Power System Control and Stability (Ieee Press Power Engineering Series)

Power System Control and Stability

IEEE Press 445 Hoes Lane Piscataway, NJ 08854

IEEE Press Editorial Board Stamatios V. Kartalopoulos, Editor in Chief M. Akay J. B. Anderson R. J. Baker J. E. Brewer

M. E. El-Hawary R. J. Herrick D. Kirk R. Leonardi M. S. Newman

M.Padgett W. D. Reeve S. Tewksbury G. Zobrist

Kenneth Moore, Director ofIEEE Press Catherine Faduska, Senior Acquisitions Editor John Griffin, Acquisitions Editor Anthony VenGraitis, ProjectEditor IEEE Power EngineeringSociety, Sponsor PE-S Liaison to IEEE Press, Chanan Singh BOOKS IN THE IEEE PRESS SERIES ON POWER ENGINEERING

Power System Protection P. M. Anderson 1999 Hardcover 1344pp 0-7803-3472-2 Understanding Power Quality Problems: Voltage Sags and Interruptions Math H. 1. Bollen 2000 Hardcover 576pp 0-7803-4713-7 Electric Power Applicationsof Fuzzy Systems Edited by M. E. El-Hawary 1998 Hardcover 384pp 0-7803-1197-3 Principles of Electric Machineswith Power ElectronicApplications, Second Edition M. E. El-Hawary 2002 Hardcover 496pp 0-471-20812-4 Analysis of Electric Machineryand Drive Systems,Second Edition Paul C. Krause, Oleg Wasynczuk, and Scott D. Sudhoff 2002 Hardcover 624pp 0-471-14326-X

Power System Control and Stability Second Edition

P. M. Anderson

San Diego, California

A. A. Fouad

Fort Collins, Colorado

IEEE Power Engineering Society, Sponsor

IEEE Press Power Engineering Series Mohamed E. El-Hawary, Series Editor

+IEEE IEEE PRESS

mWILEY-

~INTERSCIENCE

A JOHN WILEY & SONS, INC., PUBLICATION

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Contents

Preface

xiii

Part I Introduction P. M. Anderson and A. A. Fouad Chapter 1. Power System Stability

1.1 1.2 1.3 1.4 1.5

Introduction Requirements of a Reliable Electrical Power Service Statement of the Problem Effect of an Impact upon System Components Methods of Simulation Problems

3 3 4 8

10 11

Chapter 2. The Elementary Mathematical Model

2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

Swing Equation Units Mechanical Torque Electrical Torque Power-Angle Curve of a Synchronous Machine Natural Frequencies of Oscillation of a Synchronous Machine System of One Machine against an Infinite Bus-The Classical Model Equal Area Criterion Classical Model of a Multimachine System Classical Stability Study of a Nine-Bus System Shortcomings of the Classical Model Block Diagram of One Machine Problems References

13 15 16 20

21

24 26 31 35 37

45

47

48 52

Chapter 3. System Response to Small Disturbances

3.1 3.2 3.3 3.4 3.5

Introduction Types of Problems Studied The Unregulated Synchronous Machine Modes of Oscillation of an Unregulated Multimachine System Regulated Synchronous Machine

53

54

55 59 66

vii

Contents

VIII

3.6

Distributionof Power impacts Problems References

69 80 80

Part II The Electromagnetic Torque P. M. Anderson and A. A. Fouad Chapter 4. The Synchronous Machine

4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16

Introduction Park's Transformation Flux Linkage Equations Voltage Equations Formulationof State-SpaceEquations Current Formulation Per Unit Conversion Normalizingthe Voltage Equations Normalizingthe Torque Equations Torque and Power Equivalent Circuit of a Synchronous Machine The Flux Linkage State-SpaceModel Load Equations Subtransientand Transient Inductances and Time Constants Simplified Models of the Synchronous Machine Turbine Generator Dynamic Models Problems References

83 83 85 88

91 91

92

99

103 105 107

109

114 122 127 143 146 148

Chapter 5. The Simulation ofSynchronous Machines

5.1 5.2 5.3 5.4

Introduction Steady-StateEquations and Phasor Diagrams Machine Connected to an Infinite Bus through a TransmissionLine Machine Connectedto an Infinite Bus with Local Load at Machine Terminal 5.5 Determining Steady-State Conditions 5.6 Examples 5.7 Initial Conditions for a Multimachine System 5.8 Determinationof Machine Parametersfrom Manufacturers' Data 5.9 Analog Computer Simulationof the Synchronous Machine 5.10 Digital Simulationof Synchronous Machines Problems References

150 150 153 154 157 159 165 166

170

184 206 206

Chapter 6. Linear Models ofthe Synchronous Machine

6.1 6.2 6.3 6.4 6.5

Introduction Linearization of the Generator State-SpaceCurrent Model Linearization of the Load Equation for the One-Machine Problem Linearization of the Flux Linkage Model Simplified Linear Model

208

209

213 217 222

Contents

6.6 6.7

Block Diagrams State-Space Representation of Simplified Model Problems References

IX

231 231 232 232

Chapter 7. Excitation Systems

7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11

Simplified View of Excitation Control Control Configurations Typical Excitation Configurations Excitation Control System Definitions Voltage Regulator Exciter Buildup Excitation System Response State-Space Description of the Excitation System Computer Representation of Excitation Systems Typical System Constants The Effect of Excitation on Generator Performance Problems References

233 235 236 243 250 254 268 285 292 299 304 304 307

Chapter 8. Effect ofExcitation on Stability

8.1

8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9

8.10

8.11

Introduction Effect of Excitation on Generator Power Limits Effect of the Excitation System on Transient Stability Effect of Excitation on Dynamic Stability Root-Locus Analysis of a Regulated Machine Connected to an Infinite Bus Approximate System Representation Supplementary Stabilizing Signals Linear Analysis of the Stabilized Generator Analog Computer Studies Digital Computer Transient Stability Studies Some General Comments on the Effect of Excitation on Stability Problems References

309 311 315 321 327 333 338 344 347

353 363 365 366

Chapter 9. Multimachine Systems with Constant Impedance Loads

9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11

Introduction Statement of the Problem Matrix Representation of a Passive Network Converting Machine Coordinates to System Reference Relation Between Machine Currents and Voltages System Order Machines Represented by Classical Methods Linearized Model for the Network Hybrid Formulation Network Equations with Flux Linkage Model Total System Equations

368 368 369 373 374 377 378 381 386 388 390

x 9.12

Contents

Multimachine System Study Problems References

392 396 397

Part III The Mechanical Torque Power System Control and Stability P. M. Anderson

Chapter 10. Speed Governing 10.1 10.2 10.3 10.4 10.5 10.6

The Flyball Governor The Isochronous Governor Incremental Equations of the Turbine The Speed Droop Governor The Floating-Lever Speed Droop Governor The Compensated Governor Problems References

402 408 410 413 419 421 428 428

Chapter 11. Steam Turbine Prime Movers 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10

Introduction Power Plant Control Modes Thermal Generation A Steam Power Plant Model Steam Turbines Steam Turbine Control Operations Steam Turbine Control Functions Steam Generator Control Fossil-Fuel Boilers Nuclear Steam Supply Systems Problems References

430 432 435 436 437 444 446 458 461 476 480 481

Chapter 12. Hydraulic Turbine Prime Movers 12.1 Introduction 12.2 The Impulse Turbine 12.3 The Reaction Turbine 12.4 Propeller-Type Turbines 12.5 The Deriaz Turbine 12.6 Conduits, Surge Tanks, and Penstocks 12.7 Hydraulic System Equations 12.8 Hydraulic System Transfer Function 12.9 Simplifying Assumptions 12.10 Block Diagram for a Hydro System 12.11 Pumped Storage Hydro Systems Problems References

484

484 486 489 489

489 498 503 506 509 510 511 512

Contents

xi

Chapter 13. Combustion Turbine and Combined-Cycle Power Plants 13.1 Introduction 13.2 The Combustion Turbine Prime Mover 13.3 The Combined-Cycle Prime Mover Problems References

513 513 518 527 527

Appendix A. Appendix B. Appendix C. Appendix D. Appendix E. Appendix F. Appendix G. Appendix H. Appendix I. Appendix J.

529

Index

Trigonometric Identities for Three-Phase Systems Some Computer Methods for Solving Differential Equations Normalization Typical System Data Excitation Control System Definitions Control System Components Pressure Control Systems The Governor Equations Wave Equations for a Hydraulic Conduit Hydraulic Servomotors

531

545 555 582 590 614 622 631 640 651

Preface It is well over thirty years since some of the early versions of this book were used in our classes, and it is more than a quarter of a century since the first edition appeared in print. Normally, one would have expected users of the book to almost give it up as old-fashioned. Yet, until very recently the questions the authors were frequently asked explained the rationale for the added material in this edition, especially by new users: When will the Second Edition be out? Over these past thirty years the size of the systems analyzed in stability studies, the scope of the studies (including the kind of answers sought), the duration of the transients analyzed, and the methods of solution may have varied, but central to all is that the proper system model must be used. Such a model must be based on description of the physical system and on its behavior during the transient being analyzed. This book has focused on modeling the power system components for analysis of the electromechanical transient, perhaps with emphasis on the inertial transient. The one possible exception reflects the concern of the time the book came into being, namely analysis of the linear system model for detection and mitigation of possible poorly damped operating conditions. Since the 1970s, several trends made stability of greater concern to power system engineers. Because of higher cost of money and delay of transmission construction because of environmental litigations, the bulk power system has experienced more congestion in transmission, more interdependence among networks, and so on. To maintain stability, there has been more dependence on discreet supplementary controls, greater need for studying larger systems, and analysis of longer transients. Since then, additional models were needed for inclusion in stability studies: turbine governors, power plants, discrete supplementary controls, etc. Thus, the need for modeling the power system components that make up mechanical torque has become more important than ever. The authors think it is time to meet this need, as was originally planned. Now that the electric utility industry is undergoing major restructuring, the question arises as to whether the trend that started in the 1970s is likely to continue, at least into the near future. Many power system analysts believe that the answer to this question is yes. Since the revised printing of this book appeared, the electric utility industry has undergone a significant restructuring, resulting in heavier use of the bulk power transmission for interregional transactions. It is expected that new engineering emphasis will be given to what engineers refer to as mid-term or long-term analysis. We believe that in the restructured environment, this type of analysis will continue be needed because there will be greater emphasis on providing answers about system limitations to all parties involved in the various activities as well as in the interregional transactions. Modeling of mechanical torque will be important in conducting these studies. The material on the "mechanical torque" presented in Chapters 10 through 13 and in Appendices F through J are the work of author Paul Anderson and he should be contacted regarding any questions, corrections, or other information regarding these portions of the book. This material is a bit unusual to include in a book on power system stability and control, but we have recognized that a complete picture of stability and the supporting mathematical models cannot xiii

xiv

Preface

be consideredcomplete without a discussionof these importantsystem components. The models presented here can be described as "low-order" models that we consider appropriate additions to studies of power systemstability. This limits the modelsto a short time span of a minute or so, and purposely avoids the modeling of power plant behavior for the long term, for example, in the study of economics or energy dispatch.

San Diego, California Fort Collins, Colorado

P. M. ANDERSON A. A. FaUAD

Part I

Introduction

P. M. Anderson A. A. Fouad

chapter

1

Power System Stability

1.1

Introduction

Since the industrial revolution man's demand for and consumption of energy has increased steadily. The invention of the induction motor by Nikola Tesla in 1888 signaled the growing importance of electrical energy in the industrial world as well as its use for artificial lighting. A major portion of the energy needs of a modern society is supplied in the form of electrical energy. Industrially developed societies need an ever-increasing supply of electrical power, and the demand on the North American continent has been doubling every ten years. Very complex power systems have been built to satisfy this increasing demand. The trend in electric power production is toward an interconnected network of transmission lines linking generators and loads into large integrated systems, some of which span entire continents. Indeed, in the United States and Canada, generators located thousands of miles apart operate in parallel. This vast enterprise of supplying electrical energy presents many engineering problems that provide the engineer with a variety of challenges. The planning, construction, and operation of such systems become exceedingly complex. Some of the problems stimulate the engineer's managerial talents; others tax his knowledge and experience in system design. The entire design must be predicated on automatic control and not on the slow response of human operators. To be able to predict the performance of such complex systems, the engineer is forced to seek ever more powerful tools of analysis and synthesis. This book is concerned with some aspects of the design problem, particularly the dynamic performance, of interconnected power systems. Characteristics of the various components of a power system during normal operating conditions and during disturbances will be examined, and effects on the overall system performance will be analyzed. Emphasis will be given to the transient behavior in which the system is described mathematically by ordinary differential equations. 1.2

Requirements of a Reliable Electrical Power Service

Successful operation of a power system depends largely on the engineer's ability to provide reliable and uninterrupted service to the loads. The reliability of the power supply implies much more than merely being available. Ideally, the loads must be fed at constant voltage and frequency at all times. In practical terms this means that both voltage and frequency must be held within close tolerances so that the consumer's 3

Chapter 1

4

equipment may operate satisfactorily. For example, a drop in voltage of 10-15% or a reduction of the system frequency of only a few hertz may lead to stalling of the motor loads on the system. Thus it can be accurately stated that the power system operator must maintain a very high standard of continuous electrical service. The first requirement of reliable service is to keep the synchronous generators running in parallel and with adequate capacity to meet the load demand. If at any time a generator loses synchronism with the rest of the system, significant voltage and current fluctuations may occur and transmission lines may be automatically tripped by their relays at undesired locations. If a generator is separated from the system, it must be resynchronized and then loaded, assuming it has not been damaged and its prime mover has not been shut down due to the disturbance that caused the loss of synchronism. Synchronous machines do not easily fall out of step under normal conditions. If a machine tends to speed up or slow down, synchronizing forces tend to keep it in step. Conditions do arise, however, in which operation is such that the synchronizing forces for one or more machines may not be adequate, and small impacts in the system may cause these machines to lose synchronism. A major shock to the system may also lead to a loss of synchronism for one or more machines. A second requirement of reliable electrical service is to maintain the integrity of the power network. The high-voltage transmisssion system connects the generating stations and the load centers. Interruptions in this network may hinder the flow of power to the load. This usually requires a study of large geographical areas since almost all power systems are interconnected with neighboring systems. Economic power as well as emergency power may flow over interconnecting tie lines to help maintain continuity of service. Therefore, successful operation of the system means that these lines must remain in service if firm power is to be exchanged between the areas of the system. While it is frequently convenient to talk about the power system in the "steady state," such a state never exists in the true sense. Random changes in load are taking place at all times, with subsequent adjustments of generation. Furthermore, major changes do take place at times, e.g., a fault on the network, failure in a piece of equipment, sudden application of a major load such as a steel mill, or loss of a line or generating unit. We may look at any of these as a change from one equilibrium state to another. It might be tempting to say that successful operation requires only that the new state be a "stable" state (whatever that means). For example, if a generator is lost, the remaining connected generators must be capable of meeting the load demand; or if a line is lost, the power it was carrying must be obtainable from another source. Unfortunately, this view is erroneous in one important aspect: it neglects the dynamics of the transition from one equilibrium state to another. Synchronism frequently may be lost in that transition period, or growing oscillations may occur over a transmission line, eventually leading to its tripping. These problems must be studied by the power system engineer and fall under the heading "power system stability." 1.3

Statement of the Problem

The stability problem is concerned with the behavior of the synchronous machines after they have been perturbed. I f the perturbation does not involve any net change in power, the machines should return to their original state. If an unbalance between the supply and demand is created by a change in load, in generation, or in network conditions, a new operating state is necessary. In any case all interconnected synchronous machines should remain in synchronism if the system is stable; i.e., they should all remain operating in parallel and at the same speed.

Power System Stability

5

The transient following a system perturbation is oscillatory in nature; but if the system is stable, these oscillations will be damped toward a new quiescent operating condition. These oscillations, however, are reflected as fluctuations in the power flow over the transmission lines. If a certain line connecting two groups of machines undergoes excessive power fluctuations, it may be tripped out by its protective equipment thereby disconnecting the two groups of machines. This problem is termed the stability of the tie line, even though in reality it reflects the stability of the two groups of machines. A statement declaring a power system to be "stable" is rather ambiguous unless the conditions under which this stability has been examined are clearly stated. This includes the operating conditions as well as the type of perturbation given to the system. The same thing can be said about tie-line stability. Since we are concerned here with the tripping of the line, the power fluctuation that can be tolerated depends on the initial operating condition of the system, including the line loading and the nature of the impacts to which it is subjected. These questions have become vitally important with the advent of large-scale interconnections. In fact, a severe (but improbable) disturbance can always be found that will cause instability. Therefore, the disturbances for which the system should be designed to maintain stability must be deliberately selected. 1.3.1

Primitive definition of stability

Having introduced the term ....stability," we now propose a simple nonmathematical definition of the term that will be satisfactory for elementary problems. Later, we will provide a more rigorous mathematical definition. The problem of interest is one where a power system operating under a steady load condition is perturbed, causing the readjustment of the voltage angles of the synchronous machines. If such an occurrence creates an unbalance between the system generation and load, it results in the establishment of a new steady-state operating condition, with the subsequent adjustment of the voltage angles. The perturbation could be a major disturbance such as the loss of a generator, a fault or the loss of a line, or a combination of such events. It could also be a small load or random load changes occurring under normal operating conditions. Adjustment to the new operating condition is called the transient period. The system behavior during this time is called the dynamic system performance, which is of concern in defining system stability. The main criterion for stability is that the synchronous machines maintain synchronism at the end of the transient period. Definition: If the oscillatory response of a power system during the transient period following a disturbance is damped and the system settles in a finite time to a new steady operating condition, we say the system is stable. If the system is not stable, it is considered unstable.

This primitive definition of stability requires that the system oscillations be damped. This condition is sometimes called asymptotic stability and means that the system contains inherent forces that tend to reduce oscillations. This is a desirable feature in many systems and is considered necessary for power systems. The definition also excludes continuous oscillation from the family of stable systems, although oscillators are stable in a mathematical sense. The reason is practical since a continually oscillating system would be undesirable for both the supplier and the user of electric power. Hence the definition describes a practical specification for an acceptable operating condition.

6

Chapter 1

1.3.2

Other stability problems

While the stability of synchronous machines and tie lines is the most important and common problem, other stability problems may exist, particularly in power systems having appreciable capacitances. In such cases arrangements must be made to avoid excessive voltages during light load conditions, to avoid damage to equipment, and to prevent self-excitation of machines. Some of these problems are discussed in Part III, while others are beyond the scope of this book. 1.3.3

Stability of synchronous machines

Distinction should be made between sudden and major changes, which we shall call large impacts, and smaller and more normal random impacts. A fault on the highvoltage transmission network or the loss of a major generating unit are examples of large impacts. If one of these large impacts occurs, the synchronous machines may lose synchronism. This problem is referred to in the literature as the transient stability problem. Without detailed discussion, some general comments are in order. First, these impacts have a finite probability of occurring. Those that the system should be designed to withstand must therefore be selected a priori.' Second, the ability of the system to survive a certain disturbance depends on its precise operating condition at the time of the occurrence. A change in the system loading, generation schedule, network interconnections, or type of circuit protection may give completely different results in a stability study for the same disturbance. Thus the transient stability study is a very specific one, from which the engineer concludes that under given system conditions and for a given impact the synchronous machines will or will not remain in synchronism. Stability depends strongly upon the magnitude and location of the disturbance and to a lesser extent upon the initial state or operating condition of the system. Let us now consider a situation where there are no major shocks or impacts, but rather a random occurrence of small changes in system loading. Here we would expect the system operator to have scheduled enough machine capacity to handle the load. We would also expect each synchronous machine to be operating on the stable portion of its power-angle curve, i.e., the portion in which the power increases with increased angle. In the dynamics of the transition from one operating point to another, to adjust for load changes, the stability of the machines will bedetermined by many factors, including the power-angle curve. It is sometimes incorrect to consider a single power-angle curve, since modern exciters will change the operating curve during the period under study. The problem of studying the stability of synchronous machines under the condition of small load changes has been called "steady-state" stability. A more recent and certainly more appropriate name is dynamic stability. In contrast to transient stability, dynamic stability tends to be a property of the state of the system. Transient stability and dynamic stability are both questions that must be answered to the satisfaction of the engineer for successful planning and operation of the system. This attitude is adopted in spite of the fact that an artificial separation between the two problems has been made in the past. This was simply a convenience to accommodate the different approximations and assumptions made in the mathematical treatI. In the United States the regional committees of the National Electric Reliability Council (N ERC) specify the contingencies against which the system must be proven stable.

Power System Stability

7

ments of the two problems. In support of this viewpoint the following points are pertinent. First, the availability of high-speed digital computers and modern modeling techniques makes it possible to represent any component of the power system in almost any degree of complexity required or desired. Thus questionable simplifications or assumptions are no longer needed and are often not justified. Second, and perhaps more important, in a large interconnected system the full effect of a disturbance is felt at the remote parts some time after its occurrence, perhaps a few seconds. Thus different parts of the interconnected system will respond to localized disturbances at different times. Whether they will act to aid stability is difficult to predict beforehand. The problem is aggravated if the initial disturbance causes other disturbances in neighboring areas due to power swings. As these conditions spread, a chain reaction may result and large-scale interruptions of service may occur. However, in a large interconnected system, the effect of an impact must be studied over a relatively long period, usually several seconds and in some cases a few minutes. Performance of dynamic stability studies for such long periods will require the simulation of system components often neglected in the so-called transient stability studies. 1.3.4

Tie-line oscillations

As random power impacts occur during the normal operation of a system, this added power must be supplied by the generators. The portion supplied by the different generators under different conditions depends upon electrical proximity to the position of impact, energy stored in the rotating masses, governor characteristics, and other factors. The machines therefore are never truly at steady state except when at standstill. Each machine is in continuous oscillation with respect to the others due to the effect of these random stimuli. These oscillations are reflected in the flow of power in the transmission lines. If the power in any line is monitored, periodic oscillations are observed to be superimposed on the steady flow. Normally, these oscillations are not large and hence not objectionable. The situation in a tie line is different in one sense since it connects one group of machines to another. These two groups are in continuous oscillation with respect to each other, and th is is reflected in the power flow over the tie line. The situation may be further complicated by the fact that each machine group in turn is connected to other groups. Thus the tie line under study may in effect be connecting two huge systems. In this case the smallest oscillatory adjustments in the large systems are reflected as sizable power oscillations in the tie line. The question then becomes, To what degree can these oscillations be tolerated? The above problem is entirely different from that of maintaining a scheduled power interchange over the tie line: control equipment can be provided to perform this function. These controllers are usually too slow to interfere with the dynamic oscillations mentioned above. To alter these oscillations, the dynamic response of the components of the overall interconnected system must be considered. The problem is not only in the tie line itself but also in the two systems it connects and in the sensitivity of control in these systems. The electrical strength (admittance) or capacity of the tie cannot be divorced from this problem. For example, a 40-MW oscillation on a 400-MW tie is a much less serious problem than the same oscillation on a lOO-MW tie. The oscillation frequency has an effect on the damping characteristics of prime movers,

8

Chapter 1

exciters, etc. Therefore, there is a minimum size of tie that can be effectively made from the viewpoint of stability. 1.4

Effect of an Impact upon System Components

In this section a survey of the effect of impacts is made to estimate the elements that should be considered in a stability study. A convenient starting point is to relate an impact to a change in power somewhere in the network . Our "test" stimulus will be a change in power, and we will use the point of impact as our reference point. The follow ing effects, in whole or in part, may be felt. The system frequency will change because, until the input power is adjusted by the machine governors , the power change will go to or come from the energy in the rotating masses. The change in frequency will affect the loads, especially the motor loads. A common rule of thumb used among power system engineers is that a decrease in frequency results in a load decrease of equal percentage; i.e., load regulation is 100%. The network bus voltages will be affected to a lesser degree unless the change in power is accompanied by a change in reactive power.

"

e.; 3"/4

]:

~ "/2

J1

"/ 4

TIme, s

"

c o

Fig . 1.1.

Response or aIour-machine system during a transient : (a) stable system. (b) unstable system .

Power

1.4.1

System Stability

9

Loss of synchronism

Any unbalance between the generation and load initiates a transient that causes the rotors of the synchronous machines to "swing" because net accelerating (or decelerating) torques are exerted on these rotors. If these net torques are sufficiently large to cause some of the rotors to swing far enough so that one or more machines "slip a pole," synchronism is lost. To assure stability, a new equilibrium state must be reached before any of the machines experience this condition. Loss of synchronism can also happen in stages, e.g., if the initial transient causes an electrical link in the transmission network to be interrupted during the swing. This creates another transient, which when superimposed on the first may cause synchronism to be lost. Let us now consider a severe impact initiated by a sizable generation unbalance, say excess generation. The major portion of the excess energy will be converted into kinetic energy. Thus most of the machine rotor angular velocities will increase. A lesser part will be consumed in the loads and through various losses in the system. However, an appreciable increase in machine speeds may not necessarily mean that synchronism will be lost. The important factor here is the angle difference between machines, where the rotor angle is measured with respect to a synchronously rotating reference. This is illustrated in Figure 1.1 in which the rotor angles of the machines in a hypothetical four-machine system are plotted against time during a transient. In case (a) all the rotor angles increase beyond 7r radians but all the angle differences are small, and the system will be stable if it eventually settles to a new angle. In case (b) it is evident that the machines are separated into two groups where the rotor angles continue to drift apart. This system is unstable. 1.4.2

Synchronous machine during a transient

During a transient the system seen by a synchronous machine causes the machine terminal voltage, rotor angle, and frequency to change. The impedance seen "looking into" the network at the machine terminal also may change. The field-winding voltage will be affected by: I. Induced currents in the damper windings (or rotor iron) due to sudden changes in armature currents. The time constants for these currents are usually on the order of less than 0.1 s and are often referred to as "subtransient" effects. 2. Induced currents in the field winding due to sudden changes in armature currents. The time constants for this transient are on the order of seconds and are referred to as "transient" effects. 3. Change in rotor voltage due to change in exciter voltage if activated by changes at the machine terminal. Both subtransient and transient effects are observed. Since the subtransient effects decay very rapidly, they are usually neglected and only the transient effects are considered important. Note also that the behavior discussed above depends upon the network impedance as well as the machine parameters. The machine output power will be affected by the change in the rotor-winding EM F and the rotor position in addition to any changes in the impedance "seen" by the machine terminals. However, until the speed changes to the point where it is sensed and corrected by the governor, the change in the output power will come from the stored energy in the rotating masses. The important parameters here are the kinetic energy in MW·s per unit MVA (usually called H) or the machine mechanical time constant T j , which is twice the stored kinetic energy per MVA.

Chapter 1

10

When the impact is large, the speeds of all machines change so that they are sensed by their speed governors. Machines under load frequency control will correct for the power change. Until this correction is made, each machine's share will depend on its regulation or droop characteristic. Thus the controlled machines are the ones responsible for maintaining the system frequency. The dynamics of the transition period, however, are important. The key parameters are the governor dynamic characteristics. In addition, the flow of the tie Jines may be altered slightly. Thus some machines are assigned the requirement of maintaining scheduled flow in the ties. Supplementary controls are provided to these machines, the basic functions of which are to permit each control area to supply a given load. The responses of these controls are relatively slow and their time constants are on the order of seconds. This is appropriate since the scheduled economic loading of machines is secondary in importance to stability. 1.5

Methods of Simulation

Ifwe look at a large power system with its numerous machines, lines, and loads and consider the complexity of the consequences of any impact, we may tend to think it is hopeless to attempt analysis. Fortunately, however, the time constants of the phenomena may be appreciably different, allowing concentration on the key elements affecting the transient and the area under study. The first step in a stability study is to make a mathematical model of the system during the transient. The elements included in the model are those affecting the acceleration (or deceleration) of the machine rotors. The complexity of the model depends upon the type of transient and system being investigated. Generally, the components of the power system that influence the electrical and mechanical torques of the machines should be included in the model. These components are: I. 2. 3. 4. 5. 6.

The network before, during, and after the transient. The loads and their characteristics. The parameters of the synchronous machines. The excitation systems of the synchronous machines. The mechanical turbine and speed governor. Other important components of the power plant that influence the mechanical torque. 7. Other supplementary controls, such as tie-line controls, deemed necessary in the mathematical description of the system.

Thus the basic ingredients for solution are the knowledge of the initial conditions of the power system prior to the start of the transient and the mathematical description of the main components of the system that affect the transient behavior of the synchronous machines. The number of power system components included in the study and the complexity of their mathematical description will depend upon many factors. In general, however, differential equations are used to describe the various components. Study of the dynamic behavior of the system depends upon the nature of these differential equations. 1.5.1

Linearized system equations

If the system equations are linear (or have been linearized), the techniques of linear system analysis are used to study dynamic behavior. The most common method is to

Power System Stability

11

simulate each component by its transfer function. The various transfer function blocks are connected to represent the system under study. The system performance may then be analyzed by such methods as root-locus plots, frequency domain analysis (Nyquist criteria), and Routh's criterion. The above methods have been frequently used in studies pertaining to small systems or a small number of machines. For larger systems the state-space model has been used more frequently in connection with system studies described by linear differential equations. Stability characteristics may be determined by examining the eigenvalues of the A matrix, where A is defined by the equation

x=Ax+Bu

( 1.1 )

where x is an n vector denoting the states of the system and A is a coefficient matrix. The system inputs are represented by the r vector u, and these inputs are related mathematically to differential equations by an n x r matrix B. This description has the advantage that A may be time varying and u may be used to represent several inputs if necessary. 1.5.2

Large system with nonlinear equations

The system eq uations for a transient stabi lity study are usually non linear. Here the system is described by a large set of coupled nonlinear differential equations of the form

x=

f'(x, u, t)

( 1.2)

where f is an n vector of nonlinear functions. Determining the dynamic behavior of the system described by (1.2) is a more difficult task than that of the linearized system of (1.1). Usually time solutions of the nonlinear differential equations are obtained by numerical methods with the aid of digital computers, and this is the method usually used in power system stability studies. Stability of synchronous machines is usually decided by behavior of their rotor angles, as discussed in Section 1.4.1. More recently, modern theories of stability of nonlinear systems have been applied to the study of power system transients to determine the stability of synchronous machines without obtaining time solutions. Such efforts, while they seem to offer considerable promise, are still in the research stage and not in common use. Both linear and nonlinear equations will be developed in following chapters. Problems 1.1

1.2 1.3 1.4 1.5

Suggest definitions for the following terms: a. Power system reliability. b. Power system security. c. Power system stability. Distinguish between steady-state (dynamic) and transient stability according to a. The type of disturbance. b. The nature of the defining equations. What is a tie line'? Is every line a tie line'? What is an impact insofar as power system stability is concerned'? Consider the system shown in Figure P t.5 where a mass M is pulled by a driving force f(t) and is restrained by a linear spring K and an ideal dashpot B.

Chapter 1

12

Wr ite the d ifferential equati on for the system in terms of the d isplacement var iable x and determine the relative values of Band K to provide cr itical damping when J(I) is a unit ste p funct ion .

x

..

I

K f (I)

Fig. PI.5 .

1.6

Repeat Problem 1.5 but convert the equations to the sta te-spa ce form of( 1.1) .

chapter

2

The Elementary Mathematical Model

A stable power system is one in which the synchronous machines, when perturbed, will either return to their original state if there is no net change of power or will acquire a new state asymptotically without losing synchronism. Usually the perturbation causes a transient that is oscillatory in nature; but if the system is stable, the oscillations will be damped. The question then arises, What quantity or signal, preferably electrical, would enable us to test for stability? One convenient quantity is the machine rotor angle measured with respect to a synchronously rotating reference. If the difference in angle between any two machines increases indefinitely or if the oscillatory transient is not sufficiently damped, the system is unstable. The principal subject of this chapter is the study of stability based largely on machine-angle behavior. 2. 1

Swing Equation

The swing equation governs the motion of the machine rotor relating the inertia torque to the resultant of the mechanical and electrical torques on the rotor; i.e., I Jij

=

To Nvrn

(2.1)

where J is the moment of inertia in kg- rn? of all rotating masses attached to the shaft, () is the mechanical angle of the shaft in radians with respect to fixed reference, and To is the accelerating torque in newton meters (N· m) acting on the shaft. (See Kimbark [1] for an excellent discussion of units and a dimensional analysis of this equation.) Since the machine is a generator, the driving torque T; is mechanical and the retarding or load torque T, is electrical. Thus we write

a

(2.2) which establishes a useful sign convention, namely, that in which a positive T; accelerates the shaft, whereas a positive T, is a decelerating torque. The angular reference may be chosen relative to a synchronously rotating reference frame moving with

1. The dot notation is used to signify derivatives with respect to time. Thus .

dx

..

X = -,X

dt

d 2X

== - 2 ,etc.

dt

13

14

Chapter 2

constant angular velocity WR,2 (2.3) where a is a constant. The angle a is needed if Om is measured from an axis different from the angular reference frame; for example, in Chapter 4 a particular choice of the reference for the rotor angle lJ m gives a = 11"/2 and 8 = WR t + 11"/2 + Om' From (2.3) we see.that (j may be replaced by ~m in (2.1), with the result (2.4)

where J is the moment of inertia in kg-rn", Om is the mechanical (subscript m) torque angle in rad with respect to a synchronously rotating reference frame, W m is the shaft angular velocity in rad/s, and T is the accelerating torque in N· m. Another form of (2.4) that is sometimes useful is obtained by multiplying both sides by W m , the shaft angular velocity in rad/s, Recalling that the product of torque T and angular velocity W is the shaft power P in watts, we have Q

(2.5)

The quantity JW m is called the inertia constant and is denoted by M. (See Kimbark (I] pp. 22-27 and Stevenson [2], pp. 336-40 for excellent discussions of the inertia constant.) It is related to the kinetic energy of the rotating masses W k , where Wi = (1/2)Jw~joules. Then Miscomputedas Angular Momentum

= M = JOO m =

2 Wk/oo m J·s

(2.6)

It may seem rather strange to call M a constant since it depends upon w, which certainly varies during a transient. On the other hand the angular frequency does not change by a large percentage before stability is lost. To illustrate: for 60 Hz, W m = 377 rad/s, and a 1% change in W m is equal to 3.77 rad/s, A constant slip of 1% of the value of w", for one second will change the angle of the rotor by 3.77 rad. Certainly, this would lead to loss of synchronism. The equation of motion of the rotor is called the swing equation. It is given in the literature in the form of (2.4) or in terms of power, (2.7)

where M is in J. S, Om is in rad, W m is in rad/s, and P is in W. In relating the machine inertial performance to the network, it would be more useful to write (2.7) in terms of an electrical angle that can be conveniently related to the position of the rotor. Such an angle is the torque angle ,0, which is the angle between the field MM F and the resultant MM F in the air gap, both rotating at synchronous speed. It is also the electrical angle between the generated EM F and the resultant stator voltage phasors. The torque angle 0, which is the same as the electrical angle O~, is related to the rotor mechanical angle Om (measured: from a synchronously rotating frame) by (2.8)

where p is the number of poles. (In Europe the practice is to write the number of pole pairs.)

WI

o~

= POm' where p is

2. The subscript R is used to mean "rated" for all quantities including speed, which is designated as in ANSI standards ANSI Y10.5, 1968. Hence WR = WI in every case.

The Elementary Mathematical Model

15

For simplicity we drop the subscript e and write simply 0, which is always understood to be the electrical angle defined by (2.8). From (2.7) and (2.8) we write (2M/p)5 = (2M/p)w = Po W

(2.9)

which relates the accelerating power to the electrical angle 0 and to the angular velocity of the revolving magnetic field w. In most problems of interest there will be a large number of equations like (2.9), one for each generator shaft (and motor shaft too if the motor is large enough to warrant detailed representation). In such large systems problems we find it convenient to normalize the power equations by dividing all equations by a common three-phase voltarnpere base quantity S B3. Then (2.9) becomes a per unit (pu) equation (2.10)

where M, p, 0, and ware in the same units as before; but P is now in pu (noted by the subscript u). 2.2

Units

It has been the practice in the United States to provide inertial data for rotating machines in English units. The machine nameplate usually gives the rated shaft speed in revolutions per minute (r /rnin). The form of the swing equation we use must be in M KS units (or pu) but the coefficients, particularly the moments of inertia, will usually be derived from a mixture of MKS and English quantities. We begin with the swing equation in N · m

(2J jp)~ = (2Jjp)w = To N· m

(2.11)

Now normalize this equation by dividing by a base quantity equal to the rated torque at rated speed:

(2.12) where SB3 is the three-phase V A rating and nR is the rated shaft speed in r Imino' Dividing (2.11) by (2.12) and substituting f20/R InR for p, we compute (J1r 2ni / 900wRSa3)w = TalTa = Tau pu (2.13) where we have substituted the base system radian frequency WR = 21r fR for the base frequency. Note that w in (2.13) is in rad/s and Tau is in pu. The U.S. practice has been to supply J, the moment of inertia, as a quantity usually called WR 2, given in units of Ibm- ft 2• The consistent English unit for J is slug- ft 2 or 2 WR /g where g is the acceleration of gravity (32.17398 ft/s"). We compute the corresponding M KS quantity as J

=

WR

2

slug·ft2

g

1 ft·lbf·s2 1 slug- ft 2

746 W·s 550 ft ·lbf

746(WR

550 g

2

)

J'S2

or

kg'm 2

Substituting into (2.13), we write 746(WR2)1r2n~ . --------.~._..-. W = 550 g(900)Wk S ,U

Tou

pu

(2.14)

The coefficient of wcan be clarified if we recall the definition of the kinetic energy of a

Chapter 2

16

rotating body Wk =

W", which we can write as

! Jui 2

m

=

!

2

x 746(WR2) x (21r-nR)2 550 g 3600

= 2 311525 ·

x

1O- 4(WR 2)ni

J

Then (2.14) may be written as (2 W Ac / S 8 3WR)W

= Tau pu

(2.15)

We now define the important quantity

H ~ WAc/S B3 s where

(2.16)

= rated three-phase MVA of the system WAc = (2.311525 x 10-,o)(WR 2 )nl MJ

SB3

Then we write the swing equation in the form most useful in practice: (2H/WR)W

= T, pu

(2.17)

where H is in s, w is in rad/s, and T is in pu. Note that w is the angular velocity of the revolving magnetic field and is thus related directly to the network voltages and currents. For this reason it is common to give the units of w as electrical rad/s. Note also that the final form of the swing equation has been adapted for machines with any number of poles, since all machines on the same system synchronize to the same WR. Another form of the swing equation, sometimes quoted in the literature, involves some approximation. It is particularly used with the classical model of the synchronous machine. Recognizing that the angular speed w is nearly constant, the pu accelerating power P, is numerically nearly equal to the accelerating torque To. A modified (and approximate) form of the swing equation becomes (2.18)

The quantity H is often given for a particular machine normalized to the base VA rating for that machine. This is convenient since these machine-normalized H quantities are usually predictable in size and can be estimated for machines that do not physically exist. Curves for estimating H are given in Figures 2.1 and 2.2. The quantities taken from these curves must be modified for use in system studies by converting from the machine base VA to the system base VA. Thus we compute H sys

= H mach (S BJmach/ S B3sys )

S

(2.19)

The value of H mach is usually in the range of 1-5 s. Values for H sys vary over a much wider range. With SB3syS = 100 MVA values of H sys from a few tenths of a second (for small generators) to 25-30 s (for large generators) will often be used in the same study. Typical values of J (in MJ) are given in Appendix D. 2.3

Mechanical Torque

The mechanical torques of the prime movers for large generators, both steam and waterwheel turbines, are functions of speed. (See Venikov [6], Sec. 1.3, and Crary [7], Vol. II, Sec. 27.) However we should carefully distinguish between the case of the unregulated machine (not under active governor control) and the regulated (governed) case.

The Elementary Mathematical Model 10r---

---..;:--

-

-

-

- --

-

17

- -- - --

-

-,

o

500

4 .5 4.0

« ~ 3. 5 tr/2. Note that the accelerating power need not be plotted as a function of o. We can obtain the same information if the electrical and mechanical powers are plotted as a function of o. The former is the power-angle curve discussed in Section 2.7, and in many studies Pm is a constant. The accelerating power curve could have discontinuities due to switching of the network, initiation of faults, and the like.

omax

2.8.1

Critical clearing angle

For a system of one machine connected to an infinite bus and for a given fault and switching arrangement, the critical clearing angle is that switching angle for which the system is at the edge of instability (we will also show that this applies to any twomachine system). The maximum angle omax corresponds to the angle Om on the faultcleared power-angle curve. Conditions for critical clearing are now obtained (see [1] and [2]). Let

peak of the prefault power-angle curve ratio of the peak of the power-angle curve of the faulted network to PM ratio of the peak of the power-angle curve of the network with the fault cleared to PM sin-I Pm / PM < tr/2 sin-I Pm/r2PM > tr/2

Then for A I

=

8e

A 2 and for critical clearing,

=

cos-1{[1/(r2 - r1 )][(Pm/ PM)(8 m - 80)

+ r2cos8 m - r1cos80l}

(2.51)

Note that the corresponding clearing time must be obtained from a time solution of the swing equation. 2.8.2

Application to a one-machine system

The equal area criterion is applied to the power network of Examples 2.4-2.5, and the results are shown in Figure 2.15. The stable system of Examples 2.4-2.5 is illustrated in Figure 2.15. The angle at I = 0 is 21.09 and is indicated by the intersection of Pm with the prefault curve. The clearing angle Oc is obtained from the time soluThe conditions for A 2 = A I correspond to tion (see Figure 2.13) and is about 31.6 omax ~ 48°. This corresponds to the maximum angle obtained in the time solution shown in Figure 2.13. To illustrate the critical clearing angle, a more severe fault is used with the same system and switching arrangement. A three-phase fault is applied to the same bus with zero impedance. The faulted power-angle curve has zero amplitude. The prefault and 0

0

•

34

Chapter

2

2.0

6mo x 60

Fig.2 .15

90

120

150

180

Application of the equal area criterion to a stable system.

postfault networks are the same as before. For this system

'. '" 0 '2 '" 1.587/2 .222 '" 0.714

00 '" 21.09° Om '" 149.73°

Calculation of the critical clearing angle. using (2 .51), gives 0, '"

COS -I

0.26848 '" 74.43°

This situation is illustrated in Figure 2.16 .

2 .0

6 = 149.73 m

120

Fig.2.16

150

0

180

Appl icat ion of the equal area criterion to a critically cleared system.

The Elementary Mathematical Model 2.8.3

35

Equal area criterion for a two-machine system

It can be shown that the equal area criterion applies to any two-machine system since a two-machine system can be reduced to an equivalent system of one machine connected to an infinite bus (see Problem 2.14). We can show that the expression for the equal area criterion in this case is given by

°

J

612

6 120

(Pal _ Pa 2 ) dlJ HI

H2

12

=

0

(2.52)

where lJ 12 = lJ, - 2 In the special case where the resistance is neglected, (2.52) becomes _1_

Ho

2.9

J6

12

Pal dlJ 12

=

0

6120

Classical Model of a Multimachine System

The same assumptions used for a system of one machine connected to an infinite bus are often assumed valid for a multimachine system: I. Mechanical power input is constant. 2. Damping or asynchronous power is negligible. 3. Constant-voltage-behind-transient-reactance model for the synchronous machines is valid. 4. The mechanical rotor angle of a machine coincides with the angle of the voltage behind the transient reactance. 5. Loads are represented by passive impedances. This model is useful for stability analysis but is limited to the study of transients for only the "first swing" or for periods on the order of one second. Assumption 2 is improved upon somewhat by assuming a linear damping characteristic. A damping torque (or power) Dw is frequently added to the inertial torque (or power) in the swing equation. The damping coefficient D includes the various damping torque components, both mechanical and electrical. Values of the damping coefficient usually used in stability studies are in the range of 1-3 pu [9, 10, 11, 12]. This represents turbine damping, generator electrical damping, and the damping effect of electrical loads. However, much larger damping coefficients, up to 25 pu, are reported in the literature due to generator damping alone [7, 13]. Assumption 5, suggesting load representation by a constant impedance, is made for convenience in many classical studies. Loads have their own dynamic behavior, which is usually not precisely known and varies from constant impedance to constant MV A. This is a subject of considerable speculation, the major point of agreement being that constant impedance is an inadequate representation. Load representation can have a marked effect on stability results. The electrical network obtained for an n-machine system is as shown in Figure 2.17. Node 0 is the reference node (neutral). Nodes 1,2, ... .n are the internal machine buses, or the buses to which the voltages behind transient reactances are applied. Passive impedances connect the various nodes and connect the nodes to the reference at load buses. As in the one-machine system, the initial values of E" £2"'" En are determined from the pretransient conditions. Thus a load-flow study for pretransient

Chapter 2

36

n -machine system n generators

'1

f

j X ~l1

Transm ission sy ste m

r constant

impedance loads

I

,_ _ J

I I

+

1_--

1

I I I I I

1 1 -

I_ _ J' I I

I I

-T

L,

I

+

'-- --I

I -I I

I

I

+

~_J

N~aO

1L

_

I I I

Fig. 2.17 Representation or a muhimachine system (classical model).

conditions is needed. The magnitudes £;, i = I, 2, .. . , n are held constant during the transient in classical stability studies. The passive electrical network described above has n nodes with active sources. The admittance matrix of the n-port network, looking into the network from the terminals of the generators. is defined by

(2.53) where Y has the diagonal elements

Y;; 'Vij

Vii and

the off-diagonal elements

Y;j'

By definition.

Y;; IJ..n.. = driving point admittance for node i o, + j s, Y;i

&

=

a, + j s,

negative of the transfer admittance between nodes i andj

(2.54)

The power into the network at node i, which is the electrical power output of machine i, is given by P; = eRe El;* Pd

= £1 o, +

L n

j-I

EjEj Yij cos (8(i - 0;

+ OJ)

i = 1,2, . . . ,n

j~;

n

E1

c, + L

j-I j.,J;

EjEj[B;j sin (0/ -

0) +

c, cos (OJ -

OJ)]

i = 1,2, .. . , n

(2.55)

37

The Elementary Mathematical Model

The equations of motion are then given by

2H.' de.' + D,.W,. WR

dl

Pm; -

[2E; G;; + ~ ~ E;E ~j j

1=1

COS

J

(0;; - 0; + OJ)

j~;

do; dt

i = 1,2, ... ,n

It should be noted that prior to the disturbance (t

=

0-) Pm;o

(2.56) =

PeW

n

Pm;o

=

E; Gii O +

L

j=l

E;E; Y;jO cos (OijO - Ow + 0;0)

(2.57)

j~;

The subscript 0 is used to indicate the pretransient conditions. This applies to all machine rotor angles and also to the network parameters, since the network changes due to switching during the fault. The set of equations (2.56) is a set of n-coupled nonlinear second-order differential equations. These can be written in the form

x

=

f( x, X o, t )

(2.58)

where x is a vector of dimension (2n x I), (2.59) and f is a set of nonlinear functions of the elements of the state vector x. 2.10

Classical Stability Study of a Nine-bus System

The classical model of a synchronous machine may be used to study the stability of a power system for a period of time during which the system dynamic response is dependent largely on the stored kinetic energy in the rotating masses. For many power systems this time is on the order of one second or less. The classical model is the simplest model used in studies of power system dynamics and requires a minimum amount of data: hence, such studies can be conducted in a relatively short time and at minimum cost. Furthermore, these studies can provide useful information. For example, they may be used as preliminary studies to identify problem areas that require further study with more detailed modeling. Thus a large number of cases for which the system exhibits a definitely stable dynamic response to the disturbances under study are eliminated from further consideration. A classical study will be presented here on a small nine-bus power system that has three generators and three loads. A one-line impedance diagram for the system is given in Figure 2.18. The prefault normal load-flow solution is given in Figure 2.19. Generator data for the three machines are given in Table 2.). This system, while small, is large enough to be nontrivial and thus permits the illustration of a number of stability concepts and results. 2.10.1

Data preparation

In the performance of a transient stability study, the following data are needed: I. A load-flow study of the pretransient network to determine the mechanical power Pm of the generators and to calculate the values of E;& for all the generators. The equivalent impedances of the loads are obtained from the load bus data.

Chapter 2

38

18 kV

jO. 0625

1---_ loa d C

230 kV

P/2 = jO.0745

CD

:0

a

~

+

s,

S

a a

S

.

a

0

~ a

s

a

lI;

CD

0

N

0

.;-

!:!..

a " S 0

CD

a

:::

~

~

loa d A

B/2 = jO.l045

a s, :< " a

N M

13 .8 kV

0 .0 119 . jo. 1008

0 .0085 • ' 0. 072

r-,

0

230k V

a

" G)

~

e..

s

Load B

"

4

0()

t;

0

s,

~

16.5 kV

Fig. 2.18

Nine-bus system impeda nce diagram: all impeda nces are in pu on a IOO-MVA base.

100.0 18 kV 163. 0 (6,7 )

(35. 0)

230 kV -1 63 76 . 4 (9 .2 ) (- 0 . 8)

®

&L.

'"

N

~

r-,

...

a

I

-

~0 o. 996

r:::

cc

M

::;:.

'" '"

~

0 M

~

I

Lood A

( - 10 .9 )

1.032

as:

M

~ I

/ - 4. 0 0

85.0

CD

:i '" ""

r; M ~I

0

\

1.0 1 ~

CD 1. 026 0

no W 24 . 2 - 85.0 (3. 0) (15.0) (- 10 . 9)0) 60 .8 1. 025 (- 18P) / 4 ,7 0

- 75.9 -2 4 . 1 (- 10 ,7 ) (- 24 . 3)

86. 6 (- 8 .4)

~

Load C

0()

2-

~ ~

0

0

0

1 . 013 / -3,70

o- !2

Lood B

230 kv

Fig. 2.19 Nine-bus system load-flow diagram showing prefault condit ions; all flows are in MWand MVAR .

The Elementary Mathematical Model Table 2.1.

Generator Data 2

3

247.5 16.5 1.0 hydro 180 r/min 0.1460 0.0608 0.0969 0.0969 0.0336 8.96 0

192.0 18.0 0.85 steam 3600 r/min 0.8958 0.1198 0.8645 0.1969 0.0521 6.00 0.535

128.0 13.8 0.85 steam 3600 r/min 1.3125 0.1813 1.2578 0.25 0.0742 5.89 0.600

2364 MW·s

640 MW·s

301 MW·s

Generator

Rated MVA kV Power factor Type Speed xd xd xq x'q x-t(leakage)

TdO

T~O

Stored energy at rated speed

39

Note: Reactance values are in pu on a loo-MVA base. All time constants are in s. (Several quantities are tabulated that are as yet undefined in this book. These quantities are derived and justified in Chapter 4 but are given here to provide complete data for the sample system.)

2. System data as follows: a. The inertia constant H and direct axis transient reactance Xd for all generators. b. Transmission network impedances for the initial network conditions and the subsequent switchings such as fault clearing and breaker reclosings. 3. The type and location of disturbance, time of switchings, and the maximum time for which a solution is to be obtained. 2.10.2

Preliminary calculations

To prepare the system data for a stability study, the following preliminary calculations are made: 1. All system data are converted to a common base: a system base of 100 MV A is frequently used. 2. The loads are converted to equivalent impedances or admittances. The needed data for this step are obtained from the load-flow study. Thus if a certain load bus has a voltage VL, power PL, reactive power QL' and current ~ flowing into a load admittance YL = GL + jB L, then

PL + jQL

=

rLI!

=

VL[V"i(G L

-

jB L ) ]

=

Vt(G L

-

jB L )

The equivalent shunt admittance at that bus is given by

(2.60) 3. The internal voltages of the generators E;& are calculated from the load-flow data. These internal angles may be computed from the pretransient terminal voltages V I.!!.. as follows. Let the terminal voltage be used temporarily as a reference, as shown in Figure 2.20. Ifwe define 1 = I, + j/2' then from the relation P + jQ = ill* we have I, + j/2 = (P - jQ)/V. But since EIJ..' = V + jx~J: we compute EIJ..'

= (V +

Qx~/V)

+ j(PXd/V)

(2.61)

The initial generator angle 00 is then obtained by adding the pretransient voltage

40

Chapter 2

+

ElL Fig.2.20

Generator representation for computing 00.

angle a to 0', or 00

=

0' + ex

(2.62)

4. The Y matrix for each network condition is calculated. The following steps are usually needed: a. The equivalent load impedances (or admittances) are connected between the load buses and the reference node; additional nodes are provided for the internal generator voltages (nodes I, 2, ... , n in Figure 2.17) and the appropriate values of »; are connected between these nodes and the generator terminal nodes. Also, simulation of the fault impedance is added as required, and the admittance matrix is determined for each switching condition. b. All impedance elements are converted to admittances. c. Elements of the Y matrix are identified as follows: ~i is the sum of all the admittances connected to node i, and ¥;j is the negative of the. admittance between node i and nodej. 5. Finally, we eliminate all the nodes except for the internal generator nodes and obtain the V matrix for the reduced network. The reduction can be achieved by matrix operation if we recall that all the nodes have zero injection currents except for the internal generator nodes. This property is used to obtain the network reduction as shown below. Let YV

(2.63)

where I =

[-:~]

Now the matrices Y and V are partitioned accordingly to get

(2.64)

where the subscript n is used to denote generator nodes and the subscript r is used for the remaining nodes. Thus for the network in Figure 2.17, Vn has the dimension (n x I) and V, has the dimension (r x I). Expanding (2.64),

The Elementary Mathematical Model

41

from which we eliminate Y, to find In

=

(Ynn - Ynr Y;;. 1 Ym ) Vn

(2.65)

The matrix (Y nn - Y nr Y;,:' Y rn ) is the desired reduced matrix Y. It has the dimensions (n x n) where n is the number of the generators. The network reduction illustrated by (2.63)-(2.65) is a convenient analytical technique that can be used only when the loads are treated as constant impedances. If the loads are not considered to be constant impedances, the identity of the load buses must be retained. Network reduction can be applied only to those nodes that have zero injection current.

Example 2.6 The technique of solving a classical transient stability problem is illustrated by conducting a study of the nine-bus system, the data for which is given in Figures 2.18 and 2.19 and Table 2.1. The disturbance initiating the transient is a three-phase fault occurring near bus 7 at the end of line 5-7. The fault is cleared in five cycles (0.083 s) by opening line 5-7. For the purpose of this study the generators are to be represented by the classical model and the loads by constant impedances. The damping torques are neglected. The system base is 100 MVA. Make all the preliminary calculations needed for a transient stability study so that all coefficients in (2.56) are known. Solution The objective of the study is to obtain time solutions for the rotor angles of the generators after the transient is introduced. These time solutions are called "swing curves." In the classical model the angles of the generator internal voltages behind transient reactances are assumed to correspond to the rotor angles. Therefore, mathematically, we are to obtain a solution for the set of equations (2.56). The initial conditions, denoted by adding the subscript 0, are given by w;o = 0 and 0;0 obtained from (2.57). Preliminary calculations (following the steps outlined in Section 2.10.2) are: l. The system base is chosen to be 100 MVA. All impedance data are given to this base. 2. The equivalent shunt admittances for the loads are given in pu as

load A: YL5 load B: YL6 load C: YL8

= =

=

1.2610 - jO.5044 0.8777 - jO.2926 0.9690 - jO.3391

3. The generator internal voltages and their initial angles are given in pu by 1.0566/2.2717° = 1.0502/19.7315°

E,~ =

E2 &

E3~ =

1.0170/13.1752°

4. The Y matrix is obtained as outlined in Section 2.10.2, step 4. For convenience bus numbers I, 2, and 3 are used to denote the generator internal buses rather than the generator low-voltage terminal buses. Values for the generator Xd are added to the reactance of the generator transformers. For example, for generator 2 bus 2 will be the internal bus for the voltage behind transient reactance; the reactance between

Chapter 2

42

Prefault Network

Table 2.2.

Impedance

Bus no.

Generators"

No.1 No.2 No.3 Transmission lines

Shunt admittancest Load A Load B Load C

R

Admittance X

G

B

1-4 2-7 3-9

0 0 0

0.1184 0.1823 0.2399

0 0 0

4-5 4-6 5-7 6-9 7-8 8-9

0.0100 0.0170 0.0320 0.0390 0.0085 0.0119

0.0850 0.0920 0.1610 0.1700 0.0720 0.1008

1.3652 1.9422 1.1876 1.2820 1.6171 1.1551

-11.6041 -10.5107 -5.9751 - 5.5882 -13.6980 -9.7843

1.2610 0.8777 0.9690

-0.2634 -0.0346 -0.1601 0.1670 0.2275 0.2835

5-0 6-0 8-0 4-0 7-0 9-0

-8.4459 -5.4855 -4.1684

*For each generator the transformer reactance is added to the generator xd. tThe line shunt susceptances are added to the loads.

bus 2 and bus 7 is the sum of the generator and transformer reactances (0.1198 + 0.0625). The prefault network admittances including the load equivalents are given in Table 2.2, and the corresponding Y matrix is given in Table 2.3. The y' matrix for the faulted network and for the network with the fault cleared are similarly obtained. The results are shown in Tables 2.4 and 2.5 respectively.

5. Elimination of the network nodes other than the generator internal nodes by network reduction as outlined in step 5 is done by digital computer. The resulting reduced Y matrices are shown in Table 2.6 for the prefault network, the faulted network, and the network with the fault cleared respectively. We now have the values of the constant voltages behind transient reactances for all three generators and the reduced Y matrix for each network. Thus all coefficients of (2.56) are available. Example 2.7

For the system and the transient of Example 2.6 calculate the rotor angles versus time. The fault is cleared in five cycles by opening line 5-7 of Figure 2.18. Plot the angles 0 and K2 KJ K4 > O. The first of the above cr iteria st ates that the synchronizing power coefficient K, must be greater than the dem agnetizing component of electrical power. The second criterion is satisfied if the constants K2 , KJ , and K4 are positive. Venikov [4] points out that if the transmission network has an appreciable series capacitive rea ctance, it is possible that instability may occur. This would happen because the impedance factor producing the constant K, would become negative.

3.3.2

Effect of small changes of speed

In the linearized version of (3.2) we are interested in terms involving changes of power due to changes of the angle {) and its derivative. The change in po wer due to

58

Chapter 3

lJb, was discussed above and was found to include a synchronizing power component and a demagnetizing component due the change in E~ with lJb,' The change in speed. Wb, = dlJb,/dt, causes a change in both electrical and mechanical power. In this case the new differential equation becomes

2H WR

d2~b, = dt

apml Wb, _ (apt] Wb, + aptl lJb,) aw o aw "'0 alJ J6 0

J...

(3.15)

As in (3 .7) the change in electrical power due to small changes in speed is in the form of (3.16)

From Section 2.3 the change in mechanical power due to small changes in speed is also linear

Pmb,

=

oPm/owLowb,

(3.17)

where apm/owLo can be obtained from a relation such as the one given in Figure 2.3. If a transient droop or regulation R is assumed, we may write in pu to the machine base

(3.18) which is the equation of an ideal speed droop governor. The system block diagram with speed regulation added is shown in Figure 3.2.

Fig. 3.2

Block diagram representation of the linearized model with speed regulation added.

The characteristic equation of the system now becomes

-2H s 2 + WR

I ~ + -I) + (

WR

D

R

s

K1

-

(3.19)

or

(3.20)

System Response to Small Disturbances

59

Again Routh's criterion may be applied to determine the conditions for stability. This is left as an exercise (see Problem 3.2).

3.4

Modes of Oscillation of an Unregulated Multimachine System The electrical power output of machine i in an n-machine system is Pe;

= E;G;; +

n

L E;Ej Y;j cos (O;j -

oij).

j ....

j~;

E;G;;

+

n

L

i- I

sin

E;Ej(Bjj

0ij

+

Gij

cos

(3.21)

Ojj)

j~;

where

= 0; - OJ E; = constant voltage behind transient reactance for machine i

oij

~; = Gu + jBu is a diagonal element of the network short circuit admittance matrix Y ~j = Gij + jBij is an off-diagonal element of the network short circuit admittance matrix Y

Using the incremental model so that

Finally, for

sin

lJ;j =

sin

lJ;jO

cos

oij ~

cos

oijO -

r.:

cos

lJijA 0ijA

L

O;j = 0ljO

+ cos sin

=

OjjA'

fJ jjO sin fJijA ~

we compute

sin

fJ;jO

+

fJ jj A

cos

fJ;jO

fJ;jO

n

Pe;A

+

E;Ej(B;j

cos

0ijO -

j='=1

a; sin 0ijo) O;jA

(3.22)

j~;

For a given initial condition sin oijO and cos in (3.22) is a constant. Thus we write

are known, and the term in parentheses

0ijO

LP n

Pei A =

i» I

Sij lJ;jA

(3.23)

j~;

where (3.24) is the change in the electrical power of machine i due to a change in the angle between machines; and j, with all other angles held constant. Its units are W jrad or pu power jrad. It is a synchronizing power coefficient between nodes i andj and is identical to the coefficient discussed in Section 2.5.2 for one machine connected to an infinite bus. We also note that since (3.21) applies to any number of nodes where the voltages are known, the linearized equations (3.22) and (3.23) can be derived for a given machine in terms of the voltages at those nodes and their angles. Thus the concept of the synchronizing power coefficients can be extended to mean "the change in the electrical power of a given machine due to the change in the angle between its internal EM F and

60

Chapter 3

any bus, with all other bus angles held constant." (An implied assumption is that the voltage at the remote bus is also held constant.) This expanded definition of the synchronizing power coefficient will be used in Section 3.6. Using the inertial model of the synchronous machines, we get the set of linearized differential equations,

i

=

1,2, .. . .n

(3.25)

or

i = 1,2, ... .n The set (3.26) is not a set of n-independent second-order equations, since Thus (3.26) comprises a set of (n - l)-independent equations. From (3.26) for machine i,

i

=

1,2, ... , n

(3.26) ~oij

= O.

(3.27)

Subtracting the nth equation from the ith equation, we compute

d;~~11 _ d;~;11

+

2w~

n

L r.»; - 2iI L P,nj 0njl1 = 0 n-I

, J= 1

(3.28)

n )-1

j"i

Equation (3.28) can be put in the form

i = I, 2, .... .n - 1

(3.29)

Since (3.30) (3.29) can be further modified as

d 2o.

-T + L dt

n-I

j_1

aA.11

=0

;= 1,2, ... ,n-l

(3.31)

where the coefficients a ij depend on the machine inertias and synchronizing power coefficients. Equation (3.31) represents a set of n - 1 linear second-order differential equations or a set of 2(n - I) first-order differential equations. We will use the latter formulation to examine the free response of this system. Let XI' X 2, ••• ,X n _ I be the angles t5 l n4 , 02nA" •• ' t5(n-l)nA respectively, and let X n , ••• , X 2n - 2 be the time derivatives of these angles. The system equations are of the form

System Response to Small Disturbances

o o

o o

o

XI

X2

o

0

I I

61

Xn_1

-------------------4--------------I

Xn

AI ,n-I

II I II I I

A 2,n - 1 An-I,n-I

(3.32)*

Xn

o

X n+1

~

X2n-2

or (3.33) where U = the identity matrix XI = the n - I vector of the angle changes ~inA X2 = the n - I vector of the speed changes dlJinA/dt To obtain the free response of the system, we examine the eigenvalues of the characteristic matrix [11,12]. This is obtained from the characteristic equation derived from equating the determinant of the matrix to zero, as follows: : V ] -1---A ,-XV

- XV

= det M = 0

det - - [

where ~ is the eigenvalue. Since the matrix terminant of M as

I M I = I -xv I I (-~U) = {_l)n-l Xn- l

t

-~U

is nonsingular, we compute the de-

- A(_~U)-IU

-xu -

(3.34)

{_l/~)n-l

I

AI

=

t

X2V - A I

(3.35)

(See Lefschetz [12], p. 133.) The system described by I M I = 0, or I X - A I = 0, has 2(n - 1) imaginary roots, which occur in n - 1 complex conjugate pairs. Thus the system has n - I frequencies of oscillations. 2V

Example 3.1

Find the modes of oscillation of a three-machine system. The machines are unregulated and classical model representation is used. Solution

For an unregulated three-machine system, the system equations are given by 2H d 20 AI - -l - + PJl2lJl2A + PJ13~I3A = 0 WR dt?

2H 2 d 2 oA2

- - --2WR

dt

20 3 A3 - - --2-

2H d WR

*See the addendum on page 650.

dt

+ +

P ~

J21 u21A

~

PJ31 u31A

+

PJ23lJ23A

+

PJ32 032A =

=0 0

Chapter 3

62

Multiplying the above three equations by w R / 2H i and subtracting the third equation from the first two, we get (noting that Oij = - Oji) 20 d UA --2-

dt

+ -WR-

20 d 23A --2- -

dt

If we eliminate are obtained:

2H I

~

Ps12(}124

WR - - Ps21 2H2 0124

012A

R WR + (W -PSIJ + - ..

2H.

2H 3

p)

s31 0134

+ -WR2H J

Ps32 0234

=0

~ + (WR + -WR- Ps31(JUA - - Ps23 + -WR- p)~ s32 (}2J4 = 0 2H3

by noting that

2H2

0124

2H 3

+ 0234 + 0314 = 0, the

following two equations

or

The state-space representation of the above system is

Al3 A

0

0

,

~2J4

0

0

, 0

-

-,-, -

-

-al2

,I

0

I I

0

I

-a22

-a21

W23A

I

- - - - -

-all

WIJA

0

I

I

- -

°13A °234

0

Wl3A

0

W234

To obtain the eigenvalues of this system, the characteristic equation is given by

det

-A

0

0

-A

,I ,I

0 0

I I

- - -.- - - - - ""1 - - - -

-all

-al2

-a21

-a22

-A

I I I I I

0

,

- -

0

=0

-A

Now by using (3.35),

(A 2

A4 + A2

(all

(1/2)1-(all

+

+

+

+

2 a ll )( A a 22) a22)A2 + (all a22 -

a22) ±

[(all

+

a22)2 -

a12a21

=

0

all a21)

=

0

4(a lla22 -

aI2 a21)P/21

63

System Response to Small Disturbances

Examining the coefficients a;;, we can see that both values of A2 are negative real quantities. Let these given values be A = ± j{3, A = ± jl'. The free response will be in the form 0/1 = C 1 cos ({3t + 1>1) + C2 cos (1'1 + 1>2)' where C 1, C2 , (jJ" and (jJ2 are constants. Example 3.2 Consider the three-machine, nine-bus system of Example 2.6, operating initially in the steady state with system conditions given by Figure 2.18 (load flow) and the computed initial values given in Example 2.6 for E;&, i = I, 2, 3. A small 10-MW load (about 3% of the total system load of 315 MW) is suddenly added at bus 8 by adding a three-phase fault to the bus through a 10.0 pu impedance. The system base is 100 MYA. Assume that the system load after t = 0 is constant and consists of the original load plus the 10 pu shunt resistance at bus 8. Compute the frequencies of oscillation that will result from this small disturbance. Then compare these computed frequencies against those actually observed in a digital computer solution. Assume there are no governors active on any of the three turbines. Observe the system response for about two seconds.

Solution First we compute the frequencies of oscillation. From (3.24)

= Vi ~(Bij cos O;jO - Gij sin 0ijo)

PSi}

"J

Vi ~Bij cos

O;jO

From Example 2.6 we find the data needed to compute PSij with the results shown in Table 3.1. Table 3.1.

Synchronizing Power Coefficients of the Network of Example 2.6

ij

12 23 31

1.0566 1.0502 1.0170

1.0502 1.0170 1.0566

-17.4598 6.5563 10.9035

1.513 1.088 1.226

1.6015 1.1544

1.2936

Note that the oijO are the values of the relative rotor angles at t = 0-. Since these are rotor angles, they will not change at the time of impact, so these are also the correct values for t = 0+. This is also true of angles at load buses to which appreciable inertia is connected. For loads that are essentially constant impedance, however, the voltage angle will exhibit a step change. Also from Example 2.6 we know Hi = 23.64, 6.40, and 3.01 for i = 1, 2, 3 respectively. Thus we can compute the values of a ij from Example 3.1 as follows: all

a

l2 =

1o,

PiA(td = (Hi

It

Hi)

Pu(O+)

(3.61)

Thus at the end of a brief transient the various machines will share the increase in load as a function only of their inertia constants. The time II is chosen large enough

System Response to Small Disturbances

73

so that all the machines will have acquired the mean system retardation. At the same time t I is not so large as to allow other effects such as governor action to take place. Equation (3.61) implies that the H constants for all the machines are given to a common base . If they are given for each machine on its own base, the correct powers are obtained if H is replaced by HS B3/S'B' where SB3 is the machine rating and S,B is the chosen system base. Examining (3.56) and (3.61), we note that immediately after the impact PLA(i.e., at t = 0+) the machines share the impact according to their electrical proximity to the point of the impact as expressed by the synchronizing power coefficients. After a brief transient period the same machines share the same impact according to entirely different criteria, namely, according to their inertias.

Example 3.4 Consider the nine-bus, three-machine system of Example 2.6 with a small IO-MW resistive load added to bus 8 as in Example 3.2. Solve the system differential equations and plot PiA and WiA as functions of time . Compare computed results against theoretical values of Section 3.6.

11

Fig. 3.8

P IA

versus t following application of a 10 MW resistive load at bus 8.

Solution A nominal IO-MW (0.1 pu) load is added to bus 8 by applying a three-phase fault through a 10 pu resistance, using a library transient stability program. The resulting power oscillations PiA, i = I, 2, 3, are shown in Figure 3.8 for the system operating without governor action . The prefault conditions at the generators are given in Table 3.1 and in Example 2.6. From the prefault load flow of Figure 2.19 we determine that Vso = 1.016 and oso = 0.7". A matrix reduction of the nine-bus system, retaining only nodes I, 2, 3, and 8, gives the system data shown on Table 3.3.

74

Chapter 3

Table 3.3 Transfer Admittances and Initial Angles of a Nine-Bus System ij

~/jO

0.01826 -0.03530 -0.00965

1-8 2-8 3-8

2.51242 3.55697 2.61601

1.5717 19.0315 12.4752

From (3.24) we compute the synchronizing power coefficients Psile

=

~

Vie (B;1e cos DileO

-

Gi/( sin Dileo)

These values are tabulated in Table 3.4. Note that the error in neglecting the is small. Table 3.4. ik

P jik

(neglecting

Pjik

(with Gik term)

G ik)

2.6961 3.5878 2.6392 8.9231

L Psik

2.6955 3.6001 2.6414 8.9370

r.: (0+) are computed from (3.55) as Pjd(O+) = (p,jS!

where PLtl(O+)

term

Synchronizing Power Coefficients

18 28 38

The values of

Gil
"q is centered on the q axis. Flux linkage >"0 is completely uncoupled from the other circuits, as the first row and column have only a diagonal term. It is important also to observe that the inductance matrix of (4.20) is a matrix of constants. This is apparent since all quantities have only one subscript, thus conforming with our notation for constant inductances. The power of Park's transformation is that it removes the time-varying coefficients from this equation. This is very important. We also note that the transformed matrix (4.20) is symmetric and therefore is physically realizable by an equivalent circuit. This was not true of the transformation used by Park [10,11], where he let 'Odq = Q'abc with Q defined as

Q

2/3

1/2

1/2

1/2

cosO

cos(O - 21(/3)

+ 21r /3) -sin (0 + 21r/3) cos (0

-sin 0 -sin (0 - 21r/3)

(4.22)

Other transformations are found in the literature. The transformation (4.22) is not a power-invariant transformation and does not result in a reciprocal (symmetric) inductance matrix. This leads to unnecessary complication when the equations are normalized. 4.4

Voltage Equations

The generator v.oltage equations are in the form of (4.1). Schematically, the circuits are shown in Figure 4.2, where coils are identified exactly the same as in Figure 4.1 and with coil terminations shown as well. Mutual inductances are omitted from the schematic for clarity but are assumed present with the values given in Section 4.3. Note that the stator currents are assumed to have a positive direction flowing out of the machine terminals, since the machine is a generator. For the conditions indicated we may write the matrix equation v = -ri -

A+

Vn

i

~ _--------,r----a r

a

i

c

r n

l

~

L----------t---t"--r c

i

n

n

'--1.

Fig. 4.2

~

--.--------.-.-- n

Schematic diagram of a synchronous machine.

89

The Synchronous Machine or

'0

0

0

0

r,

0

0

0

-VF

0

0

0

0

0

0

0

0

0

0

0

Vo Vb Vc

-

,

0

0

0

t,

~o

0

0

0

ib

~b

0

0

0

t,

Ae

'F

0

0

iF

~F

0

'D

0

iD

AD

0

0

fQ

iQ

~Q

I I I I r, I I - - - - - - - -1- - - - - - - I 1 I I

,I I

where we define the neutral voltage contribution to

fobe

+t~]

V

(4.23)

as

ic

- R, i abc If

r, = r»

=

-

L, labe V

(4.24)

r, = r as is usually the case, we may also define Rabe

=

,U 3

n

(4.25)

where U 3 is the 3 x 3 unit matrix, and we may rewrite (4.23) in partitioned form as follows: (4.26) where

(4.27)

Thus (4.26) is complicated by the presence of time-varying coefficients in the A term, but these terms can be eliminated by applying a Park's transformation to the stator partition. This requires that both sides of (4.26) be premultiplied by

By definition (4.28) for the left side of (4.26). For the resistance voltage drop term we compute

Chapter 4

90

[:

o] [R abc

UJ

0

o ] [ii ] = R FDQ FDQ abc

~P0 o] UJ

= [PRa;P-

1

o ] [P-

[Rabc

0

0

R FDQ

o ] [i Odq ]

i

R FDQ

=

FDQ

1

0] [ii U

o ] [P

U3

0

3

O df 0 0] [ii ]

[RabC

R

FDQ

abC

]

FDQ

V

(4.29)

FDQ

The second term on the right side of (4.26) is transformed as (4.30)

We evaluate PAabe by recalling the definition (4.7), AOdq = PAabe , from which we compute AOdq = PAabe + PAabe • Then •

•

PAabe =

•

•

PAobc = AOdq

~dq -

-

'-1

PP

Aodq V

(4.31)

We may show that

o pp-I AOdq

0

x,

0

= wOO -

1

o

~d

(4.32)

o which is the speed voltage term. Finally, the third term on the right side of (4.26) transforms as follows: (4.33)

where by definition DOdq is the voltage drop from neutral to ground in the O-d-q coordinate system. Using (4.24), we compute nOdq

= PV n = -PRnP-IPiabc - PLnp-IPlabc = -PRnP-liodq - PLnP-llodq 3rn io

3L n io

o

o o

o

v

(4.34)

and observe that this voltage drop occurs only in the zero sequence, as it should. Summarizing, we substitute (4.28)-(4.31) and (4.33) into (4.26) to write q VOd ] [

vFDQ

= _

rR:bC

l0

0

J r~Odq] _ [~Odq] + [PP-1AOdf1 + A 0 J 0

RFDQ

[n

llFDQ

Odq]

V

(4.35)

FDQ

Note that all terms in this equation are known. The resistance matrix is diagonal. For balanced conditions the zero-sequence voltage is zero. To simplify the notation, let

The Synchronous Machine

R =

[~ ~]

'F

RR

0

0

0

'0

0

0

0

'Q

91

S = r-W~q]

LWAd

Then for balanced conditions (4.35) may be written without the zero-sequence equation as (4.36) 4.5

Formulation of State-Space Equations

Recall that our objective is to derive a set of equations describing the synchronous machine in the form

x=

where

x

=

U =

f

=

(4.37)

I(x, u, t)

a vector of the state variables the system driving functions a set of nonlinear functions

If the equations describing the synchronous machine are linear, the set (4.37) is of the well-known form

x=

Ax + Bu

(4.38)

Examining (4.35), we can see that it represents a set of first-order differential equations. We may now put this set in the form of (4.37) or (4.38), i.e., in state-space form. Note, however, that (4.35) contains flux linkages and currents as variables. Since these two sets of variables are mutually dependent, we can eliminate one set to express (4.35) in terms of one set of variables only. Actually, numerous possibilities for the choice of the state variables are available. We will mention only two that are common: (I) a set based on the currents as state variables; i.e., x' = lid iq iF t o iQ ], which has the advantage of offering simple relations between the voltages Vd and u, and the state variables (through the power network connected to the machine terminals) and (2) a set based on flux linkages as the state variables, where the particular set to be chosen depends upon how conveniently they can be expressed in terms of the machine currents and stator voltages. Here we will use the formulation x' = [Ad Aq AFAO AQ]. 4.6

Current Formulation

Starting with (4.35), we can replace the terms in A and ~ by terms in i and i'as follows. The ~ term has been simplified so that we can compute its value from (4.20), which we rearrange in partitioned form. Let

~~ql [>'FD~

=

[~O;q_ -1- -~~.] ~~~J L nr

I

L FDQ

~FDQ

Wb turns

where L~ is the transpose of Lm • But the inductance matrix here is a constant matrix, so we may write A = Li V, and the ~ term behaves exactly like that of a passive inductance. Substituting this result into (4.35), expanding to full 6 x 6 notation, and rearranging,

Chapter .4

92

Vo

r + 3rn

I

0

0

0:

- - - - - - - T - - - - - - - -- - - - - - - - - - - -

o o o

0

io

wL q

wkM Q

id iF

I

rOO

:I

0

rF

0:

0

0

:

0

0

rD:

0

0

I I I

-WLd

-wkM F

I I I

r

0

iq

0

0

0:

0

rQ

iQ

I

_______. 1.

o o

0

TI - - - - - - - - -

:

I

I I

1.

-wkM D

_

o o o o O' I -------,-------------------1--------o : t., kM F kM D :

L o + 3L n

I I I

o

I I

kM F

LF

MR

I I

II

kM D

MR

LD

'I

o

I I

o

I

I

I I I

0

,

I I ,

Lq

,kMQ

io

id iF

- - - - - - -- T - - - - - - -- - - - - - - - - - - - - T - - - - - - - - -

o o

iD

iD

kM Q

t,

LQ

iQ

V

(4.39) where k = ~ as before. A great deal of information is contained in (4.39). First, we note that the zero-sequence voltage is dependent only upon io and 10 • This equation can be solved separately from the others once the initial conditions on i o are given. The remaining five equations are all coupled in a most interesting way. They are similar to those of a passive network except for the presence of the speed voltage terms. These terms, consisting of wA or wLi products, appear unsymmetrically and distinguish this equation from that of a passive network. Note that the speed voltage terms in the d axis equation are due only to q axis currents, viz., i q and i Q • Similarly, the q axis speed voltages are due to d axis currents, i d, iF, and t». Also observe that all the terms in the coefficient matrices are constants except w, the angular velocity. This is a considerable improvement over the description given in (4.23) in the a-b-c frame of reference since nearly all inductances in that equation were time varying. The price we have paid to get rid of the time-varying coefficients is the introduction of speed voltage terms in the resistance matrix. Since w is a variable, this causes (4.39) to be nonlinear. If the speed is assumed constant, which is usually a good approximation, then (4.39) is linear. In any event, the nonlinearity is never great, as w is usually nearly constant. 4.7

Per Unit Conversion

The voltage equations of the preceding section are not in a convenient form for engineering use. One difficulty is the numerically awkward values with stator voltages in the kilovolt range and field voltage at a much lower level. This problem can be solved by normalizing the equations to a convenient base value and expressing all voltages in pu (or percent) of base. (See Appendix C.) An examination of the voltage equations reveals the dimensional character shown in Table 4.1, where all dimensions are expressed in terms of a v-i-t (voltage, current, time) system. [These dimensions are convenient here. Other possible systems are

93

The Synchronous Machine

FLtQ (force, length, time, charge) and M LtJJ (mass, length, time, permeability).] Observe that all quantities appearing in (4.39) involve only three dimensions. Thus if we choose three base quantities that involve all three dimensions, all bases are fixed for all quantities. For example, if we choose the base voltage, base current, and base time, by combining these quantities according to column 4 of Table 4.1, we may compute base quantities for all other entries. Note that exactly three base quantities must be chosen and that these three must involve all three dimensions, v, i, and t. Electrical Quantities, U nits, and Dimensions

Table 4.1. Quantity

Symbol

Units

Voltage Current Power or voltamperes

v i p or S

volts (V) amperes (A) watts (W) voltamperes (VA) weber turns (Wb turns) ohm (0) henry (H) second (s) radians per second (rad/s) radian (rad)

Flux linkage Resistance Inductance Time Angular velocity

A

,

LorM t

w

Angle

(J

or 0

v-i-t

Dimensions

Relationship

[v] [i]

[vi] [vt] [vii] [vtli] [1 ] [I It]

p

= vi

v = A v = ri

v = Ll

dimensionless

Choosing a base for stator quantities

4.7.1

The variables v«. vq , i«, iq , Ad, and Aq are stator quantities because they relate directly to the a-b-c phase quantities through Park's transformation. (Also see Rankin [15], Lewis [12] and Harris et al. [9] for a discussion of this topic.) Using the subscript 8 to indicate "base" and R to indicate "rated," we choose the following stator base quantities. Let Sa VB

Wa

= =

=

SR VR

WR

stator rated VA/phase, VArms stator rated line-to-neutral voltage, V rms = generator rated speed, elec rad/s

= =

(4.40)

Before proceeding further, let us examine the effect of this choice on the d and q axis quantities. First note that the three-phase power in pu is three times the pu power per phase (for balanced conditions). To prove this, let the rms phase quantities be V~ V and ItJ..A. Thethree-phasepoweris3Vlcos(a -,,)W. Thepu power P3q, is given by (4.41)

where the subscript u is used to indicate pu quantities. To obtain the d and q axis quantities, we first write the instantaneous phase voltage and currents. To simplify the expression without any loss of generality, we will assume that va(t) is in the form,

u, = Vmsin(O + a) = VIVsin(O + a) V u, = V2Vsin(lJ + a - 2tr/3) V V c = V2Vsin(lJ + a + 2tr/3) V Then from (4.5), 'Odq

=

P'abc or

(4.42)

94

Chapter 4

o

Vo

V3Vsina

V

(4.43)

V3 Vcosa

Vq

In pu (4.44)

Similarly, V qu =

V3Vucosa

(4.45)

3 V II2

(4.46)

Obviously, then 2

Vdu

+ vqu2 =

The above results are significant. They indicate that with this particular choice of the base voltage, the pu d and q axis voltages are numerically equal to V3 times the pu phase voltages. Similarly, we can show that if the rms phase current is f!J.. A, the corresponding d and q axis currents are given by,

V3/sin'Y

A

(4.47)

V3f cos 'Y

and the pu currents are given by idu =

V1fu sin 'Y

(4.48)

To check the validity of the above, the power in the d and q circuits must be the

same as the power in the three stator phases, since P is a power-invariant transformation. PJt!'

= iduvt/u + iquvqu = 3 lu Jt:(sin a sin l' + cos a cos 1') = 3 f u Jt: cos (a - 1') pu

(4.49)

We now develop the relations for the various base quantities. From (4.40) and Table 4.1 we compute the following:

t, = SBI VB =

SRI VR Arms VR/wR = L B f B Wb turn

= VBt B = R B = VBIla = VRIIR ~B

n

Thus by choosing the three base quantities SB' VB, and tB' we can compute base values for all quantities of interest. To normalize any quantity, it is divided by the base quantity of the same dimension. For example, for currents we write iu = i(A)1 f B (A) pu

(4.51 )

where we use the subscript u to indicate pu. Later, when there is no danger of ambiguity in the notation, this subscript is omitted.

95

The Synchronous Machine

4.7.2

Choosing a base for rotor quantities

Lewis [12] showed that in circuits coupled electromagnetically, which are to be normalized, it is essential to select the same voltampere and time base in each part of the circuit. (See Appendix C for a more detailed treatment of this subject.) The choice of equal time base throughout all parts of a circuit with mutual coupling is the important constraint. It can be shown that the choice of a common time base to forces the VA base to be equal in all circuit parts and also forces the base mutual inductance to be the geometric mean of the base self-inductances if equal pu mutuals are to result; i.e., M I2B = (L'BL2B)If2. (See Problem 4.18.) For the synchronous machine the choice of So is based on the rating of the stator, and the time base is fixed by the rated radian frequency. These base quantities must be the same for the rotor circuits as well. It should be remembered, however, that the stator VA base is much larger than the V A rating of the rotor (field) circuits. Hence some rotor base quantities are bound to be very large, making the corresponding pu rotor quantities appear numerically small. Therefore, care should be exercised in the choice of the remaining free rotor base term, since all other rotor base quantities will then be automatically determined. There is a choice of quantities, but the question is, Which is more convenient? To illustrate the above, consider a machine having a stator rating of 100 x 106 VAl phase. Assume that its exciter has a rating of 250 V and 1000 A. If, for example, we choose I RB = 1000 A, VRB will then be 100,000 V; and if we choose VRB = 250 V, then I RB will be 400,000 A. Is one choice more convenient than the other? Are there other more desirable choices? The answer lies in the nature of the coupling between the rotor and the stator circuits. It would seem desirable to choose some base quantity in the rotor to give the correct base quantity in the stator. For example, we can choose the base rotor current to give, through the magnetic coupling, the correct base stator flux linkage or open circuit voltage. Even then there is some latitude in the choice of the base rotor current, depending on the condition of the magnetic circuit. The choice made here for the free rotor base quantity is based on the concept of equal mutual flux linkages. This means that base field current or base d axis amortisseur current will produce the same space fundamental of air gap flux as produced by base stator current acting in the fictitious d winding. Referring to the flux linkage equations (4.20) let id = I B, iF = I FB, and ;0 = 10 0 be applied one by one with other currents set to zero. If we denote the magnetizing inductances ( t = leakage inductances) as L md ~ L d - ~d H

L mq ~ L, - ~ q H

L mF ~ L F - {F H

L mQ ~ L Q

-

~

Q

H

L mo ~ L o - {D H

(4.52)

and equate the mutual flux linkages in each winding, Amd = LmdI B

kMFIFB = kMol o o Wb

=

AmF = kMF/B = LmFIFB AmO = kMo/B

=

M R lDB Wb MRIFB = Lmolo B Wb

Then we can show that

Amq = Lmql B = kMQIQB Wb AmQ = kM Q I B = LmQIQ B Wb (4.53)

96

Chapter 4 Lmdli Lmql~

= LmFI}B = Lmo/~B = kMF/B/ FB = kMoIB/DB = MRIFBloB = kMQIBlQB = LmQl~B

(4.54)

and this is the fundamental constraint among base currents. From (4.54) and the requirement for equal SB' we compute VFB/VB = IBII FB = (L mFILmd) 1/2 = kMF/L md = LmF/kMF = MR/kM o ~ k., VOB/VB = IB/lDB = (L moILmd) I/2 = kMolL md = LmDlkMo = MR/kM F ~ k o VQol VB = 101lQO = (LmQI L mq)1/2 = k M QIL mq = LmQ/k M Q ;1 k Q

These basic constraints permit us to compute R FB = k~RB

{1

L Fo = k~LB H

ROB

=

kbR B

n

L oo = kbLB H

R QB = k~RB

(4.55)

n

L QB = k~Lo H

(4.56)

and since the base mutuals must be the geometric mean of the base self-inductances (see Problem 4.18), Moo

= koL o H (4.57)

4.7.3

Comparison with other per unit systems

The subject of the pu system used with synchronous machines has been controversial over the years. While the use of pu quantities is common in the literature, it is not always clear which base quantities are used by the authors. Furthermore, synchronous machine data is usually furnished by the manufacturer in pu. Therefore it is important to understand any major difference in the pu systems adopted. Part of the problem lies in the nature of the original Park's transformation Q given in (4.22). This transformation is not power invariant: i.e., the three-phase power in watts is given by Pahc = 1.5 (idVd + iqv q). Also, the mutual coupling between the field and the stator d axis is not reciprocal. When the Q transformation is used, the pu system is chosen carefully to overcome this difficulty. Note that the modified Park's transformation P defined by (4.5) was chosen specifically to overcome these problems. The system most commonly used in the literature is based on the following base quantities: SB = three-phase rated V A VB = peak rated voltage to neutral I B = peak rated current and with rotor base quantities chosen to give equal pu mutual inductances. This leads to the relations lFB

=

V2(L md/ M F)/o

VFB = (3/v2)(M F / L md) VB

This choice of base quantities, which is commonly used, gives the same numerical values in pu for synchronous machine stator and rotor impedances and self-inductances

as the system used in this book. The pu mutual inductances differ by a factor of vTj2. Therefore, the terms kM F used in this book are numerically equal to M F in pu as found in the literature. The major differences lie in the following:

I. Since the power in the d and q stator circuits is the three-phase power, one pu current and voltage gives three pu power in the system used here and gives one pu power in the other system.

97

The Synchronous Machine

2. In the system used here v~u + v;u = 3 V~, while in the other system v~u V~, where Vu is the pu terminal voltage.

+ v;u =

The system used here is more appealing to some engineers than that used by the manufacturers [9,12]. However, since the manufacturers' base system is so common, there is merit in studying both. Example 4./ Find the pu values of the parameters of the synchronous machine for which the following data are given (values are for an actual machine with some quantities, denoted by an asterisk, being estimated for academic study):

Rated MVA = 160 MVA Rated voltage = 15 kV, Y connected Excitation voltage = 375 V Stator current = 6158.40 A Field current = 926 A Power factor = 0.85 Ld

=

6.341

X

LQ

= 1.423

{d = {q(unsaturated) =

kM o = kM Q =

r(125°C) = 'F( 125°C) = ro =

10- 3 H

L F=2.189 H L o = 5.989 X 10- 3 H* L, = 6.118 x 10- 3 H

rQ

Inertia constant

=

=

10- 3 H* 0.5595 x 10- 3 H 5.782 x 10- 3 H* 2.779 x 10- 3 H* 1.542 x 10- 3 0 0.371 Q 18.421 x 10- 3 g* 18.969 x 10- 3 0* 1.765 kW·s/hp X

From the no-load magnetization curve, the value of field current corresponding to the rated voltage on the air gap line is 365 A. Solution:

Stator Base Quantities: SB = 160/3 = 53.3333 MVA/phase VB = 15000/v'1 = 8660.25 V f B = 6158.40 A

t8 AB

2.6526 X 10- 3 s = 8660 x 2.65 x 10- 3 = 22.972 Wb turn/phase B« = 8660.25/6158.40 = 1.406 n L 8 = 8660/(377 x 6158) = 3.730 x 10- 3 H L md = L d - -.e d = (6.341 - 0.5595) 10- 3 = 5.79 X 10- 3 H =

To obtain M F , we use (4.11), (4.16), and (4.23). At open circuit the 'mutual inductance L aF and the flux linkage in phase a are given by L aF = M F cos

f)

Aa

=

iFMFcos f)

The instantaneous voltage of phase a is va = iFwRMFsin (), where WR is the rated synchronous speed. Thus the peak phase voltage corresponds to the product iFwRMF. From the air gap line of the no-load saturation curve, the value of the field current at rated voltage is 365 A. Therefore, M F = 8660V!/(377 x 365) = 89.006 x 10- 3 H kM F = V3f2 x 89.006 X 10- 3 = 109.01 X 10- 3 H

Then k F = kM F/ L md = 18.854. Then we compute, from (4.55)-(4.57), f F8 = 6158.4/18.854 = 326.64 A

M F 8 = 18.854 x 3.73 x 10- 3 = 70.329

X

10- 3 H

98

Chapter .4

= (53.33 x 106)/326.64 = 163280.68 V

VF S

= 499.89 n

R F S = 163280.68/326.64

L FB

=

(18.845)2 x 3.73

X

10- 3 = 1.326 H

Amortisseur Base Quantities (estimated for this example): kMo/L md = 5.781/5.781

L DB = L B H ROB = R B n R Q B = R s/4 = 0.352 n

1.00

=

MOB = L B H kMQ/L mq = 2.779/5.782 = 0.5 L QB = L B/4 = 0.933 x 10- 3 H Inertia Constant: H

= 1.765(1.0/0.746)

= 2.37 kW ·s/kVA

The pu parameters are thus given by: Ld

=

6.34/3.73 = 1.70

L F = 2.189/1.326 = 1.651 Lo

=

{,d

=

= 1.605 {q = 0.5595/3.73 = 0.15 5.989/3.730

L q = 6.118/3.73

1.64

=

L Q = 1.423/0.933 = 1.526

LAo

=

L AQ =

kM D = kMF = M R kM Q = 1.64 - 0.15

= 1.70 - 0.15 =

1.55

1.49

r = 0.001542/1.406 = 0.001096

'F = 0.371/499.9

'0

'Q

The quantities 4.7.4

LAO

and

=

0.000742

= 0.018/1.406 = 0.0131

= 18.969 x 10- 3/0.351 = 0.0540 L AQ

are defined in Section 4. I I.

The correspondence of per unit stator EMF to rotor quantities

We have seen that the particular choice of base quantities used here 'gives pu values of d and q axis stator currents and voltages that are VJ times the rms values. We also note that the coupling between the d axis rotor and stator involves the factor k = V3j2, and similarly for the q axis. For example, the contribution to the d axis stator flux linkage "d due to the field current lr is kM FiF and so on. In synchronous machine equations it is often desirable to convert a rotor current, flux linkage, or voltage to an equivalent stator EM F. These expressions are developed in this section. The basis for converting a field quantity to an equivalent stator EM F is that at open circuit a field current iF A corresponds to an EMF of iFwRMF V peak. If the rrns value of this EMF is E, then iFwRMF = v1 E and iFwR kMF = V1 E in MKS units.' 2. The choice of symbol for the EMF due to iF is not clearly decided. The American National Standards Institute (A N SI) uses the sym bol E1 (16). A new proposed standard uses Eo! (17). The International Electrotechnical Commission (I EC), in a discussion of [17], favors Eq for this voltage. The authors leave this voltage unsubscripted until a new standard is adopted.

99

The Synchronous Machine

Since MF and WR are known constants for a given machine, the field current corresponds to a given EM F by a simple scaling factor. Thus E is the stator air gap rms voltage in pu corresponding to the field current i, in pu. We can also convert a field flux linkage AF to a corresponding stator EM F. At steady-state open circuit conditions AF = LFiF, and this value of field current iF' when multiplied by wRM F , gives a peak stator voltage the rrns value of which is denoted by We can show that the d axis stator EM F corresponding to the field flux linkage AF is given by

E;.

(4.58) By the same reasoning a field voltage uF corresponds (at steady state) to a field current UF/'F' This in turn corresponds to a peak stator EMF (uF/rF)wRMF. If the rrns value of this EM F is denoted by EFD , the d axis stator EM F corresponds to a field voltage UF or (4.59)

4.8

Normalizing the Voltage Equations

Having chosen appropriate base values, we may normalize the voltage equations (4.39). Having done this, the stator equations should be numerically easier to deal with, as all values of voltage and current will normally be in the neighborhood of unity. For the following computations we add the subscript u to all pu quantities to emphasize their dimensionless character. Later this subscript will be omitted when all values have been normalized. The normalization process is based on (4.51) and a similar relation for the rotor, which may be substituted into (4.39) to give UO u VB

r

+ 3'n

0

0

0

0

0

iouIB

v; VB

0

r

.st;

0

0

wkM Q

itiuIB

Vqu VB

0

-wLd

r

-wkMF

-wkM D

0

iqulo

-VFuVFB

0

0

0

0

0

iFuIFB

0

'F

0

0

0

0

'D

0

iOuIDB

0

0

0

0

0

0

'Q

iQu/QB

Lo

+ 3L n

0

0

0

Ld

0

0

0

Lq

0

0

0

kM F

0

iF

MR

0

iFuIFB

0

kM D

0

MR

Lo

0

ioul DB

0

0

kM Q

0

0

LQ

IQu l QB

0

0

kM F kM o

0

lOu/B

0

itiuIB

kM Q iquIB

(4.60)

where the first three equations are on a stator base and the last three are on a rotor base. Examine the second equation more closely. Dividing through by VB and setting w = WuWR, we have

100

Chapter 4

(4.61) Incorporating base values from (4.50), we rewrite (4.61) as

vdu = - -Rr ld. u B

-

W

u

~ .

Ls

I

qu

I QB k M . Q'Qu Va

WR

Wu

-

-

Ld La

:

- - 'du W

kM F wR/ F O

R

:

-----IF, WR Vo u

kM D wRID o

:

- - - - - ID WR VB u

pu

(4.62)

We now recognize the following pu quantities.

r, L qu

= rlR o

L du =

LdlL B

M Fu

=

MFwRIFB/VB

MDwR1Ds/VB

M Qu =

MQwRIQB/VB

M Du =

= Lq/L B

(4.63)

Incorporating (4.63), the d axis equation (4.62) may be rewritten with all values except time in pu; i.e.,

vdu

=

.

-ru1du -

Wu

L· qu1qu

-

kW M ' Qu1Qu u

L du

-

kMFu : I

:

-ldu WR

WR

kMDu : I

Fu -

The third equation of(4.60) may be analyzed in a similar way to write

WR

Du

(4.64)

(4.65) where all pu coefficients have been previously defined. The first equation is uncoupled from the others and may be written as VOu =

-

r + 3r,.. 'Ou RB

-

L o + 3L n wRL B

- (r + 3r,,)u ;ou -

: 'OU

~R (L o +

3L,,)u lou pu

(4.66)

If the currents are balanced, it is easy to show that this equation vanishes. The fourth equation is normalized on a rotor basis and may be written from (4.60) as (4.67) We now incorporate the base rotor inductance to normalize the last two terms as (4.68) The normalized field circuit equation becomes (4.69) The damper winding equations can be normalized by a similar procedure. following equations are then obtained, V Du

= O=

.

rDui Du

k M Du

:

M Ru ~

L Du

+ - - 'du + - - ' Fu + WR

WR

WR

:

' Du

The

(4.70)

The Synchronous Machine

101

(4.71) These normalized equations are in a form suitable for solution in the time domain with time in seconds. However, some engineers prefer to rid the equations of the awkward l/wR that accompanies every term containing a time derivative. This may be done by normalizing time. We do this by setting 1 d WR

dt

d

(4.72)

dr

=

where (4.73) is the normalized time in rad. Incorporating all normalized equations in a matrix expression and 'dropping the subscript u since all values are in pu, we write

r

Vd

-vF 0

vq 0

-

0

'F

0

0

,I wLq

0

0

I I I I I

0

'0

wkMQ

id

0

0

iF

0

0

to

------------------~---------

-wL d

-wkMF

-wkMo

0

0

0

I I I I

l

r

0

iq

0

'Q

'o

Ld

kM F

kM D

kMF

LF

MR

kMo

MR

Lo

0

0

I I I

I I I I

0

0

id

0

0

iF

0

0

iD

-------------~---------

0

0

I

0

I

Lq

kM Q

iq

0

I I

kM Q

LQ

iQ

I

pu

(4.74)

where we have omitted the Vo equation, since we are interested in balanced system conditions in stability studies, and have rearranged the equations to show the d and q coupling more clearly. It is important to notice that (4.74) is identical in notation to (4.39). This is always possible if base quantities are carefully chosen and is highly desirable, as the same equation symbolically serves both as a pu and a "system quantity" equation. Using matrix notation, we write (4.74) as v

=

-(R

+ wN) i - L i pu

(4.75)

where R is the resistance matrix and is a diagonal matrix of constants, N is the matrix of speed voltage inductance coefficients, and L is a symmetric matrix of constant inductances. If we assume that the inverse of the inductance matrix exists, we may write

i

=

-L- ' (R

+ wN)i - L-'v pu

(4.76)

This equation has the desired state-space form. It does not express the entire system behavior, however, so we have additional equations to write. Equation (4.76) may be depicted schematically by the equivalent circuit shown in

102

CJ d ("Q ·"ciNo}L__ -CJy: . Chapter 4

,

'F

+~.

~ :----kMF + Q~R kM~

F Y _

r

Yo = 0 _

-

r

----- L

')

'o

_ Id

L0 : : . -

a

kM

a

-

1J!),q ,

+

--I

q

,-

.q

+

lJ!~ d

-

Fig. 4.3 Synchronous generator d-q equivalent circuit

Figure 4.3. Note that all self and mutual inductances in the equivalent circuit are constants, and pu quantities are implied for all quantities, including time. Note also the presence of controlled sources in the equivalent. These are due to speed voltage terms in the equations. Equation (4.74) and the circuit in Figure 4.3 ditTer from similar equations found in the literature in two important ways. In this chapter we use the symbols Land M for self and mutual inductances respectively. Some authors and most manufacturers refer to these same quantities by the symbol x or X. This is sometimes confusing to one learning synchronous machine theory because a term XI that appears to be a voltage may be a flux linkage. The use of X for L or M is based on the rationale that w is nearly constant at 1.0 pu so that, in pu, X = wL - L. However, as we shall indicate in the sections to follow, w is certainly not a constant; it is a state variable in our equations, and we must treat it as a variable. Later , in a linearized model we will let w be approximated as a constant and will simplify other terms in the equations as well. For convenience of those acquainted with other references we list a comparison of these inductances in Table 4.2. Here the subscript notation kd and kq for D and Q reo spectively is seen. These symbols are quite common in the literature in reference to the damper windings . Table 4.2.

Comparison of Per Unit Inductance Symbols

Chapter 4 Kimbark [2J Concordia [I)

Xkdd

Example 4.2 Consider a 60-Hz synchronous machine with the following pu parameters: L, = 1.70 L, = 1.64

= LD = LQ = kMF = ,fd = LF

1.65 1.605 1.526 M R = kM D ,fq

= 0.15

kMQ = 1.49 r = 0.001096

= 0.000742 rD = 0.0131 rQ = 0.0540 H = 2.37 S rF

1.55

103

The Synchronous Machine

Solution

From (4.75) we have numerically

R + wN

0.0011

0

0

0

0.00074

0

0

0

0.0131

-

I

- - - - - - - - - - - - - - - -

-I.55w

-1.70w

I I I I I I

1.64w

1.49w

0

0

0

-,-, _.

0 0

-1.55w : 0.0011 I I

0

0.0540

1.55

0

0

1.65

1.55

0

0

1.55

1.55

1.605

0

0

0

0

0

1.64

1.49

0

0

0

1.49

1.526

0

0

0

1.70

1.55

1.55

pu

- - - - - - -

I I I I I I I I - - - - - - - - - - - - - - - - - -1- - - -- - - - - -

L =

I I I I

pu

from which we compute by digital computer

5.405

-3.414

- 1.869 7.110

-1.869

-5.060

I I

I I

I I I I

0

0

0

0

0 0 -3.414 -5.060 8.804 -- - - - - -- - - - - - - - - - - - r - - - - - - - - - - I 0 5.406 -5.280 0 0 I

L-'

0

0

0

I I -

5.280

pu

5.811

Then we may compute

-L-'(R

+ wN)

=

10- 3

-5.9269

1.3878

2.0498

-5.2785

3.7433

3.7564

44.7198: -8864.9w

-8504.1w

66.2818:·

3065.9w

2785.4w

-115.3290:

5598.9w

5086.8w

I

- - - - - - - - - - - - - - - - - - - - - - ...... 1I

9190.9w

8379.9w

-8975.2w

-8183.3w

8379.9w

I I"

-8183.3w:

_

-5.9279

284.857

5.7888

-313.534

pu

and the coefficient matrix is seen to contain w in 12 of its 25 terms. This gives some idea of the complexity of the equations.

4.9

Normalizing the Torque Equations In Chapter 2 the swing equation

J{j

=

(2J/p)w = To Nvm

(4.77)

is normalized by dividing both sides of the equation by a shaft torque that corresponds to the rated three-phase power at rated speed (base three-phase torque). The result of this normalization was found to be

Chapter 4

104

(2H IwR)w

=

To pu(3¢)

where w

=

angular velocity of the revolving magnetic field in elec rad/s

To

=

accelerating torque in pu on a three-phase base

H

=

(4.78)

WR /SB3 s

and the derivative is with respect to time in seconds. This normalization takes into account the change in angular measurements from mechanical to electrical radians and divides the equations by the base three-phase torque. Equation (4.78) is the swing equation used to determine the speed of the stator revolving MMF wave as a function of time. We need to couple the electromagnetic torque Te , determined by the generator equations, to the form of (4.78). Since (4.78) is normalized to a three-phase base torque and our chosen generator VA base is a per phase basis, we must use care in combining the pu swing equation and the pu generator torque equation. Rewriting (4.78) as

(2H / WB)W = Tm

-

T, pu(3¢)

(4.79)

the expression used for T, must be in pu on a three-phase VA base. Suppose we define Teq, =

pu generator electromagnetic torque defined on a per phase V A base Tt(N - m )/(SB/WB) pu

(4.80)

Then (4.81) (A similar definition could be used for the mechanical torque; viz., Tmt/J = 3Tm. Usually, Tm is normalized on a three-phase basis.) The procedure that must be used is clear. We compute the generator electromagnetic torque in N em. This torque is normalized along with other generator quantities on a basis of SB' VB' lB' and l B to give Teq,. Thus for a fully loaded machine at rated speed, we would expect to compute Tetl> = 3.0. Equation (4.81) transforms this pu torque to the new value Te , which is the pu torque on a three-phase basis. 4.9.1

The normalized swing equation

In (4.79), while the torque is normalized, the angular speed wand the time are given in M KS units. Thus the equation is not completely normalized. The normalized swing equation is of the form given in (2.66) (4.82) where all the terms in the swing equation, including time and angular speed, are in pu. Beginning with (4.79) and substituting (4.83) we have for the normalized swing equation 2H WB dwu = Tau diu

(4.84)

thus, when time is in pu, (4.85)

105

The Synchronous Machine

4.9.2

Forms of the swing equation

There are many forms of the swing equation appearing in the literature of power system dynamics. While the torque is almost always given in pu, it is often not clear which units of wand t are being used. To avoid confusion, a summary of the different forms of the swing equation is given in this section. We begin with ta in rad/s and t in s, (2H/wa)w = Tau. If t and Ta are in pu (and w in rad/s), by substituting tu = wat in (4.79), 2H dw = 2H dw = Tau pu wo dt dt;

(4.86)

If wand T; are in pu (and t in s), by substituting in (4.79), 2H dwu dt

=

T

au

pu

(4.87)

If t, w, and T; are all in pu, (4.88) If w is given in elec deg/s, (4.79) and (4.86) are modified as follows:

---!!- dw 180fo dt

=

T

au

pu

1rH dw = T u 90 dt au p u

(4.89) (4.90)

It would be tempting to normalize the swing equation on a per phase basis such that all terms in (4.79) are in pu based on So rather than S03. This could indeed be done with the result that all values in the swing equation would be multiplied by three. This is not done here because it is common to express both Tm and T, in pu on a threephase base. Therefore, even though So is a convenient base to use in normalizing the generator circuits, it is considered wise to convert the generator terminal power and torque to a three-phase base S03 to match the basis normally used in computing the machine terminal conditions from the viewpoint of the network (e.g., in load-flow studies). Note there is not a similar problem with the voltage being based on Vo, the phase-to-neutral voltage, since a phase voltage of k pu means that the line-to-line voltage is also k pu on a line-to-line basis. 4.10

Torque and Power

The total three-phase power output of a synchronous machine is given by (4.91) where the superscript t indicates the transpose of 'abc. But from (4.8) we may write iabc = p-l iOdq with a similar expression for the voltag,e vector. Then (4.91) becomes Pout

=

'hdq (P-I)' p-I iOdq

Performing the indicated operation and recalling that P is orthogonal, we find that

Chapter 4

106

the power output of a synchronous generator is invariant under the transformation P; i.e., (4.92)

For simplicity we will assume balanced but not necessarily steady-state conditions. Thus Vo = io = 0 and Pout =

vdid

+ vqiq (balanced condition)

(4.93)

Substituting for o, and vq from (4.36), Pout

= (id~d + iq~q) + (iq~d - id~q)W - r(i~ + i~)

(4.94)

Concordia [I] observes that the three terms are identifiable as the rate of change of stator magnetic field energy, the power transferred across the air gap, and the stator ohmic losses respectively. The machine torque is obtained from the second term, (4.95)

The same result can be obtained from a more rigorous derivation. Starting with the three armature circuits and the three rotor circuits, the energy in the field is given by 6

WAd

= I: ! k-I j-I

2

(4.96)

(ikijL kj)

which is a function of O. Then using T = awftd/ao and simplifying, we can obtain the above relation (see Appendix 8 of [I D. Now, recalling that the flux linkages can be expressed in terms of the currents, we write from (4.20), expressed in pu, ~d =

Ldid + kMFi F + kMDi D

(4.97)

Then (4.95) can be written as id

iF i D pu

(4.98)

iq iQ

which we recognize to be a bilinear term. Suppose we express the total accelerating torque in the swing equation as

Ta = Tm

-

Tet/J/3 - Td = Tm

-

T, - Td

(4.99)

where T'; is the mechanical torque, T, is the electrical torque, and Td is the damping torque. It is often convenient to write the damping torque as Td

=

Dw pu

(4.100)

where D is a damping constant. Then by using (4.81) and (4.98), the swing equation may be written as

The Synchronous Machine

107

id iF

[

~i 3r.

-

)

kMo . 3r.) q

-

---I

q

~] Tj

io iq iQ

w (4.101) where Tj is defined by (4.85) and depends on the units used for wand t. following relation between () and w may be derived from (4.6). ~

Finally, the (4.102)

1

= w -

Incorporating (4.101) and (4.102) into (4.76), we obtain

itJ iF

iD

I I I

r

iQ - - - -

A

I I I I I I I - - - - - - - - - - - - - - - - - - - - - - - - - - -1- - - - - - - -

_ LtJiq 3 Tj

o

o

I

-L-1(R + wN)

_ kMFi q

_ kMoiq

3 Tj

.LqitJ

kMQi d

3 Tj

3 Tj

3 Tj

o

I

D

: I

I I I

o

o

o

_L- 1,

t,

I

iq

W

id

,

;D iq

iQ

o

w

o

5

+ Tm Tj

-1

(4.103) This matrix equation is in the desired state-space form x = f(x, U, t) as given by (4.37). It is clear from (4.101) that the system is nonlinear. Note that the "inputs" are , and Tm • 4.11

Equivalent Circuit of a Synchronous Machine

For balanced conditions the normalized flux linkage equations are obtained from (4.20) with the row for Ao omitted.

Ad Aq

Ld

0

kMF

kM o

0

Lq

0

0

AF

kMF

0

LF

MR

0

iF

AD AQ

kM v

0

MR

Lo

0

iD

0

kMQ

0

0

LQ

iQ

We may rewrite the d axis flux linkages as

0

id

kM Q i q (4.104)

108

Chapter 4

Ad = [(Ld - ~d) + ~d) 1d + kMFi F + kMoi o AF = kMFid + (L F - {F) + ~FI iF + MRi o AO = kMoid + MRi F + (L o - {D) + {oj i o

(4.105)

where {d' {F' and {o are the leakage inductances of the d, F, and D circuits respectively. Let iF = t; = 0, and the flux linkage that will be mutually coupled to the other circuits is Ad - {did' or (L d - {d)id. As stated in Section 4.7.2, Ld - {d is the magnetizing inductance Lmd. The flux linkage mutually coupled to the other' d axis circuits is then Lmdid. The flux linkages in the F and D circuits, AF and Ao, are given in this particular case by AF = kMFid, and Ao = kMoi d. From the choice of the base rotor current, to give equal mutual flux, we can see that the pu values of Lmdid, AF, and Ao must be equal. Therefore, the pu values of Lmd, kM F• and kM o are equal. This can be verified by using (4 .57) and (4.55),

kMFu = kM F = k kM F = Lmd = Lmdu MFa (MdLmd)L a La

(4.106)

In pu, we usually call this quantity LAO ; i.e.,

LAO ~ Ld - {d

= kMF = kMo

pu

(4.107)

We can also prove that, in pu ,

LAO = L o - {o = L F - {F = Ld - {d = kM F = kM o = M R

(4.108)

Similarly, for the q axis we define

L AQ ~ L, - ~q = L Q -

t Q=

kM Q pu

(4.109)

If in each circuit the pu leakage flux linkage is subtracted, the remaining flux linkage is the same as for all other circuits coupled to it. Thus (4.110) where

AAO = iALd - td) + kMFi F + kMoi o = LAO(id + iF + io) pu

(4.111)

Similarly, the pu q axis mutual flux linkage is given by

AAQ

= (L,

- {q)iq + kMQiQ = LAQ(iq

+ iQ)

(4.112)

Following the procedure used in developing the equivalent circuit of transformers, we can represent the above relations by the circuits shown in Figure 4.4, where we note that the currents add in the mutual branch . To complete the equivalent circuit, we

Fig .4.4

Flux linkage ind ucta nces of a synchronous mach ine.

109

The Synchronous Machine t

'F

___ i

td

F

d

t+

v

td

Fig . 4.5

Direct axis equivalent circuit.

consider the voltage equations

vd = - r id

-

Ad - WA q

-rid - ,f)d - [(Ld - ,fd)id

+ kMFi F + kMoio) - W\

or

u, = -rid - ,f)d - LAO(id + IF + 10 )

-

WA q

(4 .113)

Similarly, we can show that

-v F = -rFiF - ,fFi F - LAO(id + iF + io)

(4 .114)

+ iF + io)

(4.115)

Vo = 0 = -roio - ,folo - LAD(id

The above voltage equations are satisfied by the equivalent circuit shown in Figure 4.5 . The three d axis circuits (d, F, and D) are coupled through the common magnetizing inductance LAD' which carries the sum of the currents id, iF ' and io. The d axis circuit.contains a controlled voltage source WA q with the polarity as shown. Similarly, for the q axis circuits Vq =

-

VQ =

0

+ iQ) + WAd -rQiQ - ,fQiQ - LAQ(iq + iQ)

r iq - ',fi q - LAQ(iq =

(4.116) (4 .117)

These two equations are satisfied by the equivalent circuit shown in Figure 4.6 . Note the presence of the controlled source WAd in the stator. q circuit. 4.12

The Flux Linkage State-Space Model

We now develop an alternate state-space model where the state variables chosen are Ad' AF' Ao, s.. and AQ . From (4.110)

id = (lj,fd)(Ad - AAO) but from (4.111) AAO

=

t, = (Ij,fF)(AF - AAO)

io = (I/,fO)(AO - AAO)

(4.118)

(id + iF + io) LAO' which we can incorporate into (4.118) to get ~

t

q

va = 0

Fig .4 .6

Quadrature axis equivalent circuit.

Chapter 4

110

Now define (4.119) then ~AD

=

(LMD/{d)~d

(4.120)

+ (LMO/,fF)'AF + (LMO/{O)~o

Similarly, we can show that (4.121) where we define

l/L MQ ~ l/L AQ + l/,fq + I/{,Q

(4.122)

and the q axis currents are given by

;q

== (1/{q)(A q - 'AAQ)

;Q

=

(4.123)

(l/~Q)(AQ - ~AQ)

Writing (4.118) and (4.123) in matrix form, I

0

0

-I/{d:

I/{F

0

-I/{,F:

o

l/~D

l/{d

o o

~d

I

-l/~o

>"F

o

'A o

I

I I I

>"AO

,

--------------------~-------------I

o

: 0 I

4.12.1

0

-l/-f q

I/{Q

-I/{Q

: l/{q

(4.124)

>"q

AQ AAQ

The voltage equations

The voltage equations are derived as follows from (4.36). For the d equation vd

=

-rid -

s; - wA

(4.125)

q

Using (4.124) and rearranging, ~d

= -r(Ad/~d - AAO/{d) - WA q

-

o,

or (4.126) Also from (4.36) (4.127) Substituting for iF or (4.128)

The Synchronous Machine

111

Repeating the procedure for the D circuit,

Xo

=

-(ro/{o)'A o

+ (rO/{O)'AAo

(4.129)

The procedure is repeated for the q axis circuits. For the uq equations we compute

Xq

-(r/{q}Aq

=

+ (r/{q)AAQ + wAd - uq

(4.130)

and from the q axis damper-winding equation,

XQ = -(rQ/{Q)A Q + (rQ/{Q)'AAQ

(4.131)

Note that AAD or AAQ appears in the above equations. This form is convenient if saturation is to be included in the model since the mutual inductances LAo and L AQ are the only inductances that saturate. If saturation can be neglected the AAO and AAQ terms can be eliminated (see Section 4.12.3). 4.12.2

pute

The torque equation

From (4.95) Tet/J

iqAd - idA q. Using (4.124), we substitute for the currents to com-

=

o, {q - AAQ)

+ \

\ (Ad - AAO) T•• = _ I\q {d

I\d\

1

=

- {q >"d>"AQ

I >"q>"AD + ( {q 1 - {d 1 ) >"d>"q + {d

(4.132)

We may also take advantage of the relation {q = {d (called {Q in many references). The new electromechanical equation is given by

w=

- ( AAD / {

d

3T j) Aq

+ (AA Q / rE q 3T j) Ad

- (D / Tj) W

+ Tm / Tj

( 4. 133)

Finally the equation for 0 IS given by (4.102). Equations (4.126)-(4.131), (4.133), and (4.102) are in state-space form. The auxiliary equations (4.120) and (4.121) are needed to relate AAD and AAQ to the state variables. The state variables are Ad, AF , AD, Aq, AQ ,

w, and

o.

The forcing functions, are vd , vq,vF , and Tm • This form of the equations is

particularly convenient for solution where saturation is required, since saturation affects only AAD and AAQ. 4.12.3

Machine equations with saturation neglected

If saturation is neglected, LAD and L AQ are constant. Therefore, L MD and L MQ are also constant. The magnetizing flux linkages AAD and AAQ will have constant relationships to the state variables as given by (4.120) and (4.121). We can therefore eliminate AAD and AAQ from the machine equations. Substituting for AAD' as given in (4.120), in (4.118) and rearranging, i

d

=

(1 _LrEd

Mo )

o

= - LMO rEI)

L M D AF _ L M D ~ {d {F {d rED

(1 _L{FMD) {FAF _ L{FMD AorEo Ad _ L AF + (1 _L AD {d {o {F {o {o

iF = - L MD >"d {F {d i

'Ad _

{d

+

MD

MO)

(4.134)

Chapter 4

112

(4.135) Similarly, the q axis equations are

(L

·

MQ q A = - r 1 - - -) -A

q

{q

{,q

LMQ- -AQ + +r{q 1:Q

WAd -

V

q

· L M Q Aq L M Q ) AQ ( AQ = 'o {Q {q - rQ I - {Q {Q

(4.136)

and the equation for the electrical torque is given by

= AA

T

~t/J

q

d

(

L MD -L MQ)

L A A --.!!.!L

-

{, ~

Q

d

rE q ~ Q

L + A A ---!!.!!.+ AA F {. d rtF

q

L ~

(4.137)

rt D

D red

q

The state-space model now becomes

,

-w

o

'" o o

:

o

o

o

~: ~ - ~.:) :,

o

o

o

o

)..0

0

0

Aq

I

I

(I _L!tiD) ..f.

_ !L 1:. F

I I,

'F

L!tiD

t; T;

F

-

: I

-------------------------------------r---------------- -------'"

0

:-i

0

1

I

o -

-

-

-

-

-

-

-

L!tiD x - 3TJ-fj "

o

o

o -

-

-

-

-

-

-

-

-

-

-

-

_~ x

3TJ-f,,-tF"

o

-

-

-

-

-

-

0-

_

: : _

_

_

_

_~ x 3Tj ,f,,,-lo " o

_

I

(L) Q I -

I I

~ 7;

rQ L MQ -f Q

_

-;e;

_,

: :

r L!tiQ

-e:

L MQ ).. 3Tj ,[f "

0

x

!!L (I

_ L!ti

_

_

;fQ

0

_

_

_

_

L!tiQ 3Tj-f, -!q

0

_

Q)

x d

:

: I,

_

: : oJ

J

UF

0 0

I

-e Q

_

:

: _

_

_

_!! Tj

>'d

_

)..Q

o

o

_

0

w

T",

o

~

-I

(4.138)

The system described by (4.138) is in the form = f(x, u, t). Again the description of the system is not complete since vd and vq are functions of the currents and will depend on the external load connections. The 7 x 7 matrix on the right side of (4.138) contains state variables in several terms, and this matrix form of the equation is not an appropriate form for solution. It does, however, serve to illustrate the nonlinear nature of the system.

Example 4.3 Repeat Example 4.2 for the flux linkage model. Solution From the data of Example 4.1:

The Synchronous Machine

{d = {q = 0.150 pu

rEF = 1.651 - 1.550 = 0.101

pU

= 0.055

pu

0.036

pu

{D = 1.605 - 1.550 {Q

1.526 - 1.490

=

_1_ L MD

0.15

0.055

=

0.028378

_1_

=

_1_ + _1_ + _1_ 1.49 0.t5 0.036

{d

-

LMDJ

{d/

I:

LAID =

L.

MD

{d {F {d

L

{D

=

r

--fL

L

{Q

MQ

_ L ) --E Q

L MD 2

0.000235

=

0.000349

L MD 3 {{

=

0.000642

LMQ

=

0.000980

=

0.000235

0.003743

L MD 3Tj 1:d {F

=

0.044720

!..E-

LMD

=

0.066282

(1 -

L M D)

Tj

d

D

3Tj {q,eQ

LMQ

. P2 3 TJ 1..- q

= 0.115330

= 0.308485

=

0.002049

3Tj{(J

{D

0.286058

MQ =

!!L (1

0.005927

=

rf d

~D

q

pu

{D

!..E-

-e -eQ = 0.005789 {Q {q

!..E- L M D {D {F

{F {D

pu

L MQ = 0.028378

s. (1

pu

pu

35.2381

=

!L (1 - L MD) = 0.005278 {F 1: F !L L M D = 0.003756 r L MQ

LMD L MQ

= 0.005928

= 0.001387

{F {d

0.101

+ _1_ = 35.2381

(. _ L MQ ) \1 {q

!.L.. L M D

+ _1_ + _1_

_1_ 1.55

=

=

~ {q

113

.

and we get for the state-space equation for the first six variables, with D = 0 0

"d

0

0

0

>..,

-115.330

0

0

0

>"D

0

0

-5.928

5.789

0

x,

0

0

284.854

-313.530 0

>"(2

0

W

O.OOO56T",

-5.927

2.050

3.743

~'F

1.388

-5.278

3.756

66.282

~D ~q

= 10- 3

~Q

44.720 wX

0

w

3

10

-00 X

-O.235A q -O.349A q -0.642Aq O.235Ad

4.12.4

-vd vF

103 0

Ad

O.980Ad

0

+

0 -Vq

Treatment of saturation

The flux linkage state-space model is convenient for considering the effect of saturation because all the terms in the state equations (4.126)-(4.133) are linear except for the magnetizing flux linkages >"AD and >"AQ' These are affected by saturation of the mutual inductances LAD and L A Q, and only these terms need to be corrected for saturation. In the simulation of the machine, either by digital or analog computer, this can be accom-

Chapter 4

114

'AD ' A OT- - - - -

i MO K .= -.-

I

'MS

I

I

L A OO I

Fig. 4.7

Saturation curve for hAO'

plished by computing a saturation function to adjust (4.120) and (4.121) at all times to reflect the state of the mutual inductances. As a practical matter, the q axis inductance L AQ seldom saturates, so it is usually necessary to adjust only ')..AO for saturation. The procedure for including the magnetic circuit saturation is given below (18). Let the unsaturated values of the magnetizing inductances be L AOO and L AQO' The computations for saturated values of these inductances follow. For salient pole machines, (4.139) where K. is a saturation factor determined from the magnetization curve of the machine. For a round-rotor machine, we compute, according to [16] LAO K.

=

K,L ADO

= f(')..)

L AQ

=

K.L AQO

A = (A~O

+

A~Q)1/2

(4.140)

To determine K. for the d axis in (4.139), the following procedure is suggested. Let the magnetizing current, which is the sum of ;d + ;F + ;0, be ;M' The relation between ')..AO and ;M is given by the saturation curve shown in Figure 4.7. For a given value of ')..AO the unsaturated magnetizing current is ;MO' corresponding to L AOO' while the saturated value is ;MS' The saturation function K. is a function of this magnetizing current, which in turn is a function of ')..AO ' To calculate the saturated magnetizing current ;MS' the current increment needed to satisfy saturation, ;MfJ. = ;MS - ;MO' is first calculated . Note that saturation begins at the threshold value AAOT corresponding to a magnetizing current ;MT' For flux linkages greater than ')..AOT the current ;MfJ. increases monotonically in an almost expo nential way. Thus we may write approximately (4.141) where A. and 8. are constants to be determined from the actual saturation curve. Knowing ;MfJ. for a given value of AAO, the value of ;MS is calculated, and hence K. is determined. The solution is obtained by an iterative process so that the relation >"AoK.(X AO) = LAooiMS is satisfied. 4.13

Load Equation,

From (4.103) and (4.138) we have a set of equations for each machine in the form

x = f'(x;«, Tm )

(4.142)

The Synchronous Machine

115

where x is a vector of order seven (five currents, wand {) for the current model, or five flux linkages, wand {) for the flux linkage model), and v is a vector of voltages that includes Vd , vq , and vF • Assuming that VF and T; are known, the set (4.142) does not completely describe the synchronous machine since there are two additional variables Vd and vq appearing in the equations. Therefore two additional equations are needed to relate u, and u, to the state variables. These are auxiliary equations, which mayor may not increase the order of the system depending upon whether the relations obtained are algebraic equations or differential equations and whether new variables are introduced. To obtain equations for v, and u, in terms of the state variables, the terminal conditions of the machine must be known. In other words, equations describing the load are required. There are a number of ways of representing the electrical load on a synchronous generator. For example, we could consider the load to be constant impedance, constant power, constant current, or some composite of all three. For the present we require a load representation that will illustrate the constraints between the generator voltages, currents, and angular velocity. These constraints are found by solving the network, including loads, given the machine terminal voltages. For illustrative purposes here, the load constraint is satisfied by the simple one machine-infinite bus problem illustrated below. 4.13.1

Synchronous machine connected to an infinite bus

Consider the system of Figure 4.8 where a synchronous machine is connected to an infinite bus through a transmission line having resistance R~ and inductance L~. The voltages and current for phase a only are shown, assuming no mutual coupling between phases. By inspection of Figure 4.8 we can write Va = V~a + Rsi, + Leia or Va

v.:

Vb

V:t:,h

Vc

V oo c

fa

+ ReV

i,

fa

+

LeU

t.

(4.143)

i,

ic

In matrix notation (4.143) becomes Vabc = V«obc

+ R, Ui abc + L, uiabc

(4.144)

which we transform to the O-d-q frame of reference by Park's transformation: VOdq

=

PVabc

=

PVooabc

+

Rei Odq

+

L~Piabc V

or pu

(4.145)

The first term on the right side we may call VooOdq and may determine its value by assuming that v«Jabc is a set of balanced three-phase voltages, or

Re

Le

Fig. 4.8 Synchronous generator loaded by an infinite bus.

Chapter 4

116

COS(WRt Vooabe

= V2voo

COs(wRt cos(wRt

+ +

+ a) (4.146)

a-120°)

+

a

120°)

where V is the magnitude of the rms phase voltage. Using the identities in Appendix A and using (J = WR t + 0 + 1r /2, we can show that ClO

o VooOdq

=

PVooabe

= Voo V3

(4.147)

-sin (0 - a) cos (0 - a)

The last term on the right side of (4.145) may be computed as follows. From the d~finitio~ of Park's transformation iodq = Pi ahc , we compute the derivative i Odq = Pi abe + Pi abc ' Th us

where the quantity

pp-

I

: P :labe = IOdq

-

p..

labe

= :IOdq

-

p·p-I·

is known from (4.32). Thus (4.145) may be written as

o 'Odq

(4.148)

IOdq

o

= V oo V3 -sin (0 - a)

+ Rti odq + Lti odq

ial.,

-

-iq

V or pu

(4.149)

id

cos (0 - a)

which gives the constraint between the generator terminal voltage 'Odq and the generator current i Odq for a given torque angle o. Note that (4.149) is exactly the same whether in M KS units or pu due to our choice of P and base quantities. Note also that there are two nonlinearities in (4.149). The first is due to the speed voltage term, the wLti product. There is also a nonlinearity in the trigonometric functions of the first term. The angle 0 is related to the speed by 0 = W - I pu or, in radians,

o = 00 +

I'

(w .- wR)dt

(4.150)

'0

Thus even this simple load representation introduces new nonlinearities, but the order of the system remains at seven.

4.13.2

Current model

Incorporating (4.149) into system (4.75), we may write

-Li

= (R + wN)i +

o

o

(4.151)

The Synchronous Machine

117

V3v and I' = 0 - a. Now let

where K =

tel

R=r+R e

Ld=-=Ld+L e

Lq=Lq+L e

(4.152)

Using (4.152), we may replace the r, L d , and L, terms in L, R, and N 'by wN). Thus

L, to obtain the new matrices Land (R +

R, L

d,

and

-K sin I'

-ii = (R +

o

wN)i +

(4.153)

K cos I'

o

Premultiplying 'by -

i-I and adding the equations for wand ~, -Ksin-y

I I

1 I I I I I

io

io

o

I

-i-I

o

0

K cos-y

I

I I I

- - - -- - - - - - - - - - - _. _. - - -- - - - - -- _. - -. - - I'I - - - - - -

kM,.iq -~

_Ldi, 3Tj

kM/)iq -~

o

o

L,id kMQid 3Tj ~

o

o

D

:

0

Tj

I I I

,

o

-

---

I

1

w

o

o

o

-,- - - - 0

I I I

: 0

T". -I

(4.154)

x

= f(x, u, t), The system described by (4.154) is now in the form of (4.37), namely, where x' = [i d iF ioiq 'o w 0]. The function f is a nonlinear function of the state variables and I, and u contains the system driving functions, which are VF and Tm • The loading effect of the transmission line is incorporated in the matrices i, L, and N. The infinite bus voltage Vao appears in the terms K sin')' and K cos ')'. Note also that these latter terms are not driving functions, but rather nonlinear functions of the state variable o. Because the system (4.154) is nonlinear, determination of its stability depends upon finding a suitable Liapunov function or some equivalent method. This is explored in greater depth in Part II I.

4.13.3

The flux linkage model

From (4.149) and substituting for i d and i q in terms of flux linkages (see Section 4.12.3), Vd =

- -V

+ wL t {q

R3oJ V sin . (.t u tel

-

a)

+ -s,

{d

(1

(I _L{q A_ wL.L q {q{Q MQ )

L MO - -)

MQ AQ

{d

\ I\d -

+ L.

{d

ReL

MD \ - - - I\F -

1:d1:F

(I _L{d

MD)

~d

_

ReL

MD \ - - - I\D

{d1:.D

LeL MD ~F {d{F

_

LeL MO ~D {d{D

(4.155)

118

Chapter 4

Combining (4.155) with (4.135),

[

•+

L~

(. _

~d \

LMO)~ ~d ~d ~

RL MO

_

L~LMO ~F _ L~LMO Xo = -R (1 _ LMo) Ad

c.

{,d~O

1:d{F

[Le (

L MQ ) ]

+ {d{O Ao - w • + {q • - ~

Aq

wLeL M Q

+ {q{Q

{d

_ r:;

+ RL MO AF

~d {,F

.

+ v3V.. sm(o

AQ

- a)

(4.157) Similarly, we combine (4.156) with (4.136) to get

Equations (4.157) and (4.158) replace the first and fourth rows in (4.138) to .give the complete state-space model. The resulting equation is of the form Tx

1

+ L~

(I _ L M O )

{d \

{d

Lt!LM O

- {d{F

o o

o

= ex + D

Mo -LeL -1!,d,(O

o

(4.159)

I

: I

o

: ,I

o

I I - - - - - - - - - - - - - - - - - - - - - - - - - - - - -1- - - - - - - - - - - - - - - - - - - - - - 1-'- - -

T

=

o

: 1+ :

(I _ L

MQ)

1: q

_

L~LMQ

:

{q{Q:

0

I , 0 1 I I I - - - - - - - - - - - - - - - - - - - - - - - - - - - - -1- - - - - - - - - - - - - - - - - - - - - - 1- - - -

o

I

L~

{q

:

,

,

,I ,

:I

0

I I

0

,I

10

(4.160)

The SynchronousMachine

119

and the matrix C is given by

_A(, _LII") {,

"L",o

_!L.

{,{,

.{,

'oL",o

{D{'

C =

RL MO f.d{r

{,

W[I + t (I - L{:")] 0

RL",o -f.d{O

(I _LII")

"L",o {,{o

{f

'oL",o -f.o{,

_ ts. {o

_.-

wLrL",o

(I _L-iII") o - -- - - ... -

wLrL",o - {,{o

- -i d {F

I I

,,

- W

I

[I + i (I - L;,Q) J

I

0

0

-~A 3Tj { ' {. f

~

~A

3Tj {t/{n q

0

0

0

0

0

0

0

0

0

I

,

,I

0

0

I I I

- '- I

-

.,

I

I 1

I I I I

-1- I I I I

I

I

I I

_ ,_ .. ______ . . ____________ ._ .__ .' ________ 1 _ . ___ ._____

.s. (I _LIIQ) -i, ,{, 'QL"'Q {Q{,

I

_ L",o A 37(£.3 q

I

I

I I I I

I

I

0

I

{,{Q

0

I I

,

,

wLrL"'Q

RL"'Q {" {Q

_ ts. {Q

I I

(I _LIIQ) :

0

0

0

0

I

{Q:

_1.. _

L"'Q Ad

3Trf~

I

0

I

~Ad

3Tj{q{(J

I I I I I

D TJ

0

0 0

(4.161 ) and vl3Voosin(o - a)

o D

-VlVoocos(o - a)

(4.162)

o

1fT- I exists, premultiply (4.159) by T- 1 to get

x=

T-'Cx + T-'D

(4.163)

Equation (4.163) is in the desired form, i.e., in the form of x f(x, U, t) and completely describes the system. It contains two types of nonlinearities, product nonlinearities and trigonometric functions.

Example 4.4 Extend Examples 4.2 and 4.3 to include the effect of the transmission line and torque equations. The line constants are R, = 0, L, = 0.4 pu, Tj = 2HwR = 1786.94 rad. The infinite bus voltage constant K and the damping torque coefficient D are left unspecified. Solution R = r

+

R~ =

0.001096

2.04

Chapter 4

120

Then

~

R

0.0011

0

0

0

0.00074

0

0

0

0.0131

+ wN = ~

I I I I I I I I

2.04w

1.49w

0

0

0

0

I I I I

0.0011

0

0

0.0540

-------------------~----------

~

L

-2.IOw

-1.55w

-1.55w

0

0

0

2.100

1.550

1.550

1.550

1.651

1.550

I

I I I I I I

1.550 1.605 = 1.550 ______________ J 0

0

0

0

0

0

0

0

0

0

0

0

2.040

1.490

1.490

1.526

_________

I I I I I

By digital computer we find

i-I

1.709

-0.591

-1.080

:

-0.591

6.668

-5.867

I I I

-1.080

-5.867

-7.330

:

o

I --------------------,------------

1.710

-1.669

-1.669

2.286

I I

o

:

Then 0.00187

i-1(R + wN)

-0.00044

-0.0141

3.487w

2.547w

-1.206w

0.881w

~ \

-0.00065

0.00495

-0.0769

= -0.00118

-0.00436

0.0960: -2.202w

-1.609w

- 3.590w

- 2.650w

I I I

0.00) 87

-0.09007

3.506w

2.588w

2.588w: -0.00183

0.12332

I I

________________________ 1

- 2.650w

and we compute -Ksin')' -VF

i-I

0

- 1.71 K sin 1

+ 0.591

0.591 K sin 1 - 6.67 1.08 K sin 'Y

VF

+ 5.87 vF

Kcos')'

1.71 K cos ')'

0

-1.67Kcos')'

Therefore the state-space current model is given by

VF

_

121

The Synchronous Machine -0.00187

id

0.00065

iF

0.00118

iD iq iQ W

~

-3.506w -

-

--

-

-

1.206w

O.881w

0.00436

-0.0960

2.202w

1.609w

2.650w

2.650w

-0.0019

0.0901

-

-0.00032iq

-

- - - - - -

--

-2.588w - - - - - -

--

0

id

0

0

iF

0

0

iD

0

0

iq

0

0

iQ

-0.OO0559D 0

w

0

{)

I I I I I

I

I

I

0.00183 -0.12332 - - - - - - - - - - - - -- -- -, - - -

+

I I 1

0

1.71 K sin 'Y - 0.59

-0.59 K sin 'Y

I

0.000280id

0

0

- .--

- - - - - -

VF

6.67 v F

-1.08 K sin 'Y - 5.87

+

0

I

I I I

0.OO03lid

-0.00029iq

-0.00029iq

0

- -

0

I

2.547w

0.0769

-2.588w - - -

-3.487w

-0.00495

3.590w -

0.0141

+0.00044

VF

- 1.71 K cos 'Y

1.67 K cos "y

0.000559 t; -I

The flux linkage model is of the form T~ = CA + D, where T, C, and D are given by (4.159)-(4.162). Substituting, 3.1622

-0.7478

- 1.3656

o o

1.0

o

0

1.0

o o

o o

o o

o o

I

____________________ ~ __

T=

0.3162

o o

0.2365

1.0

o

o o

__ _

I I I I I L I

o o

o o

o o

0

0

:

J

~

0

o

0

_

0

I

0

0

:

1.0

0

:

0

0

:

0

1.0

\

~

.

_

0.4319 :

o

1.0

t I I I I

o

o I

I

I I I

o

0

1.0

L

- - - -- - - -- - -

0

0

______________________I

o

I I I I I

3.1625 -2.1118

0

o o

0

I I I I I

10.3162

0.6678

I

:

0

_

o

1.0:

I --- - - - - ,-- - --- -. - -- -. - ---- "II -- -- -- -- --I

I

I I

0

I

I I

I

1

0

0

The matrix C is mostly the same as that given in Example 4.3 except that the w terms are modified.

Chapter 4

122

-5.927

2.050

3.743: -3162w

1.388

-5.278

I I I

0

0

44.720

66.282

-115.330 :

0

0:

3162w

-747.7w

o

0

3.756

2112w

: I I I

0

r ----------------------,-----------,---------

c=

r

-1366w

I I

-5.928

5.789

I I

0

0 : 284.854 -313.530 : I

10-3

I

----------------------r-----------~---------

-0.7058~q

1.046~q

o

2.954Ad: -0.5596D

-1.910~q: 0.705~d

0

0

I

10

0

I

0

II

~d

17.766

28.024

-47.733

1000w

~F

1.388

-5.278

3.756

0

0:

44.720

66.282

-115.330

o

o

~D

Aq ~Q W

&

10- 3

--------------------1000w

-236.4w

-431.8w

o

o

o

-0.706~q

-1.046Aq

o

0

667.8w

0

I I

I I

, ,

o

------------~---------

188.337 -207.529

I I

,

284.854 -313.530.:

o

---------------------,------------,---------1.910~q: 0.705~d

0: 0.316 K sin ~

2.954Ad: -0.5596D

0

0:

0

w

0

+ 0.236 Vp

o +

-0.316Kcos1'

o

0.000559 T", -1

4.14

Subtransient and Transient Inductances and Time Constants

If all the rotor circuits are short circuited and balanced three-phase voltages are suddenly impressed upon the stator terminals, the flux linking the d axis circuit will depend initially on the subtransient inductances, and after a few cycles on the transient .inductances. Let the phase voltages suddenly applied to the stator be given by cosO u,

= V2 V cos(8 - 120)

Vc

cos(8 + 120)

u(t)

(4.164)

where u(t) is a unit step function and V is the rms phase voltage. Then from (4.7) we

The Synchronous Machine

123

can show that

o V3 Vu(t) o

(4.165)

Immediately after the voltage is applied, the flux linkages AF and AD are still zero, since they cannot change instantly. Thus at t = 0+ AF

=

0

=

kMFid

+ LFiF + MRio

Therefore (4.167)

(4.168) The subtransient inductance is defined as the initial stator flux linkage per unit of stator current, with all the rotor circuits shorted (and previously unenergized). Thus by definition (4.169) where L~ is the d axis subtransient inductance. From (4.168) and (4.169) L; = L _ (kMd 2Lo + (kM o)2L F - 2kM FkM oM R d

Ld

-

Lo

+ LF

LFL o -

Mi

2L A D

(4.170) (4.171)

(LFLD/L~D) -I

where LAD is defined in (4.108). If the balanced voltages described by (4.164) are suddenly applied to a machine with no damper winding, the same procedure will yield (at t = 0+) i, Ad

where

L~

=

-(kMF/LF)id

= [Ld -

(kM F )2/ L F lid

(4.172)

=

L~id

(4.173)

is the d axis transient inductance; Le., (4.174)

In a machine with damper windings, after a few cycles from the start of the transient described in this section, the damper winding current decays rapidly to zero and tile effective stator inductance is the transient inductance. If the phase of the impressed voltages in (4.164) is changed by 90° (va = vT V sin (J), u, becomes zero and vq will have a magnitude of vJ V. Before we examine the q axis inductances, some clarification of the circuits that may exist in the q axis is needed. For a salient pole machine with amortisseur windings a q axis damper circuit exists, but there is no other q axis rotor winding. For such a machine the stator flux linkage after the initial subtransient dies out is determined by es-

Chapter 4

124

sentially the same circuit as that of the steady-state q axis flux linkage. Thus for a salient pole machine it is customary to consider the q axis transient inductance to be the same as the q axis synchronous inductance. The situation for a round rotor machine is different. Here the solid iron rotor provides multiple paths for circulating eddy currents, which act as equivalent windings during both transient and subtransient periods. Such a machine will have effective q axis rotor circuits that will determine the q axis transient and subtransient inductances. Thus for such a machine it is important to recognize that a q axis transient inductance (much smaller in magnitude than L q ) exists. Repeating the previous procedure for the q axis circuits of a salient pole machine, (4.175) or iQ = -(k M Q/LQ)iq

(4.176)

Substituting in the equation for Aq , Lqiq + kMQi Q

Aq

(4.177)

or Aq

where

=

[L q

(kM Q )2j L Q ]i q ~ L;'iq

-

(4.178)

L;' is the q axis subtransient inductance L~I = L q

-

(kM Q)2/ L Q

=

Lq

-

L~Q/LQ

(4.179)

We can also see that when iQ decays to zero after a few cycles, the q axis effective inductance in the "transient period" is the same as L q • Thus for this type of machine (4.180) Since the reactance is the product of the rated angular speed and the inductance and since in pu WR = I, the subtransient and transient reactances are numerically equal to the corresponding values of inductances in pu. We should again point out that for a round rotor machine L;' < L; < L q • To identify these inductances would require that two q axis rotor windings be defined. This procedure has not been followed in this book but could be developed in a straightforward way [21, 22]. 4.14.1

Time constants

We start with the stator circuits open circuited. Consider a step change in the field voltage; i.e., v F = VF u(t). The voltage equations are given by (4.181) and the flux linkages are given by (note that id = 0) (4.182) Again at

t

= 0+, AD

=

0, which gives for that instant iF

=

-(L D / MR)i D

Substituting for the flux linkages using (4.182) in (4.181),

(4.183)

The Synchronous Machine

125

(4.184) Subtracting and substituting for i, using (4.183), :

'0

+ 'oL F + 'FLo '.0 LFL D

M~

-

=

'/

MR

-"F-----

LFLo

-

M~

(4.185)

Usually in pu '0 » 'F, while L o and L F are of similar magnitude. Therefore we can write, approximately, :

10

+

to

2

l-» - MR/L F

.

10

=

_

'/F

"1

MR/L F L o - Mi./L F

(4.186)

Equation (4.186) shows that l o decays with a time constant

" = ------:....L o - Mi./L F rD

TdO

(4.187)

This is the d axis open circuit subtransient time constant. It is denoted open circuit because by definition the stator circuits are open. When the damper winding is not available or after the decay of the subtransient current, we can show that the field current is affected only by the parameters of the field circuit; i.e., (4.188) The time constant of this transient is the d axis transient open circuit time constant TdO, where (4.189) Kimbark [2] and Anderson [8] show that when the stator is short circuited, the corresponding d axis time constants are given by

"

Td

Td

"L"/L' = T dO d d

(4.190)

TdoLd/ t.,

(4.191)

=

A similar analysis of the transient in the q axis circuits of a salient pole machine shows that the time constants are given by (4.192)

"

=

Tq

"L"/L q q

T qO

(4.193)

For a round rotor machine both transient and subtransient time constants are present. Another time constant is associated with the rate of change of direct current in the stator or with the envelope of alternating currents in the field winding, when the machine is subjected to a three-phase short circuit. This time constant is To and is given by (see [8], Ch. 6) To

=

L 2 /r

(4.194)

where L 2 is the negative-sequence inductance, which is given by

L2

=

(L d + L q)/2

(4.195)

Typical values for the synchronous machine constants are shown in Tables 4.3, 4.4, and 4.5.

Chapter 4

126 Table 4.3. Time constant

, TdO , Td

Td" To

=

T~'

Typical Synchronous Machine Time Constants in Seconds

Turbogenerators

Waterwheel generators

Synchronous condensers

Low

Avg.

High

Low

Avg.

High

Low

Avg.

High

2.8 0.4 0.02 0.04

5.6 1.1 0.035 0.16

9.2 1.8 0.05 0.35

1.5 0.5 0.01 0.03

5.6 1.8 0.035 0.15

9.5 3.3 0.05 0.25

6.0 1.2 0.02 0.10

9.0 2.0 0.035 0.17

11.5 2.8 0.05 0.30

Source: Reprinted by permission from Power System Stability, vol, 3, by E.W. Kimbark.

©

Wiley, 1956.

Table 4.4. Typical Turbogenerator and Synchronous Condenser Characteristics Synchronous condensers

Generators Parameter

Recommended average

Range

Nominal rating 300-1000 MW Power factor 0.80-0.95 Direct axis synchronous reactance Xd 140--180 Transient reactance Xd 23-35 Subtransient reactance Xd' 15-23 Quadrature axis synchronous reactance x q 150-160 Negative-sequence reactance X2 18-20 Zero-sequence reactance Xo 12-14 Short circuit ratio 0.50--0.72 3.0--5.0 · constant H, (kW Inertia --s) - r600 rJrnin (kVA) 1800 rjmin 5.0-8.0

0.90 160 25 20 155

19

13 0.64 4.0 6.0

Recommended average

Range

50--100 MVA 220 55

170-270 45--65 35-45 100-130 35-45 15--25 0.35-0.65

40

115 40 20 0.50

Source: From the 1964 National Power Survey made by the U.S. Federal Power Commission. USGPO. Note: All reactances in percent on rated voltage and kVA base. kW losses for typical synchronous condensers in the range of sizes shown, excluding losses associated with, step-up transformers, are in the order of 1.2-1.5% on rated kVA base. No attempt has been made to show kW losses associated with generators, since generating plants are generally rated on a net power output basis and losses vary widely dependent on the generator plant design.

Table 4.5.

Typical Hydrogenerator Characteristics

Parameter

Nominal rating (MVA) Powef factor Speed (r jmin) . (kW·s) Inertia constant H, - - (kVA) Direct axis synchronous reactance Xd Transient reactance Xd Subtransient reactance Xd' Quadrature axis synchronous reactance x q Negative-sequence reactance X2 Zero-sequence reactance Xo Short circuit ratio

Small units

Large units

0-40 0.80-0.95* 70-350

40-200 0.80-0.95· 70-200

1.5-4.0

3.0-5.5

90-110

80-100

20-45

20-35 10-25 1.0-2.0

25-45 20-35

10-35

1.0-2.0

20-40 15-30

Source: From the 1964 National Power Survey made by the U.S. Federal Power Commission. USGPO. Note: All reactances in percent on rated voltage and kVA base. No attempt has been made to show kW losses associated with generators, since generating plants are generally rated on a net power output basis and losses vary widely dependent on the generator plant design. ·These power factors cover conditions for generators installed either close to or remote from load centers.

The Synchronous Machine 4.15

127

Simplified Models of the Synchronous Machine

In previous sections we have dealt with a mathematical model of the synchronous machine, taking into account the various effects introduced by different rotor circuits, i.e., both field effects and damper-winding effects. The model includes seven nonlinear differential equations for each machine. In addition to these, other equations describing the load (or network) constraints, the excitation system, and the mechanical torque must be included in the mathematical model. Thus the complete mathematical description of a large power system is exceedingly complex, and simplifications are often used in modeling the system. In a stability study the response of a large number of synchronous machines to a given disturbance is investigated. The complete mathematical description of the system would therefore be very complicated unless some simplifications were used. Often only a few machines are modeled in detail, usually those nearest the disturbance, while others are described by simpler models. The simplifications adopted depend upon the location of the machine with respect to the disturbance causing the transient and upon the type of disturbance being investigated. Some of the more commonly used simplified models are given in this section. The underlying assumptions as well as the justifications for their use are briefly outlined. In general, they are presented in the order of their complexity. Some simplified models have already been presented. In Chapter 2 the classical representation was introduced. In this chapter, when the saturation is neglected as tacitly assumed in the current model, the model is also somewhat simplified. An excellent reference on simplified models is Young (19]. 4.15.1

Neglecting damper windings-the E~ model

The mathematical models given in Sections 4.10 and 4.12 assume the presence of three rotor circuits. Situations arise in which some of these circuits or their effects can be neglected. Machine with solid round rotor [2]. The solid round rotor acts as a q axis damper winding, even with the d axis damper winding omitted. The mathematical model for this type of machine will be the same as given in Sections 4.10 and 4.12 with i D or ~D omitted. For example, in (4.103) and (4.138) the third row and column are omitted. Amortisseur effects neglected. This assumption assumes that the effect of the damper windings on the transient under study is small enough to be negligible. This is particularly true in system studies where the damping between closely coupled machines is not of interest. In this case the effect of the amortisseur windings may be included in the damping torque, i.e., by increasing the damping coefficient D in the torque equation. Neglecting the amortisseur windings can be simulated by omitting t» and it] in (4.103) or ~D and AQ in (4.138). Another model using familiar machine parameters is given below. From (4.118), (4.123), (4.120), and (4.121) with the D and Q circuits omitted, id iF

iq

(rEd - LMD)/{~

-LMD/~d{F

-LMD/rfd{F

({F - L MD)/ { }

I I I I I

0

Ad

0

~F

IIL q

~q

- - - - - - _________________ 1 ____

0

0

I I I

(4.196)

Chapter 4

128

or (4.197) We can show that£ol is given by

.c -I •

D

vu.;

(4.198)

= [ -L Ao / L;L F

Therefore, the currents are given by IIL d

(4.199)

o

o

The above equations may be in pu or in MKS units. This follows, since the choice of the rotor base quantities is based upon equal flux linkages for base rotor and stator currents. From the stator equation (4.36) and rearranging, (4.200) or from (4.199) and (4.200) ~d

=

-(r/Ld)Ad + (rLAD/LjLF)AF - WA q -

Vd

pu

(4.201 )

From (4.58) we define (4.202) and converting to pu

vlJE;u VB = wR(kMFuMFB/LFuLFB)(AFuLFBIFB)

vIJe;

or in pu

=

(k M FuAFu/ L Fu)[WR (M FBIFB / VB»)

LADAF/ L F = vlJE; pu

(4.203)

Now, from (4.201) and (4.203) we compute ~d

= -(r/Ld)'Ad + (r/L d)V3E; - WA q

-

Vd

pu

(4.204)

In a similar way we compute Aq from (4.36), substituting for iq from (4.199) to write

Xq = -(r/Lq)Aq + WA q - uq pu

(4.205)

Note that in (4.204) and (4.205) all quantities, including lime, are in pu. For the field voltage, from (4.36) VF = rFiF + ~F pu, and substituting for iF from (4.199), VF = rF[-(LAD/L;LF)Ad + (Ld/LdLF)AFl + ~F pu (4.206) Now from (4.203) (4.207) Also from (4.59) we define (4.208)

The Synchronous Machine

129

and converting to pu v'JE FDu VB = wR[(kMFuMFB/rFuRFB)VFu VFB] v'J E FDu = (k M Fu VFu / 'Fu )(WR M FB VFB / VB R r B) V3"E FD = LADVF/rF pu

(4.209)

From (4.207), (4.209), and (4.206) we compute

v'J .!.f- Em LAD

=

LAD !I i; L F

-

Ad

+ L d Is: v'JE' + V3" L F i; LAD

q

Rearranging and using L~D/ L F = L d - L d and TdO

E'q

_1, TdO

=

=

LAD

E'

q

pu

LF/rF,

(E FD _ L'L d E' + L d L'- L d ~) u - M3 P d

q

(4.210)

d

(4.2'11)

~ ~

We now define rms stator equivalent flux linkages and voltages Ad =

s,

Ad/v'J

Vd =

Aq/v'J

=

Vq =

Vd/v'J

vq / V3"

(4.212)

Then (4.204), (4.205), and (4.211) become

Ad Aq

= =

-v! Ld)A d + (r/ Ld)E;

- wA q wAd - (r/Lq)A q - Vq pu

E'q -- - ~ L" dT dO

E'

q

-

Vd pu

(4.213) (4.214)

d d _I E FD pu + L T dO,- L'dL Ad +, T dO

(4.215)

Note that in the above equations all the variables (including time) and all the parameters are in pu. Thus the time constants must be in radians, or (4.216) Now we derive the torque equation. From (4.95)

i d and iq , from (4.199) we get Teef>

= Aq (Ad/Lq - Ad/L:J

+

Ter/J

= iqAd

-

idAq. Substituting for

(LADAF/L~LF)Aq pu

(4.217)

and by using (4.203) and (4.212), T,

=

E;Aq/L d - (I/L d - I/Lq)AdA q

(4.218)

From the swing equation TjW

=

Tm

iJ

=

W

-

-

T, - Dw pu 1 pu

(4.219) (4.220)

Equations (4.213)-(4.215), (4.219), and (4.220) along with the torque equation (4.218) describe the E; model. It is a fifth-order system with "free" inputs EFD and Tm • The signals Vd and Vq depend upon the external network. Block diagrams of the system equations are found as follows. From (4.213) we write, in the s domain, (r/Ld)[1 + (Ld/r)s]A d

=

(r/Ld)E; - wA q - Vd pu

(4.221)

Similarly, from (4.214) (4.222)

130

Chapter 4

Fig.4.9

Block diagram representat ion of the

E~

model.

and from (4.215)

(Ld/Ld)[1 + TdO(L d/

Ld)sIE~

= EFD + [(Ld - Ld)/LdIA d pu

(4.223)

Now define TAd ~ Ldir, TAq = Lq/r, and Td = TdOLd/ l-« . The above equations are represented by the block diagram shown in Figure 4.9. The remaining system equations can be represented by the block diagrams of Figure 4.10. The block diagrams in Figures 4.9 and 4.10 can be combined to give the block diagram of the complete model. Note that T'; and E FDare assumed to be known and Vd and Vq depend upon the load . The model developed to this point is for an unsaturated machine. The effect of saturation may be added by computing the additional field current required under saturated operating conditions. From Ad = Ldid + L,WiF and substituting for id from (4.199),

1.0

Fig.4.10

Block diagram represent ation of(4.218)-(4. 220).

The Synchronous Machine

Fig.4 .11

131

Block diagram for generating E~ with saturation .

(4.224) then

iFL.M = AA1 - Ld/L d) + [(Ld/Ld)(LAO/LF»)AF Also, from

WR

M Fi F =

vf2E in Section 4.7.4 iFLAo =

pu

(4.225)

we can show that

YJE

pu

(4.226)

- [(Ld - L;)/Ld)A d

(4.227)

Now from (4.212), (4.203), (4.226) , and (4.225)

E

= (Ld/Ld)E~

Substituting (4.227) into (4.215) , (4.228) For the treatment of saturation, Young (19) suggests the modification of (4.227) to the form (4.229) where E A corresponds to the additional field current needed to obtain the same EMF on the no-load saturation curve. This additional current is a function of the saturation index and can be determined by a procedure similar to that of Section 4.12.4. Another method of treating saturation is to consider a saturation function that depends upon E~ ; i.e., let E A= fA (E~). This leads to a solution for E~ amounting to a negative feedba ck term and provides a useful insight as to the effect of saturation (see (20) and Problem 4.33) . Equations (4.229) and (4.228) can be represented by the block diagram shown in Figure 4.11. We note that if saturation is to be taken into account, the portion of Figure 4.9 that produces the signal E~ should be modified according to the Figure 4.11 .

Example 4.5 Determine the numerical constants of the E~ model of Figures 4.9 and 4.10, using the data of Examples 4.1 and 4.2. It is also given that L; = 0.185 pu and L d = 0.245 pu. Solution From the given data we compute the time constants required for the model.

Lv - M~/LF = 0 .00599 - (0.109)2/2.189 = 0.03046s = n .149rad

'v

0.0184

132

Chapter 4

From this we may also compute the short circuit subtransient time constant as

r;' =

T;~

L;'/L;

The fictitious time constants T)"d T)..q

= =

L;/r = Lq/r =

=

T)"d

r~(0.185/0.245) =

and

T)..q

0.023 s

8.671 rad

=

are computed as

(0.245)(3.73 x 10- 3)/1.542 x 10- 3 = 0.593 s = 223.446 rad 6.] 18 x ]0- 3/1.542 x 10- 3 = 3.967 s = 1495.718 rad

This large time constant indicates that A q will respond relatively slowly to a change in terminal conditions. The various gains needed in the model are as follows: (L d

-

L;/Ld

=

0.245/1.7

L;)/L;

=

(1.7 - 0.245)/0.245

=

1/0.245 - 1/1.64

=

4.08

I/L; - I/L q I/L;

l/T)"d =

1/0.593

0.114

=

WR

=

=

3.939

3.473

0.00447

=

Note the wide range of gain constants required.

4.15.2

Voltage behind subtransient reactance-the E" model

In this model the transformer voltage terms in the stator voltage equations are neglected com pared to the speed voltage terms [19). In other words, in the equations for vd and vq , the terms ~d and ~q are neglected since they are numerically small compared to the terms wAq and WAd respectively. I n addition, it is assumed in the stator voltage equations that W ~ WR, and L~' = L;'. Note that while some simplifying assumptions are used in this model, the field effects and the effects of the damper circuits are included in the machine representation. Stator subtransient flux linkages are defined by the equations A;

=

Ad -

L; id

A;'

=

Aq

-

L;' t,

(4.230)

where L~ and L~' are defined by (4.170) and (4.) 79) respectively. Note that (4.230) represents the more general case of (4.169), which represents a special case of zero initial flux linkage. These flux linkages produce EM F's that lag 90° behind them. These EM F's are defined by ed"a =

\ " = - Wl\q

\"

WR I\q

(4.231)

(See [8] for a complete derivation.) From (4.36) the stator voltage equations, under the assumptions stated above, are given by (4.232) Combining (4.230) and (4.232), (4.233) N ow from (4.231) and (4.233), (4.234)

The Synchronous Machine

Fig. 4. 12

133

Voltage beh ind subtransient react ance equ ivalent.

where, under the assumptions used in this model , (4 .235) The voltages e;' and e;' are the d and q axis components of the EMF e" produced by the subtransient flux linkage, the d and q axis components of which are given by (4.230). This EM F is called the voltage behind the subtransient reactance. Equations (4.234) when transformed to the a-b-c frame of reference may be represented by the equivalent circuit of Figure 4.12 . If quasi-steady-state conditions are assumed to apply at any instant, the relations expressed in (4 .234) may be represented by the phasor diagram shown in Figure 4.13. In this diagram the q and d axes represent the real and imaginary axes respectively. "Projections" of the different phasors on these axes give the q and d components of these ph asors. For example the voltage E" is represented by the phasor E" shown. It s components are E;' and Ed respectively . From the abo ve we can see that if at any instant the terminal vo ltage and current of the mach ine are known , the voltage E" can be determined . Also if Ed and E; are known, E" can be calcul ated; and if the current is also known, the terminal vo ltage can be determined. We now develop the dynami c model for the subtransient case. Substituting (4.230) into (4.134) , we compute

Ad

=

[I - L:i

-f d

(I _

LMD)]>"d + LMDL d -f d

e.e.

AF +

LMDL; -f d-f D

AD

(4.236)

We can show that

(4 .237)

Fig .4.13

Phasor diagram for the quasi-static subtransient case.

Chapter 4

134

since by definition (4.238)

Therefore we may write (4.236) as

A;

(L d' LMO/f d fF)A F + (L:; LMD/~d,fD)AD

=

Using (4.203), we can rewrite in terms of

A;'

=

E; as

E;

(L d' LMOLF/fdfFLAD) VI

+

(4.239)

u; LMD/f d ~D)Ao

(4.240)

Now we can compute the constants LMOLF fd{FL AO

L~'

K - Ld" L MoO

=

{d{O

2 -

(4.241)

I _ L"d L MD L F {d~FLAD

1_

=

X d" -

s; -

(4.242)

X-t x-t

Substituting in (4.240) and using (4.23]), we compute in pu e;'

[(x; - x-t )/(x~ - x,e )] (VI

=

E; -

AD) + AD

(4.243)

Similarly from (4.230) and (4.104),

A;'

= (Lqi q + LAQi Q) - L;' iq

=

(L q - L;') i q + LAQi Q

(4.244)

which can be substituted into (4.231) to compute (4.245)

where we define the voltage (4.246)

We can also show that

A;'

=

Aq - L;'iq

=

(4.247)

(L Ao / LO}AQ

Now from the field flux linkage equation (4.104) in pu, we incorporate (4.203) and (4.226) to compute £

From the definition of

=

E; - (x d -

x;)(id

+ ;0)/ V3

(4.248)

L; (4.174) we can show that Ld - L;

=

(4.249)

L~o/LF

We can also show that (L; - L;)/(L; - {d)2 = LF/(LFL D -L~D)

(4.250)

Then from (4.104) in pu (4.251)

Eliminating i, from (4.251),

( L~L D L) 'o. F

-

D

(4.252)

The Synchronous Machine

135

Now substituting (4.203), (4.249), and (4.250) into (4.252), - ;-:;-3 E'

V.J

q -

\

"D

=

(L d

L'd

-

-

which can be put in the form

.

'D=

x; - x:;

(x; - X,e)2

LAD) 'd.

(L~ -

{d )2 . 'D L d - L:;

-

_r l ,

,

(4.253)

.

[AD-v3Eq - (Xd - X,e ) ld]

(4.254)

In addition to the above auxiliary equations, the following differential equations are obtained. From (4.36) we write (4.255) Substituting (4.187) and (4.250) in (4.255), ·

AD

=

(x; - X-t)2

-

(Xd - x:;) Td~

.

'D

(4.256)

Similarly, from (4.36) we have 'Q

'o + AQ

0

=

which may be written as [wR'Q(LAQ/LQ)]i Q + [(WRLAQ)/LQ]XQ

=

0

(4.257)

Now from (4.246), (4.247), (4.231), (4.192), and (4.257) we get the differential equation (4.258)

The voltage equation for the field circuit comes from (4.36) (4.259)

which can be put in the same form as (4.228)

TdO i;

=

E FD

-

(4.260)

E

where E is given by (4.248). Equations (4.256), (4.258), and (4.260) give the time rate of change >---D' e;, and E; in terms of i D, ed' and E. The auxiliary equations (4.245), (4.248), and (4.254) relate these quantities to id and iq , which in turn depend upon the load configuration. The voltage e;' is calculated from (4.243). To complete the model, the torque equation is needed. From (4.95), T,q,

=

i q Ad - id >---q

By using (4.230) and recalling that in this model it is assumed that

L:;

=

L;', (4.261)

and if W in pu is approximately equal to the synchronous speed, (4.261) becomes (4.262)

If saturation is neglected, the system equations can be reduced to the following: " - - 1 (x - -1 ed " "q T qO

TqO

X ") I. q

q

(4.263)

136

Chapter 4

(4 .264)

Now from (4.243) and using K, and K 2 as defined in (4.241) and (4.242) respectively, we may write

= vTK , £; + K 2 AD

e;'

(4 .266)

To complete the description or the system, we add the inertia l equations

W = (I/Tj)T", - e;\/3Tj - id e:;/ 3Tj - Dwh

(4.267)

b=w-I

(4.268)

The currents id and iq are determined from the load equations. The block diagrams for the system may be obtained by rearranging the above equations. In doing so , we eliminate the vTrrom all equations by using the rms equivalents, similar to (4.212),

e"/vT =

£;' + j£:;

(4.269)

+ (x; - X-t)ld £FD - Kd£; + xxdld + KdA D K,£; + K 2A D

(4.270)

£" =

AD = AD/YJ

Then (4.263)-(4.266) become

+ r;'o s) £:; (I + T:;O s) AD

- (Xq

(I

(I

+ r;os)£;

e;

= =

£;

-

x;') lq

E"

~

EFD-

-

-

-

-

-.
is 0.3 pu, and zero V,6 corresponds to a terminal voltage V, of v'3 pu (or V, = 1.0).

181

Simulation of Synchronous Machines

U=l=lt FF t ... +l:fl + !:1 _.. -d+ FD-"-iirr:i't++ T l l ::t:,:. H::.tj j + HJ J J-1 , L i : Il l ·· j T l 1r . TT.:c -H: j- - --+---:l fr r: t:I:+iil '-1.. _: T r- . -roO ::Lt: LI ' T

50

-

-- _

10~

_.

4.0 ~T:;~~~~fffiI~-F":" ~: '~'~:~EtIi~}f~H+iifL j' . r IT 3.0 --r n :1T.!' -II( 1-t-t1 'II 21.0.0 --H-H-+ Add 10% T --I t=t--roO - - --t!-1-1-r 111": r' 1'1 1}' -l -t 11:-tr+t'-t -+t 1 '1 d_t -,t: ill j·J II II o H -I'II t-:! i _I !-,-1 i , I : rJl-j . , 11 j I

Imt ' -,

,I

···I·-j··i ·t ' ·.I ·1··'·1··[··~-·1·-·-.....[..·

.,

.. , .. ,. ';':"1 '-::: ~--l":::~ 'T" ' -" T

4.0 _

e¢

IL.,+,+-

I I

1

I

:L+.i:' :':ILf:.·: t'J ·.····.· ,: .J! .':' _

. +.. .

; ·1 ]': '1 ··1 .·.·I' ·.t' ;lJt r .. "-1- . ..•. ·

3.0fFl~E~±1±1~+i~fFfff±tfFfffftt+=R±tii=f1~~

.',:

~:g lT

2105.0 ~:r _ +,: .

r:

r:

'F .

IfrL

1o~. t t , i '

O. 0 1

I

.

O

-1.0 i ' ···I -105f t ! ! -2. 0 '., r.:

~.5

J. ; ,.

4 2

Fig .5 .20

.

•.l:'...•+i. L

I

J.·~i.:

; ;,

.._·l'I _-1,.·_:: ,.

1:1' T

ii :

"

i1 l..l : r l.:', IIFt' f'i·:" ' L· I + i " · ·· 1 ··· ·· ! I: '

_

~

11111fl ,. "11

!

" " " ! I" "

; :i Il :,'!,'. It:!! !il l'..

:·1

·1

I t",

'1'

; .!: :: :I:~ 1(::\ :. : Iii I

i

v tll

•

'i

I', I

Ltt:t!IL ••,::.. !:

:!.

0.25 ~.; -; ·' iT' 0.125

-_00, .12 5~ _.;.I:..;'~·~I;,._·

\ ,1+1 , ' ·, 1 1 1 1

I i I!: I i i ! ' I

1+11 I I

1-: ,.

T

~j!

i

""1-. ·-.'1 .

:,,!! '

l!ffY!: !!!lfEf tj ,I'll,.•.'IT !'! ' !!

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Response of a machine initially at 100% load (Example 5.1 cond itions) to a 10% increase in Tm followed by a 10% increase in E FD to assure stable operation .

Figure 5.19 shows the response of the loaded machine to a 20% change in EFD • The generator is initially loaded at 90% of rated load (Tm = 2.7). Note that the response to this change in EFD does not excite an oscillatory response except for a small, welldamped oscillation in w 6 • The terminal voltage responds nearly as a first-order system with a time constant of about 4 s (rdO = 5.9 s). Figure 5.20 shows the system response to 10% step changes in both T'; and EFD • The system is initially in exactly the condition calculated in Example 5.1 with computer voltages given in Table 5.2. A 10% increase in T'; is the first disturbance. This excites a well-damped oscillatory response, particularly in T., i d , v" w, and ~ (as well as other variables that are not plotted). A good degree of damping is evident. However, this

182

Chapter 5

Fig. 5.21 Response of a machine initially at 90% load to a 20% increase in Tm followed by a 20% increase in EFD to restore stability.

overload on the system results in a gradual increase in (; with time, which if not arrested will cause the machine to fall out of step. Repeated runs of the system have indicated that corrective action is required before (; reaches about 95° . The corrective action chosen was a 10% increase in EFD • This quickly restores the system to a stable operating state at about the same angle (; as the initial angle, but at a higher >"F than the initial value. Figure 5.21 is similar to 5.20 except that the increments of T; and EFD are each 20% . fhe system is initially at 9U~~ load and 90% E FD(0 .9 x 2.666 = 2.399) . Then a 20% step increase in T'; is applied. The result is a fast movement toward instability, as evidenced by the rapid increase in (; and the drop in terminal voltage. A 20% increase in Em is

183

Simulation of Synchronous Machines

TableS.2. Variable

V,

v, vq

id iq

iF AAD AAQ Ad Aq AF

Tm 0*

Comparison of Digital and Analog Computed Variables Analog computed values

Computed value pu

1.732 -1 .092 1.344 -1.925 0.667 2.979 1.634 0.994 1.345 1.094 1.935 3.004 66.995

• Angle between q axis and infinite bus

=

o-

V

pu

Percent error

68.66 -44 .12 52.63 -38.39 13.42

1.717 -1.103 1.316 -1.920 0.671

-0.90 -1.01 -2.10 0.29 0.60

48.12 30.10 39.49 33.10 19.04 29.97 33.89

1.604 1.003 1.316 1.103 1.904 2.997 67.78

-1 .84 0.94 -2.13 0.85 -1 .60 -0.10 1.17

a.

applied at about the time 0 reaches 100·, and the system is quickly restored to a stable operating state. Finally, the excess load and excitation are removed . Figure 5.22 shows a plot in the phase plane, or W A versus 0, for exactly the same disturbances as shown in Figure 5.20. The system "spirals" to the right, first very fast and later very slowly, following the 10% increase in Tm • Just prior to loss of synchronism a

Fig.5 .22 Phase-plane plot "'A versus 0 for a 10% step increase in' Tm followed by a 10% step increase in E FD (see Figure 5.20). Initial conditions of Example 5.1.

184

Fig. 5.23

Chapter 5

Phase-plane plot

£1'0 = 2.666 .

W4

versus Ii for a 10'\, step increase in Tm with init ial condit ions Tm

=

0.9.

10%, increase in Ero causes the system to return to about the original 0, following along the lower traje ctory . Figure 5.23 shows an example of a stable phase-plane trajectory. The system is initially at 90% load but with 100';" of the Example 5.1 computed value of E f D , or 2.666. A 10%, increase in Tm causes the system to oscillate and to seek a new stable value of o. A comparison of Figure s 5.22 and 5.23 shows the more rapid convergence to the target value of 0 in the stable case. 5.10

Digital Simulation of Synchronous Machines

Early efforts in solving synchronous machine behavior by digital computer were simply digital applications of the constant-voltage-behind-transient-reactance model, using a step-by -step solution method similar to that' of Kimbark [7) . As larger and faster computers became available, engineers quickly realized that the digital computer was a powerful tool for handling very large systems of differential equations. This caused an expansion in power plant modeling to include exciters , governors, and turbines. It also introduced more detailed synchronous machine models into many computer programs, usually in the form of one of the simplified models of Section 4.15. More recent research [8,9] has been aimed at finding the best machine model for system dynamic studies. All digital computer simulations must solve the differential equations in a discrete manner; i.e., the time domain is broken up into discrete segments of length 14 and the equations solved for each segment. A simple flow chart of the process is shown

Simulation of Synchronous Machines

t

Fig. 5.24

~

t

max

185

No

Flow chart of digital integration.

in Figure 5.24. There are several proven methods for performing the actual numerical integration, some of which are presented in Appendix E. Our concern in this book is not with numerical methods, although this is important. Our principal concern is the mathematical model used in the simulation. A number of models are given in Chapter 4. We shall use the flux linkage model of Section 4.12 to illustrate a digital program for calculating synchronous machine behavior in a numerical exercise. 5.10.1

Digital computation of saturation

One of the problems in digital calculation of synchronous machine behavior is the determination of saturation. This is difficult because saturation is an implicit function; i.e., hAD = j(h AD). Actually, hAD is a function of i M D = id + iF + t», which flows in the magnetizing inductance LAD. But the currents i d , iF, and i D depend upon hAD, as shown clearly in the analog computer representation of Figure 5.12. Each integration step gives us new h'S by integration. From these A's we compute i M D • From i M D we estimate saturation, which gives a new hAD, and this gives new currents, and so on. The first requirement in computing saturation is to devise some means of determining the amount of saturation corresponding to any given operating point on the saturation curve. For this procedure the saturation curve is represented by a table of data of stator EM F corresponding to given field current, by a polynomial approximation, or by an exponential estimate. The exponential estimate is often used since exponentials are easy to compute. It is based upon computing the offset from the air gap line in pu based on the field current required to produce rated open circuit voltage, shown in Figure 5.25 as i FO ' Usually it is assumed there is no saturation at 0.8 pu

Chapter 5

186

iFO iF! iF2 Fi.I d Current , iF' A or pu

Fig. 5.25 Estimating saturation as an exponent ial function.

voltage. We then compute the normalized quantities

S GI

_ iFI

-

• 'FO

iFo

_ if] - in S G2 .

'n

if] - 1.2i Fo 1.2iFo

(5 .81 )

Then any saturation may be est imated as an exponential function of the form

S G -- A Ge BGvA where VA = V, - 0.8. Since at open circuit '>-..AO tion in terms of '>-..AO'

(5.82)

= v'3v" we can also compute satura(5 .83)

This is appealing since '>-..AO = (id + iF + io)L Ao and LAo is the only inductance that saturates appreciably. If SGI and SG2 a re given, these values can be substituted into (5.82) to solve for the saturation parameters A G and BG • From (5 .81) and (5.82) we write 0.4BG 0.2BG I .2SG2 = A Ge (5.84) SGI = A Ge Rearranging, we compute In(I.2SGdAG) = OABG

(5 .85)

Then

or A G = Sbl/I.2SG2

(5 .86)

This result may be subst ituted into (5.85) to compute

BG = 51n (I.2S Gd SGd

(5 .87)

Appendix D shows a plot of SG as a function of v, . The function SG is always positive and satisfies the defined values SGI and SG2 at V, = 1.0 and 1.2 respectively. Although we define saturation to be zero for V, < 0.8 pu , actually SG assumes a very small posi-

187

Simulation of Synchronous Machines

tive value in this voltage range. The exponential function thus gives a reasonably accurate estimate of saturation for any voltage. From (5.81) we can write for any voltage level, SG

= (iF - ki FO)/ki FO

(5.88)

where iF is the field current required to produce an open circuit voltage ~, including the effect of saturation. If the air gap line has a slope (resistance) R we have ~ = RkiFo • Then, from (5.81)

= (Ri F - RkiFO)/RkiFO = (Ri F -

SG(~)

~)/~

from which we may write the nonlinear equation ~ =

Ri F

-

(5.89)

~SG(~)

where RiF is the voltage on the air gap line corresponding to field current iF. Because of saturation, the actual terminal voltage is not RiF but is reduced by an amount ~SG where SG is a function of ~. Equation (5.89) describes only the no-load condition. However, we usually assume that saturation has a similar effect under load; i.e., it reduces the terminal voltage by an amount ~SG from the unsaturated value. Example 5.7

Determine the constants A G and BG needed to compute saturation by means of the exponential definition, given the following data from the saturation curve. 1.0 pu

SGI

=

30 A

= 1.2 pu

SG2

=

120 A

~ =

~

The field current corresponding to

~ =

1.0 on the air gap line is i FO = 365 A.

Solution

From (5.81) we compute in pu SGI

= 30/365 = 0.08219

120/1.2(365)

=

0.27397

Then from (5.86) AG

=

(0.08219)2/1.2(0.27397)

=

0.0205

and from (5.87) BG

5.10.2

= Sin [1.2(0.27397/0.08219] =

6.9315

Updating the integrands

After computing the new value of saturation for each new time step, we are ready to update the integrands in preparation for numerical integration. This process is illustrated by an example. Example 5.8

Prepare a FORTRAN computer program to compute the integrands of the flux linkage model for one machine against an infinite bus using the machine data of the Chapter 4 examples. Include in the program a treatment of saturation that can be

Chapter 5

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It'

I

I I " I I I I I If' I I I I I f I I I I " I I " r , I I I' ""1' I I I , , , , , I

2.8000 __' .!. ~ !- ~ .!. .!. .!. !

I

'I

~

I

'I I I I 'I 'I , I

r I

I'

I I I t I • I , I I I ,

!- .!. .!. ~ .!. .!. .!. !.. ~ . '. . '.

,

0.5

~ ..!. ~

I

1.0 Fig.5.40

I

I

I

. . 77-;-7~77777;7T777 'I

I

'I

I

" It' I I I I t 1" " I I' I t l I '" r I I 1'1 I , , , , I I I , I , I , I I' , I' , "

.!. ..'.. .!. ~ .!. ..'.. !. .!. ~ ~

Tim., s

I

1.5

Electromagnetic torque

Tet/J.

~~

I

I

r I I ,

I

I I I I I I

I

I I I I I I

t I

I

t

I I I ,

I

I

I I I I I I , I I 1'1 I' , I I I

.!. ..'.. . '. ..'.. .~ ..!. !. ..!. .!. ..'.. ..'.. ~

I

2.0

I

I

I

1'1 I I' I I I I I' , I I , , I

~

.!. ..!.

I

2.5

Chapter 5

206 Table 5.3.

Computer Mnenomics of Output Variables

Figure 5.29 5.30 5.31 5.32 5.33

5.34 5.35 5.36 5.37 5.38 5.39 5.40

Variable

"-d

"-F

AD

AADS SGD

ia iF iD ~

" (in degrees) WA (in pu)

r.,

Computer mnemonic WD WF WKD WADS SOD fA IFF fKD VT DLD DOMU TE

Problems 5.1 The synchronous machine discussed in Examples 5.1 and 5.2 is operating at rated terminal voltage, and its output power is 0.80 pu. The angle between the q axis and the terminal voltage is 45°. Find the steady-state operating condition: the d and q axis voltages, currents, flux linkages, and the angle 4». 5.2 The same synchronous machine connected to the same transmission line, as in Examples 5.1 and 5.2, has a local load of unity power factor, which is represented by a resistance R = 10 pUt The infinite bus voltage is 1.0 pu. The power at the infinite bus is 0.9 pu at 0.9 PF lagging. Find the operating condition of the machine. 5.3 Repeat Problem 5.2 with the machine output power being 0.9 pu at 0.9 PF lagging. 5.4 I n the system ~f one synchronous machine connected to an infinite bus through a transmission line (discussed in Examples 5.1, 5.2, and 5.6) the synchronous machine is to be represented by the simplified model known as the one-axis model given in Section 4.15. Prepare a complete analog computer simulation of this system. Indicate the signal levels for the operating conditions of Example 5.1, the amplitude and time scaling, the potentiometer settings, and the amplifier gains. Note: In the load equations, assume that 5.5

Lei d

=

lei q

==

0..

Repeat Problem 5.4 using the two-axis model of Section 4.15. 5.6 Repeat Problem 5.4 using the voltage-behind-subtransient-reactance model of Section 4.15. 5.7 In the analog computer simulation shown in Figure 5.13 and Table 5.1, the time scaling is (20). I f the time scaling is changed to (10), identify the amplifiers and potentiometers in Table 5.1 that will be affected. 5.8 In Figure 5.13 the signal to the resolver represents the infinite bus voltage. If the level of this signal is reduced by a factor of 2 while the level of all the other signals are maintained, identify the potentiometer and amplifier settings that need adjustment.

References

I. IEEE Committee Report. Recommended phasor diagram for synchronous machines. 2. 3.

4. 5.

6.

IEEE Trans.

PAS-88:1593--1610,1969. Krause, P. C. Simulation of a single machine--infinite bus system. Mimeo notes, Electr. Eng. Dept., Purdue Univ., West Lafayette, Ind., 1967. Buckley, D. F. Analog computer representation of a synchronous machine. Unpubl. M.S. thesis, Iowa State Univ., Ames, 1968. Riaz, M. Analogue computer representations of synchronous generators in voltage regulator studies. AlEE Trans. PAS-75:1178--84, 1956. Schroder, D. C., and Anderson, P. M. Compensation of synchronous machines for stability. Paper C 73 313-4, presented at the IEEESummer Power Meeting, Vancouver, B.C., Canada, 1973. Electronic Associates, Inc. Handbook of Analog Computation. 2nd edt Publ. 00800.0001-3. Princeton, N.J., 1967.

Simulation of Synchronous Machines

207

7. Kimbark. E. W. Power System Stability. Vol. I. Wiley, New York, 1948. 8. Dandeno, P. L., Hauth, R. L., and Schulz, R. P. Effects of synchronous machine modeling in large-scale system studies. 1£££ Trans. PAS-92:574-·82, 1973. 9. Schulz, R. P., Jones, W. D., and Ewart, D. N. Dynamic models of turbine generators derived from solid rotor equivalent circuits. IEEE Trans. PAS-92:926-33, 1973. 10. International Business Machines. System/360 Continuous System Modeling Program Users Manual, GH20-0367-4. IBM Corp., 1967. I

chapter

6

Linear Models of the Synchronous Machine 6. 1

Introduction

A brief review of the response of a power system to small impacts is given in Chapter 3. It is shown that when the system is subjected to a small load change, it tends to acquire a new operating state. During the transition between the initial state and the new state the system behavior is oscillatory. If the two states are such that all the state variables change only slightly (i.e., the variable Xi changes from XiO to XiO + XiA where XiA is a small change in Xi)' the system is operating near the initial state. The initial state may be considered as a quiescent operating condition for the system. To examine the behavior of the system when it is perturbed such that the new and old equilibrium states are nearly equal, the system equations are linearized about the quiescent operating condition. By this we mean that first-order approximations are made for the system equations. The new linear equations thus derived are assumed to be valid in a region near the quiescent condition. The dynamic response of a linear system is determined by its characteristic equation (or equivalent information). Both the forced response and the free response are decided by the roots of this equation. From a point of view of stability the free response gives the needed information. If it is stable, any bounded input will give a bounded and therefore a stable output. The synchronous machine models developed in Chapter 4 have two types of nonlinearities: product nonlinearities and trigonometric functions. The first-order approximations for these have been illustrated in previous chapters and are outlined below. As an example of product nonlinearities, consider the product x.x., Let the state variables X; and Xj have the initial values X;o and x.«. Let the changes in these variables be XiA and XjA' Initially their product is given by XiOXjO' The new value becomes (X;O

+

X;A)(XjO

+ xjA )

= XiOXjO

+

XiOXjA

+

XjOXiA

+

X;AXjA

The last term is a second-order term, which is assumed to be negligibly small. Thus for a first-order approximation, the change in the product x.x, is given by (6.1 )

We note that xjo and X;o are known quantities and are treated here as coefficients, while and XjA are "incrementa)" variables.

XiA

208

linear Models of the Synchronous Machine

209

The trigonometric nonlinearities are treated in a similar manner as cos (00 + 0A) with cos 0A

I"'J

1 and sin OA

I"'J

=

cos 00 cos 0A - sin 00 sin 0A

0A' Therefore,

cos(oo + OA) - cos oo

I"'J

(-sinoo}oA

(6.2)

The incremental change in cos D is then (-sin Do}DA; the incremental variable is DA and its coefficient is - sin 00. Similarly, we can show that the incremental change in the term sin 0 is given by sin (00 + OA) - sin 00 6.2

I"'J

(cos OO)OA

(6.3)

Linearization of the Generator State-Space Current Model

Let the state-space vector x have an initial state xo at time rent model is used, At the occurrence of a small disturbance, i.e., after slightly from their previous positions or values. Thus

1

1

to; e.g., if the cur-

16, the states will change (6.5)

Note that xo need not be constant, but we do require that it be known. The state-space model is in the form

x=

f(x, t)

(6.6)

which, by using (6.5), reduces to (6.7)

In expanding (6.7) all second-order terms are neglected; i.e., terms of the form Xi!~XjA

are assumed to be negligibly small. The system (6.7) becomes

(6.8)

from which we obtain the linearized state-space equation (6.9) The elements of the A matrix depend upon the initial values of the state vector xo. For a specific dynamic study it is considered constant. The dynamic properties of the system described by (6.9) are determined from the nature of the eigenvalues of the A matrix. The state space may be thought of as an n-dimensional space, and the operating conditions constrain the operation to a particular surface in this n space. Being nonlinear, the surface is not flat, although we would expect it to be continuous and relatively smooth. The quiescent operating point Xo and the functions A(xo) and B(xo) are different for every new initial condition. We may also compute the A(xo) by finding the total differential dx at xo with respect to all variables; i.e., with dx XA I"'J

Chapter 6

210

where the quantity in brackets defines A(xo). We begin by linearizing (4.74), proceeding one row at a time. For the first equation (of the d circuit) we write

uc«l + Ud6 = -r(idO + id6) - (wo + w6)Lq(iqo + iqA) - (wo + wA)kM Q(iQO + i Q6) -Ld(idO + idA) - kMF(IFo + i FA) - kM D(iDO + iOA) Expanding the product terms and dropping the second-order terms, UdO

+

UdA = (-ri dO - woLqi qo - wokMQi Qo - Ldido - kMFI Fo - kMoioo) -ridA - woLqiqA - iqoLqWA - wokMQi QA - iQokMQWA -LdldA - kMFiFA - kMDioA

The quantity in parenthesis on the right side is exactly equal to UdO. Rearranging the remaining quantities, UdA

=

-ridA - woLqiqA - wokM QiQ6 - (iqoLq + kMQiQo)w A -LdldA - kM FiFA - kM DioA

(6.10)

which is equal to UdA

=

-ridA - woLqiqA - wokMQi QA - AqOWA - LdidA - kMFiFA - kMoi DA

(6.11)

Similarly, for the q axis voltage change we write UqA = wOLdidA + wokMFi FA + wokMoi DA - riql1 - LqiqA - kMQiQA

+

(idOL d + iFOkMF + iDOkMD)WA (6.12)

which is equal to Uq6 = wOLdidA + wokMFiFA + wokMoidA + AdOWA - riqA - LqiqA - kMQiQA

(6.13) For the field winding we compute -UFA = -rFi FA - kMFidA - LFi FA - MRi oA

(6.14)

The linearized damper-winding equations are given by

o= o=

-r DiD6 - kMoidA - MRiFA - LoioA -rQiQA - kM QiqA - LQi QA

(6.15) (6.16)

From (4.101) the linearized torque equation may be established as TjWA = (1/3)( - LdiqoidA - LdidO;qA - kM Fiqoi FA - kM FiFO;qA - kMDiqoi DA - kMoiooiqA + LqidOiqA + LqiqoidA

+ kMQ;dOiQA + kM QiQoidA) - DWA + TmA

(6.17)

linear Models of the Synchronous Machine

211

which can be put in the form Tmo - (1 j3)[(L diqo - Aqo)ido - (Ado - LqiJo)iqo - kM Fiqoi FO - kM Diqoi Dtl + kM QidOi Q A ) - Dwo

;jW O =

(6.18)

Finally, the torque angle equation given by (4.102) may be written as (6.19) Equations (6.11 )-(6.19) are the linearized system equations for a synchronous machine (not including the load equation). If we drop the L1 subscript, since all variables are now small displacements, we may write these equations in the following matrix form: I I I

Vd

r

0

0

-VF

0

'F

0

0

0

'D

0

woLq

wokM Q

Aqo

0

id

0

0

0

0

iF

0

0

0

0

iD

J

I I I I

I

-----------------------l-----------------~-------

Vq

0

Tm

-wokM F -wokM D

-WOLd

r

I I I

0

I I I I I

-AdO 0

, 0 0 0 0 0 0 'Q I -----------------------,-----------------,------Aqo - Ldiqo

3

0

0

0

Ld kM F

LF

0

MR

kM D

f

I

0

-Ado

I

0

kM F kM D

+

Lqido kM QidO 3 3

I I I

- kM Fiqo -kM Di qo 3 3

,

0

0

0

0

I

0

0

id

I I

0

0

iF

I I

LD I 0 0 MR 0 0 I _____________ L __________ l ______

,

t

0 0 ,I L q kM Q It 0 0 kM Q L Q I 0 0 0 0 0 I I I -------------r----------r------

0

0

0

0

0

0

0

I I I I

0

0

0

0

I

-Tj

I I I

0

I

iq iQ

I

-D

0

W

I I

-I

0

0

I I

tD ~q io

(6.20)

W

~

0

or in matrix form v = -Kx - Mi pu

(6.21 )

Note that the matrix M is related to the matrix L of equation (4.74) by L M

0

I I

---i-----I

o : I I

-;j

0

0

Assuming that M- ' exists, the state equation for the synchronous generator, not including the load equations, is (6.22)

Chapter 6

212

which is the same form as

x=

Ax

+ Du

(6.23)

Example 6./

As a preparation for later examples involving a loaded machine, determine the matrices M and K for the generator described in Examples 4.1-4.3. Let Tj = 2H WR = 1786.94 rad.

Solution

The matrix M is related to the matrix L of Example 4.2 as follows

L

I I

0

0

- _1- ____

M

0 0

I I -Tj I I I

0

0

Then we write 1.700

1.550

1.550 :

1.550

1.651

1.550

1.550

1.550

1.506 :

o

I I I

o

I

,

--------------~-----------r---------

M

,. 1.640

1.490

I

o

, I I

,I

o

I

1.490 1.526 , I

--------------,-----------r--------:

o

: -1786.94

0

I I

I I

0

o

The matrix K is defined by (6.20)

K=

0.0011

o

o

o o

0.0007

o

o 0.0131

- 1.700

-1.550

-1.550

o

o

I

,I

, , I I

1.64

1.49

o o

o

o

I ----------------------,----------------

o

'0.0011 I

~

o

0

0.0540

o o

o o

o

------

o

i ----------------------,----------------r-------

Aqo - Ldiqo -LADiqo -LADiqo 3 3 3

o

0

0

: -AdO + LqidO LAQidO : 3 ', 3 I'

_D

0

:

-I

0

0

0':

When the machine is loaded, certain terms in these matrices change from the numeric values given to reflect the impedance of the connecting system. For example, when loaded through a transmission line to a large system, r, l-«. and L, change

linear Models of the Synchronous Machine

213

to R, Ld , and i, as noted in Section 4.13. Other terms are load dependent (such as the currents and flux linkages) and must be determined from the initial conditions. 6.3

Linearization of the Load Equation for the One-Machine Problem

Equation (4.149) is repeated here for convenience: Vd =

uq

=

-K sin (0 - a) + Reid + Leid + wLeiq K cos (0 - a) + Rei q + Leiq - wLeid

(6.24)

where K = V3 VfX) and a is the angle of VfX)' The same procedure followed previously is used to linearize this equation, with the result -K cos (00 - 0')06 + R eid6 + wOLeiq/i + iqOLeW6

-K sin (00 - 0')06 + R eiq6 +

i.t., -

+ t.L,

woL eid6 - idOL eW6

(6.25)

= -rid/i - woL qiq6 - wokM QiQ6 - AqOW/i - Ldld/i - kM FiF/i - kM DiD6 -K sin (00 - O')o/i + R eiq6 + L eiq6 - woL eid6 - idOLew/i = wOLdidli + wokM FiFIi + wokM ol o« + Adowli - ri qli - Lqiq/i kM QiQIi

(6.26)

Substituting (6.25) into (6.11) and (6.12), -K cos(oo - O')o/i + R e i d6 +

i.t:

+ wOLeiq6 + iqOLew/i

Rearranging (6.26) and making the substitution Ad

=

~q

=

+ Lii, Aq + Lei q

=

L, + L,

L,

Ad

we get, after dropping the subscript

o

=

o

(6.27)

~,

-Rid - woLqiq - wokMQi Q - Ldid

-

kM FiF

-

-

~qOW + Kcos(oo - a)t5

kM DiD pu

-tu, + WOldid + wokMFi F + wokMDi D + ~dOW + K sin (00 - 0')0 - Lqiq - kM QiQ pu

(6.28)

Combining (6.28) with (6.14)--(6.16), (6.18), and (6.19), we get for the linearized systern equations o -VF

o

o o o

o

o

I I I I I

o o 'F o o 'D : ______________________ L I

o

-wokM F -wokM D:

o

o

0

0

R

~qO

-K cos (00 - a)

o

o

I

0

0

I

A

L

0:

-AdO

-Ksin(oo - a)

0

0

o

0:

0

'Q:

3

3

3

3

3

o

o

o

0

0:

I

I

_

-----------------------T-----------------r---------------Aqo - Ldi'lo -kMFi'lO -kMDiqo : -Ado + LqidO kMQidO : -D 0 I I

:

I I

-I

o

w

214

Chapter 6

i,

kM F

kM F

LF

MR

:

kM o

MR

t.,

:

kM o :

id

0

I

0

i,

;0

I

-------------J----------L-----

:

o

L,

I

: kM Q _____________ l

o

kM Q

:

LQ

:

L ,

:

:

-;j

:

0

:

0

'o

_

I

(6.29)

i~

0

I

w

0

Equation (6.29) is a linearized set of seven first-order differential equations with constant coefficients. In matrix form (6.29) becomes v = - Kx - Mx, and assuming that M- l exists,

x=

-M-'Kx - M-Iv

=

Ax + Bu

(6.30)

where A = _M- K. Note that the new matrices M and K are now expanded to include the transmission line constants and the infinite bus voltage. It is convenient to compute A as follows. Let l

M,

M

I I

0

Kil

0

---,-----,--I I I

, ,

K

0 0 t M2 ____ .J. ______ L ___ 0

, I

0

I

M3

I I

K 12

I I

K13

---t-----,---

K21 ,I K22

,

K23

I K31 , K32

I I

K33

I

___ l ____ L ___

Then Mil:

____ ,

,

0

: 0

.J

,

_

o ,M', 0 , 2 ' ----r-------""1--o : 0 : Mil M -I 1K II '1 M-IK I 12

A

- - - - - -1, 21 , I

I

I , I ' I

M-1K 1 13 _

Mil K Mil K 22 Mil K 23 ------,-------r------I M-IK M 3- ' K 31 II M-IK 3 32 I 3 33

(6.31)

Note that the only driving functions in the system (6.29) are the field voltage V F 4 and the mechanical torque Tm 4 • Initially, the machine is spinning at synchronous speed and is delivering some known power to the infinite bus. A change in either VF or T; will cause the system to seek a new operating point, and this change is usually accompanied by damped oscillations of the variables. Example 6.2

Complete Example 6.1 for the operating conditions described in Example 5.2, taking into account the load equation. Find the new expanded A matrix. Assume D = O.

linear Models of the Synchronous Machine

215

Solution From Example 5.2 we compute

R = 0.0011 + 0.020

L, =

+ 0.400

1.700

L, =

+

1.640

=

0.0211

2.100

=

0.400 = 2.040

The matrix M is given by 2.100

1.550

1.550

1.550

1.651

1.550

1.550

1.550

1.605

I I I I I I

o

o

I

- - - - - - - - - - - - - - "II - - - - - - - - - - -- I,- - - - - - - - -

M=

I I , I

o ______________

~

2.040

1.490

1.490

1.526

I I I

o

o

L

_

I I I

0

I I

I I I I

-1786.9

0

0

I I

We also compute, in pu,

XdO ~qO

+ (- 1.591 )(0.4) = 1.039 + (0.701)(0.4) = 1.430 V3(cos53.735°) = 1.025 1.676

=

=

1.150

Kcos(oo - a)

=

Ksin(oo -

= vI3(sin53.735°) = 1.397

a)

!3 (\I\qO _ L d1qO . ) 1 3" (-kMDiqO) 1

3" (- AdO +

!

3

LqidO)

= 1.150 - 1.70 x 0.701 = -0014 3 .

= =

-1.55

X

3

-(1.676

0.701

+ 1.64 3

(kMQi dO ) = 1.490( -1.591) 3

=

-0.362

X

1.591)

-1.428

-0.790

= f

The matrix K is given by 0.0211

0

0

o

0.0007

0

0

0.0131

o

K

2.040

I

1.490

1.430

I I

- 1.025

0

0

:

0

0

0

0

:

0

0

I

------------------~-------------~-----------

=

- 2.100

- 1.550

- 1.550

o

0

0

I I I

I I

0.0211

0

0

0.0540

I I I

I I

-

1.039 0

- 1.397 0

------------------~-------------~-----------

-0.014

o

-0.362 0

-0.362 0

I I I I

-1.428 0

-0.790 0

The new A matrix is given by A = -M- 1 K, or with D = 0,

I I l

I

-D -1

0 0

Chapter 6

216

- 36.062

0.439

14.142

12.472

-4.950

76.857

22.776

4.356

-96.017

-2547.01

-2444.63

1206.01

880.86

845.46

-605.68

2202.43

1608.63

1543.98

-1106.10

1776.71

2387.40

-3487.18

I

I

I I I I I

- - - - - - - - - - - - - - - _______ L _______________

A

3589.95

2649.72

2649.72

-3505.70

-2587.54

-2587.54

=

----------------------0.0078

-0.2027

0.0

0.0

90.072

-36.064 35.218

- 123.320

I

~

_______________

I I I I I

-1735.01

-- ______________ 1 _______

-0.2027

-0.7993

0.0

I I I I I

-0.4422

0.0

0.0

1751.33

0.0

1000

10- 3

- 2331.37

-------

0.0

0.0

Example 6.3

Find the eigenvalues of the A matrix of the linearized system of Example 6.2. Examine the stability of the system. Generator loading is that of Example 5.2.

Solution

To perform the computation of the eigenvalues for the A matrix obtained in Example 6.2, a digital computer program is used. The results are given below, AI

=

-0.0359 + jO.9983

As

=

-0.0016 + jO.0289

A2

=

-0.0359 - jO.9983

A6

=

-0.0016 - jO.0289

A3 = -0.0991 A7 = -0.0007 A4 = -0.1217 All the eigenvalues are given in rad/rad, Note that there are two pairs of complex eigenvalues. The pair As and A6 correspond to frequencies of approximately 1.73 Hz; they are damped with a time constant of 1/(0.0016 x 377) or 1.66 s. This complex pair and the real pole due to A7 dominate the transient response of the system. The other complex pair corresponds to a very fast transient of about 60 Hz, which is damped at a much faster rate. This is the 60-Hz component injected into the rotor circuits to balance the M MF caused by the stator de currents. Note also that the real parts of all the eigenvalues are negative, which means that the system is stable under the conditions assumed in the development of this model, namely small perturbation about a quiescent operating condition. Example 6.4

Repeat the above example for the system conditions stated in Example 5.1.

Solution

A procedure similar to that followed in ExampJes 6.2 and 6.3 gives the following results: -36.062

0.439

14.142

12.472

-4.950

76.857

22.776

4.356

-96.017

3589.95

2649.72

-3505.70

-2587.54

: -3487.18

, 'I , , I

-2547.01 : -2327.01 I I I

1206.01

880.86

2202.43

1608.63 :

958.54

804.78

- 331.50

1469.69

-605.39

- - - - - - - - - - - - - - - - - - - - __ L ______________ L ______________

A =

-------------~

2649.72

1

, -36.064 I

-2587.54 :

35.218

I - - ..- - - - - - - - ,

-0.0075

-0.1929

-0.1929

0.0

0.0

0.0

,, I I

90.071

.1 1

982.66

2257.70

-123.320 : -959.60

-2204.72

- - - - - - - -

~

I

I - - - - - - - - - - -- - - -1-

-0.8399

-0.5351

0.0

0.0

I I I

1

0.0

0.0

1000

0.0

10- 3

linear Models of the Synchronous Machine

217

and the eigenvalues are given by AI

=

-0.0359 + jO.9983

As

=

-0.0009 + jO.0248

A2

=

-0.0359 - jO.9983

A6

=

-0.0009 - jO.0248

A3

=

-0.0991

A7

=

-0.0005

A4

=

-0.1230

Note that this new operating condition has a slightly reduced natural frequency (1.49 Hz) and a greatly increased time constant (2.95 s) compared to the previous example. Thus damping is substantially reduced by the change in operating point. 6.4

Linearization of the Flux Linkage Model

We now linearize the flux linkage model of a synchronous machine, following a procedure similar to that used above for the current model. From (4.135) we can compute the linear equations

·

Adl1 = -

r

td

(L I - r

WOAq A -

MD )

AqOWA -

Adl1 + r

L

MD tdtF

AFI1 + r

L

MD tdto

ADI1 (6.32)

Ud A

(6.33) · L MD LMD AOI1 = r o t o t d Adl1 + r o tOtF AFI1

-

'D to

(

I -

L MD )

to

AOI1

(6.34)

Similarly the q axis equation (4.136) can be linearized to give

(6.35) (6.36)

(6.37)

Similarly, the swing equation becomes

218

Chapter 6

and finally (6.38) For a system of one machine connected to an infinite bus through a transmission line, the load equations are given by (4.157) and (4.158). These are then linearized to give

where

and R = r + Re and K = V3 VQO' The linearized equations of the system are (6.33), (6.34), (6.36), and (6.37)-(6.40) and 8A = WA. In matrix form we write

TA=:CA+D

(6.41 )

where the matrices T, C, and 0 are similar to those defined in Section 4.13.3 for the nonlinear model. If the state equations are written out in the form of(6.41) and compared with the nonlinear equations (4.159)-(4.162), several interesting observations can be made. First, we can show that the matrix T is exactly the same as (4.160). The matrix C is similar, but not exactly the same as (4.161). If we writeC as

wo

qQ

dFD

C. : C2

:

C3

I I

, I

C6

___ L __ -.l __

C

=

C4

,

C,

,

:

C s : C9

---J----;--

C7

(6.42)

linear Models of the Synchronous Machine

219

with partitioning as in (4.161), we can observe that C" Cs, and C9 are exactly the same as in the nonlinear equation. Submatrices C2 and C4 are exactly as in (4.161) if w is replaced by W00 Submatrices C3 , C6 , C7 , and C, are considerably changed, however, and C3 and C6 , which were formerly zero matrices, now become

- XqO

o o

C

v1 V

oo

cos (00

-

a)

o

0

6 [~dO =

V3 V",

Si: (bo -

(6.43)

a) ]

where a is the angle of V:lO and 00 is the initial angle of the q axis, each measured from the arbitrary reference. We may write matrices C7 and C, as _1_ (>"AQO _ LMO>..qo) : _ LMO>"qo : _ LMO>"qo 3 Tj'td {d I 3T/t d ~F I 3 Tjtd D I

t

I

----------,--------T-------

o

- I

-3 Totd I

( AADO

:

-

LMQAd~

-p--

'Gd

0

0

LMQAdO

I

I I I

:

3T·t t J q

(6.44)

Q

---------------~------

o

I I

where AADO and AAQO are the initial values of the new D matrix to be D

=

[0

AAD

o

and

AAQ

respectively. Finally, we note

vFa 0: 0 0: Tma/T j 0]'

(6.45)

Assuming that the inverse of T exists, we can premultiply both sides of (6.42) by

T- 1 to obtain

(6.46) which is of the form

x=

Ax

+ Bu

(6.47)

The matrices A and B will have constant coefficients, which are dependent upon the quiescent operating conditions. Note that the matrices A and B will not be the same here as in the current model. Since the choice of the state variables is arbitrary, there are many other equations that could be written. The order of the system does not change, however, and there are still seven degrees of freedom in the solution. Example 6.5

Obtain the matrices T, C, and A of the flux linkage model for the operating conditions discussed in the previous examples. Solution

Machine and line data are taken from previous examples in pu as:

Chapter 6

220

L,

=

1.700

L,

=

1.640

{d

=

{q

-tQ 0.150

=

'F

=

0.036 0.0011

=

0.00074

'Q

=

r

'D

LAD

= kM F

=

LA Q

=

kMQ

=

LF

=

1.651

LMD

0.0131 = 0.0540 = 0.02838

{F

=

0.101

L MQ

=

0.02836

=

0.020

kM D = 1.550 1.490

=

LD

=

1.605

R,

D

=

0.055

L, = 0.400

LQ

=

1.526

t

Tj =

1786.94 rad

The matrix T is independent of load and is given by -1.3656',

0

0

0

0

1.0

0:

0

0

0

0

0

I .0

0

0

0

0

-2.111810

0

3.1622

-0.7478

o

o

I

:

i

_ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _.

T=

0

I

0

0

I

,

_

I

13.1625 I

I

o 0 0: 0 1.0 : 0 0 , - - - - - - - - - - - - - - - - - - -rI - -- - - - - - - - - - -,- - - ---o 0 0 'I 0 0 I 1.0 0 I o

0

0:

0

0: 0

1.0

and T- ' is computed as 0.3162

0.2364

0.4318'

I

I

I

o 1.0 0: I o 0 1.0: ________________ L

0

, L

,0.3162

1

:

I

:

,

o

: 0

0.6678'

,

0

1.0:

- - - - - - - - - - - - - - - _,- - - - - - - - - - - 1I ,

I

o

'0 II

_

0 _

I

1 0

I'

0

'

To calculate the matrix C, the following data is obtained from the initial operating conditions as given in Example 5.2: AqO

=

1.150

AQO = 1.045 AFo

=

1.676 2.200

ADO

=

1.914

AdO =

vTv vTv

cos (00 oo sin (00

oo

= 1.025

-

a)

-

a) = 1.397

The matrix C corresponding to Example 5.2 loading is then calculated to be

linear Models of the Synchronous Machine

-114.035 1.388 44.720

39.438 - 5.278 66.282

72.022 3.756

I I I I

-3162.53

21 I 1.78 : -1430.11

0

I

I I I

0

0

-115.330 :

0

_____________________ L ______________

3162.16

C =

0

-747.76

- 1365.58

0

0

--~-------------~-~--

-1.0285 0

-0.4009

-0.7322

0

221

I I

~

1024.53

0

0

0

0

_____________

I

- 114.055

111.378 II

1039.32

284.854

-313.530 :

0

0

1.6503

0

0

I

10- 3

1396.55

- - - - - - - - - - - - - - 1I - - - - - - - - - - - - -

-1.9867

0

0

I I I I

0

0

1000

Note that some of the elements of the matrices C, and C, in this example are somewhat different from those in Example 4.4 since the resistance R is not the same in both examples. The A matrix is given by -16.422 1.388 44.720

39.848

-26.141

- 5.278 66.282

3.756

: -1000.12 I I I

-115.330 :

667.83

I

-452.26

324.00

0

0

0

0

I

0

0

I I

0

0

I

I I

- - - - - - - - - - - _________ L ______________ J _____________

A

999.88

=

0

-236.44

-431.80

0

0

I

154.147

-174.142

I I

284.854

-313.530 :'

___________________ L ______________

1.0285 0

-0.4009

-0.7322

0

I

I I

-1.9867

I I

0

1

I

I

0

1.6503 0

328.63

I I

0

0

I

0

0

I

1000

0

~

10- 3

441.59

____________

I

I

I

The eigenvalues of this matrix are the same as those obtained in Example 6.3 and correspond to the loading condition of Example 5.2. For the operating condition of Example 5.1 we obtain the same matrix T. For this operating condition the initial conditions in pu are given by AdO = 1.345~ AFO = 1.935, ADO = 1.634, Aqo = 1.094, AQo = 0.994, K cos (00 - a) = 0.5607, and K sin (00 - a) = 1.3207.

The matrix C for the operating conditions of Example 5.1 is given by -114.035

c

39.437

2111.78

72.022

-3162.53

o

o

0

0

1.388

- 5.278

3.756

44.720

66.282

- 115.330

I

I

I

I I I

:

-1361.30

560.75

o

o

0

0

- - - - - - - - - - - - - -- - - - - - - - - - - - - -1- - - - - - - - - - - - - - - - - - .l- - - - - - - - - - - - - -

=

3162.16

-747.76

o

0

-1365.58

___________________

:

-114.055

111.378:

0:

284.854

-313.530:

I I I

-1.7155

1.3246

:

0

0

~

I

-0.3816

-0.6969

o

o

o

574.48

1320.68 10- 3

0

0

I I I

o

o

:

1000

L

~

-0.9790

and the matrix A is given by

I

o

_

222

Chapter 6 -16.422

39.848

1.388

-5.278

44.720

66.282

- - -

A

-1000.12

667.83

-430.50

177.33

3.756:

0

0

0

0

- I 15.330:

0

0

0

0

-236.44

o

154.15

- 174.14:

181.76

0:

284.85

-313.53:

0

I

0

-0.3816

o

I

- - - _. -1- - - - -- .- - - -- - - - - - - - - - J.- - - - - - - - - - .- -- - -

-431.80:

.. - - - - - .- - .- -- - - _. -. - __ - ...

0.9790

I I I

-- - -- - - - - - -- - - - - - - _. - -

999.88

=

-26.141

- -. -

-0.6969

0

1_ __ _

,

I I

I

....

.

-1.7155

0:

._ L _ _. __.

1.3246

0

I

I I

0:

417.60 10- 3 . __.

0

0

0

1000

0

_

The eigenvalues obtained are the same as those given in Example 6.4 and correspond to the loading condition "f Example 5.1.

6.5

Simplified Linear Model

A simplified linear model for a synchronous machine connected to an infinite bus through a transmission line having resistance R, and inductance L, (or a reactance Xe ) can be developed (see references (I] and [2]). Let the following assumptions be made:

1. Amortisseur effects are neglected. 2. Stator winding resistance is neglected. 3. The 'Ad and Xq terms in the stator and load voltage equations are neglected compared to the speed voltage terms WA q and WAd' 4. The terms WA in the stator and load voltage equations are assumed to. be approximately equal to WRA. 5. Balanced conditions are assumed and saturation effects are neglected. Under the assumptions stated above the equations describing the system are given below in pu.

The E' equation

6.5.1

From (4.74) and (4.104) the field equations are given by (6.48)

Eliminating iF' we get

e;

E;

VF

= (rF/LF)A F + AF

-

(rF/LF)kMFid

(6.49)

N ow let = v1 be the stator EM F proportional to the main winding flux linking the stator; i.e., V3 E; = WR k M FA F/ L F. Also let EFD be the stator EM F that is produced by the field current and corresponds to the field voltage VF; i.~.,

V3EFD =

wRkMFvF/'F

Using the above definitions and TdO defined by (4.189), we get from (6.49) in the s domain E FD

where /d = i d / definition for

-v1 and s

= (1 + T;os)E; - (Xd - X;)/d

is the Laplace transform variable.

(6.50)

Also using the above

E;, we can arrange the second equation in (6.48) to give E;

/V'3 +

= w Rk M Fi F

(x, - X;)/d

=

E + (x, - X;)/d

(6.51)

223

linear Models of the Synchronous Machine

where E is as defined in Section 4.7.4. Note that (6.50) and (6.51) are linear. From (4.149) and (4.74) and from the assumptions made in the simplified model, we compute u, and uq for infinite bus loading to be Ud uq

WRLqi q

=

V1 Vex> sin (0 -

=

-

-

=

WRLdid + wRkMFiF =

a)

+ R~id + wRL~iq

V3 vex> cos(o - a) + Rti q - wRL~id

(6.52)

Linearizing (6.52),

o= o=

-R~iq6

-

R~i.1J1

+ (x, + Xt )id6 + wRkMFiF6 + [K sin (00 - (x, + X~)iqJ1 + [K cos (00 - a)] 06

-

a)]06 (6.53)

where K = V3vex> and Voo is the infinite bus voltage to neutral. Rearranging (6.51) and (6.53), -(x~

+

R~ldJ1

X~)ldJ1

+

+ (x, +

£;6

R~lqJ1 = X~)lqJ1

+ [Vex> sin (00-a)]oJ1

= [Vex> cos (00

-

(6.54)

a)loA

Solving (6.54) for Id6 and Iq6' we compute ld6] = K/ [-(X q + Xt) [ Iq6

R~

e, cos (00(X~

0-

a) - (Xq + Xt) sin (0

+ Xe ) cos (00

-

a)

+ R, sin (00

-

a)] a)

[£;6

J

Vex> 06

(6.55)

where K/

=

1/[R; + (x, + X~) (x~ + X e ) ]

(6.56)

We now substitute ld into an incremental version of (6.50) to compute E FD 4

= (1/K3 +

T~O S)E~4

+ K4 84

(6.57)

where we define (in agreement with [2])

1/ K) = 1 + K/(Xd - x~)(Xq + Xe ) K4 = VooK/(x d - x~)[(Xq + Xe)sin(oo - a) - Recos(oo - a)]

(6.58)

Then from (6.58) and (6.57) we get the following s domain relation (6.59)

[Note that (6.59) differs from (3.10) because of the introduction here of E FD rather than uF . ] From (6.59) we can identify that K) is an impedance factor that takes into account the loading effect of the external impedance, and K4 is related to the demagnetizing effect of a change in the rotor angle; i.e., 1 E~4] K -4 - K 3 ~~ EFD 6.5.2

(6.60)

= constant

Electrical torque equation

The pu electrical torque T, is numerically equal to the three-phase power. Therefore,

T,

=

(I /3)(u did

+ uqiq)

= (~Id

where under the assumptions used in this model,

+

~Iq)

pu

(6.61)

224

Chapter 6

~ = xdld + wR k M F i F / 0

~ = -xqlq

(6.62)

Using (6.51) in the second equationof (6.62), (6.63)

From (6.63) and (6.61) T,

=

(6.64)

[E; - (x, - x;)Id ] I q

Linearizing (6.64), we compute Te~

= =

+ [£;0 - (x, - xJ)Ido]Iq~ - (x, IqoE;il + EqaOlq~ - (x q - xJ) /qO/d~

Iqo£;~

xJ)Iqold~

(6.65)

where we have used the q axis voltage Eqa defined in Figure 5.2 as Eqa = E + (x d - xq)I d with E taken from (6.51) to write the initial condition

=

+ (x, -

Xq)/dO = £;0 - (x d - xd)ldo + (x d - xq)Ido E;o - (x, - xJ)ldo

EqaO = Eo

(6.66)

Substituting (6.55) and (6.56) into (6.65), we compute the incrementaltorque to be Te~ = K, V~ IEqao[R e sin (00

a

-

a) + (xJ + X e) cos (00

-

a)l

+ lqo(xq - Xd)[(X q + Xe)sin(oo - a) - Recos(oo + Krt1qo[R; + (x q + X e )2] + EqaOReJE;~ K,lJ~ + K2E;~

a)Ho~

(6.67)

Where K, is the change in electrical torque for a small change in rotor angle at constant

d axis flux linkage; i.e., the synchronizing torque coefficient K)

=

~:A ]Eq.EqO

=

K, V<Xl IEqaOlR e sin (lJo - a) + (Xd + X e ) cos (00

+ Iqo(xq - Xd)[(Xe + x q) sin (00

-

-

a)l

a) - R e cos (00

-

a)ll

K2 is the change in electrical torque for small change in the d axis flux linkage at constant rotor angle

We should point out the similarity between the constant K, in (6.67) and the synchronizing power coefficient discussed in Chapter 2 and given by (2.36). If the field flux linkage is constant, E; will also be constant and K 2 = O. The model is reduced to the classical model of Chapter 2. 6.5.3

Terminal voltage equation

From (4.41) the synchronous machine terminal voltage J!; is given by

V;

=

(I /3)(v~

or in rms equivalent variables V; = V~

+

+ v;)

V;

(6.68)

225

linear Models of the Synchronous Machine

This equation is linearized to obtain

+

~6 = (J~/O/ ~o) ~~

(6.69)

(~o/ ~o) ~A

Substituting (6.63) in (6.69), ~6 = - ( ~o/ ~O)XqlqLl

Substituting for lqtl and

ld~

+

(~o/ ~O)(X;ldLl

- (K[ Voox q VdO/ V;o)[(x;

+

I(~o/~o)[l - K,x;(xq

o

~ KS 6

+ K 6 E; 6

Xe ) cos (00

+ Xe ) ]

-

-

+ Xe)sin(oo - a)] a) + R, sin (00

-

a)]f 0Li

(~o/~o)K,xqRel£;~

(6.71)

where K, is the change in the terminal voltage constant d axis flux linkage, or

and K 6 is the change in the terminal voltage linkage at constant rotor angle, or

6.5.4

(6.70)

from (6.55),

~6 = I(K[Voox;~o/~o)[Recos(oO - a) - (x,

+

+ £;.1)

~

~

for a small change in rotor angle at

for a small change in the d axis flux

Summary of equations

Equations (6.59), (6.67), and (6.71) are the basic equations for the simplified linear model, i.e.,

K3

--~-EFD6

+ K 3 T;OS «,s, + K 2E;tl Ksoj. + K6 E; 6

-

(6.72)

We note that the constants K" K2 , K3 , K4 , Ks, and K6 depend upon the network parameters, the quiescent operating conditions, and the infinite bus voltage. To complete the model, the linearized swing equation from (4.90) is used. (6.73)

The angle 06 in radians is obtained by integrating on WA twice. In the above equations the time is in pu to a base quantity of 1/377 s, T is the total torque to a base quantity of the three-phase machine power, and Tj = 2HwR' Example 6.6 Find the constants K, through K6 of the simplified model for the system and conditions stated in Example 5.1, but with the. armature resistance set to zero. Solution We can tabulate the data from Example 5.1 as follows.

Chapter 6

226

Transmission line data:

Xe = 0.40 pu Infinite bus voltage: Veo = 0.828 Synchronous machine data:

Xd = 1.700 pu

x q = 1.640 pu

Xd = 1.700 - [(1.55)2/1.651] = 0.245 pu Also, from Example 5.1 i FO I do ~o

=

2.979

= -1.112 = -0.631

I qo = 0.385 ~o

= 0.776

Jt;, = 1.000 We can calculate the angle between the infinite bus and the q axis to be ~o - a = 66.995°. Then sin (~o - a) = 0.9205, cos (00 - a) = 0.3908. From (6.66) we compute EqaO = 1.55 x 2.979/0 - 1.112(1.70 - 1.64)

= 2.5995

Also,

I/K, = R; + (x, + Xe)(Xd + K, = 0.7598

Xt)

= 1.3162

Then we compute from (6.58) K 3 = [1

+ (1/1.3162)( 1.455)(2.04)]-1 = 0.3072

K4 = 0.828 x 0.7598 x 1.455(2.04 x 0.9205 - 0.02 x 0.3908)

=

1.7124

We then calculate K. and K 2 from (6.67). - a) + (x d' + Xe) cos (00 - a)l + Xe ) sin (00 - a) - R, cos (00 - a)]J 0.7598 x 0.828[2.5995(0.02 x 0.9205 + 0.645 X 0.3908) + 0.3853 X 1.395(2.04 X 0.9205 - 0.02 X 0.3908)]

K1 = K, Veo 1Eqao[R e sin (00

+

=

lqO(Xq

-

X d) [(x q

= 1.0755 K2 = K/IIqo[R; + (x, + Xe)2] + EqaORel = 0.759810.385[(0.02)2 + (2.04)2] + 2.5995 x 0.021

= 1.2578 K, and K6 are calculated from (6.71): Ks

=

=

(K/VeoXd~O/Jt;o)[Recos(oO

- a) - (x, + Xe)sin(oo - a)]

- (K/VeoxqVdO/JI;o)[(Xd + Xe)cos(oo - a) + Resin (00 - a)l [(0.7598) (0.828) (0.245) (0.776/ 1.0)][(0.02) (0.3908) - (2.04) (0.9205) 1 - (0.7598)(0.828)( 1.64)( -0.631/1.0)[(0.645)(0.3908) + (0.02)(0.9205)]

= -0.0409

linea r Models of the Synchronous Machine

K6 = (~o/ J!;o) [1 - KIX~(Xq =

+ Xe ) ]

227

(~o/ J!;o) KJxqR e

-

0.776[ 1 - (0.7598)(0.245)(2.04)]

+ (0.63 I ) (0.7598) ( I .64) (0.02) :::: 0.4971 Therefore at this operating condition the linearized model of the system is given by E;6 = [0.3072/( I + 1.813 s)] EFD A Te 6 = 1.0755 06 + 1.2578 E;A

[0.5261/( 1 + 1.813 s)] 06

-

= -0.04090 A + 0.497IE;A

~6

Example 6.7

Repeat Example 6.6 for the operating conditions given in Example 5.2.

Solution From Example 5.2 I qo = 0.4047 pu

i FO = 2.8259 =

-0.9] 85

~o =

0.9670 pu

~o =

-0.6628

V

CZl

=

1.000 pu

Jt;o

1.172

a

=

53.736

IdO

=

00

-

0

and sin (00 - a) = 0.8063, cos (00 - a) = 0.5915. From this data we calculate E;o and EqaO £;0

=

1.55 x 2.826/0 - 1.455 x 0.9185

E qaO

=

1.1925 - 1.395( -0.9185) = 2.4738

I/KI

=

R; + (x,

KI

Then K

3

=

= 0.7598

(I +

+ Xe)(Xd + X e)

2.04 x 1.455)-1 1.316

=

=

1.1925

1.3162

= 0.3072

K4

= 1.0 x 1.455 (2.04 x 0.8063 - 0.02 x 0.5915) = 1.805

TdO

=

1.3162

5.90 s

The effective field-winding time constant under this loading is given by K3T~0

K)

= 0.3072 x 5.9 = 1.8125 s =

(0.7598)( 1.0) 1(2.474)[(0.02)(0.8063)

+

+ (0.645)(0.5915)]

(0.4047)( 1.395)[(2.04)(0.8063) - (0.02)(0.5915)]J = 1.4479

We note that for this example the constant K) is greater in magnitude than in Example 6.6. The constant K , corresponds to the synchronizing power coefficient discussed in Chapter 2. The greater value in this example is indicative of a lower loading condition or a greater ability in this case to transmit synchronizing power. K 2 = 0.759810.4047[(0.02)2

+ (2.04)2] + (2.474)(0.02)1

=

1.3174

228

Chapter 6

K5

==

(0.7598)( 1.0)(0.245)

(~·~167720) [(0.02)(0.5915)

- (0.7598)( 1.0)( 1.64) ( -

- (2.04)(0.8063)]

~.~~;8) [(0.645)(0.5915)

+ (0.02)(0.8063)]

==

0.0294

K6 = (0.9670\ [I - (0.7598)(0.245)(2.041)] 1.172/

- (-0.6628) (0.7598)( 1.64)(0.02) J .172

=

0.5257

The linearized model of the system at the given operating point is given in pu by E;1:1

=

[0.3072/(1 + 1.813s)JEFD~ - [0.5546/(1 + 1.813s)]01:1

TeA

=

1.4479 00 + 1.3174 E'~A

~A =

6.5.5

0.02940/:1

+ 0.5257 E;/:1

Effect of loading

Examining the values of the constants K, through K6 for the loading conditions of Exam pies 6.6 and 6.7, we note the following: I. The constant K3 is the same in both cases. From (6.57) and (6.58) we note that K3 is an impedance factor and hence is independent of the machine loading. 2. The constants K K2 , K 4 , and K 6 are comparable in magnitude in both cases, " sign. From (6.58), (6.67), and (6.71) we note that these conwhile K, has reversed stants depend on the initial machine loading. The cases studied in the above examples represent heavy load conditions. Certain effects are clearly demonstrated. In the heavier loading condition of Example 6.6, K5 has a value of -0.0409, and in the less severe loading condition of Example 6.7 its value is 0.0294. This is rather significant, and in Chapter 8 it will be pointed out that in machines with voltage regulators, the system damping is affected by the constant Ks. If this constant is negative, the voltage regulator decreases the natural damping of the system (at that operating condition). This is usually compensated for by the use of supplementary signals to produce artificial damping. From Examples 6.6 and 6.7 we note that the demagnetizing effect of the armature reaction as manifested by the E;A dependence is quite significant. This effect is more pronounced in relation to the change in the terminal voltage. To illustrate the demagnetizing effect of the armature reaction, let EFD~ = 0; then

E;A

=

[K 3K4/(1 + K 3 T~OS)]O/:1

(6.74)

and substituting in the expression for Te /:1 we get, TeA

=

[K 1

-

K 2K3K4/(1 + K 3 T~OS)]OA

(6.75)

The bracketed term is the synchronizing torque coefficient taking into account the effect of the armature reaction. Initially, the coefficient K, is reduced by a factor K2K4/T~O'

Similarly, substituting in the expression for J!;A'

Jt; A

= [ Ks -

K3 K4 K6 / (I

+ K3 T ~o s)] 0A

(

6.76)

The second term is usually much larger in magnitude than K s' and initially the change in the terminal voltage is given by V,A],=o

=

-(K4K6/T~O)OA

(6.77)

229

linear Models of the Synchronous Machine 1.8

1.3 1.2

re = 0.0 xe = 0. 4

1.1

-2

0.2

1.4

0.6

1.2

0.8 1. 0

0 .6 0 .8

0. 9

~

0.8

0. 7

0.6 0.4 0.6 Real Power, P

0. 8

.

0 .4

1.0

0.1 r -re =

0.4

0.2 0.1

1.6 1.4

re = 0.0 xe = 0. 4

1.0

0.8

0.6 0. 1

>l

0.3

r. == xe

= 1.0 0.8 0.6 0.4 0.2 0. 0

0.0 0.4

0 .2 0.1 O.O''-:----:-'-,,..-_

0 .1 0 .2

Fig. 6.1

----=-'-:- _ ---,:-'-:-_ _---:-'-,:--_

0.4

0.6 Real Power, P

0.8

--:-'-:-'

1.0

Variation or parameters K I • . . .• K 6 with loading: (a) K I versus P (real power) and Q (reactive power) as parameter. (b) K 2 versus P and Q. (c) K 4 versus P and Q. (d) Ks versus P and Q. (e) K 6 versus P and Q. (e IEEE. Reprinted from IEEE Trans .. vol. PAS-no Sept .y'Oct: 1973.)

The effects of the machine loading on the constants K,• K 2 , K 4 , Ks. and K 6 are studied in reference [3] for a one machine-infinite bus system very similar to the system in the above examples except for zero external resistance. The results are shown in Figure 6.1 . 6. 5.6

Com pa rison w ith classical model

The machine model discussed in this section is a lmost as simple as the classical model discussed in Chapter 2, except for the variation in the main field-winding flux . It is interesting to compare the two models. The classical model does not account for the demagnetizing effect of the armature reaction, manifested as a change in E;. Thus (6.67) in the classical model would have K 2 = O. Also in (6 .59) the effective time constant is assumed to be very large so that E; -- constant. In (6 .72) the classical model will have K 6 = O.

230

Chapter 6

To illustrate the difference between the two models, the same system in Example 6.7 is solved by the classical model. Example 6.8

Using the classical model discussed in Chapter 2, solve the system of Example 6.7. x'

R e

X

e

d

~ = 1.0

Fig. 6.2

L2-

Network of Example 6.7.

Solution

The network used in the classical model is shown in Figure 6.2. The phasor E L! is the constant voltage behind transient reactance. Note that the angle {) here is not the same as the rotor angle 0 discussed previously; it is the angle of the fictitious voltage E. The phasors ~ and Vac are the machine terminal voltage and the infinite bus voltage respectively. For convenience we will use the pu system used (or implied) in Chapter 2, i.e., based on the three-phase power. Therefore,

E=

E = E fl- = 1 + jO.O + (0.020 + jO.645)(0.980 - jO.217) = 1.3186/28.43° The synchronizing power coefficient is given by p = Pt!] = EVac(BI2COSOO - G, 2 sin oo) sOt! h =,)0

=

1.3186 x 1.0 0.4164 (0.645

X

=

(EVac / Z 2)[(Xd + Xe)COSOO + Resinoo)J

0.8794 + 0.02 x 0.4761)

=

1.826

To compare with the value of K, in Example 6.7 we note the difference in the pu system, K, = 1.448. Thus the classical model gives a larger value of the synchronizing power coefficient than that obtained when the demagnetizing effect of the armature reaction is taken into account. To obtain the linearized equation for V" neglecting R, we get

I, = [(1.3186coso - 1.00) +jl.3186sino]/jO.645

~ = 1.000 + jO.O + jO.40 I,

Substituting, we get for the magnitude of V,

V;

2 V,o V,t!

=

(0.3798 + 0.8177 cos 0)2 + (0.8177)2 sin? 0

=

-

(0.62 sin ( 0 )

or V/~

- 0.1261 o~

o~

Linear Modelsof the Synchronous Machine

231

The corresponding initial value in Example 6.7 is given by ~~]

6.6

= -

1:0+

(K4K6/ r;0) 0A

= -0.1252o A

Block Diagrams

The block diagram representation of (6.73) and the equation for ~~ is shown in Figure 6.3. This block diagram "generates" the rotor angle 0A. When combined with (6.59), (6.67), and (6.72) the resulting block diagram is shown in Figure 6.4. In both diagrams the subscript ~ is omitted for convenience. Note that Figure 6.4 is similar to Figure 3.1. Figure 6.4 has two inputs or forcing functions, namely, E FD and Tm • The output is the terminal voltage change V" Other significant quantities are identified in the diagram, such as Tt!' w, and 0. The diagram and its equations show that the simplified model of the synchronous machine is a third-order system.

E;,

T mu

~

Fig. 6.3

6.7

elee rod

Block diagram of (6.73).

State-Space Representation of Simplified Model From Section 6.5 the system equations are given by K 3T

do t ; L1 + E;L1

=

K3EFD~

r., «,s, V'A

rjW~

6~

x,«, Tm~

WA

-

+

+

{t i~t-.

K 3K4 0/1

K2E;~

K 6 E;A

r.,

(6.78)

Eliminating V,L1 and Tt!L1 from the above equations, E;L1

wL1 5L1

=

-

=

-

(1/ K 3 T~O) E;L1 - (K4 / T~O) 0L1

(K 2/Tj ) E; A - (K 1/Tj ) 0L1

= W L1 "",f,,...

By designating the state variables as E;L1'

Fig 6.4

W L1 ,

+ (1/ T~o) E FD A

+ (I/Tj ) TmA

(6.79)

and 0L1 and the input signals as EFD~ and

Block diagram of the simplified linear model of a synchronous machine connected to an infinite bus.

232

Chapter 6

Tm A , the above equation is in the desired state-space form i

= Ax + Bu

where

x'

[£;A WA t5 A]

A

U

= [EFDA] Tm A

-1/ K) TdO

0

-K2/Tj

0

0

B

-K4/TdO -K./T j

l/Tdo

0

0

l/Tj

0

0

(6.80)

(6.81 )

0

In the above equations the driving functions E FD A and Tm A are determined from the detailed description of the voltage regulator-excitation systems and the mechanical turbine-speed governor systems respectively. The former will be discussed in Chapter 7

while the latter is discussed in Part III.

Problems 6.1

The generator of Example 5.2 is loaded to 75% of nameplate rating atrated terminal voltage and with constant turbine output. The excitation is then varied from 90% PF lagging to unity and finally to 90% leading. Compute the current model A matrix for these three power factors. How many elements of the A matrix vary as the power factor is changed? How sensitive are these elements to change in power factor? 6.2 Use a digital compute!" to compute the eigenvalues of the three A matrices determined in Problem 6.1. What conclusions, if any, can you draw from the results? Let D = O. 6.3 Using the data of Problem 6.1 at 90% PF lagging, compute the eigenvalues of the A matrix with the damping D = I, 2, and 3. Find the sensitivity of the eigenvalues to this parameter. 6.4 Repeat Problem 6.1 using the flux linkage model 6.5 Repeat Problem 6.2 using the flux linkage model. 6.6 Repeat Problem 6.3 using the flux linkage model. 6.7 Make an analog computer study using the linearized model summarized in Section 6.5.4. Note in particular the system damping as compared to the analog computer results of Chapter 5. Determine a value of D that will make the linear model respond with damping similar to the nonlinear model. 6.8 Examine the linear system (6.79) and write the equation for the eigenvalues of this system. Find the characteristic equation and see if you can identify any system constraints for stability using Routh's criterion. 6.9 For the generator and loading conditions of Problem 6.1, calculate the constants K. through K 6 for the simplified linear model. 6.10 Repeat Example 6.8 for the system of Example 6.6. Find the synchronizing power coefficient and V,~ as a function of 0A for the classical model and compare with the corresponding values obtained by the simplified linear model. References I. Heffron, W.O., and Phillips, R. A. Effect of a modern voltage regulator on underexcited operation of large turbine generators. AlEE Trans. 71:692-97, 1952. 2. de Mello, F. P., and Concordia, C. Concepts of synchronous machine stability as affected by excitation control. IEEE Trans. PAS-88:316-29,1969. 3. El-Sherbiny, M. K., and Mehta, D. M. Dynamic system stability. Pt. 1. IEE£ Trans. PAS-92:1538-46, 1973.

chapter

7

Excitation Systems

Three principal control systems directly affect a synchronous generator: the boiler control, governor, and exciter. This simplified view is expressed diagramatically in Figure 7.1, which serves to orient our thinking from the problems of representation of the machine to the problems of control. In this chapter we shall deal exclusively with the excitation system, leaving the consideration of governors and boiler control for Part III. 7.1

Simplified View of Excitation Control

Referring again to Figure 7.1, let us examine briefly the function of each control element. Assume that the generating unit is lossless. This is 'n~t a bad assumption when total losses of turbine and generator are compared to total output. Under this assumption all power received as steam must leave the generator terminals as electric power. Thus the unit pictured in Figure 7.1 is nothing more than an energy conversion device that changes heat energy of steam into electrical energy at the machine terminals. The amount of steam power admitted to the turbine is controlled by the governor. The excitation system controls the generated EM F of the generator and therefore controls not only the output voltage but' the power factor and current magnitude as well. An example will illustrate this point further. r--------... Power at voltage, V Current, I

Power set-point

REF w

Fig. 7.1

REF V

Principal controls of a generating unit.

Refer to the schematic representation of a synchronous machine shown in Figure 7.2 where, for convenience, the stator is represented in its simplest form, namely, by an EM F behind a synchronous reactance as for round rotor machines at steady state. Here 233

234

Chapter 7

'J

Torqu e........................

Fig. 7.2

x

"

Equivalent circu it or a synchronous machine.

the governor controls the torque or the shaft power input and the excitation system controls Eg , the internally generated EMF. Example 7.1

Consider the generator of Figure 7.2 to be operating at a lagging power factor with a current I, internal voltage Eg , and terminal voltage V . Assume that the input power is held constant by the governor. Having established this initial operating condition, assume that the excitation is increased to a new value Assume that the bus voltage is held constant by other machines operating in parallel with this machine, and find the new value of current l ', the new power factor cos 0: and the new torque angle I 3 ~

rings

Exc it er fi e ld rhe os tat

~ c

Gen

PT'.

I I

~---"'I

L---------'

Ma nual

control

Fig. 7.7 Main exciter with rheostat contr ol.

voltage . Two additional features of the system in Figure 7.8 are the damping transformer and current compensator. The damping transformer is an electrical "dash pot" or antihunting device to damp out excessive action of the moving plunger . The current compensator feature is used to control the division of reactive power among parallel generators operating under this type of control. The current transformer and compensator resistance introduce a voltage drop in the potential circuit proportional to the line current. The pha se relationship is such that for lagging current (positive generated reactive power) the voltage drop across the compensating resistance adds to the voltage from the potential transform er . Thi s causes the regulator to lower the excitation voltage for an increase in lagging current (increa se in reactive power output) and provides a drooping characteristic to assure that the load reacti ve power is equall y divided among the parallel machine s. The next level of complication in excitation systems is the main exciter and pilot

Exci ter

shvnt field

Compensating resista nc e

Fig.7.8

Self-excited main exciter with Silverstat regulator . (Used with permission from Electrical Transmission and Distribution Referen ce Book, 1950, ABB Power T & 0 Company Inc., 1992.)

238

Chapter

I

Pi I?t

Mai n exci te r

Comm u ta tor

rs-: excit er

~

R~t~

I

7 !' S ip

C omm u ta to r

0k\

.:

rr:

~

~ 6

~·t..:Jj~F i e~

T

: fM'.....--. I

Pilot

v::.:;

exc iter

fiel d

1

brea ker

Ma nua l

co n tro l

Fig .7 .9

Main exciter and pilot exciter system .

exciter system shown in Figure 7.9. This system has a much faster response than the self-excited main exciter, since the exciter field control is independent of the exciter output voltage. Control is achieved in much the same way as for the self-excited case. Because the rheostat positioner is electromechanical, the response may be slow compared to more modern systems, although it is faster than the self-excited arrangement. The two systems just described are examples of older systems and represent direct, straightforward means of effecting excitation control. In terms of present technology in control systems they are primitive and offer little promise for really fast system response because of inherent friction, backlash , and lack of sensitivity. The first step iii sophistication of the primit ive systems was to include in the feedback path an amplifier that would be fast acting and could magnify the voltage error and induce faster excitation changes. Graduall y, as generators have become larger and interconnected system operation more common, the excitation control systems have become more and more complex . The following sections group these modern systems according to the type of exciter [3).

Commu ta to r

Fiel d brea ker

CT

H8r~---or-...o--.....-1l:Kf--< 1G';n}--1''!'''---.......-

Ampl idyne fiel d

Slip

rings

I----~

I I

: Exc i te r fi el d

@

I

rheos ta t

(ma nua l co ntro l)

I

Amplid yne reg ulat or

I

IT

II I

II

I II: I I

II

II

r "

.. ~

Ma gnet~~

: amplifi er

i

I~

tJ

:

Gr;;n.i.t~; -, ampl ifier

L:: . . _

:

n

..

1 ·· . ··-,

e' er I. In this case the voltage rise is more "sluggish," as shown in Figure 7.25. Here the overshoot is zero, the settling time is T, (i .e., the time for the response to settle within k of its final value), a nd the rise time is T R • Reference [19) suggests testing an excitation system to determine the response, such as in Figure 7.25. Then determine the area under this curve for 0.5 sand use this as a specification of respon se in the time domain . For newer, fast systems reference [3) suggests simulation of the excitation as preferable to actual testing since on some systems certain parameters are unavailable for measurement [8,9) .

r

250

Chapter 7

o

0 .1

0 .2

0.3

0.4

0 .5

Time, s

Fig. 7.25

7.5

Response of an excitation system.

Voltage Regulator

In several respects the heart of the excitation system is the voltage regulator (Def, 2.12) . This is the device that senses changes in the output voltage (and current) and causes corrective action to take place. No matter what the exciter speed of the response, it will not alter its response until instructed to do so by the voltage regulator. If the regulator is slow, has deadband or backlash , or is otherwise insensitive, the system will be a poor one. Thus we need to be very critical of this important system component. In addition to high reliability and availability for maintenance, it is necessary that the voltage regulator be a continuously acting proportional system . This means that any corrective action should be proportional to the deviation in ac terminal voltage from the desired value, no matter how small the deviation. Thus no deadband is to be tolerated , and large errors are to receive stronger corrective measures than small errors. In the late 1930s and early 1940s several types of regulators, electronic and static, were developed and tested extensively [24, 25]. These tests indicated that continuously acting proportional control "increased the generator steady-state stability limits well beyond the limits offered by the rheostatic regulator" [24.26] . This type of system was therefore studied intensively and widely applied during the 1940s and 1950s, beginning with application to synchronous condensers; then to turbine generators; and finally , in the early 1950s, to hydrogenerators. (Reference [241 gives an interesting tabulation of the progress of these developments.)

7.5.1

Electromechanical regulators

The rather primitive direct-acting regulator shown in Figure 7.8 is an example of an electromechanical regulator. In such a system the voltage reference is the spring tension against which the solenoid must react. It is reliable and independent of auxiliaries of any kind . The response, however, is sluggish and includes 'deadband and backlash due to mechanical friction, stiction , and loosely fitting parts. Two types of electromechanical regulators are often recognized; the direct-acting and the indirect-acting. Direct-acting regulators, such as the Silverstat (2) and the Tirrell (24), have been in use for many years, some dating back to about 1900. Such devices were widely used and steadily improved, while maintaining essentially the same form. As machines of larger size became more common in the 1930s the indirectacting rheostatic regulators began to appear . These devices use a relay as the voltagesensitive element [24]; thus the reference is essentially a spring, as in the direct-acting device. This relay operates to control a motor-operated rheostat, usually connected between the pilot exciter and the main exciter, as in Figure 7.9. This regulator is limited in its speed of response by various mechanical delays . Once the relay closes, to

Excitation Systems

251

short out a rheostat section, the response is quite fast. In some cases, high-speed relays are used to permit faster excitation changes. These devices were considered quite successful, and nearly all large units installed between about 1930 and 1945 had this type of control. Many are still in service. Another type of indirect-acting regulator that has seen considerable use employs a polyphase torque motor as a voltage-sensitive element [27]. In such a device the output torque is proportional to the average three-phase voltage. This torque is balanced against a spring in torsion so that each value of voltage corresponds to a different angular position of the rotor. A contact assembly attached to the rotor responds by closing contacts in the rheostat as the shaft position changes. A special set of contacts closes very fast with rapid rotor accelerations that permit faster than normal response due to sudden system voltage changes. The response of this type of regulator is fairly fast, and much larger field currents can be controlled than with the direct-acting regulator. This is due to the additional current "gain" introduced by the pilot excitermain exciter scheme. The contact type of control, however, has inherent deadband and this, coupled with mechanical backlash, constitutes a serious handicap. 1.5.2

Early electronic regulators

About 1930 work was begun on electronic voltage regulators, electronic exciters, and electronic pilot exciters used in conjunction with a conventional main exciter [24, 25]. In general, these early electronic devices provided "better voltage regulation as well as smoother and faster generation excitation control" [24] than the competitive indirectacting systems. They never gained wide acceptance because of anticipated high maintenance cost due to limited tube life and reliability, and this was at least partly justified in later analyses [25]. Generally speaking, electronic voltage regulators were of two types and used either to control electronic pilot exciters or electronic main exciters [25]. The electronic exciters or pilot exciters were high-power de sources usually employing thyratron or ignitron tubes as rectifying elements. 1.5.3

Rotating amplifier regulators

In systems using a rotating amplifier to change the field of a main exciter, as in Figure 7.10, it is not altogether clear whether the rotating amplifier is a part of the "voltage regulator" or is a kind of pilot exciter. Here we take the view that the rotating amplifier is the final, high-gain stage in the voltage regulator. The development of rotating amplifiers in the late 1930s and the application of these devices to generator excitation systems [28, 29] have been accompanied by the development of entirely "static" voltage sensing circuitry to replace the electromechanical devices used earlier. Usually, such static circuits were designed to exclude any electronic active components so that the reliability of the device would be more independent of component aging. For example, devices employing saturable reactors and selenium rectifiers showed considerable promise. Such circuits supplied the field windings of the rotating amplifiers, which were connected in series with the main exciter field, as in Figure 7.10. This scheme has the feature that the rotating amplifier can be bypassed for maintenance and the generator can continue to operate normally by manual regulation through a field rheostat. This connection is often called a "boost-buck" connection since, depending on polarity, the rotating amplifier is in a position to aid or oppose the exciter field. The operation of a typical rotating amplifier regulating system can be analyzed by reference to Figure 7.10. The generator is excited by a self-excited shunt exciter. The

252

Chapter 7

Buc k Zero u,

>

vol ta ge regi on

am p lifie r vo lto ge

Sa t ura tio n c urve

- -

~

C>

E -;:;

r;u~l:d excita t io n vo lta g e

>

Exciter Shunt Field Current

Fig. 7.26

V·I characteristic defining boost and buck regions.

field circuit can be controlled either manually by energizing a relay whose contacts bypass the rotating amplifier or automatically, with the amplifier providing a feedback of the error voltage to increase or decrease the field current. The control characteristic may be better understood by examining Figure 7.26. The field rheostat is set to intersect the saturation curve at a point corresponding to rated terminal voltage, i.e., the exciter voltage required to hold the generated voltage at rated value with full load. Under this condition the rotating amplifier voltage is zero. Now suppose the generator load is reduced and the generator terminal voltage begins to rise. The voltage sensing circuit (described later) detects this rise and causes the rotating amplifier to reduce the field current in the exciter field . This reduces the exciter voltage, which in turn reduces iF' the generator field current. Thus the shaded area a bo ve the set point in Figure 7.26 is called the buck voltage region . A similar reasoning defines the area below the set point to be the boost voltage region . Rotating amplifier systems have a moderate response ratio, often quoted as about 0 .5 (e .g. , see Appendix D) . The speed of response is due largely to the main exciter time constant, which is much greater than the amplidyne time constant. The ceiling voltage is an important factor too, exciters with higher ceilings having much faster response than exciters of similar design but with lower ceiling voltage (see [17] for a discussion of this topic) . The voltage rating of the rotating amplifier in systems of this type is often comparable to the main exciter voltage rating, and the voltage swings of the amplifier change rapidly in attempting to regulate the system [24]. 7 .5 .4

Magnetic amplifier regulators

Another regulator-amplifier scheme capable of zero deadband proportional control is the magnetic amplifier system [6, 30, 31]. (We use the generic term "magnetic amplifier" although those accustomed to equipment of a particular manufacturer use trade names, e.g., Magamp of the Westinghouse Electric Corporation and Amplistat of the General Electric Company .) In this system a magnetic ampl ifier, i.e., a static amplifying device [32, 33], replaces the rotating amplifier. Usually, the magnetic amplifier consists of a saturable core reactor and a rectifier. It is essentially an amplifying device with the advant ages of no rotating parts, zero warm-up time, long life, and sturdy construction . It is restr icted to low or moderate frequencies , but this is no drawback in power applications.

253

Excitation Systems

a ec

Saturable core

Load

Fig . 7.27

M agn et ic ampl ifier .

Basically, the magnetic amplifier is similar to that shown in Figure 7.27 [33]. The current flowing through the load is basically limited by the very large inductance in the saturable core main windings . As the core becomes saturated, however , the current jumps to a large value limited only by the load resi st ance . By applying a small (lowpower) signal to the control winding, we control the firing point on each voltage (or current) cycle, and hence the average load current. This feature , of controlling a large o utp ut current by means of a small control current, is the essence of any amplifier . The fact that this amplifier is ver y nonlinear is of little concern . One type of regulator that use s a magnetic amplifier is shown in block diagram form in Figure 7.10 [4). Here the magnetic amplifier is used to amplify a voltage error signal to a power level satisfactory for supplying the field of a rotating amplifier. The rotating amplifier is located in series with the exciter field in the usual boost-buck connection. One important feature of this system is that the magnetic amplifier is relatively insensitive to variation s in line voltage and frequency, making this type of regulato r favorable to remote (especially hydro) locations. Another application of magnetic amplifiers in voltage regulating systems, shown in Figure 7.11 [6). has several features to distinguish it from the previous example. First, the magnetic amplifiers and reference are usually supplied from a 420-Hz system supplied by a permanent-magnet motor-generator set for maximum security and reliability. The power amplifier supplies the main exciter directly in this system . Note, however, that the exciter must have two field windings for boo st or buck corrections since magnetic amplifiers are not reversible in polarity . The main exciter also has a self-excited, rheostat-controlled field and can continue to operate with the magnetic amplifiers out of service. The magnetic amplifier in the system of Figure 7.11 consists of a two-stage pushpull input amplifier that, with l-mW input signals, can respond to maximum output in three cycles of the 420-Hz supply . The second stage is driven to maximum output when the input stage is at half-maximum . and its transient response is also about three cycles . The figures of merit [34) are about 200/cycle for the input stage and SOO/cycle for the output stage. This compares with about Soo/s for a conventional pilot exciter. The power ampl ifier has a figure of merit of ISOO/cycle with an overall delay of less than 0 .0 I s. (The figure of merit of an amplifier has been defined as the ratio of the power amplification to the time constant. It is shown in [34) that for static magnetic ampl ifier s it is equal to one-half the ratio of power output to stored magnetic energy .) Reference [6] reviews the operating experience of a magnetic amplifier regulator installation on one SO-MW m achine in a plant consisting of seven units totaling over 300 MW, only two units of which are regulated . The experience indicates that. since

254

Chapter 7

the magnetic amplifier regulator is so much faster than the primitive rheostatic regulator, it causes the machine on which it is installed to absorb much of the swing in load, particularly reactive load. In fact, close observation of operating oscillograms, when operating with an arc furnace load, reveals that both exciter voltage and line currents undergo rapid fluctuations when regulated but are nearly constant when unregulated. This is to be expected since the regulation of machine terminal voltage to a nearly constant level makes this machine appear to have a lower reactance, hence it absorbs changes faster than its neighbors. In the case under study, the machine terminal voltage was regulated to ±0.25~{)' whereas a ± 1% variation was observed with the regulator disconnected [6].

7.5.5

Solid-state regulators

Some of the amplification and comparison functions in modern regulators consist of solid-state active circuits [3). Various configurations are used depending on the manufacturer, but all have generally fast operation with no appreciable time delay compared to other system time constants. The future will undoubtedly bring more applications of solid-state technology in these systems because of the inherent reliability, ease of maintenance, and low initial cost of these devices.

7.6

Exciter Buildup

Exciter response has been defined as the rate of increase or decrease of exciter voltage when it change is demanded (see Appendix E, Def, 3.) 5). Usually we interpret this demand to be the greatest possible control effort, such as the complete shorting of the field resistance. Since the exciter response ratio is defined in terms of an unloaded exciter (Def. 3.) 9), we compute the response under no-load conditions. This serves to satisfy the terms of the response ratio definition and also simplifies the computation or test procedure. The best way to determine the exciter response is by actual test where this is possible. The exciter is operated at rated speed (assuming it is a rotating machine) and with no load. Then a step change in a reference variable is made, driving the exciter voltage to ceiling while the voltage is recorded as a function of time. This is called a "buildup curve." In a similar way, a "build-down" curve can also be recorded. Curves thus recorded do not differ a great deal from those obtained under loaded conditions. If it is impractical to stage a test on the exciter, the voltage buildup must be computed. We now turn our attention to this problem.

7.6.1

The dc generator exciter

In dealing with conventional de exciters three configurations (i.e., separately excited, self-excited, and boost-buck) are of interest. They must be analyzed independently, however, because the equations describing them are different. (Portions of this analysis parallel that of Kimbark [16], Rudenberg [20], and Dahl [35] to which the reader is referred for additiona I study.) Consider the separately excited exciter shown in Figure 7.28. Summing voltage drops around the pilot exciter terminal connection, we have

~E + Ri = up where

AE = flux linkages of the main exciter field, Wb turns

= main exciter field resistance, n i = current, A

R

up

=

pilot exciter voltage, V

(7.10)

255

Excitation Systems

Fig .7.28

Separately excited exciter.

It is helpful to think in terms of the field flux ifJ E rather than the field flux linkages. If we assume the field flux links N turns, we have N¢E + Ri

=

(7.11)

up

The voltage of the pilot exciter up may be treated as a constant [16] . Thus we have an equation in terms of i and ifJ E with all other terms constant. The problem is that idepends on the exact location of the operating point on the saturation curve and is not linearly related to uF• Furthermore, the flux ifJE has two components, leakage flux and armature flux, with relative magnitudes also depending on saturation . Therefore, (7.11) is nonlinear. Since magnetization curves are plotted in terms of uF versus i, we replace ifJ E in (7.11) by a term involving the voltage ordinate UF o Assuming the main exciter to be running at constant speed, its voltage UF is proportional to the air gap flux ifJa ; i.e., (7.12) The problem is to determine how ifJa compares with ifJE ' The field flux has two components, as shown in Figure 7.29. The leakage component, comprising 10-20% of the total, traverses a high-reluctance path through the air space between poles. It does not link all N turns of the pole on the average and is usually treated either as proportional to ifJa or proportional to i. Let us assume that ifJ-f. is proportional to ifJa (see [16] for a more detailed discussion), then (7.13) where C is a constant. Also, since (7.14)

Fig .7 .29

"'a '

"'-t.

"'E "'a

"'-t.

Armature of air gap flux leakage flux and field flux = + (Reprinted by permission from Power System Stabilit y. vol. 3, by E. W. Kimbark. @ Wiley, 1956.)

256

Chapter 7

we have

r/J£ = (I + C)r/Jo = ar/Jo

(7.15)

where a is called the coefficient of dispersion and takes on values of about 1.1 to 1.2. Substituting (7 .15) into (7.11), (7 .16) where TE = (N a / k) s, and where we usually assume a to be a constant. This equation is still nonlinear, however, as U F is not a linear function of i. We usually assume up to be a constant. In a similar way we may develop the differential equation for the self-excited exciter shown in Figure 7.30, where we have X£ + Ri = uF or (7.17)

Fig . '7 .30

Self-exc ited exciter .

Following the same logic regarding the fluxes as before, we may write the nonlinear equation (7 .18) for the self-excited case where TE is the same as in (7 .16). In a similar way we establish the equation for the self-excited exciter with boostbuck rotating amplification as shown in Figure 7 .31. Writing the voltage equation with the usual assumptions, (7.19) Kimbark [16] suggests four methods of solution for (7.16)-(7.19). These are (I) formal integration, (2) graphical integration (area summation), (3) step-by-step integration (manual), and (4) analog or digital computer solution . Formal integration requires that the relationship between uF and i, usually expressed graphically by means of the magnetization curve, be known explicitly . An empirical relation, the Frohlich equation (35)

Fig . 7.3\

Self-excited exciter with a rotating a mpl ifier (boost-buck).

257

Excitation Systems VF =

ailtb + i)

(7.20)

may be used, or the so-called modified Frohlich equation UF =

ui/(b + i) + ci

(7.21)

can be tried. In either case the constants a, b, and c must be found by cut-and-try techniques. If this is reasonably successful, the equations can be integrated by separation of variables. Method 2, graphical integration, makes use of the saturation curve to integrate the equations. This method, although somewhat cumbersome, is quite instructive. It is unlikely, however, that anyone except the most intensely interested engineer would choose to work many of these problems because of the labor involved. (See Kimbark [16], Rudenberg (20), and Dahl (35) for a discussion of this method.) Method 3, the step-by-step method (called the point-by-point method by some authors [16,35]), is a manual method similar to the familiar solution of the swing equation by a stepwise procedure [36]. In this method, the time derivatives are assumed constant over a small interval of time, with the value during the interval being dependent on the value at the middle of the interval. Method 4 is probably the method of greatest interest because digital and analog computers are readily available, easy to use, and accurate. The actual methods of computation are many but, in general, nonlinear functions can be handled with relative ease and with considerable speed compared to methods 2 and 3. In this chapter the buildup of a dc generator will be computed by the formal integration method only. However, an analog computer solution and a digital computer technique are outlined in Appendix B. To use formal integration, a nonlinear equation is necessary to represent the saturation curve. For convenience we shall use the Frohlich equation (7.20), which may be solved for i to write (7.22)

i = buF/(a - uF )

We illustrate the application of (7.22) by an example. Example 7.2

A typical saturation curve for a separately excited generator is given in Figure 7.32. Approximate this curve by the Frohlich equation (7.22). Solution

By examination of Figure 7.32 we make the several voltage and current observations given in Table 7.3. Table 7.3.

i vF

A V

0 0

1 30

Exciter Generated Voltages and Field Currents 2 60

3 90

4 5 116 134

6 7 147 156

8 164

9 172

10 179

Since there are two unknowns in the Frohlich equation, we select two known points on the saturation curve, substitute into (7.20) or (7.22), and solve for a and b. One experienced in the selection process may be quite successful in obtaining a good match. To illustrate this, we will select two pairs of points and obtain two different solutions.

258

Chapte r 7

180

160

~

f!/

'/.

~,/

140

'1/

'II

I

I..-

1/

120

Solution '1

~

(;

>

u.

>

~.

E

100

(;

> ~

'u

.s

Solution '2 80

60 Saturati on curve

40

Exciter Field Current, i, onperes

Fig.7 .32

Saturat ion curve o f a sepa ra tely excited exciter.

Solution #/

Solut ion #2

Selec t i

= 3, uF = 90

i=9,uF =I72

i = 4, o, = 116

i = 8, uF = 164

Then th e eq ua tio ns to sol ve are 90 = 3a/(3 172

=

9a/(9

+ b) + b)

116

=

164

=

+ b) 8a/(8 + b)

4a/(4

for whi ch the solut io ns are

a, = 315.9 V b, = 7.53 A

02

= 279.9 V

b,

=

5.65 A

Excitation Systems

259

Both solutions are plotted on Figure 7.32. For solution I

= 315.9i/{7.53 + i) or i = 7.53 uF/ (315.9 -

uF)

(7.23)

uF = 279.9;/(5.65 +;) or ; = 5.65 uF/(279.9 - vF )

(7.24)

UF

and for solution 2

Example 7.3

Approximate the saturation curve of Figure 7.32 by a modified Frohlich equation. Select values of; = 2, 5, and 10. Solution

;=2

60

=

5

134 179

=

; =

; = 10

=

+ b) + 2c 5al(5 + b) + 5c 10al(IO + b) + 10c

2a/(2

Solving simultaneously for a, b, and c,

a = 359

b

-21.95

=

c = 48.0

This gives us the modified formula UF =

359;/(; - 21.95) + 48;

(7.25)

Equation (7.25) is not plotted on Figure 7.32 but is a better fit than either of the other two solutions. Separately excited buildup by integration. For simplicity, let the saturation curve be represented by the Frohlich equation (7.22). Then, substituting for the current in (7.16),

+ bR vF/(a

- v F) = up

(7.26)

This equation may be solved by separation of variables. we write

Rearranging algebraically,

TEVF

dt

where we have defined for convenience, h (t - to)1T E

=

(7.27)

vF)/(av p - hv F)] dVF

= [TE(a -

= up

+ bR. Integrating (7.27),

(I 1h)( V F - UFO) - (a bR 1h2) In [(a vp - hUF)1(aUp - hUFO)]

(7.28)

This equation cannot be solved explicitly for UF, so we leave it in this form. Example 7.4

°

Using the result of formal integration for the separately excited case (7.28), compute the V F versus t relationship for values of t from to I s and find the voltage response ratio by graphical integration of the area under the curve. Assume that the following constants apply and that the saturation curve is the one found in Example 7.2, solution 2. N

=

(J

=

2500 turns 1.2

Up = 125 V

k

=

12,000

R

=

34

n

UFO = 90 V

Chapter 7

260

Solution

First we compute the various constants involved. From (7.16) TE

= No l k = (2500)(1.2)/12,000 = 0.25

s

Also, from Example 7.2 a = 279.9

!"wI

b

280

Now, from the given data, the initial voltage tion (7.22) we compute

UFO

5.65

=

is 90 V. Then from the Frohlich equa-

'-0 = 5.65(90)/(280 - 90)

=

2.675 A

This means that there is initially a total resistance of R; = 125/2.675 = 46.7 g of which all but 34 g is in the field rheostat. Assume that we completely short out the field rheostat, changing the resistance from 46.7 to 34 g at t = o. Since up is 125 V, we compute the final values of the system variables. From the field circuit, i;

=

up/R

=

125/34

=

3.675 A

Then, from the Frohlich equation the ceiling voltage is uF«J

=

0

i«J/(b

+

i«J) = 280(3.675)/(5.65

+ 3.675)

=

1 to.3 V

Using the above constants we compute the uF versus t relationship shown in Table 7.4 and illustrated in Figure 7.33. Table 7.4. Buildup of Separately Excited VF for Example 7.4 VF

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

90.00 95.85 100.12 103.18 105.35 106.87 107.94 108.68 109.19

VF

0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85

109.55 109.79 109.96 110.08 110.16 110.21 110.25 110.28 110.30

From Figure 7.33, by graphical construction we find the triangle acd, which has the same area as that under the UF curve abd. Then from (7.5) with cd = 27.9 V, as shown in the figure, the response ratio = 27.9,/90(0.5) = 0.62. Self-excited buildup by integration. For a self-excited machine whose saturation curve is represented by the Frohlich approximation (7.22), we have (7.29)

261

Excitation Systems 12

::

o>

110

z,

>

~.

E

o

>

27.9 V

d

Fig. 7.33

Buildup o f th e sepa ra tely excited excit er for Exa mple 7.4.

This is recognized to be identical to the previous case except that the term on the right side is uF instead of up' Again we rearrange the equation to separate the variables as dt =

rda - vF)duf

-=-----'---:-

(a - bR)v f

-

u}

(7.30)

This equation can be integrated from to to t with the result 1 -

10

E....l n K

(7.31)

where K = a - bR.

Example 7.5 Compute the self-excited buildup for the same exciter studied in Example 7.4. Change the final resistance (field resistance) so that the self-excited machine will achieve the same ceiling voltage as the separately excited machine. Compare the two buildup curves by plotting the results on the same graph and by comparing the computed response ratios. Solution The ceiling voltage is to be 110.3 V, at which point the current in the field is 3.68 A (from the Frohlich equations). Then the resistance must be R = 110.3/3.68 = 30 n. Solving (7.31) with this value of R and using Frohlich parameters from Example 7.4, we have the results in Table 7.5 and the solution curve of Figure 7.34. The response ratio = 15.4/90(0.5) = 0.342 for the self-excited case.

262

Chapter 7

110

V

F

,

Separate ly e xc it e d

V

F

I

Self- e xcite d

~

(;

>

u.

>

~­

Ol

E

(;

>

o Fig . 7.34

Buildup or the self-excitedexciter for Example 7.5.

Table 7.5. Buildup of Self-excited Vf for Example 7.5 0 .00 0 .05

0.10 0 .15 0 .20 0 .25 0 .30 0 .35 0.40 0.45

90 .00 91.87 93 .61 95 .23 96 .73 98 .10 99 .37 100.52 101.57 102.52

\03.38 104.15 104.85 105.47 106.03 106.52 106.96 107.36 107.71

0 .50 0 .55 0 .60 0 .65 0 .70 0 .75 0 .80 0.85 0 .90

Boost-buck buildup by integration. The equation for the boost-buck case is the same as the self-excited case except the amplifier voltage is added to the right side , or TiYF + bRvd(a - VF) = VF + v R

(7.32)

Rearranging, we may separate variables to write dt = Tt:(a - vF)dvF/(A

+ MV F - v})

(7.33)

whereA = OVR and M = a - VR - bR. Integrating (7.33), we com pute _I T£

to __ 2a - M (M - Q - 2v F)(M + Q - 2v FO) ---In - - - - - - - - - - - Q (M - Q - 2v FO)(M + Q - 2v F)

(A + MV F - v}) In - - - - - - 2 (A + Mv FO - v}o)

1

+where Q = V4A

+

M 2•

(7.34)

263

Excitation Systems

Example 7.6 Compute the boost-buck buildup for the exciter of Example 7.4 where the amplifier voltage is assumed to be a step function at t = to with a magnitude of 50 V. Compare with previous results by adjusting the resistance until the ceiling voltage is again 110.3 V. Repeat for an amplifier voltage of 100V. Solution

With a ceiling voltage of 110.3 V and an amplifier voltage of 50 V, we compute with 0, Ri'X) = V F + V R = 160.3. This equation applies as long as vR maintains its value of 50 V. This requires that i ; again be equal to 3.68 A so that R may be computed as R = 160.3/3.68 = 43.6 n. This value of R will insure that the ceiling voltage will again be 110.3 V. Using this R in (7.34) results in the tabulated values given in Table 7.6. Repeating with V R = 100 V gives a second set of data, also tabulated, in which R = 57.20.

vF

=

Table 7.6.

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70

0.75 0.80 0.85 0.90

Buildup of Boost-Buck vF for Example 7.6 90.00 94.23 97.70 100.50 102.72 104.47 105.84 106.90 107.72 108.34 108.82 109.19 109.47 109.68 109.84 109.96 110.05 110.12

110.17

90.00 96.32 100.84 103.98 106.12 107.56 108.51 109.14 109.56 109.83 110.00 110.12 110.20 110.24

110.27

110.30 110.31 110.32 110.33

These results are plotted in Figure 7.35. Note that increasing the amplifier voltage has the effect of increasing the response ratio. In this case changing V R from 50 to 100 V gives a result that closely resembles the separately excited case. In each case the response ratio (RR) may be calculated as follows: 50) = 2cd/Oa = 2(24.15)/90 = 0.537 RR (v R RR (v R = 100) = 2c'd/Oa = 2(29)/90 = 0.645

7.6.2

Linear approximations for de generator exciters

Since the Frohlich approximation fails to provide a simple VF versus t relationship, other possibilities may be worth investigating. One method that looks attractive because of its simplicity is to assume a linear magnetization curve as shown in Figure 7.36, where

264

Chapter 7 1 2 0r----------.."..----~

.f 110

v ' Boost-buc k F v = 100

~

R

U.

>

'u

.lS

v

' Se parate ly F e xcited

100

d

0.3

0.4

Tim e, s

Fig. 7.35

0.5

0. 6

OJ

Buildup of boost-buck exciters for Example 7.6 . VF

= mi + n

(7.35)

Substituting (7.35) into the excitation equation we have the linear ordinary differential equation v - (R/m)(v F

where

U

Up

-

n)

separately excited

vF self-excited uF + V R boost-buck excited

-.

Actua l cu rve

~

-0 >

Linea r approximation

U.

> ~

'" E

o

>

Excit er Fi el d Cu rren t, i, amperes

Fig .7 .36

Linear approximation to a magnetizat ion curve.

(7 .36)

Excitation Systems

265

This equation ma y be solved by conventional techniques . The question of interest is, What values of m and n, if an y, will give solutions close to the actual nonlinear solutions? This can be resolved by solving (7.36) for each case and then systematically trying various values of m and n to find the best "fit." This extremely laborious process becomes much less painful , or even fun , if the comparison is made by analog computer. In this process, both the linear and nonlinear problems are solved simultaneously and the solutions compared on an oscilloscope. A simple manipulation of two potentiometers, one controlling the slope and one controlling the intercept, will quickly and ea sily permit an optimum choice of these parameters. The procedure will be illustrated for the separately excited case. Linear approximation of the separately excited case. In the separately excited case we set v = vp so that (7.36) becomes F = k, - k 2 u, where

v

k, = (l/T £)(Vp + nR/m)

k 2 = R/T£m

(7.37)

Solution of (7 .37) gives (7 .38) Equation (7.38) is solved by the analog computer connection shown in Figure 7.37 and compared with the solution of (7.26) given in Appendix B, shown in Figure B .9.

Fig. 7.37

So lut ion of the linear equ ation .

Adjusting potentiometers k, and k 2 quickly provides the "best fit" solution shown in Figure 7.38, which is a graph made directly by the computer. Having adjusted k, and k 2 for the best fit, the potentiometer settings are read and the factors m and n computed . In a similar way linear approximations can be found for the self-excited and boost-buck connections .

Time, s

F ig.7 .38

An alog co mp uter compar iso n of linear and Frohl ich model s of the separately excited bu ildup.

Chapter 7

266

1.6.3

The ac generator exciters

As we observed in Chapter 4, there is no simple relationship between the terminal voltage and the field voltage of a synchronous generator. Including all the detail of Chapter 4 in the analysis of the exciter would be extremely tedious and would not be warranted in most cases. We therefore seek a reasonable approximation for the ac exciter voltage, taking into account the major time constants and ignoring other effects. Kimbark [16] has observed that the current in the de field winding changes much more slowly than the corresponding change in the ac stator winding. Therefore, since the terminal voltage is proportional to i, (neglecting saturation), the ac exciter voltage will change approximately as fast as its field current changes. The rate of change of field current depends a great deal on the external impedance of the stator circuit or on the load impedance. But, using the response ratio definition (see Def, 3.19, Appendix E) we may assume that the ac exciter is open circuited. In this case the field current in the exciter changes according to the "direct-axis transient open circuit time constant" TdO where (7.39) This will give the most conservative (pessimistic) result since, with a load impedance connected to the stator, the effective inductance seen by the field current is smaller and the time constant is smaller. Using relation (7.39) we write, in the Laplace domain, (7.40) where uF(s) is the Laplace transform of the open circuit field voltage and vR(s) is the transform of the regulator voltage. If the regulator output experiences a step change of magnitude D at I = 10 , the field voltage may be computed from (7.40) to be UF

=

UFO

+

KD( I

- e - Kv ' with the resulting action described by

Kt

Balanced beam

2

Raise-lower threshold

1

Ms + Bs + K

r

Qui ck rc tse - lower

Rrmn •

threshold

-f;v I

+R

x

L..-_ _

-R

Fig. 7.43

OL

1--------'

Block diagram of the combined raise-lower and quick-ra ise-lower control modes.

Excitation Systems

R

RH

X

271

> K; (quick raise)

RH + ROL

X

< K; (quick lower)

(7.46)

If we set K; > Kx ' this control mode will be initiated only for large changes in V, and will provide a fast response . Thus, although the raise-lower mode will also be operational when I x I > K v ' it will probably not have time to move appreciably before x returns to the deadband. The controller of Figure 7.43 operates to adjust the total field resistance R to the desired value. Mathematically, we can describe the complete control action by combining (7.45) -(7.46). The resulting change in R affects the solution for u, in the exciter equation (7.43) . If saturation is added, a more realistic solution results . Saturation is often treated as shown in Figure 7.44, where we define the saturation function

IB

'A

Excit er Field Cur rent, i, amperes

F ig. 7.44

Exciter saturat ion curve.

(7.47) Then we can show that

(7.48) The function SE is nonlinear and can be approximated by any convenient nonlinear function throughout the operating range (See Appendix D). If the air gap line has slope JIG, we can write the total (saturated) current as i

= GuF(I + SE) = GUF + GUFSE

(7.49)

Substituting (7.49) into (7.43) the exciter equation is

TiJ F =

up - Ri = up - RGuF - RGUFSE

(7.50)

A block diagram for use in computer simulation of this equation is shown in Figure 7.45, where the exciter voltage is converted to the normalized exciter voltage EFD• The complete excitation system is the combination of Figures 7.43 and 7.45. 7.7.2

Conti nuously regulated systems

Usually it is preferable for a control system to be a continuously acting, proportional system, i.e ., the control signal is always present and exerts an effort proportional

272

Chapter 7

Fig . 7.45

Exciter block diagram.

to the system error (see Def. 2. J2. J). Most of the excitation control systems in use today are of this type. Here we shall analyze one system. the familiar boost-buck system, since it is typical of this kind of excitation system . Consider the system shown in Figure 7. J0 where the feedback signal is applied to the rotating ampl ifier in the exciter field circuit. Reduced to its fundamental components, this is shown in Figure 7.46. We analyze each block separately .

Potential transformer and rectifier. One possible connection for this block is that shown in Figure 7.47. where the potential transformer secondaries are connected to bridged rectifiers connected in series. Thus the output voltage Vdc is proportional to the sum or average of the rms values of the three phase voltages. If we let the average rms voltage be represented by the symbol v,. we may write (7 .51) where K R is a proportionality constant and TR is the time constant due to the filtering or first-order smoothing in the transformer-rectifier assembly . The actual delay in this system is small, and we may assume that 0 < TR < 0 .06 s.

Voltage regulator and reference (comparator). The second block compares the voltage J'dc against a fixed reference and supplies an output voltage called the error voltage, which is proportional to the difference; i.e.•

v..

v.

= k(

VREF

-

J'dc)

(7 .52)

This can be accomplished in several ways. One way is to provide an electronic difference amplifier as shown in Figure 7.48. where the time constant of the electronic amplifier is usually negligible compared to other time delays in the system. There is often an objection, however, to using active circuits containing vacuum tubes, transistors, and the associated electronic power supplies because of reliability and the need for replace-

Fig . 7.46

Simplified diagram or a boost-buck system .

Excitation Systems

a .....- - - -

Fig . 7.47

273

r---------hn

Potential transformer and rectifier connection .

ment of aging components. This difficulty could be overcome by having a spare amplifier with automatic switching upon the detection of faulty operation . Another solution to the problem is to make the error comparison by an entirely passive network such as the nonlinear bridge circuit in Figure 7.49. Here the input current id< sees parallel paths t, and i 6 or id< = t, + i 6 • But since the output is connected to an amplifier, we assume that the voltage gain is large and that the input current is negligible, or i, = O. Under this condition the currents i. and i6 are equal. Then the output voltage V. is (7 .53) The operation of the bridge is better understood by examination of Figure 7.50 where the v-i characteristics of each resistance are given and the characteristic for the total resistance R L + R N seen by i, and i6 is also given. Since i. = i6 , the sum of voltage drops VL and vN is always equal to ~-

ce

0

~

·w10

~

2

£

0.8

(;

>

]

5

0.40

~

0

0

20

0 .80

F

a.

; 0 .6 , K = 0,02

T

c'

F

.~

15

"

0

2.40

1.60

Time , s

~

T

t-

~ 0 .00 0 .00

F

3 .20

; 0.6, K = 0.02

F

1. 20

~

0

c

E " 0.8

·w 10 .§

(;

>

1

5

0.40

.~

0

- 20

20 T

>-

cc

F

- 15

- 10

Reol

-5

0

...."

0.00 0.00

~

a.

; 0 .6 , K = 0 .0 3 F

T

c'

.s

;;

15

"

F

1.60

0.80

Time,s

2.40

3.20

2.40

3.20

= 0 .6, K = 0.03

F

1. 2

"

0

·w 10 .§

~ 0 .80 2

(;

>

(; 0.40

5

c

.~

0

Fig.7.58(a)

- 20

- 15

- 10

Real

-5

0

...."

0.00 0.00

0 . 80

1.60

Time,s

Effect of variation of K F on dynami c response : TF = 0.6. KF Type I excitation system.

=

0.01,0.02, and 0.03 respectively.

Solution 2 The above system is studied for different values of TF and K F with the aid of special digital computer programs. The programs used are a root-finding subroutine for polynomials to obtain the zeros of equation (7.63), a root-locus program , and a timeresponse program. Two sample runs to illustrate the effect of TF and KF are shown in Figure 7.58. In Figure 7.58(a) T F is held constant at 0 .6 while KF is varied between 0.01 and 0.03 . Plots of the loc i of the roots are shown for the three cases , along with the timeresponse for the "rated" value of K A • The most obvious effect of reducing K F is to reduce the settling time. In Figure 7.58(b), KF is held constant at 0.02 while TF is varied between 0.5 and 0.7. The root-locus plots and the time-response for the system are repeated. The effect of increasing TF is to reduce the overshoot.

282

Chapter 7

20 T

t

a

F = 0 .5, KF = 0.02

c' 0

15

.~

's

s ·w10

~

8, 0 .80

E

a> ac 'f

5

-20

20 T

z-

.g

0 '"

-15

-10

Reol

-5

0

T

Tim e , s

2 .40

= 0 .6 ,

K = 0 .02 F

0.80

1. 60

F

3.20

~

0

10

~

0.80

'" a> g 0. 40 E

'f

~

- 20

T

F

-15

=0.7,

-1 0

Reol

-5

0

0. 00 0 . 00

a.

e

.~ 1.20 0

15

Times, s

2 .40

3.20

'l"F = 0.7 , K = 0 . 02 F

's

~ 8, 0.80

10

0

I-

~

K = 0.02 F

E

a>

g

5

Fig.7.58(b)

1.60

0.80

's

20

.E

0 .00

c'

15

0

0

.! 0 .00

.2 1.20 0

5

'0,

0 .40

&.

F = 0.6, KF = 0 .02

!

t 2

'l'F = 0.5, K = 0.02 F

0

!

0

1. 20

'f

~

-20

-15

-1 0

Reol

I-

-5

0

0.40 0.00 0.00

Effect of variation ofT F on dynamic response: K F Type I excitation system .

0. 80

1.60

Time, s

;

0.02,

TF;

2.40

3 .20

0.5, 0.6, and 0.7 respectively.

From Figures 7.S8(a) and 7.S8(b) we can see that the values of TF and KF significantly influence the dynamic performance of the system. There is, however, a variety of choices of K F and T F, which gives a reasonably good dynamic response. For this particular system, T F = 0.6 and KF = 0.02 seem to give the best results.

Solution 3 An engineer with experience in s plane design may be able to guess a workable location for the zero and estimate the value of K F that will give satisfactory results. For most engineers, the analog computer can be a great help in speeding up the design procedure, and we shall consider this technique as an alternate design procedure. From Figure 7.54 we write, with V; = 0,

283

Excitation Systems

-v,

Fig. 7.59

A nalog com puter d iagram for a linea r excitat ion system with deriv at ive feed bac k.

or

(7.64) For the amplifier block of Figure 7.54 we have VR = K A V./(I rearranged as

+

TAS) ,

which may be

(7.65) Equation (7.64) may be represented on the analog computer by a summer and (7.65) by an integrator with feedback. All other blocks except the derivative feedback term are similar to (7.65) . For the derivat ive feedback we have ~ = sKFEFD/( I + TFS) , which can be rewritten as

(7.66) Using (7.64) --(7.66), we may construct the analog computer diagram shown in Figure 7.59. Then we may systematically move the zero from S = 0 to the left and check the response . In each case both the forward loop gain and feedback gains may be optimized. Table 7.7 shows the results of several typical runs of this kind . In all cases K R has been adjusted to unity, and other gains have been chosen to optimize V, in a qualitative sense. The constants in these studies may be used to compute the cubic coefficients (7.62), and the equation may then be factored . If the roots are known, a root locus Table 7.7.

Summary of Analog Computer Studies for Example 7.8

0-90%

Run

00 = -

KF

KA

Settling time, s

Percent overshoot

rise time . s

1 2 3 4 5

1.75 1.50 1.25 1.00 0 .75

0 .16 0.16 0.16 0 .16 0.16

50 50 50 50 50

1.35 1.05 1.05 2.05

9.2 8.0 22.8 42 .0 70.0

0 .37 0.30 0.25 0.215 0.20

I rF

very long

284

Chap ter 7

ri

!

I I I

L I

I

ilJ1lli . I .

1

..

I I

I

I

I

.

I

I

I

.If-+

.. .-

-

' ..

_. -

_..

I~

.-

-

_ .

.-

.. 1. , _.

.

f

-f ·

I

1

..-

f

_.

--_.. e--_. .-

-+ -t-- -I - -+--'

1- -1-- ;

__ ._

._..

1

V\

_.

j

R

..-

_. ." •.. . ...

.. .- -;

- I- I-

:= ~ .- -

-

..

- ·1-

..

~~f

.. ._.

-

-

--

r -r- -t --+_ ·-f·---t--r--t--+--+--+ ---f--·--+-· ,

_.

-

-

--.- II _ I

11

.

~f-- E --- . . f- FD - -+ ....

r- -

--

sFe,...

.- - ,--

Fig. 7.60

-

- ---t- f-+-f---- --

± ..

.r: __ - ~

; . f

1

1-

-.. --

. .

I - ~ -- 1- - - -· f-- - I -. ~ . r-r-t -r-t-r1·-

--

-

.

.-

'H-t:-n v +

..

...

_.

...

.~

VI

-

..

ve

_.

..

---.

-

..

. - _. _. ... .... -

...

_. if\"

- -!- -

_.

I

.. ..

r-. · -··f ··- .· . . f ...

.,

-I t I

i

I

III

..

..

L

V REF

,.---

/

..

-

~l~

_-. ...._-. - +--

I

f

1

I

II I

II- ..-

_.

- V3

..

'.

,

f ·· 1

1 ··· 1

1

_...

._.

..

.-

'-

I - -_. ._ . --

J~.- I=

_.

.. -

-

- ._ -

. . ....

-+ --+--+--+_._+--+-- t---+-t--+--\--+--+

~

_.

.... ,- - --_.- .-- - H-f-- --'+-1-- _... _..--

r-r-

- ..

- _.

- -.-

-

1--

- I-. - !--

- ..

-v, .-

...

- _. - -

-

-

.- !--

'-

.- . .-

_.

- f-r- - I II-

--

f- --i----+--+--f.·-f - -+-+- ,.. - +-.... --t-+---f --

,- - I--.

--1-- -

-- - -.-

._ .

1--- - .-

1- - -V 3 / s ' -1---

1- .- ._ f- -. !-. -- ---.

- I-.- f-... 1 sec

.

._.

._-

- - 1 - .-

-- -_.. I---. C---- _. f--_. .. .. ..

-

._-

_. -

.. 1-

- _.

An alog co mp uter results For Example 7.8. Solution 2.

may be plotted and a comparison made between this and the previous uncompensated solution . The actual analog computer outputs for run 2 are shown in Figure 7.60 . Onesecond timing pulses are shown on the chart. Th e plot is made so that 20 such pul ses correspond to I s of real time. This system is tuned to optimize the output v" which responds with little overshoot and displays good damping. Note, however , that this requ ires exce ssive overshoot of E FD and VR , which in physical systems would both be limited by sa turatio n. Inclusion of saturation is a practical necessity, even in linear simulation . Examples 7.7 and 7.8 are intended to give us some feeling for the derivative feedback of Figure 7.54. A study of the eigenvalues of a synchronous mach ine indicates that a first-order approximation to the generator voltage response is only approx imately true. Nevertheless, mak ing this simplification helps us to concentrate on the characteristics of the excitation system without becoming confused by the added complex ity of th e generator. Visualizing the root locus of the control is helpful and shows clearly how th e compensated system can be operated at much greater gain while still hold ing a suitable damping ratio. These studies also suggest how further improvements could be rea lized by add ing ser ies compensation, but th is is left as an exercise for the interested reader .

285

Excitation Systems

7.8

Stat.-Space Description of the Excitation System

Refer again to the analog computer diagram of Figure 7.59. By inspection we write the following equations (including saturation) in per unit with time in seconds.

~ = (KR/TR) ~ - (l/TR) V;

~

=

VR =

EFD = ~

(KF/TF) EFD

(l/TF)

-

~ V R < VRmax, VR > VR min

(KA/TA) V~ - (1IT A) V R

+ KE)ITE]EFD - V; - ~

(l/T E) VR - [(SE

= VREF +

~

(7.67)

Since SE = SE(EFD) is a nonlinear function of E FD, we linearize at the operating point to write

where we define the coefficient S~ to describe saturation in the vicinity of the initial operating point. Suppose we arbitrarily assign a state to each integrator associated with the excitation. Arbitrarily, we set xs, X9' XIO and XII to correspond to the variables VI' V3 , V~~ and EFD • In rewriting (7.67) to eliminate EFD in the second equation we observe that, when per unit time is used, the product (TFTE) must be divided by WR for the system of units to the consistent. The preliminary equations are obtained:

Vi

Xs

~

X9

0

VR

x

KA TA

KA TA

EFD

XI'

0

0

lO

TR

0

0

wRKF

wRKF (S E + KE)

TFTE

TpTE

---

TF

0

0

TA (S~

TE

+ KE)

TE

KR

T-v: R I

Xs

0

X9

+

K..

X IO

- (VREF + ~) TA

XII

0 (7.68)

In equation (7.68) the term (KRITR) ~ is a function of the state variables. (4.46) or (6.69)

V;

=

(I /3)(v~

+

v~)

From (7.69)

where vd and vq are functions of the state variables; thus (7.69) is nonlinear. If the system equations are linearized about a quiescent operating state, a linear relation between the change in the terminal voltage J-';L\ and the change in the d and q axis volt-

286

Chapter 7

ages

Vd 4

and vq4 is obtained. Such a relation is given in (6.69) and repeated here: V,4

=

(I / 3) ( -V dO Vd4

Jl;o

+

V qo ~

,0

vq A ) = d.OVd4 + qovq A

(7.70)

The linear model is completed by substituting for V d A and vq A in terms of the state variables and from (6.20) and by setting vF = (V3 'F/ LAO)EFO •

7.8.1

Simplified linear model

A simplified linear model can be constructed based on the linear model discussed in Section 6.5. The linearized equations for the synchronous machine are given by (the a subscripts are dropped for convenience)

(7.71)

From (7.71)

e;

T, = K.o + K 2E;

(7.72)

JI; = K, lJ + K6E;

(7.73)

= -(ljK)Tdo)

E; - (K4 / TdO) 0

+ (I/Tdo) EFD

(7.74)

From the torque equation (6.73) and (7.72)

w = Tm/Tj

-

(K./Tj ) 0 - (K2 / Tj ) E; - (D/Tj )

(7.75)

W

and from the definition of (&)4

o=

(7.76)

W

The system is now described by (7.68) and (7.72)-(7.76). The state variables are [E; W 0 ~ Vi VR EFO ) ' The driving functions are VREF and t; assuming that ~ in (7.68) is zero. The complete state-space description of the system is given by X'

=

E'q

E'f cd

~

~

Kl 1'; o

_ K2 1')

CAl

0 0

Tit

K. 1';0

K. Tj

0

0

K6KR

~

0

v1

VR

0

0

0

0

0

0

0

w

0

0

0

0

~

0

0

0

0

Jt;

0

Vi

0

VR

-

KsKR Tit

'1'"

0

J')

0

0

0

V"

0

0

0

EFD

0

0

0

E FD

VI

wRKF 1'F

KA

KA

1' A

1' A

TA

0

0

-1'£

+

wRKF (Si

TF TE

KE)

TFTE

0

-

(S~

+

T£

0

E'q

1'jo

K£)

E FD

Tm Tj

KA 1'..

VREF

0

(7.77)

Excitation Systems

7.8.2

287

Complete linear model

By using the linearized model for a synchronous machine connected to an infinite bus developed in Chapter 6, the excitation system equations are added to the system of (6.20). Before this is done, ~ must be expressed in terms of the state variables, using (6.25) and (7.70). These are repeated here (with the d subscript omitted), VI

= dOvd + qovq

ud

=

uq

-

= -

K cos (~o - a)~ + Reid + t.l, + wOLeiq + iqOLew K sin (~o - a)~ + Reiq + Leiq - woLeid - idOLew

(7.78)

From (7.78) and using

v'3 VXl qO ~

K cos (~o - a)

v'3V~dO ~ -Ksin(~o - a)

we get V,

=

-V3(doV~qo - qOV~dO)~

+ (doR e - qowoLe)id + (dowOL e + qoRe)i q + (doiqOL e - qOidOLe)w + doLeid + qoLei q

(7.79)

Substituting in the first equation in (7.68), VI = -(I/TR) VI - (KR/T R) VJ (do V~qO - qo V~dO) ()

+ (KR/TR) (doR e - qowoLe) id

+ (KR/TR) (dowoLe + qoRe)iq + (KR/TR)(do;qo - qO;do}Lew + (KR/TR)doLeid + (KR/T R ) qoLeiq

The remaining equations in (7.68) will be unchanged. The equations introduced by the exciter (for V:, = 0) will thus become

VI -

(KR/T R) doLei d - (KR/T R) qoLei q = (KR/T R ) (doR e - qowoLe ) id + (KR/T R) (dowoL e + qoR e ) iq + (KR/T R) (doiqo - qoidO)Lew

V3

- (KR/T R) V3(doV~qo - qoV~dO){) - (l/T R) VI = -(l/T F) V3 + (KF/TFTE) VR - [KF(SE + KE)/TFTE] E FD = -(KA/T A) VI - (KA/T A) V3 - (l/T A) VR + (KA/T A) VREF

VR EFD =

(liTE) VR

-

[(S~ + KE)/T E] E FD

(7.80)

This set of equations is incorporated in the set (6.20) to obtain the complete mathematical description. The new A matrix for the system is given by A = - M- 1 K. Note that in (7.80) the state variable for the field voltage is E FD and not uF. Therefore, the equation for the field current is adjusted accordingly. In this equation the term v F is changed to (V3rF I LAD) E FD • The matrices M and K are thus given by the defining equation" = - Kx - Mx, where i d iF

v'

=

M is given by

[0

0

iD

iq

iQ

0

0

0

w I I. I

~

Tm 0

VI

V3

0

0

VR K A VREf TA

EFD

0]

288

Chapter 7 iD

kM,..

kM o

id

t

if

kM f

t.,

Mil.

if)

kMf)

Mil.

L"

d

iq

o

0

iQ

M

w

ir

i(/

w

Lq

kM Q

kM Q

LQ

0

0

-

0

KR

0

VJ

0

0

0

0

0

VII.

0

0

0

0

0

E~,.."

0

0

0

0

0

II.

o

o

o

o 0

1,

qoL ('

VII.

e.;

o

o

o

o o

o o

0

-T

V.\

o

R d L VI - K T 0 f'

II.

VI

0

o o o

o

o o

o (7.81 )

And the matrix K becomes

i et

t,

iF

if)

R

0

0

0

iF

.. ~

I I I

-- -

_. -

.... -

-woL et

-wokM,.·

-wokM o

'»

0

0

0

I

lJ VI V) VR

E Fo

!3 (~o'I

- ..

- Letiqo)

0

_. - - - -

.-

-

_.

-

-.

- ..

0

- .. _ .. -

0

0

-

.

0 ..

0

0

-

j kMoi,o

0

0

0

..... - -

0

0

0

'.-

j kMFi,o

- - - - - ..

K 81

-

0

'o wokM Q

0

I I I " -1- -

'0 ... - -

t,

---

iq

woiq

I I

I I

w K =

0

- - - .- - - - -

- - -

I

0

'F

0

io

,

-

,- I I

I I

, I

.

-

-..-

0

0

-

R

0

I I

+ LqidO)

j kMQi

,,

, , I

I I

JO

I

0

K84

0

0

I

0

0 - - -r . I -~etO I

.

I t I I I I I I I

Xqo

0

I I

-,I I I I I

I I

0

,

0

0

0

I I I I I

I

0

I I

0

0

-D

0

.

-

K86 0

0

KS7 0

V)

VR

0

0

0

0 0

I

0

, I

I I

_ ..

0

",- , , 0 I

I

-I

VI

I I I I

.. r

-V3Vcod O

0

t

I I I

I I I

-v'3VcoqO

I

II I I I I

I

I I I

'Q

0

I

I I

I

-

!3 (-~dO

I

0

0 '.

W I

I

, I

I I I

I I I

t

I

0 0

0 0

I I

, I

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

.

0

0

1-

0

0

-r,.·

KA

/(4

TA

TA

0

0

TR

0 V'fr,.. - wRkM F

_-----------------

0

t

I I I I

0

Em

1

--------------

-------------ooRKF ooRKF (Sl + TFTE

TFTE

0

TA

S~· T,;"

+ Kl.'

r«

(7.8: where

KS1

-(KR/TR)(doR~ - qowoL e )

KS4

-(KR/TR)(qoR,

K S6

-(KR/TR)(doiqo - qO;dO) t;

K S7

= -

+ dowoL e )

(V3 KR/TR)(qo V «JdO

-

do V «Jqo)

289

Excitation Systems

Example 7.9 Expand Example 6.2 to include the excitation system using the mathematical description of (7.80). Assume that the machine is operating initially at the load specified in Example 6.2. The excitation system parameters are given by 0.01 s

=

TR

3.77 pu

=

K R = 1.0

0.05 s

TA =

KA

=

0.5 s

KE

=

-0.05

18.85 pu

=

0.715

TF =

40

=

= 188.5 pu

TE

KF

= 269.55 pu

= 0.04

Let the exciter saturation be represented by the nonlinear function SE

A EX exp(B EXEFD)

=

=

0.0039 exp( 1.555 E FD )

Solution From the initial conditions V dO =

-1.148

vqO = 1.675

ido

=

-1.59

iqo

=

0.70

V3 V oo dO =

o

E FDO = 2.529

VtO = 1.172

V oo qO

-1.397

= 1.025

do = (I /3)(v dO/ V'O) = -( 1/3)( 1.148/1.172) = -0.3264 qo

=

(1/3)(v qo/Vto)

(1/3)(1.675/1.172)

=

=

0.4762

The linear saturation coefficient at the initial operating point is

s~ = asE = 8E FD

1.555 [0.0039 exp (1.555 x 2.529)] = 0.3095

The exciter time constants should be given in pu time (radians). The new terms in

the K matrix are

Kg,

=

-(1.0/3.77)(-0.326 x 0.02 - 0.476 x 0.4)

Kg4

=

-(1.0/3.77)(-0.326 x 0.4 + 0.476 x 0.02) = 0.0321

=

0.0523

Kg6 = -(1.0/3.77)(-0.326 x 0.70 + 0.476 x 1.59)0.4 = -0.0561 Kg?

=

Kgg

=

K99

=

K9 - 10 K9- 11

= =

K IO_8 =

K IO_10

=

(1.0/3.77)(-0.326 x 1.025 + 0.476 x 1.397)

l/'R I/'F

K 2_"

=

0.0751

377/(269.5 X 188.5) = -2.967 X 10- 4 -wRKF(S~ + KE)/'TF'TE = 2.967 X 10- 4 X 0.26 = 7.7 X 10- 5 KA/'A = 40/18.85 = 2.122 = K IO-9 li'A = 1/18.85 (S~

=

0.265 0.0037

-wRKF/'TF'TE

K II - 10 = liTE K II _1I

=

=

=

-0.04

=

= 0.0053 + KE)/TE =

0rF /wR k M F

The new K matrix is given by

=

X

0.053

0.15 x 0.0053

=

0.000796

0(0.000742)/1.55

=

-0.000829

290

Chapter 7 0.021

o

o

:

2.040

o 0.00074 o : I o 0 0.013: __________________ 1 - 2.100

o

K -

- 1.550

- 1.550

0

b

I f

I

'1.490

1.430

-1.025

0

0

0

0

o

o

o

0

0

0

-0.0008

0

0

0

0

0

0

o o

0.021

o

-1.039

-1.397

0

0

0

0

0

0.054

0

0

0

0

0

0

0

0

-I

0

o o

o o

o o

o

0

0

0:

------------------1--------------+~--

-0.014

-0.362

-0.362 f -1.428

-0.780:

I

I

o 0 0: 0 0 : ------------------i-------------- 1 -

0.0523

o o o

0

0:

o

o o

o

0

0

0

:

0

0

I

0

0

I

o

o

0.032

I I I

- -

---------

0.075

0.265

o

0

o

:

o

:

o

I

-----------------------------

---------

: -0.056

I I I

_

o o

0.0037 -2.967

2.122 2.122

0

0

o 0

x 10- 4 7.7 x 10- 5

0.053

0

-0.0053

0.0008

The new M matrix is given by -0.0479 0.0211

KRdoL~/TR KRqoL~/TR

2.100

1.550

1.550

1.550

1.651

1.550'

1.550

1.550

1.605:

I I I

I I I

0

I

0,

:

--------t--------

--------------~----------

o

I I

:

0

I

2.040

1.490

I I

0

1.490 1.526

0

:

1 I --------------r-----------------,--------

M

:

o

0

I I I

:-1786.94 0

I

I I I

: I I

0

0

--------------r----------r--------T--------

0.048

0

0 : -0.021

0

:

: I

0

0

J

I

o

0

0

o o

0

0:

0

0:

0

O!

0

0

I I I

I

I I

0

,0

0

I

:0

I

I

0

0

0

0

10 0

I

0

0 0

The A matrix is given by -36.062

0.4388

12.472 -4.9503 22.776

4.3557

3590.0

2649.7

-3505.7

-2587.5

o

0

1751.3:

0

0

0

76.857:

14.142 : -3487.2 1206.0

880.86:

845.46

-605.7:

0

0

0

5.5317

-96.017:

2202.4

1608.6 ~

1544.0 -1106.1:

0

0

0

-4.8673

1776.7

0

0

0

I

____________________ '

A •

2649.7 : -36.064 I

-2547.0: -2444.6 I

.

~

90.072:

I

-:-

-2587.5:

_ - _ - _. - ... - - - - - .l - - - - - - -

2387.4:

I

- - -

- - ... - - .. - - - - -

0

-0.4904

35.218 -123.32: -1735.0 -2331.4: 0 0 0 0 - - - - - - - - - - - .. - - - - - - - .. - - - - - - - - - - ~ - -- - - - - .. -' - - .... - - - - - - - - - - . - - - - - - - - - - - - - - - - - . -0.0078 -0.2027 -0.2027 t -0.7993 -0.4422: 0 0 0 0 0 0 10-) I

0:

0

I

0:

1000

0

0

- - - - - - - - - - - - - - - - - - - -1- - - - - - - - - - - - - ~.:: - - - - - - .- - - - - -

25.394

o o

o

56.019

55.361.; I

134.50

124.15

I I

211.02 -108.65

I I

0

0

0

0.0235

0

-3.7099

0.2967

-0.077

-2122.1

-5.3052

0

0

53.052

-0.7958

0

II

0

0:

0

0

0:

0

0:

0

0 : -2122.1

0

0:

0

0:

0

0:

I I

0

-265.26

0

I

0

0

-- - - - - - - - - - - - .. - .- - - - - _. -

0

Excitation Systems

291

The eigenvalues obtained are AI

=

-0.0359 + jO.9983

A2

=

-0.0359 - jO.9983

A3

=

-0.2653

= -0.0015 +jO.0290 Ag = -0.0015 - jO.0290 A9 = -0.00125 + jO.00297

A4

=

-0.0986

AIO

=

As

=

All

=

A6

=

-0.1217 -0.0548

>"7

-0.00125 - jO.00297 -0.0037

Example 7.10

Repeat Example 7.9 for different exciters. Use the same machine loading. Tabulate the data used and the eigenvalues obtained.

Solution

For this example we will use the same machine loading of Example 5.1 and three exciters made by the same manufacturer: W TRA, W Brushless, and W Low TE Brushless. Data for the exciters and the appropriate M and K constants are given in Table 7.8. The eigenvalues obtained are tabulated in Table 7.9. Table 7.8.

Exciter Data and Elements of Matrices M and K (Loading of Exam pie 5.1)

Constants and matrix elements

IEEE type I exciter WTRA

KA

W Brushless

W low T E Brushless

VRmax VRmin

400 0.05 -0.17 0.95 0.04 1.0 1.0 0.0* 0.0027 1.304 0.0874 0.1140 3.5 -3.5

400 0.02 1.0 0.80 0.03 1.0 1.0 0.0* 0.098 0.553 0.4282 0.2368 7.3 -7.3

400 0.02 1.0 0.015 0.04 0.50 1.0 0.0* 0.0761 0.4475 0.2510 0.1123 6.96 -6.96

M g• , M g_4

3.862069 -4.753316

3.862069 -4.753316

3.862069 -4.753316

K g_, K g_4

4.9464 3.6244 -6.5741 10.2754 26.5252 0.002653 -0.000112 -0.000006 21.220159 0.053050 -0.002792 -0.000156

4.9464 3.6244 -6.5741 10.2754 26.5252 0.002653 -0.000099 0.000123 53.050398 0.132626 -0.003316 0.004101

4.9464 3.6244 -6.5741 10.2754 26.5252 0.005305 -0.014147 0.015735 53.050398 0.132626 -0.176835 0.196693

TA

KE TE

KF TF KR TR

A EX REX SEO SE

KS-6 KS•7

K g_S

K9-9

K 9-1O

K9_11 K,o-s = K'O-9 K IO-1O K".IO K 11• 11

*Where

TR

=

0.0 take

TR

=

10- 4 •

292

Chapter 7 Table 7.9.

Eigenvalues for System of Example 7.10 (Loading of Example 5.1) Exciter type

WTRA

-0.03594 -0.03594 -0.265 x -0.09804 -0.12299 - 0.02536 -0.02536 -0.00076 -0.00076 -0.00340 -0.00340

+ jO.99826

- jO.99826 102

+ jO.03912 - jO.03912 + jO.02444 - jO.02444 + jO.00249 - jO.00249

W Brushless

-0.03594 -0.03594 -0.265 x -0.07300 -0.12315 -0.07870 -0.07870 -0.00071 -0.00071 . -0.00447 -0.00447

+ jO.99826 - jO.99826 102 + jO.02139 - jO.02139 + jO.02444 - jO.02444 + jO.00185 - jO.OOI85

W low T E Brushless

-0.03594 -0.03594 -0.26525 -0.09763 -0.12302 -0.16664 -0.16664 -0.00082 -0.00082 -0.00177 -0.00177

+ jO.99827 - jO.99827 x 102 + jO.86637 - jO.86637 + jO.02468 - jO.02468 + jO.00353 - jO.00353

The results tabulated in Table 7.9 are for the same machine and loading condition as used in Example 6.4 except for the addition of the exciter models. Comparing the results of Examples 6.4 and 7.10, we note that two pairs of complex eigenvalues and two real eigenvalues are essentially present in all the results. We can conclude that these eigenvalues are identified with the parameters of the machine and are not dependent on the exciter parameters. The additional eigenvalues obtained in Example 7.10 and not previously present are comparable in magnitude except for one complex pair associated with the W Low TE Brushless exciter. For this exciter a frequency of approximately 50 Hz is obtained, which might be introduced by the extremely low exciter time constant. The same example was repeated for the loading of Example 5.2 and for the same exciters. The results obtained indicate that only one pair of complex eigenvalues change with the machine loading. This pair is one of the two complex pairs associated with the machine parameters. The eigenvalues associated with the exciter parameters did not change significantly with the machine loading. 7.9

Computer Representation of Excitation Systems

Most of the problems in which the transient behavior of the excitation system is being studied will require the use of computers. It is therefore recognized that the solution of systems can be greatly simplified if a standard set of mathematical models can be chosen. Then each manufacturer can specify the constants for the model that will best represent his systems, and the data acquisition problem will be simplified for the user. As the use of computers has increased and programs have been developed that represent excitation systems, several models have evolved for such systems. Actually, the differences in these representations was more in the form of the data than in the accuracy of the representation. Recognizing this fact, the JEEE formed a working group in the early 1960s to study standardization. This group, which presented its final report in 1967 [15], standardized the representation of excitation systems in four different types and identified specific commercial systems with each type. These models allow for several degrees of complexity, depending upon the available data or importance of a particular exciter in a large system problem. Thus, anything from a very simple linear model to a more complex nonlinear model may be formulated by following these generalized descriptions. We describe the four IEEE models below.

Excitation Systems

293

The excitation system models described use a pu system wherein 1.0 pu generator voltage is the rated generator voltage and 1.0 pu exciter voltage is that voltage required to produce rated generator voltage on the generator air gap line (see Def. 3.20 in Appendix E). This means that at no load and neglecting saturation, EFD = 1.0 pu gives exactly ~ = 1.0 pu. Table 7.10 gives a list of symbols used in the four IEEE models, changed slightly to conform to the notation used throughout this chapter. Table 7.10. Symbol

Excitation System Model Symbols

Description

Symbol

EFD = exciter output voltage IF = generator field current generator terminal voltage ~ generator terminal current I,

KA KE KF K[ Kp

Kv

SE

Vs

regulator gain

TE

= regulator amplifier time constant = exciter time constant

TF

= regulator stabilizing circuit time

TA

TFI,TF2

exciter constant related to selfexcited field = regulator stabilizing circuit gain = current circuit gain in Type 3 system = potential circuit gain in Type 1S or Type 3 system = fast raise/lower constant setting, Type 4 system = exciter saturation function = auxiliary (stabilizing) input signal

Description

TR TRH VR

constant same as T F for rotating rectifier system regulator input filter time constant rheostat time constant, Type 4 regulator output voltage

VRmax.

maximum value of

VR

VRmin

minimum value of

VR

VREF

regulator reference voltage setting field rheostat setting

V RH

Note: Voltages and currents are s domain quantities.

7.9.1

Type 1 system-continuously acting regulator and exciter

The block diagram for the Type 1 system is shown in Figure 7.61. Note that provision is made for first-order smoothing or filtering of the terminal voltage V, with a filter time constant of TR' Usually TR is very small and is often approximated as zero.

Fig. 7.61

Type I excitation system representation for a continuously acting regulator and exciter. (Ct; IEEE. Reprinted from IEEE Trans., vol. PAS-87, 1968.)

The amplifier has time constant T A and gain K A , and its output is limited by VR mu and VRmin • Note that if we have no filter and the rate feedback is zero (KF = 0), the input to the rotating amplifier is the error voltage (7.83)

294

Chapter 7

W

I I I

o LL

I

I

I

I

C)

(;

>

I

I

I

I

I

I I

I I

I

-I A

A

= B- 1

I

I

Fig. 7.62

A - 6

: SE = f(EFD) " - 6-

" E

I

~

Excite r Field Current ,i

Exciter saturation curves showing. procedure for calculating. the saturation function SE. Reprinted from lEU; Trans.. vol . PAS-l!? 1968.)

(~.'

IEEE.

and this voltage is small, but finite in the steady state. The exciter itself is represented as a first-order linear system with time constant TE' However, a provision is made to include the effect of saturation in the exciter by the saturation function SE' The saturation function is defined as shown in Figure 7.62 by the relation

Sf

= (A - B)/ B

(7 .84)

and is thus a function of Em that is nonlinear. This alters the amplifier voltage VR by an amount SEEm to give a new effective value of VR • viz., (7 .85) This altered value VR is operated upon linearly by the exciter transfer function. Note that for sufficiently small E FD the system is nearly linear (SE = 0). Note also that the exciter transfer function contains a constant K E • This transfer function (7.86) is not in the usual form for a linear transfer function for a first-order system (usually stated as 1/(1 + TS). From the block diagram we write E FO = VR/(K E + TES), and substituting (7.85) for VR we have (7.87) which includes the nonlinear function SEE FD • Equation (7.87) corresponds in the time domain to

TEEFO = -KEEFD

+

VR

-

SEE FD

(7.88)

Comparing with (7.32). for example. where we computed T iJF

=

vF

+

vR

-

bRvF!(a - v F )

with the nonlinearity approximated by a Frohlich equation, we can observe the obvious similarity. Reference [15] suggests taking

Excitation Systems

295

(7.89) which corresponds to the resistance in the exciter field circuit at t = O. Some engineers approximate the saturation function by an exponential function, i.e., (7.90) The coefficients A EX and BEX are computed from saturation data, where SE and EFD are specified at two points, usuall y the exciter ceiling voltage and 75% of ceiling. The function (7.90) is easy to compute and provides a simple way to represent exciter saturation with reasonable accuracy. See Appendix D. Finally we examine the feedback transfer function of Figure 7.61 (7.91) where K F and T F are respectively the gain constant and the time constant of the regulator stabilizing circuit. This time constant introduces a zero on the negative real axis. Note that (7.91) introduces both a derivative feedback and a first-order lag. Reference [I S] points out that the regulator ceiling VRmax and the exciter ceiling EFDmax are interrelated through S E and KE. Under steady-state conditions we compute VR

=

KEEFD + SEEFD

(7.92)

with the constraint VRmin < VR < VR m ax ' then (7.93) Thus there exists a constraint between the maximum (or minimum) values of E FDmax and VRmax (EFDm in and VR min ) . 7.9.2

Type 1S system-controlled rectifier system with terminal potential supply only

This is a special case of continuously acting systems where excitation is obtained through rectification of the terminal voltage as in Figures 7.17 and 7.18. In this case the maximum regulat or voltage is not a constant but is proportional to v" i.e., (7.94) Such systems have almost instantaneous response of their main excitation components such that in Figure 7.61 KE = I, T£ = 0, and SE = O. This system is shown in Figure7 .63. A state-space representation of the Type IS system can be derived by referring to (7.67) (written for the Type I system), setting VR = E FD and eliminating (7.65), with

Fig. 7.63

Type IS system. (© IEEE. Reprinted from IEEE Trans .. vol. PAS-87 , 1968.)

296

Chapter 7

the result VI = (KRITR) V, - (I/T R) VI

EFD V.

V. + V. -

= (KAIT A) =

VREF

EFD

(I/T A )

-

~

V3

(KdTF)

EFD -

(I/TF)

J!;

EFD < VRmax, EFD > VRmin (7.95)

V3

By using (7.79) and substituting for id and iq , we can express V, as a function of the state variables. For the linearized system discussed in Chapter 6 where the state variables

we can show that

V,

=

fiE FD +

L ];.x 7

(7.96)

k

k. I

where the f coefficients are constants. Rearranging, we write VI

Xs

, VJ

X9

EFO

l

r-~

.\:10

Til

K,K, TFT A

_ KA TA

0 (KFK A

+

T A)

_

TFTA _ KA TA

fi

Xg

KF TFT A

X9

X IO

TA

Til

0

+ 0

0

0

0

0 KF

o

TF

KA

-

TA

(

VREF

+

V,)

(7.97) where 7

V,

=

ftXIO +

L

k .1

fxXk

Note that only three states are needed in this case . 1.9.3

Type 2 system-rotating rectifier system

A nother type of system, the rotating rectifier system of Figure 7.13, incorporates damping loops that originate from the regulator output rather than from the excitation voltage (39J since, be ing brush less, the excitation voltage is not available to feed back . The IEEE description of this system is shown in Figure 7.64, where the damping feed back loop is seen to be different from that of Figure 7.61. Note that two time constants appear in the damping loop of this new system, T F 1 and Tn, one of which approximates

Fig.7 .64 Type 2 excitation system represent ation -rotating rectifier system before 1967. (© IEEE. printed from IEEE Trans., vol. PAS·87, 1968.)

Re-

297

Excitation Systems

the exciter time delay [39] and is considered "major damping," with the second or "minor damping" be ing present to damp higher frequencies . A state-space representation of this system may be derived from the following equations:

V, - (IIT R) /I; (KAIT A) V, - (I IT A) VR (KRIT R)

( 1IT E) VR - [( K E + SE)I TEl KFK A KF V VR TF1TnTA t TFiTnT A

(7.98) Rearranging, we may write as

V, VI

x, x,

PI

x

•

XII

i.;

Xu

~.

lO

T.

-

K.K, Tf1TflT A

_ K.

-

T. + K.K, TFiTnT A

_ K.

T.

T.

0

0

K.

0

-

T.

0

0

0 _

0

0

_

+

Tn

Tn

T FlTn

. :«:

Tfl Tf]T A

I

- -

0

T.

0

f.I

0

0

0

0

0

0

0

0

0

x, x,.

0

X II

0

X"

000

0

. 7

X,

0

+ 0 0 0 0

.s: T'lTn

0

o

LJ.x, -,

0

0

0

K. (V... T.

o

o (7.99)

The Type 2 excitation system representatio n is intended for use in simulating the Westinghouse Bru shless excitation system. An alternate representation developed by the manufacturer is reported to represent the physical equipment more accurately . This revised Type 2 representation is shown in Figure 7.65 [40]. Regulator

(1 + TAl , )(1 + T A2')

V

l

K ' (1 + T ' ) F F3

(l • " ,,')(1 ' '' F2' ) Domping circuit

Fig. 7.65

0

Revised Type 2 excitation system representatio n. (Used with permission from Stability Program Data Preparation Manual, Advanced Systems Technology Rept, 70-736, Dec. 1972, II::> ABB Power T & D Company Inc. , 1992.)

7.9 .4 Type 3 sys tem-stati c with te rmina l poten tial a nd current supplies Some systems use a combination of current and vo ltage intelligence as a feedback signal to be compared against the reference, e.g., the systems of Figures 7.15 and 7.16. These systems are not properly represented by Type I or IS and require special treatment , as shown in Figure 7.66. (The reactance XL is the commutating reactance of the transformer and is discussed in [41].) Here the regulator and input smoothing are similar to the Type I system. However , the signal denoted VB incorporates information fed forwa rd from V, with added information concerning both I, and IF. Thus

+ V,)

298

Chapter 7

If : A > 1, V8 = 0

Fig .7 .66

Type 3 excitation system representation -static with terminal potential and current supplies. (@ IEEE . Reprinted from IEEE Trans.. vol . PAS-l!?, 1968.)

Vc represents the self-excitation from the generator terminals.

Constants K; and

K, are proportionality factors indicating the proportion of the "Thevenin voltage,"

VTH due to potential and current information . Multiplying VTH is a signal proportional to IF, which accounts for variation of self-excitation with change in the angular relation of field current (IF) and self-excitation voltage (VTH ) (151 . Obviously, systems of this type are nonlinear. To formulate a linearized state-space representation, we may write the self-excitation components as

Vc = K 1 V, + K 2/ , + KiF

(7.100)

Then we write for the entire system

+ Vc

VB

=

VR

V3

=

KFEFDs/( I + T~)

EFD = VB/(K f + TfS) V; = K R V,/( I + TRS)

(K A/ ( I + TAS)]

V. (7.101 )

But we may 'write the terminal voltage in the time domain as

V, = ffE FD + where for brevity we let may write

Vx

L 1

j;,Xk

k-I

= ffE FD +

(7.102)

Vx

be the term on the right. Also, for the terminal current we

i,

= Mdid + Mqiq = Mdx 1 + Mqx4

(7.103)

If we define the states as in (7.68), we reduce (7.10 I )-(7 .103) to the following form:

VI

Xa

v)

x

VR

X IO

EFD

XII

9

TR 0

TF

_ K.. T..

0

Note that vx ' i, and

0

0 KF T£TF

_ K.. T ..

0

T..

KR I,

R v.• T

.!..L(fiK, - K£> T£T F

X9

KFK, FK2. + --IF J --v. +--1,

0

X IO

f

R

fjK, - K e Te

K

Xa

-T

T£

XII

R

+

K TETF

T£T F

K.. (VREF + T..

KFK TETF

•

v.)

K) . K, v + K z i + -IF TE

x

TE'

Te

(7.104) iF

are all linear functions of X I-X1_

Excitation Systems

299

Fig.7.67 Type 4 excitation system representation ---nonco ntinuously selling regulator . Note : VRH limited between VRmin and V R max : lime constant of rheost at travel = TRH '

7.9.5

Type 4 system-noncontinuous acting

The previous systems are similar in the sense that they are all continuous acting with relatively high gain and are usually fast acting. However, a great many systems are of an earlier design similar to the rheostatic system of Section 7.7.1 and are noncontinuous acting; i.e., they have dead zones in which the system operates essentially open loop. In addition to this, they are generally characterized as slow due to friction and inertia of moving parts. Type 4 systems (e.g ., Westinghouse BJ30 or General Electric GFA4 regulated systems) often have two speeds of operation depending upon the magnitude of the voltage error. Thus a large-error voltage may cause several rheostat segments to be shorted out, while a small-error voltage will cause the segments to be shorted one at a time . The computer representation of a system is illustrated in Figure 7.67, where K; is the raise-lower contact setting, typically set at 5%, that controls the fast-change mechanism on the rheostat. If V. is below this limiting value of K v , the rheostat setting is changed by motor action with an integrating time constant of TRH ' An "auctioneer" circuit sets the output V R to the higher of the two input quantities . Because the Type 4 system is so nonlinear, there is no advantage in representing it in state variable form . The equations for the Type 4 system are similar to those derived for the electromechanical system of Section 7.7.1. A comparison of these two systems is recommended . 7.10

Typical System Constants

Reference [15] gives, in addition to the system representations, a table of typical constants of physical systems . These data are given in Table 7.11 and, although typical, do not necessarily represent any physical system accurately . For any real system all quantities should be obtained from the manufacturer. Also note that the values in Table 7.11 are for a system with a response rat io of 0 .5 which, although common, is certainly not fast by today's standards . The RR of modern fast systems are often in the range of 2.0-3 .5. Note that the values of VR m a• and VRmin given in Table 7.11 are unity in column I and higher values in columns 2 and 3. This difference is due to the different choice of base voltage for VR by the different exciter manufacturers and does not necessarily imply any marked difference in the regulator ceilings or performance. Changing the base voltage of VR to VRmax affects all the other constants in the forward loop. There-

300

Chapter 7

Table 7.11. Typical Constants of Excitation Systems in Operation on 3600 r/ min Steam Turbine Generators (excitation system voltage response ratio = 0.5)

Symbol

Self-excited exciters, commutator, or silicon diode with amplidyne voltage regulators

Self-excited commutator exciter with Mag-A-Stat voltage regulator (2)

(I)

0.0--0.06 25-50* 0.06--0.20 1.0 -1.0 0.01-0.08 0.35-1.0 -0.05 0.5 0.267 0.074

TR

KA TA

V R max

VR min

KF TF

KE TE

SEmax

S E.15max

*For generators with

Rotating rectifier exciter with static voltage regulator (3)

0.0 400

0.0 400 0.02 7.3 -7.3 0.03 1.0 1.0 0.80 0.86 0.50

O.OS

3.5 -3.5 0.04 1.0 -0.17 0.95 0.95 0.22

open circuit field time constants greater than 4 s.

fore, caution must be used in comparing gains, time constants, and limits for systems of different manufacture. As experience has accumulated in excitation system modeling, the manufacturer and utility engineers have determined excitation system parameters for many existing units. Since these constants are specified on a normalized basis, they can often be used with reasonable confidence on other simulations where data is unavailable. Tables 7.12-7.15 give examples of excitation system parameters that can be used for estimating new systems or for cases where exact data is unavailable. Since the formation of the National Electric Reliability Council (NERC) a set of deTable 7.12. Symbol

Westinghouse Excitation System Constants for System Studies (excitation system voltage response ratio = 0.5) Mag-A-Stat

Excitation system type (s)

I

0.0 400 0.05 4.5 -4.5 -0.17 0.04 1.0

TR

KA TA

(s)

EFDmax EFDmin

KE

(pu)* (pu)*

KF

TF(S)

Kv

Rotating-rectifier

4

I

0.0 400 0.02 3.9 0 1.0 0.03 1.0

TRH

VRmax (pu)*

3.5 -3.5 0.95 0.22 0.95

VR min (pu)*

SEmax SE.1Smax TE (s)

3600 rjmin 1800 rjmin

7.3 -7.3 0.86 0.50 0.8

8J30 Rototrol

8.2 -8.2 1.10 0.50 1.30

I

Silverstat

I

TRA

I

0.05 200 0.25 4.28 4.5 1.70 -4.5 1.0 -0.17 0.105 1.25 0.05 20

0.02 200 0.1 4.5 0.3 -0.17 0.028 0.5

0.05 400 0.0 4.5 0.2 -0.17 0.028 0.5

3.5 8.3 1.7 -3.5 0.95 0.95 0.22 0.22 0.76 0.85

3.5 0.3 0.95 0.22 0.50

3.5 0.2 0.95 0.22 0.50

Source: Used with permission from Stability Program Data Preparation Manual, Advanced Systems Technology Rept. 70-736, Dec. 1972, e ABB Power T & D Company Inc., 1992. *Values given assume (full load) = 3.0 pu. If not, multiply * values by ,,~3.0.

"F

Excitation Systems Table 7.13.

301

Typical Excitation System Constants

Type of regulator

TR

KA

TA

VRmax

VRmin

KF/TF

Mag-A-Stat (Type I) SePT (Type 3) 8J30 (Type 4) Rototrol (Type I) Silverstat (Type I) TRA (Type I) GFA4(Type4) NAIOI (Type I) Amplidyne NA 108 (Type I) Amplidyne NA 143 (Type I) AmpJidyne < 5 kW NA 143 (Type 1) Amplidyne > 5 kW Brushless (Type 2) 3600 r/rnin Brushless (Type 2) 1800 r/rnin

0 0 20.0 0.05 0 0 0.05

400 120 0.05 200 200 400 20

0.05t 0.15 0 0.25 0.10 0.05t 0

3.5 1.2 8.3 3.5 3.5 3.5 1.0

-3.5 -1.2 1.8 -3.5 -0.05 -0.04 0

0.04 0.21 TdO 0 0.084 0.056 0.056 0

TF

1.0

TdO/ 10.0 0 1.25 0.5 0.45 1.0

0.06

*

0.2

1.0

-1.0

11.5TElKA

0.35

0

*

0.2

1.0

-1.0

4TE/KA

1.0

0

*

0.2

1.0

-1.0

4TEIKA

1.0

0

*

0.06

1.0

-1.0

8TEIKA

1.0

0

400

0.02

7.3

-7.8

0.03

1.0

0

400

0.02

8.2

-8.2

0.03

1.0

Source: Used by permission from Power System Stability Program User's Guide. Philadelphia Electric Co., 1971. "Data obtained from curves supplied by manufacturer. For typical values see Appendix D and Table 7.15. t High-speed contact setting, if known.

sign criteria has been established specifying the conditions under which power systems must be proven stable. This has caused an enlarged interest and concern in the accuracy of modeling all system components, particularly the generators, governors, exciters, and loads. Thus it is becoming common for the manufacturer to specify the exciter model to be used in system studies and to provide accurate gains and time constants for the system purchased. Table 7.14.

Typical Excitation System Constants

Type of regulator

K£

T£

A EX

BEX

M ag-A-Stat (Type 1) SCPT* (Type 3) 8J30 (Type 4) Rototrol (Type I) Silverstat (Type I) TRA (Type I) G FA4 (Type 4) 8rushless (Type 2) 3600 r Imin Brushless (Type 2) 1800 r Imin

-0.17 1.0 1.0 -0.17 -0.17 -0.17 0.051t

0.95 0.05 0.76 0.85 0.5 0.5 0.5

0.0039 0 0.0052 0.0039 0.0039 0.0039 0.00105

1.555 0 1.555 1.555 1.555 1.555 1.465

1.0

0.8

0.12

0.855

1.0

1.3

0.059

1.1

Source: Used by permission from Power System Stability Program User's Guide. Philadelphia Electric Co., 1971. "K; = 1.19 K 1 = 1.19 V8 max

=

[

. -sin (cos -

I

Fp )

+

J[

V- / E 2FDFL - F p

1.4 E FDFL

tHigh-speed contact setting, if known.

1

study M VA base generator M VA base

1

I I I I I +-H

1--. --

i--. ~-

"'-4--- -1-. -. .

~ F-

-

-

'--L.-

i-

I---f-.

Fig. 7.68

o

-

-.-

.

-t -

'I

I

_...

--

.. 1--

=

.. I--i-

. - I--- .- ... 1---

r----+

'

.

I

'

!

I

I i

:

I

l _! J

i

, --+ .. 1..-1 . ..

.L.. J. J •.i..-l. ..

':

,

'

' • I ··-t-~ I

l.

I

t

T: -

, --i". -!- '-

. : --tT~~ -,_~:J 1+ . __ -t..:d

1 .. ·,.. ....

'

!

o

~-1br;..U ' ~l..LC

:

28t: ; :~ J - I L t I

W

.-2::i...... i ! ,..

"T

v-

~ '---+--t

.

t '!rt:, ril E

l~ !I

i

I ! I ,------J

~ I i I

ii

-:: '~++++ ~'; -H

.

I A

i

;

.

/

,'

/

/

,//

I

I

I

I

/

I I,

I

// /'

.

~

-, /~ // . //-~\

°6

Full model generator respon se of 10".;. step increase in Tm and EFD • Initial loading o f Example 5.1. with no exciter and no generator sa turatio n.

.·+::: : :

.... - I--i- - - I---

-- -

.-.

-i

---

I---

t - --t ·, " t

"' ''r

-- . 1-1- .-

-::- .::: ==-:t:.t:::

.- 1- _. 1-1- ... 1--

_-

.- I-~-

-1--

1-- 1-. _. i - 1--

- - t"-

.rr.

. - 1--1---.-..

- "I---

•

1-- ..

1

I

_. - i-

I

..• - .+-+-. _..J _ _ ..

... . ,• ._ -+ · t .. .. ... . .

~_ . __

-

2 ~ L

~

,

... -'--

-.

o ~ -X

..

~

-

I

---;

--l d

-

-

.....-

II

- r--:' -

I

. - 1--. - 1-- .... -. c-

(~

1-. 1---. .C

,E

- 1-- _.. - . 1 101--+--- -

4 3 2

>-

orr:

1 II I., IJ

2

O ~T~­ 4 - .n;t ¢ 3

~1 ~

4

E FD

w o

c

'-l

~

'"0

n ::r-

r-.J

't"

-

- --.-,.---

.+ • . +• • + -+ ...

":l¢ ~

-+ .., [~.~ . t~=:= . t -_.+:.~

Fig.7.69

o --

1

--

-

_.- ._.

-- -

..-

6

W

fJ

6

Full model generator response to 10% step increase in T m and 5~~ step increase in "'REF. with initial loading of Example 5.1. Exciter par ameters (Westingho use Brushless) : K = 400. T A = 0.02. K E = 1.0. TE = 0.8. K F = 0.03. TF = 1.0. K R = 1.0. TR = 0.0. VRmax = 7.3. VRmin = - 7.3. EFDmax = 3.93; no genA erator or exciter saturation.

..

_.. - - - - - - ,I--.._- - '--i--

_. - -- I- -

- 1--"-

+ - +- t - + .. t .. r-r-

-- - -- --t--

- 1--

- I-

- t - o+-

1 tj I I I I I o .H. --+-'T -+L.LL..-L~ _ - r- r - !- -It. I e¢ = -r-t-r4 - - -- _... -- - - f-3 -2 -- - -- -- I-- - f-. 1 ...-t-. - i-- 1---- -- i--i1--- -o-i-+-C '' ' ' - - - -_. -- ----- ::~~ 2 JI'-1--- - - ~ .. .- --- ---- _.J

2

3

4~

o:=-" T-t

4 3 2 1

E

10 s

m

x

w

w

o

'"

3

'" it

-e

(.J'l

::l

o'

~

!:!.

304

Chapter 7

Table 7.15.

Typical Excitation System Constants for Exciters with Amplidyne Voltage Regulators (NAIOI, NA108, NA143)

Exciter nominal response

KE

TE

KA*

K Amax*

0.5 1.0 1.5 2.0

-0.0445 -0.0333 -0.0240 -0.0171

0.5 0.25 0.1428 0.0833

20TdO/ 3

25 25 25 25

IOTdo/3 25 TdO/ 13

25TdO/22

KAt 20TdO

10Tdo 17 TdO/3 IOTdo/3

KAmaxt

AExt

BExt

50 50 50 50

0.0016 0.0058 0.0093 0.0108

1.465 1.06 0.898 0.79

Source: Used by permission from Power System Stability Program User's Guide, Philadelphia Electric CO 1971. *For all NA 101, NA 108, and NA 143 5 kW or less. tFor NA 143 over 5 kW. [See (7.90). q

7.11

The Effect of Excitation on Generator Performance

Using the models of excitation systems presented in this chapter and the full model of the generator developed in Chapters 4 and 5, we can construct a computer simulation of a generator with an excitation system. The results of this simulation are interesting and instructive and demonstrate clearly the effect of excitation on system performance. For the purpose of illustration, a Type I excitation system similar to Figure 7.61, has been added to the generator analog simulation of Figure 5.18. Appropriate switching is arranged so the simulation can be operated with the exciter active or with constant EFD • The results are shown in Figure 7.68 for constant EFD and Figure 7.69 with the exciter operative. The exciter modeled for this illustration is similar to the Westinghouse Brushless exciter. Both Figures 7.68 and 7.69 show the response of the system to a IO~~ step increase in Tm , beginning with the full-load condition of Example 5.1. For the generator with no exciter, this torque increase causes a monotone decay in both AF and J.t; and an increase in lJ that will eventually cause the generator to pull out of step. This increase in lJ is most clearly shown in the phase plane plot. Adding the excitation system, as shown in Figure 7.69, improves the system response dramatically. Note that the exciter holds AF and J.t; nearly constant when T'; is changed. As a result, lJ is increased to its new operating level in a damped oscillatory manner. The phase plane plot shows a stable focus at the new o. Following the increase in torque the system is subjected to an increase in EFD • This is accomplished by switching the unregulated machine EFD from 100% to 110% of the Example 5.1 level. In the regulated machine a 5% step increase in VREF is made. The results are roughly the same with increases noted in AF and ~, and with a decrease in {) to just below the initial value. We conclude that for the load change observed, the exciter has a stabilizing influence due to its ability to hold the flux linkages and voltage nearly constant. This causes the change in [) to be more stable. In Chapter 8 we will consider further the effects of excitation on stability, both in the transient and dynamic modes of operation. Problems 7.1

Consider the generator of Figure 7.2 as analyzed in Example 7.1. Repeat Example 7.1 but assume that the machine is located at a remote location so that the terminal voltage Jt; increases roughly in proportion to Eg • Assume, however, that the output power is held constant by the governor.

Excitation Systems 7.2

7.3 7.4 7.5 7.6 7.7

305

Consider the generator of Example 7.1 connected in parallel with an infinite bus and operat ing with constant excitation . By means of a phasor diagram analyze the change in (), I, and 8 when the governor sett ing is changed to increase the power output by 20'i~. Note particularly the change in () in both direction and magnitude . Following the change described in Problem 7.2, what action would be required, and in what amount, to restore the power factor to its original value? Repeat Example 7.1 except that instead .of increasing the excitat ion, decrease E~ to a magnitude less than that of V" Observe the new values of 0 and 8 and , in particular, the change in 0 and O. Comparing results of Example 7.\ and Problems 7.1 -7.4 , can you make any general statement regarding the sensitivity of 0 and 0 to changes in P and E~? Establ ish a line of reasoning to show that a heavily cumulative compounded exciter is not desirable. Assume linear variations where necessary to establish your arguments . Consider the separately excited exciter E shown in Figure P7.7. The initial current in the generator field is p when the exciter voltage uF = ko. At time 1 = a a step function in the voltage UF is introduced; i.e.. uF = ko + k, U(I - a) .

-

rF

+

V

F

iF

°1

IF

Fig. P7.7

7.8

Compute the current iF' Sketch this result for the cases where the time constant very large and very small. Plot the current function in the s plane . Consider the exciter shown in Figure P7.8, where the main exciter M is excited by a pilot exciter P such that the relat ion UF = k'i», ~ ki, holds . What assumptions must be made for the above relat ion to be approximately valid? Compute the current i2 due to a step change in the pilot exciter voltage, i.e., for up = U(I) .

Ld'F is both

+

v

p

-

R,

;,

t,

Fig. P7.8

7.9

A solenoid is to be used as the sensing and amplification mechanism for a crude voltage regulator . The system is shown in Figure P7.9. Discuss the operation of this device and comment on the feasibility of the proposed design . Write the differential equations that describe the system.

Fig. P7.9

306 7.10

7.11

Chapter 7 An exc iter for an ac generator, instea d of being driven from the turbine-generator shaft. is driven by a sepa ra te motor with a large flywheel. Consider the motor to have a constant output torque and write the equations for this system . Analyze the system given in Figure P7 .11 to determine the effectiveness of the damping transformer in stabilizing the system to sudden changes. Write the equations for th is system and show that , with parameters carefully selected. a degree of stabil ization is achieved. particularly for large values of R r - Assume no load on the exciter. R

L

'o ~

+

v

ip

-

F

Rp

Lt..u. · M

rrrr' -

v

Amp

o +

V

s

+

~ I

S

Fig. P7.11 7.12

The separately excited exciter shown in Figure P7 .12 has a magnetization curve as given in Table 7.3. Other constants of interest are

N (J

k

2500

Up

12.000

UF

1.2

R

125 V 8 n in field winding 120 V (rated)

10 r

Fig. P7.12 (a)

7.13

7.14

Determine the buildup curve beginning at rated voltage; i.e.• U F 1 120 V. What are the initial and final values of resistance in the field circuit? (b) What is the main exciter response ratio? Given the same exciter of Problem 7.12, consider a self-excited connection with an amplidyne boost-buck regulation system that quickly goes to its saturation voltage of + 100 V following a command from the voltage regulator. If this forcing voltage is held constant. compute the buildup. Assume U F 1 = 40 V. un = 180 V. Assume that the constants TA' TE. TG' K E • KG. and K A are the same as in Example 7.7. Let TR take the values of 0 .00 I. 0 .01, and 0.1. Find the effect of TR on the branch of the root locus near the imaginary axis .

Excitation Systems 7.15 7.16

7.17 7.18 7.19 7.20 7.21 7.22 7.23

307

Repeat Problem 7.14 with TR = 0.05 and for values of T A = 0.05 and 0.2. Obtain the loci of the roots for the polynomial of (7.63) for TF = 0.3 and for values of K F between 0.02 and 0.10. Obtain (or sketch) a root-locus plot for the system of Example 7.8 for KF = 0.05 and rr = 0.3. Complete the analog computer simulation of the system of one machine connected to an infinite bus (given in Chapter 5) by adding the simulation of the excitation system. Use a Type I exciter. AIso include the effect of saturation in the simulation. For the excitation system described in Example 7.9 and for the machine model and operating conditions described in Example 6.6, obtain the A matrix of the system and find the eigenvalues. Repeat Problem 7.19 for the conditions of Example 6.7. Repeat Example 7.9 for the operating condition of Example 6.1. Repeat Example 7.9 (with the same operating condition) using a Type 2 excitation system. Data for the excitation system is given in Table 7.11. Show how the choice of base voltage for the voltage regulator output VR affects other constants in the forward loop. Assume the usual bases for Jt; and EFD .

References

I. Concordia, C .. and Ternoshok. M. Generator excitation systems and power system performance. Paper 31 CP 67-536. presented at the IEEE Summer Power Meeting. Portland, Oreg .. 1967. 2. Westinghouse Electric Corp. Electrical Transmission and Distribution Reference Book. Pittsburgh, Pa., 1950. 3. IEEE Committee Report. Proposed excitation system definitions for synchronous machines. IEEE Trans. PAS-88:1248-58, 1969. 4. Chambers, G. S., Rubenstein, A. S.. and Ternoshok , M. Recent developments in amplidyne regulator excitation systems for large generators. AlEE Trans. PAS-80: 1066·-72, 1961. 5. Alexanderson, E. F. W.. Edwards. M. A., and Bowman, K. K. The amplidyne generator- A dynamoelectric amplifier for power control. General Electric Rev. 43: 104-6, 1940. 6. Bobo, P.O .. Carlson. J. T.. and Horton. J. F. A new regulator and excitation system. IEEE Trans. PAS-72:175-83.1953. 7. Barnes. H. C.. Oliver. J. A., Rubenstein, A. S., and Ternoshok , M. Alternator-rectifier exciter for Cardinal Plant. IEEE Trans. PAS-87: 1189-98, 1968. 8. Whitney, E. C., Hoover, D. B.. and Bobo, P. O. An electric utility brushless excitation system. AlEE Trans. PAS-78:1821-24. 1959. 9. Myers. E. H., and Bobo, P. O. Brushless excitation system. Prot. Southwest IEEE Conj. (SWIEEECO), 1966.

10. Myers, E. H. Rotating rectifier exciters for large turbine-driven ac generators. Proc. Afn. Power Con].. Vol. 27, Chicago, 1965. II. Rubenstein, A. S.. and Ternoshok , M. Excitation systems- Designs and practices in the United States. Presented at Association des lngenieurs Electriciens de l'Institute Electrotechnique Montefiore, A.I.M., Liege, Belgium, 1966. 12. Domeratzky, L. M., Rubenstein, A. S., and Temoshok, M. A static excitation system for industrial and utility steam turbine-generators. AlEE Trans. PAS-80: 1072 77, 1961 13. Lane, L. J., Rogers, D. F., and Vance, P. A. Design and tests of a static excitation system for industrial and utility steam turbine-generators. AlEE Trans. PAS-80: 1077·85,1961. 14. Lee, C. H., and Keay, F. W. A new excitation system and a method of analyzing voltage response. IEEE Int. Conv. Rec. 12:5-14,1964. 15. IEEE Committee Report. Computer representation of excitation systems. IEEE Trans. PAS-87: 1460-64. 1968. 16. Kimbark, E. W. Power System Stability. Vol. 3. Wiley, New York, 1956. 17. Cornelius, H. A., Cawson, W. F., and Cory, H. W. Experience with automatic voltage regulation on a 115-megawatt turbogenerator. A lEE Trans. PAS-71: 184-87, 1952. 18. Dandeno, P. L., and McClymont, K. R. Excitation system response: A utility viewpoint. AlEE Trans. PAS-76:1497-1501, 1957. 19. Temoshok, M., and Rothe, F. S. Excitation voltage response definitions and significance in power systems. AlEE Trans. PAS-76:1491-96, 1957. 20. Rudenberg, R. Transient Performance of Electric Power Systems: Phenomena in Lumped Networks. McGraw-Hill, New York, 1950. (MIT Press, Cambridge. Mass., 1967). 21. Takahashi, J., Rabins, M. J., and Auslander, D. M. Control and Dynamic Systems. Addison-Wesley, Reading, Mass., 1970.

308

Chapter 7

22. Brown, R. G.\ and Nilsson, J. W. Introduction 10 Linear Systems Analysis. Wiley, New York, 1962. 23. Savant, C. J., Jr. Basic Feedback Control System Design. McGraw-Hili, New York, 1959. 24. Hunter. W. A., and Ternoshok , M. Development of a modern amplidyne voltage regulator for large turbine generators. AlEE Trans. PAS-71 :894 900, 1952. 25. Porter, F. M .• and Kinghorn, J. H. The development of modern excitation systems for synchronous condensers and generators. AlEE Trans. PAS-65: 1070·_·27, 1946. 26. Concordia. C. Effect of boost-buck voltage regulator on steady-slate power limit. A lEE Trans. PAS69:380-84. 1950. 27. McClure, J. B.• Whittlesley, S. I., and Hartman, M. E. Modern excitation systems for large synchronous machines. AlEE Trans. PAS-65:939--45, 1946. 28. General Electric Co. Amplidyne regulator excitation systems for large generators. Bull. GET-2980. 1966. 29. Harder, E. L., and Valentine. C. E. Static voltage regulator for Rototrol exciter. Electr. Eng. 64: 60\. '945. 30. Kallenback , G. K., Rothe. F. S., Storm, H. F .• and Dandeno, P. L. Performance of new magnetic amplifier type voltage regulator for large hydroelectric generators. A JEE Trans. PAS-7 \ :201·--6, \952. 31. Hand, E. W, McClure, F. N., Bobo. P.O., and Carleton, J. T. Magarnp regulator tests and operating experience on West Penn Power System. AlEE Trans. PAS-73:486-9\, \954. 32. Carleton, J. T., and Horton, W. F. The figure of merit of magnetic amplifiers. AlEE Trans. PAS71:239-45, 1952. 33. Ogle, H. M. The amplistat and its applications. General Electric Rev. Pt. I\ Feb.: Pt. 2, Aug.; Pl. 3, oei., 1950. 34. Hanna, C. R., Oplinger, K. A., and Valentine, C. E. Recent developments in generator voltage regulation. AlEE Trans. Sg:838-44, 1939. 35. Dahl, O. G. C. Electric Power Circuits. Theory and Application. Vol. 2. McGraw-Hill, New York, 1938. 36. Kimbark , E. W. Power Svstem Stability. Vol. 1. Elements ojStability Calculations. Wiley, New York, 1948. 37. Kron, G. Regulating system for dynamoelectric machines. Patent No. 2.692,967, U.S. Patent Office, 1954. 38. Oyetunji, A. A. Effects of system nonlinearities on synchronous machine control. Unpubl. Ph.D. thesis. Research Rept. ERI-71 130. Iowa State Univ., Ames, 1971. 39. Ferguson, R. W., Herbst. R., and Miller, R. W. Analytical studies of the brushless excitation system. AlEE Trans. PAS-78:1815-21. 1959. 40. Westinghouse Electric Corp. Stability program data preparation manual. Advanced Systems Technology Repl. 70-736, 1972. 41. Lane. L. J .. Mendel, J. E., Ewart, D. N., Crenshaw. M. L.. and Todd, J. M. A static excitation system for steam turbine generators. Paper CP 65-208, presented at the IEEE Winter Power Meeting. New York, 1965. 42. Philadelphia Electric Co. Power system stability program. Power System Planning Div .. Users Guide U6004-2, 1971.

chapter

8

The Effect of Excitation on Stability

8. 1

Introduction

Considerable attention has been given in the literature to the excitation system and its role in improving power system stability. Early investigators realized that the socalled "steady-state" power limits of power networks could be increased by using the then available high-gain continuous-acting voltage regulators [1]. It was also recognized that the voltage regulator gain requirement was different at no-load conditions from that needed for good performance under load. In the early J950s engineers became aware of the instabilities introduced by the (then) modern voltage regulators, and stabilizing feedback circuits came into common use [2]. In the 1960s large interconnected systems experienced growing oscillations that disrupted parallel operation of large systems [3-12]. It was discovered that the inherently weak natural damping of large and weakly coupled systems was the main cause and that situations of negative damping were further aggravated by the regulator gain [13]. Engineers learned that the system damping could be enhanced by artificial signals introduced through the excitation system. This scheme has been very successful in combating growing oscillation problems experienced in the power systems of North America. The success of excitation control in improving power system dynamic performance in certain situations has led to greater expectations among power system engineers as to the capability of such control Because of the small effective time constants in the excitation system control loop, it was assumed that a large control effort could be expended through excitation control with a relatively small input of control energy. While basically sound, this control is limited in its effectiveness. A part of the engineer's job, then, is to determine this limit, i.e., to find the exciter design and control parameters that can provide good performance at reasonable cost [14]. The subject of excitation control is further complicated by a conflict in control requirements in the period following the initiation of a transient. In the first few cycles these requirements may be significantly different from those needed over a few seconds. Furthermore, it has been shown that the best control effort in the shorter period may tend to cause instability later. This suggests the separation of the excitation control studies into two distinct problems, the transient (short-term) problem and the dynamic (long-term) problem. It should be noted that this terminology is not universally used. Some authors call the dynamic stability problem by the ambiguous name of "steadystate stability." Other variations are found in the literature, but usually the two problems are treated separately as noted. 309

310

Chapter 8

8. 1. 1

Transient stability and dynamic stability considerations

In transient stability the machine is subjected to a large impact, usually a fault, which is maintained for a short time and causes a significant reduction in the machine terminal voltage and the ability to transfer synchronizing power. If we consider the one machine-infinite bus problem, the usual approximation for the power transfer is given by p

= (~Veo/x)sino

(8.) )

where ~ is the machine terminal voltage and Veo is the infinite bus voltage. Note that if ~ is reduced, P is reduced by a corresponding amount. Prevention of this reduction in P requires very fast action by the excitation system in forcing the field to ceiling and thereby holding ~ at a reasonable value. Indeed, the most beneficial attributes the voltage regulator can have for this situation is speed and a high ceiling voltage, thus improving the chances of holding V, at the needed level, Also, when the fault is removed and the reactance x of (8.1) is increased due to switching, another fast change in excitation is required. These violent changes affect the machine's ability to release the power it is receiving from the turbine. These changes are effectively controlled by very fast excitation changes. The dynamic stability problem is different from the transient problem in several ways, and the requirements on the excitation system- are also different. By dynamic stability we mean the ability of all machines in the system to adjust to small load changes or impacts. Consider a multimachine system feeding a constant load (a condition never met in practice). Let us assume that at a given instant the load is changed by a small amount, say by the energizing of a very large motor somewhere in the system. Assume further that this change in load is just large enough to be recognized as such by a certain group of machines we will call the control group. The machines nearest the load electrically will see the largest change, and those farther away will experience smaller and smaller changes until the change is not perceptible at all beyond the boundary of the control group. Now how will this load change manifest itself at the several machines in the control group? Since it is a load increase, there is an immediate increase in the output power requirements from each of the machines. Since step changes in power to turbines are not possible, this increased power requirement will come first from stored energy in the control group of machines. Thus energy stored in the magnetic field of the machines is released, then somewhat later, rotating energy [( I /2)mv 2 ) is used to supply the load requirements until the governors have a chance to adjust the power input to the various generators. Let us examine the behavior of the machines in the time interval prior to the governor action. This interval may be on the order of I s. In this time period the changes in machine voltages, currents, and speeds will be different for each machine in the control group because of differences in unit size, design, and electrical location with respect to the load. Thus each unit responds by contributing its share of the load increase, with its share being dictated by the impedance it sees at its terminals (its Thevenin impedance) and the size of the unit. Each unit has its own natural frequency of response and will oscillate for a time until damping forces can decay these oscillations. Thus the one change in load, a step change, sets up all kinds of oscillatory responses and the system Brings" for a time with many frequencies present, these induced changes causing their own interaction with neighboring machines (see Section 3.6).

Effect of Excitation on Stability

311

Now visualize the excitation system in this situation. In the older electromechanical systems there was a substantial deadband in the voltage regulator, and unless the generator was relatively close to the load change, the excitation of these machines would remain unchanged. The machines closer to the load change would recognize a need for increased excitation and this would be accomplished, although somewhat slowly. Newer excitation systems present a different kind of problem. These systems recognize the change in load immediately, either as a perceptible change in terminal voltage, terminal current, or both . Thus each oscillation of the unit causes the excitation system to try to correct accordingly, since as the speed voltage changes, the terminal voltage also changes. Moreover, the oscillating control group machines react with one another, and each action or reaction is accompanied by an excitation change . The excitation system has one major handicap to overcome in following these system oscillations; this is the effective time constant of the main exciter field which is on the order of a few seconds or so. Thus from the time of recognition of a desired excitation change until its partial fulfillment, there is an unavoidable delay. During this delay time the state of the oscillating system will change, causing a new excitation adjustment to be made . This system lag then is a detriment to stable operation, and several investigators have shown examples wherein systems are less oscillatory with the voltage regulators turned off than with them operating [7, 12]. Our approach to this problem must obviously depend upon the type of impact under consideration . For the large impact, such as a fault, we are concerned with maximum forcing of the field, and we examine the response in building up from normal excitation to ceiling excitation . This is a nonlinear problem, as we have seen, and the shape of the magnetization curve cannot be neglected. The small impact or dynamic stability problem is different. Here we are concerned with small excursions from normal operation, and linearization about this normal or "quiescent" point is possible and desirable . Having done this, we may study the response using the tools of linear systems analysis; in this way not only can we analyze but possibly compensate the system for better damping and perhaps faster response . 8.2

Effect of Excitation on Generator Power Limits

We begin with a simple example, the purpose of which is to show that the excitation system can have an effect upon stability. Example 8.1 Consider the two-machine system of Figure 8.1, where we consider one machine against an infinite bus. (This problem was introduced and analyzed by Concordia [I].) The power output of the machine is given by P

s

[E, Ed(X, + X2 ) ] sin 0 0\

+ 02

(8.2)

t + x.

+

V

t

&

= 1 pu

~Fig . 8.1

One machine -infinite bus system .

Chapter 8

312

Ix,

Fig . 8.2

=

I

Phasor diagram for Example 8.1.

This equation applies whether or not there is a voltage regulator. Determine the effect of excitation on this equation .

Solution We now establish the boundary conditions for the problem. First we assume that X, = Xl = 1.0 pu and that JI; = 1.0 pu. Then for any given load the voltages £, and £2 must assume a certain value to hold V, at 1.0 pu . If the power factor is unity, £1 and £1 have the same magnitude as shown in the phasor diagram of Figure 8.2 . If £1 and £2 are held constant at these values, the power transferred to the infinite bus varies sinusoidally according to (8.2) and has a maximum when 0 is 90°. Now assume that £. and £1 are both subject to perfect regulator action and that the key to this action is that JI; is to be held at 1.0 pu and the power factor is to be held at unity . We write in phasor notation

Adding these equations we have

2

' on 6/ 2 Pe rf ec t regulator

0..'

~

1. 0

.£

o

90 Ang le 6, de gre es

Fig.8.3

Comparison of power transferred at unity power factor with and without excitation control.

Effect of Excitation on Stabil ity

313

E 1 = E 2 = __1_ coso/2

(8.3)

or

Substituting (8.3) into (8.2) and simplifying, we have for the perfect regulator, at unity power factor,

P = tano/2

(8.4)

The result is plotted in Figure 8.3 along with the same result for the case of constant (unregulated) E 1 and E 2 • In deriving (8.4) , we have tacitly assumed that the regulators acting upon E 1 and E2 do so instantaneously and continuously. The result is interesting for several reasons. First, we observe that with this ideal regulation there is no stability limit. Second , it is indicated that operation in the region where 0 > 90° is possible . We should comment that the assumed physical system is not realizable since there is always a lag in the excitation response even if the voltage regulator is ideal. Also, excitation contro l of the infinite bus voltage is not a practical consideration , as this remote bus is probably not infinite and may not be closely regulated . Example 8.2 Consider the more practical problem of holding the voltage E2 constant at 1.0 pu and letting the power factor var y, other things being the same. Solution Under this condition we have the phasor diagram of Figure 8.4 where we note that the locus of £2 is the dashed circular arc of radius 1.0. Note that the power factor is constrained by the relation

(8.5) where 61 = 21T - 6 and 8 = 8 1 + 82 , Writing phasor equations for the voltages, we have E,

Fig. 8.4

Phasor diagram for Example 8.2.

314

Chapter 8 3.0 r --

-

-

-

-

-

-

-

-

-

-

-

- - -::=='"-,300 E,

2.5

250

2 .0

200

~

Q.

....:

."

e 0

6

1. 5

150 II v 6>

,.j

v

a:

."

.0 ." 100 e0

1. 0

-o

P

0.5

50

o

90

Torque Ang le,

6"

degre es

Fig. 8.5 System parameters as a function of oZ ,

£,

I

+ jl =

£Z = 1 - jl =

- lsinO + jlcosO = E,ei&1

+ 1sin 8 - jl cos 8

=

Eze - j&z

(8 .6)

where 8, 0"0,, and Oz are all measured positive as counterclockwise. Noting that £ 2 = 1, we can establish that 1

=

2 sin 8.

sin 0 = 2 sin O.

s, sin 0

=

2 sin Oz

tanol = sinoz/(2 - cosoz)

(8 .7)

Thus once we establish 02, we also fix 0, I, 0, and 0" although the relationships among these variables are nonlinear. These results are plotted in Figure 8.5 where equations (8.7) are used to determine the plotted values. We also note that

P = V,l cosO

(8.8)

but from the second of equations (8.6) we can establish that 1 cos 8

P = sin Oz

=

sin Oz or (8.9)

so Oz also establishes P. Thus P does have a maximum in this case , and this occurs when Oz = 90° (£2 pointing straight down in Figure 8.4). In this case we have at maximum power

E,

=

o=

2 + jl - 45°

2.235/26.6°

1= 1.414

o=

116.6°

The important thing to note is that P is again limited, but we see that 0 may go

Effect of Excitation on Stability

315

1.0

"~. 0. 5 Q. ~

~

£

o

90

Torq ue Angle 0, deg ree .

Fig.8 .6

Vari ation of P with

o.

beyond 90· to achieve maximum power and that this requires over 2 pu £.. variation of P with 0 is shown in Figure 8.6.

The

These simple examples show the effect of excitation under certain ideal situations. Obviously, these ideal conditions will not be realized in practice. However, they provide limiting values of the effect of excitation on changing the effective system parameters. A power system is nearly a constant voltage system and is made so because of system component design and close voltage control. This means that the Thevenin impedance seen looking into the source is very small. Fast excitation helps keep this impedance small during disturbances and contributes to system stability by allowing the required transfer of power even during disturbances. Finally, it should be stated that while the ability of exciters to accomplish this task is limited, other considerations make it undesirable to achieve perfect control and zero Thevenin impedance. Among these is the fault-interrupting capabil ity. 8.3

Effect of the Excitation System on Transient Stability

In the transient stability problem the performance of the power system when subjected to severe impacts is studied. The concern is whether the system is able to maintain synchronism during and following these disturbances. The period of interest is relatively short (at most a few seconds), with the first swing being of primary importance . In this period the generator is suddenly subjected to an appreciable change in its output power causing its rotor to accelerate (or decelerate) at a rate large enough to threaten loss of synchronism. The important factors influencing the outcome are the machine behavior and the power network dynamic relations. For the sake of this discussion it is assumed that the power supplied by the prime movers does not change in the period of interest. Therefore the effect of excitation control on this type of transient depends upon its ability to help the generator maintain its output power in the period of interest. To place the problem in the proper perspective, we should review the main factors that affect the performance during severe transients. These are: I. The disturbing influence of the impact. This includes the type of disturbance, its location, and its duration . 2. The abilit y of the transmission system to maintain strong synchronizing forces during the transient initiated by a disturbance . 3. The turbine-generator parameters. The above have traditionally been the main factors affecting the so-called first-swing transients. The system parameters influencing these factors are:

316

Chapter 8

I. The synchronous machine parameters. Of these the most important are: (a) the inertia constant, (b) the direct axis transient reactance, (c) the direct axis open circuit time constant, and (d) the ability of the excitation system to hold the flux level of the synchronous machine and increase the output power during the transient. 2. The transmission system impedances under normal, faulted, and postfault conditions. Here the flexibility of switching out faulted sections is important so that large transfer admittances between synchronous machines are maintained when the fault is isolated. 3. The protective relaying scheme and equipment. The objective is to detect faults and isolate faulted sections of the transmission network very quickly with minimum disruption. 8.3.1

The role of the excitation system in classical model studies

In the classical model it is assumed that the flux linking the main field winding remains constant during the transient. If the transient is initiated by a fault, the armature reaction tends to decrease this flux linkage [15]. This is particularly true for the generators electrically close to the location of the fault. The voltage regulator tends to force the excitation system to boost the flux level. Thus while the fault is on, the effect of the armature reaction and the action of the voltage regulator tend to counteract each other. These effects, along with the relatively long effective time constant of the main field winding, result in an almost constant flux linkage during the first swing of I s or less. (For the examples in Chapter 6 this time constant K) TdO is about 2.0 s.) It is important to recognize what the above reasoning implies. First, it implies the presence of a voltage regulator that tends to hold the flux linkage level constant. Second, it is significant to note that the armature reaction effects are particularly pronounced during a fault since the reactive power output of the generator is large. Therefore the duration of the fault is important in determining whether a particular type of voltage regulator would be adequate to maintain constant flux linkage. A study reported by Crary [2] and discussed by Young [15] illustrates the above. The system studied consists of one machine connected to a larger system through a 200mile double circuit transmission line. The excitation system for the generator is Type 1 (see Chapter 7) with provision to change the parameters such that the response ratio (RR) varies from 0.10 to 3.0 pu. The former corresponds to a nearly constant field voltage condition. The latter would approximate the response of a modern fast excitation system. Data of the system used in the study are shown in Figure 8.7. A transient stability study was made for a three-phase fault near the generator. The sending end power limits versus the fault clearing time are shown in Figure 8.8 for different exciter responses (curves 1-5) and for the classical model (curve 6). From Figure 8.8 it appears that the classical model corresponds to a very slow and weak excitation system for very short fault clearing times, while for longer clearing times it approximates a rather fast excitation system. If the nature of the stability study is such that the fault clearing time is large, as in "stuck breaker" studies (15], the actual power limits may be lower than those indicated when using the classical model. In another study of excitation system representation [16] the authors report (in a certain stability study they conducted) that a classical representation showed a certain generator to be stable, while detailed representation of the generator indicated that loss of synchronism resulted. The authors conclude that the dominant factor affecting loss

317

Effect of Excitation on Stability

Gener ator : x d = 0.63 x q = 0.42 Xd = 0.21 H = 5.0 TdO = 5.0 x, = 0 . 10 Line: x r

= =

y

=

s s

Regulating system : 11{ = 20

pu pu pu

11.. = Tr =

Emax = E min =

pu

Syste m da mp ing:

0.8 ll /mi /line 0.12 ll /mi /line 5.2 x 10 - 6 mho /m i/line

Fault o n Td 11 Td12 Td 21 Td 22

System: x m = 0.2 pu H = 50.0 s Fig. 8.7

4

0.47 s 2.25 pu - 0 .30 pu

I 0 0 15

Fa ult clea red 4

3 3 18

Two-machine system with 200-m ile tran smis sion lines .

of synchronism is the inability of the excitation system of that generator, with response ratio of 0.5, to offset the effects of armature reaction. 8.3.2

Increased reliance on excitation control to improve stability

Trends in the design of power system components have resulted margins. Contributing to this trend are the following :

In

lower stability

I. Increased rating of generating units with lower inertia constants and higher pu reactances. 2. Large interconnected system operating practices with increased dependence on the transmission system to carry greater loading. These trends have led to the increased reliance on the use of excitation control as a 1.20 ,...--

-

-

-

-

-

-

-

-

-

,

-e c

~

Curve

1.05

'6c

I

2

~ 1.00 ~

0..

o.95 '---:--':-:---:--'-:-:~:_':_:___:_":~__::_':_:_---' o 0.02 0 .04 0.06 0 .08 0 .1 0 Fault Clearing Time, •

Fig. 8.8

3 4

5

6

RR 0.042 0. \7 0.68 2.70 11 .0 s Cla ssica l model

3.0 2.0 1.0 0.25 0 .10

Send ing-end power versus fault clearing time for different excitation system responses.

318

Chapte r 8

I

N

c'

.s .;0 ~

0.0

>-

s U ~

g- -Q .3 3

Sont o Bcrbere

u,

•

0 .0

1.0

(0 )

I

TIme, s

2.0

3.0

(b)

N

c'

.s

0 's

"

0

>-

0 .0

u

c

" g~

Il:

- 1. 0

0.0

2.0 Time, s

3. 0

4.0

5. 0

(e )

Fig. 8.9

Resu lts of excitat ion system stud ies on a western U.S. system : (a) One-line diagram with fau lt locat ion , (b) frequen cy dev iat ion compar ison for a four -cycle fa ult, (c) frequency de viat ion compariso n for a 9.6-cycle fault: A = 2.0 ANSI co nventio na l excitation system ; B - low time constant excitation system with rat e feedba ck; C = low time cons tant excitation system withou t rate feedback . (© IEEE. Reprinted from IEEE Trans.. vol. PAS-90, Sep l./Ocl. 1971.)

means of improving stability [17]. This has prompted significant technological advances in excitation systems. As an aid to transient stability, the desirable excitation system characteristics are a fast speed of response and a high ceiling voltage. With the help of fast transient forcing of excitation and the boost of internal machine flux, the electrical output of the machine may be increased during the first swing compared to the results obtainable with a slow exciter. This reduces the accelerating power and results in improved tr ansient performance.

319

Effect of Excitation on Stability

Modern excitation systems can be effective in two ways: in reducing the severity of machine swings when subjected to large impacts by reducing the magnitude of the first swing and by ensuring that the subsequent swings are smaller than the first. The latter is an important con sideration in present-day large interconnected power systems. Situations may be encountered where various modes of oscillations reinforce each other during later swings, which along with the inherent weak system damping can cause transient instability after the first swing. With proper compensation a modern excitation system can be ver y effective in correcting this type of problem. However, except for transient stability studies involving faults with long clearing times (or stuck breakers), the effect of the excitation system on the severity of the first swing is relatively small. That is, a very fast, high-response excitation system will usually reduce the first swing by only a few degrees or will increase the generator transient stability power limit (for a given fault) by a few percent. In a study reported by Perry et al. [18] on part of the Pacific Gas and Electric Company system in northern California, the effect of the excitation system response on the system frequency deviation is studied when a three-phase fault occurs in the network (at the Diablo Canyon site on the Midway circuit adjacent to a 500-kV bus) . Some of the results of that study are shown in Figure 8.9. A one-line diagram of the network is shown in Figure 8.9(a). The frequency deviations for 4-cycle and 9.6-cycle faults are shown in Figures 8.9(b) and 8.9(c) respectively. The comparison is made between a 2.0 response ratio excitation system (curve A), a modern, low time constant excitation with rate feedback (curve B) and without rate feedback (curve C). The results of this study support the points made above. 8.3.3

Parametric study

Two recent studies [17, 19] show the effect of the excitation system on "first-swing" transients . Figure 8.10 shows the system studied where one machine is connected to an infinite bus through a transformer and a transmission network. The synchronous machine data is given in Table 8.1. The transmission network has an equivalent transfer reactance X, as shown in Table 8.1. Xd

Machine Data for the Studies of Reference! 19J

= 1.72

Xd =

0.45 Xd = 0.33 x q = 1.68 x~ = 0.59 x~' = 0.33

pu pu pu pu pu pu

,6-B v

TdO TdO

T~O T~O

= 6.3 s = 0.033 s = 0.43 s 0.033 s

H = 4.0 s

X

xr1

Vb

r =nnnnr.....

j O. IS 3 \1 Fault

Fig .8.10

System representation used in a parametric study of the effect of excitation on transient stability. (e IEEE. Reprinted from IEEE Trans.. vol. PAS·89, July/Aug. 1970.)

Chapter 8

320

Figure 8.10 . A transient is initiated by a three-phase fault on the high-voltage side of the transformer. The fault is cleared in a specified time. After the fault is cleared, the transfer reactance X. is increased from X. h (the value before the fault) to X. a (its value after the fault is cleared). The machine initial operating conditions are summarized in Table 8.2. Prefault Operating Conditions. All Values in pu

TableS.2.

x;

V,

V"

p

Q

0.2 0.4 0.6 0.8

1.0 1.0 1.0 1.0

0.94 0.90 0.91 0.97

0.90 0.90 0.90 0.90

0.39 0.45 0.44 0.44

With the machine operating at approximately rated load and power factor, a threephase fault is applied at the high-voltage side of the step-up transformer for a given length of time. When the fault is cleared, the transmission system reactance is changed to the postfault reactance X.a , and the simulation is run until it can be determined if the run is stable or unstable. This is repeated for different values of X.a until the maximum value of X.a is found where the system is marginally stable. . Two different excitation system representa tions were used in the study: 1. A 0.5 pu response alternator-fed diode system shown in Figure 8.11. 2. A 3.0 pu response alternator-fed SCR system with high initial response shown in Figure 8.12. This system has a steady-state gain of 200 pu and a transient gain of 20 pu . An external stabilizer using a signal V. derived from the shaft speed is also used (see Section 8.7).

Fig .8.11

Excitation block diagram for a 0.5 RR alternator-fed diode system . (e IEEE. Reprinted from

1£££ Trans.. vol, PAS-89. July/Aug. 1970.)

From the data presented in [19]. the effect of excitation on the "first-swing" transients is shown in Figure 8.13, where the critical clearing time is plotted against the transmission line reactance for the case where X. a = X. h and for the two different types of excitation system used . The critical clearing time is used as a measure of relative stability for the system under the impact of the given fault. Figure 8.13 shows that for the conditions considered in this study a change in exciter response ratio from 0.5 to 3.0 resulted in a gain of approximately one cycle in critical clearing time.

Effect of Excitation on Stabil ity

321

+ ±4. 9 pu

-, Fig. 8.12

Excitat ion blo ck diagr am for a 3.0 RR alternator-fed SCR excitation system . printed from IEEE Trans.. vol , PAS-89 . July/Aug. 1970.)

8.3.4

«(I) IEEE . Re-

Reactive power demand during system emergencies

A situation frequently encountered during system emergencies is a high reactive power demand . The capability of modern generators to meet this demand is reduced by the tendency toward the use of higher generator reactances. Modern exciters with high ceiling voltage improve the generator capability to meet this demand. It should be recognized that excitation systems are not usually designed for continuous operation at ceiling voltage and are usually limited to a few seconds of operation at that level. Concordia and Brown [17] recommend that the reactive-power requirement during system emergencies should be determined for a time of from a few minutes to a quarteror half-hour and that these requirements should be met by the proper selection of the generator rating .

8.4

ERect of Excitation on Dynamic Stability

Modern fast excitation systems are usually acknowledged to be beneficial to transient stability following large impacts by dr iving the field to ceiling without delay. However, these fast excitation changes are not necessarily beneficial in damping the oscillations that follow the first swing, and they sometimes contribute growing oscillations several seconds after the occurrence of a large disturbance. W ith proper design and compensation, however , a fast exciter can be an effective means of enhancing stability in the dynamic range as well as in the first few cycles after a disturbance. Since dynamic stability involves the system response to small disturbances, analysis as a linear system is possible, using the linear generator model derived previously [II). For simplicity we analyze the problem of one machine connected to an infinite bus

10

~" 6 c '"

j

U

-0 .~

U

Fig.8 .13

4

2

o

Transient stability stu dies resulting from studies of [19J: A - 0 .5 RR diode excitation system; B ~ 3.0 pu RR SCR excitation system. «(I) IEEE. Reprinted from IEEE Trans.. vol. PAS -89, Jul y/Aug. 1970.)

322

Chapter 8

through a transmission line. The synchronous machine equations, for small perturbations about a quiescent operating condition, are given by (the subscript d is omitted for convenience)

T, =

+ K2E; E; = [K)/(l + K)TdOS)]EFD - [K)K 4 / ( 1 + K)TdOS)]O ~ = Ks~ + K 6E; TjWS

K,~

= Tm

-

T,

(8.10) (8.1 )

(8. I2) (8.13)

where TdO is the direct axis open circuit time constant and the constants K , through K 6 depend on the system parameters and on the initial operating condition as defined in Chapter 6. In previous chapters it was pointed out that this model is a substantial improvement over the classical model since it accounts for the demagnetizing effects of the armature reaction through the change in E; due to change in ~. We now add to the generator model a regulator-excitation system that is represented as a first-order lag. Thus the change in EFD is related to the change in ~ (again the subscript ~ is dropped) by

= -Kf/(l + TfS)

EFD/~

where K, is the regulator gain and 8.4.1

r,

(8.14)

is the exciter-regulator time constant.

Examination of dynamic stability by Routh's criterion

To obtain the characteristic equation for the system described by (8.10)-(8.14), a procedure similar to that used in Section 3.5 is followed. First, we obtain

T(s) = [K' _ K2K4 TdO

e

From (8.13) for T m

S2 + s(l/T

S

f

+ (I/T + K sK f

Tt ) ]O(S) K3K6Kf)/KJTdOTf]

t/K4

+ I/K3T~o) + [(I +

(8.15)

= 0, S2 0 = -(Wi/Tj)Te = -(wR/2H)Te

(8.16)

By combining (8.15) and (8.16) and rearranging, the following characteristic equation is obtained: S4 + as)

where

a {j

{3s2

+

+

"YS

7J

=0

= I/T« + I/K]Tdo = [(I + KJK6Kt)/KJTdOTt] + K,(WR/2H)

'Y = ;~ (K./T, 7J

+

+ K./K3TdO - K2K4/TdO)

= ~ [KI(l + 2H

~3K6K,}

K3TdOTf

_ K,2 K4 ~ + KsK,)] TdOTf \ K4

Applying Routh's criterion to the above system, we establish the array S4 S3 S2

s'

SO

a

{3

7J

"Y

0

a. a2 b, b2 C,

(8.17)

323

Effect of Excitation on Stability

where al = (l/a)(a~ - 1') = ~ - l'/a a2 = (1/£1)(£117 - 0) = 17

(8.18)

According to Routh's criterion for stability, the number of changes in sign in the first column (I, a, ai, b c, and CI) corresponds to the number of roots of (8.17) with positive real parts. Therefore, for stability the terms a, aJ, bv, and CI must all be greater than zero. Thus the following conditions must be satisfied. I. a = I/T + I/K]Tdo > O,andsince r, and TdO are positive, f

(8.19) K 3 is an impedance factor that is not likely to be negative unless there is an exces-

sive series capacitance in the transmission network. Even then enough to satisfy the above criterion. 2. a. = {j - 'Y/£1 > 0 (

1

+

K]K 6K f

K]TdOTf

+ K)

~) 2H

_

TdO/Tf

is usually large

K3,TdOT• ~ [K I (T. + ~3TdO) _ K2,K4]

K3TdO

+

r.

2H

K 3 TdO Tf

TdO

> 0

or (8.20)

This inequality is easily satisfied for all values of constants normally encountered in power system operation. Note that negative K, is not considered feasible. From (8.20) K, is limited to values greater than some negative number, a constraint that is always satisfied in the physical system.

3. hi [

KI

= 'Y -

(~ + r,

> 0

Ct17

{j - 'Y/ a

_I) _ K] TdO

K2K4] TdO

Rearranging, this expression may be written as K 2K4(1

+

K 3K6K f )

K 3 Td5Tf

(8.21 )

We now recognize the first expression in parentheses in the last term of (8.21) to be the positive constant a defined in (8.17). Making this substitution and rearranging

324

Chapter 8

(8.22)

The expressions in parentheses are positive for any load condition. Equation (8.22) places a maximum value on the gain K, for stable operation. 4. c. = 11 > 0 Kl

(I + K,3 K K,) K

6

3i dOi f

_ K2K4

(I

f

f(K,K

)

K4

idOif

K

Since K.K6

+ KsK

>

0

K2Ks) > K2K4

6 -

-

K 1/K3

K 2K s > 0 for all physical situations, we have

-

KE > (K2K4

-

K1/K3)/(K1K6

-

This condition puts a lower limit on the value of K

f

K2Ks)

(8.23)

•

Example 8.3

For the machine loading of Examples 5.1 and 5.2 and for the values of the constants K1 through K6 calculated in Examples 6.6 and 6.7, compute the limitations on the gain constant K using the inequality expressions developed above. Do this for an exciter with time constant if = 0.5 s. f

,

Solution

In Table 8.3 the values of the constants K, through K6 are given together with the maximum value of K, from (8.22) and the minimum value of K, from (8.23). The regulator time constant if used is 0.5 s, ;;0 = 5.9 s, and' H = 2.37 s. Case 1 is discussed in Examples 5.1 and 6.6; Case 2, in Examples 5.2 and 6.5. From Table 8.3 it is apparent that the generator operating point plays a significant Table 8.3. Computed Constants for the Linear Regulated Machine Constants

Case I (Ex. 5.1)

Kl

1.076 1.258 0.307 1.712 -0.041 0.497 2.552 0.331 2.313 0.906 0.143 0.851 -0.616 5.051 4.000 -2.3 269.0

K2

K3

K4 Ks K6 a

K 2K 3K4 T K3TdO + T K)TdOTf K 2K4 / a TdO f

f

K4K6

aKsTdO K4 T dOT f

I/T;

Kf>

«,«

Case 2 (Ex. 5.2)

1.448 •. 317 0.307 1.805 0.029 0.526 2.552 0.365 2.313 0.906 0.158 0.949 0.442 5.325 4.000 -3.2 1120.2

Effect of Excitation on Stability

325

role in system performance. The loading seems to influence the values of K, and K, more than the other constants. At heavier loads the values of these constants change such that in (8.22) the left side tends to decrease while the right side tends to increase . This change is in the direction to lower the permissible maximum value of exciterregulator gain K,. For the problem under study, the heavier load condition of Case I allows a lower limit for K, than that for the less severe Case 2.

Routh 's criterion is a feasible tool to use to find the limits of stable operation in a physical system . As shown in Example 8.3, the results are dependent upon both the system parameters and the initial operating point. The analysis here has been simplified to omit the rate feedback loop that is normally an integral part of excitation systems. Rate feedback could be included in this analysis, but the resulting equations become complicated to the point that one is almost forced to find an alternate method of analysis. Computer based methods are available to determine the behavior of such systems and are recommended for the more complex cases [20. 21] . One special case of the foregoing analysis has been extensively studied [II). This analysis assumes high regulator gain (K JK6K, » I) and low exciter time constant (T, « KJT~O) ' In this special case certain simplificat ions are possible . See Problem 8.4. 8.4.2

Further considerations of the regulator gain and time constant

At no load the angle {) is zero, and the {) dependence of (8.10) -(8 .23) does not apply. For this condition we can easily show that the machine terminal voltage V, is the same as the voltage E;. Changes in this latter voltage follow the changes in Em with a time lag equal to TdO. A block diagram representing the machine terminal voltage at no load is shown in Figure 8.14. From that figure the transfer function for V, / VREF can be obtained by inspection. (8.24) Equation (8.24) can be put in the standard form for second-order systems as

V,/ VREF =

K/(S2

+ 2fwns +

w~)

(8.25)

where K = K.lTdOT" w~ = (I + K,)/T~OT" lfwn = (I/T, + I/T~o). For good dynamic performance, i.e.. for good damping characteristics, a reasonable value of f is 1/V2. For typical values of the gains and time constants in fast exciters we usually have TdO » T, and K, » I. We can show then that for good performance K, '" T;0/2T,. This is usually lower than the value of gain required for steady-state performance. In [II] de Mello and Concordia point out that the same dynamic performance can be obtained with higher values of K, by introducing a lead-lag network with the proper choice of transfer function . This is left as an exercise (see Problem 8.5).

Fig .8 .14

Block diagram representing the machine terminal vo ltage at no load .

326

Chapter 8

8.4.3

Effect on the electrical torque

The electrical torque for the linearized system under discussion was developed in Chapter 3. With use of the linear model, the electrical torque in pu is numerically equal to the three-phase electrical power in pu. Equation (3.13) gives the change in the electrical torque for the unregulated machine as a function of the angle o. The same relation for the regulated machine is given by (3.40). From (3.13) we compute the torque as a function of angular frequency to be (8.26) The real component in (8.26) is the synchronizing torque component, which is reduced by the demagnetizing effect of the armature reaction. At very low frequencies the synchronizing torque T.. is given by (8.27) In the unregulated machine there is positive damping introduced by the armature reaction, which is given by the imaginary part of (8.26). This corresponds to the coefficient of the first power of s and is therefore a damping term. In the regulated machine we may show the effect of the regulator on the electrical torque as follows. From (3.40) the change of the electrical torque with respect to the change in angle is given by TdO

1

1) (I

2 + (S - + - - + S r, K) TdO

+

K)K 6K f )

K) TdOTf

K 2K 4(1 + TfS) + K 2K sK, [(11K ) + K 6K f ) + S2(TdO T,)] + S(TdO + T,IK)

_

_ K

-

S + (liT, + K sK,IK 4T f)

,

(8.28)

It can be shown that the effect of the terms K 2K4(1 + TfS) in the numerator is very small compared to the term K 2K sK This point is discussed in greater detail in [II]. Using this simplification, we write the expression for T~/o as f •

T~ ~ K, _

o

K2K sK [(11K) + K 6 K f) + TdO TfS2] + S(TdO + TfIK) f

(8.29)

which at a frequency W can be separated into a real component that gives the synchronizing torque T, and into an imaginary component that gives the damping torque Tdo These components are given by Ts

~

T ~ d -

K, -

K 2K sK f [( I / K 3 + K 6K f) - W2TdOTf] [(11K) + K 6K f) - W2TdOTfF + W2(TdO + TfIK J)2

------~-----------

K 2K sK,(TdO + TfIKJ)w [(11K ) + K6 K , ) - w2TdOTf]2 + W2(TdO + TfIK)2

(8.30) (8.31)

Note that the damping torque Td will have the same sign as K«. This latter quantity can be negative at some operating conditions (see Example 6.6). In this case the regulator reduces the inherent system damping. At very low frequencies (8.30) is approximately given by (8.32) which is higher than the value obtained for the unregulated machine given by (8.27).

Effect of Excitation on Stability

Fig . 8.15

327

Block diagram of a lineari zed excitation system model.

Therefore, whereas the regulator improves the synchronizing forces in the machine at low frequencies of oscillation, it reduces the inherent system damping when K, is negative, a common condition for synchronous machines operated near rated load .

8.S

Root-Locus Analysis of a Regulated Machine Connected to an Inflnite Bus

We have used linear system analysis techniques to study the dynamic response of one regulated synchronous machine . In Section 7.8, while the exciter is represented in detail, a very simple model of the generator is used . In Section 8.4 the exciter model used is a very simple one. In this section a more detailed representation of the exciter is adopted, along with the simplified linear model of the synchronous machine that takes into account the field effects . The excitation system model used here is similar to that in Figure 7.54 except for the omission of the limiter and the saturation function Sf. This model is shown in Figure 8.15. In this figure the function GF(s) is the rate feedback signal. The signal V. is the stabilizing signal that can be derived from any convenient signal and processed through a power system stabilizer network to obtain the desired phase relations (see Section 8.7). The system to be studied is that of one machine connected to an infinite bus through a transmission line. This model used for the synchronous machine is essentially that given in Figure 6.3 and is based on the linearized equations (8.10)-(8 .13). To simulate the damping effect of the damper windings and other damping torques, a damping torque component - Dw is added to the model as shown in Figure 8.16. The combined block diagram of the synchronous machine and the exciter is given in Figure 8.17 (with the subscript Ll omitted for convenience).

t--t---..--.- 6

Fig.8.16

Block diagram of the simplified linear model of a synchronou s machine connected to an infinite bus with damping added .

Fig.8.17

K.,

Combined block diagram of a linear synchronous machine and exciter .

K.

Co) to.)

00

CO ..,

"0

Q

::r

(')

00

329

Effect of Excitation on Stability

Fig .8.18

Block diagram with V, as the takeoff point for feedback loops.

To study the effect of the different feedback loops, we manipulate the block diagram so that all the feedback loops "originate" at the same takeoff point. This is done by standard techniques used in feedback control systems (22). The common takeoff point desired is the terminal voltage v" and feedback loops to be studied are the regulator and the rate feedback GF(s). The resulting block diagram is shown in Figure 8.18. In that figure the transfer function N(s) is given by N(s)

=

K3K6(2Hi + Ds + K1wR) - WRK2K3KS

(I

2

+ K3TdOS)(2Hs + Ds + K1WR) - WRK2K3K4 I

(8.33)

Note that the expression for N (s) can be simplified if the damping D is neglected or if the term containing K s is om itted (K s is usually very small at heavy load conditions). The system of Figure 8.18 is solved by linear system analysis techniques , using the digital computer. A number of computer programs are available that are capable of solving very complex line ar systems and of displaying the results graphically in several convenient ways or in tabular forms [20, 21). For a given operating point we can obtain the loci of roots of the open loop system and the frequency response to a sinusoidal input as well as the time response to a small step change in input. The results of the linear computer analysis are best illustrated by some examples. In the analysis given in thi s section , the machine discussed in the examples of Chapters 4,5, and 6 is analyzed for the loading condition of Example 6.7. The exciter data are K A = 400, TA = 0.05, K E = -0.17, TE = 0.95. K R = 1.0 and TR = O. The machine constants are 2H = 4.74 s, D = 2.0 pu and TdO = 5.9 s. The constants K , through K6 in pu for the operating point to be analyzed are

1.4479

K3

=

0 .3072

K, = 0.0294

1.3174

K4

=

1.8052

K6

=

0.5257

Example 8.4 Use a linear systems analysis program to determine the dynamic response of the system of Figure 8.18 with and without the rate feedback . The following graphical solutions are to be obtained for the above operat ing condit ions: I. 2. 3. 4.

Root-locus plot. Time response of V,to to a step change in VRE F • Bode diagram of the closed loop transfer function. Bode diagram of the open loop transfer function.

I

8

i

I

--

- -- --

II I -I ~

t

-- - -- .+- -. -- - - . .;

6

iI-;- -

tI ;

2

.- -j

,

I

I

Ii

-

J

I

I

I

II I -I I I i

I.

I

- - --

j

---

_.-

- ~

--

-

• .. .

-

..

~

t-:-

--r-

tt t 1--1:-'-IJ.

o +1T--: -8

1 ---

-- ---f-- I- ,.

..: ~

t-.

:

1

rl l ~

..-

1---:- --

---t-t+ ~~ ~~;_: -~ t

I

_L

N

I ..o

>..

c:

' 51 4

o

,§

I

!

I

1

---

- - , -,

-o

-2

Real (0) Fig .8 . 19

=V~

f- V- -

~

, .:l.l

,

-4

-6

--

-

--I I

-.

;

+l-i-~, --l--

-- - -

-

!

~- ~-, _1-

I

-. I-r +---

. f- _L ~ .

.rl~--t-'1ft

.-

-

1-1

-i-

-

-

-1- --.-.- - -- - -- --- -- ---

-

cl-

,

I

i

-

f- - t-

-8

o

-2

-4

-6

Real (b)

Ro ot locus of the system o f Figure 8.17: (a) wit ho ut rate feedback, (b) wit h rate feedback.

o

o

Time, s

2

Fig .8.20 -~ "2!'

I

1

f"\

I

i

-I

...

I

I~

!

I

I

I

i I

,

-_ !

I~ -

i

...i

!

I~

i

C i

&!.

,

I

."

\

I

I

I

:

i

i :.

I

i

I

! .

..

\

!

IW . ~D

I

i

t-

!

I 1

i

:

-I

i

~- t-.L _.-. -.- -_ I

I

-

..

\ I' I RADI ANS/ SEC.

(0 )

Fig . 8.21

I .i.,

I

--:

. --

--

."

i

i

1

iW

i

~

-- ---- -

-

:

I

~

!

~

I

---

_. ... "

!

, t---;-

1+H--I

!

j

... -..

r-r- t-:: ---'- -- - I :

-. -

-I j

:-1 -- ! -

' ''h 'I . "

\

"

i

"2

--,

;

- I-

j

"

++ ,

,

-

----

-- -~

.

i

i

~; e :~ S/

I

,

i

hJ

i

!

1

--

i

I

I

._ .0.

..~

i

I

;

."

\

I'

j

."

; i

r-, 1-

I I

I

I

:

tL

!

~ --

.... •

I ;

'---1--

I

!

\

I RADI ANS/S EC.

---

I---c- c-:

,~

=

I

~

r+

.

:

K .1

i

,,;

;.

i r~ I

:-

.~o

r::\

.--- -- -

,

!

I

i

,- t-- ~ i ~

i

I

d'S:. ~ r--: I

1----'-- ---'-'~

---

1

i__ ~

;

I

I

!\-

I

l-r- ,

1

--

i

I :".' + '--- f- ~ ---I---

."

• ..

--

f--

\

I'

."

~

( b)

Bode plots of the closed loop transfer funct io n: (a) G F

I

,

._-- _.- '- - -, - ~ - f----'--

-, !

·1

4.· ···

l!l -: -

:

~ ~_~rs: .t-

-- --- --'---- ----:- - -

i

I

.

- -...

1---- ---.

I

I

-

--

;

I

' 1 'i'

,

' I 1"--- 1----, t;-..' _ ' r

-,--e-:-- -

-!

-

i

!

-

:

~.

O,lb) GF(s) .. O.

I

_.:..(--_..:.. ; :

I

-. .J _ -- \

----

I

I

--- --

,

I

!

rrt


0.4

o

2

4 Time, s (0)

6

8

o

4 Tim e, s

2

6

8

( b)

Fig.8 .24 Time response to a step change in VREF for the system of Example 8.5: (a) D = O. (b) K s = O.

quency and its damping is inherently poor . This is an important consideration in the study of power system stabilizers. 8.6

Approximate System Representation

In the previous section it is shown that the dynamic system performance is dominated by two pairs of complex roots that are particularly significant at low frequencies. In this frequency range the system damping is inherently low, and stabilizing signals are often needed to improve the system damping (Section 8.7). Here we develop an approximate model for the excitation system that is valid for low frequencies. We recognize that the effect of the rate feedback GF(S) in Figure 8.17 is such that it can be neglected at low frequencies (s = jw -+ 0) or near steady state (t -+ 00). We have already pointed out that K, is usually very small and is omitted in this approximate model. The feedback path through K4 provides a small positive damping component that is usually considered negligible [II). The resulting reduced system is composed of two subsystems: one representing the exciter-field effects and the other representing the inertial effects. These effects contribute the electrical torque components designated Td and T. , respectivel y. 8.6.1

Approximate excitation system representation

The approximate system to be analyzed is shown in Figure 8.25 where the exciter and the generator have been approximated by simple first-order lags [II). A straightforward analysis of this system gives v, K

E'

q

(

T"""+"""TS (

_

-

-

-

-

- - G. (,)

---------l_

Fig. 8.25 Approxim ate repre sentation of the excitation system.

334

Chapter 8

(8.34) Since KJK6K~ » 1 in all cases of interest, (8.34) can be simplified to G (5) = x

I

+ [(T ~ +

K2/K6

K 3 TdO) / K 3 K 6 K f) S

~ 52

(T dO T

K2K / TdoT + [(T + K 3 T dO) / K3 T dO T 5 + K6 K K2K / TdOT K2K / TdOT f )

I

+

f

2rx

wx s + w; f

where W x is the undamped natural frequency and

I /

K6K

f )

52

f

f

52

+

f

d(s)

f /

T

dO T

I

I

(8.35)

rx is the damping ratio: (8.36)

We are particularly concerned about the system frequency of oscillation as compared to W x • The damping r, is usually small and the system is poorly damped. The function Gx(s) must be determined either by calculation or by measurement on the physical system. A proven technique for measurement of the parameters of Gx(s) is to monitor the terminal voltage while injecting a sinusoidal input signal at the voltage regulator summing junction [8, 12,23,24,25). The resulting amplitude and phase (Bode) plot can be used to identify Gx(s) in (8.35). Lacking field test data, we must estimate the parameters of Gx(s) by calculations derived from a given operating condition. It should be emphasized that this procedure has some serious drawbacks. First, the gains and time constants may not be precisely known, and the use of estimated values may give results that are suspect [10, 12,24J. Second, the theoretical model based on the constants K. through K 6 is not only load dependent but is also based on a one machine-infinite bus system. The use of these constants, then, requires that assumptions be made concerning the proximity of the machine under study with respect to the rest of the system. A procedure based on deriving an equivalent infinite bus, connected to the machine under study by a series impedance, is given in Section 8.6.2. 8.6.2

Estimate of G.(s)

The purpose of this section is to develop an approximate method for estimating K 1 through K6 that can be applied to any machine in the system. These constants can be used in (8.36) to calculate the approximate parameters for Gx(s). The one machine-infinite bus system assumes that the generator under study is connected to an equivalent infinite bus of voltage V«Ji!! through a transmission line of impedance Z~ = R~ + jX~. This equivalent impedance is assumed to be the Thevenin equivalent impedance as "seen" at the generator terminals. Therefore, if the driving-point short circuit admittance Vii at the generator terminal node i is known, we assume that (8.37) The equivalent infinite bus voltage V«J is calculated by subtracting the drop l;z~ from the generator terminal voltage P;;, where 1; is the generator current. The procedure is illustrated by an example.

335

Effect of Excitation on Stability

Example 8.6 Compute the constants K1 through K6 for generator 2 of Example 2.6, using the equivalent infinite bus method outlined above. Note that the three-machine system is certainly not considered to have an infinite bus, and the results might be expected to differ from those obtained by a more detailed simulation. Solution

From Example 2.6 the following data for the machine are known (in pu and s).

= 0.8958 x:/2 = 0.1198

Xd2

X q2 X;2

= 0.8645 = 0.1969

H2

x-t 2 = 0.0521

=

6.4

= 6.0

Td02

We can establish the terminal conditions from the load-flow study of Figure 2.19: /2~

= /r2 + j/x2 = (P2 - jQ2)IV2

= (1.630 - jO.066)/1.025 = 1.592 / - 2.339

0

pu

From Figure 5.6 ta n (

- - ' - - --

(1-u' ) +j 2C u

W

u ::W

n , = dampi ng rotio

-1 2

-18 0.01

0.02

0 .040.06

0.1

0.2

0 .4

0.6

3

4 5

u

( a)

o - 15 -30 -45

y=

-60

-8 ~

(1 -

w u =_

II -75

u') + j2eu

w

n

-90

C := damping rati o

w, and S, is small. For small damping the phase changes very fast in the neighborhood of w, (where cPlag = 90·). Many textbooks on control systems, such as (22), give curves of phase shift as a function of normalized frequency, U = w/wn , as shown in Figure 8.26. In the above example, with U = 8.8/5 .967 = 1.47 and = 0.13, it is apparent from Figure 8.26(b) that the phase lag is great.

r

8.6.3

The inertial transfer function

The inertial transfer function can be obtained by inspection from Figure 8.17. For the case where damping is present,

338

Chapter 8

-0

wR/2H

(8.38)

Where w" is the natural frequency of the rotating mass and!" is the damping factor,

vlK,WR/2H !n = D/4Hwn = D/2v12HK 1wR

Wn =

(8.39)

The damping of the inertial system is usually very low. Example 8.8

Compute the characteristic equation, the undamped natural frequency, and the damping factor of the inertial system of generator 2 (Example 2.6). Use D = 2 pu.

Solution From the data of Examples 2.6 and 8.6 we compute d(5) = 52

+ 0.1565 + 72.894

= "\1'72.894 = 8.538 rad/s t, = 2/[2(12.8 x 2.975 x 377)1/2] = 0.009

w"

d(jw) = 1 - O.0137w 2

At the system frequency of oscillation w ¢Iag

8.7

= tan "

+ jO.00214w

= Wos c =

8.8 rad/s,

[0.0183/(-0.0222 - 0.0604)]

Supplementary Stabilizing Signals

Equation (8.31) indicates that the voltage regulator introduces a damping torque component proportional to K s . We noted in Section 8.4.3 that under heavy loading conditions Ks can be negative. These are the situations in which dynamic stability is of concern. We have also shown in Section 8.6.2 that the excitation system introduces a large phase lag at low system frequencies just above the natural frequency of the excitation system. Thus it can often be assumed that the voltage regulator introduces negative damping. To offset this effect and to improve the system damping in general, artificial means of producing torques in phase with the speed are introduced. These are called "supplementary stabilizing signals" and the networks used to generate these signals have come to be known as "power system stabilizer" (PSS) networks. Stabilizing signals are introduced in excitation systems at the summing junction where the reference voltage and the signal produced from the terminal voltage are added to obtain the error signal fed to the regulator-exciter system. For example, in the excitation system shown in Figure 7.54 the stabilizing signal is indicated as the signal ~. To illustrate, the signal usually obtained from speed or a related signal such as the frequency, is processed through a suitable network to obtain the desired phase relationship. Such an arrangement is shown schematically in Figure 8.27. 8.7.1

Block diagram of the linear system

We have previously established the rationale for using linear systems analysis for the study of low-frequency oscillations. For any generator in the system the behavior

Effect of Excitation on Stability

Fig.8.27

339

Schematic diagram of a stabilizing signal from speed deviation .

can be conveniently characterized and the unit performance determined, from the linear block diagram of that generator. This block diagram is shown in Figure 8.28. The constants K, through K 6 are load dependent (see Section 8.6 for an approximate method to determine these constants) but may be considered constant for small deviations about the operating point. The damping constant D is usually in the range of 1.0--3.0 pu. The system time constants, gains, and inertia constants are obtained from the equipment manufacturers or by measurement. The PSS is shown here as a feedback element from the shaft speed and is often given in the form [II] (8.40)

The first term in (8.40) is a reset term that is used to "wash out " the compensation effect after a time lag TO, with typical values of 4 s [II] to 20 or 30 s [12]. The use of reset control will assure no permanent offset in the terminal voltage due to a prolonged error in frequency, such as might occur in an overload or islanding condition. The second term in Gs(s) is a lead compensation pair that can be used to improve the phase lag through the system from VREF to WA at the power system frequency of oscillation . Qualitatively, we can recognize the existence of a potential control problem in the system of Figure 8.28 due to the cascading of several phase lags in the forward loop. In terms of a Bode or frequency analysis (see [22J, for example) the system is likely to have inadequate phase margin. This is difficult to show quantitatively in the complete system because of its complexity . We therefore take advantage of the simplified representation developed in Section 8.6 and the results obtained in that section . 8.7.2

Approximate model of the complete exciter-generator system

Having established the complete forward transfer funct ion of the excitation control system and inert ia, we may now sketch the complete block diagram as in Figure 8.29. We note that a common takeoff point is used for the feedback loop, requiring a slight modification of the inertial transfer function using standard block diagram manipulation techniques . We also note that the output in Figure 8.29 is the negative of the speed deviation . The parameters L, W and r., w. are defined in (8.36) and (8.39) respectively . Examining Figure 8.29 we can see that to damp speed oscillations, the power system stabilizer must compensate for much of the inherent forward loop phase lag. Thus the PSS network must provide lead compensation . X

Fig.8.28

Block diagram of a linear generator with an exciter and power system stabilizer.

K.

CO

n zr: o "lJ it ...

~

o

w

Effect of Excitation on Stability

S2

v

Fig. 8.29

8.7.3

2(

...

341

w s + :v 2

.n n

n

s

Block d iagram of a simplified model of the complete system .

lead compensation

One method of providing phase lead is with the passive circuit of Figure 8.30(a) . If loaded into a high impedance, the transfer function of this circuit is (l/a)(1 + aTs) 1 + TS

Eo E; where a T

(R , + R 2)/R 2 > 1 R 1R 2C/(R 1 + R 2 )

(8.41)

The tran sfer function has the pole zero configuration of Figure 8.30(b), where the zero lies inside the pole to pro vide phase lead . For this simple network the magnitude of the parameter a is usually limited to about 5. Another lead network not so restr icted in the parameter range is that shown in Figure 8.3 1 [26]. For this circuit we co mpute

Eo E; where

TA

T8 t

c

To

K, K2

K 1 RC I R IC I

=

K 2RC2 RC F =

=

(I

+

I + (TA + T8) S T8s)[1

+

(Te

+

(8.42)

TO)S]

lead time constant noise filter time con stant « TA = lag time constant stabili zing circuit time constant « Te =

R 8/(R A + R 8 ) Ro/(R e + R o)

Approx imately , then

Eo/E;

= (I

+

TA S)/(I

+

TeS)

(I

+

aT s)/(1

+

TS)

where a c a >O

j'

R, t .I

R,

EO

1

T (a) F ig. 8.30

-~ aT (b)

Lead net work : (a) passi ve net work. (b ) pole zero configuration.

(8.43)

342

Chapter 8

I

If

R A

E.

I

11 R(

R c, B

(-

(,

R1

R O

";"

";"

Fig. lUI

EO

Acti ve lead network ,

For any lead network the Bode diagram is that shown in Figure 8.32, where the asymptotic approximation is illustrated [22). The maximum phase lead r/>m occurs at the median frequency W m , where W m occurs at the geometric mean of the corner frequencies: i.e., 10glOwm

=

(1/2)(IoglO(I/aT) + 10glO(I/T)]

=

(1/2)!oglO(I/aT 2 )

loglO(I/nla)

=

Then =

Wm

I/n/a

(8.44)

The magnitude of the maximum phase lead r/>m is computed from r/>m

=

arg[(1 + jwmaT)/(1 + jWmT)]

=

tan -'wmaT - tan-1wmT ~ x - y

(8.45)

From trigonometric identities tan(x - y)

(tanx - tany)/(I

+

(8.46)

tanxtany)

Therefore, using (8.46) in (8.45) tan r/>m

=

(wmaT -

W

mT)/[ I + (wmaT)(wmT»)

=

I)/( I +

WmT(a -

aw~ T2)

(8.47)

This expression can be simplified by using (8.44) to compute tan r/>m

=

(a -

1)/ 2Vii

(8.48)

Now, visualizing a right triangle with base 2Vii, height (a - I) and hypotenuse b, I I I

I I

I°E~O I

I

: ~:

0

db

& E. I

0

-90

Fig. 8.32

~m

log w

I I

I

I I

45'/ decode

I

I

+90

20 db/decode

- - - -1 -

I

-

I I

I I I

l/oT

'.m

,

log w

I

I/T

Bode diagram for the lead network (I

+ QTS)/( I + TS)

where Q > I.

343

Effect of Excitation on Stability

we compute b2

+

1)2

= (a -

4a = (a

+

1)2 or

sin ¢m = (a - 1)/(a + I)

(8.49)

This expression can be solved for a to compute a = (I + sin ¢m)/(I - sin ¢m)

(8.50)

These last two expressions give the desired constraint between maximum phase lead and the parameter Q. The procedure then is to determine the desired phase lead ¢m' This fixes the parameter Q from (8.50). Knowing both a and the frequency W m determines the time constant T from (8.44). In many practical cases the phase lead required is greater than that obtainable from a single lead network. In this case two or more cascaded lead stages are used. Thus we often write (8.40) as Gs(S)

= [KOTos/( 1 + TOS)][( I +

where n is the number of lead stages (usually n

=

ars)/( 1

+

(8.51 )

TS)r

2 or 3).

Example 8.9

Compute the parameters of the power system stabilizer required to exactly compensate for the excitation control system lag of 161.6° computed in Example 8.7. Solution Assume two cascaded lead stages. Then the phase lead per stage is ¢m

= 161.6/2 = 80.8°

From (8.50) a = (1

+ sin 80.8)/( I - sin 80.8)

=

154.48

This is a very large ratio, and it would probably be preferable to design the compensator with three lead stages such that ¢m = 53.9°. Then a

(I + sin 53.9)/( I - sin 53.9) = 9.42

=

which is a reasonable ratio to achieve physically. The natural frequency of oscillation of the system is from (8.44) T

Thus Gs (s)

= [

=

I/wmw = 0.037

«,ToS / (I +

T os)][

Wos c =

Wm

= 8.8 rad/s, Thus

a r = 0.3488

(I + 0.349s)/ (1 + 0.037s)P

A suitable value for the reset time constant is TO = lOs. The gain Ko is usually modest [26J, say 0.) < Ko < 100, and is usually field adjusted for good response. It is also common to limit the output of the stabilizer, as shown in Figure 8.28, so that the stabilizer output will never dominate the terminal voltage feedback. Example 8./0

Assume a two-stage lead-compensated stabilizer. Prepare a table showing the phase lead and the compensator parameters as a function of Q.

Solution As before, we assume that W m

=

8.8 rad/s,

Chapter 8

344 Table 8.5. a

5

\0

15 20 25

Lead Compensator Parameters as a Function of a

cbm

2cbm

T = I/wmv7i

WHi= liT

aT

wLO= IlaT

41 .81 54.90 61.05 64.79 67.38

83.62 109.80 \22.10 \29 .58 134.76

0 .0508 0.0359 0.0293 0.0254 0.0227

19.68 27.83 34.08 39.35 44.00

0.2541 0.3593 0.4401 0 .5082 0.5682

3.935 2.783 2.272 1.968 1.760

These results show that for a large a or large rPm the corner frequencies WHi and WLO must be spread farther apart than for small rPm . See Figure 8.32 and Problem 8.1 I. Linear Analysis of the Stabilized Generator

8.8

In previous sections certain simplifying assumptions were made in order to give an approximate analysis of the stabilized generator. In this section the system of Figure 8.28 is solved by linear system analysis techniques using the digital computer (see Section 8.5) . The results of the linear computer analysis are best illustrated by an example.

Example 8./ / Use a linear systems analysis program to determ ine the following graphical solutions for the system of Figure 8.28 : I. 2. 3. 4.

Root-locus plot Time response of W~ to a step change in VREF Bode diagram of the closed loop transfer function Bode diagram of the open loop transfer function .

Furthermore, compute these graphical displays for two conditions, (a) no power system stabilizer and (b) a two-stage lead stabilizer with a = 25:

8

8

2

-6

-4 Real (a)

Fig. 8.33

-2

o

-4

Real

o

(b)

Root locus of the generator 2 system : (a) no PSS. (b) with the PSS hav ing two lead stages with a = 25.

345

Effect of Excitation on Stability

i

1.21

N

« ., : :1 ,' \ ,:\ ,'\ r.... ~ 0 0 , " Ii, \ I '. i &. • l !\ i \J \.1 v ~

N~ I

~

10

I

3~

-1.2

]\

i

\i j,I

1"\

\

\i .

-2.4J-I-,-V~~__ ~ o

2

\ /~/ "",-""'--------.

4 TIme,s

~

-0,4 \

3~

-1 2 \

5.

6

8

0.4

o

~

\

0

I

.

iI

-2.0

\1

I

o

2

4 Time,s ( b)

(0 ) Fig.8 .34

6

8

Time response to a step cha nge in VREF : (a) no PSS, (b) with the PSS having two lead stages with a ~ 25.

Gs(s) = [IOs/(\

+ IOs»)[(1 + O.568s)/(\ + O.0227s)F

The system constants are the same as Examples 8.7 and 8.8.

Solution

The system to be solved is that of Figure 8.28 except that the PSS limiter cannot be represented in a linear anal ysis program and is therefore ignored . The results are shown in Figures 8.33-8 .36 for the four different plots. In each figure, part (a) is the result without the PSS and part (b ) is with the PSS. In the root-locus plot (Figu re 8.33) the major effect of the PSS is to separate the torque-angle zeros from the pole s, forcing the locu s to loop to the left and downward , thereby increasing the damping . The root locus shows clearly the effect of lead compensation and has been used as a basis for PSS parameter identification [27). Note that

",8

0..;

z7

-.

8 8 ;

m

~~ ~----:;;'" o

~8

~o ~8

. ..

\

\ "~\.IO

\

\'

RRDI RNS/ SEC.

(0)

Fig. 8.35

'W

7-1--:-I"'TT-rT'mr-:'l:-rrmm--r-rn:nvrC=-TTTrrt't':-:-::r"iTtTii" RRD I RNS/S EC.

(b)

Frequency resp onse ( Bode diagram) of the closed loop tr ansfer function : (a) no PSS, (b) with the PSS ha ving two lead st ages with a = 25.

346

Chapter 8 ~

-.i

~

,

8

co

08

zo

w

t-,

§g

Z'

t:l

a

!L

, ~l "'\WIO,I

~ i ~i "I\w,OJ

RRD1RNS/SE(.

~ I ~i

8 ~

"Ii

lIID-'1

~'Sllh'\ '~O,I RRDI RNS/SEC.

~'~'lh'\lIIOJ

~

i

~"h"

8

8

-~

-~

"9

~8

~

~8

~ a::~

~~ ~'

~' o

0

~~

~~

wi' en a:

wi"

:£

x:

~8

G..8

~

~+--'r"""'r""O""T'TTT'I"'-----'~TTTTnr---r--rT""1n'TTTr--'-"'T"rT-r"T'nT""-"--T-rr1"TTT1 RRDI RNS/S[C.

wIO

Si \' 'hi\.IOJ Si ~i 'hli

(b)

(0)

Fig.8.36

-- r ~ 1 ~"hi\.lO" RADIANS/SEC.

Frequency response (Bode diagram) of the open loop transfer function: (a) no PSS, (b) with the PSS having two lead stages with a = 25.

the locus near the origin is unaffected by the PSS, but the locus breaking away vertically from the negative real axis moves closer to the origin as compensation is added (this locus is off scale in 8.33(a)]. From the computer we also obtain the tabulation of poles and zeros given in Table 8.6. From this table we note that the natural radian frequency of oscillation is controlled by the torque-angle poles with a frequency of 8.467 rad/s. This agrees closely with W n = 8.538 rad/s computed in Example 8.8 using the approximate model and also checks well with the frequency of 021 in Figure 3.3. Figure 8.34 shows the substantial improvement in damping introduced by the PSS network. Note the slightly decreased frequency of oscillation in the stabilized response. Table 8.6. Condition

Poles

No PSS

With PSS a

Root-Locus Poles and Zeros

25

- 20.000 0.179 -0.102 -0.289 -0.289 -1.000 -20.000 0.179 -0.010 -0.289 -0.289 - 1.000 -0.100 -45.500 -45.500

+ jO.OOO

+ jO.OOO

+ jO.OOO + j8.533 - j8.533

+ jO.OOO

+ jO.OOO

+ jO.OOO + jO.OOO + j8.533

- j8.533 + jO.OOO + jO.OOO + jO.OOO - jO.OOO

Zeros

-0.944 -0.944 -0.452 -0.452

+ jO.955 - jO.955 + j8.467 - j8.467

-0.100 -0.941 -0.941 -0.955 -0.955 -45.000 -45.000

+ jO.OOO

+ jO.959

- jO.959

+ j7.439

- j7.439

+ j24.847

- j24.847

347

Effect of Excitation on Stability

Figures 8.35 and 8.36 show the frequency response of the closed loop and open loop transfer functions respectively. The uncompensated system has a very sharp drop in phase very near the frequency of oscillation. Lead compensation improves the phase substantially in this region, thus improving gain and phase margins. 8.9

Analog Computer Studies

The analeg computer offers a valuable tool to arrive at an optimum setting of the adjustable parameters of the excitation system. With a variety of compensating schemes available to the designer and with each having many adjustable components and parameters, comparative studies of the effectiveness of the various schemes of compensation can be conveniently made. Furthermore, this can be done using the complete nonlinear model of the synchronous machine. 8.9.1

Effect of the rate feedback loop in Type 1 exciter

As a case study, Example 5.8 is extended to include the effect of the excitation system. The synchronous machine used is the same as in the examples of Chapter 4 with the loading condition of Example 5.1. Three IEEE Type 1 exciters (see Section 7.9.1) are used in this study: W TRA, W Brushless, and W Low 'E Brushless. The parameters for these exciters are given in Table i .8. The analog computer representation of the excitation system is shown in Figure 8.37. This system is added to the machine simulation given in Figure 5.18. Note that the output of amplifier 614 (Figure 8.37) connects to the terminal marked EFD in Figure 5.18, and the terminal marked VI in Figure 5.18 connects with switch 421 in Figure 8.37. The new "free" inputs to the combined diagram are VREF and Tm • The potentiometer settings for the analog computer units are given in Tables 8.7, 8.8 and 8.9 for the three excitation systems described in Table 7.8. Saturation is represented by an analog limiter on V R in this simulation. With the generator equipped with a \V TRA exciter, the response due to a 10°/;) increase in T; and 5% change in VREF and the phase plane plot of WL\ versus OL\ for the initial loading condition of Example 5.1 are shown in Figure 8.38. The results with W Brushless and W Low 'E Brushless exciters are shown in Figures 7.69 and 8.39 respectively. Table 8.7. Pot. no.

Amp. no.

600 601

601 601

VREf

800 701

800 800

VR VR

801 703

801 801

-E FD -E FD

802 810

802 802

Vz Vz

812 803

810 803

- Vy Vx

100 50

lim 800

800

..

...

Out V REF

.

...

Potentiometer Calculations for a Type 1 Representation of a W TRA Exciter (a = 20) In

L;

50 50

REF REF

100 100

0.50 0.50

I I

- Ve VR

50 I

0.02 1.00

10 10

VR -E FD

I 10

10.00 1.00

50 50

Vy -E FD

100 10 50 40

Lo

. ..

Vz

v,

. .. .. .

., .,

C

LolL;

. .

= constant

(Lol L;)C

s-, of P60I

lnt. cap.

Amp. gain

Pot. set.

0.0250 0.4994

... ., .

I I

0.0250 0.4994

8.0 1.0

1.0 1.0

10 10

0.8000 0.1000

I/o T E = 1/(20)(0.95) = 0.05263 I K E I /aTE == 0.17/(20)(0.95) = 0.008947

0.5263 0.0089

1.0 1.0

I I

0.5263 0.0089

0.50 5.00

1/ T F = 1/1.0 = 1.0 KF/TF = 0.04/1.0 == 0.04

0.5 0.2

... ...

I I

0.5000 0.2000

2.00 1.25

I/o = 1/20 == 0.05 1/v} == 0.5773

0.1 0.7217

1.0

I I

0.1000 0.7217

. .. . ..

V R max = 3.5 pu = 3.5 v VR min = -3.5 pu = -3.5 v

I

+ (2.667)( -0.17)/400 = 0.998g

KAla T A == 400/(20)(0.05) 1/0 T A = 1/(20)(0.05) = I

= 400

...

Fig.8 .37

x 803

Qm

IT

LIM

V'A

...... 421

(40)

LIM

r

. -,

810

/'1 1

411l...l./"222

c-----"1

( >:---l I c

~

, .,...I

/

,,--,I

1. 10

1. 0

0. 9

>-

-,

",0-

.'"

a.

E

'0 0.8

>

1.0 0 .4

0 .6

Fig .8.49

0 .8

TIme , ,

1. 0

1. 4

Vol tages o f generator 2 with BBC exciter .

f\\

4 .0

I\

I \

I \

-

3 .0

I \ I I I

I \ / \// , I II ->

w~

fI

'0

>

2.0

/

/

/

~/



/' E

/ \

,

~

a.

j\v,

1. 0

0 .9

'\

Fig. 8.50

~

'0

>

0.8

Vol tage s o f generat o r 2 with T ype I 0.5 R R exciter.

362

Chapter 8

125

II

"..-

~ ~

..\

,

\

0

100

\

\

]

\

Ii. Q.

\

0

.=

\

~

~

II

~

."

-o

.f

\

\

75 : I

\

I

\

,I

N

\

\

I 501

II

0,

c

-{

/

\

I

t \

I

~

\

I

0-

....~

I

\

I I 25 I

\

/

/

/~

/

",

~

With PSS

I

~

7

/

\ ' --. / / 'Without

pss

I

I

I

I

0

Fig . 8.51

2.00

Torque angle

02 1

for a three-phase fault near generator 2. PSS with a : 25. Wos c : 8.9 rad/s.

inertial response to the accelerating torque in the rotor. the stabilizer has little effect on this first swing. On subsequent oscillations. however. the effect of the stabilizer is quite pronounced . To illustrate the effect of the PSS. some transient stability runs are made for a threephase fault near bus 7 applied at t = 0.0167 s (I cycle) and cleared by opening line 5-7

,

""_, 601 with \

3.0 ~

Q.

0 u..

w

/ /'-...., '

/

/

/

2.0

.....

..........

/

/

/

/

/

/

1.0

o

0. 25

\

\

/\ /

4.0

/

\

pss

/

/ \

/

.

\

\

\

/'",

> O. From (10.17) this means that d > Crxo, but we see from Figure 10.4 and (10.5) that this inequality always holds . Finally, note carefully that rs, acting through the spring constant K.. is in fact the speed reference. A simple manipulation of this position will cause a change in x and eventually, as the shaft responds, will cause w to seek a new steady-state value.

10.2 The Isochronous Governor The flyball governors similar to those shown in Figure 10.2 are capable of sensing changes in speed and responding by making a small change in a displacement or stroke (x) according to Equation (10.19). However, the force available to move a throttle mechanism in the x direction is small and the displacement is usually small as well. Therefore, what is needed is a force-stroke amplifier to magnify the stroke and exert a sufficient force to manipulate the valve . This is accomplished by means of a hydraulic amplifier or servomotor (see Appendix E). Consider the system shown in Figure 10.7, which consists ofa flyball governor, a spool (pilot) valve , and a piston that is capable of exerting a large linear force. * The flyball governor equation is the same as (10.19) except that a new force, the hydraulic reaction force due to the spool valve, must be added . This hydraulic reaction force, or Bernoulli force, has two components; a steady-state component that is always proportional to x and acts in a direction to close the valve orifice, and a transient component that is proportional to Xll and may be either a stabilizing (closing) or a destabilizing (opening) force [7]. Since the valve transient period is very short compared to the turbine response time, we need to represent only the steady-state hydraulic reaction force, which we write as simply (10.21) where the hydraulic reaction scale K h depends on the orifice area gradient and the pressure drop across the orifice. A detailed discussion of (10.21) is given in Appendix F, which is recommended for further reading. Adding these forces to (10.19), the governor-plus-spool valve equation can be written as

Pilot ~J!.b===::::::::::::::::==t1l Valve I

Fig.10.7

J!

The isochronous governor .

*Portions of the development here and in subsequent sections are similar to the treatment in Raven mended for further reading on the subject.

Pl. which is recom-

Speed Governing

409 (10.22)

where Kg = K, - K, + Kh • The new governor equation is basically the same as before except the X d coefficient is larger since the hydraulic reaction force is in opposition to the displacement [Fh is subtracted from the right-hand side of(10.7) since F h produces a reaction in the -Xd direction for an acceleration in the +xA direction]. The hydraulic piston moves in the +y direction as long as there is a positive x displacement of the spool valve. From Appendix J, Equation (1.53), we write

(10.23)

KqXd == alYd

where Kq is the spool valve volumetric flow per unit of valve displacement and at is the piston area. Note that the spool valve-piston combination is in fact an integrator since the output y continues to increase as long as a positive x displacement exists. Substituting (10.22) into (10.23) and solving for the piston displacement, we have . _ K~q - - - rd

Y~

Kga t

KwKq

(10.24)

- - - Wd

Kga}

and we see clearly the integrating effect of the hydraulic piston. It is convenient to normalize (10.24) on the basis of the full load rating of the generator. This is designated hereafter by a subscripted capital R. To do this, we define the per-unit (pu) quantities, with subscript u as follows.

Yd YAu== YR

pu

W Au

Wd

=-

WR

(10.25)

pu

Then (10.24) may be written in the Laplace domain as Ydu(S)==

K~qrR [

K

g!ltYRS

rd u -

KwWR

-

KsrR

W Au

]

(10.26)

The leading coefficient is interpreted as the inverse of a time constant 71 in seconds (the reader may wish to verify the dimensions). The coefficient of WAu may be simplified by performing the following conceptual test. Assume the system is initially in the steady state (j;A == 0) and at rated full load (reference) condition (r A == rR) when the load is suddenly dropped, causing a change in speed of (10.27) Substituting into (10.24) we compute 0= K,KqYR _ K."KqRWR Kgfll

(10.28)

Kgfll

Then the coefficient of WAtt in (10.26) can be determined from (10.28), with the result KwWR 1 - - == KsrR R

== C

g

(10.29)

This is the same result as that discussed in Section 2.3. Thus (10.26) can be simplified to the normalized form (10.30)

410

Chapter 10

Fig. 10.8

Block diagram of the isochronous governor.

where

Kf!1IYR T,=---

«s»,

The integrating governor system described by (10.30) is called an isochronous governor since it attempts to integrate the speed error until the error vanishes . A block diagram of the isochronous governor is shown in Figure 10.8. Note that the comparator is due to the flyball governor and the integration is due to the hydraulic servomotor.

10.3 Incremental Equations of the Turbine In order to study the performance of the governor, it is desirable to develop the incremental (linear) equations of the controlled plant, in this case, the turbine-generator system. It is not necessary here to provide a detailed analysis for large excursions since we are interested in the system behavior only in the neighborhood of the steady-state operating point. Therefore, we can estimate the behavior by taking partial derivatives in this neighborhood. As in many control system problems , it will be useful to develop the system and control equations such that a block diagram similar to Figure 10.9 can be constructed [7, 9, 10). In the preceding section we developed the equations and the block diagram for the control section corresponding to an isochronous governor . The output of this control is the "manipulated variable" M(s) = Ya(s), which corresponds to the valve position. This variable would correspond to the steam valve stroke (or valve area) for a steam turbine or the wicket gate position (or gate area) for a hydro turbine. The control transfer function and feedback function are, respectively,

Gc(s) =

-

I

TIS

and

H(s) = Cg

(10.31)

as noted in Figure 10.8. The input transfer function A(s) = 1.0 in this problem, so the command signal U(s) and the reference R(s) are identical.

Fig.IO.9 General Block Diagram ofa Control System [7,9].

Speed Governing

411

We now seek a general relationship for the plant transferfunction Gp(s) and the disturbance function N(s) for a turbine, where the output speed C(s) = w is controlledby the governor. The flow control valve in Figure 10.7 admits steam (water for a hydro turbine or fuel mixture for a combustion turbine) as a function of valve area, which in tum is a function of the valve stroke y. Usually, the valve is designed such that valve area is linearly related to stroke (see Appendix F.7, function generators). The fluid flow rate Wthroughthe valve is proportional to the product of valve area A and fluid pressure P. W= k'AP = kyP Then the incremental flow can be written as

aw

(10.32)

aw ar

(10.33)

W~=k-Yd+k~P~

ay

For the analysis in this chapter we will consider the pressure to be constant such that we may write (10.34) where ky is a positive constant. The relationship between Wand the developed mechanical torque ky is a direct one since all working fluid entering the turbine produces torque with no appreciable delay [10-12]. In a steam turbine, there is a lag associated with the control valve steam chest storage and another greater lag associated with the reheater (see Chapter 11). There are also lags in hydro turbine systems (see Chapter 12). For the purpose of this elementary model, we include a simple firstorder lag Ts for the turbinecontrol servomotor system to write T

KtWd

KtKYYd

K1Yd

:=--:=--:=--

1 + TsS

md

1 + "'sS

1 + "'sS

(10.35)

wherewe combine the two constants K, and K; into the singlepositiveconstantK I • K I wouldbe expected to have a normalized value of unity, but is approximately 0.6 in steam turbines due to valve nonlinearities [11]. Finally, we write the swing equation, from (5.78):

(10.36)

which describes the inertial behavior due to any upset in torque. The term DWd is added to account for electrical load frequency damping and turbine mechanical damping. Combining the plant equations (10.35) and (10.36) with the control equations of Figure 10.8, we can construct the systemblock diagramshown in Figure 10.10. The steady-state operationof the general control system block diagram of Figure 10.9 can be evaluated in terms of the steady-state gain of each block [7]. Suppose we define for this purpose the steady-state gain functions K;> Gc(O)

K N = N(O)

Kp := Gp(O)

K A := A(O)

KH=H(O)

(10.37)

that is, we determine the gain of each block with S replacedby zero. Then, from Figure 10.9we can write, in the steady state, Css

:=

KAKcKpuss

1 +K; fj

Tel > 1;2 > 1;1

Fig. 10.11 Steady-state operating characteristic of the isochronous governor.

Speed Governing

413 OJ

Fig. J0.12

Root locus plot for the isochronous governor .

stable for small values of the gain K but will have a sluggish response since two roots are very near the origin. We conclude that the isochronous governor has a desirable steady-state operating characteristic, is sluggish in its transient response, and becomes unstable for low values of gain . Furthermore, it the damping D is zero, the system is unstable.

10.4 The Speed Droop Governor The isochronous governor, although having good steady-state characteristics, is very nearly unstable and with sluggish response for reasonable values of gain. A better control scheme for this application is to use proportional, rather than integral, control. This can be accomplished by using mechanical feedback in the form ofa "summ ing beam," as shown in Figure 10.13. This governor is called a "speed droop governor" or a regulated governor. The mechanical feedback transforms the hydraulic integrator into an amplifier, which is used to increase the force and stroke of the governor throttle rod position. Using the notation of Section 10.3 and (10.22), we sum forces in the x direction to write

K.(xe,. + XtJ - Kxxe,. + Khxe,. + Kwwe,. = 0 or

(Ks - K, + Kh)xe,. + Ksxt" = - Kwwe,.

(10.42)

Using Kg = K, - K, + Ki, this equation can be written as

Kgxe,. + Ksxt" = -Kwwe,.

(10.43)

For the summing beam, we can write the displacement equation, for small displacements, as

a b xt" = - Ye,. - -re,.

L

L

(10.44)

where L = a + b. Substituting into (10.43) we get

Ksa Ksb Kgxe,. + LYe,. = Lre,. -K",we,.

(10.45)

For the hydraulic piston, we can again write, from (10.23),

Kqxe,. =alYa

(10.46)

Chapter 10

414 X' ~---

>1

a -----:;!~~---- b ----~

T

In;:.---I..r.....u-

lotm;::::::::::::::::::::GEJ E Flow Control Va lve

Fig. 10.13 The speed droop governor.

Combining (10.45) and (10.46) we have

Kqo l

Kq

•

Kso

Ksb

Y~ + LY~ = Lr~ -Kww~

(10.47)

This equation is normalized and rearranged to write, in the s domain

) brR K.,LwR "v s + I Y~u(s) = -r~u - -K--w~u (~ aK q 0YR 0 sYR s' ....

(10.48)

To determine the normalized coefficients in Equation (10.48) we perform two conceptual tests. The first test is conducted at full (rated) load with the system operating at steady-state rated speed . i.e., w~=o

Y~=YR

rsr r«

y~ =O

(10.49)

Substituting (10.49) into (10.47) we compute o

YR

b

(10.50)

which means that the coefficient ofr6u in (10.48) is unity. For the second test, we remove the load, allowing the speed to increase, but with the reference r~ held at the same position. The conditions for this test are, in the steady state

Speed Governing YA == Yd == 0

rs == rn

(10.51)

== RWR

W/i

415

where we recognize that the speed change in going from full load to no load is, by definition, Substituting (10.51) into (10.47) and using (10.50), we compute

RWR'

WR

==

YR

aKs KwLR

(10.52)

Thus, the coefficient of W/1u in (10.48) is Cg == llR as in the isochronous case. Dropping the

u subscript, we write the per-unit speed droop governor equation as

(10.53) where T} == atKgLlaKj(q. The governor block diagram is shown in Figure 10.14. Comparing this diagram with Figure 10.8 for the isochronous case, we see that the isochronous integrator I/T}s has been transformed into the amplifier 1/(1 + T}S) by means of mechanical feedback through the summing beam. Note that T] can be adjusted by changing the ratio aiL. In order to analyze the performance of the speed droop governor, we interface the system of Figure 10.14 with a single turbine representation using one-time lag, together with the inertial torque equations derived in the previous section. The result is the system of Figure 10.15. Note that the integral control of the isochronous case has been replaced by an additional lag in the control system. We will now examine the steady-state and transient performance of this system. The steady-state performance of the speed droop governor is analyzed from (10.37) using the factors

K A== 1 K N == -1 Kc==K} K; == lID K[-{==

Cg

(10.54)

Then W

ss

==

K}rss ----D + K}c, D + K}Cg

(10.55)

for the speed droop governor. Clearly, the steady-state speed is now a function of both the reference setting r ss and the generator load Tess. In particular, as T; is increased, the steady-state speed is reduced. The manipulated variable for this system is Tm , the mechanical torque applied to the shaft. In the steady state, we can compute Tmss to be

Fig. 10.14

Block diagram of the speed droop governor.

Chapter 10

416

I-------------~------Control Turbine I Te/1 Plant I Ir

I I

/1

1

+

Y/1

l+'t'l s

K

I

Tm /1 I + T

l+'t'ss

+ L

I

1

D+2Hs

OJ

/1

I

I I

I I I_________ Cg ----L-------I Fig. 10.15

Typical system application block diagram.

where ess is the steady-state error. This equation describes a family of parallel straight lines in the Tm-w plane, each with Tmintercept K) and with slope -KtCg . Thus, the steady-state operating characteristic may be visualized as the family of curves shown in Figure 10.16. Note that, for each setting of the reference, the steady-state speed is dependent on the shaft load T, and that the higher loads cause a greater reduction in speed. Also note, from (10.56), that the error ess is always greater than zero, whereas it was always integrated or reset to zero for the isochronous governor. A positive steady-state error signal is characteristic of a proportional control system. The characteristic of Figure 10.16 should be carefully compared with the operating characteristic shown in Figure 10.11 for the isochronous governor. The transient response of the speed droop governor may be analyzed by plotting the root locus of the open-loop transfer function (OLTF) : OLTF=

K)C g (I + TtS)(I + Tss)(D + 2Hs)

=

K (s + a)(s + b)(s + c)

(10.57)

where a = 11Th b = liT., C = D/2H, and K = K)C/2HTITs ' Note that K, b, and c are exactly the same as for the isochronous case. In most physical systems, we would expect to find T) < 'Ts , with Ts = 2T] being about typical [II]. Thus, the root locus takes the form of Figure 10.17. Compare this,plot with that of Figure 10.11 for the isochronous governor. Note that the eigenvalues of the speed droop governor have much larger negative real parts than can be achieved for the isochronous governor. This means that the system can be satisfac-

1j>r2>1j

~3 > 1;2> 1;\

Fig. 10.I6

Steady-state operating characteristic of the speed droop governor .

Speed Governing

417

m

/

-a

/

/

/

\ - b -c \

Fig, 10,17

/

\

\

o

(J

\

Root locus for the speed droop governor.

torily operated at much higher values of 'gain and with improved damping and smaller settling time. Overall, the performance of the speed droop governor is preferred because of its better transient response. The improvement in transient response is accomplished by moving the pole at the origin, for the isochronous governor case, to s = -a = -liT., which is well to the left in Figure 10.17. We can analyze the closed-loop governor behavior by writing the closed-loop transfer function for a given electromagnetic torque, T, as

K(/2HT( Ts (s + a)(s + b)(s + e) + K

WA

rs

KR

S3

+ (a + b + e)s2 + (ab + be + ea)s + (abe + K)

(10.58)

We now apply Routh's criterion to the denominator by forming the array:

(ab + be + ca)

S3 S2

(a + b + c)

(abc + K)

Sl

m

0

SO

(abc + K)

0

Then the necessary conditions for stability are found to be

a, b, c> 0 K>O m=

(a + b + c)(ab + be + ca) - (abc + K) >0 a+b+c

(10.59)

The latter of these constraints may be simplified by converting into the form

K < (a + b)[c2 + (a + b)c + ab]

(10.60)

Chapter 10

418

or, substituting gains and time constants and simplifying, we get

D2 D(Tl + Ts ) 2H ) + + -K 1Cg < (TI + Ts ) ( 2H TITs TITs

(10.61)

Since the damping D is always a stabilizing force, we examine (10.61) for the case where D = 0 to compute K1 ~ -1+ -1 ) -J

Boiler

Fig. 11.37

Block diagram of a lumped parameter drum-type boiler.

pressure and flow changes. The configuration of Figure 11.38 is recognized to be a "boilerfollowing" control arrangement. Multivariable controllers have an additional problem not usually present in single variable controllers-the consistency of results [19]. Thus, in a boiler, an increase in firing rate will always produce an increase in pressure; an increase in air flow will always decrease boiler pressure; an increase in desuperheat spray will always decrease throttle temperature, and so on. These are primary or dominant effects and their sign is always the same. Some effects, on the other hand, are opposing. Thus, an increase in fuel increases steam pressure and this tends to increase steam flow. Increased steam flow tends to decrease temperature, whereas the increase in fuel input would ordinarily increase temperature. Thus, the exact operating point plus conditions of soot, slag, etc. will effect the response and its direction.

Desired Unit Generation

Actual Unit eneration

Combustion Control

Desired Steam Pressure

Boiler Fig. 11.38

Load Anticipation Index Generator

Typical control system configuration for a drum-type boiler.

Output

467

Steam Turbine Prime Movers

One of the problems in designing an appropriate controller is that of starting with a good mathematical model of the system. This is especially difficult in boiler systems because of the difficulty in modeling a distributed parameter system and also because of the nonlinear character of steam properties. The equations of the system are those of mass flow and heat transfer in superheater and reheater tubes, and these equations'are nonlinear partial differential equations in space and time. The usual approach to the solution of these equations is to break the space continuum into a series of discrete elements and convert the partial differential equations into ordinary differential equations in the time domain [18,19]. These equations may be solved by digital computer. Models of this kind have been studied but are beyond the scope of this book. The references cited will be helpful to one who wishes to pursue the subject further. Finally, before leaving the subject of drum-type boiler control we note one type of multivariable control that has been used on both drum-type and once-through boilers. This system, shown in Figure 11.39, is called a "Direct Energy Balance Control System" [21] by its manufacturer. This kind of control is designed to perform the following operations: 1. Adjust both boiler and turbine-generator together, as required by automatic or manual controls. 2. Observe load limit capabilities of boiler, turbine, and generator. 3. Reduce operating level (runback) to safe operating level upon loss of auxiliaries. Figure 11.39 displays the major components of this type of system. Referring to the figure, the desired unit demand signal (from the automatic load control device), actual unit generation, main steam pressure, and desired steam pressure are all input quantities to the controller. Computer outputs are generated to the combustion and governor controllers. Thus, the system does not simultaneously adjust all possible variables, but it does deal with the primary variables. Compare Figure 11.39 with Figure 11.38 to see the difference between the two types of controls. The controller of Figure 11.39 is shown in block diagram form in Figure 11.40. It consists of two components: the "boiler-turbine governor" and the "unit coordinating assembly." The boiler-turbine governor produces a "required output" set point that takes into account the capa-

Desired Unit Generation--.....

Actual Unit Generation

DirectEnergy

.....--------41 Balance Control 1---.., System

Combustion Control

Desired Steam Pressure

Boiler

Main Steam Pressure

Turbine

Fig. 11.39 A multivariable control system [21].

Generator

Output

Chapter 11

468 Desired Unit Generation - - - - - -......

.-

Actual Unit Generation

;-------- -------,,, BoilerTurbine Governor

Frequency Bias (Rates of Change) (Limits) (Runbacks)

Unit Coordinating Assembly

To Combustion Control

- - - -- --- -'

Desired Steam Pressure Fig. 11.40

,, To Governor Control

Block diagram of a controller [21].

bilities of all components-boiler, turbine, and auxiliaries. It also fixes the rates of change according to a preselected setting and provides for emergency runbacks and limits. The unit coordinating assembly coordinates the combustion control with the turbine-governor control. Both of these blocks are described in greater detail below. The "boiler-turbine governor" is shown in greater detail in Figure 11.41. When operating under automatic load control, a signal is received from the load control unit. This fixes the desired generation for this unit. When not on automatic control, a selector switch provides an input signal from a manual setting, properly biased when system frequency is other than normal, For any size step change in the manual output setter, the unit automatically achieves the new setting at a preset maximum rate of change, taking limits into account as noted. The "unit coordinating assembly" is shown in greater detail in Figure 11.42. This unit compares the required output for the unit against the actual unit generation and computes an error signal from which the governor and fuel-air systems are controlled. At the same time, the measured pressure is compared against a desired pressure set point and this produces a pressure error that is used to bias both the governor and fuel-air action, but in opposite directions. This is because the governor (control) valves and fuel-air systems have opposite effects on pressure; an increase in governor setting tends to reduce the pressure but an increase in fuel-air setting tends to increase it. The overall effect of the control is to take appropriate action for changes in both load and pressure as noted in Table 11.7. In practice, the control just described may be operated in anyone of the following four modes. The operator selects the operating mode he wishes to use.

1. Base input control. In this mode, the operator adjusts the boiler inputs and the turbine governor manually. 2. Base input-turbine follow. In this mode, the governor adjusts the pressure automatically, as shown in Figure 11.3, and the turbine follows the boiler. The operator runs only the

469

Steam Turbine Prime Movers

Maximum Generation Setter Minimum Generation Setter

, Runback

Min. Fuel

Min. Air Low Deviation

Max.Fuel Max. Air Max. Feedwater Governor OpenLimit High Deviation

Required Output

To Unit Coordinating Assembly Fig. 11.41

Boiler-turbine governor control unit [19].

boiler inputs, either automatically or manually. This mode is often used during startup and certain unusual operating conditions. It frees the operator from having to watch both the boiler and the turbine. 3. Direct energy balance automatic control. This mode is the normal operating mode for this type of control and is the mode for which the system was designed. 4. Automatic control-boilerfollow. This mode is like the "conventional" mode as illustrated in Figure 11.4, except that use is made of the "required output" signal, which provides several advantages over conventional boiler-follow control, such as providing frequency bias, limiting and runback actions, and fixed rates of change. It also couples the governor and the fuel-air controls to provide an anticipatory boiler signal to accompany governor changes due to a load change. This "automatic boiler-follow mode" is shown in Figure 11.43.

11.9.2 Once-through boilers Since the late 1950s, an increasing number of large boilers installed have been of the "once-through" design. The striking difference between this type of boiler and the conventional drum-type boiler of Figure 11.35 is the absence of the drum, down comers, and waterwall risers. Instead of these features, water from the boiler feed pump passes through the economizer, furnace walls, and superheater to reach the turbine, passing from liquid to vapor along the way. See Figure 11.34 for a simple description of the two types of boilers. In the once-through boiler,

470

Chapter 11 Required Output (FromBTG)

Unit Generation

Generation Error

Desired Steam Pressure

Main Steam Pressure Pressure Error

To Turbine Governor Fig. I t.42

The unit coordinating assembly [2 t].

the pumping rate has a direct bearing on steam output as well as the firing rate and turbine governing. A simplified flow diagram of a typical once-through boiler is shown in Figure 11.44 [22]. The once-through boiler has a significantly smaller heat storage capacity than a drum-type boiler of similar rating, since it contains much less fluid. It also costs less, because of the absence of the drum, and has lower operating costs. It does, however, require a more intelligent control system. In operation, the once-through boiler is much like a single long tube with feedwater flowing in one end and superheated steam leaving at the outlet end. A valve at the discharge end can be used to control the pressure. If the pressure is constant, heat is absorbed by the fluid at a constant rate and the steam temperature is a function of the boiler throughput (pumping rate). The heat absorbed (Btu/hr) divided by throughput (lbmlhr) gives the enthalpy (Btu/Ibm). Thus, for steady-state operation, the control must equate flow into and out of the tube, holding steam tem-

Table 11.7

Steam Pressure High Low Low High

Net Control Action by the Unit Coordinating Assembly [19]

Generator Output

Action Applied To Governor

High High Low Low

Difference> Zero Difference> Decrease Difference> Zero Difference> Increase

Action Applied To Fuel and Air Inputs Sum Sum Sum Sum

Decrease Zero =: Increase =: Zero =:

=:

Steam Turbine PrimeMovers

471

Actual Unit enerauon

,......

----lPressure Error

Combustion Control

Generation Error

Desired Steam Pressure Main Steam Pressure

Boiler

Turbine Generator

Output

Fig. 11.43 Automatic boiler-follow control [21].

perature at the desired value by maintaining the correct ratio of heat input (fuel and air) to throughput (flow rate). Transient conditions are difficult to control because of the limited heat storage in the fluid. Thus, when load is increased, the pumping rate must be increased to satisfy the increased load and provide greater energy storage, and heat input must simultaneously be increased to match load and the increased storage level [23].

Heat Recovery Area

Primary Super Heater

-¥*I

L..--+----J~el ...---'----, A ir I

8--Fan

Superheater and Reheater Dampers

Fluid Path : Boiler L _ Feedpump Fig. 11.44 Fluid path for a once-through boiler [22].

JI

Chapter 11

472

Partly because of the lower storage of the once-through design, the response to sudden load changes is much faster than that of the drum-type boiler. The time required for water to pass through the boiler and be converted to superheated steam is only two or three minutes compared to six to 10 minutes for the drum-type designs [24]. Also, since the pumping rate is directly coupled to the steam produced, there is little of the "cushioning effect" that exists in drum-type boiler designs. Rigorous analysis of the once-through boiler, like the drum-type boiler, is a difficult problem, but such analysis is necessary if a control system is to be designed accurately. A common approach is to lump the spatial variation and waste heat transfer equations for each lump. This method has been used on a supercritical unit for a 191 MW unit in which the analysts divided the boiler into 14 sections or lumps [25]. Another report describes the use of 36 lumps to describe a large boiler used to supply a 900 MW generating unit [26]. Having eliminated the spatial parameter by lumping, the resulting ordinary differential equations are nonlinear. Assuming operation in the neighborhood of a quiescent point results in a linearized system of equations that may be numerically integrated by known digital techniques. Comparison of such results with field tests have generally been quite good [25, 26]. Another approach to this problem has been pursued [22] in which the boiler is lumped into 30 or so sections and the nonlinear equations for each lump are solved iteratively by digital computer. This method is more time consuming than the linearized model, but it is also more accurate for larger excursions from the quiescent point. A flow diagram of the iterative process is shown in Figure 11.45. The solutions obtained by this process, give the boiler open-loop re-

Iterated Press.

Presssure, flow rate, and density profile p Density from iterative solution of pressure drop, p continuity, pressure-temperature-density Sp !-eat steam table relations, turbine pressure, temperature and flow relations as well as p Flow ....._Rate _.... pump characteristics

Pump Speed Turbine Valve Position Spray Valve Position

.

Fluid

m

Pressure Density

Temperature Profile

Inner Fluid Energy Balance 1--------' and Transport Delays

Specific Heat

Metal Heat Storage

Flow Rate Profile

Gas to Metal Heat Flux

Profile

Gas Path Energy Balance

Radiation, Convection, Heat Transfer Fig. t t.45

Metal

Heat Transfer

Temperature Profile Firing Rate AirFlow By-Pass Damper Position

Iterative solution flow diagram [22].

Steam Turbine Prime Movers

473

sponses to step changes in turbine valve position , pump speed, spray flow, and heat flux. These results have been used in the synthesis of a control philosophy and control hardware, a portion of which is described below . The control system of Figure 11.46 is basically the direct energy balance system of Figure 11.39, but shown in block diagram form. This scheme has been used for many once-through boiler installations. Considering this control scheme , we investigate various innovations that may improve response. Referring to Figure 11.46, we examine the significance of combining MW error into the control scheme . If we let Po be the pressure set point , P.1 the pressure error, MW the megawatt level, and K v a constant proportional to the valve opening, then, from [11] MW = KvP = KrJ...Po + P.1)

or MW - K vP.1 = KvPo

(11.48)

This difference is proportional to the load level and is interpreted as the turbine valve opening. The authors of [22] present variations to the basic control scheme of Figure 11.46. Basically, the problem is to design an adaptive control system that has the ability to alter its control parameters to satisfy the changing, nonlinear needs of the system at various load levels and to do this in the shortest possible time .

11.9.3 Computer models offossil-fueled boilers From the foregoing discuss ion, it is clear that large fossil-fuel boilers are large complex systems . Detailed mathematical models of these systems have been constructed and are used by system designers and control experts . However, these large detailed models are not appropriate for use in power system stability analysis. Our interest is simply in the ability of the boiler to maintain steam pressure and flow for a few seconds or, at most, a few minutes .

Speed

Turbine Valve

Pressure Anticipatory Feed Forward Action From Desired MW Fig. 11.46

Demand For: Feedwater Firing Rate Air Flow Etc Boiler Measured an a es

Coupling of turbine load controls with boiler controls [22].

474

Chapter 11

Boiler control, on the other hand, involves the analysis of system performance over many minutes and analysis of various subsystems within the control hierarchy. These large detailed models are too detailed and too cumbersome for power system stability analysis; not that they are incorrect, but they simply are far too detailed. Their inclusion would greatly retard the solution time and the added complexity is unwarranted. However, it is also not correct to assume that the boiler is an "infinite bus" of steam supply under all conditions. Clearly, what is needed for stability analysis is a low-order model that will correctly represent the steam-supply system for up to 10 to 20 seconds. The stability analyst is not concerned with the many control loops within the boiler, but only the essential steam supply and pressure at the throttle valve. This problem has been investigated for many years and is well documented in the literature [26-37]. The IEEE Power Engineering Society has been particularly active in documenting appropriate model structures and data for proper representation and two excellent reports have been issued as a result of these efforts [29, 37]. These reports focus especially on the dynamics of prime movers and energy supply systems in response to power system disturbances such as faults, loss of generation or loads, and system separations. Figure 11.47 shows the elements of the prime mover control model that was developed by the IEEE working group. The mechanical shaft power is the primary variable of interest as it drives the generator. This variable is directly affected by the turbine control valve (CV) and intercept valve (IV), both of which admit steam to the turbine sections. Steam flow through these valves is, in tum, affected by throttle pressure, labeled PT in the figure. This pressure is directly affected by the boiler performance. Models of these system components are needed in order to provide an adequate dynamic model of the mechanical system. The relationship between the prime mover system and the complete power system are shown in Figure 11.48, where the boiler-turbine system is shown within the dashed lines. This diagram is instructive as it links the boiler-turbine systems to the controlled turbine-generator system and the external power system. It is a complex nonlinear system. There are several types of turbine systems of interest in a power system study. These generic models are described in [37]. Later, improved models of a steam turbine system, including the effects of the intercept valve, have been developed and are shown in a general way in Figure 11.48 [38], which shows how the boiler and turbine models are linked to other power system variables and controllers. The prime mover energy supply system is shown inside the dashed box in Figure 11.48. We can see that the prime mover responds to commands

p~

f

Load Demand

LD

Load Reference Boiler Turbine Controls

LR

OJ

I

tt t Speed Load Control

IV CV

Steam Mass Flow Rate, ms Fue /Air.Feedwater Fig. 11.47

Elements ofa prime mover system [37).

Turbi ne Including Reheater

T

Pmech

475

Steam Turbine Prime Movers

~

Automatic Generation Control

Jnterchange Power

-

Frequency

Electric System Generators Network Loads I

Desired Unit Generation Speed

,

Unit Electric Power

Angle

--

....

Turbine/ Generator Inertia ~

,. -- -- ---------------1 --------------------- ------ , , 'f ,, II ~ Turbine/ Turbine Generator , , Turbine/ Unit Reheater I . Valve

I

,

I

Boiler Controls I I

.."..

Speed Changer

~

1

1/

Boiler Controls

Boiler, Inputs

Dynamics

Controls I'

~

Mechanical Power

~

Boiler Pressure Dynamics

)steam Flow " Rate

[\

,, f

f f

, ,,, f

f

Main Steam Pressure

f f

f

,

f

____________________I!~il~~ !\!rp~n~ §y~t~f!l

:

Fig. 11.48 Functional block diagramof prime mover controls [38].

for generation changes from the automatic generation control system, or from manual commands issued by the control center. The turbine-boiler control also responds to changes in speed . The resulting mechanical power responds to changes in main steam pressure and turbine valve positions . The output variable of primary interest is the unit mechanical power that acts on the turbine inertia to accelerate or decelerate the inertia in accordance with Newton's law. A more detailed model of a generic turbine model is shown in Figure 11.49. The effect of intercept valve operation is that portion of the figure within the dashed box , where the intercept valve opening or area is represented by the "IV" notation. The control valve position is shown as "CV" in this figure . In many cases, these effects are modeled linearly as a first-order lag. This model is believed to be more accurate as it accounts for the valve limits . The steam turbine speed and load controls are of two types. The older units operated under a mechanical-hydraulic control system . A generic model of this type of control system is shown in Figure 11.50. The manufacturers of speed-governing equipment have their own special models for speed governors of their design , and these manufacturers should be consulted to determine the best way to model their equipment. These experts can also provide appropriate numerical data for the model parameters. In some studies it is also desirable to provide a model of the boiler. This is true of studies that extend the simulation time for long periods where boiler pressure may not be considered constant. An appropriate low-order boiler model has also been recommended by the IEEE committee responsible for the above speed -governing system model. This boiler model is shown in

Chapter 11

476

CVPr

Fig. 11.49 Generic turbine model including intercept valve effects [38].

Figure 11.51 and features a lumped volume storage of steam at an internal pressure labeled here as drum pressure, in series with a superheater, and with steam leads and their associated friction pressure drops. The energy input to the boiler represents heat released by the furnace. This heat generates steam in the boiler waterwalls at a mass flow rate of »: (note carefully the dot over the m, representing a derivative with respect to time, or a rate of mass flow). The steam generation process is a distributed one and this is approximated in the model by two lumped storage volumes for the drum, CD and the superheater, CSlJ connected through an orifice representing the friction pressure drop through the superheater and piping . The major reservoir for energy storage is in the waterwalls and the drum, both of which contain saturated steam and water . In once-thru boilers, the major storage is in the transition region. The output of the model is the steam flow rate to the high pressure turbine.

11.10 Nuclear Steam Supply Systems Nuclear power plants generate steam by utilizing the heat released in the process of nuclear fission, rather than by a chemical reaction as in a fossil-fuel boiler. The nuclear reactor controls the initiation and maintenance of a controlled rate of fission, or the splitting of the heavy uranium atom by the absorption of a neutron, in a chain reaction . In the so-called "thermal" reactors a moderator, principally water, heavy water, or graphite, is required to slow down the neutrons and thereby enhance the probability of fission .

REF

+

~

OJ

Servo Motor

Fig. 11.50 Approximate representation of control valve position control in a mechanical -hydraulic speed goveming system [38J.

477

Steam Turbine Prime Movers

HP Turbine

Drum and Water Walls

Superheater and Steam Leads (a) The Physical System

cv

Turbine Valve

+

Steam Flow Rate

Signal to Fuel and Air

(b) The System Model Fig. 11.5\

A computer model of boiler pressure effects [38].

There are several distinct types of nuclear steam supply systems that have been designed and put into service in power systems. The major systems in use are the following : I. 2. 3. 4.

Boiling water reactor (BWR) Pressurized water reactor (PWR) CANDU reactor Gas-cooled, graphite-moderated reactors

In the PWR, the reactor is cooled by water under high pressure . The high-pressure water is piped to heat exchangers where steam is produced . In the BWR, the water coolant is permitted to boil and the resulting steam is sent directly to the turbine. In Europe, gas-cooled, graphite-moderated reactors have been developed . In these reactors, the heat generated in fuel assemblies is removed by carbon dioxide, which is used to produce steam that is carried to steam generators. The CANDU reactors have been developed in Canada. These reactors use heavy water under pressure and utilize natural uranium as a fuel. Our treatment will focus on the BWR and PWR types, since they are so common in the United States .

Chapter 11

478 Pressure .1. Set Point

V

...

8 t: o

0..

o

Fig. 11.52

11.10.1

Major components of a BWR nuclear plant [39].

Boiling waterreactors

The major components in a BWR nuclear reactor are shown in Figure 11.52 [39] and these components should be included in a dynamic model. Note that the steam produced by the reactor is boiled off the water surface and fed directly to the turbines. A block diagram for the boiling water reactor is shown in Figure 11.53 [40). The variables noted in the figure are defined in Table 11.8. This is a low-order model for such a complex sys-

Fig. 11.53

Block diagram of a reduced-order BWR reactor model.

Steam Turbine Prime Movers

479

Input Signal Control Rod

~

Steam

Generator

® Reactor Fig. 11.54

Major components of a PWR nuclear reactor model [39].

tern, and was constructed for use in power system stability analysis, where it is important to keep models reasonably simple .

11.10.2 Pressurized water reactors The major components in the pressurized water reactor are identified in Figure 11.54 and the major subsystem interactions are shown in Figure 11.55. The model of the PWR nuclear reactor and turbine are rather complex . One model for the PWR is that shown in Figures 11.55 and 11.56, where the high- and low-pressure valve positions are unspecified or are unchanging. These positions are functions of the speed governor model, which is not specified here, but is similar to other speed governor models. One can also

Rod Position Rod Position Regul ator Model

a.

Fig. 11.55

Interaction of PWR subsystem models [4)].

480

Chapter 11 QB)'p.Qr

-

-'

+

Total +S team Flow )-_ _-;;~ 1+ s7; I-P"';J~=====111 I

II

H

Turbine Tailrace Fig. )2.15

Notation for changes in flow and head (all values are considered deviations from the quiescent values).

Hydraulic Turbine Prime Movers

505

For the penstock, we apply equations (12.51) and (12.57) to write h == (sech TeP)h t - Zp(tanh TeP)q - 4Jp q 1

qp == (cosh TeP)q + -(sinh TeP)h Zp

(12.68)

where qp == friction coefficient of penstock T, == elastic time of penstock Zp =

Z~o

=

normalized impedance of penstock

and all h's and q's are defined in Figure 12.15. For the turbine, we may write the following equation [8]: q

aq aq aq = - h + - n + - z = allh + aJ2n + aJ3z ah an az

(12.69)

where n = per-unit turbine speed z == per-unit gate position Also, we can write

ot;

sr;

st;

Tm == --h + --n + - - z = a21 h + a22n + a23z

ah

an

az

(12.70)

where Tm is the per unit turbine mechanical driving torque. All values defined as a's in (12.69) and (12.70) are not constants but are nearly constant for any operating quiescent point. These values will be read from curves of turbine characteristics. Also from Newton's Law, we have (12.71) where J m = per-unit mechanical inertia Tm = turbine starting time Here we assume no electrical torque as we are interested only in the relationship between the variables, not in the way the turbine accelerationis restrained by shaft load. Combining equations (12.63) and (12.65) we can write Q(s) ==-F3(s) H(s)

(12.72)

where

F1

1 +ztanh TeS F 3(s) ==

P

cPp + F) + Zp tanh TeS

(12.73)

which gives a relation between the per-unit turbine flow rate and the turbine head. We note that it depends only on the characteristicsof the penstock, surge-riser tanks, and conduit, and not on the turbine characteristics as determined by partial derivatives in (12.63) and (12.64), nor on the turbine inertia as given by (12.71).

Chapter 12

506

z

z~~ Hydraulic Supply (a)

Water Turbine

Hydraulic System (b) Hydraulic System

Hydraulic Components

Fig. 12.16 Block diagrams ofa hydraulic system.

Now, combining (12.69), (12.70), and (12.72) we get

N(s) a23(all + F3) - a13 a21 F4 (s) = = ----------Z(s) Jm(all + F3)s - a22(all + F3 ) + a12a21

(12.74)

Equation (12.74) is not yet in the desired form. Combining (12.69), (12.70), (12.72), and (12.74), we can write

H(s) Z(s) =-Fs

(12.75)

where al3

Fs = -

+ a l 2F4

-

all

- + F3

(12.76)

and (12.77) where (12.78) Finally, between (12.76) and (12.78) we deduce that

Tm(s)

F6

- - = - = F7 H(s) s,

(12.79)

In block diagram notation, we can express the hydraulic system as shown in Figure 12.16. Using equations (12.75) and (12.79), we have the representation of Figure 12.16 (a). We may, however, lump these characteristics and use only (12.78) and Figure 12.16(b).

12.9 Simplifying Assumptions It is quite apparent that the transfer functions (12.76), (12.77), and (12.78) are very difficult to work with and that some simplification would be helpful. One approach is suggested at the end of Section 12.8. In this approach, a complex hyperbolic function is represented by an infinite series and then higher-order terms can be deleted as an approximation. This is a purely mathematical approach and is quite acceptable as long as the deleted terms are small. Another approach to simplification is through a combination of mathematical manipulation and physical reasoning. This requires a certain amount of experience and intuition, and should be verified by staged tests on a physical system. Our approach is this latter method, drawing generously from the recorded thoughts of Old-

Hydraulic Turbine Prime Movers

507

enburger and Donelson, as presented in [8]. These approximations are not only those devised by experienced engineers, but tested extensively to prove their validity. The first approximation noted is that concerning the hydraulic resistance. It is noted that, although present in F., F 3, and all other factors (note cPc and cPp ) , the error in neglecting the hydraulic resistance term is negligible. Thus, the resistance head-loss term we so carefully added in equation (12.56) is not needed in the small-disturbance case. We will not bother to remove the cP term in all expressions, but note that little error would result from doing so. One possible simplification is that of neglecting the conduit portion of the hydraulic system and assume that the surge tank isolates the conduit from the penstock. Thus, in equation (12.62) we set the conduit flow to zero, i.e., cPc == O. This says that the water flow in the conduit does not change and the conduit is essentially closed. Under this condition, from (12.64) and (12.65) we have

qc == 0 == (qt + qr) + qp

Ttht = qt + qr = -qp

u,

o,

-

or

1

(12.80)

== -F)(s) = - -

TtS

(12.81) and the surge tank acts as an integrator. A second simplification involving F 3 is possible from experience with physical systems. We write

F1

l+ztanhTeS

F3(s) ==

P

cPp + F 1 + Zp tanh TeS

==

1

_

Zp tanh TeS

(12.82)

Both this assumption and the assumption on the isolation of the conduit (12.79) have been validated by experiment. We now examine certain approximations suggested by Oldenburger and Donelson [8], which provide several degrees of simplification. 1. In the simplified expression for F 3(s) from (12.82) we can set, as an approximation, (12.83) with the result (12.84) Using this approximation, we compute

F4 ==

(att a23 - a13 a2))Zp TeS + a23 J ZT 2 + (Jm + (a12 a21 - all a22)Zp Te)s - a22 mall

bls C2 S2

+ bo

p

+ CIS + Co

eS

(12.85)

Chapter 12

508

Similarlywe find that _ Fs -

(a13 C2 + aI2b IZpTe)S2 + (aUCI + aI2bOZpTe)s + aI3 cO 3 2 a}}c2ZpTeS + (C2 + at ICIZpTe)s + (CI + a. )cOZpTe)s + Co d zs2 + d.s + do

(12.86)

a2)(dzsZ + d.s + dO)(C2SZ + CIS + co)- a22(b ls + bO)(e3s3 + e2s2 + els + eo) +a23(e3s3 +e2s2+e.s + eo)(czs2 + Ct s + co) F6 = - - - - - - - - - - - - - - - - - - - - - - - - (e3s3 +e2s2 + els+ eO)(c2s2 + CIS+ co)

5th Order Polynomial 5th Order Polynomial

(12.87)

2. Simplify F 1 by letting (12.88) andF3 by

F1

F 3-

1 +-tanh reS

z

P

cPp + Zp tanh TeS

(12.89)

and, finally, with (12.90) This results in a more complex model that is undoubtedly more accurate. In this case, the function F 4 is 5th degree polynomial 6th degree polynomial

F4 = - - - - - - - -

and is much more detailed than the previous case. Experiments have indicated that, for all except the most careful experiments, such detail is not necessary. 3. If the water in the conduit is assumed to be rigid, then equation (12.62) becomes [8] h w - h, = Tc4c+ cPcqc

(12.91)

In this case, F} becomes a second order function:

reS + cPc

F 1 = - - Z- - - -

TcTtS + cPcTtS + 1

(12.92)

and the other transfer functions also become higher order. 4. All of the above should be compared to the classical water-hammer formula based on a lumped system: (12.93)

Hydraulic Turbine Prime Movers Penstock Ref

Error Signals

Servo Stroke

Hydraulic y Control Amplifier

Gate Position

Turbine Head

509 LoadTorque

t,

Z Hydraulic H Wicket System Gates Function

Shaft Speed

..--... Turbine- N Generator Rotor

Speed Governor Fig. 12.17

Block diagram of a hydro turbine speed control system.

where Tw is the so-called "water starting time" (about one second). This gives a second-order representation for F 4 . In verifying these approximations experimentally, Oldenburger and Donalson conclude that the hydraulic system consisting of conduit, surge tank, riser tank, penstock, scroll case, and draft tube can indeed be represented by a single transfer function relating Q to H as in (11.71). They verified that hydraulic resistance may be neglected without serious error. They note that a second-order representation of F 4 is adequate unless very accurate studies are to be performed. The assumption that the surge tank isolates conduit and penstock systems is also verified. Thus, although the hydraulic system is quite complicated, it may be represented adequately for control purposes by a linear model in which all transfer functions are ratios of polynomials.

12.10 Block Diagram for a Hydro System In considering the problem of controlling a hydro station, it is convenient to think of the system block diagram, which is shown in Figure 12.17. For a given steady load on the turbine Te , the electrical torque* is a constant and the speed N will be that set by the speed reference p. This would be the case in an isolated system. In an interconnected system, the speed is governed by the prevailing system frequency and the setting of the reference p determines the load that will be assumed by this machine. We can analyze the hydro system operation in a general way as follows. Any change in speed is changed by the speed governor into a change in position or displacement x, which is compared (usually mechanically) against a reference position p. Any difference in these positions produces an error signal 8), which is amplified by a control or servo amplifier to produce a servo stoke Y, proportional to 8) but having a much greater mechanical force to drive the wicket gates. This operation also usually introduces a delay or lag, which depends on the design of the servomotor. The servomotor stroke Y repositions the wicket gates to produce a new gate position Z. In hydro turbines, the gate position is fed back mechanically as a means of adjusting the droop or speed regulation. In many hydro installations, the wicket gates are very large and massive. This means that the servo amplifier must also be very large and capable of exerting large driving forces for moving such a large gate in a timely manner. *It is common to represent the torque by the symbols Tor M. We use the T here, but recognize that this symbol is also used for time constants.

510

12.11

Chapter 12

Pumped Storage Hydro Systems

The hydro systems described above assume a storage reservoir of water that is elevated in a configuration that will permit the water to be directed through a system of penstocks to hydro turbines that are situated at a lower elevation. This is true of stations that use a storage system fed by high-altitude streams , confined behind a dam. The confined water is held in storage until power output from the station is needed , at which time it is used to power hydro turbine generators. This type of system is also used for a run-of-river system, where there is a continual flow of water past the dam, some portion of which might be directed through hydro turbines to produce electric energy. In some cases, a minimum river flow might be necessary to support navigation or other uses of the water downstream, even if the generators are unavailable for some reason. A pumped storage hydro power plant is different from the run-of-river system . In the pumped-storage system there are two reservoirs, one at a high elevation into which water is pumped for release later, usually at times of high system loading . This is accomplished using a design of generator that can be operated efficiently as a motor and utilizing a turbine that can be operated as a pump. There is a cost associated with providing the pumping power, which must be performed at off-peak times when excess generation is available . Thus, there is an interesting economic tradeoff between the cost of providing the pumped storage facility and the availability of off-peak capacity to operate the pumps . Thus , the elevated water is not provided by nature , but must be created by forcing the water into the elevated storage reservoir. If the pumping energy is available at a reasonable cost, and the generation provided by the pumped-storage plant is of high value, then the overall economics of constructing such a facility may be quite attractive. The operating modes of a pumped storage system are shown in Figure 12.18. Pumped storage plants require a suitable topolog y, where an elevated pool can be built above the plant site. Aside from this physical restriction, there must be generation available for pumping that can be obtained at a cost differential that will make the entire facility operation an economic success. This requires the ability to pump power at a reasonably modest cost and a higher energy value during the generating cycle. Such a variation of energy value on a daily basis is not uncommon, since peaking load usually requires the scheduling of peaking generation with higher operating costs. Obviously, the economic parameters must be carefully evaluated in considering the construction of a pumped -storage facility.

Fig. 12.18 The two operating modes of a pumped storage power plant.

Hydraulic Turbine Prime Movers

511

Problems 12.1. Select a hydroelectric site of interest to you and record the physical features of the plant including the type of turbine, the head, the installed capacity, etc. Document the sources of your research and prepare a brief report on your findings. 12.2. Prepare a list of at least 10 hydroelectric sites, including a wide range of heads and physical features. 12.3. The system under study in [8] has the following constants:

Tee == 13 s T; == 0.25 s J m == 8 s

f4>e == 0.009 s 4>p == 0.001 s Z; == Zp == 4

The base quantities are: Torque:

40 MW at 225 rpm

Gate:

8 inches (at 80% of servomotor stroke)

Speed:

225 rpm

Head:

428 feet (headwater-tailwater)

Flow rate:

1600 ft3/ S

The turbine constants per unit are: All

== 0.57

A2 1 == 1.18

A 2 1 =-0.13

A22 == -0.35

A 1S == 1.10

A23 == 1.5

Use approximation (12.70) and compute the following:

F I == fi(s)

F3 == f3(F I , tanh TeS) F4 == f4(F3 ) 12.4. Find the transfer function of the hydraulic system shown in Figure 12.16 (b), where the hydraulic supply and water turbine transfer functions are given by (12.75) and (12.79), respectively. 12.5. Examine the effect of nonlinearity on the transfer functions F I , F 3, F s, and F 6 by using the approximation

(a) tanh(Ts) == TS

(TS)3 (b) tanh(Ts) ==TS - -3(TS)3

(c) tanh(Ts) == TS- -3-

2( TS)5

+ -15-

and finding the transfer functions for each F. Use an approximating technique to factor the truncated polynomials of (a), (b), and (c) and determine, by pole-zero plots, how the addition of extra terms in the series changes the system response. Use the data from problem 3.

512

Chapter 12

References 1. Knowlton, A. E., Standard Handbook/or Electrical Engineers, Section 10, Prime Movers, McGrawHill, New York, 1941. 2. Tietelbaum,P. D., Nuclear Energy and the US. Fuel Economy, 1955-1980, National Planning Association, Washington,D.C., 1964 3. Federal Power Commission, National Power Survey, 1964, U.S. Government Printing Office, Washington, D.C., 1964 4. Notes on Hydraulic Turbines, Los Angeles Departmentof Water and Power, Private Communication. 5. Barrows, H. K., Water Power Engineering, McGraw-Hill, New York, 1943. 6. Craeger, W. P. and 1. D. Justin, Hydroelectric Handbook, Wiley, New York, 1950. 7. Schleif, F. R., and A. B. Wilbor,The Coordinationof HydraulicTurbine Governorsfor Power System Operation, IEEE Trans. v. PAS-85, n. 7, p. 750-758, July 1966. 8. Oldenburger, R. and 1. Donelson,"Dynamic response of a hydroelectric plant," Trans. AlEE, Part III, 81, pp. 403-419, Oct. 1962. 9. deMello, F. P., Discussionof reference 8, Trans. AlEE, Part 111,81, pp. 418-419, Oct. 1962. 10. Oldenburger,R. Frequency Response, Macmillan,New York, 1956.

chapter

13

Combustion Turbine and Combined-Cycle Power Plants

13.1 Introduction Two additional types of generating unit prime movers that are growing in importance are the combustion turbine and combined-cycle units. Combustion turbine units were once considered as generating additions that could be constructed quickly and were reliable units for rapid start duty. The early units were not large, limited to-about 10 MVA, but later units have become available in larger sizes and, in some cases, may be considered a reasonable alternative to steam turbine generating units. A more recent addition to the available types of generating units is the combined-cycle power plant, in which the prime mover duty is divided between a gas or combustion turbine and a heat recovery steam turbine, with each turbine powering its own generator. The dynamic response of combined-cycle power plants is different from that of conventional steam turbine units and they must be studied carefully in order to understand the dynamic performance of these generating units.

13.2 The Combustion Turbine Prime Mover Combustion turbines, often called gas turbines, are used in a wide variety of applications, perhaps most notably in powering jet aircraft. They are also widely used in industrial plants for driving pumps, compressors, and electric generators. In utility applications, the combustion turbine is widely used as fast-startup peaking units. Combustion turbines have many advantages as a part of the generation mix of an electric utility. They are relatively small in size, compared to steam turbines, and have a low cost per unit of output. They can be delivered new in a relatively short time and are quickly installed compared to the complex installations for large steam turbine units. Combustion turbines are quickly started, even by remote control, and can come up to synchronous speed, ready to accept load, in a short time. This makes these units desirable as peaking generating units. Moreover, they can operate on a rather wide range of liquid or gaseous fuels. They are also subjected to fewer environmental controls than other types of prime movers [1]. The major disadvantage of combustion turbines is their relatively low cycle efficiency, being dependent on the Brayton cycle, which makes combustion turbines undesirable as base-load generating units. Another disadvantage is their incompatibility with solid fuels. The combination of low capital cost and low efficiency dictates that combustion turbines are used primarily as peaking units.

513

Chapter 13

514

Combustion turbines can be provided in either one- or two-shaft designs. In the two-shaft design, the second shaft drives a low-pressure turbine that requires a lower speed. However, in practice the single-shaft design is the most common [1]. The combustion turbine model presented here represents the power response of a singleshaft combustion turbine generating unit [2]. The model is intended for the study of power system disturbances lasting up to a few minutes. The generator may be on a separate shaft, in some cases connected to the turbine shaft through a gear train. The model is intended to be valid over a frequency range of about 57 to 63 Hz and for voltage deviations from 50 to 1200/0 of rated voltage. These ranges are considered to be typical of frequency and voltage deviations likely to occur during a major system disturbance. It is assumed that the model is to be used in a computer simulation in which, to obtain economical computer execution times, the timestep of the model might be one second or longer. The model is a rather simple one, but it should be adequate for most studies since the combustion turbine responds rapidly for most disturbances. Figure 13.1 shows a simple schematic diagram of a single-shaft combustion turbine-generator system with its controls and significant auxiliaries [2]. The axial-flow compressor (C) and the generator are driven by a turbine (T). Air enters the compressor at point 1 and the combustion system at point 2. Hot gases enter the turbine at point 3 and are exhausted to the atmosphere at point 4. The control system develops and sends a fuel demand signal to the main turbine fuel system, which in tum, regulates fuel flow to the burner, based on the unit set point, the speed, load, and exhaust temperature inputs. Auxiliaries that could reduce unit power capability are the

AUXILIARY POWERBUS

AUXILIARY ATOMIZING AIR SYSTEM

AUXILIARY FUEL HANDLING SYSTEM

FUEL DEMAND CONTROL SYSTEM SPEED REFERENCE

AIR IN

EXHAUST TEMPERATURE BURNER

1

SPEED FEEDBACK

Fig. 13.1

SHAFT

3 ...-------. POWER GENERATOR ' " - - - - - - ' OUTPUT

Combustion turbine schematic diagram [2].

515

Combustion Turbine and Combined-Cycle Power Plants

atomizing air and fuel handling systems shown in the figure. The atomizing air system provides compressed air through supplementary orifices in the fuel nozzles where the fuel is dispersed into a fine mist. The auxiliary fuel handling system transfers fuel oil from a storage tank to the gas turbine at the required pressure, temperature, and flow rate.

13.2. 1 Combustion turbine control Figure 13.2 shows a block diagram of a single-shaft combustion turbine-generator control system. The output of this model is the mechanical power output of the turbine. The input signal, AGCPS, is the power signal from the automatic generation control (AGC) system, in perunit power per second. The power is expressed in the system MVA base [2]. The governor speed changer position variable, noted in Figure 13.2 as GSCP, is the integral of the AGC input. An alternative input K M represents a manual input that is used if the generator is not under automatic generation control. The load demand signal shown in the diagram is the difference between the governor speed changer position and the frequency governing characteristic. The frequency governing characteristic is often characterized as a normal linear governor "droop" characteristic. Then the frequency error is divided by the per-unit regulation to determine the input demand. A nonlinear droop characteristic may be used in some cases. Typical data for the parameters shown in Figure 13.2 are provided in Table 13.1 [2]. The load demand upper power limit varies with ambient temperature according to the relation (13.1) where A = (the per-unit change in power output per per-unit change in ambient temperature) T = ambient temperature in °C T1 = reference temperature in °C

(Osys

-1

Linearor Nonlinear Frequency Governing Characteristics

f

K3

AGCPS

Off-Nominal Voltage and Frequency

Rate Limit

Effects on Power Output Nonwindup

Nonwindup Magnitude Limit

Magnitude Limit

Governor

Speed Changer Position (GSCP)

Fig. 13.2

Combustion turbine model block diagram [2].

Power

K3 ....O_ut~ PM

516

Chap~r

13

Table 13.1 Typical Combustion Turbine Model Parameters [2)

Constant

Description

Value

Manualrate, per-unitMW/s on given base Conversion, unit base/system base GSCPupper position temperature Combustion turbine time constant,s Normal regulation, per-unit freq/pu MVA Alternateregulation, see Figure 13.4

0.00278 0.11 0.25 0.04 0,01

According to (13.1), the turbine will provide 1.0 per-unit power at a reference ambient temperature of IS °C. The power limit is increased for temperatures below the reference and is decreased for ambient temperatures above the reference. The lower power limit corresponds approximately to the minimum fuel flow limit. This limit is necessary to prevent the blowing out of the flame and corresponds to zero electric power generated. There are three different off-nominal voltage and frequency effects. These are defined in the next section. Figure 13.3 shows the approximate computed response of a General Electric FS-5, Model N, single-shaft combustion turbine in response to a step change in setpoint from no load to full load, using liquid fuel [3] . The analytical model used to compute this response included the effects of the controls, the transport times, heat soak effect of turbine components in the hot gas path, and the thermocouple time constants. The turbine response will vary by several tenths of a second for other models or when using other fuels. Notice the fast response characteristic of the unit to its new power level.

1.0

-------

0.8 ..... '2

::l .... 0.6

- - - 1.0

RI

o

I I

I I Po

Rl

R2

R2

p

I

I I I

Fig. 13.4 Nonlinear governor droop characteristic [1].

All of the foregoing limiting functions apply to the limiter block on the right-hand side of Figure 13.2.

13.2.3 Nonlinear governor droop characteristic In some cases, it is desirable to include in simulations a nonlinear governor droop characteristic rather than the simple 4% or 5% linear droop characteristic often assumed . This might be necessary, for example, in providing an accurate model of the speed governor characteristic, which is not linear over a wide range, but tends to saturate for large excursions in speed or power. An example of a nonlinear droop characteristic is shown in Figure 13.4 [1, 3]. This is only one type of droop characteristic that might be examined . For example, it is not entirely clear that the slopes labeled R2 need to be equal in the high- and low-frequency ranges, nor is it clear that the center frequency in the R1 range should be exactly at the center between WI and Wz. Given adequate data, one might devise a continuous nonlinear curve to represent a range of frequencies and power responses. However, lacking better data, the droop characteristic of Figure 13.4 probably represents an improvement over the single droop characteristic so often used . Finally, it should be noted that the nonlinear droop characteristic was suggested as one device for improving the system response to very large disturbances, which create large upsets in power plants as well as loads. Some studies are not intended to accurately represent the power system under such extreme conditions, in which case the single droop characteristic may be adequate .

13.3 The Combined-Cycle Prime Mover There are a number of ways in which a combination of power cycles can be used in the generation of electricity, and power plants that use a combination of power cycles can have higher efficiencies that those dependent on a single power cycle. One typical combined-cycle turbine model is shown in Figure 13.5. This system utilizes a combination of a gas turbine Brayton cycle and a steam turbine using a Rankine cycle. The gas exhausted from the gas tur-

Combustion Turbine and Combined-Cycle Power Plants

Gas Turbine

519

Gene rator

Hot Gases Air

Generator

Condenser

Fig. 13.5

A typical combined-cycle power plant arrangement [3].

bine contains a significant amount of sensible heat and a portion of this heat is recovered in a steam generator, which in tum provides the working fluid for the steam turbine . Many combined-cycle power plants are more complex than that shown in Figure 13.5, which shows only the basic components. More practical systems are described below, but all systems can be conceptually reduced to the configuration of Figure 13.5. Figure 13.6 shows the schematic diagram for a combined-cycle power plant with a heat recovery boiler (HRG) [1]. In some designs, the steam turbine may have a lower rating than the gas turbine . In some large-system designs, supplementary firing is used, which may cause the steam turbine to achieve a rating greater than that of the gas turbine. Moreover, there may be more than one HRG, which could significantly increase the steam supply and therefore the power production of the steam subsystem. A descriptive technical paper on combined-cycle power plants has been prepared by the IEEE Working Group on Prime Mover and Energy Supply Models for System Dynamic Performance Studies [6]. Their detailed model of the combined-cycle unit is shown in Figure 13.7. Figure 13.8 shows the interactions among the subsystems of the combined-cycle system [6], and identifies the input and output variables of each subsystem and the coupling among these submodels . This structure is convenient for mathematical modeling of the combined-cycle power plant, which is described in greater detail below . The speed and load controls are described in block diagram form in Figure 13.9. The inputs are the load = \demand, VL> and the speed deviation, AN. The output is the fuel demand signal, F D'

Chapter 13

520

Combu stion Chamber

~~ Gas _

----r

r Air Compressor ~============ll ~ !

Air

T ur bime

-

Generator I

Optional Fuel --~--l Supplementary Firing System

HRB Legend: SU =Superheater B = Boiler EC = Economizer

Steam

To Stack

~S

Drum r--~---+--r U

~\.I'-~---+-l. Heat A A AY ~ B Recovery

Y

.

fA

Boiler

E

·C

)~-I Steam

i.>::= Steam Turbine

Generator 2

Condenser

.........

.... A

Feedwater Heater

......_ _J

Fig. 13.6 Schematic flow diagram ofa combined-cycle heat-recovery boiler [I] .

Combustion Turbine and Combined-Cycle Power Plants

521

Steam Turbine Generation

Heat Recovery Steam Generator

II r-

-t--

From Other HRSGs-

- -'

From OthLT

HRSGs

Fuel

Condensate Pump

To Other Gas Turbines

Gas Turbine Generation

Fig. 13.7 Two-pressure nonreheatrecovery feedwaterheatingsteam cycle generatingunit(HRSGwithinternal deaerator evaporator) (6).

Inlet Temperature

Speed

L oad Retlerence

Speed Dev iation

t

Fue) Demand Speed/Load Control

FD

I

Air Flow

Fuel and Air Fuel Controls Flow

t

Speed Deviation Gas Turbi ne

Gas Turbine

Mechan ical Powe r Gas Turbine Flow Rate

Exhaust Temperature

HRS G and Steam Turbine Fig.

13.8

Subsystems ofthe combined-cycle power plant (6).

Stea m Turb'me Mecha nical Power

522

Chapter 13 MAX

!1N Fig. 13.9

13.3.1

MIN Combined -cycle speed and control [6].

Fuel and Air Controls

The gas turbine fuel and air controls are show in block diagram form in Figure 13.10 [6]. In this control scheme, the inlet guide vanes are modulated to vary the air flow, and are active over a limited range . This allows maintaining high turbine exhaust temperatures, improving the steam cycle efficiency at reduced load. The fuel and guide vanes are controlled over the load range to maintain constant gas turbine inlet temperature. This is accomplished by scheduling air flow with the load demand F D and setting the turbine exhaust temperature reference TNto a value that is calculated to result in the desired load with the scheduled air flow at constant turbine inlet temperature. The exhaust temperature reference is calculated from the following basic gas turbine thermodynamic relations (taken from reference [6]).

( 13.5)

TE Exhaust Temp

1 + J;s

~

T,s

+

1.05

- --

1+ TRls

1.0

7;

>"

Input Temp

N 1.0

TR

Wo

F.D Fuel Demand Signal N Speed Fig. 13.10

Gas turbine fuel and air flow controls [6].

Combustion Turbine and Combined-Cycle Power Plants

523

where TR = reference exhaust temperature per unit of the absolute firing temperature at rated conditions Also (13.6) where PRO =

design cycle pressure ratio

= PRO W = isentropic cycle pressure ratio 'Y = ratio of specific heats = cp/cv

PR

We also define the following

W = design air flow per unit

TIT = turbine efficiency Tf = turbine inlet temperature per unit of design absolute firing temperature

Then the per-unit flow required to produce a specified power generation at the given gas turbine inlet temperature r:r is given by the turbine power balance equation

iw«,

(13.7)

where kW is the design output in per unit. Also 3413 . kW Ko = - - - - o WgOTfoCp

(13.8)

and where we define

kWo = base net output per unit WgO = base net flow per unit

Tj'o = turbine inlet temperature per unit of design absolute firing temperature

Cp = average specific heat T, = compressor inlet temperature per unit of design absolute firing temperature TIc = compressor efficiency

The combustor pressure drop, specific heat changes, and the detailed treatment of cooling flows have been deleted for purposes of illustration of the general unit behavior. These performance effects have been incorporated into equivalent compressor and turbine efficiency values [6]. Equations (13.7) and (13.8) determine the air flow Wand pressure ratio parameter X for a given per-unit generated power in kW, and at a specified per-unit ambient temperature T; The reference exhaust temperature TR is given by (13.6) by setting 1j.= 1.0. The air flow must be subject to the control range limits. The block identified as A in Figure 13.10 represents the computation of the desired air flow WD and the reference exhaust temperature over the design range of air flow variation by means of vane control. Desired values of WD and TR are functions of FD (the desired values of turbine output from speed/load controls) and ambient temperature T; These are determined by the solution of (13.7) and (13.8) with appropriate limits on WD and TR . The vane control response is modeled with a time constant TR and with nonwindup limits corresponding to the vane control range. The actual air flow WA is shown as a product of desired air flow and shaft speed. The reference exhaust temperature TR is given by (13.6) with Tfset equal to unity.

524

Chapter 13

The measured exhaust temperature TE is compared with the limiting value TR and the error acts on the temperature controller. Normally, T£ is less than TR, which causes the temperature controller to be at the maximum limit of about 1.1 per unit. If TE should exceed TR, the controller will come off limit and integrate to the point where the its output takes over as the demand signal for fuel Vee through the low-select (LS) block. The fuel valve positioner and the fuel control are represented as given in [7], giving a fuel flow signal Wfas another input to the gas turbine model.

13.3.2 The gas turbine power generation A block diagram of the computat ion of gas turbine mechanical power PM G and the exhaust temperature TE is shown in Figure 13.11. The equations used in the development of the gas turbine mechanical power PMG are shown in Figure 13.11. The gas turbine output is a function of the computed turbine inlet temperature Tfo which is a function of the turbine air flow Wf '

1 10 .

means that the input voltage of such amplifiers is essentially zero since the output is a Iways limited to a fi nite value (0 ft en 100 V). The multiplier. The multiplier used on modern analog computers is an electronic quarter-square multiplier that operates on the following principle. Suppose u and i are to be multiplied to find the instantaneous power; i.e., p = vi. To do this, we begin with two voltages, one proportional to v, the other proportional to r, Then we form sum and difference signals, which in turn are squared and subtracted; i.e., M = (v + i)2 - (v - i)2 = (v 2 + 2ui

+ ;2) - (v 2 - 2v; + ;2)

=

4v;

and p = (I /4)M, or one quarter of the difference of the squared signals. The symbol used for multiplication varies with the actual components present in the computer multiplier section, but in its simplest form it may be represented as shown in Figure 8.6. Note that it is usually necessary to supply both the positive and negative of one signal, say VI' The multiplier inverts and divides the result by I00 (on a IDO-V computer).

v;-------' Fig. B.6. The multiplier: VJ

= -

VI V21100 Y

= - VI V2 pu.

Other components. Most full-scale analog computers have other components not described here, including certain logical elements to control the computer operation. These specialized devices are left for the interested reader to discover for himself.

8.1.2

Analog computer scaling

Two kinds of scaling are necessary in analog computation, time scaling and amplitude scaling. Time scaling can be illustrated by means of a simple example. Consider the first-order equation

T~~

=

(B.l )

!(V,t)

where v is the dependent variable that is desired, T is a constant, and I is a nonlinear function of v and t, The constant T would appear to be merely an amplitude scale factor, but such is not the case. Suppose we write T dv = ~ = dv = dt d( t j T) dT

I(v

'

t)

(B.2)

where T = t l T, Thus replacing the constant T by unity as in (B.2) amounts to time scaling the equation. In an analog computation the integration time must be chosen so

534

Appendix B

that the computed results may be conveniently plotted or displayed. For example, if the output plotter has a frequency limit of 1.0 kHz, the computer should be time scaled to plot the results more slowly than this limit. Analog computers must also be amplitude scaled so that no variables will exceed the rating of the computer amplifiers (usually 100 V). This requires that the user estimate the maximum value of all variables to be represented and scale the values of these variables so that the maximum excursion is well below the computer rating. Actually, it is convenient to scale time and amplitude simultaneously. One reason for this is that the electronic integrator is unable to tell the difference between the two scale factors. Moreover, this makes one equation suffice for both kinds of scaling. We begin with the following definitions. Let the time scaling constant a be defined as follows: T

=

computer time

T

a = t

t = real time

computer time real time

(8.3)

For example, if a = 100, this means that it will take the computer 100 times as long to solve the problem as the real system would require. It also means that ]00 s on the output plotter corresponds to I s of real time. Also define L as the level of a particular variable in volts, corresponding to 1.0 pu of that variable. For example, suppose the variable v in (B.1) ordinarily does not go above 5.0 pu. If the computer is rated 100 V, we could set L = 20 V on the amplifier supplying v. Then if v goes to 5.0 pu, the amplifier would reach 100 V, its maximum safe value. The scaling procedure follows: I. Choose a time scale a that is compatible with plotting equipment and will give reasonable computation times (a few minutes at most). 2. Choose levels for all variables at the output of all summers and integrators. K

l.

In fntegrotor or summer

Fig. B.7.

Time and amplitude scaling.

3. Apply the following formula to all potentiometer settings (see Figure B.7):

PG where

=

KLout/aLin

(8.5)

a = time scale factor P = potentiometer setting, 0 S; P < I G = amplifier or integrator gain K = physical constant computed for this potentiometer Lout = assigned output level, V Lin = assigned input level, V

8.1.3

Analog computation

Example B.I Suppose the integrator in Figure 8.7 is to integrate - {) (in pu) to get the torque angle {) in radians. Then we write

Appendix B

535 (B.6)

Thus the constant K in Figure B.7 and (B.5) is WR. which is required to convert from pu to h in rad/s , In our example let WR = 377.

li

in

Solution

Let a = 50. Then the levels are computed as follows: 8m.. = 100" = 1.745 rad, so let L O UI = 50 Y, (1.745 x 50 < 100). Also estimate 8m.. = 1.25 pu, so let Lin = 75 Y, (1.25 X 75 < 100). Then compute

PC Since 0 setting .

~

p

~

=

KLou./aL in = (377 x 50)/(50 x 75) = 5.03

I let C = 10

=

gain of integrator and P

=

0.503

potentiometer

Example 8 .2

Compute the buildup curve of a de exciter by analog computer and compare with the method of formal integration used in Chapter 7. Use numerical data from Examples 7.4, 7.5, and 7.6. Solution

For this problem we ha ve the first-order differential equation

v

F

where

U =

=

(u -

RO/T

(B.7)

up when separately excited

uF when self-excited uR + uF when boost -buck excited

where both up and UR are constants. Thus the analog computer diagram is that shown in Figure B.8, where U FO = uF(O). Close switch for self -excited or boost-buck oper ation

Fig. B.8. Solut ion diagram for dc exciter buildup .

An alternate solution utilizing the Frohlich approximation to the magnetization curve is described by the equation (B .8) Solving this equation should exactl y dupl icate the results of Chapter 7 where this same equation was solved by formal integration.

Appendix B

536

Using numerical data from Example 7.4 we have Tf;=

0.25s

a

= 279.9

b

=

5.65

The values of R and v depend upon the type of buildup curve being simulated. From Examples 7.4, 7.5, and 7.6 we have Separatelyexcited:v=vp = 125 V Self excited : v = v,. R = 30 n Boost-buck excited: v = v,. + 50 V

n

R

34

R

43.6

n

and these values will give a ceiling of 110.3 V in all cases. Also, from Table 7.5 we note that the derivative of v,. can be greater than 100 VIs. This will help us scale the voltage level of uF• Rewriting equation (B.8) with numerical values, we have 0.25

v,.. = v

- 5.65 Rvd(279 .9 - V,.) V

(B.9)

where R and v depend on the type of system being simulated. Suppose we choose a base voltage of 100 V. Then dividing (B.9) by the base voltage we have the pu equation 0.25 v,.

=

v - 0.0565 RV,./(2.799 - V,.)

(B.10)

where v,. and v are now in pu. A convenient time scale factor is obtained by writing TeV,. = TE

d v;

dt

dv,. d o, = dUITE) = dT

or a = Til = liTE = 4.0 S-I Then the factor 0.25 in front of (B.IO) becomes unity, and 4 s on the computer corresponds to 1.0 s of realtime. The analog computer solution for (8.10) is shown in Figure B.9, and the potentiometer settings are given in Table B.l . By moving the three switches simultaneously to positions R, C, and L, the same computer setup solves the separately excited, selfexcited, and boost-buck buildup curves respectively. Voltage levels are assumed for

• REF

(100) ~

LwJ

Switch Code R = Separatel y excited C = Self-excited L = Boost-buck e xcited

( ) = Voltoge le vel of 1.0 pu

r---,-..... - REF Fig . B.9.

Solution diagram for Frohlich approximated buildup.

Appendix B

537

each amplifier and are noted in parentheses. These values are substituted into (8.5) to compute the PG products given in Table B.l. For example, for potentiometer 5

PC

=

(K/a)(Lout/Lin)

=

(1.0/4)(50/10)

=

1.25 = 0.125 x 10

or for potentiometer 7

PC = (1.92/1)(10/50)

0.384

0.384 x 1

Other table entries are similarly computed. Table B.I.

Function

K

PG

P

G

scale scale time scale initial value, v f O bR (separately) bR (self) bR (boost-buck) scale a

1.25 0.50 1.0 1.0 1.0 0.45 1.92 1.695 2.46 1.0 2.799

0.125 0.050 0.20 0.20 1.25

0.125 0.050 0.20 0.20 0.125 0.45 0.384 0.339 0.492 0.40 0.56

I I I I 10

Potentiometer

I 2 3 4 5 6 7 8 9 10 II

Potentiometer and Gain Calculations for Figure 8.9

vI'

VR

0.384 0.339 0.492 0.40 0.56

The computed results are shown in Examples 7.4,7.5, and 7.6. B.2

Digital Computer Selutien of Ordinary Differential Equations

The purpose of this section is to present a brief introduction to the solution of ordinary differential equations by numerical techniques. The treatment here is simple and is intended to. introduce the subject of numerical analysis to the reader who wishes to see how equations can be solved numerically. One effective method of introducing a subject is to turn immediately to a simple example that can be solved without getting completely immersed in details. We shall use this technique. Our sample problem is the de exciter buildup equation from Chapter 7, which was solved by integration in Examples 7.4 -7.6. Since the solution is known, our numerical exercise will serve as a check on the work of Chapter 7. However, the real reason for choosing this example is that it is a scalar (one-dimensional) system that we can solve numerically with relative ease. Larger n-dimensional systems of equations are more challenging, but the principles are the same. The nonlinear differential equation here is ·

UF =

F dU -

dt

= - I (u T£

R'I)

(8.11 )

which we will solve by numerical techniques using a digital computer. Such problems are generally called "initial value problems" because the dependent variable uF is known to have the initial value (at t = 0) of vF(O) = UFO' 8.2.1

Brief survey of numerical methods

There are several well-documented methods for solving the initial value problem by numerical integration. All methods divide the time domain into small segments ~t long

Appendix B

538

and solve for the value of v; at the end of each segment. In doing this there are three problems: getting the integration started, the speed of computation, and the generation of errors. Some methods are self-starting and others are not; therefore, a given computation scheme may start the integration using one method and then change to another method for increased speed or accuracy. Speed is important because, although the digital computer may be fast, any process that generates a great deal of computation may be expensive. Thus, for example, choosing ~t too small may greatly increase the cost of a computed result and may not provide enough improvement in accuracy to be worth the extra cost. A brief outline of some known methods of numerical integration is given in Table B.2. Note that the form of equation is given in each case as an nth-order equation. However, it is easily shown that any nth-order equation can be written as n first-order equations. Thus instead of v(n)

}(v, t)

=

(8.12)

we may write

or in matrix form

XI =

!t(v, t)

X2 =

h(v, t)

.i·n

!,,(v, t)

=

x=

f(x, t)

(B.13)

Thus we concern ourselves primarily with the solution of a first-order equation. Table B.2. Method

Direct integration, . trapezoidal rule, Simpson's rule Euler Modified Euler (Heun) Runge-Kutta Milne Hamming Crane

Some Methods of Numerical Integration of Differential Equations Form of equation v(n)

=

!(I )

Order of errors

61

Remarks

Must know n - J derivatives to solve for v(n)

= ft», I) = It». I)

(~t)2 (~1)3

Self-starting Self-starting predictor-corrector

(~t)J (~t)5

v(n)

= ft», I) = /(v, t) = ft», I)

v(n)

= [t», t)

Self-starting, slow Start by Runge-Kutta or Taylor series Imposes maximum condition on j.t for stable solution Varies size of ~ t to control error

v(n) v(n)

v(n)

v(n)

A complete analysis of every method in Table B.2 is beyond the scope of this appendix and the interested reader is referred to the many excellent references on the subject. Instead, we will investigate only the modified Euler method in enough detail to be able to work a simple problem. 8.2.2

Modified Euler method

Consider the first-order differential equation

v = ft», t)

(B.14)

Appendix B

539

Area =

v, (; t

C(v, ) (a)

f - - (;I

..

I...

(; 1 - - 1

v

(b)

Fig. B.IO. Graphical interpretation of the pred ictor -corrector routine : (a)

vversus

I.

(b) () versus I .

where v is known for I = 0 (the initial value) . Suppose the curves for v and shown in Fig . B.IO, where the time base has been divided into finite intervals Now define

uare as

!J.I

wide.

(B.15) which gives the initial slope of the v versus I curve. Next a predicted value for v at the end of the first interval is computed . If we define v = VI when I = til, we compute the predicted value VI as (B .16) which is an extension of the initial slope out to the end of the first interval, as shown in Figure B.IO(b). But votil is the rectangular area shown in Figure B.IO(a) and is obviously larger than the true area under the v versus I curve, so we conclude that P(v l ) is too large [also see Figure B.IO(b)]. Suppose we now approximate the value of VI by substituting P(v,) into the given differential equation (B.14). Calling this value P(v l ) , we compute

v

(B .17)

Now approximate the true area under the versus I curve between 0 and !J.I by a trapezoid whose top is the straight line from vo'to P(v,), as shown by the dashed line in Figure B.I O(a). Using this area rather than the rectangular area, we compute a corrected value of VI' which we call C(v l ) , (B.18)

540

Appendix B

We call (B.18) the corrector equation.

Now we substitute the corrected value of

v" C(v l ) , into the original equation to get a corrected

l\.

C(v,) = ![C(V1),dt]

(8.19)

We now repeat this operation, using C(u 1) in (B.18) rather than P(v 1) to obtain an even better value for C(v l ) . This is done over and over again until successive values of C(v l ) differ from one another by less than some prescribed precision index or until (B.20)

where k is the iteration number and E is some convenient, small precision index (10-6, for example). Once VI is determined as above, we use it as the starting point to find v2 by the same method. The general form of predictor and corrector equations is

+ V;(dt) + Hv; + P(Vi + I )]/ 21~l

P(V;+ I) = C(V;+I) =

8.2.3

Vi

V;

(B.21)

(B.22)

Use of the modified Euler method

Example B.3

Solve the separately excited buildup curve by the predictor-corrector method of numerical integration. Use numerical values from Example 7.4.

Solution

The equation requiring solution is TEV F

=

vp

-

Ri

(B.23)

where i as a function of V F is known from Table 7.3. We could proceed in two different ways at this point. We could store the data of Table 7.3 in the computer and use linear (or other means) interpolation to compute values of i for V F between given data points. Thus using linear interpolation, we have for any value of v between VI and v2 (B.24) In this way we can compute the value of i corresponding to any V F and substitute in (B.23) to find uF • An alternative method is to use an approximate formula to represent the nonlinear relationship between V F and i, Thus, by the Frohlich equation, (B.25) where a and b may be found as in Example 7.2. Let us proceed using the latter of the two methods, where from Example 7.2 we have a = 279.9

b = 5.65

Thus (B.23) becomes V•

F

V =:.J!.. TE

Rbu; _ TE(a -

vF )

(B.26)

or (B.27)

Append ix B

COMP UTE 1 ill

V OOT ~

[

W -

541

RBV 1 A=\iJ

COMP UTE PV ~ V + V OOr- DELTA PVOOT ~

..!.-[w_ARBPV 1 - pvJ

TEE

SET T ~ T + DELTA V " CV V OOT ~ CVDOT

SET

O LD ~ PV CV DO T ~ PVOOT

COMPUTE CV ~ v + 0 . 5 (VDO T • CVDOT)" DELTA

CO M PUTE 1 [ RBCV CVDor ~ TEE W- .A:-=CVJ

1

?

OLD

~

CV

YES

Fi g . B.II .

Co m p uter flow d iagram. separately e xcited case.

To avoid confusion in programming, we drop the subscript on constant W, and repl ace T by T to wr ite

v=

Uf '

represent up by a (B.28)

WjT - (RbjT)[uj(a - u)]

The data that must be input to begin the so lut io n is sho w n in Table B.3 with certain additional variables that must be defined . The computer flow diagram is shown in Figure B.11 for the separately excited case. The FORTRAN coding is given in Figure B.12 . The solution is printed in tabular form in Table B.4 for values of t from 0 to 0.8 s. Note that both U f and f are given. The derivati ve may not be needed, but it is known and can just as well be printed. The computed results agree almost exactly with the results of Example 7.4 and are therefore not plotted.

v

Appendix B

542

VDOTl(W,V) = (W-R*B*V/(A-V))/TEE READ( 1, 101)W,TEE,R,B,A,VO,OElTA,KENO,EPS FORMAT (FS.2,F4.3,FS.2,F5.3,F6.3,F5.2,FS.4,I3,F7.7) V=VO VOOT=O.O PV = 0.0 CV =-= 0.0 PVDOT=O.O CVDOT=O.O T=O.O VDOT=VDOTl (W,V) WRITE(3,110)T,V,VDOT DO 200 J = 1, KENO 105 PV=V+VDOT*DELTA PVDOT:: VDOTl (W,PV) 102 OLD=PV 103 CVDOT =PVDOT 104 CV=V+O.S*(VDOT+CVDOT)*DELTA IF(CV-OlD-EPS) 107,107,106 106 CVDOT =VDOT 1(W ,CV) OLD=CV GO TO 104 107 T = T + DEl TA V=CV VDOT=CVDOT WRITE(3,110)T,V,VDOT 110 FORMAT(' ',Fl0.3,Fl0.2,Fl0.2) 200 CONTINUE STOP END 101

Fig. B.12.

Table B.3.

FORTRAN coding for the separately excited case.

Data and Variable Symbols. Names. and Formats

Symbol up

Name W

T

TEE

a

A

at

DELTA

R b

v(O)

v v P(v i + , ) C(v i + I) P(v i + , ) C(v i + , ) t

R B

VO

KEND EPS V VDOT PVDOT CVDOT PV

Format

Constant

F5.2 F4.3 F5.2 F5.3 F6.3 F5.2 F5.4 13

x x

F7.7

F5.2 F6.2

Cy

T

F5.3

Variable

x x

x x x x x

x x x x x x x

Appendix B Table 8.4.

0.0 0.010 0.020 0.030 0.040 0.050 0.060 0.070 0.080 0.090 0.100 0.110 0.120 0.130 0.140 0.150 0.160 0.170 0.180 0.190 0.200 0.210 0.220 0.230 0.240 0.250 0.260 0.270 0.280 0.290 0.300 0.310 0.320 0.330 0.340 0.350 0.360 0.370 0.380 0.390

543

Separately Excited Results in Tabular Form

.

Vf

VF

40.00 44.50 48.93 53.30 57.60 61.83 66.00 70.09 74.11 78.05 81.92 85.71 89.42 93.06 96.61 100.08 103.46 106.76 109.97 113.10 116.14 119.09 121.95 124.72 127.41 130.00 132.50 134.92 137.24 139.48 141.63 143.69 145.66 147.56 149.36 151.09 152.73 154.30 155.79 157.20

452.90 446.55 440.10 433.50 426.75 419.84 412.78 405.57 398.20 390.69 383.03 375.23 367.29 359.21 351.01 342.68 334.24 325.70 317.05 308.32 299.52 290.65 281. 74 272.79 263.82 254.84 245.88 236.94 228.05 219.21 210.46 201.80 193.26 184.84 176.57 168.45 160.51 152.76 145.20 137.85

0.400 0.410 0.420 0.430 0.440 0.450 0.460 0.470 0.480 0.490 0.500 0.510 0.520 0.530 0.540 0.550 0.560 0.570 0.580 0.590 0.600 0.610 0.620 0.630 0.640 0.650 0.660 0.670 0.680 0.690 0.700 0.710 0.720 0.730 0.740 0.750 0.760 0.770 0.780 0.790 0.800

VF

VF

158.55 159.82 161.02 162.16 163.24 164.26 165.21 166.11 166.96 167.76 168.51 169.21 169.87 170.49 171.06 171.60 172.11 172.58 173.02 173.43 173.82 174.17 174.51 174.82 175.11 175.38 175.63 175.86 176.08 176.28 176.46 176.64 176.80 176.95 177.09 177.22 177.34 177.45 177.55 177.65 177.73

130.72 123.82 117.15 110.72 104.52 98.58 92.87 87.42 82.20 77.23 72.50 68.00 63.73 59.68 55.85 52.23 48.82 45.59 42.56 39.7\ 37.03 34.51 32.15 29.94 27.87 25.93 24.12 22.43 20.85 19.37 18.00 16.72 15.52 14.41 13.38 12.41 11.52 10.68 9.91 9.19 8.52

References Analog Computation Ashley, J. R. Introduction to Analog Computation, Wiley, New York. 1963. Blum, J. J. Introduction to Analog Computation. Harcourt, Brace and World, New York, 1969. Hausner, A. Analog and Analog] Hybrid Computer Programming. Prentice-Hall, Englewood Cliffs. N.J.. 1971. James. M. L., Smith, G. M., and Wolford. J. C. Analog and Digital Computer Methods in Engineering Analysis. International Textbook Co., Scranton, Pa., 1964. - - . Analog Computer Simulation of Engineering Systems, 2nd ed. Intext Educational Publ., Scranton, Pa.. 197'. Jenness, R. R. Analog Computation and Simulation, Allyn and Bacon, Boston, 1965. _ _ . Analog Computation and Simulation: Laboratory Approach. Allyn and Bacon, Boston, 1965. Johnson. C. L. Analog Computer Techniques. McGraw-Hili. New York, 1963.

544

Appendix B

Digital Computation Hildebrand. F. B. Introduction to Numerical Analysis. McGraw-Hill. New York. 1956. James, M. L., Smith. G. M .. and Wolford, J. C. Analog and Digital Computer Methods in Engineering Analysis. International Textbook Co .. Scranton, Pa., 1964. Korn, G. A., and Korn, T. M. Mathematics Handbook for Scientists and Engineers. McGraw-Hili, New York. 1968. Pennington, R. H. Introductory Computer Methods and Numerical Analysis. Macmillan. New York. 1965. Pipes. L. A. Matrix Methods for Engineering. Prentice-Hall, Englewood Cliffs. N.J., 1963. Stagg, G. W., and El-Abiad, A. H. Computer Methods in Power System Analysis. McGraw-Hill. New York. 196~L Stephenson, R. E. Computer Simulation for Engineers. Harcourt Brace Jovanovich, New York, 1971. Wilf, H. S. Mathematics for the Physical Sciences. Wiley, New York, 1962.

appendix

D

Typical System Data

In studying system control and stability, it is often helpful to have access to typical system constants. Such constants help the student or teacher become acquainted with typical system parameters, and they permit the practicing engineer to estimate values for future installations. The data given here were chosen simply because they were available to the authors and are probably typical. A rather complete set of data is given for various sizes of machines driven by both steam and hydraulic turbines. In most cases such an accumulation of information is not available without special inquiry. For example, data taken from manufacturers' bids are limited in scope, and these are often the only known data for a machine. Thus it is often necessary for the engineer to estimate or calculate the missing information. Data are also provided that might be considered typical for certain prime mover systems. This is helpful in estimating simulation constants that can be used to represent other typical medium to large units. Finally, data are provided for typical transmission lines of various voltages. (See Tables 0.1--0.8 at the end of this appendix.) D.1

Data for Generator Units

Included here are all data normally required for dynamic simulation of the synchronous generator, the exciter, the turbine-governor system, and the power system stabilizer. The items included in the tabulations are specified in Table 0.1. Certain items in Table 0.1 require explanation. Table references on these items are given in parentheses following the identifying symbol. An explanation of these referenced items follows. ( 1)

Short circuit ratio

The SCR is the "short circuit ratio" of a synchronous machine and is defined as the ratio of the field current required for rated open circuit voltage to the field current required for rated short circuit current [1]. Referring to Figure D.), we compute (0.1)

It can be shown that SCR "-' l/x d pu

where

Xd

(0.2)

is the saturated d axis synchronous reactance.

555

Appendix D

556 1.4

1. 2

1.0

------

I I I I

0.8

I

I

>

I

~

en

2

-0

>

0 .6

I

E

I

~

I-

u

I

I

I I

I

I

I I

I I

I

u

c v a.

I

0 .2

o

I I

-1-- ---1---

0. 4

.~

o

c ~

I

o .S

'5

I

I I

~

_ _----'-

;

-0

1. 0

~

U

I I I

~

'5

s

0 .5 ~

-L._..J...-----'-_ --'-_ _..J...---'-_- ' O

Fiel d C urren t , IF

Fig . D . !.

(2)

Op en circuit, full load . and sho rt circui t character istics ofa synchrono us generator.

Generator saturation

Saturation of the generator is often specified in terms of a pu saturation function SG' which is defined in terms of the open circuit terminal voltage versus field current characterist ic shown in Figure 0 .2. Wecompute (0 .3) where (0.3) is valid for any point v" [2, 3]. With use of this definition . It IS common to specify two values of saturation at V, = 1.0 and 1.2 pu. These values are given under open circuit conditions so that V, is actually the voltage behind the leakage reactance and is the voltage acro ss LAD. the pu saturated magnetizing inductance. Thus we can easily determine two satura tion values from the generator saturation curve to use as the basis for defining a saturation function . From Figure 0 .1 we arb itrar ily define (/0 - IA)/(of

(0.4)

(/ , - 1.2IA)/I.2IA

(0 .5)

and will use these two values to generate a saturation function.

Appendix D

557

"

0>

E

o

>

's ~

o c

o" Q.

Field Current, IF

Fig . D .2 .

C o nst ruc tio n used for computing saturation .

There are several ways to define a saturation function, one of which Section 5.10.\ where we define

IS

given in (0 .6)

where

v.\

=

v, -

0.8

(0 .7)

is the difference between the open circuit terminal voltage and the assumed saturation threshold of 0.8 pu . Since (0 .6) contains two unknowns and the quantities SG and V.\ are known at two points, we can solve for A G and BG explicitly . From the given data we write (0.8)

Rearranging and taking logarithms, In(SGlo/A G)

=

(0.9)

0.2 BG

Then, or (0.10) Example D.I

Suppose that measurements on a given generator saturation curve provide the following data: SGLO = 0.20

SGIl = 0.80

Then we compute, using (0 .10), AG

= (0.20)2/1.2(0.80) = 0.04167

BG

= Sin (1.2

x 0.8/0.20)

= 7.843

This gives an idea of the order of magnitude of these constants; A G is usually less than 0.1 and BG is usually between 5 and 10.

Appendix D

558

The value of SG determined above may be used to compute the open circuit voltage (or flux linkage) in terms of the saturated value of field current (or MMF). Referring again to Figure 0.\, we write the voltage on the air gap line as (0 .11)

Refer to Figure 0 .2. When saturation is present, current In does not give V,2 but only produces VF 1, or

V,I = V,2 - V; = Rl n - V;

=

Rln

(0.12)

where V; is the drop in voltage due to saturation. But from Figure 0 .2 tan 8 = R = V;/Un - IF!)

(0.13)

From (0.3) we write (0 .14)

Then from (0 .12) (0.15)

where SG is clearly a function of V,I' Equation (0.15) describes how v" is reduced by saturation below its air gap value Rln at no load . Usually, we assume a similar reduction occurs under load . Note that the exponential saturation function does not satisfy the definition (0 .3) in the neighborhood of V, = 0.8, where we assume that saturation begins. The computed saturation function has the shape shown in Figure 0 .3. Note that SG > 0 for any V&. The error is small, however, and the approximation solution is considered adequate in the neighborhood of 1.0 pu voltage . Note that A G is usually a very small number, so the saturation computed for V, < 0.8 is negligible. Other methods of treating saturation are found in the literature [1,2,4,5,6. 7].

-0 .8

o

0.4 1.2

Fig. D.J . The approximate saturation function, SG'

(3)

Damping

It is common practice in stability studies to provide a means of adding damping that is proportional to speed or slip. This concept is discussed in Sections 2.3, 2.4, 2.9. 4.10, and 4.15 and is treated in the literature [8-12]. The method of introducing the damping is by means of a speed or slip feedback term similar to that shown in Figure 3.4. where D is the pu damping coefficient used to compute a damping torque T,

559

Appendix D defined as

(D.16) where all quantities are in pu . The value used for D depends greatly on the kind of generator model used and particularly on the modeling of the amortisseur windings. For example, a damping of 1-3 pu is often used to represent damping due to turb ine windage and load effects (2]. A much higher value, up to 25 pu is sometimes used as a representation of amortisseur damping if this important source of damping is omitted from the machine model. The value of D also depends on the units of (D .16). In some simulations the torque is computed in megawatts. Then with the slip W 6 in pu (D .17) It is also common to see the slip computed in hertz, i.e., Ij, Hz. Then (D .17) becomes (D.18) where S03 is the three-phase MY A base . fR is the base frequency in Hz, and the slip in Hz. A value sometimes used for D' in (D .18) is

D'

=

PdlR MW 1Hz

It;

is

(D .19)

wnere PG is the scheduled power generated in MW for this unit. This corresponds to D = PGISo3 pu . (4)

Voltage regulator type

The type of voltage regulator system is tabulated using an alphabetical symbol that corresponds to the block diagrams shown in Figures D.4-D.II . Excitation systems have undergone significant changes in the past decade, both in design and in the models for representing the various designs . The models proposed by the IEEE committee in 1968 [3] have been largely superseded by newer systems and alternate models for certain older systems . The approach used here is the alphabetic labeling adopted by the Western Systems Coordinating Council (WSCC), provided through private communication . The need for expanded modeling and common format for exchange of modeling data is under study by an IEEE working group at the time of publication of this book .

v,

O ther signa ls Sta bil izer

F ig. 0.4 .

Type A-continuously acting de ro ta ting excitatio n system. Representative systems: (I) TR = 0: General Electric NA 143. NA 108: Westinghouse Mag-A-Stat. WMA ; All is Chalmers Regulux: (2) TR ... 0: General Electric NA 101: Westinghouse Rototrol, Silverstat, TRA .

Appendix D

560

Re gulator

Excit er

v

s O th er signa ls Stabi lizer

Fig. 0 .5. Type B -Westinghouse pre-1967 brush less.

Regul at or

Exci ter

v, O the r

sig na ls

Fig. 0 .6. Type C - -West inghouse brushless since 1966.

A = (0.7 8XLIFt l VTHEV)2

If : A > l, V = O B

Fig. 0 .7. Type O-SCPT system .

Appendix D

V K'

Rmox

A V

Fig . 0 .8.

If :

~ V t ~ KV ' VR = VRmo x

\'" vd v,

V R = V RH

t: V t

V =V R Rmin

-c K

1 • T RH ' Rmin

561

~

-K

v,

Type E ---no nco ntinuo usly acting rheostatic excit ati on system . Represent ative systems : General Electr ic GFA4 . We st ingh ouse BBO.

Integrat ing regulator

Exc iter

V V

Rmin

EFDm in = 0

5

O ther 'igna l'

Fig . 0 .9 .

T ype F - West inghou se continuously acti ng brush less rotating alternator excitation system.

vs

O ther ,i gnals

Fig . D .IO.

Type G -- G ene ra l Electric SCR exci tat ion system .

562

Appendix D

v,

Other

signa ls

Sta bilize r

F ig. D .I I.

T ype K - -Gener al Electric A lterrex .

Not e that the regulator base voltage used to norm alize VR may be chosen arbitrarily. Since the exciter input signal is usually VR - (Sf + Kt.)E FD, choosing a different base affects the constant Sf and KE, and also the gain KA • (5)

Exciter saturation

The saturation of de gener ator exciters is represented by an exponential model der ived to fit the actu al saturation curve at the exciter ceiling (ma x) voltage (zero field rheostat setting) and at 75,%" of ceiling. Referring to Figure 0 .12, we define the following constants at ceiling, 0.75 of ceiling and full load. S Ema>

= (A

- 8)/8

...

S E7Sma.

0

u,

i 2 a

Full load

>

= (E -

F)/F

S UL

II

II II II

"2

I,

.~

~

II

1!

'u x w

II

0

II

BE

A

& ci ter Field Current

Fig. D.12. A dc exciter satu rat ion curve .

(C - D)/ D

(0 .20)

Appendix D

563

Then in pu with EFDf L as a base (actually, any convenient base may be used), EfDmax = EFDmax(V)/EFDFL(V) =

BID pu

or B

(0.21 )

DEfDmax

We can also compute B/F = 4/3

DEFDmax / F

or (0.22)

F = O.75DEFDmax Combining (D.20)···([).22)we can write SErna..

(A - B)/ B

=

(A - B)/ DEFDrnax

SE.75rna;'

(E - F)/ F

=

(E - F)/O.75DEFDrnax

(4/3)(£ - F)I D£FDmax

(0.23)

Now define the saturation function (D.24)

which gives the approximate saturation for any £FD. Suppose we are given the numerical values of saturation at £FDmax and O.75EFDmax. These values are called SErnax and SE.75max respectively. Using these two saturation values, we compute the two unknowns A EX and B EX as follows. At E FD = EFDmax (D.25) and at E FD = O.75£FDrnax (D.26) We then solve (0.25) and (0.26) simultaneously to find A EX

(6)

=

S~.75max I S1max

(0.27)

Governor representation

Three types of governor representation are specified in this appendix: a general governor model that can be used for both steam and hydro turbines, a cross-compound governor model, and a hydraulic governor model. The appropriate model is identified by the letters G, C, and H in the tabulation. The governor block diagrams are given in Figures D.13--D.15. The regulation R is the steady-state regulation or droop and is usually factory set at 5~o for U.S, units.

564

Appendix D P

ma x

1 1 + T~S

1

~ P

e

Fig . D.I-' .

G eneral purpose go vernor block diagr am .

I

~

:: }--------------~

Fig. D .14.

C ross-co m po und go vern or blo ck dia gr am .

RP mO

J

VEl . min

I

Gote ser vo

.,

P ma x

l:C

0

'.

T S

w

p e

+

rig . D .15.

1-

1 + (T /2)


0

X·

0-

:J

(l)

-U -U

I'V

TA2

KA TA or TA J

TR

Name RR

VR type

EXCITER

D

E FDFL

SGI.2

SGJ.O

TF

WR

TqO Ta

TqO

TdO Til q Tq

TdO

Td

Td

Xo

x2

Ta x.{, or x p T2

xq

xd x" q x'q

xd

xd

Unit no. Rated MVA Rated kV Rated PF SCR

GENERATOR

pu

s s

(4) (4) s

(3)

(2) (2) (2)

n

s s s MW·s

S

s s s s s

pu pu pu pu pu pu pu pu pu pu pu

(I)

0.50 0.000 400.000 0.050 0.000

A WMA

2.000

· ..

1.120 1.920 ... 0.199 ... ... ... · .. ... ... 6.000 · .. · .. ... 1.500 · .. 992.00 ... 0.082 0.290

...

0.317 1.950

...

FI2

330.00 20.00 0.90 0.580

C

BRLS 0.50 0.000 400.000 0.020 0.000

FI3

384.00 24.00 0.85 0.580 0.260 0.324 1.798 0.255 1.051 1.778 0.0014 0.1930 0.0054 0.2374 0.1320 0.035 0.159 0.042 5.210 0.035 0.581 0.042 1.500 0.450 1006.50 0.1245 0.162 0.508 3.053 2.000

..

..

C

BRLS 0.50 0.000 400.000 0.020 0.000

0.2632 0.5351 2.7895 2.000

.

0.158 1.500 . .. 15Ut70

.. . . ..

0.2261 0.1346 . .. . .. 0.042 5.432

.

FI4

410.00 24.00 0.90 0.580 0.2284 0.2738 1.7668 0.2239 1.0104 1.7469 0.0019 0.1834

..

.

NA143A 0.50 0.000 50.000 0.060 0.000

A

0.060 0.470 0.150 1190.00 0.1357 0.0910 0.400 2.870 2.000

.

"

FI5

448.00 22.00 0.85 0.580 0.205 0.265 1.670 0.205 0.460 1.600 0.0043 0.150 0.023 0.175 0.140 0.023 1.070 0.032 3.700

..

..

..

G

ALTHYREX 1.50 0.000 200.000 0.3950 0.000

0.090 0.400 2.700 2.000

. ..

1347.20

. ..

. .. 0.480

.

3.800 . ..

.

. ..

.

.. . .. . .. . ..

0.470 1.650 0.004 0.160

.

FI6

512.00 24.00 0.90 0.580 0.200 0.270 1.700

Table 0.3 (continued)

.

BBC 0.50 0.000 30.000 0.400 0.000

A

. .. 1.230 . .. 3010.00 0.0711 0.111 0.518 3.000 2.000

. ..

. ..

3.650

"

0.013 0.167 0.112 0.030 0.550

...

FI7

552.00 24.00 0.90 0.580 0.198 0.258 1.780 0.172 0.247 1.770 0.0047

G

ALTHYREX 3.50 0.000 200.000 0.3575 0.000

FI8

590.00 22.00 0.95 0.500 0.215 0.280 2.110 0.215 0.490 2.020 0.0046 0.155 0.026 0.215 0.150 0.0225 . .. 0.032 4.200 0.0225 .. . 0.062 0.565 0.140 1368.00 0.1094 0.079 0.349 2.980 2.000

C WTA 2.00 0.000 400.000 0.020 0.000

0.134 0.617 3.670 2.000

FI9

835.00 20.00 0.90 0.500 0.339 0.413 2.183 0.332 1.285 2.157 0.0019 0.246 . .. 0.309 0.174 .. . . .. 0.041 5.690 . .. . .. 0.144 1.500 . .. 2206.40 .

G

ALTHYREX 2.50 0.000 250.000 0.200 0.000

0.090 0.402 3.330 2.000

0.160 2625.00

.,

F20

896.00 26.00 0.90 0.52 0.180 0.220 1.790 . .. 0.400 I. 715 0.001 0.135 0.019 0.135 0.130 0.035 0.596 0.032 4.300 0.035 0.298

A

BBC 0.50 0.000 50.000 0.060 0.000

0.340 1.120 3.670 2.000

2265.00

0.900

0.192 0.105 . .. . .. . .. 6.000

F21

911.00 26.00 0.90 0.64 0.193 0.266 2.040 0.191 0.262 1.960 0.001 0.154

~

w

Ul

CJ

x·

o;

;:j

(1)

» ~ ~

pu (4) pu (4) pu s (5) (5) (5) (5) pu(5) pu pu s s

TQ TQI TQI TQ2 TQ2 T()J TQ) ~Iim

PSS K Qv K QS

STABILIZER

T2 T) T4 TS F

P max TI

GOV R

pu

S

s

s s

s

(7) (7) (7) s s

(6)

S

s s s s

(6) (6) MW

TURBINE GOVERNOR

TFor TFI TF2

EFDmax EFDmin KF

B EX

AEX

SE.75max SEmax

KE TE

VR max VR min

...

... ... ...

"

.. .

... .

"

"

. .. . .. . . .. . .. . ..

.. . .. .

... .

. ..

. .. . .. ...

0.050 360.00 0.220 0.000 0.200 0.250 8.000 0.270

G

8.130 -8.130 1.000 0.812 0.459 0.656 0.1572 0.2909 4.910 0.000 0.060 1.000 0.000

.

"

0.050 8.000 0.250

OAOO

G

0.050 347.00 0.100 0.000

3JSIO -3.810 -0.170 0.950 0.220 0.950 0.0027 0.3857 4.890 -4.890 0.040 1.000 0.000

. ..

.

"

"

.

.

. ..

"

. .. . .. . .. . ..

. .. . ..

0.050 367.00 0.180 0.000 0.040 0.250 8.000 0.267

G

5.270 -5.270 1.000 0.920 0.435 0.600 0.1658 0.3910 3.290 0.000 0.030 1.000 0.000

S 0.000 4.000 10.000 0.230 0.020 0.230 0.020 0.000 0.000 0.100

G 0.050 390.00 0.100 0.000 0.300 0.050 10.000 0.150

1.000 -1.000 -0.0465 0.520 0.071 0.278 0.00\2 1.2639 4.320 -4.320 0.0832 1.000 0.000

S 0.000 26.000 3.000 0.150 0.050 0.150 0.050 0.000 0.000 0.050

G 0.050 460.00 0.150 0.050 0.300 0.160 8.000 0.270

3.840 -3.840 1.000 0.000 0.000 0.000 0.000 0.000 3.840 -3.840 0.0635 1.000 0.000

.

..

"

.

. ..

"

.

. .. . .. . .. . .. . .. . .. . ..

G

0.050 497.00 0.100 0.000 0.300 0.100 10.000 0.300

5.990 -5.990 -0.020 0.560 0.730 1.350 0.1154 0.5465 4.500 -4.500 0.050 1.300 0.000

S 0.000 24.400 3.000 0.150 0.050 0.150 0.050 0.000 0.000 0.050

0.050 553.00 0.080 0.000 0.150 0.050 10.000 0.280

G

5.730 - 5.730 1.000 0.000 0.000 0.000 0.000 0.000 5.730 - 5.730 0.0529 1.000 0.000

G

1.000 0.942 0.813 2.670 0.023 0.9475 5.020 0.000 0.030 1.000 0.000

1~.300

18.300

F 0.000 0.400 10.000 0.650 0.020 0.650 0.020 0.000 0.000 0.100

0.050 766.29 0.180 O.OJO 0.200 0.000 8.000 0.300

-

S

0.000 24.000 10.000 0.200 0.060 0.150 0.020 0.000 0.000 0.050

G 0.050 810.00 0.100 0.000 0.200 0.100 8.720 O.JOO

5.150 -5.150 1.000 0.000 0.000 0.000 0.000 0.000 5.150 -5.150 0.036 1.000 0.000

0.050 820.00 0.100 0.000 0.200 0.100 8.720 0.300

G

1.000 -1.000 -0.0393 0.440 0.064 0.235 0.00\3 1.1562 4.500 -4.500 0.070 1.000 0.000

0

0-

x·

:=J

Cl)

» ""0 ""0

tn

.......... ~

GENERATOR

VR type Name RR TR KA

EXCITER

D

SGI.2 £FDFL

'F SGI.O

Td Td TdO TdO Tq Tq TqO TqO Ta WR

Xo

x2

'2

'a x -t or x p

Unit no. Rated MVA Rated kV Rated PF SCR xd xd xd x"q x'q xq

(4) (4) s pu

MW·s 12 (2) (2) (2) (3)

S

s

S

s s s s s

s

pu pu pu pu pu pu pu pu pu pu pu

(I)

NAIOI 0.50 0.060 25.000

A

0.600 0.171 305.00 .. . 0.121 0.610 2.640 2.000

o.oso

CFI-HP 12S.00 13.80 0.85 0.64 0.171 0.232 1.680 0.171 0.320 1.610 0.0024 0.095 0.026 0.171 .. . 0.023 o.s 15 0.034 5.890 0.023 0.410

NAIOI 0.50 0.060 25.000

A

0.g5 0.64 0.250 0.369 1.660 0.250 0.565 1.590 0.003 0.140 0.020 0.250 ... 0.023 1.130 0.037 5.100 0.023 0.570 0.070 0.326 0.205 787.00 .. . 0.1122 0.433 2.640 2.000

u.so

CFI-LP 12g.00

Table D.4.

A

WMA 0.50 0.000 245.000

A

2.000

WMA 0.50 0.000 275.000

2.000

2J~40

A

2J~40

0.150 1.500 0.390 596.70 0.141 0.0982 0.4161

0.023 1.000 0.047 5.400 0.023 0.500 0.150 1.500 0.390 464.00 . .. 0.1249 0.500 2.570 2.000

.

...

WMA 0.50 0.000 245.000

A

., . 0.0905 0.345 2.500 2.000

141~.00

0.023 1.292 0.053 5.390 0.023 0.650 0.135 1.500 0.330

..

0.3~0

1.581 0.24g 0.955 1.531 0.0039 0.291 0.028 0.249

CF3-LP 221. 70 20.00 0.90 0.58 0.252

0.231 0.311 1.675 0.229 0.979 1.648 0.0043 0.304 . 0.029 0.229

0.5~

CF3-HP 27g.30 20.00 0.90

WMA 0.50 0.000 275.000

...

... 0.150 1.500 0.390 650.70 0.141 0.0982 0.4161

.

0.64 0.225 0.315 1.670 0.224 0.958 1.640 0.0036 0.186 0.028 0.224 0.101 . .. OJQO 0.043 5.000

0.~5

1~.00

CF2-LP 192.00

.

.. ..

"

CF2-HP 192.00 18.00 0.S5 0.64 0.225 0.315 1.670 0.224 0.958 1.640 0.0036 0.186 0.028 0.224 0.101 . 0.820 0.043 5.000

..

NAI43A 2.00 0.000 592.000

A

0.060 0.470 0.150 639.50 0.1357 0.0926 0.4139 2.730 2.000

.

CF4-HP 445.00 22.00 0.90 0.64 0.205 0.260 1.650 0.205 0.460 1.590 0.0043 0.150 0.022 0.175 0.140 0.020 ., . 0.032 4.800 0.020

o.tso

NA 143A 2.00 0.000 312.000

A

0.250 1.500 0.181 0.440 1.400 0.0045 0.140 0.022 0.145 0.135 0.020 . .. 0.036 g.OOO 0.020 . .. 0.070 0.410 0.110 3383.50 0.3958 0.1333 0.5555 2.560 2.000

CF4-LP 375.00 22.00 0.90 0.64

Typical Data for Cross-Compound Fossil Steam (C F) Units

r.soo

ALTHYREX 2.50 0.000 250.000

G

0.220 0.490 1.720 0.0027 0.160 0.025 0.220 0.150 0.023 0.586 0.032 3.700 0.023 0.293 0.060 0.480 0.150 633.00 0.1259 0.0866 0.410 2.900 2.000

CF5-IIP 4g3.00 22.00 0.90 0.604 0.220 0.2gS

G ALTHYREX 2.50 0.000 250.000

CF5-LP 426.00 22.00 0.90 0.645 0.205 0.285 1.750 0.205 0.485 1.580 0.0036 0.155 0.025 0.205 0.150 0.023 1.360 0.035 8.400 0.023 0.680 0.070 0.460 0.110 2539.00 0.343 0.177 0.532 2.915 2.000

»

'-J t1't

t1't

0

x·

a...

::J

(1)

-0 -0

s s pu(4) pu(4) pu s (5) (5) (5) (5) pu(5) pu pu s s

~Iim

K QS TQ TQI TQt TQ2 TQ2 TQ3 TQ3

PSS KQv

STABILIZER

F

T4 TS

T3

T2

Tl

P max

GOV R

(7) (7) (7) s s s s s s s pu

(6)

S

S

(6) (6) MW s s s

TURBINE GOVERNOR

TF2

TFor TFI

EFDmax EFDmin KF

SEX

SE.7Smax SEmax AEX

TE

V R min KE

TA2 V R max

TA or TAl

S

0.000 12.000 10.000 1.000 0.020 0.750 0.020 0.000 0.000 0.050

G 0.050 107.50 0.100 0.000 0.150 0.300 10.000 0.606

0.200 0.000 1.000 -1.000 -0.051 0.5685 0.0778 0.3035 0.0013 1.3750 3.960 -3.960 0.091 0.350 0.000

S

0.000 8.000 10.000 1.000 0.020 0.250 0.020 0.000 0.000 0.050

G 0.050 107.50 0.100 0.000 0.150 0.300 10.000 0.000

0.200 0.000 1.000 -1.000 -0.051 0.5685 0.0778 0.3035 0.0013 1.3750 3.960 -3.960 0.091 0.350 0.000

o.oso

F 0.000 0.600 10.000 0.490 0.020 . 0.490 0.020 0.000 0.000

G

0.050 172.50 0.100 0.000 0.150 0.300 4.160 0.560

0.060 0.000 0.984 -0.984 -0.0667 1.230 0.1688 0.2978 0.0307 0.5331 4.260 -4.260 0.033 0.330 0.000

F 0.000 0.600 10.000 0.455 0.020 0.455 0.020 0.000 0.000 0.080

a

0.050 172.50 0.100 0.000 0.150 0.300 4.160 0.000

0.060 0.000 0.984 -0.984 -0.0667 1.230 0.1688 0.2978 0.0307 0.5331 4.260 -4.260 0.033 0.330 0.000

S 0.000 10.000 10.000 0.250 0.020 0.400 0.020 0.000 0.000 0.050

G 0.050 267.00 0.250 0.000 0.000 0.050 12.000 0.549

0.050 0.000 2.780 - 2.780 -0.170 1.370 0.220 0.950 0.0027 1.639 3.570 - 3.570 0.040 1.000 0.000

0.000 10.000 10.000 0.700 0.020 0.450 0.020 0.000 0.000 0.050

S

G 0.050 213.00 0.250 0.000 0.000 0.300 12.000 0.000

0.050 0.000 2.780 -2.780 -0:170 1.370 0.220 0.950 0.0027 1.639 3.570 - 3.570 0.040 1.000 0.000

F 0.000 1.170 10.000 0.265 0.020 0.265 0.020 0.000 0.000 0.060

G 0.050 411.00 0.100 0.000 0.200 0.100 8.720 0.540

0.053 0.000 13.050 - 13.050 -0.591 0.512 1.094 3.048 0.0506 0.7719 5.310 -5.310 0.070 1.880 0.000

F 0.000 1.170 10.000 0.640 0.020 0.640 0.020 0.000 0.000 0.080

0.050 339.00 0.100 0.000 0.200 0.100 8.720 0.000

G

0.050 0.000 10.770 -10.770 -0.4035 1.080 0.647 2.545 0.0106 1.0891 5.030 -5.030 0.090 2.250 0.000

O.OdO 24.000 10.000 0.200 0.050 0.200 0.020 0.000 0.000 0.050

S

G 0.050 436.00 0.100 0.000 0.300 0.050 14.000 0.580

0.140 0.000 5.150 -5.150 1.000 0.000 0.000 0.000 0.000 0.000 5.150 -5.150 0.062 1.000 0.000

S

0.000 24.000 10.000 0.200 0.070 0.300 0.020 0.000 0.000 0.050

0.050 382.00 0.100 0.000 0.300 0.050 14.000 0.000

G

0.060 0.000 4.910 -4.910 1.000 0.000 0.000 0.000 0.000 0.000 4.910 -4.910 0.025 1.000 0.000

CJ

x·

CD

0....

:::s

» -0 -0

0..

"'.I

Ul

577

Appendix D Table D.5. Typical Data for Nuclear Steam (N) Units GENERATOR Unit no. Rated MVA Rated kV Rated PF SCR

xd xd xd til

'q

xq

xq

'a x-t or x p

'2

x2

Xo Td Td

(I)

pu pu pu pu pu pu pu pu pu pu pu

s

NI 76.g0 13.M 0.g5 0.650 0.190 0.320 1.660 0.120 0.470 1.5HO

0.150 0.125 0.450

0.032 4.7g0

'dO 'dO Tq

N2 145.5 14.4 0.g5 0.640 0.210 0.320 1.710 0.210 0.510 1.630 0.0032 0.125 0.025 0.160 0.110 0.230 0.03H 7.100

T

q TqO

TqO ra ",'

R

'F SGI.O SGI.2

EFDFL D

MW·s 12 (2) (2) (2) (3)

2SI.70 0.OSS7 0.3244 2.5X7 2.000

0.073 0.3XO 0.210 1136.00 0.217 0.1309 0.5331 2.730

IV3 500.00 Ig.OO 0.90 0.5~m

0.2g3 0.444 1.7S2 0.277 1.201 1.739 0.0041 0.275 0.029 0.2XO 0.035 1.512 0.055 6.070 0.035 0.756 0.152 1.500 0.310 1990.00 0.0900 0.3520 2.710 2.000

N4 920.35 Ig.OO 0.90 0.607 0.275 0.355 1.790 0.275 0.570 1.660 0.004g 0.215 0.02g 0.230 0.195

0.032 7.900

0.055 0.41 0.19 3464.00 0.0901 O.Og 16 0.3933 2.X70

N5

N6

N7

1070.00 22.00 0.90 0.500 0.312 0.467 1.933

12HO.00 22.00 0.95 0.500 0.237 0.35M 2.020 0.237 0.565

1300.00 25.00 0.90 0.4g0 0.315 0.467 2.129 OJOg 1.270 2.074 0.0029 0.151

1.144 1.743 0.360

0.2g4

6.660

3312.00

2.000

i.seo

0.0019 0.205 0.029 0.215 0.195

0.034 9.100

lVg 1340.00 25.00 0.90 0.4g0 0.2MI OJ46 1.693 0.2S I 0.991 1.636 0.0011 0.22g 0.22g

0.052 6.120

0.043 6.5XO

0.144 1.500

0.124 1.500

0.059 0.460 O.IKO 4690.00 0.0979 0.0779 0.3055 2.945 2.000

45S0.00 0.0576 0.0714 0.3100 3J40 2.000

469S.00 0.0576 0.0769 0.4\00 2.70K 2.000

A EA210 1.50 0.000 50.000 0.020 0.000 1.000 -1.000 -0.0244 0.1455 0.Og63 0.2148 0.0056 0.6g18 5.350 0.000 0.0233 0.7750 0.000

C BRLS 2.23 0.000 400.000 0.020 0.000 6.960 -6.960 1.000 0.015 0.3400 0.5600 0.0761 0.4475 4.460 0.000 0.040 0.050 0.000

C BRLS 2.00 0.000 400.000 0.020 O.OUO 6.020 -6.020 1.000 0.015 0.3900 0.5630 0.1296 0.3814 3.850 0.000 0.040 0.050 0.000

EXCITER VR type Name RR TR

KA TA

A

(4) (4)

pu

or

TAl

TA 2

s

V R min

pu (4) pu (4) pu

SE.75 max SEmax A EX

(5) (5) (5) (5)

V R mu

KE TE

B EX EFD max EFDmin KF

TFor TFI r F2

s

pu (5) pu pu

NAIOI 0.50 0.060 25.000 0.200 0.000 1.000 -1.000 -0.0516 0.579 0.0794 0.3093 0.0013 1.4015 3.881 - 3.881 0.093 0.350 0.000

A NAIOI 0.50 0.060 25.000 0.200 0.000 1.000 -1.000 -0.04g9 0.550 0.0752 0.2932 0.0016 1.6120 4.090 -4.090 0.088 0.350 0.000

A WMA 0.50 0.000 256.000 0.050 0.000 2.S58 - 2.g58 -0.170 2.150 0.2200 0.9500 0.0027 1.5966 3.665 - 3.665 0.040 1.000 0.000

A NAI43 0.50 0.000 25.000 0.200 0.000 1.000 -1.000 -0.0464 0.522 0.0714 0.27g4 0.0016 1.5330 4.310 -4.310 0.084 1.000 0.000

C BRLS 2.00 0.000 400.000 0.020 0.000 10.650 - 10.650 1.000 1.000 0.375 1.220

4.800 0.000 0.060 1.000 0.000

Appendix D

578

Table D.5. (continued) TURBINE GOVERNOR

(6) (6) MW

GOV

R

r-:

G

G

0.050 65.00

0.050 208.675

11

T2 T3

14 TS

(6)

F

G 0.050 450.00 0.250 0.000 0.000 0.300 5.000 0.320

G

0.050 790.18

0.030 0.100 0.200 6.280 0.330

G 0.050 1216.00 0.150 0.000 0.210 0.814 2.460 0.340

G 0.050 1090.00 0.180 0.000 0.040 0.200 5.000 0.300

0.050 1205.00 O.IHO 0.000 0.040 0.200 5.000 0.300

F 0.000 10.000 10.000 0.080 0.020 0.080 0.020 0.000 0.000 0.100

S 0.000 1.530 3.000 0.150 0.050 0.150 0.050 0.000 0.000 0.100

F 0.000 20.000 10.000 0.300 0.020 0.000 0.000 0.000 0.000 0.100

F 0.000 20.000 10.000 0.300 0.020 0.000 0.000 0.000 0.000 0.100

G

0.050 951.00 O.H~O

G

STABILIZER (7) (7)

PSS K Qv K Qs

TQ TOI TQI T02 TQ2 T03 1Q3

~Iim

(7)

'

',~

pu

S 0.000 0.200 10.000 1.330 0.020 1.330 0.020 0.000 . 0.000 0.100

579

Appendix D Table D.6. Typical Data for Synchronous Condensor (SC) Units GENERATOR Unit no. Rated MVA Rated kV Rated PF SCR x:J Xd Xd

x"q x'q

xq ra

x,(, or x p '2 X2 Xo

SCI

(1) pu pu pu pu pu pu pu pu pu pu pu

s

Td Td T" dO TdO T;'

T~

,

T;O TqO Ta

WR

'F SGl.O

SGl.2

E FDFL

D

s s s

25.00 13.80 0.00 0.2035 0.304 1.769 0.199 0.5795 0.855 0.0025 0.1045 0.0071 0.177 0.115

SC2 40.00 13.80 0.00 0.558 0.231 0.343 2.373

1.172 1.172 0.132

0.035

0.0525 8.000

0.058 11.600

0.0151

0.201

SC3 50.00 12.70 0.00 1.004 0.141 0.244 1.083 0.170 0.720 0.720 0.006

0.160 0.058 0.041 0.858 0.050 6.000

SC4 60.00 13.80 0.00 0.477 0.257 0.385 2.476 0.261 1.180 1.180 0.0024 0.146

0.225 0.165 0.035 0.058 12.350

S

s s s

S

MW·s

n

(2) (2) (2) (3)

30.00 0.4407 0.304 0.666 3.560

0.159 60.80 0.295 0.776 4.180

0.150 0.200 105.00 0.0631 0.0873 0.310 2.338

SC5 75.00 13.80 0.00 0.800 0.170 0.320 1.560 0.200 1.000 1.000 0.0017 0.0987 0.180 0.185 0.128 0.041 3.230 0.039 16.000 0.0473

0.188

0.235

0.290 60.60 0.274 0.180 0.708 4.224

0.288 89.98 0.279 0.150 0.500 3.730

EXCITER

VR type Name RR

(4) (4)

TR

S

KA TA or TAl TA2

VR malt VR min KE TE SE.75max SEmax

A EX HEX

EFDmax EFDmin

KF TF

or TFI

TF2

pu

s

s pu (4) pu (4) pu s (5) (5) (5) (5) pu (5) pu pu s S

A

WMA 0.50 0.000 400.000 0.050 0.000 4.407 -4.407 -0.170 0.950 0.220 0.950 0.0027 1.0356 5.650 - 5.650 0.040 1.000 0.000

A

WMA 1.00 0.000 400.000 0.050 0.000 6.630 - 6.630 -0.170 0.950 0.220 0.950 0.0027 0.6884 8.500 -8.500 0.040 1.000 0.000

A

3.85 0.000 200.000 0.050 0.000 11.540 - 11.540 -0.170 1.000 0.220 0.950 0.0027 0.3956 14.790 -14.790 0.070 1.000 0.000

A

WMA 1.00 0.000 400.000 0.050 0.000 5.850 -5.850 -0.170 0.950 0.220 0.950 0.0027 0.7802 7.500 -7.500 0.040 1.000 0.000

A

NA143 2.00 0.000 18.000 0.200 0.000 1.000 -1.000 -0.0138 0.0669 0.0634 0.1512 0.0047 0.4782 7.270 -7.270 0.0153 1.000 0.000

580

AppendixD

Table D.7.

Typical Data for Combustion Turbine (CT) Units EXCITER

GENERATOR Unit no. Rated MVA Rated kV Rated PF SCR

xd xd xd X"q

x'q xq To

X{ or

xp

'2

xl

Xo

T"d

Td TdO TdO T"q T

q

T;;O

T;'O To

WR 'F 5 G 1.O

5 G I. 2

EiDFL D

(I)

pu pu pu pu pu pu pu pu pu pu pu s s s s S S

s s s

MW·s

u

(2) (2) (2) (3)

CTI 20.65 13.80 0.85 0.580 0.155 0.225 1.850

...

...

1.740

. .. .. . . ..

.. .

... ... ...

...

4.610

...

...

...

... ...

183.30

... " "

. .

2.640

...

CT2 62.50 13.80 0.85 0.580 0.102 0.159 1.640 0.100 0.306 1.575 0.034 0.113 0.352 0.102 0.051 0.035 0.730 0.054 7.500 0.035 0.188 0.107 1.500 0.350 713.50 0.261 0.0870 0.2681 2.4348 2.000

VR type Name RR TR KA TA or TAt TA2 VRmax VR min KE TE SE.7Smax 5 Emax AEX

HEX

E FDmax EFDmin KF TFor TFt TF2

Kp

D SCPT

(4) (4)

...

s

0.000 120.000 0.050 0.000 1.200 -1.200 1.000 0.500

pu

s

s pu (4) pu (4) pu s (5) (5) (5) (5) pu (5) pu pu s

...

... ...

...

. .. 0.020 0.461

...

s

C BRLS 0.50 0.000 400.000 0.020 0.000 7.300 -7.300 1.000 0.253 0.500 0.860 0.0983 0.2972 7.300 0.000 0.030 1.000 0.000

1.19 2.32

K/' TURBINE GOVERNOR GOV R

(6) (6)

Tt T2

s

T3

s

74

s s

P max

TS

F

MW

s

(6)

Fuel:

G 0.050 17.55 0.000 0.000

G 0.040 82.00 0.500 1.250

Oil Gas 0.025 0.100

0.700

0.000 0.000 0.025 0.100 0.0 0.5

0.700 0.000 1.000

3 4

4

2 3

I

2

I I I

1 1

I

69 115 138 161 230 345 345 500 500 500 500 735 735

(kV)

Conductors per phase @ 18 in. spacing

Line-toline voltage

12 14 16 18 22 28 28 38 38 38 38 56 56

226.8 336.4 397.5 477.0 556.5 ( 1.750) (1.246) (2.500) ( 1.602) ( 1.165) (0.914) ( 1.750) ( 1.382)

15.1 17.6 20.1 22.7 27.7 35.3 35.3 47.9 47.9 47.9 47.9 70.6 70.6

(ft)

(ft)

0.465 0.451 0.441 0.430 0.420 0.3336 0.1677 0.2922 0.1529 0.0988 0.0584 0.0784 0.0456

xQ

0.3294 0.3480 0.3641 0.3789 0.4030 0.4325 0.4325 0.4694 0.4694 0.4694 0.4694 0.5166 0.5166

xd

+ xd

0.7944 0.7990 0.8051 0.8089 0.8230 0.7761 0.6002 0.7616 0.6223 0.5682 0.5278 0.5950 0.5622

xa

60-Hz inductive reactance Ujmi

Typical 60-Hz Transmission Line Data

Geometric mean distance

Table 0.8. Flat phase spacing

ACSR Conductor area (ordiam) kCM(in.) 0.1074 0.1039 0.1015 0.0988 0.0965 0.0777 0.0379 0.0671 0.0341 0.0219 0.0126 0.0179 0.0096

x~

0.0805 0.0851 0.0890 0.0926 0.0985 0.1057 0.1057 0.1147 0.1147 0.1147 0.1147 0.1263 0.1263

xd

+ xd 0.1879 0.1890 0.1905 0.1914 0.1950 0.1834 0.1436 0.1818 0.1488 0.1366 0.1273 0.1442 0.1359

x~

60-Hz capacitive reactance MU·mi

386.4 388.6 391.6 393.5 400.6 374.8 293.6 372.1 304.3 278.6 259.2 292.9 276.4

(n)

=0

Surge impedance

12 34 49 66 132 318 405 672 822 897 965 1844 1955

Surge impedance loading (MVA)

01

co

0

x·

:J CL

(1)

» -0 -0

appendix

E

Excitation Control System Definitions

There are two important recently published documents dealing with excitation control system definitions. The first [I) appeared in 1961 under the title "Proposed excitation system definitions for synchronous machines" and provided many definitions of basic system elements. The second report (2] was published in 1969 under the same title and, using the first report as a starting point, added the new definitions required by technological change and attempted to make all definitions agree with accepted language of the automatic control community. The definitions that follow are those proposed by the 1969 report. J Reference is also made to the definitions given in ANSI Standard C42.10 on rotating machines [3], ANSI Standard C85.1 on automatic control [4], and the supplement to e8S.1 [5]. Finally, reference is made to the IEEE Committee Report "Cornpu~er representation of excitation systems" [6], which defines certain time constants and gain factors used in excitation control systems.

Proposed IEEE Definitions 1.0

Systems

1.0J Control system, feedback. A control system which operates to achieve prescribed relationships between selected system variables by comparing functions of these variables and using the difference to effect control. 1.02 Control system, automatic feedback.

erates without human intervention.

A feedback control system which op-

The source of field current for the excitation of a synchronous machine and includes the exciter, regulator, and manual control. 1.03 Excitation system [I, definition 4].

1.04 Excitation control system (new). A feedback control system which includes the synchronous machine and its excitation system. 1.05 High initial response excitation system (new). An excitation system having an excitation system voltage response time of 0.1 second or less.

I. @ IEEE. Reprinted with permission from IEEE Trans., vol. PAS-88, 1969.

582

Appendix E 2.0

583

Components

2.01 Adjuster [1, definition 40]. An element or group of elements associated with a feedback control system by which adjustment of the level of a controlled variable can be made. 2.02 Amplifier. A device whose output is an enlarged reproduction of the essential features of an input signal and which draws power therefore from a source other than the input signal. .

A feedback element of the regulator which acts to compensate for the effect of a variable by modifying the function of the primary detecting element. 2.03 Compensator [I, definition 44].

Notes: I. Examples are reactive current compensator and active current compensator. A reactive current compensator is a compensator that acts to modify the functioning of a voltage regulator in accordance with reactive current. An active current compensator is a compensator that acts to modify the functioning of a voltage regulator in accordance with active current. 2. Historically, terms such as "equalizing reactor" and "cross-current compensator" have been used to describe the function of a reactive compensator. These terms are depreca ted. 3. Reactive compensators are generally applied with generator voltage regulators to obtain reactive current sharing among generators operating in parallel. They function in the following two ways. a. Reactive droop compensation is the more common method. It creates a droop in generator voltage proportional to reactive current and equivalent to that which would be produced by the insertion of a reactor between the generator terminals and the paralleling point. b. Reactive differential compensation is used where droop in generator voltage is not wanted. It is obtained by a series differential connection of the various generator current transformer secondaries and reactive compensators. The difference current for any generator from the common series current creates a compensating voltage in the input to the particular generator voltage regulator that acts to modify the generator excitation to reduce to minimum (zero) its differential reactive current. 4. Line drop compensators modify generator voltage by regulator action to compensate for the impedance drop from the machine terminals to a fixed point. Action is accomplished by insertion within the regulator input circuit of a voltage equivalent to the impedance drop. The voltage drops of the resistance and reactance portions of the impedance are obtained respectively in pu quantities by an "active compensator" and a "reactive compensator." 2.04 Control, manual (new). Those elements in the excitation control system which provide for manual adjustment of the synchronous machine terminal voltage by open loop (human element) control. 2.05 Elements, feedback. Those elements in the controlling system which change the feedback signal in response to the directly controlled variable.

584

Appendix E

2.06 Elements, forward. Those elements situated between the actuating signal and the controlled variable in the closed loop being considered. 2.07 Element, primary detecting. That portion of the feedback elements which first either utilizes or transforms energy from the controlled medium to produce a signal which is a function of the value of the directly controlled variable. 2.08 Exciter [I, definition 5]. The source of all or part of the field current for the excitation of an electric machine. 2.09 Exciter, main [1, definition 5]. The source of all or part of the field current for the excitation of an electric machine, exclusive of another exciter. 2.09.1 DC generator commutator exciter. An exciter whose energy is derived from a de generator. The exciter includes a de generator with its commutator and brushes. It is exclusive of input control elements. The exciter may be driven by a motor, prime mover, or the shaft of the synchronous machine. 2.09.2 Alternator rectifier exciter. An exciter whose energy is derived from an alternator and converted to de by rectifiers. The exciter includes an alternator and power rectifiers which may be either noncontrolled or controlled, including gate circuitry. It is exclusive of input control elements. The alternator may be driven by a motor, prime mover, or by the shaft of the synchronous machine. The rectifiers may be stationary or rotating with the alternator shaft. 2.09.3 Compound rectifier exciter. An exciter whose energy is derived from the currents and potentials of the ac terminals of the synchronous machine and converted to de by rectifiers. The exciter includes the power transformers (current and potential), power reactor, power rectifiers which may be either noncontrolJed or controlled, including gate circuitry. It is exclusive of input control elements. 2.09.4 Potential source rectifier exciter. An exciter whose energy is derived from a stationary ac potential source and converted to dc by rectifiers. The exciter includes the power potential transformers, where used, power rectifiers which may be either noncontrolled or controlled, including gate circuitry. It is exclusive of input control elements. 2.10 Exciter, pilot [1, definition 7]. The source of all or part of the field current for the excitation of another exciter. 2.11 Limiter [I, definition 43]. A feedback element of the excitation system which acts to limit a variable by modifying or replacing the function of the primary detector element when predetermined conditions have been reached. 2.12 Regulator, synchronous machine [I, definition 8]. A synchronous machine regulator couples the output variables of the synchronous machine to the input of the exciter through feedback and forward controlling elements for the purpose of regulating the synchronous machine output variables. Note: In general, the regulator is assumed to consist of an error detector, preamplifier, power amplifier, stabilizers, auxiliary inputs, and limiters. As shown in Figure 7.20, these regulator components are assumed to be self-explanatory, and a given regulator may not have all the items included. Functional regulator definitions describing types of regulators are listed below. The term "dynamic-type" regulator has been omitted as a classification [I, Definition 15].

Appendix E

585

2.12.1 Continuously acting regulator [1, definition 10). One that initiates a corrective action for a sustained infinitesimal change in the controlled variable. 2.12.2 Noncontinuously acting regulator [1, definition II]. One that requires a sustained finite change in the controlled variable to initiate corrective action. 2.12.3 Rheostatic type regulator [I, definition 12J. lating function by mechanically varying a resistance.

One that accomplishes the regu-

Note [l , Definitions 13, 14]: Historically, rheostatic type regulators have been further defined as direct-acting and indirect-acting. An indirect-acting type of regulator is a rheostatic type that controls the excitation of the exciter by acting on an intermediate device not considered part of the regulator or exciter. A direct-acting type of regulator is a rheostatic type that directly controls the excitation of an exciter by varying the input to the exciter field circuit. 2.13 Stabilizer, excitation control system (new). An element or group of elements which modifies the forward signal by either series or feedback compensation to improve the dynamic performance of the excitation control system. 2.14 Stabilizer, power system (new). An element or group of elements which provides an additional input to the regulator to improve power system dynamic performance. A number of different quantities may be used as input to the power system stabilizer such as shaft speed, frequency, synchronous machine electrical power and other. 3.0

Characteristics and performance

3.01 Accuracy, excitation control system (new). The degree of correspondence between the controlled variable and the ideal value under specified conditions such as load changes, ambient temperature, humidity, frequency, and supply voltage variations. Quantitatively, it is expressed as the ratio of difference between the controlled variable and the ideal value. 3.02 Air gap Line. The extended straight line part of the no-load saturation curve. 3.03 Ceiling voltage, excitation system [I, definition 26]. The maximum dc component system output voltage that is able to be attained by an excitation system under specified conditions. 3.04 Ceiling voltage, exciter [1, definition 24]. Exciter ceiling voltage is the maximum voltage that may be attained by an exciter under specified conditions. 3.05 Ceiling voltage, exciter nominal [1, definition 25]. Nominal exciter ceiling voltage is the ceiling voltage of an exciter loaded with a resistor having an ohmic value equal to the resistance of the field winding to be excited and with this field winding at a temperature of I. 75°C for field windings designed to operate at rating with a temperature rise of

60°C or less. 2. 100°C for field windings designed to operate at rating with a temperature rise greater than 60°C. 3.06 Compensation. A modifying or supplementary action (also, the effect of such action) intended to improve performance with respect to some specified characteristics. Note: In control usage this characteristic is usually the system deviation. Compensa-

Appendix E

586

tion is frequently qualified as "series," "parallel,' "feedback,' etc., to indicate the relative position of the compensating element. 3.07 Deviation, system. The instantaneous value of the ultimately controlled variable minus the command. 3.08 Deviation, transient. The instantaneous value of the ultimately controlled variable minus its steady-state value. 3.09 Disturbance. An undesired variable applied to a system which tends to affect adversely the value of a controlled variable. 3.10 Duty, excitation system (new). Those voltage and current loadings imposed by the synchronous machine upon the excitation system including short circuits and all conditions of loading. The duty cycle will include the action of limiting devices to maintain synchronous machine loading at or below that defined by ANSI C50.13-1965. 3.11 Duty, excitation system (new). An initial operating condition and a subsequent sequence of events of specified duration to which the excitation system will be exposed.

Note: The duty cycle usually involves a three-phase fault of specified duration located electrically close to the synchronous generator. Its primary purpose is to specify the duty that the excitation system components can withstand without incurring maloperation or specified damage. 3.12 Drift [1, definition 36]. An undesired change in output over a period of time, which change is unrelated to input, environment, or load.

Note: The change is a plus or minus variation of short periods that may be superimposed on plus or minus variations of a long time period. On a practical system, drift is determined as the change in output over a specified time with fixed command and fixed load, with specified environmental conditions. 3.13 Dynamic. Referring to a state in which one or more quantities exhibit appreciable change within an arbitrarily short time interval. 3.14 Error.. An indicated value minus an accepted standard value, or true value.

Note: ANSf C85 deprecates use of the term as the negative of deviation. accuracy, precision in ANSI C8S.I.

See also

3.15 Excitation system voltage response [I, definition 21]. The rate of increase or decrease of the excitation system output voltage determined from the excitation system voltage-time response curve, which rate if maintained constant, would develop the same voltage-time area as obtained from the curve for a specified period. The starting point for determining the rate of voltage change shall be the initial value of the excitation system voltage time response curve. Referring to Fig. E-l, the excitation system voltage response is illustrated by line ac. This line is determined by establishing the area acd equal to area abd. Notes: I. Similar definitions can be applied to the excitation system major components such as the exciter and regulator. 2. A system having an excitation system voltage response time of 0.1 s or less is defined as a high initial response excitation system (Definition 1.05).

587

Appendix E

I

Ib Ie

V

//

,..... /

,...../j

I

I I

I

I /_/_ _ _ _ _ _ _ _ _ _ _ _ Id ~

E'"

a

(;

Resporse ratio =

>

ce -

00

(oo)(oe)

(Def. 3 . 18)

Where oe = 0 . 5 . 0 0 ;;: Synch ronous maeh i ne rate d load fie ld voltage

(Del. 3.21)

o F ig. E.I .

Time , s

Exc iter o r synchronou s machine excitat ion system vo lta ge response (Def. 3.15).

3.16 Excitation system voltage response time (new). The time in seconds for the excitation voltage to reach 95 percent of ceiling voltage under specified conditions. 3.17 Excitation system voltage time response [I, definition 19]. The excitation system output voltage expressed as a function of time. under specified condit ions . Note: A similar definition can be applied to the excitation system major components: the exciter and regulator sepa ra tely.

3,18 Excitation system voltage response ratio [I, definition 23]. The numerical value which is obtained when the excitation system voltage response in volts per second, measured over the first half-second interval unless otherwise specified, is divided by the rated-load field voltage of the synchronous machine. Unless otherwise specified, the excitation system voltage response ratio shall apply only to the increase in excitation system voltage. Referring to Fig. E.I the excitation system voltage response ratio = (ce - ao)/(ao)(oe) , where ao = synchronous machine rated load field voltage (Definition 3.21) and oe = 0.5 second, unless otherwise specified . 3.19 Exciter main response ratio; formerly nominal exciter response. The main exciter response ratio is the numerical value obtained when the response, in volts per second, is divided by the rated-load field voltage; which response, if maintained constant, would develop, in one half-second, the same excitation voltage-time area as attained by the actual exciter. Note: The response is determined with no load on the exciter, with the exciter voltage initially equal to the rated-load field voltage, and then suddenly establishing circuit conditions that would be used to obtain nominal exciter ceiling voltage. For a rotating exciter, response should be determ ined at rated speed . This definition does not apply to main exciters having one or more series fields (except a light differential series field) nor to electronic exciters.

Appendix E

588

3.20 Field voltage, base (new). The synchronous machine field voltage required to produce rated voltage on the air gap line of the synchronous machine at field temperatures. I. 75°C for field windings designed to operate at rating with a temperature rise of 60°C or less. 2. 100°C for field windings designed to operate at rating with a temperature rise greater then 60°C.

Note: This defines one pu excitation system voltage for use in computer representation of excitation systems [6]. 3.21 Field voltage, rated-load [I, definition 38]; formerly nominal collector ring volt-

Rated-load field voltage is the voltage required across the terminals of the field winding or an electric machine under rated continuous-load conditions with the field winding at one of the following. age.

I. 75°C for field windings designed to operate at rating with a temperature rise of 60°C or less.

2. 100°C for field windings designed to operate at rating with a temperature rise greater than 60°C.

3.22 Field voltage, no-load [I, definition 39]. No-load field voltage is the voltage required across the terminals uf the field winding of an electric machine under conditions of no load, rated speed, and terminal voltage and with the field winding at 25°C. 3.23 Gain, proportional. The ratio of the change in output due to proportional control action to the change in input. Illustration: Y = ±PX where P = proportional gain, X = input transform, and Y = output transform. 3.24 Limiting. The intentional imposition or inherent existence of a boundary on the range of a variable, e.g., on the speed of a motor. 3.25 Regulation, load. The decrease of controlled variable (usually speed or voltage) from no load to full load (or other specified limits). 3.26 Regulated voltage, band of (I, definition 37]. Band of regulated voltage is the band or zone, expressed in percent of the rated value of the regulated voltage, within which the excitation system will hold the regulated voltage of an electric machine during steady or gradually changing conditions over a specified range of load. 3.27 Regulated voltage, nominal band of. Nominal band of regulated voltage is the band of regulated voltage for a load range between any load requiring no-load field voltage and any load requiring rated-load field voltage with any compensating means used to produce a deliberate change in regulated voltage inoperative. 3.28 Signal, actuating. ure 7.19).

The reference input signal minus the feedback signal (Fig-

3.29 Signal, error. In a closed loop, the signal resulting from subtracting a particular return signal from its corresponding input signal (Figure 7.19). 3.30 Signal, feedback. signal (Figure 7.19). 3.31 Signal, input. 3.32 Signal, output.

That return signal which results from the reference input

A signal applied to a system or element. A signal delivered by a systern or element.

Appendix E 3.33 Signal, rate (new).

input signal.

589

A signal that is responsive to the rate of change of an

3.34 Signal, reference input. One external to a control loop which serves as the standard of comparison for the directly controlled variable.

3.35 Signal, return. In a closed loop, the signal resulting from a particular input signal, and transmitted by the loop and to be subtracted from that input signal, 3.36 Stability. For a feedback control system or element, the property such that its output is asymptotic, i.e., will ultimately attain a steady-state, within the linear range and without continuing external stimuli. For certain nonlinear systems or elements, the property that the output remains bounded, e.g., in a limit cycle of continued oscillation, when the input is bounded. 3.37 Stability limit. A condition of a linear system or one of its parameters which places the system on the verge of instability. 3.38 Stability, excitation system. The ability of the excitation system to control the field voltage of the principal electric machine so that transient changes in the regulated voltage are effectively suppressed and sustained oscillations in the regulated voltage are not produced by the excitation system during steady-load conditions or following a change to a new steady-load condition. Note: It should be recognized that under some system conditions it may be necessary to use power system stabilizing signals as additional inputs to excitation control systems to achieve stability of the power system including the excitation system. 3.39 Steady state. That in which some specified characteristic of a condition, such as value, rate, periodicity, or amplitude, exhibits only negligible change over an arbitrarily long interval of time. Note: It may describe a condition in which some characteristics are static, others dynamic. 3.40 Transient. In a variable observed during transition from one steady-state operating condition to another that part of the variation which ultimately disappears.

Note: ANSI C85 deprecates using the term to mean the total variable during the transition between two steady states. 3.41 Variable, directly controlled.

sensed to originate a feedback signal.

In a control loop, that variable whose value is

References I. AlEE Committee Report. Proposed excitation system definitions for synchronous machines. AlEE Trans. PAS-80:173-180, 1961. 2. IEEE Committee Report. Proposed excitation system definitions for synchronous machines. IEEE Trans. PAS-88: 1248-58, 1969. 3. ANSI Standard C42.10. Definitions of electrical terms, rotating machinery (group 10). American National Standards Institute, New York, 1957. 4. ANSI Standard C85.1-1963. Terminology for automatic control. American National Standards Institute, New York, 1963. 5. ANSI Standard C85.1a-1966. Supplement to terminology for automatic control C85.1-1963. American National Standards Institute, New York, 1963. 6. IEEE Committee Report. Computer representation of excitation systems. IEEE Trans. PAS-87:146064, 1968.

appendix

F

Control System Components

The electrical engineer is usually acquainted with common control system components used in all-electric or electromechanical systems. Our goal here is to introduce mechanical and hydraulic components and, in some cases, to compare these with electric components that perform a similar function. * The purpose for doing this is to enable one to recognize basic functions such as summation, integration, differentiation, and amplification when performed either electrically or mechanically. Such familiarity is an obvious aid to both analysis and synthesis of control systems.

F. 1

Summation

A summer is a device that adds two or more quantities with due regard for algebraic sign. Electrically, this is easily done by adding as many connections as desired through resistors R I' R2 , • • • , R; to the input of an operational amplifier with feedback resistor R.fi as shown in Figure F.I, summing the currents entering the summing junction, where the voltage is practically zero because of the high gain A. Therefore, we can write E

o

= -

Rr ) Rr' E + -E Rf + ... + -E (R 1R 2 s.:: 2

J

n

(F.!)

A mechanical summer can be built using a "floating lever" or "walking beam" as shown in Figure F.2. The object is to sum displacements, not forces, of x and y with the displacement z being proportional to some function of x and y, or z == .f{x,y)

(F.2)

For small displacements, we assume a linear approximation z == -az

ax

I x + -az. I y == C1x + C y r

2

(~ r

(F.3)

where the bar-r notation means the derivative is evaluated at a reference position. We use linear superposition to evaluate C 1 with y fixed and C2 with x fixed. By similar triangles, we have

liz

C}= lim dz,ax~O

ax

b

==-a

+b

*Many of the ideas illustrated here are due to the late M. A. Eggenberger and his exceJlent paper "Introduction to the Basic Elements of Control Systems for Large Steam Turbine Generators" [1].

590

591

Control System Components

Feedback

RrJl'

Output Inputs

R"

I

,

SummingJunction

~

-&' PracticallyGround Potential

Fig. F.I

Electric summer using resistors and op amp.

(FA)

Therefore,

b a z=--x+--y a+b

a+b

(F.5)

For the special case where a = b we have

x+y z=-2-

(F.6)

Obviously, (F.5) and (F.6) should not be used if the beam becomes tilted, but is reasonably accurate if the tilt angle is less than 30°. In a similar way, we can use a wobble plate to add three displacements, as shown in Figure F.3. If the wobble plate is an equilateral triangle, then the sum is

z=

x+y+w 3

(F.7)

Another way of adding more than two quantities is to add them to the same beam, in which case (F.3) includes a term for each component. Unfortunately, changing one of the coefficients also changes the others, so this must be studied for each individual case .

Fig. F.2

Mechanical summer (floating lever).

Appendix F

592

x

Fig. F.3 A mechanical summerfor three variables(wobble plate).

e.g.,

Still another way of adding more than two quantities to break up the sum into partial sums, z

= u + v + x + y = (u + v) + (x + y)

(F.8)

where a separate beam is used for each partial sum and still another beam for the total. Unlike the electronic summer, the addition of mechanical hardware can cause problems of friction and backlash, which may lead to serious error. Angular addition of two quantities can be performed by a mechanical differential gear arrangement. Other electric summers include transformers, difference amplifiers, and resistance networks. Many of these schemes are described in the literature [2, 3]

F.2 Differentiation Differentiation would seem to be possible in an electric network by using the technique shown in Figure FA , where 1 Z,=Cs

(F.9)

Then, adding currents entering the summing junction we have (F.lO)

which is obviously a differentiation of E; multiplied by a negative constant. However, this circuit will not perform well due to the amplification of noise. This is due to the wide-band amplifying capability of the operational amplifier and the fact that s = 5 + jw is in the numerator. Therefore, any high-frequency noise (large w) available at the input is amplified at the output, Since all electronic equipment generates a certain amount of noise, this circuit is not practical and is usually avoided . Feedback

Rt ;?

Output

Fig. FA An electronicdifferentiator.

Control System Components

593

..--

~7

f--

II

l~

'-

r ---

0 0 0 0 0 0 0 0 ----0.-.. --0-

Y

Fig. F.5

Mechanical position differentiator (for low frequency).

Various electrical and electromechanical circuits for approximate differentiation have been proposed [2]. Usually, we can solve the system equation by integration rather than differentiation and this is recommended. One method of strictly mechanicaldifferentiation at low frequencies is the dashpot, shown in Figure F.5. The transfer function of this device is found from the differential equation

My =B(x - y) which, with M

IfTs

~

:=;

Ky

(F.ll)

0 and T = BIK becomes Y(s)

Ts

X(s)

1+ Ts

(F.12)

I Y(s) --:=;Ts X(s)

(F.13)

y(t):=; TX(t)

(F.14)

and

F.3 Integration Integration involves none of the problems of noise amplification present in the circuit of Figure FA . In fact, integration tends to smooth any input disturbances and is an operation ideally suited for electronic simulation. The usual way of doing this is by means of the circuit of Figure F.6. Adding the currents entering the summingjunction, we get

Appendix F

594

Fe edbac k

;/

Cr

,..----1

's . A

Output

ummmg J unction .

Fig. F.6

An electronic integrator.

-E.

E = --' o

(F.15)

RiCfs

This integrator is inverting, as indicated by the minus sign, and has a gain of I/R;Cf A good example of a mechanical integrator is the combination of a pilot valve and a piston, as shown in Figure F.7. Its operation is explained as follows. Suppose the pilot valve is lifted an amount XI above its neutral position. As this opens the port to the pipe connecting the pilot valve to the piston, the high -pressure hydraulic fluid will flow through this pipe and push against the piston, compressing the piston spring. Unless the piston reaches a stop, this slight movement XI will cause the piston to continue its motion, traveling at some given speed . Thus, in each increment of time dt, the piston will travel a distance Ay = Kx.dt, as shown in Figure F.8, where Kx, is the velocity. Obviously, if the pilot valve is opened a greater amount, the velocity will be increased, although not as a linear function of X, except for small displacements. By graphical integration, we have

or, in the s-domain

y(t) = K[X(t)dt o

(F.13)

KX(s) Y(s) = - s

(F.14)

Rearranging (F.14) we see that

Output Input

~

Oil Pressure (Auxiliary Powe r) Fig. F.7

Mechanical integrator.

595

Control System Components

x,y

Ii }i -- Kxdt

I

Xi ~

V

VV

,

v .'

~

~

II' .'

o

o ~I Fig. F.8

k:--dt

Graphical integration.

(F.15)

Kx(t) = y(t)

or the speed of y is proportional to the displacement of x. Another familiar example of a mechanical integrator is a rotating shaft such as a turbine. Here , the moment of inertia is the gain constant. We can write

(F.l6) where Ta is the sum of all torques acting to accelerate the shaft. Transforming (F.16) we have

O(s)

=_

T(s)

(F.17)

Js

Another example of an integrator is a steam pressure vessel in which the steam pressure in the vessel is the integral of the algebraic sum of steam flows into the vessel [I].

F.4 Amplification The amplifier is a common device in electrical technology. Using a high-gain operational amplifier, it is quite easy to produce gains over several orders of magnitude, say from 10-3 to 10+3 . The circuit for doing this is shown in Figure F.9 where

RJ

E;» -li Ej

(F.18)

I

In many cases, it is desirable to produce gain in mechanical devices. A mechanical stroke amplifier is shown in Figure F.l 0, from which we can write

b

(F.19)

Y(s) = -X(s) a

Outpu t Fig. F.9

A dc voltage amplifier.

596

Appendix F

Fig. F.IO

A mechanical stroke amplifier .

Note that the force is not amplified in this device; only the stroke or displacement. A mechanical power amplifier, which amplifies both stroke and force, is usually called a servomotor or a mechanical-hydraulic amplifier. Such a device , as shown in Figure F.11, uses hydraulic fluid, such as oil, under pressure from an auxiliary power source. This is analogous to an electronic amplifier, which also uses power from an auxiliary (+B) supply. The device in Figure F.ll will typically amplify the energy level by 1000:1 or so and can be used to drive substantial loads. The output Y follows a change in X position with a time lag. Usually, the mass of the moving parts is low compared to the force available such that the response is quite fast. The servomotor pictured in Figure F.ll is called double-acting since the two control "lands" of the pilot valve simultaneously control fluid flow to and from the opposite sides of the piston. We may analyze the system of Figure F.lI according to the block diagram of Figure F.l2 [4]. By inspection we write [2] Y(s) = G(GZG3G4 X(s) 1 + GZG3G4H3 + G3Hz + G3G4H(

(F.19)

By inspection of Figures F.8 and F. 12 we write G _ R _

£1 __ b_ a+b

(- X- X

y~o-

(F.20)

The pilot valve transfer function is (F.21)

x

a:.l< b

E

°4

Pressure Flow Rate ljI (invs/in')

Fig. F.11

A mechanical-hydraulic power amplifier or servomotor.

597

Control System Components

y

1--

Fig.F.12

-;

H

3

~--------J

block diagram of the power servomotor.

where Qo is the average flow gradient for small displacements, Qv is the valve flow in cubic inches per second, and E is the valve displacement in inches . This relationship is illustrated in Figure F.l3. The leakage coefficient of the valve is defined as the change in flow per unit change in pressure [4]. Calling this leakage coefficient L, we have, for constant E,

QL in3/s H2 = = LIlP psi

(F.22)

Transfer function G3 can be derived from the fluid compressibility equation [4]

Vo 2B s6.P(s) = Qc;(s)

(F.23)

or

G3 =

IlP

-

Qc

= -

2B

(F.24)

s Vo

where IlP is the change in pressure on either side of the piston in psi, B is the bulk modulus of elasticity of the fluid in psi, Vo is the fluid volume at zero pressure differential in in 3 and Qc is the compressibility flow.

Actual Curve

~-­

------~po..;-------~E,

/ Fig. 1'.13

Valve flow curve for a pilot valve.

inches

598

Appendix F

We find G4 from Newton's Law. Consider a force F acting on an area A with a small change in pressure M. Then (F.25)

My =F=A' dP

or (F.26) Finally, we compute HI which gives the relationship between valve displacement and piston velocity at zero feedback [4] or Ay=Qp

or (F.27) From Figures F.II and F.12, we compute,by inspection H 3 ==

F

Y ==

ac d(a + b)

(F.28)

Combining(F.20) through(F.28) we get Y(s)

X(s)

=

GIIH3

_

VoN! LM A -_-S3 + _-S2 + - - s + 1 2QoABH3 QoAH3 QoH3

(F.29)

If the mass M is small, as we have assumedhere, then Y(s) =~ Ts + 1

X(s)

(F.30)

where the servomotorgain is

G)

K= H3

bd

==-

ac

(F.3!)

and T is the servomotortime constant A

T=--=--~--

QoH3

(F.32)

WpcP (a + b)d

with A = Ap and Qo = WpcP as in Figure F.II. A servomotor can also be constructed as a "single-acting" unit, as shown in Figure F.14, where the oil force on one side of the servomotor is replacedby a strong spring. In this figure, Y has the opposite directionof X. The transfer function for this configuration is given by equation F.30, but in this case

-b

K=a

and

(F.33)

599

Control System Components

x

o o o o

0 0 0 0

Fig. F.14 A single-acting servomotor.

T=

A a

(F.34)

p

--b WPcP a+ Note carefully the difference between the force-stroke amplifier of Figure F.14 and the mechanical integrator in Figure F.7. The differen ce is clearly the presence of the mechanical feedback linkage such that the amplifier finds a new equilibrium position corresponding to a new input position x. Recall that the integrator continues to drive the piston for any pilot valve displacement until the pilot valve is returned to its neutral position . The response ofthe servomotor amplifier is given by equation F.30 and may be represented by the curves of Figure F.15. Note that this is not the response for the electronic amplifier in equation F.17, where there is no delay indicated . We may change the electronic amplifier of Figure F.9 slightly to obtain a first-order delay similar to Figure F.15. Ifwe replace the feedback resistor in Figure F.9 with a parallel R-C combin ation such that

R

(F.35)

---------71"--- ---

t

100% of step

//1 - 1-

Y

"'I - - T r----,f-,.e.----l--~---_t_63.2%

~

:

~

L\y(!~oo)

~~~y~!----:-~---t· ~ T Fig. F.15 Step response of the servomotor.

600

Input

Appendix F

El+ )

Current Amplifier

A

Torque Motor Demodulator

Excitation ('\)

H. P. Fluid

LVDT (See Section 6.2) ~§§~_I ~

~==-=~~~ Ram

Output Fig. F.16 An electrohydraulic amplifier.

then

- R/R· Z1 £o=--E.= I E R.I I + Res i

(F.36)

I

which is comparable to (F.30) Eggenberger [I] also gives an example of an electrohydraulic amplifier that can be used to drive large loads such as steam valves . Such a device is shown in Figure F.16, with the device response shown in Figure F.17. Clearly, this is a higher-order response than the first-order lag shown in Figure F.15.

/Output( y)

100% Output Step

+

-~-----r-

100% Input Step

L

In ut e

j

- - - - _.

I

o Fig. F.17 Response of the electrohydraulic amplifier.

Control System Components

601

E ;(+)

Fig. F.18

Electrical low value gate.

-----:'Ik-------~

Fig. F.19

Response of the circuit of Figure F.18 for EL < O.

--......J'k---~

Fig. F.20

E;

E;

Response of the circuit of Figure F.18 with diode reversed and EL > O.

F.5 Gating A gate is a device that makes a decision as to whether a signal should be passed or not, or that chooses between two eligible input signals to determine which , if either, should pass the gate. This can be accomplished in an electric circuit by a scheme such as that shown in Figure F.l8, which illustrates a "low-value gate" device . Here, it is assumed that E 1 is positive and E[ is negative . Then Eo will be the greater (less negative) of either E[(-) or--(R/R1)E t (+), as shown in Figure F.l9. Reversing the diode and the polarity of E[ gives the response shown in Figure F.20. Thus, it is seen that this circuit has the ability to select between E; and E[, "auctioning off' the output to the highest (or lowest) bidder.

602

Appendix F n

•

II•

•

1(+)

4'~· 11111

•

IU

i

Y( +)

1'"0 "0"" 0 0 ~

11111

~

1111

IU

(a) Mechanical Overriding Device (Single-actingrelay, X controlling)

•

(+)~~-~

Output (b) Mechanical Overriding Device (Double-acting relay, X controlling) Fig. F.21

Mechanical gating devices.

Many other gating circuits are possible and such circuits often contain diodes, Zener diodes, or some other nonlinear elements . Many references in the analog computer field give examples of such circuits, e.g., see [5] and [6]. Other circuits with characteristics similar to Figures F.l9 and F.20 are possible . In some applications, the value of E/. is fixed and the circuit is called a limiter. Another useful device is the comparator, which behaves in a certain way up to a limiting value, then changes state and acts in a different manner. Both limiters and comparators could be used as overriding gates in the sense intended here. Gating can also be accomplished using hydraulic-mechanical controls . Such a system is shown in Figure F.21, where both inputs XI and X 2 can be either control signals or limit signals. In both systems, XI can be used to control Y providing that X 2 is between its maximum and minimum limits. If X 2 is outside these limits, then XI has no control over the variable Y.

F.6 Transducers A transducer is a device that measures some quantity and produces an output that is related, in a useful way, to the measured quantity. Usually, a transducer is useful over a limited range and these limits must be compatible with the normal operating range of the quantity to be measured.

Control System Components

603

In many cases, the transducer will be designed such that its output varies linearly with the measured quantity, if within specified limits . The "output" will usually be a mechanical position or a voltage . Space does not permit an exhaustive survey of all known transducers. Here, our treatment will be confined to components used in power system control.

F.6.1 Rotational speed transducers (tachometers) It is very important to have a simple and reliable measure of the angular velocity of the generator shaft so that frequency can be closely monitored and controlled. Probably the oldest and best method know for measuring shaft speed is the flyball governor shown in Figure F.22. We can approximate the transfer function of this device, for small parameter changes, by the expression

LU

!J..n =K 1

(F.37)

Actually, the characteristic is not linear, but quadratic , as shown in Figure F.23 (also see Appendix C). However, when changes in speed are small, the error in assuming linearity is not great and the approximation of (F.37) is adequate . Moreover, the characteristic of Figure F.23 is single-valued in the range of interest (n > 0) so that the use of (F.37), even though technically incorrect, will always generate an error signal of the correct polarity . An example of an electromechanical speed transducer, which is convenient is some cases, is the permanent magnet ac generator as shown in Figure F.24. One advantage of this device is its linearity, since the generated emf (the rms value) varies directly with speed, as shown in Figure F.2S. An electromechanical scheme is the magnetic pickup device shown in Figure F.26. A com-

Limit //HAH/

Position (Output)

Speed (Input)

Fig. F.22

A mechanical speed governor.

604

Appendix F Limit +2 +1 '""' ~

'§

'-"

0 1----------:f'-~t_"1""""-'-I

-2 0.8

Fig. F.23

0.9 1.0 1.05 I.I Speed 11 (units)

Characteristics of the mechanical speed governor .

bination of these last two devices is also possible, wherein a frequency of the PM generator is sensed and converted to a voltage, as in Figure F.26 Another important speed transducer is the shaft-mounted oil pump. The oil discharge of the pump is directed through an orifice or needle valve. If a gear-type pump is used, the flow of oil will be directly proportional to speed, or (F.38) When discharged through an orifice, a square root characteristic exists between flow and pressure drop, or Q=k2 vP

(F.39)

Thus, we have the relationship between speed and pressure

P = k2n2

(F.40)

Permanent Magnet Motor

\

Speed

1=======1

E""

R

n

Fig. F.24

Permanent magnet generator speed transducer.

Control System Components

"---------------'-~

o Fig. F.25

605

n

Characteristics of permanent magnet generator speed transducer.

Magnetic Pickup Frequency-toVoltage Converter

Fig. F.26

Magnetic pickup speed transducer .

GOvERNOR PUMP

OIL

Slier/ON

SUAM VALVt:

Fig. F.27

An oil pump speed transducer used as a governor.

606

Appendix F

v

x

t":~ ~1

M~~W:t>":~j

~

Fig. F.29

Insulating Coil Form

~ ~:t>"n"j ~j

~

~

~

Secondary

Primal)'

Secondary #2 (£2)

Cross-section of a linear variable differentialtransfonner (LCDT).

607

Control System Components

~L VDT---'l>,.I...

-3>'.L Final 1

---Demodulato..-.

Ol(E':,

~

IFilterl

Output

e"

C

Primary

0.1

o

R E

Lvnr demodulator and filter [8).

Fig, F.30

for magnetic flux to link the primary and secondary coils. When the core is exactly in the center, each secondary is equally coupled to the primary and the induced voltages in the secondaries are equal, i.e., el = e 2' Moving the coil toward one end increases the coupling to one secondary and, simultaneously, reduces the coupling to the other. Thus, in Figure F.29, movement of the core to the right will result in el > e2 ' To convert the secondary voltages to de, we require a demodulator. This device, shown in Figure F.30, rectifies el and e2 with polarity such that the connection shown gives the difference, which is proportional to displacement, i.e., (FA! )

L VDT Core Position (in.)

__

ValveQp~ _

- +5

XN

e" (volts) -4

+4

-5 Fig. F.3 \

L vor transfer function.

+8

e" = -KLVDTXIl

608

Appendix F

The final stage in Figure F.30 is a low-pass filter, the output of which is loaded into a loading resistor, say lOOK, such that (F.42) Figure F.31 shows the LVDT transfer function, where X is indicated as a steam turbine valve position and shows typical values of parameters used. Note the linearity of the device and the fact that the resolution is infinite. Other translational and angular position transducers are available that utilize different principles. For example, change in resistance with strain, change in capacitance with change of plate spacing, magnetostricton, piezoelectricity, and many others. Some of these devices are useful over a very small range of displacement [3]. Our concern here has centered on devices usable over relatively large changes in displacement.

F.6.3 Pressure transducers Pressure transducers can be either mechanical or electrical, that is, the output can be either a position or a voltage . A common mechanical pressure transducer is the spring-loaded bellows shown in Figure F.32. For small displacements, the change in output Ax is proportional to the change in pressure, i.e.,

A Ax = -6.P G

(F.43)

where P is the pressure , A is the effective bellows area, and G is the spring gradient. An electrical pressure transducer makes use of the LVDT shown in Figure F.33, where the output voltage change may be written as

6.V = K6.P

(F.44)

where K is a constant depending on both the LVDT characteristic and the Bourdon tube charac-

x

Fig. F.32

A mechanical pressure transducer (for low pressure).

609

Control System Components

Demodo'>lo,

Fig. F.33

1-----'3~

An electrical pressure transducer (for high pressure) .

teristics. This transformer is very linear, down to almost zero pressure. However, it must be mounted where vibration will not produce noise in the output.

F.7 Function Generators Function generators are rather common in analog computer work, where a given nonlinear characteristic is duplicated by an electronic simulation. There are many mechanical function generators in the machines of industry. A few examples will illustrate the use of such function generators in the control scheme of a steam turbine. A cam is a function generator as it determines the position and velocity of a valve as a function of time or as a function of the control stroke. Thus , in Figure F.34, the stoke Y2 opens the valve according to the curvature of the cam . This gives the valve lift L a nonlinear characteristic, as shown in Figure F.35, and permits the linearization of steam flow using a nonlinear compensator, as shown in Figure F.36. In this particular case, the steam flow saturates for large values of valve lift. We compensate for this by opening the valve faster at large values of stroke Y2• This nonlinear function generation can also be accomplished in the feedback path, as

o o o

0 0 0

o

Valve Lift

Servomotor Output

Steam Flow Fig. F.34

A cam as a function generator.

Appendix F

610

Valve Stroke , Y2 Fig. F.35 Typical valve lift vs. stroke nonlinearity.

L L valve lift

servo stroke

Fig. F.36

steam

flow

Camshaft and valve function generators.

Y, (Input)

o

o

o

t J."LJ'nnI------.......J x

-l

Pressure Flu id

(O utp ut)

o

Fig. F.37 Mechanical function generator in feedback (intercept valve relay).

Control System Components

611 Intercept Valve

Servo Mo tor

IlIV

~y"

IlIV

2

Feedback Cam Fig. F.38

Block diagram of mechanical intercept valve flow control using a feedback function generator.

Steam Valve

LVDT

Steam Flow

RAM

Fig. F.39

Electrohydraulic valve flow control with feedback function generator.

100%

Final

Slope~

-,

Intercept Point

"

(Next Valve Starts Opening) ""'Diode Rounding

OL-----------------------' 100% o Valve Lift

Fig. F.40

Approximation of valve characteristic by electrical function generator (utilizing two slopes).

Appendix F

612 -22 V

~

E1(+)

RB

21

BIAS

--0 R1

TO SERVO VALVE

EfH INITIAL SLOPE DEMODULA TOR

Fig. FA)

RAM POSITION

Example of an electrical function generator in a feedback circuit.

shown in Figure F.3? Here, the valve is an intercept valve that is operated by stroke fj (the output in Figure F.32). As the input stroke f, increases, calling for additional output fj, the feedback position F is increased, but not linearly. In block diagram form, this situation behaves as shown in Figure F.38. The nonlinear feedback path tends to linearize the /Ltv versus fl' The notations in Figure F.38 refer to Figure F.3? Note that the feedback cam has the same nonlinear characteristic as the intercept valve . These same ideas can be used in electromechanical systems in which an electronic simulation of the nonlinearity replaces the cam. An electro-hydraulic valve controller is shown in Figure F.39, where the feedback signal is electrical rather than mechanical. Thus, the nonlinear "valve" characteristics must be simulated electrically. This is usually done using several straight line segments and nonlinear elements, such as diodes. Suppose the desired curve is similar to that shown in Figure FAO and the representation is to be as shown, where two straight lines are used to approximate the curve. There are several ways to do this electrically, but one easy way is that shown in Figure FA I. Until the voltage E[o(-) becomes as negative as the value set as the break point, all current flows through the initial slope resistance Rz. However, once the break-point voltage is reached (a negative value) the current flows through the.initial slope and final slope resistors in parallel, giving the flatter characteristic of Figure FAO. If greater accuracy is required, several break points can be incorporated so that the straight-line segments become shorter and the functional representation more precise .

References 1. Eggenberger, M. A., Introduction to 'he Basic Elements of Control Systems for Large Steam Turbinegenerators, General Electric Co. publication GET-3096A, 1967. 2. Savant, C. J., Jr. Basic Feedback Control System Design, McGraw -Hili, New York, 1958.

Control System Components

613

3. Bragge, E. M., S. Ramo, and D. E. Woolridge, HandbookofAutomation, Computation and Control, v. 3, Systemsand Components. Wiley, New York, 1961. 4. Lewis,E. E. and H. Stern, Design ofHydraulic ControlSystems, McGraw-Hill, New York, 1962. 5. Shigley, 1. E. Simulation ofMechanical Systems: An Introduction, McGraw-Hill, New York, 1967. 6. Ashley, 1. R., Introduction to Analog Computation, Wiley, New York,1963. 7. ElliottCompany, Fundamentals ofTurbine Speed Control, BulletinH-21A. 8. Westinghouse Electric Corp.,Servoactuators. Unpublished technical notes. Privatecommunication.

appendix

G

Pressure Control Systems

Pressure control systems, such as the turbine-following system of Figure 11.3, have been analyzed from a control viewpoint. * The block diagram for such a control system is shown in Figure G.l, where system variables are defined both by name and by symbols. The variables defined in Figure G.l(b) are related to physical quantities shown in Figure G.l(a). The multiplier of Figure G.l(a) will be eliminated by mathematical manipulation. The transfer functions for Figure G.1(b) will be derived. In doing so, it will be convenient to refer to a typical physical system that exhibits some of the features under discussion. Such a physical system is shown in Figure G.2. It consists of a summing beam B (see Appendix F) on which several forces act, including the pressure-sensing bellows, rjJ, the reference, Pp, the steady-state feedback, 11ISp and the temporary feedback, e; All forces are summed with the correct algebraic sign to provide an output, B, which operates the pilot valve input to the relay piston integrator (see Appendix F). This relay piston operates the force and stroke amplifier to obtain the stroke np (not shown). Feedback lever L, produces the steady-state droop by acting in opposition to B (negative feedback) with the droop adjusted by changing the lever arm as noted. Feedback lever L 2 produces a transient droop that is gradually reduced to zero by controlled leakage through a preset needle valve KN V' which equalizes the initial pressure difference. This amounts to a mechanical differentiation and is called reset control. 1. Pressure Regulator, GR Three transfer functions for pressure regulation are used: (a) Proportional control is represented by the block diagram of Figure G.3, where l/G is the time it would take the output (stroke) 111 to go through full or unit stroke if a rated pressure error is applied and with no feedback. The constant Sp is a droop constant fed back mechanically to stabilize the system. We compute, for zero reference, PP = 0 (G.l)

*This analysis follows closely that of Eggenberger and Callan, ref. 7.29.

614

Pressure Control Systems

Pressure Error

Pressure Feedback

Pressure Regulator Position

Servo Motor Position

615 Flow into the Controlled System

Equivale nt Valve Valve Area Flow +

....L.

L-

Steam Pressure

-J

(a) Identification of System Variables

+

(b) Identification of System Transfer Function Fig. G.I

A turbine-followi ng representation.

whe re K= GTR TR = l/GfJ p The temporary feedback loop in Figure G.2 is inactive for proportional control and the needle valve is open , i.e., (b) Proportional plus reset control is represented by the block diagram of Figure GA , where the system is arranged to slowly reset itself. Thus , TL is fairly large (a few seconds) and is adjusted by setting the needle valve KN V in Figure G.2. We compute

(G.2)

which we simplify to

(G.3) We have defined

Appendix G

616

Proportional Feedback

Pressure Sensor Summing Beam B

~:)(;~==~ Power Fluid

r.========:1 Pilot Valve

11 1 ~

•

Relay Output

Fig. G.2

A typical pressure regulator.

TL =

-

C

-

KN V

(0.4)

where C is a mechanical constant and KN V is the flow factor (in3/sec-psi) for the damping needle. (c) Proportional plus partial reset control is represented by the block diagram of Figure 0.5, where the transfer function is given as (0.5) Here, TL is defined as before and two new time constants are defined as follows :

Fig. G.3 Block diagram of a regulator for proportional control.

Pressure Control Systems

Fig. GA

617

Block diagram of a regulator for reset control.

(G.6)

WR'(-), wRi+) =

~[_l + G(8 + 8 2 T ±

~{[_l + 2

p) ]

i

L

TL

G(8 + i

8p )]Z_ 4G8 p TL

}I/Z

(G.7)

where the two frequencies are defined according to the choice on the sign of the second term. By proper choice of the several parameters, this type of regulator is adaptable to many applications.

2. Hydraulic Servomotor, G; The transfer function of a hydraulic servomotor of a force and stroke amplifier, is shown in Appendix F, and is defined as Gh = - - -

(G.8)

1 + Tzs

3. Steam Valve-Steam Flow, GA and GM We begin by assuming the flow through the valve is proportional to the product of the equivalent valve area and the pressure :

11t

Fig. G.5

Block diagram for a regulator with proportional plus partial reset control.

Appendix G

618

M=APlbm/s

(G.9)

This assumes that the equivalent valve area has been linearized in the valve drive cams or in the valve itself. We would like to eliminate this multiplication and to linearize equation (G.9). To do this, we write the differential

/1J.L:::::: dJ.L= ( -aJ.L ) da+ ( -aJ.L ) dt/! aa P=con st at/! A=const

(G.lO)

Since under normal operation the pressure is at nearly rated value, P = P R and the first term in (G. I0) may be evaluated at rated pressure. By definition

(~ )P=PR

=

I

Since a unit change in TJ2 produces a unit change in

(XI>

(G.I I) Therefore

GAGM(Pr ) = I

(G.12)

The change is /1J.L caused by dt/! can be introduced at the summing point as shown in Figure G.I(b).

4. Steam Volume We assume that the steam flow, J.Lj, being fed into the steam volume, is constant and is independent of pressure. The steam vessel or drum ahead of the control valves acts as an integrator. Thus , any flow in that is not balanced by flow out of the drum will increase the pressure at a rate given by the integrator gain G{ where I M, Gv = - = I

Tv

Wr

(G. 13)

where Tv is the characteristic time of the steam volume. We represent the steam-volume portion of the system by the block diagram of Figure G.6, where the feedback function H(a) is approximately equal to J.Lj, i.e., for

J.L= I;H = I J.L= 0; H = 0 and the loop time constant is

Fig. G.6

Block diagram for flow- volume-pressure relationships.

Pressure Control Systems

619

1 1+ 1;s

Fig. G.7

Block diagram for proport ional initial-pressure control with a large steam vessel (T v ~ I).

Tv T=H

(G.14)

For 0 < H < 1 the transfer function is given by

(G. IS) Reference (11 .29) points out that, in most cases, we may assume this to be an integration, or t/J

1

ILi - IL

TvS

(G.16)

--a-

Combining all of the above, the block diagram for a turbine-following system with proportional control is given by Figure G.7. Reference 11.29 solves this system using Bode diagrams with the result shown in Figure G.8 for typical values of the time constants. The quantity most easily changed is 8p • A larger regulation, 8p , makes the system more stable, but results in a greater steady-state error. Recalling that the steady-state error is defined as [26]

. . Kv=hmsKG(s)=hms 5--->0

5--->0

s(

1

Ki8 pTv

+ TR,S)(1 + Tzs )

=~ ~ upTv

r-----;:;;;;;t=-----r------r--, +90

~=6.67

spTv max

Phase M rgin

0),

Fig. G.8

radls

3 1/5

Bode diagram ofa proportional -pressure control system .

(G.17)

Appendix G

620

The system is type I [26] and has a steady-state position (pressure) error of zero. Stability depends on the gain, K v• of (G.17). If either proportional plus reset control or proportional plus partial reset control are used, the results are changed as shown in Figure G.9 and G.ID, respectively, where typical values of constants are used. These systems could also be analyzed by root locus and this method is recommended to the interested reader .

IjI

(a) Block Diagram of Proportional Plus Reset Control +40 .--- - - , -- - - - -,-- - - - - - - --,-- - - ,--- , +90 y = 550 1-..3looIO:'::::......:::...------t---

'" ~

---'''''''''oo:;:-- - - --,t-- - - t---i+45

Q) "'0

E '2 0 I - - -t -- - - --..:::p.......::::----------II-"'~--+_-- I OJ) Cu is the angle corresponding to Ru- For the force FG we have the familiar expression for the weight of an object

where g is the acceleration of gravity. The forces pennendicular to the arm L are defined as F p, where

Fp == Fe cos 4> - FG sin cP or

Fp == mLwl; sin 4> cos 4J - 2KL sin cP cos cP + 2KL sin cPu cos cP - mg sin cP

(H.10)

If the system is in equilibrium, then Fp == 0 and we compute the relationship tan

A,.

0/

==

-2KL sin cPu (mLw[; - 2KL)cos 4> - mg

(H 11)

.

which, unfortunately, is awkward to solve. If the spring is quite stiff and it overpowers the gravitational effect, then we may rewrite (H.10) as

Fp == mLwl; sin 4> cos cP - 2KL sin 4> cos 4> + 2KL sin u

(H.14)

Then (H.l5)

625

The Governor Equations

a b

Fig. H.3

Definition of the angle cPo

or, by trigonometric maniputation ) 1/2 4K 2 41(2 4 cos- 4>u ( wG - -w 2 m G +mcos 4> == 2K

wl;--

(H.14)

m

Factoring the numerator, we get 2K 2K ( wl; - -;;; ± -;;; sin cos 4> == 2K

o/u

)1/2 (HoIS)

wl;-m

or

cos 4> == [

4Kw~ 4K2 ] 1/2 Wf; - ---;;;- + m2 (I + sin? 4>m)

(H.16)

If, on the other hand, we assume that the spring has an unstressed length R; == 0 (at 4>u == 0) then this simplifiesthe equilibiumcondition for (H.I0) such that mg

(H.I?)

cos cP == mL WG2 2 KL If there is no spring at all, then K == 0 and we have g

cos r/J == - L 2

(H.18)

WG

In any case, we obtain 4> as a function of WG' From Figure HoI we note that an angular displacement 4> results in a linear displacementof the collar C. This is shown in Figure H.4, where we note that (HoI9)

x == d - (a + b)

or x == d - (L cos 4> +

VL~ -

R2)

(H.20)

where

R == L sin 4> Substituting (Ho3) for R and defining A == I.Jl. we have x == d - (L cos 4> + LVA2 - sin? 4»

(Ho21)

626

Appendix H

a

d

b

~c Fig. H.4 Relationship between 4J and x.

If L; = L this becomes

x=d-2Lcos4J Thus, the equations derived for cos small displacements

4J may

(H.22)

be used as a proportional measure of x. For (H.23)

from which we compute (H.24)

H.1.2 The dynamic equations Up to this point, we have concerned ourselves with the "static" governor equations, that is, the equations based essentially on constant speed . Actually, of course, this is a dynamic problem. Any acceleration of the mass M is governed by Newton's laws and the equations describing the system behavior are differential equations. Furthermore, we must include all forces acting on the mass M. The force (perpendicular to L), given by F p in (H.IO), is a displacement force due to the position of the mass (or the angle 4J) at any time. There is also a viscous friction force acting to retard the motion and this force is usually depicted as B~ where B is the viscous constant. Combining all forces we write the following equation, considering m to be a point mass .

m¢J = mLw{; sin

4J cos 4J - 2KL sin 4J cos 4J + 2KL sin 4J" cos 4J - mg sin 4J - B~

(H.25)

Now, suppose the turbine-generator has moment of inertia J with mechanical driving torque Tm and electrical (load) torque Te . Then, for a turbine angular velocity w we can write (H.26)

627

The Governor Equations

where

i.e.,

T; is the accelerating torque. However, there is a simple gear ratio N relating wand WG' (H.27)

From (H.22), we note that the governor stroke, x, is a function of cos cPo Thus, the mechanical torque must be proportional to cos cPo Ifwe assume an operating angle cPo at which point the torque is TmO' we write (H.28) where k> 0 is a constant. Note that, as ep increases, Tm decreases and vice versa (also note that 0 S cP :5 90°). Thus, as the speed decreases, decreasing ep, Tm is increased by the admission of more steam as shown in Figure H.5. This explanation ignores the delays in servos and steam systems. We now define a constant F as follows: (H.29) which is dependent on the load torque T; Also, for convenience, we define the angular speed in the cP direction to be f/J, i.e.,

f/J = cb

(H.30)

Combining (H.24) and (H.30) we have a normalized system of equations as follows:

. 2~ 2n f/J = n2Lw2 sin ep cos ep - - - sin l/J cos l/J + - - sin l/Ju cos l/J m

m

(R.3!)

These equations are the state equations for the system, ignoring any delays in converting governor stroke to mechanical torque. When operating at a constant load Te , the rotor speed to must be constant, thus giving constant governor speed wN and constant governor angle l/J. Thus, a state ofequilibrium exists where

cP = cPo {

t/J=4>=0 w= Wo

(H.32)

/ " Operating Point

TmO

I I I I I I

" -, -,