AC Machine Systems: Mathematical Model and Parameters, Analysis, and System Performance

Jingde Gao Linzheng Zhang Xiangheng Wang AC Machine Systems Mathematical Model and Parameters, Analysis, and System Pe...

Author: Jingde Gao | Linzheng Zhang | Xiangheng Wang

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Jingde Gao Linzheng Zhang Xiangheng Wang

AC Machine Systems Mathematical Model and Parameters, Analysis, and System Performance

Jingde Gao Linzheng Zhang Xiangheng Wang

AC Machine Systems Mathematical Model and Parameters, Analysis, and System Performance

With 158 figures

Authors Jingde Gao Dept. of Electrical Engineering Tsinghua University 100084, Beijing, China

Linzheng Zhang Dept. of Electrical Engineering Tsinghua University 100084, Beijing, China E-mail: [email protected]

Xiangheng Wang Tsinghua University 100084, Beijing, China E-mail: [email protected]

ISBN 978-7-302-19342-5 Tsinghua University Press, Beijing

ISBN 978-3-642-01152-8

e-ISBN 978-3-642-01153-5

Springer Dordrecht Heidelberg London New York Library of Congress Control Number: pending © Tsinghua University Press, Beijing and Springer-Verlag Berlin Heidelberg 2009 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: Frido Steinen-Broo, EStudio Calamar, Spain Printed on acid-free paper Springer is a part of Springer Science+Business Media (www.springer.com)

Abstract

This book is written on the basis of the authors’ achievements in teaching and research over many years in the field of ac machines and systems at Tsinghua University, and reflects the new developments in and applications of science and technology in recent years. When researching on the systems related to ac machines as main roles, analysis methods and operating performance are stressed in this book. This book is divided into six chapters. Chapter 1 makes salient-pole electric machines as the main subject of study. Considering ac machine as a circuit model composed of stator and rotor having relative motion, and starting from a single coil, the mathematic models and parameter calculation methods are discussed. Chapter 2 discusses electromagnetic relations and parameters of synchronous machines and their several operating conditions at synchronous speed, such as 3-phase sudden short circuit, sudden increase of load, opening short circuit, etc. Chapter 3 introduces some special operating modes of synchronous machine systems at constant speed and variable speed, such as two phase sudden short circuit, asynchronous operation of synchronous machines, asynchronous starting of synchronous motors, synchronous machine in connection with capacitances, cycloconverter-fed synchronous motor, ac-dc hybrid loads of synchronous machines, etc. Chapter 4 analyses oscillation, stability and excitation regulation of synchronous machine systems. Chapter 5 introduces electromagnetic relations of induction machines, some of their operating modes, such as starting characteristics, reswitching process, self-excitation during motor in series with capacitances, and combination system for polyphase induction generator with dual windings and rectified load. Chapter 6 provides description of inner asymmetric problems in the stator winding and squirrel-cage rotor of ac machines. This book is a textbook for the postgraduates in the field of electric machines and control, and also a reference book for technicists as well as for the teachers and students of relevant specialties in universities and colleges. i

Preface

With the increase of single unit capacity and broader application scope of ac machines, as well as the development and application of power electronics, computer, micro-electronic technology and control theory, the operating conditions of ac machines and systems become more complicated, their automation level is higher, and their operation demands are also stricter. Therefore, new projects and new research results are presented one by one. With development of science and technology, more attention has been paid to the performance analysis of the following systems: variable speed drive system of ac machines with power electronics and control devices, power energy system composed of ac machines and power transmission lines, electric machine system consisting of ac machines and excitation control devices. Therefore, a monographic book on ac machine systems is needed urgently to reflect the new demand and new developments. This monographic book is written mainly on the basis of the following 5 monographic books: Analysis of AC Machines and Their Systems (1st Edition) in 1993 and 2nd Edition in 2004, Fundamental Theory and Analysis Methods of Transients in Electric Machines Volumeĉin 1982 and VolumeĊ in 1983, AC Machine Transients and Operating Modes Analysis in 1963, in which the first book was awarded the First Prize of National Excellent Science and Technology Book in 1995 and the third and fourth books also won the First Prize of National Excellent Science and Technology Book in 1983, together with the Excellent Prize of National Teaching Material in 1988. The first, third, and fourth books were finished under the leadership of Professor Gao Jingde, and the fifth book was also written by Prof. Gao, a Member of Chinese Academy of Sciences, the former President of Tsinghua University, Fellow of IEEE, and the founder of transient theory of electric machines and power systems in China. For international academic exchange, this monographic book has been written in English. Professor Wang Xiangheng wrote Chapters 1 and 6 (except for Sections 6.2 and 6.3), Dr. Wang Shanming wrote Section 3.7, Dr. Wu Xinzhen iii

wrote Section 5.6, Dr. Sun Yuguang wrote Sections 6.2 and 6.3 and Professor Zhang Linzheng wrote Preface, Introduction, Appendices and Chapters 2, 3, 4 and 5 (except Sections 3.7 and 5.6) together with modifying and unifying all chapters in this book. In the process of compiling this book, some books and papers were referred to, and some books and papers written by our colleagues were also used. Main part of them have been listed in the references in order to help the readers know their origins and learn about some interesting problems in more depth. We are grateful to the authors of those books and papers, especially to our colleagues. It is very difficult to avoid defects and careless omissions in this book owing to the finite acquirements of the authors although we have made our efforts. Please send your comments and suggestions regarding this book to Department of Electrical Engineering, Tsinghua University, Beijing, China. Authors

iv

Introduction

AC machines are basic equipment of electric energy production and application. Electric energy used in industry, agriculture and daily life is mostly produced by alternators and the generated electric energy is likely utilized through ac motors. The capacity of a single unit has been increased greatly for modern alternators, the application domain is extended day by day, the automation grade of ac machine systems is heightened increasingly, and the operating condition is more complicated. Therefore, it is very important for electro-technical experts to understand and master the operating performance and the analysis method of ac machine system. The ac machine system is a relatively mature but quickly developing product of science and technology. Emergence and development of ac machines have a history of more than one hundred years, in which ac machines of various types have been built for various uses. During design, manufacture, and operation of ac machines and through researches and improvement, people have comprehended the structure principles, manufacture technique, and operating performance deeply. However, due to restraints of technical development and research conditions in the past, people paid attention to normally operating performance, outer faults and economic parallel-operation in safety for ac machines, which is natural because those problems often exist in practice. During the past decades, the power electronic technique has been widely used, the dc transmission had a great progress, ac-dc hybrid transmission systems emerged, large capacity alternators were built to bring about some new requirements, and modern control theory, micro-electronics and computer technology also progressed greatly, thus making the theory and operating performance of ac machine systems undergo a new development. For example, the analysis of inner unsymmetrical windings, the study of control system consisting of ac machines, power electronic devices, micro-electronics and computer technology, the analysis of electromagnetic harmonics effect and restraint are all the new topics. During the operation of the ac machine, due to manufacturing defect or misoperation, some fault occurs in the machine winding that creates unnecessary loss. For large hydraulic generator, its stator winding often has multi-branches. If inter-turns or inter-phases short circuit occurs or the winding is broken partly, the large fault current will damage the winding or the iron core, and the negative v

sequence magnetic field during fault may burn the rotor too. When the turbo-generator and the induction machine are working, some serious inner faults may occur in their stator winding. As far as the rotor of an ac machine is concerned, weld defects, and the breaking or short circuit of conductors may occur, too. Therefore, it is necessary to study the various state quantities during inner faults in order to understand the harm of faults, to improve the design and manufacturing technology and to put forward the corresponding protection relays. So far as the dc transmission system goes, due to the combination of alternator and rectifiers, it is quite different from the ac transmission system. As for the alternator, it is often on unsymmetrical operating mode and in the system exist serious harmonics due to nonlinearity of rectifying elements, so the corresponding special topics need studying. In addition, the ac excitation system is widely used and sometimes the excitation bars are installed in the upper positions of stator slots for the main alternator to provide excitation current for the excitation system of main alternator through the rectifiers, which also brings about the similar problems as mentioned above. Since they connect computer, micro-electronics and power electronic technique, the speed regulation and control of ac motors have been developed swiftly, and some electromagnetic processes, which were hardly measured and controlled in the past, can be regulated and controlled today through on-line measurement, computer analysis and control. With the development of computer technology and the application of numerical method, some problems of the electromagnetic field of an electric machine, which were hardly calculated in the past, can be solved today to enhance the capability of design and operating performance analysis, and the accuracy of parameters calculation greatly for ac machines. In addition, the achievements in some aspects are also gotten, such as pole-changing speed-regulation of induction motors, principles and methods of starting, the combination of capacitors and stator winding, special machine provided with rectangular waveform current, storage electric machine, etc. For a deep analysis of the ac machine systems, it is necessary to put forward new theory and methods. As you know, as far as the electromagnetic relation of ac machine is concerned, it can be considered as the combination of several coils coupled magnetically and studied according to usual circuit rules. However, due to rotor rotation, there exists relative motion between those coils. 3-phase symmetrical winding is adopted for the stator winding of an ac machine to make the air-gap magnetic field be of sinusoidal waveform and revolve synchronously when 3-phase symmetrical current passes through the 3-phase winding. In so doing, various targets of ac machines are all good. Therefore, people used to assume the air-gap flux to be sinusoidal and to merge small harmonic flux into leakage flux during the analysis of ac machines. Analyzing their operating performance, the parameters used are based on the criterion that they can reflect the total action of the air-gap magnetic vi

field, such as d- and q-axes armature reaction reactances xad and xaq for synchronous machines, magnetization reactance xm for induction machines, etc. Correspondingly, other parameters such as xd , xq , xdc , xqc , xdcc and xqcc for synchronous machines, are all some values to reflect the total electromagnetic effect. The mathematical patterns built by using the above parameters and the corresponding analytical results can meet the practice requirements; so in the text-books and references, the phase winding has been the basis and the total effect of multiphase winding has been taken to analyze ac machines for many years. As stated before, with the development of science and technology, there came to exist the inner unsymmetrical problem of electric machine winding, the new combination system of electric machine and power electronic devices and some electric machines of special structure, which make taking a single machine as the analysis object and symmetrical phase winding as the basis of theory and methods be unsuitable for the study of some topics. Thus, it is necessary to take the electric machine and the associated outer devices as a system to consider the influence of electromagnetic harmonics. Based on the aforementioned points, a single coil is adopted as the basis in this book to describe the basic electromagnetic relation of ac machines and to build the corresponding expressions. After getting the relations, they can be used to study not only the problems that are analyzed by using the existing methods but also some new topics as mentioned before. Since the basic electromagnetic relations of ac machines are based on single coil, in the light of the requirements for different research problems compose the corresponding loops to get the associated electromagnetic relations, inclusive of the electromagnetic relations obtained when taking the phase winding as the basis. Because electromagnetic harmonics are counted in the basic relations, we can study not only some topics which need to count harmonic influence but also a certain problem that can neglect harmonic effect for convenience. The ac machine parameters used in the new method are self inductances and mutual inductances for various coils, so it is not difficult to find out the parameters and the basic relations to reflect the total electromagnetic effect of ac machine winding according to the self inductances and mutual inductances if necessary. The study starts from a single coil and combines the corresponding loop equations according to the actual loops connection of ac machines to research into needful topics, which is termed the Multi-Loop Theory and is the main basis of this book. This book is divided into six chapters, referring to Contents of this book in detail. Firstly, based on a single coil and using the Multi-Loop Theory created by some authors of this book, we get various inductances and parameters which are the same as classical method, and then analyze several operating modes of synchronous machine systems, such as transformations of reference axes systems connecting the relevant mmf to form a clear concept, short circuit and vii

asynchronous operation, combination system for synchronous machine and capacitances, cycloconverter-fed synchronous motor system, synchronous generator system with ac and dc hybrid loads, oscillation, stability and excitation regulation. Also, we analyze some operating modes of induction machine systems, such as why the transient starting characteristic is quite different from steady-state starting one, why the reclosing current is larger than the starting current, discovery of new self-excitation regions during induction motor in series with capacitances, and special induction generator with dual windings and rectified load. By using the Multi-Loop Theory, we study the stator winding inner faults and squirrel cage faults, which are hardly found in other books and references. This book provides a new way for analyzing ac machine systems.

viii

Contents

Abstract ................................................................................................................ i Preface ................................................................................................................ iii Introduction ........................................................................................................ v 1

Circuit Analysis of AC Machines—Multi-Loop Model and Parameters ................................................................................. 1 1.1 Electromagnetic Relations of AC Machine Loops ................................ 2 1.2 Air-Gap Permeance Coefficient of Electric Machines .......................... 6 1.3 Air-Gap Magnetic Field Produced by Single Coil and the Corresponding Inductance .................................................................... 8 1.4 Inductances of Stator Loops ................................................................ 19 1.5 Inductances of Rotor Loops ................................................................ 23 1.6 Mutual Inductances Between Stator Loops and Rotor Loops ............. 32 1.7 Influence of Saturation on Parameters ................................................ 34 1.8 Inductances Produced by Leakage Magnetic Field ............................. 37 1.9 Electromagnetic Torque and Rotor Motion Equation of AC Machines ...................................................................................... 40 1.10 Analysis of Parameters and Performance of Single-Phase Induction Motors............................................................................... 44 References.................................................................................................... 56

2

Electromagnetic Relations and Parameters of Synchronous Machines and Analysis of Their Several Operating Conditions at Synchronous Speed ................................................................................ 58 2.1 Basic Relations of Synchronous Machines ......................................... 59 2.2 Basic Relations of Synchronous Machines in d, q, 0 Axes ................. 78 2.3 Per-Unit Systems in Synchronous Machines ...................................... 89 2.4 Basic Equations in Per-Unit, Operational Reactances and Electromagnetic Torque of Synchronous Machines.......................... 100 2.5 3-Phase Sudden Short Circuit and Transient Parameters of Synchronous Machines ......................................................................111 2.6 Voltage Dip during Sudden Increase of Load for Synchronous Machines ........................................................................................... 142 ix

2.7 Transient EMFs of Synchronous Machines ...................................... 147 2.8 Electromagnetic Torque after 3-Phase Sudden Short Circuit ............ 153 2.9 Suddenly Opening 3-Phase Short Circuit of Synchronous Machines ........................................................................................... 159 References.................................................................................................. 165 3

Some Special Operation Modes of Synchronous Machine Systems at Constant Speed and Variable Speed ....................................................... 166 3.1 Transformations of Reference Axes Systems and Their Formulas in Synchronous Machines ..................................................................... 168 3.2 2-Phase Sudden Short Circuit of Synchronous Machines................. 180 3.3 Steady-State Asynchronous Operation of Synchronous Machines ... 203 3.4 Asynchronously Starting of Synchronous Motors ............................ 214 3.5 Analysis of Combination System for Synchronous Machine and Capacitances ..................................................................................... 224 3.6 Mathematic Model and Performance Analysis of Cycloconverter-fed Synchronous Motor Systems with Field-Oriented Control............... 235 3.7 Analysis of Synchronous Generators with AC and DC Stator Connections....................................................................................... 249 References.................................................................................................. 266

4

Oscillation, Stability and Excitation Regulation of Synchronous Machine Systems ...................................................................................... 267 4.1 Small Perturbation and Linearization of AC Machine Systems ........ 268 4.2 Steady-State Small Oscillation and Torque Coefficients of Synchronous Machine Systems ........................................................ 270 4.3 Static Stability of Synchronous Machine Systems and Influence of Excitation Regulation on Static Stability .......................................... 301 4.4 Dynamic Stability and Analysis Methods of Synchronous Machine Systems ............................................................................................. 315 References.................................................................................................. 338

5

Electromagnetic Relations of Induction Machine Systems and Analyses of Some Operating Modes........................................................ 339 5.1 Basic Relations and Parameters of Induction Machines ................... 340 5.2 Basic Relations and Parameters of Induction Machines in d, q, 0 Axes ................................................................................... 354 5.3 Analysis of the Starting Process for Induction Motors...................... 363 5.4 Transients of Reswitching on Induction Motors ............................... 371 5.5 Self-Excitation when Connecting Induction Motor in Series with Capacitance and Counting the Inertia Effect in ........................ 386 5.6 Dual-Stator-Winding Multi-Phase High-Speed Induction Generator........................................................................................... 402

x

References...................................................................................................411 6

Internal Asymmetric Analysis of AC Machines ..................................... 412 6.1 Analysis of Rotor Winding Failures of Squirrel Cage 3-Phase Induction Machines ............................................................. 415 6.2 Internal Fault Analysis of Stator Winding of Salient Pole Synchronous Machines ..................................................................... 431 6.3 Internal Fault Analysis of Stator Winding of Non-Salient Pole Synchronous Machines ............................................................. 447 References.................................................................................................. 454

Appendix A Air-Gap Permeance Coefficients in Consideration of Slot Effect................................................................................ 455 Appendix B

Expressions of Effective Air-Gap .......................................... 458

Appendix C

Inductances Due to Leakage Flux of Stator Winding End........................................................................... 461

Appendix D

Heaviside’s Operational Calculus and Its Application to Transients Analysis ....................................... 474

Appendix E

Basic Relations and Equivalent Circuits of Transformers .......................................................................... 480

Index ................................................................................................................ 484

xi

1 Circuit Analysis of AC Machines—Multi-Loop Model and Parameters

Abstract The Multi-Loop Model and parameters are introduced and described in this chapter as the basis of this book, which is more convenient than the classical method for the analysis of inner faults. According to the model, an ac machine is considered as an electric circuit linked with ferromagnetic circuit. The difference between ac machines and general static circuits is the existence of relative movement between the stator and the rotor for the former. Hence, some loop inductances of ac machines are relative with the rotor position, that is, these inductances are time-variant coefficients. The salientpole synchronous machines are analyzed as structure models of ac machines, because a non-salient-pole synchronous machine is a special case of salientpole type. The single coil is taken as the fundamental unit in order to study and discuss the electro-magnetic relation of ac machines. This chapter discusses the circuit model of ac machines, inclusive of loop voltage equations and fluxlinkage equations of stator and rotor, loop parameters—mainly inductances (main inductances and leakage inductances), electro-magnetic torque and rotor motion equations, etc. Finally, as an example of using the Multi-Loop Model, the parameters and performance analysis of a single phase induction motor are presented. Evidently, the difference between the calculated values and experimental data is bigger when considering only the fundamental of air-gap magnetic field by the use of ordinary designed program. However, when counting the higher harmonics in by using the Multi-Loop Method, the calculated values are close to the experimental data.

AC machines are main equipment which convert mechanical energy into electric energy (generators) or vice versa (motors). According to varying demands, synchronous machines and induction machines with different capacities and various sorts are produced and applied. Though the structures, principles and performance of these machines are different and numerous, the stator windings of ac machines with 3 phases or single phase are connected with several coils or components; the structures of the rotor are of salient pole or non-salient pole, the rotor windings are composed of conducting bar and end-ring or connected by several coils; therefore the electromagnetic relations of the salient-pole synchronous machine with non-uniform air-gap are more complicated. From the point of view

AC Machine Systems

of the rotor structures, the synchronous machine with damper winding and excitation winding is more representative. From the point of view of electromagnetic relations, all ac machines are composed of several coils having movement with each other, hence after finding one coil’s relationship with current, flux-linkage and voltage, the corresponding loop equations are obtained based on the concrete structures and operation status, then the performance can be calculated and analyzed. In this chapter, the salient-pole synchronous machines are considered as structure models of ac machines, the single coil as the fundamental unit in order to study and discuss the electromagnetic relation of ac machines, including the inductance of single coil and the inductances of the loop composed of these coils, etc. As the basis of the Multi-Loop Theory and Method, thoroughly understanding and mastering it is the key for reading the rest of this book[1– 6].

1.1 Electromagnetic Relations of AC Machine Loops AC machines are circuit structures with iron, the difference between ac machines and general static circuits is the existence of relative movement between the stator and the rotor for ac machines, hence some loop inductances of ac machines depend upon the rotor position, i.e. these inductances are time-variant coefficients. When studying the basic electro-magnetic relations of ac machines a lot of theorems and rules about static circuits may still be applied. Taking the coupled circuit in Fig. 1.1.1 as an example, in terms of the positive directions in Fig. 1.1.1 and the right hand screw rule of the current and flux-linkage, the voltage and flux-linkage equations are as follows:

\ 1 L1i1 M 12 i2 \ 2 M 21i1 L2 i2 u1 p\ 1 r1i1 u2 p\ 2 r2 i2 These electro-magnetic relations and concepts of simple and basic circuits are often used in discussing the electro-magnetic relations of ac machines.

Figure 1.1.1 Coupled circuit with mutual inductance 2

1

Circuit Analysis of AC Machines — Multi-Loop Model and Parameters

So far as electro-magnetic relations are concerned, the salient-pole synchronous machines have more general meaning. The non-salient-pole synchronous machine with uniform air gap may be treated as a special example of the salient-pole synchronous machine with non-uniform air gap. If we need to consider the action of the eddy current in the rotor iron of the non-salient-pole synchronous machine, its action can be replaced by equivalent damper winding. Therefore, some principles and concepts drawn from salient-pole synchronous machine may be used in non-salient-pole synchronous machine under certain conditions. As for the induction machine, there exists not only a uniform air gap but its rotor is also symmetrical to arbitrary axis in electric and magnetic aspects, so it may be treated as a special example of salient-pole synchronous machine under some instances. In order to analyze the operation performance of salient-pole synchronous machines, firstly it is needed to list the voltage and flux-linkage equations of the stator and rotor loops. Generally, the stator phase winding is considered as a loop under symmetric 3 phases. But, in the inner asymmetric condition of the stator winding, it is difficult to consider the stator phase winding as a loop, so selecting the loop according to concrete problem is required. For example, when an inner fault occurs in the stator winding, the short circuit part needs to be treated as a single loop. For a normal single phase machine, although its phase winding may be treated as a loop, under some special structure, e.g. for the single phase machine with concentrated winding (q 1) , its harmonic magnetic field has to be calculated, or else more errors will be brought in. So, based on a single coil, it is significant to discuss the electro-magnetic relation of each loop. In the following, we analyze concretely the electro-magnetic relation of each loop of the salient-pole synchronous machine from two sides of the stator and rotor.

1.1.1

Stator Loops

When analyzing the performance of ac machines, it is necessary to list the loop equations. Many loops exist in the stator, in which some loop may be a phase winding, a branch of stator winding, or the special circuit connected by a coil or several coils. For example, for a loop K, the positive directions of current, flux-linkage and voltage are drawn in Fig. 1.1.2, i.e. positive current produces negative flux-linkage, positive current produces positive voltage drop along the load direction. Such regulation is consistent with the positive direction of the stator loop-K current and its axis-K direction in Fig. 1.1.3, namely, the positive direction of loop-K flux-linkage. Under the above regulation, the voltage equation of arbitrary loop K of the stator is written as

uK

p\ K rK iK

(1.1.1) 3

AC Machine Systems

where uK ,\ K , rK , iK are the terminal voltage, flux-linkage, resistance and current d is differential operator. of the loop K respectively; p dt

Figure 1.1.2 Positive direction relation of i, u ,\ of stator loop

1.1.2

Figure 1.1.3 Positive direction regulation of stator and rotor of electric machine

Excitation Loop

Selecting the direction of direct axis d and quadrature axis q as in Fig. 1.1.3, the direct axis is consistent with pole central line, the quadrature axis leads the direct axis by S / 2 electrical radians along the rotation direction of the rotor. The fluxlinkage positive direction of the excitation loop j is consistent with the positive direction of d-axis. The positive direction of the current of excitation loop j is as in Fig. 1.1.3, i.e. the positive excitation current produces positive excitation loop fluxlinkage. The positive direction of excitation voltage is as in Fig. 1.1.4, i.e. the voltagedrop positive direction is consistent with the positive direction of excitation winding Figure 1.1.4 Positive direction regulation of the voltage and current of excitation current. Based on the above positive direction winding regulation of each quantity, the voltage equation of excitation loop j is written as

u fd j

p\ fd j rfd j i fd j

(1.1.2)

where u fd j ,\ fd j , rfd j , i fd j are the terminal voltage, flux-linkage, resistance and current of excitation loop j, respectively.

1.1.3

Damper Loops

The damper bars of the salient-pole synchronous machine are all short-circuited or partly short-circuited, these bars constitute a mesh circuit in which the 4

1


selection of current loops is arbitrary. According to the requirement of the analyzed problem, the damper loops may be treated as an equivalent d-axis loop and q-axis loop on each pole, or they can be selected based on practical mesh circuit. The positive direction regulation of every quantity of the damper loops is similar to excitation loop. In this book, the damper loops will be analyzed in terms of practical mesh circuit as in Fig. 1.1.5.

Figure 1.1.5

Sketch map of damper loops of salient-pole synchronous machine

As an example, the voltage equation of damper loop i is 0

p\ i rii i rc (ii 1 ii 1 )

(1.1.3)

where \ i , ri , ii are the flux-linkage, resistance and current of damper loop i, respectively; rc is the bar resistance; ii 1 and ii 1 are the current of damper loops ( i 1 ) and ( i 1 ), respectively. The voltage Eqs (1.1.1), (1.1.2), and (1.1.3) all include the loop flux-linkage. On the basis of the circuit theorem, the relation between the flux-linkage and current (flux-linkage equation) is listed as follows: ª \ s1 º ª Ls1 « » « # « # » « « \ sm » « M sm s1 « » « «\ fd1 » « M fd1s1 « # » « # « » « «\ fdn » « M fdn s1 « \ » « M 1s1 « 1 » « « # » « # « \ » « M ¬ l ¼ ¬« ls1 or written as

"

M s1sm

M s1 fd1 #

"

# Lsm

"

M s1 fdn

M s11

"

#

"

M sm fd1

" M sm fdn

# M sm 1

" M fd1sm

L fd1

" M fd1 fdn

M fd11 "

# " M fdn sm

# M fdn fd1

"

# L fdn

"

M 1 fdn

L1

"

"

# M lfdn

# M l1

"

"

M 1sm

M 1 fd1

"

# M lsm

# M lfd1

\

LI

# M fdn 1 "

M s1l º ª is º » 1 # »« # » « » M sml » « is » »« m » M fd1l » « i fd » 1 # »« # » »« » M fdn l » «i fd » n M 1l »» « i1 » « » # »« # » » Ll ¼» ¬« il ¼»

(1.1.4) 5

AC Machine Systems

where L is self inductance; M is mutual inductance; each subscript represents the corresponding loop, where s1 , s2 ," , sm represent the sequence of stator loops, 1,2, " , l represent the sequence of damper loops, fd1 , fd 2 ," , fd n represent the sequence of excitation loops. In the above formula, the corresponding mutual inductances are reversible, i.e.,

M s1s2

1.2

M s2 s1 ,

M fdn sm

M sm fdn ,

M sml

M lsm ,

M 1l

M l1 , "

Air-Gap Permeance Coefficient of Electric Machines

The parameters of electric machines is the key for analyzing their performance, and the self inductances and mutual inductances are the most basic parameters of electric machines which represent the capability of certain loop current producing certain loop flux-linkage. Generally, the flux-linkage of certain loop depends upon the currents in the related loops, as in the formula (1.1.4), and the elements of inductance matrix L include the stator loop inductances, the rotor loop inductances and the inductances between stator and rotor loops. The flux-linkages produced by loop currents in ac machines are divided into two kinds, one is the leakage, and the corresponding inductances are considered to be independent of the rotor position, i.e. to be constant; another flux-linkage is the main part which links the stator and rotor through the air-gap. The inductances corresponding with these flux-linkages will vary when the reluctance variation occurs owing to rotor rotation. The permeance analysis method of magnetic circuit and the analysis method of magnetic field may be applied to research these inductances. Here we discuss mainly the permeance analysis method of magnetic circuit. While researching the inductances by using the permeance analysis method, the effect of iron saturation is not taken into account, considering that the reluctance of magnetic circuit does not vary with the flux density, and the effect of iron reluctance is considered by magnifying the air gap appropriately. Suitabe parameter value is applied to consider the effect of iron saturation according to the operation condition, if necessary. In addition, the effect of some secondary factors such as the hysteresis and eddy-current also will be neglected in the analysis. In this chapter, the effect of slots is represented by the Cater’s coefficient; the slot harmonic variation of the air-gap reluctance is neglected. Please see Appendix A for the inductance including the slot harmonics. The effect of iron saturation and slots may be calculated by the magnetic field analysis in which the magnetic field method should be used to count the inductances of ac machines, and obviously the time spent in the process will be greater. As there are specialized books on this topic, this book does not intend to discuss it further. 6

1


When the inductances are calculated by using the permeance analysis method, at first, the air-gap mmf and air-gap permeance are analyzed in terms of the basic concept of inductance coefficients[7]; then the air-gap magnetic field can be calculated, the self and mutual flux-linkages are found, and finally the corresponding inductances will be obtained. As mentioned above, setting out from a single coil and then to arbitrary loop composed of the coils, the inductances of electric machine loops are studied, therefore the air-gap mmf is analyzed in terms of the air-gap mmf of the single coil. Here, the air-gap mmf includes integral and fractional harmonics with the 1 2 3 orders , , ," , in which P is pole-pair number. P P P The relation between the mmf F, the permeance / and magnetic flux ) of the magnetic circuit is

F/ ) S , where S is the cross-section area l of the magnetic circuit, l is length of the magnetic circuit, and P is permeability. For magnetic flux of unit area (namely flux density) there is

For uniform magnetic circuit, there is /

P

FO

B

For uniform magnetic circuit, O

P

(1.2.1)

1 represents the permeance of unit area that l

is called the permeance coefficient. Along the different positions of the rotor coordinates system, the air-gap length of salient pole synchronous machine is different owing to non-uniform air-gap, hence the air-gap permeance /G and the permeance of unit area OG are varied, and they are relative to the mmf and rotor position. This issue is discussed further here.

Figure 1.2.1

Rotor coordinates of salient pole synchronous machine

The permeance of unit area is represented by OG ( x) which is called air-gap permeance coefficient, and here x is the abscissa on the rotor as in Fig. 1.2.1. Owing to the rotor symmetry relative to the d-axis, the air-gap lengths in the 7

AC Machine Systems

positions x and x of the d-axis are equal, hence their air-gap permeance coefficients are equal too, i.e. OG ( x) OG ( x ) . The distributing status of air-gap permeance coefficient under each pole is homologous, thus OG ( x ) OG (S r x) . Therefore, the air-gap permeance coefficient is the even function of the variable x only with even harmonics, and its general expression is

OG ( x)

O0 2

¦ O2l cos 2lx

l 1, 2,"

(1.2.2)

l

If the equivalent air-gap G ( x) of salient pole synchronous machine with non-uniform air-gap is known, then the air-gap permeance coefficient OG ( x) is

OG ( x)

P0 G ( x)

(1.2.3)

The following expression obtained according to Fourier series is

O0

4 S 4 S P OG ( x)dx = ³ dx ³ S 0 S G ( x)

(1.2.4)

O2l

4 S OG ( x) cos 2lxdx S ³

(1.2.5)

4 S P 0 cos 2lxdx S ³ 0 G ( x)

Based on the structure features of the salient pole synchronous machine, the equivalent air-gap lengths G ( x) of the pole shoe space and interpolar space may be dealt with respectively. In the pole shoe space, it is considered that all flux lines are through the air-gap along the radial direction, and the equivalent air-gap length may be given according to geometric size. But, on the basis of the results of graphic method and analysis method, in the interpolar space, the distributing status of the magnetic field is relative to the position of the corresponding mmf, hence the expressions of equivalent air-gap length are given respectively according to two statuses in which the zero value or the maximum value of air-gap mmf is on the central line of the interpolar space. The expressions of equivalent air-gap lengths G ( x) are given in Appendix B. After getting the air-gap mmf and air-gap permeance coefficient, the air-gap magnetic field can be obtained, then the loop flux-linkage and corresponding inductance are obtained.

1.3 Air-Gap Magnetic Field Produced by Single Coil and the Corresponding Inductance As mentioned above, several coils constitute each loop of the stator. No matter what connection the stator winding has or how asymmetrical the stator winding is, 8

1


the stator loops are composed of several corresponding coils. Therefore, researching and realizing the air-gap magnetic field (main magnetic field) have general meaning. Besides the air-gap magnetic field (main magnetic field), the leakage magnetic fields exist in electric machines, and the points with regard to the leakage magnetic field and the corresponding inductances will be discussed in Section 1.8. The air-gap magnetic field produced by a single coil and the related inductances are discussed as follows.

1.3.1

Air-Gap MMF Produced by Single Coil’s Current

Assume that electric current passes through the stator coil AAc , and rectangular mmf is formed as in Fig. 1.3.1. Harmonic analysis is made for the rectangular mmf along the inner circumference of the stator, and then exists F (T )

¦a

n

n

Figure 1.3.1

n cos T P

n 1, 2,"

(1.3.1)

Air-gap mmf produced by the current of stator coil AAc

The above formula has only the cosine component, because the rectangular mmf is symmetrical to the axis of the coil AAc . Therefore, there is an

1 PS §n · F (T ) cos ¨ T ¸ dT ³ P S PS ©P ¹

2 nE S iwK sin nS 2P

(1.3.2)

where P is pole-pair number, E is the ratio of actual pitch to full pitch, wK is turn number of the coil, T is the electric angle along circumferential direction of the electric machine. It is seen that in the rectangular mmf of the single coil, the fractional harmonics and lower harmonics are very strong. In formula (1.3.1), the fundamental wave of mmf is a1 cos

T

, its period is P 2PS ; the P-th harmonic of mmf is aP cosT , its period is 2S ; the KP-th harmonic of mmf is akP cos kT , its period is 2S / k . 9

AC Machine Systems

If the space wave of 2S electric radians is considered as fundamental wave according to general custom, then the formula (1.3.1) may be rewritten as

¦F

F (T )

k

cos kT

1 2 3 , , ," P P P

k

k

(1.3.3)

in which Fk

2 kES iwK sin kPS

2 iwK k yk kP S

k ES is pitch factor of the coil. 2 Owing to non-uniform air-gap of salient pole synchronous machine, the flux density produced by the same air-gap mmf is different in different rotor positions. Obviously, when the mmf is located on d-axis or q-axis, the air-gap is different, hence the corresponding air-gap magnetic field is different too. For the sake of convenience for analyzing the air-gap magnetic field, the stator mmf is divided into two components, Fdk ( x ) and Fqk ( x) , so that their positions are not varied relative to the rotor, and the corresponding air-gaps relative to the two mmf are not varied too. Therefore, it is needed to transform the air-gap mmf formula (1.3.1) written in the stator coordinates system to rotor coordinates system. Since T x J (see Fig. 1.3.1), if the d-axis position is considered as the origin of the abscissa x established on the rotor, J is the rotor position angle (it is electric angle by which the rotor d-axis leads the stator coordinates axis along the rotation direction), then there is

and k yk

sin

J

t

³ Z dt J 0

0

where J 0 is the electric angle between rotor d-axis and stator coordinates axis at t 0 , i.e. the rotor initial position angle; Z is rotor angular velocity. When the rotor rotates at uniform speed, there is J Z t J 0 , and ( x J ) is substituted into T , then formula (1.3.3) is changed to F (T )

¦F

k

cos k ( x J )

k

¦[F

dk

k ( x) Fqk ( x)]

(1.3.4)

k

where Fdk ( x)

Fk cos kJ cos kx

Fdkm cos kx

(1.3.5)

Fqk ( x)

Fk sin kJ sin kx

Fqkm sin kx

(1.3.6)

It is seen that the amplitude of Fdk ( x) is 10

1


Fdkm

2 k ES iwK sin cos kJ kPS

Fk cos kJ

That is, the amplitude is relative to the current and position angle J , and the position of the amplitude is located at x 0 . When the current and rotor position are changed, only the amplitude of component Fdk ( x ) is changed, not its position, i.e. it is a pulsating mmf component which has no motion relative to rotor d-axis. Therefore, the mmf is called the direct axis component of the k-th harmonic mmf. Similarly, the amplitude of Fqk ( x) is Fqkm

2 kE S iwK sin sin kJ kPS

Fk sin kJ

which has the same features for the rotor q-axis, hence it is called the quadrature axis component of the k-th harmonic mmf.

1.3.2

Air-Gap Magnetic Field Produced by Single Coil’s Current

Based on the air-gap mmf and air-gap permeance coefficient, the air-gap magnetic field can be obtained. The air-gap magnetic field has both integral harmonics and fractional harmonics, because the harmonic orders of the air-gap 1 2 3 mmf are k , , ," , and besides the constant component O0 / 2 , the even P P P harmonics exist in the air-gap permeance coefficient OG ( x) . As mentioned above, the air-gap magnetic fields produced by the same amplitude mmf of d- and q-axes of the salient pole synchronous machine are different, so the d- and q-axes flux densities produced by the d- and q-axes mmf should be discussed, respectively. For example, the air-gap flux density produced by the direct axis component Fdk ( x) of k -th harmonic mmf is Bd ( x)

Fdk ( x)OG ( x) §O Fdkm cos kx ¨ 0 © 2 l O Fdkm ® 0 cos kx l ¯2

·

f

¦

O2l cos 2lx ¸ ¹

1,2,3," f

¦

1,2,3,"

½ [cos(2l k ) x cos(2l k ) x]¾ 2 ¿

O2l

i.e. the direct axis k -th harmonic mmf will produce many different harmonics of air-gap flux density. Through Fourier analysis, it is seen that the amplitude Bdkjm of j -th air-gap flux density produced by the direct axis component Fdk ( x ) of k -th harmonic mmf is 11

AC Machine Systems

Bdkjm

2 PS Bd ( x) cos jxdx PS ³ 0 PS O O 2 ½ Fdkm ³ ® 0 cos kx ¦ 2l [cos(2l k ) x cos(2l k ) x]¾ cos jxdx 0 2 PS l ¯ ¿ (1.3.7)

where l 1, 2," , f. When | 2l k |z j and k z j , the integral result of the above formula is zero; and when | 2l r k | j (i.e., 2l | k r j | ) and k z j , the integral result of the above formula is Bdkjm

1 Fdkm u (O2l 2l |k j| O2l 2 1 Fdkm u (O|k j| O|k j| ) 2 Fdkm Odkj j | 2l r k |

2l |k j|

)

in which

Odkj

1 (O|k j| O|k j| ) 2

| k r j | 2l

(1.3.8)

which is called the harmonic permeance coefficient of j -th harmonic flux density produced by the d-axis k - th harmonic mmf. Here,

O|k r j|

O2l

l 1, 2,3,"

When k j and 2l k j , i.e., l k (N.B. the term corresponding to 2l k j 0 does not exist, since l does not include zero ( l z 0 )), the integral result of formula (1.3.7) is Bdkkm

1 Fdkm u (O0 O2 k ) 2

Fdkm Odkk

where

Odkk

1 (O0 O2 k ) 2

(1.3.9)

is harmonic permeance coefficient of k -th harmonic flux density produced by the d-axis k -th harmonic mmf. After comparing formula (1.3.8) with (1.3.9), it is known that formula (1.3.9) can be obtained through substituting k j into formula (1.3.8). Hence, whether k j or k z j , if only 2l | k r j | , l 0,1, 2," , the direct axis harmonic permeance coefficient can be calculated by formula (1.3.8). 12

1


Similarly, the j -th harmonic flux density produced by the quadrature axis k - th harmonic mmf and harmonic permeance coefficient Oqkj are obtained as follows: Bqkjm

Oqkj

Fqkm Oqkj 1 (O|k j| O|k j| ) 2

(1.3.10)

where

O|k r j|

O2l

l

0,1, 2, "

Based on the above results, we can observe the following facts: (i) Simple relation exists between the amplitude of k -th harmonic mmf and the amplitude of j -th harmonic flux density produced by it, i.e., Bdkjm

Fdkm Odkj ,

Bqkjm

Fqkm Oqkj

Furthermore, simple relation exists between the air-gap harmonic permeance coefficient Odkj and Oqkj as well as the air-gap permeance coefficient O2l ( l 0,1, 2," ). Hence, Odkj and Oqkj are obtained easily according to O2l . The Odkj and

Oqkj are not changed with the rotor position angle since O2l is constant which is independent of rotor position. (ii) In salient pole synchronous machine, the order j of harmonic flux density produced by k -th harmonic mmf and the order k of harmonic mmf have some relations, i.e., j | k r 2l | (l 0,1, 2,") . For example, when k 1 , then 1 1 9 11 19 , then j , , , ," j 1,3,5, 7," ; when k 5 5 5 5 5 Therefore, definite relation exists between the subscripts k and j which represent the harmonic order in Odkj and Oqkj , i.e., | k r j | 2l (l 0,1, 2,") . If | k r j |z 2l , then both Odkj and Oqkj are zero. (iii) Based on formulas (1.3.8) and (1.3.9), it is seen that the harmonic permeance coefficients are reversible, i.e.,

Odkj Oqkj

1.3.3

Odjk ½° ¾ Oqjk °¿

(1.3.11)

Self Inductance of Single Coil

As mentioned above, according to air-gap harmonic permeance coefficient and 13

AC Machine Systems

direct axis and quadrature axis k -th harmonic mmf produced by the current of single coil AAc , the direct axis and quadrature axis j -th harmonic flux densities produced by the corresponding mmfs can be obtained, i.e., Bdkjm

Fdkm Odkj

Bqkjm

Fqkm Oqkj

From the direct axis air-gap flux density, the self flux-linkage of coil AAc can be obtained as follows:

\ dkj

s2

wK ³ Bdkjm cos jxds s1

§ x · Wl d ¨ Rl ¸ dx , R is the stator inner radius, l is stator iron length, ©P ¹ S W is pole pitch. Hence,

where ds

wKW l J E2S ES Bdkjm cos jxdx S ³ J 2

\ dkj

4 wK2 W l k yk k yj Odkj i cos kJ cos jJ PS kj

(1.3.12)

Similarly, the self flux-linkage of coil AAc corresponding to the quadrature axis air-gap flux density is wKW l J E2S ES Bqkjm sin jxdx S ³ J 2

\ qkj

4 wK2 W l k yk k yj Oqkj i sin kJ sin jJ PS kj

(1.3.13)

Self inductances of the coil corresponding to the direct axis and quadrature axis flux densities are Ldkj

Lqkj

\ dkj i

\ qkj i

4 wK2 W l k yk k yj Odkj cos kJ cos jJ PS kj

(1.3.14)

4 wK2 W l k yk k yj Oqkj sin kJ sin jJ PS kj

(1.3.15)

Accordingly, based on the total of every harmonic flux density produced by every harmonic mmf, the self inductance of coil AAc by total air-gap magnetic field is 14

1

LG


¦¦ ( L

dkj

k

Lqkj )

j

k yk k yj 4 wK2 W l (Odkj cos kJ cos jJ Oqkj sin kJ sin jJ ) ¦¦ PS k j kj where k is the harmonic order of mmf, k of flux density, j | k r 2l |, l

(1.3.16)

1 2 3 , , ," ; j is the harmonic order P P P

0,1, 2,"

From formula (1.3.16), it is seen that the self inductance of single coil is a double series which represents the total effects of every harmonic, in which fractional harmonics and lower harmonics are so strong as to compare with the fundamental wave. If we calculate only the effect of the fundamental magnetic field produced by fundamental mmf, then there may be large error that is not permitted. For example, for a 550kW salient pole synchronous machine (its polepair number is P 6 ), the self inductance of single coil (constant part L0 ) is 0.106 u 102 H, but if we consider only the fundamental wave, the self inductance is only 0.884 u 104 H. It is obvious that the harmonic mmf and harmonic flux density produced by the single coil current are very strong and need to be calculated. The formula (1.3.16) can be changed to LG

k yk k yj 2 wK2 W l [(Odkj Oqkj ) cos(k j )J (Odkj Oqkj ) cos( k j )J )] ¦¦ PS k j kj (1.3.17)

Owing to | k r j | 2l , l

0,1, 2," , the above formula can be written as

LG (J )

L0G L2 cos 2J L4 cos 4J "

(1.3.18)

The above formula indicates that when the rotor position angle varies S electric radians, the self inductance of single coil will repeat once. Only the self inductance produced by the air-gap magnetic field is discussed above. After calculating the self inductance L0l produced by slot leakage flux and end leakage flux (for leakage inductances, please see Section 1.8), the self inductance of the coil AAc is L(J )

L0 L2 cos 2J L4 cos 4J "

(1.3.19)

where L0

L0l L0G 2 º 2 wK2 W l ª§ k yk · « » ( O O ) L0l ¨ ¸ ¦ dkk qkk P S 2 k «© k ¹ »¼ ¬

(1.3.20) 15

AC Machine Systems

L2

º 2wK2 W l ° ª k yk k y (2 k ) (Odk (2 k ) Oqk (2 k ) ) » 2 ®¦ « PS ¯° k ¬ k (2 k ) ¼ ª k yk k y ( k 2) º ½° 2¦ « (Odk ( k 2) Oqk ( k 2) ) » ¾ k ¬ k ( k 2) ¼ ¿°

In the summation symbol

¦ of

(1.3.21)

formula(1.3.20) and the second summation

k

1 2 3 , , ," , and in the first summation P P P k 1 2 2P 1 symbol ¦ of formula(1.3.21) there exists k , ," , . P P P k From formula (1.3.8) and formula (1.3.10), we can get

symbol

¦ of formula (1.3.21), there is k

Odkk Oqkk

O0

(1.3.22)

Odk (2 k ) Oqk (2 k )

O2

(1.3.23)

Odk ( k 2) Oqk ( k 2)

O2

(1.3.24)

Hence, formula (1.3.20) and formula (1.3.21) are simplified further as

1.3.4

L0

2 w2 W l O L0l K 2 0 PS

L2

2wK2 W l O2 PS 2

§k · ¦k ¨ kyk ¸ © ¹

2

ª k yk k y ( k 2) º °½ ° ª k yk k y (2 k ) 2¦ « ®¦ « »¾ k ¬ k ( k 2) ¼ ¿ °¯ k ¬ k (2 k ) °

(1.3.25)

(1.3.26)

Mutual Inductance Between Coils

Calculating the mutual inductance between coils of stator winding is similar to the calculation of self inductance, and the only difference is that the angle D of the two coils should be considered. If the axis of coil BBc lags behind the axis of coil AAc by D electric angle, then in the process of calculating the mutual fluxlinkage of coil BBc caused by the current of coil AAc , the upper limit and lower ES ES and J D , respectively, i.e., limit of the integral should be J D 2 2

16

1


wKW l J D ES ES Bdkjm cos jxdx S ³ J D

\ dkjAB

4wK2 W l k yk k yj Odkj i cos kJ cos j (J D ) PS2 kj

(1.3.27)

wKW l J D ES ES Bqkjm sin jxdx S ³ J D

\ qkjAB

4 wK2 W l k yk k yj Oqkj i sin kJ sin j (J D ) PS2 kj

(1.3.28)

The corresponding mutual inductances are M dkjAB

4 wK2 W l k yk k yj Odkj cos kJ cos j (J D ) PS2 kj

(1.3.29)

M qkjAB

4 wK2 W l k yk k yj Oqkj sin kJ sin j (J D ) PS2 kj

(1.3.30)

Considering the total effects of every harmonic flux density produced by every harmonic mmf, the mutual inductance between the single coil AAc and coil BBc owing to air-gap magnetic field is M ABG

¦¦ (M k

dkjAB

M qkjAB )

j

k yk k yj 2 wK2 W l ^(Odkj Oqkj ) cos[(k j)J jD ] ¦¦ PS k j kj (Odkj Oqkj ) cos[(k j )J jD ]`

(1.3.31)

1 2 3 , , ,"; j | k r 2l |; l 0,1, 2," P P P After spreading formula (1.3.31), use the following relations:

where k

M dkjAB M qkjAB M djkAB M qjkAB 4wK2 W l k yk k yj ª ( j k )D D ·º § cos «(k j ) ¨ J ¸ » ®(Odkj Oqkj ) cos PS kj ¯ 2 2 ¹¼ © ¬ ( j k )D D ·º ½ ª § cos «(k j ) ¨ J ¸ » ¾ (Odkj Oqkj ) cos 2 2 ¹¼ ¿ © ¬

(1.3.32)

17

AC Machine Systems 2

M dkkAB M qkkAB

2wK2 W l § k yk · ¨ ¸ ^(Odkk Oqkk ) cos kD P S © k ¹

D ·½ § (Odkk Oqkk ) cos 2k ¨ J ¸ ¾ 2 ¹¿ ©

(1.3.33)

where k and j are the certain values, the multiplier (k j ) or ( k j ) before the

D· § angle ¨ J ¸ is even only, and there is also 2¹ © Odkk Oqkk

1 1 (O0 O2 k ) (O0 O2 k ) 2 2

O2 k

Only when k is a positive integer, O2k has its value, otherwise O2 k 0 . The formula (1.3.31) can be written as even harmonic cosine series of the D· § angle ¨ J ¸ , i.e., 2¹ © M ABG

D· D· § § M AB 0G M AB 2 cos 2 ¨ J ¸ M AB 4 cos 4 ¨ J ¸ " 2¹ 2¹ © ©

(1.3.34)

When the axis of the rotor pole is located at the center of the two coils, i.e.,

J

D

0 , the mutual inductance reaches its extremum. 2 After considering the mutual inductance M AB 0l caused by the leakage flux, the mutual inductance between coils AAc and BBc is M AB

D· D· § § M AB 0 M AB 2 cos 2 ¨ J ¸ M AB 4 cos 4 ¨ J ¸ " 2¹ 2¹ © ©

(1.3.35)

where M AB 0

M AB 2

M AB 0l

2 º 2 wK2 W l ª§ k yk · « ¸ (Odkk Oqkk ) cos(kD ) » 2 ¦ ¨ Pʌ k «© k ¹ »¼ ¬

(1.3.36)

º 2 wK2 W l ° ª k yk k y (2 k ) (Odk (2 k ) Oqk (2 k ) ) cos(1 k )D » 2 ®¦ « PS °¯ k ¬ k (2 k ) ¼ ª k yk k y ( k 2) º °½ 2¦ « (Odk ( k 2) Oqk ( k 2) ) cos(1 k )D » ¾ k ¬ k ( k 2) ¼ °¿

(1.3.37)

Based on formulas (1.3.22), (1.3.23), and (1.3.24), the above formula can be 18

1


simplified further as 2 w2 W l O K 2 0 PS

M AB 0

M AB 0l

M AB 2

2 wK2 W l O2 PS2 2¦ k

ª§ k yk ·2 º « » k cos D ¨ ¸ ¦k « k »¼ ¹ ¬©

(1.3.38)

° ª k yk k y (2 k ) cos(1 k )D ®¦ « ¯° K ¬ k (2 k )

º ½° cos(1 k )D » ¾ k (k 2) ¼ ¿°

k yk k y ( k 2)

(1.3.39)

It is needed to point out that when the axes of coils BBc and AAc coincide with each other, i.e. D 0 , the formulas (1.3.38) and (1.3.39) become formulas (1.3.25) and (1.3.26), which accords with the physical concept. In addition, in formulas (1.3.19) and (1.3.35), the stator reference axis is the axis of coil AAc . If the reference axis is another axis, and the axis of coil AAc lags behind the new axis by D a electric angle, then J in the formulas should be replaced by ( J D a ). In the mutual inductance between two coils, the influence of fractional harmonics and low order harmonics of the air-gap magnetic field is very strong. For example, in the 550kW salient-pole synchronous machine mentioned above, its mutual inductance between two adjacent coils (merely counting the constant value part which does not vary with the rotor position) is 0.962 u 103 H ; but if only the fundamental wave is considered, then this mutual inductance will be 0.830 u 10 4 H only, the difference between the two values is very large, which is produced by very strong fractional harmonics and low order harmonics of the air-gap magnetic field. Finally, it should be pointed out that, for calculating self inductance and mutual inductance according to formulas (1.3.19) and (1.3.35), generally, taking only two items (constant item and second harmonic item) will satisfy the precision demand.

1.4

Inductances of Stator Loops

In practice, it is often needed to research some problems related to the electric machine loops. But, the constitution of the electric machine loop is different from the researched problems. Because the electric machine stator loops are composed of related coils, it is very easy to calculate the inductances of every loop based on the inductances of associated coils. At first, let us study the simple loop composed of two coils a1 , a2 as in Fig. 1.4.1. Suppose that La1 , La2 are self inductances of coils a1 and a2 , M a1a2 is mutual 19

AC Machine Systems

inductance between the two coils, then, in terms of the basic definition of the inductance, it is known that the self inductance of the whole loop-a (branch) is La

\a

\ a \ a

i

i

1

2

The flux-linkages of coils a1 and a2 are, respectively

\a

\ a a \ a a

\a

\ a a \ a a

1

2

1 1

1 2

2 1

2 2

where \ a1a1 ,\ a2 a2 are self flux-linkages, \ a1a2 or \ a2 a1 is the mutual flux-linkage of the two coils. Hence,

\ a a \ a a \ a a \ a a

La

1 1

1 2

2 1

i

Figure 1.4.1 Self inductance of the circuit composed of coils a1 and a2

2 2

La1 La2 2M a1a2

Figure 1.4.2 Mutual inductance of two loops a and b

Similarly, if the stator loop a is composed of two coils a1 , a2 , stator loop b is composed of coils b1 , b2 as shown in Fig. 1.4.2, then the mutual inductance of loops a and b is formed by the mutual inductances of these coils, i.e., M ab

M a1b1 M a2b1 M a1b2 M a2b2

If M S ( i )Q ( j ) represents the mutual inductance between i -th coil of loop S and j -th coil of loop Q of the stator, then in terms of formula (1.3.35), the mutual inductance between loops S and Q will be m

M SQ

n

¦¦ M

S (i )Q ( j )

i 1 j 1

M SQ 0 M SQ 2 cos 2(J D SQ 2 ) M SQ 4 cos 4(J D SQ 4 ) " 20

(1.4.1)

1


where m is the coil number of loop S, n is coil number of loop Q. As is the same as formulas (1.3.19) and (1.3.35), for formula (1.4.1), generally, taking only the first two items will satisfy the demand of engineering calculation. If the constant item of mutual inductance between i -th coil of loop S and j -th coil of loop Q is denoted by M S ( i ) Q ( j )0 , then the constant item of formula (1.4.1) can be calculated by m

M SQ 0

n

¦¦ M

(1.4.2)

S ( i ) Q ( j )0

i 1 j 1

As for the amplitude M SQ 2 and phase angle D SQ 2 of the second harmonic item in formula (1.4.1), it can be calculated in terms of the following approach. Suppose that the amplitude and phase angle of the second harmonic item of mutual inductance between i-th coil of loop S and j -th coil of loop Q is denoted by M S ( i ) Q ( j )2 and D S (i ) Q ( j )2 , respectively (the phase angle equals half of the included angle between the axes of these two coils), then m

M SQ 2 cos 2(J D SQ 2 )

n

¦¦ M

S ( i ) Q ( j )2

cos 2(J D S (i ) Q ( j )2 )

(1.4.3)

i 1 j 1

When 2J

0 , then m

M SQ 2 cos 2D SQ 2

n

¦¦ M

S ( i ) Q ( j )2

cos 2D S ( i )Q ( j )2

(1.4.4)

sin 2D S ( i )Q ( j )2

(1.4.5)

i 1 j 1

When 2J

S , then there is 2 m

M SQ 2 sin 2D SQ 2

n

¦¦ M

S ( i ) Q ( j )2

i 1 j 1

M SQ 2 and tan 2D SQ 2 can be calculated from formulas (1.4.4) and (1.4.5), then

D SQ 2 will be obtained. The effect of the harmonics of air-gap magnetic field on the inductances of stator branches is also very strong, especially for fractional harmonics and low order harmonics. In the same example of the 550kW salient pole synchronous machine, the self inductance of a branch and mutual inductance of adjacent branches in the same phase (merely counting the constant part independent of the rotor position) are 0.185 u 101 H and 0.134 u 103 H , respectively, but if only the air-gap fundamental magnetic field is considered, then the two inductances are 0.293 u 102 H . It is obvious that the effect of harmonics on the inductances of stator branches is very strong. 21

AC Machine Systems

It can be seen from formulas (1.3.19), (1.3.35) and (1.4.1) that the self inductance and mutual inductance of stator single coil and the inductances of stator loops are relative to the rotor position angle J , i.e. they are time-variant parameters. For non-salient pole electric machine and induction machine, there is OG ( x)

O0 2

, i.e. the above inductances are constants independent of rotor

position. Finally, it should be stated that in most instances the phase winding can be considered as a researched unit. In anordinary designed electric machine, the distributed and fractional-pitch winding is adopted generally in the phase winding, hence the harmonics of the air-gap magnetic field are weakened greatly. Please refer to details related to the problem in Section 2.1. When the inductances are calculated concretely in terms of formulas (1.3.25), (1.3.26), (1.3.38), and (1.3.39), it is needed to sum the infinite series, and it is enough to calculate the finite terms by the ordinary method. But, for the inductances of single coil, only calculating the several terms of the series is not sufficient owing to strong harmonic effect of air-gap magnetic field. The relation between the inductance M 0 ( j ) of stator single coil of a 550kW salient pole electric machine and the calculated highest harmonic order k is listed in Table 1.4.1, in which M 0 ( j ) represents the constant item of the inductance between coil j and the first coil owing to air-gap magnetic field. The stator coil pitch of the electric machine is 8, and in the table the constant items of inductances M 0 ( j ) are listed at k 7, 27,127 . It can be seen that taking 27 as the calculated highest order harmonic is sufficient for calculating single coil inductances, of course, much precise result can be obtained by taking higher harmonic order, but the effect is not obvious already. Table 1.4.1 Relation between M 0 ( j ) of a 550 kW salient-pole electric machine and harmonic order k j

M0 ( j) / H

1

2

3

4

5

6

7

7

1.059

0.962

0.794

0.657

0.505

0.353

0.214

27

1.089

0.950

0.802

0.653

0.505

0.357

0.208

127

1.096

0.950

0.802

0.653

0.505

0.356

0.208

12

30

50

k

j

M0 ( j) / H

8

k

22

9

10

11

7

0.052 4

0.068 7 0.094 2 0.086 3 0.089 2 0.088 5 0.082 8

27

0.059 6

0.083 7 0.089 0 0.088 7 0.088 6 0.088 5 0.088 0

127

0.059 7

0.087 6 0.088 8 0.088 7 0.088 6 0.088 6 0.088 0

1

1.5


Inductances of Rotor Loops

Besides excitation winding, normally the damper winding is located on the rotor of the salient pole synchronous machine. Their inductances will be analysed in the following section.

1.5.1

Inductance of Excitation Winding

In general, the excitation winding on each pole is concentrated winding, the windings on each pole may be in series or in parallel connection. The inductance of excitation winding is discussed here. When other connection is adopted for excitation winding or when internal fault occurs in it, its inductance can be obtained by a similar method, the details are not repeated here. The self inductance of excitation winding consists of two parts, i.e., L fd

L fdG L fdl

(1.5.1)

where L fdl is the inductance corresponding to end region leakage flux, leakage flux between rotor poles and leakage flux of pole surface; L f d G is the inductance corresponding to air-gap flux density. The mmf produced by the current of excitation winding is rectangular wave, the excitation mmf F ( x) and the corresponding air-gap flux density B ( x) are drawn in Fig. 1.5.1. Suppose that the turn number of excitation winding on each pole is w fd , the parallel branch number of excitation winding is a fd , total excitation current is i fd , then the branch current of excitation winding will be

i fd a fd

. Each pole excitation mmf is Ff

w fd

i fd a fd

, and it can be considered as

Figure 1.5.1 Excitation mmf and its air-gap flux density

23

AC Machine Systems

rectangular wave with a width of S radians approximately. Here, the air-gap flux density is

B f ( x)

P 0 Ff G ( x)

i fd

P 0 w fd

1 ( a fd G x)

Wl

i fd

(1.5.2)

Each pole flux is

)fd

Wl

S³

S S

B f ( x)dx

w fd

S

a fd

³

S S

P dx G ( x)

From formula (1.2.4), we can see

)fd

W lw fd i fd 2

a fd

O0

Each pole flux-linkage is

\ fd

w fd)fd

Wl 2

w2fd

i fd a fd

O0

Therefore, the self inductance of excitation winding corresponding to air-gap flux density is L fdG

1.5.2

2 P \ fd a 2fd i fd a fd

W lP a 2fd

w2fd O0

(1.5.3)

Inductance of Damper Winding

Normally, the damper winding of the salient pole synchronous machine is cage type with damping bars and damping rings; the damping bars are short-circuited by damping rings fully or partially, and it is usually symmetric to the direct axis or quadrature axis. In such cage circuit, the selection of current loop may be arbitrary. Figure 1.5.2 represents damper winding with 4 damping bars per pole; in the figure the selected damping loops are symmetric to the direct axis or quadrature axis, and loops 1d , 2d are symmetric to direct axis, loops 1q, 2q are symmetric to quadrature axis. Actual currents in the damping bars are algebraic sum of the corresponding direct and quadrature axis loop currents. When only the space fundamental of the air-gap magnetic field is considered, obvious merits exist for such selected damping loops as follows: since the space angle difference S electric radians, and in between direct and quadrature axis current loops is 2 24

1


structure the rotor is symmetric to direct axis and quadrature axis, mutual inductance and mutual resistance do not exist between the direct and quadrature axis current loops, then the actual damping winding can be considered as two imaginary damping winding groups shown in Fig. 1.5.3, namely direct axis damping winding and quadrature axis damping winding.

Figure 1.5.2

Damping winding of salient pole synchronous machine

Figure 1.5.3 Direct axis and quadrature axis damping winding of salient pole synchronous machine

In practical salient pole synchronous machine, there are many damping bars, accordingly, there are many direct axis damping windings 1d , 2d ," , nd and many quadrature axis damping windings 1q, 2q," , nq. Therefore, normally two methods 25

AC Machine Systems

are adopted for the analysis and calculation of damping winding; one is to calculate the parameters of each damping winding in terms of practical distribution of damping windings, the other is to replace all damping winding loops by a direct axis and a quadrature axis equivalent damping windings. Respective merits and shortcomings exist for these two methods: using the former the practical instance of every damping bar can be obtained, but the workload for calculating it is greater; using the latter the analysis can be simplified when the internal status of the damping winding does not need to be researched, but its equivalent parameters are difficult to be calculated and the error of calculation is difficult to be avoided. For more general instance, when strong space harmonics of air-gap magnetic field exist or internal fault of damping winding occurs, the above direct axis and quadrature axis damping windings will be relative to each other. In such an instance, it is more suitable to select damping loops based on Fig. 1.5.4. The inductances of such damping loops are discussed in the following. As for the parameter calculation relative to d-axis and q-axis damping windings or equivalent damping loops, the details are available in various books and hence do not need to be discussed here.

Figure 1.5.4 Selection of damping loops of salient-pole synchronous machine

Referring to Fig. 1.5.5, assume the damping loop 11c to lead ahead of d-axis by D1 electric angle; when a current i flows into the damping loop a rectangular wave mmf appears, using the Fourier analysis we can obtain F ( x1 )

¦a

n

n

cos

n x1 P

n 1, 2,"

(1.5.4)

Figure 1.5.5 Model for calculating inductances of salient pole synchronous machine 26

1


nE S 2 wr i sin 1 , wr is the turn number of the damping loop (normally 2P nS wr 1 ), E1 is short pitch ratio of the damping loop; x1 is the abscissa on the rotor, its origin is located on the axis of damping loop 11c . Formula (1.5.4) contains only cosine items owing to the symmetry of rectangular wave mmf to the axis of damping loop. For the convenience of representation, the wave with 2S electric radians is considered as the fundamental wave, in this instance formula (1.5.4) should be rewritten as 1 2 3 F ( x) ¦ Fk cos kx1 k (1.5.5) , , ," P P P k where 2E S 2 Fk wr i sin 1 kPS where an

When the mmf formula is transformed to the abscissa x in which the origin is on d-axis, then the corresponding mmf expression is F c( x)

¦F

k

cos k ( x D1 )

k

k

1 2 3 , , ," P P P

(1.5.6)

The air-gap flux density produced by the mmf is BG ( x)

F c( x)OG ( x)

¦F

k

k

ªO º cos k ( x D1 ) « 0 ¦ O2l cos 2lx » l ¬2 ¼

l 1, 2,"

(1.5.7)

in which the air-gap flux density caused by k - th harmonic mmf is B( x)

ªO º Fk cos k ( x D1 ) « 0 ¦ O2l cos 2lx » 2 l ¬ ¼ Fk

O0 2

O2l

cos k ( x D 1 ) ¦ Fk

2

l

cos[(k 2l ) x kD1 ]`

^cos[(k 2l ) x kD1 ]

l 1, 2,"

(1.5.8)

From the above formula, the k - th harmonic flux density in it is Bkk ( x)

Fk

O0 2

cos k ( x D1 ) Fk

Fk cos kD1

O0 O2 k

O2 k 2

cos k ( x D1 )

cos kx Fk sin kD1

O0 O2 k

2 2 Fk cos kD1Odkk cos kx Fk sin kD1Oqkk sin kx

sin kx

(1.5.9) 27

AC Machine Systems

The j - th harmonic flux density is Bkj ( x)

Fk

O|k j| 2

cos( jx kD1 ) Fk

Fk cos kD1

O|k j| O|k j|

O|k j| 2

cos( jx kD1 )

cos jx Fk sin kD1

O|k j| O|k j|

2 Fk cos kD1Odkj cos jx Fk sin kD1Oqkj sin jx

2

sin jx (1.5.10)

In the above formula, | k j | and | k j | are obtained in such a way: In formula (1-5-8), when j k r 2l , j k r2l , so there is 2l | k j | ; when j k 2l (since k and 2l are positive, j cannot equal k 2l ), then 2l | k j | . If according to formula (1.5.6), the k - th harmonic mmf is decomposed into direct axis and quadrature axis mmf s firstly, then k - th and j - th harmonic flux densities produced by the mmfs are obtained respectively, its result is the same as formulas (1.5.9) and (1.5.10). From formulas (1.5.9) and (1.5.10), it is known that j - th air-gap harmonic flux density is B j ( x)

Fj

O0 2

cos j ( x D1 )

¦

¦

Fk

O2l

2l |k j |

Fk

2l |k j |

O2l 2

cos( jx kD1 )

2

cos( jx kD1 )

2l

2, 4,"

After arranging the above formula we get B j ( x)

¦

2l |k j |

Fk

O2l 2

cos( jx kD1 )

¦

2l |k j |

Fk

O2l 2

cos( jx kD1 )

(1.5.11)

1 2 3 , , ,";| k j | 0, 2, 4,";| k j | 2, 4," P P P Suppose D1 0 in formula (1.5.11), the formula will be simplified as

where k

B j ( x)

ª O2l O2l º « Fk 2 2l |k j| Fk 2 2l |k j| » cos jx ¬ ¼ Fk Odkj cos jx | k j | 0, 2, 4,"; | k j | 2, 4,"

Here, only the d-axis mmf exists, i.e. the above formula is the expression of j - th harmonic flux density produced by d-axis k - th harmonic mmf when the axis of the damping loop11c and the rotor d-axis overlap each other. 28

1


After getting the expression of the air-gap flux density, the flux-linkage of each loop is obtained, and the inductance of each loop will be gained. Firstly, the self inductance of damping loop 11c is calculated. The self fluxlinkage caused by j -th harmonic flux density B j ( x) in damping loop 11c is wrW l E1S E S B j ( x )dx1 S ³

\j

wrW l E1S D1 B j ( x)dx ES S ³ D1

O 2 wr2W l kE S jE S i ® ¦ 2l sin 1 sin 1 cos( j k )D1 PS ¯ 2l |k j| kj

¦

O2l kj

2l |k j |

sin

k E1 S jE S ½ sin 1 cos( j k )D1 ¾ ¿

(1.5.12)

where | k j | 0, 2, 4,";| k j | 2, 4," . The corresponding self inductance is

O2l 2 wr2W l kE S jE S sin 1 sin 1 cos( j k )D1 ® ¦ PS ¯ 2l |k j| kj

Lj

¦

2l |k j |

O2l kj

sin

½ k E1S jE S sin 1 cos( j k )D1 ¾ ¿

(1.5.13)

where | k j | 0, 2, 4,";| k j | 2, 4," . After counting each harmonic of the air-gap magnetic field, the total self inductance of damping loop 11c is L

¦L

j

j

¦

2l |k j|

O2l 2 wr2W l kE S jE S sin 1 sin 1 cos( j k )D1 ¦® ¦ PS j ¯ 2l |k j| kj O2l kj

sin

½ k E1S jE S sin 1 cos( j k )D1 ¾ ¿

(1.5.14)

1 2 3 , , ,";| k j | 0, 2, 4,";| k j | 2, 4,"; normally, it is sufficient P P P to take the definite order of k , j , 2l . The mutual inductance between damping loops 11c and 22c is calculated in the following (see Fig. 1.5.5). Suppose current i flows into the damping loop 11c , then j -th harmonic air-gap flux density produced by it is expressed as formula (1.5.11), the flux-linkage of damping loop 22c is

where j

29

AC Machine Systems

wrW l E2 S E S B j ( x )dx2 S ³ 2

\ 12 j

wrW l E2 S D 2 B j ( x)dx E S S ³ 2 D 2

O kE S jE S 2 wr2W l i ® ¦ 2l sin 1 sin 2 cos( jD 2 kD1 ) PS ¯ 2l |k j| kj

O2l

¦

2l |k j |

kj

sin

½ k E1S jE S sin 2 cos( jD 2 kD1 ) ¾ ¿

(1.5.15)

where | k j | 0, 2, 4,";| k j | 2, 4," ; x2 is the abscissa on the rotor and its origin is on the axis of damping loop 22c ; E 2 is the short pitch ratio of damping loop 22c . Corresponding mutual inductance is M 12 j

O2l 2 wr2W l kE S jE S sin 1 sin 2 cos( jD 2 kD1 ) ® ¦ PS ¯ 2l |k j| kj

¦

O2l

2l |k j |

M 12

¦M

kj

sin


(1.5.16)

12 j

j

O2l 2wr2W l kE S jE S sin 1 sin 2 cos( jD 2 kD1 ) ¦® ¦ PS j ¯ 2l |k j| kj

¦

2l |k j |

O2l kj

sin


(1.5.17)

1 2 3 , , ,";| k j | 0, 2, 4,";| k j | 2, 4," P P P If D 2 D1 and E 2 E1 are substituted into formula (1.5.17), formula (1.5.14) is obtained. That is, self inductance is a special example of mutual inductance.

where j

1.5.3

Mutual Inductance Between Excitation Winding and Damping Loop

As mentioned above, suppose that current i fd flows into excitation winding, its mmf is rectangular wave, the air-gap flux density B ( x ) produced by it (refer to Fig. 1.5.1) is 30

1


B( x)

F ( x)OG ( x)

F ( x)

P0 G ( x)

(1.5.18)

From Fig. 1.5.1, we can see B( x)

B( x),

B( x)

B (S r x)

So, after the Fourier analysis, only cosine items exist in B ( x ) , and only odd order harmonics exist, its general expression is B( x)

¦B

fkm

cos kx

k 1,3,5,"

(1.5.19)

k

where B fkm

2 S B( x) cos kxdx S ³ P 2 S F ( x) 0 cos kxdx ³ S G ( x) 4 S Ff P0 cos kxdx S ³ G ( x)

(1.5.20)

Corresponding permeance coefficient is

Odk

B fkm Ff

4 S P 0 cos kxdx S ³ G ( x)

k 1,3,5,"

(1.5.21)

After comparing formula (1.5.21) with formula (1.2.5), it is known that the calculation formulas of air-gap permeance coefficient O2l of salient pole synchronous machine and the permeance coefficient Odk corresponding to harmonics flux density produced by rectangular wave excitation mmf are similar in form. Their only difference is that 2l in formula of O2l is even, and k in formula of Odk is odd. The flux-linkage of k - th harmonic air-gap flux density in damping loop 11c produced by the current i fd of excitation winding is

\ 1 fk

wrW l E1S D1 B fkm cos kxdx ES S ³ 1 D1 2 wr w fdW l i fd Odk kE S sin 1 cos kD1 S a fd k

(1.5.22)

The mutual inductance corresponding to k -th harmonic air-gap flux density is 31

AC Machine Systems

2wr w fdW l 1 Odk kE S sin 1 cos kD1 S a fd k

M 1 fk

(1.5.23)

Considering all the air-gap flux density harmonics, the total mutual inductance between damping loop 11c and excitation winding is M1 f

¦M

1 fk

k

2 wr w fdW l 1 S a fd

¦ k

Odk k

sin

k E1S cos kD1

k 1,3,5,"

(1.5.24)

If we suppose that the current flows into damping loop 11c firstly, then when we calculate the flux-linkage of excitation winding, the same result as formula (1.5.24) will be obtained. Finally, it should be pointed out that the inductances of rotor loops are independent of the rotor position, and formulas (1.5.14), (1.5.17), and (1.5.24) reflect this feature. This is in accord with the rule, because stator iron is round.

1.6 Mutual Inductances Between Stator Loops and Rotor Loops Inductances between stator loops and rotor loops are changed with the rotor position owing to the rotor rotation, i.e. they are time-variant coefficients. Firstly, the mutual inductance between stator single coil AAc and excitation winding is calculated. Suppose that the excitation winding connection is as in Section 1.5, the air-gap flux density produced by excitation current i fd is seen from formula (1.5.19). The flux-linkage in stator coil AAc owing to k -th harmonic air gap flux density is

\ fak

wKW l ES ES B fkm cos kxdT S ³ wKW l ES J ES B fkm cos kxdx S ³ J

where T x J . The expression of B fkm in formula (1.5.20) is substituted into the above formula, there is

\ fak

2 w fd wKW l i fd Odk kES sin cos kJ S a fd k

Corresponding mutual inductance is 32

(1.6.1)

1


2 w fd wKW l 1 Odk kES sin cos kJ S a fd k

M fak

(1.6.2)

When the total of all air-gap flux density harmonics is obtained, the total mutual inductance between stator coil AAc and excitation winding is M fa

¦M

2 w fd wKW l 1 S a fd

fak

k

¦ k

Odk k

sin

kES cos kJ

k 1,3,"

(1.6.3)

It may be seen that the mutual inductance depends upon the rotor position. Then, the mutual inductance between stator single coil AAc and damping loop 11c is analyzed. Referring to Fig. 1.5.5, suppose the current i flows into damping loop 11c , then j -th harmonic air-gap flux density produced by it is as in formula (1.5.11), and the flux-linkage relative to B j ( x) for coil AAc is

\ 1aj

wKW l ES E S B j ( x)dT S ³ wKW l ES J E S B j ( x )dx S ³ J

O kE S 2 wK wrW l jE S i ® ¦ 2l sin 1 sin cos( jJ kD1 ) PS ¯ 2l |k j| kj

¦

O2l

2l |k j|

kj

sin

½ k E1 S jE S sin cos( jJ kD1 ) ¾ ¿

(1.6.4)

where | k j | 0, 2, 4,";| k j | 2, 4," Considering all the air-gap flux harmonics, the total mutual inductance between stator single coil AAc and damping loop 11c is obtained as M 1a

O2l 2wK wrW l kE S jE S sin 1 sin cos( jJ kD1 ) ® ¦ ¦ PS kj j ¯ 2l |k j |

¦

2l |k j |

O2l kj

sin

½ k E1S jE S sin cos( jJ kD1 ) ¾ ¿

(1.6.5)

where | k j | 0, 2, 4,"; | k j | 2, 4," 1 2 3 , , ," In formulas (1.6.4) and (1.6.5), j P P P In formulas (1.6.4) and (1.6.5), the mutual inductance is calculated by the current flowing into the rotor. If it is calculated by the current flowing into the stator, the result is the same. 33

AC Machine Systems

Comparing formula (1.6.5) with (1.5.17), it is seen that the difference of both is to make the following substitution: wK o wr ,

E o E 2 , J o D 2

But, the difference of the two formulas in meaning is obvious: the result of formula (1.5.17) is constant, i.e. the mutual inductances between damper loops are not relative to rotor position; and the result of formula (1.6.5) is function of the angle J , i.e. the mutual inductances between damper loop and stator coil is relative to rotor position. After obtaining the mutual inductances between stator single coil and excitation winding or damping loop, the mutual inductance between stator loop and rotor loop consisting of them will be obtained easily. Finally, it should be pointed out that in the calculation of the self inductance of excitation winding, mutual inductance between excitation winding and damping loop, and mutual inductance between excitation winding and stator loop as in front, all excitation winding is considered as series or parallel connection. If other connections are adopted for excitation winding, or internal fault of excitation winding occurs, the excitation winding will be divided into several loops; at this time the inductances of excitation loops and mutual inductances between them and damper loops or stator loops need to be calculated, respectively. It is not discussed further here.

1.7

Influence of Saturation on Parameters

In discussing the inductances related to air-gap magnetic field in the above sections, the concept of air-gap permeance and harmonic analysis method are adopted, and it is supposed that mmfs are all spent in air-gap; as for iron reluctance, it is considered by enlarging air-gap length suitably. In the unsaturation state of electric machine, the iron section works in the linear part of the B-H curve, and the relation of excitation current and the no-load voltage of electric machine is also at linear part. Under this instance, compared with air-gap reluctance the iron reluctance is less, so it may be ignored, or air-gap is enlarged in terms of the ratio between iron reluctance and air-gap reluctance or according to the linear part of the no-load characteristic from experiment. Normally, an electric machine operates near saturation state, at this time the effect of iron reluctance is boosted obviously, here the influence of saturation may be considered as follows. If the electric machine operates in no-load, the saturation coefficient K may be confirmed according to the no-load characteristic of the electric machine, and the 34

1


air-gap should be enlarged to K times. As in Fig. 1.7.1, the excitation current corresponding to rating voltage U N is i f 0 ; if saturation is not considered, the excitation current is icf 0 according to air-gap line, so the saturation coefficient is K

if 0 i cf 0

df de

Figure 1.7.1 Confirmation of saturation coefficient based on no-load characteristic

Figure 1.7.2 Phasor diagram of electric machine operation on load

If the electric machine operates on load, the corresponding phasor diagram is shown in Fig. 1.7.2. Here, air-gap emf corresponding to air-gap flux density is EG . Based on the relation as in Fig. 1.7.2 and after getting EG , the saturation coefficient will be obtained in terms of EG and no-load characteristic (refer to Fig. 1.7.1) as follows: Kc

d cf c d cec

The effect of saturation is researched by using this method based on fundamental wave magnetic circuit under normal operation. If the researched problem is related 35

AC Machine Systems

to the instance with strong space harmonics of air-gap magnetic field, using this method will definitely lead to error. If needed, numerical calculation may be used to analyze the distribution of the magnetic field in terms of the loops and the magnetic field of the electric machine. When this kind of problem is researched, normally the electromagnetic field of the whole electric machine needs to be calculated. Since the air-gap of the salient pole synchronous machine is non-uniform, analysis and calculation of many electromagnetic fields are also needed. Therefore, in the instance of definite computer memory and calculation speed, sometimes it is difficult to adopt very thin mesh to reach the expected goal. Each loop parameter under saturation is calculated by the above two methods respectively on a 30 kVA simulation electric machine (6-pole salient pole synchronous machine). As an example, the mutual inductance of the first coil and j -th coil on stator is M ( j)

M 0 ( j ) M 2 ( j ) cos 2(J D ( j ))

where j 1, 2," , Z1 ( Z1 is stator slot number). When j 1 , the mutual inductance becomes self inductance. Figure 1.7.3 shows M 0 ( j ) obtained by two methods (including the inductance M 0G ( j ) related to air-gap magnetic field and leakage inductance M 0l ( j ) ). When the electromagnetic field of the whole electric machine is calculated, the cell number is taken as 3132, the node number is 1621. From the figure, we can see that the results of the two methods are the same. The pitch of the stator coil of the electric machine is 7. When j ı 8 , the overlapping part no longer exists between the j -th coil and the first coil, hence using the permeance analysis method M 0G ( j ) is constant; then adopting the numerical analysis method, the larger the j is, the less the absolute value of M 0G ( j ) is. The latter result is more rational.

Figure 1.7.3 Comparison of M 0 ( j ) by two methods 36

1


1.8 Inductances Produced by Leakage Magnetic Field Inductances produced by air-gap magnetic field (main magnetic field) are discussed in the above sections. Leakage magnetic field is not the main part of the whole magnetic field, but the leakage magnetic field should not be ignored, and it should be considered in calculating the inductances.

1.8.1

Inductance Produced by Rotor Leakage Magnetic Field

The problem has been discussed in a lot of literatures[8], here we only list the result. (1) Leakage self inductance of excitation winding of salient pole synchronous machine is Lf V

2 PP0

w2 fd a 2fd

Of l

(1.8.1)

where O f is the specific leakage permeance of excitation winding; l is rotor iron length; and § dt · §a · 0.25 ¸ 1.75 ¨ P 0.20 ¸ © cP ¹ © cP ¹

O f | 4.00 ¨

2

§h · §a · §b · 1.27 ¨ P 0.50 ¸ 1.15 ¨ m ¸ 0.44 ¨ m ¸ © l ¹ © cP ¹ © cm ¹ in which dt cP cm

bP2 4 D1 2Sdt W bP 2P ʌ (hm 2hP 2G ) W bm 2P hP G

The size of each part of the magnetic pole of the salient pole synchronous machine is shown in Fig. 1.8.1, here D1 is the inter diameter of the stator iron; G is air-gap length; hP and bP are the height and width of the pole shoes respectively; hm and bm are the height and width of the pole body, respectively. (2) Leakage self inductance of damping winding of the salient pole synchronous machine is Lc

P 0lc (Oc Ot ) 37

AC Machine Systems

where lc is rotor pole length; Oc is the specific slot leakage permeance; Ot is the specific tooth top leakage permeance; and

Oc

0.30 0.64

1 bs / d c hs 1 bs / d c bs

(for circular bar)

2

Ot

§ 2G · 1 ¨ ¸ 1 2G 2G G © bs ¹ ln arctan ʌ 2 ʌbs bs bs

The size of each part is referred to in Fig. 1.8.2.

Figure 1.8.1 Size of each part of magnetic pole of salient pole synchronous machine

Figure1.8.2 Size of damping bar hole of salient pole synchronous machine

(3) Leakage inductance of each segment of the end ring of the salient pole synchronous machine is Le

· t2 P 0 § 2le 0.25 ¸ ¨ ln 2 ʌ © re ¹

where t2 is length of each segment of the end ring; le is the interval between the end ring and rotor end surface; re is the equivalent radius of the cross-section of the end ring.

1.8.2

Inductances Produced by Stator Leakage Magnetic Field

In analyzing the inductance of each stator loop produced by the air-gap magnetic field, the harmonic magnetic field is considered already, so the effect of stator differential leakage reactance is included in this inductance. The inductances produced by slot leakage and end part leakage are discussed below: (1) Inductance produced by slot leakage flux With regard to inductance produced by slot leakage flux, detailed discussion and well known calculation method already exist. If the leakage flux is not in 38

1


saturation state and the reluctance of iron part is ignored, then the mutual inductance produced by slot leakage flux exists only between the coil bars in the same slot, and the mutual inductance produced by slot leakage flux does not exist between the coil bars in different slots. After knowing the connection of stator winding, the inductance of each stator loop produced by slot leakage flux may be calculated by adopting the method of searching coil bars in the same slots one by one. It is very convenient to do this using computer. (2) Inductance produced by end part leakage flux It is very difficult to calculate exactly the inductance produced by end part leakage flux; relatively rigorous method of calculating the end part leakage flux is numeration of the end region magnetic field. Under the state of 3-phase symmetry, the 3-phase currents are treated as equivalent current sheet. But, this method is not suitable for the inductance of stator single coil, discrete mode may be used to deal with it. Concrete method is as follows. When calculating the end region magnetic field produced by single coil current, the coil end part is divided into several current elements, and the magnetic field produced by the current elements is calculated in terms of Biot-Savart law. The magnetic field on each point of end region is the superposition of the magnetic field produced by all current elements. After getting the magnetic field, the flux-linkage of the end part of each coil can be calculated, and then the inductance produced by end region leakage flux will be obtained. In the calculation, the effect of stator iron is replaced by image current, the effect of air-gap is replaced by air-gap current; now the end region magnetic field problem produced by a single coil current becomes the problem of the magnetic field produced by coil end part current, air-gap current and their image currents altogether in uniform medium. In the calculation process, both cartesian coordinates system and cylindrical coordinates system may be adopted. The process of calculating the inductance produced by stator end region leakage flux can be found in Appendix C. The calculation results of end part inductance of stator single coil of a 630kW salient pole synchronous machine are listed in Table 1.8.1. From the table we can Table 1.8.1 Comparison of calculated results of end part inductances in two coordinates systems unit: H Slot number between two coils 0 1 2 3 4 5

Calculated results in Cartesian coordinates

0.914 u 104 0.311 u 104 0.206 u 104 0.550 u 105 0.395 u 105 0.466 u 105

Calculated results in Cylindrical coordinates

0.895 u 104 0.298 u 104 0.198 u 104 0.513 u 105 0.380 u 105 0.485 u 105 39

AC Machine Systems

see that difference between the calculated results of the two coordinates systems is not obvious. In the example, generally absolute error does not exceed 0.19u105 H , relative error is below 7%.

1.9

Electromagnetic Torque and Rotor Motion Equation of AC Machines

Electromagnetic torque and rotor motion equation are important foundation for analyzing the electric machine operation performance. The electromagnetic torque of salient pole synchronous machine is discussed[9] here. Suppose that the stator winding has m loops, the m-th loop in stator is represented by sm ; the excitation winding has n loops, the n -th loop of excitation winding is represented by fd n ; the damper winding has l loops, the l -th loop of damper winding is represented by l ; then the total magnetic field energy of the electric machine is 1 (\ s1 is1 \ s2 is2 " \ sm ism \ fd1 i fd1 \ fd2 i fd2 " 2 \ fdn i fdn \ 1i1 \ 2 i2 " \ l il )

Wm

(1.9.1)

where the flux-linkages of various loops are, respectively:

\s

Ls1 is1 " M s1sm ism M s1 fd1 i fd1 " M s1 fdn i fdn

1

M s11i1 " M s1l il

"

\ fd

1

(1.9.2)

M fd1s1 is1 " M fd1sm ism L fd1 i fd1 " M fd1 fdn i fdn M fd11i1 " M fd1l il

"

\1

(1.9.3)

M 1s1 is1 " M 1sm ism M 1 fd1 i fd1 " M 1 fdn i fdn L1i1 " M 1l il "

(1.9.4)

In the above formulas, the minus before the related items of stator is caused by the regulation that the positive current produces negative flux-linkage. Total magnetic field energy Wm of electric machine is expressed by the following matrix: 40

1

Wm


1 [is " ism i fd1 " i fdn 2 1 " M s1sm M s1 fd1 ª Ls1 « # # « # «M " L M sm sm fd1 « sm s1 « M fd1s1 M fd1sm L fd1 « u« # # # «M " M fdn sm M fdn fd1 « fdn s1 « M 1s " M 1s M 1 fd1 1 m « # # « # « M M lfd1 ¬ ls1 " M lsm

i1 " il ] "

M s1 fdn

M s11

"

# " M sm fdn

# M sm 1

"

" M fd1 fdn

M fd11 "

"

# L fdn

"

M 1 fdn

L1

#

#

M lfdn

M l1

"

# M fdn 1 " " "

M s1l º ª is º 1 » # » «« # »» M sml » « is » »« m » M fd1l » « i fd » »« 1 » # »« # » M fdn l » « i fd » »« n » M 1l » « i1 » »« » # »« # » Ll »¼ «¬ il »¼

1 i L iT 2

(1.9.5) In the above formula, the minus appears before the stator current items of the current row matrix, and the positive signs are taken for the whole elements of the inductance matrix which is different from the sign in matrix L of formula (1.1.4). Based on the virtual displacement principle when each loop current is kept as constant, there is f

wWm wg

(1.9.6) i constant

where f is broad sense force, g is broad sense displacement. If rotor mechanical angle T is considered as broad sense displacement, then broad sense force should be electromagnetic torque Te of electric machine, hence Te

wWm wT

i constant

§1 · w ¨ i L iT ¸ 2 © ¹ wT i 1 wL T i i 2 wT

constant

(1.9.7)

In formula (1.9.7), Te is consistent with rotor rotation direction, that is Te is driving torque. 41

AC Machine Systems

If P is the pole-pair number of the electric machine, J is the electric angle by which the rotor d -axis leads ahead of the stator coordinates system axis (i.e. rotor position angle), then there is

T

J /P

If according to generator convention, positive electro-magnetic torque is considered as braking torque, then the electro-magnetic torque can be written as

Te

P wL T i i 2 wJ

(1.9.8)

When the rotor motion equation of the electric machine is set up, the generator is considered as a standard, such as in Fig. 1.9.1. Here, Tm is the mechanical torque of the prime mover which is consistent with rotor general rotation direction; Te is the electro-magnetic breaking torque; surplus torque Tm Te is the accelerating torque, and the rotor motion equation can be set up as J

dZ dt

Tm Te

where J is rotary inertia; Z is mechanical angular speed;

(1.9.9) dZ is angular dt

acceleration.

Figure 1.9.1 Sketch map for setting rotor motion equation

If the mechanical angle between rotor d -axis and phase a axis is expressed dJ by J , then Z , thus formula (1.9.9) becomes dt J

d 2J dt 2

Tm Te

(1.9.10)

where the unit of J is mechanical radian; the unit of t is s ; the unit of Tm and Te is N m; the unit of J is kg m 2 . 42

1


If the unit of flywheel torque GD 2 of rotation part is t m 2 , then 1 GD 2 u 103 (kg m 2 ) 4

J

If J in formula (1.9.10) is expressed by electric radian, then the electric machine rotor motion equation is J

1 d 2J P dt 2

Tm Te

(1.9.11)

where P is the pole pair number of the electric machine. In practice, per-unit value is adopted mostly in the above formula. Generally, the base value of torque is selected as TG

S N u 103 (N m) 2Sn 60

where S N is the normal apparent power of the electric machine (kV A); n is synchronous speed (rpm). 1 Generally, the base value of time is selected as tG (s), it corresponds to 2Sf the time when the rotor of the synchronous machine rotates by one electric radian at synchronous speed. Two sides of formula (1.9.11) are divided by these base values, and we get

ª 1 GD 2 3 2 § n · º d 2J 2S f ¨ ¸ » « 2 © 60 ¹ ¼ § t · ¬ P SN d¨ ¸ © tG ¹

Tm Te TG TG

(1.9.12)

i.e., the per-unit value form of motion equation is H

d 2J dt 2

H

dZ dt

Tm Te

(1.9.13)

where H

1 GD 2 3 2 § n · 2S f ¨ ¸ P SN © 60 ¹

2S3 f

GD 2 § n · ¨ ¸ S N © 60 ¹

2

(1.9.14)

H is called as inertia constant, and simple relation exists between it and the per-unit value of the time. From Eq. (1.9.13), we can get d 2J dt 2

dZ dt

Tm Te H 43

AC Machine Systems

Suppose that the surplus torque Tm Te 1 , W is the time when the electric machine rotation speed is raised from zero to synchronous speed ( Z 1 ), then

W

³

W 0

dt

³

1 0

HdZ

H

(1.9.15)

It can be seen that while constant surplus torque with per-unit value 1 ( Tm Te 1 ) is exerted on original standstill electric machine rotor, the per-unit value of the time equals H when the rotation speed is raised from zero to synchronous speed.

1.10

Analysis of Parameters and Performance of SinglePhase Induction Motors

In the above sections, the basic electromagnetic relations of the salient pole synchronous machine are analyzed mainly, that is, related physical concept, basic voltage equations and flux-linkage equations, calculation method of each loop parameter of electric machine, electro-magnetic torque and rotor motion equation, etc. are analyzed. After getting these relations, various operation performances of the ac machine can be analyzed and calculated. In this section, as an example, the single-phase induction machine is presented for explaining the calculation method of the electric machine loop parameters, for writing voltage equations and fluxlinkage equations according to the actual loops, and for analyzing the performance of the electric machine. There are various types of induction machines, their stator windings have both symmetric distribution and asymmetric distribution, and the winding connection has various styles. In the operation of a single induction machine, normally its air-gap magnetic field is not round; besides the positive sequence rotating magnetic field, the negative sequence magnetic field exists in air-gap. The air-gap harmonic magnetic field of a single phase induction machine is very strong, and its effect on the electric machine’s performance is greater. Normally, symmetric component method is used for researching the single phase electric machine; in this method, the above factors would bring about many inconveniences, and bigger error would be caused. Adopting the Multi-Loop Analysis method for researching the single phase electric machine, writing equations and calculating parameters can be realized according to actual loops, and the effects of many factors can be considered more comprehensively. This method is suitable particularly for the condition with stronger space harmonics of air-gap magnetic field and special winding connection. In the following, the Multi-Loop Method is used to research the single phase electric machine. Figure 1.10.1(a) is the stator connection sketch of a single phase induction machine. Here, m and a represent main phase winding and auxiliary phase 44

1


winding, respectively, C is series capacitor. Figure 1.10.1(b) is the sketch of the rotor loops of a single phase induction machine.

Figure 1.10.1 Sketch map of single phase induction machine circuit (a) Stator connection; (b) Rotor loops

When parameter calculation and performance analysis of single phase induction machine are carried out, the stator phase windings can be considered as stator loops. At the same time, it is needed to calculate the inductances of stator single coil, then by combining them the inductances of stator phase winding are obtained. In the following, firstly the inductances related to air-gap main magnetic field are researched, sum of the inductances caused by the main magnetic field and leakage one are the inductances of stator phase winding.

1.10.1

Inductances of Single Phase Induction Machine

Induction machine is with uniform air-gap, its air-gap permeance is constant. Hence, formula (1.2.3) can be simplified as

OG ( x)

O0 2

P0 G

(1.10.1)

in which the effect of slot and iron reluctance is considered in choosing G . Suppose that the electric angle per stator slot is T , then the electric angle between i -th coil and j -th coil of the stator is (i j )T , according to formula (1.3.38) the mutual inductance between these two coils is M ij

ª§ k yk ·2 º 2 wK2 W l « » O cos ( ) T k i j ¨ ¸ ¦ 0 P S2 »¼ k «© k ¹ ¬

(1.10.2)

1 2 3 , , ," P P P The mutual inductance between the stator coils of the induction machine is independent of rotor position. Suppose that the stator phase winding consists of 2P series coils, each coil where k

45

AC Machine Systems

group has q coils. When the slot number per pole per phase is an integer, the serial number of i -th coil under m -th pole is (m 1)2q i , the serial number of j -th coil of the same phase coil group under n -th pole is (n 1)2q j (the serial number of the first coil under the first pole is considered as 1). Here, 2q is the slot number per pole of the single phase induction machine. It is known that the self inductance LI of stator phase winding is the sum of the inductances of all coils (including 2 Pq u 2 Pq coil) in the phase winding, that is 2P 2P

q

q

¦¦¦¦ M

LI

minj

m 1n 1 i 1 j 1

2

§k · 2 wK2 W l O0 ¦ ¨ yk ¸ (1)m n ¦¦¦¦ 2 m 1 n 1 i 1 j 1 PS k © k ¹ u cos k[(m 1) u 2q i (n 1) u 2q j ]T 2P 2P

q

q

2

§k · 2 wK2 W l O0 ¦ ¨ yk ¸ (1)m n ¦¦¦¦ 2 m 1 i 1 j 1 n 1 PS k © k ¹ u cos k[(m 1)S (i j )T ] q

2P

q

2P

(1.10.3)

1 2 3 , , ," ; M minj represents the inductance between i -th coil under P P P m -th pole and j -th coil under n -th pole. N.B. The winding direction of the coil

where k

group under the adjacent pole is opposite, so the multiple (1)m n appears in formula (1.10.3). Based on formula (1.10.3), it can be seen that if m, i and j are fixed, and n is changed from 1 to 2P , then the following summation is zero while k is fraction or even, i.e., 2P

¦ (1)

mn

cos k[(m n)S (i j )T ] 0

n 1

that is, when k is fraction or even, the formula (1.10.3) is zero. When k is odd 2P

¦ (1)

mn

cos k[(m n)S (i j )T ] 2 P cos k (i j )T

(1.10.4)

n 1

Substituting the above formula into formula (1.10.3), we get q

q

§k · (2 P ) 2 u 2 wK2 W l O0 ¦ ¨ yk ¸ 2 PS k © k ¹ 46

2

§k · 2 wK2 W l O0 ¦ ¨ yk ¸ u 2 P cos k (i j )T ¦¦¦ 2 m 1 i 1 j 1 PS k © k ¹ 2P

LI

2

q

q

¦¦ cos k (i j )T i 1 j 1

(1.10.5)

1


where k 1,3,5" Following the triangular formula qT 2 cos § D q 1T · ¨ ¸ T 2 ¹ © sin 2

sin

q

¦ cos(D iT ) i 1

(1.10.6)

we get qkT 2 cos § kiT q 1 kT · cos k (i j )T ¦ ¦¦ ¨ ¸ kT 2 © ¹ i 1 j 1 i 1 sin 2 2 qkT · § sin ¨ 2 ¸ cos § q 1 kT q 1 kT · ¨ ¸ ¨ ¸ 2 2 © ¹ ¨ sin kT ¸ 2 ¹ © q

q

q

sin

qkT § ¨ sin 2 ¨ ¨ sin kT 2 ©

· ¸ ¸ ¸ ¹

2

(1.10.7)

Therefore, qkT § § k yk · ¨ sin 2 (2 P) 2 w W l O0 ¦ ¨ ¸ ¨ kT PS 2 k © k ¹ ¨ sin © 2 2

LI

2

2 K

· ¸ ¸ ¸ ¹

2

qkT § § k yk · ¨ sin 2 (2 PqwK ) u 2W l O0 ¦ ¨ ¸ ¨ kT PS2 k © k ¹ ¨ q sin © 2 2

2

2 w12W l §k · O0 ¦ ¨ wk ¸ P S2 k © k ¹

· ¸ ¸ ¸ ¹

2

2

(1.10.8)

where k 1,3,5"; w1 2 PqwK is the series turn number per phase of stator winding; qkT 2 k yk kT q sin 2 sin

kwk

k yk k pk

is stator winding factor. Based on the inductance of single coil, the mutual inductance between main 47

AC Machine Systems

phase m winding and auxiliary phase a winding can be obtained similarly. If the turn number per coil of stator main phase winding is wKm , and the turn number per coil of auxiliary phase winding is wKa , the pitches of the main phase and auxiliary phase windings are the same, and the serial number of stator coil begins from the first coil under the first pole of the main phase winding, then the serial number of i -th coil under n -th pole of the main phase winding is (n 1)2q i , the serial number of j -th coil under nc -th pole of the auxiliary phase winding is (nc 1)2q q j. Accordingly, the mutual inductance between main phase winding and auxiliary phase winding is 2P 2P

q

q

¦¦¦¦ M

M ma

nmincaj

n 1 nc 1 i 1 j 1

2

§k · 2 wKm wKaW l O0 ¦ ¨ yk ¸ (1)n nc ¦¦¦¦ 2 PS n 1 nc 1 i 1 j 1 k © k ¹ u cos k[(n 1) u 2q i (nc 1) u 2q q j ]T 2P 2P

q

q

2

§k · 2 wKm wKaW l O0 ¦ ¨ yk ¸ (1)n nc ¦¦¦¦ 2 PS n 1 i 1 j 1 nc 1 k © k ¹ u cos k[(n nc)S (i j )T qT ] 2P

q

q

2P

(1.10.9)

1 2 3 , , ,"; M nmincaj represents the mutual inductance between i -th P P P coil under n -th pole of winding m and j -th coil under nc - th pole of winding a . When k is odd in which k

2P

¦ (1)

n nc

cos k[(n nc)S (i j )T qT ]

nc 1

2 P cos[k (i j )T kqT ] When k is fraction or even, the above formula is zero. Hence, formula (1.10.9) can be simplified further as M ma

§k · (2 P)2 u 2 wKw wKaW l O0 ¦ ¨ yk ¸ 2 PS k © k ¹

2

q

q

¦¦ cos[k (i j )T kqT ] i 1 j 1

where k 1,3," Using formula (1.10.6), we get q

q

¦¦ cos[k (i j )T kqT ] i 1 j 1

q

¦ i 1

48

qkT 2 cos § kiT kqT q 1 kT · ¨ ¸ kT 2 © ¹ sin 2

sin

1

Circuit Analysis of AC Machines — Multi-Loop Model and Parameters 2

qkT § ¨ sin 2 ¨ ¨ sin kT 2 ©

· ¸ q 1 q 1 · § kT kT ¸ ¸ cos ¨ kqT 2 2 © ¹ ¸ ¹

qkT § ¨ sin 2 ¨ ¨ sin kT © 2

· ¸ ¸ cos kqT ¸ ¹

2

Therefore, qkT § § k yk · ¨ sin 2 (2 P)2 u 2 wKm wKaW l O0 ¦ ¨ ¸ ¨ kT PS2 k © k ¹ ¨ sin © 2 2

M ma

qkT § § k yk · ¨ sin 2 2 PqwKm u 2 PqwKaW l O0 ¦ ¨ ¸ ¨ kT PS 2 k © k ¹ ¨ q sin © 2 2

2

· ¸ ¸ cos kqT ¸ ¹ 2

· ¸ ¸ cos kqT ¸ ¹

2

2 wm waW l §k · O0 ¦ ¨ wk ¸ cos kqT 2 PS k © k ¹

(1.10.10)

in which k 1,3," ; wm and wa are the series turn number per phase of the main phase winding and auxiliary phase winding, respectively; kqT is the electric angle between k -th harmonic central lines of the main phase and auxiliary phase windings. Formulas (1.10.8) and (1.10.10) are deduced according to the phase winding with single branch. If the phase winding is with multi-branches, then w1 of formula (1.10.8) is changed into w1 / a1 , and wm and wa of formula (1.10.10) are changed into wm / am and wa / aa , respectively, where a1 , am and aa are the parallel branch numbers of the corresponding phase windings respectively. As mentioned above, the self inductance and mutual inductance of stator phase winding of the single phase induction machine are deduced according to single coil inductance, as shown in formulas (1.10.8) and (1.10.10). It can be seen that the inductances of stator phase winding of the single phase induction machine are independent of rotor position owing to uniform air-gap. It should be illustrated that formulas (1.10.8) and (1.10.10) can be calculated directly too, based on main phase and auxiliary phase windings; that is, according to the air-gap mmf produced by the main phase and auxiliary phase windings, the corresponding air-gap flux density is calculated. If several different connection types of the main phase and auxiliary phase windings exist, then respective 49

AC Machine Systems

calculations are needed based on actual connection. If the single coil inductance is calculated firstly, in spite of coil connection mode it is very convenient to calculate the inductances of windings, branches, or certain several coils connected, if needed, in terms of the obtained single coil inductances. The reason why the above method is adopted is to illustrate its concrete usage. The rotor circuit of the single phase induction machine is shown in Fig. 1.10.1(b). When the inductances of the rotor loops of the single phase induction machine are researched, the formulas (1.5.14) and (1.5.17) of the rotor damper loop inductances of the salient pole synchronous machine can be cited. But, the air-gap permeance of the single phase induction machine has only the constant term O0 / 2 (harmonic terms O2 O4 " 0 ) owing to uniform air-gap of the single phase induction machine, i.e., only when 2l

k j

0 , there are values in the

two formulas, hence the self inductance of rotor i - th loop is Li

2wr2W l kE S 1 O0 ¦ 2 sin 2 r P S2 k 2 k

(1.10.11)

The mutual inductance between rotor i - th loop and j - th loop is M ij

2wr2W l kE S 1 O0 ¦ 2 sin 2 r cos k ( j i )M 2 PS 2 k k

(1.10.12)

1 2 3 , , ,"; E r is the short pitch ratio of the rotor loop; wr is the turn P P P number of the rotor loop, generally wr 1 ; i, j are the serial number of rotor loops; M is the electric angle between the central lines of rotor adjacent coils. Actually, the self inductances of all rotor loops have the same value, the mutual inductances between all rotor loops have another same value, and they are all independent of rotor position. Formulas (1.10.11) and (1.10.12) reflect this fact. In the following, still based on the inductance between stator single coil and rotor loop, the inductance between stator winding and rotor loop is calculated. Cite formula (1.6.5) of the mutual inductance between stator single coil and damper loop of the salient pole synchronous machine, and pay attention to the fact that only O0 has its value for the induction machine, then the mutual inductance between stator i - th coil and rotor j -th loop of the single phase induction machine can be obtained where k

M sij

2 wK wrW l 1 k E S k Er S O0 ¦ 2 sin sin cos k (J D ) 2 PS 2 2 k k

(1.10.13)

where J is the electric angle by which the rotor d axis leads the central line of the stator coil; D is the electric angle by which the central line of the rotor loop 50

1


leads the rotor d axis; (J D ) is the electric angle by which the central line of the rotor loop leads the central line of the stator coil. If the electric angle by which the central line of rotor 0 - th loop leads the central line of stator 0 - th coil is taken as angle J c , then formula (1.10.13) can be written as 2 wK wrW l 1 k E S k Er S O0 ¦ 2 sin sin cos k (J c iT jM ) 2 PS 2 2 k k

M sij

(1.10.14)

N.B. The serial number of stator and rotor loops is along reverse rotation direction. If the stator phase winding consists of 2P series coil groups, there are q coils in each coil group, and the first coil under the first pole is taken as the initial coil, then the total serial number of n - th coil under m -th pole is (m 1)2q n . If

M smnj expresses the mutual inductance between stator n -th coil under m -th pole and rotor j -th loop, then the mutual inductance between stator phase winding and rotor j -th loop is 2P

q

¦¦ M

M ij

smnj

m 1n 1 2P q

2 wK wrW l kE S 1 k ES O0 ¦ 2 sin sin r (1) m 1 2 2 2 PS m 1n 1 k k u cos k{J c [(m 1) u 2q n]T jM } 2P q 2 wK wrW l kE S 1 k ES O0 ¦ 2 sin sin r (1) m 1 ¦¦ 2 2 2 PS m 1n 1 k k c u cos k[J (m 1) S nT jM ]

¦¦

(1.10.15)

When k is odd 2P

¦ (1)

m 1

cos k[J c (m 1)S nT jM ] 2 P cos k (J c nT jM )

m 1

When k is fraction or even, the value of the above formula is zero. Hence, formula (1.10.15) can be simplified further as M sj

2 wK wrW l u 2 P kE S 1 k ES O0 ¦ 2 sin sin r PS2 k 2 2 k q

u ¦ cos k (J c nT jM )

k 1, 3, 5, "

n 1

Using formula (1.10.6), finally we get 51

AC Machine Systems

M sj

2(2 PqwK ) wrW l kE S 1 k ES O0 ¦ 2 sin sin r 2 PS 2 2 k k qkT sin 2 cos k § J c jM q 1T · u ¨ ¸ kT 2 ¹ © q sin 2 2w1 wrW l kE S 1 O0 ¦ 2 kwk sin r 2 PS 2 k k q 1 · § u cos k ¨ J c jM T¸ 2 ¹ ©

k 1, 3, "

(1.10.16)

q 1 · § where ¨ J c jM T ¸ is the electric angle between the central line of the 2 ¹ © stator phase winding and the central line of rotor j -th loop. It can be seen that the mutual inductance between stator phase winding and rotor loop of the single phase induction machine depends on rotor position, that is, it is time-variant parameter at rotor rotation. Similarly, formula (1.10.16) is also deduced according to single-branch phase winding; if the phase winding is with multi-branches, branch number is a1 , then w1 of the formula should be changed into w1 / a1 . Only the air-gap magnetic field is considered above, but the leakage flux should be considered in calculating stator phase winding inductances and rotor loop inductances, that is, corresponding leakage inductance terms should be added to formulas (1.10.8), (1.10.10), (1.10.11), and (1.10.12), description of which is not needed here.

1.10.2

Basic Relations of Single Phase Induction Machine

After knowing the inductance of each loop of the single phase induction machine, the voltage equations and flux-linkage equations can be written and researched. (1) Steady state expressions of stator and rotor currents Suppose that the stator power supply of the single phase induction machine is sine wave voltage, the corresponding stator current is also fundamental sine current owing to uniform air-gap of the induction machine and symmetric multi-phase winding in the rotor. The reason is that no matter what rotation speed the harmonic magnetic field has, produced by the stator fundamental current, only the corresponding harmonic magnetic field with the same rotation speed is produced by the induced current in symmetric multi-phase winding in the rotor. If rotor angular speed is Z , the rotation speeds of positive rotation and reverse rotation magnetic fields of k -th harmonic produced by the fundamental phase 52

1


current of the stator single phase winding in air gap are Z 0 / k and Z 0 / k , respectively, where Z 0 2ʌf is fundamental angular frequency, k 1,3,5," (suppose the stator is integer slot winding), then the rotation speeds of these positive rotation and reverse rotation magnetic field relative to the rotor are Z § Z · r 0 Z , accordingly, the frequencies of rotor induced currents are ¨ r 0 Z ¸ k , k © k ¹

that is, Z0 kZ and Z 0 kZ . If the stator current and rotor current are written as sine and cosine functions, then the current of stator main phase winding is im

I m cos Z0t I mc sin Z0 t

(1.10.17)

The current of stator auxiliary winding is ia

I a cos Z0 t I ac sin Z0 t

(1.10.18)

The current of rotor first loop is i1

¦ {I

1k

cos(Z0 kZ )t I 2 k sin(Z0 kZ )t I 3k cos(Z 0 kZ )t

k

I 4 k sin(Z0 kZ )t}

(1.10.19)

k 1, 3, 5, "

Considering the phase angle relation of induced currents in rotor loops produced by positive rotation and reverse rotation magnetic fields, the current of rotor n - th loop is in

¦ {I

1k

cos[(Z0 kZ )t k (n 1)M ] I 2 k sin[(Z0 kZ )t

k

k (n 1)M ] I 3k cos[(Z0 kZ )t k (n 1)M ] I 4 k sin[(Z 0 kZ )t k (n 1)M ]}

k 1, 3, 5, "

(1.10.20)

where M is the electric angle between the central lines of adjacent loops. (2) Voltage equation and flux-linkage equation of each winding of single phase induction machine When voltage equation and flux-linkage equation are written, the following positive direction regulation is adopted: the current flowing into the electric machine is considered as positive, and the positive direction of voltage drop is consistent with the current positive direction; positive current produces positive flux-linkage for both stator and rotor. And then, the voltage equation of the main phase winding is um

p\ m rm im

(1.10.21)

The voltage equation of auxiliary phase winding is 53

AC Machine Systems

ua

p\ a ra ia

(1.10.22)

The voltage equation of rotor n -th loop is un

p\ n rr in rc (in 1 in 1 )

n 1, 2," , Z 2

(1.10.23)

d is differential operator; Z 2 is rotor slot number; rr is resistance of dt rotor loop; rc is resistance of rotor bar. The flux-linkage equation of each loop is

where p

ª\ m º «\ » « a» «\ 1 » « » « # » «\ n » « » ¬« # ¼»

ª Lm «M « am « M 1m « « # « M nm « ¬« #

M ma La M 1a # M na #

M m1 " M mn M a1 " M an L1 " M 1n # # M n1 " Ln #

#

"º ªim º "»» «« ia »» "» « i1 » »« » »« # » "» « in » »« » ¼» ¬« # ¼»

(1.10.24)

In the inductances of the above formula, the inductances of stator phase windings (left-upper-angle square matrix) and the inductances of rotor loops (right-downangle square matrix) are all constant, the mutual inductances between stator phase windings and rotor loops are time-variant parameters, that is, functions of rotor position angle J c

t

³ Z dt J 0

0

Z t J 0 (at constant rotation speed), here

J 0 is rotor initial position angle. All the inductances can be obtained by formulas (1.10.8), (1.10.10), (1.10.11), (1.10.12), and (1.10.16). (3) Forming steady-state linear algebraic equation group The expressions of stator and rotor currents (1.10.17), (1.10.18), (1.10.19), and (1.10.20) are substituted into the flux-linkage equation (1.10.24), then the voltage equations (1.10.21), (1.10.22), and (1.10.23) are brought in, a group of equations will be obtained in which the amplitudes of the sine parts and cosine parts are unknown quantities, including each harmonic sine function and cosine function of Z t . Pay attention to the fact that the rotor loop of the induction machine is short circuit (un 0) , the electric source of stator winding is fundamental voltage (refer to Fig. 1.10.1(a)), then um ua

54

u U c cos Z0t U cc sin Z0t

(1.10.25)

1 ( I a cos Z 0t I ac sin Z 0t )dt C³ I Ic U c cos Z0t U cc sin Z0 t a sin Z 0t a cos Z 0t (1.10.26) Z0C Z0C

u uC

U c cos Z0t U cc sin Z0t

1


It is enough to analyze only according to a certain loop equation of rotor, because the amplitude of each loop current of the rotor is the same, and the phase angle difference of the adjacent loops is also the same. As for the current of another loop of the rotor, it is not difficult to be calculated based on equation (1.10.20). In this way, three equations of transcendental function with multi harmonics are obtained. Each harmonic of the equations can form an equation itself, if the number of the researched harmonics is k1 , then a total of 3k1 equations will be got. For each harmonic equation, two special times are selected in order to make the sine term or cosine term to be zero (for example, for fundamental order Z t 0 or Z t ʌ / 2 ), then two corresponding linear algebraic equations are got, in this way a total of 6k1 linear algebraic equations are obtained. In these equations, the unknowns are the amplitudes of the sine term and cosine term of each current, the coefficients of the equations are the inductance and resistance of each loop, the right terms of the equations are either zero or the amplitudes of the sine term or cosine term of the electric source voltage. Solving this linear algebraic equation group, all loop currents of the machine can be obtained. After getting the input current and voltage of the electric machine, the input power of the electric machine can be obtained, and various losses of the electric machine are subtracted from the input power, then the output power of the electric machine will be obtained. Electro-magnetic torque is produced under the interaction of flux and current. The electro-magnetic torque of the single phase induction machine is (refer to formula (1.9.7)) Te

1 wL T 1 i i >im ia i1 2 wT 2 ª Lm M ma M m1 «M La M a1 « am L1 w « M 1m M 1a u « # # wT « # « M nm M na M n1 « # # «¬ #

" in "@ " M mn " M an " M 1n # " Ln #

"º ªim º "»» ««ia »» "» «i1 » »« » » «# » "» «in » »« » »¼ «¬# »¼

(1.10.27)

where T is the rotor position angle (mechanical angle). Considering only the inductances between stator windings and rotor loops to be a function of the rotor position angle in inductance matrix, the formula (1.10.27) can be written as Z2

Te

im ¦ ii i 1

Z2 wM mi wM ai ia ¦ ii wT wT i 1

(1.10.28)

As an example, an electric machine for an electric fan with pole-pair number 55

AC Machine Systems

P 2 , stator winding full pitch and q 1 is calculated by using the Multi-Loop Method. The winding factor of each harmonic of the electric machine is the same as the fundamental, and the amplitude ratio of harmonic mmf to fundamental mmf is Fk F1

1 k

k 1,3,5,"

Therefore, the stator harmonic mmf of this single phase induction machine is very strong, in the calculation it is necessary to consider the effect of different harmonic air-gap magnetic field. The experimental data, the calculated values of the original designed program, and the calculated values by using the Multi-Loop Method are shown in Table 1.10.1. In the table, I m is the main phase winding current, I a is the auxiliary phase winding current, I l is the total current, P1 is input power, k is the calculated harmonic order of stator mmf. Operation conditions of the electric machine are: U 220V, n 1250rpm, phase-splitting capacitor C 1.202PF . Table 1.10.1 Comparison of operation performance of a single phase induction motor I m /A

I a /A

I l /A

I 1 /A

Experimental data Calculated values of original designed program k 1 Calculated values by using k 1, 3

0.205 6

0.176 0

0.309 0

62.56

0.230 0

0.146 9

0.357 7

74.33

0.246 2

0.149 0

0.328 3

66.21

the Multi-Loop Method

0.215 8

0.161 4

0.315 9

64.36

k

1, 3, 5

0.205 9

0.167 0

0.315 6

64.54

k

1, 3, 5, 7

0.205 0

0.169 3

0.315 2

64.41

It can be seen from the table that while only considering the fundamental of the air-gap magnetic field, the calculated values by using the Multi-Loop Method are close to the calculated values of the original designed program, but the difference is bigger compared with the experimental data. The higher the considered harmonic order of air-gap magnetic field is, the closer to experimental data the calculated results are.

References [1] Gao J D, Wang X H (1987). Multi-loop theory of AC machines. Journal of Tsinghua University (Science and Technology), 1: 1 8 (in Chinese) [2] Gao J D, Wang X H, Jin Q M (1987). Calculation of loop parameters of salient-pole synchronous machine. Journal of Tsinghua University (Science and Technology), 1: 9 19 (in Chinese) 56

1


[3] Gao J D, Wang X H, Li F H (1993). Analysis of AC Machines and Their Systems, First Edition. Beijing: Tsinghua University Press (in Chinese) [4] Gao J D, Wang X H, Li F H (2005). Analysis of AC Machines and Their Systems, 2nd Edition. Beijing: Tsinghua University Press (in Chinese) [5] Wang X H, Sun Y G, Ouyan g B, Wang W J, Zhu Z Q, Howe D (2002) Transient behavior of salient-pole synchronous machines with internal stator winding faults. IEE Proc.-Electr. Power Appl. 2: 143 151 [6] Wang X H, Chen S L, Wang W J , Sun Y G, Xu L Y (2002). A study of armature winding internal faults for turbogenerators. IEEE Trans. on IA, 3: 625 631 [7] Gao J D, Zhang L Z (1982). Fundamental Theory and Analysis Mothods of Transients in Electric Machines. Beijing: Science Press (in Chinese) [8] Danileviq Y B, Dombloski V V, Kazovski E Y (1965). Parameters of AC machines. Beijing: Science Press (in Russian) [9] Wang X H, Lo K L (1997). New development of torque analysis of synchronous machine. Electric Machines and Power Systems, 8: 827 838

57

2

Electromagnetic Relations and Parameters of Synchronous Machines and Analysis of Their Several Operating Conditions at Synchronous Speed

Abstract In this chapter, firstly study electromagnetic relations and parameters of conventional 3-phase synchronous machines by using multi-loop method, whose results are the same as classical method. Secondly, the actual a, b, c axes are converted into d, q, 0 axes according to both mmfs being equal to form a clear concept. Furthermore, two types of per-unit systems are introduced for synchronous machines, inclusive of distinct formula deduction, and then basic equations in per-unit, operational reactances and electromagnetic torque formula are also built. In order to help readers grasp how to use those basic equations in per-unit and operational equivalent circuits to solve practical problems, several symmetrical operating modes at synchronous speed are analysed, which may occur in practical operation. Among of them, 3-phase sudden short circuit will threaten synchronous machines. Thus, comprehension of the transients may provide scientific basis not only for reasonable design and reliable operation of synchronous machines but also for a careful choice of relay protection. Voltage dip during sudden increase of load for synchronous machines occurs in practice and is also studied in this chapter. Transient emfs of synchronous machines are introduced, which are proportional to the flux-linkages of rotor loops. The flux-linkage of rotor winding, especially the flux-linkage of excitation winding, under a certain condition, can be considered as a constant, which will make analysis and calculation simplified. Electromagnetic torque after 3-phase sudden short circuit, inclusive of pulsating torque and average torque, are also analysed in this chapter. Finally, suddenly opening 3-phase short circuit of synchronous machines is also paid attention to.

In Chapter 1, electromagnetic relations and parameters of ac machines have been studied according to salient-pole type, writing down voltage and flux-linkage equations and giving out inductance computation of stator and rotor loops on the basis of single coil. In this chapter, electromagnetic relations and parameters are studied for conventional 3-phase synchronous machines. During normal operation of 3-phase synchronous machine, its stator winding is symmetrical. Even if exist unbalanced loads and asymmetrical faults, the stator

2 Analysis of Several Operating Conditions of Synchronous Machines at Synchronous Speed

winding it self is still symmetrical, so it is convenient to study those problems on the basis of one phase winding and corresponding magnetic field. In electric machinery, fundamental flux in air-gap is mainly considered and harmonic flux effect can be counted in by using the differential leakage reactance, thus causing a certain deviation. In Chapter 1 we have studied the influence of airgap harmonic flux on parameters and performance of single phase induction motors. In this chapter, firstly study the effect of harmonics on 3-phase synchronous machines, and then research on their several operating conditions on the basis of one phase winding. In general, air-gap harmonics can be reduced a lot by means of special design for structure and windings, such as good pole-arc form, fractional-pitch and distributed stator winding, fractional-slot winding, etc. During small air-gap harmonics, we use coordinates conversion, especially d, q, 0 axes to analyse operating conditions of synchronous machines, which is convenient because the basic equations in d, q, 0 axes are of constant coefficients referring to Refs.[1,3,5]. Therefore, d, q, 0 axes are paid more attention to in this chapter. By using basic equations in d, q, 0 axes, we analyse some symmetrical operating modes of synchronous machines at synchronous speed, such as symmetrical steady-state operation, the transient process when rotor winding is short-circuited and voltages are suddenly applied to the stator winding, the currents after 3-phases are suddenly short-circuited, the voltage dip when the load of a synchronous machine increases suddenly, the transient emfs of synchronous machines, electromagnetic torque after 3-phases are suddenly short-circuited, opening 3-phase short-circuit, referring to Refs.[2,4,6].

2.1

Basic Relations of Synchronous Machines

In the stator of synchronous machines there are phase a, b and c windings, and in the rotor there is excitation winding besides damping winding that can be represented by equivalent d-axis damping winding 1d and q-axis damping winding 1q. According to reference directions stated in Chapter 1, various winding voltage equations in matrix form for 3-phase salient-pole synchronous machines can be written as p\ RI

U

(2.1.1)

in which

U

[ua

ub

I

[ia

ib

\

[\a

uc ic

0 0]T

u fd i fd

i1d

i1q ]T

\ b \ c \ fd \ 1d \ 1q ]T 59

AC Machine Systems

R

ª r « r « r « « « « « ¬

R fd R1d

º » » » » » » R1q »¼

and subscripts a, b, c, fd , 1d , and 1q indicate stator phase a, b and c windings, rotor excitation winding, and equivalent d-axis and q-axis damping windings, respectively, and r is stator phase resistance. In a similar way, the relations for multi-loop damping winding can be found out if necessary. Various winding flux-linkage equations in matrix form for 3-phase salient-pole synchronous machines can be expressed as

\

(2.1.2)

LI

in which

L

ª Laa « M ba « « M ca « « M fda « M 1da « «¬ M 1qa

M ab Lbb M cb M fdb M 1db M 1qb

M ac M bc Lcc M fdc M 1dc M 1qc

M afd M bfd M cfd L ffd M 1 fd 0

M a1d M b1d M c1d M f 1d L11d 0

M a1q º M b1q »» M c1q » » 0 » 0 » » L11q »¼

(2.1.3)

L with corresponding subscript indicates self-inductance for the relevant winding, and M with related subscript shows corresponding mutual inductance. The corresponding mutual inductances are reversible, namely M ab M ba , M afd M fda , M a1d M 1da , M a1q M 1qa , M f 1d M 1 fd

M bc M cb , M bfd M fdb , M b1d M 1db , M b1q M 1qb ,

M ac M ca M cfd M fdc M c1d M 1dc M c1q M 1qc

In addition, mutual inductance between q-axis damping winding 1q and d-axis damping winding 1d or and excitation winding fd is zero because winding 1q is set S electric radians apart from winding 1d and winding fd . Various self-inductances and mutual inductances in Eq. (2.1.3) are important for study of synchronous machine operation. Computation of single coil inductance 60


has been discussed in Chapter 1, by which the inductance for phase winding composed of several coils is analysed here. For convenience, ferro-magnetic saturation, eddy current and hysteresis effect, and slot influence are neglected. If it is necessary to consider slot influence, refer to Appendix A in this book.

2.1.1

Inductances of Stator Phase Windings

In Chapter 1, Eq. (1.3.31) indicates stator single coil inductance corresponding to air-gap flux, in which D is electric angle between two coils. If two coils are numbered as i c and j c , respectively, and each slot electric angle is T , then their inductance corresponding to air-gap flux is

M icj cG

k yk k yj 2 wK2 W l ^(Odkj Oqkj ) ¦¦ PS k j kj u cos[(k j )(J i cT ) j ( j c i c)T ] (Odkj Oqkj ) cos[(k j )(J i cT ) j ( j c ic)T ]`

(2.1.4)

in which

k

1 2 3 , , , "; j | k r 2l |; l P P P

0,1, 2, "

(J i cT ) is the electric angle, at which rotor d-axis leads ahead of stator ic -th coil when taking stator 0-th coil central line as stator reference axis. Supposing that stator phase winding is composed of 2P coil-groups in series, each coil-group has q coils during q integer, and the first coil of the phase winding under the first pole is taken as No.1 coil, then the number of ic -th coil of the phase winding under m -th pole is 3q (m 1) ic, the number of j c -th coil of the same phase under n -th pole is 3q (n 1) j c, and the stator phase winding self-inductance is 2P 2P

LI

q

q

¦¦¦¦ M

mi cnj c

m 1 n 1 ic 1 jc 1

k yk k yj 2 wK2 W l (Odkj Oqkj )(1)m n ^ ¦¦¦¦ 2 ¦¦ S P kj m 1 n 1 ic 1 jc 1 k j u cos[( k j )(J i cT ) j (3q (n 1) j c 3q(m 1) i c)T ] (Odkj Oqkj )(1) m n 2P 2P

q

q

u cos[( k j )(J i cT ) j (3q(n 1) j c 3q(m 1) i c)T ]` 61

AC Machine Systems

k yk k yj 2 wK2 W l ^(Odkj Oqkj )(1)mn ¦¦¦¦ 2 ¦¦ P kj S c m 1n 1 i 1 n 1 k j u cos[(k j )(J i cT ) j (3q (n m)T ( j c i c)T )] q

2P

q

2P

(Odkj Oqkj )(1) m n u cos[(k j )(J i cT ) j u 3q (n m)T ( j c i c)T ]` in which 3qT

S

If making m, i c and j c constant and letting n change from 1 to 2P in the summation term above, then the summation term equals zero during j fraction and also zero during j even number, namely, 2P

¦ (1)

mn

cos[(k B j ) (J i cT ) B j ((n m)S ( j c ic)T )] 0

n 1

During j odd number exists 2P

¦ (1)

mn

cos[(k B j ) (J i cT ) B j ((n m)S ( j c i c)T )]

n 1

2 P cos[(k B j ) (J i cT ) B j ( j c i c)T )] Hence, stator phase winding self-inductance is also written as k yk k yj 2 P u 2wK2 W l {(Odkj Oqkj ) ¦¦ 2 Pʌ kj m 1 ic 1 j c 1 k j u cos[(k j )(J icT ) j ( j c ic)T ] (Odkj Oqkj ) cos[(k j )(J icT ) j ( j c ic)T ] 2P

q

q

¦¦¦

LI

where k 1, 3, "; j | k r 2l | . By using Eq. (1.10.6) exists q

q

¦¦ cos[(k B j )(J icT ) B j ( j c ic)T ] ic 1 jc 1

qjT 2 cos ª(k B j )(J i cT ) r jicT B q 1 jT º ¦ « » j T 2 ¬ ¼ ic 1 sin 2 2 qjT · § sin ¨ 2 ¸ cos ª(k B j )J q 1 kT B q 1 jT º ¨ ¸ « » 2 2 ¬ ¼ ¨ sin jT ¸ © 2 ¹ q

62

sin


qjT § ¨ sin 2 ¨ ¨ sin jT 2 ©

2

· ¸ q 1 ·º ª § T¸ ¸ cos «(k B j ) ¨ J 2 ¹ »¼ © ¬ ¸ ¹

so we can get

LI

qjT § ¨ sin 2 k yk k yj °° 2(2 PqwK ) 2 W l ®(Odkj Oqkj ) ¨ ¦¦ Pʌ 2 kj ° k j ¨ q sin jT © 2 ¯° q 1 · § u cos(k j ) ¨ J T¸ 2 ¹ © (Odkj

qjT § ¨ sin 2 Oqkj ) ¨ ¨ q sin jT 2 ©

· ¸ ¸ ¸ ¹

2

2 ½ · ¸ q 1 · °° § T ¸¾ ¸ cos(k j ) ¨ r 2 ¹° © ¸ ¹ °¿

kwk kwj 2w2W l 2 ¦¦ Pʌ k j kj (Odkj

q 1 · § T¸ ®(Odkj Oqkj ) cos(k j ) ¨ J 2 ¹ © ¯ q 1 ·½ § Oqkj ) cos(k j ) ¨ J T ¸¾ 2 ¹¿ ©

(2.1.5)

where k 1, 3, "; j | k r 2l | Referring to Eq. (1.3.18), the formula above can also be written as LI

q 1 · q 1 · § § T ¸ L4 cos 4 ¨ J T ¸ " L0G L2 cos 2 ¨ J 2 ¹ 2 ¹ © © L0G

2 º 2w2W l ª§ kwk · (Odkk Oqkk ) » ¸ 2 ¦ «¨ Pʌ k ¬«© k ¹ ¼»

2w2W l §k · O0 ¦ ¨ wk ¸ 2 Pʌ k © k ¹ L2

(2.1.6)

2

k 1, 3, "

(2.1.7)

2w2W l 2 ®k w (Od 11 Oq11 ) Pʌ 2 ¯ 1 ª k wk k w( k 2) º ½° (Odk ( k 2) Oqk ( k 2) ) » ¾ 2¦ « k ¬ k ( k 2) ¼ ¿° 63

AC Machine Systems

ªk k º ½° 2 w2W l ° 2 O2 ®kw1 2¦ « wk w( k 2) » ¾ 2 Pʌ k ¬ k ( k 2) ¼ ¿ ° ¯°

k 1, 3, "

(2.1.8)

in which w 2 Pqwk is total number of turns for stator phase winding. The self-inductance above corresponds to air-gap flux only, so the phase winding self-inductance formula (2.1.6), after considering leakage self-inductance L0l relevant to slot and end leakage flux, is reformed as LI

q 1 · § T ¸ " L0 L2 cos 2 ¨ J 2 ¹ © q 1 · § T ¸ " ( L0l L0G ) L2 cos 2 ¨ J 2 ¹ ©

(2.1.9)

where L0G and L2 are indicated, respectively, by formulas (2.1.7) and (2.1.8). q 1 T is the included electric angle between phase In the formulas above, 2 winding central axis and 0-th coil central line, namely reference axis, so rotor d-axis q 1 · § leads ahead of stator phase winding central axis by electric angle ¨ J T ¸. 2 ¹ © The phase winding self-inductance formula above is obtained in accordance with single branch. If there are as branches in parallel, several branch currents for the phase winding are the same during symmetrical operation or external asymmetrical operation, so there is no need for distinction between single branch and several branches. However, the branch current, namely coil current, for as branches in parallel per phase, is 1/as of single branch current for the same phase current, so will the air-gap flux density. In addition, number of each branch coil-groups is 2P/as for several branches in parallel, so the flux-linkage for each branch winding is 1/a2s of single branch condition. Hence, phase winding self-inductance during several branches in parallel is 1/a2s of single branch condition, which is the only difference between them. Evidently, referring to formulas (2.1.7) and (2.1.8), the phase self-inductance components during as branches in parallel are: L0G

L2

64

2w2W l O0 Pʌ 2 as2

§k · ¦k ¨© kwk ¸¹

2

k 1, 3, "

ª k wk k w( k 2) º ½° 2w2W l O2 ° 2 k 2¦ « »¾ 2 2 ® w1 Pʌ as ¯° k ¬ k ( k 2) ¼ ¿ °

(2.1.10)

k 1, 3, "

(2.1.11)


Mutual inductance between phase a and phase b windings will be discussed as follows. Numbering stator coils and starting from the first coil of phase a winding under the first pole, then number of ic -th coil for phase a under m-th pole is [3q (m 1) i c] , and number of j c -th coil for phase b under n-th pole is [3q (n 1) 2q j c] . During single-branch and integer-slot winding, mutual inductance between phase a and phase b corresponding to air-gap flux can be written as 2P 2P

M abG

q

q

¦¦¦¦ M

mai cnbj c

m 1 n 1 ic 1 jc 1 2P 2P

q

q

k yk k yj 2wK2 W l {(Odkj Oqkj ) 2 ¦¦ kj 1 Pʌ k j

¦¦¦¦ m 1 n 1 ic 1 jc

u (1)m n cos[(k j )(J i cT ) j (3q (n 1) 2q j c 3q(m 1) ic)T ] (Odkj Oqkj )(1) m n u cos[( k j )(J icT ) j (3q(n 1) 2q j c 3q(m 1) i c)T ]} q 2P 2P q k yk k yj 2wK2 W l {(Odkj Oqkj )(1) m n ¦¦¦¦ 2 ¦¦ P kj ʌ m 1 ic 1 jc 1 n 1 k j u cos[(k j )(J i cT ) j (3q(n m)T 2qT ( j c i c)T ] (Odkj Oqkj )(1) m n cos[(k j )(J i cT ) j (3q (n m)T 2qT ( j c i c)T ]}

in which k

1 2 3 , , , "; j | k r 2l |; 3qT P P P

ʌ

M maicnbj c is the mutual inductance between ic -th coil for phase a under m-th pole and j c -th coil for phase b under n-th pole. In a silimar way as before, during j is odd number exists 2P

2ʌ ª º cos «(k B j )(J icT ) B j ((n m)ʌ ( j c ic)T ) » 3 ¬ ¼ n 1 ª § 2ʌ ·º 2 P cos «(k B j )(J icT ) B j ¨ ( j c ic)T ¸ » 3 © ¹¼ ¬

¦ (1)

mn

and the formula above equals zero during j there is

fraction or even number. Hence,

65

AC Machine Systems

k yk k yj 2 P u 2wK2 W l ¦¦¦ ¦¦ 2 Pʌ kj m 1 ic 1 jc 1 k j 2P

q

q

®(Odkj Oqkj ) ¯ 2ʌ ª º u cos «(k j )(J icT ) j j ( j c i c)T » 3 ¬ ¼ 2ʌ ª º½ (Odkj Oqkj ) cos «(k j )(J i cT ) j j ( j c i c)T » ¾ 3 ¬ ¼¿

M abG

where k 1, 3, "; j | k r 2l | By use of formula (1.10.6), there is q

q

ª

¦¦ cos «¬(k B j )(J icT ) r j ic 1 jc 1

2ʌ º B j ( j c i c)T » 3 ¼

qiT 2 cos ª(k B j )(J i cT ) r j 2ʌ r ji cT B q 1 jT º ¦ « » T j 3 2 ¬ ¼ ic 1 sin 2 2 qjT · § sin ¨ 2 ¸ cos ª(k B j )J r j 2ʌ q 1 kT B q 1 jT º ¨ ¸ « » 3 2 2 ¬ ¼ ¨ sin jT ¸ © 2 ¹ q

sin

qjT § ¨ sin 2 ¨ ¨ sin jT © 2

2

· ¸ ª q 1 · 2ʌ º § T¸r j » ¸ cos «(k B j ) ¨ J 2 ¹ 3¼ © ¬ ¸ ¹

so we can get qjT § ¨ sin 2 k yk k yj °° 2(2 PqwK ) W l ®(Odkj Oqkj ) ¨ ¦¦ Pʌ 2 kj ° k j ¨ q sin jT 2 © °¯ 2

M abG

2ʌ º q 1 · ª § T ¸ j » (Odkj u cos «(k j ) ¨ J 2 ¹ 3¼ © ¬ ½ 2ʌ º °° q 1 · ª § T¸ j » ¾ u cos «(k j ) ¨ J 2 ¹ 3¼ ° © ¬ °¿ 66

· ¸ ¸ ¸ ¹

2

qjT § ¨ sin 2 Oqkj ) ¨ ¨ q sin jT 2 ©

· ¸ ¸ ¸ ¹

2


k wk k wj 2 w2W l 2 ¦¦ Pʌ k j kj (Odkj

ª 2ʌ º q 1 · § T¸ j » ®(Odkj Oqkj ) cos «(k j ) ¨ J 2 3¼ © ¹ ¬ ¯ ª 2ʌ º ½ q 1 · § Oqkj ) cos «(k j ) ¨ J T ¸ j »¾ 2 ¹ 3 ¼¿ © ¬

where k 1, 3, "; j | k r 2l | Like the relation in formula (1.3.34), the mutual inductance between phase a and phase b windings can be written as M abG

q 1 ʌ· § M ab 0G M ab 2 cos 2 ¨ J T ¸ 2 3¹ © q 1 ʌ· § T ¸ " M ab 4 cos 4 ¨ J 2 3¹ ©

(2.1.12)

in which 2

M ab 0G

2 w2W l § k wk · 2kʌ (Odkk Oqkk ) cos ¦ 3 Pʌ 2 k ¨© k ¸¹ 2

2w2W l 2kʌ §k · O0 ¦ ¨ wk ¸ cos 3 Pʌ 2 k ¹ k ©

M ab 2

k 1, 3, "

(2.1.13)

2 w2W l 2 ®k w1 (Od 11 Oq11 ) Pʌ 2 ¯ ª k wk k w( k 2) 2ʌ º °½ (Odk ( k 2) Oqk ( k 2) ) cos(1 k ) » ¾ 2¦« 3 ¼ ¿° k ¬ k ( k 2) k k 2 w2W l 2 2ʌ ½ O2 ®kw1 2¦ wk w( k 2) cos(1 k ) ¾ k 1, 3, " 2 Pʌ k ( k 2) 3¿ k ¯

(2.1.14)

q 1 ʌº ª T » is the included electric angle between rotor In formula (2.1.12), «J 2 3¼ ¬ d-axis and the central line that is between phase a and phase b, and M abG reaches extreme value when rotor d-axis coincides with the central line between phase a q 1 ʌº ª and phase b, namely, «J T » 0. 2 3¼ ¬ In formula (2.1.13), only consider the mutual inductance component caused by air-gap flux, and leakage mutual inductance component M abl should be added in after considering slot leakage flux and end leakage flux. Obviously, M abl is a constant and independent of rotor position, which has negative value because phase 67

AC Machine Systems

a and phase b windings have a difference of 120ein space. Let M abl total mutual inductance between phase a and phase b windings is M ab

M ml , then

q 1 ʌ· § M ab 0 M ab 2 cos 2 ¨ J T ¸ 2 3¹ © q 1 ʌ· § T ¸ " M ab 4 cos 4 ¨ J 2 3¹ ©

(2.1.15)

where M ab 0G M ml

M ab 0

In a similar way, the mutual inductance components for as branches in parallel are:

M ab 0G

M ab 2

ª§ k wk ·2 2kʌ º ¦k «¨© k ¸¹ cos 3 » «¬ »¼

2 w2W l O0 Pʌ 2 as2

(2.1.16)

k wk kw( k 2) 2 w2W l O2 2 2ʌ ½ cos(1 k ) ¾ k 2¦ 2 2 ® w1 3¿ Pʌ as ¯ k (k 2) k

(2.1.17)

where k 1, 3, " Clearly, the stator phase winding inductances of salient-pole synchronous machines depend upon rotor position, namely, variable parameters with time. It should be noted that the discussion above is appropriate to number of stator slots per pole per phase q integer. If q fraction, then distribution factor and harmonic orders in self-inductance and mutual inductance formulas are different to make some accommodation.

2.1.2

Inductances Between Stator and Rotor Windings

The mutual inductance between stator single coil and rotor excitation winding has been deduced for salient-pole synchronous machines in formula (1.6.3). If taking the central line of stator 0-th coil as the stator reference axis, then the mutual inductance between stator i-th coil and rotor excitation winding is M ifd

2w fd wKW l 1 ʌ a fd

¦ k

Odk k

sin

kE ʌ cos k (J iT ) 2

k 1, 3, "

Supposing that the stator phase winding is composed of 2P coil-groups in series and each coil-group has q coils, then the mutual inductance between corresponding stator phase winding and rotor excitation winding for q integer is 68

2 Analysis of Several Operating Conditions of Synchronous Machines at Synchronous Speed 2P

q

¦¦

M I fd

m 1 i 1

u (1) Because 3qT

2 w fd wKW l 1 ʌ a fd

m 1

¦

Odk

k

k

sin

kE ʌ 2

cos k{J [(m 1)3q i ]T }

k 1, 3, "

(2.1.18)

ʌ , there is 2P

¦ (1) m 1 2P

¦ (1)

m 1

cos k{J [(m 1)3q i ]T }

m 1

cos k{J (m 1)ʌ iT }

m 1

2 P cos k (J iT )

In the light of formula (1.10.6) exists qkT 2 cos k § J q 1T · ¨ ¸ kT 2 ¹ © sin 2

sin

q

¦ cos k (J iT ) i 1

Hence, formula (2.1.18) can be reformed as

M I fd

qkT sin kE ʌ 2 cos k § J q 1T · ¦k k sin 2 ¨ ¸ kT 2 ¹ © q sin 2 2 ww fdW l 1 Odk q 1 · § ¦ kwk cos k ¨© J 2 T ¸¹ k 1, 3, " a fd k k ʌ 2 PqwK 2w fdW l 1 a fd ʌ

Odk

(2.1.19)

q 1 º ª where «J T is the included electric angle between rotor d-axis and the 2 »¼ ¬ central line of stator phase a winding. If taking the phase a central line as stator reference axis, then the mutual inductance between phase a winding and excitation winding can be written as M afd

2 ww fdW l 1 ʌ a fd

¦ k

Odk k

kwk cos kJ

k 1, 3, "

(2.1.20)

2ʌ · 2ʌ · § § If substituting ¨ J ¸ and ¨ J ¸ for J , respectively, then the mutual 3 ¹ 3 ¹ © © inductances between phase b or phase c and excitation winding can be got as: M bfd

2 ww fdW l 1 ʌ a fd

¦ k

Odk k

2ʌ · § kwk cos k ¨ J ¸ 3 ¹ ©

k 1, 3, "

(2.1.21) 69

AC Machine Systems

2ww fdW l 1 ʌ a fd

M cfd

¦

Odk

k

k

2ʌ · § kwk cos k ¨ J ¸ 3 ¹ ©

k 1, 3, "

(2.1.22)

Evidently, the mutual inductances between stator phase windings and excitation winding depend upon rotor position. The formulas above are appropriate to single branch condition. If having as branches in parallel, then the right sides of corresponding formulas above should 1 be multiplied by . For example, the mutual inductance between phase a winding as and excitation winding, referring to formula (2.1.20), should be changed to 2 ww fdW l

M afd

ʌas a fd

¦ k

Odk k

k wk cos kJ

k 1, 3, "

(2.1.23)

The mutual inductances between stator phase windings and damping loops will be discussed next. The mutual inductance between stator single coil and damping loop has been shown in formula (1.6.5). If taking stator 0-th coil axis as stator reference axis, then the mutual inductance between stator i-th coil and damping loop 11c is M i1

O2l 2wK wrW l kE ʌ jE ʌ sin 1 sin cos[ j (J iT ) kD1 ] ® ¦ ¦ 2 Pʌ kj 2 2 j ¯ 2l |k j |

¦

2l |k j |

O2l kj

sin

½ k E1 ʌ jE ʌ sin cos[ j (J iT ) kD1 ]¾ 2 2 ¿

(2.1.24)

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j

1 2 3 , , ," P P P

For single branch and integer number of slots, study the mutual inductance between stator phase winding and damping loop as follows. Taking the first coil of the phase winding under the first pole as No.1 coil, then the mutual inductance between stator phase winding and damping loop 11c is q

M I1

O2l 2 wK wrW l kE ʌ jE ʌ sin 1 sin (1) m 1 ® ¦ ¦ 2 P ʌ kj 2 2 j ¯ 2l |k j | 1

2P

¦¦ i 1 m

u cos{ j[J (( m 1)3q i )T ] kD1} O kE ʌ jE ʌ ¦ 2l sin 1 sin (1) m 1 2 2 2 l |k j | kj ½ u cos{ j[J ((m 1)3q i )T ] kD1}¾ ¿ 70

(2.1.25)


where | k j | 0, 2, 4, "; | k j | 2, 4, "; 3qT

ʌ

Analysing the summation term of formula (2.1.25), making m change from 1 to 2P, and supposing i to be constant and j to be odd number, then there is 2P

¦ (1) m 1 2P

¦ (1)

m 1

cos{ j[J ((m 1)3q i )T ] r kD1}

m 1

cos{ j[J (m 1)ʌ iT ] r kD1}

m 1

2 P cos{ j (J iT ) r kD1}

which equals zero during j fraction or even number. Thus, formula (2.1.25) can be written as M I1

O2l kE ʌ jE ʌ sin 1 sin ® ¦ 2 2 j 1,3," ¯ 2 l | k j | kj q O kE ʌ jE ʌ u ¦ cos[ j (J iT ) kD1 ] ¦ 2l sin 1 sin 2 2 i 1 2 l |k j | kj 2wK wrW l u 2 P Pʌ 2

¦

q ½ u ¦ cos[ j (J iT ) kD1 ]¾ i 1 ¿

where | k j | 0, 2, 4, "; | k j | 2, 4, " Referring to formula (1.10.6) exists

¦ cos[ j (J iT ) r kD ] 1

i 1

qjT 2 cos ª j § J q 1T · r kD º 1» ¸ « ¨ jT 2 ¹ ¬ © ¼ sin 2

sin

q

so the mutual inductance between stator phase winding and damping loop 11c is

M I1

qjT sin O2l 2 PqwK 2 wrW l ° k E1 ʌ jE ʌ 2 ¦j ®2l¦|k j| kj sin 2 sin 2 jT Pʌ 2 ° q sin 2 ¯ O2l k E1 ʌ ª § º q 1 · sin u cos « j ¨ J T ¸ kD1 » ¦ 2 ¹ 2 ¬ © ¼ 2l |k j| kj

71

AC Machine Systems

qjT ½ sin jE ʌ q 1 · ª § º° 2 T ¸ kD1 » ¾ cos « j ¨ J u sin 2 q sin jT 2 ¹ ¬ © ¼° 2 ¿ O2l kE ʌ 2 wwrW l q 1 · ª § º sin 1 kwj cos « j ¨ J T ¸ kD1 » ® ¦ ¦ 2 Pʌ 2 2 ¹ j ¯ 2 l |k j | kj ¬ © ¼ ½ O kE ʌ q 1 · ª § º T ¸ kD1 » ¾ ¦ 2l sin 1 k wj cos « j ¨ J kj 2 2 ¹ 2 l |k j | ¬ © ¼¿

(2.1.26)

in which | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " q 1 º ª «J 2 T » is the electric angle, by which rotor d-axis leads ahead of stator ¬ ¼ phase winding axis. If taking the central line of phase a winding as stator reference axis, then the mutual inductance between phase a winding and damping loop 11c is

O2l 2wwrW l kE ʌ sin 1 k wj cos( jJ kD1 ) ® ¦ ¦ 2 Pʌ 2 j ¯ 2 l |k j | kj

M a1

¦

O2l

2l |k j |

kj

sin

½ k E1 ʌ k wj cos( jJ kD1 ) ¾ 2 ¿

(2.1.27)

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " 2ʌ · 2ʌ · § § Substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in formula (2.1.27), 3 ¹ 3 ¹ © © the mutual inductances between phase b or phase c and damping loop 11c can be obtained. If stator phase winding has as branches in parallel, like the mutual inductance between stator phase winding and excitation winding, the mutual inductance between stator phase winding and damping loop should be changed from formula (2.1.26) to M I1

O2l 2wwrW l kE ʌ ª § º q 1 · sin 1 k wj cos « j ¨ J T ¸ kD1 » ® ¦ ¦ 2 Pʌ as j ¯ 2l |k j| kj 2 2 ¹ ¬ © ¼

¦

2l |k j |

72

O2l kj

sin

k E1 ʌ ª § º½ q 1 · kwj cos « j ¨ J T ¸ kD1 » ¾ 2 2 ¹ ¬ © ¼¿

(2.1.28)


where j 1, 3, "; | k j | 0, 2, 4, "; | k j | 2, 4, " Clearly, the mutual inductance between stator phase winding and damping loop depends upon rotor position. As mentioned before, an equivalent d-axis damping winding and an equivalent q-axis one are used to represent damping winding, so several parameters have to be replaced in associated mutual inductance formulas. For example, during calculation of mutual inductance between stator phase winding and equivalent d-axis damping winding, the number of turns w1d for equivalent d-axis damping winding will replace wr , the fractional pitch ratio E1d for equivalent d-axis one will replace E1 , and let D1 M I 1d

0 . Hence, there is

2 ww1dW l O kE ʌ ª § q 1 ·º ¦ ® ¦ 2l sin 21d kwj cos «¬ j ¨© J 2 T ¸¹»¼ Pʌ 2 as j ¯ 2l |k j| kj

¦

O2l

2l |k j|

kj

sin

k E1d ʌ ª § q 1 ·º ½ T¸ ¾ kwj cos « j ¨ J 2 2 ¹ »¼ ¿ ¬ ©

(2.1.29)

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " In addition, during the calculation of mutual inductance between stator phase winding and equivalent q-axis damping winding, the number of turns w1q for equivalent q-axis damping winding will supersede wr , the fractional pitch ratio

E1q for equivalent q-axis one will supersede E1 , and let D

ʌ/2 since rotor

q-axis leads ahead of rotor d-axis by ʌ/2 electric radians. Thus exists M I 1q

2 ww1qW l

O2l

k E1q ʌ

ª § q 1 · kʌ º k wj cos « j ¨ J T¸ Pʌ as j ¯ 2l |k j| kj 2 2 ¹ 2 »¼ ¬ © k E1q ʌ O ª § q 1 · kʌ º ½ ¦ 2l sin kwj cos « j ¨ J T¸ (2.1.30) ¾ 2 2 ¹ 2 »¼ ¿ 2 l | k j | kj ¬ © 2

¦® ¦

sin

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " The discussions above are targeted on stator 3-phase winding inductances and mutual inductances between stator phase windings and rotor windings. As for rotor winding inductances, the formulas (1.5.3), (1.5.14), (1.5.17), and (1.5.24) can still be used. If using an equivalent d-axis damping winding and an equivalent q-axis one, then the self-inductance of d-axis damping winding is 73

AC Machine Systems

2 w12dW l O2l kE ʌ jE ʌ sin 1d sin 1d 2 ¦® ¦ Pʌ 2 2 j ¯ 2 l |k j | kj

L11d

¦

2l |k j|

O2l kj

sin

k E1d ʌ jE ʌ ½ sin 1d ¾ 2 2 ¿

(2.1.31)

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " The self-inductance of equivalent q-axis damping winding is L11d

2 w12qW l Pʌ

¦

O2l

¦® ¦

2

j

O2l kj

2 l |k j |

¯ 2l

sin

|k j|

kj

k E1q ʌ 2

sin sin

k E1q ʌ 2 j E1q ʌ 2

sin

cos

j E1q ʌ 2

cos

( j k )ʌ 2

( j k )ʌ ½ ¾ 2 ¿

(2.1.32)

where | k j | 0, 2, 4, "; | k j | 2, 4, "; j 1, 3, " The mutual inductance between excitation winding and equivalent d-axis damping winding is M f 1d

¦ k

2.1.3

2 w1d w fdW l 1 Odk kE ʌ sin 1d ʌ a fd k 2

k 1, 3, "

(2.1.33)

3-Phase Synchronous Machine Inductances in Accordance with Fundamental Air-Gap Flux

In comparison with single coil inductances, the stator phase winding inductances for integer number of slots and 60ephase belt only comprise fundamental and odd harmonics effect except fractional harmonics and even harmonics as in single coil. In addition, the fractional pitch and distributed winding is used to make air-gap flux approximate to sinusoidal curve, so the calculation error of inductances, during only consideration of fundamental only, is usually within 10%. Considering fundamental air-gap flux only, then phase a winding self-inductance is Laa

L0 L2 cos 2J

(2.1.34)

in which J is the electric angle the rotor d-axis leads ahead of stator phase a 74


winding axis by, and 2 w2W l O0 2 kw1 Pʌ 2 as2

L0

L0l

L2

2w2W l O2 2 k w1 Pʌ 2 as2

(2.1.36)

When rotor d-axis coincides with phase a winding axis, namely J to formula (2.1.34) there are Laa

L0 L2

Laad

L0G L2

(2.1.35)

0 , referring

L0l Laad

2 w2W l (O0 O2 )k w21 Pʌ 2 as2 2

4W lP § wkw1 · Od 11 as2 ʌ ¨© P ¸¹

(2.1.37)

where Laad is phase a winding inductance caused by air-gap flux when rotor d-axis coincides with phase a winding axis. When rotor q-axis coincides with phase a axis, namely J ʌ / 2 , there exist Laa

L0 L2

Laaq

L0G L2

L0l Laaq

2 w2W l (O0 O2 )k w21 Pʌ 2 as2 2

4W lP § wk w1 · Oq11 as2 ʌ ¨© P ¸¹

(2.1.38)

where Laaq is phase a winding inductance caused by air-gap flux when rotor q-axis coincides with phase a winding axis. In the light of formulas (2.1.37) and (2.1.38) there are L0G L2

1 ( Laad Laaq ) 2 1 ( Laad Laaq ) 2

Hence, phase a winding self-inductance corresponding to fundamental air-gap flux can be written as 75

AC Machine Systems

Laa

1 1 L0l ( Laad Laaq ) ( Laad Laaq ) cos 2J 2 2 L0l Laad cos 2 J Laaq sin 2 J

(2.1.39)

2ʌ · 2ʌ · § § Substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in formula (2.1.34), then 3 3 ¹ © ¹ © phase b and phase c winding inductances Lbb and Lcc can be obtained. The mutual inductances between phase a and phase b windings are: ʌ· § M ab 0 M ab 2 cos 2 ¨ J ¸ 3¹ ©

M ab

w2W l O0 kw21 M ml Pʌ 2 as2

M ab 0

M ab 2

2w2W l O k2 2 2 2 w1 Pʌ as

(2.1.40) (2.1.41) (2.1.42)

Indicating mutual inductance between phase a and phase b by using Laad and Laaq , there is M ab

1 1 ʌ· § M ml ( Laad Laaq ) ( Laad Laaq ) cos 2 ¨ J ¸ 4 2 3¹ © 2ʌ · 2ʌ · § § M ml Laad cos J cos ¨ J ¸ Laaq sin J sin ¨ J ¸ 3 3 ¹ © ¹ ©

(2.1.43)

2ʌ · 2ʌ · § § Substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in Eq. (2.1.40), then the 3 3 ¹ © ¹ © mutual inductance between phase b and phase c and that between phase c and phase a can be found out. The mutual inductance between phase a winding and excitation winding is M afd

2 ww fdW l 1 Od 1kw1 cos J ʌas a fd

(2.1.44)

Let M afd 0

2ww fdW l ʌas a fd

Od 1kw1

(2.1.45)

then M afd

M afd 0 cos J

(2.1.46)

2ʌ · 2ʌ · § § Substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in Eq. (2.1.46), then the 3 ¹ 3 ¹ © © 76


mutual inductances between phase b winding or phase c winding and excitation winding can be got. When considering fundamental flux produced by fundamental mmf only, namely taking j 1 and k 1 in Eq. (2.1.29), we can get the mutual inductance between phase a winding and equivalent d-axis damping winding. If taking phase a winding axis as reference axis, namely ordinate, then J is substituted for q 1 · § T ¸ , so there is ¨J 2 ¹ © 2 ww1dW l E ʌ sin 1d k w1 (O0 O2 ) cos J 2 Pʌ a s 2 E ʌ§ w 8W l · w1d sin 1d ¨ kw1 ¸ Od 11 cos J 2 ʌ as 2 © 2P ¹

M a1d

(2.1.47)

Let

E ʌ§ w 8W l · w1d sin 1d ¨ kw1 ¸ Od 11 2 ʌ as 2 © 2P ¹

M a1d 0

(2.1.48)

then M a1d

M a1d 0 cos J

(2.1.49)

2ʌ · 2ʌ · § § Substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in Eq. (2.1.49), the mutual 3 ¹ 3 ¹ © © inductances between phase b or phase c windings and d-axis damping winding can be obtained. When considering fundamental flux produced by fundamental mmf only, namely taking j 1 and k 1 in Eq. (2.1.30), we can get the mutual inductance between phase a winding and equivalent q-axis damping winding after substituting q 1 · § J for ¨ J T ¸ as follows: 2 ¹ © M a1q

2 ww1qW l 2

Pʌ as

sin

E1q ʌ 2

kw1 (O0 O2 )( sin J )

E1q ʌ § w 8W l · w1q sin kw1 ¸ Oq11 sin J ¨ 2 ʌ as 2 © 2P ¹

(2.1.50)

E1q ʌ § w 8W l · w1q sin kw1 ¸ Oq11 ¨ 2 ʌ as 2 © 2P ¹

(2.1.51)

Let M a1q 0

77

AC Machine Systems

then M a1q

M a1q 0 sin J

(2.1.52)

2ʌ · 2ʌ · § § Similarly, substituting ¨ J ¸ and ¨ J ¸ for J , respectively, in Eq. (2.1.52), 3 3 ¹ © ¹ © then the mutual inductances between phase b or phase c windings and q-axis damping winding can be got. It is noted that Eq. (2.1.45) differs from Eqs. (2.1.48) and (2.1.51) because excitation winding fundamental flux is assumed to be produced by rectangular mmf and d-axis or q-axis damping winding fundamental flux is assumed to be produced by fundamental mmf.

2.2

Basic Relations of Synchronous Machines in d, q, 0 Axes

In the section above, votage and flux-linkage equations together with several inductances have been found out for 3-phase salient-pole synchronous machines, which are a set of differential equations with variable coefficients to solve them in a difficulty. As stated above, the ideal machine pattern can be used to analyse synchronous machine operation and the caused error is limited and allowed, which is discussed next.

2.2.1

Flux-Linkages and Parameters of Stator Phase Windings

As mentioned before, the flux-linkage of phase a produced by stator currents ia , ib , and ic is

\ as

Laa ia M ab ib M ac ic

(2.2.1)

Referring to Eq. (2.1.39) and (2.1.43), the self- and mutual-inductances are: Laa M ab

L0l Laad cos 2 J Laaq sin 2 J 2ʌ · § M ml Laad cos J cos ¨ J ¸ 3 ¹ © 2ʌ · § Laaq sin J sin ¨ J ¸ 3 ¹ ©

Substituting them into Eq. (2.2.1), we can get 78


\ as

ª 2 · 2 ·º § § Laad «ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » cos J 3 ¹ 3 ¹¼ © © ¬ ª 2 · 2 ·º § § Laaq «ia sin J ib sin ¨ J ʌ ¸ ic sin ¨ J ʌ ¸ » sin J 3 ¹ 3 ¹¼ © © ¬ [ L0l ia M ml (ib ic )]

(2.2.2)

Various terms in the formula above have the following meanings. After stator phase windings are passed with currents, they will constitue a mmf in air-gap each. As stated before, the mmf amplitude of the fundarnental produced by each phase current should be proportional to the phase current, so although ia , ib , and ic are current quantity, they also represent the mmf amplitudes produced by various phase currents. The mmf amplitudes lie on their own phase winding axes, so after ia , ib and ic are multiplied by cos J , 2 · 2 · 2 · § § § cos ¨ J ʌ ¸ , and cos ¨ J ʌ ¸ , respectively, i.e. ia cos J , ib cos ¨ J ʌ ¸ , 3 ¹ 3 ¹ 3 ¹ © © © 2 · § and ic cos ¨ J ʌ ¸ , they can represent the mmfs produced on rotor d-axis 3 ¹ © ª 2 · § by stator phase currents and the sum of them «ia cos J ib cos ¨ J ʌ ¸ 3 ¹ © ¬

2 ·º § ic cos ¨ J ʌ ¸» can represent the resultant mmf produced on rotor d-axis by all 3 ¹¼ © stator phase currents. Obviously, the resultant d-axis mmf acts on the permeance Od 11 , so the d-axis flux produced by all stator phase currents should be proportional to the product ª 2 · 2 ·º § § of the current «ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » and the permeance Od 11 . 3 3 ¹¼ © ¹ © ¬ Multiplying the result above by corresponding winding data, we can get d-axis flux-linkage produced by all stator phase currents. Therefore, in Eq. (2.2.2), the item ª 2 · 2 ·º § § Laad «ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » 3 3 ¹¼ © ¹ © ¬ 2

ª 4W lP § wk w1 · 2 · 2 ·º § § Od 11 «ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » ¸ 2 2 ¨ as ʌ © P ¹ 3 ¹ 3 ¹¼ © © ¬

is the d-axis air-gap flux-linkage produced by all stator phase currents. In order to get a clear conception, now we may suppose that on the stator d-axis there is a fictitious winding which has the same turns number and form as the practical stator phase winding, and then we can see that when 79

AC Machine Systems

ª 2 · 2 ·º § § «ia cos J ib cos ¨ J 3 ʌ ¸ ic cos ¨ J 3 ʌ ¸ » is multiplied by Od 11 and corres© ¹ © ¹¼ ¬ ponding winding data, i.e. by Laad , the d-axis air-gap flux-linkage of the fictitious d-axis winding may be written as ª 2 · 2 ·º § § Laad «ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » 3 ¹ 3 ¹¼ © © ¬

\ dsG

(2.2.3)

If a fictitious current i( d ) circulates through the fictitious d-axis winding, then we can know that the air-gap fulx-linkage produced in the d-axis winding by the fictitious current is Laad i( d ) . Comparing the result with Eq. (2.2.3), we can understand that when both flux-linkages are equal the fictitious current in the fictitious d-axis winding is i( d )

2 · 2 · § § ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ 3 ¹ 3 ¹ © ©

(2.2.4)

That is to say, so far as the air-gap flux density is concerned, the d-axis air-gap flux density produced by stator currents may be calculated according either to stator 3-phase currents or to fictitious current i( d ) , both results being equal. Because the air-gap flux density is distributed sinusoidally in space, the flux-linkage of phase a corresponding to d-axis air-gap flux density is \ dsG cos J . As for q-axis we have the similar explanation, i.e., ª 2 · 2 ·º § § «ia sin J ib sin ¨ J ʌ ¸ ic sin ¨ J ʌ ¸ » 3 ¹ 3 ¹¼ © © ¬ can represent the resultant mmf produced on q-axis by all stator phase currents. When multiplying it by Laaq , we can get the q-axis air-gap flux-linkage of the fictitious q-axis winding which has the same turns number and form as the practical stator phase winding. Let its value be \ qsG , then

\ qsG

ª 2 · 2 ·º § § «ia sin J ib sin ¨ J ʌ ¸ ic sin ¨ J ʌ ¸ » ( Laaq ) 3 ¹ 3 ¹¼ © © ¬

(2.2.5)

ª 2 · 2 ·º § § «ia sin J ib sin ¨ J ʌ ¸ ic sin ¨ J ʌ ¸ » 3 ¹ 3 ¹¼ © © ¬

(2.2.6)

Similarly, let i( q )

be the fictitious q-axis winding current, then the q-axis air-gap flux-linkage produced in the q-axis winding by the fictitious q-axis current is also \ qsG . 80


Multiplying the flux-linkage by sin J and taking negative sign, we can obtain the flux-linkage of phase a winding corresponding to the q-axis air-gap flux density, namely, \ qsG sin J . L0l and Lml in Eq. (2.2.2) are stator leakage self-inductance and leakage mutual inductance, respectively. Therefore, [ L0l ia M ml (ib ic )] is the leakage fluxlinkage not across air-gap produced in phase a winding by stator 3-phase currents. The resultant flux-linkage of phase a winding is equal to the sum of various flux-linkages mentioned above, and Eq. (2.2.2) can be rewritten as

\ as \ dsG cos J \ qsG sin J [ L0l ia M ml (ib ic )] ( Laad i( d ) ) cos J ( Laaq i( q ) )sin J [ L0l ia M ml (ib ic )]

(2.2.7)

In practice, we shall use d-axis and q-axis armature reaction inductances Lad and Laq to calculate some problems instead of inductances Laad and Laaq . Lad is the stator equivalent inductance when rotor d-axis coincides with the axis of resultant magnetic field produced by stator 3-phase currents, and Laq is the corresponding equivalent inductance when rotor q-axis coincides with the axis of the resultant magnetic field. There is a simple relation between Lad and Laad or between Laq and Laaq , and the simple relation can be easily found as follows. Let 3-phase balanced currents be ªia º «i » « b» «¬ ic »¼

sin Z t ª º « I m «sin(Z t 120e ) »» «¬sin(Z t 120e ) »¼

Referring to Eq. (2.2.2), the flux-linkage corresponding to air-gap flux-density produced in phase a winding by stator 3-phase currents is

\ asG

ª 2 · 2 · § § Laad « I m sin Z t cos J I m sin ¨ Z t ʌ ¸ cos ¨ J ʌ ¸ 3 3 ¹ © ¹ © ¬ 2 · 2 ·º § § I m sin ¨ Z t ʌ ¸ cos ¨ J ʌ ¸» cos J 3 3 ¹¼ © ¹ © 2 · § 2 · ª § Laaq « I m sin Z t sin J I m sin ¨ Z t ʌ ¸ sin ¨ J ʌ ¸ 3 ¹ © 3 ¹ © ¬ 2 · § 2 ·º § I m sin ¨ Z t ʌ ¸ sin ¨ J ʌ ¸» sin J 3 ¹ © 3 ¹¼ © 3 3 Laad I m sin(Z t J ) cos J Laaq I m cos(Z t J )sin J 2 2 81

AC Machine Systems

When J

Z t , there is \ asG

3 Laaq I m sin Z t 2

3 Laaq ia 2

and the corresponding equivalent inductance is

\ asG ia

3 Laaq 2

i.e., when J Z t , rotor speed is equal to the speed of revolving field produced by stator 3-phase currents and rotor q-axis coincides with the revolving field axis. Hence the corresponding equivalent inductance is Laq , namely, Laq When J

3 Laaq 2

2

3 4W lP § wkw1 · Oq11 2 as2 ʌ 2 ¨© P ¸¹

(2.2.8)

Z t ʌ / 2 exists \ asG

3 Laad I m sin Z t 2

3 Laad ia 2

and the corresponding equivalent inductance is

\ asG ia

3 Laad 2

i.e., when J

Z t ʌ / 2 , rotor d-axis coincides with the axis of revolving field. Thus, the corresponding equivalent inductance is Lad , namely, Lad

3 Laad 2

2

3 4W lP § wkw1 · Od 11 2 as2 ʌ 2 ¨© P ¸¹

(2.2.9)

Using inductances Lad and Laq to rewrite Eq. (2.2.7), there is

\ as

2 º 2 º ª ª « Lad 3 i( d ) » cos J « Laq 3 i( q ) » sin J ¬ ¼ ¬ ¼ [ L0l ia M ml (ib ic )]

(2.2.10)

Supposing

82

id

2 i( d ) 3

iq

2 i( q ) 3

2ª 2 · 2 ·º ½ § § ia cos J ib cos ¨ J ʌ ¸ ic cos ¨ J ʌ ¸ » ° « 3¬ 3 ¹ 3 ¹¼ ° © © ¾ 2ª 2 · 2 ·º § § «ia sin J ib sin ¨ J ʌ ¸ ic sin ¨ J ʌ ¸ » ° 3¬ 3 ¹ 3 ¹ ¼ °¿ © ©

(2.2.11)


and calling them to be the stator current d-axis and q-axis components, respectively, then exists

\ as

> Lad id @ cos J ª¬ Laq iq º¼ sin J [ L0l ia M ml (ib ic )]

(2.2.12)

i.e., when keeping flux-linkage the same, for convenience, we have substituted 3 inductances Lad and Laq which are amplified by a factor of for inductances 2 Laad and Laaq , and have also substituted currents id and iq which are reduced by 2 for currents i( d ) and i( q ) . 3 Out of Eq. (2.2.11) we can see that from the view-point of mathematics currents id and iq are two new variables constituted by original variables ia , ib and ic . However, if there are three original variables, then usually exist three new variables. For example, when a 3-phase machine operates asymmetrically, three original variables cannot be replaced by two new variables, so we must choose the third new variable as follows:

a factor of

i0

1 (ia ib ic ) 3

(2.2.13)

The relation between i0 and ia , ib , ic has the same form as zero sequence current in ordinary Symmetrical Components Method, but currents i0 , ia , ib and ic here are all instantaneous values and not complex phasors that represent currents changing sinusoidally as in Symmetrical Components Method. The current i0 is called zero-axis component of stator current. From the winding connection we can see that the zero-axis component is practically one third of neutral-line current. If having no neutral line, the zero-axis component of stator current will be zero. When the same single-phase current circulates through 3-phase-windings, i.e. ia ib ic I m sin Z t , obviously there is i0 I m sin Z t and the pulsating fundamental mmfs produced by currents ia , ib and ic have the same instantaneous 2 value, but their space positions differ from one another by ʌ. Therefore, the 3 resultant fundamental mmf in air-gap is equal to zero thus having no air-gap flux, but each phase winding can produce its own leakage flux. Thus, currents id , iq and i0 can be written as

83

AC Machine Systems

ªid º «i » « q» «¬ i0 »¼

ª2 « 3 cos J « « 2 « sin J « 3 « 1 « ¬ 3

2 2 · 2 2 ·º § § cos ¨ J ʌ ¸ cos ¨ J ʌ ¸ » 3 3 3 3 © ¹ © ¹» ªia º 2 § 2 · 2 § 2 ·» « » sin ¨ J ʌ ¸ sin ¨ J ʌ ¸ » « ib » 3 © 3 ¹ 3 © 3 ¹» «i » »¬ c¼ 1 1 » 3 3 ¼

Transforming the equation above inversely, we can describe currents ia , ib and ic as ªia º « » «ib » «¬ ic »¼

cos J sin J 1º ªid º ª « cos(J 120e ) sin(J 120e ) 1»» «« iq »» « «¬cos(J 120e ) sin(J 120e ) 1»¼ «¬ i0 »¼

(2.2.14)

Substituting (2.2.14) into (2.2.12) and arranging it in order, there is

\ as

( Lad Ll )id cos J ( Laq Ll )iq sin J L0 i0

(2.2.15)

where Ll

L0l M ml

(2.2.16)

is called stator leakage inductance that consists of leakage self-inductance and leakage mutual inductance for stator winding. In the discussion before, harmonic mmfs have been neglected, but during analysis of transients those harmonics can be considered by a differential leakage reactance, i.e. during calculation of leakage inductance Ll count a differential leakage inductance in. In (2.2.15), the expression L0

L0l 2 M ml

(2.2.17)

is called stator zero-axis inductance, which also consists of leakage self-inductance and leakage mutual inductance of stator winding but has a different value from the stator leakage inductance Ll . It is easily shown that L0 is stator equivalent inductance when the steady-state zero-sequence current circulates through 3-phasewindings. Supposing Ld Lq

Lad Ll ½ Laq Ll ¾¿

(2.2.18)

then exists

\ as 84

Ld id cos J ( Lq iq )sin J L0 i0

(2.2.19)


or

\ as \ ds cos J \ qs sin J \ 0

(2.2.20)

in which

\ ds Ld id \ qs Lq iq \ 0 L0i0 where Ld is called stator d-axis inductance and Lq the stator q-axis inductance. Ld and Lq have the following physical conceptions: If steady-state 3-phase currents circulate through stator phase windings and rotor d-axis coincides with the axis of revolving field produced by 3-phase currents, then referring to Eqs. (2.2.14) and (2.2.19) there are i0

0, iq

0, id

ia

id cos J , Ld

Im

\ as ia

Thus, Ld is stator equivalent inductance on the condition above. Compared with Lad , the Ld includes leakage-flux effect. Similarly, when rotor q-axis coincides with the revolving field axis mentioned above, referring to Eqs. (2.2.14) and (2.2.19) exist i0

0,

id

ia

iq sin J ,

0,

iq Lq

Im

\ as (ia )

Hence, Lq is stator equivalent inductance when rotor q-axis coincides with the revolving field axis. Lq also includes the leakage-flux effect. From Eq. (2.2.20) we can see that phase a flux-linkage produced by all stator phase currents can be indicated by three flux-linkages \ ds , \ qs and \ 0 . \ ds and

\ qs are called d-axis and q-axis components of stator flux-linkage produced by all stator phase currents, respectively. Compared with \ dsG

Lad id and \ qsG

Laq iq , the d-axis component \ ds and q-axis component \ qs all include the stator leakage flux-linkage. \ 0 is called zero-axis component of stator flux-linkage that has no magnetic connection with rotor winding. When studying phase b and phase c flux-linkages in a similar way or substituting 85

AC Machine Systems

2 · § directly ¨ J B ʌ ¸ for J in Eq. (2.2.20), we can write the flux-linkages of phase 3 ¹ © b and phase c produced by all stator phase currents as follows: ª\ bs º « » ¬\ cs ¼

ª § 2 · 2 · º § «cos ¨ J 3 ʌ ¸ sin ¨ J 3 ʌ ¸ 1» ª\ ds º ¹ © ¹ »« » « © \ qs « § 2 · 2 § · »« » «cos ¨ J ʌ ¸ sin ¨ J ʌ ¸ 1» «¬\ 0 »¼ 3 ¹ 3 ¹ ¼ © ¬ ©

(2.2.21)

The stator phase a flux-linkage produced by rotor winding currents is

\ ar

M afd i fd M a1d i1d M a1q i1q M afd 0 i fd cos J M a1d 0 i1d cos J M a1q 0 i1q sin J ( M afd 0 i fd M a1d 0 i1d ) cos J M a1q 0 i1q sin J

\ dr cos J \ qr sin J

(2.2.22)

in which

\ dr \ qr

M afd 0 i fd M a1d 0 i1d M a1q 0 i1q

Comparing Eq. (2.2.22) with Eq. (2.2.20), we can see that \ dr and \ qr are d-axis and q-axis components of stator flux-linkage produced by rotor currents. In addition, excitation winding current and d-axis damping winding current produce only d-axis component of stator flux-linkage, q-axis damping winding current produces only q-axis component of stator flux-linkage, and both cannot produce zero-axis component of stator flux-linkage. That is not difficult to understand, because currents i fd and i1d produce only d-axis flux and current i1q produces only q-axis flux. According to the results analysed above, we can describe the resultant flux-linkage produced in phase a of stator winding by stator and rotor currents as

\ a \ as \ ar \ d cos J \ q sin J \ 0

(2.2.23)

in which

\d \q \0 Similarly, there are 86

Ld id M afd 0 i fd M a1d 0 i1d ½ ° Lq iq M a1q 0 i1q ¾ ° L0 i0 ¿

(2.2.24)


ª\ b º «\ » ¬ c¼

ª § 2 · 2 · º § «cos ¨ J 3 ʌ ¸ sin ¨ J 3 ʌ ¸ 1» ª\ d º ¹ © ¹ »« » « © \q « § 2 · 2 · »« » § «cos ¨ J ʌ ¸ sin ¨ J ʌ ¸ 1» «¬\ 0 »¼ 3 ¹ 3 ¹ ¼ © ¬ ©

(2.2.25)

Inversely, from Eqs. (2.2.23) and (2.2.25) exist

ª\ d º «\ » « q» «¬\ 0 »¼

2.2.2

ª2 « 3 cos J « « 2 « sin J « 3 « 1 « ¬ 3

2 2 ·º § cos ¨ J ʌ ¸ » 3 3 ¹» © ª\ a º 2 · 2 § 2 ·» ʌ ¸ sin ¨ J ʌ ¸ » ««\ b »» 3 ¹ 3 © 3 ¹» «\ » »¬ c¼ 1 » 3 ¼

2 2 · § cos ¨ J ʌ ¸ 3 3 ¹ © 2 § sin ¨ J 3 © 1 3

(2.2.26)

Flux-linkages and Parameters of Rotor Windings

Referring to Eq. (2.1.2), the excitation winding flux-linkage is

\ fd

M fad ia M fbd ib M fcd ic L ffd i fd M f 1d i1d

According to Eq. (2.2.14) and mutual inductance relations in Eq. (2.1.46), or according to the concept that the stator d-axis ficticious winding couples electromagnetically the excitation winding, we can obtain

\ fd

M afd 0 i( d ) L ffd i fd M f 1d i1d

\ fd

3 M afd 0 id L ffd i fd M f 1d i1d 2

or (2.2.27)

The formula above shows that in excitation winding flux-linkage there are not only flux-linkage components L ffd i fd and M f 1d i1d produced by excitation winding current and d-axis damping winding current but also flux-linkage component § 3 · ¨ M afd 0 id ¸ produced by all stator phase currents. That is not difficult to © 2 ¹ understand because air-gap flux-linkage produced on d-axis by stator currents depends only upon d-axis mmf produced by stator currents or i( d ) . In Eq. (2.2.27) exists id

2 i( d ) , so the corresponding mutual inductance is multiplied by 3/2. 3 87

AC Machine Systems

In a similar way, the flux-linkage equations of d-axis and q-axis damping windings can be described as

\ 1d \ 1q

2.2.3

3 ½ M a1d 0 id M 1 fd i fd L11d i1d ° ° 2 ¾ 3 ° M a1q 0 iq L11q iq °¿ 2

(2.2.28)

Voltage Equations of Synchronous Machines

We can see that phase a voltage equation is p\ a ria p(\ d cos J \ q sin J \ 0 )

ua

r (id cos J iq sin J i0 ) cos J p\ d \ d sin J pJ sin J p\ q \ q cos J pJ

p\ 0 rid cos J riq sin J ri0 ( p\ d \ q pJ rid ) cos J

( p\ q \ d pJ riq )sin J ( p\ 0 ri0 )

where pJ Z is rotor instantaneous speed. If letting ud uq u0

p\ d \ qZ rid ½ ° p\ q \ d Z riq ¾ ° p\ 0 ri0 ¿

(2.2.29)

ud cos J uq sin J u0

(2.2.30a)

then exists ua

i.e., ua has a similar expression as ia and \ a referring to Eqs. (2.2.14) and (2.2.23). 2ʌ · § Substituting ¨ J B ¸ for J in (2.2.30a), ub and uc can be written as 3 ¹ © ub uc 88

2 · 2 · ½ § § ud cos ¨ J ʌ ¸ uq sin ¨ J ʌ ¸ u0 ° 3 ¹ 3 ¹ ° © © ¾ 2 · 2 · § § ud cos ¨ J ʌ ¸ uq sin ¨ J ʌ ¸ u0 ° °¿ 3 ¹ 3 ¹ © ©

(2.2.30b)


Making inverse transformation in Eqs. (2.2.30a) and (2.2.30b), the voltage conversion formulas, just as Eqs. (2.2.11), (2.2.13), and (2.2.26), can be described as

ª ud º «u » « q» ¬« u0 ¼»

ª2 « 3 cos J « « 2 « sin J « 3 « 1 « ¬ 3

2 2 ·º § cos ¨ J ʌ ¸ » 3 3 ¹» © ª ua º 2 · 2 § 2 ·» « » ʌ ¸ sin ¨ J ʌ ¸ » «ub » 3 ¹ 3 © 3 ¹» «u » »¬ c¼ 1 » 3 ¼

2 2 · § cos ¨ J ʌ ¸ 3 3 ¹ © 2 § sin ¨ J 3 © 1 3

(2.2.31)

Just as stator flux-linkages and currents, ud , uq and u0 are called the stator voltage d-axis, q-axis and zero-axis components, respectively. In Section 2.1, we have got excitation winding and damping winding voltage equations u fd 0 0

p\ fd R fd i fd ½ ° p\ 1d R1d i1d ¾ p\ 1q R1q i1q °¿

(2.2.32)

Formulas (2.2.29) and (2.2.32) compose voltage equations in d, q, 0 axes for synchronous machines.

2.2.4

Park’s Formulas of Synchronous Machines

Equations (2.2.29), (2.2.32), (2.2.24), (2.2.27), and (2.2.28) constitute together the basic relations of synchronous machines in d, q, 0 axes, which are named Park’s formulas and the corresponding quantities are shown in actual values. Advantages of actual values are clear concept and distinct unit dimension, but practical calculation is not convenient. Hence, per-unit systems are often adopted to analyse several problems and discussed in the following section.

2.3

Per-Unit Systems in Synchronous Machines

In sections above we have discussed electromagnetic relations of Salient-pole synchronous machines on the basis of actual values. Advantages of actual values are clear physical concept and explicit unit dimension. However, it is not convenient to calculate practical problems. Therefore, per-unit systems are often used to research into many problems of electric machines. By using per-unit systems, the bases have to be selected at first. Generally speaking, stator rated amplitudes are chosen 89

AC Machine Systems

as the stator bases of synchronous machines. However, so far as the choice of rotor bases is concerned, there are different methods that are discussed as follows.

2.3.1

Per-Unit System of Reversible Mutual Inductances

If I mG , U mG , and \ G are stator current base, voltage base and flux-linkage base that are all corresponding rated amplitudes, I fdG , U fdG , and \ fdG are excitation winding current base, voltage base and flux-linkage base, I1dG , U1d G , and \ 1dG are d-axis damping winding current base, voltage base and flux-linkage base, tG is time base and let kifd

I mG I fd G

kufd

U mG U fdG

\ G / tG \ fdG / tG

\G \ fdG

I mG I1dG U mG U1dG

\ G / tG \ 1dG / tG

\G \ 1dG

ki1d ku1d

½ ° ° ° ° ° ¾ ° ° ° ° ° ¿

(2.3.1)

we can see that the coefficients kifd , kufd , ki1d , and ku1d represent ratios of stator current base (or flux-linkage base) to rotor corresponding bases, respectively. Compared with transformers, those coefficients are similar to transformation ratio of stator winding to rotor winding. In order to keep the per-unit equations having the same form as original equations, when choosing various winding bases we must maintain the following expressions, voltage base current base u impedance base power base voltage base u current base reactance base angular frequency base u inductance base flux-linkage base inductance base u current base voltage base flux-linkage base/time base time base 1/angular frequency base Two sides of the flux-linkage equations

ª\ d º «\ » « fd » «¬\ 1d »¼ 90

ª « Ld « 3 « M afd 0 « 2 « 3 «¬ 2 M a1d 0

M afd 0 L ffd M 1 fd

º M a1d 0 » » ª id º M f 1d » «i fd » »« » » «¬ i1d »¼ L11d » ¼

(2.3.2)


are divided by \ G , \ fdG , and \ 1dG , respectively, and by use of Eq. (2.3.1) and

\G

LG I mG , after arrangement we can obtain ª\ d º « » «\ fd » « » «¬\ 1d »¼

ª Ld « « M fad « M 1ad ¬

M afd L ffd M 1 fd

M a1d º ª i d º » M f 1d » ««i fd »» L11d »¼ «¬ i1d »¼

(2.3.3)

The quantities with bottom-bars above represent corresponding per-unit values, in which ½ Ld ° Ld ° LG ° M afd 0 1 ° M afd LG kifd ° ° 3 ° M afd 0 ° 2 M fad kufd ° LG ° M a1d 0 1 ° M a1d LG ki1d ° ° 3 ° M a1d 0 ¾ (2.3.4) 2 M 1ad ku1d ° LG ° ° L ffd kufd L ffd ° LG kifd ° ° L11d ku1d L11d ° LG ki1d ° M 1 fd ku1d ° M 1 fd ° LG kifd ° M 1 fd kufd ° M f 1d ° LG ki1d ° ¿ We can see that Eqs. (2.3.2) and (2.3.3) have the same form, and various mutual inductances in Eq. (2.3.3) can be reversible or not, i.e. M afd and M fad , M a1d and M 1ad , or M f 1d and M 1 fd may be equal, respectively or not, which depends upon the choice of transformation coefficients kifd , kufd , ki1d , and ku1d at all. If those mutual inductances are made reversible, the following relations have to be maintained, namely, 91

AC Machine Systems

3 kufd 2 3 ku1d 2 kufd ki1d

1 ½ ° kifd ° 1 °° ¾ ki1d ° ku1d ° ° kifd °¿

(2.3.5)

2½ 3 °° ¾ 2° 3 °¿

(2.3.6a)

or kufd kifd ku1d ki1d

From Eq. (2.3.6a) it is clear that the products kufd kifd , and ku1d ki1d each must equal 2/3 in order to get reversible mutual inductances after stator bases are taken. It is well known that in transformers the condition making their mutual inductances reversible is ku ki 1. In synchronous machines there is an electromagnetic connection among stator d-axis winding, excitation winding and d-axis damping winding as in three-winding transformers, but here the condition making their mutual inductances reversible becomes ku ki 2 / 3 , because current id is taken as 2/3 times the fictitious d-axis winding current i( d ) that produces equivalent d-axis air-gap flux density. It is clear that Eq. (2.3.6a) is easily sufficed and coefficients kifd and ki1d (or kufd and ku1d ) can be chosen arbitrarily. The result shows that the methods of choosing rotor d-axis bases are still a lot although stator bases are already chosen and the mutual inductances are made reversible. However, once coefficients kifd and ki1d (or kufd and ku1d ) are chosen, kufd and ku1d (or kifd and ki1d ) are also determined, i.e. only one between kifd and kufd (or ki1d and ku1d ) can be chosen arbitrarily. The rotor q-axis bases can be selected in the same way and we can reach the conclusion, i.e., when mutual inductances are needed to be reversible, only one between coefficients ki1q I mG / I1qG and ku1q \ G /\ 1qG can be selected arbitrarily, in which I1qG and \ 1qG are q-axis damping winding current base and flux-linkage base, respectively, and there is ki1q ku1q 92

2 3

(2.3.6b)


In practice, when we need to choose rotor bases of a synchronous machine, kifd , ki1d and ki1q are selected at first, i.e. rotor current bases are chosen firstly. Although coefficients kifd , ki1d and ki1q are selected arbitrarily, per-unit systems used in practice have only several types. We shall discuss a per-unit system used in this book as below. In the per-unit system, current base of excitation winding is so chosen that the excitation current base produces stator no-load voltage of actual amplitude xad I mG that is evaluated according to air-gap line when a synchronous machine rotates at synchronous speed and its stator winding is open-circuited. Here xad is the synchronous reactance of d-axis armature reaction, xad ZG Lad , and ZG is angular frequency base, namely synchronous angular frequency Z s 2ʌf . On the basis of the conditions mentioned above, there is

ZG M afd 0 I fdG

ZG Lad I mG

xad I mG

or kifd

I mG I fdG

M afd 0

(2.3.7)

Lad

Substituting Eq. (2.3.7) into Eq. (2.3.4) exists M afd

M afd 0 Lad LG M afd 0

L ad

(2.3.8)

i.e., when excitation current base is chosen according to the way above, M afd in per-unit is equal to Lad in per-unit. Substituting Eqs. (2.1.45) and Eq. (2.2.9) into Eq. (2.3.7) we can get the following relation: 3 4 § wkw1 · § I mG ¨ 2 ʌ ¨© 2 P ¸¹ © as

· ¸ Od 11 ¹

w fd

I fd G a fd

Od 1

which shows that fundamental air-gap flux produced by stator 3-phase base currents is equal to that produced by excitation winding current base. Similarly, if letting ki1d

I mG I1dG

M a1d 0 , Lad

ki1q

I mG I1qG

M a1q 0 Laq

and referring to Eq. (2.3.4), then there are M a1d

L ad ,

M a1q

L aq 93

AC Machine Systems

Out of the relations above, it is clear that the per-unit system makes not only mutual inductances reversible but also some mutual inductances equal that are originally not equal to one another, so referring to Eqs. (2.3.4) and (2.3.6) we can get M afd

M fad

M a1d

M 1ad

M a1q

M 1aq

L aq

M f 1d

M 1 fd

2 M f 1d 3 LG 2

L ffd

2 L ffd 3 LG

§ I fdG · ¨ ¸ © I mG ¹

L11d

2 L11d 3 LG

§ I1dG · ¨ ¸ © I mG ¹

L11q

2 L11q 3 LG

§ I1qG · ¨ ¸ © I mG ¹

2

2

L ad ½ ° ° § I fdG I1dG · ° ¨ ¸° 2 © I mG ¹ ° ° ° ¾ ° ° ° ° ° ° ° ¿

(2.3.9)

Substituting those results into Eq. (2.3.3) and considering inductance in per-unit to be equal to reactance in per-unit corresponding to rated angular frequency, we can write the following equation: ª\ º « d» «\ » « fd » «¬\ 1d »¼

ª xd « « x ad « ¬ x ad

x ad x ffd x1 fd

x ad º ª i d º » x f 1d » ««i fd »» x11d ¼» «¬ i1d »¼

(2.3.10)

Similarly, there are

ª\ q º « » «¬\ 1q »¼

ª xq « x ¬ aq

x aq º ª i q º x11q »¼ «¬i1q »¼

(2.3.11)

Because

kufd kifd

§ U mG ¨¨ © U fdG

·§ I mG ¸¨ ¸¨ I ¹© fd G

· ¸¸ ¹

2 3

the excitation voltage base can be determined as

U fdG 94

3 U mG 2

§ I mG ¨¨ © I fdG

· ¸¸ ¹

(2.3.12)


The resistance (or impedance) base of excitation winding should be so chosen that there is no remainder coefficient in equations expressed in per-unit. When current and voltage bases are already chose, resistance (or impedance) base should be taken as Z fdG

U fdG


3 U mG 2

I fdG 3 U mG 2 I mG


· ¸¸ ¹

· 1 ¸¸ ¹ I fd G

2

3 ZG 2


· ¸¸ ¹

2

(2.3.13)

where U mG / I mG ZG xG LG ZG is impedance base of stator winding. Similarly, impedance bases of d-axis and q-axis damping windings are 2 § I mG · ½ ¨ ¸ ° © I1d G ¹ °° 2 ¾ 3 § I mG · ° ZG ¨ ¸ 2 ¨© I1qG ¸¹ °°¿

3 ZG 2

Z1dG Z1qG

(2.3.14)

When voltage equations of stator and rotor windings are described in per-unit, the equation forms are similar to those expressed in actual values. For example, ua p\ a ria for phase a, two sides of which are divided by U mG , and then exists

ua U mG

d\ a 1 ri a dt U mG U mG

d(\ a /\ G ) r ia d(t / tG ) ZG I mG

i.e.

ua

d\ a dt

r ia

(2.3.15)

In addition, u fd p\ fd rfd i fd for excitation winding, two sides of which are divided by U fd G , and then there is

u fd U fdG

d\ fd

1

dt U fdG

rfd i fd

d(\ fd /\ fd G )

U fdG

d(t / tG )

rfd

i fd

Z fdG I fdG

i.e. u fd

p\

fd

r fd i fd

(2.3.16)

95

AC Machine Systems

2.3.2

Park’s Per-Unit System of Irreversible Mutual Inductances

When rotor has only an excitation winding and no damping winding, it is also convenient to use the per-unit system of irreversible mutual inductances, by which Park formulated basic equations of synchronous machines. Without damping windings, the relations corresponding to Eq. (2.3.4) are Ld M afd

M fad L ffd

Ld LG

½ ° ° M afd 0 1 ° ° LG kifd ° ° 3 ¾ M afd 0 ° 2 kufd ° LG ° ° L ffd kufd ° LG kifd °¿

(2.3.17)

Evidently, if mutual inductances M afd and M fad are not asked to be reversible then transformation coefficients kufd and kifd are all chosen arbitrarily. Park chose the bases according to the following way. At first, excitation current base can produce stator rated no-load voltage that is evaluated according to air-gap line when a synchronous machine rotates at synchronous speed. Secondly, excitation voltage base can produce excitation current base and thirdly, excitation current base may produce flux-linkage base of excitation winding. In the light of the conditions above, there is

ZG M afd 0 I fdG

U mG

ZG\ G

ZG LG I mG

i.e. we have selcted

kifd In addition, exists \ fdG

kufd

I mG / I fd G

M afd 0 / LG

(2.3.18)

L ffd I fdG , i.e., we have chosen

\ G /\ fdG

LG I mG / L ffd I fdG

M afd 0 / L ffd

(2.3.19)

According to the results above, it is not difficult to reduce Eq. (2.3.17) to the following formulas:

96


M afd 0

M afd

LG

LG M afd 0

1

3 M afd 0 M afd 0 2 LG L ffd

M fad

L ffd M afd 0

L ffd

LG L ffd M afd 0

LG

½ ° ° 3 2 ° M afd 0 ° ° 2 ¾ LG L ffd ° ° ° 1 ° °¿

(2.3.20)

Substituting the results above into flux-linkage equation of a synchronous machine without damping winding referring to Eq. (2.3.3), the corresponding equations indicated by Park’s per-unit system are

ª \ dp º « » «¬\ fdp »¼

ª x dp 1º ª i dp º «M 1»¼ «¬i fdp »¼ fadp ¬

(2.3.21)

It is noted that the corresponding quantities of Eq. (2.3.21) are all added with subscript P to distinguish Park’s per-unit system from the per-unit system of reversible mutual inductances. As stated before, Park chose the resistance voltage drop produced by excitation current base as the voltage base of excitation winding, namely U fdG in which Td 0

L ffd rfd

rfd I fdG

rfd

I fdG L ffd

rfd

L ffd

L ffd

\ fdG

1 \ fdG Td 0

(2.3.22)

is time constant of excitation winding when stator winding is

opened. It should be noted that here U fdG z \ fd G / tG since tG

1 z Td 0 . 2ʌf

Because p\ fd rfd i fd

u fd we can get u fd U fdG

i.e.

Td 0 tG

u fdP

§\ · d ¨ fd ¸ ¨\ ¸ r i © fdG ¹ fd fd rfd I fd G § t · d¨ ¸ © tG ¹

T d 0 p\

fdP

i fdP

(2.3.23) 97

AC Machine Systems

Comparing the equation expressed by Park’s per-unit system with the original equation indicated by actual values, we can see that there is an obvious difference between them as for their forms. That is because during choice of 2 excitation winding bases we do not comply with the rules of kufd kifd and 3 \G U flux linkage base z mG . , to bring about kufd voltage base time base \ fd G U fdG If describing the corresponding equations of synchronous machine without damping windings by the per-unit system of reversible mutual inductances mentioned before, referring to Eqs. (2.3.10) and (2.3.16) we can obtain x d i d x ad i fd ½ °° x ad i d x ffd i fd ¾ ° p\ fd r fd i fd ° ¿

\d \ fd u fd

(2.3.24)

Comparing the first equation in Eq. (2.3.24) with the corresponding equation in Park’s per-unit system, we can get

\d

\ dP

½ ° i d i dP ° ¾ x d x dP ° x ad i fd i fdP °¿

(2.3.25)

The first three expressions in Eq. (2.3.25) indicate the same choice of stator bases for two per-unit systems. The last expression in Eq. (2.3.25) shows a difference of x ad times between excitation current per-unit values for two per-unit systems, because their excitation current bases are different from each other. Rewriting the second equation in Eq. (2.3.24) as the following form: x ad \ x ffd

2

fd

x ad i d x ad i fd x ffd

2

x ad i d i fdP x ffd

and comparing it with the corresponding equation in Park’s per-unit system referring to Eq. (2.3.21), we can get 2

M fadP

\ fdP 98

½ ° ° ¾ x ad \ ° x ffd fd °¿ x ad x ffd

(2.3.26)


i.e., there is a difference of x ad / x ffd times between excitation winding flux-linkage values for two per-unit systems. 2 x Because ad x d xcd , \ fdP usually has the following form: x ffd

\ fdP

( x d xcd )i dP i fdP

(2.3.27)

in which xcd is called d-axis transient reactance of synchronous machines. Rewriting the third equation in Eq. (2.3.24) as the following form: x ffd

x ad u fd r fd

r fd x ffd r fd

§x p ¨ ad \ ¨x © ffd p\

fdP

· ¸ x ad i fd fd ¸ ¹

i fdP

and comparing it with the corresponding Eq. (2.3.23) in Park’s per-unit system, in which x ffd / r fd L ffd / r fd T d 0 , we can easily know u fdP

x ad u fd r fd

(2.3.28)

i.e., there is a difference of x ad / r fd times between excitation voltages expressed by two per-unit systems. From now on, we shall use the per-unit system of reversible mutual inductances stated before. In the per-unit system, excitation current in per-unit, during no-load rated voltage on air-gap line, is equal to 1/ x ad , and corresponding excitation voltage in per-unit is r fd i fd

r fd / x ad . In many cases, machine parameters in

per-unit are known, and excitation voltage in actual value u fdx corresponding to no-load rated stator voltage and relevant excitation current in actual value i fdx are also known. If saturation effect is neglected, then according to excitation current 1 in actual value i fdx and its per-unit we can write its base as x ad I fdG

i fdx 1/ x ad

x ad i fdx

(2.3.29)

Similarly, on the basis of excitation voltage in actual value u fdx and its per-unit r fd / x ad we can describe its base as 99

AC Machine Systems

u fdx

U fdG

r fd / x ad

x ad u fdx r fd

(2.3.30)

2.4 Basic Equations in Per-Unit, Operational Reactances and Electromagnetic Torque of Synchronous Machines 2.4.1

Basic Equations in Per-Unit of Synchronous Machines1

In Section 2.2, flux-linkage equations and voltage equations in d, q, 0 axes have been found out for synchronous machines, which are named Park’s formulas. In Section 2.3, per-unit systems are also discussed and corresponding voltage equations in per-unit are: ud uq u0 u fd

0 0

p\ d \ qZ rid ½ ° p\ q \ d Z riq ¾ ° p\ 0 ri0 ¿

(2.4.1)

p\ fd R fd i fd ½ ° p\ 1d R1d i1d ¾ p\ 1q R1q i1q °¿

(2.4.2)

The flux-linkage equations in per-unit are:

\d \q \0 \ fd \ 1d \ 1q

xd id xad i fd xad i1d ½ ° xq iq xaq i1q ¾ ° x0 i0 ¿

(2.4.3)

xad id X ffd i fd X f 1d i1d ½ ° xad id X 1 fd i fd X 11d i1d ¾ ° xaq iq X 11q i1q ¿

(2.4.4)

From the equations above we can see that original quantities in a, b, c axes disappear and there are the quantities corresponding to d, q, 0 axes. Now exist three circuits that are fictitious circuits obtained by reference axis conversion. The first circuit is called stator d-axis circuit whose axis coincides with rotor d-axis and rotates synchronously with rotor d-axis. Its flux-linkage is \ d , current 1

The signs with bottom-bars indicate per-unit values, but bottom-bars are omitted from now on.

100


is id , self-inductance is xd and resistance is r. It has a magnetic connection with rotor d-axis windings and the mutual inductance is xad . The second circuit is called stator q-axis circuit whose axis coincides with rotor q-axis and also rotates synchronously with rotor q-axis. Its flux-linkage is \ q , current is iq , self-inductance is xq and resistance is r. It has a magnetic connection with rotor q-axis winding and the mutual inductance is xaq . The third circuit is called stator 0-axis circuit that has no electromagnetic connection with other windings. Its flux-linkage is \ 0 , current is i0 , self-inductance is x0 and resistance is r. From Eq. (2.4.1) it is clear that stator d-axis voltage consists of three parts. The first part p\ d is transformer emf caused by change of d-axis flux-linkage. The second part \ qZ is speed emf produced in stator d-axis winding by revolving q-axis flux-linkage and the third part rid is the resistance voltage-drop of stator d-axis winding. Their algebraic sum should be equal to terminal voltage of stator d-axis winding and according to the reference direction rules described in Chapter 1 the sum is ( p\ d \ qZ rid ). Comparing this result with Eq. (2.2.29), we can see that ud in Eq. (2.2.29) is just the value. Therefore, d-axis component of stator voltage converted by Eq. (2.2.31) gets clear physical concept now, i.e., it is just the terminal voltage the fictitious stator d-axis winding should have. Similarly, stator q-axis voltage consists of three parts too, in which p\ q is transformer emf caused by change of q-axis flux-linkage, \ d Z is speed emf produced in stator q-axis winding by revolving d-axis flux-linkage and riq is the resistance voltage drop of stator q-axis winding. Their algebraic sum ( p\ q \ d Z riq ) should be equal to terminal voltage of stator q-axis winding. Just as stator d-axis winding, the terminal voltage is also uq , the q-axis component of stator voltage converted by formula (2.2.31). So far as stator 0-axis winding is concerned, because it has no electromagnetic connection with other windings, its terminal voltage is only the algebraic sum of transformer emf p\ 0 caused by change of its own flux-linkage and resistance voltage-drop ri0 , namely ( p\ 0 ri0 ) . That is also u0 the 0-axis component of stator voltage converted by formula (2.2.31). When fundamental flux is considered only, 0-axis flux-linkage has no magnetic connection with rotor winding so that it doesn’t produce any torque. During calculation, 0-axis component can be processed alone. Comparing basic Eqs. (2.4.1) (2.4.4) of synchronous machines with those of dc machines except 0-axis component, there is some similarity between them. The fact shows that 3-phase ac machine after conversion may be replaced by a dc machine. The physical model of dc machine is shown in Fig. 2.4.1. It is a dc machine with two pairs of brushes that are set on d-axis or q-axis respectively. The armature circuit determined by d-axis brushes corresponds to stator d-axis winding of synchronous machines and the armature circuit determined by q-axis brushes corresponds to stator q-axis winding of synchronous machines. Excitation 101

AC Machine Systems

winding of synchronous machines corresponds to excitation winding set on d-axis pole of dc machine, and d-axis and q-axis damping windings of synchronous machines correspond to two short-circuited coils set on d-axis and q-axis poles of dc machine, respectively.

Figure 2.4.1 Physical pattern of 3-phase ac machine after axis conversion

Obviously, 0-axis component can be represented by a static circuit as shown in Fig. 2.4.2.

Figure 2.4.2 Static circuit corresponding to 0-axis component

2.4.2 Equivalent Circuits and Operational Reactances of Synchronous Machines After obtaining the basic equations of synchronous machines with reversible and constant mutual inductances in per-unit, we can draw the corresponding equivalent circuits. Using Heaviside’s or Laplace’s operational calculus to analyse many 102


operating modes of synchronous machines, those equivalent circuits are significant. Now, according to different types of machines, two conditions will be discussed. (1) Having Excitation Winding and No Damping Winding on Rotor Under that condition, the stator d-axis flux-linkage equation and excitation winding voltage equation can be changed into

\d

xd id xad i fd

u fd

§ xad id ¨ X ffd ©

p

½ ° R fd · ¾ i ¸ fd p ¹ °¿

(2.4.5)

According to the expressions above, the equivalent circuit is plotted in Fig. 2.4.3.

Figure 2.4.3 d-axis operational equivalent circuit when having excitation winding only on rotor

That circuit shows the relation between flux-linkages and currents. Comparing it with usual static circuits, the flux-linkage in Fig. 2.4.3 corresponds to a voltage, reactance corresponds to a resistance and the resistance corresponds to a capacitance. In Fig. 2.4.3, ( xl xd xad ) is stator leakage reactance and ( X fdl X ffd xad ) is excitation winding leakage reactance, which can be proved as follows. Referring to (2.3.9) and (2.3.7) there is X ffd

L ffd

2 L ffd § I fdG · ¨ ¸ 3 LG © I mG ¹

2

2 L ffd 3 LG

§ Lad · ¨¨ ¸¸ © M afd 0 ¹

2

and on the basis of Eqs. (1.5.3), (2.1.45) and (2.2.9) exist L ffd M afd 0

W lP a 2fd

w2fd O0 L fdl

2ww fdW l ʌas a fd

Od 1k w1 2

Lad

3 4W lP § wk w1 · Od 11 2 as2 ʌ 2 ¨© P ¸¹

so we can obtain 103

AC Machine Systems

X ffd

ª 3 1 wk w1 º « S a P Od 11 » ª º 2 1 W lP 2 s » « 2 w fd O0 L fdl » « 1 » 3 Lį «¬ a fd »¼ « « a w fd Od 1 » fd ¬ ¼ L ad

2

O0 Od 11 2 L fdl Od21 3 LG

§O O xad xad ¨ 0 2d 11 © Od 1

2

§ I fdG · ¨ ¸ © I mG ¹ · 1¸ X cfdl ¹

or X fdl

X ffd xad

§O O · X cfdl xad ¨ 0 2d 11 1¸ © Od 1 ¹

in which X cfdl is excitation winding leakage reactance component corresponding to end leakage flux, pole-to-pole leakage flux and leakage flux between two §O O · adjacent pole-surfaces, xad ¨ 0 2d 11 1¸ is excitation winding differential leakage © Od 1 ¹ reactance, and X fdl is excitation winding total leakage reactance. In many cases only stator quantities need studying, so it is preferable to delete rotor quantities from basic equations to obtain simpler equations. Using Thevenin’s Law and the equivalent circuit in Fig. 2.4.3, we can get the following simple equation:

\d

G ( p )u fd xd ( p )id

(2.4.6)

in which G ( p )u fd is the voltage when terminals a, b in Fig. 2.4.3 are opened, i.e., G ( p )u fd

u fd p u fd

xad xad X fdl

R fd p

xad p( xad X fdl ) R fd

xad u fd pX ffd R fd Hence, there is G( p) 104

xad pX ffd R fd

(2.4.7)


xd ( p ) is short circuit impedance observed from terminals a, b when the supply u fd / p is short-circuited, i.e., xd ( p)

xl

1 1 xad

xd xad xd xd xd

1 X fdl

R fd

p pxad X fdl xad R fd pX ffd R fd

pxad X ffd pxad X fdl xad R fd xad R fd pX ffd R fd pxad ( X ffd X fdl ) pX ffd R fd 2 pxad pX ffd R fd

(2.4.8)

G ( p ) is called operational conductance and xd ( p ) is called d-axis operational reactance. They are all operational expressions that are independent of machine speed. Therefore, they can be used to study any operating mode of synchronous machine at any speed. Supposing u fd 0 in Eq. (2.4.6), we can get \ d xd ( p)id . As stated above, if the source u fd / p is short-circuited, the impedance observed from terminals a, b will be operational reactance xd ( p ). Thus, this type of circuit is called equivalent circuit of stator d-axis operational reactance for synchronous machines. Without damping windings, the stator q-axis flux-linkage equation is

\q

xq iq

(2.4.9)

The corresponding equivalent circuit is shown in Fig. 2.4.4. Accordingly, this type of circuit is called equivalent circuit of stator q-axis operational reactance for synchronous machines.

Figure 2.4.4 q-axis operational equivalent circuit without damping windings 105

AC Machine Systems

(2) Having a Set of d-axis and q-axis Damping Windings besides Excitation Winding on Rotor Under that condition, the stator d-axis flux-linkage equation, excitation winding and d-axis damping winding voltage equations can be described as ª\ d º «u » « fd » « p » « » ¬ 0 ¼

ª x « d « « « xad « « « xad ¬

xad X ffd

º » »ªi º »« d » X f 1d » «i fd » » «i » ¬ 1d ¼ R1d » X 11d » p ¼

xad

R fd p

X 1 fd

(2.4.10)

The equivalent circuit of d-axis operational reactance corresponding to the matrix equation above is shown in Fig. 2.4.5.

Figure 2.4.5 d-axis operational equivalent circuit when having a d-axis damping winding besides excitation winding on rotor

Similarly, after deleting rotor currents from stator flux-linkage equation exists

\d

G ( p )u fd xd ( p )id

(2.4.11)

in which G( p)

p ( X 11d xad X f 1d xad ) xad R1d A( p)

xd ( p )

xd

B( p) A( p )

A( p )

p 2 ( X 11d X ffd X 2f 1d ) p( X 11d R fd X ffd R1d ) R1d R fd

B( p)

2 2 2 2 2 p 2 ( X 11d xad 2 X f 1d xad X ffd xad ) p( xad R1d xad R fd )

(2.4.12)

(2.4.13a)

If neglecting the mutual leakage reactance between excitation winding and d-axis damping winding, i.e., considering that there is no mutual flux that links 106


only the two windings and does not link stator winding, then we can write the mutual inductance between excitation winding and d-axis damping winding as M f 1d

M f 1dG M f 1dl

M f 1dG

2w1d w fdW l 1 E S Od 1 sin 1d 2 S a fd Referring to Eqs. (2.3.9) and (2.3.7), its per-unit can be described as M f 1d

2 M f 1d § I fd G I1dG · ¨ ¸ 3 LG © I m2G ¹ 2 M f 1d § Lad · § Lad · ¨ ¸¨ ¸ 3 LG ¨© M afd 0 ¸¹ © M a1d 0 ¹

After substituting the relations of Lad , M afd 0 , M f 1d and M a1d 0 into the equation above referring to Eqs. (2.2.9), (2.1.45), and (2.1.48) and arranging it in order, we can get M f 1d

L ad

X f 1d

xad

namely,

Accordingly, there is xd ( p)

xd

2 3 2 2 2 ( X 11d xad 2 xad X ffd xad ) p 2 ( xad R1d xad R fd ) p 2 ( X 11d X ffd xad ) p 2 ( X 11d R fd X ffd R1d ) p R1d R fd

(2.4.13b)

and the equivalent circuit in Fig. 2.4.5 can be reduced to that in Fig. 2.4.6.

Figure 2.4.6 The reduced d-axis operational equivalent circuit

In the circuit in Fig. 2.4.6, X 1dl X 11d xad is the leakage reactance of d-axis damping winding, which is proved in the same way as X fdl . Considering harmonic effect, we have to add differential leakage reactance of d-axis damping winding to leakage reactance X 1dl . 107

AC Machine Systems

The stator q-axis flux-linkage equation and q-axis damping winding voltage equation are ª\ q º «0» ¬ ¼

ª xq « « x «¬ aq

xaq

º ª iq º » R1q » « » i1q p »¼ ¬ ¼

X 11d

(2.4.14)

The equivalent circuit of q-axis operational reactance corresponding to the equations above is shown in Fig. 2.4.7.

Figure 2.4.7 q-axis operational equivalent circuit with one q-axis damping winding

In the circuit in Fig. 2.4.7, ( X 1ql X 11q xaq ) is the leakage reactance of q-axis damping winding. Considering harmonic effect, we have to add differential leakage reactance of q-axis damping winding to leakage reactance X 1ql , too. After deleting rotor current, there is

\q

xq ( p )iq

(2.4.15)

in which xq ( p )

xl

1 1 xaq

xq

1 X 1ql

R1q

2 pxaq

pX 11q R1q

(2.4.16)

p

xq ( p) is called the operational reactance of stator q-axis. So far as the non-salient pole synchronous machine is concerned, its d-axis and q-axis equivalent circuits are similar to Figs. 2.4.5 2.4.7, but xad xaq only. As for an induction machine, there is xad xaq xm owing to uniform air-gap. Since induction machine rotor is electromagnetically symmetrical with respect to any axis, its d-axis and q-axis operational reactances are equal to each other, i.e., xd ( p)

xq ( p)

x( p )

in which xs is stator leakage reactance, 108

xs

1 1 xm

(2.4.17)

1 Xr

Rr p


X r and Rr are rotor equivalent leakage reactance and resistance, respectively referring to Chapter 5. The equivalent circuit of operational reactance is shown in Fig. 2.4.8.

Figure 2.4.8 Equivalent circuit of operational reactance x( p) for induction machines

Because the rotor winding of induction machine is often short-circuited, stator d-axis and q-axis flux-linkages can be written as

\d \q

x( p )id x( p)iq

Therefore, the corresponding equivalent circuits of d-axis and q-axis operational reactances are shown in Figs. 2.4.9(a) and 2.4.9(b).

Figure 2.4.9 d-axis and q-axis operational equivalent circuits of induction machines

2.4.3

Output Power and Electromagnetic Torque in Synchronous Machines

The instantaneous output power of 3-phase synchronous machine, when expressed in actual values, is p

ua ia ub ib uc ic

(2.4.18)

The power base is PG

3UI

§ U ·§ I · 3 ¨ mG ¸¨ mG ¸ © 2 ¹© 2 ¹ 109

AC Machine Systems

in which U and I are rms values of stator rated phase voltage and current, U mG and I mG are instantaneous bases of phase voltage and current. Thus, instantaneous output power in per-unit can be described as 2 (ua ia ub ib uc ic ) 3

p

(2.4.19)

Substituting the variables in d, q, 0 axes for variables in a, b, c axes as stated before, we can write the instantaneous output power in per-unit according to d, q, 0 axes as p

ud id uq iq 2u0 i0

(2.4.20)

Substituting Eq. (2.4.1) into Eq. (2.4.20) exists p

id ( p\ d \ qZ rid ) iq ( p\ q \ d Z riq ) 2i0 ( p\ 0 ri0 )

(id p\ d iq p\ q 2i0 p\ 0 ) (iq\ d id\ q )Z r (id2 iq2 2i02 )

(2.4.21)

Observing various terms in the equation above, we can see that the first term (id p\ d iq p\ q 2i0 p\ 0 ) is total products of currents and rates of flux-linkage change. Assuming negative flux-linkage to be produced by positive current as mentioned before, the first term will represent decrease rate of stator magnetic field energy. The third term r (id2 iq2 2i02 ) indicates stator resistance loss. On the basis of power balance, the second term (iq\ d id\ q )Z should be the power delivered to stator across air-gap. Because

electromagnetic torque

the power delivered to stator across air-gap rotor speed

the electromagnetic torque of a synchronous machine is Te

iq\ d id\ q

(2.4.22)

It should be pointed out that the basic equations of synchronous machines, after converting a, b, c axes to d, q, 0 axes, are changed from differential equations with variable coefficients to those with constant coefficients, but u , i, \ , Z are still variables with time. The equations as stated before include xd ( p ), xq ( p ) and p together with time-variables u (t ), i (t ), \ (t ) and Z (t ), which are called Heaviside’s operational calculus equations and different from pure operational calculus equations with u ( p ), i ( p ) and z ( p). If necessary, refer to Reference [2] or Appendix D in this book.

110


2.5

3-Phase Sudden Short Circuit and Transient Parameters of Synchronous Machines

As mentioned before, basic equations of synchronous machines in d, q, 0 axes are differential equations with constant coefficients, by which we can study several operating modes of synchronous machines simply, especially those when stator phase windings are connected symmetrically. Among of them, 3-phase sudden short circuit will threaten synchronous machines. Thus, comprehension of the transients may provide scientific basis not only for reasonable design and reliable operation of synchronous machines but also for a careful choice of relay protection. In addition, the analysis methods and parameters derived for study of the problems above are of importance to analyse other transients, too. Therefore, this section will lay a foundation for researching further into other transients of synchronous machines. Studying 3-phase sudden short circuit, we can consider the short circuit not to have happened but the stator terminals to be suddenly applied on by 3-phase voltages that are equal but opposite to the phase voltages before short circuit. In thinking so, the problem of 3-phase sudden short circuit of synchronous machines is turned into the synthetic problem of the following two operation conditions: (i) Steady-state operation condition that is the same as the condition before short circuit. (ii) Excitation winding is short-circuited and stator terminals are suddenly applied on by 3-phase voltages that are equal but opposite to the phase voltages before short circuit. Because basic equations are linear under that condition, we can use superposition theorem to treat the problem above.

2.5.1

Original Steady-State Operation Condition

In steady-state operation, stator emfs can be described as ea eb ec

E sin J

½ ° E sin(J 120e )¾ E sin(J 120e ) °¿

(2.5.1)

When a synchronous machine is loaded, there is an angle between its terminal voltage and no-load emf due to load current. If terminal voltage lags behind no-load emf by a phase angle G, then instantaneous terminal voltages can be written as 111

AC Machine Systems

U sin(J G )

ua

½ ° U sin(J G 120e )¾ U sin(J G 120e ) °¿

ub uc

(2.5.2)

in which U is terminal voltage amplitude. By use of d, q, 0 axes, the equations above can be converted into ª ud º « » « uq » «¬ u0 »¼

ª cos J « 2« « sin J 3« 1 « ¬ 2

cos(J 120e )

cos(J 120e )º » ª ua º » ) sin(J 120e ) » ««ub »» sin(J 120e » «u » 1 1 »¬ c¼ 2 2 ¼

ªU sin G º «U cos G » « » ¬« 0 ¼»

(2.5.3)

i.e., ud and uq are constant. In the same way it can be proved that stator currents id and iq are also constant. Therefore, according to Eq. (2.4.3) the flux-linkages can be written as

\d

xad i fd xd id

\q

xq iq

E xd id

Constant

Constant ½° ¾ °¿

(2.5.4)

Substituting the expressions above into Park’s equations, we can obtain ud

\ q rid

uq

\ d riq

xq iq rid

½° ¾ E xd id riq °¿

(2.5.5)

Solving the equation above, we get id iq

rud xq ( E uq ) 2

r xd xq xd ud r ( E uq ) r 2 xd xq

rU sin G xq ( E U cos G ) ½ ° r 2 xd xq ° ¾ xd U sin G r ( E U cos G ) ° ° r 2 xd xq ¿

(2.5.6)

In practice, the stator resistance r is small and may be neglected, so the expressions above can also be simplified as id iq

E uq

E U cos G ½ ° xd xd ° ¾ ud U sin G ° ° xq xq ¿

(2.5.7)

After having current and voltage relations, we can further analyse the output power, reactive power and electromagnetic torque. 112


(1) Output power Substituting the results mentioned above into Eq. (2.4.20), we can describe output power as P

ud id uq iq ª rU sin G xq ( E U cos G ) º U sin G « » U cos G r 2 xd xq ¬« ¼» 1 r xd xq 2

ª xd U sin G r ( E U cos G ) º « » r 2 xd xq ¬« ¼»

xd xq 2 ª º 2 U sin 2G ( xq sin G r cos G )UE » « rU 2 ¬ ¼ (2.5.8)

If stator resistance r is neglected, we can obtain P

§ 1 EU 1 ·U 2 sin G ¨ ¸ sin 2G ¨x ¸ xd © q xd ¹ 2

(2.5.9)

Because d-axis and q-axis synchronous reactances are equal to each other in non-salient-pole synchronous machines, output power under the condition r 0 is P

EU sin G xd

(2.5.10)

The curve of output power P changing with angle G is termed power-angle characteristic of a synchronous machine. The stator resistance neglected, power-angle characteristic of a non-salient pole synchronous machine is shown EU and corresponding angle in Fig. 2.5.1 where the maximum output power is xd

G 90e. The power-angle characteristic of a salient-pole machine is shown in Fig. 2.5.2 after neglecting stator resistance, and the maximum output power will appear in the range of 45e 90ebecause of the second-order harmonic.

Figure 2.5.1 Power-angle characteristic of non-salient-pole synchronous machines

Figure 2.5.2 Power-angle characteristic of salient-pole synchronous machines 113

AC Machine Systems

(2) Reactive power If there is no 0-axis component, on the basis of Eq. (2.5.3) we can obtain ud2 uq2 ½° ¾ id2 iq2 °¿

U I

(2.5.11)

in which U and I are amplitudes of stator voltage and current, respectively. So the reactive power is

Q

S 2 P2 (ud2 uq2 )(id2 iq2 ) (ud id uq iq )2 (2.5.12)

uq id ud iq

in which S UI is the apparent power of a synchronous machine. Substituting Eqs. (2.5.3) and (2.5.6) into the formula above, we can obtain Q

ª º 1 U2 U2 cos 2G » ( xd xq ) «( xq cos G r sin G )UE ( xd xq ) r xd xq ¬ 2 2 ¼ 2

(2.5.13) If stator resistance is neglected, we can get Q

§1 EU 1 ·U 2 § 1 1 ·U 2 ¨ ¸ cos G ¨ ¸ cos 2G ¨x ¸ ¨ ¸ xd © q xd ¹ 2 © xq xd ¹ 2

(2.5.14)

In non-salient-pole synchronous machine, there is Q

EU U2 cos G xd xd

(2.5.15)

(3) Electromagnetic torque Substituting expressions (2.5.4) and (2.5.6) into formula (2.4.22) and arranging it in order, we can get

Te

iq\ d id\ q

r (r 2 xq2 ) 2

(r xd xq )

2

E2

xd xq

ª ( xd sin 2 G xq cos 2 G )r (r xd xq ) 2 «¬ 2

1 EU º ( xd xq r 2 )sin 2G » U 2 2 [(rxd sin G xq2 cos G )2r 2 (r xd xq )2 ¼ ( xd xq r 2 )(r cos G xq sin G )] 114

(2.5.16)


or Te

Tk Tr Ts

(2.5.17)

in which the first term r (r 2 xq2 )

Tk

(r 2 xd xq )2

E2

(2.5.18)

is the short circuit torque of a synchronous machine. It depends upon no-load emf E, is independent of terminal voltage U, and corresponds to the torque produced when excitation current is constant and stator winding is short-circuited. When excitation current is constant and stator winding is short-circuited, referring to equation (2.5.6) the corresponding stator current can be written as xq E

½ ° r xd xq ° ¾ rE ° r 2 xd xq °¿

id

2

iq

(2.5.19)

i.e. the amplitude of stator short-circuit current is id2 iq2

ik

r 2 xq2 r 2 xd xq

E

Therefore the stator loss caused by short circuit current is ik2 r

r (r 2 xq2 ) 2

(r xd xq )

2

E2

which is equal to the short circuit torque, i.e. ik2 r

Tk

(2.5.20)

Obviously, when r 0, i.e. there is no resistance loss in stator, the torque will be equal to zero. The torque Tk is an asynchronous torque produced through the interaction between the rotating field caused by excitation current and that caused by stator short-circuit current induced by excitation field. Compared with an induction machine, its slip is equal to 1, so Tk ik2 r can be obtained. The second term

115

AC Machine Systems

Tr

xd xq

1 ª º ( xd sin 2 G xq cos 2 G )r ( xd xq r 2 )sin 2G » U 2 (r xd xq ) 2 «¬ 2 ¼ 2

(2.5.21)

is the salient pole torque, which depends upon terminal voltage and is independent of no-load emf. That is the torque caused by difference between d-axis and q-axis synchronous reactances, i.e. by difference between d-axis and q-axis air-gap permeances. If air-gap is uniform, for example in a non-salient pole synchronous machine, then there are xd

xq ,

Tr

0

If stator resistance is neglected, then Tr

1§ 1 1 · 2 ¨¨ ¸¸U sin 2G 2 © xq xd ¹

(2.5.22)

The third term Ts

UE [(rxd sin G xq2 cos G )2r ( xd xq r 2 )(r cos G xq sin G )] 2 (r xd xq ) 2

(2.5.23) is the synchronous torque, which depends upon not only terminal voltage but also no-load emf. If stator resistance is neglected, then exists Ts

UE sin G xd

(2.5.24)

From the analysis above we can see that when a synchronous machine operates symmetrically in steady-state, various components of the electro-magnetic torque are all constants, i.e. there is only average torque and no pulsating torque. That is natural, because under this condition there are only synchronously rotating fields and no others. When stator resistance is neglected, the resultant electromagnetic torque is Te

UE U2 § 1 1· sin G ¨¨ ¸¸ sin 2G xd 2 © xq xd ¹

(2.5.25)

Comparing the formula above with equation (2.5.9), we can see that the output power and electromagnetic torque of a synchronous machine are equal to each other in values when r 0. That is natural, because under this condition the 116


power across air-gap to stator corresponding to the electromagnetic torque is delivered to network without any loss.

2.5.2

Current Components Caused by 3-phase Sudden Short Circuit

Before short circuit, referring to (2.5.3) there are ud

U sin G

ud 0

uq

U cos G

uq 0

so after short circuit, it is equivalent to the situation that ud 0 and uq 0 are applied suddenly to stator terminals during closed excitation winding as mentioned before, i.e. during synchronous speed, namely Z 1 exist ud

ud 0

U sin G

uq

uq 0

U cos G

p\ d \ q rid ½° ¾ p\ q \ d riq °¿

Because excitation winding is short-circuited, i.e. u fd can be reduced to

\d \q

(2.5.26)

0, flux-linkage equations

xd ( p)id ½° ¾ xq ( p )iq °¿

(2.5.27)

Substituting (2.5.27) into (2.5.26) and solving them, we can obtain id iq

[ pxq ( p ) r ]ud 0 xq ( p )uq 0

½ ° [ pxd ( p ) r ][ pxq ( p ) r ] xd ( p ) xq ( p) ° ¾ [ pxd ( p ) r ]uq 0 xd ( p )ud 0 ° [ pxd ( p ) r ][ pxq ( p ) r ] xd ( p ) xq ( p) °¿

(2.5.28)

According to machines with or without damping windings, change of those currents will be discussed respectively as follows. (1) Current change of a machine without damping windings A machine having no damping winding, its operational reactances are xd ( p)

xd

xq ( p)

xq

2 ½ pxad ° R fd pX ffd ¾ ° ¿

(2.5.29)

The corresponding current expressions are 117

AC Machine Systems

id

iq

1

½ ° ° °° ¾ 1 2 2 ° x X x u p x R rX u {( ) [( ) d ffd ad q0 d fd ffd q0 ° ap 3 bp 2 cp d ° 2 ( xd X ffd xad )ud 0 ] p (rR fd uq 0 xd R fd ud 0 )} °¿

{xq X ffd ud 0 p 2 [(rX ffd R fd xq )ud 0 ap bp 2 cp d xq X ffd uq 0 ] p (rR fd ud 0 xq R fd uq 0 )} 3

(2.5.30)

in which a b c d

2 xd xq X ffd xad xq

½ ° 2 r ( xd xq ) X ffd rxad xd xq R fd °° ¾ 2 2 xd xq X ffd xad xq ( xd xq )rR fd r X ffd ° ° xd xq R fd r 2 R fd °¿

(2.5.31)

Its characteristic equation is ap 3 bp 2 cp d

0

(2.5.32)

The general solution of the equation above is p1 p2 p3

a 3

½ ° ° 1 3 a° (u1 v1 ) ¾ (u1 v1 ) j 2 2 3° 1 3 a° (u1 v1 ) ° (u1 v1 ) j 2 2 3¿

u1 v1

(2.5.33)

in which u1

3

q q 2 m2 2 4 27

v1

3

q q 2 m2 2 4 27

q m

2a 3 ab c 27 3 a2 b 3

The calculation is tedious and it is not easy to predict the effect of machine 118


parameters on currents. Therefore, we shall find approximate solutions according to actual condition of machine parameters. Because stator and rotor resistances are much smaller than reactances, they may be neglected during approximate calculation. Now we find the approximate solutions according to two conditions as follows. (a) Neglecting excitation winding resistance Referring to equation (2.4.8), the operational reactance under this condition will be a constant, i.e. xd ( p) |R fd

0

xd

2 xad X ffd

xl

1 1 1 xad X fdl

xdc

(2.5.34)

which is termed d-axis transient reactance of a synchronous machine. When the excitation voltage and resistance of a synchronous machine without damping windings are all equal to zero, its equivalent circuit of operational reactance changes from Fig. 2.4.3 to Fig. 2.5.3.

Figure 2.5.3 Equivalent circuit of d-axis transient reactance

From Eq. (2.5.28) we can see that the characteristic equation under this condition is reduced to ( pxdc r )( pxq r ) xdc xq

(2.5.35)

0

whose solution is

p1,2

r§ 1 1· r2 § 1 1· ¨ ¸ r j 1 ¨ ¸ ¨ ¸ ¨ 2 © xdc xq ¹ 4 © xdc xq ¸¹

2

(2.5.36)

2

In practice,

1

r2 § 1 1· ¨¨ ¸¸ | 1, so 4 © xdc xq ¹

p1,2 |

1 r xdc xq rj rj c 2 xd xq Ta

(2.5.37) 119

AC Machine Systems

in which 1 2 xdc xq r xdc xq

Ta

x2 r

(2.5.38)

and x2

2 xdc xq

(2.5.39)

xdc xq

is the negative sequence reactance. (b) Neglecting stator winding resistance From equation (2.5.28) we can know that the characteristic equation under this condition is p 2 xd ( p ) xq xd ( p) xq

0

i.e. 2 § pxad ( p 2 1) xq ¨ xd ¨ R fd pX ffd ©

· ¸¸ 0 ¹

(2.5.40)

Its roots are p1,2 p3

rj

xd R fd 2 ad

xd X ffd x xd 1 xdc Td 0

xd R fd xdc X ffd

1 Tdc

in which

Tdc

xdc Td 0 xd

ª 1 « 1 « X fl 1 1 R fd « «¬ xad xl

º » » » »¼

(2.5.41)

is the time constant of excitation winding when stator winding is short-circuited whose resistance r is equal to zero referring to Fig. 2.4.3. Obviously, the roots of equation (2.5.40) p1,2 r j correspond to the roots p1,2 120

1 r j obtained above referring to equation (2.5.37) because during r Ta

0


both are equal. According to analysis of the two approximate conditions, the roots of characteristic equation (2.5.32) have the following approximate values: p1,2 p3

1 ½ r j° Ta ° ¾ 1 ° °¿ Tdc

(2.5.42)

Therefore, the characteristic equation can be written as § ·§ · 1 ·§ 1 1 ¨ p ¸¨ p j ¸¨ p j ¸ Tdc ¹© Ta Ta © ¹© ¹

0

(2.5.43)

Substituting the result above into equation (2.5.30), we can get {xq X ffd ud 0 p 2 [(rX ffd R fd xq )ud 0 xq X ffd uq 0 ] p (rR fd ud 0 xq R fd uq 0 )}

id

·§ · °½ 1 ·§ 1 1 ° § ®a ¨ p ¸¨ p j ¸¨ p j ¸ ¾ Tdc ¹© Ta Ta °¯ © ¹© ¹ °¿ iq

2 2 {( xd X ffd xad )uq 0 p 2 [( xd R fd rX ffd )uq 0 ( xd X ffd xad )ud 0 ] p

(rR fd uq 0 xd R fd uq 0 )}

° § ·§ · ½° 1 ·§ 1 1 ®a ¨ p ¸¨ p j ¸¨ p j ¸ ¾ Tdc ¹© Ta Ta ¹© ¹ ¿° ¯° ©

Because a

2 xd xq X ffd xad xq

xdc xq X ffd

we can get id

iq

° ª§ r º § r ud 0 uq 0 · ½° 1 · 2 ¸¸ ud 0 uq 0 » p ¨¨ ¸¸ ¾ ®ud 0 p «¨¨ °¯ © xq Td 0 Td 0 ¹ ¿° ¬«© xq Td 0 ¹ ¼» ª § 1 ·§ 2 2 1 ·º « xdc ¨ p ¸¨ p p 1 2 ¸ » Tdc ¹© Ta Ta ¹ »¼ «¬ © ª§ 1 º § r u · °½ r · ° 2 ¸ uq 0 u d 0 » p ¨ uq 0 d 0 ¸ ¾ ®uq 0 p «¨ Tdc ¹ ¿° © xdc Td 0 ¬© Tdc xdc ¹ ¼ ¯° ª § 1 ·§ 2 2 1 ·º « xq ¨ p ¸¨ p p 1 2 ¸ » Tdc ¹© Ta Ta ¹ »¼ «¬ ©

½ ° ° ° ° ° ° ¾ ° ° ° ° ° ° ¿

(2.5.44)

121

AC Machine Systems

Using Decomposition Theorem to solve the equation above, we can obtain

id

A Be

t Tdc

2Ce

t Ta

cos(t T )

in which § r · Tdc Tc 1 uq 0 ¨¨ ud 0 uq 0 ¸¸ | d uq 0 xdc Td 0 xd § 1 · © xq ¹ xdc Td 0 ¨1 2 ¸ © Ta ¹ °§ 1 · 2 º § r u uq 0 · ½° 1 ª§ r 1 · d0 ®¨ ¸ ud 0 «¨ ud 0 uq 0 » ¨ ¸ ¸¸ ¾ ¨ Tc Tdc ¬«¨© xq Td 0 ¸¹ ¼» © xq Td 0 Td 0 ¹ ¿° ¯°© d ¹

A

B

° 1 ª§ 1 · 2 2 1 1 º ½° c « »¾ 1 x ® d ¨ ¸ Ta2 » ° °¯ Tdc «¬© Tdc ¹ Ta Tdc ¼¿ º ª1 1 ª 1 1 1º | u u u 1 ¬« Tdc q 0 Td 0 q 0 ¼» ¬« xdc xd »¼ q 0 xdc Tdc 2 ° ª§ 1 · § r · r 1 º ª 1 1 ·§ 1 « » ud 0 « j j ¸¸ ¨ j ¸ ® ¨ ¸ ¨¨ T x T T x T d0 ¹© a q d0 » ¹ © q ¹ ¬ Ta °¯ «¬© a ¼

Ce jT

|

1 º °½ » uq 0 ¾ Td 0 ¼ ¿°

·§ 1 1· ° § 1 °½ ® xdc ¨ j ¸¨ j ¸ (2 j) ¾ Tdc ¹ ¹© Ta ¯° © Ta ¿°

( j) 2 ud 0 juq 0 2 jxd ( j)

1 (uq 0 jud 0 ) 2 xdc

2

U jG e 2 xdc

i.e. C

1 2 xdc

ud2 0 uq20

T

arctan

ud 0 uq 0

sin T cosT Therefore, 122

G

ud 0 2 d0

u uq20 uq 0 ud2 0 uq20

U 2 xdc

2 Analysis of Several Operating Conditions of Synchronous Machines at Synchronous Speed t t ½ ª§ 1 1 · T c 1 º 1 T «¨ ¸ e d » uq 0 e a ª¬ud 0 sin t uq 0 cos t º° ¼ xd »¼ xdc ¬«© xdc xd ¹ °° ¾ t t ª§ 1 ° 1 · Tdc 1 º U Ta «¨ ¸ e » U cos G e cos(t G ) ° xd ¼» xdc «¬© xdc xd ¹ °¿

id

(2.5.45)

Similarly, we can obtain t

iq

1 1 ud 0 e Ta [ud 0 cos t uq 0 sin t ] xq xq t

U U sin G e Ta sin(t G ) xq xq

Transforming them to a, b, c axes, we can get ia

t ª§ 1 u 1 · T c 1 º «¨ ¸ e d » uq 0 cos(t J 0 ) d 0 sin(t J 0 ) xd »¼ xq ¬«© xdc xd ¹ t 1§ 1 1 · Ta ¨ ¸ e [uq 0 cos J 0 ud 0 sin J 0 ] 2 ¨© xdc xq ¸¹ t 1§ 1 1 · Ta ¨ ¸ e [uq 0 cos(2t J 0 ) ud 0 sin(2t J 0 )] 2 ¨© xdc xq ¸¹

(2.5.46)

) for J 0 in the formula above, we can obtain the Substituting (J 0 B 120e expressions for ib and ic . As mentioned above, after voltages ud 0 and uq 0 are suddenly applied to stator winding, there is not only fundamental current but also aperiodic current and second harmonic current in ti. Its physical concept is mentioned as follows. Wher 3-phase symmetrical voltages are suddenly applied to stator winding, in it the corresponding 3-phase fundamental currents will be produced which can cause the flux-linkage values in stator and rotor windings changing suddenly. Therefore, the aperiodic currents in stator and rotor windings will be caused to keep flux-linkage constant. Because the rotor revolves at a synchronous speed, aperiodic current in stator will cause an ac current of basic-frequency in rotor winding, which will produce a pulsating magnetic field owing to asymmetrical rotor winding. The pulsating magnetic field can be divided into two opposite revolving fields, one of which will cause the second harmonic current in stator winding. In addition, the stator aperiodic current and second harmonic current will 123

AC Machine Systems

decay to zero at a time-constant Ta . The stator fundamental current will decay to its steady-state value at a time-constant Tdc. The time-constant Ta is referred to as the decay time-constant of stator aperiodic current, which is approximately equal to the ratio of negative sequence reactance to stator resistance. The timeconstant Tdc is called the decay time-constant of d-axis transient current, which is approximately equal to the ratio of excitation winding reactance when the stator winding having no resistance is short-circuited to excitation winding resistance. (2) Current change of a machine with damping windings The machine having damping windings, its operational reactance can be expressed as equation (2.4.13). Those expressions substituted into equation (2.5.28), the corresponding characteristic equation will become an equation of fifth degree. Solving directly the equation not only brings about complicated result but also makes it not easy to predict the effect of machine parameters on currents. Therefore, we shall use an approximate method to solve the current, i.e. at first to find the initial value and steady-state value of the current and then to search for the corresponding decay time-constants. At last, we can write the expression for the current to change with time according to those results. Under general condition, the results obtained by this approximate method are sufficiently accurate. (a) The initial value of stator current According to Initial Value Theorem in operational calculus we can see that the value of the operational expression when p f is the corresponding initial value. However, supposing directly p f in equation (2.5.28), we can only find the initial values id 0 and iq 0, and can not obtain the initial values expressed as a function of time mentioned above. Therefore, we shall use another method to find the initial values. Each inductive loop can keep its flux-linkage constant at the beginning of transients, and the superconductor loop will keep its flux-linkage constant for ever after transients. Thus, we can use the concept of superconductor loop to find the initial value of stator current. If rotor circuits are considered as superconductor loops, i.e. the rotor circuit resistances are all equal to zero, referring to equation (2.4.13b) d-axis operational reactance has the following constant, xd ( p) |R fd

R1d 0

xd xl

124

2 3 2 X 11d xad 2 xad X ffd xad 2 X 11d X ffd xad

1 1 1 1 xad X fl X 1dl

xdcc

xd ( p ) | p

f

(2.5.47)


xdcc is known as d-axis subtransient reactance, and the corresponding equivalent circuit is shown in Fig. 2.5.4. It is not difficult to understand that the equivalent circuit is similar to that of the operational reactance when U fd 0 and R fd R1d 0 as shown in Fig. 2.4.6.

Figure 2.5.4 Equivalent circuit of d-axis subtransient reactance

From equation (2.5.47) we can realize that d-axis operational reactance when rotor resistances are zero has the same result as the expression of d-axis operational reactance when p f. That is not difficult to understand. According to Initial Value Theorem we can realize that the value during p f corresponds to that during t 0, and during t 0 the passive loop has the property that its flux-linkage can not change abruptly, i.e. the property the superconductor loop has. Similarly, rotor resistances being equal to zero, q-axis operational reactance has the following constant, xq ( p ) |R1q

0

xq ( p) | p xl

f

xq

1 1 1 xaq X 1ql

2 xaq

X 11q xqcc

(2.5.48)

xqcc is termed q-axis subtransient reactance, and the corresponding equivalent circuit is shown in Fig. 2.5.5. It is not difficult to understand that the equivalent circuit is similar to that of the operational reactance when R1q 0 as shown in Fig. 2.4.7.

Figure 2.5.5 Equivalent circuit of q-axis subtransient reactance

125

AC Machine Systems

If the machine has no damping winding, the corresponding results are xd ( p ) |R fd

xq ( p)

0

xd ( p) | p

xd

2 xad X ffd

xqc

xq

f

xl

xdc 1 1 1 xad x f

(2.5.49)

which are the results of equations (2.5.34) and (2.5.29) as mentioned before. From formulas of xdcc and xqcc we can recognize that their values depend mainly upon the leakage reactances of stator and rotor windings because xad and xaq are much larger than xl , X 1dl , X fl and X 1ql and their equivalent circuits are similar to that of a transformer whose secondary is short-circuited. However, the stator current is alternating and rotor current is aperiodic, which is different from the static transformer. According to equation (2.5.28), operational formulas of currents, after stator and rotor resistances are neglected, can be reduced to the following forms: pud 0 uq 0 ½ ° xdcc ( p 2 1) ° ¾ puq 0 ud 0 ° xqcc( p 2 1) °¿

(2.5.50)

uq 0 ½ ud 0 sin t (1 cos t ) ° xdcc xdcc ° ¾ uq 0 ud 0 sin t (1 cos t ) ° ° xqcc xqcc ¿

(2.5.51)

id iq Solving them, we can obtain id iq or id

U [cos G cos(t G )] xdcc

iq

U [sin G sin(t G )] xqcc

When t 0, these currents will be equal to zero. That is because the machine has no current before voltages are suddenly applied to it. Transforming the results above to a, b, c axes, we can get

126


ia

ª uq 0 º 1§ 1 u 1· cos(t J 0 ) d 0 sin(t J 0 ) » ¨ ¸ [ud 0 sin(2t r0 ) « xqcc «¬ xdcc »¼ 2 ©¨ xdcc xqcc ¹¸ 1§ 1 1· uq 0 cos(2t r0 )] ¨ ¸ (uq 0 cos r0 ud 0 sin r0 ) ¨ 2 © xdcc xqcc ¸¹

(2.5.52)

or ia

U 2

§ 1 1· U§ 1 1· ¨¨ ¸¸ cos(t J 0 G ) ¨¨ ¸¸ cos(t J 0 G ) 2 © xdcc xqcc ¹ © xdcc xqcc ¹

U 2

§ 1 1· U§ 1 1· ¨¨ ¸¸ cos(2t J 0 G ) ¨¨ ¸¸ cos(J 0 G ) 2 © xdcc xqcc ¹ © xdcc xqcc ¹

Substituting (J 0 B 120e ) for J 0 in the formula above, we can obtain ib

ic

½ ° ° ° 1§ 1 1· ° ) uq 0 cos(2t J 0 120e )]° ¨ ¸ [ud 0 sin(2t J 0 120e ¨ ¸ 2 © xdcc xqcc ¹ ° ° 1§ 1 1· ° ) ud 0 sin(J 0 120e )] ¨ ¸ [uq 0 cos(J 0 120e ° 2 ¨© xdcc xqcc ¸¹ ° ¾ ª uq 0 º ud 0 ° sin(t J 0 120e )» cos(t J 0 120e ) « ° xqcc ¬« xdcc ¼» ° ° 1§ 1 1· ) uq 0 cos(2t J 0 120e )]° ¨ ¸ [ud 0 sin(2t J 0 120e ° 2 ©¨ xdcc xqcc ¹¸ ° ° § · 1 1 1 ° ) ud 0 sin(J 0 120e )] ¨ ¸ [uq 0 cos(J 0 120e 2 ©¨ xdcc xqcc ¹¸ ¿°

ª uq 0 º u cos(t J 0 120e ) d 0 sin(t J 0 120e )» « xqcc ¬« xdcc ¼»

(2.5.53)

(b) The steady-state value of stator current According to Final Value Theorem of operational calculus it is not difficult to find the current steady-state value. Because xd ( p) | p

0

xq ( p) | p

0

xd ½° ¾ xq °¿

(2.5.54)

taking p 0 in equation (2.5.28) we can get the steady-state value of stator current as follows: 127

AC Machine Systems

id iq

rud 0 xq uq 0 ½ ° r 2 xd xq ° ¾ ruq 0 xd ud 0 ° r 2 xd xq °¿

(2.5.55)

or ia

1 [(rud 0 xq uq 0 ) cos(t J 0 ) ( xq ud 0 ruq 0 )sin(t J 0 )] r xd xq 2

(2.5.56) The stator resistance being neglected, the steady-state value of stator current is id iq

uq 0 ½ ° xd ° ¾ u d0 ° xq °¿

(2.5.57)

or ia

uq 0 xd

cos(t J 0 )

ud 0 sin(t J 0 ) xq

(2.5.58)

(c) Time constants As mentioned before, the initial and steady-state values of stator current have been found when resistances are neglected. Those results are sufficiently accurate because resistances of practical machines are small. From those results we can see that in steady-state the aperiodic and second harmonic components of stator current will decay from their initial values to zero, and the fundamental component can decay from its initial value to steady-state value. Time constants at which those current components will decay are discussed as follows. (i) The time constant Ta Because the aperiodic and second harmonic components in stator current correspond to the fundamental component in rotor current, those components will decay at the same time constant. Obviously, the time constant depends upon parameters of stator and rotor windings. However, the rotor reactance of basic frequency is much larger than its resistance, so the rotor resistance may be considered as zero when time constant Ta is determined. Under this condition, we have xd ( p ) xq ( p ) 128

xdcc ½° ¾ xqcc °¿

(2.5.59)


so equation (2.5.28) can be reduced to id iq

( pxqcc r )ud 0 xqccuq 0

½ ° ( pxqcc r )( pxqcc r ) xdcc xqcc ° ¾ p ( xdcc r )uq 0 xdccud 0 ° ( pxdcc r )( pxqcc r ) xdcc xqcc °¿

(2.5.60)

The corresponding characteristic equation is ( pxdcc r )( pxqcc r ) xdcc xqcc

(2.5.61)

0

and its roots are p1,2

r§ 1 1· r2 § 1 1· ¨ ¸ r j 1 ¨ ¸ 2 ¨© xdcc xqcc ¸¹ 4 ¨© xdcc xqcc ¸¹

2

(2.5.62)

The corresponding time constant is 2 xdcc xqcc r xdcc xqcc

Ta

x2 r

(2.5.63)

in which x2

2 xdcc xqcc xdcc xqcc

is negative sequence reactance. Observing equation (2.5.62), we can know that when rotor resistance is neglected but stator resistance is taken into account, in currents id and iq there is no basic-frequency component but an approximate basic-frequency component whose frequency is 2

r2 § 1 1· 1 ¨ ¸ |1 ¨ 4 © xdcc xqcc ¸¹

In the corresponding stator phase currents ia , ib and ic there are no aperiodic and second harmonic components but a low frequency component and an ac component whose frequency is approximately equal to double basic-frequency. However, during calculation of these currents the difference is not taken into account. (ii) Time constants Tdcc, Tqcc and Tdc The fundamental component in stator current corresponds to the aperiodic component in rotor current, so the two components will decay at the same time constant. Of course, the time constants depend also upon the parameters of stator and 129

AC Machine Systems

rotor windings. However, the stator reactance of basic frequency is much larger than its resistance, so stator resistance may be considered as zero when time constant is determined. Under this condition, equation (2.5.28) can be reduced to pud 0 uq 0 ½ ° xd ( p)( p 2 1) ° ¾ puq 0 ud 0 ° xq ( p )( p 2 1) °¿

id iq

(2.5.64)

The corresponding characteristic equations are xd ( p )( p 2 1) xq ( p )( p 2 1)

0 ½° ¾ 0 °¿

(2.5.65)

in which the roots p1,2 r j of equation ( p 2 1) 0 correspond to the fundamental component in currents id and iq , i.e. the aperiodic and second harmonic components in stator current mentioned before, so the roots of equations xd ( p) 0 and xq ( p) 0 correspond to the aperiodic component in currents id and iq , i.e. the fundamental component in stator current. These time constants can be determined as follows. First of all, we find out the root of equation xq ( p ) 0. Because xq ( p)

2 ( X 11q xq xaq ) p R1q xq

pX 11q R1q

(2.5.66)

the corresponding time constant is Tqcc

2 X 11q xq xaq

X 11q xqcc

xq R1q

R1q xq

Tq 0

in which Tq 0

X 11q R1q

xqcc xq

1 ª 1 º X 1ql « » 1 1 R1q » « xaq xl »¼ ¬«

(2.5.67)

is the time constant of q-axis damping winding when stator

winding is open-circuited and Tqcc is the time constant of q-axis damping winding when stator winding is short-circuited whose resistance is equal to zero. Thus Tqcc is called the decay time constant of q-axis subtransient current. Now we search for the roots of equation xd ( p) 0. Because 130


xd ( p)

xd

2 3 2 2 2 [ p 2 ( X 11d xad 2 xad X ffd xad ) p ( xad R1d xad R fd )] 2 [ p 2 ( X 11d X ffd xad ) p( X 11d R fd X ffd R1d ) R1d R fd ]

(2.5.68)

xd ( p) 0 is an equation of second degree which is also tedious. In practice, an approximate method is also used. Because the damper resistance is much larger than excitation winding resistance in actual machines the aperiodic current of damping winding decays much faster than that of excitation winding. Therefore, we can approximately consider that during the initial period after voltages are suddenly applied to the stator winding, the aperiodic current of excitation winding does not decay and only that of the damping winding decays. The aperiodic current of excitation winding will not decay when its resistance is equal to zero, so we can find the time constant by R fd 0 in equation (2.5.65). The corresponding characteristic equation is xd

2 3 2 2 p ( X 11d xad 2 xad X ffd xad ) xad R1d 2 p ( X 11d X ffd xad ) X ffd R1d

0

(2.5.69)

and the corresponding time constant is Tdcc

2 2 3 2 2 xad X ffd xad xd ( X 11d X ffd xad ) X 11d xad 2 R1d ( xd X ffd xad )

Tdcc0

xdcc xdc

1 § 1 · X 1dl ¨ 1 1 1¸ R1d ¸ ¨ ¨ xad X fl xl ¸¹ ©

(2.5.70)

in which

X 11d Tdcc0

R1d

2 xad X ffd

1 § 1 · X 1dl ¨ 1 1 ¸ R1d ¨ ¸ ¨ xad X fl ¸¹ ©

(2.5.71)

is the time constant of d-axis damping winding when stator winding is opencircuited and excitation winding is short-circuited referring to Fig. 2.4.6. Tdcc is the time constant of d-axis damping winding when stator and excitation windings are all short-circuited in which r 0 and R fd 0, and it is called the decay time constant of d-axis subtransient current. After a certain time, the aperiodic current of damping winding has basically decayed out. It can be approximately considered that decay of the aperiodic current in excitation winding will determine decay of the fundamental current in 131

AC Machine Systems

stator winding and damping winding is open-circuited after that time, so xd

xd ( p )

2 pxad pX ffd R fd

and the time constant corresponding to equation xd ( p) 2 xd X ffd xad

Tdc

Td 0

xd R fd

xdc xd

(2.5.72) 0 is (2.5.73)

If damping winding resistance Rld is not very different from excitation winding resistance R fd , during estimation of time constants Tdc and Tdcc it can not be considered that R fd 0 in the initial period and Rld f after that period. Both resistances must be considered simultaneously. The corresponding time constants Tdc and Tdcc can be obtained according to the following way. Let c xad

2 xd xad xad xd

1 1 1 xad xl

(2.5.74)

be the equivalent mutual reactance between d-axis damping winding and excitation winding when stator winding is short-circuited whose resistance is zero. c T ffd

X cffd

2 X ffd xd xad

R fd

R fd xd

X fl

1 1 1 xad xl R fd

c X fl xad

(2.5.75)

R fd

is the time constant of excitation winding when stator winding is short-circuited and damping winding is open-circuited. T11c d

X 11c d R1d X 1dl

2 X 11d xd xad R1d xd

1 1 1 xad xl R1d

c X 1dl xad R1d

(2.5.76)

is the time constant of d-axis damping winding when stator winding is shortcircuited and excitation winding is open-circuited. 132


V cf 1d

1

c2 xad X cffd X 11c d

(2.5.77)

is the leakage-flux factor between d-axis damping winding and excitation winding when stator winding is short-circuited. The factors above substituted into equation (2.5.68), the equation of second degree xd ( p) 0 can be reduced to c T11c d p 2 (T ffd c T11c d ) p 1 0 V cf 1d T ffd

(2.5.78)

Solving it, we can get p1,2

c )(r q) (T11c d T ffd c T11c d 2V cf 1d T ffd

(2.5.79)

in which q

1

c V cf 1d 4T11c d T ffd c )2 (T11c d T ffd

|1

(2.5.80)

The corresponding time constants are c T11c d 2V cf 1d T ffd

Tdcc

1 p1

c ) (1 q )(T11c d T ffd

Tdc

1 p2

c T11c d 2V cf 1d T ffd c ) (1 q)(T11c d T ffd

|

c T11c d V cf 1d T ffd c T11c d T ffd

1 c ) | T11c d T ffd c (1 q )(T11c d T ffd 2

(2.5.81)

(2.5.82)

(d) The general formulas of stator current Having the initial values, steady-state values and time constants obtained above, we can easily write the general expressions of stator current as id

t t t ª§ 1 1 · T cc § 1 1 · T c 1 º 1 T «¨ ¸ e d ¨ ¸ e d » uq 0 e a [ud 0 sin t uq 0 cos t ] xd »¼ xdcc © xdc xd ¹ ¬«© xdcc xdc ¹ (2.5.83)

iq

t ª§ 1 1 · Tt cc 1 º 1 «¨ ¸ e q » ud 0 e Ta [ud 0 cos t uq 0 sin t ] xq » xqcc «¬¨© xqcc xq ¸¹ ¼

(2.5.84)

133

AC Machine Systems

or id

iq

t t t ª§ 1 1 · T cc § 1 1 · T c 1 º U T «¨ ¸ e d ¨ ¸ e d » U cos G e a cos(t G ) xd »¼ xdcc © xdc xd ¹ ¬«© xdcc xdc ¹ (2.5.85) t ª§ 1 1 · Tt cc 1 º U «¨ ¸ e q » U sin G e Ta sin(t G ) xq » xqcc «¬©¨ xqcc xq ¹¸ ¼

(2.5.86)

2.5.3 Currents after 3-phase Sudden Short Circuit of Synchronous Machines 3-phase sudden short circuit is a severe accident produced possibly in synchronous machine operation and it is also a basis for analysis of other sudden short circuit phenomena. As stated before, during analysis of 3-phase sudden short circuit, we can consider the short circuit not to have happened but the stator terminals to be suddenly applied on by 3-phase voltages which are equal but opposite to the voltages before short circuit. In thinking so, the problem of sudden short circuit is turned into the synthetic problem of the following two operation conditions: (i) Original steady-state operation condition before short circuit, referring to 2.5.1. (ii) Stator terminals are suddenly applied on by 3-phase voltages which are equal but opposite to original voltages before short circuit, referring to 2.5.2. Because the basic equations are linear under this condition, we can use Superposition Theorem to treat this problem. (1) Stator currents after 3-phases are suddenly short-circuited If the amplitudes of terminal voltage and excitation emf before short circuit are U and E respectively and the terminal voltage lags behind excitation emf by an angle G, then referring to equation (2.5.3) we can write stator terminal voltage as ud 0 uq 0

U sin G ½° ¾ U cos G °¿

(2.5.87)

According to Superposition Theorem, it is not difficult to understand that the stator current after 3-phases are suddenly short-circuited consists of the original steady-state current and transient current caused by applying the voltages (ud 0 ) and (uq 0 ) to stator winding. (a) The original steady-state current If stator resistance is neglected, according to equation (2.5.7) the currents can 134


be written as id iq

E U cos G ½ ° xd ° ¾ U sin G ° °¿ xq

(2.5.88)

(b) The transient current caused by applying the voltages (ud 0 ) and (uq 0 ) to stator winding Referring to equations (2.5.85) and (2.5.86), we can write transient currents as t t t ½ ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º U Ta e e U cos e cos(t G ) ° G «¨ » ¸ ¨ ¸ xd »¼ xdcc «¬© xdcc xdc ¹ © xdc xd ¹ °° ¾ t t ª§ 1 ° 1 · Tqcc 1 º U Ta iq «¨ ¸ e » U sin G e sin(t G ) ° ¨ xcc x ¸ xq » xqcc q ¹ °¿ ¬«© q ¼ (2.5.89) (c) Stator actual currents after 3-phase sudden short circuit Adding equation (2.5.88) to equation (2.5.89), we can write stator actual current after short circuit as

id

id

t t t ª§ 1 1 · Tdcc § 1 1 · Tdc º E U Ta e e U cos e cos(t G ) G «¨ » ¸ ¨ ¸ xd xdcc «¬© xdcc xdc ¹ »¼ © xdc xd ¹ (2.5.90) t

iq

t § 1 1 · T cc U ¨ ¸ e q U sin G e Ta sin(t G ) ¨ xcc x ¸ xqcc q ¹ © q

(2.5.91)

Having id and iq , we can transform them into ia , ib , ic . For convenience, we suppose the machine to be no-load before short circuit, and then there are G 0 U E ia

t t ª§ 1 1 · T cc § 1 1 · T c 1 º «¨ ¸ e d ¨ ¸ e d » E cos(t J 0 ) xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹ t º § 1 E ª§ 1 1· 1· e Ta «¨ ¸ cos J 0 ¨ ¸ cos(2t J 0 ) » ¨ xcc xcc ¸ 2 «¬¨© xdcc xqcc ¸¹ »¼ q ¹ © d

(2.5.92)

(J 0 120e ) and (J 0 120e ) substituted respectively for J 0 in the formula above, we can obtain the expressions for ib and ic . 135

AC Machine Systems

In nonsalient-pole machines such as turbogenerators which have strong damping windings, xdcc | xqcc, so the formula above can be reduced to ia

t ª§ 1 1 · Tdcc § 1 1 ¨ e «¨ ¸ cc c c «¬© xd xd ¹ © xd xd

t t · Tdc 1 º E Ta e E cos( t ) e cos J 0 J » ¸ 0 xd »¼ xdcc ¹

(2.5.93) If the sudden short-circuit current does not decay, according to formula (2.5.92) we can express phase a current as ia

§ 1 § 1 E 1 ·E 1 ·E cos(t J 0 ) ¨ ¸ cos J 0 ¨ ¸ cos(2t J 0 ) ¨ ¸ ¨ ¸ xdcc © xdcc xqcc ¹ 2 © xdcc xqcc ¹ 2

(2.5.94)

From here we can know that the amplitude of stator fundamental current is E / xdcc ; in comparison with the corresponding short-circuit current E / xd in steady-state the difference between them is to change xd into xdcc. The value of xdcc depends mainly upon stator and rotor leakage reactances, whose physical meaning can be stated as follows. If there were no leakage flux in rotor winding, the rotor flux-linkage would induce the same emf in stator winding as before short-circuit, so the stator fundamental current amplitude would be E / xl , i.e. the subtransient reactance xdcc would equal the stator leakage reactance xl . In fact, leakage flux exists in rotor winding and the stator armature reaction increases during sudden short-circuit. Therefore, rotor current will rise so as to keep the rotor flux-linkage unchanged, and the rotor leakage flux will increase too. Thus the mutual flux between rotor and stator will decrease, i.e. the stator emf will be less than the value E which is the emf before short-circuit. Provided that the stator emf during sudden short-circuit is ( E 'E ), the stator fundamental current amplitude will become ( E 'E ) / xl . However, it is not convenient to calculate 'E directly since 'E depends upon the increase of rotor leakage flux or the decrease of mutual flux. In practical estimation, we can suppose the stator emf after short-circuit is the same as before, but xl changes into ( xl 'x); that is to say, the corresponding current ( E 'E ) / xl can be considered to equal E /( xl 'x ), so there is xdcc xl 'x. Therefore, it is evident that xdcc depends basically upon the stator and rotor leakage reactances. From equation (2.5.92) we are clear that the stator short-circuit current is dependent on the rotor position angle J 0 at the short-circuit moment; taking non-salient pole synchronous machines as example we shall discuss the following two cases. During J 0 90e, expression (2.5.93) will be turned into 136


ia

t t ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º «¨ ¸ e ¨ ¸ e » E sin t xd »¼ «¬© xdcc xdc ¹ © xdc xd ¹

(2.5.95)

From here we can recognize that no aperiodic component exists in stator current ia in the case. That is natural because \ a is just equal to zero at the short-circuit moment, so there is no need to cause an aperiodic current to keep the original flux-linkage. The curve of stator current ia is shown in Fig. 2.5.6. It should be noted that the aperiodic component still exists in current ib or ic .

Figure 2.5.6

ia -curve during 3-phase sudden short circuit at J 0

90e

Under that condition, the stator fundamental current consists of the following three parts: E (i) Steady-state short-circuit current sin t , xd t

§ 1 1· (ii) Subtransient current ¨ ¸ e Tdcc E sin t , © xdcc xdc ¹ t § 1 1· (iii) Transient current ¨ ¸ e Tdc E sin t. © xdc xd ¹ During J 0 ia

0, formula (2.5.93) will change into t t t ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º E Ta e e E cos t e «¨ » ¸ ¨ ¸ xd »¼ xdcc «¬© xdcc xdc ¹ © xdc xd ¹

(2.5.96)

whose curve is shown in Fig. 2.5.7. On this condition, phase a fundamental current has the same three components as before except a phase-angle difference of 90e. In addition, an aperiodic component appears in stator current ia to keep the original flux-linkage of phase a unchanged suddenly. The aperiodic component will make the curve of ia asymmetrical about the abscissa. Current ia will reach its maximum in about half a cycle after short circuit, whose value is around twice as large as the maximum value during J 0 90e. Incidentally, the initial value of 137

AC Machine Systems

aperiodic current is equal but opposite to that of the fundamental current. That is natural because the current ia at t 0 must be zero.

Figure 2.5.7

ia -curve during 3-phase sudden short circuit at J 0

0

Because the stator fundamental current decays slowly, its rms value or amplitude change is significant. Obviously, the rms value or amplitude at any instant, with reference to formulas (2.5.95) and (2.5.96), can be written as t t ª§ 1 1 · T cc § 1 1 · T c 1 º «¨ ¸ e d ¨ ¸ e d » E xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹

(2.5.97)

whose curve is shown in Fig. 2.5.8.

Figure 2.5.8 Maximum-value envelope of stator fundamental current during 3-phase sudden short circuit

(2) Rotor currents after 3-phase sudden short circuit The rotor currents after short circuit can be calculated in the same way as the stator currents. It may be estimated according to the following two parts: (i) The rotor currents before short circuit, (ii) The rotor current caused by applying voltages (ud 0 ) and (uq 0 ) to stator winding. 138


If the machine operates symmetrically in steady-state before short circuit, the first part can be easily calculated, i.e. I fd 0 I1d 0

U fd R fd I1q 0

½ ° ¾ 0 °¿

(2.5.98)

The solution of the second part is also tedious, so it can be calculated by the same approximate method as the stator current, i.e. we shall estimate only the values at two or three instants and then write its expression according to the time constants obtained before. The equivalent circuits shown in Figs. 2.4.3 to 2.4.7 have given the relation between currents id , iq and rotor currents when U fd 0. Because currents id and iq have been found, rotor currents can be obtained according to those equivalent circuits. (a) The initial values of rotor currents caused by applying voltages (ud 0 ) and (uq 0 ) to stator winding During calculation of the initial values, all rotor resistances can be considered as zero, so it is convenient to use the equivalent circuits shown in Figs. 2.5.3 and 2.5.4 to solve the problem because there is no operator p in those equivalent circuits. When there is no damping winding on rotor, according to Fig. 2.5.3 we can get i fd

xad id X fl xad

xad id X ffd

xad U [cos G cos(t G )] X ffd xdc

(2.5.99)

When there is a damping winding on rotor, according to Fig. 2.5.4 we can obtain i fd

2 X 11d xad xad i 2 d X 11d X ffd xad 2 X 11d xad xad U [cos G cos(t G )] 2 X 11d X ffd xad xdcc

i1d

2 X ffd xad xad 2 X 11d X ffd xad

(2.5.100)

id

2 X ffd xad xad U [cos G cos(t G )] 2 X 11d X ffd xad xdcc

(2.5.101)

139

AC Machine Systems

From the formulas above we can see that during the same value of id the value of I fd when there is a damping winding is less than that when there is no damping winding. That is natural, because the damping winding shares part of the induced current. From the formulas above we can also know 2 X 11d xad xad 2 X ffd xad xad

i fd i1d

X 1dl X fl

(2.5.102)

In addition, according to Fig. 2.5.5 we can see i1q

xaq X 11q

iq

xaq U [sin G sin(t G )] X 11q xqcc

(2.5.103)

(b) The steady-state values of rotor currents caused by applying voltages (ud 0 ) and (uq 0 ) to stator winding Obviously, the current values are equal to zero because there is p 0 in the equivalent circuit of operational reactance shown in Fig. 2.4.6. (c) General expressions of rotor currents caused by applying voltages (ud 0 ) and (uq 0 ) to stator winding Because there are resistances in the machine, the rotor current will decay gradually from the initial value to zero. The decay time constants are similar to those of the corresponding components of stator current. Therefore, when there is a damping winding on rotor, general expressions of rotor currents caused by applying voltages (ud 0 ) and (uq 0 ) to stator winding are i fd

° ª§ X x x 2 11d ad ad ® «¨¨ 2 °¯ «¬© X 11d X ffd xad 2 § X x xad ¨ 11d ad 2 ¨ © X 11d X ffd xad

i1d

2 § X ffd xad xad ¨¨ 2 © X 11d X ffd xad

t · 1 xad 1 º Tdcc § xad ¨ e » ¸¸ ¨ cc c x X x » d ffd d ¹ © X ffd ¼

· 1 Ttc ½° ¸¸ e d ¾U cos G °¿ ¹ xdc

· U Tt ¸¸ e a cos(t G ) ¹ xdcc 2 · U Ttcc § X ffd xad xad ¸¸ e d cos G ¨¨ 2 ¹ xdcc © X 11d X ffd xad

(2.5.104) · U Tt ¸¸ e a cos(t G ) ¹ xdcc (2.5.105)

t

i1q

140

xaq U Tqcc xaq U Tta e sin G e sin(t G ) X 11q xqcc X 11q xqcc

(2.5.106)


When there is no damping winding on rotor, we can get t

t

xad U Tdc x U Ta e cos G ad e cos(t G ) X ffd xdc X ffd xdc

i fd

(2.5.107)

The rotor actual current after 3-phase sudden short circuit can be obtained by sum of the two parts as mentioned before. When there is no damping winding on rotor, we have i fd

U fd R fd

t

t

xad U Tdc x U Ta e cos G ad e cos(t G ) X ffd xdc X ffd xdc

(2.5.108)

whose curve is shown in Fig. 2.5.9.

Figure 2.5.9 Excitation winding current after 3-phase sudden short circuit and without damping winding

When there is a damping winding on rotor, the actual current in excitation winding is i fd

U fd R fd

t t ª§ X x x 2 1 xad 1 · Tdcc xad 1 Tdc º ad e e «¨ 11d ad » U cos G ¸ 2 ¨ cc c¸ X ffd xdc ¬«© X 11d X ffd xad xd X ffd xd ¹ ¼» t

2 X 11d xad xad U Ta e cos(t G ) 2 X 11d X ffd xad xdcc

(2.5.109)

whose curve is shown in Fig. 2.5.10. It is noted that the component contained in i fd and decaying at time constant Tdcc is generally negative value. Therefore, the curve of aperiodic component of excitation winding current i fd is bent down in its initial part. Because damping winding current is zero before short circuit, the damping winding actual current after short circuit is the same as expressions (2.5.105) and (2.5.106). 141

AC Machine Systems

Figure 2.5.10 Excitation winding current after 3-phase sudden short circuit and with damping winding

2.6

Voltage Dip during Sudden Increase of Load for Synchronous Machines

If 3-phase symmetrical constant loads are suddenly connected to a synchronous generator, so far as calculation of its currents is concerned, we can consider that condition as 3-phase sudden short circuit after the load resistance and reactance per phase are added to stator phase resistance and leakage reactance respectively. Provided that the load is a starting induction motor, it can also be treated as a constant impedance because it may be represented approximately by a short circuit impedance during part of the starting process. If the load reactance and resistance are X L and RL , the corresponding operational reactance and stator circuit resistance are

xdG ( p ) X L ½ ° xq ( p ) xqG ( p ) X L ¾ ° R rG RL ¿ xd ( p )

(2.6.1)

in which the subscript G represents the values corresponding to the synchronous generator. According to the concept of 3-phase short circuit and referring to equations (2.5.28) and (2.5.3), we can find the basic relations after a 3-phase symmetrical load is suddenly connected to a synchronous generator as follows:

142

ud 0

U sin G

uq 0

U cos G

0 ½° ¾ E °¿

(2.6.2)


xq ( p )uq 0

id iq

½ ° [ pxd ( p ) R][ pxq ( p ) R ] xd ( p ) xq ( p ) ° ¾ [ pxd ( p ) R]uq 0 ° [ pxd ( p ) R][ pxq ( p ) R ] xd ( p ) xq ( p ) °¿

(2.6.3)

It should be noted that deriving those relations we have supposed the speed to be constant, because the speed change is much slower than the voltage dip. It is very complicated to solve equation (2.6.3) directly, and we still use an approximate method to solve it. Because the time constant of damping winding is usually very small and the corresponding subtransient component decays very fast, studying the voltage dip we don’t take the effect of damping winding into account, i.e. we can consider the machine having no damping winding. This problem is now studied as follows. (1) The initial value of stator current Since the resistance of excitation winding is very small, evaluating the initial value of stator current we can take the resistance as zero, i.e. xd ( p ) |R fd

0 or p f

xdc

c XL xdG

(2.6.4)

xq ( p ) |R fd

0 or p f

xq

xqG X L

(2.6.5)

The values above substituted into equation (2.6.3), it is not difficult to find the initial value of stator d-axis current as follows, xq E

id

[ pxdc R][ pxq R] xdc xq

̯

xq E R 2 xdc xq

xq E

Z xdc xq ( R 2 xdc xq )

e

t Ta

sin(Z t M )

(2.6.6)

in which

Z M Ta

2 ½ R2 § 1 1· ° 1 ¨ ¸ 4 ¨© xdc xq ¸¹ ° °° arctan ZTa ¾ ° c 2 x x 1 d q ° ° R xdc xq °¿

(2.6.7)

Ta is the decay time constant of stator aperiodic current. The second term in equation (2.6.6) is the aperiodic and second harmonic components of stator 143

AC Machine Systems

current, which are neglected when the voltage dip is calculated because those components have a little effect on the voltage dip when R is small and they will decay quickly when R is large. For example, when an induction motor starts, its impedance is X L 0.3 and RL 0.1, and the decay time constant, if the generator 2 u 0.6 u 1.1 c 7.8 parameters are xdG 0.3, xqG 0.8 and r 0.005, is Ta 0.1005(0.6 1.1) 7.8 second, i.e. only about one cycle. radians or 2S u The aperiodic current and second harmonic component neglected, the initial value of stator d-axis current will be xq E

id

2

R xdc xq

(2.6.8)

Similarly, the initial value of q-axis current is RE R 2 xdc xq

iq

(2.6.9)

Because the stator aperiodic current and second harmonic component are caused by the stator transformer emf, those componnts will disappear when the two terms p\ d and p\ q are neglected in Park’s equation. That is to say, we use the voltage equations \ q Rid ½° ¾ \ d Riq °¿

ud uq

(2.6.10)

to analyse the problem, i.e. we shall neglect the stator aperiodic current and second harmonic component. For convenience, when analysing some problems, such as the voltage dip and machine stability, we often use this kind of equation instead of original Park’s equation. Equation (2.6.10) used to analyse the voltage dip, equation (2.6.3) can be reduced to id iq If xd ( p) 144

xdc and xq ( p)

xq ( p ) E

½ ° R xd ( p ) xq ( p ) ° ¾ RE ° R 2 xd ( p ) xq ( p ) °¿ 2

xq , the initial values of stator current are

(2.6.11)


id iq

xq E

½ ° R xdc xq ° ¾ RE ° R 2 xdc xq °¿ 2

(2.6.12)

that are equations (2.6.8) and (2.6.9) obtained by another method. Obviously, the two methods have the same result. (2) The steady-state value of stator current Supposing p 0 in equation (2.6.3) or (2.6.11), we can obtain the steady-state value of stator current id iq

xq E

½ ° R xd xq ° ¾ RE ° R 2 xd xq °¿ 2

(2.6.13)

(3) Time constant The time constant of stator fundamental current depends upon the effective reactance and resistance corresponding to the aperiodic current in excitation winding. The flux-linkage of excitation winding in steady-state is

\ fd

X ffd I fd xad id

X ffd I fd

xad xq E R 2 xd xq

2 § xad xq · X ¨¨ ffd ¸¸ I fd 2 R x x d q © ¹

Therefore, the corresponding time constant is Tdc

1 R fd

§ \ fd ¨¨ © I fd

· ¸¸ ¹

R 2 xdc xq R 2 xd xq

Td 0

(2.6.14)

(4) The general formulas of current and voltage After finding the initial value, steady-state value and time constant, we can write the general formulas of stator current as id

t § · xq E 1 1 Tdc 2 ¨¨ 2 ¸¸ xq Ee 2 R xd xq © R xdc xq R xd xq ¹

iq

t § · RE 1 1 Tdc 2 ¨¨ 2 ¸¸ REe 2 R xd xq © R xdc xq R xd xq ¹

R id xq

(2.6.15)

145

AC Machine Systems

According to the formulas above, we can obtain the amplitudes of stator current and terminal voltage

| ia |

2 d

2 q

i i

§R 1 ¨ ¨x © q

2

· ¸¸ id ¹

ª§ º · Ttc 1 1 1 2 «¨¨ 2 » R 2 xq2 E ¸¸ e d 2 c R x x R x x R x x d q d q ¹ d q » ¬«© ¼

|u|

(2.6.16)

X L2 RL2 | ia | ª§ R 2 xd xq º ( RL2 X L2 )( R 2 xq2 ) · t 1¸ e Tdc 1» E «¨¨ 2 ¸ R 2 xd xq «¬© R xdc xq »¼ ¹

(2.6.17)

Supposing t 0 in the formula above, we can get the initial amplitude of terminal voltage | u |t

( RL2 X L2 )( R 2 xq2 ) 0

R 2 xdc xq

E

(2.6.18)

From that formula we can see: (a) If power factor of the load is very low, i.e. R may be neglected, then the initial amplitude of terminal voltage is | u |t

0

XL E xdc

XL E c xdG X L

c id E xdG

(2.6.19)

i.e. under this condition, the initial dip of terminal voltage depends mainly upon d-axis transient reactance, and the voltage dip is severe. (b) If power factor of the load is high, i.e. the load reactance X L can be neglected, then the initial dip of terminal voltage is not large, and sometimes the terminal voltage will rise. For example, under this condition, xqG 1.0, RL 1.0, X L | rG | 0, we can get | u |t

0

1.414 E c 1 xdG

(2.6.20)

c 0.414, terminal voltage during initial period will not decrease but i.e. when xdG increase. Because q-axis synchronous reactance xq of a turbogenerator is larger c is smaller than a hydraulic generator, the and transient reactance xdG turbogenerator can undertake a larger load switched on than the hydraulic generator so far as the voltage dip is concerned. 146


2.7

Transient EMFs of Synchronous Machines

The flux-linkage of rotor winding, especially the flux-linkage of excitation winding, under a certain condition, can be considered as a constant, which will make analysis and calculation simplified. However, it is not convenient directly to use the rotor flux-linkage, so we shall use other quantities which can represent the property of rotor flux-linkage. Those new quantities are called d-axis transient emf, d-axis or q-axis subtransient emf, and the transient emf behind transient reactance, etc. Because those emfs can represent the property of rotor flux-linkage, they are important quantities when the main process of basic-frequency current is studied. We shall discuss those emfs and their applications as follows. (1) d-axis transient emf If there is only an excitation winding and no damping winding on the rotor, the flux-linkage equations can be written as

\ fd \d

xad id X ffd I fd ½° ¾ xd id xad I fd °¿

(2.7.1)

Deleting I fd , we can get

\d

§ x2 · x ¨ xd ad ¸ id ad \ fd ¨ X ffd ¸¹ X ffd © x xdc id ad \ fd X ffd

(2.7.2)

or Eqc \ d xdc id in which Eqc

(2.7.3)

xad \ fd is a quantity that is proportional to the flux-linkage of X ffd

excitation winding and represents the property of excitation winding flux-linkage. Eqc is called d-axis transient emf. (2) d-axis subtransient emf If there is not only an excitation winding but also a damping winding on rotor, the flux-linkage equations can be written as

\ fd \ 1d \d

xad id X ffd I fd xad I1d ½ ° xad id xad I fd X 11d I1d ¾ ° xd id xad I fd xad I1d ¿

(2.7.4)

147

AC Machine Systems

Deleting I fd and I1d , we can get xdccid

\d

2 2 X ffd xad xad X 11d xad xad \ \ 1d fd 2 2 X 11d X ffd xad X 11d X ffd xad

(2.7.5)

or Eqcc \ d xdccid

(2.7.6)

in which Eqcc

2 2 X ffd xad xad X 11d xad xad \ \ 1d fd 2 2 X 11d X ffd xad X 11d X ffd xad

(2.7.7)

is a quantity that is proportional to the flux-linkages of excitation winding and damping winding and represents the property of rotor d-axis flux-linkages. Eqcc is called d-axis subtransient emf. (3) q-axis subtransient emf On q-axis of a machine with damping winding, the flux-linkage equations can be written as

\ 1q \q

xaq iq X 11q I1q ½° ¾ xq iq xaq I1q °¿

(2.7.8)

Deleting I1q , we can obtain

\q

2 § · xaq xaq ¨ xq \ 1q iq ¸ ¨ X 11q ¸¹ X 11q © xaq xqcciq \ 1q X 11q

(2.7.9)

or Edcc \ q xqcciq in which Edcc

xaq X 11q

(2.7.10)

\ 1q is a quantity that is proportional to the flux-linkage of

q-axis damping winding and represents the property of q-axis damper flux-linkage. Edcc is called q-axis subtransient emf. (4) The transient emf behind transient reactance When a synchronous machine without damping windings operates symmetrically in steady-state, referring to equation (2.7.3) we can obtain the following relations: 148


\d

E xd id

\q

xq iq

ud

\ q rid

uq

\ d riq

Eqc xdc id ½ ° ° ¾ ° ° ¿

(2.7.11)

If taking d-axis as real axis, q-axis as imaginary axis and supposing uˆ ud juq , iˆ id jiq , we can plot a vector diagram on complex-number plane as shown in Fig. 2.7.1.

Figure 2.7.1 Vector diagram for transient emfs

In that figure, E c uˆ (r jxdc )iˆ ( xq xdc )iq jEqc is called the transient emf behind d-axis transient reactance or simply the transient emf behind transient reactance, and Eq E ( xd xq )id is called the emf behind q-axis synchronous reactance. Neglecting stator resistance, we have E c uˆ jxdc iˆ

In practice, because the flux-linkage of q-axis damper \ 1q decays rapidly, it can be neglected, i.e. Edcc 0. Therefore, there is only d-axis subtransient emf Eqcc and transient emf Eqc , and we can call them the subtransient emf and transient emf respectively. If xdc xq , i.e. there is no transient saliency-effect, E c will be equal to Eqc . In actual machines, the difference between transient emf Eqc and the transient emf behind transient reactance E c is very small referring to Fig. 2.7.1, and the latter can be found according to the current, terminal voltage and transient reactance. Thus, we often use E c | Eqc to analyse some problems. Similarly, we can obtain E cc uˆ (r jxdcc )iˆ | uˆ jxdcciˆ, which is called the subtransient emf behind subtransient reactance. In practice, we can consider E cc | Eqcc. Because the flux-linkage of damping winding changes much faster than that of excitation 149

AC Machine Systems

winding, E cc or Eqcc will be changed rapidly into E c or Eqc . Therefore, we often neglect E cc or Eqcc but use E c or Eqc to study many problems, i.e. Only the effect of excitation winding is taken into account. (5) Application examples of the transient emf We have mentioned definition of the transient emf as before. Now some application examples are introduced as follows. (a) Application of the transient emf to calculating the initial value of threephase short circuit current When a synchronous machine that is operating symmetrically in steady-state is suddenly 3-phase short-circuited, the initial values of aperiodic components in currents id and iq , i.e. the initial values of stator basic-frequency current, are (id 0 uq 0 / xdcc ) and (iq 0 ud 0 / xqcc) respectively referring to equations (2.5.90) and (2.5.91). After arrangement, those values are equal to 1 (iq 0 xqcc ud 0 ) xqcc

1 (id 0 xdcc uq 0 ) xdcc

Eqcc xdcc

and

Edcc respectively. That is natural, because Eqcc and Edcc represent xqcc

the emfs which are proportional to rotor d-axis and q-axis flux-linkages during short circuit; Those flux-linkages can not change abruptly at the moment of short circuit, so the corresponding emfs Edcc and Eqcc can not change abruptly, either. In addition, the reactances corresponding to 3-phase sudden short circuit are just xdcc and xqcc, so the corresponding currents are Edcc / xqcc and Eqcc / xdcc. If there is no damping winding, then the initial value of aperiodic component in current id is Eqc / xdc and that in current iq is equal to zero because there is no loop on rotor q-axis and its flux-linkage can not be kept constant. The above calculation is based on the stipulation that a synchronous machine operates symmetrically in steady-state before short circuit. However, the formulas can also be used under any operation mode before short circuit if Edcc , Eqcc and Eqc have correct values. That is because those emfs are proportional to the rotor flux-linkages and are some quantities which can not change abruptly. For example, studying sudden asymmetrical short circuit,we can use the transient emf, transient reactance and Symmetrical Components Method to solve it. (b) Application of the transient emf to estimating the voltage dip when the load of a synchronous generator increases suddenly For convenience, we have studied the voltage dip when a synchronous generator that has no load is suddenly loaded in Section 2.6. If the machine has undertaken a certain load before it is suddenly loaded, the corresponding result is different. Because the transient emf, when operation condition changes, does not change abruptly, we can use it while studying the initial voltage dip after the load of a synchronous generator increases suddenly. 150


Letting RL and X L represent the parallel resistance and reactance of the original load with the new load with reference to equation (2.5.6), the initial and steady-state currents, when the aperiodic component of stator curent is neglected, can be written as idc iqc

idy iqy

xq Eqc( i ) ½ ° R 2 xdc xq ° ¾ REqc( i ) ° R 2 xdc xq °¿

(2.7.12)

xq E

½ ° R xd xq ° ¾ RE ° R 2 xd xq °¿

(2.7.13)

E ( xd xdc )id 0

(2.7.14)

2

in which Eqc (i )

The subscript i indicates that the machine has an original load before loaded. Because the expression of decay time-constant is the same as equation (2.6.14), the formulas of current and voltage dip are similar to the corresponding expressions when a synchronous generator that has no load is suddenly loaded. However, there is a difference between them. If the machine has no load before loaded, the emf in those formulas is excitation emf E and the impedance RL , X L is a new-loaded impedance. If the machine has undertaken an original load before loaded, the emf is the transient emf Eqc (i ) before a new load is switched on, and the impedance RL , X L is the parallel impedance of original load with the new load. The corresponding formulas are id (i )

§ xq Eqc( i ) xq E · Ttdc xq E 2 ¨¨ 2 ¸¸ e 2 R xd xq © R xdc xq R xd xq ¹

iq ( i )

§ REqc( i ) · Ttc RE RE 2 ¨¨ 2 ¸¸ e d 2 c R x x R x x R xd xq d q d q ¹ ©

| ia |( i )

R 2 xq2 xq

id ( i )

(2.7.15)

R id (i ) xq

§ R2 x2 R 2 xq2 · Ttc q ¨ c E E ¸e d ¨ R 2 xdc xq q (i ) R 2 xd xq ¸ © ¹

½ ° °° ¾ R 2 xq2 ° E 2 R xd xq °° ¿

(2.7.16) 151

AC Machine Systems

RL2 X L2 | ia |

| u |( i )

ª§ R 2 xd xq Eqc( i ) º ( R 2 xq2 )( RL2 X L2 ) · Ttc d 1 e 1 E «¨ 2 » ¸¸ 2 R x x «¬©¨ R xdcc xq E » d q ¹ ¼

(2.7.17) | u |( i )( t

( R 2 xq2 )( RL2 X L2 ) 0)

R 2 xdc xq

Eqc( i )

(2.7.18)

Because the transient emf Eqc when the machine has no load is equal to the excitation emf E, those formulas (2.7.15), (2.7.16), (2.7.17) and (2.7.18) can also be used when a synchronous generator that has no load is suddenly loaded. (c) Output power expressed by transient emf The output power and electromagnetic torque are important quantities when the rotor motion is studied. The machine has a disturbance and those quantities will change accordingly. Because the flux-linkage of excitation winding changes slowly in transient process and can be considered approximately as a constant in the initial period, it is convenient that the output power or electromagnetic torque is expressed by the transient emf Eqc which represents the property of excitation flux-linkage. Neglecting stator resistance and referring to equations (2.5.9) and (2.4.22), we can write the output power and torque in steady-state as follows: P Te

EU 1§ 1 1 · sin G ¨ ¸ U 2 sin 2G ¨ xd 2 © xq xd ¸¹

(2.7.19)

In addition, there are the following relations:

\d

E xd id

\q

xq iq

ud

\ q

uq

\d

Eqc xdc id ½° ¾ °¿ U sin G

xq iq

Eqc xdc id

½° ¾ U cos G °¿

(2.7.20)

(2.7.21)

Therefore, we can get the output power or torque expressed by the transient emf Eqc as follows: P

ud id uq iq EqcU

Eqc uq xdc

ud

u d uq xq

1§ 1 1· sin G ¨ ¸ U 2 sin 2G ¨ xdc 2 © xq xdc ¸¹

152

(2.7.22)


Te

\ d iq \ q id

uq iq ud id

P

EqcU

1§ 1 1 · sin G ¨ ¸ U 2 sin 2G ¨ 2 © xq xdc ¸¹ xdc

(2.7.23)

According to equations (2.7.19) and (2.7.22), we can plot two curves A and B shown in Fig. 2.7.2, supposing U fd to be constant.

Figure 2.7.2

P - G curves for steady-state and transients

From those curves we can see that when power angle G changes very slowly, i.e. the changing process may be considered as steady-state, the output power P will change from the steady-state operation point ( P0 , G 0 ) along curve A shown in Fig. 2.7.2. However, when power angle G changes quickly, the output power P, if the excitation winding flux-linkage is constant, will change from the steady-state operation point ( P0 , G 0 ) along curve B shown in Fig. 2.7.2. Curve A is called the steady-state P - G curve, and curve B the transient P - G curve. From curves A and B we can see that the maximum value of curve A is situated at the region G 90eand that of curve B at the region G ! 90e. In addition, the maximum value of the latter is much larger than that of the first. The reason for producing the angle difference is that the salient-pole torques during steady-state and transients have different signs, because usually xd ! xq and xdc xq . From curves A and B we can also see that the ability for a machine to bear sudden disturbance is much stronger than that for a machine to bear slow disturbance.

2.8

Electromagnetic Torque after 3-Phase Sudden Short Circuit

Synchronous machines, after 3-phases are suddenly short-circuited, can produce average and pulsating electromagnetic torque. Those torques are significant for calculation of stresses in machine-parts and frames and analysis of operationstability. During analysis of those torques, it is tedious to solve strictly the basic equations. If the approximate method like in analysis of 3-phase short circuit is 153

AC Machine Systems

used, i.e. the currents and flux-linkages are found when resistances are neglected at first, then the corresponding time-constants are considered, and at last the currents and flux-linkages are substituted into the electro-magnetic torque formula, then the average torque is lost at all. However, the pulsating electromagnetic torques obtained in this approximate method are sufficiently accurate. In this section, we shall use the approximate method to analyse pulsating electromagnetic torque components; as for the average electromagnetic torque, another method is used to analyse it separately. In addition, we shall discuss only the torque after 3-phase short circuit during no-load for convenience. (1) Pulsating torque after 3-phase sudden short circuit After 3-phases of a synchronous machine are suddenly short-circuited during no-load, the currents id and iq , according to G 0, U E and equations (2.5.90) and (2.5.91), can be written as id iq

t t t ½ ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º E Ta «¨ ¸ e ¨ ¸ e » E e cos t ° xd »¼ xdcc ° «¬© xdcc xdc ¹ © xdc xd ¹ ¾ t ° E Ta e sin t ° xqcc ¿

(2.8.1)

Because the machine before short circuit is not loaded, the initial flux-linkages in phases a, b, c are

\ a0 \ b0 \ c0

E cos J 0

½ ° E cos(J 0 120e )¾ E cos(J 0 120e ) °¿

(2.8.2)

The stator resistance neglected, from Park’s equations we can see that under steadystate condition of 3-phase short circuit the flux-linkages in phases a, b, c are

\ ay \ by \ cy

0

(2.8.3)

After 3-phase sudden short circuit, the stator flux-linkages will decay from their initial values to zero. The decay time constant is Ta as mentioned before, and the general expressions for flux-linkages in phases a, b, c are

154

\a

Ee

\b

Ee

\c

Ee

t Ta

t Ta

t Ta

½ ° ° ° cos(J 0 120e )¾ ° cos(J 0 120e )° ° ¿ cos J 0

(2.8.4)


Transforming them into d, q components, we can get

\d \q

t Ta

½ cos t ° ¾ t ° Ta Ee sin t ¿

Ee

(2.8.5)

Thus, Te

iq\ d id\ q 2t

2t

E 2 Ta E 2 Ta e sin t cos t e sin t cos t xqcc xdcc t t t ª§ 1 § 1 1 · 1 · 1º «¨ ¸ e Tdcc ¨ ¸ e Tdc » E 2 e Ta sin t xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹ t t t ª§ 1 1 · T cc § 1 1 · T c 1 º 2 T «¨ ¸ e d ¨ ¸ e d » E e a sin t xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹ 2t 1§ 1 1 · 2 Ta ¨ ¸ E e sin 2t 2 ¨© xdcc xqcc ¸¹

(2.8.6)

From the formula above we can see that when the stator and rotor resistances are neglected, the electromagnetic torque after short circuit contains only the pulsating torque components and has no average torque. In addition, the pulsating torque components are the fundamental and second harmonic components. The first is very large, whose initial maximum value is inversely proportional to xdcc and decays slowly. The latter is small, especially in turbogenerators, whose maximum value depends upon the difference between xdcc and xqcc and is equal to zero when xdcc xqcc. It is not difficult to understand that there are those torque components in a synchronous machine after 3-phase sudden short circuit. The field produced by stator fundamental current has a synchronous speed, the field caused by stator aperiodic current is at a standstill and that produced by the stator second harmonic current has double synchronous speed. The field produced by rotor aperiodic current has synchronous speed and that caused by the rotor fundamental current has two components whose speeds are double synchronous speed and zero respectively. It is not difficult to know that the relative speeds between those fields are zero, Z s and 2Z s and the fundamental and second harmonic pulsating torque components can be produced. The interaction between stator and rotor fields whose relative speed is zero will produce an average torque, but the average torque is equal to zero because the stator and rotor resistances have been neglected and their field-axes coincide with each other. 155

AC Machine Systems

The electromagnetic torque after 3-phase sudden short circuit is shown in Fig. 2.8.1.

Figure 2.8.1 Pulsating torque after 3-phase sudden short circuit

(2) Average torque after 3-phase sudden short circuit After 3-phases of a synchronous machine are suddenly short-circuited, in stator winding there are the fundamental, second harmonic and aperiodic currents and in rotor winding there are the fundamental and aperiodic currents, which is the synthetic result when the aperiodic currents circulate through the stator and rotor windings simultaneously. Because the fields produced by the two groups of currents have different speeds, the average torque components corresponding to them can be studied separately. (a) The average torque component corresponding to the rotor aperiodic current and stator fundamental current The revolving fields produced by those currents have the synchronous speed and constitute average torque. Because stator winding is short-circuited, the average torque component is equal to the short circuit torque Tk referring to equation (2.5.18), i.e. the copper loss caused by the fundamental current in stator winding. That is natural, because under the condition of short circuit the energy needed according to the copper loss can be obtained only through air-gap and the value is equal to corresponding electromagnetic torque during synchronous speed. The copper loss produced by the dc current in excitation winding is supplied by an excitation source, so the loss can not produce average torque. After the 3-phases of a synchronous machine are suddenly short-circuited, the fundamental current in stator winding referring to equation (2.5.92) can be written as ia (50 )

t t ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º «¨ ¸ e ¨ ¸ e » E cos(t J 0 ) xd »¼ «¬© xdcc xdc ¹ © xdc xd ¹

(2.8.7)

According to equations (2.5.109), (2.5.108), (2.5.106) and (2.5.105), the aperiodic current in rotor winding can be written out easily. If the decay process can be considered as a series of steady-state conditions, according to the average torque 156


formula during 3-phase short circuit in steady-state, the average torque component corresponding to the rotor aperiodic current and stator fundamental current can be expressed as 2

Tecp (50 )

t t ª§ 1 1 · T cc § 1 1 · T c 1 º 2 «¨ ¸ e d ¨ ¸ e d » E r xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹

(2.8.8)

It should be noted that after 3-phase sudden short circuit, in rotor winding there is not only the dc current supplied by an excitation source but also the aperiodic current changing with time which is produced in order to keep the rotor flux-linkage constant. The copper losses caused by those currents in rotor winding are provided with the excitation source and magnetic energy reserved in rotor winding. Therefore, those copper losses can not produce average torque. (b) The average torque components corresponding to the stator aperiodic and second harmonic currents and rotor fundamental current After a dc current circulates through stator winding, the fundamental current will be produced in rotor winding. Because the rotor is electrically and magnetically asymmetrical, the latter current will cause the second harmonic current in stator winding. The revolving fields corresponding to those currents have double synchronous speed and zero speed respectively, and a certain average torque will be developed. (i) So far as the stator second harmonic current is concerned, like the stator fundamental current as mentioned before, it also causes the copper loss in stator winding and produces the average torque component which is equal to the resistance loss. That is because both are the short circuit torque, but the first is caused by the revolving field corresponding to rotor dc current and the latter is caused by the revolving field corresponding to rotor fundamental current. After 3-phase sudden short circuit, the stator second harmonic current, according to equation (2.5.92), can be written as ia (100 )

t E§ 1 1 · Ta ¨ ¸ e cos(2t J 0 ) 2 ¨© xdcc xqcc ¸¹

(2.8.9)

If the decay process can be considered as a series of steady-state conditions, the corresponding average electromagnetic torque component is 2

Tecp (100 )

2t E2 § 1 1 · Ta ¨ ¸ e r 4 ¨© xdcc xqcc ¸¹

(2.8.10)

(ii) So far as the rotor fundamental current is concerned, it also causes the copper loss in rotor winding and produces corresponding average torque. It can be proved that the copper loss in rotor winding caused by the rotor fundamental 157

AC Machine Systems

current, when expressed in terms of the stator current components id and iq , 1 [| id (50 ) |2 Rd 1 | iq (50 ) |2 Rq1 ], in which id (50 ) and iq (50 ) are the 2 fundamental components of currents id and iq , and Rd 1 and Rq1 are d-axis is equal to

and q-axis equivalent resistances, i.e. the corresponding values in the following expressions: xd ( j )

xd ( p) | p

j

xq ( j )

xq ( p ) | p

j

X d 1 jRd 1 ½° ¾ X q1 jRq1 °¿

(2.8.11)

After 3-phase sudden short circuit, the fundamental components of currents id and iq , according to equation (2.5.90), can be written as id (50 ) iq (50 )

t ½ E Ta e cos t ° xdcc ° ¾ t E Ta ° e sin t ° xqcc ¿

(2.8.12)

If the decay process can be considered as a series of steady-state conditions, the corresponding average electromagnetic torque component is Tecp (0 )

1 [| id (50 ) |2 Rd 1 | iq (50 ) |2 Rq1 ] 2 2

2

2t 2t 1 § E · Ta 1 § E · Ta ¨ ¸ e Rd 1 ¨¨ ¸¸ e Rq1 2 © xdcc ¹ 2 © xqcc ¹

(2.8.13)

(iii) The stator aperiodic current also causes the copper loss in stator winding, but the loss is provided with the magnetic energy reserved in stator winding and can not produce average torque. (c) The average electromagnetic torque after 3-phase sudden short circuit According to the results obtained above, we can write the average torque after short circuit as Tecp

Tecp (50 ) Tecp (100 ) Tecp (0 ) 2

t t ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º 2 «¨ ¸ e ¨ ¸ e » E r xd »¼ «¬© xdcc xdc ¹ © xdc xd ¹ 2 2t ª§ 1 2 Rd 1 2 Rq1 º E 2 Ta 1· « » ¨ ¸ r 2 2 e xdcc xqcc » 4 «¨© xdcc xqcc ¸¹ ¬ ¼

158

2 Analysis of Several Operating Conditions of Synchronous Machines at Synchronous Speed 2

t t ª§ 1 1 · Tdcc § 1 1 · Tdc 1 º 2 e e » E r «¨ ¸ ¨ ¸ xd »¼ «¬© xdcc xdc ¹ © xdc xd ¹

2

2t 1§ 1 1· 1 1 ª º ¨ ¸ E 2 «b 2 r (1 b)2 Rd 1 (1 b)2 Rq1 » e Ta 4 ©¨ xdcc xqcc ¹¸ 2 2 ¬ ¼

(2.8.14)

in which b

xqcc xdcc xdcc xqcc

In practice, the copper loss and torque Tecp (100 ) caused by the stator second harmonic current are very small and can be neglected. Therefore, the average electromagnetic torque after short circuit can also be evaluated according to the following formula: Tecp

t t ª§ 1 1 · T cc § 1 1 · T c 1 º 2 «¨ ¸ e d ¨ ¸ e d » E r xd »¼ © xdc xd ¹ ¬«© xdcc xdc ¹ 2t 1 2 § Rd 1 Rq1 · Ta E ¨ 2 2 ¸e 2 ¨© xdcc xqcc ¸¹

(2.8.15)

The analyses above show that the average torque Tecp after short circuit is equal to the copper losses produced by the stator and rotor ac currents, and the copper losses caused by the stator and rotor aperiodic currents are supplied by the excitation source and magnetic energy reserved in the machine. The important conclusion can be used not only to symmetrical short circuit but also to asymmetrical short circuit. At last, it should be noted that the instantaneous values of the currents and flux-linkages, according to the state-variable equation, can be estimated by digital computers, and the instantaneous values of the torque may also be evaluated according to the torque formula Te iq\ d id\ q . By the numerical method we can obtain the instantaneous values of the torque, but can not predict the harmonic components of the torque.

2.9

Suddenly Opening 3-phase Short Circuit of Synchronous Machines

It often appears that 3-phase short circuit of a synchronous machine is suddenly opened. Analysing that problem, we can consider the 3-phase short circuit not to be opened but a current to be suddenly injected into the stator which is equal but 159

AC Machine Systems

opposite to the original short-circuit current. If the machine has 3-phase short circuit in steady-state before opened, the short circuit current, when stator E and iq 0 0. Therefore, the problem, that 3-phase resistance neglected, is id 0 xd short circuit is suddenly opened, can be considered as that the current E id 0 is suddenly injected into the stator. By Superposition Theorem the xd problem will be discussed as follows. (1) Each quantity caused by the current (id 0 ) Because the flux-linkage equation of a synchronous machine is

\d

G ( p)U fd xd ( p )id

(2.9.1)

and that expression can be rewritten as id

1 pG ( p ) § U fd · \d ¨ ¸ xd ( p ) xd ( p ) © p ¹

(2.9.2)

from the equivalent circuit of d-axis operational reactance of a synchronous machine shown in Fig. 2.4.6 we can see that the xd ( p ) in the formula above is x ( p) operational self-impedance and d is operational mutual impedance. In pG ( p ) addition, according to the Reciprocity Theorem we can see that when U fd 0, i.e.

U fd p

0 we have

I fd

pG ( p) \d xd ( p)

(2.9.3)

During this time there is

\d

xd ( p )id

I fd

pG ( p)id

so (2.9.4)

As mentioned above, the problem that 3-phase short-circuit is suddenly opened can be considered as that the current (id 0 ) is suddenly injected into the stator, so we have I fd

\d 160

pG ( p )id 0 ½° ¾ xd ( p)id 0 °¿

(2.9.5)


Having those expressions, we can solve the excitation current I fd and d-axis flux-linkage \ d . According to two conditions the problem will be discussed as follows. (i) If there is no damping winding, we can get 2 ª pxad « xd R fd pX ffd «¬

\d

º » id 0 »¼ t

x2 ̯ xd id 0 ad id 0 e Td 0 X ffd

id 0 [ xd ( xd xdc )e

t Td 0

(2.9.6)

] t

I fd

pxad x id 0 ̯ ad id 0 e Td 0 pX ffd R fd X ffd

(2.9.7)

(ii) If there is a damping winding, we have

\d

2 3 2 2 2 ª 2 xad X ffd xad R fd xad p 2 ( X 11d xad ) p( R1d xad ) º x « d » id 0 2 2 p ( X 11d X ffd xad ) p( X 11d R fd X ffd R1d ) R1d R fd ¼» ¬«

(2.9.8)

For convenience, we shall use the same approximate method as in Section 2.5. (a) Instantaneous values of \ d Supposing R fd R1d 0 or p f in equation (2.9.8), we have

Supposing R fd Supposing p

xdccid 0

½ ° 0 and R1d f, we can get ° °° \ d xdc id 0 ¾ ° 0, we can obtain ° ° \ d xd id 0 °¿

\d

(2.9.9)

(b) Decay time constants of \ d According to equation (2.9.8), its characteristic equation is 2 p 2 ( X 11d X ffd xad ) p ( X 11d R fd X ffd R1d ) R1d R fd

For convenience, taking approximately R fd

0

(2.9.10)

0, we have

2 p ( X 11d X ffd xad ) X ffd R1d

0

and the corresponding time constant is 161

AC Machine Systems

X 11d Tdcc0

2 xad X ffd

(2.9.11)

R1d

f, we have

Supposing R1d

pX ffd R fd

0

and the corresponding time constant is X ffd

Td 0

(2.9.12)

R fd

(c) The general expression of \ d caused by the current (id 0 ) is

\d

id 0 [ xd ( xd xdc )e

t Td 0

( xdc xdcc )e

t Tdcc0

]

(2.9.13)

(d) The excitation-winding current caused by the current (id 0 ) : According to equation (2.9.5), we have I fd

2 p 2 ( X 11d xad xad ) pxad R1d id 0 2 2 p ( X 11d X ffd xad ) p( X 11d R fd X ffd R1d ) R1d R fd

Supposing R1d

R fd

0 or p I fd

Supposing R1d

f and R fd

(2.9.14)

f in equation (2.9.14), we can obtain

2 X 11d xad xad i 2 d0 X 11d X ffd xad

(2.9.15)

0 in equation (2.9.14), we can get I fd

xad id 0 X ffd

(2.9.16)

If p 0 in equation (2.9.14), then exists I fd 0. The time constant of I fd is the same as that of \ d because their characteristic equations are similar, so we can obtain I fd

t 2 ªx § x X x xad id 0 « ad e Td 0 ¨ ad 11d ad 2 ¨X «¬ X ffd © ffd X 11d X ffd xad

· Ttcc º ¸¸ e d 0 » »¼ ¹

(2.9.17)

(e) The d-axis damper current I1d caused by the current (id 0 ) : The quantity can be obtained by using the same method as in Section 2.5. Its expression is omitted here. 162


(2) The original short-circuit values Neglecting stator resistance r, we have id 0

id

iq

0

\d

uq

\q

u d

I fd

I fd 0

E xd 0 0

U fd R fd

xd id 0 xad

(2.9.18)

(3) Various actual values when 3-phase short circuit is suddenly opened The actual values of stator current are

in addition

id

id 0 id 0

iq

0

\q

0

0½ ° ¾ ° ¿

(2.9.19)

(a) When there is only excitation winding on rotor, we have

\d

t ª º id 0 « xd ( xd xdc )e Td 0 » «¬ »¼

I fd

I fd 0

t

xad id 0 e Td 0 X ffd

(2.9.20)

ª x xdc Tt º e d0 » I fd 0 «1 d x d ¬« ¼» t

ud

p\

uq

\d

d

1 ( xd xdc )e Td 0 id 0 Td 0

E Td 0

ª xdc Tt º «1 e d 0 » «¬ xd »¼

t ª º id 0 « xd ( xd xdc )e Td 0 » «¬ »¼ t ª § xc · º E «1 ¨1 d ¸ e Td 0 » xd ¹ «¬ © »¼

(2.9.21)

(2.9.22)

(2.9.23)

in which ud is the transformer emf and uq is the speed emf. Because ud uq , the amplitude of phase emf is

163

AC Machine Systems

|u|

2 d

2 q

u u | uq

t ª § xdc · Td 0 º E «1 ¨1 ¸ e » xd ¹ «¬ © »¼

(2.9.24)

According to equations (2.9.20) and (2.9.23), the curves of | u | and I fd are shown in Fig. 2.9.1.

Figure 2.9.1 I fd - t curve and | u |- t curve after opening short circuit of synchronous machines

From the results above we can see that after 3-phase short circuit is suddenly opened, the stator current and armature reaction are decreased abruptly to zero. Therefore, \ fd and the stator voltage are to increase suddenly, but \ fd can not increase suddenly, so the stator voltage can not be increased abruptly to its steady-state value. In the meantime, there is a negative aperiodic current in excitation winding to keep its flux-linkage not changing abruptly, so xc the excitation current I fd is decreased suddenly to d I fd 0 . However, the decrease xd of I fd results in the decrease of leakage flux in excitation winding, so the mutual flux-linkage still increases suddenly and the corresponding stator voltage will xc rise suddenly from zero to d E. In addition, because the negative aperiodic xd current in excitation winding will decay to zero according to the time constant Td 0 , the corresponding stator voltage and \ d will rise to their steady-state values according to that time constant. (b) When there is not only an excitation winding but also a damping winding on the rotor, we can get | u || \ d I fd 164

t t ª º Td 0 Tdcc0 c c cc ( xd xd )e » id 0 « xd ( xd xd )e «¬ »¼

t 2 ªx § x x X x xad id 0 « d ad e Td 0 ¨ ad 11d ad 2 ¨X © ffd X 11d X ffd xad ¬« xad X ffd

· Ttcc º ¸¸ e d 0 » ¹ ¼»


It should be noted that when t

0 we have | u || \ d

id 0 xdcc

xdcc E. Compared xd

with the machine without damping windings, its value decreases by a factor § xcc xc xdcc · of ¨ d E d E ¸ , i.e. after 3-phase short circuit is suddenly opened, the xd xdc ¹ © xd terminal voltage with damping windings is lower than without damping windings, which has a benefit to quenching electric arc of the switch-gear.

References [1] Gao J D (1963) AC machine transients and operating modes analysis (Chapter 2, 3, 4, in Chinese). Science Press, Beijing [2] Gao J D, Zhang L Z (1982) Fundamental theory and analysis mothods of transients in electric machines, Vol 1 (Chapter 1, 4, 5, in Chinese). Science Press, Beijing [3] Gao J D, Wang X H, Li F H (2004) Analysis of ac machines and their systems, 2nd Ed (Chapter 2, in Chinese). Tsinghua University Press, Beijing [4] Jackson W B, Winchester R L (1969) Direct- and quadrature-axis equivalent circuits for solid-rotor turbine generators. J IEEE Trans. PAS 88(7): 1121 1136 [5] Randin A W (1945) Per unit impedances of salient-pole synchronous machines. J AIEE Trans. Aug. Part I: 569 573, Dec. Part Ċ: 839 841 [6] Randin A W (1945) The direct- and quadrature-axis equivalent circuits of the synchronous machine. J AIEE Trans. Aug. Part I: 569 573, Dec. Part Ċ: 867 868

165

3

Some Special Operation Modes of Synchronous Machine Systems at Constant Speed and Variable Speed

Abstract As is stated in Chapter 2, by use of d, q, 0 axes can get the differential equations with constant coefficients for symmetrical faults of synchronous machines at synchronous speed, but can’t for unsymmetrical faults or variable speed. Therefore, in this chapter it is necessary to introduce other reference axes systems first, such as D , E , 0 axes; 1, 2, 0 axes; d c , qc , 0 axes; F, B, 0 axes; Fc , Bc , 0 axes. Then use D , E , 0 axes to analyse two phases sudden short circuit of synchronous machines, inclusive of stator and rotor loop parameters and currents, overvoltage, pulsating and average torques. Furthermore, use 1, 2, 0 axes to study asynchronous operation of synchronous machines to get corresponding equivalent circuits, currents, torques and negative sequence reactance that depends upon the operating modes. In a similar way, analyse asynchronously starting of synchronous motors to get starting torque and pull-in one and suggest the optimal time for applying a dc voltage to excitation winding. Analysis of combination system for synchronous machine and capacitances is also an interesting topic, in which by use of Herwitz’s, Routh’s and Mihainov’s criterions determine the salient-pole synchronous self-excitation region, asynchronous and repulsion synchronous self-excitation regions, explain the mechanism to cause those self-excitation regions, and illustrate why the sub-synchronous resonance occurs. Cycloconverters can change the source frequency to lower frequency that is under 1/3 1/2 of source frequency, and cycloconverter-fed synchronous motor systems with field-oriented control are widely used to drive slow speed and large capacity production machines because of their some advantages over the ac-dc-ac frequency converter. Therefore, their mathematic patterns and performance analysis belong to another important topic. Taking a steel-mill driven by a 4 000kW-cycloconverter-fed synchronous motor as example, analyse the steady-state and transient performance of the cycloconverter-fed synchronous motor system, simulative and experimental results being in accord with each other and thus giving a suggestion that is how to dispose damping bars. Some special power supply systems, such as those used on ships, tractions, and aircraft, are often required to provide high quality ac and dc electricity simultaneously, so analysis of synchronous

3 Some Special Operation Modes of Synchronous Machine Systems

generators with ac and dc stator connections is also distinct. The ac-dc generator has five suits of independent star-connected 3-phase windings in its stator, one is called ac winding to supply ac power and the others belong to dc windings, which are connected to four 3-phase bridge rectifiers respectively, and then the dc output sides of the four 3-phase bridge rectifiers are in parallel. By use of multi-loop model analyse the ac-dc generator to compared with experimental data, both results being in accord with each other.

In Chapter 2, we have studied some symmetrical operation modes of synchronous machines by use of d, q, 0 axes. It is evident that analysis of several problems in d, q, 0 axes is convenient especially when the stator connection of a synchronous machine is symmetrical. However, the reference axes system is not more convenient than other ones when asymmetrical operation modes of synchronous machines are researched on. Therefore, with progress in science and technology some new methods are developed for study of transients of synchronous machines. With regard to axes transformation referring to [2,3], there are the following reference axes systems to be used practically: (i) The stationary reference axes systems that lie on the stator, namely a, b, c axes, D , E , 0 axes and 1, 2, 0 axes. (ii) The revolving reference axes systems that are put on the rotor and rotate with it, namely d, q, 0 axes and F, B, 0 axes. (iii) The synchronous-speed reference axes systems that rotate at synchronous speed, namely d c , qc , 0 axes and Fc , Bc , 0 axes. Those reference axes systems have their advantages and shortcomings. The choice of either this reference axes system or that one depends upon the following conditions: (i) The needed accuracy and convenience of calculation. (ii) The conditions of studied problems, such as steady-state or transient, symmetrical or asymmetrical, speed-changing or constant, accelerative or oscillatory, and so forth. (iii) The used tools, such as mathematical analysis or experiment simulation, digital computers, analogue computers or physical simulation, etc. The same problem can often be solved by several methods, some of which are used conveniently or by which more accurate results can be obtained and the others are more tedious or not precise enough. Similarly, the same method is used easily under some condition, but it is complicated on other conditions. Therefore, the researchers should grasp various axes transformation methods and use them very well according to the concrete condition. Generally speaking, if 167

AC Machine Systems

the stator or rotor is symmetrical and the other is asymmetrical, it is more suitable to put the reference axes on the asymmetrical part. For example, when the rotor of a synchronous machine is asymmetrical and the stator is symmetrical, it is more convenient to make use of d, q, 0 axes or F, B, 0 axes. Moreover, the rotor of an induction machine is normally symmetrical and it is more suitable to take D , E , 0 axes or 1, 2, 0 axes when the operation problem about the asymmetrical stator is studied. If both the stator and rotor are asymmetrical, the basic equations of the machine expressed by any reference axes system are all equations with variable coefficients even though the machine speed is constant. Owing to the causes above, we shall discuss other reference axes systems in addition to d, q, 0 axes and the transformation relations between them in this chapter. In the meantime, we shall research into some special operation problems of synchronous machines further by using some of the reference axes systems mentioned above.

3.1

Transformations of Reference Axes Systems and Their Formulas in Synchronous Machines

The transformation of reference axes is that of variables, i.e. a set of variables is replaced by another set. Letting the original set of variables be x1 , x2 , x3 ," , xn and supposing ª y1 º «y » « 2» « y3 » « » «#» «¬ yn »¼

ªM11 «M « 21 «M31 « « # «¬M n1

M12 M13 " M1n º ª x1 º M 22 M 23 " M 2 n »» «« x2 »» M32 M33 " M3n » « x3 » #

#

M n 2 M n3

»« » # »« # » " M nn »¼ «¬ xn »¼

(3.1.1)

or

Y

CX

we can get a new group of variables y1 , y2 , y3 " , yn , in which the element M in the transformation matrix C is called the transformation coefficient which can be taken as a constant (real or complex number) or a function of time. In the linear transformation, M is independent of variables x and y. From Linear Algebra we can see that in order to fulfil the transformation and 168


have a single-value relation between the new and original variables, the determinant corresponding to the transformation matrix C in equation (3.1.1) must suffice the following condition,

M11 M12 M13 " M1n M 21 M 22 M 23 " M 2 n | C | M31 M 32 M33 " M 3n z 0 #

#

#

(3.1.2)

#

M n1 M n 2 M n3 " M nn There are many methods to choose the transformation coefficients to suffice equation (3.1.2). For example, the variables in a, b, c axes of a 3-phase synchronous machine taken as original variables, the new variables after transformation have various types. d, q, 0 axes mentioned before are an example, whose transformation matrix is

C

ªM11 M12 M13 º «M » « 21 M 22 M 23 » ¬«M31 M 32 M 33 »¼ 2 2 ª2 º cos(J 120e ) cos(J 120e )» « 3 cos J 3 3 « » 2 2 2 « sin J sin(J 120e ) sin(J 120e )» « 3 » 3 3 « » 1 1 « 1 » 3 3 ¬« 3 ¼»

Obviously, | C |z 0. It depends upon the actual need and mathematical convenience how to choose the transformation matrix and corresponding reference axes. In practice, there are already several transformation models. In the following we shall introduce other reference axes except d, q, 0 axes. It should be pointed out that in various reference axes the transformation coefficients M 31 , M 32 , M 33 are equal to 1 M31 M32 M33 , i. e. the 0-axis component has the same form in various 3 reference axes. It is also noted that in study of transient processes the three variables — current, voltage and flux-linkage must all be transformed but they have the similar transformation formula. Therefore, in the following we shall take only the current as an example. (1) D , E , 0 axes In the reference axes, the corresponding transformation matrix is 169

AC Machine Systems

ªM11 M12 M13 º «M » « 21 M 22 M 23 » «¬M31 M 32 M 33 »¼

C

ª2 «3 « «0 « « «1 «¬ 3

1 3 1

3 1 3

2 2 ª2 º ) cos(0e 120e )» « 3 cos 0e 3 cos(0e 120e 3 « » 2 « 2 sin 0e 2 sin(0e 120e ) sin(0e 120e )» « 3 » 3 3 « » 1 1 1 « » «¬ »¼ 3 3 3

º » » » 3» » 1 » 3 »¼

1 3 1

(3.1.3)

Taking the current as an example and substituting equation (3.1.3) into (3.1.1), we can get iD iE i0

i i ·½ 2§ ia b c ¸ ° ¨ 3© 2 ¹° ° 1 (ib ic ) ¾ 3 ° ° 1 (ia ib ic ) ° 3 ¿

(3.1.4)

½ ° ° 1 3 ° iE i0 ¾ iD 2 2 ° ° 1 3 iE i0 ° iD 2 2 ¿

(3.1.5)

The inverse-transformation is ia ib ic

iD i0

The relation between D , E components and d, q components is iD iE

id cos J iq sin J ½° ¾ id sin J iq cos J °¿

(3.1.6)

iD cos J iE sin J ½° ¾ iD sin J iE cos J °¿

(3.1.7)

or id iq 170


in which J is the included angle between d-axis and D-axis or a-axis. In D , E , 0 axes, the formulas of the flux-linkage, voltage, out-put power and electromagnetic torque are

\D \E

[cos J xd ( p ) cos J sin J xq ( p )sin J ]iD

½ ° [cos J xd ( p)sin J sin J xq ( p ) cos J ]iE cos J G ( p )u fd ° ¾ [sin J xd ( p) cos J cos J xq ( p )sin J ]iD ° [sin J xd ( p )sin J cos J xq ( p ) cos J ]iE sin J G ( p)u fd °¿

(3.1.8)

p\ D riD ½° ¾ p\ E riE °¿

(3.1.9)

uD iD uE iE 2u0 i0

(3.1.10)

uD uE

P

Te \ D iE \ E iD

(3.1.11)

It is noted that the flux-linkage equation (3.1.8) and votage equation (3.1.9) are the expressions described by Heaviside’s Operation, so the arrangement order of various physical quantities in the expressions can not be exchanged arbitrarily, i.e. the operational reactances xd ( p ), xq ( p ) and operational conductance G ( p ) don’t act on cos J or sin J before them but act on the functions of time written after them referring to Appendix D in this book. It is not difficult to understand that the flux-linkages \ D , \ E are some complicated functions of currents iD and iE . However, the rotor resistance neglected, the equations above can be much reduced and used easily. Studying d, q, 0 axes, we have mentioned that the currents id and iq represent the stator d-axis and q-axis mmfs which are observed from the reference axes that rotate with the rotor. Discussing D , E , 0 axes now, from equations (3.1.3) and (3.1.4) we can know that the currents iD and iE represent the stator D-axis and E -axis mmfs respectively, in which D-axis coincides with the stator a-axis S and E -axis leads ahead of D-axis by an angle of referring to Fig. 3.1.1, so those axes are static ones relative to the stator. In addition, id , iq , \ d , \ q and ud , uq in d, q, 0 axes represent the currents, flux-linkages and voltages in the fictitious circuits of the stator d-axis and q-axis respectively. Similarly, iD , iE , \ D , \ E and uD , u E in D , E , 0 axes represent the currents, flux-linkages and voltages in the fictitious circuits of the stator D-axis and E -axis referring to Fig. 3.1.1. Thus, we 171

AC Machine Systems

have another physical model of synchronous machines which is an equivalent 2-phase synchronous machine.

Figure 3.1.1 D , E , 0 pattern of synchronous machines

From equation (3.1.9) we can see that there is on speed emf in D , E , 0 axes. That is not difficult to understand, because D , E , 0 axes are static ones and

\ D , \ E don’t rotate relatively to the stator. (2) 1, 2, 0 axes The transformation matrix in the reference axes is

C

ªM11 M12 M13 º «M » « 21 M 22 M 23 » «¬M 31 M 32 M 33 »¼

ª1 «3 « «1 «3 « «1 «¬ 3

1 a 3 1 2 a 3 1 3

1 2º a 3 » » 1 » a 3 » » 1 » 3 »¼

(3.1.12)

1 3 e j120e j is the complex operator. 2 2 Taking the current as an example and substituting equation (3.1.12) into (3.1.1), we can obtain in which a

i1 i2 i0

1 ½ (ia aib a 2 ic ) ° 3 ° 1 ° (ia a 2 ib aic ) ¾ 3 ° ° 1 (ia ib ic ) ° 3 ¿

(3.1.13)

It is not difficult to understand that after transformation the original actual instantaneous currents ia , ib and ic become the instantaneous currents i1 and i2 172


expressed by complex-number, so the reference axes are sometimes called the complex-number axes, which are motionless relatively to the stator. The inverse-transformation is ia ib ic

i0 i1 i2

½ ° i0 a i1 ai2 ¾ ° i0 ai1 a 2 i2 ¿ 2

(3.1.14)

From the formulas above we can see that the basic formulas of the complexnumber axes look like those of the ordinary Symmetrical Components Method, but both are essentially different. In complex-number axes the corresponding variables are complex-number instantaneous values, but in ordinary Symmetrical Components the corresponding variables are complex-number vectors representing the quantities changing sinusoidally with time. In ordinary Symmetrical Components the operator a e j120e means that the time-vector rotates by an angle of 120e, but in complex-number axes the operator a e j120e sometimes means that the spacevector rotates by an angle of 120e. Besides, 1, 2 components in complexnumber axes are always conjugate to each other, but in Symmetrical Components Method the positive and negative components have not the same relation as above. The relations among 1,2 components, D , E components and d, q components are

i1 i2

1 ½ (iD jiE ) ° ° 2 ¾ 1 (iD jiE ) ° °¿ 2

(3.1.15)

or iD iE

i1 i2

i1 i2

½° ¾ j(i1 i2 ) °¿

(3.1.16)

1 ½ (id jiq )e jJ ° ° 2 ¾ 1 jJ ° (id jiq )e °¿ 2

(3.1.17)

or

id iq

i1e jJ i2 e jJ j(i1e

jJ

½° ¾ i2 e ) °¿ jJ

(3.1.18)

173

AC Machine Systems

In 1, 2, 0 axes, the formulas of flux-linkage, voltage, output power and electromagnetic torque are

\1 \2

1 jJ ½ e G ( p )u fd e jJ xs ( p )e jJ i1 e jJ xD ( p )e jJ i2 ° ° 2 ¾ 1 jJ e G ( p )u fd e jJ xs ( p )e jJ i2 e jJ xD ( p )e jJ i1 ° °¿ 2

(3.1.19)

in which 1 [ xd ( p ) xq ( p )] 2

xs ( p )

is the average operational reactance and 1 [ xd ( p ) xq ( p)] 2

xD ( p )

is the half-difference operational reactance.

p\ 1 ri1 ½ ¾ p\ 2 ri2 ¿

(3.1.20)

2(u2 i1 u1i2 u0 i0 )

(3.1.21)

j2(\ 1i2 \ 2 i1 )

(3.1.22)

u1 u2 P Te

It should be noted that the flux-linkage formula (3.1.19) is still expressed by Heaviside’s Operation, so the arrangement order of various quantities in the expressions can not be exchanged arbitrarily. When the speed of a machine is constant, i. e. the angular velocity Z is constant, according to Heaviside’s Shift Theorem the equation (3.1.19) can be rewritten as the following simple form:

\1 \2

1 jJ ½ e G ( p )u fd xs ( p jZ )i1 e j2J xD ( p jZ )i2 ° ° 2 ¾ 1 jJ e G ( p )u fd xs ( p jZ )i2 e j2J xD ( p jZ )i1 ° °¿ 2

(3.1.23)

Besides simplicity mentioned above, there is a property that 1, 2 components are always conjugate to each other. we can use the property to check whether the calculation results are correct. When xd ( p ) xq ( p ), 1, 2 components are independent of each other, so it is sufficient to calculate any one only. The induction machine having a symmetrical rotor falls into that condition. In practice, it is convenient to use 1, 2 components simultaneously. 174


(3) d c , qc , 0 axes The transformation matrix in the reference axes is

C

ªM11 M12 M13 º «M » « 21 M 22 M 23 » ¬«M31 M 32 M 33 »¼ 2 ª2 ) « 3 cos(t J 0c ) 3 cos(t J 0c 120e « « 2 sin(t J c ) 2 sin(t J c 120e ) 0 0 «3 3 « 1 1 « 3 3 ¬«

2 º cos(t J 0c 120e )» 3 » 2 sin(t J 0c 120e )» » 3 » 1 » »¼ 3

(3.1.24)

Taking the current as an example and substituting equation (3.1.24) into (3.1.1), we can get

idc iqc i0

2 ½ [ia cos(t J 0c ) ib cos(t J 0c 120e ) ic cos(t J 0c 120e )] ° 3 ° 2 ° ) ic sin(t J 0c 120e )]¾ (3.1.25) [ia sin(t J 0c ) ib sin(t J 0c 120e 3 ° 1 ° [ia ib ic ] ° 3 ¿


idc cos(t J 0c ) iqc sin(t J 0c ) i0

½ ° idc cos(t J 0c 120e ) iqc sin(t J 0c 120e ) i0 ¾ ° idc cos(t J 0c 120e ) iqc sin(t J 0c 120e ) i0 ¿

(3.1.26)

The relations between d c , qc and d , q components are idc iqc

id cos 'J iq sin 'J ½° ¾ id sin 'J iq cos 'J °¿

(3.1.27)

idc cos 'J iqc sin 'J ½° ¾ idc sin 'J iqc cos 'J °¿

(3.1.28)

or id iq in which 'J

t

³ 'Z dt J 0

0

J 0c

J (t J 0c ) 175

AC Machine Systems

and 'Z

Z 1

is the difference between the actual speed and synchronous one. When Z 1, then 'Z 0 and 'J J 0 J 0c , so idc , iqc are the same as id , iq

except that there is an angle of (J 0 J 0c ) between them. That is natural, because d c , qc , 0 axes are d, q, 0 axes during synchronous speed.

In d c , qc , 0 axes, the formulas of flux-linkage, voltage, output power and electromagnetic torque are

\ dc

\ qc

[cos 'J xd ( p ) cos 'J sin 'J xq ( p )sin 'J ]idc ½ ° [cos 'J xd ( p )sin 'J sin 'J xq ( p) cos 'J ]iqc ° ° cos 'J G ( p )u fd ° ¾ [sin 'J xd ( p) cos 'J cos 'J xq ( p)sin 'J ]idc ° [sin 'J xd ( p )sin 'J cos 'J xq ( p) cos 'J ]iqc ° ° °¿ sin 'J G ( p )u fd

(3.1.29)

p\ dc \ qc ridc ½° ¾ p\ qc \ dc riqc °¿

(3.1.30)

P

udc idc uqc iqc 2u0 i0

(3.1.31)

Te

\ dc iqc \ qc idc

(3.1.32)

udc uqc

(4) F, B, 0 axes The transformation matrix in the reference axes is C

ªM11 M12 M13 º «M » « 21 M 22 M 23 » «¬M31 M 32 M 33 »¼ ª « « « « « « « «¬

176

2 jJ e 3 2 jJ e 3 1 3

2 jJ ae 3 2 2 jJ ae 3 1 3

2 2 jJ º ae » 3 » 2 jJ » ae » 3 » » 1 » 3 »¼

(3.1.33)


Taking the current as an example and substituting equation (3.1.33) into (3.1.1), we can get iF

2 (ia aib a 2 ic )e jJ 3

iB

2 (ia a 2 ib aic )e jJ 3

i0

1 (ia ib ic ) 3

½ 2i1e jJ ° ° °° 2i2 e jJ ¾ ° ° ° °¿

(3.1.34)


1 2 1 2 1 2

½ ° ° ° 2i0 ) ¾ ° ° 2i0 ) ° ¿

(iF e jJ iB e jJ 2i0 ) (a 2 iF e jJ aiB e jJ (aiF e jJ a 2 iB e jJ

(3.1.35)

The relations between F, B and d, q components are iF iB

id iq

1

½ (id jiq ) ° ° 2 ¾ 1 (id jiq ) ° 2 ¿° 1

½ °° ¾ 1 j (iF iB ) ° °¿ 2 2

(3.1.36)

(iF iB )

(3.1.37)

It is not difficult to understand that the original instantaneous currents ia , ib and ic have been changed into the instantaneous currents expressed by complexnumbers iF and iB . Therefore, the components are also complex-number components. However, their axes are stationary not relatively to the stator but to rotor. Thus, discriminating them from 1, 2 components, we sometimes call them the forward and backward components. In F, B, 0 axes, the formulas of flux-linkage, voltage, output power and electromagnetic torque can be written as 177

AC Machine Systems

\F \B

1

½ G ( p)u fd xs ( p )iF xD ( p )iB ° ° 2 ¾ 1 G ( p)u fd xD ( p )iF xs ( p )iB ° °¿ 2

(3.1.38)

p\ F jZ\ F riF ½ ¾ p\ B jZ\ B riB ¿

(3.1.39)

u F iB uB iF 2u0 i0

(3.1.40)

j(iB\ F iF\ B )

(3.1.41)

uF uB P Te (5) Fc , Bc , 0 axes

The transformation matrix in the reference axes is ªM11 M12 M13 º «M » « 21 M 22 M 23 » ¬«M 31 M 32 M 33 »¼

C

ª « « « « « « « «¬

2 j(t J 0 ) e 3 2 j(t J 0 ) e 3 1 3

2 j(t J 0 ) ae 3 2 2 j(t J 0 ) ae 3 1 3

2 2 j(t J 0 ) º ae » 3 » 2 j(t J 0 ) » ae » 3 » » 1 » 3 »¼

(3.1.42)

Taking the current as an example and substituting equation (3.1.42) into (3.3.1), we can get iFc

2 (ia aib a 2 ic )e j( t J 0 ) 3

iBc

2 (ia a 2 ib aic )e j( t J 0 ) 3

i0

1 (ia ib ic ) 3

½ 2i1e j(t J 0 ) ° ° ° j( t J 0 ) ° 2i2 e ¾ ° ° ° °¿

(3.1.43)

They are also complex-number components, but their axes rotate at a synchronous speed.

178



½ ° 2 ° 1 2 ° j( t J 0 ) j( t J 0 ) aiBc e 2i0 ) ¾ (a iFc e 2 ° ° 1 (aiFc e j( t J 0 ) a 2 iBc e j( t J 0 ) 2i0 ) ° 2 ¿ 1

(iFc e j( t J 0 ) iBc e j(t J 0 ) 2i0 )

(3.1.44)

The relations between Fc , Bc and d c , qc components are iFc iBc

1

½ (idc jiqc ) ° ° 2 ¾ 1 (idc jiqc ) ° °¿ 2

(3.1.45)

or idc iqc

1

½ (iFc iBc ) ° ° 2 ¾ 1 j (iFc iBc ) ° °¿ 2

(3.1.46)

In Fc , Bc , 0 axes, the formulas of the flux-linkage, voltage, output power and electromagnetic torque are

\ Fc

\ Bc

1 j'J e G ( p )u fd e j'J xs ( p )e j'J iFc 2 e j'J xD ( p )e j'J iBc

½ ° ° °° ¾ 1 j'J e G ( p )u fd e j'J xs ( p )e j'J iBc ° ° 2 ° j'J j'J e xD ( p )e iFc °¿

(3.1.47)

( p j)\ Fc riFc ½ ¾ ( p j)\ Bc riBc ¿

(3.1.48)

u Fc iBc uBc iFc 2u0 i0

(3.1.49)

uFc uBc P Te

j(iBc\ Fc iFc\ Bc )

(3.1.50) 179

AC Machine Systems

3.2

2-Phases Sudden Short Circuit of Synchronous Machines

When a 3-phase synchronous machine operates symmetrically in steady-state, its many targets such as voltage, efficiency and torque are all good. Therefore, many synchronous machines are designed and used according to the operation condition. However, asymmetrical operation always arises, for example, unbalanced loads — electric trains, two phases short circuit and one phase open circuit, etc. So it is significant to analyse asymmetrical operation modes of synchronous machines. In this section, we shall discuss the transient process after two phases of a synchronous machine are suddenly short-circuited, in order to grasp the analysis method of asymmetrical operation modes of synchronous machines. As mentioned before, when either the stator or rotor is asymmetrical, it is convenient to put the reference axes on the asymmetrical part. For example, when studying 3-phase sudden short circuit of synchronous machines, we shall use d, q, 0 axes, because the stator is symmetrical and rotor is asymmetrical. The analysis result shows that the basic equations of synchronous machines have constant coefficients when rotor speed is constant. However, when both stator and rotor are asymmetrical, the constant coefficient equations can not be obtained whether we put the reference axes on the stator or rotor. Under this condition, the equations can be solved only approximately. But when analysing two phases sudden short circuit of synchronous machines, we can use D , E , 0 axes conveniently if neglecting the stator and rotor resistances refferring to [2,3,5,6,9]. Now we use the reference axes to analyse two phases sudden short circuit of synchronous machines under no-load condition if rotor speed remains synchronous. After two phases are short-circuited, referring to Fig. 3.2.1 there is

ub

uc

ia

0

ib ic

½ ° ¾ 0 °¿

(3.2.1)

When transforming them into D , E , 0 components, we obtain iD i0 uE

0½ ° 0 ¾ 0 °¿

(3.2.2)

After substituting them into equations (3.1.9) and (3.1.8), we have uD 0 180

p\ D

½° ¾ p\ E riE °¿

(3.2.3)


Figure 3.2.1 Two phases short circuit of synchronous machines

\D \E

[cos J xd ( p )sin J sin J xq ( p ) cos J ]iE cos J G ( p)u fd ½° ¾ [sin J xd ( p )sin J cos J xq ( p ) cos J ]iE sin J G ( p )u fd °¿

(3.2.4)

where J t J 0 ; J 0 is an angle by which the rotor direct axis leads ahead of the stator phase a axis when t 0. It is complicated to solve these equations directly. In practice, they can be solved by the following approximate method and the calculation results are sufficiently accurate. (1) Steady-state values of currents and voltages When two phases of a synchronous machine are suddenly short-circuited, corresponding aperiodic currents are induced in the stator and rotor loops in order that the flux-linkages of those loops don’t change abruptly. The stator aperiodic currents produce a standstill magnetic field. In the rotor winding a basic-frequency current is induced by that magnetic field because the rotor revolves at a synchronous speed. Since the rotor is asymmetrical, the rotor basic-frequency current produces a pulsating magnetic field relative to the rotor. If the pulsating field is divided into two revolving fields — the positive revolving field and negative revolving field, the negative revolving field is stationary relatively to stator and the positive one has twice the synchronous speed. Therefore, the voltage and current of twice the basic frequency are induced in stator winding. As a result of the asymmetrical stator, the pulsating field of twice the basic frequency produced by the stator current is also divided into the positive and negative revolving fields at twice the synchronous speed. The positive revolving field has a synchronous speed relatively to rotor and the basic-frequency voltage is still induced in rotor by the field; besides, the negative revolving field has triple the synchronous speed with respect to rotor and the current of triple the basic frequency is induced in the asymmetrical rotor winding. In doing so over and over again, at last, there are a series of even harmonics currents in stator winding and odd harmonics currents in rotor winding referring to Fig. 3.2.2(a). In addition, the rotor aperiodic current, inclusive of the excitation current, has the same condition; i. e. it causes the basic-frequency current in stator and the stator basic-frequency current induces a rotor current of twice the basic-frequency due 181

AC Machine Systems

to asymmetrical stator. The rotor current of twice the basic-frequency also causes the stator current of triple the basic-frequency as a result of asymmetrical rotor. In doing so repeatedly, at last, there are a series of odd harmonics currents in stator winding and even harmonics currents in rotor winding referring to Fig. 3.2.2(b).

Figure 3.2.2 Stator and rotor currents of various frequencies during two phases sudden short circuit

When two phases short circuit reaches steady-state, the stator aperiodic current equals zero and the rotor aperiodic current becomes the constant excitation current u fd . Therefore, a series of harmonic currents corresponding to the stator I fdo R fd aperiodic current become zero during steady-state and a series of harmonic currents corresponding to the rotor aperiodic current reach their steady-state values. Thus, in steady-state there are only odd harmonic currents in the stator and they can be written as follows: f

ib

¦

ic

f

¦

An cos nt

n 1,3,5,"

Anc sin nt

(3.2.5)

n 1,3,5,"

From formula (3.2.5) we get iE

1 3

(ib ic )

2 ª « 3 ¬n There is J

2 3

ib

f

¦

1,3,5,"

t when we choose J 0

n

¦

(3.2.6)

0. Because the steady-state excitation

voltage equals a constant u fd , we make p 182

º Anc sin nt » 1,3,5," ¼ f

An cos nt

0 in formula (2.4.12) and obtain


xad u fd

G ( p )u fd

I fdo xad

R fd

E

in which E is no-load voltage of a synchronous machine. Substituting the results above into equation (3.2.4), we can get

\E

ª¬sin txd ( p)sin t cos txq ( p) cos t º¼ iE E sin t ª sin txd ( p)sin t « 3 ¬n

2

ª u« ¬n

f

¦

º 2 cos txq ( p) cos t Anc sin nt » 3 1,3,5," ¼ f

An cos nt

1,3,5,"

n

¦

º Anc sin nt » E sin t 1,3,5," ¼ f

An cos nt

1,3,5,"

n

sin txd ( p) ® 3 ¯n

1

f

¦

¦

f

¦

An [sin(n 1)t sin(n 1)t ]

1,3,5,"

½ Anc [cos(n 1)t cos(n 1)t ]¾ 1,3,5," ¿ f

n

¦

cos txq ( p) ® 3 ¯n

1

f

¦

An [cos(n 1)t cos( n 1)t]

1,3,5,"

½ Anc [sin(n 1)t sin(n 1)t]¾ E sin t 1,3,5," ¿ f

n

¦

Because cos(n r 1)t

1 j( n r1)t e j( n r1)t ] [e 2

in steady-state exists xd ( p) cos(n r 1)t

1 xd ( p )[e j( n r1) t e j( n r1)t ] 2 1 {xd [ j(n r 1)]e j( n r1)t xd [ j(n r 1)]e j( n r1) t } 2

If neglecting the rotor resistances, we have xd [r j(n r 1)] and referring to formula (2.5.47) or Fig. 2.5.4; thus, xd ( p ) cos(n r 1)t

xdcc during n r 1 ! 1

1 xdcc[e j( n r1)t e j( n r1) t ] 2 xdcc cos(n r 1)t

sin txd ( p) Anc cos(n r 1)t sin txdcc Anc cos(n r 1)t 183

AC Machine Systems

When n 1 0, we can get xd [r j(n 1)] (2.4.13b). Therefore, there are xd ( p ) cos(n 1)t xd sin txd ( p ) Anc cos(n 1)t

xd (0)

xd referring to equation

xd A1c sin t

In doing so, when n r 1 ! 1 , we can obtain sin txd ( p) An sin(n r 1)t

sin txdcc An sin(n r 1)t

cos txq ( p ) An cos(n r 1)t

cos txqcc An cos(n r 1)t

and

When n 1 0, there is cos txq ( p ) An cos(n 1)t

cos txq A1

When n r 1 ! 1, we have cos txq ( p ) Anc sin(n r 1)t

cos txqcc Anc sin(n r 1)t

Thus, the steady-state flux-linkage equation above can be rewritten as follows:

\E

1 ª « A1cxd sin t xdcc sin t 3¬ n

f

¦

( An An 2 )sin(n 1)t

1,3,5,"

º 1 ª A1 xq cos t ( Anc 2 Anc ) cos(n 1)t » 3¬ 1,3,5," ¼ f

xdcc sin t n

¦ f

xqcc cos t

¦

º ( Anc Anc 2 )sin(n 1)t » 1,3,5," ¼ f

( An An 2 ) cos(n 1)t xqcc cos t

n 1,3,5,"

n

¦

E sin t

1 ® A1cxd sin t 3¯ n

½ 1 xdcc ( Anc 2 Anc ) >sin(n 2)t sin nt @¾ 1,3,5," 2 ¿

f

1 xdcc ( An An 2 ) > cos nt cos(n 2)t @ 1,3,5," 2

¦

f

n

¦

1 ® A1 xq cos t 3¯ n

½ 1 xqcc( Anc Anc 2 ) >sin(n 2)t sin nt @¾ E sin t 1,3,5," 2 ¿

f

1 xqcc( An An 2 ) > cos(n 2)t cos nt @ 1,3,5," 2

¦

f

n

¦

(3.2.7) 184


Neglecting stator resistance, from formula (3.2.3) we can obtain

p\E

0

Taking a derivative of equation (3.2.7), we get

p\ E

1 ® A1cxd cos t 3¯ n

f

1 xdcc ( An An 2 )[(n 2)sin(n 2)t n sin nt ] 1,3,5," 2

¦

½ 1 xdcc ( Anc 2 Anc )[(n 2) cos(n 2)t n cos nt ]¾ 1,3,5," 2 ¿ f

n

¦

1 ® A1 xq sin t 3¯ n

½ 1 xqcc( Anc Anc 2 )[(n 2) cos(n 2)t n cos nt ]¾ E cos t 1,3,5," 2 ¿

f

1 xqcc( An An 2 )[(n 2)sin(n 2)t n sin nt ] 1,3,5," 2

¦

f

n

¦

0

(3.2.8)

or 1 ª 1 1 º 1 ª 1 A1 xq xdcc ( A1 A3 ) xqcc( A1 A3 ) » sin t A1cxd xdcc ( A3c « « 2 2 2 3¬ 3¬ ¼ f 1 1 º A1c) xqcc( A1c A3c ) 3E » cos t ¦ > xdcc ( An An 2 ) 2 3 n 1,3,5," ¼

xqcc( An An 2 ) xdcc ( An 2 An 4 ) xqcc( An 2 An 4 ) º¼

n2 sin(n 2)t 2

f

1 3n

¦

1,3,5,"

ª¬ xdcc ( Anc 2 Anc ) xqcc( Anc Anc 2 ) xdcc ( Anc 4 Anc 2 )

xqcc( Anc 2 Anc 4 )

n2 cos(n 2)t 2

0

(3.2.9)

From equation (3.2.9) we can see that in order to keep this formula always being zero, the sine amplitudes at every frequency must be zero and the cosine amplitudes must also be zero. That is, 1 1 xdcc ( A1 A3 ) xqcc( A1 A3 ) 0 2 2 xdcc ( An An 2 ) xqcc( An An 2 ) xdcc ( An 2 An 4 ) xqcc( An 2 An 4 )

A1 xq

0

1 1 xdcc ( A3c A1c) xqcc( A1c A3c ) 3E 0 2 2 xdcc ( Anc 2 Anc ) xqcc( Anc Anc 2 ) xdcc ( Anc 4 Anc 2 ) xqcc( Anc 2 Anc 4 ) 0

A1cxq

in which n 1,3,5," 185

AC Machine Systems

or ( xq xscc) A1 xDcc A3

0

½ ° xDcc An 2 xsccAn 2 xDcc An 4 0 ° ¾ ( xd xscc) A1c xDcc A3c 3E ° xDcc Anc 2 xsccAnc 2 xDcc Anc 4 0 °¿

(3.2.10)

where 1 ( xdcc xqcc) 2 1 ( xdcc xqcc) 2

xscc xDcc

The first two equations in formula (3.2.10) constitute the following algebraic equations: ( xq xscc) A1 xDcc A3

0½ ° xDcc A1 2 xsccA3 xDcc A5 0 ° ¾ xDcc A3 2 xsccA5 xDcc A7 0 ° ° " ¿

(3.2.11)

They have a group of effective solutions, A1

A3

A5

" 0

The latter two equations in formula (3.2.10) constitute the following algebraic equations: ( xd xscc) A1c xDcc A3c 3E ½ ° xDcc A1c 2 xsccA3c xDcc A5c 0 ° ¾ xDcc A3c 2 xsccA5c xDcc A7c 0 ° xDcc A5c 2 xsccA7c xDcc A9c 0 °¿

(3.2.12)

If

A3c

k1 A1c, A5c

k3 A3c , A7c

k5 A5c , " , Anc 2

kn Anc ,

(3.2.13)

the third equation in formula (3.2.12) may be rewritten as the following form: k1 xDcc A1c 2k3 xsccA3c k5 xDcc A5c

0

Comparing this equation with the second equation in formula (3.2.12), we can see that k1 , k3 and k5 must all be equal. In a similar way, according to the 186


remainder equations in formula (3.2.12), it is not difficult to prove k5 k7 " kn ; that is to say, the ratios between any two neighbouring Anc and Anc 2 are all the same. Letting the ratio be –b, we obtain k1

k3

k5

" kn

b

(3.2.14)

Substituting those values into the second equation in formula (3.2.12), we get ( xDcc 2 xsccb xDcc b 2 ) A1c 0

so xDcc 2 xsccb xDcc b 2

0

Its solutions are xqcc xdcc B 2 xdcc xqcc

2 xscc r 4 xscc2 4 xDcc2 2 xDcc

b1,2

xqcc xdcc

( xqcc B xdcc )2 ( xqcc xdcc )( xqcc xdcc ) Obviously, the solution ( xqcc xdcc ) 2

xqcc xdcc

( xqcc xdcc )( xqcc xdcc )

xqcc xdcc

is not correct, because the harmonic current amplitudes can not increase with their orders. Therefore, the suitable solution is b Substituting A3c

( xqcc xdcc ) 2

xqcc xdcc

( xqcc xdcc )( xqcc xdcc )

xqcc xdcc

(3.2.15)

bA1c into the first equation in formula (3.2.12), we obtain A1c

3E xd xscc xDcc b

3E xd xdcc xqcc

(3.2.16)

After the An and Anc values are obtained as above, according to formula (3.2.5) the steady-state currents of two phases short circuit can be written as follows: ib

ic

3E xd xdcc xqcc

[sin t b sin 3t b 2 sin 5t "]

(3.2.17) 187

AC Machine Systems

On the other hand, if the two phases short circuit in steady-state is analysed by Symmetrical Components Method, its currents after neglecting resistances are Ib

3E x1 x2

Ic

3E . Comparing x1 x2 xd , we get

That is to say, the amplitude of the basic-frequency current is this expression with formula (3.2.17) and taking x1 x2

xdcc xqcc

(3.2.18)

It should be pointed out that during different asymmetrical conditions the negative sequence reactance x2 has various values and the value above is suitable for two phases short circuit. After the currents ib and ic are expressed, the expressions of flux-linkage \ D and voltages ua , ub and uc may be obtained. On the basis of formula (3.2.4), the relation of steady-state flux-linkage \ D is

\D

[cos txd ( p)sin t sin txq ( p) cos t ]

2 3

ib E cos t

Substituting current ib into the equation above and arranging it in order referring to formula (3.2.7), we obtain 2 xdcc xqcc

\D Because ia

xd xdcc xqcc

0 and i0

E (cos t b cos3t b 2 cos5t ")

0, we have

\ 0 x0i0 0 \ a \D \ 0 \D ua p\ a ria p\ a Since u0 ub 188

ua ub uc uc

1 ua 2

(3.2.19)

2 xdcc xqcc xd xdcc xqcc 0 and ub

p\ D

E (sin t 3b sin 3t 5b2 sin 5t ")

(3.2.20)

uc , we get

xdcc xqcc xd xdcc xqcc

E (sin t 3b sin 3t 5b2 sin 5t ")

(3.2.21)


Evaluating the steady-state currents before, we have taken J 0 In fact, J 0 may be an arbitrary value and J the formulas above. Thus, we have ib

ic

ub

xd xdcc xqcc 2 xdcc xqcc

\D

ua

3E

xd xdcc xqcc

uc

t J 0 must be substituted for t in

(sin J b sin 3J b 2 sin 5J ")

(3.2.22)

E (cos J b cos3J b2 cos5J ")

(3.2.23)

E (sin J 3b sin 3J 5b 2 sin 5J ")

(3.2.24)

xd xdcc xqcc 2 xdcc xqcc

0 for convenience.

xdcc xqcc xd xdcc xqcc


(3.2.25)

(2) Initial values of currents and voltages Evaluating initial values, we may consider the flux-linkages of all the closed loops to be kept constant, i. e. the resistances of all the closed loops to be zero. On this condition, we have xd ( p) xdcc , xq ( p) xqcc and

\E

1 3

(\ b \ c )

1 3

(\ b 0 \ c 0 ) \ E 0

(3.2.26)

where \ E 0 , \ b 0 , \ c 0 are the corresponding flux-linkages before short circuit. Because the synchronous machine has no load before short circuit, we may obtain \ q 0 0, \ d 0 uq 0 E and

\ E 0 \ d 0 sin J 0

E sin J 0

Substituting the results above into formula (3.2.4), in the initial period after short circuit we have

\E

\E0

[ xdcc sin 2 J xqcc cos 2 J ]iE sin J G ( p)u fd

(3.2.27)

If the excitation voltage u fd before and after short circuit is constant, we may take p

0 in equation (2.4.12) and make G ( p )u fd

E; thus, the expression

above can be rewritten as follows: 189

AC Machine Systems

\E

\E0

[ xdcc sin 2 J xqcc cos 2 J ]iE E sin J

1 [ xdcc xqcc ( xdcc xqcc) cos 2J ]iE E sin J 2

(3.2.28)

in which E is no-load voltage induced by air-gap flux produced from the original excitation current and E sin J is the flux-linkage of fictitious winding E constituted by this air-gap flux, and 1 [ xdcc xqcc ( xdcc xqcc) cos 2J ]iE 2 is the flux-linkage of winding E produced by the transient current iE . Their sum is \ E 0 . The result is obvious, because the sudden short circuit may be considered as superposition of the two conditions, i. e. the steady-state condition before short circuit and the transient condition when rotor winding is short-circuited and stator winding is applied on by a voltage equal but opposite to the original voltage before short circuit. We can conclude that the equal but opposite voltage is the cause producing the stator current iE after short circuit, because the machine has no load before short circuit and the stator current is equal to zero. Thus, in flux-linkage equation (3.2.28), the flux-linkage corresponding to current iE should be the flux-linkage caused by the equal but opposite voltage, and the flux-linkage corresponding to E is the flux-linkage caused by the excitation voltage. Solving equation (3.2.28), we can obtain iE

ib

2( E sin J \ E 0 ) ( xdcc xqcc) ( xdcc xqcc) cos 2J ic

xdcc xdcc xqcc

3E (sin J sin J 0 ) ( xdcc xqcc) ( xdcc xqcc) cos 2J

3 iE 2 3E

(3.2.29)


3\ E 0 § 1 · 2 3 ¨ b cos 2J b cos 4J b cos 6J "¸ 2 xdcc xqcc © ¹

(3.2.30)

in which b

xqcc xdcc xqcc xdcc

is the ratio between two neighbouring odd-harmonic amplitudes or two neighbouring even-harmonic amplitudes. 190


Therefore, we can see that after two phases sudden short circuit the stator current has a series of odd and even harmonics including the aperiodic component. The two kinds of harmonics can be known to be caused by the stator and rotor aperiodic currents. The stator odd-harmonic currents are produced by rotor aperiodic current and the stator even-harmonic currents are produced by stator aperiodic current referring to Fig. 3.2.2. When two phases are suddenly short-circuited at J 0 0, i.e. the resultant fluxlinkage of phases b and c is equal to zero (\ E 0 0), the stator even harmonics will disappear. It is not difficult to understand that condition, because at the moment of short circuit the flux-linkage of the closed loop constituted by phases b and c is just equal to zero and it is not necessary to have an aperiodic current in stator to cancel the flux-linkage increased abruptly. When two phases are suddenly short-circuited at J 0 90e, i.e. the resultant flux-linkage \ E 0 of phases b and c is maximum, the stator even harmonics will have their maximum amplitudes. That is because there is a maximum aperiodic current in the stator to cancel the flux-linkage increased abruptly which is caused by the ac current. After finding the current iE (or ib and ic ), we can obtain the remainder quantities. Because i0 0

\0

so

\ a \D \ 0 \D

x0 i0

0

( xdcc xqcc)sin J cos J iE E cos J

1 E cos J ( xdcc xqcc)sin 2J iE 2 ( E sin J \ E 0 )( xdcc xqcc)sin 2J E cos J ( xdcc xqcc) ( xdcc xqcc) cos 2J

ua

p\ a

E sin J ( xdcc xqcc)

b 2 sin 5J ")

2 xdcc xqcc xdcc xdcc xqcc

(3.2.32)

d ª E sin 2J « (sin J b sin 3J dt « xdcc xdcc xqcc ¬

·º 2 3 ¨ b cos 2J b cos 4J b cos 6J "¸ » xdcc xqcc © 2 ¹»¼

\ E 0 sin 2J § 1

E (sin J 3b sin 3J 5b2 sin 5J ")

4b\ E 0 (cos 2J 2b cos 4J 3b 2 cos 6J ")

Because uE

(3.2.31)

0 and u0

p\ 0 ri0

(3.2.33)

0 191

AC Machine Systems

so ub

uc

1 ua 2

xdcc xqcc xdcc xdcc xqcc


2b\E 0 (cos 2J 2b cos 4J 3b 2 cos 6J ")

(3.2.34)

From the formula above, we can see that during xdcc xqcc (i.e. b 0) the phase a voltage after two phases (b and c) are suddenly short-circuited is the same as before, i.e. ua E sin J . It can be considered that the change of the voltage ua after phases b and c are suddenly short-circuited is caused by the asymmetry of rotor. In fact, if the machine has a symmetrical rotor, the current iE (or ib and ic ) has only the aperiodic component and fundamental current and the rotor current has only the fundamental, aperiodic component and the second harmonic. The magnetic field produced by the aperiodic component of current iE is static relatively to stator and no voltage can be induced in phase a. The pulsating field produced by the fundamental of current iE is perpendicular to the axis of phase a and there is no mutual flux-linkage between them, thus not any voltage can be induced in phase a. The revolving field at synchronous speed, which is produced by the rotor extra-aperiodic current due to short circuit, is just equal but opposite to that revolving field of a synchronous speed, which is produced by the rotor second harmonic current; therefore, in phase a the voltages induced by the two fields cancel each other. The rotor fundamental harmonic current produces a revolving field of a synchronous speed relative to rotor, but the field is static relatively to stator so that no voltage can be induced in phase a. Therefore, the voltage of phase a depends still upon the rotor excitation current before short circuit. From the formula of ua , we can see that ua has its maximum amplitude only when \E 0 reaches its maximum value (\E 0 E ) , i.e. the two phases are suddenly short-circuited at J 0 90e. That is because the aperiodic currents of phases b and c have their maximum value on that condition. Under that condition, we get ua



4bE (cos 2J 2b cos 4J 3b 2 cos 6J ") From the formula above we know that during J 270e , i.e. t maximum instantaneous value for the first time, whose value is 192

(3.2.35) S , ua has its



E (1 3b 5b 2 ") 4bE (1 2b 3b 2 ")

2 xdcc xqcc

ª 1 2 b º 1 4bE E« 2 » 2 b b) 2 (1 ) (1 xdcc xdcc xqcc ¬ (1 b ) ¼ ª 1 2 b º 1 4bE (1 b) E « 2 » 2 (1 b) ¼ (1 b)2 ¬ (1 b ) E E

(1 b)(1 b ) 2 2 b (1 b )2 (1 b )2 (1 b ) 2 1 6b b 2 (1 b)2

§ xqcc · E ¨ 2 1¸ © xdcc ¹

(3.2.36)

Deriving the formula above, we have used the following equation: (1 x) 2

1 2 x 3x 2 4 x3 "

If two phases are suddenly short-circuited at J 0

0, then \ E 0

0, the maximum

xqcc S· § 90e¨ i.e. t ¸ and equals cc E. So we xd ¹ © can see that the damping winding may decrease the inner-overvoltage or operating overvoltage, because the damper makes xqcc be equal to xdcc approximately. value of voltage ua will appear at J

If the neutral point of a machine is free from earth and the fault-point is grounded referring to Fig. 3.2.3, the voltage of phase a to earth is ua 0

ua ub

3 ua 2

therefore, the maximum amplitude of voltage ua 0 during J 0 § xqcc 1 · 3 § xqcc · E ¨ 2 1¸ 3 E ¨ ¸ 2 © xdcc ¹ © xdcc 2 ¹

(3.2.37) 90ewill be (3.2.38)

Figure 3.2.3 Two phases sudden short circuit when the fault-point is earthed 193

AC Machine Systems

The initial values of the stator current and voltage have been obtained as above. In order to obtain the initial value of rotor current, we must get currents id and iq first according to the following formulas: id iq

iE sin J iE cos J

2

½ ib sin J ° ° ¾ ib cos J ° °¿ 3

3 2

(3.2.39)

then use the equivalent circuit of operational reactance ( p f, u fd 0) and can find the rotor current. The calculation method is the same as in 3-phase sudden short-circuit; such as a machine without damping winding, we have id

2 E sin 2 J 2\ E 0 sin J ( xdc xq ) ( xdc xq ) cos 2J 2 E sin J 2\ E 0 xdc xdc xq E xdc xdc xq

[1 (1 b) cos 2J b(1 b) cos 4J b 2 (1 b) cos 6J "]

2\ E 0 xdc xdc xq E

xdc xdc xq

\E0 xdc xq


[sin J b sin 3J b 2 sin 5J "]

[1 (1 b) cos 2J b(1 b) cos 4J "]

(1 b)[sin J b sin 3J b 2 sin 5J "]

(3.2.40)

in which b

xq xdc xq xdc

Therefore, the current of excitation winding is x I fd I fd 0 ad id X ffd I fd 0 194

xad E [1 (1 b) cos 2J b(1 b) cos 4J "] X ffd [ xdc xdc xq ]

xad\ E 0 X ffd xdc xq

(1 b)[sin J b sin 3J b 2 sin 5J "]

(3.2.41)


(3) Time constants The even harmonics of the stator current and voltage and the odd harmonics of the rotor current are all caused by the stator aperiodic current. Thus, those harmonics will decay to zero with the stator aperiodic current, i.e. the steadystate values of those harmonics are all zero. The odd harmonics of the stator current and voltage and the even harmonics of the rotor current are all caused by the rotor aperiodic current including the damper aperiodic current. So, these harmonics will decay with the rotor aperiodic current. However, the rotor aperiodic current cannot finally decay to zero except the damper aperiodic current but decays to the steady-state value determined by the excitation voltage. Therefore, the corresponding harmonics finally decay to some steady-state values, too. The change process of the current and voltage from their initial values to steady-state values is complicated. In practice, we can use the same approximate method as in 3-phase sudden short circuit and various time constants are discussed as follows. (a) The decay time constant Ta 2 of the stator even harmonic currents and rotor odd harmonic currents This time constant depends upon the equivalent reactance and resistance corresponding to the aperiodic current of phases b and c. Because phases b and c in 3-phase machine are equivalent to phase Ein two phases D and E machine, thus the time constant is also equal to the ratio of the equivalent reactance to resistance corresponding to the aperiodic current of phase E. The equivalent reactance can be found as follows. From equation (3.2.30), we can see that if the synchronous machine rotates at a synchronous speed, its excitation winding is closed and the aperiodic current of phase E is I 0 / 2, then the resultant current caused by the aperiodic current is iE

§1 · I 0 ¨ b cos 2J b 2 cos 4J b3 cos 6J " ¸ ©2 ¹ f 1 I 0 I 0 ¦ (b)n cos 2nJ 2 n 1,2,3,"

J

in which

b

(3.2.42)

t J0 xqcc xdcc xdcc xqcc

Referring to formula (3.2.4) and taking u fd

0 in it, we can describe the flux-

linkage of phase E as 195

AC Machine Systems

\E

f ª1 º [sin J xd ( p)sin J cos J xq ( p) cos J ] « I 0 I 0 ¦ (b) n cos 2nJ » 2 n 1,2,3," ¬ ¼ 1 1 1 ® I 0 sin J xd ( p)sin J I 0 cos J xq ( p ) cos J I 0 sin J xd ( p ) 2 2 ¯2 f

u

¦

(b)n [sin(2n 1)J sin(2n 1)J ]

n 1,2,3,"

1 I 0 cos J xq ( p ) 2 n

½ (b) n [cos(2n 1)J cos(2n 1)J ]¾ 1,2,3," ¿ f

¦

Neglecting rotor resistance and referring to equation (3.2.7), we have xd [r j(2n r 1)] xq [r j(2n r 1)]

xdcc xqcc

where n 1, 2,3,"

\E

1 1 1 ® I 0 xdcc sin 2 J I 0 xqcc cos 2 J I 0 xdcc sin J 2 2 2 n ¯ sin(2n 1)J ]

1 I 0 xqcc cos J 2 n

1 xqcc 2 n

(b) n [sin(2n 1)J

1,2,3,"

½ (b) n [cos(2n 1)J cos(2n 1)J ]¾ 1,2,3," ¿ f

¦

1 1 1 1 I 0 ® ( xdcc xqcc) ( xdcc xqcc ) cos 2J xdcc 2 ¯2 2 2 n cos(2n 2)J cos(2n 2)J cos 2nJ ]

f

¦

f

¦

(b)n [cos 2nJ

1,2,3,"

½ (b) n [cos 2nJ cos(2n 2)J cos(2n 2)J cos 2nJ ]¾ 1,2,3," ¿ (3.2.43) f

¦

Thus, the average flux-linkage, i.e. the aperiodic flux-linkage produced by the 1 aperiodic current I 0 in phase E can be written as 2

\ E cp

196

1 ª1 1 1 º I 0 « ( xdcc xqcc) bxdcc bxqcc » 2 ¬2 2 2 ¼ xqcc xdcc º 1 ª » I 0 « xdcc xqcc ( xdcc xqcc) 4 « xqcc xdcc »¼ ¬ 1 I 0 xdcc xqcc 2

(3.2.44)


We can see that so far as the aperiodic current is concerned, the equivalent reactance of phase E is xdcc xqcc x2 . Because the resistance of phase E is r, the time constant can be written as xdcc xqcc

Ta 2

r

x2 r

(3.2.45)

(b) The decay time constants Tdcc2 , Tdc2 of the stator odd harmonic currents and rotor even harmonic currents These time constants Tdcc2 and Tdc2 depend upon the equivalent reactance and resistance corresponding to the aperiodic currents of rotor. If the machine has no damping winding, then the average flux-linkage of excitation winding during steady-state two phases short circuit can be written as

\fdcp

Xffd I fdcp xad idcp X ffd I fd 0

X ffd I fd 0

xad E xd xdc xq

2 xad I fd 0

xd xdc xq

§ xdc xdc xq ¨ ¨ xd xdc xq ©

· ¸ I fd 0 X ffd ¸ ¹

Thus, the equivalent reactance of excitation winding is xdc xdc xq xd xdc xq

X ffd

The corresponding time constant is Tdc2

X ffd xdc xdc xq R fd xd xdc xq

Td 0

xdc x2 xd x2

(3.2.46)

The result is similar to that of 3-phase sudden short circuit, only the numerator and denominator are added to the x2 respectively. That is well known from the conception of Symmetrical Components. When the rotor has a damping winding besides excitation winding, the two phases short circuit can also be analysed in the same way as 3-phase sudden short circuit, i.e. considering the stator odd harmonic currents in the initial stage to decay approximately according to Tdcc2 then to Tdc2 . The first stage depends mainly on the damping winding and the latter stage on the excitation winding. On the basis of the same conception, the time constants can be written as 197

AC Machine Systems

Tdcc2

xdcc xdcc xqcc

Tdcc0

Tdc2

xdc xdcc xqcc xdc xdcc xqcc

Td 0

xd xdcc xqcc

xdcc x2 ½ ° xdc x2 °° ¾ xdc x2 ° Td 0 ° xd x2 ° ¿ Tdcc0

(3.2.47)

(4) General expressions of currents and votages Having the initial values, steady-state values and time constants mentioned above, we can write the general formulas of currents and voltages in the same way as 3-phase sudden short circuit as follows: ic

ib

ua

3F (sin J b sin 3J b 2 sin 5J ") 3\ E 0

xdcc xqcc

e

§1 · 2 ¨ b cos 2J b cos 4J " ¸ 2 © ¹

(3.2.48)

2 xdcc xqcc F (sin J 3b sin 3J 5b 2 sin 5J ") 4b\ E 0 e

ub

t Ta 2

uc

t Ta 2

(cos 2J 2b cos 4J 3b 2 cos 6J ")

(3.2.49)

xdcc xqcc F (sin J 3b sin 3J 5b 2 sin 5J ") 2b\ E 0 e

t Ta 2

(cos 2J 2b cos 4J 3b 2 cos 6J ")

(3.2.50)

in which t

t

§ 1 1 · Tdcc2 § 1 1 · Tdc 2 E ¨ ¨ ¸ Ee ¸ Ee cc c c xd x2 © xd x2 xd x2 ¹ © xd x2 xd x2 ¹

F

(3.2.51)

During two phases sudden short circuit, the excitation current of a machine without damping winding is I fd

I fd 0

t ª§ xad x 1 E Tdc 2 e ad (1 b) E «¨ X ffd xdc xdc xq X ffd «¨ xdc xdc xq ¬©

º » (cos 2J b cos 4J b 2 cos 6J ") » ¼

· tc 1 ¸ e Td 2 ¸ c xd xd xq ¹ xd xdc xq

xad (1 b) \ E 0 e Ta 2 (sin J b sin 3J b 2 sin 5J ") X ffd xdc xq

1

t

198

(3.2.52)


(5) Electromagnetic torque after two phases sudden short circuit When two phases of a no-load synchronous machine are suddenly shortcircuited, referring to formula (3.2.33) the initial flux-linkage of phase D can be written as follows: 1 E cos J ( xdcc xqcc)sin 2J iE 2 ( E sin J \ E 0 )( xdcc xqcc)sin 2J E cos J ( xdcc xqcc) ( xdcc xqcc) cos 2J

\D

2 xqccE cos J \ E 0 ( xdcc xqcc)sin 2J ( xdcc xqcc) ( xdcc xqcc) cos 2J 2 xdcc xqcc xdcc xdcc xqcc

E (cos J b cos 3J b 2 cos 5J ")

2b\ E 0 (sin 2J b sin 4J b 2 sin 6J ")

(3.2.53)

Its steady-state value referring to equation (3.2.23) is 2 xdcc xqcc

\D

xd xdcc xqcc

E (cos J b cos 3J b 2 cos 5J ")

According to the initial value, steady-state value and time constants, the general expression of flux-linkage \ D may be described as 2 F xdcc xqcc (cos J b cos 3J b 2 cos 5J ")

\D

2b\ E 0 e

t Ta 2

(sin 2J b sin 4J b 2 sin 6J ")

2 Fxqcc( xdcc xdcc xqcc ) cos J ( xdcc xqcc) ( xdcc xqcc) cos 2J

( xqcc xdcc )\ E 0 sin 2J

( xdcc xqcc) ( xdcc xqcc ) cos 2J

e

t Ta 2

(3.2.54)

in which F is evaluated with formula (3.2.51). On the other hand, from equation (3.2.48) we can obtain iE

2 3

ib

2 F (sin J b sin 3J b 2 sin 5J ")

2\ E 0 xdcc xqcc

e

t Ta 2

§1 · 2 ¨ b cos 2J b cos 4J "¸ ©2 ¹

2( xdcc xdcc xqcc ) F sin J xdcc xqcc ( xdcc xqcc) cos 2J

2\ E 0 xdcc xqcc ( xdcc xqcc) cos 2J

e

t Ta 2

(3.2.55) 199

AC Machine Systems

Substituting the results above into the torque equation, we can get Te

\ D iE \ E iD

\ D iE

1 [2 F 2 xqcc( xdcc xdcc xqcc ) 2 sin 2J [( xdcc xqcc) ( xdcc xqcc) cos 2J ]2 4 Fxqcc( xdcc xdcc xqcc )\ E 0 e

t Ta 2

cos J

2( xdcc xqcc)( xdcc xdcc xqcc ) F\ E 0 e 2( xdcc xqcc)\ E2 0 e

2t Ta 2

t Ta 2

sin 2J sin J

(3.2.56)

sin 2J ]

Referring to Appendix 2 of Reference [3] exist 1 ( xdcc xqcc) ( xdcc xqcc) cos 2J sin J cc cc ( xd xq ) ( xdcc xqcc) cos 2J d dJ

1 §1 · 2 ¨ b cos 2J b cos 4J " ¸ xdcc xqcc © 2 ¹ 1 xdcc xdcc xqcc


ª º 2( xdcc xqcc )sin 2J 1 « » [( xdcc xqcc) ( xdcc xqcc) cos 2J ]2 ¬« ( xdcc xqcc) ( xdcc xqcc) cos 2J »¼ 2b (sin 2J 2b sin 4J 3b 2 sin 6J ") xdcc xqcc

ª º 2 xqcc cos J ( xdcc xqcc)sin 2J sin J sin J « » 2 «¬ ( xdcc xqcc) ( xdcc xqcc) cos 2J »¼ [( xdcc xqcc) ( xdcc xqcc) cos 2J ] 1 (cos J 3b cos 3J 5b 2 cos 5J ") xdcc xdcc xqcc

d dJ

therefore there is

Te

2 F 2 xdcc xqcc (sin 2J 2b sin 4J 3b 2 sin 6J ") 2 F\ E 0 e

2b\ E2 0 xdcc xqcc

t Ta 2

e

2t Ta 2

(cos J 3b cos 3J 5b 2 cos 5J ") (sin 2J 2b sin 4J 3b 2 sin 6J ")

(3.2.57)

It should be noted that the formula above is obtained after neglecting the stator and rotor resistances but considering the effect of resistances on time constants. From the torque expression we can see that the electromagnetic torque after two 200


phases sudden short circuit is only the pulsating torque under this condition. The torque contains a series of odd and even harmonics. These harmonic torques are caused by a series of harmonic currents under the short circuit condition. Because both the stator and rotor are asymmetrical, the stator aperiodic current produces not only stator even harmonic currents but also rotor odd harmonic currents. Similarly the rotor aperoidic current causes stator odd harmonic currents and rotor even harmonic currents. The magnetic fields produced by stator odd harmonic currents act with the magnetic fields produced by rotor even harmonic currents to produce the even harmonic torques, and the field produced by the stator fundamental current acting with the field produced by rotor aperiodic current does not cause an average torque because we have supposed the resistances to be zero. Those even harmonic torques depend obviously upon the rotor aperiodic current and decay finally to the steady-state value, which are the first group of harmonics in the torque formula (3.2.57). The fields produced by stator odd harmonic currents act with the fields produced by rotor odd harmonic currents to cause the odd harmonic torques, and the fields produced by stator even harmonic currents act with the fields produced by rotor even harmonic currents to cause the odd harmonic torques, too. Sum of the two kinds of torques is the second group of harmonics in the torque formula (3.2.57). Obviously, the group of harmonic torques is determined by the stator and rotor aperiodic currents and decays finally to zero. The fields produced by stator even harmonic currents act with the fields produced by rotor odd harmonic currents to cause the even harmonic torques that are the third group of harmonics in torque formula (3.2.57). This group of harmonic torques depends obviously upon stator aperiodic current and decays finally to zero. In addition, during xdcc xqcc, i.e. b 0, the third group of harmonic torques will disappear, because under this condition the rotor is symmetrical, there is only the aperiodic current in stator even harmonic currents and only the basic-frequency current exists in rotor odd harmonic currents. The rotor basic-frequency current only produces the static field relative to stator because the rotor is symmetrical, so the static field does not produce the harmonic torques by acting with the field produced by stator aperiodic current. Of course, there is no average torque because we have supposed resistances to be zero. When the effect of stator and rotor resistances is considered, in the electromagnetic torques after two phases sudden short circuit there are not only the pulsating torques but also the average torque. Generally speaking, although the average torque is not large, sometimes it must be considered when we study some problems, such as the power system stability. However, the pulsating torques calculated by neglecting resistances are accurate enough. The average torque can be analysed as below. As mentioned above, the operation condition of two phases sudden short circuit is approximately considered as the condition of the aperiodic currents circulating simultaneously through the stator and rotor windings. When evaluating the average electromagnetic torque, we can study the two groups of currents separately, because the fields produced by them have different speeds. 201

AC Machine Systems

(i) The rotor aperiodic current causes the odd harmonic currents in stator and the even harmonic currents in rotor. It may be proved that the average electromagnetic torque caused by this group of currents is equal to the effective resistance losses produced by these harmonic currents; but the resistance loss of the rotor aperiodic current is provided with the excitation supply and the rotor storage magnetic-energy, so it does not produce the electromagnetic torque. Namely, their average electromagnetic torque is 2 3m

c Tecp

f

¦

| im |2 r

1,3,5,"

1 2s

f

¦

[| ids |2 sRds | iqs |2 sRqs ]

(3.2.58)

2,4,6,"

in which | im | is the stator m-th harmonic current amplitude, | ids |,| iqs | are the rotor s-th harmonic current amplitudes of the d and q axes,

s m 1, Rds , Rqs are the corresponding values of the following expressions, xd ( js )

X ds jRds

xq ( js )

X qs jRqs

The first term of the formula (3.2.58) represents the stator effective resistance losses produced by all the stator odd harmonic currents and the second term expresses the rotor effective resistance losses produced by all the rotor even harmonic currents. (ii) The stator aperiodic current causes stator even harmonic currents and rotor odd harmonic currents. It can be similarly proved that the average electromagnetic torque caused by this group of currents equals the effective resistance losses produced by these harmonic currents, but the resistance loss of the stator aperiodic current is provided with the stator storage magnetic-energy, so it does not produce the electromagnetic torque, i.e. their average electromagnetic torque is cc Tecp

2 3m

f

¦

| im |2 r

2,4,6,"

1 2s

f

¦


(3.2.59)

1,3,5,"

Therefore, the resultant average torque after two phases short circuit is Tecp

c Tecp cc Tecp 2 3m

f

¦

1,2,3,"

| im |2 r

1 2s

f

¦


1,2,3,"

[the stator effective resistance losses produced by all the stator harmonic currents] [the rotor effective resistance losses produced by all the rotor harmonic currents] 202

(3.2.60)


3.3

Steady-state Asynchronous Operation of Synchronous Machines

In practice, the asynchronous operation of a synchronous machine often arises. For example, starting of a synchronous motor, pulling a synchronous generator into step, falling out of step and periodical oscillation of a synchronous machine are all connected with the asynchronous operation. During asymmetrical operation of a synchronous machine, its stator current has not only the fundamental component but also a series of harmonics. Although the machine rotor rotates at a synchronous speed at this time, there is asynchronous operation so far as a series of harmonic currents are concerned. When a synchronous machine operates in steady-state at an asynchronous speed, its stator or rotor current is the sum of different harmonic currents of various frequencies if the terminal voltages of its stator have symmetrical and sinusoidal waves. If the supply frequency is f, the synchronous and actual speeds are Z s and Z respectively which are all constant, and the machine slip is Zs Z , then the supply voltage will produce 3-phase symmetrical currents s

Zs of frequency f in the stator. However, the synchronous revolving-field produced by the symmetrical currents has a relative speed sZ s with respect to the rotor, so a current of frequency sf can be induced in the rotor. Owing to asymmetrical rotor, that current will produce the revolving fields at speeds (Z r sZ s ), which induce the symmetrical currents of frequencies f and (1 2 s ) f in stator respectively. In addition, the rotor excitation current also induces the symmetrical currents of frequency (1 s ) f in stator. Therefore, there are harmonic currents of frequencies f, (1 s ) f and (1 2 s ) f in stator and direct current and harmonic current of frequency sf in rotor referring to [2,3]. On the basis of the deduction, the stator current may be written as follows, ia

A1 cos t +A2 cos(1 2 s )t +A3 cos(1 s )t A1c sin t +A2c sin(1 2s )t +A3c sin(1 s )t

ib

A1 cos(t 120e )+A2 cos[(1 2 s )t 120e ] A3 cos[(1 s )t 120e ] A1csin(t 120e A2csin[(1 2s )t 120e ]+A3c sin[(1 s )t 120e ]

ic

A1 cos(t 120e ] )+A2 cos[(1 2 s )t 120e c A3 cos[(1 s )t 120e ] A1sin(t 120e A2csin[(1 2s )t 120e ]+A3c sin[(1 s )t 120e ]

½ ° ° ° ° ° ¾ ° ° ° ° ° ¿

(3.3.1)

203

AC Machine Systems

If taking I1 I2 I3

1 ( A1 jA1c), 2 1 ( A2 jA2c ), 2 1 ( A3 jA3c ), 2

I1* I*2 I*3

1 ½ ( A1 jA1c) ° 2 ° 1 ° c ( A2 jA2 ) ¾ 2 ° 1 ° ( A3 jA3c ) ° 2 ¿

(3.3.2)

We obtain ia

I1e jt I2 e j(1 2s)t I3e j(1s)t I1*e jt I* e j(1 2s)t I* e j(1s)t 2

ib

3

j (1 2s)t 120e j (1 s)t 120e @ @ I1e j(t 120e) I2 e > I3 e > j (1 2s)t 120e j (1 s)t 120e @ @ I1*e j(t 120e) I*2 e > I*3e >

ic

j (1 2s)t 120e j (1 s)t 120e @ @ I3 e > I1e j(t 120e) I2 e > j (1 2s)t 120e j (1 s)t 120e @ @ I1*e j(t 120e) I*2 e > I*3e >

½ ° ° ° ° ¾ ° ° ° ° ¿

(3.3.3)

It is noted that a current changing sinusoidally with time can usually be represented by a rotating time-vector in ac Circuitry Theory, but here it is represented by two conjugate rotating vectors which have half the amplitude of that rotating time-vector. Transforming the currents ia , ib , ic above into 1, 2, 0 components according to formula (3.1.13), we can get i1 i2 i0

½ I1e jt I2 e j(1 2 s ) t I3e j(1 s ) t ° * j * j(1 2 ) * j(1 ) t s t s t I1e I2 e I3 e ¾ ° 0 ¿

(3.3.4)

If the 3-phase symmetrical terminal voltages are ua ub uc

B cos t Bc sin t

½ ° B cos(t 120e ) Bc sin(t 120e )¾ B cos(t 120e ) Bc sin(t 120e ) °¿

(3.3.5)

1 ½ ( B jBc) ° ° 2 ¾ 1 ( B jBc) ° °¿ 2

(3.3.6)

and we take U U * 204


then there are u1 Ue jt ½ ° u2 U *e jt ¾ ° u0 0 ¿

(3.3.7)

Provided that we choose J 0 at the moment t 0 , i.e. J 0 0, then during steady-state asynchronous operation, according to formulas (3.1.23) and (3.3.4) and taking Z 1 s together with J (1 s )t , we can obtain the following basic relations,

\1

\2

½ ° ° ° s D 1 2 ° [ xs ( js ) I2 xD ( js ) I1* ]e j(1 2 s )t ° ° \ 3e j(1 s )t \1e jt \ 2 e j(1 2 s )t ° ¾ 1 ª * x (0) I º e j(1 s )t ° E x I (0) s D 3 3» «2 ° ¬ ¼ ° t * j [ xs ( js ) I1 xD ( js ) I2 ]e ° ° * j(1 2 s ) t [ xs ( js ) I 2 xD ( js ) I1 ]e ° \ *3e j(1 s )t \ 1*e jt \ *2 e j(1 2 s )t °¿ ª1 * º j(1 s )t « 2 E xs (0) I 3 xD (0) I 3 » e ¬ ¼ [ x ( js ) I x ( js ) I* ]e jt

(3.3.8)

in which

\1 \

* 1

\ 2 \ 2* \ 3 \ 3*

xs ( js ) I1 xD ( js ) I2* x ( js ) I* x ( js ) I

½ ° ° s D 1 2 * ° xs ( js ) I 2 xD ( js ) I1 ° ° xs ( js ) I2* xD ( js ) I1 ¾ ° 1 E xs (0) I3 xD (0) I*3 ° 2 ° ° 1 * E xs (0) I 3 xD (0) I 3 ° ¿ 2

(3.3.9)

From equations (3.1.20), (3.3.4) and (3.3.8) we can get

205

AC Machine Systems

u1 Ue jt

u2

½ p\ 1 ri1 ° [ j(1 s )\ 3 rI3 ]e j(1 s )t ( j\1 rI1 )e jt ° ° [ j(1 2 s )\ 2 rI2 ]e j(1-2s )t ° ¾ * jt U e p\ 2 ri2 ° ° * * * j(1 s ) t ( j\1 [ j(1 s )\ 3 rI 3 ]e ° rI1* )e jt [ j(1 2 s )\ 2* rI2* ]e j(1 2 s ) t °¿

(3.3.10)

Comparing various terms having the same exponential function in the equations above and referring to formula (3.3.9), we obtain r j

jU

\1 I1

U * j

\1* jrI1*

r xs ( js ) I1 xD ( js ) I2* I1 j

½ ° ° ¾ xs ( js ) I1* xD ( js ) I2 jrI1* ° °¿ r I2 j(1 2s )

½ xs ( js ) I2 ° ° r ° * I2 ° xD ( js ) I1 s j(1 2 ) ° ¾ 2 I*2 xs ( js) I*2 ° 0 \ *2 j ° (1 2 s ) ° r * ° xD ( js ) I1 j I2 °¿ (1 2s ) 0 \ 2

(3.3.12)

r I3 j(1 s )

1 ½ E xs (0) I3 ° 2 ° r ° * I3 xD (0) I 3 ° j(1 s ) ° ¾ r * 1 0 \ *3 j I3 E xs (0) I*3 ° ° (1 s ) 2 ° r * ° xD (0) I3 j I3 °¿ (1 s )

0 \ 3

(3.3.11)

(3.3.13)

On the basis of those equations we can draw these equivalent circuits of steady-state asynchronous operation of a synchronous machine as shown in Figs. 3.3.1 and 3.3.2. In Fig. 3.3.1(a) there is an equivalent circuit of calculating or measuring the stator current component caused by the direct current in the excitation winding. Figure 3.3.1(b) shows an equivalent circuit of calculating or measuring the stator current components produced by the stator terminal voltages. Figure 3.3.2 indicates the resultant equivalent circuit. 206


Figure 3.3.1 Equivalent circuit (a) Equivalent circuit for the stator current component caused by excitation emf; (b) Equivalent circuit for the stator current components caused by stator terminal voltage

Figure 3.3.2 Resultant equivalent circuit for asynckronous operation of synchronous machines

207

AC Machine Systems

The output power and electromagnetic torque of the machine can be measured in the resultant equivalent circuit or estimated by the following formulas, 2(u2 i1 u1i2 u0 i0 ) 2[ I U * I U *e j2 st I U *e jst

P

1

2

3

I1*U I2*Ue j2 st I3*Ue jst ] Te

(3.3.14)

j2(i2\ 1 i1\ 2 ) j2[ I1\1* I1*\1 I2\ 2* I*2\ 2 I*3\ 3 I3\ 3* ] ( I*\ I \ * I*\ I \ * )e jst 2

3

3

2

3

1

1

3

( I1*\ 3 I3\1* I*3\ 2 I2\ 3* )e jst ( I1\ 2* I*2\1 )e j2 st ( I2\1* I1*\ 2 )e j2 st ]

(3.3.15)

The average electromagnetic torque of the machine is Tecp

j2( I1\1* I1*\1 I21\ 2* I*2\ 2 I*3\ 3 I3\ 3* )

(3.3.16)

Therefore, we can see that various components of the average electromagnetic torque are all constituted by the currents and flux-linkages whose frequencies are the same. That is natural, because the revolving fields caused only by them have the same speed. It should be pointed out that the equivalent circuit for the direct current of the excitation winding as shown in Fig. 3.3.1(a) is independent of the equivalent circuit for applying supply voltage to stator as shown in Fig. 3.3.1(b). That is clear according to the Superposition Theorem, because the machine has been supposed to be connected to an infinite network, so the action of the direct current of excitation winding is similar to the action of 3-phase short circuit. It is also noted that there are harmonic torques and power of frequencies 2sf and sf besides the average electromagnetic torque and power in the torque Te and output power P. That is not difficult to understand, because in this operation mode there are three kinds of revolving fields. The first is the revolving field of speed Z s corresponding to stator basic-frequency current, the second is the revolving field of speed (1–2s) Z s corresponding to the stator current of frequency (1– 2s) f and the third is the revolving field of speed (1 –s) Z s corresponding to the direct current of excitation winding or the stator current of frequency (1–s)f. The relative speeds between these fields are 2sZ s and sZ s . Therefore, the harmonic torques or power values of frequencies 2sf and sf are produced. The excitation winding current and damper current can be estimated on the basis of the equivalent circuit of operational reactance. However, first we must calculate currents id and iq by using the following formulas, 208


id iq

i1e jJ i2 e jJ ( I3 I3* ) ( I1 I2* )e jst ( I2 I1* )e jst

½ ° ° ¾ j(i1e jJ i2 e jJ ) ° * * jst * jst ° j( I 3 I 3 ) j( I1 I 2 )e j( I 2 I1 )e ¿

(3.3.17)

As for study of the rotor motion behavior, the average torque of synchronous machine under asynchronous operation condition is significant. Now we derive the average-torque formula by use of the equivalent circuit during asynchronous operation obtained above referring to Figs 3.3.1 and 3.3.2. The complex power may be described as * P UI

P jQ

6I i2 Ri j6I i2 X i

in which P 6I i2 Ri is the resultant active power through a certain point of a circuit whose voltage is U and whose current is I ; I i2 Ri is the consumed active power of the branch i; Q 6I i2 X i is the resultant reactive power through the same point of a circuit; I i2 X i is the absorbed reactive power of the branch i .However, it should be noted that X i is positive value for an inductive branch and negative value for a capacitive branch. On the basis of the concept above, by use of the equivalent circuit as shown in Figs 3.3.1(a) and 3.3.1(b) and equation (3.3.16) we can obtain the following expressions: (i) j2( I3*\ 3 I3\ *3 )

r 2 r 2º ª j2 « j I3 j I3 » 1 s ¬ 1 s ¼ 2 r 2 I3 1 s

(3.3.18)

(ii) j2[ I1 \ 1* I1*\ 1 I2\ 2* I2*\ 2 ] 2 2 2 4 I1 Rqs 2 I1 I2* ( Rds Rqs ) 4 I2 Rqs 8 I2 2 2 2 I1 I 2* Rds 2 I1 I2* Rqs 8 I2

2

2

r 1 2s

r 1 2s

(3.3.19)

in which Rds and Rqs are included in the following expressions: xd ( js )

X ds jRds

xq ( js )

X qs jRqs 209

AC Machine Systems

Impedances xd ( js ) and xq ( js ) may be measured or calculated by the equivalent circuit of operational reactance during p js. In the expressions above, both Rds and Rqs have a coefficient ( j), because the equivalent circuit of operational reactance is usually made up of the equivalent resistance and capacitance. From formula (3.3.17) we can see | id ( nep ) |2 Rds | iq ( nep ) |2 Rqs 4 | I1 I2* |2 Rds 4 | I1 I2* |2 Rqs

(3.3.20)

where | id ( nep ) | and | iq ( nep ) | are the amplitudes of ac components of currents id and iq . According to formulas (3.3.18), (3.3.19) and (3.3.20) we can get Tecp | 2 I3 |2

r 2r | 2 I2 |2 1 s 1 2s

1 [| id ( nep ) |2 Rds | iq ( nep ) |2 Rqs ] 2

(3.3.21)

Various terms in that average electromagnetic torque expression have the following physical meanings: r (i) The first term | 2 I3 |2 is the short circuit torque (or asynchronous 1 s torque), because at this time the excitation field has a relative speed (1 s ) Z s with respect to the stator, so the stator winding resistance r should be divided by (1 s). Obviously, this torque will be zero when there is no excitation source. (ii) The latter two terms are the salient-pole torque (or reluctance torque) and asynchronous torque caused by the applied voltage. When s 0.5, the current I2 will be zero, so the second term of the average torque expression will become zero. If the excitation voltage is supposed to be zero in the analysis above, that is steady-state asynchronous operation of a synchronous machine without excitation source. Because under this condition E 0 and I3 0, the average electromagnetic torque is Tecp | 2 I2 |2

2r 1 [| id ( nep ) |2 Rds | iq ( nep ) |2 Rqs ] 1 2s 2

(3.3.22)

So far as the induction machine without excitation source is concerned, there R is the relation xd ( js ) xq ( js ) X j because its rotor is symmetrical, thus in s this asynchronous operation mode we obtain I2 0 and I3 0. In addition, 210


according to formulas (3.3.20) and (3.3.22) we can get | id ( nep ) | | iq ( nep ) | | 2 I1 | Tecp

R | 2 I1 | s 2

| 2U |2

R· § ¨r ¸ s¹ ©

2

½ ° R ° ° ¾ s ° X2° °¿

(3.3.23)

That is the usual torque formula of induction machines. The minus sign appears before the formula because the basic equations are written according to generator convention. The equivalent circuit of steady-state asynchronous operation of the synchronous machine without excitation source is shown in Fig. 3.3.1(b). When r becomes infinite and one of the corresponding equivalent s 0.5, the value 1 2s circuits is shown in Fig. 3.3.3.

Figure 3.3.3 One of the equivalent circuits for Fig. 3.3.1(b) at s 0.5

It is not difficult to see that under this condition there is only the basic-frequency alternating current in stator winding and the corresponding impedance is U I1

Z

R jX

1 1 r ( Rds Rqs ) j ( X ds X qs ) 2 2

(3.3.24)

Parameters xd ( js ) and xq ( js ) or X ds , Rds and X qs , Rqs are all dependent on slip s, whose changes can be evaluated by the equivalent circuit of operational reactance referring to Section 2.4. From the equivalent circuit, we can see that when the value of slip s increases from zero to infinite, the values of reactances X ds , X qs will decrease from the synchronous reactances xd , xq to subtransient reactances xdcc , xqcc. However, because the rotor winding of a synchronous machine has a small resistance, these reactances X ds , X qs are nearly equal to the subtransient reactances xdcc , xqcc when slip s is more than a not large value. These reactances X ds , X qs change strongly with slip s when the slip s becomes small 211

AC Machine Systems

values. This property is not difficult to be proved so long as we evaluate the equivalent circuit of operational reactance of a typical machine. Of course, it can also be proved through measurement of a specified machine. The test results of a synchronous machine with an excitation winding short-circuited are shown in Fig. 3.3.4. From those curves we can see that when slip s is more than 0.2, the reactances X ds , X qs change little with the slip s and approximate to the subtransient reactances xdcc , xqcc. When 0 s 0.2, the reactances X ds , X qs change strongly with slip s.

Figure 3.3.4 Test results of a synchronous machine with an excitation winding short-circuited

Therefore, when slip s has a large value, for example s ! 0.5 , we can neglect the stator and rotor resistance for calculating currents and approximately consider xd ( js ) xq ( js )

xdcc xqcc

Under this approximate condition, one of the equivalent circuits shown in Fig. 3.3.1(b) can be reduced to Fig. 3.3.5.

Figure 3.3.5 One of the equivalent circuits for Fig. 3.3.1(b) during xd ( js ) and xq ( js ) xccq 212

xdcc


When s 2 , which is called the asynchronous braking of synchronous machines, this condition corresponds to the operation that the stator terminals are applied on by negative-sequence voltages. On this condition, the reactance measured from the stator terminals should be the negative-sequence reactance x2 . From Fig. 3.3.5 we can see that the reactance during asynchronous braking of a synchronous machine, i.e. the negative-sequence reactance, has the following value, x2

xqcc

1 1 1 xqcc 1 ( xcc xcc) d q 2

2 xdcc xqcc

(3.3.25)

xdcc xqcc

That is the negative-sequence reactance we have used in Chapter 2 to calculate time constants. This result is obtained under the condition of non-sinusoidal currents and sinusoidal voltages. If voltages are non-sinusoidal and currents sinusoidal, after neglecting the stator and rotor resistances the negative-sequence reactance corresponding to the asynchronous braking is x2

xdcc xqcc

(3.3.26)

2

Studying two phases sudden short circuit, we have obtained the negativesequence reactance which is x2 xdcc xqcc . Now under asynchronous braking condition there are other expressions referring to formulas (3.3.25) and (3.3.26). This fact shows that the negative-sequence reactance is not a constant. Its value changes with the operation modes of synchronous machines. After neglecting the stator and rotor resistances, the negative-sequence reactance values in some important operation modes are given in Table 3.3.1. From this table we can see that the negative-sequence reactance x2 is a derivative parameter of xdcc , xqcc and x0 . Table 3.3.1 Negative sequence reactance x2 x2

Operating modes 1. Two-phase short circuit at machine terminals 2. Two-phase short circuit at lines

3. Asynchronous braking during sinusoidal voltage and non-sinusoidal current

(2) 2

x

x2(~)

x2( a )

xccd xqcc xdcc xqcc 2

2 xdcc xqcc xdcc xqcc

213

AC Machine Systems Continued x2

Operating modes 4. Asynchronous braking during sinusoidal current and non-sinusoidal voltage 5. Single phase-to-neutral short circuit

x2(~)

x2(1)

6. Two phases-to-neutral short circuit

x2(2 0)

xccd xccq 2

§ cc 1 ·§ cc 1 · 1 ¨ xq x0 ¸¨ xd x0 ¸ x0 2 ¹© 2 ¹ 2 © 2 x0 xccd xccq (2 x0 xdcc )(2 x0 xqcc ) xccd xccq xdcc xccq xdcc xqcc (2 x0 xccd )(2 x0 xqcc ) 2 x0 xdcc xqcc

Notes: (1) If x0 (2) The value of

(1) 2

(2) 2

(2 0) 2

(a) 2 (a) 2 is

0, then x x and x x ; x2(~) is maximum and the value of x minimum.

When these parameters are known, the value of negative-sequence reactance can be calculated indirectly. In non-salient pole machines, we may consider xdcc | xqcc, so x2 | xdcc | xqcc | constant

3.4

Asynchronously Starting of Synchronous Motors

As 3-phase induction motors used in the electric drive system without speed regulation, the 3-phase synchronous motors are also widely used, for example, in compressors, pumps, ventilators, grinders, crushers, rolling mills, clipping machines, saw machines and the motor-generator sets. Compared with induction motors, synchronous motors have several advantages, such as high over-load ability, a good power factor, the invariable speed, the larger air-gap and the torque affected less by the supply voltage. On some conditions, especially during large power and at low speed, a good effect can be obtained by using synchronous motors. The basic equations of synchronous motors in various reference axes systems are the same as those of synchronous generators, so the theory and study methods of synchronous machines mentioned before can all be used. The torque-slip characteristic during asynchronous starting of synchronous motors will be discussed in this section. It is well known that there is the auxiliary prime-mover method, the frequency-changing method and the asynchronous starting method so far as the starting method of synchronous motors is concerned. Of all those methods the asynchronous starting method is most convenient, so the method is most 214


noticeable and widely used. There is a damping winding on the rotor of a synchronous motor for the asynchronous starting. However, the starting winding of the synchronous motor has higher resistance which is made of brass, aluminum bronze or other suitable alloys. It is developed to use massive pole-shoes for the asynchronous starting in recent years. During starting a synchronous motor there are two processes, i.e. the asynchronous operation process without excitation and the process of pulling into step with excitation. The asynchronous operation during starting of a synchronous motor is not a steady-state asynchronous operation process. However, because of the large inertia of the rotor, the rotor acceleration can be neglected so far as the electromagnetic transient process is concerned. If the electromagnetic transient process is not considered, the whole asynchronous starting process can be regarded as a series of steady-state asynchronous operation conditions with different slips. During asynchronous starting, the excitation winding of a synchronous motor is often closed through a discharge resistance whose value is about ten times of the excitation winding resistance. The discharge resistance is usually necessary because an over-voltage will exist at the terminals of excitation winding which is open-circuited during starting. Moreover, when the excitation winding is opened, the torque of pulling into step is lower, which is also adverse to the starting process. On the contrary, if the excitation winding is short-circuited during starting, the starting characteristic is bad because it has a sink at about half a synchronous speed due to the single axis phenomenon, although the over-voltage will not exist and there is higher pull-in torque. When a synchronous motor speeds up close to its synchronous speed, the synchronous motor will come into the stage of pulling into step if a dc voltage is impressed on the excitation winding. Under certain conditions, the machine will be brought into step and run normally. On some adverse conditions, the machine can not be pulled in and will operate asynchronously, which is undesirable. The process and conditions of pulling into step are complicated which fall into a nonlinear problem of electromechanical transient process, so it can only be studied approximately by analytical method in this section and will be analysed by numerical method later referring to [2,3]. (1) The torque of a synchronous motor during asynchronous starting Because the asynchronous starting process can be considered as a series of steady-state asynchronous operation conditions with various slips, some conclusions reached in Section 3.3 can also be used here. The stator resistance of synchronous motor is usually small and can be neglected. Incidentally, the 3-phase voltages can be regarded as symmetrical. Under that condition, according to equations (3.3.11) and (3.3.12) we can obtain 215

AC Machine Systems

jU

xs ( js ) I1 xD ( js ) I2* °½ ¾ xD ( js ) I1 xs ( js ) I2* °¿

0

(3.4.1)

Solving the simultaneous equations above, we can get jxs ( js )U ½ ° x ( js) xD2 ( js ) ° ¾ jxD ( js)U ° xs2 ( js) xD2 ( js ) °¿

I1

2 s

I

* 2

(3.4.2)

so with reference to formula (3.3.17), Id and Iq , namely the time vectors of d-axis current id and q-axis current iq , are Id

Iq

½ xs ( js ) xD ( js ) U ° 2 2 x s ( js ) x D ( js ) ° °° jU jU ¾ xs ( js ) xD ( js ) xd ( js ) ° ° j U U ° I1 I*2 j xs ( js ) xD ( js ) xq ( js ) °¿

I1 I*2

j

(3.4.3)

or | id ( nep ) |

j2U xd ( js )

| iq ( nep ) |

j2U xq ( js )

½ ° X R °° ¾ U ° 2 2 ° X qs Rqs °¿ U

2 ds

2 ds

(3.4.4)

in which U 2 | U | is the terminal voltage of the synchronous motor. The results above substituted into equation (3.3.22) and stator resistance r neglected, the average torque during asynchronous starting can be written as Tecp

ª Rds Rqs º 2 « 2 2 2 » «¬ X ds Rds X qs Rqs »¼

U2 2

ª 1 U2 1 º Im « » 2 «¬ xd ( js) xq ( js ) »¼

(3.4.5)

It is noted that we still adopt the same reference directions of current and torque as the synchronous generator when studying synchronous motors, so here exists 216


the negative sign. The values of xd ( js ) and xq ( js ) in the formula above with reference to Chapter 2, can be estimated as follows. Because X lld Tdcc0

2 xad X ffd

xd

Td 0

2 § xad 1 ¨ xd X lld ©

Tldc Tdcc

X lld Rld

Tld 0

Rld xd

Tdc

2 xad X ffd

· ¸ Tld 0 ¹

Td 0

X ffd R fd

2 2 3 xd ( X lld X ffd xad ) xad ( X lld X ffd ) 2 xad 2 Rld ( xd X ffd xad )

we can write out 2 p 2 ( X lld X ffd xad ) p ( X lld R fd X ffd Rld ) Rld R fd

p 2 Rld X ffd Tdcc0 pRld R fd (Td 0 Tld 0 ) Rld R fd Rld R fd ª¬Tdcc0Td 0 p 2 (Td 0 Tld 0 ) p 1º¼ , 2 3 2 2 2 p 2 [ X lld xad ] p ( xad Rld xad R fd ) 2 xad X ffd xad 2 2 p 2 [ xd ( X lld X ffd xad ) ( xd X ffd xad ) Rld Tdcc]

§T T · 2 Rld R fd ¨ d 0 ld 0 ¸ pxad ¨X ¸ © ffd X lld ¹ p 2 xd Rld R fd [Td 0Tdcc0 TdccTdc ] ª x2 º x2 pRld R fd « ad Td 0 ad Tld 0 » , X lld »¼ ¬« X ffd xd ( p)

xd

2 3 2 2 2 xad X ffd xad p 2 ( X lld xad ) pxad ( Rld R fd ) 2 p 2 ( X lld X ffd xad ) p ( X lld R fd X ffd Rld ) Rld R fd

ª x2 º x2 p 2 xd (Td 0Tdcc0 TdccTdc ) p « ad Td 0 ad Tld 0 » X lld ¬« X ffd ¼» xd 2 cc Td 0Td 0 p (Td 0 Tld 0 ) p 1 217

AC Machine Systems

ª º x2 x2 TdccTdc p 2 p «Td 0 Tld 0 ad Td 0 ad Tld 0 » 1 xd X ffd X lld xd «¬ »¼ xd u 2 Tdcc0Td 0 p (Td 0 Tld 0 ) p 1 xd

TdccTdc p 2 (Tdc Tldc ) p 1 Tdcc0Td 0 p 2 (Td 0 Tld 0 ) p 1

(3.4.6)

Similarly, there is xq ( p )

xq xq

2 pxaq

xq pxqccTq 0

pX llq Rlq

1 pTq 0

pTqcc 1

(3.4.7)

pTq 0 1

in which Tq 0

X llq Rlq

Tqcc Tq 0

xqcc xq

According to xd ( p) and xq ( p ), we can get the expressions for

1 and xd ( js )

1 as follows: xq ( js )

1 xd ( js ) 1 xq ( js )

1 (1 s 2Td 0Tdcc0 ) js (Td 0 Tld 0 ) ½ ° xd (1 s 2TdcTdcc) js (Tdc Tldc ) ° ¾ 1 1 jsTq 0 ° ° xq 1 jsTqcc ¿

(3.4.8)

It is complicated to use formula (3.4.8) for calculation and the damper parameters in that formula are not easy to be evaluated accurately. In practice, the following approximate formulas can often be adopted for estimation. Generally speaking, the time constant of excitation winding is much larger than that of the damping winding, i.e. Td 0 Tld 0 and Tdc Tldc , so equation (3.4.6) can be rewritten approximately as xd ( p )

218

TdccTdc p 2 (Tdc Tldc ) p 1 Tdcc0Td 0 p 2 (Td 0 Tld 0 ) p 1 ( pTdcc 1)( pTdc 1) | xd ( pTdcc0 1)( pTd 0 1) xd


Therefore, 1 1 § 1 1 · Tcp | ¨ ¸ d xd ( p ) xd © xdc xd ¹ 1 Tdc p § 1 1 · T ccp ¨ ¸ d cc © xd xdc ¹ 1 Tdccp

(3.4.9)

In addition, from equation (3.4.7) we can obtain 1 xq ( p )

1 § 1 1 · Tqccp ¨ ¸ xq ¨© xqcc xq ¸¹ 1 Tqccp

The results obtained from formulas (3.4.9) and (3.4.10) during p

(3.4.10) js are

substituted into equation (3.4.5), and the average electromagnetic torque can be written approximately as Tecp

U2 2

ª§ 1 § 1 1 · sTdc 1 · sTdcc ¨ ¸ «¨ ¸ 2 2 «¬© xdc xd ¹ 1 ( sTdc ) © xdcc xdc ¹ 1 ( sTdcc)

§1 1 · sTqcc º ¨ ¸ ¨ xcc x ¸ 1 ( sT cc) 2 »» q ¹ q © q ¼

(3.4.11)

According to that formula it is clear that the average electromagnetic torque during asynchronous starting consists of three parts; the first reflects the main effect of excitation winding, the second and third parts represent the main influence of starting winding. Formula (3.4.11) can give an accurate result when the slip value is larger, but there is an obvious error during small slips because the stator and rotor resistances will play a more important role at this time, and the reactance value rises rapidly from subtransient or transient reactance to synchronous reactance with the slip decreasing. However, the approximate formula can still be used since the initial starting torque Tn (during s 1), the pull-in torque Tb (during s

0.05) and the maximum torque Tm are most important

so far as the torque-slip characteristic for asynchronous starting of a synchronous motor is concerned, in which the corresponding slip values are all not very small except the slip value corresponding to the pull-in torque. The torque-slip characteristic of a synchronous motor is shown in Fig. 3.4.1, in which the effect of the stator resistance is neglected for curve a and the resistance effect is considered for curve b. Because of the asymmetrical rotor of a synchronous motor, there is the pulsating torque with a frequency of 2sf besides the average torque during 219

AC Machine Systems

starting process. Referring to the equivalent circuit in Fig. 3.3.1, we can see that during r 0 there is

\ 2 \ 2* \1

jU

\1*

jU *

0

The results above substituted into equation (3.3.15), the amplitude of pulsating torque with a frequency of 2sf during starting process of a synchronous motor can be described as Tm (2 s )

j2 u 2(U ) 2 xD ( js ) xs2 ( js ) xD2 ( js ) U2 1 1 2 xq ( js ) xd ( js ) U2 2

( X ds X qs )2 ( Rds Rqs )2 ( X ds2 Rds2 )( X qs2 Rqs2 )

(3.4.12)

The relation between the amplitude of pulsating torque with a frequency of 2sf and the slip during starting process of a synchronous motor is shown in Fig. 3.4.2. Because the rotor inertia of synchronous motors is ordinarily larger, the effect of the pulsating torque on starting process is not important and can be neglected in practical calculation. From Fig. 3.4.2 we are also clear that the amplitude of pulsating torque will rise to a large value with the slip decreasing and the corresponding pulsating

Figure 3.4.1 Torque-slip characteristic of a synchronous motor

220

Figure 3.4.2 Pulsating torque amplitude with frequency of 2sf-slip curve during the starting process of a synchronous motor


frequency (2sf )is small which is several Hz, so the speed oscillation may exist near the synchronous speed, too. According to equation (3.4.12) we can know that the pulsating torque with a frequency of 2sf depends upon the asymmetrical condition of synchronous machine rotor. When the rotor is symmetrical and there is xd ( p) xq ( p ) such as in an induction motor, the pulsating torque will be zero. During s 0, i.e. §1 1 1 · synchronous operation, the amplitude of pulsating torque will be U 2 ¨ ¸ , ¨ 2 © xq xd ¸¹ i.e. the torque amplitude at the synchronous speed will become the maximum value of the salient synchronous torque. In addition, from equation (3.4.5) we are also clear that the average electromagnetic torque Tecp (i.e. the asynchronous torque) during s 0 will be equal to zero. That is natural, because during s 0 , i.e. the steady-state synchronous operation, the supply voltage will produce the salient synchronous torque besides causing the synchronous torque with the excitation current, but no other torque exists referring to Section 2.5. (2) The calculation method of asynchronous starting torque of synchronous motors according to the d-axis and q-axis equivalent circuits Since a synchronous motor usually has a strong starting winding, the value of xd ( js ) is close to that of xq ( js ) , i.e. the value of xD ( js ) is small. Accordingly, the value of I2 is also not large referring to Fig. 3.3.1 and equation (3.3.11), so exists jU

xs ( js ) I1 xD ( js ) I2* jrI1 | xs ( js ) I1 jrI1

i.e. 2U | {[r jxd ( js )] [r jxq ( js )]}( I1 )

(3.4.13)

which corresponds to the equivalent circuit shown in Fig. 3.4.3. If the difference between xd ( js ) and xq ( js ) is small, that equivalent circuit can also be divided into two equivalent circuits as shown in Fig. 3.4.4.

Figure 3.4.3 Equivalent circuit corresponding to formula (3.4.13) 221

AC Machine Systems

Figure 3.4.4 Corresponding equivalent circuits during small difference between xd ( js ) and xq ( js )

Because xd ( js ) is ordinarily smaller than xq ( js ), Id is a little more than I1 and Iq is a little less than I1 . 2r caused by the stator resistance Therefore, when the small torque | 2 I2 |2 1 2s is neglected referring to equation (3.3.21), the average torque during asynchronous starting is 1 [| id ( nep ) |2 Rds | iq ( nep ) |2 Rqs ] 2 2[| I1 I2* |2 Rds | I1 I2* |2 Rqs ] | 2[| Id |2 Rds | Iq |2 Rqs ] 1 [| 2 Id |2 Rds | 2 Iq |2 Rqs ] 2

(3.4.14)

The equivalent circuits can be reproduced as those in Fig. 3.4.5 and then the average electromagnetic torque during asynchronous starting according to equation (3.4.14) will correspond to the average value of two starting torques of the following two induction motors: one has a symmetrical rotor with d-axis characteristic and another has a symmetrical rotor with q-axis characteristic.

Figure 3.4.5 Equivalent circuits reproduced referring to Fig. 3.4.4

Some factories have adopted the method to evaluate the asynchronous starting torque of a synchronous motor before the dc voltage is applied to excitaion winding. The error of the calculation method will increase with the values 1 of xD ( js ) [ xd ( js ) xq ( js )] and stator resistance r rising. However, the error 2 222


of this approximate method is usually not large except near the slip of 0.5. The starting characteristic of a typical synchronous motor is shown in Fig. 3.4.6 for reference, in which Ted is the torque of induction motor with d-axis characteristic, Teq is the torque with q-axis characteristic and Tecp is the resultant starting torque.

Figure 3.4.6 Starting characteristic of a typical synchronous motor

(3) The starting torque after a dc voltage is applied to the excitation winding of a synchronous motor In order to pull a synchronous motor into step, a dc source must be provided for the excitation winding when the rotor speed rises near to synchronous speed. If the transient process is neglected, the average asynchronous torque produced by excitation current, with reference to equation (3.3.21) and Fig. 3.3.1(a), is Tecp ( f ) | 2 I3 |2

r 1 s

[r 2 (1 s )2 xq2 ](1 s )r [r 2 (1 s )2 xd xq ]2

E2

(3.4.15)

Because stator resistance is small, we can take r2

r3

r4

0

Those values substituted into formula (3.4.15), we have Tecp ( f )

r §E· | ¨ ¸ 1 s © xd ¹

2

(3.4.16)

whose value is not large and not more than twenty percent of the rated torque and whose sign is opposite to that of the average torque produced by the ac supply, 223

AC Machine Systems

i.e. the torque is a braking one. Therefore, it is adverse to provide an excitation current unduly early. Besides the average torque, the pulsating torque with a frequency of sf is caused by the excitation current referring to equation (3.3.15). Neglecting stator resistance and referring to equation (3.3.13) and Fig. 3.3.1, we have I3

E , \ 3 2 xd

0, \1

jU , \ 2

0

The values above substituted into formula (3.3.15), the amplitude of pulsating torque with a frequency of sf will be Tm ( s )

2 | I2*\ 3 I3\ 2* I3*\1 I1\ 3* | E 2 jU 2 xd

EU xd

(3.4.17)

That is the well known amplitude of the synchronous torque whose value is large, so the excitation current will give rise to the serious oscillation of the power, current, voltage and speed during a larger slip. From that view point, it is also adverse to apply a dc voltage to excitation winding unduly early. However, that torque will play an important role in the pull-in process since the torque will become the synchronous one ultimately.

3.5

Analysis of Combination System for Synchronous Machine and Capacitances

In order to raise the technical and economic targets of the electrical power transmission, the capacity, voltage and distance of electrical power systems are increasing day by day, but the capacitances between long transmission lines and of lines to ground are large enough. Moreover, for the sake of raising the capacity of power transmission, the series connection of capacitors is used to compensate the inductances of long transmission lines. Therefore, the problem about synchronous machines in connection with capacitances is obviously more important. When there are capacitances in electrical power systems, some special problems may be brought about such as the self-excitation and inner over-voltage in synchronous machines referring to [2,3,10]. In the early days of 1970’s there were two severe accidents in which the main shafts of two large turbo-generators were damaged by torsion because series capacitors result in self-excitation currents of low-frequencies. Thus, it is significant to analyse these special problems when synchronous machines are in connection with capacitances and to take corresponding measures. (1) Basic equations during a synchronous machine in series with symmetrical capacitances 224


According to Fig. 3.5.1 we can write the following equations: ua ub uc

x · ½ § p\ a ¨ r c ¸ ia ° p¹ ° © x · °° § p\ b ¨ r c ¸ ib ¾ p¹ ° © x · ° § p\ c ¨ r c ¸ ic ° p ¹ ¿° ©

(3.5.1)

or pua pub puc

p 2\ a ( pr xc )ia ½ ° p 2\ b ( pr xc )ib ¾ ° p 2\ c ( pr xc )ic ¿

(3.5.2)

Figure 3.5.1 The system of synchronous machine in series with capacitances

Because ud uq

2 ½ [ua cos J ub cos(J 120e ) uc cos(J 120e ] ° ° 3 ¾ 2 ) uc sin(J 120e [ua sin J ub sin(J 120e ] ° °¿ 3

(3.5.3)

we have pud

puq

2 ½ [cos J pua cos(J 120e ) pub cos(J 120e ) puc ] ° 3 ° 2 ) uc sin(J 120e )]Z ° [ua sin J ub sin(J 120e ° 3 ¾ 2 ) pub sin(J 120e ) puc ]° [sin J pua sin(J 120e ° 3 ° 2 ) uc cos(J 120e )]Z ° [ua cos J ub cos(J 120e 3 ¿

(3.5.4)

225

AC Machine Systems

By equations (3.5.2) and (3.5.3), the formulas above can be transformed into the following forms, pud uqZ

puq ud Z

2 ½ [cos J p 2\ a cos(J 120e ) p 2\ b cos(J 120e ) p 2\ c ] ° 3 ° 2 ) pib cos(J 120e ) pic ] ° r[cos J pia cos(J 120e ° 3 ° 2 ° )ib cos(J 120e )ic ] xc [cos J ia cos(J 120e ° 3 ¾ (3.5.5) 2 ) p 2\ b sin(J 120e ) p 2\ c ]° [sin J p 2\ a sin(J 120e ° 3 ° 2 ° r[sin J pia sin(J 120e ) pib sin(J 120e ) pic ] 3 ° ° 2 xc [sin J ia sin(J 120e )ib sin(J 120e )ic ] ° 3 ¿

Because p\ d p\ q p 2\ d

2

p \q

2 [cos J p\ a cos(J 120e ) p\ b cos(J 120e ) p\ c ] \ qZ 3 2 [sin J p\ a sin(J 120e ) p\ b sin(J 120e ) p\ c ] \ d Z 3 2 [cos J p 2\ a cos(J 120e ) p 2\ b cos(J 120e ) p 2\ c ] 3 2 ) p\ b sin(J 120e ) p\ c ]Z [sin J p\ a sin(J 120e 3 Z p\ q \ q pZ

½ ° ° ° ° ° ° ¾ 2 ) p 2\ b sin(J 120e ) p 2\ c ] ° [sin J p 2\ a sin(J 120e ° 3 ° 2 ) p\ b cos(J 120e ) p\ c ]Z ° [cos J p\ a cos(J 120e ° 3 ° Z p\ d \ d pZ ¿

½ °° ¾ (3.5.6) ° °¿

(3.5.7)

we can obtain 2 ½ [cos J p 2\ a cos(J 120e ) p 2\ b cos(J 120e ) p 2\ c ] ° 3 ° p 2\ d 2Z p\ q \ d Z 2 \ q pZ ; °° ¾ 2 ) p 2\ b sin(J 120e ) p 2\ c ]° [sin J p 2\ a sin(J 120e ° 3 ° 2 2 p \ q 2Z p\ d \ qZ \ d pZ °¿ 226

(3.5.8)


In addition, there is pid piq

2 ½ [cos J pia cos(J 120e ) pib cos(J 120e ) pic ] iqZ ° ° 3 ¾ 2 [sin J pia sin(J 120e ) pib sin(J 120e ) pic ] id Z ° °¿ 3

(3.5.9)

Substituting equations (3.5.9) and (3.5.8) into (3.5.5) and simplifying them, we can get p (ud p\ d rid \ qZ ) xc id

Z (uq p\ q riq \ d Z )

(3.5.10)

p (uq p\ q riq \ d Z ) xc iq Z (ud p\ d rid \ qZ )

(3.5.11)

According to equation (3.5.1) and the transformation formula for the zero-axis component, it is not difficult to obtain u0

§ x · p\ 0 ¨ r c ¸ i0 p¹ ©

(3.5.12)

Equations (3.5.10), (3.5.11) and (3.5.12) are the basic equations when a synchronous machine is in series with symmetrical capacitances. On the basis of those equations, some problems about the system of synchronous machine in series with capacitances can be analysed. (2) Self-excitation of synchronous machines If the parameters of a synchronous machine are not a suitable match for those of electrical power system including capacitances, self-excitation may occur in the synchronous machine system at synchronous or asynchronous speed. Under this condition, regulation of the excitation current can not control the current and voltage of the synchronous machine, and those quantities depend mainly upon the parameters of the synchronous machine and power system. During self-excitation, the mechanical energy of the synchronous machine will be exchanged with the electromagnetic energy of the machine and power system, which falls into the problem of electromechanical parameter-resonance. Because the inertia constant of synchronous machine is ordinarily large, for convenience it can be taken as infinite, i.e. the speed of synchronous machine can be considered as constant during analysis of this problem. Thus, self-excitation of synchronous machines will be reduced to a problem of electro magnetic parameter-resonance. During self-excitation of synchronous machines, there is not only the current of supply-frequency but also the self-excitation current caused by parameter-resonance in the machine and power system, which will 227

AC Machine Systems

bring about the current increasing constantly or make the machine operate abnormally. Now we analyse the causes and conditions producing self-excitation so as to protect the system against it. Because the excitation voltage of synchronous machine can not be controlled during self-excitation, for convenience the excitation voltage may be supposed to be zero, i.e. U fd 0. Thus,

\d \q

xd ( p)id ½° ¾ xq ( p )iq °¿

(3.5.13)

Furthermore, supposing the speed of a synchronous machine to be asynchronous constant one and letting it be Z 1 s, we can reduce the basic equations (3.5.10) and (3.5.11) to the following forms: {[(1 s ) 2 p 2 ]xd ( p )

½ ° pr xc }id (1 s )[2 pxq ( p ) r ]iq ° ¾ (1 s )[2 pxd ( p ) r ]id ° 2 2 {[(1 s ) p ]xq ( p) pr xc }iq °¿

(3.5.14)

½ ° {[ p 2 (1 s) 2 ](1 s) xq ( p) (1 s) xc }uq º¼ A( p ) °° ¾ ª¬{[ p 2 (1 s) 2 ](1 s) xd ( p ) (1 s ) xc }ud ° ° {[ p 2 (1 s )2 ]Z d ( p) pxc }uq ¼º A( p ) °¿

(3.5.15)

pud (1 s )uq puq (1 s )ud

Solving them, we can obtain id

iq

ª¬{[ p 2 (1 s ) 2 ]Z q ( p) pxc }ud

in which A( p ) [ p 2 (1 s )2 ][ Z d ( p) Z q ( p) (1 s ) 2 xd ( p) xq ( p)] xc [ pZ d ( p ) pZ q ( p ) (1 s ) 2 xd ( p) (1 s )2 xq ( p ) xc ] Z d ( p)

pxd ( p ) r

Z q ( p)

pxq ( p) r

From the formulas above we can see that when the roots of the characteristic equation [ A( p) 0] are positive values or complex-numbers with positive real parts, currents id and iq will increase constantly, i.e. self-excitation occurs. Therefore, the problem whether there is self-excitation has become the problem whether the equation A( p) 0 has a positive real root or complex-number root with a positive real part. 228


There are many methods to identify if the equation has positive real roots or complex-number roots with positive real parts. The usual methods are Herwitz’s, Routh’s and Mihainov’s criterions. The synchronous machine having a damping winding, the order of the equation A( p) 0 is higher and the approximate digital solution can be obtained by a computer. However, in order to make the problem clear, we shall analyse only the self-excitation of synchronous machines without damping winding in the following. Under this condition, we have xd ( p)

xd

xq ( p )

xq

s

2 pxad pX ffd R fd

xdc Td 0 p xd Td 0 p 1

0

so A( p )

xdc xqTd 0 p 5 [ xd xq rTd 0 ( xdc xq )] p 4 [2 xdc xqTd 0 r ( xd xq ) Td 0 xc ( xdc xq ) Td 0 r 2 ] p 3 [2 xd xq rTd 0 ( xdc xq ) xc ( xd xq ) r 2 2rxcTd 0 ] p 2 [ xdc xqTd 0 r ( xd xq ) Td 0 xc ( xdc xq ) 2rxc Td 0 ( xc2 r 2 )] p [ xd xq xc ( xd xq ) xc2 r 2 ]

(3.5.16)

That is the fifth degree equation and it is tedious to write the corresponding discriminant. However, the practical values given, calculation is convenient. For example, a machine has the following parameters: xd

1.0,

xdc

0.30,

xq

0.6,

Td 0

1 000

According to those values, the parameter range causing self-excitation can be determined as shown by full line in Fig. 3.5.2.

Figure 3.5.2 Self-excitation region of synchronous machine 229

AC Machine Systems

The self-excitation under several special conditions is discussed as follows. a. Neglecting stator resistance (r 0) Under this condition, equation (3.5.16) is turned into the following form: A( p )

xdc xqTd 0 p 5 xd xq p 4 Td 0 [2 xdc xq xc ( xdc xq )] p3 [2 xd xq xc ( xd xq )] p 2 Td 0 [ xdc xq xc ( xdc xq ) xc2 ] p [ xd xq xc ( xd xq ) xc2 ]

(3.5.17)

From Routh’s and Herwitz’s criterions we can see that the self-excitation can be produced only under the following condition: xc xq and xq xc xd or xc xd

(3.5.18)

Therefore, when stator resistance is neglected and there is xc xd , the self-excitation will be caused. In practice, because of the stator resistance the self-excitation region is smaller than above; ordinarily the region in xdc xc xd . That conclusion can be known from Fig. 3.5.2. b. The excitation winding open-circuited (Td 0 0) Under this condition, equation (3.5.16) will be simplified. The self-excitation can be discriminated to be caused in the following region: xd xq ( xd xq ) 2 4r 2 2

xc

xd xq ( xd xq )2 4r 2 2

(3.5.19)

Obviously, the shade-line area in Fig. 3.5.2 corresponds to the self-excitation region indicated by formula (3.5.19). c. Neglecting excitation winding resistance (Td 0 f) Under this condition, the self-excitation will be caused in the following region: xdc xq ( xdc xq ) 2 4r 2 2

xc

xdc xq ( xdc xq )2 4r 2 2

(3.5.20)

According to formulas (3.5.19) and (3.5.20) we can see that there is no xd xq xq xdc or r ! . That conclusion is also known self-excitation when r ! 2 2 from the typical example in Fig. 3.5.2, in which the self-excitation region indicated by formula (3.5.20) is shown by a dotted line. That self-excitation region, i.e. the region when Td 0 f, is very close to the self-excitation region between xq and xdc in a practical machine with larger Td 0 . Because the typical generator without damping winding has large Td 0 , its self-excitation range can be discriminated by formulas (3.5.19) and (3.5.20). 230


On the basis of formula (3.5.19) it is not difficult to understand that there is no self-excitation when xd xq and the excitation winding is open-circuited, i.e. a non-salient-pole synchronous machine, when its excitation winding is open-circuited, has no self-excitation. However, when the excitation winding is short-circuited and xdc z xq , this type of machine may be self-excited referring to formula (3.5.20). That is because the salient-pole machine, when its excitation ª1 º § 1 1 · winding is open-circuited, can develop the active power « U 2 ¨ ¸ sin 2G » «¬ 2 ¨© xq xd ¸¹ »¼ due to its asymmetrical magnetic circuit to balance the stator resistance loss, but the non-salient-pole machine has no possibility. However, when the excitation winding of a non-salient-pole machine is short-circuited, it can produce an active ª1 º § 1 1· power « U 2 ¨ ¸ sin 2G » due to the asymmetrical electric circuit of its «¬ 2 ¨© xdc xq ¸¹ »¼ rotor to compensate the stator resistance loss. The first self-excitation phenomenon is called the salient-pole synchronous self-excitation, and the latter the repulsion synchronous one. The salient-pole synchronous self-excitation is still maintained when the stator flux-linkage is constant, so this self-excitation can finally reach a steady-state value and the growing rate of its current and voltage is generally slow. However, the repulsion synchronous self-excitation will grow constantly at a higher rate, or else the corresponding transient current can’t be induced in the rotor circuit and there is no parameter xdc . During the two synchronous self-excitation conditions above, stator selfexcitation current has supply-frequency and the machine has a synchronous speed. The current during this self-excitation is shown in Fig. 3.5.3.

Figure 3.5.3 Stator self-excitation current during synchronous self-excitation

During self-excitation, the oscillation period of the stator self-excitation current depends upon the inductance and capacitance in the stator circuit. Therefore, in the stator there can also be another self-excitation current whose frequency is different form the supply-frequency when the machine has a synchronous speed. This self-excitation phenomenon is called the asynchronous self-excitation one. Under the condition of asynchronous self-excitation, the speed of air-gap field produced by the self-excitation current is different from the rotor speed. Letting 231

AC Machine Systems

Z f be the speed of air-gap field, Z s the synchronous speed, i.e. the practical rotor speed and s

Z f Zs the slip, there is a current of frequency sf in the rotor Zs

circuit. If the rotor circuit is asymmetrical, the stator current will contain two components of the frequencies (1 r s) f as shown in Fig. 3.5.4.

Figure 3.5.4 Stator self-excitation current during asynchronous self-excitation

When the rotor is symmetrical as in an induction machine, the stator current contains only one component of frequency (1 s ) f and the slip s must be negative. Under this condition, the induction generator will develop an active power to compensate the stator resistance loss. As discussed above, the machine is supposed to be a synchronous generator with a synchronous speed. If the machine is a synchronous motor with a synchronous speed, the conclusions above can also be utilized. When the machine is a starting synchronous motor, i.e. its speed is below the synchronous speed, the salient-pole synchronous self-excitation or asynchronous self-excitation can also occur. However, the stator current frequency during the salient-pole synchronous self-excitation must correspond to the rotor speed. For example, when that self-excitation takes place at a synchronous speed, the self-excitation current has a frequency of 50Hz; when the self-excitation appears at half a synchronous speed, the self-excitation frequency is 25Hz. If there is an asynchronous self-excitation in the starting process, the stator self-excitation frequency will be lower, because only in this way the speed of the revolving field produced by the asynchronous self-excitation current is below the rotor speed to make the machine as an induction generator to maintain the asynchronous self-excitation. During the repulsion synchronous self-excitation, the stator self-excitation frequency also corresponds to the rotor speed. As mentioned before, this type of self-excitation will increase constantly at a higher rate, through which the self-excitation can exist. There is the self-excitation phenomenon when the d-axis and q-axis transient parameters are not equal to each other. If a pair of complex-number roots in the characteristic equation of the asynchronous self-excitation change to a pair of equal roots, the asynchronous self-excitation is turned into the repulsion synchronous self-excitation, so it is difficult to differ from them. 232


Because the salient-pole synchronous self-excitation can appear only under the larger capacitance condition, the asynchronous self-excitation is a phenomenon which often occurs in practice and must be prevented first. During the asynchronous self-excitation, the speed of the revolving filed produced by the self-excitation current is below the synchronous speed and there is an interaction between that revolving field and the synchronous rotating-field produced by the supplyfrequency current to produce the pulsating torque and cause oscillation and noise. If the frequency of the pulsating torque is equal to or near the natural frequency of the main-shaft torsion-oscillation, the mechanical resonance may occur to make the main-shaft broken or damaged, which is termed sub-synchronous resonance. As mentioned before, the main shafts of two large turbogenerators were damaged due to that cause. When a rated voltage is applied to a generator without excitation the corresponding stator current is named the charge current of a generator. Obviously, the current is I Z U N / xd 1/ xd and the corresponding apparent power is called the generator charge-capacity, which is PZ

U N2 xd

1 . xd

Similarly, the apparent power of a capacitor-load at rated voltage is called the U N2 1 . Therefore, charge capacity of a capacitor-load which is equal to PZN xc xc in order to protect the machine against any self-excitation phenomenon, it is essential to keep xc ! xd or PZ ! PZN , i.e. the generator charge-capacity must be larger than the charge capacity of a capacitor-load. (3) The transient process when a synchronous machine is suddenly connected to a capacitor-load This problem is significant in practice. For example, the synchronous machine with no-load is abruptly connected to a long-distance transmission line or the transmission line in series with capacitors is suddenly short-circuited, which all fall into that scope. Now let us discuss the transient process when the synchronous machine rotating at a synchronous speed without load is suddenly connected to a 3-phase symmetrical capacitor-load. Under this condition, we can equally consider that the generator in series with capacitors is suddenly short-circuited. By the Superposition Theorem, it may be thought that the generator is not short-circuited but suddenly applied on by a voltage which is equal but opposite to original no-load voltage. Substituting the following relations: s

0, ud

0,

uq

E

into equation (3.5.15), we can obtain 233

AC Machine Systems

id iq

( p 2 1) xq ( p ) xc

½ E ° A( p) ° ¾ 2 ( p 1) Z d ( p ) pxc ° E° A( p ) ¿

(3.5.21)

The equation can be solved by Operational Calculus, but the corresponding characteristic equation, due to capacitors, has a higher order. Therefore, it is more difficult to solve the equation. In practice, we only discuss its maximum short circuit current and maximum voltage drop in the capacitor, because those quantities are significant for use of the machine and choice of protection equipment. Solving those quantities we can also neglect the stator and rotor resistances much as in 3-phase short circuit. After obtaining id and iq , we can get the maximum value of current ia by virtue of the axes-transformation formulas and calculate the voltage drop in a capacitor according to the following formula: u xc

1 ia dt C³

xc ³ ia dt

(3.5.22)

The possible maximum value of voltage u xc is generally not more than triple the rated phase-voltage. However, the capacitor-load being asymmetrical, especially the machine having no damping winding, the operating over-voltage can be severe, whose causes are: 1) even if there is no capacitor, under the condition of asymmetrical short circuit, there are a series of harmonics in the stator current due to the asymmetrical rotor, which can result in higher instantaneous value of the voltage; 2) when there is capacitor, the capacitance and inductance of the machine and transmission line will produce resonance at a certain frequency, so the instantaneous value of the voltage will further increase. Thus, when xc n 2 xdc xq , the over-voltage is specially severe. n is positive and generally

n 3 and n 5 are most important. When a machine has damping winding, the over-voltage will obviously decrease because the damping winding makes the ratio xqcc xdcc close to 1. If there is only the q-axis damping winding, the effect is also obvious. As mentioned before, when xqcc xdcc 1, the harmonics are not easy to appear. The conclusions above are obtained on the basis of the machine operating separately. If the machine is parallel to other machines with damping windings or it has an enough resistance or induction-motor-load, the hazard of over-voltage can decrease. 234


3.6

Mathematic Model and Performance Analysis of Cycloconverter-fed Synchronous Motor Systems with Field-Oriented Control

Cycloconverters can change the source frequency to lower frequency that is under 1/ 3 1/ 2 of source frequency. Cycloconverter-fed synchronous motor systems with field-oriented control are widely used to drive slow speed and large capacity production machines such as steel-rolling machine, grinder, minehoister, etc. In comparison with ac-dc-ac frequency converter, the cycloconverter has the following features: (i) The intermediate dc filter can be saved and there is only one stage powerconverting circuit, so its efficiency is high. (ii) The source can convert current directly, so there is no need for auxiliary current-converting circuit. (iii) A cycloconverter generally consists of 36 thyristors, so its structure is complicated. In general, large cycloconverters are of voltage type and the cycloconverter-fed synchronous motor system is shown in Fig. 3.6.1, which is composed of a 3-phase rectiformer with 3 separated secondary windings for each phase, cycloconverter, synchronous motor, position and speed sensors, and field-oriented vector control system, referring to [4,7].

Figure 3.6.1 A cycloconverter-fed synchronous motor system

The output voltage of cycloconverter is pieced together by the segments of its input voltage, so there are plentiful harmonics and the harmonic frequency of its output voltage is f0h

KPfi r (2n 1) f 0

(3.6.1)

The harmonic frequency of its input current is f ih

( KP r 1) fi r 6nf 0

(3.6.2) 235

AC Machine Systems

in which fi and f 0 are the fundamental frequency of input and output respectively, k 1, 2,3," , n 0,1, 2," , and P is number of pulses for the cycloconverter. Firstly the mathematical pattern is built up for the cycloconverter-fed synchronous motor speed regulation system, then the steady-state and transient performance can be analysed for the system, and finally some useful conclusions will be drawn for design of damping winding for the cycloconverter-fed synchronous moter.

3.6.1

Mathematical Patterns of the Cycloconverter-fed Synchronous Motor System

The cycloconverter-fed synchronous motor system with field-oriented control consists of synchronous motor, cycloconverter and field-oriented control system, whose patterns will be discussed respectively as follows. (1) Mathematical pattern of synchronous machine The multi-loop pattern is adopted for the synchronous machine to consider the influence of spacial harmonics in air-gap field fully. In addition, damping winding has an important effect on the performance of cycloconverter-fed synchronous machine, especially the dynamic performance. The multi-loop pattern is to write down the corresponding equations according to actual damping winding, so the actual currents of various damping loops can be got, which is essential for design of damping winding about this type of synchronous machine. The voltage equations can be written down respectively for 3-phase stator windings and excitation winding of synchronous machine, and then take d-axis and q-axis as central lines respectively to write loop voltage equations. For example, there are 6 damping bars per pole in Fig. 3.6.2, so for d-axis exist 3 damping loops and for q-axis also exist 3 damping loops.

Figure 3.6.2 Damping loops selected

The voltage equation set of synchronous machine is

U 236

L

d I (R Z G) I dt

(3.6.3)


in which

U

[ua ub uc u fd 0 0 0 0 0 0]T

is voltage matrix and

I

[ia ib ic i fd i1d i2 d i3d i1q i2 q i3q ]T

dL and T is rotor dT position angle; L and R are inductance and resistance matrices respectively and there is is current matrix; Z is electric angular velocity of rotor; G =

ª Lss «M ¬ rs

L

M sr º Lrr »¼

where ª Laa « « M ab «¬ M ac

Lrr

ª Lrrd «M ¬ rqd

M rdq º , L Lrrq »¼ ss

M sr

ª M afd « « M bfd « M cfd ¬

M a1d M b1d M c1d

M rs

M srT

Lrrd

ª L ffd «M « f 1d «M f 2d « ¬« M f 3d

M f 1d L11d M 12 d M 13d

M f 2d M 12 d L22 d M 23d

Lrrq

ª L11q « « M 12 q « M 13q ¬

M 12 q L22 q M 23q

M 13q º » M 23q » L33q »¼

M a 2d M b 2d M c2d

M a 3d M b 3d M c3d

M ab Lbb M bc M a1q M b1q M c1q

M ac º M bc »» Lcc »¼ M a 2q M b2q M c 2q

M a 3q º » M b 3q » M c 3q »¼

M f 3d º M 13d »» M 23d » » L33d ¼»

When considering the fundamental and odd harmonic air-gap fields only, M rqd and M rdq are all zero-matrices. The resistance matrix R is

R

ª Rs « ¬

º Rr »¼ 237

AC Machine Systems

where

Rs

Rr

ª r1 º « » r1 « » «¬ r1 »¼ ª rfd « r1d r12 d « « r12 d r2 d « r13d r23d « « « « « ¬

r13d r23d r3d r1q r12 q r13q

r12 q r2 q r23q

º » » » » » r13q » » r23q » r3q »¼

In the formulas mentioned above, the subscripts a, b, c represent stator 3-phases respectively; the subscripts fd , 1d , 2d , 3d , 1q, 2q, 3q indicate excitation winding and various loops on d, q axes; r1 is stator phase resistance. The electromagnetic torque formula is Tem

I T GI

(3.6.4)

dZ dt

TL Tem

(3.6.5)

and the rotor motion equation is H

Calculation of various inductances in above formulas has been discussed in Chapter 1. (2) Mathematical pattern of cycloconverter The main circuit of 3-phase bridge-type cycloconverter with 6 pulses is shown in Fig. 3.6.3, in which the input terminals for 3 sets of current converters are supplied by a 3-phase rectiformer with 3 separated secondary windings for each phase, and each phase is composed of 12 thyristors, i.e. two 3-phase rectifierbridges controlled fully are connected in parallel inversely. For cycloconverters the cosine-intersection method is usually adopted to control thyristors conducting. The main circuit of i-th phase (i a, b, c) for 3-phase bridge-type cycloconverter with 6 pulses is shown in Fig. 3.6.4, in which various thyristors are triggered into conduction according to the sequence of 1 2 3 4 5 6 1, and the triggering time and conducting length of time can be determined for various thyristors by use of cosine-intersection method. When considering the secondary leakage inductances of 3-phase rectiformer, without circulating current the 3-phase bridge-type cycloconverter with 6 pulses has three conducting modes, namely normal two thyristors conducting mode, two 238


Figure 3.6.3 The main circuit of 3-phase bridge-type cycloconverter with 6 pulses

Figure 3.6.4 The main circuit of i-th phase (i

a, b, c) for 3-phase bridge-type

cycloconverter with 6 pulses

thyristors converting mode, and positive and negative sets interchanging mode, which are illustrated as follows. (a) Normal two thyristors conducting mode Supposing that thyristors 1 and 2 of i-th phase are conducted as shown in Fig. 3.6.5(a), then the effective voltage impressed on the cycloconverter is U iNc

U as U cs

2L f

d ii U iN dt

(3.6.6)

in which U iN is the voltage between terminal i and converter neutral point N, ii is the i-th phase load current, and L f is the phase leakage inductance of rectiformer secondary windings. 239

AC Machine Systems

Figure 3.6.5 Conducting modes of cycloconverter with 6 pulses (a) Normal two thyristors conducting mode; (b) Two thyristors converting mode.

(b) Two thyristors converting mode For rectiformer secondary windings exists leakage inductance L f , so switching-on and switching-off of thyristors can’t be accomplished in an instant, and the load current is transited from the switching-off thyristor to switching-on thyristor gradually that is two thyristors converting mode. If thyristors 1 and 2 of i-th phase are conducting, then thyristor 3 is triggered to cause current converting from thyristor 1 to thyristor 3 and in the meantime thyristors 1, 2, 3 are conducting simultaneously as shown in Fig. 3.6.5(b). The loop voltage equation for ii is U as U cs

2Lf

d d ii L f ihi U iN dt dt

where ihi is the converting loop current. The converting loop voltage equation is U bs U as

2L f

d d ihi L f ii dt dt

During ihi ii the current conversion is finished. Deleting ihi from the two formulas above, we can get the cycloconverter’s voltage equation under this mode as follows: U iNc

U as U bs U cs 2

1.5 L f

d ii U iN dt

(3.6.7)

which is the effective voltage impressed on the cycloconverter under this mode. Synthesizing formulas (3.6.6) and (3.6.7), the generalized expression of cycloconverter’s voltage equation can be got under two thyristors conducting mode and two thyristors converting mode of i-th phase as follows: 240


U iNc

L fi

d ii U iN dt

(3.6.8)

where L fi

° 2 L f ® °¯1.5 L f U iN

for two thyristors conducting mode for two thyristors converting mode U iO U NO

and subscripts N,O indicate the neutral points of cycloconverter and synchronous motor stator winding respectively. (c) Positive and negative sets interchanging mode When a phase load-current changing its direction for the cyclconverter with 6 pulses and without circulating current, the cycloconverter needs to go into the positive and negative sets interchanging stage, and in the meantime all thyristors of positive and negative stes for this phase are blockaded to achieve cycloconverter’s switching-over without circulating current, which is termed the dead-region stage and its persistence time is about 2 8ms. If i-th phase is under positive and negative sets interchanging mode, then exists ii

0

(3.6.9)

Solving the state equation, we can decrease one dimension for state variables. In the meantime, the thyristors of this phase are blocked and the output voltage of cycloconverter is determined not ty cycloconverter itself but by the load voltage impressed on this phase, so there is U iNc

U iO U NO

(3.6.10)

(3) Mathematical pattern of air-gap flux-linkage oriented control system In the mathematical pattern of air-gap flux-linkage oriented control system for cycloconverter-fed synchronous motor, the speed regulator, flux regulator and current regulator all use proportional-integral regulation, whose transfer function is W ( p)

Kp

1 TN p TN p

where K p is amplification and TN is integral time constant. Various feedback quantities in the control system all contain filtering links and the amplitude limiters are provided for various regulators. By use of the mathematical pattern above make simulative calculation and experiment for a 4.5kW-reverse-mounted cycloconverter-fed synchronous motor with field-oriented control, and the simulative results are approximate to the experimental data, which also proves the mathematical pattern above to be correct. 241

AC Machine Systems

3.6.2

Simulation and Performance Analysis of Cycloconverter-fed Synchronous Motor Systems

The cycloconverter-fed synchronous motor system with field-oriented control and speed regulation is a nonlinear, multivariable and multiloop system, whose mathematical pattern is a differential equation set with variable coefficients and can be solved by use of numerical methods. In comparison with usual synchronous motor, the cycloconverter-fed synchronous motor system has some topics to need studying, such as the harmonics in steady-state operation and the performance in transient operation. In addition, the simulative calculation in steady-state operation is also different from that in transient operation. During simulation in steady-state the rotor speed is constant, and the action of control system is to provide a standard voltage for the cycloconverter, whose frequency is equal to the operating frequency of synchronous motor and its phase angle depends upon the machine power factor and rotor initial position angle. Therefore, so long as the frequency and power factor is predetermined for steady-state operation, the action of control system can be superseded by setting a standard voltage during simulative calculation in steady-state. During simulation in transient operation, the synchronous motor has a transient process from original steady-state to another steady-state, so the control system has to join in simulative calculation. In order to shorten the simulative process, parallel calculation is used for synchronous motor and series calculation is adopted for control system. Taking a steel-mill driven by a 4 000kW-cycloconverter-fed synchronous motor as example, analyse the steady-state and transient performance of the cycloconverterfed synchronous motor system. The parameters of 4 000kW-cycloconverter-fed synchronous motor are: rated power: 4 000kW rated voltage: 1 650V rated current: 1 512A number of phases: 3 rated power factor: 0.978 rated frequency: 4Hz rated speed: 40rpm number of pole-pairs: 6 excitation voltage: 147V excitation current: 544.4A (1) Steady-state operating performance of cycloconverter-fed synchronous motor systems In steady-state, the output voltage of cycloconverter contains plentiful harmonics, and in the meantime the harmonic voltage, harmonic current, harmonic loss, harmonic torque, vibration and noise must be paid attention to. During rated operation of 4 000kW-cycloconverter-fed synchronous motor, the wave-forms of U aN , U aO , ia , Tem , i f are shown in Fig. 3.6.6 and those of 6 damping bars currents for each pole are indicated in Fig. 3.6.7. 242


Figure 3.6.6 Simulative waveforms of U aO , ia , U aN , Tem , i f during rated operation of 4 000kW-cycloconverter-fed synchronous motor

From Fig. 3.6.6 we can see, the cycloconverter’s output voltage U aN is not equal to the synchronous motor voltage U aO because the neutral point of synchronous motor is isolated from the neutral point of cycloconverter to bring about the voltage difference between them, whose value is U NO . From Fig. 3.6.6 it is also evident that the waveforms of synchronous motor phase current and electromagnetic torque are rather smooth although the waveform of its phase voltage contains ripples or burrs, because the harmonic impedance of synchronous motor is relatively large to filter current harmonics. At the same time, the excitation current of synchronous motor is not a smooth straight-line any longer and contains some harmonics. During steady-state operation, 6 damping bars currents also contain stronger harmonics as shown in Fig. 3.6.7. By use of FFT analysing program, we can evaluate the harmonic spectra of various quantities mentioned above and then analyse the relations between them and various factors. (2) Dynamic performance of cycloconverter-fed synchronous motor systems For a cycloconverter-fed synchronous motor system with field-oriented control used to drive steel-mill and during rolling period, the synchronous motor is loaded, unloaded or rotating forward or backward frequently and its load torque is also variable, so all the system is in a transient process. Thus it is significant to study dynamic performance of cycloconverter-fed synchronous motor. 243

AC Machine Systems

Figure 3.6.7 Simulative waveforms of 6 damping bars currents for each pole during rated operation

In order to improve the dynamic performance of cycloconverter-fed synchronous motor, the damping winding is installed usually for this kind of synchronous motor. There are two types of damping windings, i.e. half-damping type without full end-ring and having no connection between adjacent poles and full-damping type with full end-ring. The transient processes for this kind of synchronous motor are studied respectively for sudden load, sudden reversal-rotation and speed-up with weak field as follows. (a) Sudden load When a 4 000kW-cycloconverter-fed synchronous motor with flux-linkageoriented control operates at rated speed 40rpm and load torque is increased suddenly from 0.1 per-unit to 1.0 per-unit, the changing processes of rotor speed n, power angle G , stator current d-axis and q-axis components id and iq , stator current’s torque-component iT , electromagnetic torque Tem and excitation current i f are shown respectively in Fig. 3.6.8(a) (f) for full-damping type of synchronous motor. According to the same situation also analyse the transient processes of half-damping and no-damping synchronous motors, and the two types of synchronous motors can’t operate stably due to oscillation. From the above example it is clear that q-axis damping winding plays an important role to keep air-gap flux-linkage constant and to raise the system stability during the transient process of sudden load. (b) Sudden reversal-rotation Taking the 4 000kW-cycloconverter-fed synchronous motor as example, assuming the rotor speed and load torque to be 1.0 per-unit and during sudden 244


Figure 3.6.8 Simulative waveforms of n, G , id , iq , iT , Tem , i f during transient process of sudden load

change of rotation from backward to forward direction, the dynamic curves of rotor speed n, stator current ia , stator current d-axis and q-axis components id and iq , stator current’s torque-component iT , electromagnetic torque Tem and power angle G are shown respectively in Fig. 3.6.9(a) (f) for full-damping type of synchronous motor. In addition, also analyse the sudden reversal-rotation process of half-damping and no-damping synchronous motors, and the calculation results indicate that the two types of synchronous motors can’t operate stably. When sudden reversal-rotation is compared with sudden load, both can cause the sudden change of stator current’s torque-component iT and stator current q-axis component iq and then the rather large rotor q-axis damping winding 245

AC Machine Systems

current is produced to compensate the change of air-gap flux-linkage and to keep air-gap flux-linkage constant, but sudden reversal-rotation is more serious than sudden load. In order to improve the system dynamic performance and to raise the system stability, the cycloconverter-fed synchronous motor must have a full-damping rotor to provide a passage for rotor q-axis damping current,which is more important for the cycloconverter-fed synchronous motor used in frequent change of rotating direction. In China, some half-damping synchronous motors have burning marks on their rotor-pole dovetails during frequent change of rotating direction, which has proved the above conclusion.

Figure 3.6.9 Dynamic simulative wave-forms of n, ia , id , iq , iT , Tem , G during sudden change of rotating direction for synchronous motor 246


(c) Speed-up with weak field Taking the 4 000kW-cycloconverter-fed synchronous motor as example, keeping the load torque of synchronous motor as 0.5 per-unit, raising the rotor speed from 1.0 per-unit to 1.5 per-unit and diminishing the air-gap flux-linkage from 1.0 per-unit to 0.67 per-unit, calculate the change processes of synchronous motor speed n, stator current ia , stator current d-axis and q-axis components id and iq , stator current’s torque-component iT , electromagnetic torque Tme and power angle Gas shown in Fig. 3.6.10(a) (f). Evidently, during a certain load torque and speed-up with weak field, the stator current q-axis component changes

Figure 3.6.10 Dynamic simulative wave-forms of n, ia , id , iq , iT , Tem , G during a certain load torque of synchronous motor and speed-up with weak field 247

AC Machine Systems

a little so the rotor q-axis damping winding current caused by it is also small. However, the stator current d-axis component is double or more so the rotor d-axis damping winding current caused by it is also large. Calculation results indicate that during speed-up with weak field and even if small load torque exists serious oscillation for half-damping and no-damping synchronous motors, thus the system being unstable. Therefore, the full-damping winding is also essential even if speed-up with weak field

3.6.3

Design of Damping Winding for Cycloconverter-fed Synchronous Motors with Field-oriented Control

From the above analysis we can see, the full-damping winding has to be installed for cycloconverter-fed synchronous motor to get good dynamic performance. Design of damping winding must meet three requirements: (i) Improve the system dynamic performance (ii) Make the damping winding loss as small as possible (iii) Let the damping bars currents and losses be distributed uniformly for each pole. Therefore, it is necessary to make an overall calculation, especially dynamic-performance calculation, and then the damping winding can be disposed rationally. We still take the 4 000kW-cycloconverter-fed synchronous motor as example to discuss the damping winding current distribution as follows. Assuming the number of damping bars for each pole to be 5 and 6 and increasing load torque suddenly from 0.1 per-unit to 1.0 per-unit at rated speed, the distribution curves of various damping bars currents in rms value are shown respectively in Figs. 3.6.11(a) and (b).

Figure 3.6.11 Distribution of various damping bars currents during sudden load

From Fig. 3.6.11 it is evident that during sudden load the damping bars currents on the pole-fore are larger than those on the pole-aft for all the two situations of 5 and 6 damping bars per pole. During 5 damping bars per pole, the 248


damping bar in the middle of pole has the largest current of all damping bars. When the number of damping bars per pole is even, the damping bars currents and losses are distributed more uniformly than odd number of damping bars per pole, because during sudden load the stator current’s torque-component occurs mainly on q-axis of the synchronous motor with field-oriented control. Assuming the rotor speed and load torque to be 1.0 in per-unit, supposing the number of damping bars per pole to be 5 and 6 respectively and during sudden reversal-rotation, the distribution curves of various damping bars currents per pole are shown in Figs. 3.6.12(a) and (b) for the 4000kW-cycloconverter-fed synchronous motor, in which the damping bars currents during even number of damping bars per pole are distributed more uniformly than odd number of damping bars per pole.

Figure 3.6.12 Distribution of various damping bars currents during sudden reversal-rotation

During speed-up with weak field, the load is relatively small, so the rotor damping bars currents are also small. Therefore, it is not necessary for design of damping winding to consider the dynamic process of speed-up with weak field. To sum up, for cycloconverter-fed synchronous motors with field-oriented control used in sudden load or sudden reversal-rotation, it is better to take the number of damping bars per pole as even number.

3.7

Analysis of Synchronous Generators with AC and DC Stator Connections

Some special power supply systems, such as those used on ships, tractions, and aircraft, are often required to provide high quality ac and dc electricity simultaneously. The synchronous generator with ac and dc stator connections (ac-dc generator) is an effective way to satisfy this requirement. There are five 249

AC Machine Systems

suits of independent star-connected 3-phase windings in the ac-dc generator stator, one is called ac winding to supply ac power and the others belong to dc windings1, which are connected to four 3-phase bridge rectifiers respectively, and then the dc output sides of the four 3-phase bridge rectifiers are in parallel to supply electricity to dc load, as shown in Fig. 3.7.1.Owing to the construction with the ac and dc windings arranged in the same stator, the ac-dc generator has many obvious advantages. Firstly, this generator has small volume, light weight and low cost. In addition, there is no electrical connection between the ac winding and dc winding, but only magnetic couple between them. So the voltage harmonic distortion of ac winding caused by the commutation process of dc winding becomes small, then the influence of dc load on the voltage of ac winding decreases greatly and the electromagnetic compatibility performance is improved.

Figure 3.7.1 The schematic diagram of an ac-dc generator

The mathematic model of ac-dc generators is complicated and number of equations is high, so the simulation is used to analyze the performance, such as the voltage harmonic distortion of ac voltage of ac winding, ripple of dc voltage, power quality and electromagnetic compatibility. Moreover, the established model and simulation method can be used to analyze the low frequency power oscillation while the counter emf load is connected to dc side, and to find out the effective way to restrain the oscillation and guarantee the reliable operation. Also it can be applied to simulate the process of sudden short circuit at ac side or dc 1

In order to distinguish the ac winding with 3-phase bridge rectifier from the oridinary ac winding, the former is defined as dc winding.

250


side, and provide the reliable protection. In this section the multi-loop model is applied to analyze this system, which implies that the a, b, c coordinates system is based. So the coordinates transformation is not needed for simulation, and the state variables are the actual values, which make the simulation results directly perceived and the debugging of program easy. On the other hand, because some elements of inductance matrix are time-variant, the inverse of inductance matrix at each simulation step increases the calculation. Another advantage of using the multi-loop model is that the effects of the spatial harmonic components of air-gap magnetic field are included naturally, which improves the accuracy. On the stator of ac-dc generators there are five 3-phase windings, the slot number per pole per phase of which can not be too much because of the limited stator slot number. While the slot number per pole per phase of dc winding is equal to 1, the spatial harmonic components of air-gap magnetic field are rather strong. At last, the damper winding would influence the commutation process of rectifier, so the multi-loop model helps to simulate the commutation process with better accuracy, where the damper winding is dealt with according to its actual structure. Because of the frequent switching of rectifiers, the generator always works in an unsymmetrical state, which makes the coordinates transformation have no obvious advantage and even increase the complexity of programming. Although the application of d, q, 0 coordinates system can obtain the constant coefficient inductance matrix and save the calculation of inversing matrix, the transformation of d, q, 0 to a, b, c coordinates system must be needed to get the voltage and current in a, b, c system to judge the commutation, which makes the debugging of program more complicated. In brief, although the calculation of multi-loop model is a vast amount, the multi-loop model is suitable to simulate the performance of ac-dc generators, because the calculation of parameters is accurate, and the coordinates transformation is not necessary for the judgment of commutation. If only the fundamental component of air-gap magnetic filed is considered, the d, q, 0 coordinates system leads to a simple system equation, which is easy to make theoretical analysis and understand the system rules, referring to [1,8,11].

3.7.1

Mathematical Model of AC-DC Generators and Loads

(1) Multi-loop model of ac-dc generators Under normal condition, every stator phase winding is considered as a unit, while all of the rotor field winding is regarded as a unit. The multi-loop method enables one to deal with damper winding according to its actual structure. While only considering the odd order harmonic field, the damper winding can be divided into two groups, d-axis and q-axis windings, whose axes are orthogonal 251

AC Machine Systems

to each other. There is no electrical and magnetic link between d and q axis loops, but mutual resistance and inductance within each group. Under fault condition, the stator single coil can be dealt as a unit. If the reference direction is set using generator convention for stator circuit and motor convention for rotor circuit, positive current excites positive flux linkage, and the positive directions of d-axis and q-axis flux linkage are consistent with d-axis and q-axis positive direction, the multi-loop equation of generators can be expresses as a matrix form:

UF

LF pI F ( pLF ) I F RF I F d , the current vector is dt

where the differential operator p iF

(3.7.1)

[i fd i1d " imd iA iB iC ia1 ib1 ic1 ia 2 ib 2 ic 2 ia 3 ib 3 ic 3 ia 4 ib 4 ic 4 ]T

where the subscript fd represents the field winding, 1d to md represent the damper winding, A, B, C represent 3-phase ac winding respectively, a1, b1, c1, a2, b2, c2, a3, b3, c3, a4, b4, c4 represent four suits of 3-phase dc windings respectively. The voltage vector in equation (3.7.1) UF

[u fd 0 " 0 u A u B uC ua1 ub1 uc1 ua 2 ub 2 uc 2 ua 3 ub 3 uc 3 ua 4 ub 4 uc 4 ]T

where the voltages of damper loops are zero because the damper loops are short circuited without external supplies. The resistance matrix in equation (3.7.1) is

RF

diag(rfd , Rd , Rac , Rdc )

where the subscription fd, d, ac, dc represent the field winding, damper winding, ac winding and dc winding respectively. The resistance matrix of damper winding is

Rd

ª r11d " r1md º « » # » « # ¬« rm1d " rmmd ¼»

The self resistance matrix of ac winding is

Rac

diag(rac , rac ,rac )

The self resistance matrix of dc winding is

Rdc 252

diag(rdc ,rdc ," ,rdc )


where rac is the phase resistance of ac winding, rdc is the phase resistance of dc winding. The mutual inductance matrix in equation (3.7.1) is

ª L fdfd «L « fdd « L fdac « ¬« L fddc

LF

L fdd Ldd Ldac Lddc

L fdac Ldac Lacac Lacdc

L fddc º Lddc »» Lacdc » » Ldcdc ¼»

The calculation method of the inductance parameters is presented in detail in Chapter 1. (2) Voltage equation of ac load for ac-dc generators The voltage equation of ac load is

U ac

Lac

dǿ ac Rac I ac dt

(3.7.2)

where U ac [ua1 , ub1 , uc1 ]T is the voltage vector, I ac [ia1 , ib1 , ic1 ]T is the current vector. For 3-phase symmetrical star-connected load, the inductance matrix is Lac diag( Ll , Ll , Ll ) and the resistance matrix is Rac diag( Rl , Rl , Rl ). Ll and Rl are the inductance and resistance of ac load respectively. The delta-connected load can be equivalent to star-connected load. For the 3-phase asymmetrical load, the inductance or resistance of 3-phases can be set as different values. (3) Voltage equation of dc load for ac-dc generators The voltage equation of dc load is

udc Edc

Ldc

didc Rdc idc dt

(3.7.3)

where Ldc , Rdc , Edc are the inductance, resistance and counter emf of dc load respectively.

3.7.2

System Equation of ac-dc generator

Combining the ac-dc generator equation (3.7.1) with the load equations (3.7.2) and (3.7.3), a system circuit equation is gotten as follows:

U

LpI ( pL) I RI

(3.7.4)

where U

[U FT

U acT

U dc ],

I

[ I FT

I acT

I dc ]T 253

AC Machine Systems

ª LF L «« 0 «¬ 0 ª RF R «« 0 «¬ 0

0 Lac 0

0 º 0 »» Ldc »¼

0 Rac 0

0 º 0 »» Rdc »¼

It must be noted that equation (3.7.4) does not represent the actual loops and can not be solved. Under normal condition, the generator and loads must be connected with each other to form the actual loops which alter with the operating load condition of the ac and dc windings and the state of rectifiers. The connection can be represented by a connection transformation matrix T in which each row corresponds to an actual loop and each column to a component. If the component and loop are related to each other and their reference directions are the same, the element in the matrix is 1. If the component and loop are related and their reference directions are opposite, the element is –1. If they are not related, the element is 0.

Figure 3.7.2 The connection of ac winding and star-connected load

The connection of the ac winding and load is easy to understand. Figure 3.7.2 shows the connection between 3-phase star-connected ac winding and the star-connected load. There are two independent loops formed between them, which are represented as the 9th and 10th rows of the connection transformation matrix in Fig. 3.7.3. Every row of matrix is corresponding to one actual loop, and the total row number is the number of independent loops of system. In Fig. 3.7.3 the zero’s are not filled. The 9th row is corresponding to actual loop 1 composed of phase A and B ac windings and phase A and B loads, where the reference direction of B phase current is opposite with that of loop current. Similarly, the 10th row is corresponding to actual loop 2. The connection transformation matrix can be established similarly, if the ac winding is abnormally connected. The former 8 rows of the connection transformation matrix in Fig. 3.7.3, represent one field loop and 7 damper loops respectively. 254


Figure 3.7.3 Example of connection transformation matrix

The connection between the dc winding and dc load is more complicated because it is determined by the working state of rectifiers. While the state of rectifiers changes, the connection transformation matrix must be modified. Figure 3.7.4 shows the connection of dc windings, rectifier bridge and dc load at a certain instant, where the shaded diode is in conduction state and the diode without shade is switched off.

Figure 3.7.4 Connection of dc winding and dc load

As shown in Fig. 3.7.4, for the first 3-phase bridge rectifier there are three diodes in conduction state and the rectifier is in commutation state, so two independent loops are formed between dc windings, rectifier bridge and load. The loop 1 is composed of winding a1, diode 3 of the first rectifier bridge, dc load (E, L and R), diode 2 of the first rectifier bridge and winding c1, which is represented by the 11th row of connection transformation matrix in Fig. 3.7.3. The column corresponding to wingding c1 is –1 because the reference direction of loop is opposite to that of wingding c1. The loop 2 is composed of winding b1, 255

AC Machine Systems

diode 1 of the first rectifier bridge, dc load (E, L and R), diode 2 of the first rectifier bridge and winding c1, which is represented by the 12th row of connection transformation matrix in Fig. 3.7.3. There are two diodes in conduction state for the 2nd, 3rd , 4th rectifier bridge, and these bridges are in conduction state, so only one independent loop is formed between dc windings, rectifier bridge and load. For the 2nd rectifier bridge, the loop is composed of winding a2, diode 3 of the 2nd rectifier bridge, dc load (E, L and R), diode 2 of the 2nd rectifier bridge and winding c2, which is represented by the 13th row of connection transformation matrix in Fig. 3.7.3. The switching of diode is determined by the voltage and current of the diode. The simulation program can establish the connection transformation matrix automatically in accordance with the states of diodes. After founding the connection transformation matrix according to the relation between ac and dc windings and their loads, the voltages and currents can be transformed as U c TU I TTIc

Hence, the system equation is

Uc

LcpI c ( pLc) I c RcI c

where Lc TLT T , Rc TRT T

Therefore, the standard state equation can be expressed as pI c

AI c B

(3.7.5)

where A Lc 1( pLc Rc), B

3.7.3

Lc 1U c

Model of Rectifier Bridge

Generally, the diode of rectifier bridge is considered as ideal component and the conduction voltage drop is neglected in simulation. But it would lead to the inconsistence between simulation and experiment under certain load condition, even the divergence would happen. For example, the simulation results of 3phase rectifier bridge under short circuit state at dc side as shown in Fig. 3.7.5 show that the conduction voltage drop of diode must be considered to obtain correct result. If the short circuit occurs at dc side, after considering the conduction 256


voltage drop of diode, because the sum of voltage across the two diodes in the same arm is zero, the two diodes on the same arm of bridge would not conduct at the same time, which is different from the equivalent 3-phase short circuit when neglecting the conduction voltage drop. Figure 3.7.6 gives the measured waveform of voltages across a diode of 3-phase rectifier bridge short-circuited at dc side. It can be seen that each diode is not in conducting state for all time, and when the voltage between two terminals of a diode is negative, the diode is shut off. For the sake of this, the voltage drop is counted in and set to +0.7V.

Figure 3.7.5 3-phase rectifier bridge shorted at dc side

Figure 3.7.6 Voltage across each diode in the same arm of rectifier bridge during short circuit of dc side

The correct calculation of the commutation of diodes is the key to the simulation of all the ac-dc generator system. In the ac-dc generator, each one of the four 3-phase dc windings is connected to one 3-phase rectifier bridge independently and then they are in parallel to supply dc power at the load side. The commutation is much more complicated than general 3-phase rectifier bridge, which contains not only commutation among the diodes of single 3-phase bridge, but also that among various bridges. Once the working state of diodes alters, the topology of circuit varies accordingly, and the state variables vary too. Previous works suggested dividing all the possible state of rectifier into several groups, each of which can be expressed as an equation, called mode classifying 257

AC Machine Systems

method. Now in this system the number of diodes reaches 24, while the number of modes reaches 192, which is much more complicated than that of general 3-phase bridge. Using mode classifying method, the program must be rewritten when the number of rectifier bridges increases or decreases. In this part, the connection transformation is applied to represent the topology, which results in that only the elements of the connection transformation matrix need to be changed with topology. Because the four neutral points of 3-phase dc windings are independent of each other, the switch on of diode is determined by the line voltage and dc voltage. The diode is conducted while the line voltage is larger than or equal to the dc load voltage and voltage drop of rectifier, otherwise it is in shut-off. The time of shut-off is determined by the current of the diode, that is, the diode shutsoff when the current changes from positive to negative. In the simulation program, an array is used to record the state of diodes. Firstly, the maximum line voltage is found among the 24 line voltage of four rectifier bridges and the order number of two diode are recorded too where the maximum voltage is applied. When the maximum line voltage is larger than the sum of dc voltage and twice the voltage drop of diode, then the two diodes are set to conduction. Any one diode would be switched off once its current is less than zero.

3.7.4

Model of Automatic Voltage Regulator (AVR)

To establish the precise model of AVR is complicated. But for the analysis of ac voltage waveform under steady state, only the steady state excitation current must be calculated for specified load condition. Furthermore, only the steady-state regulation precision is concerned, while the dynamic process is not considered. So a first order PI model is applied to represent the AVR as follows: ǻu U ~ U * ǻu fd

½° ¾ K P ǻu K I ³ ǻudt °¿

(3.7.6)

where ǻu fd is the variation of excitation voltage, K P , K I are the proportional and integral coefficients of PI regulator respectively, U * , U ~ and 'u are the reference value, feedback value, and their error respectively. The feedback rms value is defined as U ~ (t0 )

1 t0 2 u (t )dt T ³ t0 T

1 ¦ u j 2 (t )'t T j

(3.7.7)

where T , 't are the period of main frequency and the sampling interval of voltage respectively, u (t ) is the instantaneous value of line voltage. 258


In order to calculate the rms value of line voltage, various instantaneous values of line voltage during the whole period ahead of the calculation instant t0 are needed. Because the variable-step Runge-Kutta integration method is used, the number of calculation intervals within one period of main frequency T varies, which leads to the complexity of saving the instantaneous values of line voltage and calculating the rms value. In simulation program the step is adjusted to calculate the line voltage in fixed step, and an array is used to record the line voltage, which would be updated during the simulation. The simulation results of excitation regulating process show that the proper coefficient of PI regulator can achieve high steady-state precision for this simple PI regulator, i.e. the steady-state error between the reference and feedback values would be less than 5‰ under rated load condition. The model of AVR is used to calculate the field current for any specified ac voltage and any load condition, which is important to the analysis of steady-state performance.

3.7.5

Motion Equation of Rotor and Speed Regulators

When studying some transient process of ac-dc generators, the variation of rotation speed must be considered, therefore the motion equation of rotor should be solved. The electromagnetic torque of generator is Te

P wL T i i 2 wJ

(3.7.8)

where i is the current vector, and L is the inductance matrix. The motion equation of rotor is d 2J dt 2

dZ dt

Tm Te H

(3.7.9)

where J is the rotor position angle, Z is the rotation speed, Tm is the mechanical torque of prime mover, H is the inertia constant. The mathematic model of speed PI regulator is 'Z 'Tm

Z Z *

½° ¾ K P 'Z K I ³ 'Z dt °¿

(3.7.10)

where 'Tm is the variation of mechanical torque of prime mover, K P , K I are the proportional and integral regulation coefficients of PI speed regulator respectively, Z *, Z and 'Z are the reference value, feedback value of speed and their error respectively. 259

AC Machine Systems

3.7.6

Simulation Procedure of AC-DC Generator System

In order that the simulation of a long time process must be finished without letup, the record file is used to record the state of the last step. In the record file, all the key information must be logged, including the loop currents, rotor position, rotation speed, the working state of all the diodes, the integral values of excitation regulator and speed regulator, and the samples of instantaneous line voltage of ac side during the latest period, which are used to calculate the rms value of line voltage of ac side. The flow chart of simulation program is shown in Fig. 3.7.7.

Figure 3.7.7 Flow chart of simulation of ac-dc generator systems

At the beginning of every simulation, the latest state must be read from the record file, while at the end of existing simulation, the record file must be updated with the new state. So the simulation can be continued from the latest step and need not simulate from the original state, by which the simulation time can be spared greatly, especially when the steady state must be reached. When the simulation is executed for the first time, the no load condition can be used to write the record file, where all of the loop currents are set as zero, except that the field current is set as the ratio of field voltage to the resistance of field winding. The Runge-Kutta integration method with variable step is applied to solve the state equation. In order to find the exact commutation instant, the step must be adjusted. Once the state of diode varies in next step, the existing step length is shortened to a half until that the step reaches the minimal one or no commutation occurs. Because of the variable step simulation, in order to plot the calculated results easily and use FFT to analyze the waveform, the simulation step must be automatically adjusted to make the output of results in fixed step. 260


3.7.7

Verification of Simulation Program

The calculated results are compared in detail with those of experiments on a model ac-dc generator as follows, in which the dimensions and data of the model ac-dc generator are shown in Table 3.7.1. Table 3.7.1 Dimensions and data model ac-dc generator Diameter of iron core Rated frequency Core length Pole arc factor Damper bar number per pole Rated power Rated voltage Rated current AC winding Power factor Winding pitch Series turns per phase

250mm 50Hz 152mm 0.72 7 15kW 390V 27.8A 0.8 10 72

Pole-pair number Stator slot number Maximum airgap length Minimum airgap length Skew of stator core Rated power Rated voltage Rated current DC winding Winding pitch Series turns per phase

2 48 1.5mm 1mm one slot pitch 3.3kW 510V 6.5A 10 64

Owing to a limited space, only one set of comparison between calculated and measured results is given below. According to the Shannon’s sampling theorem, in order to sample correctly the sinusoidal waveform whose frequency is equal to nf 0 , the sampling frequency must be greater than 2nf 0 . The lowest frequency of the high frequency components in the output dc voltage is 24 times the main frequency 50Hz, so the sampling frequency must be greater than 48 times the main frequency 50Hz, which means that the sampling period, also the maximum simulation step length, must be less than 4.17 u 104 s . The maximum simulation step length is 100ȝs , and the minimum is 0.1ȝs , which can achieve a compromise between the simulation speed and accuracy. The minimum simulation step determines the accuracy of searching the commutation instant of diodes, so it would not be too long. If the pure resistant or nearly pure resistant load is connected to the ac winding or dc winding, the variation rate of loop current would be great; while if the pure inductive or nearly pure inductive load is connected, the variation rate of voltage would be great. The step of Runge-Kutta integration method with variable step is rather sensitive to these two kinds of loads. So for these two kinds of loads, the maximum simulation step length must be decreased to improve the accuracy, which is generally less than 10ȝs . The comparisons of line voltages between the calculation and experiment results for ac winding and dc winding, while the field current I fd is 1.3A, the load current of ac winding I ac is 7.8A, power factor is 0.8, the line voltage of ac 261

AC Machine Systems

winding U ac is 180V, and the dc winding is no load, are shown in Figs. 3.7.8 and 3.7.9 respectively.

Figure 3.7.8 Calculated and measured waveforms of line voltage of ac winding (a) Calculated; (b) Measured

Figure 3.7.9 Calculated and measured waveforms of line voltage of dc winding (a) Calculated; (b) Measured

The colculated and measured harmonic components and the voltage harmonic distortion in percentage of line voltage of ac winding are compared with each other in Table 3.7.2, while those of line voltage of dc winding are shown in Table 3.7.3. The percentage of harmonic component is the ratio of amplitude of harmonic component to fundamental component. It can be seen that the results of calculation and experiment are consistent with each other. 262

3 Some Special Operation Modes of Synchronous Machine Systems Table 3.7.2 Comparison between the calculated and measured harmonic components of line voltage of ac winding Harmonic order Calculated harmonic components /ă

5 6.18

7 4.43

11 6.10

13 4.56

17 1.45

19 0.49

23 3.45

25 0.31

VHD 11.4

Measured harmonic components /ă

4.2

4.2

3.7

1.9

0.6

0.8

5.9

1.2

11.0

Table 3.7.3 Comparison between the calculated and measured harmonic components of line voltage of dc winding Harmonic order Calculated harmonic components /%

5 10.6

7 10.8

11 3.43

13 2.66

17 1.85

19 1.58

23 0.28

25 0.70

VHD 16.1

Measured harmonic components /%

9.36

11.7

2.87

2.39

2.06

1.15

0.66

0.10

15.6

More simulation results of various conditions show that when the load current is small (below half of the rated load), the good consistence is achieved between the calculated and measured results. But with the increase of load currents, the errors between calculated and measured results also increase. The errors are caused by the assumption that the multi-loop model is linear model, and the saturation of magnetic circuit is dealt with by use of the equivalent enlarged air-gap. The enlargement of air-gap is according to the fundamental magnetic field, so when the magnetic circuit is saturated, the harmonic magnetic field can not be analyzed accurately, which leads to the larger error. From the comparison, it can be seen that the calculated harmonic components are mostly greater than those of experiments, which means the calculation results are more rigorous and make the design more reliable. The accurate analysis of the effect of saturation on the harmonic field must be simulated by combination of the multi-loop model and the finite element method. From the comparison between the simulation and experiment, it can be seen that the calculation of voltage harmonics under steady state is accurate enough. Furthermore, the design optimization of ac-dc generators for different dimensions and load conditions can be conducted.

3.7.8

Simulation Research of Voltage Harmonics of AC-DC Generators

In applications of the military, aerospace, and communication, the high quality of power supply is required in order to make the high sensitivity equipment avoid the interference. The relations between the VHD of ac winding and various factors, especially the structure dimensions, are analyzed as below. 263

AC Machine Systems

(1) Influence of load current on the harmonic component and VHD The load current has great effect on the VHD of line voltage of ac winding. So the influence of currents at ac side and dc side are calculated, and six working conditions are chosen as follows: Condition 1: no load at ac side, and no load at dc side. Condition 2: no load at ac side, rated load at dc side (6.5A, pure resistance). Condition 3: half-load at ac side (14A, power factor 0.8), no load at dc side. Condition 4: half-load at ac side (14A, power factor 0.8), rated load at dc side (6.5A, pure resistance). Condition 5: rated load at ac side (28A, power factor 0.8), no load at dc side. Condition 6: rated load at ac side (28A, power factor 0.8), rated load at dc side (6.5A, pure resistance). The calculated results of harmonic components and VHD of line voltage of ac winding are shown in Table 3.7.4 and Fig. 3.7.10. Table 3.7.4 Relation of the harmonic components and VHD of line voltage of ac winding with the working conditions unit: ă Harmonic order 5 7 11 13 17 19 23 25 VHD

1 1.76 2.52 1.39h10–5 2.02h10–5 8.29h10–7 8.63h10–6 5.83h10–6 8.64h10–6 3.07

2 3.04 1.85 1.04 0.487 1.07 0.204 0.332 2.29 4.73

Working condition 3 4 2.65 2.97 3.16 3.18 1.66h10–2 1.86 6.70h10–3 0.382 1.33h10–4 0.954 –4 1.45h10 0.609 1.19h10–4 13.8 9.94h10–5 10.6 4.12 18.3

5 3.70 3.91 3.34h10–2 1.27h10–2 2.33h10–5 3.11h10–5 2.26h10–5 1.88h10–5 5.39

6 2.33 6.94 1.94 1.44 2.69 0.795 19.6 15.6 26.7

Figure 3.7.10 Relation chart between VHD of line voltage of ac winding and its working conditions 264


It can be seen that the VHD of ac windings increases with the increase of load currents at ac side and dc side. For the working condition 1, 3 and 5, the dc side is no load, and the VHD of ac winding increases slightly with the load current of ac winding. For the working condition 2, 4 and 6, the dc side is rated load, and the VHD of ac winding increases more rapidly with the load current of ac winding than the working condition 1, 3 and 5. Comparing condition 2 with 1, condition 4 with 3, and condition 6 with 5, it can be seen that when the dc load current increases, the VHD of ac winding increases greatly. It is clear that the load current of dc winding would cause greater distortion in ac voltage of ac winding because of the frequent commutation. (2) Influence of dc winding pitch on the harmonic components and VHD In the ac-dc generator, the commutation of rectifier bridge makes the ac voltage distort through the mutual inductance coupling. Although there is no restriction of VHD of line voltage of dc winding, the reasonable design of dc winding would improve the VHD of ac winding. So here the influence of dc winding pitch on the VHD of ac winding is studied. The harmonic components of ac winding are calculated and the results are shown in Table 3.7.5, while the pitch of dc winding varies from 2/3 1. Table 3.7.5 Relation of the harmonic components and VHD of line voltage of ac winding with the pitch of dc winding unit: ă Harmonic order 5 7 11 13 17 19 23 25 VHD

2/3 2.33 6.94 1.94 1.44 2.69 0.795 19.6 15.6 26.7

3/4 2.71 6.21 1.36 0.993 1.81 0.236 13.2 15.1 21.6

Short pitch ratio 5/6 4.42 2.62 1.82 0.510 0.868 1.01 8.00 9.38 13.7

11/12 2.07 5.33 1.06 0.508 1.74 1.11 7.79 4.96 11.4

1 2.71 3.93 1.34 0.521 1.90 1.26 12.5 9.64 16.9

The relation between the VHD and the pitch of dc winding is drawn as Fig. 3.7.11. The VHD of ac winding reaches the minimum while the pitch is 11. It can be concluded that a suitable pitch of dc winding can effectively make the VHD of ac winding decrease. The influence of the relative position of ac and dc windings in stator slot, the relative spatial position of ac and dc windings, the number of damper bars per pole and the open width of damper slot, on the harmonic components and VHD are also studied. But the results show that their influence is not too much. The model and simulation method established in this section can be used not only to analyze the steady state performance like the harmonic distortion of ac 265

AC Machine Systems

Figure 3.7.11 Relation curve of VHD of line voltage for ac winding with the pitch of dc winding

voltage, but also simulate the transient process, such as the low frequency power oscillation and the sudden short circuits occurring at ac side or dc side.

References [1] Franklin P W (1972) Theory of the 3-phase salient pole type generator with bridge rectifier output. J IEEE Trans. PAS-91(5): Part I, 1960 1967, Part Ċ, 1968 1975 [2] Gao J D (1963) AC machine transients and operating modes analysis (Chapter 2, 3, 4, 7, in Chinese). Science Press, Beijing [3] Gao J D, Zhang L Z (1982) Fundamental theory and analysis mothods of transients in electric machines, Vol 1 (Chapter 6 and Appendix 2, in Chinese). Science Press, Beijing [4] Gao J D, Wang X H, Li F H (2004) Analysis of ac machines and their systems,2nd Ed (Chapter 3 and Section 4.7, in Chinese). Tsinghua University Press, Beijing [5] Goodman E D (1984) Troque, voltage and current harmonics of shychronous machines. J IEEE Trans. IA-20(1): 209 215 [6] Kirschbaum H S(1945) Transient electrical torque of turbine generators during short circuits and synchronizing. J AIEE Trans. Feb. 64: 65 70 [7] Liu T, Wang X H (2000) Simulation of dynamic performance for cycloconverter-fed synchronous motor. Proc IPEMC’2000, Beijing, pp 676 679 [8] Wang S M, Wang X H, Li Y X, Su P S (2002) Steady-state performance of synchronous generators with ac and dc connections considering saturation. IEEE Trans. on Energy Conversion 17: 176 182 [9] Wong M C, Zhang L Z (1999) Analysis of asymmetric systems with harmonics and their compensation. Proc CICEM’99, International Academic Publishers, Xi’an, pp 946 949 [10] Zhang L Z, Chen L F (1991) Analysis of subsynchronous resonance in a turbogenerator and its shstem using compensation capacitors. Proc CICEM, Huazhong University of Science and Technology Press, Wuhan, pp 104 108

266

4 Oscillation, Stability and Excitation Regulation of Synchronous Machine Systems

Abstract The definition of static stability for synchronous machines is not unified in some references. Static stability is to discuss a problem whether the synchronous machine can restore its original steady-state mode or not when undergoing as small perturbation and then abolishing it, but the rate and fashion of small perturbation are not prescribed. If the small perturbation is caused by small and slow increase of loads, the excitation current has not any change and only the rotor position angle reaches a new balance position, which is the first possible case. The second possible case is that the increase rate of small loads is relatively fast to bring about change of excitation current even if having no excitation regulator. Therefore, the static stability can be divided into steady-state static stability and dynamic-state static stability in some references. By static stability refer to dynamic-state static stability in this chapter even if having no excitation regulator. Due to small perturbation, the corresponding basic equations can be linearized and small oscillation of synchronous machine systems is an example of linearization, in which according to the linearized basic equations get synchronization torque coefficient and damping torque coefficient to find out natural oscillation frequency, decay time-constant, forced oscillation produced by mechanical torque harmonics, resonant curves and influence of stator resistance on the oscillation, and then analyse the oscillation causes from physical concept in detail. Moreover, use Herwitz’s criteria to get the prerequisites of static stability for synchronous machine systems, utilize Routh’s criterion to find out the corresponding stable regions for generator operation and motor one, and then analyse the influence of excitation regulation on static stability. The dynamic stability problem of synchronous machine systems is to study whether the synchronous machine can operate synchronously after undergoing a certain or large disturbance, inclusive of basic equations corresponding to dynamic stability analysis, the influence of ferromagnetic saturation on parameters, corresponding mathematical patterns and calculation example. Through calculation it is clear to cut off the fault line in time to increase dynamic stability, but the area rule and graphic method provide only the cut-off angle G OTK , so the dividing intervals calculation should be used together with the area rule to search for the cut-off time limit tOTK.

AC Machine Systems

This chapter includes small perturbation and linearization of ac machine systems, steady-state small oscillation and torque coefficients of synchronous machine systems, static stability of synchronous machine systems and influence of excitation regulation on static stability, and dynamic stability and analysis methods of synchronous machine systems.

4.1

Small perturbation and Linearization of AC Machine Systems

In Chapter 2 and Chapter 3 we have discussed some problems of ac machines at synchronous speed or constant speed, in which the basic equations are linear ones with constant or variable coefficients, so it is not necessary to consider rotor motion. However, in practical terms the rotor motion must be counted in for some operating conditions of ac machines, where the electromagnetic transients and mechanical process have to be considered simultaneously and both influence each other. For example, the prime mover of synchronous generator is Dieselengine without uniform torque or a synchronous motor drives air-compressor, for which the electromagnetic transients and mechanical process have to be evaluated simultaneously to get correct analysis results. During a synchronous machine in parallel operation with network, the machine has to operate at synchronous speed strictly according to network frequency. However, due to some causes the small oscillation of synchronous machine rotor speed may occur, which will be studied in Section 4.2. In addition, the synchronous machine and its connected network dispatch loads or encounter some accidents to bring about a question whether they can still keep synchronous operation or not, namely the stability problem of synchronous machine system. The stability problem is a significant and interesting problem, because the main generator in a power system is out of step to make stop electricity of large area and to cause confusion. In order to answer that question above, the effect of speed change has to be taken into account for synchronous machines. For convenience, the stability problem of synchronous machines can be divided into two types, namely static stability and dynamic stability, referring to [1,2,4,6]. Static stability is to discuss a problem whether the synchronous machine can keep synchronous operation under a certain steady-state operating mode. If the machine can’t restore its original state after undergoing a small perturbation under that steady-state operating mode, it can be considered to be unstable even though the machine can still operate. On the contrary, if the machine can restore its original steady-state operating mode after sustaining a small perturbation, it can be considered to be stable under that steady-state operating mode. Under the definition of static stability mentioned above, nonlinear basic equations of 268


synchronous machines can be linearized to simplify the problem whether the machine can be stable or not. The dynamic stability problem is to discuss whether a synchronous machine can keep synchronous operation or not when undergoing a large disturbance such as change of parameters, dispatching loads or accidents. During study of the problem above, the basic equations of synchronous machines can’t be linearized any longer, so the problem will be studied only by numerical method. Generally speaking, dynamic stability limit is lower than static stability one, but a synchronous machine can’t operate normally if exceeding the static stability limit while it may operate normally if exceeding the dynamic stability limit because the accidents can’t often occur and are not so severe as calculation of dynamic stability limit even though the accidents happen. Of course, the stability limits of two types are all not permitted to be exceeded during design of power systems, but here illustrate the difference between them only. It should be pointed out, that the definition of static stability for synchronous machines is not unified in some references and bibliography. As mentioned before, static stability is to discuss a problem whether the synchronous machine can restore its original steady-state operating mode or not when undergoing a small perturbation and then abolishing it, but the rate and fashion of small perturbation are not prescribed. If the small perturbation is caused by small increase of loads and the increase rate of machine loads is very slow, the excitation current has not any change and only the rotor position angle, namely power angle, reaches a new balance position at a very low speed, which is the first possible case. The second possible case is that the increase rate of small loads is relatively fast to bring about change of excitation current due to change of rotor position angle even if the synchronous machine has no excitation regulation device. Of course, with excitation regulation device exists more change of excitation current. In the second possible case, the static stability limit is higher than the first possible case, namely constant excitation current. Therefore, in accordance with the cases above, the static stability can be divided into steady-state static stability and dynamic-state static stability in some references. There are excitation regulation devices in modern generators, so analysis of static stability has to consider influence of excitation current change on static stability. Therefore, by static stability refer to dynamic-state static stability in this book even if there is no excitation regulation device. For combination system of synchronous machine and capacitances as discussed in Chapter 3, when the rotor inertia is infinite the influence of speed change on self-excitation can be neglected, i.e. the self-excitation is considered as an electromagnetic parameter resonance. If the rotor inertia is relatively small, the effect of rotor speed change must be taken into account, i.e. the self-excitation is an electromechanical parameter resonance. Therefore, during analysis of the 269

AC Machine Systems

conditions to cause self-excitation, the influence of rotor motion should be counted in and the small perturbation and Linearization theory can be used to analyse the self-excitation phenomenon. In brief, no matter what problem such as small oscillation, static stability or self excitation it is, the basic equations of synchronous machines have been nonlinear. However, those problems have a common feature, namely study of the phenomena for synchronous machines after undergoing a small perturbation, so the behaviors after sustaining a small perturbation depend mainly upon the initial state and the corresponding basic equations can be linearized. By linearization refer to the following way: when a synchronous machine undergoes a small perturbation, the variable x can be written as ( x0 'x), in which x0 is the initial state before perturbation and 'x is an increment caused by perturbation. Substituting those variables into nonlinear basic equations and neglecting the second-and higher-order increments, the linearized equations with variable 'x can be got. Obviously, it is convenient to use the linearized equations to research into several problems such as small oscillation, static stability, self-excitation phenomenon, etc.

4.2

Steady-State Small Oscillation and Torque Coefficients of Synchronous Machine Systems

In practical operation of synchronous machines there may be an oscillation due to some adverse causes. During oscillation, the speed, current, voltage, power and torque are all not constants, and possibly reach the hazardous values. Therefore, the oscillation has an adverse effect on the machine itself and power system connected with it, which sometimes brings about users not to be capable of work, e.g. the electric lights blink and the motor speeds are unstable, which possibly makes the machine out of synchronism and causes the electric service to be interrupted, or even makes the machine and apparatus damaged. Thus, it is significant to understand the oscillation phenomenon of synchronous machine systems and to take some measures, for example fitting a damping winding, increasing the machine-set inertia and installing protection equipment. There are many causes to stimulate oscillations, e.g. the design and regulation of controllers are not suitable, the prime-mover or load has a pulsating torque and the outer load is unstable or asymmetrical, etc. On the basis of the different causes producing oscillations, they can be divided into the free oscillation and forced oscillation. During an oscillation as mentioned before, the basic equations of a synchronous machine are nonlinear, because its speed is not constant. However, here we only study the small oscillation in steady-state of synchronous machine systems, so the 270


basic equations can be linearized according to the small perturbation theory referring to [3,4,6]. Under this condition, the angle G between the rotor axis and synchronous axis changes periodically around a certain value G 0 . Let us suppose this function to be

G

G0 D

G 0 D m sin O t

(4.2.1)

in which

D D m sin O t D m is the amplitude of the small oscillation and O is the angular velocity of the machine oscillation whose value lies in the range from zero to 0.05, and 0.02 corresponds to an oscillation period per second. In practical operation of a synchronous machine, its oscillation is more complicated than mentioned above, but the oscillation, in virtue of its periodic nature, can always be analysed into corresponding harmonics, so the supposed function is still suitable as far as a certain harmonic is concerned. Because

J

t G J 0c

t J 0c G 0 D m sin O t

(4.2.2)

we can obtain

Z

dJ dt

1

dD dt

1 OD m cos O t 1 s

(4.2.3)

dD OD m cos O t is the instantaneous slip of a machine. dt If there is no oscillation in a machine, the corresponding relations, by virtue of Park’s equations, can be gotten as follows:

in which s

ud 0 uq 0

\d0 \ q0 Te 0

U sin G 0

\ q 0 rid 0 ½ ° U cos G 0 \ d 0 riq 0 ° ° E xd id 0 ¾ ° xq iq 0 ° ° \ d 0iq 0 \ q 0 id 0 ¿

(4.2.4)

The steady-state values above, due to the small oscillation, will change a little with time. Representing these small changes by 'ud , 'uq , '\ d , '\ q , 'id , 'iq and 'Te , we can write the corresponding values ud , uq , \ d , \ q , id , iq and Te as

271

AC Machine Systems

ud uq

\d \q id iq Te

ud 0 'ud ½ ° u q 0 'u q ° \ d 0 '\ d ° °° \ q 0 '\ q ¾ ° id 0 'id ° ° iq 0 'iq ° Te 0 'Te °¿

(4.2.5)

The machine having no excitation-regulation, we have U fd

constant,

G ( p )U fd

E

Substituting the values above into Park’s formulas (2.4.1), (2.4.11) and (2.4.15) and torque equation (2.4.22), we can obtain p (\ d 0 '\ d ) (\ q 0 '\ q )(1 s ) r (id 0 'id ) ½ ° p'\ d \ q 0 s\ q 0 '\ q s'\ q rid 0 r 'id ° uq 0 'uq p'\ q \ d 0 s\ d 0 '\ d s'\ d riq 0 r 'iq ° ° °° \ d 0 '\ d E xd ( p)(id 0 'id ) E xd id 0 xd ( p)'id ¾ \ q 0 '\ q xq iq 0 xq ( p)'iq ° Te 0 'Te (\ d 0 '\ d )(iq 0 'iq ) (\ q 0 '\ q )(id 0 'id ) ° ° ° \ d 0 iq 0 \ d 0 'iq '\ d iq 0 '\ d 'iq \ q 0id 0 ° \ q 0 'id '\ q id 0 '\ q 'id °¿ u d 0 'u d

(4.2.6)

Subtracting equation (4.2.4) from the above equations, we have 'ud 'uq

p'\ d s\ q 0 '\ q s'\ q r 'id ½° ¾ p' \ q s\ d 0 '\ d s'\ d r 'iq °¿ '\ d '\ q

xd ( p )'id ½° ¾ xq ( p )'iq °¿

'Te \ d 0 'iq '\ d iq 0 '\ d 'iq \ q 0 'id '\ q id 0 '\ q 'id The machine connected to an infinite source, we can get

272

(4.2.7)

(4.2.8) (4.2.9)


½ ° U sin G U sin(G 0 D ) ° ° U sin G 0 cos D U cos G 0 sin D ¾ U cos G 0 'uq U cos G U cos(G 0 D ) ° ° U cos G 0 cos D U sin G 0 sin D ¿°

U constant ud0 'ud U sin G 0 'ud uq 0 'uq

Because D is small, we can approximately take cosD the above equation and formula (4.2.4), we can obtain 'ud

U D cos G 0

'uq

U D sin G 0

1, sin D

D uq 0

½° ¾ D ud 0 °¿

(4.2.10)

D . Thus, from

(4.2.11)

Since the product of two small increments can be neglected as compared with a small increment (s is also a small value), equations (4.2.7) and (4.2.9) can be reduced to 'ud

D uq 0

'uq

D ud 0

'Te

p'\ d s\ q 0 '\ q r 'id ½° ¾ p'\ q s\ d 0 '\ d r 'iq °¿

\ d 0 'iq '\ d iq 0 \ q 0 'id '\ q id 0

(4.2.12) (4.2.13)

Referring to equations (4.2.1) and (4.2.3) and substituting the time functions of s and D and the equation (4.2.8) into formula (4.2.12), we have

D m uq 0 sin O t O\ q 0D m cos O t

½ ° pxd ( p )'id xq ( p )'iq r 'id ° ¾ D m ud 0 sin O t O\ d 0D m cos O t ° pxq ( p)'iq xd ( p)'id r 'iq °¿

(4.2.14)

which is a group of operational formulas with constant coefficients, in which 'id and 'iq seem to be produced by some emf that changes sinusoidally at an angular velocity O. Therefore, its steady-state value is also a sinusoidal value with an angular velocity O and we can use the Complex-number Vector Method to solve it. Let the complex-number time-vectors of 'id and 'iq be Id D m and IqD m respectively. According to equation (4.2.14) we have

uq 0 jO\ q 0 ud 0 jO\ d 0

jO xd ( jO ) Id xq ( jO ) Iq rId ½° ¾ jO xq ( jO ) Iq xd ( jO ) Id rIq °¿

(4.2.15)

273

AC Machine Systems

After solving it, we can get

Id Iq

[r jO xq ( jO )](uq 0 jO\ q 0 ) xq ( jO )(ud 0 jO\ d 0 ) ½ ° r 2 jO r[ xd ( jO ) xq ( jO )] (l O 2 ) xd ( jO ) xq ( jO ) ° ¾ [r jO xd ( jO )](ud 0 jO\ d 0 ) xd ( jO )(uq 0 jO\ q 0 ) ° r 2 jO r[ xd ( jO ) xq ( jO )] (l O 2 ) xd ( jO ) xq ( jO ) °¿

(4.2.16)

Having Id and Iq , we can find \ d D m and \ qD m the complex-number time-vectors of '\ d and '\ q according to equation (4.2.8) as follows:

xd ( jO ) Id ½° ¾ xq ( jO ) Iq °¿

\ d \ q

(4.2.17)

Supposing Id I

q

I d jI dc I q jI qc

\ d \ d j\ dc \ q \ q j\ qc we can obtain 'id 'iq '\ d '\ q

D m I d sin O t D m I dc cos O t D m I q sin O t D m I qc cos O t

½ ° ° ¾ D m\ d sin O t D m\ dc cos O t ° D m\ q sin O t D m\ qc cos O t °¿

(4.2.18)

Substituting the above equation into equation (4.2.13), we have 'Te

in which Ts TD

(\ d iq 0 \ d 0 I q \ q id 0 \ q 0 I d )D m sin O t

½ ° (\ dc iq 0 \ d 0 I qc \ qc id 0 \ q 0 I dc )D m cos O t ° ° TsD TD s ° ¾ ° \ d iq 0 \ d 0 I q \ q id 0 \ q 0 I d ° ° 1 ° (\ dc iq 0 \ d 0 I qc \ qc id 0 \ q 0 I dc ) O ¿

(4.2.19)

From the formulas above we can see, 'Te the additional electromagnetic torque caused by the rotor oscillation consists of two components M sD and TD s. 274


The latter is proportional to slip and the first to the angle D which the rotor leaves its balanced position by. Therefore, the corresponding coefficients Ts and TD are called the synchronization torque coefficient and damping torque one respectively. Since 'Te also changes sinusoidally, it can be expressed in a complex-number vector. Letting the complex-number vector of 'Te be TD m and according to 'Te TsD TD s TsD TD pD , we have TD m

TsD m jOTDD m

or T

Ts jOTD

\ d 0 Iq \ d iq 0 \ q 0 Id \ q id 0

[iq 0 xd ( jO ) \ q 0 ]Id [\ d 0 id 0 xq ( jO )]Iq

(4.2.20)

so Ts and TD can be found according to Id and Iq . This calculation method is convenient and we shall use it later. (1) The synchronization torque and damping torque coefficients when stator resistance r can be neglected We have stated the concept and calculation method of those coefficients as before. However, those coefficients have rather complicated forms. In order to illustrate their concepts and analyse them conveniently, we shall at first discuss the synchronization torque and damping torque coefficients when stator resistance r can be neglected. The stator resistance r neglected, equation (4.2.16) can be simplified to Id Iq Because r

0, we have ud 0

ud 0 O 2\ q 0 jO (\ d 0 uq 0 ) ½ ° (1 O 2 ) xd ( jO ) ° ¾ uq 0 O 2\ d 0 jO (\ q 0 ud 0 ) ° ° (O 2 ) xq ( jO ) ¿

(4.2.21)

\ q 0 and uq 0 \ d 0 , and according to xd ( jO )

X d O jRd O

xq ( jO )

X qO jRqO

we can obtain

Id

1 ud 0 xd ( jO )

Iq

1 uq 0 xq ( jO )

X d O jRd O ½ ud 0 ° 2 2 X d O Rd O ° ¾ X qO jRqO uq 0 ° ° X q2O Rq2O ¿

(4.2.22)

275

AC Machine Systems

Substituting the above equation into formula (4.2.20), we can get T

Ts jOTD [iq 0 xd ( jO ) ud 0 ]

X qO jRqO X d O jRd O ud 0 [uq 0 id 0 xq ( jO )] 2 uq 0 2 2 X d O Rd O X qO Rq2O

X qO jRqO 2 X d O jRd O 2 ud 0 2 uq 0 2 2 X d O Rd O X qO Rq2O

(4.2.23)

º RqO U 2 ª Rd O sin 2 G 0 2 cos 2 G 0 » « 2 2 2 O ¬« X d O Rd O X qO RqO ¼»

(4.2.24)

uq 0 id 0 ud 0 iq 0 Thus, TD

uq 0 id 0 ud 0 iq 0

Ts

Q0

X qO U 2 X d OU 2 2 sin G cos 2 G 0 0 X d2O Rd2O X q2O Rq2O

X qO U 2 X d OU 2 2 sin G cos 2 G 0 0 X d2O Rd2O X q2O Rq2O

(4.2.25)

in which

Q0

uq 0 id 0 ud 0 iq 0

U (id 0 cos G 0 iq 0 sin G 0 )

U [ I 0 sin(G 0 M 0 ) cos G 0 I 0 cos(G 0 M 0 )sin G 0 ] UI 0 sin M 0 is the output reactive power before the oscillation occurs, in which I 0

id20 iq20

and M 0 are the corresponding steady-state current and its phase-angle. Because Q0

UI 0 sin M 0

EU U2 U2 cos G 0 cos 2 G 0 sin 2 G 0 xd xd xq

EU U2 1§ 1 1 · cos G 0 ¨ ¸ (1 cos 2G 0 ) xd xd 2 ¨© xq xd ¸¹ the formula (4.2.25) can be written as another form: Ts

X qO U 2 X U2 EU U2 U2 cos G 0 cos 2 G 0 sin 2 G 0 2 d O 2 sin 2 G 0 2 cos 2 G 0 xd xd xq X d O Rd O X qO Rq2O § 1 ª X EU 1· 1º cos G 0 ¨ ¸ U 2 (cos 2 G 0 sin 2 G 0 ) « 2 d O 2 » U 2 sin G 0 ¨ ¸ xd ¬ X d O Rd O xd ¼ © xq xd ¹

276


ª X qO 1º « 2 » U 2 cos 2 G 0 2 «¬ X qO RqO xq »¼ xd xq 2 ª X 1 EU cos G 0 U cos 2G 0 « 2 d O 2 xd xd xq ¬ X d O Rd O xd

º 2 2 » U sin G 0 ¼

ª X qO 1º « 2 » U 2 cos 2 G 0 2 «¬ X qO RqO xq »¼

(4.2.25a)

From statements above we can see, the synchronization torque coefficient EU cos G 0 is the fundamental component consists of four parts, in which the first xd

that depends upon the corresponding steady-state excitation emf and terminal xd xq 2 U cos 2G 0 is the salient-pole component. Those two voltage; the second xd xq components are independent of the rotor-circuit condition, which are the synchronization torque coefficient when the rotor oscillates at a very slow speed. Letting Ts 0 represent those two components, we have Ts 0

Ts |O

0

xd xq 2 EU cos G 0 U cos 2G 0 xd xd xq

The latter two components in formula (4.2.25a) depend upon the oscillation currents in the rotor d-axis and q-axis windings respectively. Letting TsO represent them, we have TsO

ª X dO 1 « 2 2 ¬ X d O Rd O xd

ª X qO º 2 2 1º » U 2 cos 2 G 0 » U sin G 0 « 2 2 «¬ X qO RqO xq »¼ ¼

Likewise, the damping torque coefficient consists of two components, which depend upon the oscillation currents in the rotor d-axis and q-axis windings respectively. Ts 0 and TsO varying with G 0 during O 0.03 in a typical machine are shown in Fig. 4.2.1. The damping torque coefficient M D varying with G 0 in the same machine is shown in Fig. 4.2.2. On the basis of those curves it is not difficult to understand that the q-axis damping winding has a significant effect. The TD and Ts expressions can also be rewritten as the following forms to use them conveniently.

277

AC Machine Systems

Figure 4.2.1 Synchronization torque coefficient components Ts 0 -G 0 and TsO - G 0 curves during O

0.03

Figure 4.2.2

Damping torque coefficient TD - G 0 curve

(a) Without damping winding 1 xd ( jO )

X dO R j 2 dO 2 2 X Rd O X d O Rd O 2 dO

1 jOTd 0 xd jO xdc Td 0

xd O 2 xdc Td20 jOTd 0 ( xd xdc ) xd2 O 2 xdc2Td20

278

(4.2.25b)


i.e. xd xdc O 2Td20 xd2 O 2 xdc2Td20 OTd 0 ( xd xdc ) xd2 O 2 xdc2Td20

X dO 2 X d O Rd2O Rd O X Rd2O 2 dO

In addition, X qO

1 xq ( jO )

X

2 qO

R

X

2 qO

X

2 qO

j

2 qO

RqO X

2 qO

Rq2O

1 xq

i.e. X qO Rq2O

RqO Rq2O

1 xq 0

Substituting those values into equations (4.2.24) and (4.2.25a), we can get the damping torque and synchronization torque coefficients of a synchronous machine without damping winding and supposing the stator resistance r 0 as follows: ( xd xdc )TdcU 2 sin 2 G 0 xd xdc (1 O 2Tdc2 )

TD

( xd xdc )Td 0 2 2 U sin G 0 xd2 O 2Td20 xdc2

Ts

xd xq 2 ( x xc ) xc O 2T 2 EU U cos 2G 0 d 2 d 2d 2 d20 U 2 sin 2 G 0 cos G 0 xd xd xq xd ( xd O Td 0 xdc )

(b) With a d-axis and q-axis windings, (4.2.25b), and

1 has the same form as equation xd ( jO )

1 can be obtained by a similar method as follows: xq ( jO )

X qO

xq xqccO 2 Tq20

X q2O Rq2O

xq2 O 2 xqcc2 Tq20

Rq O

OTq 0 ( xq xqcc) xq2 O 2 xqcc2 Tq20

X

2 qO

2 qO

R

Therefore, in accordance with equations (4.2.24) and (4.2.25a), the synchronization 279

AC Machine Systems

torque and damping torque coefficients of a synchronous machine with a d-axis and q-axis windings, supposing the stator resistance r 0, can be written as TD

( xq xqcc)Tq 0 2 ( xd xdc )Td 0 2 2 U sin G 0 2 U cos 2 G 0 2 2 2 2 xd O Td 0 xdc xq O 2Tq20 xqcc2

Ts

xd xq 2 ( x xc ) xc O 2T 2 EU cos G 0 U cos 2G 0 d 2 d 2d 2 d20 U 2 sin 2 G 0 xd xd xq xd ( xd O Td 0 xdc )

( xq xqcc) xqccO 2Tq20 xq ( xq2 O 2Tq20 xqcc2 )

U 2 cos 2 G 0

(c) If on the rotor there is not only an excitation winding but also a d-axis and q-axis damping windings, and the excitation-winding resistance can be neglected, then the d-axis operational reactance xd ( p ), according to equations (2.5.69), (2.5.71), (2.5.47) and (2.5.49), can be reduced to xd ( p)

xd

2 ( xd xdcc ) X ffd Tdcc0 p xad

X ffd (1 pTdcc0 )

xdc xdccTdcc0 p 1 pTdcc0

According to the above formula and referring to equation (4.2.25b), we can get X dO X d2O Rd2O Rd O X Rd2O 2 dO

xdc xdccO 2Tdcc02 xdc2 O 2 xdcc2 Tdcc02 OTd 0 ( xdc xdcc ) xdc2 O 2 xdcc2Tdcc02

Therefore, according to equations (4.2.24) and (4.2.25a) and supposing the stator and excitation winding resistances r R fd 0, the synchronization torque and damping torque coefficients, when there are both excitation winding and damping winding on the rotor, can be obtained as follows: TD

( xq xqcc)Tq 0 2 ( xdc xdcc )Tdcc0 U 2 sin 2 G 0 2 U cos 2 G 0 2 2 2 2 2 2 2 xdc O Tdcc0 xdcc xq O Tq 0 xqcc

Ts

xd xq 2 ( x xdcc ) xdccO 2Tdcc02 ( xd xdc ) xdc 2 2 EU cos G 0 U cos 2G 0 d U sin G 0 xd xd xq xd ( xdc2 O 2Tdcc02 xdcc2 )

( xq xqcc) xqccO 2Tq20 xq ( xq2 O 2Tq20 xqcc2 )

U 2 cos 2 G 0

(d) If there are excitation winding and d-axis and q-axis damping windings on the rotor, but the excitation-winding resistance can’t be neglected, then xd ( jO ) has a more complicated form referring to equation (2.5.68). However, we can 280


reduce it as follows. Because X 11d Tdcc0

2 xad X ffd

R11d xd

Tdc

2 xad X ffd

xd X ffd

Td 0

R fd

,

, T1d 0

Td 0 , Tdcc

X 11d R11d

2 § · xad ¨1 ¸ T1d 0 xd X 11d ¹ © 2 2 3 ) xad ( X 11d X ffd ) 2 xad xd ( X 11d X ffd xad

T1cd

2 ) R11d ( xd X ffd xad

,

we can get 2 p 2 ( X 11d X ffd xad ) p ( X 11d R fd X ffd R11d ) R11d R fd

p 2 R11d X ffd Tdcc0 pR11d R fd (Td 0 T1d 0 ) R11d R fd R11d R fd [Tdcc0Td 0 p 2 (Td 0 T1d 0 ) p 1], 2 3 2 2 2 p 2 [ X 11d xad ] p ( xad R11d xad R fd ) 2 xad X ffd xad 2 2 p 2 [ xd ( X 11d X ffd xad ) ( xd X ffd xad ) R11d Tdcc]

§T T 2 R11d R fd ¨ d 0 1d 0 pxad ¨X © ffd X 11d

· ¸¸ ¹

ª x2 º x2 p 2 xd R11d R fd [Td 0Tdcc0 TdccTdc ] pR11d R fd « ad Td 0 ad T1d 0 » , X 11d «¬ X ffd »¼ xd ( p)

xd

2 3 2 2 2 xad X ffd xad p 2 ( X 11d xad ) pxad ( R11d R fd ) 2 p 2 ( X 11d X ffd xad ) p ( X 11d R fd X ffd R11d ) R11d R fd

ª x2 º x2 p 2 xd (Td 0Tdcc0 TdccTdc ) p « ad Td 0 ad T1d 0 » X 11d ¬« X ffd ¼» xd 2 Tdcc0Td 0 p (Td 0 T1d 0 ) p 1 ª º x2 x2 TdccTdcp 2 p «Td 0 T1d 0 ad Td 0 ad T1d 0 » 1 xd X ffd X 11d xd »¼ ¬« xd 2 Tdcc0Td 0 p (Td 0 T1d 0 ) p 1 xd

TdccTdcp 2 (Tdc T1cd ) p 1 Tdcc0Td 0 p 2 (Td 0 T1d 0 ) p 1 281

AC Machine Systems

According to this expression of xd ( p ), the values corresponding to xd ( jO ) are X dO 2 X d O Rd2O

(1 O 2Td 0Tdcc0 )(1 O 2TdcTdcc ) O 2 (Td 0 T1d 0 )(Tdc T1cd ) xd (1 O 2TdcTdcc ) 2 O 2 (Tdc T1cd ) 2

Rd O 2 X d O Rd2O

O (1 O 2Td 0Tdcc0 )(Tdc T1cd ) (1 O 2TdcTdcc )(Td 0 T1d 0 ) xd (1 O 2TdcTdcc ) 2 O 2 (Tdc T1cd )2

According to those results and supposing the stator resistance r 0, the synchronization torque and damping torque coefficients, when there are excitation winding and d-axis and q-axis damping windings on the rotor, can be obtained as follows: TD

Ts

U 2 (1 O 2TdcTdcc )(Td 0 T1d 0 ) (1 O 2Td 0T1ccd 0 )(Tdc T1cd ) 2 sin G 0 xd (1 O 2TdcTdcc )2 O 2 (Tdc T1cd )2 ( xq xqcc)Tq 0 2 2 U cos 2 G 0 xq O 2Tq20 xqcc2 xd xq 2 EU O2 cos G 0 U cos 2G 0 {(1 O 2TdcTdcc )(TdcTdcc Td 0Tdcc0 ) xd xd xq xd (Tdc T1cd )(Td 0 T1d 0 Tdc T1cd )}/{(1 O 2TdcTdcc ) 2 O 2 (Tdc T1cd ) 2 } U 2 sin 2 G 0

O 2 ( xq xqcc) xqccTq20 2 U cos 2 G 0 xq ( xq2 O 2 xqcc2Tq20 )

We have found the synchronization torque and damping torque coefficients during oscillation as above, and with them further discuss the conditions causing the oscillation in a synchronous machine. (2) The steady-state small-oscillation conditions of synchronous machines when the stator resistance r neglected As mentioned before, the electromagnetic torque during an oscillation consists of the following parts: Te

Te 0 'Te

Te 0 TD s TsD

(4.2.26)

Therefore, the rotor-motion equation during the steady-state small-oscillation can be written as H

d 2 (G 0 D ) TD s TsD Te 0 dt 2

or 2

H 282

dD TD s TsD dt 2

Tm Te 0

½ Tm ° °° ¾ ° ° °¿

(4.2.27)


in which Tm is the mechanical torque of a prime-mover or a load that may be constant or variable. In the latter, Tm varies periodically with time and can be written as Tm

Tm 0 ¦ TmO cos(O t M O )

(4.2.28)

Because the synchronous machine is in a steady-state small-oscillation state, i.e. the average value of D is equal to zero, we have Te 0 Tm 0 , or else D will have an average value. Thus, the rotor-motion equation can also be written as H

d 2D dD TD TsD 2 dt dt

¦ T O cos(Ot MO ) m

(4.2.29)

That is a linear differential equation, so the harmonics of the mechanical torque can be studied separately. So far as each of the harmonics is concerned, the corresponding equation is equal to that of the R-L-C circuit shown in Fig. 4.2.3, because the equation of the R-L-C circuit has the following form: Um

Z

sin(Z t M )

or U m cos(Z t M )

½ di 1 idt dt C ³ °° ° ¾ ° 2 di di 1 L 2 R i° dt dt C °¿

Ri L

(4.2.30)

Figure 4.2.3 Equivalent R-L-C circuit

Equation (4.2.29) compared with equation (4.2.30), there is the following relation between them, L

R

p H

p TD

1 i Um C p p p Ts D TmO

Z p

O

and we can obtain the following results. (a) When TD 0 (corresponding to R

0 in the electric circuit), even if the

mechanical torque is constant, i.e. TmO

0 (supposing Ts ! 0) , the oscillation 283

AC Machine Systems

1 in the electric ZC circuit). That phenomenon is called the no-decaying natural oscillation. Accordingly, there are the following relations: the frequency of natural 1 Ts oscillation f 0 , the angular velocity of natural oscillation O0 2Sf 0 2S H may occur and does not decay (corresponding to Z L

Ts , the period of natural oscillation T0 H

1 f0

2S

H . Ts

Thus, the conditions producing the no-decaying natural oscillation are TD

0 (corresponding to R

0)

and

O2H

Ts

§ ¨ corresponding to Z L ©

1 · Z C ¸¹

(b) When TD ! 0, the oscillation caused by some forcing-factor can not be maintained for long after the forcing-factor disappeared. The decaying rate depends upon the electromechanical parameters of the machine. It is more complicated strictly to solve the decaying process, because the machine is not in the steady-state small-oscillation state now and the corresponding TD and Ts are already not constants. However, if the decaying rate and oscillation frequency are all very low, TD and Ts can be taken approximately as constants, i.e. the whole decaying process consists of a series of steady-state small natural oscillations. Therefore, the rotor-motion equation (4.2.29) can still be used, i.e. ( Hp 2 TD p Ts )D

0

(4.2.31)

Solving it, we can obtain p1,2

TD r TD2 4 HTs 2H

(4.2.32)

Observing equations (4.2.24) and (4.2.25) we can see, TD is positive and Ts may be positive or negative if the stator resistance r is neglected. Thus, we can know: (i) When TD2 ! 4 HTs and Ts ! 0, the rotor will come back to its original position aperiodically. (ii) When TD2 4 HTs and Ts ! 0, the machine can maintain its oscillation at a certain time, but its amplitude will gradually decay to zero and the oscillation stops. Accordingly, there are the following relations, 284


4 HTs TD2

The oscillation angular velocity O The oscillation frequency

f

The oscillation period

T

The decay time-constant

Td

2H 4 HTs TD2 4SH 4SH 4 HTs TD2 2H TD

Therefore, if D m is the amplitude when the oscillation starts to decay (i.e. t

Dme

T D t 2H

0),

is the amplitude when time is equal to t, and the next neighbouring

amplitude is D m e

TD ( t T / 2) 2H

. Thus, the ratio between two neighbouring amplitudes is

' e

TD t 2H

e

TD ( t T / 2) 2H

TDT

e 4H

(4.2.33)

or ln '

TD T t D (t T / 2) 2H 2H

TDT 4H

(4.2.33a)

in which ' is called the ratio of the oscillation decay. (iii) when Ts 0, equation (4.2.31) has a positive real root. Therefore, the machine can not stop its oscillation but will aperiodically increase its rotor-position angle G to bring about out of step. This loss of synchronism is called the crawling phenomenon, because during this process the rotor-position angle G grows slowly. since Ts Ts 0 TsO and TsO is always positive referring to Fig. 4.2.1, Ts is always more than Ts 0 . If O | 0, then TsO | 0 and Ts | Ts 0 ; thus, Ts 0 0 is the condition causing the crawling phenomenon and falling out of step. (c) Now we shall further discuss the forced oscillation produced by the mechanical torque harmonic TmO cos(O t M O ). (i) According to Fig. 4.2.3, the amplitude D mO can be written as

D mO

TmO / O T · § TD2 ¨ O H s ¸ O¹ ©

TmO 2

2

2 D

O T (O 2 H Ts ) 2

(4.2.34)

When TD and Ts are all equal to zero, i.e. the stator and rotor windings are all open-circuited, the oscillation amplitude D mc O and instantaneous value D Oc have 285

AC Machine Systems

D mc O

TmO , D Oc O2H

D mc O cos(O t M O )

Thus,

[O

D mO D mc O

O2H O 2TD2 (O 2 H Ts )2

(4.2.35)

is called the oscillation amplification or resonance coefficient, which represents the amplitude amplification caused by the electromagnetic action. If O O0 , then O02 H Ts and [O has its maximum value, which is

[ mO

0

O0 H TD

(4.2.36)

D mO changing with O is shown in Fig. 4.2.4. These curves are called the resonant curves.

Figure 4.2.4

Resonant curves

In order to prevent the oscillation amplitude from too big, the oscillation angular velocity of the mechanical torque applied must be far away from the angular velocity of the natural oscillation. So far as the machine itself is concerned, the following measures during design can be used. (a) The inertia constant H changes, i.e. the resonant curve moves horizontally; (b) The damping winding is installed or strengthened to enhance the damping torque coefficient, i.e. to move the resonant curve vertically. When the damping winding designed, the leakage reactance has a little effect on the damping torque coefficient and it is a good method to change its resistances R11d and R11q . However, the cross-section of the damping bar 286


increases sometimes to bring about a difficulty of manufacture. Therefore, the choice of the damping bar material and change of the ratio between d-axis and q-axis damping resistances are also important measures. For example, we can decrease the q-axis resistance R1q with the help of increasing the d-axis resistance R1d , because the machine ordinarily operates at the angle G 0 20 ~ 30eand the q-axis damping winding has a strong influence referring to Fig. 4.2.2. (ii) The non-uniformity of the applied mechanical torque can usually be indicated by v the non-uniformity of the angular velocity caused in non-uniform mechanical torque when the stator and rotor windings are all open-circuited, in which

v

Z H G Z HM Zs

(4.2.37)

in which Z H G , Z HM and Z s are the maximum, minimum and average angular velocities respectively when all windings are opened. Ordinarily, v is smaller 1 1 . than 250 150 So far as a certain harmonic angular velocity Ois concerned, we have

vO

(Z s 'Z OG ) (Z s 'Z OG )

Zs 2'Z OG

Zs Because Z s amplitude

2

dD Oc dt

HG

Zs

2D mc O O

Zs

(4.2.38)

1, from equation (4.2.35) and the above formula we can get the

D mO

[ OD mc O

vO [ O 2O

(4.2.39)

As mentioned before, under the condition of the forced oscillation, the synchronization torque coefficient Ts is always more than Ts 0 , and during natural oscillation and Ts 0 0 the machine will crawl and lose synchronism. Therefore, there is a problem whether we can, during Ts 0 0, use a mechanical method to maintain a certain forced oscillation to reach Ts ! 0 and to prevent the machine from out of step. The answer to that problem is “no”, i.e. during Ts 0 0, even though the result Ts ! 0 can be obtained by a forced oscillation, the crawling phenomenon may still occur. That is because the forced oscillation only applies a new oscillation torque to the rotor. At a viewpoint of mathematics, the forced oscillation torque can not change the root nature of the rotor-motion characteristic equation but adds a mechanical oscillation of a new frequency referring to Fig. 4.2.3. 287

AC Machine Systems

(3) The effect of stator resistance r For convenience, as mentioned before, we have considered the stator resistance r to be neglected, in which TD is always positive. Therefore, whether the machine can operate stably or not depends upon the sign of Ts 0 . When the forced oscillation occurs owing to some cause, if Ts 0 is negative the machine will still crawl and lose synchronism even though the cause producing an oscillation disappears. On the contrary, if Ts 0 is positive, in the machine there is only the natural oscillation without decay or that with decaying to zero. In practical operation, e.g. a synchronous machine connected to the infinite network through the transmission line, sometimes the machine loses synchronism when Ts is positive. Furthermore, the out-of-step modes include not only increasing aperiodically the rotor-position angle G but also enhancing periodically the oscillation amplitude whose frequency is ordinarily low and about 1 2Hz. The latter is called the self-oscillation out of step phenomenon. There are many causes to make that phenomenon, in which the large stator-resistance is a main cause, because the damping torque coefficient TD decreases actually when the stator resistance increases. In addition, when the stator resistance is more than a certain value, TD will become negative. According to equation (4.2.32) we can see, when TD is negative, equation (4.2.31) will have positive real roots or complex-number roots with a positive real part. Under the first condition, the machine will increase aperiodically its rotor-position angle and lose synchronism. In the latter, the machine is out of step through self-oscillation. Thus, it is significant to study the effect of stator resistance on the damping torque coefficient and synchronization torque coefficient. Substituting ud 0 and uq 0 in equation (4.2.4) into equation (4.2.16), we can get

Id Iq

{[r 2 jO rxq ( jO )]iq 0 rxq ( jO )id 0 [(1 O 2 ) xq ( jO ) jO r ]\ q 0 ½ ° r\ d 0 }/{r 2 jO r[ xd ( jO ) xq ( jO )] (1 O 2 ) xd ( jO ) xq ( jO )} °° ¾ {[r 2 jO rxd ( jO )]id 0 rxd ( jO )iq 0 [(1 O 2 ) xd ( jO ) jO r ]\ d 0 ° ° r\ q 0 }/{r 2 jO r[ xd ( jO ) xq ( jO )] (1 O 2 ) xd ( jO ) xq ( jO )} °¿

(4.2.40)

Substituting the formulas above into equation (4.2.20) and arranging it, we can obtain

T

Ts jOTD Q0

1 {(1 O 2 )[\ d20 xd ( jO ) r 2 jO r[ xd ( jO ) xq ( jO )] (1 O 2 ) xd ( jO ) xq ( jO )

\ q20 xq ( jO )] jO r[\ d20 \ q20 2\ d 0 id 0 xd ( jO ) 2\ q 0 iq 0 xq ( jO ) id20 xd ( jO ) xq ( jO ) iq20 xd ( jO ) xq ( jO )] r 2 [2\ q 0 iq 0 2\ d 0 id 0 id20 xq ( jO ) iq20 xd ( jO )]} (4.2.41) 288


That expression is more complicated. For convenience, we shall take some approximate conditions. At first, although the stator resistance r can not be neglected, its value and O are all not very large. Therefore, we can neglect r 2 and higher power terms and take (1 O 2 ) | 1, so the above formula can be reduced to T

ª§ \ · \ q20 \ d20 jO r «¨ d 0 id 0 ¸ Ts jOTD | Q0 ¨ ¸ xq ( jO ) xd ( jO ) «© xq ( jO ) ¹

2

¬

§ \ q0 · ¨ iq 0 ¸ O ( j ) x © d ¹

2

º » »¼

(4.2.42)

Without damping winding, we have 1 xd ( jO )

1 jOTd 0 x xdc 1 | j d OTd 0 xdc2 xd xdc jOTd 0 xdc

1 xq ( jO )

1 xq

Substituting those values into equation (4.2.42) and supposing the imaginary part 1 of to be much less than the real part, we can obtain xd ( jO ) Ts | Q0

\ d20 xq

\ q20

(4.2.43)

xdc

2 2 2 ª§ \ · § \ q0 § \ q0 · · º 1 d 0 « » TD | 2 ( xd xdc ) ¨ r i i ¨ ¸ ¸ ¨ d0 ¸ q0 ¸ O Td 0 xdc «©¨ xq » © xdc ¹ © ¹ ¹ ¬ ¼

(4.2.44)

As mentioned before, when TD 0 there is a self-oscillation out of step and when Ts 0 0 there is a crawling and loss of synchronism. Therefore, if TD or Ts 0 is negative, then the machine will lose synchronism, i.e. TD ! 0 and Ts 0 ! 0 are the conditions under which the machine can operate in stability. Because there is a little effect of r on Ts 0 , the condition determining TD ! 0 is also that determining the critical value of r. According to equation (4.2.44) we can see, when

r!

§\ · 1 ( xd xdc ) ¨ q 0 ¸ 2 O Td 0 © xdc ¹ 2

2

§\ d0 · §\ · id 0 ¸ ¨ q 0 iq 0 ¸ ¨¨ ¸ c x x ¹ © q ¹ © d

2

rk

(4.2.45)

289

AC Machine Systems

the machine can not operate in stability, in which rk is called the critical value of the stator resistance. Obviously, TD will decrease when r increases and TD will become negative when r is more than rk . However, TD also depends upon a load, so rk will change with the load. When a certain synchronous machine has no damping winding, its critical resistance rk changes with a load, as shown in Figures 4.2.5. and 4.2.6. According to those curves we can see, the lighter the load (i.e. the smaller the value of I 0 or G 0 ) , the smaller the critical resistance rk , i.e. the more easily does the oscillation occur. During no-load, i.e. G 0 | 0, we have \ q 0 | 0 and rk | 0, so the synchronous machine without damping winding can’t be used as a synchronous condenser.

Figure 4.2.5 Critical resistance rk - I 0 curve

Figure 4.2.6 Critical resistance rk - G 0 curve

Because the value of rk is obtained according to the condition TD

Ts . Since the value of Ts does not change greatly H

accordingly we have O

O0

with r, there is O 2

Ts in equation (4.2.45), so H

O02

rk

0,

§\ · H ( xd xdc ) ¨ q 0 ¸ TsTd 0 © xdc ¹ 2

2

§\ d0 · §\ · id 0 ¸ ¨ q 0 iq 0 ¸ ¨¨ ¸ c ¹ © xq ¹ © xd

2

(4.2.45a)

It is clear that with a certain load the larger the inertia constant, the larger the critical resistance rk . Thus, the machine stability can be raised through increasing the inertia constant. By calculating in detail we can see, the damping winding, especially the q-axis 290


damping winding, can make TD and rk increase. The stronger the damping winding, the larger the value of rk . Therefore, the damping winding has an important effect on the machine stability, especially when Ois close to O0 , the damping winding will play an excellent role in limiting the oscillation amplitude. Of course, when O is much different from O0 , its effect is not so significant, which can be known from equation (4.2.34). In addition, installation of a damping winding can decrease the need of raising the inertia constant, i.e. the machine with damping winding may have a lighter rotor. (4) Some physical concepts during machine oscillation We have analysed some problems about small oscillation in detail as before. In order to understand the change of the rotor-position angle G during an oscillation, we shall further discuss it according to the following approximate assumptions: (i) A non-salient pole synchronous machine without damping winding, (ii) Only considering the transient process in excitation winding and not regarding that in stator winding. (iii) The speed of synchronous machine being synchronous speed, P | Te . That is to say, we study this problem according to the following approximate equations, ud

\ q rid

uq

\ d riq p\ fd R fd I fd

U fd

\d

xd id xad I fd

\q

xq iq

\ fd

xd id E

xd iq

xad id X ffd I fd

P | Te

E\ q xd

At first, neglecting the stator resistance r, we discuss the influence of the excitationwinding self-inductance, because its value is large which has an important effect on the excitation-winding flux-linkage or Eqc during an oscillation. Because

U fd Eqc

p\ fd R fd I fd xad \ fd X ffd

we have

U fd

X ffd xad

pEqc R fd I fd

(4.2.46a) 291

AC Machine Systems

If the machine has no oscillation, there is U fd R fd I fd 0 . When the machine oscillates and U fd is kept constant, we have

U fd

X ffd xad

p ( Eqc0 'E c) R fd ( I fd 0 'I fd )

(4.2.47)

Subtracting equation (4.2.46a) from equation (4.2.47), we can get

X ffd p'E c xad R fd 'I fd

0

(4.2.48)

or 0 Td 0 p'E c 'E in which

'E

xad 'I fd

Because

\d

xd id xad I fd

Eqc xdc id

without an oscillation we have

Eqc 0 xdc id 0

xd id 0 xad I fd 0

(4.2.49)

and with an oscillation there is

Eqc 0 'E c xdc (id 0 'id )

xd (id 0 'id ) xad ( I fd 0 'I fd )

(4.2.50)

Subtracting equation (4.2.49) from the above equation, we have

'E c ( xd xdc )'id xad 'I fd

(4.2.51)

According to the above equation and formula (4.2.48) and after deleting 'I fd , we can obtain ( pX ffd R fd )'E c R fd ( xd xdc )'id or (Td 0 p 1)'E c ( xdc xd )'id

(4.2.52)

If the machine operating in steady-state and the stator resistance r neglected, we have

292

E0

xad I fd 0

id 0

1 ( E U cos G 0 ) xd


With an oscillation, there is

'E c ( xd xdc )'id

'E

xad 'I fd

'id

1 ['E (U sin G 0 )D ] xd

(4.2.53)

1 ['E c ( xd xdc )'id (U sin G 0 )D ] xd

(4.2.54)

or

'id

1 ['E c (U sin G 0 )D ] xdc

(4.2.55)

When oscillating exist

G

G0 D

E0 'E Eqc Eqc 0 'E c E

Substituting equation (4.2.55) into equation (4.2.52), we can get

§ xd · ¨ Td 0 p ¸ 'E c xdc ¹ ©

( xdc xd )U sin G 0 D xdc

(4.2.56)

If D D m sin O t , according to the above equation we can find the corresponding 'E c as follows:

'E c

E

( xd xdc )U sin G 0 2

§x · xdc ¨ d ¸ O 2Td20 © xdc ¹ O xc arctan d Td 0 xd

½

D m sin(O t E ) ° ° ° ¾ ° ° °¿

(4.2.57)

or

§ xc · 'E c ¨1 d ¸ (U sin G 0 cos E )D m sin(O t E ) xd ¹ © If X ffd

0, then Td 0

0 and E

'E c

(4.2.57a)

0, so the above formula can be changed into ( xd xdc )U sin G 0 D xd 293

AC Machine Systems

The above formula means that when the excitation winding reactance is neglected, 'E c has a linear relation to D and a reverse sign against DThat result coincides with the physical process at all, because when G increases, id will be raised, so the excitation winding flux-linkage \ fd xad id X ffd I fd will decrease, i.e. Eqc will diminish, whose meaning is reflected by the negative sign in the formula above. However, the excitation winding reactance considered, from equation (4.2.57) we can see, the change of the flux-linkage, i.e. the change of Eqc , will lag behind the change of G, i.e. when Gincreases, Eqc can’t decrease to its steadystate value at once, and when G decreases, Eqc can’t increase to its steady-state value accordingly. Because

D m sin(O t E ) D m sin O t cos E D m cos O t sin E D cos E D m2 D 2 sin E

(4.2.58)

supposing

§ xc · A ¨1 d ¸U sin G 0 cos E xd ¹ ©

(4.2.59)

equation (4.2.57a) can be written as 'E c A[D cos E D m2 D 2 sin E ]

(4.2.57b)

Arranging it ,we have 'E c2 2 A cos E ('E c)D A2D 2

A2D m2 sin 2 E

or 2

§ 'E c · § 'E c ·§ D · § D · ¨ ¸ 2cos ¨ ¸¨ ¸¨ ¸ © AD m ¹ © AD m ¹© D m ¹ © D m ¹

2

sin 2 E

(4.2.58a)

which is an elliptical equation. Therefore, the relation between 'E c and D , due to the excitation-winding self-inductance, become an elliptical equation instead of a linear relation as X ffd 0. As mentioned before, 'E Td 0 p'E c, so 'E jOTd 0 'E c, i.e. there is a phase-angle difference of 90ebetween 'E and 'E c, whose amplitudes differ from each other by OTd 0 times. Therefore, the relation between 'E and D is similar to that between 'E c and D , i.e. the relation between 'E and D is also an elliptical equation. 294


Now we further discuss the output power change of a non-salient pole machine without stator resistance during an oscillation. In steady-state, the output power or torque of a non-salient pole machine without damping winding is

Te 0

P0

E0U sin G 0 xd

(4.2.59a)

which is shown by a steady-state P - G curve in Fig. 4.2.7. The corresponding transient emf is

Eqc 0

E0 ( xd xdc )id 0

E0

xd xdc ( E0 U cos G 0 ) xd

xdc x xdc E0 d U cos G 0 xd xd From the two expressions above, the output power or torque can be indicated by another formula as follows:

Te 0

Eqc 0U 1§ 1 1 · sin G 0 ¨ ¸U 2 sin 2G 0 xdc 2 © xdc xd ¹

P0

(4.2.59b)

When the machine has an oscillation and U is kept constant, neglecting the second-power small increment we can obtain

Te

P

E\ q xd

P0 'P

EU sin G xd

Te 0 'Te

( E0 'E )U sin(G 0 D ) xd

E0U EU 'EU sin G 0 0 cos G 0D sin G 0 xd xd xd 'Te

'P

E0U 'EU cos G 0D sin G 0 xd xd

(4.2.60a)

or

Te 'Te

P

EqcU 1§ 1 1· sin G ¨ ¸ U 2 sin 2G 2 © xdc xd ¹ xdc Eqc0U § 1 1 · 'E cU cos G 0D sin G 0 ¨ ¸U 2 cos 2G 0D 'P xdc xdc © xdc xd ¹ (4.2.61a) 295

AC Machine Systems

If D changes slowly, i.e. the transient process in the excitation winding can be neglected, we have

'E

0

'Te

'P

E0U cos G 0D xd

(4.2.60b)

or 'E c 0 'Te

'P

ª Eqc0U º § 1 1 · cos G 0 ¨ ¸U 2 cos 2G 0 » D « © xdc xd ¹ ¬« xdc ¼»

(4.2.61b)

Obviously, when the machine has a small oscillation and G changes very slowly and after the transient process in the excitation winding is neglected referring to equation (4.2.60b) or (4.2.61b), we can approximately consider that the output power or torque will change according to the tangent through point G 0 on the steady-state P -G curve as shown in Fig. 4.2.7, in other words, according to the steady-state P -G curve. If the excitation winding reactance is taken into account referring to equations (4.2.60a) or (4.2.61a), the output power or electromagnetic torque will change around angle G 0 with an elliptical curve shown in Fig. 4.2.7, whose direction is indicated by arrows in that figure. That is because Eqc or E can’t at once change to its steady-state value due to the effect of the excitation winding reactance when angle G changing by virtue of an oscillation, so there is an elliptical relation between 'E c or 'E and D as mentioned before.

Figure 4.2.7 Steady-state P -G curve

Figure 4.2.8

The ellipse moved from Fig. 4.2.7

In order to illustrate the change of the output power or torque during an oscillation, the ellipse in Fig. 4.2.7 is moved to Fig. 4.2.8. From that figure we can see, when G G 0 D m , i.e. operating at point a in Fig. 4.2.8, the machine has a surplus torque ae so an acceleration to make angle G increase. As mentioned 296


before, owing to the effect of the excitation winding reactance and D D m at this moment, 'E c become positive and the corresponding Eqc is more than Eqc0 that is the steady-state value during G G 0 D m because of Eqc Eqc 0 'E c, so the instantaneous output power is more than the corresponding steady-state value, which makes point a above the steady-state P -G curve. On the basis of the same cause, the instantaneous output power at other points is also different from the value on the steady-state P -G curve and forms an elliptical curve. When point b in Fig. 4.2.8 is reached, the torques are balanced so there is no acceleration, but this time the speed has been more than synchronous speed so the surplus kinetic energy makes angle G increase continually. After passing through point b, the speed will diminish due to the negative surplus torque. Reaching point c, the machine speed is equal to synchronous speed again, but by virtue of negative surplus torque cf the machine has a deceleration so angle G will decrease gradually. When angle G reaches point d, the torques are balanced again, but now the speed has been less than synchronous speed so angle G will go on to diminish. During a small steady-state oscillation, P and G will change so repeatedly, in which the change of the surplus kinetic energy is indicated in Table 4.2.1. Table 4.2.1 Change of suplus kinetic energy

G

The area proportional to the surplus kinetic energy

Notes

From a to b

aeba

Preservation

From b to c

bcfdb

Consumption

From c to d

cfdc

Consumption

From d to a

daebd

Preservation

From that table we can see, during a small steady-state oscillation, the preservation of the surplus kinetic energy is equal to the consumption of that, so the average energy from the prime-mover is constant. However the preserved energy aeba from a to b is less than the consumed energy bcfdb from b to c, so the difference between them must be provided with the pulsating torque of the prime-mover at the positive half-cycle, whose value is proportional to area bcdb. On the contrary, the surplus energy from c to d and from d to a has to be absorbed by the pulsating torque of the prime-mover at the negative half-cycle, whose value is proportional to area dabd. If the pulsating torque causing a small oscillation disappears, because of preserved energy aeba being less than consumed energy bcfdb or consumed energy cfdc being less than preserved energy daebd, angle G can’t change from a to c or from c back to a, so the oscillation will gradually decay to 297

AC Machine Systems

zero. From the above condition we can see, when angle Gchanges clockwise along the elliptical locus, the machine is stable. During r 0 and G 90e , the machine can suffice that condition so it is stable. Now we further discuss the condition when r z 0. Under that condition, during a small oscillation and referring to equation (4.2.46) we have id

rU sin G xd ( E U cos G ) r 2 xd2

(4.2.62)

xd U sin G r ( E U cos G ) r 2 xd2

(4.2.63)

iq

P Te

E\ q

EU ( xd sin G r cos G ) rE 2 2 2 r xd r 2 xd2

Eiq

xd

(4.2.64)

Supposing

T

arctan

r xd

and considering the stator resistance loss to be small, we have id

xd E U cos(G T ) 2 2 r xd r xd2

(4.2.62a)

2

rE 2 UE sin(G T ) r 2 xd2 r 2 xd2 UE | sin(G T ) r 2 xd2

P Te

(4.2.64a)

If the machine oscillates around angle G 0 and voltage U is maintained constant, corresponding to the steady-state operating point we can get P0

id 0

298

Te 0

UE0 r 2 xd2

sin(G 0 T )

E0U r 2 xd2

xd E0 U cos(G 0 T ) 2 2 r xd r 2 xd2 xd E0 U cos G 0c 2 2 2 r xd r xd2

sin G 0c

(4.2.65)

(4.2.66)


in which

G 0c G 0 T Substituting E E0 'E and G G 0 D into equations (4.2.62a) and (4.2.64a) and neglecting the second-power small increment, we can obtain xd 'E U sin G 0cD 2 r 2 xd2 r xd2

'id

'P

UE0

'Te

r 2 xd2

cos G 0cD

U 'E r 2 xd2

(4.2.67)

sin G 0c

(4.2.68)

Substituting equation (4.2.53) into formula (4.2.67), we can get 'id

xd U ['E c ( xd xdc )'id ] sin G 0cD 2 r 2 xd2 r xd2

or 'id

1 [ xd 'E c r 2 xd2 U sin G 0cD ] r xd xdc 2

(4.2.69)

Substituting the above formula into equation (4.2.52), we have (Td 0 p 1)'E c

xdc xd [ xd 'E c r 2 xd2 U sin G 0cD ] r 2 xd xdc

or § r 2 xd2 T p ¨ d0 r 2 xd xdc © Supposing D

· ¸ 'E c ¹

( xd xdc ) r 2 xd2 U sin G 0cD r 2 xd xdc

D m sin O t , we can obtain 'E c

( xd xdc ) r 2 xd2 U sin G 0c 2

§ r 2 xd2 · 2 2 (r 2 xd xdc ) ¨ 2 ¸ O Td 0 c r x x d d ¹ © O (r 2 xd xdc ) E c arctg Td 0 r 2 xd2

D m sin(O t E c)

(4.2.70)

This formula is similar to equation (4.2.57), so when the stator resistance r is considered there is also an elliptical relation between 'E c and D . Similarly, 299

AC Machine Systems

because of 0 Td 0 p'E c 'E referring to equation (4.2.48), there is a phase-angle difference of 90ebetween 'E and 'E c, whose amplitudes differ from each other by OTd 0 times, so there is also an elliptical relation between 'E and D . Equation (4.2.68) compared with equation (4.2.60a) we can see, the relation between output power or torque increment and power angle increment D considering the stator resistance r is similar to that without the stator resistance, but their coefficients differ a little from each other. During G 0c ! 0, i.e. G 0 ! T , the coefficients of both equations have the same sign, so the output power or torque considering the stator resistance still changes around angle G 0 along the elliptical curve shown in Fig. 4.2.7 or 4.2.8, in which the direction of change is the same as that without the stator resistance. That is to say, during G 0 ! T the small oscillation, even if the effect of the stator resistance is considered, will tend to the stable condition after a perturbation disappears. During G 0c 0, i.e. G 0 T , the sign of sin G 0c in equation (4.2.68) is opposite to that of sin G 0 in equation (4.2.60a). Therefore, although the output power or torque considering the stator resistance still changes around angle G 0 along the elliptical curve, the direction of change is counter-clockwise as shown in Fig. 4.2.9. From Fig. 4.2.9 we can see, the area adbea from a to d, which represents the preserved kinetic energy, is more than the area dfcd from d to c which represents the consumed energy, so the oscillation, if any force is not applied to, can not stop at point c, and its amplitude increases more and more to step out ultimately, i.e. during G 0 T the machine may lose synchronism. The larger the r, the biger the T . Therefore, the lighter the load, i.e. the smaller the G , more easily does the oscillation occur by virtue of the stator resistance.

Figure 4.2.9 Steady-state P - G curve and corresponding ellipse during G 0 T

r , the machine can operate stably only when r xd tanG . xd The approximate stable condition considering the effect of the stator resistance is obtained according to a non-salient pole machine, but it can also be used approximately for a salient-pole machine. However, in a salient-pole machine

Because T

300

arctan


there is xd ! xq , so r xq tanG is often used for a stable condition because it is safer to use xq .

Figure 4.2.10 Critical resistance rk - G curve

The critical resistance rk changing with angle Gis shown in Fig. 4.2.10 which is determined according to TD 0 or rk xq tanG . It is not difficult to understand the approximate grade of using r xq tanG .

4.3

Static Stability of Synchronous Machine Systems and Influence of Excitation Regulation on Static Stability

During a synchronous machine in parallel with network, the machine has to operate at synchronous speed strictly according to network frequency. Therefore, the synchronous machine and its connected network dispatch loads or encounter some accidents to bring about a question whether they can still keep synchronous operation or not, namely the stability problem of synchronous machine system. The stability problem is a significant and interesting problem, because the main generator in a network is out of step to make stop electricity of large area and cause confusion. In order to answer the above question, the effect of speed change has to be taken into account for synchronous machines, i.e. not only electromagnetic transient process but also mechanical transients have to be considered simultaneously. For convenience, the stability problem of synchronous machines can be divided into two types, namely static stability and dynamic stability (or transient stability), in which dynamic (or transient) stability will be stated in Section 4.4 and static stability is discussed in the following. Static stability is to discuss a problem whether the synchronous machine can keep synchronous operation under a certain steady-state operating mode. If the 301

AC Machine Systems

machine can’t restore its original state after undergoing a small perturbation under that steady-state operating mode, it can be considered to be unstable even though the machine can still operate. On the contrary, if the machine can restore its original steady-state operating mode after sustaining a small perturbation, it can be considered to be stable under that steady-state operating mode. Under the definition of static stability mentioned above, nonlinear basic equations of synchronous machines can be linearized to simplify the problem whether the machine can be stable or not. It should be pointed out, that the definition of static stability for synchronous machines is not unified in some references and bibliography. As mentioned before, static stability is to discuss a problem whether the synchronous machine can restore its original steady-state operating mode or not when undergoing a small perturbation and then abolishing it, but the rate and fashion of small perturbation are not prescribed. If the small perturbation is caused by small increase of loads and the increase rate of machine loads is very slow, the excitation current has not any change and only the rotor position angle, namely power angle, reaches a new balance position at a very low speed, which is the first possible case. The second possible case is that the increase rate of small loads is relatively fast to bring about change of excitation current due to change of rotor position angle even if the synchronous machine has no excitation regulation device. Of course, with excitation regulation device exists more change of excitation current. In the second possible case, the static stability limit is higher than the first possible case, namely constant excitation current. Thus, in accordance with the above cases, the static stability can be divided into steadystate static stability and dynamic-state static stability in some references. There are excitation regulation devices in modern generators, so analysis of static stability has to consider influence of excitation current change on static stability, referring to [1,2,4,5]. Therefore, by static stability refer to dynamic-state static stability in this book even if there is no excitation regulation device.

4.3.1

Basic Relations of Synchronous Machine Systems for Analyzing Static Stability

When a synchronous machine operating in steady-state, according to Park’s formulas the relations with various quantities are ud 0 \ q 0 rid 0 ½ ° uq 0 \ d 0 riq 0 ° \ d 0 E xd id 0 ¾ ° \ q 0 xq iq 0 ° Tm 0 Te 0 \ d 0 iq 0 \ q 0 id 0 ¿ 302

(4.3.1)


If the machine undergoing a small perturbation such as a mechanical torque increment 'Tm , then among various quantities exist the following relations: p (\ d 0 '\ d ) (\ q 0 '\ q )(1 p'G ) r (id 0 'id ) ½ p'\ d \ q 0 \ q 0 p'G '\ q '\ q p'G rid 0 r 'id °° uq 0 'uq p'\ q \ d 0 \ d 0 p'G '\ d '\ d p'G riq 0 r 'iq ° ° \ d 0 '\ d G ( p )(u fd 0 'u fd ) xd ( p )(id 0 'id ) ° E xd id 0 G ( p)'u fd xd ( p)'id ¾ ° \ q 0 '\ q xq iq 0 xq ( p)'iq ° Te 0 'Te \ d 0 iq 0 \ q 0 id 0 '\ d iq 0 '\ q id 0 ° ° \ d 0 'iq \ q 0 'id '\ d 'iq '\ q 'id ° 2 Hp 'G Tm 0 'Tm Te 0 'Te ¿ (4.3.2) u d 0 'u d

where the subscript “0” indicates original steady-state values. Subtracting various equations in (4.3.1) from the corresponding equations in (4.3.2) and neglecting multiplication of two small increments, there are 'ud p'\ d \ q 0 p'G '\ q r 'id ½ ° 'uq p'\ q \ d 0 p'G '\ d r 'iq ° '\ d G ( p)'u fd xd ( p)'id ° ¾ '\ q xq ( p)'iq ° 'Te '\ d iq 0 \ d 0 'iq '\ q id 0 \ q 0 'id ° ° Hp 2 'G 'Te 'Tm ¿

(4.3.3)

The above equations can be reduced further to 'ud pG ( p )'u fd ½ ° Z d ( p )'id xq ( p )'iq \ q 0 p'G ° 'uq G ( p)'u fd ° ¾ xd ( p )'id Z q ( p )'iq \ d 0 p'G ° 'Tm iq 0 G ( p )'u fd ° [\ q 0 iq 0 xd ( p )]'id [\ d 0 id 0 xq ( p )]'iq Hp 2 'G °¿

(4.3.4)

where Z d ( p) Z q ( p)

pxd ( p ) r pxq ( p) r

That is to say, through the procedures above, the nonlinear differential equation set (4.3.2) has been changed to linear differential equation set (4.3.3) or (4.3.4), by use of which we can study the static stability of synchronous machine under a certain steady-state operating mode. 303

AC Machine Systems

4.3.2

Approximate Analysis of Static Stability for Synchronous Machine Systems

In order to illustrate the analytical steps of static stability, we take a synchronous machine connected to infinite source as example referring to Fig. 4.3.1.

Figure 4.3.1 A synchronous machine connected to infinite source

Under that condition exist U sin G 0 ½ U cos G 0 ¾¿

(4.3.5)

U cos G 0 'G uq 0 'G ½ U sin G 0 'G ud 0 'G ¾¿

(4.3.6)

ud 0 uq 0 'ud 'uq

Supposing the excitation winding voltage to be constant, namely 'u fd equation (4.3.4) can be simplified as ½ 0 Z d ( p )'id xq ( p)'iq (uq 0 \ q 0 p )'G ° 0 xd ( p )'id Z q ( p )'iq (ud 0 \ d 0 p )'G ¾ 'Tm [\ q 0 iq 0 xq ( p )]'id [\ d 0 id 0 xq ( p )]'iq Hp 2 'G °¿

0, so

(4.3.7)

Deriving the equation set, we have assumed the factor of causing various quantity change for synchronous machines to be 'Tm . Therefore, if a synchronous machine can return to original steady-state operating mode when abolishing 'Tm , then the synchronous machine is stable; or it is unstable if incapable of returning to original steady-state operating mode. The above problem is not difficult to respond when knowing the root properties of characteristic equation corresponding to the differential equation set above. Supposing that there is no positive real-root or complex-root with positive real part for characteristic roots, the synchronous machine is stable; or it is unstable when there is a positive real-root or complexroot with positive real part. Thus, judgement of static stability for synchronous machines is converted into discriminating whether the real root or the real part of a complex root is negative for the characteristic equation. It should be noted that not only 'Tm but parameter and load changes also belong to the perturbation types, but the basic equation set is similar after 304


cancelling the perturbation. If taking 'Tm 0 and solving equation set (4.3.7), it is clear that the decay regularity of 'id , 'iq and 'G will change with different roots of the characteristic equation. Especially the variation of 'G expresses the varying situation of rotor position for synchronous machines, so the rotor returns to its original steady-state position directly without any oscillation if its characteristic roots are all negative real roots; or the rotor becomes out-of-step without any oscillation when its characteristic roots are all real roots in which exists one positive real root at least, i.e. the rotor will step out through the crawl phenomenon. If there is complex root but on complex root with positive real part or no positive real root, then the rotor will return to its original steady-state position through a certain oscillation. Supposing that in characteristic equation there are complex root with negative real part together with positive real root, then the rotor will leave its balance position to bring about out of step although its oscillation amplitude decays step by step. If the real part of complex root is positive, then its oscillation amplitude will increase step by step to lose synchronism finally, i.e. the rotor will step out through self-oscillation. However, it should be noted that the above conclusions are correct only under small change condition, so when the rotor position is relatively far away from its balance position, i.e. 'G is not a small quantity, the small perturbation, and linearization mentioned before can’t illustrate the rotor motion condition any longer. The characteristic equation for (4.3.7) is

D( p)

Z d ( p) xd ( p )

xq ( p ) Z q ( p)

[\ q 0 iq 0 xd ( p )] [\ d 0 id 0 xq ( p )]

(uq 0 \ q 0 p) (ud 0 \ d 0 p ) Hp 2

Hp 2 {Z d ( p ) Z q ( p ) xd ( p) xq ( p)} [\ d 0 id 0 xq ( p)]{Z d ( p )[\ d 0 p ud 0 ] xd ( p )[\ q 0 p uq 0 ]} [\ q 0 iq 0 xd ( p )]{Z q ( p)[\ q 0 p uq 0 ] xq ( p)[\ d 0 p ud 0 ]} 0

(4.3.8)

which is a high degree equation and even if there is no damping winding D( p ) 0 is still fifth-order equation thus using numerical method. In order to realize the main effect on static stability and its physical conception, it is necessary to reduce the equations further. If neglecting the stator flux-linkage change p\ d and p\ q , the equation D( p ) 0 will be degraded. In Park’s voltage equations, \ qZ and \ d Z represent the speed emfs in stator circuit, p\ d and p\ q express the transformer emfs in stator circuit, and rid and riq indicate the voltage drops in stator resistance. Therefore, Park’s voltage equations can be rewritten as 305

AC Machine Systems

u d \ qZ uq \ d Z

p\ d rid ½ p\ q riq ¾¿

(4.3.9a)

Neglecting p\ d and p\ q , it corresponds to a pure-resistance circuit, so only the speed emfs and terminal voltages cause the corresponding currents, and there is no current produced by transformer emfs. Taking 3-phase sudden short circuit as example, the former causes stator fundamental current and the latter gives rise to the stator aperiodic current and second harmonic current. Because the torques produced by the latter currents have a small influence on the rotor motion, during discussion of some static stability problems the latter can be neglected, i.e. the static stability can be studied according to the following equation, ud uq

\ qZ rid ½ \ d Z riq ¾¿

(4.3.9b)

If neglecting the stator resistance further, then Park’s equations can be reduced to ud uq

\d \q

\ qZ

½ °° G ( p )u fd xd ( p )id ¾ ° xq ( p)iq °¿

\ dZ

(4.3.9c)

Now let’s use the above equations to study the static stability problems during a synchronous machine without damping winding in parallel with infinite source. Supposing that the excitation voltage is still constant and the machine undergoes a small perturbation, then there are the following relations: ud 0 'ud \ q 0 '\ q \ q 0 p'G '\ q p'G uq 0 'uq \ d 0 '\ d \ d 0 p'G '\ d p'G \ d 0 '\ d E id 0 xd xd ( p)'id \ q 0 '\ q iq 0 xq xq 'iq Hp 'G Tm 0 'Tm \ d 0 iq 0 \ q 0id 0 '\ d iq 0 '\ q id 0 \ d 0 'iq \ q 0 'id '\ d 'iq '\ q 'id Substituting (4.3.1) into the above relations and neglecting multiplication of two small increments and \ q 0 p'G together with \ d 0 p'G , we can get 'ud 'uq 'Tm

306

uq 0 'G '\ q xq 'iq ud 0 'G '\ d xd ( p )'id [\ q 0 iq 0 xd ( p )]'id (\ d 0 id 0 xq )'iq Hp 2 'G


It should be pointed out that \ d 0 p'G and \ q 0 p'G represent the speed emfs caused when the rotor leaving synchronous speed, which will produce a certain current and asynchronous torque but are neglected only to bring about a small influence on static stability. The characteristic equation for the above equation set is xd ( p )[ xq Hp 2 (\ d 0 id 0 xq )uq 0 ud 0\ q 0 ] xq ud 0\ q 0

D( p)

0

into which substituting the following relations: xd xdc Td 0 p , 1 Td 0 p

xd ( p )

\d0

uq 0

U cos G 0 ,

\ q0 id 0

u d 0

U sin G 0

E U cos G 0 , xd

iq 0

U sin G 0 xq

the characteristic equation can be rewritten as a0 p 3 a1 p 2 a2 p a3

0

where a0 a2

xdc xqTd 0 H , a1 xd xq H xdc Td 0 (\ d 0 uq 0 \ q 0 ud 0 xq id 0 uq 0 ) xqTd 0 ud 0\ q 0

a3

ª EU º §1 § 1 1· 1 · xdc xqTd 0 « cos G 0 ¨ ¸ U 2 cos 2G 0 ¨ ¸ U 2 sin 2 G 0 » ¨x ¸ «¬ xd »¼ © xdc xd ¹ © q xd ¹ xd (\ d 0 uq 0 \ q 0 ud 0 xq id 0 uq 0 ) xq ud 0\ q 0 ª EU º §1 1 · xd xq « cos G 0 ¨ ¸ U 2 cos 2G 0 » ¨ ¸ © xq xd ¹ ¬« xd ¼»

According to Herwitz’s Criteria, the prerequisites of stable operation are (i) a0 ! 0, '1 a1 (ii) ' 2 a3 a1 (iii) ' 3 a3 0

a1 a0 a2 a0 a2 0

! 0; a1a2 a0 a3 ! 0; 0 a1 a3

a3

a1 a3

a0 !0 a2

(namely a3 ! 0)

In accordance with the coefficients of characteristic equation above, the prerequisites have the following situations: (i) The first criterion, namely a0 ! 0 and a1 ! 0, can be satisfied on any operating condition. 307

AC Machine Systems

(ii) The second criterion

' 2 a1a2 a3 a0

§ 1 1 · xdc xd xq2Td 0 H ¨ ¸U 2 sin 2 G 0 ! 0 © xdc xd ¹

can also be met under any operating mode. (iii) The third criterion ' 3 ! 0, namely a3 ! 0, can’t be satisfied for some operating modes, so it is the only criterion to discriminate the range of static stability. According to the criterion a3 ! 0, the main influence on static stability is those quantities in coefficient a3 , so the following equation

§ 1 EU 1· cos G m ¨ ¸ U 2 cos 2G m ¨x ¸ xd © q xd ¹

0

can be used to find out the limiting angle G m of static stability. During xd z xq and referring to equation (2.5.9), the output power is P

EU U2 § 1 1 · sin G ¨¨ ¸¸ sin 2G xd 2 © xq xd ¹

(4.3.10)

Therefore, the criterion for static stability limit is the same as the approximate dP 0 in Electric Machinery, and here provide a theoretical basis for criterion dG dP the approximate stable criterion ! 0. dG If needing more accurate static stability range, it is necessary to solve equation (4.3.8), but writing out its general expression is difficult. Analysing practical problems, the numerical solutions can be found out in the light of actual values of machine parameters and by use of digital computer. Now take a synchronous machine without damping winding in parallel with infinite source as example to illustrate the problem. Let the machine parameters be U 1.0, xd 1.0, xq 0.6, xdc 0.3, Td 0 1 000, H 1 000, E 1.0 or 1.5, according to which and equation (4.3.8) together with Routh’s Criterion we can get the curves as shown in Fig. 4.3.2, where the ordinate represents the stator resistance r, the abscissa indicates the power angle G, the full line corresponds to the results for E 1.0, and the dashed line corresponds to the results for E 1.5. From Fig. 4.3.2 it is clear that the synchronous generator without damping winding in parallel with infinite source can’t operate stably during larger stator resistance r and small power angle G such as G 30e. Using it as synchronous motor, its stable range is narrow and maybe it can’t operate as motor when the stator resistance exceeding a certain value such as r ! 0.6. From Fig. 4.3.2 it is also evident that the larger the stator resistance, the larger 308


the power angle G corresponding to stable operation. During zero stator resistance, the machine may be unstable even if G 90ebecause there is xd z xq here. Considering the rotor to be unsymmetrical on magnetic circuit, namely xd z xq , the output power or torque ( P Te during r 0) contains the second harmonic, so the maximum power or torque can’t occur at G 90ebut G 90e.

Figure 4.3.2 Discriminating operation stability by use of Routh’s Criterion

dP dG angle G m corresponding to the maximum power Pm can be got as Differentiating with respect to formula (4.3.10) and letting

cos G m

2

ª º 1 xq E r « » 4( xd xq ) 2 ¬« 4( xd xq )U ¼» xq E

0, the power

(4.3.11)

N.B. there are several solutions for G m , in which the solution less than 90e should be taken. On the basis of expression (4.3.11), for the demonstrated example exist G m | 65eduring E 1.0 and G m | 70eduring E 1.5. Because the synchronous generator is usually connected to outer impedance and operates at higher excitation emf E, i.e. under over-excitation condition, thus ª1 § 1 º 1 · in formula (4.3.10) the salient-pole power « ¨ ¸U 2 sin 2G » can be «¬ 2 ¨© xq xd ¸¹ »¼ neglected. That is to say, during calculation of static stability, the salient-pole synchronous generator has a synchronous reactance equal to xd and can be calculated according to the corresponding formulas of non-salient pole synchronous generators. When the stator resistance r neglected, output power in steday-state is equal to 309

AC Machine Systems

electromagnetic torque. Therefore, when the machine undergoing a small perturbation to bring about very slow change, the output power P is also approximate to electromagnetic torque Te in per-unit, so there is dTe dG

dP dG

or

'Te

§ dP · ¨ ¸ 'G © dG ¹

(4.3.12)

In addition, when the synchronous machine undergoing a small perturbation and its power angle G changing slowly, the process can also be considered as dG steady-state small oscillation but its oscillating angular velocity is O | 0. In dt Section 4.2, the additional torque during steady-state small oscillation is

'Te

Ts 'G TD

dG | Ts 0 'G dt

(4.3.13)

That is to say, when the machine undergoing a small perturbation and its power dP 'Te | Ts 0 . Hence the stable prerequisite angle G changing slowly exists dG 'G dP ! 0 is also the synchronization torque coefficient Ts 0 ! 0 referring to dG Section 4.2, which can also be illustrated in the following way. On the basis of equation (4.2.25b), during O | 0 exist X d O | xd , Rd O | 0, X qO | xq and RqO | 0. Substituting the above relations into equations (4.2.24) and

(4.2.25a) and having supposed xd

xq , we can get

TD | 0 Ts

Ts 0

Seeing that P

EU U2 U2 U2 U2 cos G 0 sin 2 G 0 cos 2 G 0 cos 2 G 0 sin 2 G 0 xd xd xq xd xq EU cos G 0 xd

EU sin G , there is xd dP dG G

G0

EU cos G 0 xd

which demonstrates the stable prerequisite synchronization torque coefficient Ts 0 ! 0. 310

Ts 0

(4.3.14)

dP ! 0 being in accord with the dG


dP ! 0 doesn’t count in the dG influence of stator resistance and speed change, in which the damping torque coefficient is always positive when neglecting stator resistance, so when dP ! 0 there is no crawl out-of-step phenomenon satisfying the stable criterion dG but maybe exists a self-oscillation out-of-step phenomenon of low frequency such as 1 2Hz. The damping winding has no effect on the synchronization torque coefficient Ts 0 , so it also has no influence on the crawl phenomenon, but there is main effect on depression of self-oscillation out-of-step phenomenon. In practical terms, the stator resistance is small and there is stronger damping winding, so the possibility of self-oscillation out-of-step phenomenon is very low for the synchronous generator without excitation regulation device. Therefore, dP ! 0 is still an important and useful criterion. dG If considering the aperiodic component of stator current, i.e. counting in p '\ d and p'\ q , then the electromagnetic torque also contains pulsating torque It should be noted that the stable criterion

whose frequency is about the basic frequency such as 50Hz, so there is also high frequency oscillation of approximate basic frequency besides crawl and low frequency self-oscillation. Taking the synchronous machine without damping winding as example, and counting in stator resistance and aperiodic component of stator current, then its characteristic equation will be of fifth-order. Therefore, besides one real root may exist two pairs of complex roots corresponding to high frequency oscillation and low frequency self-oscillation. However, aperiodic component of stator current corresponding to high frequency oscillation decays very quickly and the rotor inertia is also large, so the amplitude of high frequency oscillation is small and decays very quickly, which can’t be perceived in practical operation. In general, generators can’t operate on limiting power condition and the safe margin is necessary. So far as turbogenerators are concerned, their rated power angle G N is about 30e, and as for hydraulic generators the power angle G N is some 25e.

4.3.3

Influence of Excitation Regulation on Static Stability

The analysis of static stability mentioned above is based on the supposition that the excitation voltage keeps constant, i.e. the influence of excitation regulation on static stability is neglected. In practical terms, modern generators all have excitation regulation devices, so when the generator undergoing a perturbation the excitation regulation is put into action to increase the stable limit of 311

AC Machine Systems

synchronous machine systems and to decrease the requirements of machine parameters. As stated before, assuming the synchronous machine to be of non-salient pole and neglecting its stator resistance, its output power is P

EU sin G xG

(4.3.15)

so the output power can be increased if the excitation regulation device is put into action to regulate the excitation emf E in time. Now take a synchronous generator in parallel with infinite source as example to illustrate the problem. Suppose the power system to be shown in Fig. 4.3.3, its equivalent circuit is indicated in Fig. 4.3.4 and the phasor diagram in steady-state is shown by full lines in Fig. 4.3.5.

Figure 4.3.3 The power system comprising a synchronous generator in parallel with infinite source.

Figure 4.3.4 Equivalent circuit corresponding to Fig. 4.3.3

For the studied power system exists P

where

xc

EU sin G xG xc xT 1

(4.3.16)

1 xL xT 2 2

Evidently, if excitation current or excitation voltage is invariable, i.e. E is constant, then after the power system changing its operating condition the corresponding steady-state phasor diagram is indicated by dashed lines as shown in Fig. 4.3.5. From Fig. 4.3.5 it is clear that during increase of power angle G the phase angle of generator terminal voltage U G will change and the voltage magnitude decrease. On the contrary, when an excitation regulation device is installed at point U G of the generator, the change of terminal voltage due to variation of generator operation will cause the change of excitation current or 312


excitation voltage to restore the terminal voltage to its original value or to approximate to original terminal voltage. When U G being constant, the excitation emf E of generator will increase with power angle G as shown in Fig. 4.3.6.

Figure 4.3.5 Steady-state phasor diagram during a synchronous machine in parallel with infinite source

Figure 4.3.6 Steady-state phasor diagram during U G

constant

If plotting P-G curves for various values of E, we can get a family of sinusoidal curves as shown in Fig. 4.3.7, which are termed inner power characteristics or inner P-G curves. When an excitation regulation device is installed at point U G of the generator, due to the change of excitation emf E its 313

AC Machine Systems

P-G curve will shift from a sinusoidal curve to another sinusoidal one, and a certain point on original curve also shifts to another corresponding curve. Thus the points mentioned above will constitute a curve with an arrow as shown in Fig. 4.3.7, which is called outer power characteristic or outer P-G curve. Under this condition it can be seen that its output power P still increases with power angle G in a certain range even if G ! 90ebecause the increase of excitation emf E is more than the decrease of sin G when G is slightly more than 90ereferring to formula (4.3.16).

Figure 4.3.7 Inner power characteristics and outer power characteristic

Of course, the above fact can also be explained according to the following conception. Seeing that U G and U are all constant under this condition, output power P can also be calculated in the light of U G and U , namely P

U GU sin G c xc

(4.3.17)

where G c is the angle included between phasors U G and U . U GU Evidently, P Pm during G c 90e , so the maximum value of outer xc , not at G 90e. There is G ! G c due to power characteristic exists at G c 90e generator reactance xG referring to Fig. 4.3.5, so the maximum value of outer power characteristic appears during G ! 90e. 314


As mentioned before, excitation regulation has an important influence on static stability, which has been researched into for many years to get distinguished results. However, the influence of excitation regulation is in many ways. For example, it can regulate terminal voltage automatically and raise stable limit, but it can also cause self-oscillation or self-oscillation out-of-step phenomenon to affect the stable limit of power transmission and also to influence power system operating quality. Studying the problems above, it is necessary to write down the equations corresponding to generator, excitation regulation device and power system, to analyse them further and then to take corresponding technique measures.

4.4

Dynamic Stability and Analysis Methods of Synchronous Machine Systems

The dynamic stability problem of synchronous machine systems is to study whether the synchronous machine can operate synchronously after undergoing a certain or large disturbance. There are many disturbances, but they can be summarized into the following several types: (i) Sudden change of load (ii) Cut-off of transmission line or transformer (iii) Disconnection of generators (iv) Short circuit fault, in which the short circuit fault is the most dangerous so it has to be paid attention to firstly. For a power system with neutral point earthed may occur the following short circuit faults: (i) Single phase earthed (ii) Two phases short circuit (iii) Two phases short circuit earthed (iv) 3-phases short circuit. Statistical data about accidents indicate that the number of times for single phase earthed owns about 70% of the total short circuit faults for high voltage networks and the number of times for 3-phases short circuit only owns 6% 7% of the total short circuit faults. So far as the influence of short circuit faults on stability is concerned, 3-phases short circuit will cause the most difficulty because during 3-phases short circuit the voltage will dip quickly to weaken the connection between power stations or to lose the system stability finally. For single phase earthed or two phases short circuit the voltage dip is relatively small, especially the former, so it is of benefit to synchronous generators. For important transmission systems to maintain stability is necessary not only for unsymmetrical short circuit but also for 3-phases short circuit, because short circuit fault is the most dangerous from the viewpoint of dynamic stability. Therefore, for networks with neutral point earthed the stability calculation should be made respectively according to 3-phases short circuit, two phases earthed and single phase earthed. As for 2-phases short circuit people examine it only a few. When a synchronous generator undergoing the above disturbances, its output power will change suddenly but the prime mover’s torque can’t change abruptly due to regulator inertia. Therefore, the output power of generator can’t balance the prime mover power, thus making the generator speed variable and effecting 315

AC Machine Systems

the electromechanical transient process of whole power system. Under some adverse condition, the generator will lose synchronism. In order to discriminate whether the synchronous generator will step out during the above situation, it is necessary to study the whole electromechanical transient process. As stated before, it is a complicated nonlinear problem. In practical terms, according to the problems studied and considering the corresponding computation methods, get the reduced mathematical patterns to analyse them referring to [1,3,4,7].

4.4.1

The Equations Corresponding to Analysis of Dynamic Stability for Synchronous Machine Systems

Before analysing dynamic stability for synchronous machine systems, firstly discuss the corresponding equations. On the rotor of synchronous machine exist d-axis damping winding and q-axis one besides the excitation winding, whose basic equations have been discussed in detail in Section 2.4. and can be used to study dynamic stability of synchronous machine systems. Of course, for convenience of analysis it is necessary to reform and reduce those basic equations. Substituting formulas (2.4.3) and (2.4.4) into formulas (2.4.1) and (2.4.2) and rearranging them, the voltage equation of synchronous machines can be written as ª ud º «u » « q » «U fd » « 0 » « » ¬ 0 ¼

Z xq pxad pxad Z xaq ª (r pxd ) º « Z x » (r pxq ) Z xad Z xad pxaq d « » 0 ( R fd pX ffd ) 0 pX f 1d « pxad » « px » 0 ( R1d pX 11d ) 0 pX f 1d ad « » «¬ 0 0 0 ( R1q pX 11q ) »¼ pxaq ª id º «i » « q » « I fd » «I » « 1d » ¬« I1q ¼»

(4.4.1)

where Z is rotor speed in per-unit. In addition, there is

J

t

t

³ Z dt J ³ (Z 'Z )dt J Z t ³ 'Z dt J ³ Z dt J c Z t J c 0

0

s

0

t

s

Jc 316

0

0

t

0

s

0

s

0

0


Letting G be the angle included between d, q axes and d c , qc axes, then exist

G J

t

J J c ³ 'Z dt (J 0 J 0c ) 0 Zst G J 0 G 0

t

³ 'Z dt G 0

0

in which

G0

J 0 J 0c is the angle included between d, q axes and d c , qc axes at t 0;

'Z

Z Z s is the difference between rotor speed and synchronous speed.

Hence, the rotor speed in per-unit can also be written as

Z

dJ dt

1 pG

(4.4.2)

Tm Te

(4.4.3)

In addition, the rotor motion equation is Hp 2G

The flux-linkage equations are

\d

xd id xad I fd xad I1d

(4.4.4)

\ fd

xad id X ffd I fd X f 1d I1d

(4.4.5)

\ 1d

xad id X f 1d I fd X 11d I1d

(4.4.6)

\q

xq iq xaq I1q

(4.4.7)

\ 1q

xaq iq X 11q I1q

(4.4.8)

The electromagnetic torque equation is Te

\ d iq \ q id

[id iq I fd I1d

Taking X f 1d

ª 0 « x « d I1q ] « 0 « 0 « ¬ 0

xq 0 0 0 0

0 xad 0 0 0

0 xad 0 0 0

xaq º ª id º « » 0 » « iq » » 0 » « I fd » 0 » « I1d » »« » 0 ¼ ¬« I1q ¼»

(4.4.9)

xad , i.e. neglecting the mutual leakage reactance between

excitation winding and damping winding, the formulas of stator flux-linkages 317

AC Machine Systems

\ d and \ q can be reduced and the corresponding relations, referring to Section 2.4, can be written as

\d

G ( p)U fd xd ( p )id

(4.4.10)

\q

xq ( p )iq

(4.4.11)

in which 1 p(Td 4 Td 5 ) p 2Td 4Td 6 1 p (Td 1 Td 2 ) p 2Td 1Td 3

xd ( p)

xd

xq ( p )

xq

G( p)

1 pTd 7 xad 2 1 p(Td 1 Td 2 ) p Td 1Td 3 R fd

Td 1 Td 2 Td 3 Td 4 Td 5 Td 6 Td 7 Tq1 Tq 2

1 pTq1

(4.4.12)

(4.4.13)

1 pTq 2

(4.4.14)

X ffd

Td 0 R fd X 11d T1d 0 R1d 1 § ¨ X 11d R1d ¨© 1 § ¨ X ffd R fd ©

2 xad X ffd

· ¸¸ Tdcc0 ¹

2 · xad ¸ Tdc xd ¹ x2 · ad ¸ xd ¹

§ ¨ X 11d © 2 xad (2 xad xd X ffd ) ½° 1 ° X ® 11d ¾ Tdcc 2 R1d ¯° xd X ffd xad ¿° X 11d xad R1d 2 · xaq 1 § ¨¨ X 11q ¸ Tqcc xq ¸¹ R1q © X 11q Tq 0 R1q 1 R1d

Since Td 2 and Td 3 are much less than Td 1 , and Td 5 and Td 6 are much less than Td 4 , thus the operational reactances xd ( p ), xq ( p ) and operational conductance 318


G ( p ) mentioned above can also approximate to xd ( p)

xd

(1 pTdc )(1 pTdcc ) (1 pTd 0 )(1 pTdcc0 )

(4.4.15)

G( p)

1 pTd 7 xad (1 pTd 0 )(1 pTdcc0 ) R fd

(4.4.16)

xq ( p )

xq

1 pTqcc 1 pTq 0

(4.4.17)

The error due to simplification of time constants above doesn’t exceed 5%. In the light of formulas (4.4.15), (4.4.16) and (4.4.17), the stator flux-linkage equations (4.4.10) and (4.4.11) can be rewritten as

\d

\q

A B id id xdccid 1 pTd 0 1 pTdcc0 Gc G cc U fd U fd 1 pTd 0 1 pTdcc0

xq xqcc 1 pTq 0

iq xqcciq

A | xd xdc ½ ° B | xdc xdcc ° Tdc ° xdc xd ° Td 0 ° T cc ° xdcc xdc d Tdcc0 ° ° Tqcc ¾ cc xq xq ° Tq 0 ° Td 0 Td 7 xad ° Gc Td 0 Tdcc0 R fd ° ° Tdcc0 Td 7 xad ° G cc ° Td 0 Tdcc0 R fd ° ¿ xd xdc xc xdcc id d id xdccid 1 pTd 0 1 pTdcc0 Gc G cc U fd U fd 1 pTd 0 1 pTdcc0

(4.4.18)

(4.4.19)

(4.4.20)

\d |

(4.4.21) 319

AC Machine Systems

According to formulas (4.4.10) (4.4.14) and opening damping winding, i.e. letting R1d f and R1q f, then there is

Td 2

Td 3

Td 5

Td 6

Td 7

Tq1

0

Therefore, if neglecting damping winding, then exist 1 pTdc 1 pTd 0 xad 1 G( p) 1 pTd 0 R fd 1 pTdc xad 1 U fd xd id \d 1 pTd 0 R fd 1 pTd 0

xd ( p )

xd

Since Gc

\d

there is

Td 0 Td 7 xad x | ad Td 0 Tdcc0 R fd R fd

1 pTdc Gc U fd xd id 1 pTd 0 1 pTd 0

Referring to Section 2.7, the d-axis transient emf of synchronous machine without damping winding is Eqc \ d xdc id , so another expression for Eqc is

Eqc \ d xdc id x xdc Gc U fd d id 1 pTd 0 1 pTd 0

(4.4.22)

In addition, the d-axis and q-axis transient emfs of synchronous machine with damping winding are respectively

Eqcc \ d xdccid

(4.4.23)

Edcc \ q xqcciq

(4.4.24)

Substituting expressions (4.4.23) and (4.4.24) into formulas (4.4.19), (4.4.21) and (4.4.22), and rearranging them, we can get the following relations:

320

pEqc

1 {G cU fd ( xd xdc )id Eqc } Td 0

(4.4.25)

pEqcc

1 {G ccU fd ( xdc xdcc )id Eqcc Eqc } pEqc Tdcc0

(4.4.26)


1 {( xq xqcc)iq Edcc} Tq 0

pEdcc

(4.4.27)

Substituting expressions (4.4.23) and (4.4.24) into formula (4.4.9), another expression for electromagnetic torque is

Te

Edccid Eqcciq ( xqcc xdcc )id iq

(4.4.28)

As you know, the stator voltage equations of synchronous machine are

ud

p\ d Z\ q rid

(4.4.29)

uq

p\ q Z\ d riq

(4.4.30)

in which p\ d and p\ q represent d-axis and q-axis transformer emfs respectively, Z\ d and Z\ q indicate d-axis and q-axis speed emfs respectively, and the former will influence the aperiodic component of stator current mainly and the latter the basic-frequency component of stator current. Therefore, for convenience of analysis it is necessary to use some approximation. For example, during analysis of some transients the rotor flux-linkage can be considered as constant at initial stage of disturbance, and the aperiodic component of stator current is also neglected, i.e. p\ d and p\ q are omitted referring to Sections 2.7 and 2.6. In other cases such as a generator with excitation regulation device, its rotor flux-linkage after disturbance will change a lot to bring about main effect. In order to count the influence in, sometimes use the following approximation:

p\ d

p ( Eqcc xdccid ) | pEqcc

(4.4.31)

p\ q

p ( Edcc xqcciq ) | pEdcc

(4.4.32)

Z\ d

Z ( Eqcc xdccid ) | Z Eqcc xdccid

(4.4.33)

Z\ q

Z ( Edcc xqcciq ) | Z Edcc xqcciq

(4.4.34)

Thus, equations (4.4.29) and (4.4.30) can be reduced to

ud

pEqcc Z Edcc xqcciq rid

(4.4.35)

uq

pEdcc Z Eqcc xdccid riq

(4.4.36)

Even if using the approximate formulas above, sometimes still make a further approximation, such as: (i) In formulas (4.4.35) and (4.4.36) take Z 1; 321

AC Machine Systems

(ii) In formulas (4.4.35) and (4.4.36) neglect pEqcc and pEdcc; (iii) In formula (4.4.35) take xqcc xdcc. In practical terms, for more convenience sometimes neglect the damping winding effect, by which the results can usually meet the practical requirement, but in some cases such as relatively large change of rotor speed omit the damping winding effect to bring about large error. In order to remedy the drawback of this approximation, a damping torque is added in electromagnetic torque to consider the damping winding effect referring to Section 4.2, i.e. the rotor motion equation (4.4.3) can be replaced by Hp 2G

Tm Te TD pG

in which the damping torque coefficient TD is taken approximately as a constant. Generally TD 1 3 and for very strong damping winding TD 25. Neglecting damping winding effect or without damping winding, the corresponding relations are xd xdc xad 1 id xdc id U fd 1 pTd 0 1 pTd 0 R fd

\d

\q

xq iq

(4.4.38)

Eqc

xad x xdc 1 U fd d id 1 pTd 0 R fd 1 pTd 0

(4.4.39)

(4.4.37)

In addition exists

\d

Eqc xdc id

(4.4.40)

According to the relations above and using the approximation like formulas (4.4.31) (4.4.34), the corresponding calculation results can be got. If using the following approximation,

p\ d

p ( Eqc xdc id ) | 0

(4.4.41)

p\ q

p ( xq iq ) | 0

(4.4.42)

Z\ d

Z ( Eqc xdc id ) | Eqc xdc id

(4.4.43)

Z\ q

Z ( xq iq ) | xq iq

(4.4.44)

then equations (4.4.29) and (4.4.30) can be reduced to

ud 322

xq iq rid

(4.4.45)


xdc id Eqc riq

uq

(4.4.46)

Rearranging equation (4.4.39) exists

pEqc

1 Td 0

° xad °½ U fd ( xd xdc )id Eqc ¾ ® ¯° R fd ¿°

(4.4.47)

and the corresponding electromagnetic torque formula is

Te

{Eqc ( xq xdc )id }iq

(4.4.48)

In analysis above, the influence of ferromagnetic saturation is not considered to cause a certain error. There are several methods to count in ferromagnetic saturation, in which it is the most convenient to use invariable saturatedparameters. However, this method still causes obvious error when the operating condition of synchronous machine changes in a wide range. In the following we shall introduce a method to adjust the saturated-parameters according to operating condition change. This method is based on the supposition that the leakage flux way is unsaturated and the leakage flux also has no effect on magnetic saturation, and the no-load characteristic of synchronous machine is simulated to consider the saturation effect. It is well-known that the mutual flux-linkages \ md and \ mq on

d-axis and q-axis during linearity are

\ md

xad (id I fd I1d )

\ mq

xaq (iq I1q )

xad idt

(4.4.49)

xaq iqt

(4.4.50)

id I fd I1d ½ ¾ iq I1q ¿

(4.4.51)

in which

idt iqt

represent the total currents respectively on d-axis and q-axis and are proportional to total mmfs respectively on d-axis and q-axis. Neglecting saturation effect, the flux-linkages \ md and \ mq are proportional to the currents idt and iqt i.e. xad and xaq are invariable. Considering saturation effect,

xad and xaq will change with saturation level. For convenience, assume the ratio of xaq to xad to be invariable, namely xaq / xad

\ md \ mq

xad idt xaq iqt

K dq

constant, so exist

xm idt K dq xm iqt 323

AC Machine Systems

The saturation level depends upon the total flux produced by the total mmf which is composed of d-axis and q-axis mmfs in perpendicular in space, so the total mmf will be proportional to an equivalent current it and can be written as

idt2 K dq2 iqt2

it

(4.4.52)

Correspondingly, the saturated values of xad and xaq are

xad xaq

xm f (it ) ½ K dq xm K dq f (it ) ¾¿

(4.4.53)

In the following, we discuss how to use no-load characteristic to get the value of xm f (it ). At no-load exist idt

0, and the corresponding flux-linkage \ md

I fd and iqt

and no-load emf uq are respectively

\ md uq

xm I fd

f ( I fd ) I fd

\ md

(4.4.54) (4.4.55)

If the no-load characteristic is indicated by

uq

g ( I fd )

(4.4.56)

then there is

\ md

g ( I fd )

f ( I fd ) I fd

namely

f ( I fd )

g ( I fd ) I fd

(4.4.57)

When the synchronous machine is loaded, the equivalent excitation current is it , so exists

xm

f (it )

g (it ) it

(4.4.58)

After finding out xm , we can use the saturated reactances xm and K dq xm to supersede xad and xaq in the above formulas during neglect of saturation effect, thus obtaining accurate calculation results. 324


4.4.2

Mathematical Patterns Corresponding to Analysis of Dynamic Stability for Synchronous Machine Systems

During analysis of dynamic stability for synchronous machine systems, different mathematical patterns can be adopted in terms of concrete research topics and computation methods. Several mathematical patterns in the following are suggested for reference only. (1) Pattern 1 This is the simplest mathematical pattern of synchronous machines. Based on the pattern to study dynamic stability, use the following approximation: (i) Neglect transformer emfs in the stator voltage equations. (ii) Except the rotor motion equation, assume rotor speed to be constant, namely Z 1, in other equations. (iii) Neglect the damping winding effect. (iv) Neglect the ferromagnetic saturation effect. (v) d-axis transient emf Eqc is constant. Obviously, the above equations (4.4.45), (4.4.46) and (4.4.48) belong to the relations of this mathematical pattern. For analysis of some dynamic transients, sometimes neglect the salient-pole effect further, i.e. take xq xdc approximately. Taking this approximation exists Eqc E c, so the mathematical pattern can be reduced further to the pattern with E c constant, in which E c is the transient emf behind the transient reactance referring to Section 2.7. (2) Pattern 2 In comparison with Pattern 1, this pattern will cancel the supposition that d-axis transient emf Eqc is kept constant, so besides the above relations (4.4.45), (4.4.46) and (4.4.48) the Pattern 2 also includes relation (4.4.47) to count the change of Eqc in. (3) Pattern 3 This is the simplest mathematical pattern of synchronous machines with damping windings. Based on the pattern to study dynamic stability, suppose the following conditions: (i) Take Z 1 in stator voltage equations (4.4.35) and (4.4.36). (ii) Let pEqcc pEdcc 0 in stator voltage equations (4.4.35) and (4.4.36). xad (iii) Take G c in equation (4.4.25). R fd (iv) Take G cc 0 in equation (4.4.26). The relations of pattern 3 are (4.4.25) (4.4.28), (4.4.35) and (4.4.36). (4) Pattern 4 In comparison with Pattern 3, for this pattern G c and G cc have to be calculated 325

AC Machine Systems

according to formula (4.4.20), so the basic relations of Pattern 4 are (4.4.20), (4.4.25) (4.4.28), (4.4.35) and (4.4.36). (5) Pattern 5 This pattern can consider not only damping winding effect but also ferromagnetic saturation influence, so the basic relations of Pattern 5 are (4.4.1), (4.4.9), (4.4.51) (4.4.53) and (4.4.58). In comparison with other patterns, this pattern has considered more factors, so the calculation results are more accurate but the computation is complicated.

4.4.3

Calculation Example of Dynamic Stability for Synchronous Machine Systems

By use of mathematical Pattern 1 analyze the dynamic stability of a synchronous machine system. Therefore, firstly discuss the relation between electromagnetic torque Te and power angle G. In the light of formula (4.4.48), the electromagnetic torque expression of Pattern 1 is {Eqc ( xq xdc )id }iq

Te

During steady-state symmetrical operation, referring to equation set (2.7.21) there are

id iq

Eqc U cos G xdc U sin G xq

(4.4.59) (4.4.60)

N.B. the angle G can be realized as the angle between terminal voltage and no-load emf referring to equation (2.5.2), and is also considered as the angle included between d, q axes and d c , qc axes referring to formula (4.4.2). Substituting formulas (4.4.59) and (4.4.60) into (4.4.48), the torque formula of steady-state symmetrical operation can be written as

Te

EqcU

1§ 1 1 · sin G ¨ ¸ U 2 sin 2G 2 ¨© xq xdc ¸¹ xdc

(4.4.61)

If using an equivalent machine without transient salient-pole effect to represent actual machine, then formula (4.4.61) referring to Section 2.7 can be reformed as

Te 326

E cU sin G eq xdc

(4.4.62)


where G eq is the angle included between the transient emf E c behind transient reactance and terminal voltage U for the equivalent machine. For convenience, G eq can be written as G here, so exists

Te

E cU sin G xdc

(4.4.63)

Now let a synchronous generator be connected with infinite source to illustrate the calculation of dynamic stability as shown in Fig. 4.4.1, in which the generator passes through double transmission lines to connect infinite source.

Figure 4.4.1 A generator passes through double transmission lines to connect infinite source

As you know, if neglecting stator resistance the output power of generator during normal operation will equal the electro magnetic torque. The output power before short circuit can be written as

P

E cU sin G xc

(4.4.64)

where

xc

xdc xT 1

1 xL xT 2 2

is the effective reactance between generator and infinite source; xT 1 and xT 2 are short circuit reactances of transformer 1 and transformer 2 respectively; xL is the reactance of each transmission line. When the power system in Fig. 4.4.1 is in normal situation, its transient P - G curve is indicated by curve I as shown in Fig. 4.4.2. Now supposing that the unsymmetrical short circuit occurs at point K on the left of transmission line 2 referring to Fig. 4.4.1, then during initial stage of short circuit the equivalent circuit 327

AC Machine Systems

of power system is shown in Fig. 4.4.3, in which xK represents the equivalent reactance caused by short circuit fault at point K. By use of Symmetrical Components Method, the values of xK can be indicated in Table 4.4.1, in which x2 and x0 are corresponding negative sequence and zero sequence reactances viewed from the fault point K of the power system. It is noted that here only consider the influence of positive component on rotor motion. In some cases, especially during unsymmetrical fault near generator terminal, the negative sequence current may produce obvious braking torque, which can be calculated according to negative sequence network to get stator negative sequence current and negative sequence resistance if necessary.

Figure 4.4.2

P - G curves before short circuit and after fault at point K.

Figure 4.4.3 Equivalent circuit during unsymmetrical short circuit at point K of the power system in Fig. 4.4.1 Table 4.4.1 Values of xK for different faults Types of unsymmetrical short circuit Two phases short circuit

Values of x K x2

Single phase earthed

x2 x0

Two phases short circuit earthed

x2 x0 x2 x0

In Fig. 4.4.3, U is the constant voltage of infinite source, E c is the transient emf of generator and xdc is its transient reactance. Reaching the steady-state finally, E c and xdc will change to E and xd respectively. Although E c can’t change suddenly during short circuit, it has to decay with time. Therefore, the 328


actual output power is in accord with the transient power characteristic during E c constant only at initial stage of short circuit. With time delayed, there is a deviation of actual output power from the transient power characteristic during E c constant. Figure 4.4.4(a) indicates the unstable condition of synchronous machine system. Figure 4.4.4(b) shows the stable situation, in which the actual power is approximate to the corresponding value on the transient power characteristic during E c constant for the first oscillation period, then the oscillation decays gradually, and finally the generator operates at point C of power-angle curve during excitation emf E constant to reach a new stable operation condition. Figure 4.4.4(c) belongs to another condition, in which the actual output power is approximate to the corresponding value on the transient power characteristic during E c constant for the first oscillation period, it seems to be stable, but the synchronous generator can’t reach a new stable operating condition because the P - G curve during E constant can’t intersect the line for prime mover power P0 . Under this condition, the stability will be lost after several oscillation periods. In practical terms, if the generator after fault doesn’t lose synchronism during the first oscillation, then there is a few possibility to step out after the first oscillation; at the same time, the subtransient component of power system decays very quickly to neglect it, and the transient component decays slowly, so during discrimination of dynamic stability, i.e. within the first oscillation period, the transient component can be considered as constant, which also gives the reason why mathematical Pattern 1 can be used to calculate dynamic stability approximately. In the following we shall use mathematical Pattern 1 to analyse dynamic stability of the power system shown in Fig. 4.4.1.

Figure 4.4.4

P - G curves for transients and in steady-state

Figure 4.4.3 indicates a T-form electric circuit, so it can be converted into a Sform electric circuit as shown in Fig. 4.4.5. The reactances xE and xu in Fig. 4.4.5 are connected in parallel with E c and U respectively, so they can be neglected for calculation of generator output power. Under this condition, the output active power of generator is

P

E cU sin G xcc

(4.4.65) 329

AC Machine Systems

where xcc is the equivalent reactance of power system corresponding to the Sform electric circuit.

xcc

§x · ( xdc xT 1 ) ¨ L xT 2 ¸ © 2 ¹ xc xK

is substituted into formula (4.4.65), and the corresponding transient P - G curve is shown by curve Ċin Fig. 4.4.2.

Figure 4.4.5

S-form electric circuit obtained from a T-form circuit in Fig. 4.4.3

From here we can see, due to sudden change of network parameters during short circuit, the output power of generator changes from point a on P - G curve ĉ in Fig. 4.4.2 to point b on P - G curve Ċ . In the meantime, because of regulator lagging for prime mover, the input mechanical power of generator still maintains constant and is equal to P0 . Due to the surplus power or surplus torque effect, the generator is accelerated to make the power angle G grow. With the power angle G growing, the generator output power also increases thus reducing the surplus power or surplus torque. At point c in Fig. 4.4.2 the surplus power will be equal to zero, but the rotor speed relative to d c , qc axes is maximum. Due to rotor inertia the power angle G will continue growing to exceed point c in Fig. 4.4.2. At this time, the output power of generator has been more than the mechanical power provided by prime mover, so the surplus torque becomes negative to make the rotor decelerate. For the deceleration period, the surplus power of generator is supplied by the storage kinetic-energy during acceleration. When reaching point d in Fig. 4.4.2, the kinetic energy stored in rotor has been used up and the rotor will stop relative motion to d c , qc axes, i.e. the rotor speed is restored to synchronous speed. However, the surplus torque at point d is of braking effect, so the rotor motion relative to d c , qc axes begins in the opposite direction. During opposite motion and due to rotor inertia, the rotor passes through point c to reach the minimum value of power angle G and then stops relative motion. Later on, the acceleration starts again. In so doing, the rotor operates at point c through several oscillations with decaying amplitude, and the point c is the stable point on new power-angle curve ĊIf the rotor exceeds the angle G kp corresponding to P0 on new P - G curve Ċduring the first oscillation referring to Fig. 4.4.2, then the surplus torque will become positive again to make rotor accelerated. Later on, with the power angle G growing, the 330


acceleration torque increases constantly to bring about the stepout of synchronous generator. P - G curves in Fig. 4.4.2 provide a short-cut method to determine the maximum position angle G max and to discriminate the stable possibility of power system. It is noted that during oscillation Z | Z s so there is 'Te 'P. For the acceleration period the rotor moves from G 0 to G CT , i.e. from point a to point c in Fig. 4.4.2, and the work done by surplus torque 'Te is Ayck

G CT

³G

'Te dG

0

G CT

³G

'PdG

f abc

(4.4.66)

0

where f abc represents the area of shade-line part abc. Therefore, for the acceleration period, the kinetic energy stored in rotor is just equal to the area f abc that is named the acceleration area. When the rotor position exceeding point c on new P ~ G curve Ċthe surplus torque will become negative to make rotor decelerated. During deceleration the rotor moves from angle G CT to angle G max , and the kinetic energy stored in rotor is negative, whose value can be expressed as ATOP

G max

³G

CT

'Te dG

G max

³G

'PdG

f cde

(4.4.67)

CT

The area f cde is termed the deceleration area. During deceleration period, the kinetic energy stored in rotor is returned to power system in order to compensate the deficiency of prime mover power. When the whole kinetic energy stored in rotor during acceleration period has been used up at the deceleration stage, i.e. the work ATOP done during deceleration is balanced against the work Ayck done during acceleration, then the rotor relative speed will be equal to zero to stop relative motion and the rotor operates at the maximum position angle G max . Therefore, according to the following formula, or

Ayck ATOP f abc f cde

0 ½° ¾ 0 °¿

(4.4.68)

we can get the value of angle G max . As mentioned above, the acceleration area must be equal to the deceleration area at the maximum position angle G max , so the topic is converted into how to determine point d (see Fig.4.4.2) to suffice the above condition, which can be solved by use of graphic method. In Fig. 4.4.2 the maximum possible deceleration area is equal to f cdcc obviously. If the area f cdcc being less than the acceleration area f abc , then the generator will lose synchronism. The ratio of possible deceleration area to acceleration area f cdcc / f abc is the storage mark of power system dynamic-stability, which is termed the storage coefficient of dynamic stability. 331

AC Machine Systems

When possible deceleration area is less than acceleration area, if cutting off the fault line quickly it is still possible to get stable operation because the transmitted power by use of single transmission line is more than the power transmitted by double lines during short circuit referring to Fig. 4.4.1. After cutting off the fault line, the transient power equation is

P

E cU sin G xccc

(4.4.69)

xccc xdc xT 1 xL xT 2 where Curve ċin Fig. 4.4.6 is the transient power characteristic corresponding to equation (4.4.69). In Fig. 4.4.6, curves ĉand Ċ indicate the transient power characteristics for normal situation and short circuit fault respectively.

Figure 4.4.6 Transient power characteristics for normal situation, short circuit fault and cut-off of fault line

As stated before, during short circuit the generator output power will decrease so the rotor begins to accelerate. If cutting off the fault line at point d on curve Ċ then the operation point after cut-off will change to point e on curve ċ(see Fig. 4.4.6), thus increasing the generator output power to make the maximum possible deceleration area f cdefccc be more than the sustained short circuit situation; furthermore, the more quickly cut off the fault line, i.e. the less the angle G OTK is, the more the deceleration area grows. Therefore, the cut-off of fault line can enhance dynamic stability of power system, which is simple and effective so is widely used in practice to become an important measure for raise of dynamic stability. In Figs. 4.4.2 and 4.4.6, the method to use area to discriminate dynamic stability is called the area rule. According to Fig. 4.4.6 we can use graphic method to find out the limit value of angle G OTK , at which cut off the fault line to guarantee stable operation, i.e. in the light of the rule that the acceleration area equals the maximum possible deceleration area we can get the limit value of cut-off angle G OTK . Similarly, by use of analytical method also obtain the limit 332


value of angle G OTK , i.e. let sum of the acceleration area and the maximum possible deceleration area be zero, then exists f abcdeccc

G OTK

³G

( P0 PmĊ sin G )dG

0

³

G KP G OTK

(P0 Pmċ sin G )dG

(4.4.70)

0

where PmĊ and Pmċ are maximum values on the transient power characteristics during short circuit and after cut-off of fault line respectively. After integral on the basis of formula (4.4.70) there is

P0 (G OTK G 0 ) PmĊ (cos G OTK cos G 0 ) P0 (G KP G OTK ) Pmċ (cos G KP cos G OTK )

0

(4.4.71)

whose solution is cos G OTK

P0 (G kp G 0 ) Pmċ cos G kp PmĊ cos G 0 Pmċ PmĊ

where the angles are expressed in radians. Substituting the values of G 0 and G KP S arc sin( P0 / Pmċ ) into the above formula, then the values of cos G OTK and G OTK can be found out. However, so far as the practice is concerned, the above calculation is not enough, because it is not necessary to know the angle G OTK when demanding interruption speed from the breaker and relay, but need the time at which the generator rotor reaches the angle G OTK , i.e. the permitted time limit of cutting off short circuit, shortly cut-off time limit. The cut-off time limit can’t be obtained according to the area rule above, so we have to adopt another method for analysis of dynamic stability, namely dividing intervals calculation. The dividing intervals calculation is also a routine method for analysis of dynamic stability, whose merits are obviously to watch the short circuit process at different instants and also to consider the influence of various factors on the cut-off time limit. For example, the dividing intervals calculation can count in effects of voltage regulator and armature reaction to determine the permitted time limit of cutting off short circuit. In fact, the dividing intervals calculation is to solve the rotor motion equation by use of difference method so it is an approximate method. The rotor motion equation in expression (4.4.3) can be reformed as

H d 2G 360 f dt 2

P0 P

(4.4.72)

where H is expressed in seconds and t also in seconds, the unit of G is electric degree, P0 and P are all expressed in per-unit. 333

AC Machine Systems

In equation (4.4.72), the maximum power Pm for P Pm sin G will change with different operating conditions such as normal operation, short circuit and cut-off of fault line referring to formulas (4.4.64), (4.4.65) and (4.4.69). In the light of equation (4.4.72), solve it for G f (t ) and then get a complete picture of G changing with time to discriminate whether the generator can keep in synchronism or not. However, equation (4.4.72) is nonlinear so there is no analytical solution generally, and it is necessary to solve it by use of numerical methods. As you know, there are various numerical methods, but here only introduce difference method. By use of difference method, the whole oscillation process can be divided into a series of time interval 't , and then the angle increment 'G will be evaluated step by step for each time interval. In general, the length of time interval 't is taken as 0.05 second. At the beginning of short circuit, the surplus power is assumed to be 'P(0) P0 Pm sin G 0 and will not change within the short time interval 't , so according to the acceleration motion formula and after the first time interval 't we can get the rotor angular speed increment 'Z(1) and power angle increment 'G (1) as follows: d 2G dt 2

'Z (1)

'G (1)

d 2G dt 2

't D (0) 't

(4.4.73)

G G0

G G0

't 2 2

D (0)

't 2 2

(4.4.74)

where D represents the angular acceleration. At the beginning of short circuit, the rotor relative angular speed to d c , qc axes equals zero so after the first time interval 't the rotor relative angular speed Z (1) equals the corresponding angular speed increment 'Z (1) , namely Z (1)

'Z(1) . In addition. On the basis of equation

(4.4.72) exists

D (0)

360 f 'P(0) H

(4.4.75)

so we can get

Z (1)

'Z(1)

D (0) 't

360 f 't 'P(0) H

(4.4.76)

and

'G (1) 334

D (0) 't 2 2

360 f 't 2 'P(0) 2 H

k

'P(0) 2

(4.4.77)


where 360 f 't 2 H

k

After obtaining the angle increment 'G (1) for the first time interval, then the power angle at the beginning of the second time interval is

G (1)

G 0 'G (1)

(4.4.78)

After having the value of G (1) , the surplus power 'P(1) for the second time interval can be determined as

P0 Pm sin G (1)

'P(1)

During calculation of the angle increment 'G (2) after the second time interval, besides the effect of acceleration D (1) exists the rotor relative angular speed Z (1) at the beginning of the second time interval, i.e.

'G (2)

Z (1) 't

D (1) 't 2

Z (1) 't k

2

'P(1) 2

N.B. it is not very accurate to get the rotor relative angular speed Z (1)

(4.4.79)

D (0) 't

by use of formula (4.4.76), because the surplus power 'P(0) and acceleration

D (0) are not constant for the first time interval. If taking the average acceleration for the first time interval to calculate Z (1) , i.e. taking

D (0) D (1)

D (0) cp

2

then the result is more accurate. In so doing, the rotor relative angular speed Z (1) will become

Z (1) D (0) cp 't

D (0) D (1) 2

't

Substituting the above formula into equation (4.4.79) and referring to formula (4.4.77), there is 'G (2)

D (0) D (1) 2

D (0) 't 2 2

't 2

D (1) 't 2

D (1) 't 2

2 'G (1) k 'P(1)

(4.4.80) 335

AC Machine Systems

After getting 'G (2) we can evaluate G (2) , 'P(2) and 'G (3) as follows,

G (2)

G (1) 'G (2)

'P(2)

P0 Pm sin G (2)

'G (3)

'G (2) k 'P(2)

If cutting off the fault line at the k-th time interval referring to Fig. 4.4.7, then the surplus power will change suddenly from 'P(cK 1) to 'P(ccK 1) . When evaluating the power angle increment 'G ( K ) for the first time interval after cut-off, the surplus power can be taken as the average value of 'P(cK 1) and 'P(ccK 1) , namely

'G ( K )

'G ( K 1) k

'P(cK 1) 'P(ccK 1) 2

(4.4.81)

For the later time intervals, we can still use formula (4.4.80) to estimate the increment of power angle G. If the power angle G doesn’t begin decreasing or can’t be predicted to be infinite, the calculation must go on. The dividing intervals calculation can be used together with the area rule. Firstly, the area rule is used to find out the power angle limit G OTK of cutting off the fault line, and then the time tOTK is the cut-off time limit, referring to Fig. 4.4.8 that is obtained by use of the dividing intervals calculation.

Figure 4.4.7

P -G curves if cutting off the fault line at the k-th time interval

Figure 4.4.8

G - t curve obtained by use of the dividing intervals calculation

and the cut-off time limit 336


In order to illustrate the conception and computation of generator dynamic stability, the simplest mathematical pattern 1 is adopted in the above example. Of course, other mathematical patterns can also used if needing higher accuracy. Finally, as should be pointed out, in this section we only discussed the basic relations and mathematical patterns of synchronous machine systems, and then discriminated the dynamic stability of power system during the transient emf E c constant and according to the first oscillation period of the generator rotor for convenience. In many cases, the dynamic process of synchronous machine systems develops quickly, generally in 1 2 seconds after undergoing a disturbance we can discriminate whether the power system is stable or not. In addition, the output power of prime-mover can’t change obviously in a short time due to large inertia of speed regulator for the prime-mover, so the effect of speed regulator may be neglected temporarily to consider the output power of prime-mover to be constant. Furthermore, due to large inductance of excitation winding its flux-linkage also can’t change clearly in a short time, so the influence of excitation regulation may be neglected to consider the transient emf E c to be constant. Therefore, the method in this section, even if neglecting the effect of speed regulator and excitation regulation, still can give satisfactory results. However, it is different for long-distance transmission or weakly linking power system, which may lose dynamic stability through many times of oscillations for a long time such as several seconds or even tens seconds. Evidently, under this condition it is not suitable to discriminate the dynamic stability according to the first period of oscillation as mentioned before. As is well-known, the influence of excitation regulation and prime-mover speed regulation on dynamic stability depends upon not only the regulation speed but also cut-off time of the fault line. For modern power systems, automation is high to make the cut-off time very short, so the influence of those regulation systems on the first period of oscillation is finite. However, the effect of regulation systems will be raised with time to bring about an important effect on the generator rotor oscillation. Therefore, when studying the dynamic stability for this kind of power system, the effects of speed-regulation and excitation-regulation systems must be paid attention to. In so doing, it is necessary not only to write down accurate mathematical patterns of synchronous machine as mentioned before but also to express the corresponding relations of speed-regulation and excitation-regulation systems. Obviously, it is complicated to solve the non-linear simultaneous differential equations. In addition, when the power system undergoing a disturbance or fault, whether the synchronous machine loses synchronism is no doubt an important problem, but whether the generator oscillation can be depressed quickly is also an important problem since it influences the dynamic quality of power system. Therefore, although the calculation as stated above is complicated, it is still necessary in practice because so far as the regulation systems are concerned, increase of dynamic stability limit is not in accord with the requirement of dynamic quality improvement. 337

AC Machine Systems

References [1] Arthur R Bergen (1986) Power systems analysis. Prentice-Hall, London [2] Demello F P, Concordia C (1969) Concepts of synchronous machine stability as affected by excitation control. J IEEE Trans. PAS-88(4): 316 329 [3] Gao J D (1963) AC machine transients and operating modes analysis (Chapter 4, 5, in Chinese). Science Press, Beijing [4] Gao J D, Zhang L Z (1982) Fundamental theory and analysis mothods of transients in electric machines (Chapter 7, 8, in Chinese). Science Press, Beijing [5] Gao J D, Wang X H, Li F H (2004) Analysis of ac machines and their systems, 2nd Ed (Chapter 3, in Chinese). Tsinghua University Press, Beijing [6] Zhang L Z, Chen L F (1991) Analysis of subsynchronous resonance in a turbogenerator and its shstem using compensation capacitors. Proc CICEM, Huazhong University of Science and Technology Press, Wuhan, pp 104 108 [7] Zhang L Z, Yao L (1996) Effect of transfoemer connection banks on analysis of a synchronous generator and its system during unsymmetrical faults. Proc ICEMA’96, International Academic Publishers, Harbin, pp 369 373

338

5 Electromagnetic Relations of Induction Machine Systems and Analyses of Some Operating Modes

Abstract Induction machine systems are widely used in practice, so their operating behavior, especially transient performance, must be paid attention to. Firstly, we introduce the basic relations and parameters of induction machines in d, q, 0 axes, and then analyze the starting process of induction motors, in which dc, qc, 0 axes with synchronous speed are preferable to d, q, 0 axes due to variable rotor speed. The relations in dc, qc, 0 axes can be found out according to basic equations in d, q, 0 axes together with the conversion formulas in Section 3.1. By using the 4th order Runge-Kutta Method, we calculate the starting characteristics for different fly-wheel moments and loads, compare the transient starting characteristics with the steady-state ones, which are quite different, especially during no-load and small rotating inertia, and then illustrate why the transient starting characteristics are quite different from the steady-state ones. Furthermore, some fault is often temporary, so the production loss caused by failure in power supply can be decreased with the reswitching of the induction motors as quickly as possible after supply recovery. Transients of reswitching of the induction motors include two stages, one is the disconnection process due to source fault, and the other is the reclosing process after supply recovery, whose main problems are surge current and large electromagnetic torque; therefore we analyze them using D , E , 0 axes. As you know, the reclosing surge current is more serious than the direct starting current, which can be diminished in two ways: monitor the stator voltage phase-angle at the reclosing instant and use the extinguishing-flux method to make rotor current decay quick at the disconnecting instant. Moreover, induction motors in series with capacitance can keep the supply voltage constant, but sometimes self-excitation may occur, which is analyzed by using the D-domain partition method to get two self-excitation regions during lower compensation degree that is a new discovery, because some scientists pointed out only one self-excitation region before. In addition, a special generator is introduced as an important part of ship’s integrated power system, in which the stator has two suits of windings, one is a 12-phase power winding connected to a rectifier load and another is a 3-phase control winding connected to a static excitation regulator, and there is a squirrel-cage solid rotor. There are also the capacitors for

AC Machine Systems

self-excitation and inter-phase reactors in the system for performance improvement. It can be divided into two sub-systems: one consisting of 12-phase power winding with a rectifier load calculated by the circuit method, and another consisting of dual-stator-winding and squirrel-cage solid rotor analyzed by the electromagnetic field finite element method, the simulative results being approximate to experimental data.

As far as the structures and uses are concerned, induction machines are different from synchronous machines, but their electromagnetic relations are the same, i.e. some coils are coupled by fluxes. Therefore, the analysis methods and concepts of synchronous machines can also be used for induction machines. Furthermore, the basic equations of induction machines due to symmetrical rotor are simpler than those of synchronous machines. In this chapter, we mainly discuss the conventional 3-phase induction machines; the other types can be studied in a similar way. The voltage equations, flux-linkage equations and parameters of stator and rotor windings are the basis for studying the performance of induction machines. Learning the equations in steady-state and equivalent circuit, we can find out the relation between multi-loop method and conventional method. The basic equations and parameters in d, q, 0 axes are useful in the analysis of induction machine performance. Those voltage and flux-linkage equations can easily be converted to other axes, if necessary. Starting characteristic is an interesting topic. Moreover, the reclosing transient process of induction motors and self-excitation during motors in series with capacitors are also important, which are discussed in this chapter, refer to [1,2,3,5,10 14]. In addition, a dual-stator-winding multi-phase high-speed induction generator is also studied here, refer to [6 9].

5.1

Basic Relations and Parameters of Induction Machines

In chapter 1, the electromagnetic relations and parameters of ac machines have been found out by using a single coil. In this section, conventional 3-phase induction machines will be discussed according to the above mentioned results and reference directions in Section 1.10, i.e. the current entering the machines is positive, the reference directions of voltage drop and current are the same, and the positive currents will produce positive flux-linkages for stator and rotor. On the basis of reference directions, the voltage equations of induction machines can be written as 340

5 Analyses of Some Operating Modes of Induction Machine Systems

ua ub uc

p\ a ria ½ ° p\ b rib ¾ p\ c ric °¿

(5.1.1)

where ua , ub and uc are stator phase voltages, respectively,

ia , ib and ic are stator phase currents, respectively,

\ a , \ b and \ c are stator phase flux-linkages, respectively, d is differential operator. dt The rotor winding of induction machines may be wound-rotor type or squirrelcage one, whose number of phases may or may not be equal to stator phases.

r is stator phase resistance and p

5.1.1

Wound-Rotor Induction Machines

Phase voltage equations for wound-rotor are uac ubc ucc

p\ ac r ciac ½ ° p\ bc r cibc ¾ p\ cc r cicc °¿

(5.1.2)

The corresponding flux-linkage equations can be written as

\a \b \c \ ac \ bc \ cc

Laa ia M ab ib M ac ic M aac iac M abc ibc M acc icc ½ ° M ba ia Lbb ib M bc ic M bac iac M bbc ibc M bccicc ° M ca ia M cb ib Lcc ic M caciac M cbcibc M cccicc °° ¾ M aca ia M acb ib M acc ic Lacaciac M acbcibc M acccicc ° M bca ia M bcb ib M bcc ic M bcaciac Lbcbcibc M bcccicc ° ° M cca ia M ccb ib M ccc ic M ccaciac M ccbcibc Lccccicc °¿

(5.1.3)

in which the signs with prime belong to rotor quantities. Just as 3-phase synchronous machines, the air-gap flux of 3-phase induction machines approximates to sinusoidal wave by means of distribution, fractional pitch and skew-slot windings; so they can be considered as ideal machines. If necessary, we can count higher harmonics in by using the method stated in Chapter 1. The hypotheses of ideal machines are as follows: (i) The influences of ferromagnetic saturation, hysteresis and eddy current can be neglected, and the skin effect of iron-cores and conductors is also omitted. (ii) The rotor is cylindrical, air-gap is uniform, and the surfaces of stator and rotor are all smooth without slot effect. 341

AC Machine Systems

(iii) 3-phase coils of stator are symmetrical, and each phase winding will produce mmf and flux density distributed sinusoidally in space. (iv) The rotor winding is of multi-phase and symmetrical. Under the supposition of ideal induction machine and referring to (1.10.8), the self-inductance of stator phase winding is Laa

Lbb

2w2W l O0 k w2 L0l P S

Lcc

constant

(5.1.4)

Referring to (1.10.10), the mutual inductance between stator phase windings is M ab

M bc

2 w2W l O0 kw2 cos120e M ml PS

M ca

w2W l O0 kw2 M ml P S

(5.1.5)

constant

Similarly, the self-inductance of rotor phase winding is Lacac

Lbcbc

Lcccc

2( wc) 2 W l O0 (kwc )2 L0cl P S

constant

(5.1.6)

The mutual inductance between rotor phase windings is M acbc

M bccc

M ccac

( wc) 2W l O0 (kwc ) 2 M mlc PS

constant

(5.1.7)

In which: x L0l is the leakage self-inductance of stator phase winding. x M ml is the leakage mutual inductance between stator phase windings. x L0cl is the leakage self-inductance of rotor phase winding. c is the leakage mutual inductance between rotor phase windings. x M ml x W and W c are number of turns in series per phase for stator and rotor, respectively. x kw and kwc are stator and rotor winding-factors for the fundamental wave. respectively. Referring to (1.10.16), the mutual inductance between the stator and rotor windings is

M aac

M bbc

M ccc

2wwcW l O0 kw kwc cos J PS

(5.1.8)

where J is the included electric angle between stator phase a and rotor phase a' winding axes, J velocity of rotor. 342

t

³ Zdt J 0

0

, J 0 is the rotor initial-angle, Z is the electric angular


M abc M acb

M bcc M bcc

M cac M cca

2 wwcW l ½ O0 kw kwc cos(J 120e )° ° PS ¾ 2 wwcW l ° c e O k k cos( J 120 ) 0 w w °¿ PS

(5.1.9)

The above parameters are substituted into flux-linkage and voltage equations of stator phase a, and there are

\a

ua

§ 2 w2W l · § w2W l · O0 kw2 L0l ¸ ia ¨ O0 kw2 M ml ¸ (ib ic ) ¨ © PS ¹ © PS ¹ 2wwcW l O0 kw kwc [cos J iac cos(J 120e )ibc cos(J 120e )icc ] PS p\ a ria § 2 w2W l · § w2W l · 2 O O0 kw2 M ml ¸ p (ib ic ) k L pi ¨ ¨ 0 w 0l ¸ a © PS ¹ © PS ¹ 2wwcW l O0 kw k wc [cos J piac iac sin J pJ cos(J 120e ) pibc PS ibc sin(J 120e ) pJ cos(J 120e ) picc icc sin(J 120e ) pJ ] ria

(5.1.10)

Similarly, the flux-linkage and voltage equations of rotor phase a' are

\ ac

uac

2wwcW l O0 kw kwc [cos J ia cos(J 120e )ib PS § 2 wc2W l · )ic ] ¨ O0 kwc2 L0cl ¸ iac cos(J 120e © PS ¹ § wc2W l · O0 kwc2 M mlc ¸ (ibc icc ) ¨ © PS ¹ p\ ac r ciac 2wwcW l O0 kw kwc [cos J pia ia sin J pJ P S ) pib ib sin(J 120e ) pJ cos(J 120e cos(J 120e ) pic ic sin(J 120e ) pJ ] § 2wc2W l · O0 kwc2 L0cl ¸ piac ¨ © PS ¹ 2 § wc W l · ¨ O0 kwc2 M mlc ¸ p(ibc icc ) r ciac © PS ¹

(5.1.11)

343

AC Machine Systems

In a similar way, we can get the flux-linkage and voltage equations of stator phases b, c and rotor phases bc, cc. According to the above relations and parameters together with electromagnetic torque and motion equations, we can study various operational modes of wound-rotor induction machines. Firstly, we analyze the 3-phase symmetrical operation in steady-state. Suppose that the stator currents in steady-state are

ia

2 I sin(Z t M )

ib

2 I sin(Z t M 120e )

ic


and the rotor currents are iac

2 I c sin( sZ t M c)

ibc

2 I c sin( sZ t M c 120e )

icc

2 I c sin( sZ t M c 120e )

where M and M c are the initial phase angles of the corresponding currents, s is rotor slip, and J (1 s )Z t J 0 . The current expressions above are substituted into (5.1.10), and there exists ua

§ 3w2W l · O0 kw2 L0l M ml ¸ Z u 2 I cos(Z t M ) ¨ © PS ¹ 3wwcW l O0 kw kwc Z u 2 I c cos(Z t M c J 0 ) PS r u 2 I sin(Z t M ) ( x1 xm ) u 2 I cos(Z t M ) x12 u 2 I c u cos(Z t M c J 0 ) r u 2 I sin(Z t M )

(5.1.12)

in which x1 Z ( L0l M ml ) is the stator leakage reactance, 3w2W l O0 kw2 is the magnetization reactance of induction machines, P S 3wwcW l O0 kw kwc is the mutual reactance between stator and rotor. x12 Z PS The same way is true of voltage equations for stator phases b and c. The voltage equation of rotor phase a' is xm

344

Z


uac

3wwcW l O0 kw kwc sZ u 2 I cos( sZ t M J 0 ) P S § 3wc2W l · ¨ O0 kwc2 L0cl M mlc ¸ sZ u 2 I c cos( sZ t M c) S P © ¹ r c u 2 I c sin( sZ t M c) sx12 u 2 I cos( sZ t M J 0 ) s( x2 xm 2 ) u 2 I c cos( sZ t M c) r c u 2 I c sin( sZ t M c)

(5.1.13)

Z ( L0cl M ml c ) is rotor leakage reactance, Z 3wc2W l O0 (kwc ) 2 is the magnetization reactance referred to rotor. 2

where x2 xm 2

PS The same way is true of voltage equations for rotor phases b' and c'. wkw k wkw 1 is the , in which k Two sides of (5.1.13) are multiplied by wckwc s wck wc s

ratio of stator phase-winding effective turns number to rotor effective turns number. For a wound-rotor induction machine, its rotor winding is usually short-circuited, i.e., uac ubc ucc 0, hence 0

3w2W l O0 kw2Z u 2 I cos( sZ t M J 0 ) PS ª 3wwcW l º wk « O0 kw kwc w ( L0cl M mlc ) » wckwc ¬ PS ¼ u Z u 2 I c cos( sZ t M c)

wkw r c u 2 I c sin( sZ t M c) wckwc s

xm u 2 I cos( sZ t M J 0 ) ( x12 kx2 ) u 2 I c cos( sZ t M c) rc k u 2 I c sin( sZ t M c) s

(5.1.14)

N. B. The above formula is the voltage equation for angular frequency sZ . If J 0 0, then (5.1.12) and (5.1.14) can be changed respectively into

ua

( x1 xm ) u 2 I cos(Z t M ) x12 u 2 I c cos(Z t M c) r u 2 I sin(Z t M )

0

xm u 2 I cos( sZ t M ) ( x12 kx2 ) u 2 I c cos( sZ t M c) rc k u 2 I c sin( sZ t M c) s

(5.1.15)

(5.1.16) 345

AC Machine Systems

Let I 2c

wckwc Ic wkw

Ic k

then (5.1.15) and (5.1.16) are changed respectively into ua

( x1 xm ) u 2 I cos(Z t M ) xm u 2 I 2c cos(Z t M c) r u 2 I sin(Z t M )

0

where x2c r2c

xm u 2 I cos( sZ t M ) ( xm x2c ) u 2 I 2c cos( sZ t M c) rc 2 u 2 I 2c sin( sZ t M c) s

(5.1.17)

(5.1.18)

k 2 x2 is the rotor leakage reactance referred to stator, k 2 r c is the rotor resistance referred to stator.

Formulas (5.1.17) and (5.1.18) are voltage equations expressed as instantaneous values, which can be written as complex operator equations as follows: 2U sin(Z t M u ), then (5.1.17) and (5.1.18) can be converted into If ua Im(Ue jMu e jZ t )

Im[ j( x1 xm ) Ie jM e jZ t ] Im[ j( xm I 2c e jM c e jZ t ) Im(rIe jM e jZ t )

0

(5.1.19)

Im( jxm Ie jM e jsZ t ) Im[ j( xm x2c ) I 2ce jM c e jsZ t ]

§ rc · Im ¨ 2 I 2ce jM c e jsZ t ¸ ©s ¹

(5.1.20)

Taking away the sign of imaginary part, there are Ue jMu e jZ t

j( x1 xm ) Ie jM e jZ t jxm I 2c e jM c e jZ t rIe jM e jZ t

0

jxm Ie jM e jsZ t j( xm x2c ) I 2c e jM c e jsZ t rc 2 I 2ce jM c e jsZ t s

(5.1.21)

(5.1.22)

Formulas (5.1.21) and (5.1.22) are not simultaneous equations due to different frequencies. Multiplying (5.1.22) by e j(1 s )Z t , there exists 0

346

jxm Ie jM e jZ t j( xm x2c ) I 2ce jM c e jZ t rc 2 I 2ce jM c e jZ t s

(5.1.23)


Now (5.1.21) and (5.1.23) are simultaneous equations having the same frequency, which can be reduced to U 0

j( x1 xm ) I jxm I2c rI ½ ° r2c ¾ jxm I j( xm x2c ) I 2c I 2c ° s ¿

(5.1.24)

or U 0

(r jx1 ) I jxm ( I I2c )

½ ° § r2c · ¾ jxm ( I I 2c ) ¨ jx2c ¸ I 2c ° ©s ¹ ¿

(5.1.25)

According to (5.1.25), there exists the equivalent circuit of induction machines as shown in Fig. 5.1.1, which is familiar to us in Electric Machinery.

Figure 5.1.1 Equivalent circuit of induction machines

After getting relations and equivalent circuit of induction machines in steadystate, the corresponding phasor diagram can be drawn and various characteristics found out. Just as Electric Machinery, the referring calculation has two steps, i.e. refer different numbers of turns to the same, and then refer different frequencies to the same too, for stator and rotor. After referring calculation, the electromagnetic relations for stator and rotor can be drawn in the common circuit. Obviously, the air-gap flux and its speed produced by actual rotor current are equal to those caused by the rotor current referred to stator.

5.1.2

Squirrel-cage Rotor Induction Machines

For squirrel-cage rotor, suppose that the number of bars is Z 2 , and the number of bars per pole-pair is Z 2 / P, i.e. the number of rotor phases is K Z 2 / P, in which each phase is composed of P branches in parallel. Hence, voltage and flux-linkage equations for stator and rotor can be written as 347

AC Machine Systems

ua ub uc un

\a \b \c \n

½ ° ° p\ b rib ° p\ c ric ° ° r r ° p\ n R in c (in 1 in 1 ) P P ° Laa ia M ab ib M ac ic M a1i1 " M aK iK ¾ ° M ba ia Lbb ib M bc ic M b1i1 " M bK iK ° ° M ca ia M cb ib Lcc ic M c1i1 " M cK iK ° M na ia M nb ib M nc ic M n1i1 " M nK iK ° ° n 1, 2," , K °¿

p\ a ria

(5.1.26)

where rR is the resistance of one loop for the squirrel-cage rotor, rC is the bar resistance, \ n is rotor flux-linkage of the n-th phase and the subscript n indicates the order number of rotor-phases. The stator inductances for squirrel-cage type are the same as those for woundrotor type, refer to (5.1.4) and (5.1.5). Referring to (1.10.16), the mutual inductance between stator phase a and rotor n-th phase is

M an

kE S 2 wW l 1 O0 ¦ 2 kwk sin r cos k[J (n 1)M 0 ] 2 PS k k

(5.1.27)

where k 1, 3, " J is the electric included angle between winding axes of rotor first phase and stator phase a, 360e . M 0 is the electric angle between two adjacent rotor-phases, M 0 K During the 3-phase symmetrical operation in steady-state, the air-gap flux is approximate to sinusoidal, so (5.1.27) is changed to

M an

ES 2 wW l O0 kw sin r cos[J (n 1)M 0 ] PS

(5.1.28)

According to (1.5.14), the self-inductance of rotor phase winding is

L11

L22

" LKK

2W l P 2 S

O0

¦j j

2

sin 2

jEr S L0cl 2

(5.1.29)

On the basis of (1.5.17), the mutual inductance between two rotor-phase windings is 348


2W l P 2 S

M mn

O0

¦j j

2

sin 2

jEr S (m n)360e c cos j M mnl K

(5.1.30)

N. B. the rotor phase winding is composed of P branches in parallel, so the inductance expressions (5.1.29) and (5.1.30) have a sign P more in their c is the rotor denominators, in which L0lc is the rotor leakage self-inductance, M mnl leakage mutual inductance between the m-th phase and n-th phase, (m n)360e/ K is the electric included angle between the winding axes of rotor m-th phase and n-th phase and j 1,3," In rotor inductance expressions, there are relative strong higher harmonics for squirrel-cage type. However, during the 3-phase symmetrical operation in steadystate, the air-gap flux caused by rotor winding is still approximate to sinusoidal, which can be considered as ideal electric machine, too. Therefore, the selfinductance and mutual inductance of rotor phase-winding can be expressed as L11 M mn

ES 2W l ½ O sin 2 r L0cl constant ° 2 0 PS ° E S e 2W l ( m n )360 ° 2 r c O0 sin M mnl cos ¾ P 2 S K ° ES ° 2W l c O sin 2 r cos(m n)M 0 M mnl constant ° 2 0 PS ¿

L22

" LKK

(5.1.31)

The above inductances are substituted into the corresponding flux-linkage equations, and the flux-linkage equations are substituted into voltage equations, and then there are

\a

ua

§ 2 w2W l · § w2W l · 2 O k L i O0 kw2 M ml ¸ (ib ic ) ¨ ¨ 0 w 0l ¸ a © PS ¹ © PS ¹ E S 2wW l O0 kw sin r [cos J i1 cos(J M 0 )i2 " PS cos(J ( K 1)M 0 )iK ] p\ a ria § 2 w2W l · § w2W l · 2 O k L pi O0 kw2 M ml ¸ p(ib ic ) ¨ ¸ ¨ 0 w 0l a © PS ¹ © PS ¹ Er S 2wW l O0 kw sin p[cos J i1 cos(J M 0 )i2 " PS cos(J ( K 1)M 0 )iK ] ria

(5.1.32)

349

AC Machine Systems

Similarly, the flux-linkage and voltage equations of rotor n-th phase winding are

\n

2 w1W l ES O0 kw sin r [cos(J (n 1)M 0 )ia P S cos(J (n 1)M 0 120e )ib cos(J (n 1)M 0 120e )ic ]

ES 2W l O sin 2 r [cos(n 1)M 0 i1 " 2 0 P S c iK ) cos(n K )M 0 iK ] ( M 1cnl i1 " M Knl

un

r rR in c (in 1 in 1 ) P P 2 w1W l Er S O0 kw sin p[cos(J (n 1)M 0 )ia P S cos(J (n 1)M 0 120e )ib cos(J (n 1)M 0

0

p\ n

ES 2W l O sin 2 r p[cos(n 1)M 0i1 " 2 0 P S c iK ) cos(n K )M 0 iK ] p ( M 1cnl i1 " M Knl 120e )ic ]

r rR in c (in 1 in 1 ) P P

(5.1.33)

The same way is true of the flux-linkage and voltage equations for stator phases b, c and rotor other phases. According to those basic relations and parameters, together with electromagnetic torque and rotor motion equations, we can study various operation modes for squirrel-cage rotor induction machines. Now we discuss the 3-phase symmetrical operation in steady-state. If ia 2 I sin(Z t M ) ib


ic


If

i1

2 I c sin( sZ t M c)

then

i2

2 I c sin( sZ t M c M 0 )

iK

2 I c sin[ sZ t M c ( K 1)M 0 ]

then

and Supposing J 0

J

(1 s )Z t J 0

0, there is

J

(1 s )Z t

Substituting the above formulas into (5.1.32), there exists 350


ua

§ 3w2W l · O0 kw2 L0l M ml ¸ u 2 Ip sin(Z t M ) ¨ P S © ¹ ES wW l O0 kw sin r u 2 I cKp sin(Z t M c) PS r u 2 I sin(Z t M ) § 3w2W l · O0 kw2 L0l M ml ¸ Z u 2 I cos(Z t M ) ¨ P S © ¹ ES wW lK O0 kw sin r Z u 2 I c cos(Z t M c) PS r u 2 I sin(Z t M ) ( x1 xm ) u 2 I cos(Z t M ) x12 u 2 I c cos(Z t M c) r u 2 I sin(Z t M )

(5.1.34)

in which

ES wW lk O0 kw sin r is the mutual reactance between stator and rotor for PS squirrel-cage type. The same way is true of voltage equations for phases b and c. The voltage equation of rotor n-th phase is changed to x12

Z

0

ES 3wW l O0 kw sin r sZ u 2 I cos[ sZ t M (n 1)M 0 ] P S ES W lK 2 O0 sin 2 r sZ u 2 I c cos[ sZ t M c (n 1)M 0 ] P S § Lr 2 Lc · cos M 0 ¸ sZ u 2 I c cos[ sZ t M c (n 1)M 0 ] ¨ P ©P ¹ § r 2r · ¨ r c cos M 0 ¸ u 2 I c sin[ sZ t M c (n 1)M 0 ] ©P P ¹

(5.1.35)

Lc M (cn 1) nl M (cn 1) nl is the leakage mutual inductance between two P adjacent rotor-phases, which is zero for nonadjacent rotor-phases of squirrelcage type, Lr L0cl , P Lc is the leakage inductance of one rotor-bar, where

Lr is the leakage inductance of one rotor-loop. 351

AC Machine Systems

Two sides of (5.1.35) are multiplied by

k s

wk w s sin

Er S

, and there is

2

0

3w W l O0 kw2Z u 2 I cos[ sZ t M (n 1)M 0 ] PS ES wW lK O0 kw sin r Z u 2 I c cos[ sZ t M c (n 1)M 0 ] PS wkw ( L 2 Lc cos M 0 )Z u 2 I c cos[ sZ t M c (n 1)M 0 ] ES r P sin r wkw rr 2rc cos M 0 u 2 I c sin[ sZ t M c (n 1)M 0 ] Er S s P sin xm u 2 I cos[ sZ t M (n 1)M 0 ] ( x12 kx2 ) u 2 I c cos[ sZ t M c (n 1)M 0 ] rc k u 2 I c sin[ sZ t M c (n 1)M 0 ] s

(5.1.36)

If we write the voltage equation for rotor first phase, namely n 1, then (5.1.36) is changed to 0

xm u 2 I cos( sZ t M ) ( x12 kx2 ) u 2 I c cos( sZ t M c) rc k u 2 I c sin( sZ t M c) s

(5.1.37)

where x2 Z ( Lr 2 Lc cos M 0 ) / P is rotor leakage reactance, r c (rr 2rc cos M 0 ) / P is rotor resistance. ES K sin r I c KI c Let I 2c 3wk w 3k and then (5.1.34) and (5.1.37) are changed respectively to ua

( x1 xm ) u 2 I cos(Z t M ) xm u 2 I 2c cos(Z t M c) r u 2 I sin(Z t M )

0

352

xm u 2 I cos( sZ t M ) ( xm x2c ) u 2 I 2c cos( sZ t M c) rc 2 u 2 I 2c sin( sZ t M c) s

(5.1.38)

(5.1.39)


in which 3 2 x2c k x2 is the rotor leakage reactance referred to stator, K 3 2 r2c k r c is the rotor resistance referred to stator. K Equations (5.1.38) and (5.1.39) are identical with (5.1.17) and (5.1.18). Hence, by the use of complex operators we can write the voltage equations for squirrel-cage type as U

(r jx1 ) I jxm ( I I2c )

½ ° § rc · ¾ jxm ( I I2c ) ¨ 2 jx2c ¸ I2c ° ©s ¹ ¿

0

(5.1.40)

Similarly, the corresponding equivalent circuit can be plotted just as in Fig. 5.1.1. In order to compare the equivalent circuit with that in Electric Machinery, take rotor phase current as PI c , i.e. the rotor phase winding is composed of P bars in parallel, which belong to the same phase under P pole-pairs. Thus, there exist PI c

Ic

2sin KI c 3k

I 2c

PI c

M0

2sin

2

KPI c 6k sin

3 2 k x2 K

x2c

Er S

Er S

K PI c 6wk w

PI c ki

3 2 k Z ( LR 2 Lc cos M 0 ) / P K

Because LR 2( Lc Lk ), in which Lk is one ring-segment inductance for squirrelcage, there are x2c

3 K

w2 k w2 sin

2

Er S

Z u 2[ LK Lc (1 cos M 0 )]/ P

§ 12 Pw2 k w2 ¨ LK Z ¨ Lc M Z2 ¨¨ 2sin 2 0 © 2 § · ¨ ¸ LK ke kiZ ¨ Lc ¸ P M 2sin 2 0 ¸¸ ¨¨ © 2 ¹

· ¸ ¸ ¸¸ ¹

P

353

AC Machine Systems

r2c

§ ¨ rK 3 2 k r c ke ki ¨ rc M K ¨¨ 2sin 2 0 2 ©

· ¸ ¸ ¸¸ ¹

P

where ki

6 wk w K

ke

2 wk w

k e ki

6 Pwk w Z2

12 Pw2 k w2 Z2

The current transformation ratio ki , emf ratio ke and impedance ratio ke ki are familiar to us in Electric Machinery. In this section, considering only the fundamental flux and using the Multi-loop method stated in Chapter 1, we get the same results as in Electric Machinery for steady-state. However, during strong higher harmonic flux, special winding or inner-asymmetrical fault, it is better to use the Multi-loop method, because of accurate consideration to higher harmonics and precise results, which is clear from the example of single phase induction machine in Section 1.10.

5.2

Basic Relations and Parameters of Induction Machines in d, q, 0 Axes

In Section 5.1, we obtained the voltage and flux-linkage equations with corresponding parameters according to phase-coordinates, analyzed the steadystate operation and obtained the complex operator equations and corresponding equivalent circuit. However, for analyzing some special operation modes and transients of induction machines, it is necessary to convert coordinates just as synchronous machines. In this section, we discuss the basic relations and parameters in d, q, 0 axes for induction machines. On the basis of conversion formulas stated in Section 3.1, it is not difficult to find out induction machine relations in other axes, if necessary. Again we take the ideal-machine hypothesis and the reference directions of currents, voltages and flux-linkages in Section 5.1. The stator winding of induction machines is of 3-phases, wound-rotor winding is also of 3-phases, and the number of phases for squirrel-cage is K Z 2 / P. In general, if the number of rotor phases is K, just as in (5.1.26), we can write the stator and rotor voltage equations and flux-linkage equations in per-unit values as follows1: 1

the signs with bottom-bars indicate per-unit values, but bottom-bars are omitted from now on.

354


p\ a ria

ua

½ ° ° ° p\ c ric ° p\ n rr in ° ¾ xaa ia xab ib xac ic xa1i1 " xaK iK ° xba ia xbb ib xbc ic xb1i1 " xbK iK ° ° xca ia xcb ib xcc ic xc1i1 " xcK iK ° xna ia xnb ib xnc ic xn1i1 " xnK iK °¿ p\ b rib

ub uc un

\a \b \c \n

(5.2.1)

in which rr is rotor phase resistance, and the mutual resistance between the rotor phases is neglected and might be counted in if necessary. In (5.2.1), several reactances, namely inductances in per-unit, can be obtained from (5.1.4), (5.1.5), (5.1.27) and (5.1.31). Referring rotor parameters to stator and taking per-unit values, there exist xaa

xbb

xcc

xssc

xm xls

xab

xbc

xca

xms

x11

x22

"

xKK

constant xrrc xm xlr

xnK

xm cos(n K )M 0 xnKl

xan

xm cos[J (n 1)\ 0 ]

constant

½ ° ° ° constant ¾ ° ° °¿

(5.2.2)

in which xm is the mutual inductance between a stator phase and rotor phase, whose axes coincide with each other, xls and xlr are leakage inductances for stator and rotor phases, respectively, xnKl is the mutual inductance between rotor K-th phase and n-th phase caused by the flux that does not pass through the air-gap. As given above, the basic equations of induction machines are also a set of equations with variable coefficients. Although those coefficients vary more simply than those of the synchronous machine, it is difficult to solve the equations directly. For convenience, we make use of coordinates conversion or variables conversion just as a study of synchronous machines. Here, various rotor and stator quantities are converted into d, q, 0 axes quantities, d, q axes revolving with the rotor. Since d, q axes are of two phases, it is necessary to transform rotor currents, voltages and flux-linkages from poly-phase to 2-phase. The conversion formulas are Ydr Yqr

2 K ½ ¦ Yn cos(n 1)M0 °° Kn1 ¾ 2 K ¦ Yn sin(n 1)M0 °°¿ Kn1

(5.2.3)

355

AC Machine Systems

in which Yn represents rotor current, flux-linkage or voltage. Generally, induction machines have no neutral line, so the zero-axis component is neglected here. If necessary, we can count it in according to the following formula, Y0

1 K

K

¦Y

n

(5.2.4)

n 1

On the basis of (5.2.1), (5.2.2) and those conversion formulas, the rotor basic equations can be written as

udr uqr u0 r

\ dr

2 K

p\ dr rr idrc ½ ° p\ qr rr iqrc ¾ ° p\ 0 r rr i0cr ¿

K

¦\

n

cos(n 1)M 0

n 1

2 K 2 K cos x i n M ¦ ( n1) n n ¦ x(n 2) nin cos(n 1)M0 0 Kn1 Kn1 2 K " ¦ x( n K 1) n in cos(n K 2)M 0 Kn1 K 2 2 K ¦ xna ia cos(n 1)M 0 ¦ xnb ib cos(n 1)M 0 Kn1 K n1 K 2 ¦ xnc ic cos(n 1)M 0 Kn1 2 K xrrc idrc ¦ [ x12l xm cos M 0 ]in cos nM 0 Kn1 K 2 ¦ [ x13l xm cos 2M 0 ]in cos(n 1)M 0 " Kn1 2 K ¦ [ x1Kl xm cos( K 1)M 0 ]in cos(n K 2)M 0 Kn1 2 K ¦ xm cos(n 1)M 0 {ia cos[J (n 1)M 0 ] Kn1 ib cos[J 120e (n 1)M 0 ] ic cos[J 120e (n 1)M 0 ]} xrrc idrc

2 K ¦ [ x12l xm cos M0 ]in [cos(n 1)M0 cosM0 Kn1 2 K sin(n 1)M 0 sin M 0 ] ¦ [ x13l xm cos 2M 0 ]in Kn1 u [cos(n 1)M 0 cos 2M 0 sin(n 1)M 0 sin 2M 0 ] " xrrc idrc

356

(5.2.5)


2 K ¦ [ x1Kl xm cos( K 1)M0 ]in [cos(n 1)M0 cos( K 1)M0 Kn1 sin(n 1)M 0 sin( K 1)M 0

2 K ¦ xm cos(n 1)M0 [ia cos J cos(n 1)M0 ia sin J sin(n 1)M0 Kn1 ) cos(n 1)M 0 ib sin(J 120e )sin( n 1)M 0 ib cos(J 120e

) cos(n 1)M 0 ic sin(J 120e )sin(n 1)M 0 ] ic cos(J 120e 1 xm iqrc sin 2M 0 xm idr cos 2 2M 0 2 1 1 xm iqrc sin(2 u 2M 0 ) " xm idrc cos 2 ( K 1)M 0 xm iqrc 2 2

xrrc idrc xm idrc cos 2 M 0

K

K

n 2

n 2

u sin 2( K 1)M 0 ¦ idrc x1nl cos(n 1)M 0 ¦ iqrc x1nl sin(n 1)M 0 2 K ) ¦ xm cos 2 (n 1)M 0 [ia cos J ib cos(J 120e Kn1 2 K )] ¦ xm cos(n 1)M 0 sin(n 1)M 0 ic cos(J 120e K n1 ) ic sin(J 120e )] u [ia sin J ib sin(J 120e

xlr idrc xm idrc [1 cos 2 M 0 cos 2 2M 0 " cos 2 ( K 1)M 0 ] 1 xm iqrc [sin 2M 0 sin(2 u 2M 0 ) " sin 2( K 1)M 0 ] 2 3 K 3 K ¦ xm id cos 2 (n 1)M 0 ¦ xm iq cos(n 1)M 0 sin(n 1)M 0 K n1 Kn1

K

K

n 2

n 2

¦ idrc x1nl cos(n 1)M 0 ¦ iqrc x1nl sin( n 1)M 0

(5.2.6)

where udr , uqr , u0 r , \ dr , \ qr , \ 0 r , idrc , iqrc , i0cr are rotor voltages, flux-linkages and currents in d, q, 0 axes, respectively. Since K 2 sin 0e+ sin2M 0 sin(2 u 2M 0 ) " sin 2( K 1)M 0

1 cos 2 M 0 cos 2 2M 0 " cos 2 ( K 1)M 0

id iq

0

2 [ia cos J ib cos(J 120e ) ic cos(J 120e )] 3 2 [ia sin J ib sin(J 120e ) ic sin(J 120e )] 3 357

AC Machine Systems K

¦ ic x

qr 1nl

sin(n 1)M 0

0

n 2

so

\ dr

xlr idrc

K K 3 xm idrc xm id idrc ¦ x1nl cos(n 1)M 0 2 2 n 2

K K 3 ª º « xlr 2 xm ¦ x1nl cos(n 1)M 0 » idrc 2 xm id n 2 ¬ ¼ 3 xrr idrc xm id 2

(5.2.7)

where xrr

K K xm ¦ x1nl cos(n 1)M 0 2 n 2 K K 2 xrrc xm ¦ x1nl cos(n 1)M 0 2 n 2

xlr

is the rotor self-inductance equivalent to two phases. In a similar way, there exists

\ qr

xrr iqrc

If rotor winding is of 3-phase, namely K xrr

in which x12l xmr

(5.2.8)

3, then there is

xm x12l cos120e x13l cos 240e 2 x xrrc m x12l xrrc xmr 2 xrrc

(5.2.9)

x13l ,

x12l xm cos120eis the rotor mutual inductance between various phases.

\ 0r

358

3 xm iq 2

1 K 1 K 1 K \ n xrrc i0cr ¦ x( n 1) n in ¦ x( n 2) n in ¦ Kn1 Kn1 K n1 K K 1 1 1 K " ¦ x( n K 1) n in ¦ ia xna ¦ ib xnb Kn1 Kn1 Kn1 K 1 ¦ ic xnc Kn1 1 K xrrc i0cr ¦ [ x12l xm cos M 0 ]in K n1


1 K ¦ [ x13l xm cos 2M0 ]in " K n1 1 K ¦ [ x1Kl xm cos( K 1)M0 ]in K n1 1 K ia ¦ xm cos[J (n 1)M 0 ] K n1 1 K ib ¦ xm cos[J 120e (n 1)M 0 ] K n1 1 K ic ¦ xm cos[J 120e (n 1)M 0 ] K n1

K K 1 § · ¨ xrrc ¦ x1nl ¸ i0cr xm i0cr ¦ cos nM 0 n 2 n 1 © ¹ K K § · ¨ xlr ¦ x1nl ¸ i0cr xm i0cr ¦ cos(n 1)M 0 n 1 n 2 © ¹ K § · ¨ xlr ¦ x1nl ¸ i0cr x0 r i0cr n 2 © ¹

(5.2.10)

K

where x0 r

xlr ¦ x1nl . n 2

If the rotor winding is of 3-phase, then there exists x0 r

xlr x12l x13l xrrc 2 xmr

xlr 2 x12l

xrrc xm 2 x12l

(5.2.11)

Transforming the stator quantities to d, q, 0 axes, there exist ud

p\ d \ qZ r rs id

uq

p\ q \ d Z r rs iq

u0

p\ 0 rs i0

\d

2 K ¦ xanin cos J 3n1 2 K 2 K ) ¦ xcn in cos(J 120e ) ¦ xbn in cos(J 120e 3n1 3n1 2 K ( xssc xms )id ¦ in xm cos[J (n 1)M 0 ]cos J 3n1 K 2 ) ¦ in xm cos[J 120e (n 1)M 0 ]cos(J 120e 3n1 xssc id xms id

359

AC Machine Systems

2 K ¦ in xm cos[J 120e (n 1)M0 ]cos(J 120e) 3n1 2 K 3 ( xssc xms )id ¦ in xm cos(n 1)M 0 3n1 2 K xss id xm idrc 2

in which xss xssc xms is the stator synchronous reactance. In a similar way, we can get

\q \0 x0

K xm iqrc 2 ( xssc 2 xms )i0 x0 i0 xssc 2 xms xss iq

The basic equations deduced above can be summarized as ud uq u0

\d \q \0 udr uqr u0 r

\ dr \ qr \ 0r

360

p\ d \ qZ r rs id ½ ° p\ q \ d Z r rs iq ¾ ° p\ 0 rs i0 ¿

(5.2.12)

K ½ xm idrc ° 2 ° K c xss iq xm iqr ¾° 2 ° x0 i0 ¿

(5.2.13)

xss id

p\ dr rr idrc ½ ° p\ qr rr iqrc ¾ ° p\ 0 r rr i0cr ¿ 3 ½ xm id ° 2 ° 3 xrr iqrc xm iq ¾° 2 ° x0 r i0cr ¿ xrr idrc


3 K xm in the formula above is not equal to xm in 2 2 (5.2.13). For convenience, they can be equal after changing the current and 3 K impedance base-values of the rotor or stator, i.e. let X m xm , idr idrc , 2 3 K 3 3 3 K i0cr , Rr rr , X rr xrr , X 0 r x0 r . iqr iqrc , i0 r 3 K K K 3 After changing the base values, the basic equations in d, q, 0 axes for induction machines can be written as The mutual inductance

ud uq u0

\d \q \0 udr uqr u0 r

\ dr \ qr \ 0r

p\ d \ qZ r rs id ½ ° p\ q \ d Z r rs iq ¾ ° p\ 0 rs i0 ¿

(5.2.14)

xss id X m idr ½ ° xss iq X m iqr ¾ ° x0 i0 ¿

(5.2.15)

p\ dr Rr idr ½ ° p\ qr Rr iqr ¾ ° p\ 0 r Rr i0 r ¿

(5.2.16)

X rr idr X m id ½ ° X rr iqr X m iq ¾ ° X 0 r iq 0 ¿

(5.2.17)

Substituting (5.2.15) into (5.2.14), there exist ud uq u0

( pxss rs )id pX m idr Z r xss iq Z r X m iqr ½ ° Z r xss id Z r X m idr ( pxss rs )iq pX m iqr ¾ ° ( p\ 0 rs )i0 ¿

(5.2.18)

Substituting (5.2.17) into (5.2.16), there exist udr uqr u0 r

pX m id ( pX rr Rr )idr ½ ° pX m iq ( pX rr Rr )iqr ¾ ° ( pX 0 r Rr )i0 r ¿

(5.2.19)

361

AC Machine Systems

Equations (5.2.18) and (5.2.19) can be expressed in a matrix form as follows: ª ud º « » « uq » «udr » « » «uqr » « u0 » « » ¬«u0 r ¼»

ª pxss rs « « Z r xss « pX m « « 0 « 0 « ¬« 0

Z r xss pxss rs 0 pX m 0 0

pX m Zr X m pX rr Rr 0 0 0

Z r X m pX m 0 pX rr Rr 0 0

0 0 0 0 px0 rs 0

0 º ª id º »« » 0 » « iq » » « idr » 0 »« » 0 » « iqr » » « i0 » 0 »« » pX 0 r Rr ¼» ¬«i0 r ¼» (5.2.20)

For induction machines, their rotor windings are usually short-circuited, so udr uqr 0. For induction machines, the input power and electromagnetic torque are P

ud id uq iq 2u0 i0

(5.2.21)

Te

\ d iq \ q id

(5.2.22)

X m (idr iq iqr id )

The rotor motion equation for induction machines is H

dZ r dt

Te Tm

(5.2.23)

in which Tm is the counter-torque of production machines, the inertia constant H and time are all expressed in electric radians, namely per-unit values. In practice, substituting slip s for speed Z r , there exist

Zr

1 s

dZ r dt H

ds dt

ds dt

Tm Te

(5.2.24) (5.2.25) (5.2.26)

If we delete the rotor current in (5.2.15), then there exist

\d

\q 362

§ Xm pX m2 · ½ udr ¨ xss ¸ id ° pX rr Rr pX rr Rr ¹ ° © ° ¾ G ( p )udr x( p)id ° ° G ( p )uqr x( p)iq ° ¿

(5.2.27)


where G ( p )

Xm is the operational conductance for induction machines, pX rr Rr

pX m2 is the corresponding operational reactance. pX rr Rr In general, the rotor current base-value is so chosen that the reactances in per-unit have the following relations: x( p )

xss

xss X rr

xs X m Xr Xm

in which xs and X r represent equivalent leakage reactances per phase, respectively, for stator and rotor. According to those formulas from (5.2.15) to (5.2.17) together with the relations above, the operational equivalent circuits can be shown in Fig. 5.2.1 for induction machines.

Figure 5.2.1 Operational equivalent circuits for induction machines

After getting the relations in d, q, 0 axes above, it is not difficult to research into different problems to select appropriate axes by using the conversion formulas in Section 3.1.

5.3

Analysis of the Starting Process of Induction Motors

The starting characteristics of induction motors, inclusive of starting current and torque varying with rotor speed, are considered especially for large induction motors. On the basis of the starting characteristics, we evaluate the effect of the starting process on line voltage, estimate the minimum voltage needed for the starting process, determine the starting current and time avoided for protection relays of induction motors and service power, etc. In the past, the starting process was often analyzed by using the equivalent circuit in Electric Machinery, which is in steady-state. However, several phenomena occur during the tests of starting the motor, which cannot be explained according to the steady-state theory. For example, the maximum speed in the starting process may exceed the synchronous speed, torque and speed may oscillate near the steady-state operation point, the 363

AC Machine Systems

maximum torque in the starting process is lower than that provided by manufacturers, sometimes a 40% decrease, etc. Now, we will discuss those differences and then explain why they occur.

5.3.1

Basic Equations for the Analysis of Starting Process

In the starting process, the induction motor speed is variable, so it is convenient to select d c , qc , 0 axes with synchronous speed. On the basis of the basic equations in d, q, 0 axes together with those conversion formulas in Section 3.1, the relations in d c , qc , 0 axes can be found out. Due to symmetrical and short-circuited rotor winding, take udr uqr 0 and neglect zero-axis component, then we can get ªudcs º «u » « qcs » « 0 » « » ¬ 0 ¼

ª\ dcs º ª rsc «\ » « x qcs » p« « ss «\ dcr » « 0 « » « ¬«\ qcr ¼» ¬ sX m

xss rs sX m 0

0 Xm Rr sX rr

X m º ªidcs º « » 0 »» «iqcs » sX rr » «idcr » »« » Rr ¼ ¬«iqcr ¼»

(5.3.1)

where ª\ dcs º «\ » « qcs » «\ dcr » « » «¬\ qcr »¼

ª xss « 0 « «Xm « ¬ 0

0 xss 0 Xm

0 º ªidcs º « » X m »» «iqcs » 0 » «idcr » »« » X rr ¼ «¬iqcr »¼

Xm 0 X rr 0

in which subscript s indicates stator quantity, subscript r indicates rotor quantity, subscripts d c and qc express the quantities in d c , qc axes, xss is stator self-inductance, X rr is rotor self-inductance, X m is mutual inductance between stator and rotor, rs and Rr are stator and rotor resistances respectively, s is rotor slip, s 1 Z , Z is rotor angular velocity. Rewriting (5.3.2) and (5.3.1) as the state equation, there is

ªidcs º «i » qcs p« » «idcr » « » «¬iqcr »¼ 364

ª xss « 0 « «Xm « ¬ 0

0 xss 0 Xm

Xm 0 X rr 0

0 º X m »» 0 » » X rr ¼

1

ªudcs º °« » ° «uqcs » ® °« 0 » ° «¬ 0 »¼ ¯

(5.3.2)


ª rs « x « ss « 0 « ¬ X m

xss rs Xm

ª 0 « 0 Z « « 0 « ¬Xm

0 0 Xm

0

0

0 Xm Rr X rr 0 0 0 X rr

X m º ªidcs º « » 0 »» «iqcs » X rr » «idcr » »« » Rr ¼ ¬«iqcr ¼» 0 º ªidcs º ½ « »° 0 »» «iqcs » ° ¾ X rr » «idcr » ° »« » 0 ¼ «¬iqcr »¼ °¿

(5.3.3)

The rotor motion equation is H

dZ dt

Te Tm

(5.3.4)

The electromagnetic torque Te is Te

X m (iqcs idcr idcs iqcr )

(5.3.5)

By using (5.3.3), (5.3.4) and (5.3.5), the torque and current can be calculated for the starting process of induction motors.

5.3.2

Numerical Calculation of Starting Characteristics

In the starting process, the rotor speed is variable, so (5.3.3), (5.3.4) and (5.3.5) are non-linear differential equations, which can be solved by numerical method because it is difficult to use the analytical method. The rotor speed changes more slowly than the electromagnetic quantities, so (5.3.4) can be solved by using the first order Euler Method to make numerical integral. Start at Z 0, make integral for a step length, take the integral result as the initial value for the next step and so forth, until the calculation is finished. The integral formula of the m-th step is

Z (mh) Z[(m 1)h] (Te Tm )h / H in which h is integral step length, electromagnetic torque Te is evaluated by (5.3.5), where the d c , qc components for stator and rotor currents are given by (5.3.3), so the integral calculation needs simultaneous equations (5.3.3), (5.3.4) and (5.3.5). Making numerical calculation according to (5.3.3), we use the 4th order Runge-Kutta Method to get satisfactory results. For each discrete step, the speed Z is considered as constant, so non-linear equation (5.3.3) can be linearized for each integral step length. 365

AC Machine Systems

In general, the initial conditions for the starting process are all zero, i.e., idcs (t0 )

0, iqcs (t0 )

0

idcr (t0 )

0, iqcr (t0 )

0, Z (t0 )

0

where t0 is the starting instant. For the calculation results to accord with practical condition, we have to consider the following. (i) Studying transients, we usually select rated maximum phase-voltage as voltage base-value and rated maximum phase-current as current base-value. However, in the design of electric machines, we often take active current as current base-value, which can be estimated by machine rated output power. Because of different current base-values, the impedance, power and torque base-values are also different. Evaluating the starting process, it is necessary to convert those parameters in per-unit used in electric machine design. (ii) For deep-slot rotor or double-cage induction motors, their rotor parameters vary obviously with rotor slip. In addition, ferromagnetic saturation will influence the reactances, too. If necessary, according to the starting and operating parameters provided by manufacturers, estimate the rotor resistance for a slip-interval as follows: r2c r2c

( s se )(rstc 2 rec2 ) rec2 rec2

s ı se ½ ¾ s se ¿

(5.3.6)

in which r2c is rotor phase resistance at any slip s, se is rated slip, rstc 2 is rotor phase resistance at start, rec2 is rotor phase resistance at rated operation. The rotor leakage reactance x2c in a slip-interval can be evaluated as x2c x2c

s se ( xstc 2 xec2 ) xec2 xec2

s ı se ½ ¾ s se ¿

(5.3.7)

where xstc 2 and xec2 indicate the rotor leakage reactance at the start and rated operation, respectively.

5.3.3 Comparison and Analysis of Calculation Results in Two Ways The no-load starting process is evaluated for a 2000 kW-induction motor, whose results are shown in Fig. 5.3.1, in which continuous lines indicate T-s curve and 366


I-s curve in steady-state, and dash lines express those in the starting process. Slip s 1 Z , where Z is the rotor speed in per-unit.

Figure 5.3.1 Comparison of T-s curves and I-s curves for steady-state and starting process of induction motor

At middle speed, namely 0.7>s>0.2, two T-s curves or two I-s curves for steady-state and transients are approximate to each other. At low speed, namely s>0.7, the T-s curve and I-s curve in transients all contain strong oscillating components, but their average values are approximate to those in steady-state. In Fig. 5.3.1, the line-segments for s>0.7 cannot be drawn due to strong oscillating components. Near the critical slip sK , the torque T in the starting process is less than that in steady-state, especially for maximum Torque. In the following, we will find out the main reasons that cause this difference. Due to small voltage drop of stator resistance, rs can be neglected in (5.3.1), so the stator current in steady-state can be estimated by the following formula: ªidcs º «i » « qcs » «idcr » « » «¬iqcr »¼

ª 0 « x « ss « 0 « ¬ sX m ª B11 «B ¬ 21

xss 0 sX m 0

0 Xm Rr sX rr

ª udcs º B12 º «« uqcs »» B22 »¼ « 0 » « » ¬ 0 ¼

Xm º 0 »» sX rr » » Rr ¼

1

ª udcs º « u » « qcs » « 0 » « » ¬ 0 ¼

(5.3.8)

367

AC Machine Systems

in which exists p\

0 , and the expression 1 R ( sW X rr ) 2

B22

W

2 r

ª Rr « sW X ¬ rr

sW X rr º Rr »¼

xss X rr X m2 xss X rr

can be obtained by inversing the partition matrix. Other sub-matrices have no effect on the following analysis, so they are omitted here. Similarly, neglecting stator resistance, the stator and rotor currents in transients can be obtained by ªidcs º «i » « qcs » «idcr » « » «¬iqcr »¼

ª udcs p\ dcs º « » B12 º « uqcs p\ qcs » » p\ dcr B22 »¼ « « » p\ qcr «¬ »¼

ª B11 «B ¬ 21

(5.3.9)

Near the critical slip, p\ dcs and p\ qcs can be neglected, so (5.3.9) is changed to ªidcs º «i » « qcs » «idcr » « » «¬iqcr »¼

ª udcs º « » B12 º « uqcs » B22 »¼ « p\ dcr » « » «¬ p\ qcr »¼

ª B11 «B ¬ 21

(5.3.10)

In order to judge the difference between steady-state and transients near the ª p\ dcr º critical slip, pay attention to B22 « » , which influences the rotor currents ¬ p\ qcr ¼ idcr , iqcr and then the stator currents idcs , iqcs . s

Take an induction motor to illustrate this problem, whose parameters at 0.03 are rs 0.004 22, Rr 0.004 686, X m 3.577 4, xss 3.650 63, X rr

3.690 73. The applied voltages are udcs

0.544 021, uqcs

0.839 071 5. At speed

Z 0.97, the estimation results are p\ dcs

0.000 2,

p\ qcs

0.000 7

p\ dcr

0.006 33,

p\ qcr

0.003 64

At the same speed Z 0.97, the stator and rotor currents together with electromagnetic torque are given in Table 5.3.1 for different operation conditions.

368

5 Analyses of Some Operating Modes of Induction Machine Systems Table 5.3.1 Currents and torques evaluated for different operation conditions Calculation formula

idcs

iqcs

idcr

Estimation in transients

(5.3.3)

4.301

0.556

4.1539

Estimation in steady-state

(5.3.8)

4.115

0.402 9

3.965 2 0.558 35 2.93

Operation condition

In addition, according to the motor parameters at s B22

iqcr 0.420 3

Te

2.089

0.03 , there exists

ª 88.764 104.695º «104.695 88.764 » ¬ ¼

which is multiplied by small rates of change for rotor flux-linkage, and then the effect of transients on rotor currents is clear. The estimation result is ª p\ dcr º B22 « » ¬ p\ qcr ¼

ª 0.180 79 º « 0.985 82 » ¬ ¼

(5.3.11)

whose values 0.180 79 and 0.98582 indicate the increments of idcr and iqcr caused by small rates of change for rotor flux-linkage. The increments cannot be neglected to compare the following steady-state rotor current ªidcr º «i » ¬ qcr ¼

ª 3.965 2 º « 0.558 35» ¬ ¼

(5.3.12)

because the addition of the corresponding values in (5.3.11) and (5.3.12) represents transient rotor currents idcr and iqcr at s 0.03, as shown in Table 5.3.1. Evidently, at critical slip, the stator and rotor currents in steady-state are quite different from those in transients, which is caused mainly by the rates of change for rotor flux-linkages \ dcr , \ qcr and motor parameters. For example, the smaller the rotor resistance, i.e. the smaller the critical slip, the larger the various element values of sub-matrix B22 are, so small change rates of rotor flux-linkages influence rotor currents and then stator currents, obviously. However, increasing rotor slip, several element values in B22 are decreased distinctly, so the change rates of rotor flux-linkages have only marginal influence. Therefore, at middle speed Z 0.6, namely s 0.4, the torque and current in steady-state are close to those in transients, as shown in Fig. 5.3.1. The calculation results for a 2 000kW-induction motor are shown in Fig. 5.3.2. Clearly, starting the rotor near the steady-state operation point, the torque and rotor speed swing around the equilibrium point several times. In addition, the maximum speed in the starting process can exceed the synchronous speed at s 0.024, which cannot appear for steady-state characteristics. This phenomenon can be explained as follows. Although there is no excitation source for the rotor winding 369

AC Machine Systems

of induction motors, the starting process is accompanied with electromagnetic transient process, and the rotor current at synchronous speed cannot decay suddenly to zero, which is a temporary excitation source. Therefore, the electromechanical oscillation exists just as drawing rotor in step for a synchronous motor. Of course, the maximum speed in the starting process depends upon several factors. Increasing rotor inertia and load, the maximum speed may decrease, and the swing process will be completed quickly.

Figure 5.3.2 Speed and torque oscillation near the steady-state operation point in starting process

The starting characteristics for different rotating inertias and loads are shown in Fig. 5.3.3. During Z 0.7, namely s ! 0.3, rotating inertias and loads have a little effect on the starting characteristics, so the line-segments during Z 0.7, namely s ! 0.3, are omitted in Fig. 5.3.3. For evaluation, suppose the load to be a ventilator, which can be indicated by the formula T 0.15 0.75Z 2 and whose load-factor is 0.9. The total fly-wheel moment of motor and ventilator is GD 2 5 370kg m 2 , and the fly-wheel moment of motor itself is GD 2 870kg m 2 . Curve ċ in Fig. 5.3.3 corresponds to a fictitious condition, i.e. the load is the same but total fly-wheel moment is GD 2 870kg m 2 .

Figure 5.3.3 Effect of load and rotating inertia on starting characteristics ĉ— steady-state operation, Ċ— GD 2 ċ— GD 2

370

870kg m 2 , no-load starting, 870kg m 2 , load starting, Č— GD 2 5 370kg m 2 , load starting


Clearly, in Fig. 5.3.3, the maximum torque can increase with rotating inertia and load. If rotating inertia is large enough, T-s curve in transients is approximate to that in steady-state, which can be explained as follows. The larger the rotating inertia, the longer the starting process, thus approaching the steady-state process. From calculation and analysis above, we can get the following conclusions: (i) Using the steady-state theory to analyze the starting process brings about some errors for a certain speed region, which must be paid attention to. In the starting process, the rotor speed may exceed the synchronous speed in short time to cause oscillation, but it cannot damage motors. (ii) T-s curve and I-s curve in the starting process are different from those in steady-state, which needs to be noted, too. The maximum torque provided by the manufacturers is obtained in steady-state, so the maximum torque measured in the starting process is less than that given by the manufacturers. (iii) Load and rotating inertia have obvious effect on starting characteristics. The maximum torque increases with load and rotating inertia. When the rotating inertia is large enough, T-s curve in transients will approach that in steady-state. Therefore, the two starting characteristics can be used respectively according to different rotating inertias and loads. For example, T-s curve in steady-state is used to determine the over-load capability for steady-state operation, and T-s curve in transients is used to evaluate the starting time.

5.4

Transients of Reswitching on Induction Motors

The transient process of induction motors at variable speed is a practical topic which include the starting process, plugging operation, disconnecting process due to source fault, reclosing process due to supply recovery, etc. In this section, we analyze the transient process of reswitching on the induction motors, by which we can understand not only some problems about reswitching on but also other characteristics of induction motors at variable speed. Sometimes, temporary fault occurs for a supply system to bring about source voltage dip and then tripping away for induction motors, such as short circuit of power system, mis-action of protection relays, and switching over large motors. Source fault is often temporary, so production loss caused by failure in power supply can be decreased if we reswitch on the induction motors as quickly as possible after supply recovery, especially for some continuous production process. Transients of reswitching on the induction motors include two parts, one is the disconnection process due to source fault, and the other is the reclosing process after supply recovery. The main problems of reswitching on are surge current and electromagnetic torque, which will be discussed as follows. 371

AC Machine Systems

5.4.1

Transients of Induction Motors after Disconnection

When interrupter’s contacts are separated, 3-phase currents should be changed to zero at the same time, but it sometimes cannot be changed to zero simultaneously due to electric arc or other reasons. For example, when phase a current passes through zero and interrupter’s contacts are away from each other in the meantime, phase a current can be considered as zero, but phase b and phase c currents still exist until phase b or phase c current passes through zero to make the two phase currents change to zero simultaneously. If 3-phase currents cannot be changed to zero at the same time, the disconnecting process can be divided into two stages, the first stage is from one phase cut off to the instant of all phases cut off, i.e. from ia 0 to ia ib ic 0, and the second stage is from 3-phase zero currents to reclosing instant, in which the first stage is much shorter than the second stage, so the results in this way are approximate to those when assuming 3-phase currents to be changed to zero at the same time. Of course, the latter is much simple and used in practice. In order to understand those problems comprehensively, the two methods will be discussed as follows. (1) Analysis of 3-phase currents not to be zero simultaneously, i.e. the first method Evidently, before 3-phase currents are all changed to zero, the stator circuits are unsymmetrical, so it is convenient to put coordinates axes on the unsymmetrical stator such as D , E , 0 axes. Therefore, the relations of induction machines are converted into those in D , E , 0 axes. On the basis of (5.2.15), there are

\ d cos J \ q sin J

xss id cos J X m idr cos J xss iq sin J X m iqr sin J

Thus, referring to (3.1.6), there exists

\D

xss iD X m iD r

(5.4.1)

According to (5.2.17) and (3.1.6), we can also get

\E \Dr \ Er

xss iE X m iE r ½ ° X rr iD r X m iD ¾ X rr iE r X m iE °¿

Since p (\ cos J )

cos J ( p\ ) \ sin J ( pJ ) p(\ sin J ) sin J ( p\ ) \ cos J ( pJ )

372

(5.4.2)


on the basis of (5.2.14) and (5.2.16), there exist p\ D rs iD ½° ¾ p\ E rs iE °¿

(5.4.3)

p\ D r Z r\ E r Rr iD r ½° ¾ p\ E r Z r\ D r Rr iE r °¿

(5.4.4)

uD uE

uD r uE r

Substituting (5.4.1) and (5.4.2) into (5.4.3) and (5.4.4), there are xss piD X m piD r rs iD ½° ¾ xss piE X m piE r rs iE °¿

(5.4.5)

X rr piD r X m piD Z r X rr iE r Z r X m iE Rr iD r ½° ¾ X rr piE r X m piE Z r X rr iD r Z r X m iD Rr iE r °¿

(5.4.6)

uD uE

uD r uE r

which can be written as matrix form, U

XI Z r GI RI

U

ª uD º «u » « E », I « uD r » « » «¬u E r »¼

ª iD º «i » « E» « iD r » « » «¬iE r »¼

X

ª xss « 0 « «Xm « ¬ 0

Xm 0 X rr 0

0 xss 0

Xm

R

ª rs « « « « ¬

G

ª 0 « 0 « « 0 « ¬ X m

rs Rr

0 0 Xm 0

º » » » » Rr ¼ 0 0 0 X rr

(5.4.7)

0 º X m »» 0 » » X rr ¼

0 º 0 »» X rr » » 0 ¼ 373

AC Machine Systems

Having the relations above, together with electromagnetic torque formula and rotor motion equation, it is not difficult to analyze the situation for 3-phase currents not to be zero simultaneously. (a) The first stage As mentioned above, the first stage is from one phase cut off to the instant of all phases cut off. In the stage exists ia 0, but ib and ic are not equal to zero. Relevantly, assume the steady-state stator currents before disconnection to be ian ibn icn

½ ° ° § 2S · ° I m sin ¨ t ¸¾ ¹° © § 2S · ° I m sin ¨ t ¸° ¹¿ ©

I m sin t

(5.4.8)

which can be converted into D , E components iD n iE n

2 1ª § 2S · § 2S · º ½ ½ ® I m sin t « I m sin ¨ t ¸ I m sin ¨ t ¸ ¾° ¹ ¹ »¼ ¿° 3¯ 2¬ © © ¾ 1 ª § 2S · § 2S · º ° « I m sin ¨ t ¸ I m sin ¨ t ¸ » I m cos t ° 3¬ © ¹ © ¹¼ ¿

(5.4.9)

Substituting the formulas above into (5.4.5) and (5.4.6), and supposing rotor winding to be short-circuited, the stator and rotor relations before disconnection can be written as

uE n

xss I m cos t X m pID rn rs I m sin t ½° ¾ xss I m sin t X m piE rn rs I m cos t °¿

(5.4.10)

uD rn

0

X rr piD rn X m I m cos t Z r X rr iE rn ½ ° Z r X m I m cos t Rr iD rn ° ¾ 0 X rr piE rn X m I m sin t Z r X rr iD rn ° ° Z r X m I m sin t Rr iE rn ¿

(5.4.11)

uD n

uE rn

or

in which Z r 374

( X rr p Rr )iD rn Z r X rr iE rn

sX m I m cos t

Z r X rr iD rn ( X rr p Rr )iE rn

sX m I m sin t

1 s.


Evidently, the steady-state rotor currents iD rn and iE rn before disconnection are sinusoidal form with operating frequency, which can be indicated as complex operator form ( jX rr Rr ) ID rn Z r X rr IE rn Z r X rr ID rn ( jX rr Rr ) IE rn

jsX m I m ½° ¾ sX m I m °¿

(5.4.12)

whose solutions are s 2 X rr X m I m jsRr X m I m Rr2 ( sX rr ) 2

jsX m I m Rr jsX rr

ID rn

jIE rn

IE rn

js 2 X rr X m I m sRr X m I m Rr2 ( sX rr )2

namely iD rn iE rn

½ s 2 X rr X m I m sR X I sin t 2 r m m 2 cos t ° 2 2 Rr ( sX rr ) Rr ( sX rr ) ° ¾ 2 s X rr X m I m sRr X m I m ° t t cos sin ° Rr2 ( sX rr ) 2 Rr2 ( sX rr )2 ¿

(5.4.13)

Substituting the results above into (5.4.10), the steady-state stator voltages uD n and uE n are uD n uE n

½ ª ª s 2 X rr X m2 º sR X 2 º I cos t « rs 2 r m 2 » I m sin t ° « xss 2 2 » m Rr ( sX rr ) ¼ Rr ( sX rr ) ¼ ¬ ¬ ° ¾ 2 2 2 ª ª s X rr X m º sRr X m º ° I sin t « rs 2 I cos t ° « xss 2 2 » m 2 » m ( ) ( ) R sX R sX r rr r rr ¬ ¼ ¬ ¼ ¿ (5.4.14)

As stated before, after interrupter’s contacts are separated, there is ia

0

(5.4.15)

ib

ic

(5.4.16)

Without neutral line, there exists

Because phase b and phase c are still connected to the supply, the line voltage ubc after the disconnection of phase a is the same as before, namely ubc

ub uc

ubcn

(5.4.17) 375

AC Machine Systems

The boundary conditions above can be expressed in D , E components, namely i i · 2§ ia b c ¸ ¨ 3© 2 ¹

iD

1

uE

3

(ub uc )

0 1 3

(5.4.18)

ubc

1 3

ubcn

uE n

(5.4.19)

which are substituted into (5.4.7), thus getting ª uD º «u » « En » « 0 » « » ¬ 0 ¼

ª xss « 0 « «Xm « ¬ 0

0 xss 0 Xm

Xm 0 X rr 0

ª 0 « 0 Zr « « 0 « ¬ X m

0 0 Xm 0

0 º ª 0 º ª rs « » X m »» « iE » «« rs 0 » « iD r » « »« » « X rr ¼ ¬«iE r ¼» ¬ 0 0 ºª 0 º « » 0 0 »» « iE » X rr » «iD r » 0 »« » X rr 0 ¼ ¬«iE r ¼»

Rr

ºª 0 º »«i » »« E » » « iD r » »« » Rr ¼ ¬«iE r ¼»

(5.4.20)

in which uD

Xm

diD r dt

(5.4.21)

so (5.4.20) can be reduced to ªu E n º « 0 » « » «¬ 0 »¼

ª xss « 0 « «¬ X m

0 X rr 0

ª 0 Z r «« X m «¬ 0

X m º ª iE º ª rs « » 0 »» « iD r » «« X rr »¼ «¬iE r »¼ «¬ 0 0 X rr

Rr

º ª iE º » «i » » « Dr » Rr »¼ «¬iE r »¼

0 º ª iE º « » X rr »» « iD r » 0 »¼ «¬iE r »¼

which can be written as the standard state equation as follows: ª iE º d« » iD r dt « » «iE r » ¬ ¼

376

ª xss «« 0 «¬ X m

0 X rr 0

Xm º 0 »» X rr »¼

1

rs ª « u «(1 s ) X m «¬ 0

0 Rr (1 s ) X rr

º ª iE º « » » (1 s ) X rr » « iD r » »¼ «¬iE r »¼ Rr 0


ª xss «« 0 «¬ X m

Xm º 0 »» X rr »¼

0 X rr 0

1

ªu E n º « 0 » « » «¬ 0 ¼»

(5.4.22)

Referring to (5.4.9) and (5.4.13), the initial values of iE , iD r and iE r are indicated as Im

iE 0 iD r 0 iE r 0

½ ° sR X I 2 r m m 2 °° Rr ( sX rr ) ¾ s 2 X rr X m I m °° Rr2 ( sX rr ) 2 °¿

(5.4.23)

Referring to (5.4.14), the voltage uE n is uE n

ª s 2 X rr X m2 º x « ss » I m sin t Rr2 ( sX rr )2 ¼ ¬ ª sR X 2 º « rs 2 r m 2 » I m cos t Rr ( sX rr ) ¼ ¬

After interrupter’s contacts are separated, electromagnetic torque and rotor speed will change, and the corresponding formulas, referring to (5.2.22) and (5.2.26), can be written as Te H

X m (idr iq iqr id ) ds dt

Tm Te

X m (iE iD r iD iE r )

(5.4.24) (5.4.25)

Having the basic relations from (5.4.22) to (5.4.25), we can analyze the transients of the first stage, i.e. from phase a cut off to the instant of all phases cut off. The basic relations are non-linear differential equations, which can be solved in the following steps: (i) According to (5.4.23), we get several current initial values iE 0 , iD r 0 , iE r 0. (ii) Since phase a is broken, and phase b and phase c are still connected to the source, the line voltage ubc is the same as before, so uE will not change and can be expressed by uE n before disconnection, referring to (5.4.14), into which t 0 is substituted to get uE n (0). (iii) The change of rotor speed is much slower than that of electromagnetic quantities, so slip s is considered as constant exactly for a short time interval 'T . 377

AC Machine Systems

Thus, (5.4.22) is approximately state equations with constant coefficient s. On the basis of several initial values iE 0 , iD r 0 , iE r 0 , uE n (0), s0, and (5.4.22), by means of the Runge-Kutta Method or other numerical methods, evaluate iE ('T ), iD r ('T ), and iE r ('T ) for t 'T . (iv) Substituting t 'T into (5.4.14), we get uE n ('T ). (v) The instantaneous values of stator currents at t 'T are 3 iE ('T ) 2

ia ('T ) iD ('T ) 0, ib ('T ) ic ('T )

3 iE ('T ) 2

(vi) Referring to (5.4.24), the electromagnetic torque at t Te ('T )

'T is

X m iE ('T )iD r ('T )

(vii) According to (5.4.25) and initial values s0, Tm (0), Te (0) X m iE 0 iD r 0 , by the Runge-Kutta Method or other methods, calculate rotor slip s( 'T ) at t 'T . Using the calculation procedures above, the evaluation steps for t 2'T are: (i) According to iE ('T ), iD r ('T ), iE r ('T ), s ('T ), uE n ('T ), and state equation (5.4.22), by means of the same numerical method, calculate iE (2'T ), iD r (2'T ) and iE r (2'T ) at t 2'T . (ii) Substituting t 2'T into (5.4.14), we get uE n (2'T ). (iii) The instantaneous values of stator currents at t 2'T are ia (2'T ) iD (2'T ) ib (2'T ) ic (2'T ) (iv) Electromagnetic torque at t Te (2'T )

0

3 iE (2'T ) 2 3 iE (2'T ) 2 2'T is

X m iE (2'T )iD r (2'T )

(v) On the basis of (5.4.25) and s ('T ), Tm ('T ) and Te ('T ), by using the same method as above, estimate rotor slip s (2'T ) at t 2'T . In so doing, the iterating calculation continues until iE (k 'T ) 0, i.e. ib ( k 'T ) 0, thus completing the estimation for the first stage.

378


(b) The second stage As mentioned above, at t k 'T , there exist iE (k 'T ) 0 and ib (k 'T ) 0, which means that phase b is cut off and phase c is also open, so the second stage begins, in which there is no stator current and rotor currents decay. The initial values in the second stage are rotor currents iD r (k 'T ), iE r (k 'T ) and rotor slip s (k 'T ) , respectively. On the basis of (5.4.7), the basic relations in the second stage are ª uD º «u » « E» «0» « » ¬0¼

0 Xm 0 ºª 0 º ª xss « » « 0 0 xss X m »» « 0 » « « X m 0 X rr 0 » « iD r » « »« » 0 X rr ¼ «¬iE r »¼ ¬ 0 Xm ª 0 ª rs ºª 0 º « « »« 0 » r s » « » (1 s ) « 0 « « » « iD r » « 0 Rr « »« » « Rr ¼ ¬«iE r ¼» ¬ ¬ X m

0 0 Xm 0

0 0 0 X rr

0 ºª 0 º « » 0 »» « 0 » X rr » « iD r » »« » 0 ¼ ¬«iE r ¼»

(5.4.26) or uD uE

ª0º «0» ¬ ¼

ª X rr « 0 ¬

diD r ½ dt °° diE r ¾° Xm dt ¿°

Xm

0 º ª iD r º ª Rr « » X rr »¼ ¬iE r ¼ «¬ 0

0 ª « ¬ (1 s ) X rr

(5.4.27)

0 º ª iD r º « » Rr »¼ ¬iE r ¼

(1 s ) X rr º ª iD r º » «i » 0 ¼ ¬ Er ¼

(5.4.28)

Writing (5.4.28) as the standard state equation, there is

d ªiD r º « » dt ¬iE r ¼

ª Rr º 1 s» « X ªi º rr » « Dr » « « Rr » ¬iE r ¼ « (1 s ) » X rr ¼ ¬

(5.4.29)

Having the basic equations above, it is not difficult to analyze the transients of the second stage, i.e. from 3-phase zero currents to reclosing instant. The following are the corresponding steps: 379

AC Machine Systems

(i) On the basis of iD r (k 'T ), iE r (k 'T ), s (k 'T ) and state equation (5.4.29), by using the same numerical method, evaluate rotor currents iD r [(k 1)'T ] and iE r [(k 1)'T ] at t

(k 1)'T .

(ii) According to (5.4.24), electromagnetic torque is always zero in the second stage, i.e., Te [(k j )'T ] 0

j

0,1, 2,3,"

(iii) On the basis of (5.4.25) and s (k 'T ), Tm (k 'T ) and Te (k 'T ) 0, by using the same method, estimate rotor slip s[(k 1) 'T ] at t (k 1)'T . Following the calculation procedures above, the evaluation steps for t (k 2)'t are: (i) According to iD r [( k 1) 'T ], iE r [( k 1)'T ], s[(k 1) 'T ] and state equation (5.4.29), by using the same method, evaluate rotor current components iD r [(k 2) 'T ] and iE r [(k 2)'T ] at t (k 2)'T . (ii) On the basis of (5.4.25) and s[(k 1)'T ], Tm [( k 1)'T ] and Te [( k 1) 'T ] 0, in the same way, reckon rotor slip s[(k 2)'T ] at t (k 2)'T . In so doing, the iterating calculation continues until the reclosing instant t (k n)'T , thus completing the estimation for the second stage. (2) Analysis for 3-phase currents to be zero simultaneously, i.e., the second method As stated above, 3-phase stator currents all change to zero when interrupter’s contacts are separated for the second method, so it is simpler than the first method. For comparison with the first method, assume the disconnection instant to be t 0, and the stator currents before disconnection are expressed in (5.4.8). Using D , E , 0 axes, substitute iD iE 0 and uD r u E r 0 into (5.4.7) and refer to (5.4.26), there are uD uE

d ªiD r º « » dt ¬iE r ¼

380

diD r ½ dt °° diE r ¾° Xm dt °¿

Xm

ª Rr º 1 s» « X ªi º rr » « Dr » « « Rr » ¬iE r ¼ « (1 s ) » X rr ¼ ¬

(5.4.30)

(5.4.31)


The stator currents iD 0 and iE 0 before disconnection can be evaluated by (5.4.9), and the corresponding rotor currents iD r 0 and iE r 0 are still estimated by (5.4.23), so due to the conservation of rotor flux-linkages after stator disconnection, the rotor current initial values should be written as

iD r 0 iE r 0

X rr iD r 0 X m iD 0 ½ ° X rr ° X rr iE r 0 X m iE 0 ¾° °¿ X rr

(5.4.32)

The calculation procedures are: (i) On the basis of rotor current initial values iD r 0, iE r 0 in (5.4.32), initial slip s0 and state equation (5.4.31), by using a certain numerical method, count rotor currents iD r ('T ) and iE r ('T ) at t 'T . (ii) According to (5.4.24), electromagnetic torque is always zero, namely

Te (k 'T )

k

0

0,1, 2,3,"

(iii) By (5.4.25) and initial values s0 , Tm (0) together with Te (0) 0, using the same method as above, get rotor slip s ('T ) at t 'T . Using the calculation steps above, the evaluation procedures for t 2'T are: (i) On the basis of iD r ('T ), iE r ('T ), s ('T ) and state equation (5.4.31), by using the same method, estimate rotor currents iD r (2'T ) and iE r (2'T ) at t 2'T . (ii) According to (5.4.25) and s( 'T ), Tm ( 'T ), and Te ('T ) 0, in the same way, obtain rotor slip s (2'T ) at t 2'T . In so doing, the iterating calculation continues until the reclosing instant t (k n)'T , in which we take t (k n)'T for comparison with the first method.

5.4.2

Transients of Induction Motors after Reclosing

We still use D , E , 0 axes to analyze the reclosing transients. The basic equation (5.4.7) can be written as standard state equation

ª iD º « » d « iE » dt «iD r » « » «¬iE r »¼

ª xss « 0 « «Xm « ¬ 0

0 xss 0 Xm

Xm 0 X rr 0

0 º X m »» 0 » » X rr ¼

1

381

AC Machine Systems

rs ª « 0 u« « 0 « ¬ (1 s ) X m ª xss « 0 « «Xm « ¬ 0

0 rs (1 s ) X m 0

Xm 0 X rr 0

0 xss 0 Xm

0 º X m »» 0 » » X rr ¼

0 0 Rr (1 s ) X rr 1

0 º ª iD º »«i » 0 »« E » (1 s ) X rr » « iD r » »« » Rr ¼ ¬«iE r ¼»

ª uD n º «u » « En » « 0 » « » ¬ 0 ¼

(5.4.33)

where uD n and uE n are D , E components of stator voltage after reclosing, which can be found out as follows. Assume the reclosing instant to be the beginning time for calculation and let 3-phase stator voltages be uan ubn ucn

U m sin(t T )

½ ° § 2S · U m sin ¨ t T ¸ °° © ¹¾ 2 S § ·° T ¸° U m sin ¨ t © ¹° ¿

(5.4.34)

where T is the switching phase angle. Converting 3-phase stator voltages into D , E components, there are uD n

uE n

u ucn · ½ 2§ uan bn ¸° 3 ¨© 2 ¹° uan U m sin(t T ) ° ¾ 1 ° (ubn ucn ) ° 3 ° U m cos(t T ) ¿

(5.4.35)

Having the relations above, the reclosing transients of induction motors can be estimated as follows: (i) Taking the reclosing instant as the beginning time for calculation and according to one of the two methods above, evaluate iD r [(k n)'T ], iE r [(k n)'T ],

s[(k n)'T ], which are the corresponding initial values when reclosing, namely iD r 0 s0 382

iD r [(k n)'T ], s[(k n)'T ]

iE r 0

iE r [(k n)'T ]


In addition, there are

iD 0 (ii) Substituting t

0,

iE 0

0

0 into (5.4.35), get uD n (0) and uE n (0).

(iii) According to iD 0 , iE 0 , iD r 0 , iE r 0 , s0, uD n (0), u E n (0) and state equation (5.4.33), by using a certain numerical method, estimate several current components iD ('T ),

iE ('T ), iD r ('T ) and iE r ('T ) at t

'T .

'T , the electromagnetic torque is

(iv) By (5.4.24) and t

Te ('T )

X m [iE ('T )iD r ('T ) iD ('T )iE r ('T )]

(v) Referring to (3.1.5), the stator phase currents at t

'T are

ia ('T ) iD ('T ) ib ('T ) ic ('T )

1 iD ('T ) 2 1 iD ('T ) 2

3 iE ('T ) 2 3 iE ('T ) 2

(vi) According to s0 , Tm (0), Te (0) 0 and (5.4.25), by using a certain numerical method, evaluate rotor slip s ( 'T ) at t

'T .

Following the calculation procedures above, the evaluation steps for t

2'T

are: (i) Substituting t

'T into (5.4.35), get uD n ('T ) and uE n ('T ).

(ii) According to iD ('T ), iE ('T ), iD r ('T ), iE r ('T ), s('T ), uD n ('T ), uE n ('T ) and state equation (5.4.33), in the same way, evaluate several current components

iD (2'T ), iE (2'T ), iD r (2'T ) and iE r (2'T ). (iii) Electromagnetic torque at t

Te (2'T )

2'T is

X m [iE (2'T )iD r (2'T ) iD (2'T )iE r (2'T )]

(iv) The stator phase currents at t

2'T are

ia (2'T ) iD (2'T ) ib (2'T ) ic (2'T )

1 iD (2'T ) 2 1 iD (2'T ) 2

3 iE (2'T ) 2 3 iE (2'T ) 2 383

AC Machine Systems

(v) According to equation (5.4.25), s ('T ), Tm ('T ) and Te ('T ), in a similar way, get s (2'T ) at t 2'T . In so doing, the iterating calculation continues until the induction motor reaches steady-state at t k 'T . In this way, we can obtain stator currents ia (t ), ib (t ), ic (t ) curves, electromagnetic torque Te (t ) and slip s (t ) curves. In addition, we can get the effect of different switching phase angle T on the reclosing surge current and torque. As an additional note, let us notice that the calculation methods above are suitable not only for the analysis of electromechanical transients of reswitching on the induction motors but also for the analysis of starting electromechanical transients, whose difference is due to different initial conditions and the latter is iD 0 iE 0 iD r 0 iE r 0 0, and s0 1. Obviously, the rotor current at the reclosing instant has an important effect on impact degree for reswitching on the induction motors, which can be illustrated as follows. When the stator is disconnected from source, the rotor current plays a role of excitation current to produce air-gap flux and then to induce 3-phase voltages in stator winding. At the reclosing instant, if those induced voltages are opposite to source voltages, the surge current is more serious than the starting situation, which can be diminished by the following methods: (i) Monitor the stator voltage phase-angles at the reclosing instant. (ii) Use the extinguishing-flux method to make rotor current decay quick at the disconnecting instant. Of course, the 3-phase flux-extinguishing resistors are connected only in stator circuit for squirrel-cage type. In the following, we analyze the transient process after the extinguishing resistors are connected in stator circuit.

5.4.3

Transients of Induction Motors after Flux-extinguishing Resistors are Connected in Stator Circuit

When motor stator is disconnected from source and then 3-phase resistors RL are connected in stator circuit to extinguish flux, the stator resistance r is changed to (r RL ) and there are uD 0 and uE 0. According to (5.4.7), the corresponding matrix equation is

ª0º «0» « » «0» « » ¬0¼ 384

ª xss 0 X m « 0 x 0 ss « « X m 0 X rr « ¬ 0 Xm 0

0 º ª iD º 0 0 ª 0 « » » « X m » « iE » 0 0 0 Zr « « 0 Xm 0 » « iD r » 0 » « » « X rr ¼ «¬iE r »¼ ¬ X m 0 X rr

0 º ª iD º « » 0 »» « iE » X rr » « iD r » »« » 0 ¼ «¬iE r »¼


ª(rs « « « « ¬

RL ) 0 (rs 0 0

0 0 0 º ª iD º « » RL ) 0 0 »» « iE » 0 Rr 0 » « iD r » »« » 0 0 Rr ¼ ¬«iE r ¼»

(5.4.36)

or ª iD º « » d « iE » dt «iD r » « » «¬iE r »¼

0 ª xss « 0 xss « «Xm 0 « ¬ 0 Xm ª rs RL « 0 u« « 0 « ¬ (1 s ) X m

Xm 0

1

0 º X m »» 0 » X rr » 0 X rr ¼ 0 rs RL (1 s ) X m 0

0 0 Rr (1 s ) X rr

0 º ª iD º »«i » 0 » « E » (5.4.37) (1 s ) X rr » « iD r » »« » Rr ¼ ¬«iE r ¼»

Initial slip s0 is known, stator current initial values are iD 0 0 and iE 0 0, and rotor current initial values iD r 0 and iE r 0 are still calculated by (5.4.32). Having the relations above, transients after connecting resistors in stator circuit can be evaluated as follows: (i) According to iD 0 0, iE 0 0, and iD r 0 , iE r 0 , s0 and state equation (5.4.37), by using a certain numerical method, estimate stator and rotor current components iD ('T ), iE ('T ), iD r ('T ) and iE r ('T ) at t 'T . (ii) On the basis of (5.4.24), electromagnetic torque at t 'T is

Te ('T )

X m [iE ('T )iD r ('T ) iD ('T )iE r ('T )]

(iii) According to (5.4.25) and s0 , Tm (0), and Te (0) 0, by a certain numerical method, evaluate rotor slip s ( 'T ) at t 'T . Following the calculation procedures above, the evaluation steps for t 2'T are: (i) According to iD ('T ), iE ('T ), iD r ('T ), iE r ('T ), s('T ), and state equation (5.4.37), in the same way, evaluate stator and rotor current components iD (2 'T ), iE (2'T ), iD r (2'T ) and iE r (2'T ) at t 2'T . (ii) On the basis of (5.4.24), electromagnetic torque at t 2'T is

Te (2'T )

X m [iE (2'T )iD r (2'T ) iD (2'T )iE r (2'T )]

(iii) According to (5.4.25) and s ('T ), Tm ('T ) and Te ('T ), by using the same method, get rotor slip s (2'T ) at t 2'T . 385

AC Machine Systems

In so doing, iterating calculation continues until the reclosing instant t k 'T , in which rotor currents iD r ( k 'T ), iE r (k 'T ), and rotor slip s (k 'T ) are the initial values for the reclosing calculation, i.e., iD r 0 iD r (k 'T ), iE r 0 iE r (k 'T ), s0 s(k 'T ), iD 0 0 and iE 0 0 . On the basis of those initial values and state equation (5.4.33), by using the numerical method above, calculate transients after reclosing. Evidently, the calculation of flux-extinguishing process has to link with that of the reclosing process to identify what kind of resistors should be selected and how the flux-extinguishing resistors influence surge currents after reclosing.

5.5 Self-Excitation when Connecting Induction Motor in Series with Capacitance and Counting the Inertia Effect in In Section 3.4, we assumed speed to be constant to analyze the self-excitation of synchronous machine in series with capacitance, i.e. machine’s inertia constant to be infinite. Under that assumption, analysis and calculation are all simple. In addition, the results are accurate enough for large inertia constant and stable source voltage. However, while analyzing the self-excitation of induction motors, it is necessary to consider variable speed, because induction motors often have small inertia constant and sometimes there exist self-excitation region near rated speed, which cannot be reflected by constant speed. In addition, we cannot process the effect of source voltage and load on self-excitation region by constant speed. Considering variable speed, we can get accurate results and understand the influence factors and degree. Considering the inertia effect, self-excitation of induction motors is a resonance of electromechanical parameters. When motor parameters, such as resistance, reactance, inertia constant, etc, are not a good match for network parameters and series capacitance, a resonance of electromechanical parameters may occur to bring about self-excitation oscillation of induction motors. While counting the inertia effect in, the basic equations will be complicated. For convenience, we use synchronously rotating d c , qc , 0 axes in Section 3.1 and the corresponding small perturbation equation to analyze that problem.

5.5.1

Basic Equations of Induction Machines when Counting the Inertia Effect in

An induction machine is in series with capacitance and then connected to electric source, whose scheme can be reduced as in Fig. 5.5.1. 386


Figure 5.5.1 One-line diagram for motor-capacitance-network

For comparison with the analysis of synchronous machine in series with capacitance stated in Section 3.4, we still use generator convention to write the following voltage relations: ua ub uc

xC ½ ia p °° x ° ubc C ib ¾ p ° xC ° ucc ic ° p ¿

uac

(5.5.1)

in which ua , ub , uc are several phase voltages of electric source, respectively, uac , ubc , ucc , are phase terminal voltages of machine, respectively,

ia , ib , ic are several phase currents of stator, respectively, xC is series capacitance. Converting them into d c , qc components, there are udc

uqc

½ §i · 2 ª§ ia · )° xC «¨ ¸ cos(t J 0 ) ¨ b ¸ cos(t J 0 120e 3 ¬© p ¹ © p¹ ° ° º §i · ° )» ¨ c ¸ cos(t J 0 120e © p¹ °° ¼ ¾ § ib · 2 ª§ ia · c xC «¨ ¸ sin(t J 0 ) ¨ ¸ sin(t J 0 120e ) °° uqc 3 ¬© p ¹ p © ¹ ° ° º § ic · )» ¨ ¸ sin(t J 0 120e ° © p¹ °¿ ¼

c udc

(5.5.2)

Because

387

AC Machine Systems

§ ia · ¨ ¸ cos(t J 0 ) © p¹

§i · 1 j( t J 0 ) e j(t J 0 ) ] ¨ a ¸ [e 2 © p¹ ½ 1 1 1 [ia e j( t J 0 ) ] [ia e j(t J 0 ) ]¾ ® 2¯p j p j ¿ 1 {( p j)[ia e j(t J 0 ) ] 2( p 2 1) ( p j)[ia e j(t J 0 ) ]}

§i · sin(t J 0 ) ¨ a ¸ © p¹

§i · 1 j( t J 0 ) [e e j( t J 0 ) ] ¨ a ¸ j2 © p¹ ½ 1 1 1 [ia e j( t J 0 ) ] [ia e j(t J 0 ) ]¾ ® j2 ¯ p j p j ¿ 1 {( jp 1)[ia e j(t J 0 ) ] 2( p 2 1) ( jp 1)[ia e j(t J 0 ) ]}

(5.5.3)

there are

pudc uqc

c uqc c pudc

2 xC 1 {( p 2 1) 3 2( p 2 1)

u [ia e j(t J 0 ) ib e j(t J 0 120e) ic e j( t J 0 120e) ] ( p 2 1)[ia e j( t J 0 ) ib e j( t J 0 120e) ic e j( t J 0 120e) ]} x c uqc c C [2ia cos(t J 0 ) pudc 3 2ib cos(t J 0 120e ) 2ic cos(t J 0 120e )] c uqc c xC idc pudc puqc udc

c udc c puqc

(5.5.4a)

2 xC 1 { j(p 2 1)[ia e j( t J 0 ) 3 2( p 2 1)

ib e j(t J 0 120e) ic e j( t J 0 120e) ] j( p 2 1)[ia e j(t J 0 ) ib e j(t J 0 120e) ic e j( t J 0 120e) ]} x c udc c C [2ia sin(t J 0 ) 2ib sin(t J 0 120e puqc ) 3 2ic sin(t J 0 120e )] c udc c xC iqc puqc (5.5.4b) 388


Referring to Section 3.1, there are c udc c uqc

p\ dc \ qc ridc ½° ¾ p\ qc \ dc riqc °¿

(5.5.5)

so

pudc uqc puqc udc

( p 2 1)\ dc 2 p\ qc (rp xC )idc riqc ½° ¾ ( p 2 1)\ qc 2 p\ dc (rp xC )iqc ridc °¿

(5.5.6)

Stator flux-linkage relations are

\ dc \ qc

[cos 'J x( p ) cos 'J sin 'J x( p )sin 'J ]idc ½ [cos 'J x( p)sin 'J sin 'J x( p) cos 'J ]iqc °° [sin 'J x( p ) cos 'J cos 'J x( p )sin 'J ]idc ¾° [sin 'J x( p)sin 'J cos 'J x( p) cos 'J ]iqc °¿

(5.5.7)

in which 'J is the included angle between rotor axis and synchronous axis, 'J

t

³ 'Z d t , 0

where 'Z

Z 1 is the difference between rotor speed and

synchronous speed. x( p ) is the operational reactance of induction machines, whose equivalent circuit is shown in Fig. 5.5.2.

Figure 5.5.2 Equivalent circuit of operational reactance for induction machines

Referring to Section 5.2, there is x( p )

xss xcT0 p 1 T0 p

in which xss is stator reactance when rotor circuit is open, xss xs X m , where xs is stator leakage reactance per phase and X m is magnetization reactance of induction machine, X rr T0 is time constant of rotor winding, T0 , Rr 389

AC Machine Systems

xc is transient reactance of induction machines, whose relation to x( p ) is xc [ x( p)] p

f

and there is

xc

§ X m2 · xss ¨1 ¸ W xss xss X rr ¹ ©

in which X rr is rotor reactance when stator circuit is open, X rr

X r X m , and X r is rotor leakage reactance,

W is leakage flux coefficient between stator and rotor windings, and there is W 1

X m2 xss X rr

While counting the inertia effect in, we have to use the rotor motion equation and the torque formula, iqc\ dc idc\ qc ½ ° ¾ d 2J H 2 Tm Te ° dt ¿

Te

(5.5.8)

Equations (5.5.6), (5.5.7) and (5.5.8) constitute the basic equations for the induction machine in series with capacitance to be connected to electric source, in order to analyze self-excitation.

5.5.2

Basic Relations of Induction Machines in Steady-state

Assume source voltages to be ua ub uc

U sin(t M ) ½ ° U sin(t M 120e )¾ U sin(t M 120e ) °¿

(5.5.9)

Converting them into udc and uqc components, there are udc uqc in which T 0 390

M J 0.

U sin(M J 0 ) U sin T 0 ½ U cos(M J 0 ) U cosT 0 ¾¿

(5.5.10)


Referring to Section 3.1 and converting (5.5.7) into complex forward component Fc with synchronous constant speed, the relation between flux-linkage and current is e j'J x( p )e j'J iFc

\ Fc

(5.5.11)

in which 1

\ Fc

2 1

iFc

2

(\ dc j\ qc )

(idc jiqc )

When steady-state operation is at constant speed Z r t

t

'J

³ (Z

\ Fc

e jst x( p )e jst iFc

0

r

1)dt

³ sdt 0

1 s, there is st

(5.5.12)

x( p js )iFc

(5.5.13)

so

or

\ dc j\ qc

x( p js )(idc jiqc ) [ X RP ( p ) j X IP ( p)](idc jiqc )

(5.5.14)

in which X RP ( p ) X IP ( p )

1 ½ [ x( p js ) x( p js )] is the real part of x(p + js ) °° 2 ¾ 1 [ x( p js ) x( p js )] is the imaginary part of x ( p j s ) ° j2 °¿

(5.5.15) Because x ( p js )

xss xcT0 ( p js) 1 T0 ( p js ) xss xc( sT0 )2 ( xss xc)T0 p xcT02 p 2 (1 T0 p) 2 ( sT0 ) 2 ( xss xc) sT0 j (1 T0 p ) 2 ( sT0 )2 391

AC Machine Systems

there are X RP ( p ) X IP ( p )

xss xc( sT0 )2 ( xss xc)T0 p xcT02 p 2 ½ ° (1 T0 p) 2 ( sT0 ) 2 ° ¾ ( xss xc) sT0 ° °¿ (1 T0 p) 2 ( sT0 )2

(5.5.16)

According to (5.5.14), there exist

\ dc \ qc

X RP ( p)idc X IP ( p)iqc ½° ¾ X IP ( p )idc X RP ( p )iqc °¿

(5.5.17)

Solving (5.5.6), (5.5.10) and (5.5.17) simultaneously, there are [( p 2 1) X RP ( p ) 2 pX IP ( p ) ½ ° xC rp ]idc [( p 2 1) X IP ( p ) 2 pX RP ( p ) r ]iqc ° ¾ pU cosT 0 U sin T 0 [( p 2 1) X IP ( p ) 2 pX RP ( p) ° r ]idc [( p 2 1) X RP ( p ) 2 pX IP ( p ) xC rp ]iqc °¿

pU sin T 0 U cosT 0

During asynchronous operation in steady-state, namely p above can be reduced to U cosT 0 U sin T 0

[ X ( s ) xC ]idc [r R( s ) ]iqc ½° ¾ [r R( s ) ]idc [ X ( s ) xC ]iqc °¿

(5.5.18a)

0, the equations

(5.5.18b)

where

X (s) R( s )

xss xc( sT0 )2 ½ ° 1 ( sT0 )2 ° ¾ ( xss xc)( sT0 ) ° X IP (0) 1 ( sT0 ) 2 °¿ X RP (0)

(5.5.19)

Solving simultaneous equations (5.5.18b), there exist idc iqc

[r R( s ) ]U sin T 0 [ X ( s ) xC ]U cosT 0 ½ ° [r R( s ) ]2 [ X ( s ) xC ]2 ° ¾ [r R( s ) ]U cosT 0 [ X ( s ) xC ]U sin T 0 ° ° [r R( s ) ]2 [ X ( s ) xC ]2 ¿

(5.5.20)

According to (5.5.17), the stator flux-linkages at asynchronous operation in steady-state are 392


\ dc \ qc

X ( s ) idc R( s ) iqc ½° ¾ R( s ) idc X ( s ) iqc °¿

(5.5.21)

From the results above, it is clear that the currents and flux-linkages at steady-state asynchronous operation are constants if using synchronously rotating d c , qc axes, thus making the study of self-excitation convenient.

5.5.3

Small Perturbation Equations of Induction Machines

When an induction machine is operating in steady-state and then is perturbed, there is a small increment for each quantity. After small perturbation, there is

J

t

³ Z dt J 0

t

³ (Z

0

0

r

'Z r )dt J 0

t

Z r t J 0 ³ 'Z r dt 0

(1 s )t J 0 D

(5.5.22)

where

Zr D

1 s

½° ¾ ³ 0'Zr dt °¿

(5.5.23)

t

Z r is rotor speed before perturbation, and D is the increment of rotor positionangle, so 'J

t

t

t

0

0

0

³ 'Z dt ³ (Z 1)dt ³ (Z st ³ 'Z dt st D

r

'Z r 1)dt

t

0

r

(5.5.24)

According to (5.5.11), the relation after small perturbation is

\ Fc 0 '\ Fc

e j( st D ) x( p)e j( st D ) (iFc 0 'iFc ) e jD x( p js )e jD (iFc 0 'iFc )

(5.5.25)

in which \ Fc 0 and iFc 0 are steady-state flux-linkages and current before perturbation, respectively, '\ Fc and 'iFc are flux-linkage and current increments after perturbation, respectively. Angle D is a small increment, so there exists e r jD

cos D r jsin D | 1 r jD 393

AC Machine Systems

According to (5.5.25), we can get

\ Fc 0 '\ Fc

(1 jD ) x( p js )(1 jD )(iFc 0 'iFc ) (1 jD ) x( p js )(iFc 0 jD iFc 0 'iFc jD'iFc ) x( p js )(iFc 0 jD iFc 0 'iFc jD'iFc ) jD x( p js )(iFc 0 jD iFc 0 'iFc jD'iFc )

(5.5.26)

In the light of (5.5.11), the relation before perturbation is

\ Fc 0

x( p js )iFc 0

x( js )iFc 0

(5.5.27)

Neglecting the second-order small increment, (5.5.26) can be rewritten as '\ Fc

x( p js )'iFc jx( p js )D iFc 0 jD x( p js )iFc 0 x( p js )'iFc jiFc 0 x( p js )D jD x( js )iFc 0

(5.5.28)

namely '\ dc j'\ qc

[ X RP ( p ) jX IP ( p)]('idc j'iqc ) j(idc 0 jiqc 0 )[ X RP ( p ) jX IP ( p)]D j(\ dc 0 j\ qc 0 )D

(5.5.29)

Separating the real part from the imaginary part, there are '\ dc

X RP ( p )'idc X IP ( p)'iqc

½ ° [\ qc 0 idc 0 X IP ( p) iqc 0 X RP ( p)]D ° ¾ X IP ( p)'idc X RP ( p )'iqc ° [\ dc 0 idc 0 X RP ( p) iqc 0 X IP ( p)]D °¿

'\ qc

(5.5.30)

In accordance with (5.5.6), the voltage equations after small perturbation are p (udc 0 'udc ) (uqc 0 'uqc )

½ ° ( p 1)(\ dc 0 '\ dc ) 2 p (\ qc 0 '\ qc ) ° (rp xC )(idc 0 'idc ) r (iqc 0 'iqc ) °° ¾ p (uqc 0 'uqc ) (udc 0 'udc ) ° ( p 2 1)(\ qc 0 '\ qc ) 2 p(\ dc 0 '\ dc ) ° ° (rp xC )(iqc 0 'iqc ) r (idc 0 'idc ) °¿ 2

and the relations before perturbation are pudc 0 uqc 0 puqc 0 udc 0 394

( p 2 1)\ dc 0 2 p\ qc 0 (rp xC )idc 0 riqc 0 ½° ¾ ( p 2 1)\ qc 0 2 p\ dc 0 (rp xC )iqc 0 ridc 0 °¿


Subtracting the second set of formulas from the first set above, the increment relations are p'udc 'uqc p'uqc 'udc

( p 2 1)'\ dc 2 p'\ qc (rp xC )'idc r 'iqc ½° ¾ ( p 2 1)'\ qc 2 p'\ dc (rp xC )'iqc r 'idc °¿

(5.5.31)

According to (5.5.8), in a similar way, we can obtain 'Te 2

\ dc 0 'iqc iqc 0 '\ dc \ qc 0 'idc idc 0 '\ qc ½°

Hp D Ha0

'Tm 'Te

¾ °¿

(5.5.32)

where a0 is angular acceleration before perturbation. Formulas (5.5.30), (5.5.31) and (5.5.32) constitute the relations of several increments after small perturbation, which can be used to study the change condition of increments.

5.5.4

Determination of Self-excitation Region for Induction Machines when Counting the Inertia Effect in

For an induction machine in series with capacitance and not in good match with the electromechanical parameters, self-excitation may occur. If we connect the series capacitance before starting the motor, self-excitation may occur in the starting process, thus causing static self-excitation around a certain low speed and not reaching the normal speed. If we connect the series capacitance at rated speed, self-excitation may also cause the motor return to a certain lower speed to bring about static self-excitation. In some special cases, self-excitation may occur in the starting process, but the motor still passes through the self-excitation region and finally reaches the normal speed. When the load, connected to the network behind capacitance, increases to a large value, self-excitation may occur at high speed. In the starting process or in the sudden connection of series capacitance or during a certain large load behind capacitance, the self-excitation is always formed gradually. Therefore, whether self-excitation happens for a certain speed can be examined by applying a small perturbation to the state before self-excitation. In the starting process or in the sudden connection of series capacitance, there exists a transient component that decays quickly. Especially in the starting process, self-excitation is formed when the rotor speed increases to a certain value. Similarly, when the load behind capacitance increases to a certain value, self-excitation happens under the original static operation condition and at a perturbation. Thus, it is accurate enough to consider the motor at steady-state operation before self-excitation, that is to say, the small perturbation equations deduced from the original static asynchronous operation, as stated before, are all suitable for those 395

AC Machine Systems

conditions above. Obviously, if several increments 'idc , 'iqc and D approximate to zero after a small perturbation, self-excitation does not occur; otherwise it may occur. Therefore, according to the roots of the characteristic equation for the small perturbation relations as before, we can ascertain whether self-excitation occurs or not. For convenience, the beginning time of small perturbation is selected so that idc 0 0. In accordance with the choice above and (5.5.20) and (5.5.21), the corresponding values iqc 0 , \ dc 0 and \ qc 0 are U

iqc 0

2

[r R( s ) ] [ X ( s )

\ dc 0 \ qc 0 When taking idc 0

R( s ) iqc 0 X ( s ) iqc 0

½ ° xC ] ° ° ¾ ° ° ° ¿ 2

(5.5.33)

0 as the beginning instant, there is ia 0

idc 0 cos(t J 0 ) iqc 0 sin(t J 0 ) i0 iqc 0 sin(t J 0 )

If electric source is infinite, after small perturbation, there exist 'udc 'uqc

0½ 0 ¾¿

(5.5.34)

On the basis of (5.5.30), (5.5.31), (5.5.32), (5.5.33) and (5.5.34), there are 0

Z1 ( p )'idc Z 2 ( p )'iqc Z A ( p)iqc 0D

0

Z 2 ( p )'idc Z1 ( p )'iqc Z B ( p )iqc 0D

'Tm Ha0

½ ° ° Z 3 ( p)iqc 0 'idc Z C ( p )iqc 0 'iqc ¾° ° [ Hp 2 iqc2 0 Z 3 ( p)]D ¿

(5.5.35)

where Z1 ( p) Z 2 ( p) Z3 ( p) Z 4 ( p) Z A ( p) Z B ( p) ZC ( p) 396

(1 p 2 ) X RP ( p ) 2 pX IP ( p ) ( xC rp) ½ ° (1 p 2 ) X IP ( p) 2 pX RP ( p) r ° ° X ( s ) X RP ( p) °° R( s ) X IP ( p) ¾ ° (1 p 2 ) Z 3 ( p) 2 pZ 4 ( p ) ° 2 ° (1 p ) Z 4 ( p ) 2 pZ 3 ( p ) ° R( s ) X IP ( p ) °¿

(5.5.36)


The characteristic equation of (5.5.35) is

A( p )

Z1 ( p )

Z 2 ( p)

Z A ( p )iqc 0

Z 2 ( p)

Z1 ( p )

Z B ( p )iqc 0

Z C ( p )iqc 0 [ Hp 2 iqc2 0 Z 3 ( p)]

Z 3 ( p )iqc 0

[ Hp 2 iqc2 0 Z 3 ( p )][ Z12 ( p ) Z 22 ( p )] [ Z 3 ( p ) Z A ( p) Z B ( p ) Z C ( p )]iqc2 0 Z1 ( p ) [ Z 3 ( p ) Z B ( p ) Z A ( p ) Z C ( p )]iqc2 0 Z 2 ( p )

0

Dividing the two sides of the equation above by iqc2 0 , the characteristic equation

can be changed to another form A( p)

ªH 2 º 2 2 « 2 p Z 3 ( p) » [ Z1 ( p ) Z 2 ( p )] «¬ iqc 0 »¼ [ Z 3 ( p) Z A ( p ) Z B ( p) Z C ( p )]Z1 ( p) [ Z 3 ( p) Z B ( p ) Z A ( p ) Z C ( p)]Z 2 ( p ) 0

(5.5.37)

In the light of (5.5.37), whether self-excitation occurs for a rotor speed Z r can be judged. However, it is the twelfth order equation, whose solutions are tedious. In practice, according to concrete topics, select appropriate criterion. In general, it is convenient to use the Mihainov’s method or the D-domain partition method.

5.5.5

Effect of Inertia Constant on Self-excitation Region

Motor inertia constant H has obvious effect on the self-excitation region under some conditions. Especially for small inertia constant, if assuming rotor speed to be constant without mechanical oscillation to study self-excitation, the results are sometimes quite different from experimental data. Considering the inertia effect and analyzing the self-excitation region on the basis of (5.5.37), it is convenient to use the D-domain partition on single complex parameter plane. Thus, (5.5.37) can be reformed to HR( p) Q( p )

A( p )

H

namely in which

R( p)

1 2 qc 0

i

0

Q( p) R( p)

[ Z12 ( p ) Z 22 ( p )] p 2

(5.5.38) (5.5.39) (5.5.40) 397

AC Machine Systems

Q( p) [ Z12 ( p ) Z 22 ( p )]Z 3 ( p ) [ Z 3 ( p) Z A ( p ) Z B ( p ) Z C ( p )]Z1 ( p ) [ Z 3 ( p) Z B ( p ) Z A ( p ) Z C ( p )]Z 2 ( p )

Let p

(5.5.41)

jk , there is H ( jk )

Q ( jk ) U ( k ) jV ( k ) R jk

(5.5.42)

According to (5.5.42), use the D-domain partition on single complex parameter plane to determine inertia constants at the self-excitation region boundary points for a certain speed Z r . During the calculation, take k f ~ f to get a series of U (k ) and V (k ) values and then to plot the projection curve of the characteristic root plane imaginary axis on to U - V plane, namely H complexplane, which is the D-domain partition boundary on H-plane. A 1.6kW-motor has the following parameters: r 0.167 1, xs 0.138 7, X r 0.078 3, X m 1.720 , Rr 0.056 and series capacitance xc 0.1086. At rotor speed Z r 2 880rpm, estimate the D-domain partition boundary curve as shown in Fig. 5.5.3. Since the

Figure 5.5.3 D-domain partition boundary curve for an induction motor in series with capacitance 398


D-domain partition boundary curve is symmetrical to the real axis, namely the U-axis, it is sufficient to make the calculation for k 0 ~ f only. According to these evaluation results, draw a half of boundary curve and then plot the remainder curve on the basis of symmetrical relation. In the light of the self-excitation features of induction machines, it is enough to take k 0.1 ~ 1.0 for practical estimation. After having the D-domain partition boundary curve, move along the curve from k f to k f and plot shade lines on the left side of the curve. Obviously, if we pass through the D-domain partition boundary curve on the H-plane from the shade-line side to the no shade-line side, then a characteristic root passes through the imaginary axis from left half plane to right half plane on the characteristic equation root plane. The next step is to partition and mark regions in the following way. When passing through the D-domain partition boundary curve once from the no shade-line side to the shade-line side, region grade will be raised once. According to the principle, region ċin Fig. 5.5.3 has the highest grade and is certainly a stable region. It should be pointed out that the highest grade region is a possible region without self-excitation but sometimes it does not exist, so it is necessary to check whether the highest grade region is a stable region. However, H value for stable operation can be estimated practically to save this check. For example, during H f, there is no self-excitation region at high speed, and for small inertia constant H, there is no self-excitation region at low speed. During V (k ) 0 in (5.5.42), there is H ( jk ) U (k ) real number. Thus, when the inertia constant H is equal to U (k ) corresponding to V (k ) 0, there exists Q ( jk ) HR( jk ), namely HR( jk ) Q( jk )

A( jk )

0

(5.5.43)

which indicates that the characteristic equation A( p ) 0 has a pair of imaginary roots r jk for the inertia constant stated above that corresponds to the boundary of the self-excitation region. According to Fig. 5.5.3, U (k ) corresponding to H 124 V (k ) 0 is 124 per-unit, namely 0.395s. Therefore, during inertia 314 314 H ! 0.395s and at speed Z r 2880 rpm or s 0.04, self-excitation constant 314 does not occur because region ċhas the highest grade, which coincides with the estimation result that during infinite inertia constant there is no self-excitation region at high speed. On the basis of motor parameters and series capacitance xc 0.1086, making similar calculation at other speeds, the self-excitation region can be obtained as shown in Fig. 5.5.4, in which sign “ ” indicates the experimental datum-points H 0.354s or 1.07s at Z r 2880 rpm, thus calculation results for 314 approximating experimental data. ƻ

399

AC Machine Systems

Figure 5.5.4 Self-excitation region for an induction motor in series with capacitance

From (5.5.43) it is evident that k-value corresponding to self-excitation boundary point is stator self-excitation current frequency in d c , qc axes at that speed. If m indicates stator self-excitation current frequency in a, b, c axes, there is m 1 k . Calculation results and test data all show that m has a range of 0.1 1.0, by which we get the corresponding k to make estimation. As for rotor self-excitation current frequency n, there are n

(1 k ) Z r

sk

(5.5.44)

and k 1 m

(5.5.45)

Thus, k-value represents rotor mechanical oscillation frequency or pulsating torque frequency. H When 0.395s in Fig. 5.5.3, the corresponding k-value is 0.276, so there 314 exist stator self-excitation current frequency (1 0.276) u 50 36.2Hz and rotor mechanical oscillation frequency 0.276 u 50 13.8Hz during static self-excitation H 0.395s. at Z r 2880rpm and 314 Considering the inertia influence, the calculation of the self-excitation region for induction machines is complicated, so it is appropriate to use computers. For accuracy of evaluation, interval 'k is taken as a small value, such as 'k 0.001 or 0.0001. In the calculation results of H ( jk ) U (k ) jV (k ), only those data corresponding to U (k ) positive and V (k ) 0 are useful, and the other data need not be printed. While studying the self-excitation regions of induction motors, several speeds in the starting process have to be examined, in which rotor slip s has a range of 0 1.0 and interval 's is taken as 0.05 or 0.01 to suffice for demands. Figures 5.5.5 to 5.5.7 show the calculation results of the self-excitation regions 400


for a 1.6kW-induction motor in series with capacitances xc 0.04, 0.07 and 0.108 6, respectively. From those curves we can see the important properties, i.e. there are two self-excitation regions during lower compensation degree, the two regions are broadened with the increase of the compensation degree, and the two regions are to be connected to form a self-excitation region when the compensation degree exceeds a certain value. If assuming rotor speed to be constant to analyze the self-excitation region, only one self-excitation region exists no matter what compensation degree there is, thus bringing about serious error sometimes.

Figure 5.5.5 Self-excitation region for an induction motor in series with capacitance F c 0.04

Figure 5.5.6 Self-excitation region for an induction motor in series with capacitance xc 0.07

Figure 5.5.7 Self-excitation region for an induction motor in series with capacitance xc 0.108 6

401

AC Machine Systems

5.6

Dual-Stator-Winding Multi-Phase High-Speed Induction Generator

As an important part of ship’s integrated power systems, the dual-stator-winding multi-phase high-speed induction generator system with a rectifier load is characterized by high power density, good electric supply quality, high mechanical strength, reliable operating performance, simple maintenance, and convenient adjustment. Its main shortcoming of low compensation ability for the induction generator may be overcome by setting control winding in the generator. Therefore, the induction generator system represents the development tendency of the power generation sector in ship’s integrated power systems. To meet the requirements of minimizing the machine size, improving the mechanical strength, reducing the noise, and enhancing the electromagnetic compatibility, the stator of this new induction generator has two embedded windings. These windings include a 12-phase power winding connected to a rectifier load and a 3-phase control winding connected to a static excitation regulator, respectively. It has a squirrel-cage solid rotor. Also, there are the capacitors for self-excitation and the inter-phase reactors in the system for performance improvement. It is very difficult to analyze the dual-stator winding multi-phase high-speed induction generator with a rectifier load by general methods owing to the system’s complicated nonlinear magnetic/electric circuits, multi-loop variable-topology configurations, existence of capacitors for self-excitation and inter-phase reactors, and abrupt variation of loop currents. From the point of analyzing the mmf of the induction generator, a new idea which divides the whole system into two sub-systems is presented. The circuit method and the field finite element method are adopted to analyze the two sub-systems which are linked only by appropriate terminal variables.

5.6.1

Construction of New Induction Generator

Figure 5.6.1(a) is a connection sketch of the 12-phase winding with a rectifier load of the dual-stator-winding induction generator. The 12-phase power winding supplies active power to the load through a bridge rectifier. The 12-phase winding consists of four Y-connection windings shifted by 15eeach, electrical angle with independent neutral points, which are represented by a1b1c1, a2b2c2, a3b3c3, and a4b4c4, where phase a2 leads ahead of phase a1 by 15eelectrical angle, the same is true of phase a3 and a2 or phase a4 and a3. C and LB are self-excitation capacitance and inter-phase inductance, respectively, and RL is the load resistance. 402


Figure 5.6.1 Dual-stator-winding system of new induction generator (a) Power winding with a rectifier load; (b) Control winding with an excitation regulator

Figure 5.6.1(b) is a connection sketch of the 3-phase control winding with a static excitation regulator. The 3-phase Y-connection winding is used to regulate the reactive power of the generator for keeping the stability of the rectified load voltage. Control winding phase A leads ahead of power winding phase a1 by 22.5e electrical degrees. Solid cage structure is adopted for the high-speed rotor, which also complicates the structure of the induction generator. There exist many difficulties in analyzing such a complicated new induction generator system. Firstly, the phasor cannot be used to analyze the generator system with nonlinear and unsymmetrical circuits, and complete decoupling of the system cannot be achieved by coordinates transformation. Secondly, it is difficult to use the mode classification method in the system analysis because the system operation modes are more than one hundred under different turn-on and turn-off conditions of diodes in rectifier bridges. Thirdly, the existence of capacitors for self-excitation and inter-phase reactors increases the order of the circuit system that makes the system model more complex. Finally, it needs to use the electromagnetic field method instead of the equivalent circuit method to deal with the solid cage structure.

5.6.2

MMF of Multi-Phase Induction Generator

The expression for a desired harmonic air-gap mmf due to a specified harmonic power winding current is obtained with the current resolved into fundamental and harmonic components; consequently, the superposition method is employed to perform the mmf analysis. If the order of the harmonic current is denoted by P ( P 1, 5, 7, "), the 12-phase power winding currents decomposed by the Fourier series are written as 403

AC Machine Systems

iak (t )

¦

ibk (t )

¦

ick (t )

¦

ʌº ½ ª 2 I P sin P «Z t (k 1) » ° 12 ¬ ¼ ° 2ʌ ʌ º° ª 2 I P sin P «Z t (k 1) » ¾ 3 12 ¼ ° ¬ 4ʌ ʌ º° ª 2 I P sin P «Z t (k 1) » ° 3 12 ¼ ¿ ¬

(5.6.1)

where k 1, 2,3, 4. With the space coordinates origin located at the axis of phase a1, the pulsating harmonic mmfs of order Q (Q r1, r 5, r 7, ") due to Q - th harmonic current are f akQ ( P ) f bkQ ( P ) f ckQ ( P )

ʌº ʌº ½ ª ª FQ ( P ) sin P «Z t (k 1) » cosQ «D (k 1) » ° 12 ¼ 12 ¼ ¬ ¬ ° 2ʌ ʌº 2ʌ ʌ º° ª ª FQ ( P ) sin P «Z t (k 1) » cosQ « a (k 1) » ¾ 3 12 ¼ 3 12 ¼ ° ¬ ¬ 4ʌ ʌº 4ʌ ʌ º° ª ª FQ ( P ) sin P «Z t (k 1) » cosQ « a (k 1) » ° 3 12 ¼ 3 12 ¼ ¿ ¬ ¬

(5.6.2)

Each phase pulsating mmf is resolved into two rotating mmfs by triangular identity and the sum of 12-phase mmfs yields the resulting rotating mmf as expressed by fQ ( P )

6 FQ ( P ) sin( PZ t QD )

(5.6.3)

Let order ȝ be 1for the fundamental component of the current, and order Ȟ be 1, 5, 7, 11, 13, 17, 19, 23, 25, " , respectively, for all mmf components, substituting them into (5.6.2), the resulting rotating mmfs yield that f1(1), f23(1), f25(1), " are of determined values with f5(1), f7(1), f11(1), f13(1), f17(1), f19(1), " being equal to zeros. Repeating the above process for 5, 7, 11, 13, 17, 19, 23, 25 harmonic components of the current, " , respectively, the corresponding resulting rotating mmfs are obtained for various harmonic components of the current. In general, for symmetrical 12 phase winding, the relation between ȝ and Q is

Q

24k P

k

0, r 1, r 2, r 3, "

(5.6.4)

The relation between the harmonic order of current and mmf of the 12-phase winding is different from that of the 3-phase winding. The space harmonic mmfs of the 12-phase winding are much less than those of the 3-phase winding when the same currents flow in them. It is noted that the harmonic mmfs of order 5, 7, 11, 13, 17, 19, " are not produced in the 12-phase winding with only the fundamental current flowing, no fundamental mmf can be produced with 5 and 7 order currents flowing, etc. 404


Under the condition of the power winding current being rectangular wave and the control winding current being sinusoidal wave, it is known from mmf amplitude calculation that any other harmonic mmf amplitudes are less than 5% of the fundamental mmf amplitude produced by fundamental current. Therefore, only considering the fundamental mmf produced by the fundamental current in analyzing the generator performance is generally permitted. With only the fundamental mmf produced by the fundamental current considered, the induction generator system is divided into two sub-systems, to which the circuit method and the electromagnetic field method are applied individually. The system which consists of 12-phase power winding with a rectifier load is calculated by the circuit method, and then the emf and current of power winding under constant voltage operation condition are obtained through iterative computation. The system of dual stator winding and solid cage rotor is analyzed by the harmonic electromagnetic field finite element method, and the stator frequency and the control winding current are obtained by the optimization approach.

5.6.3

Circuit Method for the System of Power Winding with Rectifier Load

The system of the 12-phase power winding with a rectifier load of the induction generator belongs to circuit network with multi-loop and variable-topology configuration in essence, and the a, b, c phase coordinates system is used to analyze it. The circuit transients are calculated by combining circuit topology theory and loop current method, from which network equations are automatically established by the systematic approach; finally, the calculation is conducted. It is easier to deal with inductances and diode branches to form the impedance matrix by the loop current method; therefore, the method is adopted to analyze the system owing to the existence of many inductances and diode branches. It is known from Fig. 5.6.1 that every rectifier bridge system has 16 branches except load one and 9 independent loops. Since 6 diodes in a rectifier bridge cannot be turned on simultaneously in ordinary operation, only 0, 2 or 3 diodes are turned on at the same time; correspondingly, the actual branch numbers are 9, 12, or 13, respectively with the actual loop numbers 3, 4 or 5. The system of 12-phase winding with a rectifier bridge has a total of 36 independent loops with 65 branches. In the actual operation process, the independent loop l and branch b are less than 36 and 65, respectively, which are relative to the topology configuration. Suppose that the column vector of loop current is il, and the column vector of the branch current is i, then the relation of il and i is i

B T il

(5.6.5)

where the basic loop matrix B is of l u b orders. 405

AC Machine Systems

Based on the relation between voltage and current of inductances and capacitors, the circuit differential equations are handled by using the backward Euler’s method. A transient equivalent circuit composed of the resistances, inductances and capacitances is obtained in the transient analysis. Suppose that Z is the impedance matrix of b u b orders, us and is are the independent voltage and current source column vectors of b u 1 orders, respectively, and u is the branch voltage matrix of b u 1 orders. According to the relation between variables in the branch and assumed positive directions, there is u

Zi us Zis

(5.6.6)

Using KVL leads to

Bu

0

(5.6.7)

Based on the above equations, it follows that BZB T il

BZis Bus

(5.6.8)

The loop current is il

Zl 1 BZis Z l 1 Bus

(5.6.9)

where loop impedance matrix Zl is of l u l orders. The loop current il is obtained from equation (5.6.9). Substituting it into equations (5.6.5) and (5.6.6), the branch current i and branch voltage u will be obtained. In system operation, the state variation of diode is determined according to voltage uD and current iD of the diode. If the voltage uD of a turned off diode is increased to 0.7V, then the diode becomes to be in a turned on state from its turned off state, and if the current iD of a turned on diode is decreased to 0A, then the diode becomes to be in a turned off state from its turned on state. The variation of the state of the diode in the rectifier bridge changes the system topology configuration. Considering the features of the rectifier load, a method combining the fixed step and varied step is applied in the simulation. The simulation begins from a topology configuration with fixed step, after every step, the state of the system topology configuration has to be judged by the calculated result. If no change takes place, the simulation with fixed step continues, else if the system topology configuration changes, an accurate time point when the topology configuration changes has to be found by varying the step. After getting the values of some variables at different time points, the fundamental and harmonics of these periodic variables are decomposed by Fourier transformation. The FFT is adopted in order to save computation time. Rated power PN of the sample machine is 18.4kW and output dc voltage Udc at 406


the rated load is 230V. The rotational speed keeps at 1500rpm, correspondingly, the normal frequency is 50Hz. Since the stator frequency f1 is very close to normal frequency in the ordinary operation of the induction generator, f1 is supposed to be 50Hz in the simulation. The calculation begins from a supposed Ep, emf of power winding, with known variables f1 and RL. After the load voltages at different moments are obtained, the average value Udc' is compared with 230V so that a new Ep is re-supposed. The final Ep is found when the difference between Udc' and Udc is less than the set error, from which other variables are also gained. Figures 5.6.2(a) and (b) are the simulated waves and experimental waves of phase a1 winding current in a period, respectively, and Figs. 5.6.3(a) and (b) the corresponding phase voltage waves. The variation trends of simulation waves and experimental waves in Figs. 5.6.2 and 5.6.3 are basically the same without distinct difference. This shows that the analysis method is effective.

Figure 5.6.2 Current waves of power winding (a) Simulated wave; (b) Experimental wave

Figure 5.6.3 Phase voltage waves of power winding (a) Simulated wave; (b) Experimental wave

407

AC Machine Systems

The calculated value, experimental value and their error of the phase current fundamental Ip are 16.71A, 16.47A and 1.46%, respectively, those of voltage fundamental Up 99.76V, 101.1V and 1.33%, and those of power factor angle ĳt 19.46e, 19.08eand 1.78%. The calculated fundamental values and the experimental ones of the variables match well, which validates the method used in calculating the fundamentals. It also shows that considering only the fundamental mmf produced by the fundamental current as the analysis foundation is reasonable.

5.6.4

Field Method for the System of Dual-Winding-Stator and Solid Cage Rotor

According to the positive direction of the power winding variables prescribed by the generator convention, it is assumed that the phasor of phase a1 emf reaches its maximum at zero moment, so its initial phase angle is 0 and the initial phase angle of the current phasor is ĳp. Because the excitation regulator is considered as a pure capacitor when the control winding only sends reactive power to the system, the phasor of phase A current leads ahead of that of emf by 90ewith the same generator convention for the positive direction of control winding variables. Phase A emf lags phase a1 emf by 22.5eowing to the fact that phase A control winding leads phase a1 power winding by 22.5eelectric degrees in space. The resultant phasor diagram is shown in Fig. 5.6.4.

Figure 5.6.4 Resultant phasor diagram

Under the condition of known rms value of power winding emf Ep, it is easy to get that of control winding emf Ec as follows: Ec

Wc kwc1 Ep W p k wp1

(5.6.10)

where Wc and kwc1 are the total series turn number per phase and the fundamental winding factor of the control winding, Wp and kwp1 are those of the power winding. 408


The adopted method is carried out in a Cartesian coordinates system. To simplify the physical model, the assumptions are made as follows: (i) The generator is infinite in axial length in order to convert the three-dimensional calculation into a two-dimensional one. (ii) The rotating rotor is replaced by a static one in the calculation and the current with slip frequency is added on the stator winding. (iii) The flux density of stator and rotor varies with time sinusoidally. (iv) The stator slot leakage flux is considered by the stator slot leakage reactance and the reduction of air-gap fundamental flux is considered by the skew-slot coefficient. For the two-dimensional sinusoidal electromagnetic field in the rest medium, the equation with the magnetic vector potential is w ª 1 wA º w ª 1 wA º « » « » wx ¬ P wx ¼ wy ¬ P wy ¼

jZV A Js

(5.6.11)

where the magnetic vector potential is of z axial component only and it does not vary in the axial direction, Ȧ is slip angular frequency, ȝ magnetic permeability, ı conductivity. The two terms to the right of equation (5.6.11) are the induced eddy current density and the source current density, respectively. For the problem to be solved, the source current only exists in the area of power winding and control winding of the stator slots, and the eddy current only exists in the area of the conductor in rotor slots and solid rotor zone. The boundary condition of the magnetic vector potential is zero around the stator outer periphery and the rotor inner periphery as well as periodical radial boundary. The finite element method is used for calculation. At first, the conductivity and the magnetic permeability of material in different areas are given, and the source current density is added to the area of power winding and control winding in the stator slots. In order to produce rotating magnetic field with negative phase sequence relative to the rotor, the slip frequency of the current should be reverse to the actual state, e.g. the phase angle of the current in the upper layer of phase a1 power winding should be – ĳp, the phase angle of the current in the upper layer of phase A control winding should be 67.5e , then in this condition the calculated phase angle of emf of phase a1 power winding should be 0. In the calculation, by using the ANSYS harmonic field, the slip frequency f is a variable to be determined by iteration. After the flux density of each node in the air-gap center periphery is gained through post-treatment of the result by the finite element method, the amplitude and phase angle of the magnetic density fundamental will be taken out by Fourier transformation. The amplitude of flux ĭm per pole is calculated from the fundamental flux density, consequently, the rms value of fundamental emf per phase of the power winding will be obtained as follows: E cp

2ʌ f1W p k pw1)m

(5.6.12) 409

AC Machine Systems

On the basis of the waveform of fundamental flux density, the phase angle M 0c of fundamental emf of power winding phase a1 may be determined as well. The rms value of control winding current Ic and that of slip frequency f are two variables to be found. The following two equations need to be satisfied for finding out the above two variables: E cp E p ½ ¾ M 0c 0 ¿

(5.6.13)

E cp and M 0c are obtained by the finite element method, so it is impossible to express them by apparent formulae of Ic and f. Since there is no traditional method to solve the above equations, a multi-variable optimization approach is used. This optimization model is without constraint. Selecting Ic and f as optimization variables, the object function can be written as f (Ic , f )

( E cp E p ) 2 kM 0c2

(5.6.14)

where k is a positive ratio coefficient. In reality, the optimization solution of equation (5.6.14) is to get the minimum of the object function min f(Ic, f). Obviously, the minimum of the object function is 0, namely min f(Ic, f)) is equivalent to equations (5.6.13). The direct search method without derivation is applied as a multi-variable optimization approach; concretely, it is the coordinate alternant method with good stability and simple programming, and only the axial search motion for the variables is made. The optimization begins with the initial values of Ic and f. If the object function no longer decreases at a certain round of the optimization process, the steps of two variables ǻIc and ǻf should be decreased. If the step is decreased to reach the prescribed precision, the optimization process ends, and consequently, Ic and f corresponding to the optimization point will be found. Final calculated results are obtained by finite element analysis and optimization approach at the rated load operation with constant voltage. Calculated flux density amplitude is 0.7092T from which ĭm is calculated, and E cp equals 101.3V based on equation (5.6.12). Obviously, E cp is equal to Ep, and M 0c to 0, so equation (5.6.13) is satisfied, which leads to finding Ic and f. At rated load, the calculated value, experimental value and their error of the control winding current Ic are 8.38 A, 7.71 A and 8.73%, and those of the slip s –0.010 96, –0.011 74 and –6.64%, respectively. The calculated results and experimental ones are in good agreement, which shows that the adopted analysis method is correct. The reason why the error occurs is due to two points: firstly, the iron loss has not been calculated. Secondly, the influence of leakage flux produced by control winding currents on the emf in power winding has not been considered. 410


References [1] Gao J D (1963) AC machine transients and operating modes analysis, Vol 2 (Chapter 7, 8, in Chinese). Science Press, Beijing [2] Gao J D, Zhang L Z, Huang L P (1983) Induction motor reswitching transients and the methods of restraining the reswitching surge current in an induction motor (in Chinese). China Symposium on Electric Machines, Shanghai, pp 154 162 [3] Gao J D, Zhang L Z, Huang L P (1984) Study of starting characteristics for induction motors (in Chinese). J Electrotechnical Journal, Chinese Electrotechnical Society, Beijing 1: 1 6 [4] Gao J D, Wang X H, Li F H (2004) Analysis of ac machines and their systems, 2nd Ed (Chapter 1, 5, in Chinese). Tsinghua University Press, Beijing [5] Rustebakke H M, Concordia C (1970) Self-excited oscillations in a transmission system using series capacitors. J IEEE Trans. PAS-89 (7): 1504 1512 [6] Wang X H, Wu X Z (2008) Research on dual-stator winding multi-phase high-speed induction generator with rectifier load. J Sci China Ser E-Tech Sci, 51(6): 683 692 [7] Wu X Z,Wang X H (2007) Circuit analysis of power winding with rectifier system for 12-phase induction generator (in Chinese). J CSEE 27(15): 75 82 [8] Wu X Z,Wang X H (2007) Determination of control winding current and stator frequency for dual stator-winding high-speed induction generator (in Chinese). J CSEE 27(18): 23 29 [9] Wu X Z,Wang X H, Luo C (2005) Relationship between harmonic currents and corresponding harmonic magnetomotive forces of multi-phase induction machines (in Chinese). J Tsinghua Univ. (Sci&Techn.) 45(7): 865 868 [10] Zhang B D, Zhang L Z (1988) Some measures to decrease the switching-over surge current when starting an induction motor and renewal evaluation of several classical starting mothods (in Chinese). J Beijing Society of Electrical Engineering 3: 27 40 [11] Zhang B D, Zhang L Z, Wang Z R (1988) Analysis and calculation of switching-over current for starting an induction motor by use of reduction of its stator voltage (in Chinese). J Beijing Society of Electrical Engineering 3: 11 26 [12] Zhang L Z (1995) Transient theory of induction machines and its uses in some aspects. Proc CICEM’95, International Academic Publishers, Hangzhou, pp 338 342 [13] Zhang L Z, Cheang T S (1999) Further research on double fed induction motors controlled by inverters. Proc CICEM’99, International Academic Publishers, Xi’an, pp 29 32 [14] Zhang L Z, Chok S C, Cheang T F (1995) Internal faults of 3-phase induction motors. Proc CICEM’95, International Academic Publishers, Hangzhou, pp 323 325

411

6 Internal Asymmetric Analysis of AC Machines

Abstract Internal faults in ac machine stator winding will cause severe damages to the machine. In order to protect electric machine and lighten the damage owing to internal fault, the speediness and reliability of relay protection are very important which demand understanding the currents in electric machine windings under internal faults clearly. Under the situation of internal fault of electric machine winding, there are stronger space harmonics in air-gap magnetic field and strtonger time harmonics in winding currents, which make symmetric component method be not suitable for analyzing it. At the same time the ideal electric machine model is no longer suitable, accordingly, adopting d, q, 0 coordinates system based on the corresponding model will cause very large errors. Therefore it is needed to adopt Multi-loop Theory and Method based on single coil as stated in chapter 1 for analyzing it. In this chapter using the presented method the rotor failures of bar broking and end-ring broking of induction machine, the internal faults of stator windings of salient pole synchronous machine and non-salient pole synchronous machine, etc. are discussed. The short circuit current of internal fault for stator winding is large, so it is necessary to pay special attention to the protection. In addition, the transient ac component is a little more than the steady-state current not just as the short circuit of external terminals. By the way, the short circuit current depends upon not only the short circuit sorts but also the short circuit turn ratio and the space position of short circuit turn, etc. The calculation results are supported by corresponding experiments.

In above Chapters the operation problems of ac machines with symmetric 3-phase stator windings and their system are analyzed mainly. In this chapter the internal asymmetric problems in electric machine windings will be analyzed and researched. The electric machine with internal asymmetric winding can be of normal condition, or non-normal condition or fault condition. For example, the synchronous machine is an electric machine with asymmetric rotor winding, its excitation winding is on direct axis, damping effects of direct axis and quadrature axis are different too. Single phase induction machine whose application is very wide is electric machine with asymmetric stator winding, its main winding and auxiliary winding are asymmetric two phase windings.


If part coils of stator winding of ac machine are damaged, after removing the damaged coils the machine could continue operation, in this instance the electric machine stator winding is of asymmetric condition too. When internal fault in ac machine stator winding occurs, the electric machine winding is also of asymmetric condition. Here the electric machine could be damaged severely. In order to protect electric machine and lighten the damage owing to internal fault, the speediness and reliability of relay protection are very important, which demand understanding the currents in electric machine windings under internal faults clearly. Under the condition of larger unit capacity the normal current of electric machine can reach thousands of amperes, even several ten thousands amp., at this time adopting multi-branch configuration for stator winding is necessary, and higher need will be presented for the relay protection of electric machine. The internal faults of stator winding of ac machine are various, including symmetric 3-phase internal short circuit and asymmetric internal short circuit. The forms of asymmetric internal short circuit are much more too, for example, inter turn short circuit (including short circuit between coils of the same branch and short circuit between different branches of the same phase), short circuit between branches of different phases, 3-phase asymmetric short circuit and short circuit of simultaneous two faults and so on. Besides short circuit, the internal asymmetry of ac machine may be caused owing to wrong connection of branch coil group. The polarity of coil group is connected in reverse, coil is connected to wrong branch, etc. which are all actual examples. Main feature of internal asymmetry of electric machine is of very strong space harmonics in air-gap magnetic field, even if the mmf is sine distribution, the space harmonics of air-gap magnetic field produced by it are also stronger. It will cause very large error to calculate only space fundamental of air-gap magnetic field. As an example, a salient pole synchronous machine is researched, this electric machine stator is of 108 slots, and the stator winding can be connected to form 6 parallel branches. The calculated results of self and mutual inductances of stator single coil, branch and phase winding are showed in Table 6.0.1, including two conditions—only considering space fundamental magnetic field and considering space harmonic magnetic field. N.B., in the table only the constant inductance parts independent of rotor position, that is, L0 and M 0 (refer to formulas (1.3.25), and (1.3.38), etc.) are listed. From Table 6.0.1 it can be seen that for stator single coil or branch, if only consider space fundamental magnetic field, the errors of the inductances are very large, even the sign is in reverse sometimes. But for the phase winding, the errors caused by only considering space fundamental magnetic field are not large (below 8% in the example). The reason is that when the current flows in stator single coil or single branch, the fractional and lower order harmonics are very strong, some of them even exceed the fundamental magnetic field, but while the current flows in phase winding, the basic wave of air-gap magnetic field is 413

AC Machine Systems

fundamental, the fractional harmonics and lower order harmonics of mmf of various coils composing the phase winding counteract each other, so total mmf of phase winding is fundamental mainly. Table 6.0.1 Calculated results of stator winding inductances of a 550kW salient pole synchronous machine unit: H Only considering space fundamental of air-gap magnetic field

Considering space harmonics of air-gap magnetic field

0.884 u104

0.106 u102

Mutual inductance between adjacent coils

0.830 u10

4

0.962 u103

Self inductance of single branch

0.293 u102

0.185 u101

Mutual inductance between adjacent branches of the same phase

0.293 u102

0.134 u103

0.127

0.132

0.0561

0.0578

Inductance sort Self inductance of single coil

Self inductance of phase A winding Mutual inductance between phase A, B windings

This example shows that for internal asymmetric problems in the stator, only considering space fundamental magnetic field may cause not permitted error. Of course, magnitudes of the error are different owing to the difference of researched problems. For example, for the inter-turn short circuit of synchronous machine with wave winding and two poles turbo-generator, the space harmonics of air-gap magnetic field is less; but for inter-turn short circuit of synchronous machine with lap winding, the space harmonics of air-gap magnetic field are very strong, the more pole-pair number is, the larger harmonic component is. To sum up, under the situation of internal asymmetry of electric machine winding, due to strong space harmonics in the air-gap magnetic field and with different rotation speed and different rotation direction, the current harmonics in each winding are more and stronger, that is, at this time there are stronger space harmonics in air-gap magnetic field and stronger time harmonics in winding currents. Therefore in the face of the feature of internal asymmetry and various sorts of winding internal asymmetry it is needed to adopt Multi-loop Theory and Method based on single coil as showed in Chapter 1 for analyzing. In this Chapter using the method the rotor failures of bar broking and end ring broking of induction machine, the internal faults of stator winding of salient pole synchronous machine, calculation of loop parameters, etc. are discussed. As to another internal asymmetry in the winding, for example the internal faults of stator winding of induction machine, the operation of synchronous machine after 414


cutting damaged stator coils, internal faults of excitation winding, etc., they can be researched by using similar method, here do not give unnecessary details.

6.1 Analysis of Rotor Winding Failures of Squirrel Cage 3-Phase Induction Machines Rotor bar broking and end ring broking of induction machine are familiar failures for induction machine, they may be caused by the disfigurements in manufacture process or in operation. For example, thin bar or broking bar may appear in cast aluminium rotor of medium-size and small induction machine owing to technology problems; dummy jointing between bar and end ring in larger-scale induction machine may appear due to bad welding; improper operation, bad work environment and frequently starting, etc. can easily cause the failures in squirrel cage rotor. Therefore realizing the effects of rotor winding failures of induction machine on operation performance and detecting the rotor failures based on the variation of electric quantities of the machine after failures are important technical steps for avoiding the accidents in operation and ensuring safe production[1 3]. For analysis convenience, under the condition of no effect on physical essence, make the following assumptions: (i) Surfaces of stator and rotor are smooth, the effects of tooth and slots are expressed by Carter’s coefficient. (ii) Neglecting the eddy current, hysteresis and nonlinear, the effect of iron reluctance is calculated by magnifying the air-gap adequately. (iii) Air-gap of electric machine is uniform, the slot harmonics are not considered, and before failure occurring the electric machine is symmetric in both electric and magnetic aspects. (iv) Insulation exists between rotor bar and iron. In the following the rotor failures of squirrel cage induction machine are discussed according to the electric machine example showed in Fig. 6.1.1.

Figure 6.1.1 Stator windings and rotor loops of squirrel cage 3-phase induction machine 415

AC Machine Systems

Figure 6.1.1 (a) is 3-phase symmetric stator windings of squirrel cage induction machine, Y-connection (similarly treating for '-connection); Figure 6.1.1(b) is rotor loops. If the rotor slot number is n , before the rotor failure occurring the node number of rotor circuit is 2n , the branch number is 3n , hence the current number of independent loops is 3n (2n 1) n 1 . The rotor currents are showed in the figure, the loop currents are i1 , i2 ," , in , ie respectively. So the number of independent currents in stator and rotor is n 4 .

6.1.1

Basic Equations of Squirrel Cage Three Phase Induction Machine

Based on above mentioned assumptions and positive direction regulation in Fig. 6.1.1 the voltage equations and flux-linkage equations before failure occurring are listed as U

p\ RI

(6.1.1)

\

LI

(6.1.2)

where U , I , \ all are (n 4) u 1 column matrix; R, L are (n 4) u (n 4) square matrix. Above two formulas can be spread to

416

ª ua º «u » « b» « uc » « » «0» «0» « » «#» «0» « » «¬ 0 »¼

ª\ a º ª r «\ » «0 « b» « «\ c » «0 « » « \ 0 « p 1»« «\ 2 » «0 « » « « # » «# «\ » «0 « n» « «¬\ e »¼ «¬0

ª\ a º «\ » « b» «\ c » « » «\ 1 » «\ 2 » « » «# » «\ » « n» «¬\ e »¼

ª Laa «M « ba « M ca « « M 1a « M 2a « « # «M « na «¬ M ea

0 r 0 0 0 #

0 0 0 0 r 0 0 rr 0 rb # # 0 0 rb 0 0 re

0 0 0 rb rr # 0 re

" 0 " 0 " 0 " rb " 0 # " rr " re " " " " "

º ªia º » «i » »« b» » « ic » »« » » « i1 » » «i2 » »« » »«# » re » «in » »« » nre »¼ «¬ ie »¼

M ab Lbb M cb M 1b M 2b #

M ac M bc Lcc M 1c M 2c #

M a1 M b1 M c1 L11 M 21 #

M a2 M b2 M c2 M 12 L22 #

M nb M eb

M nc M ec

M n1 M e1

M n 2 " Lnn M e 2 " M en

0 0 0 re re #

M an M bn M cn M 1n M 2n #

M ae º ªia º M be »» «« ib »» M ce » « ic » »« » M 1e » « i1 » M 2e » «i2 » »« » # »«# » M ne » «in » »« » Lee »¼ «¬ ie »¼

(6.1.3)

(6.1.4)


In formula (6.1.3) rr

6.1.2

2(rb re ) is self resistance of rotor loop.

Calculation of Loop Inductances of Squirrel Cage Three Phase Induction Machine

As is the same as the inductances of single phase induction machine (refer to Section 1.10), calculation formulas of loop inductances of squirrel cage 3-phase induction machine can be written as follows. (1) Self inductances of stator windings (refer to formula (1.10.8)) are 2

Laa

Lbb

Lcc

Ls

2 w12W l §k · O0 ¦ ¨ wk ¸ Lsl 2 PS k © k ¹

(6.1.5)

Mutual inductances between stator windings (refer to (1.10.10)) are

M ab

M bc

M ca

M ba

M cb

M ac

Ms

2

2 w12W l §k · § 2S · O0 ¦ ¨ wk ¸ cos ¨ K ¸ M sl 2 PS © 3 ¹ k © k ¹

(6.1.6)

It is the result of changing qT in formula (1.1.10) to 2ʌ / 3 , in which the electric angle between central lines of phase windings. In above two formulas k 1, 3, 5, " . Lsl for formula (6.1.5) and M sl in formula (6.1.6) are leakage self inductance and leakage mutual inductance of phase windings respectively. Self inductances of rotor loops (refer to formula (1.10.10)) are

L11

L22

" Lnn

Lr

kE S 2W l 1 O0 ¦ 2 sin 2 r 2( Lb Le ) P S2 k 2 k

(6.1.7)

1 2 , ,"; 2( Lb Le ) is leakage self inductance of rotor loop. P P Self inductance of end ring loop is sum of leakage inductances of several segments of the end ring, that is in which k

Lee

nLe

(6.1.8)

Mutual inductance between i - th loop and j - th loop of rotor (refer to formula (1.10.12)) is M ij where k

kE S 2W l 1 O sin 2 r cos k ( j i )M 2 0¦ 2 PS 2 k k

(6.1.9)

1 2 , ," P P 417

AC Machine Systems

When the two rotor loops are adjacent, that is | i j | 1 or n 1 , the leakage mutual inductance Lb should be added to above formula. Mutual inductance between end ring loop and bar loop is M ek

M ke

k 1, 2, " , n

Le

(6.1.10)

Above formula indicates that the mutual inductance between end ring loop and bar loop is namely leakage inductance of one segment of end ring, because the end ring loop and bar loop are perpendicular to each other, the mutual inductance responding to air-gap magnetic field is zero. Under uniform air-gap for induction machine formulas (6.1.7) and (6.1.9) can be simplified (refer to Fig. 6.1.2). mmf produced by the current of rotor i -th loop is

F ( x)

n 1 °° n i ® ° 1 i °¯ n

M 2

x

x

M 2

M 2

or x !

M 2

where the electric angle between rotor adjacent loops is M

2PS . The air-gap n

flux density produced by this mmf is B( x)

F ( x)OG ( x)

F ( x)

O0 2

Hence the self inductance of rotor loop produced by air-gap magnetic field is Lii

\ ii i

Wl

M

iS ³

2

M B ( x )dx 2

Wl

M

n 1 O0 dx n 2 2

M S³ 2

PW l O0

n 1 n2

Considering leakage self inductance, 2( Lb Le ) should be added to above formula.

Figure 6.1.2 Air-gap mmf produced by the rotor loop current of induction machine

Mutual inductance between rotor i - th loop and j - th loop ( i z j ) is

418


M ij

\ ij i

Wl

iS ³

( j i )M

( j i )M

M 2

M

B( x)dx

2

Wl

S³

( j i )M

( j i )M

M 2

M

2

1 O0 dx n 2

PW l O0

1 n2

If i -th loop and j -th loop are adjacent, the leakage mutual inductance Lb should be added to the mutual inductance too. (2) Mutual inductance between stator winding and rotor loop (refer to formula (1.10.16)) is

M sj

2 w1W l kE S 1 2S ª º O0 ¦ 2 kwk sin r cos k «J (i 1) jM » 2 PS 2 3 ¬ ¼ k k

(6.1.11)

in which k 1,3,5,"; J is electric angle by which the central line of rotor loop leads the stator winding axis; s represents stator winding a, b, c 3-phases which correspond to i 1, 2,3 respectively; j is sequence number of rotor bar loop, j 1, 2," , n . The mutual inductance between stator winding and end ring loop can be neglected, that is M ae

M be

M ce

M ea

M eb

M ec

0

(6.1.12)

From above formulas it can be seen that in squirrel cage 3-phase induction machine the inductances of stator windings and inductances of rotor loops are all constants independent of rotor position which is easy to be understood due to the cylindrical form of stator and rotor. Only the mutual inductances between stator windings and rotor loops are time-variant parameter relative to rotor position. It is needed to point out that formulas (6.1.5), (6.1.6) and (6.1.11) are all gained for single branch stator winding. If phase winding is of multi-branches with branch number a1 per phase, then w1 in the formulas has to be changed to w1 / a1 . Also attention, if skewed slot is adopted in the electric machine, the skewed kE § kE · slot coefficient ksk sin ¨ ¸ , should be added to formula (6.1.11), here 2 © 2 ¹ E is skewed slot space electric angle.

6.1.3

Amending Basic Equations Based on Rotor Failure Condition

After various inductances are substituted into flux-linkage equation (6.1.4) and voltage equation (6.1.3), a differential equation set is obtained which is an equation under normal condition, with inductances and resistances as coefficients, and various loop currents are the unknowns in the equation set. After rotor failures occur, the equation set can be modified in order to get corresponding equations, without writing other equations again. 419

AC Machine Systems

(1) Rotor bar broking Figure 6.1.3 shows the variety condition while rotor bar j broking. Here the rotor loop current i j 1 i j , which is equivalent to combining of loop j and loop j 1 to bar j adjacent to become a new loop. Therefore accordingly basic equations (6.1.3) and (6.1.4) have to be modified as follows: (i) Elements in ( j 4)-th row of the matrix R and L are added to corresponding elements in ( j 3)-th row. (ii) Elements in ( j 4)-th column of the matrix R and L are added to corresponding elements in ( j 3)-th column. (iii) Elements in ( j 4)-th row and ( j 4)-th column of the matrix R and L are canceled. (iv) Elements in ( j 4)-th row of the column vector U , I and \ are canceled. Actually such treatment is canceling original ( j 3)-th and ( j 4)-th equations (i.e. the equations of loop j and loop j 1 of the rotor), and they are exchanged to the equation of rotor new “large” loop (it consists of original loop j and loop j 1 ). (2) Rotor end ring broking As showed in Fig. 6.1.4(a), if end ring of squirrel cage rotor is broken, obviously, there is i j 0 . Accordingly in the basic equations (6.1.3) and (6.1.4), the ( j 3)-th row and ( j 3)-th column of the matrix R and L , and ( j 3)-th row of the column vector U , I and \ should be canceled.

Figure 6.1.3 Variety of rotor currents while rotor bar j broking (a) General expression of rotor loops; (b) Bar current and end ring current corresponding to Fig. (a); (c) After bar j broking i j 1 i j ; (d) Expression of loop current corresponding to Fig. (c)

If two end rings are broken simultaneously, failure of one of the two end rings can be treated according to above mentioned method, and suppose that another 420


end ring is broken at loop K . Here the loop current ik is equal to negative value of the end ring loop current ie (refer to Fig. 6.1.4(b)), which corresponds to combining loop K and end ring loop to form a new loop. Accordingly, basic equations (6.1.4) and (6.1.3) have to be modified as follows:

Figure 6.1.4 Expression of end ring broking

(i) The negative values of the elements in ( K 3)-th row of the matrix R and L are added to corresponding elements in last row. (ii) The negative values of the elements in ( K 3)-th column of the matrix R and L are added to corresponding elements in last column. (iii) Elements in ( K 3)-th row and ( K 3)-th column of the matrix R and L are canceled. (iv) Elements in ( K 3)-th row of the column vector U , I and \ are canceled. Actually such treatment is to cancel original ( K 3)-th and last equations (namely the equations of rotor loop K and end ring loop), and they are exchanged to the equation of rotor new end ring loop (it consists of original end ring loop and loop K ). If the end ring loop is broken at several places, based on broking condition and above mentioned method it can be treated one by one. If the bar and end ring are broken simultaneously, according to above two methods it can be treated respectively.

6.1.4

Calculation of Steady State Loop Currents

Based on above mentioned method, after modifying the basic equations according to the failure situations, a new group of differential equations with variant coefficients and variables of loop currents is obtained. Then corresponding transient and steady-state conditions of the electric machine may be analyzed and researched. Calculation of the steady-state loop currents of squirrel cage 3-phase induction machine is discussed as follows. In steady-state firstly the expression of each loop current of electric machine is assumed, then the expressions are substituted into above mentioned differential equations, based on the principles that the same frequency quantities in two sides of equation are equal to each other, the differential 421

AC Machine Systems

equation group is transformed to linear algebraic equation group. Then the steady-state values of every various currents can be obtained according to the linear algebra equation group. Suppose that the power source is sine wave voltage with fundamental frequency in the steady-state operation of squirrel cage 3-phase induction machine with rotor failures, and the voltage expressions are ua ub uc

½ ° ° 2ʌ · ° § U cos ¨ Z 0 t ¸ ¾ 3 ¹° © 2ʌ · ° § U cos ¨ Z 0 t ¸° 3 ¹¿ © U cos Z 0 t

(6.1.13)

These stator fundamental voltages will cause fundamental currents in stator windings, and produce revolving magnetic field with synchronous speed and positive direction in air-gap, and produce the currents with frequency sZ 0 in rotor loops. The rotor currents will produce the revolving magnetic field with various rotation speeds and different rotation directions owing to rotor asymmetry. But the currents of stator 3-phase symmetric windings produced by fractional harmonic and high order harmonic magnetic field caused by rotor currents can be neglected in general, hence only the fundamental magnetic field due to rotor currents with positive and reverse direction rotation can induce the currents in stator windings, the former induces fundamental currents, the latter induces the currents of frequency (1 2 s )Z0 . Since stator three windings are symmetric, these stator currents of two frequencies will correspond to the rotor currents of frequency sZ 0 only, and do not cause currents with other frequency in rotor. Therefore the expressions of stator phase currents can be written as

ia ib

ic

422

½ I1 cos Z0 t I1c sin Z 0 t I 2 s cos(1 2s )Z0 t I 2cs sin(1 2 s)Z 0 t ° ° ° 2S · 2S · § § I1 cos ¨ Z0 t ° ¸ I1c sin ¨ Z0 t ¸ 3 ¹ 3 ¹ © © ° °° 2S º 2S º ª ª I 2 s cos «(1 2 s )Z0 t » I 2c s sin «(1 2s )Z 0 t » ¾ 3¼ 3¼ ¬ ¬ ° ° 2S · 2S · § § I1 cos ¨ Z0 t ° ¸ I1c sin ¨ Z 0 t ¸ 3 ¹ 3 ¹ © © ° ° 2 2 S S ª º ª º ° I 2 s cos «(1 2 s )Z0 t » I 2cs sin «(1 2 s )Z0 t » 3¼ 3¼ °¿ ¬ ¬

(6.1.14)


The expressions of rotor loop currents can be written as

i1 i2 in ie

I r1 cos sZ 0 t I rc1 sin sZ 0 t ½ ° I r 2 cos sZ0 t I rc2 sin sZ 0 t ° ° # ¾ I rn cos sZ0 t I rnc sin sZ 0 t ° ° I re cos sZ0 t I rec sin sZ0 t °¿

(6.1.15)

The expressions of these currents are substituted into formulas (6.1.4) and (6.1.5), then we can get the following formulas: n

Ls ia M s ib M s ic ¦ M aj i j

\a

j 1

n

( Ls M s )ia ¦ M aj i j j 1

( Ls M s )[ I1 cos Z 0 t I1c sin Z 0 t I 2 s cos(1 2s )Z 0 t n

I 2cs sin(1 2 s)Z 0 t ] ¦ M rs cos[(1 s )Z0 t jM )] j 1

u ( I rj cos sZ0 t I rjc sin sZ 0 t ) ½ M rs n ( I rj cos jM I rjc sin jM ) ¾ cos Z 0 t ®( Ls M s ) I1 ¦ 2 j1 ¯ ¿ ½ M n ®( Ls M s ) I1c rs ¦ ( I rj sin jM I rjc cos jM ) ¾ sin Z0 t 2 j1 ¯ ¿ M ®( Ls M s ) I 2 s rs 2 ¯ M ®( Ls M s ) I 2cs rs 2 ¯

½ cos jM I rjc sin jM ) ¾ cos(1 2 s )Z0 t j 1 ¿ n ½ ( I rj sin jM I rjc cos jM ) ¾ sin(1 2s )Z0 t ¦ j 1 ¿ n

¦ (I

rj

where for mutual inductances between stator windings and rotor loops, the effects of high order harmonics are neglected, only considering the fundamental, that is (refer to formula (6.1.11)) M aj

2 w1W l ES O0 kw1 sin r cos[(1 s )Z 0 t jM ] 2 PS 2 M rs cos[(1 s )Z 0 t jM ]

moreover 423

AC Machine Systems

ua

U cos Z0 t

p\ a ria

Z0 M rs ®Z0 ( Ls M s ) I1c 2 ¯

n

¦ ( I j 1

rj

½ sin jM I rjc cos jM ) rI1 ¾ cos Z 0 t ¿

½ ZM n ®Z0 ( Ls M s ) I1 0 rs ¦ ( I rj cos jM I rjc sin jM ) rI c¾ sin Z 0t 2 j1 ¯ ¿ n 1 ®(1 2 s )Z0 ( Ls M s ) I 2cs (1 2 s )Z0 M rs ¦ ( I rj sin M 2 j 1 ¯ ½ I rjc cos jM ) rI 2 s ¾ cos(1 2s )Z 0 t ® (1 2s )Z 0 ( Ls M s ) I 2 s ¿ ¯ n ½ 1 (1 2 s )Z0 M rs ¦ ( I rj cos jM I rjc sin jM ) rI 2cs ¾ sin(1 2s )Z 0 t 2 j 1 ¿

(6.1.16) Letting the terms of two sides of the above equation including cos Z0 t , sin Z0 t , cos(1 2 s ) Z 0 t and sin(1 2 s )Z 0 t by equal respectively, then

U

Z 0 ( Ls M s ) I1c

½ ° 1 ° c Z 0 M rs ¦ ( I rj sin jM I rj cos jM )u rI1 ° 2 j 1 ° 0 Z 0 ( Ls M s ) I1 ° ° n 1 ° Z0 M rs ¦ ( I rj cos jM I rjc sin jM ) rI c 2 ° j 1 ¾ 0 (1 2 s )Z 0 ( Ls M s ) I 2c s ° n ° 1 (1 2 s )Z 0 M rs ¦ ( I rj sin jM I rjc cos jM ) rI 2 s ° 2 j 1 ° ° 0 (1 2 s )Z 0 ( Ls M s ) I 2 s ° n ° 1 (1 2 s)Z 0 M rs ¦ ( I rj cos jM I rjc sin jM ) rI 2c s ° 2 j 1 ¿ n

(6.1.17)

For phases b and c the voltage equation similar to equation (6.1.17) can be got. The flux-linkage of rotor loop j is

\j

M ja ia M jb ib M jc ic Lr i j M ij ¦ ik Lb (i j 1 i j 1 ) Le ie k

424


M rs cos[(1 s )Z0 t jM ][ I1 cos Z0 t I1c sin Z 0 t I 2 s cos(1 2s )Z 0 t I 2c s sin(1 2s )Z 0 t ] 2S º 2S · ª § M rs cos «(1 s )Z0 t jM » ® I1 cos ¨ Z0 t ¸ 3 ¼¯ 3 ¹ ¬ © 2S · 2S º § ª I1c sin ¨ Z 0 t ¸ I 2 s cos «(1 2s )Z 0 t » 3 ¹ 3¼ © ¬ 2S º ½ 2S º ª ª I 2cs sin «(1 2 s )Z0 t » ¾ M rs cos «(1 s )Z 0 t jM » 3 ¼¿ 3¼ ¬ ¬ 2S · 2S · § § u ® I1 cos ¨ Z0 t ¸ I1c sin ¨ Z 0 t ¸ 3 3 ¹ © ¹ © ¯ 2S º 2S º ½ ª ª I 2 s cos «(1 2s )Z 0 t » I 2cs sin «(1 2 s)Z 0 t » ¾ 3 3 ¼¿ ¬ ¼ ¬ Lr ( I rj cos sZ 0 t I rjc sin sZ 0 t ) M ij ¦ ( I rk cos sZ 0 t I rkc sin sZ 0 t ) K

Lb ( I rj 1 cos sZ 0 t I rjc 1 sin sZ0 t I rj 1 cos sZ 0 t I rj 1 sin sZ 0 t ) Le ( I re cos sZ0 t I rec sin sZ 0 t ) 3 ® M rs ( I1 cos jM I1c sin jM I 2 s cos jM I 2c s sin jM ) Lr I rj ¯2 ½ M ij ¦ I rk Lb ( I rj 1 I rj 1 ) Le I re ¾ cos sZ0 t k ¿ 3 ® M rs ( I1 sin jM I1c cos jM I 2 s sin jM I 2c s cos jM ) Lr I rjc ¯2 ½ M ij ¦ I rkc Lb ( I rjc 1 I rjc 1 ) Le I rec ¾ sin sZ0 t k ¿ It is known that the voltage equation is uj

p\ j rr i j rb i j 1 rb i j 1 re ie

0

Substituting \ j into above formula, then get 0

ª3 ® sZ0 « M rs ( I1 sin jM I1c cos jM I 2 s sin jM I 2c s cos jM ) ¬2 ¯ º Lr I rjc M ij ¦ I rkc Lb ( I rjc 1 I rjc 1 ) Le I rec » rr I rj k ¼ ½ ª3 rb I rj 1 rb I rj 1 re I re¾ cos sZ0 t ® sZ0 « M rs ( I1 cos jM ¬2 ¿ ¯ 425

AC Machine Systems

I1c sin jM I 2 s cos jM I 2cs sin jM ) Lr Lrj M ij ¦ I rk k

º ½ Lb ( I rj 1 I rj 1 ) Le I re » rr I j rb I rjc 1 rb I rjc 1 re I rec ¾ sin sZ0 t ¼ ¿

(6.1.18)

where k 1, 2," , j 1, j 1," , n . Letting the terms of two sides of above formula including cos sZ 0 t and sin sZ 0 t be equal respectively, then we can obtain the following two equations: 0

0

½ ° ° ° ° ° ° ¾ ª3 sZ0 « M rs ( I1 cos jM I1c sin jM I 2 s cos jM I 2cs sin jM ) ° ° ¬2 ° º ° Lr I rj M ij ¦ I rk Lb ( I rj 1 I rj 1 ) Le I re » ° k ¼ ° rr I rjc rb ( I rjc 1 I rjc 1 ) re I rec ¿ ª3 sZ0 « M rs ( I1 sin jM I1c cos jM I 2 s sin jM I 2c s cos jM ) ¬2 º Lr I rjc M ij ¦ I rkc Lb ( I rjc 1 I rjc 1 ) Le I rec » k ¼ rr I rj rb ( I rj 1 I rj 1 ) re I re

(6.1.19)

where k 1, 2," , j 1, j 1," , n . Similarly, for end ring loop there is n

\e

¦L i

e j

nLe ie

j 1 n

¦ L (I e

rj

cos sZ 0 t I rjc sin sZ 0t ) nLe ( I re cos sZ 0 t I rec sin sZ0 t )

j 1

n

ue

0

p\ e nre ie ¦ re i j j 1

n

n

[ sZ 0 ( Le ¦ I rjc nLe I rec ) nre I re re ¦ I rj ]cos sZ 0 t j 1

j 1

n

n

j 1

j 1

[ sZ 0 ( Le ¦ I rj nLe I re ) nre I rec re ¦ I rjc ] sin sZ 0 t

(6.1.20)

or 0 0 426

n n ½ sZ0 ( Le ¦ I rjc nLe I rec ) nre I re re ¦ I rj ° j 1 j 1 ° ¾ n n sZ0 ( Le ¦ I rj nLe I re ) nre I rec re ¦ I rjc ° °¿ j 1 j 1

(6.1.21)


In the linear algebraic equation set consisting of formulas (6.1.17), (6.1.19) and (6.1.21), the unknowns are the amplitudes of sine and cosine terms of electric machine various loop currents, the total number of unknowns is 2( n 3) , the coefficients are inductances and resistances, the number of independent equations is also 2(n 3) . Hence the currents of various loops can be obtained by Gauss elimination method. Electro-magnetic torque (refer to formula (1.9.8)) is Te

P wL T I I 2 wJ

(6.1.22)

The currents of above formula are divided into stator currents and rotor currents, and the inductance matrix is written as partition matrices, then

Te

d ª º Lsr » T « ªI º dJ P »« s » Is Ir « d 2 « » ¬ I rT ¼ L « dJ rs » ¬ ¼ · P§ d d Lrs I sT I s Lsr I rT ¸ ¨ Ir 2 © dJ dJ ¹ PI s

d Lrs I rT dJ

PM rs

Z

>ia

ib

ic @

cos[(1 s )Z0 t 2M ] ªcos[(1 s )Z0 t M ] « 2S 2S ª º ª º M » cos «(1 s )Z 0 t 2M » d «cos (1 s )Z0 t u « «¬ 3 3 ¼ ¬ ¼ dt « 2 S 2 S º ª º «cos ª(1 s )Z t M » cos «(1 s )Z 0 t 2M » 0 «¬ «¬ 3 3 ¼ ¬ ¼

" cos[(1 s )Z 0 t nM ] 2S ª º " cos «(1 s )Z 0 t nM » 3 ¬ ¼ 2 S ª º " cos «(1 s )Z 0 t nM » 3 ¬ ¼

0 º ªi1 º » «i » 2 0» « » » «# » »« » i 0» « n » »¼ «¬ie »¼

The expressions of stator currents and rotor currents (6.1.14) and (6.1.15) are substituted into above formula, and it is simplified as Te

Tave Tpul 427

AC Machine Systems

where Tave is average torque, Tpul is pulsating torque, and n n ª º 3 PM rs «( I1c I 2cs )¦ I rj cos jM ( I1 I 2 s )¦ I rjc cos jM » 4 j 1 j 1 ¬ ¼

Tave

n n ª º 3 PM rs «( I1c I 2cs )¦ I rjc sin jM ( I1 I 2 s )¦ I rj sin jM » 4 j 1 j 1 ¬ ¼

n n ª 3 PM rs «( I1c I 2cs )¦ I rj cos jM ( I1 I 2 s )¦ I rjc cos jM 4 j 1 j 1 ¬

Tpul

n n º ( I1c I 2cs )¦ I rjc sin jM ( I1 I 2 s ) ¦ I rj sin jM » cos 2sZ 0 t j 1 j 1 ¼

n n ª 3 PM rs «( I1c I 2cs )¦ I rj sin jM ( I1 I 2 s )¦ I rjc sin jM 4 j 1 j 1 ¬

n n º ( I1c I 2cs )¦ I rjc cos jM ( I1 I 2 s ) ¦ I rj cos jM » sin 2sZ0 t j 1 j 1 ¼

Experiments, analysis and calculation have been carried out for a 7.5kW squirrel cage 3-phase induction machine under four conditions: rotor symmetry, one bar broking, rotor series three bars broking and end ring broking, etc. Table 6.1.1 shows stator currents, input power, electro-magnetic torque and power factor under Table 6.1.1 Comparison between calculated results and experimental data of a 7.5kW squirrel cage 3-phase induction machine Operation conditions performances

Stator current I1 / A Average electromagnetic torque Tave / N m Input power P1 / W Power factor

cos M 428

Calculated Experimental Relative error Calculated Experimental Relative error Calculated Experimental Relative error Calculated Experimental Relative error

Rotor symmetry

9.51 9.45 0.63% 56.6 55.4 2.17% 9444 9500 0.59% 0.871 0.882 1.25%

Rotor one Rotor series bar three bars broking broking

9.27 8.91 4.04% 54.7 51.9 5.39% 9111 9025 0.95% 0.862 0.889 3.04%

8.53 8.61 0.93% 47.4 47.3 0.21% 7893 8365 5.64% 0.813 0.852 4.60%

Rotor end ring broking

8.86 9.18 3.49% 51.8 52.2 0.77% 8624 9008 4.26% 0.854 0.861 0.78%


above conditions while stator phase voltage is 380V, slip is 0.03 . Therein power factor angle M is the phase angle between stator phase a voltage and phase a fundamental current ia1 , and it is gained from the calculated results directly. Input power is calculated based on formula 3U a I a1 cos M . From the table it can be seen that calculated results and experimental data are consistent basically. In order to realize the effect of rotor failures more clearly, based on the calculated values of rotor currents, the rotor current phasor diagrams under rotor winding normal condition and failure condition are drawn as showed in Fig. 6.1.5. Figure 6.1.5(a) is rotor current phasor diagram under rotor winding normal condition, and Fig. 6.1.5(b) is rotor current phasor diagram under the condition of bar 10,11,12 broking simultaneously. A light rotor failure can not bring about large effect on normal operation of squirrel cage 3-phase induction machine, normally it only causes the slip slightly to increase at constant load, accordingly the stator current, input power, etc. increase. But when severe rotor failure occurs, the starting time of the electric machine maybe obviously increases, even electric machine can not be started, and acute vibration is accompanied with it. At this time the electric machine loading capability drops obviously, which will affects the normal operation of the machine. From above analysis and calculations it can be seen that while a certain bar broking, adjacent bar currents maybe increase, and the adjacent bar broking will be accelerated. Therefore when slight rotor failure of electric machine occurs, detecting the failure in time and adopting suitable steps are very necessary. It is very important in both economics and safe production. Only fundamental component exists in the stator current of induction machine in normal operation. After rotor failure occurs, addtional component of frequency (1 2 s ) f appears in stator current, and its magnitude depends on severe degree of rotor asymmetry. So detecting the stator current component of frequency (1 2 s ) f can judge rotor failure. The current component has following specific features: (i) Generally the slip of induction machine under steady-state operation is very small, so frequency (1 2s ) f is very near to the fundamental frequency f , and it is very difficult to separate the component of frequency (1 2s ) f from the frequency f component by using filter; (ii) The amplitude of the current component with frequency (1 2s ) f is very small, and it varies with different values of s . Considering these two features it is difficult to analyze the frequency of stator currents by general harmonic analysis apparatus. One method is to memorize the current signal of the electric machine by magnetic tape recorder, then analyzing the signal using signal processor, and getting its frequency spectrum. When analyzing the signal the “by-pass petal” of main frequency should decrease as best as possible, in order to prevent the signal (1 2 s ) f from submerging in the by-pass petal. So it is needed to select suitable window function. The main petal of 429

AC Machine Systems

Figure 6.1.5 Rotor current phasor diagram of induction machine (a) Rotor current phasor diagram under rotor winding normal condition; (b) Rotor current phasor diagram under the condition of bar 10,11,12 broking simultaneously.

Hamming Window is wider, but the by-pass petal can be restrained well, hence it can be selected and used in practice. Moreover, because these two frequencies of current signals are adjacent, the sampling point number should increase as best as possible in order to improve frequency differentiation rate. Another method is that the signal uia A cos(Z t M ) Ac cos > (1 2s )Z t M1 @ , is drawn out from phase 430


a current, the signal uac A1 cos(Z t 30e ) is drawn out from voltage uac , and make A | A1 (refer to stator voltage phasor diagram Fig. 6.1.6), then these two signals are subtracted from each other as uia uac

A cos(Z t M ) A1 cos(Z t 30e ) Ac cos[(1 2s )Z t M1 ]

M §M · § · | 2 A sin ¨ 15e ¸ sin ¨ Z t 15e ¸ 2 2 © ¹ © ¹ Ac cos[(1 2s )Z t M1 ]

Figure 6.1.6 Phasor diagram of stator voltage and current

It can be seen that for the signal uia , ratio of component (1 2s ) f to ª §M ·º component f is Ac / A , but for signal uia uac , the ratio is Ac « 2 A sin ¨ 15e ¸» . ©2 ¹¼ ¬ The power factor angle M of 3-phase induction machine is near 30e generally, while analyzing the signal uia uac , the amplitude of component (1 2 s ) f increases much more relative to the component f , therefore it is more easy to detect the component (1 2 s ) f .

6.2 Internal Fault Analysis of Stator Winding of Salient Pole Synchronous Machines Internal faults of stator winding of synchronous machine are normal faults with destructive damages. The very large short circuit current of internal fault will produce destructive electro-magnetic force, and maybe also produce overheat, cause the winding and iron to be burned. The non asynchronous magnetic field produced by the fault current maybe exceeds the value permitted greatly, and causes severe damage of rotor. Therefore it is very important to research the internal faults of stator winding of synchronous machine[4 7], find the distribution and variation rules of the electric quantities at internal faults, and design protection scheme for the internal faults in order to decrease fault damage. The symmetric component method was used to analyze the internal faults of 431

AC Machine Systems

stator winding of synchronous machine before, but the effect of both air-gap magnetic field harmonics and the winding space position can not be considered by the method, and all branch currents of windings can not be obtained by it, so the results can not reflect the internal fault regulation, and also can not direct the design of internal fault relay protection. An important character of stator winding internal fault is existence of very strong space harmonics in air-gap magnetic field which make phase sequence components of symmetric component method have dependence relation, so as to lose the excellences of symmetric component method. At the same time the ideal electric machine model is no longer suitable, accordingly, adopting d , q, 0 coordinates system based on the model will cause very large errors. Phase coordinates method can consider the effect of air-gap magnetic field harmonics on the parameters[8]. Based on the phase coordinates method the parameter calculation is made by considering the phase winding as a whole, but the phase winding at stator winding internal fault is no longer a whole one, hence the phase coordinates method is not suitable for analyzing the winding internal fault either. In this section the Multi-loop Method is presented to analyze internal fault performance both for steady-state and for transients. When the internal faults occur, the electric machine circuit changes, and very strong harmonics appear in air-gap magnetic field of the electric machine, it is necessary to consider the effect of these magnetic field harmonics on inductances which is main character and difficulty of using Multi-loop Method for researching electric machine[9 13].

6.2.1

Mathematic Model of Stator Winding Internal Fault of Salient Pole Synchronous Machine

In the mathematic model of stator winding internal fault of synchronous machine according to Multi-loop Method, the electric machine is considered as the circuit consisting of many loops having relative motion, the voltage equations and fluxlinkage equations are listed in terms of actual loops of stator and rotor windings. Stator branch equations and rotor loop equations are discussed respectively in the following, then the stator equations will be modified according to stator winding fault sort. In this section the positive direction selection of electro magnetic quantities is consistent with Chapter 1: for stator loops, positive current produces negative flux-linkage, and the relation between voltage and current is defined according to generator convention, that is, positive directions of current and voltage are consistent towards load direction; for rotor loops (including excitation loop and damper loops), positive current produces positive flux-linkage, and the relation between voltage and current is defined according to motor convention, that is, positive directions of current and voltage are consistent towards winding direction. 432


(1) Stator branch equations Each winding branch without internal fault is considered as a branch and the branch voltage equation is listed. When internal fault occurs, the fault branch is divided into two branches from short circuit position. Suppose that the parallel branch number per phase of the stator winding of salient pole synchronous machine is a , and the phase number is m , then the total branch number N of stator windings is

N

For normal winding ma ° ®ma 1 For inter-turn short circuit in the same branch °ma 2 For short circuit between different branches ¯

The voltage equation of any branch in stator winding is uQ

p\ Q rQ iQ

(6.2.1)

where uQ , \ Q , rQ , iQ are voltage, flux-linkage, resistance and current respectively. The flux-linkage equation of branch Q is

\Q

N

Nd

S 1

ld 1

¦ M QS iS ¦ M Qld ild M Qfd i fd

(6.2.2)

in which iS , ild and i fd are the currents of stator branch S , damper loop ld and excitation loop respectively, M QS is mutual inductance between stator branch S and branch Q , M Qld is mutual inductance between damper loop ld and stator branch Q , M Qfd is mutual inductance between excitation loop and stator branch Q , Nd is the total number of damper bars. It should be noted that any loop (branch) current has its own contribution to the flux-linkage. Substitute formula (6.2.2) into formula (6.2.1), and obtain differential equation as follows in which the unknowns are various branch currents uQ

Nd ª N º p « ¦ M QS iS ¦ M Qld ild M Qfd i fd » rQ iQ ld 1 ¬ S1 ¼

(6.2.3)

The voltage equations of stator load side are ua ub uc

LT pia rT ia uac ½ ° LT pib rT ib ubc ¾ LT pic rT ic ucc °¿

(6.2.4)

where rT , LT , uac , ubc , ucc are the resistance, inductance and network phase voltages of the transformer referred to generator side respectively (refer to Fig. 6.2.2). 433

AC Machine Systems

In such a way the voltage and flux-linkage equations of all stator branches are obtained. (2) Rotor loop equations Excitation loop voltage equation is u fd

p\ fd rfd i fd

(6.2.5)

where \ fd , rfd are excitation loop flux-linkage and resistance respectively. Flux-linkage equation of excitation loop is

\ fd

N

Nd

S 1

ld 1

¦ M Sfd iS ¦ M ldfd ild L fd i fd

(6.2.6)

in which M Sfd is mutual inductance between stator loop S and excitation loop, M ldfd is mutual inductance between damper loop ld and excitation loop, L fd is self inductance of excitation loop. Substitute formula (6.2.6) into formula (6.2.5), then obtain the following differential equation in which the unknowns are various currents: u fd

Nd § N · p ¨ ¦ M Sfd iS ¦ M ldfd ild L fd i fd ¸ rfd i fd ld 1 © S1 ¹

(6.2.7)

According to selected loops showed in Fig. 6.2.1, the voltage equation of a damper loop gd is 0

p\ gd rgd igd rc (igd 1 igd 1 )

(6.2.8)

where \ gd , rgd are flux-linkage and resistance of the damper loop gd , rc is damper bar resistance.

Figure 6.2.1 Sketch map of damper loops

Flux-linkage equation of damper loop gd is

434


\ gd

N

Nd

S 1

ld 1

¦ M Sgd iS ¦ M gdld ild M gdfd i fd

(6.2.9)

in which M gdld is mutual inductance of two damper loops gd and ld . Substitute formula (6.2.9) into formula (6.2.8), then obtain the following differential equation of damper loop in which unknowns are various currents: 0

Nd § N · p ¨ ¦ M Sgd iS ¦ M gdld ild M gdfd i fd ¸ rgd igd rc (igd 1 igd 1 ) ld 1 © S1 ¹

(6.2.10)

In such a way all voltage equations of rotor loops are gained. All voltage equations of stator and rotor are written as matrix form: ª u1 º « # » «u » « Q» « # » « uN » «0» « » « # » «0» « # » « » «0» «u fd » «u » « a» « ub » ¬« uc ¼»

ª L1 °« # °« ° « M Q1 °« # °« M N1 °« ° « M 1d 1 °° « # p ®« °« M gd 1 °« # °« M Nd 1 °« °« M fd 1 °« 0 °« 0 °« °¯ ¬ 0

" M 1Q # " LQ # " M NQ " M 1dQ # " M gdQ # " M NdQ " M fdQ " " " " " "

" M 1N # " M QN # " LN " M 1dN # " M gdN # " M NdN " M fdN " " " " " "

M 11d # M Q1d # M N 1d L1d # M gd 1d # M Nd 1d M fd 1d " " "

" M 1gd # " M Qgd # " M Ngd " M 1dgd # " Lgd # " M Ndgd " M fdgd " " " " " "

" M 1Nd # " M QNd # " M NNd " M 1dNd # " M gdNd # " LNd " M fdNd " " " " " "

M 1 fd 0 # # M Qfd # # # M Nfd # M 1dfd # # # M gdfd # # # M Ndfd # L fd 0 LT 0 " 0 " "

0 0º # # » » # # » # # » # # » » # # » # # » # # »» # # » # # » » # # » 0 # » LT 0 » » 0 LT ¼

ª i1 º ½ ª r1 º ª i1 º ª 0 º « # »° « % » «# » «0» « »° « » « » « » « iQ » ° « » « iQ » « 0 » rQ « »° « » « » « » % « # »° « » « # » «0» « i »° « » «i » « 0 » r N « N »° « » « N» « » r1d rc rc « i1d » ° « » « i1d » « 0 » « # »° « » « # » «0» rc % % ° »u« » « » u« »¾ « « igd » ° « » « igd » « 0 » rc rgd rc « »° « » « » « » % % rc « # »° « » « # » «0» «iNd » ° « » «i » « 0 » rc rc rNd « »° « » « Nd » « » i rfd « fd » ° « » « i fd » « 0 » « i »° « » « i » «u » rT « a »° « » « a » « ac » « ib » ° « rT » « ib » «ub c » « »° « » « » « » rT ¼ ¬ ic ¼ ¬ ucc ¼ ¬ ic ¼ ¿ ¬

(6.2.11) 435

AC Machine Systems

Formula (6.2.11) is re-written as simple form: U

p ( LI ) RI B c

(6.2.12)

Owing to the relative movement between stator and rotor, the matrix L is time-variant, hence the voltage equations of stator and rotor are all differential equations with time-variant coefficients. (3) Loop selection of synchronous generator with internal fault and forming of state equations The rotor voltage in formula (6.2.12) is loop voltage, but the stator voltage in the equation is branch voltage, for convenience of treatment the loop voltage should be adopted for both stator voltage and rotor voltage, so it is needed to treat newly stator loop, that is, stator branch voltage should be converted to stator loop voltage equation.

Figure 6.2.2 Sketch map of stator circuit at normal operation

Figure 6.2.2 is sketch map of stator circuit at normal operation of electric machine, based on selected loops in the figure the transformation matrix from branches to loops is as follows:

436


H

ª1 1 º « » 1 1 1 1 « » 1 1 « » « 1 1 1 1»» « « » 1 1 « » 1 « » « » % « » 1 « » « » % « » 1 « » «¬ »¼ 1

(6.2.13)

Formula (6.2.12) pre multiplied by formula (6.2.13) is HU

HLpI HpLI HRI HBc

(6.2.14)

There is also the transformation formula of stator branch current and rotor loop current to stator loop current and rotor loop current I

HTIc

(6.2.15)

where I c is stator and rotor loop currents. Substitute formula (6.2.15) into formula (6.2.14), then the stator and rotor loop voltage equations with unknown loop currents are obtained.

Uc

HLH T pI c HpLH T I c HRH T I c HBc LcpI c RcI c HBc Lc

where Rc

(6.2.16)

HLH T pLc HRH T

The loop currents are considered as state variables, transforming formula (6.2.16), then state equation of synchronous generator is obtained as pI c Lc1RcI c Lc1 (U c HB c) AI c B

(6.2.17)

A Lc1 Rc

where B

Lc1U c Lc1 HBc

Formula (6.2.17) is the Multi-loop mathematic model of synchronous machine. When the interturn short circuit in the same branch occurs, the loop selection is showed as Fig. 6.2.3 (as an example, branch number per phase is 2). 437

AC Machine Systems

Figure 6.2.3 Sketch map of interturn short circuit in the same branch

Here the transformation matrix from branches to loops is

H

ª 1 1 º « » 1 1 1 1 « » « » 1 1 « » 1 1 1 1» « « » 1 1 1 « » 1 « » « » 1 « » % « » « » 1 « » « » % « » 1 « » «¬ »¼ 1 (6.2.18)

When the short circuit between different branches occurs, as an example the loop selection at short circuit between branches of different phases is showed in Fig. 6.2.4. 438


Figure 6.2.4 Sketch map of short circuit between branches of different phases

Here the transformation matrix from branches to loops is

H

ª 1 1 º « » 1 1 1 1 « » « » 1 1 1 « » 1 1 1 1 1 1» « « » 1 1 1 « » 1 1 « » « » 1 « » % « » « » 1 « » « » % « » 1 « » «¬ »¼ 1 (6.2.19)

The key for solving above state equations is calculation of time-variant coefficients. After getting the expressions of time-variant inductances using four-th order Runge-Kutta Method to solve the differential equation set with time-variant coefficients, then the numerical solutions of various loop currents of stator and rotor will be obtained, and the steady-state and transient currents of stator branches are also gained through the transformation of formula (6.2.15). 439

AC Machine Systems

6.2.2

Loop Parameters of Stator Winding Internal Faults of Synchronous Machine

Calculation of loop parameters (here are loop inductances mainly) is the key of analyzing stator winding internal faults of synchronous machine. The inductance parameters are mostly time-variant owing to relative movement between stator and rotor. In the following the inductances of stator circuit and rotor circuit and mutual inductances between stator and rotor circuits are calculated respectively. (1) Inductances of stator loops (branches) The inductances of electric machine circuits are produced mainly by air-gap magnetic field, furthermore by leakage field. Basic thought of calculating the inductances relative to air-gap magnetic field is: according to the basic concept of inductance M

\

, firstly the air-gap mmf produced by the current i flowing i in the loop are resolved into harmonics, and each harmonic magnetic field is obtained by using the air-gap permeance, then each harmonic flux-linkage of the loop and total flux-linkage are calculated, finally the ratio of flux-linkage to current i is the inductance of electric machine. Calculation of the inductances relative to leakage field includes two parts owing to end region leakage flux and slot leakage flux. Basic thought of calculating the inductance produced by end region leakage field is: the coil end part with current is divided into several current elements, and the magnetic field produced by the current elements is calculated in terms of Biot-Savart law. The magnetic field ata certain point of end region is superposition of magnetic fields at that point produced by all current elements. Finally the flux-linkage of coil end part and corresponding inductance produced by end region leakage flux are obtained. As to the inductance owing to slot leakage flux, if the leakage flux is not in saturation state and the reluctance of iron part is ignored, then mutual inductance produced by slot leakage flux only exists between coil bars in the same slot, and mutual inductance produced by slot leakage flux does not exist between coil bars in different slots. After knowing the connection of stator winding, the inductance of each stator loop produced by slot leakage flux can be calculated by adopting the method of searching coil bars in the same slots one by one. The total inductance is the sum of the inductance produced by air-gap magnetic field and the inductance produced by leakage magnetic field. As to the concrete calculation formulas please see Chapter 1. The inductances of stator loops of salient pole synchronous machine are time-variant parameters relative to rotor position. (2) Inductances of rotor loops Calculation formulas of the inductances of excitation winding and damper loops are showed in Chapter 1. When the slot effect is not considered, the inductances of rotor loops are constants independent of rotor position.

440


(3) Inductances between stator loops (or branches) and rotor loops Based on the inductances of stator single coil and rotor loops the mutual inductances between stator loops (or branches) and rotor loops are obtained. Corresponding calculation formulas can be found in Chapter 1. The mutual inductances between stator and rotor loops are time-variant parameters relative to rotor position.

6.2.3

Calculation of Steady State for Internal Fault of Salient Pole Synchronous Machine

After the stator and rotor voltage equations of synchronous machine are established by using Multi-loop Model and the loop parameters are calculated, a group of differential equations with time-variant coefficients is gained as formula (6.2.17), then using fourth order Runge-Kutta method to solve the differential equation set, finally get the steady state and transient values of stator and rotor currents, furthermore other electric quantities (for example, power, etc.). If only the steady-state of internal faults of electric machine stator winding is concerned, the following method can be adopted in order to save computing time. (1) Simplification of internal fault model in steady-state At first based on physical concept the frequencies of various loop currents at internal faults of electric machine stator winding are defined. When the electric machine rotates at synchronous speed, because the configuration of the windings (excitation winding and damper loops) under each pole of rotor are the same, only stator winding is internal asymmetric, so the frequencies of stator and rotor loop currents can be proved to be: Frequency of stator loop current is mZ , m 1, 3, " Frequency of excitation loop current is m1Z , m1 2, 4, " n Frequency of damper loop current is Z , n 1, 2, 3, " , P is pole-pair number P of the electric machine. Accordingly, general expressions of various loop currents can be written out, for example, stator loop Q current can be expressed as iQ

¦ (I

Qm

c sin mZ t ) cos mZ t I Qm

m 1, 3, "

(6.2.20)

m

Excitation current can be expressed as i fd

I f 0 ¦ ( I fm1 cos m1Z t I cfm1 sin m1Z t )

m1

2, 4, "

(6.2.21)

m1

The i - th damper loop current of first pole can be expressed as 441

AC Machine Systems

i1,i

§

¦ ¨© I n

in

n n · cos Z t I inc sin Z t ¸ P P ¹

(6.2.22)

Accordingly, i - th damper loop current of g - th pole can be expressed as ig , i

n n ½ (1) g 1 ¦ ® I in cos >Z t ( g 1)S@ I inc sin >Z t ( g 1)S@¾ P P ¯ ¿ n

(6.2.23) In above two formulas there is n 1, 2, 3, " The expressions of various loop currents are substituted into differential equation set (6.2.17), obtaining a transcendental equation set; then based on the equivalent principle of the components with the same frequency, for each frequency component the corresponding equation is listed. If two special times are selected in order to make the sine term or cosine term be zero (for example, for fundamental order Z t 0 and Z t S / 2 ), then two corresponding linear algebraic equations without the time t are got, in these equations the unknowns are the amplitudes of sine term and cosine term of each harmonic of each loop current, the coefficients of the equations are inductance (including amplitude and phase angle) and resistance of each loop. Solving these linear algebraic equations, the steady state currents of all loops of the machine with internal faults, furthermore other electric quantities, can be obtained. (2) Comparison between steady state calculations and experiments of stator winding internal faults of salient pole synchronous machine While the internal faults of stator winding of salient pole synchronous machine occur, the loop number of stator and rotor is more, besides fundamental, stator currents also include 3, 5, " , etc. odd harmonics, besides direct current component, excitation current also includes 2, 4, " even harmonics, the harmonics of damper loop currents are much more. But the effect of the harmonics of higher order on calculation results is less, hence generally calculated harmonic order is defined in a certain scope. If the harmonic order of stator current is taken as J1 , the harmonic order of excitation current is taken as J1 1 , the harmonic order of damper loop current is taken as J 2 , the total number of stator branches is N , damper bar number per pole of rotor is nc , then the total number of unknowns (i.e. the amplitudes of sine terms and cosine terms of various loop currents of electric machine) in the linear algebraic equation set describing the internal fault in steady state is M

( J1 1) N J1 1 J 2 u 2 Pnc

(6.2.24)

In general this high order linear algebraic equation set can be solved by Gauss elimination method. 442


In the following the calculation results and corresponding experimental data of various loop currents of stator winding internal faults of two salient pole synchronous machines are given. Here the internal short circuits of a 630kW salient pole synchronous machine occur during no-load and separate operation (island operation), the electric machine stator winding is of 6 parallel branches per phase, 5 coils per branch. Calculations and experiments are made for two internal faults: the short circuit of the point 40% of first branch a1 of phase a to the neutral point, and the short circuit of the point 40% of first branch a1 of phase a to the point 40% of first branch b1 of phase b . In Table 6.2.1 the calculation results and experimental data of the short circuit of the point 40% of first branch a1 of phase a to the point 40% of first branch b1 of phase b are listed. In the table the dc excitation current of calculation and experiment is I f 0 10.64A , various currents are rms values, I f 2 is rms value of second harmonic of excitation current. Another electric machine is a 30kW salient pole synchronous machine for simulation and experiments, its stator winding is of two parallel branches per phase. Three representative internal short circuit modes of the machine on line with load are studied: the short circuit of the point 40% to point 2% of first branch a1 of phase a , the short circuit of the point 40% of first branch a1 to the point 10% of second branch a2 of phase a , and the short circuit of the point 20% of first branch a1 of phase a to the point 10% of first branch b1 of phase b . In Table 6.2.2 the calculation and experiment results of the short circuit of the point 40% to point 2% of first branch a1 are listed. In the table the dc excitation current of calculation and experiment is I f 0 4.25A , various currents and voltages are amplitude values, I f 2 is amplitude value of second harmonic of excitation current, I 0 0 is current of connection line between two neutral points, 3U 0 is amplitude value of fundamental component of zero sequence voltage.

6.2.4

Calculation of Transients for Internal Fault of Salient Pole Synchronous Machine

As mentioned above adopting Runge-Kutta method, etc. the differential equation (6.2.17) is solved, and transient and steady state values of stator and rotor currents can be obtained. Simulations and experiments of stator winding internal faults of a 12kW salient pole synchronous machine are made. Stator of the electric machine is of two branches per phase, each branch consists of seven coils. The simulations and experiments of interturn short circuit, the short circuit of two branches of the same 443

AC Machine Systems

phase, and short circuit of two branches of different phases are compared with each others. Figure 6.2.5 is the simulation and experiment waves of the short circuit current and excitation current while the short circuit of fifth coil of second branch of phase a to neutral point occurs. Figure 6.2.6 is the simulation and experiment waves of the short circuit current and excitation current while the short circuit of sixth coil of first branch of phase a to third coil of second branch of the same phase occurs. Figure 6.2.7 is the simulation and experiment waves of the short circuit current and excitation current while the short circuit of third coil of second branch of phase a to third coil of second branch of phase c occurs. Table 6.2.1 Calculation and experiment results of short circuit of the point a1 40% to point b1 40% of a 630 kW salient pole synchronous machine Experimental

Table 6.2.2 Calculation and experiment results of interturn short circuit of the branch a1 of a 30 kW simulation electric machine

Calculated Relative J 1 3 J 2 2 error/%

Relative ExperiCalculated error/% mental

I kL

51.0

50.372

1.2

U ab / V

***

330.2

I a1 f

27.0

27.450

1.7

U bc / V

***

377.5

I b1 f

22.5

20.782

7.6

Ia / A

49.6

44.2

10.9

I a1 Ia2

***

22.947

Ib / A

50.5

47.5

5.9

1.75

1.973

12.7

I a3

10.7

13.0

21.5

5.304

4.8

Ic / A

5.57

Ia4

5.02

4.858

3.2

P / kW

14.5

14.7

1.4

I a5

4.83

4.686

3.0

I a1 / A

75.8

64.3

15.2

I a6 I b1

6.70

6.241

6.9

Ia2 / A

30.6

30.0

2.0

***

29.608

I b1 / A

52.6

44.8

14.8

Ib 2

4.75

4.678

1.5

Ib2 / A

9.0

6.6

26.7

Ib3

4.82

5.095

5.7

I c1 / A

34.0

26.1

23.2

Ib 4

***

5.360

Ic2 / A

36.7

33.8

7.9

Ib5

4.31

4.922

14.2

I kL / A

343.0

323.8

Ib6

5.6

9.56

9.657

1.0

I c1

3.478

21.3

If2 /A

4.42

0.36

0.404

Ic 2

2.30

1.553

32.5

I00 / A

8.0

6.6

Ic3

***

0.332

Ic 4

***

Ic5

12.2 17.5

3U 0 / V

***

37.0

0.321

U2 / A

***

30.2

***

0.439

I2 / A

***

20.0

Ic6

3.33

3.289

1.2

P2 / W

***

239.9

If2

2.30

2.260

1.7

Note:***express no experiment results.

Note:***express no experiment results.

444


6.2.5

Features of Internal Faults of Salient Pole Synchronous Machine

When internal faults of stator winding of salient pole synchronous machine occur, besides the fundamental magnetic field with positive rotation synchronous speed and fundamental magnetic field with reverse rotation synchronous speed (negative sequence magnetic field), other harmonic magnetic fields with different rotation speeds and different rotation directions exist in the air-gap. Except the fundamental positive rotation magnetic field, other negative sequence magnetic field and harmonic magnetic fields with relative movement to the rotor induce the currents in rotor damper loops, but only negative sequence magnetic field and odd harmonic magnetic fields cause the induced currents in the excitation winding. Therefore the variation of electric quantities of stator winding internal faults has its own features. At first, the short circuit current of internal fault is very large, as showed in Figs. 6.2.5, 6.2.6 and 6.2.7, if converted to no-load rated voltage condition the excitation current I f 0 8.24A, the steady state internal short circuit currents of the three states are 78.4A, 59.1A and 46.0A respectively, their values are 7.23, 5.45 and 4.24 times of rated branch current 21.7 / 2 10.85A respectively and exceed enormously each branch current 7.06A of 3-phase steady state short circuit. Considering so large internal fault short circuit current it is necessary to provide suitable protection devices to deduce the damages. The magnitude of internal short circuit current is relative to short circuit turn number, winding form, connection fashion and space place of winding short circuit part etc. Generally speaking, the less the short circuit turn number is, the larger the short circuit current is, but its effect on asymmetry of other branch currents without short circuits is smaller. In order to design suitable protection schedule for internal faults it is necessary to analyze and calculate each internal fault roundly and consider each factor synthetically. The harmonics of each current of stator winding internal short circuit are stronger. Besides dc component, second harmonic, fourth harmonic, etc. exist in the excitation current; for stator currents, besides fundamental component, third harmonic, fifth harmonic, etc. exist, as showed in Figs. 6.2.5, 6.2.6 and 6.2.7. In Fig. 6.2.6 the second and fourth harmonics of excitation current are 50.5% and 14% of its dc component respectively, and third and fifth harmonics of stator short circuit current are 13.8% and 4% of its fundamental component respectively. These current harmonic components are caused by air-gap negative sequence magnetic field and harmonic magnetic fields. Similar to 3-phase sudden short circuit, the transient currents of internal short circuits have both ac and dc components, the ac component will decay to steady 445

AC Machine Systems

Figure 6.2.5 Waves of short circuit currents and excitation currents at short circuit from U a 25 to neutral point (dc excitation current I f 0 1.48A ) (a) Experiments; (b) Simulations

Figure 6.2.6 Waves of short circuit currents and excitation currents at short circuit from U a16 to U a 23 (dc excitation current I f 0 3.10A ) (a) Experiments; (b) Simulations

446


Figure 6.2.7 Waves of short circuit currents and excitation currents at short circuit from U a 23 to U c 23 (dc excitation current I f 0 2.05A ) (a) Experiments; (b) Simulations

value, the dc component will decay to zero, and initial value of dc component is relative to the short circuit time. But large difference exists between the two short circuit modes: compared with the steady state current the transient current of internal short circuit is larger but not too much, which is not like 3-phase sudden short circuit as showed in Figs. 6.2.5, 6.2.6, and 6.2.7. The reason is that even if during the internal short circuit in steady-state, the induced currents are also caused in rotor excitation winding and damper loops (owing to the air-gap negative sequence magnetic field and harmonic magnetic fields with relative movement to the rotor), consequently the currents of stator side become larger.

6.3 Internal Fault Analysis of Stator Winding of Non-Salient Pole Synchronous Machines Similar to salient pole synchronous machine, the internal faults of non-salient pole synchronous machine stator winding will cause severe damages to the machine. It is necessary to analyze deeply the internal faults of non-salient pole synchronous machine stator winding and design suitable protection devices. Analysis method of internal faults of non-salient pole synchronous machine is different compared with the salient pole synchronous machine owing to its 447

AC Machine Systems

configuration features[14,15]. The internal fault analysis begins from mathematic model of non-salient pole synchronous machine as follows.

6.3.1

Mathematic Model of Stator Winding Internal Fault of Non-salient Pole Synchronous Machine

Compared with salient pole synchronous machine there are the following features for the mathematic model of non-salient pole synchronous machine. (i) The air-gap of non-salient pole electric machine is uniform, without considering the effect of slot harmonic permeance, its self inductances and mutual inductances of stator loops are all constants, the expressions of other inductances are also simpler than salient pole machine (refer to Chapter 1). (ii) Excitation winding of non-salient pole synchronous machine is distributed symmetrically relative to the direct axis, so its calculation formula of inductance is also different from salient pole electric machine. (iii) The rotor of non-salient pole synchronous machine is whole solid configuration forged mostly, so it is necessary to consider the eddy current damping effect of solid magnetic pole. Considering the features of non-salient pole synchronous machine, the calculation of corresponding parameters is discussed in the following. (1) Self inductance of excitation loop Self inductance of excitation loop consists of two components, namely L fd

L fdl L fdG

(6.3.1)

in which L fdl is inductance corresponding to end region leakage flux, inter pole leakage flux and pole face leakage flux. L fdG is the inductance corresponding to air-gap flux, its calculation method is as follows.

Figure 6.3.1 Excitation loop of non-salient pole electric machine 448


The excitation winding of non-salient pole electric machine consists of a series of concentric coils in series, as showed in Fig. 6.3.1. Therein the pitch of i - th concentric coil is E fdiW , the current in it is i fd / a fd , mmf produced by the current is Ffd ( x)

¦F

fdk

cos(kx)

(6.3.2)

k

where

i fd k E fdi S 14 sin w fd 2 kS a fd

Ffdk

w fd is turn number of one excitation coil per pole, a fd is parallel branch number of excitation winding. So the self inductance of excitation winding is N

L fdG

16 w2fdW lP P 0

N

¦¦ M

fdij

S2 a 2fd G

i 1 j 1 N

N

u ¦¦

i 1 j 1k

§ k E fdi S · § k E dfj S · 1 sin ¨ ¸ sin ¨ ¸ 2 1, 3, 5, k © 3 ¹ © 2 ¹

¦

(6.3.3)

where N is series coil number per pole. (2) Mutual inductance between damper loop and excitation loop The mutual inductance between excitation winding and damper loop 11c is 8P 0 w fd wrW l

N

M 1, fd

¦¦ M i 1

1, fdik

k

N

u¦

i 1 k

S2 a fd G

§ k E fdi S · § k E1S · 1 sin ¨ ¸ sin ¨ ¸ cos kD1 2 1, 3, 5, " k © 2 ¹ © 2 ¹

¦

(6.3.4)

where wr is damper loop turn number (generally it is 1), E1 is short pitch ratio of damper loop 11c , D1 is the angle between the central line of damper loop 11c and rotor direct axis. (3) Inductances between stator coils and excitation winding The mutual inductance between stator coil AAc and excitation winding is M fda

8P 0 w fd wKW l S2 a fd G N

u¦

i 1 k

§ k E fdi S · § k E S · 1 sin ¨ ¸ sin ¨ ¸ cos kJ 2 1, 3, 5, " k © 2 ¹ © 2 ¹

¦

(6.3.5)

449

AC Machine Systems

where wK is stator coil turn number, E is short pitch ration of stator coil and J is electric angle by which the rotor d-axis leads the stator coordinates axis. (4) Effect of eddy current in rotor iron on parameters of non-salient pole synchronous machine Solid rotor is adopted mostly for non-salient pole synchronous machine (especially for large turbo-generator), the effect of eddy current on electric machine parameters, especially damper loop parameters, is larger. Once electric machine configuration and material, etc. are defined, the eddy current effect of solid rotor will be definite, and some corresponding parameters can be used to represent it, the relation between parameters and frequencies is nonlinear. Here equivalent concentrated treatment method is introduced, the eddy current parameters under Multi-loop system are obtained from frequency response characteristic. For the internal faults in stator winding of non-salient pole synchronous machine, it can be considered that the induced eddy current in the rotor is mainly the component of 100Hz . Without considering rotor iron saturation the calculation formula of actual penetration depth corresponding to rotor eddy current effect is 2 2Sf sVP

G

Equivalent resistance and inductance are respectively Red

l

VG b

,

Led

Red 2S fs

(6.3.6)

in which l is rotor length, P is magnetic permeability, V is conductivity, f s is 100Hz, b is equivalent width of calculated eddy region, while adopting full pitch damper loop equivalent to the eddy current effect of solid rotor there is SDr b , Dr is rotor diameter. 4P The penetration depth is in reverse proportion to square root of frequency, when frequency increases, the penetration depth decreases, resistance increases and inductance decreases. It can be considered that the flux corresponding to the eddy current only links the rotor itself, hence gained equivalent inductance can be seen as leakage inductance.

6.3.2

Calculation and Experiments of Internal Faults of Non-salient Pole Synchronous Machine

The flux-linkage equations and voltage equations in internal faults of stator winding of non-salient pole synchronous machine are similar to salient pole 450


synchronous machine. While calculating internal faults of stator winding of nonsalient pole synchronous machine, its parameter expressions are substituted into corresponding flux-linkage equations and voltage equations, gaining a differential equation set with time-variant coefficients, the fourth order Runge-Kutta method or other numerical methods are adopted to solve the differential equation set, then steady state and transient values of stator and rotor currents, furthermore other electric quantities (for example, power, etc.), will be obtained. If only analysis and calculation of steady-state of internal faults are concerned, according to physical concept the frequencies of loop currents of internal faults of stator winding of electric machine are defined at first, and the expressions of various loop currents are substituted into above mentioned differential equation set, obtaining a transcendental equation set; then based on the equivalent principle of the components with the same frequency, for each frequency component the corresponding equation is listed. If two special times are selected in order to make the sine term or cosine term be zero (for example, for fundamental order Z t 0 or Z t ʌ / 2 ), then two corresponding linear algebraic equations without the time t are got, in these equations the unknowns are the amplitudes of sine term and cosine term of each current, the coefficients of the equations are inductance (including amplitude and phase angle) and resistance of each loop. Solving these linear algebraic equations, the steady state currents of all loops of the machine with internal faults, furthermore other electric quantities, can be obtained. The solving process is similar to the internal fault calculation of salient pole synchronous machine, here more describing is not needed. In order to research internal faults of stator winding of non-salient pole synchronous machine, a series of simulation calculation and corresponding experiments for internal faults of a non-salient pole generator are carried out. Main specifications of the electric machine are: S N 30kV A, PN 24kW, U N 400V, I N 43.3A,cos M N 0.8, f N 50Hz, 54, q 3, a1 2 (stator branch number per phase), wk 8 Z 30 (turn number per coil), 2 (rotor slot number/division number of rotor slots), Z 2c 42 nN

G

1000r / min, Z1

1.1mm, I f 0

0.9576A, N c

7 (damping bar number per pole).

The following three sort internal faults are researched: interturn short circuit of the same branch; short circuit between two branches in the same phase; short circuit between two branches in different phases (refer to Fig. 6.3.2). The conditions for calculation and experiments are: internal short circuits occur under single machine separate operation and no load, the dc component of excitation current I f 0 is kept constant. Figures 6.3.3, 6.3.4 and 6.3.5 are the simulation wave and experimental wave at stator short circuit between 40% of branch a1 and 2% of 451

AC Machine Systems

branch a1. Table 6.3.1 is corresponding simulation and experiment result comparison between steady state currents and line voltages. In the table ikL is current of short circuit loop, ia1 , ib1 , ic1 are all the branch currents, uab , ubc are terminal line voltages, ifd2 is second harmonic of excitation current, i0 0 is current of connection line between two neutral points.

Figure 6.3.2 Three sort internal faults

Figure 6.3.3 Transient current of branch a1 during interturn short circuit between

40% of branch a1 and 2% of the same branch (a) Experiment wave; (b) Simulation wave

Figure 6.3.4 Transient current of branch b1 during interturn short circuit between

40% of branch a1 and 2% of the same branch (a) Experimental wave; (b) Simulation wave

452


Figure 6.3.5 Transient current ikL during interturn short circuit between 40% of

branch a1 and 2% of the same branch (a) Experimental wave; (b) Simulation wave

Table 6.3.1 Comparison between simulation and experimental results of steady state quantities during interturn short circuit between 40% of branch a1 and 2% of the same branch ikL /A

ia 1 /A

ib1 /A

ic1 /A

uab /V

ubc /V

i fd 2 /A

i0 0 /A

Simulation results (effective value)

126.79

21.92

6.36

9.2

165.3

215.7

0.106

4.1

Experimental data (rms value)

126.53

19.96

6.61

10.84

152

204

***

3.33

0.2

9.8

3.8

15

8.7

5.7

Error/ˁ

23

Note: ***indicates no experiment result.

Similar to salient pole synchronous machine, for internal faults of stator winding of non-salient pole synchronous machine the following features exist: (i) The short circuit current of internal fault is very large, so it is necessary to pay special attention to the protection for internal faults. As for as the internal fault current is concerned, the difference between branch currents for the same phase is some times quite small, hence the difficulty in designing the internal fault protection increases. (ii) As the short circuit of external terminals, decaying dc component also exists in the internal fault current. But there is larger difference between ac components of both, the transient ac component of internal short circuit is a little more than the steady state current not just as the short circuit of external terminals, because at steady state internal short circuit the larger induced current is also caused in rotor winding, which affects the stator current. (iii) When internal fault occurs, the current and other electric quantities are relative to not only the short circuit sorts but also the short circuit turn ratio and the space position of short circuit turn, etc. 453

AC Machine Systems

References [1] Qiu A R, Zhang L Zo (1987) Steady Performance Analysis of Rotor Failures of Bar Broking and End Ring Broking of Squirrel Asynchronous Machine (in Chinese). Transactions of China Electrotechnical Society, 3: 7 12 [2] Zhang L Z, Qiu A R (1987) Detecting Rotor Failure of Asynchronous Machine by Frequency Spectrum Analysis (in Chinese). Transactions of China Electrotechnical Society, 4: 46 50 [3] Williamson S, Smith A C (1982) Steady-state analysis of 3-phase cage motors with rotor — bar and end-ring faults. IEE Proc. B, 3: 93 100 [4] Kinitsky V A (1965) Calculation of Internal Fault Currents in Synchronous Machines. IEEE PAS, 5: 381 389 [5] Kinitsky V A (1968) Digital Computer Calculation of Internal Fault Currents in a Synchronous Machines, .IEEE PAS, 8: 1675 1679 [6] Kazofusk E Y i, etc. ( 1969) Abnormal Operation of Large Scale Synchronous Machine (in Russian). Science Press [7] Wang Xiangheng (1987) The Internal Fault Analysis of Stator Winding of Synchronous Machine with Multi Branches on No-load (in Chinese). Proceeding of the CSEE, 5: 1 10 [8] Brameller A, Pandy R S (1974) General Fault Analysis using Phase Frame of Reference. PIEE, 5: 366 368 [9] Zhang L Z, Wang X H, Gao J D (1991) A New Approach to Analysis of Synchronous Machines with Asymmetrical Armature Windings (ĉ), Fundamental and Approach. Science in China, Series A, 7: 866 874 [10] Zhang L Z, Wang X H, Gao J D (1991) A New Approach to Analysis of Synchronous Machines with Asymmetrical Armature Windings (Ċ), Parameter Calculation. Science in China, Series A, 8: 998 1004 [11] Wang X H, Sun Y G., Ouyang B, Wang W J, Zhu Z Q ,Howe D (2002) Transient behavior of salient-pole synchronous machines with internal stator winding faults. IEE proc. -Electr. Power Appl, 2: 143 151 [12] Gao J D, Wang X H, Li F H (1993) Analysis of AC Machines and Their Systems, 1st Edition (in Chinese). Tsinghua University Press, Beijing [13] Gao J D, Wang X H, Li F H (2005) Analysis of AC Machines and Their Systems, 2nd Edition (in Chinese). Tsinghua University Press, Beijing [14] Wang X H, Chen S L, Wang W J, Sun Y G, Xu L Y (2002) A Study of Armature Winding Internal Faults for Turbogenerators. IEEE Trans. on IA, 3: 625 631 [15] Kulig T S, Buckly G W, Lambreicht D and Liese M (1990) A new approach to determine transient generator winding and damper currents in case of internal and external faults and abnormal operation. IEEE Trans. Energy Conversion, 3: 70 78

454

Appendix A Air-Gap Permeance Coefficients in Consideration of Slot Effect

If the stator inner-surface is smooth, then during consideration of rotor salientpole effect the air-gap permeance coefficient is

Or ( x)

O0 2

¦ O2l cos 2lx

(A1)

l

where l 1, 2, " and the origin of abscissa x is taken on the rotor d-axis. Supposing the rotor surface is smooth, considering the stator slot effect the air-gap permeance coefficient is

Os (D1 ) a0 ¦ aQ cosQ Q

Z D1 P

(A2)

in which v 1, 2, " , Z is number of stator slots, D1 is expressed in electric degree and the coordinate origin is taken on the central line of stator slot as shown in Fig. A1. The coordinate origin is also taken on the central line of stator tooth as shown in D 2 , and the distance between two coordinate origins is half a toothpitch D t / 2, where

Dt

Figure A1

Substituting D1 D 2

Dt 2

P u 2S Z

Stator teeth and their coordinates

into expression (A2) exists

(A3)

AC Machine Systems

§ Z · D 2 Q S¸ © P ¹ Q Z a0 ¦ (1)Q aQ cosQ D 2 P Q

Os (D 2 ) a0 ¦ aQ cos ¨Q

(A4)

where Q 1, 2, " From the above formula we can see, no matter how taking the origin on the central line of stator slot or tooth, the form of Os is unchanged but the harmonic terms change their sign. In the following we take D to express the electric angle of stator coordinates to unify them. The whole air-gap permeance coefficient can be expressed as

O ( x, D ) Or ( x)Os (D ) § Z · a0 Or ( x) ¦ aQ cos ¨Q D ¸Or ( x) © P ¹ Q

Q 1, 2, "

(A5)

in which the first term a0 Or ( x) indicates the decrease of air-gap permeance due to stator slot effect, and the second term represents the slot harmonics of air-gap permeance due to stator slot effect. If Ot indicates the slot harmonics, then there is

Ot ( x, D )

§ Z

·

aQ cos ¨Q D ¸O ( x) ¦ © P ¹ Q r

Q 1, 2, "

(A6)

Formula (A6) is the product of two infinite series, in which the first several terms are primary. If taking the first two terms for the two series, then exists § O0 · O2 cos 2 x ¸ © 2 ¹

Ot ( x, D ) (a1 cos tD a2 cos 2tD ) ¨

(A7)

Z and the relation of stator coordinates x to rotor coordinates D P (see Fig. A2) is

in which t

D

x T

Figure A2 Relation between stator and rotor coordinates

456

(A8)

Appendix A Air-Gap Permeance Coefficients in Consideration of Slot Effect

Substituting formula (A8) into (A7) exists

Ot ( x)

1 1 a1O0 cos t ( x T ) a2 O0 cos 2t ( x T ) 2 2 1 1 a1O2 cos[(t 2) x tT ] a1O2 cos[(t 2) x tT ] 2 2 1 1 a2 O2 cos[(2t 2) x 2tT ] a2 O2 cos[(2t 2) x 2tT ] 2 2

(A9)

If expressing the slot harmonics of air-gap permeance coefficient according to stator coordinates D, then there is

Ot (D )

1 1 1 a1O0 cos tD a2 O0 cos 2tD a1O2 cos[(t 2)D 2T ] 2 2 2 1 1 a1O2 cos[(t 2)D 2T ] a2 O2 cos[(2t 2)D 2T ] 2 2 1 a2 O2 cos[(2t 2)D 2T ] 2

(A10)

Based on the slot harmonics of air-gap permeance coefficient, it is not difficult to get the inductances of stator and rotor various loops corresponding to the slot harmonics of air-gap permeance.

457

Appendix B Expressions of Effective Air-Gap

Air-gap and relevant dimensions are shown in Fig. B1 for salient-pole synchronous machines. Because the effective air-gap under pole-shoe is quite different from that between adjacent poles, the air-gap can be divided into two parts, namely DS · § the part under pole-shoe ¨ 0 x ¸ and the part between adjacent poles ¹ © S· § DS x ¸ for convenience, and then get the corresponding calculating formulas. ¨ ¹ ©

Figure B1

Air-gap and relevant dimensions for salient-pole synchronous machines

(1) The part under pole-shoe The air-gap length under pole-shoe is much less than a pole-pitch in practical machines, so it can be considered that the flux passes through air-gap in radial direction and the air-gap permeance coefficient O2lP is inversely proportional to G P ( x) only. Thus, according to formula (1.2.5) write down the calculating formula of O2lP the air-gap permeance coefficient under pole-shoe as follows:

OG ( x) O2lP

P0 G P ( x) 4P 0 DS cos 2lx dx S ³ G P ( x)

(B1)

Appendix B

Expressions of Effective Air-Gap

Based upon Fig. B1, G P ( x) can be expressed as

G P ( x) R H cos

x § x· rP2 H 2 sin 2 ¨ ¸ P © P¹

(B2)

For convenience, the above formula of G P ( x) can be approximate to

G P ( x) G min (G max

§ x· sin 2 ¨ ¸ © P¹ G min ) DS · § sin 2 ¨ ¸ © 2P ¹

If using relative value G P ( x), then there is

G P ( x) 1 ( U 1) G P ( x) G min

O2lP

P0 G min

§ x· sin 2 ¨ ¸ ©P¹ § DS · sin 2 ¨ ¸ © 2P ¹

ª 4 D2S cos 2lx º dx » « ³0 G P ( x) ¼ ¬S

(B3)

P0 O 2lP G min

where

O 2lP U

4 D2S cos 2lx dx S ³ 0 G P ( x)

G max G min

(2) The part between adjacent poles The effective length of air-gap between adjacent poles is large, so the flux distribution depends upon not only the air-gap length but also the order of mmf harmonics and the position the mmf acts on. In the light of graphic and analytical results, we can get the approximate formula of O2li the air-gap permeance coefficient between adjacent poles. (i) For the mmf, whose zero-value is at the central line between two adjacent poles, such as the mmf Fdk ( x) during k odd number and mmf Fqk ( x) during k even number, the relative value of the coefficient is

O d 2li

4 S2 cos 2lx dx DS S ³ 2 G di ( x)

(B4)

in which 459

AC Machine Systems

U sinh K

G di ( x)

K

ª S x º sinh « K » ¬ D S ¼

ª1 D W º 0.768 « » ¬ 2 G max ¼

0.617

(B5)

ª1 D W º arctan « » ¬ 2 G max ¼

(ii) For the mmf, whose maximum is at the central line between two adjacent poles, such as the mmf Fqk ( x) during k odd number and mmf Fdk ( x) during k even number, the relative value of the coefficient is 4 S2 cos 2lx dx DS S ³ G qi ( x)

O q 2li

(B6)

where

G qi ( x)

ª

U «1 ¬

DS · S º § » ¨x ¸ cos G max S © 2 ¹ 2P ¼ W

(B7)

Having the permeance coefficient under pole-shoe and that between two adjacent poles on two conditions, the relative values of air-gap permeance coefficient O 2l on two conditions can be got and indicated respectively by O d 2l and O q 2l (i) When the zero-value of mmf is at the central line between two adjacent poles, there is

O d 2l

O 2lP O d 2li

(B8)

(ii) When the maximum value of mmf is at the central line between two adjacent poles exists

O q 2l

O 2lP O q 2li

(B9)

In practical terms, the difference between Od 2li and Oq 2li for the part between adjacent poles is small. In addition, neither the zero-value nor the maximum value of mmf is at the central line between two adjacent poles during harmonic order k fraction, so during calculation of air-gap permeance coefficient take the average value of both O d 2li and O q 2li , namely

460

O 2li

1 (O d 2li O q 2li ) 2

O 2l

1 (O d 2 l O q 2 l ) 2

(B10)

O 2lP O 2li

(B11)

Appendix C

Inductances Due to Leakage Flux of Stator Winding End

Calculating the leakage flux of stator winding end for 3-phase symmetrical winding, the symmetrical 3-phase currents can be processed usually by use of equivalent current sheets, but this method is unsuitable for studying inner unsymmetrical problems of synchronous machine with salient-poles based on the single coil, so the discrete method can be used to analyse those problems. When a coil end carrying current, it can be divided into several current elements, the corresponding magnetic field for one current element is calculated according to Biot-Savart law, and the magnetic field at a certain point of coil end can be got by superposition of the corresponding magnetic fields for various current elements to obtain flux-linkages of various coil ends and the corresponding inductances. The conception of ferromagnetic image and air-gap current is adopted for evaluation. The inductances due to leakage magnetic field of stator winding end are discussed respectively on Cartesian coordinates and cylindrical coordinates as follows. (1) Calculation of inductances due to leakage magnetic field of stator winding end on cartesian coordinates For convenience, the evolute part and straight-line part of stator winding end are considered to be on the same cylindrical surface, which can be developed into a plane and then the inductances due to leakage magnetic field of stator winding end are discussed on this plane. Cartesian coordinates are built as shown in Fig. C1, in which CDEFG indicates the reduced coil end on the plane xOy. The leakage magnetic field of stator winding end is of three dimensions and in the magnetic field exist various media such as air, iron, etc. The media in the magnetic field can be reduced in the following. Firstly the air-gap action is superseded by air-gap current. In air-gap there is imaginary air-gap current and the air-gap doesn’t exist any longer, so the analysed problem is changed to the problem that one side is iron and another is air. Then, according to the conception of ferromagnetic image method the ferromagnetic medium action can be replaced by image current, so the leakage magnetic field of stator coil end is simplified as the field caused by coil-end current, air-gap current and their image in the uniform air medium. Supposing current i passes 2P E through the coil end CDEFG, then the air-gap current of section AB is i 2P Ei for the outer part of coil end. BC and AG for the inner part of coil end and 2P are imaginary radial current to make current continuous, and CDcE cF cG is the

AC Machine Systems

image of coil end current CDEFG. Similarly, air-gap current and radial current have their own images, too.

Figure C1 The reduced stator coil end on Cartesian coordinates

After so simplification, the leakage magnetic field of stator winding end can be solved by use of Biot-Savart law. The passages are all short for radial currents BC and AG and then have opposite direction, so their effect can be neglected. In addition, the air-gap current and coil-end current are not on the same plane, so they can be processed separately. In the following analyse the magnetic field produced by the coil end current and its image on the coil end plane. The plane xOy is plotted separately as shown in Fig. C2, where the coordinate positions are indicated for various points of coil end.

Figure C2 462

Magnetic field produced by coil end current and its image on the plane xOy

Appendix C


According to Biot-Savart law, for the flux density at any point N ( x0 , y0 ) exists

dB

P 0 idl u r0 4S

(C1)

r2

Considering the action of each section current separately, the current element of section FG is idx, and there is

P0 idx sin T

dBFG

4S

r2

P0

§a · i ¨ y0 ¸ ©2 ¹

2 3/ 2 4S ª §a · º 2 «( x0 x) ¨ y0 ¸ » ©2 ¹ ¼» ¬«

dx

where dBFG is the flux density in z-direction produced by the current element idx for the section FG at point N assuming the positive direction of flux density B to be out of the paper. The flux density in z-direction caused by the section FG current at point N is

BFG

³

0

§a · i ¨ y0 ¸ 2 © ¹

P0

b

2 3/ 2 4S ª §a · º 2 «( x0 x) ¨ y0 ¸ » ©2 ¹ »¼ «¬

P0

dx

x0 x0 b ª « 2 2 4S a y §a · §a · 2 0 « x2 ( x0 b) ¨ y0 ¸ y 2 « 0 ¨© 2 0 ¸¹ ©2 ¹ ¬ i

º » » » ¼

(C2)

The flux density in z-direction produced by the section F cG current at point N is

BF cG

³

0 b

P0

P0

§a · i ¨ y0 ¸ 2 © ¹

2 3/ 2 4S ª §a · º 2 «( x0 x) ¨ y0 ¸ » ©2 ¹ »¼ «¬

dx

x0 x0 b ª « 2 2 4S a y §a · §a · 2 0 « x2 ( x0 b) ¨ y0 ¸ y 2 « 0 ¨© 2 0 ¸¹ ©2 ¹ ¬ i

º » » » ¼

(C3)

The flux density in z-direction caused by the section CD current at point N is 463

AC Machine Systems

³

BCD

b 0

P0

§a · i ¨ y0 ¸ ©2 ¹

2 3/ 2 4S ª §a · º 2 «( x0 x) ¨ y0 ¸ » ©2 ¹ »¼ ¬«

dx

P0

x0 x0 b i ª º « » 2 2 a 4S y §a · §a · » 2 0 « x2 ( ) y x b y ¨ 0¸ 0 0¸ » 2 « 0 ¨© 2 ¹ ©2 ¹ ¼ ¬

(C4)

The flux density in z-direction produced by the section CDc current at point N is

BCDc

³

0

b

P0

§a · i ¨ y0 ¸ ©2 ¹

2 3/ 2 4S ª §a · º 2 «( x0 x) ¨ y0 ¸ » ©2 ¹ »¼ ¬«

dx

P0

x0 x0 b i ª º « » 2 2 a 4S y §a · §a · » 2 0 « x2 ( ) y x b y ¨ 0¸ 0 0¸ » 2 « 0 ©¨ 2 ¹ ©2 ¹ ¼ ¬

(C5)

The flux density in z-direction caused by the section EF current at point N is

BEF

a ac x0 (b c) y0 2 2 i 3/ 2 4S ª a 2 4(b c) 2 º (c b) « » 2 ¬ 4(b c) ¼

P0

° ½° c P1 b P1 ® ¾ 2 2 ¯° Q1 (c P1 ) Q1 Q1 (b P1 ) Q1 ¿°

(C6) The flux density in z-direction produced by the section E cF c current at point N is

BE cF c

a ac x0 (c b) y0 2 2 i 3/ 2 4S ª a 2 4(c b) 2 º (b c) « » 2 ¬ 4(c b) ¼

P0

° c P2 b P2 ® 2 2 °¯ Q2 (c P2 ) Q2 Q2 (b P2 ) Q2

½° ¾ ¿°

(C7) The flux density in z-direction caused by the section DE current at point N is

464

Appendix C

BDE


a ac x0 (b c) y0 2 i 2 3/ 2 4S ª a 2 4(b c) 2 º (c b ) « » 2 ¬ 4(b c) ¼

P0

° ½° b P3 c P3 ® ¾ 2 2 ¯° Q3 (b P3 ) Q3 Q3 (c P3 ) Q3 ¿°

(C8) The flux density in z-direction produced by the section DcE c current at point N is

BDcE c

a ac x0 (c b) y0 2 2 i 3/ 2 2 2 4S ª a 4(c b) º (b c) « » 2 ¬ 4(c b) ¼

P0

° c P4 b P4 ® 2 2 °¯ Q4 (c P4 ) Q4 Q4 (b P4 ) Q4

½° ¾ °¿

(C9) P1 and Q1 in formula (C6) are expressed respectively as x0 P1

· a § ac y0 ¸ ¨ 2(b c) © 2(b c) ¹ 2 2 a 4(b c) 4(b c) 2 2

Q1

(C10)

§ ac · ª ·º a § ac y0 ¸ « x0 y0 ¸ » x02 ¨ ¨ 2(b c) © 2(b c) © 2(b c) ¹ « ¹» 2 2 « » a 4(b c) a 2 4(b c) 2 « » 2 2 4(b c) 4(b c) ¬ ¼

2

(C11)

In formulas (C10) and (C11), b and c are replaced respectively by ( b) and ( c) to get P2 and Q2 in formula (C7); a is superseded by ( a) to obtain P3 and Q3 in formula (C8); a, b and c are replaced respectively by ( a), ( b) and ( c) to get P4 and Q4 in formula (C9). The flux-density normal component perpendicular to the coil-end plane produced by air-gap current of section AB at point N is Bz1

a 2 a 2

³

P0 2P E 4S

2P

ix0 dy [ f ( y0 y )2 x02 ]3 / 2 2

ª a « y0 P0 2P E 1 « 2 ix0 2 2 4S 2 P f x02 « « f 2 x2 § a y · ¨ 0 0¸ « ©2 ¹ ¬

º » » 2 » §a · f 2 x02 ¨ y0 ¸ » ©2 ¹ »¼ (C12) a y0 2

465

AC Machine Systems

The air-gap current for the outer part of coil end is divided into two sections, and the flux-density normal components perpendicular to the coil-end plane produced by them at point N respectively are Bz 2

³

a 2 Pa

E

P0 E

ix0 dy 4S 2 P [ f x ( y0 y )2 ]3 / 2 2

2 0

ª a y0 1 « 2 « ix0 2 2 4ʌ 2 P f x02 « §a · 2 2 « f x0 ¨ y0 ¸ ©2 ¹ «¬

P0 E

Pa

Bz 3

³

E a 2

Pa

º » » 2 » ª º § · Pa f 2 x02 «¨ y ¸ 0» » ¬© E ¹ ¼ »¼ (C13)

E

y0

P0 E ix0 dy 4S 2 P [ f 2 x02 ( y0 y )2 ]3 / 2

Pa ª y0 1 « E « ix0 2 2 f x02 « 4ʌ 2 P ª º § · Pa « f 2 x02 «¨ ¸ y0 » «¬ ¬© E ¹ ¼

P0 E

º » » 2 » ª º a § · f 2 x02 «¨ ¸ y0 » » ¬© 2 ¹ ¼ »¼ (C14) a y0 2

In formulas (C12), (C13) and (C14), f is the perpendicular distance between the air-gap current and coil-end plane. Counting in the image of air-gap current, the values of Bz1 , Bz 2 and Bz 3 obtained above should be doubled. Summarizing the flux-density normal components produced by various section currents of coil end, air-gap current and their images at point N on the coil-end plane, we can get the total flux density at this point as follows: B

BFG BFG c BCD BCDc BEF BE cF c BDE BDcE c 2( Bz1 Bz 2 Bz 3 )

(C15)

Dividing the coil-end plane into several domains, finding out the flux-density and flux for each domain, adding the flux values of all domains and then multiplying the flux by corresponding number of turns to get flux-linkage, the flux-linkage is divided by current i to obtain self inductance corresponding to leakage flux of coil end. In a similar way, the mutual inductance due to leakage flux of coil end can also be got. During evaluation the conductor current is assumed to be concentrated on the central line, so the flux density of conductor position can’t be calculated or the result is infinite. Calculating the leakage self inductance of coil end, the conductor is considered to have circular cross-section and only the flux passing through the 466

Appendix C


area C1 D1 E1 F1G1 is counted in, the area being formed by conductor inner-surface of coil end as shown in Fig. C3. As for the inner self inductance due to inner flux-

P0

le where le is the length of coil end. 8S After finding out the inductance of single coil end, the inductances due to coil-end leakage flux can also be obtained for various stator loops. N.B. each stator coil has two ends. linkage of the conductor itself, its value is

Figure C3 Outer self inductance and inner self inductance for the coil-end conductor

(2) Calculation of inductances due to leakage magnetic field of stator winding end on cylindrical coordinates Although it is feasible to study inductances due to leakage flux of stator coil end on Cartesian coordinates, the coil end is in fact a cylindrical surface so it is more suitable to use cylindrical coordinates to study the topic. The evolute part and straight-line part of the coil end are assumed to be on the same cylindrical surface. On cylindrical coordinates the flux densities caused by the end straight-line part current, end evolute part current and air-gap current are analysed respectively as follows. (a) Flux density produced by the end straight-line part current Firstly analyse the normal flux density produced by axial current element idl for point A at point K on the same end cylindrical surface, referring to Fig. C4.

Figure C4 Analysis of the flux density caused by the end straight-line part current on cylindrical coordinates 467

AC Machine Systems

In the light of Biot-Savart law exists dB

P 0 dl u r0 4S

i

r2

The dB and dl u r0 have the same direction. Since dB is perpendicular to dl, the dB at point K sits in the lateral plane and is perpendicular to the plane of Ƹ$BK. Thus, dB is perpendicular to BK, too. The normal component of dB at point K is (dB) n

dB cos

D 2

where D is the difference between spacial angles for points A and K along the circumference and dB is

dB

P0

sin T idl 2 4S r

P0

BK idl 3 4S r

2 R sin

P0 4S

id l

D 2

2 ª 2 § D· º « AB ¨ 2 R sin ¸ » 2 ¹ ¼» © ¬«

3/ 2

If the axial coordinate position for calculating point K is z0 and the position for current element point A is Z, then there is AB z0 z and dl dz , so exists

(dB) n

P0 4S

iR

sin D 2 ª D· º § 2 «( z0 z ) ¨ 2 R sin ¸ » 2 ¹ ¼» © ¬«

3/ 2

dz

(C16)

Assuming the overhanging length to be b for the end straight-line part, then the normal flux density caused by its current at point K of the end is Bn

³

b

P0

0

4S

iR

sin D 2 ª D· º § 2 «( z0 z ) ¨ 2 R sin ¸ » 2 ¹ ¼» © ¬«

3/ 2

dz

ª « P0 z0 z0 b i « « 2 2 D 4S 2 R tan D· § 2 « z 2 § 2 R sin D · ( ) 2 sin z b R ¸ ¨ ¸ 0 2« 0 ¨ 2¹ 2¹ © © ¬

º » » » » » ¼

(C17)

In order to obtain the normal flux density produced by the end straight-line part currents and their images, totalling 4 section currents, at any point K on the end cylindrical surface, we build the cylindrical coordinates as shown in Fig. C5, in which the coordinate position at point K is ( R, D 0 , z0 ), CD and FG are the end 468

Appendix C


straight-line parts, CDc and FG c are their images respectively. The normal flux densities caused by various section currents at point K are studied respectively in the following.

Figure C5 Analysis of normal flux density produced by the end straight-line part currents on the end cylindrical surface according to cylindrical coordinates

§T · For section CD, the angle D in formula (C17) is replaced by ¨ D 0 ¸ ; ©2 ¹ §T · For section FG, the angle D in formula (C17) is superseded by ¨ D 0 ¸ ; ©2 ¹ §T · For section CDc, in formula (C17) the angle D is replaced by ¨ D 0 ¸ 2 © ¹ and z0 is replaced by ( z0 b); §T · For section F cG, in formula (C17) the angle D is superseded by ¨ D 0 ¸ and 2 © ¹ z0 is superseded by ( z0 b). Thus, various normal flux densities caused by the end straight-line part currents and their images at any point K on the cylindrical surface are BFG BF cG BCD BCDc ° P0 i ° 1 ® (T / 2) D 0 4S 2 R ° tan ° 2 ¯

ª « 2 z0 « « 2 (T / 2) D 0 º « 2 ª » « z0 «¬ 2 R sin 2 ¼ ¬

º » z0 b z0 b » 2 2 » (T / 2) D 0 º (T / 2) D 0 º » ª ª ( z0 b)2 « 2 R sin ( z0 b) 2 « 2 R sin » » » 2 2 ¬ ¼ ¬ ¼ ¼ 469

AC Machine Systems

ª « 2 z0 z0 b 1 « « 2 2 (T / 2) D 0 (T / 2) D 0 º (T / 2) D 0 º ª « 2 ª tan 2 ( z0 b) « 2 R sin 2 » » « z0 «¬ 2 R sin 2 2 ¼ ¬ ¼ ¬ º½ »° z0 b »° ¾ 2 » (T / 2) D 0 º » ° ª 2 ( z0 b) « 2 R sin » »° 2 ¬ ¼ ¼¿ (C18)

(b) Flux density produced by the end evolute part current The coil end and its image are unfolded on the end cylindrical surface as shown in Fig. C6, in which E is the angle included between the evolute part and straight-line part. If the current element i'l of the end evolute part is divided into axial and circumferential components, then there is i'l

i ('l ) z e z i ('l )M eM i'l cos E e z i'l sin E eM

(C19)

where e z and eM are axial and circumferential unit vectors respectively.

Figure C6

The map of coil end and its image unfolded on the end cylindrical surface

The actions of axial and circumferential currents will be discussed respectively as follows. The normal flux density caused by the axial current of evolute section DE at point K on the end is 470

Appendix C


P0

¦ 4S iR

Bna1

z

sin D 2 ª D· º § 2 z z R ( ) 2 sin « 0 ¨ ¸ » 2 ¹ »¼ © «¬

3/ 2

'z

(C20)

cz T D 0 and z is from b to c for summation. cb 2 The normal flux density produced by the axial components of sections DE and EF currents and their images, totalling 4 section currents, at point K is

where D

Bna

° P0 °° sin D1 ¦z 4S iR ® 2 3/ 2 °ª D1 · º § 2 ° «( z0 z ) ¨ 2 R sin ¸ » 2 ¹ »¼ © ¯° «¬ sin D 3 sin D 2 2 3/ 2 2 3/ 2 ª ª D2 · º D3 · º § § 2 2 «( z0 z ) ¨ 2 R sin ¸ » « ( z0 z ) ¨ 2 R sin ¸ » 2 ¹ »¼ 2 ¹ ¼» © © «¬ ¬« ½ ° °° sin D 4 ¾'z 3 / 2 2 ª º ° D · § 2 «( z0 z ) ¨ 2 R sin 4 ¸ » ° 2 ¹ ¼» °¿ © ¬«

(C21)

where

D1 D3

czT D0 , cb 2 czT D0 , cb 2

D2 D4

czT D0 cb 2 czT D0 cb 2

The normal flux density caused by the circumferential current of evolute section DE at point K on the end is Bnt1

P0

¦ 4S i tan E cosD z

z0 z 2 ª D· º § 2 «( z0 z ) ¨ 2 R sin ¸ » 2 ¹ ¼» © ¬«

3/ 2

'z

(C22)

where

D

cz T D0 cb 2 471

AC Machine Systems

The normal flux density produced by the circumferential components of evolute sections DE and EF currents and their images, totalling 4 section currents, at point K on the end is

Bnt

° °° P0 ( z z0 ) cos D1 ¦z 4S i tan E ® 2 3/ 2 °ª D1 · º § 2 ° «( z z0 ) ¨ 2 R sin ¸ » 2 ¹ »¼ © °¯ «¬ ( z z0 ) cos D 2 ( z z0 ) cos D 3 3 / 2 2 2 3/ 2 ª ª D2 · º D3 · º § § 2 2 «( z z0 ) ¨ 2 R sin ¸ » «( z z0 ) ¨ 2 R sin ¸ » 2 ¹ ¼» 2 ¹ ¼» © © ¬« ¬« ½ ° °° ( z z0 ) cos D 4 ¾' z 3 / 2 2 ª º ° D § · 2 «( z z4 ) ¨ 2 R sin 4 ¸ » ° 2 ¹ ¼» ¿° © «¬

(C23)

where

D1 D3

czT D0 , D2 cb 2 czT D0 , D4 cb 2

czT D0 cb 2 czT D0 cb 2

From formula (C21) to (C23), z is from b to c for summation. (c) The end flux density produced by air-gap current Air-gap current and its image total two sections, and the normal flux density produced by air-gap current of coil inner-part at point K is BnG1

P0

2i ¦ D 4S

z0 R c 2P E 'D cos D 2 2 2 2P ( z0 R Rc 2 RRc cos D )3 / 2

(C24)

where Rc is the radius for the position the air-gap current sits at, and D is from T· T· § § ¨ D 0 ¸ to ¨ D 0 ¸ . 2¹ 2¹ © © The normal flux density produced by air-gap current of coil outer-part at point K is BnG 2 472

P0

E

2i cos D ¦ 4S 2 P (z D

2 0

z0 R c 'D 2 c R R 2 RRc cos D )3 / 2 2

(C25)

Appendix C


T· T § § · where D is from ¨ D 0 ¸ to ¨ D 0 2S ¸ . 2 2¹ © © ¹ Summarizing the results of formulas (C18), (C21), (C23), (C24) and (C25), the total normal flux density at point K on the end is Bn

BFG BF cG BCD BCDc Bna Bnt BnG1 BnG 2

(C26)

After obtaining the normal flux density of coil end, the flux-linkage and corresponding inductance can be got for the coil end. As mentioned above, calculation of coil-end inductance has been discussed on the basis of Cartesian coordinates and cylindrical coordinates. When evaluating the normal flux density of coil end, the analytical expression can be got according to Cartesian coordinates, but on cylindrical coordinates the analytical expression can’t be sometimes obtained due to difficulty of integration, which obliges us to adopt summation of various pieces to bring about a tedious calculation. The calculating example also demonstrates that the evaluation results according to the two coordinates above are approximate to each other.

473

Appendix D Heaviside’s Operational Calculus and Its Application to Transients Analysis

In Laplace’s operational calculus, the transformation functions of voltages and currents are all the functions of p, which is sometimes named the pure operational calculus and usually used in analysis of linear circuits and control systems. Besides the method has existed another operational calculus that is known as Heaviside’s operational calculus. If taking a R-L series circuit as an example and writing the voltage equation as Heavisides’s operational form, we can get u (t )

Lpi (t ) Ri (t )

( Lp R)i (t )

or i(t )

1 u (t ) Lp R

1 u (t ) Z ( p)

d is termed the operator. In this method, the operational impedance dt Z ( p) is a function of p but the voltage and current are still time functions. The mathematical basis of this method is not perfect enough but it is often convenient to use it. In analysis of machine transients, the basic relations are nonlinear differential equations during a variable speed, which can be described easily by Heaviside’s operational method. However, we have to look out for neglect during application of this method. Furthermore, the method is often used only on the zero initial condition, so the unit function [1] is sometimes put behind the expression but is omitted here for convenience. When this method is used to analyse a circuit of the nonzero initial condition, it is necessary to adopt the superposition theorem to give a special treatment. Supposing that the initial condition of the circuit is zero, the applied voltage is u (t ), the transient current is i (t ), and the operational impedance is Z ( p ), we can write Heaviside’s operational equation as in which p

Z ( p )i (t ) u (t ) or i(t )

1 u (t ) Z ( p)

(D1)


It should be noted that in the formulas above Z ( p ) is considered to be applied 1 applied on u (t ). For example, the latter is considered to be not on i (t ) or Z ( p) 1 that the original function f (t ) of is multiplied by u (t ), i.e. Z ( p) 1 i (t ) z f (t ) u (t ), but that operates on u (t ). If the equation form, however, Z ( p) 1 1 is i(t ) u (t ) is not applied on u (t ) but i (t ) is , it will imply that Z ( p) Z ( p) 1 equal to u (t ) times the original function f (t ) of . Therefore, during Z ( p) application of this operational method, the sequence of various terms can’t change arbitrarily, which is also a weakness of this operational method. Supposing that the voltage u (t ) in equation (D1) is converted into the corresponding image function u ( p), the current can be expressed as the image function as well, namely i( p)

1 u ( p) Z ( p)

(D2)

which is the pure operational form, where the sequence of various terms can change at will because it has been a pure algebraic equation. After determination of the Heaviside’s operational expression of the current, the corresponding original function may be looked up from the Heaviside’s operational collating table which is found in some references and omitted here. As for several ordinary voltages, however, it is convenient to use the Heaviside’s shifting theorem and decomposition one to get the solutions of the currents, which will be introduced as follows. (a) Heaviside’s shifting theorem F ( p )u (t ) AeD t

AeD t F ( p D )u (t )

(D3)

in which A and D are all constants (real or complex). (b) Heaviside’s operation and its decomposition theorem when the applied voltage is an exponential voltage UeD t Under this condition, the Heaviside’s operational expression with reference to equation (D1) can be written as i (t )

1 UeD t Z ( p)

(D4)

In the light of Heaviside’s shifting theorem exists 475

AC Machine Systems

i (t ) UeD t

1 Z( p D)

Now the problem is how to find the original function of

(D5) 1 , through Z(p D)

which times UeD t we can get the transient current i (t ). The method of deciding 1 is the same as the pure operational calculus. the original function of Z(p D) Generally speaking, Z ( p) may be a complicated rational fraction and letting Z ( p)

Q( p) H ( p)

there will be Z(p D)

Q( p D ) H(p D)

1 Z(p D)

H(p D) Q( p D )

so

Supposing that the roots of Q( p) 0 are p1 , p2 ," , pn , the roots of Q( p D ) 0 can be written as p1 D , p2 D ," , pn D , so

H ( pk ) 1 H (D ) n e( pk D ) t ̯ ¦ Z ( p D ) Q(D ) k 1 ( pk D )Qc( pk ) n 1 1 e( pk D ) t ¦ Z (D ) k 1 ( pk D ) Z c( pk )

(D6)

where pk is each root of Q( p) 0. After substituting equation (D6) into (D5), there will be ª H (D ) D t n º H ( pk ) i(t ) U « e ¦ e pk t » k 1 ( pk D )Q c( pk ) ¬ Q(D ) ¼ ª 1 Dt n º 1 e ¦ e pk t » U« k 1 ( pk D ) Z c( pk ) ¬ Z (D ) ¼

(D7a)

which is sometimes called the Heaviside’s decomposition theorem during an exponential voltage. Taking D 0 in equation (D7a), we can get the decomposition theorem for a dc voltage as follows: 476


ª H (0) n H ( pk ) pk t º i(t ) U « e » ¦ ¬ Q(0) k 1 pk Qc( pk ) ¼ n ª 1 º 1 e pk t » U« ¦ ¬ Z (0) k 1 pk Z c( pk ) ¼

(D7b)

U is the dc constraint component and the second is a free Z (0) Q (0) is the total resistance corresponding to the dc steadycomponent; Z (0) H (0) state component which is a special condition of the operational impedance and Q( p) will be equal to Z ( p) during p 0. H ( p) The result above is used to solve the transient process of a R-L series circuit as follows. Because

where the first term

i (t )

1 UeD t Z ( p)

1 UeD t R pL

and the root of the characteristic equation R pL

0 is p

(D8)

R , according to L

the decomposition theorem in equation (D7) we can get ª º R » « 1 t 1 eD t e L » i (t ) U « § R · «R DL » ¨ D ¸ L «¬ »¼ © L ¹ §R · ¨ D ¸ t º UeD t ª «1 e © L ¹ » R D L ¬« ¼»

(D9)

(c) Heaviside’s operation and its decomposition theorem when the applied voltage is a sinusoidal voltage U m sin(Z t \ ) On that condition, the Heaviside’s operational expression will become i (t )

1 U m sin(Z t \ ) Z ( p) H ( p ) U m j(Z t \ ) e j(Z t \ ) ] [e Q( p) 2 j

(D10)

By virtue of Heaviside’s shifting theorem, there will be

477

AC Machine Systems

i(t )

ª j(Z t \ ) H ( p jZ ) j(Z t \ ) H ( p jZ ) º e «e Q ( p jZ ) Q( p jZ ) »¼ ¬ ª H ( p jZ ) º U m Im «e j(Z t \ ) Q ( p jZ ) »¼ ¬ Um 2j

(D11)

Assuming that the roots of Q( p) 0 are p1 , p2 ," , pn , the roots of Q( p jZ ) 0 can be written as p1 jZ , p2 jZ ," , pn jZ , and according to equation (D6) we may get ª H ( jZ ) j(Z t \ ) n º e j\ H ( pk ) i(t ) U m Im « e e pk t » ¦ k 1 ( pk jZ )Q c( pk ) ¬ Q ( jZ ) ¼ j(Z t \ ) j\ n ªe º e 1 e pk t » U m Im « ¦ ¬ Z(jZ ) k 1 pk jZ Z c( pk ) ¼

(D12)

in which pk is each root of Q( p) 0. The expression above is also referred to as Heaviside’s decomposition theorem for the sinusoidal voltage, in which the first term is the ac constraint component Q ( jZ ) is the complex impedance and the second the free component; Z ( jZ ) H ( jZ ) corresponding to the ac steady-state component, i.e. the complex impedance is a special condition of the operational impedance and will be equal to the operational impedance Z ( p) during p jZ . For example, in a R-L series circuit there is i (t )

and the root of R pL

1 U m sin(Z t \ ) Z ( p) 1 U m sin(Z t \ ) R pL

0 is p

(D13)

R , so according to equation (D12) we can L

obtain ª º R » j\ « t 1 e e j(Z t \ ) e L » i (t ) U m Im « R j R L Z § · « » ¨ jZ ¸ L «¬ »¼ © L ¹ R ª tº 1 e j(\ M ) e j(Z t \ M ) e L » U m Im « «¬ R 2 (Z L)2 »¼ R 2 (Z L)2

478

(D14)


or i (t )

Um R 2 (Z L) 2

[sin(Z t \ M ) sin(\ M )e

R t L

]

(D15)

in which

M

arctan

ZL R

479

Appendix E

Basic Relations and Equivalent Circuits of Transformers

Firstly discuss a single phase transformer as shown in Fig. E1, in which r1 , r2 , L1L , L2L , L1 , L2 , W1 , W2 are resistances, leakage inductances, main inductances and number of turns for the primary and secondary windings respectively, and M is the mutual inductance between the primary and secondary. Letting the transformation ratio of transformer be K W1 / W2 , then exists L1 K 2 L2 . If neglecting the magnetization current, then there is i1 i2 / K or i2 Ki1 . The voltage equations of the primary and secondary are u1 u2

di1 di di ½ L1 1 M 2 ° dt dt dt ° ¾ di di di r2 i2 L2 L 2 L2 2 M 1 ° dt dt dt °¿ r1i1 L1L

Figure E1

Substituting i2

(E1)

Single phase transformer

Ki1 into equation (E1) exist u1 u2

di1 di di L1 1 KM 1 dt dt dt di di di Kr2 i1 KL2 L 1 KL2 1 M 1 dt dt dt

r1i1 L1L

From the two equations above, we can get u1 Ku2

di1 di ( L1 2 KM K 2 L2 ) 1 dt dt di (r1 K 2 r2 )i1 ( L1L K 2 L2 L ) 1 dt

(r1 K 2 r2 )i1 ( L1L K 2 L2 L )

(E2)

Appendix E


In a similar way exists u2

1 u1 K

§ 1 · § 1 · di2 ¨ 2 r1 r2 ¸ i2 ¨ 2 L1L L2 L ¸ ©K ¹ ©K ¹ dt

(E3)

For the 3-phase transformer of Y / ' 1 connection referring to Fig. E2, letting the transformation ratio be K3 U AB / U ab , then the corresponding transformation ratio of single phase transformer is K1 K3 / 3.

Figure E2

Three-phase transformer of Y / ' 1 connection

According to Fig. E2 and equation (E2) exists u AB

U A0 U B 0 K1 (uab ubc ) (rA K12 ra )iA ( LAL K12 LaL ) (rB K12 rb )iB ( LBL K12 LbL )

diA dt

diB dt

(E4)

In a similar way, the expressions can be obtained for uBC and uCA , so the corresponding equivalent circuit referred to Y-connection side is got as shown in Fig. E3.

Figure E3 Equivalent circuit referred to Y-connection side for 3-phase transformer of Y / ' 1 connection

If the 3-phase transformer of Y / ' 1 connection is referred to the secondary referring to Fig. E2, then there is iD

ia iJ ,

iE

ib iD ,

iJ

ic iE 481

AC Machine Systems

Supposing there is no zero-sequence current for the secondary, i.e. iD iE iJ

0,

then exist

1 (ia ib ), 3

iD

1 (ib ic ), 3

iE

1 (ic ia ) 3

iJ

and iA

1 (ia ib ), 3K1

iB

1 (ib ic ), 3K1

iC

1 (ic ia ) 3K1

In addition, referring to Fig. E3 also exists u AB u BC

If there is rA rB then exists

3K1ubc (rA K12 ra )iA 2(rB K12 rb )iB di (rC K12 rc )iC ( LAL K12 LaL ) A dt di d i 2( LBL K12 LbL ) B ( LCL K12 LcL ) C dt dt 1 ia 3K1ubc [(rC K12 rc ) (rA K12 ra )] 3K1 1 ib [2(rB K12 rb ) (rA K12 ra )] 3K1 1 ic [2(rB K12 rb ) (rC K12 rc )] 3K1 1 dia [( LCL K12 LcL ) ( LAL K12 LaL )] 3K1 dt 1 dib [2( LBL K12 LbL ) ( LAL K12 LaL )] 3K1 dt 1 dic [2( LBL K12 LbL ) ( LCL K12 LcL )] 3K1 dt

rC r1 , ra

ubc

rb

rc r2 , LAL LBL LCL L1L , LaL

LbL LcL L2 L ,

· 1 1§ 1 (u BC u AB ) ¨ 2 r1 r2 ¸ (ib ic ) 3K1 3 © K1 ¹ · d(i i ) 1§ 1 ¨ 2 L1L L2 L ¸ b c 3 © K1 ¹ dt

Similarly, the expressions of uca and uab are obtained, so the corresponding equivalent circuit referred to the secondary can be got as shown in Fig. E4. 482

Appendix E


Figure E4 Equivalent circuit referred to the seconday for 3-phase transformer of Y / ' 1 connection

483

Index

A Acceleration area 331-333 Accidents 224,268,269,301,315,415 ac-dc generator 167,249-251,253,257, 259, 260,261,263,265 Air-gap flux 11,14,23,24,27,29-33,35,49, 61,64,65,67,74,75,79-81,83,92,93,190, 341,347-349,384,448 Air-gap permeance coefficient 6,8,11,13, 31,455-458,460 Another expression for electromagnetic torque 321 Area rule 267,332,336 Asymmetrical rotor 181,182,219,234 Asymmetrical stator 168,181,182 Asynchronous and repulsion synchronous self-excitation region 166 Asynchronous operation of synchronous machines 203 Asynchronously starting of synchronous motors 214 Automatic voltage regulator (AVR) 258 Average operational reactance 174 Average torque after 3-phase sudden short circuit 156

B Bad welding 415 Base value 43,361 Basic equations of induction machines when counting inertia effect in 386 Basic-frequency current 147,150,181,188, 201,208

Biot-Savart law 39,440,461-463,468 Broad sense displacement 41 Broad sense force 41 By-pass petal 429,430

C Calculation example of dynamic stability 326 Carter’s coefficient 6,415 Cartesian coordinates system 409 Characteristic equation 118-121,124,129131,161,232,234,287,304,305,307,311, 396,397,399,477 Clockwise along the elliptical locus 298 Commutation instant of diodes 261 Complete decoupling of the system 403 Complex-number axes 173 Concentrated winding 3,23 Connection transformation matrix 254-256, 258 Control winding 339,402,403,405,408-411 Cosine-intersection method 238 Coupled circuit 2 Crawling phenomenon 285,287 Critical resistance 290,301 Current base 90,92,93,96-98,363,366 Cut-off angle 267,332 Cut-off of fault line 332-334 Cut-off time limit 267,333,336 Cycloconverter with 6 pulses and without circulating current 241 Cycloconverter-fed synchronous motor system with field-oriented control 236,242,243

Index Cylindrical coordinates system 39

D Damper loops 4-6,34,252,254,432,434, 435, 440,441,445,447 Damper winding 2,3,23,24,40,251,252 Damping bar 24-26,38,166,236,242-244, 248 Damping ring 24 Damping torque coefficient 267,275,277, 278,280,282,286,288,322 d-axis operational equivalent circuit 103, 106,107 d-axis operational reactance xd ( p) 280 d-axis speed emf 321 d-axis transformer emf 321 D-domain partition boundary curve 398, 399 D-domain partition method 339,397 Dead-region stage 241 Debugging of program 251 Decay time-constant 124,151,267,285 Deceleration area 331-333 Deep-slot rotor 366 Determination of self-excitation region for induction machines 395 Direct axis (d-axis) 4,11,12,14,24-26,28, 181,412,448,449 Direct search method without derivation 410 Disconnection process due to source fault 339,371 Dispatching loads 269 Dividing intervals calculation 333,336 Double transmission lines 327 Double-cage 366 Dual-stator-winding multi-phase high-speed induction generator 340,402 Dynamic stability 267-269,315,316,325-327, 329,331-333,337

E Eddy current and hysteresis effect 61

Eddy current density 409 Electromagnetic compatibility 250 Electromagnetic torque 40,41,58,59,100, 109,110,112,114,116,117,152-159,171, 174,176,177,179,199-202,208,210,219, 221,222,238,243-245,247,259,282,310, 311,317,321-323,339,344,350,362,365, 371,374,377,378,380,381,383-385 Electromechanical parameter resonance 269 Elliptical equation 294 End leakage flux 15,64,67,104,467 Equivalent concentrated treatment method 450 Equivalent current sheet 39,461 Euler’s method 406 Extreme value 67

F Fc , Bc ,0 axes 166,167,178,179

F, B, 0 axes 166-168,176,177 Ferro-magnetic saturation 61 Fictitious d-axis winding 80,92 Fictitious q-axis winding 80 Final value theorem of operational calculus 127 Five suits of 3-phase windings 167 Flow chart of simulation of ac-dc generator systems 260 Flux-extinguishing resistors 384,386 Flux-linkage base 90,96 Fly-wheel moment 370 Forced oscillation 267,270,285,287 Four 3-phase bridge rectifiers 167,250 Fourier series 8,403 Fourier transformation 406 Fourth order Runge-kutta method 441,451 Fractional harmonics 7,9,11,15,19,21,74, 414 Fractional-pitch and distributed stator winding 59 Fractional-slot winding 59 Free oscillation 270 Frequency differentiation rate 430 485

AC Machine Systems

G

Low order harmonics

19

Graphic method 8,331

M H H complex-plane 398 Half-difference operational reactance 174 Hamming window 430 Harmonic spectrum 243 Heaviside’s or Laplace’s operational calculus 102 Heaviside’s shift theorem 174 Herwitz’s criteria 267,307 Hydraulic generator 146,311

I Ideal machine pattern 78 Inertia constant 43,227,259,286,290,291, 362,386,397,398,399 Influence of excitation requlation on static stability 267,268,311 Inner asymmetric condition 3 Inner power characteristics 314 Inner-overvoltage 193 Instantaneous slip 271 Internal asymmetric analysis of ac machines 412 Internal faults 412-415,431,432,440,441 Inter-turn short circuit 414,433 Iterative computation 405

J Judge the commutation

251

L Large disturbance 267,269,315 Leakage mutual inductance 67,84,342, 349,351,417,419 Leakage self inductance 37,417,418,466 Left-upper-angle square matrix 54 Loading capability 429 Long-distance transmission 233,337 Loop current method 405 Lose synchronism 287-289,300,305,316, 329,331 486

Magnetic vector potential 409 Main petal of Hamming window 429 Mathematical patterns corresponding to analysis of dynamic stability 325 Mihainov’s method 397 Mode classifying method 258 Multi-harmonics 55 Multi-loop method 1,44,56,251,340,354, 432 Multi-loop model and parameters 1 Multi-variable optimization approach 410 Mutual flux-linkages \md and \mq 323

N Negative sequence reactance 120,124,129, 166,188,213 No-decaying natural oscillation 284 No-load characteristic 34,35,323,324 Nonlinear problem 215,316 Non-salient pole synchronous machine 108,113,116,136,291,412,447,448,450, 451,453 Normal two thyristors conducting mode 238-240 Numerical calculation of starting characteristics 365 Numerical integral 365 Numerical method 215,242,269,305,334, 365,378,380,381,383,385,386,451

O Object function 410 Operating overvoltage 193 Operational conductance G( p) 171,318 Operational reactance of stator q-axis xq ( p) 108 Optimization approach 405,410 Oscillation amplification 286 Out of step 203,268,285,287-289,301,305 Outer power characteristic 314

Index Output reactive power

276

P P- G curves for steady-state and transients 153 Park’s formulas (Park’s equations) 89 Park’s per-unit system of irreversible mutual in ductances 96 Per-unit system of reversible mutual inductances 90,98,99 Per-unit systems in synchronous machines 89 Per-unit value 43,44,91,98,100,354,355, 362 Phase-belt 74 Phase-splitting capacitor 56 Physical pattern of 3-phase ac machine after axis conversion 102 Positive and negative sets interchanging mode 239,241 Positive direction regulation of stator and rotor 4 Power base 90,109 Power winding 339,340,402,403,405,407, 408-411 Power-angle characteristic 113 Preserved kinetic energy 300 Pull-in torque 215,219 Pulsating mmf 11 Pulsating torque after 3-phase sudden short circuit 154,156

Q q-axis operational equivalent circuit 108, 109 q-axis speed emf 321 q-axis transformer emf 321 Quadrature axis (q-axis) 4,13,14,24-26, 28,412 Quantities with bottom-bars 91

R Reactance base 90

Reactive power Q 114 Reciprocity theorem 160 Reclosing transients 381,382 Record file 260 Rectangular mmf 9 Rectified load 403 Rectiformer 235,238-240 Reluctance variation 6 Resonant curves 267,286 Right-down-angle square matrix 54 Ring-segment inductance 353 Ripple of dc voltage 250 Rotor bases 93 Rotor extra-aperiodic current 192 Rotor motion equation 1,40,42-44,238, 317,322,325,333,350,362,365,374,390 Routh’s criterion 267,308,309 Runge-Kutta method 365,378,379,441, 443,451

S Salient pole synchronous machine 7,8,10, 11,13,15,23-26,36-40,50,108,113,116, 136,291,412,413,432,440,442,443,445, 447,448,450,451,453 Salient pole torque Tr 116 Salient-pole synchronous self-excitation region 166 Saturated-parameters 323 Self-excitation of Synchronous machines 227,229 Self-excitation when connecting induction motor in series with capacitance 386 Self-oscilation out of step 288,289 Shannon’s sampling theorem 261 Short circuit between branches of different phases 438,439 Short circuit torque Tk 115,156,157,210 Short pitch ratio of damper loop 27,30,50 Simulative waveform 243-245 Single phase induction motor 56,59 Skew-slot 341,409 Skew-slot coefficient 409 487

AC Machine Systems Slot leakage flux 15,38,39,67,409,440 Small increment 273,295,299,303,306, 393,394 Small oscillation 268,270,271,291,296-298, 300,310, Small perturbation 267-271,302,303,305, 306,310,386,393-396 Small perturbation and linearization of ac machine systems 268 Solid cage structure 403 Source current density 409 Specific leakage permeance 37 Speed and torque oscillation 370 Speed-up with weak field 244,247-249 Squirrel-cage rotor induction machine 347, 350 Stable limit 311,315 Starting characteristic 215,223,339,340, 363,365,370,371 Starting process for induction motors 339, 363 Static circuits 1,2,103 Static stability 267-270,301-309,311 Statistical data about accidents 315 Stator bases 90,92,98 Stator rated amplitudes 89 Storage coefficient of dynamic stability 331 Storage kinetic-energy 330 Sub-synchronous resonance 266,338 Sudden load 244-246,248,249 Sudden reversal-rotation 244-246,249 Summation term 62,71 Superposition method 403 Surplus kinetic energy 297 Surplus torque 42,44,296,297,330,331 Synchronization torque coefficient 275, 277-279,287,288,310 Synchronous torque Ts 287

T T-form electric circuit 329 Time base 90,98 488

Ta

x2 time constant corresponding to the r

aperiodic and second harmonic components in stator current 128 Tdcc time constant of d-axis damping winding when stator winding and excitation winding are all short-circuited in which r 0 and R fd 0 131 Tdc time constant of excitation winding when

stator winding is short-circuited whose resistance r is equal to zero 132 Tqcc time constant of q-axis damping winding when stator winding is short-circuited whose resistance is equal to zero 130 Time-variant coefficients 1,2,32,436,441, 451 Time-variant parameter 54 Topology of circuit 257 Torque coefficients 268,270,275,279,280, 282,287 Transcendental function 55 Transient emfs of synchronous machines 58,59,147,320 Transient parameters of synchronous machines 111 Transient stability 301 Transients of reswitching on induction motors Triggering time 371 Turbogenerator 57,136,146,155,233,266, 311,338,454 Two phases earthed 315 Two thyristors converting mode 240,241

U Unsymmetrical short circuit 315,327,328

V Variable-step Runge-Kutta integration method 259 Variable-topolog configurations 402 Virtual displacement principle 41 Voltage base 90,94-97,366

Index Voltage dip 58,59,142-144,146,150,515, 315,371

W Weakly linking power system 337 Window function 429

Wound-rotor induction machine 341,344, 345

Z Zero-axis component 83,85,86,89,356,364 Zero-axis inductance 84

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