motor will not be able to develop the necessary torque and so
will slow down. As the speed drops however the rotor slips more
rapidly through the revolving field, so that a greater e.m.f. is
induced in the rotor conductors and the current increases, and,
at some lower speed, is again sufficient for the load.
The speed regulation however is very poor because, if the load
were decreased, less current would be required, and the motor
would automatically speed up so as to reduce the slip and
thereby reduce the rotor voltage and current. At noload,
the rotor current being then very small, the slip would be
practically zero and the motor would run at practically syn
chronous speed. This method of speed control is therefore
similar to that of a directcurrent shunt motor with resist
ance in the armature circuit and has the objection that the speed
regulation is poor and the efficiency is low.
Where good speed regulation is desired, the stator may be sup
plied with two separate windings, one of which is wound so as to
294 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, ixxvi
produce a revolving field with p poles and the other a revolving
field with pi poles and then, by using one or other of these wind
ings, the synchronous speed may be —  or . It is sel
dom possible on account of expense to put more than two such
windings on one stator.
318. Induction Generator. — Suppose that an electric ear driven
by an induction motor is operating on a road where there is a
long pull up followed by a lor^ coast down. When on the up
grade, the motor runs at about 4 per cent, less than synchronous
speed, at which speed the rotor slips fast enough through the
revolving field to cause fullload current to flow.
Starting on the down grade, the car begins to drive the motor
and the speed of the motor first becomes equal to the synchronous
speed, when the rotor current is zero, and then runs at a speed
which is greater than synchronous so that the rotor is again slip
Fia. 348. — The rotor of a self starting aynchronous motor.
ping through the revolving field. But since it is now running
faster than the field, the direction of motion of the conductors
relative to the field has been reversed, and the torque which was
a driving torque now becomes a retarding torque so that the
machine is now acting as a generator and deUverii^ power to the
mains. When the speed is about 4 per cent, above synchronous
speed, the machine will be delivering fuUload as a generator.
319. Self starting Synchronous Motors. — Polyphase syn
chronous motors have stators which are exactly the same as those
of induction motors, they may therefore be made self starting if a
squirrel cage winding is added to the rotor as shown in Fig, 348,
Art. 320] POLYPHASE INDUCTION MOTORS 295
If alternating e.m.fs. are applied to the stator of such a machine,
the field coils not being excited, the machine will start up as an
induction motor and will attain practically synchronous speed.
If the field coils are then excited, the machine will be pulled into
synchronism.
The singlephase synchronous motor cannot be made self
starting in this way because, as shown on page 308, the single
phase induction motor is not self starting.
320. Dampers for Synchronous Machines. — In addition to
making a sjrnchronous machine self starting, the squirrel cage
acts as a damper to prevent hunting, see page 259.
A machine which is hunting is moving at a speed which is
alternately slower and faster than the synchronous speed, and,
in each case, the squirrel cage rotor cuts lines of force. A driving
torque is produced as in an induction motor when the machine is
running below synchronous speed, and this tends to speed it up,
while a retarding torque is produced as in an induction generator
when the machine is running above synchronous speed, and this
tends to slow it down. When the machine is running at syn
chronous speed there is no current in the squirrel cage. The tor
que due to the squirrel cage therefore tends at all times to pre
vent oscillations in speed and therefore to prevent hunting.
CHAPTER XXXVII
INDUCTION MOTOR APPLICATIONS AND CONTROL
321. Choice of Type of Motor. — ^The synchronous motor is
the only motor whose power factor can be controlled. . Such a
machine may be overexcited and made to draw a leading current
from the line so as to raise the average power factor of the total
load connected to the line and thereby allow the use of a smaller
alternator and a smaller cross section of copper in the transmis
sion line than would be required for a load consisting entirely of
induction motors, see page 257.
The synchronous motor runs at a constant speed at all loads.
It suffers from two disadvantages, namely, that the starting
torque is small even if the motor is of the self starting type, while
direct current is necessary to excite the field magnets. Because
of these disadvantages it is used only in large sizes and only when
the starting torque required is small as, for example, for the
driving of large constant speed fans, pumps, and air compressors,
when the load can be relieved during the starting period; it is
also much used as the motor end of a motor generator set, see
page 318.
The singlephase machines described in the next chapter are
more expensive than polyphase machines of the same output and
are not used for general power work. They are used in compara
tively small sizes where polyphase current is not available.
The polyphase induction motor is used for practically all the
power work when only alternating current is available, care must
be taken however to use the proper type.
The squirrelcage induction motor takes a lagging current and
has a fullload power factor of about 80 per cent, for a 1 h.p.
motor and 90 per cent, for a 100 h.p. machine.
It is available only with the following speeds
296
Art. 324] INDUCTION MOTOR APPLICATIONS 297
)les
On 25 cycle
On 60 cycle
mains
mains
2
1500 r.p.m.
3600 r.p.m.
4
750 r.p.m.
1800 r.p.m.
6
500 r.p.m.
1200 r.p.m.
8
375 r.p.m.
900 r.p.m.
300 r.p.m.
720 r.p.m.
P
120 X 25 r.p.m.
120 X 60 r.p.m
The fullload speed is about 4 per cent, lower than the above.
The principal objection to the squirrelcage motor is that it takes
a large starting current to develop a moderate starting torque.
About 5 times fullload currwit is required to give 1.5 times full
load torque or, with reduced voltage, see page 302, 3.3 times full
load current for fullload torque. The motor is therefore not
suitable for heavy starting duty.
It is very rugged, has no sliding contacts, and is the most
satisfactory constant speed motor for loads that do not require a
large starting torque.
The wound rotor induction motor has the same running char^
acteristics as the squirrelcage motor, but has better starting
characteristics and will develop fullload torque at starting with
f uUrload current in the line.
The service for which each of these induction motors is suited
can best be illustrated by a few typical examples.
322. A line shaft should run at practically constant speed at all
loads and may be driven by a squirrelcage motor if there are not
many countershaft belts, or if loose pulleys are used to take the
load off the motor during the starting period. If the starting
torque required exceeds fullload torque then a wound rotor type
of motor must be used.
323. Woodworking machinery such as planers and saws are
driven by squirrelcage motors, except in the case of certain types
of heavy planing mills which have a large inertia and require the
large starting torque developed by the wound rotor motor.
324. Cement Mills. — The squirrelcage motor is used for the
driving of cement machinery because it has no sliding contacts;
the commutator of a directcurrent motor or the slip rings of a
wound rotor motor would wear very rapidly in a cement mill.
To obtain the high starting torque required for some of the
machines, a large resistance must be inserted in the rotor circuit,
see page 289. In the case of the squirrel cage motor this ia done
298 PRINCIPLES OF ELECTRICAL BNQINBERINO IChap.xxxvu
by making the rotor end connectors of high resistance metal.
The resistance in this case remains permanently in the circuit
and causes the efficiency at fullload to be about 5 per cent, lower
than it would be with a wound rotor motor, since this latter mar
chine is so constructed that the resistance can be cut out when the
motor is up to speed.
325, Motors for Textile Machinery. — The qutdity of the
finished product of a loom depends largely on the constancy
of the motor speed and for such service induction motors are
preferred to directcurrent motors because their speed depends
only on the frequency of the power supply and is not affected by
variations of the applied voltage.
326. Adjustable Speed Motors. — A wound rotor induction
motor with the rotor shortcircuited drops about 4 per cent, in
1„
^
^
5s
=V
G3
\
'•\
^.
^
1
\
H
P
'»
\*
2
\
'E 1
OO u
s uo
Fio. 349.
Effect of rotor tesistance c
the speed of i
Speed between noload and full load, as shown in curve a, Fig. 350.
When resistance is inserted in the rotor circuit, as shown in
Fig. 349, the motor will drop in speed so as to generate the larger
e.m.f. required to send the current through this higher resistance.
At noload however the rotor will run at practically synchronous
speed because only a small slip is required to generate enough
e.m.f. to send the noload current through the rotor. Curves
ft and c, Fig. 350, are the speed curves with different values of
rotor resistance; the larger the drop in speed required, the larger
the rotor resistance must be, and the greater the loss in this
resistance.
327. Crane motors should be able to develop a large starting
torque without taking an excessive current from the line; they
Abt. 239] INDUCTION MOTOR APPLICATIONS 299
should run at a slow speed when the load to be lifted is heavy
and at a high speed when the load is light.
Large starting torque may be obtained by inserting resistance
fai the rotor circuit. By this means fullload torque may be
obtained at starting with fullload current in the line, and twice
fullload starting torque with about twice fullload current, see
page 289.
A drooping speed characteristic such as that in curve c, Fig. 350,
is also obtained by inserting resistance in the rotor circuit.
Crane motors in small sizes are generally of the squirrel cage
type, the high rotor resistance being obtained by making the rotor
end connectors of high resistance metal. This gives a very simple
typ>e of machine but, because of the large rotor resistance and
therefore the large rotor resistance loss, the machine heats up
quickly. For outputs greater than 20 h.p. it is therefore advisable
to use the wound rotor type of machine in which case the
resistance is outside of the machine and can be varied as desired.
Crane operation by means of induction motors is not so efficient
as the operation by directcurrent series motors, since these latter
machines have the desired characteristics without the use of
additional resistance, see page 103.
328. Shears and punch presses are generally supplied with a
flywheel to carry the peak load. In order that the flywheel
may be effective, the speed of the motor must drop as the load
comes on. A standard machine drops in speed about 4 per cent,
between noload and fullload. For a larger drop in speed,
resistance must be connected permanently in the rotor circuit
so as to give a characteristic such as that shown in curve 6, Fig.
350. The motor may be a squirrelcage machine with high
resistance end connectors, but for large presses it will often be
necessary to use a wound rotor motor to obtain sufficient starting
torque to accelerate the flywheel on first starting up.
329. For adjustable speed service such as the driving of lathes
and other machine tools, the only alternatingcurrent motor at
present available is the wound rotor induction motor.
The speed of such a machine may be lowered for a given torque
by inserting resistance in the rotor circuit as shown in Fig. 349,
and the speed characteristics with different values of resistance
are shown in Fig. 350.
Because of the use of this resistance, there is a large resistance
loss which increases as the speed is decreased, so that the efficiency
300 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvii
of an induction motor at reduced speed is low. Moreover the
speed regulation is poor, as shown by curves b and c, Fig. 350.
If for example an induction motor operating at reduced speed is
driving a lathe in which a forging such as that in Fig. 129, page 108,
is being turned, then the motor will slow down when the cut is
deep and will speed up when the cut is light, so that the speed
will be very irregular.
For machine tool driving then the alternatingcurrent motor
is not such a suitable machine as the directcurrent shunt motor
controlled by field resistance, see Arts 126 and 127, page 107,
and it has been pointed out that for crane service the direct
current motor is superior to the alternatingcurrent motor, so
that, when the bulk of the load consists of adjustable speed or
crane motors, direct current should be supplied even if it is neces
sary to use a motor generator set, see page 318, to change from
alternating to direct current.
330* Resistance for Adjustable Speed Motors. — As in the case
of the directcurrent motor, the number of ohms in the controlling
rheostat and the current carrying capacity of the resistors de
pends on the service for which the motor has to be used.
If a fan and a reciprocating pump take 10 h.p. at full
speed, then at half speed the pump takes 5 h.p. since the
torque is constant, while the fan takes Vl0=2.15 h.p., see
page 111, the rotor current of the fan motor will therefore
be less at half speed than that of the pump motor. The voltage
generated in the rotor by the revolving field, however, will be the
same in each case, so that the number of ohms in the fan rheostat
must l;e greater than the number in the pump rheostat so as to
reduce the current to the lower value.
STARTERS AND CONTROLLERS
331. Switches for Alternatingcurrent Circuits. — It causes
less arcing to open a circuit in which alternating current is flowing
than to open one in which directcurrent is flowing, because, in
the former case, the current passes through zero twice every cycle..
The quick break type of switch, see page 114, is used for small
currents. For larger currents or in highvoltage circuits the
contacts are immersed in insulating oil which quenches any arc
that Torms. The oil switch in alternatingcurrent circuits takes
Art. 332]
INDUCTION MOTOR APPLICATIONS
301
the place of the magnetic blowout switches used in directcurrent
circuits.
332. Starting of Squirrelcage Induction Motors. — A squirrel
cage motor at standstill takes about 5 times fullload current
from the line with normal applied voltage, and develops about
1.5 times fullload torque.
Ranning Position
Starting Positio'n
;ioEtNl LJ L ,
starting Faiei
Bunnlng FaieB
Fig. 351. — Connections for a small threephase, squirrelcage induction
motor.
Motors with an output of 5 h.p. or less are generally
connected directly to the line without a starter, by means of a
doublethrow switch such as that shown diagrammatically in
Fig. 351, This switch is so constructed that it will stay in the
starting position only when held there against the tension of a
spring.
«—E\
pX^
Aato Transformers
Fig. 352. — Connections of a starting compensator for a threephase,
squirrelcage induction motor.
Motors with an output which is greater than 5 h.p. are
generally started up on reduced voltage by means of autotrans
formers connected as shown in Fig. 352. By this means the start
ing current is reduced, but the starting torque also is reduced;
the revolving field <i> is proportional to the applied voltage E which
produces it, see page 292, while the rotor current I2 is proportional
to the revolving field by which it is produced, so that the start
302 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvn
ing torque, being proportional to ^ X /i is proportional to <f>^
and therefore to E^.
A motor at standstill takes 5 times fullload current with normal applied
voltage and develops 1.5 times fullload torque. What must the applied volt
age be to obtain fullload torque and what will be the starting currents in the
motor winding and in the line?
(Et)* fullload torque 1
{El)* 1.5 fullload torque 1.5
therefore Et
J— X El
\1.6
» 0.815^1
or 81.5 per cent, of normal voltage
Et
The starting current in the motor » 5 (fulMoad current) —
El
B 4.1 (fullload current)
 /«, Fig. 352.
Et
The starting current in the Ime = 4.1 (fullload current) —
El
a 3.3 (full load current)
= /i, Fig. 352.
333. Starting Compensator. — The combined autotransformers
and switches constitute what is called a starting compensator.
One type is shown in Fig. 353 and consists essentially of three
autotransformers T and a doublethrow switch S by means of
which the motor is connected to lowvoltage taps for starting and
is then connected directly to the line when nearly up to full speed.
The complete connections of this compensator are shown in
diagram A, Fig. 354.
Diagram B shows the connections during the starting period;
normal voltage Ei is applied to the lines a, b, and c while a reduced
voltage Ez is tapped off from the autotransformers and is applied
to the motor.
Diagram C shows the connections during the running period.
The voltage applied to the motor is now normal and the overload
relays are connected in the circuit while the no voltage release
coil M is connected across one leg of the circuit.
The no voltage release feature is similar in principle to that
used on starters for directcurrent motors, see page 120, and
consists of a latching solenoid which holds the starting arm against
the tension of a spring. When the line voltage fails, the solenoid
INDUCTION MOTOR APPLICATIONS
Fio. 353. — Starting compensator for a threephase induction motor.
Complete connections Starting Runoing
FiQ 364, — Diagram of connections of a threephase starting compensator.
304 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvii
M is deenergized and the starting handle then returns to the
off position.
In the case of a heavy overload, the plungers of the overload
relays are raised and open the circuit of the no voltage release
M, the starting handle then returns to the off position.
The no voltage release is not supplied in all cases because an
induction motor can carry a starting current of 5 times fullload
current without injury during the short interval of time taken to
attain full speed. As a rule, the only objection to this large
current is the disturbance it causes due to the opening of circuit
breakers.
334. The stardelta method of starting is sometimes used for
threephase motors. The windings of the three phases are kept
' Ig tA Standstill
V7
at Standstill
i/o" ~S^ at Standstill
il Starting Connection BBunntng Oonnectioa
Fig. 355. — Voltage and current relations in a Ydelta starter.
separate from one another and six leads are brought out from the
machine. Under normal running conditions the windings are
delta connected as shown in diagram B, Fig. 355, and the voltage
per phase is Ei volts. During the starting period the windings
are Yconnected as shown in diagram A in which case the voltage
per phase is equal to Bi/1.73 or 58 per cent, of normal voltage.
If a delta connected motor at standstill takes 5 times fullload current with
normal applied voltage and develops 1.5 times fullload torque, what is the
starting current in the motor and also in the line, if the motor is Yconnected,
and what is the starting torque under these conditions ? (The student should
make a diagram of connections showing the doublethrow switch required to
change from Y to delta.)
The starting torque = 1.5 (fullload torque) X (0.58)*
•s 0.5 (fullload torque)
Am. 3351 INDUCTION MOTOB APPLICATIONS 305
The starting currmt in the line when delta connected — I,
The starting current in the motor when delta connected = l,l\/%
The Htarting eiUT«nt in the motor when Yconnected = (/./V3) X 0.58
= /,/3
'n>e starting cunent in the line when Yconnected = /./3
Now 7. is 5 times fnllloftd current, 80 that, when the motor is Yconneeted,
the starting current id the line is 1/3 X S or 1.67 times fuUload current.
The starting torque, however, is only 0.5 times fulllftad torque and, if tlus
is not sufficient to start the motor with the load, then a starting compensator
will be required
336. Starter for a Wound Rotfa* Motor. — To start up a motor
of this type the main switch iS, P^. 349, ia closed and then the
Fia. 356. — Siding contact type of starter for a. woundrotor induction motor.
resistance U in the rotor circuit is gradually cut out as the motor
comes up to speed.
Since the rotor is wound in three sections, see page 290, three
sets of contacts are required which for small motors are generally
mounted on a faceplate as shown in Fig. 356. This starter may
be used as a speed regulator if the resistance has sufficient
current carrying capacity.
For motors larger than about 50 h.p., the multiple switch
type of starter, see page 119, is generally preferred. Such a
starter for the motor shown diagrammatically in Fig. 357 would
consist of three doub!epole switches so interlocked that they can
be closed only in the order A, B, C. The switches are held closed
306 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxxvh
by a latch which is released by the oovoltage release magnet when
the voltage fails.
336. Automatic Starters. — Squirrelcage motors of small out
put are thrown directly on the line and the self starter for such a
motor consists of a single doublepole magnetic switch such as
that shown diagrammatically at A, Fig. 357.
If a solenoid were attached to the handle of a startii^ rheostat
such as that in F^. 356, then a sliding contact type of self starter
FiQ. 367. — Wound rotor type of motor.
Fia. 359.
FiO. 358. — Automatic starter for a woundrotor ii
duction motor
would be produced similar to the directcurrent starter shown
in Fig. 154, p^e 130. If in addition the main starter switch is
mt^netically operated then the motor may be started from a
distance by a control circuit such as that shown in Fig. 156, page
131, this circuit being connected across one of the phases and the
magnets being laminated' so as to be suitable for operation with
alternating currents.
'e of an alternating current magnet must be laminated for the same
. transformer core is laminated, see page 269, namely, to prevent
ore loss due to the alternating magnetic flux.
•Thea
Abt. 336.] INDUCTION MOTOR APPLICATIONS 307
For large motors, the multiple switch type of starter is used,
ty means of which the switches A, B and C, Fig. 357 are closed
in their proper order and the motor thereby brought up to speed
without the starting current exceeding a predetermined value.
The principle of operation is the same as for the directcurrent
starter described in Art. 158, page 133, although the mechanical
details are different.
Fig. 358 shows the control circuit for the three switches A, B
£knd C in Fig. 357. When the control switch s is closed, the line
& is excited as far as point c, and the magnet X, connected between
points 1 and 2, closes the double contactor switch A. The
motor then starts up with all the rotor resistance R, Fig. 357,
in the circuit, while about one and a half times fullload current
flows in the lines.
When switch A closes, the disc d drops, closes the contacts e and
f, and extends the excited part of line 6 as far as g. The coil t is
now excited from the points 3 and 4 and its plunger is lifted,
thereby tilting the lever j to which it is attached and removing
the support of the plunger of coil u. The line current passing
around u however holds up this plunger until the motor has
attained about onethird of normal speed and the current has
decreased to about fullload value when the plunger of u drops
and closes the contacts h and k. The coU Y is now excited from
the points 5 and 6 and closes the doublepole switch B thereby
cutting out the first step of the resistance from each leg of the
rheostat. The current then mcreases to about 1.5 times fulHoaa
current but gradually decreases as the motor speeds ^P^^^^®^'
As soon as switch B closes, the interlock discs attached to tha^
switch are lowered, and the contact Im is closed, so that coil Y i
now excited from the points 5 and 7 and the switch ^^f^**^®^®;^ .
held closed independently of the plunger of coU u. /^® J!^^^ ,
np is closed at the same instant by the lower interlock ^^^^c aaa
the excited part of line 6 is thereby extended as far as g, aa
same series of operations is repeated before switch ^^ j and t;
Fig. 359 shows the complete starter, with ^^^^^^^^^^^^ ^^^er
on the top panel and the switches A, B and C on
panels.
22
CHAPTER XXXVIII
SINGLEPHASE MOTORS
337. Single phase Induction Motors. — If one of the phases of a
twophase induction motor is opened while the motor is running,
the machine will continue to rotate and carry the load.
A twopole singlephase induction
motor at standstill is shown diagram
matically in Fig. 360. When an al
ternating current flows in the winding
Af an alternating flux 4> is produced.
Since this magnetic field is not rotat
ing, there is no tendency for the rotor
to turn, so that the machine is not self
starting. Yiq. 360.— Diagrammatic
338. Splitphase Method of Starting, repreaentation of a single
^  . , xi J J X phase mduction motor.
— One of the many methods used to
obtain a rotating field from a singlephase supply is shown dia
grammatically in Fig. 361. The motor is wound as for two
phase operation, and a resistance R is inserted in series with
the winding A. The current la therefore does not lag as much
as the current h, and the resultant magnetic field of the motor
has a rotating component which will cause the rotor to turn.
If, in addition to the resistance «, a condenser C is inserted in
series with the winding A, then the current la may be made to
lead the voltage E. The currents la and h will then be more
nearly 90 degrees out of phase, and the operation of the motor
will be more nearly that of a twophase machine.
This method of starting is called the splitphase method.
339. Running Torque of a Singlephase Motor. — If, as in
Fig. 363, two equal vectors P rotate with the same velocity but in
opposite directions, the resultant vector R alternates between the
values iZ = 2P and R = — 2P, and always lies in the line joining
these two points.
An alternating magnetic flux <t>, such as that in Fig. 362, is
therefore equivalent to two rotating fields 4>x and 4>y which are of
308
I
Art. 339]
SINGLEPHASE MOTORS
309
equal strength and rotate in opposite directions with the same
velocity. These fields tend to start the rotor in opposite direc
tions, so that the resultant starting torque is zero.
E
Resistance in series with one Resistance and condenser in
winding series with one winding
Fig. 361. — Splitphase connections for a singlephase induction motor.
Fios. 362 and 363.
FiG. 362. Fig. 363
Resolution of an alternating field into two revolving
components.
Suppose that the rotor has been started by phasesplitting and
is running in the direction of the field <t>x with a speed which is
little less than synchronous speed. The rotor bars will then be
cutting the field <l>xf and an e.m.f . will be induced which will send
310 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxviii
through these bare a current I^ of frequency sf, see page 291,
where s is the per cent, slip and / is the applied frequency. Since
this rotor frequency is low, the current lags by a very small angle,
and the torque due to this current is large, see page 288.
The rotor bare, however, are also cutting the field ^, and,
since this field is moving in a direction opposite to that of the
rotor, the frequency of the resulting rotor current /» is almost
equal to 2f. This frequency is high, and the current 7, lags by
a lai^e ai^le, so that the torque due to this current is negligible,
see p^e 288.
The resultant rotor cuiTent is the resultant of /. and /„, but
the latter current is not effective in producing torque, and the
machine runs as if it had only one field 0.. The characteristics
with respect to slip, efficiency and power factor are therefore
similar to those of a polyphase induction motor.
340. Sin^ephase Series Motor. — This machine is wound and
connected like a directcurrent series motor, but a few structural
changes are necessary to make a machine which will operate with
alternating current.
In Fig. 3&4, the current is flowing through such a machine in
one direction, while in Fig. 365, half a cycle later, the current
is flowing in the opposite direction in both armature and field
coils, but the direction of the torque is unchanged.
The characteristics of this machine are similar to those of the
directcurreot series motor. The torque is approximately pro
portional to the square of the current, since torque = K^I,
where the flux * b proportional to the current /. The torque,
however, is pulsating, being a maiumum when the current is a
maximum and zero when the current is zero, but the average
Abt. 340]
SINGLEPHASE MOTORS
311
torque is the same as would be produced by a direct current of
the same magnitude.
Since the magnetic flux in the poles is alternating, the field
structure must be laminated to keep the eddy current loss small.
Due to this alternating flux, a voltage E/ of self induction is
induced in the field coils, to overcome which the applied voltage
must have a component E/, Fig. 366.
^ ^a HotorCarr«iit
Fio. 366. — ^Vector diagram for a single phase series motor.
Since the armature is rotating in a magnetic field, a back e.m.f.
is generated in the armature in the same way as in a directcurrent
motor, see page 79. This e.m.f., however, is alternating since
it is produced by the cutting of an alternating flux <t>. It is a
maximum when the flux is a maximum and zero when the flux is
zero, so that, to overcome this back e.m.f., the applied voltage
100
""
\
. ^
:^
' /
•
2*
\
"^
/
e
H
1
\
X
f
00
/ '
^
^
^
^
80
60
40
ao
8
o
I
o
f^
26 60 75 100 126
Per Cent oi Fall Load Carrent
FiQ. 367. — Characteristics of a singlephase series motor.
must have a component Ea in phase with the flux and therefore
in phase with the current I.
In Fig. 366, then
is the alternating magnetic flux
/ is the current in the machine
Ea is the component of voltage to overcome the back e.m.f.
312 PRINCIPLES OF ELECTRICAL BNGINEERINO [Chap, xxxvui
Ef is the compooent of e.m.f. to send the current / through the
self induction of the field coila
E, the reaultant of E^ and E/, is the applied voltage
cos a is the power factor of the motor.
The lower the frequency of the supply, the smaller the value
of Ef, and the higher the speed of the motor, the larger the
value of Ea, so that low frequency and high speed are the condi
tions for h^h power factor.
If the load on such a motor is increased, the current increases
and so therefore does the flux ^ &nd the voltage Ef. Since the
Fio. 368. — Conductively compen Fia. 369. — Inductively compen
sated. Bated.
FiQS, 36S and 369. — Methods of compenaating for armature reaction in &
singlephase series motor.
applied voltage E is constant, therefore E, decreases slightly; to
generate this back e.m.f. with the larger flux 4>, the armature
must run at a slower speed. Typical curves for such a motor
are shown in Fig. 367. The speed and torque curves are similar
to those of a directcurrent series motor, and such a machine will
run equally well with either a directs or an alternatingcurrent
supply.
341. Armature Reaction. — From Figa. 364 and 365 it may be
seen that there is a cross magnetizing effect due to the armature
current just as in the direct^current machine, but this cross flux
00 is alternating and generates an e.m.f.' of self induction in the
armature coils which tends to lower the power factor, because
self induction always tends to make the current lag. This cross
flux is eliminated by means of a compensating winding connected
as shown in Fig. 368 so that its magnetizing effect at every instant
is exactly equal and opposite to that of the armature.
Art. 342]
SINGLEPHASE MOTORS
313
Another method of elimiDating the cross flux i^o is shown
diagram matically in Fig. 36d. In thb case the compensating
coils are short circuited. The flux ^o threads these coils and
generates in them an e.m.f. which, according to Lenz's law, sends
a current in such a direction as to oppose the change of the flux,
so that the cross flux is automatically Icept down to a small value.
342. The Repulsion Motor. — If, in the machine shown in Fig.
368, the armature is disconnected from the circuit and is then
short circuited, a^ shown in Fig. 370, the resulting machine is
what is called the repulsion motor.
Fio. 371. Pio, 372,
Fios. 370 to 372.— The Btarting torque of a singlephaae repulaioi
If the coils B alone are acting, as shown in Fig. 371, then
e.m.fs. are generated in the armature coib in such a direction
as to oppose the change of the flux <j>, but no current passes
through the armature winding or through the conductor x since
the voltage between a and b is opposed by that between b and c.
If the coils A alone are acting, as shown in Fig. 3V2, then a
314 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xxxvm
large current can flow, but the resultant torque is zero because,
under each pole, half of the conductors carry current in one direc
tion while the other half carry current in the opposite direction.
If both sets of coils are acting, as in Fig. 370, then the current
produced in the armature by coils A can produce a torque with
the magnetic field produced by coils B.
343. Commutation of Series and Repulsion Motors. — The
coils which are short circuited by the brushes on the commutator
are threaded by an alternating flux ^, see Figs. 364 and 370, so
that large currents are induced in these coils which currents,
flowing through the brush contacts, cause sparking. The com
mutation of these machines is therefore much poorer than that of
the directcurrent series motor.
344. Wagner Singlephase Motor. — ^Because of the poor
commutation, the singlephase repulsion motor has not come into
extensive use in this coimtry except in small sizes. The principle
of the repulsion motor, however, is used by the Wagner com
pany to obtain a large starting torque from a singlephase in
duction motor.
This machine starts up as a repulsion motor and, when it has
attained almost synchronous speed, a centrifugal device short
circuits all the commutator bars on one another and at the same
time lifts the brushes. The motor operates thereafter as a single
phase induction motor and the commutator has a long life because
it is used only during the starting period.
CHAPTER XXXIX
MOTORGENERATOR SETS AND ROTARY CONVERTERS
346. MotorGenerator Set — When it is desired to change from
one alternating voltage to another a static transformer is used,
see Chap. 34. To change from one direct voltage to another
no such simple device is available and a motorgenerator set has
to be used. This consists of two machines mechanically con
nected together one of which, running as a motor, takes power
from the source of Supply and drives the other machine as a gen
erator ; this latter machine is wound for the desired voltage.
If 10 kw. at 220 volts is required from a 110 volt line then:
the generator output = 10 kw. at 220 volts
XI. X XX 10X1000 1 ,^, ^^,^ ,^
the motor output = ^r^ X qoS = 15 horsepower at 110 volts
where the generator efficiency is taken as 88 per cent.
346. The booster set is a special t3rpe of motorgenerator set
and is shown diagrammatically in Fig. 373. The generator in
Fia. 373. — Booster motorgenerator set.
this case is connected in series with the line and its voltage can
add to or subtract from the line voltage.
If 10 kw. at 220 volts is required from a 110 volt line and a booster set
is used then:
the booster output = EI, Fig.373
= 5 kw. at 110 volts
xu X X r XI. J • • X 5 X 1000 1
the output of the drivmg motor = — =j^ — X
746
0.85
= 8 horsepower at 110 volts
and this set can perform the same duty as that performed by the motor
generator set figured out in the last article.
315
316 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxxix
347. The balancer set is a special type of motorgenerator set
used when it is desired to change from a twowire to a threewire
directcurrent system. As shown diagrammatically in Fig. 374,
it consists of two shunt wound directcurrent machines with their
armatures on the same shaft, which armatures in the particular
case shown, are wound for 110 volts.
Compare now the operation of a three vrire system without a
balancer as in Figs. 376 and 377, and with a balancer as in Figs.
Balancer
Fio. 374. — ^Balancer set.
T AQ 6 1
220
i_^
+
B
n'ivi
220
M>
<4
"lUU
B
ml
Fia. 375.
Figs. 375 and 376.
FiG. 376.
Balanced load.
110
' "^ Y T T T 'P ' ©"* '•*V Y Y T ^ I ®^^*
tt y 1 ,f l
Af JllO UBa
ni
lOtMa
1
Fig. 377. Fio. 378. Fig. 379.
Figs. 377 to 379. — Unbalanced load.
376 and 378. In the case where the load is equal on the two
sides of the system, there is no current in the neutral wire and
the balancer in Fig. 376 is running light as two motors in series
on 220 volts, while the voltages across the two sides of the
system are equal.
If now some of the lamps on the side B of the system are
switched oflf then in Fig. 377 the same total current is passing
through the lamps connected across A as is passing through the
smaller number of lamps connected across B so that each lamp B
is carrying more current than each lamp A and the voltage across
B is now greater than that across A. In the extreme case when
Abt. 348]
MOTOBrGENERATOR SETS
317
only one lamp is connected across B the voltage across that lamp
is practically 220 volts.
When the balancer is used, as in Fig. 378, the voltages tend to
take the same values as in Fig. 377, but when the voltage of B
rises and that of A falls then the machine M tries to increase in
speed and G to decrease in speed or each machine tries to run as a
motor at the speed corresponding to its voltage. But the two
machines are coupled together, so that the actual speed must be
between these two values and the one that tends to run fast will
try to increase in speed and will therefore act as a motor and
drive the other machine as a generator, so that G operates as a
generator and current passes through it from the negative to the
positive terminal while M operates as a motor driving G and
Fig. 380. — Threewire generator.
current passes through it from the positive to the negative termi
nal. The voltage across G will be (110 — hRa) volts and that
across M will be (110 + laRa) volts, where laRa is the armature
resistance drop, and the change in the voltages across the two
sides of the system is comparatively small even when the load is
considerably unbalanced.
Neglecting the losses in the balancer itself, the generator out
put and the motor input are each equal to (110 volts X half the
unbalanced current), and if care is taken to arrange that the load
is nearly balanced under all conditions, then the balancer may
be of comparatively small output.
When the load on side B of the system becomes greater than
that on side A, the current in the neutral wire flows in the opposite
direction, as shown in Fig. 379.
348. Threewire Generator. — The balancer set is eliminated
by the use of the threewire generator shown diagrammatically
in Fig. 380, which consists of a standard twowire directcurrent
318 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.xxxix
generator with a coil C of high reactance and low resistance con
nected permanently across diametrically opposite points on the
armature. The voltage between a and 6 is alternating, see page
240, so that, even with no external load, an alternating current
flows through the coil C, this current, however, is extremely
small since the reactance of coil C is large.
The center point o is alwayB midway in potential between a
and h and this point is connected to the neutral line of the three
wire system. When the loads on the two sides of the system
differ, a current flows in the neutral line and enters the armature
through the reactance coil which offers only a small resistance
to direct current. The currents in the sections M flow against
the generated e.m.f. so that these sections are equivalent to the
motor end of a balancer, while the currents in sections G flow in
the direction of the generated e.m.f. and so these sections are
equivalent to the generator end of a balancer.
349. To transform from alternating to direct currenti the
motorgenerator set may consist of either an induction or a
synchronous motor, direct connected to a directcurrent generator.
The same transformation can be made in a single macbme called
the rotary converter.
360. Rotary Converter. — ^Fig. 381 shows an alternator such as
that on page 241 connected to a directcurrent generator which has
the same number of poles. If an alternating e.m.f . is applied to
machine Af, it will operate as a synchronous motor and direct
current may then be obtained from machine 0. The two ma
chines M and may be combined to form a single machine as
shown in Fig. 382, and, if an alternating voltage is applied to the
slip rings a&, the machine will run as a synchronous motor
while at the same time direct current may be obtained from the
brushes at the commutator end. The resultant current in the
armature will be the difference between the direct current /j
drawn from the machine and the alternating current /« which the
machine takes from the power mains.
There is a fixed ratio between the voltages of the direct and
alternating current sides of the machine. The voltage between
the slip rings a and 6 is a maximum when x and y are in the
neutral position and this must be the voltage between the
brushes c and d of the directcurrent side therefore Edy the
voltage of the directcurrent end, is equal to the maximum volt
Abt. 351] MOTOR€ENERATOR SETS 319
age at the alternatii^cuirent end or to V2£o where E^ is the
effective voltage at the alternatingcurrent end.
It is impossible to change the voltage of the directcurrent side
of the machine without chan^ng the alternating applied voltage.
A change in the field excitation for example does not change the
applied voltage, but it does change the phase angle of the current
the converter takes from the line, see page 256, and a converter
can be used for power factor correctioa in the same way as a
synchronous motor, see page 257.
If it is desired to raise the voltage of the directcurrent side of
the machine then the alternating applied voltage must be raised.
Fio 382. — Diagrammatio repreaentation of a rotary converter.
One method of doing this is to supply the rotary converter through
a transformer which has taps on the secondary side by means of
which the voltage may be raised or lowered. Another method is
to insert a booster, generally on the alternatingcurrent side of
the machine. This booster consists of a small alternator, on the
same shaft as the rotary converter, the voltage of which may be
added to or subtracted from that of the alternatingcurrent power
mains.
361. MotorGenerator Sets and Rotaiy Converters. — The
principal points of difference between the rotary converter, the
320 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap.
induction motorgenerator set and the synchronous motorgen
erator set are as follows:
Starttng. — On the alternatingcurrent side, the rotary con
verter operates as a synchronous motor and must therefore
be brought up to speed and synchronized, see page 258. The
induction motorgenerator set is selfstarting.
Parallel Operation. — A synchronous motor and a rotary con
verter are liable to himt, see page 258, although this trouble is
practically eliminated by the use of dampers. Induction motors
do not hunt.
Voltage Variation on the Directcurrent Side. — ^In the case of
the rotary, a booster or a boosting transformer is required to raise
the voltage of the directcurrent side whereas the voltage of the
directcurrent side of either type of motorgenerator set can be
controlled by the field excitation.
Efficiency and Cost. — Being a single machine instead of two
separate machines, the rotary converter is cheaper than either
type of motorgenerator set and its efficiency is the highest since
it has only the constant losses (friction and iron loss) of one
machine.
Power Factor Control. — ^This is possible with the rotary con
verter and with the synchronous set but not with the induction
motorgenerator set.
For small sizes, up to 100 kw., the induction motorgenerator
set is generally preferred because it is easily started and because
the voltage of the directcurrent end can easily be regulated.
For large sizes, 500 kw. and over, a synchronous machine is
preferred because it can be used to control the power factor of the
system; if the voltage regulation is not of great importance, as in
railway work, the rotary converter is used, but if a wide variation
of voltage is required from the directcurrent side, a motor
generator set is generally preferred.
362. Polyphase Rotary Converter. — The only difiference be
tween the singlephase and the polyphase machine is that the
former is tapped at two points, as shown in Fig. 382, whereas the
latter has to be tapped as if the machine was a polyphase alter
nator of the revolving armature type, see page 241. A three
phase rotary is shown in Fig. 383; the armature is tapped at three
points, 120 electrical degrees apart.
368. Splitpole Rotary Converter. — One method of changing
the voltage of the directcurrent side of a threephase rotary con
Art. 354]
MOTOR^ENERATOR SETS
321
verter would be to change the total flux which passes between
ab, Fig. 384, without changing the flux which passes between ac.
This is accomplished by splitting each pole as shown diagram
niatically in Fig. 384 and exciting each part separately. If then
the excitation of the main pole M is unchanged, the flux threading
etc will be unchanged, but the total flux threading ah may be
increased or decreased by varying the excitation of R, and by
this means the voltage of the directcurrent side of the machine
may be raised or lowered without any change being required in
the alternating applied voltage.
Fig. 383. — Threephase rotary con Fig. 384. — Splitpole rotary
verter. verter.
con
Another method of changing from alternating to direct current
is described on page 357, the piece of apparatus by which the
transformation is made is called the mercury vapor converter.
364. Frequency Changers. — To change from one frequency
to another, a motorgenerator set must be used which consists of a
synchronous motor direct connected to an alternatingcurrent
generator. If 60 cycles have to be obtained from a 25cycle line,
then the number of poles must be in the ratio of 60 to 25, and
the smallest possible number of poles would be 10 for the motor
and 24 for the alternator; this set would run at 300 r.p.m., so
that such frequency changers cannot readily be made for small
outputs.
CHAPTER XL
ELECTRIC TRACTION
An electric car with a seating capacity for fifty persons is gen
erally equipped with four motors each of 50 h.p. Such a powerful
equipment is required to obtain the high speeds and high rates
of acceleration used in electric traction.
366. Tractive Effort. — The cycle of operations of an electric
car with a reasonable distance between stops is shown in Fig. 385.
During the time interval oa, energy is put into the motors and the
car is accelerating; during the interval a6, the motor circuit is
o a be ^stop
Fig. 385. — Cycle of operations of an electric car.
open and the car is coasting; during the interval be the brakes
are applied and the car gradually brought to rest.
If TT is the weight of the train or car in tons (2000 lb.) and a
is the acceleration in miles per hour per sec. then the tractive
effort in lb., required for acceleration
= X acceleration in ft. per sec. per sec.
9
_ (W X 2000) g X 5280
32.2 ^ 3600
= 90 TT X a
An additional accelerating tractive effort of about 10 per cent, is
required to supply the rotational kinetic energy of the gears,
motor armatures and other rotating parts so that the tractive
effort required for acceleration = 100 W X a = 100 lb. per ton
for each mile per hour per sec. of acceleration; an acceleration
of 1.5 miles per hour per sec. may be used without discomfort
to the passengers.
The tractive effort to overcome train friction (bearing, air and
322
Abt. 356] ELECTRIC TRACTION 323
rolling friction) is expressed in pounds per ton weight of train
and is given by the empirical formula
/ = T7^ + 0.03 V + 0.002 A ^ (l +^^)'
50 72
:^ + 0.03 V + 0.002 A ^
where / is the tractive efifort to overcome train friction in lb. per
ton weight of train
W is the weight of the train in tons (2000 lb.)
V is the velocity of the train in miles per hour
• A is the cross section of the car above the axle in sq. ft.
and is generally about 120
n is the number of cars; the side friction of each additional
car increases the air friction by 10 per cent.
When there is a curve in the track, an additional tractive effort
of 0.7 lb. per ton for each degree of curvature is required due to the
increased rolling friction; the number of degrees of curvature is
equal to 5730 divided by the radius of the curve in feet.
When there is a grade an additional tractive effort of 20 lb.
per ton is required for each 1 per cent, of grade. When the
grade is 1 per cent, the car has to be lifted through 1/100 of the
distance it travels and this is equivalent to lifting 1/100 of the
weight, or 20 lb. per ton, through the whole distance traveUed.
The maximum tractive effort that can be used without causing
the driving wheels to slip is about 22 per cent, of the weight on
these wheels, since maximum tractive effort = m X weight on the
driving wheels, where ^i, caUed the coefficient of adhesion, has the
following values:
for a clean dry rail = 0.28
a greasy moist rail = 0.15 or when rail sanded = 0.25
a dry snow covered rail = 0.11 or when rail sanded = 0.15
366. Speed Time Curve. — The characteristics of a traction
load can best be shown by the working out of an actual example.
A 30ton suburban car is equipped with four 50h.p. motors having the
characteristics shown in Fig. 386. The acceleration has to be 1.25 and the
deceleration 1.5 miles per hour per sec; the road is level and without curves
and the distance between stops 0.75 miles, the time of the stop 9 sec. and the
' Electric Traction by A. H. Armstrong, Standard Handbook for Electrical
Engineers.
324 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xi,
schedule speed 21 miles per hour. It is required to draw the speed time
curve.
^ ^ ^ + 003F + 0.002 X 120 X^(H^)
» 9 + 0.3 + 0.8 = 10.1 lb. per ton at 10 miles per hour
» 9 + 0.6 + 3.2 = 12.8 lb. per ton at 20 miles per hour
» 9 + 0.9 + 7.2 = 17.1 lb. per ton at 30 miles per hour
Now the effective tractive effort for an accleration of
1.25 miles per hr. per sec.
» 125 lb. per ton.
The train friction during acceleration will be approximately
«11 lb. per ton.
Therefore the total tractive effort required
30
= 136 lb. per ton = 136 X x = ^^^ ^^ P®' motor.
Corresponding to this value on Fig. 386,
the speed with 550 volts = 19 miles per hour
and the current per motor — 100 amp.
85 1400
o
e
H
ft.
o
30
26
^20
i
O
A
CO 10
^1200
a
9
^1000
e
800
I 600
H 400
aoo
\
/
\
V^
V
/
X
V
/
^^
^'■^*.
'
/>
/
<
r
X
/
aO 40 60 80
Ampem
100 lao
Fig. 386. — Characteristics of a 50horsepower, 550volt directcurrent
motor, the tractive effort and the speed being measured at the rim of the
driving wheels of the car.
The motor is so controlled that the current remains constant at 100 amp.
per motor and the acceleration has a constant value of 1.25 miles per
hour per sec. until the speed has become 19 miles per hour; the starting
resistance has then all been cut out and normal voltage is applied to the
motors.
The time taken to attain this speed = vel./accel. = 19/1.25 = 15.2 sec.
and the distance covered during this time = ^ X oggn = 211ft.
from these figures the points a, a\ and a% are plotted in Fig. 387.
The motor, running on normal voltage, will now speed up, and the current
will decrease. Let the current decrease to 60 amp. then:
The corresponding tractive effort = 500 lb. per motor and the speed « 23
miles per hour, Fig. 386.
Art. 356]
ELECTRIC TRACTION
325
The average tractive eflfort while the current is decreasing = ^
= 760 lb. per motor = 760 X 4/30 = 101 lb. per ton
The train friction at 21 miles per hour = 13 lb. per ton
The tractive effort available for acceleration = 88 lb. per ton
The average acceleration = 88/100 = 0.88 mile per hour per sec.
23 — 19
The time taken for the speed to attain 23 miles per hour = "TToo"" == 4.6 sec.
The distance covered during this time = s X ofion X 4.6 = 142 ft.
The distance covered from the start = 211 + 142 = 353 ft.
From these figures the points 6, 6i, and 62, are plotted in Fig. 387; points
c for a current of 40 amp. and d for 30 amp. are determined in a similar
way.
Now. the schedule speed is 21 miles per hour.
The distance covered is 0.75 mile.
36
32
28
24
100 A 20
I 80i
I 60 1 12
S I
140** 8
20 4
16
^ ^

_^^
^^
r/r
di
nfeet
^
_^
9"^
'''
\
, di
•x"**"
4
V
'S.
^
^^
,.y
t.
^
V
fi.
A
^
<
\
T\
Z',.^.
>
;>
y^
www* ^
2400 3
l\
^
\^
y
1800
1200
/
1
\g^i
rent
/
^
\
/
^y
"*
"""■
— —
"3
\
600
L
<
^.
\
1
2
9
41
5
6
8
7
econd
[B
8
9
1
DO 1]
LO 1:
M)
Fig. 387. — Speedtime curve of an electric car.
The time per cycle = 0.75/21 ^ 0.0356 hour = 129 sec. of which 120
sec. is the running time and 9 sec. the time of the stop.
In order to bring the car to rest in 120 sec. from the start the brakes
must be applied at some point / such that the deceleration will be 1.5
miles per hour per sec.
If the distance curve be now completed it will be found that the car
has travelled a ilistance of 4500 ft. In order that this distance be 0.75
mile, power must be taken off the car at some point g and the car allowed
to coast thereby reducing the average speed of the run. During coasting,
the speed will fall off according to the curve gh. The slope of this line is
determined as follows: the average velocity during coasting is 25 miles per
hour and the corresponding retarding force due to train friction is 15 lb.
per ton which will cause a deceleration of 0.15 mile per hour per sec. or a
decrease in speed of 1.5 miles every 10 sec.
326 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap. xi.
367. Energy Required by a Car. — The series parallel method
of control, page 126, is used for railway motors so that between
a and 6, Fig. 388, the current taken from the line is twice the cur
rent per motor while between b and c four times the current per
motor is taken from the line.
The average current per cycle = 225 amp.
225 X 550
The average power per cycle = 
The energy per cycle
1000
123,000 X 44
3600
= 123 kw.
= 1500 watthr.
It is convenient to express the energy required by a car or
fUU
w
i 300
1
i
1 200
•J
B
\
wvwv^
N
^^^5^
1
C
I 100
•v^
1
a
h
e
100
• 80
8 00
\J
f3 ^
V
" "
 —
^
i
) a
1
1
1 1
10
10 20 80 40 50 60
Seconds JMMmm UtwMB Stopi bi MOm
Fig. 388. — Current taken by an electric Fig. 389. — Energy consumed
car during one cycle. by an electric car.
train in watthours per ton mile or in kilowatthours per car mile.
These quantities are determined in the following way:
The energy per cycle = 1500 watthours
The distance travelled per cycle = 0.75 mile
The weight of the car = 30 tons
The watthours per ton mile = n 7^y^0 ~ ^^
The kilowatthours per car mile = ToTv^ X 7rf^ = 2.0.
These two quantities decrease as the distance between stops
increases, as shown in Fig. 389, because then the energy required
to accelerate the car becomes a smaller portion of the total energy
per cycle. In the extreme case of city service, the coasting period
is generally eliminated, the distance between stops being so short
that only acceleration and braking are required.
From curves such as those in Fig. 389, the power house capacity
Abt. 3581 ELECTRIC TRACTION 327
may be determined once the number of cars and their schedule
has been fixed.
368. Characteristics Desired in Railway Motors. — The series
motor is particularly suited for traction service, see page 103,
because it develops the large starting torque required with the
minimum current, see page 102, In addition, it talces light loads
at a high speed and heavy loads at a slow speed, slows down on an
up grade and speeds up on a down grade, it therefore maintains
the load on the power house more uniform than if constant speed
motors such as the directcurrent shunt motor or the alternating
current induction motor were used.
The alternatingcurrent series motor, see page 310, has the same
characteristics as the directcurrent series motor but is more
Pio. 390. — Directcurrent street railway motor.
expensive, has a lower efficiency and does not commutate so well.
The method of control is simpler however, the low voltage re
quired for starting being obtained by means of an autotrans
former instead of by the use of resistance and the seriesparallel
connection. These motors can run on direct as well as on alter
nating current circuits.
For one particular kind of service the polyphase induction
motor is specially suited, even although it is a constant speed
motor, and that is for mountain grade worlc where the grades are
long and fairly uniform. When the car is on the down grade, the
motors tend to speed up and run above synchronous speed, and,
under these conditions they become induction generators and
supply power back to the line. With a speed of 4 per cent, above
synchronous speed, the machine will deliver its fullload rating
328 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xl
Abt.3611 electric TRACTION' 329
to the line, so that brakmg with brake shoes is not necessary and
the danger of trouble on a long grade due to overheating of brake
shoes is practically eliminated.
369. Motor Construction. — To keep down the size of the motors
they must run at a high speed, see page 101, so that gears are
generally necessary. The standard construction for urban and
interurban cars is shown in Fig. 390.
Motors which are placed between the driving wheels, as are all
street car motors, are restricted in size. This restriction is most
keenly felt in the design of large horsepower slowspeed motors
for electric locomotives for freight service. One method of
providmg mcreased space for the motors is shown in Fig. 391
where two motors each of 2000 h.p. are mounted in the car
and are connected to the driving wheels through side rods.
360. Distribution to the Cars. — The three standard railway
systems are shown diagrammatically in Figs. 392, 393 and 394.^
With the singlephase system the trolley voltage may be as
high as 11,000 volts and in order to render high speed collection
reliable with long spans the catenary construction is used, the
trolley wire being carried from a steel messenger cable, the latter
having a sag while the former lies parallel with the rails.
Because of commutation troubles the standard directcurrent
voltage is not higher than 600 volts although there is now in
successful operation a trunk line with a trolley voltage of 3000.
Large currents are therefore necessary for electric trains and to
carry this current a third rail insulated from the ground is used
instead of the trolley wire if the trains run on a private right of
way. For city service the familiar trolley is used and the trolley
system is supplies by feeders from substations.
' The return circuit is through the rails in each case. This cir
cuit is made as highly conducting as possible by bridging all the
rail joints with flexible pieces of copper called bonds which are
electrically welded to the side of the rail.
361. Alternating and Directcurrent Traction. — The single
phase alternating current system has the advantage that voltages
as high as 11,000 volts may be used on the trolley so that the
current in the trolley wire is small and power may be transmitted
for long distances before the voltage drop becomes too large,
the substations therefore are few in number and contain merely
^ Taken from a paper by George Westinghouse, in the Trans, of American
See. of Mech. Eng., July, 1910.
330 PRINCIPLES OF ELECTRICAL BNOINEBRINO [Chap, i
1
h
i
i
.1^
1
i''
Si
^Ji
Art. 361]
ELECTRIC TRACTION
331
QQ
f
CO
6
332 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap. xi.
the necessary stepdown transformers. For the directcurrent
system, on the other hand, at 600 volts, substations must be closer
together to maintain the trolley voltage and these substations
contain rotary converters as well as stepdown transformers.
The principal disadvantages of the alternatingcurrent system
are that the motors are heavier than directcurrent motors of
the same horsepower, while the alternating magnetic flux link
ing the trolley causes interference with adjoining telephone
^3rstems.
362. Motor Car Trains. — ^A number of cars each with its own
motors and controller may be joined together to form a train
which, by means of the multiple unit system of control, see page
133, can be controlled by a single operator at the head of the train.
This system of operation is very flexible because cars can be
added to suit the traffic while trains can readily be split up at
junctions or single cars can be attached or dropped as required
without the necessity of stopping the whole train.
The acceleration can be rapid because all the weight is on
the drivers and a large draw bar pull can be obtained. Heavier
trains can also be run at higher speeds than by locomotives, while
the wear and tear on the track is reduced because the weight is so
well distributed. Such trains can be run in either direction
without having to be turned end for end.
363. Electric Locomotives. — When the cars of a steam road
have to be hauled into the city, through tunnels, or on mountain
grades, electric locomotives are often used. Their capacity is not
limited by grate area and boiler capacity so that they can develop
very large torques for starting and accelerating and can generally
give 50 per cent, more draw bar pull than a steam locomotive
of the same weight. The torque also is uniform, such locomo
tives can therefore give high acceleration and high schedule
speeds.
ELECTRIC HOISTING
364. Crane and Hoist Motors. — In the case of cranes, where
the load is visible at all times and where the starting and accelerat
ing of the load is a large part of the total hoisting cycle, the
directcurrent series motor is used because of its good starting
characteristics. Where only alternating current is available,
the wound rotor polyphase induction motor is generally used, but
it is not such a satisfactory machine, see page 298.
ART. 365] . ELECTRIC TRACTION 333
Motors for large cranes, such as those used in rolling mills,
take large currents and are generally controlled by magnetic
switch controllers.
When the load has to be raised a considerable distance, as in
the case of mine hoisting, the load is accelerated for only a small
portion of the total cycle and is run thereafter at a constant speed
not greater than that fixed by law. For such service the direct
current compound motor is suitable because its speed cannot
exceed a safe maximum value. Where only alternating current
is available, the polyphase wound rotor induction motor is used.
366. Braking. — The brakes used on cranes and hoisting
machines must be so constructed that they will set if there is any
interruption in the current supply. Such a brake is shown in
Fig. 395. When power is switched on to start the hoisting motor,
the solenoid lifts the lever L
and releases the brake, but as
soon as current ceases to flow,
the weight W sets the brake.
When a light load is being
lowered, power must be ap
plied to drive it down. When ^jq 395.— Brake for hoists and cranes,
the load is heavy, it may be
allowed to overhaul the motor, the speed being kept within rea
sonable limits by means of the brake. This causes jerky opera
tion, and excessive wear of the brake, and, for coal and ore un
loaders and other such hoists doing fast hoisting service, mechan
ical braking is not very satisfactory and dynamic braking is now
used.
The hoisting motors for such hoists are directcurrent compound
wound and are connected as in Fig. 396 during the hoisting period.
When being lowered, the empty bucket is allowed to overhaul the
motor, which is connected as shown in Fig. 397, the shunt coils
being separately excited while the armature is short circuited
through an adjustable resistance R. The machine then acts as a
generator driven by the descending load and sends a current I
through the resistance R, As the machine speeds up, the current
/ increases until the retarding torque due to the current / becomes
equal to the effective torque driving the armature.
To raise the lowering speed, the resistance R must be increased.
This reduces the current I and the braking torque temporarily,
and the speed increases until the greater e.m.f . generated in the
334 PRINCIPLES OF ELECTRICAL ENGINEERING . [Chap, xl
armatuie is again able to send the necessary current / through
the higher resistance. By varying the resistance R, the lowering
speed may be varied from almost zero to any desired value. To
stop the load, a mechanical brake must be applied, since dynamic
braking can only take place while the armature is moving.
I
W
W
Fio. 396. — Hoisting period. Fig. 397. — ^Lowering period, with
dynamic braking.
Figs. 396 and 397. — Connections of a compoundwound directcurrent
hoist motor.
D3mamic braking is also applied to crane motors. During
hoisting, the motor is connected as shown in Fig. 398 On the first
step of the controller about twothirds of fullload current flows
through the circuit, and this is sufficient to release the brake and
start the average load.
Brake + '
Brake
Fig. 398. — Hoisting. Fig. 399. — ^Lowering. Fig. 400. — ^Dynamic
braking.
Figs. 398 to 400. — Connection of a serieswound directcurrent crane motor.
When lowering, the machine is changed over to a shunt machine
as shown in Fig. 399 and a resistance R is inserted to limit the
flow of current in the field coil circuit to twothirds of fullload
current, which is sufficient to release the brake. The direction of
Aet. 366] ELECTRIC TRACTION 335
the field current is the same a3 during hoisting, but the armature
terminals are reversed so that the armature current is reversed
and the motor drives the load down. If the load is heavy enough
to overhaul the motor, then the motor speeds up and its back
e.m.f, increases, until finally the machine nms as a generator and
acta as a brake. The current la is now reversed and the machine
supplies its own exciting current, so that it may be disconnected
from the line as shown in Fig. 400 and the braking speed adjusted
by means of the rheostat Rb
366. Flywheel Motor Generator Sets for Mine Hoisting. — The
load on the motor of a hoist is extremely variable, as shown by
curve A, Fig. 401, and, when the motor has an output of 500
h.p. or more as in the case of motors for mine hoists, such
a load is not a very desirable one for a power company, and a
Curve A Current taken by hoist motor.
Curve B Curnat IsksD by iaduotion motor.
Fia. 401. — Characteristics of a Eioistin^ load when a flywheel motor
generator set IS used.
cheaper rate for power can generally be obtained if the load is
more uniform.
To obtain this result and at the same time to reduce the large
losses in the starting rheostat, the Ward Leonard system, see
page 110, is used. The high speed motor generator set consists
of a wound rotor induction motor which takes power from the
transmission line, and a directcurrent generator which supplies
power to the hoisting motor, while a heavy flywheel is mounted
on the same shaft as shown in Figs. 402 and 403.
The excitation of the hoisting motor is kept constant by means
of the exciter E, while the speed of the motor is controlled by the
resistance r in the generator field circuit, by means of which the
voltage El applied to the motor terminals may be varied.
As the load on the hoist motor increases, the current in the
induction motor leads tends to increase in the same ratio, but a
slight increase in the current / causes the plunger p of the sole
336 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xu
noid S to raise the plates of the water rheostat W thereby increas
ing the resistance in the rotor circuit, so that the rotor current
and torque are maintained constant. This torque is not sufficient
Fig. 402, — Flywheel motor^^erator set for the operation of mine hoista
for the load, so that the speed drops and thereby causes the fly
wheel to give up enei^.
3. — Connections for a hoist motor supplied by a flywheel motor
generator eet.
If the load on the hoist motor now decreases, the current I
tends to decrease, but a slight decrease in this current causes the
pull of solenoid S to decrease and the plates of the water rheostat
Abt. 367] ELECTRIC TRACTION 337
to drop, thereby decreasing the resistance in the rotor circuit, so
that the rotor current and the rotor torque are maintained.
This torque is greater than necessary for the load and so acceler
ates the flywheel.
By means of such a slip regulator, the power taken from the
line is maintained approximately constant, as shown in curve B,
Fig. 401.
367. Safely Devices. — The safety brake is of the type shown
in Fig. 395 and will set and hold the load as soon as current ceases
to flow in the solenoid B.
If for example the field circuit of the hoisting motor were to
open, the current in the solenoid B would be interrupted and the
brake would set.
If the cage overtravels, it closes the switch T in the hatchway
and this shunts the current from the solenoid B and allows the
brake to set. The resulting overload then opens the circuit
breaker C.
The solenoid B is also shunted when, due to an excessive over
load, the circuit breaker C opens and closes the contacts ab. A
similar pair of contacts cd ensure that the brake shall be set when
the main switch is open.
CHAPTER XLI
TRANSMISSION AND DISTRIBUTION
368. Directcurrent Stations. — When direct current is used,
the voltage can be transformed up and down only by means of
motor generator sets and, since these are expensive and require
supervision, they are seldom used, so that the connected load of
motors and lamps must operate at the power house voltage.
The 110volt lamp is practically standard because it has a
stronger filament than have lamps of higher voltage. The
use of such a low voltage necessitates the use of conductors of
large cross section to carry the current, and higher voltages are
desirable.
If 50 kw. has to be transmitted a distance of 100 yd. with a drop in vol
tage not exceeding 2.5 per cent., find the necessary cross section of the con
ductor if the voltage is 110 volts and also if it is 220 volts. The resistance
of copper is taken as 11 ohms per cir. mil foot.
At 110 volts
At 220 volts
Current in the line ....
Voltage drop in the
line.
Resistance of the line. .
Resistance of 200 yd.
of wire.
Cross section of wire in
cir. mils.
50X1000 ^^^
yTA — =455 amp,
50 XIOOO
220
= 227 amp.
= 2.5% of 110 = 2.75 = 2.5% of 220 = 5.5
volts
^ 2.75
455
^ 11 X 200 X 3
cir. mils
= 1,100,000
= 0.006 ohms
= 0.006
volts
5.5
» 0.024 ohms
227
^ 11 X 200 X 3
cir. mils
« 275,000
= 0.024
J
The cross section of the wire is inversely proportional to the
square of the voltage for the same loss in the line.
If the three wire system of distribution is used then 110 volt
lamps may be operated from 220 volt mains as shown in Fig. 376,
page 316. The neutral wire may be small in cross section if the
load is nearly balanced under all conditions of loading.
Where there is only a small Ughting load, or where the Ughts
338
Art. 369] TRANSMISSION AND DISTRIBUTION 339
are supplied from special mains, thea a voltage of 550 may be
used for the motors. Voltages greater than 550 are dangerous;
even 110 volts may prove fatal to a person who happens to mak6
unusually good contact with the mains.
389. AltematingcuTrent Ststions. — Where power has to be
transmitted a long distance or has to be distributed over a wide
area, the alternatingcurrent system is preferred because the
voltage can be transformed up and down by means of transform
ers. These require no supervision and may be installed in the
open, on poles or in manholes.
A typical highvolt^e transmission system is shown in Fig.
404. The power station contains the generators and atepup
g=jfr
transformers while the stepdown transformers are placed in a
terminal station. The load ie carried by several units operating
in parallel so that one machine may be shut down for repair
without affecting the total load.
From the terminal station, power is transmitted to a number of
substations which are conveniently located with respect to the
load. In the case of a transformer substation, the equipment
consists of the necessary stepdown transformers to reduce the
voltage to 2200 volts at which it is supplied to the feeders. Each
feeder is generally controlled by a feeder regulator, see page 281,
by means of which its voltage may be adjusted. These feeders
340 PRINCIPLES OF ELECTRICAL ENOINEERINQ [Chap.iu
supply transfonners which are placed on poles, or underground
in manholes, and step the voltage down to 110 volts for the
consumer.
When it is necessary to transform to direct current as for exam
ple to supply power to a lar^ machine shop, the substation equii>
ment is as shown in Fig. 405. Power enters the station over the
three bare wires a, h, and c which are carried through porceUin
m np
Via. 405. — ConnectioDB of ai
A.C. to D,C. motorgenerator substn^n.
wall bushii^ and are connected to the hightension bus bars rn,
n and p. Power to operate the synchronous motor is taken from
these bus bars through a deltaconnected bank of transformers by
which the voltage is reduced to 2200 volts, while half voltage
taps are also supplied so that the selfstarting motor may be
started at half voltage; the doublethrow switch for this purpose
ia not shown.
Abt. 371] TRANSMISSION AND DISTRIBUTION 341
370. The voltages used in practice are :
Direct Current:
110 volts for lighting, generally obtained from a 220volt three
wire system.
110, 220 and 550 volts for motors.
600 volts for street railway systems.
1200 volts for interurban systems.
2400 volts for trunkline electrification.
AUemating Current:
110 volts single phase for lighting and for small motors.
110, 220, 440 and 550 volts for polyphase motors up to 50 h.p.
440, 550 and 2200 volts for polyphase motors greater than
50 h.p.
13,200 volts is the highest voltage generated by alternators;
lowvoltage alternators with stepup transformers are more
reliable.
100 volts per mile of line with a maximum of 110,000 volts for
power transmission.
The tendency is to use a frequency of 60 cycles for power and
lighting work as it gives a better choice of speeds for induction
motors than does 25 cycles, see page 297. For singlephase rail
way work, however, 25 cycles are necessary because of the diffi
culty in constructing motors that will commutate satisfactorily
at a higher frequency. In the case of cement mills and other
such places where most of the induction motors are slowspeed
machines, 25 cycles may be used with advantage.
371. Comparison between singlephase and threephase
transmission.
10,000 kw. at 80 per cent, power factor has to be delivered at the end of a
singlephase 25mile line at 50,000 volts and 60 cycles. The size of wire is
No. 000 and the wires are spaced 72 in. apart. Find the voltage at the
generating station and also the power loss in the line.
the resistance of this wire = . 326 ohms per mile, page 216.
the reactance at 60 cycles = . 742 ohms per mile, page 216.
,, . . ^, ,. 10,000 X 1000 __
the current m the Ime = n » y fyO noo ~ amp.
the resistance drop in 50 miles of wire » IR
= 250 X 0.326 X 60
= 4075 volts
the reactance drop in 50 miles of wire = IX
= 250 X 0.742 X 50
= 9250 volts.
the voltage of the generating station » Eo, Fig. 406.
342 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xm
 V (giCoe a f /Jg)« f (^tain a + IX)*
 V (40,000 + 4075)« +(30.000 + 9250)*
» 59,000 volts
 250« X 0.326 X 50
= 1020 kw.
» 10.2 per cent, of the output
15,000 kw. at 80 per cent, power factor has to be delivered at the end of
a threephase 25mile line at 50,000 volts and 60 cycles. The size of wire
the power loss in the line
FiQ. 406.
NMteal \nn Cmnim N« Canmt
FiQ. 407.
is No. 000 and the wires are spaced 72 in. apart. Find the voltage in the
generating station and also the power loss in the line.
the resistance of the wire — 0.326 ohms per mile.
the reactance at 60 cycles » 0.742 ohms per mile.
15,000 X 1000
the current in the line
0.8 X 1.73 X 60,000
» 217 amp.
The problem is best solved by considering each line separately as shown in
Fig. 407 then
Et, the line voltage to neutral » 50,000/1.73 » 29,000 volts
the resistance drop in 25 miles of wire — IR
= 217 X 0:326 X 25
= 1770 volts
the reactance drop in 25 miles of wire « IX
= 217 X 0.742 X 25
= 4020 volts
the line voltage to neutral at the generating station
= V (29,000 X 0.8 + 1770)« + (29,000
X 0.6 + 4020)«
= 32,950 volts
the voltage between lines
the power loss in the line
= 32,950 X 1.73
= 57,000 volts
» 3 X 217« X 0.326 X 25
= 1150 kw.
a 7.6 per cent, of the output.
Abt. 372] TRANSMISSION AND DISTRIBUTION
343
<6
R
^HK^p
372. Lightning arresters are used to protect electrical equip
ment from lightning discharges and abnormally high voltages
of all kinds.
The current due to a lightning discharge has a high frequency
and will not pass readily through a reac
tance, so that, if, as in Fig. 408, a resis
tance path R is provided to ground with
an air gap g long enough to prevent the
flow of current under normal conditions,
and a reactance or choke coil C is placed
between the line and the equipment to be
protected, then a lightning discharge will
be held up by the choke coil and will
jump across the air gap to ground.
Once an arc is started across the gap,
however, the line current will follow
through the path abcdy and provision must be made in the
arrester to prevent this current from passing. The current
may be limited by inserting a resistance R in series with the gap,
while, if the electrodes of the air gap are made of nonarcing metal.
I
Fig. 408. — Connection of
lightning arresters.
^ '1
?
^
«
HHkBV'is
■ t ,
1
HB^^Kw^'''
1
•«
s
•ft
tt
.5>
ik
1
^
IJ
—QBp.m'ft . .'
Line
Blowout
Coll
I 6
m
Arrester
Fig. 409.— 3000 volt
multigap arrester for sta
tion installation.
Fig. 410.
"^ Ground
Lightning arrester for directKiurrent
circuits.
that is metal such as zinc which has a low boiling point, then,
on an alternatingcurrent Une, the arc will not be maintained but
will stop as the alternating current passes through zero. A single
gap of nonarcing metal will protect a 300volt line, for higher
voltages, several gaps are placed in series as in the 3000volt
344 PRINCIPLES OF ELECTRICAL ENGINEERINQ [Chap.iu
arrester shown in Fig. 409 which is really equivalent to three
arresters one with 3 gaps and a high resistance in series, an
other with 6 gaps and a lower resistance in series, and the third
with 15 gaps in series.
It is not so easy to rupture a direct current and, even when non
arcing electrodes are used, it is necessary also to supply a blowout
coil connected as shown in Fig. 410, which is excited when a power
current flows through the arrester.
Fio. 411. — Aluminium arrester.
For the protection of loi^distance highvoltage transmission
lines the aluminium arrester is generally used. An aluminium
cell consisting of two aluminium plates on which a film of hydrox
ide has been formed, when immersed in a suitable electrolyte,
will allow only a very small current to flow, until the voltage
reaches a critical value. At a higher voltage, the current that
can flow is very large but the high resistance is reestablished as
Art. 3731 TRANSMISSION AND DISTRIBUTION 345
soon as the voltage is reduced below the critical value. Such
a cell can therefore act as a safety valve and aluminium arresters
are made up in the form shown in Fig. 411, about 300 volts per
pair of plates being allowed.
Even with 300 volts between plates, a small current fiows, to
prevent which, the arrester is connected to the line through a
horn gap as shown in Fig. 404. When the arrester is disconnected
from the line, however, the hydroxide 61m dissolves, so that the
arrester must be charged daily by being connected to the line
by the closing of the horn gap for a few seconds.
Fia. 412. — Handoperated, triplepole oil switch.
373. Switches. — Oil switches such as that shown in Fig. 412
are used to rupture the current in highvoltage lines. These
switches are generally kept at a distance from the operator and
are opened and closed through a system of levers, or may be of the
remote control type operated by means of a solenoid or by a small
motor. In all cases the switch is closed against the tension of a
spring and is held closed by means of a latch. This latch may
be released by an overload relay, so as to allow the switch to open
when the current becomes excessive.
To localize trouble, important switches are generally mounted
aa shown in Fig, 413 with each pole in a separate brick or concrete
compartment.
Disconnecting switches such as that shown in Fig. 405 are not
346 PRINCIPLES OF ELECTRICAL ENOINEBRINO [Chap, xli
intended to open while current is Sowing but are used to discon
nect apparatus once the circuit has been opened by an oil switch.
Such disconnecting switches are opened and closed by a long stick
with a hook attached to the end.
374. Overhead Line ConBtruction. — For voltages up to 50,000,
wooden poles with pin insulators are used to support the line.
For higher voltages, pin insulators become veiy large and the
Fio. 413. — Motoroperated, triple^jole oil Bwitch.
stresses on the pin become excessive, so that the suspension type
of insulator has to be used and these are generally suspended from
steel towers as shown in Fig. 415.
To protect the line from lightning, it is usual to run a steel wire
parallel to the power wires, and to ground this wire at every tower.
Lightning will generally strike this ground wire and pass to the
Abt. 375] TRANSMISSION AND DISTRIBUTION
347
ground without doing injury, rather than strike the power wires
and then pass to ground through the insulators.
376. Underground Construction. — To carry current under
ground, stranded copper cable is used. The copper is insulated
with paper which is then impregnated with a compound such as
resin oil after which the cable is sheathed with lead which keeps
out moisture and at the same time protects the cable against
mechanical injury. The cable has to be flexible enough to bend
round comers because it has to be drawn into tile ducts through
manholes such as that shown in Fig. 416, which manholes are
restricted in size.
348 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xu
The necessary cross section of copper is generally fixed by the
permissible voltage drop in the case of low voltage cables, but is
always fixed by heating in highvoltage cables. A current
density of IQDO amp. per sq. in, of copper section can
seldom be exceeded, and this requires 8.7 volts per 600 ft.,
which is 2.5 per cent, of 350 volts, so that, if the voltage drop is
limited to 2.5 per cent, and the transmission distance is 600 ft.,
then, for voltages less than 350 volts, the current density must be
less than 1000 amp. per sq. in. while for voltages greater than
350 volts, the drop in the cable will be less than 2.5 per cent.
Street Lerel
Fig. 416. — Manhole.
376. Switchboards. — ^For convenience of manipulation, all the
apparatus for controlling the machines and circuits in a power
station, as well as all the measuring instruments, are assembled
as compactly as possible on a switchboard.
The method of designing a switchboard is first of all to lay
out the complete diagram of connections, then design the front
of the board placing the different pieces of apparatus in the most
convenient positions, after which the connections on the back of
the board can be laid out and any rearrangement necessary can
then be made.
Fig. 417 shows the diagram of connections for a single shunt
generator which supplies four feeders.
Fig. 418 shows the front of the board.
Fig. 419 shows the connections as they would appear if the
slate front of the board was removed. This diagram is lettered
similarly to Fig. 417.
In larger stations it is usual to provide a separate panel on the
switchboard for each machine and also for each feeder.
Fig. 420 shows a three panel switchboard for a three phase alter
nator, a three phase feeder and an exciter.
Abt. 376] TRANSMISSION AND DISTRIBUTION
349
The exciter panel is equipped with:
1 ammeter Ai.
1 handwheel for the exciter rheostat i?i.
1 switch Sj with a fuse which is on the back of the panel.
2 switches Si and Si for the station lighting and other auxiliary
circuits.
Fig. 417.
II II I I II
n
y
ircait
 ,.  . Breakers ,
voltmeter ^..^ Ammeter
ft Field Rheoitat n
Handle ^=g
ftfift
Main Switch
FiQ. 418. Fia. 419.
Figs. 417 to 419. — Switchboard for a directcurrent shunt generator,
and four feeder circuits.
The exciter voltage is indicated by the voltmeter Vi on the
swinging bracket. As the station grows in size, a second exciter
will have to be added to operate in parallel with the first but, by
means of a plug P, the voltmeter Vi may be made to indicate
the voltage of each machine.
350 PRINCIPLES OP ELECTRICAL BNOINBERINQ [Chap.Xu
The generator panel is equipped with:
1 three phase indicating wattmeter.
1 ammeter.
1 voltmeter.
1 three phase watthour meter, (called in practice a record
ing wattmeter.)
1 handwheel for the alternator field rheostat R^.
1 field switch iSi.
1 triple pole single throw (T. P. S. T.) oil switch.
1 current transformer C.
2 potential transformers V.
Switchboard Conneelion diiccini
Fia. 420. — Switchboard with an exciter panel, a threephase alternator and
a threephtise feeder panel.
As the station grows in size, additional alternators have to be
added and they must operate in parallel so that a synchroscope
must be provided. This is placed on the swinging bracket and
the necessary connections are made by plugs on the generator
panels. For a single alternator a synchroscope is not required.
The generator oil switch has no overload release; protection
against overloads is provided for by the use of automatic awitchee
on the feeder circuits.
The feeder panel is equipped with:
3 ammeters.
1 T.P.S.T. oil switch with overload release.
3 current transformers.
Abt. 377] TRANSMISSION AND DISTRIBUTION
351
Ammeter
Wattmeter
377. Instrument Transformers. — The instruments in circuits
with a voltage of 2300 volts or greater are not connected directly
in the circuit but are connected through transformers as shown
diagrammatically in Fig. 421.
r is a potential transformer and is built exactly like a standard
lighting transformer but on a
smaller scale. The voltmeter V
measures the secondary voltage
but is calibrated in terms of the
voltage of the primary side.
The series transformer has the
primary side connected directly in
the line and the secondary short
circuited by an ammeter or by the
current coils of a wattmeter.
Since the secondary ampere turns of a transformer are always
equal in number to the primary ampere turns therefore nji =
nil I or 1 2 = — /i so that the current measured by the instruments
is proportional to the current in the line.
Voltmeter.
Fia. 421. — Connections of in
strument transformers.
CHAPTER XLII
ELECTRIC LIGHTING
A hot body gives off radiant energy in the form of heat, light
and chemical energy and, to maintain the temperature of the
body, energy must be given to the body at the same rate as it is
dissipated by the body. An illuminant should have as much of
the total radiation as possible in the form of light; power is
required to maintain the other radiation but no light is obtained
from it.
As the temperature of a body is raised, the color of its light
changes from red to white, while its light efficiency increases
rapidly, so that high temperature is a necessary condition for an
illuminant of the incandescent type.
378. The carbon incandescent lamp consists of a filament of
carbon enclosed in a glass globe from which the air has been ex
hausted. When the temperature of such a filament reaches about
1600° C, the rate of evaporation of the carbon becomes excessive
and the life of the filament becomes short. The life of such a
lamp is the number of hours the lamp will bum before its candle
power drops to 80 per cent, of the original value; the decrease in
candle power is due lai^ely to evaporation of carbon from the
filament, this carbon deposits on the inside of the globe and
blackens it.
If the voltage applied to a carbon lamp increases, the current
increases, and so also does the temperature of the filament and
the efficiency of the lamp, but the life is reduced. A standard
16 candlepower, 110 volt lamp takes 60 watts, or 3.1 watts per
candlepower, with a life of about 600 hours; if the voltage is
increased 2 per cent., the efficiency is increased about 7 per cent,
and the life is reduced about 40 per cent.
379. The Tungsten Lamp has a filament of metallic tungsten.
The temperature of such a filament can be maintained at about
200Q° C, which is higher than the operating temperature of a car
bon filament, so that the tungsten lamp is the more efficient, taking
only 1.2 watts per candlepower, and gives the whiter light. The
352
Art. 380]
ELECTRIC LIGHTING
353
tungsten lamp is the more fragile of the two and the life of the
lamp is not limited by a decrease in candlepower, but by the
wear of the filament, which causes it to break after about 1000
hours of service.
A lowvoltage lamp is more robust than a high voltage lamp
of the same candlepower, because the filament is shorter and of
larger cross section, since it has to carry a larger current with a
smaller voltage drop.
One important diflference between tungsten and carbon is that,
while the temperature coeflScient of resistance of the former is
positive, that of the latter is negative. Because of this, the
tungsten lamp is less sensitive than the carbon lamp to voltage
fluctuations. If for example the voltage is increased by k per
cent., the corresponding increase of current in the tungsten lamp
will be less than k per cent., because the resistance of the filament
increases, while the increase of current in the carbon lamp will be
greater than k per cent, because the resistance decreases. The
effect of an increase in voltage is shown in the following table:
Voltage
Candlepower
Watts per candle
power
T.ifft
Normal or 100%
102%
98%
100%
111%
107%
90%
93.3%
100%
93%
96.3%
106%
103.7%
100%
60% carbon
76% tungsten
147% carbon
125% tungsten
Because of the positive temperature coeflScient of resistance,
the tungsten lamp has a much lower resistance when cold than
when hot, so that, when the lamp is switched on, the initial current
is several times as large as the normal operating current. This
result, called overshooting, reduces the life of a lamp if it is
switched off and on frequently; for sign lighting, low voltage
lamps are used since they have a stouter filament than standard
110 volt lamps.
380. Gasfilled Tungsten Lamp. — If tungsten is heated in an
atmosphere of nitrogen instead of in a vacuum, the temperature
at which evaporation becomes excessive is higher in the former
case than in the latter. Such nitrogenfilled lamps therefore have
a high eflBciency and, in large sizes, take only 0.6 watts per candle
power. Since the globe contains gas, convection currents are set
354 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xui
up when the lamp is lit, and the globe has the comparatively high
temperature of about 200° C.
To obtain high efBiciency from this lamp, it is found that the
filament must be large in cross section; a filament to carry 20
amp. at 10 volts gives 2 candlepower per watt while a thinner
filament to carry 5 amp. at 40 volts gives only 1.4 candlepower
per watt. The lamp at its best efficiency cannot be made at
present for less than about 350 candlepower, in which size it is
suitable for the lighting of large areas at present lit by arc lamps
of lower efficiency.
381. The Unit of Light. — ^The light giving power of a lamp is
expressed in candle power. This, however, has little meaning
Fia. 422. — ^Light distribution in a
vertical plane, from a 32 candlepower
tungsten lamp.
Fia. 423.— One type of
arclamp mechanism.
unless the direction of the light is specified, and curves such as
that in Fig. 422 are used for this purpose. This curve shows the
distribution of light about the vertical plane of a tungsten lamp,
and is obtained by measuring the candle power in different direc
tions and then plotting along each radius a length proportional to
the candle power in that direction.
Incandescent lamps are generally rated in mean horizontal
candle power. Thus the lamp on which the curve in Fig. 422
was taken would be rated at 32 candle power although the average
candle power in all directions, called the mean spherical candle
power (m.s.c.p.) is only about 24 m.s.cp.
382. Arc Lamps. — If two sticks of carbon connected in an
electric circuit as shown in Fig. 423 are brought into contact, a
Art. 384] ELECTRIC LIGHTING 355
current will flow through the circuit and, since the contact be
tween the carbons is poor and the resistance of the contact is
therefore high, the carbons at the contact begin to glow, while a
small quantity of carbon vapor passes between them.
If the carbon contacts are now separated by about a quarter of
an inch, it will be found that the current still flows, because the
space between the contacts is filled with carbon vapor which is
conducting. The arc so formed is a powerful source of Ught.
An arc lamp consists of two sticks of carbon, with a mechanism
which, when the voltage is applied, brings the carbons into con
tact and then separates them, and which also feeds the carbons
together as they are consumed.
One of the many types of mechanism for this purpose is shown
diagrammatically in Fig. 423. The upward pull of the shunt
coil S tends to bring the carbons together, and this pull increases
with the voltage E) the upward pull of the series coil L increases
with the current I and tends to separate the carbons. When the
main switch is closed, a large current I passes through the coil L
while the voltage E is comparatively small, so that the pull of L
is greater than that of S and the carbons are separated. As
they separate, E increases and I decreases and, when the arc
has reached the proper length, then E and I have their normal
values and the pulls balance.
383. The directcurrent open arc takes the form shown in
Fig. 425. The temperature of the positive tip is about 3700° C,
the temperatures of the arc stream and of the negative tip are
much lower. Of the total light from such an arc, 85 per cent,
comes from the crater, 10 per cent, from the positive tip and
6 per cent, from the arc stream. The distribution of light from
such an arc is shown by the polar curve in Fig. 424; directly below
the arc the illumination is practically zero due to the shadow cast
by the lower carbon.
Because of the high temperature, the light is white and the
efficiency is high, but the life of the carbons is only about 10 hours.
384. Directcurrent Enclosed Arc. — To cut down the trimming
expense, the arc is enclosed in a globe which is almost air tight,
so that after the first few seconds the arc is operating in an atmos
phere of carbonic acid gas, and the carbons are consumed more
slowly than in an open arc and have a life of about 100 hours.
The arc is operating under slight pressure and, for satisfactory
operation, the arc stream is longer than that of the open arc.
35
356 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xui
The crater is not now so pronounced, and a greater portion of the
total light now comes from the arc stream, so that the light dis
tribution in a horizontal direction is improved. Data on this
arc is given in the table on page 360.
386. Alternatingcurrent Enclosed Arc. — When operating
with alternating current, the carbon arc lamp does not go out
when the current passes through the zero value, because there
is enough heat in the carbon tips to maintain the arc stream while
the current reverses. Alternatingcurrent arcs have no crater,
and each tip is equally hot but is not so hot as the crater of a
directcurrent arc, so that the efficiency is lower, moreover the
light is not directed downward but is directed horizontally, so
that a reflector has to be supplied to deflect the light in the
direction shown in Fig. 424.
VgW
Crater
Fia. 424. — flight distribution of arc lamps.
Fia. 425. — Shape of a
direct*K5urrent arc.
386. Flame Arc Lamps. — The positive carbon of the direct
current flame arc lamp and either or both carbons of the alter
nating current flame arc are impregnated with salts which have
high selective radiation. Such salts when heated give most of
their radiation in one particular part of the spectrum. Being
selective, the portion of the total radiation which is given oflf as
light is greater than that given oflf by an ordinary incandescent
body at the same temperature, such lamps are therefore very
efficient, and a large portion of the light comes from the arc
stream. Calcium salts are often used and give a yellow light.
Barium salts give a white light, but the white flame arc is not so
efficient as the yellow flame arc.
Art. 388J ELECTRIC LIGHTING 367
The open flame arc requires daily trimming. K the arc is
enclosed so as to limit the supply of air, the life of the carbons may
be increased to 100 hours without any marked reduction in the
efficiency. The enclosing globe, however, must be kept free
from soot, by arranging that the fumes shall be carried away and
allowed to deposit in a condensing chamber and not on the sides
of the enclosing globe.
387. Luminous Arc Lamp. — This is a low temperature arc
which depends entirely on selective radiation for its efficiency.
It is essentially a directcurrent arc and has a positive electrode
of copper and a negative electrode of magnetite. Magnetite
boils at a much lower temperature than carbon, and the tempera
ture of the arc is not high enough to melt the copper or to enable
the arc to be used on alternating current; the arc goes out as the
current passes through zero.
The efficiency is not so high as that of the flame arc. The light
is white, and the magnetite electrode has a life of about 160 hours
while that of the copper electrode exceeds 1000 hours. The mag
netite arc must never be connected with the polarity reversed
because, if the copper were made the negative electrode, a copper
arc would be produced instead of a magnetite arc, since the mate
rial of the arc stream comes from the negative electrode.
For alternatingcurrent circuits, titanium arcs are being devel
oped, which have an efficiency comparable with that of the flame
arc.
388. Mercury Vapor Converter. — To obtain directcurrent
from an alternatingcurrent supply for the operation of magnetite
arcs, the mercury vapor converter is used.
The globe shown in Fig. 426 is filled with mercury vapor. It
is found that a very high voltage is required to start an arc
between a and 6 but about 14 volts is all that is required to main
tain the arc; there is a high resistance at the negative electrode
which resistance is broken down when current is flowing, but
is reestablished as soon as the current ceases to flow. Alter
nating current cannot pass between the electrodes because the
high resistance is reestablished as the current passes through the
zero value.
To operate as a converter, the globe is fitted with three elec
trodes and is connected up as shown in Fig. 428, and means are
provided whereby the resistance of electrode c is kept broken
down. At the instant shown in Fig. 428, the electrode a is posi
358 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xu;
tive and that of 6 is negative, so that current passes from a to
c but no current can pass from a to 6 or from c to 6, because of the
high resistance of negative electrode b. Half a cycle later b is
positive and a is negative, so that current can pass from 6 to c but
no current can enter a. The current in the line L therefore flows
always in one direction or is a direct current.
There is a small reactance coil S, called a sustaining coil,
placed in the line L to carry the rectifier over the point of zero
current. This coil causes the current to lag slightly behind the
e.m.f . so that there is still a small current flowing from a, for exam
ffingle PliMO Malm
B«ACtftace Ooll
single Phue Mal&i
nm
FiQ, 426. Fig. 427. Fig. 428.
Figs. 426 to 428. — The mercuryvapor converter.
pie, when the arc from b is ready to strike. The current entering
c therefore never becomes zero and the high negative resistance of
that electrode is always broken down.
Before such a converter can be started, the resistance at c must
first be broken down. To accomplish this result an additional
starting electrode is placed at w and current is passed between
w and c by tipping the globe until the mercury forms a bridge
between these two electrodes. The tube is then raised and an
arc is drawn between w and c for half a cycle, which breaks down
the resistance of c long enough to allow the arc to start from b
and thereby start the rectifier in operation.
Abt. 390]
ELECTRIC LIGHTING
359
389. Mercury Vapor Lamp. — ^The converter described above
is an arc lamp and, since the light is due to the selective radiation
of mercury vapor, it is a high efficiency lamp. Unfortunately
the color is greenish blue and gives ghastly color effects.
filngle Phase
Mains
Eeactaace
Coll
No fieflector
Fig. 429. — Mercuryvapour lamp for operation by alternatingcurrent.
The alternatingcurrent lamp is made in the form shown in
Fig. 429, which is lettered similarly to Fig. 427. To start the arc,
the lamp is tipped so that the
mercury runs down and forms a
metallic connection between the
two electrodes through which a
current flows, the lamp is then
allowed to return to its original
position, the mercury thread is
ruptured, and an arc follows.
The directcurrent lamp is
supplied with two terminals and
is started in the same way.
When a quartz tube is used
instead of a glass tube, the
lamp may be shortened, and run
at a higher temperature with a
higher efficiency.
390. Shades and Reflectors.
— The light distribution from a Fia. 430. — ^Light distribution of
«^,,««^««*.u^^^««^i^4.^i„^u«*»«^^ a 32 candlepower tungsten lamp
source can be completely changed equipped witt different reflectors.
by the use of a shade or reflec
tor. Curve A, Fig. 430, shows the light distribution from a tung
sten lamp while curves B, C and D show the distribution when
B Extenilre
Beflector
nteniire
Beflector
D Focaiing
^ Beflector
360 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xmi
diflferent types of reflectors are used. Curve B is obtained with
what is called an extensive reflector, curve C with an intensive
reflector, and curve D with a focusing reflector.
When the distance between the lamps is at least twice the height
of the lamp above the ground, the extensive type is used; it is
suitable for a small room Ut by a source placed on the centre of
the ceiling. When the distance between lamps is about one and
a half times the mounting height, the intensive type is preferred;
it is suitable for a large room Ut from several points on the ceiling.
The focusing type is used when the lamps are more closely spaced
and is suitable for the illumination of desks, show cases, etc.
391. Efficiency of Dluminants. — Since the light from a source
can be suitably directed by means of reflectors, light eflBciency
should be based on the mean spherical candle power of the lamp.
Values attained in practice are given below^
Mean
spherical
candle
power per
watt
Candle
power at
l(f angle
per watt
Available
size in
mean
spherical
candle
power
Incandescent lamps
Ordinary carbon filament
Tungsten filament
Gas filled tungsten
Arc lamps
Enclosed carbon
6.6 amp., 450 watts A.C.
6.6 amp., 480 watts D.C.
Flame carbon 500 watts, yellow
300 watts, yellow
500 watts, white
D.C. magnetite 4 amp., 300 watts
6.6 amp., 500 watts
A.C. titanium 220 watts
Mercury lamps glass tube
quartz tube
0.21
0.64
1.28
0.39
0.62
3.1
1.95
1.95
1.0
1.5
1.9
1.55
2.0
0.4
1.25
2.5
0.5
1.0
6.2
4.0
4.0
2.2
3.2
4.0
any size
any size
above 36C
175
300
1550
585
975
300
750
420
These figures take account of the loss in the reflectors, and the
candle power at 10° angle below the horizontal is that obtained
when the lamp is equipped with a reflector suitable for street
lighting.
^ Efl&ciency of Illuminants by C. P. Steinmetz, General Electric Review,
March, 1914.
Art. 392]
ELECTRIC LIGHTING
361
392. Light and Sensation. — ^The eye is able to see objects
distinctly over a range of intensity of 1,000,000 to 1 as determined
by the exposure of a photographic plate. The size of the pupil is
controlled automatically by the intensity of the light and decreases
as the light intensity increases so as to limit the amount of light
that can enter the eye. The sensibility of the optic nerve also
changes automatically with the light intensity but at !a much
slower rate; when one goes from daylight into a darkened room,
for example, objects that at first are invisible become quite
distinct after a time.
Ultra Bed
Ultra Violet
Fig. 431. — Effect of color on the light efficiency.
The Ught sensation produced by a given amount of radiant
power entering the eye depends on the color of the hght. White
light is composite and, when passed through a prism, as shown in
Fig. 431, is divided up into its constituents and produces a band
of color varying from red to violet called the spectrum.
The relation between candlepower per watt and color is
shown in Fig. 431; with a high light intensity the eye is most
sensitive to yellow light while with a low intensity it is most sen
sitive to green light. If for example a mercury vapor lamp and
a yellow flame arc burning side by side have the same brightness
at a moderate distance from the observer, then, when the lamps
are close at hand the flame arc appears the brighter since the light
362 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xmi
intensity is high, but when the lamps are a considerable distance
away the intensity is low and the green mercury lamp appears
the brighter. A yellow light is therefore the most efBicient for
high intensity illumination and a green light the most efBicient
for low intensities; these colors however are often objectionable.
There is one important exception; yellow light is used for fog
signal work because it penetrates fog better than does green light.
393. Reflection and Color. — When light strikes an opaque
body, some of the light is absorbed and some is reflected. The
color of an opaque body is the color of the light which it reflects,
a green opaque body is one which absorbs all but the green rays.
When light strikes a transparent body, the color of the body as
seen from the side away from the source is the color of the light
which is transmitted; a green glass transmits only the green rays,
whereas a transparent body transmits all the light that falls on it,
a colored globe on a lamp therefore cuts off light and reduces the
efficiency.
The color of a body depends on the color of the light that
strikes on it. An opaque body which is yellow in daylight reflects
only yellow rays; in the green light of the mercury vapor lamp
this body looks black because it absorbs the green light and there
is no yellow to reflect. It is important therefore that colors be
matched in the light in which they are going to be used.
The color of the light in a room is not always the color of the
source. If daylight enters a room which has walls of a green color,
then the bulk of the light that strikes the walls is absorbed while
the green rays are reflected and a green tint is added to the light
in the room. Dark walls and ceilings result in low efficiency of
illumination. In an ordinary room, about 70 per cent, of the
light from the source strikes the walls and, if the paper is dark,
most of this light is absorbed and lost, if the paper is light yellow
in color, a large part of the light is reflected and is useful.
Red wall papers and table cloths are to be avoided in a reading
room because, under such circumstances, a large part of the light
that enters the eye is red light and, for a given distinctness of
vision, a larger amount of energy must enter the eye if the light
is red than if yellow or white, see Fig. 431, page 361, and the
excessive amount of energy is harmful to the eye tissues.
394. Principles of Illumination. — For satisfactory illumination,
the light should be of good quality, glare should be avoided, and
the shadows should be distinct so as to give good perspective.
Art. 397] ELECTRIC LIGHTING 363
396. Quality of the Light. — The light should be as nearly white
as possible. Most of the harm done by artificial illumination is
due to the red and ultra red rays for the reason just pointed out.
Green light also is harmful under certain circumstances because
it is found that the pupil fails to respond to variations in intensity
of light of this color. Green light should therefore not be used
for high intensity illumination, as for example for the lighting of
drawing oflSces, but, as pointed out on page 362, it is particularly
suited for low intensity illumination. Ultra red and ultra violet
radiation is the most harmful and there is much of the latter in
arc lamps, so that an arc should always be enclosed in a glass
globe since glass is opaque to such radiation.
Flickering light is bad for the eye because the pupil tries to
adjust itself to the rapidly varying intensity and becomes fatigued.
Alternating current lighting at 26 cycles has not been uniformly
successful for this reason.
396. Glare. — ^The eye cannot look with comfort on near objects
which have a higher surface intensity than 4 candle power per
square inch, so that the sources of illumination should be kept
out of the range of vision, and direct reflection into the eye from
such sources should also be avoided. Side Ughts and Ught from
below are particularly objectionable since the eye is not protected
from such light.
If the source of illumination cannot be kept out of the range
of vision, then frosted incandescent lamps should be used or the
lamp supplied with a shade or globe that will keep the direct
rays away from the eye.
It is impossible to see distinctly past a bright light and for
that reason country roads should not be lit by powerful arc lamps
spaced far apart because, not only doe^ most of the light go to
illuminate the adjoining fields, but the arcs are generally hung
so low to clear the trees that it is impossible to see past them, and
driving on such roads is dangerous. Incandescent lamps of
moderate candle power, spaced closer together, give better illu
mination.
397. Shadows. — In certain cases as for example for the illu
mination of drafting rooms it is desirable to eliminate shadows.
This result is obtained by the use of a large number of light
sources, or by the use of indirect methods of lighting whereby
a reflector is used to throw all the light on to the ceiling from
which it is reflected on to the drawing tables.
364 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xlii
For other classes of work, such as street lighting, diffusion is to
be avoided since, due to the elimination of shadows, there is loss
of perspective and obstacles are not clearly seen.
398. Intensity of Illumination. — The unit of light intensity
is that produced by a source of one candlepower at a distance of
1 ft. and is called the footcandle.
In Fig. 432, the same amount of light strikes the surfaces
A, B and C, so that the intensity at B is less than that at A in
the ratio Or^/Wy or intensity is inversely proportional to the
square of the distance. Again the intensity on surface C is less
than that at B in the ratio area B/ area C = cos a.
Q 10 20 80 40 50 Candle Power
Fig. 432. — Variations of light intensity with the distance from the source,
and with the angle of incidence.
Curve D, Fig. 432, shows the light distribution curve of a 40candle
power tungsten lamp equipped with a reflector for street lighting. It is re
quired to determine the illumination at a point X, 50 ft. from the post, the
height of the lamp being 12 ft.
The candlepower in direction OX = 51
The distance OX = V50* + 12« = 51.5 ft.
The intensity on an obstacle normal to the light
51
(51.5)«
The intensity on the street surface = 0.0192 X cos a
12
:" 0.0192 ft. candle
= 0.0192 X
51.5
= 0.0045 ft.candles.
The minimum intensity on the street surface due to two lamps spaced
100 ft. apart = 2 X 0.0045 = 0.009 ft.candles. For country roads, the
minimum intensity on an obstacle normal to the light should not be less
than 0.02 ft.candles.
399. Lines of Illumination. — It is convenient to represent the
total light from a source by lines of illumination, called lumens,
Art. 400] ELECTRIC LIGHTING 365
such that the number crossing unit area placed perpendicular to
the direction of the light is made proportional to the light in
tensity. One footcandle is represented by one line per sq. ft.
If a source of one candlepower were surrounded by a sphere
of 1 ft. radius, the intensity at the surface of the sphere
would be 1 ft.candle, there must therefore be one line per sq.
ft. or a total of 4x lines.
To provide for a surface intensity of I ft.candles over an area
of A sq. ft.
The number of lines of illumination = J X A
J X A
The candlepower of the source = — 7 —
Since some of the light is absorbed by walls and ceilings
the necessary candlepower = . ,
where k = rT^\ — is less than 1 ; average values are
total light
fc = 0.6 if clear reflectors are used and the room has light
walls and ceiling.
= 0.4 with clear reflectors and dark walls and ceiling.
Detennine the candlepower required to give a light intensity of 3 ft.
candles over a room which is 40 ft. long and 30 ft. wide the room having
light walls and ceiling.^
rpu A^ 3 X 40 X 30 ._ „ . ..
The necessary candlepower = . ^ ^ ^ — = 477 candlepower. A 40
watt lamp gives 40/1.25 — 32 horizontal candlepower and 24 mean
477
spherical candlepower, see page 354, so that ot = 20 lamps of 40 watts
each are required, or a smaller number of lamps of larger candlepower. The
choice of the candlepower of each lamp, and the spacing of the lamps, is a
matter that must be left to the judgment of the individual; there are many
examples of good and of bad illumination to be found in every city.
In the above problem
The total power required = 20 X 40 = 800 watts
The watts per sq. ft. = 800/(40 X 30) « 0.75.
400. Power Distribution for Lighting. — Interior lighting is
generally carried out at 110 volts, either alternating or direct
3urrent, the lamps being connected in parallel across the circuit.
When arc lamps are connected across such a circuit, a steady
ing resistance must be placed in series with the arc as shown in Fig.
433. Suppose an arc is carrying a current I with a constant
^ A list of economical intensities for all classes of work will be found in
the American Electricians' Handbook by Terrell Croft.
366 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xlii
voltage E at the terminals. If the current / were to increase for
an instant, more carbon vapor would leave the negative electrode,
the arc stream would become more conducting, and the current
/ would increase still further, so that an arc is unstable when used
I
T
Conitant
V^
I
Fia. 433. — Directcurrent. Fio. 434. — Alternatingcurrent.
Figs. 433 and 434. — Multiple arcs.
on constant potential. If, however, a resistance is placed in
series with the arc, as shown in Fig. 433, then an increase in the
current / causes the voltage Ei = IR to increase, and therefore
the arc voltage E^ to decrease so that it cannot maintain the in
creased current across the arc.
[13
a
Q i Q Q
VWv%
ConiUmt Oamnt
Traniformet
Fig. 435. — Alternatingcurrent system.
Hercary Yapoar
CoiiTerter
Fig. 436. — Directcurrent system.
Series connection of lamps.
In the case of alternatingcurrent parallel arcs a steadying
reactance is used, because it consumes very little power and still
reduces the voltage. The same result may be obtained by sup
plying the lamp through an autotransformer since the sec
ondary voltage E^) Fig. 434, decreases with an increase of the arc
current I^.
If the lamps can be used in a series circuit, as shown in Fig.
436, then the wire has to carry the current of only one lamp and
Art. 400] ELECTRIC LIGHTING 367
may be small in cross section. This system of distribution is
largely used for street lighting, the constant current required for
the operation of the arcs being obtained by means of a constant
current transformer, see page 267.
By means of a series transformer s it is possible to connect a
circuit of tungsten lamps taking a large current in series with a
main circuit of arc lamps taking a smaller current.
For the operation of magnetite arcs, direct current is necessary
and this is obtained by means of a mercury vapor converter con
nected as shown in Fig. 436.
The voltage between electrodes should be about:
47 volts for an open carbon arc,
72 volts for an enclosed carbon arc,
45 volts for an open flame arc,
78 volts for open magnetite arc.
CHAPTER XLIII
LABORATORY COURSE
401. Protectioii of Circuits. — ^A typical circuit is shown in
Fig. 437. The lamps L take power from the mains and the
current 7 is measured by the ammeter A while the voltage E is
measured by the voltmeter V, see page 19.
All the connections should be made before the switch S is
closed, and the circuit must be protected by fuses F, see page 117,
or by an automatic circuit breaker, see page 39, set so as to
open the circuit if the current should become large enough to
injure any of the apparatus connected in the circuit.
To make sure that the instruments are reading in the proper
direction, the switch S should be closed for an instant and then
^9( ,z
t I Ammeter' I I 
F^
"" S
Fig. 437. — ^Typical circuit.
opened before any appreciable current has had time to flow.
If the needle moves in the wrong direction, the instrument
connections must be reversed.^
402. Ammeter Shtmts. — ^Most ammeters are wound with
very fine wire, see page 8, and can carry only a small fraction
of an ampere without being burnt out. To use such instriunents
for the measurement of large currents they must be connected
in parallel with shunts as shown in Fig. 437. The current in
the instrument, and therefore the deflection, will be proportional to
the line current I so that, if the shunt and instrument are always
used together, the scale of the instrument may be in terms of the
line current to be measured and not in terms of the current in the
instrument. For instruments with a range of less than 5 am
peres, the shunt is generally placed inside the case.
403. Safe Carrying Capacity of Copper Wires. — If too large
a current flows in an insulated wire, the insulation will be
* A safer method is to provide the ammeter with a short circuiting switch
that is normally kept closed and is opened when it is desired to take a
reading.
368
Abt. 404]
LABORATORY COURSE
369
damaged. The values given in the following table should not
be exceeded.
Size of wire, Brown
& Sharpe gauge
number
Diameter of wire
in inches
Maximum current in the wire in
amperes
Rubber
insulation
Other
insulation
14
0.064
12
16
12
0.081
17
23
10
0.102
24
32
8
0.128
33
46
6
0.162
46
66
5
0.182
64 ,
77
4
0.204
66
92
3
0.229
76
110
2
0.268
90
131
1
0.289
107
166
0.326
127
185
404. Control of the Current in a Circuit. — To vary the current
flowing in the coil C, Fig. 438, an adjustable resistance R is
inserted in the circuit, see page 25. This rheostat must be
able to carry the current / without overheating.
R
Fio. 438. Fig. 439.
Figs. 438 and 439. — Methods of controlling the current in a circuit.
If the resistance of the coil C is large compared with that of
the resistance R, then the current variation will not be very
large. Under such circumstances, the potentiometer connection
shown in Fig. 439 is to be preferred for experimental work. If
the movable contact a is placed at b then the voltage Ei is equal
to E, the line voltage. If contact a is placed at c, then the
voltage El is zero. By moving the contact between these two
points, the voltage Ei and the current / may have any value from
zero to Bi = ^ and I == E/Rc. If the resistance Re is small and
the contact a is close to b, as shown in Fig. 439, then the current
in ab may become dangerously large, so that the rheostat must
370 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xuii
i^^^^^
Yoltmeter Q
FlO. 440.
be watched during operation and the circuit opened if the
rheostat becomes too hot.
EXPERIMENT 1
Object of Experiment. — To determine the resistance of the
shunt field circuit of a directcurrent machine with different
values of current in the circuit.
Reference. — Pages 20 and 95.
Connections. — If the connection in Fig. 440 does not give
sufficient range of current, then use the connection shown in
Fig. 439.
Readings. — ^Volts and amperes.
Report — Describe the method of
carrying out the experiment and embody
Field ^^® answers to the following questions in
Goiu the report.
1. Find the resistance of the field coil
circuit.
2. How does temperature affect this
resistance? Show the result experi
mentally by measuring the resistance after the current has
been flowing for 30 minutes.
3. What per cent, of the output of the machine is the excitation
loss? The machine output is given on the name plate.
4. What range of instruments would you use to determine
the resistance of the shunt field circuit of a 50 kw., 240 volt
generator if the excitation loss is known to be less than 4 per
cent.?
EXPERIMENT 2
Object of Experiment. — To determine the resistance of the
armature circuit of a directcurrent machine with different values
of. current in the circuit.
Connections. — See. Fig. 438, page 369. The rheostat used
to control the current must be able to carry the fullload armature
current of the machine without injury.
Readings. — Volts across the terminals; volts across the com
mutator, obtained by attaching the voltmeter leads to the seg
ments which are under the brushes; amperes.
Report. — Describe the method of carrying out the experiment
and embody the answers to the following questions in the report:
Art. 404] LABORATORY COURSE , 371
1. Plot the armature resistance against current.
2. Plot the brush contact resistance against current.
3. What per cent, of the output of the machine are the armature
resistance loss and the brush contact resistance loss at fullload?
4. Has the pressure on the brushes much effect on the brush
contact resistance? try the experiment. What do you imagine
limits the brush pressure?
5. What range of instruments would you use to determine the
resistance of the armature circuit of a 50kw., 240 volt generator if
the loss in the complete armature circuit is known to be less than
4 per cent, at fullload?
EXPERIMENT 3
Object of Experiment. — To find how the speed of a direct
current shunt motor at noload varies with:
a. The exciting current; armature voltage being constant.
6. The armature voltage; exciting cur
rent being constant. "^^^^^
References. — Pages 85 and 89. ^ ^
Connection. Fig. 441. i
Readings. — Exciting current; arma ^ ^
ture voltage; speed. yiq, 441.
Curves a. Speed and exciting current.
6. Speed and armature voltage.
Questions. — 1. Explain the shape of the curves, without
formulas.
2. What would happen if, in starting up a shunt motor, a
starting resistance was not placed in series with the armature
circuit? How many times fullload current would flow through
the armature under these conditions? (Use the value of armature
resistance determined in the last experiment.)
3. What is the back e.m.f . of a motor and what relation has it to
the applied e.m.f.? Perform the experiment described in Art.
88, page 79, to answer this question.
4. What would happen if, during operation, the field coil
circuit were to open (do not perform this experiment) ?
5. Draw a diagram of connections of any starter in the labo
ratory, which has a novoltage release and also overload pro
tection.
EXPERIMENT 4
Object of Experiment. — To find how the voltage of a direct
current generator at noload varies with;
372 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xun
a. The exciting current; speed being constant.
6. The speed; exciting current being constant.
Reference. — ^Page 70.
Connection. — See Fig. 95, page 70.
Readings. — ^Voltage, exciting current and speed.
Curves a. Voltage and exciting current.
b. Voltage and speed.
Questions. — 1. Why is there a small voltage even with no
exciting current?
2. Explain the shape of the curves.
3. How would you reverse the polarity of the generator, t.e.,
how reverse the direction of the voltage?
EXPERIMENT 5
Object of Experiment. — To determine how the terminal volt
age of a constant speed generator varies with the load :
a. The generator being separately excited;
6. The generator being shunt excited;
c. The generator being compound excited.
References. — Pages 71 to 77.
Connections.— See Fig. 97, page 72; Fig. 98, Fig. 99.
Readings. — ^Terminal voltage, line current, exciting current
and speed.
Curves. — Terminal voltage on a line current base.
Shunt exciting current on a line current base.
Questions. — 1. Why does the terminal voltage of a separately
excited generator decrease with increase of load? How much of
the drop at fullload is due to armature resistance?
2. Why is the voltage drop greater in a shunt than in a sepa
rately excited generator?
3. What would be the efiFect of shifting the brushes further for
ward from the neutral position?
4. What is the principal advantage of the compound generator?
5. How would the terminal voltage of a compound generator
vary with increase of load if thf^ series field coils were connected
so as to oppose the shunt coils?
6. If the generator is overcompounded while flat compounding
is desired, what can be done to fix the machine?
7. If the voltage of a shunt generator builds up when the
generator rotates in a given direction, why will it not build up if
the direction of rotation is reversed? Try the experiment.
Abt. 404] LABORATORY COURSE 373
EXPERIMENT 6
Object of Experiment. — To determine the efficiency and also
the speed and torque characteristics of shunt, series, and com
pound motors, by loading the machines by means of a brake,
the applied voltage being constant.
References,— Chaps. XV, XVII and XVIII.
Connections— Fig. 107, page 86, Fig. 113 and Fig. 117,
Readings. — Applied voltage (constant), armature current,
shunt exciting current (constant), speed, brake reading.
Curves. — Speed and torque on an armature current base.
Brake horsepower and total input on an armature current base.
Efficiency on a brake horsepower base.
Questions. — 1. How would you reverse the direction of
rotation of each machine?
2. For what type of service is each machine suited?
3. The field coils of a machine become hot during operation,
what effect will this have on the noload and on the fullload
speed of each type of motor?
4. How can the speed of each type of motor be varied for a
given load?
5. Why is the speed regulation of a shunt motor poor when
the speed is controlled by a resistance in the armature circuit?
Try the experiment.
6. Explain without formulae, the shapes of the speed, torque,
and efficiency curves.
EXPERIMENT 7
Object of Experiment. — To determine the relation between
starting torque and armature current in a shunt, a series, and a
compound motor.
References. — Pages 85 and 90.
Connections. — Same as for experiment 6, with resistance in
the armature circuit to limit the current (do not use a starting
box for this purpose).
Readings. — Armature current and brake reading.
Curves. — Torque on an armature current base.
Questions. — 1. How does the starting torque compare with
the running torque, as determined in experiment 6, for the same
armature current? Does theory indicate that there should be a
difference?
2. Why is the series motor preferred for heavy starting duty?
374 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xliii
EXPERIMENT 8
Object of Experiment. — To measure the stray loss, the
armature copper loss, and the excitation loss in a shunt motor,
and calculate the efficiency from these figures.
Reference. — Chap. XVI, page 95.
The work of this experiment should be done without instruc
tion. A diagram of connections should be drawn out and the
range of the necessary instruments determined before the
apparatus is connected up.
Report. — Describe the method of carrying out the experi
ment. Plot the efficiency curve on a horsepower output base
up to 25 per cent, overload, and compare this curve with that
obtained by brake readings in experiment 6; if the results show
considerable difiference, which would you consider to be the more
reliable?
EXPERIMENT 9
Object of Experiment. To run a shunt generator in parallel
with the power house and determine the temperature rise of the
machine at fullload.
References. — Pages 21, 99 and 163.
Connection. — Pig. 179, .page 164.
Readings. — Measure the resistance of the field coil circuit
at the beginning of the test and every ten minutes thereafter, take
also readings of temperature of the field coil surface at the same
time. Find the temperature of the armature core and armature
winding immediately the generator is shut down; the heat run
should last for at least an hour and a half.
Curves. — Observed temperature rise of field coil surface, and
also the temperature rise determined by resistance measurements,
on a time base.
Questions. — 1. Explain the shape of the curves.
2. Why is the temperature rise of the field coils as determined
by resistance measurements greater than that determined by
thermometer?
EXPERIMENT 10
Object of Experiment. — To determine the voltage regulation
of a three wire system.
References.— Pages 316, 317 and 338.
Connections. — Use a balancer set, see Fig. 374, or a three wire
i
Abt. 404]
LABORATORY COURSE
375
generator. If these are not available, a singlephase rotary
converter with a suitable autotransformer may be connected
up as shown in Fig. 442 to form a three wire generator.
Readings. — Vary the loads on the two sides of the system from
perfect balance, when Ai — Ai and An is zero, to a maximum
of unbalance when the load on one side of the line is zero. Take
readings of Vi, Fj, Ai, and As; Fi + V2 to be kept constant.
Curves. — Plot Vi and Vi against
the unbalanced current A1A2.
Questions. — 1. Explain the action
of the apparatus used.
2. What are the advantages of the
three wire over the two wire system
of distribution?
3. Explain the shapes of the curves. Fia« 442. ,
VyVt
/ wv/ywv
■a
Fuse Wire
EXPERIMENT 11
Object of Experiment. — To determine the characteristics of
fuse wire.
Reference. Page 117.
Connection. Fig. 443.
Readings. — Length of we between blocks, average current,
time taken for fuse to melt after switch S is closed.
Curves. — ^Plot amperes on a time
base for four lengths of fuse wire and
from these curves plot another set
with amperes on a length base for a
fusing time of 10 sec.
Questions. — 1. Explain the shape
of the ciures.
2. If a fuse is rated at 5 amp., about what current would you
expect it to carry continuously, and what current for 30 sec. ?
EXPERIMENT 12
Object of Experilment. — To calibrate a circuit breaker.
Reference. Page 39.
Connections. — Same as for fuse testing.
Readings. — ^Amperes to open, and position of plunger.
Curve. — Draw a current scale that could be attached to the
circuit breaker.
Fio. 443.
T
■VWWNr
376 PRINCIPLES OF ELECTRICAL ENOINEERINO [Chap, xmii
Questions. — 1. Explain the construction and the operation of
the circuit breaker used.
2. What are the advantages and disadvantages of circuit
breakers and fuses, for what type of circuit is each suited?
3. If a 10 amp. fuse, and a circuit breaker set for 15 amp.;
are used to protect the same circuit, which would open first in
the case of an overload on the circuit.
4. If a circuit breaker and a switch are both in a circuit, as in
Fig. 417, page 349, which would you close first?
EXPERIMENT IS
Object of Experiment. — To determine the effect of change of
frequency on the constants of an alternatingcurrent circuit.
Reference. Chaps. XXIX and XXX.
Connection. Fig. 444.
Readings. — Vary the alternator speed and adjust the alter
B ^1 nXi ^fi^or excitation to keep Et conr
'^^^^^^^'^r^^ stant. Take readings of Ery Ei,
sr—^ — El *\* E^ Eey I, watts, and speed. Measure
n the resistance R and the resistance
i&J J of the coil Xi with direct current.
Curves. — On a frequency base,
'^* * plot the inductive reactance Xi =
Ei/If the capacity reactance Xt = Ee/I, the resistance R,
and the power factor ( rrr)
Questions. — 1. For any one frequency, draw the voltage vector
diagram to scale, and compare the value of Et so found with that
found experimentally.
2. Calculate the coefficient of self induction of the coil Xi and
the capacity of the condenser X^.
3. What is meant by resonance? At what frequency does it
occur in this experiment and how does that frequency compare
with the theoretical value / = — r=?
4. Explain the shape of the power factor curve.
5. Under what conditions can the voltage Ei be greater than
the applied voltage Et?
EXPERIMENT 14
Object of Experiment. — To predetermine the characteristics
of a given circuit and compare the results with those obtained by
actual test.
Art. 404]
LABORATORY COURSE
377
Reference. Chaps. XXIX and XXX.
Connection. — Use the same resistance; inductance, and capacity
as in the last experiment and connect as in Fig. 445.
Curves. — Predetermine the values of Er, Ex, Ii, le and I,
with Et, the normal voltage of the alternator, the same as in the
last experiment. Plot these values
against frequency, then determine the
same curves by test and compare the
results.
I
EXPERIMENT 15
E^
Fig. 446.
ITili
Fig. 446. Fio. 447.
Object of Experiment. — To deter
mine the characteristics of a transformer.
References. Chap. XXXIV, page 261.
Connectibns. — A low voltage transformer should be used, con
nected as shown in Fig. 446, or else two like transformers con
nected as shown in Fig. 447 with the high voltage leads taped up.
Readings. — Keep Ei, the secondary power factor W%IEJi^y
and the frequency all ' constant, and
^ ^Hiniild ^^^ readings of Ei, /i, Tf i, E^, U, and
jlL LtLJc — ^2 for values of I^ from zero up to 25
per cent, overload. Measure also the
resistances of the primary and the sec
ondary windings with direct current.
Curves. — Plot efficiency (F'2/F'i), voltage 72, and primary and
secondary power factors against /2.
Questions. — 1. Why is the primary power factor less than that
of the secondary particularly at light loads?
2. Why does the voltage E^ decrease with increase of load, and
why is the voltage drop greater with inductive than with non
inductive load?
3. Calculate the resistance loss in the windings at fullload.
Find also by calculation the fullload efficiency and compare the
result with that determined by test.
4. If the transformer is connected to the line for 24 hours a
day, but carries fullload for only 5 hours a day, find the all day
efficiency.
EXPERIMENT 16
Object of Experiment. — To predetermine the regulation curves
of a singlephase alternator at 100 per cent., 80 per cent, and
378 PRINCIPLES OF ELECTRICAL ENGINEERINO [CHAP.xLin
zero power factors, from noload saturation and shortcircuit
curves, and compare the result at 100 per cent, power factor
with that found by actual load test. (A lamp bank has a power
factor of approximately 100 per cent.)
Reference. XXXII, page 244.
Connections. Fig. 287, page 247.
Readings. — a. Noload saturation; armature volts and field
current at constant speed.
6. Shortcircuit; armature amperes and field current at the
same speed.
c. Load curve; terminal voltage and armature current with
constant field excitation and the same constant speed.
d. Measure the armature resistance with direct current.
Curves. — a. Noload saturation; armature voltage on a field
current base.
b. Short circuit; armature current on a field current base.
c. Armature reactance determined from curves a and b plotted
on a field current base.
d. Load curve; terminal voltage on an armature current base,
by calculation at 100 per cent., 80 per cent, and zero power
factors, and by test at 100 per cent, power factor.
Questions. — 1. Why is the power factor of a bank of in
candescent lamps approximately equal to 100 per cent.
2. Give the theory of the method used to determine the re
actance of the armature of the alternator.
3. Why does the voltage drop more rapidly at low power factors
than at high power factors?
4. Why are alternators rated in kilovoltamperes and not in
kilowatts?
5. How is the voltage of an alternator maintained constant in
practice, at all loads and power factors?
EXPERIMENT 17
Object of Experiment. — To start up a synchronous motor and
determine its running characteristics.
Reference. Chap. XXXIII, page 252, also page 296.
Connections. — Fig. 398, page 258, for single phase machines.
For three phase machines use the connection in Fig. 448.
Method of Starting. — Start up the synchronous motor by
means of the belted directcurrent motor M and adjust the speed
Art. 404]
LABORATORY COURSE
379
of the synchronous motor and its excitation until the voltage E2
is equal to Ei and the synchronizing lamps all remain dark for a
few seconds at a time. (If all the lamps do not become dark at
the same instant then interchange two of the motor leads.)
When all three lamps are dark at the same instant, then close
the three switches Si 82 and Sz at the same time. The starting
motor can then be disconnected by throwing the belt, and the
synchronous motor loaded by means of a prony brake.
Readings. — With the applied voltage E2, the frequency, and
the brake readings all constant; take readings of current / and of
watts (Wi+ W2) for different values of the exciting current // and
for three different settings of the brake.
Bynchronising
Lamp*
Alternator
Synehronoai
Motor
StarCIng
Motor
Fig. 448.
Curves. — Plot armature current, and power factor
(voltTmperes) ' ^^ ^^ ^^^^*^^S current base.
Questions. — 1. Explain the shapes of the curves.
2. What are the advantages and disadvantages of the syn*
chronous motor compared with the induction motor?
3. What efifect has the power factor of the load on the size of
the alternator and on the power loss in the transmission line supply
ing the load?
EXPERIMENT 18
Object of Experiment. — ^To determine the characteristics of a
rotary converter.
Reference. Page 318.
Connection. — On the A.C. end the machine is connected in the
same way as a synchronous motor, see experiment 17, while on the
D.C. end it is connected in the same way as a shunt generator.
380 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xlih
Method of Starting. — The machine may be started up as a
motor from the D.C. end instead of by means of a starting motor,
but it must be synchronized in the same way as a synchronous
motor, see experiment 17, before it is connected to the A.C.
mains.
Readings. — With the applied voltage Et, the frequency, and
£s /s the output of the direct current side all constant, take
readings of the current / and of the watts W for diflferent values
of the exciting current //. Take a complete set of readings for
1/4, 1/2, 3/4 and full load output from the direct current side.
Curves. — Plot armature current, and power factor
/ watts \ x X 1
I — TT ) , on an exciting current base.
wolts amperes/ ' ^
Plot applied voltage Ez and direct voltage Ez on a direct
current (/a) base.
Plot efficiency on a kilowatt output base at 100 per cent, and
at some other power factor.
Questions. — 1. Answer questions 1 and 3 in experiment 17 if
that experiment was not performed.
2. Why is the voltage of the direct current side independent
of the field current, and how can this voltage be controlled?
EXPERIMENT 19
Object of Experiment. — To determine the starting and the
running characteristics of a polyphase induction motor.
References. Chaps. XXXVI and XXXVII, page 283.
Connections. — The student should draw out a diagram of
connections and specify the range of the instruments required.
Readings. — To show the eflfect of increase in applied voltage
on the current, torque and power factor at starting (start with
a low value of applied voltage).
Also readings to show the variation of efficiency, power factor,
current, and speed with horsepower output, the applied voltage
and frequency being normal and constant.
Curves. — Draw the curves necessary to show the characteristics
clearly.
Questions. — 1. Why is the current for a given torque greater
when the motor is at standstill than when running?
2. What are the advantages and disadvantages of the squirrel
cage induction motor compared with the wound rotor induction
motor?
Art. 404] LABORATORY COURSE 381
3. What are the relative advantages and disadvantages of
the squirrel cage induction motor and the synchronous motor?
4. Explain the shape of the power factor curve. Why is the
power factor low at light loads?
5. Explain the shape of the speed curve.
6. How would you reverse the direction of rotation of an
induction motor?
7. How can the speed of an induction motor be varied for a
given load? How does the induction motor compare with the
direct current shunt motor for the operation of machine tools,
and with the direct current series motor for the operation of
cranQs?
8. What effect has an increase or decrease of applied voltage
on the speed of an induction motor at noload and at fullload?
Try this experiment.
EXPERIMENT 20
Object of Experiment. — To determine the voltage and current
relations with different transformer connections.
References.— Pages 233 to 238, also Chap. XXXV.
Connections. — a Y to 7, Fig. 322, page 276.
6 y to delta
c delta to delta
d Scott connection to obtain two phase from
3 phase.
F, St
Fig. 449.
Note. — Each transformer should be protected with fuses in
case of short circuit. If for example the third transformer in a
delta connected bank was connected as in diagram B, Fig. 449,
instead of as in diagram A, there would be a large voltage be
tween points 8i and 8z, and if they were joined together a very
large current would flow through the closed circuit.
Readings. — Measure the voltage per phase, the voltage be
tween lines, the current in each transformer and the current in
382 PRINCIPLES OF ELECTRICAL ENGINEERING [Chap, xlii
the lines, and compare the values obtained with the theoretical
values.
Question. — 1. What are the advantages and disadvantages
of the Y and delta connections?
2. Show by experiment that three phase power can be ob
tained from a delta connected bank if one transformer is removed,
but cannot be obtained from a Fconnected bank after the
removal of one transformer.
PROBLEMS
TO ACCOMPANY GRAY'S
PRINCIPLES AND PRACTICE OF
ELECTRICAL ENGINEERING
CHAPTERS 1 TO 4
1. A small room is lit by a 32candlepowery 110volt tungsten lamp which
takes 1.25 watts per candle power. Find the current taken by the lamp.
0.365 amp.
What does it cost to light the room for 4 hours if the cost of energy is
6 cents per kw. hour. 0.96 cents.
2. A 10horsepower, 110volt motor has an efficiency of 85 per cent.
Find:
(a) The motor input at full load in kw. 8.78 kw,
(6) The current taken by the motor at full load. 79.8 amp.
(c) The cost per annum to run this motor on full load for 50 hours a
week if the cost of energy is 3 cents per kw. hour. f^86.
(d) The cost per kw. year with the above schedule. %78.
8. The load on a 110volt generator consists of two hundred 32candle
power tungsten lamps that take 40 watts each, also a 10horsepower motor
which has an efficiency of 85 per cent. Find:
(a) The output of the generator. 16.78 kw.
(6) The current taken by the load. 15S amp.
(c) The horse power of the driving engine if the efficiency of the generator
is 87 per cent. £5.8 hp.
4. Find the work done in ergs and the power expended in kw. in raising
a weight of 2000 lb. through 500 ft. in 1 minute.
1.356 X 10" ergsy n.6 kw.
If the efficiency of the hoisting mechanism is 75 per cent., and that of the
motor is 88 per cent., what is the average horse power required during the
hoisting period, also the average current taken from the line if the voltage
is 110? JiO.6 hp., SIO amp.
6^ Find the power required to melt 10 lb. of ice in 5 minutes. Find also
the current taken from a 110volt line and the cost of the operation if the
cost of energy is 6 cents per kw. hour. (The latent heat of ice is 80 lb.
calories.) 6.1 kw., 46 amp.j 2.66 cents.
6. To keep a car moving with a velocity of 36 miles an hour, a tractive
effort of 700 lb. is required. Find the horse power developed by the motor,
also the current taken from the line if the line voltage is 550 and the motor
efficiency is 90 per cent. 67 hp., 101 amp.
7* A conductor 20 cm. long is moved through a uniform magnetio field
383
384 PROBLEMS TO ACCOMPANY
of 8000 lines per sq. cm. with a velocity of 30 metres per sec. Pind the
e.m.f. between the ends of the conductor. 4'S volts.
If the ends of the conductor are joined through an external circuit of such
resistance that a current of 50 amp. flows through the conductor, find the
retarding force. 800,000 dynes or 1.8 lb.
Find the power required to keep the conductor moving.
e4 X 10^ ergs per sec. '^ 240 waUs » 4.8 volts X 60 amp.
8. Is there any such unit as a watt per second?
9. Do you buy energy or power from the power company? Is your
bill in kilowatts or in kilowatthours?
10. A man claimed that his electric power cost him 10 cents a kw.
what did he probably mean?
IL A house uses 5 lamps of 40 watts each for 4 hours a day what would
be the monthly bill at 10 cents a kw.hr. ?
12. How many 40watt lamps could be safely used on a 110volt line
with &amp. fuses?
If a 5amp. electric toaster were put on the same line how many of the
lamps could now be lit at once?
18. Assuming an efficiency of 50 per cent, for the whole arrangement,
what current will be used by a 250volt electric hoist when raising 2500
lb. 200 ft. per min. 7
14. The efficiency of a 110volt water heater is 80 per cent. What
resistance must the heating elements have in order to raise the tempera
ture of 1 lb. of water from O^C. to boiling in 10 min.? What will this
heating cost when electric energy costs 9 cents per kw.hr.?
16. A 50hp., 110volt motor has an efficiency of 90 per cent.
Find:
(a) The motor input at full load in kilowatts.
(&) The current taken by the motor at full load.
(c) The cost per annum to run this motor on full load for 50 hours a week
if the cost of energy is 3 cents per kw.hr.
(d) The cost per kilowatt year with the above schedule.
16. Repeat the above problem for a 220volt motor of the same output
and efficiency and explain in as great detail as possible why 100 amp. at
200 volts costs as much as 200 amp. at 100 volts.
17. A consumer has a 50hp. motor which has an efficiency of 90 per
cent, and operates 10 hours a day. The rate is $50 per kw. year no matter
how many hours a day the motor is used. What is the annual bill and what
is the actual cost of power per kilowatthour?
18. The load on a 220volt generator consists of 1000 hp. of motors with
an average efficiency of 85 per cent. Find:
(a) The current to supply the load.
(&) The generator output.
(c) The horsepower of the engine if the generator efficiency is 95 per cent.
19. Forty wires each 10 cm. long are connected in series and made to
cut across a field having a flux density of 12,000 lines per sq. cm. at the rate
of 3000 cm. per sec.
(a) Find the e.m.f. generated in the circuit.
PRACTICE OF ELECTRICAL ENQINEERINQ 386
(&) If the resistance of the circuit is 10 ohms what current will flow and
what will be the power in watts to keep the wires moving?
(c) What is the force required to keep the wires moving?
(d) fVom your figures show that force X velocity (in watts) "* volts X amp.
CHAPTER 6
20. An electric flatiron takes 3 amp. at 1 10 volts^ what is the resistance of
the heating coil? 96,6 okma.
If the voltage applied to the heating coil is reduced to 100 volts, what will
be the approximate temperature of the iron.
8iS,6 per cent, of tke oriffinal value,
21. If ten 40watt tungsten lamps are placed in parallel with the flatiron,
what will be the total current taken from the line? Will the current in the
iron be changed by the addition of the lamps? 6,6$ afnp,y No,
22. If the ten lamps connected in parallel are now placed in series with
the flatiron across a 220volt line, what will be the current taken from the
line, the voltage drop across the lamps and that across the flatiron.
S,iS8 amp., 99 volU, IHI voUa,
If five of the lamps are switched off, what will then be the voltage drop
across the lamps and across the iron. 1S7 voUSf 83 voUa,
23. Given three resistances, A of 5 ohms, B of 11 ohms and G of 10 ohms.
Find the voltage drop across each resistance, the current in each resistance
and the total current taken from the line when A and G in parallel are con
nected in series with B across 110 volts.
Va  iS6,6, Vh  84^6, la  6,1, h  IS,6, h  7,7,
24. A hall is lit by five hundred 40watb 110volt lamps and is supplied
by a station 400 ft. distant. Find:
(a) The total current required. 182 amp,
(6) The minimum size of weatherproof wire that can safely be used.
(See table on page 369.) No, 0, B, A S,
(c) The resistance of 800 ft. of this wire at 25 deg. G. 0,08 ohms,
{d) The necessary voltage variation at the power house to keep the lamp
voltage constant at 110 with any number of lamps from to 500.
110 to 1246 voUa,
(e) The size of wire required in order that the powerhouse voltage need
not be varied more than 2.5 per cent. 560,000 cir, miU,
if) If the lamps are connected two in series across 220 volts, find the size
of wire required in order that the powerhouse voltage need not be increased
more than 2.5 per cent. IJfifiOO txr, mile,
26. It is required to deliver 100 kw. at 600 volts at the end of a halfmile
line, the voltage drop in the line not to be greater than 5 per cent, of the
terminal voltage. Find the size of wire for both copper and aluminium.
(The specific resistance of aluminium is 1.62 times that of copper.)
810,000 cir, mile, 600,000 cir, mHa.
If aluminium weighs 0.1 lb. per cu. in. and costs 30 cents per lb. while
copper weighs 0.32 lb. per cu. in. and costs 16 cents per lb., which is the
cheaper metal to use? Aluminium ie 6 per cent, cheaper,
26. If copper wire is used, what is the annual charge of the line, neglecting
386 PROBLEMS TO ACCOMPANY
supports, if the power loss in the line is charged at $25 per kw. year and
interest on the copper is charged at 6 per cent, per annum.
%47.6 irUerestf %1B6 power.
With the above rate of interest and power cost what is the most economical
section of copper wire and what is then the total annual charge of the line
and also the voltage drop? 600^000 dr. mihf %164i 3,1 per cent.
•27. Why should a voltmeter have a high resistance and an ammeter a
low resistance? What would happen to the instruments if the voltmeter
and ammeter in Fig. 20, page 18, were interchanged, E being 110 volts.
Ammeter burnt outf voltmeter reads 110.
28. Two voltmeters, A of 10,000 ohms resistance and B of 5000 ohms
resistance, are connected in series across 150 volts. What will be the reading
of each? A = 100, B = 50.
29. A 150range voltmeter has a resistance of 10,000 ohms. What addi
tional resistance must be placed in series with the instrument so that 450
volts will give a fullnscale deflection? fSO^OOO ohms.
80. A 25amp. meter has a resistance of 0.002 ohm. What is the voltage
drop across the instrument for a fullscale deflection? 0.05 volt.
Specify the shunt to be used so that this meter may be used to read 100
amperes with a fullscale deflection. 0.00067 ohm, to carry 75 amp.
81. What would happen if by accident an ammeter was used to measure
the voltage of a 150volt line?
What would happen if by accident a voltmeter was placed in a line to
measure the current flow?
82. An 8ft. sample cut from a long coil of nichrome wire (used for
heaters) has an average diameter of 0.161 in. Its resistance is found to be
0.186 ohm. What is the resistance per circ. milfoot of this wire at the same
temperature?
88. An aluminum bar with an average cross section of 0.185 sq. in. and a
length of 2.956 ft. has a resistance of 207 microhms. What is the resist
ance of aluminum per cirQ. milfoot at this temperature?
84. The resistance of a conductor increases by 31 per cent, from 23**
to 75" C. What is the temperature coefiicient of the metal?
86. The resistance of the coils of an electromagnet at room temperature
of 20" C. is 20 ohms. After the current has been on for an hour the
resistance becomes 24 ohms. What is the cause of the increase in resistance
and what is the average temperature of the winding?
If the applied voltage is 110 what is the current in the coil at the beginning
and at the end of the hour?
86. A carbon lamp and a tungsten lamp each takes 0.5 amp. from a
110volt line. Two carbon lamps in series across 110 volts take less than
0.25 amp. while 2 tungsten lamps in series across 110 volts take more
than 0.25 amp. Explain this result.
87. If a 110volt electric water heater is used on a 100volt circuit why
will it take about 20 per cent, longer time than usual to heat the water to
the desired temperature?
38. Two voltmeters whose resistances are 17,500 ohms and 5300 ohms are
connected in series across 1 80 volts. What will be the reading on each meter?
PRACTICE OF ELECTRICAL ENGINEERING 387
89. The coil of a wellknown type of moving coil instrument has a re
sistance of 1.5 ohms and will give a fullscale deflection with 50 millivolts
across the terminals.
(a) How can this meter be used as a 50 amp. range ammeter? Specify
the shunt (resistance and current capacity).
(6) How can it be used as a 150H3caIe voltmeter? Specify the series
resistance (resistance and current capacity).
40, In order to determine the value of an unknown high resistance, a 150
acale voltmeter whose resistance is 17,000 ohms is connected in series with
it across 220volt mains. If the voltmeter reads 40 volts what is the
value of the unknown resistance?
41. A lighting circuit 5 miles long is laid in an underground duct, each
wire having an insulation resistance to ground of 4 megohms per mile.
If the potential difference between the wires is 2000 volts, what is the
leakage current?
What is the insulation resistance of 4 miles of this insulated wire?
48. Two resistances of 5 and 10 ohms are connected in parallel across
a line ■ of unknown voltage. If the total current is 35 amp. what is the
current through each? What is the voltage of the line?
43. A total current of 23.5 amp. divides between three branches in
parallel, of resistance ri = 2, r2 = 5 and fs = 12 ohms respectively.
Determine the current in each branch. What is the total power supplied?
44. In the laboratory circuit shown find the voltages at points a, 6,
c and df if the resistance of the feeder is 0.4 ohms per 1000 ft. of wire
— 40 ft.
^ • 230 Voltl
F
a
30 Amp, 20 Amp. aOiMip. WA'nip.
Diagram No. 1.
46. It is desired to connect a factory to the 110volt electric light mains
located 500 ft. away. The load consists of 60 lamps of 40 watts each.
What is the smallest B. & S. gauge wire (see p. 216) that can be used with
satisfactory service? Resistivity of copper is 10.2 ohms per circ. milfoot.
(The voltage variation at the lamps should not exceed 2 per cent.)
What is the smallest conductor that could be used if there was no restric
tion to the voltage drop (see page 369).
46. In the previous problem, what size of wire could be used if the vol
tage had been 220 volts, allowing same percentage drop?
47. Three cars a, h and c are }i, 1 an^d 2 miles respectively from the
power station. Cars a and h are drawing 50 amp. and car c which is start
ing draws 200 amp. Find the voltages at each car given that the station
voltage is 600, the trolley resistance 0.42 ohms per mile of wire and the return
track resistance 0.03 ohms per mile of track. What are the voltages at
a and h when car c is at standstill drawing no current?
27
388
PROBLEMS TO ACCOMPANY
48. The ranstance of a transmission line is 1.2 ohms. What voltage is
necessary at the generating end in order to produce a current of 75 amp.
(a) When the line is shortcircuited at the receiving end; (h) when a pressure
of 500 volts must be maintained at the receiving end?
40. The opencircuit potential difference of a storage battery is 6.6
volts. This battery will maintain a current of 3 amp. at a potential
difference of 5.8 volts. What is the battery resistance?
50. What resistance must be connected in series with a storage battery
whose e.m.f . is 6.6 volts and internal resistance 0.27 ohm so that a charging
current of 5 amp. will flow when connected across 110volt mains?
What does it cost to charge this battery for 5 hours if the cost of eneigy
is 10 cents per kw.hr.?
What would it cost to charge four such batteries in series?
CHAPTER 6
61. How many ohms resistance would you use for loading the following
220volt generators: 2 kw., 10 kw., 50 kw., and 500 kw.?
Which would be the largest and heaviest rheostat and why?
58. A castiron grid rheostat was built of 100 grids arranged in two layers
of 50 each. The 50 grids per layer were connected in series and the two
layers were connected in parallel. This rheostat drew 30 amp. from a 125
volt line and the temperatures attained were ia == 90** C; fe « 116® C; U
 57** C; ti  83** C; U « 25** C. « room temperature.
c
Ul
J
\y
(U.
v^tr
s12^
i
Diagram No. 2.
Design a rheostat to absorb the load of a 30kw. 250volt generator the
temperature of the air at "a" not to exceed 120** C. the load to be put on
in 10 equal steps: The air temp, being 25** C.
63. What is the value of watts per sq. in. of gross surface of one side
of the grid, i.e., watts/xy; this value seldom exceeds 1.0.
64. The resistance of the above rheostat is 2.08 ohms; why not use
3 ft. 6 in. of No. 20 nichrome wire which has the desired resistance.
66. Why does not a rheostat for absorbing 45 kw. for 15H9ec. i>eriods
need to be as large as one to absorb 30 kw. continuously.
66. How would a rheostat to absorb 500 watts at 10 volts differ from one
for 500 watts at 100 volts? Explain in detail. Using the same size of
German silver wire in each case compare the weights of wire and arrangement
in the two cases.
67. Why is it necessary to have flowing water about the electrodes of a
rheostat for large power absorption? Allowing a temperature rise of 50** C.
FKACTWt: OF ELECTRICAL ENGINEERING ZS9
how many gallons per minute must flow by to carry away the heat when
absorbing 30,000 kw.? (1 wattsec. = 5.3 X IQr* lb. calories.)
58. How many pounds of German silver wire are needed in a rheostat
that will absorb the output of a 10kw. 110volt generator for a 5 sec.
•period? Neglect the heat radiated during this interval. For German
silver k » 250 ohms per circ. milfoot, si>ecific heat = 0.096, density ">
0.31 lb. per cu. in. A maximum temperature rise of 100° C. is permissible.
What should be the resistance of the rheostat in ohms and what should be
the cross section and length of the wire?
CHAPTERS 7 AND 8
59. A coil of 200 turns of No. 14 wire (see page 369) is wound in one layer
on a wooden annular ring which has a mean diameter of 12 cm. and a circular
cross section of 5 sq. cm. Find:
(a) The resistance of the coil at 25° C. 0.13 ohm, approx.
(6) The flux density in the core when the exciting current is 4 amp.
£6,7 lines per aq, cm,
(c) The voltage and power required to maintain the exciting current.
OM voU, B,08 watU.
60. If the wooden ring is replaced by (a) a castiron ring, (6) a cast
steel ring of the same dimensions, what will then be the flux density in the
core when the exciting current is 4 amp.? (Use the curves in Fig. 41.)
Cast iron, SI 00; cast steel, H,300 lines per sq, cm,
61. With 200 turns of wire on the caststeel ring, what is the current
required to produce a total magnetic flux of 50,000 lines in the core?
7 amp, turns per cm,; l.SB amp.
If the voltage applied to the coil is doubled, what will then be the total
flux? 65,000 lines.
If, on the other hand, the number of turns is doubled, the voltage and the
size of wire being unchanged, what will be the total flux? 60,000 lines.
If the cross section of the wire is doubled, the voltage and the number of
turns being unchanged, what wiU then be the total flux? 66,000 lines,
68. If the caststeel ring wound with 200 turns is cut and opened to form
an air gap of 0.2 cm. length, what will be the current required to produce a
total magnetic flux of 50,000 lines? (Compare with Prob. 61.)
8 amp, gap; 1.32 amp. steel; 9.32 amp, total.
Find also the ampere turns required to produce flux densities of 8000,
10,000, 12,000, 14,000 and 16,000 lines per sq. cm., and plot a curve of total
flux and amperes excitation. (This is called the saturation curve of the
magnetic circuit.) 1430, 1864, 2336, 2936, 4^30,
63. If the steel ring is cut in two halves across a diameter, what will be the
flux in the magnetic circuit and the pull required to separate the two parts
if the air gaps are each 0.1 cm. long and the exciting current is 3 amp.?
17,600 from Proh, 22; 4.9 kg.
Find also the total flux and the pull if the gaps are reduced to 0.001 cm.
each. 60,000 lines; 68 kg,
64. In the case of the nfagnet shown in Fig. 56, page 44, the length li is
made up of two pole pieces each 4 cm. long, the inner faces of which are 8 cm.
390 PROBLEMS TO ACCOMPANY
apart. On this winding space of 8 cm. the exciting coil is wound.
(a) Show that the section of wire of an exciting coil is given by
., 10.6 X mean turn in ft. X ampere turns
cir. mils = ;
volts per coil
where 10.6 is the resistance of copper per cir. mil foot.
(&) Find the cross section of the wire, the number of turns and the exciting
current for a pull of 145 kg. (take the current density in the wire to be 1
amp. per 1000 cir. mils), the applied voltage being:
(a) 10 volts. 1180 cir. mils; IAS amp,; 1700 turns,
(6) 120 volts. 100 cir. mils; 0.10 amp.; 20,000 turns.
Take the mean turn of the coil to be 17 cm.
(c) When the coil of 1700 turns is tightly wound, the external periphery
measures 22 cm. The winding space is 8 cm., so that what is called the
radiating surface of the coil » 8 X 22 = 176 sq. cm. The temperature
rise of the surface of such a coil being about 60^ C. for 0.1 watt per sq. cm. of
radiating surface, what will be approximately the temperature rise of the
above coil when the pull is 145 kg. 7 40** C.
66. The magnetic circuit of a generator such as that in Fig. 82, page 59,
is made up of:
Two caststeel poles each with a cross section of 180 sq. cm. and a mag
netic length of 25 cm.
Castiron base with a cross section of 400 sq. cm. and a magnetic length
of 40 cm.
Two air gaps each with a cross section of 450 sq. cm. and a magnetic
length of 0.5 cm.
Path through the rotating armature, of wrought iron, with a cross section
of 160 sq. cm. and a magnetic length of 15 cm.
Plot total flux against exciting current up to a value of 4 amp., the num
ber of turns in the exciting coil being 3000.
One point on curve; flux = 2,600,000; ampere turns = 900, poles; 1280,
base; 4400, gaps; 405, armature; exciting current = 2.3S amp.
66. A circular solenoid is formed by winding wire on a wooden ring
whose diameter of section is 2 cm. The inside diameter of the ring is 10
cm. There are 275 turns of wire wound on the ring, and a current of 1
amp. is maintained in the coil. Find the ampereturns per centimeter, the
flux density and the total flux.
(a) If a wroughtiron core is substituted for the wooden core in the above
problem, what would be the flux density and the total flux?
(h) If cast iron is used instead of wrought iron what will be the flux density
and the total flux?
67. If the voltage applied to the coil with the wroughtiron core is
doubled, what will be the total flux?
If the voltage and size of wire are unchanged but the number of turns is
doubled, what will be the total flux?
If the area of cross section of the wire is doubled with the voltage and
number of turns unchanged, what will be the total flux?
68. How many amperes are required to produce a flux of 50,000 lines
in the wroughtiron core?
i
PRACTICE OF ELECTRICAL ENGINEERING
391
69. If a ituiial cut is made in the wroughtiron ring of Prob. 68, and
the iron spread apart, how long an air gap must be made to double the
reluctance of the magnetic circuit if the flux remains constant?
Find the puU tending to close the gap in this case (see page 44).
70. Draw a saturation curve for the ring in Prob. 68.
(a) Without the slit.
(6) With the sUt.
For these curves we plot flux against amperes as abscissae.
71. An iron ring of 56 cm. mean diameter and cross section of 30 sq.
cm. is wound uniformly with 320 turns of wire. The exciting current is
brought to 24 amp. and then reversed by means of a switch. A ballistic
galvanometer shows a change in flux equal to 452,000 lines. What is the
reluctance of the ring and what is its permeability?
78. In the magnetic circuit shown, the useful flux passes through the
air gap between the two steel poles; a part of the flux is shunted through
the castiron part of the circuit. At low saturation a considerable part of the
total flux is shunted through the castiron part, but as the flux density increases
Oftp Area 00 ■q.cm*
G»p Length 0.25 cmrv
Gait Steel
\ry^ * ■q.em.
~~\ 7' F 1 I ^*"' '"**
*8cm^ ^^Bcmr* 12,q.cin.
7cm.dia.Cait Steel
im^
\\\
m
\ \
Diagram No. 3.
the cast iron becomes saturated, and a larger portion of the flux is deflected
into the air gap. What percentages of the total flux in the yoke are shunted
through the cast iron when the flux density in the air gap is 1000 and 7000
lines per sq. cm. respectively? The foregoing arrangement illustrates the
principle used in some practical cases when it is desired to modify the rela
tion between the flux and the magnetomotive force by providing a highly
saturated magnetic path in parallel with a feebly saturated one.
73. When the flux density in the gap is 7000 lines per sq. cm. find the flux
density in the other parts of the circuit and the total ampereturns required.
74. Diagram 4 is a part of the mechanism of a magnetic relay. For small
values of exciting current, most of the flux crossing the air gap A will pass
through L and back through the coil. This will cause S to move to the
right and close gap A. For large values of exciting current L will become
saturated, and then most of the flux crossing A will pass through gap C.
392 P^BLEMS TO ACCOMPANY
Since the magnetic pull dependa upon the tquan of the dennltf of flux, the
pull at C may become greater than that at A, though ^, is lees than 4„
niia will cause S to move to the left and cloee gap C.
For some critical or operating value of current the pulls in the two air
gaps are equal. It is desired to find the cross section of the lever L such
that the critical or operating current will occur when B, the flux density in
the main core, id 16,000 lines per sq. cm.
Not« that ALD and ACED are parallel magnetic circuits.
Use the following figures as a continuation of Fig. 41.
lines per sq. em. 20,000 : 22,500 : 22,000 : 27,S00
Amp.tum8 per cm. 275 1000 2600 4600
CHAPTERS g, 10 AND 11
78. A wire passes 40 times a sec. acrosH the pole face of a field magnet
the flux density of which is 15,000 lines per sq. cm. The pole face area ia
30 X 30 cm. What average e.m.f. is induced in the wireT
jl 3 »*«»
DiAORAif No. 4.
76. Show that BLV X 10~* will give the average voltage bduced in a
wire of length L cm. moving at right angles to a flux of density B linea per
sq. cm. with a velocity of V cm. per sec.
77. A 6pote D.C. generator has 388 conductor on the armature. The
average e.m.f. induced per conductor is 2 volts and the allowable current
per conductor is 10 amp.
(a) Find the no4oad terminal voltage and line current if the armature
has a lap or multiple cireuit winding.
(&} Find e.m.f. and line current if the winding is of the wave or two
circuit type.
78. A 4pote D.C. generator driven at a certain speed and having a
certom field excitation has an average induced voltage per conductor ot
PRACTICE OF ELECTRICAL ENGINEERING 393
2 volts. The current per conductor is limited by heating to 10 amp.
The generator delivers to the line a current of 40 amp. at an induced vol
tage of 120 volts. Does the armature have a multiple (lap) or series <wave)
winding? What is the total number of conductors on the armature?
70. A 6pole D.G. generator armature has 350 active conductors and a
wave or two circuit winding. The flux per pole is 6 X 10* lines of force
what is the generated voltage if the speed is 210 r.p.m.?
If the output of the machine is 50 kw. what is the current per conductor?
80. If this machine had a lap or multiple winding with 1050 active con
ductors what would then be the generated voltage at the same speed the flux
I>er pole being unchanged? If the output is still 50 kw. what is the current
per conductor and which machine requires the most copper?
81. An 8pole D.C. generator armature has 608 conductors. When
running at 250 r.p.m. the induced voltage is 110. What is the flux density
in the air gap if the pole area is 780 sq. cm.? Armature has a multiple
winding.
82. Draw the following winding diagrams:
(a) 6pole doublelayer lap winding with 30 slots, 2 conductors per slot
and 1 turn per coiL
(6) The same machine with 6 conductors per slot and 3 turns per coil:
Show only the first two coils.
(c) 6pole doublelayer wave winding with 29 slots, 2 conductors per slot
and 1 turn per coil.
If in cases (a) and (c) the average voltage per conductor is 10 and the
safe current per conductor is 100 find the terminal voltage, line current and
kilowatt output for the two windings.
88. Why is the magnetic circuit of an electric generator made of iron?
Why is the armature core made of laminations while the xemainder of the
magnetic circuit may be a casting? What would happen if a steel casting
were used for the armature?
Why are the armature conductors insulated from the core and why are
the commutator segments insulated from the core and from one another?
What insulating materials are generally used for this purpose?
84. In Fig. 65 would the polarity of the brushes B^ and B. be changed
if the direction of rotation was reversed? Yes,
86. Make a sketch of a fourpole compoundwound machine showing the
direction in which thd field coils are wound on the poles and the way in
which they are connected.
86. Why are the brushes of modem directcurrent generators invariably
of carbon, and why are they generally moved from the neutral position?
If the direction of rotation of a generator is reversed will it be necessary
to change the position of the brushes? How will the correct position be
determined? By trial for minimum sparking.
In the case of reversible machines, such as crane motors, what is the best
position for the brushes? On the neutral.
If the direction of rotation of an interpole generator is reversed, the
polarity of the main poles being unchanged, will it be necessary to reverse
the connections of the interpole windings? No,
394 PROBLEMS TO ACCOMPANY
If the interpole windings of a generator are disconnected what will happen
to the machine? Spark badly.
If the interpole windings are connected backward, will the commutation
be better or worse than if there were no interpoles? Worse.
87. Draw a bipolar machine similar to Fig. 85 except that the ring
winding on the armature should be complete and the field coils not connected,
and let the residual magnetism of the left hand pole be N and the direction
of rotation clockwise.
(a) Find the polarity of the brushes due to residual magnetism; connect
the shunt coils so that the machine will build up and the series coils so that
the machine will be a compound generator.
(h) Indicate the direction of the current in the armature conductors,
find the direction of the force on the conductors due to the current they
carry and then explain why the larger the current or the larger the voltage
the larger the required horsepower of engine.
88. Draw a bipolar machine similar to Fig. 91 except that a ring wind
ing on the armature and a compound winding on the field coils be shown
complete, the residual magnetism being as shown but the direction of
rotation clockwise.
Connect up the machine so that it will build up and operate as a com
pound generator. Show the brushes in the proper position for good
commutation.
89. With the rest of the machine as in the last problem add interpoles
and show how the interpole coils are connected for proper operation.
Answer Prob. 102 if possible.
90. A 4pole 1500r.p.m. machine has 80 commutator segments and the
brush is wide enough to cover 2 segments at once. Show that the current
in the short^sircuited coil is reversed in a time of Kooo sec.
9L Find the time of commutation for some of the machines in the
laboratory.
92. A laminated wroughtiron ring of 10sq. cm. crosssection area has
two coils wound upon it. Coil A has 6 turns and coil B has 8 turns. By
reversing the current in coil A the flux in the ring of 10^ lines is made to
reverse in 0.001 sec. What is the average induced e.m.f. at the terminals
of the coil B and at the terminals of coil A while the current is changing?
The former is a voltage of mutual induction and the latter a voltage of
selfinduction.
93. The flux through a shortcircuited armature coil of 8 turns changes
from 10^ lines in one direction to the same value in the opposite direction
in 0.001 sec. What is the average e.m.f . induced in the shortcircuited coil
during this interval?
CHAPTERS 12, 13, AND 14
94. Plot the magnetization curve from the following data obtained by
test on a separately excited 25kw., 120volt generator connected as shown
in Fig. 05, page 70, and run at a constant speed of 900 r.p.m. (Keep this
curve for later problems.)
PRACTICE OF ELECTRICAL ENGINEERING 395
Voltage at no load 4 40 60 80 100 120 140 volts.
Exciting current 0.0 0.67 1.03 1.50 2.07 2.94 4.35 amperes.
Kxplain why the curve bends over and why there is a small voltage with
no exciting current.
Is the polarity of this machine changed by:
(o) Reversing the direction of rotation? Yes.
(b) Reversing the connections of the exciting coils? Yes,
96. (a) If the exciting voltage is constant at 120 what must the resistance
of the field coil circuit be so that the generated voltages may be 40 volts,
100 volts, 120 volts, the speed being 900 r.p.m.? 179y 68, U ohms,
(&) If the field coils have a resistance of 33 ohms, specify the rheostat to
give a range of voltage from 40 to 120. 146 ohms to carry 2.94 omp,
(c) If the resistance of the field coil circuit is kept constant at 41 ohms,
what is the exciting current if the exciting voltage is 120, and what is the
generated voltage when the speed is 700 r.p.m., 900 r.p.m., 1100 r.p.m.?
e.94 amp; 93.6, IW, I46 volts.
96. If the above machine is now shunt connected as in Fig. 96, then :
(a) If the direction of rotation is reversed will the machine build up?
No.
(&) If in addition the residual magnetism is reversed, will the machine
then build up?
No.
(c) If the connections of the exciting coils of the original machine are
reversed will the machine then build up?
No.
(e) If the direction of rotation is reversed and also the connections of the
exciting coils, will the machine build up and will the polarity of the brushes
be changed? Yes, Yes,
97. (a) When the machine is shunt connected, find the resistance of the
field coil circuit so that the generated voltages may be 40 volts, 100 volts,
120 volts, the speed being 900 r.p.m. 60, 4S, 4t ohms.
Why do the figures differ from those in Prob. 96?
(&) If the resistance of the field coil circuit is kept constant at 41 ohms,
what is the generated voltage when the speed is 700 r.p.m., 900 r.p.m., 1100
r.p.m.? Explain why the figures differ from those in Prob. 95.
62, 120, 170 volts.
98. A generator supplied with both shunt and series field coils was tested
under load:
(a) As a shunt generator with the series coils out of circuit.
(b) As a separately excited generator with the series coils out of circuit.
(c) As a compound generator with both shunt and series coils connected.
Plot the regulation curves in each case.
Answer questions 1, 2, 3, 4, 5, 6 and 7, experiment 5, page 372, given that
the resistance of the armature circuit is 0.024 ohm.
396
PROBLEMS TO ACCOMPANY
Terminal voltace
Line current, amperes
Shunt excited
Separately excited
Compound excited
120.0
120.0
120.0
52
116.5
118.2
120.4
104
112.5
116.2
120.8
166
108.0
114.8
120.9
208 (fuU load)
102.0
112.5
120.9
260
94.0
110.0
120.6
312
84.0
107.5
120.0
99* The armature of the above generator can be spun around by hand but,
when delivering current to an external circuit, a 38hp. engine is required
to rotate the armature at the normal speed of 900 r.p.m. What is the cause
of the increased retarding torque and what is its value?
ees lb. all St.
100. The above generator when separately excited and running at 900
r.p.m. gives 120 volts at no load and 112.5 volts with a fullload current of
208 amp. The resistance of the armature circuit is 0.024 ohm. Find:
(a) The voltage drop in the armature circuit. 6 volts,
(b) The actual voltage generated at full load. i/7.5 voUs,
(c) The cause of the loss of the 2.5 volts (120— 117.5) which is no longer
generated at full load. Armature reaction reduces flux 2.1 per cent.
(cO If this machine, run as a separately excited motor, is loaded so as to
take 206 amp. from a 120volt circuit, what is the back e.m.f. of the motor
and at what speed does the machine run? 116 volte; 880 r.p.m,
(e) If this motor on no load takes such a small current that armature
reaction can be neglected, what is the back e.mi. at no load, the increase
in magnetic flux over the fullload value, the noload speed?
About 120 volte; 2.1 per cent.; 900 r.p.m,
101. (a) If a separately excited motor, which takes an armature current of
100 amp. on full load from a 120volt line, runs at 1200 r.p.m. both at no
load and at full load, what is the back e.m.f. of the motor if the armature
resistance is 0.05 ohm, and what is the reduction of flux at full load due to
armature reaction? IIS volts; 4.2 per cent.
(&) If this same machine, run as a separately excited generator, has to
deliver 100 amp. at 120 volts, what wiU then be the voltage drop in the
armature circuit, the actual voltage generated, the speed of the generator,
and the voltage at no load? 6 volts; 125 volts; 1300 r.p.m.; ISl volts,
108« If the direction of rotation of the machines in Probs. 88 and
89 is reversed what will happen and what must be done to make the
machines operate sparklessly as compound generators with the same
polarity at the brushes as existed before the rotation was changed?
103. What changes in connections will now be necessary to have the
machine run in the same direction as a cumulative compound motor with
interpoles connected to assist commutation?
PRACTICE OF ELECTRICAL ENGINEERING 397
If this interpole motor is for reversing duty how would it be connected
80 that the throwing over of a 8witch*would cause the proper connections
to be made?
104« Plot the following data for the saturation curve of a D.C. generator.
R.p.m. « 1000 (normal and constant).
Terminal volts 5 22 42 60 75 93 103 110 120
Shuntfield current. 0.1 0.2 0.3 0.4 0.6 0.8 1.0 1.5
106. With the machine running at a constant and normal speed of 1000
r.p.m. insert resistance in the fieldcoil circuit; plot the noload terminal
voltage against resistance of the field circuit for the following cases:
(a) Machine separately excited from 110volt mains.
(5) Selfexcited, shunt connected.
(c) Selfexcited, compound connected.
The resistance of the shuntfield winding is 80 ohms. Which machine is
the most sensitive to field rheostat adjustments? Why?
106. Specify the graded field rheostat (total resistance, maximum cur
rent, and minimum current carrying capacity) to change the voltage from
120 to 40 volts for the three types of connections.
107. With no field rheostat find the terminal voltage with speeds of 250,
500, 1000 and 1500 r.p.m. when the machine is
(a) separately excited from 110 volts,
(6) shunt excited.
Which machine is most sensitive to change of speed? Why?
108. Will the noload e.m.f. of the above machine (field rheostat short
circuited) be increased or decreased after a load run which has raised the
temp, of the generator field 60^ C. above room temp, of 20*^0.? What
will be the value of Eq after the load run? Can it be brought to 120 volts
again, and if so how?
109. What must be the brush setting of a separately excited D.C. generator
in order that the voltage drop under load will be (a) greater than that due
to armature resistance; (6) equal to that due to armature resistance; (c)
less than that due to armature resistance? Which would be the case in
actual practice? Why?
110. Tabulate the three causes of voltage drop in D.C. shunt generators
under load. Indicate these by curves and give their value at full load
using the values given in Prob. d8.
111. How is it possible to make a D.C. generator give (a) the. same volt
age at full load as at no load; (6) a greater voltage at full load than at no
load?
Conditions: Speed constant and brushes placed for sparkless commuta
tion. Draw the typical regulation curves illustrating this.
112. A compoimd generator having its shunt field separately excited
has its series coil reversed, i.e., it is differentially compounded.
(a) Explain why there is a limit to the current output of this machine
and compute its value. There are 1000 shunt turns per pole and 20 series
and the shunt current is 0.1 amp.
(&) Will it approach this limit closer with a high r.p.m. or a low r.p.m.?
Why? (This generator is sometimes used for automobile lighting.)
398 PROBLEMS TO ACCOMPANY
lis. A compound D.C. generator when installed was found to be too
highly compounded. What would you suggest as an easy and satisfactory
method of correcting this? The terminals of the series field are accessible
(see also Art. 197).
114. Draw two bipolar machines similar to Fig. 103 except that the arma
ture should have a complete ring winding, and let the first be a generator
and the direction of rotation be clockwise. Find the polarity of the brushes
and the direction of the force on the conductors due to the current they
carry.
Connect the second machine to a source of voltage of such polarity as
to make it rotate in the same direction as the first machine and show the
direction of the currents in the conductors.
(a) Do you find that the force on the conductors, due to the fact that
they are carrying a current and are lying in a magnetic field, retards or aids
the rotation in: (a) The generator, (5) the motor? Explain.
(6) Is the induced e.m.f. in the direction of the current or against it in
the case of: (a) The motor, (6) the generator?
(e) Will the induced e.m.f. be greater or less than the terminal voltage
in the case of: (a) The motor and (6) the generator? Explain.
(d) Place the brushes in the best position for commutation in each case
and explain why they are shifted forward in one machine and backward
in the other.
116. From a study of the formula for an induced e.m.f. in a motor
EfEa — laRa = fc* X r.p.m.
Answer the following:
(a) What effect will a change in ^ or exciting current have upon the speed,
armature current remaining the same? Why? Gould the speed of a
motor be regulated by adjusting the field rheostat? Why?
(6) In a noninterpole motor whose brushes are set for sparkless commu
tation, what will be the effect of armature reaction upon the speed? Why?
(See Arts. 9192.)
116. A D.C. generator when separately excited and running at ICXK)
r.p.m. gives 250 volts at noload and 235 volts with a fullload current of
100 amp. The resistance of the armature circuit is 0.1 ohm. Find:
(a) The voltage drop in the armature circuit.
(6) The actual voltage generated at full load.
(c) The cause of the additional voltage drop.
(d) If this machine, run as a separately excited motor, is loaded so as
to take 1(X) amp. from a 250 volt circuit, what is the back e.m.f. of the motor
and at what speed does the machine run?
(e) If this motor on noload takes such a small current that armature
reaction can be neglected, what is the back e.m.f. at noload, the increase
in magnetic flux over the fullload value, the noload speed?
117. If a separately excited motor, which takes an armature current of
100 amp. on full load from a 250volt line, runs at 1200 r.p.m. both at no
load and at full load, what is the back e.m.f. of the motor if the armature
lesistance is 0.1 ohm, and what is the reduction of flux at full load due to
armature reaction?
PRACTICE OF ELECTRICAL ENGINEERING 399
If this same machine, run as a separately excited generator, has to de
liver 100 amp. at 250 volts, what will then be the voltage drop in the arma
ture circuit, the actual voltage generated, the speed of the generator, and
the voltage at no load?
118. A D.G. shunt motor connected to a 220volt line runs at 1500r.p.m.
at no load and at 1475 r.p.m. at full load of 90 amp. The resistance of the
armature is 0.08 ohm.
(a) What is the counter e.m.f. of motor at full load?
(6) What is the per cent, reduction in flux at full load due to armature
reaction?
(c) What will be the speed and the voltage actually generated in the
armature when this machine, running as a separately excited generator with
brushes shifted for sparkless commutation is delivering fullload current
of 90 amp. at a terminal e.m.f. of 220 volts?
(d) What will be the terminal e.m.f. at no load, the speed being as found
in (c)?
119. A 4pole shunt motor whose speed is 1000 r.p.m. at fuU load of 10
hp. has its armature rewound with the same size of wire but with the wind
ing changed from lap to wave. Explain clearly and exactly how the speed
for the same torque will differ from 1000 r.p.m. How will the horsepower
output of the motor compare with the former case? Allow the same
current density in the conductors.
If the machine had been a series motor what would be the speed for the
same torque if the winding had been thus changed?
CHAPTERS 15 AND 16
120. A 30hp., 120volt, 900r.p.m. shunt motor has an efficiency of 88 per
cent. The voltage drop in the armature circuit is 4 per cent, and the ex
citing current is 1.4 per cent, of the fullload current.
(a) Find the fullload current in the line, the armature current, the resist
ance of the armature circuit. 212 amp.; 209 amp.; 0.023 ohm.
(6) Find the torque developed at the driving pulley at full load.
175 lb. at 1 ft.
(c) Specify the starting resistance to keep the starting current down to
1.25 times fullload current, and what will be the starting torque under these
conditions? 0.44 ohm to carry 260 amp.; about 220 lb. at 1 ft.
(d) Would this same starting resistance be suitable for all 120volt, 30hp.
shunt motors, no matter of what speed? Give reasons. Yes.
(e) Specify the starting resistance for a 10 hp., 120volt shimt motor to
give 1.25 times fullload torque. 1.26 ohms to carry 92 amp.
(J) What would happen if the 30hp. starter were used for the 10hp. motor
and the 10hp. starter for the 30hp. motor?
121. A 30 hp., 120volt, 900r.p.m. series motor has an efficiency of 88 per
cent. The voltage drop in the armature is 4 per cent, and in the exciting
coils is 1.5 per cent.
(a) Find the fullload current in the line, the armature current, the resist
ance of the armature and that of the field coils.
212 amp.; 212 amp,; 0.023 ohm; 0.0086 ohm.
^
400 PROBLEMS TO ACCOMPANY
(6) Find the torque developed at the pulley at full load.
176 U). ail ft.
(e) Specify the starting resistance to keep the starting current down to
1.25 times full4oad current and what will be approximately the starting
torque under these conditions.
OJiS ohm to carry tS6 amp,; about t/5 lb, ai 1 ft,
112. If the shunt motor of Prob. 120 is connected as shown in Fig. 108 and
is protected by suitable fuses (see page 117), what will happen :
(a) If the starting arm is moved over too rapidly? Fiims blow,
(b) If the field coil circuit is open and an attempt is made to start the
motor? Fuses blow,
(c) If the field coil circuit breaks while the motor is running on no load?
Motor runs away,
(d) If the starter has a novoltage release as in Fig. 109 and the field coil
circuit then breaks while the motor is running on no load?
Motor diaconneded and stops,
(e) If there is an instantaneous overload of 100 per cent.?
Motor sparks btU carries load,
(J) If there is an overload of 50 per cent, for some time? Fuses blow,
(g) If the torque of the load is increased 50 per cent., what is approxi
mately the current taken from the line, also the speed of the motor and the
output? S16 amp.; 900 r,p,m,; 45 hp,
188. If the series motor of Prob. 121 is connected up with a suitable start
ing resistance, is protected by fuses and belted to the load, what will happen:
(a) If the field coil circuit is open and an attempt is made to start the
motor? No current, nothing happens.
(h) If the field coils are shortcircuited and an attempt is made to start
the motor? Fuses blow.
(c) If the field coil circuit breaks while the motor is running?
Motor stops.
(d) If the belt breaks? Motor runs away.
(e) If the torque of the load is increased 50 per cent., what is approxi
mately the current taken from the line, also the speed of the motor and the
output? i^O amp,; 736 r,p,m,; S7 hp,
(J) If the load on the motor is increased 50 per cent, what is approximately
the current taken from the line, the torque developed and the speed in terms
of the values at full load?
1,6 F,'L, current; 2,26 F,L, torque; 0,67 F,L, speed,
124. If the shunt motor of Prob. 120 is carrying full load and a resistance
of 0.05 ohm is inserted suddenly in the armature circuit, the torque of the
load remaining constant, find:
(a) The back ejn.f. at normal load and speed. 116,2 volts,
(6) The armature current immediately after the resistance has been
inserted. 66 amp.
(c) The torque developed at the same instant. 0,S2 F,'L, torque,
(d) Explain why the speed drops, also what happens while conditions are
becoming steady.
(e) What is the final speed of the motor? 820 r,p.m9
PRACTICE OF ELECTRICAL ENGINEERING
401
126. The following data were obtained by means of a brake test on a
10hp., 120volt, 900r.p.m. shunt motor:
AppliiBd
voltage
Exciting
current
Armature
current
Speed,
r.p.m.
Torque in lb.
at 1 ft.
120
2
3.5
920
0.0
120
2
19.0
915
14.2
120
2
35.0
910
28.7
120
2
52.0
905
43.0
120
2
69.0
900
58.0 (full load)
120
2
87.0
895
73.0
Plot torque and speed on an armature current base.
Plot efficiency on a horsepower output base.
126. The same machine was supplied with series coils instead of shunt
coils and the test data obtained were as follows:
Applied voltage
Armature current
Speed, r.p.m.
Torque in lb. at 1 ft.
120
37
1550
17.6
120
54
1150
34.8
120
71
900
58.0 (full load)
120
89
780
83.0
Hot torque and speed on an armature current base.
Plot efficiency on a horsepower output base.
127. Answer the questions in experiment 6, page 373.
128. From the test data in Prob. 125 find the stray loss at 920 r.p.m. and
the resistance of the field coil circuit. 4W watts; 60 ohms,
129. The resistance of the armature circuit is 0.094 ohm. Calculate the
efficiency of this shunt motor from the losses up to 25 per cent, overload and
compare the figures with those of the test curve.
180. Draw diagrams of connections for starting the following motors;
shunt, series, compound wound and compound with interpoles. Show con
nections of armature, of shunt, series and interpole field coils, field rheostat
and starting resistance. Make clear to yourself why the starting resistance
is not placed at some other point.
181* To the above diagrams add ammeters and voltmeters for measur
ing input to the motors. Show sample readings and compute the watts
input to field and armature, and total watts input. What becomes of this
power in each case? Use figures of Prob. 120 if necessary. How would you
vary the speed of the motor in each case of Prob. 130.
182, Draw the characteristic speed curves against armature current as
abscisssD for the above motors. Explain their shape.
188. How would you reverse the direction of rotation of the motors
listed in Prob. 130?
184. What is meant by per cent, regulation of a shunt motor? Illustrate.
186. Is the speed regulation of a shunt motor with the brushes shifted
to improve commutation better or worse than one with brushes in the
neutral position?
186. What will happen if the field circuit of a shunt motor is broken while
the motor is running under load? What in the case of a series motor?
402 PROBLEMS TO ACCOMPANY
187. Draw a diagram of connections for a shunt motor which runs at
1000 r.p.m. under a certain prony brake load. What will be the effect
upon the operation of the motor (speed, commutation, etc.) if
(a) field terminals are reversed?
(6) line wires are reversed?
(c) brushes are shifted against direction of rotation?
{d) if they are shifted forward?
(e) a resistance is inserted in field circuit? In the armature circuit? Id
the line?
CO the armature is removed and the field bored out so that the air gap
is lengthened and the armature is then inserted?
{g) if the air gap is decreased by "shimming'' up the poles?
ifi) if the armature is rewound with a fewer number of turns of larger
wire?
(t) if it is rewound with a larger number of turns of smaller wire?
0) some of the field turns are short circuited?
(ib) if field is rewound with the same number of turns of wire of twice
the cross section?
(0 if there should be no residual magnetism in the field poles?
138. A 30hp., 240volt, 900r.p.m. shunt mot^has an efficiency of 88
per cent. The voltage drop in the armature circuit is 4 per cent., and the
exciting current is 1.4 per cent, of the fullload current.
(a) Find the fullload current in the line, the armature current, the
resistance of the armature circuit.
(6) Find the torque developed at the driving pulley at full load.
(c) Specify the starting resistance to keep the starting current down to
1.25 times fullload current; what will be the starting torque under these
conditions?
(d) Would this same starting resistance be suitable for a 120volt, 30hp.
shunt motor?
ie) Specify the starting resistance for a 10hp., 240volt shunt motor
to give 1.25 times fullload torque.
(/) What would happen if the 30hp. starter were used for the 10 hp.
motor and the 10 hp. starter for the 30 hp. motor? The motors are pro
tected by fuses and it is necessary to develop the fullload torque of the
motor at starting. Compare results with Prob. 120.
189. A 30hp., 240volt, 900r.p.m. series motor has an efficiency of 88
per cent. The voltage drop in the armature is 4 per cent., and in the ex
citing coils is 1.5 per cent.
(a) Find the fullload current in the line, the armature current, the
* resistance of the armature and that of the field coils.
(6) Find the torque developed at the pulley at full load.
(c) Specify the starting resistance to keep the starting current down to
1.25 times fullload current and what will be approximately the starting
torque under these conditions. Compare the results with those of Prob. 121.
140. If the shunt motor of Prob. 138 is carrying full load and a resistance
of 0.5 ohm is inserted suddenly in the armature circuit, the torque of the
load remaining constant, find:
PRACTICE OF ELECTRICAL ENGINEERING 403
(a) The back e.m.f. at normal load and speed.
(&) The armature current immediately after the resistance has been
inserted.
(c) The torque developed at the same instant in per cent, of fullload
torque.
(d) Explain why the speed drops, also what happens while conditions
are becoming steady.
(c) What is the final speed of the motor?
141. If on the other hand the exciting current is reduced suddenly so as
to reduce the flux suddenly to 80 per cent, of normal value find:
(a) Tne armature current immediately after the flux has been changed.
(6) The torque at the same instant.
(c) Explain in detail what happens while conditions are becoming steady,
also what is the final speed the horse power being kept constant.
{d) Why should the field rheostat be moved slowly when increasing
the speed? Is it as important to move it slowly when the field is being
strengthened?
142. A shunt motor is to be rewound as a series motor to give the same
fullload speed and torque. The data on the shunt machine is as follows:
40 hp., 4 poles, 120 volte^ fullload armature current 275 amp., shuntfield
current 5 amp., shunt field has 900 turns per pole.
(a) Find the number of turns for the series field.
(6) Give, briefiy, the technical reasons for the differences in the character
istic curves caused by the change in method of excitation.
CHAPTERS 17 AND 18
143. The 10hp., 120volt, 900r.p.m. shunt motor in Prob. 125 has an
efiSciency of 87 per cent., and the resistance of the armature circuit is 0.094
ohms.
(a) If the speed is reduced to 450 r.p.m. by means of a resistance in the
armature circuit, the torque being constant, what is the value of the resist
ance, the loss in the resistance, the output of the motor and the overall
efficiency? 0,86 ohms; 3.9 kw.; 5 hp.; 44 per cent,
(6) Why does the motor run hotter than on normal load?
(c) If the torque is reduced to half fullload value, the external resistance
being unchanged, what is the approximate speed of the motor? 700 r.p.m.
144. If the armature of the 10hp. motor in Prob. 143 is rewound with
twice the original number of turns, the wire being of half the original cross
section, what will be the speed on full load, the permissible armature current
and the permissible output? 4^0 r.p.m.; 345 amp.; 5 hp.
Explain why this motor with 5hp. load will be hotter than the original
machine with 10hp. load. Same copper losSy poorer ventilation.
146. If the voltage of a plant is changed permanently from 120 to 240 volts,
what change is necessary on both the shunt and the series motors so that
their characteristics may not be changed?
Twice the turns of half the section for armature and field coHs,
146. The data for the magnetization curve of a 30hp., 120volt, 900
r.p.m. motor are given in Prob. 94; plot the curve.
28
404 PROBLEMS TO ACCOMPANY
(a) Specify the additional field resistance required to obtain an increase
in speed of 50 per cent, at no load. 39 ohms,
(6) If the voltage applied to this motor is only 100 volts due to excessive
line resistance, what will be the magnetic flux as a fraction of its normal
value? 0.91.
(c) What will then be the speed of the motor and the current required
for a load of 30 hp.? Why is the commutation poorer and the temperature
rise greater than normal? 8iS0 r.p.m.; 266 amp.
(d) Show that if the voltage applied to a shunt motor is reduced, the
starting torque for a given current is reduced, but that in the case of a series
motor it is unchanged.
(e) When started cold the above motor runs at 900. At the end of 3
hours the temperature rise of the field coils is 40** C. How much is the excit
ing current reduced and what is now the speed of the motor?
To 86 per cent, of normal; 970 r.p.m,
147. A shunt motor was bought to drive a fan at 1200 r.p.m. Due to a
mistake in the pulley sizes the fan speed was 1350 r.p.m. Why did the motor
spark and run hot? 4^ per cent, overload.
148. Why is it that when the speed of a noninterpole 50hp. motor *s
doubled by field weakem'ng sparking occurs at less than 50 hp., whereas in
the case of an interpole motor the speed may be increased to three times the
normal value without sparking, the load being 50 hp.?
149. A shuntwound pump motor takes 30 amp. at 240 volts. When
tested at the factory the noload current was 2 amp. in the armature and 1
amp. in the field coil circuit, while the armature resistance was 0.4 ohm.
What is the power supplied to the pump? 6.H kw,
CHAPTERS 22, 28 AND 26
180. (a) Twelve primary cells each having an e.mi. on no load of 1.5
volts and an internal resistance of 0.25 ohm are connected in parallel to an
external circuit which has a resistance of 0.5 ohm. What is the current in
the external circuit; also that in each cell? IB.87 amp.; 0.24 amp.
(6) If the cells are connected six in series, two in parallel, what is then the
current in the external circuit and that in each cell? 7.S amp.; S.6 amp.
(c) If one of the cells in this latter case is taken out of circuit, what is then
the current in the external circuit and that in each cell? What also will be
the circulating current if the external circuit is disconnected?
6.96y S.7, S.26 amp.; 0.66 amp. circtdating.
151. (a) Taking the curves of charge and discharge given in Fig. 176, page
161, what is the volt efficiency for a lead cell and also for an Edison cell?
87 per cent.; 78 per cent.
(6) If the rating of the lead cell is 100 amp.hours at the 8hour rate,
what current can be drawn from this cell for 8 hours and what will be the volt
age of the cell at the end of this time? If the cell takes 14.5 amp. for 8
hours to recharge, what is the amperehour and the watthour efficiency?
12.6 amp.; 1.8 volts; 86 per cent,; 76 per cent.
PRACTICE OF ELECTRICAL ENGINEERING 406
(c) What current may be drawn from this cell at the 4hour rate?
About 20 amp.
162. A small isolated power plant has a day load of 50hp., at 110 volts,
and a night load of twentyfive 110volt tungsten lamps of 32 cp. and 40
watts each. The lamp load has to be carried for 12 hours by a storage
battery. Specify the battery, the charging resistance (page 171), the size
of the generator and the horse power of the engine. State the assumptions
made.
168. A small storage battery with three cells in series and a capacity of
100 amp. hours at the 4hour rate is charged from 110volt mains at the
4hour rate. What is the battery voltage at the beginning of the charge
and at the end (see Fig. 176)? About 6 and 7,5 volts.
Specify the charging resistance. S.5 ohms to carry SO amp.
What does it cost to charge the battery if tho cost of charging is 8 cents
per kw.hour? $/ .06,
What would it cost to charge four such batteries in series? $t,06,
CHAPTERS 27 AND 28
164. A 24pole alternator is driven at 300 r. p.m. What is the frequency of
the generated e.mi.? 60 cycles.
How many poles would a 25cycle alternator have for the same speed?
10 poles.
How many poles has a turbo alternator that gives 60 cycles at 3600 r.p.m.?
2 poles.
166. Draw a curve of sin* $, Show that the average value of sin* $ is equal
to .0.5 times the maximum value, and that therefore the average heating
effect of an alternating current with a maximum value of 1 amp. is the same
as that of a direct current of l/\/2 amp.
(5) Which produces the greater heating effect, 25 amp. direct current or
an alternating current which h s a value of 25 amp. as measured by an
instrument? Eqital effect,
(c) Whether will an alternating voltage of 2200 as measured by an instru
ment or a direct voltage of 2200 break down insulating material the more
readily? Alternating voltage,
156. Two 6pole alternators, A and B, rigidly connected to the same shaft,
each generate 120 volts. If the poles of machine B are displaced by 20
mechanical degrees relative to those of machine A while the stationary
armatures are in line what is the phase angle between the two voltages and
what is the resultant voltage?
(a) If the windings are in series as in Fig. 226? 60 degrees; 208 volts.
(6) If the connections of B are reversed? 240 degrees; 120 volts.
Show the result by sine curves and also by vector diagrams.
167. (a) By means of a rotating vector, construct a sine wave of voltage
having a maximum value of 100 volts. If the frequency of this voltage is
406
PROBLEMS TO ACCOMPANY
60 cycles per sec. indicate the scale of absciaBs in seconds, fractions of
a cycle, radians and in electrical degrees.
(6) Assume that the alternator generating the above e.m J. has 14 poles.
Through how many mechaniccd degrees will the shaft of this alternator turn
while the e.m.f. passes through 360 electrical degrees?
168. Draw a sine wave of current with an effective value of 6 amp.
If this be the curve obtained by an oscillograph at a in the adjoining circuit
plot the curves that you would expect to obtain by oscillographs placed at
b and c respectively.
d6
Diagram No. 5.
169. The above diagram represents a 6amp. series lighting circuit. When
the current at a has the maximum value of 8.5 amps, in one direction what
will be the value and direction of the currents at b and c at the same instant?
ISO. Find the average value of onehalf cycle of the above curve taking
at least 10 ordinates and compare this with the value given on page 197 of
text. Will the average value for a quarter cycle be the same? Of a whole
cycle? Of an eighth of a cycle?
161. The effective value of an alternating voltage is defined as
=V (average value of e*). Show by squaring the above chosen ordinates
that this is equal to ^maz. divided by \/2.
Why is \/(av. value of e*) not the same as (av. value of e)7
162. Explain clearly, as to a nontechnical man, the difference between
alternating and direct currents.
(a) What actually happens in the wire?
(6) What is the frequency of supply in this city?
(c) Why do we not see the lamps flicker?
{d) What is the frequency of flicker?
(e) What is meant by amplitude of a sine wave?
163. Find the number of poles for each of the following A.C. generators:
14,000 kw. 25 cycle, driven by steam turbine at 1500 r.p.m.
14,000 kw. 60 cycle, driven by steam turbine at 1800 r.p.m.
14,000 kw. 25 cycle, driven by low head water wheel at 58 r.p.m.
14,000 kw. 60 cycle, driven by low head water wheel at 55 r.p.m.
164. An engine builder whose engine runs normally at 650 r.p.m. wishes
a directconnected 60cycle alternator. How many poles should the
alternator have? What would you suggest to the builder?
166. Show with sketches the meter movements of a hot wire meter; a
PRACTICE OF ELECTRICAL ENGINEERING 407
soft iron repulsion type; a dynamometer type. Show that the meter reads
the effective and not the average value of voltage or current. Will the
meters read the same on A.C. and D.C. circuits?
166. Two 6pole alternators are keyed to the same shaft and connected
in series. Each machine generates 100 volts (effective). The poles of
machine A are displaced 20 mechanical degrees behind those of B. Find
the voltage at the terminals and its phase position relative to the two
generator voltages as follows:
(a) Plot to scale in their correct phase relation the waves representing
the voltages of the two machines and then find the resultant sine wave by
adding the voltages of the two machines point by point.
(d) Draw vectors representing the two voltages in their proper phase
relation, find the geometric sum and show that it corresponds in magnitude
and in phase relation with the corresponding curve found by method (a).
167. If the terminals of alternator B are reversed show by means of a
vector diagram the relative magnitudes and phase relations of voltages
A, B and the terminal voltage.
•
CHAPTER 29
168. A circuit has an inductance of 0.1 henry and a noninductive resist
ance of 10 ohms in series. If the circuit is connected across a direct voltage
of 120, what will be the voltage drop across the inductance and also that
across the resistance:
(a) At the instant the switch is closed? Vi = IW, Vr = 0,
(b) After the current has reached its maximum value?
Vi ^0,Vr = 120.
What will be the maximum value of the current ? 12 amp,
169. (a) A coil of 200 turns, wound on a wooden ring which has a mean
diameter of 12 cm. and a circular cross section of 5 sq. cm., is threaded by a
flux of 134 lines of force when the current flowing is 4 amp. (see Prob. 59).
Find the inductance of the coil. 6.7 X 10"^ henry s,
(6) When the wooden ring is replaced by a steel ring of the same dimen
sions the magnetic flux is 50,000 lines when the current is 1.32 amp. (see
Prob. 61). Find the inductance of the coil. 7.6 X 10^ Henrys.
(c) When the steel ring has an air gap of 0.2 cm. length the magnetic flux
produced by a current of 9.32 amp. is 50,000 (Prob. 62), What is the induc
tance of this coil? 1.08 X 10^ henry s,
170. A coil of negligible resistance has an inductance of 0.1 henry and
is connected across a 120volt line:
(o) What current will flow at 30, 60 and 120 cycles? 6.36; 3.18; 1.69.
(6) What is the average power taken from the line in each case? Zero.
(c) What is the maximum rate at which energy is given to the circuit dur
ing one half cycle and returned by the circuit to the line during the next
half cycle (see Fig. 231)? 760, 380, 110 waits.
{d) Explain without formulae why the current is inversely proportional
to the frequency, the voltage being constant.
408 PROBLEMS TO ACCOMPANY
171. A coil with a noninductive resistance of 10 ohms is connected across
a 120 volt line :
(a) What current will flow when the frequency is 30, 60 and 120 cycles?
liS amp,
(6) What 18 the average power taken from the line in each case?
1440 waUs,
(c) What is the maximum rate at which energy is given to the circuit
during each half cycle (see Fig. 237)? ^880 waUs,
172. An inductance of 0.1 henry and a noninductive resistance of 10
ohms are connected in series across a 120volt line:
(a) What current will flow when the frequency is 30, GO and 120 cycles?
6.6, 3,07, 1,67 amp,
(b) What is the average power taken from the line in each case?
S16, 94, 26 toaUs.
(c) What is the voltage drop across each part of the circuit at 60 cycles?
Vi = 116, Vr = S0,7,
(d) What is the power factor of the circuit in each case?
477c; 26,6%; 13%
Note from the above figures that, since the average power taken by an
inductance is zero, that taken by a circuit having resistance and inductance
in series » I*R watts.
178. A coil takes 10 amp. from a 120volt D.C. circuit and takes 3 amp.
from a 120volt, 60cycle line. What is the resistance of the coil, its induct
ive reactance, and its inductance?
12 ohms; 38,1 ohms at 60 cycles; 0,1 henry.
174. A coil takes 3 amp. and 108 watts from a 120volt, 60cycle line.
What is the resistance of the coil and its inductance? 12 ohms; 0,1 henry,
176. Given two coils: A which has a resistance of 3 ohms and an inductive
reactance of 4 ohms at 60 cycles; and B which has a resistance of 5 ohms
and an inductive reactance of 1 ohm at 60 cycles. Find the current in each
coil, the voltage drop across each coil, the power factor of each coil, the
resultant power factor and the line current:
(a) When the coils are connected in series across a 120volt, 60cycle line.
12,7 amp,; 63.8, 66 voUs; 60%; 98%; 85%,
(6) When the coils are connected in parallel across this line.
24, 23,6 amp.; 120 volts; 60%; 98%; 86%; 44,3 amp.
(c) If the two coils A and B in parallel are connected in series with a third
coil C, which has a resistance of 2 ohms and an inductive reactance of 5
ohms at 60 cycles across a 120volt, 60cycle line, what will be the current
in each coil and the voltage drop across each?
la = 8.4; h = 8.2; Ic = 16,6, F^ = F* = 42; F, « 83,
Draw the vector diagram to scale for each of the above cases.
176. (a) An alternatingcurrent motor delivers 10 hp., at 120 volts.
The efficiency of the motor is 85 per cent, and, on account of the reactance
of the motor windings, the current lags behind the voltage and the power
factor of the motor is only 80 per cent. What is the fullload current?
91 amp.
PRACTICE OF ELECTRICAL ENGINEERING 409
(6) What is the fullload current of a 10hp., 120volt, D.C. motor of
85 per cent. eflBciency? 73 amp,
(c) What is the power taken from the line in each case? 8,8 kw,
177. The load on a 120volt alternator consists of 150 incandescent lamps
each of which takes 0.5 amp., also 10 hp. of motors with an average effi
ciency of 85 per cent, and an average power factor of 80 per cent. Find
the current output of the alternator, the kilowatt output of the alternator,
the power factor of the total load (the power factor of incandescent lamps
is 100 per cent., see page 210), and the horse power of the engine if the
alternator efficiency is 90 per cent. 158 amp.; 17,8 kw,; 94%; fS6.5 hp.
178. 75 kw. at 2200 volts and 60 cycles has to be delivered at the end of
a 5mile line. Choose the size of wire so that the copper loss shall not ex
ceed 8 per cent, of the power delivered:
(a) When the power factor of the load at the end of the line is 100%.
(&) When this power factor is 79 per cent., and the current is lagging.
105,500; 167,800 cir. mils (see page £16),
If the spacing of the wires is 48 in., what is the inductive reactance of
the line, the resistance of the line, the voltage at the generating station,
and the power lost in the line?
At 100% power factor: 7,21 ohms; 5.18 ohms; 2S80 volts; 6 kw.
At 79% power factor: 6.92 ohms; 3.26 ohms; 2500 volts; 6 kw.
179. A wooden ring having a mean diameter of 12 cm. and a cross section
of 2 X 5 sq. cml is woimd with 200 turns of wire.
(a) Find the flux per ampere.
(5) Find the inductance of the coil yj 10"* henries) •
(c) Find reactance of coil at 30 and 60 cycles.
(d) Neglecting the resistance of the coil, find the current when connected
across 100 volts with frequencies as in (c).
(e) If a second coil of 600 turns is wound on top of the 200 turns what
will be the inductance of this new coil?
(/) How does the value of L vary with the number of turns of the coil?
(g) If the two coils (600 T and 200 T) are connected in series what will
be the total inductance?
(A) If the two coils are connected in series but their magnetic effects are
in opposition what will be the total inductance?
180. The wooden ring of Prob. 179 is replaced by a laminated steel ring of
the same dimensions.
(a) What is the inductance of the coil of 200 turns? Assume the current
to be less than 0.5 amp. and therefore the permeability is practically
constant. Use Fig. 41.
(6) Neglecting the resistance of the coil find the voltage required to send
0.5amp. 60cycle current through the coil.
181. If a laminated iron core is moved into an A.C. solenoid the alter
nating current is decreased in magnitude. Explain why this is so?
182. If a direct current were flowing through the above coil what effect
would the insertion of the iron core have on the value of the current? Is
there any permanent or momentary change in the current?
410 PROBLEMS TO ACCOMPANY
188. A noninductive resistance takes 10 amp. from 220yolt, 6(V«ycle
mains. What current will it take from 220volt, 30cycle mains?
184. A choke coil of negligible resistance takes 3 amp. from 220volt,
60oycle mains. What current will it take from 22(>yolt, 30cycle mains?
Will the current and voltage be in the same phase relation as before?
What is the average power taken from the line?
What is the maximum rate at which energy is given to the circuit during
one half cycle and returned by the circuit to the line during the next half
cycle for the two frequencies?
186. A resistance of 4 ohms and a reactance of 8 ohms are connected in
series across a 110 volt, GO^ycle line. Find the impedance of the circuit,
the current, the inductance of the coil. Draw a complete vector diagram
for the circuit and label fully.
For what frequency will the current be 21 amp.?
186. A resistance of 11 ohms and a reactance of 10 ohms are connected in
parallel across a 110volt, 60cycle line. Find the current in resistance,
reactance, and in the line. Draw complete vector diagram.
Repeat the problem with the frequency reduced to 30 cycles. Are any
of the currents the same as before?
187. What will be the power taken from the line in Prob. 185? Prove
that it is equal to I*R.
188. What will be the power taken from the line in Prob. 186; show that
it is equal to Ir*R where /r is the current in the resistance?
189. A coil takes 10 amp. from a 240volt D.C. circuit and takes 6 amp.
from a 240volt, 25cycle line. What is the resistance of the coil, its inductive
reactance and its inductance?
190. A coil takes 5 amp. and 240 watts from a 120volty 60cycle line.
What is the resistance of the coil and its inductance?
191. Given 2 coils: A which has a resistance of 3 ohms and an induct
ive reactance of 6 ohms at 60 cycles; and B which has a resistance of 7
ohms and an inductive reactance of 4 ohms at 60 cycles. Find the current
in each coil, the voltage drop across each coil, the power factor of each coil,
the resultant power factor and the line current:
(a) When the coils are connected in series across a 120volt, 60cycle line.
(6) When the coils are connected in parallel across this line.
(c) If the 2 coils A and B in parallel are connected across a 120volt,
60cycle line, in series with a third coil C, which has a resistance of 2 ohms
and an inductive reactance of 5 ohms at 60 cycles, what will be the current
in each coil and the voltage drop across each? Draw the vector diagram to
scale for each of the above cases.
192. Given the voltage and values of R and X draw vector diagram for
the circuit shown on page 411.
198. 75 kw. at 2200 volts and 50 cycles has to be delivered at the end of
a 5mile line. Choose the size of wire so that the copper loss shall not exceed
10 per cent, of the power delivered (see page 216).
PRACTICE OF ELECTRICAL ENGINEERING 41 1
(a) When the power factor of the load at the end of the line is 100 per cent.
(6) When this power factor is 80 per cent, and the current is lagging.
If the spacing of the wires is 48 in., what is the inductive reactance of the
line, the resistance of the liaa, the voltage at the generating station, and the
power lost in the line? Use table on page 216.
Draw a vector diagram to scale.
T
E
1
Diagram No. 6.
194. (a) A 15hp., 120 volt A.C. motor has an efficiency of 85 per cent,
and, on account of the reactance of the motor windings, the current lags
behind the voltage and the power factor of the motor is only 80 per cent.
What is the fullload current?
(6) What is the fullload current of a 15hp., 120volt D.C. motor of
85 per cent, efficiency?
(c) What is the power taken from the line in each case?
196. The load on a 120volt alternator consists of 220 incandescent lamps
of 40 watts each; also 15 hp. of motors with an average efficiency of 85 per
cent, and an average power factor of 80 per cent. Find the current drawn
from the alternator, the kilowatt output of the alternator, the power factor
of the total load (the power factor of incandescent lamps is 100 per cent.;
see page 210), and the horsepower of the engine if the alternator efficiency
is 90 per cent.
196. Explain clearly, with sketch of meter and curves, why an electro
dynamometer type of wattmeter reads "average power" when connected
in an A.C. circuit containing inductance and resistance.
197. In order to determine the power taken by a piece of apparatus " A "
it is connected in series with a noninductive resistance R and the c rrent
and three voltages Eat Er and the line voltage Ee are measured. Draw
a vector diagram and compute power taken by "A" given that Ea ^ 70;
Er — 66; ^, = 110; / — 15 amp. and frequency is 60 cycles. This is called
the "threevoltmeter method of measuring power."
198. Show that the power taken by "A" in the previous problem is
equal to
2R
199. In the above problem if we assume "A " consists of a resistance and
a reactance connected in series find the values of R and X.
200. In Prob. 198, if we assume A consists of a resistance and a reactance
connected in parallel find the values of R and X, Will they be the same
as in Prob. 199? Why?
CHAPTER 80
801. How many sheets of mica 0.1 mm. thick separating plates of 100
412 PROBLEMS TO ACCOMPANY
sq. cm. area are required to construct a condenser of 1 microfarad capacity
(the specific inductive capacity of mica is 6)? 189.
If plates are 0.1 mm. thick what are the dimensions of the condenser?
10X10 X S,8 cm,
802. A condenser of lOOmicrofarad capacity is connected across a 120
volt line:
(a) What current will flow at 30, 60 and 120 cycles? £.iSS; 45; 9 amp.
(6) What is the average power taken from the line in each case? Zero.
(c) What is the maximum rate at which energy is given to the condenser
during one ))alf cycle and returned by the condenser to the line during the
next half cycle (see page 223)? 1670, 6lfi, 1080 waUs.
{d) What is the maximum chaige in the condenser on each half cycle?
0.017 catdomh.
(e) What average current would this maintain in an external circuit for
Hso sec.? What effective current if the current changes as a cosine func
tion (compare with a)? 8.!S amp.; 9 amp.
if) Explain without formulie why the current is directly proportional
to the frequency, the voltage being constant; the reverse is the case for an
inductive circuit.
208. Two condensers, A of 1microfarad capacity and B of 2microf arad
capacity are connected across 120volt D.C. mains. What is the charge
in each condenser, the total charge and the voltage drop across each con
denser:
(a) When the condensers are in parallel?
0. = J.£ X iO*, Qt,^2.4X 10*, e  5.ff X 10"*.
(h) When the condensers are in series?
Q ^0.8X 10^, E» = 80, Eb = 40.
204. A condenser of l(X)microfarad capacity and a noninductive resist
ance of 10 ohms are connected in series across a 120volt line:
(a) What current will flow when the frequency is 30, 60 and 120 cycles?
B.SS, 4.BS, 7.2 amp.
(&) What is the average power taken from the line in each case. Explain
why this is equal to I^R. 60, 179, 620 waUa.
(c) What is the voltage drop across each part of the circuit at 60 cycles?
Vc = 112, Vr = 42.S.
{d) What is the power factor of the circuit in each case?
19%; 86%; 60%.
206. A circuit is made up of il, a noninductive resistance of 10 ohms,
B, an inductance of 0.1 henry, and C, a condenser of l(X)microfarad capacity.
Find the current in each, the total current, the voltage drop across each and
the resultant power factor:
(a) When A, B and C are connected in series across a 120volt, 120cycle
line. 1.9 amp.; Va = 19, Vh = 144, Vc = 26.3; power factor = 16.8%.
(b) When A, B and C are connected in parallel across a 120volt, 120
cycle line. la = 12, h — 169, h — 9, I ^ 141; power factor = 86%.
(c) When E and C in parallel are connected in series with A across a
120volt, 120cycle line.
/. = 6 .84, h = 1.86, /. = 7.7, y. = 68.4, Vh = 102; power factor = 68%.
PRACTICE OF ELECTRICAL ENGINEERING 413
(cO In Prob. (a) what is the frequency of resonance and what is then
the current in the circuit and the voltage drop across each part of the circuit?
What must be the value of the resistance so that the voltage drop across
the condenser at resonance shall not exceed the value of 250 for which it
was designed?
60,6 cycles; 12 amp.; 7« = 120^ Vh = S80y V^ = S80; 16,0 ohms,
206. In Prob. 176 specify the condenser to be placed in parallel with the
60cycle motor so that the resultant power factor may be 100 per cent.
What will then be the currents in the motor, the condenser and the line?
X. « 2.2 ohms, C « 1220 microfarads; 91, 64.6, 7S amp.
What would this lowvoltage condenser cost at $1 a microfarad?
$1220.
207. In Prob. 178 what would be the size of condenser required to raise
the power factor of the load from 79 per cent, to 100 per cent.? With
2200volt condensers at $10 a microfarad would it be cheaper to use con
densers and the small wire or to use the wire of larger crossHsection, the
copper loss in the line to be the same in each case? Cost of wire at 16
cents per lb.
Xa '^ 83 ohmSf C — 320 mf.; condensers = %3200; difference in cost of wire
at 16 cents per lb.  $1600.
208. A condenser of a wireless receiving set has 40 plates each of 30 sq.
cm. area and separated by air spaces of 0.1 cm. what is the capacity of the
condenser in microfarads?
If this condenser is immersed in oil which has a specific inductive capacity
of about 2.5 what is now the capacity?
What is the total capacity of two such condensers in parallel and what
when they are in series?
209* A condenser of 50 microfarad capacity is connected across a 240
volt line:
(a) What current will flow at 30, 60 and 120 cycles?
(&) What is the average power taken from the line in each case?
(c) What is the maximum rate at which energy is given to the con
denser during onehalf cycle and returned by the condenser to the line
during the next half cycle (see page 223)?
(d) What IS the maximum charge in the condenser on each half cycle?
(«) What average current would this maintain in an external circuit for
^80 sec? What effective current if the current changes as a cosine function
(compare with a)?
(/) Explain without formul® why the current is directly proportional
to the frequency, the voltage being constant; the reverse is the case for an
inductive circuit (compare with Prob. 202).
210^ A condenser of 50 microfarad capacity and a noninductive resistance
of 30 ohms are connected in series across a 120volt line:
(o) What current will flow when the frequency is 30, 60 and 120 cycles?
(&) What is the average power taken from the line in each case? Explain
why this is equal to I^R.
(c) What is the voltage drop across each part of the circuit at 60 cydes?
414 PROBLEMS TO ACCOMPANY
211. An inductive reactance of 10 ohms and a capacity reactance of 11
ohms are connected in series across a 110volt, 60cycle line. Find the
current and the voltage drop across each.
If the two reactances were connected in parallel across the same line what
would be the current in each and also the line current?
Draw vector diagrams and explain what will happen in each case if the
inductive reactance be increased to 11 ohms.
212. Given a circuit with a resistance of 10 ohms, a condenser of 0.1275
microfarad and an inductance L all connected in series across a 220volt,
500cy cle line. What will be the value of the inductance so that the voltage
drop across it has the largest possible value and what will then be the
current?
•Draw a vector diagram for the circuit and explain what happens to the
voltages as the resistance is gradually reduced to zero. (This problem is
of interest in wireless telegraphy.)
213. A circuit is made up of A, a noninductive resistance of 30 ohms,
B, an inductance of 0.1 henry, and C, a condenser of 50microfarad capacity.
Find the current in each, the total current, the voltage drop across each and
the resultant power factor.
(a) When A , B and C are connected in series across a 240volt, 120cycle line.
(&) When A, B and C are connected in parallel across a 240volt, 120
cycle line.
(c) When B and C in parallel are connected in series with A across a
240volt, 120cycle line.
(d) In Prob. (a) what is the frequency of resonance and what is then the
current in the circuit and the voltage drop across each part of the circuit?
(e) What must be the value of the resistance so that the voltage drop
across the condenser at resonance shall not exceed the value of 500 for which
it was designed?
214. In Prob. 194 specify the condenser to be placed in parallel with the
60cycle motor so that the resultant power factor may be 100 per cent.
What will then be the currents in the motor, the condenser and the line?
What would this lowvoltage condenser cost at $1 a microfarad?
216. In Prob. 193 what would be the size of condenser required to raise
the power factor of the load from 80 per cent, to 100 per cent.? What
would be the cost with 2200volt condensers at $10 a microfarad?
CHAPTER 31
216. (a) An 8pole alternator has 24 slots, is wound three phase and is
yconnected. Make a developed diagram such as that in Fig. 251 showing
the coils and connections.
(6) Show on another diagram how the coils would be deltaconnected.
(c) If the voltage between terminals is 2200 when the machine is delta
connected, and the current in each line is 1000 amp., what would be the
terminal voltage and line current if the machine were reconnected Y?
S800 voUsy 680 amp.
(d) What is the output if the power factor of the load is 80 per cent.
3050 kw.
PRACTICE OF ELECTRICAL ENGINEERING 415
S17. (a) AnewpimchiiigissuppUedwiihlOslotsforatwoithasewiiidiii^
Make a developed diagram such as Fig. 251 showing the ooib and the ood>
nections to the external circuit.
(6) Draw a vector diagram showing the voltages and ourrents if the power
factor of the load is 80 per cent.
(c) If the voltage per phase is 2400 and the current per phase is 800 amp.»
what is the output in kw., the rating of the alternator in kva. (page 251),
and the horse power of the driving engine if the alternator efficiency is ^4
per cent. 7 S060 kw.; S8S0 kva,, 4S60 hp.
(d) If the two windings were connected in series, what would be the
resultant singlephase voltage and what would be the rating of the alternator
if the current in the winding is limited to 800 amp.?
S400 rolte, trOO kva.
(e) If the power factor of the external singlephase circuit is 86 per cent.,
show on a vector diagram, the voltage of each phase, the current, and the
resultant voltage. What is the power factor of each phase considered
separately? 96,6%; t6.7%,
218. The load on a threephase 120volt alternator consists of 150 incan
descent lamps, each of which takes 0.5 amp. at 120 volts, also a 10hp. motor
with an efficiency of 85 per cent, and a power factor of 80 per cent.
(a) The lamps are connected in three groups of 50 each, must they be K
or deltaconnected to the lines? Delta,
(h) Is it necessary to know how the windings of the threephase motor
are connected?
NOf they are already properly connected by ike manufacturer,
(c) Find the current in each Hne supplying the motor, the current in each
line supplying the lamps, the readings of the voltmeter, ammeter and the
sum of the readings of the two wattmeters connected in the circuit as shown
in Fig. 271. 63 amp., 4S,S amp,, ISO volts, 91,6 amp,, 17.8 kw,
219. (a) Three heater units delta connected to a 120volt threephase line
take 10 amp. each at 100 per cent, power factor, what is the line current, the
total power given to the heaters, the resistance of each?
17jS amp., S.6 kw., OS ohms,
(6) If Yconnected heaters were used to give the same total heat, what
would be the resistance of each, the current in each and the current in the
lines? 4 ohm^, 17,3 amp., 17.3 amp,
220. What would happen :
(o) If the heaters to be deltaconnected were connected F?
Not hot enoitgh,
(6) If the heaters to be Tconnected were connected delta? Burn up,
221. An alternator has an output of 100 kva. at 2400 volts and gives 60
cycles at 1200 r.p.m. what would be the voltage, frequency and rating at 600
r.pm.? 1200 volts, 30 cycles, less than 60 kva,
222. If a 25cycle, 10pole, threephase alternator with 30 slots on the
armature has to be changed to 60 cycles, what changes must be made on the
machine if it is already running at its highest safe speed (it would
probably be cheaper to buy a new machine) ?
Supply 24 poles, 72 slot armature, new coils.
416 PROBLEMS TO ACCOMPANY
828. In a certain threephaae alternator the voltage generated in the
windings of each of the three phases is 1270. The conductors have such a
cross section that they will safely carry 100 amp.
(a) If the machine has 8 poles and 24 slots make a developed diagram
such as Fig. 251 showing the coils and their connections.
(h) Show how the three phases would be connected together to make a
yconnected alternator. Show on another diagram how the machine would
be delta connected.
(c) What is the terminal voltage, the current per phase, the line current
and the output in kilovoltamperes for the two connections?
884. If all three windings were connected in series so as to give the largest
possible singlephase voltage at the terminals what would then be the kilo*
voltampere output of the machine? Show the connections of the coils
under this condition.
885. Draw sine curves representing the voltages generated in the three
phases and label them Ay B and C.
(a) Let the machine be Fconnected and combine the curves so as to
obtain the curves representing the terminal voltages. (Subtract the
ordinates of adjacent curves.)
(6) Draw the vector diagram for the above six sine curves and show that
the terminal voltage is equal to \/3^ times voltage per phase.
(c) For the deltaconnected windings draw curves representing the
currents per phase (100 amp.) also resultant curves representing the line
current which at any instant is the difference between the currents in the
two phases connected to that line.
(a) Show from the curves and corresponding vector diagram that the
line current is \/3 times the current per phase.
886. The load on a threephase, 120volt alternator consists of 150
incandescent lamps of 40 watts each, also a 15 hp. motor with an efficiency
of 85 per cent, and a power factor of 80 per cent.
(a) The lamps are connected in three groups of 50 each, must they be
y or deltaconnected to the lines?
(6) Is it necessary to know how the windings of the threephase motor
are connected?
(c) Make a diagram of the system and find the current in each line
supplying the motor, the current in each line suppl3ring the lamps, the read
ings of the voltmeter, ammeter and the sum of the readings of the two watt
meters connected in the circuit as shown in Fig. 271. What is the rating
of the alternator?
227. (a) Three heater units deltaconnected to a 120volt threephase
line are wound noninductively and have a resistance of 10 ohms. What
is the current in each heater, the line current and the total power given
to the heaters?
(&) If Fconnected heaters were used to give the same total heat, what
would be the resistance of each, the current in each and the current in the
lines?
228. Explain in detail what would happen :
(a) If the heaters to be deltaconnected were connected F?
(}>) If the heaters to be Feonnected were connected delta?
PRACTICE OF ELECTRICAL ENGINEERING 417
CHAPTER 82
229. A singlepliafle alternator has an output of 1000 kva. at 6600 volts.
When rotating at normal speed the noload voltage with a certain field
excitation was 6600 and the current on short circuit with the same field
excitation was 380 amp. The resistance of the winding is 1.5 ohms.
(a) find the reactance of the winding. 17.4 ohms.
{h) Find the regulation at 100 per cent, power factor, also at 80 per cent,
and at zero power factor with lagging current, the fullload voltage being
6600 volts in each case. 10.8%; 30%; 40%.
230. A threephase alternator has an output of 1000 kva. at 2400 volts
and is yconnected. The stray power loss is 20 kw., the exciting current is
125 amp. at 100 per cent, power factor and 160 amp. at 80 per cent, power
factor while the exciter voltage is 120. The resistance of each phase of the
armature winding is 0.5 ohms. Find:
(a) The armature copper loss. 8.6 kw.
(&) The fullload efficiency at 100 per cent, and at 80 per cent, power
factor. 96%, 94.6%.
231. The load of a plant consists of 1000 hp. of 2200volt threephase
motors with an average efficiency of 90 per cent, and an average power factor
of 80 per cent. Find the current in the lines and the rating of the alternator
required. iS72 amp., IO40 kva.
(a) Determine the rating of the overexcited synchronous motor required
to raise the power factor to 100 per cent., also that required to raise the power
factor to 90 per cent., the synchronous motor running without load.
6B0 kva.; 226 kva.
{b) Determine the capacity of three Fconnected condensers required to
raise the power factor of the load to 100 per cent. S40 m.f. each.
Which is cheaper, condensers at $10 a microfarad or a synchronous motor
at $10 a kva. 7 Condensers $10,200; syn. motor %6200.
(c) If 500 hp. of the load may suitably be driven by a synchronous motor,
what must be the input rating of this motor so that the power factor of the
load may be 100 per cent., the efficiency being 90 per cent. ? 620 kva.
(d) What effect has the power factor of the load on the size of alternators,
the loss in the transmission Une, the size of engines and boilers?
282. The following data were taken on noload saturation and short
circuit runs of a 220kva. 2200volt, 60cycle, 600r.p.m. singlephase
alternator.
Field current
in amp.
20
No load
▼olts
740
Short circuit
amp.
65
40
1460
130
60
2100
198
80
2600
260
100
2030
325
120
3190
385
140
3380
460
160
3520
525
The resistance of the armature is 1.1 ohm.
418 PROBLEMS TO ACCOMPANY
(a) Plot the noload saturation and shortcircuit curves from the above
data.
(6) Find the reactance of the armature and plot on same sheet as the
curves in (a). (Note that the reactance is not constant when the field
circuit approaches saturation.)
(c) Assuming a constant value of reactance of 10 ohms what is the re
sistance drop and what the reactance drop with fullload current?
(d) Find the regulation at 100 per cent, power factor also at 80 per cent,
power factor with leading current and 60 per cent, power factor with
lagging current the terminal voltage being 2200 at full load in each case.
(e) If the terminal voltage has to be kept constant at all loads by means
of a voltage regulator what range of excitation must be taken care of by the
regulator for all loads from no load to full load at 60 per cent, power factor
lagging?
(/) Why is the field current of the machine so large compared with
the armature current?
(g) Why is it safe to short circuit an A.C. generator even with full field
excitation while a D.C. generator would bum up with the same treatment?
{h) If the coreloss, windage and friction of this alternator at normal
voltage are 6 kw. what are the losses at full load with 100 per cent, power
factor and with 60 per cent, power factor and lagging current and what are
the efficiencies under these two conditions of operation?
233. An alternator of 1000 kva. capacity has an efficiency of 93 per cent,
at 100 per cent, power factor and of 91 per cent, at 80 per cent, power
factor with lagging current.
(a) What size of engine would you use if the machine is used entirely
for incandescent lighting load?
(6) What size of engine would you use if it were to be used for motor load
and it was known that the power factor would never exceed 80 per cent. ?
CHAPTER 34
234. Prove that the effective value of the secondary voltage of a trans
former is given by the expression
Et = 4.44iV</>/10« (see Art. 241 for hint as to method)
where N is the number of turns, 4> is the maximum value of the flux and /
the frequency.
236. A 15kva. 2200/220volt, 60cycle lighting transformer has a core
area of 8.8 sq. in. There are 1440 turns on the primary winding. How
many secondary turns are there? Find the flux through the core. Is this
an average, effective or maximum value? What is the maximum flux
density? How many ampere turns per centimeter are required to set up the
flux (use wroughtiron curve of Fig. 41 of text) ? What will be the effective
value of the primary current required to produce this flux density if the
length of the magnetic circuit is 100 cm.? What percentage of fullload
current is this?
236. Why does the primary current increase as the resistance of the
secondary connected circuit decreases?
PRACTICE OF ELECTRICAL ENGINEERING 419
287. What is meant by the leakage reactance of a transformer and why
is it reasonable to assume that it is constant?
288* How does the flux through the magnetic circuit vary with increase
of load current from zero to full load? How do the leakage fluxes change
during the same time?
239. A transformer has 2 turns on the secondary for every one turn on
the primary.
(a) The primary resistance » 0.3 ohm; the primary reactance » 0.9
ohm; the secondary resistance "■1.2 ohms, the secondary reactance » 3.6.
The transformer on noload takes 20 amp. and 1000 watts at 200 volts.
Draw the vector diagram to scale for noload conditions.
(b) The load on the secondary is 50 amp. at 400 volts and 80 per cent,
power factor. Draw the complete vector diagram and find the applied volt
age and the primary current and power factor by scaling from the diagram.
(e) Give in tabulated form the order in which you drew the vectors and
the reasons for their direction and magnitude.
(d) Why is the primary power factor almost equal to the secondary powei
factor on full load, but is lower at light loads?
(e) Is the voltage regulation better or worse at 60 per cent, than at 80
per cent, power factor. Draw an actual vector diagram for each case if
you are not certain.
240. A 1600kva., 63,600/13,200volt, 25cycle transformer has a core loss
of 20 kw. The primary resistance » 16.2 ohms and the secondary resist
ance — 0.53 ohm. Find the efficiency of the transformer at fullload cur
rent, unity power factor. What will it be at fullload current, 80 per cent.
power factor?
If the transformer of the previous problem is operated at full load and
100 per cent, power factor for only 6 hr. of the 24, but is left connected to
the line all day, what is its allday or energy efficiency?
241. Why can an alternator be short circuited safely? Is it safe to short
circuit a transformer? What are the per cent, resistance and reactance
drops in the transformer of Prob. 239.
CHAPTER 36
242. Specify the transformers required for a 250hp., 2200volt, three
phase motor of 90 per cent, power factor and 92 per cent, efficiency, the
line voltage being 13,200 volts.
Three 75'kva. tranaformere 1S2O0/2W0 voUa; delta to delta,
248. If one of these deltaconnected transformers is cut out so that the
bank is operating open delta, to what value must the load on the motor be re
duced to prevent overheating of the transformers? AbotU 144 hp.
244. If the above threephase motor has to be supplied from a 2200volt
twophase line, specify the Scottconnected transformers required and give
the current capacity of both primary and secondary windings.
ne.5 kva.; 2200/2200 volts; 610/690 amp.
112.6 kva,; 2200/1900 volts; 610/690 amp,
246. Three transformers are connected Y on the highvoltage and delta
on the lowvoltage side. If the total output is 300 kva., the primary voltage
20
420 PROBLEMS TO ACCOMPANY
2200 and the seeondary voltage 220, what is the voltage, the current, the
rating of each transformer? Ig70/2i0 voUa; 79/4S6 amp,; 100 h>a.
249. A 5amp. 56volt incandescent lamp has to be operated from a 110
volt line by (a) a resistance in series, (6) a reactance in series, (c) an auto
transformer of 00 per cent, efficiency. Find for each case the additional
loss, the current taken from the line and the power factor of the total load.
(a) 976 watU; 6 amp,; 100%
(6) Sfruitt; 6 amp.; 60%
(c) SO vhUU; t,78 amp.; about 100%.
247. It is desired to deliver 110 kw. at a distance of 10 miles from a
power house with not more than 10 per cent, loss in I*R of line. The power
is to be used at 110 volts for lighting a laige factory (100 per cent, power
factor). Find the sise and cost of copper wire and total cost of wire and
transformers for singlephase line with: (1) 110volt transmission. (2)
2200volt transmission. Transformers cost $5 per kva. (3) 11,000volt
transmission. Transformers cost $7 per kva.
248. If 2200/110volt transformers connected FKielta deliver 300kw.
at 80 per cent, power factor find the phase and line voltages and currents.
249. A 3pha8e, 06,000volt transmission line supplies 3 loads A, B, and
C Load A consists of 400 hp. of motors, 2200 volt, 85 per cent, efficiency,
80 per cent, power factor. Load B supplies a transmission line at 13,000
volts to a substation where deltadelta transformers lower the voltage to
2200 volts. The voltage is then stepped down to 110 by deltadelta con
nected transformers for an A.C. incandescent lighting load of 500 kw. The
loss in the lines is 8 per cent.
Load C supplies 400 60 watt lamps at 110 volts for local lighting near
the power house.
The 66,000volt transformers are all connected Kdelta.
(a) Make a diagram of the system.
(&) Find currents in all line wires and all transformer windings.
(c) Find kilovoltampere ratings of all individual transformers, also
the voltage ratio.
(d) Find the current in the 66,000volt lines and the resultant power
factor. (Note that transformer losses are neglected and primary power
factor assumed equal to secondary power factor).
CHAPTERS 86 AND 87
260. The following data was obtained by test on a 5hp., 220volt, three
phase, 60cycle, squirrelcage, induction motor.
R.p.m.
1200
1102
1180
1150
1110
(a) How many poles has the motor and what is the span of the stator
coils? 6 polea; 1/6 of cvrcwmferonot.
Terminal voltase
Amperes
per line
220 (no load)
5.2
220
6.6
220
8.5
220 (fuU load)
13.2
220
19.5
220 (at standstill)
72.0
Wattmeter
Torque in
readings
lb. at 1 ft.
+ 690/ 390
0.0
+ 1280/ + 50
5.5
+ 1810/4 540
11.2
+ 2910/4 1490
23.0
+ 4260/4 2340
36.0
413000/ 1400
35.0
PRACTICE OF ELECTRICAL ENGINEERING 421
(6) If a 900 r.p.m. motor is wanted what changes must be made?
New coiU to span 1/8 of circumference,
(c) What is the noload loss and what is the input at full load?
SOO;MOO watts,
(d) Plot power factor and eflSciency on a horsepower output base.
FvUrload effic. = 8S%; power factor = 87%.
(e) Answer the questions in Experiment 19, page 380.
(/) If an autotransformer is used to reduce the starting current, what is
the secondary e.m.f . for fullload torque and what are then the current in
the motor, the current in the line, and the power factor?
180 voUs, 69 amp., 48 amp,, 4^%,
(g) What would have been the starting current for fullload torque if
the motor had been of the woundrotor type with a suitable resistance in the
rotor circuit? lS.iS amp.
(h) Since the woundrotor motor gives fuUload torque at starting with
fullload current, what must be the power factor at starting?
Same as at full load, 87%.
261. What happens when the voltage applied to a motor carrying full load
is reduced to 90 per cent, of the normal value?
(a) To the strength of the rotating field?
Reduced to 0.9 normal. See Art. 316.
(6) To the rotor current? Increases to 1.11 normal wdue.
(c) To the speed?
Reduced to give 1.11 normal current with 0.9 normal field; slip increased
to 1.24 normal.
Efficiency and power factor are not much changed.
262. Draw a diagram similar to Fig. 336 for a 6pole, threephase, 60
cycle, induction motor with one slot per phase per pole (18 total). Plot
the resulting magnetic field for the four instants A, B, C and D, Fig. 335,
and from these diagrams find the speed of the revolving field in revolutions
per minute and derive the speed formula in terms of the number of poles
and the frequency. (Be sure that you get the starts of the three phases
120 deg. and not 60 electrical degrees apart).
268. In starting a squirrelcage induction motor why is the torque very
low though the current drawn from the line is very large?
264. Why does the reactance of the rotor vary greatly between starting
and running conditions?
266. How is it possible to get a good starting torque with a polyphase
induction motor?
266. Draw speedload curves for D.C. shunt and series and A.C. syn
chronous and induction motors. What is the per cent, change in speed
between no load and full load for each motor? What is meant by syn
chronous speed?
267. On what does the torque of an induction motor depend? Why
does the torque vary as the square of the applied e.m.f.?
268. What must be the frequency of the line if a &pole induction
motor runs at 1125 r.p.m. on full load? Tabulate the various possible
422 PROBLEMS TO ACCOMPANY
synchronous speeds of induction motors used on 25cycle and 60cycle lines.
This table is of importance to a fan or pump designer.
CHAPTERS as AND 86
269. Describe the oonstniotion of a synchronous motor, naming its
principal parts. How is the field of a synchronous motor excited? Can
the speed be varied by varying field current?
260. Why 15 a single phase synchronous motor not selfnstarting? How
can it be started and made to carry load?
What three conditions must be fulfilled before closing the switch to throw
the motor on the line? In case any one of them is not satisfied what will
be the result?
261. A synchronous motor is frequently started by means of an induction
motor of relatively small capacity connected to the same shaft. How many
poles should an induction motor, used for this puri)ose, have if the normal
speed of the 60cycle synchronous motor is 720 r.p.m.?
262. In throwing a polyphase synchronous motor on the line what fourth
condition must be fulfilled in addition to the three of Prob. 2607 .It is
possible to start a polsrphase synchronous motor as an induction motor
itself without the use of a separate starting motor (see page 379, also page
294).
268. A threephase, 2200volt, 60cycle, 600 r.p.m. synchronous motor is
direct connected to a D.C. generator which delivers 500 kw. to a city railway
line. The efficiency of the set is 87 per cent, at full load. The set is started
by means of an auxiliary 40hp. induction motor on the same shaft. The
synchronous motor field is overexcited so as to draw a leading current of
90 per cent, power factor.
A group of 550volt, threephase, induction motors aggregating 600 hp.
at an average efficiency of 82 per cent, and average power factor of 80 per
cent, are supplied through deltadelta connected transformers from the
same 2200 volt mains. Neglect losses and voltage drops in transformers.
The above loads are carried by one 32pole alternator having a direct
connected exciter, the efficiency being 90 per cent.
(a) What will be the current in the 2200 volt line due to the synchronous
motor load? Due to the induction motor load? Total current per wire?
Show on vector diagram.
(&) What will be the voltages and currents in the deltadelta transformer
windings?
(c) Compute the horse power of the Corliss engine driving the alternator.
What should be its speed?
(d) Check horse power of the engine ■= horse power load X offi ~( it ) ^
1
effic. (load) '
(«) Why not use a 5hp. induction motor instead of a 40hp. one to start
the synchronous set? What constitutes the load oh this starting motor?
(/; How many poles should the auxiliary induction motor have?
INDEX
Adjustable speed operation, in Aro welding generator, 190
duction motors, 293, 298, Armature, 49 58,
300
series motors, 92
shunt motors, 89, 105112
Air blast transformers, 271
Air compressors, drive for, 296
Air gap, direct current machines, 48,
68
induction motors, 292
All day efficiency, 269
Alternating current, 48, 197
Alternator, armature reaction of,
244
characteristics, 244
construction, 229, 239
efficiency, 250
excitation, 240
inductor type, 241
magneto, 242
parallel operation of, 259
rating, 251
reactance, 246
•regulation, 245, 248
revolving armature type 240
revolving field type 229, 239
simple, 191
singlephase^ 231
threephase, 231
twophase, 230
vector diagram, 245
Aluminium arrester, 344
Amalgamation of zinc, 143
Ammeter, 8, 19, 198
Ammeter shunt, 368
Ampere, 5, 18
Amperehour, 18, 33
Amperehour efficiency, 153, 161
Ampereturn, 33
Anode, 139
Arc lamps, 354, 365
Arc light generator, 75
Armature copper loss, 95
core, 55
reaction, 67, 83, 89, 244, 312
resistance drop, 72, 79
windings, 48, 230
Arrester, lightning, 343
Automatic current regulator, 76
feeder regulator, 281
motor starter, 130« 133, 306.
switch, 183
voltage regulator, 74, 249
Automobile batteries, 149, 151, 153
Hghting, 186
Autotransformer, 279, 301
Average current, 197
Axle generator, 182
Back e.m.f., 79, 253, 261, ftll
Balanced load, 238» 316
Balancer, 316
Battery capacity, 153, 155, 175
characteristics, 152, 160
construction, 148, 158
control, 171179
efficiency, 153, 161
electromotive force, 141 142,
151, 161
primary, 140145
resistance, 141, 152, 161
storage, 146161
temperature, 155, 161
Blowout coil, 115, 123, 129
Booster, 173, 176, 180, 315
Boosting transformer, 281
Boring mill, motor for, 108
Brake, 46, 333
Brake test, 98
Braking, dynamic, 333
Brushes, 50, 58
Brushes, carbon, 66
423
424
INDEX
Brushes, resistance of, 63, 66
shifting of, 63, 83
Bucking field coils, 186
Building up of voltage, 71
CA.V. generator, 189
Cables, underground, 347
Candlepower, 354, 365
Capacity circuits, 218
Capacity of a battery, 153, 155, 175
Capacity reactance, 221
Car lighting, 181
Carbon battery regulator, 176
brushes, 66
incandescent lamp, 352
lamp regulator, 170, 182
pile rheostat, 29
switch contacts, 114
Cast iron grid resistance, 27
Cement mill motors, 297
Characteristics, alternator, 244
battery, 152, 160
d.c. generator, 7077
d.c. motor, 8594
induction motor, 290, 292
singlephase motor 310, 311
Charge of a condenser, 218
Choke coil, 343
Circuit, breaker, 39, 117, 170
f ormulie, 223
magnetic, 33, 46
Circuits, electric and hydraulic, 18
parallel and series, 22, 212, 216,
224,226
Circular mil, 20
Circulating oil method of cooling,
272
Clutch, electromagnetic, 46
Coefficient of adhesion, 323
of self induction, 206
Color of light, 362
Commutation, 62, 68, 83
limit of output, 100
of a.c. motors, 314
Commutator, 50, 58
Compensating winding, 312
Compensator, starting, 302.
Compound excitation, 61
generator, 74, 166
Compound motor, 93
starter, 120
Condensers, 218, 222
Constant current generator, 75
regulator, 76
transformer, 267
Contactor switch, 116, 132
Contacts, carbon, 114
Control, of batteries, 171179
of railway motors, 126, 133, 327
multiple unit, 133
series parallel, 126
Controllers, 122, 124
automatic, 130
crane, 122
magnetic switch, 131
master, 132
Conversion factors, 15
Converter, mercury vapor, 357
rotary, 318
Cooling of machines, 99, 100
transformers, 270
Copper losses, 95
Core depth, 49, 56
laminations, 55
Core type transformers, 279
Corkscrew law, 5
Cost of motors, 101, 108
Coulomb, 18
Coulomb's law, 1
Counter e.m.f ., 79 ^
Crane motors, 92, 103, 298, 332
Cross magnetizing effect, 67, 83, 312
Current, alternating, 48, 197, 212
direct, 48
direction of, 4
effective, 197, 212
unit, 5
Current carrying capacity of wires,
368
Current transformer, 351
Cycle, 194
Dampers, 295
DanieU cell, 141
Delta connection, 233, 236, 276, 278,
Demagnetizing effect, 68, 83
Dielectric strength, 22
Differential booster, 176
INDEX
425
Direct current, 48
Direct current generator, see Gene
rators
Direction of current, 4
ejni., 9, 11
force on a conductor, 7
maipietic field, 2
Disconnecting switch, 345
Driving force of a motor, 78
Drop in armature, 72, 79
in transmission lines, 23
Drum type controllers, 124
windings, 52
Dry cells, 143
Dynamic braking, 333
Eddy current loss, 96, 269
Edison battery, 157
Edison Lalande cell, 144
Effective current, 197, 212
Efficiency, 97
alternator, 250
battery, 152, 161
directcurrent machines, 98
illmninants, 360
induction motor, 299
rotary converter, 320
transformer, 268
values of, 98
Electric furnace, 74
hammer, 38
locomotive, 332
welder, 264
Electrical degrees, 200
energy, 14
power, 14
Electrodynamometer instruments,
199
Electrolysis, 139
Electrolyte, 139, 156, 160
Electromagnet, 5, 37
Electromagnetic brakes, 46, 333
clutches, 46
induction, 9
motor, 41
Electromotive force, back, 79
direction of, 9, 11
generation of, 9, 11
of a battery, 141, 142, 151, 161
Electromotive force, of self induc
tion, 12,
unit of, 9
Elevator motors, 103
Enclosed arc lamps, 355
Enclosed machines, 100
End cell control of batteries, 172
Energy, chemical and electrical, 140
heat and electrical, 15
mechanical and electrical, 14
required for an electric oar, 326
Equalizer connection, 166
Ewing's theory of magnetism, 36
Excitation, 59, 71
Exciter for alternators, 240
Exciting current, 60
External characteristics, 71
Fan drive. 111, 112, 296, 300
Farads, 219
Farm house lighting, 169
Feeder regulator, 281, 339
Field copper loss, 95
Field, magnetic, 15, 32
Field, reversing, 65, 68, 83
Field rheostat, 74
Flame arc lamp, 356
Flashing at switches, 12
Flat compounding, 75
Fleming's rule, 9
Float switch control, 130
Floating battery, 154, 179
Flux density, 3, 33, 44
Flywheel motor generator set, 335
Flywheel, use of, 94, 104
Force on a conductor, 6
Formula for circuits, 223
Frequency, 194, 341
natural, 226
of resonance, 226
Frequency changer, 321
meter, 195
Furnace, induction, 263, 267
Fuses, 117
Gasfilled lamp, 353
Gassing of batteries, 147
Generator, axle driven, 182
car lighting, 181
constant current, 75
426
INDEX
Generator, direetonrrenti armature
reaction^ 67
characteristics, 71
commutation, 82, 08
compound, 74
construction, 56
excitation, 60
limits of output, 100
parallel operation, 164, 166
regulation, 71
series, 75
shunt, 72
windings, 48
induction, 294
retarding force of, 78
threewire, 317
variable speed, 181
Glare, 363
Gramme ring winding, 48
Grid resistance, 27
Growth of current, 12
Hammer, electric, 38
Head and end system, 181
Heat and electrical energy, 15
Heater units, 27
Heating of machines, 99, 100, 239
of transformers, 270
Henry, 207
Hoist motor, 332
Hoisting, 332
Holding magnet, 43
Horn gap arrester, 345
switch, 116
Horsepower, 14
Horsepowerhour, 15
Hunting, 258, 295
Hydrometer, 156
Hysteresis, 36
loss, 95, 269
Ignition, make and break, 206
Illuminants, efficiency, 360
Illumination, principles, 362
intensity, 364
Incandescent lamps, 352
Inductance, 205, 209
adjustable, 210
of transmission line, 211, 216
Induction, electromagnetic, 9
furnace, 263, 267
generator, 294
motcxr, adjustable speed opera
tion, 293, 298
applications, 296
characteristics, 291
construction, 283
eflSciency, 290, 299
for raQway service, 327
power factor, 293
singlephase, 306
slip, 290
speed, 287, 290, 293
squirrel cage, 283, 296
starters, 301^07
starting torque, 287, 297
vector diagram, 291
wound rotor, 288 297
mutual, 11
i^^tor, 281
self, 11, 206
Inductive circuit, 207, 210
reactance, 208
Inductor alternator, 241
Instrument transformers, 351
InsuLiting materials, 22, 26, 32, 99
Insulation of windings, 58
Insulators, line, 346
Intensity of illumination, 364
of magnetic field, 2, 3
Intermittent ratings, 101
Internal resistance of a battery, 141^
152, 161
Interpole machines, 65, 100
Ions, 139
Iron loss, 96
Iron, magnetic properties, 32
Ironclad solenoids, 41
Isolated lighting plants, 160
Joules, 14
Kathode, 139
Kilovoltampere, 251
KUowatt, 14
Kilowatthour, 15
Knife switches, 114
INDEX
427
Lagging current, 203, 208
Lamination of armature core, 55
transformer core, 269
Lamps, arc, 354
connection of, 365
incandescent, 352
mercury vapor, 359
Lamp circuit regulator, 170, 182
Lathes, drive for, 108, 299
Lead battery, 146
Leading current, 203, 221
Leakage reactance of transformers,
264
Leclanch^ cell, 143
Left hand rule, 7
Lenz's law, 10
Lifting magnets, 43
Light, quality of, 363
unit of, 354
Lighting of cars and vehicles, 181
country roads, 363, 364
drawing offices, 363
farm houses, 169
reading rooms, 362
streets, 360, 364
Lightning arresters, 343
protection, 346
Limits of output, 100, 255, 291
Line construction, 346
Line shaft drive, 102, 297
Lines of force, 2, 3
Liquid rheostat, 29, 335
Loading back tests, 163
Local action, 143
Locomotive, electric, 332
Long shunt, 61
Losses in alternators, 250
directcurrent machines, 95
transformers, 268
transmission lines, 215
Lumens, 364
Luminous arc lamps, 357
Machine tool drive, 108, 299
Magnet, 1
alternating current, 306
electro, 5, 37
lifting, 43
permanent, 35, 59
Magnet, pull of, 40
Magnetic, brakes, 46, 333
circuit, 33, 46
clutches, 46
field, 15, 32
flux density, 3, 33, 44
hammer, 38
properties of iron, 32
separator, 47
switch controller, 131
Magnetism, molecular theory of, 36
residual, 35, 70
Magnetization curves, 34, 70
Magnetizing current, induction
motor, 292
transformer, 262
Magneto, 59, 242
Magnetomotive force, 34
Make and break ignition, 206
Manganin, 21
Manholes, 347
Master controller, 132
Maximum output of directcurrent
machines, 100
induction motors, 291
synchronous motors, 255
Maxwells' formula, 44
Mercury vapor converter, 357
lamp, 359
Mechanical losses, 95
Mil, circular, 20
Molecular magnets, 36
Motor, directcurrent, armature re
action, 83, 89
commutation, 83
compound, 93
driving force, 78
electromagnetic, 41
limits of output, 100
railway, 327
series, 90
shunt, 85
speed, 80, 82
starters for, 87, 117, 130
theory of operation, 80
variable speed, 89
induction, see induction motor
repulsion, 313
singlephase induction, 308310
428
INDEX
Motor, aingle phase aerioB, 310, 313,
327
synchronous, see synchronous
motor
Motor applications to air com
pressors, 296
boring mills, 108
cement mills, 297
cranes, 92, 103, 298, 332
crushers, 94
elevators, 103
fans, 111, 112, 296, 300
lathes, 108, 299
line shafts, 102, 297
machine tools, 108, 300
printing presses. 111, 113
pumps, 103, 296
shears and punch presses, 103,
299
textile machinery, 298
woodworking machines, 102,
297
Motor car lighting, 186
Motor car trains, 332
Motor generator sets, 315, 335
Multiple switch starter, 119, 131,
305
unit control, 133
voltage system, 109
Multipolar, machines, 51, 56, 287
Mutual induction, 11
Natural frequency, 226
Neutral line, 62
Noload saturation curve, 70
Novoltage release, 87, 125, 302, 304
Nonarcing metal, 343
Noninductive resistance, 211
Ohm's law, 20
Oil circulation for transformers, 272
Oil for transformers, 270
Oil switch, 300, 345
Open delta connection, 277
Open machines, 100
Oscillograph, 193
Overcompound generator, 75
Overhead line construction, 346
Overload relay on oil switch, 345,
350
release, 118, 125, 304
ParaUel circuits, 22, 216, 226
Parallel operation of alternators, 259
directcurrent generators, 164,
166
rotary converter, 320
P&sted plates for a battery, 149
Permanent magnets, 35, 59
Permeability, 33
Phase, relation, 203
single, two, and three, 231
Pin insulators, 346
Plants plates, 148
Polarisation, 141
Polyphase cirouits, 238
rotary converter, 320
Potential transformer, 351
Power, 14
in a capacity circuit, 223
in an inductive circuit, 209
in a resistance circuit, 21, 211
in a threephase circuit, 236, 237
measurement of, 214, 238
Power factor, 213
correction, 256, 320
of an induction furnace, 267
of an induction motor, 293
of a synchronous motor, 256
of a transformer, 262
Power station, 171, 338
Primary of a transformer, 261
Printing press drive, 111, 113
Pull of magnets, 40, 43
solenoids, 37
Pump drive, 103, 296
Punch press drive, 103, 299
Puncture of insulation, 22
Quantity of electricity, 18, 218
Quick break switch, 114, 300
Railway motors, 327
Rating of machines, 101, 251
Ratio of transformation, 262
of rotary converter, 318
INDEX
429
Reactance, capacity, 221
inductive, 208
of an alternator, 246
of a transformer, 264
of a transmission line, 211,
216
Reflectors, 359
Regulation curves, 71
of alternators, 245, 248
of directcurrent generators,
7175
of a transmission line, 215
speed, 107, 113, 293
Regulator, axle generator, 184, 187
constant current, 76
feeder, 281, 339
lamp circuit, 170, 182
voltage, 74, 249
Reluctance, 33
Repulsion motor, 313
Residual magnetism, 35, 70
Resistance, 19
battery internal, 141, 152, 161
brush contact, 63, 66
control of batteries, 171
drop in armature, 72, 79
for adjustable speed motor,
112, 300
power loss in, 21, 211
specific, 20
starting for motor, 85, 87, 89,
118, 305
steadying, 365
temperature coefficient of, 21
Resistors, 26
Resonance, 224, 227
Reverse current circuit breaker, 170
Reversing drum, 129
Reversing field, 65, 68, 83
Revolving field alternator, 192, 229
239
of an induction motor, 284
Rheostats, 25
carbon pile, 29
castiron grid, 27
field circuit, 74
liquid, 29, 335
Right hand rule, 9, 192
Ring winding, 48
Rosenberg machine ,188
Rotary converter, 318
Rotor, 229
induction motor, 283, 288
Safety devices, 337
Saturation curve, 70
magnetic, 46, 71
Scott connection, 274
Searchlight generator, 190
Secondary of a transformer, 261
Selfcooled transformer, 270
Self excitation, 60, 71
Self induction, 11, 206
Selfstarting synchronous motor
294, 296
Semienclosed machines, 100
Separately excited machines, 60, 71
Separator, magnetic, 47
Series arc lighting, 366
booster, 180
circuits, 22, 212, 224
excitation, 60
generators, 75
motors, 90, 310
parallel control, 126
shunt, 77, 167
Shades for lamps, 359
Shadows, 363
Shears, motor for, 103, 299
Shell type transformers, 279
Shifting of brushes, 63, 83
Shortcircuit curve, 246
Short shunt, 61
Shunt excitation, 60
generators, 72, 162, 164
motors, 85, 105
Singlephase, 231
motors, 308314
railway system, 329
Sliding contact starter, 117
Slip of induction motors, 290, 293
Slipping belt generator, 185
clutch generator, 186
Slow speed drive, 113
Solenoid, 32, 37, 41
brake, 333
starter, 130, 306
Sparking, at a commutator, 63
430
INDEX
Sparking, at a switch, 12
limit of output, 100
Specific gravity of electrolyte, 156,
157
inductive capacity, 219
resistance, 20
Speed, adjustable, 89, 92, 105112,
293, 298, 300
Speed of a motor, 8083, 89, 101
of an induction motor, 287, 290,
293, 298
of a series motor, 92, 312
of a shunt motor, 89, 105112
of a synchronous motor, 252,
296
regulation, 107, 113, 293
regulators, 121
synchronous, 252
variation by armature control,
106, 112
by field control, 107
Speed time curve, 323
Spider, 59
Splitphase method of starting, 308
Splitpole rotary converter, 320
Squirrelcage motor, 283, 296
Stardelta starter, 304.,
Starters, automatic, 130, 133, 306
handoperated, 87, 117121,
301305, 340
Starting compensator, 302
Starting torque of an induction
motor, 287, 297
of a series motor, 90
of a shunt motor, 85
of a synchronous motor, 252,
294, 296
Starting resistance, 85, 87, 89, 118,
305
Stator, 229, 283
Steadying resistance, 365
Stiff field, 69
Stone generator, 185
Storage battery, see battery, 146
161
Stray loss, 96
Street car controller, 126
Street lighting, 360, 364
Substation, 339
Sulphation, 147
Sum of alternating voltages, 203
Suspension insulator, 346
Switches, automatic battery, 183
carbon break, 114
contactor, 116, 132
disconnecting, 345
float, 130
magnetic, 131
oil, 300, 345
quickbreak, 114
sparking at, 12
Switchboards, 348
Symbols for alternating currents, 198
Synchronizing, 258, 378
Synchronous motor, 252259, 294
296
applications, 296
dampers for, 295
hunting, 258, 295
power factor, 256
selfstarting, 294, 296
speed, 252, 296
startmg torque, 252, 294, 296
synchronizing, 258, 378
vector diagram, 254
Synchronous speed, 252
Synchroscope, 258, 350
Temperature coefficient of resistance,
21
of a battery, 155, 161
of motors, 99, 100, 102
Terminal station, 339
Textile mill, drive, 298
Threephase alternator, 231
connection of load, 237
connection of transformers, 276
transformers, 279
Threewire generators, 317
system, 316, 338
Torque of a motor, 82
of an induction motor, 287, 297
of a series motor, 90, 310
of a singlephase motor, 308, 310
of a shunt motor, 85, 88
of a synchronous motor, 252,
294, 296
Traction, 322332
INDEX
431
Tractive effort, 322
Train lighting, 181
friction, 322
Transmission line, drop, 23
inductance, 211, 216
insulation, 346
losses, 215 .
regulation, 215
single and threephase, 341
Transmission, underground, 347
voltages, 338, 341
Transformation ratio, 262
Transformer, boosting, 281
connections, 273279, 381
constant current, 267
cooling, 270
efficiency, 268
instrument, 351
leakage reactance, 264
lighting, 273
losses, 268
theory of operation, 261
vector diagram, 263, 265
Trolley wire, 329
Tungsten lamp, 352
Twophase alternator, 230
connection of transformers,
273
Underground cables, 347
Unit of current, 5
e.m.f., 9
light, 354
pole strength, 1
power, 14
work, 14
V<5onnection, 277
Variable speed generators, 182
Variable speed operation, induction
motors, 293, 298, 300
series motors, 92
shunt motors, 89, 105112
Vector diagram for alternators, 245
induction motors, 291
Vector diagram for parallel circoit,
217, 227
series circuit, 213, 225
series motor, 311
synchronous motor, 254
transformer, 263, 265
transmission line, 215, 342
Vector sum, 203
Vector representation, 201
Vehicle lighting, 181
Vibrating contact regulator, 187
Volt, 9
Volt efficiency, 152, 161 "
Voltage, average, 197
drop in transmission lines, 23
effective, 212
for arc lamps, 367
for traction, 329
for transmission, 338, 341
of a battery, 151, 160
regulators, 74, 249
Voltages used in practice, 341
Voltameter, 139
Voltmeter, 19, 198
Wagner motor, 314
Ward LeonarcLsystem, 110, 335
Water cooled transformers, 271
Water rheostat, 29, 335
Watts, 14
Wattmeter, 214
Wave form, 193
Welder, electric, 264
Windings, 48, 229
Wire, current carrying capacity,
338, 348, 368
Woodworking machinery, motors
for, 102, 297
Work, unit of, 14
Wound rotor motor, 288
Yconnection, 233, 235, 237, 276,
278
Ydelta starter, 304
Yoke of machine, 56, 58
V
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