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Electrical Drives, Machines, and Power Systems / Theodore Wildi Electrical Machines, and Power Systems Drives, Fifth Edition Theodore Wildi Professor Emeritus, Laval University Prentice Hall Upper Saddle River, New Jersey Columbus, Ohio Library of Congress Cataloging-in-Publication Data page 136 by Weston Instilments; pages 204, 239, 251, 312, by ABB; page 207 by Hammond; pages 209, 232, 777, 778, 796, 797 by 339, 344, 370, 583, 584, 626, 700, 701, 782 Theodore. Wildi, Electrical machines, Theodore Wildi.— 5th p. and drives, power systems / ed. cm. Includes bibliographical references and index. ISBN 1. 3. 0-1 3-093083-0 (alk. Electric paper) machinery. Electric driving. I. 2. Electric power systems. Title. TK2182.W53 2002 621.31' 042— dc21 2001051338 Editor in Chief: Stephen Helba Assistant Vice President and Publisher: Charles E. Stewart, Jr. Production Editor: Alexandrina Benedicto Wolf Design Coordinator: Diane Ernsberger Cover Designer: Thomas Borah Cover Art: Corbis Stock Market Production Manager: Matthew Ottenweller set in Times Roman and Helvetica by Communications, Ltd. It was printed and bound by R.R. Donnelley & Sons Company. The cover was printed by Phoenix Color Corp. This book was Carlisle Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson Pearson Photo Education Ltd., London Education Australia Pty. Limited, Sydney Education Singapore, Pte. Ltd. Education North Asia Ltd., Hong Kong Education Canada, Ltd., Toronto Education de Mexico, SA. de C.V. Education-Japan, Tokyo Education Malaysia, Pte. Ltd. Education, Upper Saddle R iver, New Jersey credits: Pages 21, 86, 87, 88, 107, 115, 371, 643, 640, 651, 652, 691, 701, 702, 708, 713, 731, 741, 743, 762, 773 General Electric; pages 1 1 7, Westinghouse; pages 232, 250 by Ferranti-Paekard; pages 704 by Montel, Sprecher and Schnh; page 235 by American Superior Electric; pages 252, 386, 644, 668, 669, 670, 689, 691, 703, 706, 708, 764, 765 by Hydro-Quebec; pages 265, 354, 392, 628, 737, 839, 840 by Lab Volt; pages 2(57, 301 by Brook Crompton-Parkinson Ltd.; page 290 by ElectroMecanik; pages 294, 295, 315, 354, 441, 442, 452, 547, 744 by Siemens; pages 300, 301, 391, 404 by Gould; page 304 by Reliance Electric; pages 337, 338, 340, 645 by Marine Industrie; pages 342, 344, by Allis-Chalniers Power Systems, Inc.; page 345 by Air France; pages 424, 425, 433 by Pacific Scientific, Motor and Control Division, Rockford, IL; pages 425, 426 by AIRPAX Corporation; pages 440, 442, 447, 450, 452, 458, 785 by Square D; pages 440, 441, 448, 449 by Klockner-Moeller; page 441 by Potter and Brumfield; pages 443, 451 by Telemeeanique, Group Schneider; page 454 by Hubbel; pages 477, 493, 500 by International Rectifier; page 595 by Robicon Corporation; page 627 by Carnival Cruise Lines; page 645 by Les Ateliers d'Ingeniere Dominion; page 647 by Tennessee Valley Authority; page 649 by Novenco, Inc.; page 694 by Pirelli Cables Limited; page 650 by FosterWheeler Energy Corporation; page 651 by Portland General Electric; pages 653, 654 by Electricity Commission of New South Wales; page 658 by Connecticut Yankee Atomic Power Company Georges Betancotirt; page 659 by Atomic Energy of Canada; page 667 by Canadian Ohio Brass Co., Ltd.; page 674 by IREQ; page 700 by Canadian General Electric; pages 703, 704, 714 by Dominion Cutout; pages 703, 704, 705, 707 by Kearney; page 731 by Sangamo; page 735 by Gentec Inc.; page 738 by Service de la C.I.D.E.M., Ville de Montreal; page 758 by GEC Power Engineering Limited, England; page 759 by Manitoba Hydro; page 762 by New Brunswick Electric Power Commission; pages 763, 764 by United Power Association; page 772 by EPRI; page 289 by Services Electromecaniques Roberge; page 830 by Fluke Electronics Canada, Inc.; pages 842, 843, 844, 849 by Oinron Canada Inc.; pages 851, 852, 853, 855 by St. Lawrence Stevedoring; pages 854, 855, 856, 857 by Schneider Electric; page 133 by Leroy Somer and Emerson Electric; page 436 by 233, 09, 100, 264, 300, 402, 59 1, 605, 619 by 289 by H. Roberge; pages by Baldor Electric Company; — Emerson Electric. Copyright © 2002, 2000, 1997, 1991, 1981 by Sperika Enterprises Ltd. and published by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Prentice Hall 10 98765432 ISBN 0-13-093083-0 Preface This fifth edition was prompted in part by the great in- is crease of computers in industrial controls and au- in tomation, which has produced computer programs can simulate relays and relay contacts. These that no longer pertinent isolation to discuss dc and ac machines because wherever they are being age. Consequently, the term drive motor alone but the now on/off discrete controls have eliminated the wiring the and installation of hardware components torque and speed of the machine. This virtual relays a keyboard. and contacts The devices that that in favor of can be programmed on perform these operations are called Programmable Logic Controllers (PLCs), or simply programmable were initially controllers. These devices stand-alone computers that controlled a specific robot or manufacturing operation. However, with the advent of the Internet, they have now been integrated with the overall manufacturing process, leading seamlessly to integration with sales, procurement, and consumer satisfaction. The 20 is management, relay control of machines covered now supplemented by coverage Chapter 3 in of PLC controls Chapter 3 of PLCs and shows, by way of example, how they used . 1 covers the basic principles are running the activities of a large service enter- in prise. 1 This new chapter illustrates computer-based setting how these trend- activities involving controls and automation are being integrated with other business activities, including e-commerce. As I mentioned in the last edition, similar up- heavals have occurred in ply amazing influence on the involves not entire unit that directs the way power technology. It is simof power elec- to witness the entrance tronics into every facet of industrial drives. Thus, it having a is machinery electrical courses are being taught. How has this dramatic change come about? mainly due to the high-power solid state It is switching devices, such as insulated gate bipolar transistors (IGBTs), which can operate at frequencies of up to 20 kHz. The change has also been driven by tors and gate turn-off thyristors (GTOs) is thyris- that handle currents of several thousand amperes ages of up to 5 kV. Another key element Chapter in direct in- an electronic control forms part of the pack- stalled, can at volt- the com- puting power of microprocessors that can process signal data in real time with incredible speed. The high switching frequencies of IGBTs permit the use of pulse-width-modulation techniques in power converters. This, in turn, enables torque and speed control of induction motors down to zero was not feasible in rectangular-wave converters that were employed only a few years ago. Most industrial drives are in the fractional horsepower to the 500 hp range. That is precisely the range now available for control by IGBTs. The result has speed. This been an explosion in the retrofitting Lower maintenance costs, of existing drives. higher efficiency, and PREFACE economically attractive. pedagogical quality. As a made such changeovers greater productivity have Thus, dc drives are being placed by induction motor drives, which require less • Every sector of is • industrial this revolutionary con- verter technology. Electric elevators, electric tives, electric transit vehicles, and ing, ventilating air locomo- • is — an power distribution of electric Most I distorted wave. are wave in Power Research (EPR1) Institute in Palo Alto, California, in collaboration with several electrical manufacturers, has also resulted in the cre- ation of high-power static switches, thyristor-eon- and converters trolled series capacitors, that electric power. method that enables into its Once they know how harmonic components, harmonics quickly in a unravel a to their interest rises. at all. All the important changes first introduced in Important development work, carried out by the Electric they affect the behav- students to calculate the harmonic content seeing large rotating machines, such as synchronous have no moving pails how also devised a simple condensers and frequency changers, being replaced by solid-state converters that have added a new chapter I generated and and the quality of and we importantly, ics are industry that has been relatively stable for over 50 years. Here, im- this ior of capacitors, inductors, cables, transformers, new technology. utilize this also affecting the transmission make on harmonics. Chapter 30 reveals how harmon- servomechanisms, heat- conditioning systems, fans, being modified to The change to portant topic easier to understand. compressors, and innumerable industrial production lines are their solutions Chapter 7 on Active, Reactive, and Apparent Power was completely revised and commercial activity by therefore being affected The end-of-chapter problems and were revised and double-checked. maintenance while offering equal and often superior dynamic performance. more than 20 result, percent of the pages were altered. re- can fill previous editions have been kept tion. in this fifth edi- Thus, the writing of circuit equations, the discussion of higher frequency transformers, and the equivalent circuit diagram of the single-phase induction motor have all been retained. the role of phase-shift transformers. These new methods of power flow FACTS by the acronym (Flexible control, AC known Systems) will permit existing transmission and bution lines to carry more demand creasing extremely bilize a power for electricity. to On that It is sta- is may suddenly be menaced by an in electric Most students in many formulating them. I particularly easy to follow. Readers will be re- minder of the circuit-solving procedure. on all rest • motor drives is Chapter panded similar to ers. employed to control the flow of power in elecAs a result, everything falls neatly and coherently into place. The teaching and learning of electric machines, drives, and power systems are 1 1 on Special Transformers was ex- to include higher The reader is frequency transform- guided through the reasoning that behind the design of such transformers, and tric utilities. they thereby made much ex- dis- glad to refer to this section as a convenient base. In other words, the converter tech- nology used 2. such equations, but close an ac/de circuit-solving methodology that account of their remarkable that these innovations common to solve perience difficulty unexpected disturbance. a section covering the writing of circuit know how distri- can also A new equations was added to Chapter meet the ever-in- fast response, the converters network • Transmission become smaller why as the frequency increases. High-frequency transformers are directly related to the higher frequencies encountered in switch- ing converters. easier. The following changes have been made in the • Chapter 16 on Synchronous Generators has been expanded fourth and fifth editions: to show why an increase in size in- evitably leads to higher efficiencies and greater • Every page of the original work was examined for clarity of expression and reviewed as to its outputs per kilogram. This fundamental aspect of machine design will interest many readers. PREFACE • A new section was added Chapter to 1 phase induction motor. many worked-out problems, to solve the circuit, presents a rigorous, yet It which permits a better under- Chapter 2 1 , Electronics, Fundamental Elements of Power was revised and expanded to in- modulation (PWM) techniques. how made they can be to converters and generate almost any motors operating major addition Electric Utility to Part It power sags, swells, It power becomes become visit students made may in to establish how the lines. book requires user-friendly treatment of even topics, this book broad range of readers. will First, meet it is the needs of a appropriate for students following a two-year electrical community versities. in colleges, technical institutes, and uni- Owing to its very broad coverage, the text can also be incorporated program. program Many for their electric in a 4-year technology universities have adopted the book power service courses. wealth of practical information that can be di- rectly applied to that greatest laboratory electrical industry itself. actual use. at close hand the The photographs help convey the 1 chapters, a conscious effort was coherence, so that the reader can see various concepts fit together. For example, the to those found lines, in turn, And reactive transmission in bring up the question power is an important aspect in electronic converters. Therefore, knowledge it in is one sector applied is strengthened and broadened in another. As a result, the learning of electrical machines, drives, and HPwer sy^ems be- comes a challenging, thought-prov^^f^experience. In order to convey the real-w|ftd aspects df machinery and power systems, particular. attentio|i has been paid to the inertia of revolving masses, the physical limitations of materials, ai^l .the problems created by heat. This appro^kfjal^^^ multidisciplinary programs of many fftfe and eo^lle^es technical institutes. Instructors responsible for industrial training will find a in the transmission and distribution of Transmission when complex the impor- not have had the opportunity to machines are similar increasingly important. gained its book shows terminology and power equations for synchronous and some trigonometry. to the an industrial plant or to see only a background in basic circuit theory, algebra, Owing and Websites articles, by diagrams and pictures, showing illustrated Throughout the 3 also as regards a reality, these subject matter covered in this will also invited to con- is various stages of construction or in of reactive power. The also avail- magnificent size of these devices and machines. methods of controlling the quality of electricity will is end of most chapters at the electrical energy. harmonics, and brownouts. As dereg- ulation of electric electronic power Manual Industrial Application prob- of books, technical equipment used explains the electronically. discusses the quality of electric appear The Reference section toward the end of the book. Some vector control. IV dealing with Power Systems. practical, inter- tance given to photographs. All equipment and sys- them technologies that are being developed to control the flow of electric that tems are at Chapter 29, Transmission and Distribution represents a lems A Solutions A quick glance through special section explains the PWM drives and flux basics of — appeal to hands-on users. The reader in the Chapter 23, Electronic Control of Alternating A the end of each chapter are di- at end of the book. sult the list /^Current Motors, was greatly expanded to cover variable speeds. exercises mediate, and advanced. Furthermore, to encourage the waveshape and frequency. the properties of induction The vided into three levels of learning able for instructors. illustrates the It oflGBT its particularly suitable the reader to solve the problems, answers are given at clude switching converters and pulse width extraordinary versatility is for self-study. programmed standing of this ubiquitous single-phase machine. • being de- is voted to continuing education, this book, with motor. Hand-held computers can be • effort velop the equivalent circuit diagram of a single- simple approach, based on the 3-phase induction • when much Finally, at a time 8 to de- v of all — the In summary, I employ a tlicM>Bk&$ ipfefeAal, multidisciplinary approach to give a broad under- standing of modern electric power. Clearly, longer the staid subject it was considered it is to no be PREFACE vi some years ago. There this is dynamic, expanding good reason to believe that field will open career op- and Bernard Oegema of Schneider Canada; Carl Tobie of Edison Electric Institute; Damiano Esposito portunities for everyone. and Vance make a final remark concerning the As mentioned previously, power technology has made a quantum jump in the past Scott Lindsay of Daiya Control I would like to use of this book. Belisle, machines, drives, and power systems, there will now be a long period of consolidation during which existing machines and devices will be replaced by newer Systems; Louis and Jean Lamontagne of Lumen; Benoit Arsenault and Les Halmos of Allen Bradley. eight years, mainly on account of the availability of fast-acting semiconductors. In the field of electrical E. Gulliksen of Carnival Cruise Lines; extend a special note of thanks to Professor I Thomas Young of Rochester the Technology, to Dr. Robert University, College, and to Jean Anderson of Lab- Volt Ltd. for models. But the basic technology covered herein will having extensively reviewed and commented not change significantly in the foreseeable future. on various aspects of this book and Consequently, the reader will find that this book can of Peros of Seneca Professor Martin to Institute H. Alden of McMaster T. their valued viewpoints. also I in depth for having offered want to acknowledge also be used as a valuable long-term reference. the contribution of Professor Stephane Montreuil of Acknowledgments end-of-chapter problems and the solutions manual. CEGEP Levis-Lauzon for having gone over all the the want I In preparing this edition and previous editions of my acknowledge book, 1 would like to the impor- tant contribution of the following persons. Consultant; David Krispinsky, Rochester Institute of Technology; Athula Kulatunga, Southeast Missourri State University; Rick Miller, Ferris State University; Nehir, Montana State University; Martin University; Chandra James E, Sekhai; University; Gerald Sevigny, Southern Maine Technical College; Philippe Viarouge, Laval University; Stacy Wilson, Western Kentucky Rochester Institute A& M and University; Sri R. Kolla, University; Thomas of Technology; Dr. P Texas Ted James, Pasadena City College; Bowling Green State University. Commercial, industrial and institutional contributors: Andre Dupont, Raj Kapila, G. Linhofer, Katherine Sahapoglu of ABB; Roger Bullock, Gerry McCormick, James Nanney, Darryl J. Van Son, and Roddy Yates of Baldor Electric Company; Jacques Bedard, Guy Goupil, and Michel Lessard of Lab-Volt Ltd.; Richard B. Dube of General Electric Company; Abdel-Aty Edric and Ashock Sundaram of Electric Power Research Institute; Neil H. Woodley of Westinghouse Electric Rene Poulin of Inc. in of Lawrence in the application and to of pro- photographer Hughes also also hereby acknowledged. is want E. Stewart, Editor; the Centre de Robotique Industrielle reviewing and describing the essential features PLCs and Jr., to to express my appreciation to Charles Publisher; to Delia Uherec, Associate Alexandrina B. Wolf, Senior Production Editor, of Prentice Hall, for planning, coordinating, and administrating As to in this text. previous editions, my provide his valuable help art, in son Karl continued preparing the line photographs, and word processing of this latest edition. My thanks also go to ing supported me thor, consultant, Goyette, Jim Corporation; Maurice Larabie, Jean-Louis Marin, controllers, St. providing industrial ex- in Chicoine for his work. The important contribution of Young, Enjeti, know-how perience and grammable M. Roach, Bob Jones Purdue appreciation to Jean- Serge and Giles Campagna of Electric, I Peros, Seneca College; my Stevedoring for their help Professors and reviewers: Robert T. H. Alden, McMaster University; Ramon E. Ariza, Delgado Community College; Fred E. Eberlin, Educational M. H. to express Lamirande of Omron, Pierre Juteau of Schneider I tors in my my wife, Rachel, for hav- continuing vocation as au- and teacher. also wish to voice my gratitude to the instruc- and students, practicing engineers, and techni- cians who asked questions and made suggestions by e-mailing their messages to wildi@wildi-theo.com. You are cordially invited to do the same. Theodore Wildi PART I. FUNDAMENTALS Distinction between sources and 2.2 loads 1. UNITS 3 l.O Introduction 3 LI Systems of units 3 1.2 Getting used to SI 4 1.3 Base and derived units of the SI 4 \A Definitions of base units 5 1. Definitions of derived units 5 1.6 Double-subscript notation for units 7 Conversion charts and their use 8 Per-unit system with one base l.ll Per-unit system with Sign notation for voltages 2.6 Graph of an 2.7 Positive and negative currents 18 19 Sinusoidal voltage 2.9 Converting cosine functions into sine 2. 10 2. 1 two bases 10 1 19 Effective value of an ac voltage 20 Phasor representation 2 2.12 Harmonics 23 2.13 Energy in 14 Energy in a 2.15 Some an inductor 25 capacitor 25 useful equations 26 1 12 ELECTROMAGNETISM 2. MAGNETISM, AND CIRCUITS 15 2.0 Introduction 2. Conventional and electron current flow 15 17 alternating voltage 2.8 1 6 FUNDAMENTALS OF ELECTRICITY, 1 17 2.5 2. l.IO Questions and Problems 17 functions 20 The per-unit system of measurement 9 1. 2.4 Multiples and submultiples Commonly used 1. Sign notation voltages of SI units 7 1. 16 2.3 15 Magnetic density 2.17 2. 1 8 2.19 field intensity H and B 27 B-H curve B-H curve of f ACULTAD vacuum 27 DE q r ~n of a magnetic material" 27 Determining the relative permeability 28 * WIN A _ - - BlBi-iO I fcw* CONTENTS i 2.20 Faraday's law of electromagnetic 3.9 induction 29 a conductor 30 2.21 Voltage induced 2.22 Lorentz force on a conductor 3 2.23 2.24 2.25 Kinetic energy of rotation, inertia in 3.10 Torque, inertia, and change Direction of the force acting on a 3.11 Speed of a motor/load system 57 straight conductor 3 3.12 Power flow Residual flux density and coercive in a mechanically coupled system 58 force 32 3.13 Motor driving a load having Hysteresis loop 33 3.14 Electric motors driving linear motion Hysteresis loss 33 Hysteresis losses caused by 3.15 Heat and temperature 60 rotation 33 3.16 Temperature scales 61 Eddy Eddy 3.17 currents 34 3.18 Eddy-current losses in a revolving core 35 2.31 Current in an inductor 36 Transmission of heat 62 Heat transfer by conduction 62 3.20 Calculating the losses by convection 63 Kirchhoffs voltage law 40 2.33 Kirchhoffs voltage law and double- 3.22 Heat transfer by radiation 64 3.23 Calculating radiation losses 64 Questions and Problems 65 40 2.34 Kirchhoffs current law 41 2.35 Currents, impedances, and associated PART II. voltages 41 2.36 Kirchhoffs laws and ac 2.37 KVL and 2.38 Solving ac and dc circuits with sign 2.39 circuits 4. 44 Circuits and hybrid notation ELECTRICAL MACHINES AND TRANSFORMERS 43 sign notation 43 45 Questions and Problems 46 DIRECT-CURRENT GENERATORS 4.0 Introduction 71 4.1 Generating an ac voltage 71 4.2 Direct-current generator 72 4.3 Difference between ac and dc generators 73 FUNDAMENTALS OF MECHANICS AND HEAT 50 4.4 Improving the waveshape 73 4.5 Induced voltage 75 Neutral zones 76 3.0 Introduction 50 4.6 3.1 Force 50 4.7 Value of the induced voltage 76 3.2 Torque 51 4.8 Generator under load: the energy 3.3 4.9 3.5 Mechanical work 5 Power 52 Power of a motor 52 3.6 Transformation of energy 53 3.7 Efficiency of a machine 53 4.11 commutation 78 Commutating poles 79 3.8 Kinetic energy of linear motion 54 4.12 Separately excited generator 79 3.4 58 temperature Heat transfer by convection 63 3.21 2.32 notation to raise the 3.19 CIRCUITS AND EQUATIONS subscript notation Heat required of a body 6 currents in a stationary iron core 35 2.30 inertia loads 59 2.26 2.28 in speed 57 2.27 2.29 moment of 54 conversion process 77 4. 10 Armature reaction 77 Shifting the brushes to improve 71 CONTENTS 4. 13 No-load operation and saturation 5.18 curve 79 4.14 Shunt generator 80 5. 4.15 Controlling the voltage of a shunt 5.20 1 9 generator 81 4.16 Dynamic braking and mechanical constant 1 Armature reaction 5.21 1 1 Flux distortion due to armature reaction Equivalent circuit 82 1 1 Commutating poles 113 Compensating winding 114 Separately excited generator under 5.22 load 82 5.23 Basics of variable speed control 4. 18 Shunt generator under load 83 5.24 Permanent magnet motors 4.19 Compound 4.20 Differential 4. 1 7 Questions and Problems generator 83 compound 1 1 Load 4.22 Generator specifications 84 84 6. EFFICIENCY AND HEATING OF ELECTRICAL MACHINES 120 120 Introduction 6.1 Mechanical losses Field 84 6.2 Electrical losses 4.24 Armature 85 6.3 Losses as a function of load 4.25 Commutator and brushes 86 6.4 Efficiency curve 123 4.26 Details of a multipole generator 88 6.5 Temperature 125 4.27 The The 6.6 Life expectancy of electric 6.7 Thermal 4.28 ideal commutation process 91 practical commutation process 92 1 20 20 1 rise DIRECT-CURRENT MOTORS 96 Introduction 96 classification of 126 6.8 Maximum ambient temperature and 6.9 Temperature 6.10 Relationship between the speed and Counter-electromotive force 5.2 Acceleration of the motor 97 Mechanical power and torque 98 5.4 Speed of rotation 100 5.5 Armature speed control 101 rise size of a machine 1 Shunt motor under load 103 ACTIVE, REACTIVE, POWER 134 7.0 Introduction 7.1 Instantaneous power 7.2 Active power 136 7.3 Reactive power 137 7. 102 Series motor Series 5.10 Applications of the series motor 104 motor speed control 105 1 5.11 Compound motor 5.12 Reversing the direction of rotation 5.13 Starting a shunt 5.14 Face-plate starter 108 5.15 Stopping a motor 109 06 106 1 07 7.4 motor 108 5.16 Dynamic braking 109 5.17 Plugging 110 7.6 3 134 134 Definition of reactive load and reactive source 7.5 ! AND APPARENT Field speed control 5.7 5.8 27 30 Questions and Problems 5.6 5.9 1 by the resistance method 129 (cemf) 96 5.3 23 insulators hot-spot temperature rise 5.1 1 equipment 126 Questions and Problems 93 5.0 1 1 6.0 CONSTRUCTION OF DIRECT-CURRENT GENERATORS 4.23 1 17 generator 84 4.21 characteristics tin 1 138 The capacitor and power 139 reactive Distinction between active and reactive power 140 CONTENTS Combined active and apparent power 141 7.7 reactive loads: Relationship between 7.9 7.10 Power Power 7.11 Further aspects of sources and loads 7.12 7. 3 1 1 4 Q, and S 141 8.20 Varmeter 177 8.2 A remarkable 1 144 7. 1 6 17 Systems comprising several loads 146 9. 148 Solving AC circuits using the method 148 Power and vector notation power Rules on sources and loads (sign 154 Rules on sources and loads (double subscript notation) Introduction 9. Voltage induced 1 Introduction 83 Elementary transformer 9.4 Polarity of a transformer 9.5 Properties of polarity marks 9.6 Ideal transformer at no-load; voltage 1 84 1 85 1 86 186 187 ratio 9.7 Ideal transformer under load; current 9.8 Circuit 188 symbol for an ideal transformer 191 158 Polyphase systems 158 9.9 Single-phase generator 159 9. Impedance 10 Power output of a single-phase ratio 191 Shifting impedances from secondary to primary and vice versa 192 Questions and Problems 60 Two-phase generator 160 Power output of a 2-phase 8.5 1 9.3 8.1 1 a coil Applied voltage and induced 8.2 generator in 9.2 ratio 8.4 80 1 183 9.0 154 THREE-PHASE CIRCUITS 158 8.3 178 THE IDEAL TRANSFORMER 183 voltage 151 Questions and Problems 155 8.0 single-phase to 3-phase 144 notation) 7. 3-phase, Questions and Problems triangle 7.15 in 177 transformation Reactive power without magnetic fields 7. P, 143 triangle Power measurement 4-wire circuits 7.8 factor 19 8. 10. 195 PRACTICAL TRANSFORMERS 197 generator 16 10.0 Introduction 8.6 Three-phase generator 162 10. Ideal transformer with an imperfect 8.7 Power output of generator 1 a 3 -phase 10.2 8.8 Wye Voltage relationships connection 164 8.10 Delta connection Power transmitted by line 8.12 8.13 97 199 Primary and secondary leakage 10.3 167 reactance 200 a 3-phase 1 0.4 168 Equivalent circuit of a practical transformer 202 Active, reactive and apparent 3-phase circuits 1 197 Ideal transformer with loose coupling 165 8. 1 core 62 8.9 1 1 power in 1 Construction of a power 0.5 169 transformer 203 Solving 3-phase circuits 170 10.6 Standard terminal markings 204 8.14 Industrial loads 171 10.7 Polarity tests 8.15 Phase sequence 174 10.8 Transformer taps 205 10.9 Losses and transformer rating 206 8. 1 6 Determining the phase sequence 8. 1 7 8. 1 8 Power measurement Power measurement 3-wire circuits 176 in ac circuits in 3-phase, 1 75 1 76 204 10.10 No-load saturation curve 206 10.11 Cooling methods 207 10. 1 2 Simplifying the equivalent circuit 209 CONTENTS Voltage regulation 211 10.14 Measuring transformer transformers 260 impedances 212 Questions and Problems 260 10.15 Introducing the per unit method 215 1 11. 0. 6 1 Impedance of a transformer 2 Typical per-unit impedances 216 10.18 Transformers 1 THREE-PHASE INDUCTION MOTORS 13.0 Introduction 263 13.1 Principal 13.2 Principle of operation 13.3 The TRANSFORMERS 225 Introduction 225 1 1 1 .2 Autotransformer 226 1 .3 Conventional transformer connected 1 1 .5 1 1 .6 Number 3.5 1 1 cage motor 273 228 Acceleration of the rotor-slip 274 13.7 Motor under load 274 and slip speed 274 Voltage transformers 230 13.8 Current transformers 23 13.9 Slip 13.10 Voltage and frequency induced Opening the secondary of a CT can be rotor Toroidal current transformers 234 .7 11.8 Variable autotransformer 235 11.9 High-impedance transformers 236 1 1 . 1 0 1 1 . 1 1 1 2. 1 Active power flow 278 Torque versus speed curve 28 13.15 Effect of rotor re s stance 282 13.16 Wound-rotor motor 284 Introduction 243 13.17 Three-phase windings 285 Basic properties of 3-phase 13.18 1 2.4 Wye-delta connection 247 12.6 Open-delta connection 248 Three-phase transformers 249 12.8 Step-up and step-down 1 2. 1 1 0 1 12. 12 19 3.2 1 i Sector motor 288 Linear induction motor 289 Traveling waves 291 Properties of a linear induction motor 291 13.22 Wye-wye connection 248 12.7 3. 13.20 244 1 1 an induction 13.14 Delta-wye connection 246 2. in 13.13 2.3 1 Estimating the currents High-frequency transformers 238 1 12.9 13.12 Questions and Problems 24 Delta-delta connection 2.5 Characteristics of squirrel-cage motor 277 transformer banks 243 1 13.11 Induction heating transformers 237 2.2 1 in the 275 induction motors 276 THREE-PHASE TRANSFORMERS 243 12.0 of poles-synchronous Starting characteristics of a squirrel- 3.6 dangerous 233 1 264 265 rotating field speed 271 1 as an autotransformer components 263 Direction of rotation 270 13.4 transformer 225 11. 4 263 Questions and Problems 221 in parallel Dual-voltage distribution 1 . Polarity 219 SPECIAL 1 13. 1 10.17 11.0 12. marking of 3-phase 10.13 12.13 Magnetic levitation 293 Questions and Problems 295 autotransformer 25 SELECTION AND APPLICATION OF THREE-PHASE INDUCTION Phase-shift principle 253 MOTORS 14. Three-phase to 2-phase 14.0 transformation 254 1 4. 1 4.2 1 Calculations involving 3-phase trans- 299 Introduction 299 Standardization and classification of induction motors 299 Phase-shift transformer 256 formers 258 xi Classification according to environ- ment and cooling methods 299 CONTENTS Classification according to electrical 14.3 16. 16.0 Introduction 335 16. Commercial synchronous 335 14.4 Choice of motor speed 303 14.5 Two-speed motors 303 14.6 Induction motor characteristics under 16.2 Number various load conditions 305 16.3 16.4 Main Main Field excitation and exciters 342 14.7 Starting an induction 1 generators 335 motor 308 of poles 335 336 features of the stator features of the rotor 340 14.8 Plugging an induction motor 308 16.5 14.9 Braking with direct current 309 16.6 Brushless excitation 343 14.10 Abnormal conditions 310 Mechanical overload 310 16.7 Factors affecting the size of 16.8 No-load saturation curve 345 14.11 1 4. 1 2 Line voltage changes 3 Single-phasing 310 14.14 Frequency variation 311 4. 1 5 synchronous generators 344 1 14.13 1 circuit of an ac generator 346 Induction motor operating as a 16.10 Determining the value of generator 3 16.11 Base impedance, per-unit 1 Complete torque-speed characteristic of an induction machine 314 14.17 Features of a wound-rotor induction Start-up of high-inertia loads 315 14.19 Variable-speed drives 315 Frequency converter 3 s 348 s 349 350 1 6. 1 2 Short-circuit ratio 1 6. 1 3 Synchronous generator under 16.14 14.18 X X load 350 motor 315 4.20 Synchronous reactance-equivalent 16.9 14.16 1 SYNCHRONOUS GENERATORS and mechanical properties 301 Regulation curves 352 1 6. 1 5 Synchronization of a generator 353 1 6. 1 6 Synchronous generator on an infinite bus 355 1 Questions and Problems 3 16.17 1 Infinite bus-effect of varying the exciting current 355 15. EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR 322 15.0 15.1 16.18 Infinite bus-effect of varying the mechanical torque 355 Introduction 322 The wound-rotor induction motor 322 Power relationships 325 16.19 Physical interpretation of alternator behavior 357 1 6.20 Active power delivered by the motor 326 1 6.2 Control of active power 359 15.4 Breakdown torque and speed 327 16.22 5.5 Equivalent circuit of two practical 1 15.2 1 1 5.3 Phasor diagram of the nfl, i generator 358 iction 1 6.23 motors 327 15.6 15.7 5.8 15.9 Power transfer between two sources 361 Calculation of the breakdown 16.24 Efficiency, power, and size of torque 328 electrical Torque-speed curve and other Questions and Problems 364 characteristics 1 Transient reactance 359 machines 362 329 Properties of an asynchronous 17. SYNCHRONOUS MOTORS generator 330 17.0 Introduction 369 Tests to determine the equivalent 17.1 Construction 370 circuit 33 l Questions and Problems 333 1 7.2 17.3 Starting a synchronous Pull-in torque 372 369 motor 372 CONTENTS ] 1 Motor under load-general 7.4 7.5 18.19 Deducing xiii the circuit diagram of a description 372 single-phase motor 411 Motor under load-simple Questions and Problems 4 1 calculations 373 Power and torque 376 17.7 Mechanical and 17.8 Reluctance torque 378 19.1 Elementary stepper motor 417 Losses and efficiency of a 19.2 Effect of inertia 1 7.9 19. 377 electrical angles 1 synchronous motor 379 17.10 1 7. 1 1 power 380 Excitation and reactive Power 1 9.0 418 Effect of a mechanical load 4 9.3 19.5 Start-stop stepping rate 19.6 Slew speed 421 420 V-curves 382 17.13 Stopping synchronous motors 383 19.7 Ramping 422 The synchronous motor versus induction motor 385 Synchronous capacitor 385 19.8 Types of stepper motors 422 14 7. 17.15 the Motor windings and associated 424 19.10 High-speed operation 427 Modifying the time constant 428 19.11 19.12 Bilevel drive 428 19.13 Instability and resonance 434 19.14 Stepper motors and linear drives 434 Questions and Problems 434 19.9 drives Questions and Problems 388 SINGLE-PHASE 1 8. MOTORS 391 Introduction 39 18.0 Construction of a single-phase 1 induction motor 39 18.2 Synchronous speed 393 18.3 Torque-speed characteristic 394 18.4 Principle of operation PART III. 394 ELECTRICAL AND ELECTRONIC DRIVES Locked-rotor torque 396 1 8.5 1 8.6 Resistance split-phase motor 396 1 8.7 Capacitor-start motor 398 1 8.8 Efficiency and 1 1 Torque versus current 420 19.4 factor rating 38 Introduction 41 17.12 1 18. STEPPER MOTORS 417 17.6 8.9 18.10 1 8. 1 1 8. 12 1 power 20. factor of single- BASICS OF INDUSTRIAL CONTROL MOTOR 439 phase induction motors 399 20.0 Introduction 439 Vibration of single-phase motors 40! 20.1 Control devices 439 Capacitor-run motor 402 20.2 Normally-open and normally-closed contacts 443 Reversing the direction of 403 20.3 Relay Shaded-pole motor 403 20.4 Control diagrams 445 rotation coil exciting current 443 18.13 Universal motor 404 20.5 Starting methods 446 18.14 Hysteresis motor 405 20.6 Manual across-the-line starters 18.15 Synchronous reluctance motor 407 20.7 Magnetic across-the-line 18.16 Synchro drive 408 20.8 Inching and jogging 450 EQUIVALENT CIRCUIT OF A SINGLE-PHASE MOTOR Reversing the direction of 20.9 rotation 451 1 8. 1 7 18.18 Magnetomotive force Revolving motor 410 mmfs in distribution a single-phase 409 20.10 20. 1 1 20.12 447 starters Plugging 453 Reduced-voltage starting 454 Primary resistance starting 454 448 CONTENTS xiv 20. 1 3 20.14 Autotransformer starting 458 21.17 Power gain of Other starting methods 460 21.18 Current interruption and forced 20.15 Cam 20.16 Computers and controls 462 a thy ristor 494 commutation 495 switches 461 21.19 Basic thyristor power circuits 496 2 .20 Controlled rectifier supplying a 1 ELECTRIC DRIVES passive load (Circuit 7 Fundamentals of electric drives 462 1 8 Typical torque-speed curves 463 1 9 Shape of the torque-speed 20. 1 20. 20. curve 464 20.20 Current-speed curves 466 20.21 Regenerative braking 467 21 D) Controlled rectifier supplying an ac- 21 .22 Line-commutated inverter (Circuit tive load (Circuit 2, 21.0 2 .23 1 472 Potential level Voltage across some circuit .2 21 .25 2 1 .4 2 1 .5 Battery charger with series .7 6, Table 2 D) 502 1 Delayed triggering-rectifier 21.29 Delayed triggering-inverter mode 507 21.30 Triggering range 508 21.31 Equivalent circuit of a 2 1 .32 Currents in a 3-phase, 6-pulse converter 5 Single-phase bridge rectifier 480 481 Filters 2 .9 Three-phase, 3-pulse diode 21.10 Three-phase, 6-pulse rectifier 485 21.11 Effective line current, fundamental 1 Power factor 511 21.34 Commutation overlap 514 21.35 Extinction angle 514 21.33 483 489 Distortion power line current 21.12 Three-phase, 6-pulse controllable mode 505 476 Battery charger with series rectifier Table converter 509 21.8 1 5, Three-phase, 6-pulse rectifier feeding 1 inductor 478 21 Cycloconverter (Circuit an active load 504 The diode 475 Main characteristics of a diode 476 resistor Table 2 .27 THE DIODE AND DIODE CIRCUITS .6 4, Basic principle of operation 503 1 1 switch (Circuit 500 21 .26 2 .28 2 3, 21D) 501 elements 474 21.3 static converter (Circuit 21.1 1 AC 2 ID) Introduction 472 2 Table 2 ID) 497 Table 2 ID) 498 2 1 .24 FUNDAMENTAL ELEMENTS OF POWER ELECTRONICS 472 Table 21.21 Questions and Problems 468 21. 1, 496 factor 490 Displacement power factor, total power factor 490 21.14 Harmonic content, THD 491 21.13 DC-TO-DC SWITCHING CONVERTERS 21.36 DC-to-DC switching converter 21.38 Rapid switching 519 21.39 Impedance transformation 522 2 .40 Basic 2-quadrant dc-to-dc 21.41 Two-quadrant electronic 2 1 .42 Four-quadrant dc-to-dc 1 converter 522 THE THYRISTOR converter 525 AND THYRISTOR CIRCUITS 21.15 Thethyristor 492 21.16 Principles of gate firing Semiconductor switches 515 21.37 converter 526 492 21.43 Switching losses 528 5 1 CONTENTS Dc-to-ac rectangular wave ELECTRONIC CONTROL OF ALTERNATING CURRENT MOTORS 575 converter 529 23.0 Introduction 575 Dc-to-ac converter with pulse-width 23.1 Types of ac drives 575 modulation 530 23.2 DC-TO-AC SWITCHING CONVERTERS 2 .44 1 2 .45 1 23. 21.46 Dc-to-ac sine wave converter 532 21.47 Generating a sine wave 533 PWM pulse train Synchronous motor drive using current-source dc link 577 Synchronous motor and 23.3 cycloconverter 580 534 21.48 Creating the 21.49 Dc-to-ac 3-phase converter 535 21.50 xv Cycloconverter voltage and frequency 23.4 control Conclusion 537 580 Squirrel-cage induction motor with Questions and Problems 537 23.5 ELECTRONIC CONTROL OF DIRECT- 23.6 cycloconverter 582 22. CURRENT MOTORS Squirrel-cage motor and static voltage controller 541 quadrant speed control 541 22.1 First 22.2 Two-quadrant control-field reversal 544 SELF-COMMUTATED INVERTERS 23.8 Self-commutated inverters for cage 23.9 Current-source self-commutated motors 592 Two-quadrant control-armature 22.3 reversal 545 frequency converter (rectangular Two-quadrant control-two 22.4 wave) 593 converters 545 Four-quadrant control-two converters 22.5 23.10 Two-quadrant control with positive 22.6 23. torque 549 22.7 Four-quadrant drive 549 22.8 Six-pulse converter with freewheeling 23. 1 1 1 2 diode 551 Half-bridge converter 556 22.10 Detraction 558 Motor drive using a dc-to-dc 22. 12 Introduction to brushless dc 22. Commutator replaced by reversing Recovering power in a wound-rotor 23.14 Pulse-width modulation and ind modulation 602 motors 604 Synchronous motor 22. 1 1 5 6 U^'COi" TORQUE AND SPEED CONTROL as a brushless dc OF INDUCTION MOTORS machine 568 22. wound- Review of pulse-width switches 566 22. 14 motor 597 23.13 motors 565 3 control of a rotor induction PULSE-WIDTH MODULATION DRIVES switching converter 560 1 wave) 594 Chopper speed induction motor 599 22.9 1 Voltage-source self-commutated frequency converter (rectangular with circulating current 546 1 590 Introduction 54 22.0 22. 589 Soft-starting cage motors 23.7 BGi'l. VAO Q[ ^!3 L fTRM Standard synchronous motor and 23.15 Dc motor and brushless dc machine 569 23.16 Slip speed, flux orientation, Questions and Problems 571 604 23.17 Features of variable-speed controlconstant torque mode 607 A aBdBwiOiECA torque 605 Practical application of a brushless dc motor 569 flux orientation CONTENTS xvi 23. 8 1 Features of variable-speed controlconstant horsepower 23.19 24. Features of variable-speed control- 1 Induction motor and its 24. 1 2 Equivalent circuit of a practical Volts per hertz of a practical Speed and torque control of induction motors 614 Carrier frequencies 615 23.25 Dynamic Condenser 650 24.15 Cooling towers 650 24. 6 Boiler- feed 24. 7 Energy flow diagram for a steam 24. Thermal 1 8 stations and the environment 652 616 Principle of flux vector control 23.27 Variable-speed drive and electric NUCLEAR GENERATING STATIONS 618 Principal 23.29 Operating mode mode Composition of an atomic nucleus; 24.20 The source of uranium 655 isotopes 655 of the 3-phase converter 622 Operating 24. 9 1 components 621 23.28 of the single-phase converter 624 Conclusion 629 Questions and Problems 629 IV. pump 65 plant 651 control of induction 23.26 PART thermal generating 648 Turbines 650 motors 615 23.31 646 24.14 1 23.24 23.30 installations 24.13 1 traction of a hydropower plant 644 Makeup of a station motor 613 23.23 Makeup Pumped-storage THERMAL GENERATING STATIONS equivalent motor 612 23.22 0 24.11 circuit 61 23.2 1 mode 610 generator 23.20 Types of hydropower stations 643 24.9 mode 610 ELECTRIC UTILITY POWER 24.21 Energy released by atomic fission 656 24.22 Chain reaction 656 24.23 Types of nuclear reactors 657 24.24 24.25 Example of a light-water reactor 658 Example of a heavy-water reactor 659 24.26 Principle of the fast breeder reactor SYSTEMS 24.27 660 Nuclear fusion 661 Questions and Problems 661 24. GENERATION OF ELECTRICAL ENERGY 635 25. 24.0 Introduction 635 24. Demand 1 TRANSMISSION OF ELECTRICAL ENERGY of an electrical system 635 24.2 Location of the generating station 637 24.3 Types of generating 24.4 Controlling the power balance 24.5 between generator and load 638 Advantage of interconnected stations 25.0 25. 1 637 Conditions during an outage 641 Frequency and electric clocks 642 components of a power 664 25.2 Types of power 25.3 Standard voltages 667 25.4 Components of a line 24.7 lines HV 665 transmission 667 25.5 Construction of a line 668 25.6 Galloping lines 669 Corona effect-radio interference 669 Pollution 669 Lightning strokes 670 HYDROPOWER GENERATING STATIONS 25.7 Available hydro power 642 25.9 25.8 24.8 Principal distribution system systems 639 24.6 664 Introduction 664 CONTENTS 25.10 Lightning arresters on buildings 671 25.11 Lightning and transmission lines 671 25.12 Basic impulse insulation level 26 A] Low-voltage distribution 709 PROTECTION OF MEDIUM-VOLTAGE DISTRIBUTION SYSTEMS (BIL) 672 25.14 Ground wires 673 Tower grounding 673 25.15 Fundamental objectives of a 25.13 26. 2 1 Coordination of the protective devices 714 26. transmission line 675 1 3 26.14 25.16 Equivalent circuit of a line 676 25.17 Typical impedance values 676 25.18 Simplifying the equivalent circuit 678 25.19 xvii Fused cutouts 7 1 Reclosers 716 26.15 Sectionalizers 716 26.16 Review of Voltage regulation and power- MV protection 717 LOW-VOLTAGE DISTRIBUTION transmission capability of 26.17 transmission lines 680 25.20 Resistive line 680 25.21 Inductive line 681 25.22 Compensated inductive 25.23 25.24 25.25 25.26 25.27 25.28 25.29 26. line 683 1 8 LV distribution system Grounding 717 electrical installations 7 Electric shock 26.20 Grounding of 120 V 240V/120V and systems 720 Inductive line connecting two systems 685 26.21 Equipment grounding 721 Review of power transmission 686 line voltage 687 Methods of increasing the power capacity 689 Extra-high-voltage lines 689 Power exchange between power centers 692 Practical example of power exchange 693 26.22 Ground-fault circuit breaker 723 26.23 Choosing the Rapid conductor heating: 2 I t factor 724 26.24 The 26.25 Electrical installation in role of fuses 725 buildings 725 26.26 Principal components of an electrical 725 installation Questions and Problems 727 Questions and Problems 695 THE COST OF ELECTRICITY 729 26. DISTRIBUTION OF ELECTRICAL Introduction 729 ENERGY 698 Tariff based upon energy 730 Tariff based upon demand 730 26.0 Introduction 698 Demand meter 730 SUBSTATIONS Tariff based 26.1 Substation equipment 698 upon power factor 732 Typical rate structures 733 Demand 698 26.2 Circuit breakers 26.3 Air-break switches 702 26.4 Disconnecting switches 702 26.5 Grounding switches 702 26.6 Surge arresters 702 1 719 26.19 Power ^giiTAD Of ILF controllers 733 factor correction Measuring \ 737 electrical energy, the watthourmeter 740 27.9 Operation of the watthourmeter 741 26.7 Current-limiting reactors 705 27.10 Meter readout 742 26.8 Grounding transformer 706 27.11 Measuring three-phase energy and 26.9 Example of a substation 707 power 743 26.10 Medium-voltage distribution 709 Questions and Problems 743 CONTENTS xviii 28. DIRECT-CURRENT TRANSMISSION 746 Why Distribution system 785 Compensators and 28.0 Introduction 746 28. Features of dc transmission 746 29. 10 1 28.2 Basic dc transmission system 747 28.3 Voltage, current, and Power fluctuations 29. power 1 1 characteristic 28.6 Power on a dc line 29.13 752 28.9 Power reversal 755 Components of a dc transmission line 30. 755 Inductors and harmonic 28.12 Converter transformers 756 28. 13 Reactive power source 757 compensator: principle of Conclusion 796 Harmonic filters on the ac side 757 28.15 Communications link 757 28.16 Ground electrode 757 28. 7 Example of a monopolar converter station 757 28. 8 Thyristor converter station 758 28.19 Typical installations 760 30.0 Introduction 799 Harmonics and phasor diagrams 799 Effective value of a distorted wave 800 Crest factor and total harmonic 30.3 distortion 30.5 Displacement power factor and power Non-linear loads 804 Generating harmonics 805 30.8 Correcting the power factor 807 30.9 Generation of reactive power 808 EFFECT OF HARMONICS 30. 10 30. 29. Thyristor-controlled series capacitor 1 1 30.12 Harmonic current in a capacitor 809 Harmonic currents in a conductor 810 Distorted voltage and flux in a 810 Harmonic currents coil 30.13 in a 3-phase, 4-wire distribution system 812 (TCSC) 769 29.2 Vernier control 771 29.3 Static 29.4 Eliminating the harmonics 776 29.5 Unified power flow controller Harmonics and resonance 813 Harmonic filters 818 30.16 Harmonics in the supply network 819 (UPEC) 776 30.17 29.6 Static synchronous compensator 773 total 804 30.6 SOLID-STATE CONTROLLERS 768 Introduction 768 factor circuits 30.7 TRANSMISSION AND DISTRIBUTION 29.0 802 Harmonics and Questions and Problems 765 TRANSMISSION POWER FLOW CONTROLLERS (THD) 801 30.4 1 1 799 30.1 on the 28. 14 1 series HARMONICS 30.2 filters dc side (6-pulse converter) 756 29. The 754 Bipolar transmission line 754 1 principle of Questions and Problems 797 control 753 Effect of voltage fluctuations 1 The shunt compensator: operation 793 28.7 28. 29. 12 75 28.8 28. 10 circuit operation 787 Typical rectifier and inverter 28.5 784 analysis 787 relationships 748 28.4 PWM converters? 29.8 29.9 30.14 30.15 Transformers and the K factor 821 frequency changer 780 HARMONIC ANALYSIS DISTRIBUTION CUSTOM POWER PRODUCTS 30. 29.7 1 8 Procedure of analyzing a periodic Disturbances on distribution wave 823 systems 782 Questions and Problems 827 CONTENTS 31. PROGRAMMABLE LOGIC 31.18 Getting to CONTROLLERS 831 31.19 Linking the PLCs 853 Introduction 83 31.20 Capacity of industrial PLCs 83 31.21 Programming the PLCs 853 The transparent enterprise 855 Elements of system 832 31.0 3 1 . 3 1 .2 3 1 .3 3 l .4 1 a control 31.10 31.11 Conventional control circuits and 31.6 3 1 .7 3 1 .8 3 1 .9 circuits Appendixes 865 AXO Conversion Charts 865 AX1 Properties of Insulating Materials 869 AX2 PLC Mechanical and Thermal Properties of Some Electrical, Common Conductors (and Insulators) 870 Security rule 847 31.13 Programming the PLC 847 Programming languages 847 31.14 References 859 844 31.12 31.15 851 Questions and Problems 856 Examples of the use of a PLC 835 The central processing unit (CPU) 838 Programming unit 838 The I/O modules 839 Structure of the input modules 839 Structure of the output modules 840 Modular construction of PLCs 84 Remote inputs and outputs 84 31. 5 know PLCs AX3 Properties of Round Copper Conductors 871 Advantages of PLCs over relay Answers cabinets 848 MODERNIZATION OF AN INDUSTRY PLCs 850 31.16 Industrial application of 31.17 Planning the change 850 to Problems 873 Answers to Industrial Application Problems 877 Index 879 xix 1 To Rachel Part One Fundamentals Chapter 1 Units example, Introduction 1.0 in measuring length some people use the inch and yard, while others use the millimeter and an important role Units play everything effect, we buy and thing we familiar that sell is Some means of units. we our daily how lives. In meter. Astronomers them become for granted, they started, or deal with the rod and chain. But these units of length so can be compared with great accuracy because the seldom why is based upon the speed of light. Such standards of reference make it possible to compare the units of measure in one country, or in standard of length they were given the sizes they have. was defined as the length of 36 barleycorns strung end and the yard was the distance from the tip Centuries ago the foot to the end of we have come measure more now based upon This way precisely. in improvement in in in defining Most units are which terms of the speed of our standards of measure has hand with the advances in 1.1 Systems Over the years systems of units of units have been devised to A may be described as one in which the units bear a direct numerical relationship to each technology, other, usually expressed as a whole number. Thus the English system of units, the inch, foot, the other. are related to each other by the Although the basic standards of reference are recall any meet the needs of commerce, industry, and science. system of units and the one could not have been achieved without ognized by in the anchors that tie together the units used in the and time by the duration of atomic vibrations. gone hand measure Standard units of length, mass, and time are world today. and reproducible. Thus the me- and yard are measured light, a long specialty, with the units of other. of King Edgar's nose the physical laws of nature, are both invariable ter one his outstretched hand. Since then our units of to end, employ the parsec, physicists some surveyors still have to use the angstrom, and measured and compared by of these units have often take stopping to think in see and feel and every- The same countries of the world, the units of numbers 1 in and yard 2, 3, and 36. correlation exists in metric systems, except that the units are related to each other by everyday measure are far from being universal. For multiples of ten. 3 Thus the centimeter, meter, and FUNDAMENTALS 4 kilometer are related by the numbers 1 00 000. 1 00, 1 000, and 4. It centimeters than to convert yards into feet, and this decimal approach scientist, the layman, thereby blending the theoretical and one of the advantages of the is can be used by the research technician, the practicing engineer, and by the therefore easier to convert meters into It is the practical worlds. metric system of units.* Despite these advantages the SI Today the officially recognized metric system the International System of Units, for which the is SI. The SI was formally universal abbreviation introduced 1960, in the at "Systeme international d to everything. In specialized areas of not the answer and even in atomic physics, The used to official introduction of Units, and its be measure will continue to plane angles in degrees, even though the SI unit radian. Furthermore, 1 day and hour will is the be used, still unites. despite the fact that the SI unit of time 1.2 Getting may day-to-day work, other units more convenient. Thus we General Eleventh Conference of Weights and Measures, under the official title is is SI Base and derived units 1.3 of the International System adoption by most countries of the is the second. of the SI The foundation of the International System of Units rests upon the seven base units listed in Table A. l world, did not, however, eliminate the systems that were previously employed. become habits, units cannot readily from yards to let go. lates to the It is because long familiarity with a an idea of magnitude and how its (particularly in the electrical it it re- growing importance of SI the and mechanical know necessary to TABLE 1A BASE UNITS Quantity Symbol Unit And physical world. Nevertheless, makes we not easy to switch overnight meters and from ounces to grams. this is quite natural, unit gives us Just like well-established a part of ourselves, which Length meter m Mass Time kilogram kg second s Electric current ampere Temperature kclvin A K Luminous candela cd mole mol fields) the essentials of this Amount intensity of substance measurement system. Consequently, one must be able to convert ple, from one system unambiguous way. to another in a sim- From In this regard the reader will these base units we derive other units to discover that the conversion charts listed in the express quantities such as area, power, force, mag- Appendix are particularly netic flux, and so on. There number of units The SI possesses a tures shared helpful. number of remarkable by no other system of fea- we can is really derive, but no limit to the some occur so frequently that they have been given special names. units: Thus, instead of saying that the unit of pressure 1 . 2. It is It a decimal system. employs many dustry and pere, kilogram, 3. It is units commerce; commonly used some of the most ships in electricity, mechanics, The metric Canada am- that have special names are listed in Table l B. and watt. a coherent system that expresses with star- tling simplicity * in in- for example, volt, unit of length is the official spelling heat. spelled either meter or metre. In is metre. TABLE 1B DERIVED UNITS basic relation- and is newton per square meter, we use a less cumbersome name, the pascal. Some of the derived units the Quantity Unit Symbol Electric capacitance farad F Electric charge coulomb C Electric conductance Siemens S UNITS A TABLE 1 B 5 tuned to the resonant fre- epiartz oscillator, (continued) quency of cesium atoms, produces a highly accuQuantity Symbol Unit and stable frequency. The ampere (A) is that constant current which, if maintained in two straight parallel conductors of inrate Electric potential volt V Electric resistance ohm tt Energy joule J finite length, Force newton N placed Frequency hertz Hz tween these conductors a force equal newton per meter of Illumination lux lx iicni y n Luminous flux lumen lm Magnetic flux weber Wb Magnetic flux density tesla T 1 1 1 LI LI<_ LuIlLC Plane angle radian The kelvin ature, watt W pascal Pa steradian in units illustrate the ated with this water, ice, italics is sr and water vapor definition. The The candela definitions of the SI base extraordinary precision associunits. The text in explanatory and does not form part of the by light in the length of the path travelled vacuum during a time interval of 1/299 until A coexist equal point is to called the is 273.16 equal kelvins, to 0.01 de- temperature ofO °C is therefore (cd) 1983 the speed of light was defined 792 458 m/s exactly. to he 299 the luminous intensity, in a is given direction, of a source that emits monochromatic radiation of frequency 540 X 10 12 hertz and that has a radiant intensity in that direction of 1/683 watt per steradian. is the amount of substance of system that contains as many elementary a entities as there are atoms in 0.012 kilogram of carbon 12. Note: 792 458 of a second. /// is triple The mole (mol) is cooled cell is equal to 273.15 kelvins, exactly definition: The meter (m) of water. triple point an evacuated point of water and triple base units modern system of thermodynamic temperthermodynamic begins to form. The resulting temperature where ice gree Celsius (°C). official length. the fraction 1/273.16 of the is temperature of the by The following 10~ 7 X to 2 rad Power 1.4 Definitions of vacuum, would produce be- in (K), unit of Pure water Pressure Solid angle of negligible circular cross-section, and meter apart 1 tities When the mole is used, the elementary en- must be specified and may be atoms, mole- cules, ions, electrons, other particles, or specified The kilogram (kg) is the unit of mass; it is equal to the mass of the international prototype of groups of such particles. the kilogram. The international prototype of the kilogram is a 1.5 Definitions of derived units particular cylinder of platinum-iridium alloy that is preserved in International a vault at Sevres, France, by the Bureau of Weights and Measures. Duplicates of the prototype exist in all important standards laboratories in the world. The platinum- iridium cylinder (90 percent platinum, 10 percent iridium) is about 4 cm high and 4 cm The second (s) is the in diameter. duration of 9 192 63 1 770 periods of the radiation corresponding to the transition state between the two hyperfine levels of the ground of the cesium- 33 atom. 1 Some of the more important derived units are de- fined as follows: The coulomb (C) transported in (Hence 1 l is the quantity of electricity second by a current of coulomb = / The degree Celsius (°C) and is used in l ampere. ampere second.) is equal to the kelvin place of the kelvin for expressing Celsius temperature (symbol t) defined by the equa- = T — Ta where T is the thermodynamic perature and Tn = 273.15 K, by definition. tion t tem- FUNDAMENTALS 6 The farad (F) is the capacitance of a capacitor between the plates of which there appears a ence of potential of volt I when quantity of electricity equal to force coulomb. farad coulomb per volt) The henry (H) is the inductance of a closed circuit in which an electromotive force of volt is produced when the electric current in the circuit varies uniformly at a rate of ampere per second. (Hence = 1 (I / = I volt tromotive force. (Hence second per ampere.) angle with that force is which gives kilogram an acceleration of second per second. (Hence I newton = to I a angle with kilogram mass and an acceleration, defined in The terms of a Exponent form 24 000 000 000 000 000 000 000 000 1() 000 000 000 000 000 000 000 10 000 000 000 000 000 000 000 000 000 000 000 10 l I I I in length to the radius. is the unit of electric conduc- vertex at the center of a sphere and en- tesla (T) 000 000 10 I 000 00 1 l() 0.1 0.01 2 IN 0.000 000 001 0.000 000 000 001 6 3 1() 10 2 10' 10-' fi l()" 9 10 10" 0.000 000 000 000 000 001 10 i2 15 Symbol yotta Y zetta Z E peta P tera T tr't- mana G mega M kilo k hecto h dec a da deci d centi c milli m micro nano n pico P fern to f alto a ~ lx 10" 21 10 SI 2 10 10 0 000 000 000 000 000 000 000 001 Prefix 12 uf 0.000 000 000 000 001 0.000 000 000 000 000 000 001 UNITS exa 0.001 0.000 001 SI ' l() 10 l in length to the radius. the unit of magnetic flux density ,s 000 000 000 000 000 000 000 I is equal to one weber per square meter. also applies to sta- it Multiplier I its PREFIXES TO CREATE MULTIPLES AND SUBMULTIPLES OF TABLE 1C (S) of a square with sides equal is ampere. ) closing an area of the spherical surface equal to that meter per second squared. Although the newton J volt per ohm. (The Siemens was formerly named the mho. ) The steradian (sr) is the unit of measure of a solid meter per I ohm = tance equal to one reciprocal newton meter) 1 I vertex at the center of a circle and sub- The Siemens 1 The newton (N) its tended by an arc equal 1 mass of am- The pascal (Pa) is the unit of pressure or stress equal to one newton per square meter. The radian (rad) is the unit of measure of a plane I I I pere, this conductor not being the source of any elec- The hertz (Hz) is the frequency of a periodic phenomenon of which the period is second. The joule (J) is the work done when the point of application of newton is displaced a distance of meter in the direction of the force. (Hence J joule = every application where a to 1 1 henry and involved. points, produces in this conductor a current of 1 I is The ohm (il) is the electric resistance between two points of a conductor when a constant difference of potential of volt, applied between these two charged by a is it tionary objects differ- 24 zepto z yocto y UNITS the difference of electric poten- electricity. They contain notes between two points of a conducting wire carry- the reader who The volt (V) tial is ing a constant current of (Hence I = volt is power the is (Hence I = watt The weber (Wb) ing a circuit of one motive force of uniform rate in J joule watt. 1 in Quantity second. {Hence I an electro- it reduced is it to zero at a weber = I volt second. Multiples 1.6 and submultiples of SI units by adding appropriate prefixes Thus prefixes such as to the units. mega, nano, and centi kilo, multiply the value of the unit by factors radian rad square meter nr Energy (or work) joule J Force newton N Length meter m Mass Power kilogram kg watt W Pressure pascal Pa Speed meter per second m/s 1 rad/s Nm Volume Volume cubic meter m liter L rotation . Although the radian ID, IE, and IF 2. Most countries, including COMMON UNITS 3. units some common IN A The newton is 4. The pascal 5. In this is units 6. in spelled liter or litre. ( 1 rad/s = 9.55 The W joule per (kilogram kelvin) J/kg-Kor J/kg °C Temperature kelvin K Temperature difference kelvin or degree Celsius Thermal conductivity watt per (meter-kelvin) Kor °C W/m-Kor W/m°C is organi- 1 N/nr. Canada It is litre. Note Specific heat K some r/min). official spelling in J I we 57.3°). mainly used for liquids and gases. is Symbol SI unit exactly equal to a temperature difference of I °C. I 2 The °C is l I a recognized SI unit and, in practical often used instead of the kelvin. Thermodynamic, or absolute, temperature °C. is This unit of volume watt is ~ the revolution per minute (r/min) to joule it (as well as a very small pressure equal to book we use Heat temperature difference of rad (I a very small force, roughly equal to the Thermal power calculations, 2. Canada book THERMODYNAMICS Quantity 1. in this force needed to press a doorbell. encountered in mechanics, thermodynamics, and TABLE 1E 6 meter. 10 watts. list 5 the SI unit of angular measure, is designate rotational speed Tables 4 radian per second 6 Commonly used 1.7 3 zations in the United States), use the spelling metre instead of seconds, 10 megawatt — 1 newton meter amperes, -9 = Note 2 Torque use the degree almost exclusively /^ampere = 1000 nanosecond Symbol SI unit Area J 1 MECHANICS listed in Table 1C. For example, 1 IN Angle Speed of Multiples and submultiples of SI units are generated UNITS per second. ) produces turn, COMMON TABLE 1D joule per sec- the magnetic flux that, link- is volt as 1 1 1 that gives rise to the production of energy at the rate of ond. equal to watt per ampere.) I The watt (W) particularly useful to not yet familiar with the SI. ampere, when the power 1 between these points dissipated is 1 The absolute temperature T is is expressed in kelvins. On related to the Celsius temperature t the other hand, the temperature of objects by the equation T= t + 273. 1 5. is usually expressed FUNDAMENTALS 8 COMMON TABLE 1F UNITS ELECTRICITY AND IN MAGNETISM Quantity Symbol SI unit Capacitance farad F Conductance Siemens c Electric charge c Electric current coulomb ampere F nprtrv joule j Erecjuencv hertz Hz Inductance henry Potential difference volt i i A Figure 2 watt Resistivity ohm ohm Magnetic ampere Drawn from and Conversion Charts" by Theodore Enterprises Ltd. All rights reserved. n "Metric Units Om meter for units of length. Conversion chart adapted and reproduced with permission. Copyright © 1991, 1995 by Sperika W Power 1.1 Conversion chart H V Resistance field strength Note Wildi. A/m IEEE Press, Piscataway NJ, 08855-1331. 3 per meter Magnetic flux weber Wb Magnetic flux density tesla T 4 Magnetomotive force ampere A 5 1 . 2. 3. 4. 5. listed in descending order of size, and the Formerly called mho. of the connected units: the yard Hz = cycle per second. A/m = ampere turn per meter. T - Wb/nr. What was formerly called an ampere than the inch, the inch I I ampere: I A I ampere turn is now simply simple method. Conversion charts and their use Unfamiliar units can be converted to units is an arithmetic process that often leaves us wondering if from yard move downward in the Appendix eliminate this problem because they show the relative size of a unit by the position it occupies on the page. The charts in the in units are we reach we want to to two arrows millimeter. convert from millime- we move up- ward against the direction of the arrows until we start at millimeter and we apply the following rules: 1 If, in . traveling move in the from one unit to another, direction of the arrow, we we multi- ply by the associated number. between. Conversely, 2. The if to mil- we have ters to yards, largest unit is at the top, the smallest at the bottom, and intermediate units are ranked direction of the (36 and 25.4) until Conversely, from yards in Fig. 1.1, reach yard. In making such conversions our calculations are correct. The conversion to convert limeters. Starting we know well by using standard conversion tables. But this strictly we can turn. Suppose we wish 1.8 36 times larger 25.4 times larger than the convert from one unit to any other by the following I called is is millimeter, and so on. With this arrangement 1 1 1 lines join- them bear an arrow that always points toward the smaller unit. The numbers show the relative size ing connected by arrows, each of which if we move against the arrow, we divide. bears a number. to the smaller hence, its The number is the ratio of the larger of the units that are connected and, value is always greater than row always points toward the smaller In Fig. I.I, for unity. The ar- unit. example, five units of length the mile, meter, yard, inch, and millimeter — are Because the arrows point downward, this means when moving down the chart we multiply, and when moving up, we divide. Note that in moving from one unit to another, we can follow any path we that please; the conversion result is always the same. UNITS The rectangles bearing SI units extend toward the Each rectangle bears from other units. for the unit as ENERGY slightly of the chart to distinguish them left name well as the the 9 TNT kilotonne of J symbol 1.167 x 10 6 of the unit written . out in full. | I kilowatt hour . kW-h | | 3.6 mega joule MJ| | Example I- 1 1000 Convert 2.5 yards to millimeters. British thermal unit Btu | | | 1.055 Solution kJ fkilojoule | Starting (Fig. from yard and moving toward millimeter we move downward 1.1), the arrows. 1000 in the calorie direction of | 1 | 4.184 We must therefore multiply the numbers joule | associated with each arrow: N-m |n ewton-meter 2.5 - yd 2.5 ( = 2286 X 36) X ( 25.4) millimeters van second mm I I 6.24 x 10 18 Example 1-2 eV [electronvolt | Convert 2000 meters into miles. Figure 1.2 Solution Starting move See Example from meter and moving toward mile, we and then against, the direction of with, first the arrows. Consequently, 2000 meters = 2000 X 1 ( = 2000 X we 1 995 by Sperika Enterprises Drawn from "Metric Units and Conversion Charts" by Theodore Wildi. IEEE Press, sion. Copyright© 1991, Ltd. All rights reserved. obtain .0936) (- 1760) miles Piscataway, NJ, 08855-1331. 1.0936 1760 = 1-3. Conversion chart adapted and reproduced with permis- The per-unit system of measurement 1-9 1.24 mi The SI units just described enable us to specify the Example 1-3 magnitude of any quantity. Thus mass Convert 777 calories to kilowatt-hours. kilograms, volts. Solution Referring to the chart on moving from calorie travel downward to ENERGY (Fig. 1.2) kilowatt-hour, we and first (with the arrow 4.184) and then upward (against the arrows 1000, 1000, and Applying the conversion rule, we power in watts, and However, we can often size of expressed in get a better idea of the something by comparing thing similar. In effect, is electric potential in it to the size of some- we can create our own unit and specify the size of similar quantities compared to this arbitrary unit. This concept gives rise to the per-unit 3.6). find method of expressing the magnitude of a quantity. For example, suppose the average weight of 777 calories = 777 (X 4.184) (- 1000) adults in 1000) (- 3.6) New any individual = 9.03 X 10" 4 kW h York is 130 lb. Using this arbitrary weight as a base, we can compare the weight of a in terms of person weighing 160 lb this base weight. Thus would have a per-unit FUNDAMENTALS 10 weight of 160 weighing mb/ 130 I The lb/ lb = 130 lb - Another person 1.23. would have 15 lb 1 n 3500 weight of a per-unit a 4800 0.88. measurement has per-unit system of vantage of giving the size of a quantity ± the ad*2 terms of in 450 xv 3000 n n a particularly convenient unit, called the per-unit in reference to our previ- a football player has a per-unit base of the system. Thus, ous example, weight of 1.7 if we immediately know above average. Furthermore, far 1.7 X 130 Note = that 221 1 .7 per-unit, To where it is Figure 1.3 weight is Conventional are given, they would be absurd weighs state that the football player is weight 1 .7 lb. to His weight the selected base unit generalize, a per-unit system of consists of selecting one or is 130 lb. Fig. 1 .3, composed of several we them. In this book decide to use an impedance of base, the per-unit measuring *,(pu) and impedance. The base may be power is 40 ratings of 25 hp, hp, The 1 the base = 3. is of 15 hp. In 40 hp/50 hp - 0.8 and world in this per-unit and 150 hp/ 15 hp - ^50 n 4800 n if expressed _ " = 1.67, 40 hp/ 15 hp 3.2 3000 a 1500 ft in vector notation shown per-unit values. in Fig. 1 is same impedances are We can solve this other circuit. For example, used, the per-unit circuit is that .5. power - 2.33(pu) 3.2(pu) -nrrv 2.67, 10. therefore important to value, the actual values of per-unit _ ~ real circuit, but the we would any R 2 (pu) know the magnitude If we do not know the quantities we are 0.30 i dealing with cannot be calculated. The _ 0.30 ~ 150011 of the base of the per-unit system. its 12 1500 and 3 pu, respectively. well have selected a base case the respective per-unit rating this 2.33 50 hp, the three motors have would be 25 hp/ 15 hp It is 0.5; ratings of 0.5, 0.8, We could equally = per-unit circuit (Fig. 1.4) contains the now Thus, = elements as the circuit as of 50 hp. The corresponding per-unit ratings hp XL (pu) said to have PH where power a 3500 a power, a voltage, and 150 hp. Let us select an arbitrary base power = = Xc (pu) a current, or a velocity. For example, suppose that 50 hp/50 hp = R 2 (pu) one quantity as our system stick, the per-unit are then 25 hp/50 we 150011 system with one base three motors have If as the impedances are as follows: more convenient mea- select the size of only a single base. 500 ohms are particularly interested in current, power, torque, we 1 measurement selecting convenient measuring sticks for voltage, 1.10 Per-unit resistors, capacitors, and inductors having the impedances shown. suring sticks and comparing similar things against If circuit. lb. whenever per-unit values always pure numbers. Thus are his his actual method can also be applied to im- pedances. Consider, for example, the circuit in Figure 1.4 Per-unit circuit. 2(pu) UNITS 2.33 3.2 1 1 In order to understand the significance of this re- j sult, the reader should study the two following ex- amples. The bases are the same as before, namely 4kV 0.30 500 Figure 1.5 = /B kW ZB = A 125 32 ft Example 1-4 Per-unit circuit with notation. i A 400 II resistor carries a current of 60 A. above base values, a. system with two bases Per-unit 1.11 b. c. when two bases particularly useful are bases are usually a base voltage £" B P B Thus power 4 becomes used. The electrotechnology the per-unit system In . kV and d. e. The per-unit resistance The per-unit current The per-unit voltage across the resistor The per-unit power dissipated in the resistor The actual E and P of the resistor and a base the selected base voltage may Solution be a. the selected base power 500 kW. The per-unit resistance The two base values can be selected quite indeb. unit The per-unit current system is that it automatically establishes a cor- />w For example, is base power PB base voltage EH impedance Z B and the base if — is d. base voltage The the base voltage / 4 is 500 kW, the base current ZB = £ b /'b = In effect, system impedance. per-unit is kV and the base is 1 The 0.48 is V/125 A = 32 ft also get a base current and a base Consequently, the so-called = 0.48 = 6 = X X tf(pu) 12.5 is E(pu) = 6 - 2.88 X X /(pu) 0.48 is E — E B X E(pu) = 4 kV X 6 = 24 kV is 4( )00 /(pu) actual voltage across the resistor A 25 = power P(pu) by selecting the voltage/power per- we The n = ^b/^b = 500 000/4000 = The base impedance system. 12.5 per-unit voltage across the resistor EH — base current e. 'b per-unit = / B is £(pu) unit ft/32 ft = 60 A/ 125 A = /(pu) Thus c. base current power is interesting feature of the voltage/power per- responding base current and base impedance. the = 400 tf(pu) pendently of each other. One Using the calculate: 2-base system really gives us a 4-base per-unit The actual power dissipated in the resistor i$^;^'as'' P = PB x ^(pu) = 500 kW X = 1440kW BCOLTAD 2.88 Df. RU,]?\ vlii TA A ©IBUCUfcCA FUNDAMENTALS 12 Example 1-5 A 7.2 /|.(Pu) . = 2.844 kV source delivers power to a 24 II resistor and a 400 kW electric boiler (Fig. 1.6). Draw equivalent per-unit circuit diagram. Use the base values as Example in the same /i(pu) /,(pu) = 0.444 = 2.4 1-4. R{pu) boiler [ 0.75 j Calculate The The The The The The a. b. c. d. e. f. per-unit E(pu), /?(pu), PCpu) per-unit current P(pu) 0.8 / 2 (pu) per-unit line current (pu) l { power absorbed by the resistor power absorbed by the resistor per-unit actual Figure 1.7 Per-unit version of Figure 1 .6. actual line current The per-unit line current /t (pu) ft 24 n r 400 kW The d. per-unit is = /,(pu) + / 2 (pu) = 0.444 + 2.4 = 2.844 power P(pu) Figure 1 .6 See Example /L in the resistor is = = E(pu) = 4.32 X X 1.8 / 2 (pu) 2.4 1-5. The e. actual power in the resistor is Solution a. The per-unit line voltage £,(pu) P 2 = P a X «pu) = 500 kW X 4.32 is - 7.2kV/4kV = 1.8 = 2160 kW The per-unit resistance is The f. R(pu) = 24 n= fi/32 actual line current = 12 The per-unit power of the P(pu) We b. can The boiler = is = 400 kW/500 kW = now draw c. The = £(pu)//?(pu) = 1.8 - 2.4 - - 1 /, 125 Name /, X (pu) 2.844 = 355.5 A the seven base units of the International 0.75 1 -2 1-3 per-unit current X Questions and Problems is 1 / 2 (pu) IB 0.8 the per-unit circuit (Fig. 1.7) per-unit current f2 is 0.75 Name System of Units. five derived units of the SI. Give the symbols of seven base units, paying is particular attention to capitalization. /,(pu) = P(pu)/E(pu) - 0.444 = 0.8/1.8 1-4 Why are some derived units given special UNITS 1-5 What ergy, 1-6 1-58 revolution 1 -60 oersted 1-59 degree 1 -6 ampere are the SI units of force, pressure, en- power, and frequency? Give the appropriate prefix for the following multipliers: 100, 1000, 10 1/1000, 10" 6 , Make the following conversions using the conver- 1/10, 1/100, , io'-\ 1-62 10 square meters to square yards 1-63 250 1-64 1645 square millimeters Express the following SI units in symbol form: 1-7 megawatt 1-21 millitesla 1-8 terajoule 1-22 millimeter 1-9 millipascal turn sion charts: 10" 9 , 6 1 1 1-23 1 revolution 1-10 kilohertz 1-24 megohm 1-11 gigajoule 1-25 megapascal 1-12 milliampere 1-26 millisecond -65 1 3 MCM to square millimeters 000 circular mils to square millimeters 1-66 640 acres 1-67 81 1-68 to square inches square kilometers to 000 watts to Btu per second 33 000 foot pound-force per minute to kilowatts 1-13 microweber 1-27 picofarad 1-14 centimeter 1-28 kilovolt 1-15 liter 1-29 megampere 1-16 milligram 1-30 kiloampere 1-17 microsecond 1-31 kilometer 1-69 1 -70 1-71 1-18 millikelvin 1-19 milliradian 1-20 terawatthour 1-32 1-33 1 -34 flow I -38 density 1-39 power 1-36 plane angle 1-40 temperature 1-41 microjoules 10 pound-force to kilogram-force 60 000 -72 1 -73 1 -74 50 ounces 1 -75 76 oersteds 1 -76 5 1 -77 1 lines per square inch to teslas kilogauss .2 teslas to to kilograms to 000 meters amperes per meter to miles and frequency magnetic flux to meters milliliter 1-35 1-37 0 foot pound-force 1 symbol: rate of 1 feet to cubic nanometer State the SI unit for the following quantities write the 250 cubic mass 80 ampere hours 1-78 25 pound-force 1-79 25 pounds 1-80 3 tonnes to 1-81 1-82 Give the names of the SI units that eorresponcl to 1 00 000 0.3 to to coulombs newtons to kilograms pounds lines of force to webers pounds per cubic inch kilograms per to cubic meter the following units: 1-83 1-42 Btu 1-51 bar 1-43 horsepower 1-52 pound-mass 1-44 line of flux 1-53 1 -84 1 -85 pound-force 2 inches of mercury to millibars 200 pounds per square inch to pascals 70 pounds-force per square inch to newtons per square meter 1-45 inch 1-54 kilowatt-hour 1-46 angstrom 1-55 gallon per 1-47 cycle per second 1-48 gauss 1-49 line per 1-50 °F 1 1 square inch 1-57 mho -87 1-88 pound-force per 1 square inch 1 5 revolutions per minute to radians per second minute 1-56 -86 -89 A temperature of 20 °C to kelvins A temperature of 200 °F to kelvins 1 A temperature kelvins difference of 1 20° Celsius to 14 I -90 FUNDAMENTALS A resistance of 60 ft is Industrial application selected as the base resistance in a circuit. If the circuit contains 1 -94 A motor has an efficiency of 92.6%. What is the efficiency in per-unit? A variable-speed motor having a nameplate three resistors having actual values of 100 fl, 3000 II, and 20 O, calculate the per-unit 1-95 value of each resistor. 1-91 A power of 25 kW rating of 15 hp, and a voltage of 2400 are selected as the base voltage of a V power and base power system. Calculate 890 r/min develops a torque of 25 newton meters at 1260 r/min. Calculate the per-unit values of the torque, the speed, and power. value of the base impedance and the base 1-96 Three resistors have the following ratings: current. 1 -92 A resistor has a per-unit value of 5.3. base power 12 is 470 is 250 If the resistor resistance A 10012 kW and the base voltage ohmic value of V, calculate the the resistor. 1-93 A length of 4 m is selected as a base unit. b. the per-unit length of 1 mile C 300 II the per-unit length of 1 foot of resistors 1 d. e. the per-unit value of a the magnitude of the base area (in m f. ft W W 40 W 24 75 and voltage values of the resistance, power, m2 3 the magnitude of the base volume (in m c. 50 Using resistor A as a base, determine the per-unit Calculate a. B power 3 the per-unit value of an area of 2 square A rating C, respectively. 30 hp cage motor has the following cur- rent ratings: ) volume of 6000 -97 B and ) FLA: full-load current LRA: locked NLA: 36 A rotor current 2 8 1 A no-load current 14 A. miles Calculate the per-unit values of LRA and NLA. Chapter 2 Fundamentals of Electricity, Magnetism, and Circuits Introduction 2.0 positive ^ negative ( terminal (+) -fli-ftr This chapter briefly reviews some of t ermina the funda- mentals of electricity, magnetism, and circuits. We assume the reader already familiar with the is dry cell including the solution of electric circuits. basics, However, a review useful because is it focuses on those items that are particularly important in technology. Furthermore, used throughout this Some currents. it book power establishes the notation and to designate voltages Figure 2.1 of the topics treated here will also Dry cell, provide the reader with a reference for subjects covered in later chapters. electron current flow Conventional and electron 2.1 current flow Consider the dry cell positive ( shown + and one difference of potential volts) is tive terminal If ( — ) 1 , having one terminal. The between them (measured in to an excess of electrons at the nega- compared we connect tential in due in Fig. 2. negative ) to the positive terminal. a wire across the terminals, the po- difference causes an electric current to flow is composed of a steady come out of the negative the circuit. This current stream of electrons that terminal, move along the wire, and reenter the cell Figure 2.2 by the positive terminal (Fig. 2.2), Electron flow. 15 ' <-) FUNDAMENTALS 16 Before the electron theory of current flow was fully understood, scientists of the 17th century arbi- decided trarily that current in a from the positive terminal to the negative terminal This so-called conventional current flow (Fig. 2.3). used today and is still conductor flows is the accepted direction of power technology. book we use the conventional current current flow in electric In this flow, but flow is it is worth recalling that the actual electron opposite to the conventional current flow. In order to establish a general rule, consider two black boxes A and B ally changing between sources / that is in direction (Fig. 2.4). drop along the wires assumed is that are connected to be zero. some way in A2 to the external ter- . Under such highly are also continually changing. is how can we tell To answer sometimes important and loads in an electric to identify the sources circuit. By definition, a source delivers electrical power whereas a load absorbs it. Every electrical device (motor, resistor, thermocouple, battery, capacity, generator, How can we tell the A or B the question, suppose we have appro- instantaneous polarity the terminals ( + )( — ) determine the of the voltage across and the instantaneous direction of conventional current flow. The following rule then applies: etc.) that carries a current can be classified as either a source or a load. whether a source or a load? priate instruments that enable us to is Each and B h B 2 A variable voltage exists across the terminals, and its magnitude and polarity minals A], and loads It continu- The voltage box contains unknown devices and components variable conditions, 2.2 Distinction connected by a pair of that are wires carrying a variable current • A device is a source whenever current flows out of the positive terminal. one from the other? • A device is a load whenever current flows into a positive terminal. conventional current flow If the instantaneous polarities and instantaneous current flow are as from the load. rule that However, if shown box A is in and box The above is A the follows B is a box B would become load. rule for establishing a source or load it the current should reverse while the polarity remains the same, the source Fig. 2.4, a source and box is whether a device very simple, but it has impor- tant applications, particularly in alternating current circuits. Figure 2.3 Some devices, such as resistors, can behave only as loads. Other devices, such as photocells, can act only as sources. However, many Conventional current flow. either as sources or as loads. / Figure 2.4 between a source and a Distinction load. devices can behave Thus when a battery FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS power, delivers electric flows out of the recharged, (4 ) ( + ) it when being is it acts as a load (current flows into the it The acts as a source (current terminal); ties • on a system, but they can briefly behave generators When is thing when is like £AB = + tween 100 V, which reads: The voltage be- A and B is 100 V, and A is ( the capacitor it + is acts as a source On terminal. ) charging up, and current flows into the + ( ) it • positive with the £ BA = 100 V, which reads: The voltage between A and B is 100 V, and B is negative with respect to A. acts As another example, terminal. if we know ator voltage in Fig. 2.6 has a value that the gener- E2 = — \ 100 V, we know that the voltage between the terminals is 00 V and that terminal 2 is negative with respect to Sign notation 2.3 relative polari- true of capacitors. discharging and current flows out of the as a load and the can be designated by the respect to B. The same a capacitor other hand, B and the electromechanical conditions are if appropriate. A double-subscript notation, as follows: terminal). Similarly, electric motors usually act as loads potential difference of terminals 1 1 In arithmetic we scribe addition mechanics, direction ) to de- etc., to indicate the compared + that the direction This interpretation of ( + to an ar- the speed if —400 r/min, 100 r/min to of rotation has reversed. ) and — ( terminal 2.5 Sign notation for voltages Although we can represent the value and the polarity of voltages by the double-subscript notation (£, 2 £ AB in the signs is fre- we etc.), £2 V etc.) ( + sign. ) in marked with to G having a positive terminal terminal B. Terminal A is positive minal B. Similarly, terminal spect to terminal A. tive a positive ( + ) by sign. is itself: it is Note B is A and The other terminal 2 shows a a negative with respect to ter- negative with re- that terminal A is not posi- only positive with respect to B. Figure 2.6 = 100 If E 21 terminal V, terminal 2 is a arbitrarily describe a system of notation that enables us indicate the polarity of voltages. Fig. 2.5 source (£,, For example. Fig. 2.7 shows which one of the terminals Double-subscript notation for voltages We now symbol , It and identifying one of the terminals by a chapters that follow. source £, 2.4 often prefer to use the sign notation. consists of designating the voltage by a , ) , positive met quently 1. and of an electric current, of a mechanical motor changes from means ) In electricity meaning the ( chosen direction. For example, bitrary it we broaden + and — ( and subtraction. of a rotational speed, force, of a use the symbols negative with respect 1. Figure 2.5 Figure 2.7 Double-subscript notation to designate a voltage. Sign notation to designate a voltage. to FUNDAMENTALS 18 unmarked, but is is automatically assumed to be neg- ative with respect to the With • If ( we this notation the + = + state that £, ) indicated in the diagram. + ( ) sign that the The terminal bearing then actually positive and the is other terminal is negative. Furthermore, the magnitude of the voltage across the terminals Figure 2.9 islOV. Solution of Conversely, • means 10 V, this of the terminals corresponds to that real polarity the terminal following rules apply: if the terminals £, = - shown on diagram. The terminal bearing the ( + ) sign actually negative, and the other terminal is The magnitude of the voltage across tive. terminals is the is the 10 V. circuit of Fig. 2.8 consists of three sources 7 V V2 and V? , ,, with a positive in series to — each ( + ) — having a terminal marked The sources are connected using jumper wires A, B, C, sign. a resistor R, Determine the actual value and polarity of the voltage across each source, V, spectively. In effect, point V2 = + 10 V, and knowing V? = -40 that V = ues and polarities are as in — That is a negative sign. 2.6 Graph of an alternating voltage In the chapters that follow, whose voltages change alternating voltages voltage at 2. 1 may be 0). we encounter sources polarity periodically. The Such represented by means vertical axis indicates the each instant, while the horizontal axis in- dicates the corresponding time. Voltages are positive when they when are E2 above the horizontal axis and nega- they are below. Figure 2.10 shows the \ produced by the generator of Fig, 2.6. find that the true val- shown in Fig. 2.9. However, jumper A, it seems im- 2 directing our attention to possible that ( re- negative with respect to point C. voltage we only { V. Solution stated, is B and positive with respect why A carries both a positive and tive Using the rules just A It jumpers B and C, to point of a graph (Fig. and D. -4 inherently positive nor inherently negative. has a polarity with respect to posi- Example 2-1 The 2-1 10 V, the real polarity of the reverse of that is Example ). it can be both positive However, we must remember Figure 2.8 Example Circuit of 2-1. ( + ) and negative that A is neither Figure 2.10 Graph of an alternating voltage having a peak of 1 00 V. FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS from zero, Starting taining + 100 V falls to zero E2] gradually increases, after 0.5 second. spect to terminal because 1 E2 During the interval from 1 spect to terminal polarities onds are 1 . 2.7 Positive is insets at 0.5, I, IT, E2X is negative with re- The instantaneous of the generator shown by positive. to 2 seconds, negative; therefore, terminal 2 0 this positive with re- is is \ at- then gradually It end of one second. During at the one-second interval, terminal 2 1 and voltages and .5, and TIT of Fig. 2.10. 2. 1 7 sec- and negative currents 0 We also indicate 1 make use of positive and negative signs to the direction of current flow. The signs are *- time >v X | allocated with respect to a reference direction given on the circuit diagram. For example, the current a resistor (Fig. 2.11) Y to X. One be positive may flow from X of these two directions + ( is in Y or from considered to and the other negative ) to ( — Figure 2.13 and the corresponding graph Electric circuit The arrow ). of current. indicates the positive direction of current flow. Solution According zero to Because may is the interval flow from X to Y or from Y to X. interval from from +2 A to in the resistor. to l it to l second. from B to A of the arrow). During the 2 seconds, the current decreases zero, but circulates it still Between increases from zero to ative, from 0 positive, the current flows in the resistor (direction Figure 2.11 Current it from to the graph, the current increases +2 A during from B to A 2 and 3 seconds, the current —2 A and, because it is neg- really flows in a direction opposite to that of the arrow; that is, from A to B in the resistor. Figure 2.12 Circuit element showing positive direction of current flow. 2.8 Sinusoidal voltage The The positive direction means of an arrow (Fig. 2 A flows from X to Y, and if is it is 2. 1 shown 2). flows Thus, of the arrow), from is it if by a current of +2 Y to X (direction opposite to that designated by the symbol graph a resistor in shown of this graph. in Figure 2. R 1 —2 A. varies according to the 3. It may therefore be expressed by the equation e = Em cos {lirft + 6) (2.1) A. Conversely, Example 2-2 The current ac voltage generated by commercial alternators very nearly a perfect sine wave. in the positive direction designated by the symbol current flows arbitrarily is Interpret the meaning where e — Em — / = instantaneous voltage [V] peak value of the sinusoidal voltage [V] frequency [Hz t = time 6 = a fixed angle [radj [s] FUNDAMENTALS 20 The expression lirft and 6 are angles, expressed However, it is often more convenient to The voltage in radians. express the angle in e ab degrees, as follows: e = Em cos (360 ft * = £ ni cos + In these (2.3) 0) equations the symbols have the same sig- s is 100 cos 488 622° X 50 X 27.144 + 30°) V moment at this terminal a Note 21 AAA- 100 cos (360 -20.8 Thus, + = = or <<|> t = (2.2) 6) at the voltage is V —20.8 and negative with respect to terminal b. is that an angle of 488 622° corresponds to 488 622/360 = 357 complete cycles plus 0.2833 cycle. The latter corresponds to 0.2833 X 360° = 102°, and 100 cos 102° = -20.8v 1 nificance as before, and the time-dependent angle (= 360 ft) is Example 2-3 The sine wave _ in Fig. 2.14 represents the voltage £. lb across the terminals a operates at 50 Hz. and b of an ac motor Knowing that 6 100 V, calculate the voltage 27 .1 44 <j) also expressed in degrees. at t 2.9 Converting cosine functions into sine functions that = 30°, and E m = = 0 and at t = We can convert a cosine function of voltage or current into a sine function Em Solution The voltage at t e, xh = At this is moment 0 Em is = E m cos (360 ft + 9) - 100 cos (360 X 50 X = 100 cos 30° - a by adding 90° to the angle 6. Thus, s. 86.6 Similarly, 0 + 30°) +86.6 V and terminal B + = 90°) we can convert a sine function /m is 0) (360 ft cos (360 ft + 6) (2.4) into a cosine function by subtracting 90° from the angle V the voltage + (360 ft sin / m sin therefore positive with respect to terminal b. + cos (360 ft 6. Thus, = + 6-90) (2.5) 2.10 Effective value of an ac voltage Although the properties of an ac voltage are known when its frequency and peak value Em are specified, it is much more common to use the effective value £e(j . For a voltage E between cfi - that varies sinusoidally, the relationship and £ni is given by the expression Ecn = EJ,2 (2.6) - The effective value of an ac voltage is some- times called the RMS the voltage. a measure of the heating effect of It is (root compared the ac voltage as mean square) value of to that of an equivalent dc voltage. For example, an ac voltage having an fective value of 1 ing effect in a resistor as does a dc voltage of Figure 2.14 Sinusoidal voltage having a peak value of 100 expressed by e ab = Em cos (360 ft + 30°). V and ef- 35 volts produces the same heat1 35 V. The same remarks apply to the effective value of an ac current. Thus a current that varies sinusoidally and whose peak value is / m possesses an effective value /C given by | t FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS U n 2 (2.7) Most alternating current instruments are brated to rent show c. e cali- the effective value of voltage or cur- and not the peak value (Fig. 2.15). value of an alternating voltage or current When given is 339 it understood that it the is effective E^-and Furthermore, the subscript in Owing E given by sin 21 600 t t to the phase lag of 30°, the current is value. given by /eir is dropped / and the effective values of voltage and current are simply represented by the symbols is = E m sin 360 ft = 339 sin 360 X 60 the d. is Let us assume the voltage 21 and /. = /m sin (360 ft - 30) = 14.1 sin (21 600 f = 14.1 sin (<J> - 30) 30) Example 2-4 A 60 Hz source having an effective voltage of 240 V delivers an effective current of 10 A to a circuit. The current lags the voltage by 30°. Draw the waveshape for E and Em = E b. / In The peak voltage x 2 m = / 1 Phasor representation most power studies the frequency we simply is = 240 x 2 val- 1 take it is fixed, for granted. Furthermore, and so we n 2 is magnitudes and phase angles. = 10x2= 14.1 are not particularly concerned with the instantaneous = 339 V voltages and currents but more with their The peak current / The waveshapes giving the instantaneous ues of e and are shown in Figure 2. 6. 2.1 /. Solution a. e. A ages are measured in And because the RMS volt- terms of the effective values E Figure 2.15 Commercial voltmeters and ammeters are graduated ing up to 2500 A and 9000 V. (Courtesy of General Electric.) in effective values. This range of instruments has scales rang- FUNDAMENTALS 22 \ 16.67 \ 30° 14 • A \ \ (a) ms A 1 '* 9' <b) 1 t E h- 1.39 E / V 339 I ms Figure 2.17 current phasor The / and voltage phasor E are in phase. Figure 2.16 Graph showing the instantaneous values of voltage and current. The current lags 30° behind the voltage. The effective voltage is 240 V and the effective current is 10 A. E and interested in E m we peak values rather than the are really only , Figure 2.18 Phasor / lags behind phasor 0. E by an angle of 0 degrees. This line of reasoning has given rise to the phasor method of representing voltages and currents. The basic purpose of phasor diagrams sense that it A phasor is phasor show represents. the same angle If a phasor bears an arrow, and its length is The angle between two phasors to the electrical phase angle propor- 3. is it equal between the quantities. 1 . Two phasor E is / is phasors are said to be are parallel to each other them in phase when they and point in the same Two one The phase angle between direction as phasor said to lead phasor /. same Fig. 2. is 1 8, it / lags behind make then Conversely, a E if phasor make has to be rotated counterclockwise to point in the /, it direction. Thus, referring to clear that phasor £ by we could E leads phasor / equally well say that 0 degrees. then zero. is phasors are said to be out of phase they point in different directions. gle to be rotated clockwise to same by 6 degrees. But 4. 2. rotate line up. said to lag behind phasor rules apply to phasors: direction (Fig. 2.17). we sweep through to make them to E has point in the phasor / The following we have similar to a vector in the tional to the effective value of the voltage or current it Consequently, whether /. phasor or the other, between voltages the magnitudes and phase angles and currents. to is between them is point in the same The phase the angle through one of the phasors has when to be rotated to Referring phasor / now to Fig. 2. 1 9 we could rotate (3 to make it clockwise by an angle an- which make it direction as the other. Thus, referring to Fig. 2. 1 8, phasor / has to be rotated counterclockwise by an angle 6 to in the same phasor E has gle 6 to make it point direction as phasor E. Conversely, make to be rotated it clockwise by an an- point in the same direction as Figure 2.19 Phasor / leads £ by B E by degrees. (3 degrees. But phasor / also lags FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS same point in the direction as phasor E. could then say that phasor 23 We E by leads phasor / P degrees. But this is the same as saying that phasor / lags phasor E by 0 degrees. In practice, we always select the smaller phase angle between the two phasors to designate the lag or lead situation. 5. common Phasors do not have to have a may be shown that in Fig. 2.20. £| same from each entirely separate By phase with in is applying rule 3, we can because they point l see in the x direction. Furthermore, sor £, origin but other, as by 90°, and E2 phasor / 2 leads lags behind /2 by 1 pha- 35°. Figure 2.21 Different ways of showing the phase relationships between three voltages that are mutually displaced at 120°. Solution To draw trary equivalent to 240 V. Phasor Figure 2.20 Phasors do not have to start from a show their the phasor diagram, common origin to magnitudes and phase relationships. it we direction for phasor E, E with a length lags 30° behind (Fig. 2.22). / is select any arbimaking its length then drawn so that Knowing equivalent to frequency that the is 1 0A 60 Hz, between the positive peaks the time interval is given by In the in Fig. same way, £bc the three phasors 2.2 1 a can be rearranged as without affecting the shown , and £ca in Fig. 2.2 1 = 360 ft 30 = 360 X 60 = 1.39 ms 9 b phase relationship between t £ab in Fig. 2.21b still points in the as Eab in Fig. 2.21a, and the same is them. Note that same direction true for the Fig. 2.2 three 240 V other phasors. 1 c shows still t another arrangement of the phasors that does not in any way alter their magnitude or phase relationship. The angle 0 between two phasors is a measure peak positive val- of the time that separates their ues. Knowing the frequency, we can calculate the of the voltage and current given in Figure 2.16. time. Example 2-5 Draw the phasor diagram of the voltage and current in Fig. 2. 16. Calculate the time interval between the positive Figure 2.22 Phasor diagram peaks of E and /. 2.12 Harmonics The voltages and currents in a power The quently not pure sine waves. circuit are freline voltages FUNDAMENTALS 24 usually have a satisfactory waveshape but Fig. 2.23. This distortion distorted, as netic saturation in the cores of transformers or switching action of thyristors or IGBTs In order to understand the distorting effect of a the cur- shown in can be produced by mag- sometimes badly rents are by the in electronic drives. harmonic, us consider two sinusoidal sources in series (Fig. 2.24a). quencies are respectively 60 Hz and 1 Their c, fre- 80 Hz. The V and 20 V. The fundamental (60 Hz) and the third harmonic 80 Hz) voltages are assumed to pass through zero at the same time, and both are perfect sine waves. Because the sources are in series, the terminal corresponding peak amplitudes are 100 ( 1 voltage —4 let and e 2 connected is equal to the sum of the instantaneous The resulting wave (Fig. 2.24b). voltages produced by each source. — A 0 4- V terminal voltage is a flat-topped - Thus, the sum of a fundamental voltage and a har- monic voltage yields a nonsinusoidal waveform whose degree of distortion depends upon the mag- — nitude of the harmonic (or harmonics) 60 1 20 1 300 240 80 it contains. 420 360 Figure 2.23 This severely distorted 60 Hz current obtained on an electronic drive contains the following harmonics: funda- mental (60 Hz) = 59 A; fifth harmonic (300 Hz) - 15.6 A; seventh harmonic (420 Hz) = 10.3 A. Higher harmonics are also present, but their amplitudes are small. (Courtesy of Electro-Mecanik.) The to the distortion of a voltage or current can be traced harmonics it contains. A harmonic age or current whose frequency is is any volt- an integral multi- ple of (2. 3, 4. etc., times) the line frequency. Consider a set of sine waves in which the lowest frequency is/ and multiples of / the lowest the other By all frequency waves other frequencies are integral definition, the sine is wave having called the fundamental and are called harmonics. For example, waves whose frequencies are 20, 40, and 380 Hz is said to possess the following a set of sine 1 00. components: (b) fundamental frequency: 20 Hz (the lowest frequency) Figure 2.24 second harmonic: 40 Hz (2 fifth harmonic: l()0Hz(5 X 20 Hz) a. X 20 Hz) Hz ( 19 X 20 Hz) sinusoidal sources having different frequen- cies connected b. nineteenth harmonic: 380 Two in series. A fundamental and third harmonic voltage can gether produce a flat-topped wave. to- FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS We can produce a periodic voltage or current of any conceivable shape. All gether a fundamental we have to do component and an of harmonic components. For example, to is add to- power we can distorted gen- wave having an amplitude of 00 V and a frequency of 50 Hz by connecting the following sine wave sources in series, as shown in Table 2A. In 1 100 V SQUARE WAVE waveshapes Freq. [VI [Hz| that are rich in mental voltage harmonics. produce together power. This fundamental power motor a is fundamental power the useful and an arc furnace to rotate The product of heat up. Amplitude in ac circuits the fundamental current and funda- the corresponding Harmonic produced whenever voltages electronic circuits. All these circuits produce that causes a TABLE 2A are also and currents are periodically switched, such as arbitrary set square erate a They circuits. 25 to harmonic voltage times harmonic current also produces a Relative amplitude harmonic power. The latter is usually dissipated as heat in the ac circuit and, consequently, does no 127.3 50 third 42.44 150 1/3 fifth 25.46 250 1/5 fundamental useful 1 work. Harmonic and currents voltages should therefore be kept as small as possible. It should be noted that the product of a funda- mental voltage and a harmonic current yields zero seventh 18.46 350 1/7 ninth 14.15 450 1/9 net power. Harmonics are covered in greater detail in Chapter 30. 2.13 Energy in an inductor lh I27 1.00 6350 1/127 A energy coil stores carries a current /. W 127.3/n A tal square wave wave and an is thus infinite 50 n 1/n W= composed of a fundamennumber of harmonics. The and they are consequently less wave. and pointy corners of the square In practice, square waves are not produced by adding sine waves, but the example does l - field when it given by , LI (2.8) energy stored L = inductance of / = in the coil [J] the coil [H| current [A] important. However, these high-frequency harmonics produce the steep sides = is where higher harmonics have smaller and smaller amplitudes, magnetic in its The energy show that any waveshape can be built up from a fundamental wave and an appropriate number of harmonics. we can decompose a distorted periwave into its fundamental and harmonic components. The procedure for decomposing a distorted wave is given in Chapter 30. Harmonic voltages and currents are usually undesirable, but in some ac circuits they are also unavoidable. Harmonics are created by nonlinear If the current varies, the stored energy rises and falls in step with the current. Thus, whenever the current increases, the coil absorbs energy and current The cussed falls, energy is properties of an inductor are in Section 2.3 whenever the released. more fully dis- 1 Conversely, odic loads, such as electric arcs and saturated magnetic 2.14 Energy in A capacitor stores ever a voltage energy is E a capacitor energy in its electric field appears across its when- terminals. The given by W='c£ 2 (2.9) FUNDAMENTALS 26 W= where W= energy stored C— E= capacitance of the capacitor [F] in the capacitor fJ] 8 The energy Example 2-6 having an inductance of 10 in series with a 100 capacitor. jjlF current in the circuit is 40 A voltage across the capacitor energy stored this X 1/2 10 10" 3 X X 40 2 J stored in the capacitor is voltage [V) W= A coil 1 Ml LI = in the electric mH = is connected 2 1/2 32 CE = 1/2 X 100 X 10" 6 X 800 2 J The instantaneous Some 2.15 and the instantaneous 800 V. Calculate the and magnetic fields at is useful equations We terminate this tions (Table moment. 2B) section with a list of useful equa- that are frequently required solving ac circuits. The equations when are given without proof on the assumption that the reader already pos- Solution The energy stored in the coil sesses a is knowledge of ac circuits in general. IMPEDANCE OF SOME COMMON AC CIRCUITS TABLE 2B Impedance Circuit diagram XL = Equation (2-I0) 2TTJL (2-ll) 2irfC Z = \R 2 + X — o o 1| o-m—it—— Ac H Z= \ R~ 2 (2-12) + Xc r (2- 1 3) XL Z = VR- + (X} - XC Y (2-14) RX, a* (2-15) \R 2 + X : ( RX C (2-16) Vr 2 + Xc 2 Xc \ R +_XL 2 1 2 VR + (X L - Xcf (2-17) FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS ELECTROMAGNETISM X by def10~ 7 or approximately 1/800 000. This enables us to write In the SI, the inition. Magnetic 2.16 and Whenever a component, H field intensity B magnetic flux exists in a it field intensity is due H, given by H=U/I (2.18) in the approximate form: //= 800 000 3 magnetic field intensity a flux density of H— U= / = magnetic field intensity f rubber, and air have length of the that component [m] The resulting magnetic flux density is 1) A/m of 800 produces millitesla. I Nonmagnetic materials such A/m] magnetomotive force acting on the component [A] (or ampere turn) (2.2 The B-H curve of vacuum is a straight line. A vacuum never saturates, no matter how great the flux density may be (Fig. 2.25). The curve shows that a where fixed, is has a numerical value of 4tt body or presence of a magnetic to the It Eq. 2-20 flux density cf> magnetic constant 27 B-H as copper, paper, curves almost identical to of vacuum. mT 2.0, given by B = $IA r . — 1 , 1 (2.19) where B = flux density [T] § = A = flux in the There component WbJ [ cross section of the is density (B) component [m 2.17 In ] a definite relationship between the flux and the magnetic field intensity any material. This relationship graphically 2 by the B-H curve B-H curve of is (H) of usually expressed Figure 2.25 of the material. B- /-/curve of of nonmagnetic materials. vacuum vacuum, the magnetic flux density B is directly 2.18 expressed by the equation B = The il 0 H (2.20) B-H curve of a magnetic material proportional to the magnetic field intensity H, and is vacuum and flux density in a magnetic material also de- pends upon the magnetic is subjected. value Its is field intensity to which it given by where B = B = flux density [T] H= magnetic |x 0 — field intensity where [A/m magnetic constant [= 4tt X 7 I0~ l* B, |jl before, and 0 , and (jl,. is H have ijL oK /y the (2.22) same significance as the relative permeability of the material. The value of ijl,. is not constant but varies with the flux density in the material. Consequently, the Also called the permeability of vacuum. The complete expression for (jl 0 is 4 tt X K) 7 henry/meter. relationship between this makes Eq. 2.22 B and H is not linear, and rather impractical to use. We FUNDAMENTALS 28 prefer to show the relationship by saturation curve. Thus, Fig. 2.26 means of a B-H shows typical saturation curves of three materials used in electrical and cast steel. field intensity of 1.4 sity T commonly machines: silicon iron, cast iron, ial, it is The curves show that a magnetic of 2000 A/m produces a flux den- in cast steel but only 0.5 T jjl,- ~ 800 000 (2.23) fl/// in cast where 2.19 Determining the relative permeability tio easy to calculate the relative permeability using the approximate equation iron. The would be produced in vacuum, under the same magnetic field intensity H. Given the saturation curve of a magnetic matersity that relative permeability of the flux density in B = flux density in the magnetic material [TJ H corresponding magnetic field intensity [A/ml — Example 2-7 jjl,. of a material is Determine the permeability of the ra- flux density of the material to the flux den- 1 silicon iron .4 T. teslas 0 1000 2000 3000 4000 *H Figure 2.26 B-H saturation curves of three magnetic materials. 5000 6000 A/m ( 1 %) at a FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS the magnetic materials saturate Solution Referring to the saturation carve (Fig. 2.26), that a flux density of intensity jlil.. 1 .4 see a magnetic field eventually all the B-H more and more and B-H curve curves follow the of vacuum. of 1000 A/m. Consequently, = 2.20 Faraday's law of 800 000 B/H = 800 000 X At T requires we 29 1.4/1000 this flux density, silicon iron is more permeable than vacuum (or electromagnetic induction =1120 1120 times air). shows the saturation curves of a broad from vacuum to Permalloy®, one most permeable magnetic materials known. Fig. 2.27 In 1831, while pursuing his experiments, Michael Faraday made one of the most important discoveries in electromagnetism. Now known as Faraday's it revealed a range of materials law of electromagnetic induction, of the fundamental relationship between the voltage and Note that as the magnetic field intensity increases, flux in a circuit. Faraday's law states: Figure 2.27 Saturation curves of magnetic curve of vacuum where H is and nonmagnetic materials. Note high. that all curves become asymptotic to the B-H FUNDAMENTALS 30 function of time, a voltage tween its of the induced voltage tional to the rate of By 1 definition, when units, its E = propor- is and according terminals. Consequently, V if induced be- is is The induced voltage (2.24) in A/ where In = — At coil [Wb] fixed in space. is common) (although case, it is induction with relative is induced ac- in this special easier to calculate the induced voltage with reference electromagnetic move The in the flux linking the cording to Faraday's law. However, [s] of generators, the coils coils and, consequently, a voltage time interval during which the flux law many motors and motion produces a change in the coil change of flux inside the Faraday's induced a conductor respect to a flux that induced voltage [V] changes zero as soon as the flux falls to ceases to change. 2.21 Voltage A<I> A<J> X1/10 1000 given by N~— = number of turns 3 the flux varies in- side a coil of /V turns, the voltage induced /V in is = 60 V system of the flux inside a loop varies at the rate of E= AO = 2000 X N— At to the SI 1 change takes place uniformly this 1/10 of a second (A/), the induced voltage change of flux. weber per second, a voltage of tween induced be- is terminals. The value 2. Because linking a loop (or turn) varies as a If the flux 1. to the conductors, rather than with reference to the coil itself. In effect, established the basis of operation of transformers, whenever a conductor cuts a magnetic field, a voltage is induced across its terminals. The value of the generators, and alternating current motors. induced voltage opened the door to a host of practical applications and is given by E= Example 2-8 A coil of 2000 turns surrounds a flux of 5 mWb produced by a permanent magnet net is (Fig. 2.28). The mag- suddenly withdrawn causing the flux inside the coil to drop uniformly to 2 second. What is mWb in 1/10 of a the voltage induced? flux E— B= = / induced voltage [VJ flux density [Tl active length of the conductor in the magnetic change AO (5 v is mWb - N= 2 mWb) 3 (2.25) where Solution The Blv — field [m] relative speed of the conductor fm/s| mWb Example 2-9 The stationary conductors of a 2000 an active length of 2 N S \ 0.6 teslas. ,> moving at m large generator and are cut by a have field Calculate the voltage induced in each conductor. 5 4> 2 mWb =2mWb Solution A/ = 1/10 s According to Eq. 2-25. we find Figure 2.28 Voltage induced by a moving magnet. See Example 2-8. E= Blv = 0.6 X of a speed of 100 m/s (Fig. 2.29). 2 X 100 - 120 V FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS 120 3 V Figure 2.30 Force on a conductor. Figure 2.29 Voltage induced Example in a stationary conductor. See 2-9. - on a conductor 2.22 Lorentz force When conductor a current-carrying magnetic field, it is is placed force is of fundamental importance because stitutes the and of tors, it con- basis of operation of motors, of genera- many nitude of the force electrical instruments. The force is parallel to it = 0. The mag- when greatest the conductor perpendicular to the field (Fig. 2.30) and zero is Figure 2.31 Force depends upon the orientation of conductor with respect to the direction of the field. it we electromagnetic force, or Lorentz force. This call the 1 N in a subjected to a force which — (Fig. 2.3 l ). is when Calculate the force on the conductor BII tremes, the force has intermediate values. tor is straight perpen- it is Solution Between these two ex- The maximum force acting on a if dicular to the lines of force (Fig. 2.30). X 0.5 X 200 = 300 N 3 conduc- given by 2.23 Direction of the force acting F= BII on a straight conductor (2.26) where Whenever F= force acting on the conductor [N] B = flux density of the field [T] a conductor carries a current, rounded by a magnetic into the field. For page of this book, the circular have the direction shown in it a current lines Figure 2.32a. of force = active length of the conductor [m] figure / - current in the conductor A] N, S poles of a powerful permanent magnet. [ the magnetic field created The magnetic shape shown Example 2-10 A conductor 3 m long carrying a current of 200 A is placed in a magnetic field whose density is 0.5 T. in field sur- The same / shows is flowing between does not, of course, have the the the figure because lines of force never cross each other. What, then, the resulting field? is the shape of FUNDAMENTALS 32 N // H cross-section . Xi - - ct> I . I (a) Y turns length - / Figure 2.33a Method of determining the B-H properties of a mag- netic material. curve oa (b) a value Figure 2.32 a. Magnetic field due to magnet and conductor. b. Resulting magnetic field pushes the conductor downward. If in Bm Figure 2.33b. The flux density reaches for a magnetic field intensity the current the flux density curve, but To answer the question, we observe that the lines of force created respectively by the conductor and permanent magnet the act in the above the conductor and low in same direction opposite directions be- Consequently, the number of lines above the it. B we the magnetic H nv gradually reduced to zero, does not follow the original moves along oa. In effect, as sity, now is a curve ab situated above reduce the magnetic field inten- domains that were lined up under Hm tend to retain their origiphenomenon is called hystereConsequently, when H is reduced to zero, a sub- the influence of field nal orientation. This sis. stantial flux density remains. It is called residual flux density or residual induction (B r ). conductor must be greater than the number below. The resulting magnetic field therefore has the shape given in Figure 2.32b. Recalling that lines of flux act like stretched elastic bands, upon it is easy to visualize that a force acts the conductor, tending to push it downward. 2.24 Residual flux density and coercive force Consider the coil of Figure 2.33a, which surrounds a magnetic material formed in the shape of a ring. A magnetic field intensity current source, connected to the coil, produces a current whose value and direction can be changed from zero, we gradually increase /, Figure 2.33b H Residual induction and coercive force. at will. Starting so that and B increase. This increase traces out // FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS we wish If to eliminate this residual flux, 33 we have to reverse the current in the coil and gradually H increase in the opposite direction. we move along curve As we do so, be. The magnetic domains gradually change their previous orientation until the becomes zero flux density point at The magnetic c. required to reduce the flux to zero field intensity is called coercive force In also B reducing the flux density from have to furnish overcome the The energy supplied A to zero, magnetic in orientation. dissipated as heat is we used to is frictional resistance of the domains as they oppose the change material. r energy. This energy the in very sensitive thermometer would in- dicate a slight temperature rise as the ring is being Figure 2.34 Hysteresis loop. If B expressed is amperes per meter, the area demagnetized. dissipated per cycle, in in and teslas of the loop is H in the energy joules per kilogram. 2.25 Hysteresis loop Transformers and most electric motors operate on such devices the flux alternating current. In iron tion. changes continuously both The magnetic domains in depends upon the frequency. Thus, pressed if at a rate that the flux has a in J/nr ) is equal to the area (in T-A/m) of the hysteresis loop. value and direc- are therefore oriented one direction, then the other, first in in the To reduce hysteresis losses, we select magnetic materials that have a narrow hysteresis loop, such as the grain-oriented silicon steel used in the cores of alternating-current transformers. frequency of 60 Hz, the domains describe a com- every 1/60 of a second, passing succes- plete cycle through peak flux densities +Z? m and —B m the peak magnetic field intensity alternates be- sively as tween + // m and — // m as a function of H, . we we If v B obtain a closed curve called hysteresis loop (Fig. 2.34). B and coercive force plot the flux density H c The residual induction have the same signifi- cance as before. 2.27 Hysteresis losses caused by rotation Hysteresis losses are also produced for example, an armature volves in AB, made of S (Fig. 2.35). The magnetic domains armature rotates, the describing a hysteresis loop, the flux and +# m . + 5 m +B n , 0, and a, of moves — B m —B n , corresponding respectively b, c, d, e, f, energy that the is 0, to points a, Figure 2.34. The magnetic material absorbs energy during each cycle this iron, that re- in the armature field, irrespective of the position of the armature. Consequently, as the 2.26 Hysteresis loss successively from a piece of a field produced by permanent magnets N, tend to line up with the magnetic In when iron rotates in a constant magnetic field. Consider, dissipated as heat. We and can prove amount of heat released per cycle (ex- first sal toward A and N poles of the domains point then toward B. A complete rever- occurs therefore every half-revolution, as can be seen in Fig. 2.35a and 2.35b. Consequently, the magnetic domains cally, in the armature reverse periodi- even though the magnetic field is constant. Hysteresis losses are produced just as they are in an ac magnetic field. FUNDAMENTALS 34 Figure 2.37 Concentric conductors carry ac currents due to ac flux <1>. currents are progressively smaller as the area of the (b) loops surrounding the flux decreases. In Fig. 2.38 the ac flux passes Figure 2.35 Hysteresis losses due to rotation. through a solid metal plate. It is packed of rectangular conductors touching set basically equivalent to a densely each other. Currents swirl back and forth inside Eddy currents 2.28 Consider an ac flux that links a rectangular-shaped <E» conductor (Fig. 2.36). According to Faraday's law, an ac voltage If the E is { tor to heat up. If a the first, induced across conductor alternating current is /, current U short-circuited, a substantial second conductor is is conduc- placed inside induced because centric it is the power disshows four such conloops carrying currents I 7 2 7 3 and / 4 The is less than penetrated by an ac flux can this regard, special care /, and so, too, x , , , The flux c|) in Fig. be increasing. As a the flux <1> in trans- 2.37 and 2.38 result, eddy currents flow in is assumed to due to the Lenz's law, such a way as to oppose the increasing flux. . conductor Figure 2.36 has to be taken is hot. In sections of the enclosing tanks to overheat. eddy currents An ac become very formers so that stray leakage fluxes do not cause Consequently, the short-circuit sipated in this loop. Fig. 2.37 in the figure. These so-called eddy currents (or Foucault currents) can be very large, due to the low resistance of the plate. Consequently, a metal plate that terminals. its will flow, causing the a smaller voltage links a smaller flux. shown the plate, following the paths V — metal plate Figure 2.38 induces voltage Ev Large ac eddy currents are induced in a solid metal plate. FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS Eddy currents 2.29 iron 35 a stationary in core insulation The eddy -current problem becomes particularly important when iron has to carry an ac flux. This the case in all ac 2.39a shows a coil carrying A large of the core. ally become red hot (even due to these at a its Figure 2.39b Eddy currents are reduced by splitting the core in half. in two The voltage of what it was length, taking care to insulate the from each other induced in each section (Fig. 2.39b). is one half before, with the result that the eddy currents, and corresponding losses, are considerably reduced. we If continue to subdivide the core, we composed of stacked laminations, usually of a millimeter thick. Furthermore, a small silicon is find that decrease progressively. In practice, the core the losses ity, /'•-•'' core could eventu- frequency of 60 Hz) can reduce the losses by splitting the core two along is currents eddy-current losses. sections the Eddy up as shown and they flow throughout the entire length We i.-A an ac current that pro- duces an ac flux in a solid iron core. are set is motors and transformers. Figure alloyed with the steel to increase thereby reducing the losses still more a fraction amount of its (Fig. 2.39c). The cores of ac motors and generators are fore always laminated. A eddy current in one lamination resistiv- there- Figure 2.39c Core built up of thin, insulated laminations. thin coating of insulation covers each lamination to prevent electrical contact between them. The stacked laminations are tightly held in place by bolts and appropriate end-pieces. For a given iron core, the eddy-current losses decrease in proportion to the square of the laminations. number of 2.30 Eddy-current losses in a revolving core The stationary field in direct-current motors and gen- erators produces a constant dc flux. This constant flux induces eddy currents To understand how they in the revolving armature. are induced, consider a solid cylindrical iron core that revolves between the poles of a magnet (Fig. 2.40a). As lines and, duced along Owing it its its is in- length having the polarities shown. to this voltage, large core because resistance eddy currents flow is These eddy currents produce are immediately converted is turns, the core cuts flux according to Faraday's law, a voltage in the very low (Fig. 2.40b). large FR into heat. losses which The power loss proportional to the square of the speed and the square of the flux density. To reduce Figure 2.39a Solid iron core carrying an ac flux. the eddy-current losses, we laminate the armature using thin circular laminations that are FUNDAMENTALS 36 rotation (b) (b) Figure 2.41 Armature a. Figure 2.40 Much b. up of thin laminations. a revolving armature. a. Voltage induced b. Large eddy currents are induced. in built smaller eddy currents are induced. of current. However, known and we want insulated from each other. The laminations tightly stacked with the flat side are running parallel to the flux lines (Fig. 2.41). rent /. We To often happens that e knowledge of advanced mathemat- get around this problem, we can use a graphical solution, called the volt-second method. 2,31 Current in It is well known an inductor yields the that in an inductive circuit the volt- known Ai = L~~ how = instantaneous voltage induced in the circuit | V] L = inductance of the A/7 A/ = E /, time . e, later, after carries a current an interval At. circuit [H] /, at a time From Eq. 2-27 we the rate of change = 1 eAt L of change of current [A/s] when we know which a We want to determine the current a very short Ai rate in appears across an inductance L. can write This equation enables us to calculate the instanta- neous voltage = response to in applied voltage. Suppose the inductance / e the current in an induc- and decreases with time, variable voltage where It and has the advantage of en- Consider, for example, Fig. 2.42, (2.27) At results abling us to visualize a e same tor increases age and current are related by the equation is can use the same equation, but the solu- tion requires a ics. it to calculate the resulting cur- which means that the short interval At is change given by in current Ai during a FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS Volt-seconds are gained (and it -« voltage 1 is when a lost) 37 variable applied across an inductor. Figure 2.42 Variable voltage applied across sulting change in current. Initial an inductor and current re- is A,. mental changes — (f 2 f|). As U = initial A = /, in current A/ during the long period we a result current find current /, + + e 2 At 2 (A/, + A/ 2 I 2 at + time + A/ 3 . t . 2 .) average voltage e during the interval A/ - X duration Af of the interval I L 2 = + | + /, (^Af, - = <? 3 I curve during the interval .) . . .) . / algebraic sum of all + between and t A under t 2 the voltage -seconds across the induccurve between tance during the interval At The values of is tive ( + + Af) = initial current + A/ e2 or negative ) eas AAj, (f, and t f { Therefore, the current in the inductance after the in- /at time little areas under the voltage curve net area volt the I 4 = Am x Af . 2 \ terval + Af 3 + AA 2 + AA 3 + (A/4, / area A/\ under the voltage A/ + AA 2 The sum of , A/\ 3 these ( and so on may be posi- , ) and, therefore, the and so on may be , ( — + ) 2 and ( — ) ( + ) little ar- or ( — ). values of the A/\s gives the net area under the voltage curve between /, + AA fj are usually current at a time ter f, f2 (Fig. 2.43). more , 2 . in Fig. 2.44 the net area/\ after time inter- val Tis equal to (A interested in calculating the when We f Thus, L We and f 2 is many Af ize, the , — A2 ) volt-seconds. To general- current after an interval Tis always given by intervals af- then have to add the incre- / = /, + AIL (2.28) FUNDAMENTALS 38 volts Figure 2.44 The T is A and A 2 net volt-seconds during interval algebraic sum areas of the equal the to . A where /( = current at start of interval T / = current after lime interval T\A] A — under the volt-time curve dur- net area T ing lime /. — [ V-s] inductance [HI Consider, for example an inductor L, having negli- gible resistance, connected to a source whose volt- age varies according to the curve of Fig. 2.45a. the initial current is zero, the value at instant =A F t\ If Figure 2.45 is /L l As time goes However, the current reaches time /-> because at this its moment maximum value at the area under the any more. Beyond becomes negative and, consequently, vol age curve ceases to increase l t, the voltage the net area begins to diminish. ample, the net area equal to is the corresponding current f=(A + i At instant /4 , At instant ^, for ex+ A 2 — A$) and (/\, stant is / the negative area 04 3 words, it the current is to is A is ex- + A 2 The net 4- 4) ). l the inductor. onds during the interval from 0 to t 2 . As it becomes direct proportion to the volt-seconds received. during the discharge period from t 2 in Then to /4 the inductor loses volt- seconds and the current decreases accordingly. An inductor, therefore, behaves very a capacitor. in known much like However, instead of storing ampere-secstores volt-seconds. a capacitor having a capacitance that the voltage E across its C For it is terminals is given by also zero. After inin other where changes direction. at in charged up with volt-seconds, the current increases well becomes negative; Another way of looking inductor stores volt-seconds. example, A 2 - A y )IL zero and so the current 4, Current onds (coulombs), an inductor is actly equal to the positive area (A area An b. in- and so does the current. progressively creases under the curve by, the area a. the situation (Fig. 2.45), consider that the inductor accumulates volt-sec- in E is } the initial voltage and Q . L is the charge coulombs (ampere seconds, positive or negative) the capacitor received during a given interval. FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS In the same way, for an inductor having an in- ductance L, the current / it carries Example 2-11 The voltage across given by is 39 the terminal s of an inductor of 2 H varies according to the curve given in Fig. 2.46. a. Calculate the instantaneous current cuit, where in /] is the initial current and knowing the cir- / in that the initial current is zero. 1 QL is the "charge' b. Repeat the calculations for an current initial of 7 A. volt-seconds (positive or negative) that the in- ductor received during a given interval. It 1 interesting to note that is volt-second. turns coil of 1 weber-turn 600 turns of 20 milliwebers has stored a flux netic Thus a 12V-s/3H tained equal to mag- a total 12 000mWb 12 volt-seconds. If the inductor has an induc- tance of 3 henries, Fig. it X 20mWb = charge of 600 turns = in is that surrounds it is carrying a current of Q { /L = =4 A. the voltage of Fig. 2.45a an inductance of 100 H. The and the current rises to a maximum initial is applied to current of 6.9 zero, is A before again dropping to zero after a time interval of Important Note: an interval T is value to all the If the 27 s. current at the beginning of not zero, we simply add the initial ampere values calculated by the volt-second method. Figure 2.46 See Example 2-11. a. Interval from zero to 3 s: During this interval the area in volt-seconds increases uniformly and progressively. Thus, area A and so is 4 V forth. s; after after one second, the two seconds 8 it is Using the expression = / V s; AIL, the current builds up to the following respective 2.45b shows the instantaneous current ob- when Solution values: 2 A, 4 A, and so on, attaining a final value of 6 Interval A after three from 3 s to 5 seconds. s: The area continues to increase but at a slower rate, because voltage is smaller than before. When = t 5 surface starting from the beginning therefore the current Interval from 5 s to by 4 squares, which is 7 is 16 V-s/2 s: s, is H = The surface equivalent to 8 E the total 16 V s; 8 A. increases V s. FUNDAMENTALS 40 means that the sum of the voltage rises is equal to sum of the voltage drops. In our methodology it Consequently, the current increases by 4 A, thus reaching 12 A. Note that the current no the is age rise" or a "voltage drop/' is not constant during this interval. from 7 Interval s to 8 We The voltage sud- s: denly changes polarity with the result that the V 8 during this interval subtract from the s ously. The from the beginning net area therefore 24 V - s 8 V = s Consequently, the current interval 7=16 is Interval voltage from 8 is s at the = V-s/2 77 to 10 s: V 16 ence. from end of 2.33 Kirchhoff's voltage law and double-subscript notation we assumed is may elements zero in which six circuit The s: — at t 14 is Beyond zero. current of +7 s, this point be sources or loads, and the conI to 4. we have A, to add we can start with any node cw or ccw direction until we starting point. In so doing, we in either a come back to the This ordered voltage The new current wave is simply 7 A above the curve shown in Fig. 2.46. Thus, the = 1 1 s is CIRCUITS When + 6 7 2.7. We 1 set of labels subscripts. We is used to establish the write the voltage sub- scripts in sequential fashion, following the it is order as the nodes cw around in presume the reader same meet. loop ABCD, we successively encounter nodes 2-4-3-1-2. The resulting KVL equation is therefore written essential to fol- E24 + £43 + E31 + E l2 = based upon the voltage rules that are we For example, starting with node 2 and moving 3 A. AND EQUATIONS and current notations covered and = writing circuit equations, low certain going elements A, E, and D, and move ously. at t In encounter the labeled nodes one after the other. each of the currents calculated previ- current elements around a circuit loop, such as the loop involving negative volt- equal to the positive area, and so the net current A to Consider Fig. 2.47 A, B, C, D, E, and F are connected together. The the current changes direction, 7 by the sign notation. later nections (nodes) are labeled the negative area initial followed this 8 A. seconds continue to accumulate and With an a matter of individual prefer- will begin with the double-subscript nota- Because the terminal that 10 s to 14 in The choice tion, coil resistance). Interval can be expressed that voltages is a "volt- is second area does not change and neither does (remember We is s. zero during this interval, the net volt- the current have seen either double-subscript or sign notation. of one or the other volt-seconds that were accumulated previ- b. not necessary to specify whether there longer follows a straight line because the volt- age 0 Sections 2.4, 2.5, familiar with the is solution of such equations, using linear and vector method will review only the writing of these equations, using Kirchhoff s algebra. Consequently, our voltage law (KVL) and By following a current law (KCL). few simple rules, solve any circuit, ac ordc, no matter it is possible to how complex. We begin our explanation of the rules regarding voltages. 2.32 Kirchhoff's voltage law Kirchhofrs voltage law sum states that the algebraic of the voltages around a closed loop Thus, in a 3 is zero. closed circuit, Kirchhoff's voltage law Figure 2.47 Rule for writing notation. KVL equations using double-subscript FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS If we select loop CEF and start with node 4 and move ccw, we successively encounter nodes 4-2-3- 4. KVL equation The resulting + £42 The set usually 0 ac or dc. If they are ac, the voltages will to it is essential to equate all we have done so far and We do not recommend attempts equations to zero as continue to do. equate voltage rises to voltage drops. In £, set even represent instantaneous values. order to prevent errors, will and so be expressed as phasors, having certain of voltages can KVL = -10 V KVL equa- magnitudes and phase angles. In some cases the In = ~E ]2 ~~ E 2 $ = -E X2 + E yl = -40 + 30 £31 of voltages designated by the may be tions Transposing terms, is + em £23 = +10 V 3 indicating that terminal is 1 positive with respect to terminal 3 and that the voltage between the two is 10 V. 2.34 Kirchhoff s current law finding the solution to such double-subscript equations, pressed as it useful to is EXY remember that a voltage ex- can always be expressed as — E YX Kirchhoff s current law sum means zero. This rents that leave Example 2-12 shows two sources connected in series, having terminals (nodes) 2, and 3. The magnitude 2.48 1 +40 V and £ 32 Fig. 2.49 and is equal to the sum of the cur- shows five currents arriving at a The sum of common the currents flowing , E l2 and E^ 2 are specified as E i2 = = +30 V. We wish to determine the I equal to is the currents that it. terminal (or node). into the is (I 2 + node I4 + is (/, 7 S ). magnitude and polarity of the voltage between the open terminals point at a sum of that the flow into a terminal and polarity of states that the algebraic of the currents that arrive , and vice versa. Fig. 41 /, + / 3 ), while the sum that leaves it Applying KCL, we can write + /3 = I2 + I4 + /, 3. Solution In writing the loop equation, let us start at terminal and move ccw The resulting we till again come back KVL equation £, 2 to terminal 1 1 is + £ 23 + E M = 0 Figure 2.49 ?3 1t Rule for writing KCL equations. 2.35 Currents, impedances, and associated voltages Consider an impedance 2 Z carrying a current A connected between two terminals marked Figure 2.48 (Fig. 2.50). See Example 2.12. will A 1 and 2 voltage E, 2 having a magnitude , /Z, appear across the impedance. The question of FUNDAMENTALS 42 Let us write the circuit equations for Fig. 2.5 Loop 2312, starting with node 2 and moving cw: =0 + /4 Z4 + E 3I -/,Z, Figure 2.50 E12 = + /Z. Voltage /4 Z4 is preceded by a + ( ) sign, going around the loop we are moving now polarity The arises: Is question is across an impedance direction as the current flow voltage IZ or — IZ? /, Z in write £, 2 = +/Z the the associated Conversely, moving across an impedance against ative sign. Thus, or dc, and the ductive In ( jX, most ), E2] the direction resistive (R), in- (—jXc ). impossible to predict the or capacitive circuits it is On the other hand, voltage I{Z Loop 3423, starting with + /3Z3 Voltages /3Z3 and because in Fig. 2.5 and Zj, 1 in , I2 node 3 and moving ccw: Z2 + /4 Z4 = 0 /4 Z4 are preceded by a ( + sign, ) the direction of the respective currents. Voltage is preceded by a negative sign because ing against current Loop 242, I2 are Z2 mov- . starting with £24 we in /2 node 2 and moving cw: ~ h^i = 0 Consider for example the circuit of which two known voltage sources E, 3 E24 are connected to four known impedances Z 2 Z3 and Z4 Because the actual directions of , pre- going around the loop we are moving actual direction of current flow in the various circuit elements. is X against the direction of/,. in ~IZ. The current can be ac impedance can be . in when of current flow, the voltage /Zis preceded by a neg- = tion of /4 because in the direc- ceded by a negative sign because we are moving preceded by a positive sign. Thus, is we Fig. 2.50, + /Z equal to resolved by the following rule: When moving same £ l2 1 KCL node at 2: /s = U +h /. . unknown, we simply assume arbitrary directions as shown in the figure. It is a remarkable fact that no matter what directions are assumed, the final outcome after solving current flows are presently the equations (voltages, currents, polarities, phase angles, power, etc.) is always correct. KCL node at 3: h+ /. = h Example 2-13 Write the circuit equations and calculate the currents flowing in the circuit of Fig. 2.52, that £AD - + 108 V and ECD = + 48 knowing V. Solution We first and /3 select arbitrary directions for currents and write the 6 108 Figure 2.51 Writing KVL and KCL V-—- See Example a Ail 12 Figure 2.52 equations. 2.13. /,, circuit equations as follows: Q ^ 48 V /2 , FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS For the loop DABCD composed of the two sources we and the 6 fi and 4 fl resistors, For the loop DCBD Applying 1 composed of the (cw) 48 V source 2 II resistors: E DC + 47 2 + -48 + 4/ 2 + KCL at 12/ 3 = 12/, = 0 (ccw) Figure 2.53 See Example Solution a. We conclude and 73 72 To solve = -3 A we 73 that the directions flows from To = +5 A establish an arbi- first we assumed assumed because Thus, suppose between points a and write the circuit equation, let / b. move cw us c. This yields for I\ 2 is bears a nega- /2 left to right around the loop, starting from terminal However, the actual direction of opposite to that we find ECd + were correct because both currents carry a positive sign. the circuit, trary direction of current flow. *3 Solving these simultaneous equations, = +8 A 2.14. 0 node B, we get h + h = /, 763 obtain £da + 6'i - 4/ 2 + ECD = 0 -108 + 67, - 47 2 + 48 - 0 and the 4 il and 16 43 7(16 + j Substituting the values of tion + £bc = 63) Eac and Ehi and combining the terms in . we /, 0 in this equa- get tive sign. -200 2.36 Kirchhoff s laws and ac circuits The same basic rules of writing double-subscript equations can be applied to ac circuits, including 3-phase circuits. sistive sistive, elements The only difference in all Solving b. 765 A. + 75.8° we 100 find that I = L 1 150° .9 Z. = 0 20.5°. To determine £. lb we write the following equation, moving cw around the loop: , that the re- £ua + £ ;ib + £bc = 0 Transposing terms, three. Furthermore, the volt- magnitudes and phase angles. is + this equation, ages and currents are expressed as phasors, having sor equations 120° dc circuits are replaced by reor capacitive elements, or a inductive, combination of is A. The more time-consuming, but equations themselves can be written by inspection. Let us consider = £ac — £ bc = 200 L 120° - solution of phathe down almost 100 l_ 150° Using vector algebra, we find two examples. E ilb = 123.9 L 96.2° Example 2-14 In the circuit ate the of Fig. 2.53, two sources A, B gener- 2.37 KVL and Voltages E, c sign notation following voltages: = 200 L E bc = 120° 100 L 150° in ac and dc circuits are frequently indi- cated with sign notation and designated by symbols such as E. v e m for such circuits, Calculate a. The value of the current b. the value of £ ab and its / in the circuit phase angle , and so on. To write the equations we employ As we cruise around a larity (+ or — ) of the first the following rule: loop, we observe the po- terminal of every voltage FUNDAMENTALS 44 (£,, E. v e m etc.) , we meet. of the voltage source minal If only the ( + terminal ) marked, the unmarked is ter- 2.38 Solving ac and dc circuits with sign notation taken to be negative. This observed polar- is (+ or — precedes the respective voltages as we write them down in the KVL equation. The following example illustrates the application of this rule. ity In circuits using sign notation, wc treat the IZ volt- ) ages in same way the subscript notation. In other words, the IZ voltage across an impedance sign Example 2-15 In Fig. 2.54, it is known wish age Z preceded by a positive is whenever we move across the impedance in the direction of current flow. Conversely, the IZ , 1 determine the value and polarity of the volt- to Ec EA and £ B E A = + 37 V and E R - - 5 V. We given the polarity marks of that as in circuits using double- voltage move is preceded by a negative sign whenever The following example across the open terminals. to we against the direction of current flow. illustrates the procedure be followed. Solution First, we assign an arbitrary polarity minal voltage Ec We . then proceed + to the tercw around the ) ( loop in Fig. 2.54, starting with voltage EA . This yields the following equation: Example 2-16 The circuit of Fig. E = 1600 /L 60°. 2.55 is powered by an ac source The values of the respective im- pedances are indicated -EA + Ec -E B = (cw) 0 Ec Calculate a. b. T2 T1 in the figure. The current flowing in each element The voltage Ex across the 72 ohm capacitive reactance. Solution a. To solve to flow this problem, the currents are assumed in the arbitrary directions shown. We then write the following equations. Moving cw around - E- 40) Figure 2.54 Rule Note for writing KVL that the sign sponds equations using sign notations. preceding each voltage correof the terminal that was to the polarity encountered in going cw around first the loop. Transposing terms. Ec ~ E A + E H = + 37 = +22 V Thus, the magnitude of of terminal Tl is to is 22 V, and the polarity indeed positive with respect to minal T2. The polarity happens Ec 5 we assumed have been the correct one. at ter- the outset Figure 2.55 See Example 2.15. the loop - 7,(30) BDAB, we + I2 (- j obtain 37) = 0 (cw) FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS moving ccw around Then, we the second loop 45 ABCA, obtain /2 (-j Finally, - 37) (- /3 KCL at applying + /, /2 j - 72) = 21 / 3 0 (ccw) N node A, we have + h = o Upon solving these equations, we obtain the fol- lowing results: = /, 44.9 zl 215° = /3 b. I2 = 30.3 Figure 2.56 L 40° See Example 2.16. 14.9/1 24° We can think of E x as being a voltmeter connected across the capacitor. As a result, Solution the 'Voltmeter" and capacitor together form a closed loop for which we can To meet write ing a circuit equation, as for Thus, in traveling we any other loop. cw around requirement, this KVL we which equations, write the followthe should reader verify: the loop write -/,(-j72) + E x = £ l2 + Eb - E, = £23 + Ec - Eh = E M + £a - E c = 0 Thus 0 0 0 Transposing terms, we obtain Ex = = /3 (-j72) 14.9 zl 24° (- j E 12 = Ea - Eb = 26 72) 45 and so £23 = Eb Ex = 1073 zl Ec = 26 £ 31 - Ec ~ E = iX 45 and hybrid notation 2.39 Circuits In tation We employ both sign noand double-subscript notation as shown in the some circuits it useful to is Ll I notation). We wish to determine the values of £ 23 and £ 31 (double-subscript , notation). E ]2 * 20° -26/1 240° - 26 0° ing the loop created by E Therefore E, N = x = 240° Z_ = 210° can even express the sign notation in and terminals N terms of in follow- and 1 , we KVL equation £ N! + Example 2-17 Fig. 2.56 shows a 3-phase system in which E — 26 L 0°, Eb = 26 zl 20°, and Ec = 26/1 240° (sign 1 L L - 90° double-subscript notation. For example, can write the following example. zl 26 120° -30° zl 45 -66° -26/1 0° zl zl E, E N = — £,,, which , 0 can be expressed as E. v This completes our review on the writing of dc and ac circuit equations. 46 FUNDAMENTALS Questions and Problems 2-6 What 2-1 Three dc sources G h G 2 and generate voltages as follows: , £34 = -100 V = -40 V £56 = +60 V £,2 Show the actual polarity minals 2-2 In in G3 (Fig. 2.57) c. d. Magnetomotive force b. 2-7 Problem ( + )( — ) of the ter- 2-8 2-3 if 2-9 the show the voltage and the actual polarity of the generator minals 2-4 at instants A conductor 2 m 1 , 2, 3, long and moves 60 km/h through a magnetic 2-10 is moved, and the coil falls to 1.2 Figure 2.57 2-3. to A. Draw the force on the moving N pole. N pole act in the direction as the direction of rotation? the waveshape of a sinusoidal V and a frequency of 5 Hz. at a b. speed of If the voltage is = 5 voltage at field t having a the flux linking mWb in 0.2 the average voltage induced. Figure 2.58 Calculate the force on the Does same a. carries a cur- 2-11 A sinusoidal zero at t = 0, what is the ms? 75 ms? 150 ms? current has an effective value of 50 A. Calculate the peak value of current. The magnet See Problem T mm. Calculate the force on the conductor. b. 4. A coil having 200 turns links a flux of 3 mWb, produced by a permanent magnet. 2-1 produce a flux density of 0.6 voltage having a peak value of 200 voltage. See Problems to gap having a length of 8 c. ter- flux density of 0.6 T. Calculate the induced 2-5 air mmf required. Conductor AB in Figure 2.29 rent of 800 A flowing from B a. Terminals 1-4 and 3-6 Terminals 1-3 and 4-6 Referring to Figure 2.58, want an at 0.2 T, 0.6 T, T. Calculate the determine the voltage following terminals are connected together. a. Terminals 2-3 and 4-5 c. We in and polarity across the open terminals b. of cast iron tive permeability 2-1, if the three sources are in series, Referring to Figure 2.26, calculate the rela- and 0.7 each case. connected the SI unit of is Magnetic flux Magnetic flux density Magnetic field intensity a. and 2-2. s. Calculate 2-12 A sinusoidal voltage of 120 V is applied to a resistor of 10 O. Calculate a. the effective current in the resistor b. the peak voltage across the resistor FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS 2-13 c. the d. the power dissipated by the resistor peak power dissipated by the resistor A distorted voltage contains an th 1 2- 8 1 fre- 2- 1 9 quency of the fundamental. 2-14 The current lags in a 60 Hz ( — ) , and curthe is each case, indicate which phasor is ( ) ) { 36 degrees behind the voltage. 2-20 The alternating voltage e 2 in Fig. 2.24a is given by the expression angle between the following phasors and, c. is B 2 which box The resistance of the conductors joining the two boxes in Figure 2.4 is zero. If A, is + with respect to A 2 can B be — with respect to B 2 ? Referring to Fig. 2.59, determine the phase b. terminal A, A 2 to , positive peaks of voltage and current. a. if from ( single-phase motor Calculate the time interval between the 2-15 Figure 2.4, source? har- monic of 20 V, 253 Hz. Calculate the In rent flows 47 e2 = 20 cos (360 ./f- 9) in If 6 lagging. = 1 In d it stria I 2-2 1 and/ = 50° value of e 2 /, and / 3 li and / 3 E and /, at t = 0, 1 80 Hz, calculate the and at / = 262.37 s. Appl icatio n In Fig. 2.60 write the KVL circuit equa- tions for parts (a), (b), (c), and (d). (Go cw around the loops.) 6 A Figure 2.59 See Problem 2-15. 2- 1 6 The voltage applied to an ac the current is / = 20 magnet is E — 60 sin - 60°), all sin given by the expression cj> 1 and an- (c|> gles being expressed in degrees. a. Draw the phasor diagram for E and /, Draw the function of c. 1 a. / as In Fig. 2.6 1 write the for parts (a), (b), <\>. and KCL circuit equations (c), and determine the true direction of current flow. peak negative power in the circuit. 2-23 leads of the third if cw around the harmonic source are In Fig. 2.62 write the equations for parts Referring to Fig. 2.24, draw the wave- shape of the distorted sine wave, b. a Calculate the peak positive power and the 2- 7 2-22 waveshape of E and 2.21. us- ing effective values. b. Figure 2.60 See Problem 2-24 An KVL and KCL circuit (a), (b), (c), and (d). (Go the loops.) electronic generator produces the out- shown reversed. put voltage pulses Calculate the peak voltage of the result- this voltage is applied across a 10 fl resis- ing waveshape. tor, calculate in Fig. 2.63. If 1 FUNDAMENTALS 48 a. the fundamental frequency of the 2-25 current Repeat the calculations of Problem 2-24 for the b. the peak power, in watts c. the energy dissipated per cycle, in 2-26 In Fig. 2.65 write the power per cycle the average e. the value of the dc voltage that in the — the effective value of the voltage in the figure o 1 0 the average voltage 8 4 V + 100 resistor 7A y*4A 2 in parts the loops.) would produce the same average power g. KVL and KCL circuit (Go cw around (a) to (g). d. in Fig. 2.64. equations for the ac circuits shown joules f. waveshape shown A A 3 A / ^ 2 4 6 8 s Figure 2.63 See Problem 2.24. ^4_A A 1' + 100 (b) (a) v (c) 0 4 - Figure 2.61 See Problem 2.22. 2 -100 V Figure 2.64 See Problem 'l 2 5 LI R 11 /1 | (b) (a) 6" I — T 4 LI 1 7Q I— ',1 T (c) Figure 2.62 See Problem 2.23. CSJ 6 1 0 2.25. seconds FUNDAMENTALS OF ELECTRICITY, MAGNETISM, AND CIRCUITS E l2 = 100 20 0° >50U 12 tf (a) (b) (c) £-13 £ ab = 49 £ B = 50 150_° 101.30° E A = 20 = 30 1-3 0° Eba =100L0 c [45_° 7 £ 3 = 100(0° 12 J 40 U 'l A 60 12 ) i 30 * (^O 3' 1 24 12 ^ 40 12 ^ "1 (d) Figure 2.65 See Problem 2.26. (e) (g) 30 hr 12 Chapter 3 Fundamentals of Mechanics and Heat 3.0 Introduction In order to get a thorough grasp of electrical technology, is it mechanics and large motors is essential to power know something about For example, the starting of heat. determined not only by the magnitude of the torque, but also by the inertia of the revolving And parts. the overload capacity of an alternator determined not only by the size of also by the temperature that its its the — — is we know exert a muscular effort to is overcome tional force that continually pulls the force of lift it a stone, we the gravita- downward. There are other kinds of forces, such as the force by exploding dynamite. All these forces are ex- windings can safely And we could mention many more cases where comprehensive approach familiar force For example, whenever we exerted by a stretched spring or the forces created of the conductors as by the currents they chanical/thermal approach gravity. is And the stringing of a transmission line is determined as much by the ice-loading and mechancarry. The most conductors, but withstand. ical strength Force 3.1 pressed in terms of the newton (N), which is the SI unit of force. The magnitude of the force of gravity depends upon the mass of a body, and is given by the ap- proximate equation the electrical/me- F = 9.8w essential to a thor- (3. 1) ough understanding of power technology. For this reason, this introductory chapter covers certain fundamentals of mechanics and heat. The top- ics are not immediately essential to an understanding of the chapters which follow, but they constitute a may wish Consequently, we rec- valuable reference source, which the reader to consult ommend from time a quick to time. first reading, followed by a closer study of each section, as the need arises. where F= m = 9.8 = force of gravity acting on the body [N] mass of the body [kg] an approximate constant that applies when objects are relatively close to the surface of the earth (within 30 km). FUNDAMENTALS OF MECHANICS AND HEAT Example 3-1 51 F Calculate the approximate value of the force of on a mass of gravity that acts 1 2 kg. Solution The force of gravity m = = 17.6 N F= 9.8 is X 9.8 = 12 Figure 3.1 Torque T= 17.6 newtons 1 Fr. 1 When using the make to the equal to 0.453 hand, a between the pound a distinction pound-force A pound (lbf). 592 37 kg, pound-force is On exactly. and (lb) T= F= r = the other X equal to (9.806 65 0.453 592 37) newtons exactly, or about 4.448 N. If the pulley around Example 3-2 (3.2) where mass a unit of is T = Fr we have English system of units, is torque |N m| force [N] radius [mj move, free to it will begin to rotate axis. its Calculate the approximate value of the force of gravity that acts on a mass of 140 newtons and sult in Express the lb. re- Example 3-3 A in pounds-force. motor develops a starting torque of 150 pulley on the shaft has a diameter of Solution the braking force needed to prevent the Using the conversion charts in Appendix massif 140 Eq. 3. 1 - lb 9.8 = 140 (- 2.205) the force of gravity F= m = 9.8 X AXO, = 622.3(^4.448) = 1 = 63.5 622.3 39.9 pound-force = 39.9 lbf 1 mass of 140 entirely different The N F = lb. 139.9 is N lbf. radius T/r action is from a mass of 140 al- lb. 3.3 Mechanical it rotate. Mechanical work pose a string /• is (Fig. 3. 1 Torque If we pull on the cl is done whenever a force in the direction the pulley will tend to rotate. The torque exerted is F moves of the force. The work Fd (3.3) where W= F= cl = work ( J] force [N] distance the force moves [m] a ra- string with a force on the pulley by the tangential force be is of the force. For example, sup- wrapped around a pulley having ). N would given by between the axis of rotation and the point of application required. If the ra- work W= on a body, tending to make is sufficient to prevent rotation. equal to the product of the force times the perpen- F = 300 N almost However, produced when a force exerts a twisting dicular distance motor from m; consequently, a braking force 50/0.5 1 dius were 2 m, a braking force of 75 is dius 0.5 is = Torque Torque If a turning. a distance 3.2 m. 63.5 kg. Using though the numbers are nearly the same, a force of is N m, calculate Solution Note that the force of gravity of exactly equal to the a is Again using the conversion charts, a force of 622.3 140 lbf l given by Example 3-4 A mass of 50 kg 3.2). is lifted to a height Calculate the work done. of 10 m (Fig. 1 FUNDAMENTALS 52 500 kg Figure 3.2 W= Work Fd Figure 3.3 Power P - Wit Solution The force of gravity acting on the 50 kg F= m = 9.8 The work done X 9.8 50 mass is F= = 490 N The work done X 490 10 W= = 4900 J The power Power Power is 9.8 X 500 = 4900 N is is W= Fd = 3.4 m= 9.8 P = the rate of doing work. It is given by the = Fd = 4900 X 30 = 147 000 J is Wit 147 000/12 = 12 250 W= 12.25 kW equation: Expressed P= Wit (3.4) in horsepower, P= 12 250/746 = 16.4 hp where P = power [W] W = work done = t The unit of power is the watt (W). corresponds is use the sometimes expressed to the is motor equal average power output height of 30 m a mass of lifts in power developed by 12 s 500 kg through a the motor, in kilowatts and in Solution ity in the acting on the cable mass is The mechanical power output of a motor depends upon its rotational speed and the torque it develops. The power is given by where P = T— n = 9.55 = (Fig. 3.3). Calculate the horsepower. The tension equal to the force of grav- that is being motor in Example 3-5 electric of a to of a dray horse. An Power 1000 W. The One horsepower units. [s] We often equal to is power output of motors horsepower (hp) It [J] time taken to do the work kilowatt (kW), which 746 W. 3.5 lifted: mechanical power [WJ torque [N-m] speed of rotation [r/min] a constant to take care of units (exact value = 30/tt) We can measure the power output of a motor by means of a prony brake. It consists of a stationary flat belt that presses against a pulley mounted on the motor shaft. The ends of the belt are connected to two spring scales, and the belt pressure is adjusted FUNDAMENTALS OF MECHANICS AND HEAT V (Fig. 3.4). As the motor turns, by tightening screw 53 3.6 Transformation of energy we can increase or decrease the power output by adpower developed by the motor into heat When register The mechanical tension of the belt. justing the by the the entirely converted is it not running, the spring scales equal pulls and so the resulting torque (as P 2 The re- does in Fig. 3.4), pull Thermal energy (heat released by ley is therefore a radius ters. late f>, exceeds pull — P 2 newtons. If the pulley has T = (P — P 2 )r newton me- (/>, Knowing a stove, by by the sun) Chemical energy (energy contained 3. in dyna- mite, in coal, or in an electric storage battery) ) the net torque r, . friction, on the circumference of the pul- in mov- ing car) is However, when the motor turns clockwise sultant force acting one of the following forms: a coiled spring or the kinetic energy of a 2. zero. exist in Mechanical energy (potential energy stored 1. rubbing against the pulley. belt motor is Energy can produced by Electrical energy (energy 4. a gen- x the speed of rotation, we can erator, or calcu- the power, using Eq. 3.5. by lightning) Atomic energy (energy released when 5. cleus of an atom is the nu- modified) Although energy can be neither created nor destroyed, it can be converted from one form to an- other by means of appropriate devices or machines. For example, the chemical energy contained in coal can be transformed into thermal energy by burning The thermal energy contained the coal in a furnace. in steam can then be transformed into mechanical energy by using a turbine. Finally, the mechanical energy can be transformed into electrical energy by means of a In the generator. above example, the furnace, the turbine, Figure 3.4 and the generator are the machines Prony brake. ergy transformation. that Unfortunately, whenever energy the output Example 3-6 During a prony brake test on an electric motor, the 25 spring scales indicate (Fig. 3.4). turns at 0.1 N and 5 N, respectively Calculate the power output if the motor 1700 r/min and the radius of the pulley is always transformed, because all ma- chines have losses. These losses appear in the form of heat, causing the temperature of the machine to rise. Thus, the electrical energy supplied to a motor is partly dissipated as heat in the windings. Furthermore, is some of its mechanical energy m. is less than the input do the en- is also lost, due to bear- ing friction and air turbulence created by the cooling Solution fan. The torque T= = The power The mechanical losses are also transformed into put of a motor Fr (25 - 5) X 0.1 = 2 Nm is less than the electrical input. 3.7 Efficiency of a machine is The efficiency of a machine P = ai mechanical power out- heat. Consequently, the useful is 779.55 = 1700 X 2/9.55 = 356 W The motor develops 356 W, or about 0.5 hp. is given by the equation P t\ = p X 100 (3.6) FUNDAMENTALS 54 where Kinetic energy — Tl P0 = The energy power input \/2mv power of the machine fW] output efficiency is is machine to the particularly [ W] low when thermal from 25 the efficiency of steam turbines ranges that of internal gines (automobile engines, diesel motors) we must remember it lies how low that a having an efficiency of 20 percent loses, of heat, 80 percent of the energy Ek = kinetic energy [JJ m = mass of the body [kg] v = speed of the body fm/s] to be- A bus having a mass of 6000 kg moves at a speed of machine 100 km/h. If it carries 40 passengers having a total mass of 2400 kg, calculate the total kinetic energy of the loaded vehicle. What happens to this energy in the form receives. efficiently. Example 3-8 these Electric motors transform electrical energy into mechanical energy much more when the bus Solution pending on the size of the motor. Total 150 kW percent when mass of it motor has an efficiency of 92 operates at full-load. The speed + 2400 = 8400 = 100 km/h v 100 refers to the The mechanical output power = 163 kW To stop the is is \/2mv 3 2 1/2 245 928 J X 8400 X 3.25 MJ 27. 2 bus, the brakes are applied and the re- sulting frictional heat is entirely produced at the ex- pense of the kinetic energy. The bus will finally come losses are 150 27.8 m/s kinetic energy Ek = = P 0 = 150kW 163 m s mechanical The 150/0.92 kg X 1000 3600 The 150 kW rating always power output of the motor. The input power is The 6000 is is Solution { to a stop? Calculate the losses in the machine. P = PJi] = braked the loaded bus m = _ electric is Their ef- ficiency ranges between 75 and 98 percent, de- Example 3-7 (3.7) combustion en- tween 15 and 30 percent. To realize efficiencies are, 2 where converted into mechanical energy. Thus, 40 percent, while A a form of mechanical energy efficiency [percent] — Pi is given by the equation 13 to rest when all the kinetic energy (3.25 MJ) has been dissipated as heat. kW Considering the high efficiency of the motor, the losses are quite moderate, but they be enough to heat a large home in the would still winter. 3.8 Kinetic 3.9 Kinetic A energy of linear motion A falling stone or a swiftly sess kinetic energy, which moving automobile posis energy due to motion. energy of moment middle of Its rotation, of inertia revolving body also possesses kinetic energy. magnitude depends upon the speed of rotation and upon (he mass and shape of the body. The netic energy of rotation on page 56. is ki- given by the equation FUNDAMENTALS OF MECHANICS AND HEAT TABLE 3A MOMENT OF INERTIA J AROUND AN AXIS OF ROTATION 0 Figure 3.5 Mass m revolving at a distance J r around axis 0. = mr (3.9) Figure 3.6 Solid disc of mass m and radius r. (3.10) Figure 3.7 Annular ring of mass J m having a rectangular cross-section. = -(Rf + R 2 I (3.11) ) Figure 3.8 Straight bar of mass m pivoted on its center. J (3.12) 12 Figure 3.9 Rectangular bar of mass J m revolving around = -(Rf + /V axis 0. (3.13) 55 1 FUNDAMENTALS 56 5.48 X 10"'V/7- (3.8) The b. Ek = = where E k = kinetic energy [J] J = moment of inertia [kg-rrrl = rotational speed [r/minj X 1()~ 3 = constant to take care of units 2 Lexact value = tt /1800] // 5.48 The moment of inertia./ (sometimes simply called depends upon the mass and shape of the body. inertia ) may be value Its calculated for a shapes by using Eqs. 3.9 to 3. 1 3 given in Table 3 A. If body has a complex shape, the number of simple can always be bro- it ken up into two or more of the simpler shapes given in the table. are then The individual Js of these simple shapes added to give the kinetic energy = Note as 5.48 X 5.48 X 3. 1 is l0"-V/r X (3.8) 175 X 1800 2 MJ that this relatively small flywheel possesses much kinetic energy as the loaded bus Example tioned in men- 3-8. Example 3-10 A flywheel having the shape given in Fig. 3.1 1 is composed of a ring supported by a rectangular hub. The ring and hub respectively have a mass of 80 kg and 20 kg. Calculate the moment of inertia of the flywheel. J of the body. total Inertia plays a very important part in rotating machines; consequently, it is worth our while to solve a few problems. 80 kg Example 3-9 A 1 1400 kg solid m flywheel has a diameter of steel and a thickness of 225 mm (Fie. 225 3. 10). mm 1400 kg V 1800 r/m Figure 3.11 Flywheel in Example 3.10. Figure 3.10 Flywheel in Example 3.9. Calculate Its b. The at Solution moment a. of inertia kinetic energy For the when ring, the flywheel revolves J I800r/min ] Solution a. Referring to Table 3 A, inertia we find the moment of = m (Rr + R 2 2 )/2 = 80 (0.4 2 + 0.3 2 )/2 = (3.1 1) 10 kg m 2 For the hub, is J2 J (3.10) 1400 X 0.5 The 2 total = mL 2 /\2 = 20 X (0.6) 2 /12 = moment = 175ks-m 2 J = J (3. 12) 0.6 kg m 2 of inertia of the flywheel X + J2 = 10.6 kg m is 2 k, FUNDAMENTALS OF MECHANICS AND HEAT 3.10 Torque, inertia, volving body, and that a change the speed of a to is to subject The given period of time. depends upon the Speed 1 In electric way only one is 3-1 speed in There and change it re- to a torque for of change of speed rate inertia, as well as on the torque. A = A/7 9.55 power technology, a system there are three erted by the load, and the speed. (3.14) T= The load a shaft (Fig. 3. 2). interval of time during torque J 9.55 is — moment = constant applied creasing or Suppose the T [sj . Y to take care = in a We now explain exerts a constant torque counterclockwise direction. TM developed by the can be varied by in- the torque acts clockwise, and decreasing system it the current electric rest is initially at and that /. 7M = Because the torques are equal and opposite, the net torque acting on the shaft 2 zero, and so is it has ] no tendency to rotate. of units Load mm 30/ttJ the torque acts in the direction of rotation, the speed rises. Conversely, tion which the of inertia fkg-m [exact value If motor [Nm] torque = Ar speed [r/min] in to consider: they interreact. 1 change main factors the torque developed by the motor, the torque ex- Tu that always acts On the other hand, = often happens that Consider a load coupled to a motor by means of TMJ where A/? it system an electric motor drives a mechanical load. In such how simple equation relates these factors: of a motor/load 57 if of rotation, the speed acts against the direc- it The term An may falls. therefore represent either an increase or a decrease in speed. ' M Example 3-11 The to fly wheel increase torque of 20 of Fig. 3.11 turns Nm. at 60 r/min. We wish Motor 600 r/min by applying a For how long must the torque be speed its to Figure 3.12 applied? Shaft is stationary TM 71- Solution The change in A/? speed - The moment of (600 - inertia J As is = 60) = 540 a result of the twists and r/min becomes shaft deformed, but other- wise nothing happens. Suppose we want is 10.6 kg- opposing torques, the slightly /z, To do so, we increase the motor current so TM exceeds T The net torque on the shaft acts speed 2 that . . x clockwise, and so Substituting these values in Eq. 3.14 the load to turn clockwise at a it begins to rotate clockwise. The speed increases progressively with time but as soon 9.55 7Af/./ A/7 (3.14) as the desired 540 = 9.55 x 20 A/7 10.6 TL The . yielding speed motor current so 30 s is TM reached, is let us reduce the again exactly equal to net torque acting on the system and the speed Ar /z, that any more (Fig. /z, is now zero will neither increase or decrease 3. 13). FUNDAMENTALS 58 Load Load Figure 3.13 Shaft turns Figure 3.14 cw 7M Shaft turns 71- Whenever This brings us to a very important conclusion. The speed of a mechanical load remains constant when the torque TM developed by the motor is equal and opposite to the torque T L exerted by the load. At first, this we accept, because when TM = 7^ rather difficult to is the opposite to a motor remains constant motor torque is exactly equal and load motor/load system is torque. In the effect, then in a state of dynamic equilibrium. With the load now running clockwise suppose we reduce TM so that it is at a speed TL less than . on the shaft now acts counterclock- net torque wise. Consequently, the speed decreases and will continue to decrease as long as T imbalance between the enough, the speed we then reverse. If TM — TL when motor torque 7M and load torque and opposite, the speed will change. The rate of change depends upon the inerof the rotating parts, and this aspect tia more is covered in detail in Section 3. 13. Power flow 3.12 our reasoning (and reality) shows. the the are not exactly equal { T, will eventually exceeds TM and TM . If long lasts become zero and control the motor torque so that the reverse speed reaches a value n 2 , in a mechanically coupled system The speed of repeat: whenever The TL . are inclined to believe that the system should simply stop. But , this is not so, as We conclusion ccw TM = TL Returning again to Fig. 3. 13, we torque TM acts in the torque T acts opposite to speed n see that motor same direction (clockwise) as speed n lt This means that the motor delivers mechanical power to the shaft. On the other hand, load { x the load receives mechanical . Consequently, power from the shaft. We can therefore state the following general rule: When the torque developed by a motor acts in same direction as the speed, the motor delivpower to the load. For all other conditions, the motor receives power from the load. the ers In Fig. 3.14, for example, the power from the load because 7M motor receives acts opposite to n 2 . the system will continue to run indefinitely at this Although new speed brief periods in electric trains and electric hoists. (Fig. 3. 14). In conclusion, torques Figs. 3.12, 3.13, TM and T { are identical in and 3.14, and yet the shaft may be turning clockwise, counterclockwise, or not The actual steady-state whether TM was greater or speed less than depends 7L this is an unusual condition, it occurs for The behavior of the motor under these conditions will be examined in later chapters. at all. upon 3.13 Motor driving a load for a certain having inertia period of time before the actual steady-state condi- was reached. The reader should ponder moments over this statement. tion a few When a motor drives a mechanical load, the speed is usually constant. In this state of dynamic equilibrium. FUNDAMENTALS OF MECHANICS AND HEAT the torque TM developed by the motor is exactly equal T imposed by the load. and opposite to the torque The inertia of the revolving parts does not However, play under these conditions. torque speed is raised so that it increase, will come if we have already when less than that of speed drops. The increase or decrease in speed (An) motor torque is TM torque Tis now replaced by the net torque (TM — 7L = A/7 - T )AtIJ 9.55 (rM x stays con- paper remains must be greater than the load torque It . We in have given by the Eq. 3.14, except that is still Nm) in the order for the speed to increase. seen. Conversely, because the tension unchanged. Let the required motor torque be exceeds the load torque, the as the speed increases from 120 r/min to 160 stant motor the As r/min, the load torque (5400 into the load, the the b. x 59 - A/7 ): J (3.15) At where 160 - = 4500 = 5 s 120 m kg r/min - TL )At 9.55 (7\, An = = 40 2 J = change in speed [r/minj TM = motor torque [N-m] 7L = load torque [N m] At = time interval during which TM and TL are acting [s] J = moment of inertia of all A/7 _ 40 - 9.55 4500 Thus, TM - 5400 = 3770 rM = 9170 revolving parts [kg-irrj The motor must Example 3-12 A large reel torque of 9170 of paper installed machine has a diameter of moment of and a 1 inertia of at the end of a paper kgm 2 . It is therefore develop a constant Nm during the acceleration period. m, a length of 5.6 m, .8 4500 5400) 5 The mechanical power of the driven reel motor 160 r/min at accelerating is by a directly coupled variable-speed dc motor turn- The paper ing at 120 r/min. tension of a. 160 r/min c. in 5 c. from 120 r/min to seconds, calculate the torque that power of the this interval. motor after it As soon 7 = Fr = 6000 X 1 reel .8/2 The power developed by the P = nT 120 P = is is N m). The power nT _ ~ 160 X 5400 9.55 9.55 90.5 kW (equivalent to 121 hp) is = 5400 N m reel motor 3.14 Electric motors driving linear motion loads is Rotating loads such as fans, pumps, and machine X 5400 9.55 kW 160 r/min) therefore reduced to tools are well suited for direct mechanical coupling to electric motors. 67.85 ( equal to the load torque (5400 has (3.5) 9.55 kW (equivalent to 206 hp) as the desired speed of the motor Solution The torque exerted on the 9.55 reached, the motor only has to develop a torque reached the desired speed of 160 r/min. a. 160X9170- - 153.6 the reel speed of 120 r/min. motor must develop during Calculate the - power of the motor when the speed has to be raised the nT P = 6000 N. Calculate the If kept under a constant 9.55 turns at a constant b. is (about 91 hp) move in On the other hand, loads that a straight line, such as hoists, trains, wire- FUNDAMENTALS 60 drawing machines, must be equipped with a etc., where motion converter before they can be connected to a rotational speed [r/min] T= torque [N-m] F= v = 9.55 = we seldom converters are so utterly simple that = n The motion converter may be a rope-pulley arrangement, a rack and pinion mechanism, or simply a wheel moving over a track. These rotating machine. force [N] linear speed [m/s| a constant [exact value = 30/tt] think of the important part they play. Straight-line motion involves a linear speed v and a force while rotary motion involves a rota- speed n and a torque tional tities F, related when T. How are these quan- a motion converter is used? Consider a jack driven by a motor that rotates T (Fig. a linear speed plied in raising the load is v. motive turns the other hand, the needed is at 1 to pull an electric train The motor on board the loco- 200 r/min. Calculate the torque de- at F Solution The power sup- given by power kN This nT= 9.55fY 1200 T = X P a = Fv On of 25 a speed of 90 km/h. veloped by the motor. 3. 15). causes a vertical ram to exert a powerful force at A force at a speed n while exerting a torque while moving Example 3.13 9.55 (3.16) 25 000 j = 4974 N-m = input to the jack 5 X (90 000/3600) kN m is given by = nT P: 3.15 Heat and temperature (3.5) 1 9.55 Assuming verter, Whenever there are no losses in the motion con- we have the SI unit p. 1 1 heat applied to a body, is mal energy. Heat is ' energy? Consequently, nT= 9.55Fv a body receives this type of atoms of the body vibrate more First, the intensely. Second, we can receives ther- the joule. What happens when = po it therefore a form of energy and is its temperature increases, a fact verify by touching it or by observing the (3.16) reading of a thermometer. For a given amount of heat, the increase tem- in perature depends upon the mass of the body and the material of which 100 kJ of heat to is it made. For example, if by 24°C. The same amount of heat supplied of copper raises we add kg of water, the temperature 1 rises to temperature by 263°C. its therefore obvious that heat and temperature are kg 1 It is two quite different things. If we remove heat from a body, its temperature drops. However, the temperature cannot low a lower zero. or It limit. corresponds — 273.1 5 °C. At This limit to a fall be- called absolute is temperature of 0 kelvin absolute zero all and the only motion Figure 3.15 tions cease Converting rotary motion into linear motion. that of the orbiting electrons. atomic vibra- that subsists is FUNDAMENTALS OF MECHANICS AND HEAT 1806 T iron melts T 450 K i 1083 aluminum melts 933 660 1220 lead melts 600 327 621 water boils water freezes 373 273 100 212 32 0 -273.15 ® ® 1981 -459.67 Fahrenheit scale Celsius scale Kelvin scale 2791 810°F 450 °C 1, 356 copper melts — T 1533 61 Figure 3.16 Temperature scales. 3.16 The Temperature scales given specific heat capacity of several materials Table in AX2 in the is Appendix. The kelvin and the degree Celsius are the SI units of temperature. Fig. 3. 6 1 tionships between shows some interesting rela- Kelvin, the Celsius, and Fahrenheit temperature scales. For example, iron melts at 1 806 K or 1 533°C or 279 °F 1 Example 3-14 Calculate the heat required to raise the temperature of 200 L of water from 10°C tank perfectly insulated (Fig. 3.17). is heat capacity of water 3.17 Heat required to raise the temperature of a body The temperature heat it terial. receives, rise its weighs 1 kg. of a body depends upon the mass, and the nature of the ma- The relationship between these quantities is given by the equation Q = mcAt (3. 17) where Q = quantity of heat added to (or m = c = removed from) mass of the body = body [JJ specific heat capacity of the material At a |kg] making up the body [J/(kg-°C)l Figure 3.17 change Electric in temperature [°C] water heater. is to 70°C, assuming the The specific 41 80 J/kg °C, and one liter n FUNDAMENTALS 62 Solution quired convection convection The mass of water is 200 kg, and so the heat re- is Q ~ mcAt = 200 X 4180 X = 502 MJ (70 - 10) Referring to the conversion table for Energy (see Appendix), we find that 50.2 MJ is equal to 13.9 kW-h. 3.18 Transmission of heat Many problems in electric power technology lated to the adequate cooling of devices are re- and ma- knowledge of the mechanism by which heat is transferred from one body to another. In the sections that follow, we Figure 3.18 Heat transmission by convection, conduction, and ra- diation. chines. This, in turn, requires a briefly review the elementary physics of heat trans- We mission. also include some simple but useful Referring to Fig. 3.19, we can equation equations, enabling us to determine, with reason- P = able accuracy, the heat loss, temperature rise, and so on of electrical equipment. its we - — atoms (Fig. 3. 18). This A = — atomic vibration ] t 2) — end of the bar. atom is to the next, to the other at d = thickness of the body [m] Consequently, the end opposite the flame also warms up, an observation one time or another. ferred along the bar we have In effect, heat is all trans- by a process called conduction. The rate of heat transfer depends upon the thermal conductivity of the material. Thus, copper is a better thermal conductor than steel is, and plastics and other nonmetallic materials are especially poor conductors of heat. The SI unit of thermal conductivity per meter degree Celsius conductivity of several in Tables AX 1 and AX2 [W/(m common in the °C)J. is the watt The thermal materials Appendix. is given 2 body [m difference of temperature between surface area of the opposite faces [°C] transmitted from one made (3.18) [W/(m-°C)J bring a hot flame near one end of an iron bar, temperature rises due to the increased vibration its ] P = power (heat) transmitted [W] X = thermal conductivity of the body (t of - where 3.19 Heat transfer by conduction If calculate the ther- mal power transmitted through a body by using the Figure 3.19 Heat transmission by conduction. ] FUNDAMENTALS OF MECHANICS AND HEAT Example 3-15 63 tank The temperature difference between two sides of a mica sheet of 50°C is cm" and thickness ing 3 is through the sheet, (Fig. 3.20). If mm, its area is 200 calculate the heat flow- in watts. Sol ui ion AX1, According to Table of mica ducted is, W/m 0.36 is the thermal conductivity The thermal power con- °C. therefore. - \A(t P t l 2) (3.18) 0.36 d Figure 3.21 X Convection currents 0.02 (120 70) 120 in oil. W 0.003 The warm 0.02 m 2 it comes in contact with the cooler tank, X = 0.36 chills, nal tank. 120' oil becomes heavier, sinks to the bottom, and moves upward again to replace the warmer oil now moving away. The heat dissipated by the body is, therefore, carried away by convection to the exter- mica The tank, in turn, loses convection to the surrounding x its heat by natural air. 70°C 3 3.21 mm Calculating the losses by convection Figure 3.20 Mica sheet, The Example 3-15. heat loss by natural convection in air is given by the approximate equation P = 3.20 Heat transfer by convection In Fig. 3.18 the air in contact with the hot steel bar 3/Uf, - t 2 y (3.19) where warms up and, becoming a chimney. As the bar, smoke moves upward, it is which, in turn, also warms the hot air placed by cooler air A continual lighter, rises like current of air removing its is P — A — f, = t2 = in re- up. therefore set up around heat by a process called nat- heat loss by natural convection [W| surface of the body [m] surface temperature of the body [°C] ambient temperature of the surrounding air l°CJ ural convection. The convection process can be accelerated by employing a fan to create a rapid circulation In the case of forced convection, such as that produced by a blower, the heat carried away is of given approximately by fresh in air. Heat transfer by forced convection is used P = 1280 Va {t 2 most electric motors to obtain efficient cooling. when oil. The Natural convection also takes place body is immersed contact with the currents in a liquid, such as (3.20) where oil in body heats up, creating convection which follow the path shown /,) a hot in Fig. 3.21. P — heat loss by forced convection [W] — volume of cooling air |m Vs] V. x FUNDAMENTALS 64 = i\ temperature of the incoming (cool) air 3.22 Heat transfer by radiation [°C| f — 2 We temperature of the outgoing (warm) have air|°C] same when hydrogen, Surprisingly, Eq. 3.20 also applies a much lighter gas, basked all is used as the cooling medium. properties Example 3-16 area of 1.2 nr. When operates it at full-load, 60°C surface temperature rises to 20°C in the an ambient of (Fig. 3.22). Calculate the heat loss by natural as energy light, P = = diate when only converted to heat is have discovered even those heat, as the physi- that that all bodies ra- very cold. The are amount of energy given off depends upon the tem- perature of the body. On L25 - 3A{t t2 } the other hand, all bodies absorb radiant en- 3 X (60 1.2 ergy absorbed depends upon the temperature of ) - 20) L25 W = 362 the surrounding objects. There tinual ial 362 i f ) \ W H bodies, as convection body the is same when in as that of body then radiates a body as the temperature of a surroundings. The its much energy is hotter than is tinually lose heat in therefore, a con- each were a miniature sun. if Equilibrium sets \ is, exchange of radiant energy between mater- and the net radiation zero. its On as it receives the other hand, environment, by radiation, even if it it will is if con- located vacuum. 3.23 Calculating radiation losses Figure 3.22 Convection and radiation losses a in totally enclosed motor. The heat that a body loses by radiation is given by the equation P = kA Example 3-17 kW fan rated at 3.75 through a 750 kW motor the inlet temperature is 3 1 is blows 240 rrrVmin of to carry 22°C and away air away by P = A = T, = T2 = the outlet temper- the circulating air. lx = 1280 (t 2 - k = X 240/60 heat radiated (31 - 22) (3.21) = 46 kW IW] surface area of the body \m'\ absolute temperature of the body [K] absolute temperature of the surround- a constant, ture of the f,) (approximate) ) ing objects IK] Consequently, the losses are P = 1280 V - T2 4 where °C. estimate the losses in the motor. losses are carried 4 (7, the heat. If Solution The passes readily it ergy from the objects that surround them. The en- Solution ature and meet a solid body, such earth. Scientists convection. A the and living things on the surface of the cal objects enclosed motor has an external surface totally warmth produced by through the empty space between the sun and the earth. Solar the sun's rays A in the sun's rays. This radiant heat energy possesses the Table 3B which depends upon the na- body surface gives the values of k for surfaces monly encountered in electrical equipment. com- FUN DA MENTA LS OF MFC HA NfCS A NI) HFA I 3-3 RADIATION CONSTANTS TABLE 3B Give the SI unit symbol Type of surface polished silver bright oxidized copper 3 aluminum paint 3 Nichrome 2 tungsten 2 oxidi/ed iron 4 insulating materials paint or nonmetallic perfect emitter 5 enamel 5 5.669 (blackbody) ) X l()" X 10 s X 10~ s X 10 s X 10" s X 10 8 X 10'"* X 10" s X 10 s X 10'"* 1 force work pressure area mass temperature thermal energy thermal power 3-4 200 force of An tion, coated with a non- knowing surrounding objects are that all mechanic exerts a end of a wrench hav- automobile engine develops a torque of at an at a speed of power output 4000 r/min. Calculate watts and in horse- in power. by radia- lost at the ing a length of 0.3 m. Calculate the torque The motor is N power he exerts. the Example 3-16 electrical In tightening a bolt, a Example 3-18 in energy electrical 600 N-m enamel. Calculate the heat mechanical power mechanical energy 3-5 metallic and the corresponding SI for the following quantities: 4 s 0.2 copper oxidized W/(m 2 -K Constant k 65 3-6 ambient temperature of 20°C. A crane lifts 200 ft in and in 15 a mass of 600 lb to a height of s. Calculate the power watts in horsepower. Solution An 3-7 = = 60°C or (273. 15 + 60) = 333 K T2 = surrounding temperature = 20°C or (273. 15 4- 20) - 293 K 7, surface temperature From Table 3B. k The power = 5 X 10 by radiation lost s W/(m 2 -K 4 is, line 4 = 5 - 7\ 10" 8 X (7, X 3-8 (3.21) ) 4 - 293 20 kW. c. A large flywheel has a it (hp] J lb-fr. Calculate rotates at its moment of inertia of kinetic energy when 60 r/min. 4 The rotor of an induction motor has a moment of inertia of 5 kg-nr. Calculate the 3-9 ) W (approximate) = 296 kW from the 20 The power output of the motor [kW] and The efficiency of the motor The amount of heat released Btu/h] b. therefore, (333 1 to ). 4 1.2 motor draws and has losses equal Calculate a. 500 P = kA electric energy needed to bring the speed It is interesting to note that the almost as much motor dissipates heat by radiation (296 W) as it does by convection (362 W). 3- 1 0 Questions and Problems 3-11 3- 1 A cement is from zero b. from 200 r/min c. from 3000 r/min Name ried Practical level What What block has a mass of 40 kg. the force of gravity acting on it? 3-2 is needed How much to energy to the three to 400 r/min to 400 r/min ways whereby heat from one body A motor develops a cw 50 N-m. If this situation persists for some time, will the direction of rotation eventually be needed to lift a sack of flour weighing 75 kg to a height of 4 m? car- torque of 60 N-m, lift it? is is to another. and the load develops a ccw torque of a. force 200 r/min a. b. What value of motor torque the speed constant? is cw needed or to ccw? keep FUNDAMENTALS 66 3-12 A motor drives r/min. Nm, and 15 Nm. 12 of cw speed of 1000 a cw torque of exerts a ccw torque a load at 3- 9 1 The motor develops the load a. Will the speed increase or decrease? b. If this situation persists for some 1 negligible, calculate the following: what time, in a. direction will the shaft eventually rotate'? 3-13 Referring to Fig. 3.12, what 3-14 is the if Referring to Fig. 3.13, =50 TM = 40 N power delivered by if the b. m, power 3-20 motor? TM = 40 N m r/min, calculate the = 50 Calculate the heat [MJJ required to raise the temperature of 100 kg of copper from 20°C deliv- 3-2 1 Repeat Problem 3-20 for 1 00 kg of alu- minum. Referring to Fig. 3.14, n2 The torque developed by the motor [N-m] The force opposing the motion of the bus [N] to 100°C. and ered by the motor. 3-15 The electric motor in a trolley bus develops a power output of 80 hp at 200 r/min as the bus moves up a hill at a speed of 30 miles per hour. Assuming that the gear losses are if TM = 40 N m and r/min, calculate the 3-22 power received by the motor. The motor in Fig. 3.23 drives a hoist, raising a mass m of 800 kg at a uniform rate of 5 m/s. The winch has a radius of 20 cm. Calculate the torque [N-m] and speed Intermediate level [r/min] of the motor. 3- 6 1 During a prony brake test on a motor (see and speed Fig. 3.4), the following scale readings were noted: P2 = 5 n = 1 If P, lbf =28 lbf 160 r/min the diameter of the pulley calculate the 12 inches, is power output of the motor in kilowatts and in horsepower. 3- 7 1 A motor drives a flywheel having a moment of inertia of 5 kg-nr. The speed increases ij from 1600 r/min to 1800 r/min in 8 Motor: s. Calculate a. The torque developed by b. The energy c. The motor power [W] d. The power in the input the flywheel [W] at at motor [N-m] 1800 r/min Figure 3.23 |kJ| to the flywheel at A 3-23 dc motor coupled velops 120 hp at a constant speed of 700 The moment of 2 parts is 2500 lb-ft ing a. | inertia of the revolv- 1 Problem 3-22 m/s, calculate the r/min] and torque [ft- lbf] is re- new speed of the motor. i Calculate the torque [N-m] developed by the 3-24 How many Btus are required to raise the temperature of a 50 gallon (U.S.) reservoir of water from 55°F to Calculate the motor torque IN-m] needed so that the {Note; to Industrial appl cat ion . motor. b. Tf the hoisting rate in duced to a large grinder de- r/min. Problem 3-22. 1750 r/min 3-18 Electric hoist, 1600 r/min speed will increase to The torque exerted by mains the same.) 750 r/min in 5 the grinder re- s. the tank will it is take 1 80°F, assuming that perfectly insulated. if the tank electric heater? is How heated by long a 2 kW FUNDAMENTALS OF MECHANICS AND HEA T 3-25 A large indoor transformer using an is painted a non- It is proposed to refurbish aluminum paint. Will this affect metallic black. it the temperature of the transformer? If so, will 3-26 An it electrically heated perature is m X cement floor covers 30 m. The surface tem- 25 °C and the ambient tempera- ture is 23 °C. Approximately heat is given the point of is off, in The cable and other electrical how much kilowatts? Note: from view of heat radiation, cement considered to be an insulator. components inside a sheet metal panel dissipate a total of 2 kW. A blower inside the panel keeps the inside temperature at a uniform level throughout. The panel run hotter or cooler? an area of 100 3-27 67 high, and 2 Assuming tion ft is 4 ft wide, 8 ft deep, and totally closed. that heat is radiated by convec- and radiation from all sides except the bottom, estimate the temperature inside the panel if the ambient temperature The panel enamel. is is 30 °C. painted with a nonmetallic Part Two Electrical Machines and Transformers Chapter 4 Direct-Current Generators This Introduction 4.0 We begin our study of rotating machinery with the direct-current generator. generators are not as common because direct current, when are discussed. as they used to be, required, is mainly without current any using moving important because tion to the and it is 4.1 motor and vice versa. Owing similar construction, the reason fundamental properties of erator after we We show when operates meant by the neutral fine what how the induced voltage termines it at is its is point. it is (ac) generator. The that the voltage generated in any dc gen- inherently alternating and only becomes dc is has been rectified by the commutator. 1 begin with the basic principles the importance of brush position the study of a direct- Fig. 4. shows an elementary ac generator composed of a coil that revolves at 60 r/min between the N, S poles of a permanent magnet. The applied to a dc motor. chapter may seem, edge of the alternating-current to their iearn about a dc generator can be di- of a 2-pole generator it current (dc) generator has to begin with a knowl- generators and motors are identical. Consequently, In this Generating an ac voltage Irrelevant as built same way; consequently, any dc generator can we actual including multipole designs. many dc Commercial dc generators and motors are rectly their voltage-regulation characteristics. physical construction of direct-current machines, brief periods. anything next. of dc generators represents a logical introduc- behavior of dc motors. Indeed, operate as a commutating poles and The chapter ends with a description of the industry actually operate as generators for in for We then discuss the major types parts. Nevertheless, an understanding of dc generators The need problem of pole-tip saturation are covered the produced by electronic rectifiers. These rectifiers the followed by a study of the behavior of the current flow, and the importance of armature reaction Direct-current can convert the current of an ac system into direct motors is generator under load. Mechanical torque, direction of is due to an external driving force, such motor (not shown). The coil is connected to two slip rings mounted on the shaft. The slip rings are connected to an external load by means of two no-load. rotation and de- as a We show generated and what de- stationary brushes x and magnitude. 71 y. 1 ELECTRICA L MA CHINES AND TRANSFORMERS 72 rotation (Fig. 4.2). 60 r/min The waveshape depends upon We shape of the N, S poles. designed The assume the were the poles generate the sinusoidal wave shown. to coil in our example revolves uniform at speed, therefore each angle of rotation corresponds to makes a specific interval of time. Because the coil one turn per second, the angle of 360° 4.2 cor- in Fig. responds to an interval of one second. Consequently, we can also represent the induced voltage as a func- tion of time (Fig. 4.3). v 1 +20 cycle - N \ F Al> c an elementary ac generator of 1 revolution per second. \ V y_ \ \ \ / \ / \ f \ / \ turning at \ / \ 1 \ / / \ Figure 4.1 Schematic diagram \ / \ \ / \ \ / \ s 1 As the coil rotates, a voltage A is induced (Eq. 2-25) between its between the brushes and, therefore, across the load. terminals The voltage is time 1.25 / / and D. This voltage appears / - 1 cycle generated because the conductors of the coil cut across the flux produced by the N, S poles. The induced voltage (20 V, say) when the coil is zontal position, as shown. coil is momentarily quently the voltage is No maximum therefore momentarily flux is in the hori- cut when in the vertical position; at these instants feature of the voltage is that its is zero. Figure 4.3 Voltage induced as a function of time. the conse- Another polarity changes 4.2 Direct-current generator If the brushes in Fig. 4.1 could be switched from every time the coil makes half a turn. The voltage can one therefore be represented as a function of the angle of about to change, slip ring to the other every time the polarity was we would obtain a voltage of con- stant polarity across the load. V Brush x would always be positive and brush y negative. We can obtain this result by using a commutator (Fig. 4.4), A commu- + 20 tator in that is its simplest form composed of is cut in half, with each a slip ring segment insulated from t-'.A I) the other as well as \ 180 - 360 degrees 450 angle connected from to coil-end the shaft. The commutator revolves with age between the segments tionary brushes x and One segment A and the other to coil-end is the coil and the volt- picked up by two The voltage between brushes nating voltage in the coil Figure 4.2 in the ac generator as a function of the angle of rotation. sta- y. x and y pulsates but never changes polarity (Fig. 4.5). Voltage induced is D. is rectified The alter- by the com- mutator, which acts as a mechanical reversing switch. DIRECT-CURRENT GENERATORS Due 60 r/min to the constant polarity 73 between the brushes, Hows in The machine represented in Fig. the current in the external load always the same 4.4 is direction. called a direct-current generator, or 4.3 Difference dynamo. between ac and dc generators The elementary ac and dc generators in Figs. 4.1 and 4.4 are essentially and an ac voltage chines only differ same way. built the between the poles of case, a coil rotates is induced in the way We commutator Elementary dc generator machines which carry both is simply an ac generator rectifier called a commu- (Fig. 4.6a). tator (Fig. 4.6c). generators carry while dc generators require a Figure 4.4 equipped with a mechanical each magnet The ma- the coil. in In the coils are connected to the external circuit (Fig. 4.6): ac slip rings (Fig 4.6b) a sometimes build small slip rings and a commu- Such machines can function si- multaneously as ac and dc generators. tator. 4.4 Improving the Returning +20 to the waveshape dc generator, we can improve the pulsating dc voltage by using four coils and four segments, as shown shape but it is given never The resulting waveThe voltage still pulsates in Fig. 4,7. in Fig. 4.8. falls to zero; it is much closer to a steady dc voltage. By increasing the number of coils and segments, we can obtain a dc voltage that is very smooth. 0 0 90 180 270 360 degrees Modern dc generators produce The angle 9 ripple of less than 5 percent. Figure 4.5 voltages having a coils are lodged in the slots of a laminated iron cylinder. The elementary dc generator produces a pulsating dc the cylinder constitute the voltage. The coils The percent ripple is the ratio of the RMS value of the ac component of voltage to the dc component, expressed (b) (a) in percent. ( C ) Figure 4.6 The three armatures rings or and armature of the machine. (a), (b), and (c) have identical windings. a commutator), an ac or dc voltage is obtained. Depending upon how they are connected (to slip ELECTRICAL MACHINES AND TRANSFORMERS 74 rotation rotation A Figure 4.7 Schematic diagram of a dc generator having 4 and 4 commutator bars. See Fig. 4.9. Figure 4.9 coils The actual physical construction of the generator shown in Fig. 4.7. The armature has 4 slots, 4 coils, and 4 commutator bars. v A schematic diagram such as Fig. 4.7 where the tips, and so on. But we must remember coil sides (a h a 2 ; b,, b2 ally located at 180° to side as Figure 4.7 ol 0 I I 90 180 I I |_ I 360 270 degrees *0 Figure 4.8 The voltage between the brushes than in Fig. that 4.5. one of Fig. 4.7, because we in will be using similar drawings of dc machines. The four coils the figure are identical to the coil At the instant shown, coil A is shown in Fig. 4. 1 not cutting any flux The reason is that the coil sides of these two coils are midway between the poles. On the other hand, coils B and D are cutting flux coming and neither is coil C. from the center of the N and S poles. Consequently, the voltage induced in these coils possible value (20 V, say). That is at its is maximum also the voltage across the brushes at this particular instant. each other and not side by seems to indicate. coil side is at the is at a, is in important to understand the physical meaning to explain the behavior that the of each coil are actu- etc.) The actual construction of this armature is shown in Fig. 4.9. The four coils are placed in four slots. Each coil has two coil sides, and so there are two coil sides per slot. Thus, each slot contains the conductors of two coils. other It is ; For reasons of symmetry, the coils are wound so more uniform is us between the poles, under the poles, near the cated: pole tells coil sides of the individual coils are lo- the top of slot bottom of bottom of a slot 3. The 1, while coil side a 2 coil the is in connections to the com- mutator segments are easy to follow armature. and the slot the top. For example, in Fig. 4.7 coil side The reader should compare in this simple these connec- tions with those in Fig. 4.9 to verify that they are the same. Note also the actual position and schematic position of the brushes with respect to the poles. the position of the coils when moved the armature has through 45°. The sides a,, A are now sweeping past pole tip and a 2 of coil Fig. 4.10 shows 1 pole same tip 4. The sides of coil C are experiencing the flux because they are in the A. Consequently, the voltage e.A same slots as coil induced in coil A is DIRECT-CURRENT GENERATORS rotation 15 rotation Figure 4.10 when Position of the coils rotated the armature of Fig. 4.9 has through 45°. Figure 4.11a Physical construction of a dc generator having 12 exactly the same as the voltage e c induced in coil C. A C while coil coils, 12 slots, and 12 commutator moving downward, moving upward. The polarities of e a however, that coil Note, is bars. is rotation and ec are, therefore, opposite as shown. The same reasoning leads us and? are equal and opposite Ll + + + = conclude that e h to in polarity. This means at all times. Consequently, no current will flow in the closed loop formed by the that e.d eh ec This four coils. is circulating current ed 0 most fortunate, because any such would produce 2 I R The voltage between the brushes c c (or e a + c\\) at the instant minimum to the shown. shown voltage losses. is equal to e b It in Fig. 4.8. The armature winding we have just discussed called a lap winding. the It is + corresponds most common is type of winding used in direct-current generators and motors. 4.5 Figure 4.11b Schematic diagram Induced voltage induced Figures 4. 1 and la 4. 1 show lb a more in When the armature rotates, the voltage E induced each conductor depends upon the flux density which it cuts. This fact is E= Because the density point to point, coil depends based upon the equation Blv in the air (2.25) gap varies from the value of the induced voltage per upon its of the armature and the voltages the 12 coils. realistic ar- mature having 12 coils and 12 slots instead of only 4. in instantaneous position. Consider, for example, the voltages induced armature when Fig. 4.11. it in the occupies the position shown The conductors in slots in and 7 are ex- 1 actly between the poles, where the flux density is zero. The voltage induced coils lodged in slots 1 and 7 is, in the therefore, zero. two On the other hand, the conductors in slots 4 and 10 are directly under the center of the poles, greatest. where the The voltage induced in flux density the two is coils ELECTRICAL MACHINES AND TRANSFORMERS 76 lodged in Finally, due duced in as that slots induced in the coils lodged and in slots 5 0V same the is rotation in- 1 coil A / 1 shows the instantaneous voltage ineach of the 2 coils of the armature. They 18, and 20 V, respectively. Note that the in 4. 1 lb 1 are 0, 7, brushes short-circuit the coils is maximum. therefore, is, magnetic symmetry, the voltage to the coils lodged in slots 3 and 9 Figure duced these which the voltage in momentarily zero. Taking polarities into account, we can see that between the brushes the voltage 18 + to brush 7) = 70 and brush x V, is + (7 is 18 + 20 + positive with respect y. This voltage remains essentially con- stant as the armature rotates, because the number of coils between the brushes always the same, is irre- spective of armature position. Note 4.11b straddles two that brush x in Fig. commutator segments Consequently, the that are brush connected However, since the induced voltage momentarily zero, no current brush. sitioned on the to brush If That we were the case in Figs. 4. is (0 + 7 + 1 1 is a momen- and 4. 1 1 b. 18 + 20 + 18) = 63 continually short-circuit coils that generate 7 V. Large currents will flow in the short-circuited coils and brushes, and sparking with coils that are momentarily age between the brushes and the at same time sparking occurs, there is poor commutation. a neutral zone. 4.7 Value of the induced voltage The voltage induced winding is in a dc generator having a lap given by the equation E0 = E0 = Z= n = (J> = Z/?*/60 (4J) voltage between the brushes [V] total number of conductors on the armature speed of rotation [r/min] tlux per pole |Wb] This important equation shows that for a given generator the voltage only holds true is zones Neutral zones are those places on the surface of the armature where the tlux density is zero. When the generator operates at no-load, the neutral zones are located exactly between the poles. No voltage is in- is directly proportional to the flux per pole and to the speed of rotation. tion. If the 4.6 Neutral in We in contact will result. Thus, shifting the brushes off the neutral position reduces the volt- When through the neutral zone. brushes so they are where V. decreases. Furthermore, in this position, the brushes said to be try to set the between the brushes would be- Thus, by shifting the brushes the output voltage causes sparking. in a coil that cuts always brush yoke by 30° (Fig. to shift the 4.12), the voltage come which y, The brushes are neutral position when they are pocommutator so as to short-circuit duced coil B. those coils in which the induced voltage tarily zero. A. coil in this coil is flow through the will The same remarks apply momentarily short-circuits said to be in the to coil A. short-circuits Figure 4.12 Moving the brushes off the neutral point reduces the output voltage and produces sparking. if the brushes are Example 4-1 The armature of a slots. Each Wb. equation brushes are shifted off neutral, the equivalent to reducing the 0.04 The on the neutral posieffect number of conductors Z _ 6-pole, coil has 600 r/min generator, has 90 4 turns and the tlux per pole is Calculate the value of the induced voltage. DIRECT-CURRENT GENERATORS The current delivered by Solution Each turn corresponds to two conductors on the armature, and The coils arc required to fill the number of armature conductors total Z = 90 90 coils X 4 turns/coil X 90 slots. flows through 2 conductors/turn The speed we in the same discover direction in The same is N mo- true for conductors that are mentarily under a S pole. However, the currents unis n = 600 der the r/min N pole flow the opposite direction to in those under a S pole. Referring to Fig. Consequently, En = Z//4V60 = 720 X 600 X = 288 V The voltage between the brushes fore always flows we would those conductors that are momentarily under a pole. = 720 generator also the the armature conductors. If could look inside the machine, that current is all 77 at 4. 13, the ar- mature conductors under the S pole carry currents 0.04/60 that flow into the page, away from the reader. Conversely, the armature currents under the no-load is there- N pole flow out of the page, toward the reader. Because the conductors 288 V, provided the brushes are on neutral. lie in a magnetic field, they are subjected to a force, according to Lorentz's 4.8 Generator under load: the energy conversion process When a direct-current generator is under load, some fundamental flux and current relationships take place that are directly related to the mechanical-electrical energy conversion process. Consider for example, a 2-pole generator that is while delivering current / driven counterclockwise to a load (Fig. 4. 1 3). law (sections 2.22 and 2.23). find that the individual forces all we examine F on the di- we act clockwise. In effect, they the conductors produce a torque which the gen- that acts opposite to the direction in erator is we must being driven. To keep the generator going, exert a torque on the shaft to overcome this opposing electromagnetic torque. The resulting mechanical which rotation If rection of current flow and the direction of flux, how is power is converted into electrical power, delivered to the generator load. That is the energy conversion process takes place. torque due to F 4.9 Armature reaction Until now, we have assumed motive force (mmf) acting to the field. in a that the only dc generator However, the current flowing ture coils also creates a powerful that distorts we we and field weakening takes place reaction. the impact of the armature mmf. return to the generator under load (Fig. 4.13). If consider the armature alone, magnetic chanical torque. magnetomotive force mmf is called armature To understand Figure 4.13 The energy conversion process. The electromagnetic torque due to Fmust be balanced by the applied me- due arma- both motors and generators. The effect produced by the armature load that in the and weakens the flux coming from the poles. This distortion in magnetois at right field as shown it in Fig. 4.14. will produce a This field acts angles to the field produced by the N, S poles. The intensity of the armature flux depends upon its mmf, which in turn depends upon the current carried by the armature. Thus, contrary armature flux is to the field flux, the not constant but varies with the load. ELECTRICAL MACHINES AND TRANSFORMERS 78 armature flux rotation neutra zone , neutral zone Figure 4.14 Magnetic field produced by the current flowing armature conductors. We can in the immediately foresee a problem which the armature flux will produce. Fig. zone flux in the neutral is shows 4. 14 that the Figure 4.15 Armature reaction S poles. may sparking by the brushes. As will The occur. depend upon a result, severe intensity of the sparking the armature flux the load current delivered that is it distorts the flux mmf and field whose shape is mmf in the direction The all The neutral still another effect: the higher flux density in pole tips 2. 3 causes saturation to set in. Consequently, the increase in flux decrease in flux under pole tips 2, 3 is less than the under pole tips 1, 4. As a result, the total flux pro- duced by the N, S poles less than may be 1 as . was when induced voltage given as 10 percent. in flux under load, we could zone when the gener- move the brushes to re- For generators, the brushes are shifted to the new zone by moving them tation. For motors, the brushes are shifted against in the direction of ro- the direction of rotation. As soon However, We as the brushes are improves, tion forth it For large machines, the decrease much ro- neutral no-load. This causes a cor- at in the does not duce the sparking. rises responding reduction 4. to the shift in the neutral is the is generator was running by Eq. Due ator of rotation of the dc generators. flux distortion produces it improve commutation to produced by the illustrated in Fig. 4.15. zones have shifted in space; 4.10 Shifting the brushes produces a magnetic field armature. This occurs in important to note that the orientation of the with the armature. tate the armature combination of the armature poles. In effect, the It is armature flux remains fixed and hence upon by the generator. The second problem created by mmf N, no longer zero and, con- sequently, a voltage will be induced in the coils that are short-circuited produced by the distorts the field and if meaning moved, the commuta- there is less sparking. the load fluctuates, the armature falls and so the neutral zone shifts mmf back and between the no-load and full-load positions. would therefore have to move the brushes back and forth to cedure not practical and other is obtain sparkless commutation. This pro- means are used to DIRECT- CURRENT GENERA TORS resolve the problem. ever, the to For small dc machines, how- brushes are set ensure reasonably in an intermediate position good commutation at all loads. mmf of the the commutating poles is greater than the armature mmf. This flux in the neutral zone, which aids made 19 slightly creates a small the commuta- tion process (see Section 4.28). Commutating poles 4.11 Fig. 4. 16 shows how the commutating poles of a 2-pole machine are connected. Clearly, the direc- To counter the of armature effect reaction in medium- and large-power dc machines, we always place a set of commutating poles* between the main These narrow poles carry wind- poles (Fig. 4. 16). ings that are connected with the armature. in series The number of turns on the windings that the equal and opposite mmfa of the armature. the designed so As exactly bucking each other ing the armature mmf flowing through the windings in- opposite to the mmf of the commutating poles acts mmf of the armature and, therefore, neutralizes effect. is its restricted to the However, the neutralization narrow brush zone where com- mutation takes place. The distorted flux distribution under the main poles, unfortunately, remains the same. the load current varies, rise at all in this between the main poles we no longer have mmfc magnetomotive force the to two magnetomotive forces space is poles develop a magnetomotive force tion of the current dicates that the and fall times. together, By way, the flux is 4.12 Separately excited generator nullifyin the always zero and so to shift the brushes. In practice, Now that we have learned some basic facts about we can study the various types and dc generators, their properties. magnets Thus, instead of using permanent to create the magnetic field, we can use a shown pair of electromagnets, called field poles, as O (+) in Fig. 4. 17. When generator supplied by an independent source is the dc field current in such a (such as a storage battery or another generator, called an exciter), the generator is said to be sepa- rately excited. Thus, in Fig. 4. 7 the dc source con1 nected to terminals a and b causes an exciting current / x to flow. If the armature is or a diesel engine, a voltage £0 brush terminals x and driven by a motor appears between y. O (-) Figure 4.16 Commutating poles produce an the mmf, of the mmf c that opposes armature. Commutating poles are sometimes called mierpoles. Figure 4.17 Separately excited 2-pole generator. The N, S field poles are created by the current flowing field windings. in the 1 ELECTRICAL MACHINES AND TRANSFORMERS 80 How does the saturation curve relate to the induced £n ? If we drive the generator at constant speed, 4.13 No-load operation voltage and saturation curve E0 is directly proportional to the flux When a separately excited dc generator runs at no- load (armature circuit open), a change the excit- in ing current causes a corresponding change in the in- We now examine duced voltage. whose shape 4. 8a. 1 which increases the If we f|> plot as a function of I X1 per pole. is 1 is mmf of the obtain the satu- ration curve of Fig. 4. 8a. This curve whether or not the venerator obtained turning. identical to the saturation curve of Fig. result is shown in Fig. 4. 1 8b; rated voltage of a dc generator above the knee of the curve. usually a is In Fig. 4. example, the rated (or nominal) voltage varying the exciting current, called the it is we can is 1 1 8b. for 20 V. By vary the in- duced voltage as we please. Furthermore, by re- versing the current, the flux will reverse and so, too, will the polarity of the induced voltage. Induced voltage vs speed. For a given exciting current, the induced voltage increases in direct pro- rated flux portion to the speed, a result that follows from Eq. 4. A . obtain a curve little flux we Consequently, we Let us gradually so that the , is , no-load saturation curve of the generator. The raise the exciting current / x field increases, The as a function of 7 X the relationship between the two. Field flux vs exciting current. Ea by plotting If we of the induced voltage also reverses. However, ity we 1 reverse the direction of rotation, the polar- reverse both the exciting current and if the direc- tion of rotation, the polarity of the induced voltage remains the same. Figure 4.18a Flux per pole versus exciting current. When the flux the exciting current relatively small, is small and the iron in the machine is saturated. Very little mmf is needed un- is to establish the flux through the iron, with the result that the mmf developed by the field coils is available to drive the flux through the air gap. Because the permeability of air 3 A almost entirely is constant, the Figure 4.18b Saturation curve of a dc generator. flux increases in direct proportion to the exciting current, as shown by the linear portion 0a of the 4.14 Shunt generator saturation curve. However, as we continue current, the iron in the field to saturate. A to raise the exciting and the armature begins large increase in the quired to produce a small increase mmf is now in flux, as by portion be of the curve. The machine to is re- shown now said be saturated. Saturation of the iron begins to be important when we reach the so-called "knee" ab of the saturation curve. A shunt-excited generator shunt-field winding is is connected a machine whose in parallel with the armature terminals, so that the generator can be self-excited (Fig. 4. 19). this connection is that it The principal advantage of eliminates the need for an external source of excitation. How is generator self-excitation achieved? is When started up, a small voltage is a shunt induced in i DIRECT-CURRENT GENERATORS 81 field rheostat y Figure 4.20 Controlling the generator voltage with a field rheostat. rheostat is a resistor with an adjustable To understand how suppose that p is in tact Ea is 1 the output voltage varies, V when 20 movable contact the we move the conresistance R between the center of the rheostat. If toward extremity m, the points A sliding contact. { p and b diminishes, which causes the excit- ing current to increase. This increases the flux and, £tv On consequently, the induced voltage hand, if (b) we move the other R the contact toward extremity n, l increases, the exciting current diminishes, the flux Figure 4.19 b. diminishes, and so Self-excited shunt generator. a. Schematic diagram field is of one designed a shunt generator. to be connected in A We shunt shunt (alter- know points the armature, due to the remanent flux in the poles. produces a small exciting current This voltage the shunt field. The resulting small mmf acts / x in in the same direction as the remanent flux, causing the flux per pole to increase. which increases /x , The increased flux increases which increases more, which increases E0 the flux even more, and so This progressive buildup continues until a maximum E0 forth. reaches value determined by the field resistance and the degree of saturation. See next section. p and R E0 if we of the shunt field circuit between { We draw b. ing to the slope of a straight line correspond- R and superimpose on the it { uration curve (Fig. 4.2 sat- This dotted line passes 1). through the origin, and the point where it intersects the curve yields the induced voltage. Ea still will fall. the saturation curve of the generator and the total resistance nate term for parallel) with the armature winding. £0 can determine the no-load value of For example, 50 12 if the shunt field has a resistance of and the rheostat R = 50 12. { The line is m, then R must E = 50 V, / = pass set at extremity corresponding to { through the coordinate point I A. This line intersects the saturation curve where the voltage is 150 V (Fig. 4.21). That is the maximum voltage the shunt generator can produce. By changing 4.15 Controlling the voltage of a shunt generator to It is easy to control the induced voltage of a shunt- excited generator. rent We simply vary the exciting cur- by means of a rheostat connected in series with the shunt field (Fig. 4.20). the setting of the rheostat, the total resistance of the field circuit increases, causing decrease progressively. For example, creased to 120 12, A\ is E0 in- the resistance line cuts the satu- ration curve at a voltage we continue if Ea of 120 V. a critical value will be reached where the slope of the resistance line is If to raise A*,, ELECTRICAL MACHINES AND TRANSFORMERS 82 Figure 4.22 Equivalent circuit of a dc generator. revolving conductors. Terminals l , 2 are the external armature terminals of the machine, and F,, F 2 are the field now winding terminals. Using study the more common we this circuit, will types of direct-current generators and their behavior under load. 4.17 Separately excited generator under load Figure 4.21 The no-load voltage depends upon the resistance the shunt-field of Let us consider a separately excited generator that circuit. driven equal to that of the saturation curve region. When unsaturated in its this resistance is attained, the induced voltage suddenly drops to zero and will remain so for any R v greater than this critical value. In Fig. 4.2 the critical resistance corresponds to 200 at a battery (Fig. 4.23). and so is The When load, terminal voltage E0 resistance whose field the E {2 zero. is constant machine operates is However, is excited by The induced voltage £0 is at no- equal to the induced because the voltage drop is is exciting current the resultant flux. therefore fixed. voltage il. constant speed and if in the armature we connect a load 4.16 Equivalent circuit across the armature (Fig. 4.23), the resulting load We R Kr Terminal voltage E l2 is now less than the induced voltage £ As we increase the load, the terminal current / produces a voltage drop across resistance set have seen that the of identical coils, all armature winding contains a of which possess a certain re- () sistance. which The exists machine the armature resistance R^ total between the armature terminals when is stationary. It is measured on the com- mutator surface between those segments that der the ally ( + ) that is and ( — ) The brushes. resistance lie is . voltage diminishes progressively, as 4.24. shown as a function of cun e of the generator. load current is called the load y un- usu- very small, often less than one-hundredth of an ohm. Its value depends mainly upon the voltage of the generator. circuit, we can represent one of the brushes. If To simplify machine has resistance of these windings The equivalent circuit posed of a resistance R„ (Fig. 4.22). The is in series latter is the with interpoles, the included of a generator in series power and the generator R 0 as if it were the in R ir is thus com- with a voltage voltage induced Ea in the in Fig. The graph of terminal voltage Figure 4.23 Separately excited generator under load. DIRECT-CURRENT GENERATORS V tions 100 produce corresponding changes in 83 the genera- tor terminal voltage, causing the lights to flicker. Compound generators eliminate this problem. A compound generator (Fig. 4.25a) is similar to a shunt generator, field coils connected except that in series has additional it with the armature. These series field coils are composed of a few turns of heavy wire, big enough to carry the armature cur0 10 5 rent. A The 4.25b showing the shunt and Figure 4.24 of the series coils total resistance fore, small. Figure When In practice, ration the induced voltage Ev also decreases with increasing load, because pole-tip satu- slightly tends to decrease the field flux. Consequently, voltage the terminal E l2 falls off more rapidly than can be attributed to armature resistance alone. the series coils series field connections. As the generator off more sharply with increasing load than that of a separately excited generator. field The reason current in a separately excited constant, whereas citing current falls in a self-excited load to full-load is drop coils acts in the is said to 15% and 10%, Compound its raises the value of original Ea . If practically constant from no-load it respectively. generator prevent voltage of a dc generator from de- we can usually tolerate a reasonable drop in terminal voltage as the load increases, this has a serious effect on lighting circuits. For example, the distribu- tion system of a ship supplies power to both dc ma- chinery and incandescent lamps. The current mmf no-load value, properly designed, the terminal voltage remains regulation creasing with increasing load. Thus, although flows developed the series coils are voltage from no- to now direction as the machine remains The voltage however, Consequently, the field flux un- generator the ex- The compound generator was developed the terminal same /c mmf The that the whereas for a separately excited generator usually less than 10 percent. be field. coils. about 15 percent of the full-load is 4.19 in coils, loaded, the terminal volt- as the terminal voltage drops. For a self-excited generator, the voltage, is is through the series field der load rises above The terminal voltage of a self-excited shunt generator The shunt zero. age tends to drop, but load current which falls is flux, just as in a standard self-excited shunt generator. by these Shunt generator under load there- diagram carry exciting current / x which produces the field of the shunt 4.18 is, the generator runs at no-load, the current Load characteristic of a separately excited generator. in a schematic is Figure 4.25 delivered by the generator fluctuates continually, in a. Compound response to the varying loads. These current varia- b. Schematic diagram. generator under load. to full-load. The ELECTRICAL MACHINES AND TRANSFORMERS 84 rise in the induced voltage compensates for the ar- Load characteristics 4.21 mature IR drop. In the some cases we have to compensate not only for armature voltage drop, but also for the IR drop in between the generator and the load. the feeder line The generator manufacturer then adds one or two extra turns on winding so the series that the terminal ma- voltage increases as the load current rises. Such chines are called over-compound generators. compounding placed is too strong, a low resistance can be with the series in parallel the current in the series field as reducing the If the field. This reduces The load characteristics of some shunt and compound generators are given in Fig. 4.26. The voltage of an over-compound generator increases by 1 value of the diverter resistance is if the equal to that of the series field, the current in the latter is reduced by half. compound flat-compound generator remains constant. tor 15 percent is compound generator the mmf of the As a re- terminal voltage falls drastically with in- We can make such a generator by simply reversing the series field of a standard com- pound generator. Differential were formerly used in below its the is 30 percent lower. 4.22 Generator specifications The nameplate of a generator indicates the power, voltage, speed, and other details about the machine. ratings, or nominal characteristics, are the following information compound nameplate of a 100 Power 100 Voltage 250 20 generators kW V A 50° C Exciting current Temperature rise These specifications tell dc arc welders, because they tended to limit the short-circuit current and to stabi- punched on is the kW generator: Speed 1200r/min Type Compound B Class us that the machine can power of 100 deliver, continuously, a lize the arc On no-load value, while that of a differential-compound generator ple, the series field acts opposite to the shunt field. creasing load. applied, whereas that of a values guaranteed by the manufacturer. For exam- generator sult, the is other hand, the full-load voltage of a shunt genera- These In a differential full-load and has the same effect number of turns. For example, 4.20 Differential when percent kW at a volt- age of 250 V, without exceeding a temperature rise during the welding process. The voltage regulation of the differential compound generator in Fig. 4.26 is (no-load — full00-70)770 = 42.9%. load )/full- load = ( of 50°C. can therefore supply a load current of It 100 000/250 = 400 A. and the current in the It possesses a series winding, shunt field is 20 A. In practice, 1 the terminal voltage its % 100 rating of overcom pound power from compound ceed 1 separate excitation class B used in the 80 shunt 60 differential 00 250 V. is adjusted to a value close to We may draw any amount of the generator, as long as kW and the current is it does not ex- 400 A. The less than designation refers to the class of insulation machine. compound 40 CONSTRUCTION OF 20 DIRECT-CURRENT GENERATORS We have 0- 50 100 described the basic features and properties % of direct-current generators. We now look at the Load current mechanical construction of these machines, direct- Figure 4.26 ing our attention to the field, the armature, the Typical load characteristics of dc generators. mutator, and the brushes. com- DIRECT-CURRENT GENERATORS 85 4.23 Field The field chine. It produces the magnetic flux is composed of side in the ma- basically a stationary electromagnet a set of salient poles bolted to the in- of a circular frame (Figs. 4.27 and 4.28). Field mounted on the poles, carry the dc exciting coils, current. The frame is usually stacked iron laminations. In solid cast our discussions so far 2-pole generators. ator or poles. some generators the created by permanent magnets. flux is In made of whereas the pole pieces are composed of steel, we have considered only However, motor may have in practice a 2, 4, 6, or as many The number of poles depends upon flux dc generas 24 Figure 4.29 Adjacent poles of multipole generators have opposite magnetic polarities. ical size of the machine; the bigger the phys- poles frame will have. it By using is, it the more a multipole design, we can reduce the dimensions and cost of large mafield chines, and also The improve field coils their performance. of a multipole machine are con- nected together so that adjacent poles have oppo- commutator site magnetic polarities (Fig. 4.29). The shunt composed of coils are field several hundred turns of wire carrying a relatively small current. The coils armature are insulated from the pole pieces to prevent short- circuits. Figure 4.27 mmf The Cross section of a 2-pole generator. developed by the coils produces a magnetic flux that passes through the pole pieces, the frame, the armature, is the short space pieces. It and the air gap. The air gap between the armature and the pole ranges from about 1 erator rating increases from 1 Because the armature and .5 to kW mm as the gen- 5 to 100 kW. field are composed of magnetic materials having excellent permeability, most of the mmf produced by the field is used to drive the flux across the air gap. Consequently, by reducing its shunt field made fect length, coils. we can becomes too must be enough so Figure 4.28 Cutaway view does not overheat when It has 3 field, the coils are top of the shunt-field coils. tor size a 4-pole shunt generator. gap cannot be great. generator has a series wound on of air too short otherwise the armature reaction ef- If the brushes per brush set. diminish the size of the However, the large rent of the generator. it The conduc- that the winding carries the full-load cur- ELECTRICAL MACHINES AND TRANSFORMERS 86 4.24 Armature The armature the rotating part of a dc generator. is It consists of a commutator, an iron core, and a set of The armature coils (Fig. 4.30). keyed is revolves between the field poles. composed of slotted, to form a to a shaft The and iron core is iron laminations that are stacked solid cylindrical core. The laminations are individually coated with an insulating film so that they do not As come in electrical contact with each other. The a result, eddy-current losses are reduced. are lined up to provide the space needed slots to insert the armature conductors. The armature conductors carry the load current They are insulated from the iron core by several layers of paper or mica and delivered by the generator. are firmly held in place by fiber slot sticks. If the ar- mature current is below 10 A, round wire is Figure 4.31 Armature lamination with tapered but for currents exceeding 20 A, rectangular con- ductors are preferred because they make slots. used; iron teeth fiber slot stick better use of the available slot space. The lamination of a small armature tion view of the is shown slot in Fig. 4.31. A of a large armature cross sec- is shown in Fig. 4.32. Figure 4.32 Cross-section of a slot containing 4 conductors. 4.25 Commutator and brushes The commutator is composed of an assembly of ta- pered copper segments insulated from each other by mica sheets, and mounted on the shaft of the ma- chine (Fig. 4.33). The armature conductors are con- nected to the commutator in a manner we will ex- plain in Section 4.26. Great care Figure 4.30 Armature of a dc generator showing the commutator, stacked laminations, slots, and shaft. (Courtesy of General Electric Company, USA) is taken in building the commutator because any eccentricity will cause the brushes bounce, producing unacceptable sparking. to The sparks burn the brushes and overheat and carbonize the commutator. DIRECT-CURRENT GENERATORS yoke that permits the entire brush tated through an angle position. In going 87 assembly to be and then locked ro- in the neutral around the commutator, the suc- cessive brush sets have positive and negative polarities. Brushes having the same polarity are connected together and the leads are brought out to one positive and one negative terminal (Fig. 4.34b). The brushes are made of carbon because it has good electrical conductivity and its softness does not score the commutator. To improve the conductivity, a small amount of copper is sometimes mixed with the carbon. The brush pressure is set by means of adjustable springs. friction tator A of is too great, the and brushes; on the other hand, if commuit is too a dc machine. 2-pole generator has two brushes fixed dia- metrically opposite to slide pressure weak, the imperfect contact may produce sparking. Figure 4.33 Commutator If the produces excessive heating of the each other (Fig. 4.34a). They on the commutator and ensure good electrical contact between the revolving armature and the sta- tionary external load. Multipole machines possess as they have poles. The brush of one or that many brush sets, in turn, are more brushes, depending upon sets as composed the current has to be earned. In Fig. 4.35c, for example, brushes mounted side-by-side make up the brush The brush sets are spaced at two set. equal intervals around the commutator. They are supported by a movable brush + (a) 9 6 (b) Figure 4.34 a. Brushes of a 2-pole generator. b. Brushes and connections of a 6-pole generator. Figure 4.35 Carbon brush and ultraflexible copper lead. b. Brush holder and spring to exert pressure. c. Brush set composed of two brushes, mounted on a. rocker arm. (Courtesy of General Electric Company, USA) 88 ELECTRICAL MACHINES AND TRANSFORMERS The pressure is usually about 15 kPa 2 lb/in"), and the permissible current density mately 10 A/cm 2 65 A/in is In order to get a better 2 ). Thus, atypical brush cm X cm of 4.5 N having a cross section of 3 1.2 in 1 X lb) and can the construction of a modern 0.4 in) exerts a pressure 1 carry a current of about 30 A. Fig. 4.36 shows 4-pole dc generator. In 4.26 Details of a multipole generator approxigenerators, was 1 2- the schematic diagram is of such a machine having 72 slots on the armature, 72 segments on the commutator, and 72 coils. The armature has a lap winding, and the reader should note erator that understanding of multipole us examine the construction of a pole machine. Fig. 4.38a order to appreciate the progress that has been made. Fig. 4.37 shows a gen- let how similar it is 2-pole machine (Fig. built in 1889. Figure 4.36 Sectional view of a 100 kW, 250 V, 1750 r/min 4-pole dc generator. (Courtesy of General Electric Company, USA) to the 4. 1 1 schematic diagram of a b). Coils A and C are mo- DIRECT-CURRENT GENERATORS 89 The voltage generated between brushes x and y sum of the equal to the connected coils commutator segments to brush sets are similarly generated by five ( form the + + -2, 2-3, ) ( coils. brush sets are connected together The terminal. ) ( — connected to form the larly I and 5-6. The voltages between the other 3-4, 4-5, The is voltages generated by the five to brush sets are simi- ) — ( terminal. These ) connections are not shown on the diagram. For similar reasons of clarity, that are placed Fig. diagram. Coil is B stalled in 1889 250 A at a voltage of 1 1 0 It deliv- Other prop- V. pioneering machine include the following: Speed Total first in- to light the streets of Montreal. ered a current of erties of this Thompson generator was 1 300 r/min 2390 kg weight mm 330 mm 292 Armature diameter Stator internal diameter Number of commutator Only A has its coil sides in slots B are in slots is connected the three 4 and 10. I and 7, while Furthermore, coil is poles. segments 3 and B The voltage I, 4. A are between the poles. Consequently, induced the coil sides of to shown, the coil-sides of coil the neutral zone no voltage Figure 4.37 y. C are shown so as not to complicate the In the position This direct-current the interpoles connected to commutator segments 72 and while coil in show between brushes x and A, B, and those of coil A not 4.38b gives a detailed view of the armature coils lying coils we do between the N, S poles. A. in coil On the other hand, are directly under the in coil B is maximum N and S at this mo- ment. Consequently, the voltage between adjacent commutator segments 3 and 4 is maximum. The voltage in coil C is also zero because its coil sides are sweeping across the neutral zone. Note that the positive circuit coils and negative brushes each short- having zero induced voltage. 76 bars #4 Armature conductor size # Shunt field conductor size 14 Example 4-2 The generator in Fig. 4.38 generates 240 V between adjacent brushes and delivers a current of 2400 A modern generator having the same power and speed weighs 7 times less and occupies only 1/3 the floor Calculate space. a. b. mentarily in the neutral zone, while coil coming from the flux The coil width (known as coil pitch) the coil sides tween poles l. B and the center of pole of coil is cutting A is such that coming from adjacent N, Thus, the coil sides of coil center of pole 2 B c. The current delivered per brush set The current flowing in each coil The average voltage induced per coil the center of the poles. the coil sides cut the flux S poles. A to the load. lie under the 3. Similarly, Solution a. A current of 2400 A flows out of the + ( and back into the ( There are 12 brush The — ) sets, 2. 3. / terminal 6 positive and 6 negative. current per brush set is are in the neutral zones be- 2 and poles ) terminal of the generator. 2400/6 = 400 A 90 ELECTRICAL MACHINES AND TRANSFORMERS Figure 4.38b Closeup view of the armature coils between adjacent brushes. DIRECT-CURRENT GENERA TORS b. Each positive brush coils to the right / c. - 400/2 each in 200 coil is A 71 There are six coils between adjacent brush The average voltage per Eavge The 4.27 coil 240/6 a generator | { The commutation 72 I currents flowing the If 80 A, the shown carry all 40 versal takes place other. left. If In Fig. 1 , the current is how commutation 4.39a the brush and the 40 short distance, now is A from brush unite is The to U re- | (c) 80 di- we right A and output. 1 40 40 contact with segment 2, 1 . Owing the total current, 72 | 1 | only one-fourth of namely 0.25 X 80 V | 3 | 4040 7 (e) 80 between the only one-fourth of the total contact area, and so is I to the commutator is proportional to the conThe area in contact with segment 2 is from segment 2 40 ©| |© 71 while 75 per- contact resistance, the conductivity 40 40 V. and 25 percent of the brush surface contact with segment the current 3 | middle of seg- on the give the 80 in tact area. © ,2 | to the changes takes place, in the the coils in brush and N 40 40 4.39b the commutator has moved a Fig. 1 called commutation. brush produces a voltage drop of about is | j The contact resistance between the segment and cent 72 4.39a to 4.39e. the left of the In 40 40 during the millisecond interval that rection in this brief interval ment I means move from one end of the brush To understand | (b) © 71 on the right-hand side of the The process whereby refer to Figs. 3 I 80 0 40 A. current in these coils must reverse. a coil takes to © 2 1/ T moving from are 40 40 in Fig. brush will soon be on the left-hand side. This that the | flow toward and the the right coils commutator segments right to left, the coils 3 | 6020 the armature in that the currents in the coils the load current is 2 I © V | under load, the individual coils coming both from the brush, / 20 40 40 Note 1 = 40 V windings next to a positive brush are 4.39a. ® (a) on the armature carry one-half the load current carried by one brush. © 40 sets. 71 is 40 is process When 72 | 40 ideal 40 40 40 to the left of the brush. Consequently, the current = from the set gathers current and 9 = 20 A. By Figure 4.39 Commutation of the current in coil 1. are neglected and current reversal brush contact resistance. is Inductive effects caused by the ELECTRICA L MA CHINES A ND TRA NS FORMERS 92 same token, the brush is we now If the current X 80 - 60 0.75 from segment cover that the current flowing and 2 now are the ities means dis- must be 20 A. 1 A to dropped from 40 and the brush area is in coil X 1/10 a little in L = the currents are equal. This is zero Segment 2 left. is now still far- contact with 75 in A from from segment 2 and 20 If coil I again 20 A, but is the opposite direction to now understand how what it segments In Fig. plete slide e and the current in coil 2 is 1 is com- about to be reversed. it is important (amperes per square centimeter) remains the same at the brush face. Thus, the heat produced by the con- resistance is every point across spread uniformally across the brush surface. Unfortunately, such ideal commutanot possible in practical machines, and tion is now investigate the reason why. induced voltage [V] change of current [A/s] rate of LA//A/ = It we The practical X 6 10 X 5.75 + 40 - [ X V the presence of this induced voltage (attribut- is that flow in coil 1 coil is considered. when We ues for these currents 1, and the currents creasing or decreasing. tance does not come that that it takes place very short time; consequently, the current can- The reason is the armature coils have inductance and it it should. strongly opposes a rapid change in current. in Fig. 4.39 has 72 bars and that the armature turns at 600 r/min. in 1/10 One revolution that the is, therefore, The currents in Fig. 4.39. the middle of seg- is in in the coils are neither in- As a result, the coil induc- into play. 4.40b the current in coil 1 is changing due However, the in- value of 20 A. Suppose the coil current 35 A. From Kirchhoff s current law, the currents flowing from segments then respectively 75 20 A. Note completed of a second and during this short period 72 form over the brush touches segment into the brush are 5 A, instead of face. 2, 60 A and no longer uniis low where and high where it 1 In Fig. 4.40c the brush 1 is The density segment cally placed as regards rent in coil and 2 1 A and that the current density the brush touches commutator Suppose, for example, val- order to determine the re- 4.40a the brush In Fig. ment currents have assumed plausible in should be compared with those is not reverse as quickly as new the self-inductance of the sulting current flows in the brush. to its ideal is 40)] duced voltage e prevents the current from dropping process in a - ( 3 10 able to L), that opposes the change in current. In Fig. commutation The problem with commutation in- is to the contact resistance effect. 4.28 given by (4.2) Figs. 4.40a to 4.40e illustrate the in coil commutation process, is LM&t 1.39 over the brush. to note that the current density tact = 100 We the brush contact resis- 4.39e the current reversal In this ideal self-induction = seg- tance forces a progressive reversal of the current as the only inductance of the coil [H| Hows did before! it is has an inductance of, say, 100 fxH. the 1 Applying Kirchhoff" s current law, we find 1. that the current in coil can = duced voltage percent of the brush, and so the currents divide ac- A 1 at this instant. moved 4.39d the commutator has cordingly: 60 in = e 1 in coil which 1 the same. Consequently, the conductiv- same and so past the brush. Thus, the or 1.39 ms! s The voltage induced by A//A/ In Fig. 1/720 e contact with segments in that the current in coil ther to the ment - 1/72 20 A. moved 4.39c the commutator has In Fig. further, we contact with the brush, the cur- in rent in this coil has commutator bars sweep time available to reverse the current apply Kirchhoff s current law, Thus, by coming to the 1 A. is momentarily symmetri- segments 1 and 2. has not fallen to zero, and But the curis still, say. DIRECT-CURRENT GENERATORS 30 A. As a 40 40 40 40 40 while that sity © 71 72 | segment 2 2 3 | f still 35 40 2 / | 3 I I has left- moved beyond it in coil 1 the has has a value of 20 A, from segment to the brush is is The 1 resulting high current den- sity causes the brush to overheat 720 coils are being <t>) at the tip. Because commutated every second, overheating raises the brush 75 5 71 therefore, 7 40 © V | A despite the fact that the contact area getting very small. 72 1 Assuming not reversed. now 60 A, | is, midpoint of the brush and the current the current flowing 71 70 is 1 0 A. The current den- will tend to overheat. 4.40d segment In Fig. (a) 80 40 1 on the left-hand side of the brush hand side of the brush 4040 40 only is times greater than on the right-hand side. The vl /l | segment result, the current in in 93 tip to the this incandescent point and serious sparking will result. In 80 designing dc motors and generators, every is made to reduce the self-inductance of the One of the most effective ways is to reduce the number of turns per coil. But for a given output voltage, this means that the number of coils must be increased. And more coils implies more effort coils. 40 40 40 30 40 © 71 | 72 1^1/2 | 3 I 1 commutator 7010 (c) 80 bars. Thus, in practice, direct-current generators have a large number of mutator bars — not so much the output voltage but to coils and com- to reduce the ripple in overcome problem of the commutation. 20 40 40 40 40 Another important factor © 71 72 | J 1 | 2 7 is made 3 | that the (d) 80 zone. As © | 72 1 | x 2 , | the brush 3 I 4 40 4C (e) 80 tance opposes the reversal always is commutation induced is 1 .5 V. the brush . The of current. in the to the self- measures, the composition of carefully chosen. It affects the brush V to as much This drop occurs between the surface of and the commutator surface. A large brush drop helps commutation, but unfortunately 1 mmf. created in the neutral coil. creases the losses. of the current in coil is voltage drop, which can vary from 0.2 as Figure 4.40 Commutation is armature the which opposes the voltage due In addition to these 71 than this tlux, a voltage inductance of the 40 40 aiding commutation the coil side undergoing sweeps through coil 40 greater Therefore, a small tlux 60 20 40 slightly in commutating poles 4 | tht\ 40 mmf of the As a result, the it in- commutator and brushes become hotter and the efficiency of the coil induc- generator is slightly reduced. ELECTRICAL MACHINES AND TRANSFORMERS 94 Questions and Problems resistance the Practical level Sketch the main components of a dc gener- 4-1 ator. 4- 2 1 4-2 Why are the brushes of a de ways placed at machine al- 100 is <}, calculate the machine operates a. At no-load b. At full-load Fig. 4. 1 mmf when rated voltage at 8b shows the no-load saturation curve of a separately excited dc generator the neutral points? when it revolves 1500 r/min. Calculate at 4-3 Describe the construction of a commutator. the exciting current needed to generate 4-4 How 120 is the induced voltage of a separately excited dc generator affected a. the speed increases? b. the exciting current How 4-5 do we adjust 4- 3 if 1 is reduced? tion the voltage of a shunt is 4. 10, the induced voltage momentarily 18 V, in the posi- shown. Calculate the voltages induced C A, B, and same at the Referring to Fig. 4. 1 age induced A when decreases with increasing load. Explain. has rotated by 90°; by 120°. over- 4- 5 1 load and differential compound generators as to construction b. as to electrical properties 1 instant. lb, calculate the volt- the armature positive with respect to brush y 1 b. Show Does the polarity of each of the polarity reverse when a coil turns through 180°? 4-16 The generator of Fig. 4.38 revolves at r/min and the flux per pole is 20 960 mWb. Calculate the no-load armature voltage if each armature coil has 6 turns. Intermediate level separate excited dc generator turning at 4-17 How many a. 127 V. The armature resistance machine delivers is brush sets are needed for the gen- erator in Fig. 4.38? 1400 r/min produces an induced voltage of the is in coil the 12 coils. a. A Brush x in Fig. 4, Explain the difference between shunt, compound, 4-9 330 r/min. The terminal voltage of a shunt generator increases. 4-8 1 D in coils 4-14 Explain why the output voltage of an compound generator increases as the 4-7 at Referring to Fig. in coil generator? 4-6 V If the b. 2 12 and machine delivers a total load current of 1800 A. calculate the current flowing a current of 12 A. armature in each coil. Calculate Advanced a. The terminal voltage |V] b. c. The heat dissipated in the armature [W] The braking torque exerted by the armature A separately excited dc generator pro- 4-18 [Nm| 4-10 duces a no-load voltage of 115 V. segments 3 and 4 must be greater than 40 V? What happens if a. The speed is increased by 20 percent? b. The direction of rotation is reversed'.' c. The exciting current is increased by 10 percent? d. The polarity of the field is reversed? level The voltage between brushes x and y is 240 V in the generator shown in Fig. 4.38. Why can we say that the voltage between 4- 9 1 Referring to Fig. ity of £ xv when 4. 1 0, determine the polar- the armature turns counter- clockwise. 4-20 a. In Fig. 4.38 determine the polarity of tween commutator segments 3 and 4- 1 1 Each pole of a 1 00 kW, 250 V flat-compound generator has a shunt field of 2000 turns and a series field of 7 turns. If the total shunt-field ing that the armature b. At the same instant, is 4. £ 34 be- know- turning clockwise, what segment 35 with respect is to the polarity of segment 34? DIRECT-CURRENT GENERATORS 4-2] The armature shown 5.4 (Chapter 5) in Fig. has 81 slots, and the commutator has 243 segments. lap It will be winding having flux per field pole is wound / n du stria I 4-24 to give a 6-pole 30 mWb, The induced voltage at a calculate the is 60 ohms and 2 The I R loss in are 15 mm wide and commutator 4-22 A 200 W, 1 20 is V, that the diameter of the 1 4-23 dc generator has a winding on the armature. The rated armature current The The I total losses in the 2 R effi- is is 5 A. 0.023 pu. machine losses in the armature The generator in Problem 4-24 weighs 2600 lb. Calculate the output in watts per 4-26 In Problem 4-24 calculate The full-load current of the generator The current carried by the armature coils the torque re- quired to drive the generator (The shunt field is at 1750 r/min. powered by a separate source.) 4-27 A 4-pole 2 8 A. dc generator delivers a current of The average brush voltage drop on each of the four brush sets Calculate b. the rated current the armature a. 1 a. r/min separately kilogram. equal to 1.33 ms. A 4-pole 250 kW, 750V lap 4-25 800 r/min dc generator The brush width is such as to cover 3 commutator segments. Show that the duration of the commutation is 1 b. c. 450 mm. has 75 commutator bars. process V 750 Calculate The average flux density per pole The time needed to reverse the current in each armature coil, knowing that the brushes c. 500 speed of 1200 r/min b. A 240 kW, ciency of 94%. The shunt field resistance following: a. Appl tea t ion excited dc generator has an overall turn per coil. If the 1 95 is found to be 0.6 V. Calculate the total brush loss in the machine, neglecting friction loss. 1 Chapter 5 Direct-Current Motors Today, 5.0 Introduction this general statement can be challenged because the availability of sophisticated electronic Now thai erators, we have we can a good understanding of dc gen- motors transform Direct-current drives has begin our study of de motors. energy electrical hoists, fans, cars. vari- ment has given it service and force (cemf) has to drive, and this require- rise to three basic types Direct-current motors are built the of motors: erate either as a trate, Series motors 3. Compound motors motor or means of a switch all electric utility sys- electric trains, it is in steel mills, As soon form the alternating current der to use dc motors. flows mines, and sometimes advantageous low. to trans- is To connected to a dc source (Fig. 5. 1 ). illus- which the armature, E The armature has is s by a re- created by a set in The as the switch is closed, a large current the armature because its resistance individual armature conductors are is very imme- diately subjected to a force because they are im- into direct current in or- The reason as of permanent magnets. tems furnish alternating current. However, for speapplications such as in sistance R, and the magnetic field Direct -current motors are seldom used in ordinary because as a generator. consider a dc generator initially at rest, is industrial applications same way generators are; consequently, a dc machine can op- Shunt motors 2. mersed that the torque- in the magnetic field created by the perma- nent magnets. These forces add up to produce a speed characteristics of dc motors can be varied over a in Counter-electromotive 5.1 The torque- speed characteristic of the motor must be adapted to the type of the load cial still thousands more are being produced every year. definite torque-speed pump or fan) or a highly able one (such as a hoist or automobile). . possible to use alternating current pumps, calendars, punch-presses, and These devices may have a characteristic (such as a 1 it there are millions of dc motors mechanical energy. They drive devices such as into made motors for variable speed applications. Nevertheless, wide range while retaining high efficiency. powerful torque, causing the armature 96 to rotate. DIRECT-CURRENT MOTORS motor 5.2 Acceleration of the The net voltage acting in the armature circuit in Fig. (£ s —£ The resulting armature current 5.2 is / is limited only by the armature resistance R, and so () volts. ) = / (£ s - EJ/R (5.1) When the motor is at rest, the induced £0 = 0, and so the starting current is Figure 5.1 Starting a line. The On the other hand, as gins to turn, a cut a in soon as the armature be- second phenomenon takes place: the We know generator effect. duced that a voltage Ea is in- the armature conductors as soon as they magnetic field (Fig. 5.2). This is always true, The value and the same as those no matter what causes the rotation. polarity of the obtained induced voltage are when the machine operates The induced voltage the E0 is as a generator. therefore proportional to speed of rotation n of the motor and to the flux per pole, as previously given by Eq. 4. tf> £0 in or the circuit-breakers to the type the (4. polarity . is equal to It £ (1 is force (cemf) because always acts against the source voltage acts against the voltage in the sense that the net voltage acting in the series circuit equal to (£ s — £ () ) volts and not (£ s of Fig. 5.2 + £n ) they are consequent rapid acceleration of the armature. As the speed increases, the counter-emf creases, with the result that the value of (£ s diminishes. It follows from Eq. 5. ) armature that the 1 £0 in— £u current / drops progressively as the speed increases. maximum volts. is to accelerate until () voltage £ s . voltage (£s to act ical In effect, if £0 were reaches a def- less than the source equal to — E0 would become ) the current The driving /. £ s , the net zero and so, too, forces would cease on the armature conductors, and the mechan- drag imposed by the fan and the bearings would immediately cause the motor to slow down. As the speed decreases the net voltage (Es and so does the current fall it speed. At no-load this speed pro- duces a counter-emf £ slightly would case of a motor, the induced voltage called counter-electromotive £s the armature and Z if ductors produce a powerful starting torque and a 1 a constant that de- of winding. For lap windings However, trip. mo- blow Although the armature current decreases, the number of armature conductors. In the its Z is the fuses to absent, the large forces acting on the armature con- motor continues number of turns on be 20 to 30 times would cause tor. In practice, this 1 Z//4V60 the case of a generator. pends upon the may current starting greater than the nominal full-load current of the inite, As voltage = EJR / dc motor across the 97 as soon as the torque current is /. — £0 The speed ) increases will cease to developed by the armature equal to the load torque. Thus, motor runs when a no-load, the counter-emf must be at £ slightly less than s? to flow, sufficient to so as to enable a small current produce the required torque. Example 5-7 The armature of a permanent-magnet dc generator has a resistance of when 1 H is and generates a voltage of 50 500 r/min. If the armature is V con- nected to a source of 150 V, calculate the following: Figure 5.2 Counter-electromotive force (cemf) the speed in a dc motor. a. The starting current "1 ELECTRICAL MACHINES AND TRANSFORMERS 98 b. The counter-emf when the motor runs at 1000 Mechanical power and torque 5.3 r/min. At 1460 r/min. c. The armature current at 1000 r/min. At 1460 The power and torque of r/min. motor are two of a dc most important properties. its We now derive two sim- ple equations that enable us to calculate them. Solution a. moment of start-up, At the tionary, so rent E0 = 0 V 1 the armature (Fig. 5.3a). The is . sta- According to Eq. 4. wound armature starting cur- £ = limited only by the armature resistance: is (t - EJR = / 150 V/l n= A 150 the 1 cemf induced in a lap- given by is Z/i$/60 (4.1) Referring to Fig. 5.2, the electrical power P. supx b. Because the generator voltage r/min, the cemf of the 1000 r/min and 146 c. The V motor at is 50 V will be at 100 V at E s P* However, drop s 150 - 100 E in the = EJ s is equal to the sum of E0 E = Ea + s = (E S = 50/1 When the A (Fig. 5.3b) = (E0 + = EJ + motor speed reaches 1460 r/min, the cemf will be 146 V, almost equal to the source voltage. Under these conditions, the armature current is The 2 I R term IR)I 2 I R only electrical = = (Es 4 - EJ/R = (150 - 146)/1 (5.4) represents heat dissipated in the ar- mature, but the very important term / (5.3) EJ P* 50 IR follows that - EJ/R - plus the IR is It I (5.2) armature: = 50 V The corresponding armature current I: 1000 is E - Eu = equal to the supply voltage is multiplied by the armature current 1460 r/min. net voltage in the armature circuit at r/min 500 plied to the armature power that is EJ is the converted into mechanical power. The mechanical power of the motor therefore exactly equal to the product of the A is cemf multiplied by the armature current and the corresponding motor torque smaller than before (Fig. 5.3c). is much P = EJ (5.5) DIRECT- CURRENT MOTORS where where P — E0 - mechanical power developed by the T motor [W] Z induced voltage in = total torque (N-mJ number of conductors on the armature [Wb|* - effective flux per pole <& = / = armature current A| 6.28 = constant, to take care of units the armature (cemf) IV] / [ current supplied to the armature [A] [exact value 2. 99 Turning our attention to torque that the mechanical power P is T, = 2tt 1 we know given by the Eq. 5.6 shows that motor expression either we can raise the torque of a by raising the armature current or by raising the flux created by the poles. P = n 7/9.55 where n is (3.5) Example 5-2 The following the speed of rotation. Combining Eqs. /?779.55 3.5, 4.1, and 5.5, we obtain - EJ = Z/i3>//6() hp), 250 V, details are given on a 225 k\V (~ 300 1200 r/min dc motor (see Figs. 5.4 and 5.5): 243 armature coils turns per coil and so 1 type of winding T = Z*//6.28 The torque developed by a lap-wound motor therefore given armature slots 81 commutator segments 243 is field poles by the expression T = Z$//6.28 lap (5.6) 6 diameter of armature 559 axial length of armature 235 The effective flux is niven by <I> mm mm - 60 EJZn. Figure 5.4 225 kW, 250 V, 1200 r/min. The armature core has a diameter and an axial length of 235 mm. It is composed of 400 stacked laminations 0.56 mm thick. The armature slots and the commutator has 243 bars. (H. Roberge) Bare armature and commutator of a dc motor rated of 559 has 81 mm ELECTRICAL MACHINES AND TRANSFORMERS 00 1 Figure 5.5 Armature a. of Fig. 5.4 in the process b. One c. Connecting the d. Commutator connections ready of the 81 coils coil of ready to be placed ends to the being wound; coil-forming machine gives the coils the desired shape. in the slots. commutator bars. for brazing. (H. Roberge) T= Calculate a. The b. The number of conductors per c. The rated armature current slot 9.55 Pin = 9.55 X 225 000/1200 = 1791 N-m flux per pole The flux per pole is Solution a. = 6.28 TiZI nearly equal to the applied voltage (250 V). = (6.28 The = 25.7 We can assume that the induced voltage rated armature current / - F/E0 = E0 d> is is 225 000/250 5.4 = 900 A Speed When b. Each coil is made up of 2 conductors, gether there are 243 X 2 = 486 so alto- conductors on Conductors per slot Coil sides per slot The motor torque 1790)/(486 is = = 486/81 6 = IR drop due always small compared On 900) a dc motor drives a load between no-load and full-load, the equal to 6 X of rotation to armature resistance to the supply voltage This means that the counter-emf the armature. c. X mWb E s £0 is is E s . very nearly . the other hand, may be expressed by we have already seen that E0 the equation En = Z/7*/60 (4.1) DIRECT-CURRENT MOTORS 101 (variable) ! motor armature motor field (fixed) O ( j O 3-phase motor Figure 5.6 Ward-Leonard speed control system. Replacing E0 by £ we s 3-phase obtain , as Zn*/60, the line. This method of speed control, known Ward-Leonard system, is found in steel and paper 'A mills, high-rise elevators, mines, That RKTR is 60£ v (approx) A 'SiaJU3$6^A modern In installations the generator placed by a high-power electronic converter that changes the ac power of the electrical utility to dc, The Ward-Leonard system = speed of rotation [r/min] E = armature voltage fVJ Z = total number of armature armature of a dc motor. s shows This important equation is conductors speed of the that the directly proportional to the armature supply and inversely proportional voltage We will now study more than is simple way of applying a variable dc voltage // pole. mills. often re- by electronic means. where motor is to the flux per how this equation is applied. It just a to the can actually force the mo- tor to develop the torque and speed required by the load. For example, suppose slightly higher than the Current will then flow 5.6, E is s E cemf i} adjusted to be of the motor. in the direction and the motor develops shown in Fig. a positive torque. The armature of the motor absorbs power because Armature speed control 5.5 According to Eq. 5.7, constant with Now, suppose we reduce Es by reducing the gen<1> G As soon as E s becomes less than £u current / reverses. As a result, the motor torque speed depends only upon the reverses and (2) the armature of the motor delivers if the flux per pole (permanent magnet fixed excitation), the E By armature voltage s . In practice, we can tion or field fall in vary E s of the motor s, the proportion. by connecting the is G (Fig. 5.6). The field excita- kept constant, but the generator can be varied from zero and even reversed. to maximum The generator output voltage f can therefore be varied from zero s mum, with £ kept M to a separately excited variable- voltage dc generator excitation / x field <I> is raising or lowering motor speed will rise and motor armature to either positive or negative maxi- polarity. Consequently, the motor speed can be varied from zero to maximum generator is / flows into the positive terminal. in either direction. Note erator excitation . ( , 1 ) power to generator G. In effect, the dc motor suddenly becomes a generator and generator G suddenly becomes a motor. The electric power that the dc motor now netic its the delivers to G is derived at the expense of the connected mechanical load. Thus, by reducing motor is £ s, suddenly forced to slow down. What happens to the dc power received by genG? When G receives electric power, it operates as a motor, driving its own ac motor as an asynchronous generator!* As a result, ac power is fed erator that the driven by an ac motor connected to a ki- energy of the rapidly decelerating armature and The asynchronous generator is explained in Chapter 14. 1 ELECTRICAL MACHINES AND TRANSFORMERS 02 back into the P = EJ = 380 X 3000 = normally feeds the ac motor. line that The tact that power can be recovered this way makes the Ward-Leonard system very efficient, and constitutes another of Example 5-3 A 2000 kW, 500 by a 2500 kW control system generator, using a shown T = 9.55/Vw motor V, variable-speed in Fig. 5.6. is driven Ward-Leonard The total resis- tance of the motor and generator armature circuit The motor turns when E0 is 500 V. 10 mi}. r/min, at a is nominal speed of 300 The motor torque and speed when The speed of of this (400 = n with the armature (Fig. 5.7). The current It is E s , in the which subtracts yielding a smaller lot below its nominal recommended for small motors beof power and heat is wasted in the rheoonly and the overall efficiency is is low. Furthermore, poor, even for a fixed setting increases as the armature current increases. This produces a substantial drop motor armature speed with increasing in mechanical load. is P = EJ = 380 X 2000 - 760 The motor speed to control to place a rheostat in se- is 380)/0.01 A to the connected me- of the rheostat. In effect, the /R drop across the rheo- - stat The power motor the speed regulation is s its electromechanical braking torque. cause a stat, = (E - EJ/R = motor and Rheostat Speed Control Another way speed. Solution 2000 the ables us to reduce the speed s The armature current kN-m supply voltage across the armature. This method en- E = 350 V and £0 = 380 V -= 47.8 140 000)/228 1 from the fixed source voltage The motor torque and speed when / = X chanical load will rapidly drop under the influence ries s a. (9.55 rheostat produces a voltage drop E = 400 V and E0 = 380 V b. = the speed of a dc Calculate a. kW 140 Braking torque developed by the motor: advantages. its 1 kW is X 300 = 228 (380 V/500 V) The motor torque r/min is T = 9.55P/R b. Because - (9.55 = 31.8 X 760 000)/228 kN-m E a — 380 V, the motor speed is still 228 r/min. The armature current / is = (E EJ/R = = - 3000 A s The current is (350 - Power returned by 38())/0.01 it flows in re- motor torque also reverses. the and the 10 mi} resistance: 5,6 Field According negative and so verse; consequently, the Figure 5.7 Armature speed control using a rheostat. motor to the generator a dc speed control to Eq. 5.7 we can also vary the speed of motor by varying the keep the armature voltage numerator in Eq. 5.7 is field flux E s <t>. Let us now constant so that the constant. Consequently, the motor speed now changes in inverse proportion to DIRECT-CURRENT MOTORS the flux drop, cr>: if we increase the flux the speed will This method of speed control frequently used is when the called base speed. To control the flux (and hence, motor has the speed), to run we connect above a rheostat its R rated speed, in series To understand at constant speed. less method of speed this motor in Fig. 5.8a is The counter-emf running slightly ZTn is £ due s, we suddenly to the increase the resistance of the rheostat, both the exciting current /x and the flux <J> reduces the cemf jump a to to will diminish. This immediately causing the armature current much is again almost equal to E / higher value. The current . £ will accelerate until It same E0 with Clearly, to develop the motor must turn raise the motor speed above shunt-wound motors, £0 and s () s. flux, the troducing a resistance E motor develops a the field, greater torque than before. with control, sup- initially than the armature supply voltage IR drop in the armature. If Despite the weaker value depends its between the very small difference { the field (Fig. 5.8a). pose that the changes dramatically because upon and vice versa. 103 its weaker a can therefore nominal value by with the in series this We faster. in- For field. method of speed control enables high-speed/base-speed ratios as high as 3 to 1. Broader speed ranges tend produce to instability and poor commutation. Under abnormal conditions, the certain exciting current of a shunt motor dentally, the only flux manent magnetism that the motor has remaining in the poles.* is is may flux drop to dangerously low values. For example, if the interrupted acci- due that This flux to the re- is so small dangerously high to rotate at a speed to induce the required cemf. Safety devices are introduced to prevent such runaway conditions. (a) 5-7 Shunt motor under load Consider a dc motor running chanical load at no-load. If a suddenly applied is me- to the shaft, the small no-load current does not produce enough torque to carry the load and the motor begins to slow down. This causes the cemf to diminish, p.u. —/ Tvsn 2 — "-7 / — / / / / higher torque. * Tvs motor I is When the torque developed by the exactly equal to the torque imposed by the mechanical load, then, and only then, will the speed / remain constant (see Section t 3.1 1). To sum up, as the mechanical load increases, the armature current •rated load (b) re- sulting in a higher current and a corresponding rises and the speed drops. "—7 The speed of / stant <L 2 1 ^ speed a shunt from no-load motor stays only drops by 10 to 15 percent when it is p. u. / The term residual magnetism Figure 5.8 also used. However, the is IEEE Standard Dictionary of Electrical and a shunt motor including the a. Schematic diagram b. Torque-speed and torque-current characteristic of induction; a shunt motor. will be less than the residual induction." of full-load n armature current field relatively con- to full-load. In small motors, rheostat. Terms states. . netie eireuit, the if . If Electronics there are no air gaps ... in the inag- remanent induction there are air gaps . . . will equal the residual the remanent induction "1 1 ELECTRICAL MACHINES AND TRANSFORMERS 04 applied. In big machines, the drop in part, to is even less, due By the very low armature resistance. A 51 ad- justing the field rheostat, the speed can, of course, be kept absolutely constant as the load changes. Typical torque-speed and torque-current charac- of a shunt motor are shown in Fig. 5.8b. The speed, torque and current are given in per-unit values. The torque is directly proportional to the arteristics mature current. Furthermore, the speed changes only from from 0 pu pu 1.1 pu as the torque increases to 0.9 to 2 pu. Example 5-4 A shunt motor rotating 120 V 1500 r/min at and the shunt-field resistance mature resistance a. b. c. is fed by a source (Fig. 5.9a). The line current is 0. is II. If The = Figure 5.9 See Example 120 V/120 P = 6000 - 250 - 5750 = 7.7 motor (equivalent to 5750/746 = / b. 5 The voltage across W O= A 1 = 50 A I the armature E= and in armature iron losses. 120 5.8 Series is A series motor motor identical in construction to a is shunt motor except for the V Voltage drop due to armature resistance nected = 50X0.1 = The cemf generated by 5 V field. The field con- is This series field is armature current (Fig. 5.10a). composed of a few turns of wire having a cross section sufficiently large the armature is with the armature and must, there- in series fore, carry the full IR hp) is - I is The actual mechanical output is slightly less than 5750 because some of the mechanical power is dissipated in bearing friction losses, in windage losses, The armature current 5.4. W the field current (Fig. 5.9b) is /x A Mechanical power developed by the armature The current in the armature The counter-emf The mechanical power developed by Solution: a. 51 the ar- calculate the following: ii, 1 120 is to carry is the current. E0 = - 120 5 = V 115 Although the construction ties c. The total power supplied to the motor is similar, the proper- of a series motor are completely different from is those of a shunt motor. In a shunt motor, the flux F P, = El = 120 Power absorbed by X = 6120 51 W per pole field the armature is is constant P n = El = 120 X 50 = 6000 W armature is IR 2 = 50 2 X 0. 1 in a series motor When the current is large and vice versa. Despite these is differences, the P = But depends upon the armature current and. hence, upon the load. large, the flux in the loads because the shunt line. is the flux per pole Power dissipated at all connected to the - 250 W same basic principles and equations apply to both machines. DIRECT-CURRENT MOTORS 105 series field (b) Figure 5.10 a. Series motor connection diagram. b. Schematic diagram of a series motor. When a series motor operates at full-load, the per pole flux identical same as that of a shunt starts up, the armature current 3 p.u. motor of power and speed. However, when motor series the is speed /; armature current the / higher is Figure 5.11 than normal, with the result that the flux per pole also greater than normal, torque of a series motor ft is is follows that the starting of ing the On T versus / curves of Figs. 5.8 and the other hand, if the than full-load, the pole are 5. 1 Conversely, the speed motor operates necting an external resistor at less ture The weaker field same way as it would smaller than normal. shunt motor with a weak shunt the load current of a series For ex- field. motor drops ample, if half normal value, the flux diminishes by half and its so the speed doubles. Obviously, if the load is to ate at no-load, ft tends to run away, and the resulting could tear the windings out of the centrifugal forces and the sistor and field. The field reduces be lowered by con- in series total with the arma- IR drop across the armature the voltage, and so the speed must re- supply fall. Typical torque-speed and torque-current characteristics are ferent shown in Fig. 5. 1 1 . They are quite dif- from the shunt motor characteristics given in Fig. 5.8b. small, may rise to dangerously high values. For reason we never permit a series motor to oper- speed this may 1 armature current and the flux per causes the speed to rise in the for a and current-torque characteristic a series motor. considerably greater than of a shunt motor. This can be seen by compar- that the Typical speed-torque Example 5-5 A 15 hp, 240 V, 1780 r/min dc series motor has full-load rated current of teristics are 54 A. Its a operating charac- given by the per-unit curves of Fig. 5.11. armature and destroy the machine. Calculate 5.9 When Series motor speed control a series motor carries a load, its speed a. The current and speed when 24 N m b. The efficiency under may the load torque is these conditions have to be adjusted slightly. Thus, the speed can be low resistance increased by placing a with the series field. smaller than before, and an increase in The field in parallel current which produces a drop speed. is then in flux Solution a. We first establish the base power, base speed, and base current of the motor. They correspond to the full-load ratings as follows: ELECTRICAL MACHINES AND TRANSFORMERS 106 PH = // = hp 15 X 746 = 15 11 190 W are also used in electric cranes and hoists: light loads are lifted quickly = 1780r/min H = 54A /B Compound motor 5.11 The base torque therefore, is, A compound dc P vh 9.55 Tv = = X 9.55 11 780 190/1 mmf of the Fig. 5. N-m torque of 24 corresponds to a per- two runs = T(pu) = 24/60 2 1 at Referring to Fig. 5. tained at a speed of From the X /j(pu) T vs / 1 .4 1 winding 0.4 1 , a torque of 0.4 pu pu. Thus, the speed nB = 1.4 shows low and the is However, is at- by current machine: is As X 1780 /x it curve, a torque of 0.4 pu re- is The total current b. X /(pu) - /B 0.6 To calculate the and X 54 = away like a shunt no-load. at mmf of the series field remains flux per falls with increasing load and the to full-load is generally between 10 percent and 30 percent. A we have efficiency, to run to know P 32.4 x P0 = /2779.55 = 7776 = 2492 X () W 24/9.55 = 6263 W = PJP, = 6263/7776 = 0.805 or 80.5% 5.10 Applications of the series motor Series motors are used on equipment requiring a high starting torque. They which must run series motor is at are also used to drive devices high speed at light loads. The particularly well adapted for traction purposes, such as in electric trains. Acceleration rapid because the torque is high at is low speeds. Furthermore, the series motor automatically slows down as the train goes up a grade yet turns at top speed on flat ground. The power of a series motor tends to be constant, because high torque is accom- panied by low speed and vice versa. Series motors Figure 5.12 Connection diagram of a dc compound motor. b. Schematic diagram of the motor. a. is fully excited Pr P = EI = 240 X T| 32.4 is therefore greater under load than at no-load. speed drop from no-load = the series of the series field mmf of the shunt field mmf (and the resulting The motor speed is the motor / in and so the motor behaves the load increases, the constant. pole) When the shunt field does not tend increases but the r/min mmf quires a current of 0.6 pu. Consequently, the load / and the connection and schematic no-load, the armature current negligible. - 2492 fields add. diagrams of a compound motor. unit torque of = a series field compound motor, the The shunt field is always stronger than the series field. = 60 N-m // motor carries both a shunt field. In a cumulative "is A load and heavy loads more slowly. DIRECT-CURRENT MOTORS If 1.6 the series field the shunt field, we is connected so increasing load. 1.2 creases, and this it The speed may opposes compound obtain a differential motor. In such a motor, the total 1.4 that 107 mmf decreases with rises as the lead to instability. load in- The differ- -r-+ential 1.0 j a Fig. 5. a x> compound motor differential 0.8 compound compound a> 0) a of shunt, 1 3 shows the typical torque-speed curves compound and basis. Fig. 5. 14 w has very few applications. shows series motors on a per-unit a typical application of dc 0.6 motors in a steel mill. 0.4 5.12 Reversing the direction 0.2 of rotation 0 0 0.2 0.4 0.6 0.8 Torque 1.2 1.0 1.4 1.6 (per-unit) the direction of rotation of a dc motor, must reverse either (2) both the shunt Figure 5.13 Typical To reverse speed versus torque characteristics of various and l ) in we the armature connections or series field connections. terpoles are considered to The change dc motors. ( connections form is The in- part of the armature. shown in Fig. 5. 15. Figure 5.14 Hot strip finishing mill to the runout table composed (left of 6 stands, each driven by a 2500 kW dc motor. The wide steel foreground) driven by 161 dc motors, each rated 3 kW. (Courtesy of General Electric) strip is delivered "1 1 ELECTRICAL MACHINES AND TRANSFORMERS OS (-) (+) (-> (+) <+> (0 (b) (a) | I Figure 5.15 compound a. Original connections of a b. Reversing the armature connections to reverse the direction of rotation. c. Reversing the field motor. connections to reverse the direction of 5.13 Starting a shunt motor [f we apply full rotation. 5.14 Face-plate starter voltage to a stationary shunt motor, Fig. 5. 16 shows the schematic diagram of a manual the starting current in the armature will be very high face-plate starter for a shunt motor. Bare copper and we run the contacts are connected to current-limiting resistors risk of /?,, Burning out the armature; a. Damaging b. the commutator and brushes, due to * and R4 Conducting arm . when it of insulated handle is 1 sweeps across pulled to the right by means 2. In the position shown, the M heavy sparking; c. R2 the contacts Overloading the feeder; d. Snapping off the shaft due e. Damaging the driven to mechanical shock; equipment because of the sudden mechanical hammerblow. arm touches dead copper contact and the motor circuit is open. As we draw the handle to the right the conducting arm first touches fixed contact N. The sii; y voltage Es immediately causes full field current / x to flow, but the armature current / is limited by the four resistors in the starter box. The All dc motors must, therefore, be provided with a means to limit the starting current to values, usually between rent. One solution is to 1 .5 reasonable and twice full-load cur- connect a rheostat in series motor begins to turn and, as the cemf E0 the armature current gradually falls. speed ceases builds up, When any more, the arm the motor is pulled to the next contact, thereby removing resistor R from the to rise } with the armature. The resistance is duced as the motor accelerates and eliminated entirely, full when the gradually reis eventually machine has attained Today, electronic methods are often used to control. and to provide speed circuit. The current immediately jumps to higher value and the motor quickly accelerates to next higher speed. we move speed. limit the starting current armature arm When the speed again levels off to the next contact, and so finally touches the last contact. netically held in this position net 4, which is in a the forth, until the The arm is mag- by a small electromag- series with the shunt field. DIRECT-CURRENT MOTORS 109 Figure 5.16 Manual face-plate starter for a shunt motor. If the supply voltage the field excitation is suddenly interrupted, or electromagnet releases the arm, allowing to it to return dead position, under the pull of spring its 3. This prevents the motor from restarting un- safety feature expectedly if should accidentally be cut, the when the supply voltage is reestablished. When the tion of the cemf Ea motor are as shown armature IR drop, If is running normally, the direc- armature current £ () is we suddenly open motor continues I x and the polarity of the Neglecting the in Fig. 5.17a. equal to E s. the switch (Fig. 5.17b), the to turn, but its speed will gradually drop due to friction and windage losses. 5.15 Stopping a motor One inclined to believe that stopping a dc other hand, because the shunt field induced voltage is almost a simple, always trivial, not a heavy inertia load, the true. system to come is same it a generator a large dc motor may to a halt. For many coupled is take an hour or more Ea whose armature is motor under these circumstances, is motor is by simple mechanical armature is suddenly connected to reasons such the external resistor (Fig. 5. 7c). Voltage way we stop a car. this current motor original current same A more elegant method consists of electrically. flows in the brake the circulating a reverse current in the armature, so as to brake the mediately produce an armature current a braking friction, in the Two methods are /,. It ( nected to a source nected to the throw switch. £ s, whose field is directly and whose armature is either the line or to an external resistor R (Fig. very smooth stop. concon- 5. 17). im- However, /2 . is The reverse torque brings the machine to a rapid, but l same source by means of a doubleThe switch connects the armature to . developed whose magnitude depends upon Dynamic braking Consider a shunt motor E0 will opposite direction to the dynamic braking and (2) plugging. 5.16 /2 follows that a reverse torque em- ployed to create such an electromechanical brake: now Let us close the switch on the second set of contacts so that the we must apply One way to at the is for often unacceptable and, torque to ensure a rapid stop. the on open-circuit. 1 a lengthy deceleration time On excited, continues to exist, falling rate as the speed. In essence, the operation. Unfortunately, this When is to motor is still Figure 5.17a Armature connected to a dc source Es . 1 1 ELECTRICAL MACHINES AND TRANSFORMERS 0 Figure 5.17b Armature on open circuit generating a voltage E0 \x Time . Figure 5.18 Speed versus time curves for various braking methods. reversing the armature current by reversing the ter- minals of the source (Fig. 5.19a). Under normal motor conditions, armature rent is /, /, Dynamic braking. where In practice, resistor R R0 cur- given by is = (E s - £0 )//? 0 the armature resistance. If we suddenly reverse the terminals of the source, the net voltage is chosen so that the initial acting on the armature circuit becomes (E 0 + E s The so-called counter-emf E0 of the armature is no ). braking current rent. The initial is about twice the rated motor cur- braking torque is then twice the nor- longer counter to anything but actually adds to the mal torque of the motor. As E0 in E supply voltage the motor slows down, the gradual decrease produces a corresponding decrease in /2 becoming zero when finally the arma- The speed drops quickly This net voltage would produce greater than the full-load armature current. This current tor, ture ceases to turn. . . Consequently, the braking torque becomes smaller and smaller, s an enormous reverse current, perhaps 50 times would initiate an arc around the commuta- destroying segments, brushes, and supports, at first and then more slowly, as the armature comes even before the line circuit breakers to a halt. The speed decreases exponentially, somewhat like the voltage across a discharging capacitor. Consequently, the speed decreases by half intervals of time Ta To dynamic braking, . in equal illustrate the usefulness Fig. 5. 18 of compares the speed- time curves for a motor equipped with dynamic braking and one that simply coasts to a stop. 5.17 Plugging We a can stop the motor even more rapidly by using method called plugging. It consists of suddenly Figure 5.19a Armature connected to dc source Es . could open. DIRECT-CURRENT MOTORS value. However, much is it draw the easier to speed-time curves by defining a new time con- T0 which stant the time for the speed to de- is crease to 50 percent of its original value. There is a direct mathematical relationship between the conventional time constant T0 constant T and the half-time given by It is . Ta = 0.6937 We Figure 5.19b is Plugging. can prove that this given by Jnr Tn = (5.9) ° To prevent such a catastrophe, we must by introducing a reverse current resistor R in series As in dynamic with the reversing circuit (Fig. 5. 9b). 1 braking, the resistor U braking current With oped even effect, at when Ta = the armature has E0 = = moment J devel- = is braking motor P = cuit, otherwise interruption it the armature cir- will begin to run in reverse. Circuit mounted on the motor The curves of Fig. 5.18 enable us dynamic braking plugging and for the to completely after an interval if dynamic braking the its is 2T0 On . it same initial is still 25 per- original value at this time. Nevertheless, more popular in motor to the braking resistor a constant (exact value = a constant This equation is braking effect pated is in the exact value = log c 2] due that the energy dissi- resistor. In general, the motor entirely braking to the subjected to an extra braking torque due to windage and friction, and so the braking time be less than that given by Eq. hp), 250 V, 1 280 r/min dc motor mechanical time constant drives a large flywheel and the total We can therefore speak of a mechanical much same way we speak of time constant T the electrical time constant of a capacitor that dis- in the charges into a resistor. essence, motor T is it is speed is its initial connected to a 210 V is kW. moment 177 kg of m2 . It in- The dc source, and 1280 r/min just before the armature switched across a braking resistor of 0.2 its is il. Calculate The mechanical time constant T0 of the braking system takes for the speed 36.8 percent of of the flywheel and armature motor a. the time to fall to will 5.9. has windage, friction, and iron losses of 8 ertia of the | Dynamic braking and braking. [W] = most applications. We mentioned that the speed decreases exponentially with time when a dc motor is stopped by dynamic In the based upon the assumption is Example 5-6 A 225 kW (- 300 5.18 starts [r/min] power delivered by 2 0.693 comparative simplicity of dynamic braking ren- ders shaft compare the other hand, used, the speed motor (30/77) /log c 2] shaft. braking current. Note that plugging stops the motor cent of = 131.5 usually controlled by an automatic is null-speed device of inertia of the rotating initial l we must immediately open stops, fall to |sj speed of the motor when initial i as the previous value lkg-nr] /? which its parts, referred to the to a stop. In = EJR, As soon value. its initial come 0, but I2 is time for the motor speed to one-half to limit the initial reverse torque circuit, a zero speed, about one-half designed 131.5 7, where about twice full-load current. to plugging this is limit the (5.8) mechanical time constant b. The time for the motor speed to drop to 20 r/min ELECTRICAL MACHINES AND TRANSFORMERS c. The time for the speed to drop to only braking force friction, due that is 20 r/min to the if The stopping time increases the windage, 20 r/min and iron losses Solution a. We ; note that the armature voltage the speed When the armature 210V. The the resistor P, 210 V is (276/10) = 28 min This braking time switched to the brak- initial dynamic braking very is still power delivered to X 60 = 1656 2 /R 2 = 210 /0.2 = 220 500 The time constant Ta s Ta = 7//, 177 131.5 - 10 28 times longer than when used. to a is dynamically complete stop. In practice, however, we can assume that the machine stops W is 2 is is a motor which Theoretically, braked never comes ter an interval equal to 5 If the /(131.5 Py) (5.9) motor af- seconds. plugged, the stopping time has a is 2 X 1280 X 220 500 Tn definite value given by ts = 2T0 (5.10) where s — ts b. = and is = E approximately is 1280 r/min. is ing resistor, the induced voltage close to is proportion to the in time constant. Consequently, the time to reach The motor speed drops by 50 percent every The speed versus time curve follows the se- 10 s. stopping time using plugging Tn — time constant as given in [ sj Eq. 5.9 [sj quence given below: speed (r/min) time Example 5-7 The motor in Example 5-6 (s) 1280 0 640 320 10 ing resistor 30 Calculate 80 40 a. 40 20 50 b. an interval of 60 60 to 20 r/min The initial braking The stopping time after The We ro - = .///, />, /(131.5 P, X = 276 s = (177 The initial /, The , initial power is s braking current - EiR = initial is 420/0.4 braking power = 1050 = 8000 According to Eq. 5.9, Tn A is X { 220.5 kW has the same value as before: is T0 = 10 s ) 2 1280 )/(131.5 4.6 current and braking P = EJ = 210 X 1050 = have 1280 The new time constant 2 so that the fl, as before. E = E + E = 210+ 210 - 420 V s. of the braking time. = same net voltage acting across the resistor ( x the Solution The initial windage, friction, and iron losses are 8 kW. These losses do not vary with speed in exactly the same way as do the losses in a braking resistor. However, the behavior is comparable, which enables us to make a rough estimate n is 20 160 The speed of the motor drops c. plugged, and the brak- increased to 0.4 is braking current . . is min X 8000) The time to come ts to a complete stop = 2T0 = 20 s is DIRECT-CURRENT MOTORS Armature reaction 5.19 now we have assumed Until ing in a dc motor current flowing is that due the flux that the only However, to the field. N mmf actthe armature conductors also cre- in the magnetomotive force ates a 113 that distorts and weakens coining from the poles. This distortion and zone field weakening takes place in motors as well as We recall that the magnetic action mmf is called armature reaction. generators. armature in of the Figure 5.20 Flux distribution a motor running in at no-load. due 5.20 Flux distortion to armature reaction When motor runs a at no-load, the small current flowing in the armature does not appreciably affect the flux when coming from <2>, the the poles (Fig. 5.20). But armature carries its normal current, it duces a strong magnetomotive force which, would create a acted alone, superimposing and <1>| increases under the we 2, (Fig. 5.2 if it By 1). obtain the resulting Figure 5.21 Flux created by the full-load armature current. our example the flux density (Fig. 5.22). In flux <J> 02 flux pro- half of the pole and left de- it ceases under the right half. This unequal distribution produces two important effects. First the neutral zone shifts toward the rotation). The left result is (against the direction of poor commutation with sparking at the brushes. Second, flux density in pole tip A, due to the saturation higher sets Consequently, the increase of flux under the hand side of the pole the right-hand side. is Flux 4> 3 than flux slightly less less than the <F, at chines the decrease in flux percent and therefore no-load. For large may left- decrease under at full- load is be as much zone in. Figure 5.22 Resulting flux distribution a motor running at full- main the as 10 case of a dc generator, these narrow poles de- velop a magnetomotive force equal and opposite causes the speed to increase with load. it in load. the mmf of the armature to so that the respective mag- Such a condition tends to be unstable; to eliminate the problem, we sometimes add netomotive forces a series field of rise and fall one current varies. In practice, the or two turns to increase the flux under load. Such tating poles motors are said to have a stabilized-shunt winding. Commutating poles 5.21 set of improve commutation, we always place a commutating poles between the main poles of medium- and large-power dc motors made (Fig. 5.23). As slightly greater than that of the armature. Consequently, a small flux subsists in the region of the commutating poles. The flux is signed to induce To counter the effect of armature reaction and thereby is together as the load mmf of the commu- in the coil tion a voltage that is equal and opposite to the self- induction voltage mentioned in Section 4.28. result, commutation de- undergoing commuta- is As a greatly improved and takes place roughly as described in Section 4.27. DIRECT-CURRENT MOTORS Figure 5.24 Six-pole dc motor having a compensating winding distributed in slots in the main poles. The machine also has 6 commutating poles. (Courtesy of General Electric Company) base speed. In so doing, the rated values of armature current, armature voltage, and field flux must not be may be we assume an exceeded, although lesser values In rately making our E. v is negligible (Fig. 5.25). the armature current exciting current /f in per-unit values. £a happens /a is that it 1 . The advantage of the Thus, the per-unit torque used. ideal sepa- unit flux <I> f to , The armature /a , the flux and the speed n are Thus, be 240 if all <t> is T is given by the per- times the per-unit armature current T=$> r I ll /., (5.11) voltf, the expressed the rated armature voltage By the voltage Ea same reasoning, is 1 shunt field flux O, has a per-unit the per-unit armature equal to the per-unit speed n times the per-unit flux O, = V and the rated armature current 600 A, they are both given a per-unit value of Similarly, the rated per-unit approach renders the torque-speed curve universal. excited shunt motor in which the armature re- sistance age analysis, value of The (5.12) logical starting point of the torque-speed curve (Fig. 5.26), is the condition where the motor 1 1 6 ELECTRICAL MACHINES AND TRANSFORMERS (T= develops rated torque 'a The rated speed ) — speed (n at rated I ). order to reduce the speed below base speed, In we l often called base speed. is gradually reduce the armature voltage to zero, and while keeping the rated values of at their per-unit value of l . 3> r constant Applying Eq. (5. 7=1 corresponding per-unit torque X 1 1 the ), = 1 1. Furthermore, according to Eq. (5.12), the per-unit Figure 5.25 £a = n X £a /a voltage Per-unit circuit diagram the state of operation, , = n. Figures 5.27 and 5.28 show ] and known <£>, during this phase of motor mode. as the constant torque Next, to raise the speed above base speed, alize that the armature voltage anymore because The only solution T already at it is is to keep means = base speed, the per-unit flux OH 1 0 ' 2.0 1.0 ^ speed n is Eq. (5.12), 1. 1 this Thus, above \/n. equal to the recipro- of the per-unit speed. During this operating cal mode, the armature current can be kept of level T= Figure 5.26 = and so 1, rated level of its flux. Referring to { re- E,d at its rated level of and reduce the that n<$ we cannot be increased ct» 1. = / r a Recalling Eq. (5.11), (\/n) X it at its rated follows that -\ln. Consequently, above base 1 speed, the per-unit torque decreases as the reciprocal of the per-unit speed. motor 1 is clear that since the per- and armature voltage are both unit armature current equal to It is during this phase, the power input to the equal to 1 . Having assumed an chine, the per-unit mechanical equal to is why 1, which corresponds the region ideal power output is maalso to rated power. That above base speed is named the constant horsepower mode. We conclude that the ideal dc shunt motor can operate anywhere within the limits of the torque- speed curve depicted in Fig. 5.26. In practice, the actual torque-speed fer considerably from that shown curve may in Fig. 5.26. dif- The curve indicates an upper speed limit of 2 but some machines can be pushed to limits of 3 and even 4, by reducing the flux accordingly. However, speed is raised lems develop and centrifugal forces <H ' 0 1.0 1.25 >- speed n L2.0 dangerous. When the ventilation when the above base speed, commutation probthe motor runs becomes poorer and tends to rise above its may become below base speed, the temperature rated value. Consequently, the armature current must be reduced, which reduces the Figure 5.28 torque. Eventually, when the speed is zero, all forced DIRECT-CURRENT MOTORS ventilation ceases and even the field current must be 117 magnet motors 5-24 Permanent reduced to prevent overheating of the shunt field coils. As a result, the permissible stalled torque only have a per-unit value of 0.25. tical torque-speed curve The ishes The resulting pracspeed dimin- can be largely overcome by using an external of air, no matter what It delivers a constant stream the speed of the Under these conditions, approaches that shown We have seen and a that shunt-field field current to motors require coils produce the flux. consumed, the heat produced, and in Fig. 5.29. drastic fall-off in torque as the blower to cool the motor. to be. shown is may motor happens the torque-speed curve large space taken up by the vantages of a dc motor. By The energy the relatively field poles are disad- using permanent mag- nets instead of field coils, these disadvantages are overcome. The result is a smaller motor having a higher efficiency with the added benefit of never risking run-away due to field failure. in Fig. 5.26. A further advantage is that the effective air The reason that that the is many increased times. magnets have a permeability nearly equal to that of is mature sible is of using permanent magnets gap air. As a result, the ar- mmf cannot create the intense field that is pos- when soft-iron pole pieces are employed. Consequently, the field created by the magnets does not become distorted, armature reaction is as shown in Fig. 5.22. Thus, the reduced and commutation is im- proved, as well as the overload capacity of the motor. 0 0.2 0.4 0.6 0.8 1.0 — A further advantage is that the long air gap reduces the 2.0 speed inductance of the armature and hence n much more quickly to changes in it responds armature current. Permanent magnet motors are particularly advan- Figure 5.29 Torque-speed curve of a typical dc motor. tageous in capacities below about 5 hp. The magnets Figure 5.30 Permanent magnet motor rated slots 20; 1 .5 hp, commutator bars: 40; turns per 0.34 a. (Courtesy of Baldor Electric Company) 90 V, 2900 coil: 5; r/min, 14.5 A. conductor Armature diameter: 73 mm; armature length: 115 mm; 17 AWG, lap winding. Armature resistance at 20°C: size: No. ELECTRICAL MACHINES AND TRANSFORMERS are ceramic or rare-earth/cobalt alloys. shows PM motor. and tia fast Its the series winding, per pole. 5.30 Fig. 2900 r/min elongated armature ensures low iner- the construction of a response hp, .5 when used The only drawback of tively high cost 1 has a V, motors motor of the magnets and the inability to a. b. obtain higher speeds by field weakening. 5-12 Questions and Problems Name 1 make Explain what is a 5- 3 1 meant by the generator What determines ef- ity in If the line, calcu- following: full-load when V the armature is connected 200 1 to a source. Calculate the armature voltat 1500 The following details are known about a 250 hp, 230 V, 435 r/min dc shunt motor: A nominal full-load current: 862 the magnitude and polar- of the counter-emf V r/min. At 100 r/min. fect in a motor. 5-3 23 A. separately excited dc motor turns at 115 three types of dc motors and is field and the age required so that the motor runs sketch of the connections. 5-2 A The shunt 15 (1, 1 connected to a 230 mmf per pole at mmf at no-load The The r/min Practical Level 5- is late the the rela- is of total resistance nominal armature current servo applications. in PM 90 H insulation class: a dc motor? weight: 3400 kg The counter-emf of a motor 5-4 is always mm external diameter of the frame: 915 slightly less than the applied armature volt- length of frame: 1260 mm age. Explain. and efficiency a. Calculate the total losses Name two methods 5-5 that are used to vary the speed of a dc motor. 5-6 why Explain ing current a starting resistor motor up needed to bring a percent of the total losses Show one way to reverse the direction rotation of a compound motor. 5-8 A 230 V 5-9 20 per- Calculate the value of the armature resistance knowing as well as the counter-emf. speed? to field excit- the shunt field causes if cent of the total losses. c. is approximate shunt b. Calculate the the armature current of a shunt motor decreases as the motor accelerates. Why 5-7 at full-load. of that at full-load are 50 due to armature resistance. d. If we wish to attain a what should be shunt motor has a nominal arma- the speed of 1 100 r/min, approximate exciting current? ture current of tance 0.15 11, If the armature c. The mechanical power developed by [kW] and the armature if the motor across the 230 b. V the mo- inal is initial starting shown The value of 20 hp, 240 V, 400 nom400 A, calculate in Fig. 5. 17. If the is the braking resistor to limit the 125 percent of maximum its R if we braking current to nominal value line, to limit the initial current to 1 b. The braking power [kW] when the motor has decelerated to 200 r/min, 50 r/min, 0 r/min. a. The motor 15 A. 5-15 in Problem 5- 1 4 is now stopped by using the plugging circuit of Fig. 5.19. The compound motor of Fig. 5.12 has 1200 turns on the shunt 1 armature current want directly connected Intermediate level 1 wish to stop a the following: a. Calculate the value of the starting resistor needed 5-1 We ing circuit [WJ [hp] a. In Problem 5-9 calculate the current 5-14 r/min motor by using the dynamic brak- The counter-emf [V| The power supplied to tor, resis- calculate the following: b. a. 5-10 is 60 A. winding and 25 turns on Calculate the the maximum new braking resistor braking current is R so that 500 A. DIRECT-CURRENT MOTORS Calculate the braking power [kW] b. motor has decelerated to when the 5-20 200 r/min, 50 r/min, ing: 0 r/min. Compare c. the braking power developed at 200 r/min to the instantaneous power dissipated in resistor R. Advanced 5-16 5-2 1 tor has a 225 kW, a diameter of 559 length of 235 mm. 1 The The c. at a The value of c. The the armature the counter-cmf at full flux per pole, in milliwebers A standard 20 hp, 240 V. 1 load [mWb] 500 r/min self- it 200 r/min Calculate the following: ing. total kinetic ings and commutator is the decided to cool the machine by the air by tor equal to the J calcu- in- blower and channeling means of an The highest 30°C and exits the mo- air duct. the temperature of the air that J of the wind- from 1500 r/min without overheat- expected ambient temperature energy of the revolving parts if Ft is to stalling an external turns at 1200 r/min speed of 600 r/min, requirement has arisen whereby the mo- tor should run at speeds ranging kinetic energy of the armature alone when The number of conductors on b. A 200 r/min mo- mm and an axial The approximate moment of inertia, knowing 3 that iron has a density of 7900 kg/m b. a. cooled dc motor has an efficiency of 88%. level The armature of a. Referring to Fig. 5.30, calculate the follow- is should not exceed 35°C. Calculate the capacity of the blower required, in cubic lated in (a) feet per minute. (Hint: see Section 3.21.) 5-17 If we reduce the normal exciting current of a practical shunt motor by 50 percent, speed increases, but it 5-22 the A 250 hp, nominal never doubles. 500 V dc shunt motor draws field current of 5 A a under rated The field resistance is 90 12. Calculate ohmic value and power of the series re- load. Explain why, bearing in mind the saturation the of the iron under normal excitation. 5- 8 1 The speed of a series motor drops with ing temperature, while that of a shunt ris- mo- tor increases. Explain. Industrial 5-19 5-23 magnetism per 100°C increase The motor runs at 3% in of its tempera- a no-load speed of 2500 r/min when connected to a 150 A5 500 hp dc motor draws a In each case, calculate the V field as a What source in an ambient temperature of 22°C. room where that the field current drops the shunt field and resistor V source. field current of A when the field is connected to a 150 V source. On the other hand, a 500 hp motor draws a field current of 4.3 A when the field is connected to a 300 V dc source. magnet motor equipped with Estimate the speed A when 0.68 cobalt-samarium magnets loses ture. needed so are connected to the A pplication A permanent sistor to 4.5 if the motor is placed the ambient temperature is in a 40°C. power required for the percentage of the rated power of the motor. conclusions can you draw from these results? Chapter 6 Efficiency and Heating of Electrical Machines 6.0 Introduction Whenever a one form loss. The 6.1 machine transforms energy from to another, there is loss takes place in the causing ( duction in efficiency. 1 ) an increase in Mechanical losses are due machine bearing friction, to brush friction, and windage. The friction losses always a certain depend upon itself, temperature and (2) a Mechanical losses the speed of the machine and upon the design of the bearings, brushes, commutator, re- and Windage depend on the speed and design of the cooling fan and on the tur- slip rings. losses From the standpoint of losses, electrical machines may be divided into two groups: those that bulence produced by the revolving have revolving parts (motors, generators, sence of prior information, those that do Electrical (transformers, not etc.) reactors, and mechanical losses are produced and tests of these mechanical losses. In this chapter chines, but the we analyze same internal fan the losses in dc losses are also found in mamost in us a clue as to We how is they important because may be it determine the value mounted on the motor shaft. It cool air from the surroundings, blows the windings, and expels machines operating on alternating current. The study of power losses parts. In the ab- usually conduct Rotating machines are usually cooled by an machines. in stationary itself to etc.). in ro- tating machines, while only electrical losses are produced on the machine we it draws it over again through suitable vents. In hostile environments, special cooling methods are sometimes used, gives reduced. as illustrated in Fig. 6.1. also cover the important topics of tempera- ture rise and the service life of electrical equipment. We show that both are related to the class of insula- tion used and have been that these insulation classes 6.2 Electrical losses standardized. Electrical losses are 120 composed of the following: EFFICIENCY AND HEATING OF ELECTRICAL MACHINES 1. Conductor / R losses (sometimes called copper R = losses) 2. Brush losses 3. Iron losses 1. Conductor Losses The in resistance, in turn, depends upon the length, cross section, resistivity, and tempera- A = rent ture it carries. of the conductor. The following equations determine the resistance able us to ture resistance and the square of the cur- The at at) (6.2) which losses in a conductor de- R = L = its (6.1) P A = pod + p pend upon 121 en- p resistance of conductor [il] cross section of conductor length of conductor [m] = resistivity [ m2 ] of conductor at temperature conductor at 0°C / any tempera- and for any material: = a = resistivity of Po m] temperature coefficient of resistance 0°C t [il = at l/°C] [ temperature of conductor [°C] The values of p and a for different materials are listed in Appendix AX2. In dc motors and generacopper losses occur tors, shunt field, the armature, the series in the field, the commutating poles, and compensating winding. These I~R losses show the up as rise conductor temperatures heat, causing the to above ambient temperature. Instead of using the 2 I R we sometimes equation, number prefer to express the losses in terms of the of watts per kilogram of conductor material. The losses are then given by the equation P c = l00Oy 2 p/£ (6.3) where Pc — J = Figure 6.1 Totally tor for enclosed, water-cooled, 450 kW, 3600 r/min mo- use machine is in Warm a hostile environment. air inside Westinghouse nameplate. After releasing its water-cooled pipes, the cool air reenters the chine by pipes located diagonally as cooling-water inlet on the heat exchanger serve and outlet respectively. (Courtesy of Westinghouse) (W/kg| ] p £ = density of the conductor [kg/m 1000 = constant, to take care of units conductor [nll-m) 1 ] heat to a maway of two rectangular pipes leading into the end bells. The cooling air therefore moves in a closed circuit, and the surrounding contaminated atmosphere never reaches the motor windings. The circular capped set of [A/mm loss 2 resistivity of the the above the current density power = blown upward and through a water-cooled heat exchanger, situated immediately specific conductor According mass is density. For tween 1 to this equation, the loss per unit proportional to the square of the current .5 copper conductors, we use densities be- A/mm 2 losses vary and 6 A/mm 2 from 5 W/kg to . The corresponding 90 W/kg (Fig. 6.2). The higher densities require an efficient cooling system to prevent an excessive temperature rise. ELECTRICAL MACHINES AND TRANSFORMERS 122 9.6 W/kg cm 1 carbon brush 2 200 80° C copper conductor commutator Figure 6.2 Copper losses may be expressed in watts per Figure 6.3 Brush contact voltage drop occurs between the brush face and commutator. kilo- gram. 2. Brush Losses The 2 I R losses in the brushes are negligible because the current density A/mm", which 0.1 per. is only about used and eddy currents, as previously explained in Sections 2.27 and 2.30. Iron losses depend upon the cop- magnetic flux density, the speed of rotation, the However, the contact voltage drop between the quality of the steel, and the size of the armature. is far less than that brushes and commutator may produce in significant pending on the type of brush, the applied pressure, They typically range from 0.5 W/kg to 20 W/kg. The higher values occur in the armature teeth, where the flux density may be as high as .7 T. The and the brush current (Fig. losses in the armature core are usually losses. 3. The drop varies from 0.8 V to 1.3 V, de- 6.3). Iron Losses Iron losses are produced in the ar- mature of a dc machine. They are due to hysteresis 1 The losses can be much lower. minimized by annealing the steel (Fig. 6.4). Figure 6.4 kW electric oven is used to anneal punched steel laminations. This industrial process, carried out in a conatmosphere of 800°C, significantly reduces the iron losses. The laminations are seen as they leave the oven. (Courtesy of General Electric) This 150 trolled EFFICIENCY AND HEATING OF ELECTRICAL MACHINES Some iron losses are also They faces. are due produced in the pole to flux pulsations created as suc- and cessive armature teeth sweep across slots the Strange as chanical drag it may seem, on the armature, producing the same mechanical effect as impose a me- iron losses power the line to continue to rotate. This no-load overcomes the friction, windage, and iron losses, and provides for the copper losses 2 pole face. 123 The I R losses in commutating field load current in the shunt field. the armature, series field, seldom more than 5 percent of is and are negligible because the nothe nominal full-load current. friction. As we load machine the current increases the PR in Example 6-1 the armature circuit. Consequently, the Adc machine turning at 875 r/min carries an armawinding whose total weight is 40 kg. The cur- the armature circuit (consisting of the armature and ture A/mm 2 rent density is 5 ture is 80°C. The amount to 1 1 and the operating tempera- total iron losses in the armature 00 W. all the other windings in series with the other hand, the no-load losses losses in will rise. it) On mentioned above remain essentially constant as the load increases, unless the speed of the machine changes appreciably. It follows that the total losses increase with Calculate load. a. b. The copper losses Because they are converted into heat, the tem- perature of the machine rises progressively as the The mechanical drag [N inJ due to the iron losses load increases. However, the temperature must not exceed Solution a. Referring to Table sistivity p of copper = p0 = 15.88 = 21.3 ( at in the 80°C Appendix, the re- maximum used is the + 1 AX2 at) machine. Consequently, there in the power that the + 0.00427 X 80) The specific power loaded beyond is loss 8890 kg/m deliver. is a limit to This temper- inal or rated nllm The density of copper machine can power enables us to establish the nompower of the machine. A machine ature-limited (1 the allowable temperature of the insulation heat. 3 The its nominal rating inevitably shortens the service will usually over- more insulation deteriorates life rapidly, which of the machine. is If a machine runs intermittently, it can carry heavy 2 P L = 100O/ p/£ = 1000 X 5 2 X (6.1) 21.3/8890 overloads without overheating, provided that the operating time rating of = 60 W/kg 1 is copper loss is is P = 60x 40 = 2400 W The braking torque due to iron losses is it can be for P = «779.55 = 875 T= 12 A ( 3.5) 779.55 N-m or approximately 8.85 ft-lbf Losses as a function of load physically impossible for a generator kW to deliver an output of 100 kW, even 6.4 Efficiency curve The efficiency of ful output power it must absorb some power from machine is the ratio of the use- to the input = P - - X power P- (Section x plus the losses p. ti However, a power PG Furthermore, input power dc motor running at no-load develops no useful power. kW for one millisecond. 3.7). 6.3 2 from calculated 1100 1 for higher loads the capacity limited by other factors, usually electrical. For in- stance, rated at 10 b. Thus, a motor having a nominal However, short periods. Total short. kW can readily carry a load of 0 100 We = is equal to useful can therefore write P ° X 100 (6.4) ELECTRICAL MACHINES AND TRANSFORMERS 124 The where = T| ful output power [W] P = input power [W] zero is losses the 25 percent of rating, the nominal its copper losses we have V, in the 2 = 50 r/min, 230 loaded is to armature curof 1/4) its full- the following: to calculate the efficiency of a dc machine. Example 6-2 A dc compound motor having motor approximately 25 percent (or is square of the current, 1 no-load because no use- load value. Because the copper losses vary as the |W1 The following example shows how 1 at developed by the motor. 25 percent load When rent x = is efficiency [%] P0 — p efficiency power armature circuit W X 595 = 37 (1/4) no-load losses a rating of 10 kW, 50 A, has the following losses W = 830 at total losses full load: = bearing friction loss brush friction loss windage loss mechanical losses (1) total (2) iron losses (3) copper loss in the shunt field copper losses at in the b. in the series field c. in the total (4) commutating winding copper W 50 W 200 W 290 W 420 W 120 W = = = 500 load at full — load at 25, 50, 75, 100, 595 Pa = at no-load of the machine. Draw a graph showing 10 kW X power supplied P, 830 In the (1/4) to the = 867 W = 2500 at motor W 25 percent same way, we At 50 percent load 2 (6.2) 100 = 74% find the losses at 50, 75, 100, and 150 percent of the (1/2) W is = (PJPJ X 100 = (2500/3367) X efficiency as a function of mechanical load (neglect 3.35 hp) is = 2500 + 867 - 3367 and the efficiency W and 150 percent of the nom- + is T! Calculate the losses and efficiency inal rating W 25 W 70 W 37 Useful power developed by the motor loss in the armature circuit and 40 full load: armature a. = = = = = = nominal load: the losses are X 595 + 830 = 979 W the losses due to brush contact drop). At 75 percent load Solution 2 No-load The copper losses (3/4) in the At 100 percent load sum of the mechanical losses (1), the iron losses (2), 595 W the losses are + 830 = 1425 W and the shunt-field At 150 percent load losses (3): 2 (1.5) no-load losses 1165 armature circuit are negligible at no-load. Consequently, the no- load losses are equal to the the losses are X 595 + 830 = = 290 + 420 + 120 - 830 the losses are X 595 + 830 = 2169 W W The efficiency calculations for the various loads 6A and These losses remain essentially constant as the load are listed in Table varies. graphically in Fig. 6.5. the results are shown EFFICIENCY AND HEATING OF ELECTRICAL MACHINES 125 Figure 6.5 Losses and efficiency as a function of mechanical power. See Example It is LOSSES AND EFFICIENCY OF A DC TABLE 6A when Total Output Load losses power P m [Wl [W] 75 1 100 1 l r /r] 830 0 74 979 5 000 5 980 83.6 165 7 500 8 665 86.5 10 15 000 1 000 1 selecting a motor to 425 87.5 17 170 roughly equal to the load We 3 367 425 The efficiency curve mature rises sharply as the load in- fall. This is typical of the efficiency curves of all electric motors, both ac dc. Electric motor designers usually and try to attain the above calculation of efficiency we could have included the losses due to brush voltage drop. Assuming a constant drop, brush loss at full-load brushes = say, amounts losses, V per brush, the to 0.8 V X 50 A X 2 of 0.8 80 W. At 50 percent load, the brush loss would be 40 W. These losses, do loads the Consequently, a particular job, we W ( 1 6.5 it has to drive. maximum our example (830 + at that load where the ar- copper losses are equal to the no-load this corresponds to a = 660 W, 830) 1 an output of 1 total 8 1 may wish to check these Temperature The temperature rise 1 The 5.8 hp) and an efficiency of 87.68 percent. results. rise of a machine or device difference between the temperature of its is the warmest accessible part and the ambient temperature. It may be measured by simply using two thermometers. peak efficiency at full-load. In the circuit losses. In loss of 87.4 over a broad range of power, and then slowly begins to at light poor. can prove that the efficiency of any dc ma- chine reaches a reader creases, flattens off is Efficiency P- [Wj 0 2 169 150 power t) motor should always choose one having a power rating 500 830 867 50 Input 2 0 important to remember that efficiency of any MOTOR 25 6.2. when added modify the efficiency curve only to the other slightly. However, due to the practical difficulty of placing a to the really warmest spot inside method is seldom used. We usuupon more sophisticated methods, de- thermometer close the machine, this ally rely scribed in the following sections. Temperature rise has a direct bearing on the power rating of a machine or device. It also has a di- ELECTRICAL MACHINES AND TRANSFORMERS 126 rect bearing on its temperature rise useful service is life. Consequently, In crystallizing, organic insulators and a very important quantity. brittle. ical vibration will expectancy of equipment 6.6 Life electric a Apart from accidental electrical and mechanical is its insulation: higher the temperature, the shorter made on many its insulating materials have cause them to break. life. The Low 1 200°C for the same length of time. temperatures are just as harmful as high temperatures are, because the insulation tends that tors their flexibility at temperatures as life means ture of 105°C, will it have a service years at a temperature of 115°C, of 125°C, and of only one year The at factors that contribute at life a tempera- 6.7 of only four two years 6.6). Because of these lation tallize it upon their ability to withstand heat. correspond to the (6) time (Fig. maximum slowly begins to crys* rapidly as the temperature rises. Such as IEEE. Underwriters Laboratories. Canadian Standards Association. humidity high temperature chemicals fungus (UU so dust noxious gases rodents Figure 6.6 Factors that may These classes temperature levels shorten the service life of an of: 105°C, 130°C, 155°C, 180°C. and 220°C (formerly and the transformation takes place more time that set standards* have grouped insulators into five classes, depending factors, the state of the insu- changes gradually; Thermal classification Committees and organizations to the deteriora- and retain — 60°C. of insulators tion of insulators are (1) heat, (2) humidity, (3) vi- bration, (4) acidity, (5) oxidation, low as at 135°C! most have been developed, however, which motor has a that if a expectancy of eight years to freeze and crack. Special synthetic organic insula- approximately by half every time the temperature increases by 10°C. This that On the other synthetic polymers can withstand temper- atures as high as Tests shown some does not exceed 00 °C. the service life of electrical apparatus diminishes normal Under normal expectancy of eight to ten years provided life hand, expectancy of electrical apparatus limited by the temperature of hard conditions of operation, most organic insulators have their temperature failures, the life become Eventually, the slightest shock or mechan- insulator. vibration EFFICIENCY AND HEATING OF ELECTRICAL MACHINES represented by the letters A, B, thermal classification (Table 6B) F, is H, and R). This 40°C, This standardized temperature was estab- a cornerstone in lished for the following reasons: design and manufacture of electrical apparatus. the 1 6.8 Maximum ambient temperature and hot-spot temperature rise maximum ambient TABLE 6B Class l()5°C A temperature, which Illustrative B F when immersed H be included life at 220°C R 240°C S above 240°C C in this class if such as and paper when suitably impregnated or silk, oil. by experience or accepted Other materials or combinations of materials tests shown they can be shown to may be included have comparable thermal life at to have comparable substances. Other materials or combinations of materials shown to have comparable may life at with suitable bonding etc., in this class if by experience 130°C. Materials or combinations of materials such as mica, glass fiber, asbestos, with suitable bonding etc., be included in this class if by experience 155°C. Materials or combinations of materials such as silicone elastomer, mica, glass fiber, asbestos, etc., with suitable bonding substances such as appropriate silicone resins. Other materials or combinations of may they can be shown to have Materials or combinations of materials which by experience or accepted tests can be shown to have shown to have shown to have be included life at in this class if by experience or accepted tests 180°C. the required thermal life at 200°C. Materials or combinations of materials which by experience or accepted tests can be the required thermal life at 22()°C. Materials or combinations of materials which by experience or accepted tests can be the required thermal life at 240°C. Materials consisting entirely of mica, porcelain, glass, quartz, and similar inorganic materials. Other materials or combinations of materials they can be implies that electrical sive, performance guarantees. I05°C. shown to continuously at may be included have the required thermal The above insulation classes indicate a normal ated to give Materials or combinations of materials such as mica, glass fiber, asbestos, materials N likely to encounter. enables them to standardize the size of their substances. Other materials or combinations of materials comparable 200°C It machines and in a dielectric liquid or accepted tests they can be 80°C machines are examples and definitions or accepted tests they can be I55°C 2. Materials or combinations of materials such as cotton, coated or may 130°C enables electrical manufacturers to foresee their usually is It CLASSES OF INSULATION SYSTEMS thermal 1 . the worst ambient temperature conditions that Standards organizations have also established a equipment insulated with 105°C. Note that life life at in this class if expectancy of 20 000 h a class A this classification insulation system assumes by experience or accepted tests temperatures above 240°C. to 40 000 h at the stated temperature. This would probably that the insulation system last for is 2 to 5 years if oper- not in contact with corro- humid, or dusty atmospheres. For a complete explanation of insulation classes, insulation systems, and temperature indices, see the 127 companion IEEE Standards Publications Nos. 96. 97, 98, 99, and 101. See also IEEE Std Underwriters Laboratories publication on insulation systems UL 1446, 1978. 1 IEEE Std 1-1969 and 17-1974 and ELECTRICAL MACHINES AND TRANSFORMERS 128 The temperature of a machine point, but there are places warmer motor, relay, and so forth, he intends to put on the varies from point to where the temperature market. Thus, for class is than anywhere else. This hottest-spot tem- perature must not exceed the maximum To show how allowable and Figure 6.7 shows the hot-spot temperature limits F. and H insulation (curve 1). built a 10 test the They are the temperature limits previously mentioned maximum insulation, the is (130 - 40) kW motor using class B motor he places it in temperature of 40°C and loads in 10 The maximum ambient temperature of 40°C is also shown (curve 3). The temperature difference between curve and curve 3 gives the maxSection 6.7. kW insulation. it up until This limiting temperature delivers detectors, located at strategic points inside the ma- chine, record the temperature of the windings. After the temperatures have stabilized (which rise to establish the physical size may take is called the hot-spot temperature. If the hot- Class say is, H 180°C Class F 155°C 165°C r / Class / © © 105°C // // B 130°C A / J / / j 145°C / hot-spot 120°C temperature rise by // J/ J 100°C embedded average temperature by the thermocouple rise resistance method © 40°C 40°C limiting ambient temperature Figure 6.7 Typical limits of Shows Shows Shows the the noted, and of the spot temperature so recorded Class is enables the this manufacturer (3) it permissible temperature rise for each insula- tion class. (2) To of mechanical power. Special temperature several hours), the hottest temperature (1) 90°C. a constant ambient 1 imum = the temperature rise affects the size of a machine, suppose a manufacturer has designed temperature of the particular class of insulation used. for class A, B, B allowable temperature rise some dc and ac industrial machines, according to the insulation class: maximum permissible temperature of the insulation to obtain a reasonable maximum permissible temperature using the resistance method the limiting ambient temperature service life 147°C, the EFFICIENCY AND HEATING OF ELECTRICAL MACHINES manufacturer would not be permitted to The reason product. - (147° = 40°) On B the hottest-spot tempera- if is ( 100° The manufacturer immediately 6()°C. ceives that he can still per- make rise limits. For instance, he can reduce the conducthe hot-spot temperature rise 90°C. Obviously, close to this very is reduces the weight and cost of the windings. But the manufacturer also realizes that the him ables turn, reduces the amount of iron. ing the with a now reduced conductor size reduce the size of the to By en- thus redesign- motor, the manufacturer ultimately ends up machine The hot-spot temperature measure because that operates within the permissible it rise is rather difficult to has to be taken at the very inside of a winding. This can be done by embedding a small temperature detector, such as a thermocouple or thermistor. However, this direct method of measuring hot-spot temperature is costly, and only is justified for larger machines. To simplify This, in slots. Temperature rise by the resistance method 6,9 per- remain within the permissible temperature tor size until the size product. — more economical design a The manufacturer could reduce standards. of the motor and thereby market a more competitive insulation. only 100°C, the temperature rise = and for class the other hand, ture is 40°) maximum 1()7°C exceeds the 90°C missible rise of his sell that the temperature rise is 129 matters, accepted standards permit a second method of determining temperature rise. It is based upon the average winding temperature, measured by resistance, rather than the hot-spot temper- The maximum allowable average winding ature. temperature rise limits and has the smallest possible temperatures for the various insulation classes are physical size, as well as lowest cost. In practice, it is shown in not convenient to carry out per- formance of 4()°C. tests in a is usually loaded to rated ca- its much lower (and more comfortable) am- pacity in bient temperatures. by tablished testing controlled ambient temperature The motor Toward this end, it bodies for that, purposes, the ambient temperature ture rise is recorded as before. under these conditions 90°C than is B (for class allowed to sell his temperature to standards-setting may is If the tempera- insulation), the manufacturer product. is assumed in the to correspond to a hot-spot - rise of ( 120° - 40°) - 80°C 40°) = is rise assumed of ( 30° 1 90°C. The average temperature of by the resistance method. It a winding is A 75 kW known winding temagain when the machine is hot. For example, if the winding is made of copper, we can use the following equation (dethe winding resistance at a perature, and measuring it F, operates If U = (125° (6.5) The motor F 234 = = 93°C permissible hot-spot tem- insulation average temperature of the winding when rise is 32°) to Fig. 6.7, the perature rise for class 15°C. - 234 where Solution 1 /,) 125°C, does the motor meet the The hot-spot temperature = U = ^(234 + the hot- mpe rat Lire stand ard s ? According av- at full- R, ambient temperature of 32°C. spot temperature is te its erage temperature: motor, insulated class load in an found consists of measuring rived from Eqs. 6.1 and 6.2) to determine Example 6-3 case average winding tempera- correspond to a hot-spot temperature lie equal to or less 120°C For example, 2, Fig. 6.7. insulation, an temperature of 130°C. Consequently, an average has been es- anywhere between 10°C and 40°C. The hottest-spot temperature ture of curve B of class is (155° - 40°) easily meets the temperature R2 = R = ] t\ — hot f°C] a constant equal to 1/a = 1/0.004 27 hot resistance of the winding cold resistance of the winding temperature of the winding [°ci when cold 1 ELECTRICAL MACHINES AND TRANSFORMERS 30 Knowing winding temperature by the the hot we can immediately sistance method, Alternatively, re- insulation. calculate the temperature rise If this within the permissible limit (80°C for class lation), the product is B falls insu- acceptable from a standards when performance point of view. Note that tests are carried out using the resistance method, the ambient temperature must again If the lie winding happens wire, Eq. 6.3 can must be changed between 10°C and 40°C, to be be used, still made of aluminum but the number 234 to 228. A dc motor be rewound using class F a very last resort, A word of final may caution: temperature rise stan- also on the type of apparatus (motor, transformer, relay, etc.), the type of construction (drip-proof, totally enclosed, etc.), and the of application of field the apparatus (commercial, industrial, naval, be consulted before conducting a heat-run machine or device test been idle for several is days found to in an il. The motor then opwhen temperatures have staresistance is found to be 30 il. The maximum Although bilized, the field basic physical size depends corresponding ambient temperature built with class B is 24°C. If the allowable temperature rise power shown in Fig. 6.8. Suppose we have to build another generator having the same power and the winding, at full- To generate dards we either have the to same voltage at half the speed, double the number of conductors on the armature or double the flux from the Consequently, we must practice, The average temperature of the shunt full-load t2 increase both. We conclude that for a (R 2 /R i ) true for both ac - 234 + 19) - 234 + (234 (30/22) (234 /,) The average temperature its rise at full-load is c. - 24° = 87°C. The maximum allowable temperature resistance for class B insulation is rise (120° - kW, 2000 r/min motor by 40°) 80°C. Consequently, the motor does not meet the standards. Either its rating will have be reduced, or the cooling system improved, can be put on the market. is machine depends uniquely torque. Thus, a 100 111° it always and dc machines. Basically, the size of a upon b. before is bigger than a high-speed machine (Fig. 6.9). This is = = we given power output, a low-speed machine field at = 111°C = poles. either increase the size of the armature, or increase the size of the poles. In Solution a. volt- age, but running at half the speed. load The full-load temperature rise by the resistance method Whether the motor meets the temperature stan- its upon the power and Consider the 100 kW, 250 V, 2000 r/min generator The average temperature of es- rating of a machine, speed of rotation. insulation, calculate the following: c. a and size of a machine have a tablishes the nominal b. on (Fig. 6.10). shunt-field resistance of 22 a. etc.). Consequently, the pertinent standards must always erates at full-load and, is size 6.10 Relationship between the speed that has ambient temperature of 19°C, motor its dards depend not only on the class of insulation, but specific Example 6-4 may have to be increased. corresponding temperature rise by subtracting the ambient temperature. As it to Figure 6.8 100 kW, 2000 r/min motor; mass: 300 kg. EFFICIENCY AND HEATING OF ELECTRICAL MACHINES // 1 te nn eel la te A 6-9 1 3 level V dc motor connected to a 240 line pro- duces a mechanical output of 160 hp. Knowing 6-10 A 1 V 5 1 kW, that the losses are 12 late the input power and calcu- the line current. dc generator delivers 20 1 A to a load. If the generator has an efficiency of power 81 percent, calculate the mechanical needed 6- Figure 6.9 1 1 it [hp]. Calculate the full-load current of a 250 hp, 230 100 kW, 1000 r/min motor; mass: 500 kg. to drive V dc motor having an efficiency of 92 percent. 6-12 same physical size as a 10 kW motor 200 r/min because they develop the same A machine having class B insulation attains has about the a temperature of running at torrid torque. a. Low-speed motors are therefore much more costly than high-speed for b. it is often cheaper to use a small 6- 1 3 high-speed motor with a gear box than to use a large low-speed motor directly coupled to its load. Practical level 6-15 What causes the losses in a dc motor. iron losses Explain and how can they why the temperature of a What determines If the power electric tor in ma- the vents in a motor, its by always low of 6-16 it delivers an output of motor driving m a skip hoist with- of minerals from a 1.5 metric tons deep every 30 seconds. Thermocouples power must be reduced. Explain. per- the mo- in kilowatts. are used to measure the winding temperature of insulated class runs at full-load, what If the 94 power output of horsepower and kW ac motor, out- nomi- its hp. An cent, calculate the machine rating of a so, if Calculate the efficiency of the motor in nal hot-spot we cover up put is hoist has an overall efficiency of chine? 6-5 the temperature rise? efficiency of a motor trench 20 increases as the load increases. 6-4 machine running too hot and, draws be reduced? 6-3 Is the The 1 6-2 8()°C. is Example 6-2 when Questions and Problems Name 1 What when it operates at 10 percent nal power rating. Explain. 6-14 6-1 resistance) in a how much? motors of equal power. Consequently, low-speed drives, 208°C (by ambient temperature of is the F. If a inter- 1200 the motor maximum tem- perature these detectors should indicate in an 6-6 If a motor operates may we load it in above a cold environment, its rated ambient temperature of 40°C? 3()°C? 14°C? power? 6-17 Why? 6-7 Name some A 60 hp ac motor with class F insulation has a cold winding resistance of 12 of the factors that contribute to 23°C. the deterioration of organic insulators. When it runs at rated load in 11 at an am- bient temperature of 3 °C, the hot winding 1 6-8 A motor is built with class H insulation. What maximum hot-spot temperature can withstand? resistance it is found to be 17.4 il. a. Calculate the hot winding temperature. b. Calculate the temperature rise of the motor. 1 ELECTRICAL MACHINES AND TRANSFORMERS 132 Could the manufacturer increase c. plate rating of the 6-18 An motor has a normal electric when eight years is 30°C. If it is 6-24 life of is service A No. 1 6-25 of the life Knowing The current density [A/mm b. The specific copper losses is the conductor No. 4 A/mm at Express the current density is its in the rise its is losses. 6-27 On the reasonably is is tion 1 is What is its rewound using F ft cable dc circuit to carry 48 A. Assuming a The power maximum loss, in watts, in the The approximate voltage 6-28 In Problem 6-27, if 2-conductor at the 243 V when rying a current of 60 A, what insu- load end is if V. the voltage drop in the cable must not exceed 10 motor having class B insula- Assume would normally have a service life of h, provided the winding tempera- of 70°C. by resistance does not exceed 120°C. By how many hours duced RW 75. A 420 V a max- No. 6 gauge cop- the voltage at the service panel expected class Code allows in a cable situ- 20 000 ture in it is car- minimum conductor size would you recommend? kW ac 1 that the following: a. lation? An found operating temperature of 70°C, calculate deliver at a temperature rise of two years. A being used on a 240 b. it By 25°C. at it is AWG wire size, and Electrical current of 65 a current of 6.5). ated in a particularly hot location has a ser- if ohms of 56 per conductor, type nominal rating its electromagnet (insulated class A) span 27 meters long. is of a 4-pole dc motor has a Determine the The National imum Based on these facts, if a 20 kW motor has a fullload temperature rise of 80°C, what life field calculate the weight of the wire per pole, range between 50 percent example, Fig. it AWG that The shunt inches. in circular mils of a motor efficiency and 150 percent of power can of 105°C? 6.2, calculate the resistance un- the bare copper wire has a diameter of 0.04 120°C, calcu- fW/kg]. roughly proportional to (see, for be as high as 70°C. kilograms. The temperature other hand, AWG conductor in scraping off the insulation, b. constant may total resistance a cur- 2 conductor temperature it is der these conditions of a 2-conductor cable ] per ampere. 6-23 the prop- lists of commercially available copper an area where the operating temperature of 105°C, cal- conductor operates If the life pounds. in AX3 Appendix proposed to use a No. 4 [W/kgl a. vice a temper- at that the 6-26 An aluminum An ohms 35 long level late the specific losses 6-22 in Using Eq. a. rent density of 2 1 1 conductors. In an electrical installation, temperature of the conductor 6-2 The Table erties culate the following: 6-20 0. weight of the conductor 60°C, 210m 0 round copper wire carries a current of 12 A. Advanced of No. 2/0 single copper conductor reel ature of 25°C. Calculate the approximate motor? 6-19 A has a resistance of installed in a location new probable the is Industrial application the ambient temperature where the ambient temperature what name- the motor? Explain. if the is motor runs ature (by resistance) of the service life re- 6-29 A dc a maximum operating temperature busbar 4 inches wide, 1/4 inch thick, and 30 feet long carries a current of 2500 A. Calculate the voltage drop for 3 h at a temper- temperature of the busbar 200°C? the power loss per meter? is 1()5°C. if the What is EFFICIENCY AND HEATING OF ELECTRICAL MACHINES 6-30 Equation 6.3 gives the resistance/temperature relationship of brush pressure: copper conductors, 1 33 1.5 lbf brush contact drop: 1 .2 V namely, coefficient of friction: 0.2 t2 = R 2 IR (234 X + t x ) - 234 Calculate the following: Using the information given AX2, deduce minum 6-3 1 in Appendix a. a similar equation for alu- b. conductors. c. The commutator of a .5 hp, 2-pole, 3000 r/min dc motor has a diameter of 63 mm. the contact voltage drop 1 d. Calculate the peripheral speed in feet per minute and 6-32 in miles per hour. e. The following information is given on the brushes used in the motor of Problem 6-3 newtons) number of brushes: 2 1 5 3/4 in long. resistivity The frictional brushes A (The 5/16 in when energy expended by the two the commutator makes one revo- lution (in joules) brush dimensions: 5/8 in wide, 5/16 in thick, contact with the The total electrical power loss (in watts) due to the two brushes The frictional force of one brush rubbing against the commutator surface (in lbf and in 1 f. current per brush: The resistance of the brush body in ohms The voltage drop in the brush body The total voltage drop in one brush, including X 5/8 in area commutator) of brush: 0.0016 Il.in g. is in h. The power loss due of 3000 r/min The total brush loss motor rating to friction, given the speed as a percent of the 1 .5 hp Chapter 7 Active, Reactive, and Apparent Power 7.0 Introduction The 7.1 The instantaneous power supplied concept of active, reactive, and apparent power plays a major role in electric power Instantaneous power nology. In effect, the transmission of electrical energy across and the behavior of ac machines are often easier that to encouraged to The terms The reader pay particular attention active, reactive, is which the positive value always expressed is means may be that in The A positive or negative. power flows into the de- vice. Conversely, a negative value indicates that in power is flowing out of the device. to describe transient-state be- we apply them to dc circuits. Our study begins with an analysis of the instantaneous power in an ac circuit. We then go on to define the meaning of active and reactive power and how to identify sources and loads. This is followed by a definition of apparent power, power factor, and the power triangle. We then show how ac circuits can be solved using these power concepts. In conhavior, nor can clusion, vector notation active and reactive it. instantaneous power to this chapter. and apparent power voltages and currents are sinusoidal. They cannot be used is terminals times the instantaneous current watts, irrespective of the type of circuit used. therefore apply to steady-state alternating current circuits its flows through Instantaneous power understand by working with power, rather than dealing with voltages and currents. to a device simply the product of the instantaneous voltage tech- power is in Example A sinusoidal voltage having a peak value of 162 and a frequency of 60 Hz is V applied to the terminals of an ac motor. The resulting current has a peak value of 7.5 a. A and b. 134 in terms of the <J>. Calculate the value of the instantaneous voltage and current circuit. lags 50° behind the voltage. Express the voltage and current electrical angle used to determine the an ac 7-1 at an angle of 120°. ACTIVE, REACTIVE, c. Calculate the value of the instantaneous p = power ei = AND APPARENT POWER X 140.3 = + 989 7.05 1 35 W at 120°. d. Plot the Because the power curve of the instantaneous power deliv- positive, is flows it at this instant into the motor, ered to the motor. d. a. In order to plot the power, Solution Let us assume that the voltage starts at zero We increases positively with time. and can therefore we for angles ranging Table 7A curve of instantaneous repeat procedures (b) and (c) above lists from (J) = 0 to <J) = 360°. part of the data used. write e = Em sin § = 1 62 sin TABLE 7A (f> = / b. At 50°, consequently, = /m (J) e / c. = sin (<(> - G) = 162 sin 120° = 140.3 = 7.5 sin (120° 7.5 = 7.05 Voltage 7.5 sin (c|> 162 sin - degrees 50°) = X 162 - 50°) = 0.866 7.5 sin 70° 0.94 A The instantaneous power at 120° p USED TO PLOT in an ac circuit. (]j Power Current 7.5 sin <t|j - volts amperes 50°) P watts 0 -5.75 0 -3.17 -218 50 124.1 0 75 156.5 3.17 497 0 115 146.8 6.8 1000 155 68.5 7.25 497 180 0 5.75 0 3.17 -218 205 -68.5 230 -124.1 is power AND 68.5 Figure 7.1 Instantaneous voltage, current and i, 25 0 V X Angle can write we have 120° = = we e, FIG. 7.1 The current lags behind the voltage by an angle 0 VALUES OF (See Example 7-1 .) 0 0 ELECTRICAL MACHINES AND TRANSFORMERS 136 The of power voltage, current, and instantaneous The power plotted in Fig. 7.1. attains a positive + 000 W and a negative peak of - 2 1 ative power means that power is 1 are peak W. The neg- 8 actually flowing 0-50°, 180° -230°, 360° -410°. and Although a power flow from a device considered to be impossible, it happens often in We is given The simple to be 20 power cycle may seem The the sections that follow. quency of s. 120 Hz, which the voltage and current is to an ac generator. and as we would expect E and / The line E and a reading To / are effective values. the instantaneous values of voltage did power * con- the sinusoidal curves of The peak values are and V2/ amperes because, as (Fig. 7.2d). ^2E volts ously, E and we we will give watts (Fig. 7.2c). we have drawn circuit, / it get a better picture of what goes on in such a produce the and respec- /, are in phase (Fig. 7.2b). If P = El re- effective in a resistive circuit, nect a wattmeter (Fig. 7.3) into the line, twice the fre- that quite normal: the voltage and current are designated phasors This means that the frequency of is is always twice the ac circuit of Fig. 7.2a consists of a connected ac circuits. also note that the positive peaks occur at in- tervals of 1/1 the in is power* 7.2 Active tively, reason phenomenon frequency. sistor a load to a device considered to be a source this from the load (motor) to the source. This occurs during the intervals power. Again, frequency of ac power flow in Section 7.1, we E respectively stated previ- By multiplying and current as obtain the instantaneous in watts. Many persons refer to active power as real power or true power considering it to be more descriptive. In this book we use the term active power, because conforms to the IEEE designation. it Figure 7.2 a. An ac voltage E produces an ac current / in this re- Figure 7.3 Example of a high-precision wattmeter rated 50 V, 200 V; 1 A, 5 A. The scale ranges from 0-50 sistive circuit. c. Phasors E and / are in phase. A wattmeter indicates El watts. d. The b. tive active power power pulses. is composed V, 100 W to of a series of posi- 1000W. (Courtesy of Weston Instruments) 0- ACTIVE, REACTIVE, The power wave consists of a from zero pulses that vary to a = 2EI = 2P value of The fact that X (V2/) power is always positive reveals watts. that always it flows from the generator to the resistor. This pow properties of what although er: imum, it 137 series of positive maximum (i2E) of the basic AND APPARENT POWER is is one called active pulsates between zero and max- never changes direction. The direction of it power flow is shown by an arrow P (Fig. 7.2c). is clearly midway between The average power IP and zero, and so That watts. is = EI power indicated by the value its precisely the P = is 2E//2 wattmeter. The two conductors leading to the resistor in Fig. 2EJ power. However, unlike cur- 7.2a carry the active rent flow, power does not flow down one conductor Power flows over both conconsequently, as far as power is con- and return by the other. ductors and, cerned, line, In as we can replace the conductors by shown in Fig. 7.2c. general, the line represents any transmission number of conductors The generator an active load. and the unit megawatt is is it may / Reactive power consists of a series of positive and lags 90° behind E. are frequently used multiples of back and forth in this manner is power (symbol Q), to distinguish it from the unidirectional active power mentioned before. The reactive power in Fig. 7.4 is also given by the product EL However, to distinguish this power from active power, another unit is used the van Its identical to the resistive is 7.2a) except that the resistor by a reactor X E . x As is a result, current / now re- therefore again is twice the line frequency. Power that surges drawn the waveforms for in such a E and / power circuit, we and, by again we obtain (Fig. 7.4c). This power p consists of a series of identical positive and negative pulses. instantaneous the reactor The positive waves correspond power delivered by the generator to and the negative waves represent instan- power delivered from generator. to the reactor to the The duration of each wave corresponds The one quarter of a cycle of the line frequency. megavar (Mvar). Special instruments, called vanneters, are available to measure the reactive multiplying their instantaneous values, curve of instantaneous — multiples are the kilovar (kvar) and lags 90° (Fig. 7.4b). To see what really goes on to Phasor frequency of the power wave behind the voltage taneous b. c. called reactive circuit (Fig. the this in- The symbol for active power is P The kilowatt (kW) and The circuit of Fig. 7.4a have / in negative power pulses. Reactive power placed current circuit. have. the watt. 7.3 An ac voltage E produces an ac ductive an active source and the resistor the watt (W). (MW) Figure 7.4 a. connecting two devices, irrespective of the line is a single (C) 7.5). A line voltage sin 8 if E (where B reading power in a circuit (Fig. varmeter registers the product of the effective is times the effective line current is the phase angle only obtained between when E and / / times E and /). A are out of phase; they are exactly in phase (or exactly 180° out of phase), the varmeter reads zero. Returning pulse is to Fig. 7.4, the dotted area under each the energy, in joules, transported in direction or the other. Clearly, the energy is one deliv- ered in a continuous series of pulses of very short duration, every positive pulse being followed by a ELECTRICAL MACHINES AND TRANSFORMERS 138 By definition*, a reactor is considered to be a re- active load that absorbs reactive power. Example A 7-2 reactor having an inductive reactance of 4 connected of a 120 to the terminals V 0 is ac generator (Fig. 7.6a). a. Calculate the value of the current in the reactor b. Calculate the power associated with the reactor c. Calculate the power associated with the ac generator d. Figure 7.5 Varmeter with a zero-center scale. or negative reactive power flow up It to Draw the phasor diagram for the circuit indicates positive 100 Mvars. , 4 Q 30 A| negative pulse. The energy flows back and forth between the generator and the inductor without ever being used up. What tive is A the reason for these positive and nega- , energy surges? The energy flows back and forth because magnetic energy + rr^> 120V alternately being stored is 3.6 kvar positive, the magnetic field is inside the coil. A moment is when later L l 4J 30 A up and released by the reactor. Thus, when the power / building up the power negative, the energy in the magnetic field is is E de- 120 V creasing and flowing back to the source. We now tive have an explanation for the brief nega- power pulses in Fig. 7.1. In effect, they repre- 30 A (c) sent magnetic energy, previously stored up in the motor windings, that is being returned to the source. Figure 7.6 See Example 7.4 Definition of a reactive load 7.2. and reactive source Solution a. Current in the circuit: Reactive power involves real power that oscillates back and forth between two devices over a transmission line. Consequently, is it whether the power originates at /,= impossible to say one end of the line assume that b. or the other. Nevertheless, some devices generate absorb it. In other it is useful to reactive power while others E = XL 120 4 Power associated with Q= EI = 120 V = 30 A f1 the reactor: X 30 - 3600 var = 3.6 kvar words, some devices behave like reactive sources and others like reactive loads. This definition is in agreement with IEEE and 1EC conventions. ACTIVE, REACTIVE, power This reactive is AND APPARENT POWER equal to the current is Because the reactor absorbs 3.6 kvar of reactance times the voltage across ries power, the ac generator must be supplying Consequently, the generator power: delivers 3.6 kvar. it The flows therefore in the direction reactive shown power Q The = EIC 120 reactive pressed car- it namely V X A= 30 power delivered by in vars = 3600 var 3.6 kvar the capacitor power or kilovars. Reactive is ex- Q now flows from the capacitor to the reactor. We This phasor diagram applies (the reactor) and the reactive source (the ac genera- tor) as Q = (Fig. 7.6b). The phasor diagram is shown in Fig. 7.6c. Current // lags 90° behind voltage E. d. terminals, its it. a source of reactive is 39 source of reactive power. The reactive power deliv- absorbed by the reactor. ered by the capacitor c. 1 to the reactive load have arrived very important conclusion: at a a source of reactive power. a reactive power source whenever it is a capacitor is It acts as part of a sine-wave-based, steady-state circuit. well as the line connecting them. Let us take another step and remove the reactor 7.5 The capacitor and reactive power from the circuit in Fig. 7.7a, yielding the circuit of Fig. 7.8a. now Suppose that we add a capacitor having a reac- tance of 4 11 to the circuit of Fig. 7.6. This yields the circuit of Fig. 7.7a. pacitor is /c expect, it = The current 120 V/4 il = 30 /L . drawn by A and, as the ca- we would leads the voltage by 90° (Fig. 7.7b). rent of capacitor is now alone, connected to It still 30 A, leading the voltage 7.8b). Consequently, the capacitor E carries a cur- by 90° still this power go? The answer power delivers reactive is to the (Fig. acts as a source of reactive power, delivering 3.6 kvar. does The vector sum of I L and / c is zero and so the ac is no longer supplying any power at all to The the terminals of the ac generator. Where that the capacitor very generator to generator the circuit. not 30 However, the current changed; consequently, AX 1 Where can only V = 20 is it in the reactor has continues to absorb 3.6 kvar of reactive power. power coming from? capacitor, which acts as this reactive come from the c It 3.6 kvar a -4/ 30 A (a) 'c 30 A (b) 120 V varmeter 30 A (c) LTi capacitor 120 V (b) Q = FJ C Figure 7.8 30 A Figure 7.7 See Example 7.3. a. Capacitor connected to an ac source. b. Phasor c. Reactive power flows from the capacitor to the l c generator. leads Eby 90°. 1 ELECTRICA L MA CHINES A ND TRA NS FORMERS 40 which it is connected! For most people, this takes a time to accept. little How, we may ask, can a passive device like a capacitor possibly produce any power? The answer energy is that reactive power really represents a pendulum, swings back and forth that, like without ever doing any useful work. The capacitor acts as a temporary energy-storing device repeatedly accepting energy for brief periods and releasing as a reactor does, a capacitor stores electrostatic en- we connect 7.8c), var, it that reactive from the capacitor power is to the generator. El = -3600 The generator X = 400 3.5 1 prefer to call a receiver of reactive power, which, of same thing. In cerned the generator acts as a load. The active power absorbed by active power = 1 circuit W) power (1304 power (812 var). tive 7.9). The power associated with L, C respective elements is a basic difference remember is between active and most important re- thing one cannot be converted and reactive powers function independently of each other and, consequently, in electric a burden on the transmission line that whereas active power eventually 3./ produces a tangible result (heat, mechanical power, light, etc.), reactive 14 A }= A that oscillates 4 -3.5./ power only represents power back and forth. All ac inductive devices such as magnets, trans- Q formers, ballasts, and induction motors, absorb 20 A re- power because one component of the current they draw lags 90° behind the voltage. The reactive power plays a very important role because it proactive Figure 7.9 duces the ac magnetic field See Example a source of ac- they can be treated as separate quantities carries them, but, -+* 16.12 that the into the other. Active Both place - is and a receiver of reactive circuits. the generator. 2Q 1=3 a source of between active and reactive power There carry the currents shown. Calculate the active and reactive is 7.6 Distinction to connected to a group of R, elements (Fig. it 304 W. active power, and perhaps the is the resistors must be supplied by the generator; hence summary, a ca- pacitive reactance always generates reactive power. Example 7-3 An ac generator G var is we sometimes like reactive load, but course, amounts to the 2 indeed flowing now behaving it 20 In conclusion, the ac generator a varmeter into the circuit (Fig. will give a negative reading of showing XC = 1 ergy (see Section 2.14). If 3.5 il capacitor generates reactive power: 2 1 The R, L, C circuit generates a net reactive power of 400 - 588 = 8 12" var This reactive power must be absorbed by the generator; hence, as far as reactive power is con- it However, instead of storing magnetic energy again. The Qc = 7.3. A building, in these devices. shopping center, or city may be con- sidered to be a huge active/reactive load connected an electric Solution to The two tain resistors absorb active P = 2 1 2 power given by + (16.12 2 X 784 + 520 = 304 W R = (14 X 4) 1 The 3 II reactor absorbs reactive Q L = 1% = l4 2 X 3 = 588 var power: 2) = utility system. Such load centers con- thousands of induction motors and other elec- tromagnetic devices that draw both reactive power (to sustain their (to magnetic fields) and active power do the useful work). This leads us to the study of loads that absorb both active and reactive power. AND APPARENT POWER ACTIVE, REACTIVE, 7.7 Combined and reactive loads: apparent power The active flow in the the arrows in Fig. P Loads that absorb both active power power tance ple, Q may be considered to be and reactive made up of a the circuit actor are of Fig. 7. 1 Oa in which a resistor and p , while the reactor draws a current / q According to our definitions, the resistor active load while the reactor Consequently, behind. I p is in is phase with The phasor diagram an a reactive load. E while (Fig. 7. the resultant line current / lags . is 1 /t] lags 90° Ob) shows that behind Furthermore, the magnitude of E / is and a varmeter by an an- 7. into the circuit, the readings will both be positive, indicating £/q re- connected to a source G. The resistor draws a current / gle 6. resis- and an inductive reactance. Consider, for exam- power components P same direction, as shown by 10c. If we connect a wattmeter active and reactive Q both and 141 P = £V p Q= watts and vans, respectively. Furthermore, if we connect an ammeter into the will indicate a current of / line, it sult, we amperes. As a plied to the load is equal to EI watts. But viously incorrect because the an active component (watts) is nent (vars). For this reason the product EI Apparent power is ob- this is composed of and a reactive compopower called is apparent power. The symbol for apparent power given by re- power sup- are inclined to believe that the expressed neither in is 5*. watts nor but in voltamperes. Multiples are the kilo- in vars, voltampere (kVA) and megavoltampere (MVA). 7.8 Relationship between P Q, 3 and S Consider the single-phase circuit of Fig. 7. 1 1 a com- source] posed of a source, a load, and appropriate meters. Let us assume that (b) E • the voltmeter indicates • the • the wattmeter indicates • the varmeter indicates ammeter Knowing indicates / that volts amperes +P watts +Q vars P and Q are positive, we know that the load absorbs both active and reactive power. Consequently, the line current an angle Current nents I p / lags behind E. lb by 8. / and rature, with values of / can be decomposed into two compo/C] , respectively in phase, and phasor and / L| E (Fig. 7.1 lb). in quad- The numerical can be found directly from the in- strument readings / l Figure 7.10 a. Circuit consisting of a source feeding an active and reactive load. Phasor diagram of the voltage and currents. c. Active and reactive power flow from source to load. q = PIE = QIE (7.1) (7.2) Furthermore, the apparent power 5 transmitted over the line b. P is given by S El, I = from which S/E (7.3) 1 ELECTRICAL MACHINES AND TRANS EORMERS 42 Figure 7.11 Instruments used to measure £, P, and Q in a circuit. The phasor diagram can be deduced from the instrument readings. a. /, b. Referring to the phasor diagram (Fig. 7.1 lb), it is Example 7-5 A wattmeter and varmeter are connected into a obvious that r- 2 = / p single-phase line that feeds an ac motor. 2 + / q spectively indicate 1 W and 960 800 20 V 1 They re- var. Consequently, Calculate -> P E s E That 2 Q + The in-phase and quadrature components a. E and c. in 0 2 (7.4) d. line current which = P = Q= S We Solution apparent power [VA] Referring to Fig. power [Wl reactive power [var] active tor, can also calculate the value of the angle 0 be- cause the tangent of 6 Thus, is obviously equal to / // p / a. / . b. p q = arctan / // q p = arctan QiP 1 1 , where the load is now = PIE = 1 = QIE = 960/120 800/120 = - 15 8 A A From the phasor = V7 2 + Example 7-4 alternating-current motor absorbs 40 diagram we have (7.5) / tive 7. / 2 = Vl5 2 + 8 2 = 17A kW of ac- power and 30 kvar of reactive power. Calculate power supplied to the motor. c. The apparent power is the apparent S = EI= 120 X 17 = 2040 VA Solution S = VP^Vq 2 ~ V40 r + 30 2 = 50 kVA d. a mo- we have we have 0 An / q The line current / The apparent power supplied by the source The phase angle between the line voltage and b. is, 2 2 S = P + / The phase angle 6 between E and / is (7.4) 6 = arctan = 28.1° QIP = arctan 960/1800 (7.1) (7.2) ACTIVE, REACTIVE, The power Example 7-6 A voltmeter and ammeter connected tive circuit into the induc- of Fig. 7.4a give readings of 140 V and AND APPARENT POWER factor of a resistor cause the apparent power tive power. On it equal to the ac- is the other hand, the an ideal coil having no resistance 20 A, respectively. 100 percent be- is draws 143 power factor of zero, because is it does not consume any active power. Calculate To sum The apparent power of the load The reactive power of the load a. b. is parent The active power of the load c. up, the power factor of a circuit or device simply a way of stating what fraction of power is real, In a single-phase circuit the The apparent power S and current. Thus, referring is Q = factor also is to Fig. 7.11, PIS = El p IEI is = v = = cos B El = 2800 would give a reading of 2800 factor X 20 140 var power = 2.8 kvar Consequently, vanneter were connected into the circuit, If a power = = EI= 140 X 20 = 2800 VA = 2.8 kVA The reactive power b. ap- a measure of the phase angle 6 between the voltage Solution a. its or active, power. power it factor = cos 6 = P/S (7.7) var. where The active power c. is zero. power If a wattmeter were connected into the circuit, factor 0 but because the current voltage, 7.9 it is is also equal to is 2800 VA, 2800 voltage and current is the ratio of the active power S. It is power P to the ap- said to be lagging factor = leading (7.6) power factor is said to be the current leads the voltage. in Example 7-5 and the phase angle between the line voltage power delivered or absorbed by circuit or device [W] active apparent Power factor a if Calculate the power factor of the motor and = the current lags behind the Example 7-7 PIS where S if voltage. Conversely, the given by the equation power P = phase angle between the If we know the power factor, we automatically know the cosine of the angle between E and / and, hence, we can calculate the angle. The power factor var. The power factor of an alternating-current device or parent = 90° out of phase with the Power factor circuit is factor of a single-phase circuit or device would read zero. To recapitulate, the apparent power = power it is power of the circuit or device line current. the Solution [VA] power factor expressed as a simple number, or as = PIS = 1800/2040 - 0.882 or 88.2% percentage. Because the active power apparent power S, it P can (lagging) never exceed the follows that the power factor can never be greater than unity (or 100 percent). cos 0 = 0.882 therefore, 0 = 28.1° ELECTRICAL MACHINES AND TRANSFORMERS 144 Example 7-8 A single-phase 20 V, 60 Hz 1 motor draws a current of 5 A from a The power factor of the motor is line. 65 percent. Calculate a. b. The The power absorbed by the motor power supplied by the line active reactive Solution a. The apparent power drawn by = 5m EI = 120 X 5 the motor is - 600 VA ^(390 W) The power absorbed by active the motor is P m = S m cosQ = 600 X b. The 0.65 - 390 power absorbed by reactive (7.7) Figure 7.12 Power triangle is 4. the motor VSI^ the that power from amount of nonpro- Power Active power is considered P absorbed by to Active power device is P a circuit or device be positive and that is drawn hori- Reactive power vice is delivered by a circuit or considered to be negative and drawn horizontally 3. is Q upwards is to the left when is drawn circuit is Example 7-8 and Q shown in The power is rules. look like phasors, but they The concept of the power as convetriangle is solving ac circuits that comprise sev- and reactive power components. Further aspects of sources and loads 1 Let us consider Fig. 7.13a in which a resistor and capacitor are connected to a source. The circuit similar to Fig. 7. 10 except that the capacitor As power flows from is a is re- a result, reactive power flows G while active to the source the source active and reactive G to the resistor. power components The therefore flow in opposite directions over the transmission line. A wattmeter connected into the circuit give a positive reading ter will absorbed by a circuit or de- considered to be positive and vertically S, P, from the capacitor zontally to the right 2. triangle for However, we can think of them nient vectors. useful by a downwards accordance with these active source. convention, the following rules apply: . are not. 7.1 triangle 2 2 2 The S = P + Q relationship expressed by Eq. 7.4, brings to mind a right-angle triangle. Thus, we can show the relationship between S, P, and Q graphically by means of a power triangle. According to 1 in eral active ductive power. 7.10 2 1 that is delivered considered to be negative and vertically components the line than active power. This burdens the line with a relatively large is The power Fig. 7. motor draws even more reactive Q Reactive power drawn (7.4) = V600 2 - 390 2 = 456 var Note a motor. See Example 7-8. W or device Qm = of G P = EI p power Q. Thus, Q— ElLy The source power P but receives reactive give a negative reading delivers active will watts, but a varme- G is simultaneously an active source and a reactive load. ACTIVE, REACTIVE, AND APPARENT POWER 145 device (or devices) connected to the receptacle. If the device absorbs active power, the receptacle will provide G (a) E (source) if it; the device delivers active power, the receptacle will receive 7 p simple receptacle outlet 7 g Y liver tive — or accept power nected to Q — In other words, a it. is at all either active times ready to de- power P or reac- accordance with the devices con- in it. The same remarks apply to any 3-phase 480 V service entrance to a factory or the terminals of a (b) kV high-power 345 Example A 50 jjiF transmission line. 7-9 paper capacitor terminals in Example is placed across the motor 7-8. Calculate a. (c) b. c. d. The reactive power generated by the capacitor The active power absorbed by the motor The reactive power absorbed from the line The new line current Solution a. Figure 7.13 a. The impedance of Source feeding an active and reactive (capacitive) Xc = load. b. Phasor diagram c. The active of the circuit. and reactive powers flow in opposite di- the capacitor is 1/(2 77/C) = 1/(217 = 53 (2.11) x 60 x 50 x 1()" 6 ) il rections. The current It in may seem unusual to we must line, but P not the is / have two powers flowing opposite directions over the remember that active power a reactive power Q and that again same as The reactive each flows independently of the other. tacle in a home, also deserves our such outlets are ultimately alternators that power Qc active also 20 V attention. All connected may seem, an re- source (as we would as it expect), but behave as an active or reactive load. factors huge to the can act not only as an active or Odd determine whether way or the other? It all it will behave A £/ - q - 120 X the capacitor 2.26 271 var recep- the electrical transmission and distribution systems. electrical outlet 1 2.26 is 120/53 power generated by Speaking of sources and loads, a deceptively simple electrical outlet, such as the capacitor = EIXC = = same transmission in the it may What in one depends upon the type of b. The motor continues to draw the same power because it is still fully loaded. active Consequently, P m = 390 The motor also draws the W same reactive power as before, because nothing has taken place to change its magnetic field. Qm = Consequently, 456 var is ELECTRICAL MACHINES AND TRANSFORMERS 146 The motor draws 456 var from c. the line, but the same power drawn from the capacitor furnishes 271 var to the line. The line net reactive is, therefore, Ql = Q m - Q c = 456 = The active 185 var power drawn from P\. d. 271 the line = 390 W = The apparent power drawn from - V390 + - 432 VA 2 The new 1 the line /W390 W) is Figure 7.14 Power triangle of a motor and capacitor connected an ac line. See Example 7-9. 2 85 432/120 nected in 7. 15a). We wish A to 3.6 A by This represents a big improvement because the line smaller and the operation of the motor has factor of the line 4), = /ys L = = <|> L = 390/432 We concerned) of active and reactive power flow arcos 0.903 = delivers reactive arrow active and reactive active powers power P. We rather complex inductive, cir- Consider, for example, a group of loads con- On to the system. it ar- the other represents a capacitor, power it The 16 kvar in independent) nature of the powers enables us to add all the a circuit to obtain the total active same way, we can add the reactive power Q. The reapparent power S is then found by In the to obtain the total reactive S The concept of active and reactive power enables us some C distinct (and Systems comprising simplify the solution of is directed therefore toward the source. is The powers the line. several loads to A the source to the load. 25.5° sulting total cuits. Thus, because load hand, because load observe the effect of the capacitor on the ap- 7.12 to simply draw a block diagram of the individual row flows from 0.903 or 90.3% power supplied by ab- absorbs reactive power; consequently, the 5 kvar The power triangle is shown in Fig. 7,14. The repower Qc generated by the capacitor is drawn vertically downward. By comparing this power triangle with that in Fig. 7.12, we can visuparent power Using the power approach, we do not have is is active ally V source (Fig. worry about the way the loads are interconnected. (Fig. 7.15b). cos 380 to a loads, indicating the direction (as far as the source not been changed in the least. The new power way to calculate the apparent by the source. placing the capacitor in parallel with the motor. is a very unusual sorbed by the system as well as the current supplied Thus, the line current drops from 5 current to line current is = SJE = = 3.6 A /j is recall that = \ P2 -\ when adding Q 1 (7.4) reactive powers, we assign a positive value to those that are absorbed by the system and a negative value to those that are generated (such as by a capacitor). In the same way, AND APPARENT POWER ACTIVE, REACTIVE, Reactive power absorbed by the system: 2. 380 V> 14kW D A 8 kvar 9 kvar (a) 0 + (5 + 7 Q 2 = (-9 Q Net reactive power 4. - -25 16) Q = + 20 kvar kvar absorbed by the system: = 25) ( 16 kvar = +20 8) Reactive power supplied by the capacitors: 3. 5 kvarf = G, 2kW 147 5 kvar kW 8 Apparent power of the system: 5. S = \ P2 7Q - 24.5 kVA V Because the 380 6. = 1 V24^+7~ 5? source furnishes the appar- ent power, the line current = I 7. The power cos = 4> L power, but it tem of the netic PIS = = 500/380 = 24/24.5 64.5 A is 0.979 (leading) kW source delivers 24 of active receives 5 kvar of reactive power. This power flows reactive comes = 24 factor of the system V The 380 (b) S/E is into the local distribution sys- company, where electrical utility available to create magnetic fields. fields may be it be- The mag- associated with distribution transformers, transmission lines, or even the elec- 8 kvar tromagnetic relays of customers connected to the Figure 7.15 a. b. Example of active and reactive loads connected to a 380 V source. All loads are assumed to be directly connected to the same distribution system. The power triangle for the system 7.15c. It is we Thus, starting with the 5 kvar load, 380 V receptacle. is shown move from one progressively device to the next around the system. While so doing, we draw the magnitude and we assign a positive value to active powers that are absorbed and a negative value to those that are generated (such as by an Note that usually powers we cannot add power total apparent their power factors are Let us I. now Active the apparent various parts of a circuit to obtain the in S. We can only add them power absorbed by if meet. 7. 1 5: (2 + 8 + 14) right) of point, direc- each power vector, tail accordance with the power of each device When we can starting point to the end the selection draw a power vector from the is complete, which yields the inclined vector having a value of 24.5 kVA. The horizontal component of the right, 24 we know kW and, because that by the system. The it represents vertical it is this vec- directed to power absorbed component of 5 kvar is the system: directed P = left, tor has a value of identical. solve the circuit of Fig. to head, in we alternator). down, tion (up, in Fig. the graphical solution to our problem. = +24 kW tive downward: consequently, power generated by it the system. represents reac- 1 48 ELECTRICAL MACHINES AND TRANSFORMERS 5 kvar starting point Figure 7.15c Power triangle of the system. 7.13 Reactive magnetic We with the rapidly operating switch rather than with power without the resistor fields sometimes encounter situations where loads absorb reactive power without creating any magnetic field at all. This can happen in electronic power circuits when the current flow is delayed by means of a rapid switching device, such as a itself. Nevertheless, reactive power consumed just as surely as in the circuit. This switching circuit will be cussed in detail in if is a reactor were present dis- Chapter 30. 7.14 Solving AC circuits using the power triangle method thyristor. Consider, for example, the circuit of Fig. 7. 16 in We have seen that active and reactive powers can be 60 Hz source is connected to a resistive load of 10 (2 by means of a sy nchronous mechanical switch. The switch opens and closes its rather contacts so that current only flows during the latter tation. which a 100 V, We can some to vector (j) no- calculate the active and reactive powers associated with each circuit element and deduce the corresponding voltages and currents. The following see, if we connected a wattmeter and varmeter between the source and the (sometimes called displacement power factor) of 84.4 percent. The reactive power example demonstrates the usefulness of power this triangle approach. W would respectively read +500 and corresponds to a lagging power fac- + 318 var. This tor draw a phasor diagram or resorting We to solve ac circuits without ever having to almost by intu- behind the voltage. Indeed, switch, they complex forced delay causes the current to lag part of each half cycle. ition, that this added algebraically. This enables us is associated Example 7-10 In Fig. 7.17a, the voltage is 60 V. between terminals 1 and 3 ACTIVE, REACTIVE, AND APPARENT POWER 149 0)^ 7.07 A (eff 5ft (a) 12H R 60 V T ©6 (a) 141 V S 14.1 © — A 9- 700 = r VA S --^ *l = 780 VA 1 r^rs_+_ I (b) 60 V Figure 7.16 a. Active and reactive power flow in a switched resis- tive load. b. Figure 7.17 The delayed current flow is the cause tive power absorbed by the system. of the reac- a. Solving ac circuits by the power triangle method. b. Voltages and currents in the circuit. See Example 7-10. from which the active power absorbed Calculate a. The current b. The voltage between terminals c. The impedance between terminals in each circuit element I P = and 2 l S Solution We know the impedances of the elements and (Fig. 7. 60 V exists between terminals 3 and We now proceed in logical steps, as follows: a. The current in the capacitor Ic = 60/5 = The current 12 = X 60 - -720 60/12 = 7b). The current 5 A /, = must, therefore, be = S/E 3] The voltage across A in the resistor is /R = VP + Q = V300 + (-720) 2 = 780 VA /1 from which the reactive power generated Qc = 1 that E 23 = IXL 780/60 = A 13 the inductive reactance =13X8= 104 is V is The var terminals 1-3: 2 1 2 is 12 W The apparent power associated with and 2 I X 60 = 300 5 is reactive reactance power absorbed by the inductive is G L = E23 X = +1352 /, = var 104 X 13 1 ELECTRICAL MACHINES AND TRANSFORMERS 50 The power absorbed by total reactive Qi. + Qc = = +632 var Q= the circuit 1.25 Mvar is 1352 0.2 MW T - 720 ]— substation The total active power absorbed by the circuit 15 is 12.47 kV 3 The apparent power absorbed by 2.8 MW 0.75 Mvar 2 Mvar 2 The voltage of £2 = , c. is C 10.03 kV 289 A MW = \P + Q 2 = V300 2 + 632 2 = 700 VA S b. the circuit load Q 2,4 W P = 300 L2 the line = SII { 10.03 kV therefore is 700/13 = V 53.9 The impedance between terminals 2-1 Z= E 2\U\. = 12.47 kV 53.9/13 = 289 A is 4.15 II 289 A Figure 7.18 Voltages, currents and power. See Example 7-1 1 Example 7-11 A kV single-phase 12.47 transmission line several C from kilometers long feeds a load (Fig. 7.18). The reactance of 15 Instruments at the inputs to 6 = arccos 0.833 power dissipated b) Active MW and 2 Mvar, respectively. P = RI2 = = Calculate the line current and phase angle with respect its X 0.2 = 33.6° in the line: 2.4 Y a) at the substation: substation in- power dicate that the active and reactive the line are 3 Phase angle between the voltage and current has a resistance of 2.4 il and a line il. a substation 10 6 X 289 2 = 0.2 MW Active power absorbed by the load: to the line voltage at the substation b) the active — power absorbed by the load power absorbed by the load c) the reactive rL ^sub = 2.8 MW - 0.2 MW 3 MW d) the line voltage at the load e) the and phase angle between the voltage at the load c) Q L = XJ = Solution a) Reactive power absorbed by the 2 that at the substation 15 X 289 2 = 1.25 X line: 10 6 = 1.25 Reactive power absorbed by the load: Apparent power delivered VP S to the line: V3 + + Q2 = 2 3.60 2 2 Qc 2 = Qsub - Gl = = 0.75 Mvar 2 Mvar - 1.25 Mvar MVA d) Apparent power at the load: Line current: / = S = 3 600 000 E Power Mvar 12 470 VA Sc = V 289,4 Voltage factor at the substation FP = P S = 3 3.6 MW MVA = 0.833 = - + Ql = \fl& 2.90 at the Ec= 0.75 2 MVA load end of the line: Sc / 2.90 = MVA 289 A 10.03 kV AND APPARENT POWER ACTIVE, REACTIVE, Power factor 1 5 load end of the line: at the P, 2.8 S, 2.90 FP MW MVA = 0.965 ou 96.5% p e — s= E.dh r *|b Phase angle between the voltage and current at the rest of circuit j load: (a) = 9C It = arccos 0.965 15.2° follows that the phase angle between the voltage the substation at 15.2°) = We that at the load is — (33.6° 18.4°. Fig. 7. sis. and summarizes the 8 1 S = + E\I* results of this analy- could have found the same values using However, on account of its simpower method of solving this problem vector algebra. the plicity, is (b) very appealing. 7.15 If rest of circuit Power and vector notation vector notation is used to solve an ac can readily determine the active with associated We sources. component, any current that P is the active value for rest of circuit or Q (c) The vector product it.' power S in terms of P + jQ, power and Q the reactive P E4 I* E power absorbed (or delivered) by the component. A positive S=- the the conjugate (/*) of the flows through £/* gives the apparent where including simply multiply the phasor voltage component by across the we circuit, and reactive power means same sequence ab com- that the Figure 7.19 Method of writing power equations. Z associated with The apparent power S (not ba). is therefore written ponent absorbs active or reactive power. Negative values mean reactive component that the power. When It calculating the £7* vector product, procedure very important to follow a standard der to obtain the correct result. plies to circuits that tion or the is culate the active nal a Z We to 7. The procedure ap- 1 9a in which a circuit element of circuit." We want to cal- and reactive power associated with note that current terminal Consequently, the in or- would be incorrect In Fig. 7. that b, when i.e. in / flows from termithe sequence ab. calculating the product subscripts of voltage E must be written in the a current has a value /^B, its conjugate /* = IL~ B. S = £ ba /.* to write 9b, sign notation current enters / Z is by used, and the Consequently, the apparent power S element In the case of Fig. 7. 19c, we want to, ) is seen given by is a ( + ) ( + the in Fig. 7. 1 we Zby write the ( — sign because ) terminal of S = — £4 /* be) terminal. we can determine power associated with from b ( it terminal. Z. cause the current enters If + = +£,/* The £]/* product is preceded by current / is shown as entering by Thus, 'If 1 use the double subscript nota- part of a larger "rest element is sign notation (see Sections 2.4 and 2.5). Consider Figure Z it = £ah /* S delivers active or the apparent the "rest of circuit" (roc). 9a, because the current circulates to a in the roc, we would write: 1 ELECTRICAL MACHINES AND TRANSFORMERS 52 we would Similarly, in Fig. 7.19c, write Let us illustrate the procedure by a few examples. Example 7-12 In the circuit 2 of Fig. the following values are 7. 19c, given Figure 7.20 E4 = 70 Z25° See Example = I 7-13. 4 Z40° Calculate the active and reactive power associated with element Z. The voltage across Solution We have / = 4 Z40°; therefore I* Since the current flows into the power equation must bear 5 a — ( — ( ) = 4 the capacitor E y2 + /(— IO7) = Z-40° -£4/* ~ = -7()Z25° X 4 Z.-40 0 = -280 Z-15° Current = terminal +7 28()(cos(-15°) = -270.5 +7 sin (-15°)) in the We active 10./ x = 24.6 Z(-47° + = 24.6 Z43° 2.46 Z -47° 90°) capacitor flows from terminal 2 to with the capacitor is 72.5 S = W and Q ~ -270.5 1 Consequently, the power associated 3. = P+.iQ Thus P 0 10 J sign: ) given by = = £32 terminal, the is = +72.5 var W that element Z delivers 270 .4 of power and absorbs 72,5 var of reactive conclude £23 /* = -24.6 Z43° X = -60.5 Z90° 2.46 - -60.5 (cos 0 - 60.5 7 + 90° Z47° 7 sin 90°) = power. = P +JQ Example 7-13 Given the circuit . of Fig. 7.20 which in E l2 = 30 Z78°, determine the power associated with the capacitor whose reactance is Hence P — 0 and Q = —60.5; Consequently, the acpower associated with the capacitor is zero, and tive it delivers 60.5 var of reactive power. 10 O. Example 7-14 Solution Going cw around the circuit, we can write (see The circuit in Fig. 7.21 sistor Sections 2.32 to 2.39) connected actance. E2] ~ 7 1(1 E2{ _ - 10/ ~ = -2.46 Z - 10 7) 12.5 133° - 30 Z 0 a 45 i! rere- The source generates by the phasor a voltage described E ab = 159Z65°. 78° Z -55° = + composed of with a 28 (1 inductive is in series 2.46 Z-47° Calculate a. The magnitude and phase of the current / AND APPARENT POWER ACTIVE, REACTIVE, 4512 a Voltage across the reactance Ecb = = 159 [65_° £\ib ' 28 j28 c. is / X 3^33.11° j28 = 84^(33.11° + = 84^123.11° n 153 90°) The conjugate /* of the current / is = 3^-33.11° /* The apparent power associated with the resistor is Figure 7.21 S r = E.J* Solving an ac circuit using vector notation. = (135^33. 11°) (3^-33.11°) = 405^0° b. and across the reactance the resistor c. The active and reactive power associated with Applying Kirchhoff 's voltage law (see Section 2.32), we because there sin 0°) jO) £ba + E ac + Ecb = -£ab + -159^65° + we phase angle / = + - j28 159 53 L 45/ + j28/ = 0 /(45 + j28) - 0 / - S r = 159 L 45 45 + j28 45 = 252/190° - 252 (cos 90° = 252 - + (0 + j sin 90°) jl) j252 Thus, the reactance absorbs only reactive power (252 var). = 65° 3 / (65° 31.89° £ ac = 45 (84 Z_123. 11°) (3^-33.11°) The apparent power associated with - 53^31.89° Z_ the reac- 65° + 28 = 53 = arctan 28/45 = 31.89° \ W) . = = Voltage across the resistor = in 2 2 ~ ~EJ A 31.89°) = -477^(65° is / X 3^33.11° 135zl33.ll E\vJ''~ the source * = -(159^65°) (3^-33.11°) = 3^33.11° = power (405 real 0 obtain amplitude hence 45 no j component is Transforming the denominator into polar coordinates, is The apparent power associated with obtain tance b. + (1 j Thus, the resistor absorbs only Solution and so + (cos 0° = 405 and the source the resistor, the reactance, a. = 405 - 405 The magnitude and phase of the voltage across 0 33.11°) = -477^31.89° = -477 (cos 31.89° - -477 (0.849 = -405 - j + 252 j + j sin 31.89°) 0.528) is ELECTRICAL MACHINES AND TRANSFORMERS 154 The active and reactive powers are both negative, which proves that the source delivers an active power of 405 and a reactive power of 252 var. W on sources and loads 7.16 Rules (sign notation) We are often interested in determining whether a de- vice an active/reactive source or an active/reactive is load without such as sis, making a complete mathematical analy- we performed Section in To enable 7. 15. Figure 7.22 us to positively identify the nature of the source or load, consider Fig. 7.22 in line current /. The device The voltage between which a device is part of a circuit. the terminals + A carries a ( is £, and that are respectively parallel to, to E. Let / E. p be the component of out of phase with E. behind or 90° ahead of The 1 . A voltage b. the line current + ) A device when double /, re- is is an active source subscript notation used. Consider Figure 7.23 in which a device ries a current / flowing in the direction b is enable us to state 3. / is shown in phase and A device is shown. The £ab The . voltage b. line current / shown is in phase and as entering terminal Otherwise, the device is an active source. A device a. is current / is an active source. rule also applies: a reactive load when lags 90° behind voltage E. xh rule also applies: a reactive load /c] when lags 90° behind voltage line current / is shown E and as entering the ( +) terminal. Otherwise, the device Based on these rules, tionships in Fig. 7.22, active load because A is / p is a reactive source. and observing the phasor we deduce is in that device A relais an phase with E. Also, device a reactive source because / q is 90° ahead of E. Figure 7.23 Same These rules are in fol- an active load when: £ab and component / p are a. The following as entering the agreement with IEEE and IEC conventions. is A car- lowing rule applies: 4. component b. whether a device or active load terminal. The following a. We can also tell rule applies*: Otherwise, the device 2. on sources and loads (double subscript notation) an active load a. ( and rela- /. voltage between terminals a and when E and component / are is between Eand 7.17 Rules an active load or an active is The following device in E. E between whether a device source. angles diagram, together with the phasor circuit lationships at right / that is parallel to phase with, or 180° Similarly, / can be either 90° be either will therefore It active/reactive source or depending upon the phasor and one sign. ) A may be an active/reactive load tionship The phase angle between E and / can have any value. As a result, / can be decomposed into two components, I p and / of the terminals bears a Device circuit as in Fig. subscript notation is 7.22 except that double- used. and a. ACTIVE, REACTIVE, b. shown line current / is nal as entering by termi- c. Otherwise, the device a reactive source. is we deduce tionships in Fig. 7.23, active source because A Also, device 90° behind Eab is I p is 1 that device A an is 80° out of phase with E, lb a reactive load because / q e. 7-1 1 Using the rules given 7. 17, . lags Sections 7.16 and in determine which of the devices in Fig. 7.24a through 7.24f acts as an active (or power active) . 7-12 Questions and Problems 55 The peak power output of the reactor The duration of each positive power pulse d. Based on these rules, and observing the phasor rela- 1 The reactive power absorbed by the reactor The apparent power absorbed by the reactor The peak power input to the reactor a. b. a. AND APPARENT POWER re- source. A single-phase motor draws a current of 2 A at a power factor of 60 percent. 1 Calculate the in-phase and quadrature comPractical level 7-1 What ponents of current power? reactive the unit of active is power? apparent power? 7- 7-2 7-3 A capacitor of 500 kvar 7-5 placed 1 3 in parallel A 240 V, 60 Hz line. A 1 wattmeter line gives a reading of the Name motor and a static device that can generate reac- power. Name 7- 14 a static device that absorbs reactive If a the reactive in parallel Problem 7-13, What a. is the approximate power factor, in The current in What is the power large The apparent power of The line current e. The power 7- 5 1 motor absorbs 600 kW at a power is An a. b. is connected across a 1 20 A 10 12 reactance 60 Hz line. is V, connected to a 120 V, Calculate: induction motor absorbs an apparent at a power factor of 80 percent. Calculate: generates. 10 (2 resistor and without drawing any power of 400 kVA 60 Hz source. Calculate the reactive power A 14) circuits in Fig. 7.25. connected to a 240 V, 60 Hz source. Calculate: a. The active power absorbed by the resistor b. The apparent power absorbed by the resistor c. The peak power absorbed by the resistor d. The duration of each positive power pulse 7. phasor diagrams, find the impedance of the 7-16 machine. it factor of the motor/capacitor Using only power triangle concepts (Section 90 percent. Calculate the apparent power and reactive power absorbed by the p,F capacitor the ac line combination factor of A 200 calculate: active d. motor? Intermediate level A is with the motor of power reading of the wattmeter The total reactive power absorbed by the capacitor and motor The b. a single-phase motor lags power factor of the power it absorbs. capacitor having a reactance of 30 (2 connected power. factor of the 7-10 a 2765 W. Calculate 50° behind the voltage. 7-9 to connected into the c. 7-8 with respect apparent power of the group. candescent lamp? 1-1 / single-phase motor draws a current of A from percent, of a capacitor? of a coil? of an in- 7-6 and with an inductor of 400 kvar. Calculate the tive 7-4 is / the line voltage. c. 7-17 A power absorbed by the motor power absorbed by the motor What purpose does the reactive power serve The The active reactive circuit composed of a 1 2 ft resistor in series with an inductive reactance of 5 (2 carries an ac current of 10 A. Calculate: a. The active b. The reactive c. d. power absorbed by the resistor power absorbed by the inductor The apparent power of the circuit The power factor of the circuit (0 (b) (a) G E F / / (d) (e) Figure 7.24 See Problem 7-11. Hf- 1 40 V[ ^ ion 4I2( ] en 5A J X (c) (b) (a) Figure 7.25 See Problem 7-18 7-15. A coil having a resistance of 5 ft and an ductance of 2 H 7-2 in- carries a direct current of 20 A. Calculate: a. The active power absorbed b. The reactive power absorbed Advanced 7-19 7-20 A ply voltage is connected 200 is ft. If the sup- V, calculate: a. The reactive power absorbed by The reactive power generated by the coil the capacitor Problem a 7- 3, if 1 in parallel we d. 7_22 power absorbed by The The apparent power of the c. The power The power V the system is 0.6 lagging (Fig. 7.26). calculate: sy stem factor of the system 156 coil factor at the terminals of a source Without using phasor diagrams, with the motor, calculate: a. The active power dissipated by the The apparent power of the circuit 120 place a capacitor of b. total active c. a. The value of E b. The impedance of the load re- in parallel with b. level motor having 500 var having a reactance of 1011 and a a capacitive reactance of 10 power factor of 0.8 absorbs an active power of 1200 W. Calculate the reactive power drawn from the line. In A coil sistance of 2 ft Z ACTIVE, REACTIVE, AND APPARENT POWER 157 29. E= 120 V »E= 120 V Figure 7.26 /= 5 A See Problem 7-22. 7-23 In Figs. 7.27a and 7.27b, indicate the mag- 150° (b) (a) nitude and direction of the active and reactive power flow. and / and , q treat (Hint: Decompose / into / 5 them independently.) Figure 7.27 See Problem Industrial application 7-24 A single-phase A p 7-23. capacitor has a rating of 30 kvar, 480 V, 60 Hz. Calculate its capacib. tance in microfarads. The active and reactive power consumed by the line 7-25 In Problem 7-24 calculate: c. a. The peak voltage across the capacitor when it is connected to a 460 V source b. The active, reactive and apparent power ab- sorbed by the load d. resulting energy stored in the capacitor 7-28 at that instant, in The joules The voltage across A2 hp, 230 V, 1 the load 725 r/min 60 Hz single- phase washdown duty motor, manufactured 7-26 Safety rules state that one minute after a capacitor is disconnected from an ac voltage across discharge that is it must be 50 V or less. done by means of a by Baldor Electric Company, has the line, the The Full load current: resistor permanently connected across the is fol- lowing characteristics: efficiency: capacitor terminals. Based on the discharge power 1 1.6 A 75.5% factor: 74% curve of a capacitor, calculate the discharge resistance required, in ohms, tor in Problem 7-24. Knowing the tance is when weight: 80 lb for the capaci- a. resis- sorbed by this machine subjected to the service voltage the capacitor is in operation, calculate b. 7-27 A 3.2 kV, 1 60 Hz single-phase line con- (1. The metering equipment at the substation indicates that the line voltage is 12.5 kV and that the line of active power and 2 is drawing 3 Mvar of reactive power. Calculate: a. The current flowing operates at If a 40 microfarad capacitor is connected current feeding the motor. The c. MW Will the presence of the capacitor affect the temperature of the motor? has a resistance of 2.4 II and a reac- tance of 12 it across the motor terminals, calculate the line nects a substation to an industrial load. line when full load. wattage rating. its Calculate the active and reactive power ab- 7-29 A single-phase heater absorbs 4 kW on a 240 V line. A capacitor connected in parallel with the resistor delivers 3 kvar to the line. a. Calculate the value of the line current. b. If the capacitor in the line new is line current. removed, calculate the Chapter 8 Three-Phase 8.0 Introduction 8.1 Electric power We is tributed in the generated, transmitted, and dis- form of 3-phase power. Homes and small establishments are wired power, but this power ferred over single-phase is can gain an immediate preliminary understand- piston pre- is comparable to a single-phase the other hand, a 2-cylinder engine for several impor- is Three-phase motors, generators, and transformers are simpler, cheaper, and more down move efficient b. Three-phase transmission lines can deliver c. more power for a given weight and cost The voltage regulation of 3-phase transmission inside identical They in unison. power to deliver rather than at the running engine and a A cuits knowledge of 3-phase power and 3-phase is, cir- the basic different times. circuit will see that most 3-phase circuits we identical, can be reduced to time. the reader is As this the reader may know produces a smoother much smoother output torque. a 3-phase electrical system, the As a result, the total one phase may be used behavior of power at power flow is all to represent the three. Although we must beware of carrying analogies elementary single-phase diagrams. In this regard, we assume not way as very smooth. Furthermore, because the phases are techniques used to solve single-phase circuits can be directly applied to 3-phase circuits. Furthermore, do cylinders, but they three phases are identical, but they deliver therefore, essential to an understanding of power technology. Fortunately, a and to the shaft in successive pulses same lines is inherently better in In move up are staggered in such a from personal experience, Similarly, to 6-cylinder engine could be called a 6-phase machine. 6-cylinder engine identical pistons a. machine. On comparable The more common a 2-phase machine. tant reasons: common A single-cylinder engine having one gasoline engine. merely represents a tap-off from the power Polyphase systems ing of polyphase systems by referring to the for single-phase basic 3-phase system. Three-phase Circuits familiar with the previous too chapters dealing with ac circuits and power. far, system 158 the is above description reveals basically that a 3-phase composed of three single-phase THREE-PHA SE CIRCUITS systems that operate in sequence. much fact is realized, Once this basic A winding stator 59 1 of the mystery surrounding 3-phase systems disappears. Single-phase generator 8.2 Consider a permanent magnet stant NS revolving con- at speed inside a stationary iron ring (Fig. 8.1). The magnet is driven by an external mechanical source, such as a turbine. The Figure 8.2 ring (or stator) re- duces the reluctance of the magnetic circuit; consequently, the flux density in the air if having terminals a, insulated ductors, A were absent. the ring from one in it. 1 is Each each Ea1 = At this instant gap is flux does not cut greater than multiturn rectangular coil mounted 0 because the the conductors of winding A. inside the ring but voltage E. xi maximum when is position of Fig. turn corresponds to two con- 8. the poles are in the because the flux density 1 On greatest at the center of the pole. slot. the voltage when zero is is the other hand, the poles are in the position of Fig. 8.2 because flux does not cut the conductors winding stator A at this If tion, moment. we plot E. xl as a function of the angle of rota- and provided the N, S poles are properly shaped, we obtain the sinusoidal voltage Suppose Fig. 8.3. * shown in the alternating voltage has a peak value of 20 V. Machines that produce such voltages are called alternating-current generators or syn- chronous generators. The particular machine shown Figure 8.1 in Fig. 8. embedmaximum ( + ). A single-phase generator with a multiturn ded in two slots. At this instant Ea1 is called a single-phase generator is 1 coil v + 20 As the magnet turns, it sweeps across the con- inducing a voltage in them according / / / —y \ X \ \ 10 ductors, / V to the / / \ equation: / \ £al = B/v (2.25) wherein - / KJO \— — 90 0 10 2 ( B = instantaneous flux density cutting across conductors I = = the \ \ \ \ \ \ 3(i0 degrees 4S >0 / \ \ — \ f / / \ \ - \ / \ / 20 / in the slots [T] Figure 8.3 length of conductors lying in the magnetic field v V] \ / \ instantaneous voltage induced in the coil '0 \ / igle \ = / N Voltage induced in winding A. [m] peripheral speed of the revolving poles [m/s] The poles shown The sum of the voltages induced in all the ductors appears across the terminals. con- The terminal voltage in Fig. 8. composed of negative pulses. 1 would generate an alternating rather brief t'lat-toppped positive and ELECTRICAL MACHINES AND TRANSFORMERS 160 Power output of a single-phase generator 8.3 If a resistor rent will is connected across terminals The current /,, is in 1 a cur- phase with the voltage and, con- sequently, the instantaneous series of positive pulses, as average power power trical a, flow and the resistor will heat up (Fig. 8.4). is is power is composed of a shown in Fig. 8.5. The one-half the peak power. This elec- derived from the mechanical power provided by the turbine driving the generator. As a result, the turbine ergy in pulses, to must deliver mechanical en- its match the pulsed electrical output. This sets up mechanical vibrations whose frequency is twice the electrical frequency. Consequently, the generator will vibrate and tend to be noisy. 8.4 Two-phase generator Using the same single-phase generator, a second winding (B) on the let us mount Figure 8.5 Graph of the generator stator, is voltage, current, and power when the under load. identical to voltage E. tl becomes zero and voltage maximum therefore out by curves load on phase Note A in Fig. £ aj that positive value 8.6b and by phasors Eh2 because before E h2 does. stator Figure 8.4 resistor. its is it in Fig, 8.6c. reaches its peak called a two-phase generator, windings are respectively called phase A and phase Single-phase generator delivering power to a attains value. leads This machine and the E b2 The two voltages are of phase by 90°. They are represented positive B. Example 8-1 The generator shown in Fig. 8.6a rotates at 6000 r/min and generates an effective sinusoidal voltage winding A, but displaced from it by a mechanical of 170 V per winding. angle of 90° (Fig. 8.6a). As the magnet rotates, sinusoidal voltages are in- Calculate each winding. They obviously have the a. same magnitude and frequency but do not reach maximum value at the same time. In effect, at b. duced in their moment when the magnet occupies the position shown in Fig. 8.6a. voltage passes through its c. positive value, whereas voltage zero. This is conductors Eh2 is an- gle of 90° the maximum The peak voltage across each phase The output frequency The time interval corresponding to a phase Solution a. The peak voltage per phase is because the flux only cuts across the in slots 1 after the rotor has and a at this instant. made one However, quarter-turn (or 90°), £ ni = \2E = = 240 V 1.414 X 170 (2.6} THREE-PHASE CIRCUITS b. One cycle makes one is completed every time the magnet The period of one cycle turn. T= = = s = Power output now Let us 0.01 will flow in phase with /= c. A phase terval is \/T= means Hz 100 1/0.01 angle of 90° corresponds to a time in- Consequently, phasor behind phasor A and B each E. {i (Fig. 8.7a). Currents and /., /h They are respectively in The currents are, therefore, resistor. and Eh2 - 90° out of phase with each other (Fig. 8.7b). This of one quarter-revolution, or 10 ms/4 2.5 ins. connect two identical resistive loads across phases s 10 ins The frequency of a 2-phase generator 1/6000 min 60/6000 8.5 is 161 £a] E h2 lags 2.5 = that /., reaches period before now produces /b its maximum value one quarter- does. Furthermore, the generator a 2-phase power output. The instantaneous power supplied ms to each resis- tor is equal to the instantaneous voltage times the . instantaneous current. This yields the two power waves shown of phase in Fig. 8.8. A is maximum, Note that that when the power B is zero, and of phase we add the instantaneous powers of we discover that the resultant power is and equal to the peak power P m of one vice versa. If both phases, constant (a) 1 load on (a) phase \ -h (b) 0- 90. if i 360 21 —t an jle of 450 _ r ot ation 6 i load on phase B (c) n (b) Figure 8.6 a. Schematic diagram of a 2-phase generator. b. Voltages induced c. Phasor diagram of the induced voltages. in a 2-phase generator. Figure 8.7 Two-phase generator under load. b. Phasor diagram of the voltages and currents. a. A ELECTRICAL MACHINES AND TRA NSEORMERS 62 1 phase.* In other words, the total 2-phase generator also constant. is benefit, it peak power = it is to drive the less noisy. 'I 1 instantaneous power of phase in size, i except for the addition of an extra winding. Three-phase generator A 3-phase generator is similar to a 2-phase generator, except that the stator has three identical windings stead of two. placed at The three windings a-1, b-2, 120° to each other, as When magnet the position in- maximum Voltage in Fig. 8.9a, E b2 Consequently, only voltage will reach its Ea is in , is The The i i i i ; 180 270 360 the *- angle of rotation Total instantaneous power output it is used to current is one-third of a turn). Similarly, voltage designate different things. has to be read in context to be understood. ways the in Figure 8.8 Power produced by a 2-phase generator. which out of phase with the voltage (relets to pha- tain its positive peak Ec3 will at- after the rotor has turned through 240° (or two-thirds of a turn) from Consequently, the three stator voltages three phases of a transmission line (the three conduc- tors of the line) £b2 and , The phase-to-phase voltage 4. The phase sequence (the line voltage) (the order in Ec3 They 120°. 3. its ini- position. tial sor diagram) 2. 90 0 at its positive peak after The following examples show some of the word phase is used. . r i positive value. The icrm phase 1 i windings have the the rotor has turned through an angle of 120° (or :;: i in Fig. 8.9a. At the moment when the magnet shown i Instantaneous power of phase B and c-3 are effective values, but the peaks occur at differ- ent times. Ib = rotated at constant speed, is the voltages induced in the three same Eb . pov rer = .peak /MA, i shown A | j 8.6 Pm = fa A 2-phase generator does i without any increase Ea every instant. As a As an important produces twice the power output not vibrate and so added power output of the at mechanical power needed result, the generator same the is — are and as phasors — al , are respectively out of phase by shown as sine waves in Fig. 8.9b, in Fig. 8.9c. which the phasors follow each other) 5. The burned-out phase (the burned-out winding of a 3-phase 8.7 machine) 6. The 3-phase voltage (the line voltage of a 3-phase system) 7. The 3-phase currents are unbalanced (the currents in a 3-phase 8. line or machine are unequal and not displaced The pliase-shifi transformer (a device that at 120°) (a short-circuit between two line and E52 conductors) Phase-to-ground fault six wires to deliver phase loads (Fig. The phase-to-phase fault (a short-circuit between a line or /c , 8. 1 power to the individual singleThe resulting currents /.„ / b 0a). , are respectively in phase with voltages and EcV Because the currents have the same 1. The phases are unbalanced (the line voltages, or the line £al , resistors are identical, the effective values, but they are winding and ground) 1 to can change the voltage) 10. Let us connect the three windings of the generator three identical resistors. This arrangement requires phase angle of the output voltage with respect to the input 9. Power output of a 3-phase generator mutually out of phase by 1 20° (Fig. 8. 1 that they are out of other) reach their positive peaks The 0b). phase simply means currents, are unequal or not displaced at 120° to each fact that they at different times. THREE-PHASE CIRCUITS The instantaneous power supplied to each resisis again composed of a power wave that surges tor between zero and a (a) 163 maximum power peaks in the the same time, due to P nv However, value do not occur the three resistors at the phase angle between the we add the instantaneous powers of all resistors, we discover that the resulting power voltages. If three is T 1 Eai constant, as in the case of a 2-phase generator. However, the 1 ^c3 SN \ put \ (b) 0 740 3iiO 50 \/ J output of a 3-phase generator has P m Because is 1 the electrical out- . constant, the mechanical power required generator does not vibrate. Furthermore, the power erator to the load, is line, connecting the gen- constant. Example 8-2 The 3-phase generator shown in Fig. nected to three 20 12 load resistors. voltage induced in each phase 120° (c) I 120° a. c. a. d. Three-phase generator. in a 3-phase generator. b. Voltages induced c. Phasor diagram of the a. Each resistor to behaves as a single-phase load in each resistor (b) Three-phase, 6-wire system. Corresponding phasor diagram. 120 V. calculate an effective voltage of 120 V. The power dissipated b. con- The power dissipated in each resistor The power dissipated in the 3-phase load The peak power P lu dissipated in each resistor The total 3-phase power compared to P lu connected Figure 8.10 is the effective Solution induced voltages. a. is 8.10a If the following: b. Figure 8.9 to also constant, and so a 3-phase flow over the transmission -N X is drive the rotor t J total a magnitude of 1.5 is, therefore, 1 64 ELECTRICAL MACHINES AND TRANSFORMERS I i u * U 'b AI 360 240 0 480 "\ t x I (a) Figure 8.11 a. Three-phase, 4-wire system. b. Line currents in a 3-phase, 4-wire system. P = E 2 IR = = 720 b. The (all total l20 The peak power /2() W power dissipated three resistors) 2 in the 3-phase load d. 3 The ratio of X 720 is Pm to 8.485 is 2 60/1440 1 = 1.5 Thus, whereas the power The peak voltage across one resistor sates is tal E m = <2E = V2 X - 169.7 V The peak current /m X 69.7 * is absolutely constant from instant to instant. c. PT PT /P m = = 2160 This power each resistor = £ m /m = = 1440W is P v = 3P = in in 120 each resistor = EJR = = 8.485 A is 169.7/20 between 0 and a power for all in maximum each resistor of 1440 three resistors is W, pul- the to- unvarying and equal to 2 160 W. Wye 8-8 The connection three single-phase circuits of Fig. 8.10 are electrically we independent. Consequently, can connect the three return conductors together form a single return conductor (Fig. to 8.1 la). This reduces the number of transmission line conduc- The tors from 6 tral conductor (or simply neutral), carries the sum to 4. of the three currents (/ a + /b + /c ). At first it seems that the cross section of this conductor should be three times that of lines and diagram of Fig. 'c return conductor, called neu- a, b, 8,1 lb clearly c. However, the shows that the sum of the three return currents is zero at every instant. For example, at the instant corresponding to = / max and I h = / a = -0.5 / max making + 4 = 0- We arrive at the same result (and Figure 8.12 240°, I c Three-phase, 3-wire system showing source and load. 4+ A> , THREE-PHASE CIRCUITS I much more simply) by taking the sum of the phaOb. The sum is clearly (/., + / b + I c ) in Fig. 8. sors 1 zero. We can, therefore, together without in remove the neutral wire al- any way affecting the voltages one stroke or currents in the circuit (Fig. 8.12). In we accomplish a great saving because the number of conductors line However, the loads in order to remove not identical, the from drops six three! to 8.11a must be in Fig. identical the neutral wire. If the loads are absence of the neutral conductor produces unequal voltages across the three loads. The circuit of Fig. —composed and load — 8.12 generator, transmission line, n 3-phase, 3 -wire system. The generator, connected load, are said to be as the of the is called as well in wye, be- cause the three branches resemble the letter Y. For equally obvious reasons, to some people prefer use the term connected In star. The circuit of Fig. 8.1 la 4-wire system. system is The usually the than the line is called a 3-phase, neutral conductor in such a same size or slightly smaller conductors. Three-phase, 4-wire sys- tems are widely used to supply electric power to commercial and industrial users. The line conductors are often called phases, which is the same term applied to the generator windings. 8.9 Voltage relationships Consider the wye-connected armature windings of a 3-phase generator (Fig. 8. 13a). in The induced voltage each winding has an effective value sented by the length of each phasor Fig. 8. are 1 3b. Knowing represented tion is, in the repre- diagram in that the line-to-neutral voltages by phasors E ail , Eblv and Ecn the quesE. lb Ehc and what are the line-to-line voltages £ca ? Referring E lN to Fig. 8.13a, lowing equations, based on we , , can write the fol- Kirchhoff s voltage law: Eab = Ean + E nh = ^an ~ £bn £ bc = £ bn + E nc = E hn - Ecn Eca = Ecn + E na = Ecn - Ean (8.1) (8-0 Figure 8.13 Wye-connected a. (8.2) Line-to-neutral voltages of the generator. c. Method d. Line voltages to determine line voltage £ab £ bC) and £ca placed at 120°. (8.3) 3-phase gen- b. (8.2) (8.3) stator windings of a erator. , £ab . are equal and dis- ELECTRICAL MACHINES AND TRANSFORMERS 166 Referring we draw phasor Eq. 8.1, first to ex- E.db a actly as the equation indicates: The diagram shows that line by 30° (Fig. 8.13c). Using resulting phasor voltage E Ean leads ilb simple trigonometry, and based upon the fact that the length of the line-to-neutral phasors is £j N we , have the following: E length x £ah = = of phasor 2 X 2 X £ LN V3/2 £, N cos 30° Figure 8.14 Voltages induced in a wye-connected generator. - V3 £ LN Calculate The line-to-line therefore V3 voltage (called line voltage) is times the line-to-neutral voltage: - V3£ LN a. b. (8.4) c. The line-to-neutral voltage The voltage induced in the individual windings The time interval between the positive peak A and voltage of phase where phase E = x £ LN — V3 = Due d. effective value of the line-to-neutral Solution voltage [V| a. = a constant [approximate value to the equal to V3 E t N . The of truth this = can be seen by referring to Fig. 8.13d, which shows three phasors: £. l]v drawn according The line £" hc , and phasors are and 8.3, respec- . is The windings 1 3 V 800 are connected in wye; conse- quently, the voltage induced in each winding all £ca The to Eqs. 8.1, 8.2, line voltage line-to-neutral voltage symmetry of a 3-phase system, we con- is peak of £ln = ^i/^ 3 = 23 900/V3 b. tively. The 1.73] clude that the line voltage across any two generator terminals The peak value of the effective value of the line voltage [V] the positive B 13 c. 800 is V. One complete cycle (360°) corresponds to 1/60 s. Consequently, a phase angle of 120° corresponds voltages are equal in magnitude and to an interval of mutually displaced by 120°. To further clarify these results. Fig. the voltages 8. between the terminals of generator whose line-to-neutral voltage The line voltages are all equal to 14 is 3-phase system, but the voltage between b, c\ b and n, etc.) is Example 8-3 3-phase 60 Hz generator, connected in wye, gen- erates a line (line-to-line) voltage of 23 900 V. 1 X 1/180 s 60 5.55 ms positive voltage peaks are, therefore, sepa- rated by intervals of 5.55 ms. lines a, b, c, consti- nevertheless an ordinary single-phase voltage. A The 100 V3, or tute a b and = 100 V. The voltages between 120 360 a 3-phase 173 V. any two lines (a and T= shows d. The peak line voltage is E m = V2E L = 1.414 X = 33 800 The same voltage 23 900 (2.6) relationships exist in a wye- connected load, such as that shown in Figs. 8.1 THREE- PHASE CIRCUITS and 8. 1 2. In other words, the line voltage is 1 67 V3 times the line-to-neutral voltage. Example 8-4 The generator in Fig. 8.12 generates a line voltage of 865 V, and each load resistor has an impedance of 50 fl. Calculate The voltage across each resistor The current in each resistor The total power output of the generator a. b. c. Solution The voltage across each a. resistor is £ L n = £| A/3 = 865/V3 = 500 V The current b. in / each resistor = ELN /R = (8.4) is 500/50 10A All the line currents are, therefore, equal to 10 Power absorbed by each c. resistor P = ElN I = 500 X 5000 A. (b) is 10 W The power delivered by the generator to all three resistors is P = 3 X 5000 = 15 kW Figure 8.15 Impedances connected in delta. b. Phasor relationships with a resistive a. 8.10 Delta connection A is 3-phase load said to be voltages are equal and the This corresponds to three balanced when the line line currents are equal. identical impedances connected across the 3-phase line, a condition that is usually encountered in The three impedances (as we The 3-phase may the line; consequently, resistor currents £bc , and ECiV h /2 , and /3 £ab law, the line currents are given by be connected 8. -h wye in 1 (8.5) 5a). (8.6) h - (8.7) 12 shown). Let us determine the voltage tionships in and current rela- such a delta connection,* assuming a resistive load. The resistors are , Furthermore, according to Kirchhoff's voltages are produced by an external gen- erator (not / are in phase with the respective line voltages circuits. already have seen) or in delta (Fig. line load. connected across The connection letter A. is so named because it resembles the Greek ELECTRICAL MACHINES AND TRANSFORMERS 68 1 Let the current each branch of the delta-connected in which corresponds load have an effective value the length of phasors /2 , /3 . Furthermore, let to the have an effective value / L which corresponds to the length of phasors /.,, / b /c Referring line currents , , equation shows . we draw phasor exactly as the indicates. The resulting phasor diagram Eq. 8.5, first to that / a /., leads ple trigonometry, /, by 30° (Fig. we can now 8. 1 5b). Using sim- write A — 10 lx = 2 X I, cos 30° = 2 X I, V 3/2 = V3/ c o— Figure 8.15c See Example y . The line current is therefore V3 8.5. times greater than the current in each branch of a delta-connected load: /,. = V3/Z (8.8) = l(W3 = Iy b. 5.77 A The voltage across each impedance is 550 V. Consequently, where = — /, /, Z= - effective value of the line current [AJ effective value of the current in one EIIy = 550/5.77 95 ft branch of a delta-connected load [A] — "V3 a constant [approximate value The reader can = 1.73] position of phasors /b and /c and thereby observe that 20°. , the three line currents are equal and displaced by Table 8A summarizes 1 the basic relationships bein by a 3-phase magnitude and readily determine the tween the voltages and currents wye-connected and delta-connected loads. The relationships are motor winding, generator wind- ing, etc.) as long as the elements in the three line The apparent power supplied by is a single-phase equal to the product of the line voltage line current /. The question now arises: line E times the What is the apparent power supplied by a 3-phase line having line voltage valid for any type of circuit element (resistor, capacitor, inductor, Power transmitted 8.11 If we 8. 16a, E and a a line current I? refer to the the apparent wye-connected load of power supplied to Fig. each branch is phases are identical. In other words, the relationships in Table 8A apply to any balanced 3-phase load. The apparent power supplied is Example 8-5 Three identical impedances are connected across a 3-phase a. b. is 550 V apparent power in delta line (Fig. 8.15c). If the line S = E , X / X 3 = Solution The current :!: In 3-phase balanced circuits, factors. If the in each impedance is \ 3 El \3 10 A. calculate the following: The current in each impedance The value of each impedance [ft] is we can add ers of the three phases because they a. the . total current to all three branches obviously three times as great. * Consequently, power the apparent pow- have identical power factors are not identical, the apparent powers cannot be added. THREE-PHASE CIRCUITS VOLTAGE AND CURRENT RELATIONSHIPS TABLE 8A Wye IN 169 3-PHASE CIRCUITS Delta connection connection 7/1.73 Figure 8.16a Impedances connected • The current current • • in in Figure 8.16b Impedances connected wye. each element is equal to the line • The voltage across each element line voltage E divided by V 3. The voltages across is equal to the • the elements are 120° out of in the elements are 120° out of • • case of a delta-connected load (Fig. the apparent power supplied S = E X each branch to is equal to the line The voltage across each element The The 8. 1 6b), is equal to the voltages across the elements are 120° out of currents in the elements are 120° out of power / is the same Consequently, the as for a total in The relationship between active power P, reactive power Q, and apparent power S is the same for bal- wye-connected apparent power and apparent 3-phase circuits 8.12 Active, reactive, is V3 is load. also the anced 3-phase We circuits as for single-phase circuits. therefore have 2 S = \'P + same. We each element phase. phase. which in divided by V3. phase. The currents In the / delta line voltage E. phase. • The current current /. in therefore have Q 2 (8. JO) and S = V 3 El cos 6 (8.9) = PIS (8. where where S = total apparent power delivered by a 3-phase line E— / = V3 = [VA] effective line voltage [V] effective line current [A) a constant [approximate value = 1.73] = total 3-phase apparent power VA| = P total 3-phase active power [W| Q = total 3-phase reactive power |var| cos 0 = power factor of the 3-phase load 0 = phase angle between the line current S | and the line-to-neutral voltage |°| 1 1 ELECTRICAL MACHINES AND TRANSFORMERS 70 Example 8-6 A 3-phase motor, connected power a line current of 5 A. If the motor a. is The 440V to a line, factor of the The total active The total reactive 60 Hz power power absorbed by Solution The = V3 = V3 X 440 X EI = 3811 VA = 3.81 kVA total active See Example is O 1 R power = 3.05 total reactive is X 3.81 The The = \ S current in each line is 8 1 V = also 3. resistance of each element is is A 3. 15 1 5 A. 0.80 R = - P2 = 2 each resistor 1 kW power in = P/E= 000 W/3 / EII = - 318/3.15 101 12 Example 8-8 is In the circuit Q = 8-7. The current P = ScosO = The JT 5 b. c. O Figure 8.17 apparent power total S b. w line machine the 3000 = 9" V 550 3-phase apparent power total c. The A o 80 percent, calculate the following: b. a. P draws \ 3.81 2 - 3.05 2 a. b. 2.28 kvar of Fig. 8. 1 8, calculate the following: The current in each line The voltage across the inductor terminals Solution 8.13 Solving 3-phase circuits A a. balanced 3-phase load may be considered composed of three identical Consequently, the easiest cuit is to consider examples way to be single-phase loads. method 4 { to is composed of an inductive reactance fl in series with a resistance R= 3 II. Consequently, the impedance of each branch Z to solve such a cir- only one phase. The following illustrate the Each branch X = ~~~ V4- 3^ = 5 The voltage across each branch a (2.i: is be employed. ELN = ELN3 = 440 V/V3 - 254 V Example 8-7 The current Three identical 3000 550 V W resistors dissipating a total are connected in wye in / across a 3-phase (50.8 A is - E LN /Z - The current The value of each in each line resistor 440 Solution a. The power dissipated by each resistor V 3-phase line is P = 3000 W/3 = 1000W The voltage across the terminals of each resistor is Figure 8.18 E — 550 VA/3 -318 V See Example 254/5 = 50.8 also the line current.) Calculate b. is power of line (Fig. 8. 17). a. each circuit element 8-8. is A THREE-PHASE CIRCUITS The voltage across each inductor b. E = IX = 50.8 X = 203.2 V V In a is wye connection 1 impedance per phase the 7 is understood to be the line-to-neutral impedance. The 4 voltage per phase is simply the line voltage divided by V3. Finally, the current per phase equal to the is line current. Example 8-9 A identical capacitors If Hz 3-phase 550 V, 60 the line current is line is connected connected to three in delta (Fig. 8.19). 22 A, calculate the capacitance of each capacitor. The assumption of a wye connection can be made not only for individual loads, but for entire load centers such as a factory containing motors, lamps, heaters, furnaces, and so forth. We assume in that the load center is connected simply wye and proceed with the usual calculations. Solution The current / each capacitor in = / L /V3 = 22 A/V3 Voltage across each capacitor Capacitive reactance X Xc = EJI = is c - Example 8-10 A A = 550 V of each capacitor 550/12.7 - The capacitance of each capacitor C= 12.7 1/(2tt = 61.3 V (line-to-line) 8.20a). If the plant is power factor total of 415 kVA 3-phase line (Fig. is 87.5 percent lag- ging, calculate the following: 43.3 il b. The impedance of the plant, per phase The phase angle between the line-to-neutral c. The complete phasor diagram a. is voltage and the line current 1/2tt/Xc = manufacturing plant draws a from a 2400 X 60 X 43.3) (2.11) for the plant Solution |jlF a. We assume a wye connection composed of impedances Z(Fig. 8.20b). three identical The voltage per branch E= = The Figure 8.19 See Example 8-9. 8.14 Industrial loads In is connected = in delta or in capacitors, and so on, often have only three external and there is no way to tell how the inter- connections are made. Under these circum- we simply assume (A wye connection is connection stances, that the wye. slightly easier to handle than a delta connection.) (8.9) 000/(2400 V 3 = A 100 = b. E/I = ) is 1386/100 13.9 12 The phase angle 0 between the line-to-neutral voltage 386 V) and the corresponding line current ( 100 A) is given by ( nal is S/(EV3) Z= wye. For ex- V = 415 ample, 3-phase motors, generators, transformers, terminals, 1386 The impedance per branch most cases, we do not know whether a particular 3-phase load 2400/V3 current per branch / is 1 is in cos 6 0 = power factor = = 29° 0.875 (8.11) ELECTRICAL MACHINES AND TRANSFORMERS 172 4000 V 3 phase 2400V y 1 ?\ 3- phase line —1((a) 1800 kvar 'Tan / a = 100 1386 A V Figure 8.21 Industrial motor and capacitor. A delta-connected capacitor bo- is See Example bank rated also connected to the line. If the at 8-1 1 1800 kvar motor produces an output of 3594 hp at an efficiency of 93 percent and a (b) power factor of 90 percent (lagging), calculate the following: c. The The The d. The apparent power supplied by a. b. power absorbed by the motor power absorbed by the motor reactive power supplied by the transmisactive reactive sion line the transmis- sion line e. f. g. The transmission line current The motor line current Draw the complete phasor diagram for one phase Solution a. Figure 8.20 a. Power input to a See Example wye connection b. Equivalent c. Phasor diagram The factory. current in P 2 = 3594 X 8-10. and is 0.746 equivalent to = 268 kW 1 Active power input to motor: of the factory load. of the voltages Power output of 3594 hp currents. = p 2^ = 2681/0.93 = 2883 kW each phase lags 29° behind the b. (3.6) Apparent power absorbed by the motor: line-to-neutral voltage. Sm c. The complete phasor diagram is shown in Fig. 8.20c. In practice, we would show only one phase; for example, Ean , /.„ Q m = \Si= A 1 P- n = V3203 T^r 2883 2 395 kvar . 5000 hp wye-connected motor 4000 2883/0.90 Reactive power absorbed by the motor: and the phase angle between them. Example 8-11 = PJcos 0 = = 3203 kVA V (line-to-line), 3-phase, 60 is Hz connected to a line (Fig. 8.2 1 ). c. Reactive power supplied by the capacitor bank (see section 7.5): THREE-PHASE CIRCUITS Qc = -1800kvar Total reactive cause the capacitors are connected power absorbed by assumed a wye connection the load: in delta, 1 and 1800 + 1395 we try to This is is pacitor bank. an unusual situation because reactive being returned to the the capacitor most cases line. In bank furnishes no more than Qm The Active power supplied by the line P L = P m = 2883 5L e. = VPl + Qi = V2883 2 + = 2911 kVA Transmission line current h. f. Motor /m g. The to recognize that is in wye same reactive power), the line current would lead £ LN by 90°. Consequently, A 90° ahead of £, N That is the correct position for phasor /c no matter how the capacitor /c . is bank is connected internally. Phase angle 0 L between the transmission current and line is ( - 405) £ LN cos6,, 2 line-to-neutral voltage = PJS L = 0.99 eL = 2883/291 8° (8.9) 4000) 4000) 462 A is £LN = 4000/V3 = 2309 V 420 A 462 A 420 A 462 A 420 A 462 A Phase angle 9 between the motor current and the line-to-neutral voltage 4000 is: V 3-phase 6 line is: is = 5 m /(£L V3) = 3 203 000/(V3 X = 462 A cos 0 /'/ (while gen- is = SJ(E^3) = 2911 000/(V3 X = 420 A line current of 260 kW Apparent power supplied by the solution were connected erating the we draw kilovars of reactive power, d. if follow the actual currents inside the ca- the capacitors power we for the motor. This can create unnecessary phase-angle complications QL = Qc + Q m = = -405 kvar 73 = power factor = = 25.8° 0.9 260 260 260 A A A (The motor current lags 25.8° behind the voltage, as shown in Fig. 8.22a.) Line current drawn by the capacitor bank /c = Gc/(£lV3) = 800 000/(V3 X - 260 A 1 (b) is —IH 1800 kvar 4000) Figure 8.22 Phasor relationships a. for one phase. See Example 8-11. the Where should phasor current /c be located on phasor diagram? The question is important be- b. Line currents. Note that the motor currents exceed the currents of the source. "1 1 ELECTRICAL MACHINES AND TRANSFORMERS 74 The line current (420 A) leads Eln by 8° be- cause the kvars supplied by the capacitor bank ex- ceed the kvars absorbed by the motor. The phasor diagram one phase tor is shown in Fig. 8.22a. The circuit diagram and shown in Fig. 8.22b. current flows are We want to emphasize the ing a tual wye importance of assum- connection, irrespective of what the ac- connection nection for all may By assuming a wye conwe simplify the be. circuit elements, Figure 8.23 The sequence are observed in the Figure 8.24 The letters are observed in the sequence a-c-b. in the letters a-b-c. calculations and eliminate confusion. As a final remark, the reader has no doubt no- ticed that the solution of a 3-phase problem in- volves active, reactive, and apparent power. The impedance value of devices such as resistors, mo- tors, and capacitors seldom appears on a nameplate. This is to be expected because most industrial loads involve electric motors, furnaces, lights, and so on, which are seldom described in terms of resistance and reactance. They are usually represented as de- draw a given amount of power vices that power The situation is somewhat 3-phase transmission sistances fixed. at a given factor. lines. different in the case of Here we can define re- and reactances because the parameters are The same remarks apply to equivalent circuits describing the behavior of individual machines such as induction motors and synchronous machines. In conclusion, the solution of may L, 3-phase circuits involve either active and reactive power or R, C elements — and 8.15 sometimes both. Phase sequence Figure 8.25 The In addition to line voltage letters are observed sequence a-c-b. and frequency, a 3-phase system has an important property called phase sequence. Phase sequence is important because it de- termines the direction of rotation of 3-phase motors and whether one 3-phase system can be connected in parallel with another. Consequently, in 3-phase systems, phase sequence quency and voltage is as important as the fre- understanding of phase sequence by considering the following analogy. Suppose the letters a, b, c are printed at 20° 1 tervals on a slowly revolving disc in- (Fig. 8.23). If the disc turns counterclockwise, the letters appear in the sequence a-b-c-a-b-c. Let us call this the posi- are. Phase sequence means the order three line voltages We can get a quick intuitive in which become successively the positive. tive sequence. It can be described three ways: abc, bca, or cab. in any one of THREE-PHASE CIRCUITS If the same disc turns clockwise, the sequence (Fig. 8.24). We call this becomes a-c-b-a-c-b . . . and the negative sequence, it can be described by rule: When 175 using the double-subscript notation, the sequence of the first subscripts corresponds to the phase sequence of the source. any one of three forms: acb ? cba, or bac. Clearly, there a difference between a positive sequence is and a negative sequence. In Fig. 8.17, the Suppose we interchange any two in Fig. 8.23, the result is letters on the disc while retaining the same counterclock- wise rotation. the letters If as shown becomes c-b-a-c-b-a a and c . . , The sequence now which is the same as the negative sequence generated by the disc in Fig. 8.24. We to phase sequence of the source be A-C-B. Draw sequence can be converted into a negative sequence by simply interchanging two the line voltages. Solution The voltages follow the sequence A-C-B, which is the same as the sequence AC-CB-BA-AC .... ECB -E BA and shown in Fig. 8.27. is We can reverse the phase line by interchanging any two se- let- quence of a 3-phase conductors. Although this verted into a positive sequence by interchanging change, it may appear to be a trivial can become a major problem when large letters. Let us now busbars or high-voltage transmission lines have to consider a 3-phase source having ter- minals a, b ? c (Fig. 8.26a). Suppose the line voltages £ AC - is the corresponding phasor diagram Similarly, a negative sequence can be con- any two is the phasor diagram of Consequently, the line voltage sequence conclude that for a given direction of rota- tion a positive ters. known are interchanged, in Fig. 8.25. . Example 8-12 that £ab E hc E , , Cil revolving phasors sweep past be interchanged. In practice, measures are taken so are correctly represented shown in by the As they Fig. 8.26b. be such drastic mechanical changes do not have to made at the last minute. major distribution systems The phase sequence of all is known in advance, and the horizontal axis in the conventional any future connections are planned accordingly. counterclockwise direction, they follow the se- quence If Eab -E bc -Eca -Eab -E bc .... we direct our attention to the first letter in each subscript, we find that the sequence The source shown sequence a-b-c. in Fig. 8.26a is We can, therefore, is a-b-c-a-b-c . sequence . said to possess the state the 8.16 Determining the phase following Special instruments are available to indicate the phase sequence, but ing we can also determine two incandescent lamps and a it by us- capacitor. three devices are connected in wye. If The we connect the circuit to a 3-phase line (without connecting the neutral), 3-phase one lamp source der: (a) — — ^/^^^^ Determining the phase sequence of a 3-phase £"cb source. b. always burn brighter than (b) Figure 8.26 a. will The phase sequence is in the following bright lamp dim lamp capacitor the other. Phase sequence depends upon the order in the line voltages reach their positive peaks. which Figure 8.27 See Example 8-12. £ ac or- ELECTRICAL MACHINES AND TRANSFORMERS 176 3-phase (a) line lamp Figure 8.29 Method of connecting a single-phase wattmeter. moves upwhen the connections between source and load made as shown in Fig. 8.29. Note that the ± cur- In single-phase circuits the pointer (b) scale are rent terminal nal. Figure 8.28 a. When is connected to the the wattmeter upscale reading Determining phase sequence using two lamps and means supply terminals 1, is that ± potential termi- connected power is way. an this flowing from 2 to load terminals 3, 4. a capacitor. b. Resulting phasor diagram. 8,18 Suppose, for example, cuit is connected to a 3-phase 8.28a. If the brightly, the lamp connected phase sequence ages follow each other which is to say in the line, as to is phase shown C sequence is CB line volt- CB-BA-AC, E BA EAC The £" , . , shown in in Fig. burns more C-B-A. The sequence in the corresponding phasor diagram 8.28b. in Fig. In a 3-phase, 3-wire system, the active power sup- may be measured by two shown in Fig. equal to the sum of the two single-phase wattmeters connected as 8.30. The total power is wattmeter readings. For balanced loads, is less than 1 if the power 00 percent, the instruments give different readings. Indeed, Power measurement in 3-phase, 3-wire circuits plied to a 3-phase load factor 8.17 Power measurement that a capacitor/lamp cir- if the will power factor is ac circuits Wattmeters are used measure active power to in single-phase and 3-phase circuits. Owing is built, to its external a wattmeter meter and ammeter combined Consequently, rent terminals. it way connections and the may be considered in it to be a volt- same box. the has 2 potential terminals and 2 cur- One of the potential terminals one of the current terminals bears a signs are polarity marks that ± sign. and The ± determine the positive when same time or negative reading of the wattmeter. Thus, the ± voltage terminal as current is is entering the positive at the ± current terminal, then the wattmeter will give a positive (upscale) reading. The maximum voltage and current the instrument can tolerate are shown on the nameplate. Figure 8.30 Measuring power in a 3-phase, 3-wire two-wattmeter method. circuit using the THREE-PHASE CIRCUITS less than 50 percent, one of the wattmeters will give We must then reverse the con- 177 Power measurement 8.19 a negative reading. in 3-phase, 4-wire circuits nections of the potential coif so as to obtain a reading of this negative quantity. In this case, the of the 3-phase circuit is power equal to the difference be- The two-wattmeter method gives the active balanced or un- is single-phase three that the ± ± as current terminal potential terminal. shown When is total in Fig. 8.3 power. 1 . Note again connected to the the wattmeters are con- nected this way, an upscale reading means that active balanced. Example 8-13 A full-load test on a 10 hp, 3-phase following results: P, the current in line circuits, wattmeters are needed to measure the The connections are made tween the two wattmeter readings. power absorbed whether the load 4-wire 3-phase, In voltage motor yields the = +5950 W; P 2 = + 2380 W; each of the three lines 600 is V. Calculate the 10 A; and the is power factor of power is flowing from source A, B, C, N to the load. The total power supplied to the load is equal to the sum of the three wattmeter readings. The threewattmeter method gives the active power for both balanced and unbalanced loads. Some wattmeters, such as those used on switch- boards, are specially designed to give a direct read- the motor. out of the 3-phase power. Figure 8.32 shows a Solution megawatt-range wattmeter Apparent power supplied to the motor ^3EI = V3 X 600 X S = = Active 10 390 power is circuit that a generating station. The measures the current trans- formers (CT) and potential transformers (PT) step 10 down VA power supplied in the line currents and voltages to values com- patible with the instrument rating. to the motor is SSI P = 5950 + 2380 = 8330 W cosO = P/S= 8330/10 390 = 0.80, or 80 percent p. Example 8-14 > When the the line motor Example 8-13 runs in current drops to 3.6 readings are P, A at p and the wattmeter - +1295 W; P 2 = Calculate the no-load losses no-load, and power LOAD . .. • : p> 845 W. factor. Solution Apparent power supplied to motor ) S = V3 EI = V3 X 600 X = 3741 VA 3.6 Figure 8.31 Measuring power in a 3-phase, 4-wire circuit. No-load losses are P = P, + P2 = = 450 1295 - 8,20 Varmeter 845 A varmeter indicates W It is Power factor = P/S = 450/3741 = 0.12 = \2% built the the reactive same way power as a wattmeter ternal circuit shifts the line voltage in a circuit. is, but an in- by 90° before it ELECTRICAL MACHINES AND TRANSFORMERS 178 Figure 8.32 Measuring active power is in applied to the potential employed tions in the control a high-power coil. circuit. Varmeters are mainly rooms of generating and the substations of electrical utilities load resistance (Fig. 8.33). Furthermore, given and the In 3-phase, 3-wire balanced We circuits, the we can cal- two wattmeter simply multiply the differ- ence of the two readings by V 3. For example, two wattmeters indicate = 6 76 vars. Note that 1 surement is if is this (5950 — 2380) , phase sequence of the 1 it is line voltages £ l2 connected as indicated. capacitive and If the in- ductive reactances are interchanged, the 3-phase system becomes completely unbalanced. X method of var mea- only valid for balanced 3-phase circuits. Example 8-15 A 800 phases kW 1 = 440 Z single-phase load connected between is and 2 of a 440 V, 3-phase 0, E23 = 440 Z - line, E ]2 wherein E3l = 440 Z 120, 12(X Calculate the load currents and line currents 8.21 It A remarkable single-phase to 3-phase transformation sometimes happens unity power 3-phase line. that a large single-phase a. This can create a badly unbalanced it three phases perfectly is possible to balance the by connecting the single-phase load b. When is connected alone line balancing reactances are added across the remaining lines, as shown in Fig. 8.34 Solution a. The resistance of the single-phase load a capacitive reactance and an inductive reactance across the other two lines. When on the 3-phase factor load has to be connected to a system. However, , essential that the three impedances be the +5950 W and +2380 W repower spectively, the reactive V3 -2-3- , power from readings (Fig. 8.30). l £23 E3l large industrial consumers. culate the reactive impedances V 3 times greater than the value of the sta- The reactances must each have R E2 P = 800 000 0.242 n is THREE-PHASE CIRCUITS The current in the 1 79 load and in two of the three lines is The current in the third line 3-phase system b. By is and so the zero, is badly unbalanced, introducing capacitive and inductive reac- tances having an impedance of 0.242 we 0.4 19 fl, V3 = obtain a balanced 3-phase line, as demonstrated below. Taking successive loops around the respective Section 2.32), E ]2 - elements in Fig. 2.38 £ 3] - j obtain the following results: = 0 /. /, = 4A3E l2 = X 440Z0 = 1817Z0 4.13 £23 + we 0.242/, £12 2.38 circuit and using Kirchhoff's voltage law (see 8.34, = 0 j 0.419 U X 440Z(-120 + 0.419 /3 - 0 = j 2.38 £ 23 = 1047^-30 U = .'. 90) /3 X 440Z(120 + 90 - = £ 31 = = 1047Z30 -j 2.38 180) £23 Applying Kirchhoff's current law and Figure 8.33 A single-phase resistive load can be transformed a balanced 3-phase load. 3, we to nodes 1 , 2, obtain into U = /, ~ /3 = 1817Z0 - 1047Z.30 = 1817 - 907 - j 523 = 1047/1-30 /b Ic = = = = = h ~ h 1047Z-30 - 1817Z0 907 - j 523 - 1817 -907 — j 523 1047Z210 = h ~ h = 1047Z30 - 1047Z.-30 = 907 + j 523 - 907 + j 523 =1047j = 1047Z90 Figure 8.34 Thus, / A / B /c make up a balanced 3-phase system because they are equal and displaced at 120° See Example 8-14. to , , each other (Fig. 8.35). 180 ELECTRICAL MACHINES AND TRANSFORMERS Tm e. 8-6 The ohmic value of each a. What b. Could we reverse is resistor the phase sequence in Fig. 8.10? it by changing the direction of rotation of the magnet? 8-7 A 3-phase draws a motor connected line current to a 600 V line of 25 A. Calculate the apparent power supplied to the motor. 8-8 Three incandescent lamps rated 60 W, 120 are connected in delta. needed so £ 23 8-9 8-14. b. fl resistors are new power Questions and Problems 8-10 If one cut, is Practical level 8- 1 wye-connected generator V in each of its windings. 8-11 voltage of 100 terminals is in Fig. 8.9 V What is 1, a to a 8-3 8-12 at 240° each of these b. If c. R = is £al Could we ahead of£ al ? . Eb2 is 120° lines a-b-c of Fig. 15 (2, what resistors are line voltage is 13.2 is a. b. c. d. the line current if the resistors are in delta? known to be connected in wish to apply full-load 100 kVA, to a each resistance the elements are connected wye a. In b. In delta The windings of a 3-phase motor nected in delta. If the resistance two terminals is 0.6 II, what is are con- between the resis- the voltage across each resistor? is the current in tance of each winding? each line? Calculate the power supplied to the 3-phase Three the resistors are also say that 8-14 Three 24 fl resistors are connected across a load. 8-5 is if wye? the resistors are If We if 1 What the line current in kW when V, 3-phase line. load. Calculate the value of The voltage between 8.12 is 620 V. a. What 208 4 kV, 3-phase generator using a resistive 8- 3 8-4 is wye, calculate the resistance of each. instants? Referring to Fig. 8.9c, phasor is c. the instantaneous value of the volt- behind phasor Eh2 240°, and 330°. each of these instants? is connected the polarity of terminal a with re1 at What connected b. at 0°, 90°, 120°, age across terminals 2, b same conductor of a 3-phase line the load then supplied by a single- heater dissipates 15 generates a peak per phase. spect to terminal c. 3-phase load? supplied to the load. A 3-phase Calculate the instantaneous voltage between What to the in delta burns out, calculate the line connected a. The generator b. connected in- Calculate the line voltage. a. line V is phase voltage or a 2-phase voltage? A 3-phase duces 2400 8-2 one the fuse in If line voltage lamps burn normally? on a 208 V, 3-phase line. a. What is the power supplied Figure 8.35 See Example Three 10 that the What connected kV and in delta V, 3-phase line. Calculate the resistance of three elements connected in in delta. If the wye the line current 1202 A, calculate the following: The current in each resistor The voltage across each resistor The power supplied to each resistor The power supplied to the 3-phase load 600 8-15 that A 60 hp would dissipate the 3-phase motor absorbs 50 from a 600 V, 3-phase b. kW line. If the line cur- is 60 A, calculate the following: The efficiency of the motor The apparent power absorbed by the motor rent a. same power. THREE-PHASE CIRCUITS The reactive power absorbed by The power factor of the motor c. d. 8-16 Three 15 voltage If the line 530 is motor Q reac- in Fig. 8.18. The 8-24 V, calculate the An industrial plant 2.4 kV 8- 1 7 The voltage across each W lamps and a Two 60 connected in wye. The to the terminals outlet. nal Y, What a. 1 resistor b. 0 |jlF circuit capacitor are 8- 1 8 connected is V X-Y-Z of a 3-phase 20 1 8-25 wye 1 Two kW, the phasor diagram for the line voltages. b. The 0 jxF capacitors are connected An In Problem to terminal 8-20 Hz line. Three 1 5 If the power generated 1 is X, which lamp will be brighter? connected resistors to a in in 3 8-27 absorb 60 3-phase line. If a. b. H resistors (R) and three 8 fl reline. Without draw- 8-28 R and X in series, connected in wye R and X in parallel, connected in delta R connected in delta and X connected in wye 8-23 In is in wye and that the motor R with an inductive reactance X. Calculate the values of R and X. power output. energy does the motor consume h? The The load power factor; line current if the line voltage 630 is V. A 20 H resistor is connected A and B of a 3-phase, 480 V between line. lines Calculate Two 30 il resistors are AB and BC connected between of a 3-phase 480 V line. Calculate the currents flowing in lines A, B, and C, respectively. is 8-30 A 1 50 kW, 460 V, 3-phase heater stalled in a hot water boiler. that each branch can be represented by a resistance a. motor has an efficiency of 85 percent, balanced, calculate the following: phases 50 Hz instead of 60 Hz. Problem 8-15, assume connected V from a 600 respectively. 8-29 In Fig. 8.19, calculate the line current if the frequency A line. the currents that flow in lines A, B, and C, rent for each of the following connections: 8-22 the load Industrial application ways ing a phasor diagram, calculate the line cur- c. .5 1 16 A, is The wattmeters in Fig. 8.30 register +35 kW and —20 kW, respectively. If the load is wye, calculate the absorbed. across a 530 V, 3-phase b. kW and motor having a cos 0 of 82 per- How much connected actors (X) are connected in different a. line indicate 3.5 calculate the mechanical 8- 7, if the capacitor new power 1 V Calculate the active power supplied to the a. Calculate b. Three delta-connected kW when deter- 8, motor. they are reconnected 8-2 electric 3-phase in c. 8-19 1 respectively. If the line current The apparent power The power factor of b. line current reactive 8. the resistance and reactance. cent draws a current of 25 across a 2300 V, 60 The can be represented by calculate the following: a. the following: a. that the plant wattmeters connected into a 3-phase, 3-wire 220 level Three a imped- the equivalent line-to-neutral mine the values of 8-26 Advanced kVA from factor of 80 percent an equivalent circuit similar to Fig. the phase sequence? is is Assuming b. The capacitor is connected to termiand the lamp that burns brighter is What Draw draws 600 power ance of the plant? connected to terminal X. a. line at a lagging. and apparent power sup- active, reactive, plied to the 3-phase load b. the phase angle between the line cur- is and the corresponding line-to-neutral voltage? following: a. What b. rent and three 8 ft resistors connected as shown tors are the 181 does in series 8-3 1 it produce Three 5 if the line voltage il resistors are across a 3-phase 480 V connected line. is in- What power is in 470 V? wye Calculate the 1 82 ELECTRICA L MACHINES AND TRANSFORMERS has a full-load efficiency of 93.6% and a current flowing in each. If one of the resistors is that flows in the 8-32 One power disconnected, calculate the current remaining two. of the three fuses protecting a 3-phase 200 kW, 600 V is removed so as to reduce the heat produced by the boiler. What power does the heater de- factor of 83%. Calculate the following: b. The active power drawn by the motor; The apparent power drawn by the motor; c. The a. full-load line current; electric heater rated at A 450 kW, duces 1 300 a. 575 lb V, 3-phase steam boiler pro- 8-34 is 612 produced if A 40 hp, 460 V, TEFC, premium 1 1 Company V, 3-phase, Assuming the is 32 in, 450 kg Square D used to drive a 1600 hp, 60 Hz squirrel-cage motor. motor has a minimum 96% effi- and 90%, What is the reactive power drawn from the line at full-load? 80 r/min, 3-phase, 60 Hz manufactured by Baldor Electric X in delivered by the controller. b. efficiency induction motor X 24 respectively, calculate the full-load current the line V. in ciency and power factor of of steam per hour. Estimate the quantity of steam voltage A 92 motor controller 2400 velop under these conditions? 8-33 8-35 c. What is the phase angle neutral voltage and the between the line current? line-to- Chapter 9 The Ideal Transformer 9.0 Introduction frequency/ The transformer ful electrical is probably one of the most use- devices ever invented. It can raise or lower the voltage or current in an ac circuit, isolate circuits from each other, and it it can can increase or decrease the apparent value of a capacitor, an inductor, or a resistor Furthermore, the transformer enables us to transmit electrical energy over great distances and to distribute it safely in factories and homes. We will study transformers some of the basic properties of in this chapter. It will help us under- stand not only the commercial transformers covered in later chapters but also the basic operating princi- ple of induction motors, alternators, and synchro- nous motors. All these devices are based upon the Figure laws of electromagnetic induction. Consequently, a. we encourage 9.1 b. covered here. Voltage induced in a coil A itive 9.1 voltage able the reader to pay particular attention to the subject matter A is induced in a coil when coil of Fig. 9. la, links) a variable flux soidally at a The which surrounds (or links a vari- sinusoidal flux induces a sinusoidal voltage. and negative peaks $ max . The induces a sinusoidal ac voltage Consider the it flux. effective value is alternating flux in the coil, whose given by flux alternates sinu- E= frequency/ periodically reaching pos- 183 4.44/7V<D max (9. 1) 1 ELECTRICAL MACHINES AND TRANSFORMERS 84 where 9.2 Applied voltage E = effective voltage induced [V] /= frequency of the flux [Hz] = number of turns /V ^max = peak value of = 4.44 It on the and induced voltage and draws a current / m [Wb] the flux a constant [exact value = is It even by an ac current that flows — N is As in any inductive is in <l> The Thus, derived from Faraday's law equation which AcjVAr A(|)/Ar in of flux and c b, 1 when and so the voltage AcJ)/Ar is is change the instantaneous induced voltage. is in Fig. 9. the flux the rate of is the flux time, the rate of change Ac|)/Ar is is is increasing with greater than zero positive. Conversely, when decreasing with time, the rate of change less than zero; consequently, the voltage negative. Finally, when the flux is neither in- creasing nor decreasing (even for one microsecond), the rate of change Act>/A/ voltage given by is circuit, / m lags 90° behind phase with the current (Fig. 9.2b). detailed behavior of the circuit can be ex- c|> density sinusoidal current , which Consequently, I is peak flux Bm . ]X is which, use the peak rms value? The reason m induces an effective voltage of the coil, whose value E E must be is cIJ milx given by Eq. is The . E a and flux 9. 1 . On the the induced identical because they appear be- tween the same pair of conductors. Because we can <J>. across the terminals other hand, the applied voltage voltage a sinusoidal sinusoidal flux called the magnetizing cur- £u = = 4.44/MP max from which we obtain is (9.2) 4.44 JN determines the level of saturation. Example 9-1 The coil in Fig. . 9.1 possesses 4000 turns and links an ac flux having a peak value of 2 quency is mWb. If the fre- 60 Hz, calculate the effective value and (a) frequency of the induced voltage E. Sol til ion 4.44/W* inax (9.1) 4.44 X 60 X 4000 X 2131 V 0.002 (b) The induced voltage has an of 2131 age is V effective or RMS Figure 9.2 value a. and a frequency of 60 Hz. The peak volt- 2131 V2 = 3014 V. £, write proportional to the peak flux in iron cores, m produces rent.The peak value of this ac flux zero, and so the why do we also arises: / in turn creates a zero. max instead of the that the is The mmf /V7 m E, is The question flux Xm of the coil plained as follows: Eq. 9.1 is If the resistance . negligible, the current and in the coil itself. e to a si- 2 tt/V2) be created by a moving magnet, a nearby ac coil, or connected coil of /V turns coil does not matter where the ac flux comes from: may shows a Fig. 9.2a nusoidal ac source £„. The coil has a reactance The voltage E induced plied voltage b. Eg in a coil is equal to the ap- . Phasor relationships between £fg! E, / m and , <l>. THE IDEAL TRANSFORMER This equation shows that for a given frequency and a given to the number of turns, (I> lllax varies in proportion applied voltage E„. This means that if £g is kept constant, the peak flux must remain constant. For example, suppose we gradually a. b. c. d. The The The The 185 peak value of flux mmf peak value of the inductive reactance of the coil inductance of the coil insert an Solution iron core into the coil while keeping E„ fixed (Fig. The peak value of the ac flux will remain ab- solutely constant during this operation, retaining original value $ max even when the core is its com- pletely inside the coil. In effect, if the flux increased (as we would also increase. at But this is we said, E„ is is therefore the same. netizing current core is flux, a The peak current is lm is much Wik) = ^2/ = = kept fixed. For a given supply voltage E„, the ac flux and 9.3 b. E would impossible because E — E„ (9.2) 90) expect), the induced voltage every instant and, as 9.2 = EJ(4MJN) = 120/(4.44 X 60 X = 0.005 - 5 mWb <D max a. 9.3). in Figs. The peak However, the mag- smaller when the iron U= same - inside the coil. In effect, to produce the smaller magnetomotive force is in Fig. 9.3 is much X 4 A 5.66 is NI m = 90 X 5.66 509. 1 needed with The an iron core than with an air core. Consequently, the magnetizing current mmf U >/2 smaller than c. in Fig. 9.2. flux is equal to 5 mmf is the coil 509. 1 mWb at the The inductive reactance d. The inductance when is X m = EJI m = = 30 instant ampere-turns. 120/4 SI is L = XJ2irf = 30/(2tt (2.10) X 60) - 0.0796 = 9,3 79.6 mH Elementary transformer In Fig. 9.4, a coil having an air core an ac source E„. The resulting current (b) a total flux around the <1>, which we coil. If Figure 9.3 The a. is flux in the coil remains constant so long as Eg first, flux. constant. An second Phasor relationships. b. the it A coil Hz having 90 turns is in produces the space bring a second coil close to ac voltage and its E2 is therefore induced in the value can be measured with a The combination of the two coils is The coil connected to the called a transformer. is connected to a 120 V, 60 source. If the effective value of the magnetizing current dispersed excited by /m will surround a portion 4> ml of the total coil voltmeter. Example 9-2 is is 4 A, calculate the following: source is called the primary winding (or primary) and the other one (or secondary). is called the secondary winding 1 ELECTRICAL MACHINES AND TRANSF ORMERS 86 Figure 9.5 Terminals having the Figure 9.4 marked Voltage induced <I> m1 leakage ; in a secondary winding. Mutual flux is <I> f1 flux is . value voltage exists only between primary terminals is and secondary terminals 3-4, respectively. voltage exists between primary terminal ondary terminal trically isolated The flux up into two <I> 3. The secondary is No and sec- 1 therefore elec- from the primary. parts: a mutual flux 4> ml , which links the same does. Suppose, during instant as apart, the mutual flux total flux <t>, two we coils is is very small compared to the then say that the coupling between weak. We two coils closer together. two coils touch, the ) by bring- However, even mutual flux will small compared to the total flux pling is weak, voltage E2 is <J>. When still be the cou- relatively small and, when still, load connected across the secondary terminals. In most it collapses almost completely industrial transformers, the and is a primary and sec- ondary windings are wound on top of each other positive with respect to sec- (Fig. 9.5). Terminals 1 and 3 are terminal 1 and another large dot beside secondary The dots are called polarity marks. The polarity marks in Fig. 9.5 could equally terminal 3. well be placed beside terminals 2 and 4 because, as the voltage alternates, they too, become simultaneously positive, every half-cycle. Consequently, the polar- marks may be shown beside terminals beside terminals 2 and 1 and 3 or 4. 9,5 Properties of polarity A transformer is marks usually installed in a metal enclo- sure and so only the primary and secondary terminals are accessible, together with their polarity marks. But although the transformer may lowing rules always apply to not be visible, the fol- to polarity marks; improve the coupling between them. A current entering a polarity- marked terminal produces a mmf that acts in a "positive'' direc- 9.4 Polarity of a transformer tive" direction* (Fig. 9.6). Conversely, a current 1 . tion. In Fig. 9.4 fluxes On and 3> ml are both stant. It They also pass through zero at the follows that voltage E2 will reach same its in- peak As a result, it produces a flux in the "posi- flowing out of a polarity-marked terminal pro- produced by magnetizing current /nr Consequently, the fluxes are in phase, both reaching their peak values at the same instant. 1 that if worse is primary terminal then said to possess the same polarity. This sameness ity E2 bring the secondary right up to the primary so that the secondary terminal 3 ondary terminal 4 can obtain a better cou- pling (and a higher secondary voltage ing the that positive with respect to primary terminal 2 which links only the turns of the primary. If the coils are far we at the can be shown by placing a large dot beside primary created by the primary can be broken turns of both coils; and a leakage flux the instantaneous polarity are one of these peak moments, A 1-2 same with a dot. * "Positive" and "negative" are cause we can shown in quotation marks be- rarely look inside a transformer to see in which direction the flux is actually circulating. THE IDEAL TRANSFORMER 187 9.6 Ideal transformer at no-load; voltage ratio Before undertaking the study of practical, commer- we cial transformers, shall examine the properties of the so-called ideal transformer. By definition, an ideal transformer has no losses and nitely permeable. Furthermore, by the primary ondary, is its core is infi- any flux produced completely linked by the sec- and vice versa. Consequently, an ideal transformer enclosure transformer has no leakage flux of any kind. current entering a polarity-marked terminal pro- duces a have properties which ap- Practical transformers Figure 9.6 A flux in a "positive" direction. proach those of an ideal transformer. Consequently, our study of the ideal transformer will help us understand the properties of transformers Figure 9.8a shows an ideal transformer duces a mint and flux in the "negative" direction. Thus, currents that respectively flow into and out 2. turns. The primary of polarity-marked terminals of two coils pro- E„ and The If one polarity-marked terminal is momentarily positive, then the other polarity-marked terminal to momentarily positive (each with respect is its in primary and secondary respectively possess duce magnetomotive forces that buck each other. in general. is 4 the and N2 connected to a sinusoidal source the magnetizing current / U1 creates a tlux flux is <t> m it is a mutual flux. flux varies sinusoidally, and reaches a peak value ) mrtx . According to Eq. 9. 1, we can therefore write: other terminal). This rule enables us to re- late the *m-.^. phasor voltage on the secondary side ti with the phasor voltage on the primary side. For example, with phasor in Fig. 9.7, £ ab phasor £dc is in phase . (a) (a) t: g , A', (b) (b) 4>rr Figure 9.7 a. Instantaneous polarities b. Phasor rent is increasing. relationship. when the magnetizing cur- Figure 9.8 The ideal transformer at no-load. Primary and secondary are linked by a mutual flux. b. Phasor relationships at no-load. a. . completely linked by the primary and sec- ondary windings and, consequently, The which /V, 1 ELECTRICAL MACHINES AND TRANSFORMERS 88 = £, 4.44/JV,<l> lllilx (9.3) Calculate: a. The effective voltage across the secondary and terminals E2 = From 4.44./yV2 we deduce these equations, the voltage ratio O max (9.4) b. The peak voltage across the expression for c. The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V and turns ratio a of an ideal trans- former: the secondary terminals Solution: a. The turns ratio N2 /N where £| The secondary voltage N2 = numbers of turns on voltage induced secondary [V] in the on the primary turns 25 times more £ 2 = 25 X E Furthermore, because the primary <F m they are necessarily in phase. The phasor diagram at no load is given is in If the transformer has fewer turns b. The voltage on the sec- As is £, shorter any inductor, current / m lags 90° behind applied voltage Zs The phasor reprethan phasor £",. obviously phase with magne- senting flux <P m is which produces However, because magnetic circuit this is is . in varies sinusoidally; consequently, P cak)= V 2£ = V 2 X 3000 =4242 V c. The secondary voltage at e 2 = 25 permeable and so diagram of such a transformer 9.8b except that phasor / m is is identical to Fig. infinitesimally small. primary and 2250 turns on the secondary source. tween the primary and secondary is 4 A. x 37 E x V Pursuing our analysis, I 2 will I2 con- Does The coupling be- swer is is perfect, but the let us connect a load Z across A the secondary of the ideal transformer (Fig. 9.9). transformer having 90 turns on the magnetizing current = e current ratio secondary current Hz when X 37 = 925 V Example 9-3 nected to a 120 V, 60 25 times greater than 9.7 Ideal transformer under load; required to produce the is is every instant. Consequently, it. Thus, under no-load conditions, the phasor A not quite ideal is: an ideal transformer, infinitely no magnetizing current $m V E 2 = 3000 V . tizing current I m flux 90/2250 in tT the apply Eq. 9.5: l = the peak secondary voltage 2( E2 we can /E 2 =N /N 2 \20/E2 180° out of phase) as indicated by the polarity marks. 120 in Fig. phase with phasor £, (and not ondary than on the primary, phasor ] which again yields , E2 X 25 Instead of reasoning as above, equal to the ratio of the is and secondary voltages are induced by the same 9.8b. Phasor = Ei -3000 V and secondary voltages turns. Consequently: turns. turns ratio number of therefore 25 times greater is than the primary voltage because the secondary has the secondary This equation shows that the ratio of the primary mutual (9.5) voltage induced in the primary [V] numbers of = -2250/90 x = 25 = E2 = N\ = a is: E2 immediately flow, given by: = E2 IZ we connect the load? To anwe must recall two facts. First, in change when this question, an ideal transformer the primary and secondary windings are linked by a mutual flux <I> m and by no , other THE IDEA L TRANSFORMER same the <*W: when time. Thus, maximum + is be E To N2 N, { mark on when effect, on the primary /,, ) /, lags behind behind Ee , we can now draw facts, polarity a resistive-inductive load, current E2 by an angle Flux 6. but no magnetizing current produce the phasor under load (Fig. ideal transformer Assuming 9.9b). must flow out of the the secondary side (see Fig. 9.9a). Using these to ( flows into a polarity mark /[ side, I 2 diagram of an I2 maximum + phase. Furthermore, in order to produce the in bucking (a) I 2 is 89 In other words, the currents must ). ( goes through zero, /2 goes through zero, and when 1 this flux because this is <I> / lags 90° m m is needed an idea trans- former. Finally, the primary and secondary currents (b) According are in phase. to Eq. 9.6, they are related by the equation: Figure 9.9 a. b. AT, under load. The mutual mains unchanged. Phasor relationships under load. Ideal transformer (9.7) flux re- where f = primary flux. In other words, an ideal transformer, by defini- tion, has no leakage flux. Consequently, the voltage ratio under load is the same /Vj as at no-load, namely: the if E„ the supply voltage is E induced voltage primary E2 E2 also remains fixed. remains fixed whether a load Let us remains fixed. { now examine the is It mmf would produce in current not. the mutual flux change under <L> m But we just saw . load. only remain fixed We if conclude If it acted that <E> in m does not <I> m can the primary develops a mmf which exactly counterbalances Thus, a primary current /j that flux N2 I 2 at every must flow so AVi = N 2 I 2 To obtain . profound change instant. f and I2 what we gain and vice versa. This is see that the in voltage, the primary E2 l 2 we lose consistent with the Ef to x must equal the apparent power output power inputs and outwould mean that the transformer itself absorbs power. By definition, this of the secondary. If the puts were not identical, is impossible in it an ideal transformer. Example 9-4 ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a (9.6) 200 the required instant-to-instant bucking ef- fect, currents ratio. In effect, we the inverse of the volt- is requirement that the apparent power input An that: 9.5 and Eq. 9.7, transformer current ratio age mmf N2 I2 a turns ratio Comparing Eq. We conclude that magnetomotive forces current I 2 produces a secondary — connected or created by the primary and secondary windings. First, alone, this turns on the secondary kept fixed, then Consequently, mutual flux <P m also remains fixed. follows that number of on the primary x a Second, secondary current [A] = number of turns N2 = - N /N2 £,/£ 2 = I2 current |A) must increase and decrease at V, 50 Hz source. The load across the secondary draws a current of 2 cent lagging (Fig. A at 9. 10a). a power factor of 80 per- 90 1 ELECTRICAL MACHINES AND TRANSFORMERS Figure 9.10 See Example 9-4. b. Phasor relationships. a. current in the secondary. Therefore Calculate: a. The b. The instantaneous effective value of the primary current current in the 100 mA, primary when /, 1/ 100 is mA c. The peak d. Draw flux linked by the secondary winding the phasor diagram c. a. The is The peak turns ratio = ?s / — '2 instantaneous = 25 X 0. = 2.5 A the same as that linking the primary. flux in the secondary = <P max 1/25 current ratio is = E^/(4MfN ) = 200/(4.44 X 50 X = 0.01 ] therefore 25 and because the 10 primary has fewer turns, the primary current 25 times greater than the secondary current. 25 X 2 d. by means of Eq. N,/, 90/, /, b. To draw = 50 A Instead of reasoning as above, the current mWb the phasor diagram, lows: Secondary voltage = we can The instantaneous current E2 is in Phase angle between in the power primary is al- ways 25 times greater than the instantaneous as fol- 25 X 200 phase with £, indicated by the polarity marks. For the same reason, 2 we reason is E2 = 25 X Ei = = 5000 V calculate 9.6. = N 2 l2 - 2250 X = 50 A 90) is Consequently: /, is is: ^=yv,/yv 2 -90/2250 The = In an ideal transformer, the flux linking the sec- ondary Solution: l2 is: instantaneous 1 the instantaneous current in the secondary when /j is in E 2 and l 2 is factor = cos 6 0.8 = cos e 9 = 36.9° phase with I2 . THE IDEA L TRA NS FORMER Figure 9.11 Symbol for an b. Symbol for an a. ideal transformer ideal transformer and phasor diagram using sign notation. and phasor diagram using double-subscript The phase angle between £, and /, is also 36.9°. The mutual tlux lags 90° behind E„ (Fig. 9.10b). In and for an 1 To highlight the bare former, it is 9.1 lb), ideal transformer best to /, la, draw in it symbolic form. Thus, in- and the mutual tlux ing primary 4> m , we (Fig. 9. 1 1). to indicate the direction of current flow as well as the polarities of E voltages into , and £2 F° r example, - a current one polarity-marked terminal panied by a current is ity-marked terminal. Consequently, ways in { flowing always accom- /, and U are al- phase. /N 2 = a, we and E2 are always in phase, and so are double-subscript notation £ab /2 and Ecd are always (which in used (Fig. is phase and so are . may sometimes the nature of the load be a source) connected to the secondary side. 9.9 Impedance ratio Although a transformer is generally used to trans- form a voltage or current, it also has the impor- tant ability to transform an impedance. Consider, for example, Fig. 9. 12a in which an former T load Z. The is ideal trans- connected between a source E„ and a ratio of transformation is a, and so we can write Furthermore, N /[ flowing out of the other polar- I2 x simply show a box hav- and secondary terminals marks are added, enabling us Polarity and E The angle a depends upon essentials of an ideal trans- stead of drawing the primary and secondary windings notation. / 2 .* If the symbol 9.8 Circuit 9 an ideal transformer, and specifically referring to Fig. 9. /, 1 if we let the ratio of transformation obtain £, * = a£ 2 Some texts show the respective voltages and currents as being 180° out of phase. This situation can arise depending upon how the behavior of the transformer is described, or how the By us- voltage polarities and current directions are assigned. and ing the methodology /, = / 2 /a never any doubt as to we have adopted how the phasors in this book, there should be drawn. is 92 1 ELECTRICAL MACHINES AND TRANSFORMERS impedance across the actual tical to the secondary terminals multiplied by the square of the turns ratio. The impedance transformation is real, and not illusory like the image produced by a magnifying An glass. (a) ideal transformer can of any component, be ductor. For example, modify the value a resistor, capacitor, or in- it a 1000 12 resistor if placed is across the secondary of a transformer having a pri- mary secondary turns ratio of to across the primary as X (1/5)" = 40 12. if it Similarly, a reactance of 1000 fl ondary, it 1 :5, will appear it had a resistance of 1000 is if a capacitor having connected to the sec- appears as a 40 il capacitor across the primary. However, because the reactance of a capacitor — (X c Figure 9.12 a. Impedance transformation using a transformer. b. The impedance seen by the source differs from is inversely proportional to l/2ir/'C), the the primary terminals actual value. Z We its capacitance apparent capacitance between is 25 times greater than its can therefore artificially increase (or decrease) the microfarad value of a capacitor by means of a transformer. EJE, impedances from secondary to primary and vice versa 9.10 Shifting and LIU = As far as the ance source Z x between is l/a concerned, it sees an imped- the primary terminals given by: Zx = EJ1 As a further illustration of the impedance-changing properties of an ideal transformer, consider the circuit of Fig. 9. 1 3a. It is composed of a source E„, a ! transformer T, and four impedances Z, to On the other hand, Z given by the secondary sees an impedance transformer has a turns ratio We can progressively to 9.l3e. Zx can be expressed ^ = a£\ in another way: / 2 /a 2 Z As the impedances are shifted x 1 3b way, the impedances are transferred a~. to the pri- A, mary side, the ideal this position the a~Z (9.8) This means that the impedance seen by the source times the real impedance (Fig. ideal transformer has the 9. in this transformer ends up treme right-hand side of the circuit (Fig. Z = 2 secondary imped- impedance values are multiplied by If all Consequently, a shift the the circuit configuration remains the same, but the shifted a~£\ a /, . ances to the primary side, as shown in Figs. Z = E 2 /I 2 However, Z4 The a. amazing 9. 1 is 2b). Thus, an ability to increase is iden- 1 3d). In secondary of the transformer is on open-circuit. Consequently, both the primary and secondary currents are zero. move In der We can therefore re- the ideal transformer altogether, yielding the equivalent circuit or decrease the value of an impedance. In effect, the impedance seen across the primary terminals at the ex- 9. comparing how a circuit shown Figs. 9. 1 in Fig. 9.l3e. 3a and 9. which contains a 1 3e, real we may wonT transformer THE IDEAL TRANSFORMER Z, a 2 193 Z3 (e) Figure 9.13 a. The and actual circuit showing the actual voltages currents. Impedance Z 2 b. is c. Impedance Z3 Impedance Z4 is rents All in /2 . change E3 and in /3 Note . Note .The cur- shifted to the primary side. the corresponding e. E 2 and in shifted to the primary side. is the corresponding d. Note shifted to the primary side. the corresponding changes T are now change E4 and in /4 zero. the impedances are now transferred to the mary side and the transformer pri- no longer needed. is can be reduced to a circuit which has no transformer any meaningful relationship at all. In effect, is there between the two circuits? The answer is yes — there is a useful relationship between the real circuit of Fig. 9. 13a and the equivalent circuit of Fig. 9.l3e. The reason that the voltage is £ across each element in the secondary side becomes a£ when the element shifted to the primary side. Similarly, the current each element the element On in the is secondary side becomes account of this relationship, simply reduce in Fig. 9. rents. by I 1 it Z4 easy to solve is it shown in Fig. 9. to the equivalent 3e and solve for 3a. the voltages and cur- all /a and by a, which yields the actual voltages illustrate, in Fig. 9. 14 in the suppose that the is £4 volts and amperes. Then, secondary that the real current it cuit, the voltage across the a"Z4 impedance £4 X is /4 a volts. On through the impedance side. real voltage across through to 1 form shown These values are then respectively multiplied and currents of each element To when //a shifted to the primary side. a real circuit such as the one We is / in in the equivalent ciris equal the other hand, the current is equal to /4 ^ a amperes ELECTRICAL MACHINES AND TRANSFORMERS 194 Figure 9.14 Actual voltage and current in Z4 impedance a 2 . Z4 a£ 4 Figure 9.15 Equivalent voltage and current (c) . z, 1? words, whenever an impedance (Fig. 9. 15). In other is Z4 in transferred to the primary side, the real voltage across the impedance increases by a factor a, while the real current decreases by the factor a. general, In ferred whenever an impedance from one side of a transformer the real voltage across the turns ratio. If the side it changes impedance is is trans- to the other, transferred to the where the transformer voltage is higher, the voltage across the transferred impedance will also be higher. Conversely, ferred to the side if the J proportion to in impedance is Figure 9.16 a. where the transformer voltage is b. lower than the the ratio of the In real voltage — again, of course, in c. some cases it is useful to shift is Impedance Z 1 is The source is impedances secondary side (Fig. 9.16a). The procedure in to the is the d. and in £-, A,. transferred to the secondary side. in £g Note . also T are zero. All the impedances and even the source are now on the secondary side. The transformer is no longer needed because its currents are zero. that the currents in from the primary side and transferred to the secondary side. Note the corresponding change number of turns. the opposite way, that actual circuit, showing the real voltages Note the corresponding change lower, the voltage across the transferred impedance is The currents on the primary side. trans- THE IDEA L TRA NS FORMER same, but all vided by a 2 impedances so transferred are now (Fig. 9.16b). We can even source E„ to the secondary side, where source having a voltage is now Eg /a. The it Z- becomes a ideal transformer \ R 2 + (X L - Xc = V4 2 + primary of circuit (Fig. 9.16c). In this position the circuit in Fig. 9. 18 95 is shift the located at the extreme left-hand side of the the transformer + \ 16 - (5 - 2 (2. 17) ) 2 2) 9 on open-circuit. Consequently, is both the primary and secondary currents are zero. before, The impedance of the di- 1 we can remove = As the transformer completely, The s n current in the circuit is leaving us with the equivalent circuit of Fig. 9.16d. = E/Z = / Example 9-5 The voltage across Calculate voltage Fig. 9.17, E knowing and current / in the circuit that ideal transformer primary to secondary turns ratio of 1 : T £7 100 has a 100. The the impedances to solve this to the problem is , = 8V therefore, 100 - 800 V Questions and Problems impedance values are or 10 000. Voltage E becomes £7100, but current / remains unchanged because 9- 1 The already on the primary side (Fig. 9. 1 500 turns and coil in Fig. 9.2a has a re- actance of 60 il but negligible resistance. it it is X is, is primary side of the trans- turns than the secondary, the 2 E A to shift all former. Because the primary has 100 times fewer divided by 100 8 2 2X4 = //? actual voltage E= way easiest = = the resistor of Solution The 10/5 8). is connected to a 120 V, 60 Hz source If £u , calculate the following: a. The effective value of the magnetizing cur- rent / m 9-2 b. The peak value of / ni c. d. The peak and mm!' produced by The peak flux 4> m:ix In Problem to 40 9-3 , if the voltage E„ What coil is new reduced is mmf developed and the peak flux (I) I1UIX . meant by mutual flux? by leakage flux? Figure 9.17 See Example 1 V, calculate the by the 1:100 9- the coil 9-5. 9-4 The ideal transformer in Fig. 9.9 has turns on the primary and secondary. of 12 Figure 9.18 Equivalent circuit of Fig. 9.17. a voltage a resistance Calculate the following: The voltage E 2 b. The c. e. The current /, The power delivered to The power output from the secondary In Problem the impedance seen d. 9-5 11. Z is 500 turns on the The source produces E„ of 600 V, and the load a. 300 current I 2 9-4, by the source what is the primary [ WJ W] | 196 9-6 ELECTRICAL MACHINES AND TRANSFORMERS In Fig. 9. 17, calculate the voltage across the capacitor through 9-9 and the current flowing Two coils are set up as shown in Fig. 9.4. Their respective resistances are small and may be it. 1, neglected. The having terminals coil 2 has 320 turns while the coil having ter- Industrial application minals 9-7 The nameplate on a 50 kVA transformer shows a primary voltage of 480 V and a sec- when 1 is mine the approximate number of turns on the primary and secondary windings. Toward this wound around the end, three turns of wire are and a voltmeter external winding, across this 3-turn coil. then applied to the 1 connected is V winding, voltage across the 3-turn winding found to How many turns are there on the 480 V and 120 V windings (approximately)? A coil with an air core has a resistance of is connected to a 42 V, 60 14.7 fl. When it Hz ac source, it draws a current of 1 .24 A. Calculate the following: a. b. c. The impedance of the coil The reactance of the coil, and its inductance The phase angle between the applied voltage (42 V) and the current ( 1 .24 A). Hz It is voltage is found that applied to ter- voltage across terminals 3-4 V. Calculate the peak values of <)>, <(>, ,, . |jlF, 600 V paper capacitor we need one having jjlF. It is is available, a rating of about proposed to use a transformer modify the 40 fxF so that it appears as fxF. The following transformer ratios are available: 120 V/330 V; 60 V/450 V; 480 V/1 50 V. Which transformer is the most 300 be 0.93 V. 9-8 1-2, the 4> ml A 40 but to and the is ^ 300 A voltage of 76 V is 20 22 and 4 has 160 turns. a 56 V, 60 minals We wish to deter- ondary voltage of 20 V. 3, appropriate and what is the reflected value of the 40 fxF capacitance? To which side of the transformer should the be connected? 40 p,F capacitor Chapter 1 Practical Transformers 10.0 Introduction In we Chapter 9 discovered real its with an imperfect core studied the ideal transformer and basic properties. However, in the The world transformers are not ideal and so our happens account. Thus, the windings of practical transform- have resistance and the cores are not iron infinitely is not completely captured by the sec- And losses, a practical we primary which con- The transformer can be described by an resistance current is in is phase with E current its £p that 1 0. 1 a). The produces a x (Fig. 10.1b). Xm a is measure of the permeability of the transformer core. Thus, age regulation and the behavior of transformers that also used to illustrate with the primary To furnish these drawn from the line. This The magnetizing reactance cir- The per-unit method mode of application. We represents the iron losses and /, the permeability in parallel. Rm losses a small current developed from fundamental concepts. This connected in parallel excited by a source is enables us to calculate such characteristics as volt- are low? the resulting heat they produce. equivalent circuit comprising an ideal transformer cuit is rather is voltage £], discover that the properties of and resistances and reactances. The equivalent and eddy-current hysteresis whose permeability R m and X m elements tribute to the temperature rise of the transformer. In this chapter What replaced by an terminals of the ideal transformer (Fig. finally, the iron cores pro- duce eddy-current and hysteresis having core is can represent these imperfections by two circuit ondary. Consequently, the leakage flux must be taken into account. the previous in such a perfect core if losses and permeable. Furthermore, the flux produced by the primary transformer studied ideal chapter had an infinitely permeable core. simple analysis must be modified to take this into ers Ideal transformer 10.1 /m is flowing X m is relatively through X m represents low, if The mag- low. the netizing current needed to create the flux 3> m in the is core. This current lags 90° behind 197 E . } 1 ELECTRICAL MACHINES AND TRANSFORMERS 98 = 0 /1 Jo /o=0 A 5 120 V 60 Hz A An ideal X T Figure 10.1a An imperfect core represented by a reactance a resistance Rm Xm and Figure 10.2a See Example 10-1 . Xm The values of the impedances R m and can be found experimentally by connecting the trans- transformer is shown former to an ac source under no-load conditions and <t> measuring the active power and reactive power absorbs. m = is The peak value again given by Eq. 9.2: E f(AMfN x x (9.2) ) it The following equations then apply: R m = E* 2 /P m Xm = Er/Q m in Fig. 10.1b. of the mutual flux <£ m Example 10-1 (10-1) A (10.2) exciting current 70 of 5 large transformer operating at no-load A when draws an the primary is where a wattmeter test R [u = Xm = resistance representing the iron losses [Cl] equal to 180 magnetizing reactance of the primary Calculate winding Ei = Pm — Q in = The [flj primary voltage [V] c. iron losses reactive power needed IvarJ total / in . [W] flux <t> m to set up the mutual It is ally a small is equal to the phasor called the exciting current 7 0 . <t> m at no-load for The reactive power absorbed by The value of R m and X m The value of 7f 7 m and 7 sum of 7f It is usu- , , the core t) The apparent power supplied Sm in percentage of the full-load current. The phasor diagram W.* Solution a. current needed to produce the flux an imperfect core and a. b. it is = E,7C = 120 to the core X is 5 = 600 VA The iron losses are P m = 80 W this less-than-ideal 1 The reactive Qm = power absorbed by ^^~Pi = 572 Figure 10.1b Phasor diagram of a practical transformer at no-load. con- Hz source (Fig. 10.2a). From known that the iron losses are nected to a 120 V, 60 var the core = V600 r^780I is PRACTICAL TRANSFORMERS The impedance corresponding b. 99 1 to the iron losses is 2 Rm = E = { 120 /180 The magnetizing reactance c. 2 } = 0 is 2 /Q m = 120 /572 25.2 ft The current needed If 1 a so Xm = E / 2 = IP m supply the iron losses to = E\/R m = 1.5 = is 120/80 Figure 10.3 A Transformer with The magnetizing current /m = E IX m = = 4.8 A across the primary 120/25.2 The exciting current - V/? 5 is x $m its h = infinitely + /t) i peak value Vl.5 2 + 4.8 2 cause it A E2 to the current is given in Fig. 10.2b. Ep and it sets up a mutual flux given by ^> nikl is Ep and = £ p /(4.44 /TV,). infinitely is permeable and be- /, = 0. E2 = (N2 /N E p Owing zero, no mmf is available to has no losses, the no-load current The voltage The phasor diagram at no-load. This flux lags 90° behind a in the core. Because the core is = II permeable core is is given by being drive flux through the air; { ) . consequently, there is no leakage flux linking with the primary. If = 1.5 A 120 V Let us now connect a load Z across the sec- ondary, keeping the source voltage A'L^Si = 4.8 /m 10.4). /« ~ 5 A Ep fixed (Fig. This simple operation sets off a train of events which we list as follows: Figure 10.2b Phasor diagram. 10.2 Ideal transformer with loose coupling We have just seen how an ideal transformer behaves when it has an imperfect core. We now assume a transformer having a perfect core but rather loose coupling between ings. We also its primary and secondary wind- assume that the primary and sec- ondary windings have negligible resistance and the turns areN ,N2 Consider the transformer to a source E„ and operating Figure 10.4 Mutual fluxes and leakage fluxes produced by a trans- . l in Fig. 10.3 at no-load. connected The voltage former under load. The leakage fluxes are due to the imperfect coupling between the coils. ELECTRICAL MACHINES AND TRANSFORMERS 200 1 Currents . and /, 2. I2 flow to = N2 /N hence : { produces an mmf /V,/,. equal and /V,/| mmf N2 I 2 = N2 I2 while - /| produces an These magnetomotive forces are in direct when opposition because flows into the polarity-marked terminal flows out of polarity-marked terminal The mmfiV 2 / 2 produces 3. portion of C I> ( 2 ( <t>, . mmf /V]/, A portion Flux <I>, of (<I> t 2 - produces a (<I> ml is , ) called the The magnetomotive the magnetic field ) does we can Figure 10.5 not. A $ ni links with the sec- P n ) does analyze this and /, In general, 1 of two parts; a same the new mutual flux (The mutual flux <!>,-,. <I> inl total flux posed of a mutual flux <I> A voltage E n . composed and we combine is £n A voltage E 2. produced by mutual flux <I> m ni , and (Fig. 10.5). This /2 com- is is we flux <I> n is mutual flux is NJ phase with 2 1 . <I> n and induced by mutual flux x (10.5) 4> in and = 4.44./N <I> l Ep = six basic facts, (10.6) m applied voltage E^. we now proceed to note that the primary leakage flux / lr and leakage flux O l2 10-3 Primary E We can better identify the four induced voltages E2 £ n and £ t2 by rearranging the transformer cir- induced composed of two A voltage E [2 , , s in the secondary is cuit as parts: induced by leakage flux winding <I> I2 and secondary leakage reactance is in . Fifth, the voltage actually Using these . I primary in the =4.44.^,*,-, Sixth, induced voltage cre- h while the secondary leakage created by /V 2 / 2 Consequently, leakage flux is in induced parts: develop the equivalent circuit of the transformer. created by phase with are not in phase. Ep 3> m2 into a single ated by the joint action of the primary and sec- Fourth, (10.4) given by . ondary mmfs. (J>,-, E2 #m given by not m2 and a leakage flux <£ (2 <t> 4.44/7V2 induced by leakage flux £, Third, m and and a leakage (p mJ in Fig. 10.4 as ( F mla in Fig. 10.3.) Second, the E [2 composed of two reason as follows: is tp induced by mutual flux Similarly, the voltage is /] E2 E2 = how is, situation? we voltage upset /2 that existed in the core be- produced by flux. given by ( new First, the total flux A 2. c ( primary leakage flux. forces due to |. Referring to Fig. 10.4, flux transformer possesses two leakage fluxes and a ac flux total was connected. The question fore the load A . ^ m 2) links with the primary ondary winding, while another portion not. 3. mutual Similarly, the 4. t called the secondary leakage flux. 2 is <I>, / 1, 1 2 a total ac flux 4> 2 winding while another portion Flux in by the ideal-transformer equation related /,// 2 immediately begin I2 and secondary windings. They are the primary and 10.6. to . 4.44/N 2 <I\2 (10.3) by two fluxes, clearly <J> r2 and This rearrangement does not change the value of the induced voltages, but Er Thus, the secondary show even more that the jV 2 turns are linked $m given by shown in Fig. is drawn twice age stand out by itself. it Thus, does make each it becomes volt- clear that PRACTICAL TRANSFORMERS 201 Figure 10.6 Separating the various induced voltages due to the mutual flux and the leakage fluxes. Figure 10.7 Resistance and leakage reactance E l2 is of the primary and secondary windings. drop across a reactance. This really a voltage X secondary leakage reactance - { 2 is given by Example 10-2 The secondary winding of 180 turns. X = Ef2 /I2 (1 l2 0.7) When a transformer possesses the transformer is under load, the secondary current has an effective value of 8 A, 60 1 The primary winding also is shown twice, to separate £, from E n Again, it is clear that E n is simply a voltage drop across a reactance. This pri. mary leakage reactance Xn Xn is =£ fl shown in Figure 10.7. We (10.8) // l tive which, of course, act windings. mWb. The m has a peak value of secondary leakage flux <I>, 2 has a mWb. Calculate a. in series and b. with the respec- c. /?, The voltage induced by have also added the primary and secondary winding resistances Ri, 20 peak value of 3 <I> given by The primary and secondary leakage reactances are Hz. Furthermore, the mutual its in the secondary winding leakage flux The value of the secondary leakage reactance The value of E 2 induced by the mutual ELECTRICAL MACHINES AND TRANSFORMERS 202 —Oi Eo 1 'lm Figure 10.8 Complete equivalent /f X a practical transformer. The shaded box T circuit of If Solution a. The effective voltage induced by the secondary leakage flux an is we add ideal transformer. elements circuit sent a practical core, we Xm and Rm to repre- obtain the complete equiv- alent circuit of a practical transformer (Fig. 10.8). is In this circuit E l2 = = 4.44/yV 2 f2 X 60 X 4.44 = O 143.9 X 180 T is an ideal transformer, but only the (10.3) primary and secondary terminals 0.003 cessible; I -2 and 3-4 are ac- other components are "buried" inside transformer the V all However, by appropriate itself. we can find the values of all the circuit elements that make up a practical transformer. Table 10A shows typical values of /?,, R 2 X n X r2 X m and R m for transformers ranging from kVA tests b. The secondary leakage reactance is X = EnJI 2 (10.7) {2 = 143.9/18 , , I , 400 MVA. The nominal primary and secondary £np and £ ns range from 460 V to 424 000 V. The corresponding primary and secondary currents / np and / ns range from 0.4 7 A to 29 000 A. The exciting current /0 for the various transto = 8ft c. The voltage induced by voltages the mutual flux is 1 E2 = 4.44^2*. = (10.4) X 60 X 4.44 1 X 80 0.02 formers also shown. is is It always much smaller than the rated primary current = 959 V Note that in where S n is the rated /, EnpInp = En J ns — each case 5n power of the transformer. 10.4 Equivalent circuit TABLE 10A of a practical transformer The circuit of Fig. 1 0.7 and inductive elements is composed of (/?,, R 2 X n X i2 , , resistive , Z) cou- m which links the primary and secondary windings. The leakagepled together by a mutual flux free <t> in same properties and obeys For example, we can primary (Ay/V 2 ) side 2 , as we by shift multiplying did before. I 10 2400 2400 1 400000 100 1000 2470 69000 13800 V A A 460 347 600 6900 424000 0.4 17 4.17 8.02 14.5 29000 2.17 28.8 167 145 943 a 58.0 5.16 11.6 27.2 0.0003 pos- R, fi 1.9 0.095 0.024 0.25 0.354 X~n a 32 4.3 39 151 0.028 Chapter Xa xm Q. 1.16 0.09 0.09 1.5 27 150000 460 Km A» 52.9 impedances their 'ns kVA V same It the rules as the ideal transformer discussed in 9. /„p the dotted actually an ideal transformer. is sesses the Ens , magnetic coupling enclosed square £n P ACTUAL TRANSFORMER VALUES to the values by a 200000 29000 400000 51000 220000 505000 432000 A 0.0134 0.0952 0.101 0.210 317 , PRACTICAL TRANSFORMERS 10.5 Construction of a power Winding resistances 2 to reduce the J transformer R loss R and R 2 : 1 0.9a Power transformers are usually designed so that a characteristics approach those of an ideal ondary are wound on one their transformer. Thus, to attain high permeability, the made of The resulting magnetizing current / m is at least 5000 times smaller than it would be if an air core were used. Furthermore, to keep the iron losses down, the core core is is iron (Fig. 1 0.9a). laminated, and high resistivity, high-grade silicon steel is used. Consequently, the current supply the iron losses than /m is /r needed to usually 2 to 4 times smaller power transformer and secondary leg. In practice, the amount of copper. For built up so as Figure 1 to is 0.9b shows how 2. 10a). small transformer are stacked to build up the core. Figure 1 0.9c shows the primary winding of a voltages. them as closely together as insulation considera- The coils are carefully insulated from each other and from the core. Such coils is means that the tight when It also guarantees a load is connected secondary terminals. cou- secondary almost exactly equal to times the primary voltage. voltage regulation 1 the laminations of a ondary coils on top of each other, and by spacing voltage at no-load the not square (as shown) but be nearly round. (See Fig. ondary windings depends upon between the sec- primary reason, in larger transformers the cross section Leakage reactances X n and X r2 are made as small as possible by winding the primary and sec- pling ensure much bigger transformer. . tions will permit. to a simplified version of coils are distributed over both core of the laminated iron core is is which the primary and in legs in order to reduce the same are kept low, both and resulting heat and high efficiency. Figure 203 N2 /N i good to the The number of turns on the primary and sectheir than a low-voltage winding. current in a HV winding is On the other hand, the much smaller, enabling us to use a smaller size conductor. amount of copper windings respective A high-voltage winding has far more turns is in As a result, the the primary and secondary about the same. In practice, the outer coil (coil 2, in Fig. I0.9a) length per turn is greater. weighs more because the Aluminum ductors are used. laminated iron core Figure 10.9a Figure 10.9b Construction of a simple transformer. Stacking laminations inside a coil. or copper con- ELECTRICAL MACHINES AND TRANSFORMERS 204 subtractive polarity additive polarity x2 « Figure 10.10 Additive and subtractive polarity H cation of the 1 -X additive polarity 1 depend upon the lo- terminals. when terminal H, is diagonally opposite terminal X,. Similarly, a transformer has subtractive polarity when terminal Hj terminal X, (Fig. 10.10). If we know is adjacent to that a power transformer has additive (or subtractive) polarity, we do not have to identify the terminals by symbols. Subtractive polarity is standard for all single- phase transformers above 200 kVA, provided the high-voltage winding is rated above 8660 V. All other transformers have additive polarity. Figure 10.9c Primary winding 290 A. of a large transformer; rating 128 kV, 10.7 Polarity tests To determine whether (Courtesy ABB) a transformer possesses ad- ditive or subtractive polarity, A ther transformer is reversible in the sense that ei- winding can be used as the primary winding, 1 where primary means the winding that is we proceed as follows (Fig. I0.ll): . connected Connect the high-voltage winding to a low- voltage (say 120V) ac source E„. to the source. 2. 10.6 Standard terminal saw in markings Section 9.4 that the polarity of a trans- a jumper J between any two adjacent HV and LV terminals. 3. We Connect Connect a voltmeter adjacent HV and LV Ex between the other two terminals. former can be shown by means of dots on the primary and secondary terminals. This type of marking on instrument transformers. On power is used j transformers, however, the terminals are designated by the symbols H, and H 2 for the high-voltage (HV) winding and by X and X 2 for the low-voltage (LV) winding. By con( vention, H, and Xi have the Although the H2 bols H,, , polarity X,, and power transformers the four dard X2 it is is same polarity. known when the sym- are given, in the case of common practice to mount terminals on the transformer tank in a stan- way so that the transformer has either additive or subtractive polarity. A transformer is said to have Figure 10.11 Determining the polarity of a transformer using an ac source. PRA CTICA L TRA NS FORMERS Connect another voltmeter 4. winding. polarity Ex If Ep across additive. This tells us that H, is HV the gives a higher reading than Ep ing. , the and X, to the On the other hand, if £x than E the polarity is subp are diagonally opposite. gives a lower reading jumper J In this polarity text, E secondary voltage the Ep voltage tracts s E E s with the primary terminal of the voltmeter is Ex = Ep + £ or Ex = the polarity. We can now see is marked H, . 10,8 Transformer taps Due to age in a particular may consistently be lower than normal. Thus, a dis- voltage drops in transmission lines, the volt- s region of a distribution system tribution transformer having a ratio of may be connected voltage the polarity test, an ordinary 120 V, 60 Hz source can be connected 2400 V/120 to a transmission line V where the never higher than 2000 V. Under these con- is its to the HV winding, nominal voltage may be several During a polarity test on a 500 kVA, 69 kV/600 motors may 1 1 V stall 10. 13). this problem taps are provided on the Taps enable us 8 V, Ex = We percent. 119 V. Determine markings of the terminals. is additive because HV may be 4'/>, 9, Ex LV is greater than voltage is only 2076 would use terminal (or H 2 and and Some X,). Figure 10.12 shows another circuit that may used to determine the polarity of a transformer. source, in series with an open switch, LV winding a voltage A dc connected of the transformer. The trans- former terminal connected is is to the positive side marked X,. A dc voltmeter is of the Determining the polarity /,, 9, ratio so or 13'/> is 1 V 1 0. (instead of and tap 1 3, if the line 2400 V), we 5 to obtain 120 V whenever above or below a preset the secondary voltage level. Such tap-chang- ing transformers help maintain the secondary volt- age within ±2 percent of its rated value throughout the day. connected 2 O- Figure 10.13 of a transformer using a dc on transformers are designed to change the taps automatically HV terminals. When the switch is closed, is momentarily induced in the HV wind- Figure 10.12 source. be 1 or 13'/2 percent below normal. Thus, the secondary side. source by 4 secondary voltage, even though the primary voltage terminals con- Consequently, the across the change the turns can therefore maintain a satisfactory nected by the jumper must respectively be labelled to the to referring to the transformer of Fig. Solution X2 electric under moderate overloads. as to raise the secondary voltage the following readings 10.11), Ep = were obtained: H, and cook food, and electric stoves take longer to To correct transformer (Fig. polarity consider- primary windings of distribution transformers (Fig. Example 10-3 the polarity is ably less than 120 V. Incandescent lamps are dim, hundred kilovolts. . H2 marked ditions the voltage across the secondary making even though Ep ) the terms additive and subtractive originated. In The + ( and the other either adds to or sub- In other words, , the pointer of the voltmeter upscale, the transformer terminal connected effectively connects in series Consequently, . from Ep — Es depending on how , and terminals H, and X, are adjacent. tractive, moment, this If, at moves 205 Distribution transformer with taps at 2184 V, and 2076 V. 2400 V, 2292 V, ELECTRICAL MACHINES AND TRANSFORMERS 206 Losses and transformer 10.9 rating Like any electrical machine, a transformer has They losses. are composed of The nameplate of a distribution transformer indicates 250 kVA, 60 Hz, primary 4160 V, secondary 480 V. the following: Calculate the nominal primary and secondary a. 2 1. 2. 3. I R __ Example 10-4 losses in the windings currents. we apply 2000 V to the 4 60 V primary, can we still draw 250 kVA from the transformer? If b. Hysteresis and eddy-current losses in the core 1 Stray losses due to currents induced in the tank and metal supports by the primary and sec- Solution ondary leakage fluxes a. The duce l) form of heat and pro- losses appear in the an increase efficiency. temperature and 2) a drop in Under normal operating conditions, efficiency of transformers is very high; it may Nominal current of L np nominal S = 99.5 percent for large power transformers. — ns The heat produced by the iron losses depends upon the peak value of the mutual flux <E> m which in turn depends upon the applied voltage. On the , E np 4160 b. we ... apply 2000 V to the primary, the flux rent should not we must set limits to at an accept- two limits determine the nominal voltage / np E np The power rating of a transformer is product of the nominal voltage times the nominal current of the However, the primary or secondary result is winding. not expressed in watts, be- cause the phase angle between the voltage and current may have any voltage and equal to the value at all, depending on the na- ture of the load. Consequently, the capacity of a transformer is power-handling expressed in voltam- S = 2000 V X is directly related to the through it. apparent power This means that a 500 will get just as hot feeding a as a 500 kW kVA that flows transformer 500 kvar inductive load resistive load. The rated kVA, frequency, and voltage are always shown on the nameplate. In large transformers the corresponding rated currents are also shown. A= 120 kVA Ep on the pri- mary of a transformer, with the secondary open-circuited. creases As the voltage rises, the mutual flux in direct Om in- proportion, in accordance with Eq. 9.2. Exciting current / u will therefore increase but, when the iron begins to saturate, the magnetizing current /m has to increase very steeply to produce /Q , of a transformer 60 Let us gradually increase the voltage voltamperes (MVA), depending on the size of the rise using this far lower 10.10 No-load saturation curve the required flux. If The temperature cur- nominal value, other- is peres (VA), in kilovoltamperes (kVA) or in mega- transformer. However, the load its maximum power output the of the transformer winding (primary or secondary). exceed and and wise the windings will overheat. Consequently, both the applied voltage and the current drawn by the load. These nominal current A 521 . the iron losses will be lower than normal Consequently, keep the transformer temperature is n the core will be cooler. to = 60 A s other hand, the heat dissipated in the windings de- able level, winding is S — 250 X 1000 — — — £ 480 £ns pends upon the current they carry. V 480 n.. nominal If the nominal S — / winding 250 X 1000 Nominal current of the reach V 4160 Sn ~ £p nominal in the we pass we draw a graph of Ep versus see the dramatic increase in current as the normal operating point Transformers are usually designed peak flux density of about 1 .5 T, (Fig. we 10.14). to operate at a which corresponds roughly to the knee of the saturation curve. Thus, when nominal voltage is applied to a transformer, the corresponding flux density is about 1.5 T. We can exceed the nominal voltage by perhaps 10 percent, but if we were to apply twice the nominal PRA CTICA L TRA NS FORMERS 207 kV 20 18 16 - no 'mat oper atincjpoi it 14 En 12 10 — nc>min al cu rrem 8 6 4 2 0 0 0,5 6 5 4 1 A exciting current /Q Figure 10.14 No-load saturation curve of a 167 kVA, 14.4 kV/480 V, 60 Hz transformer. Figure 10.15 Single-phase dry-type transformer, type AA, rated voltage, the exciting current could become even greater than the nominal full-load current. The nonlinear shows Xm relationship between that the exciting branch in Fig. 10. la) is effect, although Rm not as constant as is Ep and /0 (composed of R m and it Xm de- creases rapidly with increasing saturation. However, and so R m and X m terials inside close to rated voltage, remain essentially constant. 10.11 Cooling To prevent rapid at methods deterioration of the insulating ma- kVA air. The metallic housing is fitted the can be di- with ventilating may flow over windings and around the core (Fig. it away in a to the tank, dissipated by radiation and convection to Oil 1 consequently, is; 0. 1 6). is a much it is better in- invariably used on high-voltage transformers. As the power 1 0. 15). filled tank (Fig. rating increases, external radiators 10.17). Oil circulates around the transformer windings and moves through the radiators, air. where For the heat still is again released to surrounding higher ratings, cooling fans blow over the radiators (Fig. For transformers may be Such dry-type transformers are used inside build- oil hostile atmospheres. is are and enclosed oil sulator than air Larger transformers can be built the same way, but away from mineral in the outside air (Fig. forced circulation of clean air must be provided. ings, for are added to increase the cooling surface of the oil- by the natural flow of the surrounding louvres so that convection currents immersed steel tank. Oil carries the heat a transformer, adequate cooling of the Indoor transformers below 200 200 kVA Distribution transformers below usually where windings and core must be provided. rectly cooled V, appears. In reasonably constant, most transformers operate at 60 Hz, insulation class 150°C indoor use. Height: 600 mm; width: 434 mm; depth: 230 mm; weight: 79.5 kg. (Courtesy of Hammond) 15 kVA, 600 V/240 10. in the 1 air 8). megawatt range, cooling effected by an oil-water heat exchanger. Hot drawn from the transformer tank heat exchanger where it is pumped to a flows through pipes that are ELECTRICA L MA CHINES AND TRANSFORMERS 208 Figure 10.16 Two single-phase transformers, type OA, rated 75 60 Hz, 55°C temperature rise, pedance 4.2%. The small radiators at the side inkVA, 14.4 kV/240 V, im- Figure 10.17 crease the effective cooling area. Three-phase, type OA grounding transformer, rated 1900 kVA, 26.4 kV, 60 Hz. The power of this transformer is 25 times greater than that of the transformers in contact with cool water. Such a heat exchanger is very effective, but also very costly, because the water itself used. Thus, a transformer 18 000/24 000/32 . is still self-cooled. Note, the transformer much room as itself. method of cooling may have a triple rating of 000 kVA depending on whether by the natural circulation of it The type of transformer cooling is designated by the following symbols: air (AO) ( 1 8 AA-dry-type, self-cooled 000 AFA-dry-type, forced-air cooled or OA-oil-immersed, self-cooled by forced-air cooling with fans (FA) (24 000 kVA) 3. it cooled kVA) 2. 10.16, but big transformers are designed to have a multiple rating, depending on the 1 in Fig. has to be continuously cooled and recirculated. Some is shown however, that the radiators occupy as OA/FA-oi I- immersed, or by the forced circulation of forced-air cooling (FOA) oil (32 air accompanied by 000 kVA). self-cooled/forced- cooled A O/FA/FOA-oil -immersed, self-cooled/forced- air cooled/forced-air, forced-oil These elaborate cooling systems are nevertheless cooled economical because they enable a much bigger output from a transformer of a given size and weight The temperature (Fig. 10.19). transformers is rise by resistance of oil-immersed either 55°C or 65° C. The tempera- PRA CTICA L TRA NS FORMERS 209 Figure 10.19 Three-phase, type OA/FA/FOA transformer rated 36/48/60 MVA, 225 kV/26.4 kV, 60 Hz, impedance Figure 10.18 7.4%. The circular tank enables the Three-phase, type FOA, transformer rated 1300 MVA, the temperature rises 24.5 kV/345 kV, 60 Hz 65°C temperature oil in ance: 1 1 imped- contact with air. Other nuclear power generating station, The is one forced-oil circulating below the cooling (Courtesy of Westinghouse) just of the largest weight of core and turc must be kept low By fans. weight of to preserve the quality coils: 37.7 t weight of tank and accessories: 28.6 pumps can coil (44.8 Total weight: 104.5 oil. details: .5%. This step-up transformer, installed at a units ever built. be seen rise, oil to expand as and reduces the surface of the m3 ): 38.2 t t t of the contrast, the temperature rise of a dry-type transformer may be as high as 1 80°C, depending on the type of insulation used. 10.12 Simplifying the equivalent circuit N, The complete equivalent circuit of the transformer as shown in Fig. 0.8 gives far more detail than is needed in most practical problems. Consequently, 1 let us try to simplify the circuit former operates I . l ) at At no-load (Fig. 10.20) because T is when no-load and 2) 1 2 is /n flows a transformer at no-load. the trans- drop across them zero and so These impedances are so small circuit of at full-load. current in is /, an ideal transformer. Consequently, only the exciting current Figure 10.20 Complete equivalent N2 in R and } Xn . that the voltage is negligible. Furthermore, the R 2 and X [2 is zero. We can, therefore, neglect these four impedances, giving us the much simpler tio, a = N /N2 ] circuit of Fig. 10.2 , is 1 obviously equal . The turns ra- to the ratio of ELECTRICAL MACHINES AND TRANSFORMERS 210 primary the to secondary voltages Ep/E meas sured across the terminals. 2. hi full-load I p Consequently, least is at we can 20 times larger than sponding magnetizing branch. The resulting shown cuit is cuit may /() a£ s in Fig. 10.22. This simplified when be used even the load is a cir- 2 Z J cir- only 10 percent of the rated capacity of the transformer. We 1 . neglect /0 and the corre- Figure 10.23 Equivalent circuit with impedances shifted to the can further simplify the circuit by shift- mary pri- side. ing everything to the primary side, thus elimi- nating transformer T (Fig. nique was explained in 10.23). This tech- Section 9.10. Then, by summing the respective resistances and reacwe obtain the circuit of Fig. 10.24. In tances, this circuit Rp = fl t + crR 2 Xn = X n + a 2 (10.9) Xr (10.10) . where Rp = total transformer resistance referred to the total of a large transformer is mainly reactive. primary side Xp = Figure 10.24 internal impedance The transformer leakage reactance referred to the primary side The combination of R p and X constitutes the totransformer impedance Z referred to the prip mary side. From Eq. 2. 2 we have tal 1 \Rcn VR- + X; L, Impedance Z p a:1 N, N2 when one of the important parameters of is the transformer. It produces an internal voltage drop the transformer is loaded. Consequently, Zp affects the voltage regulation of the transformer. Figure 10.21 Transformers above 500 Simplified circuit at no-load. Xp reactance Rp *f2 *2 as . In such transformers voltages tance Xp 10.25). a transformer at full-load. we can neglect Rp , is as far and currents are concerned.* The is thus reduced to a simple reac(Fig. quite remarkable that the relatively circuit of Fig. simple reactance Figure 10.22 possess a leakage between the source and the load It complex kVA that is at least five times greater than equivalent circuit Simplified equivalent circuit of (I0.ll) 10.8 can be reduced to a in series with the load. PRACTICAL TRANSFORMERS Solution a. Rated primary current /np = 5 n /Enp = 3 000 000/69 000 = 43.5 A Rated secondary current Ais Figure 10.25 The internal impedance b. of a large transformer = SJEns = 3 000 000/4160 = 721 A Because the transformer exceeds 500 kVA, the windings have negligible resistance compared to is their leakage reactance; mainly reactive. we can therefore write Zp = X p = 127(1 Referring to Fig. 10.26a, the approximate imped- ance of the 2000 10.13 Voltage regulation An important attribute of a transformer is its Z = E~IP = 4160 = 8.65 il volt- age regulation. With the primary impressed voltage held constant at its rated value, the tion, in percent, is voltage regula- Load impedance defined by the equation: a voltage regulation = — E l — E - X 100 2 Z= / p 2 secondary voltage X (69/4. 16) /2 secondary side 000 000 8.65 = 2380 12 we have 28.95 A no-load [V] at a£ at full-load s [V] 2 (a Z) / p = 2380 X 28.95 = 68 902 V The voltage regulation depends upon the power power factor factor of the load. Consequently, the must be specified. load voltage If the load may exceed is capacitive, the no- the full-load voltage, in which case the voltage regulation E = s negative. at 3000 kVA, 69 X Because the primary voltage it = (4.16/69) is 4154 held constant follows that the secondary voltage 4160 is 68 902 at at single-phase transformer rated kV/4. 6 kV, 60 1 of 127 (1, Hz has a total internal impedance V. x9 = 127 n Ij_ Zp referred to the primary side. \ kV Calculate a. b. I The rated primary and secondary currents The voltage regulation from no-load to fullload for a 2000 kW resistive load, knowing that the primary supply voltage is fixed . a = t-t^ = at 4.16 69 kV c. The primary and secondary ondary is currents if accidentally short-circuited. the sec- Figure 10.26a See Example 10-7. 16.58 V 69 kV, no-load Example 10-5 A is = 69 000/V127 2 + 2380 2 = secondary voltage 2 referred to primary side: Referring to Fig. 10.26b (10.12) where £ NL = £ hl = kW load on the 1 z T is ELECTRICAL MACHINES AND TRANSFORMERS 2 2 1 Voltage regulation is; X voltage regulation 4160 ^ 100 41 54 X (10.12) 100 4154 0.14% The voltage regulation c. is excellent, Referring again to Fig. 10.26b, is accidentally short-circuited, / p = E p/Xp = = 543 A if the secondary aEs = 0 and so Figure 10.27 Open-circuit test and determination of 69 000/127 The corresponding current /s turns Rm Xm , , and ratio. on the secondary and 10.24 by means of an open-circuit and a side: short- circuit test. = /s a/ p = (69/4.16) X 543 During the open-circuit = 9006 A test, rated voltage and current plied to the primary winding /0 , is ap- voltage E p and active power P m are measured (Fig. 10.27). The secondary open-circuit voltage E s is also mea, These sured. test results give us the following in- formation: active power absorbed by core = apparent power absorbed by core power absorbed by core reactive where Figure 10.26b See Example short-circuit currents in both the primary secondary windings are rated values. 1 2 The I R 1 Rm R m = Ep losses are, therefore, 12.5 2 or imme- diately to prevent overheating. Very powerful elec- Turns a ratio (10.1) X m = Ep 2 /Q m (10.2) Xm is a = NJNo 56 times greater than normal and, unless the wind- may be During the short-circuit is or torn apart. is For a given transformer, values of Xm , /? , less than 5 percent much lower / sc should be less than its than of rated voltage) The primary nominal value to prevent overheating and, particularly, to prevent a we can determine m R p and the secondary winding applied to the primary (Fig. 10.28). current impedances test, Ep/E, short-circuited and a voltage E„ normal (usually 10.14 Measuring transformer is is tromagnetic forces are also set up. They, too, are ings are firmly braced and supported, they 2 /P m Magnetizing reactance 56 times greater than normal. The circuit-breaker damaged P~m corresponding to the core loss and 2.5 times greater than the or fuse protecting the transformer must open 1 - = S m = Ep Ia = Qm 10-7. Resistance The Qm = f* in X p shown the actual in Figs. 10.21 rapid change in winding resistance while the test being made. is PRACTICAL TRANSFORMERS The voltage E sc current / sc , measured on the primary side , and power (Fig. 10.28) P sc are Transformer impedance referred to the primary and the Zp = £ sc //sc = - 650 (2 following calculations made: Total transformer mary side impedance referred to the pri- is RJtJ = Rp 1 is 2600/4 Resistance referred to the primary (10.13) is 2400/16 = ison Total transformer resistance referred to the primary side 2 is Leakage reactance referred to the primary is (10.14) primary side V650 2 - 150 2 Xp Total transformer leakage reactance referred to the - 632 a is VZ; Example 10-6 (10.11) __ During a short-circuit test on a transformer rated 500 kVA, 69 kV/4. 6 kV, 60 Hz, the following 1 age, current, and Terminals X,, 2600 V £\ ( volt- power measurements were made. were in short-circuit (see Fig, X2 Figure 10.29 See Example 10-6. 10.28): 2600 V 4A 2400 Example 10-7 W An Calculate the value of the reactance and resistance of the transformer, referred to the HV side. Solution shop, a 69 Referring to the equivalent circuit of the trans- former under short-circuit conditions (Fig. 10.29), we was conducted on the transExample 10-6. The following results were obtained when the low-voltage winding was excited. (In some cases, such as in a repair open-circuit test former given in kV voltage may not be available and the open-circuit test has to be done by exciting the winding LV at its rated voltage.) find the following values: E = 4160 V /n K Using this acteristics a. = 2A 5000 W information and the transformer char- found in Example the values of X m and Rm 10-6, calculate: on the primary side (Fig. 10.21) b. the efficiency of the transformer plies a load of 80 % when 250 kVA, whose power it (lagging). Solution a. Applying Eq. 10.1 to the secondary side: Figure 10.28 Short-circuit test to winding resistance. determine leakage reactance and = 4160 2 /5000 = 3461 (1 sup- factor is ELECTRICAL MACHINES AND TRANSFORMERS 214 The apparent power S m = E 5m s = 4160 X I {) 2 and R p Let us assume = 8320 VA load is 4 60 6650 The load current 11 /? Xm m and 000/4 160) 2 11 referred the primary side = 275 times greater. The Rm = X 2602 275 275 X = II = 3461 fl 715 X 10 952 X 10 3 turns ratio = 715 kH (1 - 952 k(l the time. Thus, when we = with cos 6 is about 250 about that loads and voltages fluctuate b. Industrial 0.8, it all state that a load is is kVA and power The total swer, even in if iron loss is = 3.62 X is 150 1966 same the rated voltage on the is LV as that measured at side of the transformer. Total losses are certain it. = 6966 The circuit of the transformer and 5000 bosses Knowing this, assumptions that make it much able to give active + 1966 W = 7 kW power delivered by the transformer its add the The values of R p known, and so we only have to magnetizing branch. To simplify the calcu- lations, we is Rp 2 is calculating efficiency, there we were The equivalent Xp Il = The is easier to arrive at a solution. load A 3.62 arriving at a precise mathematical an- we can make and - P iron = 5000W Consequently, in 60/16.59 2 = about 69 kV. no point = is: copper loss (primary and secondary) 250 kVA, factor the primary side /2 /a Copper the Furthermore, the primary voltage 0.8. = 16.59 if understood that the load that the is The current on /, would have been found primary had been excited at 69 kV. These are the values 000/4160 = 69W/4I60 V - a (1 3 is The values on the primary side are therefore: Xm = calculate the efficiency of = SIE, = 250 = 60 A 12 = 4160 2 /6650 = 2602 will be (69 Xp the transformer. = V8320 2 - 5000 2 The values of We now V. 1 greater than that the voltage across the . ^s^pI Qm = much cause these impedances are is: to the load is represented by Fig. 10.30. PG = S cosG = 250 X are already shifted input terminals 1 1 1, 50f2 X m and R m from points 3, 4 to the 2. 632 This change a is /, justified be- = 200 The active power received by 60 A Note efficiency that in is 7 therefore = PJP - 200/207 = 0.966 or 96.6% X making the calculations, sider the active power. transformer and 10-7. is 3 ti Figure 10.30 the transformer = + bosses = 200 + = 207 kW The See Example 0.8 kW its ciency calculations. The reactive we only con- power of the load does not enter into effi- PRACTICAL TRANSFORMERS The 10.15 Introducing the per-unit best approach calculate is often encountered when dealing with transformers and other electrical machines. reason is relative rents employ to The that per-unit values give us a feel for the its ohmic value and use For example, former listed in _ E ns _ " L ~ and kilowatts, we simply work volts we don't have to carry when per-unit values are used. The per-unit method as applied to transformers V 8 A 28 ' 12.012 Using value of the secondary resistance who are easy to understand. However, readers useful to read Sections 1.9 to 1.11 in Chapter 1 tual values R of { , at Table be- 1 OA which , , , transformers ranging from scanning through the table, 1 kVA we is 400 to ances vary from 505 000 17 to 0.0003 MVA. In 12, a range over all that the various voltages, currents and impedances are expressed in actual values using volts, amperes and ohms. ACTUAL TRANSFORMER VALUES /„s *i Ri *n Q Q n n 4.17 / np Using this load V A 2400 , impedance 576 as a reference, the rela- value of the primary resistance R\ /?,(pu) is 5.16(2 = 0.0090 576 The l 1000 100 10 12 and R 2 (pu) are pure relative values R\(p\i) numbers because they are the ratio of two quantities that bear the same unit. Circuit elements on the primary side are always compared with the nominal load impedance Z np on 400000 2400 2400 12470 69000 13800 460 347 600 6900 424000 0.417 4.17 8.02 14.5 29000 167 145 943 .6 27.2 0.0003 17 28.8 58.0 5.16 1.9 0.095 0.024 0.25 0.354 32 4.3 39 151 0.028 0.09 0.09 1.5 27 2. 1. 16 1 1 a Q 200000 29000 150000 505000 460 Rm 400000 51000 220000 432000 317 A, A 0.0134 0.0952 0.101 0.210 52.9 R\,R 2 X n X l2 X tn and R m in ohms, we could express them relative to another ohmic value. The question is: what value should we , , on the secondary side are compared with the nominal load impedance Z ns on the secondary Proceeding of the 10 kVA values X, |(pu), The relative in this side. for the other impedances we transformer, obtain the relative m (pu), etc. displayed in Table 10B. impedances of the other transformers /? are calculated the tive way same way. In nominal load impedances as the reference impedances. each case, the respec- Z np and Z ns are chosen Using the rated voltage and power of the transformer, they are given E _np Instead of expressing pri- 12 the primary side. Similarly, circuit elements kVA /np E no recognizable pattern to the values; they are impedance on the is: in is TABLE 10A 12.017 ~ np tive is 0.0079 (pu) of five excess of a billion to one. Furthermore, there is side R2 0.095 12 2 re- see that the imped- as a reference, the relative Similarly, the nominal load displays the ac- It R 2 Xn X (2 X m and R m map. The reason R mary Let us begin by looking ohmic value it fore proceeding further. produced here for convenience. this not yet familiar with per-unit calculations will find V V A A trans- 347 with numbers. Consequently, Ens can is along units £„p kVA case of the 10 in the We as a reference. Table 10A, the nominal load im- pedance on the secondary side and powers. Thus, instead of dealing with ohms, amperes, the it magnitudes of impedances, voltages, curZ,ls is the nominal load (voltage and current) of the transformer. method Per-unit notation is 215 *«P. - E by: 1 (HU5a) sa , En, (10.15/;) choose as a basis of comparison? s tt ELECTRICAL MA CHINES AND TRANSFORMERS 2 6 1 symbols #|(pu), 7\? 2 (pu), R2 X values of/?,, In practice, the relative < tl , etc., and are designated by the are called per-unit values X n (pu), The etc. quantities Example 10.8^ A transformer rated 250 kVA, 4 60 V/480 V, 60 Hz 1 has an impedance of 5.1 %. Calculate: used as references are called base quantities. Thus, Z np Z ns S n E , a. £ ns . , , / / ns listed in , Table 10B are impedance on the base base quantities. all examining Table OB. the reader I will note that former referred for a given transformer, the values of X, ,(pu) and X r2 (pu) kVA V V A 2400 100 1000 400000 2400 12470 69000 13800 460 347 600 6900 424000 0.417 4.17 8.02 14.5 29000 17 28.8 167 145 943 5760 576 1555. 4761 0.4761 211.6 12.0 3.60 47.61 449.4 R\ (pu) 0.0101 0.0090 0.0075 0.0057 0.00071 /? 2 (PU) 0.0090 0.0079 0.0067 0.0053 0.00079 X X rj X ni (pu) 0.0056 0.0075 0.0251 0.0317 0.0588 (pu) 0.0055 0.0075 0.0250 0.0315 0.0601 (pu) 34.7 50.3 96.5 106 966 /? ni Znp = E p - 69 2 /S n = 4160 2. 69.4 (pu) 0.032 A,<P») 88.5 141.5 0.023 0.013 90.7 0.015 666 0.0018 2 is /250 000 tt Base impedance on the secondary side 2 n Q zns Base impedance on the primary side Table 10A. in 10 I A A,s z„ P ri of the trans- Solution a. PER UNIT TRANSFORMER VALUES TABLE 10B Aip Zp primary side to the AVpu) and are nearly the same. This pat- does not show up tern of similarity £„» and sec- are nearly the same. Similarly, the values of /? 2 (pu) t„p impedance the total internal b. In the primary ondary side Zns = E /S n = 480 = 0.92 n 2 s b. The actual value of Zp = Zp % X Z np = 5.1 is /250 000 on the primary side 0.051 X 69 (1 = is: 3.52 11 10,17 Typical per-unit impedances We have seen ative that we can get a better idea of the rel- magnitude of the winding resistance, leakage actance, etc., re- of a transformer by comparing these im- pedances with the base impedance of the transformer. There is even a similarity between the per-unit whose values of transformers ferent. ratings are quite dif- For example, the R\(pu) of the former (0.0101 ) is as the /?,(pu) of the I kVA trans- of the same order of magnitude 1000 kVA transformer despite the fact that the latter is (0.0057) 1000 times more powerful and the voltages are vastly different. Clearly, the per-unit method offers insights that In making the comparison, circuit elements located on the primary side are compared with the primary base impedance. Similarly, circuit elements on the secondary side are compared with the secondary base impedance. The comparison can be made either on a percentage or on a per-unit basis; ter. kVA transformers ranging from 3 example, the table shows would otherwise not be evident. we shall use the lat- Typical per-unit values are listed in Table IOC for to 100 MVA. For that the per-unit resistance of the primary winding of a transformer ranges from 10,16 Impedance of a transformer The total internal impedance Zp of a transformer was defined in Section 10. 12 and highlighted in Fig. 10.24. In power and distribution transformers its value However, is it always indicated on the nameplate. is expressed as a percent of the nominal load impedance. Thus, 3.6 %. the per if the nameplate unit value of Zp is 0.036. is marked 0.009 to and 100 0.002 for all MVA. Over the per-unit resistance power this 7?, ratings between 3 kVA tremendous power range, of the primary or secondary windings varies only from 0.009 to 0.002 of the base impedance of the transformer. Knowing the base im- pedance of either the primary or the secondary winding, we can readily estimate the order of magnitude of the real values of the transformer impedances. Table IOC is, therefore, a useful source of information. PRACTICAL TRANSFORMERS TABLE 10C 217 TYPICAL PER-UNIT VALUES OF TRANSFORMERS Figure 10.31 Equivalent circuit of a transformer. Typical per-unit values Circuit element (see Fig. /?, 0.3 1 3 1 orR 2 . X n orX n kVA to 250 kVA l MVAio I00MVA 0.009-0.005 0.005-0.002 0.008-0.025 0.03-0.06 20-30 50-200 20-50 100-500 0.05-0.03 0.02-0.005 Example 10-9 Using the information given the approximate 0.35 SI 1.7 SI Table 10C, calculate in 23 ma 4.6 mSl impedances of a real values of the 250 kVA, 4160 V/480 V, 60 Hz distribution trans- former. Solution We first determine the base impedances on the mary and secondary Example 10-8, we From side. the results pri- of Figure 10.32 have See Example Z np = Z lls = 69 0.92 n This example shows the usefulness of the per- We now calculate the real impedances by multiplying Znp and Z ns by the per-unit values given in Table IOC. This yields the following results: = 0.005 X 69 R2 = Xn = X r_ = Xm = Rm = 0.005 X 0.025 X 69 0.025 X /?, 30 50 (2 0.92 tt 10-9. tt = 0.35 12 H= = 0.92 (2 1 = 4.6 mil alent circuit of the range of transformers. .7 12 23 m() X 69 fl = 2070 12 = X 69 n - 3450 (2 - method of estimating impedances. The equiv250 kVA transformer is shown in Fig. 10.32. The true values may be 20 to 50 percent higher or lower than those shown in the figure. The reason is that the per-unit values given in Table 10C are broad estimates covering a wide unit 2 kil 3.5 kI2 Example 10.10 The 500 kVA, 69 kV/4160 V, 60 Hz transformer shown in Fig. 10.30 has a resistance R of 150 (2 ELECTRICAL MACHINES AND TRANSFORMERS 218 and a leakage reactance unit a. X p of 632 f2. Using the per- G(pu) method, calculate: = V0.5 2 when the voltage regulation kVA tween zero and 250 = VS 2 (pu) - P 2 (pu) the load varies be- = The 80% b. the actual voltage across the c. the actual line current I 250 kVA R The Solution clear that the pres- is it R p and Xp voltage drop across magnetizing branch does not affect the voltage We now draw regulation. in To determine fer all voltages, HV (69 kV) terminals 1 is We assume the voltage between 69 kV, and that it remains fixed. The base power P H The base voltage E is is ti kVA 69 kV 500 Consequently, the base current 'b = Pb/Ek = 300 = 7.25 A and the base impedance Z B = E B /I H = 69 The will re- impedances, and currents to the side. 2 , we the voltage regulation, per-unit value of Rp Figure Note terminals per-unit value of Xp (pu) The Xp it 4 of the circuit 250 = 4 The Z ]2 (pu) = primary Figure 10.30. The per-unit im- is 2/136.87° 1.2 0.0158 + 1.6 + j(1.2 + 1, 0.0664 0.0664) is = 69 000/69 kV = A-p(pu) flp(pu) 1.0 j 0.0664 power absorbed = 250 kVA/500 kVA = A'.(pu) 0.5 j per-unit value of the active the load 3.33 power absorbed by is P(pu) = 5(pu) cos 9 = 0.5 X 0.8 - 0.4 Figure 10.33 The per-unit value of the reactive by the load is is 5(pu) The 2 = 1.616 + j 1.266 = 2.053Z 38.07° is E i2 in impedance between terminals per-unit is = not X j 3.33 + j 3.33 = 1.6+j per-unit value of the apparent by the load 3, 2.50 Z34 (pu) 0.0158 The shown equivalent circuit diagram.) is 632/9517 is does not enter into the calcula- is ft shown (These terminals are not accessible; they exist only in the = 9517 3.333 The magnetizing branch pedance between terminals 000/7.25 = Q is 0.3 that the load appears across the 3, 000/69 000 per-unit value of voltage E, 2 (pu) Uf_ the equivalent per-unit circuit R p (pu) = 150/9517 = 0.0158 The corresponding to v " 0(pu) 10.33. shown because tions. 0.4 £2 (pu) _ X L (pu) Consequently, the , X 2 2.50 ~ P(pu) P is corresponding to 1.0 ) (pu) v per-unit load reactance ence of the magnetizing branch does not affect the /?, £ 2 (pu _ . Fig. 10.30, 0.3 per-unit load resistance load { examining 2 lagging power factor at a of In 0.4 is power absorbed Per-unit equivalent circuit of a feeding a 250 kVA load. 500 kVA transformer PRACTICAL TRANSFORMERS The per-unit current I 219 ensure proper load-sharing between the two trans- is { formers, they must possess the following: /i(pu) .0 - Z 12 (pu) L 2.053 a. 38.07° b. The same primary and secondary voltages The same per-unit impedance = 0.4872^-38.07° Particular attention must be paid to the polarity The per-unit voltage £34 (pu) = /j(pu) £34 across the load of each transformer, so that only terminals having is 10.34). - (0.4872/1-38.07°) (2^36.87°) = same the X Z34 (pu) circuit as 0.9744/1 -1.20° polarity An soon as the transformers are excited. In order to calculate the currents flowing in each transformer The per-unit voltage regulation E 34 (pu) at £34 (pu) is must — £34 (pu) no-load at Consider 0.0263 is former referred therefore 2.63 is first mary voltage 0.9744 The voltage regulation when we they are connected in parallel, determine the equivalent circuit of the the equivalent circuit when a single transformer feeds a load Z, (Fig. 10.35a). The 1^ -j0.9744_ a. first system. full-load full-load at connected together (Fig. are error in polarity produces a severe short- £p pri- and the impedance of the trans- to the primary side is Zpl . If the ratio %. 2 We can now calculate the actual values of the volt- O- age and current as follows: Voltage across terminals 4 3, A is O- £ 34 = EM (pu) X £ B = 0.9744 X 69 000 = b. = £34 X 67.23 is B (4160/69 000) X 10 3 X >H 2 Actual line current /, Figure 10.34 Connecting transformers a .. in parallel to share a load. is = /,(pu) = 0.4872 = 3.53 X X /B 7.246 A 10.18 Transformers When X 1# 0.0603 = 4054 V c. * 2# . 2 kV 67.23 Actual voltage across the load £ S6 = *H 1 in parallel growing load eventually exceeds the power rating of an installed transformer, connect a second transformer we sometimes in parallel with it. To Figure 10.35a Equivalent circuit of a transformer feeding a load ZL . ELECTRICAL MACHINES AND TRANSFORMERS 220 of transformation that shown in Fig. is 1 a, the circuit can be simplified to 0.35b, a procedure we are already portional to the respective we want kVA ratings. Consequently, to fulfill the following condition: familiar with. If is a second transformer having an impedance connected circuit the in parallel becomes with the shown that first, spectively /, and /2 . (I0.18) 2 in Fig. 10.35c. In effect, impedances of the transformers are The primary currents Z the equivalent in the From Eqs. and 10.18 10. 17 can readily be it in parallel. proved transformers are re- Because the voltage drop £, 3 is the same, we can write across the impedances that the desired condition is met if the trans- formers have the same per-unit impedances. The following example shows what happens when the per-unit impedances are different. (10.16) Example 10-11 that is. A 1 00 kVA transformer an existing 250 (10.17) The of the primary currents is connected in parallel with supply a load of 330 kVA. The transformers are rated 7200 V/240 but the ratio is kVA transformer to 1 V, kVA unit has an impedance of 4 percent 250 kVA transformer has an impedance of 00 therefore de- while the termined by the magnitude of the respective primary impedances —and not by the ratings of the two 6 percent (Fig. 10.36a). trans- formers. But in order that the temperature rise be the Calculate same a. for both transformers, the currents must be pro- The nominal primary current of each trans- former c. The impedance of the load referred to the primary side The impedance of each transformer referred to d. The b. the primary side actual primary current in each transformer Solution a. Nominal primary current of former the 250 kVA trans- is Figure 10.35b Equivalent circuit with all impedances referred to the = 250 000/7200 34.7 A primary side. X t H, - 1:250 kVA -d H 7 1 > 240 j: V z a Z y , <\ H, X, 100 kVA: Z p2 d :H 2 = 4% X2 Figure 10.35c Equivalent circuit of two transformers a load Zv All impedances in parallel feeding referred to the primary side. load ;= X 2 b- : 7200 6% ::|Zpr Figure 10.36a Actual transformer connections. b V 330 kVA PRACTICAL TRANSFORMERS Nominal primary current of the 100 kVA former = and A = 46 - side, is im- 0.36b, we justified be- 46 find that the Load impedance referred primary side to the is = A 28.8 17.2 which is A seriously overloaded is 25 percent above unit is its A 28.8 20.7) carries a primary current of it The 250 kVA cause the transformers are fairly big. + 20.7/(12.4 The 100 kVA transformer because are considered to be enis = 46 X /, that transformer This assumption tirely reactive. A primary to the Note Zp2 13.9 of the two transformers circuit in Fig. 10.35c. Zpl = 100 000/7200 and the load, referred pedances 1 1 load current divides in the following way: The equivalent given Referring to Fig. d. is / n2 b. trans- 22 17.2 A, rated value of 13.9 A. not overloaded because carries a current of 28.8 A versus its it only rated value of 34.7 A. Clearly, the two transformers are not carry2 2 = E p /S| olld = 7200 = 157 n The approximate 2 /330 000 ing their proportionate share of the load. The 100 kVA transformer of load current its c. = JE p S hr = 330 000/7200 = 46 A The base impedance of the kVA 207 unit its Transformer impedance referred tends to carry proportionate share of the load. If the is percent impedances were equal, the load would be shared between the transformers il power their respective to the in proportion to ratings. primary is Zpl = X 207 = 0.06 Base impedance of the 100 Questions and Problems 12.4(1 kVA Practical level unit is 1 Znp2 = 7200 2 /10() 000 = 518 Transformer impedance referred side to the transformer (6 percent). A low-impedance transformer always more than 250 kVA Z npt = 7200 2 /250 000 - side overloaded because is impedance of the 250 A. is low impedance (4 percent), compared 0- 1 12 to the Name the principal parts of a transformer. primary 10-2 Explain how a voltage is induced in the secondary winding of a transformer. is Z p2 = 0.04 X 518 = 10-3 20.7 il The secondary winding of a transformer has twice as many turns as the primary. Is the secondary voltage higher or lower than the primary voltage? /„, /, = 34.7 A = 28.8 10-4 A Which winding is connected to the load: the primary or secondary? ma a z pi = Z pl = 20.7 46 A 10-5 State the voltage and current relationships between the primary and secondary wind- 7200 V n 157 I 2 = 17.2 /n , A 7 = 13.9 A Figure 10.36b Equivalent circuit. transformer is Calculations show seriously overloaded. that the ings of a transformer under load. The primary and secondary windings have /V| a 100 kVA and N2 turns, respectively. 10-6 Name 10-7 What purpose does the no-load current of a transformer serve ? the losses produced in a transformer. 222 10-8 ELECTRICAL MACHINES AND TRANSFORMERS Name in three conditions that must be met order to connect two transformers 10-15 What the purpose of taps is on a trans- Name 10-16 three methods used 10-17 The primary of a transformer is connected to a 600 V, 60 Hz source. If the primary has 1200 turns and the secondary has 240, the peak flux a 60 in Hz ac supply voltage 10-18 The transformer by a 20 1 V, fixed. is 10.37 in Fig. is excited 60 Hz source and draws a no- load current / 0 of 3 A. The primary and secondary windings respectively possess low-voltage winding 200 and 600 turns. If mary linked by the secondary, cal- V is excited by a source, calculate the voltage across the A 6.9 kV HV winding. transmission line a transformer having is a. connected to b. 1500 turns on the c. If d. the load across the secondary has an im- pedance of 5 H, calculate the following: a. The secondary voltage b. The primary and secondary currents The primary of a transformer has twice as many turns as the secondary. The primary voltage is 220 V and a 5 17 load is con- as the primary Figure 10.37 See Problem 10-18. is 40 percent of the pri- the transformer, as well and secondary currents. The voltage indicated by the voltmeter The peak value of flux 4> The peak value of <t> m Draw the phasor / tr 4> m 10-19 In Fig. 1 , and 0.38, diagram showing when 600 V is applied to V is measured H2 across terminals X,, a. What and is E h £2 the voltage , 80 X2 . between terminals H, X2 ? b. If terminals H,, X, are connected together, calculate the voltage across terminals c. , <£, terminals H] and nected across the secondary. Calculate the power delivered by tlux culate the following: primary and 24 turns on the secondary. 10-14 why Explain in the core. The windings of a transformer respectively have 300 and 7500 turns. If the 2400 10-13 Problem 10-11, calculate the peak transformer remains fixed as long as the calculate the secondary voltage. 10-12 In value of the tlux to cool trans- formers. 10-11 nominal Intermediate level former? 10-10 a ratio of to 2.4 kV. Calculate the current of each winding. parallel. 10-9 A 3000 kVA transformer has 60 kV in Does H2 X2 , the transformer have additive or sub- tractive polarity? . PR A CTICA L TRA NS FORMERS ^ = 300 /, TV, 223 = 1200 (96 A) Figure 10.38 See Problem 10-20 10-19. what would happen a. Referring to Fig. 10.34, if we reversed terminals H, and H2 of trans- Would X2 The primary if H h H2 terminals of transformer B were and Advanced 10-27 level Referring to Fig. 10.39, calculate the peak value of flux Explain leg reversed? Explain. 10-21 wound on one is other. the operation of the transformer bank be affected X,, 10-33. and the secondary on the former B? b. Figure 10.39 See Problem why the secondary voltage of a in the supplied by a 50 is core Hz if the transformer source. practical transformer decreases with in1 creasing resistive load. The impedance of a transformer 0-28 as the coupling 10-22 10-23 What is meant by the following terms: a. Transformer impedance b. Percent impedance of a transformer The transformer in 0-24 A 2300 V the im- is R\ = 18 12 R 2 = 0.005 xn = connected to terminals 40 10-25 A 66.7 kVA, calculate the following: a. The percent impedance of the transformer The impedance [12] referred to the sec- d. The percent impedance in the f. transformer un- Calculate the losses and efficiency the transformer delivers 66.7 when MVA to a load having a power factor of 80 percent. 10-26 If the transformer shown were placed ture rise Explain. in a tank of would have to The total copper losses at full load The percent resistance and percent reac- tance of the transformer der these conditions. b. referred to the sec- ondary side e. Calculate the losses referred to ondary side percent. a. [12] c. an efficiency when it delivers full power load having a power factor of 100 of 99.3 percent to a The transformer impedance the primary side connected across the secondary MVA transformer has a s 1 b. is 12 E p = 14.4 kV (nominal) £ = 240 V (nominal) x n_ = o.oi n the transformer has a nominal rating of 75 and 4 in Fig. 10.13. Calculate the following: a. The voltage between terminals X, and X 2 b. The current in each winding, if a 12 kVA load given for the is transformer circuit of Fig. 10.22. If line The following information 10-29 pedance 1(2] referred to: a. The 60 kV primary b. The 2.4 kV secondary 1 increases reduced between the primary and secondary windings. Explain. Problem 10-15 has an impedance of 6 percent. Calculate is in Fig. oil, 10.15 the tempera- be reduced to 65°. 10-30 During a short-circuit test on a 10 66 kV/7.2 kV transformer (see the following results were obtained: Eg = 2640 V / sc - 72 A R = 9.85 kW . ( MVA, Fig. 10.28), ELECTRICAL MACHINES AND TRANSFORMERS Calculate the following: a. The Industrial application and the total resistance actance referred to the 66 b. total leakage kV primary The nominal impedance of re- 10-34 side A transformer has a rating 200 kVA, 400 V/277 14 the transformer The high-voltage windWhat is the V. ing has a resistance of 62 H. referred to the primary side c. The percent impedance of V approximate resistance of the 277 the transformer winding? 10-31 In Problem 10-30, if rated voltage are 35 the iron losses at kW, calculate the full- load efficiency of the transformer power factor of the load is if 10-35 The primary winding of the transformer Problem 10-34 the is wound with No. 1 in gauge 1 AWG wire. Calculate the approximate 85 percent. cross section (in square millimeters) of the 10-32 a. The windings of a transformer operate current density of 3.5 made of copper and ture of A/mm 2 . If at a conductors 1 operate 0-36 tempera- at a If 75°C, calculate the copper loss per aluminum windings were 0-33 If cording to Fig. 10.39, it would have very poor voltage regulation. Explain propose a method of improving why and it. kVA weighs 1 transformer rated whereas a 100 18 kg, of the same kind weighs in watts per kilogram in each case. 10-37 a transformer were actually built ac- 10 445 kg. Calculate the power output same condi- tions. 1 secondary winding. oil-filled distribution kVA transformer used, calculate the loss per kilogram under the An at kilogram. b. in the they are The transformer shown in Fig. 10. 13 has rating of 40 kVA. If 80 V is applied between terminals X, and will X 2, a what voltage appear between terminals 3 and 4? If a single load is applied between terminals what is the 3 and 4 current that can be maximum drawn? allowable Chapter 1 Special Transformers windings, each rated 11.0 Introduction connected Many transformers are designed to meet specific we industrial applications. In this chapter at 1 20 V. The windings and so the in series, total tween the lines is 240 V while that between the lines and the center tap is 20 V (Fig. ). The center tap, called neutral, is always connected to study 1 some of the special transformers that are used in dis- tribution systems, neon signs, laboratories, induction furnaces, and high-frequency applications. they are special, they 1 . 2. The ampere-turns of I . I on the high-voltage winding to the neutral terminal is usu- of the secondary windings are connected that both to ground. the transformers are under load: The voltage induced in a winding is proportional to the number of turns, quency, and the flux in the core. bonded winding so a result, the following approximations can be made when H2 Terminal ally of the standard transformers discussed in Chapter 10. As I ground. Although possess the basic properties still are voltage be- The nominal rating of these distribution trans3 mounted on poles of the fre- pany (Fig. 1 1 .2) to kVA 500 kVA. They are comsupply power to as many as 20 formers ranges from directly to the electrical utility customers. the primary are equal and The load on opposite to the ampere-turns of the secondary. distribution transformers varies greatly throughout the day, depending on customer 3. 4. The apparent power input to the transformer equal to the apparent power output. The exciting current may be in the is demand. In residential districts a peak occurs morning and another peak occurs primary winding noon. The power peaks never neglected. in the in the late after- last for more than one or two hours, with the result that during most of the 24-hour day the transformers operate far below 11.1 their Dual-voltage distribution transformer normal rating. tem, every effort Transformers that supply electric power to dential areas generally have Because thousands of such transformers are connected to the public small. This resi- two secondary is is made utility sys- keep the no-load losses achieved by using special low-loss silicon-steel in the core. 225 to ELECTRICAL MACHINES AND TRANSFORMERS 226 — I A Figure 11.3 Autotransformer having N2 turns A/-, turns on the primary and on the secondary. H2 11.2 Autotransformer Figure 11.1 a. Distribution transformer with ondary. b. Same The 1 central conductor 20 V/240 V secis distribution transformers the neutral. turns, mounted on an ing connected reconnected to give only 120 V. N Consider a single transformer winding having is iron core (Fig. 1 1 to a fixed-voltage ac .3). { The windand source the resulting exciting current /(> creates an ac flux <& m in the core. the flux is As in Suppose a tap there are any transformer, the peak value of fixed so long as E\ N2 C is is fixed (Section 9.2). taken off the winding, so that turns between terminals A and C. Because the induced voltage between these terminals is proportional to the number of turns, E2 is given by E 2 = (N2 /N0 X £, (ll.l) Clearly, this simple coil resembles a transformer having a primary voltage £, and a secondary voltage E 2 However, . A and the A are no longer isolated from the primary terminals B, secondary terminals C, common each other, because of the If we connect terminal A. a load to secondary terminals CA, the resulting current I2 immediately causes a pri- mary current /] to flow (Fig. 1 .4). The BC portion of the winding obviously carries current /,. Therefore, according to Kirchhoff s 1 CA portion carries a current (I 2 — /]). Furthermore, the mmf due to /, must be equal and opposite to the mmf produced by (7 2 — /]). As current law, the Figure 11.2 a result, we have Single-phase pole-mounted distribution transformer rated: 100 kVA, 14.4 kV/240 V/120 V, 60 Hz. (N I ] l - N2 = ) (I 2 - I,)N2 SPECIAL TRANSFORMERS 221 B O— : (N,-N 2 c ) (h -fx) E2 A -o- load Figure 11.4 Autotransformer under load. The currents flow which reduces opposite directions in 3.6 to I ] assuming N = I2 N 2 that both the transformer losses b. and exciting current are negligible, the apparent c. power drawn by the load must equal the apparent power supplied by the source. Consequently, 1 1 . 1 , 1 1 1 1 .3 CA and Solution = E 2 I2 and .2, connected across the secondary, is The secondary voltage and current The currents that How in the w inding The relative size of the conductors on windings BC (U.3) a. Equations load (ll. 2) ] £,/, kW the upper and lower windings. calculate: a. Finally, in The secondary voltage are identical to those of a standard transformer having a turns ratio N /N2 is E2 = 80% X 300 = 240 V . i However, in winding actually part of the primary winding. In effect, is this autotransformer the secondary separate secondary winding. As difference in size cal a result, autotrans- becomes power and 2. On isolation windings is a E = PIE 2 = 3600/240 = The current supplied by ] /E 2 The /, important lies = = PIE, the current in between between the primary and secondary serious drawback in Autotransformers are used to some start secondary applications. induction mo- voltage of transmission lines, and, in general, to transform voltages to is 15 A the source (Fig. 1 1.5). is the current in = A winding BC = 12 A winding CA = 15 — 3600/300 12 12 = 3 A the other hand, the absence of electri- tors, to regulate the mary output. particularly the ratio of transformation b. current and cheaper than lighter, standard transformers of equal 0.5 I2 an autotransformer eliminates the need for a formers are always smaller, when The secondary ratio Example 11-1 The autotransformer is close to in Fig. when the pri- The conductors in the secondary winding can be one-quarter the size of those ing BC because the current (see Fig. winding 1 1.5). BC is However, is in CA wind- 4 times smaller the voltage across equal to the difference be- tween the primary and secondary voltages, I 11.4 has an 80 per- cent tap and the supply voltage £, c. is 300 V. If a namely (300 - 240) = 60 V. Consequently, winding CA has four times as many turns as winding BC. Thus, the two windings require essentially the same amount of copper. ELECTRICAL MACHINES AND TRANSFORMERS 228 -12A /, B -O 2 = / G : 15A 300 V 1 3A A 240 <\ V Figure 11.5 Autotransformer of Example 11-1. 11.3 Conventional transformer 6. The voltages add when terminals of opposite X2 polarity (H, and connected as an autotransformer conventional changed into may add to, is in series. Depending made, the secondary volt- or subtract from, the primary voltage. basic operation and behavior of a transformer unaffected by a mere change in is external connections. Consequently, the following rules apply whenever a conventional transformer is connected as an auto- transformer: 1 . 2. 3. in If nominal voltage 1 1 .6 We Hz. wish to reconnect in three different age a. shown in kVA, 600 V/l 20 V, 60 ways it as an autotransformer to obtain three different volt- ratios: 600 600 20 1 to 720 V V secondary primary V primary to 480 V secondary V V primary to 480 maximum secondary load the transformer can carry in each case. rating. rated current flows in one winding, rated cur- ing (reason: 4. If The ampere-turns of the windings always equal). rated voltage exists across one winding, rated voltage automatically exists across the other winding (reason: The same mutual flux links 2 both windings). 5. If the i current in a winding flows from Hj to the current in the other X 2 to X, and 5 1 rent will automatically flow in the other wind- are H2 single-phase transformer has a rating of a Calculate the rating. The voltage across any winding should not exits Fig. c. any winding should not exceed nominal current ceed (or are connected together. ) Example 11-2 The standard b. The current its and Xj) are con- an autotransformer by connecting the upon how the connection age X2 two-winding transformer can be primary and secondary windings The H2 or when H, and X, voltages subtract A , nected together by means of a jumper. The H2 winding must flow from and vice versa. — 120 v-^o o , Figure 11.6 Standard 15 kVA, 600 V/120 V transformer. SPECIAL TRANSFORMERS 229 Solution Nominal current of the 600 V winding = SIE = /, = 000/600 15 { is Nominal current of the 120 V winding I2 a. = = S/E 2 To obtain 480 15 000/120 = A 25 is 125 A secondary voltage (120 V) V, the X2 between terminals X,, must subtract from we the primary voltage (600 V). Consequently, connect terminals having the same polarity shown gether, as in Fig. ing schematic diagram Note .7. 1 The correspond- given that the current in the 120 same the 1 is as that in the load. to- in Fig. V 1 1 winding Because this Figure 11.7 Transformer reconnected as an autotransformer to .8. give a ratio of 600 V/480 V. is wind- ing has a nominal current rating of 125 A, the maximum load can draw a S., The . A X 25 480 V= 60 kVA currents flowing in the circuit at full-load shown are 1 1 power. If in Fig. we assume from X, 25 A to 1 1 Note the following: .8. that the current X2 in the of 1 25 A flows winding, a current of H2 must flow from to H, in the other winding. All other currents are then found by applying Kirchhoff 's current law. 2. The apparent power supplied by is S b. = 100 To obtain a = 720 V. site polarity gether, as same 600 V/720 = V, the of Fig. 1 1 .7 showing voltages and secondary primary voltage: 600 to the Figure 11.8 Schematic diagram current flows. V = 60kVA 600 + Consequently, terminals of oppo- (H, and X2 in the must be connected ) in Fig. 1 1 secondary winding load 125 is is again and therefore the is again 125 A. 720 V = 90kVA load current to- .9. as that in the load, maximum maximum Sb add shown The current the A X ratio of voltage must 120 the source equal to that absorbed by the load: The now A X The previous examples show ventional transformer is that when a con- connected as an auto- Figure 11.9 Transformer reconnected to give a ratio of 600 V/720 V. ELECTRICAL MACHINES AND TRANSFORMERS 230 69 kV line primary \C c/__ 25 A distributed \^ capacitance voltmeter Oto 150 V secondary grounded Figure 11.10 Transformer reconnected to give a ratio of 120 V/480 V. Figure 11.11 on a 69 kV line. Note between the windings. Potential transformer installed transformer, it can supply a load far greater As than the rated capacity of the transformer. mentioned earlier, this is the distributed capacitance one of the advantages of using an autotransformer instead of a con- most exactly ventional transformer. However, this is The nominal secondary voltage ways example the case, as the next part of our not al- be. To obtain we the desired ratio of 120 to 480 X,X 2 is now connected same transmission 600 V winding maximum ilar to that load current cannot exceed 25 A. AX and Sc = 25 This load is less than the 480 the isolate V = 12kVA full line nominal rating (15 winding voltage on the In this regard, is HV to withstand the one terminal of the secondary always connected to ground ing these three autotransformer connections. to be isolated The temperature rise of the transformer is the same in each case, even though the loads are respectively 60 kVA, 90 kVA, and 12 kVA. The pacitance between the two windings that the currents in the windings and to eliminate when touching one of the secondary leads. Although the secondary appears from the primary, the distributed ca- visible connection between voltage ground. the secondary By grounding one of nals, the highest voltage and so the losses are the same. lines and ground The nominal usually less than insulation is is makes an in- which can produce a very high the flux in the core are identical in each case 11.4 Voltage transformers sim- side. the danger of a fatal shock is is between the primary and secondary windings must be particularly great kVA) of the standard transformer. We want to make one final remark concern- reason metering of conventional transformers. However, the insulation therefore, is, to lines (Fig. 11.11). The construction of voltage transformers is as that in the load; consequently, the The corresponding maximum load to measure or monitor the voltage on lines equipment from these in the ir- may This permits standard instruments and relays ers are used to a), to terminals (Fig. 11. 10). This time, the current usually 115 V, be used on the secondary side. Voltage transform- V, again connect H, and X, (as in solution but the source the V is respective of what the rated primary voltage shows. c. phase with the primary voltage. in winding and the secondary termi- between the secondary limited to 115 V. rating of voltage transformers 500 VA. As a result, the often far greater than the volume of Voltage transformers (also called potential trans- formers) are high-precision transformers the ratio of primary voltage to a known load."' constant, in which secondary voltage which changes very little Furthermore, the secondary voltage steel. is with is copper or al- In the case of voltage transformers and current transformers, the load is called burden. is volume of SPECIAL TRANSFORMERS Voltage transformers installed on ways measure the line-to-neutral eliminates the need for two one side of the primary is late the kV lines al- to ground. For transformer shown 11.12 has one large porcelain bushing to iso- HV line from the grounded case. The latter houses the actual transformer. The basic impulse insulation (BIL) of Current transformers 1 1 .5 This bushings because connected example, the 7000 VA, 80.5 in Fig. HV HV voltage. 23 650 kV expresses the transformer's ability to withstand Current transformers are high-precision transform- which the ers in rent a is of primary to secondary cur- ratio known constant that changes very little with the burden. The phase angle between the pri- mary and secondary current is very small, usually much less than one degree. The highly accurate current ratio and small phase angle are achieved by keeping the exciting current small. Current transformers are used to measure or lightning and switching surges. monitor the current ondary in a line and to isolate the equipment connected tering and relay me- to the sec- The primary is connected in series with shown in Fig. 1. 3. The nominal sec- side. the line, as 1 ondary current mary current 1 usually 5 A, irrespective of the pri- is rating. Because current transformers (CTs) are only used for measurement and system protection, their power rating small is and 200 VA. As —generally in the formers, the current ratio to the number of ondary windings. tio of 150 A/5 A between 15 VA case of conventional transis inversely proportional turns on the primary and sec- A current transformer having a rahas therefore 30 times more turns on the secondary than on the primary. For safety reasons current transformers must ways be used when measuring currents mission lines. The insulation between in HV the primary and secondary windings must be great enough withstand the line surges. stand is full line-to-neutral The maximum voltage the CT can line load' primary C :-j-:c / distributed capacitance Figure 11.12 7000 VA, 80.5 kV, 50/60 Hz potential transformer having an accuracy of 0.3% and a BIL of 650 kV. The primary terminal at the top of the bushing is connected to the HV line while the other is connected to ground. The secondary is composed of two 115 V windings each tapped at 66.4 V Other details: total height: 2565 mm; height of porcelain bushing: 1880 mm; oil: 250 L; weight: 740 kg. Figure 11.13 (Courtesy of Ferranti- Packard) Current transformer installed on a 69 kV secondary secondary grounded to voltage, including always shown on the nameplatc. 69 kV al- trans- - line. with- ELECTRICAL MACHINES AND TRANSFORMERS 232 the As in the case of voltage transformers (and for same reasons) one of the secondary terminals is always connected to ground. ing serves to isolate the HV 1 kV line line. from The A current large bush- the ground. The terminals that are connected in series with the The line current Hows HV current c. CT a typical installation is shown By way of comparison, former shown because it is in Fig. shown is 1 1 . 17 in Fig. 11.15 is is 50 in- a. The current in Fig. trans- The turns ratio = 400/5 N^/Nn 11,17 has a rating I2 = 80 is smaller, mainly It is imped- ratio is and The secondary current of 50 VA, 400 A/5 A, 36 kV. 60 Hz. total the transmission-line Solution insulated for only 36 kV. Example 1 1-3 The current transformer If il. The secondary current The voltage across the secondary terminals The voltage drop across the primary /,// 2 VA current much 2 280 A, calculate in Fig. 11.16. the 1. the up the bushing and out by the other terminal. The of a a. line. bushing, through the primary of the transformer, then ternal construction 3. wires on the secondary side possess a b. one terminal, down into 1 ance (burden) of CT is housed in the grounded steel case at the lower end of the bushing. The upper end of the bushing has two manner similar to that shown in The ammeters, relays, and connecting 4.4 kV, in a Fig. II. Figure 11.14 shows a 500 VA, 100 A/5 transformer designed for a 230 into an ac line, having a line-to-neutral voltage of = = 1/80 is 280/80 = 3.5 A connected ! Figure 11.14 500 VA, 100 A/5 A, 60 Hz current transformer, insulated for a 230 kV line and having an accuracy of 0.6%. (Courtesy of Westinghouse) N Figure 11.15 Current transformer in the final (Courtesy of Ferranti-Packaro) process of construction. SPECIAL TRANSFORMERS M " 233 Mm ' Figure 11.17 Epoxy-encapsulated current transformer rated 50 VA, A, 60 Hz and insulated for 36 kV. (Courtesy of Montel, Sprecher & Schuh) 400 A/5 /, Current transformer 220 kV, 3-phase in one phase series with line inside The voltage across c. IR = X 3.5 The secondary voltage The primary voltage is E, = 4.2/80 = is 1 .2 = 4.2 V than normal. is = 52.5 a miniscule voltage drop, the 14.4 kV cle. mV The the is compared to line-to-neutral voltage. Opening the secondary of a CT can be dangerous and falls <t» much s first falls, for but it most of remains at a the time. The thing happens during the second half-cycle. During these saturated intervals, the induced volt- age across the secondary winding cause the flux changes very little. is negligible be- However, during tremely high primary flowing is accidentally opened, the primary current in the circuit. If the sec- to- primary current the unsaturated intervals, the flux changes is is half cycle, flux <P secondary circuit of a current transformer while ondary higher of every half cy- II. 18, as the core also rises and in the 1 so large that the core during the fixed, saturation level secondary may be 00 Every precaution must be taken to never open the current to that normal exciting cur- core reaches peaks flux Referring to Fig. rises same 11-6 to the tally saturated for the greater part /, This 200 times greater than rent, the flux in the therefore 4.2 V. 0.0525 no further bucking effect due ampere-turns. Because the line current is to E2 = compared the exciting current of the transformer because there a substation. the burden negligible is of the electrical load. The line current thus becomes of a is b. continues to flow unchanged because the imped- ance of the primary Figure 11.16 rate, at an ex- inducing voltage peaks of several hundred volts across the open-circuited secondary. This is a dangerous situation because an unsuspect- ELECTRICAL MACHINES AND TRANSFORMERS 234 ing operator could easily receive a bad shock. voltage is The particularly high in current transformers having ratings above 50 VA. view of In ondary must the above, first if CT circuit of a a meter or relay in the sec- we has to be disconnected, secondary winding and short-circuit the then remove the component. Short-circuiting a current transformer does no harm because the primary current remains unchanged and the secondary current can be no greater than that determined by the turns ra- winding may be re- tio. The moved short-circuit across the after the secondary circuit is again closed. 11-7 Toroidal current transformers Figure 11.18 When the line current exceeds 100 A, we can some- times use a toroidal current transformer. It consists Primary current, CT is flux, and secondary voltage when a open-circuited. of a laminated ring-shaped core that carries the sec- ondary winding. The primary is composed of a sin- gle conductor that simply passes through the center of the ring (Fig. conductor is 1 1 . 1 9). The unimportant as long as it is secondary possesses less centered. If the the ratio of transformation is having a ratio of 1000 A/5 200 turns position of the primary more N N. Thus, a toroidal A or turns, CT has 200 turns on the secondary winding. Toroidal widely used (MV) CTs are indoor installations. They are also incorporated in circuit-breaker (Fig. simple and inexpensive and are low-voltage (LV) and medium-voltage in 1 bushings to monitor the line current 1.20). If the current limit, the CT causes the Figure 11.19 Toroidal transformer having a ratio of nected to measure the current in a 1000 A/5 A, con- line. exceeds a predetermined circuit-breaker to trip. Example 11-4 A potential transformer rated 14 400 V/l 15 current transformer rated 75/5 V and a A are used to measure the voltage and current in a transmission line. If the voltmeter indicates 1 1 V 1 and the ammeter reads 3 A, calculate the voltage and current in the line. Solution cm The voltage on E= The current 111 the line x is (14 400/115) = 13 900V Figure 11.20 in the line is Toroidal transformer surrounding a conductor inside a /= 3 X (75/5) = 45 A bushing. SPECIAL TRANSFORMERS 11.8 Variable autotransformer 1 1.22). a fixed A variable autotransformer is often used when we wish to obtain a variable ac voltage from a fixedvoltage ac source. a single-layer The transformer toroidal iron core. A composed of movable carbon brush in slid- winding serves as a variable ing contact with the tap. is winding wound uniformly on a The brush can be set in 235 The input voltage £, is usually connected to 90 percent tap on the winding. This enables E 2 to vary from 0 to 1 10 percent of the input voltage. Variable autotransformers are efficient and provide good voltage regulation under variable ondary line loads. The sec- should always be protected by a fuse or circuit-breaker so that the output current /2 never ex- ceeds the current rating of the autotransformer. any position between 0 and 330°. Manual or motorized positioning may be used (Figs. 11.21 and 11.23). As the brush slides over the bared portion of the winding, the secondary voltage portion to the Figure 11.21 Cutaway view number of of a E2 turns increases in pro- swept out (Fig. manually operated 0-140 V, 15 A showing (1) the laminated variable autotransformer toroidal core; (2) the single-layer winding; (3) the mov- able brush. {Courtesy of American Superior Electric) Figure 11.23 200 A, 0-240 V, 50 A, 120 V units, connected in series-parallel. This motorized unit can vary the output voltage from zero to 240 V in 5 s. Dimensions: 400 mm x 1500 mm. {Courtesy of American Superior Electric) Variable autotransformer rated at 50/60 Hz. Figure 11.22 Schematic diagram ing a fixed 90% tap. of a variable autotransformer hav- It is composed of eight 236 ELECTRICAL MACHINES AND TRANSFORMERS 11.9 High-impedance transformers The primary winding P is connected to a V ac source, and the two secondary windings 240 The transformers we have studied so far are alt designed to have a relatively low leakage reactance, ranging perhaps from 0.03 to 0. S are connected tube. per unit (Section 1 <T> 10. 1 However, some 3). applications require and commercial industrial much higher reactances, some- across the long neon the secondary voltage of the transformer (Fig. circuit voltage (20 following in the in series to the large leakage fluxes E2 tp., falls rapidly as typical applications: kV) 1 soon as the neon tube current is 1 .24c). The high open- initiates the discharge, but lights up, the automatically limited to 15 secondary mA. The electric toys arc welders corresponding voltage across the neon tube fluorescent lamps electric arc furnaces to neon signs reactive oil and with increasing current, as seen in the regulation curve times reaching values as high as 0.9 pu. Such high- impedance transformers are used b , Owing falls The power of these transformers ranges from 50 VA to 1500 VA. The secondary voltages power regulators 1 5 kV. burners Let us briefly examine these special applications. 1 . A toy transformer is often accidentally short- by children circuited, but being used ther practical nor safe to protect Consequently, the transformer that its leakage reactance is is it it nei- is with a fuse. designed so so high that even a permanent short-circuit across the low- voltage secondary will not cause overheating. The same remarks apply to some bell trans- formers that provide low-voltage signalling power throughout a home. If a short-circuit oc- curs on the secondary side, the current is auto- matically limited by the high reactance so as not to burn out the transformer or damage the fragile annunciator wiring. 2. Electric arc furnaces and discharges in gases possess a negative E/I characteristic, meaning that once the arc struck, the current increases is as the voltage falls. To maintain a uniform discharge, ance in series ance may a steady arc, or we must add an imped- with the load. The series imped- be either a resistor or reactor, but prefer the latter because it consumes very we little active power. However, the load, it is if a transformer usually is used to more economical porate the reactance in the transformer designing it itself, have a high leakage reactance. to typical example shown in Fig. 1 is 1 the neon-sign transformer .24. 0 supply to incor- by A Figure 11.24 a. Schematic diagram — 15 of a 30 mA neon-sign transformer. b. Construction of the transformer. c. Typical E-I characteristic of the transformer. SPECIAL TRANSFORMERS range from 2 kV to 20 kV, depending mainly ings are very loosely coupled. mary windings upon the length of the tube. Returning to Fig. 1 1 .24a, we note that the center of the secondary winding (typically is only one-half the voltage across the neon tube. As a result, less insulation is have properties similar to neon-sign HV 1 .25). The controller permits more or less sec- causing the leakage flux to flow, A change in the flux produces a corresponding active power absorbed by leakage change in transformer, incorporated in a static var improve the power factor of the pensator, is the re- The com- the transformer. transformers. Capacitors are usually added to total circuit. line 765 kV) while windings (typically 6 kV) to vary accordingly. Fluorescent lamp transformers (called ballasts) 1 ondary current needed for the high-voltage winding. kV and between 230 are connected to an electronic controller (Fig. This ensures that the secondary line-to-ground voltage three pri- are connected to the the three secondary grounded. is The 237 further discussed in Section 25.27. Oil-burner transformers possess essentially the same about 10 A kV the oil jet. 3. Some oil Q secondary open-circuit voltage of The M '4 arc continually ignites the va- while the burner is in operation. between two carbon taining an intense arc A relatively low secondary voltage leakage flux elecis used tertiary and the large secondary current is ^ primary electric furnaces generate heat by main- trodes. kV Q M two closely immediately above creates an arc between spaced electrodes situated porized 3-phase primary input 230 characteristics as neon-sign trans- formers do. winding limited by the leakage reactance of the transformer. Such trans- formers have ratings between 100 MVA. In kVA and 500 secondary leakage flux very big furnaces, the leakage reactance of the secondary, together with the reactance of the conductors, usually sufficient to provide is the necessary limiting impedance. 4. Arc- welding transformers are also designed to have a high leakage reactance so as to stabilize the arc during the welding process. The open- about 70 V, which facilitates circuit voltage striking the arc is when the electrode touches the work. However, as soon as the arc lished, the secondary voltage 15 V, a value that the arc 5. As and the is falls to about we mention the units that absorb reactive transmission line. enormous 3-phase power from a 3-phase These transformers are tionally designed to 11.10 Induction heating transformers welding current. a final example of high-impedance trans- formers, leak- estab- depends upon the length of intensity of the Figure 11.25 Three-phase static var compensator having high age reactance. inten- produce leakage flux and, consequently, the primary and secondary wind- High-power induction furnaces also use the former principle to produce high-quality other alloys. The induction stood by referring to Fig. frequency 500 that Hz 1 trans- steel and principle can be under1.26. ac source is A relatively high- connected to a coil surrounds a larse crucible containing molten ELECTRICAL MACHINES AND TRANSFORMERS 238 primary molten iron Figure 11.26 Coreless induction furnace. The currents in the molten metal. the reactive The iron. coil is the primary, itself. O produces eddy capacitor furnishes power absorbed by the coil. and the molten iron secondary turn, short-circuited acts like a single upon flux The Consequently, carries a very large it secondary current. This current provides the energy that keeps the iron scrap metal as 15 it is melting other in a liquid state, added to the pool. Such induction furnaces have ratings between kVA and 40 000 kVA. The operating frequency becomes progressively lower power as the Figure 11.27 Channel induction furnace and is required to drive the flux through the molten iron and through the must remember far air. In this regard, that the temperature of above the Curie point, far as permeability is and so it we molten iron The magnetizing is like air as why these Capacitors are installed close to the coil to sup- In power it known as a channel fur- nace, a transformer having a laminated iron core made to link with a channel of in Fig. 1 1 ted to the .27. The channel Hows molten iron, as Hz is fit- coil is channel and through the molten iron in the crucible. In effect, the channel lent to a single turn short-circuited coil. is On is the large because the sec- obviously not tightly coupled to the Nevertheless, the power factor II. 26, and 80 percent. As a is higher being typically between 60 result, a smaller capacitor bank required to furnish the reactive power. Owing to the very high ambient temperature, the primary windings of induction furnace transformers are always made of hollow, water-cooled copper on is itself. aluminum, copper, and other metals, 1 1.28 shows as well as iron. a very special application of is source, and the secondary current in the liquid primary is than that in Fig. Figure shown a ceramic pipe that bottom of the crucible. The primary excited by a 60 12 is other hand, the leakage flux ondary turn low because the flux is permeable iron core. conductors. Induction furnaces are used for melting absorbs. another type of furnace, current to a highly is furnaces are often called coreless induction furnaces. ply the reactive confined is behaves concerned. That water-cooled rating 60 Hz is used when the power exceeds about 3000 kVA. The power factor of coreless induction furnaces is very low (typically 20 percent) because a large magincreases. Thus, a frequency of netizing current its transformer. equiva- the induction heating principle. 11.11 High-frequency transformers In electronic power supplies there is often a need to from the input and to reduce the isolate the output weight and cost of the such as in aircraft, unit. In there is other applications, a strong incentive to SPECIAL TRANSFORMERS 239 order to illustrate the reason for this phenomenon, limit we how avoid a cumbersome theoretical analysis, take a practical transformer and observe when haves the frequency Consider Fig. tional 20 V/24 1 ing of 36 in the II. 29, be- 1 .5 T. a rat- 0.5 kg The flux core attains a peak of 750 fjiWb. The lami- nated core is about made of 1 ordinary silicon steel having mm (12 mils) and the W. The and the primary 1 .5 A current rating silicon A >~ 1 t core mA for 12 mil 1.5 600 total 300 cm mA 120 V 60 Hz is for the secondary. core: 6 x 5 x 2.5 300 it which shows a conven- peak flux density of at a to will raised. 60 Hz transformer having V, a thickness of 0.3 is is VA. This small transformer weighs and operates loss we our discussion to transformers. Furthermore, 120 24 V t T . J P = 36 VA B=1.5T 0max = 75OAWb core loss = 1 W Figure 11.29 Figure 11.28 Special application of the transformer effect. This picture of shows one stage in a steam-turbine generator. the diameter of a 5 t It consists of expanding coil-retaining ring. A coil of as- its temperature up to 280°C expansion enables the ends, where it ring to in h. The main resulting be slipped over the rise of the large this it in- mass. (Courtesy of ABB) means that the can be increased E= = = which 4.44,/TV] 400 Hz, while in electronic to to, typically power supplies quency may range from 5 kHz An is the at a fre- same peak ^ max will re- to Eq. 9.3, corresponding primary voltage to the fre- 50 kHz. increase in frequency reduces the size of such devices as transformers, inductors, and capacitors. In <I> (9.3) max X 6000 X 600 X 750 X 000 V 12 is 6 l() 100 times greater than before! The sec- becoming 2400 frequency Assuming 4.44 by using a relatively high frequency compared in aircraft the it 00 times higher than 750 |xWb. However, according at ondary voltage 60 Hz. Thus, 1 follows that the flux minimize weight. These objectives are best achieved say, for. is coil- clean and produces a very uniform is was designed it flux density, the ring, bringing about 3 cools and contracts. This method of duction heating temperature in to the transformer, us consider the effect of operating what is which induces large eddy currents let quency of 6000 Hz, which wound around the ring and connected to a 35 kW, 2000 Hz source (left foreground). The coil creates a 2000 Hz magnetic field, bestos-insulated wire Without making any changes the construction of the rotor will likewise be 100 times greater, The operating conditions are The primary and secondary currents remain unchanged and so the power of the transformer is now 3600 VA, 00 times greater than shown in Fig. V. 11.30. 1 in Fig. 1 1 .29. Clearly, raising the a very beneficial effect. frequency has had ELECTRICAL MACHINES AND TRANSFORMERS 240 However, the advantage because at 6000 Hz 700 W), due to the increase in is eddy current and hys- Thus, the transformer teresis losses. not feasible because To it seems enormous (about it in Fig. we of 12 mil silicon cording to 1 .5 T to 0.04 T. As be a result, ac- Eq. 9.3, the primary and secondary volt- ages will have to be reduced to 320 spectively. as reduction in steel, this requires a from the flux density The new power of P = 320 X 0.3 = 96 V and 64 V, re- the transformer will VA (Fig. 11.31 ). This is al- most 3 times the original power of 36 VA, while same temperature rise. By using thinner laminations made of re- it is special possible to raise the flux density above 0.04 T while maintaining the same core losses. Thus, if we replace the original core with this special material, the flux density to 0.2 T. = cl> max of 750 |xWb X (0.2 T/1.5 T) means primary voltage can be raised to that the 100 £ = 4.4477V, * nlllx = 4.44 X 6000 X 600 X = 1600 V core: 6 x 5 x 2.5 12 mi silicon 300 120/ silicon 12 600 Figure 11.31 interested, of course, in maintaining the 1 V to 24 V. This 20 read- is number of turns on the primary will be reduced to 600 t X (120 V/1600 V) = 45 turns, while the secondary will have only 9 turns. Such a drastic reduction in the wire the Bearing former can size mind in is still number of turns means be increased that the capacity of the trans- 480 VA, it follows that the rated pri- mary current can be raised to 4 A while that secondary becomes 20 A. This rewound same size in the trans- 11.33) has the special core (Fig. its that significantly. and weight as the one in Fig. 11.29. Furthermore, because the iron and copper losses same are the both cases, the efficiency of the in high frequency transformer It is now obvious is better. that the increase in has permitted a very large increase pacity of the transformer. It in the frequency power ca- follows that for a given A cm special core J ' = 3600 VA 300 mA 1 ff=1.5T 2400 V ^max^SOjUWb 1600^ core loss = 700 6 kHz ^ W 600 r 120 5 p = 480 VA B = 0.2T A t+ t «W= lOO^Wb 320 V r core: 6 x 5 x 2.5 cm core loss = 1 W t 120/ cm special core 1.5 . is .32). from 600 mil mA 6 kHz 1 Figure 11.32 core: 6 x 5 x 2.5 32*0V^ 1 power output a high frequency transformer is much smaller, cheaper, more efficient, and lighter than a 60 Hz transformer. Figure 11.30 300 VA (Fig. 480 core: 6 x 5 x 2.5 — / I(T A= the cm 15A 600 X 100 .5 achieved by rewinding the transformer. Thus, I mA 12 kV 6 kHz ^Wb, which V, ily can be raised This corresponds to a peak flux 1 former with taining the nickel-steel, V X We are 320 original voltage ratio of same 320 is and so the enhanced capacity of the transformer is Based upon the properties 1.29. 1 .30 can reduce the flux density so that the core losses are the in Fig. 1 1 will quickly overheat. get around this problem, they were The corresponding secondary voltage not as great as is the core loss 64 A vh Y P = 96 VA # = 0.04T 0max = 2OAiWb core loss = 1 W 4 20 A A B= 120~\T 6 kHz 1 45/ Figure 11.33 9/ 24 V T 0.2 T <^max = lOO^Wb core loss = 1 W SPECIA L TRA NS FORMERS Questions and Problems Calculate the voltage across the secondary a. winding Practical level 11-1 What is What the difference between an auto- is former? 1 1 -3 Why 1 -4 11-5 Calculate the voltage drop the transformer c. If the produces on the Of a is looped four times through the toroidal opening, calcu- current transformer? new late the current ratio. Industrial application why the secondary winding PT must be grounded. Explain CT or A toroidal of a 11-12 The nameplate of a small transformer dicates 50 VA, 120 V, 12.8 V When 1 current transformer has a ratio of How many turns does have? it current transformer has a rating of 10 VA, 50 A/5 A, 60 Hz, 1 V 8.8 would 2.4 kV. Calculate V 13.74 If V were 120 at no-load available, what Why the secondary voltage be? this voltage the nominal voltage across the primary in- applied to the primary, the is voltage across the secondary is A line conductor. primary conductor must we never open the secondary of 1500 A/5 A. 11-6 ammeter has an impedance the b. the purpose of a voltage trans- a current transformer? 1 if of 0.1 5 n. transformer and a conventional transformer? 11-2 24 is higher than the indicated nameplate voltage? winding. 11-13 Intermediate level 11-7 A single-phase transformer has a rating of 100 kVA, 7200 V/600 V, 60 Hz. If re- it is connected as an autotransformer having a ratio it 1 1-8 In can carry. Problem 11-14 1 1-7, how should the trans- H2 , X,, X2 ) be con- in Advanced 11-10 A of 6.6 k V/600 how is Problem 1 1-7 is V What load can it 120 V winding wound Many airports use series lighting systems in rent winding, or vice versa? sion line it where is installed numin 60 Hz kept constant at 20 A. The secondary A to a incandescent lamp. a. Calculate the voltage across each lamp. b. The 0.07 resistance of the secondary winding 0 while that of the primary Knowing It that the is is 0.008 U. magnetizing current and the leakage reactance are both negligible, has a primary to secondary capacitance If is 100 W, 6.6 should the connections be made? of 250 pF. large windings are individually connected carry 100 VA, 2000 A/5 A, 60 Hz, 138 kV. calculate the voltage across the primary on a transmis- winding of each transformer. the line-to-neutral voltc. If 140 lamps, spaced 138 kV, calculate the capacitive vals, are leakage current that flows to ground (see 11-11 V the 12.8 current transformer has a rating of is il. Is the upon source. In one installation, the primary cur- recon- level age 0.306 the pri- 5.2 11 and that of the secondary 1 ber of current transformers are connected nected again as an autotransformer having a and mary series across a constant current, The transformer ratio However, the resistance of which the primary windings of a nected? 1-9 epoxy and cannot be seen. is of 7800 V/7200 V, calculate the load former terminals (H,, 1 Referring to Problem 11-12, the windings are encapsulated in at in series power The a temperature of 105°C. toroidal current transformer of Fig. 1 000 A/5 A. The conductor carries a current of 600 A. line in inter- using wire, calculate the Fig. M.13). 11.19 has a ratio of every 50 No 14 minimum voltage of the Assume the wire operates at connected 11-15 A source. no-load test on a 1 5 kVA. 480 V/l 20 60 Hz transformer yields the following V, 242 ELECTRICAL MACHINES AND TRANSFORMERS saturation curve data winding is The primary a. Draw is If the the 1 20 V c. known to have 260 Draw the saturation curve at flux in 60 Hz (peak mWb versus current in mA). At what point on the saturation curve does turns. uration the saturation curve (voltage versus become important? sat- Is the flux dis- torted under these conditions? mA). current in b. when excited by a sinusoidal source. experiment were repeated using a 50 Hz source, redraw the resulting saturation curve. E 14.8 31 A) 59 99 49.3 144 66.7 210 90.5 430 110 120 130 136 142 V 700 1060 1740 2300 3200 mA Chapter 12 Three-Phase Transformers connected 12.0 Introduction is distributed throughout North America by means of 3-phase transmission lines. In order to transmit this power efficiently and economically, the Power vice versa. kV 3.8 power to that has to be transmitted and the distance is also A it uniform, ranging from systems to 600 V 1 20/240 shift from one this re- ratio of the the primaries and secon- enables us to change the number of if there were a 12-phase system. practical application for it, we could even convert a 3-phase system into a 5-phase can be achieved by using three system by an appropriate choice of single-phase transformers and interconnections. form a 3-phase transformer bank. In making tant to When how into a 2-phase, a 6-phase, or a Indeed, single-phase transformers connected together to 12.1 The amount of phase depends again upon the turns phases. Thus, a 3-phase system can be converted windings mounted on a 3-legged core. However, result between the 3-phase input voltage and shift feature level to another. The transformers may be inherently 3-phase, same shift daries are interconnected. Furthermore, the phase- having three primary windings and three secondary the are connected. transformers, and on quires the use of 3-phase transformers to transform the voltages may wye, or 3-phase transformer bank can also produce a the 3-phase output voltage. V single-phase 3-phase systems. Clearly, in a result, the ratio of the 3-phase input upon how they phase the appropriate voltage levels used in factories and homes. These are fairly As and the secondaries only upon the turns ratio of the transformers, but 765 kV) depend upon the amount of has to be carried. Another aspect in delta voltage to the 3-phase output voltage depends not voltages must be at appropriate levels. These levels (1 ways. Thus, the primaries in several be connected the various connections, observe transformer polarity may produce polarities. it is An imporerror in a short-circuit or unbalance Basic properties of 3-phase transformer banks the line voltages three single-phase transformers are used to former banks can be understood by making the The transform a 3-phase voltage, the windings can be and currents. basic behavior of balanced 3-phase trans- lowing simplifying assumptions: 243 fol- ELECTRICAL MACHINES AND TRANSFORMERS 244 1 . 2. 3. The exciting currents nected are negligible. The transformer impedances, due to the resis- is Terminal H, of each trans- H2 connected to terminal of the next X2 tance and leakage reactance of the windings, transformer. Similarly, terminals X, and are negligible. cessive transformers are connected together. The total tual physical layout of the transformers apparent input power to the trans- former bank Fig. I2.l. is put power. Furthermore, when single-phase transformers are connected into a 3-phase system, they retain their basic single-phase properties, voltage ratio, and flux ratio, polarity marks X,, X2 in and Hj, between primary and secondary that Ex x is in Fig. phase with EH the core. H2 is , Given the C AO the phase shift Ht X Ho X; -O-i 1 2 balanced BOH . three-phase CO R of to a level appropriate for the outgoing transmission line is The schematic diagram is drawn in such a way to show not only the connections, but also the phasor H2 X; Hi X, Ho X2 load I transform the voltage of the incoming transmission line A, B, line l, 2, 3. The incoming connected to the source, and the outgoing connected to the load. The transformers \ 3/„ line are con- Figure 12.1 Delta-delta connection of three single-phase trans- formers. The incoming lines (source) are A, B, the outgoing lines (load) are V3/ 1 , 2, 3. s load Figure 12.2 Schematic diagram is in Fig. 12.2. zero, in the sense three single-phase transformers P, Q, and 12.1 The corresponding schematic diagram all such as current 12.2 Delta-delta connection The is of suc- The acshown in equal to the total apparent out- given is in delta-delta. former of a delta-delta connection and associated phasor diagram. C and THREE-PHASE TRANSFORMERS relationship between the primary and secondary c. drawn d. corresponding primary winding to e. voltages. Thus, each secondary parallel to the which is it winding coupled. Furthermore, duces voltages £AB £ BC ECA , , if is G source in- current in the HV current in the LV lines lines currents in the primary and secondary windings of each transformer pro- according to the The The The 245 f. The load carried by each transformer dicated phasor diagram, the primary windings are same way, phase by phase. For example, the primary of transformer P between lines A and B is oriented horizontally, in the same direction Solution oriented the £AB as phasor a. The apparent power drawn by S . Because the primary and secondary voltages EH and Hn phase, Ex En (secondary voltage of b. E AB (primary £2 3 s m phase transformer P) must be in phase with of the same transformer). Similarly, with £ BC , and E3i ECA with i- tween the respective incoming and outgoing mission lines are If in small. is c. connected to lines 1 The -2-3, the incoming = (24.4 = 102 in the / V3 times greater than the respective and /s ondary windings the transformer flowing (Fig. bank in the The rating of I2 single transformer. Note tutes that arrangement, each transformer, considered alone, acts as phase H2 circuit. in the rent / s if it were placed Thus, a current primary winding flowing from X2 to is X| / p MVA. is (8.9) X 10 )/(V3 X 138 000) A LV lines is = 5/(V3 E) = (24 A 6 X 10 )/(V3 X 4160) = 3386 A although the transformer bank consti- a 3-phase also 24.4 line 6 current in the three times the rating of a is is HV primary and sec- The power 12.2). HV line current in each /,=5/(V3£) d. currents MVA follows that the apparent power fur- in any delta connection, the line produces balanced line currents A-B-C. As It nished by the trans- phase. a balanced load lines 24.4 (7.7) The transformer bank itself absorbs a negligible amount of active and reactive power because the 1~R losses and the reactive power associated resulting line currents are equal in magnitude. This currents are 21/0.86 = is with the mutual flux and the leakage fluxes are , such a delta-delta connection, the voltages be- In P/cos 0 of a given transformer must be in Xi follows that it = = the plant e. Referring to Fig. mary winding in a single- flowing from H] to / p 1 2.2, the current in each pri- is = 1 02/V 3 = 58.9 A associated with a curin the The secondary. current in each secondary winding /s = 3386/V3 = 1955 is A Example 12-1 Three single-phase transformers are connected delta-delta to step down a line voltage of 4160 V to supply power The plant draws 21 MW 1 38 kV in to to a manufacturing plant. at a lagging power factor f. Because the plant load is balanced, each trans- former carries one-third of the 24.4/3 The - 8.13 total load, or MVA. individual transformer load can also be obtained by multiplying the primary voltage of 86 percent. times the primary current: Calculate a. b. The apparent power drawn by the plant The apparent power furnished by the HV S = line = E p Ip = 8.13 138 000 MVA X 58.9 ELECTRICA L MACHINES AND TRANSFORMERS 246 Note that we can do not know how the fect, the plant load 3-phase load (shown composed of hundreds of which are connected as a is connected. In ef- box in Fig. individual loads, in delta, we others 12.2) is some of wye. in Furthermore, some are single-phase loads operating at much lower voltages than 4160 V, powered by smaller transformers located inside the plant. sum total of these The calculate the line currents and the currents in the transformer windings even though The relative values of the currents in the trans- former windings and transmission Fig. C 1 are The 2.4. Thus, the line currents in V 3 times the currents line currents in in the phases lines are given in phases A, B, and primary windings. 2, 3 are the l, same A shift delta-wye connection produces a 30° phase between the of the incoming and line voltages outgoing transmission lines. Thus, outgoing line loads usually results in a reasonably well-balanced 3-phase load, represented by the box. AO- 12.3 Delta-wye connection When the transformers are connected in delta-wye, the three primary windings are connected the way as in Fig. 1 2. 1 . together, creating a BO same H, X H2 M X;2 X f 1 — C—-— balanced 2 : I H, Xt H2 X; H, X H2 x 2 |— -O( • three-phase load However, the secondary windings are connected so that all the common X2 terminals are joined neutral N CO (Fig. 12.3). In 3 -O--I such a delta- wye connection, the voltage across each primary winding age. is equal to the incoming line volt- However, the outgoing line voltage is a/3 the secondary voltage across each transformer. times as the currents in the secondary windings. Figure 12.3 Delta-wye connection of three single-phase transformers. load Figure 12.4 Schematic diagram of a delta-wye connection and associated phasor diagram. (The phasor diagrams on the mary and secondary sides are not drawn to the same scale.) pri- THREE-PHA SE TRANSFORMERS 375 A 3932 A 247 1 80 kV 90 MVA 13.2kV« 375 A 3932 A Figure 12.5 See Example E i2 voltage £ AB , 12-2. is 30° ahead of incoming as can be seen The voltage across line voltage from the phasor diagram. The voltage between outgoing line feeds an isolated group of loads, the phase shift creates no problem. But, if and 3 the outgoing be connected in parallel with a line comfrom another source, the 30° shift may make such a parallel connection impossible, even line voltages are One that is it s b. 1, 2, The load C is kV 139 carried by each transformer S = 90/3 is = 30 MVA wye conThe reduces the amount of insulation needed inside the transformer. The to the therefore, the outgoing lines E = 80 V3 = otherwise identical. of the important advantages of the nection if is, is line has to ing the secondary 80 kV. If the current in the primary winding HV winding has / be insulated for only 1/V3, or 58 percent of the The p is = 30 MVA/ 13.2 kV = 2273 A current in the secondary winding is line voltage. /s Example 12-2 c. Three single-phase step-up transformers rated MVA, a 13.2 a 90 1 3.2 kV kV/80 kV at The current If in each incoming /= they feed MVA load, calculate the following: The 2273 V 3 b. c. The secondary line voltage The currents in the transformer windings The incoming and outgoing transmission line A, B, = 3937 A current in each outgoing line / a. A 375 40 are connected in delta-wye on transmission line (Fig. 12.5). = 30 MVA/80 kV = = 375 1 , 2, 3 is A 12.4 Wye-delta connection line The currents and voltages currents in a wye-delta connection are identical to those in the delta-wye connection of Solution The easiest way Section to solve this the windings of only former problem is one transformer, say, trans- P. The voltage across ously 13.2 kV. 2.3. The primary and secondary connec- tions are simply interchanged. In other words, the H 2 terminals are connected together to create a neuand the X X 2 terminals are connected in delta. tral, a. 1 to consider the primary winding is obvi- [ , Again, there results a 30° phase shift between the voltages of the incoming and outgoing lines. ELECTRICAL MACHINES AND TRANSFORMERS 248 ac source | Figure 12.6 Wye-wye connection ac source with neutral of the primary connected to the neutral of the source. ' Figure 12.7 Wye-wye connection using a 12.5 tertiary winding. Wye-wye connection to a delta-delta connection, except that former When cial transformers are connected in wye-wye, spe- precautions have to be taken to prevent severe One way distortion of the line-to-neutral voltages. to prevent the distortion is to connect the neutral of the primary to the neutral of the source, usually way of the ground (Fig. 12.6). Another way is by to provide each transformer with a third winding, called tertiary winding. three 1 2.7). The tertiary transformers are connected They windings of the in delta (Fig. often provide the substation service is delta connection is is ample, if two 50 kVA transformers are connected bank as it is obviously 2 may seem, it X 50 = 100 kVA. But, strange can only deliver 86.6 The open-delta connection emergency situations. Thus, if mainly used is b n^-Mj Q I possible to transform the voltage of a 3-phase The open-delta arrangement is in identical Figure 12.8a Open-delta connection. o [h x~| in three transformers A O- 12.6 Open-delta connection open-delta. kVA before the transformers begin to overheat. 0—4 It is in open-delta, the installed capacity of the transformer transformer. system by using only 2 transformers, connected only 86.6 per- cent of the installed transformer capacity. For ex- Note that there is no phase shift between the incoming and outgoing transmission line voltages of wye-wye connected trans- seldom used because the load capacity of the transformer bank voltage where the transformers are installed. a one absent (Fig. 12.8). However, the open- THREE-PHASE TRANSFORMERS are connected in delta-delta and comes defective and has ble to feed the load to one of them be- be removed, it is 249 Thus, the ratio possi- maximum on a temporary basis with the kVA 300 kVA 260 load installed transformer rating two remaining transformers. = Example 12-3 Two single-phase 150 formers are connected maximum V kVA, 7200 V/600 in 0.867, or 86.7% 12.7 Three-phase transformers trans- open-delta. Calculate the 3-phase load they can carry. A transformer bank transformers former Solution may (Fig. 1 composed of three single-phase be replaced by one 3-phase trans- 2.9). The magnetic core of such a transformer has three legs that carry the primary and Although each transformer has a rating of 50 k VA, 1 two together cannot carry a load of 300 kVA. The following calculations show why: secondary windings of each phase. The windings are the The nominal secondary current of each former connected internally, either in wye the result that only six terminals or in delta, with have to be brought trans- outside the tank. For a given total capacity, a 3-phase is transformer = /s 1 50 kVA/600 V= 250 A is single-phase always smaller and cheaper than three transformers. Nevertheless, single- phase transformers are sometimes preferred, partic- The current ceed 250 A mum /s in lines 1, 2, 3 (Fig. cannot, therefore, ex- 12.8b). Consequently, the maxi- load that the transformers can carry when a replacement unit one 3-phase 5000 (8.9) 259 800 Figure 12.8b Associated schematic and phasor diagram. VA is essential. For ex- kVA. To guarantee continued service we can is S = ^3EI = V3 X 600 X 250 = = 260 kVA ularly ample, suppose a manufacturing plant absorbs 5000 ond one as kVA a spare. Alternatively, we can single-phase transformers each rated plus one spare. install transformer and keep a sec- at install three 1667 kVA, The 3-phase transformer option is 250 ELECTRICA L MACHINES AND TRANSFORMERS more expensive (total capacity: 2 X 5000 = 10 000 kVA) than the single-phase option (total capacity: 4 X 1667 = 6667 kVA). Fig. 12.10 shows successive stages of construction of a 3-phase 1 10 MVA, 222.5 kV/34.5 changing transformer/ Note that 1 three A main legs, the ' in kV tap- addition to the tional lateral legs. They enable the designer to re- duce the overall height of the transformer, which simplifies the problem of shipping. In effect, when- ever large equipment has to be shipped, the designer is faced with the problem of overhead clear- ances on highways and rail lines. magnetic core has two addi- 1 £ lap-changing transformer regulates the secondary voltage by automatically switching from one tap to another on the pri- mary winding. The tap-changer is a motorized device under the control of a sensor that continually monitors the voltage that hns to be held constant. 3 I 1—t w V Figure 12.10a Core of a 1 1 0 MVA, 222.5 kV/34.5 kV, 60 Hz, 3-phase transformer. By staggering laminations of different widths, the core legs can be made almost circular. This reduces the coil diameter to a minimum, resulting in 2 less copper and lower l R losses. The legs are tightly bound to reduce vibration. Mass of core: 53 560 kg. Figure 12.9 Three-phase transformer for an electric arc furnace, rated 36 MVA, 1 3.8 kV/1 60 V to 320 V, 60 Hz. The secondary voltage is adjustable from 1 60 V to 320 V by means of 32 taps on the primary winding (not shown). The three large busbars in the foreground deliver a current of 65 000 A. Other characteristics: impedance: 3.14%; diameter of each leg of the core: 71 mm; overall height of core: 3500 mm; center line distance between adjacent core legs: 1220 mm. (Courtesy of Ferranti- Packard) Figure 12.10b Same transformer with coils windings are connected in in place. wye and The primary the secondaries in Each primary has 8 taps to change the voltage in steps of ±2.5%. The motorized tap-changer can be seen in the right upper corner of the transformer. Mass of copper: 1 5 230 kg. delta. THREE-PHASE TRANSFORMERS The 34.5 kV windings (connected in delta) are to the core. The 222.5 kV windings (connected in wye) are mounted on top of the 34.5 kV mounted next windings. A space of several centimeters separates the two windings cool 251 to ensure good isolation and to allow flow freely between them. The oil to ings that protrude nected to a 220 bushings are kV much HV bush- from the oil-filled tank are con- line. The medium voltage (MV) smaller and cannot be seen in the photograph (Fig. 12.10c). 12.8 Step-up and step-down autotransformer When the voltage of a 3-phase line has to be stepped up or stepped down by a moderate amount, nomically advantageous to it is eco- use three single-phase transformers to create a wye-connected autotrans- The former. in Fig. 1 agram is 2. 1 actual physical connections are 1 a, given shown and the corresponding schematic in Fig. 12. lib. The respective di- line- to-neutral voltages of the primary and secondary are obviously in phase. Consequently, the incoming Figure 12.10c Same has been subjected to a 1050 kV impulse test on the HV side and a similar 250 kV test on the LV side. Other details: power rating: 110 MVA/146.7 MVA (OA/FA); total mass transformer ready including oil: 158.7 for shipping. t; It overall height: 9 m; width: and outgoing transmission phase. tral, The neutral is line connected voltages are to the otherwise a tertiary winding must be added to prevent the line-to-neutral voltage distortion mentioned previously (Section 12.5). 8.2 m, length: 9.2 m. {Courtesy of ABB) A B O2 -o H Figure 12.11a Wye-connected autotransformer. in system neu- TN Figure 12.11b Associated schematic diagram. load 252 ELECTRICAL MACHINES AND TRANSFORMERS Figure 12.11c Single-phase autotransformer (one of a group of three) connecting a 700 kV, 3-phase, 60 Hz transmission line to an existing 300 kV system. The transformer ratio is 404 kV/173 kV, to give an output of 200/267/333 MVA per transformer, at a temperature rise of 55°C. Cooling is OA/FA/FOA. A tertiary winding rated 35 MVA, 1 1.9 kV maintains balanced and distortion-free line-to-neutral voltages, while providing power for the substation. Other and windings: 132 t; tank and accessories: 46 t; oil: 87 t; total 265 t. BIL rating is 1950 kV and 1050 kV on the HV and LV side, respectively. Note the individual 700 kV (right) and 300 kV (left) bushings protruding from the tank. The basic impulse insulation (BIL) of 1950 kV and 1050 kV expresses the transformer's ability to withstand lightning and switching surges. (Courtesy of Hydro-Quebec) properties of this transformer: weight of core weight: THREE-PHASE TRANSFORMERS For a given power output, an autotransformer former (see Section 1 1 .2). This is the ratio of the incoming line voltage lies between 0.5 and particularly true winding rated 1 1 .9 kV. kV an existing 300 kV The cerned) = 3 66.3 with a tertiary MVA. basic transformer rating (as far as size is considerably less than MVA. This lies between 0.5 and kV to kV transmission 230 kV line has to be supply a load of 200 MVA. totransformers are to be used. Calculate the basic power and voltage rating of each transformer, assuming they are connected as shown in Fig. Solution 199 stations ing is and in - 199 H2 between H, and 230/V3 to I, SN3E = (200 X = 335 lines 1 is and lx power flow over trans- the phase shifting principle, con- between phases B and 12. 12). As we slide contact A o £ kV AP P A is = 66 kV — — o to B : rheostat line C pri- 66 kV. line is (8.9) 6 1() )/(V3 X 345 000) output A voltage The power associated with winding X,X 2 S converter special electric controls. Phase shift- QMne a voltage each phase of the outgoing = Such multi- is kV 133 secondary voltage rating of 133 in line. in large electronic also used to control output = 133 H2 This means that each transformer has an effective The current phase an- shifting en- mission lines that form part of a power grid. line mary .5) kV The voltage of winding X,X 2 between = 1 us consider only between X, and = 345/V3 line-to-neutral voltage £, A = say). line-to-neutral voltage £AN = let shift the Such phase systems from an ordinary 3-phase phase systems are used of a 3-phase line (Fig. To simplify the calculations, The (345/230 ables us to create 2-phase, 6-phase, and 12-phase sider a rheostat connected Em = keeping with the 2.0. 3-phase system enables us to To understand 12.11b. The con- 12.9 Phase-shift principle A Three single-phase transformers connected as au- one phase (phase A, is load-carrying system. a 3-phase, stepped up to 345 in fact that the ratio of transformation gle of a voltage very simply. Example 12-4 The voltage of is its part of a 3-phase trans- It is former bank used to connect a 700 line to basic rating of the 3-phase transformer bank X 22.1 capacity of 200 2. Figure 12.11c shows a large single-phase auto- transformer rated 404 kV/173 is if outgoing line voltage to The is smaller and cheaper than a conventional trans- 253 = 66 000 X 335 - 22. 1 Winding H,H 2 has the same power is MVA rating. sic rating of each single-phase transformer fore 22.1 MVA. The is ba- there- Figure 12.12 EAP can be phase -shifted by means of a potentiometer. Voltage with respect to EAC C P ELECTRICA L MA CHINES AND TRANSFORMERS 254 from phase B toward phase C, voltage both amplitude and phase. in moving from one end of shift in to the other. changes a 60° phase the potentiometer we move from B Thus, as E AP We obtain C, voltage to EAP gradually advances in phase with respect to £AB At the same time, the magnitude of E AP varies slightly, from E (voltage between the lines) to 0.866 . E when the contact Such circuits draws is in the middle of the rheostat. a simple phase-shifter can only be used in where the load between terminals few milliamperes. a If a IR drop plied, the resulting pletely changes the voltage A and P heavier load ap- is in the rheostat com- and phase angle from what they were on open-circuit. we connect a mulautotransformer between phases B and C 12.13). By moving contact P, we obtain the To get around titap (Fig. this nals shifts as be- but this time they remain essentially un- changed when a load A and P. Why is is connected between termi- this so? The reason flux in the autotransformer fixed. mains As is is that the E BC fixed because (both in magnitude and phase) whether the autotransformer delivers a current 3 tapped autotransformers con- nected between lines A, B, and C. Contacts P,, P 2 P 3 move in obtain a maximum tandem as we switch from one set We now to source of 60° as ABC. We we move the autotransformers to the some practical applications of the phase-shift principle. 12.10 Three-phase to 2-phase transformation 2-phase system are equal but dis- in a placed from each other by 90°. There are several ways to create a source. One 2-phase system from a 3-phase of the simplest and cheapest single-phase autotransformer having taps cent and 86.6 percent. We two phases of a 3-phase line, If the A discuss shift , of taps to the next. This arrangement enables us to cre- line phase from one extremity of other. The voltages shows , to the load or not. Fig. 12. 14 P h P 2 P 3 whose phase angle ate a 3-phase source changes stepwise with respect is a result, the voltage across each turn re- fixed shifter. problem, same open-circuit voltages and phase fore, Figure 12.14 Three-phase phase connect as voltage between lines A, is to use a at 50 per- between any it shown B,C is in Fig. 100 1 2. 1 5. V, voltages EAT and £ NC are both equal to 86.6 V. Furthermore, they are displaced from each other by 90°. This relationship can be seen by referring to the phasor dia- gram 1 . (Fig. 1 Phasors 2. 1 5c) and reasoning as follows: EAB £ BC , , and ECA are fixed by the source. 2. U^A-^^^A/->^^^A/'•^l O line B 1 1 3 4 b b Figure 12.13 Autotransformer used as a phase-shifter. 7 n line Phasor E AN cause the is in phase with phasor same ac £ AB be- flux links the turns of the au- < totransformer. 3. Phasor same E AT is reason. in phase with phasor £ AB for the THREE-PHASE TRANSFORMERS and the other an 86.6 percent tap on tap the 255 primary The transformers are connected as shown in 12.16. The 3-phase source is connected to termi- winding. Fig. nals A, B, C secondary and the 2-phase load windings. The ratio connected to the is of transformation (3-phase line voltage to 2-phase line voltage) by EAB /E l2 . The is given Scott connection has the advantage of isolating the 3-phase and 2-phase systems and provid- ing any desired voltage ratio between them. load Except for servomotor applications, 2-phase 1 systems are seldom encountered today. Example 12-5 A 2-phase, kW (10 hp), 240 V, 60 Hz motor has 7.5 an efficiency of 0.83 and a power factor of 0.80. is to It be fed from a 600 V, 3-phase line using a Scott- connected transformer bank (Fig. 12.16c). <b) Calculate a. b. c. The apparent power drawn by the motor The current in each 2-phase line The current in each 3-phase line Solution a. -NC The power drawn by active the The apparent power drawn by Figure 12.15 a. Simple method to obtain a 2-phase system from a 3-phase line, using a single transformer winding. b. Schematic diagram of the connections. c. Phasor diagram of the voltages. From KirchhofPs ECA = Loads I E AN + £ NC + phasor £ NC must have and direction shown = S = The ratio b. The current in 1 00/86.6 - the fixed and given by EAH /EAV = to is 9036/0.8 VA 295/2 is = 5648 VA each 2-phase line is c. The transformer bank tle itself consumes very lit- active and reactive power; consequently, the 3-phase line supplies only the active and reac- 1.15. Another way 295 = motor f=S/E = 5648/240 = 23.5 A other, of transformation (3-phase voltage to is 1 <|> the in the figure. and 2 must be isolated from each 2-phase voltage) 1 5=11 such as the two windings of a 2-phase induction motor. P/cos The apparent power per phase voltage law, Consequently, 0. the value is P = PJt) = 7500/0.83 = 9036 W (c) 4. motor produce a 2-phase system Scott connection. It consists of two is lo use identical single-phase transformers, the one having a 50 percent power absorbed by the motor. The power furnished by die 3-phase therefore, 295 VA. tive total ap- parent line 1 1 is, ELECTRICAL MACHINES AND TRANSFORMERS 256 86.6% phase AO 1 OB Figure 12.16c See Example 12-5. The 3-phase = / line current 5/(V3 E) = l 1 is 295/(V3 X 600) = 10.9A Figure 1 2. line voltages 12.1 1 1 6c shows the power circuit and the and currents. Phase-shift transformer A phase-shift transformer is a special type of 3-phase autotransformer that shifts the phase angle between incoming and outgoing the lines without changing the voltage ratio. Consider a 3-phase transmission to the terminals (b) b. Scott connection. Phasor diagram (Fig. incoming Figure 12.16 a. former of the Scott connection. 1 A, B, 2. 1 7). C The transformer line voltages however, changing their 1 , 2, 3 are shifted connected twists a magnitude. The through an angle that all the voltages of the line line of such a phase-shift transall the without, result is outgoing transmission with respect to the voltages of the incoming line A, B, C. The angle may be lead- THREE-PHASE TRANSFORMERS 257 Example 12-6 phase-shift A phase-shift transformer is designed transformer MVA on a 230 kV, 3-phase ± variable between zero and a. to control 50 1 The phase angle line. is 15°. Calculate the approximate basic power rating of the transformer. Calculate the line currents b. outgoing transmission Tap changer incoming and in the lines. Solution a. Figure 12.17a The basic power Phase-shift transformer. ST Note 0.025 S L a max = 56 The X 0.025 (12.1) X 150 15 MVA power rating is much that the transformer carries. feature of b. is = = that the power rating less than the This a is autotransformers. all line currents are the same both lines, be- in cause the voltages are the same. The line current is I Figure 12.17b Phasor diagram showing the range over which the phase angle of the outgoing line can be varied. SL = (150 = 377 Fig. 12,18a ing or lagging, and is usually variable between zero and ±20° The phase angle is sometimes varied in discrete means of a motorized tap-changer. The basic power rating of the transformer (which determines power its size) depends upon the ap- carried by the transmission line, and upon the phase shift. For angles less than 20°, it is given by the approximate formula ST = 0.025 S L a max leg. PN = basic power rating of the 3-phase 000) A an example of a 3-phase transformer with a tap brought out shift of, say, a, 3. S, 0.025 = The incoming line and the outgoing result Similarly, is £2 n ECN l is connected to terminals line to terminals 1, A, B, a£ s C £, N lags 20° behind £ AN 20° behind E HN and £ 1N lags that . , (Fig. 12.18c). principle of obtaining a phase shift connect two voltages in E ]b apparent power carried by the trans- mission line [VA] generated by phase A. The values of E,> N and B is is to series that are generated ated by phase an approximate coefficient A and 2, 3. by two different phases. Thus, voltage shift |°] has one The wind- transformer bank VA] a max = maximum transformer phase 20 ings of the three phases are interconnected as shown. [ = A terminal at a second winding having terminals The basic Sv X 230 The transformer has two windings on each 20° behind where (8.9) h 10 )/(V3 Thus, the leg associated with phase winding The (I2.l) is X could be used to obtain a phase degrees. steps by parent that N3E = connected in series selected so that the output voltage is gener- E PN E ]b are with equal to the in- put voltage while obtaining the desired phase angle _\-SS ELECTRICAL MACHINES AND TRANSFORMERS coming terminals are A, B, C; the outgoing terminals Figure 12.18c Phasor diagram are 1,2,3. a transformer that gives a phase- of shift of 20°. - E EPN = \.\4E = 0.40E In practice, the internal circuit phase-shift transformer However, it rests upon the The purpose of such transformers just discussed. will of a tap-changing, much more complex. basic principles we have is be covered in Chapter 25. 12.12 Calculations involving 3-phase transformers The behavior of a 3-phase transformer bank is calculated the same way as for a single-phase transformer. In making the calculations, we proceed as Figure 12.18b Schematic diagram follows: of the transformer in Fig. 12.18a. 1. We assume that the primary and secondary windings are both connected between them. In our particular example, line-to-neutral voltage of the incoming if E is the line, the re- spective voltages across the windings of phase A are in wye, even if they are not (see Section 8.14). This eliminates the problem of having to deal with delta-wye and delta-delta voltages and currents. THREE-PHASE TRANSFORMERS 2. We consider only one transformer (single phase) of this assumed 3. The primary voltage of former is coming 4. wye-wye transformer 259 Z T (pu) MP") = tO.115 bank. this hypothetical trans- the line-to-neutral voltage of the in- line. The secondary voltage of this transformer is the line-to-neutral voltage of the outgoing line. 5. The nominal power rating of this transformer is one-third the rating of the 3-phase transformer bank. 6. Figure 12.19 The load on transformer this is See Example one-third the 12-7. load on the transformer bank. This Example 12-7 The 3-phase step-up transformer shown 10.18 (Chapter 10) is rated 1300 impedance kV, 60 Hz, MVA, a. kV 11.5 percent. Fig. in It steps The power a S, power 810 therefore, j 0.115 The voltage = 81 0/3 = 270 1 2. 9. 1 MVA £, across the load is ter- HV side of the transformer MVA at 370 kV with a lagging minals when the delivers is, The equivalent circuit is shown in Fig. b. The power of the load per phase is circuit of this trans- Calculate the voltage across the generator almost entirely reactive. is impedance ZT (pu) = former, per phase. b. per-unit up line. Determine the equivalent a very large transformer; consequently, the 24.5 kV/345 the voltage of a generating station to 345 is transformer impedance £, The = 370 kV/V3 = 213.6W per-unit power of the load is factor of 0.90. MVA = 5 L (pu) = 270 MVA/433.3 0.6231 Solution a. First, we By note that the primary and second- ary winding connections are not specified. We don't need this information. However, we assume in selecting £, as the reference phasor, the per-unit voltage across the load that both £ L (pu) = windings are connected We this shall use the per-unit We problem. the secondary select the method EB = The to solve £B /.(pu) L F = 199.2 kV The power /u Ratio of transformation a = per-unit current in the load = is 345/24.5 = 5,(pu) 11 £ L (pu) is 345/V3 = is 0.6231 £L = 0.581 1300/3 = I 1.0723 is 0.9. Consequently. by an angle of arcos 0.90 = 25.84°. Consequently, the amplitude and phase of the 14.08 is given by rating of the transformer will be used as the base SB = factor of the load lags behind per-unit load current The nominal power kV 0 1.0723Z.0 nominal voltage of winding as our base voltage, The base voltage 213.6 kV/1 99.2 = wye. is power 5 B Thus, /, . 433.3 MVA The (pu) - per-unit voltage 0.581 1Z. -25.84° £ s (Fig. 12.19) is ELECTRICAL MACHINES AND TRANSFORMERS 260 E s (pu) = EJpu) + / L (pu) X ZT (pu) Hi = 1.0723^0° + (0.5811/1 -25.84°) X (0.1 15/190°) = 1.0723 + 0.0668Z.64.16 = 1.0723 + 0.0668(cos 64.16° j X1 0 + sin 64.16°) = 1.1014 = 1.103/13. 12° + j 0.0601 Therefore, E = s The 1.103 X 345 kV = per-unit voltage on the primary side Ep = The 381 kVZ.3.12° 1. is also 103/13. 12° effective voltage across the terminals of the generator is, therefore, £g = = = £p(pu) X £B 1.103 X 27.02 kV ( primary) 24.5 kV Figure 12.20 Polarity 12.13 Polarity marking marking of 3-phase transformers. of 3-phase transformers The HV terminals of a 3-phase transformer are marked H h H 2 H 3 and the LV terminals are marked X h X 2 X 3 The following rules have been stan- ondary dardized: are , . , 1 . If the wye-wye terminals on the or delta-delta, the voltages between similarly-marked terminals are in phase. Thus, is in phase with E xx £ HiH| is in phase with Exx EH is in phase with Ex £H H H x If the , HV side LV side. Thus, E HH leads £ X|Xi by 30° EH H| leads £ XiX| by 30° £H Hi leads Ex 30° x by Fig. 12.20 shows two ways of representing the delta-wye terminal markings. primary and secondary windings are con- nected internal connections and so on. and so on. 2. The so that the voltages on the always lead the voltages of similarly-marked primary windings and secondary wind- ings are connected line voltages. made in wye-delta or delta-wye, there results a 30° phase shift between the primary and sec- 3. These rules are not affected by the phase sequence of the line voltage applied to the primary side. THREE-PHASE TRANSFORMERS Questions and Problems 1 2-7 In order to 261 meet an emergency, three single-phase transformers rated Practical level 1 2- 1 that the transformer terminals H2 have polarity marks H,, X,, , X2 , make wye-delta on a 3-phase 18 a. What Delta-wye b. Open-delta 1 2-2 250 kVA, 7200 V/600 in V, at load 450 kVA, is In the a. 12-9 incoming and outgoing transmis- In the of 36 MVA, in Fig. 12.9 13.8 nominal currents ondary 12-4 kV/320 in the has a rating V Calculate the lines. in Fig. mode ates in the forced-air 225 current during the 600 to is 1, V. kV 6.9 2, and 3 Then, in a P are by mistake con- Determine the voltages measured between lines 1-2, 2-3, Draw the and 3-1. new phasor diagram. Three 1 50 kVA, 480 V/4000 V, 60 Hz sin- is if the primary line volt- kV and 12-11 when The exciting the transformers are The core loss in a 300 kVA distribution transformer the primary line is 3-phase estimated to be 0.003 pu. The copper losses are 150 A. in line. operating at no-load. 0.0015 pu. the transformer overloaded? If the time, and the cost of electricity 4.5 cents per kWh, is calculate the cost of the no-load operation in the course of connected? Calculate the line currents for a 600 load. the transformer operates effectively at no-load 50 percent of Problem 12-2 are V line to 7.2 kV. How must they be kVA load the bank current has a value of 0.02 pu. Calculate 600 c. balanced and equal on a 4000 V, 3-phase 10.19 oper- used to raise the voltage of a 3-phase b. maximum lines the line current The transformers a. the Industrial application Calculate the currents in the sec- ondary lines 12-6 is A-B-C b. morning peaks. Is What gle-phase transformers are to be installed The transformer shown b. 400 kVA. and the voltage between a. Intermediate level is 250 kVA, the transformers overloaded? ings of transformer 12-10 age at are connected in open- Referring to Figs. 12.3 and 12.4, the line Calculate the nominal currents in the pri- a. V nected in reverse. mary and secondary windings of the transformer shown in Fig. 10.18, knowing that 12-5 transformer bank? to the similar installation the secondary wind- primary and sec- the windings are connected in delta-wye. line. the outgoing line voltage? transformers rated Are kV load that can be voltage between phases primary and secondary windings The transformer maximum can carry on a continuous basis? is 12-3 is kV/600 b. sion lines b. Two calculate the following currents: the delta to supply a load of 60 Hz, are con- wye-delta on a 12 470 V, 3-phase line. If the a. 2-8 2.4 Three single-phase transformers rated nected What b. a. is connected nections: are connected in schematic drawings of the following con- 1 kV 100 kVA, 13.2 kV/2.4 Assuming 12-12 one year. The bulletin of a transformer manufacturer kVA, 230 V/208 Calculate the corresponding primary indicates that a 150 and secondary currents. 60 Hz, 3-phase autotransformer weighs V, 262 ELECTRICAL MACHINES AND TRANSFORMERS 310 lb, whereas a standard 3-phase trans- 12-14 former having the same rating weighs 1220 12-13 1b. Why 1 kVA, 480 V/l 20 5 in delta to formers on a 600 H2 X h X2 , V, source. 60 Hz are con- 12-15 Then can be drawn from the 600 calculate the maximum that the autotransformer can carry. You wish to operate a 40 hp, 460 V, 3-phase motor from a 600 V, 3-phase sup- V ply. The full-load current of the motor 42 A. Three 5 kVA, 20 V/480 V, 3-phase line. The H,, marks appear on the is 1 single-phase transformers are available. the transformers should be How would you connect them? Are they connected. able to furnish the load current b. Determine the 3-phase voltage output the c. Determine the phase of the transformer. shift between the 3-phase voltage output and the 600 V, 3-phase input. V load function as autotrans- polarity Show how (kVA) at metal housing. a. Problem 12-13 calculate the maximum line current that this difference 9 Three single-phase transformers rated nected In motor without overheating? drawn by Chapter 13 Three-Phase Induction Motors that ranges 13.0 Introduction Three-phase induction motors are the motors most frequently encountered and easy are simple, rugged, low-priced, They run tain. to constant speed from The speed frequency-dependent consequently, these is motors are not supports main- at essentially zero to full-load. and, They in industry. to control the the 3-phase induction its We the basic principles of its behavior. general construction and the We way use two types of rotor windings: (1) conven- 3-phase windings made of insulated wire and then cage induction motors (also called cage motors) the and wound-rotor induction motors. A squirrel-cage is composed of bare cop- into the slots. The opposite ends are welded two copper end-rings, so that all the bars are short-circuited together. The entire construction pushed thousand horsepower permit the reader to see that same rotor per bars, slightly longer than the rotor, which are motors ranging from a few horsepower to several operate on the The type of winding gives rise to two main classes of motors: squirrel- Squirrel-cage, wound-rotor, and linear induction all circumference of the lam- (2) squirrel-cage windings. windings are made. they A number of evenly spaced slots, internal of rotor slots to provide space for the rotor winding. motor and develop the funda- mental equations describing discuss that tions. speed of commercial induction we cover frame made up of The rotor is also composed of punched laminaThese are carefully stacked to create a series fre- tional chapter steel core inations, provide the space for the stator winding. motors. In this hollow, cylindrical stacked laminations. quency electronic drives are being used more and more a punched out of the easily adapted to speed control. However, variable mm to 4 mm, depending on the from 0.4 power of the motor. The stator (Fig. 13.2) consists of a to basic principles. (bars and end-rings) resembles a squirrel cage, from which the name Principal 13.1 A 3-phase components induction motor (Fig. parts: a stationary stator rotor is and 1 3. 1 ) small and in- aluminum, molded The show progressive 13.3a). Figs. to form an 13.3b and 13.3c stages in the manufacture of a squirrel-cage motor. 263 In are die-cast tegral block (Fig. a revolving rotor. derived. made of has two main separated from the stator by a small air gap is medium-size motors, the bars and end-rings ELECTRICAL MACHINES AND TRANSFORMERS 264 A wound rotor has a 3-phase winding, similar the one on the stator. The winding tributed in the slots and is is uniformly usually connected in 3- wire wye. The terminals are connected to three rings, which revolving turn with the rotor (Fig. slip-rings and 13.4). associated slip- The stationary brushes enable us to connect external resistors ries to dis- in se- with the rotor winding. The external resistors are mainly used during the start-up period; under normal running conditions, the three brushes are short-circuited. 13.2 Principle of operation Figure 13.1 Super-E, premium efficiency induction motor rated 10 hp, 1760 r/min, 460 V, 3-phase, 60 Hz. This to- tally-enclosed fan-cooled motor has a full-load current of tor of 1 2.7 A, efficiency of 91 .7%, and 81%. Other power fac- characteristics: no-load current: 5 A; lockedrotor current: 85 A; locked rotor torque: breakdown torque: 3.3 pu; service factor 90 kg; over-all length including shaft: 491 mm; overall height: 279 mm. (Courtesy of Baldor Electric Company) 2.2 pu; 1 .15; total weight: The operation of a 3-phase induction motor is based upon the application of Faraday's Law and the Lorentz force on a conductor (Sections 2.20, 2.2 1, and 2.22). The behavior can readily be understood by means of the following example. Consider a series of conductors of length /, whose extremities are short-circuited by two bars A and B (Fig. 3.5a). A permanent magnet placed above this conducting ladder, moves rapidly to the 1 right at a speed r, so that across the conductors. its magnetic field B sweeps The following sequence of events then takes place: Figure 13.2 Exploded view of the cage motor of Fig. 13.1, showing the stator, rotor, end-bells, cooling and terminal box. The fan blows air over the stator frame, which is ribbed to improve heat (Courtesy of Baldor Electric Company) fan, ball bearings, transfer. THREE-PHASE INDUCTION MOTORS moving magnet field is replaced by a rotating windings, as in the stator 13.3 is produced by the 3-phase currents The we now will 265 field. The flow that explain. rotating field Consider a simple stator having 6 salient poles, each of which carries a coil having 5 turns (Fig. 1 3.6). Coils that are diametrically opposite are connected in series by means of three jumpers connect terminals a-a, b-b, and that respectively c-c. This creates AN, BN, CN, that are mechanically spaced at 120° to each other. The three identical sets of windings Figure 13.3a Die-cast aluminum squirrel-cage rotor with integral cooling fan. (Courtesy of Lab-Volt) 1 A . E— voltage while it Blv is induced in each conductor being cut by the flux (Faraday's law). is The induced voltage immediately produces a /, which flows down the conductor un- 2. current derneath the pole-face, through the end-bars, and back through the other conductors. Because the current-carrying conductor 3. the magnetic field of the lies in permanent magnet, it experiences a mechanical force (Lorentz force). 4. The force always acts in a direction to drag the conductor along with the magnetic field (Section 2.23). If the celerate move, conducting ladder is toward the However, as right. free to it it will ac- picks up speed, the conductors will be cut less rapidly by the moving magnet, with age E and the current the force acting If the ladder magnetic the result that the induced volt/ will diminish. Consequently, on the conductors were field, the to move at the will also decrease. same speed and the force dragging the ladder along would come as the induced voltage E, the current all /, be- zero. In an itself to Figure 13.3b (1), induction motor the ladder is form I a squirrel-cage (Fig. closed upon 3.5b) and the in the manufacture of stator and Sheet steel is sheared to size punched (3), blanked (4), and Progressive steps rotor laminations. blanked punched (2), (5). (Courtesy of Lab-Volt) ELECTRICAL MACHINES AND TRANSFORMERS 266 upper Figure 13.3c Progressive steps a. in Molten aluminum the injection molding of a squirrel-cage is poured into a cylindrical cavity. rotor. The laminated rotor stacking is firmly held between two molds. b. Compressed rams the mold assembly into the cavity. Molten aluminum is forced upward through the and into the upper mold. Compressed air withdraws the mold assembly, now completely filled with hot (but hardened) aluminum. The upper and lower molds are pulled away, revealing the die-cast rotor. The cross section view shows that the upper and lower end-rings are joined by the rotor bars. (Lab-Volt) air rotor bar holes c. d. two coils in each winding produce magnetomotive forces that act in the The same three sets of windings are connected in thus forming a perfectly common symmetrical always flow neutral N. Owing arrangement, the wye, to the line-to- bers, suppose that the peak current per phase Thus, when /a = +7 pere-turns balanced 3-phase system. Because the current ings. The displaced /.„ /h , and /c will currents will have the in flow In sume B, C, flux. It is this flux we same value but in turn, are interested will be create a we corresponding a as- by the arrows) A will is 1 0 turns = 70 amof value positive, the flux is flux. directed upward, according to the right-hand As time goes by, we rule. can determine the instanta- neous value and direction of the current each in winding and thereby establish the successive flux patterns. Thus, referring to Fig. rent /a has a value of in. order to follow the sequence of events, that positive currents (indicated vertically the wind- time by an angle of 120°. These currents produce magnetomotive forces which, magnetic in and 10 A. is A, the two coils of phase mmf of 7 A X together produce an we connect a 3-phase source to terminals A, to Furthermore, to enable us to work with num- neutral impedances are identical. In other words, as If line to neutral. line. regards terminals A, B, C, the windings constitute a alternating currents windings from in the Conversely, negative currents flow from neutral direction. A X 1 0 turns 1 3.7 at instant 10 A, whereas fb and 1 /c , cur- both —5 A. The mmf of phase A is = 00 ampere-turns, while the mmf have a value of 10 + 1 Figure 13.4b of the slip-ring end of the rotor. (Courtesy of Brook Crompton Parkinson Ltd) Close-up 267 ELECTRICAL MACHINES AND TRANSFORMERS 268 length / Figure 13.5a Moving magnet cutting across a conducting ladder. ength Figure 13.5b Ladder bent upon of phases B and C rection of the B / form a squirrel-cage. itself to are each 50 ampere-turns. mmf depends upon current flows and, using the right-hand rule, that the direction of the resulting shown in Fig. I3.8a. Note The di- the instantaneous magnetic we find is as that as far as the rotor is field concerned, the six salient poles together produce a magnetic having essentially one broad north field pole and one broad south pole. This means that the 6-pole stator actually produces a 2-pole field. combined magnetic At instant tains a 2, peak of field points one-sixth cycle — 10 A, while /a The upward. later, and current /b / L at, both have a value of +5 A new has the same shape as before, except that it field has (Fig. 1 3.8b). moved clockwise by an We discover that the angle of 60°. In other words, the flux makes 1/6 of a turn between instants l and cycle, we find that the magnetic plete turn during field one cycle (see Figs. makes one com1 3.8a to I3.8f). The rotational speed of the field depends, therefore, upon the duration of one cycle, which in turn depends on the frequency of the source. 2. Proceeding Figure 13.6 Elementary stator having terminals A, B, C connected to a 3-phase source (not shown). Currents flowing from line to neutral are considered to be positive. in this instants 3, 4, 5, 6, and way 7, for each of the successive quency is 60 Hz, the resulting separated by intervals of 1/6 in 1/60 s, that is, field If the fre- makes one turn 3600 revolutions per minute. On Figure 13.7 Instantaneous values of currents and position of the flux 269 in Fig. 13.6. ELECTRICAL MACHINES AND TRANSFORMERS 270 A A g O I'- N N Figure 13.8d Figure 13.8c Flux pattern at instant Flux pattern at instant 3. A 4. A g N Figure 13.8e Figure 13.8f Flux pattern at instant 5. Flux pattern at instant the other hand, if the frequency were 5 Hz, the field would make one turn in 1/5 s, giving a speed of only 300 r/min. Because the speed of the rotating field is change any two of the the new phase sequence necessarily synchronized with the frequency of the the same we find that the field source, it is called synchronous speed. produces a speed field that rotates clockwise. If line positive crests of the currents in Fig. in the we connected to the will be interstator, A-C-B. By following of reasoning developed in Section in the opposite, therefore, reverse The lines now revolves at 1 3.7 follow order A-B-C. This phase sequence its 3.3, or counterclockwise direction. will, direction of rotation. Although early machines were poles, the stators of 1 synchronous Interchanging any two lines of a 3-phase motor 13.4 Direction of rotation each other 6. built with salient modern motors have internal di- THREE-PHASE INDUCTION MOTORS ameters that are smooth. Thus, the salient-pole stator of Fig. 13.6 as shown is 1 3.6, the are replaced Note that two each coils of phase A (Aa and An) slots coil in Fig. coil pitch is covers more duces more flux per turn, terminal shown A to the neutral coils of phases 1 80° of the circum- A current only 60°. because efficient /a pro- it flowing from N yields the flux distribution are identical to those Fig. 120° to each other. The netic field at due 1 3.9b, they are resulting mag- to all three phases again consists of In practice, instead shown in Fig. two, three or more of using a single coil per pole 13.9a, the coils coil lodged in is subdivided into adjacent slots. The staggered coils are connected in series and constitute what known as a phase group. Spreading the coil way over two or more slots tends to create a is in this sinusoidal flux distribution per pole, the in 5 suc- in Fig. 13.20. of poles Soon after the invention of the induction motor, it was found that the speed of the revolving flux could be reduced by increasing the number of poles. To construct a 4-pole stator, the coils are distributed as shown in Fig. I3.l0a. The four identical A now span only 90° of the stator cir- cumference. The groups are connected in such a way that adjacent tomotive forces acting when other words, winding of phase two poles. as be placed in series to synchronous speed groups of phase B and C of phase A and, as can be seen in displaced Number 13.5 13.9a. in the figure. The shown is on the inner surface of the ference whereas the coils in Fig. 13.6 cover The 180° gered coils connected cessive slots and 13.24a. two by the two coils shown are lodged in stator. replaced by a smooth stator such in Figs. 13.2 In Fig. They now 271 which improves performance of the motor and makes it less noisy. A phase group (or simply group) composed of 5 stag- A in in series and groups produce magneopposite directions. In a current flows /,, (Fig. 13. 10a), it in the stator creates four al- N-S poles. The windings of the other two phases ternate cal but are displaced are identi- from each other (and from When phase A) by a mechanical angle of 60°. wye-connected windings are connected source, a revolving field having four poles (Fig. 13. 10b). This field rotates speed of the 2-pole will shortly explain field why shown at the to a 3-phase is created only half the in Fig. 13.9b. We this is so. group phase Figure 13.9a Phase group 1 is composed of a single coil lodged two slots. Phase group 2 is identical to Phase group 1 The two coils are connected in series. In practice, a phase group usually consists of two or more staggered coils. in . (/ c 1 C = - 5 A) Figure 13.9b Two-pole, magnetic A and / b full-pitch, = /c lap-wound stator and resulting when the = -5 A. field current in phase A = +10 ELECTRICAL MACHINES AND TRANSFORMERS 272 phase group group 1 group 1 phase B 1 Figure 13.10a The phase groups four pole magnetic group of phase A produce a 4- field. 1 group 1 -5 A. rent flow in the three phases, let us restrict our at- tention to phase A. In Fig. 13.11 each phase group covers a mechanical angle of 360/8 = 45°. Suppose the current in phase A is at its maximum positive value. The magnetic flux is then centered on phase A, and the N-S poles are located as shown in Fig. 13.12a. One-half cycle later, the current in phase tive value. Figure 13.10b fore, Four-pole, full-pitch, lap-wound stator magnetic field when /a = +10 A and and /b = The A will reach its maximum except that all N the poles will resulting lc = -5 A. S poles and vice versa (Fig. 13.12b). In comparing the two figures, it is netic field has shifted clear that the entire by an angle of 45° this gives us the clue to finding the We we can increase the number of poles as much as please provided there are enough slots. Thus, Fig. 13. II shows a 3-phase, 8-pole stator. phase consists of 8 groups, and the groups of phases together produce an 8-pole rotating When connected to a 60 like the Hz spokes of a wheel, Each all the field. source, the poles turn, at a synchronous speed of 900r/min. How will can we tell what the synchronous speed be? Without going into all nega- same as behave become flux pattern will be the the details of cur- tion. cles mag- — and speed of rota- The flux moves 45° and so it takes 8 half-cy(= 4 cycles) to make a complete turn. On a 60 Hz system the time fore 4 X 1/60 = 1/15 to s. make one turn is there- Consequently, the flux r/s or 900 r/min. The speed of a rotating field depends therefore upon the frequency of the source and the number of poles on the stator. Using the same reasoning as turns at the rate of 15 above, we can prove that the synchronous speed always given by the expression is THREE-PHASE INDUCTION MOTORS phase group phase group 1 maximum 1 Figure 13.12b Figure 13.12a Flux pattern 273 when the current in phase A is at when Flux pattern its maximum positive value. in Fig. the current negative value. 13.12a but it The phase A in pattern is is the at its same as has advanced by one pole pitch. ing field created by the stator cuts across the rotor bars and induces a voltage in P This where is cut, in rapid succession, = synchronous ns /= speed fr/min] the creases with frequency and decreases with the N is pole followed by a number of N and S poles that sweep across always equal is at rest, it a is frequency of the source. to the in- Because the rotor bars are short-circuited by the num- end-rings, the induced voltage causes a large cur- ber of poles. rent to flow — usually bar in machines of Example 13-1 duction motor having 20 poles when it is connected several hundred amperes per medium power. The current-carrying conductors Calculate the synchronous speed of a 3-phase in- Hz by a conductor per second; when the rotor This equation shows that the synchronous speed a 50 of them. S pole. The frequency of the voltage depends upon frequency of the source [Hz] p = number of poles to all an ac voltage because each conductor the flux created by the stator, are in the path of consequently, they all experience a strong mechanical force. These forces source. tend to drag the rotor along with the revolving Solution In ns = 120///7 = 120 X = 300 r/min 50/20 1 . field. summary: A revolving magnetic field 3-phase voltage is is set up when a applied to the stator of an induction motor. 13.6 Starting characteristics of a squirrel-cage 2. motor Let us connect the stator of an induction motor The revolving field induces a voltage in the ro- tor bars. to a 3-phase source, with the rotor locked. The revolv- 3. The induced voltage rents which flow creates large circulating cur- in the rotor bars and end-rinizs. ELECTRICAL MACHINES AND TRANSFORMERS 274 The current-carrying 4. magnetic the 5. rotor bars are by the field created immersed in they are stator; equal to the load torque. therefore subjected to a strong mechanical force. constant The sum of the mechanical forces on motor only turns which tends tor bars produces a torque the rotor along in the volving same the ro- all to drag direction as the re- at —slip load. The moment this state of equilibrium as the rotor released, is it rapidly acceler- ates in the direction of the rotating field. up speed, As picks it the relative velocity of the field with re- spect to the rotor diminishes progressively. This causes both the value and the frequency of the duced voltage to decrease because the rotor bars are more slowly. The cut first, in- will continue to increase, but never catch up with the revolving it rotor bars field. In effect, if same speed as the field (synflux would no longer cut the and the induced voltage and current fall to to produce a current overcome tor bars sufficiently large to usually less than machines That ( motors field (called slip), 0.1% of synchronous in speed is small: motor will begin to slow is initially down and at greater motor torque. running at no-load. If the revolving field will rate. The in- at synchronous speed, they sometimes called asynchronous machines. never actually turn and 13.9 Slip The state are speed slip slip s of an induction motor is the difference be- tween the synchronous speed and the rotor speed, expressed as a percent (or per-unit) of synchronous speed. The per-unit slip = = — s ns n The given by the equation is motor comes slip synchronous speed [r/minl rotor speed fr/minj zero slip is practically equal to l 100%) when (or no-load and at the rotor is is locked. . motor hp, 6-pole induction phase, 60 Hz source. excited by a 3- is the full-load speed If is 1 140 r/min, calculate the slip. Solution The synchronous speed of ns producing a greater and The question is, for how long can this go on? Will the speed continue to drop un- the seldom exceeds 5%. the motor is resulting current in the bars will increase progressively, No; it constant speed machines. However, because they A 0.5 a higher and higher duced voltage and the the and more) rarely speed. apply a mechanical load to the shaft, the motor cut the rotor bars til ). induction motors are considered to be Example 13-2 the kW 000 (1 kW and less), 10 why is the ro- in 13.8 Motor under load we 1 motors run very loads, induction exceeds 0.5% of synchronous speed, and for small the braking At no-load the percent difference between the rotor and Suppose 1 zero. chronous speed so as torque. upset, the is 3. close to synchronous speed. Thus, at full-load, the Under these conditions the force acting on the rotor bars would also become zero and the friction and windage would immediately cause the rotor to slow down. The rotor speed is always slightly less than synwould is will the rotor did turn at the chronous speed), the torque its rotor current, very large at decreases rapidly as the motor picks up speed. The speed change (Section will start to slip for large As soon when constant speed exactly equal to the torque exerted by the mechanical Under normal 13.7 Acceleration of the rotor a at very important to understand that a rate. It is motor speed field. When this state is reached, the speed will cease to drop and the motor will turn The = = \20flp = 120 X 60/6 (13.1) 1200 r/min difference between the synchronous speed of to a halt? motor and the mechanical load the revolving flux and rotor speed is the slip speed: will reach a of equilibrium when the motor torque is exactly ns - n = 1200 - 1140 = 60 r/min THREE-PHASE INDUCTION MOTORS The slip is c. s = (« s = 0.05 or - 500 r/min at 60/1200 (13.2) d. 5% Motor turning 2000 r/min at From Example The voltage and frequency induced in the rotor both slip. They are given by the follow- depend upon the a. At K- = s (13.3) sf (approx.) = motor speed n 0. slip is = n)/n s - (1200 0)/1200 the induced current) h in b. When = = 4' X 60 = 60 Hz 1 motor turns the same in the motor speed n field, the frequency of the source connected to s the stator [Hz] = E2 = Eoc = slip at rest would be induced bars were disconnected from the £oc is c. When = tf= the = direction as the positive. - (1200 the end-rings. In n in the motor speed = —500. The - (-500)1/1200 should be noted that Eq. 13.3 always holds (1200 + 500)/ 1200 but Eq. 13.4 = 1.417 valid only if the revolving flux (expressed in webers) remains absolutely constant. A slip However, between zero and full-load the actual value of E2 is Hz negative; thus, is [1200 — 35 slip is = is 1 opposite direction = /V 3 times the voltage between the open- circuit slip-rings. It 500)/ 200 X 60 = 0.583 (n s 1 slip is is — is The 0.583 motor turns to the field, the s voltage true, = n)/n s in the rotor bars wound-rotor motor the open-circuit the case of a 700/1200 fi the voltage that if = [V] cage motor, the open-circuit voltage In a (« s the rotor current) open-circuit voltage induced in the ro- when - = is The frequency of the induced voltage (and of voltage induced in the rotor at slip s tor 1 is the rotor [Hz] s = The frequency of the induced voltage (and of (13.4) frequency of the voltage and current synchronous speed of the 13-2, the standstill the where /= direc- 1200 r/min. is Consequently, the ing equations: f2 = same Solution motor E 2 = sEoc in the tion as the revolving field 13.10 Voltage and frequency induced in the rotor f2 = opposite in the direction to the revolving field = n)/n, Motor turning 215 /?)//? s greater than = 1700/1200 implies that the motor 1 is operating as a brake. only slightly less than the value given The frequency of the induced voltage and by the equation. current f2 = Example 13-3 rotor is f= s X 60 = 1.417 85 Hz The 6-pole wound-rotor induction motor of Example 13-2 the is excited by a 3-phase 60 Hz d. source. Calculate frequency of the rotor current under the follow- ing conditions: The motor speed turns in the same n = +2000. The s a. At b. Motor turning 500 r/min tion as the revolving field in the same direc- positive because the rotor direction as the field: slip is = (n s = (1200 standstill at is - n)ln s - 2000)/ 1200 = -800/1200 = -0.667 ELECTRICAL MACHINES AND TRANSFORMERS 276 A negative slip implies that the motor actually is 1. operating as a generator. The frequency of current the induced voltage means negative frequency reversed. Thus, is in the rotor wind- stator quency is mutual flux is can say that the frequency is a transformer (Fig. 1 lists I kW simply 40 Hz. no-load therefore low; is machines efficiency that the current rents are is the full-load current is compared to it. is to it. and all Finally, the base it to 0.05 for large the is is zero. under load, mmf which tends m This sets up an opposthe stator. The opposing mmfs of change the mutual flux 4> . in a transformer. are created, in The total m power needed to produce these three fluxes is slightly greater than when the motor is operating at no-load. However, the active power (kW) absorbed addition to the mutual flux <f> (Fig. 13.14). reactive the syn- is to create within ac- machines. The motor mmfs of the secondary and primary As a result, leakage fluxes <J> n and other torques are speed it the rotor and stator are very similar to the opposing other cur- all 3). needed zero because the output power ing current flow in and Similarly, the base torque the full-load torque and compared to The base torque are expressed in per-unit values. current 1 links both the similar to the is Motor under load. When 2. power range between in the and 20 000 kW. Note m chanical tolerances will permit. the typical properties of squirrel- cage induction motors 3. is the current in the rotor produces a Table I3A <E> it is made as short as meThe power factor at ranges from 0.2 (or 20%) ceptable limits, the air gap we Characteristics of squirrelcage induction motors 1 in flux the revolving field and, in order to keep for small 13.1 The consequently rotor; Considerable reactive power concerned, a negative frequency gives the same reading as a positive frequency. Consequently, and the that sup- friction losses in the rotor plus the iron losses in the stator. the phase sequence itive, windage and plies the A-B-C when the frequency is posis A-C-B when the frenegative. As far as a frequency meter is rotor voltages component ing flux 3> m and a small active the phase sequence of the if (of of a magnetizing component that creates the revolv- phase se- that the the motor runs at no between 0.5 and 0.3 pu full-load current). is quence of the voltages induced ings When at no-load. The no-load current is similar to the exciting current in a transformer. Thus, it is composed and rotor f2 = sf= -0.667 X 60 = -40 Hz A Motor load, the stator current lies chronous speed of the motor. The following explanations will clarify the meaning of the values given by the motor increases in in the table. the mechanical load. follows that the power factor TABLE 13A size — No-load * Small Current Torque Slip (per-unit) (per-unit) (per-unit) means under factor Big* Small Big Small Big Small Big Small Big l 1 1 1 0.03 0.004 0.7 0.96 0.8 0.87 to to to to 0.9 0.98 0.85 0.9 -0 0 0 0.2 0.05 1 0 0 0.4 0.1 0.5 rotor Power Efficiency Small* Full-load Locked almost direct proportion TYPICAL CHARACTERISTICS OF SQUIRREL-CAGE INDUCTION MOTORS Loading Motor It 0 0.3 0 5 4 to to to to 6 6 3 1 11 kW (15 hp); big 1 .5 0.5 means over 1 1 120 kW (1500 hp) and up to 25 000 hp. to THREE-PHASE INDUCTION MOTORS Figure 13.13 Figure 13.14 At no-load the flux tual flux tive (t> power the motor in m To create . is this flux, is mainly the mu- At full-load the mutual flux decreases, but stator considerable reac- needed. of the motor improves dramatically as the mechanical load increases. At full-load it 98% 3. and rotor leakage fluxes are created. The reactive power needed is slightly greater than in Fig. 13.13. where ranges from 0.80 for small machines to 0.90 for large machines. ciency 211 at full-load is particularly high; it The / = Ph = E= 600 = effi- can attain for very large machines. full-load current [A] output power [horsepower) rated line voltage (V) empirical constant Locked-rotor characteristics. The locked-rotor current the I The 2 is R 5 to 6 times the full-load current, losses 25 to rotor 36 times higher than normal. must therefore never remain locked for more than a few seconds. Although the mechanical power is needed Recalling that the starting current that the no-load current we can 0.3 pu, lies is 5 to 6 pu and between 0.5 and readily estimate the value of these currents for any induction motor. at standstill is motor develops a strong torque. The power zero, the factor making low because considerable reactive power to produce the leakage flux stator windings. and Example 13-4 a. 3-phase induction motor having a rating of because the stator and 500 hp, 2300 windings are not as tightly coupled (see Section 10.2). Calculate the approximate full-load current, locked-rotor current, and no-load current of a These leakage fluxes are much larger than in a transformer the rotor in the rotor is V. b. Estimate the apparent power drawn under c. State the nominal rating of this motor, locked-rotor conditions. expressed 13.12 Estimating the currents in an induction motor Solution a. The full-load current of a 3-phase induction may be motor means of the following ap- = 600 P h /E (13.5) calculated by proximate equation: I in kilowatts. The full-load current / is = 600 P h /E - 600 X 500/2300 = 130 A (approx.) (13.5) ELECTRICAL MACHINES AND TRANSFORMERS 278 The no-load current = /0 = 0.3/ X 0.3 1 energy 30 39 A starting current = / LR (approx.) 6/ - X 6 The remaining the watts, it and not of this fol- power stator. f js is owing active power P v gap and transferred Due dissi- windings. Another portion dissipated as heat in the stator core, P Due 00 kVA power of a motor always (8.9) is relates to the motor expressed in expressed car- is to the rotor jr to the PR losses in the rotor, a third portion dissipated as heat, and the remainder in kilo- The nominal SI units is is, rating therefore, windage and bearing-friction tain PL , the mechanical is finally power P m By . f v representing losses, we finally ob- subtracting a small fourth portion mechanical output to the electrical input. P available in the form of mechanical (approx.) , power available at the shaft to drive the load. The power flow diagram of Fig. 13.15 enables P = 500/1.34 = 373 kW (see Power Appendix AX0) to identify conversion chart in ties Voltages, currents, and phasor diagrams enable us understand the detailed behavior of an induction Figure 13.15 Active power flow in a 3-phase induction motor. and Efficiency. motor is us to calculate three important proper- of the induction motor: (1) power, and (3) L 13.13 Active power flow to is { to the by electromagnetic induction. EI = V3 X 2300 X 780 When electrical flows through the ma- in the iron losses. 1 it pated as heat ried across the air 3 how copper losses, a portion 130 is = easier to see to the stator tions = V3 is flows from the line into the 3-phase The apparent power under locked-rotor condi- S c. it converted into mechanical energy by chine. Thus, referring to Fig. 13.15, active Pc is = 780 A (approx.) b. is lowing the active power as = The motor. However, is its its efficiency, (2) its torque. By definition, the efficiency of a the ratio of the output power to the input power: efficiency Cn) - PJP e (13.6) THREE-PHASE INDUCTION MOTORS R losses in the rotor. It can rotor rR losses P-. are related 2. / power P T by Motor be shown * that the 4. to the rotor input motor „ the equation The torque Tm developed by torque. any speed at 9.55 279 is the given by Pm (3.5) Pjv = sP n (13.7) T 9.55 where rotor ]V — s PR losses [W] P r 9.55 transmitted to the rotor T Equation 13.7 shows that as the PR A nous speed = (s slip increases, the 0.5) dissipates in the locked is ted to the rotor is (s = 1), all /n s it the (13.9) s P power transmitted r "s 9.55 When = = power transmit- The Mechanical power. The mechanical power P m is equal to the power transmitted to the rotor minus its PR losses. Thus, developed by the motor any [W] to the rotor synchronous speed [r/minj multiplier to take care of units [exact value: 60/2tt| dissipated as heat. 3. at speed [N-ml form of heat receives. PJn torque developed by the motor rotor turning at half synchro- 50 percent of the active power r where consume a larger and larger propower P r transmitted across the air to the rotor. the rotor [ W] losses portion of the P therefore. slip P = power gap -v) s) 9.55 P = rotor (1 « s (l actual torque 7, slightly less than overcome the available Tm due , windage and at the shaft friction losses. most calculations we can neglect in is to the torque required to However, this small difference. Pm = P 1 1 = P r Equation 13.9 shows that the torque J'* sP r r (13.7) proportional to the active tor. whence The actual mechanical the load is needed to losses. In (13.8) power available P m due to drive power overcome the windage and friction most calculations we can neglect this slightly less than , must absorb a large amount of active power. The form of latter is dissipated in the to the small loss. mechanical electromagnetic power heat, consequently, the temperature of the rotor rises very rapidly. Example 13-5 A 3-phase induction motor having a synchronous speed of 1200 r/min draws 80 power output directly to the ro- Thus, to develop a high locked-rotor torque, the rotor s)P T is power transmitted kW from a 3-phase electrical X speed of flux losses transferred electromagnetic torque p, of rotor to rotor Pm - P 1 9.55 in rotor i 9.55 but from Eq. 3.5 _ (iii) p, (i) 1 rotor speed Pm ~ Tm must but the mechanical torque X mechanical torque the electromagnetic torque 9.55 Timc equal . Thus Hence, rm = Tmia _ nT^ Also from Eq. 3.5 we can write tn) Substituting Py = (ii), (iii), SP, (iv) and (iv) in (i). we find ELECTRICAL MACHINES AND TRANSFORMERS 280 feeder. tor The copper losses and iron losses in the stato 5 kW. If the motor runs at 52 r/min, amount 1 e. The efficiency is = PJPC = 1 T) calculate the following: 70/80 0.875 or 87.5% a. b. c. d. The active power transmitted to the rotor The rotor ER losses The mechanical power developed The mechanical power delivered to the load, knowing that the windage and friction losses are equal to 2 e. kW The efficiency of Example 13-6 A 3-phase, 8-pole squirrel-cage induction motor, connected line, possesses a synchronous and iron losses the copper motor the 60 Hz to a speed of 900 r/min. The motor absorbs 40 kW, and 5 kW and 1 kW, in the stator amount to respectively. Calculate the torque developed by the motor. Solution a. Active power to the rotor Solution is The power transmitted across the = b. The - 80 = 5 75 kW Pr slip is = s = (n s - Tm = n)/n s - (1200 Rotor I R Note torque) ]v c. v X 75 = 3 The mechanical power developed P^n d. 0.04 = P\ = 75 ~ - 2 I 3 kW windage rotor P 9.55 X = 361 N-m T P>„ - P, P mt due = is 72 13-7. this problem (the at a standstill or running at power P transmitted v full to the equal to 34 kW, the motor develops a torque Example 13-7 to the load to the friction 2 to independent of the speed of rotation. The and A 3-phase induction motor having a nominal rating of 100 hp (-75 = 70 kW is kW) and a synchronous speed of 1800 connected to a 600 V source (Fig. 13.16a). The two-wattmeter method shows a HP(^75kW) 1783 r/min Figure 13.16a (13.9) 34 000/900 solution the that is 100 See Example kW fn s losses. = is of 361 N-m. r/min Pi 1 9.55 motor could be rotor P losses in - 72 kW x slightly less than 5 speed, but as long as the is The mechanical power P delivered is to the rotor 0.04 losses are P = sP = ~ Pi - = 34 gap Pjs = 52)/ 1200 11 = 48/1200 2 = Pc ~ = 40 - air total power con- THREE-PHASE INDUCTION MOTORS sumption of 70 kW, and an ammeter indicates a measurements give current of 78 A, Precise line b. The slip is a rotor s known about stator iron losses windage and P = between two resistance kW 2 r = (1800 = 0.0205 the motor: friction losses Px = .2 1 kW stator terminals Rotor = 0.34 (2 2 J R b. Rotor I~R losses c. Mechanical power supplied Mechanical power developed = to the load, in Py ~ P)v = e. Torque developed Solution Power supplied to the stator P c = 70 is d. kW R 0.34/2 wye con- e. Torque 63.5 kW = 62.3 kW = = 83.5 hp - /\ 63.5 62.3 - X 1.2 1.34 (hp) is 62.3/70 - 0.89 or 89% 1763 r/min: at = T0.17 ft 9.55 R = = 3.1 kW P = 2 ]s Iron losses { Power supplied = Pc = (70 - X 3 2 (78) X 0.17 y in s = 9.55 X 64 900/1800 The torque developed by to the rotor: ^js 3.1 - summarized a Pi' - in motor depends upon speed, but the relationship between the 2) Figure 13.16b Example calculations are Fig. 13.16b. 13.14 Torque versus speed curve kW 13-7. its two cannot be expressed by a simple equation. Consequently, = 64.9 3.1 in P = 344 N m The above 2 3 I Power flow - = PJP e - = 1.33 losses are P = Pr is is R = 2 kW to the load: Efficiency of the motor T] Stator resistance per phase (assume a PL 63.5 = 1763 r/min at - "4.9 Mechanical power P, Efficiency Stator f 1.33 to the rotor d. nection) = 64.9 r horsepower a. 763)/ 1800 P = sP = 0.0205 X c. Power supplied 1 losses: ]r Calculate a. n)fn s (n s speed of 1763 r/min. In addition, the following characteristics are 281 kW kW we 2kW prefer to show the relationship in the form of a ELECTRICAL MACHINES AND TRANSFORMERS 282 curve. Fig. 13.17 shows inal full-load torque T and 1.5 torque) the is is following example illustrates the changes that occur. starting torque is the it breakdown torque. motor runs at a speed is rings. is Figure accelerat- ing from rest to the 10 n. If motor torque torque. As soon motor will as the two torques breakdown if torque), the motor Small motors (15 hp and breakdown torque at 20 breakdown torque ( at develop of about /z d 1 98% that it from no-load to full-load, N-m (-73.7 fHbf). The full-load current is is 100 A. The ro- This can be achieved by using a material of in Figure 1 3. 1 8b. It can be seen that the is start- ing torque doubles and the locked-rotor current de- 00 A to 90 A. The motor develops its breakdown torque at a speed /Vd of 500 r/min, compared to the original breakdown speed of 800 r/min. creases from If we becomes imum es- is motor having a syn- and the locked-rotor current shown of syn- rotor resistance of a squirrel-cage rotor V and end-rings. The new torque-speed curve of 13.15 Effect of rotor resistance sentially constant 380 higher resistivity, such as bronze, for the rotor bars chronous speed. The A 2.5. 500 hp and more) about 8a shows the torque-speed curve of a Let us increase the rotor resistance by a factor of T their 80% 1 tor has an arbitrary resistance R, lower will quickly stop. less) a speed synchronous speed. Big motors attain their of 100 are in balance, the the load torque exceeds 2.5 3. 1 again equal to the load turn at a constant but slightly speed. However, (the is 1 kW (13.4 hp), 50 Hz, chronous speed of 000 r/min and a full-load torque me- the chanical load increases slightly, the speed will drop until the and end- minimum The torque developed by the motor while At full-load the in the rotor bars torque (called breakdown Pull-up torque T. nom- The torque-speed curve is greatly affected by such a change in resistance. The only characteristic that remains unchanged is the breakdown torque. The T. maximum 2.5 aluminum, or other metals the torque-speed curve of a conventional 3-phase induction motor whose of 70 except A increases with temperature. Thus, the resis- 1 again double the rotor resistance so that 5 R, the locked-rotor torque attains a value of 250 A (Fig. N-m it max- for a corresponding current 13.18c). further increase in rotor resistance decreases tance increases with increasing load because the both the locked-rotor torque and locked-rotor cur- temperature rent. rises. For example, if the rotor resistance is in- In designing a squirrel-cage motor, the rotor resis- creased 25 times (25 R), the locked-rotor current tance can be set over a wide range by using copper, drops to 20 A, but the motor develops the same breakdown torque 2.5 T 2 T , | locked-rotor torque 1.5 Ti o ^pull-up 0.5 full torq ue nominal T T 20 60 40 — p~ rotational speed Figure 13.17 Typical torque-speed curve of a 3-phase squirrel-cage induction motor. 80 100 % load Nm N-m Figure 13.18 Rotor resistance affects the motor characteristics. 283 ELECTRICAL MACHINES AND TRANSFORMERS 284 Nm), torque (100 starting as locked-rotor current was 100 In summary, because it relatively A it when did the 1 . a high rotor resistance is The locked-rotor current can be rotor torque will low squirrel-cage motor. Unfortunately, it current (Fig. resistors in series with the rotor. Nevertheless, the locked- desirable produces a high starting torque and a starting drastically re- duced by inserting three external (Fig. 13.1 8d). 13.18c). also produces a rapid fall-off in 2. still be as high as that of a The speed can be varied by varying the exter- speed with increasing load. Furthermore, because the slip at rated torque The losses are high. and the motor tends is high, the efficiency is R therefore low 3. it is preferable to have up and the slip We is is small. Consequently, the effi- high and the motor tends to run cool. which require is a diagram of the circuit used to start nected to three wye-connected external resistors by set of slip-rings and brushes. Under locked-rotor (LR) conditions, the variable resistors and a low running resistance by designing the rotor bars in a special However, are set to their highest value. way (see Fig. 14.5, Chapter 14). up, the resistance the rotor resistance has to be varied if over a wide range, it may a long time to bring wound-rotor motor. The rotor windings are con- means of a can obtain both a high starting resistance ideally suited to accelerate high- to speed. Fig. 13.19 at rated torque is The speed de- less with increasing load, a ciency The motor inertia loads, a low rotor resistance (Fig. 13.18a). much I nal rotor resistors. to overheat. Under running conditions creases motor 2 load speed be necessary to use a is is As the motor speeds gradually reduced until full- reached, whereupon the brushes are By properly selecting the resistance we can produce a high accelerating torque short-circuited. wound-rotor induction motor. Such a motor envalues, ables us to vary the rotor resistance at will by means of an external with a stator current that never exceeds twice full- rheostat. load current. To 13.16 Wound-rotor motor We explained the basic difference between a squirrel-cage motor and a wound-rotor motor in Section 1 3. more than 1 . Although a wound-rotor motor costs a squirrel-cage motor, it start large motors, we offers the fol- lowing advantages: a large thermal capacity. A liquid electrolyte. To vary its resistance, trodes. The is com- a suitable we simply vary large thermal capacity of the electrolyte Figure 13.19 to the three slip-rings of in the level of the electrolyte surrounding the elec- speed controller connected use liquid rheostat posed of three electrodes immersed starting rheostat External resistors often rheostats because they are easy to control and have a wound-rotor induction motor. and THREE-PHASE INDUCTION MOTORS limits the temperature rise. For example, in conjunction bring a large synchronous machine up We used one ap- kW wound-rotor motor to plication a liquid rheostat with a 1260 in is to speed. can also regulate the speed of a wound-rotor motor by varying the resistance of the rheostat. As we increase the resistance, the speed will drop. This method of speed control has the disadvantage of heat lot ciency is that a dissipated in the resistors; the effi- is therefore low. Furthermore, for a given rheostat setting, the speed varies considerably if the rating of a self-cooled wound-rotor motor depends upon the speed at Thus, for the same temperature rise, 40 which it operates. a motor that can lOOkWat 800 r/min can deliver only develop kW at 900 r/min. it 60 lodged coils, However, the if can deliver 50 motor is in 60 slots. total The X of (4 and are staggered intervals (Fig. The 13.20). X coils in each are connected in series may 3 at 5) group one-slot coils are identical possess one or more turns. = The width and of each coil is called the coil pitch. Such a distributed winding is obviously more costly to build than a concentrated winding having only one coil per group. However, starting torque it improves the and reduces the noise under running conditions. from a the stator windings are excited 3-phase source, a multipolar revolving field duced. The distance between adjacent poles the pole pitch. It is is is pro- called equal to the internal circumfer- about ence of the stator divided by the number of poles. For cooled example, a 12-pole stator having a circumference of 1 by a separate fan, group must have a When mechanical load varies. The power coils per 285 kW at 900 r/min. 600 mm has a pole-pitch of 600/12 or 50 mm. In practice, the coil pitch is between 80% and 100% of the pole pitch. The coil pitch is usually made less than the pole pitch in order to save copper 13,17 Three-phase windings named and to Nikola Tesla invented the 3-phase induction motor. The shorter coil width reduces the cost and weight In 883 a 27-year-old Yugoslav 1 His first model had a similar to the one scientist salient-pole stator winding shown in Fig. 1 3.6. Since then the design of induction motors has evolved consider- modern machines ably; are built with lap windings distributed in slots around the stator. A lap winding consists of a set tribution improves the torque during much To Fig. 13.21a. = poles X coils is it follows that the equal to the number of groups. A stator must have at least 2, 3. or more coils per The number of tion. coils 12 slots. group and in coils are held upright, with one at least of 12 coils. many 5 coils per group has coils. Consequently, a 4-pole, 3-phase designers have discovered that us 4-pole, Furthermore, in a lap winding the stator has as it let shown at least minimum number 3-phase stator must therefore have slots as laid out flat as 4X3=12 phase groups. Because a group must have coil, The 24 is phases Thus, a 4-pole, 3-phase stator must have one and easier to insert in the slots. get an overall picture of a lap winding, The number of groups groups start-up, 2-pole machines, the shorter pitch also makes the suppose a 24-slot stator given by the equation the air gap. often results in a quieter machine. In the case of evenly distributed around the stator circumference. is in of the windings, while the more sinusoidal flux dis- coils of phase groups improve the flux distribution it is However, motor preferable to use rather than only one. slots increases in propor- For example, a 4-pole. 3-phase stator having 5 Figure 13.20 The five coils are connected phase group. in series to create one ELECTRICAL MACHINES AND TRANSFORMERS 286 coil side set in each the windings are slot. If now laid c. down so that all the other coil sides fall into the slots, we obtain the classical appearance of a 3-phase lap winding having two The coils are coil sides per slot (Fig. 1 3.2 b). 1 Number of groups poles d. The pole one for each phase. Each wind- ing consists of a One ber of poles. (say) to slot 13. around the circumference of the rically distributed The following examples show how stator. e. this is done. of = 40 4- = 10 4. pitch corresponds to pole pitch number of groups equal to the numThe groups of each phase are symmet- = number per phase 10 Coils per group connected together to create three identical windings, — = - = slots/poles 120/10 12 slots pole pitch extends therefore from slot 1 The coil pitch covers 10 slots (slot to slot The percent coil pitch = 10/12 = 83.3%. 1 The next example shows in greater detail 1 1 ). how the coils are interconnected in a typical 3-phase sta- Example 13-8 The stator of a possesses 120 tor winding. 3-phase, 10-pole induction motor slots. If a lap winding is used, calcu- a. b. c. d. e. The total number of coils The number of coils per phase The number of coils per group The pole pitch The coil pitch (expressed as a percentage of the pole pitch), 1 to slot if Example 13-9 A late the following: the coil width extends from slot stator having 24 slots has to be 1 . 2. The connections between The connections between are standing upright, with We bution for phase Solution a. A b. Coils per phase 120 - 3 - the coils the phases coils. one Assume coil side in that they each slot will first determine the coil distri- A and then proceed with the con- nections for that phase. Similar connections will 120-slot stator requires 120 coils. = with a Solution The 3-phase winding has 24 (Fig. 13.22). 1 wound 3-phase, 4-pole winding. Determine the following: then be 40. made for phases B and C. Here is the line of reasoning: a. The revolving p field creates 4 poles; the motor 4 groups per phase, or 4 therefore has phase groups in all. Each rectangle X 3 in Fig. = 1 13.22a I represents one group. Because the stator contains lln iVr 1/2 20 21 22 23 24 3 4 5 6 24 7 / slot number coils, each group consists of 24/12 2 b. The groups (poles) of each phase 24 must be uni- The group distribution for phase A is shown in Fig. 13.22b. Each shaded rectangle represents two Figure 13.21a in 2 con- secutive coils. formly spaced around the Coils held upright = stator slots. stator. upright coils connected in series, producing the two terminals shown. Note that the me- chanical distance between two successive 20 21 22 23 24 1 2 3 4 5 6 groups always corresponds to an electrical 7 phase angle of 180°. Figure 13.21b Coils laid down c. to make a typical lap winding. Successive groups of phase site magnetic polarities. A must have oppo- Consequently, the four ^each each group one group of one phase coils are ot two composed of coils in series AAA Figure 13.22a The 24 is grouped two-by-two to make 12 groups. BtJtfMtzftlMt (—180° Figure 13.22b The four groups of (electrical)— phase A are selected so as start of phase to be evenly spaced from each other. A Figure 13.22c The groups of phase A are connected start of in series to create alternate phase , start of N-S poles. phase C 1 1 cil :b; 120° 240°— Figure 13.22d The start of phases B and A C C begins 120° and 240°, respectively, after the start of phase A. '4 Figure 13.22e When all phase groups are connected, only six leads remain. 287 ELECTRICAL MACHINES AND TRANSFORMERS 288 Figure 13.22f The phase may be connected A are groups of phase wye in connected duce successive N-S-N-S poles Phase A now 1 A2 B and C same way around the stator. and three leads are brought out delta, pro- f. are spaced (Fig. The groups series in the in phases B and same way C A are A,A 2 , . sulting 3 wires corresponding to the 3 phases are brought out to the terminal box of the ma- chine (Fig. 13.22f). In practice, the connections made, not while the coils are upright shown) but only in the slots. according to Fig. 450 after they first coil of 1 and connections 3.22e. and 13.24b show the kW (600 hp) and coil induction motor. procedure used kW (50 hp) in stator. are connected in as those of phase B|B 2 and C)C 2 They may be connected either in wye or in delta inside the machine. The re, Thus, the shortened the first and sixth slots winding a smaller 37.5 (Fig. 13.22e). This yields six terminals: are in Fig. 13.25 illustrates the (Fig. 13.22d). e. to slot 6). 1 lodged Figs. 13.24a A of phase ( suit stator of a located at 120° and 240° (electrical) with re- A is span of to a may be 3.23). All the other coils 1 follow However, the the terminal box. 6 slots, the coil pitch A phase . starting terminals B, and C, are respectively spect to the starting terminal = to 5 slots (slot ter- to Because the pole pitch corresponds 24/4 3.22c). has two terminals, a starting The phase groups of phases the in in series to (Fig. minal A, and a finishing terminal d. or have been (as laid down 13.18 Sector motor Consider a standard 3-phase, 4-pole, wye-connected motor having a synchronous speed of 1800 r/min. Let us cut the stator is left in in half, so that half the winding removed and only two complete (per phase). Next, let N and S poles are us connect the three phases wye, without making any other changes to the ex- isting coil connections. Finally, nal rotor gap above we mount this sector staton leaving a (Fig. 13.26). the origi- small air THREE-PHASE INDUCTION MOTORS should be reduced to half the stator winding nal now 289 original value because its has only one-half the origi- number of turns. Under these conditions, this re- markable truncated sector motor still develops about 20 percent of The moves original rated power. its sector motor produces a revolving field that the at same peripheral speed as However, the original 3-phase motor. making the flux in instead of a complete turn, the field simply travels continuously from one end of the stator to the other. 13.19 Linear induction motor It is flat, obvious that the sector stator could be laid out without affecting the shape or speed of the field. Such a flat stator produces a field moves at constant speed, in a straight line. Using the same reasoning as in Section 3.5, we can magnetic Figure 13.24a that Stator of a 3-phase, Hz induction motor. 108 preformed One coil side 450 kW, 1 1 80 r/min, 575 V, 60 The lap winding is composed of coils having a pitch from slots falls into the bottom of a slot other at the top. Rotor diameter: 500 460 mm. (Courtesy mecaniques Roberge) length: mm; 1 to 15. and the 1 prove that the flux travels at a linear synchronous speed given by axial vs = 2wf (13.10) of Services Eiectro- where \\ = linear vr — width of one pole-pitch /= Note synchronous speed |m/s] frequency [Hz] that the linear speed does not depend upon the number of poles but only on is [in] the pole-pitch. Thus, moving at the stator (say), If a flat flat stator, along with same speed as that of a 6-pole linear provided they have the same pole-pitch. squirrel-cage winding is brought near the the travelling field drags the squirrel cage it use a simple (Section 13.2). In practice, aluminum we generally or copper plate as a rotor (Fig. 13.27). Furthermore, to increase the power and duce the reluctance of the magnetic path, two tors are usually sides of the Figure 13.24b Close-up view of the preformed coil in Fig. 13.24a. it possible for a 2-pole linear stator to create a field to re- flat sta- mounted, face-lo-face, on opposite aluminum plate. a linear induction motor. The combination The is direction of the called motor can be reversed by interchanging any two stator leads. If we connect the stator terminals to a 3-phase, 60 Hz source, the rotor will again turn 1800 r/min. To prevent saturation, at close to the voltage In many practical applications, the rotor is sta- tionary while the stator moves. For example, in some high-speed trains, the rotor is composed of a Figure 13.25 50 Stator winding of a 3-phase, carrying 48 coils connected a. Each composed coil is in hp, 575 V, 60 Hz, 1764 r/min induction motor. of 5 turns of five No. 15 copper wires connected with a high-temperature polyimide insulation. Five No. 15 wires b. c. One coil fore, from Each coil side threaded stator possesses 48 slots into slot 1 (say) in parallel. in parallel is and the other side goes The wires are covered equivalent to one No. 8 wire. into slot 12. The coil pitch is, there- to 12. 1 side side placed is The wye. in and 4 empty fills the does not touch the second coil shows 3 empty and uninsulated slots a composition paper liner. The remaining 10 slots each carry one coil a slot and same slot. Starting from the top, the photograph half slots insulated with is covered with a paper spacer so that it side. d. A varnished cambric cloth, cut in the shape of a triangle, provides extra insulation between adjacent phase groups. (Courtesy of Services Efectromecaniques Roberge) 290 THREE-PHASE INDUCTION MOTORS 13.20 Traveling We sometimes are 291 waves with the impression that left when the flux reaches the there must be a delay before more at end of the beginning. This a linear stator, returns to restart once it is not the case. ear motor produces a traveling wave of The lin- which flux moves continuously and smoothly from one end of shows how the flux moves from left to right in a 2-pole linear motor. The flux cuts off sharply at extremities A, B of the stator to the other. Figure 13.28 Figure 13.26 Two-pole sector induction motor. However, the stator. pears at the right, it N as fast as a or S pole disap- builds up again at the left. linear rotor Properties of a linear induction motor 13.21 (aluminum, copper or iron plate) The properties of a linear induction motor are al- most identical to those of a standard rotating ma- chine. Consequently, the equations for slip, thrust, power, L etc., are also similar. Slip Slip, is defined by = s (v s - (I3.ll) v)/v s where s \\ Figure 13.27 Components of a 3-phase linear induction motor. aluminum thick tending over the stator is plate fixed to the full v ground and ex- length of the track. The linear bolted to the undercarriage of the train and straddles the plate. Train speed is varied by chang- ing the frequency applied to the stator (Fig. 1 3.3 2. 1 slip synchronous linear speed [m/s] speed of rotor (or stator) [m/s] Active power flow. With reference to Fig. active way it 3.7, power flows through a linear motor and 13.8 apply tween consecutive phase groups of phase mm, A is 300 calculate the linear speed of the magnetic field. (13.6) (\ -s)P is given by: F= P pitch vs is 300 mm. Consequently, = 2vv/ = 2 = 45 m/s or X X 1 75 62 km/h T (13.12) /v s where (13.10) 0.3 (13.8) r thrust or force developed by a lin- ear induction motor Solution The pole The 13.6, (13 .7) T Pm = Thrust. that the to both types of machines: = PJP c P}r = sP 3. same Consequently, Eqs. flat. t\ Example 13-10 The stator of a linear induction motor is excited from a 75 Hz electronic source. If the distance be- 3. 15, 1 in the does through a rotating motor, except and rotor are stator 1 ). — = = F = thrust [Nl P = power transmitted to the rotor W] v = linear synchronous speed |m/s| v s ( 292 ELECTRICAL MACHINES AND TRANSFORMERS Example 13-11 An overhead crane in a factory is driven horizon- by means of two linear induction motors tally whose two rotors are the I-beams upon which steel The 3-phase, 4-pole the crane rolls. linear stators (mounted on opposite sides of the crane and facing the respective pitch of 8 webs of the I-beams) have a pole cm and are driven by a variable frequency electronic source. During a test tors, the on one of the mo- following results were obtained: Hz stator frequency: 15 power kW to stator: 5 copper loss + iron loss in stator: 1 kW crane speed: 1.8 m/s Calculate a. Synchronous speed and b. Power c. I d. Mechanical power and thrust 2 R slip to the rotor loss in rotor Solution a. Linear synchronous speed = 2wf - 2 X 0.08 X = 2.4 m/s vs The Power = (v s = 0.25 - v)/v = (2.4 - 1.8)/2.4 to the rotor /> js r = c. ! R s (13.11) is P = Pc - 5 - 2 15 slip is j b. (13.10) - P f (see Fig. 13.15) 1 4kW loss in the rotor is Figure 13.28 Shape of the magnetic field created by a 2-pole, 3-phase linear stator, over one complete cycle. The successive frames are separated by an interval of time equal to 1/6 cycle or 60%. Pir = *P = 0.25 X = kW (13.7) r 1 4 THREE-PHASE INDUCTION MOTORS d. Mechanical power cause the flux density is The (Fig- 13.15) }r sulting value induced current reaches nnn and 1 time. This current, re- and creates magnetic 3, shown as sss the re- maximum its Fig. in 13.29. is According F = PJv s = 4000/2.4 = same at virtually the turning by conductors poles thrust greatest at the center of magnet moves very slowly, the pole. If the Pr-P 4-1 = 3kW Pm = = is 293 1667 (13.12) to the laws of attraction and repulsion, while the rear half N= 1.67 magnet the front half of the attracted is repelled upward is downward. Because the distribution of the nnn and sss poles kN (-375 is sym- lb) metrical with respect to the center of the magnet, the vertical forces of attraction and repulsion are 13.22 Magnetic levitation In we saw Section 13.2 moving permanent that a magnet, sweeping across a conducting ladder, tends to drag the ladder along with the magnet. now show We that this horizontal tractive force accompanied by a is also to Referring to Fig. 13.29, suppose that conduc- 1,2,3 are three conductors of ladder. The center of the N pole of the stationary the sweeping across the top of conductor age induced in this conductor is 2. magnet is The volt- maximum be- But suppose now rapidly. magnet (stationary) \ s s s s / / that the nil. magnet moves very its maximum con- value a fraction of a after the voltage has attained Consequently, by the time the current its in maximum. conductor 2 maximum, the center of the magnet is already some distance ahead of the conductor (Fig. 13.30). The current returning by conductors and 3 again is 1 creates nnn and sss poles; however, the is now directly N pole of the above an nnn pole, with the low result that a strong vertical force tends to magnet upward.* This effect push the called the principle is of magnetic levitation. 777TTT\ n n ss is to its inductance, the current in speed N conducting ladder Owing ductor 2 reaches magnet :: force vertical only a horizontal tractive is force. second push the magnet away from the ladder. tors Consequently, there will which tends vertical force, and the resulting equal, \ 1 Magnetic n n n 0' levitation is used in some ultra-high- speed trains that glide on a magnetic cushion rather than on wheels. A powerful electromagnet fixed un- derneath the train moves above a conducting ducing currents Figure 13.29 ladder. Currents and magnetic poles at low speed. in the rail in the The force of levitation same way is rail in- as in our always accompa- nied by a small horizontal braking force which must, of course, be overcome by the linear motor magnet high that propels the train. See Figs. 13.31 and 13.32. speed N The current is always delayed (even terval of time A/, m of the rotor. This delay the is low speeds) hy an the in- IJR time constant so brief that, at low speeds, the maximum at virtually the same lime and voltage does. On the other hand, at high speeds, current reaches place as the at which depends upon its same delay At produces a significant shift in space be- Figure 13.30 tween the points where the voltage and current reach Currents and magnetic poles at high speed. respective maximum values. their Figure 13.31 This 17 t track. is driven by a linear motor. The motor consists of a stationary rotor and a flat stator undercarriage of the train. The rotor is the vertical aluminum plate mounted in the center of the electric train fixed to the The 3-tonne stator is varied from zero to 105 Hz. energized by a 4.7 MVA electronic dc to ac inverter whose frequency can be The motor develops a maximum thrust of 35 kN (7800 lb) and the top speed 200 km/h. Direct-current power at 4 kV is fed into the inverter by means with 6 stationary dc busbars mounted on the left-hand side of the track. means of a brush assembly in is contact a superconducting electromagnet. The magnet is 1300 magnet are maintained at a 2 temperature of 4 K by the forced circulation of liquid helium. The current density is 80 A/mm and the resulting flux density is 3 T. The vertical force of repulsion attains a maximum of 60 kN and the vertical gap between the magnet and the reacting metallic track varies from 100 mm to 300 mm depending on the current. Electromagnetic levitation nun long, 600 mm wide, is obtained by and 400 mm deep, of and weighs 500 kg. The coils of the , (Courtesy of Siemens) 294 THREE-PHASE INDUCTION MOTORS 295 superconducting electromagnet conducting track guide and support wheels linear motor (stator) conducting plate (rotor) brush assembly for power input, from 4 kVdc source Figure 13.32 Cross-section view of the main components of the high-speed train (Courtesy of Siemens) Questions and Problems shown in Fig. 1 3.31 Give two advantages of a wound-rotor 13 -8 motor over a squirrel-cage motor. Practical level 13-1 Name the principal components of an Both the voltage and frequency induced 13-9 in in- the rotor of an induction duction motor. motor decrease as the rotor speeds up. Explain. 13-2 Explain how a revolving field is set up in 13-10 a 3-phase induction motor. 1 3-3 If we double the number of poles on stator of an induction motor, will its A 3-phase, connected the syn- chronous speed also double? 20-pole induction motor to a a. What is b. If the voltage 600 V. is 60 Hz source. the synchronous speed? is reduced to 300 V, will the synchronous speed change? 13-4 The rotor of an induction should never be c. locked while to the stator. 1 3-5 Why full voltage is 13- Explain. 13-7 1 1 groups are there, per phase? Describe the principle of operation of a linear induction motor. does the rotor of an induction motor turn slower than the revolving field? 13-6 How many being applied 13- 12 Calculate the approximate values of the What happens to the rotor speed and rotor current when the mechanical load on an starting current, full-load current, and induction motor increases? 575 Would you recommend using a 50 hp induction motor to drive a 10 hp load? Explain. no-load current of a 150 horsepower, 13- 3 1 V, Make 3-phase induction motor. a drawing of the magnetic field cre- ated by a 3-phase, 12-poIe induction motor. 296 ELECTRICAL MACHINES AND TRANSFORMERS 13-14 How wc change can the direction of rota- the motor? tion of a 3-phase induction mmf developed by the windings, the resulting mmf point in a direction Does c. intermediate between the Intermediate level 13-15 sponding Calculate Ihe synchronous speed of a a. 1 3-phase, 12-pole induction motor that b. cited by What is load a 3-2 1 ex- is if produces the slip at full synchronous speed a when connected to number of source. Calculate the 6 percent? is slots of 900 r/min nominal speed mmf s corre- and 4? 3-phase lap- wound stator possessing 72 60 Hz source, the A to instants 3 60 Hz a coils per phase group as well as the probable 1 3- 1 6 A 3-phase 6-pole induction motor Hz nected to a 60 induced supply. in the rotor bars is con- The voltage is 4 V when the is locked. If the motor turns in the 13-17 its 13-22 The 3-phase, 4-pole stacking (axial length) of 200 At 1000 r/min maximum At 1500 r/min calculate the following: a. Calculate the approximate values of b. a. full- The kW. 4000 V, 3-phase. 1 9 A 900 b. c. 3-phasc, 75 hp, 440 V 2 percent. 1 3-23 when the stator induction motor A large 4000 3-phase, A and a 385 and a power factor of 83 percent. when Calculate the nominal current per phase. sponding speed open-circuit voltage of is 240 V appears total active found to is wye and tween two stator terminals total iron losses are windage and motor turns same direction as the r/min. in the same direction as the rotating field b. At 900 rotating field c. a. of b. field Referring to Fig. 13.7, calculate the instan- taneous values of I5() /.,, /h , and /c for an angle u . Determine the actual direction of current flow in the three phases at this instant 10 friction losses are 12 il. The kW. c. d. The load mechanical power [kW|, torque [kN-m], and efficiency 1 At 3600 r/min. opposite to the rotating is 0. kW and the 23.4 The power factor at full-load The active power supplied to the 2 The total I R losses in the rotor a. b. in the At 600 r/min, is the resistance be- Calculate the following: open-circuit voltage and frequency across a. kW corre- be 709.2 r/min. The stator duction motor when the rotor is locked. The stator has 6 poles and is excited by 60 Hz source. If the rotor is driven by a a The accurately measured and in the dc squirrel- power of 2344 operating at full-load. connected if bars in the rotor 60 Hz V, across the slip-rings of a wound-rotor in- the slip-rings 0.7 T, connected to a is The peak voltage induced The pole-pitch has a full-load efficiency of 91 percent An is a the cage induction motor draws a current of variable-speed dc motor, calculate the 13-20 flux density per pole If 60 Hz source Calculate the nominal full-load speed and is mm and mm. peripheral speed |m/s] of the revolving field induction motor, 60Hz torque knowing that the slip 3- stator of Fig. 13.25 has an internal diameter of 250 frequency: c. r/min, 1 Fig. 13.22. b. current of a 75 con- coil ro- load current, starting current, and no-load 13-18 complete same direction as the flux, calculate the approxi- mate voltage induced and a. At 300 r/min the nection diagram, following steps (a) to (f) in tor Draw coil pitch. and calculate 3-24 If we rotor ( \ 7c\ slightly increase the rotor resistance of an induction motor, what effect does this a. have (increase or decrease) upon Starting torque b. Starting current c. Full- load speed d. Efficiency THREE-PHASE INDUCTION MOTORS e. Power f. Temperature 13-30 factor of the motor rise rated at its power output 1 3-25 13-26 voltage Explain the principle of magnetic levitation. Advanced In Fig. 13.5a the The permanent magnet has and moves at 30 m/s. mm flux density in the air gap is 0.5 T 1 3-3 1 and the effective resistance per rotor bar is mil. Calculate the current 1 3-27 If the conducting ladder in Fig. 3.5a 1 is pulled along with a force of 20 N, what 1 3-28 A 5000 3-phase, hp, 6000 V, 60 Hz 1 What r/min. I~R losses 1 3-29 at The motor at 0.0073 3-28 has the 1 = 13-33 17°C = V 1600 = 6000 V = 100 A line-to-line stator voltage 6. active 7. windage and 8. iron losses in the stator 9. locked-rotor current 2207 power supplied to motor at 450 3. 1 9). r/min, calculate the Voltage between the slip rings Rotor resistance (per phase) and the tolal Approximate rotor current, per phase The train shown in Fig. 13.31 moves at 200 km/h when the stator frequency is By supposing a negligible slip, motor [mm]. A 3-phase, 300 kW, 2300 V, at = The motor has load efficiency and power 1 V = 2760 terminal voltage rises to kW effect (increase or decrease) 1800 A to stator with rotor locked = kW a full- factor of 92 per- cent and 86 percent, respectively. kW 39 6000 5 used to drive a compressor. motor operates = friction losses 60 Hz, is no- kW power a speed of 1 develop a torque of 20 1780 r/min induction motor no-load stator current, per phase 10. active at the linear 5. 91 kN-m to calculate the length of the pole-pitch of = 17°C 4. = motor has 105 Hz. rings with rotor locked load inserting resis- fol- open-circuit voltage induced between slip- 3. Problem 13-29 by in If the c. Problem 11 at wish to control the speed of the motor at 13-32 dc resistance between rotor slip-rings 2. We are the approximate rotor dc resistance between stator terminals 0.112H Torque developed by the rotor power dissipated lowing characteristics: 1. Mechanical power output e. a. 2- rated load? in the stator in d. b. pole wound-rotor induction motor turns 594 Active power supplied to the rotor following: is on the magnet? the braking force exerted I~R losses c. tors in series with the rotor (see Fig. tractive force. 1 (locked-rotor) conditions: b. given and the / LR in full- Reactive power absorbed by the motor a. level a width of 100 Referring to the motor described Problem 13-29, calculate under 297 at full-load, V If the while the determine the upon a. Mechanical power delivered by the motor b. Motor torque c. Rotational speed d. Full-load current e. Power f. Starting torque factor and efficiency Calculate a. Rotor and stator resistance per phase 75°C (assume b. it turns at at g. wye connection) Voltage and frequency induced when c. a 200 r/min and Breakdown torque i. Motor temperature in the rotor at Starting current h. rise 594 r/min j. Reactive power absorbed by the motor to k. Flux per pole Exciting current create the revolving field, at no-load 2 d. I at e. 1. R losses in the stator when the Iron losses motor runs no-load (winding temperature 75°C) Active power supplied to the rotor at no-load 13-34 A 3-phase, 60 Hz linear induction motor has to reach a top no-load speed of 12 m/s 298 ELECTRICAL MACHINES AND TRANSFORMERS and must develop a it standstill thrust of Calculate 10 kN. Calculate the required pole-pitch and the minimum 2 I R a. b. loss in the rotor, at c. standstill. d. Industrial application 1 3-35 A 1 e. 0 hp, 575 V, 1 1 60 r/min, 3-phase, 60 Hz induction motor has a rotor made of alushown in Fig. minum, similar to the rotor 3.3a. The end-rings are trimmed 1 f. is 1 3-37 What making them effect will this a. The The starting torque The temperature rise c. full hp, standstill less thick. 1 1 83 r/min, 575 V, 3-phase, 60 320 V is 3-36 The stator of a 600 that the RMS about 0.6 V. Estimate the no-load speed of the motor. 1 3-38 The rotor of a 60 hp, 1 760 r/min, 60 Hz at full-load induction motor has 1 known It is brush voltage drop load speed of the motor Hz at between the open-circuit lines of the rotor. have on the following: b. A 25 density T 0.54 wound-rotor induction motor produces in a lathe, cutting off the cooling fins and also a portion of the rings, The number of coils on the stator The number of coils per phase The number of coils per group The coil pitch (in millimeters) The area of one pole The flux per pole if the average flux hp, 1 1 60 r/min, 575 V, 3-phase, 60 Hz induction motor has 90 slots, an internal diameter of 20 inches, and an axial length of 16 inches. ter of 1 1 1 17 bars and a diame- inches. Calculate the average force on each bar (in newtons) motor is running at full-load. when the Chapter 14 Selection and Application of Three-Phase Induction Motors be replaced by that of any other manufacturer, 14.0 Introduction When purchasing a 3-phase induction motor we a particular application, several types can have to make fill The for shaft height, or the type of coupling. often discover that selection is establishes limiting values for electrical, generally simplified because the manufacturer of the lathe, fan, pump, and so forth indicates the type of motor best suited to drive the load. Nevertheless, ful to know something about it that is is use- must satisfy also cover some to motors Motors are grouped Standardization and classification of 1. * Manufacturers Motors and Generators. Standards publication in Canada Drip-proof motors. The frame motors of one manufacturer can (NEMA) which they have We limit our discussion to five important and solid particles which in a MG-I fall at drip-proof liquid drops any angle between vertical. The means of a fan directly couCool air, drawn into the motor are cooled by pled to the rotor. Standards in the United Slates are governed by National Electrical in 0 and 15 degrees downward from the motors under 500 hp have standardized dimensions. Thus, a 25 hp, Hz motor categories, de- into several motor protects the windings against induction motors* 725 r/min, 60 environment and pending upon the environment in general. to operate. 1 rise. cooling methods classes. industrial to starting 14.2 Classification according special appli- interesting devices will enable the reader to gain a better understanding of induction all requirements as in- nous generators and frequency converters. These The frames of minimum locked-rotor current, overload capacity, and temperature cations of induction machines, such as asynchro- 14.1 mechan- and thermal characteristics. Thus, motors the basic construction and characteristics of the various types of 3-phase we stan- ical, torque, duction motors that are available on the market. In this chapter The dardization covers not only frame sizes, but also we the need. Consequently, a choice. without having to change the mounting holes, the governed by Canadian Standards Association (CSA) publication titled are similarly 299 C 154. The two standards arc essentially identical. ELECTRICAL MACHINES AND TRANSFORMERS 300 through vents in the frame, is and then expelled. The ings temperature rise resistance) ing blown over the wind- 4. maximum allowable and high-power motors (measured by the change may be 6()°C in wind- 80°C 105°C 125°C, depending on the type of insulation used the windings. Drip-proof motors (Fig. 14.1 used 2. in ) ribbed motor frame (Fig. in a splash- proof motor protects the windings against liquid fall at any angle be- tween 0 and 100° downward from the temperature rise These motors are mainly used 3. vertical, similar to that in drip-proof motors and maximum the to the shaft, in is An external air over the 1 4.3). A concentric outer shield prevents physical contact with the fan and serves to channel the airstream. drops and solid particles that is coupled in can be air. blows usually cooled by an external blast of fan, directly Medium- enclosed are that are totally or most locations. Splash-proof motor. The frame Cooling Totally enclosed, fan-cooled motors. perature rise 5. is the same The permissible tem- as for drip-proof motors. Explosion-proof motors. Explosion-proof mo- tors are used highly inflammable or explosive in surroundings, such as coal mines, grain elevators. They oil refineries, and are totally enclosed (but not also the same. wet locations. Totally enclosed, nonventilated motors. These motors have closed frames that prevent the free exchange of air between the inside and the outside They are designed for very wet and dusty locations. Most are rated below 10 kW beof the case. cause it is difficult to get rid of the heat of larger machines. The motor losses are dissipated by natural convection and radiation from the frame. The permissible 1 temperature rise is 65°C, 85°C, 10°C, or I30°C, depending on the class of insu- lation (see Fig. 14.2). Figure 14.2 Two enclosed nonventilated (TENV) 2 hp, 1725 cage motors are shown in foreground and two 30 hp, 1780 r/min totally enclosed blower-cooled motors (TEBC) in background. These 3-phase, 460 V motors are intended to operate at variable speeds totally r/min ranging from a few revolutions per minute to about 3 times rated speed. The 2 hp motors have a full-load current of 2.9 A, and power factor of 76 per- efficiency of 84 percent cent. Other characteristics: no-load current: locked-rotor current: pu; tal 26 1 .7 A; A; locked-rotor torque: 4.2 breakdown torque: 5.0 pu; service factor: 1.0; to- weight: 39 kg; overall length including shaft: mm; overall height: The 30 hp motors have a efficiency of 377 235 mm. 34 A, 84 per- full-load current of 93 percent, and power factor of cent. Other characteristics: no-load current: 12 A; tion Energy efficient drip-proof, 3-phase squirrel-cage inducmotor rated 230 V/460 V, 3 hp, 1750 r/min, 60 Hz. locked-rotor current: 214 A; locked rotor torque: 1 .6 pu; breakdown torque: 2.84 pu; service factor: 1 .0; total weight: 200 kg; overall length including shaft: 834 mm; overall height: 365 mm. {Courtesy of Gould) (Courtesy of Baldor Electric Figure 14.1 Company ) SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS 301 14.3 Classification according to electrical and mechanical properties enclosures just mentioned, In addition to the various 3-phase squirrel-cage motors can have special electrical and mechanical characteristics, as listed below. Motors with standard locked-rotor torque (NEMA Design B). Most induction motors belong to this group. The per-unit locked-rotor torque decreases as the size of the motor increases. Thus, it ranges from .3 to 0.7, as the power increases from 20 hp to 200 hp (15 kW to 150 kW). The corre/. l Figure 14.3 Totally enclosed fan-cooled induction motor rated 350 440 (Courtesy of Gould hp, 1760 r/min, V, sponding locked-rotor current should not exceed 6.4 times the rated full-load current. These general- 3-phase, 60 Hz. purpose motors are used to drive fans, centrifugal pumps, machine 2. and so tools, forth. High starting-torque motors (NEMA Design C). These motors are employed when Pumps and are difficult. that have to start starting conditions piston-type compressors under load are two typical applica- tions. In the range from 20 hp to 200 hp, the locked- rotor torque is 200% of full-load torque, which cor- responds to a per-unit torque of The locked-rotor 2. current should not exceed 6.4 times the rated full- load current. equipped with In general, these motors are double-cage rotor. double-cage rotor lowing Figure 14.4 a. Totally enclosed, fan-cooled, explosion-proof motor. Note the particularly rugged construction of this type b. facts: The frequency of the rotor as the motor speeds up A conductor that of motor. (Courtesy of Brook Crompton-Parkinson Ltd (cage ) airtight) the and the frames are designed enormous pressure that may to an internal explosion. the flanges on the end-bells are made extra long Furthermore, in order to cool any escaping gases generated by such may permissible temperature rise tally the is enclosed motors (see Fig. 1 be initiated by same 4.4). lies close to the rotor surface has a lower inductive reactance than The as for to- The conductors of cage 1 are When the motor is connected build up inside the motor due an explosion. Such explosions ) much 2) smaller than those of cage 2 to withstand the spark or a short-circuit within the windings. 1 current diminishes one buried deep inside the iron core (cage c. a The excellent performance of a (Fig. 14.5) is based upon the fol- to the line with the rotor at standstill, the frequency of the rotor current is equal to line frequency. Owing ductive reactance of squirrel-cage to the high in2, the rotor cur- rent flows mainly in the small bars of cage effective motor resistance is essentially equal to that of cage high starting torque is 1. The therefore high, being 1. developed. Consequently, a ELECTRICAL MACHINES AND TRANSFORMERS 302 Figure 14.5 Typical torque-speed curves of NEMA design B, C, and D minimum NEMA Hz squirrel- motors. Each curve corresponds to the values of locked-rotor torque, pull-up torque, and breakdown torque of a 3-phase 1800 r/min, 10 hp, 60 cage induction motor. The cross-section As falls, the of the respective rotors indicates the type of rotor bars motor speeds up, the rotor frequency with the result that the inductive reactance of tors are usually windings that the reactance of both The rotor current tance of cage 1 so low (typically is is Hz) I negligible. then limited only by the resis- is and cage 2 operating in parallel. The rated speed is 1 , the effective rotor resistance at much lower than at standstill. For this reason the double-cage rotor develops both a high starting torque and a low slip at full-load. Despite their high torque. Design recommended not reason is that punch holes for starting high-inertia loads. 3. in cage 1 . Owing to its small may melt. The start- size, tends to overheat and the bars High-slip (NEMA motors rated speed of high-slip, Design lies between 85% and 95% These motors are used in sheet metal. ates the operation, a clutch When the Punching a hole requires a tremendous amount son is that the punching energy fraction of a second. punch does The energy delivered in a is is furnished by the in a itself. As the work, the speed of the flywheel lot of kinetic very short time. The speed of the motor also drops considerably, along with that of the fly- wheel. However, the Class D will not The high-remade of brass, and the mo- its drops immediately, thus releasing a energy the drawn from exceed As soon of synchronous speed. to accelerate high -inertia initi- engages the flywheel, of power, sometimes exceeding 1000 hp. The rea- the current motors usually worker is that causing the punch to descend and pierce the sheet. Design D). The loads (such as centrifugal dryers), which take a rel- motor its design ensures that rated value. as the hole is D the line at the lower speed is pierced, the only load on the flywheel, which is now gradually brought back up to speed. During the acceleration atively long time to reach full speed. period, the sistance squirrel cage thus restoring the energy is machine tools flywheel rather than by the motor motors are most of the rotor tR losses during up are concentrated it C to large drop in speed with increasing load also ideal to drive impact-type Because the conductors of cage 2 are much larger than those of cage designed for intermittent duty prevent overheating. both squirrel-cage windings diminishes. At rated speed the rotor frequency used. motor delivers energy it lost to the flywheel, during the impact. A SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS powerful motor will quickly accelerate the motor, and fly- wheel, permitting rapid, repetitive operation of the punch is press. low, a On much the other hand, if 2. it and and D it the characteristics of motors. The (as a percentage of full-load Fig. 14.5 enable us to NEMA rotor construction of equal power. Design B, C, is if increased (by using brass instead of copper or aluminum), the locked-rotor torque creases, but the speed at rated torque is By way also shown, obtained by changing the rotor design. For example, is always greater torque) than that of a similar low-speed motor can be seen that the distinguishing properties are the rotor resistance in- two 10 motor speed The choice of motor speed synchronous speeds. The difference would is rather limited because by quantum jumps, depending upon the frequency and the number of poles. For example, sible to build a and running ing an acceptable efficiency say, it impos- is conventional induction motor hav- of 2000 r/min on a 60 Hz supply. at a speed, totally en- in price justify the use of a high-speed alone motor and a 900 to drive a load operating at, say, speeds (100 r/min or r/min. has to operate at very low a gearbox less), mandatory. is The gears are often an integral part of the motor, making for a very compact unit (Fig. 14.6). synchronous speed of induction motors changes the 60 Hz, hp, 3-phase, closed, fan-cooled induction motors having different gearbox lower. of example. Table 14A compares the properties of When equipment 14.4 Choice of factor are The locked-rotor torque of a high-speed motor is The torque-speed curves of power will only take longer to bring the flywheel up to speed. compare efficiency and higher. the repetition rate smaller motor will suffice; its 303 A gearbox is when equipment also mandatory has to run above 3600 r/min. For example, in one industrial application a large gear unit is used to 5000 r/min centrifugal compressor a 3560 r/min induction motor. drive a 1200 hp, coupled to Such a motor would require 2 poles and a corresponding synchronous speed of 3600 r/min. The 2000)/3600 = slip of (3600 - 0.444 means that 44.4% of the power 14,5 The Two-speed motors stator of a squirrel-cage induction supplied to the rotor would be dissipated as heat. designed so that the motor can operate (See Section ent speeds. 1 3. 13.) The speed of a motor the speed of the for machine low-speed machines, a high-speed rectly is it obviously determined by it has to drive. However, is motor and a gearbox instead of . di- coupling a low-speed motor to the load. There are several advantages 1 often preferable to use to using a is to wind the pumps. One way stator with that only one winding only half the copper gearbox: is in Power being utilized. However, special windings have been invented whereby the speed TABLE 14A is changed by simply changing The synchronous COMPARISON BETWEEN TWO MOTORS OF DIFFERENT SPEEDS Synchronous Power speed factor drill obtain two operation at a time and so the external stator connections. of a low-speed to two separate windThe problem is in the slots is high-speed motor less than that differ- ings having, say, 4 poles and 6 poles. For a given output power, the size and cost of a is two Such motors are often used on presses, blowers, and speeds motor can be at Efficiency Locked-rotor Mass torque Price (2002) hp kW r/min % 9c % kg U.S. $ l() 7.5 3600 89 90 150 50 650 10 7.5 900 75 85 1 70 2000 25 1 ELECTRICAL MACHINES AND TRANSFORMERS 304 ac source Figure 14.6 Gear motor rated at 2.25 kW, 1740 r/min, 60 Hz. The output torque and speed are respectively 172 N-m and 125 r/min. (Courtesy of Reliance speeds obtained Electric) usually are speed the in (3600/1800 r/min, 1200/600 r/min, etc.). ratio 2:1 The lower produced by the creation of consequent is poles. Consider, for example, one phase of a two-pole, 3-phase motor (Fig, 14.7a). connected in series to a flows into terminal of terminal As 2. When and current 1 two poles the are 60 Hz ac source, current a result, one N I2 (= l\) I { tlows out pole and one S pole Figure 14.7 Two a. are created and the flux has the pattern shown. The synchronous speed = ns is 120f/p that stator This 120 X 60/2 percent of the pole-pitch. shown rent /, minal now in Fig. As a windings. When the coils are connected motor is produced. Two Because every in parallel, as 14.7b. In this case, at the instant cur- result, two S pole, the connect the two poles flows into terminal 2. series produce a of the in parallel, a 4-pole poles are conse- each pole covers only one-quarter of the achieved by using a coil pitch equal to 50 Let us in r/min circumference instead of the usual one-half. is connected quent poles. = 3600 Note b. - short-pitch coils two-pole motor. 1 N , current I2 flows into ter- poles are created by the it two nious N pole must be accompanied by a follows that two S poles will appear between N poles. The way south poles created in this inge- are called consequent poles. nection produces 4 poles in speed is all, The new con- and the synchronous 1800 r/min. Thus, we can double the number of poles by simply changing the stator connections. is upon this principle that 2-speed motors are built. It SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS now again produces 4 poles per phase, but possess the same polarity (Fig. Two-speed motors have a 1 they all 4.8b). relatively lower effi- ciency and power factor than single-speed motors They can be designed do. develop to both (at speeds) either constant power, constant torque, or variable torque. that has to The choice depends upon The 2-speed motors described so tios of 2: l . the load be driven. If the motor drives a The reason big a change in speed. far have pole fan. this is may that the ra- be too power of a fan varies as the cube of the speed. Consequently, if the speed is reduced by one-eighth, which To overcome power drops half, the to often too low to be of interest. is this problem, some 3-phase wind- ings are designed to obtain lower pole ratios such as 8/10, 1 4/ 6, 26/28, 10/ 4, 1 1 larly useful in driving and 38/46. These pole PAM, motors amplitude modulation, or 2-speed fans horsepower range and more. moderate reduction in fan PAM are particuthe in hundred motors enable a power by simply recon- necting the windings to give the lower speed. motor characteristics under 14.6 Induction various load conditions The complete torque-speed curves displayed 1 Figure 14.8 a. High-speed connection of a 3-phase stator, yielding 4 poles. b. Low-speed connection of same motor yielding 4.5 are important, but most of the time a motor runs Figure 1 4.8 shows the stator connections for a 2-speed, 4-pole/8-pole, ing. to I to 6, are 3-phase motor. Six leads, brought out from the stator wind- For the high-speed connection, power is applied terminals 1-2-3, and terminals 4-5-6 are open. resulting delta connection produces having two Tn two S poles connected (Fig. . It so happens that between circuiting terminals 1-2-3 The 1 4.8a). Note is made by and applying power resulting 4.9). is essentially a The slope of pends mainly upon the rotor resistance the line de- — the lower the resistance, the steeper the slope. In effect, the slip tance R s, it can be shown that torque T, at rated line voltage E, frequency, and rotor resis- are related by the expression 4 poles per phase = kTR/E 2 (14. 1) that where k in series. The low-speed connection minals 4-5-6. 1 s N and the four poles are The that close to synchro- nous speed, supplying a torque that varies from zero to full-load torque straight line (Fig. numbered at these limits the torque-speed curve 8 poles. in Fig. must be recognized it shortto ter- double-wye connection is a constant that depends upon the con- struction of the motor. This expression enables us formula showing how to establish a simple the line voltage and rotor ELECTRICA L MA CHINES A ND TRA NS FORMERS 306 Figure 14.9 The torque-speed curve is resistance affect the behavior of the load. In effect, motor once between the no-load and rated torque operating essentially a straight line we know motor under the characteristics of a for a given load condition, we can predict its speed, torque, power, and so on, for any other load condition. These quantities are related by the formula an accuracy of better than 5 percent which cient for most suffi- Example 14-1 A V 3-phase, 208 induction motor having a syn- chronous speed of 1200 r/min runs to a V 215 line stant torque load. Calculate the A- is practical problems. when connected Ax points. increases to 240 at 1 140 r/min and driving a con- speed if the voltage V. where n = Solution subscript referring to the initial, conditions (the given conditions spond to the = subscript referring to the s = slip E= may The slip at 215 Vis corres nominal rating of the motor) x T= R = or given load new load conditions = (n s = (1200 = 0.05 n)ln s - 11 40)/ 1200 torque [N-m] When the voltage rotor resistance rises to stator voltage [V] applying the formula, the only restriction . V, the load torque we can write is new torque Tx must not be greater than Tn (EJEn ) 2 Under these conditions Eq. 14.2 yields that the 240 and rotor resistance remain the same. Consequently, in applying Eq. 14.2, In — .v x = sn = 0.04 (E n /Ex ) 2 = 0.05 (215/240) 2 - SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS The slip speed Example 14-3 therefore, is, X 1200 = 48 0.04 A 3-phase, 4-pole r/min rating of The new speed nx = 240 at 1200 V - 48 = 1152 r/min Under is Hz line. The it wye across the Nm speed of 1000 r/min. at a N Calculate the speed for a torque of 400 b. Calculate the value of the external resistors so kW at 200 1000)/ 1800 = 0.444 motor develops 10 that the m. r/min. Solution to a. The given conditions are machine has run for several after the Tn = 300 N hours. .v Calculate The hot in is initial 23°C. The speed drops connected these conditions the motor de- a. com- 8-pole induction motor driving a connected to a fixed 460 V, 60 rotor resistance in terms of the cold re- n torque of 400 The approximate hot temperature of the bars, knowing they are made of copper m - (1800 - All other conditions being fixed, sistance b. wound-rotor induction motor has a kWJ 760 r/min, 2.3 kV, 60 Hz. Three ex- velops a torque of 300 cold rotor temperature a. 0 1 rotor slip-rings. pressor runs at 873 r/min immediately after 864 r/min 1 ternal resistors of 2 f i are is Example 14-2 A 3-phase, 307 we have for a N m the following: rotor -v x = - sn (TJTn ) = 0-444 (400/300) 0.592 Solution a. The synchronous speed ns The = 120 initial = ///; and 120 X = 900 slip speed (900 - 873)/900 = 0.03 sx = (900 - 864)/900 = 0.04 speed change is in rotor resistance. x = 0.04 = .s due entirely .v n to the change b. is The hot R, 734 r/min 1066 kW at 200 is 9.55 Pin 9.55 X (3.5) 10 000/200 = 478 N-m ) The greater than the rated torque 7Vaicd = is 9.55 Pin - R (234 9.55 X 110 000/1760 = 597 N m is + T ) - 234 (6.5) Because Tx is less than r, atcd , we can apply x Eq. 14.2. \ = 1066 r/min. is = rotor temperature t,= = to 10 r/min cold rotor resistance. b. 1800 load. = 33% - = The torque corresponding Tx = 1.33 /?„ hot rotor resistance - 1800 734 r/min with increasing therefore write (RJR n X speed drops from 1000 r/min to that the (R x /R n ) 0.03 Rx = Note are fixed; consequently, We can 0.592 r/min n = = Consequently, the motor speed final slips are The voltage and torque The 60/8 n .v the The is: 1.33 (234 = 108°C + 23) - 234 The slip is sx = (1800 - 200)/ 1800 = 0.89 ELECTRICAL MACHINES AND TRANSFORMERS 308 we All other conditions being fixed, have, from Rule 1 The - heat dissipated in the rotor during the starting period (from zero speed to final rated Eq. 14.2: sx = *n speed) (TJTn ) (RJR n ) all 0.89 = 0.44 (478/300) (RJ2) is equal to the final kinetic energy stored This rule holds true, irrespective of the stator volt- age or the torque-speed curve of the motor. Thus, and so in the revolving parts. if a motor brings a massive flywheel up to speed, and rx = Three 2.5 ft n wye-connected tor circuit will 10 2.5 if the energy stored in the flywheel joules, the rotor will have dissipated resistors in the ro- enable the motor to develop kW at 200r/min. the form of heat. Depending upon the tor and its is then 5000 5000 joules in size of the ro- cooling system, this energy could easily produce overheating. 14.7 Starting an induction motor 14.8 Plugging an induction motor High-inertia loads put a strain on induction motors some because they prolong the starting period. The starting In current in both the stator and rotor and is high during this that the revolving field lem. For motors of several thousand horsepower, a prolonged starting period may even is installed. many loads. To The line voltage may fall overload the motor riod, the It below normal chanical relieve the problem, induction motors are rotor. often started on reduced voltage. This limits the During its speed to power P m is fall. The associated me- entirely dissipated as heat in the Unfortunately, the rotor also continues to re- P r the line voltage drop as well as the heating rate of the which windings. Reduced voltage lengthens the start-up Consequently, plugging produces time, but this is whether the start-up time remembering is long or short, the following rule for a it is motor also from the heat /"/? (Fig. is plugged, the rotor 2 l stator, 14. 10). losses in the may is locked. melt the rotor bars or overheat the stator winding. In this regard Figure 14.10 a 3-phase induction motor as even exceed those when the rotor high rotor temperatures is not loaded mechanically: When dissipated Motors should not be plugged too frequently because worth that is rotor that usually not important. However, oppo- in the plugging pe- acts as a brake. ceive electromagnetic power the motor, and consequently reduces this absorbs kinetic energy from the still-revolving load, causing seconds, thus affecting other connected power drawn by suddenly turns site direction to the rotor. transmission line feeding the plant where the motor for motor industrial applications, the induction load have to be brought to a quick stop. This can be done by interchanging two stator leads, so becomes a major prob- interval so that overheating its R losses are very high. it is worth SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS remembering the following tions for a Rule 2 motor that The - is heat dissipated in the rotor during is three times the original kinetic energy of the revolving parts. all ergy dissipated original it produces In effect, the en- only equal to the in the rotor is energy stored kinetic that is does plugging. far less heat than the plugging period (initial rated speed to zero speed) The advantage of dc braking rule for plugging opera- not loaded mechanically: 309 in the revolving masses, and not three times that energy. The energy dissipated in the rotor is independent of the magnitude of the dc current. However, a smaller Example 14-4 A 100 kW, 60 Hz, dc current increases the braking time. The dc cur175 r/tnin motor 1 is coupled to a flywheel by means of a gearbox. The kinetic energy of is 300 kJ when the motor The motor is plugged to a stop the revolving parts all runs at rated speed. and allowed to run up to 1 175 r/min in the reverse two or three times the rated current of Even larger values can be used, prothe stator does not become too hot. The rent can be the motor. vided that braking torque proportional to the square of the is dc braking current. direction. Calculate the energy dissipated in the rotor if the flywheel is the only load. Example 14-5 A Solution During the plugging period, the motor speed drops from 175 r/min to zero. The heat dissipated 1 rotor is 3 erates to X 300 kJ = 900 nominal speed kJ. The motor then kJ. this + 300 = 1200 is The period By reversing the speed this way, the total dissipated in the rotor from start to finish 900 accel- in the reverse direction. energy dissipated in the rotor during 300 in the 25 kg 62 A. 24 heat therefore a. 14.9 Braking with direct current high-inertia load can also be brought to a quick stop by circulating dc current in the stator winding. Any two having a is The dc 0.32 We moment of inertia of between two stator ter- and the rated motor current 12, want total resistance to stop the is motor by connecting a battery across the terminals. Calculate b. its V . is kJ. induction motor and m2 minals c. An 50 hp, 1760 r/min, 440 V, 3-phase induction mo- tor drives a load The dc current in the stator The energy dissipated in the rotor The average braking torque if the stopping time is 4 min Solution a. The dc current stator termi/ is — EIR — 24/0.32 = 75 A nals can be connected to the dc source. The in direct current produces stationary N, S poles This current The number of poles created is equal number of poles which the motor develops to the how the motor ter- When the rotor an ac voltage is sweeps past the stationary induced in the rotor bars. The field, volt- age produces an ac current and the resulting rotor 2 I R losses are dissipated at the expense of the kinetic energy stored in the revolving parts. motor finally comes to rest when all b. The kinetic energy in the rotor and load at 1760 is 5.48 X 10" 3 5.48 X 10" 3 - 424 kJ Ek = = Jn 2 X 25 X 1760 (3.8) 2 The the kinetic en- ergy has been dissipated as heat in the rotor. is short. r/min minals are connected to the dc source. slightly higher than the rated not overheat, because the braking time normally. Thus, a 3-phase, 4-pole induction motor produces 4 dc poles, no matter is current of the motor. However, the stator will the stator. Consequently, the rotor absorbs 424 kJ during the braking period. ELECTRICAL MACHINES AND TRANSFORMERS 310 c. The average braking torque 7" can be calculated An = 1760 - factor 1.15. The allowable temperature 10°C higher than from the equation tors operating at 9.55 TAt/J 9.55 T= 19.2 T X (3.14) X (4 vided. This is because even Abnormal conditions 14.10 in the stator, to internal overheating of the bearings, etc.) or to external conditions. External problems may be caused by any of the following: 1 . not much as 125 percent, as if recommended the external frame perature of the windings Abnormal motor operation may be due problems (short-circuit overloads as to carry long as supplementary external ventilation N-m may pro- is for long periods is cool, the tem- be excessive. 14.12 Line voltage changes The most important consequence of a line voltage change is its effect upon the torque-speed curve of the motor. In effect, the torque at any speed Mechanical overload mo- normal load. During emergencies a drip-proof motor can be made 60)/25 then rise is that permitted for drip-proof pro- is portional to the square of the applied voltage. Thus, 2. Supply voltage changes 3. Single phasing 4. Frequency changes the stator voltage decreases by if voltage drop to the We will examine the nature of these problems in the sections that follow. According motor shall ±10% of the nominal voltage, and for any frequency within of the nominal frequency. If the voltage and frequency both vary, the sum of the two percentage changes must not exceed 10 percent. Finally, designed tors are up to 1000 to the mo- much On when sea level. At higher altitudes the the permissible limits poor cooling afforded by the thinner due air. Mechanical overload as twice their rated power for short periods, As may the line. less than its rated value. the motor is is too high running, the flux per pole will be at full-load, this current, with the result that the temperature increases slightly and the power factor is slightly reduced. the 3-phase voltages are unbalanced, they can produce a serious unbalance of the three rents. line cur- This condition increases the stator and rotor losses, yielding a higher temperature. Although standard induction motors can develop as much drawn from the other hand, if the line voltage A voltage un- 3.5% can cause the temperature to increase by 15°C. The utility company should be notified whenever the phase-to-phase line voltages differ by more than 2 percent. balance of as 14.11 often produced during start-up, due increases both the iron losses and the magnetizing If may exceed is starting current above normal. For a motor running to operate satisfactorily at altitudes m above temperature all heavy at A line a result of the lower voltage, the starting torque be to national standards, a operate satisfactorily on any voltage within ±5% 10%, the torque every speed will drop by approximately 20%. little as they should not be allowed to run continuously be- yond their rated capacity. heating, Overloads cause over- which deteriorates the insulation and duces the service life of the motor. In practice, the overload causes the thermal overload relays starter fore its box re- to trip, bringing the motor in the to a stop be- temperature gets too high. Some drip-proof motors are designed to carry a continuous overload of 15 percent. This overload capacity is shown on the nameplate by the service 14.13 Single-phasing If one or if line of a 3-phase line a fuse ning, the is accidentally opened, blows while the 3-phase motor machine is run- will continue to run as a single- phase motor. The current drawn from the remaining two lines will almost double, and the motor will be- gin to overheat. motor will The thermal eventually trip relays protecting the the circuit-breaker, thereby disconnecting the motor from the line. SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS % 311 breakdown torque 250 'X 200 3-phase P" ^ 150 i ^pull-up torq ue nominal torque T 100 full — load i\ i 50 sirigle- phase — i i i \ \ \ \ \ ! \ i 40 20 80 60 100 % speed \ Figure 14.11 Typical torque-speed curves when a 3-phase squirrel-cage motor operates normally and when it operates on single-phase. The torque-speed curve a 3-phase down seriously affected torque decreases to about value, all. is when motor operates on single phase. The break- 40% and the motor develops no of its original starting torque at Consequently, a fully loaded 3-phase motor simply stop if one of its lines is resulting locked-rotor current normal 3-phase enough LR current. may suddenly opened. The is about It is to trip the circuit breaker or to 90% of the therefore large blow the 1 4. II each other closely single-phase until the breakdown is 50 Hz may cause problems when they nected to a 60 torque approaches the in some to gear 1 large distribution system, except during a is major dis- However, the frequency may vary electrical signif- energy generated by diesel engines or gas turbines. The in a hospital, the electrical system on a ship, and the generators camp, are examples of this is in a lumber type of supply. The most important consequence of change may 20% not be acceptable we either have motor speed or supply an expen- 50 Hz source. well on a 60 Hz line, but terminal voltage should be raised to 6/5 (or its of the nameplate rating. The 20%) torque is new breakdown then equal to the original breakdown torque and the starting torque is only slightly re- remain satisfactory. A 60 Hz Important frequency changes never take place on a emergency power supply this duced. Power factor, efficiency, and temperature torque. on isolated systems where and A 50 Hz motor operates rise icantly the are con- system. Everything runs applications. In such cases down sive auxiliary 14.14 Frequency variation turbance. Hz faster than normal, fuses. shows the typical torque-speed curves of a 3-phase motor when it runs normally and when Note that the curves follow it is single-phasing. Fig. Machine tools and other motor-driven equipment imported from countries where the frequency a frequency the resulting change in motor speed: if the frequency drops by 5%, the motor speed drops by 5%. but its motor can also operate on 83%) of its nameplate and fore, a 50 Hz line, terminal voltage should be reduced to 5/6 (or value. The breakdown torque same as be- starting torque are then about the and the power ture rise factor, efficiency, and tempera- remain satisfactory. 14.15 Induction motor operating as a generator Consider an electric train cage induction motor that wheels. As powered by a is directly the train climbs up a hill, squirrel- coupled the to the motor will ELECTRICAL MACHINES AND TRANSFORMERS 312 Figure 14.12 The makes electric train the round trip be- tween Zermatt (1604 m) and Gornergrat (3089 m) in Switzerland. The drive is provided by four 3-phase wound-rotor induction motors, rated 78 kW, 1470 r/m, 700 V, 50 Hz. Two aerial conductors constitute phases A and B, and the rails provide phase C. A toothed gear-wheel 573 mm in diameter engages a stationary rack on the roadbed to drive the train up and down the steep slopes. The speed can be varied from zero to 14.4 km/h by means of variable resistors in the rotor circuit. The rated thrust is 78 kN. (Courtesy of ABB) run than synchronous speed, devel- at slightly less overcome both oping a torque sufficient to and the force of At the top of the gravity. level ground, the force of gravity friction hill, to overcome the The motor runs at and the motor has only into play friction of the rails light load and the and very close What happens when to air. move begins it to rotate is to a it gasoline engine (Fig. as the engine speed exceeds the synchronous speed, the motor becomes a generator, delivering active it is power P to the electrical system connected. However, to create its to mag- in- speed. Thus, a higher engine speed produces a mo- develops a counter torque that opposes the same As soon effect as a coupled to the speed. However, as soon as this takes place, the tor to above synchronous crease in speed. This torque has the and coupling line 14.13). motor has to absorb reactive power power can only come from the ac line, with the result that the reactive power Q flows in the opposite direction to the active power P (Fig. 14. 3). The active power delivered to the line is directly proportional to the slip above synchronous the train begins to accelerate and because the motor phase which synchronous speed. downhill? The force of gravity causes the train wheels, on no longer comes We can make an asynchronous generator by connecting an ordinary squirrel-cage motor to a 3- netic field, the Q. This 1 brake. However, instead of being dissipated as heat, mechanical braking power the 3-phase line in the is 3-phase system returned to the form of electrical energy. An in- P duction motor that turns faster than synchronous speed acts, therefore, as a generator, it ^ gasoline converts the engine mechanical energy and this chine is energy receives into electrical energy, returned to the induction line are rarely motors Such a ma- a motor In cranes, for to run running off it to the line. the G squirrel-cage induction motor applications that above synchronous speed. motor receives power from f a example, during the lowering cycle, "load" and returns ooo QHill used to drive trains (Fig. 14. 12), there are several industrial may cause the line. called an asynchronous generator. Although 3-phase is it mechanical Figure 14.13 Gasoline engine driving an asynchronous generator connected to a 3-phase opposite directions. line. Note that Pand Qflow in SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS 313 Solution a. The apparent power drawn by when it operates as a motor is = = = S 1.73 X 440 X 31.2 kVA = for put reached is very small at 3%. The reactive power may be supplied by a group of capacitors connected to the terminals of the moWith this is slightly When least r/min /= produces pni\2{) = 4 a at frequency a speed of slightly X 2400/120 = 80 The terminal voltage of with the capacitance, but its less the less connected reactive when power as 2400 than Consequently, the capaci- /c phase is = QIE = 5700/440 = 13 A magnitude is limited by is The capacitive reactance per phase insuffi- Xc ~ Ell = = 34 il at least as the machine normally ab- is 440/13 The capacitance per phase must be at least operating as a motor. l/2ir/X c = 1/(2tt We = 78 [xF 40 hp, 1760 r/min, 440 V, 3-phase squirrel-cage induction motor as an asynchronous to use a The motor 84%. rated current of the and the full-load power factor is is 41 A, Figure 14.15 shows connected. Note that tive a. The voltage in delta. Example 14-6 generator. 2 because the capacitors are C= wish 26.2 the generator increases capacitor bank must be able to supply much 2 5.7 kvar per phase. 440 tive current per generator voltage will not build up. The sorbs is V is Hz. saturation in the iron. If the capacitance cient, the power absorbed machine operates as an asynchronous 17^-3 = per phase than that corresponding to the speed of rotation. Thus, a 4-pole motor driven reactive generator, the capacitor bank must supply at arrangement we can supply a 3-phase The frequency generated kW 1 load without using an external 3-phase source (Fig. 14. 14). is 0.84 = V31.2 2 = 7 kvar slips, typically less than tor. 26.2 Q = \ S2 - P However, the rated out- 41 power absorbed active P = S cos B = 31.2 X The corresponding greater electrical output. machine V3 EI The corresponding Figure 14.14 Capacitors can provide the reactive power for any asynchronous generator. This eliminates the need a 3-phase external source. the Calculate the capacitance required per phase X 60 X how if 34) the generating system is the load also absorbs reac- power, the capacitor bank must be increased to if provide it. the capacitors are connected in delta. b. At what speed should the driving engine run generate a frequency of 60 Hz? to b. The driving engine must turn at slightly more than synchronous speed. Typically, the slip ELECTRICAL MACHINES AND TRANSFORMERS 314 14.16 Complete torque-speed characteristic of an induction machine 30 = = Q 7kVar kW t ^ OOP 4D= 1840 r/min I ^>25 kW On j^r B We have seen that a 3-phase squirrel-cage induction motor can also function as a generator or as a brake. — motor, generator, These three modes of operation and brake — merge into each other, as can be seen from the torque-speed curve of Fig. curve, together with the adjoining Figure 14.15 See Example 14-6. grams, 14. 16. power flow illustrates the overall properties This dia- of a 3-phase squirrel-cage induction machine. should be equal to the full-load machine operates as slip = when the a motor. Consequently, 1800 = 40 The engine should slip - 1760 r/min We see, for example, that when the shaft turns in same direction as the revolving field, the induction machine operates in either the motor or the generator mode. But to operate in the generator the mode, the therefore run at an approxi- must turn mate speed of n shaft must turn faster than at less than synchronous speed. Finally, in order to operate as a brake, the shaft - 800 + 40 = 1840 r/min 1 must turn in the opposite direction to the revolving flux. V- u BRAKE synchronous speed. Similarly, to operate as a motor, the shaft n m r | speed stator rotor + 2 h i rotor stator T = torque developed stator n = speed of rotation by the machine Figure 14.16 Complete torque-speed curve of a 3-phase induction machine. n s = synchronous speed of the revolving field SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS 14.17 Features of a wound-rotor induction motor sult, it imum possible for the motor to develop is far, we have directed our attention cage induction motor and generator and brake. motor is the its to the squirrel- that this type is of in industry. However, the wound-rotor induction motor has cer- trial 1 applications. make attractive in special indus- it These may be listed as follows: final We have already seen that for a given load, an in- crease in rotor resistance will cause the speed of an induction motor to we Thus, by varying the exter- fall. wound-rotor motor we can obtain want, so long as nous speed. The problem Frequency converter We now examine in the shortest 14.19 Variable-speed drives any speed Variable-speed drives 3. speed can be reached possible time. nal resistors of a Start-up of very high-inertia loads . 2. max- related properties as a The reason one most frequently used tain features that its torque during the entire acceleration period. Thus, the So 315 as heat in the resistors it is that the is makes for a very inefficient system, which becomes too costly these applications. below synchro- power dissipated when the motors have ratings of several thousand horsepower. 14.18 Start-up of high-inertia loads around this problem by connecting the We get slip-rings to an electronic converter. The converter changes the We recall that whenever a load is brought up to speed by means of an induction motor, the energy dissipated in the rotor is equal to the kinetic energy power at system (Fig. this 14. 17). imparted to the load. This means that a high-inertia speed control system load will produce very high losses in the rotor, caus- that become excessively ing it the wound-rotor motor in the to is hot. itself is dissipated is is a result, such a variable- very efficient, is lost in the form of in the sense heat. that the external resistors can be varied as the motor picks up speed. 14.20 Frequency converter A conventional remains cool. Another advantage power As The advantage of that the heat external resistors connected to the slip-rings. Thus, the rotor little into power at line frepower back into the 3-phase low rotor frequency quency and feeds As a re- wound-rotor motor may be used as a frequency converter to generate a frequency different from that of the utility company. The stator of Figure 14.17 The water supply in the City of Stuttgart, Germany, provided by a pipeline that 11 0 km long. Constance is in is 1 .6 m in is diameter and The water is pumped from Lake the Alps. The pump in the background driven by a wound-rotor induction motor rated at 3300 kW, 425 to 595 r/min, 5 kV, 50 Hz. The variable speed enables the water supply to be varied according to the needs of the city. The enclosed motor housing seen in the foreground contains an air/water heat exchanger that uses the 5°C water for cooling purposes. During connected to the slipup to speed the sliprings are connected to an electronic converter which feeds the rotor power back into the line. (Courtesy of Siemens) start-up, liquid rheostats are rings, but when the motor is ELECTRICAL MACHINES AND TRANSFORMERS 316 R n R R wound-rotor induction "motor" Figure 14.18 Wound-rotor motor used as a frequency converter. Figure 14.19 Power flow in a frequency converter when the output frequency machine the wound-rotor line, and the rotor by a motor is M (Fig. is driven 14. 1 connected at rotor supplies power E 2 and to the 3 -phase load at a voltage both of which depend upon the slip. frequency / 2 Thus, accord- Calculate a. , b. we have ing to Eqs. 13.3 and 13.4, greater than the line frequency. an appropriate speed The 8). is to the utility The The is turns ratio of the stator to rotor windings when the rotor same direction as rotor voltage and frequency driven 720 r/min at in the the revolving field h = (13.3) sf c. (13.4) In general, the desired times that of the 13.3, in utility frequency is company. According be positive and greater than 1 . As is to Eq. rotor voltage and frequency driven a. The turns ratio a result, the shaft a ing flux. the frequency converter is when to the the rotor revolving Solution must be rotated against the direction of the revolv- The operation of 720 r/min opposite at field two or three order to attain this frequency the slip must The b. The slip at is = E]/Eoc = 2300/500 = 4.6 720 r/min is - (1 then identical to that of an induction motor operating as a brake. However, the dissipated as heat in the rotor, power P ]n usually is now .v = available to supply power to the load. The converter acts as a generator, and the active in Fig. 14.19. power flow Note how similar is as shown this is to the power flow when an induction motor runs = The (;? s n)ln, = 0.6 rotor voltage at 720 r/min E 2 = sEoc = = 300 V as a 800 - 720)/l80() 0.6 is X 500 brake (Fig. 14.16). The Example 14-7 A3-phase wound-rotor induction motor has of 150 hp (—110 kW), 1760 r/min, 2.3 kV, 60 Hz. ro- The ro- tor voltage tor is between the slip-rings is 500 V. driven by a variable-speed dc motor. is f2 = sf= 0.6 X 60 = 36 Hz a rating Under locked-rotor conditions, the open-circuit rotor frequency c. The motor speed is considered to be negative ( — ) when it turns opposite to the revolving field. The slip at —720 r/min is SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS = (n - n)ln = (1800 - (-720))/1800 = (1800 + 720)/ 1800 s s The converter must s at a speed of 1800 r/min. The negative sign indicates that the rotor must run opposite to the re- volving 1.4 The therefore be driven 317 field. The induction motor driving the converter must, therefore, have a synchronous rotor voltage and frequency at speed of 1800 r/min. 720 r/min b. are E2 = = sE„ c 1 X 500 .4 The rotor delivers an output of corresponds to Ppated in the rotor, = 700 V f2 = sf= 1.4 X 60 = 84 Hz P useful power delivered The power P r transferred is ]V to a load (Fig. 14.20). from the stator to the rotor P, 60 kW. This but instead of being dissi- = P vh = 20 is 60/3 (13.7) kW Example 14-8 We wish rotor 60 to use a 30 kW, 880 motor as a frequency converter kW at an The power 60 Hz wound- r/min, converter (F) to generate c. frequency is 60 Hz, calculate the following: a. The speed of the induction motor (M) b. The Fig. 14. 19 active power delivered to the stator of the c. d. Will the frequency converter overheat under Solution To generate 1 80 Hz the slip the small cop- Fig. 14.20, we can see how must be rotor receives 20 kW of power from the stator and 40 kW of mechanical power from the driving motor M. The rotor converts this power into 60 kW of electrical power at a frequency of 180 Hz. Induction motor M must therefore have a rating of 40 kW, 60 Hz, 800 r/min. summary, the 1 f2 = (13.3) sf 180 X 60 s load from which f s The stator is still = = 60 Hz 3 fed from the 60 Hz line, con- sequently, the synchronous speed of the converter ven is at a 900 r/min. The converter must be dri- speed n given by s = (/? s 3 = (900 - h)//? s - (13.2) frequency converter driving «)/900 motor from which Figure 14.20 n = -1800 r/min the into (and out of) the con- electrical these conditions a. kW plus the stator. verter. In The power of the induction motor (M) and power flows active frequency converter in The remaining power input to the rotor amounting to (60 - 20) = 40 kW, is derived from the mechanical input to the shaft. By referring to that dri- ves the frequency converter equal to 20 per and iron losses approximate frequency of 180 Hz (see Fig. 14.18). If the supply-line input to the stator of the frequency is See Example 14-8. 3 1 d. ELECTRICA L MA CHINES A ND TRA NSEORMERS 8 The because the 20 its 14-10 stator of the converter will not overheat kW it absorbs much is nominal rating of 30 kW. The rotor overheat either, even though The increased power arises from the the voltage induced in the rotor higher than at standstill. The is and the rotor stator how much 60 kW. fact that 14-11 more at A is By breakdown torque and are the nected to a 520 180 V line. Explain is con- how the following parameters are affected: twice effective, a. Locked-rotor current b. Locked-rotor torque c. No-load current d. No-load speed e. Full-load current f. Full-load iron losses in the stator are normal. Questions and Problems g. 14-12 Practical level 50 r/min 1 line. 3-phase squirrel-cage induction motor 60 Hz, consequently, the is 1 having a rated voltage of 575 V, iron losses in the is V, locked-rotor torque reduced? probably not overheat. The will frequency is 60 Hz, will not rotor will be high because the frequency normal speed, the cooling 50 hp, 440 connected to a 208 V, 3-phase three times Hz; however, because the rotor turns standard squirrel-cage induction motor less than delivers it A rated at a. power factor Full-load efficiency Referring to Fig. 14.6, if we eliminated the gearbox and used another motor directly 14-1 What is between a drip-proof the difference coupled motor and an explosion-proof motor? 14-2 What is the approximate life expectancy of a motor? 14-3 Explain a NEMA Design D unsatisfactory for driving a 14-4 Identify the b. 14-13 why to the load, motor How many Draw poles would the motor have? the pump. motor components shown 14.5). in Give the values of the LR, pull-up, and breakdown torques and the correflow of active power 3 -phase induction a. b. 14-6 As As a motor when it operates 14-14 motor A 300 b. 14-8 14-9 to start If the line on such a line? b. c. 14-15 relates to induction motors. can bring an induction motor to a quick stop either by plugging from method produces the citing the stator in the 60 Hz it motor? Explain. Which amount of 2 I R the losses. voltage then drops to 1944 V, The new speed, knowing We The new power output 2 The new I R losses in the heat that the load rotor wish to make an asynchronous generator using a standard squirrel-cage induction tor rated at 40 (Fig. 14. 14). or by ex- a dc source. least 590 r/min. Calculate torque remains the same for the following applications: Give some of the advantages of standard- We V, 3-phase, calculate the following: A saw in a lumber mill A variable speed pump it 2300 if type of ac motor would you recom- ization as and r/minl. approximate value of the rotor a. a. hp, full-load speed of Will a 3 -phase motor continue to rotate What mend lbf squirrel-cage induction motor turns at a a brake motor be able [ft in a one of the lines becomes open? Will the 14-7 squirrel-cage induction motor, rated at 30 hp, 900 r/min (see Fig. sponding speeds Show power its the typical torque-speed curve of a NEMA Design C is Fig. 14.3. 14-5 what would output have to be [hp]? hp, 208 V, mo- 870 r/min, 60 Hz The generator is driven at 2100 r/min by a gasoline engine, and the load consists of three 5 12 resistors connected in wye. The generator voltage builds up when three 100 fxF capacitors are connected in SELECTION AND APPLICATION OF THREE PHASE INDUCTION MOTORS wye is a. across the terminals. 520 c. d. The tor e. voltage 14-19 is A 30 000 hp— Advanced 14-20 best suited to drive the 60 Hz hp, 13.2 kV, 3-phase, A plant. The motor runs at a. an it has an efficiency of 98. LR are respectively 0.7 1 comThe motor power drives the speed of 4930 r/min. % and a c. d. e. 14-21 in Which A The is of the two rotors will be the hottest, 60 Hz, drives rotor is in speed? a belt conveyor. connected the slip rings in wye and a. b. is 530 V. Calculate the The The rotor winding resistance per phase resistance that must be placed in series with the rotor (per phase) so that the motor knowing 14-22 A 150 hp, 40 hp at a speed of 600 r/min, that the line voltage water tempera- is load, close to its is running The motor and compressor in Problem at 1.3 14-16 are started on reduced voltage, the acceleration period compressor has a The squirrel-cage 2 000 lb -ft a J of 18 How What long will is clocked ( 1 .2 pu), is equal to calculate the b. The magnitude of the plugging torque The moment of inertia of the rotor In Problem 14-22 calculate the energy a. motor rotor alone has is that the torque exerted 14-23 . it dissipated in the rotor during the plugging take to bring the motor and compressor up b. no- following: of inertia of lb ft~ referred to the Assuming the starting torque 0.25 pu. The is moment s. during the plugging interval and the average starting torque during a. at synchronous speed of reversed, and the stopping time shaft. kV 1200 r/min. The stator leads are suddenly exchanger. 130 000 2.4 165 r/min, 440 V, 60 Hz, 1 3-phase induction motor ture as the water flows through the heat 14-18 the following; cooled by 350 gallons (U.S.) of water through the heat exchanger per minute. has the shortest acceleration nominal open-circuit voltage between circulating Calculate the increase motor and, 3-phase, wound-rotor induction motor 2.3 kV, pu and 4.7 pu. Problem 14-16 D class .4 this having a rating of 150 hp, 1760 r/min, torque and current The full-load current The total losses at full load The exact rotor TR losses if the windage and friction losses amount to 62 kW The LR current and torque The torque developed at the compressor shaft The motor 1 1800 r/min. Could to Which motor will deliver 14-17 inertia load of after reaching the no-load Calculate the following: a. B induction hp. squirrel-cage. Design 1 time from zero to 1200 r/min? b. by means of a gearbox, The motor deliver if so, exact full-load speed of 1792.8 r/min and factor of 0.90. [hp] can the motor be replaced by a a turbo compressor in a large oxygen- at a should be used, and what level kgnr, from 0 air-to-water cooled induction motor drives pressor line. motor accelerates an manufacturing kW, without overheating? following gasoline engines are which one 10 at to be con- approximate speed of the motor? What power b. Hz line voltage will be the generator? b. What a. stator current If the induction motor rated nected to a 60 bank available— 30 hp, 100 hp, and 150 14-16 A 3-phase 1450 r/min, 380 V, 50 Hz has The approximate frequency generated active power supplied to the load reactive power supplied by the capaci- The The b. If the line V, calculate the following: to speed, at interval. no-load? the energy dissipated in the rotor during the starting period [Btu]? 14-24 A 3-phase, rating of 8-poIe induction motor has a 40 hp, 575 V. 60 Hz. It drives a 320 ELECTRICAL MACHINES AND TRANSFORMERS steel flywheel having a diameter of 3 inches and a thickness of 7.875 1 .5 The in. Industrial application 14-27 torque-speed curve corresponds to that of D a design a. in Fig. 14.5. of inerlia d. Calculate the rated speed of the motor and Motor B: 75 Draw [ft-lbf]. hp, 900 day. How erates about 6 hours per day. In b. Using Eq. time. r/min; lubricate Motor A runs continually, 24 hours per Motor B drives a compressor and op- at 0, 180, Problem 14-24 calculate the average a. lubricate every 10 000 hours of running time. 360.540. 720, and810r/min. 14-25 to be 40 hp the torque-speed curve of the motor and give the torques [N-m] motor have Motor A: 75 hp, 3550 r/min; every 2200 hours of running |lb-ft~|. Calculate the loeked-rotor torque a its the corresponding torque [ft-lbf]. c. in following schedule applies to two motors: Calculate the mass of the flywheel and moment b. motor given The bearings greased regularly, but not too often. The often should the bearings of each motor be torque between 0 and 180 r/min. 3. r/min, c. greased per year? 14 calculate the time required to accelerate the flywheel from 0 to 180 14-28 mium the flywheel at 180 r/min. 540 that this N-m in 14-29 A standard 40 hp motor, similar mass of line current The energy required to to is 14-30 the av- if A constant has windings similar to those shown 60 kg minals to make tion the min] when is The resistance V in Fig. is and 2 1 measured between 12 il. in the ter- What high-speed connec- resistance would you expect to measure between terminals 4 and Assuming energy horsepower, 2-speed induction motor rated 2 hp, 1760/870 r/min, 460 Gornergrat |MJ] | mo- climb from Zermatt The minimum time required trip g. one the energy sav- ings that accrue to the high-efficiency 14.8. f. to the Problem 14-28, costs $1723 82%. Calculate factor of at full-load the loaded train erage weight of a passenger e. in tor during the 3-year period. The approximate transmission when the motors are operating total is and has an efficiency of 90.2% and power The speed of rotation of the gear wheel when the train moves at 9 miles per hour The speed ratio between the motor and the The at full- $0.06/kWh. cost of energy described gear wheel d. $2243, runs during a 3-year period, knowing that the Calculate the following: c. at Calculate the cost of driving the motor addition to The train in Fig. 14. 12 has a mass of 78 500 lb and can carry 240 passengers. b. and an efficiency of 93.6%. load 12 hours per day, 5 days a week. the flywheel load. a. pre- time the load exerts a fixed counter-torque of 300 14-26 86% The motor, priced knowing r/min. V, 3-phase, energy induction motor has a power factor of d. Calculate the time required to accelerate the flywheel from 0 to 1780 r/min, 460 60 Hz, drip-proof Baldor Super E assuming no other load on the motor. Using Eq. 3.8 calculate the kinetic energy in A 40 hp, that 80 percent of the the train is going uphill and that 80 percent of the mechanical energy verted to electrical energy 6 electrical converted into mechanical energy is 14-31 A in the low-speed connection? 150 hp 7 1 175 r/min, 460 V, 3-phase, 60 Hz induction motor has the following recon- properties: when going downhill, calculate the total electrical en- ergy consumed during a round trip [kW-h]. no-load current: 71 A full-load current: 183 A SELECTION AND APPLICATION OF THREE-PHASE INDUCTION MOTORS locked-rotor current: 1 550 A full-load torque: 886 breakdown 2552 A a. ft-lbf b. torque: ft-lbf A power factor: from the main panelboard motor, 850 elboard is ft away. The voltage 480 V ture of the cable to the at the mo- the motor is impedance is purely re- approximate current started up across the line. Estimate the resulting starting torque. d. Compare it with the rated starting torque, percentagewise. pan- and the average temperais the cable c. 3-conductor 250 kcmil copper cable stretches Assuming when 32% circuit of the under locked-rotor conditions. sistive, calculate the locked-rotor torque: 1205 ft-lbf locked-rotor Determine the equivalent tor 321 estimated to be 25°C. i -32 In Problem 14-3 1 express the currents and torques in per-unit values. Chapter 15 Equivalent Circuit of the Induction Motor 15.0 Introduction secondary windings cal —one The preceding three chapters have shown that we single primary winding can describe the important properties of squir- we want if pensible. In this chapter'" circuit and observe tor Finally, We indis- We low-power and high-power mo- we develop its proper- is the same in equivalent cir- as that of a transformer, Chapter 10, Fig. 10.20. assume a wye connection E„ = r\ — stator a-, = stator leakage reactance x2 = rotor leakage reactance r2 — rotor winding resistance is very simi- for the stator The and circuit para- identical who winding resistance externa] resistance, effectively connected slip-ring and the neutral of the rotor primary windings and 3 identi- This chapter can be skipped by the reader source voltage, line to neutral between one construction to a 3-phase transformer. Thus, the to ) meters, per phase, are identified as follows: the equivalent circuit of an 3-phase wound-rotor induction motor have lime 1 acts exactly like its their basic differences. The wound-rotor induction motor motor has 3 5. the rotor, and a turns ratio of 1:1. Rx = lar in 1 previously developed under load. 15.1 A cuit (Fig. then analyze the asynchronous generator and determine ties it a conventional transformer, and so we develop the equivalent from basic principles. characteristics of a is On consider a and a single secondary wind- When the motor is at standstill, to gain even a better understanding of the properties of the motor, an equivalent circuit diagram we can ing in analyzing the behavior of the motor. rel-cage and wound-rotor induction motors without using a circuit diagram. However, each phase. set for account of the perfect symmetry, Xm = magnetizing reactance Rm = resistance does not iron losses whose losses correspond to the and windage and friction losses study the more theoretical aspects of induction T= motor behavior. 322 ideal transformer having a turns ratio of 1 : EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR 323 Figure 15.1 Equivalent circuit of a wound-rotor induction motor at standstill. Figure 15.2 Approximation of the equivalent for 3-phase trans- the case of a conventional In acceptable circuit is load current true: / 0 may / . p compared negligible is However, in a motor this be as high as 40 percent of the air gap. Consequently, we cannot to the no longer is shown we in Fig. can 1 E2 U , on the primary and secondary side of in the secondary winding will become these new to Ei (the turns ratio This greatly simplifies the equa- induced is Fig. 1 5.2 motor when the rotor How when affected the Suppose the motor runs the rotor speed is ns ( I — motor starts a slip at s), s, where meaning /7 S is The frequency locked. is turning? that the syn- chronous speed. This will modify the values of be used. hp the exact circuit of Fig. 1 5. 1 if the slip is v, l ; is sfand I be equal motor were sta- the actual voltage sEi this changes the imped- ance of the secondary leakage reactance from j.\ 2 to jsx 2 Because resistors are not frequency-sensitive, . the values of r2 and the two together tance For motors under is shows 5.3 E2 would ) E2 = a true rep- is resentation of the is it is But because the compromising accuracy.* / Directing our attention to the secondary side, the amplitude of the induced voltage tions that describe the behavior of the motor, with- out 1 where operating conditions. tionary. it sfi the frequency of the source E„. Fig. to the input terminals, as shift 5.2. and the ideal transformer T. Furthermore, the frequency because of p eliminate the magnetizing branch. However, for motors exceeding 2 hp, I x we would be justified in removing the magnetizing branch composed of jX m and R m because former, the exciting current /0 motors above 2 hp. R2 , R K remain to the same. Let us form a single secondary lump resis- given by should R, = (I5.l) ELECTRICAL MACHINES AND TRANSFORMERS 324 external resistance frequency frequency f sf Figure 15.3 Equivalent circuit of a wound-rotor motor the stator in The is details of the Fig. I5.4a, it is running at a secondary circuit are shown and the resulting current ^~ = when But the frequency of the voltages and currents f. sE, Z_ a-£, „ A- = • R2 VRj + +js.x 2 slip s. in The frequency the rotor of the voltages in I 2 is ^ - B /U'2 (15.2) 2 Cvx,) where total (3 = resistance arctan sx 2 IR 2 ( The corresponding phasor diagram Fig. 5.4b. 1 and currents is sf. diagram It is 1 5.3) circuit shown is of rotor in important to realize that this phasor relates to the frequency sf Consequently, frequency = sf it cannot be integrated into the phasor diagram on the primary there is side, fect, the . /, and /, (frequency /) absolute value of that of U. /2 is a direct relationship between in the rotor £, and where the frequency Nevertheless, I2 (frequency Furthermore, the phase angle is exactly the same (a) sf) in the stator. In ef- exactly the is /, / as that (3 same as between between E2 and This enables us to draw the phasor diagram for £, and /, as shown in Fig. 15.5. To summarize: 1 . The effective value of tive value of 2, is /, equal to the effec- even though their frequencies (b) are different. 2. The effective value of tive value of 3. E2 E equal to the effec- is { divided by the slip The phase angle between as that between E 2 and I 2 E . and ] /] Figure 15.4 a. ,s. is the same b. Equivalent circuit of the rotor; E2 and / 2 have a frequency sf Phasor diagram showing the current lagging behind the voltage by angle (3. EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR we can Thus, on the primary side = U = /, R2 + 325 write (15.4) jsx 2 Therefore, (15.5) R2 Zn jx2 S The impedance Z2 seen between nals 1, 2 of the ideal transformer the primary termiis, therefore, Figure 15.6 Equivalent circuit of a wound-rotor motor referred to the primary (stator) side. Z2 = 1 = + jx 2 (15.6) /, As a result, we can This equivalent circuit of a wound-rotor induc- simplify the circuit of Fig. motor The leakage reactances jx Jx 2 can now be lumped together to create tion a single total leakage reactance jx. motor 15.3 to that shown in Fig. 15.6. it is is so similar to that of a transformer that not surprising that the wound-rotor induction { total It is equal to the leakage reactance of the motor referred to the stator side. is sometimes called a rotary transformer. The equivalent circuit of a squirrel-cage induction motor is the same, except that R 2 is then equal to the equivalent resistance r2 of the rotor alone referred to the stator, there being 15.2 no external resistor. Power relationships some power relationships for the 3-phase induction motor. The following equations The equivalent circuit enables us to arrive at basic electromechanical can be deduced by visual inspection of the equiva- V P = lent circuit of the wound-rotor motor (Fig. 15.7): (sx 2 h arctan sx 2 IR 2 Figure 15.5 voltage and current in the stator are separated by the same phase angle p, even though the frequency The is different. The final equivalent circuit of the wound-rotor induction motor is shown in Fig. 15.7. In this di- agram, the circuit elements are fixed, except for the resistance R 2 /s. Its value depends upon the slip and hence upon the speed of the motor. Thus, the value of R 2 ls will vary from motor goes from start-up speed (s = 0). (.v R 2 to infinity as the = 1) to synchronous Figure 15.7 The primary and secondary leakage reactances x-, and x2 are combined to form an equivalent total leakage reactance x. ELECTRICA L MACHINES AND TRANSFORMERS 326 1 . Active power absorbed by the motor Efficiency of the motor 10. is PJP n = 2. Reactive power absorbed by the motor Note: The preceding quantities are "per phase"; is some must be Q = E^IX m + Irx 3. motor Apparent power absorbed by the motor = S 2 \ P + Q is: multiplied by 3 to obtain the total quantities. is 15,3 Phasor diagram 2 of the induction 4. Power factor of the motor If cos 6 = PIS we use current /, in Fig. 5.7 as the reference phawe obtain the complete phasor diagram of the 1 sor, wound-rotor motor shown in Fig. 5.8. In this diagram (and also in future calculations) it is useful to define an impedance Z, and angle a as follows: 5. Line current 6. Active power supplied to the rotor 1 is is r Power cuit dissipated as I R losses in the rotor cir- V>~ +jc 2 a — arctan xlr In these equations r, is { ]V 2 I R 2 -sP Because stator. v and Mechanical power developed by the motor is a are fixed, r, and jx are than r and so the angle x = 9. P, ( 1 - lation of Torque developed by the motor j 9.55 - Pm 111 9.55 P - I ns n _ 9.55 Pl\ ^ is (\ s)' s) (1 5.7b) motor referred fixed, is to the follows that it 1 000 hp, jx is a approaches In the equivalent circuit .V) 5.7a) Z of Fig. 1 much larger 90°. 5.7, the calcu- mechanical power, torque, and speed de- and R 2 is. The magnetizing branch pends upon r jx, R m and jX m does not come into play. Consequently, { for these calculations the equi valent circuit and cor- responding phasor diagram can be simplified to that shown in Figs. 15.9 and 15. 10. v Figure 15.9 As far as mechanical power, Figure 15.8 Phasor diagram of the voltages and currents in Fig. 1 5.7. The power factor of the motor is cos tt. { irrespective of the speed of the motor. motors above In large (1 the stator resistance and jx the total leakage reactance of the is P = 8. 2 = Z, P = ICR 2 /s 7. motor is torque, and speed are concerned, we can neglect the magnetizing branch Xm Rm This yields a simpler motor behavior. and sis of . circuit for the analy- EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR 1 Phasors . AB angle 2. BC and have the same length and between them the angle CAB = angle 327 is ( 80 1 ACB = — a)°. a/2 Consequently, /t hR^s /,r, Figure 15.10 Phasor diagram of the circuit of Fig. 1 5.9. Note that phasor / Z is always a degrees ahead of phasor 1 15.4 Breakdown torque and speed We have seen that the torque developed by the motor is given by T = 9.55 P /n where P is the power der livered to the resistance to a basic imum power s v R 2 is transfer theorem, the (and therefore the torque the value of R 2 /s is is breakdown torque slip at = R 2 IZ sh The current is (15.9) X breakdown torque at the is According power is max- maximum) when = /, b £,/(2Z, cos a/2) The breakdown torque (I5.10) is equal to the absolute value of im- pedance Z|. Thus, for maximum R 2 /s = Z Under these circumstances, voltage drop across Z, We (Fig. 15.9). The = EJ2 cos ^ /,Z, /-,. 1 is torque - (15.8) l the magnitude of the R 2 /s. We note that the magnitudes of both the breakdown Tb and torque circuit resistance However, the R2 pends upon The phasor diagram corresponding to condition is shown in Fig. 5. It is a 1 1 . breakdown current the / ib are fixed, sense that they are independent of the rotor in the can therefore write (I5.ll) % (4Z, cos- a/2) S /? equal to that across 1 9.55 Th = . R2 . breakdown torque de- slip at the Indeed, if R2 = Z,, the breakdown torque coincides with the starting torque because s b this special special case of the phasor diagram of Fig. 15.10. Simple geom- etry yields the following results: then equal to l . These conclusions are all is borne out by the torque-speed curves in Fig. 13. 18 (Chapter 13). In the case of squirrel-cage motors, the resistance R 2 becomes equal to r 2 , which is the resistance of the rotor alone reflected into the stator. In practice, the angle a lies correspond tors. In between 80° and 89°. The larger angles to medium- and high-power cage mo- R 2 /Z such machines the ratio as 0.02. Consequently, the at slips as small as 2 percent. 15.5 Equivalent circuit of practical Figure 15.11 Phasor diagram when the motor develops its maximum torque. Under these conditions R2 ls = Z,. can be as low } breakdown torque occurs two motors The impedances and resulting equivalent circuits of two squirrel-cage motors, rated 5 hp and 5000 hp are given in Figs. 1 5. 1 2 and 1 5. 1 3, together with the ELECTRICAL MACHINES AND TRANSFORMERS 328 motor The motors ratings. wye and are both connected in 3. a/2 = 4. The slip at impedances are given per phase. the 15.6 Calculation of the We now will calculate the 2 l + .5 6 2 = 6. = 76° 1 8 nb for arclan .v/r, arctan 6/1.5 l. 2/6. The - n s (\ = 1450r/min .v 440V/V 3 0.194) is = a/2 6900 V/V'3 Motor rating: 60 Hz, 1 800 r/min, 440 full-loiid cuitciu: 7 V, 5000 3-phase A lneked-rotor current: 39 A hp, 60 Hz. 600 r/min, 358 r2 rotor resistance j.x - total 1 1 .5 12 .2 12 reactance 1 6 6 1 A r2 rotor resistance 0.080 (2 j.x = total = magnetizing reactance 46 leakage reactance 2.6 12 12 no-load losses resistance 600 (2 (The no-load losses include the iron losses plus The no-load windage and 15 friction losses.) 1 stator resistance 0.083 fi Rm = no-load losses resistance 900 12 3-phase = /X m 10 (1 V, = /*, leakage reactance 6 12 6900 A locked-rotor current: stator resistance jXm = magnetizing rating: full-load current: = Rm = b) - 1800(1 2 Z, cos 5 hp, 0.194 is = = breakdown current at /,K Motor = 8 1 O 6. a is breakdown hp motor. V I. at breakdown torque Th /, b 0.788 ] The speed n b 5. and the corresponding speed n h and current the 5 breakdown = R 2 /Z = sb breakdown torque - 38°, cos a/2 losses of 26.4 kW for windage and kW (per phase) consist of friction and 1 1 .4 kW for the iron losses. Figure 15.13 Figure 15.12 Equivalent circuit of a 5 hp squirrel-cage induction Because there R2 ^ r2 motor. rotor, . is no external resistor in the Equivalent circuit of a 5000 hp squirrel-cage induction motor. Although this powerful than the motor gram remains the same. motor in Fig. is 1000 times more 15.12, the circuit dia- EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR - 7. A3 — = 2X6.18X0.788 The power to the rotor 2 P = = I r { R 2 ls = 2 26. X 26.1 A 2 I\ Z 6.18 9.55 P 9.55 X 4210 W is r 200 0 800 600 400 The total is torque r/min SPEED the torque developed is, therefore, 3 X 22.3 per phase. = 67 N-m. Figure 15.14 Torque-speed curve The same 15.7 Torque-speed curve We can determine the complete torque-speed curve of hp motor by selecting various values of slip and solving the circuit of Fig. 15. 12. Table 15 A, and the curve in is The given 5 hp, 440 V, r/min, 5000 hp 1 5. 15 to the breakdown torque. These characteristics are typical for large squirrel-cage induction motors. in Fig. 15.14. TORQUE-SPEED AND LOAD CHARACTERISTIC 5000 60 Hz squirrel-cage induction hp, 6900 V, 600 r/min, 60 Hz squirrel-cage induction motor /, IWj [A] s T n [N-ml [r/min J Torque IkN-ml 0.0125 2.60 649 3.44 1777 5.09 1243 6.60 1755 0.026 5.29 1291 6.85 1753 0.6 0.05 9.70 2256 12.0 1710 0.4 2 1 1.49 / p 4.9 1617 0 1616 8.2 23.4 1614 1610 1120 360 10.6 40.8 2921 480 17.7 64.7 1593 26.8 6095 540 30.8 80.4 1535 42.1 10114 570 51.7 89.5 1 0.03077 47.0 11520 581.5 68.2 93.1 1133 0.02 43.1 10679 588 79.8 95.1 878 0.0067 19.9 5000 596 90. 96.6 358 0.0033 10.2 2577 598 85.1 95.4 198 7.39 0.2 26.4 4196 22.3 1440 0.1 0.4 33.9 3441 18.3 1080 0.05 0.6 36.6 2674 14.2 720 0.8 37.9 2150 11.4 360 38.6 1788 0.026. -377 -600 6.3 14.4 = [A| [9*1 0 0.2 0 | 240 1620 9.49 [r/min [hpl 0 18.8 at s ElTcy 500 3547 developed cos a 2.98 17.2 s Speed 4.95 0.1 The rated power of 5 hp Total power 0.025 1 for the results are listed motor s made the results and Fig. the near-synchronous speed from no-load right up TORQUE-SPEED CHARACTERISTIC 1800 lists shows the torque-speed curve. Note the relatively low starting torque for this large motor, as well as TABLE 15B TABLE 15A a 5 hp motor. of calculations are motor. Table 15B and other characteristics the 5 1000 1200 1400 1600 1800 N-m 1800 that this N-m X = 4210 22.3 Note 22.3 (per is The breakdown torque Tb 8. breakdown torque phase ) 440 329 363 ELECTRICA L MA CHINES A ND TRA NSEORMERS 330 breakdown torque = 47 kN m (per phase) 254 V j110 Figure 15.16 Equivalent circuit of a 5 hp motor operating as an asynchronous generator. Note that a negative resis- tance SPEED Figure 15.15 Torque-speed curve 1 . is reflected into the primary circuit. Net resistance of branch R n = -48 + of a 5000 hp motor. 2. Z = \Rl + asynchronous generator have already learned tion motor can act as a generator if lent circuit for the 5 hp motor, it - it is driven above 3. Now power = V that a squirrel-cage induc- synchronous speed. that Current its = -46.5 ( - 46.88 /, other prop- is H is .r 46.5) 2 + 6 2 0 branch 1-2-3-4 in we have the equivawe can calculate the can generate, together with -2-3-4 Impedance of branch 1-2-3-4 15-8 Properties of an We I.5 l is - EIZ = 254/46.88 - 5.42 A erties as a generator. Let us connect the motor to a 440 V, 3-phase line and drive it at a speed of 1 845 r/min, which 4. Active power delivered to the rotor P = IcR 2 /s = is is 5.42- (-48) { 45 r/min above synchronous speed. The s = (/? s = (1800 - slip is = -1410 n)ln s - This negative power means that 1410 1845)/ 1800 flowing from the rotor - -0.025 5. The value of /? 2 Av in the equivalent circuit W is, The 2 l R there- P - Ifr 2 = ]r = 6. The negative resistance indicates that power is flowing from the rotor to the stator rather than from the stator to the rotor. Referring to Fig. 15.16 make the following calculations: we 35.2 5.42" P v X 1.2 W The mechanical power equal to is losses in the rotor are fore, R 2 ls = 1.2/(- 0.025) = -48 n W to the stator. input to the shaft is plus the losses P-]r in the rotor: p = p + p. = 1410 + = 1445 W 35.2 EQUIVALENT CIRCUIT OF THE INDUCTION MOTOR 7. The 2 I R 2 F = l, js = = r, 44. 5.42 2 X 1.5 W 1 = PJS = 1294/1502 0,861 = 86.1% cos 0 losses in the stator are 16. The efficiency of the asynchronous generator useful electric 8. The windage and iron plus mechanical input 9. The to the line feeding rotor to stator 1 . 3 = phases 3 9.55 = X 1 X 3 1445 2 = /, = 176 var a- = 5.42 2 X No-load 2 1 5,6, this /X m = 254 at /110 power absorbed by the motor X nv R m The power p that the value of /, is negligible Xm R m , at at no- R 2 /s is compared very togh to / 0 Thus, . . Their values can be de- no-load, as follows: Measure the stator resistance r, R u between any is generator terminals A, 762 2 VA '"I Run the motor is £ NL load current / NI = RlA.ll no-load using rated line-to(Fig. 15. and the 1 7). total Measure the no- 3-phase active . The following calculations of total apparent power total reactive power Q N] are then made: S NL and 1502/254 factor at the generator terminals at line voltage, power P Nl p an induction motor runs exceedingly small. Referring to Fig. means value of b. = S/E= = 5.91 A in the no-load the circuit consists essentially of the power 762 vars 2 line current / and x two terminals. Assuming a wye connection, the VP 2 + Q 2 = V1294 + 1502 , by means of is a. at the , termined by measuring the voltage, current, and Qi / When is magnetizing branch var is The test load, the slip 2 r2 tests. and so current Apparent power = N-m the equivalent circuit following is Total reactive S 22.3 6 Reactive power absorbed by the magnetizing B - is: 854 the equivalent circuit can be found + Qi = 176 + 586 = 15. hp The approximate values of Q = 14. X P 9.55 is 1445/746 Torque exerted by the driving motor X 1294 - 3882 W) is = 586 3. 5.81 X 3 Reactive power absorbed by the leakage reac- Q2 = E 1 = = tance reactance 12. 3 to drive the generator 15,9 Tests to determine 2, 1 = n W (P c for the 10. The horsepower needed T= v 1294 8. 1 = P - P]S - P - P v = ]410 - 44.1 - 71.7 = 7. 1 minus losses v = 89.5% 0.895 is Pc = power delivered from ir 1445 power delivered active motor the 1294 W 71.7 is Pc P power friction losses are P f + P v = E2 /R m = 254 2 /900 = 331 is ML ELECTRICAL MACHINES AND TRANSFORMERS 332 Figure 15.17 A Figure 15.18 no-load test permits the calculation of Xm and tfm of the magnetizing branch. A locked-rotor test permits the calculation of the total leakage reactance From these P + Pv windage, t The Rm resistance representing circuit of and iron losses friction, P + r /\ results xand the total resistance + r2 we can determine the equivalent the induction motor. Hence, is /WO 'ud ~ The magnetizing reactance ). More is: r, elaborate tests are conducted on large ma- chines, but the above-mentioned procedure gives adequate results that are Locked-rotor Under test motor the rotor of an induction current I almost is p Furthermore, the slip that r2 /s is s is locked, the stator is times six its rated value. equal to one. This means equal to r2 where r 2 , the rotor reflected into the stator. is the resistance of Because greater than the exciting current 70 , / is p we can 1 5.9, composed of the leakage reactance and the reflected rotor x, the stator resistance r h sistance R 2 /s = r 2 /\ mined by measuring — re- r2 Their values can be deter. the voltage, current, and most cases. Example 15-1 A no-load test conducted on a 30 hp, 835 r/min, 440 V, 3-phase, 60 Hz squirrel-cage induction motor yielded the following results: much neglect the magnetizing branch. This leaves us with the circuit of Fig. in when rated line voltage, No-load voltage (line-to-line): No-load current: 14 440 V A No-load power: 1470 W Resistance measured between two terminals: 0.5 (2 power The locked-rotor under locked-rotor conditions, as follows: test, conducted at reduced volt- age, gave the following results: a. Apply reduced 3-phase voltage that the stator current is to the stator so about equal to its rated Take readings of total E l R (line-to-line), 3-phase power P K (Fig. ] /, R and the , Locked-rotor current: JS' LR are then = F LR I 60 A 15. 18). Determine the equivalent The following calculations V W Locked-rotor power: 7200 value. b. Locked-rotor voltage (line-to-line): 163 circuit of the motor. made: Solution R [ \v J 3 Assuming the stator windings are connected wye, the resistance per phase = Cir/3/ lr 3/T R (r, + r 2 = P LR r, * ) From - 0.5 the no-load test a/2 we = find is 0.25 II in EQUIVALENT CIRCUIT OE THE INDUCTION MOTOR Rx = = ' 10 669 P m = 1470 2 5- 1 e Il /p nl = 440 = i46a 2 /(1470 ^lr'lr V3 /10 568 -3 X 14 2 X 0.25) A we /, R X 60V 163 = 60 A 2 - 7200 _ G,.r 15 c. - 0.67 (2 = 0.25 H = 0.67 - The value of Z, and the angle a The speed when the breakdown torque is The current I \ at the breakdown torque (see The value of a. Problem 15-2, draw the equivalent circuit if the motor runs at 950 r/min in is / 2 breakdown torque [N-m] ator? Calculate the torque of the machine. is „= 7200/(3 X 60 2 , the In the same direction as the revolving flux. Does the machine operate as a gener- b. r/3 V, calculate the d. 1.42(2 2 Total resistance referred to stator + ^ = 346 reached 2 333 X 60 3 3/iLR is 5 12 and the line- is Fig. 15.9) Total leakage reactance referred to stator — voltage to- neutral 15-3 x leakage reactance following: Pl R = V16 939 333 var the 12. If find b. 15 and wye-connected squirrel-cage motor equivalent rotor resistance of 0.5 a. = to the text, explain the the impedances, currents, having a synchronous speed of 900 r/min VA VS V °LR in Fig. 15.19. voltages in Fig. 15.2. W 2 shown is Without referring total 939 because has a stator resistance of 0.7 il and an the locked-rotor test 16 circuit meaning of 15-2 7200 r2 15. 1.) Questions and Problems 10 568 var x m = ^\[/Q\l = 440 = 18.3 n From seeEq. The equivalent W 1 R, 0; VA V10 669 2 - 1470 2 - R2 = (In a squirrel-cage motor, 440 X 14V 3 333 Draw runs ) flux. the equivalent circuit if the motor 950 r/min opposite to the revolving Does the machine operate as a genera- at tor? Calculate the torque. r { 1 0.25 0.25 = 0.42 5-4 n Q A 550 780 r/min, 3-phase, 60 Hz squirrel-cage induction motor running no-load draws a current of 12 A and a total power of 1500 W. Calculate the V, value of 1 Xm and Rm at per phase (see Fig. 15.2). 15-5 440 18.3 #n 146 f3 Q The motor rent of X 30 in Problem 15-4 draws A and when connected a power of 2.43 to a 90 a cur- kW V, 3-phase line under locked-rotor conditions. The resis- tance between two stator terminals is Figure 15.19 0.8 fl. Calculate the values of r h r 2 and Determining the equivalent x and induction motor (see , circuit of Example 15-1). a squirrel-cage the locked-rotor torque [N rated voltage. m| at 334 1 5-6 ELECTRICAL MACHINES AND TRANSFORMERS If motor in 6200 V, calculate the line voltage for the Fig. 15.15 the dropped new breakdown to / nd us trio I 15-9 torque and starting 5-7 Consider the 5 hp motor whose equivalent circuit torque. 1 a. A 440 V, 3-phase, motor has the following characteristics: If the = 1 x = 6 .2 il c. n 15-8 In compare magnetizing branch can be neglected, connected in series if a 4.5 fl resis- with each Problem 15-7 calculate the is connected in series line. starting torque and the breakdown torque reactor Calculate the 50 to obtain the calculate the value of the starting torque is Determine the values of the leakage reacat a frequency of 50 Hz. and the breakdown torque tor in Fig. 15.12. tance and the magnetizing reactance 1.5 fi x r2 shown ing reactances. b. = r is Calculate the values of the inductances (in millihenries) of the leakage and magnetiz- 800 r/min squirrel-cage 1 application if a 4.5 ft with each 15-10 The 5 V line-to-neutral voltage with the voltage at 60 Hz. hp motor represented by the equiva- lent circuit 503 it Hz same magnetizing current and of Fig. 1 5. 1 2 is connected (line-to-line), 3-phase, to a 80 Hz source. The stator and rotor resistances are assumed to remain the same. a. Determine the equivalent circuit when the line. motor runs b. at 2340 r/min. Calculate the value of the torque [N-m] and power [hp] developed by the motor. Chapter 1 Synchronous Generators some 16.0 Introduction other source of motive power. 3-phase voltage Three-phase synchronous primary source of all generators consume. These machines are the verters in the world. are the the electrical energy largest energy con- into electrical energy, in powers ranging up will study the construction In this chapter we and characteristics of these are based upon large, modern in view this materia] upon the dc exciting The frequency of the speed and the number of used when the power output However, generators. may wish rotates, a the speed of rotation and voltage depends upon the for greater outputs, and more practical the elementary principles cov- Section 8.6, and the reader ered it poles on the field. Stationary-field generators are 1500 to As induced, whose value depends current in the stationary poles. They convert mechanical energy MW. They upon we is A to re- to is less than 5 employ a revolving dc tor field. revolving-field synchronous generator has a stationary armature called a stator. before proceeding further. kVA. cheaper, safer, is it winding is The 3-phase sta- directly connected to the load, with- out going through large, unreliable slip-rings and 16.1 A Commercial synchronous brushes. generators insulate the windings because they are not sub- stationary stator also makes jected to centrifugal forces. Fig. 16. Commercial synchronous generators are built with The The field is excited by a dc generator, mounted on the same shaft. Note that the brushes on the commutator have to be connected to cut by a re- another set of brushes riding on slip-rings to feed as a dc generator. which is usually The armature possesses a 3-phase winding whose terminals are connected to three slip-rings mounted on the shaft. A set of volving armature. the dc current I x into the revolving field. 16.2 brushes, sliding on the slip-rings, enables the armature to is Number of poles an external 3-phase load. The number of poles on a synchronous generator de- driven by a gasoline engine, or pends upon the speed of rotation and the frequency be connected The armature a schematic alternator. A stationary -field synchronous generator has the salient poles create the dc field, is easier to diagram of such a generator, sometimes called an either a stationary or a rotating dc magnetic field. same outward appearance 1 it to 335 ELECTRICA L MA CHINES A ND TRA NSEORMERS 3 36 ils pilot exciter 25 kW 3-phase alternator 500 MW, 12 kV, 60 Hz Figure 16.1 Schematic diagram and cross-section view of a typical 500 synchronous generator and its 2400 kW dc exciter. The dc exciting current lx (6000 A) flows through the commutator and two slip-rings. The dc control current lc from the pilot exciter permits variable field control of the main exciter, which, in turn, controls /x MW . we wish to produce. Consider, for conductor that is p = example, a stator successively swept by the N and S 120.//// = 20 X 60/200 1 poles of the rotor. when an N ilar If a positive voltage is induced pole sweeps across the conductor, a sim- negative voltage is crosses the conductor, the induced voltage goes through a complete cycle. The same other conductor on the stator; that the alternator is true for every we can therefore de- frequency is given by From an (,6I) 120 is frequency of the induced voltage [Hz] that carry a is is always connected connected to . The voltage per phase is between the the voltage 200 r/min nected to a synchronous generator. voltage has a frequency of 60 Hz, If is con- the induced how many poles in 16.3). wye and the A wye connection is only 1/V3 or lines. between a line voltage. amount of We can 58% of This means that stator is only conductor 58% of the therefore reduce the insulation in the slots which, in turn, enables us to increase the cross section of the conductors. A larger conductor permits us to in- crease the current and, hence, the power output Solution Eq. 16. ground. and the grounded stator core at composed of 3-phase lap winding (Figs. 16.2, the highest voltage hydraulic turbine turning It is preferred to a delta connection because 1 Example 16-1 From S poles identical to that of a 3-phase induction motor (Section 13. 17). The winding p = number of poles on the rotor n = speed of the rotor [r/min] does the rotor have? N and electrical standpoint, the stator of a syn- chronous generator neutral A 8 pairs of 16.3 Main features of the stator where /= 1 a cylindrical laminated core containing a set of slots pn f= 36 poles, or induced when the S pole speeds by. Thus, every lime a complete pair of poles duce = 1 , we have of the machine. SYNCHRONOUS GENERATORS 337 Figure 16.2a Stator of a 3-phase, 500 MVA, 0.95 power factor, 15 2350 mm; 378 effective axial length of iron stacking: kV, 60 Hz, 200 r/min generator. Internal diameter: 9250 mm; slots. (Courtesy of Marine Industrie) 2. When a synchronous generator voltage induced in and the waveform distortion is is under load, the each phase becomes distorted, is no longer sinusoidal. The mainly due to an undesired third harmonic voltage whose frequency that is three times of the fundamental frequency. With a wye connection, the distorting line-to-neutral harmonics do not appear between the lines because they effectively cancel each other. Consequently, the line voltages remain sinusoidal under conditions. Unfortunate! v. when all load a delta connec- tion is used, the harmonic voltages but add up. Because the delta is do not cancel, closed on itself, they produce a third-harmonic circulating current, which increases the The nominal erator depends greater the However, exceeds 25 its losses. kVA synchronous gen- rating. In general, the rating, the higher the voltage. nominal line-to-line voltage seldom kV because takes up valuable space conductors. R line voltage of a upon power the I the increased slot insulation at the expense of the copper Figure 16.2b The copper bars connecting successive 19 250 A per phase. stator poles are designed to carry a current of 3200 A. The total output is (Courtesy of Marine Industrie) Figure 16.2c The stator is built up from toothed segments silicon-iron steel laminations (0.5 an insulating varnish. The mm slots are 22.3 mm deep. The salient poles of the much thicker (2 mm) thick), of high-quality covered with mm wide and 169 composed of These laminations rotor are iron laminations. are not insulated because the dc flux they carry does not vary. the The width of the poles from tip to tip is 600 mm and gap length is 33 mm. The 8 round holes in the face air of the salient pole carry the bars of 338 a squirrel-cage winding. Figure 16,3 MVA, 3600 r/min, 19 kV, 60 Hz steam-turbine generator during the construction phase. The windings are water-cooled. The stator will eventually be completely enclosed in a metal housing (see background). The housing contains hydrogen under pressure to further improve the cooling. {Courtesy of ABB) Stator of a 3-phase, 722 339 ELECTRICA L MA CHINES A ND ERA NS FORMERS 340 16.4 Main features of the rotor made ing, the field coils are of bare copper bars, with the turns insulated from each other by strips of Synchronous generators are rotors: salient-pole rotors rotors. built with two types of mica and smooth, cylindrical Salient-pole rotors are usually driven by have to turn at Most hydraulic rotors. (Fig. turbines dc field winding, in series, embedded 1 6.6). Under normal we in the often add pole-faces conditions, this winding does not carry any current because the rotor turns at generator changes suddenly, the rotor speed begins is directly coupled to required, a large number of poles to fluctuate, 50 Hz producing momentary speed variations above and below synchronous speed. This induces a are re- voltage in the squirrel-cage winding, causing a large quired on the rotor. Low-speed rotors always pos- current to flow therein. sess a large diameter to provide the necessary space magnetic for the poles. connected maximum power from the waterwheel, and because a frequency of is coils are synchronous speed. However, when the load on the a waterfall. Because the rotor Hz The low speeds (between 50 and 300 r/min) in order to extract the or 60 6.5). a squirrel-cage winding, are driven by high-speed steam turbines. Salient-pole 1 In addition to the low-speed hydraulic turbines, and cylindrical rotors /. (Fig. with adjacent poles having opposite polarities. The large circular steel salient poles are frame which ing vertical shaft (Fig. 1 6.4). is mounted on a dampen fixed to a revolv- The field of the stator, the oscillation of the rotor. For this reason, the squirrel -cage winding To ensure good cool- current reacts with the producing forces which is sometimes called a damper winding. Figure 16.4 This 36-pole rotor is being lowered into the stator shown a 330 V, electronic rectifier. Other details are: mass: 600 (Courtesy of Marine Industrie) in Fig. t; 16.2. moment The 2400 A dc of inertia: 41 40 exciting current tm 2 ; air gap: 33 is supplied by mm. Figure 16.5 tor is made width of 89 Figure 16.6 a 250 MVA salient-pole generaof 18 turns of bare copper bars having a This rotor winding mm for and a thickness of 9 Salient-pole of a 250 MVA generator showing 12 slots t0 carry the squirre |. cage winding, mm. Figure 16.7a Rotor of a 3-phase steam-turbine generator rated 1530 MVA, 1500 r/min, 27 kV, 50 Hz. The 40 longitudinal slots are being milled out of the solid steel mass. They will carry the dc winding. Effective axial magnetic length: 7490 mm; diameter: 1800 mm. (Courtesy of Allis-Chalmers Power Systems Inc., West Mis, Wisconsin) 34 ELECTRICAL MACHINES AND TRANSFORMERS 342 Figure 16.7b Rotor with rent of 1 1 its .2 4-pole dc winding. Total mass: 204 kA is supplied by a 600 V dc Inc., lines, even anced load conditions. 2. Cylindrical rotors. ; It is well known than low-speed turbines. The same is that high- true of high- speed synchronous generators. However, to generfrequency we cannot air gap: 120 of the N main mm. The dc exciting cur- shaft. use less than to and S poles. The high speed of rotation produces strong cenwhich impose an upper trifugal forces, speed steam turbines are smaller and more efficient ate the required 2 and retained by high-strength end-rings, * serve create the due to unbal- the line currents are unequal 85 t-m the end of inertia: West Allis, Wisconsin) also tends to maintain bal- anced 3-phase voltages between the when moment brushless exciter bolted to (Courtesy of Allis-Chalmers Power Systems The damper winding t; diameter of the at rotor. In the 3600 r/min, the limit on the case of a rotor turning elastic limit of the steel requires the manufacturer to limit the diameter to a maxi- mum of erful m. .2 l 1000 On the other hand, to build the MVA to 1500 MVA pow- generators the vol- 2 poles, and this fixes the highest possible speed. ume On high-power, high-speed rotors have to be very long. a 60 Hz system speed is it is 3600 r/min. The next lower It follows that 1800 r/min, corresponding to a 4-pole ma- chine. Consequently, these steam-turbine generators possess either 2 or The of the rotors has to be large. 4 poles. rotor of a turbine-generator The dc is cylinder which contains a series of longitudinal slots milled out of the cylindrical mass (Fig. I6.7). wedged field excitation and exciters of a large synchronous a long, solid steel Concentric field coils, firmly 16.5 Field excitation into the slots generator is * 1. See Fig. 1 an important part of 28 (Chapter II). its overall design. SYNCHRONOUS GENERATORS stationary field —o A rrrr air 3-phase alternator -OB -oC gap 343 terminals pole bridge rectifier / pilot exciter ^.exciting coil 3\ 3-phase rotor 3-phase stator winding r—T main exciter y alternator Figure 16.8 Typical brushless exciter system. The reason is must ensure not only that the field stable ac terminal voltage, but to sudden load changes tem order to maintain sys- in Quickness of response stability. a must also respond one of the is voltage may have as as little 300 to to rise to twice normal value its 400 milliseconds. This in represents a very quick response, considering that the power of the exciter may be several thousand kilowatts. important features of the field excitation. In order to attain it, two dc generators and a pilot citer no rotating parts The main slip-rings. voltage lies main ex- at all are also involve employed. exciter feeds the exciting current to the of the synchronous generator by field and are used: a exciter. Static exciters that way Under normal conditions between 125 V and 600 V. of brushes /c , produced by the 1000 kVA kW alternator exciter needed (2.5% of its 1 6. 1 ). whereas a 2500 kW exciter suffices for an alternator of (only 0.5% of its rating). 500 Under normal conditions the excitation is varied MW automatically. It responds to the load changes so as to maintain a constant ac line voltage or to control the active power delivered re- to the electric utility system. serious disturbance on the system may produce a A sud- den voltage drop across the terminals of the alternator. The exciter must then react very quickly to keep the ac voltage from falling. wear and carbon dust, we constantly and replace brushes, slip-rings, and commutators on conventional dc excitation sys- systems have been developed. Such a system con- to excite a rating) to clean, repair, regulated It is pilot exciter (Fig. is to brush have tems. To eliminate the problem, brushless excitation The power rating of the main exciter depends upon the capacity of the synchronous generator. Typically, a 25 Due the exciter manually or automatically by control signals that vary the current 16.6 Brushless excitation For example, the exciter sists of a 3-phase stationary-field generator whose ac output is rectified by a group of output from the rectifiers rectifiers. The dc fed directly into the field is of the synchronous generator (Fig. 1 6.8). The armature of the ac exciter and the are mounted on the main shaft and turn rectifiers together with the synchronous generator. In comparing the excitation system of Fig. we can see they are 1 6.8 with that of Fig. 16. 3-phase rectifier replaces the commutator, rings, and brushes. (which is really a by an electronic In other slip- words, the commutator mechanical rectifier. 1, except that the identical, rectifier) is replaced The result is that the brushes and slip-rings are no longer needed. The dc control current /c from the regulates the main exciter output /v . pilot exciter as in the case of ELECTRICA /. MA CHINES A ND ERA NS FORMERS 344 Figure 16.9 This brushless exciter provides the dc current for the rotor of 4 1 is* shown in Fig. 16.7. The exciter consists kVA generator and two sets of diodes. a 7000 Each set, corresponding respectively to the posiand negative terminals, is housed in the circular rings mounted on the shaft, as seen in the center of the photograph. The ac exciter is seen to the right. The two round conductors protruding tive from the center of the shaft (foreground) lead the exciting current to the 1530 MVA generator. (Courtesy of Allis-Chalmers Power Systems Inc., West All is, Wisconsin) The frequency of a conventional dc exciter. main exciter is the generally two to three limes the syn- chronous generator frequency (60 Hz). The crease in frequency is in- obtained by using more poles on the exciter than on the synchronous generator. Fig. 1 shows 6.9 the rotating portion of a typical brushless exciter. Sialic exciters that involve no rotating parts at all employed. are also 16.7 Factors affecting the size of synchronous generators The prodigious amount of energy generated by elecutility companies has made them very con- trical scious about the efficiency of their generators. For example, station if the efficiency of a improves by only 1 %, 1 it 000 MW generating represents extra rev- enues of several thousand dollars per day. gard, the size of the generator tant because as the power its is In this re- particularly impor- efficiency automatically improves increases. For example, if a small l kilowatt synchronous generator has an efficiency of 50%, l() a larger, but similar model having a capacity of MW inevitably has an This improvement in efficiency of about efficiency with size why synchronous is MW and 000 up possess efficiencies of the order of 99%. Another advantage of large machines is son generators of 1 power output per kilogram increases increases. For example, if a l 20 kg (yielding 1000W/20 kg 90%. the rea- Figure 16.10 Partial view of 87 MVA, 428 a 3-phase, salient-pole generator rated r/min, are water-cooled. that the as the power kW generator weighs = 50 W/kg). a lOMW and the use 50 Hz. Both the The high rotor resistivity of of insulating plastic tubing water to be brought into direct parts of the machine. {Courtesy of ABB) and stator pure water enables the contact with the live S YNCHRONO US GENERA TORS 345 generator of similar construction will weigh only ticated cooling techniques (Figs. 16.10 20 000 kg, thus yielding 500 W/kg. From a power Other technological breakthroughs, such as better standpoint, large machines materials, weigh relatively less than small machines; consequently, they are cheaper. Section 16.24 why at the end of this chapter explains Everything, therefore, favors the large machines. in size, we run into seri- ous cooling problems. In effect, large machines herently produce high power in- losses per unit surface area (W/m~); consequently, they tend to overheat. To prevent an unacceptable temperature must design efficient cooling we become rise, systems that ever more elaborate as the power increases. For ex- ample, a circulating cold-air system cool synchronous generators but between 50 the 1000 hollow, point MW is MW range have to be equipped with water-cooled conductors. Ultimately, exceeds the savings made elsewhere, and upper limit to the a reached where the increased cost of cooling is To sum 1). in modifying the design of early ma- chines (Fig. 16. 12). As regards speed, low-speed generators are al- power. Slow-speed bigness simplifies the cooling problem; a good air-cooling system, completed with a heat exchanger, usually suffices. For example, the large, slow-speed 500 generators installed are air-cooled 500 MVA, 200 r/min in a typical 1 to plant whereas the much smaller high-speed MVA, 800 r/min units installed have synchronous hydropower in a steam plant be hydrogen-cooled. adequate to whose rating is below and 300 MW, we have hydrogen cooling. Very big generators to resort to in 1 ways bigger than high-speed machines of equal However, as they increase MW, 16. and novel windings have also played a major part the efficiency and output per kilogram increase with size. 50 and this fixes size. 16.8 No-load saturation curve Fig. 16. 3a shows a 2-pole synchronous generator It is driven at constant speed by a turbine (not shown). The leads from the 3-phase, wye-connected stator are brought out to terminals A, B. C, N, and a variable exciting current /x up, the evolution of big alternators has mainly been determined by the evolution of sophis- 1 operating at no-load. produces the flux in the air gap. Let us gradually increase the exciting current while observing the ac voltage £ 0 between terminal Figure 16.11 The electrical- 200/1 15 the is V, energy needed on board the Concord 12 000 r/min, 400 Hz. Each generator enormous power developed by used to cool the generator and (Courtesy of Air France) is is aircraft is supplied by four 3-phase generators rated 60 kVA, driven by a hydraulic motor, which absorbs a small portion of the turboreactor engines. The then recycled. The generator hydraulic fluid streaming from the hydraulic motor itself weighs only 54.5 kg. 346 ELECTRICAL MACHINES AND TRANSFORMERS Figure 16.12 was This rotating-field generator system. The alternator was first in 1888. It was used in a 1000-lamp street lighting steam engine and had a rated output of 2000 V, 30 A at a frewhich represents 26 W/kg. A modern generator of equal speed and power installed driven by an 11 00 in North America r/min quency of 1 10 Hz. It weighed 2320 kg, produces about 140 W/kg and occupies only one-third the A, say, and the neutral N. For small values of / x the , floor space. Fig. 16. 13c ator ing current. However, as the iron begins to saturate, phases on the the voltage rises /x . If we much less for the plot the curve of £u same increase versus the no-load saturation curve of the generator. It /x , we a schematic diagram of the gener- the revolving rotor and the three stator. in obtain 16.9 synchronous Synchronous reactance equivalent circuit of an ac generator similar to that of a dc generator is is showing voltage increases in direct proportion to the excit- (Section 4. 13). Fig. 16.13b curve of a 36 shows the actual no-load saturation Consider a 3-phase synchronous generator having MW, 3-phase generator having a terminals A, B, nominal voltage of 12 kV (line to neutral). about 9 kV, the voltage increases in Up put of 12 kV, but if 1 age rises only to 15 kV. is driven by a turbine (not proportion to shown), and is chine and load are both connected in wye, yield- 00 A produces an out- the current feeding a balanced 3-phase load (Fig. 16. 14). the current, but then the iron begins to saturate. Thus, an exciting current of C The generator to doubled, the volt- its is excited by a dc current Ix . The ma- ing the circuit of Fig. 16.15. Although neutrals N, and N2 tential are not connected, they are at the because the load is same po- balanced. Consequently, SYNCHRONOUS GENERATORS \ I; B : • I i 347 | C i i ! i Figure 16.13c Electric circuit representing the generator of Fig. 16.13a. is an alternating-current machine, the inductance X manifests itself as a reactance X = s , given by 277/L s where X = s = E — /' synchronous reactance, per phase [H] generator frequency [Hz] apparent inductance of the stator winding, per phase HJ [ The synchronous reactance of ternal Figure 16.13 Generator operating a. b. impedance, just like its a generator is an in- internal resistance R. The impedance is there, but it can neither be seen nor touched. The value of X s is typically 10 to 100 times greater than R; consequently, at no-load. No-load saturation curve 3-phase generator. of a 36 MVA, 21 kV, neglect the resistance, unless we we can always are interested in efficiency or heating effects. We can simplify the schematic diagram of 16.16 by showing only one phase of the we could connect them together (as indicated by the fect, the two other phases Fig. stator. In ef- are identical, except that short dash line) without affecting the behavior of the their respective voltages (and currents) are out of voltages or currents in the circuit. phase by The field carries an exciting current which pro- duces a flux O. As the duces in the stator three field revolves, the flux in- equal voltages Eu that are 120° out of phase (Fig. 16.16). Each phase of the stator winding possesses a resistance R and a certain inductance E. Because this 1 20°. Furthermore, tance of the windings, cuit of Fig. 1 6. 1 7. we if we neglect the resis- obtain the very simple fore be represented by an equivalent of an induced voltage E0 in circuit O which induces composed series with a reactance In this circuit the exciting current / x flux cir- A synchronous generator can thereX s . produces the the internal voltage E<r For a 348 ELECTRICAL MACHINES AND TRANSFORMERS / A \ B load C load alternator Figure 16.14 Generator connected to a load. Figure 16.17 Equivalent only circuit of a 3-phase generator, showing one phase. E given synchronous reactance, the voltage E0 terminals of the generator depends upon E0 load Z. Note that ages and / is E and are line-to-neutral volt- 16,10 Determining the value of X We can determine the unsaturated value of X by the During the open-circuit at rated test the sponding exciting current En are recorded. The excitation age is / xn is is driven raised until The attained. is s s test. generator speed and the exciting current the rated line-to-line voltage Electric circuit representing the installation of Fig. 16.14. the the line current. following open-circuit and short-circuit Figure 16.15 at and the corre- and line-to-neutral volt- then reduced to zero and the three stator terminals are short-circuited together. With the generator again running exciting current value / xn The tor is at rated gradually raised to speed, the its original . resulting short-circuit current / sc in the sta- windings is measured and X s is calculated by us- ing the expression X = £ n // s (16.2) Sc where X = synchronous reactance, per phase En — rated open-circuit line-to-neutral voltage s [11]* LV| Figure 16.16 Voltages and impedances its connected load. :;: in a 3-phase generator and This value of reactance. behavior. It X is h corresponds widely used to the direct-axis to describe synchronous synchronous machine SYNCHRONOUS GENERATORS / sc = When short-circuit current, per phase, using the same exciting current En required to produce The synchronous reactance / Xll that the terminals are short-circuited, the only impedance limiting was [A] to the the current flow X - EJ1 = - 5n is its heavily saturated, the value of X s may the iron be only half unsaturated value. Despite this broad range usually take the unsaturated value for due that 4000/800 s When is synchronous reactance. Consequently, not constant, but is varies with the degree of saturation. 349 X s The synchronous reactance per phase we because fore 5 it yields sufficient accuracy in most cases of interest. The equivalent b. is there- il. phase circuit per shown is in Fig. 16.18a. The impedance of the Example 16-2 A 3-phase synchronous generator produces an open-circuit line voltage of citing current is 6928 50 A. The ac terminals = are then short-circuited, and the three line currents are be 800 A. a. Calculate the synchronous reactance per phase. b. Calculate the terminal voltage The resistors are connected if wye in current / three 12 il across the = EJZ= minals. E= 1 X; \ 2 + 5 (2.12) 2 n 3 is The voltage across ter- 2 \ 12 = found to \R Z V when the dc ex- circuit is 1R = 4000/13 = A 308 the load resistor X 308 1 2 is - 3696 V Solution a. The line-to-neutral induced voltage The line voltage under load E, E0 = E L /V3 = 6928/V3 (8.4) = 4000 V - V3 E = V3 X 3696 - 6402 V The schematic diagram of x s = 5 is is sualize a what 16.11 We is happening Fig. 16.18b helps us viin the actual circuit. Base impedance, recall that first select when using a base voltage per-unit we EH power per phase as the base power.* lows that the base impedance Z B is given by the rated 1 line voltage = = we In the use the rated line-to-neutral voltage as the base voltage t s the per-unit system and a base power. case of a synchronous generator, Z, X ^ It and fol- (16.3) 6394 V alternator :;: Figure 16.18 a. See Example 16-2. b. Actual line voltages and currents. In many power studies the base power is selected to be equal power of the generator and the base voltage is to the rated the linc-to-linc voltage. This yields the base impedance. same value Z n for the ELECTRICAL MACHINES AND TRANSFORMERS 350 where ZB = base impedance (line-to-neutral) of the generator [ft] EB = Note base voltage (line-to-neutral) fV] [ The d. for other may X s (pu) lies between Note P = nous reactance of 60 Hz ac generator has a synchro1 .2 from are 2 / line to (pu) R(pu) pu and a resistance of 0.02 2 l X = 0.02 0.02 that at full-load the per-unit value equal to Example 16-3 5 kV, = be expressed as = of I is 1 The copper 1 7.5 per-unit copper losses at full-load are = depending upon the design of the machine. A 30 MVA, X 0.02 that the generator possesses. . 2, ZB = impedance values that all P(pu) impedances a per-unit value of ZB In general, and 0.15 ft used as a basis of comparison is Thus, the synchronous reactance 0.8 0.02 = neutral. S B — base power per phase VA] The base impedance = losses for 0.02 S B 600 = all 0.02 3 phases are X 30 = 0.6 MW kW pu. 16.12 Short-circuit ratio Calculate a. b. c. d. The base voltage, base power and base imped- Instead of expressing the synchronous reactance as ance of the generator a per-unit value of ZB The The The sometimes used. the ratio of the field current / x! actual value of the synchronous reactance actual needed winding resistance, per phase age copper losses total full-load EH a. is SH Eq. 000/V3 15 = 30MVA/3 = = 10 7 VA 10 is 1 6.2. Thus, if X as defined in s X s is l .2, the 1/1.2 or 0.833. is many = (16.3) types of loads, but they can all be reduced to two basic categories: 7.5 ft The synchronous reactance is X = X (pu) X ZB = 1.2ZB = 1.2 X = 9(2 s The resistance per phase 7.5 is Figure 16.19 R = The short-circuit. exactly equal to the the per-unit value of is volt- produce The behavior of a synchronous generator depends upon the type of load it has to supply. There are ti s |// X 2) to is 16.13 Synchronous generator under load MVA Z = E H 2 /S H - 8660 2 /10 7 c. on a sustained short-circuit ratio is The base impedance b. , armature needed reciprocal of the per-unit value of E B = £,/V3 = = 8660 V The base power the short-circuit ratio to the field current I x2 short-circuit ratio (7 X The base voltage , to generate rated open-circuit rated current / B Solution It is R(pu) X ZB Equivalent circuit of a generator under load. SYNCHRONOUS GENERATORS 1 . 2. The Isolated loads, supplied by a single generator The infinite bus diagram resulting phasor Note 16.20. E that t) given is E leads by 5 begin our study with isolated loads, leaving greater than the terminal voltage, as the discussion of the infinite bus to Section 16. 16. In Consider a 3-phase generator that supplies power to a load having a lagging power some cases the load is somewhat that current / leads the terminal voltage factor. Fig. 16. 19 represents the equivalent circuit for one phase. In or- What effect The answer does . E0 in Fig. capacitive, so by an angle The 16.21. across the synchronous reactance the following facts: list Current / lags behind terminal voltage E by 0. voltage EK to the pha- sum of £ and Ex However, the terminal voltage is now greater than the induced voltage £0 which is a sor angle 0. 90° ahead of is still £0 is again equal the current. Furthermore, an is have on the phasor diagram? this found is Fig. we would expect. der to construct the phasor diagram for this circuit, 1 in degrees. Furthermore, the internally-generated voltage We we 35 . , 2. 3. Cosine 6 Voltage = power Ex 4. EK Voltage the phasor 5. Both the £0 by 90°. / - jIX s E0 very surprising result. In effect, the inductive reac- across the synchronous reactance leads current sion factor of the load. It is given by the expres- sum of £ Ex plus £x O is equal to <I) rent / x is that s enters into partial resonance with the capacit may appear we something for nothing, the higher terminal more power. voltage does not yield any load is entirely capacitive, a very high ter- . minal voltage can be produced with a small excitare voltages that exist inside synchronous generator windings and cannot Flux X reactance of the load. Although If the ing current. However, in later that such under-excitation be measured directly. 6. itive are getting . generated by the flux and tance is chapters, we will see undesirable. produced by the dc exciting cur- Example 16-4 . A 36 MVA, 20.8 kV, 3-phase alternator has a syn- chronous reactance of 9 1 Cl and a nominal current of kA. The no-load saturation curve giving the En tionship between If the excitation is age remains fixed and / x is given rela- in Fig. 16.13b. adjusted so that the terminal voltat current required and 21 kV. calculate the exciting draw the phasor diagram for the following conditions: Figure 16.20 Phasor diagram for a lagging power factor load. a. No-load b. Resistive load of 36 c. Capacitive load of 12 MW Mvar Solution We / shall immediately simplify the circuit to show only one phase. The line-to-neutral terminal voltage for all cases is fixed at E= E a. 20.8/V3 At no-load there is = 12 kV no voltage drop chronous reactance; consequently, Figure 16.21 Phasor diagram for a leading power factor load. E = £ = tl 12 kV in the syn- ELECTRICAL MACHINES AND TRANSFORMERS 352 Q= E,E 0 1 2/3 - 4 Mvar X 10 712 ^12kV The Figure 16.22a Phasor diagram line current = Q/E = - 333 A / at no-load. is The voltage across The exciting current /x b. = The / 36/3 100 - (see Fig. I6.l3b) £ x =JIX = 1000 X ,/ This voltage The voltage £ generated by () phasor sum of sor diagram, £o = \ E and £ x its value . 1000 A is equal to the / x is kV Figure 16.22c Phasor diagram with a capacitive E0 generated £ The voltage £<, 1 6.22c). /x kV by E » Note that £0 is = kV 12 load. / x is equal to the . 12 + (-3) 70 A is (see Fig. 16.13b) again less than the terminal volt- age £. The required exciting current is The phasor diagram = 200 A /x c. kV/!90 o The corresponding exciting current given by 15 x = £ + £x = = 9kV Referring to the pha- T~£; = Vl2 2 + 9 2 = 3 » kV^90° 9 - E0 9 kV phasor sum of £ and = 9 i 90° ahead of/. is is Ex the terminal voltage. 9 00Q (333 A is The current is in phase with The voltage across Xs is S s ( leads / by 90° (Fig. 6.22a. 3 X K/712 000 = 12 1 As before £x MW 12 full-load line current = PIE = X £x = JIX,=j333 X is The phasor diagram is given in Fig. With a resistive load of 36 MW: The power per phase is P = 4 (see Fig. 16.13b) The phasor diagram is given in Fig. 16.22b. With a capacitive load of 12 Mvar: The reactive power per phase is given for this capacitive load is in Fig. 16.22c. 16.14 Regulation curves When a single load, we voltage The synchronous generator feeds a variable are interested in £ changes relationship knowing how the terminal as a function of the load current between £ and / is called the /. regula- tion curve. Regulation curves are plotted with the field excitation fixed and for a given load Fig. MVA, E 1 1 Figure 16.22b Phasor diagram with a kA unity 12 power kV factor load. 1 power factor. 6.23 shows the regulation curves for the 36 21 kV, 3-phase generator discussed in Example 16-4. They are given for loads having unity power factor, 0.9 power factor lagging* and 0.9 power factor leading, respectively. These curves were derived using the method of Example 16-4, SYNCHRONOUS GENERATORS povver factor "-^ The percent regulation x 0.9 lacigincJ % . regulation 353 is = X 100 rated load 10 Lr\/ mnn a / y lea dine (\5- = 12) X 100 - 25% 12 I We note that the percent regulation of a synchro- nous generator erator. much is The reason is greater than that of a dc gen- the high impedance of the syn- chronous reactance. 750 500 250 Load current 1250 1000 16.15 Synchronization of a generator / Figure 16.23 We often Regulation curves of a synchronous generator at three different load power E 0 was kept fixed instead of E. In each power requirements of connected starting point for all the curves Later, terminal voltage 000 A). The change rent load ( ( was set so that the was the rated line2 kV) at rated line cur- 1 is voltage between no-load and full- expressed as a percent of the rated terminal voltage. load. The percent regulation is given by the a large utility system to the when system to provide the extra power. power demand falls, selected gen- from the sys- until the power again builds up the following day. Synchronous generators are therefore regularly being connected and disconnected from a large grid in response to En " regulation many X 100 power customer demand. Such a grid said to be an infinite bus because equation % in For example, as erators are temporarily disconnected tem 1 in common build up during the day, generators are successively of the three cases, the value of E0 to- neutral connect two or more generators to factors. the except that have parallel to supply a it is contains so generators essentially connected in parallel that neither the voltage nor the frequency of the grid can be altered. where Before connecting a generator to an = EH = (or in parallel with another generator), no-load voltage VJ [ synchronized. rated voltage [VI nized 1. Example 16-5 power factor curve in Fig. 16.23. it A generator meets all is it bus must be said to be synchro- the following conditions: The generator frequency is equal to the system frequency. Calculate the percent regulation corresponding to the unity when infinite 2. The generator voltage is equal to the system voltage. Solution The rated line-to-neutral voltage at full-load is 3. The generator voltage is in phase with the sys- tem voltage. EB = 12 kV 4. The no-load terminal voltage 15 is kV The phase sequence of the generator same as that of the system. To synchronize an alternator, is we proceed as the follows: ELECTRICAL MACHINES AND TRANSFORMERS 1 . Adjust the speed regulator of the turbine so that the generator frequency is 3. ment has frequency. 2. En is E0 and E (Fig. 16.24). by This instru- a pointer that continually indicates the phase angle between the two voltages, covering Adjust the excitation so that the generator volt- age Observe the phase angle between means of a synchroscope close to the system the entire range equal to the system voltage E. from zero to 360 degrees. Although the degrees are not shown, the dial has a when the voltages are in when we synchronize an alter- zero marker to indicate phase. In practice, nator, the pointer rotates slowly as it tracks the phase angle between the alternator and system voltages. If the generator frequency is slightly higher than the system frequency, the pointer rotates clockwise, indicating that the generator has a tendency to lead the system frequency. Conversely, if the generator frequency is slightly low, the pointer rotates counterclockwise. bine speed regulator is The A fi- that the pointer barely creeps across the dial. nal check is still 4. The made to see that the alternator voltage equal to the system voltage. Then, moment Figure 16.24 Synchroscope. is the pointer crosses the zero line circuit tur- fine-tuned accordingly, so breaker is at the marker . . closed, connecting the generator to the system. {Courtesy of Lab- Volt) Figure 16.25 This floating r/min, oil derrick provides 60 Hz supply all its own energy needs. Four board are thyristor-controlled dc motors. (Courtesy of Siemens) diesel-driven generators rated the electrical energy. Although ac power is 1200 kVA, 440 V, 900 all the motors on generated and distributed, ~ SYNCHRONOUS GENERATORS In is modern generating stations, synchronization usually done automatically. If will We seldom have except parallel in isolated locations (Fig, As mentioned previously, to it is in 16.25). An bus own its many alternators is system so powerful a that its terminals. Once con- nous generator becomes part of a network comprising hundreds of other generators that deliver thousands of loads. It is power impossible, therefore, to specify the nature of the load (large or small, resistive or capacitive) connected to the terminals of this What, then, determines the particular generator. tion, = (E0 - E)/X s Because the synchronous reactance £x the current lags 90° behind (Fig. is 1 therefore 90° behind E, which is inductive, 6.26b). The machine delivers? To answer we must remember means that that both the value we were an induc- it we Consequently, when tive reactance. a synchronous generator, to the infinite bus. The it over-excite power power increases as supplies reactive reactive we raise the dc exciting current. Contrary we might expect, is impossible to make a it tor deliver active now Let us power by raising its ques- and the can vary only two genera- excitation. decrease the exciting current so that Ea becomes smaller than E. As a result, phasor E = £0 — £" becomes negative and therefore points to the left (Fig. 6.26c). As always, current / = EJX lags 90° behind E x However, this puts / 90° ahead of E, x S . which means it that the alternator sees the system as if were a capacitor. Consequently, when we under- excite an alternator, machine parameters: what to 1 this frequency of the terminal voltage across the generator are fixed. Consequently, if it nected to a large system (infinite bus), a synchro- the therefore circulate in the circuit the generator sees the system as voltage and frequency upon any apparatus connected to power E £o it. infinite imposes to / current to / will s much more common bus) that already has (infinite current s X given by given by connect a generator to a large power system connected E experience a difference of potential A connect only two generators to increase the exciting current, the volt- increase and the synchronous reactance = Synchronous generator on an infinite bus 16.16 we now Ea will age 355 it draws reactive power from the system. This reactive power produces part of the 1 . 2. The exciting current magnetic /x The mechanical torque exerted by Let us see fects the how der the turbine field required by the machine; the remain- supplied by exciting current is — 16.18 Infinite bus effect of varying the mechanical torque — 16.17 Infinite bus effect of varying the exciting current Let us return to the situation with the synchronous generator floating on the connect is equal it after we synchronize to an infinite bus, the to, and E of the system in a generator and induced voltage £ () phase with, the terminal voltage (Fig. 1 6.26a). There is no difference of potential across the synchronous reactance and, consequently, the load current the generator no power; it is is . a change in these parameters af- performance of the machine. Immediately /x / is zero. connected to the system, said to float on the line. Although it delivers and in phase. If we open line, the E0 and E being equal steam valve of the tur- bine driving the generator, the immediate result an increase in mechanical torque (Fig. 1 is The 6.27a). E0 will atmaximum value a little sooner than before. £0 will slip ahead of phasor E, leading by rotor will accelerate and, consequently, tain its Phasor a phase angle 8. same it Although both voltages have the value, the phase angle produces a difference Figure 16.26a Generator floating on an infinite bus. Figure 16.26b Over-excited generator on an infinite bus. Figure 16.26c Under-excited generator on an infinite bus. turbine Figure 16.27 a. Turbine driving the generator. b. Phasor diagram showing the torque angle 5. 356 SYNCHRONOUS GENERATORS E = £ — E across the of potential x () synchronous re- actance (Fig. 16.27b). A current / will flow (again lagging 90° behind £J, but this time it is almost in phase with E. It fol- power delivered by the generator also increases. To understand the physical meaning of the diagram, let us tor will continue to accelerate, the angle 8 will con- tinue to diverge, system the and the power delivered to up. However, as soon electrical will gradually build power delivered to the system is power supplied by the turwill cease to accelerate. The generator examine the and position of the currents, fluxes, poles inside the machine. Whenever 3-phase lows that the generator feeds active power into the system. Under the driving force of the turbine, the ro- currents flow in the stator of a generator, they produce a rotating magnetic field identical to that in an induction motor. In a synchrothis field rotates at the same speed same direction as the rotor. Furthermore, same number of poles. The respective nous generator and in the has the as the electrical it equal to the mechanical fields bine, the rotor stationary with respect to each other. again run will angle 8 between It is synchronous speed, and the Torque at Ea and £ will remain constant. important to understand that a difference of potential is created when two equal voltages are out of phase. Thus, in Fig. 16.27, a potential difference of 4 kV exists between Ea and E, although both produced by the rotor and When tween them. the stator current / veloped. The only 16,19 Physical interpretation of alternator behavior ator (by admitting it the phase angle between value of E x E0 and E increases, the increases and, hence, the value of / in- creases. But a larger current means that the active may be set up be- the generator floats on the line, zero and so no forces are de- is flux that created is induces the voltage If a 6.27b shows that when Depending on hand and the rotor poles on the other hand, powerful forces of attraction and repulsion and 1 stator are, therefore, the relative position of the stator poles on the one voltages have a value of 12 kV. The phasor diagram of Fig. 357 £ mechanical torque () is (Fig. 1 by the rotor, 6.28a). applied to the gener- more steam to the turbine), the rotor accelerates and gradually advances by a mechanical angle a, compared sition (Fig. begin 1 to its original po- 6.28b). Stator currents immediately to flow, owing to the electrical 5 between induced voltage E 0 and phase angle terminal volt- age E. The stator currents create a revolving field I: I / Figure 16.28a The N poles of the of the stator. rotor are lined up with the S poles Figure 16.28b The N poles of the the stator. rotor are ahead of the S poles of ELECTRICAL MACHINES AND TRANSFORMERS 358 and a corresponding set of N and S poles. Forces P = of attraction and repulsion are developed be- - sin 8 (16.5) tween the stator poles and rotor poles, and these magnetic forces produce a torque that opposes where P — En = E= X = 8 = mechanical torque exerted by the turbine. the When the electromagnetic torque is equal to the mechanical torque, the mechanical angle will no longer increase but will remain at a constant active power, per phase There is a direct relationship a and chanical angle between the me- the torque angle 8, given b = pa/2 by (16.4) [ V ] terminal voltage, per phase [V] synchronous reactance per phase s value a. [W] induced voltage, per phase torque angle between En This equation can be used under tions, including the case when and all E [il] [°] load condi- the generator is con- nected to an infinite bus. where To understand 8 = torque angle between the terminal voltage E0 E and the excitation voltage meaning, suppose a generator infinite generator is bus having a voltage E. kept constant so that The term E0 E/X S is then power which the alternator the generator mechanical angle between the centers vary directly with sin of the stator and rotor poles [mechanical its connected to an Furthermore, suppose that the dc excitation of the [electrical degrees) p — number of poles on a ~ is degrees] we admit more Thus, as and so, too, will the active is constant. delivers to the bus will 8, the sine gle. £0 fixed, and the active of the torque an- steam, 8 will increase power output. The relatwo is shown graphically in Note that between zero and 30° the tionship between the Example 16-6 The rotor poles shift by 1 . Fig. of an 8-pole synchronous generator 0 mechanical degrees from no-load to full- load. gle of 30°. Calculate the torque angle between a. 16.29. power increases almost linearly with the torque angle. Rated power is typically attained at an an- E0 and the E at full-load. E or £u is leading? terminal voltage Which b. voltage, , Solution a. The torque angle 8 b. When is: = pa/2 = = 40° 8 X 10/2 a generator delivers active power, ways leads E0 al- E. 16.20 Active power delivered by the generator We can prove (see Section 16.23) that the active power delivered by a synchronous generator given by the equation is Figure 16.29 Graph showing the relationship between the active power delivered by a synchronous generator and the torque angle. SYNCHRONOUS GENERATORS = However, there is an upper limit to the active power the generator can deliver. This limit is reached when 8 is 90°. The peak power output is then P m AX — E„E/X . . S If we the infinite bus. The tion is trip as flow in try to exceed and X (3 13.3) = 4()MW the stator. In practice, this condi- soon as synchronism 16.21 Control of active large, pulsating cur- never reached because the circuit breakers power is, this limit rotor will turn faster than the is lost. We resynchronize the generator before liver the alternator therefore, and lose synchronism with rotating field of the stator, rents will MW The peak power output of (such as by admitting more steam to the turbine), the rotor will accelerate 13.3 359 it then have to can again de- When synchronous generator a system, speed its is connected to a kept constant by an extremely is sensitive governor. This device can detect speed changes as small as 0.01%. system sensitive to the grid. power An automatic control such small speed changes im- to mediately modifies the valve (or gate) opening of Example 16-7 the turbine so as to maintain a constant speed and A constant 36 MVA, 21 kV, 1800 r/min, 3-phase generator On connected to a power grid has a synchronous reactance of 9 (1 per phase. If the exciting voltage is kV (line-to-neutral), and the system voltage 17.3 kV (line-to-line), calculate the following: 12 is in b. before We network have is is done as more elaborate systems under the control of a computer. individual overspeed detectors are always ready to particularly a generator, for if station so that and transmission of energy efficiently as possible. In the entire stations. communicate with each other modify the power delivered by each In addition, Solution a. station operators the generation of step (loses synchronism) falls out it power delivered by advance between the various generating The The active power which the machine delivers when the torque angle 8 is 30° (electrical) The peak power that the generator can deliver output. each generator depends upon a program established to a. power a big utility network, the respond to a large speed change, one reason or another, should suddenly become disconnected from the Ea = E= 8 = kV 12 17.3 system. Because the steam valves are kV/V3 = 10 kV attain a 30° onds. The active power delivered P = (E„EIXS) = The total 6.67 X 10/9) X grid is X centrifugal forces at synchronous speed any excess speed can quickly create a very situation. Consequently, steam valves gencies. At the - 20 all three phases is burners must be shut MW sin 90 10/9) X same is attained time, the pressure build-up off. 16.22 Transient reactance synchronous generator connected to a system subject to unpredictable load changes that 1 in the steam boilers must be relieved and the fuel A X may to 5 sec- must immediately be closed off during such emer- (E0 EIX S ) (12 4 0.5 The maximum power, per phase, when 8 = 90°. P = = The in wide are already close to the limit the materials can with- dangerous MW 6.67) speed 50 percent above normal stand, so 8 power delivered by (3 b. (12 sin power to the still open, the generator will rapidly accelerate and times occur very quickly. In is some- such cases the simple ELECTRICAL MACHINES AND TRANSFORMERS 360 X' short- direct bearing circuit at the on the capacity of the circuit breakers generator output. In effect, because they must short-circuit a interrupt cycles, it three in to six follows that they have to interrupt a very normal high current. load On the other hand, the low transient reactance simplifies the voltage regulation problem when the load on the generator increases rapidly. First the drop due ternal voltage -short-circuit- load would be ing. if to X' ci the synchronous reactance X Second, f in- smaller than is X s were below stays at a value far X it act- for a s sufficiently long time to quickly raise the exciting current /x . which helps •V'd - / Raising the excitation increases E0 . to stabilize the terminal voltage. • Example 16-8 A 250 MVA, time 25 kV, 3-phase steam-turbine gener- ator has a synchronous reactance of 1.6 pu and a Figure 16.30 transient reactance X' d of 0.23 pu. Variation of generator reactance following a short- rated output at a circuit. circuit power factor of suddenly occurs on the It delivers 100%. line, A its short- close to the generating station. equivalent circuit flect the shown in Fig. does not 16. 17 re- behavior of the machine. This circuit only valid under steady-state conditions or is when the load changes gradually. tance X' X s X f varies when a generator is suddenly short-circuited. Prior to the short-circuit, the synchronous reactance is X simply However, s much lower value X' d creases gradually until upon 100 T. it is kVA it only in lasts a fraction the 1000 It . again equal to The duration of the the size of the generator. For machines the circuit breakers should fail to a. The base impedance of then in- X interval s after a depends it The sy 2 = 25 000 /(250 = 2.5 fl nchronous reactance X 10 6 ) is may last as X =X s = long s (pu) 1.6 X ZB 2.5 = 4il The reactance X' d of the alternator. It is may called the transient reactance be as low as synchronous reactance. short-circuit current to the is machines below of a second, but for MVA range the generator at as 10 seconds. sponding open Solution immedi- the instant of short-circuit, the reactance time interval The induced voltage E0 prior to the short-circuit The initial value of the short-circuit current The final value of the short-circuit current if varies as a function of time. shows how ately falls to a c. must be replaced by another reac- whose value Fig. 16.30 a. b. For sudden load current changes, the synchronous reactance Calculate is 1 5 percent of the Consequently, much higher the The initial rated line-to- neutral voltage per phase E = 25/\3 = 14.4 kV than that corre- synchronous reactance JVS . This has a The rated load current per phase is is SYNCHRONOUS GENERATORS 36 rated —-load —I— short-circuit kA ,47.3 -i Change Figure 16.31 See Example 16-8. current in 5/V3 E 6 = 250 X X 10 /(1.73 which 5 6 s short-circuit occurs across See Example drop Ex shows and during the is 23.1 T of terval = 5774 X IX S = EJX, = = 6.8 kA 16-8. 27.2/4 only 1.2 times rated current. is Fig. 16.32 internal voltage £x = = 1 1 1 4 25 000) = 5774 A The when a the terminals of a generator. / = 1 2 3 * time 1 Figure 16.32 5780 A / 1 1 0 4 We assume to, a time in- 5 seconds. Note that in practice the cir- would cuit breakers kV the generator current prior short-circuit. certainly trip within 0. 1 s after the short-circuit occurs. Consequently, they have to The current is in phase with E because power factor of the load is unity. Thus, interrupt a current of about the ring to the phasor diagram (Fig. 16.31), 47 kA. refer- E0 is 16.23 Power transfer between E = i} \ E2 + E\ 2 V14.4 + The 2 The kV 27.2 b. two sources 23.1 transient reactance circuit of Fig. because it is 0.23 6.33a is particularly important in the study of generators, synchronous motors, and transmission is we such circuits = 1 encountered X In lines. are often interested in the active power transmitted from a source A to a source B vice versa. The magnitude of voltages £, and E 2 2.5 < = The 0.575 initial short-circuit n or as well as the phase angle between them, are quite ar- current bitrary. Applying Kirchhoff s voltage law to this is = EJX = 27.2/0.575 circuit, we obtain the equation l1 = 47.3 If kA we assume angle 6 and which c. If is 8.2 times rated current. the short-circuit tion is is sustained and the excita- unchanged, the current E that / lags leads ] E2 behind £\ by an arbitrary by an angle phasor diagram shown (Fig. leads / 1 <X we obtain the 6.33b). Phasor IX by 90°. The active power absorbed by B is will eventually level off at a steady-state value: P = E2 I cos (16. 6) ELECTRICAL MACHINES AND TRANSFORMERS 362 The active power alway s flows from the lagging voltage. In Fig. leads E2 ; 1 6.33, it hence power flows from is the leading to obvious that E { left to right. Example 16-9 Referring to Fig. 16.33a, source age age A generates a volt- E = 20 kV Z 5° and source B generates a E 2 — 15 kV Z 42°. The transmission line volt- { con- necting them has an inductive reactance of 14 11. Calculate the active power that flows over the line and specify which source is actually a load. Solution The phase angle between the two sources is 42° — 5° = 37°. The voltage of source B leads that of source A because its phase angle is more positive. Consequently, power Hows from B to A and so A is actually a load. The active power is given by: E,E2 X Figure 16.33 Power flow between two voltage sources. . - sin o 20 kV X (16.8) kV 15 sin 37° 14 From the sine law for triangles, we have 20 000 /X/sin 8 E,/sin = £,/sin(90 = EJcos Consequently, Substituting ( / cos 0 = P = i\f + 9) 12.9 Note we E = E2 = ] 8 = X= active active (16.7) E2 - sin 8 (16.8) power transmitted [W] 1 The physical [VJ phase angle between E and x E2 f°l reactance connecting the sources may seem, power flows ( 15 kV) to cost, [11 The magnitude off E x and does not have is is E2 machines machine has a pro- power output, relaand temperature rise. The following its why efficiency, these characteristics are ] Let us consider a small ac generator having the equal to that : determined by the angle 6 be- to be specified. inti- mately related. following characteristics: power P received by B between it size of an electrical found effect upon analysis reveals / 6 16.24 Efficiency, power, and size of tive and 10 the one having the higher voltage (20 kV). find delivered by A, because the reactance consumes no tween that, strange as from the source having the lower voltage voltage of source 2 [V] the phase angle X 12.9 1 X voltage of source active power. = MW electrical P = 000 B where The 15 14 £, sin h/X 16.7) in Eq. 16.6, X 0.602 = kW power output I rated voltage 120 V, 3 phase rated current 4.8 rated speed 1 A 800 r/min SYNCHRONOUS GENERATORS efficiency 73% input torque 7.27 morhent of 0.0075 kg*m inertia Under these conditions, we can N-m of the generator as erties 2 its size For example, suppose that m m 80 predict the prop- is increased. dimen- the linear all The volume 363 external diameter 0. external length 0.15 mass 20 kg mass power output/mass 50 W/kg Using we can The mass of the bigger machine will therefore be 27 X 20 kg = 540 kg. The losses will rise to 27 X 0.37 kW = lOkW. The slots are 3 times wider and 3 times deeper. As a result, the cross section of the conductors is 9 times greater which means they can carry 9 times more current. The larger machine can therefore de- this information, 1 sions are tripled. crease by a factor of 3 P" = - X calculate the losses 100 kW x 73 eq. 3.6 100 liver a current As power input = P, - losses 1.37 kW is kW - 1.37 1 .0 losses comprise the kW that kW 0.37 PR losses in the and eddy-current losses the hysteresis and the windage and windings, in the iron way that actly the same proportion, while keeping materials throughout. Thus, iron lamination type is used was used in the larger of insulation is if such in dimensions are raised linear its the the same keep the same current densities (A/itT) as in the original machine. las) in the flux densities (tes- various parts of the magnetic circuit (core, air gap, stator teeth, etc.). As a result, the losses per cm 3 R will be original machine. and 2 I It losses per cm 3 B is flux density in the larger windage and assume unchanged. However, before. it recall machine length the / is the has it. same tripled. As has the same number of conductors and be- as before cause they are connected the same way, the genera- X produce a voltage of 9 tor will 120 V = 1080 V. Thus, by tripling the linear dimensions, the voltage and current both increase by a factor of 9. This power output increases 9X9 = 81 times. The power output of the new generator is therefore 81 X kW = 81 kW. The power input needed to drive the ac generathat the 1 tor is P, : = 81 kW + losses kW. The new efficiency = is 81 kW + 10 kW = therefore: in the 71 = ° vol- 81 friction 91 its X 100 eq. 3.6 P; = kW x kW 0.89 100 - 89% that the ( left the speed at We Blv. the length of the conduc- / is The is E = which the flux cuts across v follows that the copper losses same way. number of slots, conductors and interconnections remain the same as before and that the speed of rotation 800 r/min) is further the flux density, and losses also increase the We 43.2 A. and the iron everywhere the same as that the A= times because the diameter of the rotor has tripled. 91 iron losses will increase in proportion to ume. Let's assume 4.8 regards the generated voltage per conductor, means same X creases by a factor of 9. Because the larger generator nuts and bolts. will of 9 same magnifying everything, including the bearings, We as also used, thereby duplicating and will also maintain the so, too, will a result, the voltage generated per conductor also in- machine. The same type we 27 and ex- in a particular type of in the stator, In this larger generator 27. Consequently, the Furthermore, the peripheral speed v has increased 3 friction losses. Let us increase the size of the machine a = a factor of determined by equation (2.25) tor The by the losses. of the machine: r\ will increase 3 will therefore in- 1 The which the efficiency has increased from is a dramatic improvement. power output has increased 73% to The reason 89% is that 81 times, while the ELECTRICA L MA CHINES AND TRANSFORMERS 364 losses increased only 27 times. Consequently, the bound to ciency of the machine was effi- increase with size. about 20()°C. Consequently, the cooling of large machines a very important matter. is The original machine produced an output of 50 W/kg. The larger machine has a mass of 540 kg and produces 81 kW. Consequently, it produces 81 regarding physical size, power output, efficiency, kW/540 kg = 150 W/kg which including ac and dc motors and transformers. 3 times greater is In conclusion, the general principles covered here temperature rise and so forth, apply to machines, all than before. The proof, generator larger if eighty-one produce 81 kW, X 20 therefore is relatively Questions and Problems and cheaper than the smaller machine. As lighter = kg kW 1 their 1620 kg. Practical level generators were used to combined mass would be 81 This generating center would 1 6- 1 Why obviously be more costly and take up more floor space than the single 81 As another matter of moment of mass and inertia ,/ kW of a rotor the square of we is proportional to Hence, when linear dimensions are = nir — 243. The moment of therefore 243 The X inertia tripled, X 27 2 3 J = 3 5 = 1.8 kgm 2 16-3 is the original 1 kW 1080 V, rated voltage 43.2 rated current 3 A 1800r/min 89% moment of inertia Nm 1.8 kg-nr external length 0.45 1 m m linear dimensions are temperature rise. the losses increase 27 times. Hence, the power dis- sipated per square meter increases by a factor of better prevent damage is directly-coupled genera- If the must generate a frequency of 60 Hz, The number of poles on The exaci turbine speed An the rotor isolated 3-phase generator produces a a load If is cooling hound maintain the same line voltage? What conditions must be met before a gener- Calculate the number of poles on system? the genera- means to the insulating materials, the 3. are Calculate the number of poles on shown in Fig. 16. 6-8 A 3-phase generator turning Hz. How fected to its if 200 r/min will the terminal voltage be af- terminals? Resistive load tem- b. Inductive load c. Capacitive load of 1 1 the following loads are connected a. maximum at the air1 generates a no-load voltage of 9 kV, 60 To to be hotter. perature rise has to be limited to a 6-7 craft generator 1 unless 6-6 When tripled, the heat-dissipating used, the larger machine found close to tor in Fig. 16.12 using the information given. 150W/ks is is it 350 r/min. at ator can be connected to a 3-phase surface area of the machine increases 9 times but Consequently, 6-5 540 kg big problem site, should turn to 1 The one analyzing a hydropower connected to the machine, must the exci- 1 mass power output/mass the larger? tation be increased or decreased in order 483 0.54 is having a lagging power factor of 0.8 rated speed external diameter For a given power output, which of no-load line voltage of 13.2 kV. phase efficiency input torque In a. 16-4 kW main differences between steam- that the turbines b. 81 wye? calculate the following: are in striking contrast to machine. power output State the tors . characteristics of the larger generator are summarized below. They synchronous generators? in large the stator always connected in these machines = of the larger machine 0.0075 kg-nr is ators. its will are the advantages of using a stationary turbine generators and salient-pole gener- recall that the radius (see Table 3A). its increase by a factor of J 16-2 generator. interest, What armature SYNCHRONOUS GENERATORS 16-9 In Problem 16-8, if the field current Calculate is kept constant, calculate the no-load volt- a. age and frequency b. the speed if is The synchronous impedance Z per phase The total resistance of the circuit, per s lOOOr/min a. phase 5 r/min. b. 365 The c. total reactance of the circuit, per phase Intermediate level 16-10 What meant by the synchronous is tance of a 3-phase generator? 16-11 reac- Draw nator meaning of all the parameters. State the advantages of brushless excita- Using a schematic how circuit diagram, the rotor in Fig. 16.7 16-17 show needed a a. 24.2 12.1 A 3-phase £0 16-18 Ev kV 16 generator possesses a synchro- is 100 (1, calculate the value of per phase. , The generator in Fig. is 3 kV connected to an Q and the excitation per phase to load having a lagging 1 6.2 has a synchro- nous reactance of 0.4 H, per phase. nous reactance of 6 voltage KVA, factor of 0.8. If the synchronous re- actance kV kV generator rated 3000 2400 KVA, power to generate a no- load line voltage of b. A 3-phase 20 kV, 900 r/min, 60 Hz delivers power excited. is the volt- age across the load Referring to Fig. 16.13, calculate the exciting current 16-13 The phase angle between E0 and f. the tion systems over conventional systems. 16-12 h. e. equivalent circuit of a generator and explain the g. The line current The line-to-neutral voltage across the load The line voltage across the load The power of the turbine driving the alter- d. It is bus having a infinite line voltage of 14 kV, and the excitation volt- (ref. Fig. age is adjusted to 1 . 14 pu. 16.19). Calculate the line-to-neutral volt- age E for a resistive load of 8 (2 and draw Calculate the phasor diagram. a. 16-14 a. In Problem 16-13, draw the curve of E for the following resistive loads: infin- sus / ity, 24. 12, 6, 3,0 ohms. power P per phase b. Calculate the active b. c. in inside stator circumference) corresponding each case. c. Draw to this the curve of E versus value of load resistance a is P. the For what power output 16-19 Referring to Fig. 1 6.2, calculate the length of one pole-pitch measured along the 16-16 on the 500 [in]. M VA alternator of Fig. 16.2 yields the following results: maximum? ternal displacement angle A test taken 1. 16-15 The torque angle 8 when the generator delivers 420 The mechanical displacement angle a The linear pole shift (measured along the MW ver- circumference of the in- 2. stator. The 3-phase generator shown Open-circuit line voltage is 15 kV for a dc exciting current of 1400 A. Using the same dc current, with the armature short-circuited the resulting ac line current is 21 in Fig. 000 A. 16.16 has the following characteristics: Calculate E0 = 2440 V X = s R = load impedance Z= a. 144(1 17 12 175 (2 (resistive) The base impedance of the generator, per phase c. The value of the synchronous reactance The per-unit value of X s d. The b. short-circuit ratio 366 ELECTRICAL MACHINES AND TRANSFORMERS Advanced 16-20 16-25 level b. d. tween going incoming the cool air, if the air flow air is of 500 A. 1 .3 b. be- 280 nrVs 16-26 gap (See Section 16-22 2. 1 the line circuit breakers suddenly trip, cal- second 1 later, assum- By how many mechanical degrees do 1 how many degrees? A 400 Hz electrical By alternator has a 2-hour rating of nal diameter of mmf re- length of 9.5 80 1 V, 7). and has an slots 22 inches and an in. The rotor for a field current of 3 Referring to Fig. 16.17, the following the -second interval? position) during the stator possesses quired for the iron portion of the magnetic circuit. . 80 percent power factor (Fig. 16.34a). The gap length no-load. Neglect the at lb kg nr. 75 kVA, 1200 r/mim 3-phase, 450 inches, calculate the flux density in the air 6 poles advance (with respect to their normal and warm out- that the air 10 10 culate the speed of the generating unit (tur- and carries a dc current Knowing X 6 driven is whose moment of 2 ft The rotor has a J ing that the wicket gates remain wide open. Referring to Fig. 16.4, each coil on the rotor has 21 .5 turns, If X 54 bine and alternator) The torque developed by the turbine The average difference in temperature c. is of 4.14 a. Problem 16-20 in a hydraulic turbine inertia is The total losses in the machine The copper losses in the rotor a. 16-21 The generator by The synchronous generator in Fig. 16.2 has an efficiency of 98.4% when it delivers an output of 500 MW. Knowing that the dc exciting current is 2400 A at a dc voltage of 300 V, calculate the following: 1 is A at inter- axial designated 1 1 5 V. in- Calculate formation is given about a generator: c. The number of poles on the rotor The number of coils on the stator The number of coils per phase group on d. The a. Ea = E= 12 kV 14 kV x = 2 b. the stator s Ea leads a E e. a. length of one pole pitch, measured along the circumference of the stator by 30° Calculate the total active power output of The tor resistance of the dc winding on the ro- and the power needed to excite it the generator. 1 6-23 b. Draw c. Calculate the power factor of the load. the phasor diagram for one phase. The steam-turbine generator shown 16.3 has a The excitation voltage pu and the machine is E0 is 16-27 in Fig. synchronous reactance of 1 .3 pu. adjusted to 1.2 connected to an hid List rial a pplica tion A 33.8 kVA, 480 operate at a bus of 19 kV. If the torque angle 5 16-24 active b. The line current c. Draw is given: 83.4% Weight: 730 Wk 2 lb (moment of Insulation: class the phasor diagram, for 80 percent. The is power output The factor of diesel- designed to infi- 20°, calculate the following: a. power 60 Hz is following additional information Efficiency: nite V, 3-phase, driven emergency alternator one phase In Problem 16-23. calculate the active power output of the generator if the steam valves are closed. Does the alternator receive or deliver reactive power and how much? inertia) : 15.7 2 lb. ft B Calculate a. The minimum horsepower rating of the diesel engine to drive the generator b. The maximum allowable temperature of windings, using the resistance method the SYNCHRONOUS GENERATORS 16-28 A 220 MVA, 500 0.9 power r/min, 13.8 kV, factor, water-turbine 50 Hz, Efficiency nous generator, manufactured by Siemens, Insulation class: mass of Total (t = stator: Unsaturated synchronous reactance: 1.27 pu Runaway speed 158 2 mass of rotor: Static excitation t current 270 in generator mode: 890 r/min metric ton) Total factor: Transient reactance: 0.37 pu F of inertia: 525 t*m power 98.95% synchro- has the following properties: Moment at full-load, unity 367 is is used and the excitation 2980 A under an excitation volt- age of 258 V. t Figure 16.34a Rotor and stator of a 75 kVA, 1200 r/min, 3-phase, 450 driven by a 100 hp, 1200 r/min synchronous motor. V, 400 Hz alternator for shipboard use. The alternator is Figure 16.34b Stator and rotor of the 100 hp, 1200 r/min, 60 serves as a base for the alternator. The Hz synchronous motor. The rotor is duction motor. (Courtesy of Electro-Mecanik) equipped with a squirrel stator is mounted on a bedplate that also cage winding to permit starting as an in- 368 ELECTRICAL MACHINES AND TRANSFORMERS The generator also designed to oper- is 16-29 In industry application Problem 16-28, ate as a motor, driving the turbine as a calculate the following: pump. Under these conditions, a. The horsepower b. when it runs as a pump motor The kinetic energy of the rotor when develops an output of 145 the motor MW. Both the stator and rotor are water- at rated cooled by passing the water through the c. hollow current-carrying conductors. The The is treated so that less than 5 (JiS/cm. through the stator its conductivity a rate of 8.9 second and through the rotor at liters runs when kinetic energy of the rotor its maximum it allowable runaway is speed The pure water Hows at it speed reaches water rating of the generator d. per The time to reach the runaway speed event that a short-circuit occurs 5.9 liters erator is delivering its when rated load, in the the gen- and assum- per second. Given the above information, ing that the water continues to flow unchecked make through the turbine (gates wide open) a. the following calculations: The rated active unity power power output, in MW factor and at 0.9 lagging at power factor b. c. d. The rated reactive power output, in Mvar The short circuit ratio The value of the line-to-neutral synchronous reactance, per phase e. 16-30 In Problem 16-28 calculate the power dis- sipated in the rotor windings and the power loss per pole. Knowing water flow and that the is inlet the rate of temperature 26°C, calculate the temperature of the water flowing out of the rotor windings. minimum the total losses of the generator at full load What and unity power factor the circulating water? is the resistivity (H-m) of Chapter 17 Synchronous Motors 17.0 Introduction motor speed stays constant, irrespective fixed, the of the load or voltage of the 3-phase The synchronous generators described in the previous chapter can operate either as generators or as motors. When they run operating as motors (by con- field. The speed of Most synchronous motors synchronism with the revolving rotation is 150 therefore tied to the frequency of the source. Because the frequency constant speed but because they possess We will study these features in this chapter. synchronous motors. As the name implies, synchroin at other unique electrical properties. necting them to a 3-phase source), they are called nous motors run However, line. synchronous motors are used not so much because kW (200 hp) and speeds ranging from is 1 15 are rated between MW (20 000 hp) and turn at 50 to 800 r/min. Consequently, 1 these machines are mainly used in heavy industry Figure 17.1 Three-phase, unity power factor synchro- nous motor rated 3000 hp (2200 kW), 327 r/min, 4000 V, 60 Hz driving a compressor used in a pumping station on the Trans- Canada pipeline. Brushless excitation provided by a 21 kW, 250 fier, which is V mounted on the shaft the bearing pedestal and the main {Courtesy of General 369 is alternator/recti- Electric) between rotor. ELECTRICAL MACHINES AND TRANSFORMERS 370 (Fig. we in 1 7. 1 ). At the other end of the power spectrum, find tiny single-phase synchronous motors used They control devices and electric clocks. cussed 17.1 in Chapter are dis- rent is fed into the winding from an external exciter. Slots are also punched out along ing similar to that in a 3-phase induction motor. This 18. damper winding serves to start the motor. Modern synchronous motors often employ Construction less excitation, similar to that Synchronous motors are identical in construction to salient-pole ac generators. The stator is composed of a slotted magnetic core, which carries a 3-phase lap winding. Consequently, the winding is also identical to that of a 3-phase induction motor. The rotor has by a dc current connected a set of salient poles that are excited (Fig. in series to the circumference of the salient poles. They carry a squirrel-cage wind- 17.2). two The exciting slip-rings, coils are and the dc cur- used in brush- synchronous generators. Referring to Fig. 17.3, a relatively small 3-phase generator, called fier are mounted at dc current / x from the salient-pole exciter, and a 3-phase one end of the motor rectifier windings, is recti- shaft. The fed directly into the without going through brushes and slip-rings. The current can be varied by controlling the small exciting current Ic that flows in the stationary field winding of the exciter. Fig. 17.4 Figure 17.2 Rotor of a 50 Hz to 1 6 2/3 Hz frequency converter used to power an electric railway. The 4-pole rotor at the left is associated with a single-phase alternator rated 7000 kVA, 1 6 2/3 Hz, PF 85%. The rotor on the right is for a 6900 kVA, 50 Hz, 90% PF synchronous motor which drives the single-phase alternator. Both rotors are equipped with squirrel-cage windings. (Courtesy of ABB) Figure 17.3 Diagram showing the main components chronous generator. of a brushless exciter for 1 - dc control source 2 - stationary exciter poles 3 - alternator (3-phase exciter) 4 - 3-phase connection 5 - bridge rectifier 6 - dc 7 - rotor of synchronous 8 - stator of 9 - 3-phase input to stator line a synchronous motor. motor synchronous motor It is similar to that of a syn- SYNCHRONOUS MOTORS shows how mounted The the exciter, rectifier, in a and kW synchronous salient poles are motor. rotor and stator always have the ber of poles. the 3000 As number of in the same num- poles determines the synchronous Example 17-1 Calculate the number of salient poles on the rotor of the synchronous motor shown in Fig. 17.4a. Solution 120 / P = 120,///? frequency of the source [HzJ poles s 200 = p = speed fr/min] p — number of at 200 r/min; (17.1) /z /= 60 Hz and runs consequently, where = motor at The motor operates - s J case of an induction motor, speed of the motor: // 37 The rotor possesses 1 (120 X 60)//? 36 poles 8 north poles and 1 8 south poles. Figure 17.4a Synchronous motor rated 4000 hp (3000 kW), 200 r/min, 6.9 kV, 60 Hz, 80% power factor designed to drive an ore crusher. The brushless exciter (alternamounted on the overhung tor/rectifier) is shaft and is rated 50 kW, 250 (Courtesy of General V. Electric) Figure 17.4b Close-up of the 50 kW exciter, showing the armature winding and 5 of the 6 diodes used to rectify the (Courtesy of General ac current. Electric) ELECTRICAL MACHINES AND TRANSFORMERS 372 synchronous motor 17.2 Starting a A synchronous motor cannot quently, the rotor is When slightly is can it the stator start of stator conse- up as an induction connected is motor accelerates line, the itself; axis of S pole usually equipped with a squirrel- cage winding so that motor. by start axis of N pole of rotor until to the 3-phase reaches a speed it below synchronous speed. The dc excitation suppressed during While this starting period. the rotor accelerates, the rotating flux cre- ated by the stator sweeps across the slower salient poles. moving Because the coils on the rotor possess number of turns, a high voltage is rotor winding when it turns at low a relatively large induced in the speeds. This voltage appears between the slip-rings Figure 17.5 The poles of the rotor are attracted to the opposite poles on the stator. At no-load the axes of the poles coincide. and decreases as the rotor accelerates, eventually it becoming negligible when synchronous speed. To we prove the starting torque, slip-rings or connect the rotor approaches and limit the voltage, them either short-circuit the to an auxiliary resistor power capacity of we sometimes have ited, to the stator. As in the the supply line is lim- reduced voltage to apply case of induction motors, we use either autotransformers or series reactors to Chapter 20). Very limit the starting current (see synchronous motors (20 large sometimes brought up called a tor, lations the more) are speed by an auxiliary mo- to pony motor. MW and some Finally, in motor may be brought up to big instal- pull-in torque of a right moment. For example, motor will immediately slow breakers will chronous speed, the rotor trip. In This produces alternate N moment happen at this tion is set on the to at close to syn- excited with dc current. and S poles around the circumference of the rotor (Fig. site polarity 1 strong magnetic attrac- up between them. The mutual attraction locks the rotor and stator poles together, and the rotor is literally field. yanked the circuit ment when excitation should be applied. The motor with the revolving Once is carries the induced no field. motor turns at synchronous speed, no in the squirrel-cage winding and so current. Consequently, the behavior of a is entirely different from that of into step with the revolving The torque developed because of the magnetic attraction between the poles of the rotor and the opposite poles of the To reverse the direction of rotation, we interchange any two lines connected to the stator. simply stator. 7.5). If the poles be facing poles of oppo- stator, a down and practice, starters for synchro- an induction motor. Basically, a synchronous motor running is re- then pulls automatically and smoothly into step rotates is magnetic nous motors are designed to detect the precise mo- voltage 17.3 Pull-in torque motor should happen that stator, the resulting synchronous motor as the it pulsion produces a violent mechanical shock. The it As soon if is at the emerging N, S poles of the rotor are opposite the N, S poles of the speed by a variable-frequency electronic source. synchronous motor powerful, but the dc current must be applied the during the starting period. If the The im- to at this propriately called the pull-in torque. moment is ap- 17.4 Motor under load general description When a synchronous motor runs at no-load, the ro- tor poles are directly opposite the stator poles and their axes coincide (Fig. 1 7.5). However, ply a mechanical load, the rotor poles if we ap- fall slightly SYNCHRONOUS MOTORS 373 a bigger mechanical load, and the increased power can only come from the 3-phase ac source. 17.5 Motor under load simple calculations We can get a better understanding of the operation of a synchronous motor by referring to the equiva- shown lent circuit in Fig. 1 7.7a. It phase of a wye-connected motor. represents one It identical to is the equivalent circuit of an ac generator, because both machines are built the same way. Thus, the Figure 17.6 The rotor poles are displaced with respect to the axes of the stator poles when the motor delivers mechani- flux <£> rent cal power. /x As . E0 Consequently, a between the poles increases progressively as increase the load (Fig. more powerful torque is to the and the motor develops an ever field, But there we Nevertheless, the 17.6). magnetic attraction keeps the rotor locked revolving to The mechanical angle E0 is in motor comes step creates a away from to a halt. the stator poles circuit breakers and A motor that pulls out of major disturbance on the immediately trip. line, and the This protects the motor because both the squirrel-cage and stator windings overheat rapidly when the machine ceases to run at The Under 7.7b). E0 = If, we in addition, adjust the motor "floats" on the E, the line current / practically zero. In ef- is is to supply the small friction losses in the motor, and so it is negligible. the shaft? if we The motor apply a mechanical load to will begin to causing the rotor poles to poles by an angle a. E0 reaches its Due fall to this maximum electrical degrees mechanical value a behind E. slow down, behind the stator E is i} E0 tial Ex than now The mechanical placement a produces an electrical phase between shift, little later before. Thus, referring to Fig. I7.7c, 8 dis- shift 5 and E. The phase upon the magneto- these conditions, in- phase with the line-to-neutral only current needed windage and synchronous speed. pull-out torque depends 1 What happens ceeds the pull-out torque of the motor, the rotor the and the fect, the as the angle increases. a limit. If the mechanical load ex- poles suddenly pull voltage £(Fig. excitation so that line varies with the excitation. already mentioned, the rotor and stator poles duced voltage turn at synchronous speed. in depends on the dc exciting cur- are lined up at no-load. behind the stator poles, but the rotor continues £0 created by the rotor induces a voltage the stator. This flux shift produces a difference of poten- across the synchronous reactance X s given by motive force developed by the rotor and the stator poles. The mmf of the dc excitation /x , rotor poles while that of the stator depends upon the ac current flowing pull-out torque is in E x — E — E0 depends upon the The nom- the windings. usually 1.5 to 2.5 times the Consequently, a current j'X, inal full-load torque. The mechanical angle a between stator poles has a direct bearing As to the rotor and / must flow given by = E* from which on the stator current. the angle increases, the current increases. This is be expected because a larger angle corresponds to / = -jEJX = -j(E - s E„)/X x in the circuit, 374 ELECTRICAL MACHINES AND TRANSFORMERS Example 17-2a A 500 hp, 720 r/min synchronous motor connected to a 3980 age V, 3-phase line generates an excitation volt- £n of 1790 ing current II is V (line-to-neutral) when the dc excit- 25 A. The synchronous reactance and the torque angle between Ea and E is 22 30°. is Calculate a. b. Figure 17.7a c. Equivalent circuit of a synchronous motor, showing d. one phase. The value of £ x The ac line current The power factor of the motor The approximate horsepower developed by the motor e. E Eo The approximate torque developed at the shaft Solution This problem can best be solved by using vector no- Figure 17.7b Motor tation. at no-load, with EQ adjusted to equal E. a. The voltage E (line-to-neutral) applied to the motor has a value Ex — E - E0 E = EjV3 = 3980/V3 4 = 2300 V Let us select E E as the reference phasor, whose angle with respect to the horizontal axis sumed is as- to be zero. Thus, E = 2300Z0 0 It follows that E0 is given by the phasor Ea = 1790/1-30° Figure 17.7c Motor under load 17.7b, but it The equivalent EQ has the same value as lags behind E. circuit per phase Moving clockwise around The current lags 90° behind E x because Xs is inductive. The phasor diagram under load is shown in Because / is nearly in phase with E, the motor absorbs active power. This power is entirely transformed into mechanical power, except for the relatively small copper and iron losses In practice, the excitation voltage to in the stator. E0 is adjusted be greater or less than the supply voltage E. value depends upon the and the desired power power output of factor. the given the circuit plying KirchhofTs voltage law Fig. I7.7c. is in Fig. 17.8a. in Fig. -E + E + E0 = x EX = E- write 0 E0 = 2300^0° - 1790Z-30 0 = 2300 (cos0° + ;sin0°) 1790 (cos -30° = 2300 - 1550 = 750 + j 895 Its motor and ap- we can - 1168Z50 0 +j +7 sin 895 -30°) SYNCHRONOUS MOTORS 375 Ex 1168 V Figure 17.8a Equivalent circuit of a synchronous motor connected to a source E. Thus, phasor leads phasor b. The Ex has a value E by 50°. of 1 1 68 V and it 1790 Figure 17.8b fl2I = Ex _ ~ 1 See Example A 50 ° A 90° 22 Approximate torque: e. 9.55 T= A and Thus, phasor / has a value of 53 40° behind phasor E. The power motor factor of the it = is current E across /. Hence, power X 9.55 3715 X 1 10 3 720 N-m Example 17-2b The motor in Example 17-2a has a stator resistance of 0.64 Q. per phase and possesses the following factor = cos 6 = = 0.766, or losses: cos 40° 2 I 76.6% R losses in the rotor: 3,2 Stator core loss: The power 280. given by the motor terminals and the the X P n lags cosine of the angle between the line-to-neutral voltage 17-2. 168 = 53^-40° c. V given by line current / is factor is lagging because the current Windage and 3.3 friction loss: 1 .5 kW kW kW lags behind the voltage. The complete phasor diagram is shown in Fig. 17.8b. d. Calculate a. Total active power input to the stator: b. c. Pi = 3X - 3 L Solution cos 40° a. W= 142 280. Power input to the stator lkW Stator/ Neglecting the shaft E, N / cos 6 X 2300 X 53 X = 280 The actual horsepower developed The actual torque developed at the The efficiency of the motor 2 I R 2 /? losses - 3 X is 53 2 280. X Total stator losses = power transmitted across is 280. kW. Power transmitted to the rotor stator, the electrical the airgap to the rotor 1 = 271.4 + kW Q = 5.4 kW = 8.7 kW = 280. - 8.7 0.64 losses and iron losses in the 5.4 1 3.3 1 kW Approximate horsepower developed: P - 280.1 X 3 10 /746 = 375 hp The power at the shaft minus the windage and is the power to the rotor friction losses. The rotor ELECTRICAL MACHINES AND TRANSFORMERS 376 IR losses are supplied by an external dc source and so they do not affect the mechanical power. chanical power. by Power available at the shaft: P0 = 271.4 - 269.9 X l(r - Example The corresponding torque is: value calculated in b. T - X P 9.55 P= 361.8 hp H - therefore expressed by " sin 8 (17.2) of the motor, per phase [Wl X 269.9 10* 720 // E E P = mechanical power 17-2a. X 9.55 is where very close to the approximate is synchronous motor kW 269.9 746 This power a the equation = 1.5 is available in the form of meThe mechanical power developed across the air gap En = E— line-to-neutral voltage of the source [V] X = synchronous reactance per phase s = 3580 N-m = 8 line-to-neutral voltage induced by torque angle between £ (> and [V] /x E [electrical degrees] c. = Total losses Total Total 4- 3.3 4 3.2 4 1.5 = 13.4 power input = 280.1 4- 3.2 = 283.3 power output = 269.9 kW Efficiency Note 5.4 = = 269.9/283.3 = 0.9527 that the stator resistance of 0.64 compared small to 95.3 Q Consequently, the true phasor diagram close to the phasor diagram of Fig. kW is % very of 22 reactance the kW 1 is Q. This equation shows that the mechanical power increases with the torque angle, and its maximum value is reached when 8 is 90°. The poles of the rotor are then midway between the N and S poles of the stator. The peak power P max (per-phase) very far as torque tional to the Power and torque speed When a synchronous motor operates under load, draws active power from given by the same equation The power the line. we t> As in the s ) is sin 8 voltage E, the excitation voltage PR power is This is If we and iron losses to the is directly propor- fixed. The torque is derived from Eq. 3.5: where T= P = (16.5) power £ () , and the phase neglect the relatively in the stator, all the transmitted across the air gap to the rotor. analogous it is torque, per phase [N-m] mechanical power, per phase [W] ns = synchronous speed [r/min] 9.55 = a constant [exact value absorbed by the motor depends upon the supply small concerned, previously used for case of a generator, the active angle 8 between them. is mechanical power because the rotor it the synchronous generator in Chapter 16: P = (£ £/X given by 7.8b. As 17.6 is The maximum torque = 60/2 tt| motor can develop is called the pull-out torque, mentioned previously. It occurs when 8 = 90° the (Fig. 17.9).* power P r transmitted across the air gap of an induction motor (Section 13.13). tor PR However, losses in are a synchronous motor, the roentirely source. Consequently, all supplied by the dc the power transmitted The remarks rotors. in ihis section apply to motors having Most synchronous motors have case the pull-out torque occurs at smooth salient poles: in this an angle of about 70°. S YNCHRONO US MOTORS P T 8 90 800 100 120 86.6 693 150 50 400 180 0 0 These values are plotted 377 in Fig. 17.9. The torque curve can be found by applying Eq. b. 17.4: T= Figure 17.9 Power and torque per phase as a function of the torque angle 8. Synchronous motor rated 150 kW (200 hp), 1200 r/min, 3-phase, 60 Hz. See Example 17-3. 1 1 If 9.55 /VI 200 = /VI 25 the output: 7mux = 800 N-m r/min. 460 synchronous V, 3-phase motor has a synchronous reactance of 0.8 phase. = The pull-out torque TnvAX coincides with c. maximum power Example 17-3 A 50 kW, 200 9.55 Pin s the excitation voltage En is fixed at 12, per 300 V, The the per phase, determine the following: actual pull-out torque N-m) because this is a 3 times as great is (2400 3-phase machine. Similarly, power and torque values given in Fig. 1 7.9 must also be multiplied by 3. Consequently, this 150 a. b. c. The power versus 8 curve The torque versus 8 curve The pull out torque of the motor maximum motor can develop a or about 400 kW output of 300 kW, hp. Solution a. The line-to-neutral voltage 17.7 Mechanical is As corresponding values of is a and the number of poles p. It given by (17.2) selecting different values for 8, late the case of synchronous generators, there a, the torque angle 8 is sin 5 = (266 X 300/0.8) = 99 750 sin 8 [W] = 100sin8[kWl By in the precise relationship between the mechanical angle is P = (E0 E/XJ electrical angles E = EjV3 = 460/ V3 = 266 V The mechanical power per phase and 8 = /x*/2 (1 7.5) sin 8 Example 17-4 A 3-phase, 6000 we P and can calcuT, per phase. tor P T PI [kW] [N*m 0 0 0 30 50 400 60 86.6 693 ? 4 kV, 180 r/min, 60 Hz mo1.2 ° 1 from At their no-load position. If the line-to-neutral excitation late the £0 = 2.4 kV, calcu- mechanical power developed. Solution The number of poles (continued) (1. full-load the rotor poles are displaced by a me- chanical angle of 5 kW has a synchronous reactance of \20fln s is = 120 X 60/180 = 40 ELECTRICAL MACHINES AND TRANSFORMERS 378 The axis of electrical torque angle is N pote of rotor 8 Assuming a motor to the = pa/2 wye = X (40 )/2 1 = axis of S pole ! of stator ; 20° connection, the voltage E applied is E = E,/V3 =4kV/V3 = 2.3 kV - 2309 V and the excitation voltage is Ea = 2400 V Figure 17.10a The mechanical power developed per phase P = (E E/X = power = Total (y s) (2400 X = 1 = 1573 X 3 5 sin The flux produced by the stator gap through the salient poles. is flows across the air (17.2) 2309/1.2) sin 20° axis of N pole of rotor 573 300 J axis of S pole ! of stator \ kW 1573 = 4719 kW (-6300 hp) 17.8 Reluctance torque If we gradually reduce the excitation of a synchro- nous motor when that the it is running motor continues to no-load, at speed even when the exciting current reason is that the flux to cross the short produced by the gap between the much the stator rather than the tween the poles. In other Fig. 1 7. 10a. On is find is zero. The stator prefers is less in the axis this of the phenomenon, in the tract the salient poles in 1 7. 10c). tains a mechanical load is applied to the shaft, the have the shape shown maximum Fig. 17,11 rotor poles will fall behind the stator poles, and the stator flux will its in Fig. 1 7. 1 0b. positive value at 8 maximum is namely at 8 = 90°. the reluctance torque as a func- The torque reaches a maximum = 45°. For larger angles negative value at 8 = it attains 135°. Obviously, Thus, a considerable reluctance torque can be de- to run as a reluctance-torque motor, the angle veloped without any dc excitation lie at all. The reluctance torque becomes zero when the midway between the stator poles. The reason is that the N and S poles on the stator atrotor poles are is zero where the regular torque T at- value, shows tion of the angle 8. a opposite directions (Fig. Consequently, the reluctance torque precisely at that angle motor develops a reluctance torque. If salient poles are attracted to the stator poles, thus producing a reluctance torque. and salient poles concentrated as shown account of Figure 17.10b The longer air gap be- words, because the reluc- tance of the magnetic circuit salient poles, the flux we run at synchronous must between zero and 45°. Although a positive torque still developed between 45° and 90°, this is an un- stable region of operation. The reason angle increases the power decreases. is that as the S YNCHRONO US MOTORS 379 Figure 17.12 Figure 17.10c The reluctance torque is zero when the are midway between the stator poles. salient poles In a synchronous motor the reluctance torque plus (1) the smooth-rotor torque (2) produce the resultant torque (3). Torque can be seen is due (2) is to the dc excitation in Fig. 17.12. of the rotor. However, the difference not very great, and for this reason we shall con- tinue to use Eqs. 17.2 and 17.5 to describe synchro- nous motor behavior. Losses and efficiency synchronous motor 17.9 Figure 17.11 of a Reluctance torque versus the torque angle. In order to give the reader a sense of the order of As magnitude of the pull-out torque, resistance, reaccase of a conventional synchronous in the tance, motor, the mechanical power curve has exactly the and losses of a synchronous motor, we have drawn up Table 7A. It shows the characteristics of a 2000 hp and a 200 hp synchronous motor, respectively labeled Motor A and Motor B. 1 same shape sence as the torque curve. Thus, in the ab- of dc Does 8 at = mechanical the excitation, reaches a peak power 45°. power The answer is the saliency of the poles modify the and torque curves shown in Fig. 17.9? The following points should be noted: 1. yes. In effect, the curves shown in Fig. 1 7.9 are those 27° and 37°. of a smooth-rotor synchronous motor. The torque of a salient-pole rotor motor is 1 7. 1 1 . It at full-load corresponds ranges between to the electrical an- gle 8 mentioned previously. equal to the component and nent of Fig. sum of the smoothreluctance-torque compo- The torque angle the 2. (4.2 Thus, the true torque curve of a syn- chronous motor has the shape (3) The peak reluctance torque given is of the peak smooth-rotor torque. the in Fig. 17. 12. As is about 8 per- 3. The is to excite the 2000 hp motor only about twice that needed for 200 hp motor per-unit a result, the cent greater than that of a smooth-rotor motor, as kW) larger the about 25 percent peak torque of a salient-pole motor The power needed (2. 1 kW). In general, the synchronous motor the smaller power needed total losses to excite of Motor four times those of Motor A (38 kW) B (9.5 is the it. are only kW) despite the ELECTRICAL MACHINES AND TRANSFORMERS 380 CHARACTERISTICS OF TWO TABLE 17A SYNCHRONOUS MOTORS MOTOR A NAMEPLATE RATING power [hp] power [ k W 2000 hp 492 k W 1 ] 4000 V 220 A line voltage line current speed MOTOR B 440 V A 208 90()r/min 60 Hz 3 3 .0 l pull-out torque (pu) dc exciter power 17.10 Excitation and reactive power Consider a wye-connected synchronous motor connected to a 3-phase source whose line voltage fixed (Fig. 17.13). voltage wye wye kW gap V 25 10 mm V 25 1 mm 6 stray losses stator/ rotor I 2 2 kW kW kW 3.5 kW 2 kW 9.5 kW 97.5% 94.0% 1 /? R total losses efficiency IMPEDANCES AND VOLTAGES stator X s XJR 0.0638 l 2 phase voltage fact that E E0 is therefore created by the U v E. <I> induces a line-to- in the stator. If in the stator, However, because E com- . is it fixed, we neglect the follows that E.d <I> is = also fixed, as case of a transformer (see Section 9.2). mmf needed to create the constant flux may be produced by both. either by the <4> stator or the rotor or If the rotor exciting current / x is zero, all the Ex. 0.0262 254 285 V V ten times as powerful. This another property of large motors: the more horsepower they develop, the smaller the tive losses are. As in power. Compare the effi+ two motors: 97.5% versus 94.0%. The synchronous reactance much rela- a result, the efficiencies im- prove with increase ciencies of the 4. produce On the other 23 V 2873 V is line currents / O H 0.62 il 2309 Motor A The l 122 S phase voltage is Rs is x (line-to-neutral values) 7.77(2 stator resistance ratio 1 total flux very small IR drop in the kW kW 4 kW 10.3 kW 4.2 kW 38 kW 8.5 friction also fixed. neutral voltage The stator core loss is E follows that the line-to-neutral Pursuing our reasoning, flux kW 2.1 The r. LOSSES windage and E It bined action of U.d and 2.2 27° 1 U .0 36.7° 4.2 dc exciter voltage l .4 1 at full-load connection air motor hand, the rotor produces a dc magnetomotive force factor torque angle as far as s a magnetomotive force U.d in the stator. LOAD CHARACTERISTICS power R concerned. 49 k 60 Hz phases the effect of is 200 hp 1 I800r/min frequency ways neglect performance X s per phase is 122 times larger than R As s . a result, X 6- Figure 17.13 larger than the resistance of the stator winding. Note that for the 2000 hp motor 6 s we can total flux 4> is due to the mmf produced by the ro(U ) plus the mmf produced by the stator (Ua ). For The is tor al- a given r EL , the flux <E> is essentially fixed. S YNCHRONO US produced by the flux has to be The stator must power from the stator. then absorb considerable reactive we 3-pfiase line (see Section 7.9). But if rotor with a dc current / x mmf helps the rotor , duce part of the flux O. Consequently, power drawn from is the ac line. If The ical active power absorbed power of the motor. is MOTORS 38 equal to the mechan- excite the pro'p less reactive we gradually raise the excitation, the rotor will eventually pro- duce the required flux all by The itself. stator then draws no more reactive power, with the power the What happens tive motor becomes unity factor of the critical level? we if The of reactive power, just as if it motor this we can cause power to correct the same time the at agram the current q , and motor designed p / line 36.87°. This cur/ p and power P q is = 0.8/ s (17.7) = 0.6/ s (17.8) given by P = £ab / p = The = broken up into two components / active The in Fig. 17.14. clear that power facme- power for a power equal factor of 0.8 (leading). power reactive 0.8 £ab / power delivered by Q = £ab /q = 0.6 the £ab / (17.9) s motor is (17.10) s to can supply A factor of 0.8 can deliver 75 percent of kW rated its me- motor shown 75% X 3000 = 2250 in kvar to same time as it develops its rated mekW. Motors designed to opcrate at leading power factors are bigger and more costly than unity power factor motors are. The reathe line at the chanical output of 3000 is is rating chanical load. Thus, the 3000 son it motor factor motor develops the same me- leads E. lb by arcos 0.8 /s rent can be / as the It power, they are usually designed to op- erate at a full-load Fig. 17.4 power chanical as they furnish Most synchronous motors are designed to operate at unity power factor. However, if they also have to de- reactive shows an 80% power 15 to the load they are driving. Power factor liver reactive 7. 1 also operating at full-load. The 17.11 di- at full-load. Fig. to the source important property, synchronous motors are sometimes used chanical like a absorb or deliver reactive power. to either of a plant this were a capacitor. Thus, by varying the dc excitation Because of .0). Figure 17.14 Unity power factor synchronous motor and phasor of absorbing reac- The motor then behaves line. 1 motor above excite the stator, instead ( power, actually delivers reactive power 3-phase tor result that that for a given horsepower rating, both the dc Figure 17.15 80 percent power factor synchronous motor and phasor diagram at full-load. It follows from Eqs. 17.9 and 17. 10 that exciting current and the stator current are higher. Q= This can be explained as follows. Fig. 17. 14 is the schematic power factor motor operating to-neutral voltage The active is E,db and the is, of rated mechanical output line- was as line current power absorbed per phase P = Eay Jp The P — 15% diagram of a unity at full-load. 0.75 is / p stated previously. . If we compare therefore, I p with /s , we find that / s = i .25 / p . Thus, for the same mechanical power output, a mo(17.6) tor designed for a leading power factor of 80% has ELECTRICAL MACHINES AND TRANSFORMERS 382 25% to carry a line current that is power that operates at unity greater than one ,'s2 factor. 's2 \36.9° 17.12 V-curves 200 P = 800 Suppose a synchronous motor rated mechanical load. We havior as the excitation is in wish is operating to examine power remains power 1 6. will We assume we reduce If S = /x power. 1000 kVA. As a is (b) shown Figure 17.18 in 200 A but Field excitation raised to a. with same me- chanical load. Motor delivers reactive power to the unity, line. Fig. b. = 00 A and P = 800 kW. Phasor diagram shows current leading the voltage. 1 the excitation to draw reactive power from to the active mechan- factor thus yielding the phasor diagram 7. S = 1000 kVA (a) fixed. Let us begin by adjusting the excitation I x so that the 1 its be- its Because a change varied. excitation does not affect the speed, the ical at A kW We assume 70 A, the motor the line in addition that S increases to result, the line current will in- crease from ponent of / sl p to / sl (Fig. 1 phase with in because the motor is still 7. 1 7). £. lb is Note the that the same com- as before developing the same me- chanical power. Current E b o —v I M y P = 800 kW S = 800 kVA A 4' 100 A /s) lags behind £"ab motor the rotor smaller than before, but the apparent is is power S absorbed by If (b) (a) Figure 17.16 Synchronous motor operating at unity power factor with a mechanical ioad of 800 kW. Field excitation is 100 A. b. Phasor diagram shows current in phase with the voltage. we lagging. the stator is field current I x in greater. increase the excitation to motor delivers reactive power is a. and so the power , The factor of the /x = 200 A, to the line to the which connected (Fig. 17.18). The apparent power again greater than We assume S comes /s2 and = in the unity power 1000 kVA. The it is factor case. line current be- However, the in-phase it leads ELXh component of /s2 is still equal to I p because the mechanical power is the same. . By varying the excitation this way, we can power of the synchronous motor the apparent plot as a function of the dc exciting current. This yields a V-shaped curve Ix = 70 A 36.9° P = 800 kW S = 1000 kVA (a) tive (b) the b. reduced to 70 A but with same meMotor absorbs reactive power from Field excitation chanical load. line. Phasor diagram shows current lagging behind the voltage. The V-curve the V-curve corresponds to full-load. V-curve Figure 17.17 a. (Fig. 17.19). is always displayed for a fixed mechanical load. In our case, is power The no-load also shown, to illustrate the large reacthat can be absorbed or delivered by simply changing the excitation. Example A 1 7-5 4000 hp (3000 kW), 6600 synchronous motor operates power 1 1 (2, V, 60 Hz, 200 r/min at full-load at a leading factor of 0.8. If the synchronous reactance calculate the following: is SYNCHRONOUS MOTORS 383 kVA ^ 1000 full-load x O+ ' source E yrio-load 400 O- 200 Figure 17.20 synchronous motor connected to a source E. Note the arbitrary (+) polarity marks and arbitrary direction of current flow. See Example 17-5. Circuit of a 200 120 80 40 -'x Figure 17.19 No-load and full-load V-curves nous motor. E = 3815^0° of a 1000 hp synchroIt c. The apparent power of the motor, per phase The ac line current The value and phase of E0 d. Draw e. Determine the torque angle 8 a. b. the phasor = given by 328/136.9° Writing the equation for the circuit -E+jlX + E0 = diagram s shall find 0 thus immediately change the given values correspond The to E0 = E-jIX = 3811/10° - j (328/136.9°) 11 = 3811^0° - 3608/1(36.9° + 90°) = 3811 (cos 0° + y sin0°) - active to one phase of a wye-connected motor. power per phase P= 3000/3 is = lOOOkW The apparent power per phase 3608 (cos 126.9° line-to-neutral voltage (8.11) d. Consequently, is EQ lags 26° behind E, complete phasor diagram E= The line current I / EjV3 = leads the value , \ 1 1 V e. The torque angle 8 is is shown and the in Fig. 17.21. 26°. 17,13 Stopping synchronous motors = Owing 36.9°. to the inertia of the rotor and its load, a large synchronous motor may take several hours and phase of the excita- £ we draw (> 38 1250 X 1000/3811 an angle of arcos 0.8 To determine tion voltage = C is = S/E= = 328 A E by 6600/ V3 sin 126.9 3811 + 2166 -; 2885 = 5977 -y2885 = 6637/1-26° is = 1250kVA The +; = S = P/cosd = 1000/0.8 c. we s We b. / is / Solution a. follows that after the equivalent circuit time, being disconnected from the we line. to stop To reduce the use the following braking methods: of one phase (Fig. 17.20). This will enable us to write the circuit equations. Furthermore, lect E as the reference phasor and so we se- I. Maintain full short-circuit. dc excitation with the armature in ELECTRICAL MACHINES AND TRANSFORMERS 384 e. The time required for the speed 600 r/min to 150 r/min to fall from Solution a. In Fig. 17.22a the motor has just been discon- nected from the line and generator is now operating as a The speed frequency is 60 Hz. in short-circuit. r/min, and the is still Consequently, the impedance per phase Z=VR 2 + Xi \ 0.2 6637 V = 2 600 is (2.X 12) + 16 2 16 ft Figure 17.21 See Example The current per phase 17-5. / is = EJZ= 2400/16 = 150A 2. Maintain full dc excitation with the armature connected to three external 3. Apply mechanical braking. In methods it I and 2, the The power resistors. r/min P= motor slows down because functions as a generator, dissipating the resistive elements of the circuit. its energy is Mechanical usually applied only after the motor has reached half-speed or less. A lower speed prevents undue wear of the brake shoes. Example 17-6 A 1500 kW, 4600 2 3I R = 13.5 in b. braking dissipated in the 3 phases at 3 X 150 E0 2 X is fixed, the in- 0.2 kW Because the exciting current duced voltage is proportional to the speed. Consequently, when the speed has dropped to 150 r/min, Ea = 2400 X 600 r/min, 60 Hz synchronous motor possesses a synchronous reactance of The frequency 16 ft and a stator resistance of 0.2 ft, per phase. and so = 600 V (150/600) V, is also proportional to the speed, The excitation voltage E 0 is 2400 V, and the moment of inertia of the motor and its load is 275 kg-irr. We wish to stop the motor by short-cir- The synchronous reactance cuiting the armature while keeping the dc rotor the frequency; consequently, /= 60 X (15/60) = is 15 Hz proportional to current fixed. 16U Calculate a. The power dissipated in the armature at 600 in the armature at 150 r/min b. The power dissipated r/min c. d. The The 600 is kinetic energy at 600 r/min kinetic energy at 150 r/min Figure 17.22a Motor turning at 600 r/min (Example 17-6). S YNCHRONO US 4 P = LI 13.5 150 A 600 V whence 0.2 Q. 150 r/min Note that the t MOTORS 3 85 Wit (3.4) = 508.6// = 37.7 s motor would stop much sooner if external resistors were connected across the stator terminals. Figure 17.22b Motor turning at 150 r/min (Example 17-6). The synchronous motor 17.14 versus the induction motor X, = X 16 = (15/60) 4 0 We Referring to Fig. 17.22b the phase at 150 r/min new impedance per But is have already seen that induction motors have 600 excellent properties for speeds above at r/min. lower speeds they become heavy, costly, and have relatively low power factors and efficiencies. Z = V0.2 2 + 4 2 = a 4 Synchronous motors are particularly attractive for low-speed drives because the The current phase power factor can is always be adjusted /= £0 /Z = 600/4 = A 150 to 1 .0 Although more complex and the efficiency is high. weight and to build, their cost are often less than those of induction motors of Thus, the short-circuit current remains un- changed r/min to 150 r/min. The power dissipated 3 phases therefore the is equal power and speed. This motor decelerates from 600 as the P = same 13.5 is particularly true for speeds below 300 r/min. in the as before: A synchronous motor can tor kW of a plant Furthermore, its while improve the power carrying starting torque its rated fac- load. can be made consid- The erably greater than that of an induction motor. c. The kinetic energy at 5.48 d. The A 10 -3 = 5.48 = 542.5 kJ X 10 " is reason Jrr (3.8) X 275 X of the squirrel -cage wind- at 5.48 X 10 3 150 2 motors in decelerating from - = 542.5 = 508.6 kJ The time in and kilns, A synchronous capacitor is essentially a synchronous motor running armature for the speed to drop is ultra-low speeds. Thus, huge motors 33.9 lost as heat in the r/min to 150 r/min at MW range drive crushers, rotary 17.15 Synchronous capacitor kl is the 10 variable-speed ball mills. W = E -Ek2 This energy rating. very low frequencies enable us to run synchronous 33.9 kJ The loss in kinetic energy 600 r/min to 150 r/min is motor and same nominal The biggest difference is in the starting torque. High-power electronic converters generating is X 275 X effi- synchronous speed. Figure 17.23 compares a synchronous motor having the kinetic energy at 150 r/min tance. that the resistance ing can be high without affecting the speed or ciency 600- is the properties of a squirrel-cage induction E k2 = = e. X 600 r/min given by resis- from 600 at no-load. Its only purpose is to ab- sorb or deliver reactive power on a 3-phase system, in order to stabilize the voltage (see Chapter 25). machine acts as an The enormous 3-phase capacitor (or ELECTRICAL MACHINES AND TRANSFORMERS 386 I i i s ynchrcDnous motor 97 induction > motor /J 96 rf // // // o c a) it 95 !y if if if if (a) if 94 93 92 91, 25 50 *- 75 100 125 % mechanical power Figure 17.24a % Three-phase, 16 250 rated -200 Mvar kV, 900 r/min synchronous capacitor (supplying reactive power) to +300 Mvar (absorbing reactive power). It is used to regulate the voltage of a 735 kV transmission line. Other characteristics: mass of rotor: 143 t; rotor diameter: 2670 mm; axial length of stator iron: 3200 mm; air gap synchronous motor 200 = 150 length: 39.7 mm. (b) 100 induction motor ^ 50 60 40 20 80 100 % speed Figure 17.23 Comparison between the torque (b) of efficiency (a) synchronous motor, both rated r/min, 6.9 kV, 60 Hz. inductor) and starting a squirrel-cage induction motor and a whose reactive at 4000 hp, 1800 power can be varied by changing the dc excitation. Most synchronous capacitors have ratings that range from 20 Mvar to 200 Mvar and many are hydrogen-cooled (Fig. 17.24). They are started up like synchronous motors. However, if the system cannot furnish the required starting power, a pony motor is used to bring them up to synchronous speed. For example, in one installation, a 160 Mvar mm Figure 17.24b Synchronous capacitor enclosed in its steel housing containing hydrogen under pressure (300 kPa, or" 2 about 44 lbf/in ). {Courtesy of Hydro-Quebec) SYNCHRONOUS MOTORS synchronous capacitor is and brought up started speed by means of a 1270 387 to kW wound-rotor motor. Example 17-7 A synchronous capacitor 16 kV, actance of 0.8 pu and is Calculate the value of EQ a. Absorbs 160 Mvar b. Delivers 120 It 160 Mvar, rated at is 1200 r/min, 60 Hz. has a synchronous connected to a 16 kV re- line. so that the machine Q= 120 Mvar<feMjl Mvar / Solution a. i 4335 The nominal impedance of the machine Zn = En 2 /S n = 16 000 2 /(160 X = (16.3) 10 6 5550 V 9250 V 14 800 V ) a 1.6 A is Figure 17.25b Over-excited synchronous capacitor delivers reactive The synchronous reactance per phase X = X s (pu)Zn = s = The i.28 0.8 X is power (Example 1.6 From Fig. 17-7). 7.25a 1 n we can -£+ 77X line current for a reactive load of 160 Mvar s write + £o = 0 hence is E„ /n = 5 n /(V3£n - 160 X c X 10 7( .73 1 16 000) = 5780 A The drop across the synchronous reactance is b. Ex = IX S = 5780 X = E- jIX 1.28 Note that the excitation voltage less than the line voltage The load current when ing 120 IU line-to-neutral voltage E = EjV3 = = 9250 V Selecting E as E= The current machine is / This time we have 0° lags 90° behind the absorbing reactive power; conse- quently, / = 5780^-90° 850 V) 10 7(1.73 machine X / E by leads is deliver- 16 000) 90° and so = 4335Z90 0 Fig. 17.25b we can write En = E-jIX = 9250^0° - 4335 X 1.28/180° = (9250 + 5550)Z()° s = 14 800/10° is (9250 V). ) ( X / From E because 1 90°) is = Q/(V3EU 120 the ( - = 4335 A 16 000/1.73 9250,/ Mvar - is the reference phasor, 1.28^(90° much = 7400 V The s = 9250^0° - 5780 X = 1850^0° ) ELECTRICAL MACHINES AND TRANSFORMERS 388 5780 motor [hp] knowing A it has an efficiency of 95%. 17-8 A synchronous operates at a motor driving a pump power factor of 100%. What happens if the dc excitation is in- creased? 17-9 mm^ Q= 160 Mvar 7400 A 3-phase, to a 4 kV, 60 Hz current of 320 A and line draws a absorbs 2000 kW. E V y > I 9250 Calculate V a. 1850 V b. c. 5780 225 r/min synchronous motor connected A d. Figure 17.25a Under-excited synchronous capacitor absorbs reactive power (Example 17-7). 17-10 The apparent power supplied to the motor The power factor The reactive power absorbed The number of poles on the rotor A synchronous 3-phase motor draws line. If the 1 50 A from exciting current raised, the current drops to 140 A. a is Was the motor over- or under-excited before the was changed? excitation The excitation voltage (14 800 V) is now con- siderably greater than the line voltage (9250 V). Intermediate level 17-11 a. Calculate the approximate full-load current of the 3000 hp motor Questions and Problems in Fig. 17.1, if it has an efficiency of 97%. Practical level 1 1 7- 1 7-2 Compare 7-3 the construction of a synchro- 17-12 7-4 Referring to Fig. 17.2, how Explain a synchronous motor starts up. 17-13 should the dc excitation be applied? Why does the Name some What A 3-phase a. What b. If is factor. speed of a synchronous motor 1 .8 is we used for? meant by an under-excited syn- 17-14 does 1 7-7 over-excite a synchronous motor, its power line voltage at motor draws 2000 factor of the approximate 90% leading. Calculate the power kV, but the exciting current remains how the following a. Motor speed and mechanical power output b. Torque angle 8 c. Position of the rotor poles d. Power e. Stator current factor A synchronous motor has the following parameters, per phase (Fig. 17.7a): kVA at power developed by unity suddenly drops to mechanical power output increase? A synchronous a 60 Hz operates quantities are affected: to a squirrel- chronous motor? The unchanged. Explain of the advantages of a syn- it synchronous motor rated 800 hp, 2.4 kV, meant by a synchronous capacitor is and what 7-6 what speed must quencies? cage induction motor. 1 at a squirrel -cage induction motor. chronous motor compared 17-5 the value of the field resistance? is the rotor turn to generate the indicated fre- remain constant even under variable load? 1 What nous generator, a synchronous motor, and When 1 b. E= x = s / 2.4 kV; 2 a = 900 A. Ea = 3 kV SYNCHRONOUS MOTORS Draw the phasor diagram and determine: 17-19 nous motor rated 400 hp, 2300 b. Active power, per phase r/min, 80 A, c. Power The d. Reactive power absorbed (or delivered), factor of the a. In motor is new the mechani- citation its power excitation is is adjusted 17-20 0.6 increased without making any is is the effect The synchronous capacitor (I, phase upon the per phase. 0.007 is to a stop, it by the motor dc excitation initial line motor the tor line current reactive power absorbed level 10 H, per phase. The stator is a. at full-load b. (4000 hp) with a leading power factor of 0.89. If the efficiency is 97%, calculate c. the following: c. The apparent power The line current The value of E0 per phase The mechanical displacement of the poles The their no-load position total reactive electrical f. power supplied power b. A so that the its is rated value, or 1600 V, at d. The total braking power and braking torque at 900 r/min The braking power and braking torque at 450 r/min The average braking torque between 900 r/min and 450 r/min The time for the speed to fall from 900 r/min to 450 r/min, knowing that the moment of inertia of the rotor is 1 .7 6 X 1() lbfr. 17-21 A 500 hp, 3-phase, 2200 V, unity power factor synchronous motor has a rated current of 103 A. It can deliver its rated out- put so long as the air inlet temperature 1 7- 7 1 we wish 40°C or less. The manufacturer is states that to adjust the the output of the motor must be decreased factor to unity. by 1 percent for each degree Celsius above 40°C. Calculate a. 250 voltage across the resistors system The approximate maximum power the motor can develop, without pulling out of Problem fixed at H wye. The Industrial application to the step [hp] In is in r/min. , from e. connected to three large 0.6 Calculate connected wye, and the motor operates machine coasts in Fig. 17.4 possesses a synchronous reactance of d. is 900 kV motor shown hp, 6.9 resistance per the braking resistors connected (or delivered) one-tenth of The 4000 The 11. If der to shorten the stopping time, the sta- The torque angle power absorbed by 17.24 will run for about 3 h. In or- d. active in Fig. possesses a synchronous reactance of unity. If the ex- c. a. to synchronism The The The b. adjusted , c. following: in is iy The value of Xs and of £ 0 per phase The pull-out torque [ft-lbfj The line current when the motor is about a. b. power absorbed reactive factor other change, what Advanced E hp synchronous motor drives a so that the a. synchronous reactance of Calculate pull out of compressor and b. 450 the line current if (or delivered) by the motor, per phase. 17-18 stator has a 0.88 pu, and the excitation suddenly removed, b. Calculate the A 500 V, 60 Hz, drives a compressor. to 1.2 pu. Problem 17-14 calculate cal load 17-17 factor synchro- Torque angle 8 and the new torque angle 5 17-16 power unity a. per phase 17-15 A 3-phase, 389 The exciting voltage £0 The new torque angle required, per phase If the air inlet 46°C, calculate the motor current. temperature maximum allowable is 390 1 7-22 ELECTRICAL MACHINES AND TRANSFORMERS An 8800 kW. 6.0 kV, 1 500 r/min, 3-phase, 50 Hz, 0.9 power factor synchronous motor manufactured by Siemens has the fol- sing the above information, calculate the fol- wing: a. 1. 962 Rated current: A Rated torque: Pull-out torque: 4. Locked-rotor current: 5. Excitation voltage: 56.0 b. kNm 2. 3. c. 1.45 pu 160 V 387 A Full-load efficiency, excluding excitation losses: 9. Temperature 11. f. 520 kg-m~ Maximum g. 12. 13. 14. its in gallons (U.S.) total moment which the motor can of inertia (in pull into syn- The The total losses of the motor total efficiency at full-load of the motor at full- moment 2 The reactive power delivered by the motor full-load If the iron losses are equal to the stator cop- per losses, calculate the approximate resis- 465 L/min permissible external 1370 kg-m The maximum at 25°C of cooling water: 32°C Flow of cooling water: inertia: motor including load of inertia of rotor: rise e. 97.8% to 10. the chronism d. Excitation current: Moment mass of The flow of cooling water lb-ft'), 4.9 pu 7. 8. total per minute 6. system The enclosure, in metric tons lowing properties: of tance between two terminals of the stator. h. Calculate the resistance of the field circuit. Mass of rotor: 6. 10 t (t = metric Mass of stator: 7.50 Mass of enclosure: 3.97 t t ton) Chapter 1 Single-Phase Motors 18.0 Introduction Fig. 18.2 shows the progressive steps in wind- ing a 4-pole, 36-slot stator. Starting with the lami- Single-phase motors electric are the most familiar of all motors because they are used in nated iron stator, paper insulators home ers — are first inserted in —called the slots. r/min, 60 Hz slot lin- The main appliances and portable machine tools. In general, employed when 3-phase power they are is not available. There are many kinds of single-phase motors on the market, each designed to cation. However, we meet a specific appli- will limit our study to a few basic types, with particular emphasis on the widely used split-phase induction motor. 18.1 Construction of a singlephase induction motor Single-phase induction motors are very similar to 3-phase induction motors. They are composed of a squirrel-cage rotor (identical to that in a 3-phase motor) and a stator (Fig. 1 8. 1). main winding, which creates a The set stator carries a of N, S poles. It also carries a smaller auxiliary winding that only when the motor auxiliary winding has the same num- operates during the brief period starts up. The Figure 18.1 Cutaway view of a 5 hp, 1725 phase capacitor-start motor. ber of poles as the main winding has. (Courtesy of Gould) 391 single- ELECTRICAL MACHINES AND TRANSFORMERS 392 Figure 18.2a Bare, laminated stator of a 1/4 hp squirrel-cage rotor is (1 87 W), single-phase motor. The 36 3-phase motor. slots are insulated with a paper liner. The identical to that of a Figure 18.2c Figure 18.2b Four poles of the main winding are inserted in the slots. Four poles of the auxiliary winding straddle the main winding. (Courtesy of Lab-Volt) winding 1 is then laid the in 8.2b). Next, the auxiliary that (Fig. its 1 slots winding is (Figs. I8.2a, embedded so poles straddle those of the main winding 8.2c). The reason be explained shortly. for this arrangement will Each pole of the main winding consists of group of four concentric (Fig. 1 8.3a). coils, connected Adjacent poles are connected so as produce alternate N, S polarities. the center of each pole The empty (shown as a in series to slot in a vertical dash SINGLE-PHASE MOTORS 393 one pole pitch —(180°) —H 10 20 25 30 turns No. 16 wire Figure 18.4 Main and auxiliary windings motor. iary The winding opens mounted on the nous speed. line) in stationary contact and the when shaft, a 2-pole single-phase series with the auxil- in the centrifugal switch, reaches 75 percent of synchro- partially filled slots on either side of are used to lodge the auxiliary winding. The has only two concentric coils per pole (Fig. Fig. 1 shows a 2-pole 8.4 8.3c). 1 main stator; the large winding and the smaller auxiliary winding are at right 18.2 Synchronous speed in the speed of dis- angles to each other. placed As it latter case of 3 -phase motors, the synchronous all single-phase induction motors is given by the equation ns = 120/ 1 (17.1) P where ns 90°-*! center of main winding = /= center of auxiliary p winding The synchronous speed [r/min] frequency of the source [Hz] = number of poles rotor turns at slightly less than synchronous speed, and the full-load slip 5 percent for fractional Figure 18.3 Main winding of a 4-pole, 36-slot motor showing the number of turns per coil. b. Mmfs produced by the main winding. a. c. laid out typically 3 percent to flat, Position of the auxiliary winding with respect to the main winding. is horsepower motors. Example 18-1 Calculate the speed of the 4-pole single-phase shown in percent. The tor Fig. 18.1 if the slip at full-load line frequency is 60 Hz. is mo3.4 ELECTRICAL MACHINES AND TRANSFORMERS 394 Solution produces an ac to the stator. the resulting current / s The motor has 4 = ns poles, consequently, 120///? = (120 X flux // is s . The flux pulsates back and forth but, unlike the flux in a 3-phase stator, no revolving field 60)/4 duced. The flux induces an ac voltage = 1800r/min The speed <£> is pro- in the station- ary rotor which, in turn, creates large ac rotor currents. In effect, the rotor given by: behaves like the short-circuited secondary of a transformer; consequently, the motor s = (n s 0.034 = (1800 n = 1739 r/min - n)!n s - (17.2) has no tendency to h)/1800 However, the other, As spin. 18.3 Torque-speed characteristic if we start by itself (see Fig. 1 8.5 a schematic diagram of the rotor and main is will continue to rotate in the direction of it ates until tor reaches a speed slightly below synchro- it winding of a 2-pole single-phase induction motor. turn. Fig. when is locked. If an ac voltage is applied that the develops a positive torque as soon as Suppose the rotor 8.6). a matter of fact, the rotor quickly acceler- nous speed. The acceleration indicates Fig. 1 spin the rotor in one direction or 8.6 1 the shows the is mo- begins to typical torque-speed curve main winding starting torque it is zero, the excited. Although the motor develops a power- rotor current ful torque as it approaches synchronous speed. 18.4 Principle of operation The principle of operation of a single-phase induc- tion motor is quite complex, and by the cross-field theory As soon E 120 V, 60 Hz is may be explained * as the rotor begins to turn, a speed induced in the rotor emf conductors as they cut the ac source stator flux <D S (Fig. 18.7). This voltage increases as Figure 18.5 Currents The is in the rotor bars when resulting forces cancel the rotor is locked. each other and no torque produced. *6 ac source Figure 18.7 Currents induced They produce a in the rotor bars due to rotation. flux <I\ that acts at right angles to the stator flux speed * Figure 18.6 Typical torque-speed curve of a single-phase motor. The double revolving Held theory (diseussed 1 8. 18) is in Section also used to explain the behavior of the single- phase motor. SINGLE-PHASE MOTORS = r/4 t / 395 = 3774 Figure 18.8 Instantaneous currents and flux in a single-phase momain winding excited. The duration of one T seconds, and conditions are shown at suc- tor with the cycle is cessive quarter-cycle intervals. a. Stator current b. Stator current however, c. d. e. /s is is maximum, rotor current zero, rotor current is smaller than <I> S is zero. is /r maximum; . maximum, but negative. Rotor current is maximum, but negative. After one complete cycle (t = T) the conditions Stator current is re- peat. f. Resulting flux <b in the air gap rotates ccw at synchronous speed. Its amplitude varies from a maximum of <I> S to a minimum <I> r the rotor speed increases. flow currents produce an ac flux same time behind does not reach /, to r which acts at right its is the maximum value at the as 3> s does. In effect, <I\ lags almost 90° due <t> s , to the The combined volving magnetic motor. <l> Equally important angles to the stator flux <!>,. causes currents It bars facing the stator poles. These in the rotor fact that . inductance of the action of 4\ and 4> r rotor. produces a field, similar to that in a The value of <I> r re- 3-phase increases with increasing speed, becoming almost equal to 3> s why speed. This explains in part at synchronous the torque in- creases as the motor speeds up. We can understand duced by referring the currents tor /s = + 10 A r = T and sume how the revolving field to Fig. 18.8. It and fluxes created respectively by stator, at that the successive intervals of time. motor is is pro- gives a snapshot of the ro- We as- running far below synchronous ELECTRICAL MACHINES AND TRANSFORMERS 396 speed, and so much is <!>,. smaller than <J) s . is obvious that the combination of <I> S and a revolving field. Furthermore, the flux izontally and shown wise it in Fig. 18.8f. The strong hor- Thus, the elliptic pattern flux rotates counterclock- iary rotor. As motor approaches syn- the tioned previously) is and standstill torque T— = = /s Fig. 1 8.9. sina */ a / s (1 8. 1) locked-rotor torque [N mJ locked-rotor current in the auxiliary in the main ] locked-rotor current winding [A| a starting torque in a single-phase motor, When the main at low speeds. The locked-rotor at winding [A create a revolving field. This (men- <!>,. result, <P.d strength- where 18.5 Locked-rotor torque we must somehow weak. As a T= produced. done by adding an auxiliary winding, as shown during the a given by is /., To produce is <I> the rotor flux thereby producing a powerful torque both ens chronous speed, CI\ becomes almost equal to <P S and a nearly perfect revolving field when acceleration period direction as the rotor. Furthermore, speed of the immediately see that the auxil- will winding produces a strong flux synchronous speed, irrespective of the ac- rotates at tual vertically. low speed follows the same in the weak relatively field strength at it produces is The reader By observ- ing the flux in the successive pictures of Fig. 18.8, is a = phase angle between in k = and auxiliary windings are and /s /., [°| a constant, depending on the design of the motor connected to an ac source, the main winding pro- duces a flux duces a flux <P S , the If two fluxes so that <1\ either lags or leads set up. manner To while the auxiliary winding pro- <E> S , The 2-phase revolving are out of phase, a rotating field field is obtain the desired phase shift between (and hence between pedance is created in a similar to the revolving field of a 3-phase motor (see Section /., Z in and 4> a ), The rise to various types many cases pedance in the self, as is incorporated the desired im- auxiliary winding it- explained below. special switch is also connected in series with the auxiliary winding. when and depending upon the desired starting torque. of split-phase motors. In A /s an im- resistive, inductive, or capaci- The choice of impedance gives 13.3). we add series with the auxiliary winding. impedance may be tive, <J\ the chronous It disconnects the winding motor reaches about 75 percent of synspeed. A speed-sensitive switch mounted on the shaft purpose (Fig. 1 is centrifugal often used for this 8. 10). 18.6 Resistance split-phase motor i i The main winding of a single-phase motor is always 2 made of relatively large wire, to reduce the I R losses (Fig. 18. la). The winding also has a relatively large number of turns. Consequently, under 1 o- Figure 18.9 Currents and fluxes ac source locked-rotor conditions, the inductive reactance at standstill when the main and An elliptical revolv- auxiliary windings are energized. ing field is «-o produced. high and the resistance locked-rotor current applied voltage /s E (Fig. is low. As a result, is the lags considerably behind the 18.11 b). SINGLE-PHASE MOTORS 397 Figure 18.10 a. Centrifugal switch position. b. The in Centrifugal switch position. Due the closed, or stopped, stationary contact in is closed. the open, or running, to centrifugal force, the rectan- gular weights have swung out against the straining tension of the springs. This caused the plastic collar to move re- has to the left along the shaft, thus opening the stationary contact series with the auxiliary winding. In a resistance split-phase centrifugal motor (often simply called split-phase motor), the auxiliary winding has switch ) E in ) ) auxiliary winding 70 turns per pole No. 22 wire a relatively small resistance number of higher and is turns of fine wire. Its reactance lower than that its of the main winding, with the result that the lockedrotor current / a is more nearly in phase with E. The a between /a and / s produces resulting phase angle the starting torque. The main winding /s 120 turns per pole and line current /.,. At / L is start-up, it equal to the phasor is sum of usually 6 to 7 times the nominal current of the motor. No. 16 wire Owing to the small wire used on the auxiliary winding, the current density is high and the winding heats up very quickly. If the starting period lasts for a= more than 25° 5 seconds, the and may burn winding begins out, unless the motor is to smoke protected by a built-in thermal relay. This type of split-phase tor is well mo- suited for infrequent starting of low- inertia loads. Example 18-2 A resistance split-phase motor is rated at 1/4 hp (187 W), 1725 r/min, 115 V, 60 Hz. When the rotor is locked, a test at reduced voltage on the main and (b) auxiliary windings yields the following results: Figure 18.11 a. b. Resistance split-phase motor (1/4 hp, 115 V, 1725 r/min, 60 Hz) at standstill. Corresponding phasor diagram. The current in the auxiliary winding leads the current winding by 25°. in the main main auxiliary winding applied voltage current active power E= / = V 4 A 23 60 W winding £= /, = P., = 23 1 .5 30 V A W ELECTRICAL MACHINES AND TRANSFORMERS 398 Calculate v s 2 - Pi G = s a. b. The phase angle between / a and / The locked-rotor current drawn from s = V92 2 - 60 2 = the line at V 115 = VSJ - Oa 69.7 var Pi Solution 17.0 var We first calculate the phase E of the main winding. a. The apparent power = 5S - £V S angle <$> s between Is and The Q= is X 23 4 = The The power cos factor <|> s + Ga 69.7 + 17.0 = - PJS = - 60/92 S S 0.65 = VP 2 + Q 2 V = 4> s We now calculate the phase angle and E of the auxiliary winding. = The power cos EI, = factor = 4> a = /, <T> a between r SIE -V$6J* = 125 V V current at 23 = 125/23 - = 5.44 X - (115/23) is 5.44 The locked-rotor current drawn /, The apparent power W The locked-rotor 49.6° lags 49.6° behind the voltage E. 5a is is thus, /s is 86.7 var apparent power absorbed total motor the Qs = VA 92 power absorbed by total reactive at A 1 V 15 is A 27.2 is 23 X 1.5 = 34.5 VA Due to their low cost, resistance split-phase in- duction motors are the most popular single-phase motors. They are used where a moderate starting is PJS.A = 30/34.5 = torque 0.87 is required and where the starting periods are infrequent. They drive fans, chines, oil burners, small thus, pumps, washing ma- machine tools, and other devices too numerous to mention. The power rating = *a /., 29.6° usually lies between a = = cj) s - <\>. A = /s and 49.6° - W and 250 W (1/12 hp to 1/3 hp). lags 29.6° behind the voltage. The phase angle between 60 Ia is 29.6° 18.7 Capacitor-start motor The 20.0° capacitor-start motor is identical to a split-phase motor, except that the auxiliary winding has about as b. To determine we first caloiP and Q drawn by both the total line current, culate the total value windings and then deduce the power S. The total total apparent many turns as the series with the auxiliary The capacitor active power absorbed is main winding has. Furthermore, a capacitor and a centrifugal switch are connected in 80°, which is is winding chosen so that (Fig. 18. 2a). 1 /., leads /s by about considerably more than the 25° found in a split-phase motor. Consequently, for equal starting = The reactive powers 60 Q auxiliary windings are s + 30 = 90 W and Q.A of the main and torques, the current in the auxiliary winding about half that in a split-phase motor. It is only follows that during the starting period the auxiliary winding of a capacitor motor heats up less quickly. Furthermore, SINGLE-PHASE MOTORS voltage, electrolytic capacitors are centrifugal much 399 smaller and cheaper than paper capacitors. However, elec- can only be used for short peri- trolytic capacitors ods ac circuits whereas paper capacitors can op- in on ac erate indefinitely. Prior to the development of electrolytic capacitors, repulsion-induction had motors be used whenever a high starting torque was to Repulsion-induction required. motors possess a commutator and brushes that require considerable maintenance. Most motor manufacturers special have stopped making them. when a high They are built in sizes kW (~I/6 hp to 10 hp). Capacitor-start motors are used starting torque ranging from 1 is required. 20 W to 7.5 Typical loads are compressors, large fans, pumps, and high-inertia loads. Table motor 1 8A gives the properties of a capacitor-start having a rating W(l/3 250 of 1760 r/min, 115 V, 60 Hz. Fig. 18. 1 hp), shows 3 the torque-speed curve for the same machine. Note that during the acceleration phase (0 to 1370 r/min) the main and auxiliary windings together produce a very high starting torque. When the rotor reaches 1370 r/min, the centrifugal switch snaps open, causing the motor to operate along the torque-speed curve of the main winding. The torque suddenly drops from 9.5 (b) Nm to 2.8 Nm, but the motor continues to accelerate Figure 18.12 until a. Capacitor-start motor. b. Corresponding phasor diagram. reaches 1760 r/min, the rated full-load speed. it and power factor of single-phase induction motors 18,8 Efficiency the locked-rotor line current /L cally is smaller, being typi- 4 to 5 times the rated full-load current. Owing to the high starting torque and the relatively low value of /a the capacitor- start motor to applications is well suited involving either frequent or prolonged starting periods. Although the starting characteristics of this motor are better than those of a split-phase tor, mo- both machines possess the same characteristics under load. The reason identical circuit is that the main windings and the auxiliary winding when the is motor has come up no longer are in the is full-load a 186 a direct W motor (1/4 hp) has an efficiency and power factor of about 60 percent. The low power factor is mainly due to the large magnetizing current, which ranges between 70 percent and 90 percent full-load Consequently, even current. at these motors have substantial temperature The relatively these motors to speed. The wide use of capacitor-start motors The efficiency and power factor of fractional horsepower single-phase motors are usually low. Thus, at horsepower is rises. low efficiency and power a consequence of ratings. Integral of no-load factor of their fractional horsepower single- result of the availability of small, reliable, low-cost phase motors can have efficiencies and power fac- electrolytic capacitors. For given capacitance and tors above 80 percent. ELECTRICAL MACHINES AND TRANSFORMERS 400 1370 r/min a auxiliary S winding' in circuit c 300 r/min ;entrifuga svvitch oper s \ centrifuga : \ I i clos es S\/vitch 2.8 no minal ill Figure 18.13 Torque-speed curves ^*"* " 400 600 ki. full load /bu r/mi n \\ '///// V////. 800 1000 > speed n of a capacitor-start motor, rated 1/3 1200 1400 hp (250 W), 1760 r/min, 1800 r/min 1600 1 15 V, 60 Hz, class A insulation. CHARACTERISTICS OF A CAPACITOR -START MOTOR TABLE 18A Rating: N-m V//// 200 0 ^ tore m 1.35 ji 250 W, 1760 r/min, 115 V, 60 Hz, Insulation Class 105°C No-load Full -load 115 V voltage 115 power 250 W current 4.0 V A current 5.3 A losses 105 W RF. 64% voltage Locked 63.9% efficiency speed 1760 r/min torque 1.35 Nm 115 23 7a current 7 L torque 3.4 current Nm 1600 r/min speed % 13 A V A 19 A 29 A voltage current 7 S current Breakdown rotor torque capacitor 6Nm 320 [jlF * SINGLE-PHASE MOTORS 18.9 Vibration of single- phase motors If we we touch the stator of a single-phase motor, note that it vibrates rapidly, whether full-load or no-load. operates it These vibrations do not at exist in 2-phase or 3-phase motors; consequently, single- phase motors are more noisy. What causes electric to the fact the deceleration intervals coincide with the negative peaks. Consequently, the acceleration/deceleration power whereas it properties given in Table 18A. is 5.3 draw A and the it + 1000 the W The We line. it is current, power P supplied find that 218 W. When P we to the we can mo- oscillates between power positive the motor receives energy from the when full-load current lags 50° behind the line voltage. If 18. 14). W and mechan- motor having the waveshapes of voltage and plot the instantaneous tor (Fig. is It delivers constant power. Consider the 250 ical due power is positive, negative, or zero, the mechanical power delivered is a steady 250 W. The motor will slow down during the brief periods when the electric power it receives is less than 250 W. On the other hand, it will accelerate whenever the electric power exceeds the mechanical output plus the losses. The acceleration intervals coincide with the positive peaks of the power curve. Similarly, motor always receives pulsating this vibration? that a single-phase 40 line. is Conversely, negative the motor returns energy to the However, whether the instantaneous electric occur twice per cycle, or 120 times per sec- intervals ond on 60 Hz system. As a a and rotor vibrate The stator at vibration and noise. is 18.15). both the stator are transmitted to the in turn, generates additional vibrations mounting base which, motor result, twice the^line frequency. To eliminate the problem, the often cradled in a resilient mounting (Fig. It consists of two soft rubber rings placed between the end-bells and a supporting metal bracket. Because the rotor also vibrates, a tubular rubber isolator is sometimes placed between Figure 18.14 The instantaneous power absorbed by a single-phase motor varies between +1000 output is constant at 250 W; consequently, vibrations are produced. W and -218 W. The power the ELECTRICAL MACHINES AND TRANSFORMERS 402 shaft and the mechanical load, particularly when the load a fan. is Two-phase and 3-phase motors do not because the total from phases all the is vibrate power they receive instantaneous constant (see Section 8.7). 18,10 Capacitor-run motor The capacitor-run motor is essentially a 2-phase motor that receives its power from a single-phase source. It has two windings, one of which is directly connected to the source. The other winding is also connected to the source, but capacitor (Fig. 18. Figure 18.15 Single-phase capacitor-start motor supported in a re- and noise silient-mount cradle to reduce the vibration transmitted to the mounting surface. Motor rated at 1/3 hp, 1725 r/min, 230 V, 60 Hz has a 3.0 A, efficiency of 60 percent, full-load current of and power factor of 60 percent. Other characteristics: no-load current: 2.6 A; has a large compared 1 with a paper in series The capacitor-fed winding 6). number of turns of relatively small wire, connected winding. to the directly This particularly quiet motor is used to drive fixed loads in hospitals, studios, and other places where silence is important. It has a high power factor on account of the capacitor and no centrifugal switch is locked-rotor current: 13 A; locked rotor torque: 3.6 pu; breakdown torque: 3.0 pu; service factor: required. 1 278 mm; 232 mm. (Courtesy of Baldor Electric Company) The motor acts when it operates at low. Capacitor-run motor having a NEMA rating Corresponding phasor diagram at full load. of 30 millihorsepower. motor only full-load (Fig. 18.16b). * Figure 18.16 b. is as a true 2-phase these conditions, fluxes (a) a. starting torque .35; total weight: 10 kg; overall length including shaft: overall height: However, the ct> s a and <t s created Under by the SINGLE-PHASE MOTORS two windings are equal and out of phase by 90°. The motor is The shaded-pole motor is very popular for ratings below 0.05 hp (—40 W) because of its extremely below 500 W. simple construction (Fig. Reversing the direction winding order to reverse the direction of rotation of the In we have discussed so far, we have to inter- change the leads of either the auxiliary winding or main winding. the However, if a single-phase motor with a centrifugal switch, versed while the motor is its is equipped same In the case running because both windings are times. In the case of very small motors, the ro- in Fig. 1 8. 1 7. In in position 1, while winding winding B is in A is When the switch is thrown is directly across the line, series with the capacitor. connection the motor turns clockwise. switch is the circuit such a motor, the main and auxiliary windings are identical. this in can be reversed by using a double-throw switch shown to position 2, the role reversed and the motor will <J> a that lags behind tor is as and <!>!, <t> 2 , <J> 3 all in , come <X> 3) <£> and starts the and 2 <J> a 3 This current produces a flux . x Consequently, The combined . motor. The <I> + (<J>i / b in <I> 3) <1> 2 . and As is from the (I> b the produces rotation and then run up to speed in the opposite direction. copper ring (auxiliary winding) /b main winding r\ %J • q n n o \J xJ ac source Figure 18.17 Reversible single-phase motor using a 2-pole switch Figure 18.18a and capacitor. Fluxes in a shaded-pole motor. — sJ - right. and the a revolving field that drives the rotor clockwise. o + which before, the of the windings to a halt (<t> 2 field, the ring, With When also lags up by the pole on the is set b lags behind bined action of a shaded (ring side) of the pole. Flux 4> 2 induces a current sulting flux <I> action of direction of rotation to the similar torque links the pole, inducing a produces a weak revolving unshaded side A <J> phase. Flux on the left-hand <& at all tation nents /.,. 18. 16) a copper ring surrounding The main winding is a simple coil connected to The coil produces a total flux <£> that may be considered to be made up of three compo- behind can be changed while the mo- basically a is the ac source. rather large current of a capacitor-run motor (Fig. It which the auxiliary in a portion of each pole. the main wind- If direction. the direction of rotation composed of rotation cannot be re- ing leads are interchanged, the motor will continue to turn in the is short-circuited ring running. 18. 18). small squirrel-cage motor of rotation motors 18.12 Shaded-pole motor then essentially vibration-free. Capacitor- run motors are usually rated 18.11 403 re- comweak ELECTRICAL MACHINES AND TRANSFORMERS 404 TABLE 18B Properties of a Shaded-Pole Motor, having 2 poles, Rated 6 W, 115 V, 60 Hz. No-load input A 0.26 current power W 15 3550 r/min speed Locked rotor input A 0.35 current power torque 24 W 10 mN-m Full-load input power 2900 torque 2600 r/min breakdown speed power 18.13 Universal motor starting torque, efficiency, and factor are very low, the simple construction and absence of a centrifugal switch give this marked advantage in 8. 1 9. entire very similar to basic construc- motor is is shown in Fig. laminated to re- low-power applications. The it fixed by the position of the copper rings. Table 18B gives the The is The magnetic circuit tion of a small universal motor direction of rotation cannot be changed, because is universal motor a dc series motor (Section 5.8). 1 a mN-m 21 V, The single-phase Although the mN-m W 6 breakdown torque 5 millihorsepower, 115 r/min 19 mechanical power at W 21 speed Figure 18.18b Shaded-pole motor rated 60 Hz, 2900 r/min. (Courtesy of Gould) A 0.33 current ac flux typical properties of a 2-pole shaded- pole motor having a rated output of 6 W. Example 18-3 Calculate the full-load efficiency and slip of the shaded-pole motor whose properties are listed in Table USB. Solution The efficiency is (PJP,) X X (6/21) 100 (3.6) 100 28.6% (/7 S — (3600 0.194 n)/n s - 2900)/3600 = 19.4% Figure 18.19 Alternating-current series motor, also called universal motor. SINGLE-PHASE MOTORS duce eddy-current losses. Such a motor can operate on either ac or dc, and the resulting torque-speed about the same in each case. That is why it is is called Series motors are built in motors formerly used When the motor is connected to an ac source, the ac current flows through the armature and the series field. The field in produces an ac flux <E> electric locomotives. performance curves of a 8000 r/min, universal motor 115 V, The hp. different sizes, very large traction to some Fig. 18.20 gives the ac a universal motor. many from small toy motors starting full-load current 405 175 is rated at 1/100 mA. that reacts with the current flowing in the armature to produce 18.14 Hysteresis motor a torque. Because the armature current and the flux reverse simultaneously, the torque always acts in the same in this direction. No revolving field is produced type of machine; the principle of operation is same as that of a dc series motor and it possesses the same basic characteristics. The main advantage of fractional horsepower the universal motors ing torque. is speed and high their high They can start- therefore be used to drive high-speed centrifugal blowers in vacuum cleaners. The high speed and corresponding small size for a given power output is also an advantage in driving portable tools, such as electric saws and drills. Noload speeds as high as 5000 to 5 000 r/min are pos- To understand sis motor, the operating principle of a hystere- us let first consider Fig. 1 8.2 1 . It shows a stationary rotor surrounded by a pair of N, S poles that can be rotated mechanically rection. ial The rotor is in composed of of high coercive force. Thus, magnet material whose As in it resistivity of an insulator. Consequently, up eddy currents a such a it is a clockwise di- ceramic materis a permanent approaches that impossible to set rotor. the N, S field rotates, it magnetizes the rotor; consequently, poles of opposite polarity are contin- uously produced under the moving N, S poles. In effect, the revolving field is continuously reorient- 1 sible but, as in any series motor, the speed drops ing the magnetic individual rapidly with increasing load. domains in the rotor. Clearly, the domains go through a complete cycle (or hysteresis loop) every time the field makes one complete revolution. Hysteresis losses are therefore produced in the rotor, proportional to the area of the r/mm hysteresis loop (Section 2.26). 14 000 These losses are dis- sipated as heat in the rotor. Let us assume that the hysteresis loss per revo- 12 000 lution % is Eh joules and that the field rotates 50- 10 000 efficie ncy mA 40- 8000 \ Spe 400 o stationary rotor 30-|- 300 6000 / CI rrent ^ 5= LU 4000 2000 — - rate j 20- 200 10- 100 torque 010 30 20 40 mN m Torque Figure 18.20 Characteristics of a small tor 1 15 having a full-load rating of 60 Hz universal mo1/100 hp at 8000 r/min. V, Figure 18.21 Permanent magnet revolving field. rotor and a mechanically-driven at ELECTRICAL MACHINES AND TRANSFORMERS 406 n revolutions per minute. the rotor per minute The energy The corresponding power nE h (dissipated as heat) Ph = Wit = the mechanical the N, S poles. This power P = Because P [W] in the rotor power used can to drive given by is /z779.55 = P h we is (3.4) /z£ h /60 However, the power dissipated come from hysteresis motor W= only dissipated in is (3.5) have , = /z779.55 speed Figure 18.22 Typical torque-speed curves of two capacitor-run /?£ h /60 motors: whence T= £ h /6.28 a. Hysteresis motor b. Induction motor (18.2) where T = Eh — torque exerted on the rotor [N-m] hysteresis energy dissipated in the ro- per turn tor, = 6.28 [J/r] = constant [exact value 2tt] Equation 18.2 brings out the remarkable feature that the torque 18.21) is tion. In needed to drive the magnets other words, whether the poles just barely creep around the rotor or whether they move speed, the torque exerted on the rotor is same. It (Fig. constant, irrespective of the speed of rota- is this basic at high always the property that distinguishes hys- Figure 18.23 teresis motors from all other motors. In practice, the revolving field 3-phase stator, or auxiliary winding. When inside such a stator, it is by a single-phase it Single-phase hysteresis clock motor having 32 poles produced by a stator a hysteresis rotor is is essentially constant as (a) in Fig. 18.22. This is until Thanks it the curve entirely different squirrel-cage induction motor, toward zero as shown by from a whose torque falls approaches synchronous speed. to the fixed frequency of large distribution systems, the hysteresis motor is employed ferrite rotor. placed immediately accelerates reaches synchronous speed. The accelerating torque and a having an in electric clocks, and other precise timing devices (Fig. 1 8.23). It is also used to drive tapedecks, turntables, and other precision audio equipment. In such devices the constant speed looking for. is, However, of course, the feature the hysteresis ticularly well suited to drive of their high inertia. Inertia motor we is are par- such devices because prevents many synchro- nous motors (such as reluctance motors) from coming up to speed because to reach synchronism, they have to suddenly lock in with the revolving field. SINGLE-PHASE MOTORS No such abrupt transition occurs motor because to in the hysteresis When c. develops a constant torque right up it at synchronous speed. In some W= The power dissipated function as a vibration-free capacitor-run motor. is accelerating, its full torque P = available to carry the mechanical load and to overcome Once inertia. the motor runs reaches synchronous it speed, the rotor poles are still There d. magnetized and so is = Wit no energy motor runs at to the rotor. therefore, 0.8 = 180 J in the rotor is 180/60 = 3 W loss in the rotor when the synchronous speed because the magnetic domains no longer reverse. an ordinary permanent-magnet like is, X 225 moves The en- the rotating field stalls, ergy loss per minute enhanced by designing the motor to While the motor motor 225 r/min with respect turntable audio equipment these fea- tures are further is the 407 sychronous motor. The rotor poles will lag behind the stator poles by a certain angle, whose magni- 18.15 tude depends upon the mechanical torque exerted Synchronous reluctance motor by the load. We Example 18-4 A small 60 Hz poles. In hysteresis clock motor possesses 32 making one complete turn with respect to to 0.8 b. c. the motor when when rotor losses the motor is stalled motor runs the at syn- it pull-in up as a standard salient poles lock with the re- and so the motor runs at synchronous speed. Both the pull-in and pull-out torques are to those of a hysteresis motor of accelerate high-inertia loads to synchronous speed. The and pull-out torques are about T= £ = poles s - = 120///; = 225 r/min The maximum power P = /? 779.55 = 3W = 0.8/6.28 at stator poles are X (18.2) 60/32 is (225 (or 3/746 = X 0.127)/9.55 1/250 hp) 1 ). a rate that corresponds to the slip. If the is 120 18.22. corre- slipping past the rotor Nm 0.127 The synchronous speed /? = h /6.28 18.24 approaches syn- sponding to full-load torque (operating point equal in a hysteresis motor: b. starts when The reason can be seen by referring to Fig. Suppose the motor has reached a speed //, Solution The field, weak, compared rotor losses the stator. Fig. equal size. Furthermore, reluctance motors cannot chronous speed a. Such a reluctance motor volving number poles must be equal rotor milled out to create four salient poles. chronous speed, the The pull-in and pull-out torques The maximum power output before The The number of poles on shows a squirrel-cage motor but, stalls d. number of J. Calculate a. of salient poles. The to the the revolving field, the hysteresis loss in the rotor amounts can build a synchronous motor by milling out a standard squirrel-cage rotor so as to create a Figure 18.24 Rotor of a synchronous reluctance motor. ELECTRICAL MACHINES AND TRANSFORMERS 408 rotor so is to in the lock with the revolving field, time it this interval {At), it chronous speed // s will , sweep It is particularly well adapted to variable-frequency electronic speed control. Inertia is then no problem because the speed of the revolving field always never be achieved. The tracks with the speed of the rotor. Three-phase re- is going from speed that in is must do not achieved during past a rotor pole. If pull-in problem it takes for one stator pole to /?, to syn- the kinetic energy of the re- motors of several hundred horsepower luctance have been using this approach. built, volving parts must increase by an amount given by Eq. 3.8: AE = k 5.48 X \0~'J(n 18.16 Synchro drive 2 - s 2 /;, (18.3) ) In where ./ is moment of the inertia. Furthermore, the time interval At = 60/(/z s - is some remote-control systems we may have move the position of a small given by meters away. This problem (18.4) /?,)/? flexible shaft. But if must develop an accelerating power P d of at tor P = AEJAt X 1.8 10 4 n s (n s - n\Y Jp power P will never pull into step. In essence, a reluctance motor can only synchronize when the small and the moment How does of inertia J is slip speed is impractical. Two rotors are also connected in parallel a single-phase source. about this arrangement is and energized The remarkable that the rotor other. Thus, if we slowly turn rotor A clockwise B will move clockwise through 17°. Obviously, such a system enables us to control a rheostat from a remote 'location. receiver a synchro system. feature on one ma- chine will automatically track with the rotor on the »/b of in phases of the respective ~U Figure 18.25 then rheostat motors whose 3-phase stators are connected from is cheaper than any other type of synchronous motor. Components and connections We knob and through 17°, rotor low. Despite this drawback, the reluctance motor away, the becomes 100 such a shaft work? parallel (Fig. 18.25). demanded by the load. If the sum of P u + P L exceeds the power capacity of the motor, it by using a m is Consider two conventional wound-rotor induction x easily solved electrical shaft to tie the (approx) Furthermore, the motor must continue to supply the together. (18.5) Ll = least employ an one or two the rheostat flexible-shaft solution Consequently, to reach synchronous speed, the mo- is to is rheostat that SINGLE-PHASE MOTORS Two One 409 miniature wound-rotor motors are required. the stator voltages are again in balance (phase by coupled to a control knob, phase), and the torque-producing currents disappear. (the transmitter) is and the other (the receiver) is coupled Synchros are often employed to the rheo- to indicate the po- The 5-conductor cable (conductors a-b-c-1-2) linking the transmitter and receiver constitutes the with the result that the torque requirements are flexible electrical shaft. small. stat. The behavior of system this selsyn or synchro control Assume explained as follows. is that the sition of an antenna, a valve, a gun turret, Such transmitters and receivers and so on, are built with watch-like precision to ensure that they will track with as error as possible. little transmitter and receiver are identical and the rotors are in identical positions. cited, they behave When the rotors are ex- like the primaries of two EQUIVALENT CIRCUIT OF A SINGLE-PHASE MOTOR trans- formers, inducing voltages in the respective stator windings. The voltages induced in the three stator windings of the transmitter are always unequal Chapter In (Fig. we developed 15 5.6) for 1 the equivalent circuit one phase of a 3-phase induction mo- because the windings are displaced from each other tor. by 120°. The same exception that the magnetizing branch has been in the stator is true for the voltages induced moved of the receiver. Nevertheless, no matter what the respective stator voltages of the transmitter and receiver are identical in both the rotors may be, they synchros (phase by phase) when occupy the same position. The stator volt- ages then balance each other and, consequently, no The current flows in the lines connecting the stators. rotors, however, carry a small exciting current Now if we This circuit /0 They will reproduced with the in Fig. 18.26, between to the technically correct position points 1 and 2. The reason for the change is that most single-phase motors are fractional horsepower machines for which the exact needed this circuit diagram to get reasonably accurate results. we now develop model, is Using a similar equivalent circuit for a single-phase motor. . turn the rotor of the transmitter, three stator voltages will change. is its 18.17 Magnetomotive force no longer distribution balance the stator voltages of the receiver; consequently, currents necting the /.„ / b , / L , will flow in the lines con- two devices. These currents produce a torque on both rotors, tending to line them up. Since the rotor of the receiver with the transmitter. is free to As soon move, it will line up as the rotors are aliened. In order to optimize the starling torque, efficiency, power tor, tor and noise factor, level of a single-phase mo- magnetomotive force produced by each stapole must be distributed sinusoidally across the the pole face. That number of is the reason for using the special turns (10, 20, 25, and 30) on the four concentric coils shown Let us examine the poles when I 8.3(a). the concentric coils carry a peak current 2 amperes. Table of, say, mmf, using of the in Fig. mmf created by one of the four 1 8C shows the distribution numbers as a measure of the slot distance along the pole. For example, the 25-turn coil lodged in slots these slots an 2 and 8 (Fig. mmf of 25 X 8.27). produces 1 2 = between 50 amperes (or am- pere-turns). Similarly, the 10-turn coil in slots 4 and 6 2 The Figure 18.26 Equivalent circuit of one phase of a 3-phase cage motor referred to the produces between these primary (stator) side. Fig. 1 8.27. the pole slots distribution of these is an mmf of 20 A. mmfs is illustrated in The total mmf produced in the middle of 60 + 50 + 40 + 20 = 70 A and it drops 1 ELECTRICAL MACHINES AND TRANSFORMERS 4 0 1 soidal, but the TABLE 18C only 0.4 1-9 slot slot 3-7 slot 4-6 9 30 2-8 slot Mml' Turns Coil pitch 25 2 20 2 X 10 40 - 10 mmf in 85 turns the center of the pole will be = the current reverses and mmf X 30 = 60 A X 25 = 50 A X 20 A X 20 equal to, say, — 1 ,2 A, the mmf will However, the still be distributed sinusoidally but with a peak value the center of A We A l .2 A X in 102 A. conclude that the ac current produces a pul- mmf, which is distributed each pole and whose amplitude 3 - phase stator. the mmf sinusoidally across varies sinusoidally mmf Thus, unlike the time. in = - 85 turns sating 170 ampere turns 85 turns will also reverse. 34 A. Subsequently, when is does not rotate but remains fixed 18.18 Revolving produced by a of a single-phase stator mmfs in place. a single- in phase motor It can be proved mathematically that a stationary pul- mmf sating having a peak amplitude placed by two volving in M can be re- mmfs having a fixed amplitude Mil re- opposite directions synchronous speed. at Referring to our previous example, a 4-pole pulsat- mmf that reaches positive and negative peaks of A at a frequency of 60 Hz can be replaced by two 4- pole mmfs having a constant amplitude of 85 A roing 123456789 - - 1 1 ^ pole pitch 70 tating in opposite directions at 1800 r/min. The re- mmfs are also distributed sinusoidally in space. As the oppositely moving mmfs take up successive positions, the sum of their magnitudes at any volving Figure 18.27 Distribution of the pole when current magnetomotive force across one is 2 A. point in space is equal to the pulsating point. This can be seen mmf at that by referring to Fig. 18.28, which shows a portion of the forward and backward off in steps on either side of center. Adjacent poles have the same magnetic mmf distribution but with opposite polarities. We have superposed upon this figure a smooth mmf having a perfectly sinusoidal distribution. It reveals that the stepped mmf produced by the four concentric coils tracks the sine Indeed, soidal we could mmf without The wave very replace the stepped closely. mmf by a sinu- introducing a significant error. current flowing in the four coils alternates sinusoidally (in time) at the line frequency of 60 Hz. Consequently, as the current varies, the in proportion. For example, mentarily 0.4 A, the when mmf varies the current mmf distribution is mo- remains sinu- revolving fields (mmfF and mmfB the stationary but pulsating sweeping ), past mmf. The revolving mmfs respectively produce the same effect as the revolving mmf created by a 3-phase stator. Consequently, we would expect the circuit diagram of a single-phasor motor to resemble that of a 3-phase motor. However, since the mmfs slip .v the rotor has a if with respect to the forward-moving mmf, will automatically spect to the The ing on rotate in opposite directions, their effect the rotor will be different. Thus, slip of (2 s) with it re- backward-moving mmf. circuit mmf have a diagram as regards the forward-mov- having a slip s is shown in Fig. 18.29a. SINGLE-PHASE MOTORS Figure 18.29 50 mmfB ^ r' a. /\ v \^ ~ mmf f Equivalent circuit as regards the forward-moving mmf. b. Equivalent circuit as regards the backward-moving mmf. -90° 0 6 . +90° Similarly, the circuit diagram for the backward- ~ revolving Fig. 1 mmf having 8.29b. For the the physical to ^0A -90° /0\ /t= +90° — a slip (2 moment we meaning of r,, r 2 , Vj, s) is shown in will not define .\ 2 , etc., except say that they are related to the stator and rotor resistances and reactances. How should wc merge these two diagrams into a single diagram to represent the single-phase motor? 4" Figure 18.28 pulsating mmf having a peak amplitude of 170 A can be represented by a forward and backward revolving mmf having a fixed amplitude of 85 A. Shown are successive positions of mmf F and mmf B and the corresponding amplitude of the stationary, pulsating mmf. 18.19 Deducing the circuit diagram of a single-phase motor The First, the / jF we know that the oppositely rotating same magnitude. Consequently, and / cuits can , B are identical, which be connected means in series. mmfs have the stator currents that the two cir- Second, the forward ELECTRICAL MACHINES AND TRANSFORMERS 412 2 2jx 2 2./.V, r, 2 Figure 18.31 Equivalent circuit of a single-phase motor at standstill. In practice we assume x = x 2 . x The above analysis ances r,,.Vj, etc., shown Thus, Equivalent circuit of a single-phase motor. of r E¥ is associated with EB backward voltage Because the mmfK associated is circuits are in series, the voltages must be equal to the voltage stator. It , while the with mmfB sum of r h r2 , -V], .v 2 , etc., which case the meaning of the E applied to the circuit suppose the motor slip s = 1 . is Example 18-5 A test motor behaves like a (the represent the following physical elements: Draw stator resistance 2r 2 = rotor resistance referred to the stator 2/.X-, = stator leakage reactance 2jx 2 = rotor leakage reactance referred to the 2jX m — and iron resistance corresponding to the and iron losses magnetizing reactance 4 il 3 11 600 losses: O 60 il the equivalent circuit diagram and determine the factor of Solution circuit the values of the listed friction, single- results: power output, efficiency, and power motor when it turns at 1725 r/min. The equivalent stator windage, 725 r/min 2 11 magnetizing reactance: the 2R m = 1 n tion, reveals that the parameters etc., = 60 Hz, V, resistance corresponding to the windage, fric- sim- r, 2r, 3 in short-circuit. , 20 rotor leakage reactance referred to the stator: cir- is in .V| 1 stator leakage reactance: the rotor) , on a 1/4 hp, phase motor reveals the following stationary, in which the secondary winding It imped- the equivalent circuit. stator resistance: of Fig. 18.30 therefore reduces to that shown ple transformer in 10 ohms, the value is forth, for the other rotor resistance referred to the stator: Under these conditions, Fig. 18.31. In essence, the ohms, and so parameters forward and backward circuits are identical. The cuit in the stator resistance 5 . these follows that the equivalent circuit of the sin- interpret the if is { ances gle-phase motor can be represented by Fig. 18.30 To 18.29 to 18.31 are in Figs. equal to one-half of the actual physical quantities. Figure 18.30 stator voltage indicates that the imped- The slip is s = (1800 We first determine circuit between points diagram (Fig. 18.32) shows impedances divided by .two. the 1 725)/ 1800 = 0.0417. impedance of the forward 1, 3: SINGLE-PHASE MOTORS 4 1 1.5 J 1 1 + j . = 1 = 14.89 + + 1.5 j + j 300 30 + 13.89 j 1 .5 19.53 j 21.03 The impedance of the backward points 3, 2 + 48 between circuit 120 V is 1 + 1 1.5 j 1 1 1 + j = + 1 = j 1.93 The current / = 1 + j + 0.93 + j + 1.02 j 1.5 .45 1 2.95 in the stator is £/(Z F / .5 300 30 + ZB = = +j 16.82 12()/( ) 23.98) Figure 18.32 120/(29.29^54.95) See Example = 4.097Z -54.95 The forward voltage between points 1, 3 18-5. is E v = IZ ¥ = (4.097Z -54.95) X (14.89 + = 4.097/1-54.95 X 25.77Z54.7 Backward rotor current: 21.03) j l In = I I = 105.6Z-0.25 The backward voltage between points 3, 2 E B = IZ H = 4.097^-54.95 X (1.93 + = 4.097^-54.95 X 3.52Z56.8 = is j l 2.95) 14.42/1 1.85 = 3.89 ' 1 30 4.097 300 Z + 48 1 .5 .5 1 (0.93 + j 1 .45) Z - 54.9 5 X 1,72 Z 57 .32 1.81 Z 55.78 + j Pv = 1.5 54.95 ( 1 Z- 54.95 X 3.89 + 23.96 48.02/1 1.79 2.16 to rotor: 1.5 j 1 9.53) 2 lv 9.55 P. Z 54.58 X Forward torque 1 2.044/1 j j Z- 53.4 Forward power j 48.02/1 1.79 4.097 + + 1 1 48 .02 .02 L81Z55.78 Forward rotor current: = 1 Z - 54.95 4 .097 4.09 7 'f I 300 30 j /i 48 7, = 1.02 2 X 48 = W 200.5 : X 9.55 s X 2.044 1 Backward power I pj'~ H = to rotor = 200.5 = 1.064 N m 800 3.89 PB 2 X : 1.02 = 15.4 W 1 ELECTRICAL MACHINES AND TRANSFORMERS 414 Backward torque Tn Pn 9.55 // motors? X 9.55 - 15 such a mounting necessary on 3-phase : 15.4 - 0.082 N-m 1 800 1 s What 1 - Tu = 1.064 - 0.082 - 0.982 8-8 N-m X 1725 0.982 177 ~ 9.55 b. W c. 9.55 d. Horsepower: e. 177 f. 0.24 hp 746 Active power input to = X 120 is in this best suited to drive the follow- A small portable drill A 3/4 hp air compressor A vacuum cleaner A 1/100 hp blower A 1/3 hp centrifugal pump A 1/4 hp fan for use in a hospital g. An h. A ward electric timer hi-fi turntable stator: = 4.097 cos 54.95 282.3 W Intermediate level 1 Power main advantage of a capacitor- ing loads: a. nT _ e the Which of the motors discussed chapter Mechanical power output P: £/cos is run motor? Net torque: 7V 8-7 8-9 factor: Referring to Fig. 1 8. 1 1 , the effective im- pedance of the main and auxiliary wind- cos 54.95 - 0.57 ings under locked-rotor conditions are = 57% given as follows: Efficiency: 177 0.627 = 62.7% Effective Effective resistance reactance 282 Main winding 4 Auxiliary winding 7.5 il fl 4 7.5 11 il Questions and Problems If the line Practical level 18-1 A 6-pole to a single-phase motor 60 Hz source. What connected a. synchronous b. is is its speed? 1 8-2 What is the purpose of the auxiliary wind- c. d. The power the rotation of such a State the 18-10 What 8-5 main difference between a 18-6 The palm of the human hand can just how 1/4 a shaded-pole motor the properties hp motor is 64°C and advan- a. Can b. Is a person keep his the Referring to Fig. 18. 13, if the stant at havior of the motor some single-phase motors resilient mounting? Is hand on the frame? motor connected to a load whose torque Why equipped with a an ambient tem- motor running too hot? tages of a universal motor. are in perature of 76°F, 18-11 some of If split- are their relative advantages? Explain briefly List factor under locked-rotor con- barely tolerate a temperature of 130°F. operates. 1 /s the full-load temperature of the frame of a phase motor and a capacitor-start motor. 18-4 /,, ditions motor? 8-3 119 V, calculate the is The magnitude of and / s The phase angle between /u and The line current I L ing in a single-phase induction motor? How can we change 1 voltage following: on the is is con- 4 N-m, explain the resulting be- line. when it is switched SINGLE-PHASE MOTORS 18-12 a. A single-phase moior vibrates quency of 100 Hz. What * power the ' b. c. a. motor does not have set in a resilient mounting. to be b. Why? at c. 1600 r/min. Calculate the hys- Referring to the 6 Table a. b. c. d. W shaded-pole motor 18-18 in and the voltage at the The [N m] starting torque A 3 hp, 1 725 r/min 230 V, totally-enclosed, fan-cooled, capacitor-start, capacitor- run, single-phase motor manufactured by The rated power output in millihorsepower The full-load power factor The slip at the breakdown torque The per unit no-load current and locked-ro- Baldor Electric Company has the follow- ing properties: A no-load current: 5 tor current 18-14 resistance of the transmission line starting current Industrial appl ication 8B, calculate the following: 1 the Appendix, calcu- in following: when in-lb teresis loss per revolution [J], 18-13 The The AX3 1 motor terminals 60 Hz single-phase hysteresis motor develops a torque of 6 running late the line? A capacitor-run A 4-pole, Using Table at a fre- the frequency of is 4 locked-rotor current: 90 Referring again to Fig. 18.13, calculate full -load current: A A 15 the following: a. The locked-rotor torque b. The per-unit value of the c. The starting torque winding d. The e. How is locked-rotor torque: 30 lbfft (t't-Ibf | LR torque when only the main breakdown torque: 20 full-load breakdown torque are the lorque-speed curves affected the line voltage falls from 1 V 15 to 100 if power 18-15 Table In service factor: 1.15 V? a. mass: 97 1 9 lbf The voltage across 1b Using the above information, calculate the the capacitor under following: The corresponding phase angle between and 18-16 ft 8 A, calculate the following: locked-rotor conditions b. 87% factor: full-load torque: level lbfft excited per-unit Advanced 79% full-load efficiency: a. /s The per-unit values of locked-rotor torque, locked-rotor current, and breakdown torque /., Referring to Fig. 18.16, if b. the capacitor- The full-load torque expressed in run motor operates at full-load, calculate c. The capacitor could be added across that the following: the stator so that the full- load a. The line current /, rises b. c. d. 18-17 The power factor of the motor The active power absorbed by each winding The efficiency of the motor The motor described in Table 8A has an LR power factor of 0.9 lagging. It is installed in a workshop situated 600 ft from a home, where the main service entrance is located. The line is composed of a 2conductor cable made of No. 12 gauge copper. The ambient temperature is 25°C newton- meters 18-19 A from 87% to power factor 90% 3/4 hp, 1725 r/min, 230 V, totally- enclosed, fan-cooled, capacitor-start, single-phase motor manufactured by 1 and the service entrance voltage is 122 V. Baldor Electric Company has the fol- lowing properties: no-load current: 4.4 A locked-rotor current: f u 1 1 - load c u rre nt : 5 3 . locked-rotor torque: 30 A A 9.5 lbf ft 416 ELECTRICAL MACHINES AND TRANSFORMERS locked-rotor full- load power factor: efficiency: breakdown torque: 58% long and fed from the service entrance is where the voltage 66% is 230 V ±5%. Using the above information, determine 6. 1 lbfft the following: full-load power service factor: factor: 1 68% a. b. full-load torque: mass: 29 The motor ft We wish motor to lb starting torque (newton-meters), a cable temperature of to raise the 90% pacitor across power at full-load its 25 n C factor of the by installing a ca- terminals. Calculate the approximate value of the capacitance, is copper cable Code 2.25 lbf The lowest assuming .25 fed by a 2-conductor No. 12 that has a National Electrical rating of 20 A. The cable is 240 feet microfarads. in Chapter 19 Stepper Motors writers, tapedecks, valves, Stepper motors are special motors used that are In this chapter when motion and position have to be precisely controlled. As their name implies, stepper motors rotate in discrete steps, a pulse that is each step corresponding supplied to one of Depending on its its 18°, or by as little cover the operating princi- will and By varying the pulse rate, the Elementary stepper motor A very simple stepper motor is made of a 2-pole rotor Stepper motors can turn clockwise or counter- in Fig. 19. 1 soft iron. The windings can means of three switches A, B, C. When the switches are open, clockwise, depending upon the sequence of the pulses that are applied to the windings. any position. However, The behavior of a stepper motor depends greatly upon the power supply that drives it. The power sup- sulting magnetic field created ply generates the pulses, a which microprocessor. The pulses while counterclockwise (ccw) pulses are (-). number of steps is known exactly follows that the number of revolutions ( rotate As the rotor can take up switch will line A is closed, the re- by pole up with pole 60°, Next, if 2. In we open we now so doing, it will B and siwill turn ccw switch 3. in 60° steps by closing and opening the switches in this we can make the rotor advance the sequence A, B, C, A, B, 417 If ccw Clearly, step. will attract time lining up with pole by an additional 60°, This permits the motor to be used as a precise posi- 1 up as shown. multaneously close switch C, the rotor a is al- ccw by if and simultaneously close switch B, the rotor will line ) times. accuracy of one A + at all to an open switch it are result, the net ways precisely known the rotor and so in turn are usually counted and stored, clockwise (cw) pulses being It shown consists of a stator having three salient poles and be successively connected to a dc power supply by r/min. by will also discuss 19.1 It a time, or to rotate stepwise at speeds as high as initiated We limitations. the types of drives used to actuate these machines. as a fraction advance very slowly, one step 4000 plotters, type- printers. stator windings. motor can be made to we and more common stepper motors, together with their properties to of a degree per pulse. at ple of the design, a stepper motor can ad- vance by 90°, 45°, X-Y tioning device in machine tools, 19.0 Introduction C Furthermore, ELECTRICAL MACHINES AND TRANSFORMERS 418 opposite (cw) direction. Picking up speed, again overshoot the center line of pole upon will it where- 2, the magnetic field will exert a pull in the ccw direction. The rotor will therefore oscillate like a pendulum around the center line of pole 2. The oscillations will gradually die out because of bearing friction. Fig. 1 9.2 shows the angular position of The rotor starts at the rotor as a function of time. 0° (center of pole Figure 19.1 rotor which each step moves the in switches to a halt (at 3 ms). line at reverse by operating the rotation the reverse sequence A, C, B, A, C, in the B the last switch that was closed in a switching se- quence must remain closed. This holds the rotor its last position and prevents it in from moving under the influence of external torques. In this stationary state the motor will remain locked provided the external torque does not exceed the holding torque moving from one tion of the rotor will position to the next, the mo- be influenced by the inertia and the frictional forces that amine / > come into play. We now ex- amplitude in rotor has a low ing friction. inertia It is at facing pole l. Let this cor- respond to the zero degree (0°) angular position. At the moment switch A opens and switch B closes, rotor will start accelerating ccw toward pole the 2. It rapidly picks up speed and soon reaches the center line of pole 2, where However, the rotor able speed and it does so, the it is it should now moving come to rest. with consider- will overshoot the center line. magnetic field of pole 2 will pull As it in the opposite direction, thereby braking the rotor. The rotor will come to a halt comes to rest The reader will note that in Fig. 19.2 we have drawn the instantaneous speed of the rotor as a function of time. The speed can be given in revolutions per second, but for stepper motors it is more meaningful to speak of degrees per second. The also speed momentarily zero is at t = and becomes permanently zero is ter line greatest of pole whenever the 2. 3 ms, 5 ms, 7 ms, at / > 10 ms. The rotor crosses the cen- Clearly, the oscillations last a rel- down. atively long time before the rotor settles Without making any other changes, suppose we when no-load and that the and a small amount of bear- initially until the rotor 10 ms. shaft. We discover that both the pe- riod and the amplitude of the oscillations increase 19.2 Effect of inertia motor operates rotor 4 ms. wheel on the the The reverse and again crosses the center in increase the inertia of the rotor by mounting a fly- the nature of these forces. Suppose line overshoots the center line oscillations continue this way, gradually di- minishing at speed of the motor. = t The .... In order to fix the final position of the rotor, In by 30° before coming by 60°. we can and reaches 60° (center It now moves Simple stepper motor l) of pole 2) after 2 ms. and start moving in the the inertia increases. In Fig. 19.3, for exam- time to reach the 60° position has ple, the creased from 2 ms to in- 4 ms. Furthermore, the am- plitude of the oscillations has increased. also takes a longer time to settle down The (20 rotor ms in- stead of 10 ms). The oscillations can be For example, the friction. damped by if raised sufficiently, the oscillations 19. 3 increasing the bearing friction shown is in Fig. can be suppressed so as to give only a single overshoot, shown damping accomplished by using an eddy-current is in Fig. brake or a viscous damper. fluid such as oil 19.4. In practice, the A viscous damper uses a or air to brake the rotor whenever STEPPER MOTORS deg , ing effect 60 zero 30 proportional to speed; is when the rotor 10 — *- 12 14 16 19.3 Effect of a mechanical load 18 overshoot. Let us return to the condition shown where the rotor has low and a small amount inertia of viscous damping due to bearing rotor is coupled moving, the effect would expect, it deg w) to I angu lar p os i tic ms 4 )n in Fig. shown is t 8 10 12 14 16 18 ms ms in Fig. greater. 19.2, except that the iner- in Fig. The overshoot is greater and the rotor 9.2 with is the mechanical load and the The oscillations down. order to obtain fast stepping response, in its load) should be as pressed by using a viscous damper. The time to move from one position to the next can also be reduced by increasing the current deg in the winding. However, thermal limitations due to l~R 90 -antjular pos lion losses dictate the 60 CO 1 small as possible and the oscillations should be sup- takes longer to settle down. g is damped more are oscillations the inertia of the rotor (and Figure 19.3 tia is it As we also prolong the time before the rotor settles Therefore, conditions as 9.5. Furthermore, the overshoot inertia increase the stepping time. —» time 8. 19.5). summary, both In ! Same 1 quickly. pee( 0 W in Fig. takes longer for the motor to attain smaller and the 60 19.2, friction. If the 90 » in Fig. mechanical load while to a the 60° position (compare 2 » therefore is at rest. is time Figure 19.2 In moving from pole 1 to pole 2, the rotor oscillates around its 60° position before coming to rest. The speed is zero whenever the rotor reaches the limit of ? that the brakit speed 0 its moving. Viscous damping means it is angular position 90 419 ~* maximum current that can be used. speed 30 Returning to Fig. I9.l, 1 . - 0 8 i 10 12 —* time 14 16 18 ms Same shows and the in- stantaneous position of the rotor (as well as its 1 Figure 19.4 9.6 conditions as in Fig. 19.3, except that viscous We assume and that 90 at the ngul ar po sitio n c V) to it have /.,, /h that the driving a mechanical load. Note that is is zero —" time 14 16 18 ms a duration of 8 ms. Consequently, the step- Same conditions as coupled to is, 0.048 is 1 in Fig. 19.2, except that the rotor a mechanical load. s to 000/8 = 125 steps per second. complete one therefore, minute. Figure 19.5 is = beginning and at the In this figure the pulses revolution requires 6 steps, and so 12 /c makes one-half revolution. stepper motor has some inertia end of each pulse. ping rate 10 , the motor the speed of the rotor deg a. the current pulses when speed) damping has been added. us excite the wind- let ings in succession so that the motor rotates. Fig. 60/0.048 However, start-stop the = turn. 1 takes 6 One + 25 1 The average speed 250 revolutions per stepper motor rotates jumps and not smoothly motor would. it in as an ordinary 420 ELECTRICAL MACHINES AND TRANSFORMERS current 71 a, 10A S777777777Z7A -fr77777777777i^. J777777, i 10A deg 240 — / © 180 « 120 60 0 settling time H 6 - ms 8 0 24 16 - ms time 24 16 time Figure 19.6 Graph of current pulses, angular position, and instantaneous speed (24 ms) produce one half-revolution. of rotor during the When the motor continue to flow is first four steps. at rest, a in the last Three steps holding current must winding that was excited so that the rotor remains locked in place. 19,5 Start-stop stepping rate When motor inches along the stepping stop fashion shown in Fig. 1 9.6, there is in the start- an upper limit to the permissible stepping rate. If the pulse rate of the current in the windings is too fast, the rotor is unable to accurately follow the pulses, and steps will be - lost. current This defeats the whole purpose of the motor, which Figure 19.7 Graph of pull-over torque versus current motor; diameter: 3.4 inches; length: 3.7 to correlate of in; 19.4 Torque versus current this As mentioned rate motor depends upon the current. shows the relationship between the two stepper Fig. 19.7 for a typical stepper motor. 8 A, the motor develops the torque that the When a torque of 3 N-m. This it is pulses. In order to maintain is settle down before ad- that the interval at least between successive 6 ms, which means limited to a maximum of 1 9.6, = 167 steps per second (sps). Bearing in mind what was clear that the said in Section 19.2, maximum number of steps it is second depends upon the load torque and the called the 1 steps that the stepping 000/6 is motor can exert while moving from one position to the next, so pull-over torque. the current ) to the next position. Referring to Fig. means must be previously, the torque developed by — synchronism, the rotor must vancing is instantaneous position (steps) with the number of net ( + and a stepper weight: 5.2 Ibm. a its tia is pet- iner- of the system. The higher the load and the greater the inertia, the lower will be the allowable of steps per second. number STEPPER MOTORS deg 1 mode start-stop 300 42 J 240 CD / C slew CO mg nnode 180 slew curve I 120 / 60 / 0 4 2 8 6 10 time speed start-stop k 0 6 4 2 10 8 ms 16 14 12 *- s,ewspeed ms 16 14 12 time 800 sps 400 500 600 200 0 Figure 19.9 speed »- Angular position versus time curve when the stepper motor operates in the start-stop mode and the a. Figure 19.8 and slewing characteristic of a typical stepper motor. Each step corresponds to an advance of 1 .8 slewing mode. Stepping rate is same the in both Start-stop cases. Instantaneous speed versus time curve b. degrees. curve 1 curve