iiiiiliilii ESSENTIALS OF / ELECTRICAL ENGINEERING A TEXT BOOK FOR COLLEGES AND TECHNICAL SCHOOLS BY JOHN FAY WILSON, B. S., E. E. Instructor in Electrical Engineering at the University of Michigan 282 ILLUSTRATIONS NEW YORK D. VAN NOSTRAND COMPANY 25 PARK PLACE 1915 r K COPYRIGHT, 1915, BY D. VAN NOSTRAND COMPANY Stanbopc ipress f, H. CILSON COMPANY BOSTON. U.S.A. PREFACE The widely prevalent belief that continuous and alternating cur- rents are not subject to the same general laws, is entirely errone- ous. The principles and laws which relate to the flow of continu- ous currents also govern the flow of alternating currents. This volume, which is offered as a text for the use of students pursuing either electrical or non-electrical engineering courses, is the result of the writer's class-room experience, and seeks to em- phasize the fact that continuous and alternating currents are gov- erned by the same laws. To this end the fundamental laws of the electric circuit are fully developed before any study of machines is attempted. With a thorough knowledge of the electric circuit as a foundation, the student should have little trouble in compre- hending the physical phenomena taking place in the more common types of electrical apparatus. The student is expected to be familiar with trigonometry, and a knowledge of calculus will be found advantageous but not indis- pensable. The mathematical developments of the formulae for the calculation of inductance and capacitance have been placed in appendices at the back. These and other portions of the text may be omitted when, for lack of time or for any other reason, it is necessary to shorten the course. The fact that the ideas advanced in this volume have developed with the science of Electrical Engineering, and may be regarded as the common property of the science, would make any attempt to give specific credit burdensome (and often impossible), but the writer wishes to specifically acknowledge his indebtedness to both standard and current literature, particularly to those works listed on page 333. Students desiring a more detailed discussion of particular subjects are referred to this list. The writer also wishes to express his obligation to the following men, each of whom read all or part of the manuscript, and offered valuable suggestions for its improvement: Professor C. M. Jan- 337873 iv PREFACE sky, of the University of Wisconsin; Professor H. H. Higbie, Professor A. H. Lovell, Mr. A. H. Stang and Mr. W. L. Bice, of the University of Michigan. Some idea of the practical construction of electrical machinery and instruments is given by means of a limited number of illustra- tions of actual apparatus. These illustrations are offered through the courtesy of the manufacturers. J. F. W. ANN ARBOR, MICH. June, 1915. TABLE OF CONTENTS CHAPTER PAGES I. THE ELECTRIC CIRCUIT 1-28 1 II. MAGNETISM AND MAGNETIC INDUCTION 29-45 III. PRACTICAL CONSTRUCTION OF THE DYNAMO 46-59 IV. THE CONTINUOUS-CURRENT GENERATOR 60-78 V. THE CONTINUOUS-CURRENT MOTOR 79-92 VI. LOSSES, EFFICIENCIES AND RATINGS OF CONTINUOUS-CURRENT DYNAMOS 93-106 VII. POLYPHASE ALTERNATING CURRENTS 107-122 VIII. THE ALTERNATING-CURRENT GENERATOR 123-148 IX. THE SYNCHRONOUS MOTOR 149-160 X. CURRENT-RECTIFYING APPARATUS 161-177 XI. THE TRANSFORMER 178-195 XII. TRANSFORMER CONNECTIONS . 196-202 XIII. THE INDUCTION MOTOR 203-225 XIV. SINGLE-PHASE COMMUTATING MOTORS 226-234 XV. ELECTRIC LAMPS 234-243 XVI. CURCUIT-LNTERRUPTING APPARATUS '. 244-25! XVII. METERS 252-267 XVIII. POWER TRANSMISSION AND DISTRIBUTION 268-298 XIX. THE STORAGE BATTERY 299-304 APPENDIX A. HARMONIC QUANTITIES 305-310 APPENDIX B. INDUCTANCE 311-318 APPENDIX C. CAPACITANCE 319-327 APPENDIX D. THE COMPLEX QUANTITY. ADMITTANCE, CONDUCTANCE AND SUSCEPTANCE 328-330 APPENDIX E. RESUSCITATION FROM ELECTRIC SHOCK 33 I- 332 REFERENCES.. 333 NOTATION* */ A = area. v B = magnetic flux density. b = susceptance. v C = capacitance (electrostatic capacity). r D = electrostatic flux density. ^ e electromotive force, instantaneous alternating electromotive force. */ Em = maximum alternating electromotive force. * E = continuous electromotive force, effective alternating electromotive force. -v / = frequency, force. 5~ = magnetomotive force. ^ F = electrostatic field intensity. -v g = conductance. T E = magnetic field intensity, magnetizing force. i = instantaneous current. - Im = maximum alternating current. / = continuous current, effective alternating current. K = dielectric constant. kva = kilovolt-amperes. kw = kilowatts. L = inductance. / = length. m = unit magnet pole. n = speed (revolutions per second). N = number of armature conductors, number of turns in a winding. P = power in continuous-current circuit, average power in alternating-current circuit. p = number of poles on a dynamo, instantaneous power in alternating-current circuit. P' = number of paths into which an armature winding is divided. q, Q = quantity of electricity. R = resistance. $1 = reluctance. s = slip of an induction motor. t F= time. T = torque, temperature. V = velocity, volume. W = work. X = reactance. Y = admittance. Z = impedance. a = an angle. * Based on the report of the Standardization Committee of the American Institute of Electrical Engineers (1915). vii viii NOTATION /3 = an angle. = magnetic flux, an angle. = an angle. ^ = dielectric flux. p permeability. co = angular velocity (radians per second). p = resistivity. Essentials of Electrical Engineering CHAPTER I THE ELECTRIC CIRCUIT 1. Introduction. The ultimate nature of electricity has never been discovered but it is generally conceived to be a medium, without weight or form, by means of which the energy of heat or motion may be transferred from one point to another. By means of this medium the energy of motion, as developed by the electric generator, may be transferred and caused to reappear at some dis- tant point in the form of motion (the electric motor), or as heat (the electric lamp). This conception of electricity is upheld by the similarity of an electric and a hydraulic system. Contrary to a widely prevalent belief, electricity is not erratic in its action but is governed by simple and well-defined laws. In the following pages the fundamental laws of the electric circuit are de- veloped, and the physical phenomena which take place in the more common types of electrical apparatus explained. 2. The electric current. The elec- tric circuit (Fig. i) is analogous, in many respects, to a hydraulic system, consisting of the pump and pipe connections shown in Fig. 2. The rotary pump indicated in Fig. 2a produces a continuous and , __ ^ uniform flow of water in the direction indi- cated by the arrows; the piston pump indicated in Fig. 2b produces a flow which is always in the direction indicated by the arrows but which is not uniform, i.e., when the piston Generator FIG. i. The Electric Circuit. FIG. 2a. Hydraulic Anal- speed is reduced and its motion reversed at the ogy for Continuous end of the stroke, the flow of water slackens or ceases altogether; the valveless pump indi- cated in Fig. 2c produces a flow which is not constant, either in direction or in value, but surges first in one direction and then in the other through the system. ENGINEERING Similarly the construction of the generator determines the char- acteristics of the electric current flowing in the circuit. Electric currents may be divided into s ^T x three classes: (a) continuous, (b) pulsating, (c) alternating. (a) Continuous currents. When the current in a given circuit flows continuously in one direction and the rate of flow is uniform, it is said to be a " continuous current" and is Piston .-' Pump FIG. 2b. Hydraulic Analogy for Pulsating Currents. commonly called a " direct current." The voltaic cell and certain forms of the electric generator cause a unidirectional current to flow at an approximately constant rate and are, there- fore, continuous-current appa- ratus. (b) Pulsating currents. If the current in a given cir- cuit flows in one direction but at a momentarily chang- ing rate, it is a "pulsating current." Pulsating currents Valveless Pump FlG. 2C. Hydraulic Analogy for Alternating Currents. are largely used in telephone work and in telegraphy. (c) Alternating currents. When the current in a given circuit, starting at zero, increases to maximum, decreases to zero, increases to maximum in the opposite direction, and again decreases to zero (the process being repeated periodically), it is an "alternating cur- rent." The wave forms produced by commercial alternators vary greatly, but comparisons and mathematical calculations, unless specifically stated otherwise, are based on the harmonic or sine wave ( form.* While this wave form is seldom or never attained, its as- sumption is usually sufficiently accurate for practical purposes. Calculations for the actual form of the wave may become extremely complicated. 3. Manifestations of the electric current. The presence of an electric current is made manifest in a number of ways. The prin- * See Appendix A. , THE ELECTRIC CIRCUIT 3 cipal effects of the electric current with which the electrical engi- neer is concerned are: (a) chemical, (b) magnetic and (c) heating. (a) Chemical effect. When water containing a small quantity of acid forms part of a circuit carrying a unidirectional current, bubbles may be seen to rise through the liquid. When the gases causing these bubbles are collected, they are found to be oxygen and hydrogen, the elements which enter into chemical combination to form water. The same action takes place in a circuit carrying alternating current, but each half cycle destroys the chemical effect of the preceding half cycle so that the net effect is zero. Hence, alternating currents are not used when chemical effects are desired. (b) Magnetic effect. A current-carrying conductor exhibits all the characteristics of a magnet in that it attracts pieces of iron, and is attracted to or repelled from a magnet or another current- carrying conductor. Attraction or repulsion takes place accord- ing to the polarity of the magnet, or the relative directions of the currents in the conductors. This magnetic effect is intensified if the conductor is in the form of a spiral. It is still further increased if the spiral is wound about an iron core. (c) Heating effect. When an electric current flows in a wire, heat is liberated. The heating may be so slight as to be unnoticed as in the ordinary electric bell circuit, or so great as to heat the conductor to incandescence as in the electric glow lamp. Light is not a direct manifestation of the electric current, but is due to the high tempera- ture to which the current heats the conductor. 4. Series circuits. A series circuit, represented in Fig. 3a, is one in which the entire current in the circuit flows successively through each piece of apparatus. The commercial application of the series circuit is somewhat limited, the most common use being in connection with street and other outdoor lighting. n A A e-L Load A A e-Load Co) (b) FIG. 3. Series and Parallel Circuits. 5. Parallel circuits. -7- A parallel or multiple circuit, represented in Fig. 3b, is one in which the current divides and flows through two or more branches. The current in each branch of a parallel circuit is independent of that in the other branches. Motors are always connected in parallel. 4 ESSENTIALS OF ELECTRICAL ENGINEERING 6. Electric units.* - - The units of the electric system are : (a) the ohm, (b) the ampere, (c) the volt, (d) the coulomb, (e) the joule and (/) the watt. (a) The ohm. The opposition to the flow of an electric current is measured in ohms. The standard ohm is represented by the opposition offered to an unvarying electric current by a column of mercury at a temperature of melting ice, 14.4521 grams in mass, of a constant cross section, and of the length 106.3 centi- meters. (b) The ampere. The rate at which electricity is transferred is measured in amperes, analogous to miner's inches or gallons per second, and the ampere is represented by the unvarying current which, when passed through a standard solution of nitrate of silver in water, deposits silver at the rate of 0.001118 gram per second. (c) The volt. The force which causes or tends to cause an elec- tric current to flow is termed an " electromotive force," the unit of which, the volt, is the electromotive force that, steadily applied to a conductor whose resistance is one ohm, produces a current of one ampere. (d) The coulomb. Quantity of electricity is expressed in cou- lombs, analogous to gallons or cubic feet, and one coulomb is the quantity of electricity transferred in one second by a current of one ampere. (e) The joule. The unit of energy or work in the electric system is the joule, and is the energy expended in one second when a current of one ampere flows in a circuit having a resistance of one ohm. The energy equivalent of one joule has been experimentally deter- mined to be == 0.24 calorie. = 0.00095 B.T.U. = 0.737 foot-pound. = 10,000,000 ergs. (/) The watt. The watt, which is the unit of electrical power, is represented by the expenditure of one joule in one second. Since * There is a definite relation between the C.G.S. units (the so-called absolute units) and the ohm, the ampere and the volt, but these units are to be regarded as arbitrary units, similar to the foot and the pound, which have been adopted by international agreement, and legalized by statutory enactment. THE ELECTRIC CIRCUIT 5 the energy equivalent of one joule is 0.737 foot-pound, and 550 foot-pounds per second equal one horse power, h.p. = -$&- o-737 = 746 watts. The kilowatt (kw.) is a commonly used unit of power and is equal to 1000 watts. 7. Resistance. Resistance is the inherent property of a ma- terial which opposes the flow of an electric current, and by virtue of which electrical energy is converted into heat. It is analogous, in many respects, to mechanical friction, and has been experimen- tally determined to be directly proportional to length, and in- versely proportional to cross-sectional area. d) area The unit of length commonly used is the foot; that of area, the circular mil. The constant p is, then, the resistance in ohms of a section of the material one foot long and having a cross-sectional area of one circular mil. The average value of p for commercial copper may be taken as 10.8 at ordinary temperatures (25 C.). Materials whose resistivity is low, such as copper, silver, alumi- num, etc., are known as conductors; those whose resistivity is high, such as glass, rubber, mica, porcelain, etc., are known as insulators. 8. Circular mil. The circular mil is an arbitrary unit of area much used in electrical calculations. Since electrical conductors are largely circular in cross section, it is convenient to have a unit of area that bears a simple relation to the diameter of a circle. The area of a circular cross section, in circular mils, is obtained by squar- ing the diameter in thousandths of an inch (mils). Hence, the area in circular mils is to the area in square measure as 4 is to TT. area in circular mils 4 / \ area in millionths of a square inch TT 9. Temperature coefficient. An increase in the temperature of a copper wire causes its resistance to increase. The ratio of the change in the resistance of a conductor, per degree change in ESSENTIALS OF ELECTRICAL ENGINEERING its temperature, to its resistance at its initial temperature, is its temperature coefficient. The temperature coefficient of most materials is not constant, but changes with the temperature. By reference to Table I it will be found that the temperature coeffi- cient of copper decreases as the temperature increases. TABLE I TEMPERATURE COEFFICIENTS FOR COPPER i a = 234-5 + T T a T a r a T a 0.00427 13 o . 00404 26 o . 00383 39 o . 00366 i 0.00425 14 o . 00403 27 0.00381 40 o . 00364 2 O . 00424 15 o .00401 28 o . 00380 4i 0.00362 3 0.00422 16 o . 00399 29 0.00379 42 0.00361 4 0.00420 17 0.00397 30 0.00378 43 o . 00360 5 0.00418 18 o . 00396 31 0.00377 44 o . 00360 6 0.00416 19 o . 00394 32 0.00375 45 0.00358 7 0.00414 20 0.00393 33 0.00374 46 0.00356 8 0.00412 21 0.00391 34 0.00373 47 0.00355 9 0.00411 22 o . 00390 35 0.00371 48 0.00354 10 o . 00409 2 3 o . 00388 36 0.00370 49 0.00352 ii o . 00408 24 0.00387 37 o . 00369 5 0.00351 I 2 o . 00406 2 'C o 0038^ 38 o 00367 v^ ^ w o^0 o w . w^) v^ i R t = R(i db at). R = the resistance at temperature T C. R t = the resistance at temperature (T =t /). a = the temperature coefficient at temperature T C. / = the change in temperature. The resistance of some materials, of which carbon is an example, decreases as the temperature increases. Glass, at ordinary tem- peratures, has a very high resistivity, but in the liquid state its re- sistivity is comparatively low. These materials are said to have negative temperature coefficients. Certain metallic alloys have a practically constant resistance over a wide range of temperature. 10. Ohm's Law. The electromotive force required to overcome the opposition due to the resistance of any circuit is proportional to the current flowing in the circuit. When the electromotive force (e) is expressed in volts and the current (i) in amperes, the proportionality factor (R) is in ohms. e = Ri. (3) THE ELECTRIC CIRCUIT 7 The experimental fact stated above is the first fundamental law of the electric circuit and was first demonstrated and enunciated by Dr. G. S. Ohm. It applies equally to continuous, pulsating and alternating currents. ii. Inductance. That property of a body which tends to main- tain any existing state or condition of motion of the body is termed its inertia. The force (/) required to neutralize the effect of inertia is proportional to the rate at which the speed or velocity of the body changes, and the proportionality factor (M) is the mass of the body. When the velocity of a body is changed from V' to V" in / seconds , M (V - V") av./=- ^, (4) and / = M X rate at which V changes. (5) Suppose the speed of a steam engine is increased from 100 r.p.m. to 200 r.p.m. in 10 seconds. A force is required to overcome the inertia of the flywheel and other rotating parts, but the energy used in causing the speed of the engine to increase is stored as kinetic energy in the rotating parts, and is returned to the system when the engine speed decreases to its original value. Therefore, the net expendi- ture of energy due to the inertia of the moving parts of a machine is zero when the final state of motion is the same as the initial state. An electric circuit has a property, similar to that of inertia, which tends to maintain any state or condition of current flowing in the circuit. The electromotive force (e) required to overcome the effect of this property and cause the current to increase or decrease, is proportional . to the rate at which the current changes, and the proportionality factor (L) is the inductance of the circuit. The unit of inductance is the henry. When the current in a circuit is changed from i f to i" in / seconds T > /,-\ av.|0 = L - j (6) t and e L X rate at which i changes. (7) It is evident that the energy expended in causing a current to increase is equal and opposite to that expended in causing it to decrease to its original value. Therefore, the net expenditure of energy due to the inductance of an electric circuit is zero when the final current is equal to the initial current. 8 ESSENTIALS OF ELECTRICAL ENGINEERING 12. Capacitance. In the system shown in Fig. 4, consisting of a pump, pipe connections, and a cylinder across which is stretched an elastic membrane, the pressure set up by the pump stresses the membrane until the reaction due to the stress is equal to the pressure of the pump, and the quantity of water displaced or stored is proportional to the pressure. gallons = constant X pressure. (8) FIG. 4 . Mechanical Anal- If the pressure due to the pump decreases, ogy of Capacitance. u iT i the stress in the membrane becomes less, and energy which tends to maintain the original pressure set up by the pump, is returned to the system. An electric condenser consists, essentially, of two conductors separated by an insulating material, or dielectric. In the system shown in Fig. 5, the generator sets up a pressure which stresses the dielectric separating the plates PP until the reaction equals the electromotive force of the generator, and the quantity of electricity displaced or stored is pro- FlG * 5 ' Electric Ca P acitance - portional to the electromotive force (e) of the generator. The proportionality factor (C) is termed the capacitance of the con- denser. The unit of capacitance, is the farad. q = Ce. (9) When the applied electromotive force is changed from e' to e" in / seconds ,e'-e" Resistance av. i = C do) and i = C X rate at which e changes. If the electromotive force of the generator decreases, energy is re- turned to the system and tends to maintain the electromotive force of the system at its former value, the stress in the dielectric being reduced proportionately. The energy returned to the system when the condenser is completely discharged is equal to the energy stored, the condenser being so constructed that the losses are negligible. THE ELECTRIC CIRCUIT 9 13. Alternating-current circuits containing resistance only. From equation (3) e = Ri (12) = RI m smut, (13) i.e., the current and the electromotive force in an alternating- current circuit containing resistance only, are in phase as shown in Fig. 6a. 14. Alternating-current circuits containing inductance only. - Since the electromotive force required to overcome the effects of inductance is proportional to the rate at which the current changes, the electromotive force required to overcome the inductance of an alternating-current circuit in which the current varies harmonically is <> ' e e = uLI m sin (/ + 90),* (14) i.e., the current and the electromotive ~|*' CO force in an alternating-current circuit containing inductance only, are 90 Fl . <*' Vectors P of Current M and Electromotive Force in a Non- degrees out of phase, and the current inductive Circuit. lags behind the applied electromotive FIG. 6b. Vectors of Current and force as shown in Fig. 6b. Electromotive Force in an In- . i f . / . \ ductive Circuit. The maximum value of sin (/ + 90 ) FlG> 6c> Vectors o Current and equals I , and Electromotive Force in a Capaci- E m = wLI m . (15) tive Circuit. The quantity col, is termed the inductive reactance of an alter- nating-current circuit and is expressed in ohms. (Symbol X L .) 15. Alternating-current circuits containing capacitance only. Since the quantity of electricity displaced in a condenser circuit is proportional to the applied electromotive force, the current (rate of displacement) in the circuit is proportional to the rate of change in the electromotive force. When the electromotive force varies harmonically i = uCE m sin (co/ + 90),* (16) i.e., the current and the electromotive force in an alternating-current circuit containing capacitance only, are 90 degrees out of phase, and the current leads the applied electromotive force as shown in Fig. 6c. * See Appendix A, Section 7. 10 ESSENTIALS OF ELECTRICAL ENGINEERING The maximum value of sin (/ + 90) equals i, I m = uCE m ' (17) and ; IS.-f*. ^ || ( I8 ) The quantity is termed the capacitive reactance * of an alter- coC na ting-current circuit and is expressed in ohms. (Symbol X c *) 16. Reactance of alternating-current circuits. The reactance of an alternating-current circuit is, from equations (15) and (18), the ratio of the current flowing in the circuit and the electromotive force required to overcome the combined effects of inductance and capacitance. 17. Alternating-current circuits containing resistance and in- ductance in series. In a series circuit containing resistance and inductance, the applied voltage required to cause a given current to flow in the circuit must be equal to that required to overcome the combined effects of resistance and inductance. e = e R + e L (19) = Ri + uLi (20) = RI m sin ut + (tiLIm sin (/ + 90). (21) The right-hand member of equation (21) is the surri of two harmonic electromotive forces 90 degrees out of phase. The applied electro- motive force is, therefore, a harmonic quantity, the maximum value of which is the geometric sum of the maximum values of the quad- rature electromotive forces. E m = (RIm) 2 + (uLIm) 2 ( 22 ) = Im VR 2 + (COL) 2 (23) and the current lags behind the applied electromotive force by the angle = tan- 1 ( 2 4) * Capacitive reactance is to be considered a negative quantity. THE ELECTRIC CIRCUIT II 1 8. Alternating-current circuits containing resistance and ca- pacitance in series. In a series circuit containing resistance and capacitance, e = e R + e c (25) = Ri + ~dC (26) r> T i Im sin (ait 00) . / \ = RI m sin co/ + - * 5 (27) therefore, "0* (28) and the current leads the applied electromotive force by the angle (30) 19. Alternating-current circuits containing resistance, induc- tance and capacitance in series. In a series circuit containing resistance, inductance and capacitance, e = e R + e L + e c (31) = Ri -f coLi + ^ (32) coC = RI m sin f + L7, sin (/ + 90) + /m sin ( ~ 9 ) . (33) From equation (33) E m = y jR 2 / w 2 + (L/ m + ^)' (34) = 7 m ^ + coL + (35) and the current leads or lags behind the applied electromotive force, as the quanti of lead or lag being force, as the quantity f w + ~;J is negative or positive, the angle XV (36) 12 ESSENTIALS OF ELECTRICAL ENGINEERING 20. Voltage triangle. When an alternating current flows in a circuit containing resistance and reactance in series, the applied electromotive force is the resultant of quadrature electromotive forces, and the voltages of the circuit may be represented graphi- cally by means of a right triangle. Fig. ya. Because of the impossibility of eliminating resistance from a reac- tance, just as it is impossible to FIG. 7. Voltage Triangles. . ... eliminate friction from a machine having moving parts, the more common form of the voltage tri- angle is an obtuse triangle, the hypotenuse and the sides of which are, respectively, the applied electromotive force, the voltage be- tween the terminals of the resistor, and that between the terminals of the reactor. Fig. yb. "N 21. Impedance and the impedance triangle. Dividing equation (35) by I m} and representing the quantity coL -\ by X coC f? = VFTT 2 . ( 37 ) J-m The quantity VR 2 -f X 2 is termed the impedance of an alternating- current circuit, and is the ratio of the applied electromotive force and the current flowing in the circuit. The unit of impedance (Z) is the ohm. The impedance, the resistance and the reac- tance of an alternating-current circuit form FlG - 8 - Impedance respectively, the hypotenuse, the base, and the altitude of a right triangle, and may be represented graphically as in Fig. 8. Example. A series circuit has the following: E m = 220 VOltS. / = 60 cycles. R = 30 ohms. L = o.i henry. C = 0.00013 farad. Find: (a) The reactance; (b) the impedance; (c) the current. THE ELECTRIC CIRCUIT 13 Solution. XL = 377 X o.i = 37.7 ohms. = 20.4 ohms. 377 X 0.00013 X = XL -f- Xc = 17.3 ohms. Z = V( 3 o) 2 + (i7-3) 2 = 34-7 ohms. r E m 220 I m = - = -- = 6.3 amperes. ^ 34-7 22. Alternating-current circuits containing resistance and induc- tance in parallel. At any instant the total current supplied to a parallel system is the algebraic sum of the currents flowing in the branches. i = *B + *i. (38) By definition i R = (/ m )^sinoj/. (39) From equation (14) IL = (Im) L sin (co/ - 90). (40) Substituting in equation (38), * = (Jm)fl sin w/ + (7 m ) L sin (/ - 90). (41) The right-hand member of equation (41) is the sum of two -harmonic currents 90 degrees out of phase. The total current supplied to the system is, therefore, a harmonic quantity, the maximum value of which is the geometric sum of the maximum values of the currents flowing in the branches. /. = \/CW+(/~)L 2 (42) and the total current lags behind the applied electromotive force by the angle =k. (43) 23. Alternating-current circuits containing resistance and capac- itance in parallel. When resistance and capacitance are con- nected in parallel * = ** + c> (44) * = (/.)* sin /. (45) 14 ESSENTIALS OF ELECTRICAL ENGINEERING From equation (16) ic = (Im)c sin (co/ -f 90). (46) Substituting in equation (44), i = (Im) R sin co/ + (7 m ) c sin (co/ + 90), (47) Im = V(Im) R 2 + (Im) C 2 (48) and the total current leads the applied electromotive force by the angle "T^T- (49) 24. Alternating-current circuits containing resistance, induc- tance and capacitance in parallel. When resistance, inductance and capacitance are connected in parallel * = i R + iL + ic, (5) ** = () sin co/, (51) iL = (7 m )z, sin (co/ - 90), (52) ic = (Im)c sin (co/ + 90). (53) Substituting in equation (50), i = (I)R sin co/ + (I m ) L sin (co/ - 90) + (7 m ) c sin (co/ + 90), (54) fi 2 + [(7)z, ~ (^m)cJ 2 (55) and the total current leads the applied electromotive force if [(7 m )L (7m) c] is negative, and lags behind the applied electro- motive force if [(7 m )z, (I m )c\ is posi- tive, the angle of lead or lag being tm ^ VJ (56) FIG. 9. Current Triangles. {***)& 25. Current triangle. The current relations of a parallel cir- cuit are shown graphically by means of a right triangle. Fig. 9a. Because of resistance in the inductive or capacitive branch the currents in parallel circuits are seldom 90 degrees out of phase, and Fig. 9b shows the usual relations of the component and the resultant currents. THE ELECTRIC CIRCUIT 15 26. Joule's Law. The work done in any electrical circuit is proportional to the square of the current flowing in the circuit and to the time the flow continues. When the electromotive force (e) is expressed in volts, the current (i) in amperes and the time (/) in seconds, the unit of work is the joule (A), and the proportionality factor is in ohms. A oc . ( 57 ) Joule's Law is the second fundamental law of the electric circuit and, like Ohm's Law, has been shown, experimentally, to be uni- versal. 27. Work done in an electric circuit. From equation (57) A R = Ri^t (58) when current flows in a circuit containing resistance only; A L = coL* 2 / (59) when current flows in a circuit containing inductance only; A c =^ ' (60) when current flows in a circuit containing capacitance only. 28. Power in an electric circuit. Power is, by definition, the rate of doing work. Therefore, work ,, ^ Power = -T- (61) time and p R = RP watts (62) when current flows in a circuit containing resistance only; p L = uL? watts (63) when current flows in a circuit containing inductance only; *2 pc = -7; watts (64) o?C when current flows in a circuit containing capacitance only. From equation (3) Ri = e R . (65) From equation (14) uLi = CL. (66) 1 6 ESSENTIALS OF ELECTRICAL ENGINEERING From equation (16) (6 7 ) Therefore, the instantaneous power (watts) in any electrical cir- cuit is equal to the product of the applied electromotive force (volts) and the current (amperes) flowing in the circuit. p = ei. (68) In a continuous-current circuit, e and i are constant and P = EI watts. (69) In an alternating-current circuit containing resistance only p R = E m sin. utl m sin co/ (70) = E m lm sin 2 co/. (71) From Appendix A, Section 9, the average value of sin 2 co/ during one complete cycle is ^. Therefore, av . p R = ^^ watts. (72) 2 In an alternating-current circuit containing inductance only p L = E m sin co// m sin (co/ - 90) (73) = E m l m sin co/ cos co/, (74) but the average value of sin co/ or cos co/ during one complete cycle is zero.* Therefore, av. p L = o. (75) In an alternating-current circuit containing capacitance only pc = E m sin (co/ + 90) co// w sin co/ (76) = E m l m cos co/ sin co/, (77) which is identical with equation (74) and av. pc = o. (78) In any alternating-current circuit p = E m sin co/ I m (sin co/ 0) (79) = E m / w sin co/ (sin co/ cos =b cos co/ sin 0) (80) = E m l m (sin 2 co/ cos sin co/ cos co/ sin 0). (81) * See Appendix A, Section 8. THE ELECTRIC CIRCUIT 17 From Appendix A, Section 9, av. sin 2 co/ = \ (82) and av. sin co/ cos co/ sin = o. (83) Therefore, E m l m cos , ^ ay. p = - - watts, (84) i.e., the average power * (watts) in any alternating-current circuit is equal to one-half the product of the maximum electromotive force (volts), the maximum current (amperes) and the cosine of the phase angle. 29. Effective current and electromotive force. The steady current (or electromotive force) which, acting in a circuit of constant resistance and zero reactance, transfers energy at the average rate of transfer when an alternating current (or electromotive force) acts in the same circuit, is the elective value of the alternating current (or electromotive force) . From equation (62) P = R ? (85) = RI m 2 sm 2 wt. (86) Then av. p = RI m z av. (sin 2 /)t / (87) = ~- (88) For continuous currents P=RI 2 . (89) Therefore, /?/" 2 RP = &*- (go) and 2 _j = 0.707 I m .^ (92) Similarly, ^ = ^ R 2 R X * A watt-meter indicates the average power in an alternating-current circuit. t See Appendix A, Section 9. 1 8 ESSENTIALS OF ELECTRICAL ENGINEERING and = ^ (94) V 2 = 0.707 E m . (95) Therefore, the effective value of a harmonic alternating current (or electromotive force) is equal to its maximum value divided by the square root of 2, and effective values may be substituted in any of the above equations containing maximum values. Effective values are indicated by ammeters and voltmeters, and are always to be assumed in alternating-current problems unless the values given or required are specifically stated to be maximum or instantaneous. 30. Power factor. The cosine of the phase angle (cos 0) is termed the power factor of an alternating-current circuit. The power factor may also be expressed as the ratio of the resistance to the impedance, n COS = ~ , (96) or by the ratio of the watts to the product of volts and amperes, cos0 = |^. (97) 31. Resistances in series. The resistance of a series circuit is the arithmetical sum of the resistances of its parts. E = Ei + E 2 (98) = 7 (^ + 2). (99) Dividing by I R = 1 + Ri. (100) 32. Reactances in series. The reactance of a series circuit is the algebraic sum of the reactances of its parts. E = Ei 4- E 2 (101) = /(X 1 + Z 2 ). y (102) Dividing by 7 X = Xi + X 2 . (103) THE ELECTRIC CIRCUIT 33. Impedances in series. The impedance of a series circuit is the geometric sum of the impedances of its parts. From equation (35) E = Er 2 + E x * (104) = / y/(Ri + Rtf + (X l + X 2 y. (105) Dividing by / Z = (R, + R 2 )* + (Xi + X 2 ) 2 . (106) 34. Resistances in parallel. In a parallel system the branches of which contain resistance only, the reciprocal of the resistance of the system is the arithmetical sum of the reciprocals of the resist- ances of the branches. / = /I + /2, (I0 7 ) Dividing by E 35- Reactances in parallel. In a parallel system the branches of which contain reactance only, the reciprocal of the reactance of the system is the algebraic sum of the reciprocals of the reactances of the branches. / = /l+/2, (HO) - = +. Dividing by E 36. Resistance and reactance in parallel. In a parallel system consisting of a branch containing resistance only and a branch con- taining reactance only, the reciprocal of the impedance of the system is the geometric sum of the reciprocal of the resistance and the reciprocal of the reactance of the branches. From equation (55) / = V/! 2 + /2 2 , (113) E /E\ 2 , ^ (x) (I14) 20 ESSENTIALS OF ELECTRICAL ENGINEERING Dividing by E '~ f^r I . ("5) 37. Impedances in parallel. Any impedance containing both resistance and reactance (Fig. 10) may be replaced by a parallel FIG. 10. Resistance and Inductance FIG. n. Resistance and Inductance in Series. in Parallel. system the branches of which contain only resistance or reactance (Fig. n). Referring to Fig. 10 let E = the applied electromotive force, / = the current, Z = the impedance, cos = the power factor. Then I r = I cos (= power component of current), I x = I sin ( = wattless component of current), R s = Z cos , X 8 = Z sin 4>. Referring to Fig. n, let E = the applied electromotive force, / = the line current, I r the current in the resistance, I x = the current in the inductance, R p = the resistance, X p = the reactance, Z = the impedance. i.e., the applied electromotive force, the current, the impedance and the power factor of one system are made equal to those of the other system, and the systems are, therefore, equivalent. R, = f- (6) * r (it?) COS Z cos Z 2 ELECTRIC CIRCUIT 2 I 71 (119) (o) (iai) (122) (123) (124) (125) sin < Z 2 Z sin< Z 2 From equation (115) '\ + ^k (126) (127) Therefore, for a parallel system, Z V VZl 2 ' 7 2 ' ^ 2/ r ^ 2 "T 'Z 2 + ^"J ( I2 &) I? 7-2 / -^1 I -^2 , -Vn \ / \ ^ = Z l^ + ^l + ^l)' ( I2 9) (130) Example. Let the circuit shown in Fig. 12 have: RI = 6 ohms, i?2 = 8 ohms, J? 3 = 10 ohms, J"i = 5 ohms, J\T 2 = 4 ohms, -Xs = 3 ohms. Then, Z 2 2 = 8 2 + 4 2 = 80, | x - f- | x ' Z 3 2 = io 2 + ^ 2 = 100, FlG - I2 ' Impedances in ParalleL 22 ESSENTIALS OF ELECTRICAL ENGINEERING and = v++ + + Z ' \6i 80 ice/ \6i 80 109 = V( .098 +0.1+0.092)2 + (0.082 + 0.05 + 0.028) 2 = 0.331. Z = - - = 3.02 ohms. R = (3-02) 2 (0.098 + o.i + 0.092) = 2.65 ohms, X = (3.02)2 (0.082 + 0.05 + 0.028) = 1.46 ohms. Also, tan = 0-082+0.05,+ 0.028 = 0.159 = Q 0.098 + o.i + 0.092 0.289 cos = 0.876. R = Z cos = 3.02 X 0.876 = 2.65 ohms. sin = 0.482. X = Z sin = 3.02 X 0.482 = 1.46 ohms. 38. Resonance of alternating-current circuits. In a series circuit the electromotive force required to overcome the effect of inductance leads the current by 90 degrees ; the electromotive force required to overcome the effect of capacitance lags behind the cur- rent by 90 degrees. These two harmonic forces are, then, 180 degrees out of phase, and tend to neutralize. When the inductive electromotive force is equal, numerically, to the capacitive elec- tromotive force, the circuit is said to have voltage resonance. I =o (131) Example. Find the capacitance required to neutralize an in- ductance of o.i henry in a 6o-cycle alternating-current circuit, the capacitance to be connected in series with the inductance. Solution. From equation (131) r _ _!_ */ nT 7T 2 X 4 X 3600 X o.i = 0.007042 farad. In a parallel circuit, the wattless component of current in an inductive branch lags 90 degrees behind the applied electromotive THE ELECTRIC CIRCUIT 23 force; the wattless component of current in a capacitive branch is 90 degrees ahead of the applied electromotive force. These wattless components of current are, then, 180 degrees out of phase, and tend to neutralize. When the leading wattless component is equal, numerically, to the lagging wattless component, the cir- cuit is said to have current resonance. /iag sin 0' = /ie a d sin 0" (132) when /i ag = the current flowing in the inductive branch, /lead = the current flowing in the capacitive branch, 4>' = the angle between the vector of electromotive force and that of the current in the inductive branch, 0"= the angle between the vector of electromotive force and that of the current in the capacitive branch. Example. The impedance of a 6o-cycle inductive circuit is 22 ohms, the current flowing in the circuit is 10 amperes, and lags 30 degrees behind the applied electromotive force. Find the capacitance to be connected in parallel with the impedance, the electromotive force and the resultant current to be in phase. Solution. The voltage across the condenser is EC = 22 X 10 = 220 VOltS, and the wattless component of current is I w = 10 X 0.5 = 5 amperes. From equation (18) C = s 377 X 220 = 0.00006 farad. 39. Use of electrical measuring instruments. A voltmeter is connected in parallel with that part of a circuit in which it is desired to measure the drop of poten- tial. Fig. 1 3 a. An ammeter is connected ^^\ <&> in series with the circuit in which it is desired to measure FIG. 13. Voltmeter, Ammeter and Watt- the current, the entire cur- rent, or a constant proportion of it, going through the coils of the instrument. Fig. i3b. A wattmeter consists of two current-carrying coils, a voltage 24 ESSENTIALS OF ELECTRICAL ENGINEERING coil connected in parallel with the load and a current coil con- nected in series with the load. The reaction set up by the coils is proportional to the product of the currents flowing in the coils, i.e., to the power in the circuit, and causes the deflection of a movable element. Permanent-magnet instruments indicate on continuous-current circuits only; induction instruments on alternating-current circuits only; dynamometer instruments on either continuous- or alternat- ing-current circuits. CHAPTER I PROBLEMS 1. Find the area in circular mils of: (a) a circle \ inch in diameter, (b) a rec- tangle i X i inch. 2. Find the diameter of a circle, the area of which is: (a) 211,600 circular mils, (6) 52,240 circular mils, (c) 10,400 circular mils, (d) 1000 circular mils. 3. Find the side of the square, the area of which is: (a) 211,600 circular mils, (b) 52,240 circular mils, (c) 10,400 circular mils, (d) 1000 circular mils, 4. Find the resistance, at 20 C., of 1000 feet of copper wire: (a) o.i inch in diameter, (b) 0.25 inch in diameter, (c) 0.6 inch in diameter, (d) i inch in diameter. 5. Find the resistances of the wires specified in Problem 4 at a temperature of 45 C. 6. Find the electromotive force required to cause a current of 10 amperes to flow in a circuit, the resistance of which is: (a) i ohm, (b) 5 ohms, (c) 18 ohms, (d) 25 ohms, (e) 40 ohms. 7. Find the average electromotive force of self-induction in a circuit, the inductance of which is i henry, when the current changes: (a) from i ampere to 5 amperes in o.i of a second, (b) from 10 amperes to zero in o.oi of a second, (c) from 25 amperes to zero in 0.2 second. 8. Find the capacitance of a circuit when 5 coulombs of electricity are dis- placed by an electromotive force which changes: (a) from 25 volts to zero in i second, (b} from 50 volts to 100 volts in \ of a second, (c) from zero to 10 volts in o.oi of a second. 9. The resistance of an electric circuit is 10 ohms. Find the maximum value of the current when the maximum value of the alternating electromotive force applied to the terminals of the circuit is: (a) 25 volts, (b) 40 volts, (c) 90 volts, (d) 150 volts, (e) 500 volts. 10. The inductance of a circuit is o.i henry. Find the electromotive force required to cause an alternating current of 100 amperes (maximum) to flow in the circuit when the frequency of the alternating current is: (a) 10, (b) 25, (c) 40, (d) 50, (e} 60, (/) 100. 11. Find the voltage required to charge a 6o-microfarad condenser with 5 coulombs of electricity. 12. The charge on a condenser changes from i coulomb to 2 coulombs in THE ELECTRIC CIRCUIT 25 o.oi of a second. Find the average current flowing in the circuit during the period of change. 13. An electric circuit has an inductance of o. i henry. ' Find the capacitance required to cause resonance when the frequency of the alternating electro- motive force is: (a) 25, (b) 40, (c) 60, (d) 100. 14. Find the impedance of an electric circuit having a resistance of 8 ohms and a reactance of 6 ohms connected in series. 15. Find the reactance of a circuit having an inductance of 0.15 henry when the frequency of the applied electromotive force is: (a) 15, (6) 25, (c) 40, (d) 60, (e) 100. 16. Find the reactance of a lo-microfarad condenser when the frequency of the applied electromotive force is: (a) 15, (b) 25, (c) 40, (d} 60, (e) 100. 17. The resistance of an electric circuit is 2 ohms. Find: (a) the total heat developed in the circuit when 50 amperes continuous current flows steadily in the circuit for 10 minutes, (b) the maximum value of the alternating-current required to develop the same quantity of heat in the same length of time, (c) the power (watts) in the circuit. 1 8. The continuous electromotive force applied to an electric circuit is 220 volts. This electromotive force produces a current of 20 amperes. Find: (a) the resistance of the circuit, (b) the power used in the circuit. 19. The effective value of an alternating current is 20 amperes, the applied electromotive force (effective) is 220 volts, and the power factor of the circuit is 0.85. Find: (a) the impedance of the circuit, (b) the resistance of the circuit, (c) the reactance of the circuit, (d) the power used in the circuit. 20. A rheostat and a condenser are connected in series to a no-volt, 25- cycle alternating-current generator. The value of the resistance is 10 ohms and the capacitance of the condenser is o.i microfarad. Find: (a) the reactance of the circuit, (b) the impedance of the circuit, (c) the power factor of the circuit, (d) the current flowing in the circuit, (e) the power component of electromotive force, (/) the wattless component of electromotive force. 21. Three impedances, the values and power factors of which are indicated below, are connected in series. Find: (a) the impedance of the circuit, (b) the power factor of the circuit. Zi = 10 ohms. p. f. = 0.9 (lagging). Z 2 = 15 ohms. p. f. = i. Z 3 = 20 ohms. p. f. = 0.6 (lagging). 22. The impedances specified in Problem 21 are connected in parallel. Find: (a) the impedance of the circuit, (b) the resistance of the circuit, (c) the reactance of the circuit, (d) the power factor of the circuit. 23. Find the resistance of 1000 feet of copper wire, 102 mils in diameter and at a temperature of 40 C. 24. Find the length of copper wire, 204 mils in diameter, the resistance of which, at the same temperature, is equal to the resistance of the wire specified in Problem 23. 25. The resistance of the field windings of a shunt motor was found to be 70 ohms at 20 C. After the motor had been in operation for three hours, it was 26 ESSENTIALS OF ELECTRICAL ENGINEERING found that the resistance of the windings was 10 per cent greater than when the first measurement was made. Find the temperature of the field windings. 26. Two loads A and B are connected in parallel to alternating-current mains. The power factor of A is 0.9 and I a 100 amperes; the power factor of B is 0.3 and Ib 50 amperes. Find: (a) the current i;i the line, (b) the power factor of the system. 27. Measurements taken on the series system indicated in the accompany- ing sketch were as follows: I = 100 amperes, E = 2300 volts, E\ = 1800 volts, E 2 = goo volts. 22(7 V. t 200 Ft. -*p 200 'Ft >K- ZOO Ft M 1 -X ' I flains : | 1 J215 Volts 1 | 25Amp& -i-b 2/5 Vo//s 15 Amperes 2/5 V&//S lOAmperes Find: (a) the power factor of the circuit, (b) the resistance of the circuit, (c) the reactance of the circuit, (d) the impedance of the circuit, (e) the power used in the circuit. 28. The continuous-current distributing system indicated in the accom- panying sketch is connected to 2 20- volt supply mains. Determine: (a) the sizes of wires required for the dif- ferent parts of the circuit so that 215 volts may be de- livered at the terminals of each load, (b) the resistance of the distributing system (not including the load), (c) the power (watts) loss due to the resistance of the wires, (d) the total output of the generator. R pV\^-. 29. Calculate the resistance of the circuit indicated in VWH - the accompanying sketch when: *-M/\r R s RI = 10 ohms, Rz =5 ohms, R z = 8 ohms. 30. Find the impedance of the system specified in Problem 29 when a re- actance of 2 ohms is connected in series with R z . 31. The capacitance of each of three condensers is 10 microfarads. Find the capacitance of the circuit when: (a) the condensers are connected in series, (b) the condensers are connected in parallel. 32. Find the current flowing in a 25-microfarad condenser when connected to a 2 20- volt, 6o-cycle, alternating-current system. 33. Three impedances A , B and C are connected in series to 6o-cycle mains. R a = loohms, Rb = 5 ohms, R c = 12 ohms, L a o.i henry, Lb = 0.08 henry, L c = 0.2 henry. A current of 10 amperes (effective) flows in the circuit. Find the voltage be- tween the terminals of: (a) impedance^!, (b) impedance B, (c} impedance C, (d~) the circuit (the applied voltage). 34. The impedances specified in Problem 33 are connected in parallel and to no- volt, 25-cycle mains. Find: (a) the impedance of the system, (b) the resistance of the system, (c) the reactance of the system, (d) the current in each parallel branch, (e} the total current flowing in the circuit (line) . THE ELECTRIC CIRCUIT 27 35. An induction motor is operated in parallel with 100 incandescent lamps. The motor takes 50 amperes and has a power factor of 0.8; the lamps each re- quire 0.4 ampere and their power factor is unity. Find: (a) the current sup- plied to the combination, (b) the power factor of the circuit, (c) the power component of line current, (d) the wattless component of line current. 36. The current in a line supplying two induction motors is 100 amperes, and lags 45 degrees behind the electromotive force. One motor has a power factor of 0.85, and takes 60 amperes. Find the power factor of the other motor, and the current supplied to it. 37. The maximum value of an alternating current flowing in a circuit, the impedance of which is 25 ohms and the ratio =0.75, is 100 amperes. Find: JK. (a) the effective value of the current, (b) the power in the circuit, (c) the power factor of the circuit. 38. Find the size wire required to supply 10 amperes to lamps 200 feet from the generator, when the' allowable resistance of the wires is 0.2 ohm. 39. Find the current flowing in the circuit when the impedances of Problem 33 are connected to 25-cycle mains, the voltage of which is equal to the voltage found in part (d) of Problem 33. 40. Three impedances are connected in parallel. fa 73 and lags 30 degrees. behind the electromotive force, Ib = 150 and lags 45 degrees behind the electromotive force, I c = 225 and lags 15 degrees behind the electromotive force. Find: (a) the current in the supply line, (b) the power factor of the system. 41. The resistances and reactances indicated in the accompanying sketch have the following values: Ri = 5, & = 3, ^s = 6, R 4 = 10, X l = 8, X z = 4, X, = 3. Find: (a) the impedance of the circuit, (b) the power factor of the circuit. T 42. Referring to the accompanying sketch: EI = 100, 2 = 150, / = 10, EI and E z are 60 degrees : out of phase. E, j Find: (a) E, (6) Z, (c) R, (d) X, (e) power factor, (/) Li- power in the circuit. 43. Find the impedance, the resistance, the reactance and the power factor of the following circuit: 44. Find the impedance, the resistance and the reactance of the circuit in Problem 43 when Xi is a capacitance equivalent to 5 ohms. 28 ESSENTIALS OF ELECTRICAL ENGINEERING 45. Referring to the accompanying figure: R r E = 220 volts, R = 40 ohms, r = 0.5 ohm, L = o.i henry, C = 0.00012 farad, / = 60 cycles. Find: (a) the current, (&) 22i, (c) 2 , (d) E 3 , (e) power in the circuit, (/) power factor. Draw vector diagram representing the current and the voltages in the circuit. 46. A voltmeter, the resistance of which is 3,000 ohms (reactance negligible), is connected in series with a condenser between 25o-volt, oo-cycle alternating- current mains. The indication of the meter is 1 50. Find the capacitance of the condenser, and the voltage between its terminals. 47. A current of 8 amperes flows in an impedance connected between 240- volt, 6o-cycle alternating-current mains. The power absorbed in the imped- ance is 1280 watts. Find: (a) the power factor of the circuit, (b) the resistance of the circuit, (c) the reactance of the circuit, (d) the inductance of the circuit. Draw vector diagram of the voltages and the current. 48. When an impedance (resistance and inductance in series), the power factor of which is 0.707, is connected to 6o-cycle, 2 50- volt alternating-current mains, 25 amperes flow in the circuit. Find: (a) the capacitance required to be connected in the circuit to cause voltage resonance, (b) the current in the resonant circuit, (c) the voltage across the resistance, (d) the voltage across the inductance, (e) the voltage across the capacitance. 49. The impedance in Problem 48 is connected to 6o-cycle, 25o-volt alter- nating-current mains in parallel with a condenser. The resistance of the con- denser circuit is negligible. Find the capacitance when the power factor of the circuit is: (a) unity, (b) 0.90 lagging, (c) o.oo leading. 50. Same as Problem 49 except the resistance of the condenser circuit is 5 ohms. 51. Show that when the applied voltage and the resistance of a series cir- cuit are constant, and the reactance of the circuit varies from zero to infinity, the locus of the current vector is a semi-circle, the diameter of which makes zero" angle with the vector of applied electromotive force. 52. Show that when the applied voltage and the reactance of a series cir- cuit are constant, and the resistance of the circuit varies from zero to infinity, the locus of the current vector is a semi-circle, the diameter of which makes an angle of 90 degrees with the vector of applied electromotive force. CHAPTER II MAGNETISM AND MAGNETIC INDUCTION 1. Magnetism. A body which has the power to attract a piece of iron is termed a magnet, and the property by virtue of which attraction takes place is termed magnetism. The ultimate nature of magnetism, like that of electricity, has never been determined, but electricity and magnetism are intimately associated. 2. Magnetic field. The space in and around a magnet is a magnetic field. A magnetic field is represented graphically by means of lines (lines of magnetic force or magnetic induction) which pass from the north pole to the south pole of a magnet and form complete loops or circuits. Fig. 14. The total number of lines of magnetic induction leaving a north pole and entering a south pole is a magnetic flux. FlG - X 4- 3. Production of a magnetic field. The only known means for the production of a magnetic field is the electric current (electricity in motion). It has been proved experimentally that the space surrounding a current-carrying con- ductor is a magnetic field, that the lines of mag- netic induction are concentric circles,* and that their direction is clockwise or counter-clockwise as the current in the conductor flows away from or toward the observer. The direction of a magnetic flux is assumed to be that indicated by the north-seeking pole of the magnetic needle. The magnetic field around a current-carrying wire is repre- sented, in direction and in intensity, in Fig. 15- 4. The solenoid. If a current-carrying FI(J i6 conductor is wound into a helix, the lines of magnetic induction are no longer concentric circles but pass through the helix as indicated in Fig. 16. If the helix is long in * When the conductor is isolated and uninfluenced by other magnetic fields. 29 30 ESSENTIALS OF ELECTRICAL ENGINEERING comparison to its diameter, the magnetic flux is uniformly dis- tributed over its cross-sectional area except near the ends where the lines begin to diverge. 5. The electromagnet. The magnetic effect of a current-carry- ing wire is concentrated by winding it into a helix, and greatly in- creased if the helix is wound on an iron core. A current-carrying coil wound on an iron core is an electromagnet, and is used for the pro- duction of the intense magnetic fields required for modern electrical apparatus. When once magnetized, iron retains a part of the mag- netic effect of the current-carrying coil. This permanent effect is termed residual magnetism, and varies with the quality of the iron. The polarity of an electromagnet is easily determined from the direction of the current in its exciting coil. If the direction of current around the coil is that of the rotating motion of a right- handed screw, the direction of magnetic flux is that of the longi- tudinal movement of the screw. 6. Magnetomotive force. Magnetomotive force is that prop- erty by virtue of which a magnetic flux is established or maintained. 7. Reluctance. The opposition offered to the establishment or to the maintenance of a magnetic flux is termed the reluctance of the magnetic circuit. The reluctance of a magnetic circuit is directly proportional to its length and inversely proportional to its cross-sectional area. 8. Permeability. Some materials, notably iron and many of its alloys, offer less opposition to the establishment or the main- tenance of a magnetic flux than do others. The ratio of the flux established or maintained in a given length and cross section of a material by unit magnetomotive force, to the flux established or maintained by unit magnetomotive -force in the same length and cross section of vacuum is termed the permeability of the material. 9. Magnetic units. The magnetic units are: (a) Unit pole. Unit magnet pole is one which repels with a force of one dyne a similar and equal magnet pole placed at a dis- tance of one centimeter hi air. (Symbol m.) (b) Field intensity. A magnetic field has unit intensity (one line per square centimeter of cross-sectional area) when it reacts on a unit pole, placed in the field, with a ^force of one dyne. (Symbol #.) MAGNETISM AND MAGNETIC INDUCTION 31 (c) Flux density. By flux density is meant the number of lines of magnetic force or induction per unit of cross-sectional area of a magnetized material. (Symbol B.) Note. When a magnetic flux is propagated in air, or other non- magnetic material, the flux density is equal to the field intensity. (d) Magnetic flux. The total flux in a magnetic circuit is equal to the product of the average flux density and the cross-sectional area of the circuit. The unit is the maxwell. (Symbol <.) (e) Reluctance. Unit reluctance (the oersted) is that opposition offered by a cubic centimeter of vacuum,* each face of which is one centimeter square, to the passage of a magnetic flux between its parallel faces. (Symbol 91.) (/) Magnetomotive force. The magnetomotive force of a circuit is equal to the work in ergs done when a unit magnet pole is moved around the circuit against the force, and is measured in gilberts. (Symbol cF.) Unit difference of potential exists between two points when one erg is required to transfer a unit pole from one point to the other. (g) Magnetizing force. Magnetizing force is the ratio of the magnetomotive force of the circuit to the length of the circuit. Magnetizing force is numerically equal to field intensity, and is indicated by the same symbol. (ti) Permeability. The permeability of a material is the ratio of the flux density and the field intensity, or the magnetizing force. (Symbol /*.) 10. Law of the magnetic circuit. The flux in any magnetic circuit is directly proportional to the magnetomotive force and in- versely proportional to the reluctance of the circuit. ; :? *-! <> when = the magnetic flux in maxwells, eF = the magnetomotive force in gilberts, SH = the reluctance in oersteds. 11. Faraday's Law. An electromotive force is induced in any closed circuit when the magnetic flux linking with the circuit changes in value, and the magnitude of the electromotive force induced is propor- * Air has practically the same specific reluctance as vacuum. 32 ESSENTIALS OF ELECTRICAL ENGINEERING tional to the rate at which the magnetic flux linking with the circuit changes. 12. Lenz* Law. The direction of an induced current is such that its reaction opposes any change in the value of the magnetic flux linking with the closed circuit, in which the induced current flows. Lenz' Law is simply a special application of the general principle of mechanics that action and reaction are equal and opposite. 13. Induced currents. From Faraday's Law it is evident that an electric conductor forming part of a closed circuit and lying partly or wholly within a magnetic field has an electromotive force induced in it when: (a) the relative position of the conductor and the magnetic field changes, (b) the intensity of the magnetic field varies. (a) Induction by motion. The following propositions are self- evident or are easily demonstrated by experiment: (1) The relative position of the conductor and the magnetic field is changed by the movement of either the conductor or the field. (2) The induction of an electromotive force occurs only during the period of motion. (3) The induced electromotive force is proportional to the number of lines of magnetic induction across which the conductor cuts per unit of time. (4) The direction of the induced electromotive force is reversed when the direction of motion is reversed. - Let Fig. 17 represent a cross section of a magnetic field, the flux in which flows toward the observer, and a copper conductor A which may be moved to the right or to the left, the circuit being completed through the rails B, C and D. When the conductor A is moved to the right an electromotive force is induced, the magnitude of which is proportional to FlG I7 the rate of motion, and the direction of the current in the circuit is that indicated by the arrows. If the wire is moved to the left at the same speed, the same value of current flows in the wire but its direction of flow is reversed. In the above case the induced electromotive force is proportional to the speed of the conductor, the length of the conductor and the density of the magnetic field being constant. In any construction, MAGNETISM AND MAGNETIC INDUCTION 33 the number of lines of magnetic induction cut per unit of time de- pends on: (Y) The length of conductor lying within the magnetic field. (2') The rate at which the conductor moves across the magnetic field. (3') The density of the magnetic field (the number of lines of magnetic induction per unit of cross-sectional area). The length of the conductor, the speed at which the conductor or the magnetic field moves, and the flux density may be expressed in any desired units, provided the length of the conductor and its speed are in the same unit, and the unit of area is the square of this linear unit. The unit on which the electromotive force of a circuit is based is that difference of potential found to exist when a con- ductor one centimeter long moves* at a uniform speed of one centi- meter per second across a uniform magnetic field having a flux density of one maxwell (one line of magnetic force per square centi- meter). This unit of electromotive force is inconveniently small and the commercial unit (the volt) is taken as 100,000,000 times the fundamental unit (the abvolt). Then e = I X V X B X io~ 8 , (2) when e = the electromotive force in volts induced in the conductor, / = the length of the conductor, V = the relative speed (per second) of the conductor and the magnetic field, B the density of the magnetic field (maxwells per unit of cross-sectional area),. i.e., an electromotive force of one volt is induced in a closed circuit when the magnetic flux linking with the circuit changes at the rate of 100,000,000 maxwells per second. As pointed out above, a reversal of the direction of motion causes a reversal of the direction of current flow. There is, then, a definite relation between the direction of motion, the direction of induced electromotive force and the direction of magnetic flux. By refer- * The direction of motion and the axis of the conductor are here assumed to be at right angles to each other and to the direction of the lines of magnetic force. When this condition does not exist, the effective length of the conductor is equal to / times the sine of the acute angle between the axis of the conductor and that of the magnetic field, and the effective velocity is equal to V times the sine of the acute angle between the direction of motion and the axis of the magnetic field. 34 ESSENTIALS OF ELECTRICAL ENGINEERING t&x: FIG. 18. ence to Fig. 17 it will be observed that any one of these three quantities is at right angles to each of the other two. A simple rule for determining the relations of these quantities was deduced by Dr. J. A. Fleming and is known as Fleming's finger rule. If the thumb, index and middle fingers of the right hand are placed as indicated in Fig. 18, the thumb indicates the direction of motion across the magnetic field, the index finger the direction of flux, and the middle ringer the direction of induced electromotive force. If a ring or closed loop is rotated in a magnetic field, one side of the loop cuts the flux in one direction and the opposite side cuts it in the other direction. Hence, a current flows around the loop but its direction is peri- odically reversed, and the induced electromotive force passes through successive values from zero to maximum and then from maximum to zero. Let Fig. 19 represent a loop of wire and a uniform magnetic field, the loop being rotated at a constant angular velocity in the direction indicated by the arrow. Starting from the position shown, the electromotive force increases to maximum, decreases to zero, increases to maximum in the opposite direction, and again decreases to zero for each revolution of the loop. The flux passing through (link- ing with) the loop decreases from maximum to zero, or increases from zero to max- imum, in one quarter of a revolution, and the electro- motive force induced in the loop is proportional, at any instant, to the sine of the angle through which the loop has rotated, i.e., the electromotive force is harmonic as indicated in the rectangular representation where the abscissas are angular distances (degrees) passed through by the rotating loop, and the ordinates are the instantaneous values of electromotive force in- duced in the loop. Since the electromotive force induced in one FlG * MAGNETISM AND MAGNETIC INDUCTION 35 side of the loop tends to produce a current in the same direction around the loop as the electromotive force induced in the other side of the loop, the electromotive force of the circuit is twice as great as if only one side of the loop were in the magnetic field. (b) Induction by varying the flux density. If two coils of in- sulated wire are wound on an iron core as indicated in Fig. 20, and coil A is supplied with a current of con- stant value (continuous current) no cur- rent will flow in coil B. If the current in coil A is made variable, the following effects may be noted: (1) A current flows in coil B when the current in coil A either increases or decreases. (2) A current flows in coil B only when the value of the current in coil A is changing. (3) The electromotive force induced between the terminals of coil B is proportional to the rate at which the flux in the iron ring changes. Within certain limits, the change in the flux is approxi- mately proportional to the change in the current flowing in coil A. (4) The direction of current in coil B is reversed by reversing the direction of current in coil A. (5) The direction of current in coil B is always such that it opposes the change in flux which produced it. From the above considerations it is evident that electrical energy may be transferred from one circuit to another, provided the cur- rent in the supply circuit is of varying value, i.e., either pulsating or alternating. 14. Reaction between a magnetic field and a current-carrying conductor. Let / = the force required to move a conductor across a uniform magnetic field at a velocity of V centimeters per second, B = the density of the magnetic field in maxwells, / = the length of the conductor in centimeters lying in the magnetic field and perpendicular to both the direction of motion and the direction of the magnetic flux, i = the current in amperes flowing in the conductor, e = the electromotive force in volts induced in the conductor (= eX io 8 c.g.s. units), 36 ESSENTIALS OF ELECTRICAL ENGINEERING ds = the distance through which the conductor moves, dt = the time required for the conductor to move through the distance ds. Then / (ds) = ei (dt) joules (3) = ei (dt) X io 7 ergs (4) IBVi (dt) -^ergs (5) 'ds IBVi ,,. and ; '*-^' < 6 > But |=F. (7) Therefore, / = dynes (8) io IBi , ^ 22 B f N = - -pounds. (io) 981 Xio 4 ^ From the above considerations it is evident that: (a) A current-carrying conductor lying within a magnetic field tends to move across the field, the direction of motion being at right angles to the axis of the conductor and to the direction of the flux. (b) The direction in which the conductor tends to move is re- versed by reversing either the direction of the magnetic flux or the direction of the current flowing in the conductor. (c) The force tending to move the conductor across the magnetic field is proportional to the product of the current flowing in the conductor, the density of the magnetic field, and the length of conductor lying in the magnetic field. The relations between the direction of the mag- netic flux, the direction of the current and the direction of the resultant force on the current-carry- ing conductor are shown in Fig. 21 by the index finger, the middle finger and the thumb of the left hand. 15. Magnetic flux and field intensity due to unit pole. From the definitions of unit pole and field intensity given above, it MAGNETISM AND MAGNETIC INDUCTION 37 follows that the field intensity due to unit pole is unity at any point on the surface of a sphere one centimeter in radius, and of which the unit pole is the center. But the area of such a sphere is 4 TT square centimeters. Therefore, the total flux leaving or entering a unit pole equals 4 TT maxwells, and the field intensity due to an isolated magnet pole is inversely proportional to the square of the distance x of a surface from the pole. *-$ () 1 6. Field intensity produced by an electric current. As stated in Section 3, the space surrounding a current-carrying con- ductor is a magnetic field. The intensity of the magnetic field at the center of a circular loop, and that surrounding a long straight wire, are of particular interest. (a) Field intensity at the center of a loop of r centimeters radius, in which flows a current of i amperes. From equation (n), the intensity of the field set up by a magnet pole m placed at the center of the loop is and the force with which the current-carrying conductor and the magnetic field react is, from equation (8), , m 2 irri , f >, dynes. (14) If H" is the field intensity at the center of the loop due to the current in the loop, = mH" 10 r (16) J TT// O.2 TTl / \ and H" = -- (17) r (b) Field intensity x centimeters distant from the axis of a long straight wire in which flows a current of i amperes. Let m magnet poles per centimeter length of the wire be uniformly distributed along a line parallel to and x centimeters distant from the axis of 38 ESSENTIALS OF ELECTRICAL ENGINEERING the current-carrying wire. The 4 TT lines which emanate from each unit pole radiate like the spokes of a wheel, and are in a plane perpendicular to the axis of the current-carrying wire. Therefore, the intensity of the field at the wire is (18) 2irX 2m , . (19) 10 x and the force exerted on each centimeter length of the conductor is , 2mi, , , / = - dynes. (20) IO OC If H" is the field intensity at the poles due to the current-carrying wire, / = mH." dynes, (21) -r-fft 0.2 mi / N mH" = - (22) 1 TT// O.2 I , \ and H" = - - - (23) 00 17. Relation of the gilbert to the ampere-turn. Let a current of i amperes flow in a coil of n turns. When the coil is rotated about a unit pole, or a unit pole is moved around a path threading the coil, the work done is TT7 f ^ W = - (24) 10 4 wni f \ = - -ergs. (25) 10 The magnetomotive force due to the current-carrying coil is, by definition, / = m* (26) = 1.257 m gilberts. ( 2 7) The quantity ni is termed " ampere- turns," and the ampere- turn is commonly used as a unit of magnetomotive force. 18. The C. G. S. unit of electric current. The c. G. s. unit of current is that current which, when flowing in a wire in and at right angles to a magnetic field having a density of one maxwell MAGNETISM AND MAGNETIC INDUCTION 39 per square centimeter, causes a mechanical force of one dyne to be exerted on each centimeter length of the wire. From equation (8), the c. G. s. unit is equal to ten amperes. 19. Force of magnetic traction. Lines of magnetic force act like rubber threads under tension and tend to shorten, and the tractive force or pull between two surfaces in contact is propor- tional to the square of the flux density of the magnetic field within which the surfaces lie. Let the flux passing from surface X (Fig. 22) be divided into equal parts, and enter poles m' and m" on surface Y. Pole m' lies in the magnetic field of pole m" and pole m" lies in the magnetic field of pole m', and there is a reaction between each pole and one-half the total flux. By definition, Therefore, /' = ^ X f ( 29 ) B , (30) , T> Similarly, /" = ~- dynes. (31) Therefore, /=/'+/" (32) = ^ dynes. (33) But in a uniform field , , = AB (34) / AB *J f \ / = dynes (35) AB 2 AB 2 , . (37) when A = the area of the surface in square centimeters, B = the flux density in maxwells per square centimeter. AB 2 / = - - pounds, (38) 72,134,000 ESSENTIALS OF ELECTRICAL ENGINEERING when A = the area of the surface in square inches, B = the flux density in lines per square inch. TABLE II PERMEABILITY OF CAST IRON, CAST STEEL AND SHEET STEEL Cast iron Cast steel Sheet steel B M B - B M 10 5000 500 9,800 980 12,500 I25O 20 6600 330 13,400 670 16,700 835 30 7200 24O 14,400 460 18,000 600 40 7750 194 15,200 380 19,000 480 50 8150 15.700 3*4 19,800 396 6o 8500 142 16,150 269 20,500 342 70 8800 126 16,550 236 21, IOO 301 80 9075 114 16,925 211 21,600 270 90 9325 IO4 17,275 192 22,000 244 IOO 9600 9 6 17,600 I 7 6 22,300 223 The quality of iron varies greatly, and the above are to be considered simply as representative values. 20. Counter-electromotive force. When the reaction between a current-carrying conductor and a magnetic field causes the con- ductor to move across the field, an electromotive force which opposes the flow of the current is induced in the conductor, and the current flowing in. the conductor is proportional to the geometrical dif- ference of the applied and the induced (counter) electromotive forces. 21. Magnetic leakage. Unlike the electric current, a mag- netic flux cannot be confined to a definite path, except under con- (a) Cb) FIG. 23. Magnetic Leakage. ditions which are not applicable to commercial electrical apparatus. The total flux in any magnetic circuit may be regarded as flowing in two or more parallel branches, the flux in each branch being inversely proportional to the reluctance of that branch. In Fig. 23 that part of the flux which flows through the air is termed the leakage flux. The ratio of the total flux set up by a magneto- MAGNETISM AND MAGNETIC INDUCTION motive force to the useful flux is the leakage coefficient. The leakage coefficient of a dynamo depends on its size and design, and varies from i.i to 1.5. In any magnetic circuit containing iron, the leakage flux is in- creased by: (a) Increasing the total flux in the circuit. Since the permeability of iron is a function of the flux density, the reluctance of the path 150 700 650 600 550 500 450 400 550 .100 150 200 250 AMPERE-TURNS PER INCH FIG. 24. Magnetization Curves. 500 550 through the iron increases as the total flux set up by coil A in- creases, the reluctance of the path through the air remains con- stant, and a larger part of the total flux is diverted through the air. Fig. 2$&. (b) Increasing the length of the air gap between the metallic parts of the magnetic circuit. Increasing the air gap increases the reluc- tance of one branch of the magnetic circuit without affecting that ESSENTIALS OF ELECTRICAL ENGINEERING of the other, and diverts a larger proportion of the flux through the leakage path. Fig. 230. (c) An auxiliary coil which sets up an opposing magnetomotive force. The counter-magnetomotive force set up by coil B acts as an additional opposition to the flow of the magnetic flux through that part of the circuit, and increases the relative quantity of flux diverted through the air. Fig. 23 c. 22. Magnetization curves. Typical magnetization curves are shown in Fig. 24 for cast iron, cast steel and annealed stampings. It will be observed that for low flux densities the magnetization curve is an approximately straight line, that it bends sharply to the right when the magnetizing force is increased beyond a certain defi- nite value, and again becomes an approximately straight line if the magnetizing force is sufficiently increased. From these curves it is evident that the reluctance of a magnetic circuit in iron is not con- stant, but increases with increasing flux density. 23. The elementary dynamo. A simple generator producing an alternating electromotive force may be constructed by rotating an open loop of copper wire or other conducting material between the poles of a magnet, and providing means for connecting the termi- nals of the loop to an outside circuit. This arrangement is illus- trated in Fig. 25 where the terminals of the loop are connected to insulated copper rings mounted on the shaft. Stationary conduc- tors (brushes) press against these rings and through them connec- tion is made to an outside circuit. Unidirectional currents are obtained in the outside circuit by the arrangement shown in Fig. 26. Instead of connecting the terminals MAGNETISM AND MAGNETIC INDUCTION 43 of the loop to separate rings, connection is made to a single ring which is divided, the segments being insulated from each other. By placing the brush mid-way of the pole face, it passes from one segment to the other when the loop is in the position where it is generating zero electromotive force, and the current in the outside circuit always flows in the direction indicated by the arrows. The magnitude of the electro- motive force varies with the position of the loop, the values during one revolution of the loop being plotted in Fig. 27. If, instead of a single loop, a number of symmetrically spaced loops are con- nected in series the electromotive force of the system is the sum of the instantaneous electromotive forces induced in the individual loops, and the potential difference between the stationary brushes is approximately constant. Fig. 28 shows the electromotive forces induced in each of three symmetrically spaced loops, and the resultant electromotive force due to the combined action of the three loops. The greater the number of loops in the series, the more nearly the resultant approxi- mates a straight line. Structurally the electric generator and the electric motor are identical, and the same machine may be used for the production of electrical energy or for its conversion into mechanical energy. CHAPTER II PROBLEMS 1. Find the magnetomotive force set up by a winding of 100 turns when a current of 5 amperes flows in the coil. 2. Determine the permeability of the iron around which the coil in Problem i is wound if the length of the coil is 10 inches, and the flux density in the iron is 10,000 lines per square centimeter. 3. The reluctance of a magnetic circuit is 10 oersteds. Find the flux set up by a magnetomotive force of 150 gilberts. 4. Find the flux density when the force (pull) between an electromagnet and its armature is 60 pounds per square inch. 5. A copper wire is moved across a uniform magnetic field at the rate of 500 feet per minute. Find the electromotive force, per unit length, induced in the wire if the density of the magnetic field is 75,000 lines per square inch. 6. A closed loop rotates in a uniform magnetic field at the rate of 200 r.p.m. and the maximum flux linking with the loop is 1,000,000 lines. Find: (a) the 44 ESSENTIALS OF ELECTRICAL ENGINEERING effective value of the electromotive force induced in the loop, (b) the current flowing around the loop if the resistance of the loop is o.oi ohm. Neglect any inductance. 7. Find the force acting on a current-carrying wire 10 inches long, lying wholly within a uniform magnetic field (B = 45,000 lines per square inch), and carrying a continuous current of 1000 amperes. 8. Find the average force acting on the wire in Problem 7 when an alter- nating current of 1000 amperes (effective) flows in the wire. 9. A series magnetic circuit is composed of: 6 inches of cast iron, density 35,000; 8 inches of cast steel, density 75,000; j-inch air, density 50,000. Find the ampere-turns required in the winding. Use curves in Fig. 24 for the iron and the steel. 10. A magnetic circuit 10 inches in length is made up of sheet-iron punchings (laminations), and excited by 1000 turns. Find the exciting current when the average flux density is 40,000, and 90% of the cross-sectional area is iron. 11. A conductor 1 5 inches long moves across a magnetic field having a uni- form density of 48,000 lines per square inch. Find the electromotive force induced in the conductor if the conductor moves at the rate of 4500 feet per minute. 12. If the magnetic field in Problem n consists of poles 9 inches long and separated by a distance of 3 inches, find the average electromotive force induced in the conductor. 13. A continuous-current armature is 18 inches long and 30 inches in di- ameter. The electromotive force induced in each conductor as it passes under an interpole 15 inches long is i volt when the armature rotates at a speed of 600 r.p.m. Find the flux density in the air gap under the interpole. 14. Show that the field intensity in a long solenoid is 15- Show that the ampere- turns in the exciting coil of an electromagnet are NI = eJL when B = the maxwells per square centimeter, / = the length of the magnetic circuit in centimeters. 1 6. Show that the ampere- turns in the exciting coil of an electromagnet are when B = the maxwells per square inch, / = the length of the magnetic circuit in inches. 17. An electromagnet lifts 5 tons (10,000 pounds). The mean length of the magnetic circuit = 60 inches; cross-sectional area = 20 square inches; /z = 1000. Find: (a) the total flux, (6) the flux density, (c) the ampere-turns in MAGNETISM AND MAGNETIC INDUCTION 45 the exciting coil, (d) the magnetomotive force (gilberts), (e) the field intensity, (/) the reluctance of the circuit. 1 8. A hollow wrought-iron cylinder (outside diameter =12 inches, inside diameter = 6 inches) 6 inches long forms part of a magnetic circuit, the flux in which is parallel to the axis of the cylinder. Average density = 61,000 lines per square inch, /z =20. Find: (a) ampere-turns required, (b) flux density in the iron, (c] field intensity, (d) reluctance of the cylinder. 19. A point lies on the axis of a circular loop of wire in which flows a cur- rent of i amperes. The radius of the loop is r centimeters, and the point is x centimeters distant from the plane of the loop. Show that the field inten- sity at the point is TT 0.2 -n-r-i ti = - 20. The currents in two long parallel wires are equal but flow in opposite directions. Show that the field intensity at any point on a line joining the centers of the wires is Tr o.2i . o.2i a = h D-x when D = the distance in centimeters between the axes of the wires, x = the distance in centimeters of the point from the axis of one wire. 21. A long solenoid has n turns per centimeter of its total length. Show that the field intensity at any point inside the solenoid, except near the ends, is H = 0.4 irni when i amperes flow in the windings. CHAPTER III PRACTICAL CONSTRUCTION OF THE DYNAMO i. Parts. The principal parts of a commercial dynamo are: (a) frame, (b) field poles, (c) field windings, (d) armature core, (e) armature winding, (/) yoke, (g) commutator, (h) collector rings, (i) brushes, (j) brush holders. FIG. 29. Typical Continuous Current Dynamo. General Electric Co. FIG. 30. Alternating-current Dy- namo. General Electric Co. (a) Frame. The frame of a dynamo is the supporting structure, and includes the base and the supports for the bearings in which the armature shaft rests, as well as the yoke. It is usually made of cast iron or cast steel. (b) Field poles. The field core is a body of iron around which the field winding is placed. Its function is to reduce the reluctance of the magnetic circuit (increase the flux for a given field winding and current), and to supply a mechanical support for the field wind- ing. It may be made of cast iron, of cast steel, or be built up of stampings bolted or riveted together. Fig. 32 shows a typical form 4 6 PRACTICAL CONSTRUCTION OF THE DYNAMO 47 of field pole. The part next the armature is usually spread out to give a larger cross-sectional area as it is desirable to have a smaller flux density (flux per unit of cross-sectional area) in the air gap than in the body of the pole. This enlarged part is termed the pole shoe, and is sometimes made as a separate piece and bolted to the core. The field poles and the yoke of small machines are some- times cast in a single piece. When made separately, they are bolted together, the joint being made as close as possible in order to reduce the reluc- tance of the magnetic circuit. (c) Field windings. The field winding of a dynamo is that part which produces the magnetic field or flux across which the electrical conductors move, or which moves relatively to the conductors. It consists (Fig. 33) of a coil of insulated cop- per wire, the exciting current being supplied by the dynamo itself (self-excited) or from some outside source (separately excited). (d) Armature core. The armature core is FIG. 32. Typical Field that part of the dynamo over or around which Core for Continuous the armature conductors are placed. It serves FIG. 31. Frame and Field Structure of Tri- umph Continuous Current Dynamo. Current Dynamos. Western Electric Co. as a mechanical support for these conductors and as a path through which the magnetic circuit is completed. Armature cores are of two classes: (i) ring, (2) drum. (1) Ring cores. The armature cores of early dynamos were iron rings through which the conductors were threaded, as shown in Fig. 34a. Because of mechanical and electrical deficiencies, the ring construction is seldom used in present day machines. (2) Drum cores. The drum core consists of an iron cylinder (Fig. 34b) on the surface of which the armature conductors are placed. Armature cores are built up of thin stampings (laminations) of ESSENTIALS OF ELECTRICAL ENGINEERING soft iron or steel. The purpose of this construction is to reduce the eddy currents which flow in the body of the core and cause it to heat. Laminations vary in thickness from o.oi inch to 0.04 inch. Early dynamos were built with smooth cores, the armature conductors on their surface being held in place by binding wires. In later types, the arma- ture conductors are placed in slots as shown in Fig. 37, and retained by means of wooden or fiber wedges. Different slot forms are used by different manufacturers, from an entirely closed slot, through which the armature conductors are threaded, to the rectangular (open) slot. FIG. 33. Typical Field Coils (Westinghouse). FIG. 34a. Ring Armature. FIG. 34b. Drum Armature. (a) Core. (b) Lamination. FIG. 35. Armature Core and Lamination for Crocker- Wheeler Continuous Current Dynamo. (e) Armature windings. The armature winding consists of electrical conductors which move relatively to the magnetic field. PRACTICAL CONSTRUCTION OF THE DYNAMO 49 Complete Armature for Crocker- Wheeler Continuous Current Dynamo. Retaining Hedqe For machines of the usual commercial voltages, the winding con- sists of several hundred insulated copper wires or bars, divided into groups, the wires of each, group being connected in series and the groups in parallel. The number of parallel groups is never greater than the number of poles on the dynamo (simplex windings), and there may be only two parallel groups on a continuous-current armature, re- FKJ gardless of the number of poles. The armature conductors of an alternator are usually all connected in series (single-phase alter- nator). A typical armature coil having one turn is shown in Pig- 38- Two or more coil sides are usually placed in one slot, one side of each coil being placed in the bottom of a slot and the other side in the top of a slot under an adjacent pole. This FIG. 37. Section of arrangement makes all the coils on an armature Slotted Armature ... . . ,. - . , , Core similar, and gives the finished armature a sym- metrical appearance. Continuous-current armature windings. The principal types of armature windings used on continuous-current dynamos are: (i) lap, (2) wave. (i) Lap or parallel windings. -The lap or parallel winding is used when the armature conduc- tors are to be divided into as many groups as there are poles on the machine. Starting at any com- mutator segment, the armature conductor passes under one pole, across the rear end of the core, and back under an adjacent pole (which is of an opposite polarity), then to the commutator segment adjacent to the one from which the coil started. The second coil starts from the commutator segment at which the first ends, the third where the second ends, etc., until the winding closes on itself FIG. 38. Typical Armature Coil. Triumph Electric Co. 50 ESSENTIALS OF ELECTRICAL ENGINEERING at the commutator segment from which the first coil started. Fig. 39- Dynamos having lap- wound armatures are provided with as many brush sets as there are poles. If, for any reason, such as the wearing of the bearings, the elec- tromotive forces generated in the different parallel branches of a lap-wound armature are unequal, heavy currents circulate in the winding. These currents cannot be prevented but their effects are minimized by connecting, through heavy " equalizer " rings, such points of the winding as are normally of the same potential. (2) Wave or series windings. In the wave or series winding, the armature conductors are connected into two parallel groups. Start- ing at any commutator bar, the conductor passes under one pole, across the rear end of the core, and back under an adjacent pole s FIG. 39. Elementary (Lap) Armature FIG. 40. Elementary (Wave) Armature Winding. Winding. (of opposite polarity) to a commutator bar which is separated from the one from which the coil started by a distance slightly greater or v 1-4.1 i ^ total number of commutator bars Tr slightly less than - If a dynamo number of pairs of poles has four poles, the total number of commutator bars required is C = 2F I , (i) when Y is the number of commutator bars between the terminal con- nections of any given armature coil. If Y = 9 (terminals of first coil connected to commutator bars i and 10), the total number of commu- tator bars must be either 17 or 19. By reference to Fig. 40, it will be seen that the winding closes on itself, as in the lap winding, by mak- ing a final connection to the commutator bar from which it started. PRACTICAL CONSTRUCTION OF THE DYNAMO 51 Since the wave winding offers only two paths for the current flow, only two brush sets are required and their proper position is shown in Fig. 40. Additional brushes may be used, the total brush equip- ment acting like two brushes having a total contact area equal to the sum of the areas of the individual brushes. The additional brushes do not affect the voltage to any appreciable extent. The relative positions of the brushes on a multi-polar wave- wound armature make it especially adapted for street car and other enclosed motors which are accessible from one side only. Alternating-current armature windings. The armature winding of an alternator does not differ, essentially, from that used in con- tinuous-current machines, but the winding does not usually form a closed circuit. An elementary | winding for alternators is shown in Fig. 41 . If the arma- ture rotates, the terminals are connected to insulated copper rings mounted on the shaft; if the armature is stationary, the terminals are connected to insulated blocks from which connection is made to the switchboard. Alternator armature wind- ings in common use are: (Y) FlG 4I Elementary Armature Winding chain, (2') basket. for Alternators. (V) Chain windings. In the chain winding one coil side only is placed in a slot. This winding is, therefore, especially adapted to high-voltage machines, the coils being easily insulated. The objection to the chain winding (but this objection is not serious) is that the coils are of different shapes and the individual conductors of different lengths. Fig. 42 shows a simple chain winding. (2'} Basket windings. In the basket winding two coil sides per slot are used. The coils are, therefore, similar, their form and arrangement being shown in Fig. 43. Basket windings are largely used for low- and medium-voltage alternators, and for induction motors. In both chain and basket windings all the conductors are con- nected in series (single-phase windings) so that the current flow is N ESSENTIALS OF ELECTRICAL ENGINEERING confined to a single path, i.e., the terminals of one coil are connected to the terminals of adjacent coils. (/) Yoke. The yoke is that part of the frame connecting the FIG. 42. Chain Armature Winding. General Electric Co. FIG. 43. Basket Armature Winding. Triumph Electric Co. pole pieces, and serves the double purpose of a mechanical support for the field poles and their windings, and of a path for the flux to pass from the south to the north pole. Fig. 44 shows cross sections of several common forms. The ridges on c are added to give additional mechanical strength. (a) (b) (c) FIG. 44. Cross Sections of Typical Yokes. (g) Commutator. -- The commutator is a cylindrical structure made up of copper bars insulated from each other and from the supporting structure. It is fastened to the armature shaft with which it rotates. Its purpose is to rectify (make unidirectional) the alternating current which flows in the armature winding of a contin- uous-current generator, or to periodically reverse the direction of the current flowing in the armature coils of a continuous-current motor. The best insulation obtainable is used to insulate the bars from each other and from the supporting frame. Mica is generally used for this purpose. A cross section and a sectional elevation PRACTICAL CONSTRUCTION OF THE DYNAMO 53 of a commutator are shown in Fig. 45, and a typical commutator in Fig. 46. (ti) Collector rings. In an alternator with rotating armature, FIG. 45. Commutator Structure. FIG. 46. Commutator. Tri- umph Electric Co. the terminals of the armature winding are connected to insulated copper rings mounted on the shaft. In an alternator with rotating field, similar rings (but not always of copper) are provided for the terminal connections of the field windings. Through these rings continuous current is supplied to the field. Fig. 47- (i) Brushes. The brushes are those parts of a dynamo which make sliding contact with the commutator or the rings, and through which current is taken from or supplied to the rotating armature, or supplied to the rotating field of an alternator with stationary armature. Brushes are made of: (i) carbon, (2) copper. (1) Carbon brushes. Carbon brushes are in very extensive use on commutating machines at the present time as they offer a material help in the prevention or the reduc- tion of sparking. Also, being set radially, the armature may be rotated in either direction without danger of injury to the brushes. Carbon brushes are sometimes coated with a deposit of metallic copper to give a reduced contact resistance between the brush and the line connection. (2) Copper brushes. Various forms of copper brushes are still used on alternating-current apparatus where no commutation takes FIG. 47. Collector Rings. General Electric Co. 54 ESSENTIALS OF ELECTRICAL ENGINEERING place, and where it is undesirable to have the large contact resist- ance characteristic of carbon brushes. A common form of copper brush consists of a number of thin leaves of spring copper soldered together at one end. (j) Brush holders. A brush holder is the arrangement by means of which the brush is supported and held in contact with the commutator or the ring. It is usually connected to a rocker by means of which the angular position of the brushes may be changed. A typical brush holder for carbon brushes is shown in Fig. 48. 2. Classes of dynamos. Dynamos are divided into two general classes according to the nature of the current they pro- duce or use: (a) continu- FIG. 48. Carbon Brush and Brush Holder. QUS current (ft) alternating Triumph Electric Co. current. (a) Continuous-current dynamos. Continuous-current dynamos are sub-divided into four divisions according to the character of the field excitation: (i) shunt, (2) series, (3) compound, (4) sepa- rately excited. (i) Shunt dynamos. In the shunt dynamo, the field is excited by means of a winding composed of a large number of turns of insu- lated copper wire connected to the terminals of the armature circuit so that the field winding and the load circuit are in parallel when the dynamo is operated as a generator, and the field winding and the armature circuit are in parallel when the dynamo is operated as a motor. The field current is only a small percentage of the rated current capacity of the dynamo, and is controlled by means of an adjustable resistance or field rheostat connected in series with the field winding. Since, for a given magnetization, it is required that the product of the current (amperes) flowing in the winding and the number of turns in the winding be constant, the larger the number of turns the smaller is the required current. Fig. 49 shows the schematic and the conventional wiring diagrams of a shunt dynamo. PRACTICAL CONSTRUCTION OF THE DYNAMO 55 -FieJd Resistance (2) Series dynamo. The field of a series dynamo is excited by a few turns of heavy wire through which flows the entire armature current, or a constant part of this current. The field excitation, instead of being approximately con- stant, as in the shunt dynamo, is proportional to the current flowing in the armature, i.e., to the load on the dynamo. The field flux, how- ever, may not increase or decrease in proportion to the change in the excita- FIG. 49. Wiring Diagrams for Shunt Dynamo. tion because of properties of the iron parts of the magnetic circuit. The schematic and conventional wiring diagrams of a series dy- namo are shown in Fig. 50. The resistance, shown in the conven- tional diagram as shunting the field winding, is for the purpose of varying the field excitation for a given armature current. This resistance is made of German silver or other high- resistance material, and determines the characteristic of the dynamo. It may not be changed while the dynamo is in operation, as may the resistance in the field circuit of a shunt dynamo, but is permanently connected to the field terminals and is changed only when the characteristic of the dynamo is to be altered. < to> Conventional FIG. 50. Wiring Diagrams for Series Dynamo. : b) SHORT SHUNT (conventional) LONG SHUNT < FIG. 51. Wiring Diagrams for Compound Dynamo. (3) Compound dynamo. The compound dynamo is, as the name implies, a dynamo having both a shunt and a series field winding. When the shunt field winding is connected to the terminals of the armature circuit, it is termed a " short shunt" compound dynamo; when the shunt field winding is connected so that the voltage be- tween its terminals is the line voltage, it is termed a "long shunt" compound dynamo. Conventional and schematic diagrams are shown in Fig. 51. 56 ESSENTIALS OF ELECTRICAL ENGINEERING (4) Separately excited dynamo. If the field windings of a genera- tor are supplied from some other source than its own armature, or the field and armature of a motor from different sources, it is said to be " separately" excited. Continuous-current dynamos are seldom separately excited while alternators are always excited in this way. The field winding of a separately excited dynamo does not differ from that of a shunt dynamo, and any dynamo may be separately excited by connect- m g its neld winding to a con- tinuous-current circuit of the proper voltage. Fig. 52. (b > Condon*, Q) The dternator . The FIG. 52- Wiring Diagrams for Separately ^ essential differe nce be- Excited Dynamo. tween an alternator and a continuous-current dynamo is the substitution of continuous cop- per rings for the commutator, since alternating electromotive forces are induced in a continuous-current armature. These rings serve as a means of connecting the armature winding and the outside or load circuit. The alternator is not self-exciting. However, in most alternators of the present day, the field instead of the armature is the rotating part, the armature winding being placed in slots on the inside of a cylindrical iron core, as shown in Fig. 43. In this type of alternator, rings are provided for connecting the rotating field winding to its continuous-current supply. The field excitation of an alternator is often supplied by a small continuous-current generator, the armature of which is mounted on or belted to the alternator shaft. In large installations, the fields of all the alternators are supplied from one or more (usually not less than two) continuous-current units, each of which is driven by its own prime mover. Attempts have been made to provide alternators with series (com- pound) windings by rectifying a portion of the armature current. The operation of such alternators (composite alternators) is not entirely satisfactory * and their manufacture has been discontinued. 3. Speed and frequency of an alternator. The speed, the fre- quency and the number of poles of an alternator have a fixed rela- * The degree of compounding changes with the power factor, and excessive spark- ing takes place unless the brushes are adjusted so as to pass from one segment to another at the instant the current is passing through its zero value. PRACTICAL CONSTRUCTION OF THE DYNAMO 57 tion to each other. An alternating-current cycle means the passage of the current, or the electromotive force, through all its values, both positive and negative, i.e., starting at zero the value rises to maximum, decreases to zero, rises to maximum in the opposite direc- tion, and again decreases to zero. To complete one cycle, a con- ductor must pass across a north and a south pole, or through 360 electrical degrees.* Then, for any given frequency, the speed is inversely proportional to the number of field poles, p and for any given speed, the frequency is directly proportional to the number of field poles. ': /-^, (3) when / = the frequency of the alternating current or electromo- tive force, n = the speed (revolutions per second), p the number of field poles. In America two frequencies have become standard twenty-five cycles for exclusive power service, sixty cycles for exclusive lighting service or for service which supplies both lamps and motors. In Europe a frequency of fifteen is largely used. The lower frequen- cies, while preferable in many re- spects for motor operation, are en- tirely unsatisfactory for lamps. If the frequency of an alternating current supplied to incandescent lamps is reduced to about forty, a distinct variation in the light intensity (which tires the eye) becomes noticeable. The same trouble is experienced with arc lamps. 4. The inductor alternator. An alternating electromotive force may be induced in a conductor without moving either the con- ductor or the exciting coil of the field magnet. In Fig. 53 let A be a cylindrical body of laminated iron similar to the armature struc- ture of any rotating field alternator, except the central part is cut * An electrical degree is the 36oth part of the angle subtended, at the axis of the machine, by radial lines through the centers of alternate field poles. FIG. 53. Schematic Diagram of Inductor Alternator. 58 ESSENTIALS OF ELECTRICAL ENGINEERING away to receive a coil B of insulated wire, and C is a body of iron having radial projections as shown. When coil B is supplied with continuous-current, a magnetic flux is set up as indicated by the heavy lines in the cross-sectional view. This magnetic flux is distributed in tufts over approximately one- half the inner surface of part A and moves around the cylinder as part C rotates. The conductors on the surface of A are cut by the flux in the same manner as if C were a permanent magnet. The frequency of an inductor alternator is twice that of a rotating field alternator having the same number of poles and operating at the same speed. This construction, while very simple, fails to show either good regulation or high efficiency. 5. Field discharge resistance. When the field circuit of a dynamo is opened there is induced in the field windings an electromotive winding force, due to the rapid decrease in the flux I QOQOQOQOOOOOOQQQp J ' . _ threading the windings. This induced electro- FIG. 54. Field Discharge r , , ,, Resistance motive force may be so large as to rupture the insulation unless an auxiliary resistance, in which the energy stored * in the magnetic field may be dissipated, is provided. Such a resistance is known as a " field discharge" re- sistance, and is commonly used with machines having a capacity greater than 100 kw. Fig. 54 shows the essential arrangement and operation of a discharge resistance. CHAPTER III PROBLEMS 1. Find the resistance of a 6-pole lap- wound armature, the winding of which consists of 3000 feet of No. TO copper wire (= 10,400 circular mils). Note. The resistance of any armature winding is equal to the resistance of the total length of conductor on the armature, divided by the square of the number of parallel paths into which the conductors are connected. 2. The armature in Problem i is to be wave wound, the same total weight of copper to be used and the armature to have the same number of series con- ductors between positive and negative brush contacts. Find: (a) the size and the length of the wire required, (6) the resistance of the armature winding. 3. Plot a development t of a 4-pole, wave- wound armature having 35 slots, 105 commutator bars and 630 conductors. * See Appendix B, Section 6. t Armature windings should be studied by means of wooden models on which the different windings may be placed. PRACTICAL CONSTRUCTION OF THE DYNAMO 59 4. Plot a development of a 4-pole, lap-wound armature having 44 slots, 88 commutator bars and 352 conductors. 5. The frequency of an alternator is 60. Find the speed when the number of poles equals: (a) 2, (b) 4, (c) 6, (d) 8, (e) 10, (/) 25, (g) 40. 6. The frequency of an alternator is 25. Find the speed when the number of poles equals: (a) 2, (b) 4, (c) 6, (d) 8, (e) 10, (/) 25, (g) 40. CHAPTER IV THE CONTINUOUS-CURRENT GENERATOR 1. The fundamental equation. The voltage induced in the armature windings of a continuous-current generator is directly proportional to: (a) the flux entering or leaving the armature at each pole, (b) the number of poles p in the field structure, (c) the total number of conductors N on the armature, (d) the speed n (revolutions per second) at which the armature rotates. It is in- versely proportional to the number of parallel circuits p' into which the armature conductors are connected. The equation representing the above statement is known as the fundamental^ equation of the continuous-current generator. <{>Nnp .. ( ^ E = Y (l) The constant (io 8 ) represents the ratio between the volt and the c.G.s. unit (the abvolt) of electromotive force. 2. Voltage characteristic. The voltage characteristic of a gen- erator is a curve showing the relation between the voltage of the generator and the load or the armature current, and is: (a) ex- ternal, (b) internal. (a) External characteristic. The external characteristic shows the relation between the terminal voltage of a generator, and the current flowing in the load circuit. This is the experimental curve, and is the one usually referred to when the term " voltage character- istic" is used. (b) Internal characteristic. The internal or total characteristic shows the relation between the total voltage induced in the armature winding, and the armature current. When current flows in the ar- mature circuit, the resistance of the circuit makes the terminal voltage of a generator less than that induced in the winding. The total voltage is, then, the terminal voltage plus the voltage drop in the armature circuit and that in the series-field windings, if the dynamo is either series or compound. + RJ, + E b> (2) 60 THE CONTINUOUS-CURRENT GENERATOR 6 1 when E a = the total electromotive force induced in the armature winding, E t = the terminal electromotive force, I a = the armature current, I s = the current in the series-field circuit, R a = the resistance of the armature circuit, R 8 = the resistance of the series-field circuit, E b = brush contact drop.* The total current flowing in the armature circuit is, evidently, the sum of that in the load circuit and that in the shunt field circuit. It may be measured directly or calculated by adding these two quantities. The current in the series field circuit of a compound generator is equal to that in the armature circuit or to that in the load circuit, as the generator is long or short shunt. 3. Voltage regulation. When the speed of the armature, the resistance of the armature winding,! and the resistance of the field circuit are constant, the voltage regulation of a generator is the ratio of the maximum deviation of the actual characteristic, between full load and no load, from the ideal characteristic, which is always a straight line, and the full-load (rated) voltage. 4. Building up. " Building up " is that process by which the field flux of a self -excited generator is increased to its normal value. When the iron in the magnetic circuit is once magnetized it retains some of its magnetic property (residual magnetism), and when the armature is rotated in this weak magnetic field a small electro- motive force is induced in the armature windings. This electro- motive force causes a current to flow in the field windings and, if the windings are properly connected, the flux is increased. The * It has been experimentally determined that the voltage drop between the brushes and the commutator is a function of the current density in the area of contact. The drop per pair of ordinary carbon brushes is represented, with sufficient accuracy for general calculations, by the formula: 6 = 0.8 + 0.2!), (3) when D = the current density (amperes per square centimeter) in the contact area between the brush and the commutator. Present practice allows five to six amperes per square centimeter of brush contact area at rated load. For densities less than one ampere per square centimeter, the values obtained by equation (3) are too large. A close approximation is obtained by assuming the brush-contact drop equal to the current density. f The brush-contact resistance is not constant. 62 ESSENTIALS OF ELECTRICAL ENGINEERING increased flux causes a larger current to flow in the field windings, and increases the flux still further. A self-excited generator can not build up if the flux set up by the field windings opposes the residual magnetism of the iron in the magnetic circuit. If the magnetism due to the field windings opposes the residual magnetism, the proper relations are obtained by: (a) reversing the field connections, (b) reversing the direction of armature rotation. (a) Reversing the field connections. If the terminal connections of the field circuit are reversed, the direction of the current flow in the field coils is reversed, and the two fluxes are no longer opposed. (b) Reversing the direction of armature rotation. If the direction of armature rotation is reversed, the polarity of the armature ter- minals is changed, causing the direction of the current in the field coils to reverse, and the two fluxes are no longer opposed. 5. Commutation. Commutation is the process of rectifying the alternating currents which flow in the armature conductors. Un- like the elementary (single loop) generator described in Chapter 2, Section 23, the conductors of commercial armatures may carry maximum current even though no electromotive force is being in- duced in the conductors themselves. Also, since the current in a given coil periodically reverses in direction, the current in the con- ductor must be reduced to zero and a current of like value estab- lished in the opposite direction during the very short time that the brush is in contact with the two commutator segments to which the terminals of the armature coil are connected. The inductance of the circuit tends to maintain any current that may be flowing in the coil at the time the brush short-circuits the coil, and to pre- vent the establishment of a current in the opposite direction. The current flowing in the coil may be reduced to zero and a cur- rent in the opposite direction established during the time the brush is in contact with two commutator segments by: (a) changing the relative resistances of the paths through which the current may flow, (b) causing the conductors to generate an electromotive force opposite to that which established the original current. (a) Resistance commutation. Since the resistance of an armature coil, even when composed of several turns, is very low, a compara- tively small additional resistance reduces the current very materi- THE CONTINUOUS-CURRENT GENERATOR 63 ally. Inherent properties of the circuit itself are used for this purpose. The contact resistance between a carbon brush and the commu- tator is many times the resistance of the armature coil and increases as the area of contact decreases. Let Fig. 55 represent the rela- tions of the coils, the commutator segments and the brush of a two- pole generator just before the brush makes contact with segment 4. One- half of the line current flows through coils a, b and c, unites with the other half flowing in coils d, e and /, and passes to the brush through the radial connection and segment 3 . An instant later the brush makes contact with segment 4, the current in coils , e and / naturally seeks the shorter path through segment 4, and current is diverted from coil d. As stated above, the inductance of coil'd, as well as the relative resistances of the two paths now pro- vided for the passage of the current to the brush, prevents an in- stantaneous diversion of all the current from the coil. As the armature continues to rotate, the area of the brush in contact with segment 4 increases, and the area in contact with segment 3 de- creases, changing the relative resistances of the two paths, so that current in coilJ has decreased to approximately zero when the area of the brush in contact with segment 4 is equal to the area in con- tact with segment 3. As the area of contact between the brush and segment 3 decreases still further, the resistance of this path increases, that through seg- ment 4 decreases, and a reversed current is gradually established in coil d. When contact between the brush and segment 3 is broken, the entire current should be flowing through coil d } and no sparking occur. (b) Voltage commutation. Since the electromotive force induced in an armature conductor is zero only at one point (the neutral), if the brushes are advanced in the direction of armature rotation, commutation is delayed until an opposing electromotive force is induced in coil d. This opposing electromotive force tends to reduce the current flowing in the coil, and to establish a current in the opposite direction. The magnitude of the opposing electro- FIG. 56. Commutating Pole Field Struc- ture. Crocker- Wheeler Co. 04 ESSENTIALS OF ELECTRICAL ENGINEERING motive force depends on the angular advance of the brushes. Therefore, to produce sparkless commutation the position of the brushes must be changed as the load on the generator increases or decreases. To make the commutating flux proportional to the current in the armature coil, the construction shown in Fig. 56 is used. The small poles (commutating or inter- poles) are excited by means of a winding connected in series with the load. The flux is, therefore, always of the proper value to pro- duce sparkless commutation, and no shifting of the brushes is nec- essary for changing load. In nearly all dynamos of re- cent design, a combination of (a) and (b) is used to produce good commutation, high - resistance carbon brushes being used in connection with an angular advance of the brushes or with interpoles. Interpoles are extensively used in present day dynamos, but satisfactory commutation, from no load to 25 per cent overload, is accomplished without their use and without changing the position of the brushes. 6. Armature reaction. When current flows in an armature wind- ing, a magnetic flux is set up, as indicated in Fig. 57. The magneto- motive force of the armature is proportional to the current in the winding, and the direction of the flux is at right angles to the polar axis. At one tip of the pole, the magnetism thus set up decreases the flux due to the field winding, and increases it at the other tip, its main effect being to change the symmetrical distribu- tion of the flux in the air gap* shown in Fig. 58, to the unsym- metrical distribution shown in Fig. 59, and to shift the neutral (the FIG. 57. Magnetic Field Due to Armature Current. * The effect of armature teeth on the distribution of the flux is here neg- lected. THE CONTINUOUS-CURRENT GENERATOR position in which zero electromotive force is induced in a conductor) in the direction of armature rotation. To produce good commutation the brushes must be moved fore- ward,* as indicated in Fig. 60. This movement of the brushes pro- L. of Pole CLafPole (a) (b.) FIG. 58. Symmetrical Distribution of Flux in Air Gap. C.L.ofPole FIG. 59. Unsymmetrical Distribution of Flux in Air Gap Due to Armature Currents. duces a corresponding change in the direction of the flux set up by the armature winding, and the armature magnetomotive force may be resolved into two components at right angles to each other. One of these components is proportional to the sine of the angle of brush advance and tends to set up a flux opposite in direction to that produced by the field windings, hence the term "demagnetizing action." The other component of armature magnetomotive force is proportional to the cosine of the angle of brush advance, and has the distorting effect described above. The effect of armature reaction is, then, twofold: (a) Cross-magnetization or distortion which is pro- portional to the current flowing in the armature conductors and to the cosine of the angle of brush advance. Because the angle of brush advance never exceeds a few degrees, for which the cosine is * The brushes of dynamos having commutating poles are set on the geometrical neutral (midway between the pole tips), and the effects of armature reaction are re- duced to a minimum. FIG. 60. Components of Arma- ture Reaction. 66 ESSENTIALS OF ELECTRICAL ENGINEERING approximately equal to unity, and the reluctance of the air gap between the pole tips is high, the distortion in any given generator may be assumed to be proportional to the armature current. (b) Demagnetization which is proportional to the armature current and to the sine of the angle of brush advance. Let N = the number of armature conductors on any given arma- ture, p = the number of poles on the dynamo, / = the current flowing hi each armature conductor. Then the armature ampere-turns per pole = ( ) But the conductors are uniformly disturbed over the surface of the armature, and the magnetomotive force set up by the winding is, therefore, proportional to the average cosine (over 180 degrees). ^ = ^7 av ' cos I " 9 o (5) 2 P J 90 ^I.257X0.636Ar/* ^x 0.4^7 .,, f ^ = * gilberts. (7) 7. The shunt generator. The terminal voltage of a shunt gen- erator decreases as the load (armature current) increases, the speed of the armature and the resistance of the field circuit remaining constant. This decrease in voltage is due to: (a) armature and brush-contact resistance, (b) armature reaction, (c) decreased field current. (a) Armature and brush-contact resistance. According to Ohm's Law, the resistance drop in any current-carrying conductor is equal to the product of the current and the resistance of the conductor. Assuming a constant temperature, the resistance of the armature winding is constant, and the resistance drop is proportional to the armature current. The drop due to brush-contact resistance is calculated by means of equation (3). At no load, when only the field current flows in the armature circuit, the armature current is negligibly small, and the terminal voltage is equal to the electromo- * See Appendix A, Section 8. THE CONTINUOUS-CURRENT GENERATOR 6 7 live force induced in the armature. As the current in the armature increases, the resistance drop in the armature circuit becomes ap- preciable, and the terminal voltage decreases. (b) Armature reaction. As explained in Section 6, armature reaction neutralizes part of the flux produced by the field winding at no load, and changes the distribution of the flux in the air gap. The total electromotive force induced in the winding, and, there- fore, the terminal voltage, decreases as the armature current increases. (c) Reduction of field current. When the resistance of the field circuit is constant, the current flowing in the circuit is proportional to the electromotive force at the terminals of the armature, and the decreased voltage due to resistance in the armature circuit and to 20 40 60 80 100 .PER CENT OF RATED CURRENT FIG. 61. Shunt Generator Characteristics. armature reaction, causes a decrease in the field current, and a further decrease in the terminal voltage .of the generator. The ideal characteristic of a shunt generator is a horizontal straight line (ab in Fig. 61), while the actual characteristic rises as the load decreases (cd in Fig. 61). The percentage regulation of a shunt generator is, therefore, * per cent regulation when E = the full-load (rated) voltage, EQ = the no-load voltage. (8) The regulation of the average shunt generator is too large for the satisfactory operation of lamps when the load varies greatly, al- though it may be entirely satisfactory for motors. To give satis- faction, incandescent lamps must have approximately constant voltage applied between their terminals. Since the heating of a conductor is proportional to the square of the current flowing in it, 68 ESSENTIALS OF ELECTRICAL ENGINEERING & 100 80 : 60 40 20 the light given by an incandescent lamp is approximately propor- tional to the square of the applied voltage. The inherent regulation of a shunt generator, therefore, cannot be depended on to maintain the proper voltage as the load fluctuates, and manipulation of the field rheostat must be resorted to when incandescent lamps form all or part of the load. Automatic adjustment of the field resistance is made by the Tirrill and other regulators, which are largely used for the maintenance of constant voltage either at the generator terminals or at some center of distribution. 8. The series generator. - The voltage characteristic of a series generator is, prac- tically, the magnetization curve for a combined iron and air circuit. Since the load current, or a constant part of it, flows through the series - field windings, the field excitation increases as the load increases. For low excitations, the flux is very nearly proportional to the field current, but as the ex- citation increases, the flux increases at a constantly decreasing rate until the iron becomes "saturated." Be- yond the point of saturation, the permeability of iron, is little greater than that of air, and a large increase in field current produces only a small increase in flux. The voltage of a series generator is, there- fore, approximately proportional to the armature current over a considerable range, the characteristic bending to the right, as indi- cated in Fig. 62. Armature, brush-contact and field resistances make the terminal voltage less than the induced voltage, while armature reaction re- duces the total flux set up by a given field excitation and causes an unsymmetrical distribution of the flux in the air gap as in other types of generators. 20 40 60 80 100 170 140 160 180 ?00 PER CENT OF RATED CURRENT FIG. 62. Series Generator Characteristics. THE CONTINUOUS-CURRENT GENERATOR 69 Since the series generator is used almost exclusively as a constant- current generator, its voltage regulation is of little consequence. The term "regulation," when applied to a series generator, means the ratio of the maximum deviation of the current, between rated load and short circuit, from the rated current at full load, and the rated current. When used as a constant-current generator, the series dynamo is provided with an automatic regulator which causes the terminal voltage to increase or decrease in proportion to the increase or de- crease in the resistance of the load circuit, so that the current is maintained at an approximately constant value. One of the simplest of these automatic regulators is that used on the Brush arc- light generator. The field winding is shunted by a carbon pile, the resistance of which varies inversely as the pressure between the discs. The pressure between the discs is varied by means of an electro- magnet, the winding of which is connected in series with the load. When normal current flows in the circuit, the current divides, part flowing in the field windings and part through the carbon pile. If the current falls below normal, by reason of an increase in the resist- ance of the load circuit (an increase in the number of lamps in the circuit), the pressure on the carbon pile is decreased. The de- creased pull of the magnet increases the resistance of the carbon shunt, and causes a larger current to flow in the field windings. This increased field excitation causes a larger voltage to be induced in the armature windings, and the load current rises to its normal value. Since no current flows in the field coils of a series generator on open circuit, it can " build up" only when the load circuit is closed. 9. The compound generator. The inherent tendency for the terminal voltage of a shunt generator to decrease as the load in- creases is counteracted by the addition of a series-field winding so proportioned that the total flux increases as the current in the armature winding increases. If the effect of the series- field wind- ing is just sufficient, at full load, to compensate for the decrease in voltage due to armature resistance and armature reaction (i.e., if the no-load and the full-load voltages are equal), the generator is flat compounded; if the effect of the series-field winding is such that the full-load voltage is greater than the no-load voltage, the generator is over compounded. The characteristic curves of a flat- 70 ESSENTIALS OF ELECTRICAL ENGINEERING compounded generator and of an over-compounded generator are shown in Fig. 63. When armature speed, field resistance and armature resistance are constant, the compounding of a generator is the ratio of the in- crease in voltage, between no load and full load, to the no-load voltage. Per cent compounding = when EQ = the no-load voltage, E = the full-load voltage. (E - EQ) 100 (9) It is impracticable to build a compound generator, the voltage characteristic of which is a straight line. From the definition given in Section 3, the regulation of a compound generator is the maximum deviation e (Fig. 63), between no load and full load, of the char- 20 40 60 ftO 100 120 140 PER CENT OF RATED CURRENT FIG. 63. Compound Generator Characteristics. acteristic from the straight line connecting the no-load and the full- load voltage points, divided by the full-load voltage. The maximum deviation is measured perpendicularly to the axis of abscissa, and only for the flat-compound generator is it perpendicular to the ideal curve. Per cent regulation = , (10) when E = the full-load voltage. A compound generator builds up in the same manner as a shunt dynamo since the excitation due to the series-field winding is negli- gible at no load. If the series winding is improperly connected, the two windings oppose each other magnetically, and the terminal volt- age drops very fast as the resistance of the load circuit is decreased. THE CONTINUOUS-CURRENT GENERATOR 71 The degree of compounding of a given generator may be changed, without altering its structure, by: (a) changing the resistance of the shunt around the series-field winding, (b) changing the speed. (a) Changing the resistance of the series-field shunt. The effect of changing the resistance shunting the series-field winding is to cause a greater or a less percentage of the total load current to flow in the series-field coils, thus increasing or decreasing the field excitation for a given armature current. This change in field excitation causes the full-load voltage to be increased or decreased. (b) Changing the speed. If the no-load voltage of the generator remains constant, the effect of an increase in the speed of its arma- ture is to increase the full-load voltage. The effect of a decrease in the speed of a given generator, the no-load voltage of which is constant, is to decrease the full-load voltage. The truth of these statements is evident from a study of the oper- ating principles of the electric generator. The rated full-load volt- age of a flat compounded generator is 100. At no load, 1,000,000 magnetic lines pass into the armature from each north pole, and this flux is increased, at full load, to 1,100,000 lines by the magnetic action of the series-field winding. The electromotive force induced in the armature at full load is, therefore, no, and 10 volts are re- quired to compensate for armature resistance and armature re- action. If the speed of the armature is increased 20 per cent above its rated value, the flux required to produce the no-load voltage is re- duced to 800,000 lines, while the effect of the series winding does not decrease. Assuming the magnetic effect of the series winding to be the same at all speeds, the induced electromotive force, at the higher speed, is 112.5 an d the terminal voltage, at full load, is 102.5, an over compounding of 2.5 per cent. But the compounding is greater than the above because of the fact that as the shunt-field excitation is reduced, the flux produced by a given series-field excitation increases. The last statement will be made clear by an examination of the magnetization curve of a generator. Referring to Fig. 64, it is found that 500 ampere turns are required to produce a flux of 1,000,000 lines, and that 850 ampere- turns are required to produce 1,100,000 lines. Conse- quently, the shunt-field winding must consist of 500 ampere-turns and the series-field winding of 350 ampere- turns. To produce ESSENTIALS OF ELECTRICAL ENGINEERING 800,000 lines requires only 260 ampere-turns in the shunt-field winding, but the series-field winding still produces, at full load, 350 l.200,000r ^ 1,000.000 2 800.000 o I 600.000 o 400,000 Z 00.000 200 400 600 800 1000 1200 1400 MAGNETOMOTIVE FORCE IN AMPERE TURNS. FIG." 64. Characteristic magnetization curve showing the large increase in magnet- izing force for a small increase in magnetic induction after passing the "knee" of the curve. ampere-turns or a total of 610 ampere- turns at full load. This ex- citation sets up a flux of 1,030,000 lines, the elec- tromotive force induced in the armature at full load is 128.5, an d the terminal voltage is 118.5, an over compounding of 18.5 per cent. 10. The Tirrill regu- lator. An elementary .Compenaattfty Resistance Shunt FIG. 65. Diagram of Connections for Tirrill Regulator. diagram of the Tirrill regulator, as applied to a shunt or a compound generator, is shown in Fig. 65. By means of electromagnetically operated contacts, which open and close as the electromotive force THE CONTINUOUS-CURRENT GENERATOR 73 rises above or falls below a certain value at the generator terminals, or at some center of distribution, the field rheostat is periodically short-circuited. The winding of the main contact magnet is connected directly to the bus bars, and is so proportioned that it opens the main con- tacts, against the action of a spring, when the voltage rises above the desired value. The relay is differentially wound, one coil being connected directly to the bus bars and the other in series with the main contacts. The relay contacts are connected to the terminals of the field rheostat. The function of the condenser is to prevent excessive sparking when the relay contacts open. The regulator operates as follows : If the main contacts are open, the left-hand coil of the differentially-wound relay is energized and the relay contacts opened. This action increases the resistance of the field circuit by opening the short circuit over the field rheo- stat, the voltage drops below the required value, and the spring closes the main contacts. Closing the main contacts energizes the right-hand coil of the relay and closes the relay contacts, thus raising the voltage above the value at which the main contacts open. In practice, the two sets of contacts are continually opening and clos- ing, and the voltage varies from the desired value by only a very small amount. When it is desired to maintain constant voltage at some distant center of distribution, a differential series winding is added to the main control magnet, thus increasing the bus-bar voltage at which the main contacts are opened. The drop in the transmission sys- tem may thus be compensated for. n. Parallel operation of generators. It is often desirable to supply an electric system from two or more generators instead of from one because: (a) of the greater efficiency of a machine when operated at or near its rated load, (b) continuity of service can be more easily maintained. (a) Efficiency. Because certain losses of a generator are practi- cally constant and these losses are roughly proportional to the size of the generator, the ratio of the total output to the total input (all-day efficiency) is greater for a generator running at full load than for one operating at half its rated capacity, the total output being constant. (b) Continuity of service. It is practically impossible to main- tain continuous service from a single generator because adjustments 74 ESSENTIALS OF ELECTRICAL ENGINEERING and repairs, which require the machine to be shut down, must be made from time to time. With two or more generators, these matters may be attended to during periods of light load when all the machines are not required. It is thus possible to give con- tinuous service and, at the same time, keep the machinery in repair. Parallel operation of generators, then, means dividing the total load between two or more generators, and the operation of such a number of generators at a time as will cause each generator to operate at its highest efficiency. This necessitates connecting and disconnecting generators and the load circuit as the load varies. Before two continuous-current generators are connected to the same load circuit, their terminal voltages should be approximately equal and similar terminals, either positive or negative, must be con- nected together. If the voltages are not approximately equal, an undesirable surge of current takes place in the system when the switch is closed. When dissimilar terminals of the generators are connected together, conditions identical with a short-circuit exist and an excessive current, which either operates the protective de- vices or overheats the armatures, flows around the circuit formed by the two armature windings. Shunt generators. The connections for the operation of two shunt generators in parallel are indicated in Fig. 66. If generator A is carrying the load and it is desired to divide the load between A and B, I drive generator B at its rated speed and regulate its ' field to give a voltage equal to or slightly greater than FIG. 66. Parallel Operation of Shunt Generators, f^at of A When the volt- age across the open switch is zero, the switch may be closed, after which the field rheostats should be manipulated until the volt- age of the system is the rated or required value, and the genera- tors divide the total load in proportion to their ratings.* After being properly regulated, two or more generators having similar characteristics will automatically divide the load in proportion to their ratings as the total load on the system varies. * The load on any one of two or more continuous-current generators operating in parallel is reduced by reducing either its field excitation or its speed. THE CONTINUOUS-CURRENT GENERATOR 75 In connecting two generators as described above, a voltage across the switch indicates that the terminal connections of generator B should be reversed. When any voltage is indicated across this switch, it is equal to approximately twice the voltage of each gen- erator, so that in testing, care should be taken that the voltmeter is not injured. A shunt generator, operating in parallel with other generators, should be disconnected from the load circuit by opening the line switch or circuit-breakers after reducing either its speed or its field excitation until the current flowing in its armature is a minimum. Compound generators. The connections for the parallel opera- tion of compound generators are shown in Fig. 67, the only change from the shunt diagram being a third connection between the two generators. This connec- tion is called the equal- izer, and is made between the brush and the series- field winding of each gen- erator. The effect of the equal- izer is to divide the load properly between the two generators. Consider the Field Resistance Field FIG. 67. Parallel Operation of Compound Generators. action of two compound generators without an equalizer connection. If, for any reason, the speed of one generator increases slightly, its voltage increases and it takes a greater part of the total load. If the total load on the system remains constant, the increased load on one generator causes a larger current to flow in its series-field wind- ing and reduces the current flowing in the series-field winding of the other generator. This change in field excitation disturbs the equilibrium of the generators and will, by its cumulative effect, cause one generator to take the entire load and to operate the other as a motor. Compound generators connected to the same circuit and operating without equalizer connections are, therefore, in unstable equilibrium. With an equalizer connection, the same increase in the speed of generator A causes its voltage and its current output to increase, but instead of all the increased current flowing through the series- field winding of generator A, it divides at the brush, part flowing 76 ESSENTIALS OF ELECTRICAL ENGINEERING through the series-field winding of each generator so that the voltage of each is equally increased, and their equilibrium is not disturbed. Equalizer connections should be large as they are often required to carry heavy currents, and an appreciable resistance in this con- nection between generators tends to disturb their equilibrium, and causes an unequal division of the load. Fuses or other circuit in- terrupting devices should not be placed in an equalizer circuit. The process of disconnecting a compound generator, operating in parallel with other generators, from its load circuit, is similar to that for a shunt generator. 12. Connection to load circuits. The load units supplied by shunt and compound generators are connected in parallel, i.e., the terminals of each unit, as a lamp or a motor, are connected directly to the mains leading from the terminals of the generator or to branches of these mains. The shunt generator is in common use under the following con- ditions: (a) Where the load is practically constant. (b) Where the load changes slowly and infrequently. (c) Where the load consists exclusively of motors which do not require close voltage regulation. (d) Where an attendant is employed to manipulate the field rheo- stat and maintain constant voltage. (e) Where the voltage is controlled by a Tirrill regulator or other automatic device. Compound generators are in common use under these conditions: (a f ) Where an approximate compensation is to be automatically made for the internal drop of the generator. (b') Where the drop in a long transmission line or in a long feeder is to be automatically compensated for, i.e., where the voltage at the terminals of a load some distance from the generator is to be automatically maintained at an approximately constant value as the load varies. (c') Where the terminal voltage is to increase with the load. The series generator is used almost exclusively for supplying cur- rent to series arc-lamp systems, the generator being provided with a regulator which automatically maintains the current in the system at an approximately constant value as the load changes. The terminals of the load units in a series circuit, instead of being THE CONTINUOUS-CURRENT GENERATOR . 77 connected to the mains running from the generator, are connected to each other, one terminal of the first and one terminal of the last unit in the series being connected to the mains. Consequently, the same current flows in each part of the circuit and the voltage changes as the load changes. The series generator for the operation of arc-lamp circuits, has been largely replaced by alternating-current apparatus. CHAPTER IV PROBLEMS 1. The electromotive force of a shunt generator is 118 volts when operated without load, and the rated output is n kw. at no volts. Calculate: (a) the voltage regulation, (6) the resistance of the armature circuit, if one-half the drop in terminal voltage between no load and full load is due to armature resistance. 2. The terminal voltage of a series generator is 450 when the armature cur- rent is 10 amperes. Calculate the electromotive force induced in the armature winding if the resistance of the armature circuit is i ohm and that of the field winding is 1.5 ohms. 3. A shunt generator requires, at zero load, a field current of 2 amperes to induce the voltage at which it is rated; at full load (= 100 amperes) the field current required to give rated terminal voltage is 2.33 amperes. The shunt winding on each pole consists of 4000 turns. Find the number of turns re- quired in a series-field winding to make the generator flat compounded, 57 per cent of the load current to flow in the series-field windings. 4. A 4-pole, i25-volt, lap-wound (drum) armature has 400 conductors. The flux per pole is 2,500,000. Find the speed at which the armature rotates to induce rated voltage in the armature conductors. 5. A i6-pole, lap-wound generator operates at 80 r.p.m., has a flux per pole of 7,500,000, and 2304 conductors. Find the voltage induced in the armature. . 6. Find the number of conductors required on a wave-wound armature to be used in the generator of Problem 4, the rated voltage to remain the same. 7. Compare the resistance of a 6-pole, lap-wound armature with that of a 6-pole wave-wound armature, the rated voltages, the flux per pole, the kw. outputs, and the current densities in the armature conductors being the same. 8. Allowing 500 circular mils per ampere, find the area of the armature conductors in: (a) a 6-pole lap-wound dynamo, the armature current of which is 100 amperes, (b) a 6-pole wave-wound dynamo, the armature current of which is 100 amperes. 9. Find the voltage induced in an 8-pole lap- wound armature when: r.p.m. = 300, = 2,000,000, N = 784. 10. A 55o-kw. generator has a terminal voltage of 550 at no-load and is 5 per cent overcompounded. Find: (a) the size of wire required to transmit 100 amperes a distance of 600 feet, the voltage between the terminals of the load 78 ESSENTIALS OF ELECTRICAL ENGINEERING apparatus to be 550, (b) the watts lost in the line, (c) the total output of the generator. 11. The rated output (line) of a compound generator =150 amperes, and its no-load voltage is 220. The current is transmitted over a line, the resistance of which is 0.166 ohm, and the voltage at the load terminals is the same at full load as at no-load. Find: (a) the terminal voltage of the generator at full load, (b) percentage overcompounding, (c) watts lost in line, (d) watts output of gen erator. 12. A 4-pole, wave-wound armature has 105 commutator bars. Each armature coil consists of two turns of No. 10 double cotton-covered wire. Find the flux (per pole) required when the induced voltage is 125, and the speed = looo r.p.m. 13. A 4-pole, lap- wound armature has 88 commutator bars. Each arma- ture coil consists of four turns of No. 1 1 double cotton-covered wire. Find the voltage when the flux per pole = 1,000,000 lines and the speed of the armature = 1000 r.p.m. CHAPTER V THE CONTINUOUS-CURRENT MOTOR i. The fundamental equation of the motor. Rotation of the armature of a motor induces in the armature winding a counter- electromotive force which is dependent on the same quantities as is the induced electromotive force of a generator. Counter e.m.f. = , \ volts, (i) p icr when < = the total flux entering or leaving the armature at each pole, N = the total number of conductors on the surface of the armature, n = the speed of the armature (revolutions per second), p = the number of poles, p r the number of parallel paths into which the armature conductors are connected. The counter-electromotive force is the difference between the applied electromotive force and the resistance drop in the armature circuit. Therefore, Nn , . E " RJa Eb ' (2) when E = the applied electromotive force, R a = the resistance of the armature winding, I a = the current flowing hi the armature circuit, Eb = the drop due to the so-called contact resistance be- tween the brushes and the commutator.* * As noted in Chapter 4, Section 2, the voltage drop per pair of carbon brushes may be calculated by the formula: Eb = 0.8 + 0.2 D, (3) when D the current density (amperes per square centimeter) in the contact area between the brush and the commutator. 79 8o ESSENTIALS .OF ELECTRICAL ENGINEERING Transposing the quantities in equation (2), kE c since />, p r and N are constant for any given motor. Equation (4) is the usual form of the fundamental equation for the continuous- current motor. 2. Torque. The force tending to rotate the armature of a motor is termed its torque and is usually expressed in pounds at one foot radius, i.e., in foot-pounds. It is proportional to the product of the flux in the air gap and the current flowing in the armature circuit. Let T = the torque in foot-pounds developed in the armature of any motor, n = the speed (revolutions per. second) at which the armature rotates, EC = the counter-electromotive force (volts) induced in the arma- ture conductors, I a = the current (amperes) flowing in the armature circuit. Then MI.BJ. ( 5 ) , , and T= ' 8 The quantity is constant for any given motor. The torque delivered at the pulley of the motor is less than that indicated by equation (7), because part of the torque developed in the armature is required to overcome the windage, friction and iron losses (stray power) of the motor itself. 3. Speed-torque characteristic. The operation of a motor is represented graphically by a speed-torque curve, which shows the THE CONTINUOUS-CURRENT MOTOR 8l relation between the speed of the armature and the torque developed in the armature, or that delivered at the pulley. The relations between the speed, the torque, and the armature current of a continuous-current motor are such that a change in one of these quantities produces a change in one or both of the others. In any given motor, the speed is automatically adjusted to the value which allows the required current to flow in the armature conductors. If the armature current is larger than that necessary to produce the required torque, the speed of the armature auto- matically increases until equilibrium is established; -if the current is too small, the speed of the armature decreases, allowing a larger current to flow and a larger torque to be exerted. 4. Commutation. The phenomena of commutation occur in a motor just as in a generator, since the direction of the current in each armature coil is periodically reversed. 5. Armature reaction. The direction of current in the arma- ture conductors of a motor is opposite, for a given field polarity and direction of armature rotation, to that in the conductors of a generator. The field- is, therefore, distorted as indi- cated in Fig. 68, and the brushes must be shifted backward, or opposite the direction of armature rotation, to ob- tain good commutation. FlG> 68 - . Flux Distribution in the s ,. iiT-,,1 Air Gap of Motor. 6. The shunt motor. With con- stant applied voltage, the field current and the armature flux of a shunt motor are constant, neglecting armature reaction which is small in a well-designed motor, and the torque developed in the motor is proportional to the current flowing in the armature circuit. T oc I a . (8) Characteristic. As the load on a shunt motor increases, the counter-electromotive force must decrease so that the required cur- rent may flow in the armature circuit. Armature reaction reduces the flux of a shunt motor slightly, but if the magnetic circuit is properly designed this reduction is small, and may be neglected.* * Any decrease in the flux of a shunt motor by reason of armature reaction, tends to increase its speed and improve the regulation of the motor. 82 ESSENTIALS OF ELECTRICAL ENGINEERING The speed of a shunt motor must, therefore, decrease as the load increases, and the speed-torque characteristic is a slightly drooping curve as indicated in Fig. 69. Well-designed shunt motors operate with only a small change in speed between no-load and full (rated) load, and are termed con- 20 40 60 &0 100 120 PER CENT OF FULL LOAD( RATED) TORQUE. FIG. 69. Speed-torque Characteristic of Shunt Motor. 140 stant-speed motors. They are largely used where a small variation of speed is of little consequence. Regulation. The regulation of a shunt motor is the ratio of the difference between the no-load and the full-load (rated) speeds, and the full-load speed. i * Per cent regulation = n) ioo n (9) when HQ = the speed at no load, n = the speed at full (rated) load. The resistance of the armature circuit is the controlling factor in the regulation of a shunt motor. If the armature resistance is small, the variation of the product, RJ a , between no load and full load is small, and the required variation of the counter-electromotive force is correspondingly small. But the counter-electromotive force is proportional to the speed of the armature. Therefore, a small armature resistance produces, or tends to produce, good regulation. Speed control* Speed regulation is an inherent property of a motor; speed control is the variation in speed obtained by exter- nal means. The speed of a shunt motor is varied, within certain limits, by: (a) changing the resistance of the armature circuit, (b) changing the * For an extended discussion of motor speed control, see " Electric Motors " by Crocker and Arendt. Field Resistance THE CONTINUOUS-CURRENT MOTOR 83 resistance of the field circuit, (c) changing the reluctance of the magnetic circuit, (d) changing the electromotive force between the armature terminals. (a) Changing the resistance of the armature circuit. The speed of a motor is changed by varying a resistance connected in series with the armature as indicated in Fig. 70. A resistance, the current-carrying capacity of which is equal to or greater than the maxi- mum current that will flow in the circuit, must be used. This method is effective, cheap in first cost, and easily applied, but the power wasted in heating the rheostat FlG ' 7a s P eed Contro1 by Armature Resistance, makes it very inefficient. As pointed out above, a small change in load causes a large change in speed, and the motor acts as if the resistance of the armature winding were greatly increased. (b) Changing the resistance oj the field circuit. The resistance of the field circuit is changed by manipulating the field rheostat. In- creasing the resistance of the field circuit decreases the field cur- rent and increases the speed; decreasing the resistance of the field circuit increases the field current and decreases the speed.* This method is cheap, both as to first cost and as to operation, but its range of operation is limited. Very low speeds are unattain- able because of the fact that above a certain point a large increase in field excitation produces only a small increase in flux. High speeds are also unattainable because, with a weak field, armature reaction becomes so pronounced as to cause excessive sparking which burns, and would ultimately destroy, the commutator. Armature reaction and commutation difficulties are materially reduced by means of interpoles which provide a local commutat- ing flux, and make possible the operation of shunt motors having very weak fields. Interpole motors having a maximum speed six times the minimum speed, and controlled by manipulation of the field rheostat, are in successful operation. (c) Changing the reluctance of the magnetic circuit. With con- stant field excitation, the flux in the magnetic circuit of a dynamo is inversely proportional to the reluctance of the circuit.f A number * This statement is evident from equation (4). t See Chapter 2, Section 10. ESSENTIALS OF ELECTRICAL ENGINEERING Field Resistance of variable-speed motors have been designed in which the reluctance of the magnetic circuit is varied by changing the length of the air gap between the pole and the armature. The operation of such motors is satisfactory, but they are expensive in construction and more or less complicated in operation. (d) Changing the electromotive force applied to the armature terminals. This method of speed control may be sub-divided into two parts: (i) when the change in voltage is made in steps, (2) when the range of voltage is continuous. (i) Voltage changed in steps. This method of speed control necessitates the use of several supply mains having dif- ferent voltages, and a generating system capable of supplying current at these FIG. 71. Multi-voltage Speed different voltages. The field winding is permanently connected across one pair of the supply mains and the flux is, therefore, approximately con- stant. Fig. 71. The operating characteristics of this system are good and the efficiency high, the objections being its high first cost and the fact that the speed changes by fixed steps. The latter fault is remedied by combining this method with the field rheostat method. (2) Voltage range continuous. This method, devised by Ward Leonard and generally known by his name, requires the use of three machines a vari- able-speed motor, a con- stant-speed motor* and a generator, as shown in Fig. 72. The constant- speed motor drives the generator, and the gen- erator, in turn, supplies current to the armature of the variable-speed motor, the field of which is connected to constant potential mains. By varying the field excitation of the generator, any desired voltage is delivered at the terminals of the motor armature, and a continuous variation of speed obtained over the widest possible range. Very * It is not necessary that this be an electric motor. Generator Constant speed Hohr Variable Speed Motor FIG. 72. Ward Leonard Speed Control System. THE CONTINUOUS-CURRENT MOTOR simple operating mechanism is required. The efficiency of this system is the combined efficiency of three machines, the individual efficiencies of which are usually low. Its high first cost and low operating efficiency prohibit its use except where these factors are secondary considerations. 7. The series motor. The series motor, unlike the series generator, has a very large commercial application. It is used, practically to the exclusion of all other types of motor,* for street 180 i fifi i 250 225 H- z uJ 200 or K 3 U I25< 100^ u- 75 1 UJ UJ CL 25 \ \ \ t f 140 100 ao 60 n x 4 ^ J r l ^peed f \ x^ \ X 1 \ ^, / \ ,/ ' ( sU r/ -o -jf \ ^ M | \ X 1 *, / <^ s^ . ^x 's ^ X "S^ ~j ^ jj / ^ ^N ^ v ~ ^^ ^ / *, a / t / / 40 80 120 160 200 340 PER CENT OF FULL LOAD (RATED) TO ROUE FIG. 73. Series Motor Curves. 280 railway work, for hoisting and for all purposes which require large starting torques, and do not require close speed regulation. Torque. Since the field excitation of a series motor is propor- tional to the armature current, the torque developed in its armature is, theoretically, proportional to the square of the armature current. roc/ a 2 . (10) This equation holds good for low values of armature current, but the upper part of the current-torque curve is found to be an approxi- mately straight line. Fig. 73. For low excitations, the flux ispro- * Alternating-current motors are used to a limited extent for these purposes. 86 ESSENTIALS OF ELECTRICAL ENGINEERING portional to the field current but for larger loads the iron of the magnetic circuit approaches saturation, and the flux in- creases more slowly than does the current in the field winding. The torque developed is, therefore, less than that indicated by equation (10). Speed. The speed of a series motor varies over wide limits as the load changes, the speed increasing as the load is reduced as indi- cated in Fig. 73. The full line shows the operating range of the motor, the dotted portion near the axis of ordinates indicates speeds above the safe operating limits, and excessive heating takes place if the motor is operated over the dotted portion at the right. Because of the excessive speed attained by the armature of a series motor when the load is small, a series motor should never be used where the load may, accidentally or otherwise, be reduced below the safe minimum value, or the motor may be wrecked. PER CENT OF FULL LOAD (RATED) SPEED o o f^ < ' , ^ ^. x_ ^ta 1 T / "- "^ \ L \ & & / N RAIlNGS J >', \ '<'. 101 which is the equation of a straight line, and may be plotted as in Fig. 87, after the iron losses at different speeds but constant field excitation have been determined. The hysteresis loss at any given or required speed n is equal to the product of the ordinate k h ' and the speed n\ the eddy current loss for any given or required speed n is equal to the product of the speed and the ordinate k e 'n corre- sponding to the given or required speed. 5. Efficiencies. The efficiency of an electrical machine is the ratio of the output to the input. SPfED FIG. 87. Separation of Hysteresis and Eddy Current Losses. Per cent efficiency = output X ICQ input (21) The efficiency of a dynamo is usually computed from stray power and resistance measurements rather than from an actual load test for the following reasons: (a) A considerable quantity of energy must be wasted in making a load test. (b) It is often inconvenient or impossible to supply the electrical energy required to run a large dynamo as a motor or to absorb its output when operated as a generator. (c) Calculated efficiencies are more reliable than those determined by load test. The equation for the efficiency of a generator, then, becomes rr OUtpUt X ICO . , N per cent efficiency = *- - (22) output + losses and that for a motor ~ . (input losses) 100 , , per cent efficiency = * c . (23) input The efficiencies of a dynamo are represented graphically by means of an efficiency curve, per cent efficiency being used as ordinates and per cent of full (rated) load as abscissas. The shunt dynamo. Very simple measurements are sufficient for the calculation of the efficiency of a shunt dynamo. The copper loss in the armature is proportional to the square of the current 102 ' ESSENTIALS , O.F .ELECTRICAL ENGINEERING flowing in the armature circuit and may be calculated for any given or required current after the resistance of the armature winding has been determined as explained above. The field loss is approximately constant, the change in value due to the decrease in field * current as the load on a generator increases is usually neglected, and is determined by measuring the current in the field circuit and the voltage across its terminals when the excita- tion is such as to give rated voltage when operated as a generator, or rated speed when operated as a motor. The stray power of a shunt dynamo is approximately constant, and the no-load determination is used in the calculation of efficien- cies. The dynamo is operated as a motor, without load and at rated speed, the input to the armature measured, and the calculated armature and brush contact resistance losses subtracted, as explained above. The armature and brush-contact resistance losses are small at no-load and may usually be disregarded without serious error. The fact that the stray power of a shunt dynamo is not constant will be appreciated from the following: In a shunt generator operating at constant speed, the field current falls off as the load increases. Therefore, the flux in the magnetic circuit is reduced, and the armature current changes the distribu- tion of the flux in the air gap. In the shunt motor, the speed falls off and armature reaction distorts and reduces the field. Decreased field excitation, decreased speed, and armature demag- netization cause the iron losses of a dynamo to decrease; field dis- tortion causes the iron losses to increase. These two effects, there- fore, tend to neutralize each other. No accurate and easily applied method for determining the change in the stray power of a dynamo as the load changes has been devised and, since it is usually small, no correction need be attempted. By differentiating the equation for the efficiency of a shunt dy- namo it may be shown that the efficiency is a maximum when the constant losses equal the variable losses, i.e., the efficiency is a maximum when the armature resistance losses are equal to the sum of the field resistance loss and the stray power. The series dynamo. Since the field excitation of a series gener- ator, and the field excitation and the speed of a series motor vary * If the terminal voltage of a shunt generator remains constant, the field current must be increased as the load increases. LOSSES, EFFICIENCIES AND RATINGS 103 over wide limits, the stray power of a series dynamo also varies greatly. Therefore, the stray power must be determined for the particular speed and field excitation for which it is desired to calculate the efficiency. By separately exciting the field and regu- lating the voltage applied to the terminals of the armature, the stray power of the dynamo for any speed and field excitation may be determined. Since the same current flows in the field as in the armature of a series dynamo, the copper losses for any load are equal to the square of the line current multiplied by the combined resistance, in series, of the two windings. The losses due to brush-contact resistance should be calculated as for a shunt dynamo. The compound dynamo. Since the voltage of a compound gener- ator and the speed of a compound motor may vary widely from their no-load values, the efficiency may be accurately calculated only by determining the losses at the voltage, or the speed, at which the dynamo operates when carrying the specified load. The approximate efficiency of a compound dynamo obtained by assuming the shunt-field loss and the stray power constant, is often sufficiently accurate, particularly if the dynamo is flat or only slightly overcompounded. As a generator. Given the voltage characteristic of a compound generator, very close approximations of the losses at different loads may be made from no-load measurements. Knowing the ter- minal voltage at any condition of loading and the resistance of the shunt-field circuit, the shunt-field loss is readily calculated. The resistance losses in the series field and in the armature are calcu- lated in the same way as for the series dynamo. By separately exciting the series-field winding, the stray power of a compound generator may be readily determined for any re- quired speed and field excitation. If the stray power is determined for different excitations, but constant speed, a curve may be plotted, using stray power as ordinates and per cent of rated load as ab- scissa. From this curve the stray power at any load is obtained. As a motor. When the long shunt compound dynamo is oper- ated as a motor from constant potential mains, the shunt-field loss is constant and the resistance loss in the armature and that in the series-field circuit may be calculated from the resistances of the windings and the armature current. The shunt-field loss in a short- 104 ESSENTIALS OF ELECTRICAL ENGINEERING shunt compound motor, operating from constant potential mains, decreases slightly as the armature current increases because of the drop in the series-field winding which reduces the voltage between the terminals of the shunt-field circuit. Stray power should be de- termined for speeds corresponding to different field excitations, and a curve plotted as for a generator. 6. Ratings. Aside from the greatly increased copper losses in a dynamo which materially reduce its efficiency when excessive current flows in the armature windings, the output is limited by: (a) heating, (b) sparking, (c) voltage considerations. (a) Heating. According to Joule's Law the heat liberated in a current-carrying conductor is proportional to the square of the cur- rent flowing in the circuit. The heat due to hysteresis and eddy currents also tends to raise the temperature of the armature. Insulations used on commercial dynamos will not stand an in- definite rise in temperature without injury to the insulating quali- ties. The output, then, must be kept below the value which causes the insulation of the conductors to be injured. The generally recognized safe limit of temperature is 65 to 70 degrees (Centigrade).* If the atmospheric temperature is taken as 25 degrees (an average value), the allowable temperature rise is 40 to 45 degrees. (b) Sparking. Excessive armature currents, by reason of their reaction on the magnetic field, cause sparking at the brushes which burns and ultimately destroys the commutator, and thus limits the output of a given dynamo. (c) Voltage limitations. An examination of the magnetization curve of iron shows that the flux in a given magnetic circuit cannot be increased beyond a fixed value (magnetic saturation), without increasing the exciting ampere-turns beyond the economical limits. The alternative is to increase the speed, but this is limited, by mechanical considerations, to a peripheral velocity of about six thousand feet per minute. Commutation difficulties increase as the voltage increases, and the commutation of voltages above 1200 to 1500 is practicable only on large sized machines. * This temperature is that determined by a thermometer, and is from 5 to 15 less than the maximum internal temperature. LOSSES, EFFICIENCIES AND RATINGS 105 CHAPTER VI PROBLEMS 1. A 5oo-kw., 55 the current in the line to which coils A and C are connected is maximum when co/ = 120 or 300,* and the current in the line to which coils B and C are connected is maximum when co/ = 180 or o. Therefore, (/i) = (/) sin 60 - (I P ) m sin 300 (24) = 0.866 (/ p ) m + 0.866 (I P ) m (25) = 1.732 (I P ) m (26) = V3(U (27) and /, =>/3 /,. (28) 112 ESSENTIALS OF ELECTRICAL ENGINEERING i.e., the effective line current in a balanced delta-connected system is equal to the effective current in each winding multiplied by the square root of 3. Since the line-to-line voltage and the phase current in a balanced delta-connected system supplying a non-reactive load circuit attain their maximum values at the same instant, it follows that the line current and the line-to-line voltage are 30 degrees out of phase. (c) V -connection. With the open-delta or V-connection, a three- phase system is obtained by the use of only two windings. Fig. 98. The connection is essentially that of the delta with one coil omitted. FIG. 98. Three-phase V-connection. FIG. 99. Vector Diagram of V-connection . Obviously, the line-to-line voltage is that of the winding, and the line current must flow in the winding. The vector diagram for a V-connection is shown in Fig. 99. Let the load on a delta-connected system be 300 kw. at unity power factor. It is obvious that each winding carries one-third of the total load or ioo kw. If the same load is carried by an open- delta system, each winding must carry one-half the total or 150 kw., but the current and the electromotive force in c the windings are 30 degrees out of phase and the current-carrying capacity (rating) of each open- " delta winding must be 1.73 times the current- carrying capacity of each delta winding. The ' weight of copper required for an open-delta con- FIG. ioo. Three-phase nection is, therefore, 15 per cent greater than that required for a delta connection. (d) T-connection. A three-phase system is obtained if two coils, the electromotive forces in which are in quadrature, are connected as indicated in Fig. ioo. Referring to Fig. ioo, let EAB = EBC = ECA ( 2 9) EEC \ and ECD == EBD = ~~ * \3w 2 POLYPHASE ALTERNATING CURRENTS 113 The voltages E AB , EBD and E AD are, by construction, the hypothe- nuse, the base and the altitude of a right triangle. Fig. 101. Therefore, _ (31) = 0.866 E AB , (32) i.e., a three-phase system having equal voltages between lines is produced by two windings connected as in Fig. 100, if the elec- tromotive forces in the windings are in quadrature and that in AD is 0.866 times that in BC. Let the current in winding AD be in phase with the voltage E AD . Then * A = / m sinco/, (33) is = I m sin (/ - 120), (34) . m i c = I m sin (/ - 240). (35) But e BC = (E B c)m sin (co/ - 90). (36) Therefore, at unity power factor, the current in one-half of the wind- ing BC leads the electromotive force, and that in the other half lags behind the electromotive force, as shown in Fig. 101. ,/ 4. Comparison of star and delta connec- tions. The star-connected three-phase system FIG.IOI. Vector Dia- has the following advantages over the delta- g ram of T-conneo connected system: (a) The availability of a neutral point to which a ground wire, a meter connection or a load may be connected. (b) Circulating currents cannot flow in the windings of a star- connected system. (c) For a given line-to-line voltage, the number of turns required in a star-connected armature is only 58 per cent of the number required in a delta-connected armature. (d) The electromotive-force wave of a star-connected armature winding is more nearly harmonic than that of a delta-connected winding.* * A non-harmonic electromotive force wave may be resolved into a fundamental sine wave and the odd harmonics of this fundamental. By reason of their phase relations, some of these harmonics, particularly the third, cause currents to flow in the circuit formed by delta-connected coils, but neutralize each other in star- connected coils and do not appear in the line-to-line voltage wave. 114 ESSENTIALS OF ELECTRICAL ENGINEERING 5. Unbalanced three-phase system. When the load on a three- phase system is unequally divided between the three phases, the system is said to be unbalanced, and the currents flowing in the different lines are no longer equal. The voltages between lines are also unbalanced to a slight degree. When the line currents are unequal, the algebraic sum of the instantaneous line currents may not be zero and an equalizing current, IN in Fig. 102, flows in the neutral of a star-connected system, or distorts the current triangle FIG. 102. Vector Diagram of Unbalanced FIG. 103. Vector Diagram of Unbalanced Star-connected System. Delta-connected System of a delta-connected system, as indicated in Fig. 103. If the star- connected system is not provided with a neutral connection, the current relations are still further distorted. 6. Power factor of a polyphase system. The term " power factor," when applied to a polyphase system, can have no rational meaning since each component circuit may have a different power factor. 7. Power in a polyphase system. It has been shown* that the power in a single-phase alternating-current circuit is equal to the product of the electromotive force, the current and the cosine of the angle of phase difference. P 1 = p / p cos0. (37) The power in a balanced two-phase system is, therefore, twice that in one of the component circuits, P 2 = 2 p / p COS0, ( 3 8) and that in a balanced three-phase system is three times that in one of the component circuits. Wz = 3 E P I P cos <, (39) when E p = the phase voltage, IP = the phase current, cos = the power factor of each component circuit. * See Chapter i, Section 28. POLYPHASE ALTERNATING CURRENTS 115 The line quantities of a polyphase system are often more easily measured than the phase quantities, and it is convenient to have a power equation into which the line quantities may be substituted. From equation (15), the line-to-line voltage of a star-connected system is equal to A/3 times the phase voltage; from equation (28), the line current in a delta-connected system is A/3 times the phase current. Therefore, P 3 = V3 -Ei/i cos 0, (40) when EI = the line-to-line voltage, 1 1 = the line current, cos c/> = the power factor of each component circuit. 8. Constancy of power in a balanced polyphase system. One of the advantages of the polyphase system, when compared with the single-phase system, is its constancy of power, i.e., the torque of a single-phase motor is pulsating; that of a polyphase motor is constant. Let the electromotive force and current relations in one phase of a two-phase system be such that e f = E m sin co/, (41) *' = I m sin (co/ - 0). (42) Then the relations in the second phase are e" = E m sin (co/ - 90) (43) = E m cos co/, (44) i" = Im sin (* - 90 - ) (45) = I m cos (co/ - 0), (46) and p 2 = e'i' + e"i" (47) = E m l m [sin co/ sin (co/ 0) + cos co/ cos (co/ 0)]. (48) Expanding equation (48) p 2 = E m l m cos (sin 2 co/ + cos 2 co/). (49) But sin 2 co/ + cos 2 co/ = i. (50) Therefore, /> 2 = m/m cos t/>, (51) which is a constant. In a similar manner it may be shown that the power in a three- phase system is constant, and equal to pa = 1.5 Emlm cos . (52) n6 ESSENTIALS OF ELECTRICAL ENGINEERING 9. Power measurements. The power in a single-phase system may be measured by means of a wattmeter connected as indicated in Fig. 104. Since a polyphase system is composed of two or more single-phase circuits, the power in a polyphase system is the sum of the indications of wattmeters con- nected in the component single-phase circuits. .-Current Coil Voltage Coil FIG. 104. Single-phase. In the two-phase system a wattmeter connected in either phase indicates half the total power, if the system is balanced; if the Wattmeter FIG. iosa. Two-phase Four-wire. FIG. io5b. Two-phase Three-wire. Co) system is unbalanced, a wattmeter must be connected in each phase and the sum of the readings taken as the total power of the system. The connections for a four-wire sys- tem, either independent or intercon- nected, are shown in Fig. io5a, and those for a three-wire system in Fig. io5b. The power in a balanced three- phase system may be determined from the indication of one wattmeter connected as shown in Fig. io6a. The current coil of the wattmeter is connected into any one of the three lines, and the potential coil is con- nected between this line and the neu- tral of a star-connected system. In a delta-connected system, or in a star- connected system the neutral of which Resistance Cb) Waff-meter Wattmeter FIG. 106. Three-phase. is inaccessible, a so-called "artificial neutral" may be constructed by using two resistances, each of which is equivalent to the resist- ance of the potential circuit of the wattmeter. When connected as shown in Fig. io6b, the resistances and the potential circuit of POLYPHASE ALTERNATING CURRENTS 117 the meter form a star connection, and the meter indicates, for balanced load, one-third the total power of the system. A more generally used method for determining the power in a three-phase circuit makes use of two wattmeters connected as indi- cated in Fig. io6c. The algebraic sum of the indications of the meters is the total power in the system, whether the system is balanced or unbalanced.* The current coils of the meters are con- nected into any two of the three lines, and the potential coils be- tween these two lines and the third line. The wattmeter indica- tions are equal only when the power factors of the load circuits are unity and the load is balanced. In using two wattmeters for measuring the power of a three- pliase system, the sum of their indications is to be taken if the power fac- tors of the load circuits are greater than 50 per cent and their dif- ference if the power factors are less than 50 per cent. The power factors of induction motors running at light loads are often very low. To determine whether the power factor of a circuit is greater or less than 50 per cent, disconnect the potential coil of wattmeter No. i from line B and connect it to line A. Fig. io6c. If the direction of torque (the direction in which the pointer deflects) is not reversed, the power factor is greater than 50 per cent. Referring to Fig. 107, let OA, OB and OC represent the equal line to neutral voltages of a balanced three-phase system,! the power factor of which is 50 per cent (angle of lag equals 60). The line currents, then, are represented by the FIG. 107. Vector Dia- vectors I A , I B and Ic and the line voltages by Deters AB, BC and AC. If the currents in the coils of a wattmeter remain constant, the indication of the meter is proportional to the cosine of the angle of their phase difference. Therefore, wattmeter No. 2, with a phase difference of 90 degrees, indicates zero, and wattmeter No. i, with a phase difference of 30 degrees, indicates the total load in the three-phase system. If the power factor of the system is greater than 50 per cent, the indication of wattmeter No. 2 is positive and * An unbalanced three-phase system with "grounded" neutral is essentially a four- wire system, and three wattmeters should be used if accurate determination of the power is required. t It is immaterial whether the system is star- or delta-connected. Il8 ESSENTIALS OF ELECTRICAL ENGINEERING must be added to the indication of wattmeter No. i to obtain the total power in the system; if the power factor of the system is less than 50 per cent, the direction of current in one coil of wattmeter No. 2 must be reversed to bring the indication on the scale, and its indication regarded as negative. . From the above considerations it is evident that PI = E AB I A cos (30 - 0) (53) and P 2 = E B cIc cos (30 + 0), . (54) when PI = the indication of wattmeter No. i, PI = the indication of wattmeter No. 2, EAB = EEC = the line-to-line voltage, I A Ic = the line currents, = the power-factor angles. It was shown in Section 7 that the total power in a balanced three-phase system is P 3 = VsEj/jcos*. (55) But Pi + P 2 = E l l l [cos (30 - 0) + cos (30 + 0) ] (56) = EJi (cos 30 sin + sin 30 sin 4- cos 30 sin - sin 30 sin <) (57) = 2 Eil i cos 30 cos (58) = A/3 EJ i cos 0. (59) Therefore, the total power of a three-phase system is the algebraic sum of the indications of two wattmeters connected as shown in Fig. io6c, and the indications of the meters are equal only when the phase angle $ is unity, i.e., when the load circuits are non-reactive. The power factor of a balanced three-phase system may be cal- culated from the wattmeter readings and the following formula: COS = * r , (60) when n = -= PI = the indication of that wattmeter which is always positive, Pz = the indication of that wattmeter which may be either positive or negative. POLYPHASE ALTERNATING CURRENTS 1 19 10. Comparison of two- and three-phase systems. The state- ment was made in Section 2 that the three-phase system is superior to the two-phase system. This superiority lies chiefly in the smaller weight of copper required in the three-phase system for the trans- mission or the distribution of a given power, the line voltage and the power lost in the two systems being equal. Let E = the greatest line-to-line voltage in each system, 7 2 = the current in each line of the two-phase system, 7 3 = the current in each line of the three-phase system, R 2 = the resistance of each of the four conductors in the two- phase system, R 3 the resistance of each of the three conductors in the three-phase system. Then, 2EI 2 = V^EI S (61) and 7 2 = . (62) The copper loss in the two-phase system is P 2 = 4#2/2 2 (63) and that in the three-phase system is PS = 3 *3/ 3 2 . (64) Since the copper losses in the two systems are equal 4#2/2 2 = 3^3/3 2 . (65) Substituting the value of 7 2 from equation (62) 3 *3/3 2 (66) 4 and R 2 = R 3 , (67) i.e., the resistance of each conductor in the two-phase system is equal to the resistance of each conductor in the three-phase system, and the total weight of conductor in the two-phase system is one- third greater than that in the three-phase system. When the three-wire two-phase system having a phase voltage equal to the line-to-line voltage of a three-phase system, is compared with the three-phase system, the weight of copper in the two-phase system is found to be 2.8 per cent less than that required in the 120 ESSENTIALS OF ELECTRICAL ENGINEERING three-phase system. This small saving in copper is more than offset by the increased insulation required by reason of the 40 per cent larger voltage between two of the conductors. n. Equivalent single-phase system. In problems involving polyphase quantities, it is often convenient to consider the poly- phase system replaced by an equivalent single-phase system, i.e., by a single-phase system, the voltage of which is equal to the greatest voltage between lines in the polyphase system, and in which the total power, the losses and the power factor are equal to those in the polyphase system. Equivalent of the two-phase system. Since the two-phase system is composed of two single-phase circuits, the voltage of the equivalent single-phase system is the phase voltage of the two-phase system,* and the equivalent single-phase current is twice that in each of the two-phase conductors. The copper loss in the two-phase system is P 2 = 2 2 / 2 2 , (68) when jR 2 = the resistance of each phase circuit, / 2 = the current in each line. The copper loss in the equivalent single-phase circuit is P 1 =R l (2l 2 ) 2 . (69) But Ri (2 7 2 ) 2 = 2 2 / 2 2 (70) and *i = ^> (71) 2 i.e., the resistance of the equivalent single-phase system is equal to one-half the resistance of each circuit of the two-phase system. Equivalent of the three-phase system. Assuming unity power factor, the power in a three-phase system is P 3 = V 3 / 3 (72) and that in the equivalent single-phase system is Pi = Eh. - (73) Therefore, - /i = V 3 7 3 . (74) tfi/i 2 = 3*3/3 2 (75) and #1 = -Rs, (76) * Unless the phases are interconnected to form a three-wire system. POLYPHASE ALTERNATING CURRENTS 121 when RI = the resistance of the equivalent single-phase system, R s = the resistance of each phase of the three-phase system, /i = the current in the equivalent system, 7 3 = the current in each line of the three-phase system, i.e., the resistance of the equivalent single-phase system is equal to the resistance of each phase of the three-phase system. It is evident from inspection that the equivalent resistance of a star-connected system is one-half the resistance as measured by con- tinuous-current methods between any two lines of the three-phase system. It may be shown that this is also true for the delta-con- nected system. CHAPTER VII PROBLEMS 1. A balanced 3-wire, 2-phase system delivers 50 kw. at" a voltage (phase) of 220 and a power factor of 0.9. Find: (a) the current in each line wire, (b) the voltage between the "outside" wires. 2. The current in the common wire of Problem i flows in the current coil of a wattmeter, and the voltage coil of the wattmeter is connected between the common wire and one of the "outside" wires. Find the wattmeter indica- tion. 3. The current in the common wire of Problem i flows in the current coil of a wattmeter and the voltage coil of the wattmeter is connected between the "out- side" wires. Find the wattmeter indication. 4. A 3-phase, star-connected system has a line-to-line voltage of 6600. Find the phase voltage (line to neutral). 5. The line current in a balanced delta-connected system is 150. Find the phase current. 6. Find the power in the circuit of Problem 5 when the phase voltage is 240, and the power factor of the load circuit is 0.85. 7. The current coil of a wattmeter is connected in one line of a balanced 3- phase system, and the voltage coil of the wattmeter between the other two lines. The kw. output, of the system is 100, the voltage 220 and the power factor unity. Find the indication of the wattmeter. 8. A non-inductive star-connected load and a non-inductive delta-connected load are operated from the same 3-phase system. The phase current in each load circuit is 100 amperes. Find the current in each line of the supply system. 9. Same as Problem 8 except the power factor of the star-connected loads is 0.866. Find the line currents. 10. The indications of two wattmeters connected as in Fig. io6c are 5000 and 2500. The system is balanced. Find the power factor of the load circuits. 11. The phase currents in a delta-connected 3-phase system are 50, 30 and 75 amperes, the load circuits being non-inductive. Find the current in each line. Solve graphically. 122 ESSENTIALS OF ELECTRICAL ENGINEERING 12. The phase currents of a 2-phase, 3-wire system equal 160 amperes. The power factor of the load connected to one phase is 0.85, that of the load con- nected to the other phase is unity. Find the current in the common wire. 13. Determine the single-phase equivalent of a balanced 3-phase system delivering 500 kw. at 2300 volts, and having a power factor of 0.92. 14. The line current in a mesh-connected 2-phase system (balanced) is 100 amperes. Find the current in each coil (Fig. 92). 15. Find the power in the 2-phase system in Problem 14 if the voltage be- tween alternate wires (Fig. 92) is 2300. 16. Three resistances, the values of which are 10, 15 and 20 ohms, are star- connected to a 3-phase system, the line-to-line voltage of which is 220. Find: (a) the current in each line, (b) the voltage between each line and neutral, (c) the power input to the resistances. 17. 100 kw. are delivered to a balanced 3-phase system, the power factor of which is 0.85, and the voltage between lines is 2200. Find the current flowing in each line. CHAPTER VIII THE ALTERNATING-CURRENT GENERATOR i. Voltage. The voltage of an alternating-current generator is affected by the same quantities that affect the voltage of a con- tinuous-current generator, and in addition, by the relative posi- tions of the armature conductors. The equation for the voltage of an alternating-current generator may be written E a = knpN io- 8 , (i) when E a = the effective electromotive force induced in the arma- ture winding, k = a constant, the value of which depends on the rela- tive positions of the armature conductors, = the total flux passing between each pole and the arma- ture, (a) Stator (armature) (b) Field Structure FIG. 108. Revolving Field Alternator. Triumph Electric Co. n the speed of the armature or of the rotating field in revolutions per second, p = the number of poles in the field structure, N = the number of series conductors on the surface of the armature. 123 124 ESSENTIALS OF ELECTRICAL ENGINEERING Since the product of the speed and the number of field poles of an alternator is equal to twice the frequency, equation (i) may be written E a = 2 jkfN io- 8 , (2) which is often a useful form of the expression. 2. Concentrated armature windings. Concentrated armature windings are those in which all the conductors under any given pole are placed in one slot, as indicated in Fig. ioga. It is evident that the same electromotive force is induced in each of the con- ductors, and that the maximum values are attained at the same instant. Therefore, the total electromotive force induced in the winding is that in- duced in each conductor multiplied FIG. iooa. Concentrated Anna- , . , i e ^ ^ ture Winding PX tne num ber of conductors connected in series. Since the quantity npN io~ 8 is the average electromotive force induced in the armature winding, the value of k for a con- centrated winding is the ratio of the effective to the average value. For a sine wave = I- II- (4) 3. Distributed armature windings. Distributed armature windings are those in which the conductors under a given pole are divided into groups, and each group is placed in a different slot, as indicated in Fig. logb. With this arrangement of the armature conductors, the electromotive force induced in each conductor is the same, but the maxi- mum values are not attained at the same instant in the different groups, i.e., the electromotive force induced in each group is out of phase with the electromotive FlG , 109b ' ! st , ributed Armature Winding. forces induced in the other groups. There- fore, the electromotive force of an alternating-current generator, having a distributed armature winding, is the geometric sum of the electromotive forces induced in the several groups into which the armature winding is divided. THE ALTERNATING-CURRENT GENERATOR 125 ii the armature winding consists of two similar groups of conduc- tors per pole as in Fig. 1090, the maximum in group B occurs one- quarter of a cycle (90 electrical degrees) later than the maximum in group A . The electromotive forces of the two groups are, therefore, 90 degrees out of phase, and the resultant electromotive force is \/2 times that induced in one group. For any number of groups into which the armature conductors may be divided, the values and the phase relations of the electromotive forces in the different groups may be plotted, and the resultant electromotive force found graphically. Fig. no. In calculating the effective electro- motive force induced in the armature of an alternating-current generator, it should be remembered that each group is a concen- trated winding, and that the effective elec- tromotive force induced in it is the average electromotive force multiplied by I.H, as explained for the concen- trated winding. From Fig. no it is evident that FIG. no. Vector Diagram of Distributed Winding. . k = i.nOd f Ob + Oc + Od 1. 1 1 (chord of the group arc) sum of chords between groups (s) (6) For an infinite number of groupings of the conductors, the sum of the chords becomes an arc, as indicated by the dotted arc in Fig. no. TABLE III VALUES OF k FOR DISTRIBUTED ARMATURE WINDINGS No. of slots Degrees subtended by the winding per pole 180 135 90 60 45 2 0.784 0.922 1.025 1.072 1.087 3 o-739 0.893 1. 012 1.065 1.084 4 0.724 0.882 1.007 1.063 1.083 Infinite 0.707 0.876 I .OOO I .060 1.082 126 ESSENTIALS OF ELECTRICAL ENGINEERING 4. Effects of distributed armature windings. The effects of distributing the armature winding of a alternator are: (a) to de- crease the induced electromotive force, (b) to reduce the induc- tance* of the armature circuit and improve the regulation, (c) to make the induced electromotive force more nearly harmonic, (d) to distribute the heating over the armature surface, (e) to reduce the size of the armature. The effect of a distributed winding on the shape of the electro- motive-force wave is clearly shown in Fig. in. The electromotive- A A A A A A A / v v v / V V V FIG. ma. Voltage Wave with Concen- FIG. nib. Voltage Wave~with Dis- trated Armature Winding. tributed Armature Winding. force wave of the concentrated winding (Fig. ma) shows a very pronounced third harmonic which is entirely absent from the wave of the distributed winding (Fig. nib). The harmonics of the different groups of a distributed winding tend to neutralize and eliminate the harmonic from the resultant electromotive-force wave. 5. The oscillograph. The shape of an alternating current or electromotive-force wave is determined by means of the oscillo- graph. A very fine wire or strip is bent into a loop and placed between the poles of a powerful electromagnet, as indicated in Fig. 112. Attached to the loop is a small mirror M on which may be focused a beam of light. When current flows in the looped conduc- FIG - II2 ; **** D , ia " tor, the loop tends to move so that its plane gram of the Oscillograph. is perpendicular to the lines of magnetic flux passing from N to S. This movement of the loop causes the mirror to be deflected, the deflection is proportional to the current flowing in the loop, and the beam of light reflected from the mirror acts as a pointer. The moving parts of an oscillograph are made very light so that the deflection of the mirror is always in time- * The inductance of the armature is proportional to the square of the number of conductors per slot. THE ALTERNATING-CURRENT GENERATOR 127 phase with the current flowing in the current-carrying loop, and the position of the mirror indicates the instantaneous value of the current flowing in the loop. If the beam of light reflected from the mirror M is directed onto a photographic film having a uniform motion at right angles to the movement of the beam of light, records similar to those in Fig. in are obtained. When it is not desired to make photo- graphic records a second mirror, operated by a small synchronous motor, is rotated or vibrated in such a manner that the reflection of the beam of light on a stationary screen indicates the wave form of the current or electromotive force. 6. The single-phase alternating-current generator. In the single-phase alternating-current generator, the armature conduc- tors are all connected in series. Fig. 113. It is evident from the considerations discussed in Section 3, that the value of k is low if the armature conductors are distributed over the entire arma- ture surface. The terminal volt- age of a single-phase alternator is only slightly greater when the ' conductors cover the entire arma- ture surface than when they cover three-fourths of the surface, but the weight of copper and the arm- ature resistance are proportional to the surface covered. It is, therefore, common practice to distribute the armature conductors over only a portion of the surface of the armature core. Because of this excess of materials, a single- phase alternator has a greater weight and is less efficient than a polyphase alternator of the same voltage and output. Consider an armature having four slots per pole and so wound that the average electromotive force induced in the conductors in each slot is 10 volts. The arc subtended by this winding is 180 degrees, the value of k is 0.724, and the effective electromotive force induced in this section of the armature winding is E a = 4 X 10 X 0.724 = 28.96 volts. FIG. 113. Elementary Single-phase Armature Winding. 128 ESSENTIALS OF ELECTRICAL ENGINEERING If three of the four slots in the above armature core are used, the arc subtended is 135 degrees, the value of k is increased to 0.893, and the effective electromotive force induced in the coils is E a = 3 X 10 X 0.893 = 26.79 volts. Thus by using only three of the four slots per pole, the electro- motive force induced in the armature winding is reduced only 7.5 per cent, while the weight of copper and the resistance of the armature circuit are each reduced 25 per cent. The reactance of the armature is also materially reduced. 7. The two-phase alternating-current generator. In the two- phase alternator two armature windings are wound on the same core, the conductors being so placed that the two induced elec- tromotive forces are in quadrature. As each armature winding covers one-half the surface of the core, the arc subtended is 90 degrees, the value of k is materially higher than for the single-phase alterna- tor, the material is used more ad- vantageously, and a larger output is obtained from a given weight. 8. The three-phase alternating-current generator. In the three-phase alternator three armature windings are wound on the same core and so placed that the induced electromotive forces are 1 20 degrees out of phase. Each winding covers one-third of the surface of the core, and the arc subtended by the winding is 60 degrees. The windings of three-phase alternators are always intercon- nected, the star connection being more largely used than the delta.* By connecting a single-phase load between any two terminals, a star-connected three-phase alternator may be operated as a single- phase generator in which the active armature conductors cover two-thirds of the armature surface. * In a delta-connected system harmonics cause a current to circulate around the delta and heat the conductors. Also, for a given electromotive force a delta-connected armature requires 73 per cent more conductors than does a star-connected armature. FIG. 114. Elementary Two-phase Armature Winding. THE ALTERNATING-CURRENT GENERATOR 129 9. Armature reaction. As in the continuous-current dynamo, the armature current of an alternator affects both the value of the flux in the air gap and its distribution. If the current flowing in the armature windings is in phase with the induced electro- FIG. lisa. Elementary Three-phase Armature Winding (Delta Con- nected) . FIG. iisb. Elementary Three-phase Armature Winding (Star Con- nected). motive force, the flux is distorted only; if the current and the electromotive force are not in phase, the total flux is increased or decreased as the current leads or lags behind the electromotive force. Referring to a single-phase two-pole alternator with concentrated winding, let N = the number of series conductors on the armature, = the angle by which the current leads or lags behind the electromotive force. Then and e = E m sin co/, i = I m sin (ut ). (7) (8) The direction of the flux set up when current flows in the arma- ture winding is at right angles to the plane of the coil, and its value is proportional to the product of the armature current and the number of turns in the winding. Let the instantaneous magneto- 130 ESSENTIALS OF ELECTRICAL ENGINEERING motive force in ampere-turns due to the armature winding be represented by m. Then t \ m=-- (9) ^_ JV7 m sin(co/dr 2 Ra ed Volt a^ - d 36V S ^ 1 / / s> > 14 ^2000 t- / / /\ 'L in j - ,fior / ' / j '/ K ' > '6 ^ /* s V } '/ b ^ ^10-^ / g ^ ^ ^ * o / / // S X. ! / 1 // V" ^ -^ 5 ' / / J V / f s^ f 100 /^ "v / / ^ ^ \ & c I D Z I 4 5 6 1 ) FIELD AMPERES OA open-circuit saturation curve. BC approximate zero power factor curve. BC' experimental zero power factor curve. BD full-load saturation curve at unity power factor. OF short-circuit curve. EG synchronous reactance curve. Ice (b) The open-circuit saturation curve. The saturation curve shows graphically the relations between the electromotive force in- duced in the armature winding and the field excitation. The data THE ALTERNATING-CURRENT GENERATOR 135 from which the saturation curve is plotted are obtained by meas- uring the terminal voltage of the armature with the armature circuit open, for different field currents, the frequency being main- tained at the rated value. Fig. 119. (c) The "short-circuit" curve. The short-circuit curve shows graphically the relations between the current in the short-circuited armature winding and the field excitation. Taking into considera- tion the rated current of the armature, short circuit the armature winding, as indicated in Fig. 120, through a suitable ammeter, and determine the field current required to cause different current Field Rheostat- , , Field Rheostat flfo To Lucifer (a) 5 ingle Phase (b) Three Phase FIG. 1 20. Connections for Short-circuit Test. values to flow in the armature circuit, the rotating parts of the alternator being driven at rated speed. Plot the curve as in Fig. 119. The impedance of the armature circuit is assumed to be constant and the short-circuit curve is, approximately, a straight line. (d) The zero .power factor saturation curve. The zero power factor saturation curve of an alternator shows the relation between the terminal voltage and the field excitation when the power factor of the load circuit is zero, and rated current flows in the armature circuit. A curve approximating the zero power factor curve may be constructed from data obtained when the load on an alternator consists of idle-running under-excited synchronous motors, the power factor of which is very low. An approximate zero power factor curve may also be con- structed from the open-circuit saturation curve and the short- circuit curve. Draw the curve BC, in Fig. 119, parallel to the open-circuit saturation curve, beginning at the abscissa which represents the field excitation required to cause rated current to flow in the short-circuited armature. In alternators having high reactance, high saturation, and large magnetic leakage, the zero power factor curve may lie before BC, as indicated by the line BC, and its exact location must be determined by test. 136 ESSENTIALS OF ELECTRICAL ENGINEERING Regulation by the electromotive-force method. In calculating the regulation of an alternator by the electromotive-force method, it is assumed that the electromotive force induced in the armature is the vector sum of two quadrature electromotive forces, one equal to the product of the armature current and the total resistance of the circuit, including that of the armature; the other equal to the product of the armature current and the total reactance of the circuit. E a = i a VR* + x 2 (26) cos + Ralrf + (E sin < + XJ a )\ (27) when E a = the electromotive force induced in the armature, E = the terminal electromotive force of the generator, Ra = the resistance of the armature circuit, X a = the reactance of the armature circuit (the synchronous reactance* of the alternator armature), I a = the current flowing in the armature circuit, cos $ = the power factor of the load circuit. The synchronous reactance of the armature circuit, which can- not be measured directly, must be calculated. The voltage drop in the armature is the vector sum of the resistance voltage and the reactance voltage. ES = V(Rj a y + (xjaY- (28) Dividing equation (28) by I a , z a = VRJ + XJ, (29) and X a = Vz a 2 - R a 2 . (30) A consideration of the open circuit and the zero power factor saturation curves, shows that the synchronous reactance of an al- * The synchronous reactance of an alternator armature is the combined effect of : (a) the inductance of the armature windings, (6) the flux set up by the current-carrying armature conductors. As shown in Section 9, the armature currents of a polyphase alternator set up a flux which is fixed in position, and across which the armature conductors move. The electromotive force induced in the conductors by the armature flux reduces the terminal voltage of the alternator and makes the regulation poorer in much the same manner as does the inductance of the windings. A similar effect takes place in the single-phase alternator because of the varying value of the flux threading the armature coil. THE ALTERNATING-CURRENT GENERATOR 137 ternator armature is not constant, but decreases as the excitation increases. It is also affected by the phase relation of the current and the electromotive force induced in the armature winding, i.e., by the power factor of the load circuit. Substituting in equation (27) the full- (rated) load armature cur- rent, the no-load voltage is calculated and , , . (E a E) 100 f x Per cent regulation = - - (31) E Example. A single-phase alternator has the following: E = 2300, I a = 100, Ra = I, X a = 10. Find the per cent regulation when the power factor of the load circuit is unity. Solution. = V( 23 oo + ioo) 2 + (iooo) 2 = 2600 volts. -p. , ,. 2600 2300 Regulation = - 2300 = 13 per cent. Regulation by the magnetomotive-force method. In calculating the regulation of an alternator by the magnetomotive force method it is assumed that the voltage induced in the armature windings is due to quadrature fields, one equal to that required to produce the total non-inductive drop, the other equal to that required to produce the total wattless component of electromotive force. From the saturation curve find the field current // required to induce in the armature winding an electromotive force equal to the total non-reactive drop (E cos < + RJ a ) of the circuit at rated load; from the same curve find the field current //' required to induce in the armature winding an electromotive force equal to the total reactive drop (E sin + XJ a ) of the circuit at rated load. The field current // required to produce, simultaneously, the ter- minal voltage E and the current I a in a circuit the power factor of which is cos <, is It = V(//) + (/,")'. (32) , From the saturation curve find the electromotive force E a in- 138 ESSENTIALS OF ELECTRICAL ENGINEERING duced in the armature winding when current // flows in the field windings. Per cent regulation = ^-- ? . (33) h, Example. Find, by the magnetomotive force method, the reg- ulation of the 2 300- volt, single-phase alternator, specified above. Solution. From Fig. 119, // = 44.5 amperes, and //' =12.5 amperes. Therefore, _ I f = V(44. 5 ) 2 + (i2.5) 2 = 46.22 amperes. The induced electromotive force is, from the open-circuit satura- tion curve, 2430 volts, and the regulation is 2300 = 5.2 per cent. Regulation by the A.I.E.E. method. The electromotive force induced in the armature winding of an alternator is the vector sum of the terminal voltage and the voltage drop in the armature circuit. Therefore, the drop in the armature, at zero power factor and rated armature current, is the difference between the ordinates of the open-circuit and the zero power factor saturation curves. For field excitation, Oc, the induced electromotive force is ca, and the drop in the armature is ba. Fig. 119. For any other power factor, and the same field excita- tion, the terminal voltage is the vector difference of the induced electromotive force and the internal drop. Fig. 121. By calculating the terminal electro- motive forces for different field exci- tations but constant power factor, the load saturation curve BD, in Fig. 119, may be plotted. From this curve and the open-circuit saturation curve, the electromotive force induced in the armature when the alternator is operating at rated voltage and armature current, and with THE ALTERNATING-CURRENT GENERATOR 139 the specified power factor, is determined, and the regulation cal- culated. From Fig. 119, the induced voltage is 2490 when rated current flows in the armature circuit, the terminal voltage is 2300, and the power factor of the load circuit is unity. The regulation is, therefore, = !&- X 100 / 2300 = 8.3 per cent. 12. Limits of regulation. From the above numerical examples it is evident that the electromotive force method and the magneto- motive-force method do not give the same result, these calculations serving only to establish the limits within which the actual regu- lation lies. If the saturation curve was a straight line, the actual regulation would be determined by either of these methods. Since the no-load voltage, as calculated by the electromotive-force method, is larger than that obtained by an actual load test, this method is termed " pessimistic. " The no-load voltage, as obtained by the magnetomotive force method, is smaller than that obtained by an actual load test, and this method is termed "optimistic. " It is worthy of note that, in a well-designed alternator, the optimistic method gives a closer approximation to the actual regulation than does the pessimistic method. The A.I.E.E. method for the calculation of alternator regula- tion is essentially empirical, but gives results which approximate very closely those obtained by test, and its use is recommended in preference to either the electromotive-force or the magnetomotive- force methods. 13. Graphical determination of term- inal voltage. In Fig. 122, using as a center and with a radius propor- tional to the rated electromotive force of the alternator, strike an arc ECD. Through draw the current vector I a . Lay off OA proportional to the full- FlG - I22 - Alternator Regulation, load resistance drop RJ a of the armature winding, and AB propor- tional to the full-load reactance drop XJ a of the armature cir- cuit. Draw OC to its intersection with the arc ECD, making the angle COI such that its cosine is equal to the power factor of the 140 ESSENTIALS OF ELECTRICAL ENGINEERING load circuit. BC is proportional to the generated or no-load voltage. With B as a center and a radius equal to BC, strike a second arc FCG. Divide OB into any number of equal parts and lay off equal spaces beyond O, the spaces representing percentages of the rated capacity (armature current) of the alternator. From these points draw lines, parallel to OC, to their intersection with the arc FCG. The lengths of these parallel lines are proportional to the terminal voltages at the given percentages of rated load, and from them the voltage characteristic may be constructed, or the regulation cal- culated. 14. The Tirrill Regulator. The operation of the Tirrill regu- lator, when applied to an alternator, is essentially the same as for the Main Contacts A.C. field Rheostat A - c - Generator FIG. 123. Tirrill Regulator for Alternators. continuous-current generator.* An elementary diagram showing the application of a Tirrill regulator to an alternator is shown in Fig. 123. The field rheostat of the exciter is periodically short- circuited by the closing of the main contacts, the length of time during which the short circuit exists depending on the position of the main contact. The position of this contact is controlled by the alternating-current magnet, i.e., by the voltage of the alternating- current system. The function of the compensating winding is to increase the terminal voltage as the load increases, the effect of the compensating winding being proportional to the line current. Assuming the main contacts to be open, the relay contacts are held open by the differentially wound relay magnet, and the voltage of both the exciter and the alternator falls off. The re- duced voltages allow the main contacts to close. Closing the * See Chapter 4, Section 10. THE ALTERNATING-CURRENT GENERATOR 141 main contacts demagnetizes the relay, closes the relay contacts, and short circuits the exciter-field rheostat. The increased field excitation increases the electromotive force between the terminals of the control magnets, and causes the main contacts to open. 15. The losses in an alternating-current generator. The losses in an alternator are: (a) armature copper losses, (b) field copper losses, (c) windage and friction, (d) iron (core) losses, (e) stray load losses. (a) Armature copper losses. The copper loss in the armature winding is that due to the resistance of the armature conductors, and increases as the square of the armature current. (b) Field copper losses. The field copper losses are those due to the resistance of the field winding and are proportional to the square of the field current. If the terminal voltage is to remain constant, the field excitation of an alternator must be increased as the load increases to compensate for armature reaction and resist- ance drop, and the field losses increase with the load. (c) Windage and friction. The windage and friction losses are those due to: (i) friction between the shaft and the bushings, (2) friction between the brushes and the rings, (3) the resistance offered by the air to the movement of the rotating parts. These losses are constant for any given speed. (d) Iron losses. The iron losses are due to: (i) hysteresis in the iron of the armature core, (2) eddy currents in the armature core. The iron losses do not vary greatly and are usually considered constant. (e) Stray load losses. The stray load losses, so-called for lack of .a better name, include all the losses not included in (a), (), (c) and (d) and such changes in (c) and (d) as take place when the alter- nator is loaded. The stray load losses are primarily additional iron losses which are due to distortion of the magnetic field as the arma- ture current increases; to eddy-current losses in the armature con- ductors, etc.; and are directly proportional to the armature current. Determination of the losses in an alternating-current generator* The losses of an alternator are determined as follows: Armature copper losses. The copper loss in the armature con- ductors is equal to the product of the resistance of the armature circuit and the square of the armature current. Pa = RaV. fo) * See the Standardization Rules of the American Institute of Electrical Engineers. 142 ESSENTIALS OF ELECTRICAL ENGINEERING Field copper loss. . P/ = */[(//)' + (/"/)*!, (35) when Rf = the resistance of the field circuit, // = the field current (taken from the saturation curve) re- quired to produce rated voltage at no load. I/' = the field current (taken from the short-circuit curve) required to produce a given current in the short- circuited armature. Windage and friction. Drive the alternator at its rated speed but without field excitation, and determine the input to the driving motor. Determine the losses in the motor and subtract them from the total input. The difference is the loss due to windage and friction of the alternator. Pw&f = Motor input motor losses. (36) Iron losses. Drive the alternator at its rated speed and with such field excitation as gives rated voltage at the armature terminals. The total input to the driving motor is the sum of the losses in the motor, the windage and friction of the alternator, and the iron losses of the alternator. Pi = Motor input motor losses W w &/. (3 7) Stray load losses. Drive the alternator at its rated speed with the armature winding short-circuited, and with such field excitation as produces the desired current in the armature circuit. The in- put to the driving motor is equal to the sum of the motor losses, the windage and friction of the alternator, the copper losses in the alternator armature winding, and the load losses. Pi = Motor input motor losses R a l the vertical position, i.e., the number of revolutions per second made by the loop must be equal to the fre- quency of the alternating current, and the change in the direction of the current must be made when the conductors are approximately midway of the arc between the poles. Under these conditions, the relations between the field flux and the current in the moving con- ductor are fixed, and the torque is exerted in one direction only. Therefore, if the frequency of the supply circuit is constant, the speed of a synchronous motor is constant, but the motor is not self-starting. 3. Torque-load adjustment. It is an easily demonstrated experimental fact that when -the driving torque of an alternating- current generator, opera ting, in parallel with other generators, is reduced to zero, its moving parts continue to rotate at the same speed, and that the direction of the current in the armature windings is reversed. When the load on a continuous-current shunt motor increases, the speed of the armature decreases until the counter- electromotive force is reduced. to such a value that the current necessary to produce the required torque flows in the armature circuit. Since the speed of a, synchronous motor is fixed by the frequency of the system to which it is connected, the increased armature current (torque) must be produced by other means. A synchronous motor, like a continuous-current motor, generates a counter-electromotive force which is proportional to the speed, and is dependent on the field excitation. The current in the arma- ture of a synchronous motor may, then, be changed by changing the field excitation, but it would be practically impossible to oper- THE SYNCHRONOUS MOTOR 151 ate, commercially, a motor the field current of which must vary with the load. Consider the conditions when the counter-electro- motive force of the motor is equal to the applied electromotive force. As long as the phase difference of these equal electro- motive forces is 180 degrees, no current flows in the armature of the motor, and no torque is developed in it. Consequently, the rotating parts tend to stop, and the angle of phase difference becomes greater than 180 degrees. Let the phase rela- tions of the two equal electromotive forces be as indicated in Fig. 130. A current proportional to the resultant elec 7 tromotive force E r flows FlG - T 3 - in the armature of the Mechanical and .Vector Relations of an Alternator and a Synchronous Motor. motor, and the angle between the resultant electromotive force and the current is j3, the tangent of which is equal to --v 5 = tan- 1 - when .Y = the synchronous reactance of the armature circuit, R a = the resistance of the armature circuit. Since R a is always small as compared to X a , the angle ft is usually greater than 80 degrees but can never equal 90 degrees. The input to the motor under the above conditions is ~D J7 T ( \ If the load increases, the angle between the applied and the counter- electromotive forces will, evidently, increase until equilibrium is reestablished, i.e., with constant field excitation, the torque of a synchronous motor is dependent on the phase relations of the ap- plied and the counter-electromotive forces, and the angle between them is automatically adjusted as the load changes. 4. Starting. A synchronous motor is not self-starting because the periodical reversal of the direction of the current in the station- 152 ESSENTIALS OF ELECTRICAL ENGINEERING ary armature conductors produces a torque which periodically reverses, and tends to produce rotation of the field structure first in one direction, then in the other. The commercial methods of starting a synchronous motor are: (a) by means of an auxiliary motor, (6) as an induction motor. (a) By an auxiliary motor. The starting motor is a small motor, either shunt or induction, direct-connected or belted to the shaft of the synchronous machine. When a synchronous motor is started in this way, it is a generator and must be syn- chronized * before it is connected to the supply circuit. After the synchronous motor armature is connected to the alternating-cur- rent supply circuit, the circuit of the starting motor is opened. In case the synchronous machine is provided with a separate exciter and a source of continuous current is available, the exciter may be used as a starting motor. An induction motor is, however, more often used. This method requires the use of synchronizing apparatus and the installation of a starting motor, but causes minimum distur- bance in the alternating-current system during the starting period. (b) As an induction motor .f If the field circuit of a synchro- nous motor is opened and an alternating current supplied to the armature, the changing magnetism set up by the armature currents induces currents in the field pole shoes. These currents, reacting with the magnetism set up by the armature winding, produce a small torque, if the motor is polyphase, and the unloaded motor starts without the use of auxiliary devices. When the motor attains approximately synchronous speed, which is indicated by a violent swinging of the pointer of an ammeter connected in the alternating-current supply line, the field circuit is closed and the motor pulls into step with the supply circuit. The starting torque of a synchronous motor, when started in this way, is increased by means of a "squirrel cage" field structure similar to that shown in Fig. i28b. This construction also tends to prevent " hunting," as explained in Section 8. * See Chapter 8, Section 17. . f This method of starting a synchronous motor depends on the fundamental prin- ciples underlying the action of the induction motor, and will not be understood until those principles have been studied. The statements made here are to be taken simply as a mechanical process by which the synchronous motor may be started. THE SYNCHRONOUS MOTOR 153 Because of the low power factor and the large current required during the starting period, this method of starting may cause un- desirable disturbances in the system from which it is supplied. These disturbances are particularly objectionable when a single motor forms a considerable portion of the total load on the supply system and is stopped and started frequently, or when motors are operated in parallel with incandescent lamps. 5. Stability. An engine or motor is in stable equilibrium as long as an increased load automatically produces a corresponding intake of power. As noted in Section 3, an increased load on a synchronous motor causes the angle between the applied and the counter-electromotive forces to increase. Assuming the applied FIG. 131. voltage to be constant, the operation of the motor is stable as long as the product of the current and the power factor increases; but it is evident that, as the angle between the two electromotive forces increases, a point is finally reached where the power factor decreases faster than the current increases. Beyond this point the motor is in unstable equilibrium, drops out of step, and stops. The above statement is illustrated by Fig. 131, in which OA is the vector of applied electromotive force, and OB the vector of counter-electromotive force. OC is the vector of the resultant elec- tromotive force, and OD the vector of current flowing in the arma- ture circuit. The impedance of the circuit is approximately con- 154 ESSENTIALS OF ELECTRICAL ENGINEERING stant, and OD is proportional to OC. The power input to the motor is proportional to the projection of the current vector on the vector of applied electromotive force, i.e., to the product of the current vector and the cosine of angle AOD. For the relation of the vectors shown in Fig. 13 la, the power in- put to the motor is proportional to OE. As the load on the motor increases, the angle increases until the vector of the counter- electromotive force reaches the position indicated in Fig. 13 ic. If the angle 6 increases still further, the projection OE decreases and the power input to the motor decreases correspondingly, as indicated in Fig. 13 id. Therefore, if a synchronous motor is loaded beyond a certain point, it cannot develop sufficient torque to carry the load, and will stop. In the commercial motor, the armature current usually be- comes excessive and overheats the motor before it reaches the point of instability. 6. Maximum load. That the maximum intake of a synchro- nous motor for any given counter-electromotive force (field ex- citation) occurs when 6 (3 = 180 degrees is evident from the following considerations: Referring to Fig. 1310 let 6 - = 180. (3) If the angle 6 is either increased or decreased, both the projection of the current vector on the vector of the electromotive force and the input to the motor are decreased. 7. Efficiency. The efficiency of a synchronous motor is deter- mined : (a) by a brake test, (b) from the losses. (a) Brake test. Measure the input and the output, and compute the efficiency from the equation ~ . output X 100 t , Per cent efficiency = ~ (4) input (b) From the losses. The losses in a synchronous motor are the same as those of an alternating-current generator and are deter- mined in the same way.* 8. Hunting. The phenomenon known as " hunting " in a synchronous motor, or a rotary converter, consists of a periodical * See Chapter 8, Section 15. /THE SYNCHRONOUS MOTOR 155 variation of the speed of the rotating parts, the speed being alter- nately too fast or too slow, and is indicated by the swinging of the ammeter pointer, and by a humming noise peculiar to this condition. Hunting may be caused by: (a) a sudden change in the load, (b) irregularities in the speed of the prime mover driving the generator from which the motor is supplied, (c) faulty design of the motor. The inertia of the rotating parts of the motor tends to damp out any oscillations that may be set up, the addition of a heavy fly- wheel adding to this damping effect. Hunting is often guarded against, in the design of a synchronous motor, by wedging bars of copper between the tips of the pole shoes, or by a " squirrel cage" structure similar to that shown in Fig. 128. As long as the angular velocity of the rotating parts of the motor is constant, the copper bars have no effect, but when hunting takes place the oscillatory motion causes a relative movement of the bars and the flux set up by the armature winding. This move- ment induces an electromotive force, and currents flow in the bars. The reaction between the currents in the bars and the flux set up by the armature winding tends to damp out the oscillations, hence the name "magnetic dampers" which is sometimes applied to this construction.* 9. Phase characteristic. The phase characteristic, commonly called V-curve, of a synchronous motor shows the relations between the armature current and the field excitation, the load remaining constant. Referring to Fig. 132, let OA be the vector of the applied electromotive force, OB be the vector of the counter-electromotive force, OC be the vector of the resultant electromotive force, OD be the vector of the armature current, ft be the angle between the current and the resultant electromo- tive force, < be the power factor angle, 7 cos be constant, i.e., the locus of the current vector is a straight line perpendicular to the vector of the applied electro- motive force. * The action of copper bars in the prevention of hunting will be more apparent after the induction motor has been studied. 156 ESSENTIALS OF ELECTRICAL ENGINEERING When = o (current and applied electromotive force in phase), the relations in the circuit are represented in Fig. 1320; when = 30 degrees (lagging), the rela- tions in the circuit are represented in Fig. i32a; and when $ = 30 degrees (leading), the relations in the circuit are represented in Fig. I32C. A comparison of the counter- electromotive forces in Fig. 132, shows that the power factor of a synchronous motor is dependent on its counter-electromotive force, i.e., on its field excitation. The power factor of the motor is, therefore, changed by changing its field excitation. Fig. 133 shows phase characteristics for different loads. For a con- stant load, the power factor of the motor is computed as the ratio of the minimum armature current to the armature current at the given or required field excitation. Because of distortion of the wave shape, unity power factor is seldom or never attained. 10. The circle dia- gram. In any alter- nating - current circuit having a constant applied voltage, a COTi- Maximum \potver factor L eadinq FIELD AMPERES FIG. 133. Phase Characteristics. stant reactance and a variable resistance, it may be shown that the locus of the current vector is a semicircle. The phenomena of the synchronous motor may, therefore, be represented by means of a circle diagram, when it is assumed that the synchronous reactance of the motor is constant.* * This assumption simplifies the treatment, and the error introduced is not great. THE SYNCHRONOUS MOTOR 157 Lay off AO and OD (Fig. 134) proportional to the applied elec- tromotive force, making the angle AOD such that , (5) when X a = the synchronous reactance of the motor armature, R a = the resistance of the motor armature. With D as a center and a radius proportional to the counter- electromotive force at any given or required field excitation, draw a semicircle. The semicircle is the locus of the current vector 07, and the react- ance drop OC, the resistance drop BC, and the counter electromotive force AB may be drawn in their correct phase rela- tions and magnitudes as indicated in the ^^ circle Diagram. figure. ii. The synchronous phase-modifier. As shown in Fig. I32C, a synchronous motor, when over excited, takes a leading current from the supply system. The wattless component of a leading current opposes and neutralizes an equal wattless component in a transmission line or a distributing system * due to a parallel induc- tive load such as an induction motor. When used for the purpose of improving the power factor of a circuit, a synchronous motor is termed a "synchronous phase-modifier," although it may deliver mechanical power at the same time. Example. Two ioo-h.p., 440 volt induction motors operate at 85 per cent power factor and 87 per cent efficiency. Find the saving in transmission line copper when one of the induction mo- tors is replaced by a synchronous motor, the efficiency of which is 85 per cent when over excited so as to compensate for the lagging current of the other induction motor. Also find the rated output of the synchronous motor, and its power factor. Solution. T = 200 X 746 " 0.85 X 0.87 X 440 = 465 amperes for two induction motors, * See Chapter i, Section 38. 158 ESSENTIALS OF ELECTRICAL ENGINEERING , _ 100 X 746 100 X 746 0.87 X 440 0.85 X 440 = 194 + 199 = 393 amperes for one induction motor and one synchronous motor, R' ( 4 6 5 ) 2 = R" (393) 2 i.e., the weight of copper is approximately 40 per cent greater for two induction motors than for one induction motor and one syn- chronous motor. P.P. of snchronous motor = (233 X 0. 5 2) 234 = 0.85. Rating of synchronous motor = - 0.85 = 119 h.p. (use 125 h.p.). 12. Load division of alternators in parallel. Let A and B be two similar alternating-current generators operating in parallel, the field excitation, terminal voltage and armature current being the same in each machine. If the power input to the prime mover driving alternator B is reduced, the load on the alternator tends to stop its rotating parts, the angle between the electromotive forces becomes greater than 1 80 degrees, and a current flows between the armatures, i.e., alter- nator B acts simultaneously as an alternating-current generator and as a synchronous motor, the current for its motor action being received from alternator A. Therefore, the load division of alter- nating-current generators operating in parallel is changed by chang- ing the power input to their prime movers. Since the input to a prime mover, e.g., a steam engine, is constant at a given speed, changing the field excitation of alternator B does not change its output, although it causes a current to flow between the armatures. Since the resultant electromotive force is in phase with that induced in alternator A and the angle is nearly 90 de- grees, it is evident that this current is wattless except for the small THE SYNCHRONOUS MOTOR 159 power component due to the resistance of the circuit, that it leads the electromotive force of alternator B and lags behind the electro- motive force of alternator A. This quadrature current can pro- duce no torque in either machine and the load distribution of alternators operating in parallel is, therefore, not disturbed by a change in field excitation. Since the current flowing between the armatures leads the elec- tromotive force of alternator B and lags behind the electromotive force of alternator A, its effect is to equalize the voltages induced in the armatures by reducing the flux set up by one field winding and increasing that set up by the other. CHAPTER IX PROBLEMS 1. Show, by means of the circle diagram, that the power factor of a synchro- nous motor is changed by changing its field excitation, the load remaining constant. 2. The synchronous impedance of an alternator is 1.5 ohms. When oper- ated as a synchronous motor from 22o-volt mains and without load, the arma- ture current is 10 amperes, and the power factor is 0.866 leading. The angle ft 85 degrees. Find the counter-electromotive force of the motor. 3. Find the counter-electromotive force of the motor in Problem 2 when it power factor is unity. 4. Find the power factor of the motor in Problem 2 when the applied volt- age is reduced 10 per cent. 5. Calculate the armature resistance of the motor in Problem 2. 6. Draw the vector diagram for the three-phase alternator of Problem n, Chapter 8, when operated as a motor at unity power factor, and rated armature current. 7. From the diagram obtained in Problem 6, calculate the torque developed in the armature. 8. A 3-phase, 23oo-volt, 5oo-kv-a. synchronous motor is overexcited so that its power factor is 0.6 leading. Calculate the capacitance to which it is equivalent when operated from 6o-cycle mains, and full-load (rated) current flows in the armature circuit. 9. A 3-phase, 22oo-volt induction motor having a lagging power factor of 0.85 takes zoo amperes (line), and is connected in parallel with an over- excited synchronous motor. When the power factor of the system is unity the current (line) supplied to the synchronous motor =125 amperes. Find: (a) the power factor of the synchronous motor, (b) the line current, (c) the input (watts) to each motor. Draw a diagram representing the electromotive fc and the currents in the system. 160 ESSENTIALS OF ELECTRICAL ENGINEERING 10. Plot the circle diagram for the alternator in Problem n, Chapter 8, when operated as a synchronous motor, and from this diagram construct the phase characteristic (V-curve) for an input of: (a) 500 kw., (6) 100 kw. n. Same as Problem 10 except the resistance of the armature is 2 ohms. 12. Same as Problem 10 except the resistance of the armature is 0.5 ohm. 13. The alternator in Problem 2 is used as a phase modifier. The armature current is 100 amperes and the power factor o.i. Find: (a} the power input to the armature, (6) the wattless component of lagging current that it will neutralize. 14. The alternator hi Problem 2 is driven as a synchronous motor from 220 volt mains. The counter-electromotive force is 220 volts. Find: (a) the cur- rent input to the motor at no load, (6) the power factor at which the motor operates, (c) the phase angle between the applied and the counter-electromotive forces. CHAPTER X CURRENT-RECTIFYING APPARATUS SINCE electrical energy is most economically and satisfactorily transmitted by means of alternating currents, and many commercial applications require unidirectional currents, it is necessary to pro- vide means for the rectification or conversion of alternating cur- rents. Alternating currents may be made unidirectional by: (a) a mechanically-driven commutator, (b) a motor-generator, (c) the mercury arc rectifier, (d) the synchronous converter. (a) The Commutator 1. A commutator driven by a small synchronous motor would appear to be a cheap and an efficient means for the rectification of alternating currents. This method is, however, of little commercial importance because of its limited capacity, excessive sparking, with a consequent destruction of the commutator, taking place whenever either the current or the electromotive force increases above very limited values. (b) The Motor-Generator 2. The motor-generator consists of a motor, either synchronous or induction, which drives a continuous-current generator. The apparatus differs in no way from that described elsewhere in this volume. Motor-generators are in extensive use where a wide variation of the continuous voltage is required. Because it con- sists of two machines, the first cost of a motor-generator is high, and the operating efficiency low. (c) The Mercury Arc Rectifier 3. Construction. The mercury arc rectifier consists of a highly exhausted glass tube having two or more iron or graphite terminals to which the alternating-current supply circuit is connected, and a mercury terminal to which the load circuit is connected. Fig. 135. 161 162 ESSENTIALS OF ELECTRICAL ENGINEERING It is also provided with an auxiliary mercury terminal which is used for starting purposes only. The circuits for a single-phase rectifier are shown in Fig. i36a, and those for a three-phase rectifier in Fig. 1360. The mercury arc rectifier depends, for its operation, on the fact that an electric current may flow from iron or graphite terminals to mercury, but may not flow in the reverse direc- tion. The flow of current in the rectifier circuit is similar to that of water in the hydraulic sys- tem shown in Fig. 137, where check valves VV FIG. 135. Mercury Arc permit water to flow only in the direction indi- RectifierTube. West- cate( j ty t h e arrows> i n t he single-phase rec- tifier, current flows from terminal A to the mercury during one-half cycle, and from terminal B to the mercury during the other half cycle. The current in the load circuit is, therefore, a unidirectional (pulsating) current. FIG. 136. Wiring Diagrams for'Mercury FIG. 137. Hydraulic Analogy for Mer- Arc Rectifier. cury Arc Rectifier. 4. Starting. The mercury arc rectifier cannot be started simply by connecting the iron, or graphite, terminals to an alternating- current supply circuit, because of the high initial resistance of the tube. The auxiliary mercury terminal is, therefore, necessary to break down this high resistance. When the starting switch is closed, and the bulb tilted slightly so that the mercury terminals are connected by the liquid mercury in the tube, an electric current flows through the mercury and fills the tube with vapor. The mercury vapor reduces the resistance of the main circuit so that CURRENT-RECTIFYING APPARATUS 163 current may flow between the iron terminals and the mercury. The tube may now be allowed to return to its normal position and the starting switch opened, the load circuit having been closed through a suitable resistance, since the arc will " break" and the rectifier cease to operate if the current falls below a certain min- imum value. 5. Operation. Because the arc " breaks " when the load current falls below a certain minimum value, which varies from two to four amperes, an inductance must be placed in the load circuit so that the decay of the current established during one- half cycle is delayed until that from the other half cycle can be established * making the form of the wave as shown in Fig. 138. Without this inductance, which may be the secondary of the transformer supplying the rectifier, the current decreases to zero at the end of each half cycle and the operation of the arc is interrupted. Because of the overlapping of the waves in the dif- ferent phases of a polyphase system, no inductance is required in the load circuit of a rectifier using polyphase currents. The voltage E dc of the load circuit bears a constant ratio to the voltage E ac of the supply circuit. 7,; Zeio L >- Current Wore in I PtoiHye Electrode Current Wavem H fbstrivt Electrode Wave of Rectified m Current Wave of Impressed IV FIG. 138. Wave Forms Mercury Arc Rectifier. E d , = E ac k, IT d) when k is the counter-electromotive force of the tube which varies, in different tubes, from thirteen to fifteen volts. Because of the high temperatures at which they operate, the tubes of commercial rectifiers are usually immersed in oil for the better dissipation of the heat. The mercury arc rectifier delivers constant electromotive force or constant current, as the transformer T is constant potential or constant current. 6. Efficiency. From equation (i), it is evident that from thirteen to fifteen volts are lost in the rectifier tube, whatever the applied voltage may be. Therefore, the efficiency is greater * This statement applies to single-phase rectifiers only. 164 ESSENTIALS OF ELECTRICAL ENGINEERING at high voltages than at low, but is constant at all loads. At low voltages, such as are commonly used for motors, the efficiencies of mercury arc rectifiers do not compare favorably with other types of rectifying apparatus. 7. Limitations and use. Because of its low efficiency at the voltages commonly used for continuous-current apparatus, and its poor power factor, the use of the mercury arc rectifier is restricted, at the present time, to special purposes. It is largely used for supplying unidirectional current to arc lamp systems where the high voltage and the small current required make it very satis- factory when operated in connection with constant-current trans- formers. It is also used, in small units, for charging storage batteries. (d) The Synchronous Converter 8. Construction. The synchronous, or rotary, converter does not differ essentially from a shunt or compound (continuous-cur- rent) dynamo with the addition of symmetrical connections from the armature winding to two or more continuous rings mounted on the arma- ture shaft in the same manner as for an alternating-current generator with rotating armature. Fig. 139. Practically, the converter which has more than two rings differs from the continuous - current machine of the same rating by having a much larger FIG. i 39 . Synchronous Converter commutator, and smaller yoke and Armature. Westinghouse. ,, , , , field poles. 9. Operation. A synchronous converter may be operated as: (a) A continuous-current motor. (b) A continuous-current generator. (c) A synchronous motor. (d) An alternating-current generator. (e) A double-current generator, i.e., simultaneously as a contin- uous-current generator and an alternating-current generator. (/) A converter, i.e., simultaneously as a synchronous motor and a continuous-current generator. When so operated, the arma- ture must rotate at synchronous speed, and the speed of any given CURRENT-RECTIFYING APPARATUS 165 converter is determined by the frequency of the supply circuit. The converter exhibits the same phase characteristic, power factor and line-current changing with the field excitation, as the synchro- nous motor. (g) An inverted converter, i.e., simultaneously as a continuous current motor and an alternating-current generator. When so operated, the speed and, consequently, the frequency of the alter- nating current, varies with the field excitation, unless operated in parallel with other synchronous apparatus which controls the fre- quency of the circuit. Since the armature reaction of an alternat- ing-current generator, the load circuit of which is inductive, tends to demagnetize the field, the speed of an inverted synchronous con- verter supplying such a circuit may become excessive, and means should be provided for preventing these high speeds which may wreck the machine. Two methods are employed to limit the speed of an inverted converter to safe values: (i) separate excita- tion, (2) speed-limit switches. (1) Separate excitation. If an inverted converter, instead of being excited from the supply mains, has its field windings supplied from a separate exciter direct connected, or belted, to the armature shaft, an increase or decrease in the speed of the armature causes a corresponding increase or decrease in the excitation of the converter, and tends to keep the speed constant. (2) Speed-limit switches. A speed-limit switch is attached to the shaft of the converter, and operates circuit breakers in the con- tinuous-current supply mains when the speed reaches a prede- termined value, thus stopping the machine. 10. Voltage relations. The voltage between the brushes of a continuous-current generator is the algebraic sum (constant) of the instantaneous electromotive forces induced in the different coils con- nected in series between the brushes; the maximum alternating voltage induced in the same coils is equal to the geometric sum of the maximum electromotive forces induced in the separate coils. But the geometric sum of the maximum alternating-electro- motive forces induced in the distributed windings of a synchronous converter, between adjacent continuous-current brushes, is the alge- braic sum of the instantaneous electromotive forces. Therefore, the continuous voltage of a synchronous converter is equal to the i66 ESSENTIALS OF ELECTRICAL ENGINEERING maximum alternating-electromotive force induced in the coils con : nected in series between adjacent continuous-current brushes. Let an armature have twelve coils symmetrically placed, as indi- cated in Fig. 140. The maximum alternating electromotive forces induced in the coils, and their phase relations, are represented by a twelve-sided polygon, and the maximum alternating voltage be- tween continuous-current brushes is equal to the geometric sum of the maximum electromotive forces induced in any six coils, i.e., to the diameter of the circumscribed circle. Fig. 141. The sum of the instantaneous electromotive forces induced in the coils between continuous-current brushes is also the diameter of the circumscribed circle. If the armature winding is connected to two rings, the maxi- mum alternating voltage between rings is equal to the voltage be- tween the continuous-current brushes, and the effective alternating voltage is equal to the continuous voltage divided by the square root of 2. Fig. 141. FIG. 140. Single-phase Rotary Converter. Six King = (2) (3) FourR'mq < Similarly, if the armature is provided with three rings, the maximum alternating voltage between rings is the geometric sum of the maximum voltages induced in four coils, or the side of the inscribed triangle; if four rings, the geometric sum of the maximum voltages induced in three coils, or the side of the in- scribed square; if six rings, the geometric sum of the maximum voltages induced in two coils, or the side of the inscribed hexagon. Therefore, the effective alternating voltage between adjacent rings of an w-ring synchronous converter required to give a speci- fied continuous voltage E cc is 1 80 sin Eac == (4) CURRENT-RECTIFYING APPARATUS 167 It will be remembered that the phase voltage of a four-ring (two- phase) converter is the voltage between alternate rings. Hence, the phase voltage of a four-ring converter is equal to the phase voltage of a two-ring converter, the continuous voltage remaining constant. TABLE IV E ac for a 2-ring converter = 0.707 times the continuous voltage 3- =0.612 4- =0.500 6- =0.354 From the above considerations it is evident that the ratio be- tween the alternating and the continuous voltages of a synchronous converter is fixed, i.e., the voltage at the continuous-current brushes can be changed only by changing the voltage between the alter- nating-current rings. The theoretical ratios developed above are obtained, within very narrow limits, in the commercial converter. The practical considerations which affect the ratios are: (i) the resistance of the armature windings, (2) the shape of the alternat- ing electromotive-force wave. (1) Armature resistance. The theoretical ratios of voltages given above are changed by virtue of the resistance of the armature windings but this resistance is always small, and the ratios are never seriously affected. (2) Wave shape. When the electromotive force *wave is not harmonic, the effective value is not equal to the maximum divided by the square root of 2. For example, for, a flat-topped wave the effective value is greater than for a sine wave having the same maximum, while for a peaked wave, the effective value is less. In a well-designed converter, the wave shape is not seriously distorted. ii. Current ratios. The ratio of the current in the alternating- current circuit to that in the continuous-current circuit of a con- verter is easily calculated when the losses in the converter are neglected, i.e., when the output is considered equal to the input. In an w-ring two-pole armature, the alternating voltage between the neutral and any ring is -% (5) 2 V 2 168 ESSENTIALS OF ELECTRICAL ENGINEERING and *^ = EJ CCJ (6) 2 V 2 when the power factor of the converter is unity. Therefore, the line current and the current hi the armature conductors , , _ V2l cc IOC ' ~ wsin n 12. Starting. The preferable way to start a synchronous con- verter is as a continuous-current motor, or by means of a small starting motor, after which the alternating-current end is synchro- nized * with the system from which the converter is to be supplied. When no continuous-current energy is available, the converter may be started as an induction motor, as described for the synchronous motor.f The same disturbances are caused in the alternating-cur- rent system as with the synchronous motor, and the polarity of the continuous-current brushes must be determined before connection is made to a storage battery, or other apparatus through which the current flows in a given direction. The polarity of the continuous- current brushes is fixed by the last pulsation of the alternating cur- rent as the armature pulls into step with the supply circuit. 13. Compounding. Since an increased field excitation of a converter, as of a synchronous motor, is neutralized by an increased wattless component of armature current, compounding of a synchronous converter is not possible, in the sense that the term is used in connection with continuous-current dynamos. The only way to increase the continuous voltage of a converter is to increase the alternating voltage applied at the rings. A series or com- pound field winding may be made to increase the continuous voltage, provided the alternating-supply circuit is inductive. If the alternating-current circuit supplying a converter is inductive, the inductive drop in the line makes the voltage at the rings less than it would be if the circuit were non-inductive. * See Chapter 8, Section 17. t See Chapter 9, Section 4. CURRENT-RECTIFYING APPARATUS 169 When a synchronous motor is over excited, a leading current flows in the alternating-current system,* and the circuit is equivalent to a series circuit containing resistance, inductance and capacitance. The inductive line drop is thus neutralized, partially or completely, and the voltage at the rings of a converter, and consequently, the continuous voltage, is increased. Because its average power factor is inherently low, the operation of a compound-wound converter is unsatisfactory. 14. The regulating (or split) pole converter. For automati- cally increasing the continuous-voltage as the load on the con- verter increases, the regulating pole (the so-called "split" pole) converter has been designed. In this type of converter, the field pole is divided into two parts, each part has its own winding, and may be excited to any required degree independently of the other part. If the regulating pole is unexcited, the converter operates in the usual manner and the continuous voltage of a two-ring con- verter is equal to the maximum in- duced alternating : electromotive force. Let Fig. 142 represent a two-ring converter having two sets of field poles, the smaller (regulating) poles being placed midway between the larger (main) poles. When the con- verter is in operation, the maximum C. counter-electromotive force (OA , Fig. 1 43 a) induced in the armature wind- FIG. 142. The Regulating Pole Converter. ings is equal, neglecting resistance drop, to the maximum applied electromotive force. But the counter-electromotive force is the resultant of two quadrature electromotive forces, as shown in Fig. Cb) FIG. 143. 1 43 a, one induced in the winding as it passes under the main poles, the other induced in the winding as it passes under the regulating poles, i.e., this construction is electrically equivalent to two arma- * See Chapter 9, Section 9. 1 70 ESSENTIALS OF ELECTRICAL ENGINEERING ture windings in series, the induced electromotive forces in- which are 90 degrees out of phase. The continuous voltage of a synchronous converter is the alge- braic sum of the instantaneous electromotive forces induced in the armature conductors connected in series between adjacent brushes (Section 10). The sum of the instantaneous electromotive forces induced in the armature winding in cutting the flux of the main pole is proportional to OB (Fig. 143 a), and the sum of the instan- taneous electromotive forces induced in the armature winding in cutting the flux of the regulating pole is proportional to OC (Fig. i43a). If the main and the regulating poles between brushes are both north or both south, the continuous voltage is proportional to the arithmetical sum of the lines OB and OC; if the main and the regulating poles between brushes are not of the same polarity, the continuous voltage is proportional to the arithmetical difference of the lines OB and OC. In practice, the regulating poles are not placed midway between the main poles. Under this construction, the phase angle between the components of the counter-electromotive force is less than 90 degrees, as shown in Fig. Msb. From the above considerations, it is evident that the continuous voltage of a regulating pole converter is changed by varying the field excitation. Converters of this type are in practical operation where the maximum continuous voltage is 150 per cent of the minimum value, the alternating voltage remaining constant. During the development of the regulating pole converter, fears were expressed that its alternating electromotive-force wave would be so badly distorted as to make its use undesirable. These fears have been proven entirely groundless. 15. The Booster-Converter. The booster-converter consists of an alternating-current generator and a synchronous converter, the armatures of which are connected in series. By means of the field rhetostat and a field reversing switch, the alternating voltage supplied to the converter armature may be either increased or decreased, and the direct voltage varied over a range propor- tional to twice the alternating electromotive force induced in the generator armature. 1 6. Hunting. A synchronous converter, because of the light weight of its rotating parts, is more susceptible to oscillatory dis- CURRENT-RECTIFYING APPARATUS 171 turbances than is a synchronous motor. The same things tend to cause hunting in a converter as in a motor, and the same methods are used to prevent oscillations, or to damp them out when once started. 17. Efficiency. The efficiency of a synchronous converter, which is usually higher than that of either a motor or a generator having the same rating, is easily determined by loading, since both input and output are electrical quantities which are easily and accurately measured. The approximate efficiency may be calculated from the rating, the input at no load, measured from either the continuous-current or the alternating-current end, and the resistance of the armature winding as measured between continuous-current brushes. Per cent efficiency = (E + l + w > fe) when E = the rated continuous voltage, Eb = brush-contact resistance drop.* / = the continuous- current output for which it is desired to calculate the efficiency, W = the no-load input, R a = the resistance of the armature winding as measured between continuous-current brushes, k = the ratio of the copper loss in the armature conductors to the armature copper loss when the converter is operated as a continuous-current generator with the same output (see Section 18). The value of k depends on the number of rings with which the converter is provided and may be taken from Table V. From the above, it is evident that the efficiency of a converter increases as the number of alternating-current rings is increased. Converters having more than six rings are seldom used, because the increased efficiency is not sufficient to justify the increased ex- pense and complications. The values of k given in Table V assume that the converter is operated at unity power factor. When the power factor is less than unity, the armature copper losses are increased (see Section 19), and the efficienty is decreased. * See Chapter 4, Section 2. 172 ESSENTIALS OF ELECTRICAL ENGINEERING TABLE V No. rings k 2 i-39 3 0.56 4 0-37 6 o. 26 8 0. 21 1 8. Armature reaction. It was shown in Chapter 4, Section 6, that the magnetomotive force set up by the current in the armature of a two-pole continuous-current generator is NI = av. cos NI ampere-turns. 2 7T do) The alternating current flowing in the coils of an w-ring two-pole converter, the continuous-current output of which is 7, is, from equation (8) , , , V~2l , . (12) and the ampere-turns per ring are NIJ -v/2 NI 2 n 9 . TT 2 n 2 sin - n Therefore, the maximum magnetomotive force per ring is NI av. cos . TT sin- n n (14) TTH ampere- turns. When the current and the electromotive force of an w-ring arma ture are in phase, the armature magnetomotive force is in quadra ture with the field flux,* and is equal to * See Chapter 8, Section 9. CURRENT-RECTIFYING APPARATUS 173 NI M n = ampere- turns,* (16) 2 7T i.e., when the continuous-current brushes of a polyphase converter are set midway between the pole tips, the armature reaction due to the continuous current is equal and opposite to that due to the power component of the alternating current. The armature reaction in a polyphase converter is, therefore, that due to the current required to supply the losses in the converter, and to the wattless component of current when the power factor is less than unity. From equation (15) it is evident that the maximum armature magnetomotive force of a two-ringf converter is NI M 2 = ampere-turns, (17) i.e,, the maximum value of the armature magnetomotive force of a two-ring converter, due to the power component of the alternating current, is twice that due to the continuous current. The armature magnetomotive force of a two-ring converter, therefore, varies NI NI from -f- - : to , the frequency of the variation being twice 2 7T 2 7T that of the alternating-current circuit.}; The shifting of the armature flux across the pole faces causes energy losses (hysteresis and eddy currents), and tends to cause sparking at the continuous-current brushes. Therefore, commuta- tion in a two-ring converter is inherently poor, and is not improved by an angular advance of the brushes. 19. Armature heating. Since a synchronous converter acts simultaneously as a continuous-current generator and a synchro- nous motor, or as a continuous-current motor and an alternating- current generator, the currents actually flowing in the armature conductors must be resultants of these two actions. The instan- taneous current flowing in an armature coil of a converter is, then, the algebraic sum of the constant continuous current and the in- stantaneous alternating current. * Equation (16) is the resultant of n magnetomotive forces each of which has the maximum value ampere-turns, and having a displacement of degrees in both space (direction) and time. f The two-ring or single-phase converter has a very limited commercial application. t See Chapter 8, Section 9. 174 ESSENTIALS OF ELECTRICAL ENGINEERING Let the position of the armature of a two-ring two-pole converter be such that the angle co/ is zero. Fig. 140. Assuming unity power factor, the current in the alternating-current system is maximum and its direction of flow opposite to that of the current flow in the continuous-current circuit. The current flowing in the armature windings is, therefore, the algebraic sum of the constant continuous current and the maximum alternating current. When the armature has advanced thirty degrees, the alternating current has decreased in value and the continuous current in coils a and g has reversed in direction, the coils having passed to the opposite sides of the brushes. The current flowing in coils a and g is, therefore, the arithmetical sum of the constant continuous current and the instantaneous value of the alternating current, and is greater than in the other coils. Similarly, when the armature has rotated through another thirty degrees, the direction of the continuous current in coils b and h has reversed, and the current in coils a, b, g and h is the arithmetical sum of the constant continuous current and the instantaneous alternating current. When the armature has rotated through ninety degrees, the value of the alter- nating current is zero and the current flowing in any coil is one-half that in the continuous-current (armature) circuit. During the next quarter revolution the alternating current increases in the opposite direction, and coils e, f, I and m carry the larger instantaneous current values. From the above it is evident that: (a) The armature coils of a two-ring'converter are not uniformly heated. (b) The heating decreases as the distance from the point of con- nection to the ring increases. The conditions shown to exist in a two-ring converter may be shown to exist in one having a larger number of rings, an increase in the number of rings making the heating more nearly uniform and decreasing the total quantity of heat liberated in the armature for a given value of load (continuous) current. Taking as the axis the line AB in Fig. 140, the instantaneous current in any armature coil of an w-ring two-pole converter is 2 I COS (eo/ =b a) I , , I = > (10) . 7T 2 wsm- n CURRENT-RECTIFYING APPARATUS 175 and . 2 = - r?d(<*) (19) 7T J (20) when 7 = the continuous-current output, n = the number of rings on the armature, a = the angular displacement of the coil from a position midway between the connections to two adjacent rings. But the heating of the coil due to the continuous current alone is 72 proportional to . Therefore, 4 l6cosa , \ 7T . 7T " j (2l) HIT sin - n } n I (22) when ^ ; = the ratio of the average rate at which heat is liberated in the coil of an w-ring converter to the rate at which heat is liber- ated in the same coil of the armature when operated as a continuous- current generator with the same current output. Equation (22) is independent of the number of poles for which the armature is wound, and is, therefore, applicable to multipolar machines. Integrating equation (22) with respect to a from a. = o to <* = - n (24) when k is the ratio of the average heat liberated in the armature winding of an w-ring converter to the heat liberated in the same 176 ESSENTIALS OF ELECTRICAL ENGINEERING armature when used as a continuous-current generator, with the same current output. 20. Converter ratings. From the above considerations it is evident that the current output (rating) of a given armature when used as a synchronous converter is to the current output (rating) when used as a continuous-current generator as i is to Vk (25) (26) 7T when KW = the rating as a continuous-current generator, K'W' = the rating as a synchronous converter. When the power factor of the converter is less than unity, the alternating current is increased for a given continuous-current output, the inequality of the heating of the armature is~ increased, the heating of the individual armature coils becomes more pro- nounced, and the wattless component of current tends to magnetize or to demagnetize the field as the current leads or lags behind the electromotive force. CHAPTER X PROBLEMS 1. The continuous voltage of a synchronous converter is 550. Find the alternating voltage (neglecting resistance drop) between adjacent rings when the converter is provided with: (a) v rings, (b) 3 rings, (c) 4 rings, (d) 6 rings. 2. Calculate the current in the alternating-current mains when the output of the converter in Problem i is 1000 amperes, the efficiency 95 per cent and the power factor 96 per cent. 3. The power factor of a 2-ring converter is unity and the continuous- current output 250 amperes. Determine the maximum (instantaneous) cur- rent flowing in a conductor: (a) adjacent to the connection from the armature winding to the ring, (b) mid-way between the connections to the rings. 4. Same as Problem 3 except the power factor of the alternating-current circuit is 0.866. 5. The continuous voltage of a 4oo-kw., 3-ring converter is 600, the re- sistance of its armature (measured between continuous-current brushes) is 0.027, and the no-load input when operated as a continuous-current motor without load (at rated speed and normal field excitation) is 12.5 kw. Calculate the efficiencies for 25, 50, 75, 100 and 125 per cent of rated output and plot the efficiency curve (using per cent of rated output as abscissas). CURRENT-RECTIFYING APPARATUS 177 6. The converter in Problem 5 is provided with 6 rings. Determine its rating, calculate the efficiencies, and plot the efficiency curve as in Problem 5. 7. A compound-wound 25-cycle, 3-ring converter has an inductance con- nected in each alternating-current supply line. When the input to the conver- ter is 100 kw., the power factor of the converter is unity and the voltage between continuous-current brushes is 600. When the input to the converter is 10 kw. the power factor of the converter is 0.5 and the voltage between continuous- current brushes is 550. Find: (a) the value of the inductance in the alternating- current lines, (b) the voltage of the alternating-current supply system. 8. The no-load continuous voltage of a split-pole converter is 550. The regulating pole is excited by means of a series winding, and the flux pro- duced by the regulating pole at full load is 30 per cent of the flux produced by the main winding which is constant. The distance from the center line of the main pole to the center line of the regulating pole is 45% of the distance between the center lines of adjacent main poles. Find the full-load continuous voltage, the applied electromotive force at the rings remaining constant. 9. A 6-pole, lap-wound armature has 720 conductors. The commutator has 360 bars. Determine the commutator bars which must be connected to a given ring in: (a) a 2-ring converter, (b) a 3-ring converter, (c) a 4-ring con- verter, ((/) a 6-ring converter. CHAPTER XI FIG. 144. Schematic Diagram of Transformer. THE TRANSFORMER A TRANSFORMER consists of a single magnetic circuit linking with two independent electric circuits, as shown schematically in Fig. 144. Since iron offers a path of small reluctance for the magnetic flux, the coils of commercial transformers are always wound on iron cores. i. Fundamental physical action. When the armature conductors of a con- tinuous-current motor move across the magnetic field, a counter-electromotive force which opposes the flow of current is induced in the conductors, and the armature rotates at a speed which establishes equilibrium in the system. Similarly, when an alternating current flows in the windings of coil P (Fig. 144) the changing value of the flux induces * in the coil an elec- tromotive force which opposes the flow of current, and the current is of a value which establishes equilibrium in the sys- tem. The induced electromotive force is equal to the geometrical difference of the applied electromotive force and the drop due to the resistance of the coil. The resistance drop in the primary windings of commercial transformers FlG * I45 operating without load (secondary cir- cuit open) is so small as to be negligible, and the applied and the counter-electromotive forces are, therefore, equal. Referring to Fig. 144, let m = the maximum flux produced in the iron core by coil P, f = the frequency of the alternating-current supply circuit, N p = the number of series turns in coil P, N t = the number of series turns in coil 5, * See Chapter 2, Section 13. 178 Transformer in Posi- tion on Pole. THE TRANSFORMER 179 The flux in the iron must decrease from its maximum m to zero in one-fourth of a cycle. The average rate of flux change is, there- fore, d) 4? and the average electromotive force induced in coil P is the prod- uct of the average rate of change in the flux and the number of series -turns in the coil. E av = 4mfN P io- 8 . (2) Since the ratio of the effective electromotive force to the average electromotive force is i.n, E p = 4.44 m fN P io- 8 . (3) Assuming that the flux set up by coil P is confined to the iron core, the electromotive force induced in coil S is E, = 444 m fN, io- 8 . (4) If the terminals of coil 5 are connected through a resistance as shown in the figure, current flows in the windings of the coil. The current in coil 5 tends to establish in the iron core a flux opposed to that set up by coil P * and the counter-electromotive force in- duced in coil P is reduced. The equilibrium of the system is thus destroyed, and the current in coil P increases until it neutralizes the magnetic effect of coil 5. As the current in coil P increases, the resistance drop is no longer negligible and equilibrium is reestablished with a counter- electromotive force less than the applied electromotive force. Equilibrium in a transformer is, then, maintained by an automatic change in the current flowing in the primary coil. The conditions are analogous to those in a shunt motor, the speed of which decreases as the load increases, so that the torque developed in the motor armature is automatically adjusted to equal the counter torque due to the load. 2. Magnetic leakage. In Section i it is assumed that all the flux set up by coil P passes through coil S. Such is not the case in a transformer since the flux is not confined to the iron.f That part * See Chapter 2, Section 14. t See Chapter 2, Section 21. l8o ESSENTIALS OF ELECTRICAL ENGINEERING of the flux which threads one coil but does not thread the other is termed " leakage " flux, and has the same effect as a series inductance, i.e., it reduces the electromotive force induced in coil S, and causes the primary current to lag behind the applied electromotive force. 3. Electromotive force relations. If the coils had no resistance and there were no magnetic leakage, it is evident, from equations (3) and (4), that the ratio of the voltage applied at the terminals of coil P to the voltage delivered at the terminals of coil S would be that of the number of turns in the coils. , , E a ~ 8 This ratio is found to hold very closely when the secondary circuit is unloaded. Because of resistance and magnetic leakage, the secondary terminal voltage decreases as the load increases, the applied voltage remaining constant. 4. Current relations. Neglecting the magnetizing current and the losses, which are small, the input to a transformer is equal to its output, and the product of the electromotive force and the current in the primary circuit, is equal to the product of the electromotive force and the current in the secondary circuit. E P T P = EJ 8 . (7) Therefore y* = f. (8) 1, tL v N. . i.e., the ratio of the primary and the secondary currents is equal to the inverse ratio of the number of series turns in the coils. 5. Vector diagram. The electromotive forces and currents in a transformer may be represented graphically by means of a vector diagram. Fig. i46a is the diagram of an unloaded transformer. The applied electromotive force is represented by OA, the secondary electromotive force by OB and the flux, 90 degrees behind OA and 90 degrees ahead of OB, by OC. Because of hysteresis in the iron core, the no-load current is not harmonic,* and cannot, * See Appendix B, Section 8. THE TRANSFORMER 181 therefore, be represented by a rotating vector. It is represented in transformer diagrams by what is termed the " equivalent harmonic current," i.e., a harmonic current having the same frequency and effective value as the actual current. The error due to this assump- - FIG. i46a. Vector Diagram of Unloaded Transformer. FIG. 1466. Vector Diagram of Loaded Transformer. tion is usually negligible. If the resistance of coil P were zero and there were no iron (hysteresis and eddy current) losses, the no-load current would be in phase with the flux. Because of these losses, which are practically all iron losses, there is a power component of the no-load current in phase with the applied electromotive force OA, and the vector of the no-load current is less than 90 degrees behind OA. That the electromotive force induced in the coils of a transformer is 90 degrees out of phase with the flux set up by the primary wind- ing is evident from the following consideration: Let the flux in the iron core (Fig. 144) vary harmonically. The electromotive force induced in coils P and 5 is proportional to the rate of change in the flux threading the coils.* But the rate of flux variation is maximum when the flux is passing through its zero value, f When the secondary circuit is loaded, the transformer quantities are represented in Fig. i46b. Let E p = the applied electromotive force, E a = the secondary (terminal) electromotive force, IP = the current in the primary circuit, I 8 = the current in the secondary circuit, * See Chapter 2, Section 13. t See Appendix A, Section 7. 182 ESSENTIALS OF ELECTRICAL ENGINEERING R p = the resistance of the primary circuit, R s = the resistance of the secondary circuit, X p = the reactance of the primary circuit, X s = the reactance of the secondary circuit, cos = the power factor of the primary circuit. Draw OA = E p , OD = Ip, (f> = AOD. The applied voltage E p is equal to the geometric sum of the resis- tance drop R p lp, the reactance drop X P I P and the counter-electro- motive force. Draw AG = Xplp perpendicular to the current vector OD, GE = RpI P parallel to the current vector. Then OE is the vector of the counter-electromotive force induced in the primary winding. . The load component of the primary current OD' is the geometric difference of OD and the no-load current. The secondary current I s is in phase opposition to the primary load current OD' . The electromotive force induced in the secondary winding is in phase opposition to OE. Draw OK = I s , OF = OE- s , N P FH = XJ S perpendicular to OK, BH = RJ 8 parallel to OK. OB is the secondary terminal voltage E s . 6. Types and construction of transformers. Commercial con- stant-potential transformers differ in the mechanical arrangement of the iron core and the windings, and are of two general types: (a) core, (6) shell. (a) Core type. In the core type transformer the windings are placed around the legs of the core and cover a large part of the iron. Fig. i4ya. I THE TRANSFORMER (b) Shell type. In the shell type transformer the windings are placed around the middle leg of the core as shown in Fig. 1470. Iron, evidently, encloses the greater part of the coils. There is no marked superiority of one type over the other, both types being largely used. Economical considerations usually FIG. i47a. Core Type Transformer (Case FIG. i47b. Shell Type Transformer (Case Removed). Maloney Electric Co. Removed). General Electric Co. result in the selection .of the shell type when the voltages are low and the currents relatively large, and of the core type when the voltages are high and the currents relatively small. Transformer cores are built up of thin stampings (laminations) to reduce eddy-current losses. After the core and the windings are assembled they are placed in an iron case for mechanical protection. 7. Transformer losses.* The losses in a transformer are: (a) iron losses, (b) copper losses. (a) Iron losses. The iron losses of a transformer are those due to: (i) hysteresis, (2) eddy-currents. (i) Hysteresis loss. The hysteresis loss in an iron core f is equal to P, = i,F/B. M , (10) * See Standardization Rules of the American Institute of Electrical Engineers, t See Appendix B, Section 7. 184 ESSENTIALS OF ELECTRICAL ENGINEERING when rj = the magnetic (hysteretic) constant of the iron used in the core, V = the volume of the iron, / = the frequency of the applied electromotive force, B m = the maximum flux density in the iron. Since, in any given transformer, 77, V and / are constant, and B m is proportional to the applied electromotive force, equation (10) may be written P = *". (n) (2) Eddy-current losses. The eddy-current loss in an iron core * (12) when b = a constant proportional to the reluctivity of the iron used in the core, V = the volume of the iron, / = the frequency of the applied electromotive force, / = the thickness of the laminations of which the core is built. B m = the maximum flux density in the iron. Equation (12) may be written p. = k e &. (i 3 ) Determination of total iron loss. The total iron loss in a trans- former is determined, for any applied voltage, by connecting the - - transformer as in Fig. 148, and impressing the required voltage across its terminals. The watt- _ . meter indicates the total iron FIG. 148. Connections for Transformer J oss pj us the copper loss due to Iron Losses. ^ no _ load curren t. The no-load copper loss is usually so small as to be negligible, but the correction is easily made after the resistance of the primary winding has been determined. Separation of hysteresis and eddy-current losses. The iron loss of a transformer is separated into its components by the following graphical method: From equations (10) and (12) * See Chapter 6, Section i. Supply THE TRANSFORMER 18* when B is constant, i.e., when the ratio is constant. Dividing equation (14) by/ -; = k h ' + kSf, (15) which is the equation of a straight line, and may be plotted as ab, Fig. 149. The ordinate oc is the value of the constant k h '. k e f 10 20 30 40 50 60 70 FREQUENCY (f ) FIG. 149. Separation of Hysteresis and Eddy-current Losses. The power lost in hysteresis at any frequency is obtained by multi- plying the ordinate oc = k h f by the frequency; and the power lost in eddy currents is obtained by multiplying the ordinate k e 'f, corre- sponding to the required frequency, by the frequency.* (b) Copper losses. The copper losses in a transformer are those due to the resistance of the windings, and are equal to the product of the resistance and the square of the current. PC = R P I P 2 + RJ*, (16) * Compare with the separation of hysteresis and eddy-current 1 current machinery as outlined in Chapter 6, Section 4. in continuous- l86 ESSENTIALS OF ELECTRICAL ENGINEERING when P c = the total copper loss in the transformer, R p = the resistance of the primary winding, R 8 = the resistance of the secondary winding, IP = the primary current, I 8 = the secondary current. The resistance of a transformer winding is determined by con- necting it, in series with a suitable rheostat, to a source of contin- uous current, and measuring the current in the circuit and the voltage between the terminals of the winding. The resistance of the winding is, from Ohm's Law, Direct measurement of the copper losses may be made by con- necting a transformer as indicated in Fig. 150, where the low- voltage winding is short-circuited and the high-voltage winding connected to a source of alternat- ing current, and the voltage be- tween the terminals of the winding regulated to give any required FIG. 150. Connections for Transformer Current in the windings. The volt- Copper Losses. age required to produce full-load current in the windings of a short-circuited transformer is usually only a few per cent of the rated voltage of the coil. The indication of the wattmeter is the sum of the total copper loss in the trans- former and the iron loss in the core. Since the iron loss in a transformer operating at rated voltage, is seldom more than 3 per cent of its rated output, the iron loss at the reduced voltage required for the short-circuit test is negligible, i.e., the indication of the wattmeter is the total copper loss in the windings. 8. Equivalent resistance. The equivalent resistance of a transformer is the resistance which, when multiplied by the square of the current in the circuit, gives the total copper loss of the trans- former. It will, evidently, have different values as the current used is that of the primary or of the secondary circuit. Let R e = the equivalent resistance of a transformer, R p = the resistance of the primary winding, R a = the resistance of the secondary winding, IP = the primary current, THE TRANSFORMER 187 I s = the secondary current, P p = the copper losses in the primary winding, P s = the copper losses in the secondary winding, P c = the total copper losses in the transformer. Then P c = Pp + P* (18) + RJ*. (19) From equation (9) ''-' (20) and /.-/, () Therefore, From equation (22) ' (24) when referred to the primary circuit. From equation (23) -Y + R., (25) when referred to the secondary circuit. 9. Equivalent reactance. The equivalent reactance of a trans- former is the ratio of the voltage drop due to the inductance of the windings and to leakage flux, and the current flowing in the circuit. It is determined from the short-circuit test and the equivalent resistance of the windings. when X e = the equivalent reactance of the transformer, referred to the primary or to the secondary circuit as the values of E, I and R are primary or secondary values, l88 ESSENTIALS OF ELECTRICAL ENGINEERING E = the value of the primary, or the secondary, electro- motive force required to cause current / to flow in the primary circuit, or in the short-circuited windings. / = the current flowing in the primary circuit, or in the short-circuited windings. R e = the equivalent primary, or secondary, resistance of the windings. 10. Cooling. The cases of small transformers are filled with oil which helps dissipate the heat liberated in the windings and in the core. In large transformers additional means of cooling are required, the heat being dissipated by means of an air blast, or by means of a system of pipes through which cold water is forced, and around which the oil circulates. It is cheaper to cool large transformers by mechanical means than to build them of such "shape and dimensions that they are self-cooling. 11. Efficiency. The efficiency of a properly designed trans- former is high, and varies from 95 per cent in small transformers to above 99 per cent in those of large size. If the applied electromo- tive force is constant, the iron loss is approximately constant, the copper loss is proportional to the square of the load and the effi- ciency may be calculated from the following equation: Per cent efficiency = - ~~JT^J* ' ( 2 ?) E 8 I S cos + P + R e l 8 when E a = the rated secondary voltage, I 8 = the current output at which it is desired to calculate the efficiency, P = the no-load input (= iron losses), R e = the equivalent secondary resistance of the windings, cos < = the power factor of the load circuit. Transformer efficiencies are usually calculated for unity power factor, under which condition cos in equation (27) is equal to i. Example. Find the full-load efficiency of a 2200: 2 20- volt, 15- kv-a. transformer, the no-load input to which is 200 watts, R P 2 ohms, R s = 0.02 ohm. THE TRANSFORMER 189 E a 220 volts, 15,000 = 6g x eres 220 P = 200 watts, R e = 0.02 -f- 2 X - - = 0.04 ohm, 100 RJ* = 0.04 X (68.i) 2 = 185.5 watts > COS (j> = I . ~ . 15,000 X 100 Per cent efficiency = 15,000 -f- 200 + 185.5 I5* 000 ~ 15,385-5 = 97-5- The "all-day efficiency" of a transformer is the ratio of the total output during twenty-four hours to the total input during the same period. ~, . output during 24 hours / ox All-day efficiency = - (28) input during 24 hours output during 24 hours output during 24 hours + losses Example. Calculate the all-day efficiency of the above trans- former, when operated at full load for 8 hours each day. rr 15,000 X8 X 100 Per cent efficiency = - 15,000 X 8 -f 200 X 24 + 185.5 X 8 _ 120,000 126,248 = 95- 12. Regulation. The regulation of a transformer is the ratio of the increase e in the secondary terminal voltage, between rated load and no load, and the secondary voltage E 8 at rated load, the primary voltage remaining constant. Per cent regulation ^-^ t -= (30) The secondary terminal voltage decreases as the load increases because of: (a) the resistance of the windings, (b) the reactance of 190 ESSENTIALS OF ELECTRICAL ENGINEERING the windings. Since the vector sum of the secondary terminal voltage, the resistance drop and the reactance drop is constant, the regulation of a transformer becomes larger (poorer) as the power factor of an inductive load circuit decreases. The resistance and the reactance of a transformer are determined as explained in Sections 8 and 9, and the regulation calculated by means of the following equation: Per cent regulation V \L S CQS r , , , T . ---, a ^, ~jg -XlOO, ^i; when E s = the full-load (rated) secondary voltage, I 8 = the full-load (rated) secondary current, R e = the equivalent secondary resistance, X e = the equivalent secondary reactance, cos = the power of the load circuit. . Example. Determine the regulation of the transformer in Section n, if no volts must be applied to the 2 200- volt winding to cause rated current to flow in the short-circuited secondary winding. E 5 = 220 volts, I 8 = 68.1 amperes, R e = 0.04 ohms, //H\2 X e = Y ( j + (o.o4) 2 = 0.16 ohms, cos = i, RJ 8 = 0.04 X 68.1 = 2.7 volts, X e I 8 = 0.16 X 68.1 = 10.9 ohms. -n i -L- V(22O + 2.7) 2 + (lO.o) 2 22O Per cent regulation = * - X 100 220 22O = 1.4. 13. The auto-transformer. When a single continuous winding forms both the primary and the secondary of a transformer THE TRANSFORMER the arrangement is termed an auto- transformer. Fig. 151. The auto-transformer, like the independent coil transformer, is rever- sible and may be used to increase or to decrease the applied voltage. The ordinary (two-winding) transformer may be used as an auto-transformer by interconnecting the windings as indicated in Fig. 152. Let 1 = 220 volts, N t = 10, 10 amperes. FIG. 151. Conventional Dia- gram of Auto-Transformer. When the coils are interconnected as indicated in Fig. i52a, the voltage of the load circuit is the arithmetical sum of the applied voltage and the electromotive force induced in the secondary winding; when connected as indi- cated in Fig. i52b, the load voltage is the arithmetical difference of the applied voltage and the electromo- tive force induced in the secondary winding. The above statements are evident from the fact that the wind- ings may be interconnected so that the applied and the secondary (in- duced) electromotive forces are either in phase or in phase opposition. The ratio of the currents in the windings of an auto-transformer is FIG. 152. The Auto-Transformer. the same as if the transformer were operating in the usual manner, i.e., the currents in the windings are inversely proportional to the number of turns in the windings, while the current in the supply mains is the algebraic sum of the currents in the two windings. Assuming the load circuit to be non-inductive, the power delivered by the transformer when con- nected as indicated in Fig. i52a, is P = 242 X 10 = 2420 watts. The same power, neglecting the losses, is supplied to the 19 2 ESSENTIALS OF ELECTRICAL ENGINEERING transformer, but the voltage of the primary circuit is only 220. Therefore, 22O = 11 amperes, and the current flowing in the primary winding is i ampere. Similarly, for the connection indicated in Fig. 1520, the power delivered is P = 198 X 10 = 1980 watts ii 220 = 9 amperes, and the current flowing in the primary winding is i ampere. The copper losses in an auto-transformer are less than in a two- circuit transformer having the same output and the same primary and secondary voltages, but the interconnection of the windings makes its use inadvisable except where the difference between the supply voltage and that of the load circuit is small. One of its principal uses is to reduce the voltage applied to an alternating- current motor during the period of acceleration. 14. The constant-current transformer. A transformer having a counterbalanced movable coil delivers approximately constant current to a load circuit of variable impedance, when the voltage between the terminals of the primary winding is constant. When an alternating current flows in coil P the magnetic flux set up divides between the iron and the air as indicated in Fig. 153. That part of the flux which threads coil S induces in it an electro- motive force which causes a current to flow in the closed secondary circuit. The flux which passes through the air forms a magnetic field, the direction of which is at right angles to the current-carry- ing conductors of coil 5. Since the current-carrying conductors of coil S lie in a magnetic field, the coil tends to move and the direction of its motion is upward. The magnetic circuit is designed for large magnetic leakage, so that the number of magnetic lines threading coil 5 is materially reduced as the coil moves upward. Both the induced electromotive force and the current flowing in the coil are reduced, and the force producing, or tending to produce, motion becomes less. THE TRANSFORMER 193 FIG. 153. The Constant-current Transformer. By partially counterbalancing coil S as shown in Fig. 153, the secondary comes to rest in that position which establishes equilib- rium in the system, i.e., coil S moves upward until the reaction of the magnetic field and the current- carrying conductors is just suffi- cient to neutralize the effect of the unbalanced part of the coil. If the impedance of the load circuit is increased, the current in coil 5 decreases, the upward force acting on the coil decreases, and the coil moves downward until equilibrium is restored, i.e., until the flux thread- ing the secondary coil increases sufficiently to restore the original value of current flowing in the secondary circuit. If the impedance of the load circuit is reduced, the current in the secondary circuit increases, increasing the up- ward force on the coil, and the coil rises until equilibrium is re- established. As stated in Section 2, leakage flux has the same effect as a series inductance, i.e., it causes the current to lag behind the applied electromotive force. The power factor of a constant-current transformer is, therefore, low when the magnetic leakage is large, i.e., when the transformer is operating at partial loads. In fact, the primary current of a constant-current transformer is approximately constant, and the change in load is accomplished by a change in the power factor of the primary circuit. It is, there- fore, highly undesirable that these transformers operate at loads much below their rated output. To make the operation of constant-current transformers at partial loads commercially feasible, taps are often provided on the secondary coil, by means of which the number of current-carrying conductors in the coil may be changed. The relative position of the coil for a given load and, consequently, the leakage flux, may thus be changed, and the power factor of the system maintained at approximately full-load value.* * The average power factor of constant-current arc lamp circuits is about 70 per cent when the transformers operate at full load. 194 ESSENTIALS OF ELECTRICAL ENGINEERING The constant-current transformer is largely used to supply current to arc and other lamps connected in series. In large transformers the coils may be divided, and the secondary (movable) coils balanced against each other. With divided coils the size of the auxiliary weight W is re- duced to a minimum, that necessary to give the required range of current. 15. Polyphase transform- ers. Polyphase transformers are constructed by interlink- ing the magnetic circuits of two or more transformers which are to be connected to a polyphase system. If the flux set up by coil A is in FIG. 154. Constant-current Transformer (Case quadrature with that set up Removed). General Electric Co. by coil B, the weight of the iron cores is approximately 15 per cent less than the weight of the cores of two equivalent single-phase transformers, the maximum flux density being the same in each construction. Polyphase trans- formers are extensively used in Europe but have not attained great popularity in America. CHAPTER XI PROBLEMS 1. The iron losses (measured at no-load) in a 5-kv-a. 22oo:22o-volt trans- former are 75 watts. The resistance of the primary winding is 7.5 ohms; that of the secondary winding is 0.07 ohm. Calculate the full-load efficiency: (a) when the power factor of the load circuit is unity, (&) when the power factor of the load circuit is 0.85. 2. The transformer in Problem i is operated during 24 hours as follows: 4 hours at full load and unity power factor, 6 hours at half load and 90% power factor, 14 hours without load. Calculate: (a) the all-day efficiency, (ft) the yearly loss charge against the transformer if energy is produced for i cent per kw.-hour. 3. A 20: i -transformer, the regulation of which is 2 per cent, has a terminal voltage of 115 at full load. Determine the primary voltage at full load. THE TRANSFORMER 195 4. The hysteresis loss in a transformer is 125 watts when operated on do- cycle mains. Calculate the hysteresis loss when the transformer is operated on a 25-cycle circuit, the voltage of which is equal to that of the oo-cycle circuit. Note. From equation (3) f 5. The eddy-current loss in a transformer is 100 watts when operated on 60- cycle mains. Calculate the eddy-current loss when the transformer is operated on a 25-cycle circuit, the voltage of which is equal to that of the oo-cycle circuit. 6. 20 per cent of the iron loss in a 6o-cycle transformer, when operated at rated frequency and voltage, is due to eddy currents. Calculate the per- centage increase or decrease in the total iron loss when the transformer is operated at 25 cycles and rated voltage. 7. The hysteresis loss in a oo-cycle, 2300- volt transformer is 500 watts. Calculate the hysteresis loss when operated on a 6o-cycle, ii5o-volt circuit. 8. The eddy-current loss in a 6o-cycle, 23oo-volt transformer is 400 watts. Calculate the eddy-current loss when operated on a 6o-cycle, 1150- volt circuit. 9. Two transformers have the same volume of iron of the same quality, but are so wound that the ratio of the flux densities is 2 13. Calculate the ratios of the iron losses. 10. Calculate the equivalent resistance of the transformer in Problem i : (a) referred to the primary circuit, (6) referred to the secondary circuit. 11. Compare the iron losses in three similar transformers star-connected to a balanced 3 -phase system, with the iron losses in the same transformers delta- connected to the same system. 12. Compare the copper losses in three similar transformers connected as in Problem n, the line current in the star-connected system being the same as the line current in the delta-connected system. 13. An auto-transformer has a ratio of 2:3. Determine the relative size of the conductors to be used in the windings, the current densities to be equal. 14. The iron losses in a i5-kv-a. 22oo:22o-volt transformer are 135 watts (measured at no load). Resistance of primary winding = 2 ohms; resistance of secondary winding = 0.018 ohm. Find: (a) the copper losses at full load, (6) the efficiency at full (rated) load and 90 per cent power factor, (c) the copper losses when the efficiency is maximum. 15. The maximum efficiency of a i2.5-kv-a. 22oo:22o-volt transformer is 97.5 per cent, and occurs when the output is 80 per cent of the rated value. Find: (a) the copper losses, (b~) the iron losses, (c) the equivalent resistance re- ferred to the 2 200- volt circuit, (d) the equivalent resistance referred to the 220- volt circuit. 16. A 22oo:22o-volt transformer is rated at 150 kv-a. and has a no-load input of 1.2 kw. Primary resistance = 0.16 ohm; secondary resistance 0.0018 ohm. Find: (a) the total copper loss at full load, (b) the efficiency at full load, (c) the output at maximum efficiency. CHAPTER XII TRANSFORMER CONNECTIONS 1. Protection. In connecting transformers to a supply circuit they should be protected from excessive currents by means of fuses of proper size in the primary circuits. A fuse is a short piece of soft wire which melts and opens the circuit whenever an excessive current flows in the circuit. Fuse wire is rated in amperes carrying capacity, the capacity varying, approximately, as the cross-sectional area. 2. For single-phase circuits. Connections for single-phase cir- cuits are shown in Fig. 155, where P is the primary winding and FIG. i55a. Primary and Secondary Coils FIG. I55b. Secondary Coils in Parallel, in Parallel. FIG. I55C. Primary Coils in Parallel. FIG. issd. Primary and Secondary Coils in Series. Single-phase Transformer Connections. 5 is the secondary winding. In Figs. i55a and i55b, the halves of the secondary winding are connected in parallel, and the volt- age of the load circuit is that of each coil. In making a parallel connection care must be taken that similar terminals of the two coils are connected together, or a short circuit may be formed. In Figs. i55c and i55d the secondary coils are connected in series and the voltage of the load circuit is twice that of Figs. i55a and i55b, the primary voltage being constant. It is impossible, in the series 196 TRANSFORMER CONNECTIONS 197 connection, to form a short circuit but if the coils are improperly connected their voltages oppose and neutralize each other, and the voltage of the load circuit is zero. Since, for a given load, the product of the current and the elec- tromotive force is constant, it is evident that the secondary current in the parallel connections (Figs. i$5a and 1550) is twice that in the series connections (Figs. 1550 and i55d). A three-wire system is formed by connecting a third wire to the junction of the series-connected coils, as indicated in Figs. i55c and i55d. The voltage between A and B and that between B and C is one-half that between A and C. This connection offers two avail- able voltages and is much used, the load being connected between any two of the three wires. If the load connected between A and B is not equal to that connected between B and C, the halves of the transformer are unequally loaded. With a properly-distributed load, the current in line B is materially less than that in either A or C. For this reason B is usually made smaller than A and C. Two or more transformers may be connected so that they supply the same load circuit. In making such connections the same conditions must exist as for the parallel operation of alterna- tors. Since the primary coils are connected to the same supply circuit, the frequencies of the secondary circuits are the same, but the voltages may be in phase or in phase opposition. If the secondaries are connected so that their electromotive forces are not in phase opposition, a short circuit is formed and an excessive current flows in the windings. This large current heats the trans- formers to such a degree as to seriously damage or destroy the insulation, unless the protective devices work promptly. Transformers having different characteristics should not be operated in parallel as they will not divide the load properly. For example, if two transformers of the same rating and no load voltage, but having regulations of 2 and 5 per cent, are operated in parallel, the one having the better (smaller) regulation will carry the greater part of the load. When the total load is equal to their combined ratings both transformers operate at a reduced efficiency, one being overloaded and the other underloaded. The wave shapes, or the ratios of transformation of different transformers, may be such as to make their parallel operation impossible. 198 ESSENTIALS OF ELECTRICAL ENGINEERING 3. For two-phase circuits. Transformers may be operated on each phase of a two-phase system, all the connections given for single-phase circuits being available. In addition, the primaries, or the secondaries, or both, may be interconnected as explained for the polyphase system in Chapter 7. 4. For three-phase circuits. Single-phase transformers may be connected between any two lines of a three-phase system, or between any line and the neutral FIG. 1^6. Delta-delta Connection. r , of a star-connected system, and operated as from a single-phase circuit. The primaries of three single-phase transformers may be connected to a three-phase supply system in either star or delta; the secondaries of the transformers may be connected to the load circuit in either star or delta. The voltage and current relations are those given for the polyphase system in Chapter 7. FIG. i57a. Correct Delta Connections. FIG. 1570. Incorrect Delta Connections. Instantaneous Voltages Balanced. Instantaneous Voltages Unbalanced. In making three-phase transformer connections the relations in the primary coils adjust themselves. If the secondaries are to be connected in delta, the voltage relations must be as indicated in Fig. i57a, i.e., the sum of the instantaneous electromotive forces around the delta must be zero. If any one of the coils is reversed the equilibrium of voltages is destroyed, and an excessive current flows around the delta. Fig. isyb. If the secondaries are to be star-connected, FIG. 158. Star-star , ..... ., , , ., M j Connection a snort circuit is impossible because the coils do not form a closed circuit, but the voltages be- tween lines may be unequal. The correct relations are indicated in Fig. i5Qa, and the voltage between any two lines is equal to that between any other two lines. If the terminals of any one of the coils is reversed, the relations shown in Fig. TRANSFORMER CONNECTIONS 199 exist and are indicated by the fact that the voltage between two of the lines is correct, while that between either of these and the third line is equal to the voltage in the coil, i.e., to the voltage between line and neutral. 60' njr/ N c FIG. i5Qa. Correct Star Connections. FIG. i5gb. Incorrect Star Connections. AB = AC = BC. AB AL> = LJJ = -=. V or open-delta connection. If one transformer of a delta connec- tion is omitted, a three-phase system may be operated with only two transformers. This is the V or open-delta connection. Fig. 161. While the number of transformers required for this connection is FIG. 1 60. Delta-star (or Star-delta) Connection. FIG. 161. Open-delta (or V) Connection. oooooooooooo P OGOGOOOOO less than in the delta connection, its use is not advised, except as a temporary expedient for the continuation of service, because of the phase relations of the current and the electromotive force in the windings of the transformers.* 5. Phase transformation. It is some- times desirable to change two-phase cur- rent to three-phase, or three-phase to two-phase. This may be accomplished by the "T" or Scott connection in which two transformers are connected as indi- cated in Fig. 162, i.e., the two-phase coils are independent of each other while one terminal of one three- phase coil is connected to the middle point of the other three- phase coil. The terminals A, B and C are connected to the three-phase lines. * See Chapter 7, Section 30. FIG. 162. Scott (orT) Connection. 200 ESSENTIALS OF ELECTRICAL ENGINEERING The voltage of the coil AC is that between the lines to which it is connected. The voltage of the coil BD is a component AC of the voltage AB (or BC), the other component being - - ( = AD 2 = DC). Fig. 163. B Since ABC is, by construction, an equilateral triangle and AD = DC, ADB is a right triangle and V(ABy - (ADY to 0.866 (AB). (2) FIG. 163. Vector Diagram of T- The voltage between the terminals of BD is, connection. therefore, 86.6 per cent of the voltage between the terminals of AC, and two similar transformers do not induce equal electromotive forces in the two-phase windings when the transformers are supplied from a balanced three-phase system. The voltages induced in the two-phase windings are equal if the ratio of the number of turns in BD to the number of turns in AC is made equal to 0.866. No. turns in coil BD _ , . , No. turns in coil AC 6. For synchronous converters. Transformer connections for two-, three- and four-ring converters differ in no way from those connections already described. Because of the increased output of a given converter armature when provided with six rings, six- ring converters are desirable. For supplying current to six-ring converters, the primary windings of transformers may be con- nected either star or delta to three-phase mains, but the secondary windings must be connected so as to produce six-phase currents. These connections are: (a) diametral, (6) double delta, (c) double star, (d) hexagonal. (a) Diametral. In this connection the terminals of trans- former A are connected to rings i and 4 of the converter, the terminals of transformer B to rings 2 and 5, and the terminals of transformer C to rings 3 and 6. Fig. i64a. (b) Double delta. The windings of the transformers are con- nected so as to form two separate deltas, as indicated in Fig. i64b. The parallel sides of the deltas are formed by the coils of one trans- former. TRANSFORMER CONNECTIONS 2OI (c) Double star. The six coils of a transformer may be con- nected into a double star, as indicated in Fig. i64c. In this con- nection the coils of a given transformer are diametrically opposed. 000000000. .OOQOOOOOO poooooooo. 0000000(50 WOOWoW 000000000 A 9 c 4 2 ' 53 6 FIG. 1643,. Diametral. FIG. i64b. Double Delta. FIG. 1640. Double Star. FIG. 1640!. Hexagonal. (d) Hexagonal. The six coils may be connected as indicated in Fig. i64d, connection being made to the rings of the converter from the junction of each two coils. As in the double delta, par- allel sides of the hexagon are formed by the coils of one transformer. From the voltage relations given above it is a simple matter to make any of the above connections when three identical trans- formers are used. When the transformers are not similar, the relative direction of the electromotive forces in the coils must be determined. CHAPTER XII PROBLEMS 1. The primaries of three similar 20:1 transformers are star-connected to 23oo-volt mains. Find the voltage of the secondary circuit when the second- aries are: (a) star-connected, (6) delta-connected. 2. Same as Problem i except the primaries are delta-connected. 3. A 5 kv-a. transformer, the regulation of which is 2 per cent is operated in parallel with another 5 kv-a. transformer, the regulation of which is 4 per cent. When the total load on the circuit is equal to 10 kv-a. it is equally divided be- tween the transformers. Determine the load division when the total load is (a) 7.5 kv-a., (6) 5 kv-a., (c} 2.5 kv-a. Assume that the voltage characteristics are straight lines. 202 ESSENTIALS OF ELECTRICAL ENGINEERING 4. Three similar transformers have their primaries star-connected to balanced three-phase mains. The voltages between the terminals of the star-connected secondary windings are AB = 127, BC = 127, AC = 220. i State what the trouble is and how it may be remedied. 5. Two 100 kv-a. transformers are V-connected. Find the load on each transformer when the balanced load on the system is equal to 150 kw., and the power factor is: (a) unity, (b) 0.866. 6. The primaries of two similar 10:1 transformers are connected to 2300- volt 2-phase mains. The secondaries are T-connected. Find the voltages between the 3 -phase (secondary) terminals. 7. Draw a vector diagram showing the relations of the current and the elec- tromotive force in the 3 -phase coils of a T-connection when the power factor of the system is 0.866 and the load is balanced. 8. The primaries of three similar 10 : i transformers are delta-connected to 23oo-volt mains. Determine the voltage of a single-phase circuit when the three secondaries are connected in series. 9. A transmission line is connected to 66oo-volt generators through i : 10 step-up transformers delta-connected to the generators, and star-connected to the line. Find the line-to-line voltage. 10. A 3-phase line supplies an induction motor delivering 500 brake horse power through a bank of transformers. The primaries are star-connected; the secondaries are delta-connected. Ratio 60 : i. Voltage between lines at trans- former (primary) terminals = 45,000. Transformer efficiency = 98.7 per cent, motor efficiency = 92 per cent, power factor of motor = 87 per cent. Find: (a) the voltage impressed on the motor, (V) the current per line delivered to the motor, (c) the kw. supplied to the transformers. CHAPTER XIII THE INDUCTION MOTOR i. Construction. The essential parts of an induction motor are: (a) the stator, (b) the rotor. (a) The stator. The stator of an induction motor is the station- ary part, and its structure is essentially that of the armature of a rotating field alternator. A slotted core (Fig. 166) is built up of laminations (Fig. 167), and the stator conductors, which are connected to the supply mains, are placed in the slots so as to form a distributed winding. (b) The rotor. Induction motors are differentiated by the construction of the rotor, and are: (i) squirrel cage, (2) slip-ring. (i) Squirrel-cage rotor. A squirrel-cage rotor consists of cop- , . per bars or rods placed in slots on the surface of a cylindrical laminated iron core, the ends of the conductors being connected to cop- per rings, placed at each end of the core. Since the conductors are connected in parallel, the resistance of such a winding is small. A squirrel-cage rotor is shown in Fig. i68a. (2) Slip-ring rotor. A slip-ring or wound rotor is one in which distinct windings, similar to those on the stator, are intercon- nected, and terminals brought out to slip rings mounted on the 203 FIG. 165. Induction Motor with Wound Rotor. Triumph Electric Co. FIG. 166. Stator Core and Frame. Crocker- Wheeler Co. 2O4 ESSENTIALS OF ELECTRICAL ENGINEERING shaft. Through these slip rings the rotor winding is connected to an external rheostat, by means of which the resistance of the rotor circuit may be varied. Since the conductors of a wound rotor are connected in series, its resistance is much greater than that of a squirrel - cage rotor. Fig. i68b shows a wound rotor with three rings mounted on the shaft. The rheostat for regulating the resistance of the rotor circuit is sometimes placed in the rotor struc- ture of small motors. In this case no rings are required, but a rod, by means of which the resist- ance of the rotor circuit is changed, projects through the hollow shaft. 2 . Rotating flux and its produc- tion. It was shown in Chapter 8, Section 9, that the magneto- motive force due to the rotating armature of a two-phase alterna- tor is constant in value and fixed in direction. When the arma- FIG. 167. Stator Lamination. Crocker-Wheeler Co. FIG. i68a. Squirrel-cage Rotor. Triumph Electric Co. ture is stationary, the magneto- motive force of the armature winding is constant in value, but is momentarily changing in direc- tion, and the flux ro- tates at a speed pro- portional to the fre- quency of the current flowing in the arma- ture conductors. Consider two elec- tromagnets placed at FIG. i68b. Slip-ring Rotor. Crocker- Wheeler Co. right angles to each other as shown in Fig. 169, and excited from a two-phase alter- nating-current system. It is evident that the flux due to winding A is maximum when that due to winding B is zero. As flux A THE INDUCTION MOTOR 205 decreases, flux B increases. During the quarter cycle required for flux A to decrease to zero and for flux B to increase to maximum, a resultant field is produced, the instantaneous value and direc- tion of which is represented by the vector sum of the instanta- FIG. 169. FIG. 170. neous values of the quadrature fluxes.* The flux due to each winding of Fig. 169, and the resultant flux is represented in Fig. 170 when flux A is maximum, 30, 60 and 90 degrees later. It is evident that the maximum value of the resultant flux is con- stant and equal to the maximum flux produced by each winding separately, and that its rate of rotation is proportional to the frequency of the supply circuit. It may be shown in a similar manner that a rotating flux, the maximum value of which is 1.5 times the maximum flux set up by each phase winding, is produced by connecting properly designed coils to a three-phase supply system. The magnetic effect when the stator windings of a polyphase in- duction motor are connected to a circuit supplying polyphase alter- nating current, is essentially that described, the difference being due to the arrangement of the windings, which is purely a mechanical detail, and to the reaction of the currents set up in the conductors of the rotor. 3. Generator action. The rotating flux set up by the stator windings of an induction motor cuts across the conductors of the rotor, induces in them an electromotive force, and causes a current to flow in the rotor circuit. 4. Motor action. The rotating flux set up by the stator wind- ings reacts with the currents induced in the rotor conductors to produce a torque,f and the direction of the torque is the same as the direction of flux rotation. * The fluxes are in quadrature as regards both time and space (direction), f See Chapter 2, Section 14. 206 ESSENTIALS OF ELECTRICAL ENGINEERING In a two-phase motor, the direction of flux rotation is reversed by reversing the terminal connections of either phase winding; in a three-phase motor, the direction of flux rotation is reversed by reversing any two line connections. 5. Speed of rotor.*- -The torque produced by the reaction of the stator flux and the rotor currents causes the rotor to revolve. Because the rate at which the flux cuts the rotor conductors de- creases as the speed of the rotor increases, both the induced electromotive force and the rotor current decrease, and the rotor runs at a speed which establishes equilibrium in the system. If the load increases, the speed decreases until the reaction between the flux and the increased current produces the required torque; if the load decreases, the speed increases until the current decreases to a value which reestablishes equilibrium in the system. The speed of the rotor can never attain the speed of the rotating flux because at this speed (the synchronous speed) no current flows in the rotor windings, and no torque is exerted to compensate for the frictional and other losses of the motor. 6.' Slip of the rotor. The difference between the speed of the rotating flux and that of the rotor is termed the "slip" of the rotor. The slip is approximately proportional to the load over the normal working range of the motor. At overloads the slip increases faster than the load until maximum torque is reached after which both speed and torque decrease very rapidly. The slip of an induction motor is easily measured, unless it be- comes excessive, by the stroboscopic method. On the end of the shaft or the pulley, mark as many equally-spaced radial lines as there are pairs of poles on the motor, and illuminate these lines by means of an arc lamp connected to the circuit from which the motor re- ceives its current. When the motor is in operation, the radial lines' appear to rotate in a direction opposite to that of the rotor. The speed of this apparent rotation is proportional to the slip of the rotor. The stroboscopic method of slip measurement depends on the fact that the light from an alternating-current arc lamp is pulsa- ting. If the rotor moved at synchronous speed, the radial lines * The induction motor is, inherently, an approximately constant-speed machine. Large variations in speed are obtained only by a sacrifice of efficiency or of mechanical simplicity. THE INDUCTION MOTOR 207 would advance through the angular distance of one pole pitch for each pulsation of the light, and successive pulsations would show the lines in the same relative positions. But the angular advance of the lines is less than the angular pitch of the poles, and each successive pulsation of the light shows the . lines in a position slightly behind that which it occupied at the previous pulsation. Example. A four-pole induction motor is operated from 60- cycle supply mains. Determine the slip of its rotor when 127 radial lines pass through the field of vision in one minute, ^ - = 1.76 per cent. 60 X 60 X 2 Another simple method for the determination of slip is to connect a contact maker to the shaft of the motor so that it closes, once in each revolution of the rotor, the circuit of a voltmeter connected across the mains supplying the motor. The voltmeter pointer swings back and forth, the rate of the swing being proportional to the slip of the motor. If an electrodynamometer type of volt- meter is used, the rate at which the pointer swings is twice as great as if the voltmeter is of the permanent magnet type. 7. Torque. From the above it might be supposed that the maximum torque of an induction motor is exerted at zero speed, since both the induced electromotive force and the rotor currents are then maximum. But the frequency of the rotor currents is proportional to the slip of the rotor, and the rotor current, therefore, lags behind the induced electromotive force by a constantly increasing angle as the slip increases. The lagging current tends to set up a flux which is opposed to that set up by the stator windings.* When the slip becomes large, this demagnetizing action is excessive, and the flux decreases faster than the rotor current increases. Therefore the speed-torque curve of an induction motor is not a straight line, but has the general shape, for a rotor circuit of constant resist- ance, shown in Fig. 171. Starting slightly below synchronous speed (the speed of the ro- tating flux), the torque increases as the speed decreases until the point of maximum torque is reached. At speeds less than that at which maximum torque is developed, the torque decreases very * See Chapter 8, Section 9. 208 ESSENTIALS OF ELECTRICAL ENGINEERING rapidly. Consequently, when an induction motor is loaded beyond the point of maximum torque, it stops. An examination of the speed-torque curve of an induction motor shows that the characteristic, Over the working range of the motor, I 70 JO 10 50 100 150 200 250 300 PER CENT OF FULL LOAD(RATED)TORQUE FIG. 171. Speed-torque Characteristic for Squirrel-cage Induction Motor. is similar to that of a shunt (continuous-current) motor, i.e., the speed drops slightly as the load increases. When the applied volt- age is that at which the motor is rated, the maximum torque of an induction motor is usually from two to three times its rated load torque; its starting torque is from one and one-half to two times rated value, and the starting current is from five to six times that at rated load.* The relations between the slip and the torque of an induction motor may be derived as follows: Let E r = the electromotive force induced in the rotor circuit at zero speed, I r the rotor current, R r = the resistance of the rotor circuit, X r = the reactance of the rotor circuit at zero speed, s = the slip of the rotor expressed as a fraction of the synchronous speed, * The Standardization Rules of the American Institute of Electrical Engineers require that a motor intended for continuous service shall develop a maximum run- ning torque at least 75% greater than the normal torque at rated load. THE INDUCTION MOTOR 209 But and n' = the synchronous speed of the rotor in revolutions per minute, n" = the actual speed of the rotor in revolutions per minute, cos (f> = the power factor of the rotor circuit. Rotor input = E r l r cos <. (i) sE r 100 80 70 1, COS s?55^ VRS + s 2 x r 2 ' Rr ~- ^ ^^ \ . 'A ~ - ^ ^ * 2 M . 4 ^~ , V 1 -' T 01 x 2 M 1 1 1 1 1 1 1 1 1 1 1 f ^-^ * , '~ 1 V s^ x X * K ^ K^ \ ^ X, ^ ^ "^ 1 , Pi ^-> ^ ^ i^ s i^ ^ S^ * ^ ^ > x ^ \ ^ N ^ s >> 4> *** ix^ s k^ \^ X, *x ^ s s, ^ ^ N x k v s SN ^ s, \ s^ x, Xj ^ V S S, ^ k^ s V ' s^ s ^v Xj y 1 \ s, s f i \ Trt \yn s, f \ s. \ s / ^ / 7O s^ ' ^ S ^ \ k^ \ } *\ 1 10 s s j s / t ^ / \ ^ s s 1 / \ / / y (3) SO 100 \50 200 150 300 PERCENT OF FULL LOAD (RATED) TORQUE FIG. 172. Speed-torque Characteristics for Slip-ring Induction Motor. Resistance of Rotor Increased by Steps. Substituting in equation (i), Rotor input = Rotor output = watts, = sE r 2 R r (i - s) R 2 + s 2 X 2 Rr 2 + S 2 X 2 watts, Torque 33,000 sE 2 R r (i -s) n'(R 2 +s 2 X 2 ) r -^- foot-pounds. (4) (S) (6) (7) (8) (9) 210 ESSENTIALS OF ELECTRICAL ENGINEERING The above formulae disclose the following operating character- istics of the induction motor: (a) Maximum torque is developed when Rr = sX r (10) (b) Maximum torque and is independent of rotor resistance. (c) Starting torque 7-4Er 2 Rr ( x -n'(R* + X?Y is proportional to the rotor copper losses, and is maximum when Rr = X r . (13) (d) The copper losses in the rotor circuit are equal to the product of the rotor input and the slip. (e) At speeds near synchronism, the reactance of the rotor cir- cuit is negligible, the torque is directly proportional to the slip, inversely proportional to the resistance of the rotor circuit, and (/) At maximum torque, the power factor of the rotor circuit = 0.707. (15) (g) At constant speed, the torque is directly proportional to the square of the electromotive force induced in the rotor circuit at zero speed, and, therefore, to the square of the applied voltage. (ti) At constant torque, the slip of the rotor is inversely propor- tional to the square of the electromotive force induced in the rotor circuit at zero speed, and, therefore, to the square of the applied voltage. 8. Starting. The polyphase induction motor is self-starting when supplied with alternating current of the proper voltage, fre- quency, and number of phases. The squirrel-cage motor is usually started by supplying the stator with current at a voltage less than that at which the motor is rated, the line voltage being reduced by means of an auto-transformer or other step-down device. Fig. 173. After the rotor has attained a considerable speed, the line voltage is THE INDUCTION MOTOR 211 ^Compensator Auto-transformer) applied and the starting device automatically disconnected from the line. Starting at a reduced voltage minimizes the line disturbances and reduces the heating of the motor windings, but decreases the starting torque. The slip-ring induction motor is started by applying rated volt- age to the stator windings, after resistance has been inserted in the rotor circuit. As the rotor speeds up, the resistance of the rotor circuit is gradually reduced until the windings are short-circuited. Because of their large starting currents and low power factors, induction motors often cause undesirable fluctuations of the line voltage dur- ing the starting period. These fluctuations are particularly objectionable when lamps and motors are operated in parallel. 9. Power factor. Since the air gap of an induction motor introduces considerable reluc- tance into the magnetic circuit, the mag- netizing current, as compared to the load FlG< I73 ' Com P ensat <* . . . . Connections, current, is relatively large at small percentages of the rated load. The power factor of an induction motor is, therefore, low at no-load but increases as the load increases. The frequency of the currents induced in the rotor circuit is pro- portional to the slip of the rotor. The reactance of the rotor circuit is, then, proportional to the slip of the rotor. "V f fT / /-\ A r =2 irSjLy (lO) when XT' = the reactance of the rotor circuit, at slip s, s = the slip of the rotor expressed as a fraction of the synchronous speed, / = the frequency of the supply circuit, L = the inductance of the rotor circuit. When the slip is small the reactance of the rotor circuit is neg- ligible, the impedance of the circuit is practically equal to its resistance, and the rotor current is in phase with the induced elec- tromotive force. As the slip increases, the reactance of the rotor circuit increases, the impedance becomes greater than the resistance, and the current lags behind the induced electromotive force. The power factor of an induction motor will increase as long as 212 ESSENTIALS OF ELECTRICAL ENGINEERING the power component of the induced voltage increases faster than the reactive component. When the slip becomes so great that the reactive component of the voltage increases faster than the power component, the power factor decreases. 10. The losses.* The losses in an induction motor are: (a) copper losses, (b) stray power. (a) Copper losses. The copper losses of an induction motor are those due to the resistance of the stator windings and that of the rotor circuit. The losses in the stator windings are easily calculated for any given value of stator current when the resistance of the windings is known. The resistance is determined by continuous- current methods. The copper losses in a slip-ring rotor are as easily determined as are those of the stator. Those in a squirrel-cage rotor cannot be determined directly, since neither the resistance of the circuit nor the value of the current flowing in it can be measured. The rotor copper loss should be calculated by means of the following formula: T g output X slip ,v Krlr " i - slip ' U7J Let E r = the electromotive force induced in the rotor circuit at zero speed, R r = the resistance of the rotor circuit, X r = the reactance of the rotor circuit at zero speed, s = the. slip of the rotor expressed as a fraction of the syn- chronous speed, 7 r = the current in the rotor circuit at slip s. , induced e.m.f. lr = . impedance sE r (19) Squaring equation (19) and multiplying by R r (*>) = ks. (21) * See the Standardization Rules of the American Institute of Electrical Engineers. THE INDUCTION MOTOR 2I 3 (b) Stray power. The stray power of an induction motor in- cludes windage, friction, and iron losses, and is approximately con- stant over the working range of speeds. It is evident that windage, friction, and stator iron losses decrease as the slip (load) increases, and that the iron losses in the rotor increase. The stray power of an induction motor is usually taken as the no-load input to the 180 25 50 75 100 125 150 PERCENT OF RATED OUTPUT (H.R) FIG. 174. Induction Motor Performance Curves. ZOO motor minus the stator copper losses, the rotor copper losses at no- load being so small as to be negligible. Stray power = no-load input stator RP. (22) ii. Performance curves. The performance of an induction motor is indicated by curves such as those shown in Fig. 174 data for which may be derived from: (a) a brake test, (b) the losses, (c) the circle diagram. (a) The brake test. A brake test is usually unsatisfactory be- cause of the large number of readings required, and the difficulty 214 ESSENTIALS OF ELECTRICAL ENGINEERING experienced in holding a brake load constant. The brake test is, therefore, little used in induction motor testing. (b) The losses. The set-up for the determination of the losses is the same as that used in a brake test, but the load is applied by means of an electric generator or a blower, and no measure- ments of output need be taken. The stray power is calculated from the no-load input and the stator copper losses as indicated above. Output = input stray power stator RP rotor RP. (23) But from equation (21) the rotor copper losses are proportional to the slip. Hence, Output = (input stray power stator RP) (1 5). (24) (c) The circle diagram. It may be proved both experimentally and mathematically, that the locus of the vector of the current in the rotor circuit is a semi-circle, i.e., the value of the rotor current and FIG. 175. Circle Diagram for Induction Motor. the power factor of the circuit are so related that, as the current varies, the end of the vector is always on a semi-circle as indi- cated in Fig. 175. Therefore, the semi-circle is determined when two points are determined, the diameter being parallel to the axis of abscissa. The two points usually selected for experimental deter- mination are: (i) when the motor is running without load (no-load test), (2) when the rotor is blocked to prevent its rotation (blocked rotor test). (i) No-load test. The motor is supplied with current at rated voltage and run without load. Measure the current and the watts input to the stator. THE INDUCTION MOTOR 215 t/3 = the phase angle between E r and 7 r . Then . ( 26 ) But r = tan^ (27) K r and R r = -^- (28) tan<*> = sX r COt 0. (29) Substituting the value of R r from equation (29) in equation (26), * T sEr / \ Ir = = (30) $XrV(* - = ^- .';,.". (32) Equation (32) is a polar equation of the circle.f * In plotting the circle diagram for a polyphase motor it is convenient to use the quantities for one phase, in which case the watts, torque, output, losses, etc., must be multiplied by the number of phases to obtain the quantities for the motor, or to reduce the experimental quantities to the "equivalent single-phase" system, as ex- plained in Chapter 7, Section n. t The proof given above is strictly true only when the resistance of the stator wind- ings and the magnetic leakage are negligible, but over the operating range of an in- duction motor very small differences are found between the theoretical and the actual locus of the current vector. 218 ESSENTIALS OF ELECTRICAL ENGINEERING THE SINGLE-PHASE INDUCTION MOTOR* Structurally, the single-phase induction motor is essentially the same as the polyphase motor, but the stator windings are connected to a single-phase supply system. The chief operating difference between the single-phase and the polyphase motor is the fact that the former, is not inherently self-starting. 14. Transformer action. When the windings A A (Fig. lyya) are connected to a single-phase alternating-current circuit, the flux set up by the windings periodically reverses in direction, and the changing value of flux induces an electromotive force in the conduc- tors of the squirrel-cage structure. But the directions of the cur- FIG. FIG. i;7b. rents in the conductors are such that the torque exerted on one conductor is equal and opposite to that exerted on another conductor. There is, therefore, no tendency for the conductors to rotate, the effect of the rotor currents being to largely neutralize the flux set up by the windings A A, the action being identical with that in a transformer, the secondary of which is short-circuited. 15. Generator action. If the structure supporting the rotor conductors is rotated by some external power, an electromotive force is generated in the conductors by reason of their movement across the magnetic field set up by windings AA.\ The current-carrying conductors on the rotor structure set up a quadrature flux, as indicated in Fig. lyyb. During the half- cycle in which the current in AA flows in the direction indicated by the arrows, the flux set up by A A is constant in direction, though * For a mathematical discussion of the single-phase induction motor, the student is referred to " Electric Motors " by Crocker & Arendt. t See Chapter 2, Section 13. THE INDUCTION MOTOR 219 varying in magnitude, and the direction of the flux set up by the rotor winding is that indicated m the figure. During the next half-cycle the flux due to the windings AA is reversed, since the current direction in the windings is reversed. The reversed direc- tion of the field across which the rotor conductors move, reverses the direction of the rotor flux. Therefore, the flux set up by the rotor winding of a single-phase induction motor is in quadrature, both in time and in space, with the flux set up by the stator windings, and alternates in direction, the frequency of the flux reversal being the same as that of the current reversal in the windings AA.* It is evident that the generated electromotive force and, there- fore, the quadrature flux, is proportional to the speed of the rotor; and that the magnetomotive force of the rotor, when rotating at synchronous speed, is equal to the magnetomotive force of the stator windings. 1 6. Rotating flux. While the simultaneous existence of quad- rature magnetic fluxes in the same material is entirely imaginary, the effects are real and are due to the actual flux resulting from quadrature components. There is, then, set up in a single-phase induction motor, when once started, a rotating flux the value of which is constant only when the rotor revolves at synchronous speed. Since no electromotive force could be induced in the rotor conductors at synchronous speed (field and conductors rotating FlG g ' Rotatino . at the same speed), the operating speed of the Flux of Single- rotor is always less than synchronous speed, and phase Induction the rotating field of a single-phase induction motor Mo1 is an ellipse, as indicated in Fig. 178, the short axis being proportional to the speed of the rotor. 17. Starting. To make single-phase induction motors com- mercially practicable, auxiliary starting devices are required. Two methods of producing a starting torque in single-phase induction motors are in general use, and will be described: (a) shading coils, (b) split-phase windings. * Because of the reversal of the quadrature flux, and the passage of the rotor con- ductors across the quadrature field, the currents flowing in the rotor conductors of a single-phase induction motor, are due to the combined action of four electromotive forces. 22O ESSENTIALS OF ELECTRICAL ENGINEERING (a) Shading coils. A shading coil is a closed winding placed around a portion of the pole, as indicated in Fig. 179. As the flux in the pole alternates, by reason of the reversal of the electromotive force of the supply circuit, an electromotive force is induced in the shading coil. The effect of the induced electromotive force is to oppose any change in the magnetic conditions existing in the space enclosed by the coil. This opposi- tion causes the flux threading the coil to lag behind the flux in the unshaded part of the pole, and thus reach its maximum value at a later period. A flux which moves (shifts) from the unshaded to the shaded part of the pole is thus produced , and a small starting torque obtained. Shading FIG. 179. Single-phase In- coils are commonly used in fan and other duction Motor. (b) Split-phase windings. When a single-phase induction motor is to be started by the split-phase method, an auxiliary stator wind- ing must be provided and the motor is, structurally, a polyphase motor. The connections for starting a two-phase motor from a S*ifct> Running Starting Resistance FIG. i8oa. Split-phase (Two-phase Winding). FIG. i Sob. Split-phase (Three-phase Winding). single-phase circuit are shown in Fig. i8oa, and those for a three- phase motor in Fig. i8ob. Fig. 181 is a vector diagram, with stationary rotor, of the circuits represented in Fig. i8oa. The impedance of circuit A is greater than that of circuit B because of the resistance connected in series THE INDUCTION MOTOR 221 with the motor winding; the power factors of the two circuits are different, and the currents are out of phase as indicated.. The fluxes set up by the windings are, therefore, out of phase, and produce a resultant rotating (or shifting) field. It is impracticable to produce a quadra- ture displacement of fluxes by means of split- phase connections, but a displacement sufficient to produce a small starting torque is easily obtained. The direction of rotation is reversed FIG. 181. Vector Dia- by reversing the terminal connections of either gram SpUt " Phase . . Motor. winding. 18. Speed-torque curves. Fig. 182 shows the speed-torque relations of a single-phase induction motor with variable rotor resistance. Both the maximum torque and the speed at which it . \n 50 KX) 150 200 250 PERCENT OF FULL LOAD (RATED) TORQUE FIG. 182. Speed- torque Characteristics of Single-phase Induction Motor with Variable Rotor Resistance. is developed, are reduced by increasing the resistance of the rotor circuit. THE FREQUENCY CHANGER 19. The frequency of a transmission line is often less than that required for circuits supplying lamps. To operate lamps from such a system it is necessary to raise the frequency. As indicated in Section 9, the frequency of the current in the rotor circuit of an in- duction motor is proportional to the slip of the rotor. If, then, the 222 ESSENTIALS OF ELECTRICAL ENGINEERING slip-ring rotor of an induction motor is driven at the proper speed a current of any desired frequency may be obtained from the rotor windings. When so used the induction motor is termed a frequency changer.* The voltage at the terminals of the rotor circuit is proportional to the slip of the rotor. This is easily demonstrated by considering the voltage at synchronous speed and at zero speed. At synchro- nous speed, the voltage must be zero since there is no relative move- ment between the flux and the rotor conductors. At zero speed, the stator and the rotor act as the primary and the secondary of a trans- former, and the voltages are to each other as the number of turns in the windings. The voltage of the rotor circuit at any given speed is, then, obtained by multiplying the voltage of the supply circuit by the ratio of the number of turns in the windings and by the slip of the rotor from synchronism. Er = fo.,t (33) when E r = the electromotive force induced in the rotor winding, k = the ratio of the number of turns on the rotor to the number of turns on the stator, s = the slip of the rotor expressed as a fraction of the synchronous speed, E s = the voltage of the supply circuit at the stator terminals. The ratio k of the number of turns in the windings may be deter- mined by voltage measurements at 100 per cent slip (standstill). The rotor of an induction motor, when used as a frequency changer, is driven by another motor (either induction or synchro- nous) supplied from the same system as is the stator of the frequency changer. In case the driving motor is synchronous, its fields may be over excited to compensate for the lagging current of the fre- quency changer, and the power factor of the supply circuit kept at a high value. At zero speed, it is evident that no power is required from the driving motor, and that a certain electromotive force is induced in the rotor circuit by transformer action. If the rotor is driven at * The required change in frequency may be affected by means of a motor-generator set, e.g., a twenty-five cycle motor (synchronous or induction) driving a sixty-cycle alternator. t This equation is only approximate when current flows in the windings, because of magnetic leakage and stator resistance. THE INDUCTION MOTOR 223 synchronous speed backwards, and the current maintained at a con- stant value, the voltage of the circuit is doubled. But the in- creased voltage is due to the actual movement of the rotor (i.e., to generator action), and the increased power is supplied through the motor. Since there is no fixed relation between the number of phases supplied to the stator and the number for which the rotor may be wound, the induction machine may be used to change the number of phases as well as the frequency of a system. The chief objections to the induction frequency or phase changer are its poor regulation and low power factor, due to the large air- gap leakage reactance. THE INDUCTION GENERATOR 20. When the rotor of an induction motor is driven above syn- chronism, the counter-electromotive force of the motor becomes greater than the applied electromotive force, and the power compo- nent of current in the supply circuit is reversed, i.e., the motor acts as a generator and delivers current to the supply system. When so operated, an induction machine is not self-exciting, and must be operated in connection with synchronous apparatus from which it may receive its magnetizing current. If an induction motor, the rotor of which is driven above syn- chronism by an independent prime mover, is connected to a system supplied by an alternating-current generator (synchronous alter- nator), the following effects are noted when the driving torque sup- plied to the alternator is reduced to zero: (a) The alternator operates as a synchronous motor. (b) If the field excitation of the synchronous machine is constant, the electromotive force of the system decreases as the load increases. (c) With constant rotor speed, the frequency of the system de- creases as the load increases. (d) The frequency of the system is proportional to the speed of the synchronous machine. (e) The electromotive force of the system is increased by in- creasing the field excitation of the synchronous machine. (/) The load on the system is proportional to the speed of the rotor above synchronism, i.e., above the speed of the synchronous machine. 224 ESSENTIALS OF ELECTRICAL ENGINEERING 21. Parallel operation of induction generators. The relative speeds of two or more synchronous machines connected to the same system must remain constant, the phase relations of the generated electromotive forces determining the ratio of load division.* The division of the load between two induction generators operating in parallel is proportional to the ratio of the deviation of the actual speeds of the rotors from the synchronous speed. Since the speeds of two induction machines operating in parallel need not have a constant ratio to each other, induction generators need not be synchronized but simply brought up to approximately synchronous speed, and the switch connecting the incoming machine to the system closed. After closing the switch, the load division is adjusted, as in the case of synchronous machines, by manipulating the speed governing apparatus of the prime mover. 22. Commercial applications. The induction generator has not, up to the present time, come into any extensive use f and must be considered as still in the stage of development. Theoretically it offers decided advantages when driven by water wheels, gas engines or other prime movers, the speed regulations of which are not good. CHAPTER XIII PROBLEMS 1. The maximum torque of an induction motor occurs when the slip is 18 per cent. Find the ratio of the rotor resistance and the resistance that must be added to the rotor circuit so maximum torque is developed at starting. Note. The starting effort of an induction motor is expressed in "synchro- nous watts" or "synchronous horse power," and is equal to the power that would be developed in the rotor if it were operating at synchronous speed and developing a torque equal to that developed at starting. 2. Find the frequency of the rotor current in a lo-pole, 6o-cycle induction motor when operated at a speed of: (a) 720 r.p.m., (6) 600 r.p.m., (c) 400 r.p.m., (d) 100 r.p.m., (e) o r.p.m. 3. Compare the efficiencies of two induction motors, the full-load slip of one being 4 per cent and that of the other 8 per cent. 4. A 4oo-horse-power, 440- volt, 6o-cycle, 3-phase induction motor was tested and the following data obtained: Stator curren t Watts input per phase per phase Without load 160 4000 With rotor blocked 2250 225000 Stator resistance per phase = 0.018 ohm. Construct the circle diagram. * See Chapter 9, Section 12. t The 59th Street power house of the Interborough Rapid Transit Co. (New York) , has an installation of induction generators. So far as the writer has been able to learn, this installation has been entirely satisfactory. THE INDUCTION MOTOR 225 5. From the circle diagram of Problem 4 plot the following curves: (a) speed-torque, (6) efficiency, (c) power factor. 6. From the circle diagram of Problem 4 determine the following when the output of the motor is 400 brake horse power: (a) current input, (b) torque developed in the rotor, (c) power factor, (d) speed, (e) "added" copper losses in stator, (/) rotor losses. 7. From the circle diagram in Problem 4 determine: (a) the stator current required to produce maximum starting torque, and the value of this starting effort in synchronous horse power, (b) the stator current and the torque when the power factor is maximum. 8. Determine the synchronous speed of a 6o-cycle induction motor having: (a) 2 poles, (b) 4 poles, (c) 6 poles, (d) 10 poles, (e) 16 poles, (/) 24 poles. 9. Determine the synchronous speed of a 25-cycle induction motor having poles as indicated in Problem 8. 10. Determine the voltages induced in the rotor circuit of the motor in Prob- lem 2 when the ratio of the stator turns to the rotor turns is one, and the applied voltage is 230. n. A 6- and a lo-pole induction motor are operated in cascade, the 6-pole motor being connected to a oo-cycle alternating-current circuit. Find the syn- chronous speed of the combination. Note. When two induction motors are to be operated in cascade or con- catenation, their rotors are connected to the same shaft; the stator of one motor is supplied directly from the line, and the stator of the other motor from the rotor of the first motor. The motors may be connected in direct or in differential cascade, and the synchronous speed of the combination is that of a single motor having the sum or difference of the number of poles on the two motors, and operated from a line of the same frequency as that to which the first motor in the cascade combination is connected. 12. Find the voltage supplied to the stator windings of the lo-pole motor in Problem n. 13. Find the frequency of the current in the stator windings of the lo-pole motor in Problem n. 14. A 6o-cycle alternator is direct-connected to a 6-pole, 25-cycle induction motor. Find the number of poles on the alternator. 15. An induction motor develops a starting torque of 250 foot-pounds when the applied voltage is 60 per cent of the rated voltage. Determine the starting torque that would be developed by the motor if rated voltage were applied to the stator windings. Note. Since the fundamental equation of the induction motor is the same as that of the transformer (E = 4.44/4>Afio~ 8 ) with the addition of a winding constant, both the flux and the rotor current are proportional to the applied voltage. Therefore, the torque of an induction motor is proportional to the square of the applied voltage, the slip of the rotor (the frequency of the rotor currents) remaining constant. CHAPTER XIV SINGLE-PHASE COMMUTATING MOTORS i. Action of the shunt motor. When alternating current is supplied to a shunt motor, two very noticeable effects take place: (a) the motor develops very little torque, (b) excessive sparking occurs at the brushes. (a) Torque. It is evident that if the current in the armature of a shunt motor supplied from a single-phase alternating-current system and the flux set up by the field winding are in phase, i.e., attain their maximum and minimum at the same instant, a torque varying from zero to maximum but always in the same direction is produced. The average torque of an alternating-current shunt motor is proportional to the average product of the armature current and the flux in the air gap, as in the continuous-current motor. When the field winding of a shunt motor is excited from an alter- nating-current system, the current lags behind the electromotive 'e.m.f force by a very considerable angle, the field circuit being highly inductive, and the flux has a correspond- ing lag behind the electromotive force. The power Fields factor of the armature circuit being higher than that of the field circuit, the vector of armature current FIG. 183. leads the vector of field flux, as shown in Fig. 183. Because of the phase relations of the armature current and the field flux, their product is negative during part of the cycle and positive during the remainder of the cycle. The torque, therefore, tends to rotate the armature first in one direction and then in the other, and the net torque producing, or tending to produce, rotation is the algebraic sum of the average positive and negative torques during one complete cycle. If the angle equals 90 degrees, the sum of the instantaneous torques is zero, and the armature has no tendency to rotate. The torque of a shunt motor may be improved by connecting the armature to one phase, and the field to the other phase of a two- 226 SINGLE-PHASE COMMUTATING MOTORS 227 FIG. 184. phase system. Such a machine is seldom used because of the com- plications involved, and the low power factor at which the phase supplying the field circuit must operate, and the fact that more satis- factory apparatus has been devised. (b) Sparking. Commutation in an alternating-current motor is more complicated than in a continuous-current dynamo. Let the position of the armature of an alternating-current motor be that shown in Fig. 184, the armature coil c being short-circuited by the brush. The changing value of flux in the armature core causes the coil c to act as the short-circuited second- ary of a transformer in which the current may be many times the normal current flowing in the coil. This large current flowing in the local circuit, composed of the coil, the two commutator segments to which the coil is con- nected, and the brush, causes excessive heating of the armature as well as destructive sparking when the brush passes from one com- mutator segment to another. 2. The series motor. When a continuous-current series motor is operated with alternating current, the phase displacement between the armature current and the field flux largely disappears, since the same current flows in both wind- ings and the inductance of a series field winding is much less than that of a shunt field winding, but commutation difficulties still exist. The commutation of a series motor operating with alternating current is improved by: (a) re- ducing the flux density of the magnetic circuit, (b) reducing the number of series turns per arma- ture coil, (c) reducing the frequency of the supply circuit, (c) special devices. (a) Flux density. For a given frequency, the average rate at which the flux changes is proportional to its maximum value, and the electromotive force induced in the short-circuited armature coil is proportional to the rate of change.* The impedance of the * See Chapter n, Section i. FIG. 185. The Series Motor. 228 ESSENTIALS OF ELECTRICAL ENGINEERING short-circuit remaining constant, the current in the local circuit is proportional to the induced electromotive force. (b) Series turns. Since the electromotive force induced in the secondary of a transformer is proportional to the number of series turns,* the smaller the number of series turns per armature coil, the less is the induced voltage. It might appear that reducing the number of turns in the coil reduces the impedance of the circuit in the same ratio, and that the current, therefore, remains constant; but the resistance of the brush, brush contact, commu- tator bar, and connecting leads must be taken into account in calculating the impedance of the local circuit. (c) Frequency. The rate of flux change is proportional to the frequency, i.e., the higher the frequency the shorter the time in which the flux must change from zero to maximum. The electro- motive force induced in a short-circuited coil is, therefore, pro- portional to the frequency of the supply system, and low frequencies tend to reduce commutation troubles. Series alternating-current motors are seldom used on circuits having a frequency greater than twenty-five cycles per second. (d) Special devices. While the above considerations of design and operation materially reduce both heating and commutation troubles, special devices have been found necessary to bring them within practical operating limits. The simpler (and more satis- factory) of these devices are: (i) resistance leads connected be- winding tween the terminals of the armature coils and the commutator segments, (2) balanced choke coils. (i) Resistance leads. If resistances Comma fa/or . . are connected as indicated in Fig. i86a, FIG. i86a. Resistance Leads the local current is reduced correspond- A. C. Series Motor. ingly since j t must flow through the coil and two resistances connected in series. The load current must, also, flow through these resistances and this would, apparently, decrease the efficiency by increasing the resistance losses. It is an experimental fact that the introduction of resistance leads increases the efficiency, the increased loss due to the load current flowing in the added resistance being less than the decreased loss due to the smaller current flowing in the local circuit. * See Chapter n, Section 3. SINGLE-PHASE COMMUTATING MOTORS 22Q Armature Winding Brush FIG. i86b. Balanced Choke Coils. Winding (2) Choke coils. The connections for the use of choke coils are shown in Fig. i86b. The windings on each core are so related that their inductance is cumulative to the short-circuit current, but differen- tial to the load current. This arrange- ment is not entirely satisfactory because the coils balance for the load current only when the current is equally divided between two coils, and for certain positions of the armature, the coils neutralize each other, and become ineffective so far as the short- circuit current is concerned. 3. Compensation for armature inductance. The inductance of the armature winding may be neutralized by means of a " com- pensating" winding connected as shown in Fig. 187. The com- pensating winding may be sup- plied with current : (a) inductively, (b) conductively. (a) Current supplied inductively. - If the compensating winding is closed on itself (short-circuited) as indicated in Fig. i8ya, the alternating field flux induces in the winding an electromotive force, and the magnetic effect of the winding is approximately equal and opposite to that of the armature winding. The magnetic fields of the two windings, therefore, neu- tralize each other. (b) Current supplied conductively. If the compensating winding is connected in series with the armature winding as indicated in Fig. i8yb, the same current flows in the windings. By properly proportioning the compensating winding, and connecting it so that its magnetic effect is opposite to that of the armature winding, the magnetic fields of the two windings neutralize each other. 4. Comparison of series motors. The alternating-current and the continuous-current series motors differ in the following re- spects : (b) Conductive Compensation FIG. 187. Field Winding 230 ESSENTIALS OF ELECTRICAL ENGINEERING rmafure Drop. Drvp in Compensating Winding FIG. 1 88. Clock Diagram, Series A. C. Motor. (a) The weight of the alternating-current motor is from one and one-half to two times that of the continuous-current motor develop- ing the same torque. (b) The alternating-current motor has the larger number of armature coils. (c) The alternating-current motor has the larger commutator, and the larger number of segments. (d) The entire magnetic circuit of the alternating-current motor is laminated. 5. The repulsion motor. When a continuous-current armature is placed in a magnetic field produced by an alternating current as indicated in Fig. 189, the field wind- ing acts as the primary and the armature winding as the second- ary of a transformer, and current flows between the short-circuited brushes for any position of the brushes except that shown in Fig. i8gb, when the algebraic sum of the voltages induced in the coils between brushes is zero. { FIG. 189. Repulsion Motor. For the position indicated in Fig. i8ga, maximum current flows between the brushes. This current tends to set up a flux dia- metrically opposed to that set up by the field windings, and the magnetic effect of the field windings is largely neutralized. The torque produced by the reaction between any current-carrying conductor and the field flux is neutralized by an equal and oppo- SINGLE-PHASE COMMUTATING MOTORS 2 3 I site torque produced by some other armature conductor and the same field flux. The armature has, therefore, no tendency to rotate. This equality of opposite torques is destroyed, and the armature caused to rotate, by moving the brushes in either direction, the direction of rotation being the same as the movement of the brushes. It has been determined, experimentally, that maximum torque is developed when the angle equals approximately 45 degrees. While the repulsion motor is little used commercially,* its good starting torque offers means by which the single-phase induction motor may be made self-starting. The Wagner Electric Mfg. Co. make a motor of this type in which the rotor structure, at starting, is essentially that described above. When the speed reaches a predetermined value, a centrifugal device removes the brushes from the commutator, and short-circuits the commutator bars so that the armature conductors form a squirrel-cage structure. 6. The compensated repulsion motor. - - The so-called compen- sated repulsion motor is, mechan- ically, a series motor with the addition of short-circuited brushes placed in quadrature with the main brushes, as indicated in Fig. 190. While both series and repulsion motors have series characteristics, i.e., the speed tends to rise to an infinite value as the load approaches FIG. 190. Compensated Repulsion Motor. zero, the compensated repulsion motor has shunt characteristics, and its principles of operation are materially different from either the series or the repulsion motor. The short-circuited brushes of this type of motor cause the arma- ture winding to produce a magnetic field which opposes and largely neutralizes the flux set up by the field winding. The current which flows between the short-circuited brushes is produced by transformer action, as described for "the repulsion motor when the brushes are in the position indicated in Fig. iSpa. The current flowing between the main (series) brushes sets up a * Commutation is inherently poor. ESSENTIALS OF ELECTRICAL ENGINEERING field at right angles to the line joining the short-circuited brushes. This flux, reacting with the current flowing in the armature con- ductors by reason of the short-circuited brush connection, produces the greater part of the torque of the motor, although some torque is doubtless produced by a reaction between the series-field flux and the current between the main (series) brushes. As the armature speed increases, the counter-electromotive force induced in the conductors by reason of their motion reduces the current flowing between the short-circuited brushes and the torque becomes less. Therefore, the flux is approximately constant, the current varies inversely with the speed of the armature and the motor has shunt characteristics. 7. The General Electric RI motor. - The connections of the General FIG. 191. Diagram of General Electric Company's Type RI motor Electric "RI" Motor. i are shown in Fig. 191, and typical per- formance curves in Fig. 192. From the curve sheet, it will be Typical Performance Curves Type RI - 4 Pole 1600 ff. P. M. Repulsion Induction Motor 73 100 Percent Load FIG. 192. Typical Performance Curves of Type RI Motor observed that the power factor is high for loads above 50 per cent of the rated output. 8. The Wagner BK motor. The Unity Power Factor (so- called by the manufacturers) single-phase motor of the Wagner Electric Mfg. Co. is a combination of the compensated repulsion SINGLE-PHASE COMMUTATING MOTORS 233 motor and the single-phase induction motor. The armature has a squirrel-cage, as well as a commutated winding, as indicated in Fig. 193. The electrical connections are indicated in Fig. 194. Retaini Commuted Winding (3) Magnetic Separator Squirrel Ca^e (4) FIG. 193. Section Through Slot of Wagner "BK" Motor. FIG. 194. Wiring Diagram for Wag- ner " BK " Motor. At starting, the main field winding i induces, by transformer action, currents in the commutated winding 3, and these currents flow between the short-circuited brushes 5 and 6. Similarly, cur- rents are induced in the squirrel-cage winding as described for the single-phase induction motor.* The series currents flowing in 14 V BK MOTOR H P.60~.1 PHASE, 4 POLE 220 VOLTS. 18OO R.P.M. "' 123456789 10 H.P. FIG. 195. Performance Curves of Wagner "BK" Motor. the commutated winding 3 set up a flux, the direction of which is at right angles to the flux set up by winding i. The currents in the commutated and in the squirrel-cage windings react with this quadrature flux to produce a torque, and start the motor. As the speed of the motor increases, the squirrel-cage winding sets up a quadrature flux of its own, and develops a corresponding torque. * See Chapter 13, Section 14. 234 ESSENTIALS OF ELECTRICAL ENGINEERING The torque of this motor is, then, a resultant of two torques, one of which is maximum at starting and decreases as the speed in- creases, while the other is zero at starting, increases to maximum as the speed increases, and then decreases as the speed increases still further. The action of the auxiliary winding 2, is to improve the power factor of the motor. The switch 9 is open at starting, but is auto- matically closed by a centrifugal device at a predetermined speed. The performance curves of this motor are shown in Fig. 1 95 . The power factor will be observed to be 70 per cent leading at zero load, and approximately unity from 75 to 150 per cent of rated load. It will also be noted that the speed is very nearly synchronous at rated load, rises slightly above synchronism at no-load, and drops below synchronism as the load is increased above the rated capacity. CHAPTER XV ELECTRIC LAMPS* i. Arc lamps. Arc lamps are largely used for lighting streets and other areas where a few large units are preferable to a larger number of smaller units. The basic principle of the arc lamp is the arc which is established when an electric circuit is interrupted. The heat of the arc causes one or both electrodes to be consumed, and the incandescent par- ticles emit a very intense light. Arc lamps may be classified as : (a) carbon arcs, (b) magnetite arcs, (c) flaming arcs. Carbon and flaming arc lamps may be operated on either alternating or unidi- rectional current; magnetite lamps on unidirectional current only. Arc lamps may be operated either in parallel or in series. In parallel operation, the voltage is constant and the regulating mech- anism must maintain the proper value of current; in series oper- ation, the current is automatically maintained at a constant value, and the lamp mechanism controls the voltage between the terminals of the lamp. (a) Carbon arc lamps. In this type of lamp the electrodes are made of finely ground carbon mixed with a suitable binder, pressed into the form of pencils and baked. In the early lamps, the arcs were "open," i.e., air currents circulated freely about the arc. The light from such a lamp is unsteady and the electrodes are consumed very rapidly. By enclosing the arc in a glass globe to which a very limited quantity of air is admitted, the flickering of the arc is materi- ally reduced, and the electrodes are consumed at a much slower rate. Because of these facts and the reduced fire hazard of the enclosed arc, open arc lamps are no longer manufactured. Fig. 196 shows the circuits of a lamp designed for parallel opera- tion. Regulation is accomplished by means of an electromagnet, * Problems in illumination are beyond the scope of this volume. For information regarding methods of calculating light densities, etc., the reader is referred to " Illum- ination and Photometry," by W. E. Wickenden, and " Electrical Illuminating Engineer- ing," by W. E. Barrows, Jr. 235 236 ESSENTIALS OF ELECTRICAL ENGINEERING Line Resistance \ : : Series Coil r Carbon Clutch *T T 'heat Sere Carbon en the windings of which are connected in series with the arc. When the circuit is open, the core of the magnet drops to its lowest position and the clutch releases the upper electrode, which then rests on the lower electrode as indicated. Closing the lamp circuit, causes the magnet to attract its core, raise the upper electrode, and ." strike" the arc. The resistance of the circuit becomes greater as the distance between the elec- trodes increases, and the magnet is so proportioned that its pull balances that FIG. 196. Schematic Diagram o f a spring, when rated current flows in of Connections for Parallel ^ ^^ The ^ therefo moves (Carbon) Arc Lamp. upward until this equilibrium, which corresponds to a fixed distance between the electrodes, is estab- lished. As the electrodes burn away, the length of the arc increases, the current in the circuit decreases, the pull of the magnet no longer balances that of the spring, and the core moves downward, increasing the current and decreasing the length of the arc, until equilibrium is restored. When the core has de- scended to its lower limit, the clutch releases the upper electrode which drops into contact with the lower electrode, the resistance of the circuit is reduced, and the increased Line current causes the magnet to separate the electrodes again. The result of this process is the periodical feeding of the upper electrode through the clutch by an amount equal, approximately, to the length of the arc. Since multiple arc lamps are usually connected to no-volt mains, and the _ FIG. 197. Schematic Diagram of arc requires only from seventy to eighty connections for Series (Carbon) volts, the lamp is provided with a ballast Arc Lamp. (Differential Con- by means of which the line voltage is re- duced. In continuous- current lamps the ballast is resistance; in alternating-current lamps, reactance. The series lamp (Fig. 197) has, in addition to the series magnet of the parallel lamp, a magnet connected in parallel with the arc and so arranged that its action opposes that of the series magnet. When Heat Screen .Carbon ELECTRIC LAMPS 237 current flows through the lamp, the series magnet raises the upper electrode as in the parallel lamp, but as the length of the arc in- creases, the current in the windings of the shunt magnet increases, and equilibrium is established at the length of arc at which the pull of the series magnet (constant) minus the pull of the shunt magnet, balances the pull of a spring. As the electrodes burn away, the effect of the shunt magnet increases, and equilibrium is main- tained by the changing position of the series magnet core. At the lower limit of the movement of the core, the clutch releases the upper electrode and it is fed downward as in the parallel lamp. In the series lamp, a bypass must be provided for the passage of the current in case the carbons burn out, or for any other reason the arc circuit becomes opened. This may be provided by means of a shunt coil armature which closes the circuit through an auxiliary resistance whenever the voltage across the arc reaches a given value, which is always higher than that reached in normal operation. Should the arc circuit be re-established, the voltage between the terminals of the shunt winding is reduced, the armature of the shunt magnet is released and the auxiliary circuit broken. (b) Magnetite (or "luminous") arc lamps. The magnetite arc lamp is one of the results of the researches of Dr. C. P. Steinmetz of the General Electric Co., and has a fixed upper electrode * of copper and a movable lower electrode composed of . Line a mixture of magnetite (one of the oxides of iron) and titanium oxide encased in an iron tube. This lamp is economical in power consumption and gives a brilliant " white" light which is well distributed. Fig. 198 shows the circuits of a series magnetite lamp, and its operating mechan- ism. Unlike the carbon lamp, the arc circuit FlG Ig8 . Schematic Diagram is open when the lamp is not in operation.! of Connections for Series When the current is turned on, the starting magnet closes the arc circuit by raising the lower electrode into contact with the copper rod. The series mag- net then attracts its armature and opens the circuit of the starting Snitch Resistance ' Copper Electrode 'Magnetite Electrode Magnetite (Luminous) Arc Lamp. * In the Westinghouse magnetite lamp the lower electrode is fixed, t If the arc circuit is not kept open the terminals weld or stick together, making starting difficult or impossible. 238 ESSENTIALS OF ELECTRICAL ENGINEERING magnet, allowing the lower electrode to drop and "strike" the arc. As the electrode burns away, the voltage over the arc increases until the shunt magnet attracts its armature, closes the circuit of the starting magnet, and feeds the lower electrode upward by a definite step. In case the arc circuit becomes open, the shunt magnet closes the circuit of the starting magnet and the current flows through the starting magnet and its series resistance, so that the other lamps in (a) Complete. . (b) Casing Removed. FIG. 199. G. E. Series Luminous (Magnetite) Arc Lamp. the circuit are unaffected by the burning out of an electrode, or by failure to feed properly. The mechanism of the magnetite lamp, when intended for parallel operation, is similar to that of the series lamp, with the omission of the shunt magnet. The starting magnet raises the lower electrode and the series magnet breaks the circuit of the starting magnet, which allows the lower electrode to fall and strike the arc as in the series lamp. When the electrode has burned away to such an ex- tent that the series coil is no longer able to hold its armature, the circuit of the starting magnet is closed, and the electrode fed upward. (c) Flaming-arc lamps. The light of the flaming-arc lamp is due to the luminescent vapor of metallic salts, the salts of calcium (yellow light) and of titanium (white light) being those com- monly used. The electrodes consist, essentially, of a mixture of carbon, the metallic salt, and an alkaline salt. The purpose of the alkaline salt is to steady the arc and to prevent the accumulation of slag. ELECTRIC LAMPS 239 High efficiency is obtained only by rapid consumption of the electrodes, the great length (14 to 24 inches) and small diameter (J to f inch) of which make vertical mounting impracticable. The electrodes are, therefore, usually placed as indicated in Fig. 200, and fed downward by gravity, the mechanism holding them being re- leased by an electromagnet connected in parallel with the arc, when the voltage across the arc reaches a predetermined value. The electrical circuits of one of the simpler regulating mechanisms are shown in Fig. 200, and are similar to those of the differential carbon lamp described above. The electrodes are normally separated. When the circuit is closed, the shunt magnet pulls the points of the electrodes together (lateral move- ment of one electrode is provided for), closes the arc circuit, and energizes the series magnet. The series magnet separates the electrodes, thus striking the arc, and equilibrium is established when the arc is from ij to 2 inches in length. As the electrodes FIG. 200. Wiring burn away, the voltage over the shunt coil increases, Diagram Flam- while the current in the series coil tends to decrease mg Arc Lamp ' (parallel operation), and equilibrium is maintained by an inward movement of the movable electrode. When the electrode reaches the limit of its movement, the shunt magnet causes the mechanism holding the electrodes to be released, and they are fed downward, reducing the length of the arc. The increased current in the wind- ings of the series magnet causes the electrodes to be separated and the arc is restored to its normal length. The short life (10 to 18 hours) of flaming-arc electrodes led to the development of the regenerative flaming-arc lamp, which is an adaptation of the flaming-arc principle to the enclosed lamp, and increases the life of the electrodes to seventy hours or more. Tubes through which the gases circulate conserve the heat, produce a more perfect combustion, and increase the efficiency of the lamp. The electrodes used in regenerative lamps, being shorter and of greater diameter than those used in the flaming-arc lamps, may be arranged vertically as in the carbon and magnetite lamps. The flaming-arc is the most efficient of artificial illuminants, and gives approximately three candle power per watt. Arc lamps are always provided with air dash pots or some similar device for damping the movements of the electrode, which would otherwise be jerky and irregular. 240 ESSENTIALS OF ELECTRICAL ENGINEERING 2. Instability of the electric arc. The electric arc is inherently unstable when operated from constant potential mains, i.e., when the arc is connected directly between the mains. Suppose an electric arc to be established between constant voltage mains. Any slight decrease in the value of the current flowing in the circuit causes the cross-sectional area of the arc to decrease, increases the resistance of the circuit, and still further decreases the current; any slight increase in the value of the current flowing in the circuit causes the cross-sectional area of the arc to increase, decreases the resistance of the circuit, and still further increases the current. The effects of any momentary change in the value of the current flowing in the circuit are, therefore, cumulative, and either cause the arc to " break," or the current to increase to an excessive value. Stable operation of the multiple arc lamp is established by means of the ballast (resistance in the continuous- current lamp and react- ance in the alternating-current lamp) referred to above, the opera- tion of which is as follows: If the current decreases slightly, the drop in the ballast decreases, and the voltage between the terminals of the arc increases; if the current increases slightly, the drop in the ballast increases, and the voltage between the terminals of the arc decreases. The ballast thus produces a compensating change in the voltage between the terminals of the arc for any change in the resistance of the circuit (area of the arc). 3. Power factor of the alternating-current arc. It is an experimental fact that the power factor of the alternating-current arc is less than unity, its average being about 0.85. This low FIG. 201. Distorted Pwer factor is not due to a time lag between Current Wave of the the current and the electromotive force, but to Electric Arc. distortion of the current wave. As pointed out in Section 2, the resistance of the electric arc is a function of the current flowing in the circuit, and the resistance of an alternating- current arc depends on the instantaneous value of the current. Fig. 20 1 is a reproduction of oscillograms of the current and electromotive force waves in an alternating-current arc. 4. Incandescent lamps. The incandescent lamp, which is universally used for interior lighting, consists of a hair-like filament enclosed in a highly exhausted and hermetically sealed glass globe. The light of the lamp is an indirect effect, and is due to the fact that ELECTRIC LAMPS 241 the electric current raises the temperature of the filament to a "white" heat. The purpose of the globe, in addition to the me- chanical protection which it gives, is to retard oxidation of the fila- ment and thus increase the life of the lamp. Connection between the filament and the line conductors is made through short wires embedded in the glass. For more than twenty-five years after the incandescent lamp became a commercial article, carbon was practically the only sub- stance of which filaments were made, but within a few years carbon filaments have been largely displaced by those of metallic tungsten. The light produced by the tungsten lamp is decidedly superior to that of the carbon lamp, and approximately three times as much light is available from a given expenditure of energy. The per- fection of the tungsten lamp, which is sold under the trade name " Mazda," has greatly reduced the cost of electric light, and thereby largely increased its use. While the efficiency of an incandescent lamp increases as the applied voltage is increased, the life of the filament is shortened. Lamps are rated at that voltage which has been found to give the most satisfactory results, taking into account both efficiency and length of life, and should be operated at this voltage. The temperature, and therefore the efficiency, at which a tung- sten filament may be operated is materially increased when it is surrounded by an atmosphere of nitrogen. The nitrogen-filled tungsten lamp is now a commercial article and has an efficiency, in large units, as low as | watt per candle power. The incandescent lamp is a constant potential lamp, and is used, primarily, for parallel connection between constant voltage mains, although series lamps are used to a considerable extent in street lighting. 5. The Nernst lamp. The luminous element (glower) of the Nernst lamp is a hollow cylinder composed of oxides of some of the rarer elements (zirconia, yttria, etc.). The glower is a non- conductor at ordinary temperatures, and is heated to a conducting temperature by a coil of platinum wire connected in parallel with the glower. The circuit of the heating coil is automatically opened when current begins to flow in the glower circuit. The negative temperature coefficient of the glower makes the operation of the lamp unstable unless a ballast is connected in series 242 ESSENTIALS OF ELECTRICAL ENGINEERING with the glower. The ballast coil is made of iron, the temperature coefficient of which is positive. The glowers, of Nernst lamps are always enclosed, usually by alabaster globes which conserve the heat and prolong the life of the glowers. The color effects of this lamp are good and the efficiency is high. 6. The mercury vapor lamp. The mercury vapor lamp con- sists of a highly exhausted glass tube containing a small quantity of mercury, which may be vaporized by an electric current. The light, which is due to the luminescence of the mercury vapor, is entirely devoid of red rays. This absence of red rays causes colors to be distorted, but the light is very acceptable in draughting rooms, packing rooms, warehouses and other places where its peculiar color effects are not objectionable. There is now on the market a phosphorescent reflector which adds red rays to the light emitted by the mercury lamp. Like the mercury rectifier, to which it is analogous, the mercury vapor lamp, because of its high resistance when cold, must be started by an auxiliary device of some kind. Starting may be effected by tilting the tube until the terminals are connected by liquid mer- cury through which the current flows, heating the mercury and fill- ing the tube with mercury vapor. The tube is tilted by hand, or by means of an electromagnet which is automatic in its action. Starting may also be accomplished by breaking an auxiliary circuit, the inductive kick from which breaks down the high initial resist- ance of the tube. Because of its low power factor, the use of the alternating-cur- rent mercury vapor lamp is objectionable. 7. The quartz lamp. The fundamental principle of the quartz lamp is the same as that of the mercury vapor lamp in that its light is due to luminescent mercury vapor. The pressure of the mercury vapor in the quartz lamp is materially higher than in the mercury vapor lamp, the temperature and the luminous intensity are cor- respondingly greater, and the light is not entirely devoid of red rays. The quartz lamp emits a considerable quantity of ultra-violet rays which are decidedly harmful to living organisms, and should not be used unless it is enclosed by a protecting globe. 8. The Moore tube. An application of the Geissler discharge to commercial lighting is made by means of the Moore tube, which ELECTRIC LAMPS 243 consists of a highly exhausted glass tube, of any length up to about 200 feet, the luminous properties of which are due to an electric discharge through rarified gases introduced into the tube. The quality of the light produced varies with the medium through which the electric discharge takes place; carbon dioxide gives a light approximating daylight, nitrogen an orange tinted light, air a pinkish light. The operation of the tube depends essentially on the maintenance of the proper pressure of the rarified gas in the tube, the electric discharge forming a solid precipitate which reduces the pressure in the tube. As the pressure decreases the conductivity increases, and the increased current operates a valve which admits gas to the tube, and restores the required pressure. CHAPTER XVI CIRCUIT-INTERRUPTING APPARATUS THE circuit-interrupting apparatus of an electric plant is the means by which the generators and the load are connected and dis- connected, and includes not only the switches, but fuses and other circuit-breaking devices, and the auxiliary apparatus for their man- ipulation. The satisfactory operation of a power plant depends, in no small degree, on the proper selection, installation, and opera- tion of the switching gear and the protective apparatus. 1. Fuses. A fuse is a short piece of lead and tin alloy, of such cross section and so connected in the circuit that it is melted and the circuit opened when an excessive current flows. Fuses are either "open'' or " enclosed." Enclosed fuses are to be preferred since the arc and the hot metal incident to the opening of the circuit are confined, thus reducing the fire hazard. The current at which a fuse will open a circuit can be determined only approximately because of external conditions, such as the temperature of the air, area of contact, etc. 2. Switches. Switches are for the specific purpose of con- necting and disconnecting the generator and the load apparatus, and may be divided into : (a) air-break switches, (b) carbon-break switches, (c) oil-break switches. (a) Air-break switches. An air-break switch consists of one or more blades of copper hinged at one end and making contact at the other end with spring clips which form part of the circuit. As operating switches, i.e., for opening current-carrying circuits, air- break switches are used on small apparatus only, because of the burning of the contacts when the switch is opened; for completely isolating other apparatus from a "live" line, they are universally used. (b) Carbon-break switches. To protect the copper terminals of an air break switch, a circuit having carbon terminals is connected in parallel with the one having copper terminals. In operating the switch, the copper terminals open first, without an arc, thus shunt- 244 CIRCUIT-INTERRUPTING APPARATUS 245 ing the current through the carbon circuit and protecting the copper contacts from injury. The carbon terminals, which may be re- newed, are not so easily burned as are copper terminals. (c) Oil-break switches. For the manipulation of high voltage circuits or those carrying large currents, switches having their con- tacts immersed in oil are always used. The arc formed when the contacts are separated is promptly smothered by the oil without appreciable damage to the contacts. Since a fuse cannot be depended on to open a circuit promptly when the current exceeds a specified value, and must be renewed each time it operates, automatically opening switches have been devised for the protection of elec- trical apparatus. An automatic switch is closed against the pressure of a spring, and is held by a latch. When the cur- rent exceeds a predetermined value, an electromagnet trips the latch, and the spring opens the switch. The coils of the tripping magnet are excited: (i) by the line current, the winding of the magnet being connected in series with the load, (2) from an auxiliary source, the circuit through the coils of the magnet being closed by a relay connected in the secondary circuit of a series transformer, as indicated in Fig. 205. The first method is applicable to either continuous- or alternating-current circuits; the second, to alternating-current circuits only. The current at which an automatic switch opens is adjusted by changing the length of the air gap in the magnetic circuit of the electromagnet, or the relay, or by changing the tension of a spring. In many cases, particularly in motor operation, both automatic switches and fuses are placed in the circuit. The fuses are rated slightly higher than the current at which the switch is set to operate, and their purpose is to protect the motor in case the operating mechanism of the switch becomes deranged. By the addition of an air dash-pot or other mechanical device, FIG. 202. Triple-pole, Single-throw Non-automatic Oil Switch. Gen- eral Electric Co. 246 ESSENTIALS OF ELECTRICAL ENGINEERING the tripping mechanism of an automatic switch is made less sensi- tive to momentary overloads or surges, i.e., the switch opens only after the overload has been maintained for a definite length of time, which may be made inversely proportional to the overload. Such devices are known as time limit relays. 3 . Switch operation. Switching operations, in small plants may be done manually, but in larger plants hand operation becomes unsatisfactory and often impossible, because of the size of the moving parts, as well as danger- Section through Typi- ous, because of high voltages. Elec- trical power is, naturally, used for the operation of large switches. The FIG. 203. cal Bus Bar and Switch Com- partments. Westinghouse. two methods of switch operation in general use are: (a) solenoid, (b) motor. (a) Solenoid-operated switches. Solenoid-operated switches re- quire the use of two solenoids, one for closing the switch, the other D. C. Buses Red Lamp lighted when oil srv. is close Closing Co itact FIG. 220. Voltmeter Connections, (a) Multi- plier. (6) Transformer. 4. High voltage measurements. The above types of instru- ments are not adapted to the direct measurement of voltages above seven hundred volts. For measuring voltages higher than this there must be used: (a) a multiplier, (b) a potential trans- former, (c) a special voltmeter. (a) The multiplier. A multiplier is a resistance connected in series with the current-carrying coil of a voltmeter, or the volt- meter element of a watt- meter, and serves simply to reduce the voltage that would otherwise be applied to the terminals of the coil. The resistance of the multiplier should bear a simple ratio to the resistance between the terminals of the voltmeter with which it is to be used, so that the voltage of the circuit is some multiple of the meter indication. The multiplier is applicable to continuous- or to alternating-current circuits. Fig. 22oa. (b) The potential transformer. The voltage of an alternating- current circuit may be stepped Po . nter down by means of a small potential transformer, the voltage of the primary circuit being the indication of the voltmeter multiplied by the ratio of transformation. Fig. 22ob. (c) Special voltmeters. The volt- age of a high potential system is determined by means of: (i) the .electrostatic voltmeter, (2) the spark gap. (i) The electrostatic voltmeter. The principle on which the electro- static voltmeter operates is the attraction between two oppositely charged bodies. * The essential parts of the Westinghouse electrostatic voltmeter are shown schematically in Fig. 221. A A * See Appendix C, Section 2. FIG. 221. Schematic Diagram of Con- nections for Westinghouse Static Voltmeter. METERS 257 are curved metallic plates connected, through condensers, to the conductors, the difference of potential between which it is desired to measure; and BB are hollow cylinders to which a pointer is attached. The position of the hollow cylinders changes as the voltage between lines changes, and the scale is calibrated to read volts. 2& 26 24 22 20 18 16 14 270 240 210 180 120 90 30 \ 4Ss I 10 IZ 14 2466 INCHES FIG. 222. Sparking Distances. Standard Conditions: oo sewing needles. 25 Centigrade. 760 mm. barometer. &o% humidity. Sinusoidal voltage wave form. (2) The ^>ar& gap. The voltage required to break down the resistance of the air between two needle points is approximately constant. If the distance between points at which the voltage of a given line breaks down the intervening air is determined, the voltage of the line may be determined by reference to a table or a curve. Fig. 222. ESSENTIALS OF ELECTRICAL ENGINEERING 5. Wattmeters. A wattmeter is a combination of a voltmeter and an ammeter operating on the same movable element, so that the deflection of a pointer is proportional to the product of the current in the circuit and the voltage between the terminals of the voltage coil. Two types of wattmeters have been developed: (a) electrodynamometer, (b) induction. (a) Electrodynamometer wattmeter. The electrodynamometer wattmeter consists of a fixed coil of large wire (ammeter element) connected in series with the load, and a movable coil of fine wire (voltmeter element) connected in parallel with the load. The ammeter element sets up a flux with which the voltmeter element reacts to cause a deflection of the fine FIG. 223. Schematic Diagram w j re C oil,* and the deflection is pro- of Dynamometer Wattmeter. portional to ^ p roduct of ^ currents in the coils, i.e., to the power in the circuit. This type of watt- meter indicates the power in either continuous- or alternating- current circuits. Fig. 223. Line Movable (Shunt) <\ c !l Fixed (.Series) Coils FIG. 224. Weston Wattmeter (Cover Removed). FIG. 225. Movable Element of Weston Wattmeter. (b) Induction wattmeter. The induction wattmeter is identical in principle with the induction ammeter and voltmeter described above, i.e.j deflection of a pointer is caused by the reaction between a shifting flux and the currents induced in a disc or drum. The ammeter element, which is wound with a few turns of heavy wire, has negligible inductance; the voltmeter element, which is wound with many turns of fine wire, has large inductance. The relations * See Chapter 2, Section 14. METERS 259 of the fluxes are as represented in Fig. 226a, i.e., the flux due to the parallel winding lags behind that due to the series winding by the angle 0. The angle in Fig. 226a is increased to 90 degrees by means of an auxiliary or compensating winding which is a short-circuited winding similar to a "shading coil," * except that it surrounds the entire pole. The compensating coil acts as the short-circuited secondary of a transformer, and sets up a flux the magnitude and phase i relations of which are represented by OC in ||%^^ 5H ^ Fig. 226b. The flux in the magnetic circuit ^| of the parallel winding is, then, the geometric If * .;-- ^uxdueto shunt 5 /AS : B Winding difference of that due to the shunt coil and \$^3v ,-, r ,1 , j. C*-"""" ''Resultant Flux. that set up by the compensating winding. ^ By properly proportioning the compensating FIG. 226. Vector Dia- winding, a flux in exact quadrature with that grams of Fluxes in the set up by the series winding is produced. Induction Wattmeter. The resultant of these quadrature fluxes is a shifting or rotating flux, and a torque which is proportional to the product of the voltage of the circuit and the load amperes. Like other induction apparatus, the induction wattmeter is applicable to alternating- current circuits only. That a properly designed wattmeter indicates the power in an alternating-current circuit when the load is either non-inductive or inductive is evident when the torque relations during a cycle aje considered. In a non-inductive single-phase circuit, the torque is constant in direction but varies in value from zero to maximum, and the wattmeter indication is proportional to the average torque. In an inductive circuit, the direction of the torque is not constant, and the wattmeter indication is proportional to the algebraic sum of the average positive and negative torques. The current and volt- age remaining constant, it may be proved both mathematically and experimentally, that the average net torque is proportional to the cosine of the phase angle.f The deflection of the movable system of a wattmeter is, therefore, proportional to the product of the current, the electromotive force, and the cosine of the angle by * See Chapter 13, Section 17. t See Chapter i, Section 28. 260 ESSENTIALS OF ELECTRICAL ENGINEERING which the current leads or lags behind the electromotive force, i.e., to the power in the circuit. 6. Polyphase wattmeters. For the measurement of power in polyphase alternating-current circuits, polyphase wattmeters have been designed. The polyphase wattmeter is a combination of two or more single wattmeter elements acting on the same movable part, and each pair of coils is connected as if it formed an inde- pendent instrument. 7. Watt-hour meters. A watt-hour meter is a small motor, the speed of the rotating parts of which is proportional to the power in the circuit to which it is connected. The movable parts of such motors are made very light, and friction is reduced to a minimum by the use of jewel bearings. Watt-hour meters are divided into two classes: (a) commutator meters, (b) induction meters. (a) Commutator meters. Commutator watt-hour meters are similar, in principle and in action, to the continuous-current shunt motor, and are applicable to either direct- or alternating-current circuits. The armature is wound of many turns of fine wire and is connected in series with a resistance, between the supply lines. The field, which is produced by the series windings, is propor- tional to the load current, the magnetic circuit being in air. With constant electromotive force between the terminals of the armature winding, the torque of the motor is proportional to the current flowing in the field windings. Therefore, to make the speed proportional to the power in the circuit, the load on the motor (counter torque) must be proportional to the speed. This is ac- complished by means of an eddy-current brake, consisting of a disc of copper or aluminum attached to the shaft of the motor, and rotated between the poles of a permanent magnet. When current flows in the coils of the meter, a torque is produced, and the movable parts of the meter increase in speed until the driving torque is balanced by the retarding torque of the permanent magnet and the eddy currents induced in the rotating disc. Since the torque producing motion is proportional to the power in the circuit to which the meter is connected, and the retarding torque is propor- tional to the speed of the rotating parts, the speed of the rotating parts is proportional to the power in the circuit. By means of a train of gears, the total number of revolutions made by the arma- METERS 261 ture is recorded, the calibration being such that the meter reads in either watt-hours or kilowatt-hours. For a given load, the speed of the rotating parts of a meter is changed by changing the position of the permanent magnet, which is made adjustable. Friction between the moving parts of the meter and the support- ing bearing, which would disturb the torque-speed relations at light loads to a very considerable extent, is compensated for by the addition of an auxiliary field winding connected in series with the armature winding, and so adjusted that when zero cur- rent flows in the series windings a slight jar causes the armature Series Field 'Series Field Schematic Diagram of Con- nections for Commutating Watt-hour Meter. to rotate. If the field set up by this auxiliary winding is too FIG. 227. strong, the meter " creeps," i.e., the armature rotates when the load circuit is disconnected; if the field set up by the auxiliary winding is too weak, the meter does not register on light loads. The electrical circuits of a commutator watt-hour meter are shown in Fig. 227. (ft) Induction meters. The induction watt-hour meter is, in principle, a two-phase squirrel-cage induction motor, the quadrature fluxes being produced in the same manner as described for the induc- tion watt-meter. The squirrel-cage element of the induction meter is an aluminum disc or cup which also serves as the movable part of the eddy current brake, thus making the movable system of an induction FIG. 228. Schematic Diagram of Con- me ter lighter than that of the COm- nections for Induction Watt- meter rr\ . and Watt-hour Meter. mUtat r ^ The tOr( ! Ue P Cr Umt weight is also greater. The neces- sity for friction compensation is, therefore, reduced, but is provided by means of a " shading coil"* on the shunt winding. An induction watt-hour meter is shown schematically in Fig. 228. * See Chapter 13, Section 17. 262 ESSENTIALS OF ELECTRICAL ENGINEERING Because of the principle on which it operates, the induction watt- hour meter is not applicable to continuous-current circuits. 8. Recording or graphic meters. It is often desirable to have a continuous record of the momentary fluctuations in the values of such electrical quantities as voltage, current, power, etc. Such records are obtained by means of a pen or pencil, the position of SHUNT COIL L*6HT LCA&-* ADJUSTMENT MAGNET BALANCE LOOP DISK POWER FACTOR /I ADJUSTMENT FIG. 229. Elements of Polyphase Induction Watt-hour Meter. Westinghouse Elec. & Mfg. Co. which is controlled by the magnitude of the quantity to be re- corded, and a uniformly moving paper on which the pen or pencil traces a record. A detailed description of the mechanism of such instruments is beyond the scope of this work. 9. The synchroscope. The synchroscope or synchronism indi- cator is a device by means of which the phase relations of two electromotive forces and their relative frequencies are indicated. The synchronism indicator used by the General Electric Co. is, structurally, a small synchronous motor the field winding of which is connected to the bus bars. Fig. 230. The movable coils are connected to the incoming machine through a phase- METERS 263 Movable Coils / Resistance splitting device. There is, then, superimposed on the field set up by the movable coils, an alternating flux due to the station- ary winding, and the movable parts assume a fixed position which is dependent on the phase relation between the electromotive force of the incoming machine and that of the bus bars, the frequencies of the two electromotive forces be- ing the same. Correct phase rela- tions for parallel connection of al- ternators (phase opposition) exist when the stationary (" shadow ") and movable pointers coincide. If the frequencies of the electro- motive forces are not the same, FIG. 230. Schematic Diagram of Con- a torque which causes the arma- nections . for General Electric s y nchr - . . nism Indicator. ture to rotate is produced. The direction of rotation is clockwise or counter clockwise according as the speed of the incoming machine is too fast or too slow, the rate at which the pointer rotates indicating the difference in the fre- quencies of the electromotive forces. The Weston synchro- scope operates on the elec- trodynamometer principle, the movable element vibrat- ing instead of rotating. By reference to Fig. 231, it will be seen that the fixed coils are connected to the bus bars in series with a slightly inductive resistance, and the movable coil to the incoming machine in series with a condenser. The relative values of the capacitance, the resistance and the inductance are such that the currents are in exact quadrature when the voltage of the incoming machine is in phase, or in phase opposition, with that of the bus bars. Under this condition no torque is exerted between the coils, and the pointer is at rest in the middle of the scale. When the electromotive forces are not in phase or in phase oppo- sition, a torque, proportional to the phase displacement, causes the FIG. 231. Schematic Diagram for Weston Synchroscope. 264 ESSENTIALS OF ELECTRICAL ENGINEERING movable coil to be deflected, and the pointer is moved to the right or to the left as the electromotive force of the incoming machine leads or lags behind that of the bus bars. If the frequencies are not the same, the phase displacement, and consequently the position of equilibrium, changes momentarily, and the pointer swings back and forth across the scale. The lamp, which illuminates the dial of the instrument, is so connected that it is dark when the voltages are in phase, and light when they are in phase opposition. Consequently, the pointer seems to rotate either clockwise or counter clockwise, the direction of apparent rotation indicating whether the incoming machine is too fast or too slow, and the rate of apparent rotation, the amount by which the frequencies differ. 10. Power-factor meters. If the stationary coils of a synchro- scope are connected in series with the load apparatus and the mov- able coils between the supply mains, the position of the movable element depends on the phase relations of the currents in the two coils as explained in Section 9. Since the phase relations of the currents in the coils are dependent on the power factor of the load circuit, the scale may be calibrated to read power factors. The Weston power-factor meter, The movable element of the Weston power- factor meter differs from that in the synchroscope in that it has two coils, the planes of which are at right angles. The stationary coils are connected in series with the load, or to the secondary terminals of a series transformer, and the movable coils across different phases of a polyphase system or are FIG. 232. Schematic Diagram of provided with a phase-splitting device. Connections for Weston Power- Fig. 232. If the current in the sta- factor Meter. ^ tionary coils is in phase with that in either of the movable coils, the plane of one movable coil coincides with the axis of the fixed coils; if the current in the fixed coil is not in phase with that in either of the mov- able coils, the movable system is deflected, and the angle of deflection is equal to the angle between the current in the sta- tionary coil and that in one of the movable coils, i.e., to the power-factor angle. METERS' 265 Iron Vane Westinghouse power-factor meter. The essential parts of a Westinghouse power-factor meter are: (a) a soft iron vane or armature of the shape indicated in Fig. 233, (b) two or more angu- larly displaced stator coils which are connected, through series transformers, with a polyphase sys- tem, (c) a stationary potential coil, the axis of which coincides with that of the iron vane. The stator coils set up a rotating flux which, when unaffected by other forces, causes the armature to rotate. FlG - 2 33- Schematic Diagram of ,. , ., , . ,, Westinghouse Power- factor Meter. The potential coil polarizes the arma- ture and, at unity power factor, it assumes a position at right angles to the stator coil, the current in which is in phase with the current in the potential coil. When the power factor of the system is not unity, the current in the potential coil is not in phase with that in the stator coil, and the armature is deflected through an angle equal to the angle of phase displacement. Polyphase power-factor meter. The indications of a split-phase meter are, obviously, affected by the frequency of the circuit to which it is connected. This difficulty is overcome by connect- ing the movable coils to a polyphase system, thus utilizing the inherent voltage displacement of such a system. The operation of such an instrument is in no other way different from that of the split-phase meter. ii. Frequency meters. Frequency meters indicate the frequency at which a system is operating, and are of two types: (a) vibrating-reed meters, (b) split-phase meters. FIG. 234. Principle of Operation of Vibrat- ( a ) Vibrating-reed meters. ing Reed Frequency Meter. Jf a ^^ of ^ ^^ d j f _ fering in length and in inertia, are arranged in front of an alternating-current magnet, as indicated in Fig. 234, those strips whose natural period of vibration corresponds to the frequency of the alternating-current system are thrown into violent vibration, while the others are affected only slightly or not at all. Steel Reeds . \ *Solder Weight Electro-ma'qnet AC. Supply 266 ESSENTIALS OF ELECTRICAL ENGINEERING (b) Split- phase meters. Split-phase frequency meters are essentially differential induction voltmeters. One coil is con- nected to the mains in series with a non-inductive resistance and its current is, therefore, practically independent of frequency; the A. C. Line r Inductance Resistance > I'l'l'l'l 1 Zl 22 23 25 26 27 26 FIG. 235. Section of Vibra ting-Reed Fre- quency Meter Scale Showing Method of Indication. 1 Aluminum Disc \ n FIG. 236. Schematic Diagram of Con- nections for Induction (Split-phase) Frequency Meter. other coil is connected to the same mains through an inductive resistance which causes the current to vary as the frequency changes. At normal frequency, the torque of one coil is equal and opposite to that of the other; as the frequency of the system in- creases or decreases the torques are no longer equal, and equilibrium is restored by a deflection of the movable element. Fig. 236. In the Weston frequency meter two stationary coils are inter- connected with resistances and reactances so as to form a Wheat- stone bridge which, at normal frequency, is balanced. As the frequency increases or decreases, the bridge is unbalanced, and a movable element of soft iron caused to deflect. The electrical FIG. 237. Westinghouse (Split-phase) Frequency Meter. connections and the movable element of this meter are shown in Fig. 238. 12. Ground detectors. A " ground " is a connection between a current-carrying conductor and the earth, which materially reduces the normal insulation resistance of the line. A ground METERS 267 cannot be stated to exist when the resistance between the con- ductor and the earth falls below a certain fixed minimum, because of the different insulation requirements of different systems what is good insulation under one condition may be a very serious ground under other circumstances. A ground on one line of a system does not, in itself, cause any trouble, but if two lines become grounded, a serious leakage occurs, Pointter Line FIG. 238. Schematic Diagram for Western Frequency Meter. FIG. 239. Schematic Diagram of Static Ground Detector. or a short-circuit is established. Since a line may accidentally become grounded at any time, it is good practice to have on the switchboard, instruments which indicate the fact whenever the resistance between any line and the earth becomes seriously re- duced. Such instruments are known as "ground detectors." The static ground detector operates on the same principle a. the electrostatic voltmeter, i.e., the attraction between two oppo- sitely charged bodies. Let A and C in Fig. 239 be curved plates connected to current-carrying conductors, as indicated; B a plate which is free to move about the pivot and which is connected, through a negligible resistance, to the earth. If the lines are equally insulated, equal and opposite forces tend to deflect plate B * from its position midway between plates A and C. If the lines are not equally insulated, the equilibrium of forces acting on B is destroyed, and B moves toward the plate (A or C) which is con- nected to the line having the higher (better) insulation, and the deflection is proportional to the relative resistances between the lines and the earth. * See Appendix C, Section 2. CHAPTER XVIII POWER TRANSMISSION AND DISTRIBUTION* THE advantages of power transmission by means of the electric current are such that this method has practically superseded all other methods except for very limited distances. Within the last few years a number of large hydroelectric plants have been com- pleted, the power from which is transmitted one hundred miles or more at voltages up to 140,000. The transmission line is the connecting link be- tween the generators and the distributing network, and is usually run over a private right of way, the current- carrying conductors being supported by wooden poles or steel towers. Fig. 240. Conductors of distributing systems are carried on poles or are placed in conduits underground. i. Conductors. - - Copper wires or cables are almost universally used for the trans- mission and the distribution of electric power, although FIG. 240. aluminum is used to some extent because its light weight makes the number of supporting structures required a minimum. The sizes of commercial conduc- tors are expressed by their areas in circular mils, or by gauge * For an extended discussion of transmission problems, the details of pole line design, etc., the reader is referred to "Overhead Electric Power Transmission " by Alfred Still, and to "Elements of Electrical Transmission" by O. ]. Ferguson. 268 POWER TRANSMISSION AND DISTRIBUTION 269 numbers. The American Wire Gauge (A. W. G.), often termed Brown & Sharpe (B. & S.), is universally used in the United States. A characteristic of the A. W. G. which makes it easy to ap- proximate the area corresponding to any given gauge number without the use of a wire table, is the fact that No. 10 is approx- imately o.i inch in diameter and has an area of approximately 10,000 circular mils, and that the area halves or doubles, approx- imately, for each three gauge numbers, i.e., No. 7 has an approx- imate area of 20,000 circular mils while No. 13 has an approximate area of 5000 circular mils. Gauge numbers and areas of wires, with their weights and resistances, are given in Table VI. 2. Insulation. For many purposes it is required that current- carrying wires be covered with some insulating material. The amount and the quality of this insulation depends on the use for which the conductor is intended, and may consist of: (a) a cotton braid, (b) a cotton braid impregnated with a liquid compound which dries and becomes hard, (c) a coating of vulcanized rubber over which, as a mechanical protection for the rubber, is wound a braid having a polished surface. Rubber insulation is required for all interior wiring by a rule of the National Electric Code (N. E. C.), which has been adopted by the national engineering societies as well as by the National Board of Fire Underwriters. Wires and cables intended for underground use are thoroughly insulated, then covered with a (a) Suspension Insulator. (b) Pin Insulator. FIG. 241. Line Insulators. The Ohio Brass Co. continuous moisture-proof lead sheath over which may be wound a braid, or other mechanical protection. For insulating aerial conductors from their supporting structures glass or porcelain insulators are used. Two types of line insulators are in general use, illustrations of which are shown in Fig. 241. 270 ESSENTIALS OF ELECTRICAL ENGINEERING Pin- type insulators are generally used for voltages up to 66,000; suspension type for voltages greater than this. 3. Carrying capacity of conductors. The current which an insulated conductor may safely carry is limited by the temperature at which the insulation softens, or by reason of which its insulating properties deteriorate. The maximum current allowed by the N. E. C. is given in Table VII for rubber and for other insulations. 4. Inductance. A magnetic field which opposes* any change in the value of the current is set up around any current-carrying con- ductor.f Alternating-current circuits are, therefore, subject to an inductance which varies with the distance between the conductors and with their size. Inductances for different sizes of wires and for different spacings are given in Table VIII, the values being those obtained by the following formula: J 2 D L = 0.03028 + 0.282 logio > (i) when L = the inductance in millihenries per 1000 feet of two- wire circuit (2000 feet of conductor), D = the distance between wires in inches, d = the diameter of the wires in inches. The inductance of each wire in a three-phase system is one-half that of the loop formed by any two of the three conductors. 5. Capacitance. Two or more metallic conductors separated by air or other insulating material, form a condenser, the effect of which is to cause a leading current to flow in an alternating- current circuit. The value of this leading current, termed the charging current, is proportional to the applied voltage, to the capacitance of the circuit, and to the frequency. I e (single phase) = 2 w/EC io~ 6 , (2) /i \ 4 irfEC IO" 6 f N I e (three phase) = * J > (3) V 3 when E = the line-to-line voltage, C = the capacitance in microfarads, / = the frequency of the supply circuit, I c = the charging current. * Because of the electromotive force induced in the conductor when the magnetic field changes. t See Chapter 2,' Section 3. J See Appendix B, Section i. See Appendix C, Section 9. POWER TRANSMISSION AND DISTRIBUTION 271 The capacitance of a transmission line varies with the distance between the wires, with their diameter, and with the nature of the insulating material between the conductors. The capacitances given in Table IX were calculated by means of the following for- mula: * C-VS&1, (4) lo glo when C = the capacitance in microfarads per 1000 feet of two-wire circuit (2000 feet of conductor), D = distance between wires in inches, d = diameter of conductors in inches. 6. Skin effect. When an alternating current flows in a con- ductor, the current density over the cross-sectional area is not uni- form, but increases as the distance from the center of the conductor increases. This is the "skin" effect, and increases the resistance losses in the conductor. The increase in the resistance of a con- ductor to alternating currents as compared with its resistance to continuous currents is a function of the product of the area of the conductor and the frequency of the alternating current. The skin effect is negligible when the product of the area of the con- ductor in circular mils and the frequency does not exceed 30,000,000, or, for commercial frequencies, when the diameter of a conductor is not greater than three-quarters of an inch. 7. Line calculations. The inductance and the capacitance of a transmission line are made up of an infinite number of induc- tances and capacitances uniformly distributed over the system. An exact solution of such a system involves equations too compli- cated for practical use, but an approximate solution giving fairly accurate results is easily made when the capacitance is assumed to be concentrated, e.g., one-half at each end of the line. Let EI = the voltage at the terminals of the load, E d = the voltage drop in the line, E = the voltage at the terminals of the generator, I I = the load current, I c = the charging current due to the capacitance at the load end of the line, * See Appendix C, Section 9. 272 ESSENTIALS OF ELECTRICAL ENGINEERING I = the line current, L = the inductance of the line, R = the resistance of the line, cos < = the power factor of the load circuit, ' = the angle between EI and 7, The relations between E h I t and 7 C are known, and their vectors may be plotted as shown in Fig. 242, and the value of 7 and that of the angle ' determined. ^r I = V7 i 2 cos 2 4>+(7,sin0-7c) 2 , (5) (6) 1 1 COS also, FIG. 242. Vector Diagram for Trans- ^ T /7^ T ^ T ,-r, L d = 1 V K -\- mission Line. * Adding, vectorially, the line drop and the load voltage, the volt- age at the generator terminals is obtained. E = V[E t + E d cos (e - 0')] 2 + E 2 sin 2 (0 - 0')- (8) Problem Single phase. 500 kw. are to be transmitted a dis- tance of 20 miles, the voltage at the load end (primary terminals of step-down transformers) is 25,000, the frequency is 25, and the power factor of the load circuit is 86.6 per cent. Find: (a) the drop in the line, (b) the voltage at the generator terminals. C = 0.00155 X 5.28 X 20 = 0.1637 microfarad. I c = 2 TT X 25,000 X 0.1637 X 25 X 10 = 0.64 ampere. L = 0.694 X 5.28 X 20 X 10 = 0.073 henry. wL 0.073 ^ X 57 = 11-26 ohms. R = 0.3258 X 20 X 2 = 13.03 ohms. r 500,000 1 1 = - ^ = 23.1 amperes. 25,000 X 0.866 I = ^(23.1 X Q.866) 2 + (23.1 X 0.5 - o.64) 2 = 22 ' 8 am P eres - E d = 22.8 v / (i3.o3) 2 + (n.26) 2 = 392 volts. 6 = tan" 1 0.864 = 40 50'. 1 1. 6 0.64 20 =0.548 = 28 45 = 12 5 '. "^(25,000 + 392 X 0.977)2 +'(392 X 0.209)" = 25,338 volts. POWER TRANSMISSION AND DISTRIBUTION 273 If greater accuracy is required, the charging current may be taken as that due to the mean voltage on the line, and a second approximation made. This is, however, seldom necessary as it is only at very high voltages, or over long distances that the capaci- tance of transmission lines becomes very marked, and, in many cases, it may be disregarded entirely as having no appreciable effect on the voltage regulation of the line. Problem Three phase. 2000 kw. are to be transmitted over a distance of 20 miles, the conductors used are ooo spaced 48 inches apart, the voltage at the load terminals (line to line) is 25,000, the frequency is 25, and the power factor of the load circuit is 86.6 per cent. Assume the capacitance to be concentrated at the ends of the line. Find: (a) the drop in the line, (b) the voltage between the generator terminals. C = 0.00155 X 5.28 X 20 = 0.1637 microfarad. L I 4*_X 25 X 2 5 ,ooo_X 0.1637 X io-* = 2 V 3 r 0.604 X <;.28 X 20 X io~ 3 L = - - = 0.037 henry. coZ, = o.o 37 X 157 = 5.75 ohms (per line). R = 0.3258 X 20 = 6.5 ohms (per line). 2,000,000 1 1 = - ~^= = 53-5 amperes. 25,000 X 0.866 X V3 I = ^(53-5 X Q.866) 2 + (53-5 X 0.5 - Q-37) 2 = 53-3 amperes. E d = 53-3 v-S) 2 + (5-75) 2 = 454 volts. 6 = tan" 1 0.8847 = 4 l0 3' <' = tan" 1 0.569 = 29 40'. 6 - 0' = 11 50'. E = V f^P 2 + 454 X o. 97 8) 2 + ( 4 54 X o. 2 o 5 ) 2 \ V 3 / = 14,895 volts from line to neutral 25,768 volts from line to line. 8. Distributing systems. The distributing systems in common use at the present time are: (a) series, (b) parallel, (c) series- parallel, (d) multiple wire. (a) The series system. In the series system the entire current flows successively in each piece of apparatus, the voltage varying as the load changes, and the current remaining approximately constant. It is largely used in alternating current street lighting 274 ESSENTIALS OF ELECTRICAL ENGINEERING installations in connection with a constant-current transformer, either with or without a mercury arc rectifier. (b) The parallel system. The parallel system is the one com- monly used in supplying motors and incandescent lamps. The current divides before reaching the lamps or the motors, so that the current flowing in one piece of apparatus may be greater or less than that flowing in another piece. (c) Series-parallel system. It is undesirable to increase the voltage of a distributing system above that required to operate about one hundred arc lamps connected in series. Therefore, in large installations, the lamps are connected into series groups, and these groups are connected in parallel, each group, if the supply is from alternating-current mains, being supplied through a constant- current transformer. (d) Multiple-wire systems. When the power in a system is con- stant, the current decreases as the voltage increases. If the line __^ rrrrr ^ oss ^^ * s constant > tne resistance WIT of the conductors is four times as itra I \\\\\ f an ce -I 11 great when the voltage of the system I Converfer -HI i s doubled, e.g., the cross section (or FIG. 243- The Rotary Converter we i g h t ) o f w j re used in a 22O -volt as a Three- wire Balancer. . . system is only one-quarter that used in a no-volt system, the line loss and the total power delivered be- ing the same in each case. This fact has led to the development of multiple- wire systems which effect a large saving in copper without increasing the voltage between terminals of the load apparatus. The only one of these multiple-wire systems in extensive use is that using three wires, a greater number of wires adding undesirable complications in both the generating and the distributing system. The simplest application of the three-wire system is in connection with an alternating-current system where the third or neutral wire is connected to the middle point of the secondary winding of a transformer, and the load is connected between each of the main wires and the neutral.* The application of the three-wire principle to continuous-current distribution requires the use of: (i) a rotary converter and induc- tance coils, (2) a special generator, (3) a motor-generator balanc- ing set. * See Chapter 12, Section 2. POWER TRANSMISSION AND DISTRIBUTION 275 A rmature , Winding I POLE Brush, Positive Wire - Neutral Wire n Meaa five Wire Auxiliary Winding FIG. 244. Schematic Diagram of Crocker- Wheeler Three-wire Generator. Field Rheostat (1) Rotary converter. If the rings of a rotary converter are connected through suitable inductance coils, as indicated in Fig. 243, the converter armature acts as a balance, and the voltage between either main and a neutral connected to the junction of the inductance coils is approx- imately equal to one-half that between the mains, whatever may be the relative value of the currents in the commutator. - mains. (2) Three-wire generators. Three-wire generators are essentially rotary converters, the neutral connection to the armature being made through an auxiliary winding or through inductance coils. The armature of the Crocker- Wheeler three-wire generator carries three auxiliary windings which are star-connected to the main winding and to a single collector ring, as shown in Fig. 244. The Triumph Electric Company's three-wire gen- erator makes use of star-connected external inductance coils. (3) The motor-generator balancer. If two similar shunt dynamos are mounted on the same shaft and con- nected as indicated in Fig. 245, a three-wire system is produced. The action of a balancer set is as follows: As long as the load is balanced, i.e., equal currents flow in the main wires, zero current flows in the neutral, and the currents in the armatures of the balancer set are equal. When the currents in the mains are un- equal, their difference flows in the neutral, and the current in the armature connected between the neutral and the main carrying the smaller current increases. This increased armature current causes the speed of the armature to increase, and the other dynamo is driven as a generator. The speed of the balancer set increases until equilibrium is reestablished, and the voltage between each main and the neutral is automatically maintained at a value equal, approximately, to one-half that between the mains. Neg- lecting the losses, the armatures of the balancer set are required FIG. 245. Motor-generator Balancer Set. 276 ESSENTIALS OF ELECTRICAL ENGINEERING to carry the current flowing in the neutral wire, and this current divides inversely as the voltages between the neutral and the main wires. The voltage regulation of a three-wire system is improved by the use of compound balancers, the series field windings being so connected that the motor is differentially AAiiiiiii wound and the generator cumulatively < ) wound. 9. Voltage regulation of parallel sys- tems. For the satisfactory operation of incandescent lamps the electromotive force between the terminals of the lamps I }' } {* r I { { { mus t t> e maintained at an approximately (c) ' constant value. It is not difficult to design a wiring system for a group of FIG. 246. Parallel Distri- , , , , . , -f 11 bution lamps that is operated as a unit, ^.e., all the lamps in the group are either lighted or not lighted, but to design a wiring system for the same group of lamps, any part of which may be in use at a given time, may be a rather complicated problem. The general problem is to determine the center of distribution (the center of gravity) of the group, at which the voltage should be maintained constant by means of feeder regulation, and to propor- tion the wiring between this center and the individual lamps, so that the variation of voltage at the most disadvantageously sit- uated lamp never exceeds the allowable limits. To meet the requirements of a distributed load, various wiring schemes have been devised, their object being to minimize the dif- ference in voltage between the terminals of different lamps in the group. Some of these schemes are indicated in Fig. 246. 10. General wiring formulae. The following expressions will be found useful in the solution of wiring problems: p I = = - for single-phase or continuous-current circuits. E X cos (9) p I = - for two-phase circuits. (10) 2 E X cos p I = - for three-phase circuits. (n) V 3 E X cos POWER TRANSMISSION AND DISTRIBUTION 277 21.6 X D X / X k , Ed = - - - for continuous, single-phase or two- \^s 1VJL phase circuits. (12) 18.7 X D X / X k , ^ Ea = - - for three-phase circuits. (13) when C.M. = the circular mil area of the conductor used, D = the distance of transmission in feet, E = the line voltage at terminals of the load, Ed = the volts lost in the line. / = the line current, k = the impedance factor / im P edanceN ) . (Tables XV or \ resistance / XVI.) Cos = the power factor of the load circuit, P = the total power (watts) delivered to the load circuit. Example. 30 k.w. are to be transmitted over a distance of 500 feet, the voltage at the terminals of the load is 440, and the fre- quency is 60, single-phase. Find the drop in the line when No. 4 wires spaced 18 inches apart are used, and the power factor is 0.85. *- - = 80.2 amperes. 440 X 0.85 ,-, , 21.6 X 500 X 80.2 X 1. 12 E a ' = - - = 23.2 volts. Example. 100 k.w. are to be transmitted a distance of 300 feet at a voltage of 440, a frequency of 60, a power factor of 0.85, three- phase. Find the size of wire required, the line drop not to exceed 4 per cent. T 100,000 V 3 X 440 X 0.85 = 155 amperes. From Table VIII the minimum size wire allowable is No. i the area of which is 83,690 c.m. Substituting in equation (13) Ed _ 18.7 x 390x155x1.37 . volts> 83,690 which is within the allowable limits and should be used. 2 7 8 ESSENTIALS OF ELECTRICAL ENGINEERING FIG. 247. Mershon's Diagram. ii. Mershon's diagram. Graphical solution of alternating- current distribution problems may be made by means of Mershon's diagram.* Referring to Fig. 247 let the radius of the smallest arc represent the voltage at the end of the feeder, or at the center of distribution; AB represent the resistance drop in the line, expressed as a per- centage of the delivered voltage; BC represent the reactance drop in the line, expressed as a per- centage of the delivered voltage. * Originated by Ralph D. Mershon, Past-President of the American Institute of Electrical Engineers. POWER TRANSMISSION AND DISTRIBUTION 279 Lay off AB beginning at a point A where the arc cuts the verti- cal representing the power factor of the load circuit; lay off BC perpendicular to AB, as indicated in the diagram. The generator voltage is indicated by the position of the point C and may be read directly from the diagram. The power factor of the feeder circuit is determined by the point C where a line drawn from C to the origin cuts the arc. 12. Feeder regulation. When several feeders take current from the same bus bars, it is desirable to be able to regulate, independently, the feeder voltages. Feeder regulation may be accomplished by the use of: (a) boosters, (b) auto- transformers, (c) induction regulators. Auto-transformers and induction regulators are applicable to alternating-current circuits only; boosters are used on continuous- current circuits. (a) Boosters. A line booster consists of a series generator driven by a shunt motor, or other constant-speed engine. The armature Switch \Auto- trans former ,-Jj FIG. 248. Diagram of Booster Connections. FIG. 249. Diagram of Connections for Auto-transformer Feeder Regulator. and the field of the generator are connected in series with the load, as indicated in Fig. 248. The voltage of the generator is propor- tional to the current flowing in the feeder and is added to the bus-bar voltage, and any desired increase in the voltage may be obtained by properly proportioning the booster field. Because of the addition of two rotating machines, the efficiency of the system is reduced and the operating complications greatly increased. (b) Auto-transformer. If an auto-transformer is connected as indicated in Fig. 249, the number of effective turns in the secondary depends on the position of the switch. The Stillwell regulator operates on this principle, and is provided with a reversing switch so that the secondary voltage may either "boost " or "buck " that of the bus bars. This type of regulator changes the'voltage by definite steps. 280 ESSENTIALS OF ELECTRICAL ENGINEERING (c) Induction regulators. The induction regulator is, essentially, an auto-transformer in which the angular relations of the primary and secondary coils may be changed. (c) Complete. FIG. 250. Six-phase, Motor-operated Induction Regulator. General Electric Co. Structurally, the polyphase induction regulator resembles the polyphase induction motor with a wound rotor. The primary windings, equal in number to the number of phases in the system POWER TRANSMISSION AND DISTRIBUTION 281 on which it is to operate, are symmetrically placed in slots on the surface of a movable laminated iron drum (Fig. 25oa), and are connected between the lines of a polyphase system. The secondary windings are symmetrically placed in the slots of a stationary core (Fig. 25ob), and are connected in series with the load apparatus. The primary windings set up a rotating flux which induces a con- stant electromotive force in the secondary windings. The feeder voltage is the vector sum, or difference, of the bus-bar voltage and the voltage induced in the secondary winding of the regulator. Since the phase relation of FIG. 251. Clock Diagram of the bus-bar voltage and the induced Polyphase Induction Regu- secondary electromotive force depends on lator Volta s es - the angular position of the movable coils, the bus-bar voltage may be raised or lowered by an amount equal to the secondary voltage of the regulator. Fig. 251. The single-phase induction regulator differs, both in principle and in structure, from the polyphase regulator. A schematic diagram of the single-phase regulator is shown in Fig. 252. The voltage induced in the secondary coil is proportional to the cosine of the angle between the axis of the primary coil and that of the secondary winding. The in- duced electromotive force is, there- fore, .zero when the coils are at right FIG. 252. Schematic Diagram of angles. Since the flux set up by the Smgk -phase Induction Regu- pri m ary co il passes through the sec- ondary coil in one direction when the angle is less than 90 degrees, and in the opposite direction when the angle /3 is greater than 90 degrees, the induced electro- motive force is in phase with, or in phase opposition to, the bus-bar voltage as the angle j3 is greater or less than 90 degrees. The short-circuited winding which is placed at right angles to the primary coil reduces the reactance of the secondary winding, and thus improves the power factor of the feeder circuit. By reas >n of their angular relations, the effect of the short-circuited coil increases as that of the primary winding decreases. Feeder regulators are operated either by hand or by motor. 282 ESSENTIALS OF ELECTRICAL ENGINEERING When motor operated, the regulation may be made automatic by means of a contact-making voltmeter which causes the motor circuit to be closed when the voltage becomes either too high or too low. 13. Regulating effect of a constantly-excited synchronous motor. - It was shown in Chapter 9 that the power factor of a synchro- nous motor depends on the relative values of the applied and the counter-electromotive force. Let the field excitation of a synchro- nous motor at the end of a feeder be such that the power factor of the feeder circuit is unity. If the load increases, the voltage at the terminals of the motor drops, the current leads the electromotive force, and the leading current tends to increase the voltage to its former value; if the load decreases, the voltage at the terminals of the motor increases, the current lags behind the electromotive force, and the lagging current tends to decrease the voltage to its former value. The tendency, therefore, of a constantly-excited synchro- nous motor, when located at the end of a feeder, is to maintain constant voltage at the center of distribution. 14. Voltmeter compensation. Series (differential) ^^ _ When feeder regulators are used it is -shunt winding required that the voltage at the end of the feeder, or at the center of distribu- tion, be indicated by an instrument at FIG. ,53. Voltmeter Compensa- ft ' tf For ' continuous . current tion (C. C. Circuits). feeders, it is only necessary that a prop- erly proportioned series winding be added to the voltmeter. Fig. 253. The series winding is so connected that it opposes the effect of the shunt coil, and the voltmeter indication is reduced by an amount equal to the drop in the line. Since the drop in an alternating-current feeder is not a simple ohmic drop, but the combined effect of the resistance and the react- ance of the conductors, the voltage at the end of an alternating- current feeder is indicated by a voltmeter only when the voltage and current relations in the instrument circuit are identical with those in the feeder circuit. This condition is effected by properly proportioning the resistance and the reactance of the secondary cir- cuit of a series transformer, and connecting the voltmeter as indi- cated in Fig. 254. Commercial compensators are made so that their resistance and reactance may be varied, thus making the same ap- paratus applicable to different circuits. POWER TRANSMISSION AND DISTRIBUTION 283 15. Lightning arresters. The effect of a lightning discharge on electrical apparatus may be either direct or indirect. The direct effect is the destruction of the insula- tion; the indirect effect is the estab- lishment of a low-resistance circuit which may be maintained by the nor- mal voltage of the system. A satis- factory lightning arrester must, then, divert or dissipate the energy of the discharge, and promptly interrupt any low-resistance circuit that may be es- Feeder i. V Series Transforme Reactance *~z* SL Resistance tablished for the purpose of diverting FlG . 254 . schematic Diagram or dissipating this energy. of Connections for Voltmeter The fundamental operation of light- Compensator (A. c. Circuits). ning arresters will be explained by reference to Fig. 255 in which G is an air gap between the conductor and a low-resistance ground connection and A is an inductive coil con- nected in series with the line. Under normal operating conditions the impedance of A to ftie flow of the load current, either con- tinuous or alternating, is small, while the FIG. 255. Elementary resistance o f the air gap G is so large that Lightning Arrester. . tne leakage is negligible. Since the impe- dance of A is directly proportional to the frequency, the extremely high-frequency lightning (oscilla- tory) discharge is " choked" back, breaks down the resistance of the air gap and discharges to earth. It is a well-known fact that an electric arc, when once estab- lished, is maintained by a much smaller voltage than is required to establish it. Consequently the arc established by the lightning discharge may be maintained by the normal voltage of the system FlG * 256 * unless means are taken to suppress it. This is done by: (a) the use of " non-arcing" metals, (6) the horn gap. Westinghouse Choke Coil (Air Cooled). 284 ESSENTIALS OF ELECTRICAL ENGINEERING (a) Non-arcing metal arresters. When the air-gap terminals are made of zinc or cadmium or of their alloys, it is found that the arc is not maintained after the passage of the lightning discharge. Arresters are, therefore, made of short cylinders of brass separated by a small air gap. Arresters having part of the gaps shunted by resistance have come into extensive use. Their construction is based on the theory that, for a given voltage, high-frequency discharges require a larger number of gaps than do discharges of low frequency. Fig. 257 shows a 2300- volt General Electric arrester having a high resistance and a low-resistance shunt. Low-frequency discharges pass through the high resistance (long carbon rod) and two air gaps; medium frequency discharges through the low resistance (short carbon rod) and FIG. 257. G. E. four air gaps; while high-frequency discharges Lightning Arrester. ^ through ^ ^^ ^^ Q ^ ^ Westinghouse practice differs from the above in that there is a shunt and a series resistance, as indicated in Fig. 258. '.Shunt Resistance ; Series Resistance Ground Horn\ f ftp V Line Ground FIG. 258. Schematic Diagram of West- inghouse Lightning Arrester. FIG. 259. Schematic Diagram of Horn Gap Arrester. ; (b) Horn gap arrester. The horn gap arrester has the appear- ance shown in Fig. 259, and is connected to ground in series with resistance. The lightning discharge breaks down the air gap and forms an arc between the horns. Air currents formed by the heat of the arc carry the arc itself upward, increasing the length of the arc until it can no longer be maintained by the normal voltage of the system. During the past few years there has been developed an electrolytic cell (Fig. 260), which replaces the resistance used in connection with POWER TRANSMISSION AND DISTRIBUTION 285 the horn gap. This cell consists of a series of cone-shaped aluminum elements, the spaces between which are partly filled with electrolyte, and the whole is immersed in oil. The purpose of the oil is to in- crease the insulation, prevent evaporation of the electrolyte, and dissipate the heat liberated during discharge. The action of the cell is valve-like in that, at a definite critical voltage, the resistance breaks down, and a small increase in voltage above this critical value causes a large current to flow. When the voltage drops below the critical value, which is about 40 per cent above the normal voltage of the system, the high-resistance prop- erty of the cell is restored and the horn gap promptly quenches the arc formed by the discharge. Choke Coil Ground FIG. 260. Cross-section of General Electric Electrolytic Cell. FIG. 261. An Approved Connection for Three - phase Electrolytic Lightning Arrester. The high resistance of the aluminum cell is due to a film which is deposited on the aluminum elements when a current flows through the cell. Since this film tends to dissolve, the cell is maintained in operating condition by periodically bridging the horn gaps and allowing a momentary discharge through the cell. The electrolytic cell is, by construction, a condenser in which, for a constant applied voltage, the current is directly proportional to the frequency (7 = 2 irfCE). A high-frequency lightning discharge is, therefore, diverted through the cell to the ground. The air gap between the horns prevents the small leakage current that would flow if the cell were connected directly to the line. 286 ESSENTIALS OF ELECTRICAL ENGINEERING TABLE VI DIMENSIONS, WEIGHTS AND RESISTANCES OF COPPER WIRES Diameter = when n = gauge number. circular mils Weight (pounds) per 1000 feet Resistance (ohms) per 1000 feet 33 iQ.354 circular mils No. A.W.G. (B. & S.) Diameter, inches Area, C.M. Weight-pounds Resistance, 20 C. looo feet MUe 1000 feet Mile oooo 0.460 211,600 640.5 338i o . 04893 0.2583 ooo 0.407 167,800 508.0 2682 0.06170 0.3258 oo 0.365 133,100 402.8 2127 0.07780 0.4108 o 0.325 105,500 319.5 1687 0.09811 0.5180 I 0.289 83,690 253-3 1337 0.1237 0.6531 2 0.258 66,370 200.9 1062 0.1560 0.8237 3 0.229 52,630 159-3 841.1 0.1967 1.0386 4 0.204 41,740 126.4 667.4 o . 2480 I.3094 5 0.182 33.100 IOO.2 529-0 0.3126 1.6516 6 o. 162 26,250 79.46 4I9-5 o . 3944 2.0824 7 0.144 20,820 63.02 332.7 0-4973 2-6257 8 0.129 16,510 49.98 263.9 0.6271 3-3III 9 o. 114 13,090 39.63 209.2 o . 7908 4-1754 10 O. IO2 10,380 31-43 166.0 0.9972 5-2652 ii 0.091 8,234 24-93 131.6 1-257 6.637 12 O.oSl 6.530 19.77 104.4 1.586 8-374 Resistance of aluminum wire = 160 per cent of copper wire. Weight of aluminum wire = 30 per cent of copper wire. POWER TRANSMISSION AND DISTRIBUTION 287 TABLE VII CURRENT CARRYING-CAPACITY OF COPPER WIRES National Electric Code A.W.G., (B.&S.) Sectional area, circular mils Rubber insula- tion, amperes Other insula- tion, amperes 18 1,624 3 5 16 2,583 6 8 14 4,107 12 16 12 6,530 17 23 IO 10,380 24 32 8 16,510 33 46 6 26,250 46 65 5 33-100 54 77 4 41,740 65 92 3 52,630 76 no 2 66,370 QO 131 I 83,690 107 156 o 105,500 127 i85 oo 133,100 ISO 220 qco 167,800 177 262 oooo 211,600 2IO 312 200,000 2OO 3OO 300,000 27O OW 4OO 400,000 0/v 330 *T WV i?OO 500,000 oo w 3QO O^*" 7 coo 600,000 ov 450 ov w 680 700,000 *TO coo 760 800,000 O vv eeo 840 000,000 00 600 W*f.W O2O 1,000,000 650 yw IOOO Current-carrying capacity of aluminum wire = 75 per cent of copper wire. 288 ESSENTIALS OF ELECTRICAL ENGINEERING ll II Q ^ 8 rJ-O 00 M vo ON . ON O tM t~ t^OO OO - 00 00 00 00 NO NO 00 OO M VO ON tO OO ON ON ON O O cs cs 00 cs NO O ON ON ON ON o o o o O O O O O O O O O O O O $ ^ Tj- TfNO rt-00 N ON O t^ t^ t^OO dodo 00 00 O O O Tt ON tO CM CO -^O CO oo oo oo o d d d . i>- t^ r^ d d d d \O NO OO OO Td-oo ON <^>X) ON O . t^ t>. t^ O O CM cs VO ON to t^ NO t^ ON O t^ i^ r^oo 0000 O O O O 0000 O O O O vQ OO OO O O ^-00 fO O r^oo O es cs -^ TJ- t^. M vo ON M CO Tf VO NO NO oo oo CO t-- M VO I>.00 O M 8O CM CM ^00 CM co ^- vo t^ M 0000 O O O O 0000 O O O O VO VO IO VO vo OO 00 O O t^ M NO O NO OO ON M vo vo lOO (M M rj- rj- Tj-00 CS NO ? ^33.% 3%<3 i?^?^S 8888 8888 8888 8888 dddd dddd dddd dddd * OOOlO uoio^-Tj- Tl-^rorO fO^^M 8888 8888 8888 8888 OOOO OOOO OOOO OOOO Ol OO ^" O O W O\ *O ^ ^O <*O O OO *O * ^~ ^* O 1 ^ O^ o^oo oo t^ r>o vo *O *o *o l o ^ ^ ^j* co 8888 8888 8888 8888 ddoo oood oood ddoo N H O O O OOO OO t^. t~>. t-^-O O 10 u-> 10 Tf 8888 8888 8888 8888 OOOO OOOO OOOO OOOO <0 ro ro ^" O O V 1 OOO C"O r^ ^ t^ w OO to o^ OOTt-ro NCSMO OO OOO 00 t^ l^O 8888 8888 8888 8888 oood dddd dddd ddoo '4 OOOO MNco^' too t~oO O O M OOO M M M 289 2 bjQ C/5 290 ESSENTIALS OF ELECTRICAL ENGINEERING CS CO Tj- to S 8 g; O 0^8 oooo OOOO 0000 oooo oooo oooo 3 00 ON O IH 00 ON M CN to too o 0000 M CO "^ to CO rf too O t- ON O o o o ^* oooo M CS CO ^ > CN CO Tf IO OOOO oooo oooo oooo oooo N co ^h too to to too oooo M CS CO 1 O ^o *o ^o ^o O t^^OO ON O IH * : OOOO j o o o d 5 to to io to OOOO O O O O M Ol CO ^J" tOO t^OO to to to IO OOOO o o d o tOVQ VO VO oooo d d d d O IH OO ON O tOO t^ ON M cs co HT O IH CN CO oooo oooo > > > > ^ ^0 " ^0 oooo oooo oooo OOOO 00 III! too 1^-00 00 ON O M Tf ^" IO to oooo ON O IH - CN CO ^ IO O OO ON O ? oooo d d d d o o o d OOOO o o o o o" o" o" o d d d 6 j?O .N>8 co^ 888 o o M CS CO -* o fcj ^^ POWER TRANSMISSION AND DISTRIBUTION 2QI fc ^ H Ix! u PQ u SR 8. oo I jj Tf ON CM Tf t^. \O 4 s * r^ r^* 1 d d d d d d d d dodo o o o d 1 ,8 co co co Tf ONO M ON COO &\ HH Tf Tf Tf 10 O N O> to to to too to t>- O 5 co O O t I s - ! d d d d 0000 o o o o 6 d d d 1 M OO >O M OO O COO CM co co CO 00 Tf M 00 oo M Tfo Tf HI t^. Tf Tf 10 10 O - O co co co Tf oo to M oo . co O ** IOOO HI CO CO O O fO ON M Tf O co ON I s * ON N Tf W (N CO CO t-- O w 10 PO Tf Tf Tf O O O O O O O O 0000 o o o o a ^ &> O O . ^ c?8> S" M H? co O t- O co 10 HI CM C\l (S co ONO co 00 O coO CM CO CO CO d o o o O O O O 0000 o o o o to M OO to HI 00 Tf M M coO O\ o o o" o" JiJcri" o d o o d o o o 6 d d d o o o d 4 :S O O O M N CO Tf w .^i ONO M 10 O rS cs o s s 8 M Tt O ON ONOO M r^ r^oo O M O O M M to o -- O I s - S"82S O O O O O NO M ^J- O O\oo r- Tf^O O 10 M M CN CM 6 6 6 6 to t^ to ^ O O co M M O O CO co Tl- iO\O o o o o CO O Tj- ON Tj- O ON t^ ON O >OOO t^. O CN lo Q ^H M M s OO CN ON COOO Tj- ON ON t>- ON ON >OOO t"^- ON CN IO O O O O O O O O o o o o M M 5 0 O O O O 8 CO H M O CM M CM O O CO CO 'sj- lOO o o o o H OO CN t^ ^J- Q\ Q\ f^ ON ON IOOO t^ ON CN IO O O M t-i R I tO O\ ON CO 10 TJ- t-^. 10 t^OO ON M O O O M O O O O 0 COO O >0 M M (N M 6 6 6 6 ON -^- -OOO t^. ON - Tj- 10 ON VOO O CO O l^ O\ O O O M OO t^ M CO 0 . Tf ON O M N O O 1 O o o o o \O t^ - IOO ONOO M ON ON H M Q O O co O O , * M M M M dodo M M i^ K>( N 10 X U Mf^voco CMMOO OOOO JJ, VOCOCMM MOOO OOOO OOOO O vo VONO t^NO OO ^" OO ON ON f^ OO vo vo ON M NO VONO O CO ^ t~< O COOO vo co CN) M O M vo ^*NO M r> ^" CO CMMQO OOOO bL bfl VOCOCMH MOOO OOOO OOOO c 'a VOOO OO ON ^-NO CMrt" vot-.CM^- VOCO^ON ^JJ COOO ^ COOO VOON ONCMOOVO COCMMQ CO CM VO O NO Tj" CN MMOO OOOO MM rocMM MOOO OOOO OOOO fl^d) 0) O^OO OO *O t^ M l> O *O O CO 'O *H O ^"0 OO ^O fO ^ M M O ^- O ^O ^~ fO CSHH OOOO OOO MMO OOOO OOOO OOO M O co O> to I s - "2 NO SfoB c^.7^S M CO ON O NO OO vN M ON to O-OO ^2 : ? tO ONNO OO to co w M M O 5 CM"^ CM OO to PO M NO 00 NO CO CM Tt ONNO CM t-l O O O 1 ON <-> *$ CO O l-l Tf 10 ON Tf O t^ >O M CM NO CO M to N tO CO M >H PO co fO O ONO O CO M I^N. I/} f"O \f) O HH M M Tf ONNO CM M O O O O O O 1 d * I^NO OO to O ^OO O OO CO rT M CM Q\ Q\ CN >O CO CM M !!! O COOQ 10 SS88 8 I OO NO NO rf r~. Tt M to NO CM ONNO CM M M Tj- CO W O 00 NO O CM ^ Tf CO CM M M NO CM CO ON >0 COOO ONNO ^ CM o o o o CO M ON CM oo CM r-~ to 5388 S tO' CO M CM O HI M 00 IO M OO >0 O f*5 tN NO ON co CO O O 00 ON CO Tf CM M M M OO CM CM CM t>. o o -228 o o o o 00 M S 5 ON'S 8 00 HI ^ co co O t^ >o 00 l^ Tj- to M l^. 1000 t>. to r^. M CO CM M M o" o 5 o 5 o tO ONNO ^ 1-1 o o o o o o o 2 O ON- *O ON N oo NO T- t^ O >o O M NO M CM CM M 10 O CO CM M HH 00 tO ON ON NO Tt CM M 0000 O NO NO O COOO to CO 0808 . = ut (2) and Oa = OA sin co/, (3) when a) = the angular velocity (radians per unit of time) at which the vector rotates, / = the time during which the vector has rotated, zero time coinciding with zero value of the angle . 2. Rectangular representation. A sine wave may be plotted to rectangu- lar coordinates as shown in Fig. 269, the curve there plotted showing a succes- sion of values for one complete revolution (360 degrees). A representation of greater values of ut would be a repetition of those values already plotted. 3. Polar representation. It is usually unnecessary to draw rectangular representations of harmonic quantities, the magnitudes of the quantities and their relative posi- FIG. 270. Vector tions being su ffi c i ent> Such a diagram (Fig. 270) is called or Clock gram Dia- a vector or clock diagram, and the vector rotates in a counter-clockwise direction. 4. Combination of harmonic quantities. Two or more harmonic quantities, having the same angular velocities, may be replaced by a single harmonic * This velocity is only approximately harmonic when the connecting rod is of finite length. 305 ESSENTIALS OF ELECTRICAL ENGINEERING quantity, the vector of which is the geometric sum of the vectors of the two quantities. Let OA (Fig. 271) represent the maximum value of one harmonic quantity and OB the maximum value of another harmonic quantity having the same angular velocity. Then Oa = OA sin / (4) and Ob = OB sin (ut ). (5) FIG. 2yia. Combination of Vectors. FIG. 27 ib. Combination of Sine Waves. Combining these two vectors graphically, the vector OC is obtained, for which the mathematical expression is (7) (8) - 0). (6) The mathematical proof that the resultant of two harmonic quantities is a harmonic quantity having the same angular velocity is as follows: Let a = A sin w/, b = sin(co/-0). Then a + b = A sin at + B sin (wt ). Expanding equation (9) a + b = A sin ut + B sin wt cos < B cos o>/ sin = (A + B cos 0) sin at B cos co/ sin = C cos and 5 sin = C sin ! = tan " = T COS0 yl Substituting in equation (n) a + b = \/(A z + B 2 2 AB cos <) sin ( o>/ From equation (15) the maximum value of the resultant is C = (9) (10) (n) (12) d3) (14) V /J B*-2ABcos and the angle through which the vector has rotated is . B sin < ft = tat tan" 1 . , 4 4- cos (16) (17) APPENDIX A 307 5. Phase difference. The phase (time) difference between two harmonic quantities having the same angular velocities, is the time required for the vector of either to pass through an angle equal to that between the vectors of the two quantities. The phasa angle, or angle of phase difference, is the angle AOB in Fig. 27ia. 6. Resolution of harmonic quantities. If two harmonic quantities may be combined into a single harmonic quantity having the same angular velocity, it follows that any harmonic quantity may be resolved into harmonic components having any desired phase difference. In alternating-current problems, the resolution of harmonic electromotive forces and currents into components having a phase difference of 90 degrees, is a common and a useful ex- pedient. 7. Rate of change in the instantaneous value of a harmonic quantity. An inspection of a sine curve such as that shown in Fig. 269, shows that the rate at which its instantaneous value changes is not uniform, and that the rate of change is greatest when the quantity is passing through its zero value. This may also be shown by differentiating the expression for the instanta- neous value a = A sin ut (18) with respect to /. The rate at which a changes is -~ = uA cos ut. (19) But cos o>/ is maximum when sin o>t is zero. 8. Average value of the sin w/. The average value of the sin wt for a com- plete cycle is zero, since for each positive value there is an equal negative value. For any half cycle beginning with o>/ = o or ut = T, all the values are either positive or negative and the average may be determined by adding together the values given in Table XIX, and dividing by the number of values. This gives av. sin w/ = 0.637. ( 2 ) The same result is obtained by integrating sin w/ between the limits <>/ = and co/ = o, and dividing by *-. j ~w/ = f av. sinw/=-| sinco/(f(aJ) (21) 7T J w <=0 9. Value of average sin 2 /. Let the ordinates of a curve be equal to the squares of the instantaneous values of a harmonic quantity. y = <** (23) = A* sin 2 co/. (24) 308 ESSENTIALS OF ELECTRICAL ENGINEERING TABLE XIX NATURAL SINES, COSINES, TANGENTS AND COTANGENTS A Sin Cos Tan Cot O . OOOOO I . OOOOO O . OOOOO Infinity 90 I 0.01745 0.9998 0.01745 57.2900 89 2 0.03490 0.9994 0.03492 28.6363 88 3 0.05234 0.9986 0.05241 19.0811 87 4 0.06976 0.9976 0.06993 14.3007 86 5 0.08716 0.9962 0.08749 11.4301 85 6 0.10453 0-9945 o. 10510 9-5I44 84 7 o. 12187 0.9925 o. 12278 8 . 1443 83 8 0.1392 o . 9903 0.1405 7-U54 82 9 0.1564 0.9877 0.1584 6.3138 81 10 0.1736 o . 9848 0.1763 5-6713 80 ii o. 1908 0.9816 0.1944 5 H46 79 12 0.2079 0.9781 o. 2126 4.7046 78 13 0.2250 0.9744 0.2309 4.3315 77 14 s 0.2419 0.9703 0.2493 4.0108 76 15 0.2588 0.9659 0.2679 3-7321 75 16 0.2756 0.9613 0.2867 3-4874 74 17 0.2924 0.9563 0.3057 3.2709 73 18 0.3090 0.9511 0.3249 3.0777 72 19 0.3256 0.9455 o . 3443 2.9042 7i 20 0.3420 0.9397 o . 3640 2-7475 70 21 0.3584 0.9336 0.2839 2.6051 69 22 0.3746 0.9272 o . 4040 2-4751 68 23 0.3907 0.9205 0.4245 2-3559 67 24 0.4067 0.9135 0.4452 2 . 2460 66 25 0.4226 0.9063 o . 4663 2 1445 65 26 0.4384 0.8988 0.4877 2.0503 64 27 o . 4540 0.8910 0.5095 I .9626 63 28 0.4695 0.8829 0.5317 I . 8807 62 29 o . 4848 0.8746 0.5543 I . 8040 61 3 o . 5000 0.8660 0.5774 I.732I 60 3i 0.5150 0.8572 o . 6009 1-6643 59 3 2 0.5299 o . 8480 0.6249 I . 6003 58 33 o . 5446 0.8387 o . 6404 1-5399 57 34 0.5592 0.8290 0.6745 1.4826 56 35 0.5736 0.8192 o . 7002 i .4281 55 36 0.5878 0.8090 0.7265 1.3764 54 37 0.6018 o . 7986 0.7536 1.3270 53 38 0.6157 o . 7880 0.7813 1.2799 5 2 39 0.6293 0.7771 0.8098 i 2349 Si 40 0.6428 o . 7660 0.8391 1.1918 50 4i 0.6561 0.7547 o . 8693 1.1504 49 42 0.6691 o.743i o . 9004 i .1106 48 43 0.6820 0-73U 0.9325 1.0724 47 44 0.6947 0.7193 0.9657 1-0355 46 45 0.7071 0.7071 I .0000 I. 0000 45 Cos Sin Cot Tan A For a more complete table of functions see any standard text on trigonometry. APPENDIX A 309 Integrating equation (24) between the limits co/ = TT and co/ = o, and dividing by *, av . y = f" 1 'sin 2 co/ d (tat) (25) -f ; (*) i.e., the average value of sin 2 co/ equals . The same result may be deduced without recourse to calculus. From trigonometry sin 2 w/ + cos 2 co/ = i, (27) and av. sin 2 co/ -f av. cos 2 co/ = i. (28) As co/ varies from o to 90 degrees, the sine passes through all values from o to i and the cosine passes through all values from i to o. Therefore, av. sin 2 co/ = av. cos 2 co/ (29) and av. sin 2 co/ = i. (30) 10. Frequency. The frequency of a harmonic quantity is the number of complete revolutions made by its vector per unit of time. During one cycle a harmonic quantity passes through all possible instantaneous values, both posi- tive and negative, i.e., starting at zero the quantity increases to maximum, decreases to zero, increases to maximum in the opposite direction and again decreases to zero. n. Value of . Since the distance passed through by a rotating vector is expressed in radians, the angular velocity (radians per second) is equal to the frequency (number of revolutions the vector makes in one second) multiplied by 2 TT. co = 2*/. (31) APPENDIX A PROBLEMS 1. Find the resultant of two harmonic quantities, each of which has a maxi- mum value of 1000, when the angle between their vectors is: (a) 30 degrees, (b) 45 degrees, (c) 60 degrees, (d) 90 degrees, (e) 120 degrees. 2. The maximum value of a harmonic quantity is 600, and the angle between its (two) components is 90 degrees. Find the maximum values of the compo- nents when: (a) the angle between the quantity and one component is 30 degrees, (b) the maximum values of the components are equal, (c) the maximum value of one is twice the maximum value of the other. 3. Find the angles of phase difference between the quantity in Problem 2 and its components. 4. Find the angular velocity of a rotating vector when the frequency is: (a) 20, (b) 25, (c) 30, (d) 40, (e) 50, (/) 60, (g) 100. 5. The maximum value of a harmonic quantity is 100. Find the rate at which the quantity is changing when co/ is equal to: (a) o, (b) 30 degrees, (c) 45 degrees, (d) 60 degrees, (e) 75 degrees, (/) 90 degrees. 310 ESSENTIALS OF ELECTRICAL ENGINEERING 6. Find the average value of a harmonic quantity, the maximum value of which is: (a) 40, (6) 75, (c) 100, (d) 250, (e) 800. 7. Find the average square of a harmonic quantity, the maximum value of which is: (a) 40, (b) 75, (c) 100, (d) 250, (e) 800. 8. Find the frequency when is equal to: (a) 125.6, (b} 157, (c) 188.4, (d) 251.2, (e) 314, (/) 376.8, (g) 400. APPENDIX B INDUCTANCE In Chapter i, Section n, inductance is denned as the proportionality factor between the electromotive force set up in a circuit by reason of a change in the value of the current flowing in the circuit, and the rate at which the current changes. From Chapter 2, Section 13 Therefore, L = (3) and L = *> (4) when = the total flux linking with the circuit (if the conductor has more than one turn or loop, is the product of the flux linking with one turn and the number of turns), * = the current flowing in the circuit, in c.g.s. units, e = the electromotive force of self-induction, in c.g.s. units, L = the inductance of the circuit, in c.g.s. units. i. Inductance of each of two parallel wires. Let A and B be two parallel cylindrical conductors in which the currents are equal but flow in opposite directions. The current in each conduc- tor, if uninfluenced by that in other con- ductors, would set up concentric circles of flux around the axis of the wire (Chapter 2, Section 3). The mutual effect of the currents is to produce the FlG - 2 ? 2 - Magnetic Field Between Two flux distribution indicated in Fig. 272. Current <***&* Conductors. The inductance of the conductors is due to the flux which passes between the axes of the wires, the flux which encircles A at a greater distance than D being neutralized by an equal and opposite flux set up by B. The total flux encircling the axis of each wire may be divided into two parts: (a) that in the body of the wire, (6) that in the insulating material between the wires. 3" 312 ESSENTIALS OF ELECTRICAL ENGINEERING (a) The flux in the body of the wire. The flux in any elemental zone dx (Fig. 273) within the wire is due to the current inside of zone dx, and this current is ^2 * FIG. ,73- = P' (7) Therefore, the flux in zone dx is 2 9ci = - dx c.g.s. units. (8) Since this flux surrounds only that part of the current which flows in the area of the conductor bounded by the zone dx, it is equivalent to *'-Jx^ = the flux in the iron core. Then ei(df) = Ni eit = Ni (45) (46) (47) But and Therefore, APPENDIX B = AB HI Ni = 47T Wh = ( A-JT J - +B HdBc.g.s. units, 317 (48) (49) (50) (51) i.e., the energy expended in carrying a volume of iron through a magnetic cycle is proportional to the area of the hysteresis loop, and the average power loss due to hysteresis in the iron is i_ n (52) when / = the number of magnetic cycles per second through which the iron passes. Dr. C. P. Steinmetz has shown that the area of a hysteresis loop is approxi- mately proportional to the 1.6 power of the maximum flux density attained during a magnetic cycle. The loss per cycle per cubic centimeter of iron due to magnetic hysteresis is, then, (53) when ij = the hysteretic (magnetic) constant and is dependent on the physical properties of the iron. Table XX. TABLE XX HYSTERETIC CONSTANTS (r) Kind of iron Constant Best annealed sheet o 0015 Good annealed sheet o 003* Ordinary annealed sheet o 004 Soft annealed cast iron o 008 Soft machine steel O.OI Cast steel 0. 12 Cast iron o. 16 Hardened steel 0.25 * Largely used for dynamo armature punchings. 8. Distortion of current wave due to hysteresis. From Section 7 it is evi- dent that the magnetizing current in a coil having an iron core in which the flux varies harmonically cannot be harmonic, but has a distorted wave shape as indicated in Fig. 277. 9. Growth and decay of current in an inductive circuit. When an electro- motive force of constant value is applied to a circuit containing both resist- ance and inductance, the current does not immediately rise to its final value E , because of the counter-electromotive force induced in the circuit by the 318 ESSENTIALS OF ELECTRICAL ENGINEERING increasing flux; when a circuit containing both resistance and inductance is disconnected from a source of constant electromotive force, and the circuit closed, the current does not immediately fall to zero, but is maintained by the electromotive force induced in the circuit by the decreasing flux. Let E = the applied electromotive force, R = the resistance of the circuit, L = the inductance of the circuit, = the rate at which the current flowing in the circuit changes. For an increasing current, * (54) Transposing equation (54) and multiplying by R, - R (di) - R (dl) E-Ri L ' and _ Ki Li R ^ - (58) <-|(, -,-*) For a decreasing current, 1 = o, (59) (6o) and i=|r^. (6 3 ) APPENDIX C CAPACITANCE 1. The dielectric field. When the potential of a body is greater or less than zero (the surface of the earth is assumed to be at zero potential), the body is said to be electrified or charged. The energy required to produce electrification is stored in the surrounding medium, and there is set up in the medium a stressed condition termed a dielectric or electrostatic field. The dielectric flux emanating from a surface is normal to the surface as in- dicated in Fig. 278, and is represented by lines which termi- nate at another surface. 2. Properties of electric charges. The following properties of electric charges have been established by experiment: (a) Like charges repel; unlike charges attract. (6) When a charge is produced on any body, an equal and opposite charge is produced on the same or on some other body. IG * 2 ^ * (c) The force exerted between two charges of electricity is directly propor- tional to the product of the charges, and inversely proportional to the square of the distance separating them. 5 ' - , /= ^' (I) 3. Electrostatic units. The units of the electrostatic system, and the equivalent values in practical units are: . (a) The unit of quantity (charge) is that quantity of electricity which repels with a force of one dyne a similar and equal quantity of electricity placed at a distance of one centimeter in air. To reduce electrostatic units of quantity to coulombs divide by 3 X io 9 . (b) Unit current is that which conveys unit quantity past a given point on a conductor in one second. To reduce electrostatic units of current to amperes divide by 3 X io 9 . (c) Unit electrostatic force or unit difference of potential exists between two points when one erg of work is expended in transferring unit quantity of elec- tricity from one point to the other against the force of the electrostatic field. To reduce electrostatic units of potential to volts multiply by 300. (d) Unit capacitance is that which produces unit difference of potential when charged with unit quantity of electricity. To reduce electrostatic units of ca- pacitance to microfarads divide by 900,000. (e) The specific inductive capacity or dielectric constant of a substance is the ratio of the capacitance of a condenser having that substance as a dielectric, to 319 320 ESSENTIALS OF ELECTRICAL ENGINEERING the capacitance of the same condenser using dry air, at o C. and a pressure of 76 centimeters, as the dielectric. (Table XXI.) TABLE XXI DIELECTRIC CONSTANTS (K) Material Constant Air I .O Oil (transformer) 2. I Shellac 2.75 Paraffin 2 3 Rubber 2 1< Paper 2 to 4 Gutta percha 3 to "? Glass . ... 3 to 10 Mica . 4 to 8 Varnished cambric 4 to 6 Pure water 80 (/) The intensity of a dielectric field is the ratio of the total flux, when propagated in air, to the area of the surface from which it emanates, and is equal to the force in dynes at the point.* (Symbol F.) (g) The density of a dielectric field is the product of the field intensity and the dielectric constant.* (Symbol D.) 4. Flux due to unit charge. The force with which a unit charge acts on another unit charge distant r centimeters in air is (from 2 c and 3 a) f= ^ dynes (2) (3) and the intensity of the dielectric field is F = - 2 lines per square centimeter. But the surface of a sphere, the diameter of which is 2 r centimeters is square centimeters. Therefore, the total dielectric flux emanating from unit charge is ^=1X4^2 (4) = 47rlines. (5) 5. Potential difference between points. The potential difference between points is, by definition, equal to the work in ergs done when unit quantity of electricity is transferred from one point to the other against the force of the electrostatic field, and is, therefore, the line integral of the field intensity be- tween the points. (6) * Compare with the corresponding terms used for the magnetic circuit (Chapter 2). APPENDIX C 3 2I 6. Capacitance. Capacitance has been denned (Chapter i, Section 12) as the ratio of the quantity of electricity displaced (the charge), to the electro- motive force producing the displacement. C= (7) 7. Capacitance of parallel plates. Determine the capacitance of two par- allel plates (Fig. 278) A square centimeters in area, and separated by / centi- meters of dielectric, the constant of which is K. Let Q be the charge on each square centimeter of the positive plate and Q the charge on each square centi- meter of the negative plate. Then (8) (9) (10) F = lines per square centimeter, K y K electrostatic units, and (ix) TT = - electrostatic units per square centimeter (12) 47T/ A IF = electrostatic units = j- 900,000 microfarads, (13) (14) i.e., the capacitance of a plate condenser is directly proportional to the product of the area of plates and the dielectric constant, and inversely proportional to the distance between the plates. Commercial condensers are made up of a large number of sheets of tinfoil, to FIG. 279. The Plate Condenser. FIG. 280. obtain the required area of the plates, connected as indicated in Fig. 279, and separated by sheets of paraffined paper. 8. Capacitance of concentric cylinders. Determine the capacitance of two concentric cylinders (Fig. 280), the radii xi and x 2 of the cylinders, their length /, and the dielectric constant K being given. Let the charge per centimeter length of the positive cylinder be Q and the charge per centimeter length of the 322 ESSENTIALS OF ELECTRICAL ENGINEERING negative cylinder be Q. Then the field intensity at any point distant r centi- meters from the axis of the cylinders is = - lines per square centimeter. (16) The potential difference between the two cylinders is found by integrating equation (15) between the limits r x\ and r = x% E = ^J n ~r (I7) ^log,' 5 (18) .A. Xv and C = U (19) *r electrostatic units per centimeter length of cylinders. (20) 9. Capacitance of parallel wires. Determine the capacitance of two parallel wires (Fig. 281), their radii r, the distance between their centers D, their length / and the dielectric constant K being given. Let the charge per centi- meter length of the positive wire be Q and the charge per centimeter length of the negative wire be Q. Then the field in- x .............. p "p-;;;::;-;~:| tensity at any point distant x centimeters from K ___ I ____ /j>\ the axis of conductor A is and the potential difference between A and B is found by integrating equation (21) between the limits x = D r and x = r. * The expression log e - - may be replaced by log e without appreciable error when the ratio is large. APPENDIX C 323 K r = electrostatic units per centimeter 4log e ^ length of circuit. (25) 10. Capacitance of each conductor in a three-phase circuit. The capacitance of a conductor in a three-phase system depends on its position relatively to the other conductors in the system. ^JL* There will be considered here only the two commonly used arrangements: (a) when the conductors are placed p''/ \' D . at the vertices of an equilateral triangle, (b) when the conductors are in the same plane. (a) When the conductors are placed at the vertices of an equilateral triangle. From equation (22) the poten- p IG> 2 82a. tial difference between any conductor (Fig. 28 2a) and the neutral plane halfway between the conductor and either of the other conductors is ' and the capacitance of each conductor is c=^L electrostatic units per centimeter length of conductor, (28) i.e., the capacitance of each conductor in a three-phase system, when the con- ductors are placed at the vertices of an equi- K~~ 0. >*<- D *! Q A -~-'-JB @- c lateral triangle, is twice the capacitance of the P , loop formed by any two of the conductors. (b) When the conductors are in the same plane. From equation (22) the potential difference between A or B (Fig. 282b) and their neutral plane is ,-,20,7) r /N l = -^loge . (29) the potential difference between A or C and their neutral plane is fi-'log., (30) 324 ESSENTIALS OF ELECTRICAL ENGINEERING and the potential difference between B or C and their neutral plane is 3 = flog^. . (30 Under the common practice of transposing the conductors of polyphase systems, the capacitances of the conductors are sensibly equal, and may be calculated by means of the following equation: c = D-r , 2O, 2D-r . +f log '-7- (33) D-r . . 2D-r h 2 log e K electrostatic units per centimeter length of conductor.* (34) n. Charging current in a three-phase system. From equation (16), Chapter i, the current due to the capacitance of a single-phase line is IC=2TT/CE. ( 35 ) when the capacitance is concentrated. But the voltage between the terminals of a condenser star-connected to a three-phase system is ~ times the line-to- V 3 line voltage. Therefore, the charging current due to the capacitance of a three-phase line is r 4 iffCE / ,\ Ic = J .- , (36) ^3 and is -^= (= 1.15) times the charging current flowing in the circuit formed ^3 by any two of the conductors when the voltage between lines is equal to the line-to-line voltage of the three-phase system. 12. Energy stored in the dielectric field. When the intensity of a dielectric field increases or decreases, a current flows to or from the charged body. and q = Ce. ' ' (38) Therefore, *' = C T/ <39> at * Equation (33) gives values only slightly gr sater than those given by equation (27) . APPENDIX C 325 But Wc = pdt (40) (41) (42) * . ergs (43) f eidt Cfede when the applied voltage varies from e to zero. : Substituting in equation (43) the value of e obtained in equation (9), and that of C obtained in equation (13) (44) (45) IT But F = | (46) and Al=V cubic centimeters. , (47) Therefore, We = ^ X V ergs (48) 7} 17 = - ergs per cubic centimeter (49) 8 T 7-J2 = - ergs per cubic centimeter. (50) 8 7TA 13. Dielectric hysteresis. It has been experimentally shown that the energy (heat) dissipated in the dielectric of a condenser is greater with alter- nating current than when a constant difference of potential exists between the condenser terminals, the effective value of the alternating electromotive force being equal to the constant electromotive force. This additional loss is due to so-called dielectric hysteresis. When a condenser is connected to an alternating-current circuit, the chang- ing value of the dielectric flux induces alternating electromotive forces in any conducting particles that may have become imbedded in the dielectric. Di- electric hysteresis is, therefore, more nearly analogous to eddy currents than to magnetic hysteresis. Dielectric hysteresis losses are always small, and are usually neglected. 14. Charging and discharging current in a condenser circuit. When an electromotive force of constant value is applied to a circuit containing both resistance and capacitance, a charging current flows in the circuit; when a circuit containing both resistance and capacitance is disconnected from a 326 ESSENTIALS OF ELECTRICAL ENGINEERING source of constant electromotive force, and the circuit closed, a discharging current flows in the circuit. Let E = the applied electromotive force, R = the resistance of the circuit, C = the capacitance of the circuit, -^ = the rate at which the capacitance charges or discharges. For a charging current, j~ (Si) (52) q q-CE-~J t RC' (S3) -CE\ _^ - CE I RC (54) (55) But CE = Q (56) and * = J'- (57) n I -- M Therefore * = lL\ e RC). RC V For a discharging current, q = Qe'RC ^ ( 6 3) and n ~RC *=-RC< 15. Corona. The phenomena known as corona * is an electrostatic (leak- age) discharge between wires of different potential, and takes place when the * See The Transactions of the American Institute of Electrical Engineers, Vol. XXX, pages 1889-1965 and Vol. XXXI, pages 1051-1092, "Law of Corona and the Di- electric Strength of Air," by F. W. Peek, Jr., Vol. XXXI, pages 1035-1049, "Corona Losses Between Wires at High Voltages," by C. Francis Harding. APPENDIX C 327 potential difference between the wires exceeds a certain "critical" value, de- pending on the diameters of the wires and their distance apart. In an alter- nating-current system, the power losses due to corona are proportional to the frequency and to the square of the increase in voltage above the critical value, but are usually negligible for voltages up to 45,. At voltages materially higher than the critical value, a visible halo-like envelope surrounds the con- ductor, the diameter of the envelope increasing as the voltage increases. APPENDIX D THE COMPLEX QUANTITY ADMITTANCE, CONDUCTANCE AND SUSCEPTANCE 1. The complex quantity. It is evident that the equation a = Vb* + ? (i) may be written a = b jc, (2) when it is specified that j indicates that b and c are at right angles, and are to be combined geometrically. j may be defined as the complex operator, the effect of which is to rotate a vector to which it is applied, 90 degrees forward in relation to the reference line. The effect of j Xj or j* is to rotate the vector 180 degrees in relation to the reference line, and the algebraic value of f i. (3) Therefore, j= V i. (4) Addition, subtraction, multiplication and division of equations involving the complex quantity are essentially algebraic operations, and are governed by the laws of such processes. The use of the complex quantity greatly simplifies the solution of some alternating-current problems, particularly those relating to transmission lines and distributing networks. 2. Examples. (i) Find the line current when the currents in the parallel branches of a circuit are: /i = 6 j 3, and / 2 = 8 j 2. = 14.86 amperes. (2) Find the applied electromotive force, the power component, the wattless component, and the power in a series circuit when: / = 5 + ^*2 and =20+.; 5. E = ZI = (20+75) (5 +72) = 100 +.725+.; 40 +/ 10 = 90+^65 (since/ = - i) = V( 9 o)*+(6 5 ) 2 = in volts (applied). EI = 90 volts (power component). 328 APPENDIX D 329 1 E 2 = 65 volts (wattless component). El = (90 +./ 65) (5 -K; 2) = 320+^505 P = 320 watts. (3) Find the power in an electric circuit when: I = 10 -f j 4 and E = 150 j 24. El = (150 - j 24) (10 +j 4) / 1 = 1 500 j 240 + j' 600 j 2 96 - = 1596 +.7360, P = 1596 watts. 3. Admittance, conductance and susceptance. The value of the current in any circuit is expressed by the equation '-f RjX Multiplying the right-hand member of equation (7) by R =F jX (6) (7) The value of the current in the circuit may also be expressed by the equation 7 = YE (10) = (g*jQE, (11) when F = and is termed the admittance of the circuit, g Y cos and is termed the conductance of the circuit, b = Y sin and is termed the susceptance of the circuit. From equations (n) and (9) i.e., the conductance of a series circuit is equal to its resistance divided by the square of its impedance * = Compare equations (13) and (14) with equations (124) and (129), Chapter I. 330 ESSENTIALS OF ELECTRICAL ENGINEERING and the susceptance of a series circuit is equal to its reactance divided by the square of its impedance. Admittances, like impedances, must be combined geometrically; conduc- tances or susceptances are combined algebraically. APPENDIX D PROBLEMS 1-6. Solve Problems i to 6, Chapter 18, using Complex Quantity. 7-14. Solve Problems 22, 30, 34, 35, 36, 40, 41 and 42, Chapter i, using ad- mittance, conductance and susceptance values. * Compare equations (13) and (14) with equations (120) and (125), Chapter i. APPENDIX E RULES RECOMMENDED BY COMMISSION ON RESUSCITATION FROM ELECTRIC SHOCK REPRESENTING The American Medical Association The National Electric Light Association The American Institute of Electrical Engineers DR. W. B. CANNON, Chairman DR. GEORGE W. CRILE Professor of Physiology, Harvard Professor of Surgery, Western Reserve University University DR. YANDELL HENDERSON MR. W. C. L. EGLIN Pr of essor of Physiology, Yale Univer- Past-President, National Electric sity Light Association DR. S. J. MELTZER DR. A. E. KENNELLY Head of Department of Physiology Professor of Electrical Engineering, and Pharmacology, Rockefeller In- Harvard University stitute for Medical Research DR. ELIHU THOMSON DR. EDW. ANTHONY SPITZKA Electrician, General Electric Com- Director and Professor of General pany Anatomy, Daniel Baugh Institute of MR. W. D. WEAVER, Secretary Anatomy, Jejferson Medical College Editor, Electrical World Copyright, 1912, by NATIONAL ELECTRIC LIGHT ASSOCIATION Follow these instructions even if victim appears dead. I. IMMEDIATELY BREAK THE CIRCUIT With a single quick motion, free the victim from the current. Use any dry non-conductor (clothing) rope, board, to move either the victim or the wire. Beware of using metal or any moist material. While freeing the victim from the live conductor have every effort also made to shut off the current quickly. H. INSTANTLY ATTEND TO THE VICTIM'S BREATHING i. As soon as the victim is clear of the conductor, rapidly feel with your finger in his mouth and throat and remove any foreign body (tobacco, false teeth, etc.). Then begin artificial respiration at once. Do not stop to loosen the vic- tim's clothing now; every moment of delay is serious. Proceed as follows: 332 ESSENTIALS OF ELECTRICAL ENGINEERING (a) Lay the subject on his belly, with his arms extended as straight forward as possible and with face to one side, so that nose and mouth are free for breath- ing (see Fig. i). Let an assistant draw forward the subject's tongue. FIG. 1. Inspiration; pressure off. (6) Kneel, straddling the subject's thighs, and facing his head; rest the palms of your hands on the loins (on the muscles of the small of the back), with fingers spread over the lowest ribs, as in Fig. i. (c) With arms held straight, swing forward slowly so that the weight of your body is gradually, but not violently, brought to bear upon the subject (see Fig. 2). This act should take from two to three seconds. (d) Then immediately swing backward so as to remove the pressure, thus returning to the position shown in Fig. i. (e) Repeat deliberately twelve to fifteen times a minute the swinging for- ward and back a complete respiration in four or five seconds. (/) As soon as this artificial respiration has been started and while it is being continued, an assistant should loosen any tight clothing about the subject's neck, chest, or waist. 2. Continue the artificial respiration (if necessary, two hours or longer), without interruption, until natural breathing is restored, or until a physician arrives. If natural breathing stops after being restored, use artificial respira- tion again. FIG. 2. Expiration; pressure on. 3. Do not give any liquid by mouth until the subject is fully conscious. Give the subject fresh air, but keep him warm. III. SEND FOR NEAREST DOCTOR AS SOON AS ACCIDENT IS DISCOVERED REFERENCES "Alternating Currents," F, Bedell and C, A. Crehore. "Alternating Currents," A. Russell. "Alternating Currents and Alternating Current Machinery," D. C, and J. P Jackson, "Alternating Current Motors," A. S. McAllister. "Alternating Current Transformer," J. A. Fleming. "Alternating Current Windings," C. Kinzbrunner. "Continuous Current Armatures," C. Kinzbrunner. "Dynamo-electric Machinery," S, P. Thompson. " Electrical Conductors," F. A, C. Perrine. "Electrical Meters," C. M. Jansky. " Electrical Illuminating Engineering," W. E. Barrows, Jr. "Electric Lighting," F. B. Crocker. "Electric Motors," F. B. Crocker and M. Arendt. "Electric Power Conductors," W. A. del Mar. "Electric and Magnetic Calculations," A. A. Atkinson. "Electric Transmission of Energy," A. V. Abbott. "Elements of Electrical Engineering," C. P. Steinmetz. "Elements of Electrical Transmission," O. J. Ferguson. " Experimental Electrical Engineering," V. Karapetoff. " Foster's Electrical Engineers Handbook." "Illumination and Photometry," W. E. Wickenden. " Polyphase Apparatus," M. A. Oudin. "Polyphase Alternating Currents," S. P. Thompson. "Principles of Electrical Engineering," H. Pender. " Secondary Batteries," E. J. Wade. "Standard Handbook for Electrical Engineers." "'Synchronous Motors and Converters," A. E. Blondel. "The Electric Circuit," V. Karapetoff. "The Induction Motor," B. F. Bailey. "Transformer Practice," W. T. Taylor. 333 INDEX Admittance, 329. Abvolt, 33. Air-break switches, 244. All-day efficiency, 73, 189. Alternating-current circuits, 9, 10, n, 13, 14, 270. resonance in, 22. Alternating current, hydraulic analogy for, 2. Alternating-current series motor, 227. Alternator efficiency, 143. Alternator, frequency of, 56. Alternator, inductor, 57. Alternator losses, 141. Alternator ratings, 146. Alternator regulation, 134. Alternator, single-phase, 127. speed and frequency of, 56. three-phase, 128. two-phase, 128. voltage of, 123. Alternators in parallel, load division of, 144, 158. Alternators, parallel, operation of, 143. synchronizing of, 144. voltage characteristics of, 133. American wire gauge, 269. Ammeters, 252. Ammeter shunts, 254. Ampere, 4. Ampere-turn, relation of the gilbert and the, 39. Armature core, 46. Armature inductance, compensation for, 229. Armature reaction, 64, 81, 129. Armature windings, basket, 51. chain, 51. distributed, 124. lap, 49. wave, 50. Arc lamps, carbon, 238. flaming, 238. magnetite, 237. Automatic starting rheostats, 90. Auto- transformer, the, 190. Average value of the sine, 307. Average value of the sin 2 , 307. Balanced three-phase system, 109. Batteries, 299-304. Benchboards, 251. BK motor, Wagner, 232. Boosters, 303. Booster converter, 170. Brown and Sharpe wire gauge, 269. Brush advance, 64. Brushes, carbon, 53. copper, 53. 335 336 INDEX Brush-contact losses, 94. Brush holders, 54. Building up, 61, 70. Capacitance, 8, 320. of concentric cylinders, 321. 'of parallel plates, 320. of parallel wires, 322. of three-phrase circuits, 323. of transmission lines, 270. Carbon-break switches, 244. Carrying capacity of wires, 270. C. G. S. unit of current, 39. Circuits, alternating-current, 9, 10, n, 13, 14, 270. parallel, 3. series, 2. Circle diagram for induction motor, 214-217. for synchronous motor, 156. Circular mil, 5. Charge, electric, 319. Charging current, 270, 324. Coefficient, leakage, 41. temperature, 5. Commutation, continuous-current dynamo, 62-66, 81. series alternating-current motor, 227. sparkless, 64. , Commutation, resistance, 61. voltage, 61. Commutating flux, 64. Commutator, 52, 161. Comparison of series motors, 229. Comparison of star- and delta-connected systems, 113. Comparison of two- and three-phase systems, 119. Compensated repulsion motor, 231. Compensation for armature inductance, 229. Compensation, voltmeter, 282. Complex quantity, 328. Compound dynamo, 55. generator, 69. motor, 86, 87. Compounding, 70, 71. Concatenation of induction motors, 225. Concentric cylinders, capacitance of, 321. Conductance, 329. Conductors, 268. Connection to load circuits, 76. Constant-current transformer, the, 192. Constancy of power in polyphase systems, 115. Constants, hysteretic, 317. Construction of transformers, 183. Continuous-current dynamo, 54. Continuous-current, hydraulic analogy for, 2. Control, speed, 82. Cooling of transformers, 188. Copper loss in transformers, 186. Core, armature, 46. Corona, 326. Coulomb, 4. Counter- electromotive force, 40, 79, 150, 178. Cross-magnetizing magnetomotive force, 65, 131. Cumulative compound motor, 86, 103. Current, charging, 270, 324. C. G. s. unit of, 39. INDEX 337 Current, density in carbon brushes, 61. effective, 17. induced, 32. relations in transformers, 180. the electric, i. triangle, 14. transformer, 255. wattless, 22, 23, 173. , Decay of current in an inductive circuit, 318. Delta-connected three-phase system, in. Demagnetizing magnetomotive force, 66. Detector, ground, 266. Determination of stray power, experimental, 98. Diagram, Mershon's, 278. Dielectric field, 319. energy stored in, 324. flux, 319. hysteresis, 235. Differential compound motor, 87. Direction of magnetic flux, 29, 30. Distributed armature windings, 124. effects of, 126. Distribution of flux in air gap, 64. Discharging current in condenser circuit, 326. Distributing systems, 3, 273. Distortion of current wave due to hysteresis, 316. Dynamo, compound, 55. elementary, 42. losses, 93. parts of, 46-51. series, 55. separately excited, 54. shunt, 54. Economy of the three-phase system, 119. Eddy current brake, 260. Eddy currents, 95. Edison battery, 301. Electric arc, instability of the, 240. power factor of the, 240. Electric charges, properties of, 319. Electric current, field intensity produced by the, 37. manifestations of, 2. ^Electric currents, i. hydraulic analogy for, 2. Electric circuit, power in an, 15. Electric shock, resuscitation from, 331. Electric units, 4. Electrical degree, 125. Electrolytic cell, 284. Electrodynamometer, 253. Electromagnet, the, 30. Electrostatic units, 319. voltmeter, 256. Effective current, 17. electromotive force, 17. Effects of distributed armature winding, 126. Efficiency, 101. all-day, 73. alternator, 143. synchronous motor, 154. transformer, 188. 338 INDEX Electrical degree, 125. Electricity, nature of, i. Electromotive force, 4, 33, 60, 79, 123, 165, 180. Electromotive force, counter, 40, 79. effective, 17. Elementary dynamo, 42. Elimination of harmonics, 126. Energy stored in a dielectric field, 324. magnetic field, 315. Equalizer, 50, 75. Equivalent reactance, 187. resistance, 186. single-phase system, 120. Experimental determination of stray power, 90. Faraday's Law, 31. Feeder regulation, 279. Field discharge resistance, 58. Field intensity, magnetic, 30. dielectric 320. due to unit pole, 37. produced by an electric current, 37. Field, magnetic, 29. poles, 46. windings, 46. Flux, commutating, 64. density, 31. dielectric, 319. due to unit charge," 320. pole, 37. leakage, 41, 180. magnetic, 31. Force, magnetizing, 31. magnetomotive, 31. of magnetic traction, 39. Frame, 46. Frequency, 57. changer, 221, Frequency meters, split phase, 266. vibrating reed, 265. Frequency of alternator, 56. Friction and iron losses, separation of, 100. Frictional losses, 97. Fundamental equation of the generator, 60. motor, 79. Fundamental physical action in the transformer, 178. Fuses, 244. Gap, spark, 257. General Electric RI motor, 232. Generator, compound, 69. fundamental equation of the, 60. induction, 224. series, 68. shunt, 66. Generators, parallel operation of, 73, 143. Gilbert and the ampere-turn, relation of the, 39. Graphic meters, 262. Ground detector, 266. Growth of current in an inductive circuit, 318. Harmonic quantities, 305. combination of, 305. examples of, 305. INDEX 339 Harmonic quantities, instantaneous values of, 305. rate of change in, 305. resolution of, 307. Harmonics, elimination of, 126. Heating, 104. High-voltage measurements, 256. Holders, brush, 54. Horn gap, 284. Hot wire instruments, 252. Hunting, 154. Hydraulic analogy for alternating currents, 2. continuous currents, 2. pulsating currents, 2. Hysteresic constants, 317. Hysteresis, dielectric, 325. distortion of current wave due to, 316. loop, 316. loss, 95, 316. Incandescent lamps, 240. Inductance, 7. of a coil, 314, 315. of a conductor with earth return, 314. of a three-phase system, 313. of parallel wires, 311. of transmission lines, 270. Impedance, 12. Impedances in parallel, 20. series, 19. Induced currents, 32. Induction by varying flux density, 35. instruments, 253. generator, 224. motor action, circle diagram, 214. construction, 203. generator action, 206. performance curves, 213. slip of rotor, 206. slip-ring rotor, 203. squirrel cage rotor, 203. starting, 210. synchronous speed of, 206-7. torque, 207. regulator, 280'. Impedance, 12. triangle, 12. Inductor alternator, 57. Instability of the electric arc, 240. Instruments, use of, 23. electrodynamometer, 253. hot wire, 252. induction, 253. permanent magnet, 252. soft iron, 253. Insulation, wire, 269. Insulators, pin, 269. suspension, 269. Interpoles, 64. Inverter converter, 165. Iron losses, 95. in transformers, 184. separation of, TOO, 184. separation of friction and, 100. 340 INDEX Joule, 4. Joule's Law, 15. Kilowatt, 5. Lamps, arc, 235. incandescent, 240. nitrogen-filled, 241. Nernst, 241. mercury vapor, 241. regenerative, 239. quartz, 242. Law, Faraday's, 32. Joule's, 15. Lenz', 32. of the magnetic circuit, 31. Ohm's, 6. Leakage coefficient, 41. magnetic, 40, 180. Lightning arresters, 283. Limitations of the single-phase system, 107. Limits, regulation, 139. Line calculations, 271. Load circuits, connection of generators to, 76. Load division of alternators, 144, 158. Losses, brush-contact, 94. determination of, 98, 141, 184, 186. dynamo, 93. eddy current, 95. frictional, 97. in alternator, 141, iron, 95. pole-face, 97. hysteresis, 95, 141, 183, 316. Magnetic circuit, law of the, 31. dampers, 158. field, 29. production of a, 29. energy stored in a, 315, flux, 31. due to unit pole, 37. leakage, 40, 180. neutral, 63, 64. traction, force of, 39. units, 30. Magnetism, 29. Magnetization curves, 42. Magnetizing force, 31. Magnetomotive force, 30, 31. Manifestations of the electric current, 2. Maximum load of synchronous motor, 154. torque of induction motor, 216, Measurements, high-voltage, 256. Mercury arc rectifier, construction, 161. efficiency, 163. limitations, 163. operation, 163. starting, 162. use, 164. Mercury vapor lamp, 241. Mershon's diagram, 278. Meters, graphic, 262. INDEX 341 Meters, power factor, 264. recording, 262. Mil, circular, 5. Moore tube, 241. Motor, alternating-current series, 227. compensated repulsion, 231. fundamental equation of, 79. generator, 161. General Electric RI, 232. operated switches, 249. repulsion, 230. series, 85. shunt, 81. starting rheostats, 87. Wagner BK, 232. Multiple- wire systems, 274. Multiplier, 256. Nernst lamp, 241. Neutral in three-phase system, 109. Neutral, magnetic, 63, 64. Nitrogen-filled lamp, 241. Oil-break switches, 245. Ohm, 4. Ohm's Law, 6. Oscillograph, the, 126. Parallel circuits, 3, 274. Parallel operation of alternators, 143. generators, 73. induction generators, 224. Parallel wires, capacitance of, 322. inductance of, 311. Parts of the dynamo, 46-51. Permanent magnet instruments, 252. Permeability, 30, 31. Phase characteristic, 155. difference, 307. transformation, 199. Pole-face losses, 97. Poles, field, 46. Polyphase armature reaction, 131. Polyphase systems, constancy of power in, 115. power factor of, 114 power in, 114. Polyphase transformers, 194. wattmeters, 260. Potential difference between points, 319. transformer, 256. Properties of electric charges, 319. Power in an electric circuit, 15. in a polyphase system, 114. factor, 18. factor meters, 264. factor of polyphase system, 114. factor of the electric arc, 240. measurements, 116. Production of a magnetic field, 29. Proof of the circle diagram, 217. Pulsating current, hydraulic analogy for, 2. Protection of motors, 245. 342 INDEX Protection of transformers, 196. Quarter-phase system, 107. Quartz lamp, 241. Ratings, 104, 146. Reactance, 10. equivalent, 187. synchronous, 136. Reactances in parallel, 19. series, 18. Reaction, armature 64, 81, 129. of magnetic field and current-carrying conductor, 36. Recording meters, 262. Regulation, alternator, 134. feeder, 279. limits, 139. speed, 82. transformer, 189. voltage, 61. Regulating-pole converter, 169. Regulator, induction, 280. Tirrill, 72, 140. Relation of the gilbert and the ampere-turn., 39. Reluctance, 30, 31. Repulsion motor, 230. Resistance, 5. and reactance in parallel, 19. effect of temperature on, 5. equivalent, 186. field discharge, 58. Resistances in parallel, 19. series, 18. Resonance, 10. in alternating-current circuits, 22. Reversing direction of induction motor rotation, 206. motor armature rotation, 36. Rheostats, motor-starting, 87. RI motor, General Electric, 232. Rotating field in single-phase induction motor, 219. flux, 204. Saturation curve for alternator, 134. Separately excited dynamo, 56. Separation of friction and iron losses, 100. Series circuits, 3, 273. dynamo, 55. generator, 68. motor, 85. on alternating-current circuit, 227. torque of, 85. motors, comparison of, 229. Series-parallel starting, 87. Series transformer, 255. Short-circuit curve for alternator, 135. Shunts, ammeter, 254. Shunt dynamo, 54. generator, 66. motor, 81. on alternating-current circuit, 226. Sine, average value of, 307. Sin 2 , average value of, 307. Single-phase alternator, 127. INDEX 343 Single-phase armature reaction, 129. induction motor, 218. generator action, 218. rotating flux, 219. speed- torque curves, 221. starting, 219. transformer action, 218. system, equivalent, 120. limitations of, 107. Skin effect, 271. ' Slip of induction motor, 206. Solenoid-operated switches, 246. Solenoid, the, 29. Soft iron instruments, 253. Spark gap, 257. Sparking, 87, 104. Sparkless commutation, 64. Speed of alternator, 56. Speed-torque relations, 80. Speed control, 82. regulation, 82. Split-pole converter, 169. Stability of synchronous motor, 153. Star-connected three-phase system, 109. Starting of synchronous motors, 151. Storage battery, internal actions of, 299. Stray power, experimental determination of, 98. Stroboscopic method of slip measurement, 206. Structure of synchronous motor, 149. Susceptance, 329. Switchboards, 249. Systems, distributing, 273. Switches, air-break, 244. automatic 245. carbon-break, 244. oil-break, 245. motor-operated, 249. solenoid-operated, 246. Switch operation, 246. Synchroscope, 262. Synchronous converter, armature reaction, 172. compounding, 169. construction, 164. current relations, 167. efficiency, 171. heating, 173. hunting, 171. operation, 164. rating, 176. regulating-pole, 169. starting, 168. voltage relations, 165. Synchronous motor, efficiency of, 154. maximum load of, 154. stability of, 153. starting of, 151. structure of, 149. torque-load adjustment of, 150. Synchronous phase modifier, 157. Synchronous speed of induction motor, 206-7. Synchronizing of alternators, 144. T-connected three-phase system, 112. 344 INDEX Temperature coefficient, 5. Terminal voltage, 139. Three-phase alternator, 128. system, balanced, 109. capacitance of, 323. delta-connected, in. economy of, 119. star-connected, 109. Three-phase system, T-connected, 112. unbalanced, 114. V-connected, 112. Tirrill regulator, 72, 140. Torque equation for induction motors, 209. Torque-load adjustment of synchronous motors, 150. Torque of series motor, 85. shunt motor, 81. Traction, force of magnetic, 39. Transformation, phase, 199. Transformer, fundamental physical action in a, 178. connections for single-phase circuits, 196. synchronous converters, 200. three-phase circuits, 198. two-phase circuits, 198. constant-current, 192. current, 255. losses, 184, 186. potential, 256. Transformers, construction of, 183. current relations in, 180. cooling of, 188. efficiency of, 188. polyphase, 194. protection of, 196. series, 255. types of, 183. Vector diagrams for, 180. Transmission lines, capacitance of, 270. inductance of, 270. Triangle, current, 14. impedance, 12. voltage, 12. Tube, Moore, 241. Two-phase system, 107. Types of transformers, 183. Unbalanced three-phase system, 114. Units, electric, 4. electrostatic, 319. magnetic, 30. Unit of current, C. G. S., 39. pole, 30. field intensity due to, 37. magnetic flux due to, 37. Use of instruments, 23. V-connected three-phase system, 112. Vector diagrams for transformers, 180. Volt, 4, 33. Voltage characteristics of alternators, 133. compound generators, 70. series generators, 68. shunt generators, 67. Voltage of alternators, 123. INDEX 345 Voltage of continuous-current generators, 60. regulation, 61. terminal, 139. triangle, 12. Voltmeters, 252. Voltmeter compensation, 282. electrostatic, 256. Wagner BK motor, 232. Watt, 4. Wattless current, 22, 23, 173. Watt-hour meter, commutator, 260. induction, 261. Wattmeter connections, 116. Wattmeters, electrodynamometer, 258. induction, 258. polyphase, 260. Windings, armature, basket, 51. chain, 51. lap, 49. wave, 50. field, 46. Wires, carrying capacity of, 270. Wire gauge, American, 270. Wiring formulae, 276. Work, 15. Yoke, 52. Y-connected three-phase system, 109. LIST OI 7 WORKS ON ELECTRICAL SCIENCE PUBLISHED AND FOR SALE BY D. 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Third Edition. 84 Illustrations. 8vo., cloth, 160 pp. Net, SI. 50 MORGAN, ALFRED P. Wireless Telegraph Construction for Amateurs. Third Edition, 167 illustrations. 12mo., cloth, 236 pp Net, $1.50 LIST OF WORKS ON ELECTRICAL SCIENCE. 9 NERZ, F. Searchlights, Their Theory, Construction and Application. Trans- lated by C. Rodgers. 47 Illustrations. 6x8, cloth, 145 pp.. . .Net, $3.00 NIPHER, FRANCIS E. Theory of Magnetic Measurements. With an Appendix on the Method of Least Squares. Illustrated. 12mo., cloth, 94 pp.$l .00 OHM, G. S. The Galvanic Circuit Investigated Mathematically. Berlin, 1827 Translated by William Francis. With Preface and Notes by the Editor, Thos. D. Lockwood. Second Edition. Illustrated. 16mo., cloth, 269 pp. (No. 102 Van Nosrrand's Science Series.) 50 cents OLSSON, ANDREW. Motor Control as used hi Connection with Turret Turning and Gun Elevating. (The Ward Leonard System.) 13 illustrations. 12mo., paper, 27 pp. (U. S. Navy Electrical Series No. 1.) 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Vol. I.: Direct-Current Machines. Ninth Edition, completely rewritten. Illustrated. 12mo., cloth, 281 pp Net, $2.50 Vol. II.: Alternating-Current Machines: Tenth Edition, rewritten. 12mo., cloth, 353 pp Net, $2.50 Electric Traction and Transmission Engineering. 127 illustration. 12mo., cloth. 317 pp Net, $2 .50 SLOANE, T. O'CONOR. Standard Electrical Dictionary. 300 Illustrations. 12mo., cloth, 682 pp $3.00 Elementary Electrical Calculations. A Manual of Simple Engineering Mathematics, covering the whole field of Direct Current Calculations, the basis of Alternating Current Mathematics, Networks, and typical cases of Circuits, with Appendices on special subject. 8vo., cloth. Illustrated. 304 pp Net, $2 . 00 LIST OF WORKS ON ELECTRICAL SCIENCE. 11 SMITH, C. F. Practical Alternating Currents, and Alternating Current Testing. Third Edition. 236 illustrations. 5^x894, cloth. 476 pp Net, $2.50 SMITH, C. F. 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Lessons in Practical Electricity: Principle Experi- ments, and Arithmetical Problems. An Elementary Text-book. Fifteenth Edition, enlarged with a chapter on electric lighting. 404 illustrations. 12mo., cloth, 462 pp Net, $2.00 THIESS, J. B. and JOY, G. A. Toll Telephone Practice. 273 illustrations. 8vo., cloth, 433 pp Net, $3.50 THOM, C., and JONES, W. H. Telegraphic Connections, embracing recent methods in Quadruplex Telegraphy. 20 Colored Plates. 8vo., cloth, 59 pp. .$1 .50 THOMPSON, S. P., Prof. Dynamo-Electric Machinery. With an Introduction and Notes by Frank L. Pope and H. R. Butler. Illustrated. 16mo., cloth, 214 pp. (No. 66 Van Nostrand's Science Series.) 50 cents Recent Progress in Dynamo-Electric Machines. Being a Supplement to " Dynamo-Electric Machinery." Illustrated. 16mo., cloth, 113 pp. (No. 75 Van Nostrand's Science Series.) 50 cents TOWNSEND, FITZHUGH. Alternating Current Engineering. Illustrated. 8vo., paper, 32 pp Net, 75 cents UNDERBILL, C. R. Solenoids, Electromagnets and Electromagnetic Windings. Second Edition. 218 Illustrations. 12mo.,cloth, 345 pp Net, $2.00 URQUHART, J. W. Electroplating. Fith Edition. lUustrated. 12mo., cloth, 230 pp $2.00 Electrotyping. Illustrated. 12mo., cloth, 228 pp $2.00 VOSMAER, A. Ozone. Its Manufacture and Uses. Illustrated. In Press WADE, E. J. Secondary Batteries: Their Theory, Construction, and Use. Second Edition, corrected 265 illustrations. 8vo., cloth, 501 pp. Net, $4.00 12 LIST 01 WORKS ON ELECTRICAL SCIENCE."* 3 WADSWORTH, C. Primary Battery Ignition. A simple practical pocket guide on the construction, operation, maintenance, and testing of primary batteries for automobile, motorboat, and stationary engine ignition service. 26 Illustrations. 5x7, cloth, 79 pp Net, .50 WALKER, FREDERICK. Practical Dynamo-Building for Amateurs. How to Wind for any Output. Third Edition. Illustrated. 16mo., cloth, 104 pp. (No. 68 Van Nostrand's Science Series.) 50 cents. WALKER, SIDNEY F. Electricity in Mining. Illustrated. 8vo., cloth, 385 pp. $3.50 WATT, ALEXANDER. Electroplating and Refining of Metals. New Edition, rewritten by Arnold Philip. Illustrated. 8vo., cloth, 704 pp. Net, $4.50 Electro-metallurgy Fifteenth Edilion. Illustrated 12mo., cloth, 225 pp. $1.00 WEBB, H. L. A Practical Guide to the Testing of Insulated Wires and Cables. Fifth Edition. Illustrated. 12mo., cloth., 118 pp $1 . 00 WEYMOUTH, F. MARTEN. Drum Armatures and Commutators. (Theory and Practice.) A complete treatise on the theory and construction of drum- winding, and of commutators for closed-coil armatures, together with a full re'sume' of some of the principal points involved in their design, and an exposition of armature reactions and sparking. Illustrated. 8vo., cloth, 295 pp Net, $3 . 00 WILKINSON, H. D. Submarine Cable-Laying, Repairing, and Testing. Second Edition, completely revised. 313 illustrations. 8vo., cloth, 580 pp.Net, |6.00 WILSON, J. F. Essentials of Electrical Engineering. 300 illustrations. 6x9, clot.n. 255 pp Net, $2.50 WRIGHT, J. Testing, Fault Localization and General Hints for Linemen. 19 illus- trations. 16mo., cloth, 88 pp. (Installation Serious Manuals.) . Net, 50 cents YOUNG, J. ELTON. Electiical Testing for Telegraph Engineers. Illustrated. 8vo., cloth, 264 pp Net, $4.00 ZEIDLER, J. and LUSTGARTEN, J. Electric Arc Lamps Their principles, con- struction and working. 160 illustrations. 8vo., cloth, 200 pp Net, $2.00 A 96-page Catalog of Books on Electricity, classified by subjects, will be furnished gratis, postage prepaid, on application. 2 M-ll-15 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 3Mar58MF REC'D LD MAR -* 1938 General Library LD 21A-50m-8,'57 University of Calif orni a (C8481slO)476B Berkeley YC 19586 UNIVERSITY OF CALIFORNIA LIBRARY