Introduction toElectrical Power and Power ElectronicsIntroduction toElectrical Power and Power ElectronicsIntroduction toElectrical Power and Power ElectronicsMukund R. PatelMukund R. PatelPatel6000 Broken Sound Parkway, NW Suite 300, Boca Raton, FL 33487711 Third Avenue New York, NY 100172 Park Square, Milton Park Abingdon, Oxon OX14 4RN, UKan informa businesswww.taylorandfrancisgroup.comMosttraditionalpowersystemstextbooksfocusonhigh-voltagetransmission. However, the majority of power engineers work in urban factories, buildings, or industrieswherepowercomesfromutilitycompaniesorisself-generated. IntroductiontoElectricalPowerandPowerElectronicsisthefrstbookofits kind to cover the entire scope of electrical power and power electronics systems in one volumewithafocuson topics that are directly relevantinpower engi-neers daily work.Composed of 17 chapters, the book is organized into two parts. The frst part intro-duces aspects of electrical power that most power engineers are involved in during their careers, including the distribution of power to load equipment such as motors via step-down transformers, cables, circuit breakers, relays, and fuses. For engineers working with standalone power plants, it also tackles generators. The book discusses how to design and operate systems for economic use of power and covers the use of batteries in greater depth than typically covered in traditional power system texts.Thesecondpartdelvesintopowerelectronicsswitches,aswellastheDCDC converters, ACDCACconverters,andfrequencyconvertersusedinvariable-frequencymotordrives.Italsodiscussesquality-of-powerissuesinmodern powersystemswithmanylargepowerelectronicsloads.Achapteronpower converter cooling presents important interdisciplinary design topics. Thistimelybookdrawsontheauthors30yearsofworkexperienceatGeneral Electric,LockheedMartin,andWestinghouseElectricand15yearsofteaching electricalpowerattheU.S.MerchantMarine Academy.Designedforaone-se-mester or two-quarter course in electrical power and power electronics, it is also ideal for a refresher course or as a one-stop reference for industry professionals.Electrical EngineeringK15426_COVER_final_revised.indd 1 11/13/12 11:24 AMIntroduction toElectricalPowerand PowerElectronicsBoca RatonLondonNew YorkCRC Press is an imprint of theTaylor & Francis Group, an informa businessIntroduction toElectricalPowerand PowerElectronicsMukund R. PatelCRC PressTaylor & Francis Group6000 Broken Sound Parkway NW, Suite 300Boca Raton, FL 33487-2742 2013 by Taylor & Francis Group, LLCCRC Press is an imprint of Taylor & Francis Group, an Informa businessNo claim to original U.S. Government worksVersion Date: 20121203International Standard Book Number-13: 978-1-4665-5661-4 (eBook - PDF)Thisbookcontainsinformationobtainedfromauthenticandhighlyregardedsources.Reasonableefforts havebeenmadetopublishreliabledataandinformation,buttheauthorandpublishercannotassume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint.Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmit-ted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, includingphotocopying,microfilming,andrecording,orinanyinformationstorageorretrievalsystem, without written permission from the publishers.Forpermissiontophotocopyorusematerialelectronicallyfromthiswork,pleaseaccesswww.copyright.com(http://www.copyright.com/)orcontacttheCopyrightClearanceCenter,Inc.(CCC),222Rosewood Drive,Danvers,MA01923,978-750-8400.CCCisanot-for-profitorganizationthatprovideslicensesand registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged.Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe.Visit the Taylor & Francis Web site athttp://www.taylorandfrancis.comand the CRC Press Web site athttp://www.crcpress.comtoSarla,my friend and my wife,and to her late parents,Kantaben and Shantilal Patel,for inspiring her to be an electrical engineerin 1950s rural India with no electricity.viiContentsPreface...................................................................................................................... xvAcknowledgments ..................................................................................................xviiThe Author ..............................................................................................................xixAbout This Book .....................................................................................................xxiSystems of Units and Conversion Factors ............................................................ xxiiiPART A Power Generation, Distribution, and UtilizationChapter 1AC Power Fundamentals ...................................................................... 31.1Current Voltage Power and Energy ........................................... 31.2Alternating Current ................................................................... 41.2.1RMS Value and Average Power ................................... 51.2.2Polarity Marking in AC ................................................ 61.3AC Phasor .................................................................................. 71.3.1Operator j for 90 Phase Shift ...................................... 91.3.2Three Ways of Writing a Phasor ................................ 101.3.3Phasor Form Conversion ............................................ 111.4Phasor Algebra Review ........................................................... 111.5Single-Phase AC Power Circuit ............................................... 141.5.1Series RLC Circuit ................................................. 151.5.2Impedance Triangle .................................................... 191.5.3Circuit Laws and Theorems ....................................... 211.6AC Power in Complex Form ................................................... 231.7Reactive Power ........................................................................ 271.8Three-Phase AC Power System ............................................... 281.8.1Balanced Y- and -Connected Systems ..................... 281.8.2Y Equivalent Impedance Conversion ..................... 31Further Reading ................................................................................. 37Chapter 2Common Aspects of Power Equipment ............................................. 392.1Faradays Law and Coil Voltage Equation .............................. 392.2Mechanical Force and Torque ................................................. 412.3Electrical Equivalent of Newtons Third Law ......................... 432.4Power Losses in Electrical Machines ...................................... 432.5Maximum Effciency Operating Point .................................... 442.6Thevenin Equivalent Source Model ........................................ 46viii Contents2.7Voltage Drop and Regulation .................................................. 482.8Load Sharing among Sources .................................................. 512.8.1Static Sources in Parallel ............................................ 512.8.2Load Adjustment ........................................................ 542.9Power Rating of Equipment ..................................................... 542.9.1Temperature Rise under Load .................................... 552.9.2Service Life under Overload ...................................... 552.10Temperature Effect on Resistance ........................................... 56Further Reading ................................................................................. 60Chapter 3AC Generator...................................................................................... 613.1Terminal Performance ............................................................. 623.2Electrical Model ...................................................................... 633.3Electrical Power Output .......................................................... 643.3.1Field Excitation Effect ................................................ 683.3.2Power Capability Limits ............................................. 693.3.3Round and Salient Pole Rotors ................................... 703.4Transient Stability Limit .......................................................... 723.5Equal Area Criteria of Transient Stability .............................. 743.6Speed and Frequency Regulations .......................................... 783.7Load Sharing among AC Generators ...................................... 803.8Isosynchronous Generator ....................................................... 823.9Excitation Methods .................................................................. 843.10Short Circuit Ratio ................................................................... 863.11Automatic Voltage Regulator .................................................. 86Further Reading ................................................................................. 90Chapter 4AC and DC Motors ............................................................................. 914.1Induction Motor ....................................................................... 924.1.1Performance Characteristics ...................................... 964.1.2Starting Inrush Kilovolt-Ampere Code .................... 1004.1.3TorqueSpeed Characteristic Matching ................... 1024.1.4Motor Control Center ............................................... 1044.1.5Performance at Different Frequency and Voltage .... 1044.2Synchronous Motor ............................................................... 1064.3Motor Horsepower and Line Current .................................... 1104.4Dual-Use Motors ................................................................... 1114.5Unbalanced Voltage Effect .................................................... 1134.6DC Motor ............................................................................... 1164.7Universal (Series) Motor AC or DC ...................................... 1184.8Torque versus Speed Comparison ......................................... 119Further Reading ............................................................................... 122ix ContentsChapter 5Transformer ...................................................................................... 1235.1Transformer Categories ......................................................... 1245.2Types of Transformer ............................................................ 1255.3Selection of Kilovolt-Ampere Rating .................................... 1285.4Transformer Cooling Classes ................................................ 1295.5Three-Phase Transformer Connections ................................. 1305.6Full- and Open- Connections ........................................... 1315.7Magnetizing Inrush Current .................................................. 1325.8Single-Line Diagram Model .................................................. 1345.9Three-Winding Transformer ................................................. 1365.10Percent and Perunit Systems ................................................. 1375.11Equivalent Impedance at Different Voltage .......................... 1415.12Continuous Equivalent Circuit through Transformer ............ 1435.13Infuence of Transformer Impedance .................................... 145Further Reading ............................................................................... 149Chapter 6Power Cable ...................................................................................... 1516.1Conductor Gage ..................................................................... 1516.2Cable Insulation ..................................................................... 1536.3Conductor Ampacity ............................................................. 1556.4Cable Electrical Model .......................................................... 1586.5Skin and Proximity Effects ................................................... 1606.6Cable Design .......................................................................... 1626.7Cable Routing and Installation .............................................. 171Further Reading ............................................................................... 173Chapter 7Power Distribution ............................................................................ 1757.1Typical Distribution Scheme ................................................. 1757.2Grounded and Ungrounded Systems ..................................... 1767.3Ground Fault Detection Schemes .......................................... 1797.4Distribution Feeder Voltage Drop ......................................... 1807.4.1Voltage Drop during Motor Starting ........................ 1817.4.2Voltage Boost by Capacitors .................................... 1827.4.3System Voltage Drop Analysis ................................. 1847.5Bus Bar Electrical Parameters ............................................... 1857.6High Frequency Distribution ................................................. 1867.7Switchboard and Switchgear ................................................. 1907.7.1Automatic Bus Transfer ............................................ 1917.7.2Disconnect Switch .................................................... 192Further Reading ............................................................................... 194x ContentsChapter 8Fault Current Analysis ..................................................................... 1958.1Types and Frequency of Faults .............................................. 1958.2Fault Analysis Model ............................................................. 1968.3Asymmetrical Fault Transient ............................................... 1988.3.1Simple Physical Explanation .................................... 1998.3.2Rigorous Mathematical Analysis .............................2008.4Fault Current Offset Factor ...................................................2008.5Fault Current Magnitude ....................................................... 2028.5.1Symmetrical Fault Current ....................................... 2028.5.2Asymmetrical Fault Current .................................... 2038.5.3Transient and Subtransient Reactance ...................... 2078.5.4Generator Terminal Fault Current ............................ 2128.5.5Transformer Terminal Fault Current ........................ 2128.6Motor Contribution to Fault Current ..................................... 2138.7Current-Limiting Series Reactor ........................................... 2158.8Unsymmetrical Faults............................................................ 2158.9Circuit Breaker Selection Simplifed ..................................... 216Further Reading ............................................................................... 220Chapter 9System Protection ............................................................................. 2219.1Fuse ........................................................................................ 2229.1.1Fuse Selection ........................................................... 2239.1.2Types of Fuse ............................................................ 2249.2Overload Protection ............................................................... 2279.3Electromechanical Relay ....................................................... 2289.4Circuit Breaker ...................................................................... 2309.4.1Types of Circuit Breaker .......................................... 2329.4.2Circuit Breaker Selection ......................................... 2369.4.3Standard Ratings of LV Breaker .............................. 2389.5Differential Protection of Generator ..................................... 2389.6Differential Protection of Bus and Feeders ........................... 2399.7Ground Fault Current Interrupter ..........................................2409.8Transformer Protection .......................................................... 2419.9Motor Branch Circuit Protection ........................................... 2419.10Lightning and Switching Voltage Protection ........................ 2439.11Surge Protection for Small Sensitive Loads .......................... 2479.12Protection Coordination ........................................................ 2499.13Health Monitoring ................................................................. 2499.14Arc Flash Analysis ................................................................ 250Further Reading ............................................................................... 253xi ContentsChapter 10Economical Use of Power ................................................................ 25510.1Economic Analysis ................................................................ 25510.1.1Cashfow with Borrowed Capital ............................. 25510.1.2Payback of Self-Financed Capital ............................ 25610.2Power Loss Capitalization ..................................................... 25910.3High-Effciency Motor .......................................................... 26010.4Power Factor Improvement .................................................... 26310.4.1Capacitor Size Determination .................................. 26710.4.2Parallel Resonance with Source ............................... 27010.4.3Safety with Capacitors ............................................. 27110.4.4Difference between PF and Effciency ..................... 27210.5Energy Storage ...................................................................... 27310.6Variable-Speed Motor Drives ................................................ 27310.7Regenerative Braking ............................................................ 27410.7.1Induction Motor Torque versus Speed Curve ........... 27510.7.2Induction Motor Braking .......................................... 27610.7.3DC Motor Braking ................................................... 27910.7.4New York and Oslo Metro Trains ............................ 281Further Reading ............................................................................... 285Chapter 11Electrochemical Battery ................................................................... 28711.1 Major Rechargeable Batteries ............................................... 28911.1.1LeadAcid ................................................................ 28911.1.2NickelCadmium ..................................................... 29011.1.3NickelMetal Hydride .............................................. 29111.1.4Lithium Ion ............................................................... 29111.1.5Lithium Polymer ....................................................... 29211.1.6Sodium Battery......................................................... 29211.2Electrical Circuit Model ........................................................ 29211.3Performance Characteristics ................................................. 29311.3.1Charge/Discharge Voltages ...................................... 29411.3.2c/d Ratio and Charge Effciency ............................... 29411.3.3Round-Trip Energy Effciency .................................. 29411.3.4Self-Discharge and Trickle Charge .......................... 29611.3.5Memory Effect in NiCd ............................................ 29711.3.6Temperature Effects ................................................. 29711.4Battery Life ............................................................................ 29711.5Battery Types Compared .......................................................30011.6More on LeadAcid Battery ..................................................30011.7Battery Design Process.......................................................... 30111.8Safety and Environment ........................................................ 303Further Reading ............................................................................... 306xii ContentsPART BPower Electronics and Motor DrivesChapter 12Power Electronics Devices ............................................................... 30912.1Diode ..................................................................................... 31412.2Thyristor (SCR) ..................................................................... 31812.3Power Transistor .................................................................... 32012.4Hybrid Devices ...................................................................... 32312.5di/dt and dv/dt Snubber Circuit .............................................. 32412.6Switching Power Loss ........................................................... 32712.7Device Application Trends .................................................... 33012.8Device Cooling and Rerating ................................................ 331Further Reading ............................................................................... 333Chapter 13DCDC Converters .......................................................................... 33513.1 Buck Converter ...................................................................... 33513.2 Boost Converter ..................................................................... 34213.3 BuckBoost Converter ........................................................... 34313.4 Flyback Converter (Buck or Boost) ....................................... 34513.5Transformer Coupled Forward Converter ............................. 34613.6 PushPull Converter.............................................................. 34713.7Inductor Coupled Buck Converter ......................................... 34713.8Duty Ratio Control Circuit .................................................... 34813.9Load Power Converter ........................................................... 34913.10 Power Supply ......................................................................... 350Further Reading ............................................................................... 353Chapter 14ACDCAC Converters ................................................................... 35514.1ACDC Rectifer ................................................................... 35514.1.1Single-Phase Full-Wave Rectifer ............................. 35514.1.2Ripples in DC and Ripple Factor ............................. 36014.1.3Harmonics in AC and Root Sum Square.................. 36014.1.4Harmonic Distortion Factor ..................................... 36114.1.5Three-Phase Six-Pulse Rectifer............................... 36414.2ACAC Voltage Converter .................................................... 36814.2.1Single-Phase Voltage Converter ............................... 36814.2.2Three-Phase Voltage Converter ............................... 36914.3DCAC Inverter .................................................................... 37014.3.1Single-Phase Voltage Source Inverter ...................... 37014.3.2Single-Phase Pulse with Modulated Inverter ........... 37414.3.3Three-Phase Six-Pulse Voltage Source Inverter ...... 37714.3.4Three-Phase Six-Pulse Current Source Inverter ...... 37814.4Frequency Converter ............................................................. 38014.4.1DC-Link Frequency Converter................................. 381xiii Contents14.4.2Single-Phase Cycloconverter .................................... 38214.4.3Three-Phase Cycloconverter .................................... 38414.5Thyristor Turnoff (Commutation) Circuits ............................ 38514.5.1Line Commutation ................................................... 38514.5.2Forced (Capacitor) Commutation ............................. 38514.5.3Resonant Commutation ............................................ 38614.5.4Load Commutation ................................................... 38614.5.5ZCS and ZVS ........................................................... 38714.6Other Power Electronics Applications .................................. 38714.6.1Uninterruptible Power Supply .................................. 38714.6.2Static kVAR Control ................................................. 38714.6.3Static Switch and Relay ............................................ 38914.7Common Converter Terms .................................................... 38914.8Notes on Converter Design .................................................... 390Further Reading ............................................................................... 395Chapter 15Variable-Frequency Drives .............................................................. 39715.1Pump Performance Characteristics ....................................... 39815.2Pump Energy Savings with VFD ..........................................40015.3VFD on Ships and in Oil Refneries ......................................40415.4VFD for Medium-Size Motor ................................................40415.5Constant V/f Ratio Operation................................................. 40715.6Commutation and Control Methods ...................................... 41315.7Open-Loop Control System ................................................... 41415.8Vector Control Drives ............................................................ 41515.9Twelve-Pulse VFD Design .................................................... 41715.10Special VFD Cables .............................................................. 41715.11Variable-Voltage DC Motor Drive ......................................... 42315.12VSD in Metro Trains ............................................................. 42415.13VFD as Large Motor Starter ................................................. 42415.14Converter Topologies Compared ........................................... 42615.15Notes on VFDs ...................................................................... 427Further Reading ............................................................................... 429Chapter 16Quality of Power .............................................................................. 43116.1Power Quality Terminology .................................................. 43116.2Electrical Bus Model ............................................................. 43316.3Harmonics ............................................................................. 43616.3.1Harmonic Power ....................................................... 43816.3.2THD and Power Factor .............................................44016.3.3K-Rated Transformer ................................................ 44316.3.4Motor Torque Pulsations .......................................... 44516.3.5Harmonic-Sensitive Loads ....................................... 44816.4Power Quality Studies ........................................................... 449xiv Contents16.5Harmonic Reduction .............................................................. 45116.5.1Harmonic Filter Design ............................................ 45116.5.2Very Clean Power Bus .............................................. 45616.5.3-Connected Transformer ........................................ 45716.5.4Cable Shielding and Twisting ................................... 45816.5.5Isolation Transformer ............................................... 45916.6IEEE Standard 519 ................................................................ 46016.7International Standards ......................................................... 461Further Reading ...............................................................................464Chapter 17Power Converter Cooling ................................................................. 46517.1Heat Transfer by Conduction ................................................. 46517.2Multiple Conduction Paths .................................................... 46717.3Convection and Radiation ..................................................... 47017.4Thermal Transient ................................................................. 47117.5Water Cooling ........................................................................ 47317.5.1Cooling Tube Design ................................................ 47417.5.2Pressure Drop ........................................................... 47517.5.3Cooling Water Quality ............................................. 480Further Reading ............................................................................... 482Appendix: Symmetrical Components ................................................................ 483xvPrefaceThe United States, Canada, the United Kingdom, and many other countries are pres-entlyexperiencingashortageofpowerengineers.Theshortageisexpectedtoget worseintheUnitedStates,whereabout45%oftheutilitypowerengineerswill becomeeligibletoretirebefore2020.Theywouldrequireabout7000newpower engineering graduates to replace them and about an equal number in the supporting industries. At the same time, many countries are making huge investments in build-ing new power plants and power grids that will require even more power engineers to serve the industry. It is for this trend that university students are now once again getting attracted to the electrical power programs.Noothertechnologyhasbroughtagreaterchangeintheelectricalpower industryandstillholdingthepotentialofbringingfutureimprovementsthan powerelectronics.Thepowerelectronicsequipmentpriceshavedeclinedtoabout 1/10 since the late 1980s, fueling a rapid growth in their applications throughout the power industry. Large cruise ships power plant ratings approach 100 MWe since they use electric propulsion with power electronics in abundance. The navies of the world have started changing from mechanical propulsion to electric propulsion with power electronics for the numerous benefts they offer. With high-power combat weapons onboard, they will need even larger power plants in size comparable to those on land.Until now, there has been no single book available that covered the entire scope of electrical power and power electronics systems. Traditional books on power systems focused on high voltage transmission, in which only a handful of power engineers getinvolved.Avastmajorityofpowerengineersworkincities,towns,largefac-tories,steelmills,commercialandinstitutionalbuildings,refneries,datacenters, railways,shippingports,commercialandnavyships,etc.Theymostlydealwith power taken from the utility company or self-generated mostly for running motors andotherloadsviatransformers,cables,andprotectiondevicessuchasfusesand circuitbreakers.Theyalsoroutinelydealwithbatteriesofvariouselectrochemis-tries, power electronics converters of all kinds, variable-frequency motor drives, etc. This book is the frst comprehensive volume of its kind that focuses on all of these topics, which are directly relevant to most power engineers on a daily basis. In addi-tion,engineersworkingintherenewableenergytechnologieswouldalsofndthis book useful since the electrical power system for extracting energy from wind and ocean currents is the exact mirror image of the variable-frequency motors driving pumps, fans, and ship propellers.It is hoped that the book is a timely addition to the literature and a one-volume resource for students of electrical power and power electronics at various universi-ties around the world and for a range of industry professionals working as electrical, mechanical, and chemical engineers, managers, and supervisorsactually, anyone who uses or deals with electrical power.xviiAcknowledgmentsA book of this nature incorporating all aspects of electrical power and power elec-tronics requires help from many sources. I have been extremely fortunate to receive fullsupportfrommanyorganizationsandindividualsinthefeld.Theynotonly encouraged me to write the book on this timely subject but also provided valuable suggestions and comments during the development of the book.AttheU.S.MerchantMarineAcademy,KingsPoint,NY,IamgratefultoDr. David Palmer, engineering department head, and Dr. Shashi Kumar, academic dean, for supporting the research work that led to my writing this book. I have benefted from many students at the Academy, both the undergraduate seniors and the gradu-ate students, who contributed to my learning by pointed questions and discussions based on their professional experience.My special gratitude goes to Dr. Robert Degeneff of Utility Systems Technology Inc., Niskayuna, NY, a colleague from the General Electric Company and a retired professorofelectricalpowerengineeringattheRensselaerPolytechnicInstitute, Troy, NY, and Dr. Anurag Srivastava of Washington State University, Pullman. Both professors reviewed the book proposal and provided valuable comments.My grandchildren Sevina, Naiya, Dhruv, and Rayna and my wife Sarla cheerfully contributed good time they would have otherwise spent with me.I heartily acknowledge the valuable support and encouragement I received from all.Mukund R. PatelYardley, PennsylvaniaxixThe AuthorMukundR.Patel,PhD,PE,isaprofessorofengineeringattheU.S.Merchant Marine Academy in Kings Point, NY. He has over 45 years of hands-on involvement in research, development, and design of state-of-the-art electrical power equipment and systems. He has served as a principal engineer at the General Electric Company in Valley Forge, PA; fellow engineer at the Westinghouse Research and Development Center in Pittsburgh, PA; senior staff engineer at the Lockheed Martin Corporation in Princeton, NJ; development manager at Bharat Bijlee (Siemens) Limited, Bombay, India;and3MMcKnightDistinguishedVisitingProfessorattheUniversityof Minnesota, Duluth.Dr.PatelobtainedhisPhDdegreeinelectricpowerengineeringfromthe Rensselaer Polytechnic Institute, Troy, NY; MS degree in engineering management fromtheUniversityofPittsburgh;MEdegreeinelectricalmachinedesignfrom Gujarat University; and BE degree from Sardar University, India. He is a fellow of theInstitutionofMechanicalEngineers(U.K.),associatefellowoftheAmerican Institute of Aeronautics and Astronautics, senior life member of the IEEE, registered professional engineer in Pennsylvania, chartered mechanical engineer in the United Kingdom,andanelectedmemberofEtaKappaNu,TauBetaPi,SigmaXi,and Omega Rho.Dr. Patel is an associate editor of Solar Energy, the journal of the International Solar Energy Society, and a member of the review panels for the government-funded research projects on renewable energy in the state of California and the emirate of Qatar. He has authored fve books, three of which have been translated into Chinese, and major chapters in two international handbooks. He has taught 3-day courses to practicing engineers in the electrical power industry for over 15 years, has presented and published over 50 papers at national and international conferences and journals, holdsseveralpatents,andhasearnedNASArecognitionforexceptionalcontribu-tion to the power system design for the Upper Atmosphere Research Satellite. The high-voltage, high-frequency cable he developed for the International Space Station was nominated by NASA for an Industrial Research-100 award in 1987. He can be reached at patelm@usmma.edu or patelm30@gmail.com.xxiAbout This BookThisbookhasevolvedfromtheauthors30yearsofworkexperienceatGeneral Electric,LockheedMartin,andWestinghouseElectric,and15yearsofteaching electricalpowerattheU.S.MerchantMarineAcademyatKingsPoint,NY.The book has 17 chapters, divided into two parts:Part A: Power Generation, Distribution, and UtilizationPart B: Power Electronics and Motor DrivesPart A of the book focuses on all aspects of electrical power that most power engi-neers get involved in during their professional careers. It covers the generation and dis-tribution of power to load equipment such as motors via step-down transformers, cables, circuit breakers, relays, and fuses. Some engineers working on standalone power plants get involved with generators. All these topics are covered in Chapters 1 through 9.Chapter10coversvariouswaysofdesigningandoperatingpowersystemsfor economic utilization of power and basic methods of quantifying proftable energy-saving opportunities. The regenerative braking that converts the kinetic energy of a moving mass into electricityas is done in hybrid automobilesand energy-saving benefts of variable speed motor drives are dealt with in detail in Chapter 10.The battery is always an integral part of power system design for providing emer-gency power to essential and critical loads and power to dc control circuits and com-puters.ItiscoveredinChapter11indetailnotfoundintraditionalpowersystem books.PartBofthebookstartswithpowerelectronicsswitchesinChapter12.The next three chapters cover dcdc converters, acdcac converters, and frequency con-vertersusedinvariable-frequencymotordrives.Chapter16discussesthequality-of-powerissuesinmodernpowersystemswithmanypowerelectronicsloadsof relatively large size.Chapter 17 covers power converter cooling by air and also by water, which pres-ents important interdisciplinary design topics relevant to power engineers.The unique features of this book are as follows:This is the only book available that covers all aspects of electrical power and power electronics under one cover that will otherwise require several book.Many examples, exercise problems, and concept questions appear at the end of each chapter.Long visible suffxes in equations make their use fuent to save time.BothinternationalandBritishsystemsofunitsareusedinthebooktopresent data, as they came from various sources. An extensive conversion table connecting the two systems of units is therefore given in the next section for ready reference.xxii About This BookThe book can be used for a one-semester course for electrical majors in four-year degree programs or for two-quarter courses split between power systems and power electronics as needed. For professional engineers working in the industry, the book would be ideal for a refresher course or as a single-source reference book for all top-ics on electrical power and power electronics.xxiiiSystems of Units and Conversion FactorsBoththeinternationalunits(SIorMKSsystem)andBritishunitsareusedinthis book. The table below relates the international units with the British units commonly used in the United States.Category Value in SI Unit = Factor Below Value in English UnitLength m 0.3048 ft.mm 25.4 in.m 25.4 milkm 1.6093 mi.km 1.852 nautical mi.Area m20.0929 ft.2m2506.7 circular milVolumeL (dm3) 28.3168 ft.3L 0.01639in.3cm316.3871 in.3m3/s 0.02831 ft.3/hL3.7853 gal. (U.S.)L/s0.06309 gal./minMass kg0.45359 pound masskg14.5939 slug massDensity kg/m316.020 lb. mass/ft.3kg/cm30.02768 lb. mass/in.3Force N 4.4482 lb. forcePressure kPa 6.8948 lb./in.2 (psi)kPa 100.0 barkPa 101.325 std atm (760 torr)kPa 0.13284 1mm Hg at 20CTorque N m 1.3558 lb.-force ft.PowerW 1.3558 ft. lb./sW 745.7hpEnergy J 1.3558 ft. lb.-forcekJ 1.0551 Btu internationalkWh 3412 Btu internationalMJ 2.6845 hp hMJ 105.506 thermTemperature C (F 32)5/9 FK (F + 459.67) 5/9 R Heat W 0.2931 Btu (international)/hkW 3.517 Ton refrigerationxxiv Systems of Units and Conversion FactorsW/m23.1546 Btu/(ft.2 h)W/(m2 C) 5.6783 Btu/(ft.2 h. F)MJ/(m3 C) 0.0671 Btu/(ft.3 F)W/(m C) 0.1442 Btu in./(ft.2 h F)W/(m C) 1.7304 Btu ft./(ft.2 h F)J/kg 2.326 Btu/lb.MJ/m30.0373 Btu/ft.3J/(kg C) 4.1868 Btu/(lb. F)Velocity m/s 0.3048 ft./sm/s 0.44704 mi./hKnot m/s 0.51446 knotMagnetics Wb 108lineWb/m2 (T) 0.0155kiloline/in.2PREFIXES TO UNITS Micro 106m mili 103k Kilo 103M mega 106G Giga 109T tera 1012OTHER CONVERSIONS1 nautical mi. = 1.15081 mi.1 bar pressure = 14.50 psi = 100 kPa= 29.53 in Hg = 10.20 m water = 33.46 ft. water1 cal (CGS unit) = 4.1868 J1 kg cal (SI unit) = 4.1868 kJ1 hp = 550 ft.-lb./s1 T magnetic fux density = 1 Wb/m2 = 10,000 G (lines/cm2)Absolute zero temperature = 273.16C = 459.67FAcceleration due to earth gravity = 9.806 7 m/s2 (32.173 5 ft./s2)Permeability of free space o = 4 107 henry/mPermittivity of free space o = 8.85 1012 F/mENERGY CONTENT OF FUELS1 tip of match-stick = 1 Btu (heats 1 lb. water by 1F)1 therm = 100,000 Btu (105.5 MJ = 29.3 kWh)1 Quad = 1015 Btuxxv Systems of Units and Conversion Factors1 ft.3 of natural gas = 1000 Btu (1055 kJ)1 gal. of LP gas = 95,000 Btu1 gal. of gasoline = 125,000 Btu1 gal. of no. 2 oil = 140,000 Btu1 gal. of oil (U.S.) = 42 kWh1 barrel = 42 gal. (U.S.)1 barrel of refned oil = 6 106 Btu1 barrel of crude oil = 5.1 106 Btu1 ton of coal = 25 106 Btu1 cord of wood = 30 106 Btu1 million Btu = 90 lb. coal, or 8 gal. gasoline, or 11 gal. propane1 quad (1015 Btu) = 45 million tons of coal, or 1012 ft.3 of natural gas, or 170 million barrels of oil1 lb. of hydrogen = 52,000 Btu = 15.24 kWh primary energyWorlds total primary energy demand in 2010 was about 1 quad (1015 Btu) per day = 110 1016 J/day = 305 1012 kWh/dayAbout 10,000 Btu of primary thermal energy input at the power generating plant produces 1 kWh of electrical energy at the users outlet.Part APower Generation, Distribution, and UtilizationPart A of the book focuses on all aspects of electrical power that most power engi-neersgetinvolvedinduringtheirprofessionalcareers.Avastmajorityofpower engineers work on distribution and utilization of electrical power in cities and towns, factoriesandoilrefneries,commercialandinstitutionalbuildings,electrictrains, shippingports,electriccruiseships,etc.Theireverydaytaskisrelatedtothedis-tribution of power to load equipment such as motors and process heaters via step-downtransformersandcables.Thesystemprotectionisanintegralpartoftheir work, which requires circuit breakers, relays, and fuses. Some engineers working on cogenerating power plants get involved with generators. All these topics are covered in Chapters 1 through 9. Only a small percentage of power engineers deal with high-voltage power transmission, which is therefore excluded from this book.Chapter10coversvariouswaysofeconomicutilizationofpowerforgettingthe required work done with the minimum expense of electrical energy. It includes the regen-erative braking that recovers the kinetic energy from a moving mass and converts it into electricity, as is done in hybrid automobiles. The motors use almost 60% of all electrical energy in the world; hence, the energy-saving benefts of variable speed motor drives are also covered briefy in this chapter and then again in greater detail in Chapter 15.The battery is covered in Chapter 11 in such detail that is not found in traditional power system books. It forms an integral part of the power system design for provid-ing (1) backup power to essential and critical loads in case of the main power failure and(2)powertothelowvoltagedccontrolcircuitsandcomputers.Moreover,the battery performance depends greatly on many variables in a nonlinear manner, as covered in the chapter.31AC Power FundamentalsThomas Edisons Pearl Street low-voltage direct current (dc) power generating sta-tion opened in 1882 to serve the New York City market. Its low-voltage power had to be utilized in the vicinity of the generating station to keep the conductor I 2R loss to an economically viable level. Many more dc power stations were built by Edison and his competitors in the neighborhood of local users. Then came Westinghouses alternating current (ac) power systems, with large steam and hydro power plants built where economical, such as on Niagara Falls. The high-voltage ac power was brought to the load centers, which was stepped down by transformers before feeding to the end users. The ac system proved to be much more economical and fexible and soon drove away the dc competitors. New Yorkers paid an infation-adjusted price of about $5/kWh for dc power in 1890 versus the average of about $0.12/kWh we pay for ac powertodayintheUnitedStates.WiththetransformerandNicolaTeslasinduc-tion motor for which Westinghouse acquired the patent, ac soon became universally adopted for electric power all around the world.The fundamentals of power fow in ac circuitsusually covered in an undergrad-uate course in electrical engineeringare reviewed in this chapter. A clear under-standing of these fundamentals will prepare the student for the chapters that follow and is also essential for working in the electrical power feld.1.1CURRENT VOLTAGE POWER AND ENERGYThe basis of electricity is an electric charge (measured in coulombs) moving between two points at different electrical potentials, either absorbing or releasing the energy along its way. The basic electrical entities in power engineering are defned below with their generally used symbols and units.Current I (amperes) = fow rate of electrical charge (1 A = 1 C/s)Voltage V (volts) = electrical potential difference, that is, energy absorbed or released per coulomb of charge moving from one point to another (1 V = 1 J/C)Power P (watts) = rate of energy fow (1 W = 1 J/s)Therefore,JoulesSecondJoulesCoulombCoulombP = = ssSecondWatts = V I (1.1)With time-varying voltage and current, the instantaneous power p(t) = v(t) i(t) andenergy=powertimeduration.Withtime-varyingpower,theenergyused between 0 and T seconds is given by the integral (i.e., area under the power versus time curve)4 Introduction to Electrical Power and Power ElectronicsEnergy d watt seconds (joules) = v t i t tT( ) ( )0.(1.2)Inversely, the power is the time differential of energy, that is,Powerdd(Energy) =t.(1.3)Example 1.1The electrical potential of point A is 200 V higher than that of point B, and 30 C of charge per minute ows from A to B. Determine the current and power ow from A to B and the energy transferred in 1 min.Solution: Current ow from point A to B = 30 60 = 0.50 APower ow from A to B = Voltage Current = 200 0.5 = 100 WEnergy transferred in 60 s = 100 60 = 6000 W s (J).It is worth repeating here that 1 W = 1 J/s or 1 J = 1 W s. The commercial unit of electrical energy is kilowatt-hour (kWh); 1 kWh = 1000 W 3600 s = 3,600,000 Ws = 3.6 MJ. The utility bill is based on kilowatt-hour electrical energy used over the billing period. An average urban U.S. customer using 1000 kWh per month pays about$0.15/kWh,outofwhichabout40%isforgeneration,5%fortransmission, 35% for distribution, and 20% for account service and the utility companys proft.For example, if a heater consumes 1500 W for 4 h and 750 W for 8 h every day in a winter month of 30 days, then the electrical bill at 15 cents/kWh tariff will be (1.5 4 + 0.750 8) kWh/day 30 days/month $ 0.15/kWh = $54 for that month.1.2ALTERNATING CURRENTAC is used all over the world for electrical power. It varies sinusoidally with time t asi(t) = Imsin t or i(t) = Im cos t.(1.4)Although its representation by either the sine or cosine function is called sinusoi-dal, the cosine function shown in Figure 1.1 is more common, where Im is the maxi-mum value, or amplitude, or peak value of the sinusoid (in amperes); = 2f is the angular speed (called angular frequency) of the alternations (in radians per second); T is the period of repetition (in seconds per cycle); and f = 1/T = /2 is the frequency [in cycles per second (called hertz)].5 AC Power FundamentalsIn sine or cosine representation, the ac current completes one cycle of alternation in t = 360 or 2 radians. For this reason, one cycle of ac is customarily displayed with respect to t and not with time t. As an example, for sinusoidal current i(t) = 170 cos 377t,Peak value (amplitude) Im = 170 A (front number)Angular frequency = 377 rad/sec (number in front of t)Numerical frequency f = /2 cycles/sec (Hz)Period of repetition T = 1/f = 2/ sec/cycle.We recall that the dc needs only one number to specify its value, but we see in Equation 1.4 that the ac needs three numbers to specify its value at any instant of time t, namely, the peak value Im, the angular frequency , and time t. This makes the mathematics in ac circuits complex.1.2.1RMS VALUE AND AVERAGE POWERAlthoughacvariessinusoidallywithtimewithnorealfxedvalue,weoften speak in terms of a fxed ac value, for example, ac current of 10 A or ac voltage of 120 V. We see below what a fxed number in ac means, taking an example of resistor carrying current i(t). The power absorbed by the resistor at any time t is given by p(t) = i(t)2R, which varies in time as a square function of the current. Itisalwayspositiveevenwhenthecurrentisnegativeduringone-halfcycle. Therefore,theaverageofi2isalwaysapositivenonzerovalue,althoughthe average of sinusoidal current is always zero. To fnd the average power, we must therefore use (average of i2), not (average of i)2, carefully noticing the placement of the parentheses in each case. We also note that since iavg is always zero, (iavg)2 isalsoalwayszero.However,sincei2isalwayspositiveinbothpositiveand negativehalfcycles,(i2)avgisneverzero,unlessi(t)=0atalltimes.Therefore, Pavg = {i(t)2}avg R. The square root of {i(t)2}avg is called the root-mean-squared (rms) value of the current i(t).i(t) = Im cost t 360 or2 radians0 Im+ Im+ ImOne cyclePeriod T = 1/f = 2/FIGURE 1.1Sinusoidal ac over one cycle represented by cosine function.6 Introduction to Electrical Power and Power ElectronicsInallpracticalapplications,itistheaveragepowerthatmatters.Forexample, we are mostly interested in how much a room heater, a fan motor, or a pump motor producesattheendofanhouroranyothertimeduration.Sinceacrepeatsevery cycle, the average power over one cycle is the same as that over 1 min or 1 h or 1day, aslongasthepowerison.Therefore,theeffectivevalueofthecurrentfordeter-miningtheaveragepoweristhesquarerootof(averageofi2)overonecycle,that is,I i I Ieff root.mean.squared rmsMean of = = =2. It is the equivalent dc value that would result in the same average power. For any wave shape in general, that is, cosine, sine, square, rectangular, triangular, etc., Ii t tTTrmsd=20( ). (1.5)For a sinusoidal current, II t tTIpkTpkrmsd=( )=cos 202,and similarly, for a sinusoidal voltage,VV t tTVpkTpkrmsd=( )=cos 202.(1.6)The divisor 2 above is for the sinusoidal ac only. It is different for different wave shapes. Using the basic calculus of fnding the rms value, the student is encouraged to derive the divisor 2 for a sinusoidal wave, 1.0 for a rectangular wave, and 3 for a triangular wave.1.2.2POLARITY MARKING IN ACAlthough ac circuit terminals alternate their + and polarities every one-half cycle, we still mark them with + and polarities, as if they were of fxed polarities, as in dc. Such polarity marking in ac has the following meanings:With multiple voltage sources (generators and transformers) in parallel, all + marks are positive at the same time, and all marks are negative at the same time. This information is needed to connect multiple voltage sources in parallel to share a large load.With multiple voltage sources or loads in series, the ( +) and ( +) sequence indicatesanadditivevoltagepair,whereas(+)and(+)sequenceindi-cates a subtractive voltage pair.In a single-voltage-source circuit, the + terminal is usually connected to the load, and the terminal is connected to the ground.7 AC Power FundamentalsTo eliminate such technical contradiction in using the + and marks, the modern polarity marking is sometimes done with dots for positive terminals and no mark for the negative terminals.1.3AC PHASORThe ac current i(t) = Imcos t can be represented by an arm of length Im rotating at an angular speed (frequency) rad/sec as shown in Figure 1.2. The actual instanta-neous value of i(t) at any time t is Imcos t, which is the projection on the reference axis. It alternates between + Im and Im peaks, going through zero twice at t = 90 and 270 every cycle. Since the actual value of i(t) depends on the phase angle of the rotating arm, the rotating arm is called the phasor. Decades ago, it used to be called vector since the algebra dealing with it works like the vector algebra.The ac voltage and current phasors, in general, can have phase difference between their peaks, that is, their peaks can occur at different instants of times. Voltage V and current I phasors shown in Figure 1.3a have the phase difference of angle , with the current peak lagging the voltage peak by angle . Two phasors of the same frequency with phase difference between their peaks will also have the same phase difference between their zeros. The wavy hat sign ~ on V and I signifes the sinusoidal vari-ations with respect to time.In the electrical power industry, since the voltage is given by the generator or the utility company, the power engineer always takes the voltage as the reference pha-sor and then designates the current as leading or lagging the voltage. In all practical power circuits, the current lags the voltage, so we say that the current lags in most practical power circuits.Drawing a neat sine wave by hand is diffcult. Power engineers generally circum-vent this diffculty by drawing the phasor diagram as shown in Figure 1.3b, where the V and I phasors are rotating at the same angular speed , keeping their phase dif-ference fxed. The actual instantaneous values of both V and I are their respective projections on the reference axis at any given instant of time.The phasor diagram can be drawn using the arm length equal to the peak value orthermsvalue.Sincethepowerengineerisalwaysinterestedintheaverage Reference axis (t = 0)Im cos t Im = tFIGURE 1.2Rotating phasor representing ac.8 Introduction to Electrical Power and Power Electronicspower, he or she always draws the phasor diagram using the rms values. The rms values are customarily implied in the power feld, and we will do the same in this book as well.For average power, we recognize that the voltage and current that are out of phase would produce less average power than those in phase with their peaks occurring at the same time. If V and I are in phase (i.e., when = 0), their product P = V I is always positive even when both V and I are negative during one-half cycle. However, when 0, the instantaneous power is negative when either V or I is negative and the other is positive. If positive power means the power fowing from the source to the load, then the negative power means the power fowing backward to the source from the energy stored in the load inductance or capacitance. The average power in such a case is always less than the maximum power V and I would produce if they were in phase. The average power of voltage V and current I lagging the voltage by phase angle is given by the time-average of p(t) = v(t) i(t) over one cycle of period T, that is,PTV t I t tV Iavg pk pkTpk pk= =120cos( ) cos( ) cos d == V Irms rmscos .(1.7)It(a) Time diagramV0(b) Phasor diagramIVFIGURE 1.3Two sinusoidal phasors out of phase by angle .9 AC Power FundamentalsThevoltageandcurrent,ifinphasewith=0,wouldproducethemaximum possibleaveragepowerequaltoVrmsIrms.Whennotinphase,theyproduceless averagepower.Thepowerreductionfactorcosiscalledthepowerfactor(pf). Obviously, pf = 1.0 (unity) when = 0, and pf = 0 when = 90.Example 1.2A circuit element has sinusoidal voltage of 300 cos 314t volts across its terminals anddraws80cos(314t25)amperes.Determinetheaveragepowerdelivered to the element.Solution:The current is lagging the voltage by 25, so the pf of this element is cos 25. For the average power, we use rms values and pf, that is,Pavg = (300 2) (80 2) cos 25 = 10,876 W.1.3.1OPERATOR J FOR 90 PHASE SHIFTAnuppercaseletterwithawavyhatsign(e.g.,=I)inthisbookrepresentsa sinusoidally varying ac phasor with rms magnitude I and a phase difference of angle with respect to a reference sinusoidal wave (usually the voltage). If in Figure1.4 representsanyvoltageorcurrentphasor,thenanotherphasor

Bthatisofthe samemagnitudeasbutwith90phaseshiftinthepositive(counterclockwise) directioncanbewritteninlonghandas B A = withaddedphaseshiftof+90,or B A Aj jA = 90 = orin short hand where the operator j represents the phase shift of by +90 in the positive (counterclockwise) direction.Shifting

Bfurtherby+90,wegetanotherphasor C jB j jA j A = = = ( )2.We graphicallyseeinFigure1.4that

C j A A = = 2andthereforededucethatj2=1 orj = 1.Thus,jisanimaginarynumbergenerallydenotedbyletteriinthe + 90 phase shift of Aresults in jAAC = jB = jjA = j2A = AB = jA FIGURE 1.4Operator j representing 90 phase shift in positive (counterclockwise) direction.10 Introduction to Electrical Power and Power Electronicsmathematics of complex numbers (we use letter j to avoid confusion with current i in electrical circuits).Thus, in mathematical operations, j = + 1 90 represents phase shift. (1.8)From j 2 = 1 or 1 = j 2, we get 1jj = , which is a useful relation to remember.1.3.2THREE WAYS OF WRITING A PHASORAlternative ways of writing a phasor are depicted in Figure 1.5.In polar form (also called -form), we write phasor = Im, where Im = mag-nitudeofthephasor(canbepeak,butrmsmagnitudeisusedinthisbook,asis customary in power engineering), and = phase angle of the phasor. Here, the hat sign ~ signifes a sinusoidal phasor. In routine use, we often drop the hat sign ~ and write I = 3 20, meaning an ac current of 3 A rms value lagging the voltage by 20.In rectangular form (also called j-form), we write = x + jy = Im cos + jIm sin , wherex andy arethe phasorsrectangular components on the real and imaginary axes, respectively.Inexponentialform(alsocallede-form),weuseEulerstrigonometricidentity e j= cos + j sin to write the phasor in yet another form, that is, = Im cos + j Imsin = Im (cos + j sin ) = Im e j.The three alternative ways of representing a phasor are then summarized below: = Im = Im cos + jIm sin = Ime j, where = t.(1.9)Real axis or reference axisIm0Imaginary axis or j-axisI = Im y = Im sin x = Imcos FIGURE 1.5 Polar and rectangular components of phasor .11 AC Power FundamentalsItisimportanttotakeanoteherethattwophasors=Am1=x1+jy1and

B B x jy = = +m2 2 2areequalif,andonlyif,boththeirmagnitudesandphase angles are equal, that is, Am = Bm and 1 = 2, or x1 = x2 and y1 = y2, that is, when both their real and imaginary components are individually equal.1.3.3PHASOR FORM CONVERSIONThe form we choose to represent various phasors depends on the algebraic operation required on a given set of phasors. Certain algebraic operations require the phasors in certain form, as seen in the next section. Therefore, converting the phasor from one form to another is often necessary and is done using the trigonometry of Figure 1.5.Polar to rectangular ( to j) conversion: Consider a phasor given in -form, that is, = Im, where Im and are known. To convert it in j-form, we write using the rectangular components, that is, = x + jy = Im cos + jIm sin thereforex = Im cos andy = Im sin .(1.10)Rectangular to polar (j to ) conversion: Consider a phasor given in j-form, that is, = x + jy where x and y are known. To convert it in -form, we write using the polar components, that is,

I I I x yyxm= = + =mtan where and2 2 1.(1.11)1.4PHASOR ALGEBRA REVIEWAC power engineers are routinely required to perform six basic mathematical opera-tions on phasors, namely, to add, subtract, multiply, divide, differentiate, and inte-grate phasors. These operations are normally covered in books on algebra of complex numbers and also on ac circuits. This section is a brief summary of such operations.Consider two phasors and

B given by A A x jy A e B B x jy B ej j= = + = = = + =m m m mand 1 1 112 2 22.We add or subtract two phasors in the rectangular form, that is, A B x jy x jy x x j y y + = + + + = + + + ( ) ( ) ( ) ( )1 1 2 2 1 2 1 2.(1.12) A B x jy x jy x x j y y = + + = + ( ) ( ) ( ) ( )1 1 2 2 1 2 1 2.(1.13)We multiply or divide two phasors in the polar and exponential forms, that is,12 Introduction to Electrical Power and Power Electronics AB A B Ae Be ABe Aj jj = = = =+ ( )m m m m m m 1 21 21 2mm mB + ( ) 1 2 (1.14)

ABABAeBeABeABjjj== = = ( ) mmmmmmmm 12121 2 ( ) 1 2.(1.15)Wedifferentiateorintegratephasorwithrespecttotimetintheexponential form in the time domain, that is, = Am = Amej = Amejt. Then, dddd

At t Ae j Ae j Amj tmj t= = = (1.16)

A t Ae t AejAjmj tmj td d = = = .(1.17)The summary of above phasor operations in words follows:To add two phasors, add their x and y components separately.To subtract two phasors, subtract their x and y components separately.To multiply two phasors, multiply their magnitudes and add their angles.To divide two phasors, divide their magnitudes and subtract their angles.To differentiate a phasor, multiple it by j, that is, d/dt = j.To integrate a phasor, divide it by j, that is,d / t j j = = 1 .Tip-to-tail method is the graphical method of adding or subtracting two phasors as illustrated in Figure 1.6. To add and

B in Figure 1.6 (top), frst draw . Then, AA + BBBA + BBBA FIGURE 1.6Tip-to-tail method of adding and subtracting two phasors.13 AC Power Fundamentalsat the tip of , place the tail of

B and draw

B. The end point from the origin is then A B + . To subtract

B from in Figure 1.6 (bottom), frst draw . Then, at the tip of , place the tail of

B and draw

B (i.e.,

B in the negative direction.) The end point from the origin is then A B .Example 1.3Given two phasors, = 6030 and

B = 40 60, determine (1) A B + , (2) A B , (3) A B , and (4) AB. Express each result in j-form and also in -form.General note:For simplicity in writing, all angles we write following signs in this book are in degrees, whether or not expressly shown with superscripts ().Solution:For adding and subtracting and

B, we must express phasors in j-form, that is, j j B = + = + = 60 30 30 51 96 30 40 and (cos sin ) . , (cos

660 60 20 34 64 + = + j j sin ) .Then, A B j j + = + + + = + = ( . ) ( . ) . . . 51 96 20 30 34 64 71 96 64 64 96 733 41 93 . A B j j = + = = ( . ) ( . ) . . . 51 96 20 30 34 64 31 96 4 64 32 30 8 26 ..For multiplying and dividing, we must use and

Bin -form as given, that is, A B = = + = =60 30 40 60 60 40 30 60 2400 902400 ((cos sin ) 90 90 0 2400 + = + j j AB = = = = 60 30 40 60 60 40 30 60 1 5 301 5 ( ) .. ( (cos sin ) . . 30 30 1 3 0 75 = jExample 1.4Determine 21 5 j . in -form.14 Introduction to Electrical Power and Power ElectronicsSolution:All ac voltages, currents, impedances, and powers are phasors, that is, complex numbershavingx+jycomponentsorthermsmagnitudeandphaseangle. When we write 2 in ac, it really means x = 2 and y = 0, or magnitude 2 and = 0. Also, when we write j1.5, it really means x = 0 and y = 1.5, or magnitude 1.5 and = 90.Thealgebraofdividingtwophasorsrequiresrstconvertingthemin-form and then dividing the two as follows: 21 52 00 1 52 01 5 9021 50 90 1 33 90jjj . . . .. =++== = .Recognizing that a real number alone always means = 0, and an imaginary numberalonealwaysmeans=90,wecanskiptheformalj-formto-form conversion, and write directly as 21 52 01 5 9021 50 90 1 33 90j . . .. == = or, recalling that 1/j = j = 90, we can simplify the algebra further as 21 521 51 33 1 33 90jj j. .. . = = = .Understanding all of the above three ways of arriving at the same results adds uency in the complex algebra of phasors in ac power circuits, which is required for professional electrical power engineers.1.5SINGLE-PHASE AC POWER CIRCUITIn analyzing any power circuit (ac or dc), we use the fowing basic circuit laws:Kirchhoffs voltage law (KVL): In any closed loop, phasor sum of all source voltages=phasorsumofallvoltagedropsinloadelements.Itappliesin every loop, covering every segment along the loop, in ac or dc. In a volt-age driven circuitusually the case in power engineeringit is KVL that determinesthecurrent,thatis,thecircuitdrawscurrentthatwillsatisfy KVL in every closed loop of the circuit.Ohms law: It basically gives the voltage drop across two terminals of R, L, or C element as summarized in Table 1.1. It is local; it applies only between two terminals of R, L, or C (but not of a source).Kirchhoffs current law (KCL): At any junction node (point or a closed box), phasor sum of all currents going inward = phasor sum of all currents going outward. The KCL is even more local than Ohms law; it applies only at a junction node.15 AC Power Fundamentals1.5.1SERIES RLC CIRCUITThebasicpowercircuitismadeofoneormoreofthethreebasicloadelements, namely, the resistor R, inductor L, and capacitor C, powered generally by a voltage source (e.g., the utility grid). Since the R, L, and C elements are distinctly different inunitsandalsoinbehavior,theycannotbecombinedwitheachotherusingany series-parallelcombinationformulasuntiltheyareconvertedintotheirrespective ohm values with phase shifts.ConsideraseriesRLCcircuitofFigure1.7excitedbyasinusoidalvoltage source

Vs. Since the driving voltage in this circuit is sinusoidal, the current must also be sinusoidal of the same frequency, say, = I where I = rms magnitude (note that we now drop the suffx m from Im and imply that I without the hat sign means the rms magnitude).In the series loop of Figure 1.7, we write KVL using Ohms law relations given in Table 1.1 and Equations 1.16 and 1.17:

V V V V RI LItI tCRI Lj IICs R L cddd= + + = + + = + +jj.TABLE 1.1Voltage Drop, Current, and Energy Storage Relations in R, L, and C Elements in AC or DC Circuits Parameter between Two Terminals of the ElementResistance R of Wire (ohm)Inductance L of Coil (henry)Capacitance C between Two Conductors (farad) Voltage drop in the element in dc or ac (Ohms law or its equivalent)v = R iv Lit=ddvQCi tCi Cvt= == ChargedorddVoltage drop in steady state dc (i.e., when di/dt = 0 and dv/dt = 0) V = R I Vdrop = 0 (i.e., coil in steady dc behaves like an ideal wire shorting the coil) Icap = 0 after fully charged (i.e., capacitor in steady dc behaves like open circuit)Energy stored in the elementNone (dissipates energy in heat, which leaves the circuit)1/2 LI 2 (stores energy in magnetic fux through the coil) 1/2 C V 2 (stores energy in electrical charge on the capacitor conductors)Energy conservation Dissipates the absorbed energy into heatTends to hold on stored energy by retaining its current ITends to hold on stored energy by retaining its voltage V16 Introduction to Electrical Power and Power ElectronicsRecalling that j = 1j, we rewrite the above equation as V RI j LIjC I RI j LjCIs = + = + (1.18)where j = 1 = 90 = 90 phase shift in the positive direction. We note here the following phasor relations in the R, L, and C elements: V RIR = , that is, VR is in phase with I with = 0, meaning that a pure R always draws current at pf = 1.0 (unity), absorbing average power that is not zero in R. V j LIL = ,thatis,VLleadsIby90,orIlagsVLby90,meaningthata pure L always draws current at pf = 0 lagging, giving zero average power absorbed in L. V jC IC = 1, that is, Vc lags I by 90 or I leads Vc by 90, meaning that a pureCalwaysdrawscurrentatpf=0leading,againgivingzeroaverage power absorbed in C.Thevoltmetermeasuresonlythermsmagnitudewithoutthephaseangle. Therefore, three voltmeter readings across R, L, and C (i.e., VR, VL, and VC) will not add up to the total source voltage magnitude Vs. We must add the three load voltages taking their phase differences into account (called the phasor sum) to obtain the total source voltage. Figure 1.8 shows the phasor sum using the tip-to-tail method, where

V V V V Vs R L c s= + + = .(1.19)We next write Equation 1.18 as V R j LCI Z Is+

_,

,, 1(1.20)+Vs+VRC faradsIL henries R ohms+VL+VCFIGURE 1.7Series RLC circuit powered by sinusoidal source voltage

Vs.17 AC Power Fundamentalswhere

Z R j LCR jX = + = + 1ohms.TheRandXbothhaveunitsofohmsandsodoes

Z.Therefore,Equation1.20 gives ac voltage as a product of ampere and ohm, as in dc. Equation 1.20 is Ohms law in ac, where

Z is called the total circuit impedance. The

Z is generally written as

Z Z R jX = = + , where X = (L 1/C) is called the reactance, which comes from L and C. Since the reactance has two parts, we write X = XL XC, whereXL = L = reactance from inductance (ohms)XC = 1/(C) = reactance from capacitance (ohms).The contributions of L and C in the total reactance X are subtractive, that is, they tend to neutralize each other. As a result, X can be zero in some circuits even if L and C are not individually zero.The KVL in a closed loop in any power circuit with multiple voltage sources and impedances in series can be expressed in a simple form as follows:

Iloopphasor sumof all source voltagesphasor=ssumof all impedancessource=

VZ .(1.21)The current lags the voltage if the phase angle of all load impedance combined is positive (i.e., when the loop is more inductive than capacitive). Thus, the overall pf of a circuit is determined by the nature of the load. The pf has nothing to do with the source voltage.(a) Individual phasors , VR, VL, and VCIVL IVLVC (b) Tip-to-trail addition VR + VL + VC = Vs VRVsVCVRFIGURE 1.8Phasors

VR,

VL, and

VCand their phasor sum

Vs in series RLC circuit.18 Introduction to Electrical Power and Power ElectronicsExample 1.5A one-loop circuit has source voltage of 120 V and load impedance of 10+84.3 . Determine the current and average power delivered to the load. Can you guess about the nature of the loadcould it be a heater, capacitor, motor, etc.?Solution:The 120 V with no angle stated implies that it is the reference phasor at 0. It is also implied in power engineering that it is the rms value. The current phasor in this loop will then be

IVZlooplooploop= == 120 010 84 312010 .00 84 3 12 84 3 A = . . .The 12 A is the rms value since we used 120 Vrms to derive the current, which lags the voltage by 84.3, giving the pf = cos (84.3) = 0.10 lagging. The average power delivered by the source and absorbed by the load is given by Pavg = Vrms Irms cos = 120 12 0.10 = 144 W (versus 1440 W 120 V and 12 A could produce if they were in phase with = 0). This is an example of extremely poor pf.As for the nature of the load, since it draws current with a large lag angle close to 90, it must have a large inductive reactance with small resistance. Therefore, it is most likely to be a large coil.Example 1.6Determine the average real power drawn by a coil having inductance of 10 mH and resistance of 1.5 connected to a 120-V, 60-Hz source.Solution:For a 60-Hz source, = 2 60 = 377 rad/s, hence the coil impedance

Z j j = + = + = 1 5 377 0 010 1 5 3 77 4 057 68 3 . . . . . . .Taking 120 V as the reference phasor, the current drawn by the coil isI = 1200 4.05768.3 = 29.5868.3 A,which lags the voltage by 68.3therefore, real power P = 120 29.58 cos 68.3 = 1312.45 W.Alternatively, the average real power is absorbed only by the circuit resistance, and the power absorbed by the inductance or capacitance always averages out 19 AC Power Fundamentalsto zero over one cycle. Thus, the average real power can also be derived simply from the following:real power P = I2R = 29.582 1.5 = 1312.45 W, Example 1.7Convert the time-domain circuit shown in gure (a) below in the phasor domain with impedances in ohms. Then, determine the current phasor I and the average power.75 F12 340 cos(314t)80 H12 240 Vrms j42.46 j25.12 (a)(b) Solution:First, the voltage phasors rms value = 340 2 = 2400 V (reference for angle).Angular frequency = 314, which gives f = /2 = 50 Hz.The loop impedance of series RLC circuit in gure (a) is Z jj= + = + 12 314 0 0801314 75 1012 25 12 46.( . 22 46 12 17 34 21 09 55 3 . ) . . . = = j .Thephasor-domaincircuitisshowningure(b),fromwhichwederivethe circuit current IVZ= == =240 021 09 55 324021 090 55 3. . .( . ) 111 38 55 3 . . + A.The 11.38-A current phasor leads the voltage by 55.3, meaning that the capac-itance dominates the inductance in this circuit.Average power P = 240 11.38 cos 55.3 = 1555 W.1.5.2IMPEDANCE TRIANGLEAs seen in Equation 1.20, we write the total circuit impedance in the form Z = R + jX, where the operator j = 1 in algebra or 90 phase shift in the phasor diagram. 20 Introduction to Electrical Power and Power ElectronicsForengineersandmanagersnotfullyversedwiththeoperatorj,thereasonfor assigningtheinductivereactanceXwith90phaseshiftfromRcanbesimply explained as follows in view of the circuit shown Figure 1.9a. The voltage drop in R is IR, which always subtracts from the source voltage. The voltage drop in the inductor, however, is equal to L di/dt. This drop is positive for one-half cycle of rising current from point A to B under a sinusoidal current as in Figure 1.9b when the inductor is absorbing the energy that causes the voltage to drop. It is negative for the other one-half cycle of falling current from point B to C when the induc-tor is supplying the energy that adds into the circuit voltage. The negative voltage drop means it subtracts from the positive voltage drop in the resistance. Therefore, for one-half cycle of rising current, the total voltage drop is IR + IX as in Figure 1.9c,whereasforthefollowingone-halfcycleoffallingcurrent,thenetvoltage drop is IR IX as in Figure 1.9d. We graphically show the effective rms value of BAIRIRIXVs60 HzITotal voltage drop in R and XVL LoadRX = L(a) RL circuitC(b) One cycle of ac(c) Drops add during A to Bunder rising currentIX(d) Drops subtract during B to Cunder falling current(e) Eective rms drop IZover one cycleIXIRIZ(f ) Eective impedance ZXRZ FIGURE 1.9Simple explanation of voltage drop in RL circuit and impedance triangle.21 AC Power Fundamentalsthe voltage drop over one cycle as in Figure 1.9e not positive nor negative, but in an in-between position with 90 phase shift. Analytically, the effective rms value of the total voltage drop isTotaldrop.rmsVIR IX IR IXIR IX =+ + = + =( ) ( )( ) ( )2 22 22II R X I Z2 2+ = whereZ R X = +2 2,whichhastheright-angletrianglerelationasshownin Figure1.9f, which is known as the impedance triangle. Therefore, the circuit imped-ance, in general, is expressed as a phasor

Z R jX = + , as drawn in Figure 1.9f using the tip-to-tail method, where cos is the circuit pf.1.5.3CIRCUIT LAWS AND THEOREMSAllcircuitlawsandtheoremslearnedindcapplyinthesamemannerinacas well,exceptthatallnumbersinacarephasors(complexnumbers),eachwith magnitudeandphaseangle,makingtheacalgebracomplex.Alldcformulas remain valid in ac if we replace R in dc with

Zin ac. For example, if the circuit has more than one impedance connected in a seriesparallel combination, then the total equivalent

Zcan be determined by using the corresponding dc formula after replacing R with Z and doing the phasor algebra of the resulting complex numbers, that is,

Z Z Z ZZ Zseries paralleland = + + =+ +1 21 211 1.......... (1.22)Example 1.8A 120-V, 60-Hz, single-phase source shown below powers a 0.10-mF capacitor in parallel with 20-mH coil that has 10- winding resistance. Determine the current, pf, and power delivered by the source. (Note: Practical capacitors have negligible resistance, but the inductors have signicant resistance due to long wires needed to wind numerous turns.)120 V 60 Hz C = 0.10 mF R = 10 L = 20 mH Z1 Z2 I22 Introduction to Electrical Power and Power ElectronicsSolution:We rst determine the circuit impedances (all angles are in ):capacitor impedance Z1 = j/C = j/(260 0.10 103) = j26.52 = 26.5290 coil impedance Z2 = R + jL = 10 + j260 20 103 = 10 + j7.54 = 12.5237 .Z1 and Z2 are in parallel, so they give ZZ ZZ ZZ ZTotal =+=+11 11 21 21 2.Notice that, the second term Z1 Z2 (Z1 + Z2) is simple to write and easy to rememberastheproductdividedbythesum,althoughitisvalidonlyfortwo impedances in parallel (it does not extend to three impedances in parallel). Therefore,TotalZj= + +26 52 90 12 52 3726 52 10. .. 77 54332 5321 45 62 215 48 9 2. . .. . == IV== 12015 48 9 27 75 9 2. .. . A.Power factor of the circuit = cos 9.2 = 0.987 laggingPavg = 120 V 7.75 A 0.987 = 918 W.Thenodalanalysis,meshanalysis,superpositiontheorem,andTheveninand Norton equivalent source models are all valid in ac as they are in dc. The maximum power transfer theorem, however, has a slight change. For maximum power transfer in ac, we must have Z ZLoad Th=*, where ZTh = Thevenin source impedance at the load points, and superscript * denotes the complex conjugate defned below.Forphasorimpedance

Z Z R jXTh Th Th Th= = + ,itscomplexconjugate

Z Z R jX*Th Th Th Th = = ,whichisobtainedbyfippingthesignofj-partinthe rectangular form or by fipping the sign of angle in the polar form. Thus, the com-plex conjugate of a phasor is its mirror image in the real axis as shown in Fig ure1.10. Withloadimpedance Z ZLoad Th= (calledtheloadmatching),themaximumpower that can be transferred to the load remains the same as in dc, that is, PVRmax =Th(rms)Th4.(1.23)23 AC Power Fundamentals1.6AC POWER IN COMPLEX FORMForageneralformofacpowerdeliveredfromasourceorabsorbedinaload, we consider Vv and Ii phasors shown in Figure 1.11. The ac power is given bytheproductofthevoltageandthecomplexconjugateofthecurrentpha-sor.Therefore,thepowerisalsoaphasor,acomplexnumber,hencethename + jXTjXT0ZT+Z*TRTFIGURE 1.10Complex conjugate of phasor is its own mirror image in reference axis.IivReference axisVFIGURE 1.11Product of

V and * gives complex power in ac.24 Introduction to Electrical Power and Power Electronicscomplex power. For average power, we use the rms values along with their phase angles, that is,complex powerrms v rms i

S V I = { } (1.24)where * denotes the complex conjugate, that is, the mirror image of on the refer-ence axis. Therefore,rms v rms i rms rms v i

S V I V I = = ( ))= V Irms rms(1.25)where = v i = phase difference between

V and phasors. In retrospect, if were used instead of * in the power product, would be equal to v + i, giving an incor-rect pf.In j-form, Equation 1.25 for the complex power becomes

S V I jI V I jV = +( ) = +rms rms rms rms rms rmcos sin cos ss rms = + I P jQ sin (1.26)where the frst term is the power on the real axis, called real power P = Vrms Irms cos ,whichhastheunitofwatts,orkilowatts,ormegawatts,dependingonthe power system size. The second term is the power on the imaginary axis, going in and out of the reactance X, hence called the reactive power Q = Vrms Irms sin , whichhas the unit of volt-ampere reactive (VAR, or kVAR, or MVAR). The pf cos of the load has no units. The complex power

S, being the product of volts and amperes, has the unit of volt-ampere (VA, or kVA, or MVA).Therefore, complex power S = P + jQ is a phasor, with Q out of phase by +90 from P as shown by the power triangle in Figure 1.12. The trigonometry of power triangle gives S P Q = +2 2Q(kVAR)P (kW)S (kVA)FIGURE 1.12Power triangle of real power P, reactive power Q, and apparent power S.25 AC Power Fundamentalsor kVA kW kVAR = +2 2and pfkWkVA= =PSorkW = KVA pf.(1.27)It is noteworthy that the lagging pf of the load circuit produces positive Q and vice versa. Also, in complex power

S S V I = = rms rms , the S = Vrms Irms without is known as the apparent power. With ,

S is known as the complex power. Both

Sand S have units of volt-amperes or kilovolt-amperes, or megavolt-amperes.The real power P (watts) delivered by the generator requires fuel (real source of power) to drive the generator engine. The reactive power Q (VARs) on the imaginary axisisthepowergoinginandoutofLandCcharginganddischargingLand Ceveryhalfcyclewithzeroaveragepower.Itdoesnotrequirefueltodrivethe generatorengine,althoughitloadsthegeneratorvoltageandcurrentcapabilities. Therefore, VAR adds in the capital cost of power equipment but does not add in the fuel cost for running the prime mover.The ac power absorbed in load impedance

Zcan also be derived as follows. The complex power absorbed in

Zis given by (voltage across Z) (complex conjugate of I in Z), that is, S V I IZ I I Z I R jX I R jI X P j = = ( )= = + = + = +2 2 2 2( ) QQ, wherereal powerP = I 2R, reactive powerQ = I2X, andapparent powerapparent power S I Z P Q = = +2 2 2(1.28)which is the same power triangle relation as in Figure 1.12.We take note here, that in impedance

Z R jX = + , the real power P is absorbed only in R, and the reactive power Q is absorbed only in X.26 Introduction to Electrical Power and Power ElectronicsExample 1.9Determinethecurrentandpowerinthefollowingcircuit,anddrawthephasor and time diagrams.Time t1125 VA (c) Time diagram51.3VI 51.3 878VAR51.3 703.4 W (d) R = 8 120 V60 HzC = 300 FL = 50 mH(a) Circuit120 V9.375 A (b) Phasor diagram51.3ISolution:We rst convert the RLC values into the total impedance

Zin ohms:

Z R j L j C = + 1/,where = 2f = 2 60 = 377 rad/s Therefore, /

Z j j = + ( )= +8 377 0 050 1 377 300 10 86. jj108 10 10 8 12 8 51 32 2 . = +( ) = tan ( ) . . /In a one-loop circuit,current (reference)Loop LoopI V Z = = 120 0 ( . . 12 8 51 3 ), which gives

I= (120 12.8)051.3 = 9.37551.3 A.The current lags the reference voltage by 51.3 as shown in phasor diagram (b) and time diagram (c).Since 120 V is the implied rms value, 9.375 A is also an rms value. The complex power in any element (source or load) is

S V I = * volt-amperes, where

V = volt-age across and

I* = complex conjugate of current through the element.Thus, in this case,

Ssource+ = = + 120 0 9 375 51 3 120 9 375 0 51 . . ( . ) ( .3 3 1125 51 31125 51 3 51 3 703 4= = + = +) .(cos . sin . ) . j jj878 VA.27 AC Power FundamentalsApparent power drawn from the source S = 1125 VA.Real power P = 703.4 W (absorbed in R and leaving the circuit as heat).Reactive power Q = 878.0 VAR (charging and discharging L and C, but remain-ing in the circuit).The power triangle is shown in (d). The following is another way to determine the ac powers:Apparent power = I2Z = 9.3752 12.8 = 1125 VA; real power = I2R = 9.3752 8= 703.4 W; and reactive power = I2X = 9.3752 10 = 878 VAR, all matching with the above values.Example 1.10Two parallel loads draw power from a 480-V source as shown in the gure below, where Load-1 draws 15 kW and 10 kVAR lagging, and Load-2 draws 7 kW and 5kVARleading.DeterminethecombinedpfandthetotalkVAdrawnfromthe source.Load-1Load-2480 V sourceSolution:In power engineering, the lagging kVAR has, a + sign, and the leading kVARhas, a sign. The parallel loads add up to make the total, and their real and reactive powers are added individually to make the total real and reactive powers. Therefore,Total

S S S j j j = + = + + = +1 215 10 7 5 22 ( ) ( ) 55 22 56 12 8 = . . + .Total kVA drawn from the source = 22.56, and the combined pf = cos 12.8 = 0.975 lagging since the combined power has + 5 kVAR, with the + sign indicating the lagging pf (or the + sign with 12.8 in total complex power

S comes from the inductive load, which draws lagging current).1.7REACTIVE POWERIf the current lags the voltage by 90 as shown in Figure 1.13as in a pure induc-torthe instantaneous power shown by a heavy curve is positive in the frst one-half cycleandnegativeinthesecondone-halfcycle.Thepowerisfrstpositive(from source to inductor) and then negative (from inductor to source). Therefore, the aver-age power in an inductor over one full cycle is always zero, although wires are occu-pied with current at all times, leaving less room for the real power to fow from the source to the load. Similarly, the average power is also zero over one full cycle in a pure capacitor, which draws current leading the voltage by 90.28 Introduction to Electrical Power and Power ElectronicsOn the other hand, if the current and voltage are in phaseas in a pure resistorthe power is always positive even when both the voltage and the current are negative in one-half cycle. This results in the average power having the same positive value in both one-half cycles and also over one full cycle.1.8THREE-PHASE AC POWER SYSTEMThe ac power is now universally adopted all over the world because of its easy con-version from one voltage level to another using energy-effcient power transformers. From a large ac power system (power grid), large power users are catered to at high voltage,mediumpowerusersatmediumvoltage,andlowpowerusersatsafelow voltage of 120 or 240 V.In a single-phase (1-ph) power circuit, the instantaneous power varies between the maximum and minimum values every cycle, with the average power positive. In a bal-ancedthree-phase(3-ph)powercircuit,whentheinstantaneouspowerinonephase decreases, the power in remaining phases increases, making the sum of power in all three phases a steady constant value with no time variations. For having such a smooth power fow, the three-phase ac system has been universally adopted around the world. Since all three phases are balanced and identical, the balanced three-phase power cir-cuitisgenerallyshownbyaone-linediagramandanalyzedonasingle-phase(per-phase) basis. Then, average three-phase power = 3 average single-phase power.The three-phase source or load can be connected in Y or . The basic voltage and currentrelationsbetweentheline-to-neutral(calledphase)valuesandtheline-to-line (called line) values are reviewed below.1.8.1BALANCED Y- AND -CONNECTED SYSTEMSIn a balanced three-phase system, the magnitudes of line voltage VL, line current IL, andthree-phasepower intermsofthemagnitudesofphasevoltageVphandphase VoltageAvg power in rst-half cycleInstantaneous powerAvg power in second-half cycleCurrent lagging by 90FIGURE 1.13Voltage, current, and instantaneous and average powers in inductor.29 AC Power FundamentalscurrentIpharederivedasfollows,wherethetermphasevoltage(Vph)alsomeans line-to-neutral voltage (VLN), and the term line voltage (VL) also means line-to-line voltage (VLL).The Y-connected system shown in Figure 1.14a gives the voltage of line a from lineb,

V V Vab an bn= ,whichisshowninFigure1.14bbyaheavysolidline.The geometry would give the line-to-line voltage magnitude VLL = 3 Vph. We note that the line voltage

Vab leads the phase voltage

Van by 30.The line current is obviously the same as the phase current, i.e., L = ph.Three-phase power P3-ph = 3 P1-ph = 3 Vph Iph pf = 3 (VLL/3) IL pf.IaLine BLine CIcLine AVanVcnPhase CLine current IANVbnIbPhase B(a) Phase and line voltage and current denitions Line-to-line voltage VAB30VaVABVbVcVb(b) Line-to-line voltage phasor VAB = Va VbFIGURE 1.14Phase and line voltages in balanced Y-connected source or load.30 Introduction to Electrical Power and Power ElectronicsTherefore, P3ph = 3 VLL IL pf watts and S3ph = 3 VLL IL volt-amperes.(1.29)The -connected system shown in Figure 1.15a gives the line current A = a c, which is shown by a heavy line in Figure 1.15b, the geometry of which leads to IL = 3 Iph.The line-to-line voltage here is obviously the same as the phase voltages, that is, VLL = Vph.Three-phase power P3-ph = 3 P1 = 3 Vph Iph pf = 3 VLL (IL/3) pf.Therefore, P3-ph = 3 VLL IL pf watts and S3-ph = 3 VLL IL volt-amperes.(1.30)Note that the expression for three-phase power comes out to be the same in both Y and .In a three-phase, four-wire, Y-connected system, the neutral current is the sumof three line currents. In a balanced Y system, the current in neutral wire (if used) is always zero because it is the phasor sum of three equal line currents out of phase by 120 from each other. For this reason, a balanced three-phase system performs exactly the same with or without the neutral wire, except in unbalanced load condition.In a three-phase, -connected system, no wire exists for the return currents, so thephasorsumofalllinecurrentsisforcedtobezeroregardlessofbalancedor Phase BLine BLine CLine ALine current IA = Ia IbIaVanIcVcnVbnIb(a) Phase and line voltage and current denitionsLine-to-line voltageIa30IAIcIcIb(b) Line-current phasor IA = Ia IbFIGURE 1.15Phase and line currents in balanced -connected source or load.31 AC Power Fundamentalsunbalanced load. With an unbalanced three-phase load, however, this imposed zero on the return current produces unbalanced line-to-line voltages.Example 1.11AY-connectedgeneratorhastheline-to-neutralvoltagephasor

Van = 100 0. Determine the line-to-line voltage phasors

Vab,

Vbc, and

Vca with their phase angles with respect to

Van.Solution:The phasor relations between the line and phase voltages are shown in the gure below, the trigonometry of which gives the following:

Vab+ = 3 100 30,

Vbc = 3 100 90, and

Vca or + = 3 100 210 3 100 150 .It means that with reference to the phase voltage

Van, the line voltage

Vab leads

Van by 30,

Vbc lags

Van by 90, and

Vca lags

Van by 210 or leads

Van by 150.nb c a Vbn VcnVca bc Vab VanPositive direction of phase rotationVV1.8.2Y EQUIVALENT IMPEDANCE CONVERSIONAlmost all three-phase generators are Y-connected, but some transformers and loads maybe-connected.Evenso,wealwaysanalyzethepowersystemassumingall equipment to be Y-connected since it is much easier to determine the phase values in each phase of the generator and then convert the results into three-phase values. Forthispurpose,ifaloadis-connected,wefrstc