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
Copyright © 2003 by Institute of Electrical and Electronics Engineers, Inc. All rights reserved.
No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the] 976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright
Clearance Center, 222 Rosewood Drive,Danvers, MA 01923, (978) 750-8400, fax
(978) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (20]) 748-60] 1, fax (201) 748-6008.
Limit of Liability/Disclaimerof Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representationor warrantieswith respect to the accuracy or completenessof the contents of this book and specificallydisclaim any implied warranties of merchantabilityor fitness for a particular purpose. No warranty may be created or extended by sales representativesor written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professionalwhere appropriate.Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental,consequential,or other damages. For general informationon our other products and services please contact our Customer Care Department within the U.S. at 877-762-2974,outside the U.S. at 317-572-3993or fax 317-572-4002. Wiley also publishes its books in a variety of electronic formats, Some content that appears in print, however, may not be available in electronic format. Library of Congress Cataloging in Publication Data is available.
ISBN 0-471-23862-7 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1
To Our Families
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 <,
C 1800 _r/mi n ----L..
~ 3. 0
::;
£ 2.5 2.0
nuclea r
C 3600 r/ min fossi l
I ! 1 ' I ! I I ~ I_-L_J_._----,l~,_J:_:,.__J-.-__'::_::___c~~ 800 1000 1200 1400 1600 1800 2000 2200 Gener ntor Ro ling , MVA 6OQj..J.'
(b)
Fig .2. 1
Inertia constants for large steam turbogenerators: (a) turbogenerators rated 500 M VA and below 1201. (b) expected future large turbogenerators. (@ IEEE . Reprinted from IEEE Trans.. vo l. PAS-90. No v.jDec. 1971.)
13. p.
2.3.1
Unregulated machines
For a fixed gate or valve position (i.e ., when the machine is not under active governor control) the torque speed characteristic is nearly linear over a limited range at rated speed, as shown in Figure 2.3(a). No distinction seems to be made in the literature between steady-state and transient characteristics in this respect. Figure 2.3(a) shows that the prime-mover speed of a machine operating at a fixed gate or valve position will drop in response to an increase in load . The value of the turbine torque coefficient suggested by Crary (7) is equal to the loading of the machine in pu . This can be veri fied as follows. From .the fundamental relationship between the mechanical torque
4 .5 4
« > ~
A ~ 450 - 5 14 r/m ln ~
200 - 400 r/m ln
c~
138 - 180 r/m !n
8
3
o ~ 80 1 L -_
o
Fig. 2.2
- L_ _"--_-L_ _.L-_--'-_
20
40
60
80
100
_
120 r/ min - L_
120
_
L
140
Genera to r Rating, MVA
Inertia constants of large vert ical-type waterwheel generators. including allowance of 15% for waterwheels. (® IEEE. Repr inted from Electr Eng.. vol. 56. Feb . 1937).
Chapter 2
18
o
1.0 Speed, pu (0)
2.0
-- - - - - - - \
\ Slope =
z~ Tm01---.-u·
~
~
0-
-
Tm6
T _ --i'
- l/R
-----\
_
m
o
WR W Spee d, raql, (b)
Fig .2.3
Turbine torque speed characteristic: (a) unregulated machine. (b) regulated machine.
T; and power Pm Tm = Pm/w N vrn
(2.20)
we compute, using the definition of the differential,
at;
=
:~: ar; +
00:
dw Nvrn
(2.21)
Near rated load (2.21) becomes d'I'; = (ljwR)dPm - (PmRjw~)dw N· m
If we assume constant mechanical power input, dPm
=
(2.22)
0 and
d'T; = -(PmRjw'R,)dw Nvm This equation is normalized by dividing through by TmR
(2.23)
PmRjWR with the result
d'I'; = -dw pu
(2.24)
where all values are in pu . This relationship is shown in Figure 2.3(a). 2.3.2
Regulated machines
In regulated machines the speed control mechanism is responsible for controlling the throttle valves to the steam turbine or the gate position in hydroturbines, and the
The Elementary Mathematical Model
19
mechanical torque is adjusted accordingly. This occurs under normal operating conditions and during disturbances. To be stable under normal conditions, the torque speed characteristic of the turbine speed control system should have a "droop characteristic"; i.e., a drop in turbine speed should accompany an increase in load. Such a characteristic is shown in Figure 2.3(b). A typical "droop" or "speed regulation" characteristic is 5% in the United States (4/~ in Europe). This means that a load pickup from no load (power) to full load (power) would correspond to a speed drop of 5% if the speed load characteristic is assumed to be linear. The droop (regulation) equation is derived as follows: from Figure 2.3(b), Tm = Tmo + Tm~, and TmtJ. = -wtJ./R, where R is the regulation in rad/ Nvrn-s. Thus
T;
=
Tmo - (w - wR)/R Nv m
(2.25)
Multiplying (2.25) by WR, we can write Pm Let Pmu
=
~
TmWR = PmO -
(wR/R)w~
(2.26)
pu mechanical power on machine VA base
Pmu ~ Pm/S B = PmO/S B
-
(wR/SsR)W4
or (2.27) Since Pm~
=
P; - PmO' (2.28)
where the pu regulation R; is derived from (2.28) or R" ~ SsR/w~ pu
(2.29)
As previously mentioned, R" is usually set at 0.05 in the United States. We also note that the "effective" regulation in a power system could be appreciably different from the value 0.05 if some of the machines are not under active governor control. IfLSB is the sum of the ratings of the machines under governor control, and L SSB is the sum of the ratings of all machines, then the effective pu regulation is given by (2.30)
Similarly, if a system base other than that of the machine is used in a stability study, the change in mechanical power in pu on the system base PmAs" is given by PmtJ.s"
=
-
(SswtJ.u/SsBR,,) pu
(2.31 )
A block diagram representing (2.28) and (2.31) is shown in Figure 2.4 where K
= SS/SsB
The droop characteristic shown in Figure 2.3(b) is obtained in the speed control system with the help of feedback. It will be shown in Part III that without feedback the speed control mechanism is unstable. Finally, we should point out that the steadystate regulation characteristic determines the ultimate contribution of each machine to a change in load in the power system and fixes the resulting system frequency error.
20
Chapter 2 p
mOu
p mu
1.0
w
Fig .2.4
U
Block diagram representation of the droop equation .
During transients the discrepancy between the mechanical and electrical torques for the various machines results in speed changes. The speed control mechanism for each machine under active governor control will attempt to adjust its output according to its regulation characteristic. Two points can be made here:
I. For a particular machine the regulation characteristic for a small (and sudden) change in speed may be considerably different in magnitude from its overall average regulation . 2. In attempting to adjust the mechanical torque to correspond to the speed change, time lags are introduced by the various delays in the feedback elements of the speed control system and in the steam paths; therefore, the dynamic response of the turbine could be appreciably different from that indicated by the steady-state regulation characteristics. This subject will be dealt with in greater detail in Part III. 2.4
Electrical Torque
In general, the electrical torque is produced by the interaction between the three stator circuits, the field circuit, and other circu its such as the damper windings. Since the three stator circuits are connected to the rest of the system, the terminal voltage is determined in part by the external network, the other machines, and the loads. The flux linking each circuit in the machine depends upon the exciter output voltage, the loading of the magnetic circuit (saturation), and the current in the different windings. Whether the machine is operating at synchronous speed or asynchronously affects all the above factors . Thus a comprehensive discussion of the electrical torque depends upon the synchronous machine representation. If all the circuits of the machine are taken into account, discussion of the electrical torque can become rather involved. Such a detailed discussion will be deferred to Chapter 4. For the present we simply note that the electrical torque depends upon the flux linking the stator windings and the currents in these windings. If the instantaneous values of these flux linkages and currents are known, the correct instantaneous value of the electrical torque may be determined. As the rotor moves, the flux linking each stator winding changes since the inductances between that winding and the rotor circuits are functions of the rotor position. These flux linkage relations are often simplified by using Park's transformation . A modified form of Park's transformation will be used here (see Chapter 4). Under this transformation both currents and flux linkages (and hence voltages) are transformed into two fictitious windings located on axes that are 90· apart and fixed with respect to the rotor. One axis coincides with the center of the magnetic poles of the rotor and is called the direct axis. The other axis lies along the magnetic neutral axis and is called the quadrature axis. Expressions for the electrical quantities such as power and torque are developed in terms of the direct and quadrature axis voltages (or flux linkages) and currents.
The Elementary Mathematical Model
21
A simpler mathematical model, which may be used for stability studies, divides the electrical torque into two main components, the synchronous torque and a second component that includes all other electrical torques. We explore this concept briefly as an aid to understanding the generator behavior during transients. 2.4.1
Synchronous torque
The synchronous torque is the most important component of the electrical torque. It is produced by the interaction of the stator windings with the fundamental component of the air gap flux. It is dependent upon the machine terminal voltage, the rotor angle, the machine reactances, and the so-called quadrature axis EM F, which may be thought of as an effective rotor EM F that is dependent on the armature and rotor currents and is a function of the exciter response. Also, the network configuration affects the value of the term inal voltage. 2.4.2
Other electrical torques
During a transient, other extraneous electrical torques are developed in a synchronous machine. The most important component is associated with the damper windings. While these asynchronous torques are usually small in magnitude, their effect on stability may not be negligible. The most important effects are the following. I. Positive-sequence damping results from the interaction between the positive-sequence air gap flux and the rotor windings, particularly the damper windings. In general, this effect is beneficial since it tends to reduce the magnitude of the machine oscillations, especially after the first swing. It is usually assumed to be proportional to the slip frequency, which is nearly the case for small slips. 2. Negative-sequence braking results from the interaction between the negative-sequence air gap flux during asymmetrical faults and the damper windings. Since the negative-sequence slip is 2 - s, the torque is always retarding to the rotor. Its magnitude is significant only when the rotor damper winding resistance is high. 3. The de braking is produced by the de component of the armature current during faults, which induces currents in the rotor winding of fundamental frequency. Their interaction produces a torque that is always retarding to the rotor.
It should be emphasized that if the correct expression for the instantaneous electrical torque is used, all the above-mentioned components of the electrical torque will be included. In some studies approximate expressions for the torque are used, e.g., when considering quasi-steady-state conditions. Here we usually make an estimate of the components of the torque other than the synchronous torque. 2.5
Power-Angle Curve of a Synchronous Machine
Before we leave the subject of electrical torque (or power), we return momentarily to synchronous power to discuss a simplified but very useful expression for the relation between the power output of the machine and the angle of its rotor. Consider two sources V = VL2 and E = Eli. connected through a reactance x as shown in Figure 2.5(a),3 Note that the source V is chosen as the reference. A current 3. A phasor is indicated with a bar above the symbol for the rrns quantity. For example if I is 'the rms value of the current, 1 is the current phasor. By definition the phasor T is given by the transformation (J> where 1 ~ le J8 = I(cos 8 + j sin 8) = CP (V2 I cos (wt + 8)}. A phasor is <1 complex .number related to the corresponding time quantity ;(t) by ;(t) = (R-e(VIle Jwt ) = VII cos (wt + 8) = (J> -I'(/e J8 ) .
22
Chapter 2
,&
§
1&
ry-yy-, k
~va
(0)
fY/.
o
n/ 2
n
(b )
Fig .2 .5
A simple two-m achine system : (a) schematic represent ation. (b) power-angle cur ve.
1 = IL!!.. flows
between the two sources. We can show that the power P is given by P
= (£V/x) sin b
(2 .32)
Since E, V. and x are constant. the relation between P and b is a sine curve. as shown in Figure 2.5(b). We note that the same power is delivered by the source E and received by the source j7 since the network is purely reactive. Consider a round rotor machine connected to an infinite bus. At steady state the machine can be represented approximately by the above circuit if V is the terminal voltage of the machine. which is the infinite bus voltage; x is the direct axis synchronous reactance; and E is the machine excitation voltage, which is the EM F along the quadrature axis. We say approximately because such factors as magnetic circuit saturation and the difference between direct and quadrature axis reluctances are overlooked in this simple representation. But (2.32) is essentially correct for a round rotor machine at steady state. Equation (2 .32) indicates that if E, V, and x are constant, Et/]» is a constant that we may designate as PM to write P = PM sin 0; and the power output of the machine is a function only of the angle 0 associated with E, Note that E can be chosen to be any convenient EM F, not necessarily the excitation voltage; but then the appropriate x and 0 must be defined accordingly .
2.5 .1
Classical representation of a synchronous machine in stability studies
The EM F of the machine (i .e.., the voltage corresponding to the current in the main field winding) can be considered as having two components: a component E' that corresponds to the flux linking the main field winding and a component that counteracts the armature reaction. The latter can change instantaneously because it corresponds to currents, but the former (which corresponds to flux linkage) cannot change instantly .
The Elementary Mathematical Model
23
When a change in the network occurs suddenly, the flux linkage (and hence E') will not change , but currents will be produced in the armature; hence other currents will be induced in the various rotor circuits to keep this flux linkage constant. Both the armature and rotor currents will usually have ac and de components as required to match the ampere -turns of various coupled coils. The flux will decay according to the effective time constant of the field circuit. At no load this time constant is on the order of several seconds, while under load it is reduced considerably but still on the order of one second or higher. From the above we can see that for a period of less than a second the natural char acteristic of the field winding of the synchronous machine tends to maintain constant flux linkage and hence constant E' . Exciters of the conventional type do not usually respond fast enough and their ceilings are not high enough to appreciably alter "this picture. Furthermore, it has been observed that during a disturbance the combined effect of the armature reaction and the excitation system is to help maintain constant flux linkage for a period of a second or two. This period is often considered adequate for determin ing the stability of the machine . Thus in some stability stud ies the assumption is commonly made that the main field flux linkage of a machine is constant. The main field-winding flux is almost the same as a fictitious flux that would create an EMF behind the machine direct axis transient reactance. The model used for the synchronous machine is shown in Figure 2.6, where x; is the direct axis transient reactance.
x'
d
-
+
v/..Q.
Fig. 2.6 Representation or a synchronous machine by a constant voltage behind trans ient reactance.
The constant voltage source Ell. is determined from the initial conditions, i.e.• pretransient conditions . During the transient the magnitude E is held constant, while the angle 0 is considered as the angle between the rotor position and the terminal voltage V. Example 2.1
For the circuit of Figure 2.6 let V operating at P = 0.8 pu at 0.8 PF. Solution Using
Vas reference. V
=
=
1.0 pu, x d = 0.2 pu. and the machine initially
I.OLQ
10 = 1.01-36.9° = 0.8 - jO.6 E = Ell. = 1.0 + jO.2(0.8 - jO.6) = 1.12 + jO.16
=
1.1314 /'8.13°
The magnitude of E is 1.1314. This will be held constant during the transient, although 8 may vary. The initial value of 8. called 80 , is 8 .13° .
24
Chapter 2
During the transient period, assuming that V is held constant, the machine power as a function of the angle 0 is also given by a power-angle curve. Thus P = (EVjXd) sin 0 = PMsino
For the example given above PM 2.5.2
(2.33)
1.1314/0.2 = 5.657.
=
Synchronizing power coefficients
Consider a synchronous machine the terminal voltage of which is constant. This is the case when the machine is connected to a very large power system (infinite bus). Let us assume that the machine can be represented by a constant voltage magnitude behind a constant reactance, as shown in Figure 2.6. The power is given by (2.32). Let the initial power delivered by the machine be Po, which corresponds to a rotor angle 00 (which is the same as the angle of the EM F E). Let us assume that 0 changes from its initial value 00 by a small amount 0A; i.e., 0 = 00 + 0A' From (2.32) P also changes to p = Po + PI1' Then we may write
Po + PA
=
PM sin (00
+
0A) == PM(sin 00 cos 0A
If 011 is small then, approximately, cos 0A Po + P A
"J
"J
)
and sin 0A
PM sin 00
~
+ cos 00 sin 0A)
(2.34)
0A' or
+ (PM cos 00)0~ (2.35)
The quantity in parentheses in (2.35) is defined to be the synchronizing power coFrom (2.35) we also observe that
efficient and is sometimes designated P
J
r,
~
•
~PJ ao
PM COS 00 =
(2.36) b." 60
Equation (2.35) is sometimes written in one of the forms
PA
La
=
PJ OA
La
ap OA
ao
= -
La
(2.37)
(Compare this result with dP, the differential of P.) In the above analysis the appropriate values of x and E should be used to obtain PM' In dynamic studies and the voltage E' are used, while in steady-state stability analysis a saturated steady-state reactance X d is used. If the control equipment of the machine is slow or inoperative, it is important that the machine be operating such that o ~ 0 ~ 1r /2 for the operating point to be stable in the static or steady-state sense. This is the same as having a positive synchronizing power coefficient. This criterion was used in the past to indicate the so-called "steady-state stability limit."
x;
2.6
Natural Frequencies of Oscillation of a Synchronous Machine
A synchronous machine, when perturbed, has several modes of oscillation with respect to the rest of the system. There are also cases where coherent groups of machines oscillate with respect to other coherent groups of machines. These oscillations cause fluctuations in bus voltages, system frequencies, and tie-line power flows. It is important that these oscillations should be small in magnitude and should be damped if the system is to be stable in the sense of the definition of stability given in Section 1.2.1.
The Elementary Mathematical Model
25
In this section we will illustrate the inherent oscillatory nature of a synchronous machine by the following example.
Example 2.2 A two-pole synchronous machine is connected to an infinite bus with voltage V through a reactance x as in Figure 2.5(a). The voltage E remains constant, and a small change in speed is given to the machine (the rotor is given a small twist); i.e., W = Wo + fU(t), where u(t) is a unit step function. Let the resulting angle change be 06. Let the damping be negligible. Compute the change in angle as a function of time and determine its frequency of oscillation. Solution
From (2.10) we write M~/SB3 + P, = Pm. But we let fJ = 00 + fJ A such that ~ = ~A and P, = PeO + Pe6 ; Pm is constant. Then M~6/SB3 + PeA = Pm - PeO = 0 since 50 = O. From (2.37) for small 06 we write Pe6 = Ps0 6 , where from (2.36) P, is the synchronizing power coefficient. Then the swing equation may be written as
M5 6 /SB3
RE
which has the solution of the form
8~(t) =
+ P.r;06
=
0
V
sin PsSB3 1M t elect rad (2.38) PsSB3 Equation (2.38) indicates that the angular frequency of oscillation of the synchronous machine with respect to the rest of the power system is given by vi PsS B3 / M. This frequency is usually referred to as the natural frequency 'of the synchronous machine. E
--
I t should be noted that PJ is a function of the operating point on the power-angle characteristic. Different machines, especially different machine types, have different inertia constants. Therefore, the different machines in a power system may have somewhat different natural frequencies. We now estimate the order of magnitude of this frequency. From (2.6) and (2.16) we write M / S8) = 2H /w m or f~SB3/ M = P.~wm/2H where P, is in pu, W m is in rad/s, and H is in s. Now P, is the synchronizing power coefficient in pu (on a base of the machine three-phase rating). If the initial operating angle 0 is small, P, is approximately equal to the amplitude of the power-angle curve. We must also be careful with the units. For example, a system having P,f/SB3 = 2 pu, H = 8, Wos c
Jose
= v(2 x 377)/(2 x 8) =
6.85/21r
=
=
6.85 rad/s
1.09 Hz
If MKS units are used, we write
lose where
I
=
S83 =
H
P,
= =
=
(1/21r) V 1rf( PJ /
SB3
H)
(2.39)
system frequency in Hz three-phase machine rating in MVA inertia constant in s synchronizing power coefficient in MW /rad
Next, we should point out that a system of two finite machines can be reduced to a single equivalent finite machine against an infinite bus. The equivalent inertia is J 1J2/(J1 + J 2 ) and the angle is Ol~ - fJ 2A •
26
Chapter
2
Thus we conclude that each machine oscillates with respect to other machines, each coherent group of machines oscillates with respect to other groups of machines , and so on. The frequencies of oscillations depend on the synch ro nizing power coefficients and on the inertia constants.
2.7
System of One Machine against an Infinite Bus- The Classical Model
An infinite bus is a source of invariable frequency and voltage (both in magnitude and angle). A major bus of a power system of very large capacity compared to the rating of the machine under consideration is approximately an infinite bus. The inertia of the machines in a large system will make the bus voltage of many high-voltage buses essentially constant for transients occurring outside that system. Consider a power system consisting of one machine connected to an infinite bus through a transmission line. A schematic representation of this system is shown in Figure 2.7(a).
v
(a)
Fig. 2.7
E~
(b)
One machine connected to an infinite bus thr ou gh a tr an smi ssion line: (a) one -line diagram. (b ) equ ivalent circuit.
The equation of motion of the rotor of the finite machine is given by the swing equation (2.7) or (2.10). To obtain a time solution for the rotor angle, we need to develop expressions for the mechanical and the electrical powers. In thi s sect ion the simplest mathematical model is used. This model, which will be referred to as the classical model, requires the following assumptions: I. The mechanical power input remains constant during the period of the transient. 2. Damping or asynchronous power is negligible. 3. The synchronous machine can be represented (electrically) by a constant voltage source behind a transient reactance (see Section 2.5.1). 4. The mechanical angle of the synchronous machine rotor coincides with the electrical phase angle of the voltage behind transient reactance. 5. If a local load is fed at the terminal voltage of the machine, it can be represented by a constant impedance (or admittance) to neutral. The period of interest is the first swing of the rotor angle 0 and is usually on the order of one second or less. At the start of the transient, and assuming that the impact initiating the tr ansient creates a positive accelerating power on the machine rotor , the rotor angle increases. If the rotor angle increases indefinitely, the machine loses synchronism and stability is lost. If it reaches a maximum and then starts to decrease, the resulting motion will be oscillatory and with constant amplitude. Thus according to this model and the assumptions used, stability is decided in the first swing. (If damping is present the amplitude will decrease with time, but in the classical model there is very little damping.)
The Elementary Mathematica l Model
Fig. VI
27
Equivalent circuit for a system of one machine against an infinite bus.
The equivalent electrical circuit for the system is given in Figure 2.7(b) . In Figure 2.7 we define
V=
V,
=
terminal voltage of the synchronous machine
VLQ = voltage of the infinite bus, which is used as reference
Xd = direct axis transient reactance of the machine Z TL series impedance of the transmission network (including transformers) Z, = equivalent shunt impedance at the machine terminal, including local loads if any
By using a Y-~ transformation , the node representing the terminal voltage V, in Figure 2.7 can be eliminated. The nodes to be retained (in addition to the reference node) are the internal voltage behind the tran sient reactance node and the infinite bus. These are shown in Figure 2.8 as nodes I and 2 respectively . Also shown in Figure 2.8 are the admitt ances obtained by the network reduction. Note that while three admittance elements are obtained (viz., YI2, YIO , and Y20), Y20 is omitted since it is not needed in the analysis . The two-port network of Figure 2.8 is conveniently described by the equation (4.40) The driving point admittance at node I is given by YII = YI I Lfu = YI2 + Y IO where we use lower case y's to indicate actu al adm ittances and cap ital Y' s for matrix element s. The negative of the transfer admitt ance Jil2 between nodes \ and 2 defines the admittance matrix element (1,2) or YI 2 = Y12 !J...u.. = - Y 12' From elementary network theory we can show that the power at node \ is given by PI = ffi.eEfr or d
Now define Gil = YII PI
COSOIi
and 'Y = 012
-
rr/2, then
= E 2G II + EVYI 2 sin (0 - 'Y) = Pc + PM sin (0 - 'Y)
(2.41)
The relation between PI and 0 in (2.4\) is shown in Figure 2.9. Examining Figure 2.9, we note that the power-angle curve of a synchronous machine connected to an infinite bus is a sine curve displaced from the origin vertically by an amount Pc, which represents the power dissipation in the equivalent network, and horizontally by the angle 'Y, which is determined by the real component of the transfer admittance Y12 • In the special case where the shunt load at the machine terminal V, is open and where the tran smission network is reactive , we can easily prove that Pc = 0 and 'Y = O. In this case the power-angle curve becomes identical to that given in (2.33).
28
Chapter 2 p
I
I I I
- -- - - , - - - - -- I
Fig .2.9
Power output of a synchronous machine connected to an infin ite bus.
Example 2.3
A synchronous machine is connected to an infinite bus through a transformer and a double circuit transmission line, as shown in Figure 2.10. The infinite bus voltage V = 1.0 pu. The direct axis transient reactance of the machine is 0.20 pu, the transformer reactance is 0.10 pu, and the reactance of each of the transmission lines is 0.40 pu, all to a base of the rating of the synchronous machine. Initially, the machine is delivering 0.8 pu power with a terminal voltage of 1.05 pu. The inertia constant H = 5 MJ jMV A. All resistances are neglected. The equation of motion of the machine rotor is to be determined.
Fig . 2.10
System of Example 2.3.
Solution The equivalent circuit of the system is shown in Figure 2.11. For this system:
YI2
=
IjjO.5
=
-j2 .0
YIO = 0 f. 2 = j2.0
Y;. = -j2 .0 0" = -1rj2 012 = 1rj2
therefore, Pc = 0 and "I = O. The electrical power is given by P,
= PI = Pc +
EVY'2sin(~
- "I) =
EVY'2sin~
2Esin~
Since the initial power is p.o = 0.8 pu, then E sin ~o = 0.4.
Fig .2 .11
Initial equivalent circuit of the system of Example 2.3 .
The Elementary Mathematical Model
29
To find the initial conditions, we solve the network of Figure 2.11. We have the terminal condition
v= To find the angle of
~ = 1.05 L!!., pu
1.0 ~ pu
V;, we write, since resistance is zero, PeO
sin (}'o
= 0.8 =
(VV,/x)sinlJ,o
=
0.8/3.5
()'o = 13.21 The current is found from
t
P, = 0.8 pu
=
= (1.05jO.30)sinlJ,o
0.2286
0
V; = zT + V, or
(Y; - V)jZ = (1.05 L!ld!: - 1.0LQ)/jO.3 = (1.022 + jO.240 - 1.000)/jO.3 = 0.800 - jO.074 = 0.803 / - 5.29
=
0
Then the internal machine voltage is E 11- = 1.05/13.21 + (0.803 / - 5.29 )(0.2/90 0
0
0 )
1.022 + jO.240 + 0.0148 + jO.160 = 1.037 + jO.400 = 1.111 /21.09 pu 0
Thus E = 1.111 is a constant that will be unchanged during the transient, and the initial angel is 00 = 21.09 = 0.367 rad. We also may write 0
P, = [(1.111 x 1.0)/0.50) sin 0 = 2.222 sin 0 Then the swing equation is given by 2H d WR
20
dt?
=
Pm - P,
or 2
d 0 = 377 (0.8 _ 2.222 sin 0) rad/s' dt? 10 From this simple example we observe that the resulting swing equation is nonlinear and will be difficult to solve except by numerical methods. We now extend the example to consider a fault on the system. Example 2.4
Develop the equation of motion of the system of Figure 2.11 where a fault is applied at the sending end (node 4) of the transmission line. For simplicity we will consider a three-phase fault that presents a balanced impedance of jO.1 to neutral. The network now is as shown in Figure 2.12, where admittances are used for convenience. Solution By Y-Ll transformation we compute
YI2 and since ~2 is now
=
= -
-YI2, then Y;2
j (3.333 x 5)/18.333] = - jO.909 =
jO.909. The electrical power output of the machine
P, = (0.909 x 1.111)sin I) = 1.010sin 0
30
Chapter 2
CD
-j 3. 333
8)
- j 5.0
CD
Fig.2 .12 Faulted network for Example 2.4 in terms of admillances .
From Example 2.3 the equation of motion of the rotor is
d 20
dl 2 = 37.7(0.8 - 1.0lOsino) rad/s?
At the start of the transient sin 00 =0.36, and the initial rotor acceleration is given by
d d[2 20
= 37.7[0.8 - (1.010 x 0.368)J = 16.45 rad/s!
Now let us assume that after some time the circuit breaker at the sending end of the faulted line clears the fault by opening that line. The network now will have a series reactance of jO.70 pu, and the new network (with fault cleared) will have a new value of transfer admittance, Yl2 = j 1.429 pu . The new swing equation will be
d 20
dl 2 = 37.7(0.8 - 1.587 sin 0) rad/s?
Example 2.5 Calculate the angle 0 as a function of time for the system of Examples 2.3 and 2.4. Assume that the fault is cleared in nine cycles (0.15 s).
Solution The equations for 0 were obtained in Example 2.4 for the faulted network and for the system with the fault cleared. These equations are nonlinear; therefore, time solutions will be obtained by numerical methods. A partial survey of these methods is given in Appendix B. To illustrate the procedure used in numerical integration, the modified Euler method is used in this example. This method is outlined in Appendix B. First, the swing equation is replaced by the two first-order differential equations: ~
= w(l)
-
WR
(2.42)
The time domain is divided into increments called t::.l . With the values of 0 and wand their derivatives known at some time I, an estimate is made of the values of these variables at the end of an interval of time t::.l, i.e., at time 1 + t::.l. These are called the predicted values of the variables and are based only on the values of 0(1), w(l), and their derivatives. From the calculated values of 0(1 + t::.l) and w(l + t::.l), values of the derivatives at 1 + t::.l are calculated. A corrected value of 0(1 + t::.l) and w(l + t::.l) is obtained using the mean derivative over the interval. The process can be repeated until a desired precision is achieved. At the end of this repeated prediction and correction a final value of 0(1 + t::.l) and w(l + t::.l) is obtained. The process is then repeated for the next interval. The procedure is outlined in detail in Chapter 10 of [8J. From Example 2.4 the initial value of 0 is sin - I 0.368, and the equation
The Elementary Mathematical Model
31
50
o
0 .2
0.4
0.6 0.8 Time, s
1.0
1.2
1.4
Fig.2 .13 Angle-ti me curve for Example 2.5.
for w is given by w
= 37.7(0.800 - 1.010sino) = 37.7(0.800 - 1.587 sin 0)
o~
t < 0.15
t ~ 0.15
The results of the numerical integration of the system equations, performed with the aid of a digital computer, are shown in Figure 2.13. The time solution is carried out for two successive peaks of the angle o. The first peak of 48.2° is reached at t = 0.38 s, after which 0 is decreased until it reaches a minimum value of about 13.2° at t = 0.82 s, and the oscillation of the rotor angle () continues. For the system under study and for the given impact, synchronism is not lost (since the angle 0 does not increase indefinitely) and the synchronous machine is stable.
2.8
Equal Area Criterion
Consider the swing equation for a machine connected to an infinite bus derived previously in the form (2.43) where P, is the accelerating power. From (2.43) 20
d dt?
=~
P 2H·
(2.44)
32
Chapter 2
Multiplying each side by 2(do/dt),
20 2 do d = (~ P.)(2 dO) dt dt? 2H dt
~~(dO)2J = WR dt ~ dt
d
p do H· dt
[(~~r] = W; P.do
(2.45) (2.46) (2.47)
Integrating both sides,
(2.48) or
(2.49) Equation (2.49) gives the relative speed of the machine with respect to a reference frame moving at constant speed (by the definition of the angle 0). For stability this speed must be zero when the acceleration is either zero or is opposing the rotor motion. Thus for a rotor that is accelerating, the condition of stab ility is that a value Om.. exists such that p.(Om.. ) ~ 0, and
(2.50) If the accelera ting power is plotted as a function of 0, equation (2 .50) can be interpreted as the area under that curve between 00 and Om.. ' This is shown in Fig-
+ P (I = 0 ) o
-
(0)
P
o
(b )
Fig.2.14
Equ al area cr iteria : (a) for sta bility for a stable system, (b ) for an unstable system.
The Elementary Mathematical Model
33
ure 2.14(a) where the net area under the P, versus 5 curve adds to zero at the angle since the two areas A I and A 2 are equal and opposite. Also at 5max the accelerating power, and hence the rotor acceleration, is negative. Therefore, the system is stable and
omax
is the maximum rotor angle reached during the swing. If the accelerating power reverses sign before the two areas A I and A 2 are equal, synchronism is lost. This situation is shown in Figure 2.14(b). The area A 2 is smaller than A I' and as 0 increases beyond the value where P, reverses sign again, the area A 3 is added to A I. The limit of stability occurs when the angle omax is such that Pa(omax) = 0 and the areas A I and A 2 are equal. For this case omax coincides with the angle Om on the power-angle curve with the fault cleared such that P = Pm and () > 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 <5 t , <5 2 , and <5 3 and their difference versus time. Solution
The problem is to solve the set of equations (2.56) for n = 3 and D = O. All the coefficients for the faulted network and the network with the fault cleared have been determined in Example 2.6. Since the set (2.56) is nonlinear, the desired time solutions for <5 1, <5 2, and <5) are obtained by numerical integration. A brief survey of numerical integration of differential equations is given in Appendix B. (For hand calculations see [I] for an excellent discussion of a numerical integration method of the swing equa-
- j4.1684
I
2 3 4 5 6 7 8 9
j8.4459
-j8.4459
j5.4855
-j5.4855
2
j4.1684
-j4.1684
3
I
j8.4459
Node
- j5.4855
-j4.1684
- j8.4459
2
j5.4855
2 3 4 5 6 7 8 9
I
I
j8.4459
3
- j4.1684
3
Node
- j5.4855
2
j4.1684
-j8.4459
I
2 3 4 5 6 7 8 9
I
Node
3.3074 - j30.3937 - 1.3652 + j 11.6041 -1.9422 + j 10.5107
j8.4459
4
5
- 1.2820 + j5 .5882
- 1.3652 + j 11.6041 2.6262 - j 11.8675
5
- 1.2820 + j5.5882
4.1019 - j16.1335
-1.9422 + j 10.5107
6
J.6171 - j 18.9559 -1.6171 + j13.6980
j5.4855
7
7
2.8047 - j24.93 11 -1.6171 + j13.6980
Y Matrix of Network with Fault Cleared
- 1.2820 + j5.5882
4.1019 - j16.1335
-1.9422 + j10.5107
6
j5.4855
7
- 1.1876 + j5.9751
Y Matrix of Faulted Network
- 1.3652 + j 11.6041 3.8138 - j17.8426
5
Table 2.4.
- 1.1876 + j5.9751
4.1019 - j16.1335
- 1.9422 + j 10.5107
6
Y Matrix of Prefault Network
- 1.3652 + j 11.6041 3.8138 - j 17.8426
Table 2.5.
3.3074 - j30.3937 - 1.3652 + j 11.6041 -1.9422 + j 10.5107
j8.4459
4
3.3074 - j30.3937 - 1.3652 + j 11.6041 - 1.9422 + j 10.5107
j8.4459
4
Table 2.3.
-1.6171 + j13.6980 3.7412 - j23.6424 - I. I551 + j9. 7843
8
3.7412 - j23.6424 - 1.1551 + j9.7843
8
- 1.6171 + j 13.6980 3.7412 - j23.6424 - 1.1551 + j9.7843
8
- l.1551 + j9.7843 2.4371 - j 19.2574
- I .2820 + j5 .5882
j4.1684
9
-1.1551 +j9.7843 2.4371 - j19.2574
- 1.2820 + j5.5882
j4.1684
9
-1.1551 +j9.7843 2.4371 - j19.2574
- 1.2820 + j5.5882
j4.1684
9
Chapter 2
44 Table 2.6.
Type or network
Reduced Y Matrices
Node
2
3
Prefault
I 2 3
0.846 - j2 .988 0.287 + j 1.513 0.210 +j1.226
0.287 + j 1.513 0.420 - j2.724 0 .213 + j 1.088
0.210 + j 1.226 0.213 + j 1.088 0 .277 - j2.368
Faulted
I 2 3
0 .657 - j3 .816 0.000 + jO.OOO 0.070 + jO.631
0.000 + jO.OOO 0 .000 - j5.486 0.000 + jO.OOO
0.070 + jO.631 0.000 + jO.OOO 0.174 - j2 .796
Fault cleared
1 2 3
1.181 - j2 .229 0.138 +jO .726 0.191 + j 1.079
0 .138 +jO .726 0.389 - j 1.953 0.199 + j l.2 29
0.191 + j 1.079 0.199 + jl.229 0.273 - j2 .342
tion. Also see Chapter 10 of [81 for a more detailed discussion of several nume rical schemes for solving the swing equation.) The so-called transient stabil ity digital computer programs available at many computer centers include subroutines for solving nonlinear differential equations . Discussion of these programs is beyond the scope of th is book . Numerical integration of the swing equations for the three-generator. nine-bus system is made by digital computer for 2.0 s of simulated real time. Figure 2.21 shows the rotor angles of the three machine s. A plot of 021 = 02 - 0, and 0JI = oJ - 0. is shown 400
v>~
360
u
/.
'" .!:
/ j
I ' III 6.1/
'0
320
~
Z
v -s
280 .f I
/I~J
I
I I
240 Ii
.s ~
00' -
II
I I
I I
200
I I I
I 1 I
16
I
I I I I
120
6.
/
80
I
I /
--
.-/
/ /;,/ '
I I
1/ ;/ II
/ / '
63
40 :/ /
)/ I
0
I
0
0 .5
1. 5 Time, s
Fig, 2.2\
Plot or 01.02. and OJ versustime.
I
2.0
45
The Elementary Mathematical Model
('
I \ I \
I
\
I I
\ \
I
\
\
/
V o o
20
40
0.5 Time.,. s
Fig. 2.22
~
1. 5
2.0
Plot of 0 differences versus time .
in Figure 2.22 where we can see that the system is stable. The maximum angle difference is about 85° . This is the value of 021 at t = 0.43 s. Note that the solution is carried out for two "swings" to show that the second swing is not greater than the first for either 021 or 0JI ' To determine whether the system is stable or unstable for the particular transient under study, it is sufficient to carry out the time solution for one swing only. If the rotor angles (or the angle differences) reach maximum values and then decrease, the system is stable. If any of the angle differences increase indefinitely, the system is unstable because at least one machine will lose synchronism. 2.11
Shortcomings of the Classical Model
System stability depends on the characteristics of all the components of the power system . This includes the response characteristics of the control equipment on the turbogenerators, on the dynamic characteristics of the loads, on the supplementary control equipment installed , and on the type and settings of protective equipment used. The machine dynamic response to any impact in the system is oscillatory . In the past the sizes of the power systems involved were such that the period of these oscillations was not much greater than one second . Furthermore, the equipment used for excitation controls was relatively slow and simple. Thus the classical model was adequate. Today large system interconnections with the greater system inertias and relatively weaker ties result in longer periods of oscillations during transients. Generator control systems, particularly modern excitation systems, are extremely fast. It is therefore
46
Chapter 2
questionable whether the effect of the control equipment can be neglected during these longer periods. Indeed there have been recorded transients caused by large impacts, resulting in loss of synchronism after the system machines had undergone several oscillations. Another aspect is the dynamic instability problem, where growing oscillations have occurred on tie lines connecting different power pools or systems. As this situation has developed, it has also become increasingly important to ensure the security of the bulk power supply. This has made many engineers realize it is time to reexamine the assumptions made in stability studies. This view is well stated by Ray and Shipley [14]: We have reached a time when it is appropriate that we appraise the state of the Art of Dynamic Stability Analysis. In conjunction with this we must:
I. Expand our knowledge of the characteristic time response of our system loads to changes in voltage and frequency-develop new dynamic models of system loads. 2. Re-examine old concepts and develop new ideas on changes in system networks to improve system stability. 3. Update our knowledge of the response characteristics of the various components of energy systems and their controls (boilers, reactors, turbine governors, generator regulators, field excitation, etc.) 4. Reformulate our analytical techniques to adequately simulate the time variation of all of the foregoing factors in system response and accurately determine dynamic system response.
Let us now make a critical appraisal of some of the assumptions made in the classical model: I. Transient stability is decided in the first swing. A large system having many machines
will have numerous natural frequencies of oscillations. The capacities of most of the tie lines are comparatively small, with the result that some of these frequencies are quite low (frequencies of periods in the order of 5-6 s are not uncommon). It is quite possible that the worst swing may occur at an instant in time when the peaks of
some of these nodes coincide. It is therefore necessary in many cases to study the
transient for a period longer than one second.
2. Constant generator main field-winding flux linkage. This assumption is suspect on two
counts, the longer period that must now be considered and the speed of many modern voltage regulators. The longer period, which may be comparable to the field-winding time constant, means that the change in the main field-winding flux may be appreciable and should be accounted for so that a correct representation of the system voltage is realized. Furthermore, the voltage regulator response could have a significant effect on the field-winding flux. We conclude from this discussion that the constant voltage behind transient reactance could be very inaccurate. 3. Neglecting the damping powers. A large system will have relatively weak ties. In the spring-mass analogy used above, this is a rather poorly damped system. It is important to account for the various components of the system damping to obtain a correct model that will accurately predict its dynamic performance, especially in loss of generation studies [8]. 4. Constant mechanical power. If periods on the order of a few seconds or greater are of interest, it is unrealistic to assume that the mechanical power will not change. The turbine-governor characteristics, and perhaps boiler characteristics should be included in the analysis. 5. Representing loads by constant passive impedance. Let us illustrate in a qualitative manner the effect of such representation. Consider a bus having a voltage V to which a load PL + jQL is connected. Let the load be represented by the static ad-
The Elementary Mathematical Model
Fig .2 .23
47
A load represented by pas sive ad mitt a nce.
mittances GL = Pd V2 and BL = Qd V2 as shown in Figure 2.23 . During a transient the voltage magnitude V and the frequency will change. In the model used in Figure 2.17 the change in voltage is reflected in the power and reactive power of the load, while the change in the bus frequency is not reflected at all in the load power. (n other words, this model assumes PL 0( V2, QL 0( V2, and that both are frequency independent. This assumption is often on the pessimistic side. (There are situations, however, where this assumption can lead to optimistic results. This discussion is intended to illustrate the errors implied .) To illustrate this, let us assume that the transient has been initiated by a fault in the transmission network. Initially, a fault causes a reduction of the output power of most of the synchronous generators. Some excess generation results, causing the machines to accelerate, and the area frequency tends to increase. At the same time, a transmission network fault usually causes a reduction of the bus voltages near the fault location. In the passive impedance model the load power decreases considerably (since PL 0( V2), and the increase in frequency does not cause an increase in load power. In real systems the decrease in power is not likely to be proportional to V2 but rather less than this . An increase in system frequency will result in an increase in the load power. Thus the model used gives a load power lower than expected during the fault and higher than normal after fault removal. From the foregoing discussion we conclude that the classical model is inadequate for system representation beyond the first swing. Since the first swing is largely an inertial response to a given accelerating torque, the classical model does provide useful information as to system response during this brief period. 2.12
Block Diagram of One Machine
Block diagrams are useful for helping the control engineer visualize a problem . We will be considering the control system for synchronous generators and will do so by analyzing each control funct ion in turn. It may be helpful to present a general block diagram of the entire system without worrying about mathematical details as to what makes up the various blocks. Then as we proceed to analyze each system , we can fill in the blocks with the appropriate equations or transfer functions . Such a block diagram is shown in Figure 2.24 [15). The basic equation of the dynamic system of Figure 2.24 is (2.18); i.e., (2.66) where ;5 has been replaced by ciJ and J has been replaced by a time constant Tj , the numerical value of which depends on the rotating inertia and the system of units. Three separate control systems are associated with the generator of Figure 2.24. The first is the excitation system that controls the terminal voltage. Note that the excitation system also plays an important role in the machine's mechanical oscillations, since it affects the electrical power, p•. The second control system is the speed control or governor that monitors the shaft speed and controls the mechanical power Pm .
48
Chapter 2
Fig. 2.24
Block d iagr am of a synchronous generator con trol system.
Finally, in an interconnected system there is a master controller for each system . This sends a unit dispatch signal (UDS) to each generator and adjusts this signal to meet the load demand or the scheduled tie-line power. It is designed to be quite slow so that it is usually not involved in a consideration of mechanical dynamics of the shaft. Thus in most of our work we can consider the speed reference or governor speed changer (GSC) position to be a constant. In an isolated system the speed reference is the desired system speed and is set mechanically in the governor mechanism. as will be shown later. In addition to the three control systems, three transfer functions are of vital importance. The first of these is the generator transfer function. The generator equations are nonlinear and the transfer function is a linearized approximation of the behavior of the generator terminal voltage V, near a quiescent operating point or equilibrium state. The load equations are also nonlinear and reflect changes in the electrical output quantities due to changes in terminal voltage v,. Finally. the energy source equations are a description of the boiler and steam turbine or of the penstock and hydraulic turbine behavior as the governor output calls for changes in the energy input. These equations are very nonlinea r and have several long time constants. To visualize the stability problem in terms of Figure 2.24 . we recognize immediately that the shaft speed w must be accurately controlled since this machine must operate at precisely the same frequency as all others in the system. If a sudden change in woccurs. we have two ways of providing controlled responses to this change. One is through the governor that controls the mechanical power Pm. but does so through some rather long time constants. A second controlled response acts through the excitation system to control the electrical power p. . Time delays are involved here too. but they are smaller than those in the governor loop. Hence much effort has been devoted to refinements in excitation control.
Problems 2.1 2.2
Analyze (2 .1) dimensionally using a mass, length, time system and specify the units of each quantity (see Kimbark [I D. A rotating shaft has zero retarding torque T. = 0 and is supplied a constant full load accelerating torque; i.e.• Tm = Tn . Let T( be the accelerating time constant. i.e.• the time required to accelerate the machine from rest to rated speed WR' Solve the swing equation to find T( in terms of the moment of inertia J, WR, and Tn . Then show that T( can also be related to H, the pu inertia constant.
The Elementary Mathematical Model 2.3 2.4
49
Solve the swing equation to find the time to reach full load speed WR starting from any initial speed Wo with constant accelerating torque as in Problem 2.2. Relate this time to T c and the slip at speed woo Write the equation of motion of the shaft for the following systems: (a) An electric generator driven by a dc motor, where in the region of interest the generator torque is proportional to the shaft angle and the motor torque decreases linearly with increased speed. (b) An electric motor driving a fan, where in the region of interest the torques are given by Tmotor
=
0 -
b8
T fan
= c0 2
where 0, b, an.d c are constants. State any necessary assumptions. Will this system have a steady-state operating point? Is the system linear? 2.5 In (2.4) assume that T is in N· m, ~ is in elec. deg., and J is in lbm -ft2. What factor must be used to make the units consistent? 2.6 In (2.7) assume that P is in Wand M in J. s/rad. What are the units of lJ? 2.7 A SOO-MV A two-pole machine is to operate in parallel with other U .S. machines, Compute the regulation R of this machine. What are the units of R? 2.8 A 60-MVA two-pole generator and a 600-MVA four-pole generator are to operate in parallel with other U.S. systems and are to share in system governing. Compute the pu constant K that must be used with these machines in their governor simulations if the system base is 100 MVA. 2.9 Repeat problem 2.8 if the constant K is to be computed in MKS units rather than pu. 2.10 In computer simulations it is common to see regulation expressed in two different ways as described below: (a) Pm - Pmo
where
Pm = mechanical power in pu on SsB = initial mechanical power in pu on Sso f = system base frequency in Hz Rsu steady-state speed regulation in pu on a system base = RuSso/So s = generator slip = (WR - w)/21r Hz
PmO
(b) Pm - Pmo
where
s/fRsu
=
Pm PmO K, ~W
=
K,~w
pu.
turbine power in pu on Sso initial turbine power in pu on Sol RuWR Sso speed deviation, rad/s
SsB
Verify the expressions in (a) and (b). 2.11 A synchronous machine having inertia constant H = 4.0 MJ /MVA is initially operated in steady state against an infinite bus with angular displacement of 30 elec. deg. and delivering 1.0 pu power. Find the natural frequency of oscillation for this machine, assuming small perturbations from the operating point. 2.12 A solid-rotor synchronous generator is driven by an unregulated turbine with a torque speed characteristic similar to that of Figure 2.3(a). The machine has the same characteristics and operating conditions as given in Problem 2.1 t andis connected to an infinite bus. Find the natural frequency of oscillation and the damping coefficient, assuming small perturbations from the operating point. 2.13 Suppose that (2.33) is written for a salient pole machine to include a reluctance torque term; i.e., let P = PM sin lJ + k sin 2~. For this condition find the expression for PA and for the synchronizing power coefficient. 2.14 Derive an expression similar to that of (2.7) for an interconnection of two finite machines that have inertia constants M, and M 2 and angles 0, and ~2' Show that the equations for such a case are exactly equivalent to that of a single finite machine of inertia
M and angle 0,2
=
M, M2/(M, + M 2)
0, - 02 connected to an infinite bus.
50 2.15
2.16
Chapter 2 Derive linearized expressions (similar to Example 2.2) that describe an interconnection of three finite machines with inertia constants M I , M z, and M ) and angles 15•• 15 z, and 15). Is there a simple expression for the natural frequency of oscillation in this case? Designate synchronizing power between machines 1 and 2 as Ps l Z, etc . The system shown in F igure P2 .16 ha s two finite synchronous machines, each represented by a constant voltage behind reactance and connected by a pure reactance. The reactance x includes the transm ission line and the machine reactances . Write the swing equation for each machine, and show that this system can be reduced to an equivalent one machine aga inst an infinite bus. Give the inertia constant for the equivalent machine, the mechanical input power, and the amplitude of its power-angle curve. The inertia constants of the two machines are HI and Hz s.
"62 H,
2.17
roT"
~
-
Fig. P2.16
E·A· H.
The system shown in Figure P2 .17 comprises four synchronous machines. Machines A and Bare 60 Hz, while machines C and Dare 50 Hz; Band C are a motor-generator set (frequency changer). Write the equations of motion for this system . Assume that the transmission networks are reactive .
Fig. P2 .1?
2.18
The system shown in Figure P2.18 has two generators and three nodes. Generator and transmission line data are given below. The result of a load-flow study is also given . A three-phase fault occurs near node 2 and is cleared in 0.1 s by removing line 5.
CD
eY1
®
3 6
1M Fig. P2.18
4
S
0
fa
(a) Perform all preliminary calculations for a stability study . Convert the system to a common 100-M VA base, convert the loads to equivalent passive impedances, and calculate the generator internal voltages and initial angles. (b) Calculate the Y matrices for prefault, faulted, and postfault conditions . (c) Obtain (numerically) time solutions for the internal general angles and determine if the system is stable.
The Elementary Mathematical Model
51
Generator Data (in pu to generator MVA base)
xrt
Xd
Generator number
(pu)
(pu)
(MW-s/MVA)
H
Rating (MVA)
I 3
0.28 0.25
0.08 0.07
5 4
50 120
tXT:: generator transformer reactance
Transmission Line Data (resistance neglected) Line number:
Xpu to 100-MVA base
3
4
5
6
0.08
0.06
0.06
0.13
Load-Flow Data Generator
Load
Voltage
Bus no.
Magnitude pu
Angle"
MW
MVAR
MW
MVAR
I 2 3
1.030 1.018 1.020
0.0 -1.0 -0.5
0.0 50.0 80.0
0.0 20.0 40.0
30.0 0.0 100.0
23.1 0.0 37.8
2.19
2.20
Reduce the system in Problem 2.18 to an equivalent one machine connected to an infinite ... bus. Write the swing equation for the faulted network and for the network after the fault is cleared. Apply the equal area criterion to the fault discussed in Problem 2.18. What is the critical clearing angle? Repeat the calculations of Example 2.4, but with the following changes in the system of Figure 2.11. (a) Use a fault impedance of 2, = 0.01 + jO pu. This is more typical of the arcing resistance commonly found in a fault. (b) Study the damping effect of adding a resistance to the transmission lines of R L in each line where R L = 0.1 and 0.4 pu. To measure the damping, prepare an analog computer simulation for the system. Implementation will require computation of VII' Y12' the initial conditions, and the potentiometer settings.
(c) Devise a method of introducing additional damping on the analog computer by adding a term Kd~ in the swing equation. Estimate the value of Kd by assuming that a slip
of 2.5% gives a damping torque of 50% of full load torque. (d) Make a parametric study of changes in the analog simulation for various values of H. For example, let H = 2.5,5.0,7.5 s. 2.21 Repeat Problem 2.20 but with transmission line impedance for each line of R L + jO.8, where R L = 0.2, 0.5, 0.8 pu. Repeat the analog simulation and determine the critical clearing time to the nearest cycle. This will require a means of systematically changing from the fault condition to the postfault (one line open) condition after a measured time lapse. This can be accomplished by logical control on some analog computers or by careful hand switching where logical control is not available. Let V = 0.95. 2.22 Repeat Problem 2.21 using a line impedance of 0.2 + jO.8. Consider the effect of adding a "local" unity power factor load R LD at bus 3 for the following conditions: 00
Case I: PL D = 0.4 pu P + jQ = 0.4 ± jO.20 pu Case 2: PLD = value to give the same generated power as Case 1 P'; +jQoo =O+jO pu Case 3: PL D = 1.2 pu P + jQ = -0.4 ~ jO.2 pu 00
00
00
00
(a) Compute the values of R L D and E and find the initial condition for lJ for each case.
52
Chapter 2 (b) Compute the values of )ill and V12 for the prefault, faulted, and postfault condition. if the fault impedance is ZF = 0.01 + jO. Use the computer for this, writing the admittance matrices by inspection and reducing to find the two-port admittances. (c) Compute the analog computer settings for the simulation. (d) Perform the analog computer simulation and plot the following variables: Tm , T, Ta , W A , 0, 012 - o. Also, make a phase-plane plot of w'" versus o. Compare these results with similar plots with no local load present. (e) Use the computer simulation to determine the critical clearing angle.
References 1. 2. 3. 4. 5. 6.
7. 8. 9. 10. II. 12. 13. 14.
15.
Kimbark , E. W. Power System Stability. Vol. I. Wiley, New York, 1948. Stevenson, W. D. Elements of Power System Analysis. 2nd ed. McGraw-Hill, New York, 1962. Federal Power Commission. National Power Survey. Pt. 2. USGPO, Washington, D.C., 1964. Lok ay, H. E., and Thoits, P. O. Effects of future turbine-generator characteristics on transient stability . .IEEE Trans. PAS-90:2427- 31,1971. AI EE Subcommittee on Interconnection and Stability Factors. First report of power system stability. Electr. Eng. 56:261··82, 1937. Venik ov. V. A. Transient Phenomena in Electrical Power Systems. Pergamon Press. Macmillan, New York, 1964. Crary, S. B. Power System Stability. Vol. 2. Wiley, New York, 1947. Stagg, G. W., and El-Abiad, A. H. Computer Methods in Power System Analysis. Mct.iraw-Hill, New York, 1968. Concordia, C. Effect of steam turbine reheat on speed-governor performance. ASM E J. Eng. Power 81:201-6,1959. Kirchmayer, L. K. Economic Control ofInterconnected Systems. Wiley, New York. 1959. Young. C. C.• and Webier. R. M. A new stability program for predicting the dynamic performance of electric power systems. Proc. AnI. Power Con]. 29: 1126-39. 1967. Byerly, R. T.• Sherman, D. E., and Shortley, P. B. Stability program data preparation manual. Westinghouse Electric Corp. Repl. 70--736. 1970. (Rev. Dec. 1971.) Concordia, C. Synchronous machine damping and synchronizing torques. A lEE Trans. 70:731-37, 1951. Ray, J. J., and Shipley. R. B. Dynamic system performance. Paper 66 CP 709-PWR, presented at the IEEE Winter Power Meeting. New York. 1968. Anderson, P. M., and Nanakorn, S. An analysis and comparison of certain low-order boiler models. ISA Trans. 14:17-23,1975.
chapter
3
System Response to Small Disturbances
3.1
Introduction
This chapter reviews the behavior of an electric power system when subjected to small disturbances. It is assumed the system under study has been perturbed from a steady-state condition that prevailed prior to the application of the disturbance. This small disturbance may be temporary or permanent. If the system is stable, we would expect that for a temporary disturbance the system would return to its initial state, while a permanent disturbance would cause the system to acquire a new operating state after a transient period. In either case synchronism should not be lost. Under normal operating conditions a power system is subjected to small disturbances at random. It is important that synchronism not be lost under these conditions. Thus system behavior is a measure of dynamic stability as the system adjusts to small perturbations. We now define what is meant by a small disturbance. The criterion is simply that the perturbed system can be linearized about a quiescent operating state. An example of this linearization procedure was given in Section 2.5. While the power-angle relationship for a synchronous machine connected to an infinite bus obeys a sine law (2.33), it was shown that for small perturbations the change in power is approximately proportional to the change in angle (2.35). Typical examples of small disturbances are a small change in the scheduled generation of one machine, which results in a small change in its rotor angle 0, or a small load added to the network (say 1/ 100 of system capacity or less). In general, the response of a power system to impacts is oscillatory. If the oscillations are damped, so that after sufficient time has elapsed the deviation or the change in the state of the system due to the small impact is small (or less than some prescribed finite amount), the system is stable. If on the other hand the oscillations grow in magnitude or are sustained indefinitely, the system is unstable. 'For a linear system, modern linear systems theory provides a means of evaluation of its dynamic response once a good mathematical model is developed. The mathematical models for the various components of a power network will be developed in greater detail in later chapters. Here a brief account is given of the various phenomena experienced in a power system subjected to small impacts, with emphasis on the qualitative description of the system behavior. S3
Chapter 3
54
3.2
Types of Problems Studied
The method of small changes, sometimes called the perturbation method [I, 2, 3], is very useful in studying two types of problems: system response to small impacts and the distribution of impacts. 3.2.1
System response to small impacts
If the power system is perturbed, it will acquire a new operating state. If the perturbation is small, the new operating state will not be appreciably different from the initial one. In other words, the state variables or the system parameters will usually not change appreciably. Thus the operation is in the neighborhood of a certain quiescent state Xo. In this limited range of operation a nonlinear system can be described mathematically by linearized equations. This is advantageous, since linear systems are more convenient to work with. This procedure is particularly useful if the system contains control elements. The method of analysis used to linearize the differential equations describing the system behavior is to assume small changes in system quantities such as ~b.' Vb., Pb. (change in angle, voltage, and power respectively). Equations for these variables are found by making a Taylor series expansion about Xo and neglecting higher order terms [4, 5, 6]. The behavior or the motion of these changes is then examined. In examining the dynamic performance of the system, it is important to ascertain not only that growing oscillations do not result during normal operations but also that the oscillatory response to small impacts is well damped. If the stability of the system is being investigated, it is often convenient to assume that the disturbances causing the changes disappear. The motion of the system is then free. Stability is then assured if the system returns to its original state. Such behavior can be determined in a linear system by examining the characteristic equation of the system. If the mathematical description of the system is in state-space form, i.e., if the system is described by a set of first-order differential equations,
x=
Ax
+ Bu
(3.1)
the free response of the system can be determined from the eigenvalues of the A matrix. 3.2.2
Distribution of power impacts
When a power impact occurs at some bus in the network, an unbalance between the power input to the system and the power output takes place, resulting in a transient. When this transient subsides and a steady-state condition is reached, the power impact is "shared" by the various synchronous machines according to their steady-state characteristics, which are determined by the steady-state droop characteristics of the various governors [5, 1~. During the transient period, however, the power impact is shared by the machines according to different criteria. If these criteria differ appreciably among groups of machines, each impact is followed by oscillatory power swings among groups of machines to reflect the transition from the initial sharing of the impact to the final adjustment reached at steady state. Under normal operating conditions a power system is subjected to numerous random power impacts from sudden application or removal of loads. As explained above, each impact will be followed by power swings among groups of machines that respond to the impact differently at different times. These power swings appear as power oscil-
System Response to Small Disturbances
55
lations on the tie lines connecting these groups of machines. This gives rise to the term "tie-line oscillations." In large interconnected power systems tie-line oscillations can become objectionable if their magnitude reaches a significant fraction of the tie-line loading, since they are superimposed upon the normal flow of power in the line. Furthermore, conditions may exist in which these oscillations grow in amplitude, causing instability. This problem is similar to that discussed in Section 3.2.1. It can be analyzed if an adequate mathematical model of the various components of the system is developed and the dynamic response of this model is examined. If we are interested in seeking an approximate answer for the magnitude of the tie-line oscillations, however, such an answer can be reached by a qualitative discussion of the distribution of power impacts. Such a discussion is offered here. 3.3
The Unregulated Synchronous Machine
We start with the simplest model possible, i.e., the constant-voltage-behind-transient-reactance model. The equation of motion of a synchronous machine connected to an infinite bus and the electrical power output are given by (2.18) and (2.41) respectively or 2H d 20 WR dt?
- - = Pm - Pe P, Letting 0
=
=
Pc + PMsin(o - 1')
(3.2)
00 + OA, Pe = PeO + PA , P; = PmO and using the relationship
sin (0 - 'Y)
sin (00 - 'Y + OA)
=
"J
sin (00 - 1') + cos (00 - 'Y)O/i
(3.3)
the linearized version of (3.2) becomes 2H d 20 A - - + Pso A = 0 WR dt?
-
(3.4)
where P,
=
dPeJ do
PMcos(oo - ,,)
(3.5)
60
The system described by (3.4) is marginally stable (i.e., oscillatory) for Ps > O. Its response is oscillatory with the frequency of oscillation obtained from the roots of the characteristic equation (2H/WR)S2 + P, = 0, which has the roots (3.6)
If the electrical torque is assumed to have a component proportional to the speed change, a damping term is added to (3.4) and the new characteristic equation becomes (2H/WR)S2
+
(D/WR)S
+ Ps = 0
(3.7)
where D is the damping power coefficient in pu, The roots of (3.7) are given by s
=
(3.8)
56
Chapter 3
UsuaJly (D/WR)2 < 8HPs / WR, and the roots are complex; i.e., the response is oscillatory with an angular frequency of oscillation essentially the same as that given by (3.6). The system described by (3.7) is stable for P, > 0 and for D > O. If either one of these quantities is negative, the system is unstable. Venikov [4) reports that a situation may occur where the .machine described by (3.4) can be unstable under light load conditions if the network is such that ~o < "Y. This would be the case where there is appreciable series resistance (see [4], Sec. 3.2). From Chapter 2 we know that the synchronizing power coefficient P, is negative if the spontaneous change in the angle ~ is .negative. A negative value of P, leads to unstable operation. 3.3.1
Demagnetizing effect of armature reaction
The model of constant main field-winding flux linkage neglects some important effects, among them the demagnetizing influence of a change in the rotor angle o.. To account for this effect, another model of the synchronous machine is used. It is not our concern in this introductory discussion to develop the model or even discuss it in detail, as this will be accomplished in Chapter 6. Rather, we will state the assumptions made in such a model and give some of the pertinent results applicable to this discussion. These results are found in de Mello and Concordia [8] and are based on a model previously used by Heffron and Phillips [9]. To account for the field conditions, equations for the direct and quadrature axis quantities are derived (see Chapter 4). Major simplifications are then made by neglecting saturation, stator resistance, and the damper windings. The transformer voltage terms in the stator voltage equations are considered negligible compared to the speed voltage terms. Linearized relations are then obtained between small changes in the electrical power PeA, the rotor angle 0A' the field-winding voltage VF~' and the voltage proportional to the main field-winding flux E~. For a machine connected to an infinite bus through a transmission network, the
following s domain relations are obtained,
(3.9) (3.10)
where K , is the change in electrical power for a change in rotor angle with constant flux linkage in the direct axis, K 2 is the change in electrical power for a change in the direct axis flux linkages with constant rotor angle, TdO is the direct axis open circuit time constant of the machine, K 3 is an impedance factor, and K 4 is the demagnetizing effect of a change in the rotor angle (at steady state). Mathematically, we write
K1
=
PtA/OAlE'A-O
K2
=
PeA/E~llJA·O
final value of unit step vF response K4
= -
.1- lim EA(t)] K3
1-00
=
VF~=O
lim
1-00
E~(t)ts
A
.0
(3.11)
lJA=U(I)
The constants K K 2 , and K 4 depend on the parameters of the machine, the external network, and"the initial conditions. Note that K 1 is similar to the synchronizing power coefficient Ps used in the simpler machine model of constant voltage behind
57
System Response to Small Disturbances
Fig . 3.1
Primitive linear ized block diagram representation or a generator model.
transient reactance . Equations (3 .9) and (3 .10) , with the initial equat ion (3 .2), may be represented by the incremental block d iagram of Figure 3.1 . P,IJ. =
For the case where
v FIJ. =
(K
K
1 -
2K JK 4 )
I + KJT~OS
{) IJ.
+
K 2K3
I + KJT~OS
(3 .12)
VFIJ.
0,
(3 .13) where we can clearly identify both the synchronizing and the demagnetizing components . Substituting in the linearized swing equation (3.4), we obtain the new cha racterist ic equation , (with D = 0)
or we have the third-order system 5
J
K1wR + -I- 5 2 + -5 +
KJT;O
2H
W
~
2H
I --,(K 1 KJTdO
-
K 2KJK4) =
0
(3.14)
Note that all the con stants (3 .11) a re usually positive. Thus from Routh 's criterion [10] this system is stable if K, - K2 KJ K4 > 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 =
Then
x2 =
= (wR/2)(Ps12/ H l + Ps 13/ H 1 + Ps31/ H 3 ) = 104.096 (w R / 2)( Ps 32 / H 3
-
Psn / HI)
(w R/2)(Ps31/ H3 - P s21/H2) = (wR/2)(P s21/H2
= 59.524
= 33.841
+ P s23/H2 +
P s32/H3 )
=
153.460
-(1/2)[ -(all + a22) ± v(all + a 22 )2 - 4(alla22 - al2a2.)] -(1/2)[ -257.556 ± (66336 - 55841)112] = -77.555, -180
64
3
Chapter
Now we can compute the frequencies and periods shown in Table 3.2. T able 3.2.
Frequencies of Oscillat ion of a N ine-Bus System
Qu ant ity
x
w
rad/s
JHz Ts
Eigenv alue I
Eigenvalue 2
±j 8.807 8.807 1.402 0.713
±j13.416 13.416 2.135 0.468
Thus two frequencies , about 1.4 Hz and 2.1 Hz, should be observed in the intermach ine oscillations of the system . Th is can be approximately verified by an actual solution of the system by d igital co mp uter. The results of such a solution are shown in Figure 3.3, whe re absolute angles are given in Figure 3.3(a) and angle differences relative to ~I are given in Figure 3.3(b) . As might be expected, neither of the computed frequencies is clearly observed since the response is a combination of the two frequencies. A rough measurement of the peak-to-pe ak periods in Figure 3.3(b) gives per iods in the neighborhood of 0.7 s. Methods have been devised [3, II) by which a system such as the one in Example 3.2 can be transformed to a new frame of reference called the Jordan canonical form . In Jo rdan form the different frequencies of oscillation are clearly separated . In the form of equations norm ally used, the va riables 512 and 513 (or other a ngle differences) contain 35. 0
20.0
24. 0
19 .0
13. 0
18.0
2.0
17.0
-9 . 0
16 .0
-tl
;
--:11
~ - 20. 0 ~.
~ -0(
~
0." -." ,£ 0
G
-e
~
~ .i
-
-:.E
15.0
0."
,£ ."~
."
~ 14.0
3~
C 0
2
~.
~
-42 .0
§'" \: ~
1!CD
--.
- 31 .0
~
0: '0
1' 13 . 0
I."
12.0
-75.0
I I
11.0
- 86. 0
I
9.0
I
8 .0
~o
-0(
I ...
- 53. 0
I.§
-64 . 0
-9 7. 0
6 21
I I
0 .0
0 .500
1. 000
1.500 Time , s
{oj
Fig. 3.3
2 .000
2. 500
~
Q.
~--:
"-0
1'0
, "-
18 I.... I
I I 1
0 .0
0 .500
1.000
1.500
2 .000
2. 500
TIme ,s
(b)
Unregulated response of the nine-bu s system to a sud den load a pplicatio n at bus 8: (a) a bsolute angles, (b) ang les relati ve to 15 I .
System Response to Small Disturbances
65
"harmonic' terms generally involving all fundamental frequencies of oscillation. Hence
we have difficulty observing these frequencies in measured physical variables.
Example 3.3
Transform the system of Example 3.2 into the Jordan canonical form and show that in this form the system frequencies of oscillation are clearly distinguishable.
Solution The system equations for the three..machine problem are given by
o
: I
o
:0 I
----------~-_.
or
x=
I
-all
-a 12
:
-a21
-a 22
~
I
o
A x, where x is defined by
and the a coefficients are computed in Example 3.2. We now compute the eigenvectors of A, using any method [1,3, II] and call these vectors E E2 , E3 , and E4 • We then use these eigenvectors to define a matrix E. " jO.06266 : - jO.06266; 0.14523' -0.14523 I
E = [E I E2 E) E4 ]
=
I
I
jO.07543 : - jO.07543 : -0.13831:
0.13831
0.83069:
0.83069:
1.00000
1.00000:
1.00000: -0.95234 : -0.95234
I
1
I
I
I
.
1.00000:
I
where the numerical values are found by a suitable computer library routine. We now define the transformation x = E y to compute i = E Y = A x = A E y y = E-'AEy = DywhereD = diag(X1,X2,X 3,X4) . Performing the indicated numerical work, we compute
E-l
D=E-IAE=
- j3.5245
-j3.7008
0.2659
0.2792
j3.5245
j3.7008
0.2659
0.2792
-jI.9221
j 1.5967
0.2792
-0.2319
j 1.9221
-jl.5967
0.2792
-0.2319
-j13.2571
0.0
0.0
j13.2571
0.0
0.0
0.0
0.0
0.0
0.0
0.0
-j6.8854
0.0
0.0
0.0
j6.8854
or
66
Chapter 3
Substituting into
y
Dy, we can compute the uncoupled solution y; =
C;e>";1
i = 1,2,3,4
where C; depends on the initial conditions. This method of computing the distinct frequencies of oscillation is quite general and may be applied to systems of any size. For very large systems this may not be practical, however, since the eigenvector computation may be too costly. Finally, we note that the simple model used here assumes that no damping exists. In physical systems damping is usually present; therefore, the oscillatory response given above is usually damped. The magnitude of the damping, however, is such that the frequencies of oscillation given by the above equations are not appreciably affected. 3.5
Regulated Synchronous Machine
In this section we examine the effect of voltage and speed control equipment on the dynamic performance of the synchronous machine. Again we are interested in the free response of the system. We will consider two simple cases of regulation: a 'simple voltage regulator with one time lag and a simple governor with one time lag. 3.5.1
Voltage regulator with one time lag
Referring to Figure 2.24, we note that a change in the field voltage V F41 is produced by changes in either VREF or~. If we assume that VREF 41 = 0 and the transducer has no time lags, V F41 depends only upon ~~, modified by the transfer function of the excitation system. Analysis of such a system is discussed in Chapter 7. To simplify the analysis,. a rather simple model of the voltage regulator and excitation system is assumed. This gives the following s domain relation between the change in the exciter voltage V F41 and the change in the synchronous machine terminal voltage J!;~: (3.36)
where
K, = regulator gain Tt
= regulator time constant
To examine the effect of the voltage regulator on the system response, we return to the model discussed in Section 3.3 for a machine connected to an infinite bus through a transmission network. These relations are given in (3.9) and (3.10). To use (3.36), a relation between ~~, lJ 41 , and E~ is needed. Such a relation is developed in reference [8] and is in the form (3.37) where
K,
=
~A/lJ~]EA
K6 =
VtA/E~]6~
= change in terminal voltage with change in rotor angle for constant E' = change in terminal voltage with change in E' for constant {)
The system block diagram with voltage regulation added is shown in Figure 3.4. From (3.36) and (3.37) VFA
-[Kt/(l
=
+
Tts)](K5lJ~
+
(3.38)
K6E~)
Substituting in (3.10), we compute
E~
==
L I +K.
KJ I + K 3T; OS [
TfS
(K S0 4 + K6 E J _
J
KJK4
I + K 3TdoS
04
System Response to Small Disturbances
67
Fig.3 .4 System block diagram with voltage regulation .
or, rearranging,
K4 T' dO
(I
(3.39)
S2+ S _ + T.
From (3.39) and (3.9)
(3.40)
Substituting in the s domain swing equation and rearranging, we obtain the following characteristic equation:
(3.41) Equation (3.41) is of the form (3.42) Analysis of this fourth-order system for stability is left as an exercise (see Problem 3.7).
68
Chapter 3
3.5.2
Governor with one time lag
Referring to Figure 2.24, we note that a change in the speed w or in the load or speed reference [governor speed changer (GSC)] produces a change in the mechanical torque T« . The amount of change in F; depends upon the speed droop and upon the transfer functions of the governor and the energy source. For the model under consideration it is assumed that GSC.\ = 0 and that the combined effect of the turbine and speed governor systems are such that the change in the mechanical power in per unit is in the form
r.:
where
K,
=
W4
-[K,/(I + T,S)]-
= gain constant = 1/R
T, =
(3.43)
WJl
governor time constant
The system block diagram with governor regulation is shown in Figure 3.5. Then the linearized swing equation in the s domain is in the form (with WR
in rad/s)
(2H/WR)s2SIJ,.(s)
=
-[Kg/{l
+
TgS)] s8<1(s) - P e<1(s) WR
(3.44)
The order of this equation will depend upon the expression used for P,.\(s). If we assume the simplest model possible. P,.\(s) = Ps6.\(s), the characteristic equation of the system is given by (2H/WR)S2 + [K,/(1
+ T,S)]S +
P,
=
0
(3.45)
or (3.46) The system is now of third order. Applying Routh's criterion, the system is stable if K, > 0 and Ps > O. If another model is used for P,.1(s), such as the model given by (3.9) and (3.10), the system becomes of fourth order. as shown in Figure 3.5. Its dynamic response will change. Information on stability can be obtained from the roots of the characteristic equation or from examining the eigenvalues of its characteristic matrix.
Fig . 3.5
Block diagram of a system with governor speed regulation .
69
System Response to Small Disturbances
Fig. 3.6
Bloc k di agr am of a sys tem with a go vern o r and volt age regulato r.
If both speed governor and voltage regulation are added simultaneously, as is usually the case, the system becomes fifth order, as shown in Figure 3.6 . 3.6
Distribution of Power Impacts
In this section we consider the effect of the sudden application of a small load Pes at some point in the network . (See also [7,5].) To simplify the analysis, we also assume that the load has a negligible reacti ve component. Since the sudden change in load P u creates an unbalance between generation and load , an oscillatory transient results before the system settles to a new steady-state condition . This kind of impact is continuously occurring during normal operation of power systems. The oscillatory transient is in fact a "spectrum" of oscillations resulting from the random change in loads. These oscillations are reflected in power flow in the tie lines. Th us the scheduled tie-line flows will have "random " power oscillations superimposed upon them. Our concern here is to make an estimate of the magnitude of these power oscillations. Note that the estimates made by the methods outlined below are only approximate, yet they are quite instructive . We formulate the problem mathematically using the network configuration of Figure 3.7 and the equations of Sections 2.9 and 3.4 . Referring to the (n + I)-port net work in Figure 3.7 , the power into node i is obtained from (3.21) by adding node k , P;
e.a, + L '"
=
n
i- I
E;Ei(B;;sin Oii
+
c.;cos 0;;)
+ EY k ie, sin Oik + G ik cos oid
j .,li.k
For the case of nearl y zero conductance n
Pi ~
L s.e,s; sin s., + t: VAB
j - I j .,u(
iA
sin
OiA
(3.47)
70
n_
Chapter 3
Cn + 1l- pe r! netw ork
Tn
Fig . 3.7
Network with power impact at node k ,
and the power into node k (the load bus) is
• r, = L
(3.48)
VkEjBkjsin Okj
j-I
j",k
Here we assumethat the power network has a very high XI R ratio such that the conductances are negligible. The machines are represented by the classical model of constant voltage behind transient reactance. We also assume that the network has been reduced to the internal machine nodes (nodes 1,2, . . . .n of Figure 2.17) and the node k, where the impact PL A is applied. The immediate effect (assuming the network response to be fast) of the application of PLA is that the angle of bus k is changed while the magnitude of its voltage Vk is unchanged, or V k IOko becomes V k IOko + ou. Note also that the internal angles of the machine nodes 01, 02' ... ,h. do not change instantly because of the rotor inertia . 3.6.1
Linearization
The equations for injected power (3.47) and (3.48) are nonlinear because of the transcendental functions . Since we are concerned only with a small impact PL A , we linearize these equations to find and determine only the change variables PiA and Pu . The transcendental functions are linearized by the relations
+ OkjA) '" sinokjO + (COSOkjO)OkjA = cos (OkjO + OkjA) '" cos OkjO - (sin OkjO)OkjA
sinokj = sin(okjo cos Okj
(3.49)
for any k,j. Note that the order kj must be carefully observed since Okj = -Ojk' Substituting (3.49) into (3.47) and (3.48) and eliminating the initial values, we compute the linear equations PiA
=
•
L
(E;EjB;jcOSO;jO)O;jA
j-I
j
"'i.k
•
r., = L
j -I
+
•
(VkE;B;kCOSO;kO)OikA
=L
j-I
j -I
P,ijO;jA
+
P,ikO;kA
j"'i.k
•
(VkEjBkjCOSOk jO)OkjA
L
P,kjOkjA
These equations are valid for any time t following the application of the impact.
(3.50)
System Response to Small Disturbances 3.6.2
71
A special case: , = 0+
The instant immediately following the impact is of interest. In particular, we would like to determine exactly how much of the impact PLA is supplied by each generator PiA,; = 1,2, ... .n. At the instant t = 0+ we know that OiA = 0 for all generators because of rotor inertias. Thus we can compute (with both i andj indicating generator subscripts)
Thus (3.50) becomes PkA (0+)
=
L n
j-I
PskjOkA (0+)
(3.51)
Comparing the above two equations at t = 0+, we note that at node k PkA(O+)
= -
n
L
(3.52)
PiA(O+)
i-I
This is to be expected since we are assuming a nearly reactive network. We also note that at node; PiA depends upon B« cos OikO' In other words, the higher the transfer susceptance Bik and the lower the initial angle OikO, the greater the share of the impact "picked up" by machine i. Note also that PkA = - P LA, so the foregoing equations can be written in terms of the load impact as
L n
PLA(O+) = -
Psk;DkA(O+)
i-I
From (3.52) and (3.53) we conclude that ou(O+)
=
-Pu(O+) /
PiA(O+) = (P'ik/
t
=
L n
PiA(O+)
i- I
t r.;
P'i k) PLA(O+)
(3.53)
(3.54) i
= 1,2, ... , n
(3.55)
It is interesting that at the instant of the load impact (i.e., at t = O~\ the source of energy supplied by the generators is the energy stored in their magnetic fields and is distributed according to the synchronizing power coefficients between i and k. Note that the generator rotor angles cannot move instantly; hence the energy supplied by the generators cannot come instantly from the energy stored in the rotating masses. This is also evident from the first equation of (3.51); PiA depends upon Psik or Bik , which depends upon the reactance between generator; and node k. Later on when the rotor angles change, the stored energy in the rotating masses becomes important, as shown below. Equations (3.52) and (3.55) indicate that the load impact PLA at a network bus k is immediately shared by the synchronous generators according to their synchronizing power coefficients with respect to the bus k. Thus the machines electrically close to the point of impact will pick up the greater share of the load regardless of their size. Let us consider next the deceleration of machine; due to the sudden increase in its output power PiA' The incremental differential equation governing the motion of machine; is given by
72
Chapter 3
2H; dW;4 -dt- + P/4 (I ) -_ WR and using (3.55) i
w/A t
2H dd WR
+ (PSilc!
i: j-I
0
i = I, 2, ... , n
P$Jk)PLA(O+) = 0
i
(3.56)
= 1,4, ... .n
Then if PLA is constant for all t, we compute the acceleration in pu to be
_I dWiA WR dt
=
_
PSi/( (PL4(0+)! 2H i
i:
P.)
j-I
/
; = 1,2, ... .n
(3.57)
Obviously, the shaft decelerates for a positive load P LA. The pu deceleration of machine i, given by (3.57), is dependent on the synchronizing power coefficient Psi/c and inertia H;. This deceleration will be constant until the governor action begins. Note that after the initial impact the various synchronous machines will be retarded at different rates, each according to its size H; and its "electrical location " given by p s;/( .
3.6.3
Average behavior prior to governor action (I = 'I)
We now estimate the system behavior during the period 0 < t < t" where I, is the time at which governor action begins. To designate this period simply, we refer to time as I I, although there is no specific instant under consideration but a brief time period of no more than a few seconds. Looking at the system as a whole, there will be an overall deceleration of the machines during this period. To obtain the mean deceleration, let us define an "inertial center" that has angle ~ and angular velocity w, where by definition, (3.58) Summing the set (3.57) for all values of i, we compute
.1L WR
dd (H i W i4 ) I
!!.WA dt WR
=
= Pk4
-Pu(O+)
=
-PL4(0+)
(3.59)
2H;
(3.60)
It i-I
Equation (3.60) gives the mean acceleration of all the machines in the system, which is defined here as the acceleration of a fictitious inertial center. We now investigate the way in which the impact PL4 will be shared by the various machines. Note that while the system as a whole is retarding at the rate given by (3.60), the individual machines are retarding at different rates. Each machine follows an oscillatory motion governed by its swing equation. Synchronizing forces tend to pull them toward the mean system retardation, and after the initial transient decays they will acquire the same retardation as given by (3.60). In other words, when the transient decays, dWit~/dl will be the same as dw4/dt as given by (3.60). Substituting this value of dW;4/dl in (3.56), at I = I, > 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<
t
P'jS) Pu(O+)
= 10.0 MW nominally. The results of these calculations and the
actual values determined from the stability study are shown in Table 3.5. Table 3.5. (I)
;
I
2 3 LPi~
Initial Power Change at Generators Due to 10-MW Load Added to Bus 8 (2) Pi~
(neglecting G;k)
3.021 4.021 2.958 10.000
(3)
r.,
(with
G ik)
3.016 4.028 2.956 10.000
(4)
(5)
(6)
Pi~
Pi~
Pi~
(computer study)
(91% of(2»)
(91% of(3»)
2.8 3.6 2.7 9.1
2.749 3.659 2.692 9.100
2.745 3.665 2.690 9.100
Note that the actual load pickup is only 9.1 MW instead of the desired 10 MW. This is due in part to the assumption of constant voltage VI< at bus 8 (actually, the voltage drops slightly) and to the assumed linearity of the system. If the computed PiA are scaled down by 0.91, the results agree quite well with values measured from the computer study. These values are also shown on the plot of Figure 3.8 at time t = 0+ and are due only to the synchronizing power coefficients of the generators with respect to bus 8. The plots of Pi~ versus time in Figure 3.8 show the oscillatory nature of the power exchange between generators following the impact. These oscillations have frequencies that are combinations of the eigenvalues computed in Example 3.2. The total, labeled "L,Pi A , averages about 9.5 MW.
75
System Response to Small Disturbances
Time, ,
o
0 . 1 0.2 0.3 0.4 0 .5 0 .6 0.7 0.8 0. 9 1.0
1.1 1. 2 1.3
1. 4 1. 5 1.6
1. 7 1.8
1. 9 2.0
-0 .02 -0 .0 4 -0. 06 N
r
i
J
-0 .08 -0 .10 -0 .12 -0 .14
dil at
>: -
0 .09 Hz/ ,
- 0. 16 -0 . 18 -0. 20
Fig. 3.9 Speed deviation following application of a 10 MW resistive load at bus 8.
Another point of interest in Figure 3.8 is the computed values of P;t.(tl) that depend entirely on the machine inertia. These calculations are made from
P;t.(td
(HJ"LH;)PLt. = IOH;/(23.64 + 6.40 + 3.01) = IOH;/33 .05 1.94 MW
= I i = 2
0.91 MW
i
7.15MW
i
=
3
and the results are plotted in Figure 3.8 as dashed lines. It is fairly obvious that the P;t.(t) oscillate about these values of P;t.(td. It is also apparent that the system has little damping and the oscillations are likely to persist for some time . This is partly due to the inherent nature of this particular system, but the same phenomenon would be present to some extent on any system. The second plot of interest is the speed deviation or slip as a function of time, shown in Figure 3.9. The computer program provides speed deviation data in Hz and these units are used in Figure 3.9. Note the steady deceleration with all units oscillating about the mean or inertial center. This is computed as
PLt. 0.10 2'LH; 2(23 .64 + 6.40 + 3.01) - 1.513 X 10- 3 puis = -0.570 rad/s! = -0.0908 Hz/s --- =
The individual machine speed deviations Wit. are plotted in Figure 3.9 and show graphically the intermachine oscillations that occur as the system slowly retards in frequency . The mean deceleration of about 0.09 Hz/s is plotted in Figure 3.9 as a straight line. If the governors were active. the speed deviation would level off after a few seconds to a constant value and the osc illations would eventually decay . Since the governors have a drooping characteristic. the speed would then continue at the reduced value as
76
Chapter 3
long as the additional load was present. If the speed deviation is great, signifying a substantial load increase on the generators, the governors would need to be readjusted to the new load level so that additional prime-mover torque could be provided .
Example 3.5 Let us examine the effect of the above on the power flow in tie lines . Consider a power network composed of two areas connected with a tie line. as shown in Figure 3.10 . The two areas are of comparable size, say 1000 MW each . They are connected with a tie line having a capacity of 100 M W. The tie line is carry ing a steady power flow of 80 MW from area I to area 2 as shown in Figure 3.10 . Now let a load impact Pu. = 10 MW (1% of the capacity of one area) take place at some point in area I, and determine the di stribution of this added load immediately after its application (t = 0+) and a short time later (t = t l ) after the initial transients have subsided. Because o'f the proximity of the groups of machines in area I to the point of impact, their synchronizing power coefficients are larger than those of the groups of machines in area 2. If we define :LPsidareal = Psi, LP'idarea2 = P s2' then let us assume that P si = 2Ps2 '
---
80MW
P = 10 M W LA
Fig .3 .10
Two areas connected with a tie line.
Solution Since PsI = 2Ps2 , at the instant of the impact 2/3 of the IO-MW load will be supplied by the groups of machines in area I, while 1/3 or 3.3 MW will be supplied by the groups of machines in area 2. Thus 3.3 MW will appear as a reduction in tie-line flow. In other words, at that instant the tie-line flow becomes 76.7 MW toward area 2. At the end of the initial transient the load power impact Pes will be shared by the machines according to their inertias. Let us assume that the machines of area I are
80 .0
cE
76 .7
~ ~
£
~
::; v ;::
T ---
73 . 3
I I
I I I I
o Fig .3.11
t =
0 Time, s
Tie-line power oscillations due to the load impact in area I.
77
System Response to Small Disturbances
predominantly hydro units (with relatively small H), while the units of area 2 are of larger inertia constants such that L:H;].re.2 = 2L:H;]are.1 where all H's are on a common base . The sharing of the load among the groups of machines will now become 6.7 MW contributed from area 2 and 3.3 MW from area 1. The tie -line flow will now become 73.3 MW (toward area 2). From the above we can see that in the situation discussed in this example a sudden application of a 10-MW load caused the tie-line flow to drop almost instantly by 3.3 MW, and after a brief transient by 6.7 MW. The transition from 76.7-MW flow to 73.3-MW flow is oscillatory, and power swings of as much as twice the difference between these two values may be encountered . This situation is illustrated in Figure3.11. The time t I mentioned above is smaller than the time needed by the various controllers to adjust the system generation to match the load and the tie-line flow to meet the scheduled flow .
Example 3.6 We now consider a slightly more complex and more realistic case wherein the area equivalents in Figure 3.10 are represented by their Thevenin equivalents and the tieline impedance is given . The system data are given in Figure 3.12 in pu on a IOOO-MVA base. The capacity of area I is 20,000 MW and that of area 2 is 14,000 MW . The inertia constants of the machines in the two areas are about equal. (a) (b) (c) (d)
Find the equations of power for PI and Pz • Find the operating condition when PI = 100 MW. This would correspond ap proximately to a 100-MW tie-line flow from area I to area 2. Find the synchronizing power coefficients. Consider a sudden load addition to area 2, represented by the resistive load P4 tJ. at bus 4. If this load is 200 M W (1.43% of the capacity of area 2), find the distribution of this load at I = 0+ and I = II.
® El~
+
+
F.el!t.
1.~
1.0&
Area 1 equivalent
Tie line
Area 2 equivalent
Fig. 3.12 Two areas connected by a tie line.
Solution Consider the system as a two-port network between nodes 1 and 2. Then we compute Zl2 =
0.450
+ jl.820
jil2 =
I/Z I2
=
0.533 /-76.112° == 0.128 - jO.518 pu
= - YI2
=
0.533/103.888°
Yl2
Gil =
0.128
=
1.875 /76.112° pu
glo ==
gzo == 0
Chapter 3
78 Gil
=
-0.128
B I2 = 0.518
+
+ Bl2sinol2) - VrG t2 = 0 + 1.0(-0.128cosol + 0.518sinol) + 0.128 = 0.128 + 0.533 sin (01 - 13.796°) P2 = V~g20 + VI V2 (G 12 COS021 + Bl2sin021) - V~G21
(a)
PI
=
V~glo
VI Vl(GI2COSOI2
+ 1.0(-0.128cosol - 0.518sinol) + 0.128 = 0.128 - 0.533sin(ol + 13.796°) =
(b) Given that PI
0
= 0.1 pu
0.100 = 0.128 + 0.533sin(0, - 13.796°) (c)
Psl2
=
VI V2(B 12cos 0120 - G I2 sin 0120)
1.0(0.518 cos 10.784° + O.128sin 10.784°) = 0.533 Ps21 = VI V2( B 21cos 0210 - G21sin 0210)
I.O[O.518cos(-IO.784°) + 0.128sin(-IO.784°») = 0.485
(d) Now add the 200-MW load at bus 4; P44 = 200/1000
=
0.2 pu.
To complete the problem, we must know the voltage 'V:. at t = 0-. Thus we compute
(V; - ~ )/ZI2 = (1.0 /10.784° - 1.0 LQ)/ 1.875 /76.112° = 0.100 / 19.280° = £2 + (0.100 + jO.012)/12 = 1.009 + jO.004 = 1.009 /0.252° 040 = 0.252° 0140 = 010 - 040 = 10.532° 0240 = 020 - 040 = -0.252°
112(0 -
) =
~(O-)
From the admittance matrix elements -0.103 + jO.533 -1/z24 = -9.858 + jl.183
Yl4 = - Yl4 = -
Y24 =
-Y24 =
1/z14
=
we compute the synchronizing power coefficients Ps l 4 = V, V4(B 14 cos 0,40 - G I4sin 0140)
=
Ps24
(1.009)(0.533 cos 10.532° + 0.103 sin 10.532°)
=
0.548
= V2 V4(B 24 cos 0240 - G 24 sin 0240) = 1.009[1.183cos(-0.252°) + 9.858sin(-0.252°)] = 1.150
Then the initial distribution of P4 4 is P I 4(0+)
=
PsI4(0 .2)/ (PsI4 + Ps24)
P2A(0+)
=
Ps24(0.2)/(PsI4
+
Ps24)
= (0.323)(0.2) = 0.0646 pu = (0.677)(0.2) = 0.1354 pu
The power distribution according to inertias is computed as P'A(t.)
P2A(t , )
= 0.2[20,000H/(20,OOOH + 14,000H)] = 0.11765 pu = O.2[14,OOOH/(20,000H + 14,000H)] = 0.08235 pu
In this example the synchronizing power coefficients Ps14 is smaller than P s24' while the inertia of area 1 is greater than that of area 2. Thus, while initially area 1 picks up only about
System Response to Small Disturbances
79
one third of the load P4~' at a later time t = t1 it picks up about 59% of the load and area 2 picks up the remaining 41%. In general, the initial distribution of a load impact depends on the point of impact. Problem 3.10 gives another example where the point of impact is in area I (bus 3). In the above discussion many factors have been neglected, e.g., the effect of the network transfer conductances, the effect of the reactive component of the load impact, the fast primary controllers such as some of the modern exciters, the load frequency and voltage characteristics, and others. Thus the conclusions reached above should be considered qualitative and as rough approximations. Yet these conclusions are basically sound and give a good "feel" for what happens to the machines arid to the tie-line flows under the influence of small routine load changes. If the system is made up of groups of machines separated by tie lines, they share the impacts differently under different conditions. Hence they will oscillate with respect to each other during the transient period following the impact. The power flow in the connecting ties will reflect these oscillations. The analysis given above could be extended to include governor actions. Following an impact the synchronous machines will share the change first according to their synchronizing power coefficients, then after a brief period according to their inertias. The speed change will be sensed by the prime-mover governors, which will act to make the load sharing according to an entirely different criterion, namely, the speed governor droop characteristic. The transition from the second to the final stage is oscillatory (see Rudenberg [7], Ch. 23). The angular frequency of these oscillations can be estimated as follows. From Section 3.5.2, neglecting Pt!A' the change in the mechanical power Pm A is of the form (3.62)
where R is the regulation and T$ is the servomotor time constant. The swing equation for machine i becomes, -in the s domain,
2H·sw·/t. , + ,~
WR
The characteristic equation of the system is given by S2
+ (I/T s;)s + 1/2H;R;T$;
= 0
(3.63)
from which the natural frequency of oscillation can be estimated. It is interesting to note the order of magnitude of the frequency of oscillation in the two different transients discussed in this section. For a given machine (or a group of machines) the frequency of oscillation in the first transient is the natural frequency with respect to the point of impact. These frequencies are determined by finding the eigenvalues ~ of the A matrix by solving det (A - ~U) = 0, where U is the unit matrix and A is defined by (3.1). For the second transient, which occurs during the transition from sharing according to inertia to sharing according to governor characteristic, the frequency of oscillation is given by Vf2 ~ 1/2H;R i T ,; . Usually these two frequencies are appreciably different.
80
Chapter 3
Problems 3.1
3.2 3.3 3.4
3.5 3.6 3.7 3.8 3.9
A synchronous machine is connected to a large system (an infinite bus) through a long transmission line. The direct axis transient reactance xj = 0.20 pu. The infinite bus voltage is 1.0 pu. The transmission line impedance is Zline = 0.20 + jO.60 pu. The synchronous machine is to be represented by constant voltage behind transient reactance with E' = 1.10 pu. Calculate the minimum and maximum steady-state load delivered at the infinite bus (for stability). Repeat when there is a local load of unity power factor having R10ad = 8.0 pu. Use Routh's criterion to determine the conditions of stability for the system where the characteristic equation is given by (3.14). Compute the characteristic equation for the system of Figure 3.1, including the damping term, and determine the conditions for stability using Routh's criterion. Compare the results with those of Section 3.3.1. Using 04 as the output variable in Figure 3.2, use block diagram algebra to reduce the system block diagram to forward and feedback transfer functions. Then determine the system stability and possible system behavior patterns by sketching an approximate rootlocus diagram. Use block diagram algebra to reduce the system described by (3.45). Then determine the system behavior by sketching the root loci for variations in Kg. Give the conditions for stability of the system described by (3.20). A system described by (3.41) has the following data: H = 4, rdO = 5.0, T f = 0.10, K I = 4.8, K 2 = 2.6, K) = 0.26, K 4 = 3.30, K, = 0.1, and K 6 = 0.5. Find the maximum and minimum values of K, for stability. Repeat for K s = -0.20. Write the system described by (3.46) in state-space form. Apply Routh's criterion to (3.46). The equivalent prefault network is given in Table 2.6 for the three-machine system discussed in Section 2.10 and for the given operating conditions. The internal voltages and angles of the generators are given in Example 2.6. (a) Obtain the synchronizing power coefficients Ps12, P s 13, P s 23 , and the corresponding coefficients aij [see (3.31)] for small perturbations about the given operating point. (b) Obtain the natural frequencies of oscillation for the angles Ol2A and 0134' Compare
.with the periods of the nonlinear oscillations of Example 2.7.
3.10
Repeat Example 3.6 with the impact point shifted to area 1 and let before. 3.11 Repeat Problem 3.10 for an initial condition of PL4 = 300 MW.
PL4
= 100 MW as
References I. Korn, G. A., and Korn, T. M. Mathematical Handbook for Scientists and Engineers. McGraw-Hili, New York, 1968. 2. Hayashi, C. Nonlinear Oscillations in Physical Systems. McGraw-Hili, New York, 1964. 3. Takahashi, Y., Rabins, M. J., and Auslander, D. M. Control and Dynamic Systems. Addison-Wesley, Reading, Mass., 1970. 4. Venikov, V. A. Transient Phenomena in Electric Power Systems. Trans. by B. Adkins and D. Rutenberg. Pergamon Press, New York, 1964. 5. Hore, R. A. Advanced Studies in Electrical Power System Design. Chapman and Hall, London, 1966. 6. Crary, S. B. Power System Stability, Vols, I, 2. Wiley, New York, 1945, 1947. 7. Rudenberg, R. Transient Performance of Electric Power Systems: Phenomena in Lumped Networks. McGraw-Hili, New York, 1950. (MIT Press, Cambridge, Mass., 1967.) 8. de Mello, F. P., and Concordia, C. Concepts of synchronous machine stability as affected by excitation control. IEEE Trans. PAS-88:316-29, 1969. 9. Heffron, W. G., and Phillips, R. A. Effect of a modern amplidyne voltage regulator on underexcited operation of large turbine generators. A lEE Trans. 71 (Pt. 3):692-97, 1952. 10. Routh, E. J. Dynamics of a System of Rigid Bodies. Macmillan, London, 1877. (Adams Prize Essay.) II. Ogata, K. State-Space Analysis of Control Systems. Prentice-Hall, Englewood Cliffs, N.J., 1967. 12. Lefschetz, S. Stability of Nonlinear Control Systems. Academic Press, New York, London, 1965.
Part /I
The Electromagnetic Torque
P. M. Anderson
A. A. Fouad
chapter
4
The Synchronous Machine
4.1
Introduction
In this chapter we develop a mathematical model for a synchronous machine for use in stability computations. State-space formulation of the machine equations is used. Two models are developed, one using the currents as state variables and another using the flux linkages. Simplified models, which are often used for stability studies, are discussed. This chapter is not intended to provide an exhaustive treatment of synchronous machine theory. The interested reader should consult one of the many excellent references on this subject (see [I ]-[9]). The synchronous machine under consideration is assumed to have three stator windings, one field winding, and two amortisseur or damper windings. These six windings are magnetically coupled. The magnetic coupling between the windings is a function of the rotor position. Thus the flux linking each winding is also a function of the rotor position. The instantaneous terminal voltage v of any winding is in the form,
v = ±l:,ri ± L~
(4.1)
where A is the flux linkage, r is the winding resistance, and i is the current, with positive directions of stator currents flowing out of the generator terminals. The notation ±2: indicates the summation of all appropriate terms with due regard to signs. The expressions for the winding voltages are complicated because of the variation of A with the rotor position. 4.2
Park's Transformation
A great simplification in the mathematical description of the synchronous machine is obtained if a certain transformation of variables is performed. The transformation used is usually called Park"s transformation [10, II]. It defines a new set of stator variables such as currents, voltages, or flux linkages in terms of the actual winding variables. The new quantities are obtained from the projection of the actual variables on three axes; one along the direct axis of the rotor field winding, called the direct axis; a second along the neutral axis of the field winding, called the quadrature axis; and the third on a stationary axis. Park's transformation is developed mathematically as follows.' I. The transformation developed and used in this book is not exactly that used by Park [10, II] but is more nearly that suggested by Lewis [12], with certain other features suggested by Concordia (discussion to (12)) and Krause and Thomas [13). 83
84
Chapter 4 a axi s
~
d a xis
o i
Directio n
of Rotatio n
.-----1 fb n'
I I I
sc
b
(
sb
b a x is
Fig. 4.1
q e x is
c axis
Pictorial representation of a synchronous mach ine.
We define the d axis of the rotor at some instant of time to be at angle 0 rad with respect to a fixed reference position, as shown in Figure 4.1 . Let the stator pha se currents ia , i b , and i, be the currents leaving the generator terminals. If we "p roject " these currents along the d and q axes of the rotor, we get the relations
(2/3)[ia sin 0 + ib sin (0 - h 13) + i, sin (0 + h 13») (2/3)[iacosO + ibcos(O - h/3) + iccos(O + h/3»)
iq u is idu
is
(4.2)
We note that for convenience the axis of phase a was chosen to be the reference position, otherw ise some angle of displacement between phase a and the arbitrary reference will appear in all the above terms . The effect of Park's transformation is simply to transform all stator quantities from phases a, b, and c into new variables the frame of reference of which moves with the rotor. We should remember, however, that if we have three variables i a , i b , and i., we need three new variables. Park 's transformation uses two of the new variables as the d and q axis components. The third variable is a statio na ry current, which is proportional to the zero-sequence current. A multipl ier is used to simplify the numerical calculations. Thus by definition (4.3) where we define the current vectors
(4.4)
and where the Park's transformation P is defined as
/V 2 P
v 2/ 3
[
1/V2
I/VTl
cosO
cos (8 - h 13)
cos (8
+ h 13)
sin 8
sin (0 - h 13)
sin (8
+ 21r13)
(4.5)
The main field-winding flux is along the direction of the d axis of the rotor. It produces an EMF that lags this flux by 90·. Therefore the machine EMF E is primarily along the rotor q axis . Consider a machine having a constant terminal voltage V. For generator
The Synchronous Machine
85
action the phasor E should be leading the phasor V. The angle between E and V is the machine torque angle () if the phasor V is in the direction of the reference phase (phase a). At t = 0 the phasor V is located at the axis of phase a, i.e., at the reference axis in Figure 4.1. The q axis is located at an angle 0, and the d axis is located at 8 = {) + 1r /2. At t > 0, the reference axis is located at an angle WR t with respect to the axis of phase a. The d axis of the rotor is therefore located at (4.6)
where WR is the rated (synchronous) angular frequency in rad/s and () is the synchronous torque angle in electrical radians. Expressions similar to (4.3) may also be written for voltages or flux linkages; e.g., VOdq
= PVabl'
(4.7)
If the transformation (4.5) is unique, an inverse transformation also exists wherein we may write iabc = p-l i Odq (4.8) The inverse of (4.5) may be computed to be
Vf73
p-l
1/\/2
cos fJ
sin fJ
1/V'2 cos(fJ - 21r/3) sin (0 - 21r/3)
(4.9)
1/\/2 cos(fJ + 21r/3) sin(fJ + 21r/3)
and we note that p-l = P', which means that the transformation P is orthogonal. Having P orthogonal also means that the transformation P is power invariant, and we should expect to use the same power expression in either the a-b-c or the O-d-q frame of reference. Thus p =
vaia
+
vbib
= V~dq(P-I)'
+, Veil' =
p- t iOdq
= V~dq iOdq = voio
4.3
+
v~bciabe = (P-IVOdq)'(P-liOdq)
= V~dq pp- t i Odq vdid
+
(4.10)
Vqiq
Flux linkage Equations
The situation depicted in Figure 4.1 is that of a network consisting of six mutually coupled coils. These are the three phase windings sa-fa, sb-jb, and sc-fc; the field winding F-F'; and the two damper windings D-D' and Q-Q'. (The damper windings are often designated by the symbols kd and kq. We prefer the shorter notation used here. Phase-winding designations sand f refer to "start" and "finish" of these coils.) We write the flux linkage equation for these six circuits as stator
Aa
Lao
Lab
Lac
L aF
Lao
L aQ
t,
Ab
u,
~bb
t.;
.L bF
LbO
L bQ
ib
Lea
t.,
Lee
L eF
LeO
L cQ
t.
Ae
AF rotor {
----------------------L Fa
L Fb
L Fe
L FF
L FO
L FQ
iF
AD
Loa
LOb
L De
L DF
L DD
L DQ
iD
AQ
L Qa
L Qb
L Qe
L QF
L Qo
L QQ
iQ
Wb turns
(4.11)
86
Chapter 4
where
L j le = self-inductance whenj = k mutual inductance whenj
¢
k
and where L j le = Li, in all cases. Note the subscript convention in (4.11) where lowercase subscripts are used for stator quantities and uppercase subscripts are used for rotor quantities. Prentice [14] shows that most of the inductances in (4.11) are functions of the rotor position angle (J. These inductances may be written as follows 4.3.1
Stator self-inductances
The phase-winding self-inductances are given by Lao
=
L hb
=
L, + L mcos 20 H L, + L m cos 2(0 - 21r/3) H
L cc
=
L, + L mcos 2(fJ + 2tr/3) H
(4.12)
where L, > L m and both L, and L m are constants. (All inductance quantities such as L, or M, with single subscripts are constants in our notation.) 4.3.2
Rotor self-inductances
Since saturation and slot effect are neglected, all rotor self-inductances are constants and, according to our subscript convention, we may use a single subscript notation; i.e., (4.13) 4.3.3
Stator mutual inductances
The phase-to-phase mutual inductances are functions of (J but are symmetric, Lob = L ha = -Ms L hc L eh -M. f
-
Lmcos2«(J + 7r/6) H Lmcos2(O - tr/2) H
Lea = Lac = -Ms - Lmcos2(O + 51r/6) H
(4.14)
where I M, I > L m • Note that signs of mutual inductance terms depend upon assumed current directions and coil orientations. 4.3.4
Rotor mutual inductances
The mutual inductance between windings F and D is constant and does not vary with fJ. The coefficient of coupling between the d and q axes is zero, and all pairs of windings with 90° displacement have zero mutual inductance. Thus L FQ = L QF = 0 H
4.3.5
L DQ = L Q D = 0 H
(4.15)
Stator-to-rotor mutual inductances
Finally, we consider the mutual inductances between stator and rotor windings, all of which are functions of the rotor angle O. From the phase windings to the field winding we write L oF = L Fa = MFcos(J H L bF = L Fb
L eF
=
MFcos(O - 27r/3) H
= L Fe = MFcos(8 + 2tr/3) H
Similarly, from phase windings to damper winding D we have
(4.16)
The Synchronous Machine
87
LaD = L Da = M DCOS 8 H Lbo = L Db LcD
=
MDcos(fJ - 21r/3) H
= L Dc = MDcos(fJ + 21r/3) H
(4.17)
and finally, from phase windings to damper winding Q we have L aQ
='
L Qa = M Q sin (J H
L bQ = L Qb = M Q sin «(} - 21r/3) H L cQ = L Qc = MQsin«(} + 21r/3) H
(4.18)
The signs on mutual terms depend upon assumed current directions and coil orientation. 4.3.6
Transformation of inductances
Knowing all inductances in the inductance matrix (4.11), we observe that nearly all terms in the matrix are time varying, since (} is a function of time. Only four of the off-diagonal terms vanish, as noted in equation (4.15). Thus in voltage equations such as (4.1) the ~ term is not a simple Li but must be computed as ~ = Lf + li. ,We now observe that (4.11) with its time-varying inductances can be simplified by referring all quantities to a rotor frame of reference through a Park's transformation (4.5) applied to the a-b-c partition. We compute (4.19) where
Loa = stator-stator inductances L aR , L Ra = stator-rotor inductances L RR = rotor-rotor inductances
Equation (4.19) is obtained by premultiplying (4.11) by
where P is Park's transformation and U3 is the 3 x 3 unit matrix. operation indicated in (4.19), we compute ~o
Lo
0
0
~d
0
Ld
0
~q
0
0
Lq
0
kM F
0
')...F
I I I I I I I I
0
0
0
10
0
id
0
kM Q
iq
MR
0
iF
MR
Lo
0
iD
0
0
LQ
iQ
kM F kM o 0
------------,-------------
~o
0
kM o '
0
')...Q
0
0
kM Q
, LF I
I
I I I I
Performing the
Wb turns
(4.20)
where we have defined the following new constants,
L d = L, + M, + (3/2)L m H Lo = L, - 2Ms H
L q = L$ + M, - (3/2)L m H k
=
V372
(4.21)
88
Chapter 4
In (4.20) >"d is the flux linkage in a circuit moving with the rotor and centered on the d axis. Similarly, >"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-
-
-
-
-
-.<
Fig.4.14
Block diagram for the E" model.
The Synchronous Machine E'd -
-
-
137
--,
T e
I
q q
1.0-
Fig.4.15
'v
+
E"
- ---i.
Block diagram for computation of torque and speed in the E " model.
where we have defined (Xd - Xd)(X; - x;J)
x:; - x{
x; - ·X {
(x; - x"y
(Xd - Xd)( X:! - x{)
(4.271)
x; - x{
The block diagram for (4.270) is shown in Figure 4.14. The remaining equations are given by (D + T/)W
==
Tm
-
(E~'lq
so
+ E:!ld)
==
W- I
(4.272)
The block diagram for equation (4.272) is given in Figure 4.15. Also the block diagram of the complete system can be obtained by combining Figures 4.14 and 4.15 . If saturation is to be included , a voltage increment EtJ., corresponding to the increase in the field current due to saturation , is to be added to (4.248), E
==
E~
+ EtJ. -
(x d
-
X;)(id
+
i D )/ V3
(4.273)
Example 4.6 Use the machine data from Examples 4.1-4.5 to derive the time constants and gains for the E" model.
Solution
The time constant T:!O == 0.03046 s == 72.149 rad is already known from Example 4.5. For the E" model we also need the following additional time constants. From (4.192) the q axis subtransient open circuit time constant is T~~ == LQ/rQ ==
1.423 x 10- 3 / 18.969 x 10- 3
==
0.075 s
==
28.279
rad
which is about twice the d axis subtransient open circuit time constant. We also need the d axis transient open circuit time constant. It is computed from (4.189) .
TdO
==
LF/rF == 2.189/0 .371 == 5.90 s == 2224.25
rad
Note that this time constant is about 30 times the subtransient time constant in the d
Chapter 4
138
axis. This means that the integration associated with T~o will be accomplished very fast compared to that associated with r;o. To compute the gains, the constant or L~ is needed. It is computed from (4.174):
x;
L~
= L, -
= 1.70 - (1.55)2/1.651 = 0.245 pu
L~D/LF
We can now compute from (4.271) K
1
K2
= x; x~
=
d
X xd
Xt
I - K1
K = (x d
=
- x.f.
-
-
=
0.185 - 0.15 0.245 - 0.15
0.632
xj)(xj - x;)
=
(x~ - X.f,)2
=
(x d
-
= 0.368
x~)(x:;
,
Xd -
(1.70 - 0.245)(0.245 - 0.185) (0.245 - 0.150)2 (1.70 - 0.245)(0.185 - 0.150) 0.245 - 0.150
- x-e)
x.f,
=
=
9.673 0536 .
From (4.179) we compute
L;' = L, - L~Q/ L Q = 1.64 - (1.49)2/1.526 = 0.185 pu Then, from (4.270), we compute the gain, x q
4.15.3
Neglecting "A" and
'A, for a
x;' = 1.64 - 0.185
-
= 1.455
pu.
cylindrical rotor machine-the two-axis model
In the two-axis model the transient effects are accounted for, while the subtransient effects are neglected [18]. The transient effects are dominated by the rotor circuits, which are the field circuit in the d axis and an equivalent circuit in the q axis formed by the solid rotor. An additional assumption made in this model is that in the stator voltage equations the terms ~d and ~q are negligible compared to the speed voltage terms and that w " J WR = 1 pu. The machine will thus have two stator circuits and two rotor circuits. However, the number of differential equations describing these circuits is reduced by two since ~d and Xq are neglected in the stator voltage equations (the stator voltage equations are now algebraic equations). The stator transient flux linkages are defined by \' I\q
A \ = I\q -
L"q lq
(4.274)
and the corresponding stator voltages are defined by
ed,
A
\'
\'
(4.275)
= -W I\q = -WRl\q
Following a procedure similar to that used in Section 4.15.2,
(4.276) or e~ = vd
e;
=
+ r id +
x~iq
vq + r iq
-
+ (x; -
x~) iq
X~id
(4.277) (4.278)
Since the term (x; - xd)iq is usually small, we can write, approximately,
ed
"J
Vd
+ r id + Xdiq
(4.279)
139
The Synchronous Mach ine
.'
d
Fig . 4. 16 T ra nsie nt equi valent circ uit o f a generator .
The voltages e; and e; are the q a nd d components of a voltage e' behind tr ansient reactance. Equations (4.279) and (4 .278) ind icate that during the transient the machine can be represented by the circuit diagram sho w n in Figure 4 .16 . It is interesting to note that since and a re d and q axis stator voltages, they represent v'3 times the equivalent stator rms voltages. For example, we can verify that = v'JE;, as given by (4 .203). Also, in this model the voltage e' , which corresponds to the transient flux linkages in the machine, is not a constant. Rather, it will change due to the changes in the flux linkage of the d and q axi s rotor circuits. We now develop the differential equations for the voltages and The d axis flux linkage equation s for thi s model are
e;
e;
e;
e;
e;.
(4 .280) By eliminating i, and using (4 .174) and (4. 203) ,
V3 E;
Ad -
=
Ldid pu
a nd by using (4 .275),
e'q
(4.281)
Similarly, for the q axis
Aq
Eliminating
+ LAQi Q pu
Lqiq
=
'o- we compute Aq
-
(4 .282)
=
(L AQ/ LQ)AQ
(L, - L~Q/ LQ)iq pu
(4 .283)
by defining (4.284)
L; = L, - L~Q/LQ pu
and by using (4.284) and (4.275) we get
ed
~
v'J Ed
= -(L AQ/ LQ)AQ pu
(4.285)
We also define
v'3 E =
eq
= LAD iF pu
(4.286)
=
(4.287)
We can show that [8), E
From the (4 .286) ,
+
xd1d
E~
+
x~ld
Q circuit vo lta ge equation
T;oEd =
rQiQ + dAQ/dt -
Ed -
where, for uniformity, we adopt the notation
(x, -
=
0, and by using (4 .282) with
x; )/q
(4.288)
Chapter 4
140
Fig .4.17
Block diagram representation of the two-axis model.
, =
T qO
(4 .289)
" = L Q/ rQ
TqO
Similarly, from the field voltage equation we get a relation similar to (4.228)
.
E~
I
= -,- (E FD TdO
-
(4.290)
E)
Equations (4.288), (4.290), and (4.287) can be represented by the block diagram shown in Figure 4.17 . To complete the description of the system, the electrical torque is obtained from (4.95), T.~ = )l.iq - )l.qid , which is combined with (4.274) and (4.275) to compute
(4.291)
Example 4.7 Determine the time constants and gains for the two-axis model of Figure 4.17, based on the machine data of Examples 4.1-4.6. In addition we obtain from the manufacturer's data the constant x~ = 0.380 pu.
Solution
Both time constants are known from Example 4.7. The gains are simply the pu reactances Xq -
x~
= 1.64 - 0.380 = 1.260 pu
Xd - Xd
= 1.70 - 0.245 = 1.455 pu
The remaining system equations are given by Tj
W = Tm <5 =
w -
Dw - [Edld +
-
E~/q
-
(L~
- Ld)/iq) (4.292)
I
The block diagram for (4.292) is shown in Figure 4.18. By combining Figures 4.17 and 4.18, the block diagram for the complete model is obtained . Again saturation can be accounted for by modifying (4.287), E = E~ - (Xd - Xd)/d
+ EIJ.
(4 .293)
where Eli is a voltage increment that corresponds to the increase in the field current due to saturation (see Young (19». The procedure for incorporating this modification in the block diagram is similar to that discussed in Section 4.15.2 .
The Synchronous Machine
141
T e Ui
E'q - - - - - '
+
1.0-_0-{
Fig .4 .18
4.15.4
Block diagram representation of (4.292) .
Neglecting amortisseur effects and
Xd and Xli terms-the one-axis model
This model is sometimes referred to in the literature as the one-axis model. It is similar to the model presented in the previous section except that the absence of the Q circuit eliminates the differential equation for Ed or ej (which is a function of the current iQ ) . The voltage behind transient reactance e' shown in Figure 4.16 has only changing by the field effects according to (4.290) and (4.293) . The the component component ej is completely determined from the currents and Vd • Thus, the system equations are
e;
(4 .294) The voltage Ej is obtained from (4.36) with ~d = 0, and using (4.274) and (4.275), (4 .295) The torque equation is derived from (4.95), T,.. = Adiq - Aqid • Substituting (4.274) and EFO - - - - - - - - ,
E' I--"--_q
T
e
T
m
Fig .4 .19
Block d iagram representation of the one-axis model.
Chapter 4
142
noting that, in the absence of the Q circuit, Aq
=
L, i q ,
T, = E;lq - (L q - L;)/d1q pu
(4.296)
Thus the remaining system equations are TjW
=
Tm
-
Dw - [E;lq - (L q - L;)/d1ql pu
0
=
w- I
pu
(4.297)
The block diagram representation of the system is given in Figure 4.19. 4.15.5
Assuming constant flux linkage in the main field winding
From (4.228) we note that the voltage E;, which corresponds to the d axis field flux linkage, changes at a rate that depends upon T;O. This time constant is on the order of several seconds. The voltage EFD depends on the excitation system characteristics. If E FD does not change very fast and if the impact initiating the transient is short, in some cases the assumption that the voltage (or e;) remains constant during the transient can be justified. Under this assumption the voltage behind transient reactance E' or e' has a q axis component or that is always constant. The system equation to be solved is (4.296) with the network constraints (to determine the currents) and the condiis constant. tion that The next step in simplifying the mathematical model of the machine is to assume that E; and E' are approximately equal in magnitude and that their angles with respect to the reference voltage are approximately equal (or differ by a small angle that is constant). Under these assumptions E' is considered constant. This is the constant voltage behind transient reactance representation used in the classical model of the synchronous machine.
E;
e;
E;
E;
Example 4.8
The simplified model used in Section 4.15.2 (voltage behind subtransient reactance) is to be used in the system of one machine connected to an infinite bus through a transmission line discussed previously in Section 4.13. The system equations neglecting saturation are to be developed. Solution F or the case where saturation is neglected, the system equations are given by (4.263)-(4.268). This set of differential equations is a function of the state variables e:J, AD, w, and 0 and the currents id and i q • Equation (4.266) expresses as a linear combination of the variables and ADo For the mathematical description of the system to be complete, equations for id and iq in terms of the state variables are needed. These equations are obtained from the load constraints. From the assumptions used in the model, i.e., by neglecting the terms in ~J and ~q in the stator voltage equations (com pared to the speed voltage terms) and also by as-
E;,
e;'
E;
r
E"
Fig.4.20
R
x"
e
+
vt
-
X
~ i
+
v.
-
Network representation of the system in Example 4.8.
The Synchronous Machine
143
suming that w ~ WR, the system reduces to the equivalent network shown in Figure 4.20. By following a procedure similar to that in Section 4.15.2, equations (4.234) are given by V:tjd
=
-
RId
-
X" I + q
E:;
V2l q
(4.298)
where (4.299) and Vood = - Voo sin (~ - a)
=
Vooq
Voo cos
(~
- a)
(4.300)
From (4.298) I d and I, are determined Id = Id
1
(Ri +
=-
A
(R)
2
(" (1")2 [ - A R Vcod - Ed) 1
+
A"
(X)
2
A" (Vood [X
-
+ X'A,(Vooq
-
Eq )
"A + R(Vooq
-
E q)]
Ed)
"] "
(4.301)
Equations (4.147) and (4.301) along with the set (4.263)-(4.268) complete the mathematical description of the system. 4.16
Turbine Generator Dynamic Models
The synchronous machine models used in this 'chapter, which are in common use by power system engineers, are based on a classical machine with discrete physical windings on the stator and rotor. As mentioned in Section 4.14, the solid iron rotor used in large steam turbine generators provides multiple paths for circulating eddy currents that act as equivalent damper windings under dynamic conditions. The representation of these paths by one discrete circuit on each axis has been questioned for some time. Another source of concern to the power engineer is that the value of the machine constants (such as
L~, L~',
etc.) used in dynamic studies are derived from data ob-
tained from ANSI Standard C42.10 [16). This implicitly assumes two rotor circuits in each axis-the field, one d axis arnortisseur, and two q axis amortisseurs. This in turn implies the existence of inductances L d, L d, L;, L q , L;, and L;' and time constants Tdo' T;O' T;O' and T;~, all of which are intended to define fault current magnitudes and decrements. In some stability studies, discrepancies between computer simulation and field data have been observed. It is now suspected that the reason for these discrepancies is the inadequate definition of machine inductances in the frequency ranges encountered in stability studies. Studies have been made to ascertain the accuracy of available dynamic models and data for turbine generators [21-25J. These studies show that a detailed representation of the rotor circuits can be more accurately simulated by up to three discrete rotor circuits on the d axis and three on the q axis. Data for these circuits can be obtained from frequency tests conducted with the machine at standstill. To fit the "conventional" view of rotor circuits that influence the so-called subtransient and transient dynamic behavior of the machine, it is found that two rotor circuits (on each axis) are sometimes adequate but the inductances and time constants are not exactly the same as those defined in IEEE Standard No. ItS. The procedure for determining the constants for these circuits is to assume equiva-
Chapter 4
144
lent circu its on each axis made up of a number of circuits in parallel. The transfer function for each is called a n operat ional inductance of the form (4 .302)
L(s) = [N(s)/ D( s») L
where L is the synchronous reactance, a nd N( s) and D( s) are polynomia ls in s , Thus for the d axis we write Ld(s)
L (I d (I
=
+ als)(1 + bls)(1 + CIS) + a2s)( I + b2s)( I + c2s)
(4.303)
and the constants L d, a, . a 2• b, b2, CI • and C2 are determined from the frequency domain response. If the operational inductance is to be approximated by quadratic polynomials, the constants can be identified approximately with the transient and subtransient parameters. Thus , for the d axis, LAs) becomes L ( ) _ L [I d
s -
d
+ (L;/ L d )r;osJ( I + (L~ / L;)r;os) (I + r;os)(1 + rdO s)
(4 .304)
The time constants in (4 .304) are d ifferent from those associated with the exponential decay of d or q axi s open circuit voltages, hence the discrepancy with IEEE Standard
NO .115 .
An example of the data obtained by standstill frequenc y tests is given in [24] and is reproduced in F igure 4.21. Both third-order and second-order polynom ial representa tions are given . Machine data thus obtained differ from standard data previously obtained by the manufacturer from short circuit tests. Reference [24] gives a comparison between the two sets of data for a 5S5-M VA turbogenerator . T his compa rison is given in Table 4 .6.
2 .0 ~
a. - 0"
1=:=====--......:::-----__ 1 76 (I • 0.69. ) (I • o.on.) (l ~
1. 5
.....
1. 0
~
0 8 • 1 81 (1 • 1. 28. ) (I • 0 .0 16.) • (1. 7 .7 5s) (1 . 0 .022.) 0.6
o
.....-0 ~
g
o U ~
-0
.E
(I
1 81 ( 1 • 1.69 . )( 1 • 0 . 1&)(1 ' 0 . 0038. ) 0 .3 ' (1 + 8. 57.) (1 , 0 . 24< ) (1 , 0.0047.)
• 0 . 004. ) 1. 23. ) (1 . 0 . 162.) (1. 0 . 0070)
"-
0.5
o. ~
+
"
Li ')
-,
1 76 (1 • 0. 3 1< ) (l + 0 .031<) . (I ' 0. 9OS) (1 . 0 . 074.)
-,
-,
-,
-,
'<,
0. 2
Frequenc y response plot. 555 . 5 - MVA unit - - Test re sult s
----
- - Ad juste d re su lts fo r simu la ti on o f
two ro tor wind ings in ea c h a xis
Fig. 4.21
Frequency response plot for a 555-M VA turboalternator. (@IEEE. Reprinted from IEEE Trans.. vol. PAS -93. May/June 1974.)
The Synchronous Machine
145
Comparison of Standard Data with Data Obtained from Frequency Tests for a 555-MVA turboalternator
Table 4.6.
Constants
Standard data
Adjusted data
Ld pu
1.97 0.27 0.175 1.867 0.473 0.213 0.16 4.3 0.031 0.56 0.061
1.81 0.30 0.217 1.76 0.61 0.254 0.16 7.8 0.022 0.90 0.074
L~
L:;
Lq
L'q L"q
L{,
TdO
"
'dO
';0
,~~
pu pu pu pu pu pu s s s s
Source: © IEEE. Reprinted from IEEE Trans., vol. PAS-93, 1974.
The inductance versus frequency plot given In Figure 4.21 is nothing more than the amplitude portion of the familiar Bode plot with the amplitude given in pu rather than in decibels. The transfer functions plotted in Figure 4.21 can be approximated by the superposition of multiple first-order asymptotic approximations. If this is done, the break frequencies should give the constants of (4.304). The machine constants thus obtained are given in the third column of Table 4.6. If, however, the machine constants obtained from the standard data are used to obtain the breakpoints for the straight-line approximation of the amplitude-frequency plots, the approximated curve does not provide a good fit to the experimental data. For example, the d axis time constant 'da of the machine, as obtained by standard methods, is 4.3 s. If this is used to obtain the first break frequency for log [1 j( I + T~OS)], the computed break frequency is Ij,~o
= Ij4.3 = 0.2326 radjs = 0.00062 pu
(4.305)
The break point that gives a better fit of the experimental data corresponds to a frequency of 0.1282 radjs or 0.00034 pu. Since the amplitude at this frequency is the reciprocal of the d axis transient time constant, this corresponds to an adjusted value, denoted by T~6, given by '~6
= I j O. I282 = 7.8 s
(4.306)
Reference [24] notes that the proper ajustment of T~o, T;O' and L; are all particularly important in stability studies. A study conducted by the Northeast Power Coordinating Council [26] concludes that, in general, it is more important in stability studies to use accurate machine data than to use more elaborate machine models. Also, the accuracy of any dynamic machine model is greatly improved when the so-called standard machine data are modified to match the results of a frequency analysis of the solid iron rotor equivalent circuit. At the time of this writing no extensive studies have been reported in the literature to support or dispute these results. Finally, a comparison of these results and the machine models presented in this chapter are in order. The full model presented here is one of the models investigated in the NPCC study [26] for solid rotor machines. It was found to be inferior to the more
146
Chapter 4
elaborate model based on two rotor windings in each axis. This is not surprising since the added detail due to the extra q axis amortisseur should result in an improved simulation. Perhaps more surprising is the fact that the model developed here with F, D, and Q windings provided practically no improvement over a simpler model with only F and Q windings. Furthermore, with the F-Q model based on time constants 7~0 and 7;0' larger digital integration time steps are possible than with models that use the much shorter time constants Td~ and T;~, as done in this chapter. As a general conclusion it is apparent that additional studies are needed to identify the best machine data for stability studies and the proper means for testing or estimating these data. This is not to imply that the work of the past is without merit. The traditional models, including those developed in this chapter, are often acceptable. But. as in many technical areas, improvements can and are constantly being made to provide mathematical form ulations that better describe the physical apparatus. Problems 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
Park's transformation P as defined by (4.5) is an orthogonal transformation. Why? But the transformation Q suggested originally by Park [10, II] is that given by (4.22) and is not orthogonal. Use the transformation Q to find voltage equations similar to (4.39). Verify (4.9) by finding the inverse of(4.5). Verify (4.12) by sketching the stator coils as in Figure 4.1 and observing how the inductance changes with rotor position. Verify the following equations: (a) Equation (4.13). Can you explain why these inductances are constant? (b) Equation (4.14). Why is the sign of M s negative? Why is I M.~ I > L m? (c) Explain (4.15) in terms of the coefficient of coupling of these coils. Verify (4.16)-(4.18). Explain the signs on these equations by referring to the currents given on Figures 4.1 and 4.2. Verify (4.20). Explain the signs on all terms of{4.23). Why is the ~ term negative? Consider a machine consisting only of the phase winding sa-fa shown in Figure 4.1 and the field winding F. Sketch a new physical arrangement where the field flux is stationary and coil sa-fa turns clockwise. Are these two physical arrangements equivalent? Explain. For the new physical machine proposed in Problem 4.8 we wish to compute the induced EM F in coil sa-fa. Do this by two methods and compare your results, including the polarity of the induced voltage. (a) Use the rate of change offlux linkages X. (b) Compute the Blv or speed voltage and the transformer-induced voltage. Do the results agree? They should! Verify (4.24) for the neutral voltage drop. Check the computation of Pp-I given in (4.32). The quantities 'Ad and 'A q are given in (4.20). Substitute these quantities into (4.32) and compute the speed voltage terms. Check your result against (4.39). Verify (4.34) and explain its meaning. Extend Table 4.1 by including the actual dimensions of the voltage equations in an M LtJ,L system. Repeat for an FLtQ system. Let va(t) = Vm cos (WR t + a) Vb(t) = Vmcos(wRt + a -21r/3) vc(t) = Vmcos(wRt + a + 21r/3) (a) For the pu system used in this book find the pu voltages Vd and vq as related to the rms voltage V. (b) Repeat part (a) using a pu system based on the following base quantities: Sa = threephase voltampere and VB = line-to-line voltage. (c) For part (b) find the pu power in the d and q circuits and id and iq in pu.
The Synchronous Machine
147
Using the transformation Q of (4.22) (originally used by Park) and the MKS system of units (volt, ampere, etc.), find: (a) The d and q axis voltages and currents in relation to the rms quantities. (b) The d and q axis circuit power in relation to the three-phase power. 4.17 Normalize the voltage equations as in Section 4.8 but where the equations are those found from the Q transformation of Problem 4. t • 4.18 Show that the choice of a common time base in any coupled circuit automatically forces the equality of VA base in all circuit parts and requires that the base mutual inductance be the geometric mean of the self-inductance bases of the coupled windings; i.e.,
4.16
Sl8 =
4.19 4.20
4.21 4.22
4.25
=
Imax
ib
(L I8L28 ) ' / 2
=
- ( 1/2) I max
ic
=
- (
1/ 2) I max
Plot the M M F as positive when radially outward + t, enters sa I and + ib enters sb, but +i c enters jc., Assume the MMF changes abruptly at the center line of the slot. The M M F wave should be a stepwise sine wave. Is it radially outward along d or q? Verify (4.138). Derive formulas for computing the saturation function parameters As and B, defined in (4.141), given two different values of the variables AAD' i M O , and i M S ' Compute the saturation function parameters A J and BJ given that when AAD
0,
AAD
1.20, (i M S
(i MS -
where i M S and i MO correspond to 1.2
4.26
=
Show that the constraint among base currents (4.54) based upon equal mutual flux linkages is the same as equal M M F's in each winding. Show that the I/wR factors may be eliminated from (4.62) by choosing a pu time T = wRt rad. Develop the voltage equations for a cylindrical rotor machine, i.e.. a machine in which the inductances are not a function of rotor angle except for rotor-stator inductances that are as given in (4.16)-(4.18). Consider a synchronous generator for which the following data are given: 2 poles, 2 slots/pole/phase, 3 phases, 6 slots/pole, 12 slots, 5/6 pitch. Sketch the slots and show two coils of the phase a winding, coil I beginning in slot 1 (0°) and coil 2 beginning in slot 7 (180°). Label coil t sal-fa, (start a, and finish at) and coil 2 sa2-ja 2. Show the position of Nand S salient poles and indicate the direction of pole motion. Now assume the machine is operating at 1.0 PF (internal PF) and note by + and · notation, looking in at the coil ends, the direction of currents at time to, where at to
t,
4.23 4.24
M 128
S28
0. Let AADT
=
0.8
0.
iMO)/i MO
AAD =
-
=
0.13
1.2iMO)/1.2i M o
vIJ and
=
0.40
i M S is the saturated current at
AAD
=
Compute the saturation function KJ at AAD = 1.8, using the data and results of the previous problem. 4.27 The synchronous machine described in Examples 4.2 and 4.3 is connected to a resistive load of R L = 1.0 pu. Derive the equations for the state-space current model using vF and Tm as forcing functions. Use the current model. 4.28 Repeat Problem 4.27 using the flux linkage model. 4.29 Derive the state-space model for a synchronous machine connected to an infinite bus with a local load at the machine terminal. The load is to be simulated by a passive resistance. 4.30 Repeat Problem 4.29 for a local load simulated by a passive impedance. The load has a reactive component. 4.31 Obtain the state-space model for a synchronous machine connected to an infinite bus through a series resistance, inductance, and capacitance. Hint: Add two state" variables related to the voltage (or charge) across the capacitance. 4.32 Incorporate the load equations for the system of one machine against an infinite bus (shown in Figure 4.8) in the simplified models given in Section 4.15: (a) Neglecting damper effects.
148
4.33
Chap~r4
(b) Neglecting ~d and ~q for a machine with solid round rotor. (c) Neglecting damper effects and the terms Ad and Aq • Show that the voltage-behind-subtransient-reactance model of Figure 4.14 can be rearranged to give the model of Schulz (20) given in Figure P4.33 , if the rotor has two circuits on the q-axis.
X'd- Xc -X d t
r-- - - - - - - l X E'
q
Fiel d Sta tor
Fi el d current
\d I F
E'
~
d
1 +
q
(X ' - X" ) IX - X' )
q
q
q
(X' - X )2 q t
q
( X~ - X'~) (X - X'ql
q
(X' -
X/
(X" - X l (X - X' ) q I q q X' - X q C
Fig. P4.33
4.34 4.35 4.36
Using the third-order transfer functions for Ld(s) and Lq( s) given in Figure 4.21, sketch Bode diagrams by making straight-line asymptotic approximations and compare with the given test results . Repeat Problem 4.34 using the second-order transfer functions for LAs) and Lq(s) . Repeat Problem 4.35 using the second-order transfer functions of (4.304) and substituting the standard data rather than the adjusted data.
References I. Concord ia, C. Synchron ous Machines. Wiley, New York, 1951. 2. K irnbark , E. W. Power System Stability. Vols. 1,3. Wiley, New York, 1956. 3. Adkins, B. The General Theory of Electrical Machines . Chapman a nd Hall, London, 1964.
The Synchronous Machine
149
4. Crary,S. B. Power Svstem Stabilit y, Vols. 1,2. Wiley, New York, 1945, 1947. 5. Lynn, T. W., and Walshaw, M. H. Tensor Analysis of a Synchronous Two-Machine System. lEE (British) Monograph. Cambridge Univ. Press, London. 1961. 6. Taylor, G. D. Analysis of Synchronous Machines Connected to Power Network. lEE (British) Monograph. Cambridge Univ. Press, London, 1962. 7. Westinghouse Electric Corp. Electrical Transmission and Distribution tceference Book. Pittsburgh, Pa.. 1950. 8. Anderson, P. M. Analysis of Faulted Power Systems. Iowa State Univ. Press, Ames, 1973. 9. Harris, M. R.. Lawrenson, P. J .. and Stephenson, J. M. Per Unit Systems: With Special Reference to Electrical Machines. lEE (British) Monograph. Cambridge Univ. Press, London, 1970. 10. Park, R. H. Two reaction theory of synchronous machines, Pt. I. AlEE Trans. 48:716-30,1929. II. Park, R. H. Two reaction theory of synchronous machines. Pt. 2. AlEE Trans. 52:352-55.1933. 12. Lewis, W. A. A basic analysis of synchronous machines. Pt. l. AlEE Trans. PAS-77:436-55, 1958. 13. Krause, P. C., and Thomas. C. H. Simulation of symmetrical induction machinery. IEEE Trans. PAS84:1038-52, 1965. 14. Prentice. B. R. Fundamental concepts of synchronous machine reactances. AlEE Trans. 56 (Suppl. I): 716- 20.1929. 15. Rankin. A. W. Per unit impedances of synchronous machines. AlEE Trans. 64:569-72,839-41.1945. 16. IEEE. Test procedures for synchronous machines. Standard No. 115, March, 1965. 17. IEEE Committee Report. Recommended phasor diagram tor synchronous machines. IEEE Trans. PAS-88:1593-1610.1969. 18. Prabhashankar, K., and Janischewskyj, W. Digital simulation of multimachine power systems for stability studies. IEEE Trans. PAS-87:73-80, 1968. 19. Young. C. C. Equipment and system modeling for large-scale stability studies. IEEE Trans. PAS91:99- 109. 1972. 20. Schulz, R. P. Synchronous machine modeling. Symposium on Adequacy and Philosophy of Modeling: System Dynamic Performance. IEEE Pub!. 75 CH 0970-PWR. 1975. 21. Jackson. W. B.. and Winchester. R. L. Direct and quadrature axis equivalent circuits for solid-rotor turbinegenerators. IEEE Trans. PAS-88:1121---36. 1969. 22. Schulz. R. P.. Jones. W. D.. and Ewart. D. N. Dynamic models of turbin-e generators derived from solid rotor equivalent circuits. IEEE Trans. PAS-92:926-33.1973. 23. Watson. W.. and Manchur. G. Synchronous machine operational impedances from low voltage measurements at the stator terminals. I EEl:" Trans. PAS-93:777-- 84. 1974. 24. Kundur, P.. and Dandeno, P. L. Stability performance of 555 MVA turboalternators----Digital comparisons with system operating tests. IEEE Trans. PAS·93:767 76. 1974. 25. Dandeno, P. L.. Hauth. R. L.. and Schulz. R. P. Etfects of synchronous machine modeling in largescale system studies. 1l::EE Trans. PAS-92:574- 82. 1973. 26. Northeast Power Coordinating Council. EtTects of synchronous machine modeling in large-scale system studies. Final Report. N PCC-IO. Task Force on System Studies. System Dynamic Simulation Techniques Working Group. 1971.
chapter
5
The Simulation of Synchronous Machines 5. 1
Introduction
This chapter covers some practical considerations in the use of the mathematical models of synchronous machines in stability studies. Among these considerations are the determination of initial conditions, determination of the parameters of the machine from available data, and construction of simulation models for the machine. In all dynamic studies the initial conditions of the system are required. This includes all the currents, flux linkages, and EM F's for the different machine circuits. The number of these circuits depends upon the model of the machine adopted for the study. The initial position of the rotor with respect to the system reference axis must also be known. These quantities will be determined from the data available at the terminals of the machine. The machine models used in Chapter 4 require some data not usually supplied by the manufacturer. Here we show how to obtain the required machine parameters from typical manufacturer's data. The remainder of the chapter is devoted to the construction of simulation models for the synchronous machine. Both analog and digital simulations are discussed. 5.2
Steady-State Equations and Phasor Diagrams
The equations of the synchronous machine derived in Chapter 4 are differential equations that describe machine behavior as a function of time. When the machine operates in a steady-state condition, differential equations are not necessary since all variables are either constants or sinusoidal variations with time. For this situation phasor equations are appropriate, and these will be derived. It is common to tacitly assume all machines to be in a steady-state condition prior to a disturbance. The socalled "stability study" examines the system behavior following the disturbance. The phasor equations derived here permit the solution of the initial conditions that exist prior to the application of the disturbance. This is a necessary part of any stability investigation. From (4.74) at steady state all currents are constant or, mathematically, (5.1 )
Then from (4.74) (5.2) 150
151
Simulation of Synchronous Machines
or at steady state iD
= iQ = 0
(5.3)
Using (5.1) we may write the stator voltage equation from (4.74) as (5.4) From (4.5) with balanced conditions, Va
=
Vo
= O. Therefore, from (4.9) we may compute
V273(VdCOSO
+ vq sin 8)
(5.5)
where by definition 0 = WRt + 0 + 11"/2. Then from (5.4) and (5.5) Va = V273[-(ri d
wLqiq)cos(wRt + 0 + 11"/2)
+ () + V273[-(ri d + wLqiq)cos(wRt + 0 + 11"/2) + (-riq + wLdi d t kMFwiF)coS(WRt + 0)] +(-riq
+
+
wLdid + kMFwiF)sin(U.'Rt
At steady state the angular speed is constant, w noted as reactances, or
= WR,
11"/2)] (5.6)
and wL products may be de(5.7)
From (4.226) we also identify (5.8)
where E is the stator equivalent EMF corresponding to iF. Using phasor notation,' the V2 multiplier of (5.6) is conveniently used to define the rms voltage phasor
where the superior bar indicates a total phasor quantity in magnitude and angle (a complex number).
By using the relation j
v..
=
=
1Min (5.9),
-r(~ IE. + j ~
fl) -
jX q
~
fl +
Xd
~
fl + Efl
(5.10)
Note that in this equation Va and E are stator rms phase voltages in pu, while id and iq are de currents obtained from the modified Park transformation. The choice of this particular transformation introduced the factor 1/V3 in the equation. To simplify the notation we define the rms equivalent d and q axis currents as
t,
~ id /V3
t,
~ iq /V3
(5.11)
The stator current t, expressed as a phasor will have the two rectangular components I q and I d • Thus if the phasor reference is the q axis, Ia
=
(Iq + jId) e j~
(
5.12)
I. We define the phasor A = Ae j a as a complex number that is related to the corresponding time domain quantity a(/) by the relation a(l) = <Jl.e, (V2Ae JCI1 t ) = V2A cos (wI + a).
152
Chapter 5 q a xis
Fig. 5.1
Phasor diagram representing (5.14).
Substituting (5.12) and (5.11) in (5.10) and rearranging,
E& =
and by using
E = EjJ, t, = IqlJ., E=
v..
+
-t, + jxqlq /J. -
Xdld L2.
(5.13)
t, = jldjJ, v.. + -t. + jX/q + jXdId
and
(5.14)
The phasor diagram representing (5.14) is shown in Figure 5.1 [I] . Note that the phasor jxqIq leads the q axis by 90·. The phasor jXdld makes a 90· angle with the negative d axis since Id is numerically negative for the case illustrated in Figure 5.1. To obtain Vd and vq from (5.4), we compute the rms stator equivalent voltages "
II ·
.
Vd ~ Vd/v'3
=
Vq ~ vq/V] = -rlq + Xdld + E
(5.15)
Note that Vq and Vd are the projection of Va along the q and d axes respectively. Also note that in the phasor diagram in Figure 5.1 both Vd and I d are illustrated as negative quantities. Thus the magnitude of rid is subtracted from xqlq to obtain the magnitude of Vd . This situation is shown in Figure 5.1 since lagging current (negative I d ) is commonly encountered in practice. Examining Figure 5.1 and (5.15), we note that if the angle 0 is known the phasor diagram can be constructed quite readily . If the position of the q axis is not known but the terminal conditions of the machine q ax is
Fig.5.2
Location of the q axis from a known terminal current and voltage .
153
Simulation of Synchronous Machines
are given (i.e., if Va' fa' and the angle between them are known), construction of the phasor diagram requires some manipulation of (5.15). However, an alternate procedure for locating the position of the q axis is illustrated in Figure 5.2, where it is assumed that Va' fa' and the power factor angle are known. Starting with J:: (used here as reference) the voltage drop rl; is drawn parallel to t.. Then the voltage drop jxqTa is added (this is a phasor perpendicular to Ta). The end of that phasor (£qa in Figure 5.2) is located on the q axis. This can be verified by noting that the d axis component of the phasor jXq~ is x q which is similar to that shown in Figure 5.1. Its q axis component however is xqTd , which is different from that shown in Figure 5.1. Thus to locate the phasor E in Figure 5.2, we add the phasor (x, - x q )Id to the phasor
4,
e.;
5.3
Machine Connected to an Infinite Bus through a Transmission Line
To illustrate more fully the procedure for finding the machine steady-state conditions, we solve the simple problem of one machine connected to an infinite bus through a transmission line. Although this one-machine problem is far simpler than actual systems, it serves well to illustrate the procedure of finding initial conditions for any machine. As we shall see later, this simple problem helps us concentrate on concepts without becoming engulfed in details. The differential equations for one machine connected to an infinite bus through a transmission line with impedance Z, = R, + jwRL e is given by (4.149). Under balanced steady-state conditions with zero derivatives, (4.149) becomes Vd = vq
vI3 V~ sin (0
= V1v~
- a)
cos(o - a)
+ Reid + wLei q
+ Reiq - wLei d
(5.16)
Substituting for Vd and vq from (5.4) into (5.16), -rid - wLqiq
= -
vl3vao sin(o -
a)
-riq + wLdi d + kMFwi F = vl3Vaocos(5 - a)
+
+ Reid + wLei q Reiq - wLeld
By using (5.7) and (5.11) and rearranging the above equations, we compute E
=
o=
Vao cos (0 - a) + (r + Re)/q - (x, + Xe)l d -
Vao sin(o - a) + (r + Re)ld + (x,
+ Xe}l q
(5.17)
where X e = wL e • Equations (5.17) represent the components of the voltages along the q and d axes respectively. The phasor diagram described by these equations is shown in Figure 5.3, where the phasor representing the infinite bus voltage Vao , with the q axis as reference, is given by (5.18)
Note that Figures 5.1 and 5.2 can be combined since the same q and d axes, the same EM F E, and the same current I, are applicable to both. Thus in Figure 5.3 the machine terminal voltage components Vd and ~ can be obtained using (5.15). An alternate procedure would be to start with the phasor Vao in Figure 5.3, then add the voltage drop ReIq - Xeld in the q axis direction and the voltage drop ReId + Xel, in the d axis direction to obtain the phasor ii:,. Again remember that in Figure 5.3 both I d and V~d are shown as negative quantities. The remarks concerning the location of the q axis starting from V~ and I, are also applicable here.
Chapter 5
154
Fig . 5.3
5.4
Phasor diagram or (5.17).
Machine Connected to an Infinite Bus with Local Load at Machine Terminal
The equations that relate the infinite bus voltage V. to the stator equivalent EM F
E are given by (5 .17) . Note that this form of the equations does not give the machine
terminal voltage explicitly. Since the terminal voltage is a quantity of considerable interest, we seek a solution in wh ich Vd and Vq are given explicitly. One convenient method is to add a local load at the machine terminals, as shown in Figure 5.4. For the system shown in Figure 5.4, the steady-state equations for the machine voltages, EM F's, and currents are the same as given by (5.14), (5.15), and (5.12) respectively. Equations (4.149), which at steady-state conditions are the same as (5 .16), are still applicable except that the currents id and iq should be replaced by the currents i'd and i,q . These are the d and q axis components of the transmission line current i,. In other words, with the q axis as a reference, (5.19) where we define (5.20) The transmission line equations are then given by
R• L e
e
.. I
Fig.5.4
t
vee
One mach ine with a local load connected to an infinite bus through a transmission line.
155
Simulation of Synchronous Machines
V1 V~ sin (0 V1 V cos (0 -
Vd = -
vq
=
+ Rei'd + wLei,q
- a)
+ Rtitq
a)
oo
(5.21 )
wLei'd
-
which can be stated in the form Vd
=
-
Voo sin (0 - a) + Reltd + Xel,q + Rel rq - XeI'd
(5.22)
Vq = Voo cos(o - a)
Ta , we refer to Figure 5.4. By inspection we
To obtain a relation between I, and can write the phasor relations
(5.23) where we define
ZL
= R L + jX L . Separating the real and imaginary components, (I q - l,q) R L
-
(Id - I td ) X L
=
Vq
(L, - l'd)R L
+ (/ q - l,q)X L
=
Vd
(5.24)
From (5.24) we can solve for l,q and l u(5.25) The equations for the q and d axis voltage drops can then be obtained from (5.25), (5.15), and (5.22). 5.4.1
Special case: the resistive load,
For this case XL
=
It = Rt +
iO
O. From (5.25)
(5.26) Substituting (5.26) into (5.22), Vd
=
-
Voo sin (0 - a) + Re(/d
Vq = Vco cos(o - a) + Re(lq
Vd/ R L ) Vq/R L ) -
-
-
+
Xe(/ q - Vq / R L )
Xe(/ d - Vd/R L )
or Vd(1 + Rt>/RL ) + Vq(Xe/R L )
sin(o - a) + ReId + Xel q Voo cos(o - a) + Relq - Xel d
= - Vao
- Vd(Xe/R L ) + Vq(1 + Rl'/R L } =
(5.27)
Substituting (5.15) into (5.27) and rearranging, Xl'
RL
E__ V sin. (~ -
co
+ (x q
u -
a
) + (R
R L R+ R e +
(I + ;:)E = V.,cos(o -
L
a) -
+ RL - XeXd) - ld RL RL
+
r
r RXl'
+
x) I
L
(X
L + (R e + r Re R+L R
Now define
s,
e
e
-
+
e
r ::
XeX ) R;q
+ I
q
q
Xd
RL :
L
R e) t, (5.28)
156
Chapter 5 q axis
E,
.
X1
q q
d oxis
.
R 1
q q
Fig. 5.5 Phasor diagram of a synchronous machine connected to an infinite bus with local resistive load.
R
R,
+ RL
X,xJ If':
R~ J
=
i;
= X,(I + rfRd + xq(1 + R,fRd
,+ r
R
L
-
A
/(q
=R
XJ =
+
,
r R, + R L R
_
L
X,x q R L
X,(I + rfRd + xJ(1 + R,fRd
Then (5.28) can be written as (I
(X,f RdE = - V", sin (<5 - a) + k.t, + Xqf q = V",cos(<5 - a) - XJf J + k.t,
+ R,fRdE
Let us define a phasor
(5.29)
£1 : £1 =
(I
+ R,fRdE + j(X,fRdE
(5.30)
£1 makes an angle 'Y with the q axis
where the phasor
'Y
=
arctan/X,f(R, + Rdl
(5.31 )
The phasor diagram for (5.29)--(5.31) is shown in Figure 5.5. 5.4.2
The general case: It arbitrary
For 2 L arbitrary the equations are more complicated . (5.22) and rearranging,
VJ
(I +
RLR,
Substituting (5 .25) into
+2 XLX,) + Vq (RLX, - 2 XLR,) -- _ V Sin . (~u a ) + R , IJ + X , Iq
ZL
_ Vd (R LX, - 2 XLR,) + Vq ZL.
ZL
ee
(I + RLR, ZL+ XLX,) = V 2
(~
'" COS u -
I a ) + R' q
-
X , Id (5.32)
or
VAl + A.) + VqA 2 = -V", sin (<5 - a) + R,IJ + X,lq -VJA2 + Vq(1 + AI) = V",cos(<5 - a) - X,l d + R,lq
(5.33)
where AI
= (RLR, + XLX,)jZr
Combining (5.33) and (5.15),
(5.34)
157
Simulation of Synchronous Machines
A2E = - V<X) sin(o - a) + [Re + r(1
(I
+
+ AI)E
=
(Xl'
+ xq(l +
+
AI) - xdA2]ld
AI) + rA 2 ]l q
V<X) cos(o - a) + [-Xl' - rA2 - xd(1 + AI)]ld
+
[R e
XqA2 + r(1 + AI)]/q
-
(5.35)
Again, by defining £1 i (1 + AI)E + jA 2E, '"
~
'"
~
Rd
R, + r(1 + AI) - X q A2 Xq i Xl' + xq(l + AI) + rA2
+ AI) - xdA2 + Xd( I + AI) + rA 2
'" R,
R, + r(1
=
X d = Xl'
Ii.
=
(5.36)
we may write (5.35) in the form (1
+
A2E
=
- V<X)
sin (l5 - a) + RdI d + XqI q
AI)E = V<X) cos (l5 - a) - Xdl d
+ k.t,
(5.37)
Since (5.37) is of the same form as (5.29), it can be represented by the same phasor diagram in Figure 5.5. 5.5
Determining Steady-State Conditions
The most common' boundary conditions are the terminal voltage Va and either the current I, and the power factor F, or the generated power P and the reactive power Q (per phase). In either case Va' la, and 11 (the power factor angle) are assumed to be known. Resolving Ta into components with v:z as a reference, we write
fa = I, + jI x
(5.38)
where I, is the component of Ta in phase with ii:, and / x is the quadrature component (which carries its own sign). We also define the power factor F, as Fp = cos 11
where 11 is the angle by which I a lags
(5.39)
Then
~.
(5.40)
The phasor Eqa in Figure 5.2 is given by -
Eqa
Ii.
=
=
-
-
Va + (r + j X q ) / a
+ (/, + j Ix) (r + j X q ) (Va - xqI x + rIr) + j(xqI, + rIx) =
~
(5.41)
The angle between the q axis and the terminal voltage ~ (i.e., the angle 0 - (3 in Figures 5.1 and 5.2) is given by fJ - {3
= tan - I [(x q I r + r I x ) / (Va + r I, -
X q / x )]
(5.42)
The above relations are illustrated in Figure 5.6. Then we compute (5.43)
and Vd and vq can then be determined from their relationship to Vd and Vq given by (5.15). The currents are obtained from (5.44)
and the rotor quantities i d and i q can be determined from (5.11).
The remaining
158
Chapter 5 d a xis
v
q I
I I I
Vol Ix
Fig . 5.6
r1 -r]
- '1' (
x
x 1 qq
Phasor diagram illustrating (5.41) and (5.42).
currents and flux linkages can readily be determined once these basic quantities are known . In the case of a synchronous machine connected to an infinite bus the same procedure is followed if the conditions at the machine term inals are given. The voltage of the infinite bus is then determined by subtracting the appropriate voltage drops to the machine terminal voltage v.,. If the terminal conditions at the infinite bus are given as the boundary conditions, the position of the q axis is determined by a procedure similar to the above . The machine d and q axis currents and voltages and the machine terminal voltage can then be determined . This is illustrated in Examples 5.1 and 5.2 . 5.5.1
Case 1:
Machine connected to an infinite bus with local load V~ ,
E, and the machine load an¥le 0 - ex are known .
In this case I d and I q can be determined directly from (5.37). Then from (5.15) we can determine Vd and Vq • The three-phase power of the machine can be determined from the relation P3~ = 3( Vdld + VJq). The terminal current I, is determined from (5.25), and knowing V~ we can also determine the power and power factor at the infinite bus. . Case 2: Machine terminal conditions Va. l«, and power factor are known . From la, Va. and the power factor the position of the quadrature axis is determined (see Figure 5.2). From this information Id • Vd , Iq , and Vq can be found. Also E can be calculated from (5.13). From (5.36) and (5.37) the phasor £1 can be constructed . The infinite bus voltage can then be determined by drawing Rdld + Xqlq parallel to the d axis and Rqlq parallel to the q axis, as shown in Figure 5.7. Thus j7" and the angle 0 - a are found, from which we can determine V.,d and V.,q . The current I, is determined from (5.25), and the power at the infinite bus is given by 3(V"d/'d + V.,q/,q).
x.t,
Case 3: Conditions at infinite bus are known. From j7,.. T" and Z. the machine terminal voltage Va is calculated. Then from V and ZL we can determine TL • From TL and 7,. T" is found . Now the conditions at the terminals of the machine are known and the complete phasor diagram can be constructed.
Simulation of Synchronous Machines
159
E,
Fig.5 .7
5.6
Construction of the phasor diagram for Case 2.
Examples
The procedures described are illustrated by several examples where different initial conditions are given.
Example 5./ The machine described in Examples 4.1, 4.2, and 4.3 is to be examined at rated power and 0.85 PF lagging conditions (nameplate loading) . The terminal voltage is 1.0 pu. Calculate the steady -state operating conditions. If this machine is connected by a transmission line of 0.02 + j0.40 pu impedance to a large system, find the infinite bus voltage . Solution From previous examples and the prescribed boundary conditions the following data are available:
v. = 1.000
=
1.700 pu 1.640 pu
R.
=
0 .001096 pu
L. =
Xd
=
xq
r
F; = cos cP = 0.850
Z.
=
=
pu 0.02 pu 0.4 pu 0.4005/87.138°
From the given power, power factor , and voltage we compute I a = 1.0/0.85 = 1.176 pu
The angle cP is computed from F, as cP = I, = I, cos cP = 1.000
COS-I
0.85 = 31.788° . Then from (5.40)
I, = I, sin cP = -0.620
From (5.42) and Figure 5.6 1.00 x 1.64 - 0.001096 x 0.620 (0 - (3) = arctan 1.000 + 0 .620 x 1.64 + 1.00 x 0 .001096
arctan 0 .8126
=
39.096 °
and 0 - {3 + cP = 31.788 + 39.096 = 70.884° = angle by which Ia lags the q axis. Then from (5.44)
Chapter 5
160 f q = fa
t,
cos (0 - (3 + 11) = 0.385 pu
0.667 pu
=
and f d = -lasin(o - (J
+
=
id
-1.112 pu
=
-
1.925 pu
From (5.43) ~ =
~
cos 3~.09° = 0.776 pu sin 39.09
~ = - ~
0
=
1.344 pu
Vq =
vd
-0.631 pu
=
-1.092 pu
From Figure 5.1 by inspection E
= =
+
rl, - xd1d 0.776 + 0.001096 x 0.385 + 1.70 x 1.112 ~
= 2.666 Now using (5.8) in pu, iF Then iF
=
E FD at steady state [from (4.209) and (5.8)]
v1 E/LAD where, from Example
=
= (v1 x 2.666)/1.55
=
4.1, LAD
=
1.55 pu.
2.979 pu
The currents iD and iQ are both zero. The flux linkages are given in pu by Ad
Ldid + k M FiF = 1.70( - 1.925) + (1.55)(2.979) = 1.345 (id + iF) k M F = (2.979 - 1.925)( 1.55) = 1.634
=
AAD =
Aq = Lqiq = 1.64 x 0.667 = 1.094 k M Qiq = 1.49 x 0.667 = 0.994
AAQ =
AF
=
kMFid
AD
=
k M Did
+ LFiF = 1.55(-1.925) + (1.651)(2.979) = 1.935 + M Rif = 1.55(2.979 - 1.925) = 1.634 = AAO
AQ
=
kMQiq
=
0.994 = AAQ
As a check we calculate the electrical torque Te , which should be numerically equal to the three-phase power in pu. Tet/J
= =
iq Ad - id Aq
0.667 x 1.345 + 1.925 x 1.094 = 3.004
Then T, = 1.001 pu. If we subtract the three-phase 12r losses, we confirm the generated power to be exactly P = T, .- ,[~ = 1.000. We also calculate the infinite bus voltage for this operating condition. We can write V = ~ - Ze J:. Let ~ = ~ I.J!.. = 1.0 I.J!... Then lXl
~
V
lXl
L!!.
=
[al{3 - ¢ = 1.176/{3 - 31.788
=
I.O/p" - (0.4005 187.138°)(1.176 1,8 - 31.788°)
{3
=
0
or Voo la
-
1.0 - 0.4712/55.349 = 0.828/-27.899° pu 0
Thus we have Voo = 0.828 pu, and {3 - a = 27.899 = the angle by which ~ leads Voo ' The angle between the infinite bus and the q axis is computed as 0
o-
a = (0 - (3) + ({J - a) = 39.096
+ 27.899 = 66.995
0
Simulation of Synchronous Machines
161
Example 5.2 Let the same synchronous machine as in Example 5.1 be connected to an infinite bus through a transm ission line hav ing R, = 0.02 pu, and L, = X, = 0.4 pu . The infinite bus voltage is 1.0 pu. The machine loading rema ins the same as before (P = 1.0 pu at 0.85 PF) . The boundary conditions given in this example are "mixed"; i.e., the voltage is known at one point (the infinite bus), while the power and reactive power are known at a different po int (the machine terminal) . A slight modification of the procedure of Example 5.1 is needed . Solution
A good approximation is to assume that the power at the infinite bus is the same as at the machine terminals by neglecting the ohmic power loss in the transmission line (since R, is small) . A better approximation is to assume a power loss in the transmission line based on some estimate of current (say 1.0 pu current). Let I~ R, = (1.00)2 (0.020) = 0.02 pu. Then the power at the infinite bus is 0.980 pu and the component of the current in phase with V~ is I, = 0.980 pu . The angle 8 between I, and V", is given by tan 8
=
Ix/I,
=
1.020 I,
The angle 13 between Va and V", is given by an equation similar to (5.42), viz., tan {3
=
XiI, + R,Ix V", - XlIx + R,I,
The power factor angle at the machine terminal rjJ = {3
+8=
COS -I
0 .392 + 0.02/x 1.020 - OAlx rJ>
is given by
0.85 = 31.788°
These angles are shown in Figure 5.8, with V", used as reference; i.e., a = O. Then tan rJ> = tan (COS -I 0.85) = 0.620 . Using the identity tan
rJ> =
(tan (3 + tan 8)/( I - tan (3 tan 8)
we compute 0.620 = -1.0201 x + (0.392 + 0.02l x)/(1.020 - OAlx ) 1+ [1.020(0.392 + 0.02l x)/x]j(1.020 - OAl x )
from which we get I, = -0.217 pu . q axis
d axis
Fig. 5.8
Phasor diagram of Va and V",.
162
Chapter 5
From the known value of I, we can now determine {3. {3 = tan-I [(0.392 - 0.004)] = 19.310° (1.020 + 0.082)
Also
o=
cP = 19.310 + 12.483 = 31.793°
tan- I(0.213jO.980) = 12.483°
which is a good check (see above). The terminal voltage Jt;, is given by ~
= (Vel) - x J, + Rt'J,) + j(X~/, + R~/x) = 1.106 + jO.388 = 1.172 L.!.2l.r pu
The generator phasor current is
I,
0.980 - jO.21? = 1.003/-12.48° pu
=
and P = ~/Qcos(jJ = 1.0001 pu (on a three-phase basis). The position of the q axis can be determined from an equation similar to (5.41). With a = 0, () =
tan'
(x q
+ X,J/, +
Vel) - (x, + X~)Jx
+ R~)/x + (r + R~)/,
(r
=
53.7360
The currents, voltages, and flux linkages can then be calculated as in Example 5.1. The results are given below in pu: .
id = -1.591
Ad
=
1.676
iq = 0.701
AD =
iF = 2.826
Aq = 1.150
E = 2.529
AQ
vd = - 1. 148 vq = 1.675
AAD
=
= AAQ =
Tt'q,
=
3.004
T,
=
1.001
1.914 1.045
In steady-state system studies (often called load-flow studies) it is common to specify the generator boundary conditions in terms of generated power and terminal voltage magnitude, i.e., P and J!;. (Both ~ and ~ are commonly used for the terminal voltage and both are used in this book.) In studies of large systems these boundary conditions are satisfied by iterative techniques, using a digital computer. For the one machine-infinite bus problem the system may be solved explicitly. We now consider the one machine-infinite bus problem with a local load connected to the ~ bus consisting of a shunt resistance R L and a shunt capacitance C L , representing the transmission line susceptance. The system of generator, local load, and line may be conveniently described as a two-port network (Figure 5.9) for which we write, with Vel) as reference (a = 0),
(5.45) The apparent power injected at node 1 may be computed as
81 =
PI + jQI
= V;~* = viY~ + V;Vi~*2
(5.46)
163
Simulation of Synchronous Machines
--- CD
r
[l =-1o t
Fig. 5.9 One-machine system as a two-port network.
Then we may compute
( 5A7) where we define Vk m = Gk m + jB k m for all k and m. In (5A7) PI' v" and V", are specified, while Gil' G12 , and B I 2 are known or computed system parameters. Thus we may solve (5A7) for the angle (3 . In doing so, it is convenient to define a constant angle)' related to the adm ittance element VI2 = Y12!:L Then from (5A7) we define
F = cos {)' - (3) = {PI - GII V;)/Y 12V,V",
(5A8)
from which {3 can be found . Obviously, there are lim its on the magnitude of PI that can be specified in an y ph ysical situatio n, as the co sine function is bounded in (5A8) .
Example 5.3 Compute the stead y-state co nditi o ns for the system of Exa m ples 5.1 and 5.2, where the given boundary condition s in pu are
P
=
1.0
v,
(on a three-phase ba sis)
= 1.17
V", = 1.00
and where the loc al load is given in pu as
RL
100
=
Solution For the numerical data and boundary conditions given, we compute R, + jX,
= 0 .02 + jOA = 0.4005 /87.138° pu - Jil2 = - 1(Z, = Y I 2 -0.1247 + j2.4938 = 2.4969 /92 .862 ° pu
Z,
V12
tx.
or)' = 92 .862 ° We are also given that RL = 100pu and BL = 0.01 pu. Thus the admittance from node I to reference is JiIO = 0 .01 + jO.OI pu . We then compute
VII
=
f lO + Jil2
=
Gil
+ jB II
=
0 .1347 - j 2.4838 pu
We now compute the quantity F defined in (5.48) as 0.2792 Then )' - {3
or
V,
=
= COS - I
1.17 /19.074°.
F
=
73 .788 °
{3
=
92.862 - 73.788
19.074 °
Chapter 5
164
To find the currents, we note from Figure 5.9 that I:
T:
T, + ~.
=
= 4 + Ie = (V,/R L )!.!!.. + (V,/X c ) / f3 + 90° = 0.0072 + jO.O 149 pu
Now
We also write
I, = (~ - V~)/~ [Rl'( V, cos f3
=
-
+ Xl'v, sin {3] + j [Rl' V; sin {3 - Xl'( V, cos (j -
V~)
0.9667 - jO.2161 pu
Then, noting that
Z;
=
V~ )]
0.99056 /- 12.60 or - 0.2199 radians
t, lies at an angle () from V~ (Figure 5.8), I, = T, + t, = t, L!!.. = 0.9739 - jO.2012 0.9945 / - 11.672° pu
=
We may now compute, as a check,
P + jQ
=
v;r:
1.000 + jO.595 1.164/30.746° pu =
The power factor is
Fp = cos 30.7460
=
0.859
The quantity Eqa of Figure 5.2 may be computed as a means of finding 0 we compute, as in Figure 5.6,
o.
Thus with
a =
s; = e; Ii = v, 11.. + n, L!!.. + jxql
a
=
and ()
=
L!!..
2.446 /54.024° pu
54.024°. Then we com pute () - {3
= 34.950°
1J
= ()
+ {3
=
30.746
() - {3
0
+ 1J
=
65.696°
With all the above quantities known, we compute d-q currents, voltages, and flux linkages in pu as in Example 5. 1, with the result id = -1.570 iq = 0.709
Ad AAD
vd vq
=
1.661
AAQ
E
=
2.500
AF
i, = 2.794
Tl'~
= -
1.161
Aq
=
1.662
= AD = 1.897 =
1.163
= AQ =
=
1.056
2.180
= 3.003 P, = 1.000
Example 5.4
The same machine at the same loading as in Example 5.1 has a local load of 0.4 pu power at 0.8 PF. It is connected to an infinite bus through a transmission line having R, = 0.1 pu and Xl' = 0.4 pu. Find the conditions at the infinite bus.
Solution The internal machine currents, flux linkages, and voltages are the same as in Example 5.1. Thus, in pu,
165
Simulation of Synchronous Machines
ld = -1.112 ~
t,
=
0.385
- 13
=
39.096
Vq
= 0.776
Vd = -0.631
E
=
2.666
0.4/(1.0 x 0.8)
=
0.5 pu
0
From the local load information IlL
I
=
Therefore lL = 0.4 - jO.3 pu. We can also determine that, in pu, RL
=
1.6
ZL
=
2.0
Thus we compute from (5.34) AI = (1.6 x 0.1
A2
=
+ 1.2 x 0.4)/(2.0)2 = 0.16
(1.6 x 0.4 - 1.2 x 0.1)/(2.0)2
=
0.13
Then
Rd =
+ 0.001096 x 1.16 - 0.13 x 1.7 = -0.1197 Rq = 0.1 + 0.001096 x 1.16 - 0.13 x 1.64 = -0.119 Xd = 0.4 + 1.7 x 1.16 + 0.001096 x 0.13 = 2.372 Xq = 0.4 + 1.64 x 1.16 + 0.001096 x 0.13 = 2.303 0.1
From (5.37) Vood
= =
Voo sin (8 - a)
=
-(-1.112)(-0.1197) - (0.385)(2.303)
-0.673
V oo q = VC() cos (0 - a) = (-1.112)(2.372) - (0.385)(-0.119) =
0.501
VC() = [(0.673)2
+
(0.13)(2.666)
+ (1.16)(2.666)
+ (0.501)2]1/2 = 0.839
From (5.25)
I'd = - 1.\\2 + 0.776 x 1.2 ; 0.63\ x 1.6 = -0.6268
t.; =
0.385 _ 0.776 x 1.6 ~ 0.63\ x 1.2 = 0.2639
The power delivered to the infinite bus is
P:c
=
(-0.673)( -0.6268) + 0.2639 x 0.501
=
0.554 pu
The power delivered to the local load is PL = 0.4 pu. Then the transmission losses are 0.14 pu, which is verified by computing R;/:.
5.7
Initial Conditions for a Multimachine System
To initialize the system for a dynamic performance study, the conditions prior to the start of the transient must be known. These are the steady-state conditions that exist before the impact. From the knowledge of these conditions we can assume that the power output, power factor, terminal voltage, and current are known for each machine. If they are not specifically known, a load-flow study is run to determine them. Assume that a reference frame is adopted for the power system. This reference can
Chapter 5
166
be chosen quite arbitrarily. Once it is chosen, however, it should not be changed during the course of the study. In addition, during the study it will be assumed that this reference frame is maintained at synchronous speed. Consider the ith machine. Let its terminal voltage phasor Va; be at an angle 13; with respect to the arbitrary reference frame, and let the q axis be at an angle 0; with respect to the same reference. Note that 13; is determined from the load-flow study data, while 0; is the desired initial angle of the machine q axis, which indicates the rotor position. The difference between these two angles (0; - 13;) is the load angle or the angle between the q axis and the terminal voltage. From the load-flow data we can determine for each machine the component I, of the terminal current in phase with the terminal voltage and the quadrature component Ix. By using an equation similar to (5.42), we can determine the angle 0; - P; for this machine. Then by adding the angle P;, we get the angle 0;, which is the initial rotor angle of machine i. From Va; and 0; we can determine Iq;, Id ;, Vd ;, and Vq;, which can be used in (5.14) or (5.15) to determine E;. Then from (5.7) iF; can be determined. The flux linkages can also be calculated once the d and q components of fa are known. S.8
Determination of Machine Parameters from Manufacturers' Data
The machine models given in Chapter 4 are based upon some parameters that are very seldom supplied by the manufacturer. Furthermore, the pu system used here is somewhat different from the manufacturer's pu system. It was noted in Section 4.7.3 that the pu self-inductances of the stator and rotor circuits are numerically equal to the values based on a manufacturer's system, but the mutual inductances between rotor and stator circuits differ by a factor of vT(f.. We shall attempt to clarify these matters in this section. For a more detailed discussion see Appendix C. Typical generator data supplied by the manufacturer would include the following. Ratings:
Three-phase MV A Frequency and speed
Stator line current Power factor
Stator line voltage Parameters: Of the several reactances supplied, the values of primary interest here are the so-called unsaturated reactances. They are usually given in pu to the base of the machine three-phase rating, peak-rated stator voltage to neutral, peak-rated stator current, and with the base rotor quantities chosen to force reciprocity in the nonreciprocal Park's transformed equations. This is necessary because of the choice of Park' transformation Q (4.22) traditionally used by the manufacturers. The following data are commonly supplied. Reactances (in pu):
Synchronous d axis =
Xd
Synchronous q axis = x q Transient d axis = Transient q axis =
Subtransient d axis
x~
x;
= x~'
Subtransient q axis = Negative-sequence =
x;' X2
Zero-sequence = X o Armature-leakage = x.(.
167
Simulation of Synchronous Machines
Time constants (in s):
Field open circuit
=
Subtransient of amortisseur (d axis) = Subtransient of amortisseur (q axis) =
TdO
T;'
T;'
Resistances (in n):
Stator resistance at 25°C Field circuit resistance at 25°C Other data:
Moment of inertia in lbrn- ft2 or WR2 (sometimes separate data for generator and turbine are given) No-load saturation curve (at rated speed) Rated load saturation curve (at rated speed) Calculations: The base quantities for the stator are readily calculated from the rat-
ing data:
S8
V A rating/phase V A
V8
stator-rated line-to-neutral voltage V stator-rated current A
10 Wo
211'"
X
rated frequency rad/s
The remaining stator quantities follow: l/w o S VotD Wb turn
Also the stator pu inductances are known from the corresponding reactance values. Th us Ld , L;, L;', t.; L;, L;', L 2 , t.; and {d are known. Rotor base quantities: If {d in pu is known, then LAD in pu is determined from LAD = L, - {d' the corresponding value of LAD in H is then calculated. The mutual field-to-stator inductance M F in H is determined from the air gap line on the no-load saturation curve as VIVo = woM FiF, where iF is the field current that gives the rated voltage in the air gap line. The base rotor quantities are then determined from (4.55) and (4.56); the base mutual inductance M FB is calculated from (4.57). Rotor per unit quantities: Calculation of the rotor circuit leakage inductances is made with the aid of the equivalent circuits in Figure 5.10. The field-winding leakage inductance {F is calculated from Figure 'i. IO(a) by inspection: (5.49)
which can be put in the form (5.50)
Chapter 5
168
(0)
t
L
F
d
L"
~d
(b)
Fig.5.10
Equivalent circuit for d axis inductances: (a) transient inductance, (b) subtransient inductance.
Similarly, by inspection of Figure 5.IO(b), L d"
1 l/L,w + l/{D + l/{F
V
= 'l,d +
(5.51 )
from which we can obtain fD
= LADfF(L;
- fd)/[LADf F - LF(L'd - f d ) ]
.
(5.52)
The self-inductances of the field winding L F and of the amortisseur L D are then calculated from (5.53)
The same procedure is repeated for the q axis circuits. (5.54)
L AQ = L, - {q
where {q
{d and {Q is determined from Figure 5.11 by inspection:
L;' from which we can obtain
=
tQ =
t
+
+
L AQ)
(5.55)
LAQ[(L;' - {q)/(L q - L;')]
(5.56)
q
{QLAQ/({Q
and the self-inductance of the q axis amortisseur is given by
LQ
=
LA Q +
t
Q
(5.57)
Resistances: The value used for the stator winding resistance should be that which corresponds to the generator operating temperature at the rated load. If this data is not available, a temperature rise of 80-1 OO°C is usually assumed, and the winding resistance
Fig. 5.11 Equivalent circuit of the q axis subtransient inductance.
169
Simulation of Synchronous Machines
is calculated accordingly. Thus for copper winding the stator resistance for 100°C temperature rise is given by '125 =
'25 [(234.5
+ 125)/(234.5 + 25)]
n
(5.58)
The same procedure can be used to estimate the field resistance at an assumed operating temperature. However, other information is available to estimate the field resistance. From (4.189) we compute (5.59)
where T~O is given in pu time. The damper winding resistances may be estimated from the subtransient time constants. From (4.187) and (4.190) the d axis subtransient time constant is given by ( 5.60)
Since all the inductances in (5.60) are known, (4.192) and (4.193) rQ can be found,
T;'
=
'D
can be determined. Similarly, from
(L;' 1Lq}(LQI 'Q) pu
(5.61)
Again note that T~' and T;' are given in pu. Finally, data supplied by the manufacturer may not be available in the complete form given in this section. We should also differentiate between data obtained from verified tests and those obtained from manufacturers' quotations. The latter are usually estimated for a machine of given size and type, often long before the machine is fabricated. This may also explain apparent inconsistencies that may be found in a given set of data. This section illustrates the procedure that can be used to determine the parameters of the machine. When some of the data is not available, the engineer may find it convenient to assign values for this data from typical data available in the literature for machines of the same size and type. We should always ascertain that the parameters thus calculated are self-consistent. Actual values for several existing machines are given in Appendix D. Example 5.5
The data given by the manufacturer for the machine of Example 4.1 are given below. The machine parameters are to be calculated and compared to those obtained in Example 4. t. x d = L d = 1.70 pu x q = Lq = 1.64 pu x~ = L~ = 0.245 pu x'q L~ = 0.380 pu
x:
L:
=
=t q =
x-t =
td
TdO =
5.9 s
,
T: =
0.023 s
= 0.075 s r. = 0.24 s
T;~
0.185 = L"q pu
Solution
We begin by calculating the pu d axis mutual inductance LAD =
1.70 - 0.15
=
1.55
This is also the same as kM F , kM D , and M R - Similarly,
0.15 pu
Chapter 5
170
LA Q
=
kM Q
=
1.64 - 0.15
=
1.49 pu
Now, from (5.50)
t
F =
LF
=
1.55[(0.245 - 0.15)/(1.70 - 0.245)] 0.101 + 1.55
=
0.101 pu
1.651 pu
=
From (5.52)
t
(1.55)(0.101)(0.185 - 0.15)
D
LD
= (1.55)(0.101) _ (1.651)(0.185 _ 0.15) = 0.055 pu =
1.550 + 0.055
=
1.605 pu
Also, from (5.56) {o = 1.49[(0.185 - 0.150)/( 1.640 - 0.185)] = 0.036 pu LQ
= I.490 + 0.036 = 1.526 pu
From the open circuit time constant T~o
= 5.9 s
=
2224.25 rad
We compute from (5.59)
'F
= 1.651/2224.25 = 7.423 x 10-4 pu
and from (5.60)
'D From
T;~
=
( I .605 x 1.65I - 1.55 x 1.55)(0. 185) (1.651)(0.023 x 377)(0.245)
=
0.0131 pu
= 0.075 s we compute
7; = (L;IL;) 7:;0 =
8.46 ms
=
3.19 rad
Then from (5.61) '0 =
(1.526/3.19)(0.185/1.64)
=
0.054 pu
These values are the same as those calculated in Example 4.1. 5.9
Analog Computer Simulation of the Synchronous Machine
The mathematical models describing the dynamic behavior of the synchronous machine were developed in Chapter 4. The remainder of this chapter will be devoted to the simulation of these models by both analog and digital computers. We begin with the analog simulation. Note that the equations describing the machine are nonlinear. For example (4.154) and (4.163) have two types of nonlinearities, a product nonlinearity of the form xixj (where Xi and xj are state variables) and the trigonometric nonlinearities cos l' and sin 1'. These types of nonlinearities can be conveniently represented by special analog computer components. Also, the analog computer can be very useful in representing other nonlinearities such as limiters (in excitation systems) and saturation (in the magnetic circuit). Thus in many ways the analog computer is very well suited for studying synchronous machine problems. A brief description of analog computers is given in Appendix B.
Simulation of Synchronous Machines
171
To place the matter in the proper perspective, recall that the state-space model of a synchronous machine connected to an infinite bus is a set of seven first-order, nonlinear differential equations. When the equations for the excitation system (for VF) and the mechanical torque (for Tm ) are also added, the system is typically described by 14 differential equations. Complete representation of only one synchronous machine with its controls would occupy the major part of a large-size analog computer. Thus while the analog computer is well adapted for the study of synchronous machine dynamics, it is usually limited to problems involving one or two machines with full representation or to a small number of machines represented by simplified models [2,3,4,5]. The model most suited for analog computer representation is the flux linkage model. Thus the equations developed in Section 4.12 are used for the analog simulation. The differential equations will be modified, however, to avoid differentiation. For example the state-space equation of the variable Xi is Xi
=
(5.62)
J;(x,u,t}
where xj ' j = 1,2, ... , n, are the state variables, and u., k ing functions. For analog computer simulation (5.62) is written as Xi
=
WB
a
l' 0
I, (x, u, t) dt + x;(O)
where a is the computer time scale factor and (see Appendix B). 5.9.1
WB
1,2, ... .r, are the driv-
(5.63)
is required if time is to be in seconds
Direct axis equations
From (4.126) (5.64) From (4.128) (5.65) and from (4.129) (5.66) The mutual flux linkage AAO is computed from (4.120) AAO
=
LMO(Ad/td
+ AF/tF + AD/tD)
(5.67)
Then from (4.118) the d axis and field currents are given by
id
=
(1/ td)(A d - AAO)
iF
=
(l/tF)(A F
-
AAD)
(5.68) (5.69)
The analog representation of the d axis equations is shown in Figure 5.12. Note that all integrand terms are multiplied by WB to compute time in seconds and divided by the time scaling factor Q.
172
Chapter
5 L MD
"' B
"0 'A D -vd
-' F
- Ad
~
- Ad
~ C'q
- 'D
I
' AD v F
-A F
~
-A d AA D
".,~
' AD
I
- 'D
- AD
5 .9.2
~
' AD
"'B' D/ 1 DO
Fig.5.12
~
- 'F
iF
Analog repr esentation of the d axis equations.
Quadrature ax is equations
From (4.130) (5 .70)
and from (4.131) (5.71 )
The mutu al flux linkage is compu ted from AAQ = LMQ(Aq /
t
q
+ AQ/ t
(5 .72)
Q)
Then the q axis current is given by, from (4.123),
t, =
(5 .73)
(l/tq)(A q - AAQ)
The analog simulation of the q axis equ ation s is shown in Figure 5.13. "B
-a 'A Q -----;,~-{ - v ---'<--=f q
-A
q
, u' d --~
\
AQ
-' Q
-\---=< - ' Q, - - - - - {
"o'Q ~ 'B' Q / l Oo
Fig. 5.13
-'Q
-,
q
' AQ
1
~ r:q
Analog simulatio n of the q axis equatio ns.
i q
Simulation of Synchronous Machines
173
Fig. 5.14 One machine--infinite bus system with local resistive load .
5 .9 .3
°
Load equations
In (4.149) a
id iq
=
=
will be used for convenience. Therefore,
l' ~ l' ~
al.,
0
al. ,
0
[0 V'" sin 0 + Vd - R,id - wL,iq]dt + iAO)
(5.74)
t- VJ v'"
(5.75)
cos 0 +
Vq
- R,iq + wL,idl dt + iq(O)
Equations (5.74) and (5.75) are useful in generating the voltages vd and vq • However, if they are used directly, different iation of id and iq will be required, which should be avoided in analog computer simulation. To generate vd and vq , the following scheme, suggested by Krause [2), is used. The machine is assumed to have a very small resistive load located at its terminal, as shown in Figure 5.14. This load is represented by a large resistance R. From Figure 5.14 the machine terminal vo ltage and current for phase a are given by Va
=
(5.76)
(ia - i,a)R
where t; is the phase a current to the infinite bus .
-i d (L ) j i
td
-,.
q
0 .1 M
) o
(L
0 .1 M (L ) i
(l.) ,
l..---JlI'h-----:l~
0 .1 M
vd
P = L;I RL0
0. 1 M
. 0. 1 M 'tq (L.) I
Fig.5.15 Analog simulat ion of the load equat ions.
174
Chapter 5
-T04>
Fig. 5.16
Simulation of the electrical torque
T.~.
Following a procedure similar to that used in Section 5.4, the current it can be resolved into d and q axis components i d, and i q , given by (5.74) and (5.75). The currents id and i q are given by (5.68) and (5.73). The ud and uq signals are obtained from Figure 5.14 by inspection, Ud = (id - i,d)R
uq =
(5.77)
(iq - i,q)R
where i'd and t; are obtained from (5.74) and (5.75) respectively, with subscript t added as required by Figure 5.14. The analog computer simulation of the load equations is shown in Figure 5.15. 5 .9.4
Equations for wand {)
From (4.90) and (4 .99), with W<1
=
win pu and
2Hw B dW<1u = 2H dW<1u = T ~
~
Tj
=
2HwB' we can write
- T - DW<1
m.
(5.78)
pu
u
where T. = (iq>"d - id>"q)j3 . Equation (5.78) is integrated with time in seconds to compute, with zero initial conditions and with a time scale factor of a, W<1u
= -1-
2Ha
I. 0
(Tm - T. - DW<1U>dt
(5.79)
pu
Note that the load damping signal used is proportional to W<1 (pu slip), requiring appropriate values of D. Most analog computers require that 0 be expressed in degrees to find sin 0 and cos {) (6). Therefore, since iJ = WB(W u - I) = WBW<1 pu, we compute
s=
180 WB 'Ira
f
t
0
W<1 dt
+ 180 0(0)
e1ec deg
(5 .80)
'Ir
The analog computer simulation of (5.78)-(5 .80) is shown in Figures 5.16 and 5.17. The generation of the signals - wand - 0 is shown in Figure 5.17 . The analog repre-
-6
D
1.0
Fig. 5. t 7 Simulation of W<1. w, and
o.
Simulation of Synchronous Machines
175
sentations shown in Figures 5.1 2, 5.13, and 5.15-5 .17 generate the basic signals needed to simulate a synchronous machine connected to an infinite bus through a transmission line. However , other auxiliary signals are needed. For example to produce the signals WA q and WAd shown in Figure s 5.12 and 5.13, additional multipliers are needed. To produce the signals V ~ sin 0 and V " cos 0, an electronic resolver is needed. The complete analog representati on of the system is shown in Figure 5.18. It is important to -100
Fig .5.18
Analog computer patching for a synchro no us mach ine connected to an infinite bu s through a transmission line .
Chapter 5
176
note that signals are added by using the appropriate setting for the potentiometers associated with the various amplifiers and integrators scaled to operate within the analog computer rating. This scaling is best illustrated by an example, and in Example 5.6 the scaling is given in detail for the simulation of the synchronous machine. The initial conditions may be calculated from the steady-state equations (as in Examples 5.1--5.3), and these values may be used to initialize the integrators. However, the analog computer may be used to compute these initial conditions. To initialize the system for analog computation, the following procedure is used. The integrator for the speed is kept at hold position, maintaining the speed constant. The integrators for the flux linkages are allowed to operate with the torque Tm at zero. This builds the flux linkages to values corresponding to the no-load conditions. The load T; is then applied with the speed integrator in operation. The steady-state conditions thus reached correspond to initialization of the system for transient studies. Example 5.6 The synchronous machine discussed in Examples 4.1-4.3, 5.1, and 5.2 is to be simulated on an analog computer. The operating conditions as stated in Example 5.1 represent the steady-state conditions. The system response to changes in vF and Tm is to be examined. Solution
The data for the synchronous machine and transmission line in pu is given by: L, = 1.700
LMD
=
0.02838
Lq
L MQ
=
0.02836
=
1.640
r = 0.001096
L D = 1.605
LQ
'F
=
1.526
LAD =
1.550
'D
=
0.0131
LA Q
=
1.490
'Q =
LF
=
1.651
R
0.0540 100.0
d
=
{,q
t
t
F =
{D
t
Q
L('
=
=
=
0.00074
R, = 0.02
0.101 0.055
= 0.036 =
0.150
=
0.400
H T~o
2.37
s
= 5.90
s
=
Vex> = 0.828
The additional data needed is T; = 1.00 pu and EFD = 2.666. Note that EFD = E in the steady state. This value of EFD with the proper scaling is introduced into the integrator for ~F' As explained in Section 5.9.5, the analog computer is made to initialize itself by allowing the integrators to reach the steady-state conditions in two steps. In the first step EFD is applied with T; = 0 and W = WR = constant. Then Tm is switched on with all integrators, including the W integrator, in operation. The basic connection diagrams for the analog simulation are given in Figures 5.12 5.17. The overall connection diagram is shown in Figure 5.18. In that figure the analog unit numbers and the scaling factors for the various signals are given; e.g., the scaling factor for ~F is 10, which is given in parentheses. The time scaling used is 20. The settings of the various potentiometers and the scaling are listed in Table 5.1.
30
30
30
30
000
000
000
001
001
201
201
201
002
002
002
003
003
100
200
300
001
101
201
302
301
002
102
202
003
103
AAD
1.0
20
20
30
30
30
I -A o I -Ad
30 I 1.0
30 I 1.0
30 I 0.6667 I -Ad
30 I 0.6667 I AAO
I -A F
10
10 I 3.0
E FO
1.0 -A F
AAD
-Ad
0.3333
1.0
AAO
-Ad
-wAq
1.0
10
10
10
30
30
10
30
1.0
1.0
30 I 24 I 1.25
-Vd
Input
LolL;
20
377
(0.0010965)(377) (0.15)(20)
20
ta
1
{d
I
{d
=
0.1892
0.028378378 0.15
6.667
6.667
0.5121
0.028378378 0.055416667
I
0.2804
0.1383
0.01564
0.1383
4.456
4.456
0.1378
18.85
18.85
0.1378
C
(eonst.)
0.028378378 0.101202749
0.15
=
=
LMD {O LMD
=
{F
LMD
{~
T£WB
'FWB =
(0.74236 X 10- 3)(377) (0.101203)(20) {~ 3)(377) wa v'3rF = v'3(0.74236 X 10( 1.55)(20) LAOa
'DWB _
(0.013099)(377) {Oa - (0.055417)(20) 'DWB _ (0.013099)(377) {oa - (0.055417)(20)
-taD
'WB
a
WB 377 -=-
a
WB =
_
taD -
TWa
Constant
4.444
4.444
0.1892
0.5121
0.8412
0.1383
0.01564
0.04609
4.456
4.456
0.1378
23.56
14.14
0.1378
(LoIL;)C
Potentiometer and Gain Settings for Synchronous Machine Simulation by Analog Computer (a
30 I 40 I 0.75
30
30
000
000
L;
no.
Lo
Amp.
Pot. no.
Table S.I. 20)
10
Amp. gain
10.0
10.0
10
10
0.1
0.1
10
1.0 10.0
10
10 1.0
0.1
0.1 I 100
0.1 I 100
0.1
Int. cap.
=
0.4444
0.4444
0.1892
0.5121
0.8412
0.1383
0.1564
0.4609
0.4456
0.4456
0.0138
0.2356
0.1414
0.0138
Pot. set.
3·
""""
(1) Vt
~
=:
Q
~ n
cVt
~
o
O
zr:
n
~
-e
(J)
~
o
~
o·
[
c
(J)
30 I 0.6667 I
30 I 0.6667 I - Aq
30
30
30
30
30
30
20
20
210 I DID
211 I 010
011
011
012
012
013
013
202 202
401
011
111
012
412
013
413
700
402
702
30
30
I 1.25
0.75
-A Q -Aq
1.0
1.0
6 6
16 I 1.25
20
I -wi,q
d
I 1.667 I - Aqi Adi 1.667 q
AAQ
-A Q
AAQ
-Aq
wAd
-Vq
1.0
1.0
10 10
30
30
30
I 30 I 1.0
I 24
40
AAQ
0]0
1.0
110
30
30
010
Input
010
LolL;
no.
L;
Lo
Amp.
Pot. no.
(continued)
I a =
a
WB
1.0 1.0
ta
I
t
I 0.15
0.028357472 0.15
.{a =
LMQ
{, Q
0.028357472 = 0.035809
L MQ
18.85
1.000 1.000
6.667
6.667
0.1890
0.7919
28.40
(0.053955)(377) {Qa = (0.035809)(20)
rQwB
{Qa
rQwB
28.40
=
0.1378
18.85
]8.85
0.1378
C
(const.)
(0.053955)(377) (0.035809)(20)
rWa {.aa
a
Wa _ 377 a 20 Wa
rWa _ (0.001 0965) (377) {aa (0.15)(20)
Constant
Table 5.1.
23.56
1.667 1.667
4.444
4.444
0.1890
0.7919
28.40
28.40
0.1378
23.56
14.14
0.1378
(LoIL;)C
100
10
Amp. gain
0.1
0.1
0.1
0.1 I
100
10
10
10
10
100
100
10
0.1 I 100
0.1
0.1
Int. cap.
0.2356
0.1667 0.1667
0.4444
0.4444
0.1890
0.7919
0.2804
0.2804
0.0138
0.2356
0.1414
0.0138
Pot. set.
zr:
l.n
""'
~
"'0
Q
...
()
'" 00
20
20
20
20
40
400
400
400
400
412
400
401
500
501
112
0.5
80 80
211
212 212
410
212 213
Witd lId - id
0.2
1.25
2.0
100
16
50.0
0.001
0.80 0.16
10
500
100 500
20
2.0
vq
0.5
40
20
-itq
1.0
20
W~u
1.0
W~u
-Ta
itq - iq
- Vet) cos {)
i td
1.0
20
Vd
Vet) sin {)
0.5
0.2
100
40
Note: In this table {a is used for either {,d or {,q.
500
210
303
HG
413
40
20
401
503
113
20
401
502
HG
20
401
403
1.0 1.0
7ra
180 WB
I 6Ha
-=
I
=
100(40)
=
11"(20)
( 180)(377)
1 6(2.37)(20)
P = L.jRLo
20
20 P = L;/ RLo = 100(40)
a
wB
Lea
V3WB
Lea
WB
Lea
RewB
-
0.8000 0.1600
1.0
1.0 1.0
0.1758
1.0
10- 3
-
-
0.1
0.1
0.1
0.1
0.1
0.1
0.1
1.080
X
-
-
23.56
16.32
23.56
0.9425
0.9425
16.32
23.56
.080
3.516
0.005
0.005
18.85
81.62
47.13
0.9425
0.9425
RewB = (0.02)(377) (0.4 )(20) Lea
47.13 81.62
377 (0.4)(20)
V3WB = V3(377) (0.4)(20) Lea
Lea
~=
-
I
I
1
10
-
100
100
100
10
10
100
100
0.8000 0.1600
0.1080
0.1758
0.005
0.005
0.2356
0.1632
0.2356
0.0943
0.0943
0.1632
0.2356
3' [
o
'" "0
en
:] (1)
zr:
Q
~ n
C en
:]
o
;:s-
n
-e:]
(J)
~
o
:]
s
c
(J)
180
Chapter 5
~:g J~o, tH t:~ rl l l i l ! 1 2.0 1 0 'a
N. ; l I'
r
li -i-i!
I
i , '"
Add 20% E : I i I il!1 !,fi jO : --j II
! , ,;
I; :I :ll; l:i I: ,
I
iii
I
i I
:
'f :
"
,
;
;
i
Oro 20% E ' " FD' , ! I ,
i
, " P"
.
"
;
,
,
j
ii
Ii! I
'
i I
4. 0 3. 0 2. 0 1. 0 0
''0' ',
"I '
,,'e¢f ;'"
'
,
'I "
II' I ill I' i ! iii: :ii i i i III
II Illil l I I
I
I I
"
! ""
i
!
Ii
II " Ii
I! ! I i " I i I I ~ : -1
I
,
I
! i ii I
IiI
I I ' I
2.0
1 .5 1; ( 0
Fig .5.19
I
Response of a machine initially at tion .
90~~
load and 90% excitation to a 20% step change in excita-
The steady-state conditions reached by the analog computer are listed in Table 5.2. They are compared with the values computed in Example 5.1. Figures 5.19-5.21 show the following analog computer outputs: the change in the exciter voltage Em, the mechanical torque Tm, the electromagnetic torque T,, the field flux linkage AF , the stator d axis current i d , the terminal voltage error V'6' the angular velocity error w 6 , and the rotor angle o. The results of the simulation are shown in Figures 5.19-5 .23, where all plotted quantities are given in pu. Example 5.1 is used as a base for the computer runs. Thus a 10% change in EFD is 0.2666, which is 10% of the nominal value computed in Example 5.1. Sim ilarly, 10% Tm 4> 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 !'! ' !!
I I · [. Q l l i l f:j Ll t '··
·JtJi·ttti1l H IILlt i.··':';
r.~- _r , _"I.'. ~~::t~t:t· ( t::t:+··J:::1:":. t't.::!:..: :-:..\":: .I-
_.•l i'
-'
.
'. ~ ·X · I ~-·:i' ~iiii.i~iT.. =tFf.rq-r'LF:;l-~r.::i.
1.•;. '. :.•"
:b"~H-++-j,b+t+-+-+t++r.rf:.~:±.p·+ -tJ :tt~;.. .t .. i , ,'--j'+ r TT+++ ..+-- H·_t-t-t-"· "!'·L .-I·j· l " ·' ·
'j ..i
i
.!
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
188
*••• CCNTINUGUS
SY~TE~
MODELING PROGRAM••••
••• VERSION 1.3
*.*
IN ITt AL
MACqn
sn={;E.N~AT(~I\Il
C; (, : A GSAT *~:
x~ ( K t: )C. P )
FNDM~C
INITIAL
CONSTANT
• •
PJ=J.141~~?~~.qLO=lOn.OtCLO=O.Ol ~ P F elF Y (; F.' .~ t:. ~ A T0 a;, Pow r ~ • P (; EN A ~t f)
~US T
V TAN LJ
•
T~I F t NIT E H US VOL TAG E.
v T"! F
GE I\JE. .~ A r 0 H 1 F JJ ~ 1 1\' ~ L
\}(' L TAr; F
P(;F~=l.UO
'If=I.17
=
v I "IF 1 • U 0 U TITLE SAT u RAT E'" n S Y N C H ~ (1 N 0 U S
rUN~T
rONC\T CON5T PAHA~
GF ", FHA T(, ~ ,,, J T ~ nUT E)(CIT L:P l) • () , f.lK V = 1 5 • 0 • R P F 0 • H S • X n =: 1 • 1 U _ T f)" "'.1 1 • '" 4 • )(, l ~ ? 4 r, • x 0'" p =() • 1 ij 5 , X\11 e P 0 • 1 f.i c; • ). L A U • 15 t ,~ An H M (t Q F o... ~ =0 • i' ':J 1 8 7 1 "'j ~ 4 • I F L lJ =3 6 ~ • 0 • TD P P =• 0 ? 3 • T (JP P =CJ • nn-" ~L t' U 1 h 5 () )(F"=0.4 NP =2
)(()=
• 0 , He =? •.i
=() •
PAHA~
H[=n.o2
PARAJ.1
ExrON=O.Ol
CO"lST PAHA'4
DA~"~
'f • ~ M V A
=1"
=
=
=
S~T~10=O.OH~~.S~TGl?=O.3t~8 TSTAFolT=O.~
Krv=o.o,KE.F=l.U
FIXED KKK
•
5 " T' JHAT I ON F U!" c r I UN F 0 ~ (,E"" ERA1 C'R PHOrEOl'Ht\t Af;SA T,H~SAT::Sl\ rtJ~ (Sf\TG10.SATGli?l IFf<;AT~lO.f::U.O.O) (;0 TO ~n
..
AG~AT=(SATG1~~*?)/(1.~*SATGl?)
=
H(; ~ A T 5 • 0 .. " L ()G ( ( 1 • i!;. S,'\ Tb 1 ? ) / S 1\T
20 30
ENOPRll
GO Tn 3U
Ac;~AT=f).fl ~l;SA1=O.O
C\lNT {NUt.
• •
•
l'\(\ )
COMPUTE:
INITtAL cu-o r r i ows
LJ\=xLA LAU=XD-LA
L" f~= )(.)-1. A Lt==L AI..," ( XOP-LA) / (L AO-xnp+L") LFF=LAD+LF LCON="O~P-LA
rn"t'
LKD=LF o,~ C()~t~L 1\01 (LF *l.AO-LFF 01. L K U =L A(J" ( )( 4J~ tJ - LA) I ( L A(J - (J P to> + L f\ ,
= 1 • ()/
( " .....' I ,. tl U • ) ~ T3 S H ~ ~ E 4 M V 1\~ 1 l) u 0 ,l '1 0 • \J/ 3 • ,) 1 H A Sf:
x
v~ " c;F.: s =.~ K V.. 1 () 0 () • \) / I
=
~~~ES=SHASr IVHJ\~t-, ~
~H"~ES=VHf\SfS/IH"5fS ~HA~E~=V~AS~~~iH~~E
L~~~ES=~PAS~~UrH~~E
Ll)H=XDuLBASt'"_ t; l A~=XLI\~L~A~t::S L~OH=Lf)H"Lt\H HT~=SQ~T(2.U)
MFH=~12°VHAS~S*lHASE/IFLU I\~FH=J.lT3°MFH/~r2
Iff,
OH=T[;ASE~"l.AI)rl/t(~F'"
Fig. 5.26 CSM P program for computing initial conditions.
=
="•
'#
•
0 () 1 1 1 3
Simulation of Synchronous Machines TFLDd
VFI
rH~=C;HASf::/
RFt.
nH=VFLI)~1 [FLIH~
LFl M F"~
189
rH-l=~FlIJH" rHASE
=
A S E ~ I) 4
T ( '- HAS F.. S" LF LD~ )
wr-l.r"~=VrLI)HUTHA~F.
~"=(~AnHMo]~~.~)/(~H"~F:~"2~C:I.5)
LMl>=l.OI « 1.II/LAn> l.MCJ= 1 .0/ ( ( 1 • O/LAf.l) L ~·=Xl:.
**
+ (I.U/LA) + + ( 1 • OIl. A) +
(l.f')/LF) + (l.O/Lt(Il) (1. O/LKf) )
PRnvInF LnGIC f()~ CASE wHERE TDPP Af\.If) Tn~p AJJE MISSlr-rr; p~OCF:nU~AL ~KD .~K')=TFtlf\J (I..KD.LK(J,n~H.TI)PP, TnPFJ.L Al>.LI\'J.I..J\ ,l.F)
IFCTOPP .EQ. o ,.o: (;0 TO If)4 ~ t( n=l K f) I ( (l,~ H .. 1 f) ~ P) + (L!\ I) "'l. AU LF ) I ( I) MR .. T r) '" P" ( t, A(H~ L " + L " n * L t· + L L\ it L F) ) G,l ro lor;
104 ~",n=1.0t:.+H 10 !i I" (T ()fJ P • E (J. (,0 In~
lIn
{}.
0)
«;0
10
~t«(J=LKnl (nM~4"'T()PP)'+
TO 110
Q~,Q=l.Ot: +~
()~U=l .•
1 () b (LAQ*LA) I
(nr"p"TI1P~"
(LA()+Lf\»
o
F"'DP~n
* p ~ n VI nE" L nf,; TC f ()'~ c At.CUl ATIN ~ 8 f:. S T ~ n ~ C\ T~ LE P~OCE()U~~'- ~F=T~FD(TU()PtL"'F'LFLf)8,~FLUH.~F-nr'M) IFCTOOP.EcJ.O.O) GO To 12U
~F
~F::(LFF"l.Fl.f.)~)/(T()()~*RFLUH)
GO TO
1 ? 0 ~ fo
12';
FNDJ.lQO
l~c;
=( ~ F 0 ~. M.. 3 ~ ~ • "
C(lt-.IT
I NlI~_
) I ( l~ F L LJ H 4. £.~ ~ • 5 )
~T3=S(J~T(].U)
()P~=lHU.o/"'I
OM~=l?O.o"tJJ ZE~=HE.~-Uo~+)(F**2 G12=-~t:/It:.':)
Hl~=xE/7f:S AAM~A=PI+ArAN(HltIGl~)
GIG=l.O/J.?LO
G11=G1<,-(i12 y 11 5 fJ.~ T (C; 1 ? .... ~ + H 1 ~ N l, E 1\1- (1 1 1" V T .~ .. 2 I>r N = Y 1 ~ .. v r u-v I t>J f-'
=
'-I =.. , (.;
F At =NUM II
*
0
2)
)t-~N
1. -F- AC""2) ZfT t\=~ T f\'\J (IJUNl/F AC) l"\E"TA=GJ\MMA-/ETA f)\lM=S(~~ T (
rOS~L
=cos
(HE r.'\)
SIN~L=srN(~rlA)
I l. p~. = VT 0 ( (C C' ~ 1\ L I f~ LD) - ( S I f\I J\ l. .. ci » ) IL 1~=VT4Jo ('jTf'JAl../HLI) + (COSl\l.~rLO» 1 TQ t (f~ t:. 0 ( v r ..c() 5 rd. - VI" If-) + Y. F .. V r 6 ~ TN AI ) / 7 f.. (-,
=
tTT~=(~~"VTosrNAL-x.t:*(Vl*cnStaL-VTNF»/Z,"~ IAr..lE=IL~E+ r~~~
r t ~ t M= TL t M + I ·r t M
THETA=ATAN(IAIM/IA~E)
TA =e:, (~ ~ T ( I A ~ t-. ~ .. 2 .. 1 A I M<0.. 2 )
E (J R E =v r" ens f1 L + 4
A .,. I A ..
Cf) S ( T H F.: r " ) - ~ ,") .. J A .,.SIN
(
r H F T 1\ )
EUI~=VT*srNAL+QA~lAostN(rH~rA)+xn~tA*C0~(rHFTA) E Q () =c;o 4 r (E (J ~ E:. ~ 2 + F: (J I ~, .. 2 )
*
TECHK=1"PC;Ff'J" ~iU-J.)A"l A""?
*
01.=4 TAN (E (~I '1/E'}~f) PH 0 ~ ErH I ~ F VD • V Q , to. T() • wU • \IIn , wAf) ~ • w~ ,')S t S Gn • S G t) • t~· U'" • I MII • T E i) r: I..• l) r. I. 11l. •••• T F. • ~ N =F lJ N C ( it L • V 1 , I A , R f. T A • THE r A • ~ 1 3 t ~ A • L AIJ , LA. L F , L A'1 , T F- C~ ~ k )
NN=Q
Fig.5.26
(continued)
190
Chapter 5
100
VI)=-RT:l"VT~StN(lJL-~t:TA)
V (J=~ T 3 * V T ~ Co c; ( U L - H t: 1 A ) I U=-~ T 3* T A *S IN (OL -1 HI-:' 1 ~) T:J=4TJ*IA~CvS (1J,--T~fTA)
..
UNSATlJqA ffD F IE'lD ClH.1~E"'iT :: IFlJN I F H"', =( (V U + R A.. I (J ) L Af} ) - ( (L A + L~ n ) .. In)
I
•
*
".4"
I
= I t> + I F u N
1,-" 0
Wl\nC;=LAn*lMu
SATUJ.(ATEO La-AX I C; FLUX L INKAu",.S
SGf1=<;t.NSAT(WAflS) wD=tAoTl'+'IIAOC;
C;ATU~I\ rEO (J-A.~ I'i FLUx.. LINK£\GES WA o Z L ~ (J" I (J WA() S I ~ J.'L ( ."A (J I. , 0000 1 ,G A(.) S ) s(;6::G~ NSI\ r (W IH3$)
= =
o•
6A~5=lAQ~IQ/(1.O+SG~)
'II(J
=I. A" T(J .. ~ A(J S
Tl:.::WO*Tu-wO"TI> TEnFL=Tf..-TECHK DEI I1L e n , 1 .. (T EI>EI 11 ECHr<) ~.1
AN
IFCOfLDL.LT.-tJ.O(}Otll)
TO 200
IF(flf:LOL.t;T.O.tlOOlll)
200
GO T" <;1)
GO TO 400
oL=nL+OF.LIJI. t\JN=NN + l IFC~~.~T.~O)
GO TO 100
«()L) £00
GOrO 400
4QO CoNTINUE: F~up~n
•
FINAL tNITlf\L COM"'UfATIO"JS YFF=IFlJN+SC;O*IMU
wF=LF*YFF+wAIJS
WKf1=wAOS
W~Q=wJ'!}~
VF=f.lF*'FF
Ul.U=UJ.lR*UL n~,=()MH'
OOMU=O.U T T'" A(; So ~ T ( TT r~ t: *..? + J T I ~,"" .. 2 )
=
P51=ATAN(1'IM/IT~E)
lOT =- Hr', ct I T'·1Ae,;
Tl.OO=IO-TDT
ILOu=lfJ-tur TM=Tf::
TA=O.·O
VTCH~=(1./k13)*SQRT(VO**2+v~ .. n2)
w()l=wD
~Fl=wF
WKOl='-JKU wn7=~U
WKOl=wt
DOMZ=DOMtJ
l>l...l=OL
IUTZ=lDT JeJTl=IQT
TMZ=TM VOl=\lO
vQl.=V(,)
KFn=w130RF/LAO f:Fn=vF/r
• "IOSO~T
EFnl=f:~Fll
Fig. 5.26
(continued)
Simulation of Synchronous Machines
191
executed pr ior to integration at each time step . Include a local load on the generator bus in the computation. Use the Continuous System Modeling Progr am (CSM P) [10] for solving the equations and plotting the results. Solution
An essential part of the computer program is a routine to compute the initial conditions . As noted in Examples 5.1-5 .3. this computation depends upon the bound ary conditions that are specified. The boundary conditions chosen for this example are those of Example 5.3, viz., P a nd V, at the generator terminals. The FORTRAN coding for this section of the program is included in the portion of the program listing in Figure 5.26 called INITIAL. Note that the statement of the problem does not give any explicit numerical boundary condition. This is one of the advantages of a. computer program ; once it is written and verified. problems with different boundary conditions but of the same type can be solved with ease. The boundary conditions specified in Figure 5.26 give P = 1.00 (PG EN), V, = 1.17 (YT), and V", = 1.00 (YIN F). I. Make a preliminary estimate of hAD (hAD is named WADS in the program ; W being
used for hand S meaning "saturated").
(5.90)
2. Compute the new currents. From the equ ations id
=
(hd - AAD)/f..d
lo = (AD - hAD)!f..D
i, = P'F - AAD)/f..F
i MD = id + if + i D
(5.91 )
we compute an estimate of the new currents. This estimate is not exact because the value of AAD used in (5.91) is the value computed at the start of the last dt, whereas the flux linkages Ad . Af, and AD are the integrated new values. Thus i MD computed by (5.91) does not correspond to point A of Figure 5.27, but to some new point B. Since hAD is a function of the currents and of saturation, we must find the correct new AAD iteratively . We do this by changing our estimated hAD slightly until i MD agrees with AAD on the saturation curve, or until points A and B of Figure 5.27 coincide. 3. To estimate the new AAD. we compute the saturation function SGD = f(hAD) in the
Fig .5.27
Saturation curve for the magnet izing ind ucta nce
LAD.
Chapter 5
192 nYNA~IC
MOSORT
• •
Cit""FQA TO~ Cl,HqF.:"Il S \"AOS(t=~Af)~
w~ns=l'J1PI. ("AI)~o,n.OO"l.FWM»
rp=(W(1-WAUS)/l.A IFF=fwf-WAVS)/LF I t< LJ wto- I)- \01 A0 ~ ) II." U I"1U=IO+ IFF+ P'd)
=(
Sc;n=Gf N~A T (w ,',,)S)
(;An~=LJ\I)*I~U/fl.(l+SG[)
FIN AII= wA"~"
WAfJc;O=WAU<.)
rr- "(}S-WAn~)
of::XCON
\IIIA(JS=IM"'L (.,"()S().().~)I)(Jl,FwA() TfJ W(J - .. l\ (J S ) / LA
=(
It«J= (WK'J-"'''\J~)/'-I
f,I\QS=L/HJ(tlftll(J/ (l.n+~l;U) FW"U=~A(JS.. (\;I\IJS-','IIHJ'i) oFXr.I)"J
~UHT
• •
T'l~OIJF T~. =~Oo
• •
• •
• •
T,,)-I-/IJ" Tl>
TA=T,..-Tt: C;"E.Fn
JGIH)V,=TAI (h.(}OH<':) DOM,,= IN' I;~L (·)('·... l. 1(,IH)14) OPlU=UOMU"l.1l o tw· I) M U *') ,..,~
=
AN(;t E
I om. =np.1b 0I)O""J nl=TNT'HJL (Ull. IbPL) 1J1.L>=LJJJk*fJL
t. nc at, LOAIl I ~VI'= (Of'AH/ClP) * ( Tf)- I or-
(Vn/~l P)
- ("r-t.,0fLD*vfJ) )
VD=JNTC;~L(vf)l.I(;VI)
!(;V(J=
• •
Tf.l~"'SM
•
, (j
•
• •
, 1 G I UT )
T:: ( 0"" ~ I L f ) .. ( V(" - ~ 1 3 *
O-AXTS F"ll')( L(N~"r,fS I GIJ=(lMH* (-VI)- (~A" I ()) WI>= 1 r-Jl (;I~l (111(11.
VF =KFO*t:. F!J
t 0 S (fJ L ) -
H~
1(;1)
* J (J I .. n M 11* L" ° 11 r )
-()"'~U*WQ)
J~F=O~H."(VF-kf*IFF) wF=INT~~L(WFl,I~F)
Ir,KIJ=OMH.*(-Ht
t)-AJCtS
FLlJ~
=()
t.TNKAHFS
I (J)
•
I (; K (~ :: f)'~ H" ( - WK (J - ( ~ u ) ~ K lJ J N 1 (H~ L ( ,.tt', 'J l • I GK (J
,
IC'H J
\IIU=
•
• •
Q
(O~'JoCLLJoVI»)
).(V()"~T3*~IN(nL)-JJFoJ(lT-op.\IJ*Lf.JOT)
=[f\1 T(~~ L ( J P Tl
I()T=INTr,~L('lJTL.Ir,IfJT)
• •
I
.. (IU-I'JT- (V(J/IotLf) ..
(\It,ll. I (:'V'J)
I ~~ JON t.. PIt:.
I 01
.. •
rnt~L
Tc,Ior=(OMH/Lf
• •
(H~H/Ct.r»
VIJ= t f\.1
fA H-
( - v (J -
I N Tf~~L
( J:tA0
(W'iIJl,
=
TE~~JNAL
1(,'»
OP'- u. \~ 0 )
-?.
VOLTAG~
Vl=«Vf)"*r""U*·~)o*O.~)/RT3 1A J n* I () ... 2) .... 0 • ~) / Q T 3 p~
O~IVTNG
=((
=tH'll*' t l J. U
FIINCTTO'4S
T~=T~1+~TM·T~l/lO.ooSTt:.P(TS1A~T)
E~n=EFOl
• NUSORT
.. t
Fig, 5.28 CSM P program for updating integrands.
I
193
Simulation of Synchronous Machines TE~r.4 J~'AL
DEI T flUTPUT _n PAGf G~nIlP=(1.C;5.1.''') nUTPUT wF' PAGf G~nllP=(2.1.?3) t'uTPljT WI(f) C»AGE n~r",p.(I.~."'O) ouTPUT wAnS PAGE GROlJP=(l.~,~.O) nUTPIJT S~f') J:?AN~E
PAGE
G~n"P=(O.12,().2.)
PAGE
G~nIIP= (O.<.1~,
nuTPUT 16
ouTPUT IFF'
(H~nlJfJ= (J.
PAGf:
nvTPLlT 1"f) PAG":
1.lH)
u, J. 7.4)
C;~OlJD=(-U.02.0.(}4)
OUTJ.llIT PAGF.
r,~"'JP=(
VT
PA6E
G~ntlJ.'=(~O,"'2)
OuTPIlT OLI)
1.1';,
l.l~)
nu Tf'IJT IJnMU PI\GE
r,~nllP=(-O.OOlf,.O.Ofl?)
nUT~UT
PAGE
TI~F.~
FND
IF.
~~f)I'P=(2.~.3.A)
FINTIM=?'j,nUTOtL=O.OC;.,.lELT=O.OOOl
~lUP
Fig. 5.28 (continued)
usual way, using (5.83). Then we compute AO and AN, defined in Figure 5.27, Then the error measured on the air gap line is AE sured on the saturation curve is approximately AA
=
=
AN - AO' and the error mea-
AE/(I + SGD)
Now define a new AAD to be GAO' defined as GAD = AAD + AA' Then we compute GAD = AAD
+
(AN - Ao)/(I
+ SGo)
= LADiMD/(1
+ SGD)
4. Now we test GAD to see if it is significantly different from AAD; i.e., we compute
I GAD
- AAD
I~
f
where f is any convenient precision index, such as 10- 4 • If the test fails, we estimate a new AAD from new AAD ~ FAD
=
AAD - h(G AD - AAD)
where h is chosen to be a number small enough to prevent overshoot; typically, h = 0.01. Now the entire procedure is repeated, returning to step I with the AAD = FAD, finding new currents, etc. As the process converges, we will know both the new current and the new saturated value of AAO' The second part of the program computes the integrands of all equations in preparation for integration (integration is indicated in the program by the macro INTGTL). The computer program for updating the integrands is shown in Figure 5.28. The computed output of several variables to a step change in Tm and EFD is shown in Figures 5.29-5.40. Computer mnemonics are given in Table 5.3. In both cases, the step input is applied at t = TST ART = 0.2 s.
1.7023-
Response to a 10% step increase in Tm I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~.
1.6826...
.. t
•
, ,
• I
...
..
• I
I
I
ttl' I • 1 I •
I
fl'
••
I
I
If I I
••• I I •
" "
I I I
,
I
I
I
• I
I
I.
.~~~~~~~~~~~~-~~--~~
I I I "
I'
"
+ .. .. .. ........ I t
I I I I '" , • t. ' I I , t I " I • , f , •
I I ,
I
I I I I
I I I ,
It' I I I , t , I , , • , I I I I t I I , , I I , , , •• , f
I t " , I I I I I I t t I' t , , It' I I t , t. t i t I It' I I , • I' I I , I If' • , t •• I I t I •• I
I
t
~~~~~~~~~~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~~~-~~
, ,
1•6629- .. I , , I ,
......
•• I I I , • , ttl "'11
I I
I I
I , I It'
I I
I I
I I
1'1 • I'
I ,
, II II I
'I I I
I t
: : ~ ; ; ~ : : ~ : ; : ~ ~ ~ : ~ ; : : ~ ~ ; ;: t I I' t I 1'1' t , I It' I I I I' t I I I I t i t ' t t l I , , , , , I t ' , "1 I " t t , • , , t , " , , , , "'l t " , I I I I __
.... . -
_
~
...
• , ,
_
.-
'"-
I I t
•
I t
__
I ,
I' I'
..............
• I ,
,
,
t
,
t ,
t •
..........
ttl' tit'
~
_
••
_
,
, ,
~
...
I' , , • , Itt'
..-.
,.-
.-
.....
~
1" I , ,
--..
til' , t, I
'--
..-
~
::::; It. I I "1" I tit I
tit It' ,
........
1 I
I
~
,...
~
,
lit , t ,
...
......,
"--
_
t-..
.........
_
....
t ,
I I I I I I
I t t , t I
, ,
.-.
'" , I
t
• ,
t
t i t t t l , "
I
t I Itt
,
"
,t, It' I I
'"
I
I
I
,
I
I
7-7~7-~~77777777~777----77--7~--77~-~77-7-----~7~ I "I t' r I "1 I I r , I I I 'I' , , I t , t
I I' I "1'" , • I ' I I t ' t I , •• , , , ,t, '" '" I I I I Itt.' , : I t "" " '" "'" ttl t I '" "I I "" t , '" '" I I I I , I I
,
I
t I I "1 I.
,
I I
I I • I t I t', , I It' " I t
,
ttl I I ,
I , I • , Itt'
•
I
I
,
I
,
t" "
'" t t, I "
I '" '" '" It'
t.
I
I " "'"
,
t
I
I
, •
, t
, I
, t
I
,
I
I
I
, I
I
I ,
'I
,
f7iit~T7T7777777i777-777,7f77;ii-7i-tf7fit-----i7-I • f
, 1 , ,
t
I I , ,
I I t,
,
,
,.
,
t
"
" , 1 t' I'
'" f'"
I I '" I I I "'" I ' , I" • " , , I , , , . , 1.1 I I' " , , , , • , I , , •• , •• '" 1 I I I , f " "" I t I " , I I I t If"" f , I " "" 1 1 , , "'" I " t 1 , t , "" "'" I , , I I I '" , " , , t I' "1 "" 1 I I I ' , 1ft I '" 1 1 , 1 I' t 1 1 t , , , " , I' t t l ' , , , 1 t t t
, I " I I I'
I
I I
" • I
, t , I I t . t " f f • f f I " It'" 1 • 1 1 I 1 I "I" 1 , , II """ t , , t 'I I' , I It " " I t I' 1 ,
• I
'f'
f
I I 1"
t
I I
'-~'7-----"7---7~77--'-7i'----~~'ii-'i---'~--'-77-, ,
1 "
• ,I
I
, ,
, , I'
'I' I"
., , •, t i t t , "I I '" , , 1
• t
1
"
•
•
•
"I
,
,
•
'" • I " ,., t i t ' t '" 1'" t f t t , " • " I '"
, t, t. , t , 1 II I f
"
"' " " I
tit"
"
"
"
"" ""
I"
I
I I , ,
t
•I,
" , I " , 1 I " f It' 1 'I'
,11
1'1'
,
,
"1 • 'f I'"
I I I ,t" I ,
, " I I , .. ,
I I 'I I
1 I' I I I t "
,.
"t'
I
---~i--~·~-777i7-~-'i--~--'7~7---"i-7iii~7-7-tf7'~I t , , , , 1 ,
,
t ,
, , I ,
" "
, , I
, , • t'
, , " •
, I
t , ,
" I'
••
'f
• ,
•
"I " I I "
" I'
" t I ••
""
" . ,
t.t
r
" "
" " t ••
I
I
,
I I ,
, , ,
II' I I'" I "1
'" I I
I
" t • I I
I
I
I
s.:»: i s.:»: ~.!.!..!...!.!.. ~!...!.!..!...!..!...!...!..!..!.!...!...!..!...!.!.. ~!....!..!...!..!.!....!..!..!.!-..!..!...!..!. ~!..!.. I I I I I I
1.5500 -.!.
,
,
,
,
1
,
I
o
1.6629
,
t,
,
,
I
•
,
,
I'
0.5
.,,
-.,, I
I
,
1.0
Response to a
I
,
5~~
,
,
,
I'
"
•
I
•
,
It.
1.5
TIme, s
,
I'
1
I
2.0
II
2.5
step increase in E FD
1
~~~~~:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1.6406
•,
,•,
1
-,
,
,
I
,
I I t
I
..
, ,
,
,
I't'
,
,
"
+
f , ,
I
••
,
I
,
, •
~~~~~~~~~~~~~~.~~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
,
,
,
1.6182
,
,
"
•
f
I
,
,
" ,
f
t " , , I lit 1 I' -'1"1"
,
, It
,
f'
,
"
I
""
III
I' I' , I I I II I I I' I'll
,
I
f
,
,
•
,
•
!..~~~!..~~~-----!....!...!...!..!.!.._-----.!..!..~---------~-----------'I " I I "
"" '" 1 II" 1'" "'I
'I
1.5959
,
t
-t ,
t..
t,I 'I ',
••
t
t ,
, I 1 , 1
,
'I I 1'1
t
"
"t
1 I I 1 1
" " " I' I 1
I'
.""
I
I
1 I
, 1 1 , I
"
II " I"
I
I
, 1 , ,
,
tit t
,
,
,
t
"
I
,
I
~~~~~~~~~~~~~~~~~~~~.~~.~~~~~~~~~~~~~~~~~.~~~~~~.~~
, , 1 t , , , , t • , • , , • I , , t , I' , II' I I' , 1 t t , I' , 1 t ., t i l ' II 1 " ' 1 ' •• 1 1 1 1 t , • , , , , I I I " I '" , I I I" " " , , I t I , , 1 '" , , • 1 III I 1 1 I lit , , , , , , , 1 1 , , I , , , I' t I , • I , t t , 1 lit t I I , • I _" I I I,. "" 1" t i l . I" t • I It. I , • I I • I 1 f I I • , • ,
••• '" • I'
1
1.5735
,
,
, , , , , • 1 • , , , I ' ,-. 1 • 1 , , tit "1 I' It. t I I '" , , • , 1 '" t,.,. • 1 I • , • t ,
+ • , 1 •
,
f'
•
,
, • , 1 , , • I ,
t
,
I , , I t 1 , , II I , I , • I' • lit' I' 1 , , 1 , , • I I I I' t I" I' , 1 • , , •
•• , ,
, I
, ,
t t , •
,
,
• I I + It' 1'1' .. I ,. I + 1 I , 1 , ,
i77777777f77~77~f7t7--f-7t'~'7~~""'7~~ii77771777777i
t t t • t, , , t t, " • tit 1 t. , It 1 f II I' I " 1 " • I ' I t' "t, I I 1'" , " , t. "1 t '" I " t t 'I" , 1 lit, If' , , , lilt I' ,. 1 I" '" I I 1 lit 1 " " t 1 , I' III It' I' • III I t , , 1 t, I f ' , I I , I II I'" til' '" 1 II' Itt t , " • , I It' , • ,. , . II '" 1 _""1'1111"""""'11""111", f
, .550\"
,
,
It,
,
,
,
t t , f , ••• 1 , , • t • , , ,t , , t t • , t I I • , • , t • t , , It' , I' , , • t t , " 1 1" t , til' , t , t " I I I 1 I , t. I 1 I , • I • t 'I I , t , , , , , ••• I " , t " , , , t t , • , , , •• , " " " " " " ' 1 '
T.. . . . . . . - - . . . . . . . . . 1-.. . . . - - . . . . . . . . . -r . . --.. . . . . . . . . -,- -.. . -.. . . . . . . . -.. . r-.. . . . . . -.. --,
o
0.5
1.0
Time, s
1.4
Fig. 5.29 d axis flux linkages "-d.
2.0
2.5
2.2548 -
+ +
...
~
I
-~~-~-~~~~~~~~~~-~~~~~~~~~M~~~+
.<
..
Response to a 10%step increase in Tm
I
,
,
........
_
~
I-
..........
~
""""'" r -
..-
~
..-.
•
to-
.....
..-
I I
~
,
I
• ,
I
I
I
+~""'~"""""10-4""~""'~"-0-4""""
.. to-
I I I I
• +
+ • , I
,
I
I I,
I"
I ,
"" f
"
,
I I
...........
..-
too-
to-
..-.
_
~
.....
...-....
.......
~
~~
to--
to-
..........
I
fI
I"
I
1 I
to-
................
..-
"-
'-
~.
....
.....
.....
t-
to-
..... ..... ...... ....
~
~
..-.
-.
.....
I I I I
, I 1
2.2026
I
-ot,,++.
I
I I I
,
,• 1
t--
to-
I I I I If 1 I I , , " I I I I I'
t-
I , 1 I I I I I
t - ,,-
..- .............
III I , I I r r '" I , , I I I I , r I I I I , I
1 , 1 , I
I-
~
, , 1 ,
.-
.....
.- .-
I I " If "
1 " 1 " I r • " I " I " I I I I 'II'
1',
~
...... .....
.....
....
to-
If' " I I I I I I I r I I I I I I I I I I I
t--
_
to-
" I I I r I I 'I " " I r I I
.-
......
I
...- ........... ..-
..-
.....
.....
....
I
.... .......
I ~
I
I
.....
, I , " I I I I , " I , I " I " t i l I
~
~
I
I " I I I , I I ,
I I ' I
~~~~~~~p~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~~~~~
I I I I I r II' I I r I I I 1 , I 1 I I I I r III , I , I I r , , , III I I , I II l i t I "" ,
, I I I ,
I I , , ,
1
I I I
"1 I I , I I I
I I "
I I
, I I , ,
, I ,
• I f 1 I' I' .,
" I I
I I
" " 1 I 'I " I
1
1
1
1
I " I
'"
" "
...... ., 7 .. ., .,. ., ,. .,. -. I "" I I I 1 , If 1 I I " , I , I I I I'
I
"
I 1 " "1
.
I
I
I I I
"
I I I f I I ,
,. .,
--
I I
I
1 l I I I 1 I I I I "
,
I
~
0-
--
,
,
I
I I , I I I t
~-~~~~
I 1 t I I I I , I 1 I I I 1 I I I
i
~
..... ....
I
.-
-
~
_.....
....
" 1 I " I I I'
" • "1
I I
I"
I
,
I
,
I
I
I
I
I
If
,
I
I
I
I
If
,
I
•
I
7777777~~~i~~~~~-~~~~~~~~~~~~~~~~t~~~~~~~~~~~~~~~7~~
, I
2.1000
I
"
'"
I
I"
II' 1 • I I " I I I I I I _I"
,
I
'" I" 'I I I I I '" ".
I
I 1 I I I ,.11
I
r- _ --
o
I I I I I 111111111111111111"",'1
-l --- _
11111'1111
0.5
1.0
-r -- --
---
~-l
2.0
1.5
Time, s
I ... II
2.5
Response toa Yj;) step increase in EFD
2. 2026 -. .. .. .. .. .. •
~~~~~~~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~--~
III I I I
I 1
I I
I "1"
I I
I I
I 1
I '.1
+
2.1888 -:I :t :' ": : :I :I :I :I :I ;I •I ::) ~
I I I I I I I I III I I I I I I I 777.,.~7
u.. 2. 1750-:
to<
I , I I I
:I :I :I :, :, " I I I
2.1611--!..!. I I
I I I I' I I
I
I I I
• I I
..........
, I I , .... I I , " I , , • , I , I ~-7--~-
:f
•
.. _ I
I
.. _·~_
. . . . _.. . . . . . __ . .
•
, , . , f I f I , , I , I I
, I I , I I I I I I I , I I I I , If' 1""
I I I • I
I I • I ,
I I " I I
t
f
f
I
,
,
I
, I
, I , I 1
I I I I I
I , I I ,
, I I I ,
If' I I I , , If' I f I I f I 1'1" I I I I I I I I I • I , f I
I I
f
If'
,
, ,
I I
, 1
, I
I I I I I
, I I I I
f 1 I I I
, , I , I If' " I I "'" I , I ,
I I I I I I If' I 1 f , I
I
I
-7
f I
I
I
•
I I
I I
.. -.+ .. I I I I
I I I I
1-0
........ I f I " I I I I I , , I I I
2. 1473-:I :I :, :I :I :I :, :I :I ': :, :, :I :, :I :I I: t: t: :I :I :, " : :' :": :':I I: t: .: :t :I I: :I :I :, :, I: :I :, :,
i7111,-7i7,,--,-I
I
I
I
1"" I I I I I I I
I t f I , J I I , I I I I I I I • I 1'1
I
I
"
I
I
I
I
I I I I ,
I
I
'"
I
I " I " "
I
"
I
,
I
I
I
I
I
I
,
,
I
I
I
---I
I
I
I
I
I
~
,
I
I I I , I I I , II I I I , I I If' , "" I t I I I t I " f I
I
~~---~-
+ .... ot + I
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~.!. ~ ~.!. ~!...!..!-.!.!.. ~.!.!...!.!-- ~ ~..!...!. ~
I , I If' I I II I "1 1" I
..
~--------~-~
I I I I I I I
I
I
I
I
-I
I
I
I
'"
- I
,
I
;t t; l; ;I I
I
I
,
,t
""1"
"
I
I I I I I I '"
I
I
I
.....
,--I
,
,
I
f , I I I I I I
,
~777~7~~~~7~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~·7~~~~77i
"" I '" ,.,
I
, I I
, I
I I
I I
I I f
1.,', I , I
I I
, I I
I
,
I I
&----- . . ---f ... 0.5
2.1000-: : : : : :
I
,
I
I
:
,
I
I
I
I
,
I
I
~-----~
Fig.5.30
. . . I-- . . I
I
,
1.0
I
,
, " "
I
" , I
I I
, I
I I
I ,
I I
:
:
:
:
"'l
I I
------,---------i . . -.. ---- . . . .2.5 ., 1.5 2.0 ,
I
I
,
I
Time, s
Field flux linkages AF.
I
I
I
I
,
I
I
I
t I
I
I
I
I
:
I
I
I
Response to a IOc%, step increase in Tm 1.9~3--""-""---~~---~----------""-----------""~""---""---~~;; ..
....
'"
1•9193 __.-
0-
>-
- . .-
.- ~
-
.. .. ....
.. I
,
" I ,
I ,
I
I I I , I I I ,
"1' I I , , "1 t : I , I 1" I I I , I I I I I I I I I
I I I I I I I " I
I I I I , I I
,
I
I
1.8982 - -; :- ";-; ";-; : I 7 , ";" i' , ";' .. i' , I , , , I I I I
"1 I I I , I I , I I I I I I' , I I I I I I
I I , , I I I •
I I , " I I I "
,
t ,
, I I
, I , , I I • ,
• I I I , , , , , "" I I , I I I I I I I I I
I , I
I I I I
,
:-
:
.... I I I I
I I I
I I I
It , I I
"" I " ,
I
I , I I , I I I
I I , I I I I I
I
I
I
I I I
I , I
I' "I I I I I I ,
I I , •
I
.. ..... I I I I I I I I I
, ,
,
,
t t
I I I I I
, I
I I
I I I I I
I , I • '" I I I I
I , '"
""
, I I I
I
I
-
I
I
I
,
-
I
I I I " I I '"
I
I
I
I
I
I I I I I. I I I I I I I I I I
I
, , I I I I
I •• I I , , t I I I I I I I '" I I I I I I
-
I
II
-
I
I
I
I
I " I I
I
I I
I
I
II
' " I I
I
I
-
,
I I I I I I
-
I
I
,
,
,
,
I "
I
, ,
I I I I I
I I I I I
I I I , I I 1., I '" I I I
"
I I I I I
,
I
,I , I I
,
I
,
I
, I
, I
I I I '" '" I I I , , I I , , 1'1 I I I I I I I I I I I I 1" ,.,., " , , I I
I
I
I
I
";" .. i I .. i' ........ 7 7 .... 7 ; 7 r ,
,
t
1" I I I I , I I " I I I I I I
I I • I I I
I I I I I I I I
I
I
I I I I I I I I t
I I I I I I I I I
I
,
r r ~ .. I I • '" 'I' " t • I • I I I I I I " I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
, •• • " I " I I I I I I I " '" , I I , I I
.-
I I I
r r r ";' ........... , ";' t I I I I I
.. .. I I I I I I
,I
I I I I I I , , I
I
, ,• I
I I I I I
• •
I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
,
I
I
• I
I I I I I ,
1.8000 .- -
-
I
-
-
I
o ;•
1.0
Response to a
, I I • , I ,
I I I I I • "
I
I I , , I
, I I 1'1 , , I , I I ,.,
I I , I I I I I I 1'1 I I I
I I I I I
I I I I I
I
I
,
I
I I
I , , I ,
I , • " , '" 1., " ,
------- ..... ----1.8679--1I :I :I :, :I :I :I :I
,
I
I'
I
,
I
I
, , , I I , I , I ,., , I ,
"
..... I I I
~
, I •• I I I I I I
•••
I
" " " I I
"
I
I
_
I
.. I I , I I' • I
I
"" '" I I I "" '"
I I
'"
:: I I I
-0
I " I
I
I
>-4
5~'~
- .-. -
~
I
0.5
1.8982 ~
_
----
--
- - --
I
-
I
I
I
..
_
,., " I I I I I " I "
I I
I
I
- ....
I
2.5
step increase in E FD
_-_ ............. _............ _------_ .... _.... _- ....
I I
-
2.0
1.5
Time, s
I
I I I I I I , I
" " I " I
"
I
• ,
_
..
, I '" , I"
'""4
.-.
_
~---------
_
4
-
....
, I
I
1"
, I '" I I I I " "'" "1 I I I I I , , I I , , , , " I I , " I I I " I I .. I '" "'" I I I I , , """ , , " I I , , I I " , I , , I I , '" ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~~~~~~~~~~~.~~~~.~~
1.8375-: : :: , , , I , , I
, " I " , " I " , 'I , " '"
, , , , , , ,
• •• , " • •• 1" I " • I I t "
u n uu : ""'" I I 'I I I , '" I '" I I , , , , '" I I '" I I I I I , I I , I " I I "1 I I """1'1
.
~::::
I
I
I
I
I
"
,
1"
I
,771-"1~,.,, -.~--------~ • I I I I I I. '1'1 "." • I • , I "1 I " ,
I , I I I ,
I I I I , I I
"I' • " I 'I , •• I I I I ., , "
, • I I I I I
" I I " ' " I I I I I I I I I " I I " I I , , " " t I I I I "1'1"
. . -.. -------77-.-.. . " I " '" • " I " I' "
I
I I , I I I I
I
::
, I I I I 1" I I I I I I I I I It'
-~77-77'" I • I 'I "" I I I I •• I I I I " I I "
· '-r--- -----l---.. . ---- . . -r--------l-.. . - ------r . . -------l
18000
""
o
""""""""""""" 0.5
1.0
Time, s
"
1.5
Fig. 5.31 d axis amortisseur flux linkages >"D. 196
I I
"1"""""'" 2.0
2.5
Response to a 10%step increase in Tm
1.9404 - -
"'"41
-
--
~
-
-
-
-
:»
......
0..
o 1.9193 - - ....... - - - ............ - _ ............ _ ....
~
~~
-.
I I
I
I I I I I
I I I
I I
I
I I
, I
, I
-
--
-
-
-
-
I ,
";
I
, ,
-;
:
•
I I I , 1'1 I I I I I I
, , I I I
I I I I I
I I I I I
I I
I I
I I
I I
, I 'I
I I I
+ • I
-;
I I I I
+ + ,
I
I
I
I
I
,
,
,
I I
,
I I
,
I
I
I
...... - .... ~........ - - I'" -, .............. - ........ -- -- . . -- ..........
: ~ :: , I I
I
I
........ ': '; 1'1
..
~
-. -
::: : :
:
I
, I I , I I
I I I I I
I I I I I I
1"
I
I I I I I
I I I I I
I I I I I
::: I
I I I I I
I
I I , ,
I
I
I I I I
I
I
I I I I
-~-~~+.~~~~-~~~~~~~~~~~~~~~-~~~~~-~~~-~~~~~-~~~~~~~
1•8982 - ;I
~
I I , ,
I I I I
I
~
I
,
I
I
I
I
I
~
I
I
I
I
I
I
I
I
I
I
I
I
I
I
:, I I I ,
I
I I I I
I
I
I , I I
~
I
I
, , I I
I
I
I
I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~~~~~
I I I I I I I , ,
I I 1'1 , I I I I I I I I I I I I I
I I I I I I I I I
I I I I I I I I I
~~~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~
, , I
I , I
I I
I , I
I I I
I I I'
I
I I
I'
I
_.. . _.. . . . . ~_ . . _~-~~~~~~~_ . . _--~~~~!~-~~_ . . . . I I
,
I'
,
I
I' , I I I , 1" , I I
I ,
I I I I , , I
I
I I
I
I I
.
I
I I
I
,
I I I I
I I I
I
I
I I
I
I I
I I
I I
I I
-~~--------~~-
~
...- . . . -. . . . . . . . . . . .
too-
.....
I , I
I I
I
-.. ........................
--'
I
................. ~
I ~
......, ......
I
~
...........
I ~
...............................
.....,.-f...- ...... - ......
~
...... -.. .............
~
I
I I , , , 'I"
1 •8000 '-I -
o
,
--
I I
1'1 I 1'1 I I I I
I I
I I
I I I
T
I I I
, ,
I , ,
I I ,
I I I
I ,
0.5
I
I
I
I
I
,
,
r , I
I I
I I I
1.0
Response to a
I' ::
I ,
>-
1.8982-t I I I I I I ,
I I I
,
,, I' I
I I
I I
I
, ..I 'i..-7......., ... . . ., . . . I
I' "
I I I
5~~
I I I
I I
I I' I I
I I
I I
I I
I I
-1-
I' I I , • Itt I 0-
0-
1" t I -
I
, ,
-
1.5
Time, s
I I' I I
-I
I I
I' , Itt
~
-.
, , I
I I
, I
I I
2.0
I ,
,
2.5
step increase in EFD
.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I
I
I
I'
R.
I
~
•
__ 1.8679-:: ': .... o ::, I : I : : .. ; • ~ ~~~~~~~~~~~~~~~~~~~~+~~~~.~~~.~~~~~~~~~~~~~~~~~~~~~
r<
I I I I , I I
,
I
•
I' I' " I I " I I I I
,
I'
,
,
I I I I I ' I I I I I I I '
,
,
I
I
1'1
I
,
I
,
"I , "" I" I "I I "I I I' I I I I 'I
,
I
I
I
I
I
I
, , 1" • I , I , , I I , I
I'
I
I'
,
I'
I
I
I
I I , I , I I •• I I I I I
I'
,
,
, I I I , I I I
,
I
I
,
I
,
,
I
I
I
I
I'
•
,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.~~~~.~~~~+~~
,
,
1.8374-:
:
I
I , I I I
~
I
"1
I
I
I
I
I
I
,
- ..'
I'
~
-I -
I
-I -'I
I
I
I
,
I
I
I I 'I " I I I I
I
·
, , I
'I I , I It'
1'1 It' I' I
-jH- ---
1 8000
o
, I
, I
-I I I
, ,
1'1 , , , I , , I'
0• .5 Fig.5.32
, I
, ,
I , ,
-
• I I 1'1'1 It' , , ,
I I ,
I I I
I I I
, I ,
, , ,
I I ,
I I I
t , I
I t I
i -------l---
1.0
Time,s
, I
I" t , , t
1.5
I I' It'
t ,
I I ,
,
,
I
7~·
I I
t
,
•
I' I I I I
,
I
I
, , , I t
I I I' , 1'1
I I
I
,
: : :
I'
I I , I
I
I
::
~
,
I I I ,
I
,
I I
I I I
2.0
I , I , I
I 7,1 I" 1'1'" 'I I I , ,
I' I'
'I
I I 1'1 I I I
I I I
- , ---I •
I
I , I I I
, t
I
I I I
, I' • I' 1'1 I I I I I' , .. ,
,
2.5
Saturated d axis mutual flux linkages )..ADS. 197
0.21119-
Response to a IQ~~ step increase in Tm
• + •
• + "
I I
•
,
, , I , '• ," I ' , ,,
~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~-~~~~++~~~-~~-~~~-
"
I'
I'
+ •
I'
o.19283 -
o (!)
V')
..-
.- -
•
+ ::
::
~
0-
0-
1 , " , 1 1 1 1111 It 1 t 1 I I I 1 t 1 t " III' , , .......... 'I
• -I I' II II
I'
~
to-
..................... .....-....
•
•
~
,
,
, ,
c. 17446 _.
""""4 ....
a--
, I
, I ,
...............
,
,
I' I'
:1 :1
I , 1 1
-
,
, ,
~
~
~
I'
" "
I 1'" 1 I I' II' t
--
I' 1 1 'I " "
I'
~
' "
"
"
to-
1 , ,
1 1
::
1
:
I" _
..-
"
" I' ".................... '" 1 a.-.
.-.
, .........
"
~
,
I
....., _
,
....
1 ..... t _,
,
,
,
: :: : 1 II'
:I
:,
:,
I
, , ,
I
I
1 1 1
_
1 t 1 , 1 1
,
I ' ' " I '
I ,
1
I' 'I 'I I 1 1 I 'I
I " , , I I""
:I :' l: l: :1
, I I
l l
-
I' t, 1 , til 1 1 1 1 1 III' I' t I' , I I I I I 1 1 1 III' , , 'I" t , 1 t t l 1 I I I t I 'I , I' 1 1 .... '-- ................. .....: .... ..- ,..... •
,
"
" " 'I " II " 'lit • tIl
, ,
, 1 ,
, I II'
I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I , , , , 1 1 1 1
, 1 , I , , I I , , I' I' 1 I , I
I I I ' , , I I ' , I" , " , , , , , It' " , I I I " , , , , Itt , , , 1'1" I , I " , , t , 1'1 I , I' " t ,., , . , , , I 1 , t I , 1 II " 1 II' "1111 " , III I' 'I "'" 1 ,t
" , , " 'I I'" , I " I ' I , , , 'I " I I' , I I 'I I , I I I I " t 1 It' 1 I III'" 1 I ' 1 , I I II 1 II' t I I'
t,
I , , , , 1 1 1 1
"'" I 1 "". ,. "" I " "'I I " It'" " I' tit I 1 "1' til 'III 1 • 1 ." 1 til
, • I , , 1 t , II
77-'7-i77ii'ii~'ii7''''',~4i'''''77i~i'ii''17-7i'ii-''--7--" I 1 I' 1 1 'I " 1 I I'
t i l ' "1 I' 1 'I t".", I I I I II I II" 1 I II'" 1 'I I' '" I II' 1 't '" 1 , 1 " " I 1111 'I I I' , I , "
11I1 ,t" "I 1 "t' I" 1 I I'" ""
t"
I"
I I I I I 1 I t
1 I , , I 1 t' I I 1 , I I , ,
'I I' '" t III 1 "I 1 III' I" 1 II' 1 ""
"lit 1 t' I , , , I' III' 1 1 , , 1 1 II'" 1 "I I 1 I '" I , 1 "I" I
I 1 I I 1 1 I' I' 1 1 I' "
"1 I 1 , '1111 , 1 11I1 , It' t i l 1 I I I I I II' I , , t " t , 1 "I I , , ,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~.~~~~~~~~~~~~
...................... . . -
.,..,
1 1 1
, ,
t"
I
"
"
t
po-
, 1
'" , 'I' , t
,
I ,
~
I"
..., ;
~
-e
~
.~
t
I
~
~
-
.-
.....-
I
It'
~
.-.-
...., ...., ....
II
"II " l l I 1'"
,
,
I
t
I
,
t
,
,
t
I
I
,
I
,
,
,
,
t
,
I ,
,
,
Itt
,
,
,
,
,
1 ,
II "
O.12000 _!.. .!. ~ ~ ~ ~ ~ !. ~ ~ .!. ~ ~ !.. !.,.!. ..!. ~ !.. ~: .!. .!. !. !. ~ .!. .!. !.. !. ~ !. .!. ~ .!. .!. !.. !.. .!. .!. ~ !. ~ !. ~ .!. !.. .!. ..!. .!. ..!. !.
I
I
o
I'~ ~, ;, ~, ,• I I
I I II
1.0
I
I
1.5
Time, s
I
2.0
2.5
Response to a 5(\ step increase in E FD
0.17446-; I I
I
0.5
II II
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
111'1' , I' I' , I I , , , I , , I I' , 1 I , , , , , I , • I , t I , , I t I I I , , 1 ,
o
I~
, ,
, I
I I
I ,
'" , I
I
•
•• .1 I I I •
7i777i777-----i-7-~------------~-----------~-------
O.15346 -: : : : : : : : :
V I ' '" '" t • 1 I I , , , , I II' I , I , I , 1 , til'
I I I I , I I I , I 1 1
t • , I , "
I I , I I
:::
I I " " '" I I 1"" 1 ,
1 1 I
,
I + 1'1
"
••• , I I I , + , "
,
,
+ I
7777~-,~~7-----------;7~;,777-;-------------------,
I ,
"
I
" t " I' , " " ' ", 1 1 , I , 1 I
I 'I , I I I
'"
1 ,
" " '1" I ,
" I I"I I I • , I I • , I I 1 I ,
I' I I
, ,
, ,
I ,
, ,
It' , I I
t ,
I
, I
1, 'I "I , 1 I t 1 I , I
I
I I
I I
I ,I I , 1 ,
I I
I
I I 1" , I I I I ,
I
,
1 I
I , I
I ,
, I
II , I
1 1 , , I ,
I 1 I •
I ,
, I
til I 1 ,
I • I I
+
I I I I I 1 1
, I I I I 1 I " t 'I I , , , '" 1 II' I , I , t ,
O.13245 -.!., .!.., ..!.t .!.• ~, ~, .!., .!.I ..!., .!.I !., .!." .!. ~I ~t .!.I ~, !.., ~I !.., ~I ~I .!., .!.., ~I .!.I .!." ~ .,'.. ..!.I .!.I .!.., ~, !..I ~I ~' "!.. .!. !..I .!.I .!..I.!.t '!.. i' I. 1'1' , , 1 , It, , , , , , '" 1 I , , I I 1 I , 1 , , , , I , , I I' , , , ,
I I
1" 1'1 1 " , t I I I I t ' t '" I I 1 I' , , t , , , , I' t , I' I I 1'1 1'1' , , , , I , t I I' , , I t " , I , , I I I' , , , , "1 II' , , I " ,
, I I I I I I' t I II' II' I , , , , I' , , I , , , t , I It' , I Itt I , , t , , , t , , II I I , I 1" I 1'1'
, , I , 1 I "1 I I I lit I " l l I 1'1 I' , , , , "" , , 1 I I I I I , , I 1 I , , I I , It 1 I I I " I' , I " , , "" 1'1' , I , It 1 I 1
I I , I I I ,
1 1 t I , I ,
+
•
+ + 1 1
I
I I II II
~ .!. ..!. ~
,
0.1~-.!..!.~.!.~.!..!..!..~.!..~.!..!..!.~.!..!.!.!.~~~~.!..~~~~~!.~~~~~~~.!...!.!..~~~~.!..~.!..!..!.~.!.
I
o
I
0.5
I
1.0
I
Time, s
Fig. 5.33 d axis saturation function 198
1.5 SGD-
I
2.0
I
2.5
Response to a 10% step increase in Tm
1.0014 -- - _ o 9945 - +
-- _
-
0-1
.+.+
--
+ +++'
.-.
• - _ - _ + '+ • • • • ,
R.. :, :::: I: : :: ::::::: r , , , t , ;c: : : : : : : ~ :::::::::::: ~ ::::::::::::::::::::::::::::::: .... 09800 I
I
: : : :
I
I
I
I
I
I
"
Itt
t
""1
1"
I
I
I
I
I
t'
I
I
: : ::
I
,
:::::::
,
I
,
,
I
I
,
I
I
,
I'
I
1"1'
I
""1
I
II
I
II
I
I
I
I
,
,
,
I
III
I
I
III
I
,
I
,
t
I
, -.. . . . . . . - . . - . . . . T.. . . . . . . . . -- . . . . . . . . . . r.. . . . . . . . . . . . ---1- . . . . . - . . - - . . -. -I........ - ............. - - . . T
•
ttl'
o
II
~
0.5
1.0
t
1.5
Time,
I
I"
2.0
2.5
Response to a 5% step increase in EFD
1.1586
-~
r _
~ ~
~ ~
~ ~
~
~
~
~ ~ ~ ~ ~
~ ~
+ I I , I I I I I
..
I I I I I I I
~--_ .... _-~-----_ .... _ - ...._ ...._+I ....- - _ .... _ - - - - - - - - - - - - - - .... _ - - - ............ I I I
r
, I
1 • 1176
-
~
-.. . . . ,
iI I I I I I I I ,
I I I I I I I I I
I I , I I I I I
I I I I I I I I
~ ~
,,
•
I
II I I I I I I I I
I I
~
I I I I I I , ,
..-I
I
~
7+ :
~ ~
I
,
I I I I I I I I I I I' I I I I
. . , to-
to-
..,J
,
,
I
p..
• I
~~~~~~~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I
I I
I
1.0766-
I
"
I
,
I I ,
I , I
,
"
I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I
I ,
I
-----
• "
I
+
,
_--------- ---------------------------------I
.....
,
,
, , I
• I
I
I I I , I I r I r , I I '.1' I I I t I I I I I
I
r
, I " I' I I t I • I I I , I
I
•
I
,
I
,
I I , I I I I I ,
" I. I I I I I I I' I I I I I ,
.....
I
I
I
,
I
I
I I " 'I I I I I " I I I I I I
I
I I I , " I" I I I I I I , I I 1'1 I I' + , , I
I
•
•
I
•
•
•
" I I t I I I I'
,rI
I
I I I
I I
,
•
•
I
I
•
•
• I I I I I I • 1'1 I I " I , • , Itt I I I I I r I 1'1 I I I I I I I I I I I " I I I I I I I I I I I • I I I I I I I , 'I' I I I I , I ,
I'
I
I
~~~~~~~~~~~~~~~~~+~~~~~~~~~~~~~~~~~~~~+~~~~~~~~~~~~
I' I t I I I' I I I I I I " "
1.0355-
I I I I I I I' I I I I I I I I I I
I I I I
I I I I t I I I I I I I I I I I I I I I' f I f I f I I I " I t I • , I I I I t I I I I I I I " I I I I I I I I' " ' 1 I I I I I If' I f f I I I
I I I I I I I I I I I' I I I I I I
,
I I I , I I I I ,
,.1 ., I '1" I I I I I I I I I I I I I I I I
I I I I I I I I I I t I I I t I I I I I I Itt
I
t
r,
I I
,, I I
, I
I
"
I"
t
I
I'
t
I
If'
f
I
I
I
I
t
I
I
I
I
I
If'
I
I
I
,
I I
It'
I
I
I
~~~~~~~~~~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
t I
0.9945-:
·
O 980
"1 I , ,
'" , " I I I
I ,
I I I I tit'
I I
I I
I I
I I
I' I , , f It' I
I I
I I •
I' , t I ,
I I
I I
t
:
:
I 'I I f I I
'I I I' I , I
t
I I
, I I 1'1'1 "I" I
I
:
I I
I I
, I
t I
" I ,
I ,
I I
I I I
'I , I I I '" I
I I
I ,
, 1'1 t I I
I I
O~---------l----------r--------l----------r--------i o
0.5
1.0
Fig.5.34
Time, s
1.5
2.0
2.5
Line current i a •
199
3.1063--_
Response to a 10%step increase in Tm _--_ _ ..-____
3.0816-
•
.., , ,, +
_
:..:..!...!.!.!...!!..
,
" " "" " ""
,,
.. .. 't
,,t ,, ,
t.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~.~~~~~~~~~~~.~~~~~~~~~~~
,
"" ,, , ,,. " , "
:» Q.
.
"
t
t
3.0569_ ............. _ ................ --_ ........ _--~~~~~~~~~~~!.~~~~~!...~~~~~~!._!.!.!.~~~~~ '" '" '" """ '" '" '" ,t, '" '" I I ,
'I '"
I
I
• , , I
"
3.0322-
• "
•
I I
I I
I
'" I"
I
'"
" "
"
I
• I I"
,
'"
,
It'
I
'" I"
, ,
• I' '" " '" '" , -----_ ..... ~~~~~!.~~~~!.~~~!.~~!.~~~~~~!.!.~!.~!.~~~~_ .... ~!..!.~~~~ , 1" 1'1'1' , " 'I " , I " • , I I I I I • ,., , I' t, " , I " " 1 ' , , I , I I , , 1 . 1 I I I I ' • I I I I I I , • I 1" I , I I " • I
I I
I I I I
I I
, I
I I , ,
i : : i i !i
, ,
, ,
f
, I
, I I I , ,
• I
I I I ,
I ,
, ,
, ,
, I
, ,
, ,
, ,
, ,
, ,
, I
, ,
I ,
, ,
, I
, I
i : iii : : : iii : i i : i i : i : : : : : i
3. 0075 -~ : ~ : r : : i ::: 30000 " 1 " 1 " " ' 1 " " " " " " " " ' 1 " 1 1 " ' 1 " ' "
•
-,'"
- T-
o
- --
., ,.
I I
0-4
0.5
-1- -1.0
::
, - - - -- - T
_ -,- - - -1.5
.,
2.0
2.5
Time, s
.
3.2240-
Response to a
5~/~
step increase in EFD
·" , ·,, ,, , I
I I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I I
I I I I
~~~~~~~~~7;7~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I
I
I
,
I
I
'"
3. 1699-
0-
, , ,
, , ,
I I I
, , I I I
I' I' , , , , I I
I ,
, ,
ll , , !, ~, _ _ _
_
, , I , I I
, •
1"
, ,
!...
, I I , I ,
,
,
,
, ,
.
_
_
__
....
I I I I , I , I
~-~~~~~~71~'~~~~~~~-"~~~~~~~~~~~~~~~~~~~~·~~~--~~~~
" ""
'f "'I
I' "
5.
3. 1158--
-
""
f' "" ,, " , ~~~_~~~~~~~~~~~~~~~_~_~~~~~~
" "
"
• I .,
'I
"I
T
I
"
,~
~
,,
,
• f
"
,
T ,:, ,
I '"
".
,
I'
•
,
,
I'
,
I
I'
,
•
+•
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I'
•
,
,
,
I
, I , •• , I I I I I I I , I' , • , , I , I I • , , I I" 'I I 1'1 I I I I " ,., I I , ttl ,
•
•
•
•
I' • 1.1 I' , , , , I' • I I , , , , , I • I' , , • , , , , 1'1 I , I I , , , I I , , ,
•
, , , , I' , I • , • , I' • I ""
• I
I
I'
, I I ,
I ,
,
, I
I.
I' , , , t I I , I ,
, + ,
•
I
I ,
, t , I I , I + , I
-~-------7777~7-----~77777----~777777---""--77777----
3.0616I
1'1' , , , •• I I I I' , , , I I I I 1'1 I I' • I , , , I 1" I , , I I I I I 1'1 I ,
,
I I , , , , , ,
I I I , I , I , I I I' , , , , I I , , , , I , , t , I I ' ••
, I , , , , , ,
, I I I , t I I
+ ,
I , , I I' I I , I' , I I' I I I , I , I , ,
, I , , " I' t, , I , I , ,
I I , , I I , I 1" I , 1.1 I I , I I I I I , • I I 1'1 I
,
r
I , I I , I • I I
~~~~~~~~~~-~~~.~~~.~~~~~~~~~~~~-~~~~~~-~~~.~~~-~~~~~
I' , , I I I I , 1.1'" , I , , I I , , t • I I I' , , , , • I' I I I , , I I I , , , I , • , I I • • , I I , I' • I , I I "1 I I ,
I I 1'1'1' I I , , I I' , I I' , I , , I I' , 1'1 I I t , I I It' I I f I I I • I I , I' , 1., I 1'1 I 1'1'
I I' , • t , , I I I I I I I , I , I I
, I' , " , • , I • , • , t I I , I I I f , , , • I I I • , I , I ,
, , • I I I I I " I I • , I I
______ !~~~~~~~~~_!~~~~~~~~ I , I I •
....
~~~~~~~~~
.~~~~~l~-_
1'1'1'1" , , I , , If" • , I' , , I I I I I ' , , I I" I , I I I , I I'
, f I If" I I I 'I I I I I
o
: : : :I :, :, :I :• :, '" "1 1'1'1"
0.5
: : : : : :I :I :,
I I I I , 1'1""
1.0
I'
Time,
'"
I
I I' , , I I , , I " • , I I' , , I I •• I
I , I • I +
: : : : : : : : : : :, :I :, :I :• :I :I :, :I :I :I '•
1'1 1 ' 1 ' 1 ' , II I 1 ' 1 ' 1 I"
1.5
Fig. 5.35 Field current ir-
200
__
1"'1'1 , I • I , , , I' , , I I , , , , "1 I , , , , I I I 1'1 If' , • I , I , I I , I I , 1'1
-i . . . . . . . . . . -.. . . . . . r.. . . . . . -.. -.. . . . . . . .1- - . . . . - . . - . . -1- . . -.. . . . . . -.. . . . 1.. . . . . . -.. . -.. . . . - f"
3.0075 -.• •I •I "•• 30000 I I I I "
·
, It' I • I I , , I I I I I I' , I • , I • , I
I , I I I , , , • I , I I , , , , , " , I I I I , , I I I I 1'" , , I • ,
I 1'1 f
2.0
I
fl'
I
I
2.5
Response to a IO~~ step increase in Tm
0.000017 -; ; ; ; ;
,, &. -0.00343. -, , .....o
ttl'
,
I
I I
-0.00687
•
_I
1 ,
I
+ •
+
• + + + I,
I I
,,
,
,
I
I
I
"I ,
I
I
I
+ I I
I I
I I I
,
,
,
I
" "
I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-0.02000
,
I
I
I
o
0.5
~~~~~~~~~~~~~~
1.0 TIme,
t
1
1.5
2.5
2.0
Response to a 5% step increase in E FD
0.03075-
+
, I I
I
II -----~-------------------~-~---~-------------~----1 I , I II I I II I I I I I I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I I I • I I I
I I I I I , , I I
1 , , , , I I I ,
+ + I
I
~~~·~~~~~~~~~~~~~~~~~~~M~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I
1
I
1
I t t
.. 0.00657-
....o
I I I
I I I I
I I I ,
,
,
~~~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I I I
, I
I I
I
I' I t
I I 1 I I
t
I I ,
I I II I I I I
+++++ , I
I I I I I
I
" I " I ,
I I I I II , , I ,
I I , I
I
I I 1 I I I I I I I
I
+
I
I I II I , I I
,
,
I
+
I
,
~~~~~~~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I I '" " I I
I I , I
I I I I I
I I I I I
1'1 I I I I I I I I I I I ,
I
I I
I
,
I I I ,., I I I I I I I I t I I I I I I
I I , I I I I
I I I I I t ,
I I I 1.1 I I I I I I " , I I I , I I
.
I I I 1 II I I I I
I I I I I
I I I I I
I I I I I
I I I I I
I I I I I I I I I ,., I I I
I
:
:
:
:
:
:
-0.00553-:I :I I: :I I
t , , I
I I I I I
,
I
I I I I I • I
" • I I • I I • I , , I I I ttl I I I
I
: I: :, :I :I •
•
I
•
•
:I
,.
"
•
"
I
I
,
•
,
I
I
,
•
",7777777,7----7 .... ,--------------,-----, .... --------.I I I I I I I I I I I I " I I I I I " I I I I
7 7 7 ..... 7 .....7 I' 7 ... 7 - - - . . . I I I '" III t I I ttl
-0.01762-: : :
:
:
:
;
I
I
I
I I
I I I I
I
>-4
....
>-4
I
I
I
I ........
I
I
I
I
~
....
:
:
+
I -
:
I
I
I
>- -- ....
-
:
:
I
:
I -
I
I
• • I I I I I I I I I I I I I I ,
-- >- ........ -
I
:
tit
....
I
I
-
-
:
....
,
:
I
I
I
I
>- ............ -
:
:
I
I
I
I
I
I
. - ............ -
I
lit:
:
I I t I I I 1
....
-4
:
:
-OO~O-~~~!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
.
I
o
I
0.5
I
1.0
I
TIme, s
1.5
I
2.0
I
2.5
Fig.5.36 d axis amortisseur current i o201
Response to a IO~;., step increase in Tm
..
1.1826 -
.... _!..!..
•
+ + I I
I
I
I I I I I I I I
• I I
---~---~-----------------~ •• :~!..~~~~~~~~~~~~~~~~~~~ I I I I " I"
1.1763-
t
•
I I,
I I
• •
I I, I I I
,
t,
,
t,
t
I I t t
I
,
--~~-~--------+~~!..~!..!..!..~!..!..~~!..!..~~!..~!..!..!..!!..~~!..~!.. •
• I I
I I I
I I I
•
"
I
'"
,
.;. .... -;- ';" 'j" ...~ ';' 'i" .... - - ~,- .. .!. - - .!.
" " "
,,
t
.!. - ~ .!. - --- ....
• : .!. ...... - - -
I I I I
I I I I
'" '" , ' ,, , t
1• 1700- :
I I I
, I
I I , I "1"
'" I
I
"
, I' I I I I
I , I I
, I I I
, , I I
I , I I
"1'1 , I I I I I I I I I I '" I
, , '" I I I I I , I , I , , I , I I I I I I t "'" I , I I
I , I I I
I I , I I , I I
I I I I I I I I
I I I I I I I I
I I I I I ""1 I " , " I I I I " I I I , I I I I I f I I I I I , I I I I
,
I I I I
t I I
,
,
,
I
,
,
I I
,
I I
,
,
" " I t ' "
" t
" t "
t I
I
!.. - __ ... - !- - .!. .... - - ........ - ... - .!. .. !. _ .... .!. _ !... .!. !.. ! !.. .!.!.. ~ .!. .!. !.. .!. !.. .!. !.. .!. !- !... .!. !.. _ !.. !.. .!. !... !.. 0-
, " '" , I I I I I I I " I '" '" I I I
, I I I
I ,
, I I I , I I I
'" I I I I I I '" " I I I I
I I I I
" I I I
I I I
t I I I
" I I '" I I I I '., I I I I
I I I
f
"
, I
, I : I
,
'" I I I I I I I I I I I I I I
I , , " I I I I I I I I I " I I I , " "'" I I I I I I , I '" I I I I I
I I '" I I I I I I I I I I '"
I I
I
I
I
I I " I I
I I I I
I I I
I
I I
I I I I " " I I " I' "
I I
f
I
.!. .!. .!. !.. !- .!. !.. .!. . . . !.. .!. ..!. !.. . . . .!. .!. . . . ..!. .!. !.. ..!. ..!. . . . .!. .!. ..!. ..!. ..... .!. . '. .!.. ~ !.. ~ ~ .!..!. I I I I
I I I " I
I I I
" I I I I
, , I , , , , ,
I I , I " I I I I , , , I '"
I
I I " " I
I
I I I
I I I I
I I I I
I I
I I I I
" " " I "
I
I. I I , I ,
I
.. 7 .... 7 .... - .. - " 'i" ., ...... - ........... - .... ; .... --- .... --...... I I , I I I I
I I I , I I I I
I I I " " " " I
I I I
I , I I I I I I
I
I I I t " ,t I I I I "1"
I
I I I , I I I I
'" , , I I , I I I I , I I I I
I I I I I , I I
0-
....
I " I I " " I " "
I I I , I I I •
I I I , I I I •
I
"
I , , I I I I
I I t , I I I I
'" '" I I '" I I I I I " I t
I " " I I '" " " '" , "
I I I
I
0• .5
1.0
Response to a
I I I I I I I " " '" I '" "
I I I
t
'" 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 I I
I I
I
I t
I '
I I I I I " I t
I I '" I " '" I " " I I I "
I
I " I I '" t I I I " I " I I
I I I
I I I I I
I
1.5
Time, s
I
"I I I" I
~~
.... - - - -- ~ ..
I I I I
'"
I
I
I
I
I
I
,
I I I I I I ,
I
t
I
I
'"
"
I
I I
I" I
I I
I
"
2.0
2.5
step increase in E FD
5~~
+ •
I
.!.
I
',1 -;- '; - ., '; ...- - .- - - - - -- -
-
I
.~~~~.-----------------------------_
1.1708-
I
"
I
T------.. . --~---- ----,------ . . --1---------1
J• 1500-.!. .!. .!. !. .!. !. .!.
I o
, I I
I
7.7.7, ..7.7--,••-.7,,-.777,-7.-7-------7 . . -
7'-~i-''''''''' I I •• I I " I I I I , ,
I I I I I , '" "" I I I "" "" I I I
I I I
f
..!. ........
I I I
I I I I I I '" I I '"
I
.... _----~----
'"
I
'"
I I
•
!~~~~~~.!.!!..~~~!..~~. " ., , I
I
, I
I I
1.1642-:: : t I
I I " '"
, I I
'" I I "
!l l l l I I
I I
I I
, , I
I ,
:: : i
I I
'" I I
,
.!.~.!.!!..~
.!.~
__ ~~~~..!.l~
,. I I t I I I I' '" '" • t
, , I " I I • I
I , • , I , I I
,t,
,,," "
I I I I I
,
I
I
I
I , I I I I I
, , " I , ., I I I I I I I I , I " It' I , • " I I , I
I
I I I I , I I I I I "'" I I I , I I I t '" " I I I "t I I """ t I • I I
~_~~
I I I I
"'" I I I " I , , t I , I I " I I ,., I I , I I I I , , , • I '" " I I , t I I
I
I I I , I I I I " "'" 1" " "'" I I I t I I I , " I I I ., I , t I I
.- .... _---------- .... ---+
,
,
I I " I I I I
I I , •
I
I
•
•• I I I , I •
+ • I , , I I ,
• I t
• I ,
• t
+
1. 1576-::;
; : : ::
'" •• t t. I " I ,t,
,., t
•
I I • " , " •
I I I
I
, , , "" " I '" I I ,
I I' I I I I
I
I
I
I •
I , • I • I I ,
"
'"
I I'
I
:;
"
I
t, •• 'I " "
I
., ,
I
I ,
I I I , I I I I
I I I " I '" t I I I I' '" '" It' • I Itt I
I
I
,
I
.... "
'"
t
"
'"
I
I
I
I
, • • , ,
'"
I
I I I t ,
I
I
~7-7---7-----7---- 7--~----------
·
,
1 1500 '" I
I
, I
t ,
, ,
I I
, ,
, ,
I I
" I •
,
I , I I •
,
I
,
I
I
I I
, I
I I
, .
" ,.
I I , ""
•
t
I I
I I
I ,
• , I It'
I •
" •
t
, I
, ,
, I
~.
t
I
I
I I ,
" • I " t I . , 't
I
.,
,
I ,
-i---------i-M~-------i--------i------0.5 1.0 1.5 o TIme, Fig. 5.37 Terminal voltage
"'"
I t ' •
,
"
I ,
~~
'" It' ", • ,., I ••• , t I , t I I ., '" t " I I •• '" I "" ., .".
It"
.... ---7 -- ..... 777-~777
t' , ,
..
u r u u u u u u u u u u ; : :: ::; ~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~
"
.,
"
•
"
I
,---I '" ., I
2.0
I
""
. , •• I "1 ,. " I I , , "" It •• It' I '" I
---i
2.5
Response to a 10°"step increase in Tm
53 • 975 -., •, •I •,
, ,, , , 52.645 -, ,
I
+ I + I •, I I I
I
t
+ I + + I , I +
I
+ + + + 1'1'++++++.
I
I
I
-
....
_
..-.-
.....
,........
..-
-
...........
_
....................
..-4
~4
"-lI
...................
.--
I'
........................
I"
I " ' I
I
I
I I
I
I
I
-...
................
.--
.......
...-
,
......
,
,
,
~
.......
.-..
I
I
I
,
I
I
I
--e
....
,
I
...-
I
I I
I
+ + + + + •
I
1'1
.......... _
"
I
I
........ to-
I'
.............
•
I
......
~
"" "
I I ' I
I I
r,
I
r r
I
I ~
" I " I
I
" I
.......
'" I " 'I I
"
" I
IJ.-4
'I
I
I
I
I
' I I t
,
'I
I
~~-~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~~~tJ~_~~~~~~~~~~~~
,
'
I ,
I I ,
,
t
, ,
,
I
,
II
I
I
, , I
, I I'
-y
I
I
o
-•
I
- T.-.
I
I I
Itt'
I
I'
,
,
-- ,
I' ~
0.5
61.326-
r ,
,
I
,
I
1.0
--
0- 1- -- 0- - --
I It' 0-
I'
,
,
I'
,
I
f
I
.-4
--
1.5
TIme, s
t
Itt
I' , ,
, I I
I
,
I I
,
,
50.000
,
I
I
1
, ,
I' , •
,
t.
It'
,
" I, I1'" ' ,
f I
,
,
I
•••
,
,
,
, •
T
,
,
,
,
-0
2.0
. 2.5
Response to a 5"" step increase in EFD
,
+
t
+ ,
,
t
-------~---_ •
•
•
....
•
,
_-~-----+_
. . _---_ . . _-- . . • +
•• I
•
I'
•I
+ I
•
I• ,
I • 'I
•
, It. ttl I •
•
•
•t t.
tI
••
. . . --~7777-- .......... ---77777---- . . . - •• 777-0--t
I t '
~.4~-~------~777;---t •
_--------~------
I
•
I •
•
" , I , It' I , Itt " t I Itt t I , t I
t
I •
I I
•
,
,
"
,
I I t t I I I , , , t "1 •
tit' , I I , t I I I , t I I I , I I
•
I' , t
I I I I I
•
I
,
"
•
""1 I I I
II I' I I
,
II , , I I •
I
,
,
t I , t , , t •• I' I I I I I I , I I I I I I
,
I "
I
t , I I ,
,
• I
I t , I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-~~~~-~~~~--~~~--~~~-
I
I I I I I
,
I
,
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~._~~~~~~~~~
I, , ,
,, I ,
t
"
'I
I
,
I ,
+ I
I
I
I
I
I 1 I I
, , ,
I' I I
"I I I I I I
"
, , I
I
I
I
I' It
I
I
I'
I
I t '
I
I
I
I
I
I
I
I
It'
I
II
I
I
III
I
I
I
,
I
~~~~~~.~~~~~,-~~~~~~~~~~-~~~~~~~~~~~~~~~~-~-~~-~~-~~
55.806.-
, 'I I I I I
I t ,
I ,
"
I I
,
I I I' "
I I t
'"
I
lit I I t
• I
,
It'
I •
t'
I " '"
I ' "I
"
1'" '" I
I'
III' I I , I It'
I
I
I
I I I.
t,
I I
I I
" "I I I
" , I I'
I I I 'I' I I I I I I I t ' t III I I 1'1 I I
I
I' 1 , I I
I I
III
I I
I
I , I
, I , I
t
I
I
,
I
I
Itt
~~~~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
III , , I
I • I
•I ,I ,,
53.966 -: : ,
•
I
I
I
I
I
I
r :: :: ': I
I
I I
"I
I
I
I
1'1
1'1
I
I
I
I
I
I
I
I I I'
•I
I
:: I
' I
I
I
•
t
'I I I
,
I
I
I 1.,
I
I
II
I
I
: I
, I
•
,I
I
::
I 1'1
I I
,1
I
I
I
I
:
1 I
I
: ,
I
777ii,7~-----------7---~-----""~~-------------""'7--71 1 I I t I I I , I III
I t I I , I I t I' ' I t I I I I I I' • I I I I I I I 1'1 I • II I t I I I , II I 1'1
I
1
I
I
I
I
I I 1 I I 1 "1
1"
I
I
I
I
I
I
,
It'
,
I
I
It'
I
I
1
I t , I 'lit'
I
I
I
I t I I , ,
1
~7~77777-----------7-""-~""'-----7--7--------~-------I'll I I I I II I It, II.I
1
II'
,
,
I
I •
I ,
I
I
Itt
I I
tit
I
,
I
I 1'1
,
I
I' I
I
I
I
,
I
I
t
I
t I
I
I I
I
, I
' I
I 1
I II
I
t
I
I
I
I
I
I
I I
,
I
• I I '
I I
' I
1 I
I
,
It'
I
' I 1 I
1., 1'1
I
1
I
I
I111
I
I
I 1
I
I
I
I
ii~~i7iii7-~-----""---~'---i----7~77--7--~i-""'------i7 •
'I
I I , I I , I . I I I I , , til I I' I 1'1' t I I I I I t " I I III' t i l I I t " t I I " I I I 1 ' I I I I I I' t t, "t I I , ' I I I 1 I , , t ., II I I I I 1 I I II • I 1 I I I t I' I 1 II I I I I l i t t • , , I I II I I I l i t I I I II I I I I II I I I I I I I' I , t t , I I ' , • I , • I 1 1"'11'1111111"11111'1"11111'11111111"1111111111
· I . . . . . . . . . . . - - . . . . . . . . , . . . . . . . -.. . . . . --.. . -r.. . . . . . . . _.. . . . . . . . . ,.. . . . . . . . . . -.. .
50000
o
0.5
1.0 Tfme, s Fig.5.38
1.5
Torque angle ~ in degrees.
-< .....
-
.... , ............ -
2.0
....... -
-
....
T
2.5
Response to a 10':<) step increase in Tm -0.00002 - ; : : : : : ; • • I I
I I
I I
I
I
I
, I I
I I I
, I
r
I I
I I
I I
I I
,
I I I I
I I I I
I I
I I I
I I
I I
I I
I I I I I I , I
I I I I
• • : ; ; ; ; •••••• ; : ; ; : ; : ••• ; ; ; : : ; ; : ; : ; : ; ; ; ; I
,
,
I I
I I
I I
I I
, I
, , I I
I r '" I I I I
I I
I I
I I
I I
I I
I I I I "'"
I
I I I I
I '" I I I I I I '" I I I I I I I I I I I I '" I I I I I
I
I I
I I
I I
I I
I I
I I
I I
I I
, "
I I '" I I rI II I I I I I I
I
I I I I I I
I I , I I I I I
I I I
I , I
, , I
I I I
I " I
I I I I
, I I
I I I I
I I I I
I I I
,
I ,
I I
I " I
I
I
I I I
I " I
I
I I
"" , I "'"
I
I I
I I
I I
I I
I I
, I
7~~7~7~777~7~7~~7~~~77~7-~7~~7~777~-77-7~------~--, "
r r
I
~
I I I
r
I I I I I I
t
r
r
r
I I I I
r
I I
I r I I I I I '" '" I . r
r
r
I I I
I I
I I
r r
I I
I
I I
I I
r r r r t I r r
I I I I '"
,
r
I
I I
r
I
r I
I I I
" I I I
r
I I
7~;777~7~7~~~~-~~~-~~---
I
, , I I I I
I
, ,
I
I I r I I I I I '"
I
I , I , I
"
I " I I I "
I
I
I
I " '" I I '" I I '"
I I t
I
I ,
,
I
I
,
,
,
,
,
,
I
I I I I I I
'" r , , I I I r I I I I I I I '" I I I I 1 III
I I I I I I I
I r '" I , t I I I I I I I
,
""
"
-0.00081 ~ .'.. .~ .!. !.. .!. .-. .'.. .!. !- .!. !.. .!. !- .'.. .!. !- .!. .!. .!.!.. !- .!. .'.. ,
"'" " r I " " , r I 1" I I I I I I I I I I I I , I I '"
'"
""
~ ~ .~
I
"
"'" ,t t , I
r
I I
7~- ; ,
-4
,
I
I
I
I
I '"
I I
I I '" '"
I
I
I
I I
I I
I I I
'"
"'"
I I I
~OO160-.!.!..!..!..!..~~
•
'"
'" "
I
I
'"
I t
I "
I
I ,
, ,
"'"
""
I I I I I I 1
I
I I
I
I
"
I
I
I
I
"1'
,
I I
I I
I I
I I
I I
, I
'"
t
"
'" I "" "'" 1 I I I
I I
I
I
t
I I
I I
" I " I
, I , '"
,
t
I
I , , ,
"
I
"
r,
I
" " I
I
" I
I
r, r, r I
r, I
!.. .!.
I I I
r
I I I I I I
1 I I
,
'" , I
,
" I
I
I I
I I I I
, I I I
I I I I
I I I I
I I I I
I
It' , I , I
I I
r
I
I I I I t I
I I I " " I
, I I
t "
I
I
.!. !- .!. .!. !.. .!. .!. .!.!.. .!.
I
I
'" I I I r I I , " I ,
I
I
~
'"
r '"
I I , I I I I , , • I I I I I "'" , , I "
,
,
r
I "
,
,
- .... - - - ... - --- - - - .-
I I
I ,
,
7 .......... t - ..., ~ ~ .I
"
,
,
I
I
"
"" -4 . -
~ ~
I ...
-.
I
I
~ ~
" .-
,
~ ~
I ,
,
t
, , I
.!.!..!..!...!.!..--!..~!..!..!...!.!..!..!..!...'..!..!..!..!..!..!..!..!..!..!..!..~!..!..!...!.!..!..~!..!..!..!...!.-
t t l
t o
0.5
.,
0.(X)lS5 -
, •
I I r I , I '" I I
7 7 .... 7 7 - 7 7 7 .. 7 7 7 .. ,. , -; 7 ., ~ I
t I
"1" I I I I I I '" I I 1 I
'"
I
I ,
,
"
r r , , "
I I I I
, r I I """ I I r I r I r. I I I I I I I I I I I I I I " , I "" I II'
I
r
, I I I I ,
I "
'"
7~~~-~~~~-~~-~~-~
I
I I
t
1.0
I
1.5
Time, s
I
2.0
2.5
Response to a 5°<, step increase in E FD
I
I
•
t
I
~~~~~~~~.~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I'
I' I I
,, ,, I
I
,
I
", ,I
,
,,
,,
,,
•
,
~~~~~~~~-~~~~~~~~~~.~~~~~~~~~~~~~~~-~~~~~~~~~~~~~~~
,, , ,
0.00102 .....
~1""""4
, ,
,
....
,
I
I , , I I , I I ,
I I , I I I I I I
II
,"
, ,
, I I
,
, ,
I I
, I
.......
~-
......
_~
"
.....
- . .....
I , I I I I
I I I , , ,
I
"
I I
, ,
I I
.-.~.--
....
-.04
....................
I I I , • I , ,
,
-
.....
I I I I I I I , ,
,
_-,....
...............
-~'-'"
I I I I I I I I I
......
-
,
...........
,
-..-
......
....,
................
,
--~~~~~~~~.~~~~-~-~~~~~~~~~~~~~~~~-~~~~~~~~~-~~~-~~~
I
,
,
, , I
I ,
I
•
I I I
, , I
,
0.00019
__
•
,
I I
....
._
"
I I
...-
I
t I
....
, ,
!.. .!. ,
,
1
~
.....
I ,
I ,
,
I
~
, I I I I ,
, I
P-
I
I
..........
I
I I
..-
I
I I I I I I I I I
.......
I
I I I.' I I I I
I I
I I
I
"
I
I
~
I
I I
I
I "
I
_
I ,
I
I
,
,
-0.00064
,
••••
I ,
,
"
to-
...............
.-.
I
I
I
" ' " I I 1" I I " " " , I I I
..................
I I I
I
rI
-
'" I I
I
I I
" I
-" I I" t' " I t" I "" , • , I , I I I I • I' , , • , • t It' , I I I I I I I I I I 1 I I "
I' t I
,,, " , , I I ,
..-
-
~
I I I I I , I
__
,
. . - t - t - .........
I
~
.....
r
10-
I I , , I I I I I I II I I t I I I t-~
'I
r
I I
I I ,t
I
I
"
, I
..-.
I I
,
I I I I I I I I I
.....
,
I I
I I ,
•
t
I
r
I
I I
I I
'" 1 "I' 1 ' I I' 1 , , , " , I t " , , I , I , , , , , I , I I , , , I , r "1 I
I
I I I
-
...-~
I
..-
_
..-
I
"
I
I I
__
-
I
, I
r
I
...
t- .............
." I I I .....
-
I
I
..- .....
.,
I
t-
'"
I
'I
I
"
-
-
too-
---
I
....
I
.-. .............
....-
,
.-.-
•• , I , ,
I'
I I I I
I
"
.....,
I I
I I '"
I
I
-~
, , I I I , t
, I I
I
I
I
-
I
I
" ,
I
I I
I
I'
.!.I . I. . .!., .!., .............. - •I .!.I !..I .!., -, .'~t .!.I .....I- t ..... - :. .!. .!. , ' I t I
, I
'" I'
I "
I
I I
, r
,",' 1 , ,1 , ,"I "I , , I I , , I "
I
I " I I, I"
, ,
I
,
I ,
,
I I
I I
,
I I
,
I I
I I
I
I
'" I I
I "
,
I
,
I I
, ,
I' , I t , , , , , "'I'
•
I I
I
, I
I I
t
t
" t r" I r" I " I r I I I I I I I
,
I I
-
I I , I I I I I I I I I II I I I I
I I I
r,
............... - ............... ~
-.
I I "
r,
r , , ,
~
"
!.., -I !~I -, ..: ~ !.. .!. ~ - - - - .!. !.. .!. .!. I , • , , I I • I I
"1 I I I
.-
I I
"I
I
I
.-
I
I ,
,
,
I' "
I
I
" '"
,
I
~7;~~--~~~~~~~~~~~7~~~~~~~~'7777~7;~~;~~~7-~~~~~~777
,
,
'
, , • , ,
I I , , ,
,
I
"
-", '"
I I '" I I '" I I I I "" I , I 'I I • , , , ""1 I I "1" I I I I I I I
,
"
, . " I I I ' " •
,
t
I
,
•
" 1 ,
I
I
I , I I I
I I I
I , I
""
I
I
I I I I I " I I , '" , I I "1 "'" " I I , I , "
, I
1
I' I
I
•
I
I " I' I I
, " I I I I , " I r I '" "" '" I
I
I I
,
I
" I
1
I
I
' I " " I I " "
"
lit"
I I " I ,
,
"
I
I
"
I , , , I
I
, I I I
I
,.,77777.77-------77777.,.-777777, . . ,777 •• ,.77-.7777 -0.00147
-0.00160
I , , ,
" " " "
• "
,t
•• "
I I I
, I " , , ., , , I I " "'" I' ""1' " , , I I I
,.1 ,r I
r'
t
r "'1 I I '" I I
I I I I
-t.!..!..!..!..!.~.!.!...!.i.!..!..!..!.!..!...!.!...!..!.t!.!.!...!..!.!...!..!.~.!..!..!.
o
0.5
1.0 Time,
Fig.5.39 Speed deviation
1.5 w~
in pu.
1" I I I '" I I , "1" I "'" I I ""1 '" '" "'" I
, I I • I I I
I I , I
I I " I
I
I I
I t
I I
r "
I I
.!..!.!...!.!..!..f!...!..!.
2.0
I I , I , I I , I I ,.". "'" I I I , , "'" '" I I
!~~~t
2.5
Response to a
step increase in Tm
10~'~
3.0000-; ;
,, I' I
,
,,,• 2.9000-; ,
, ,,
I
I I I
•• ,
I
•,
,
I
~~~~~~-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I I I
,
,
I
I'
t
t
,
I
,
I
1'1'
,
I
,
1'1
,
,
,
,
,
I
,
,
,
,
,
,
,
,
,
I'
,
I'
I
,
•
I
Itt'
t
••
t
2.~0-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ I J
I
o
3.5638-
I
I
0.5
1.0
Response to a
.,,• ,.,,• ,. - TTTT
5~~)
I
1.5
Time,s
2.0
2.5
step increase in E FD
---------~---------------------------------~------I' ,, , 3.4236 -- -
_
•,
,
TT
,
t
,
, I t
, t i t • I t t , t • t
• I t . I
, ,
I ,
, ,
, , •
.,
,,,, ...
~
~
i
>- ............
~
I , I
....
~
, ,
!-. .... .!. !. .................... _ •
~
_ : T•
+
r-' r-'
~
_
_
,,
, •
t
I
~
~
-
,
,
I I
I I
..-+'"-
-.. - - , ,
-•
. . . !-. .!. !.. ............. _ ............. !-. !. _ !- .!. .... .... .... ... .... .!. .... !- ._ !.., .~ :
, ,,
I I
I ~
, ....
~
......
....-
'"-
...-~
.......
~
.......
..-
~
...................
~
-..
.-
......
~
I I I
I I I I I I
~
......
•
............
_
,_
......
~
~
I I I
+
.....
~
............
I I ,
, , I , , I
~~~~~~-~~~~~~~~~~~~~~~~~
.....
"
3 • 0030 - +, ,
I I
I I '"
I
I I I 1'1
I I
I I
I
I ,
I ,
I I
I
,
,
,
,
I
f
I
• • • + : : : : : : : : : : : : : : : : I ,
r ,
I I I I I 1'1' , , , , I'
I I ,
I I I
, I ,
I I I "1 I I I
I I I "1 , I I
I I I I I "1'1 , I I I I
I'
,
I
t
: : : :
I I I 1'1 I I "
I I
t I
I
..-
f
~,
I
.. ---
I
~-
.... -
~~~~~~~~~~~~~~~~~~~~~~~~~~
I I
,
I
--~~-7177771717,,7--~--~·~-~ I ,
1'1
+ .......... ......, . . - ............ -
I I I I I ,
f
, ,
• I I I ' I I I f I I I "
, I I , I '"
I ,
I I I I I
.... --~~- . . . 7-,77177-,tt717- . . I
'"
I
I
I
I
1"
: : :
I r I I I'
t
I
I I
t
+
, ,, I I
I
to-
_ - ...
, I I , I I I
+
~
.....
t. • I • ,
..- - - -
~
I I 'I I' I I I I I' "
t
3. 2834 - .........
I I
I 'I
, , ,
-
,
I ,
I'
: : :
r ,
I
I I
I I I
I I I
I
t
I
,
: : : I I I
r
I I
I I ,
I ,
,
I I
I ,
I
I
I I
I I
: : : : :
I , I
I I I
I
I
I I I
, I I' f I
I' , I
I
I
I
I I
I I
I ,
I ,
I ,
I
I I
: : : : : : : : : :
, , I
I I I
I I
r
I I I
I , I
I , I
, f I
I I I
I I I
I , I
,
,
I
: :
, , I
I , I
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~!..~~~~~~~~~~~~~~~~~~ 1.
•
t I I It" 'I I " t "" I
,t " ,
,
I
I
I
I
'I I I I I
I'
'1'
I I I
I I I ".
I I I I 'I II II
I I I
I I " '" ' I "I
I f
I
I
I
I
I , , I,
I
I
I
(
I
I
,r
I I I
I
77~77777-777777-777-------~----I
t t , I • I
I.
I I I I 'I I I " I I
"1
,
I I I I I I I t 'I , ,
I
o
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
><:
1.6
0.4
1.0
0.2
Q = 0.0
Q =
0.0
- - - - - - - -- __
Q = 0. 01-_
0.0
-
1.2
0.4
(0.1,0.0)
I.
0.6 0.8
-0.05
-
-
--,
re = 0.0 xe = 0.4
0 .05
0.2
xe = 0.4
1.0
~.1
0.8 0.6 0.4
0.6
0.2 0. 1 0.2
0.4
0.6
0.8
Q
0.5
1.0
Reol Power, P
0.4
>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 <5 ~
E; .
Solution
This problem without numbers may be solved by sketching a phasor diagram . Indeed, considerable insight into learning how the control system functions is gained by this experience. The in itial operating condition is shown in the phasor diagram of Figure 7.3. Under the operating conditions specified, the output power per phase may be expressed in two ways : first in terms of the generator terminal conditions
P = VI cos 0 and second neglected,
In
(7.1 )
terms of the power angle, with saliency effects and stator resistance P
=
(EgVjX) sin
(7 .2)
<5
In our problem P and V are constants. Therefore, from (7 .1) / cos 0
= k,
(7.3)
where k, is a constant. Also from (7 .2)
Eg sin
<5
(7.4)
k2
where k 2 is a constant. E 9
VI I < ,
Fig . 7.3
1\
1
1
I
I
. . . . . . .( J 90°
Phasor diagram or the init ial condition .
235
Excitation Systems
I
l---k'i
i I I
I I
E
9
I
Fig .7.4
Phasor d iagram showing control constraints.
Figure 7.4 shows the phasor diagram of Figure 7.3, but with k, and k2 shown graphically. Thus as the excitation is increased, the tip of Eg is constrained to follow the dashed line of Figure 7.4, and the tip of I is similarly constrained to follow the vertical dashed line . We also must observe the physical law that requires that phasor IX and phasor TIie at right angles. Thus we construct the phasor diagram of Figure 7.5, which shows the "before and after" situation . We observe that the new equilibrium condition requires that (I) the torque angle is decreased, (2) the current is increased, and (3) the power factor is more lagging; but the output power and voltage are the same. By similar reasoning we can evaluate the results of decreasing the excitation and of changing the governor setting. These mental exercises are recommended to the student as both interesting and enlightening. E
9
Fig. 7.5
E' 9
Solution for increas ing Eg at constant P and V.
Note that in Example 7.1 we have studied the effect of going from one stable operating condition to another. We have ignored the transient period necessary to accomplish this change, with its associated problems-the speed of response, the nature of the transient (overdamped, underdamped , or critically damped), and the possibility of saturation at the higher value of Eg • These will be topics of concern in this chapter.
7.2
Control Configurations
We now consider the physical configuration of components used for excitation systems . Figure 7.6 shows in block form the arrangement of the physical components in
236
Chapter 7 Output voltage & current
Input torque
from
prime mover
Exciter power source
Fig. 7.6
Arrangement of excitation components.
any system. In many present-day systems the exciter is a de generator driven by either the steam turbine (on the same shaft as the generator) or an induction motor. An increasing number are solid-state systems consisting of some form of rectifier or thyristor system supplied from the ac bus or from an alternator-exciter. The voltage regulator is the intelligence of the system and controls the output of the exciter so that the generated voltage and reactive power change in the desired way. In earlier systems the "voltage regulator" was entirely manual. Thus the operator observed the terminal voltage and adjusted the field rheostat (the voltage regulator) until the desired output conditions were observed. In most modern systems the voltage regulator is a controller that senses the generator output voltage (and sometimes the current) then initiates corrective action by changing the exciter control in the desired direction. The speed of this device is of great interest in studying stability. Because of the high inductance in the generator field winding, it is difficult to make rapid changes in field current. This introduces considerable "lag" in the control function and is one of the major obstacles to be overcome in designing a regulating system. The auxiliary control illustrated in Figure 7.6 may include several added features. For example, damping is sometimes introduced to prevent overshoot. A comparator may be used to set a lower limit on excitation, especially at leading power factor operation, for prevention of instability due to very weak coupling across the air gap. Other auxiliary controls are sometimes desirable for feedback of speed, frequency, acceleration, or other data [I].
7.3
Typical Excitation Configurations
To further clarify the arrangement of components in typical excitation systems, we consider here several possible designs without detailed discussion.
7.3.1
Primitive systems
First we consider systems that can be classified in a general way as "slow response" systems. Figure 7.7 shows one arrangement consisting of a main exciter with manual or automatic control of the field. The "regulator" in this case detects the voltage level and includes a mechanical device to change the control rheostat resistance. One such directacting rheostatic device (the "Silverstat" regulator) is described in reference [2] and consists of a regulating coil that operates a plunger, which in turn acts on a row of spaced silver buttons to systematically short out sections of the rheostat. In application, the device is installed as shown in Figure 7.8. In operation, an increase in generator output voltage will cause an increase in de voltage from the rectifier. This will cause an increase in current through the regulator coil that mechanically operates a solenoid to insert exciter field resistance elements. This reduces excitation field flux and voltage, thereby lowering the field current in the generator field, hence lowering the generator
Excitation Systems
237
Commutator
Field
1--......Q~-.-b....~.:-1<w>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<\ncel
'~ and vq lto ge ~ Com p e nsoto r :
----fM;;gneti;;:
U ~.n~ ..
T-.. -'
J
L .. ILi~it e; ·l~"""" __..JJ
L..-l~ I.. "l ~-,~n'in~ .J L'~pl ifi ~~_1--L~:n.in9_:
1~_ . : ' : _ R~u.!.?~ E.0~':... Fig.7.10
.. J
.. .., : -.1
Sta b i Ii zer
L- .. _
r - 'Oth er
M M ~
=-= _J_-~
; 0- 52 r-...
~
I IOther IStat inputs ion
J~~~~~ory
Excitat ion control system with dc generator-commutator exciter . (@ IEEE. Reprinted from IEEE Trans .• vol. PAS-llS. Aug. 1969.) Example: General Electric type NA 143 amplidyne system 141.
Excitation Systems Commutator
~~~~\
Field br eak er
~d~,,-..,
fie[1d'
field \!:Y rheosta t (manual control) _
_
I ~~~-A-Sto~~-r= .~!j 'L J L __ powe r amplifier
I
I -- -I '--
,
Sllp n ngs
1
Fig .7.11
=pr ,
I
Type WMA
_ _
_
r egu lo t~r
rw:r~~ ~.:---- fVoTt~;l
tStObtll;e ru : _.& vo l toge,~ Compenso tor
-
. d'
t
•
L- _:. __ : ~~~ ~
~':'~~
---J
r-.--.~ I
[SigSI .
----;1 mixing '..
Li m i t er
~p~if~,er~n~ ~ng
,.. ' _ - --
r-- - -,
' a~
i
h
I t er
--'
Regulati ng s ysf en el ements
- -- - - -- - -t_ Oinptheutsr
'_ ..
se nsi ng .L - __ --.J
1
IL -
CT
}----..,...-- ----,.-
1:1:-, Exciter
90- 41 ) ) Regulat or tronsf±Ie
I
G~n
239
Regulato r
power
_
MG se t
- - - ----
1 Station
=J-=~ uX il iOry power
Excitation control system with dc generator-commutator exciter. (" I EEE . Reprinted from IEEE
Trans.. vol. PAS-88. Aug . 1969.) Example: Westinghouse type WMA Mag-A-Stat system [61 .
7.3 .2
Excitation control systems with dc generator-commutator exciters
Two systems of U.S. manufacture have de generator-commutator exciters . Both have amplifiers in the feedback path; one a rotating amplifier, the other a magnetic amplifier. Figure 7.10 [3] shows one such system that incorporates a rotating amplifier or amplidyne [5] in the exciter field circuit. This amplifier is used to force the exciter field in the desired direction and results in much faster response than with a self-excited machine acting unassisted . Another system with a similar exciter is that of Figure 7 .11 where the amplifier is a static magnetic amplifier deriving its power supply from a permanent-magnet generator-motor set. Often the frequency of this supply is increased to 420 Hz to increase the amplifier response . Note that the exciter in this system has two control fields. one for boost and one for buck corrections. A third field provides for self-excited manual operation when the amplifier is out of service.
7 .3.3
Excitation control systems with alternator-rectifier exciters
With the advent of solid-state technology and availability of reliable high-current rectifiers, another type of system became feasible. In this system the exciter is an ac generator, the output of which is rectified to provide the de current required by the generator field. The control circuitry for these units is also solid-state in most cases, and the overall response is quite fast [3]. An example of alternator-rectifier systems is shown in Figure 7.12. In this system the alternator output is rectified and connected to the generator field by means of slip rings . The alternator-exciter itself is shunt excited and is controlled by electronically adjusting the firing angle of thyristors (SCR's). This means of control can be very fast
240
7
Chapter
CT
Fiel d~ breok er }--''''-.---'~;--f'-----<>.......
C T& PT
P
~~~ere".;; -"=l~
I'
IL..=C!~C=tUilT~
I ,
C
~
.
an d de
I
vo lta ge
Slip d ngs
PT 's
--1
Regu lot on
I
' (manual'
,L,::X·.: ~
lQ
ac._}-_ ...,..; _ Ge
ra "l l\!ar
c antra ll
.
~~p litie.r_ j
, I
J .fI - - -- - - - 'Regulat ar powe r _ _ Regul at a r , - - _
_
Vol ta ge I buildup ; e l em~
~
trans fer
l
'--~--+'-l Tr;~'!'tar
Auxi lia ry
po wer for start-up
I
I
amplIfier
, I'
SCR regula tor
I
S't~bilize~
L
· L.......; I
I I
,.
-. -
,. . . .'
r-ifec iifier . .,
L-c.r{~irt
..
'
rR~fere";; ~
,
L~e.n~
vo ltage
I
~·_ t
":
, "
i----:.,~~';,'
.,
rr : - _ . . :--: Compe nsa tor '
Limi ter
:
-~_
.
~ ._-_._-_. --- ----____ _. J
L- . _ ~
L
..I
.. _ _ . . J
Excitation control system with alternator-rectifier exciter using stationary nonconlrolled rectifiers . (0<' IE EE . Reprinted from IEEE Trans .• vol. I'AS-88 . Aug . 1969.) Example: General Electric Alterrcx excitation system [71.
Fig. 7.12
since the firing angle can be adjusted very quickly compared to the other time constants involved. Another example of an alternator-rectifier system is shown in Figure 7.13 . This system is unique in that it is brush less; i.e., there is no need for slip rings since the alternator-exciter and diode rectifiers are rotating with the shaft. The system incorporates a pilot permanent magnet generator (labeled PMG in Figure 7.13) with a permanent magnet field to supply the (stationary) field for the (rotating) alternator-exciter. Thus all coupling between stationary and rotating components is electromagnetic. Note, however, that it is im possible to meter any of the generator field quantities directly since these components are all moving with the rotor and no slip rings are used . Rot ating el ement>
[
---8 !- -~ ! 1
1
5
CT
ac Gen
PT',
\.. cilQt ion
Jtbreak ~
Exci tat ion pow er
Man ua l
1 I ,--
c an lT~\
fM'- .!
~
T
ru': .r.
Lr;.:~r;,:JJ --. . __
-. -- -
I
Bo,e
o diu ste r
!
I
_ _ _ _ _ Rt~~~~:r Regulator power
Fig .7.13
Sta bi li zer 1_.'-=r J
IL .L --- ,-
I
I
I SI.9f'!O.
" .
_ W_TA_,: _ _g u_'_ a t_ar_ _ - - - -
IRe"f~rence I
I
- --,,
-
" - " -'
i
& vo ltage :..J, Compensat ar ' I_,.e.nsi ng
\-
i I
;~i;~r '~ .
. . - - - .-
-
I
J _ .._ _..
-.-...
l lml.ter
,enslng
.:
,-.J .
... ' -
===\
Other • t sensing L .. _ _'. •
-- - -
~~~~~~.] I
-
-
-
-
--'
Regulating s ys tern e lements _
I~
Other
i nputs
Excitation control system with allernator-rectifier exciter employing rotat ing rectifiers, Reprinted from IE£E Trans ., vol. PAS-88, Aug . 1969.) Example: Westinghouse type WTA Brushless excitation system [8,91.
(@ IEEE.
Excitation Systems
241
Exciter
f iel d breake r
CT PT' ,
j 'j ~ '!:rr- · i -' t- fRei~r~-1
LJ~~~F.
I
Auxili a ry powe r for sta rt -u p
':L~~fta~e : 5_e~i ng..l
II L
I
- SCR- -
excite r regul ato r
SCR
r egu la t or
F ig.7 .14
- - _ . _ -- - -
"
r ' :
b
•
•
-l :tr~i~t _J
I
IR;;cmr;r
.-_ .-I :
.
_.-_ ..
I Ref erence an d v~ ltage r---f(:cm pensct ar : , e nSlng L-' -r-- -
L ..---- J r-·--''----·CJ Limiter !
,en Slng
.
' L!C)t~ ··I. L.._ . F6ther input> L ____ . . I~n~I.rl2...._i - ---- - -
Excitation control system with alt ernato r-SCR exciter system . (It.. IEEE . Reprinted from IEEE Trans .. vol. PAS·88. Aug. 1969.) Example: General Electric Althyrex excitation system
IIIJ.
The response of systems with alternator-rectifier exciters is improved by designing the alternator for operation at frequencies higher than that of the main generator . Recent systems have used 420-Hz and 300-Hz alternators for this reason and report excellent response characteristics [8,10).
7.3.4
Excitation control systems with alternator-SCRexciter systems
Another important development in excitation systems has been the alternator-SCR design shown in Figure 7.14 [3). In th is system the alternator excitation is supplied diLi nea r reactor
Fiel d
Fie ld
rv~i tage -i rec tifie r : buildup [,: lemen tsJ
* ....
;0:. . . . . .
"T'-
---
PI's
Aux iliar y powe r for
sto rt-up
IV~ l tage
L I
SC R
re gu la tor
co nt ro l ( ma n ua l
1.
:~n t ra IL_
I
l
,
I,
Fig .7 ,15
---- -~
Excitation control system with compound-rectifier exciter. (© IEEE. Reprinted from IEEE Trans .. vol. PAS-88 . Aug . 1969.) Example: General Electric SCTP static excitation system
(12.131·
242
Chapter 7
rectly from an SCR system with an alternator source . Hence it is only necessary to adjust the SCR firing angle to change the excitation level, and this involves essentially no time delay. This requires a somewhat larger alternator-exciter than would otherwise be necessary since it must have a rating capable of continuous operation at ceiling voltage . In slower systems, ceiling voltage is reached after a delay, and sustained operation at that level is unlikely . 7.3.5
Excitation control systems with compound-rectifier exciter systems
The next classification of exciter systems is referred to as a "compound-rectifier" exciter , of which the system shown in Figure 7.15 is an example [3). This system can be viewed as a form of self-excitation of the main ac generator. Note that the exciter input comes from the generator electrical output terminals, not from the shaft as in previous examples. This electrical feedback is controlled by saturable reactors, the control for which is arranged to use both ac output and exciter values as intelligence sources . The system is entirely static, and this feature is important. Although originally designed for use on smaller units [12, 13), this same principle may be applied to large units as well. Self-excited units have the inherent disadvantage that the ac output voltage is low at the same time the exciter is attempting to correct the low voltage . This may be partially compensated for by using output current as well as voltage in the control scheme so that (during faults, for example) feedback is still sufficient to effect adequate control. Such is the case in the unit shown in Figure 7.15. 7.3.6
Excitation control system with compound-rectifier exciter plus potentialsource-rectifier exciter
A variation of the compound-rectifier scheme is one in which a second rectified output is added to the self-excited feedback to achieve additional control of excitation .
Auxilia ry
r Vol t~g;j buil d up
power in put ..... 1
for sta rt-up
L ~ lemen!s..J Excitation p o wer
I~
CT
I~
,.a;:~~:a
rings
Powe r rec ti fi e r r-- ~
Exci ta tion
= PT's
powe r
t~6~~rd~~r
L....,_--J
Exci ta ti on ( Tri n istot breake r powe r ::.mpli fie r
I L=1!c u!!!rJ I'
, I Ga;;-l ' Iii
---.
if:
I~ ~dJus ter ~r= - (ma nua l !: ~ '- --l I--- ----J control ) I : Stabi lizer -i Regulato r ~se
I
I!L . -
1-------+1. t rans fer
t
-S~g;;i ml x rng
Regula tor --,---,-p ower
rR; ie~;-l &
L_~ens tng . -~
t---.
re gu la tor
I
I r--:..r:.-·,: '' ---- ··1 ---_=_1_1-:- - - _ •
' - - - - - ----"---loiL Fig . 7.16
se nsi ng
l
r?;~ tifiel
I r~~~ '
I
O th e r
sensi ng
C-----l
'-T _ .
:-' Compe nsa tor .
! - -Limite r '-
L~~P l i fi '~ _ ~L
WTA
va l~ag.
..J
.J~----' 1
-~
V~ it;;g;l adj uste r : L._---J
I
I
I I
~ O the r
:--'1~------- -- inputs _ __I
Excitation control system with compound -rectifier exciter plus potential-source-rectifier exciter. (,ll IEEE. Reprinted from IEEE Trans., vol. PAS-88, Aug. 1969.) Example: Westinghouse type WT A-peV static excitation system [14J.
243
Excitation Systems
This scheme is depicted in Figure 7.16 [3J . Again the basic self-excited main generator scheme is evident. Here, however , the voltage regulator controls a second rectifier system (called the " Trinistat power amplifier" in Figure 7.16) to achieve the desired excitation control. Note that the system is entirely static and can be inherently very fast, the only time constants being those of the reactor and the regulator. 7.3.7
Excitation control systems with potential-source-rectifler exciter
The final category of excitation systems is the self-excited main generator where the rectification is done by means of SCR 's rather than diodes . Two such systems are shown in Figure 7.17 and Figure 7.18 [3J . Both circuits have static voltage regulators that use potential, current, and excitation levels to generate a control signal by which the SCR gating may be controlled. This type of control is very fast since there is no time delay in shifting the firing angle of the SC R's . 7.4
Excitation Control System Definitions
Most of the foregoing excitation system configurations are described in reference [3], which also gives definitions of the control system quantities of interest in this application . Only the most important of these are reviewed here. Other definitions, including those referred to by number here, are stated in Appendix E. All excitation control systems may be visualized as automatic control systems with feed forward and feedback elements as shown in Figure 7.19. Viewed in this way, the excitation control systems discussed in the preceding section may be arranged in a general way, as indicated in Figure 7.20 and further described in Table 7.1. Note that the synchronous machine is considered a. part of the "excitation control system," but the control elements themselves are referred to simply as the "excitation system ." The type of transfer function belonging in each block of Figure 7.20 is discussed in reference [lSJ. The reference to systems of Type I, Type 3, etc., in the last column of Table 7.1 also refers to system types defined in that reference. This will be discussed in greater detail in Section 7.9. Our present concern is to learn the general configuration Auxil ia ry power f\k it~gel i n pu t~ build up : for sto rt -o p ~~_e~
Fi eld b reo ker
-----
o--~Br---<
Slip ,j ngs
CT
(
Exc itation pow er
po te ntial t ra nsfo rme r
~~~ I rectifier
I ...rPower
1-
--
- ---,
, Gate ci rcuitr y : I
L:m~I~1--
-..::J
R~~~, ~ ;a~ ----r-; -- _,
f
l:Jru ns fer
Wr;-: _.--
I ' - - ---1 I'
~ -- .
J ~ ----- -- II 1:
( ma nua l co ntro l)
I :
SCR re g u la tor
Fig . 7.17
_ f"R'e-ctifi e r !
L:~~~e~_J n_ji;!t".~_J
r"Reference I ' I I a nd • Tronstst or: dC vo ttag e , ~p~lf2!!....J I
Sen!I~9~ de Regulator
PT' s
Regu la tor power
_·W fR;f~ renc e
TranSIStor . , an d ,ofi ..-.-I cmpu tr er \' 'vot~~ :
I I
r-
__ --.,
....-.! Compe nsotor l- _. .----J
I
~:'~I~
I
~i;nrt.;;:---t ~ • 'sens ing : , -- --
I :
_
'--- -~
L : iSthe;"l
I O the r i....-:; _ se!!slna........:----r--_ in puts
J
Excitation control system with potential-source-rectifier exciter. «<' IEEE. Reprinted from IEEE Trans., vol . PAS-88 . Aug . 1969.) Example: Genera l Electric SCR static excitation system [141 .
244
Chapter 7
CT
=
VJ
Exc ! to t i on
pow er po ten t ia l tr a nsform er
Tri n h ta t powe r amplifier
-8'
,
J
G ot ~
~cui !.2.......J ~
I
Regu la tor po wer
IBo5~
~
od j u.--...-J te r L.-
Regulat or
I
trcru fe r
','- -
,
( manual
conlro l)
-
I
-Ir lRe"i;r;n~ r- -,--, S tObi l i z~ 1 & vo ltoge l-t c
I
,L.e: .. , _ _ ,_ _ ,,_n~
lSi ~n~ 1
rci ~ i~er
t
't
-
I
;----t se ns ing : L:!'lifi~r 'L ,.--- ----'
WTA
mi x ing
T
re gu l a tor
-+-';r;;;,;tiiie";l c~~~~t : L __ --.J
:;l
IV Vo-' t-og -e-',
om_p_e-+_~:_ _.o_---l o_r_,+-_!_O_ j Us_t er L~d...J --';
, L.. __ -.J,
'--- - f - ':
L..-
PT' .
--
r;~h~ ;j
: se nsing
L-. _.
::
Regulati ng system elements
---I
I'
, -+-O:-th_er
'------- _ _ _ _ _ _ _ __ J I
I
i npuh
Fig .7 .I B Excitatio n co ntro l system with potenti al-sour ce-rect ifier exciter. C'" IEEE. Reprinted from II:"EE Tran s.. vo l. PAS -88. Aug . 1969.) Exa mple: West ingh ou se type WTA-Trini st at excitat ion sys tem .
of modern excitation control systems and to become famili ar with the language used in describing them . 7.4.1
Voltage response ratio
A n im p o rta nt definition used in des cribing excitati on control systems is that of the
response ratio defined in Appendix E, DeL 3.15 --3.19 . This is a rough measure of how
fast the exciter open circuit voltage will rise in 0 .5 s if the excitation control is ad justed suddenly in the maximum increase direction . In other words , the voltage reference in Figure 7.20 is a step input of sufficient magnitude to drive the exciter voltage to its ceiling (DeL 3.03) value with the exciter operating under no-load conditions. Figure 7.21 shows a typical response of such a system where the voltage vF starts at the rated load field voltage (DeL 3.21) that is the value of U F, which will produce rated Refe re nce
Actua tin g
i n pu t
si gno l
~
signa l
....
---'-~...,.---< : }-~':-7,-:-::,.--...j
( Der 3 .3 4)
., Dh e c t! v co ntro l led vo riob le ( Def 3 . 4 1)
( Der 3.2 8)
Fee dback sig na l (Def 3. 30)
Fig. 7.19
Essential elemen ts of an a uto ma tic feedback cont rol system (Def. 1.02). C<' I EE E. Reprinted from IEEE Tram .. vol, PAS -88, Aug . 1969.) N ot e: In excitat ion co ntro l system usage the actuating signa l is co mmo nly ca lled the error signal (D ef. 3.29). (See Appendix E for definitions.)
Magnetic. thyristor Thyristor
Thyristor
Compoundrectifier exciter
Compoundrectifier exciter plus potential-source rectifier exciter
Potential-source rectifier (controlled) exciter Exciter output voltage regulator. Compensated input to power amplifier
Compensated input to power amplifier
Self-excited
Exciter output voltage regulator
,
See note 6
Signal modifiers
6
MG set. synchronous machine shaft
Synchronous machine terminals
Synchronous machine terminals
Synchronous machine terminals
Alternator output
Synchronous machine terminals
Synchronous machine terminals
Synchronous machine terminals
Synchronous machine shaft
Synchronous Synchronous machine machine shaft shaft. MG set. alternator output
MG set See note 8
Excitation system stabilizers
Exciter
Power sources Regulator
8
7
Figs. 7.17 & 7.18
Fig.7.16
Fig.7.15
Fig. 7.14
Figs. 7.12 & 7.13
Figs. 7.10 & 7.11
System diagram reference
3
I
IS
Similar to 3
I
I
Type of computer representation*
'C)
Source: IEEE. Reprinted from IEEE Trans .. vol. PAS-88. Aug. 1969. 2. Primary detecting element and reference input: can consist of many types of circuits on any system including differential amplifier. amplifier-turn comparison. intersecting impedance. and bridge circuits. 3. Preamplifier: Consists of all types but on newer systems is usually a solid-state amplifier. 6. Signal modifiers: (-A) Auxiliary inputs-i-reactive and active current compensators: system stabilizing signals proportional to power. frequency. speed. etc. (8) Limiters--maximum excitation. minimum excitation. maximum V 1Hz. 8. Excitation control system stabilizers: can consist of all types from series lead-lag to rate feedback around any element or group of elements of the system. *IEEE committee report (IS].
Thyristor
Alternatorrectifier (controlled) exciter
Com pensa ted input to power arnplifier. Selfexcited field voltage regulator
Self-excited or separately excited exciter
Manual control
Power amplifier
See note Rotating. 3 magnetic. thyristor
5
Components Commonly Used in Excitation Control Systems
4
Rotating. thyristor
See note 2
2 3 Primary detecting Preelement & amplifier reference input
Alternatorrectifier exciter
dc Generatorcommutator exciter
Type of exciter
I
Table 7.1.
m
x
~
In
tV
V'
3
(l)
-cV'
o»
s='
~.
n
246
Chapter
7
I
Input
I
I
I Powe r !iystem
Inpu t>
- --
- - --
- - Regulator (De f 2 .1 2)
- - - - -t__
Excitati on system (Def 1.03) -
-
- - --
-
Exciter
=:r
----------
- - -- - Excitation c ontTol sy>tem (Def
Sync hronous
mach ine
1. 04 )---------------~
N ote : The numeral s on this diog ran refer to the columns in Tobie 7. 1 ..
Fig.7 .20
Excit ation cont rol systems. (-<) IEEE . Reprinted from IE EE Trans ., vol , PAS-88 , Aug. 1969.) Note: The numerals on this diagram refer to the columns in Table 7.1. (See Appendix E for definitions .)
generator voltage under nameplate loading. Then, responding to a step change in the reference, the open-circuited field is forced at the maximum rate to ceiling along the curve abo Since the response is nonlinear, the response ratio is defined in terms of the area under the curve ab for exactly 0.5 s. We can easily approximate this area by a straight line ac and compute Response ratio = cdl(Oa)(O.5) pu VIs (7.5) Kimbark [16) points out that since the exciter feeds a highly inductive load (the generator field), the voltage across the load is approximately v = k d¢ldt. Then in a short time At the total flux change is A¢
= -I J~I vdt k
0
area under buildup curve
To ce i li ng
vo lto g e
I >"-
a
5 o u,
w
o Fig. 7.21
1
//
I
Rated loa d f iel d vo lta ge
I I
~-l-----id Time, s
Defini tion or a volt age response ratio .
(7.6)
Excitation Systems 4.0
System attaining 95 % ce iling in 0 .1 s & having linear response
247
-+//
M
o
M
~ 3.0
E ~
~
.=
0>
b'..-'---S)"tem atta ining 95 % ceiling in 0.1 s & having on exponential
3 2.0
response
1.0
V---Synchr onaus machine rated lood field vol tage (Def 3.21 )
o
1
2
3
4 5 6 7 8 9 10 11 12 Response Ratio (Def 3.18)
Fig. 7.22 Exciter ceiling voltage as a function of response ratio for a high initial response excitation system. « IEEE. Repr inted from IEEE Trans.. vol. PAS-88. Aug. 1969.)
The time D.l = 0 .5 was chosen because this is about the time interval of older "quickresponse" regulators between the recognition of a step change in the output voltage and the shorting of field rheostat elements. Buildup rather than build-down is used because there is usually more interest in the response to a drop in terminal voltage, such as a fault condition. In dynamic operation where the interest is in small, fast changes, build-down may be equally important. Equation (7 .5) is an adequate definition if the voltage response is rather slow, such as the one shown in Figure 7.21. It has been recognized for some time, however, that modern fast systems may reach ceiling in 0.1 s or less, and extending the triangle aed out to 0.5 s is almost meaningless. This is discussed in reference [3), and a new definition is introduced (Def', 1.05) that replaces the 0.5-s interval Oe in Figure 7.21 by an interval Oe = 0.1 s for "systems having an excitat ion voltage response time of 0.1 s or less" [the voltage response time (Def. 3.16) is the time required to reach 95% of ceiling). A comparison of three systems, each attaining 95~~ ceiling voltage in 0.1 s, is given in Figure 7.22 (3) and shows how close the O.I- s response is to the ideal system, a step function .
7.4.2
Exciter yoltage ratings
Some additional comments are in order concerning certain of the excitation voltage definitions . First, it may be helpful to state certain numerical values of exciter ratings offered by the manufacturers (see [2) for a discussion of exciter ratings) . Briefly, exciters are usually rated at 125 Y for small generators, say 10 MY A and below. Larger units usually have 250- Y exciters, say up to 100 MY A; with still larger machines being equipped with 350-Y, 375-Y , or 500-Y exciters. The voltage rating and the ceiling voltage are both important in considering the speed of response [I, )7). Reference [I) tabulates the pattern of ceiling voltages for various response characteristics in Table 7.2, which shows the improved response for higher ceiling voltage ratings (and the lower ceiling voltage for solid-state exciters). It is reasonable that an exciter with a high ceiling voltage will build up to a particular volt-
248
Chapter 7
age level faster than a similar exciter with a lower ceiling voltage simply because it saturates at a higher value. This is an important consideration in comparing types and ratings of both conventional and solid state exciters as shown in Table 7.2. Table 7.2. Typical Ceiling Voltages for Various Exciter Response Ratios Response ratio
Per unit ceiling voltage conventional exciters"
0.5 1.0 1.5 2.0 4.0
1.25-1.35 1.40-1.50 1.55-1.65 l. 70-1.80
SCR exciters
1.20 1.20-1.25 1.30-1.40 1.45-1.55 2.00-2.10
• Based on rated exciter voltage.
In adopting a pu system for the exciter, there is no obvious choice as to what base voltage to use. Some possibilities are (also see [2]): (A) exciter rated voltage, (8) rated load field voltage, (C) rated air-gap voltage (the voltage necessary to produce rated voltage on the air gap line of the main machine in the case of a dc generator exciter), and (D) no-load field voltage. The IEEE [3] recommends the use of system B, the rated load field voltage. Consider, as an example, an exciter rated at 250 V. For this rating some typical values of other defined voltages are given in Figure 7.23. The pu system A 300 1.:..2 0_ -
200
-
1:~
_-
.3;.33_ Typical ceiling
!..0.Q. _ _
1:..25_ _ _ J...zS_ Rated voltage
£·~
LqQ __
2.22
Rated lood fie Id voltage
0.45
1.00
Rated air gop voltage
.l!
'0
>
100 0.36
90
- -
-
-
- -
_ _-.a.
O~
A
Fig. 7.23
8 pu System
- -
-.a..._ _
c
Per unit voltages for a 250- V exciter.
of Figure 7.23 has little merit and is seldom used. System 8 is often used. System C is often convenient since, with rated air gap voltage as a base, pu exciter voltage, pu field current, and pu synchronous internal voltage are all equal under steady-state conditions with no saturation. System D is not illustrated in Figure 7.23 and is seldom used.
7.4.3
Other specifications
Excitation control system response should be compared against a suitable criterion of performance if the system is to be judged or graded. System performance could be measured under any number of forcing conditions. It is generally agreed that the quantity of primary interest is the exciter voltage-time characteristic in response to a step change in the generated voltage of from 10 to 20% [18,19). The problem is how to state in words the various possible slopes, delays, overshoots, damping, and the like. One useful description, often used in control system specification, is that based on the
249
Excitation Systems
Time , s
Fig . 7.24
Time do ma in specificatio ns [221 .
curve shown in Figure 7.24 . Here the curve is the response to a step change in one of the system variables, such as the terminal voltage. This response, based on that of a second-order system, is a reasonable one on which to base time domain specifications since many systems tend to exhibit two "least-damped" poles that give a response of this general shape at so me value of gain [20,21) . Three quantities describe this response: the overshoot, the rise time, and the settling time . The overshoot is the amount that the respon se exceeds the steady-state responsein Figure 7.24, at pu. The rise time is the time for the response to rise from 10 to 90% of the stead y-state response. The settling time is the time required for the response to a step function to stay with in a certain percentage of its final value. Sometimes it is given as the time requ ired to arrive at the final va lue after first o vershooting this va lue . The first defin ition is preferred . The damping ratio is that value for a second-order system defined by in the expression
r
(7.7) and is related to the values a, and a2 of two successi ve overshoots [231 . The natural resonant frequency W n is also of interest and may be given as a specification . In the case of the second-order system (7.7), the response to a step change of a driving variable is
c(t )
=
1 - e-r"n1Icosw,t
+ [r/(l - r2») sin w, t l
(7.8)
where
r
(7 .9)
r r
When = 0 , the system is oscillatory; when = 0 .7, it has very little overshoot (about 5%) . Critical damping is said to occur when = 1.0. In dealing with an exciter being forced to ceiling due to a step change in the voltage regulator control, the system is often "overdarnped"; i.e. , > 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 -
(7.41)
This linearized result does not include saturation or other nonlinearities, but does include the major time delay in the system. An ac exciter designed for operation at a few hundred Hz could have a very reasonable iTdo' much lower than that of the large 60-Hz generator that is being controlled. 1.6.4
Solid-state exciters
Modern solid-state exciters, such as the SCR exciter of Figure 7.14, can go to ceiling without any appreciable delay. In systems of this type a small delay may be required for the amplifiers and other circuits involved. The field voltage may then be assumed to depend only on this delay. One way to solve this system is to assume that vF changes linearly to ceiling in a given time delay of t d s, where td may be very small. This is nearly the same as permitting a step change in U F • For such fast systems the time constants are so much smaller than others involved in the system that assuming a step change in V F should be fairly accurate. 1.6.5
Buildup of a loaded de exciter
Up to this point we have considered the response characteristics of unloaded exciters, i.e., with iF = O. If the exciter is loaded, the load current will affect the terminal voltage of the exciter vF by an amount depending upon the internal impedance of the exciter. In modern solid-state circuits this effect will usually be small, amounting to
Excitation Systems
267
essentially a small series iFR drop . In rotating dc machines the effect is greater, since in addition to the iFR drop there is also the brush drop , the drop due to armature reaction, and the drop due to armature inductance. (Dahl [35) provides an exhaustive treatment of this subject and Kim bark [16) also has an excellent analysis.) We can analyze the effect of load current in a dc machine as follows. First, we recognize that the armature inductance is small, and at the relatively slow rate of buildup to be experienced this voltage drop is negligible. Furthermore, if the machine has interpoles, we may neglect demagnetizing armature reaction . However, we do have to estimate the effect of cross-magnetizing armature reaction, which causes a net decrease in the air gap flux. Thus, the net effect of load is in the resistance drop (including brush drop) and in the decrease in flux due to cross-magnetizing armature reaction. To facilitate analysis, we assume the load current iF has a constant value. This means the iFR drop is constant, and the armature reaction effect depends on the value of current in the field, designated i in our notation. The combined effect is determined most easily by test, a typical result of which is shown in Figure 7.39. To the load
::
0
> u,
>
oJ0>
E
0
> ~
'u
.s
~
~ ~
Excite r Field Current , i , amperes
Fig.7 .39
No-load and load sat ur ation c urves. (Reprinted by permission from Power System Stability, vol. 3, by E. W . Kimbark . '" Wiley , 1956.)
saturation curve is added the resistance drop to obtain a fictitious curve designated "distortion curve." This curve shows the voltage generated by air gap flux at this value of iF as a function of i, and it differs from the no-load saturation curve by an amount due to armature reaction . The magnitude of this difference is greatest near the knee of the curve . Kimbark (16) treats this subject thoroughly and is recommended to the interested reader . We will ignore the loading effect in our analysis in the interest of finding a reasonable solution that is a fair representation of the physical device. As in all engineering problems, certain complications must be ignored if the solution is to be manageable. 7.6.6
Normalization of Exciter Equations
The exciter equations in this book are normalized on the basis of rated air gap voltage, i.e., exciter voltage that produces rated no-load terminal voltage with no saturation. This is the pu system designated as C in Figure 7.23. Thus at no load and with no saturation, Em = 1.0 pu corresponds to V, = 1.0 pu .
268
Chapter 7
The slip ring voltage corresponding to 1.0 pu EFD is not the same base voltage as that chosen for the field circuit in normalizing the synchronous machine. From (4.55) we have V FB
=
VBIB/I FB
=
SB/IFB
V
This base voltage is usually a very large number (163 kV in Example 4.1, for example). The base voltage for EFD , on the other hand, would be on the order of 100 V or so. Simply stated, the exciter base voltage and the synchronous machine base for the field voltage differ, and a change of base between the two quantities is required. The required relationship is given by (4.59), which can be written as
EFD
=
(L A D / v'3rF )
VF
pu
EFD = (w B k M F/ v'3rF ) vF V
(7.42)
Thus any exciter equation may be divided through by VFB to obtain an equation in v Fu and then multiplied by L A D / v'3r F to convert to an equation in EFDu • For example, for the dc generator exciter we have an equation of the form TEV F = !(v F ) V. Dividing through by VF8 we have the pu equation TEV Fu = !(v Fu ) . Multiplying by L A D /V3rF , we write the exciter equation TEE FDu = j(EFDu ) . It is necessary, of course, to always maintain the "gain constant" V3rF/ LAD between the exciter EFD output and the v F input to the synchronous machine. This constant is the change of base needed to connect the pu equations of the two machines.
7.7
Excitation System Response
The response of the exciter alone does not determine the overall excitation system response. As noted in Figure 7.20, the excitation system includes not only the exciter but the voltage regulator as well. The purpose of this section is to compute the response of typical systems, including the voltage regulators. This will give us a feel for the equations that describe these systems and will illustrate the way a mathematical model is constructed.
7.7.1
Noncontinuously regulated systems
Early designs of voltage regulating schemes, many of which are still in service, used an electromechanical means of changing the exciter field rheostat to cause the desired change in excitation. A typical scheme is shown in Figure 7.40, which may be explained as follows. Any given level of terminal voltage will, after rectification, result in a given voltage V c across the regulating coil and a given coil current i.. This current flowing in the regulating coil exerts a pull on the plunger that works against the spring K and dashpot B. Thus, depending on the reference screw setting, the arm attached to the plunger will find a new position x for each voltage V,. High values of V, will increase the coil voltage V c and pull the arm to the right, reducing x, etc. Note that the reference is the mechanical setting of the reference screw. Now imagine a gradual increase in V, that pulls the arm slowly to the right, reducing x until the lower contact L is made. This causes current to flow in the coil L, closing the rheostat motor contact and moving the rheostat in the direction to increase R H • This, as we have seen, will reduce V,. Note that there is no corrective action at all until a contact is closed. This constitutes an intentional dead zone in which no control action is taken. Once control action is begun, the rheostat setting will change at an assumed constant rate until the maximum or minimum setting is reached. Mathematically, we can describe this action as follows. From (7.16) we have, for the separately excited arrangement,
269
Excita tion Systems
t Ol
OR
I
;-IL I,Reference set
I I
~~~~~~
Regula ting
ca il
400--Quic k raise &. lower contacts ~ Ti m e del ayed rais e & lower contacts " - Operating coils
Fig. 7.40
A non cont inuou s regu lat or fo r a sepa ra te ly excit ed system. The scheme illustr ated is a simp lified sketch sim ilar to th e Wes tinghouse type BJ system [2J.
(7.43) and in th is case the regulating is acco m plished by a change in R . But R ch anges as a function of time whenever the arm position x is greater than so me threshold va lue Kr This condition is shown in F igure 7.41 where the choice of curve depends on the magnitude of x be ing gre ater than the dead zone ±Kx ' Note that any change in x from the equ ilibrium position is a measure of the error in the te rm inal vo ltage magnitude. This control action is design ated the " raise-lo wer mode" of operation. It results in a slow excitation change, responding to a change in V, large enough to exceed the threshold Kx • where the rheostat motor stea dily changes the rheostat sett ing. A block diagram of this control action is shown in Figure 7.42 . The balanced beam responds to an accelerat ing force
Fa = K(x o + {,) - Fe = Mx + Bi + K x
(7.44)
where X o is the reference po siti on ; {, is the un stressed length of the spring; Fe is the plunger force; and M, B, and K a re the mass, damping, and spring constants respectively. If the beam mass is negl igibl e, the right side of (7.44) can be simplified . In operat ion the beam pos ition x is ch anged co ntinuo usly in response to va riatio ns R
c:
H
,,'o e
.~
s E
2
12
R
ma x
RO R min
OL-- - - - - - - - - - - Time, s
Fig. 7.4 1 R H versus I for the cond ition
Ix I
> Kx >
o.
270
Chapter 7 Balanced
beam
Fig . 7.42
Block diagram of the rai se-lower control mode.
in V" Any change in V, large enough to cause I x I ~ K, results in the rheostat motor changing the selling of R H • As the rheostat is reset, the position x returns to the threshold region I x I < K, and the motor stops, leaving R H at the value finally reached . At any instant the total resistance R is given by R QR R QR
+ Ro + Ro +
KMt
(raise)
KMt
(lower)
(7045)
Thus the exact R depends on the integration time and on the direction of rotation of the rheostat motor. In (7 .45) and Figure 7AI, Ro is the value of RH retained following the last integration . This value is constrained by the physical size of the rheostat so that for any time t, R min < (R o ± KMt) < R max • The foregoing d iscussion pertains to the raise-lower mode only. Referring again to Figure 7040, a second possible mode of operation is recognized . If the x deflection is large enough to make the QL or QR contacts, the fixed field resistors R QL or R QR are switched into or out of the field respectively, initiating a quick response in the exciter. This control scheme is shown in Figure 7043 as an added quick control mode to the original controller. The quick raise-lower mode is initiated whenever I x I > 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 ~<, the applied voltage. If we choose the nonlinear elements carefully, the operation in the neighborhood of VREF is essentially linear; i.e., a deviation VA above or below VREF results in a change i A in the total current, which is also displaced equally above and below i REF • Note that the nonlinear resistance shown is quite linear in this critical region. Thus we may write for a voltage deviation VA' (7.54) where k L > k N • Combining (7 .54) and (7.53), we compute
V. But for a deviation
VA'
Vd<
= -(k L
VREF
+
VA'
-
kN)V A = -k VA
(7 .55)
which may be incorporated into (7.55) to write (7.56)
We note that (7 .56) has the same block diagram representation as the difference amplifier shown in Figure 7.48(b), where we set T = 0 for the passive circuit.
Difference amplifier
Fig . 7.48
+
Electronic difference amplifier as a comparator: (a) circuit connection, (b) block diagram .
274
Chapter 7
---... ide
-f i
e
0
~
+ V
-l
Fig.7.49
Input to
amplifier
e
Nonl inear bridge co mpariso n circuit.
A natural question to ask at this point is, What circuit element constitutes the voltage reference? Note that no external reference voltage is applied . A closer study of Figure 7.50 will reveal that the linear resistance R L is a convenient reference and that two identical gang-operated potentiometers in the bridge circuit would provide a convenient means of setting the reference voltage. The nonlinear bridge circuit has the obvious advantage of being simple and entirely passive. If nonlinear resistances of appropriate curvature are readily available, this circuit makes an inexpensive comparator that should have long life without component aging. The amplifier. The amplifier portion of the excitation system may be a rotating amplifier, a magnetic ampl ifier, or conceivably an electronic amplifier. In an y case we will assume linear voltage amplification K A with time constant T A , or VR
=
K A v./( I
+
(7.57)
T AS)
As with any amplifier a saturation value must be specified, such as VRmin < VR < VRmu These conditions are both shown in the block diagram of Figure 7.51.
'
The exciter. The exciter output voltage is a function of the regulator voltage as derived in (7.50) and with block diagram representation as shown in Figure 7.45 . The major difference between that case and this is in the definition of the constant K E • Since the exciter is a boost-buck system, we can write the normal ized equ at ion E FD = (VR - EFDSE)/(K E
+
TES)
(7.58)
where K E = RG -
The generator.
1
(7 .59)
The generator voltage response to a change in vF was examined in
Fig. 7.50 T he u versu s i characteristics for the nonl inear bridge .
275
Excitation Systems V
e
Fig. 7.51
Block diagram or the regulator amplifier.
Chapter 5. Looking at the problem heuristically , we would expect the generator to respond nearly as a linear amplifier with time constant T;O when unloaded and T; when shorted, with the actual time constant being load dependent and between these two extremes. Let us designate this value as T G and the gain as KG to write, neglecting saturation, (7.60) In the region where linear operation may be assumed, there is no need to consider saturation of the generator since its output is not undergoing large changes . If saturation must be included, it could be done by employing the same technique as used for the exciter, where a saturation function SG would be defined as in Figure 7.44.
Example 7.7 I. Construct the block diagram of the system described in Section 7.7.1 and compute
the system transfer function . 2. Find the open-loop transfer function for the case where T" TE
= 0.1 = 0.5
TG TR
1.0 0.05
KE K" KG
= -0.05 = 40 = 1.0
3. Sketch a root locus for this system and discuss the problem of making the system stable.
Solution I The block diagram for the system is shown in Figure 7.52. If we designate the feed-forward gain and transfer function as KG and the feedback transfer function as H, the system transfer function is (23]
v,/ VREF
=
KG(s)/[ I
+
KG(s) H(s)]
where, neglecting saturation and limiting, we have
Fig . 7.52
Block diagram or the excitation control system.
276
Chapter 7
or
and the system is observed to be fourth order.
Solution 2
The open-loop transfer function is KGH, or
KGH =
KAKGKR (I + TAS)(K E + TEs)(1 + TGs)(1 + TRS)
Using the values specified and setting K = 400 KAKRKG, we have
KGH =
K
(s + 10)(s - O.I)(s + I)(s + 20) (amp)
(exc)
(gen)
(reg)
Solution 3
Using the open-loop transfer function computed in Solution 2, we have the rootlocus plot shown in Figure 7.53, where we compute (22)
o
-15
Fig. 7.53
K
~
0 .0 5 a t arig in
Root locus for the system of Figure 7.52.
= (L,P - 'LZ)/(HP - HZ) = -(30.9 - 0.0)/4 (2) Breakaway points (by trial and error):
(I) Center of gravity
left breakaway at -16.4: right breakaway at -0.43:
= -7.75
1/3 .6 = 1/6.4 + 1/15.4 + 1/16.5 0.278 "" 0.281 1/19 .57 + 1/9.57 + 1/0 .57 = 1/0 .53 1.91 "" 1.89
(3) Gain at jw axis crossing: From the closed-loop transfer function we compute the characteristic equation
4>(s) = S4 + 30.9s3 + 226.9s2 + 177s + K' where K' = 400K - 20 and K = KAKRKG = 40K R.
277
Excitation Systems
Then by Routh's criterion we have S4
I
226.9
S3
30 .9
177
S2
221 .2
K'
Sl
177 - 0.14K'
0
K'
K'
SO
For the first column we have: From row SO K'
From row
=
400 K - 20 > 0
K > 0.05
Sl
K'
=
= 1266
400K - 20 < (177/0.14)
K < 3.21
We may also compute the point of jw axis crossing from the auxiliary polynomial in S 2 with K' = 1266, or
221.2 S2
+
1266
=
0
S2 =
-
5.73
s = ±j2.4
An examination of the root locus reveals several important system characteristics. We note that for any reasonable gain the roots due to the regulator and amplifier excite response modes that die out very fast and will probably be overdamped. Thus the response is governed largely by the generator and exciter poles that are very close to the origin. Even modest values of gain are likely to excite unstable modes in the solution. This can be improved by (a) moving the exciter pole into the left half of the s plane, which requires that R in (7 .59) have a greater value; (b) moving the generator pole to the left, which would need to be done as part of the generator design rather than afterwards; and (c) adding some kind of compensation that will bend the locus to a more favorable shape in the neighborhood of the jw axis. Of these options only (c) is of practical interest. Excitation system compensation. Example 7.7 illustrates the need for compensation in the excitation control system. This can take many forms but usually involves some sort of rate or derivative feedback and lead or lead-lag compensation. (It is
V3
Fig.7 .54
Block diagram of a typical compensated system .
278
Chapter 7
interesting to note that Gabriel K ron recognized the need for this kind of compensation as early as 1954 when he patented an excitation system incorporating these features (37).) This can be accomplished by adding the rate feedback loop shown in Figure 7.54, where time constant T F and gain K F are introduced. Such a compensation scheme can be adapted to bend the root locus near the jw axis crossing to improve stability substantially. Also notice that provision is made for the introduction of other compensating signals if they should be necessary or desirable. The effect of compensation will be demonstrated by an example.
Example 7.8 I. Repeat Example 7.7 for the system shown in Figure 7.54 . 2. Use a digital computer solution to obtain the "best" values for mize the rise time and settling time with minimum overshoot. 3. Repeat part 2 using an analog computer solution.
TF
and KF to mini-
(0)
(b)
+
-
KAK G
E
VI
(I + TA, )( K + TE')(I + TG' ) E
KF' ( I + T ) K G' R + KG (1 + TF') ~,
(e)
Fig . 7.55
Excitation system with rate feedback neglecting Sf and limiter: (a) original block diagram, (b) with rate feedback take-off point moved to V" (c) with combined feedback .
279
Excitation Systems
Solution I The system transfer function can be easily computed for SE = 0 and with limiting ignored. Figure 7.55(a) shows a block diagram of the system with SE = 0 and without the limiter. By using block diagram reduction , the takeoff point for the rate feedback signal is moved to v" as shown in Figure 7.55(b), then the two feedback signals are combined in Figure 7.55(c) . The forward loop has a transfer function KG(s) given by KG(s)
KAKG TATET G (s
I
= - - --....,.------:'.,..----~~
+
I/TA)(s
+
KE/TE)(S
+ I/Td
and the feedback transfer function H(s) is given by (KFTG/KGTF)S(S
H(s)
+
I/Td(s
+
I/T R)
+
(KR/TR)(S
+
I/T F)
= ----"-'--'--'----'---'----'---'---'-'------'--
(s
+ I/TF)(S + I/TR)
The open loop transfer function is thus given by KG H
KAK
s(s
TATETF (s
Substituting the values TA and K R = 1.0, KG H
+
I/TG)(s
+
+ (KRKGTr/TRTGKF)(S + __:_..~ I/TF) + I/TG)(S + I/TF)(S + I/TR)
I/T
R) F _ _---'--=--_ _..:-.c"--_-=---=-~..:..:..-==--...:...:....:.= -_
+
= 0 .1,
I/TA)(S
TE
+
= 0.5 ,
KE!Td(s
= 0.05,
TR
TG
= 1.0 K E = -0.05, KG = 1.0,
s(s + I)(s + 20) + 20(Tr/K F)(s + I/T F ) = 20 K A -K F _'---'..:-.c-'--_---'---'-__ TF (s + lO)(s - O.I)(s + I)(s + I/TF)(S + 20)
(7 .61)
A given TF fixes all poles of (7.61). Then the shape of the locus depends on the location of the zeros. Thus we examine the zeros of (7 .61). From the numerator we write s(s
o=
I
+
+
I)(s
+ 20) +
20(TF/KF)(s s(s + I)(s
20(Tr/KF)(s
+ I/TF) + 20)
=
I
+
+ I/TF) = 0
s(s
K(s + 0) I)(s + 20)
(7.62)
+
where we let K = 20(Tr/KF) and o = I/T F. The locus of the roots of (7.62) , which gives the zeros of (7 .61), depends upon the value of 0 = I/TF' There are three cases of interest (note that 0 > 0): Case I, o < 0 < I; Case II, I < 0 < 20; and Case III , 0 > 20. These cases are shown in Figure 7.56 where -m is the location of the asymptote. Case I is sketched in Figure 7.56(a), where a zero falls on the negative real axis at -0, which is between the origin and - I. The locus therefore falls between the origin and -0 . This means that (7.61) would have a zero on the real axis near the origin. Thus the open loop transfer function of (7.61) will have a pole at 0.1 and a zero on the real axis at -0 . The locus of the roots for this system will have a branch on the real
; I
-x---+--+-x-ox-
-20
tl ~I
Case I: 0 < a < 1 - 10 .5 < m < - 10
Fig. 7.56
-x~l~J -20 ~ ill ~ -20
o-x-x
Case II: 1 < a < 20 - 10 < m < - 0 .5
-I
1
Case III : a > 20 - 0.5 < m
Locus of zeros for the open loop transfer function of (7.62).
Chapter 7
280 I
I I I
x ----O---OX - - o - x
-z.,
- 10
X
X
- Z. -en -I
I I
I Cose I B
Ccse 1 A
~J : ~ x--+__xl z" V1 I
I
x------0-- X
-20
z;
X-X
-1 0
-0 :
-1 Z. Z,
X
-20
I
Cose II A
- 10
I z,
Cose II B
I
I
X --o-X -0 Z, - 20
X-10
I
-X -1
i' I I
Z. Z,
X
X--o-X -0 z, - 20 Cose III B
I
Cose III A
I
Z,
I I
I I
:-
I
-
-X -1
X
Z,
. (K r] Tr )[S(S + I)(s + 20) + 20(s + all Root loci of KGH = 20K A .:.....;.-'------"--'----'------'---(s + 20)( 5 + 10)(s + I)(s - 0.1)(5 + 0)
F ig. 7.57
axis near the origin, and the system dynamic performance will be dominated by this root. Its dynamic response will be sluggish . Cases II and III are shown in Figures 7.56(b) and (c) . In both cases, the root-locus plots of (7 .62) have branches that. with the proper choice of the ratio K, give a pair of complex roots near the imaginary axis. Again, these are the zeros for the system described by (7.61). However, in Case II the loci approach the asymptotes to the left of the imaginary axis, while for Case 111 the loci approach the asymptotes to the right of the origin . The position of the roots of (7.62) and hence the zeros of (7.61), are more likely to be located further to the left of the imaginary axis in Case II than in Case III. A further examination of the possible loci of zeros in Figure 7.56 reveals that for the three zeros, two may appear as a complex pair. Thus there are two situations of interest: (A) all zeros real and (B) one real zero and a complex pair of zeros. Furthermore, both conditions can appear in all cases. Figure 7.57 provides a pictorial summary of all six possibilities . In all but two cases the system response is dominated by a root very near the origin . Only in Cases liB and IIIB is there any hope of pulling this dominant root away from the origin ; and of these two, Case liB is clearly the better choice. Thus we will concentrate on Case liB for further study . (Also see [38] for a further study of this subject.) From (7.61) the open loop tran sfer function is given by KGH
= 20K K f
where I < I/TF < 20.
A
$3
Tf (s
+ 21s2 + 20(1 + TF/KF)s + 20/KF + lO)(s - O.I)(s + I)(s + 20)(s + I/TF)
(7.63)
281
Excitation Systems
20 T
F
~
a. c'
0 .0 1 ; 0 . 6, K F;
T
F
; 0 .6, K ; 0 .01 F
.~ 1.20
15
"
>-
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
<
1 i g---
!
L I
~
I
j
I
I "~,
I , ~ -11
."
:
: ,
I t
i
I'
I
I
iIJI i ! V- q), _i_ i N. I i
;
i
!
i
i
-
1
r I !
,
jfa.+-'- Ii
! ,
I ~~/ Ii'~
II
I
m I
i -!
I __o j
1--
IV"'
II
111
! I
;
i
-0.
i
.. l.
I
. 1.-
=
- --: - - --1 ---
1
, -I i ! i ·· ~ .- i I i I i +-! I t·· i.- N. i -j --- -I I I -I l l + I i i N" + ~i .I !- I- " ! ! ~t-J- K I - I 1 I - i I, ~ ~ I th i 1 I -I i
I
B
S
(b)
Time response to a step change in VR EF : (a) G F(S)
I :..j
! I
6
4 Time,
(0)
--
loU--I-J
OJ.-,......L.j......L..J~--J---------+------'---IL-L-+-~
O. (b) G F .. O.
f-~
I
I
~ -': i
-- --
I
_. -"
\ I "h '
i
Effect of Excitotion on Stability I
8
I
I
i
8
i
I
! I
o
j
I
f
j
I
I I
!
Ilf ' - 1-!
-
I
dO
-- f-.
-
-
~
.
I
-0-
--
......
.- f--;-
!
..
"
s
-
_._. .-
•...
I
, -- - --
I
i
I
-
··
-.
ri,
51
_. --
'Y
....
-- -_.. -- - -~ ..
. ID
.. ,.
-
,
,
;
..
.
l:
1\
i
A ~ DI ~ N S / S [ C .
'"
+
I
;
.
......... '-
,
,
--
!
!
! I '-
,! I I
1
'-l I I
..I
i
I I
I
'f- j i
I~
l ..
.. -]: I ' , ; .+ ..-1-..
I
! -t-
'\(.
...
I
_
. ID
-- -
--
, : " -;,1
~
'" -
--- -- .. _._- -_ ..
-
r-L
!
...
!
.-
F:
-8 - ~-~ ~ . . -+-+-'-t--'-t--'---h~--r-r-.. ,. . ,. ..,..... I ! : i "'8+-t--t---1-=-+-"",--+- r-f-t-+-f-H-I-+-IH-+-!++ !
I-
_I
'
.. --
-
-- I--
,
!
1-- - . I . I ---.- ~ -~ -I
;i i' -
...
'
I
r= ---'-
..
I
,
1 I -I -! ,
~f pi~
_. - . . .. _. -,
k
.
.
j
I
r
i
i
I
.~-
I i! I
r
~i i
! -, ! r-,
.iT
;
331
...
-j-
:. ~
i..
II
,: 1···
+ "'·1..' :i· I I
....;... ...~ ..
'" ...
I, + .•...
'i·
-+ .:::;::: rt' \
. ~
R~D
I ~NS / S [C .
( b)
(0 )
Fig.8 .22
Bode plots of the open loop transfer funct ion: (a) GF = O.(b) GF
""
O.
Compute these graphical di splays for two cond itions: (a) GF(s) = 0
(b) G~s)
= sKF/(l +
T~), with
KF
= 0.04,
and TF
=
1.0 s
Solution
The results of the computer analysis are shown in Figures 8. t9-8 .22 for the different . plots. In each figure, part (a) is for the result without the rate feedback and part (b) is with the rate feedback . Figures 8.19--8.20 show clearly that the system is unstable for this value of gain without the rate feedback. Note the basic p roblem discussed in Examp le 7.7 . With GF(s) = 0, the system dynamic response is dominated by two pairs of complex roots near the imaginary axis . The pair tha t causes instability is determined by the field Table 8.4. Condition (a) K F
(b) K F
=0
=
0.04
Root-Locus Poles and Zeros of Example 8.4 Zeros
-0.21097 + j10.45130 -021097 - j10.45130
-1 .19724 - 1.19724 -0.40337 -0.40337
+ jO.83244 - jO.83244
+ jlO.69170 - jlO.69170
Poles
- 0.27324 -20.00000 -0.17894 -0.35020 - 0.35020 -20.00000 -0.17894 -0.27324 -0.3502\ -0.35021 -1 .00000
+ jI0.72620
- j 10.72620
+ jI0.72620 - jI0.72620
332
Chapter 8
winding and exciter parameters. The effect of the pair caused by the torque angle loop is noticeable in the Bode plots of Figures 8.21 --8.22. These roots occur near the natural Irequency ,», = (1.4479 x 377/4.74)1/2 = 10.73 rad/s. The rate feedback modifies the root-locus plot in such a way as to make the system stable even with high amplifier gains. The poles and zeros obtained from the computer results are given in Table 8.4. Example 8.5 Repeat part (b) of Example 8.4 with (a) D = 0 and (b) K, = O. Solution
(a) For the case of D = 0 it is found (from the computer output) that the poles and zeros affected are only those determined by the torque angle loop . These poles now become -0.13910 ± jI0.72550 (instead of -0.35021 ± jI0.72620). The net effect is to move the branch of the root locus determined by these poles and zeros to just slightly away from the imaginary axis. (b) It has been shown that K, is numerically small. Except for the situations where K, becomes negative, its main effect is to change W n to the value w2n
(wR/2H)(K I
=
-
K 2K s/Kd
The computer output for K, = 0 is essentially the same as that of Example 8.4. The root-locus plot and the time response to a step change in VREF for the cases of D = 0 and K, = 0 are displayed in Figures 8.23--8.24. The examples given in this section substantiate the conclusions reached in Section 7.7 concerning the importance of the rate feedback for a stable operation at high values of gain. A very significant point to note about the two pairs of complex roots that dominate the system dynamic response is the nature of the damping associated with them. The damping coefficient D primarily affects the roots caused by the torque angle loop at a frequency near the natural frequency w n • The second pair of roots, determined by the field circuit and exciter parameters, gives a somewhat lower fre. . .._..
8 6
-- -~.,. -- -. _-- ---
~ ir=
- ~- --
I I
-- - --- l-r-- .- -, - f- I -
~
I-
L
:
,-
2 1i 1\ -
o
I H
-8
-
~f -
;
i ,-
-- - f- - - -
-- -- - --j- r- - -
r• ,- I -+ I; ;
,
l -I - --
- - ~- -
,-
I
-
I-
,-
i
, ,
,-
'-
i r-
.!
I
I
.- 0-
i
I
I-
;
:
i
,- ,-
I
I
..
:
-.
;
,-
;
,
i, i ,-
" 1'
-- ' -I -.-- ---.-
h
f----_..- ..! --__+ -0 !.,.- --.,.. .
f,-
I
-
--
.-
, -., ._,----- --
i
,
i
...-
,-
!' ~
I I i' ll: I I' -,-
--.. _--
_
- ~-
,-
,
!
.
.. , -H, - ";' .- .i
±
-.,
-:
~
;
I
I
IN l I 1\ -I: -l - -t - -H +-'1 + + + I ii I
i..
,- ' I I' I
-4
Reel
-~
'"
I
I
I
I-I
-2
o
:
I i
(0) Fig. 8.23
Root locus of the system of Example 8.5: (a) D
= 0, (b) K s = O.
333
Effect of Excitation on Stability
1.2
.
0.5
~
> 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 (<5 20
(32)
-
020 -
But from the load flow {32
020 -
(32
+
= V2/ {32 /2 = 12/-(820
I. 272
{32 = 5 I. 8I 8 0 ,
=
61.098
0
54.156 and 0
<1>2 =
V2
X q 2/ x 2) =
-
= 51.818 + 9.280
020
Then
x q 2/'21(V2 0
9.280
=
=
= 1.025 /-51.818 = Vq2 + jVd2 = 0.634 - jO.806 pu /32 + cP2) = 1.592 /-54.156 = Iq2 + j / d2 = 0.932 - jl.290 pu 0
~20
0
-
Neglecting the armature resistance, r = 0, E qaO = Vq 2 £20
=
~2
-
xq2ld2 =
-
Xd2 1d2
1.749 pu
= 1.789 pu
From Table 2.6 the driving-point admittance at the internal node of generator 2 is given by
fn
= 0.420 - j2.724 pu
The terminal voltage node of generator 2 had been eliminated in the reduction process. can be obtained However, since it is connected to the internal node by Xd2' by using the approximate relation Ze = 1I Y22 - jXd2' The exact reduction process gives
z.
Zt! = 0.0550 + jO.2388 = 0.2450 /77.029
0
pu
Then we compute from (6.56) K/
=
I/[R;
I/K 3 = 1
+
+ (x, + Xt!)(Xd + Xe)]
K/(Xd - Xd)(X q
K 3=O.3177 We can compute the infinite bus voltage
+
Xe)
=
=
1/0.39925 = 2.5084
3.1476
336
Chapter 8
v~
= V~fs5.. = ~2 - Ze/2 = 1.025 /9.280° - (0.2450 /77.029°)( 1.592/6.941 = 0.9706 - jO.2226 = 0.9958/-12.914°
0 )
The angle required in the computations to follow is l'
K.
=
a
020 -
=
61.098 - (-12.914)
=
74.012°
K/ V~ IEqao[R e sin l' + (Xd + X e ) cos 1'] + lqo(xq - Xd )[(x q + X e ) sin l' - R, cos 1']1 = 2.4750
=
K2 = K/IReEqaO + Iqo[R; + (x q + X e )211 = 3.0941 K4 = V~K/(Xd - xj)[(x q + Xe)sin')' - Recos')'] = 2.0265 K s = (K/V~/~o)lxj~o[Recos')' - (x, + Xe)sin')'l - x q VdO[(xj + X e ) cos l' + R, sin 1'H = 0.0640 K6 =
(~o/~o)[1
- K/xj(xq + X e ) ]
Summary:
K , = 2.475
-
(Vdo/V,o)K/xqR e K) = 0.318
=
0.5070
K, = 0.064
K 2 = 3.094 K 4 = 2.027 K 6 = 0.507 Note that these constants are in pu on 100-MYA base whereas the machine is a 192MYA generator. The constants K. and K 2 should be divided by 1.92 to convert to the machine base. Example 8.7
The exciter for generator 2 of the three-machine system has the constants K, = 400 and r, = 0.95 s. Compute the parameters of Gx(s). For the system natural frequency (see Example 3.4) calculate the excitation control system phase lag. (Here again we emphasize the need for actual measurement of the system parameters. Lacking such measurement, a judgment is made as to which parameters should be used. We use the regulator gain and the exciter time constant. It is judged that the latter is important at the low frequencies of interest. This point is a source of some confusion in the literature. It is sometimes assumed, erroneously, that the regulator time constant is to be used when the excitation system is represented by one time constant. This is not valid for low frequencies.) Solution From (8.36) we have Wx =
v(0.507 x 400)/(6.0 x 0.95) = 5.967 rad/s
!x
(0.95
=
+ 0.318
x 6.0)/(2 x 5.967 x 0.318 x 6.0 x 0.95) = 0.132
and the excitation system is poorly damped. From Example 3.4 the dominant frequency of oscillation is approximately 1.4 Hz or Wosc ~ 8.8 rad/s. At any frequency the characteristic equation of Gx(s) is obtained by substituting s = jw in the denominator of the first expression in (8.35): d(jw)
=
At the frequency of interest (w d(jwosc)
1 - 0.0281 w 2 =
+ jO.0443w
8.8 rad/s) we have
1.1761 + jO.3898 tan-I (0.3898/-1.1761)
= -
(jJlag =
=
161.661
0
337
Effect of Excitation on Stability
12
v
6
>
- - ' - - --
(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
-tc n
8=
-120
~
I - u'
-1 35 - 150
-165 -18 0 0 . 01
0.3
0 .6
1
3
6
10
30
60 !OO
u (b)
Fig .8.26
Characteristics of a second-order transfer function : (a) amplitude. (b) phase shift.
The excitation system phase lag in Example 8.7 is rather large, and phase compensation is likely to be required (see Section 8.7). The phase lag is large because Wost > 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 <,
O---J I 7..Q.3
~
Pi
I I RfJ{) ...,
I
Analog computer represen tation of a Type I excitation system .
(~ )
v
LC = l evel cho nge
( ~)
-v e
i(
1
V
I -FD I <,
+EFD (10)
To
P302
w
IX)
-0
o it ..,
:r
()
~ IX)
349
Effect of Excitation on Stability
Table 8.8.
Potentiometer Calculations for a Type 1 Representation of a W Brushless Exciter (a = 20)
Pot. no.
Amp. no.
Out
Lo
In
L;
LolL;
600 601
601 601
VREF VREF
50 SO
REF REF
100 100
0.50 0.50
800
800
VR
I
- Ve
50
0.02
KA/aT A = 400/(~0)(0.02)
701
HOD
VR
I
VR
I
1.00
801 703
801 801
-E FD -E FD
10 10
VR -E FD
1 10
10.00 1.00
802 810
802
~m2
Vz Vz
50 50
Vy -E FD
100 10
812 803
810 803
- V.v Vx
50 40
lim
800
800
100 50
. .. ..
... .,
.
Vz
v,
., . .. .
.
"
.
. ..
Amp. gain
(LoIL;)(,
Int. cap.
0.0252 0.5033
... ..
I I
0.0252 0.5033
20.0
0.1
100
0.2000
I [a T A = 1/(20)(0.02) = 2.5
2.5
0.1
10
0.2500
1/(20)(0.8) = 0.0625 K E/a T E = 1/(20)(0.8) = 0.0625
0.6250 0.0625
1.0 1.0
I I
0.6250 0.0625
0.50 5.00
1/ T F = I/ 1.0 = 1.0 KF/TF= 0.03/1.0 = 0.03
0.50 0.15
...
...
I I
0.5000 0.1500
2.00 1.25
1/ vJ = 0.5773
1/0= 1/20=0.05
0.10 0.7217
1.0 ...
1 1
0.1000 0.7217
.
=
constant
s-, of P601
I + 2.667/400
:=
1.0066
= 1000
I [a T E
=
Pot. set.
VRmal( = 7.3 pu = 7.3 v VRmin = -7.3pu = -7.3v
..
.
"
C
Comparing the responses shown in Figures 8.38, 7.69, and 8.39 with that of Figure 5.20, we note that without the exciter the slow transient is dominated by the field winding effective time constant. The terminal voltage, the field flux linkage, and the rotor angle are slow in reaching their new steady-state values. From Figures 8.38, 7.69, and 8.39 we can see that the steady-state conditions are reached sooner with the exciter present. At the same time, the response is more oscillatory. 8.9.2
Effectiveness of compensation
A detailed study of the effectiveness of four methods of compensation is given in [28J, by comparing the dynamic response due to changes in the mechanical torque T; and the reference excitation voltage VREF at various machine loadings. The dynamic response comparison is based on observing the rise time, settling time, and percent overshoot of either Pe~ or Vt~ in a given transient. For example, a 10~~ increase in the reference torque is made, and the change in electrical power output Pea is observed. The machine data and loading are essentially those given in the Examples 8.4 and 8.5. Table 8.9. Pot. no.
Amp.
600 601
Potentiometer Calculations for a Type I Representation of a W Low T E Brushless Exciter (a = 20)
Out
Lo
601 601
VREF VREF
50 50
REF REF
800
gOO
VR
I
- Ve
no.
In
Lj
Lo/L;
100
100
0.50 0.50
5".. of P601 1 + 2.667/400"'" 1.0066
50
0.02
KA/a T A = 400/(20)(0.02) I/o T A
C = constant
= 1/(20)(0.02) = 2.5
701
800
VR
I
VR
I
1.00
801
-E FD
10
VR
I
10.00
703
801
-E FD
10
-E FD
10
1.00
= '/(20)(0.015) = 3.3333 KE/aTE = 1/(20)(0.015)
802 810
802 802
Vz Vz
50 50
Vv -'E FD
100 10
0.50 5.00
I/TF= 1/0.5=2.0 KF/TF = 0.04/0.5 = 0.08
812 803
810 803
- Vy Vx
50
2.00 1.25
I/v] = 0.5773
lim 800
800
..
.
.. .
100 50
...
.. .
Vz
v,
. ..
...
4t;
. .. . ..
"
.
. ..
l/aTE
= 3.3333
1/0= 1/20=0.05
VRmax = 6.96 pu = 6.96 v V Rmin = -6.96 pu = -6.96
Int. cap.
Amp. gain
0.0252 0.5033
... ...
I 1
0.0252 0.5033
0.1
100
0.2000
20.0
= 1000
801
(Lo/L j ) (,
v
Pot. set.
2.5
0.1
10
0.2500
33.333
0.1
100
0.3333
3.3333
0.1
10
0.3333
DAD
1.00
... ...
10 I
0.1000 0.4000
0.10 0.7217
...
1.0
I 1
0.1000 0.7217
-'---......-...-
I
I
, , ,
I
-
+-
l-
f-
~
1-1-
"
o
1
-
f- .
-r-t-
L-L-t--r
iii
Fig . 8.38
I
J==btH=H=
I
':r=u:::w:::;:::
AF
2ft
0,
1
2 L -L...l--.1f---t-T
J
-
_
-YJ ..L
L .1-
I
.Li..
.i.r;
4 t -~~-r++ttQ 3
o
I
1 -H I
I"'"
I l J -j I I
II I I t l
3 2 1
4
, E 4~ c FD T ITin
System response to a step change in Tm and VREF • generator equipped with a W TRA exciter.
o
.25
-2 -3
-1
o
I
.
00
"iii..,
Q
-:T
n
w
<.ro
o
r"
Il-
~
'"
1
•
~
o
1
2
I
o
1
2
4 3
~
tAr-I-
I
Ir-
~b ~
.
n:tJ
T¢"
+-+-
o
1
2
3
4
1\
Fig.8.39
- t---r--_. t ·
I
4 2
6 -
-1--
I
I
,
I
6°
..l H"'
I
I
I
~
I
I·
X
,
'"~~ ~~
Iii
~~
- t--+
-'---'- ill
,I
.2 Ervt~
- .2
;
;
,
I
I
. ;='d
I
I
I
I
I
_.
,
I
r
I
I I
10
'-..•__•!
I
I
,
I
I
!
I
I
I
-r-r-
. ~~
til'
II
.
System response to a step change in T", and VR EF • generator equ ipped with a Low r e Brush less exciter .
+-~
I
+--;
-+; -+-;+ ; -
;-+;-+;-+-;--+-;-+-; -+-;
.
;105 ..'
+-1 . +---<1
o-t-h-t-I--1+-
1
2
43
E ~ c FD
352
Chapter 8
However, the machine is fully represented on the analog computer. The excitation system used is Type 2, a rotating rectifier system (see Section 7.9.3). The data of the exciter are: K A = 400 pu
K F = 0.04
TA
= 0.02
TE
=
KE
=
= 0.05 T R = 0.0 K R = 1.0
S
0.015 1.0
TF
S
S
SErnax
= 0.86
SE.75rnin
= 0.50
VR rna x
8.26
VR rnin
-8.26 4.45
EFD rnax
The methods of compensation used are: Rate feedback: sKF/(1 + TFS' Bridge- T filter with transfer function: C/ R
= (S2 + rn WnS +
n Power system stabilizer:
C T
=
+ nwns +
w~)
2 t rad/s 2 r = 0.1
Wn
Ii =
w~)/(S2
Ks
1+
TS
3.0 s
(1 + T1S)2 1+
TZS
i, =
0.2
S
T2
=
0.05 s
A sample of data given in reference [28] is shown in Table 8.10 for the initial operating condition of Tmt/J = 3.0 pu at 0.85 PF lagging. Table 8.10.
Comparison of Compensation Schemes p~~
Case
Uncompensated Excitation rate feedback Bridged-T only Bridged-T, two-stage lead-lag and speed Power system stabilizer
~~
Rise time
Settling time
Overshoot %
Rise time
Settling time
Overshoot %
0.06 0.06
0.22 0.22
86.6 80.0
0.20 0.98
0.60 4.20
10.0 60.0
0.05 0.04
0.23 0.21
100.0 73.4
0.21 0.28
0.56 0.37
33.0 5.0
0.05
0.21
82.6
0.23
0.42
5-10
Source: Schroder and Anderson (28).
Other valuable information that can be obtained from analog computer studies is the response of the machine to oscillations originating in the system to which the machine is connected. This can be simulated on the analog representation of one machine connected to an infinite bus by modulating the infinite bus voltage with a signal of the desired frequency. This is particularly valuable in studies to improve the system damping. When growing oscillations occur in large interconnected systems, the frequencies of these oscillations are usually on the order orO.2-0.3 Hz, with other frequencies superimposed upon them. Thus it is important to know the dynamic response of the synchronous machine under these conditions.
353
Effect of Excitation on Stability
2v/L
~ 2·~-~~~_L~~ ; .:L; ~-~~.:-.~_. - --_~ . _ : _. ~
.' " .." ."' . .,. -l] _ . --~ ~ ~ .: -~ _ ..: ..~ ': . .:. ...... . ,
:
;
(
'.
;
..
- _. . .. .
"T
"
L',
2v/L
2v/L
'I: : .. _...:.+ ~"-
.
-. T" --:- - -~ -r-:-
1; ' I
·
! . _. -l-o ••
_
-~.:.- . ~-_ . .
. I!
I
- ~--
-
.-. .
t.
.. : ; l ;. _
. 0 .
;"
'-
O.2v/L
O.2v/L
_
lv/L
lv/L
2v/L
Fig.8.40
Synchronous machine with PSS operating against an infinite bus whose voltage is being modulated at one-tenth the natu ral frequency or the machine,
A sample of this type of st ud y. taken from [28]. is shown here. The same machine discussed above . but operating under the heavy loading condition of Example 5.1. has its bus voltage modulated by a frequency of one-tenth the natural frequency . The modulating signal varies the infinite bus voltage between 1.02 and 0.98 peak . Fig ure 8.40 shows the effect of the PSS under these conditions. At time A the modulating signal of 2.1 rad/s is added . The PSS is removed at B. causing growing oscillations to build up especially on Pet.. which would simulate tie-line oscillations. Note also that the frequency of these oscillations is near the natural frequency of the machine. When the stabilizer is reinstated at point C. the oscillations are quickly damped out. At point D the modulation is removed .
8.10
Digital Computer Transient Stability Studies
To illustrate the effect of the excitation system on transient stability , transient stability studies are made on the nine-bus system used in Section 2.10 . The impedance diagram of the system (to 100-MY A base) and the prefault conditions are shown in Figures 2.18 and 2.19 respectively . The generator data are given in Table 2.1. The transient is initiated by a three-phase fault near bus 7 and is cleared by opening the line between bus 5 and bus 7. In this study the loads A, B, and C are represented by
354
Chapter 8 PSS V
a max
K,o
(1 + -
r
Kia
( 1 + 1<,0 K.
-r
A
, )(1 +
~
K. o
')( 1 + -
K, Kt a
'b')
-r ,)
B
'-- - - - - - - - - - - - - - --J K. o Typical values : Va m ax = I VRmax = 11 .85 V Rm in = 4.4 5
Em max = 4 .S5
Fig. 8.41
K1
=
10
K 2 = 0.1 K 20 K IO = \00
K 20 = 200/K 30
K 30= S-23 K40 = 0. 2
TO
= \0
Th =
0.1
TE=0 .B1 5
TA = \ T8 = 0
The Brown Boveri Co . alternator diode exciter. (U sed with permission of Brown Boveri Co .)
constant impedances; generators I and 3 are represented by classical models, i.e., constant voltage behind transient reactance. For generator 2, provision is made for the excitation system representation. A modified transient stability program was used in this study. (It is based on a program developed by the Philadelphia Electric Co ., with modifications to include the required new features.) When the excitation system is represented in deta il, the model used for the synchronous machine is the so -called " o ne-axis model" (see Section 4.15.4) with provision for representing saturation. When the machine EM F E (corresponding to the field current) is calculated , an additional value £6 is added due to saturation Table8.II. Parameter
KA K£ KF KR Kp K1 VRma.
VRmin V8 m a.
TA T£ TF TR
A,
B,
Excitation Systems Data
Amplidyne
Mag-A-Stat
SCPT
25 -0.044 0.0805 1.0
400 -0.17 0.04 1.0
1.0 -1.0
3.5 - 3.5
120 1.0 0.02 1.0 1.19 2.62 1.2 -1.2 2.78 0.15 0.05 0.60 0
0.20 0.50 0.35 0.06 0.0016 1.465
0.05 0.95 1.0 0 0.0039 1.555
Note: See Figure 8.41 for BBC exciter parameters.
355
Effect of Excitation on Stability 130
90
120
!
j!
,I/;
110
i
~;;~, /;:Y-~,' '\-
80
~0· " '\ \ \
'\\\\\ "\\
\
\\~,
100
\' .
U
x
~
-£ .~
.g 90
-"'-------
.J!'
C>
C
4: ~ ~
E! 80
....a
BBC exclter - constant E FD Type 1 - 0 .5 RR Ma g- A-S ta t BBC exci ter - 2 .0 RR Classical madel
70
\ \
\\\
\\ \
\\ . \\
70
60 C
u
.~
e U
50
.d'
,,'
C,
c
4:
40
" ~
E!
..'1
30
Fig.8 .42
0 21
for various exciters with a three-cycle fault.
effect and based on the vo lta ge beh ind the leakage reactance Es , Th is is given by EA = A g exp [Bg(E -t - 0.8)]
(8.52)
The con st ants Ag and Bg a re pr ov ided for several exciters [see (4.141 )]. The types of field represent at ion used with generator 2 are:
I. 2. 3. 4. 5.
Cl assical mod el. I EEE T ype 1,0.5 pu resp on se, a mp lidy ne N A 101 exciter (see Figure 7.61). I EEE Type I, 2.0 pu resp onse, M ag -A-Stat exciter (see Figure 7.61). I EEE T ype 3, SCPT fast exciter, 2.0 pu respon se (see Figure 7.66) . Brown Boveri Company (BBC) a ltern ato r diode exciter (see Figure 8.41).
The excitat ion system data ar e given in Table 8.11 .
8.10.1
Effect of fault duration
Two sets of runs were made for the same fault location and removal , but for different fault durations. The breaker clearing times used were three cycles and six cycles. For a t h ree-cycle fault , the results of generator 2 data are shown in F igures 8.42-8 .46. Similar res ults for a six-cycle fault are shown in Figures 8.47-8 .50.
Chapter 8
356 1.2 ,----
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-,
v!.-:: -==
~----
1. 0-
....::::-=- - -
~--===~~----=._---
BBC excite r - 2.0 RR Mag-A - Stat Type! -0 . 5RR
/
r::/
0.2 f-
I 0. 2
I 0.1
o Fig. 8.43
V, and
E~
I 0.4
I 0 .3
Time.
I
0.5
0.6
5
for var ious exciters with a three-cycle fault.
Results with three-cycle fault clearing. Figure 8.42 shows a plot of the first swing of the angle 021 for different field representations. Note that the classical run gives the angle of the voltage beh ind transient reactance, while all th e others give the position of the q ax is. A run with constant EFD is also added . We conclude from the results shown in Figure 8.42 that for a three-cycle clearing time the classical model gives appro ximately the same magnitude of 0 21 for the first swing as the different exc iter representations . When the exciter model was adjusted to give con stant Em. however, a large swing was obtained. From Figure 8,43 we conclude that the slow exciter give s the nearest simulation of a constant flux linkage in the main field winding (and hence constant E~) a nd minimum variation of the terminal voltage after fault clearing. The action of the exciter and the armature reaction effects are clearly displayed in Figure 8.44 . It is interesting to note that the actual field current, as seen by the EM F E, is hardly affected by the value of Em for most of the duration of the first swing after the fault is cleared. The effect of the armature reaction is dominant in this period. Figure 8.45 shows a time plot of P2 for this transient. Again it can be seen that the different model s give essentially the same power swing for this generator. We note, however, that the minimum swing is obtained with the slow exciter while the maximum swing is obtained with the classical model. In Figure 8.46 the rotor angle 021 is plotted for a period of 2.0 s for the classical model , a slow IEEE Type I exciter, and a relatively fast exciter with 2.0pu response . The plot shows that the first swing is the largest , with the subsequent swings slightly reduced in magnitude. Figures 8.42 -8.46 seem to indicate that for this fault the system is well below the stabil ity limit, since the magnitude of the first swing is on the order of 60·. All generator
357
Effect of Excitation on Stability
4 .0 r-
~
Q.
W
o LL
-e C
o
w
1.0 1-
7
e:.
I
I
I
I
I
0
0. 1
Fig .8.44
Em and E for various exciters with a three-cycle fault.
0.2
0 .3
Time , s
0 .4
0.5
0 .6
2 models give approximately the same magnitude of rotor angle and power swing and period of oscillation. Results with six-cycle fault clearing. For the case of a six-cycle clearing time. the plot of the angle 021 is shown in Figure 8.47 for the classical model and for two different types of exciter models . The swing curves indicate that this is a much more severe fault than the previous one, and the system is perhaps close to the transient stability limit. Here the sw ing curves for the generator with different field representations are quite different in both the magnitude of swings and periods of oscillation . The effect of the 2.0 pu respon se exciter is pronounced after the first swing. The effect of the power system stabilizer on the response is hardly noticeable until the second swing. The magnitude of the first swing for the cases where the excitation system is represented in detail is significantly larger than for the case of the classical representation . The Type I exciter gives the highest swing. Comparing Figures 8.46 and 8.47. we note that for this severe fault the rotor oscillation of generator 2 depends a great deal on the type of excitation system used on the generator. We also note that the classical model does not accurately represent the generator response for this case.
358
Chapter 8 200 ,..-
-
-
-
- --
-
-
-
-
-
-
-
-
-
-
-
-
-
--,
180
C lcsslcol model BBC exciter - 2.0 RR Mog-A-Stat Type I - 0 .5 RR ~
~
120
c,
100
80
o Fig. 8.4 5
0.2
0 .1
Time, s
0.3
0.4
0. 5
Output power P2 for vario us exciters with a three-cycle faul t.
o .~ 60
a
v
Type I - 0.5 RR BBC - 2.0 RR Clossical
o
0 .2
0.4
Fig . 8.46
0 .6
Rotor angle
20
0 .8
021
1.0
Time, s
1. 2
1.4
1.6
for var ious exciter s with a three-cycle fault.
1. 8
2.0
359
Effect of Excitation on Stability
120
160
140
80
120
1!
'l h
'u x ~
-" 100 'j
;
I!
~
/1;
80
."
U ~
Qj
,
40
~
8
.r..
a ~
I!
Ol ~
."
~
Qj
;;
00
0> c
~
<{
.~
u
~
00"
0> c
0
40
<{
Classical Type 1 - 0.5 RR BBC - 2 .0 RR BBC with PSS
~
~
0
...
!: 0
20
Fig. 8.47
Rotor angle 021 for various exciters with a six-cycle fault.
The output power of generator 2 is shown in Figure 8.48 for different exciter representations. While the general shape of these curves is the same, some significant differences are noted . The excitation system increases the output power of the generator after the first swing . The generator acceleration will thus decrease, causing the rotor swing to decrease appreciably . This effect is not noticed in the classical model. It would appear that for slightly more severe fault s the classical model may predict different results concerning stability than those predicted using the detailed representation of the exciter. Figures 8.49 and 8.50 show plots of the various voltages and EMF 's of generator 2 for the case of the 2.0 pu exciter and the Type I exciter respectively. The curves for show that although the fault is near the generator terminal, the flux linkage in the main field winding (reflected in the value of E;) drops only slightly (by about 5%); and for the duration of the first swing it is fairl y constant. The faster recovery occurs with reaches a plateau at about 1.1 s and stays fairly constant the 2.0 pu exciter, and thereafter . For the Type I exciter recovers slowly and continues to increase steadily. The oscillations of terminal voltage V, are somewhat complex . The first swing after the fault seems to be dominated by the inertial swing of the rotor, with the action of the exciter dominating the subsequent swings in v,. Thus after the first voltage dip, the swings in V, follow the changes in the field voltage Em with a slight time lag . Again the recovery of the terminal voltage is faster with the 2.0 pu exciter than with the Type I exciter. We also note that the excitation system introduces additional frequencies of oscillation, which appear in the V, response .
E;
E;
E;
Chapter 8
360
r \
I \ I \
250
I
200
;.,
/'
'\
~,
I ~(\
I
(
I I
/\ \ 1/' \I \.
~
I
\~
/
'\\;---/II
~1 50
: I~
I
I ,
<:»
I \ ,
\ \ \ \
I
II
I
\
\
\
\
\
\
\,
(
\ \
<>
I
\
\\ \
J
II
\
\\ II \ \ I, \
\\ \//t I
( /
1.6
Fig . 8.48
....
r\ \
II
'
\
o
\
I
I I
\ '
50
\\ \ \
\
Clossico l Type 1 - 0.5 RR BBC - 2 . 0 RR
I \\
I
I
\
100
\ \
!I I \\\"-.
I \ I \
I
1.8
Outpu t pow er P2 for var ious exciters with a six-cycle fault.
The plots of E clearly show the effect of the armature reaction . In the first 0 .7 s, for example, the changes in Em are reflected only in a minor way in the total internal EM F E. The component of E due to the armature reaction seems to be dom inant because the field circuit time constant is long. The general shape of the EMF plot, however, is due to the effects of both Em and the armature reaction , From the data presented in this study we conclude that for a less severe fault or for fast fault clearing, the excitation representation is not critical in predicting the system dynamic responses . However, for a more severe fault or for studies involving tong transient periods, it is important to represent the excitation system accurately to obtain the correct system dynamic response. 8 .10.2
Effect of the power system stabilizer
For large disturbances the assumption of linear analysis is not valid, However, the PSS is helpful in damping oscillations caused by large disturbances and can be effective in restoring normal steady-state conditions. Since the initial rotor swing is largely an
Effect of Excitation on Stability
/
4 .0
~
I
'\
I \ I \ I \
/ I
/
/
/
/
/'
/
\EFD
I
I
I
,
\
\ \ \
v
/
/\ \ I
\
\I / \ \
,
\
I\~
I
\
q
I
/ v/ i\ \
\
\
\ )(\ . . ""'-<.
/ ' ' ' - .'
I ,,---,
\ j E'
-;/
r-; r \
\
/-\ /
,
\
I
\/
i \v, \
/
I
~,
"
/ \
I
'0
/
~
/ ,
\
:; 3. 0
>
\, I I
I \ I/ \ I I \ / ~ / ~ I \ i> f I I/\V " ,-/ 'hi
361
\ '> , .,...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
/
/ \
/
.
\
\
\
/'",
><"
\
With
\
,,
-. """/ -, ' - - - ,
\ \
\
\
pss
//
I
\
\
-, <;
/
-,
/
-.
.....
\ :-.
/ "<, .. .
-6.,
/
.....
-,
............../
,
Without PSS
<:»
"
--"
2 . 00
0.50 Time, s
Fig.8.52
(
'
U-j A ' A
\
/
Exciter voltage E FD with and without a PSS.
Effect of Excitation on Stability
363
at t = 0.10 s (6 cycles). Generator 2 is equipped with a Type I Mag-A-Stat exciter with constants similar to those given in Table 8.11. The PSS constants are the same as in Example 8.12 (a = 25) with a limiter included such that the PSS output is limited to ±O.IO pu. Stability runs were made with and without the PSS. From the stability runs, data for the angle l5 21 and the voltage EFD are taken with and without the PSS. The results are displayed in Figures 8.51 and 8.52. From the plot of 021 in Figure 8.51 note that while the change in the first peak (due to the PSS) is very small, the improvement in the peak of the second swing is significant. The comparison in EFD , shown in Figure 8.52, is interesting. Note that this exciter is not particularly fast (RR = 0.5), and the response tends to be a ramp up and then down. The phase of E FD changes when the PSS is applied to produce a field voltage that is almost 180 out of phase with 021' This results in a delayed EFD ramp as 0 swings downward, which tends to limit the downward 0 excursion by retarding the building in Te . The improvement in the angle 021' defined as 021~ = 021 (no PSS) - 021 (PSS)' has been investigated for different PSS parameters. It is found that this angle improvement is sensitive to both the amount of lead compensation and to the cutoff level of the PSS limiter. A comparison of several runs is shown in Table 8.12. 0
Table 8.12. a
021 Improvement at Peak of Second Swing
Limit = ±O.IO
Limit
=
±O.05
25
16
8.11
Some General Comments on the Effect of Excitation on Stability
In the 1940s it was recognized that excitation control can increase the stability limits of synchronous generators. Another way to look at the same problem is to note that fast excitation systems allow operation with higher system reactances. This is felt to be important in view of the trends toward higher capacity generating units with higher reactances. For exciters to perform this function, they need high gain. Series compensation makes it possible to have a high dc gain and' at the same time have lower "transient gain" for stable performance. Modern exciters are faster and more powerful and hence allow for operation with higher series system reactance. Concordia [17], however, warns that "we cannot expect to continue indefinitely to compensate for increases in reactance by more and more powerful excitation systems." A limit may soon be reached when further increases in system reactance should be compensated for by means other than excitation control. The above summarizes the situation regarding the so-called steady-state stability or power limits. Regarding the dynamic performance, modern excitation systems play an important part in the overall response of large systems to various impacts, both in the so-called transient stability problems and the dynamic stability problems. The discussion in Section 8.3 and the studies of Section 8.10 seem to indicate that for less severe transients, the effect of modern fast excitation systems on first swing transients is marginal. However, for more severe transients or for transients initiated by faults of longer duration, these modern exciters can have a more pronounced effect. In the first place, for faults near the generator terminals it is important that the synchronous machine be modeled accurately. Also, if the transient study extends beyond the first swing, an accurate representation of the field flux in the machine is needed. If the excitation system is slow and has a low response ratio, optimistic results
364
Chapter 8
Shaft spee d
Fig . 8.53
K(', 1)(1 + ' 01)(1 + ' ,I ) (1
+'1 1)(1 +', 1)(1 +'51)
Block diagram of the PSS for the BBC exciter with a 2.0 RR : KQI : K QJ : K Q4 : 0, K Q2 : I, 10. T2 : 0.5, T) : 0.05, T 4 : 0.5. TS : 0.05, limit: ±0.05 pu .
Tl :
may be obtained if the classical machine representation is used . Transient studies are frequently run for a few swings to check on situations where circuit breakers may fail to operate properly and where backup protection is used . It should be mentioned that several transients have been encountered in the systems of North A merica where subsequent swings were of greater magnitudes than the first, causing eventual loss of synchronism . This is not too surprising in large interconnected systems with numerous modes of oscillations. It is not unlikely that some of the modes may be superimposed at some time after the start of the transient in such a way as to cause increased angle deviation . As shown in Section 8 .10, the effect of excitation system compensation on subsequent swings (in large transients) is very pronounced . This has been repeatedly demonstrated in computer simulation studies and by field tests reported upon in the literature [8,9, 13, 23,29,30,31). For example, in a stability study conducted by engineers of the Nebraska Public Power District, the effect of the PSS on damping the subsequent swings was found to be quite pronounced, while the effect on the magnitude of the first swing was hardly noticeable. The excitation system used is the Brown Boveri exciter shown in Figure 8.41. The PSS used is shown schematically in Figure 8.53, and the swing curves obtained with and without the PSS (for the same fault) are shown in Figure 8.54. Voltage regulators can and do improve the synchronizing torques . Their effect on damping torques are small; but in the cases where the system exhibits negative damping characteristics, the voltage regulator usually aggravates the situation by increasing the negative damping. Supplementary signals to introduce artificial damping torques and to reduce intermachine and intersystem oscillations have been used with great success. These signals must be introduced with the proper phase relations to compensate for the excessive phase lag (and hence improve the system damping) at the desired frequencies (32). Large interconnected power systems experience negative damping at very low frequencies of oscillations. The parameters of the PSS for a particular generator must be adjusted after careful study of the power system dynamic performance and the generator-exciter dynamic response characteristics. As indicated in Section 8.6, to obtain these characteristics, field measurements are preferred. If such measurements are not possible, approximate methods of analysis can be used to obtain preliminary des ign data, with provision for the adjustment of the PSS parameters to be made on the site after installation . Usually the PSS parameters are optimized over a range of frequencies between the natural mode of oscillation of the machine and the dominant frequency of oscillation of the interconnected power system.
365
Effect of Excitation on Stability 165 , - - --
- - -- - - - - - - - - - - - - - -- -- -- -- - - -- - - ,
ISO Wit ho ut PSS in op era tio n 135 with PSS in opera tio n
~120
e ~
Z'
." 105 ~.
e;, c
« ~
90
r 75
>-
60
45 30
L -_.l--_...J-_-l..._
o
Fig.8.54
12
24
36
--L_
48
----l_ _L -_
60
72
..L.-_
84
-l..._ - L_
96
108
-.l._
120
---lL...-_
132
.l--_
144
...J-_- L_--J
156
168
180
Time, c ycles
Effect of the PSS on tran sient stabi lity. (Obtained by private communication and used with permission.)
Recently many studies have been made on the use of various types of compensating networks to meet different situations and stimuli. Most of these studies concentrate on the use of a signal derived from speed or frequency deviation processed through a PSS network to give the proper phase relation to obtain the desired damping characteristic. This approach seems to concentrate on alleviating the problem of growing oscillations on tie lines [II, 13, 14, 24, 26, 30, 33-39). However, in a large interconnected system it is possible to have a variety of potential problems that can be helped by excitation control. Whether the stabilizing signal derived from speed provides the best answer is an open question . It would seem likely that the pr inciple of "optimal control" theory is applicable to this problem. Here signals derived from the various "states" of the system are fed back with different gains to optim ize the system dynamic performance. This optimization is accomplished by assigning a performance index . This index is minimized by a control law described by a set of equations . These equations are solved for the gain constants . This subject is under active investigation by many researchers [40-44) .
Problems 8.1 8.2 8.3 8.4
Construct a block diagram for the regulated generator given by (8.10)-(8.14). What is the order of the system? Use block diagram algebra to reduce the system of Problem 8.1 to a feed-forward transfer function KG(s) and a feedback transfer function H(s) , arranged as in Figure 7.19. Determine the open loop transfer function for the system of Problem 8.2, using the numerical data given in Example 8.3. Find the upper and lower limits of the gain K, for (a) Case 1 and (b) Case 2. Repeat the determination of stable operating constraints developed in Section 8.4 .1, with the following assumptions (see [II J):
366
Chapter 8
Recompute the gain limitations, using the numerical constants K, through K6 given in Table 8.3. 8.5 The block diagram shown in Figure 8.14 represents the machine terminal voltage at no load. The S domain equation for ~/ VREF is given by (8.24). It is stated in Section 8.4.2 that a higher value of regulator gain K, can be used if a suitable lead-lag network is chosen. If the transfer function of such a network is (I + Tts)/(I + T2S), choose T, and T2 such that the value of the gain can be increased eight times. 8.6 In (8.30) and (8.31) assume that K6K » I I K 3 , and TdO » Tfl K 3 • For each of the cases in Example 8.3, plot T, and Td as functions of w between W = 0.1 rad/s and W == 10 rad/s (use semilog graph paper). 8.7 Compute the constants K, through K6 for generator 3 of Example 2.6. 8.8 Determine the excitation control system phase lag of Example 8.7 if a low time constant exciter is used where K, == 400 and T f == 0.05 s. 8.9 Compute the open loop transfer function of the system of Figure 8.28 both with and without the stabilizer. Sketch root loci of each case. 8.10 Analyze the system in Figure 8.29 for a stabilizing signal processed through a bridged Tfi Iter: f
Gs 8.11 8.12
8.13 8.14 8.15 8.16
= (s2
+
rnwns
+ w~)/(S2 +
nwns
+ w~)"
where W n is the natural frequency of the machine, n == 2 and r == 0.1. Sketch Bode diagrams of the several lead compensators described in Example 8.10. Use a linear systems analysis program (if one is available) to compute root locus, time response to a step change in VREF , and a Bode plot for Example 8.11 with (a) A dual lead compensator with a = 15. (b) A triple lead compensator with a == 10. Perform a transient stability run, using a computer library program to verify the results of Section 8.10. Plot E~ and Jt; as functions of time and comment on these results. Modify the block diagram of Figure 5.18 showing the analog computer simulation of the synchronous machine to allow modulating the infinite bus voltage. With the help of the field voltage equation (VF = r rir + ~F), discuss the plots of E FD , E, and E~ shown in Figures 8.43 and 8.44. Explain why the curve for constant EF lJ in Figure 8.42 shows a larger swing than the other field representation.
References 1. Concordia, C. Steady-state stability of synchronous machines as affected by voltage regulator characteristics. AlEE Trans. PAS-63:215-20, 1944. 2. Crary S. B. Long distance power transmission. AlEE Trans. 69, (Pt. 2):834-44, 1950. 3. Ellis, H. M., Hardy, J. E., Blythe, A. L., and Skooglund, J. W. Dynamic stability of the Peace River transmission system. IEEE Trans. PAS-85:586-600, 1966. 4. Schleif, F. R., and White, J. H. Damping for the northwest-southwest tieline oscillations-An analog study. IEEE Trans. PAS-85: 1239-47, 1966. 5. Byerly, R. T., Skooglund, J. W., and Keay, F. W. Control of generator excitation for improved power system stability. PrOf. Am. Power Con! 29: 1011-1022, 1967. 6. Schleif, F. R., Martin, G. E., and Angell, R. R. Damping of system oscillations with a hydrogenating unit. IEEE Trans. PAS-86:438-42,1967. 7. Hanson, O. W .. Goodwin, C. J., and Dandeno, P. L. Influence of excitation and speed control parameters in stabilizing intersystem oscillations. IEEE Trans. PAS-87: 1306-' 3, 1968. 8. Dandeno, P. L., Karas. A. N .. McClymont, K. R., and Watson, W. Effect of high-speed rectifier excitation systems on generator stability limits. IEEE Trans. PAS-87:190-201, 1968. 9. Shier, R. M., and Blythe, A. L. Field tests of dynamic stability using a stabilizing signal and computer program verification. IEEE Trans. PAS-87:315-22, 1968. 10. Schleif, F. R., Hunkins, H. D., Martin, G. E., and Hattan, E. E. Excitation control to improve power line stability. IEEE Trans. PAS-87: 1426-34, 1968. 11. de Mello, F. P., and Concordia, C. Concepts of synchronous machine stability as affected by excitation control. IEEE Trans. PAS-88:316-29,1969. 12. Schleif, F. R., Hunkins, H. D., Hattan, E. E., and Gish, W. B. Control of rotating exciters for power system damping: Pilot applications and experience. IEEE Trans. PAS-88: 1259-66,1969.
Effect of Excitation on Stability
367
13. Klopfenstein. A. Experience with system stabilizing controls on the generation of the Southern California Edison Co. IEEE Trans. PAS-90:698-706,1971. 14. de Mello. F. P. The effects of control. Modern concepts of power system dynamics. IEEE tutorial course. IEEE Power Group Course Text 70 M 62-PWR, 1970. 15. Young, C. C. The art and science of dynamic stability analysis. IEEE paper 68 CP702-PWR. presented at the ASM E-I EEE Joint Power Generation Conference, San Francisco, Calif., 1968. 16. Ramey, D. G., Byerly, R. T .. and Sherman, D. E. The application of transfer admittances to the analysis of power systems stability studies. IEEE Trans. PAS-90:993-1,OOO. 1971. 17. Concordia. C .. and Brown, P. G. Effects of trends in large steam turbine generator parameters on power system stability. IEEE Trans. PAS-90:22 I 1--18, 1971. 18. Perry. H. R.• Luini, J. F.• and Coulter. J. C. Improved stability with low time constant rotating exciter. IEEE Trans. PAS-90:2084--89. 1971. 19. Brown. P. G .. de Mello. F. P.. Lenfest, E. H., and Mills, R. J. Effects of excitation. turbine energy control and transmission on transient stability. IEEE Trans. PAS-89: 1247-53.1970. 20. Melsa, J. L. Computer Programs for Computational Assistance in the Study of Linear Control Theory. Mcfiraw-Hill, New York. 1970. 21. Duven, D. J. Data instructions for program LSA P. Unpublished notes. Electrical Engineering Dept., Iowa State University, Ames. 1973. 22. Kuo. Benjamin C. Automatic Control Svstems. Prentice-Hall. Englewood Cliffs, N.J .. 1962. 23. Gerhart, A. D.. Hillesland, T .. Jr .. Luini, J. F., and Rockfield, M. L., Jr. Power system stabilizer: Field testing and digital simulation. IEEE Trans. PAS-90:2095-2101. 1971. 24. Warchol, E. J .• Schleif, F. R., Gish, W. B. and Church, J. R. Alignment and modeling of Hanford excitation control for system damping. IEEE Trans. PAS-90:714·-25, 1971. 25. Eilts. L. E. Power system stabilizers: Theoretical basis and practical experience. Paper presented at the panel discussion "Dynamic stability in the western interconnected power systems" for the IEEE Summer Power Meeting, Anaheim. Calif., 1974. 26. Keay, F. W.. and South. W. H. Design of a power system stabilizer sensing frequency deviation. IEEE Trans. PAS-90:707--14. 1971. 27. Bolinger. K., Laha, A., Hamilton, R.. and Harras, T. Power stabilizer design using root-locus methods. IEEE Trans. PAS-94: 1484 88, 1975. 28. Schroder. D. C .. and Anderson. P. M. Compensation of synchronous machines for stability. IEEE paper C 73-313-4. presented at the Summer Power Meeting, Vancouver, B.C., Canada. 1973. 29. Bobo, P.O., Skooglund, J. W., and Wagner, C. L. Performance of excitation systems under abnormal conditions. IEEE Trans. PAS-87:547-53.1968. 30. Byerly. R. T. Damping of power oscillations in salient-pole machines with static exciters. IEEE Trans. PAS-89: 1009--21. 1970. 31. McClymont. K. R.. Manchur. G .. Ross. R. J .. and Wilson. R. J. Experience with high-speed rectifier excitation systems. IEEE Trans. PAS-87: 1464-70. 1968. 32. Jones. G. A. Phasor interpretation of generator supplementary excitation control. Paper A75-437-4. presented at the IEEE Summer Power Meeting. San Francisco, Calif., 1975. 33. El-Sherbiny, M. K.. and Fouad, A. A. Digital analysis of excitation control for interconnected power systems. 1£££ Trans. PAS-90:441-48. 1971. 34. Watson. W.. and Manchur, G. Experience with supplementary damping signals for generator static excitation systems. IEEE Trans. PAS-92:199--203. 1973. 35. Hayes. D. R.. and Craythorn. G. E. Modeling and testing of Valley Steam Plant supplemental excitation control system. IEEE Trans. PAS-92:464--70, 1973. 36. Marshall. W. K.. and Smolinski. W. J. Dynamic stability determination by synchronizing and damping torque analysis. Paper T 73-007-2. presented at the IEEE Winter Power Meeting. New York, 1973. 37. El-Sherbiny, M. K.. and Huah, Jenn-Shi. A general analysis of developing a universal stabilizing signal for different excitation controls. which is applicable to all possible loadings for both lagging and leading operation. Paper C74-I06-1. presented at the IEEE Winter Power Meeting, New York, 1974. 3~L Bayne, J. P.. Kundur, P.. and Watson. W. Static exciter control to improve transient stability. Paper T74-521-1, presented at the IEEE-ASM E Power Generation Technical Conference, Miami Beach, Fla .. 1974. . 39. Arcidiacono. V.. Ferrari. E., Marconato, R.. Brkic, T .. Niksic, M.. and Kajari, M. Studies and experimental results about electromechanical oscillation damping in Yugoslav power system. Paper F75-460-6 presented at the IEEE Summer Meeting, San Francisco, Calif., 1975. 40. Fosha, C. E.. and Eigerd. O. I. The megawatt-frequency control problem: A new approach via optimal control theory. IEEE Trans. PAS-89:563-77.1970. 41. Anderson, T. H. The control of a synchronous machine using optimal control theory. Proc. IEEE-59:25-35, 1971. 42. Moussa. H. A. M., and Yu, Yao-nan. Optimal power system stabilization through excitation and/or governor control. IEEE Trans. PAS-91: 1166-- 74, 1972. 43. Hurnpage. W. D .. Smith, J. R .. and Rogers. G. T. Application of dynamic optimization to synchronous generator excitation controllers. Proc. lEE (British) 120:87-93, 1973. 44. Elrnetwally. M. M.. Rao, N. D. and Malik, O. P. Experimental results on the implementation of an optimal control for synchronous machines. IEEE Trans. PAS-94: 1192-1200. 1974.
chapter
9
Multimachine Systems with Constant Impedance Loads 9. 1
Introduction
In this chapter we develop the equations for the load constraints in a multimachine system in the special case where the loads are to be represented by constant impedances. The objective is to give a mathematical description of the multimachine system with the load constraints included. Representing loads by constant impedance is not usually considered accurate. It has been shown in Section 2.11 that this type of load representation could lead to some
error. A more accurate representation of the loads will be discussed in Part III of this
work. Our main concern here is to apply the load constraints to the equations of the machines. We choose the constant impedance load case because of its relative simplicity and because with this choice all the nodes other than the generator nodes can be eliminated by network reduction (See Section 2.10.2). 9.2
Statement of the Problem
In previous chapters, mathematical models describing the dynamic behavior·of the synchronous machine are discussed in some detail. In Chapter 4 [see (4.103) and (4.138)] it is shown that each machine is described mathematically by a set of equations of the form (9.1) where x is a vector of state variables, v is a vector of voltages, and T'; is the mechanical torque. The dimension of the vector x depends on the model used. The order of x ranges from seventh order for the full model (with three rotor circuits) to second order for the classical model where only wand () are retained as the state variables. The vector v is a vector of voltages that includes ud , uq , and UF. If the excitation system is not represented in detail, uF is assumed known; but if the excitation system is modeled mathematically, additional state variables, including vF, are added to the vector x (see Chapter 7) with a reference quantity such as VREF known. In this chapter we will assume without loss of generality that vF is known. Consider the set of equations (9.1). In the current model developed in Chapter 4, it represents a set of seven first-order differential equations/or each machine. The number of the variables, however, is nine: five currents, wand b, and the voltages vd and uq • Assuming that there are n synchronous machines in the sy.stem, we have a set of Tn differential equations with 9n unknowns. Therefore, 2n additional equations are 368
Multimachine Systems with Constant Impedance loads
369
needed to complete the description of the system. These equations are obtained from the load constraints. The objective here is to derive relations between vd; and Vq;, i = 1, 2, ... , n, and the state variables. This will be obtained in the form of a relation between these voltages, the machine currents i q; and i d; , and the angles 0;. i = 1,2, ... , n. In the case of the flux linkage model the currents are linear combinations of the flux linkages, as given in (4.124). For convenience we will use a complex notation defined as follows. For machine i we define the phasors V; and ~ as (9.2)
where Vq; ~
t;
vq;/vr iq;/vr
~
vd;/vr ld; ~ id;/vr
Vd; ~
(9.3)
and where the axis q; is taken as the phasor reference in each case. Then we define the complex vectors V and I by
V
~
T~
Vq 1 + j Vd 1
VI
Vq 2 + j Vd 2
il2
Vqn + j Vd n
~
lq,
+ j/dl
t,
Iq2
+
j/d2
t;
t.; +
j1dn
T"
(9.4)
Note carefully that the voltage ~ and the current ~ are referred to the q and d axes of machine i. In other words the different voltages and currents are expressed in terms of different reference frames. The desired relation is that which relates the vectors V and T. When obtained, it will represent a set of n complex algebraic equations, or 2n real equations. These are the additional equations needed to complete the mathematical description of the system. 9.3
Matrix Representation of a Passive Network
Consider the multimachine system shown in Figure 9.1. The network has n machines and r loads. It is similar to the system shown in Figure 2.17 except that the machines are not represented by the classical model. Thus, the terminal voltages Vi' i = 1,2, ... , n, are shown in Figure 9.1 instead of the internal EMF's in Figure 2.17. Since the loads are represented by constant impedances, the network has only n active sources. Note also that the impedance equivalents of the loads are obtained from the pretransient conditions in the system. By network reduction the network shown in Figure 9.1 can be reduced to the n-node network shown in Figure 9.2 (see Section 2.10.2). For this network the node currents respecand voltages expressed in phasor notation are 1;, 1;, ... , T" and ~. V;, ... , tively. Again we emphasize that these phasors are expressed in terms of reference frames that are different for each node. At steady slate these currents and voltages can be represented by phasors to a com-
v;,
370
Chapter 9
--
lL
In
Vn
1
n
•
--
..
Tran smissi on syste m
•
I.
--I
0
V.
Lr
V,
Fig. 9. 1.
Mult imachin e system with con stant impedance load s.
mon reference frame . To distingui sh these phasors from tho se defined by (9.2), we will use the symbols t, and Vi' i = I. 2• . .. , n. to designate the use of a common (network) frame of reference . Similarl y, we can form the matrices i and V. From the network steady-state equations we write where
~
i
[n
(9 .5)
(9.6)
V. [:.:]
InJ
Vn
and V is the short circuit admittan ce mat rix of the network in Figure 9.2. 9 .3 .1
Network in the transient state
Consider a branch in the reduced network of Figure 9.2. Let this branch, located between any two nodes in the network. be identified by the subscript k . Let the branch 1 ____ n
•• •
1.___
n 2
1..0+
~_Vn
+
?_V. F ig. 9.2.
?-"
I 0
-
Reduced n-port network .
Multimachine Systems with Constant Impedance loads
371
resistance be 'In its inductance be {k' and its impedance be Zk' The branch voltage drop and current are v, and i k • In the transient state the relation between these quantities is given by
k
=
1,2, ... , b
(9.7)
where b is the number of branches. Using subscripts abc to denote the phases abc, (9.7) can be written as
k
=
1,2, ... , b
(9.8)
This branch equation could be written with respect to any of the n q-axis references by using the appropriate transformation P. Premultiplying (9.8) by the transformation P as defined by (4.5),
(9.9) Then from (4.31) and (4.32)
o (9.10)
Substituting (9.10) in (9.9) and using (4.7). (9.11 )
which in the case of balanced conditions becomes
(9.12) It is customary to make the following assumptions: (1) the system angular speed does not depart appreciably from the rated speed, or w "'" WR and (2) the terms t i are negligible compared to the terms wti. The first assumption makes the term wtl
(9.) 3)
Equation (9.13) gives a relation between the voltage drop and the current in one branch of the network in the transient state. These quantities are expressed in the q-d frame of reference of any machine. Let the machine associated with this transformation be i. The rotor angle 0; of this machine is given by (9.14) where 0; is the angle between this rotor and a synchronously rotating reference frame.
372
Chapter 9
,
d.
ReFerence frome
(mov ing 01 synchronous speed )
Fig . 9 .3.
Position of axe s of rotor k with respect to reference frame .
From (9 .13) multiply both sides by I /~; and using (9 .3), (9 .15) where the subscript; is added to indicate that the rotor of machine; is used as reference . Expressing (9.15) in phasor notation,
ilkli}
VqkliJ
+ j Vdklil
(r k Iqklil - x, Im iJ)
+ jerk Idk1i) + Xk Iqk1iJ)
v. + jXdUq k + j/dd
or
k
1,2, . . . , b,
=
(9 .16)
Equation (9.16) expresses, in complex phasor notation, the relat ion between the voltage drop in bran ch k and the current in that branch. The reference is the q axis of some (hypothetical) rotor; located at angle 0; with respect to a synchronously rot ating system reference, as sho wn in Figure 9.3 . 9 .3.2
Converting to a common reference frame
To obtain general network relationships, it is desirable to express the various branch quantities to the same reference. Let us assume that we want to convert the phasor V; = Vq; + j Vdi to the common reference frame (moving at synchronous speed). Let the same voltage, expressed in the new notation, be V; = VQ; + j VD; as shown in Figure 9.4. From Figure 9.4 by inspection we can show that
VQi
+ j VD ;
= (Vq; cos 0; -
Vd; sin 0;)
+ j( Vq; sin 0; + Vdi cos 0i)
or (9.17) Now convert the network branch voltage drop equation (9.16) to the system reference frame by using (9 .17) . -
Vke
-j~ .
zJke-j~;
I
or
Vk
ZJk
k
=
1,2, . .. , b
(9 .18)
where b is the number of branches and Zk is calculated based on rated angular speed. Comparing (9.18) and (9 .5) under the assumptions stated above, the network in the transient state can be described by equations similar to those describing its steady-state
Multimachine Systems with Constant Impedance loads
373
di
--- --- - --
V Di
--
.i- >:
qi
Fig. 9.4 .
Two frames of reference for phasor quantities for a voltage Vi'
behavior. The network (branch) equations are in terms of quantities expressed to the same frame of reference , conveniently chosen to be moving at synchronous speed (it is also the system reference frame) . Equation (9.18) can be expressed in matrix form (9.19)
where the subscript b is used to indicate a branch matrix. The inverse of the primitive branch matrix Zb exists and is denoted rb, thus (9 .20)
Equation (9 .20) is expressed in terms of the primitive admittance matrix of a passive network . From network theory we learn to construct the node incidence matrix A which is used to convert (9.20) into a nodal admittance equation
i
=
(A'YbA)V ~ YV
(9.21 )
where Y is the matrix of short circuit driving point and transfer admittances and I if current in branch p leaves node q - I if current in branch p enters node q o if branch p is not connected to node q with p = I, 2, . .. , band q Since Y - 1 ~ Z exists,
(9.22)
1,2 , .. . , n . (9.23)
where Z is the matrix of the open circuit driving point and transfer impedances of the network. (For the derivation of (9 .21)-(9.23), including a discussion of the properties of the Y and Z matrices, see reference [I]. Chapter II .) 9.4
Converting Machine Coordinates to System Reference
Consider a voltage "ab ri at node i. We can apply Park's transformation to this voltage to obtain "dqi' From (9 .2) this voltage can be expressed in phasor notation as V;, using the rotor of machine i as reference. It can also be expressed to the system reference as Vi' using the transformation (9 .17).
Chapter 9
374
Equation (9.17) can be generalized to include all the nodes. Let ej 6 1
0
0
0
ej 62
0
0
0
ejhn
T=
V
VOl
+ jVD I
V0 2
+ jVD 2
V=
(9.24)
Vql
+ jVd1
Vq2
+ jVd2
(9.25)
Then from (9.2), (9.14), (9.17), and (9.25)
V
=
TV
(9.26)
Thus T is a transformation that transforms the d and q quantities of all machines to the system frame, which is a common frame moving at synchronous speed. We can easily show that the transformation T is orthogonal, i.e., Therefore, from (9.26) and (9.27)
T:" = T*
(9.27)
v = T*V
(9.28)
Similarly for the node currents we get
i 9.5
=
TI
I =
r-t
(9.29)
Relation between Machine Currents and Voltages
From (9.22) i
=
YV." By using (9.29) in (9.22),
rr
=
YTV
(9.30)
Premultiplying (9.30) oy T:" (9.31) where (9.32) and if M -I exists, (9.33) Equation (9.33) is the desired relation needed between the terminal voltages and currents of the machines. It is given here in an equivalent phasor notation for convenience and compactness. It is, however, a set of algebraic equations between 2n real voltages Vq" Vd1, ... , Vqn, Vdn, and 2n real currents Iq1, Idl , ... , Iqn, Jdn. Example 9.1
Derive the expression for the matrix M for an n-machine system.
Multimachine Systems with Constant Impedance loads
375
Solution
The matrix Y of the network is of the form
Y=
Y II ej lJ11
Y 12ej tJ 12
Y21 e j 82 1
Y ej (J22
Y2n e j (J 2n
Y
Y ej (Jn2
Ynn e j 6nn
nl
ej(Jnl
22
n2
(9.34)
and from (9.24) e -jh l
(9.35)
T From (9.34) and (9.35)
YT and premultiplying by T- ', we get the desired result YII e
j8 11
Y12e
j(tJ 12- 6 12)
Y21 e
j(821 -6 2 1)
Y22 e
j 822
(9.36)
To simplify (9.36), we note that Yi/ce j( (Jik -
hi/c) _
-
(
Gik cos uik ! ' t) .( Bik cos ui/c t + B ik sin ui/c + J
-
G '!) ik sin "t«
Now define FG+B(Oi/C) =
FG + B = Gik cos
FB-G(Oi/C)
F B- G =
Bile
Oi/c
cos Oi/c
+ -
sin Oi/c Gilc sin 0ik Bi/c
(9.37)
Then the matrix M is given by M
=
H + jS
(9.38)
where Hand S are real matrices of dimensions (n x n). diagonal terms are given by S;i
Example 9.2
e,
Their diagonal and off(9.39)
Derive the relations between the d and q machine voltages and currents for a twomachine system.
Chapter 9
376
Solution From (9 .31) and (9 .38)
r
= (H + jS) [Vql Vqn =
(HV q
-
~
j VJ] = (H +
jsxv, + jVd )
+ j Vdn
SVd ) + j(SVq + HVJ )
(9.40)
For a two-machine system the q axis currents are given by /ql] [ I q2 =
[Gil
FG+ B(021)
(ol2 )l [VqIJ
FG+ B G22
and the d axis currents are given by
J Vq2
-
I l
f"B _G(OI2)] Vql] +
8 22
Vq2
[
8\1
FB_ G(021)
l
Gil
FG+ B(021)
FG+B(OI 2~ [VdIJ G22
J Vd2
We note that a relation between the voltages and currents based upon (9 .33) (i.e., giving Vql, Vq2• Vdl• and Vd2 in terms of t.; I q2, I dl , and I d2) can be easily derived . It would be analogous to (9.40) except that the admitt ance parameters are replaced with the parametersof the Z matr ix of the network . Example 9.3. Derive the complete system equations for a two-machine system . The machines are to be represented by the two-axis model (see Section 4 .15 .3). and the loads are to be represented by constant impedances. Solution The transient equivalent circuit of each synchronous machine is given in Figure 4.16. A further approximation, commonly used with this model, is that x;""'" x; ~ x' . The network is now shown in Figure 9.5. The representation is similar to that or the classical model except that in Figure 9 .5 the voltages E; and Ei are not constant. The first step is to reduce the network to the "internal" generator nodes 1 and 2. Thus the transient generator impedances + jx; and r, + jxi are included in the network Y (or Z) matrix. The voltages at the nodes are E; = E;I + jE;, and Ei = E;2 + jE;2. and the currents are 7; = I ql + j/dl and T;. = I q2 + j/J 2' The relation between them is
'1
I;
o
Fig . 9.5.
Network of Example 9.3.
377
Multimachine Systems with Constant Impedance Loads
given by an equation similar to (9.40). The equations for each machine, under the assumption that x~ ~ x;, are the two axis equations of Section 4.15.3. , E"
T qO;
,
-
di
EO..,
T dO;
E~i
- (X q ;
£FDI - E;i
qi
Tm ;
Ti;W i
iJ;
+ (Xdi - x;> ldi + Iq;£ ;;) - Diwi i = 1,2
(ldiE~i
-
W;
x:) Iq ;
-
1
(9.41 )
Equations (9.40), with Vi replaced with l:, and (9.41) completely describe the system. Each machine represents a fourth-order system, with state variables £;i' £~i' Wi' and 0i' The complete system equations are given by -[1 - (x ql - x;) B II
, EO,
r q02
d2
, E',
TdOI
ql
°
]
£~I
- (x ql - x;) GII £ ; ,
- (x q I - x;) F G+ B ( 12) E; 2
°
+ (x q 1 -. X;) F B - G( 12) £;2
-[ 1 - (X q2 - X~) B 22] £~2 - (X q2 - X~) Gn E; 2 - (X q2 - X~) F C +B(021) £;1 + (X q2 - X~) F B- G(02l) £;1 E FDI - [I - (Xdl - x;) BId £;1
+ (X dl
- X;)[G II £~I
+
F B- G(ol2 ) £;2
+
FG+ B(ol2 ) £;2]
£FD2 - [I - (Xd2 - X~) B22] £;2
+
(X d2
Tm,
+
-
- X~)[FB-G(021) £;1
FG+B(0'2)(E~IE~2
Tm2
+
FG +B (021) £~I
+
G22£~2]
D I WI - [GIl (£~f + £;f) + FB-G(0I2)(£~1 £;2 -
D2 W 2
-
+ E~IE~2)]
[G22(Ed~ + E;~)
-
£;, £;2)
+ FB _ G(021) (£;1 £;2 - £;2 Edt>
+ FG+B(02I)(E~IE~2 + E;I£~2)] WI -
b2
I
=
W2 -
1
(9.42)
The system given by (9.42) is not an eighth-order system since the equations are not independent. This system is actually a seventh-order system with state variables E;" Ed" £;2' Ed2, WI' W2' and 0 12 , The reduction of the order is obtained from the last two equations Furthermore, if damping is uniform: i.e., if DI/T;t = D2/ T;2 = D/T; (or if damping is not present) then the system is further reduced in order by one, and the two torque equations can be combined in the form WI2 =
9,6
Tm , Til
-
m2
T
T i2
+ 1(£;"
£~I' £;2' £~2' on> - ~ WI2 T;
System Order
In Example 9.3 it was shown that with damping present the order of the system was reduced by one if the angle of one machine is chosen as reference. It was also pointed out that if damping is uniform, a further reduction of the system order is achieved. We now seek to generalize these conclusions. We consider first the classical model with zero transfer conductances. We can show that the system equations are given by
378
Chapter 9
L n
Tj;W;
+ D;wi =
.i=1
E;EjBi;(sin 0;; - sin o~;)
j-;.;
i
=
1,2, ... ,n
(9.43)
where the superscript s indicates the stable equilibrium angle. tor x, the vector CT, and the function f by
h(
x'
(J
k)
= [WI' W2' ••• ,Wn'
(01
-
O'i), (02
-
(
= EpEqBpq [sin ( (Jk + O;q) - sin O;q]
Defining the state vec-
2), ... , (on -
O~)]
k = I, 2, ... , m
m
=
n(n - 1)/2
and (1 = C x where C is a constant matrix. The system (9.43) may then be written in the form x=Ax-Bf(u)
(9.44)
where A and B are constant matrices. The order of the system (9.44) is determined by examining the transfer function of the linear part (with s the Laplace variable) W(s) = C(sl - A)-I B
(9.45)
This has been done in the literature [2, 3]. Expanding (9.45) in partial fractions and examining the ranks of the coefficients obtained, the minimal order of the system is obtained. It is shown that the minimal order for this system is 2n - 1. For the uniform damping case, i.e., for constant D;/T;;, the order of the system becomes 2n - 2 (see also [4]). The conclusions summarized above for the classical model can be generalized as follows. If the order of the mathematical model describing the synchronous machine i is k;, i = 1,2, ... , n, and if damping terms are nonuniform damping, the order of the system is (L7= I k, -1). However, if the damping coefficients are uniform or if the damping terms are not present, a further reduction of the order is obtained by referring all the speeds to the speed of the reference machine. The system order then becomes (L?= 1 k; - 2). The above rule should be kept in mind, especially in situations where eigenvalues are obtained such as in the linearized models used in Chapter 6. Unless angle differences are used, the sum of the column of o's will be zero and a zero eigenvalue will be obtained (see Section 9.12.4). 9.7
Machines Represented By Classical Methods
In the discussion presented above, it is assumed that all the nodes are connected to controlled sources, with all other nodes eliminated by Kron reduction (see Chapter 2, Section 2.10.2). The procedure used to obtain (9.31) assumes that all the machines are represented in detail using Park's transformation. For these machines we seek a relation, such as (9.3 l ), between the currents I and the voltages V. The former are either among the state variables if the current model is used, or are derived from the state variables if the flux linkage model is used (see [5]). If some machines are represented by the classical model, the magnitudes of their internal voltages are known. If machine r is represented by the classical model, the angle (), for this machine is the angle between this internal voltage and the system reference axis. In phasor notation the voltage of that node, expressed to the system refer-
379
Multimachine Systems with Constant Impedance loads
ence, is given by ~
= VQ, +jVD, = E,coso, +jE,sino,
(9.46)
At any instant if 0, is known, VQ, and VOr are also known. Since the voltage E, is considered to be along the q axis of the machine represented by the classical model, we can also express the voltage of this machine in phasor notation as
ii:
= E, + jO
r
=
1,2, ... , C
(9.47)
where c is the number of machines represented by the classical model. (4.93) on a per phase base
Also from
Dividing both sides by three changes the base power to a three-phase base and divides each voltage and current by W, converting to stator rms equivalent quantities. Thus we have
and using (9.47), (9.48)
Note that E, is in per unit to a base of rated voltage to neutral. Assuming that the speed does not deviate appreciably from the synchronous speed, then T, '" P, and from the swing equation (4.90) on a three-phase base
5 r
= wr -
1
(9.49)
A machine r represented by the classical model will have only w, and D, as state variables. In (9.49) E, is known, while lqr is a variable that should be eliminated. To do this we should obtain a relation between I q, and the currents of the machines represented in detail. Similarly the voltages ~i and ~i of the machines represented in detail should be expressed in terms of the currents lq; and Id; of these machines and the voltages E, of the machines represented classically. To obtain the above desired relations, the following procedure is suggested. Let m be the number of machines represented in detail, and c the number of machines represented by the classical model; i.e., m
Let the vectors I and
T=
+
cAn
V be partitioned as
lqm
+ j1dm
---- - - - - -
I qm+ I +
j/dm+ I
lqn + jldn
~
[~~]
v=
~m + j~m
--------
Em+ 1 +jO En + jO
~
~:]
(9.50)
380
Chapter 9
Then from (9.50) and (9.31) (9.51)
where in (9.51) the complex matrix M is partitioned. Now since Mill exists, (9.51) can be rearranged with the aid of matrix algebra to obtain (9.52)
Equation (9.52) is the desired relation between the voltages of the machines represented in detail along with the currents of the machines represented classically, as functions of the current variables of the former machines and the known internal voltages of the latter group. We note that the matrices Mil' M 12 , M 2 1 , and M 22 are functions of the angle differences as well as the admittance parameters. Example 9.4 Repeat Example 9.2 assuming that machine 1 is represented in detail by the twoaxis model and machine 2 by the classical model. Solution From (9.37) and using
Y12
=
M
=
f 21 and
btl =
[~~I~j~1~
Y12e j(912+o 12)
-021'
r_ ~~~J~~~6~2-J
(9.53)
Y22 e J922
I I
and from (9.53) by inspection
M-I II
=
_1_ e- j 8 11 fit
Y I 2 e j(8 12 -8 11 +6 12) Y I 2 e j ( 6l r Y II
b12)
Yt2
YII
Yn ej8n _ Yf2 e j (28 12 - 6 11) YII
eH2612-811)
(9.54)
From (9.50) and (9.52) _1_ e- j8 11
r.,
: '_
~ ej( 8 12 - 8 11- ol2 )
Y II ----------+----------------
~ e j ( 61r Yl1
I
I
6 11 +612) :
I
Yn e j 622
_
Yf2 'e j (2812 - 6 11)
r.,
(9.55)
Multimachine Systems with Constant Impedance loads
381
or ~I
(9.56 ) Note that the variables needed to solve for the swing equations are only and I q 2 •
~I' ~"
Example 9.5
Repeat Example 9.3, with machine I represented mathematically by the two-axis model and machine 2 by the classical model. Solution Again the nodes retained are the "internal" generator nodes, and the transient impedances of both generators are included in the network Y (or Z) matrix. The equations needed to describe this system are (9.41) for generator 1, (9.49) for generator 2, and an additional set of algebraic equations relating the node currents to the node voltages. Since the two-axis model retains £; and £; as state variables, it is convenient to use (9.51). For the two-machine system this is the same as (9.40), with E~ replacing VI and E~ = £2 + jO replacing V2 • The system is now fifth order. The state variables for this system are £;" £;" WI' W2' and 0'2' The complete system equations are given by
, e:dl
TqO I
, i:
TdOI
ql
TjlW I
~i2W2
hl2
9.8
[B,,(x q ,
-
1]£;, - (x q l
x~) -
£FDI
+ [BII(X d ,
WI
W2
-
x;)[GIIE;, - F G +B(OI2)£ 2]
+ (Xd l - XD[GIIE~I + FB - G (o,2)E2 ] Tm, - Dv», - [G'I(E~r + E;f) + FB- G(ol2 ) £; ,E2 + F G+B(OI2)E;I E2] Tm2 - D 2W 2 - E 2[FG+ B (021) £;1 - F8_G(021)E~1 + G22E2 ] -
-
x;) - I]E;,
(9.57)
Linearized Model for the Network
From (9.26) V = TV, where T is defined by (9.24) and V and V are defined by (9.4) and (9.17). Also from (9.31) T = MV, where M is given by (9.32). Linearizing (9.31), (9.58)
382
Chapter 9
where Mo is evaluated at the initial angles 0;0' i = 1,2, ... , n, and Vo is the initial value of the vector V. Let 0; = 0;0 + 0;4' Then the matrix M becomes Y"e
Y12 e j (812 - h 120 - h 12.1)
j 811
YlneH81n -0 InO- h Inu)
M
(9.59) Yn1ej(lI n, -OnlO-Onl.l)
Y
j(lJ n2 - 0n20 - 0n2~)
n 2e
The general term m., of the matrix M is of the form
m..
=
'J
Using the relation cos
O;jA
"J
I, sin
mij
YnnejlJnn
Yije
j(8ij-6ijO-6;jA),
thus
y..e j(8 ij-6ijo) e --j6ij 4 'J
0;;4
"J
Oij4,
we get for the general term
Yij e j(8ij+6ijO) (I - jO;jA)
"J
(9.60)
Therefore the general term in M A is given by (9.61) Thus M A has off-diagonal terms only, with all the diagonal terms equal to zero.
o
~O
~O y
nl
e
o
j(8,,1- 6,,10) ~ U"IA
~O
k-'
(9.62)
"
L
k.1
and the linearized equation (9.58) becomes IIA
/2A
Y II e y
21 e
j 8 11
j(821 - 6210)
~ j(81,.-&1,,0) I"e
Y
2n e
j(82,.-&2,.O)
"
V.4
k.1
P;A
-j
InA
Y:
"Ie
j(6,. 1- 6n 10)
Y ,." e
j 8""
L,. L
1e.1
~
Y
~
y e j(8 21e - '21eO)0 21e 2leA
kO
leO
n
~A
L
1e.1
~ leO
y
lk
nk
ej(8Ik-6IkO)
e
j(8 nle- &nkO)
0
IleA
0
"leA
(9.63)
Multimachine Systems with Constant Impedance loads
383
The set of equations (9.63) is that needed to complete the description of the system. A similar equation analogous to (9.63) can be derived relating Y.1 to 1.1 and IJijti. The network elements involved in this case are elements of the open circuit impedance matrix Z. We now formulate (9.63) in a more compact form. From (9.24) let T = To + T.1 to compute (9.64)
Similarly, we let T- I i N
=
No + N4 to compute (9.65)
Note carefully that T- I show, however, that (TO)-I
M o + M A = (No +
rr.r '
o
T l + T~I and that :/= (T-I)A = N A. We can = (T-I)o = No- Thus from M = M o + M A we compute NA)Y(T o + T A). Neglecting second-order terms, :/=
(9.66)
From matrix algebra we get the following relations,
YIn e j 8 1n
Ynl e j l1n l
=
~Y
lie
j(6 11- lJlO)
Ynn e j 6nn
'
...
Y",e j(8"I-lJ"O)
r l l e j 8 11
r
21
e j(621-6210) Ynne j8 nn
Also
(9.67)
384
Chapter 9
(9.68) From (9.66), (9.67), and (9.68) MA
-j[tlAM o - Motlt.\]
(9.69)
and the network equation is given by I~
=
MoVA - j[oAMo - MOoA]VO
(9.70)
Note that (9.70) is the same as (9.63). To obtain a relation between VA and lA, we can either manipulate (9.70) to obtain VA
=
MoirA - j[OA - Mo'oAMo]Vo
(9.71)
or follow a procedure similar to the above. Define We can then show that
Q : M- ' = T-1y-IT
(9.72) (9.73)
Example 9.6
Derive the relations between VA and
t,
for a two-machine system.
Solution From (9.53) we get for M o
(9.74)
(9.75)
Multimachine Systems with Constant Impedance loads
385
(9.76)
(9.77)
] = [ r II e j011 Vq I A + J. Y II e j 8 11 Vd I A + dIA [Iql + J' 1 A
u
Iq2~ + J'1d2~
Y 12e j( 8 12 -
Y e j(8 12+ o I20 ) V . Y j(812+ 6120> l2 qlA + J 12 e VdlA
j Y 12 e j(8 12-
cS
120) (
Vq 20
+ j Vd 20 )
]
[ -J. Y 12 e j(8 12+o120) (VqlO + J. VdID )
6
J
120 > V q 2A
+ J. Y Il e j(812 -!S120) Vd26 Y j 8 22 V . j822 + 22 e q2A + JY22 e Vd2~
'
(9.78)
~12A
By separating the real and the imaginary terms in equation (9.78), we get four real equations between I q 16 , Id I A , I q 26 , and I d 26 and ~16' V d 16, V q2A, Vd2~' and 012A. These are given below:
Iq t 6 Id l A
=
G"
+
B"
Vqt 6 -
+
Vdt6
Y)2[sin(OI2 - 0)20)
B I I Vq l A + Y 12 cos (8 1l
-
Id 2A
+ +
+ cos (0 12
0120}
-
( 120) Vd2~
-
0120) V d 26
(120) V d20]012A
(120) V q 2A
+ Y12 cos (012
+ sin (0 12 - 0120) V d20]012A Y 12 sin (8 12 + OIlO) Vd l A + G22 V q 2A + cos (012 + 0120) Vd lO] 0 12 A
Vq lD
8 22 V d2 A
-
+ Y I2 cos (8 12 + 0120) Vd l A + B 22 Vq2~ + + 0 120) Vq 10 - sin (Oil + 0 120 ) Vd 10 ] 0 126
0120)
+ Y 12 [cos (0 12
-
-
Y'2 sin ( 0 12
0 120) V q 20
0120) V q l A -
Y12 [sin(0I2 +
Y I2 sin (0 12
V q20
Vd l A + Y I2 sin (012
Gil
+ Y12 [ -cos(OI2 I q 2A
Y12cos(0I2 - (120) Vq2~ -
Vq l A
G 22 Vd 2A
(9.79)
Example 9.7
Linearize the two-axis model of the synchronous machine as given by (9.41) and the classical model as given by (9.48).
Solution
From (9.41) we get T~oEdA TdOE;A Tj W t1 06
=
- Ed A £FDt1 -
Tm 6 wt1
-
(x, - x')/ q6
£;6 + (x, - X')/ d 6 DWA -
(ldOE d A
+ Iqo E ; 6 +
EdO/d~
+
E;olqA)
(9.80)
From (9.48) we get (9.81 )
386
Chapter 9
Example 9.8
Linearize the two-machine system of Example 9.5. One machine is represented by the two-axis model, and the second is represented classically. Solution
From (9.79), (9.80), and (9.81) and dropping the a subscripts for convenience, , E·' [(Xql - x:)B II ~ I]E dl - Gil (Xql - X~)E~1 TqOI dl - [(xql - x~) Yl2 E2 sin (8 12 - 0120)] 012 , E·' E FD I + [(Xdl - x;)B II - I]E;1 + (Xdl - X;)GII£dl T dOl ql - [(Xdl - x:) Y 12E2cos (012 - 0120)]012 T ml -
DIWI
1(£dIOBIl + E;IO G II + Iqlo)£;1 + (EdIOG II - £;loB II + IdIO)Edl + Y 12 E2[E; 10 sin (812 - 0120) - E dlOCOS (812 - 0120)] 0121
-
Tm2 - D 2W 2
-
+ 0120)] £;1 - [Y 12 sin (012 + 0120)] Edl - YI2[£;lOsin«(J12 + 0120) + EdIOCOS«(J12 + 0120)]0121
E 2HYI2 cos (012
(9.82)
Equation (9.82) is a set of five first-order linear differential equations. It is of the form Bu, where
x = Ax +
u
£;1 (Xlii -
xi)B II
E~,
I
-
(Xlii -
T~OI -(Xql -
Tdol
xi )G II
(Xql -
-(£~10811
+ E;IOGII + l q lo )
-
I
-(£dIOGII - £;loB II + I l110 ) Tjl
TJl
-E2YI2COS(012
xi)B II
T;OI
TqOI
A=
WI
xi)G II
+
~120)
£2 Y 12
sin (0\2 + 0120)
TJ 2
T/2
0
0
0
0
0
0
-DlfT/,
0 -I
~12
W2 -(Xdl -
xI)Y 12E 2cos(8 12
-
~12()
0
-Dd T/ 2 -I
YI2£2(E~IOsin«(J12
+ 0120) +
r»
E~IOCOS(lJ'2
+ 0120»)
o (9.84)
From the initial conditions, which determine £dIO, E~IO' E2 , 1;10, Idlo, and 0120 and from the network Y matrix all the coefficients of the A matrix of (9.84) can be determined. Stability analysis (such as discussed in Chapter 6) can be conducted. We note again (as per the discussion in Section 9.6) that the order of the mathematical description of machine I is four, that of machine 2 is two. The system order, however, is 4 + 2 - I = 5. If the damping terms are not present, the variables WI and W2 can be combined in one variable WI2. 9.9
Hybrid Formulation
Where a combination of classical and detailed machine representations exists, a hybrid formulation is convenient. Let m machines be represented in detail, and c machines represented classically, m + c = n. Then from (9.58),
Multimachine Systems with Constant Impedance loads
~~
VA =
Vm~
387
~{~~~}
0
From (9.70) (9.85) where the subscript m indicates a vector of dimension m. By comparing (9.85) and (9.63),
(9.86)
where Km(~A) is an (m x I) vector and Kc(6A) is a (c x 1) vector. From (9.85) and (9.86)
t:~j [~::~ -i- -~:~[~~j -~~~::] =
Therefore
I mA =
MOil
j
Vm~
from which we get
-
rc~ =
(9.88)
jKm(6~)
t.,
+ jM o l \ Km(t5~) M 021 M,o,', t., + jM o21 M OIII Km(cl A)
Vm~ = M OI II
(9.87)
-
jKc(cl A )
(9.89)
Example 9.9 9.4.
Obtain the linearized hybrid formulation for the two-machine system in Example
Solution From Example 9.2
Mo =
-
[rr e 11
12e
i 811 j(812 +6120)
(9.90)
388
Chapter 9
Substituting in (9.89) -
1
-Y e
Vl~ =
- j 8 11 /-
II
1
.
+ J -y e
14
-v: Y
-j 8 11
20
II
l2e
j(812- cSI20) .t
u12~
or (9.91)
and 1q24 + J0/d24
YI2 = -Y II
e
8 cS .t ] + J°1) dlA + JO(Vq20 + J°Vd20 )Y'l2e j( 12 - I20 > ul2A
j(8 12-8 11+&I20>[(1
ql~
j(8 12+ 6120).t - JO(Vql0 + J°V)Y dlO l2 e u12A
or 1q2A + J°1d2A
=
YI 2
-Y e II
/) + J· dlA
j(812-811+cSI20)(1 ql4
ry2 1 2 j(28 12-8 1l)( V . V) + J· l-y e q20 + J d20
Y e j(8 12 + cS I20>( V ° V )] .t 12 qlO + J dlO ul2A
-
II
(9.92)
Equations (9.91) and (9092) are the desired relations giving i7;A and and bI2~. These complex equations represent four real equations:
~11l =
-y'
cosOlllqlLl
II
+ _yl
sinOllldl1l
II
+
[sin(OIl - 012 +
yYl2 II
in terms of IIA
/21l
0120)Vq20
- cos (011 - 012 + 0120) Vd20]012~ Vd l A
=
I
I
-Y sin 8 11l q l A + -y
-
°
II
II
cosOll/dl4
YI2
+ -y 11
[COS(O'1 -
012 + 0120) ~20
+ sin (0 II Iq24
Yl2
-
Yll
COS {012 -
011 +
~t2 [sin (20
+ {-
22
Yl2
0120)/q 1A -
12 -
( 22 )
Vq20 .
-
Yll
.
sm (8 12
+ cos (20 12
Id 2A = -y sln(OI2 II
+{
°
011
n2 [cos (20 Y II
Y12 + - cos (012 Yll
+ 0120)/q 1A
12 -
011) Vq 20
-
sin(20 12
- YI2 [COS (0 12 + 9.10
0120)
-
811 +
+
0120)
Vd20 ] Ol2A
0120)ld1A
011) Vd 20 1
-
+ Y12 [sin (012 + YI 2
-
0 12
-
0120)
-
Vq l O + cos (0 12 +
011
0120)
VdIOl }012L1
+ 0120)/d 1A
011) VdlOl
VqIO - sin (6 12 +
0120)
VdIO l}O I2A
(9.93)
Network Equations with Flux Linkage Model
The network equation for the flux linkage description is taken from (9.33) and (9.72). (9.94)
This is a complex equation of order n, or 2n real equations. If the flux linkage model is used, I q and I d for the various machines are not state
Multimachine Systems with Constant Impedance loads
389
variables. Therefore, auxiliary equations are needed to relate these currents to the flux linkages. These equations are obtained from Section 4.12. For machine i we have I. q'
=
..!.(1 ,{ q
A. _ L
L MQ) ,{ q
MQ
A Q.
{q { Q
q'
'
1 (1L-Mtd D) L I di = td A di - tdtF AFi MD
-
L td{D A Di MD
i = 1,2, ... .n
(9.95)
Equations (9.94) and (9.95) are the desired network equations. Together with the machine equations they complete the description of the system. While the above procedure appears to be conceptually simple, it is exceedingly complex to implement. This is illustrated below. To simplify the notation, (9.95) is put in the form I q;
= (Jqi A qi
I di
=
(JdiAdi
+ +
(J Q; A Qi (JFiA Fi
+
i
(JDiAD;
= 1,2, ... .n
(9.96)
The complex vector I thus becomes I q l + j1dl
I
I q2
=
=
+ jld 2
[O"qlAql
.~.O"QIAQIJ
(Jqn A qn
+
+j
(J Qn A Qn
[O"diAdl (Jdn A dn
+
0"~1~F1
+
(JFn A Fn
+ +
O"DIADlJ
(9.97)
(J DnADn
Now the matrix Q in (9.94) is of the form
Q=
Z lI e
j BII
••.
[Z
"Ie
j('nl-6",)
«; = [ 2"1 cos (~;I
2 1" sin ~8.,~ -
- o.d
o'")l
X""
QR + jQ,
(9.98)
Expanding (9.94),
vq +
jV d
+ jQ/)(Iq + jld )
=
(QR
=
(QR1q - Q/ld) + j(Q/lq + QRld)
(9.99)
and substituting (9.97) into (9.99),
v
q
J
= (J qnAqn
+
(JQnAQn
(9.100)
390
Chapter 9
+ (J dnAdn
+
(J
FnAFn
+
(J
OnAOn
(9.101) Equations (9.100) and" (9.101) are needed to eliminate Vqi and Vdi in the state-space equations when the flux linkage model, such as given in (4.138), is used. The above illustrates the complexity of the use of the full-machine flux linkage model together with the network equations. Much of the labor is reduced when some of the simplified synchronous machine models of Section 4.15 are used. For example, if the constant voltage behind subtransient reactance is used, the voltages E;i and Ed; become state variables. The network is reduced to the generator internal nodes. This allows the direct use of a relation similar to (9.31) to complete the mathematical description of the system model. This has been illustrated in some of the examples used in this chapter. The linearized equations for the flux linkage model are obtained from (9.97), which is linear, and (9.73). Following a procedure similar to that used in deriving (9.100) and (9.101), we expand (9.73) into real and imaginary terms as follows: ·V~
=
Vq~ + jVd~
=
(QRo + jQ/o)(lq~ + jld~) - j [O~(QRO + jQIO) - (QRO + jQIO)~~)(lqo + jIdo)
=
[QRoIq~ - Q/old~
+
j[Q/olq~
+
+ (OAQ/O
- Q/o~A)Iqo
+
(~~QRO - QRo~A)Ido]
QRoId~ - (O.1QRO - ·QRo~~)lqo
+ (O.1Q/O -
Q/o~A)Ido]
(9.102)
The terms in Iq~, IdA' I qo, and Ido are substituted for by the linear combinations of the flux linkages given by (9.97). 9.11
Total System Equations
From (4.103) for each synch ronous machine and hence for each node in Figure 9.2, the following relations apply ik
= -
Wk =
l,
Lkl(R k
(1/3T j k ) ( -
= Wk -
1
+ WkNk)ik - Lkl"k A dk ;qk
k
=
+
Aqk;dk -
I, 2, ... .n
3Dk W k + 3Tmk ) (9.103)
where i k = [idkiF/ciDkiqkiQk]', "k = [Vdk -VFk 0 Vqk 01' and the matrices Rk , L, and N, are defined by (4.74). The whole system is of the form (9.104) (see [5, 6, 7, 8, and 9]). Assuming that V Fk and Tmk , k = 1,2, ... , n, are known, (9.104) represents a set of 7n nonlinear differential equations. The vector x includes all the stator and rotor currents of the machines, and the vector \' includes the stator voltages plus the rotor voltages (which are assumed to be known). The set (9.31) provides a constraint between all the stator voltages and currents (in phasor notation) as functions of the machine angles. These equations are also nonlinear.
Multimachine Systems with Constant Impedance loads
391
By examining (9.103) and (9.31) we note the following: The differential equations describing the changes in the machine currents, rotor speeds, and angles are given in terms of the individual machine parameters only. The voltage-current relationships (9.3 J) are functions of the angles of all machines. This creates difficulties in the solution of these equations and is referred to in the literature as "the interface problem" [10]. The nature of the system equations forces the solution methods to be performed in two different phases (or cycles). One phase deals with the state of the network, in terms of node voltages and currents, assuming "known" internal machine quantities. The other phase is the solution of the differential equations of (9.103) only. The solution alternates between these two phases. This problem is mentioned here to focus attention on the system and solution complexities. This problem will be discussed further in Part III of this work. Finally, if the flux linkage model is used (for the case where saturation is neglected), the system equations will be (4.138), (9.100), and (9.10 t). Again the "interface problem" and the computational difficulties are encountered. Example 9./0
Give the complete system equations for a two-machine system with the machines represented by the voltage-behind-subtransient-reactance model and the loads represented by constant impedances. Solution The network constraints for this system are given in complex notation in (9.31) or in real variables in (9.40), and the machine equations are given in Section 4.15.2. The machine equations are obtained from (4.234) and (4.270). They are
= KI;E;; +
E;~
~; = -
f;ldi - Iq;X!, + Ed;
~; = - r;lq;
and
-
T;~;E;J;
T:!o;A. D; T~o;E;;
E~:
KuA D;
- (x q; -
+ IdiX:' +
E;~
x;~)/q;
=
E;; - AD; + (x~; - X;f.;)ldi
=
E FD; - (I
1jiW; =
b; =
Tm ;
W; -
-
+
(9.105)
Kd;)E;i
Iq;E;: - Id;E;J; 1 i = I, 2
+ Xd;!di + Kd;ADi (9.106)
The network constraints are obtained from (9.40). The system has ten differential equations, six auxiliary machine equations, and four algebraic equations for the network (or two complex equations). As per the discussion in Section 9.6, some differential equations can be eliminated by using ~I - ~2 and WI - W2 as state variables. Some of the computational labor can be reduced if the subtransient reactances of the generators are included in the network Y matrix (or Z matrix). The network equations would then give relations between the currents I q; and I di, i = I, 2, and the voltages E;~ and Ed;, i = 1,2. The auxiliary equations for ~i and Jt;,i can be omitted. Also in (9.40), E;: and E;J; should replace ~i and ~;.
392
Chapter 9
9.12
Multimachine System Study
The nine-bus system discussed in Section 2.10 is to be examined for dynamic stability at the initial operating point given in Section 2.10. Linearized machine equations are to be used. The loads are to be simulated by constant impedances based on the initial operating conditions. The system under study comprises three generators and three loads. A one-line impedance diagram is given in Figure 2.18. The initial operating system condition, indicating the power flows and bus voltages, is given in Figure 2.19. Data for the three generators are given in Table 2.1 (some of which are repeated below for convenience). The synchronous machine models to be used are as follows: classical model for generator 1, and the two-axis model for generators 2 and 3.
9.12.1
Preliminary calculations
Let the generator terminal voltage be V 1Ji, and the q axis be located at angle o. All angles are measured from reference. The generator current flags the terminal voltage by the power factor angle e. The following relations, derived in Section 5.5, are used (r ,-...; 0) to obtain the data in Table 9.1:
I;
+
j/x
=
I /-
l/J =
(P - jQ)/V
Table 9.1. Quantity
H(MW. s/IOO MVA) Tj = Xd -
xq
-
T~O T~O
2HwB
xd xd
Unit
s
pu pu pu pu
s
I qo I do VqO VdO
pu pu pu pu pu pu pu
E'
pu
TdO
E~o
EdO
s,
9.12.2
= xqlr/(V -
xqlx)
Three-Machine System Data
s
TdO
tan(a - (3)
elec deg
Generator I (classical)
Generator 2 (two-axis)
Generator 3 (two-axis)
23.6400 17824.1400 0.0852 0.0361 0 0 8.9600 3377.8404
6.4000 4825.4863 0.7760 0.7447 0.5350 201.6900 6.0000 2261.9467 0.7882 -0.6940 0.9320 -1.2902 0.6336 -0.8057 61.0975°
3.0100 2269.4865 1.1312 1.0765 0.6000 226.1900 5.8900 2220.4777 0.7679 -0.6668 0.6194 -0.5615 0.6661 -0.7791 54.1431 °
0.6780 -0.2872 1.0392 -0.0412 2.2717° 1.0566
Linearized network equations
The network is assumed to include the transient reactances of the generators. The network is reduced to the generator internal nodes. At these nodes the voltages are E~, E~, and E;. From (9.63) with V replaced with E' and for a three-machine system (using ~'2
= - ~2" ~13 = - ~31)'
393
Multimachine Systems with Constant Impedance loads YI 2eJ(' I2 - ' 120)
Yllcj('1J -6no)
Yn ej' 22
Y23eJ(' 23 - '230)
Y32eH'J2 - 6320)
Yn ej' )3
-
jEio Y12ej (' 12-'120)
-
jEio YJ)e H' IJ- 6 IJo)
j £;0 Y2Ie j (' 21-'210)
0
0
jE:o Y31e j( 831- '310)
(9.107)
With generator I represented classically, £;A = 0 and E~ £,; and with node I as the arbitrary reference node = £1 + jO = E, (a constant). Substituting in (9.107) and using 023 = 013 - 012'
E:
~A
Yl2 e j (812 - 6 120'
Yl3 e j( 813 - 6130)
7;/1
Y22e j822
Y23e j( 8 n
1;A
Yn e j( 823 +6230)
- jE20 Y12e j( 8 12- 6 120)
- jEio YlJ e j (8 13 - 6 130)
j[E I Y2Ie j( 8 12+6 120)
- j Eio Y23e j( 823 - 6230)
- 6230)
+
-
YBej833
£;0
E~A EiA
Y23e j( 823 - 6230)]
jE~o Y23e j( 823 +6 230)
°12A
j[EI Ynejl813 +6130)
013A
E~o y 23eH823 + 6230)]
+
(9.108)
Separating real and imaginary parts and dropping the subscript Ll for convenience,
If'
i.;
1'2
£;2
£:12
Y12COS(012 - 0120)
- Y12 sin (0 12 - h120)
Y12sin «(J12
-
0120)
Y1 2cos(012- 0'20)
-8 22
G22
Ed)
£;J Y" cos (Ou - ~uo)
Yu sin (0" -
tS DO)
Y2J cos(On - 02)0)
- Y.) sin (01)-
~I)
~'2 ~I)o)
Yucos(O'J - ouo) - Y2)sin(On - 0no)
Y12[£ ,i20 cos (0 12 - ~120)
YuIE,iJocos(Ou - ~IJO)
+ E;20sin (012 -
+
~120)1
£~JOsin(Ou
YIZ!£d20 sin (0 12 - 0.20)
YIJI£';JO sin (0" -
- £;20COS(0I2 - ~l2o»)
- £;JOcos(Of) - 0,,0»)
-e, YI2 sin (812 + 0 20) 1
- £';10 YncOS(02J - 02)0)
£;2
- 01)0» tS uo)
Y2JI£ dJO cos (82J - °no)
£:12
£;]
+ £;losin (Ou - 0130)]
- £;JOY2J sin (02)- 0230) I d2
8 12
G22
YHsin(OH - 02JO)
YZJcos(02J - 6B o )
E, Y'2 COS(OIZ + 0 120)
Yn[EdJOsin(82J - 0210)
- E,;JO Yn sin (02J - OlJO)
- £;30cos (021 - 0130)]
EdJ
+ £;10 Y2Jcos(8 2J - 02JO) If J
Yncos(02J + 0])0)
- Y2J sin (82J + 0BO)
GJ)
-B 13
Y2J[Ed20COS(OZJ + 02JO)
-£1 YlJsin(OIJ + 0110)
+ £;20sin(OB + °2Jo)1
- Y2JE';20COS(On + 02JO)
Yn[Ed 20sin(01J + ono)
£1 YUcos(fJll + OUO)
- £;zocos(8 H + (hJo)J
- £:110 YH sin (82) + ovo)
hl1
- Y2J£;20sin(fl2J + 02J0) I dJ
Y21sin(OB + 0210)
Y2Jcos(01J + 0210)
BJ)
GJ]
Ou
+ £;20 Y2.lCOS(OlJ + 02JO)
(9.109)
Equation (9.109) is the desired linearized network equation. It relates the incremental currents to the incremental state variables E;2' £:12' £;3' £:13' 012' and 013. 9.12.3
Generator equations
From Example 9.7 we obtain the following generator equations (again the subscript Ll is omitted): Generator I (classical) (9.110)
Chapter 9
394
Generators 2 and 3 (two-axis model)
r;o;Ed; = -Ed; Tdo;E;; = EFDi Tj;W; &i
= Tm; =
-
(x q;
x;)Iq;
-
E;i +
(Xd; -
D,», -
xI )I d;
IdioEdi -
IqioE;i -
E~iOld; -
E~ioIqi
i = 2,3
Wi
(9.111 )
Again we recall that, to obtain an independent set, the last equations in (9.110) and
(9.111) are combined to give
s;
= Wt -
Wi
;
= 2,3
(9.112)
By using (9.109), t.; t.; Iq2' Id2, I qJ, and IdJ are eliminated from (9.110) and The resulting system comprises nine linear first-order differential equations. The state variables are E;2, E:n , E;3, EdJ~ WI' W2, WJ, 012, and 013'
(9.111).
9.12.4
Development of the A matrix
The Y matrix of the network, reduced to the internal generator nodes and including the generator transient reactance, is given in Table 2.6 as the prefault Y matrix. It is repeated here in Table 9.2. Data for the terms in (9.109) are calculated and given in Table 9.3. Table 9.2.
Reduced
V Matrix for
2
3
0.2871 + jJ.5129 = 1.5399 /79.25°
0.2096 + j 1.2256 1.2434 /80.30° 0.2133 + j 1.0879 = 1.1086 /78.91° 0.2770 - j2.3681
Node
0.8455 - j2.9883 0.2871 + jl.5129 = 1.5399 /79.25°
2 3
=
a Three-Machine System
=
0.4200 - j2.7238
0.2133 + j 1.0879 = 1.1086/78.91
0.2096 + j 1.2256 1.2434 /80.30°
0
The coefficients of (9.109) and (9.111) are then calculated. The main system equations are given below. The incremental currents I q; and I di are calculated from
(9.1.09).
t.;
-1.1458
-1.0288
-0.8347
-0.9216
1.6062
1.2642
ldt
1.0288
-1.1458
0.9216
-0.8347
0.1891
0.0265
E:n
lq2
0.4200
2.7239
0.3434
-1.0541
-1.1484
0.5805
£;3
I d2
-2.7239
0.4200
1.0541
0.3434
2.4914
-0.9666
E d3
i.,
0.0800
-1.1058
0.2770
2.3681
0.8160
-1.4414
012
I d3
1.1058
0.0800
-2.3681
0.2770
-0.8305
1.9859
013
£;2
(9.113)
The generator differential equations are: Generator 1 (classical) 5.6104 x 10-sTmt
WI
=
~I
= WI
-
5.6104 x 10-sDIWt - 5.9279 - lO-5lqt
395
Multimachine Systems with Constant Impedance loads Table 9.3.
Preliminary Calculations for Three-Machine System 1-·-2
1-3
2-3
1.5399 79.2544 - 58.8259 138.0802 -1.1458 1.0288 20.4285 1.4431 0.5375
1.2434 80.2952 -51.8714 132.1666 -0.8347 0.9216 28.4238 1.0935 0.5919
1.1086 78.9084 6.9545 71.9540 0.3434 1.0541 85.8629 0.0800 1.1058
Nodes
Yij Oij oijO Oij - oijO Yij cos (Oij Yijsin(Oij Oij + oijO Yij cos (Oij Yij sin (Oij
- oijo) - oijo)
+ oijo) + oijo)
Generator 2 (two-axis) E:J2
=
-4.9581 x 10-3Ed2 - 3.6923
£;2 = 4.4210 x 10-4EFD2 W2 2.0723 X 10- 4T -
+ 2.6736
m2
x 10-4£d2
10-3/q2 4.4210 x 10- 4£ ; 2 + 3.4307 x 10-4/d2 2.0723 X 10- 4D2W2 - 1.9314 X 10- 4£;2 X
+ 1.4383
X
10-4/ d2 - 1.6334
X
10-4/q 2
~2 W2 Generator 3 (two-axis)
£:13 £;3
WJ
=
53
-4.4210 x 10-3Ed3 - 4.7592 X 10-J/q3 4.5035 x 10-4E FD3 - 4.5035 x 10- 4£ ; 3 + 5.0944 x 10-4/ dJ 4.4063 X 10- 4Tm J - 4.4063 X 10- 4DJW3 + 2.4741 X 10- 4E dJ - 2.7292 x 10- 4£ ;3 + 2.9380 x 10-4/ dJ - 3.3836 X 10-4/ q3 (9.114)
W3
By using (9.113), the currents are then eliminated in (9.114). Combining terms and using the relation oij = 0; - OJ' we obtain the linearized differential equations for the three-machine system. The results are shown in (9.115), which is of the form WI
£;2
£;2
W2
£;J
£;J
WJ
6 11
61)
WI
-0.56IOD,
0.6793
0.6099
0
0.4948
0.5463
0
-0.9520
-0.7494
WI
£;2 £:12
0
-13.7658
1.4409
0
3.6163
1.1781
0
8.5472
-3.3161
£;2
0
-15.5076
-150.1554
0
- 12.6793
38.9205
0
42.4023
-21.4333
£;2
W2
0
- 6.5352
-1.1714
-2.0723b 2
0.9552
2.2156
0
5.4592
-2.3385
W2
0
5.6334
0.4076
0
-16.5675
1.4111
0
-4.2309
10.1170
£;3
0
-38.8349
68.5987
£;3
-5.2010
10.7116
WJ
i;) i;n
,.. 10- 4
0
- 3.807 J
52.6270
0
-13.1829
-156.9117
w)
0
2.9781
3.9766
0
- 10.6238
-4.7247
~12
10000
0
0
-10000
0
0
0
0
0
612
~I)
10000
0
0
0
0
0
-tOOOO
0
0
~I)J
-4.4063D)
0.5610T""
4.421O£'D2 0
2.0723T"'2
+ 10-4
4.5035£'D3 0
4.4063 T"'3 0
0
(9.115)
396
Chapter 9
x where
=
Ax + Bu
x' = [WI E~2 E:J2 W2 E~3Ed3 W3 Oil 013] u': = [Tm , E FD2 Tm2 E FD 3 Tm3 ]
The eigenvalues of the A matrix are obtained for the case of D I 1.0 pu, using a library computer program. They are
x,
-0.002664 + jO.034648
A2 A3 A4 As
=
D2
-0.002664 - jO.034648
A6 A7
-0.000455
-0.000622 + jO.022984
As
-0.000199
-0.000622 - jO.022984
A9
-0.000199 - jO.000129
-0.010373
+ jO.000129
-0.016644
All the eigenvalues have negative real parts, and the system is stable for the operating point under study. The dominant frequencies are about 2.1 Hz and 1.4 Hz respectively. These frequencies are the rotor electromechanical oscillations and should be very similar to the frequencies obtained in Example 3.4. Thus if we plot Pl2 from the data of Figure 3.8, we find that the dominant frequency is about 1.4 Hz, which checks with the data obtained here. A similar run was obtained for the same data except for D, = D2 = D3 = O. The eigenvalues are
AI A2
+ jO.022983
-0.000458
-0.000529
-0.000281
-0.000529 - jO.022983
A3
-0.010366
A4
-0.016659
-0.002459 + jO.034636 -0.002459 - jO.034636
As
=
0
Since this is a special case of uniform damping (D/Tj = 0), the system order is reduced by one. The frequencies corresponding to the electromechanical oscillations are almost unchanged, while the long period frequency has disappeared. Problems 9.1 9.2
9.3 9.4 9.5 9.6 9.7 9.8
If the Y matrix of the network, reduced to the generator nodes, is such that Oij = 90°, i ~ j, derive the general form of the matrix M. For the conditions of Problem 9.1, obtain the real matrices for Iq and Id in terms of Vq and Vd. Compare with (9.40) for a two-machine system with Gil = G21 = O. Repeat Example 9.3, using the synchronous machine model called the one-axis model (see Section 4.15.4). Repeat Example 9.5, neglecting the amortisseur effects for the synchronous machine represented in detail (Section 4.15.1). Linearize the voltage-behind-subtransient-reactance model of the synchronous machine. Repeat Example 9.8, using the results of Problem 9.5. Develop (9.89) for a three-machine system with zero transfer conductances. For the nine-bus system of Section 2.10 the dynamic stability of the postfault system (with line 5-7 open) is to be examined. The generator powers are the same as those of prefault conditions. a. From a load-flow study obtain the system flows, voltages, and angles. b. Ca~culate the initial position of the q axes; lqo, ldo, VqO, VdO, E~o, and E dO for each machine; and the angles 0120 and 0130. c. Obtain the A matrix and examine the system eigenvalues for stability.
Multimachine Systems with Constant Impedance loads
397
References I. Anderson. Paul M. Analysis of Faulted Power Systems. Iowa State University Press, Ames, 1973. 2. Pai, M. A., and Murthy, P. G. New Liapunov functions for power systems based on minimal realizations. Int. J. Control 19:401-15, 1974. 3. Willems, J. L. A partial stability approach to the problem of transient power system stability. Int. J. Control 19:1-14, 1974. 4. Pal, M. K. State-space representation of multimachine power systems. IEEE Paper C 74 396-8. presented at the Summer Power Meeting, Anaheim, Calif, 1974. 5. Prabhashankar, K., and Janischewskyj, W. Digital simulation of multimachine power systems for stability studies. IEEE Trans. PAS-87:73-80, 1968. 6. Undrill, J. M. Dynamic stability calculations for an arbitrary number of interconnected synchronous machines. IEEE Trans. PAS-87:835-44, 1968. 7. Janischewskyj, W., Prabhashankar, K., and Dandeno, P. Simulation of the nonlinear dynamic response of interconnected synchronous machines (in two parts). IEEE Trans. PAS-91:2064-77,1972. 8. Van Ness, J. E., and Goddard, W. F. Formation of the coefficient matrix of a large dynamic system. IEEE Trans. PAS-87:80--84, 1968. 9. Laughton, M. A. Matrix analysis of dynamic stability in synchronous multi-machine systems. Proc. lEE (British) 113:325-36, 1966. 10. Tinney, W. F. Evaluation of concepts for studying transient stability. IEEE Power Engineering Society Tutorial, Spec. Publ. 70 M62-PWR, pp. 53-60, 1970.
Part 1\1 The Mechanical Torque Power System Control and Stability P. M. Anderson
chapter
10
Speed Governing
Prime mover governors, especially centrifugal flyball governors, have been in use since the late 1700s. James Watt first applied a centrifugal governor to a steam engine in about 1788. There is evidence that he considered a patent application for his governor and probably decided against it because of earlier patents for similar centrifugal devices used to regulate the speed of water wheels and windmills in the milling industry [1, 2]. During the 19th century interest in speed governing intensified and a number of scholarly papers were written on the subject. Over 100 references on the subject are given in the Royal Society of London Catalogue of Scientific Papers, 1800-1900 [3]. Many of the prominent engineers and scientists of that era made contributions to the description and analysis of governors. These include C. W. Siemens, J. C. Maxwell, W. Thompson (Kelvin), 1.B. L. Foucault, and I. Vyshnegradski. Pontryagin [4] refers to the work of the Russian engineer Vishnegradski (published in 1876) as of "complete clarity and simplicity" and credits him as being the originator of automatic controls (in Russia). Hammond [5] notes that J. C. Maxwell, writing in 1868, identified the instability of an early governor design as being due to a positive eigenvalue [6]. The mechanical flyball governor of Watt and Boulton came into wide use during the early 19th century and easily outstripped competing devices, such as the float valve regulator of French design. Watt's governor is extensively treated in the literature of that era and even some elementary quantitative analysis is evident in works prior to 1850 [2]. However, the control dynamic problems inherent in feedback systems were not recognized until the second half of the 19th century. The dynamic problems associated with speed governing almost certainly provided the incentive for establishing the mathematical theory underlying automatic control. Mayr [2] lists the earliest contributors to this quantitative theory as G. B. Airy (1840/51), 1. C. Maxwell (1868), I. I. Vyshnegradskii (1876), E. 1. Routh (1877), A. M. Lyapunov (1892), A. Stodola (1893/94), and A. Hurwitz (1895). Mostly, these works consisted of attempts to solve the differential equations by classical methods and did not present a generalized theory of feedback control. By the end of the 19th century, the dynamic speed control problem had been thoroughly documented in the technical literature, was presented in textbooks and handbooks, and was even the subject of historical studies [2]. The treatment in this book is therefore the restatement of a very old problem, but it is placed in a modem setting and is attacked with the tools of the control engineer developed in this century. In a steam or hydraulic turbine-generator system, the governing is accomplished by a speed transducer, a comparator, and one or more force-stroke amplifiers. Figure 10.1 shows the system block diagram for a steam turbine generator. The speed governor in the figure is a speed transducer, the output of which is typically the position (stroke) of a rod that is therefore pro-
401
402
Chapter 10
Reference Position
Position Error
Steam 1---3l~Valves & Turbine
+ ~ILJ'---3l~
Speed f--~..O ... )--~
Speed Governor Position Fig. 10.1
Block diagram of steam turbine control system from [II] with permission,
portional to speed . This stroke is mechanically compared to a preset reference position to give a position error proportional to the speed error. The force that controls this position error is small and must be amplified in both force and stroke. This is the purpose of the two amplifiers labeled speed relay and servomotor. This same figure also describes a hydraulic turbine control system if the valve position is changed to a gate position and the steam valve block is considered the wicket gate and hydro turbine system, including the penstock. The speed transducer is the heart of the governor system and may be a mechanical, hydraulic, or electrical device. It must measure shaft speed and provide an output signal in an appropriate form (position, pressure, or voltage) for comparison against the reference, and the subsequent amplification of the error. The centrifugal flyball governor has historically been used for this purpose . Figure 10.2 shows three examples of flyball governors as conceived by different designers. All three have the same essential components : the flying weights (flyballs), the restrain ing spring (speeder spring), and a mechanical linkage that changes a shaft or collar position as the speed changes. An example ofa hydraulic governor is shown in Figure 10.3 (also see Figure A.27 of Appendix A). In Figure 10.3, a main oil pump supplies high-pressure hydraulic fluid that flows through an orifice to the governor oil pump. The amount of governor oil flow is determined by the pressure produced by the governor oil pump, the output pressure of which is only one-fifth or so that of the main pump pressure. However, the governor pump pressure varies as the square of the speed. This controls the pressure downstream from the orifice, which is used to control the throttle setting through a hydraulic control system . Speed sensing may also be done electromechanically by coupling a small generator to the shaft, the output voltage or frequency of which is speed dependent. Examples are given in AppendixF. Such devices are not widely used for central station speed governors . The newest governor designs use high-speed electronic logic to control e1ectrohydraulic force-stroke amplifiers. These e1ectrohydraulic systems have high sensitivity and fast response. The analysis followed here is based largely on the mechanical flyball governor. This approach is used because mechanical devices are easy to understand and analyze, and because they are still widely used. In most cases, similar equations can be derived for other types of governors. Our motive is not to present any given system as being superior to others but to derive a typical mathematical model that will increase our understanding of the governor as a control system component and allow us to analyze systems similar to that of Figure 10.1. In writing the governor equations, it will be convenient to use several of the control system component descriptions and formulas given in Appendix A.
10.1
The Flyball Governor
Consider the flyball governor shown in Figure 10.4 [4, 6]. If we assume that the gravitational force is negligible compared to the centrifugal force FCo then there are two forces acting
Speed Governing
403
Arms Spring
Rotating Shaft -----:~ Steam Valve -----;;.
(a)
Turbine Shaft Worm Gear Speeder~
Rod
(b)
(c) Fig. 10.2
Examp les of centrifugal f1yball governors.
404
Chapter 10 Main Oil Pump
Governor Oil Pump
-
Oil Reservoir
Oil Ejector Check Valve Fig. 10.3 A hydraulicgovernor.
on the flyball-erankarm system : an outward centrifugal force Fe acting on the masses , and a downward spring force Fs acting on the throttle rod. The reference position r is adjusted to correspond to the desired speed. The total outward force Fe on the two flyballs depends on the mass m, the peripheral velocity v, the downward force of the spring, and the radial displacement R of the mass m:
';;---R--~
rrJ-rJ:;~::;-.d-~~~
F. ~ s
't' 0
B:======-) Throttle Control Rod
Fig. 10.4 1\ tlyballgovernor.
W=
Turbine Speed
405
Speed Governing Fc =
2mv2
-
(l 0.1)
-
R
Using the familiar relation between peripheral velocity v and shaft angular velocity can write
v =Rl/J
l/J we (10.2)
Now, relating the governor speed to the turbine shaft speed through the gear ratio N,
l/J =Nw
(10.3)
we can write (10.1) as
(lOA) where Fe is in Newtons, w is in radians per second, m is in kg, and R is in meters . By simple geometry, we can relate R to x, and x as follows:
b x, = R - d = - - x = - Crx a
(l0.5) where C,. = bla is the lever ratio constant. Then the ballhead force Fe may be written in terms of x as
Fc = 2mN2(d- Cr x)w2
(10.6)
Now, using Figure 10.5, we sum the forces on the ballhead using Newton's law:
F 2
F' 2
.. =k '" Forces= - c - H'x , - - s mXI
(10.7)
where F/; is the force due to the spring and Bx, is the force required to overcome friction , both applied at the ballhead . Equating moments about the pivot, we can relate F/; to Fs, the actual spring force, as follows :
1
1
iF/;b sin a = iFsa sin a
(10.8)
Fc: 2
T x
Fig. 10.5 Crank ann geometry and forces.
406
Chapter 10
and solve for F; with the result
,_ a _ I _ K~(r-x) F s - -Fs - - Fs - - - b
c.
c,
(10.9)
where K; is the spring constant of the speeder spring. Substituting (10.6) and (10.9) into (10.7) we have
..
.
2mx} + 2Bx}.-
K;(r-x) 2 2 C = 2mN (d - Crx)w
(10.10)
r
Now, from (10.5) Xl = -Crx and the entire equation can be written in terms of the variables
x and w with the result
2mNl
Ksr - --c(d - Crx)w 2 = 2mx + 2Bi + KsX
(10.11)
r
where we define an effective spring constant K, = K~/C;. Equation (10.11) shows clearly the nonlinearities of the system. Not only are the products and quadratic terms nonlinear, but the coefficients, particularly K, and B, can not be expected to be linear over a large range of x and x. Furthermore, there may be backlash in the gears and dead zones in the pivots or other mechanical connections, which introduces nonlinearities that are not continuous functions of x, to, and their derivatives. In order to gain a better grasp of governor behavior, we linearize (10.9) about a steady-state operating point (subscript 0) from which we will study small deviations (subscript ~). This is justified since the speed will deviate from its rated value only by small amounts in normal operation. Thus, we write
W
= Wo + W a
rr r.s r, At the quiescent points, with x., = Xo escent condition
(10.12)
= 0, (10.11) must still be satisfied. This gives the qui-
2mNl
Ksro - --c(d - Cr xo)w5 = KsXo
(10.13)
r
Now, substituting (10.12) into the system equation (10.11) and using (10.13) to simplify the result, we compute (10.14) which is a linear differential equation in the variables rd' xd , and Wt>.. The ballhead force Fe acts in the Xl direction (see Figure 10.5) on the total mass 2m. This force creates an equivalent force in the -x direction, which we shall call From Figure 10.5 we readily compute
Fe.
Fe = CrFe = 2mN2Cr(d - Crx)w2
Clearly,
Fe is a function of both x and ta. Upon linearization we can write
(10.15)
(10.16)
407
Speed Governing
where we define thepositive constants
aF' ] == -2mN2C2W1l Kx == _c ax 0 r 0
The Ballann Scale
aF~] Kw - -
The BallheadScale
_
a£O
0
_4mN Cr(d - Crxo)£Oo -
2
(10.17)
Using these definedconstants in (10.14)the systemequationbecomes (10.18) Takingthe Laplacetransform of this linearequationwe can visualize the computation of X1.\ from the block diagramof Figure 10.6. The variable X1.\' which relatesto the throttlerod motion, can be applied directly to the throttle valve or, more commonly, applied first to a force-stroke amplifierthat drives the throttle. The linearequationof motion of the governoris a second-order equation. We would expect a response that is probablyoscillatory when a step changeis made in r s or £01.\, or a well-"tuned" governor may respond in a critically damped mode. In any event, the frequency of oscillation and the damping ratio are determined by the coefficients on the right-hand side of(IO.I8). Since the governoris physicallysmall and it controls a massiveturbine, we know that the solutionof (10.18)will reach steady state much faster than the turbineshaft. We are interested primarily in the motion of the turbine shaft. Therefore, we will neglect the governoroscillatory behaviorto write, as an estimate, (10.19) which will be sufficiently accurate in the longertime span of interest. This equationis algebraic and specifies that a reduction in speed results in an immediate increase in X1.\. Since a reduction in speedwould normallyaccompany an increase in load on the turbine,the increase in x1.\ should be in a direction to furtheropen the throttlevalve. The linearequation (10.19)is commonly used to representgovernorbehaviorin powersystem simulations. The assumption of linearity is justified since deviations from synchronous speed are small, even for large disturbances such as faults on the generatorterminals. The spring constantK, is an important parameterin governor design. It determines the natural frequency of oscillation of the governorfrom (10.18), from which we compute w == n
x JKs-K 2m
(10.20)
Furthermore, it is obvious that the system is unstable when K, < K, and K; is always positive. Therefore, a minimum spring stiffnessexists for satisfactory operation.
1
2m/ +2Bs+(Ks - Kx )
Fig. 10.6
Block diagram of a linear speed governor.
Xt\
Chapter 10
408
Also note that the system is designed for correct operation with K", > 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;<;
+
(10.38)
Chapter 10
412
D+2Hs
Fig. 10.10 System block diagram for the isochronous governor.
Now, for the isochronous governor K)
K; = lim
s-+O TIS(l
+ TsS)
= 00
(10.39)
Since K, is infinite, the error B must be zero for steady-state operation . Indeed, this is the unique characteristic of any integral control system. This means that, following any deviation in speed, the controller will drive the system until r/i and C gW/i are equal, or the steady-state speed is independent of load torque. For the system of Figure 10.10, the steady-state performance equation for zero error becomes (10.40) and the steady-state W is a constant for any Te • Another view of the steady-state operating characteristic of the isochronous governor is shown in Figure 10.11, where the manipulated variable Tm is plotted against the controlled output w. For each setting of the reference, W ss is constant from (10.40), even if the load torque T; changes . This is a desirable steady-state characteristic, but the transient response also needs to be considered. The transient response of the isochronous governor can be evaluated by plotting the roots of the open-loop transfer function or OLTF on the complex plane. For the isochronous system we can write OLTF=
KICg T[s(l + T,s)(D + 2Hs)
=
K s(s + b)(s + c)
(10.41)
where we define the constants b = 1/Ts' C = D/2H, and K = KICi2HT)Ts' The root locus plot is sketched in Figure 10.12 for a typical small value of c and a larger value for b. The system is
fj
1j > r2> 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 ) -<2
R
Tl
r,
(10.62)
Now 'Ts and H are fixed positive constants. The gain K} is a function of the control valve and turbine design and is fixed for a given system, although it may vary slightly with the operating point. The quantities Rand TI vary with the lever ratio aiL since we define, from (10.47) and (10.50),
KsYR ) a R= ( KwWR L
(10.63)
Thus, increasing aiL increases R and decreases 'TJ which increases the stability margin. From Figure 10.13, we note that increasing the ratio aiL moves the flyball connection with the summing beam to the right. This increases the negative feedback, increases the droop, and reduces the governor time constant. In the root locus plot of Figure 10.17, this increase in aiL moves the pole at s = -a farther to the left. Finally, we compute the response of the system to a step increase in reference r~ (or a step decrease in Te~)' From (10.58) we have, with r ~ = AIs, w~=
KRA/s
S3
+ (a + b + e)s2 + (ab + be + ea)s + (abc + K)
(10.64)
From the final value theorem we write (10.65) or, if D = 0, as a limiting case (10.66) The response to a step increase in the reference r~ is shown in Figure 10.18 for two different values of the regulation R (ignoring any oscillatory behavior). Because of the change in speed that takes place with changes in load, the speed droop governor does not hold the frequency exactly constant, but as the load cycles up and down, the net error is usually small. Frequency corrections can be made by adjusting the reference thumbscrew T, shown in Figure 10.13. This thumbscrew is usually driven by a governor speed changer (GSC) electric motor. Each new setting of the reference moves the torque-speed curve (labeled rl' r2' or r3) to a new position in Figure 10.16. The droop or slope of the locus is rarely changed in operation. The speed droop governor is widely used for governing steam turbines and combustion turbines. Hydro turbines often use a special kind of speed droop governor discussed in Section 10.7.
419
Speed Governing
--- ---------------------
o Fig. 10.18
Step response of the speed droop governor.
10.5 The Floating-Lever Speed Droop Governor Another speed droop governor design is the floating lever governor shown in Figure 10.19(a). Here, the mechan ical feedback acts directly on the servomotor pilot valve rather than on the speeder spring. However, the effect is the same as the design of Figure 10.13. The equations of motion of the floating lever governor are determined as follows. The force FG acting at point G on the walking beam is that produced by the governor and is positive for a drop in speed or an increase in the reference position. This results in a positive change in y~ with its associated hydraulic reaction force of the pilot valve acting on the point P. A positive movement in y~ produces an upward force F R due to the hydraulic piston acting at R. These forces are computed in the usual way to write
FG = Kk/:; -x/:;) +
«», -Kww/:;
Fp=KhY~
FR =Pa ,
(10.67)
where P is the pressure of the hydraulic supply. Summing moments about R in the clockwise sense, we write, with L = a + b, -LFG + bFp = 0
(10.68)
or, substituting from (10.67),
sx,
Ksf/:; -Kww/:; = (K, -Kx)xA + TY~
(10.69)
Now we can also write the summing beam displacement equation and the hydraulic servomotor equations in the usual way, that is, ,_ b L
a
Y/:; - -x/:; - -Y/:; KqY~ =
a,y/:;
L
(10.70)
Combining (10.69) and (10.70) we get
K,
J<7s
K;
f/:; -
J<7w/:; = Y/:; + s
K~al . bK K' Y/:; q
s
(10.71)
420
Chapter 10
a
b
Flow Control Valve
(a) Schematic diagram
a
p
b
(b) Free body diagram of the walking beam Fig. 10.19
The floating-lever speed droop governor.
where
Equation (10.69) is normalized in the usual way to write rfJ,u -
Cg(J)fJ,u
=
(l
+ T\ S)YfJ,u
(10.72)
Speed Governing
421
where
and Cg = llR.
Equation (10.72) is identical with (10.53). Note, however, that the time constant 7") is defined differently for the two governor designs.
10.6 The Compensated Governor Another important governor design that is widely used, particularly in the control of hydraulic turbines, is the compensated governor shown in Figure 10.20. This governor incorporates an added feedback that gives it a unique operating characteristic. We have observed that the speed regulation provided by proportional(drooping) control is important in providing good response and also contributes to the stability of the prime mover
~~=ff):::::='= = = = = = = : : f . p d
c
xi b
a
ui Supply
Press-lie
Ps
~J
l
Ji
Increase Flow Decrease How 'Y
Flow Control Val ve Fig. 10.20
The compensated governor.
Chapter 10
422
system. Still, it would be desirable to have the governor hold nearly constant speed (frequency) if possible. This is particularly important on isolated systems where only one, or a very few, machines control the frequency. This need is satisfied by the "compensated governor," which is a governor with two values of regulation. The principle of operation is to provide a given (relatively large) droop in response to fast load changes. The resulting speed deviation is gradually removed by slowly correcting the speed back to a second (relatively low) value of droop. Thus, the larger droop provides stability and the smaller droop provides good speed regulation in the long term. If the smaller value of droop is zero, the governing is a stable isochronous operation. The two values of droop are called the "temporary" and "permanent" droops and are both adjustable within certain limits. The time required to change from the temporary to the permanent droop is also adjustable. These objectives are met in the compensated governor design of Figure 10.20. The mechanical feedback provided by the summing beam c-d provides a temporary droop exactly as in the design of Figure 10.13. The added feedback involves a floating lever system a-b connecting the speeder rod (x), the pilot valve (u), and a receiving piston of area a3' which is held in its steady-state position by a spring. As long as the piston location z remains at its steady-state equilibrium position, the flyweights must also be in their equilibrium position if the pilot valve is held closed. This means that, following a disturbance, the ballhead would return to the same position when the receiving piston (z) returns to equilibrium, if there were no permanent droop through lever c-d. Thus, without lever c-d the compensated governor would act isochronously, but it would do this in a special way. Suppose that walking beam c-d were disconnected. Then, an increase in load would cause the governor to respond to positive displacements in x, u, and y. As piston al moves in the +y direction, it causes transmitting piston a2 to be displaced downward. Since the hydraulic fluid in the chamber connecting pistons a2 and a3 is trapped, this will cause receiving piston a3 to move upward, pushing against its spring, tending to close the pilot valve. Note, however, that the hydraulic chamber also contains a needle valve that will allow hydraulic fluid to move in or out of the chamber slowly, the speed of entry or escape depending on the needle valve orifice area. The compressed spring on piston a3 will slowly force this piston downward, increasing the turbine power gradually and restoring the flyweights to their normal positions. Thus, the governor provides a temporary droop characteristic, but is isochronous in the long term. This gives the governor both a permanent and a temporary droop characteristic, each of which is adjustable. To analyze the compensated governor, it is helpful to break the system into subsystems and write the force and displacement equations for each subsystem. In doing this, it is essential that the defined positive directions of all variables be used in summing forces or moments. The first subsystem is the flyball governor system shown in Figure 10.21. Using the methods developed in previous sections, we can write equations for the forces acting at G and G' as functions of the displacements x and x', and of the speed to. Thus the force acting at G can be written as
FG = -Ks(x6, + x~) + K~6, - K ww6,
(10.73)
The force at G' is equal and opposite to this force, or
F/; = Ks(x6, + x~) -
K~6,
+ K ww6,
(10.74)
The second subsystem is the upper summing beam shown in Figure 10.22(a), for which we write both a displacement and a force equation. For incremental displacements, we can write x~
c
d
LI
LI
= -y6, - -r6,
(10.75)
423
Speed Governing
Fig. 10.21 The flyball governor subsystem.
where L( = c
+ d. Summing moments about R in the clockwise sense, we compute
a
'2.MR = 0 = cF + L]Fs = CKs(XA
+ xa)
+ CKwWA + L1Fs
- CK,xA
(10.76)
For the pilot valve beam of Figure 10.22(b) we can write, for incremental displacements
b a X A - -ZA L2 L2
(10.77)
UA = -
where L2 = a + b. Then summing moments about G in the clockwise direction we have (10.78)
or (10.79)
rAUF
R
c
R
G'
~tx'
d
s
F'G
(a) Upper Summing Beam
(b) Pilot Valve Summing Beam Fig. 10.22
(c) Compensator Summing Beam
Mechanical beams of the compensated governor.
424
Chapter 10
The compensator beam of Figure 10.22(c) is nothing but a lever for which we can write the displacement equation (10.80) and, summing moments in the clockwise sense about N, (10.81) where Ps is the supply pressure behind the hydraulic ram and a) is the ram area . The compensator system is shown in Figure 10.23 on an enlarged scale. Here, we write the equations for the forces acting at Band E as F B = a3PA -s;»,
FE =
P
a2 A
(10.82)
The equation for the volumetric discharge rate of fluid through the needle valve is
CdP A = a0'~ - a3zA
(10.83)
where P A is the incremental pressure change in the chamber in lbf/ft", Cd is the needle valve constant in ft5/S lbf or in3/s psi, and other quantities are as previously defined. The final subsystem is the hydraulic piston or ram shownin Figure 10.24. Since the available force F s is usually much greater than the load F v, and the load mass is small compared to this force, we write only the integrator equation ( 10.84) for this subsystem. If load force and mass are important considerations, the complete equations for the piston should be written (see Appendix E). This completes the subsystem equations. We now collect the equations necessary to describe the total system behavior. From (10.75) and (10.76) we compute
cs;
.«; =
- 2- rA- - L) L1
c c2 Ks -(Ks -Kx)xA + --YA -Fs L, L)
(10.85)
But F s may be calculated from (10.81) and (10.82) with the result (10.86)
Fig. 10.23 The compensator system.
Speed Governing
425
s
s
v
Fig. 10.24
The hydraulic piston subsystem .
Substituting into (10.85) and rearranging, we have r
d L)Kw rs-: - - Wu c .x,
= Yu
+
a)L\Lz(Ks - KJ . aL)(Ks - KJ eazL IKz Yil + z Z6 bcK,Kq u«, - c'fa3Ks
(10.87)
From compensator equation(10.83) and beam equation (10.80) we compute CdP6 =
aze .
jY6
(10.88)
-a3 z6
Also, from (10.82) and (10.79) we can write (10.89) and using (10.84) we compute a3 P6 -KzZ 6
aa jKh . LzKq
(10.90)
=---Yil
from which we can find P6 as a function of Z6 and Y6 ' Substituting this result into (10.88) and simplifying we have eaZ a3 aatKh . a3 . (10.91) c K + L K r Y6 = Z6 + C'" Z6 fi d z z tr'Z tP"- z which is the desired equation for the compensator. Note that (10.91) may be written in the form
(10.92) where we define a3 TZ = CdKz _ eaja3LzKq + ajatCdKh T3 z
(10.93)
fa3LzC~~q
But T3 may also be written as _ (eaza3LzKq + afalCdKh)Tz f a 5L zK q
~-
where 5' is defined as the coefficientmultiplying Tz .
E5~
(10.94)
426
Chapter 10 We may also define, from (10.87)
aL1L2(Ks - Kx )
1'}=
bcKj(q
aL}(Ks - Kx )
K=
ea2LIKz ---C2ja3Ks
u«,
(10.95)
Then the system equations (10.87) and (10.92) may be summarizedas d L}Kw - rA - - - W a =
c
cKs
YA
•
+ 7lYa + Kza
8'1'2YA = ZA + T2Za
(10.96)
Equation (10.96) can be normalized in the usual way to write drR
L}KwWR
CYR
c JlR
-rau -
K
Wau
.
Kz R
= YAu + 7lYAu + - Z a u YR
~'1'iVR . --Y~u = Z~u ZR
.
+ T2Z~u
(10.97)
The coefficients of (10.96) are determined from full-load and no-load steady-state tests. In performing these tests, we note from (10.96) that wheneverY~ = 0, then we also have z~ = z~ = oas well, and that this always holds in the steady state. ' At full (rated) load and rated speed at steady state, equation (10.96) becomes d r
c
LIKw r» - - . 0 = YR + 0 + 0 cK s
(10.98)
or YR
(10.99)
d
and the r ~u coefficient of (10.97) is unity. Now, if the load is removed and the reference is held at rR the speed will reach W A = RWR at steady state and (10.96) becomes d
-rR-
L}K
RWR
.«, w
c
=0+0+0
(10.100)
Using (10.99) in (10.100) we compute YR
L}Kcfi
(10.101)
and the coefficients of WAu in the normalized equation (10.97) becomes Cg = I/R as before. Now, if we arbitrarily let ZR = YR, then (10.96) may be written as
rau - Cgw Au = YAu + 'flY Au + KZz Au 8'T2Y~U = ZAu + T2ZAu
(10.102)
Equation (10.102) may be written. in a slightly improvedform by defining a new variable (10.103) If we multiply the compensator equation by K and define 8 = K5', where 8' is given by (10.94), we can write
427
Speed Governing y~
Fig. 10.25
Block diagram of the compensated governor.
rliu - C gWliu - vliu = Yliu Vliu
+ Tz Vliu =
+ TIYli u
8Tzyliu
(10.104)
This is the desired system description . If (10.104) is written in the s domain, the system block diagram is that given in Figure 10.25. The block diagram helps clarify the role of the compensation feedback and the derivative effect of the temporary droop 8. Note that the signal Vii will always return to zero in the steady state and the system tends toward the speed droop governor similar to Figure 10.14 in the long term. Another form of the compensated governor derived by Ramey and Skoogland [13, 14] is shown in Figure 10.26. This form of representation is instructive as it directly parallels the permanent (R) and temporary (R8) droop factors and also shows the integrating effect of the servomotor in the absence of droop. To analyze the performance of the compensated governor, we again apply the governor as the controller in the system of Figure 10.15. The result is the composite system shown in Figure 10.27. The steady-state performance of the system shown in Figure 10.26 is analyzed using (10.37) with the result W
ss
=
KJrss
D+K[Cg
- _-..:::::::...--
D+K\Cg
(10.105)
This is exactly the same result obtained for the speed droop governor with no compensation. This result was anticipated as the compensation signal Vii goes to zero in the steady state. The transient performance of the compensated governor is not easily analyzed using the manual root locus or Routh techniques because of the added compensation. A computer root 10-
Fig. 12.26 Alternate form of compensated governor representation [13,14).
428
Chapter 10
r-----------------,----------l Control I TetJ. Plant I
I
I I r, I I I I I I I 1
ytJ. l+'t"l s
~ TmtJ. 1+'t"3s
I
+
Tal>
1
D+2Hs
vtJ. &2 S
1+'t"2S
- ....- - - - - - - - - -
I
»,
I I I I I I I I I ,
Fig. 10.27 Typical system application block diagram for the compensated governor .
cus method can be used for numerical results, but this requires a cut and try procedure to optimize the variable parameters in the system. As an instructive alternative, one can use an analog computer or digital simulator to determine suitable values for all parameters and then examine the behavior in the s plane for further insight into the design optimization.
Problems 10.1 Verify the development of equation (10.11). Give a physical explanation for the resulting effective spring constant of K, = K;/C~. 10.2 Verify that the dimension of the leading coefficient on the right-hand side of(10.26) is in inverse of a time constant in seconds. 10.3 From Appendix E we find the mathematical statement in (C.32) that cos
cPo =
F
g
- = -----=--,------:k >7"..., 2KL ( 1 sin-cPu) L1Y-WO--m sin cPo
Based on this premise , find the expression for stability of the system . lOA Evaluate the function 1 - (sin cP./sin cPo) for values of cPu = 10,20, and 30 degrees, and for various positive values of cPo between 0 and 75 degrees . Plot the results . 10.5 Perform a computer simulation of the isochronous, speed droop, and compensated governors. Use the following constants for all simulations. 7'1=0.1S
7's=O.3s
2H=4.74s D=2.0pu
Cg=20pu
Determine suitable settings for the gain K1 in all governors and for the parameters 8 and 7'2 in the compensated governor.
References 1. Dickinson , H. W. and Rhys Jenkins, James Watt and the Steam Engine, Oxford , 1927.
2. Mayr, Otto, The Origins of Feedback Control (translation of Zur Friihgeschite der Technischen Regelungen), MIT Press, Cambridge, MA, 1970. 3. Royal Society of London, Catalog ofScientific Papers, 1800-1900, Subject Index, v. II, Mechanics, Cambridge , 1900, pp. 136-137. 4. Pontryagin, L. S., Ordinary Differential Equations, Addison-Wesley, Boston, 1962. 5. Hammond, P. H., Feedback Theory and its Applications, Macmillan, New York, 1958.
Speed Governing
429
Maxwell,1. C., On Governors, Proc. RoyalSociety of London, v. 16, 1868, pp. 270-283. Raven, FrancisH., AutomaticControlEngineering, McGraw-Hill, New York, 1968. Merritt, HerbertE., Hydraulic ControlSystems, Wiley, New York, 1967. Takahashi, Yasundo, Michael 1. Rabins, and DavidM. Auslander, Controland DynamicSystems, Addison-Wesley, Boston, 1970. 10. Anderson, P. M., Modeling Thermal Power Plantslor DynamicStabilityStudies, ProjectReport, Pacific Gas and Electric Company, San Francisco, 1972. 11. Eggenberger, M. A., A simplified analysis of the no-load stability of mechanical-hydraulic speedcontrol systems for steam turbines, Paper 60-WA-34, ASME WinterAnnual Meeting, New York, N.Y., November 27-December2, 1960. 12. Eggenberger, M. A., Introduction to the basic elements of control systems for largesteamturbine-generators, GET-3096B, General ElectricCo., 1970. 13. Ramey, D. G., Hydro unit transfer functions, IEEE Tutorial Course, The Role of Prime Movers in System Stability, IEEE pub. 70M29-PWR, 1970, pp. 34-39. 14. Ramey, D. G. and J. W. Skoogland, Detailed hydrogovernor representation for systemstability studies, SixthPICA Conf. Proc. May 1969, pp. 490-501. 6. 7. 8. 9.
chapter
11
Steam Turbine Prime Movers
11.1 Introduction We begin this chapter with some general considerations of prime movers and how they are controlled. Following this general overview of prime movers, we concentrate on steam turbines and develop models that can be used to represent this type of machine in computer studies of the power system. Other types of prime movers are discussed in Chapters 12 and 13. Figure 11.1 shows on overview of a large power system and the generation control structure. The system control center measures the power produced by all generators and the interchange power with neighboring systems. It compares the tie line flows with their scheduled values, and these flows are coordinated with neighboring utilities. The control center receives measurements of all generator outputs and compares these values with desired values, which are based on the economic dispatch of generation considering individual unit generation costs. Then, as the system load varies, the control center can change the generation dispatch to economically meet the demand in the most efficient manner, while still maintaining prudent reserves to assure adequate generation if unforeseen unit outages should occur. Note that the control center does not measure the system loads. The measurement of system frequency is used to assure adequate total generation to meet load and maintain rated speed, thereby assuring constant long-term system frequency. The system dispatch computer sets the governor input signal to control the mechanical torque of the prime mover, computing a unit dispatch signal (UDS), as shown in Figure 11.2. The governor compares the speed reference or load control signal against the actual speed and drives the governor servo amplifiers in proportion to this difference, which can be interpreted as a speed error. The servomotor output is a stroke or position YSM ' which indicates the position of the turbine control or throttle valves. Note that this control is different on an isolated system, where the governor input is set to hold constant speed or frequency. The fast dynamics of the generation of each unit is the solution of Newton's law, which we write per unit as (11.1) where 7j = a time contant related to the unit moment of inertia in seconds w = shaft angular velocity in radians per second Tm = the mechanical torque output of the turbine in per unit T; = the electromagnetic torque or load of the generator in per unit Ta = the accelerating torque in per unit 430
431
Steam Turbine Prime Movers
Other { Generators
SYSTEM
Tie Lines
Other Systems
TRANSMISSION
Raise/Lower Commands
NETWORK I
Load/Speed Actual Control Speed
p
,
I
•
I
I
I
I
Tie Line Power
'--y--J
Generator Output Control Signals
System Loads
Unit Generated Power
System Frequency SYSTEM CONTROL CENTER
t
system} Frequency ~ Reference
Fig. I I.!
t
Tie Line} Power ~ Schedule
Power system generation control.
The excitation system is used primarily as a voltage controller and acts much as a single-input, single -output system with VI as the output. There exists a cross-coupling to the torque output Te , but this effect is secondary. The system dispatch computer determines the desired generator output and sets the governor input signal to control the mechanical torque of the prime mover. The governor compares the speed reference or governor speed changer (GSC) signal against the actual speed and drives the servomotor amplifiers in proportion to this difference, which can be interpreted as a speed error. The servo motor output is a stroke or position YSM' which indicates the position of the turbine control or throttle valves. Finally , the prime mover term in Figure 11.2 is a transfer function that relates the turbine control valve position to the mechanical (shaft ) torque . In some cases, this block can be represented by a constant and in others it may be a simple first-order lag. In general, if the system is to be studied over a long time period , the turbine should be represented in greater detail as an energy source transfer function . In some modem thermal units, for example , the energy source controller receives feedback signals from several points , including the generated power (or load control signal) and the turbine throttle pressure, to control simultaneously the turbine valve position, the boiler firing rate, and the condensate pumping rates.
Chapter 11
432
; - - EX~itaii;n-Sy;t;m- - -: , PTs and 1-ooI!E-""-----, I
,
:
Rectifiers:
:+
,, ~ - - - - - - - - --- - - -'
Tie Line Flows
I
I
Fig. 11.2
F
v,
Block diagram of a generating unit.
11.2 Power Plant Control Modes The controls of the steam generator and turbine in a power plant are nearly always considered to be a single control system. This is true because the two units, generator and turbine, operate together to provide a given power output and, since limited energy storage is possible in the boiler-turbine system, the two subsystems must operate in unison under both steady-state and transient conditions . In this section, the different control modes commonly used by the industry are presented and compared .
11.2.1
The turbine-following control mode
The control system shown in Figure 11.3 is usually called the "turbine-following" control, although it is sometimes referred to as "base boiler input" and "admission pressure control" systems (the latter mostly in Europe) . In this control mode, a load demand signal is used to adjust the boiler* firing rate and the fluid pumping rate. As the boiler slowly changes its energy level to correspond to the demand signal, the pressure changes at the throttle (the turbine control valves). Then a back-pressure controlon the turbine changes to hold the throttle pressure constant. This back-pressure control is very slow, even for a rapidly responding boiler. Thus the system response is very slow, monotonic, and very stable. Turbine following may be used on a base-load unit, where the unit will respond only to changes in its own firing and pumping rates. It is often used in start-up or initial stages of unit operation . Turbine following is also used in some modem complex systems when the boiler capability is limited for some reason, such as a fan or pump outage . In general, turbine following is seldom used because of its slow response and its failure to use the heat storage capability of the boiler in an optimal manner to aid in the transition from one generator load level to another. "The tenn "boiler" used here should be taken in a general way to indicate a steam generator and that receives its thermal energy from either a fossil fuel or nuclear energy source.
Steam Turbine Prime Movers
433
Pumping & Firing Rate Control
Fig. 11.3 The turbine-following unit control system [I] .
11.2.2 The boiler-following control mode A more conventional mode of boiler control is called "boiler-following" mode. This control mode is shown in Figure 11.4. This control mode is sometimes called the "conventional mode" or (in Europe) the combustion control mode. This control scheme divides the control function such that the governor responds directly to changes in load demand. The response is an immediate change in generator load due to a change in turbine valve position and the resulting steam flow rate. The boiler "follows" this change and must not only "catch up" to the new load level, but also must account for the energy borrowed or stored in the boiler at the time the change was initiated. This type of control responds quickly, utilizes stored boiler energy effectively, and is generally stable under constant load [1]. Boiler-following control has the disadvantage that pressure restoration is slow and the control is nonlinear. There also may be troublesome interactions between flow, pressure, and temperature variations. If a change in demand exceeds the boiler stored energy, the result may be an oscillation in steam flow and electric power output until the pressure reaches a final, stable value. Boiler-following control is widely used as the normal control mode of many thermal generating units, particularly the older drum-type boiler units. Many newer units employ a more complex control system in which all control functions are integrated into one master control, but even in these more complex controllers, boiler following is offered as an optional control mode that may be required if there are limitations in turbine operation.
11.2.3 The coordinated control mode Most modem thermal generating units employ a control scheme that is usually called an integrated or coordinated control system . This type of system simultaneously adjusts firing rate,
Fig. 11.4 The boiler-following unit control mode [t] .
434
Chapter 11 Pumping & Firing Rate Control
I
Boiler
Fig. 11.5 The coordinated control mode.
pumping rate, and turbine throttling in order to follow changes in load demand. Such a coordinated control mode is shown in Figure 11.5. In this type of control, both pressure and generated output are fed back for the control of both boiler and turbine. In this manner, it is possible to achieve the stable and smooth load changes of the turbine-following mode and still enjoy the prompt response of the boiler-following mode . This is accomplished by making maximum use of the available thermal storage in the boiler. Both pumping and firing rates are made proportional to the generation error so that these efforts are stabilized as the load approaches the required value . Pressure deviation is controlled as a function of both the thermal storage and the generation error. A comparison of the three control methods described above is shown in Figure 11.6
I MW GENERAT ION ---------~ - - 1I ----~.~-~-:;::?"'----:':~-------.... _-----~.,..--
,r;:::· 1/
BOILER·FOLLOWING SYSTEM
" - TURBINE·FOLLOWING SYSTEM
/,
1
COORDINATED CONTROL SYSTEM
-~~-~----- -----------------
THROTILE PRESSURE ,.
.-<# . . . . . . . _ - ... - ..
2 Fig. 11.6
Set
Point
..
"
3
4
Time in minutes
5
6
Comparison of the results of different control methods [2). .
7
435
Steam Turbine Prime Movers
11.3 Thermal Generation The most universal method of electric power generation is accomplished using thermal generation, and the most common machine for this production is the steam turbine. In the United States over 85% of all generation is by powered by steam-turbine-driven generators [1]. The size of these generating units has increased over time, with the largest units now being over 1200MW. The prevalence of thermal energy production in the generation mix of the United States is shown in Table 11.1, which summarizes data compiled by the U.S. Department of Energy for the years 1997 and 1998. A more descriptive way to compare these results is by plotting the numerical values, as shown in Figure 11.7. Here, it is clear that coal is by far the largest energy source used in the United States, at least for the time period represented. As coal becomes depleted or more costly to extract, this could change. The second largest in order of size is nuclear generation. The role of hydro generation is rather small taken on a national basis; however, hydro is very important in certain regions, such as the Pacific Northwest, which is more dependent on this energy source. This is true in many parts of the world, where the predominant energy generation depends on available local natural resources. The steam used in electric production is produced in steam generators or boilers using either fossil or nuclear fuels as primary energy sources. Fossil generation uses primarily coal, natural gas, and oil as fuels. Nuclear generation uses fission reactors that operate by breakup of high-mass atoms to yield a high energy release that is much greater than that produced from chemical reactions such as burning. Fossil fueled plants generate the majority of the electrical energy, but this may gradually change as the sources of fossil fuels are depleted or become more expensive to recover and process than nuclear fuels. By "thermal" generation we usually mean a system that operates on the physical principle of the vapor power cycle or Rankine cycle. Usually, variations of the straight Rankine cycle are used, with two important innovations being the reheat cycle and the regenerative cycle. We will not belabor these concepts here as our primary motive is to study the system operation and control, but a thorough understanding of this important subject is available through many fine refer-
Table 11.1 Net Generation, U.S. Electric Power Industry by Energy Source in GWh
Energy Source Coal (1) Petroleum (2) Natural gas (3) Nuclear Hydro, conventional Other (4) Pump storage (5) Other (6)
1997, GWh 1,843,831 92,727 497,430 628,644 358,949 73,763 -4,040 3,137
1998, GWh
1997, Percent
1998, Percent
1,872,186 129,104 544,765 673,702 328,581 72,867 -4,478 2,905
53.76 2.65 14.23 17.99 10.27 2.11 -0.12 0.09
51.72 3.57 15.05 18.61 9.08 2.01 -0.12 0.08
(I) Includes coal, anthracite, culm, coke breeze, fine coal, waste coal, bituminous gob, and lignite waste. (2) Includes petroleum, petroleum coke, diesel, kerosene, liquid butane, liquid propane, oil waste, and tar oil. (3) Includes natural gas, waste heat, waste gas, butane, methane, propane, and other gas. (4) Includes geothermal, biomass (wood, wood waste, peat, wood liquors, railroad ties, pitch wood sludge, municipal solid waste, agricultural byproducts, straw, tires, landfill gases, and fish oils), wind, solar, and photo voltaic. (5) A more complete designation of this source is hydro pumped storage. (6) Includes hydrogen, sulfur, batteries, chemicals, and purchased steam.
Chapter 11
436
. ..t::
~ 1.5x10
o
Coal Petroleum Natural Gas Nuclear Hydro Geothermal, etc Pumped Storage Hydrogen, etc.
1
-
6
2 3
.S
4
"0
~... 0
1.0
-
0.5
_.
5 6 7 8
....
I:
o
>.
~
I
I:
U.l
i---
0.0
2
3
4
5
6
7
8
Fig. 11.7 Net generation by type of energy source, 1998 (top line) and 1997.
ences on the subject [2-5]. Our objective here is to study the physical design of thermal power plants with the intention of understanding how these plants work and respond to controls .
11.4 A Steam Power Plant Model Steam power plants are of two general types : those fueled by fossil fuels such as natural gas or coal, and those fueled by nuclear energy produced in a thermal reactor. The overall unit control is largely independent of the source of energy, as both types of plants must have a means of controlling the power output as well as the frequency . Figure 11.8 shows a block diagram of the controls for a thermal power plant, in which the source of thermal energy is a steam generator
Steam
Conden sate Control Center
j
/P
, SCT + I~ f
, j
/ j
PG£N
j j -
-
-
-
-
-
-
-
-
-
_.
Set
Point
W
Fig. 11.8 The control system for a thermal generating unit.
V,
Steam Turbine Prime Movers
437
that could utilize either fossil or nuclear fuel. The term "boiler" is used here to designate any type of steam generator. The boiler control inputs are the unit demand signal (UDS), the generated power (P GEN) , and the speed or frequency (w). The unit demand signal is set by the system dispatch computer based on the method of dispatch and on the level of load to be served. The generated power of the unit is fed back to the control center so that any error in generated power can be corrected, The unit speed is used by the speed governor as a first-order control on this parameter. The speed governor acts as a continuous, proportional controller to make fast, automatic adjustments to unit speed in response to a speed error. This mechanism is much faster than the governor speed changer (GSC) adjustment of the boiler controller. The input from the dispatch computer is optional and is not used when the unit is on local control. In that case, the UDS is hand set by the plant operator. Note also that the boiler controller can be turbine following (adjusting firing rate according to desired power), boiler following (adjusting firing rate to hold throttle pressure), or a completely integrated or coordinated control that does both simultaneously. The degree of detail required for computer simulation of the power system depends on the length of time required in the simulation. Studies of system performance of a few seconds, for example, need consider only those system components with response times of a few seconds, such as the generator, exciter, and speed governor. Studies of several minutes would usually require some consideration of the steam generator and steam system controls, and may require some consideration of the dispatch system. Thus, it is seen that the longer the desired simulation, the more system components that might enter into consideration. For very long periods of interest, the fastest responding components might be represented in a very simple manner and may not be required at all. In transient stability studies of 1-10 seconds duration, it is common to consider the generator, network, and the steam turbine and turbine controls. If there is interest in extending the studies to several minutes, then it is probably necessary to add at least a simple boiler model to the simulation, and it may be necessary to consider the dispatch computer as well. The general block diagram of Figure 11.8 would be applicable to these longer-duration studies.
11.5 Steam Turbines A large portion of the conversion of thermal to electrical energy occurs in steam turbines. This is due to the many advantages of the steam turbine over reciprocating engines. Among these advantages are the balanced construction, relatively high efficiency, few moving parts, ease of maintenance, and availability in large sizes. Internally, the steam turbine consists of rows of blades designed to extract the heat and pressure energy of the steam, which is usually superheated, and convert this energy into mechanical energy. To accomplish this goal, high-pressure steam is admitted through a set of control valves and allowed to expand as it passes through the turbine, to be exhausted, usually to a condenser, at relatively low pressure and temperature. Thus, the type and arrangement of turbine blading is important in extracting all possible energy from the steam and converting this energy into the mechanical work of spinning the turbine rotor and attached electric generator. Two types of turbine blading are used; impulse and reaction blading. In impulse blading, the steam expands and its pressure drops as it passes through a nozzle, leaving the nozzle at high velocity as shown in Figure 11.9 (a). This kinetic energy is converted into mechanical energy as the steam strikes the moving turbine blades and pushes them forward. Reaction blading operates on a different principle, as illustrated in Figure 11.9 (b). Here the "nozzle" through which the steam expands is moving with the shaft, giving the shaft a torque due to the unbalanced forces acting on the blade intake and exhaust surfaces. A somewhat more realistic picture of the combined impulse-reaction blading is shown in Figure 11.10. The two moving stages on
Chapter 11
438
Nozzle Mounted on Wheel Stationary Nozzle
(a) Impulse Blading
(b) Reaction Blading
Fig. 11.9 Two types of turbine blading.
the left of the figure are impulse stages, whereas those on the right are reaction stages. In many turbines, impulse stages are used at the high-pressure, high-temperature end of the turbine and reaction blading at lower pressures. This is because there is no pressure drop across impulse stages and hence there is little tendency for the high-pressure steam to leak past these stages without doing useful work. As the steam expands in passing through the turbine, its volume increases by hundreds of times. At the lower pressures, reaction blading is used. Here, the steam expands as it passes through the blading and its pressure drops. The steam velocity increases as it passes through fixed blading as shown in Figure 11.10, but it leaves the moving blades at a speed about equal to the blade speed. The impulse stage nozzle directs the steam into buckets mounted on the rim of the rotating disk and the steam flow changes to the axial direction as it moves through the rotating disk. In reaction blading, the stationary blades direct the steam into passages between the moving blades and the pressure drops across both the fixed and moving blades . In impulse blading, pressure drops only across the nozzle . In the velocity compound stages, steam is discharged into two reaction stages . The velocity stage uses a large pressure drop to develop a high-speed steam jet. Fixed blades then tum the partially slowed steam before it enters the second row of moving blades, where most of the remaining energy is absorbed. Because of the tremendous increase in the volume of steam as it passes through the turbine, the radius of the turbine is increased toward the low-pressure end. In many turbines, the steam flow is divided into two or more sets of low-pressure (reaction) turbines. Figure 10.11 shows several typical tandem compound configurations and Figure 11.12 shows several typical crosscompound designs . In some designs, the steam is reheated between stages to create a reheat cycle, as noted in the figures, which increases the overall efficiency. In other designs, a portion of the steam is exhausted from the various turbine pressure levels to preheat water that is entering the boiler, which is called a regenerative cycle system . The various valves that control the turbine operation are shown in Figure 11.12 and will be discussed in the order encountered by the steam as it moves through the system. Steam leaves the main steam reheater of the boiler at high pressure and is superheated, in most cases, to high superheat temperature. For example, a large fossil fuel unit uses superheated steam at 2400 psi and 1000°F for a 1.0 GW unit [15]. A modem 750 MW nuclear design uses 850 psi saturated (0.25 percent moisture) steam [16]. The steam heaters contain steam strainers
Steam Turbine Prime Movers
439
.1".
1
Moving
Fig. 11.10 Combined impulse and reaction blading [6].
to catch any boiler scale that could damage the turbine. A typical steam generator and turbine system is shown in Figure 11.13 [7]. The main stop valve or throttle valve (#2 in Figure 11.13) is one means of controlling the steam admitted to the turbine. It is often used as a start-up and shut-down controller. During startup, for example, other inlet valves may be opened and steam admitted gradually through the stop valve to slowly bring the turbine up to temperature and increase the turbine speed to nearly synchronous speed, at which point the governor can assume control of the unit. This mode of control is known as full-arc admission. The main stop valve is also used to shut off the steam supply if the unit overspeeds. The unit may be under automatic or manual control, but is usually controlled automatically through a hydraulic control system. . A typical example of the several valves controlling a large steam unit is presented in Figure 11.13 [7]. This system is typical of many large steam power plants, having both superheater and reheater boiler sections and three separate turbines, representing high pressure (HP), intermediate pressure (IP), and low pressure (LP) units. The admission or governor valves, also known as control valves (#3 in the figure), are located in the turbine steam chest and these valves control the flow of steam to the high-pressure turbine. In large units there are several of these valves, and the required valve position is determined by the governor (D in the figure). An overview of the turbine control for a typical steam power plant is shown in Figure 11.14. Steam is admitted through the main stop valves to a set of control valves and admission of steam into the high pressure turbine is regulated by a set of nozzles distributed around the periphery of the first stage of turbine blading. If only a few of the control valves are open, the
Chapter 11
.440
Single-Casing Single-Flow
Two-Casing Double-Flow
Reheater
Reheater
Single-Casing Opposed-Flow
Reheater Two-Casing Double-Flow-Reheat
Three-Casing Tripple-Flow-Reheat
Four-Casing Quadruple-Flow-Reheat
Fig. 11.11 Typical tandem compound steam turbine designs with single shaft [6].
steam is said to be admitted under partial arc of the first stage rather than through all 360 degrees of the circumference. This is called "partial arc admission." Two types of overspeed protection are provided on most units. The first is the normal speed control system, which includes the control valves and the intercept valves. The second type of overspeed control closes the main and reheat stop valves, and if these valves are closed, the unit is shut down. Two types of control valve operation are used. In one type, the control valves are opened by a set of adjustable cam lifters, as shown in Figure 11.15. In this arrangement, the valves can be opened in a predetermined sequence as the cam shaft is rotated. In response to a load increase, the flow ofsteam to one input port may be increased and a closed port may simultaneously be cracked
441
Steam Turbine Prime Movers Reheater
Reheater
Two-Casing Double-Flow
Four-Casing Quadruple-Flow-Reheat
Two-Casing Double-Flow-Reheat
Reheater
Four-Casing Quadruple-Flow-Reheat
Six-Casing Sextuple-Flow-Double-Reheat
Five-Casing Sextuple-Flow-Reheat
Six-Casing
Octuple-F low-Reh eat
Fig. 11 .12 Typical cross-compound steam turbinedesigns with multiple shafts [6].
442
Chapter 11
A- MAN G<M:RNOR
B- AUXILIARI' GOVERNOR C-CHECK ~VE
0- GCNERNOR CONTROL MECHANISM E-MAIN PUMP r - GCl'Vt:RNOR IM"£U..ER G-INTERC£PTOR CONTROL t.AECHANISM H-ORIFICE K-Al/TO STOP VALVE - - - - GCl'Vt:RNOR CO/IITROL.
H
== ~ GCM:RNOR CONTROL
~
-
CM:RSPEED CONTRa.. HIGH PRESSURE: 011.. 5Um.V
LP
GO£RATOR l-RE!.D" '-"LVE
2-1lf'lOTTLE \lA.LVE
3-GOVERNOR "'LVE 4-INTERCEPTOR \IlLVE 5-SY-F\4,SS
e- DU.4P
=
w..vt
l '~11
'-"LVE STEAM ut-£S
Example of a large boiler configurat ion showing major system component s and controls [7].
Fig. 11.13
Steam Generator
1-------
c=====l' ~ontrol 1I
I
I
I
Overspeed Trip
Main Stop Valve
Crossover
Speed
I
Intermediate Pressure Turbine
tL ,------_
Generator
Load Intercept Valve
Reheater
Fig. 11.14 A reheat turbine flow diagram.
Condenser
Steam Turbine Prime Movers
Fig. 11.15
443
Cam lift steam turbine control valve mechanism .
open. This distributes the steam around the periphery of the first stage, assuring a uniformtemperaturedistribution and controllingthe power input.The cam shaft is controlledby the governor actingthrougha power servomotor, as shown in Figures 11.13 and 11.14. The other type of steam admission control is called the "bar lift" mechanism. This type of valve control is shown in Figure 11.16; each valve in a line of valves is lifted using a bar, but each valve is a differentlength so that the valves open sequentially. As load is added to the turbine, the bar is raised and steam flow is not only increased to the first-opening valve, but additional valves are also opened. The separate valves feed steam to different input ports around the peripheryof the first-stage blading and thus increase the power input to the turbine. The bar lift is actuatedby the governorservomotorthrough a lever arrangement.
Valve Lift Bar
Steam Chest
\.......-'I.......-lI.-.L--H .,;41~--L_.L-..+--.
Nozzle,...,
Chest Balance Rod" L-_",--_-' Fig. 11.16
Bar lift steam turbine control valve mechanism [2].
444
Chapter 11
The high-pressure turbine receives steam at high pressure and high temperature, and converts a fractionjofthe thermal energy into mechanical work. As the steam gives up its energy, it expands and is cooled. Steam is also bled from the turbine and piped to feedwater heaters. This has proven economical in reducing the boiler size and also reducing the size required at the low-pressure end of the turbine. The turbine extraction points vary in number from one to about eight, the exact number being dictated by design and economy. In the reheat turbines, shown in Figure 11.14, the steam exhausted from the HP (high-pressure) turbine is returned to the boiler in order to increase its thermal energy before it is introduced into the intermediate-pressure (IP) turbine. This reheat steam is usually heated to its initial temperature, but at a pressure that is somewhat reduced from the HP steam condition. Following the reheater, the steam encounters two valves before it enters the IP turbine, as shown in Figures 11.13 and 11.14. One of these is the reheatstop valve and serves the function of shutting off the steam supply to the IP turbine in the event the unit experiences shut-down, such as in an overspeed trip operation. The second valve, the interceptvalve, shuts off the steam to the IP turbine in case of loss of load, in order to prevent overspeeding. It is actuated by the governor, whereas the reheat stop valve is actuated by the overspeed trip mechanism. The IP turbine in Figure 11.13 is similar to the HP turbine except that it has longer blades to permit passage of a greater volume of steam. Extraction points are again provided to bleed off spent steam to feedwater heaters. The crossover, identified in Figure 11.14, is a large pipe into which the IP turbine exhausts its steam. It carries large volumes of low-pressure steam to the low-pressure (LP) turbine(s). Usually, the LP turbine is double or triple flow as shown in Figures 11.11 and 11.12. Since a large volume of steam must be controlled at these low pressures, doubling or tripling the paths available reduces the necessary length of the turbine blades. The LP turbines extract the remaining heat from the steam before exhausting the spent steam to the vacuum of the condenser. It is desirable to limit condensation taking place within the turbine, as any water droplets that form there act like tiny steel balls when they collide with the turbine blades, which are traveling at nearly the speed of sound. We previously specified that the HP turbine extracts a fractionjofthe thermal power from the steam. Then the IP and LP turbines extract the remaining 1 - j of the available power to drive the shaft. Usually,jis on the order of 0.2 to 0.3. For example, in a certain modem 330 MW turbine,fis determined to be 0.24. This is a rather typical value.
11.6 Steam Turbine Control Operations The controls for a steam turbine can be divided into those used for control of the turbine and those used for the protection of the turbine. It is difficult to sketch a "typical" control system for a steam turbine since these controls depend on the age of the unit and the type of controls available at the time of unit installation. Since power plants operate for many years, there are likely to be many different controls, using different technologies, on any given power system. However, we can summarize the most common controls as being either "traditional" or "modem," with those terms also having a somewhat variable meaning due to the steady advance in control technology. The control operations that are usually considered to be "traditional" are listed in Table 11.2. These are controls that have been required for many years and that require only the very basic technologies for their operation. It is apparent that plant control systems become more complex due to the demands of interconnected operation and the availability of more modem methods of control. The newer controls provide many functions that were not considered necessary for older units, and some that were not available due to limitations of the available technology at the time of manufacture.
Steam Turbine Prime Movers
445
Table 11.2 Traditional and ModemSteamTurbine Generator Controls
Traditional Controls Speedcontrol, near rated speed Overspeed protection Loadcontrol-manual or remote Basiccontrol and protection Initialpressure Vacuum Vibration Others, as needed
ModemControls All traditional controls and protections Long-range speed(zero to rated speed) Automatic line speedmatching Loadcontrol; automatic load setback Admission modeselection Automatic safetyand condition monitoring On-line testing of all safetysystems Fast or earlyvalveactuation Interface to the plant computer Interface to area generation control system
Many of the plant controls are hydraulic, using high-pressure oil supplied by a shaftmounted main oil pump. These high pressures are practical for the operation of power servomotors for control purposes. For example, many control valves are actuated by hydraulic means. In modem plants, many systems also use electric controls as well. The control functions for the turbine include the servomotor-driven control or governing valves and the intercept valves, which control the amount of steam admitted to the turbine. Positioning intelligence for these valves comes primarily from the speed governor, the throttle pressure regulator, or from an auxiliary governor. There is also an interlocking protection between the control and intercept valves so that the control valves cannot be operated open when the intercept valves are closed. The protective controls include the main stop valve (throttle valve) and the reheat stop valve. The reheat stop valve is always either fully open or fully closed, and is never operated partially open. The main stop valve may operate partially open when used as a startup control. Both valves are under control of a device that can rapidly close both valves, shutting down the turbine on the occurrence of emergency conditions such as overspeed trip, solenoid trip, lowvacuum trip, low bearing oil trip, thrust bearing trip, or manual trip. During normal operation, both of these stop valves are completely open. A primary function of the main stop valve is to shut off the steam flow if the unit speed exceeds some predetermined ceiling value, such as 110% of the rated value. Steam turbine blading experiences mechanical vibration or oscillation at certain frequencies. The turbine designer assures that such oscillations occur above or below synchronous speed, with a generous margin of safety. Also, with the longer blades traveling at nearly the speed of sound, destructive vibration levels may be reached if the speed is permitted to increase substantially beyond rated speed. Thus, speed control on loss of load is very important and is a carefully designed control function. [9]. The operation of a steam turbine on loss of load is approximately as shown in Figure 11.17. It is assumed that the generator breaker opens at t = 0 when the unit is fully loaded. On loss of load, the turbine speed rises to about 109% in about one second. As the speed increases, the control valves and intercept valves are closing at the maximum rate and should be completely closed by the time the speed reaches 109% of the rated value, at which time the turbine speed begins to drop. At about 106%, the intercept valves begin to reopen so that a no-load speed of 105% might be achieved. If the speed changer is left at its previous setting, the unit will continue to run at 105% speed on steam stored in the reheater. There is usually sufficient steam for one to three minutes of such operation. Once the reheater steam supply is exhausted, the speed will drop to near 100% and the governor will reopen the control valves. The definition of what constitutes an emergency overspeed [10] is a figure agreed upon by
446
Chapter 11
110
Intercept Valve Starts to Reopen
109 108 13 107 & 106 en i:: 105
Time to Blowdown Reheater
~
8 104 ~
Time to Blowdown
I
-<
103 102 101
Reheater
>-
I
~NoLoad
5% Auxiliary Load Remaining on Generator
-E-
Remaining on Generator
l00------------~-------------1.-~
o
Fig. 11.17
1
Time in minutes
2
Estimated speed versus time following sudden reduction from a maximum load to the values noted.
turbinemanufacturerand purchaser, but may be in the region of 110to 120%of the rated value. If the speed reaches this range, an emergency overspeedtrip device operates. Usually the overspeed trip mechanismdepends on centrifugal force or other physicalmeasurements that are not dependent on the retention of power supply. Some devices include an eccentric weight or bolt, mounted in the turbine shaft, with the weight being balanced by a spring. At a predetermined speed, such as III %, the centrifugal force overcomes the spring force and the bolt moves out radially far enough to strike a tripper,which operates the overspeedtrip valve.
11.7 Steam Turbine Control Functions We now investigate the transfer functions that describethe operationand control of a typical steam turbine.* The system under investigation is the reheat steam turbine of Figure 11.13, with controls as describedin the precedingsections. The block diagramfor this system is shown in Figure 11.18 [10], with controls as described in the preceding paragraphs. Our immediate concern is with the thermal system betweenthe control valves, with input 1]2 and turbine torque 'T. The symbols used in Figure 11.17 represent per-unit changes in the variables, as defined in Table 11.3. For the present, we will accept the transfer functions of governor and servomotor and reserve these for later investigation. Let us examine the functions between 1]2 and 'T in Figure 11.18 more carefully. The control valve transfer function is nearly a constant and would be exactly 1.0 were it not for nonlinearvariations introduced by control valve action. This is due to a combination of nonlinearities. First of all, the steam flow is not a linear function of valve lift, or displacement, as shown by the right-hand block of Figure 11.19. It is, in fact, quite nonlinear, exhibitinga definite saturationas the valve opening increases. One way to counteractthis nonlinearity is to introduce a nonlinearity in the valve lifting mechanism, as shown in the left block of Figure 11.19. This is accomplished with a cam lift mechanism, as shown in Figure 11.20. Here, the cam acts as a function generatorprovidingan output 'We follow closely here the excellent reference by the late M. A. Eggenberger [10] who did significant work in this field. The authors are indebted to Mr. Eggneberger for having shared his work, some of which is unpublished .
447
Steam Turbine Prime Movers
f Pa
+
1
171
1 + 1;s
1 1 + 1;s
~
172
lL
J1v
1
1
1 + 1;s
lfIR 1 1+ TRs
T4s
+
1- f
a
T/P + LP
L . - - - - - - - - - - - - - - 1 Cg Fig. 11.18
Block diagram of mechanical reheat turbine speed control [10].
(11.2)
L = / (1)2, L)
in which the output L is a function not only of 1)2 but also of 1. In this way, the transfer function of the two blocks taken together are nearly linear for any given valve. Still, a small nonlinearity exists in the overall transfer function, as shown in Figure 11.18, due to "valve points," as this phenomenon is known in the industry. This refers to the point at which one valve, or set of valves, approaches its rated flow and a new valve (or valves) begins to open.
Table 11.3
Per Unit Change Variable
Definition of Per-Unit Change Variables
DefiningEquation Nf!,.
Remarks
Speedof rotation
a =-
Developed torque
'T
Load torque
A= -
Steamflow
IL
Servomotor stroke
Tl2 =
Speedrelay stroke
Ylf!,. TIl = -y
YI R = Speed relay stroke for full load
Speed/load reference
Rf!,. P=RR
RR = Reference positionat rated load and rated speed
Speedgovernorstroke
t> X
Speed error signal Valve steam flow HP turbine torque Reheatpressure IP + LP torque Accelerating torque
NR
NR = Rated speed
r:
TmR = Rated full load torque
r:
TeR = Rated electrical torque
= -
TmR TeR Qf!,.
=-
QR Y2f!,. Y2R IR
XI!. R
B
ILv 'T/{?
!/JR 'T/P&LP 'Ta
QR = Rated steam flow in Ib/sec Y2R =
Servomotor position for steady rated load
X R = Speed governorstroke for 5% speed change Speed relay input Control valve output HP turbine output variable Reheateroutput variable IP + LP turbine torque
448
Chapter 11 L
L servo stroke
valv e
steam
lift
flow
Fig. 11.19 Block diagram for camshaft and valve function generators [10].
This causes the transferfunction to consistof a series of small curvedarcs, as shown in Figure IUS. To compute the transfer function of steamflow versus servomotor stroke, we write K) = /Lv 1/2
(11.3)
If it were not for valve points, the curve expressing the function K) would be a constant with valueof unity,with the incremental regulation at the operating point the sameas that of the governor (usually 5%). If we define incremental regulation R, as [10] da
R; = dP
(11.4)
where a is the per-unitspeed, P is the per-unitpower,and R, is evaluated at the operating point. Ifwe let Rs be the steady-state regulation or droop No-N R =N- S
(11.5)
r
o o o
0 0
0
Control Valve
Valve Lift
Fig. 11.20 Mechanical function generator (cam-operated control valve).
Steam Turbine Prime Movers
449
then we have
K
3
Rs R
=-
(11.6)
I
Eggenberger [10] points out that R; is often between 0.02 and 0.12 over the range of valve strokes and may be taken as 0.08 as a good approximate value . Using this value, we would compute for a typical case 0.05
K3 = 0.08 = 0.625
(11.7)
From Figure 11.8, we see that the steam is delayed in reaching the turbines by a bowl delay
T3, expressed in terms of servo stroke and turbine flow parameters as JLr
K3
112
1+ T3s
(11.8)
where T3 is the time it takes to fill the bowl volume VB (ft') with steam at rated initial conditions, with specific volume initially of v (Ibm/sec) , or [I 0] T3 = -
VB
vQr
seconds
(11.9)
Typical values of T3 are given as 0.05 to 0.4 seconds . For a straight condensing turbine with no reheat, the torque versus servomotor stroke is given by (11.8). This situation is illustrated in Figure 11.21 and is accomplished mathematically by replacing JLr in (11.8) by T . This is equivalent to setting the fractionjoftorque provided by the HP turbine to unity. For a reheat turbine , there is a large volume ofsteam between the HP exhaust and the IP inlet. This introduces an additional delay in the thermal system . From Figure 11.18 with elementary reduction , we have [10] T
112
K 3(1
+fF&S)
(I + T1s )(1 + T&S)
~ o 0 o
g
V
Fig. 11.2\
0 0
B
Torque production as controlled by servomechanism stroke.
(11.10)
450
Chapter 11
wherefis the fraction of the total power that is developed in the high-pressure unit and is usually between 0.2 and 0.3. The parameter TR is the time constant of the reheater and is defined in a manner similar to (11.9) or
(ll.ll) where VR = volume of reheater and piping, ft3 QRr = full load reheater steam flow, Ibm/sec V R = average specific volume of steam in the reheater, ft3/lbm Since the reheat temperature is not constant, computation of TR involves taking averages, but it is usually in the neighborhood of 3 to II seconds. This long time constant in the reheater causes a considerable lag in output power change following a change in valve setting. In HP turbines, there may be a delay of up to 0.5 seconds, depending upon control valve location. A much larger delay occurs in the IP and LP sections, however. This is due to the large amount of steam downstream of the control valves , and this steam must be moved through the turbines and reheater before the new condition can be established. These delays are both shown in Figure 11.22, where the control valve is given a hypothetical step change and the power output change is plotted [10]. A five second value for TR is assumed. The speed-torque transfer function is given in Figure 11.18 as [10] T
(ll.l2)
T4s
The time constant T4 is the total time it would take to accelerate the rotor from standstill to rated speed if rated torque, Tm , is applied as a step function at t = O. At rated speed, the kinetic energy in the rotating mass is
(ll.l3)
Control Valve Position
70%
1_----
,:
1
IP & LP Turbines
Turbine Power Output
I I
I
I
60%
I
I
o
I
....!-
_ _ _ 1...1
2
3
4
5
~
6
7
Time, seconds Fig. 11.22 Reheat turbine response to a control valve change.
8
9
Steam Turbine Prime Movers
451
and the differential equation of motion is ]w = T; == a constant
(11.14)
where we take (11.15) the rated value of torque. Solving (11.14) for constant torque gives T4 =
wRJ
-
TmR
wlJ
== -
r. -
(11.16)
since TmR = P /WR' From (11.16) and (11.13) we can compute 2Wk
T4 = - - seconds Pr
(11.17)
where the units must be consistent. We usually compute Wk =
±( ~~; )( 2::, y( ~:~) Joule 0.83(WR2)N;
= 3600 x 106 MWs so that T
4
--
(WR2)N;. d - - - -9)P - secon s (2.165 X 10 r
(11.18)
where Pr = rated power in MW WR2 = rotor inertia in lbm-ff NR = rated speed in rpm Another useful constant is the so-called specific inertia of the turbine-generator [10]: J.~p = (
2
WR )( N )2 ---p;3600 x 10-3Ibm-ftlMW r
(11.19)
and this is convenient since it usually turns out to be nearly unity. In terms of this constant, T4 = 5.98 J sp seconds
(11.20)
Actually, as the turbine speed increases, the load torque increases and the loss torque varies as some power of the speed. Eggenberger [10] shows that this can be accounted for by replacing the single block in Figure 11.18that relates a to l' by a feedback system wherein a portion of the speed increase is fed back as a negative torque [10]. However, as the losses are very small, this is usually neglected. A set of typical constants for all values shown in Figure 11.18 is given in [10] and is valuable for making comparisons of the various system lags under consideration. These constants are shown in Table 11.4. Additional insight into the control of the steam turbine system is gained through an evaluation of system performance by the root locus method [12]. Referring to Figure 11.14 and equations (11.3) through (11.12), we may write the open-loop transfer function as KG(s) ==
+ I/fTR ) s(s + lIT})(s + IIT2)(s + IIT3)(s + IITR) K(s
(11.21)
Chapter 11
452
Table 11.4 Typical Values of Constants Used in Steam Turbine Analysis
Parameter Cg T.
T2 K3
T3 TR
f
T4
Normalized speed governorconstant(5% regulation) Speedrelay time constant Servomotor time constant Valve gain at no-loadpoint Valve bowl time constant Reheatertime constant Load on HP turbineper unit Turbinecharacteristic time
Non-reheat Turbine
Reheat Turbine
20 0.08 to 0.14 s 0.15 to 0.25 s 0.625 0.05 to 0.3 s
20 0.08 to 0.18 s 0.15 to 0.30 s 0.6 to 0.8 0.05 to 0.4 s 3 to 11 s 0.2 to 0.3
6 to 12 s
5 to 12 s
where
K=
fK3Cg
T]T2T3T4
Considering the range possible for each variable as shown in Table 11.4, we have a range of pole-zero locations and gains as shown in Table 11.5. The range of values shown in Table 11.5 has some influence on system behavior, as shown in Figure 11.23, where poles of a nonreheat turbine are plotted as a band of values rather than as a point in the s plane. It is obvious that, since the system response depends on these pole locations, this system may be designed with a wide range of response characteristics. This is especially true for the valve bowl delay, which may vary from 0.05 to 0.3 seconds [10]. Other component values affect the response as well, especially the servomotor pole, which may be quite close to the origin. A similar plot for the reheat turbine is shown in Figure 11.24. Here, the four poles due to the inertia, servomotor, speed relay, and valve bowl are far enough from the origin to be offscale for the scale chosen for this figure. This means that the reheater pole and zero will always be relatively close to the origin and will, therefore, have a great influence on the system dynamic response, even for small disturbances. For large disturbances, the problem is greatly complicated because the reheater should then be treated as a nonlinear model to account for the spatial distribution of flow and pressure in both reheater and piping. A convenient method of analyzing steam turbine systems is to use the root locus technique [12]. Two examples, one for the straight condensing (nonreheat) turbine and one for the reheat turbine will illustrate the method.
Table t 1.5
Range of Values for Poles, Zeros, and Gains
Pole/Zero Pole
Zero Gain
Reheat
Nonreheat
Item Symbol
Minimum
Maximum
Minimum
Maximum
lIT, llT2 lIT] l/TR l/fTR
7.15 4.00 3.33
12.50 6.67 20.00
46.3
5340
5.55 3.33 2.50 0.091 0.303 9.27
12.50 6.67 20.00 0.333 1.667 1600
K
Steam Turbine Prime Movers
453 ill
...
+5
....00
E
ll) 0= ll)_
IE -20
I.-
~~
t:
ll)
.5
~I~
>1
valve bowl delay
CT
0
-5
-10
-15
.€0=
0
"0»
-5 Fig. 11.23
s Plane plot of poles for the nonreheat turbine .
Example 11.1 Prepare a root-locus plot for a nonreheat turbine with the following constants :
T3 = 0.0667 s Determine the damping ratio and undamped natural frequency for the two least damped roots if K3 = 0.625 and Cg = 20. Solution The block diagram for this system is that shown in Figure 11.25. The open-loop transfer function is KG(s) =
K s(s + 5)(s + 10)(s + 15)
K = ---:---=--~-3 2 S4
+ 30s + 225s + 750s
(11.22)
For the constants given in this example, we can compute the gain K as K
=
K 3Cg
T,T2T3T4
= 937.5
(11.23)
0.5 Zero Range
1< -2
-1.5
-1
~I -05
U ole Range
Fig. 11 .24
Pole and zero for the reheater.
o
-0.5
454
Chapter 11
Fig. 11.25
Block diagram for the nonreheat turbine.
We also compute the following constants , which are required in order to construct the root locus plot: I. The excess of poles over zeros = P - Z = 4 - 0 = 4 2. The asymptotes lie at angles of
Ok=
(2k+ 1)180° p -z
±45°, ±135°
(11.24)
lP-lz -30 ==-7.5 P-Z 4
(11.25)
=
3. The center of gravity is located at
e.G. = 4. Write the polynomial
D(s)
+ KN(s) = 0
(11.26)
In our case, we have
s(s + 5)(s + 10)(s + 15) + K=S4 + 30s3 + 275s2 + K
(11.27)
From (11.27), we construct the Routh's table [13] to find the critical value of gain and the point of the w-axis crossing:
S4 S3 s2
1 30 740
275 750
s'
55500- 9K K
0
SO
K 0
3K
For the first column in this array to be positive, we require that
K:5 6167 The auxiliary polynomial [13] is
740s2 + 3(6167) = 0 or
s = ±j5
(11.28)
5. The locus "breaks away" from the negative real axis at points k, and k2 defined by the equations
SteamTurbine Prime Movers
455
1
1
1
1
~
5 -~
10-~
15 -~
- =--+---+--1
1
1
1
15 - k 2
k 2 - 10
k2 - 5
k2
--- =---+--+-
(11.29)
We solve (11.29) by trial and error to find
k, == 1.91 (actually -1.91)
(11.30)
k 2 = 15 - 1.91 = 13.09
(11.31)
and, by symmetry,
6. Incorporating information accumulated in equations (11.24) to (11.31), we construct the root locus diagram shown in Figure 11.26. We can also locate the point corresponding to the assumed gain of 937. With this value of gain, the damping ratio is ~=
0.7
(11.32)
/ / /
/
1: =0.7 (~1
•
= 2.2
/ / /
"
speed relay
-10 /
/ "
,,/
/
/ t"
/ /
' - - - Asymptotes
servo motor
--5
C.G.=-7 ,5 -,
/ /
/
"
"
~ -,
/ / /
/
Fig. 11.26
Root locus for a nonreheat turbine system.
Chapter 11
456
and the undamped natural frequency is Wn
= 2.2 radians/s
(11.33)
These values are indicated in Figure 11.26. Also note in the root locus plot that the poles are labeled to remind us of the reason for their existence. They can be moved by changing the appropriate design parameters. We now recognize the significance of the solution just obtained. Note that, corresponding to a gain of 937, there are actually four solutions, indicated by the dots on the locus. Two of these solutions correspond to responses that are very quickly damped out, being located at approximately -13.5 in the negative-real direction. By comparison, the least damped roots are located at -{Wn
= -1.54
(11.34)
and we can neglect the quickly damped solutions with very little error. Thus, our system will respond approximately as a second-order response [14]: (11.35) where
k=vT="f k
cf> = tan-- 1{
io.> kio; u(t) = unit step function
This response is a damped oscillatory response and this is, generally speaking, what we would like. We would hope to have the damping factor {be fairly large for good damping and to prevent an overshoot or too long an oscillation. Certainly, ~ ~ 0.2 is desirable as this corresponds to about 50% overshoot (actually 52.6%). In our case, with {= 0.7 there is practically no overshoot and the system is very well damped. If some oscillation can be tolerated, this system could be operated at a higher gain. Figure 11.27 shows a typical second-order response for values of l* of 0.2 and 0.7. Note that when l* = 0.7 there is very little overshoot, but with l* = 0.2 the overshoot is about 50% (actually 52.6%) and oscillations ring down for almost four seconds. If some oscillation can be tolerated, this system could be operated at a higher gain.
Example 11.2
If the system of Example 11.1 is a reheat system, the fraction f of power generated by the HP turbine and the reheater time constant TR must be specified. Suppose we let
1=0.2 TR = 5 s Then the open-loop transfer function becomes
KG(s) =
K(s + 1) s(s + 5)(s + 10)(s + 15)(s + 0.2)
---------
(11.36)
and the normal value of K is (11.37)
457
Steam Turbine Prime Movers 1.6 , - - - , - - - -, --
--,------ , - --
--.------,---
---,
1.4 1.2
0.8 0.6 0.4 0.2
o -0.2
~---'-----'----"-----I.---L..---..l------l
o
5
2
Fig. I 1.27
6
7
Step response of a second-order system.
The block diagram for this new system is shown in Figure 11.28. The root locus plot is shown in Figure 11.29. From this plot, we observe that for a gain of about 187, the damping ratio is about 0.4, corresponding to an overshoot of about 25%, and the undamped natural frequency is about 0.5 radians per second. Thus the product -{Wn = - 0.2
(11.38)
is much less than for the straight condensing turbine . Note also, however, that the system gain could be increased substantially with practically no change in {up to a frequency of about 1.5 or 2.0, which would improve the product by a factor of three or four and the oscillations would decay much faster as we see from the exponent of (11.35) . The block diagram of a more detailed dynamic model of a reheat steam turbine system is shown in Figure 11.30. This more detailed model consists of high-pressure, intermediate-pressure, and low-pressure turbines on a single shaft, driving a generator and excitation system, as shown in Figure 11.14. The principal dynamic components that effect the time lag of delivered mechanical power are the speed relay, control valves, steam bowl, the drum, and the feedwater heaters. In normal operation, the intercept valve is fully open, but the control valve may be only partially open, depending on the scheduled generation output of the unit. These dynamic components are connected in the system diagram of Figure 11.30 by solid lines.
Fig. 11.28
Block diagram of a reheat turbine system.
458
Chapter 11
,
-,
-,
1;=0 .4
-,
-,
-,
-,
<,
-. -,
bowl del ay
-,
-,
<,
speed rela y
-15
/
<,
"-
- 10
/
/
/
/
/
/
/
/
/
/
/
/
/
/
"- " / / servo .... "- / motor /
/
"-
"-
-5
"-
"-
"-
"- -,
"-
"- -,
"""-
"-
"-
Fig. 11.29 Root locus for a reheat steam turbine system.
The dashed lines in Figure 11.30 show the connection of an overspeed protection system. This system will initiate fast turbine control and intercept valve closure in the event of a load rejection. The control logic operates by comparing the turbine power, which is determined by measuring cold reheat pressure, and the generated power, measured by the generator current. This protection will operate if the difference between these measured power values becomes greater than a preset value, typically about 40% of full load , and the rate-of-change in generator current is also greater than a set point value. This provides overspeed protection for the generating unit that might follow a loss of load. 11.8
Steam Generator Control
The expansion of power system interconnections has necessitated more precise control in order to hold the frequency stable and to control disturbances. It has also introduced a new class of stability problems that are not so much concerned with system recovery following major impacts, such as faults, as with the control and damping of sustained oscillations over periods of several minutes duration. Thus, system components that are usually thought of as quite slow in response must be investigated for poss ible behavior that might be detrimental to system damping . The steam generator is such a component. Steam generators can be either fossil or nuclear fuel systems, but here we shall concentrate on fossil -fueled boilers. The recovery time of boiler pressure following a sudden change in turbine control valve setting is measured in minutes for systems of conventional design . During this period, the boiler-turbine system is operating with
.....
0'1 '0
LRM
~ynarmcs
Generator Current
>- l- -L":'o-d-""" a Unbalance --~~I Logic
I LRM=Load Reference Motor I
L-.
I-
Governor Dead Band
L __ JL~
: Position
I I
(i)HP
I
Fig. I 1.30
"a
>
Control Valve
1~1;s
1
>
> "a
ee
~o
I
Wr
~
'" en
E
tt:
o
~
~
IX
Steam Bowl
1 + :Z;S
~
_1_ I WH
Typical turbin e control dynamic for a reh eat steam turb ine system.
Speed Relay
1 +1;s
1
>
<)
Q.,
o
.s.;;; c
Drum
+ l~
1 -(J(H+ K / )
IP FW Heaters
HP FW Heaters
1 +7;s
KH(l+~S)
HPSteam Flow, pu
+
+
460
Chapter 11 Table 11.6 Normal Boiler Single VariableControls
Independent Variable
Controlled Variable
Desuperheating spray Firingrate Burnertilt Feedwater flow
Main steam temperature Output(drum)pressure Reheattemperature Drumlevel
its open-loop gain changingand possibly oscillating slowly. How these low-frequency oscillations will affect the overall system behavioris not always clear, but they can hardly be considered to be beneficial. The introduction of the once-through boiler in the late 1950salso focused attentionon boiler control. This type of boiler, because of its thermal design, requires a more sophisticated control. This increasedinterestin boiler controlhas affected later designsfor drum-type boilerstoo, with the result that faster response and more precise controlare being realized. Traditionally, thecontrolsystemfora boilerhas beenaccomplished by usinganalogdevices, which respondto an error in a singlevariable. Any response to such an error will, in most cases, cause errors to appear in other variables. For example, in most boilers,the usual single-variable controlsare those shownin Table 11.6[15].With this type of system,a step changein any of the independent variable references or in load will cause a readjustment of all variables, each respondingin its ownway.Thus,a chainreactionof controlled responses follows the changein one error and may unbalance the systemfor severalminuteswhile all systemsreadjustthemselves. One alternative to this situation is the use of one multivariate controller [15, 16], so that several input variables can actuatea numberof actuators simultaneously, as indicated in Figure 11.31. In this kind of control, the outputs x are related to all inputs m by a matrix G(s) in the equation x(S) = (;(s)mB(s)
(11.39)
Each elementof G(s) may be foundby settingall inputs m to zero except one. The outputx corresponding to this component of m determines one column of the transfer function G. Repeating for other components of m determines G completely. This kind of systemmodel causes cross couplingbetweenvariables, as shown in Figure 11.31. The size of the off-diagonal terms, Gij(S), i =I: i, is an indication of the cross coupling that exists in the system. Such controllers should force the system toward the new steady-state position in a much more optimal manner. However, the design of a multivariable controller requires the use of an accurate model of the
m(s)
Inputs
Fig. 11.31
Outputs
Block diagram of a coupledtwo-variableprocess.
461
Steam Turbine Prime Movers Fuel
-
Throttle Pressure
-
........
~
Main Steam Temperature_ ---. Boiler Reheat Steam Temperature -___ - Turbine Drum Level --. -. System Steam Flow Rate
Air
-
Tilt
Spray
-
-
Feedwater
~
-
.-
-
Excess Air
Turbine Valve_
~
~
Fig. 11.32 A multivariableprocess.
controlled plant and this is not available for many problems. Applying this concept to a steam generator system, we can construct the system model as shown in Figures 11.32 and 11.33.
11.9 Fossil-Fuel Boilers As the technology has evolved, two distinct types of fossil-fueled steam generators have been designed and are widely used; drum-type boilers and once-through boilers. A simple comparison of these two types of boilers is illustrated in Figure 11.34. As suggested by its name, the drum boiler employs a large drum as a reservoir for fluid that is at an evaporation temperature. The once-through (or once-thru, as it is often called) design has no drum and the fluid passing through the system changes state into steam and then into su-
Pressure Fuel Air Tilts
Process
Trottle Temp
Matrix
Reheat Tern
Including
Spray Feedwater
Actuators
ms Ji•• t1° 2 Controller Matrix
U
t1TR /iTT liP
Fig. 11.33 Multivariablecontrol.
SP
Chapter 11
462
, :0,
:t:
,~ c;t:
1~i:I~' s, I
tl :: t :- - I
I I I I I
T
t
-c:llfl1""
:::EE "'f5 .-::,.....
G- -~-I
we
-~~-T
q) FP
t
Drum-Type Boiler
Line Types - - - Water Steam ~ - -- Flue Gas ---;;:...
Once-Thru Boiler
T S E D FP WC
o
Legend Tube Waterwall Sections SuperheaterSection Evaporator Section Drum Feed Pump Water Circulating Pump Steam Output to Turbine
Fig. 11 .34 Drum and once-through boiler configurations. Figuresadapted from similar items in Power Station Engineering and Economy, G. Bernhardt, A. Skrotski, and W. A. Vopat, McGraw-Hili, NewYork, 1960.
perheated steam. The once-through design contains less fluid than the drum-type design and generally has faster transient response. 11.9.1
Drum-type boilers
A simplified sketch of the working fluid path in a drum-type boiler is given in Figure 11.35. In such a system, the drum serves as a reservoir of thermal energy that can supply limited amounts of steam to satisfy sudden increases in demand. It also serves as a storage reservoir to receive energy following a sudden load rejection. Since the fuel firing and pumpingsystems lag behind the drum demand by several seconds, the drum serves as a buffer between the turbine-generator system and the boiler-firing system. It is, however, a very elastic connection as the drum is not an "infinite bus" of thermal energy. Some of the major control systems for the drum-type boiler are the following [16]: (a) Combustion control-fuel and air control (b) Burner and safety control (c) Boiler temperature control-s-bumertilt, gas recirculation (d) Feedwatercontrol (e) Superheatertemperature control-desuperheating (t) Reheat temperature control-gas recirculation
Steam Turbine Prime Movers
Air For Combustion
463
.....,;;......- Air Inlet
Feedwater Healer Feedwater Inlet
Valve Feedwater Pump
Regenerative Fccdwater Heaters
Condensate Pump
Fig. 11.35
A drum-type boiler arrang ement.
Some other control systems are: (a) Feedwater heating system control (b) Air heater temperature control (c) (d) (e) (f)
Fuel oil temperature control (in an oil fired boiler) Turbine lubricating oil temperature control Bearing cooling water temperature control Mill temperature control (in a coal burning boiler)
464
Chapter 11
These controls are usually single-variable control loops. In order to apply advanced control concepts, it is necessary to have an adequate mathematical model of the process . Some valuable work [17-19] has added to our knowledge of boiler behavior as an element in a dynamic system . One boiler representation [20] considers the drum as a lumped storage element as shown in Figure 11.36 (a) and is easily studied by means of an electric analog as shown in Figure 11.36 (b). This simplified model assumes that feedwater effects can be neglected and that the feedwater control satisfies the drum requirements. It also ignores the geometry of the boiler , which is actually a huge distributed parameter system. Still, it should provide at least a rough idea of the system behavior and permit us to study various control arrangements without becoming burdened by system complexity. Such is the approach presented in [20]. A certain mass of steam is stored in the boiler and any change in this mass affects the boiler pressure. Such changes result from transient effects wherein the steam generated and the steam demanded by the turbine are unbalanced. Thus , boiler pressure depends on steam flow. We also recognize that the pressure at the drum is not the same as pressure at the control valves because of the pressure drop across the superheater, which varies as the square of steam flow rate. If we linearize about a quiescent operating point, however, the change in pressure drop is proportional to the change in flow rate and we are justified in using the linearized model of Figure 11.36 (b) Referring to the linear circuit of Figure 11.36 (b), we define the following analogous quantities:
VR T = throttle pressure Vc = drum pressure
I, = steam generated /2 = steam flow to turbine R = friction resistance of the superheater
RT = resistance of the turbine at a given valve opening
Drum Pressure
Steam Generatio n >
(a) Schematic of Boiler-Turbine System
(b) Electric Analog of BoilerPressure Phenomenon Fig. I 1.36 A simplified boiler-turbine representation [20].
(11.40)
Steam Turbine Prime Movers
465
In this model, a change in control valve opening is represented by a change in Rr . We may then write Vc = RI2 + R TI2 Vc o + VCA = R(I2o + 12A) + (RT O + R TA)(I2o + 12A)
(11.41)
and solving for 12A we get 12A ==
R TtiI20 R +R ro
VCA -
(11.42)
and the throttle pressure VTR will experience a drop proportional to RTd , the change in valve opening. The value of R is a function of the quiescent point of operation (the load level). In terms of system quantities,we write the pressure drop from drum to throttle as PD (in lb-mass)or, at constant firing rate: (11.43) where K is the friction coefficient and Q is the steam flow rate in lbm/s. Then, for small perturbations, we can write PD A = (2KQo)QA
(11.44)
where Qo is the steady-state flow rate and Qd is the change in flow rate. In the analog, R = 2KQo
(11.45)
and is a function of Qo as noted. The steam flow to the turbine, Q, is a function of the throttle pressure, P t» and a coefficient K; proportionalto the valve opening, i.e., (11.46) Linearizing, we write
(11.47) where K vo is a function of load level. The steam generated by the boiler is proportional to the heat released in the furnace, but lags behind this heat release by 5 to 7 seconds, as an estimate [20]. If we let Qw be the flow of steam from the boiler, then we can think of the generated steam as being delayed by a time constant Tw, the waterwall time constant. The boiler storage effect is an integration with capacitance (or thermal inertia or time constant) C. This gives the needed relationship between the net unbalance in boiler steam flow to the drum pressure. Finally, the fuel system dynamics can be represented by a delay and dead time. The delay time constant TF is typically about 20 seconds and the dead time Td depends on the type of fuel system, and may be anything from zero to about 30 seconds [20]. All of the above relationships, linearized about a quiescent operating point, may be represented by the lumped parameter model shown in Figure 11.37. To study the control of the boiler dynamics,the system can be arranged as shown in Figure 11.38.With this configuration, it is possible to investigate the nature of the control system and also to optimize the effect of both
466
Chapter 11
.=< ~v
.=
:1' oVj
~ ~
=
=' ~I~
I
I I I I I I
Go) .w rJ)
o~
&
I.F;~
~
e- Td S
IFF6
l+TF s I
Fuel
f-E-& Air
System
I I I I I :>1"<
~
0
~ £e
---
1 l+Tws
t::
=
tl1
\)
v
~
0
0
W6
fi:
e5 v
2
~
OJ:
.£
e
o ='
~
(/J
~ ~
E-4~
+
I I I I I I
>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~
......,
Position
P eal
Impulse Chamber
HP Turb ine
+
ST16
Reheater
Turbine Model
I~
Fig. 11.56 PWR reactor and turbine model [41].
model the turbine bypass system [41], but that option is not pursued here and the total bypass flow is assumed to be a zero input in the reactor model. Several other PWR models have been presented and these are recommended for study [42-46]. Problems 11.1. Verify the results of Example 11.1 by working through each step of the problem and plotting the root locus diagram. Locate the points for which the gain is approximately 937. Repeat for a longer bowl delay using T) = 0.25. 11.2. Examine the stability of the open-loop transfer function of Example 11.1 by performing a Bode plot. What is the gain margin? The phase margin?
Table 11.8 Variable Identification, per Unit
LD = Load demand P T = Throttle pressure Ks = Steam flow pressure drop factor T = Oscillation period, s ( = Oscillation damping factor TJ = Oscillation rate TC, s Tp = Power response TC, s
LR = Turbine load reference PR = Reactor Pressure
M T = Turbine Steam flow M B = Bypass steam flow Ms = Total steam flow Rn = Speed regulation Aw = Speed error
481
Steam Turbine Prime Movers
11.3. Prepare a Nyquist diagram for the system of Example 11.1 and find the gain margin and phasemargin. Compare these results with those of the previous problem. 11.4. Verifythe resultsof Example 11 .2 by working through each step of the problem and plotting the root locus diagram. Locate the points for whichthe gain is about 187. 11.5. Examine a turbine control system similar to that of Example 11.1 except that, instead of the shortbowl delayused in the example, use a long bowl delayof T3 = 0.25 s. Sketch the root locus and find the normal operating point for K3 and Cg as given in Example 11.1 . 11 .6. Find the state-space model for the governor and boiler system shown in the following figure. Initial
Power
K; (1+ I; s) (I + 7;s)(1 + 1;s) Auxiliary Signal
Governor
(I+K2 Ts s)
YI
(1+ 1;s)(1 +
"""
Y2
Tss)
---1
Shat
Power
Ste am System Dynamics
A governor, boiler, and reheat steam turbine system
11.7. Examine the pressure control systems of Figures B.7, B.8, B.9, and B.I0 of Appendix B by root locus, using the values given for the various parameters. References I. McGraw-Hill Encyclopedia ofScience and Technology, 7th Edition, McGraw-Hill, New York, 1992. 2. Skrotzki, A. H. and W. A. Vopat, Power Station Engineering and Economy, McGraw-Hill, New York,1960. 3. Zerban, A. H. and E. P. Nye, Power Plants, International Textbook Co., Scranton, PA, 1964. 4. Potter, Philip 1., Power Plant Theory and Design, Ronald Press, New York, 1959. 5. Power Station web site, for example : http://www.firstgov .gov/, then under "search EIA using" enter "power station" and hit GO. 6. Skrotzki, B. G. A. (Associate Editor), Steam Turbines, a Power Magazine special report, June 1961. 7. Reynolds, R. A., "Recent development of the reheat steam turbine," from "Reheat Turbines and Boilers," American Society of Mechanical Engineers Publication, September 1952, pp. 1-7, reprinted from Mechanical Engineering, January and February, 1952 and the May 1952 Transactions of the ASME. 8. J. Kure-Jensen, "Control of large modem steam turbine-generators, " paper 83T12, General Electric Company, 1983. 9. ASME Power Test Codes, "Overspeed trip systems for steam-turbine generator units," ASME, Power Test Codes 20.2, 1965. 10. Eggenberger, M. A., Introduction to the Basic Elements of Control Systems for Large Steam Turbine Generators, General Electric Company publication GET 3096A, 1967. II . IEEE Report, "Recommended specification for speed governing of steam turbines intended to drive electric generators rated 500 MW and larger," IEEE Publication 600, IEEE, New York, 1959. 12. Evans, W. R., "Graphical analysis of control systems," Trans. AlEE, 67, pp. 547-551 , 1948. 13. Brown, R. G. and 1. W. Nilsson, Introduction to Linear Systems Analysis. Wiley, New York, 1962. 14. Savant, C. J., Jr., Basic Feedba ck Control System Design, McGraw-Hill, New York, 1958. 15. deMello, F. P., "Plant dynamics ofa drum-type boiler system," Trans. IEEE, PAS-82, 1963.
482
Chapter 11
16. Stanton, K. N., "Computer control of power plants," paper presented at the Fourth Winter Institute on Advanced Control, University of Florida, Gainesville, Florida, February 20-24, 1967. 17. Federal Power Commission, National Power Survey, U.S. government Printing Office, Washington, D.C., 1964. 18. Thompson, F. T., "A dynamic model ofa drum-type boiler system," IEEE Trans., PAS-82, 1963. 19. deMello, F. P., "Plant dynamics and control analysis," IEEE Trans., PAS-82, 1963. 20. deMello, F. P. and F. P. Imad, "Boiler pressure control configurations," IEEE paper 31PP67-12, presented at the IEEE Winter Power Meeting, Jan. 29-Feb. 3, 1967, New York. 21. Bachofer, 1. L. C. Jr. and D. R. Whitten, "The application of Direct Energy Balance Control to Unit 2 at Portland Station," paper presented at the 6th National ISA Power Instrumentation Symposium, Philadelphia, PA, May 13-15, 1963. 22. Ahner, D. 1., C. E. Dyer, F. P. deMello, and V. C. Summer, "Analysis and design of controls for a once-through boiler through digital simulation," paper presented at the Ninth Annual Power Instrumentation Symposium, Instrument Society of America, Detroit, Michigan, May 16-18, 1966. 23. Kenny, P. L., "Once-through boiler control," Power Engineering, January 1968 and February 1968. 24. Scutt, E. D., "An integrated combustion control system for once-through boilers," Proc. American Power Conference, XXI, 1959. 25. Adams, 1. D. R. Clar, 1. R. Louis, and 1. P. Spanbauer, "Mathematical modeling of once-through boiler dynamics," IEEE Trans., PAS-84, February 1965. 26. Concordia, C., F. P. deMello, L. Kirchmayer, and R. Schulz, "Effect of prime-mover Response and Governing Characteristics on System Dynamic Performance," Proc. American Power Conference, 28, 1966. 27. Littman, B. and T. S. Chen, "Simulation of Bull-Run Supercritical Generating Unit," IEEE Trans., PAS-85, 7, July 1966. 28. IEEE Working Group on Power Plant Response to Load Changes, "MW response of fossil-fueled steam units," IEEE Trans. on Power Apparatus and Systems, PAS-92, 1973. 29. IEEE Committee Report, "Dynamic models for steam and hydro turbines in power system studies," IEEE Trans. on Power Apparatus & Systems, 92, 6, Nov/Dec. 1973, pp. 1904-1915. 30. Schulz, R. P., A. E. Turner, and D. N. Ewart, "Long Term Power System Dynamics," EPRI Report RP90-7, v. 1, June 1974 and v. 2, Oct. 1974. 31. Morris, R. L. and F. C. Schweppe, "A technique for developing low order models of power plants," IEEE Paper 80SM598-3, presented at the IEEE Power Engineering Society Summer Meeting, Minneapolis, July 13-18, 1980. 32. IEEE Committee Report, "Bibliography of literature on steam turbine-generator control systems," IEEE Trans. on Power Apparatus and Systems, PAS-109, 9, 1983. 33. Kundur, P., R. E. Beaulieu, C. Munro, and P. A. Starbuck, "Steam turbine fast valving: Benefits and teclmical considerations," Canadian Electrical Association, Position Paper ST 267, March 24-26, 1986. 34. IEEE Task Force on Stability Terms and Definitions, "Conventions for block diagram representation," IEEE Trans., PWRS-l, 3, August 1986. 35. Younkins, T. D. et. al., "Fast valving with reheat and straight condensing steam turbines," IEEE Trans. on Power Apparatus and Systems, PWRS 2,2, May 1987. 36. IEEE Committee Report, "Update of bibliography of literature on steam turbine-generator control systems," IEEE Trans. on Energy Conversion, Ee-3, 1988. 37. IEEE Committee Report, "Dynamic models for steam and hydro turbines in power system studies, IEEE Trans., 92,6, Nov.lDec. 1973, pp. 1904-1915. 38. IEEE Committee Report, "Dynamic models for fossil fueled steam units in power system studies," IEEE Trans., PWRS-6, 2, May 1991. 39. Inoue, T., T. Ichikawa, P. Kundur, and P. Hirsch, "Nuclear plant models for medium to long-term power system stability studies," IEEE Paper 94 WM 187-5 PWRS, presented at the IEEE Power Engineering Society Meeting, January 30-February 3, 1994, New York.
Steam Turbine Prime Movers
483
40. Younkins, T. D., "A reduced order dynamic model of a boiling water reactor," paper presented at the IEEE Symposium on Prime Mover Modeling, IEEE Power Engineering Society, Winter Meeting, New York, January 30, 1992. 41. Van de Meulebroeke, F., "Modelling of a PWR unit," paper presented at the IEEE Symposium on Prime Mover Modeling, IEEE Power Engineering Society, Winter Meeting, New York, January 30, 1992. 42. Ichikawa, T., and T. Inoue, "Light water reactor plant modeling for power system dynamic simulation," IEEE Trans. on Power Systems, PWRS-3, May 1988, pp. 463-71. 43. Inoue, T., T. Ichikawa, P. Kundur, and P. Hirsch, "Nuclear plant models for medium- to long-term power system stability studies," IEEE Paper 94 WM 187-5 PWRS, presented at the IEEE Power Engineering Society Meeting, Jan. 30-Feb 2, 1994, New York. 44. Kundur, P. and P. K. Dar, "Modeling of CANDU nuclear power plants for system performance studies," paper presented at the IEEE Symposium on Prime Mover Modeling, IEEE Power Engineering Society, Winter Meeting, New York, January 30, 1992. 45. Culp, A. W., Jr., Principles ofEnergy Conversion, McGraw-Hill, New York, 1979. 46. Schulz, R. P. and A. E. Turner, "Long term power system dynamics, phase II final report," Project EL-367, Electric Power Research Institute, Palo Alto, CA, February 1977. 47. Di Lascio, M. A., R. Moret, and M. Poloujadoff, "Reduction of program size for long-term power system simulation with pressurized water reactor," IEEE Trans. on Power Apparatus and Systems, PAS102,3, March 1983. 48. Kerlin, T. W., E. M. Katz, 1. G. Thakkar, and J. E. Strange, "Theoretical and experimental dynamic analysis of the H. B. Robinson nuclear plant," Nuclear Technology, 30, September 1976.
chapter
12
Hydraulic Turbine Prime Movers
12.1 Introduction The generation of hydroelectric power is accomplished by means of hydraulic turbines that are directly connected to synchronous generators. Four types of turbines or water wheels are in common use. The three most common are the impulse or Pelton turbine, the reaction or Francis turbine, and the propeller or Kaplan turbine. A fourth and more recent development is the Deriaz turbine, which combines some of the best features of the Kaplan and Francis designs. All of these types make use of the energy stored in water that is elevated above the turbine. Water to power the turbines is directed to the turbine blading through a large pipe or penstock and is then discharged into the stream or tailrace below the turbine. The type of turbine used at a given location is based on the site characteristics and on the head or elevation of the stored water above the turbine elevation.
12.2 The Impulse Turbine The impulse or Pelton wheel is generally used in plants with heads higher than 850 feet (260 meters), although some installations have lower heads. One plant, at Bucks Creek in California, has a static head of 2575 feet (785 m) and another in Switzerland has a head of over 5800 feet (about 1800 m). Impulse turbines are often installed on a horizontal shaft with the generator mounted beside the turbine. Some designs have two turbines on a shaft with a generator between them and are called "double-overhung" units. The turbine wheel is spun by directing water from nozzles against the wheel paddles and using the high momentum of the water to drive the wheel. Figure 12.1 shows a double-overhung unit with a single nozzle for each wheel. Occasionally, several nozzles are directed toward each wheel. A stripper, also shown in Figure 12.1, is used to clear water from the bucket as it moves upward, thereby increasing the efficiency of the unit. Speed regulation of the impulse turbine is accomplished by adjusting the flow of water through the nozzle by means of a needle that can be moved back and forth to change the size of the nozzle opening. This arrangement is shown in Figure 12.2 (a) and is seen to be similar to the familiar garden hose nozzle. This needle adjustment is used to make small, steady changes in water flow and power input. However, since the impulse wheel is used in plants having high heads and long penstocks, it is not advisable to use the nozzle to cut off the water jet abruptly. The reason for this is that a sharp cut-off in flow causes a pressure wave to travel back along the penstock causing possible damage due to water hammer. Thus, another means must be found to divert the water stream away from the wheel while the nozzle is closed slowly. One way this is accomplished is by mechanically deflecting the water stream by means of a jet deflector as 484
485
Hydraulic Turbine Prime Movers Water Wheel Housing Buckets
Generator
/-
Nozzle Body Nozzle and Needle
Base Frame and Wheel Pit Liner
Stripper Fig. 12.1
A double-overhung impulse wheel.
shown in Figure 12.2(b). Thus, the governor of an impulse wheel will control the nozzle for normal changes, but must recognize a load rejection by quickly moving the jet deflector. In an impulse turbine, the total drop in pressure of the water occurs at the stationary nozzle and there is no change in pressure as the water strikes the bucket. All of the energy input to the shaft is in the form of kinetic energy of the water, and this energy is transformed into the mechanical work of driving the shaft or is dissipated in fluid friction . Ideally then, the water veloc-
Bushing /
I
/
.Needle Stem
" Governor Connecting Rod
,; Nozzle Tip
"
Flow Straightening Vanes (a) Needle Nozzle with Jet Deflector
(b) Diverting Action of the Jet Deflector Fig. 12.2
Impulse wheel nozzle and deflector arrangements.
486
Chapter 12
ity is reduced to zero after it strikes the turbine buckets. Actually, a small kinetic energy remains and is lost as the deflected water is directed downward to the exit passageway. The power available at the nozzle is given by the formula
p == WHQ hp n 550
(12.1 )
where P; == power availble at the nozzle, hp W == weight of one cubic foot of water == 62.4 lbm/ft' Q == quantity of water, ft 3/s H == static or total head, ft Recall that 550 lbm/s is equal to one horsepower. If 1Jt is the turbine efficiency, the shaft power may be written as
p == HQ1Jt h
8.8
s
P
(12.2)
where the maximum efficiency is usually 80 to 90% [1]. The quantity of water depends on the water velocity, the head, and a nozzle coefficient. It is also restricted by the mean river or stream flow, which is dictated by nature. For a given design, we can compute Q ==AVft3/s
(12.3)
where A == jet area, ft2 V == jet velocity, ft/s Then
V == C
V2iii ft/s
(12.4)
where g == 32.2 ft/s? h == net head at nozzle entrance, ft C == nozzle coefficient, usually = 0.98 If we assume that h =kH for a given situation, where k is a constant, then we may write
Ps == k1H 3/2
(12.5)
12.3 The Reaction Turbine In the impulse turbine, the high pressure in the penstock at the nozzle is changed to momentum so that no pressure drop is experienced at the turbine. In the reaction turbine, however, there is only a partial pressure drop at the nozzle, the remainder taking place in the rotating runner. Thus, water completely fills the cavity occupied by the runner, flows across this pressure drop, and transfers both pressure energy and kinetic energy to the runner blades. Since so much of the turbine blading is active in this energy transfer, the diameter of the reaction turbine is smaller than an impulse turbine of similar rating. Most reaction turbines in use today are of a radial inward-flow type known as the "Francis" turbine after James B. Francis, who designed the first such water wheel in 1846. In these turbine designs, water under pressure enters a spiral case surrounding the moving blades and flows through fixed vanes in a radial inward direction. The water then falls through the runner, exert-
Hydraulic Turbine Prime Movers
487
ing pressure against these movable vanes and causing the runner to tum. The generator is usually directly connected to the runner shaft as shown in Figure 12.3. Reaction turbines are classed as radial flow, axial flow, or mixed flow according to the direction of water flow. In radial flow, the water flows perpendicular to the shaft. In axial flow the stationary vanes direct the water to flow parallel to the shaft . Mixed flow is a combination of radial and axial flow. Reaction turbines are installed either in a horizontal or vertical shaft arrangement, with the vertical turbines being the most common. It is a versatile design, being applicable to installations with heads as high as 800 feet (244 m) and as low as about 20 feet (= 6 m). The control for a reaction turbine is in the form of movable guide vanes called wicket gates through which the water flows before reaching the runner . Positioning these vanes can cause the water to have a tangential velocity component as it enters the runner. For one such position, usually at 80 to 90% of wide open, the runner will operate at maximum efficiency. At any other wicket gate setting, a portion of the energy is lost due to less efficient angling of the water streamline. Although the wicket gates are close-fitting, they usually leak when fully closed and subject to full penstock pressure . Thus, a large butterfly valve is often installed just ahead ofthe turbine case for use as a shut-down valve . The draft tube is an integral and important part of the reaction turbine design. It serves two purposes. It allows the turbine runner to be set above the tailwater level and it reduces the discharge velocity, thereby reducing the kinetic energy losses at discharge. The large tube with the 90° bend just below the runner in Figure 12.3 is the draft tube. The importance of the draft tube is evident when the energy of water leaving the runner is considered. In some designs, this energy may be as high as 50% of the total available energy . Without the draft tube, this kinetic energy would be lost. With the draft tube constructed air-tight, however, a partial vacuum is formed due to the fast-moving water. This low pressure tends to increase the pressure drop across the turbine blading and increase the overall efficiency. One of the important empirical formulas used in waterwheel design is the specific speed formula.
Ns =
NYPs IF/ rpm
(12.6)
4
Tail
Water
Fig. 12.3
A typical vertical shaft reaction turbine arrangement.
Chapter 12
488 Table 12.1
Typical Specific Speeds for Waterwheels
Type of Wheel Impulse Reaction Propeller Deriaz
oto 4.5 10 to 100 80 to 200 10 to 100
max V, 10 150 250
where N = speed in rpm H = head in feet p s = shaft power in hp This quantity is the speed at which a model turbine would operate with a runner designed for one horsepower and at a head of one foot. It serves to classify turbines as to the type applicable for a certain location. As a general guide, then, we say that the specific speeds given in Table 12.1 are applicable. Under this classification, an impulse turbine is a low-speed , low-capacity (in water volume) turbine and the reaction turbine is a high-speed, high-capacity turbine. The same formulas (12.1) to (12.5) used in conjunction with the impulse turbine also apply for the reaction turbine. For (12.4), the value of C is about 0.6 to 0.8 and this value usually decreases for turbines with higher values of Ns. The control of a reaction turbine is through the movable wicket gates. These are deflected simultaneously by rotating a large "shifting ring" to which each gate is attached. The force required to move this assembly is very large and two servomotors are often used to rotate the ring, as shown in Figure 12.4.
Fig. 12.4 Wicket gate operating levers and position servomotors. Figure courtesy F. R. Schleif, Electric Power Branch, Bureau of Reclamation, U.S. Department of the Interior. USBR photo by C. W. Avey.
Hydraulic Turbine Prime Movers
489
The machine shown in Figure 12.4 is one of the generators at the Grand Coulee Dam Powerhouse in Washington State. It shows the wheel pit of a 165,000 horsepower turbine generator. The two rods are connected to power servomotors and operate to rotate the shifting ring, thereby changing the wicket gate position of all gates. A second control device used in reaction turbines is a large bypass valve, which is actuated by the shifting ring. If load is rejected and the wicket gates are dri yen closed very quickly by the governor servomotor, the pressure regulator is caused to open and does so very rapidly. This prevents the large momentum of penstock water from hammering against the closed wicket gates. The pressure regulator then closes slowly to bring the water gradually to rest.
12.4 Propeller-Type Turbines The propeller-type turbine is really a reaction turbine since it uses a combination of water pressure and velocity to drive the shaft. It employs water velocity to a greater extent than the Francis turbine. It also has a higher specific speed, as indicated in Table 12.1. Three types of propeller turbines can be discussed. The fixed blade or Nagler type was developed in 1916 by F. A. Nagler. It operates at a high velocity and operates efficiently only for fixed head and constant flow applications. A few years later, in 1919, Kaplan developed the adjustable blade propeller turbine shown in Figure 12.5. This design has the advantage of fairly high efficiency over a wide range of head and wicket gate settings. Adjustments of wicket gate setting and blade angle can both be made with the unit running. This permits optimization of turbine efficiency over a wide range of head and load conditions. Kaplan turbines are used at locations with heads of 20 to 200 feet (about 15 to 150 m). Compared to the Francis turbines, the Kaplan units operate at higher speeds for a given head and the water velocity through the turbine is greater, leaving the runner with a fast swirling motion. Thus, the draft tube design is important in Kaplan turbine applications.
12.5 The Deriaz Turbine The Deriaz turbine is a more recent development in reaction turbine design and incorporates the best features of the Kaplan and the mixed-flow Francis designs. It is essentially a propeller turbine with adjustable blades. The blades are contoured similar to the Francis blading and are set at 45 degrees to the shaft axis rather than 90 degrees as in the Kaplan turbines. These differences are illustrated in Figure 12.6, where the blades are identified by the letters A and the direction of water flow by the letter W. Wicket gates are generally not used with a Deriaz turbine and control is maintained by blade adjustment only. The Deriaz turbine has the capability of operating at high turbine efficiency over a wide range of loadings, as shown in Figure 12.7. Thus, this design is well suited for situations requiring large variations in loading schedules.
12.6 Conduits, Surge Tanks, and Penstocks It is assumed that any hydroelectric generation site has a supply of elevated water from which water may be drawn to power the turbine. The selection of sites and construction of dams, spillways, and the like are important, but are beyond the scope of this text. Many excellent references are available that discuss these important items [5, 6]. We will assume that a reservoir of water exists and is large enough in capacity that, during periods of interest for control analysis, the head is constant. That is to say, the water source is an infinite bus. From the reservoir, water is drawn from an area called the forebay into a couduit or large pipe, and flows to the turbine as shown in Figure 12.8. In some cases, a relatively level section
490
Chapter 12
Pit Liner Blade
Scromotor Pit Liner
Wicket Gate s
Fig. 12.5 The Kaplan propeller turbine .
of pipe, called the conduit, is necessary to move the water to a point where it begins a steep descent through the penstock to the turbine. As the water flows through this conduit and penstock at a steady rate, a head loss develops , similar to the voltage drop in a nonlinear resistor. The hydraulic gradient in Figure 12.8 represents the approximate profile of the head, measured in feet, as a function of distance from forebay to turbine. Under steady-flow conditions, this head loss at the turbine is
he = H - h = kQ" where hL = head loss, feet H = static head, feet h = effective head at the turbine , feet k = a constant corresponding to pipe resistance
(12.7)
Hydraulic Turbine Prime Movers
491
w
w
(a) The Francis Runner
w
w
(b) The Kaplan Runner
w
(c) The Deriaz Runner Fig. 12.6
Comparison of reaction turbine runners.
Q = flow rate, ft3/S
n = a constant, where 1 :5 n :5 2
Thus, when the flow is steady, the head loss will be directly proportional to the length of pipe, as indicated in the figure. One of the serious problems associated with penstock design and operation is that of water hammer. Water hammer is defined as the change in pressure, above or below normal pressure, caused by sudden changes in the rate of water flow [6]. Thus, following a sudden change in load, the governor will react by opening or closing the wicket gates. This causes a pressure wave to travel along the penstock , possibly subjecting the pipe walls to great stresses. Creager [6] gives a graphic example of this phenomena as shown in Figure 12.9. Suppose the load on the turbine is dropped suddenly. The turbine-governor reacts to this change by quickly moving the wicket gates toward the closed position and, because of the momentum built up by the penstock water, the hydraulic gradient to changes from the normal full load gradient A-C, to the positive water-hammer gradient, A- D. This supernormal pressure is not stable, and once the wicket gate movement stops, gradient A-D swings to A-E and oscillates back and forth until damped by friction to a new steady-state position.
Chapter 12
492 100
-.....;..: .. ._%
~Deriaz ------Impulse /'
80
/ Kaplan
/
Francis .:60
"
~
'
... .'
..... /'
Francis /
/
-:
/
//
N, =50 N, = lOO
/
-:
,,/
/
-----~:.:: ~
'. 7'''-;-: • //
.
/ /
/
Fixed / ' Propeller
40
o
.. .•. •.
- . / -~ :-~
o
20
40
60
80
% of Full Load
100
Fig. 12.7 Turbine efficiency as a function ofload.
A sudden increase in load, accompanied by wicket gate opening has just the opposite effect. Thus, not only must the penstock be well reinforced near the turbine , but it must be able to withstand these shock waves all along its length. Examining this phenomenon more closely, reveals that it is much like the distributed parameter transmission line. The (closing) wicket gate can be thought of as a series of small step changes in gate position. Each step change causes a positive pressure wave to travel up the penstock to the forebay and, upon reaching this "open circuit," it is reflected back as a negative
_ _ Forebay
----- -
r-~_
Static Hydraulic Gradient
~
~-S''''"y_S"" I~'d G . . ~- - ~-~~ ,,",,",
i
1
I
h
Conduit
Wicket Gates Fig. 12.8
A typical conduit and penstock arrangement.
H
Turbine Tailrace
Hydraulic Turbine Prime Movers
493
r_ G{\\die~ - _
_
...,I
--
~ \\te{
D
I
\\\\~e
\,os~:'- - -
I
--
A - - Static Hydraulic Gradient I B -----I-~~-~--~-~~~~~-Sffi~ LOadG~~-1 -_
Forebay
- - __
----
S W'
-----II c la dleut
F=::~\==:::::=-----=-=~·- -!.nR.....0[-A-n- ,----:\----=::::::::::= Conduit
Wick et Gates
-
-l
E
I
Turbine Tailrace
Fig. 12.9 Hydraulic gradient following a loss ofload.
pressure wave of almost the same magnitude. The time of one "round trip" of this wave is called the critical time, p" which is defined as
2L
p, = -
a
seconds
(12 .8)
where L = length of penstock, feet a = pressure wave velocity, ftls For steep pipes, the wave velocity is approximately 4675 a = 1 + (d/IOOe) ft/s
(12 .9)
where d = pipe diameter, inches e = pipe wall thickness, inches Pressure wave velocities of 2000 to 4000 feet per second are not uncommon. The change in head due to water hammer produced by a step change in velocity has been shown to be [6] (12.10) where h6, = change in head, feet V6, = change in velocity, ftls g = acceleration of grav ity, ft/s? and a is the pressure wave velocity as previously defined. Equation 12.10 is the fundamental equation for water hammer studies. Note that to keep water hammer to a low value, V6, must be
Chapter 12
494
kept small either by using a pressure regulator or by introducing intentional time lag in the governor. The introduction of time lags are particularly troublesome for interconnected operation as this contributes to tie-line oscillation [7]. Usually, the time for closure ofthe wicket gates of a hydraulic turbine is much greater than JL of equation (12.8). Suppose, however, that the gate is opened by only a small amount, such that it can be closed in a time JL. In such a case, the pressure rise can be greater than that due to closure from full gate to zero. For this reason, JL is usually considered the critical governor time. From the above, we see that water hammer, both positive and negative, can be a serious problem in penstock design. It may require that penstocks be built with much greater strength than would ordinarily be necessary. It may also cause violent pressure oscillations, which can interfere with turbine operation. The pressure regulator is helpful in controlling positive water hammer as it provides relief for the pressure buildup due to closing of the gates. However, it is of no help in combating negative water hammer. A device often used to relieve the problems of both positive and'negative water hammer is the surge tank, a large tank usually located between the conduit and penstock, as shown in Figure 12.10. To be most effective, the surge tank should be as close to the turbine as possible but, since it must also be high enough to withstand positive water hammer gradients without overflowing, it is often placed at the top of the steep-descent portion of the penstock, as shown in the figure. Sometimes an "equalizing reservoir" is constructed to serve as a surge tank for large installations and may actually be cheaper and more beneficial. This is due to the general rule that the larger the tank area, the smaller the pressure variation [6]. Surge tank dimensions are important. The tank must be high enough so that in no case is air drawn into the penstock. Letting y denote the maximum surge up or down in feet (measured from the reservoir level for starting, from a distance below this equal to the friction head for stopping) we have [5]
)1/2 aLvi y= ( - - +P gA
(12.11)
where a = conduit area, ft2 L = conduit length, ft
A Forcbay
Surge Tank
- ---
--:::..---- - - - -
.... ===-- --- -Conduit
Turbine Tailrace Fig. 12.10
Condu it and penstock with a surge tank.
495
Hydraul ic Turbine Prime Movers == velocity change , ft/s g == 32.2 ft/s 2 F == friction head, ft A == area of surge tank, ft2
V.i
Barrow [5] also gives a formula for the time interval that elapses between turbine load change and the occurrence of the maximum surge as
t ==
7T
2
22 (LA + c v A2 ) 1/2 ga
(12.12)
a2
where c == coefficient of friction cv 2 = q = flow in ft3/S The factor F in (12.10) is important since it represents the friction that eventually damps out oscillations following a sudden change . Since damping is desirable, it is sometimes advantageous to add hydraulic resistance at the surge tank opening to produce a choking effect. This is done in two ways: by placing a restricted orifice between the tank and the penstock, or by constructing a "differential surge tank ." The differential surge tank, shown in Figure 12.11, consists of two concentric tanks : an inside riser tank of about the same diameter as the penstock and an outer or surge tank of larger diameter with a restricted passage connecting it to the penstock. Because of this restriction, the water level in the outer tank is independent of the accelerating head and the head acting on the turbine. These heads are determined by water in the riser tank, which acts like a simpler surge tank with small diameter. The diameter of the differential surge tank is about one-half that of a simple surge tank . The riser diameter is usually the same as that of the penstock. The damping effect due to the added friction of the differential surge tank is shown in Figure 12.12, where the surge is compared for two types of tank design [6]. Note the relatively long period (about 300 seconds , or five minutes) of the surge. This surge would be due to a sudden increase in load, where the turbine wicket gates are opened at time t = O. Note that an accelerating head is created , which increases steadily for about 80 to 85 seconds , at which time the flow
Riser Tank
A
Foil
Forebay
I I
Lo~ amd;': 1~
.!~;~~~~~ll
I
Conduit
II H
I
Wicke t Gates Fig. 12.11
The differential surge tank.
Turbine Tailrace
Chapter 12
496
o Differential :
5
10
Final Le vel
20
25
o
50
100
150
200
250
300
350
Time in seco nds Fig. 12.12
Comparison of surges in simple and differential surge tanks.
of water from that tank ceases. In the differential tank, the accelerating head is established very fast, but not so fast as to prevent the governor from keeping up with the change . In the discussion of a technical paper [8], deMello suggests a lumped parameter electric analog of the hydraulic system, including conduit, surge tank, penstock, and turbine [9]. Figure 12.13 shows this analog, where head is analogous to voltage, volumetric flow is analogous to current, and the turbine is represented by the variable conductance, G. With water being considered incompressible, the inertia of water in the penstock and conduit are represented by inductances L) and L 2 , respectively (series resistance could be added to represent hydraulic resistance) . If the effect of water wheel speed on flow is neglected, the turbine can be simulated by G or Gt:., where a change in gate setting is under consideration. The surge tank behaves much like a capacitor as it tends to store water (charge) and release it when the head (voltage) at the turbine falls. (How could a differential surge tank be represented?)
Conduit
i;
v
Penstock L1
Sur ge Tank
Fig. 12.13 Electric analog of the hydraulic system.
Hydraulic Turbine Prime Movers
497
If linearized equations about a quiescent operating point are written we have, for the head at the reservoir described in the s domain,
_ . (S(L
Vd - -lid
1
+ L 2)(1 + LCs2) S 1+L
)
2C2
(12.13)
where
Also
. _ i 10 ( Vl~ ) G~. + -110 2 VIO Go
'lA- -
(12.14)
From the square root relationship between flow and head
Q = aVii
(12.15)
we write (12.16) Combining, we get
.
'tA
=
2(G A/GO) VIO 2vo s(L 1 + L2)(1 + LCs2) - + - - - - -2- 1 + L2C2S io
(12.17)
Now, assume a change in turbine power at constant efficiency or
(12.18)
When the surge tank is very large, C is large and (12.18) reduces to the so-called waterhammer formula
(12.19)
where VIO
Ro = -.10
(12.20)
Then (12.19) may be written as
Ga
Po-(l- TwS) Go
(12.21)
Chapter 12
498
where [9]
Tw == water starting time == 1 second
(12.22)
Furthermore, as pointed out by deMello [9], when the tunnel inertia is great, or L2 is large, then (12.19) becomes
p/).==
1 LIC -+CS+_-S2 2Ro n;
(12.23)
These results are not greatly changed by consideringthe conduit and penstock as a distributed parameter system.
12.7 Hydraulic System Equations The hydraulic system and water turbine transfer functions have been thoroughly analyzed by Oldenburger and Donelson [8]. This excellent description is based on a rigorous mathematical analysis and is supported by substantial experimental evidence to testify to its validity. As shown in the previous section, the flow of water through a conduit is analogous to an electric transmissionline in which head is analogousto voltage and volumetric flow rate is analogous to current. This is easily seen when the partial differential(wave) equations for a uniform pipe with negligible friction are examined. For the uniform pipe, we write
au
ah
au
ah
-==-0-
ax
at
-=-g-
at
ax
(12.24)
where u == water velocity, fils x == distance along pipe, ft h == head, ft a
= a constant = P~~ +; )
p = density of fluid
g = accelerationof gravity K == bulk modulus of elasticityof fluid
r ~ internal pipe radius
f == pipe wall thickness
E.= Young's modulus for the pipe Equation (12.24) should be compared to the equations of the transmission line, which can be written as follows:
av
ai
- - =L- +Ri
ax
ai
at
av
--=cax at +Gv The similarity for the lossless case should be obvious.
(12.25)
Hydraulic Turbine Prime Movers
499
Now, let us define the following: H = H(s, x) = L[h(!, x)] U
= U(s, x) = L[u(t, x)]
(12.26)
We may write the Laplace transform of(12.24) with the result, assuming zero initial conditions,
au ==-asH
-
ax
on ==--su 1
-
ax
(12.27)
g
The solution of (12.27) may be shown to be
U = Kle-sxla + K2e+sxla H
= K3e-sxla + K4e+sxla
(12.28)
This result can be written in hyperbolic form as sx sx U= C 1 cosh - + C2 sinha a sx H = C3 cosh a
sx
+ C4 sinh -
(12.29)
a = vg;a = wave velocity
(12.30)
a
where
These results may be simplified by eliminating of the arbitrary constants subscripted by 3 and 4. With this simplification, we have [8]
(12.31) or sx sx U == C 1 cosh - - C2 sinh a a
H == -
sx
C1
C2
sx
•
r - cosh - - - - SInh vag a ~ a
~
(12.32)
Note we may apply (12.31) or (12.32) to any cross section of pipe such as I or II of Figure 12.14, or any arbitrary cross section i. Thus, in (12.31) and (12.32) we may subscript all x's with a numeral (I, II, or i) to indicate the particular section under study. This helps in evaluating the constants C t , Cb K 1, and K 2 as they depend on boundary conditions. For example, we may write
s
S
C 1 == U/cosh -X/+ ~sinh -XI a a
(12.33)
Chapter 12
500
Fig.12.14
A view of an arbitrary pipe section selected for study.
We may then write (12.32) as, for the section at 11,
s a
s a
s a
s a
- vagHI sinh -Xu cosh -Xl - U, sinh -Xl sinh -Xu
(12.34)
Now, let XI=O
Xu = L = length of pipe
(12.35)
Then, (12.33) and (12.34) become
C1 = U, C2=-vagHI
(12.36)
and Ul/ = U, cosh T~ - agll, sinh T~
Ul . Hu = - sinh T~ ag
+ HI cosh T~
(12.37)
where
L
Te = - = elastic time
(12.38)
q=AU
(12.39)
a
Now, since where q = volumemetric flow rate, tt31s A = pipe cross sectional area, t12
(12.40)
then we may write
Q(s, X) = AU(s, X)
(12.41)
Q=AU
(12.42)
or, simply and this applies at any section such as I or 11. Thus, we convert the U equation to a Q equation and rewrite (12.3 7) as
501
Hydrau1ic Turbine Prime Movers 1
QIl == Q/ cosh T.,s - Zo sinh T.,s HII = -ZoQI sinh TeS + HI cosh T.s
(12.43)
where Zo=
1 . . r - = the "characteristic" impedance
Avag
(12.44)
From the time-domain translation theorem of Laplace transform theory we write
e-bsp(s) = L[u(t - b)f{t - b)]
(12.45)
We readily conclude that the Laplace transform of the following differential equation may be written: for T, > write
°
°
L[(sinh TeP)f(t)] = F(s) sinh Tes
andf(t) = when t < T, and where we use the notationp
°
(12.46)
== dldt. Similarly, we also
L[(cosh TeP)f(t)] = F(s) cosh Tes
(12.47)
for f(/) = when t < Teo From these relations, we conclude that the second item in (12.43) is the Laplace transform for f(/) when t < T; We can see that (12.43) is the Laplace transform of the equations
1 qll = (cosh TeP)q/- Zo (sinh TeP)h/
hlI = -Zo(sinh TeP)qI + (cosh TeP)h I
(12.48)
where
qlO, t) = hlO, t) == 0 for t > T, Now note that (12.46) can be rearranged and hyperbolic identities used to write
1 Q/ = Qll cosh T,» + Zo H ll sinh Tes HI = ZOQIl sinh Tes + HIl cosh Tes
(12.49)
and in the time domain this equation pair becomes
1 q/ == (cosh TeP)qll + Zo (sinh TeP)hll hI = Zo(sinh TeP)qu + (cosh TeP)h II where
q/,(L, I) = hIlL, t) =
°
for t <
(12.50)
T;
Now, we rearrange (12.49) and subsequently (12.50) to write the hybrid equation pair
1 q/ = (cosh TeP)qll + Zo (sinh TeP)hIl (12.51 )
Chapter 12
502
Equations (12.51) may be evaluated by expanding the hyperbolic differential operators in an infinite series. We recall that (12.52) and if this series converges rapidly, we may write approximately
e-TePf(t) == (1 - TeP)f(t)
(12.53)
or, if more accuracy is require, we may add more terms. In a similar way, we may expand the hyperbolic terms by the expansions U
cosh U =: 1 + -
2
2!
u3
sinh u = u + -
4!
US
3!
If these sequences in u (12.51)
u4 + - + .. ·
+ - + . ·. 5!
= TeP converge rapidly, we may write for the first of equations
qj
1
.
= qIl + Zo TehIl
(12.54)
We also note that equations (12.51) are linear in both q and h such that, if we define
q[=qO+qd h/=ho+h d
(12.55)
and write new equations in terms of the d-quantities, the new equations will be identically the
same as (12.51).
The head loss due to friction has been shown to be proportional to q2. Thus, the head equation is, from (12.51) and including a friction-loss term
hJl== (sech TeP)h,-Zo(tanh Tep)qJl-k~qh
(12.56)
This nonlinearity is removed by the approximation (12.55), or
hJ[d = (sech TeP)h 1d - Zo(tanh TeP)ql/d - k2ql/d
(12.57)
k2 = 2k~q 110
(12.58)
where
We may also write (12.51) and (12.57) in per-unit terms by dividing through by a base quantity. Let Base q
=:
qo
Base h = ho Then, in per-unit terms, (12.51) becomes
1 qJ == (cosh TeP)qJ[ + -(sinh TeP)hl/ Zo
hlJ = (sech TeP)h I
-
Zn(tanh TeP)qJl
(12.59)
Hydraulic Turbine Prime Movers
503
where we define
hI per unit hI == h o .
per unit hll ==
h/I h;;
. qI per unit qI = -
qo
. q/I per unit qII == -
qo
. Zoqo per unit Zj v Z;> ~
(12.60)
We need not use any special symbol to indicate whether these are per-unit or system quantities as the equations are identical (except for Zo and Zn)' In what follows, we will assume: 1. All flows and heads are deviations from the steady state, but we will avoid using the subscript for brevity.
t1
2. All values are per unit.
12.8 Hydraulic System Transfer Function We now apply the equations of Section 12.7 to typical hypothetical situations and derive transfer functions for the hydraulic system. In so doing, we are interested in dynamic oscillations about some quiescent operating point. Partial derivatives of nonlinear relationships are assumed to be derived at the quiescent point or Q-point. The results of this section and the assumptions made have been verified for at least one physical case as recorded in [8]. Verification was checked by the frequency-response method [8, 10], wherein the wicket gates are oscillated at a range of frequencies and measurements taken to determine the system Bode diagram. We will not dwell on this technique except to acknowledge that experimental verification has been checked by others. It has been observed in physical situations that when the wicket gates are oscillated at low frequencies, the levels in the riser tank and surge tank are practically the same. Also, when the frequency of oscillation is high, the levels in both tanks are practically constant as the water inertia prevents it from responding to rapid changes. Thus, we assume that the levels in riser and surge tanks are identical, or (12.61) where h, == surge tank head, per unit h, = riser tank head, per unit Experimental runs verify this assumption [8]. From (12.57) applied to the conduit (from forebay to surge tank) we have
h, == (sech Teep)h w - Ze(tanh Teep)qe - cPeqe where Tee == elastic time for the conduit h w == forebay head, per unit
(12.62)
504
Chapter 12
h;- = norma I'tzed conduiU1t impedance
Zc = Zocqo
qc = conduit flow rate near surge tank, per unit cPc = friction coefficient for conduit If we assume that the reservoir is large, we may write
hw=O
(12.63)
since there will be no change in head at the forebay. We now observe that, from Figure 12.15, that the per-unit flow rate at the surge-tank end of the conduit is
qe = qt + qr + qp
(12.64)
We can further describe the flow into the two tanks by the differential equation
T/lt=qt+qr
(12.65)
where T, = surge tank riser time. Combining (12.62) and (12.64) and taking the Laplace transform with zero initial conditions, we have (12.66) where
F 1(s) =
cPe + Z, tanh TecS 1 + cPeTrs + ZeTtS tanh TecS
(12.67)
---'-''------''----=.:...--
This equation is especially interesting since it indicates that the relationship between surge tank head, ht, and penstock flow rate, qp, depends only on the conduit and surge-riser tank characteristics and not on the characteristics of any component following the surge tank. In other words , the hydraulic system up to the penstock is completely described by (12.66) .
A Forebay
-- - - -
>=====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
0.
.5 ....
0.4
0
0..
0.2
o
o
0. 1
0.2
0.3
0.4
0.5
Time in seconds Fig. 13.3
CT response to a step change in setpoint from no load to rated load [3).
Combustion Turbine and Combined-Cycle Power Plants
517
13.2.2 Off-nominal frequency and voltage eHects The power supply for the governor system is usually provided by the station battery that can provide power for at least 20 minutes and is, therefore, unaffected by the voltage and frequency of the ac power system [3]. The shaft-driven main fuel and lubrication oil systems can be consideredas unaffectedby ac system voltage deviations. If the power demand exceeds the power limit, the combustionturbine power output capability decreases as the frequency drops. A basic characteristic of the combustionturbine is that the air flow decreases with shaft speed and the fuel flow must also be decreasedto maintainthe firing temperature limit. The amount of the air flow decreaseis on the order of 2% in output capability for each 1% drop in frequency. This is shown in equation (13.2), which represents the limitingmultiplieron power demandwhen the unit is running on an exhaust temperature limitation. RPFE= I-B}(DPF)(wBP- W Sy s )
= Reducedpower frequency effect multiplier
(13.2)
where
0 when power demand < power limit B I = { 1 when power demand> power limit
DPF = per-unit change in unit output per-unit change in frequency
= 0 if data not available, bypasses the multipliereffect wsys = system frequency WBP = system frequency when unit exceeds its power limit
The RPFE is one of the possible limiting effects noted by the limitationblock on the righthand side of Figure 13.2The invocationof this limitation depends on the initial power level of the generating unit and the change in frequency during the transient. For example, if the frequency declines 3 Hz or 5% on a 60 Hz system, then the power capabilityof the unit will be reduced by 2% for each 1% reduction in speed after the power limit is exceeded. A unit operating initially at full load would reach the power limit immediately and the output of the unit would be decreasedby 10%. Off-nominal voltage and frequency both have an effect on the system auxiliaries, such as the fuel system, heaters, and air handling equipment. These effects vary depending on the unit design,the particularinstallation limitations, the utility practice, and the site variables. This represents another limiting function that is referred to in the literature as the auxiliary equipment voltage effect, or AEVE [2]: AEVE:= 1 - max[DPV(VBP - Vr), 0]
(13.3)
where
DPV:= per-unit change in unit output per unit change in voltage VBP = voltage level above which there is no reduction in unit output Vt == generatorterminalvoltage
[2]
Anotherunit limitationis based on a reductionin system frequency. This limit in definedas AEFE == Auxiliary equipmentfrequency effect == I-max[DPA(wBP- wsys ) , 0]
(13.4)
where DPA is the per-unit change in unit output due to a per-unit change in frequency from the base point frequency WBp.
Chapter 13
518 Aeo, pu
ro, pu
R2
----r--+-------t -- --1 -------L----l- ~-t>
- - - 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<2 Tf = TCD + -W'W where
K2 =
=
X-I] T; [ 1 + - +W-j(2 TIc
(13.9)
W
TdT = per-unit combustor temperature rise fO
TCD = compressor discharge temperature per unit of absolute firing temperature Wr= design air flow per unit The gas turbine exhaust temperature TE is determined by equation (13.6), substituting TE for TR and using (13.7) for the computation of X. The mechanical power PM G is a function of the turbine inlet temperature and the flow rate of combustion products Wa + Wf
K
K+ - ' •
Fig. 13.11
1+
I;s
Gas turbine mechanical power and exhaust temperature model [6].
T,'E
Combustion Turbine and Combined-Cycle Power Plants
525
13.3.3 The steam turbine power generation The heat recovery steam generator (HRSG) system responds to changes in the exhaust flow from the gas turbine W and its exhaust temperature Te. This heat is delivered to the high- and low-pressure steam generators, which can be approximated. The exhaust gas and steam absorption temperatures through the HRSG are indicated in Figure 13.12. The transient heat flux to the high- and low-pressure steam generation sections can be approximated using the relations for constant gas side effectiveness , and are computed as follows [6]. 7]g l =
7]g2 =
Tex - T'
(13.10)
t; - r: T' - T"
(13.11)
T' - T m2
where T' and T" are the gas pinch points shown in Figure 13.12. Temperatures Tml and Tm2 are the average metal temperatures in the HP and IP evaporators, respectively. The gas heat absorption by the HRSG section can be computed as follows [6]. (13.12) (13.13) where (13.14) (13.15)
(econ2 = ~ w + 7JeciT" - Ti n)
and where Qeconl, Qecon2, and Q 'econl are the HP and IP economizer heat fluxes.
HP Superheater
HP Evaporator HP Economizer
Qed
IP Evaporator
IP Evaporator
STEAM ENERGY
Qec2
Heat Absorption, % Fig. 13. 12 Steam energy exhaust gas tempe rature versus heat absorption (6).
100
Chapter 13
526
The economizer heat absorption is approximated using the constant effectiveness expressions, as follows [6]: Q econ2 = T/eciT" - Tf W)mLP Qeconl = T/ecl(T' - tecon2)mllP Q;conl = T/eciT" - Trw)mJIP tecon2 = tfw + T/ec2(T" - Tf n)
(13.16)
Then equations (13.11) through (13.17) are solved to find the temperature and heat flux profiles . The steam flows, mHP and mLP are computed by the pressure/flow relationship at the throttle and admission points as follows: mHP=KrPHP mlfp
+ m/p =
(13.17)
K 'P/p
where K r = throttle valve flow coefficient K' = admission point flow coefficient Steam pressures P HP and PLP are found by integrating the transient energy equations, which are given as DllP?HP = QgHP - hhpmllP + hjWmHP + hjWmllPjW DLP?LP = Q gl.P - hLPmLP + hjWmLPjW
(13.18)
The HP and LP metal temperatures Tm l and Tm2 are determined by integration of the gas and steam side heat flux as shown in Figure 13.13. The steam turbine power in kilowatts is computed as
kWg -
mIlP . AEIlP + mLP . AELP
3413
(13.19)
HP Evap . Metal Temp
T.III
Exhaust Flow
IP Evap. Meta l Temp
IP Admiss ion Pressure Fig. 13.13 Steam system model.
Combustion Turbine and Combined-Cycle Power Plants
1
(1 + sTM Xl + st; ) Fig.13.14
527
PMS
A simplified steam power response model [6].
where AE1f P and AELP are the steam actual available energies [6]. The dynamic relations for the HRSG and steam turbine are shown in Figure 13.13. Note that the heat transferred from the high pressure boiler QGI is a function of the exhaust gas temperature TE , the HP evaporator metal temperature TMI' and the IP evaporator metal temperature TM2' It is noted in reference [6] that the total contribution to mechanical power from the two pressure boilers can be approximated with a simple two-time constant model. The gain between the gas turbine exhaust energy and the steam turbine output will, in general, be a nonlinear function that can be derived from steady-state measurements through the load range, or from design heat balance calculations for rated and partial load conditions. These simplifications will result in a low-order model as shown in Figure 13.14 [6]. Such a low-order model would be very simple to implement in a computer simulation, and may be quite satisfactory for may types of studies, especially studies in which the major disturbance of interest is far removed from the combined cycle power plant. Moreover, this simple model could be "tuned" by comparing it against the more detailed model of Figure 13.13. The detailed model should be considered for studies of disturbances in the vicinity of the combined-cycle plant. From [6] the values of the time constants for this simrlified model are given as
TM = 5 S
TB = 20 s
Problems 13.1
The combustion turbine presented in Figure 13.1 is a single-shaft design. Other combustion turbines are designed to employ two different shafts. Sketch how such a two-shaft unit might be configured and compare with the single-shaft design. What are the advantages of a two-shaft design? Hint: Consult the references at the end of the chapter, if needed. 13.2 The single-shaft combustion turbine shown in Figure 13.1 is called a "direct open cycle" design since it exhausts its hot exhaust to the atmosphere. A different design is called a "closed-cycle" system, which recycles the exhaust back to the air input port. Make a sketch of how such a closed-cycle system might be configured. 13.3 It has been noted that the ideal cycle for the gas turbine is the Brayton cycle. Explore this cycle using appropriate references on thermodynamic cycles and sketch both the P-V and the T-S diagrams for this cycle.
References 1. EI-Wakil, M. M., Powerplant Technology, McGraw-Hill, New York, New York, 1984. 2. Turner, A. E. and R. P. Schulz, Long Term Power System Dynamics, Research Project 764-2, User's Guide to the LOTDYS Program, Final Report, Electric Power Research Institute, Palo Alto, CA, April 1978.
528
Chapter 13
3. Bailie, R. C., Energy Conversion Engineering, Addison-Wesley, Reading, MA, 1978. 4. Pier, J. B. and S. Bednarski, "A simplified single shaft gas turbine model for use in transient system analysis," General Electric Company Report, 72-EU-2099, 1972. 5. Schulz, R. P., A. E. Turner, and D. N. Ewart, Long Term Power System Dynamics, volume 1, Summary and Technical Report, EPRI Report 90-7-0 Final Report, June 1974. 6. IEEE Working Group on Prime Mover and Energy Supply Models for System Dynamic Performance, F. P. deMello, Chairman, "Dynamic models for combined cycle plants in power system studies," IEEE Transactions Power Systems, 9, 3, August 1994, p. 1698. 7. Rowen, W. I., "Simplified mathematical representations of heavy-duty gas turbines," Trans. ASME, 105 (1),1983, Journal ofEngineeringfor Power, Series A, October 1983, pp. 865-869.
A
appendix
Trigonometric Identities for Three-Phase Systems In solving problems involving three-phase systems, the engineer encounters a large number of trigonometric functions involving the angles ± 120 Some of these are listed here to save the time and effort of computing these same quantities over and over. Although the symbol (0) has been omitted from angles ± 1200, it is always implied. 0
•
sin (fJ ± 120) = -1/2 sin fJ ± v1/2 cos fJ
(A.I)
cos(fJ ± (20) = -1/2cosfJ
(A.2)
1=
sin? (fJ ± 120) = 1/4 sin? fJ + 3/4 cos? fJ 1/2 + 1/4cos2fJ cos2(fJ ± 120)
1=
v'J/2 sin fJ =f=
V]/2 sin fJ cos fJ
(A.3)
v1/4sin2fJ
=
1/4 cos 2fJ + 3/4 sin 2fJ ± VJ/2 sin (J cos (J
=
1/2 - 1/4cos2fJ ± VJ/4sin2fJ
sin fJsin (fJ ± 120)
=
(A.4)
-1/2sin 2fJ ± VJ/2 sin fJcosfJ 1/4 + 1/4 cos 2fJ ± v'J/4 sin 2fJ
(A.5)
cos Ocos(fJ ± 120) = -1/2cos 2fJ 1= -YJ/2 sin fJcosfJ = -1/4 - 1/4 cos 2(J =f= VJ/4 sin 2fJ
(A.6)
= -
sinOcos«(J ± 120)
=
-1/2sin(Jcos(J =F V]/2sin -1/4sin2fJ ± -YJ/4cos20 =F v3/4 2(J
(A.7)
cos (J sin (fJ ± 120) = - 1/2 sin {} cos fJ ± VJj2 cos? fJ (A.8)
-1/4 sin 2fJ ± VJ/4 cos 20 ± VJ/4 sin(O + 120)cos(O + 120)
=
-1/2sinOcosO - -YJ/4cos 20 + VJj4sin 2(J
= -1/4sin2{}- VJ/4cos28
(A.9)
sin(8 + 120)cos«(J - 120)
=
sinfJcos(J - VJ/4
=
1/2sin20 -
v3/4
(A.IO)
sin(fJ - 120)cos(fJ + 120)
=
sin {}cosfJ + VJ/4
=
1/2sin28 + -YJj4
(A.II)
sin (fJ - 120) cos({} - 120)
1/2 sin () cos {} + 0/4 cos? 0 - V]j4 sin? 0
=
-
=
-1/4sin2fJ + 0/4 cos 2fJ
sin({} + 120)sin({} - 120) = 1/4sin 20 - 3/4cos 20 ~ -1/4 - 1/2cos2{}
(A.12) (A.13) 529
Appendix A
530
cos (0 + 120) cos (0 - 120)
=
1/4 cos? (J
sin (20 ± 120)
=
cos (28 ± 120)
= -
-
-
3/4 sin? (J
=
-1/4
+ 1/2 cos 20
1/2 sin 20 ± vJ/2 cos 2fJ
(A.IS)
vJ/2 sin 20 + 120) = 0
(A.16)
1/2 cos 28
=F
sinO + sin(O - 120) + sin(O
(A.I?)
cosO + cos(O - 120) + cos(O + 120) = 0 sin 2 tJ
+ sin 2 (0
cos tJ + cos 2
2(O
(A.14)
(A.18)
+ sin 2(O + 120) = 3/2 - 120) + cos 2(O + 120) = 3/2
- 120)
sin 8 cos 0 + sin (8 - 120)cos (0 - 120) + sin (0 + 120)cos (0 + 120)
(A.19)
(A.20) =
0
(A.21)
In addition to the above, the following commonly used identities are often required: sin 2 (J + cos 2 0 = 1 sin (J cos (} = 1/2 sin 2fJ 28 cos - sin 20 = cos 20
cos 28 = (I + cos20)/2 sin 2 8 = (I - cos 28)/2
appendixJ
Hydraulic Servomotors
The hydraulic servomotor, such as the mechanical integrator described in Appendix A, is a class of control devices that are used to move large loads with precision and speed. The newer designs incorporate electromechanical elements to improve the speed and accuracy. These devices have two main mechanical components: a control valve and a piston. The purpose of this appendix is to write the basic equations that describe the behavior of these two components and of the servomotor system.
H.l Control Valve Flow Equations The control valve or spool valve is usually described in terms of the number of spools or lands and the number of ways the hydraulic fluid can enter or leave the valve. All valves require at least a supply line, a return line, and a line to the load-a three-way configuration. Many valves, such as the valve shown in Figure J.l, are four-way valves. All are analyzed in a similar way. Our analysis follows closely that of Merritt [1], which is recommended for further study. Consider a three-land, four-way spool valve shown in Figure J.l. This valve is described by four sets of equations that describe the flow and pressure relationships. The flow past the spool orifices are given by Bernoulli's equation*
QI
=C~IJ;(PS-PI)
Qz = c~zJ ;(PS - Pz)
Q3=C~3J;PZ
Q4 = C~4J ;P
1
(J.l)
where Q= volumetric flow rate, ft3/S Cd = dimensionless discharge coefficient A = orifice area, ft2 *Dimensions of all quantities are given in a consistent set of units, often using the ft-Ibm-s system. Actual devices might be analyzed using different dimensions for convenience, e.g., using A in square inches or metric units.
640
Hydraulic Servomotors
Q,
Suppy
s
i; Rerun
J~ ~
U
-
641
L1
i;
Q,
Fig. J.I
A three-land, four-way spool valve [I].
and
P = pressure, Ibf/ftz
p = mass density of fluid , lbm/ft? or Ibf-s z/ft4 The flow to the load can be written as (1.2)
and these relationships are readily verified by examining the Wheatstone bridge equivalent of the spool valve in Figure 1.1. The orifice area in each case is a function of the displacement x. Thus, we can write
AI = A )(x) A z = Az(-x) A) = A)(x)
A 4 =A 4 (- x)
(J.3)
Finally, we note that the pressure drop across the load is given by
(J.4) These four equations, 1.1-1.4, with appropriate simplifications, must be solved simultaneously to give QL as a function of x and Pu i.e., QL = QL(X, PL). The first simplification is to assume matched symmetrical valve orifices: Matched:
A) = A) A z = A4
Symmetrical:
(J.5)
A )(x) = Az(- x) Aix) = A 4(-x)
(1.6)
We also define the neutral position area (1.7)
642
Appendix J
Usually, we assume that orifice area varies linearly with valve stroke so that only one defining equation is required, i.e.,
A =wx
(1.8)
where w is the width of the slot in the valve sleeve in ft 2/ft (or in2/in). Now, for matched symmetrical valves
Ql =Q3 Q2= Q4
(1.9)
From the first equality, and using (J.5), we write
CeYllJ f(ps-p == CeYllJ fp 2 t)
or
PS=P1+P2
(1.10)
Combining (J.I0) with (J.4), we compute
PS+PL Pl=--2(1.11 ) These relationships are shown graphically on a pressure scale in Figure J.2. From (J.2) we also compute
QL = QI-Q4
=
C,p4tJ;(Ps-P C,p42J;P JPrP JPS+PL 1) -
=C~l
L
P
-C~2
1
P
Drop Across 1
Ps/2 ~:----Ir------
f} =0
(Drain) Fig. 1.2
Graphical illustration of pressure division for matched symmetric orifices.
(J.12)
643
Hydraulic Servomotors
Also, from Figure J.l,
(J.13)
(J.14) For a symmetrical valve, we can write
(J.15) Thus, for any x we can write (J.16) Now, our goal is to determine a linear equation for QL' We can use a Taylor's series expansion to write (1.17) Thus (J.18) where
QL
a- ] Kq == the flow gain == ax
0
QL K c == the flow-pressure coefficient == - a -- ] aPL 0
(J.19)
Equation (J.18) is the desired relationship and will be used in evaluating the small-signal behavior of the system. There are obvious limitations that should be kept in mind, however, as equation (1.16) is obviously not linear, even though much of the operating range is reasonably linear.
J.2 Control Valve Force Equations The equations giving the forces acting on the spool valve are developed for either a steadystate or a transient condition. Consider the spool valve shown in Figure 1.3, where the spool is displaced a small amount in the +x direction. Continuity requires that
QI = Qz = Cc00
J
;(PI - P z) = CcCvAo
J
;(P 1 - P z)
(J.20)
644
Appendix J
P2
t
x
Fluid Element
QI
Vena Contracta
PI
F I < .._-_.__ ... Face a
-,
- - --
F2
-
1<
L
Face b
---'-F,
- ----
Fig. J.3 Flow forces on a spool valve due to flow leaving the valve chamber . From Hydraulic Control Systems. by Herbert E. Merritt, © 1967 by John Wiley & Sons, Inc.
where we have defined the discharge coefficient as the product
Cd = CeCv
(J.2l)
where we have defined C; = contraction coefficient (0.6 < C; < 1.0) C, = velocity coefficient == 0.98 Also, we have devined Ao to be the orifice area. The effective area, due to flow contraction is given by [lJ (J.22) Thus, we write
QI = Q2 = CvA2J
~(PI - P2)
(1.23)
The steady-state force acting on the spool valve is given by Fr=Ma
.
= prj
Q~ ) = PQ~ A
Y\A 2 V
(J.24)
2
which is a force normal to the plane of the vena contracta. The force normal to the spool is given by
Fs = Flcos () = 2CeC~O(P2 - PI) cos ()
(1.25)
Using (J.l5) to express Ao as a linear function of x, we write, for small x ~=~~
~~
This is a steady-state (Bernoulli) force that always acts in a direction to close the orifice, or in the -x direction in Figure 1.3. The transient flow force is derived by considering the forces produced by accelerating the element of fluid shown in Figure J.3 in reacting with the face area of the spool. If the fluid element is accelerated in the direction of flow, the pressure on the left must exceed that on the right, or the pressure at face a exceeds that at face b. The direction of this force tends to close the valve. The magnitude is given by
Hydraulic Servomotors
F =Ma = LA d(Q/A) = L dQI t P dt P dt Using
645 (1.27)
QI from (1.20) with the area expressed as a linear function of x, we compute (1.28)
where P A = PI - P2 • Merritt [1] observes that the first term on the right side of (1.28) is the more significant as it represents a damping term. The second term is usually neglected. The quantity L is called the damping length and is the axial length of fluid between incoming and outgoing flows. In power system control analysis, it is customary to ignore the transient force (J.28). This is simply in recognition of the fact that the valve transient period is very short compared to the load transient period.
J.3 The Hydraulic Valve Controlled Piston A hydraulic valve controlled piston or linear servomotor is shown in Figure J.4. This is similar to the mechanical-hydraulic integrator described in Appendix F and reference 2. In our analysis, we assume that the valve orifices are matched and symmetrical, that equal pressure
~~ Supply Fig. J.4
Return
A hydraulic-valve-controlled piston [1].
Appendix J
646
drops exist across the valves, that the valves have equal coefficients, and that the supply pressure, P s, is constant. Then, from (J.18), for small deviations, (J.29) where P L == PI - P2 is the pressure drop across the load or across the piston. We can also write a continuity equation for the weight flow rate in and out of the contained volume. If we consider a contained volume V of mass m and density p, we can write the continuity equation dm
I Win - I Wout = Wstored = g"dt
(J.30)
where W = weight flow rate, lbf/s? g = acceleration of gravity, ft/s? p == density, lbm/ft" (or lbf-svft") v = volume, ft3 From (J.30) we can write dV
IWin-IWout=gpdi
dp
+gVdi
(1.31)
But we can also write the weight flow rate as
W=gpQ
(1.32)
Then (J.31) can be written as
IQin - IQout=
dV
V dp
dt + p di
(1.33)
Now, at constant temperature
Po P = Po + - p f3e
(J.34)
where Po is the density at zero pressure, f3e is the effective bulk modulus (lbf/ft") and P is the pressure. Thus, (J.33) may be written as
IQin - IQout=
dV
dt +
V dP f3e
dt
(J.35)
which is a convenient form of continuity equation for this problem [1]. For the piston chambers, we write the continuity relations dVI
V J dP J
dV
V2 dP2 f3e dt
QI-C;P(P1-P2)-Ce,J'1 = dt + f3e C;p(P, - P2)- Ce,J'2 - Q2 =
dt2 +
dt (J.36)
where VJ == total volume of forward chamber including valve, connecting line, and piston volume, ft3 V2 = total volume of return chamber, ft3 C ip == internal cross port leakage coefficient of piston, ft5ls-1bf Cep = extemalleakage coefficient of piston, ft 5ls-lbf
647
Hydraulic Servomotors
Now, let
where
VI = VOl +ApY V2 = V02 + ApY
(J.37)
VOl = V02 == Vo
(J.38)
Ap = piston area, ft2 VOl, V02 = initial volumes, ft3
and assume that [1]
Also note that the total volume, Vt, is constant, i.e., Vt = VI + V2 = 2Vo
(J.39)
Taking derivatives of (J.37) and substituting into (J.36) we get dy
VI
QI - Cip(P t - P2) - Ce,JJt = Apdi + f3e
ar, dt
dy V2 dP2 CiP(P 1- P2) - Ce,JJ2- Q2 = -A pdi + f3e dt
(J.40)
Now, we subtract these equations and divide by two to write
QI + Q2 2 PL
=
_
(c. + C2
ep )
Ip
P _ P = A dy + Vo (dP l (I 2) p dt 2f3e dt
_
dP2 ) + ApY (dP l + dP2 dt 2f3e dt dt
)
(J.41)
Using (J.ll), we can show that the last term on the right side of (J.40) is zero. Also, using PI - P2 , (J.4l) can be written as
QL =
dv V. dP Q +Q I 2 = C p + A _'.I' + _ 0 _ L 2 tp" L p dt 2f3e dt
(J.42)
where we define _
Cep
Ctp-CiP + 2
(J.43)
We now apply Newton's law to the forces acting on the piston to write Mty = -Ky - BpY -FL + A,JJL
(1.44)
where M( = total mass of piston and load, lbf-svft Bp = viscous damping coefficient of piston and load, lbf-s/ft K = spring constant, lbf/ft F L = load force, lbf In summary, then, we have three equations that describe the servomotor behavior. In the sdomain, these equations are QL = Kqx - KePL QL = ApSY + Ctp(1 +
~S)PL f3e Ctp
MtS 2y + BpSY + Ky + FL = A,JJL
(J.45)
Appendix J
648
These equations are easily combined to write
x,
Kce( Vt ) A/ - A; 1 + 4(3)(ces FL
y= VIMt
4f3"Al
3
+ (KceA1t + Bp~ ) 2 + A; 4f3"A; s
(1 + B;<'ce + KK + Kc)( A~ 4f3"A; s A~ t
)
(J.46)
where we define the new coefficient (J.47)
to
Equation (J.45) can be arranged in the block diagram form shown in Figure J.5. In most applications, the spring force is missing and K = O. This changes the form of (J.46)
(J.48)
where we have incorporatedthe assumption that [1] (J.49)
B;<'ce ~A; We also have defined the following parameters: T =
w~
~ 2Y = lag time constant f3eL")..ce
4f3eA 2
= --p ~Mt
x;
(h = A
p
=
hydraulic natural frequency
J----v; f3eA1/
s, V t-: fiji;
+ 4A
(1.50)
p
Note that (J.48) has a pure integration, which is not present in the system (J.46) where the spring was included. The block diagram for this system is the same as Figure J.6, but with K = O. In some systems, the mass M, of the piston and load is negligible, i.e., the time constant is small, or
1
M,s 2 + BpS +K
Fig. J.5
Y
Block diagram of servomotor position y as a function of control valve position x and load force FL.
Hydraulic Servomotors
Fig. J.6
649
Servomotor with negligible load mass and small lag time constant.
When this assumption holds, the output transfer function in Figure 1.5 becomes simply an integration. If we also assume that time constant 'T is small , the system reduces to that of Figure J.6. Many practical systems, such as the speed governor servomotor for a steam turbine can be modeled as a system similar to Figure J.6. Another assumption that is commonly made is that the load force FL is small compared to the piston force F p , i.e.,
FL -s A~L
(1.52)
In this case, the load force can be neglected entirely and the transfer function for the servomotor becomes
Kqx
y = -
ApS
(1.53)
or the entire system becomes an integrator with integrating time A,JKq • This is the form often assumed for the power servomotor. It should be noted that (J.53) may not be an adequate mathematical model if the piston load is massive. For example, the intercept valve for a large steam turbine may weight three or four tons. In such a case, it may not be a good assumption to write (J.53) unless the piston area Ap and pressure drop PL are both very large such that the acceleration can be very fast compared to the turbine response . In summary, the following assumptions have been used in deriving (1.52):
K=O VI ~ 4(3)(ce
FE.
«r,
MI~Bp
B~ce ~ AJ
(1.54)
and when these assumptions hold , the valve-controlled piston is approximated as an integrator.
References 1. Men-itt, Herbert E., Hydraulic Control Systems , Wiley, New York, 1967. 2. Eggenberger, M. A., Introduction to the Basic Elements ofControl Systems, General ElectricCompany Publication GET-3096 B, 1970.
Addendum
Page 61, general formula for the A's in Eq. 3.32:
where n is the number of machines and a machine n is the reference.
650
appendix
B
Some Computer Methods for Solving Differential Equations The solution of dynamic systems of any kind involves the integration of differential equations. Some physical systems, such as power systems, are described by a large number of differential equations. Hand computation of such large systems of equations is exceedingly cumbersome, and computer solutions are usually called for. Computer solutions fall into two categories, analog and digital, with hybrid systems as a combination of the two. The purpose of this appendix is to reinforce the material of the text by providing some of the fundamentals of computer solutions. This material is divided into two parts: analog computer fundamentals and digital computer solutions of ordinary differential equations. A short bibliography of references on analog and digital solutions is included at the end of this appendix. B.l
Analog Computer Fundamentals
The analog computer is a device designed to solve differential equations. This is done by means of electronic components that perform the functions usually required in such problems. These include summation, integration, multiplication, division, multiplication by a constant, and other special functions. The purpose of this appendix is to acquaint the beginner with the basic fundamentals of analog computation. As such it may be a valuable aid to the understanding of some of the text material and may be helpful in attempting an actual analog simulation. It should be used as a supplement to the many excellent books on the subject. In particular, the engineer who attempts an actual simulation will surely need the instruction manual for the computer actually used. B.l ,1
Analog computer components
Here we consider the most important analog computer components. Later, we will connect several components to solve a simple differential equation. We discuss these components using the common symbolic language of analog computation and omit entirely the electronic means of accomplishing these ends. The summer. The first important component is the summer or summing amplifier shown in Figure B.I, where both the analog symbol and the mathematical operation are indicated. Note that the amplifier inverts (changes the sign) of the input sum and multiplies each input voltage by a gain constant k, selected by the user. On most computers k; may have values of 1 or 10, but some models have other gains available. Usually V4 is limited to 100 V (10 V on some computers).
531
532
Appendix B
The integrator. It is necessary to be able to perform integration if differential equations are to be solved. Fortunately, integration may be done rapidly and very reliably by electronic means, as shown in Figure B.2, where Vo is the initial value (at t = 0) of the output variable V4 • Gain constants k, are chosen by the operator and are restricted to values available on the computer, usually I and 10. The output voltage is limited, usually to 100 V.
The potentiometer. The potentiometer is used to scale down a voltage by an exact amount as shown in Figure 8.3, where the signal is implied as going from left to right. Potentiometers are usually 10-turn pots and can be reliably set to three decimals with excellent accuracy.
Fig. 8.3.
The potentiometer: V2
=
k VI, 0
~
k
~
1.
The function generator. The function generator is a device used to simulate a nonlinear function by straight-line segments. Function generators are represented by the "pointed box" shown in Figure B.4 where the function / is specified by the user, and this function is set according to the instructions for the particular computer used. This feature makes it possible to simulate with reasonable accuracy certain nonlinear functions such as generator saturation. The function/must be single valued.
Fig. 8.4.
~
The function generator; V 2
=
f( VI)'
The high-gain amplifier. On some analog computers it is necessary to use high-gain amplifiers to simulate certain operations such as multiplication. The symbol usually used for this is shown in Figure 8.5, although it should be mentioned that this symbol is not used by all manufacturers of analog equipment. Note that the gain of the amplifiers is very high, usually being greater than 104 and often greater than 106 • This
533
Appendix B
Fig. B.5. The high-gain amplifier: V2
=
-A
VI'
4
A > 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)<
Hydr oturb ine go vernor block d iagr am .
Appendix D
565
V' li m
1 + TiS 0,
~
0,
Fig. D.16.
(7)
Power system stabilizer block diagram . Stabilizer types : (I) Vx ; rotor slip = quency deviation ; f~ , (3) Vx = accelerating power = Pa .
w~,
(2) Vx ; fre-
Power system stabilizer
The constants used for power system stabilizer (PSS) settings will always depend on the location of a unit electrically in the system, the dynamic characteristics of the system, and the dynamic characteristics of the unit. Still there is some merit in having approximate data that can be considered typical of stabilizer settings . Values given in Tables D.2-D .5 are actual settings used at certain locations and may be used as a rough estimate for stabilizer adjustment studies . The PSS block diagram is given in Figure D.16. D.2
Data for Transmission Lines
Data are provided in Table D.8 for estimating the impedance of transmission lines. Usually, accurate data are available for transmission circuits, based on actual utility line design information . Table D.8 provides data for making rough estimates of transmission line impedances for a variety of common 60-Hz ac transmission voltages. References I. Fitzgerald . A. E.. Kingsley . C.. Jr .. and Kusko . A. Electric Machinery . 3rd ed. McGraw Hill. New York. 1971. 2. Byerly. R. T .. Sherman. D. E.. and McCauley. T. M. Stability prcgram data preparation manual. Westinghouse Electric Corp. Rept . 70 736. 1970. (Rev. Dec. 1972.) 3. IEEE Working Group. Computer representat ion of excitation systems . IEEE Tram . PAS-ll7: 1460 64. 196H. 4. Prubhashank ar . K.. and Janischewdkyj . W. Digital simulation of mult i-machine power systems for stability studies . IEEE Trans . PAS-H7:73··ll0. 1968. 5. Crary. S. B.. Shildneck . L. P.. and March . L. A. Equivalent reactance of synchronous machines. Electr. Eng . Jan .: 124 32: discussions. Mar. : 4ll4· 88: Apr .: 603 7.1934 . 6. Kingsley . c.. Jr . Saturated synchronous reactance . Electr. Eng. Mar .: 300 305.1935 . 7. Kilgore . L. A. Effects of saturation on machine reactances. Electr. Eng. Ma y: 545..50.1935. 8. Concordia. C. Effect of steam -turbine reheat on speed-governor performances. ASME J. Eng . Power . Apr. : 201 -6. 1950. 9. Kirchmayer. L. K. Econ omic Control of Interconnected Systems. Wiley. New York. 1959. 10. Young . C. c.. and Wehler. R. M. A new stability program for pred icting dynamic performance of electric power systems. Proc. Am . Power Con] , 29: 1126· 39. 1967. II. Crary. S. B. POII'er Svstem Stab tlitv, Vol. 2. Wiley. New York . 1947. 12. Concord ia. C. Synchronous machine damping and synchron izing torques. AlEE Trans . 70:731 -37. 1951.
Appendix D
566 Table D.I.
Definitions of Tabulated Generator Unit Data EXCITER (continued)
GENERATOR Unit no. Rated MVA Rated kV Rated PF SCR
·\'d xd xd x"q x'q
xq '0
x-t.orxP
'2 x2 xo
Td
Td T:1o
TdO Tq Tq TqO T
qO
TO
WR
'F
SGI.O SGI.2
EFDFL D
Arbitrary reference number Machine-rated MVA: base MVA for impedances Machine-rated terminal voltage in kV: base kV for impedances Machine-rated power factor (I) Machine short circuit ratio Unsaturated d axis subtransient pu reactance Unsaturated d axis transient reactance pu Unsaturated d axis synchronous pu reactance Unsaturated q axis subtransient pu reactance pu Unsaturated q axis transient reactance pu Unsaturated q axis synchronous reactance Armature resistance pu pu Leakage or Potier reactance Negative-sequence resistance pu pu Negative-sequence reactance Zero-sequence reacta nee pu d axis subtransient short circuit time s constant d axis transient short circuit time s constant d axis subtransient open circuit time s constant d axis transient open circuit time s constant q axis subtransient short circuit lime s constant q axis transient short circuit time s constant q axis subtransient open circuit time s constant q axis transient open circuit time s constant Armature time constant s MW·s Kinetic energy of turbine + generator at rated speed in MJ or MW·s U Machine field resistance in U (2) Machine saturation at 1.0 pu voltage in pu Machine saturation at 1.2 pu voltage (2) in pu (2) Machine full load excitation in pu (3) Machine load damping coefficient
EXCITER VR Type Name
(4)
RR
(4)
TR
s pu
TA
s s
KA
or TAl T A2
Excitation system type Excitation system name Exciter response ratio (formerly ASA response) Regulator input filter time constant Regulator gain (continuous acting regulator) or fast raise-lower contact selling (rheostatic regulator) Regulator time constant (# I) Regulator time constant (#2)
VR max
pu(4)
VR min
pu (4)
A.'E
pu
TE
s
S£.75 max
(5)
SEmax
(5)
A EX
(5)
HEX
(5)
EFD max
pu
t'FDmin
pu pu
1(1' TForTFI
s
TF2
s
(5)
Maximum regulator output. starting at full load field voltage Minimum regulator output. starting at full load field voltage Exciter self-excitation at full load field voltage Exciter time constant Rotating exciter saturation at 0.75 ceiling voltage. or I( / for SC PT exciter Rotating exciter saturation at ceiling voltage. or A.'p for SC PT exciter Derived saturation constant for rotuting exciters Derived saturation constant for rouuing exciters Maximum field voltage or ceiling voltage. pu Minimum field voltage Regulator stuhilizing circuit gain Regulator stabilizing circuit time constant (# I) Regulator stabilizing circuit time constant (#2)
TURBINE-GOVERNOR GOV
(6)
R
(6)
TI
r-;
MW s
T2
s
T3
s
T4
s
T5
s
F
(6)
Governor type: G == general. C == cross-compound. /I = hydraulic Turbine steady-state regulation selling or droop Maximum turbine output in M\\' C ontrol time constant (governor delay) or governor response time (type 1/) Hydro reset time constant (type (j) or pilot valve timc (type II) Servo time constant (type G or C). or hydro gate time constant (type G) or dashpottime constant (type H) Steam valve bowl time constant (zero for type G hydrogovernor) or (Tw12 for type H) Steam reheat time constant or 1/2 hydro water starting time constant (type C or G) or minimum gate velocity in MW Is (type H) pu shaft output ahead of reheater or - 2.0 for hydro units (types Cor G). or maximum gate velocity in MW Is (type II)
STABILIZER PSS
(7)
I(Qv
(7)
I( QS
(7)
TQ TQl TQI T{)2 TQ2
T(}3
TQ3
VS li m
s s s s s s s pu
PSS feedback: F == frequency. S = speed. P == accelerating power PSS voltage gain. pu PSS speed gain. pu PSS reset time constant First lead time constant First lag time constant Second lead time constant Second lag time constant Third lead time constant' Third lag time constant PSS output limit selling. pu
567
Appendix D Table D.2. Typical Data for Hydro (H) Units GENERATOR Unit no. Rated MVA Rated kV
Raled P.... SCR
xd xd xd
r" 'q
x~
'\q
ra
.v-t or-,"p '2 x2 ""0
Td Td T;jO TdO Tq' Tq TqO TqO
(t) pu pu pu pu pu pu pu pu pu pu pu
HI
H2
H3
9.00 6.90 0.90 1.250 0.329 0.40H 0.911
17.50 7.JJ O.HO 0.330
25.00 13.20 0.95 2.2XO 0.310
1.070
1.020
0.660 0.660 0.003 0.310 0.030 0.490 0.200 0.035 1.670
0.650 0.650 0.0032 0.924 0.030 0.150 0.035 2.190
114 35.00 IJ.HO 0.90 LI67 0.235 0.260 1.000 0.204 0.620 0.620 0.004 0.170 0.040 0.270 0.090 0.035 2.300
5.400 0.035 0.X35
7.200 0.035 1.100
7.100 0.035 1.150
0.5HO 0.5HO .
4.200
OAoO
H5 40.00 !J.MO 0.90 I.IXO 0.2HH 0.31H 0.Y90 0.J06 0.615 0.615 0.0029 0.224
H6 54.00 13.~W
0.90 I. \X 0.340 0.3g0 1.130 0.J40 0.6HO 0.6HO 0.0049 0.2100
1/7 65.79 13.g0 0.95 \., 75 0.240 0.260 0.900 0.540 0.540 0.0022
Hg 75.00 13.g0 0.95 2.36 0.140 0.174 0.495 0.135 0.331 0.004\ 0.120
0.297 0.125
0.340 O.IHO
0.014 0.260 0.130
0.130 0.074
1.700
3.000
1.600
I.HSO
5.300
8.500
5.500
8.400
H9 ~6.00
n.so
0.90 r. Ig 0.25M 0.320 1.050 0.306 0.670 0.670 0.0062 0.140 0.060 0.312 0.130 0.044 2.020 0.051 4.000 0.017 0.033
Ta
Ut'R
'F SGI.O
SGI.2
EFD FL
D
MW·s 11 (2)
(2) (2)
(3)
O.IHOO 23.50
117.00
no.oo
0.160 0.446 2.mW 2.000
0.064 1.0lH 2.130 2.000
0.064 I.OIH 2.130 2.000
E
E AJ23 0.5 0.000 0.050 20.000 0.000 5.940 1.210 1.000 0.760 0.220 0.950 0.0027 1.91H5 3.050 1.210 0.000 0.000 0.000
E Gt-'A4
254.00 0.064 1.0lH 2.130 2.000
107.90 0.269 0.194 0.6H5 2.030 2.000
16g.00 O.JOI 0.3127 0.7375 2.320 2.000
176.00 0.199 0.lg27 0.507 1.904 2.000
524.00 0.155 O.I 70 0.440 1.460 2.000
0.2X6 233.00 0.332 0.245 0.770 2.320 2.000
A NAIOg 0.5 0.000 65.200 0.200
A REGULUX
A WMA I.X5 0.000 37.300 0.120 0.012 1.410 -1.410 -0.137 0.560 0.328 0.6H7 0.0357 1.1507 2.570 - 2.570 0.055 1.000 0.000
A NAIOX OJ 0.000 IHO.OOO 1.000 0.000 3.000 - 3.000 -0.150 2.000 0.623 1.327 0.0645 1.1861 2.550 - 2.550 0.150 1.000 0.000
A NAI43 0.5 0.000 242.000 0.060 0.000 5.320 - 5.320 -0.1219 2.700 0.450 1.500 0.0121 1.3566 3.550 - 3.550 0.100 1.000 0.000
EXCITER VR type Name RR
(4)
KA
pu
TR
TA
(4)
or TAl
TA 2 V R rnax V R rni n
/(E
pu (4) pu (4) pu
TE SE.75 max SErnax
AEX
BEX
EFDmax EFDmin
KF
T For T FI T F2
(:')
(5)
(5) (5) pu (5) pu pu
RHEO O.HH 0.000 0.050 20.000 0.000 4.320 0.000 1.000 2.019 0.099 0.3gS 0.0017 1.7412 3.120 0.000 0.000 0.000 0.000
0.5 0.000 0.050 20.000 0.000 4.390 0.000 1.000 1.970 0.096 0.375 0.0016 1.7059 3.195 0.000 0.000 0.000 0.000
t: WMA 0.5 0.000 0.050 20.000 0.000 5.940 1.210 1.000 0.760 0.220 0.950 0.0027 1.9185 3.050 1.210 0.000 0.000 0.000
o.ooo
2.607 - 2.607 -0.111 1.930 0.176 0.610 0.0042 0.94XH 5.240 - 5.240 0.120 1.000 0.000
OJ
0.000 25.000 0.200 0.000 1.000 -1.000 -0.057 0.646 0.OHg5 0.34~0
0.0015 1.573M 3.4MO - 3.4g0 0.10.1 1.000 0.000
Appendix D
568 Table D.2 (continued) TURBINE-GOVERNOR GOV
R P max
(6) (6)
MW
TI T2 TJ T4
TS F
(6)
G
0.050 8.60 48.440 4.634 0.000 0.000 0.579 - 2.000
G
0.050 14.00 16.000 2.400 0.920 0.000 0.300 -2.000
G
0.050 23.80 16.000 2.400 0.920 0.000 0.300 - 2.000
G
0.050 40.00 16.000 2.400 0.920 0.000 0.300 -2.000
G
0.056 40.00 0.000 0.000 0.500 0.000 0.430 -2.000
G
0.050 52.50 0.000 0.000 0.000 0.000 0.785 -2.000
G
0.050 65.50 25.600 2.800 0.500 0.000 0.350 -2.000
G
0.050 90.00 20.000 4.000 0.500 0.000 0.850 -2.000
G
0.050 86.00 12.000 3.000 0.500 0.000 1.545 -2.000
STABILIZER
PSS K Qv K Qs
(7) (7) (7)
TQ
TOI
TQI T{)2 TQ2 TQ3 TQJ
Vslim
pu
F
0.000 1.000 30.000 0.500 0.030 0.500 0.030 0.000 0.000 0.100
F 0.000 4.000 30.000 0.700 0.100 0.700 0.050 0.000 0.000 0.100
F 0.000 3.150 10.000 0.75g 0.020 0.758 0.020 0.000 0.000 0.095
Table 0.2. tcont.)
569
Appendix 0
Table D.2 (continued) GENERATOR
Unit no. Rated MVA Rated kV Rated PF (I) SCR
xd
xd xd til
'q
xq
xq
'a
.r -t or x p
'2 .t2 .\'0
pu pu pu pu pu pu pu pu pu pu pu
HIO 100.10 13.~W
0.90 1.20 0.280 0.314 1.014 0.375 0.770 0.770 0.0049 0.163 0.326
HII 115.00 12.50 0.85 1.05 0.250 0.315 1.060 0.287 0.610 0.610 0.0024 0.147 0.027 0.269 0.161
H12 125.00 13.HO 0.90 1.155 0.20S 0.300 1.050 0.221 a.6H6 0.686 0.0023 0.21H 0.008 0.211 0.150
1d
0.035
1:10
0.039 6.550
2.260 0.040 8.680
0.071
0.080
0.27H Mw -s 312.00 12 0.332 (2) 0.219 (2) 0.734 (2) 2.229 (3) 2.000
0.330 439.00 0.156 0.178 0.592 2.200 2.000
392.09 0.379 0.200 0.612 2.220 2.000
A
A
1d
I.~IO
TdO T
q
1q Tq'O 1qO Ta WR 'F SGI.O SGI.2
E FDFL
D
1.940 6.170
H13 131.00 13.HO 0.90 1.12 0.330 0.360 1.010 0.330 0.570 0.570 0.004 0.170 0.330 0.150 0.030 2.700 0.030 7.600 0.030 0.040 0.180 45H.40 0.IH2 0.113 0.478 1.950 2.000
H14 145.00 14.40 0.90 1.20 0.273 0.312 0.953 0.402 0.573 0.573 0.280
0.041 7.070
0.071
469.00 0.220 0.725 2.230 2.000
HIS 15K.00 13.tW 0.90 0.220 0.300 0.920 0.290 0.510 0.510 0.002 0.130 0.045 0.255 0.120 0.024 1.600 0.029 5.200 0.028 0.034 0.360 502.00 0.206 0.1642 0.438 1.990 2.000
HI6 231.60 13.80 0.95 1.175 0.245 0.302 0.930 0.270 0.690 0.0021 0.340 0.258 0.135 0.020 3.300 0.030 8.000 0.020
HI7 250.00 18.00 0.85 1.050 0.155 0.195 0.995 0.143 0.568 0.56H 0.0014 0.160 .
9.200
H18 615.00 15.00 0.975 0.230 0.2995 0.8979 0.2847 0.646 0.646 0.L396
7.400
0.060 0.200 7H6.00 0.181 0.120 0.400 1.850 2.000
1603.00 0.0769 0.282 1.88 2.000
3166.00 0.180 0.330 2.000
EXCITER VR type Name
(4)
RR
(4)
TR KA
pu
TA or TAl
TA 2
V R max V R min
K£ 1£
S £.75 max S£max
A EX
BEX
EFDmax EFDmin
KF 1For TFI TF2
pu (4) pu (4)
pu
(5) (5)
(5) (5)
pu (5) pu pu
A
WMA J .0 0.000 400.000 0.050 0.000 4.120 -4.120 -0.243 0.950 0.484 1.308 0.0245 1.0276 3.870 - 3.870 0.040 1.000 0.000
WMA NA 143A 1.5 1.5 0.000 0.000 54.000 276.000 0.105 0.060 0.011 0.000 1.960 3.X50 - 1.960 - 3.850 -0.062 -0.184 0.732 1.290 0.270 0.410 0.560 1.131 0.0303 0.0195 0.5612 1.1274 5.200 3.600 - 3.600 - 5.200 0.140 0.0317 \.000 0.4~0 0.000 0.000
G
SCR
A
WMA 1.0 0.5 0.000 0.000 272.000 400.000 0.020 0.050 0.000 0.000 4.120 2.730 - 2.730 -4.120 1.000 -0.24.' 0.000 0.950 0,480 0.000 0.000 1.310 0.000 0.0236 0.000 1.0377 2.730 3.870 - 3.870 0.000 0.0043 0.040 0.060 1.000 0.000 0.000
A
A
NA143 0.5 0.000 17.XOO 0.200 0.000 0.710 -0.710 -0.295 0.535 0.333 0.533
SIEMEN 1.0 0.000 50.000 0.060 0.000 1.000 -1.000
0.0~12
0.6303 2.9H5 - 2.985 0.120 1.000 0.000
-o.oso
0.405 0.200 0.407 0.0237 0.9227 3.080 - 3.0XO 0.0648 1.000 0.000
A
ASEA 1.0 0.000 100.000 0.020 0.000 5.990 - 5.990 -0.020 0.100 0.127 0.300 0.0096 1.1461 3.000 - 3.000 0.000 0.000 0.000
J
0.000 200.000 0.020 0.000 7.320 0.000 1.000 0.000 0.000 0.000 0.000 0.000 7.320 0.000 0.010 1.000 0.000
Appendix D
570 Table D.2 (continued) TURBINE·GOVERNOR GOV R PmaJl
(6) (6)
MW
TI.
T2 T) T4 T5
F
(6)
G
0.030 133.00 52.100 4.800 0.500 0.000 0.498 -2.000
-2.000
G 0.050 111.00 31.00 4.120 0.393 0.000 0.515 -2.000
F 0.000 0.300 10.000 0.431 0.020 0.431 0.020 0.000 0.000 0.100
0.000 8.000 30.000 0.600 0.100 0.600 0.040 0.000 0.000 0.100
G 0.051 115.00
G 0.050 120.00 27.500 3.240 0.500 0.000 0.520 -2.000
G 0.038 160.00 65.300 6.200 0.500 0.000 0.650 -2.000
G 0.050 155.00
-2.000
G 0.050 267.00 124.470 8.590 0.250 0.000 0.740 - 2.000
G 0.050 250.00 30.000 3.500 0.520 0.000 0.415 -2.000
G 0.050 603.30 36.000 6.000 0.000 0.000 0.900 -2.000
F 0.000 4.000 55.000 1.000 0.020 1.000 0.020 0.000 0.000 0.090
F 0.000 10.000 15.000 0.000 0.053 0.000 0.053 0.000 0.000 0.050
F 0.000 5.000 10.000 0.380 0.020 0.380 0.020 0.000 0.000 0.050
STABILIZER PSS J(QV J(QS TQ TO' TQI T{>2 TQ2
(7) (7) (7)
Ten
TQ) V-flim
pu
F
0.000 1.000 10.000 0.700 0.020 0.700 0.020 0.000 0.000 0.050
F
TA or TAl
KA
TR
VR type Name RR
EXCITER
SGI.2 E FDFL D
SGI.O
rF
TdO Tq Tq Tfo TqO Ta WR
Td Td T"dO
Xo
x2
r:
x-t or x p
ra
xq
xd x"q x'q
xd xd
Unit no. Rated MVA Rated kV Rated PF SCR
GENERATOR
(4) (4) s pu s
(2) (2) (2) (3)
{}
MW·s
S
s s
S
pu pu pu pu pu pu pu pu pu pu pu s s s s s
(I)
E
BJ30 0.50 0.000 0.050 20.000
Fl
25.00 13.80 0.80 0.80 0.120 0.232 1.250 0.120 0.715 1.220 0.0014 0.134 0.0082 0.120 0.0215 0.035 0.882 0.059 4.750 0.035 ... 0.210 1.500 0.177 125.40 0.375 0.279 0.886 2.500 2.000
.
A
NAI43A 0.50 0.000 57.140 0.050
0.210 0.805 3.000 2.000
...
154.90
"
...
... · .. · ..
5.500
...
0.118 0.077 ... ...
...
1.372 · .. ...
· ..
...
F2 35.29 13.80 0.85 0.80 0.118 0.231 1.400
A
WMA 1.50 0.000 400.000 0.050
F3
51.20 13.80 0.80 0.90 0.105 0.209 1.270 0.116 0.850 1.240 · .. 0.108 · .. 0.105 0.116 · .. 0.882 . .. 6.600 ... . .. · .. . .. · .. 260.00 0.295 0.2067 0.724 2.310 2.000 E
GFA4 0.50 0.000 0.050 20.000
F4
..
NAIOI 0.50 0.060 25.000 0.200
A
0.092 0.300 0.140 498.50 0.215 0.0933 0.4044 2.292 2.000
.
F5
100.00 13.80 0.80 0.90 0.145 0.220 1.180 0.145 0.380 1.050 0.0035 0.075 0.020 0.095 0.065 . .. . .. 0.042 5.900 . ..
A
NAIOI 0.50 0.060 25.000 0.200
0.1026 0.4320 2.220 2.000
. ..
0.023 1.280 0.033 8.970 0.023 0.640 0.070 0.500 0.390 596.00
...
F6
125.00 15.50 0.85 0.90 0.134 0.174 1.220 0.134 0.250 1.160 0.004 0.078 0.017 0.134
. ..
A
WMA 0.50 0.000 175.000 0.050
0.057 0.364 2.670 2.000
. ..
0.218 1.500 0.470 431.00
F7 147.10 15.50 0.85 0.64 0.216 0.299 1.537 0.216 0.976 1.520 0.0034 0.133 0.0284 0.216 0.093 0.035 . .. 0.0484 4.300 0.0072
Typical Data for Fossil Steam ~F) Units
75.00 13.80 0.80 1.00 0.130 0.185 1.050 0.130 0.360 0.980 0.0031 0.070 0.016 0.085 0.070 . .. . .. 0.038 6.100 . .. . .. 0.099 0.300 0.140 464.00 0.290 0.100 0.3928 2.120 2.000
Table 0.3.
.. .
A
NAIOI 0.50 0.060 25.000 0.200
0.033 5.900 . .. . .. 0.076 0.540 0.240 634.00 0.370 0.1251 0.7419 2.680 2.000
.,
.
F8 160.00 15.00 0.85 0.64 0.185 0.245 1.700 0.185 0.380 1.640 0.0031 0.110 0.016 0.115 0.100
. ..
NAIOI 0.50 0.060 25.000 0.200
A
0.105 0.477 2.640 2.000
F9 192.00 18.00 0.85 0.64 0.171 0.232 1.651 0.171 0.380 1.590 0.0026 0.102 0.023 0.171 . .. 0.023 0.829 0.033 5.900 0.023 0.415 0.078 0.535 0.254 634.00
C BRLS 0.50 0.000 250.000 0.060
0.141 1.500 0.420 960.50 . .. 0.0987 0.303 2.580 2.000
0.248 0.143 0.350 0.950 0.0437 5.140
FlO 233.00 20.00 0.85 0.64 0.249 0.324 1.569 0.248 0.918 1.548 0.0016 0.204
A
BBC 0.50 0.000 30.000 0.400
0.500 0.297 1115.00 0.166 0.125 0.450 2.300 2.000
0.140 0.060 0.027 0.620 . .. 4.800
Fll 270.00 18.00 0.85 0.6854 0.185 0.256 1.700 0.147 0.245 1.620 0.0016 0.155
»
x·
0'1 'J
0
0-
::3
CD
-0 -0
(5) pu (5) pu pu s s
(5)
s pu (4) pu(4) pu s (5) (5)
~lim
K QS TQ TQI TQI TQ2 TQ2 TQ3 TQ3
KQV
PSS
STABILIZER
F
P max TI T2 T3 T4 TS
R
GOV
pu
(7) (7) (7) s s s s s s s
(6) (6) MW s s s s s (6)
TURBINE GOVERNOR
TF2
EFDmax EFDmin KF TFor TFI
BEX
SE.1Smax SEmax AEX
TE
KE
TA2 VR malt VR min
"
"
· ..
·
., . · .. · ..
·
· .. .. .
...
...
· ..
G
0.050 22.50 0.200 0.000 0.300 0.090 0.000 1.000
0.000 6.812 1.395 1.000 0.700 0.414 0.908 0.0392 0.8807 3.567 1.417 0.000 0.000 0.000
"
.
· · .. · ..
.,
., . ...
...
· .. . .. .. . · ..
G 0.050 36.10 0.200 0.000 0.300 0.200 0.000 1.000
0.000 1.000 -1.000 -0.0445 0.500 0.0684 0.2667 0.0012 1.2096 4.500 -4.500 0.080 1.000 0.000
F 0.000 0.700 10.000 0.300 0.020 0.300 0.020 0.000 0.000 0.100
53.00 0.200 0.000 0.300 0.090 0.000 1.000
0.07~
G
0.000 0.6130 -0.6130 -0.0769 1.370 0.1120 0.2254 0.0137 0.6774 4.130 -4.130 0.040 1.000 0.000
. .. . .. ... .. .
... . ..
. ..
'"
. .. .,. . ..
G 0.050 75.00 0.090 0.000 0.200 0.300 0.000 1.000
0.000 4.380 0.000 1.000 1.980 0.0967 0.3774 0.0016 1.7128 3.180 0.000 0.000 0.000 0.000
"
. .. ... ... . .. .. . .. . ... . .. . .. .
G 0.050 105.00 0.090 0.000 0.200 0.300 0.000 1.000
0.105 0.350 0.000
-3.43~
0.000 1.000 -1.000 -0.0582 0.6544 0.0895 0.349 0.0015 1.5833 3.438
. .. . .. . .. ... ... .. . . .. . ..
. .. .. . . ..
G
0.050 132.00 0.083 0.000 0.200 0.050 5.000 0.280
0.000 1.000 -1.000 -0.0601 0.6758 0.0924 0.3604 0.0016 1.6349 3.330 - 3.330 0.108 0.350 0.000
"
., . . .. . ., . .. . .. .
. ..
. .. . .. . ..
. ..
G 0.050 121.00 0.200 0.000 0.300 0.090 10.000 0.250
0.000 3.120 -3.120 -0.170 0.952 0.220 0.950 0.0027 1.4628 4.000 -4.000 0.030 1.000 0.000
.
"
... . ..
. .. . .. . .. . .. . .. . .. . .. . ..
G 0.050 142.30 0.100 0.000 0.200 0.050 8.000 0.300
0.000 1.000 -1.000 -0.0497 0.560 0.0765 0.2985 0.0013 1.3547 4.020 -4.020 0.0896 0.350 0.000
S 0.000 15.000 10.000 1.000 0.020 0.750 0.020 0.000 0.000 0.050
G 0.050 175.00 0.083 0.000 0.200 0.050 8.000 0.271
0.000 1.000 -1.000 -0.0505 0.5685 0.0778 0.303 0.0013 1.3733 3.960 -3.960 0.091 0.350 0.000 G 0.050 210.00 0.150 0.000 0.100 0.300 10.000 0.237
0.000 4.420 -4.420 1.000 0.613 0.010 0.270 0.000 3.7884 3.480 0.000 0.053 0.330 0.000 G 0.050 230.00 0.100 0.000 0.259 0.100 10.000 0.272
0.000 4.590 -4.590 -0.020 0.560 0.730 1.350 0.1154 0.7128 3.450 -3.450 0.050 1.300 0.000
'J
01
}>
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<:----v~
Fig. F.28
A potentiometer used to indicate position.
which we can linearize for small changes. A typical oil-pump governor arrangement is shown in Figure F.27 .
F.6.2 Position Transducers It is often desirable to convert a mechanical position into an electrical signal. There are many ways of doing this. One common way is to use a potentiometer, as shown in Figure F.28. This techn ique can be used to indicate translational or rotational position and can be linear or nonlinear, depending on the potentiometer design . If the potentiometer is wire-wound, the resolution is finite and this may be a problem for some applications. In this case, the transfer function is not a straight line as in Figure F.28, but is a stair-step function . The main advantage of this type of device is its simplicity and low cost. Another useful position transducer is the linear variable differential transformer (LVnT) shown in cross-section in Figure F.29 [8]. This device consists of a primary winding, two secondary windings, and a movable magnetic core. The windings are concentric about the cylindrical core, which moves axially, as indicated in the figure. The magnetic circuit is excited by the primary winding, which is located in the center (axially). The movable core provides a flux path
I I
fj:~~~~:fJ W~~:+:~:~J ~:t)~~:i
Position to be Controlled
I
I
Core
~I
~% :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) C
::E
~
~
"'0
d
'§
o ::E C
---& -20 1-- --II--- - - - - t - - -- - - - --=k - - -+---1-45
--40L - _--L! : . 0.33
...I-
1.0
~
00,
o
_ _---="_""'''__'''_'''--'
rad/s
(b) Bode Diagram for Proportional Plus Reset Control Fig.0.9 Proportional plus reset pressure control.
'" .c C
c,
Pressure Control Systems
621
(a) Block Diagram of Proportional Plus PartialResetPressure Control Diagram +90
+40 y=54°
:€l
+20
+45 "~
0
o '§l
"'"cf"
of
-e
a
'&, ::;'"
::;'"
-20
1
TL
-40
-45
1 R2
1.0
1; " 5.0 T
"
10.0
"gj
E:
-90
co, radians/s
(b) Bode Diagram of Proportional Plus PartialResetPressure Control Diagram Fig. G. I0
Proportional plus partial reset pressure control system.
appendix
H
The Governor Equations
Considerable literature exists on governors, some of it quite elementary [1-7]. Only a few references provide a more rigorous analytical treatment [8, 9]. This appendix explores the governor equations in greater detail than is usually needed for linearized control. It is presented as a background for the material for Chapter 10 and forms a basis from which simplifying assumptions may be made for physical systems.
H.1 The Flyball Governor· Consider the flyball governor shown in Figure H.l, where two flyballs are held to a rotating shaft by rigid arrns L 1 and L2 and further restrained by a spring K. As the angular velocity of the shaft increases, the balls are thrown out, such that the collar C slides upward on the shaft. Thus, the vertical position of the collar C from some stationary reference is a measure of the angular velocity and a mechanical linkage attached to C could be used to control the throttle of the prime mover, providing the force available is sufficient to move the throttle lever.
H.1.1 The equilibrium equations To analyze the forces acting on one of the flyballs of Figure H.I, refer to the sketch in Figure H.2. As the flyball rotates at angular velocity lJJG (radians/second), the ann of length L, holding the ball of mass M, swings out to radius R and assumes an angle c/J with the vertical rotating shaft. Under these conditions, the ball is acted upon by three forces:
= Gravitational Force (weight) F~ = Centrifugal Force (weight) Fs = Spring Force due to Spring K Fg
The difference between centrifugal force and spring force is the net outward force Fe or (H.1)
From classical dynamics we write
mv' F'=-c R *The development here follows closely that ofPontryagin [8].
622
(H.2)
The Governor Equations
Fig. H.I
623
A simple flyball governor.
Fe cos
\
\ Fe \
\
./
\
./
./
./
./
./
Fe sin
\
\
\
\
\
./
./
./
./ Fe cos
Fe Fig. H.2
Forces acti ng on the flybal l,
r;
Appendix H
624
where v is the peripheral velocity of the ball. In terms of the angle cP we note that
R == L sin 4J
(H.3)
Fe = L sm· = L sin· w2; = (mL sin
(H.4)
or mv!
mR2
Also, writing v as a function of Rand WG' v
=.RwG
(H.5)
we have F~ =
mLwl; sin cP
(H.6)
For the spring force, we can write (H.7)
where R; is the unstressed length of the spring. Combining (H.1), (H.6), and (H.7) we get
Fe == mLwl; sin cP - 2KL sin cP + 2KL sin cPu
(H.8)
where 4>u 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
(H.12)
for which the equilibrium condition is tan4J=
-2K sin cPu 2 2K mWG-
(H.I3)
This can be viewed as a right triangle as shown in Figure H.3, where we define
a =mwl;-2K b == -2K sin 4>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
" -, -,
<,
o " ' " - - - - - + - - - - - + - - - - + - - - - - - +......' ~,--+-----t--~ •• <1> <, <1>0
Fig. H.5
1t
2
......
.......
..................
Mechanical torque as a function of angle
1t
Appendix H
628
From (H.3!), we learn more about the state of equilibrium by setting the left-hand side to zero and substituting (H.32).
o== l/Jo o== LN2 W6 sin cPo cos cPo - g sin cPo 2KL
- - - (sin m
cPo - sin cPu)cos cPo (H.33)
From (H.33) we compute cos
cPo == F/k 2KL) 2KL 5- --;;;cos cPo + --;;;- sin cPu cos cPo
g == ( LN2 W
.po == 0
(H.34)
We now linearize (H.31) by the substitution
cP = cPo + cPa l/J == .po + l/Ja W== Wo + Wa
(H.35)
to write
(H.36) where we have defined
I( Wa, cPa, cPu, l/Ja) == LN2( Wo + wa)sin( cPo + l/Ja)cos( l/Jo + l/Ja) 2KL - - - sine cPo + cPa)cos( cPo + cPA) m
(H.37) Equation (H.36) must be examined for higher-order terms, such as those involving squared variables, etc., which may reasonably be neglected. Also (H.34) may be incorporated to give the result (H.38) where
A21 == A 23 =
g sin 2
cos
2g sin
Wo
cPo
~
%
cPo
2KL sin
cPu
sin
'+'0
- - - ---:-;: cos
m
4KL.
.
2
cPo
+ --(sIn cPo - SIn cP,Jcos cPo
wom
(H.39)
The result is a linear system that is restricted to small deviations from the initial states.
629
The Governor Equations
It is instructive to examine the stability of the linear system (H.38). We call the system matrix A and compute P(A) = det A - A1
=0
(H.40)
where 1 is the unit matrix. Thus we have the result _
3
B
kA23
2
•
P(A) - A + -A -A 2 1A + - - slncPo m J
(H.41)
or, by definition (H.42) Note that a3 :> 0, therefore, by Routh's criterion, we require that, not only must all a's be positive, but also, if stability is to be assured, (H.43) where these coefficients are defined above. This is the sufficient condition for stability [8]. Rearranging (H.42) and incorporating(H.33) we compute BJ _ 2KNBJ m w5m2N2
(1 _ sin cPu ) > 2F sirr' cPo
Wo
(H.44)
where F is proportionalto the load torque, T; Now, the right-hand side of (H.44) correspondsto a particular operatingpoint on the torque speed characteristic of the prime mover. Recall from (H.28) that F is a constant for a given value of W00 These incrementalchanges on the torque-speedcurve are referred to as the "incremental regulation" (incrementaldroop) of the prime mover, defined by
aw
R·= I
aF
(H.45)
This corresponds to the slope at a given point (wo, To) on the torque-speed curve as shown in Figure H.6. Since the slope is usually negative, the incremental regulation computed by (H.45) is a positive quantity. The derivative (H.45) may be computed from (H.34) with the result -cPwo Wo K ( sin cPu ) R, = cPTo = 2To - woT~m 1 - sin3 cPo
OJ
-----------------~T Fig. H.6
Location of the operating point on a torque-speed curve.
(H.46)
Appendix H
630
from which we compute 2F Wo
1
1[
2K (
sin
= R; - R; wfjN2m 1- sin3
(H.47)
By factoring BJ/m from the left side of (H.44), we have the result 1 BJ ->m R;
(H.48)
This is an importantresult and is the sufficientconditionfor stability. From (H.48) we may summarizeour findings as follows: 1. B (friction) is essential for stability 2. Large J (inertia) is beneficial to stability 3. Large m (flyball mass) is detrimental to stability 4. Stability is increasedby increasing the regulationor droop of the torque-speed characteristic
For a given system with fixed B, 1, and m, the only control we have on stability is through the regulation. As seen from (H.46), this depends on the values of K and
appendix
I
Wave Equations for a Hydraulic Conduit
The purpose of this appendix is to derive the equations for head and velocity of fluid in an elastic conduit. The resulting equations are very similar to the familiar wave equations used by electrical engineers to describe the voltage and current at any point along a transmission line. In hydraulicapplications,these equations are often called the "water hammer" equations,since they describe mathematicallythe traveling pressure waves in a conduit. The derivation used here follows closely that ofParmakian [1], which is recommended for further reading on the subject. All variables used in this derivation, together with the variable names, are given in Table 1.1. It will be convenient to recognize that
..rp = g, the acceleration of gravity I.1 Dynamic Equation of Equilibrium The dynamic equilibrium condition (F = ma) for an element of water dx long may be derived as follows. Consider two faces or sections along the conduit labeled Band C in Figure 1.1. If the fluid at face B has area A, then the fluid at face C has area A + dA, where JA dA = -dxft2
(1.1)
ax
The pressure also is different at the two faces. At B the pressure is
PB = y(H - Z) lbf/ft?
(1.2)
where H is the head and Z the height at B. At C the pressure is Pc = y[(H + dB) - (Z + dZ)] lbf/ft?
(1.3)
JH dH= -dxft
(1.4)
dZ = -sin a dx ft
(1.5)
where we compute
ax
and Substituting into (1.3), we get
r; =
{(H-
Z) + (
a;
+ sin
a}ix] thf/ft
2
(1.6)
631
Appendix I
632
Table I.1
Variab le Names
Dimenson
Variable
Symbol
ft2 ft ft ft ft Ibfft2 Ibfft2 Ibfft3 lbm/ft? or lbf-s?fft2 ft/s2 radians ft ft ft ft/s Ibfft2 Ibfft2 Ibffft2
Pipe insidearea Pipe insideradius Pipe inside diameter Pipe length Pipe wall thickness Modulusof elasticityof pipe material Bulk modulusof water Specificweight of water Fluid (water) mass density Acceleration of gravity Angle of slope of conduit Distancealong pipe in directionof flow Head at any point x and at any time t Height above conduit outlet or gate Velocityof fluid Longitudinal stress in pipe wall Circumferential stress in pipe wall Pressureat any point x Force of fluid
A R D L e E
K g p
g a
x H= H(x, t) Z V UI
U2
P F
lbf
Finally, we analyze the forces at faces Band C caused by the pressure acting over a given area. At face B
FB = yA(H- Z) Ibf and at C Fe =
~A + : dx)[(H - Z) + (
a;
(1.7)
+ sin a
)dx] Ibf
(1.8)
Hydraulic Gradient for Gate Closure
--
w.s. -
-.t~(JH dx ox
-----,
--_]/~x~~------------l
lJ --,-.1-__ [-
I I I .!.-_..L'
-_
I
F;
I
ex
Center Line of Gate Fig. I.I
n;
Con trol -.;>Gate
Sketch of conduit showing element of length dx between faces Band C [1].
+x
633
Wave Equations for a Hydraulic Conduit
The quantities computed in (1.2)-(1.8) are summarizedin Table 1.2. In addition to the forces due to fluid pressure, there is also a force due to gravity, as indicated in Figure 1.1, which acts on the center of gravity of the element. Calling this downward force F g we compute (1.9)
Fg= yAcgdx where we take the area at the center of gravity to be A + (1/2)dA. Then _/ 1 aA ) Fe = '\A + 2" a.; dx dx lbf
(1.10)
of which a fraction, F g sin a acts to the right, along the pipe longitudinalaxis. Thus, the accelerating force may be written as
(1.11 ) It is often assumed that
aH
aA ax
(1.12)
Fa = -yA -dx lbf
(1.13)
A-
ax
~(H-Z)-
so that we can write the approximate solution
aH
ax
But
Fa = (mass) x (acceleration) yA dV =-dxg dt or
(1.14)
aH 1 dV --=--
ax
Finally then
aH = ax
(1.15)
g dt
_.!.( dV + vdV) g
dt
(1.16)
dx
Table 1.2 Area, Pressure, and Force Quantities on a Differential Length dx of Fluid
Quantity
Value at Face B
Value at Face C
Area, ft2
A
aA A+-dx
Pressure, Ibf/ft2
y(H -Z)
Force,lbf
yA(H-Z)
ax
i(H -Z) + ( : + sin
~A + : dx)[(H - Z) + ( :
a)dx] + sin
a)dxJ
Appendix I
634
which is one of the wave equations for the conduit and is derived from the equat ions of dynamic equilibrium for an element of water.
1.2 The Continuity Condition The second equation relating H, V, x, and t is derived from the continuity condition. This condition requires that all space inside the boundaries of the element be occupied by water at all times . Consider the element of water dx long as shown in Figure 1.2. The element boundaries are Band C at time t as shown in (a) but have moved to D and F, respectively, at time t + dt. Thus, in time dt, B moves to D and C moves to F. At time t + dt we may compute the velocity at face D as
VD= VB+dVB av av = V+ -dx+ -dt ax at av av =V+BD-+-dt ax at
(1.17)
and the velocity at face F is
VF= Ve+dVe av aVe aVe = V+ -dx+ -dx+ - d t ax ax at = V + av dx + !..-.(v + av dx)CF + !-.(v + av dx)dt ax ax ax at ax These velocities are shown in Figure 1.2.
c
B
(a) At timet FJ
C F
f)
t=---~ ---
; av
JV: V + -BD+ -cIt dX at (b) At time t +dt
~:-3
V: JV dX+i.(V+ dV cIX)CF dX ax dX
av (
av )
+- V+-dt dt at
ax
Fig. 1.2 The change in length of dx in time dt [I).
(1.18)
Wave Equations for a Hydraulic Conduit
635
The change in length of the element is
dL =BD-CF
(I.l9)
where we note that, if dL > 0, the element becomes shorter or compresses because of the way that dL is defined. Now, the averagevelocity of face B in moving to D in time dt is 1 av + -dt av) Vave = -(VB + VD) = -1 ( V+ V+ -BD 2
=
1
av
2
1
ax
av
at
V+ --BD + - - d t 2 ax 2 at
(1 .20)
The distancethat face B moves in time dt is
1 av ) I av BD= ( V+ --BD + --dt dt 2 ax 2 at
(1.21 )
Then, we can compute,neglectinghigher-orderterms
av
dL =BD - CF= --dxdt
ax
(1.22)
The change in length computedby (1.22) is caused by two factors. 1. The change (increase) in pressure causes the pipe shell to expand and causes dx to shrink
in order to contain the same volume of water. 2. Since the water is compressible, a change(increase)in pressurecausesa change(decrease) in the volume of water within the element,causing a furtherchange (decrease) in length. Note that these two effects are additive.
1.2.1
Deformation ofthe shell
A small segmentof the pipe shell is shown in Figure 1.3. If'we define 0') as the longitudinal stress, 0'2 as the circumferential stress, and I.t as Poisson's ratio, then the change in radius may be computed as (1.23)
I
I
I
I
.R~,/ / // 1,/
/ // k/
I
1,,;/
/
(11=Circumferential Stress
- 1-<: dX~-'----- -Center line axis of pipe
Fig.I.3
A sequent of the pipe shell of length dx [2].
Appendix I
636
where el2 is negligible relativeto R. We may also compute the changein lengthdue to stressing of the pipe material as (1.24) In both equations (1.23) and (1.24), the d quantities are changes in stress due to a change in pressure. Knowing fiR and S~ we may compute the new volumeof the elementas New Volume = 1T(R + 1iR)2(dx + Sx)
(1.25)
If we define the change in length due to changein stress as dLm then we can write new volume- old volume dL = - - - - - - U old area
'TT(R + 1iR)2(dx + Sx) - 1TR2dx 1TR2
fiR =Sx+2-dx R
(1.26)
with higher-order terms neglected. Expanding (1.26) by incorporating (1.23) and (1.24), we get dl., ==
dx
E [(1 -
2J..L) Au l + (2 - J-t) d U2]
(1.27)
The exact solutionof (1.27) depends on exactlyhow the pipe is anchored. Three cases that are sometitnes of interestare shown in Table1.3. It is apparentthat, in any case, we may write yDdH
(1.28)
dLu==Cl~dx
where 5
- - J.L
Case 1
1 - J.L2
Case 2
4
C] ==
(1.29)
1-!!... Case 3 2
Parmakian [1] gives examples to show that the results are nearly the same for all values of Ct. For example, with IL == 0.3 for steel pipes, we compute C 1 to have values of 0.95, 0.91, and Table 1.3 Evaluationof dl., for Three Cases of Interest
Case 1. Pipe anchored at one end, free at the other 2. Pipe anchored throughout its entire length 3. Pipe with expansion joints throughout its entire length
~O"l
~0"2
')'DdH
')'DdH
4e
2e
JJ.-~0"2
0
')'DdH
2e ')'DdH
2e
ai, yDdH ( 2. _JL
Ee
4
)dx
')'DdH - - ( 1 - Il})dx
Ee
yDdH ( 1 _ JL
Ee
2
)dx
Wave Equations for a Hydraulic Conduit
637
0.85 for the three cases. Thus, in general, we could take C t to be a constant somewhat less than unity, or about 0.9.
1.2.2 Compressibility of the water The change in volume of the original length dx of water due to water compressibility under pressure change vdl! is fJ. V = (force)dx = (area x pressure)dx = (1TR2)( ydH)dx ft3 KKK
(1.30)
This change in volume causes a change in length dLK equal to (1.31 )
Then the total change in length is dL
== ==
==
dLK + dl.;
ydH dx K
C1yDdHdx
+---eE
.l ~ + _C1_D )dH dx 1'\ K eE
(1.32)
But H is a function of both x and t so that
aH on aH dx aH ( aH aH) dH== -dx + -dt== --dt+ -dt== + V - dt ax at ax dt at at ax
(1.33)
Then we may write
dL ==
on + VaH) dt dx at ax ~ 1 + -C1D- )( --
K
eE
(1.34)
Since the change in length is also computed in (1.22), we can set the two expressions equal and write
CID)(aH av dL == y ( -1 + - -- + VaH) - dtdx==--dtdx K eE at ax ax
(1.35)
or
'Y(~ + K
C1D)( aH + v eE at
aH
ax
) == _av ax
(1.36)
Now define
K==(~+CID) 1 K eE
(1.37)
Using this expression,we can write (I. 36) as
on aH 1 av -+v-==--at ax K 1 ax
(1.38)
Appendix I
638
which is the second of the wave equations, this one being derived from the continuity of water inside the pipe. It is sometimes convenient to write (1.38) in a slightly different way. Suppose we let (1.39) where
C1D) p-+--
a=
1
(K
ftls
(1.40)
eE
Then we can write
aH
on
a2 av
-+v-=---
at
ax
ax
g
(1.41 )
The wave equations then may be written as
av
av
aH
at + Va; =-ga;
on
on
a2
av
-+v-=---
at
ax
g
ax
(1.42)
The solution to these equations is well known and may be thought of as two waves traveling in the +x and -x directions at a velocity of a feet per second. This being the case, we may write x
= ±at + k
(1.43)
This simple relationship helps us analyze the second terms on the left side of (1.42). We compute
av ax
V
av at
v-=±--
aH
a
vaH
v-=±--
ax
a
at
(1.44)
Now, the constant "a" may be evaluated for a given physical system and will typically have a value of from 2000 to 4000. This is 100 times or so the value expected for V, so both quantities (1.44) have multipliers VIa that are very small. We conclude that
av at
av ax
-~v-
aB -aB ~ vat ax
(1.45)
and we can neglect the second terms on the left side of (1.42) to write
av
aH
-=-g-
at ax 2 all a av -=:--at g ax
(1.46)
Wave Equations for a Hydraulic Conduit
639
This is the more familiar form of wave equation and corresponds to a lossless transmission line. The solution maybe thought of as an incident wave f+ and a reflected wave t: or H-
n, = f+(t - ~ )+ f-(t + ~ )
v- Vo = ~~+(t- ~ )+f-(t + ~)J Reference 1. Parmakian,1., Waterhammer Analysis, Prentice-Hall, New York, 1955.
(1.47)
appendixJ
Hydraulic Servomotors
The hydraulic servomotor, such as the mechanical integrator described in Appendix A, is a class of control devices that are used to move large loads with precision and speed. The newer designs incorporate electromechanical elements to improve the speed and accuracy. These devices have two main mechanical components: a control valve and a piston. The purpose of this appendix is to write the basic equations that describe the behavior of these two components and of the servomotor system.
H.l Control Valve Flow Equations The control valve or spool valve is usually described in terms of the number of spools or lands and the number of ways the hydraulic fluid can enter or leave the valve. All valves require at least a supply line, a return line, and a line to the load-a three-way configuration. Many valves, such as the valve shown in Figure J.l, are four-way valves. All are analyzed in a similar way. Our analysis follows closely that of Merritt [1], which is recommended for further study. Consider a three-land, four-way spool valve shown in Figure J.l. This valve is described by four sets of equations that describe the flow and pressure relationships. The flow past the spool orifices are given by Bernoulli's equation*
QI
=C~IJ;(PS-PI)
Qz = c~zJ ;(PS - Pz)
Q3=C~3J;PZ
Q4 = C~4J ;P
1
(J.l)
where Q= volumetric flow rate, ft3/S Cd = dimensionless discharge coefficient A = orifice area, ft2 *Dimensions of all quantities are given in a consistent set of units, often using the ft-Ibm-s system. Actual devices might be analyzed using different dimensions for convenience, e.g., using A in square inches or metric units.
640
Hydraulic Servomotors
Q,
Suppy
s
i; Rerun
J~ ~
U
-
641
L1
i;
Q,
Fig. J.I
A three-land, four-way spool valve [I].
and
P = pressure, Ibf/ftz
p = mass density of fluid , lbm/ft? or Ibf-s z/ft4 The flow to the load can be written as (1.2)
and these relationships are readily verified by examining the Wheatstone bridge equivalent of the spool valve in Figure 1.1. The orifice area in each case is a function of the displacement x. Thus, we can write
AI = A )(x) A z = Az(-x) A) = A)(x)
A 4 =A 4 (- x)
(J.3)
Finally, we note that the pressure drop across the load is given by
(J.4) These four equations, 1.1-1.4, with appropriate simplifications, must be solved simultaneously to give QL as a function of x and Pu i.e., QL = QL(X, PL). The first simplification is to assume matched symmetrical valve orifices: Matched:
A) = A) A z = A4
Symmetrical:
(J.5)
A )(x) = Az(- x) Aix) = A 4(-x)
(1.6)
We also define the neutral position area (1.7)
642
Appendix J
Usually, we assume that orifice area varies linearly with valve stroke so that only one defining equation is required, i.e.,
A =wx
(1.8)
where w is the width of the slot in the valve sleeve in ft 2/ft (or in2/in). Now, for matched symmetrical valves
Ql =Q3 Q2= Q4
(1.9)
From the first equality, and using (J.5), we write
CeYllJ f(ps-p == CeYllJ fp 2 t)
or
PS=P1+P2
(1.10)
Combining (J.I0) with (J.4), we compute
PS+PL Pl=--2(1.11 ) These relationships are shown graphically on a pressure scale in Figure J.2. From (J.2) we also compute
QL = QI-Q4
=
C,p4tJ;(Ps-P C,p42J;P JPrP JPS+PL 1) -
=C~l
L
P
-C~2
1
P
Drop Across 1
Ps/2 ~:----Ir------
f} =0
(Drain) Fig. 1.2
Graphical illustration of pressure division for matched symmetric orifices.
(J.12)
643
Hydraulic Servomotors
Also, from Figure J.l,
(J.13)
(J.14) For a symmetrical valve, we can write
(J.15) Thus, for any x we can write (J.16) Now, our goal is to determine a linear equation for QL' We can use a Taylor's series expansion to write (1.17) Thus (J.18) where
QL
a- ] Kq == the flow gain == ax
0
QL K c == the flow-pressure coefficient == - a -- ] aPL 0
(J.19)
Equation (J.18) is the desired relationship and will be used in evaluating the small-signal behavior of the system. There are obvious limitations that should be kept in mind, however, as equation (1.16) is obviously not linear, even though much of the operating range is reasonably linear.
J.2 Control Valve Force Equations The equations giving the forces acting on the spool valve are developed for either a steadystate or a transient condition. Consider the spool valve shown in Figure 1.3, where the spool is displaced a small amount in the +x direction. Continuity requires that
QI = Qz = Cc00
J
;(PI - P z) = CcCvAo
J
;(P 1 - P z)
(J.20)
644
Appendix J
P2
t
x
Fluid Element
QI
Vena Contracta
PI
F I < .._-_.__ ... Face a
-,
- - --
F2
-
1<
L
Face b
---'-F,
- ----
Fig. J.3 Flow forces on a spool valve due to flow leaving the valve chamber . From Hydraulic Control Systems. by Herbert E. Merritt, © 1967 by John Wiley & Sons, Inc.
where we have defined the discharge coefficient as the product
Cd = CeCv
(J.2l)
where we have defined C; = contraction coefficient (0.6 < C; < 1.0) C, = velocity coefficient == 0.98 Also, we have devined Ao to be the orifice area. The effective area, due to flow contraction is given by [lJ (J.22) Thus, we write
QI = Q2 = CvA2J
~(PI - P2)
(1.23)
The steady-state force acting on the spool valve is given by Fr=Ma
.
= prj
Q~ ) = PQ~ A
Y\A 2 V
(J.24)
2
which is a force normal to the plane of the vena contracta. The force normal to the spool is given by
Fs = Flcos () = 2CeC~O(P2 - PI) cos ()
(1.25)
Using (J.l5) to express Ao as a linear function of x, we write, for small x ~=~~
~~
This is a steady-state (Bernoulli) force that always acts in a direction to close the orifice, or in the -x direction in Figure 1.3. The transient flow force is derived by considering the forces produced by accelerating the element of fluid shown in Figure J.3 in reacting with the face area of the spool. If the fluid element is accelerated in the direction of flow, the pressure on the left must exceed that on the right, or the pressure at face a exceeds that at face b. The direction of this force tends to close the valve. The magnitude is given by
Hydraulic Servomotors
F =Ma = LA d(Q/A) = L dQI t P dt P dt Using
645 (1.27)
QI from (1.20) with the area expressed as a linear function of x, we compute (1.28)
where P A = PI - P2 • Merritt [1] observes that the first term on the right side of (1.28) is the more significant as it represents a damping term. The second term is usually neglected. The quantity L is called the damping length and is the axial length of fluid between incoming and outgoing flows. In power system control analysis, it is customary to ignore the transient force (J.28). This is simply in recognition of the fact that the valve transient period is very short compared to the load transient period.
J.3 The Hydraulic Valve Controlled Piston A hydraulic valve controlled piston or linear servomotor is shown in Figure J.4. This is similar to the mechanical-hydraulic integrator described in Appendix F and reference 2. In our analysis, we assume that the valve orifices are matched and symmetrical, that equal pressure
~~ Supply Fig. J.4
Return
A hydraulic-valve-controlled piston [1].
Appendix J
646
drops exist across the valves, that the valves have equal coefficients, and that the supply pressure, P s, is constant. Then, from (J.18), for small deviations, (J.29) where P L == PI - P2 is the pressure drop across the load or across the piston. We can also write a continuity equation for the weight flow rate in and out of the contained volume. If we consider a contained volume V of mass m and density p, we can write the continuity equation dm
I Win - I Wout = Wstored = g"dt
(J.30)
where W = weight flow rate, lbf/s? g = acceleration of gravity, ft/s? p == density, lbm/ft" (or lbf-svft") v = volume, ft3 From (J.30) we can write dV
IWin-IWout=gpdi
dp
+gVdi
(1.31)
But we can also write the weight flow rate as
W=gpQ
(1.32)
Then (J.31) can be written as
IQin - IQout=
dV
V dp
dt + p di
(1.33)
Now, at constant temperature
Po P = Po + - p f3e
(J.34)
where Po is the density at zero pressure, f3e is the effective bulk modulus (lbf/ft") and P is the pressure. Thus, (J.33) may be written as
IQin - IQout=
dV
dt +
V dP f3e
dt
(J.35)
which is a convenient form of continuity equation for this problem [1]. For the piston chambers, we write the continuity relations dVI
V J dP J
dV
V2 dP2 f3e dt
QI-C;P(P1-P2)-Ce,J'1 = dt + f3e C;p(P, - P2)- Ce,J'2 - Q2 =
dt2 +
dt (J.36)
where VJ == total volume of forward chamber including valve, connecting line, and piston volume, ft3 V2 = total volume of return chamber, ft3 C ip == internal cross port leakage coefficient of piston, ft5ls-1bf Cep = extemalleakage coefficient of piston, ft 5ls-lbf
647
Hydraulic Servomotors
Now, let
where
VI = VOl +ApY V2 = V02 + ApY
(J.37)
VOl = V02 == Vo
(J.38)
Ap = piston area, ft2 VOl, V02 = initial volumes, ft3
and assume that [1]
Also note that the total volume, Vt, is constant, i.e., Vt = VI + V2 = 2Vo
(J.39)
Taking derivatives of (J.37) and substituting into (J.36) we get dy
VI
QI - Cip(P t - P2) - Ce,JJt = Apdi + f3e
ar, dt
dy V2 dP2 CiP(P 1- P2) - Ce,JJ2- Q2 = -A pdi + f3e dt
(J.40)
Now, we subtract these equations and divide by two to write
QI + Q2 2 PL
=
_
(c. + C2
ep )
Ip
P _ P = A dy + Vo (dP l (I 2) p dt 2f3e dt
_
dP2 ) + ApY (dP l + dP2 dt 2f3e dt dt
)
(J.41)
Using (J.ll), we can show that the last term on the right side of (J.40) is zero. Also, using PI - P2 , (J.4l) can be written as
QL =
dv V. dP Q +Q I 2 = C p + A _'.I' + _ 0 _ L 2 tp" L p dt 2f3e dt
(J.42)
where we define _
Cep
Ctp-CiP + 2
(J.43)
We now apply Newton's law to the forces acting on the piston to write Mty = -Ky - BpY -FL + A,JJL
(1.44)
where M( = total mass of piston and load, lbf-svft Bp = viscous damping coefficient of piston and load, lbf-s/ft K = spring constant, lbf/ft F L = load force, lbf In summary, then, we have three equations that describe the servomotor behavior. In the sdomain, these equations are QL = Kqx - KePL QL = ApSY + Ctp(1 +
~S)PL f3e Ctp
MtS 2y + BpSY + Ky + FL = A,JJL
(J.45)
Appendix J
648
These equations are easily combined to write
x,
Kce( Vt ) A/ - A; 1 + 4(3)(ces FL
y= VIMt
4f3"Al
3
+ (KceA1t + Bp~ ) 2 + A; 4f3"A; s
(1 + B;<'ce + KK + Kc)( A~ 4f3"A; s A~ t
)
(J.46)
where we define the new coefficient (J.47)
to
Equation (J.45) can be arranged in the block diagram form shown in Figure J.5. In most applications, the spring force is missing and K = O. This changes the form of (J.46)
(J.48)
where we have incorporatedthe assumption that [1] (J.49)
B;<'ce ~A; We also have defined the following parameters: T =
w~
~ 2Y = lag time constant f3eL")..ce
4f3eA 2
= --p ~Mt
x;
(h = A
p
=
hydraulic natural frequency
J----v; f3eA1/
s, V t-: fiji;
+ 4A
(1.50)
p
Note that (J.48) has a pure integration, which is not present in the system (J.46) where the spring was included. The block diagram for this system is the same as Figure J.6, but with K = O. In some systems, the mass M, of the piston and load is negligible, i.e., the time constant is small, or
1
M,s 2 + BpS +K
Fig. J.5
Y
Block diagram of servomotor position y as a function of control valve position x and load force FL.
Hydraulic Servomotors
Fig. J.6
649
Servomotor with negligible load mass and small lag time constant.
When this assumption holds, the output transfer function in Figure 1.5 becomes simply an integration. If we also assume that time constant 'T is small , the system reduces to that of Figure J.6. Many practical systems, such as the speed governor servomotor for a steam turbine can be modeled as a system similar to Figure J.6. Another assumption that is commonly made is that the load force FL is small compared to the piston force F p , i.e.,
FL -s A~L
(1.52)
In this case, the load force can be neglected entirely and the transfer function for the servomotor becomes
Kqx
y = -
ApS
(1.53)
or the entire system becomes an integrator with integrating time A,JKq • This is the form often assumed for the power servomotor. It should be noted that (J.53) may not be an adequate mathematical model if the piston load is massive. For example, the intercept valve for a large steam turbine may weight three or four tons. In such a case, it may not be a good assumption to write (J.53) unless the piston area Ap and pressure drop PL are both very large such that the acceleration can be very fast compared to the turbine response . In summary, the following assumptions have been used in deriving (1.52):
K=O VI ~ 4(3)(ce
FE.
«r,
MI~Bp
B~ce ~ AJ
(1.54)
and when these assumptions hold , the valve-controlled piston is approximated as an integrator.
References 1. Men-itt, Herbert E., Hydraulic Control Systems , Wiley, New York, 1967. 2. Eggenberger, M. A., Introduction to the Basic Elements ofControl Systems, General ElectricCompany Publication GET-3096 B, 1970.
Addendum
Page 61, general formula for the A's in Eq. 3.32:
where n is the number of machines and a machine n is the reference.
650
Index
Acceleration, mean value, 72 Admission valve, 439 Admittance matrix: defined, 36,40, 370 primitive, 373 reduction, 40-41 Air gap line, 248, 584 A matrix, 11,65,209,212,214,219,221-222,232,386, 394--95. See also Eigenvalues including excitation system, 287, 290, 307 American National Standards Institute (ANSI), 14, 98, 143,318,581 Amplidyne, 239, 251-252 Amplification, 595 Amplifier: as analog computer component, 532-533 defined, 451 figure of merit, 253 magnetic, 239, 252-254 rotating, 239, 251-252 transfer function, 274 Analyzing steam turbine systems, 452 Analog computer simulation: differential equations, 531-537 excitation control system, 302-304, 307, 347-353 excitation system, 257, 265, 282-284, 535-537 synchronous machine, 170-184 Anderson, P. M., 125, 352 Armature reaction, demagnetizing effect, 56-57, 228, 229-230,326 Automatic control, 401 Backlash, in voltage regulator, 238, 250-251 Bar lift, 443 Base quantity, choice, 93, 95-96, 104, 147, 167, 550 Bode plot: compensated excitation system, 329-331, 334, 339, 344--346,366 lead compensator, 342, 366 machine inductance, 144--145
regulated synchronous machine, 329-331, 334 Boiler, 233-234 configuration (large), 442 -follow control, (automatic), 471 -following mode, 433 -turbine representation (simplified), 464 storage effect, 465 Boiling water reactors, 478 Boost-buck, 250-251,268-271,274-277,305 Braking: dc, 21 negative sequence, 21 Brown Boveri Corp., 354--356, 358, 361, 364 Brown, P.G., 321 Buildup, exciter voltage, 247 Bypass valve, for hydro turbines, 489 Cam lift, steam turbine control, 443 Ceiling voltage, exciter, 23, 247-248, 260-261, 263, 266,295,311,562,584 Centrifugal flyball governor, 402 Centrifugal governors, 401 Classical model: defined, 26 multimachine system, 35-37, 316-317 shortcomings, 45-47, 316-317 synchronous machine, 22-24, 55-56, 355, 358-359 Classical stability study, nine-bus system, 37-45 Clearing angle, critical, 33-34 Clearing time, critical, 33, 320-321 Combined cycle power plant, 519 Combined cycle prime mover, 518 Combined cycle units, 513 Combustion turbine control, 515 Combustion turbine units, 513 Combustion turbine schematic diagram, 514 Compensated governor, 421, 422 analysis of, 422 principle of operation, 422 permanent droop, 422
651
652
Index
Compensated governor (continued) temporary droop, 422 Compensation, See also Bode plot, Root locus current, 237 excitation system, 277-284,321,341,584 excitation system, lead network, 339, 341, 344, 363, 366 linear analysis, 344-347 Compensator, 451 Compressibility of water, 637 Computed response, of combustion turbines, 516 Computer methods, differential equations, 531-544 Concordia, C., 56, 83, 102, 106,311,321,325,363 Conduits, 489, 494 Constant flux-linkage assumption, 23, 46 Constant voltage behind transient reactance, 142. See also Classical model Continuous system modeling program (CSMP), 188-193 Continuity conditions, in hydraulic conduits, 634 Control: generating unit, 234 optimal, 365 system, 581 Control system: for a boiler, 560 components, 590 Control valve, 411,439 Control valve flow equations, 640 Control valve force equations, 643 Control valve operation, 440 Control valve position control. 476 Coordinated control mode, for a thermal unit, 433 Crary, S. B., 316 Critical time, of a hydro unit, 493 Current compensation, excitation system, 237 Dahi, O.G.C., 254, 257, 267 Damping: critical, 249 effect, of a hydro unit, 495 effect on system order, 377-378, 527 excitation system, 297 generator unit oscillation,S, 46, 558-560 positive sequence, 21 ratio, 249, 334, 336, 337-339 system oscillation, 309-310 torque (D~, 21,35,46, 106,326-27,339,558-560 Damping transformer, as excitation stabilizer, 237, 306 Deadhand, in voltage regulator, 250-251, 268, 311 Deformation of the shell, in hydro conduits, 635 Delta: maximum value, 32 mechanical (q;,), 14 de Mello, F. P., 56, 325 Deviation, 585 Differential surge tank, 495 Differentiation, in control systems, 592 Digital computer simulation: differential equations, 537-543
excitation system, 257 synchronous machine, 184--206 transient stability, 353-363 Dimensions, machine equations, 92-93 Direct axis, 20, 22, 23, 84-85 Dispersion, coefficient, 256 Distortion curve, 267 Disturbance, 53-80, 584 Double-overhung hydro units, 484 Draft tube, for hydro units, 487 Drift, 585 Droop, 10,58,563 Droop characteristic, 19-20 Drum-type boilers, 461 Duty, excitation system, 585 Dynamic, 454 Dynamic equation of equilibrium, for a hydraulic conduit, 631 Dynamic equations of governors, 626 Dynamic system performance, 5,46,325 E (EMF proportional to iF)' defined, 98 E FD (EMF proportional to VF), defined, 99, 129 E~
(EMF proportional to
Eqa , defined, 152-153
~),
defined, 99, 128
Economic control, 10 Eigenvalues: A matrix, 11,54,61,79,209,216-217,222,232, 284,396 A matrix with linear exciter, 291-292, 307 effect ofunifonn damping, 378 Eigenvectors, A matrix, 65 Electric analog of a hydro system, 496 Electrical angle (Be), 15 Electrical load frequency damping, 411 Electrohydraulic systems, in steam turbine control, 402 Emergency trip conditions, in steam turbines, 445 Equal area criterion, 31-35 two-machine system, 35 Equal mutual flux linkage, 95-96, 547-548 Equivalent circuit, synchronous machine, 107-109 Equivalent stator, pu d-q quantities, 129. See also Stator equivalent Error, 585 Euler method, modified, 30, 532-534 Excitation control: alternator-rectifier system, 239-241, 583 altemator-SCR systems, 241-242 brushless, 240, 296, 304 compound rectifier, 242, 583 compound rectifier plus potential source rectifier, 242-243 configurations, 236, 244 de generator-eommutator systems, 239, 583 potential source rectifier, 243, 583 rheostatic, 236-237, 268-271,584 Routh's criterion, 11, 57, 59, 68,232,271,277, 322-327 simplified view, 233-235
653
Index Excitation control system, 236-244. See also Excitation control analog computer solution, 282-284 boost-buck response, 275-277 comparison with classical representation, 316-317 complete linear model, 287-291 definllion~243-249,581-589
linear analysis of compensation, 344-347 linear numerical example, 288-291 simplified linear model, 286 Excitation systems, 431 approximate representation, 333 compensation, 277-284. See also Bode plot; compensation, computer representation, 292-299 Types A, B, 559 Types C, D, 560 Types E, F, G, 591 Type K, 562 Type 1,293-295,304,307,316,347-349, 355-356,359 Type 1S, 295-296 Type 2, 296-97, 307 Type 3, 297-298, 355 Type 4,299 damping, 297 defined, 243, 246, 581 duty,605 effect on power limits, 311-315 effect on stability, 309-366 high initial response, 247, 581 normalization, 248, 267-268, 299 primitive, 236-238 rate feedback, 277-284, 325, 352 response, 268-284, 585-586 continuously regulated, 271-284 noncontinuously regulated, 268-271 rheostat, 236-239, 247, 268 self-excited, 237 saturation, 271, 294-295, 307, 562-563 separately excited, 238, 305-306 stabilizer, 237, 306, 338-344 state-space description, 285-297 thyristor, 239, 241--244, 266 typical constants, 299-304 voltage response, 585-586 Exciter: boost-buck transfer function, 274-275 ceiling voltage, 23, 247-248, 260-261, 263, 266, 295, 311,562,584 voltage rating, 247-248 Exciter build-down, 254 Exciter buildup, 247, 254-268 ac generator exciter, 266 analog computer solution, 535-537 de generator exciter, 254-265, 306 digital computer solution, 540-552 formal integration, 256, 259-263 linear approximation, 263-264 loaded exciter, 266-267
response ratio, 268-284, 585-586 solid-state exciter, 266 Faults, effect on transient stability. 3, 16-17, 355-360 Feedback, 19,244,309,329,352 Feedback control system, 244 Field voltage: base, 248, 587 rated load, 248, 587 Figure of merit, amplifiers, 253 Filter, bridged T, 352, 366 First swing stability, 35-37,46,315,320 Flux-linkage: equations, synchronous machine, 85-88 network, 388-390 mutual, 95-96, 416-17 subtransient, 132, 134 transient, 138 Flyball governor, 401, 402, 408, 622, subsystem, 423 Fossil-fueled boiler computer models, 473 low-order model, 474 prime mover control model, 474 block diagram of prime mover controls, 475 Fossil-fueled steam generators, 461 drum-type boilers, 461 once-through boilers, 461 FORTRAN, 187,541-542 Frame of reference, (abc, Odq), 84 Frequency: natural resonant (undamped), 249 oscillation, 24-25, 310 Friction head, in a hydro penstock, 495 Frohlich equation, 256-261, 263, 265, 294, 535-536, 540-541 Fuel and air controls, in combustion turbine unit, 522 Fuel system dynamics, in steam turbines, 465 Function generators, 609 Gain, 587 Gas turbine power generation, 524 Gating, in control systems, 601 General Electric Co., 238, 240-41, 243, 252, 299, 560-561 Generation control, 430 isolated system, 430 network system, 431 Generation mix, in U.S. power systems, 435 Generating unit block diagram, 432 Governor, 10,48,68,233-234 analysis: Ballarm scale, 407 compensator system, 424 block diagram, 427 transient performance, 427 behavior, 406 closed-loop, 417 block diagram, 546
654 Governor (continued) computerrepresentation,563 -564 droop, 10, 58 characteristic,nonlinear, in combustionturbine system, 518 equations, 622 equilibriumequations, 622 linear synchronousmachine, 68-69 values, in steam turbines, 439 Grand Coulee Dam, 489 H, change of base, 16
estimatingcurves, 17 typical values, 126 Harris, M. R., 93 Head, change in, 493 Head loss, 490 Heffron, W. G., 56 Hybrid formulation, linear n-machinesystem, 386-88 Hydraulicgradient, 492, 493 Hydraulicreaction force, in governors,408 Hydraulicservomotor: general description,640 transfer function, 618 Hydraulicsystem equations,498 Hydraulicsystem transfer function, 503 Hydraulicturbine prime movers ,484 adjustableblade propeller turbine, 489 Deriaz turbine, 484, 489 Francis turbine, 484, 486 impulse turbine, 484 Kaplan turbine, 484, 489, 490 Nagler turbine, 489 Pelton turbine, 484 propeller type turbine, 484, 489 reaction turbine, 484, 487, 489 Hydraulicvalve controlledpiston, 645 Hydro system, block diagram, 509
Ideal transformer,546-547 Impact: distribution,54, 69-79 effect, 8-10 large vs small, 6, 53 Impedance,characteristic,501 Impedancematrix, n-port, 373, 383 Incrementalvariables, 208 Inductance,synchronousmachine leakage, 108, III defined, 86-87, 122-124 negative sequence, 125 magnetizing, 108 table, 126 transient and subtransient, 123, 143 Odq, defined, 87 Inertia constant: effect on stability. 317 H, defined, 14. See also H M, defined, 14 units, 15, 16 Inertial center, 72
Index Infinite bus, 26, 115-116 Initial conditions: examples, 159-165 stability study, ISO, 165-166 Institute of Electricaland ElectronicsEngineers(IEEE), 143-145,238-248,292-297,316,319,321,347, 355-356,581 Integration,in control systems, 593 Integrator,analog computercomponent,532 Interceptvalve, in steam turbine systems, 444 Interface,between system differentialand algebraic equations,522 International Electrotechnical Commission(lEe), 98 Instability,dynamic,46 Isochronousgovernor,408-413 Jordan canonicalform, 64-65 Kimbark,E. W., 13, 14,48, 102, 125-26, 184,246, 254-255,256-257,266-267 Kinetic energy of rotating mass, Wk, 14, 16 Krause, P. C., 83, 173 Kron, Gabriel, 278 Kron reduction, 378. See also Matrix reduction LAD and LAQ , defined, 108 Ld,defined,87 Limitingeffects, in combustionturbines, 517 Lmd and Lmq , defined, 108 L MD and L MQ , defined, 110 L q , defined, 87 defined, 87 Leakage inductance,synchronousmachine, 108, III Lefschetz,S., 61 Lewis, W. A., 83, 93, 95 Liapunov,M. A., 117 Limiter,defined, 583, 587 Linear analysis, dynamic stability, 53, 321 Linearization, nonlinearequations, 53, 70, 208-210, 381-385 Linearizedsystem equations, 10-11, 60, 386 Load equations: infinite bus form, 115-116 linear current model for one machine, 213-217 synchronousmachine, 114-122 Load-flowstudy, 35, 37-38, 162 nine-bus system, 38 Load representation, constant impedance,35, 46, 368
u;
Main exciter, 237, 250, 255, 305, 583. See also Exciter; Excitation Main stop valve, 439, 445 Mathematicalmodel, elementary, 10, 13-52. See also Classicalmodel Matrixreduction, 40-41, 378 Mechanicalflyball governor, 402-408 Moment of inertia, 9, 13, 15 Multimachine systems, 35-37, 165-166, 368-397 Multiplier,analog computercomponent, 533 Multivariable control system, in governor control, 467
Index National Electric Reliability Council (NERC), 300-301 Nebraska Public Power District, 364 Network equations: based on flux-linkagemodel. 388-390 linearized n-machine form, 381-384 n-machine system, 36-37, 369-370 Nine-bus system: defined, 37-39 linearized solution, 392-396 load-flow study, 38 oscillation, 61-66 stability simulation, 353 swing curve, 44-45 n-machine system, 35-37 hybrid formulation, 386-88 network equations, 369-70, 381-384 system equations, 377-78, 386, 396 Node incidence matrix, 373 Nonlinearities,machine equations, 107, 116-117, 119, 170,185,208 Nonlinear system equations, 11 Nonreheat turbine block diagram, 454 Normalization: comparison of pu systems, 96-98, 550-551 guidelines, 545 swing equation, 15, 103-105 synchronous machine equations, 92-96, 99-103, 545-554 time, 101 torque equations, 103-105 Northeast Power Coordination Council, 145 n-port network, 370 Nuclear steam supply systems, 477 Numerical methods, integration, 537-540 Nyquist, H., 11 Off-nominal frequency and voltage effects, in combustion turbine systems, 517 Omega: mechanical (ltln), defined, 14 rated (~), defined, 14 One machine-infinite bus solution, 26-31, 115-122, 153--157,311-315 Once-throughboiler, 469 Order, system equations, n-machine system 377-378, 386,396 Oscillation: generator unit, 5, 46, 558-560 modes, 59-66, 364 natural frequencies, 24-25, 310 system, 309-310 three-machine,nine-bus system, 61-66 tie-line, 7, 55 Overshoot, 249 Overspeed protection, in steam turbines, 440 Overspeed trip, 445 Pacific Gas and Electric Co., 319 Park, R. H., 20 Park's transformation,20, 83-85, 88, 115, 146-147,371
655
Penstocks, 489, 494 Perry, H. R., 319 Perturbation method, 54 Per unit: comparison of various systems, 96-98, 102, 550-551 conversion, 92-96 torque, 103-107 Phasor: defined, 21, 151 diagram, synchronousmachine, 152-165 equations, in d-q reference frame, 374--376 reference frame definition, 370, 372-373 reference frame transformation, 372-374 relation between system d-q quantities, 374--377 Philadelphia Electric Co., 301, 304, 354 Phillips, R. A., 56 Pilot exciter, 238, 250, 255, 305, 583. See also Exciter; Excitation P matrix, 84 Position transducers, in control systems, 614 Potential transformer, transfer function, 272 Potentiometer,analog computer component, 532 Power: accelerating,15,32, 33 factor, 157 invariance, 85,414 limits, effect of excitation, 311-315 synchronizing,24. See also Synchronizingpower coefficient Power-anglecurve, 21-22, 33-34 Power system components, in stability study, 10 Power system stabilizer (PSS), 338-339, 343, 345-346, 352,387,359-365,562-563,584 Predictor-correctormethod, 30, 534 Prentice, B. R., 86 Pressure regulator, in control systems, 614 Pressure regulator, typical, 616 Pressure transducers, 608 Pressurized water reactors, 479 Prime mover governors, 401 Proportionalplus partial reset pressure control, 620 Proportionalplus reset pressure control, 620 Pumped storage hydro systems, 510 Quadrature axis, 20, 84-85 Quiescent operating point, 209, 311 Rankin, A. W., 93 Rankine cycle, 435 Rate feedback, excitation system, 277-84, 325, 352 Ray, J. 1., 46 Reactance, direct-axis: synchronous,xd,22 transient, Xd , 23 Reaction hydro turbine, 486 Reactive power, emergency demand, 321 Reference frame: phasor, 370, 372-374 synchronously rotating, 13, 371-372 Regenerativevapor cycle, 566
656 Regulator. See also Voltage regulator continuously acting, 250, 271,584 proportional control, typical, 616 synchronous machine, 66-69, 583, 585 Reheat steam turbine system block diagram, 459 Reheat stop valve, 444 Reheat turbine flow diagram, 442, 457 Reheat turbines, 444 Reheat vapor cycle, 435 Reliability, 3-4 Reset control, block diagram, 617 proportional plus reset control, 617 Response ratio, 244-246, 248, 260-261, 263, 299, 306, 316-317,320-321,357,363 Rheostat, excitation system, 236-239, 247, 268 Rise time, defined, 249 Riser tank, in hydro systems, 495 Root locus, 11 compensated excitation system, 276, 281-282, 306-307,327-333,344,366 Rotor angle, 13-14 Rotational speed transducers, 603 Routh's criterion, II, 57, 59, 68,232 applied to excitation control system, 271, 277, 322-327 Rudenberg, R., 79,254,257 Saturable reactor, 251-252 Saturation: computer representation of exciters, 294-295 de generator exciter, 255, 257-60, 267, 271 digital calculation, 185-187 excitation systems, 271, 294-295, 307, 562-563 exponential function, 114, 186, 593-594, 563 linearized exciter, 285 synchronous machines, 20, 113-114,355,556-558 Scaling, analog computer solutions, 533-534 Schroder, D. C., 352 Schulz, R. P., 148 SCR, 239, 320-321. See also Thyristor Servomotor, 402 Settling time, 249 Shipley, R. B., 46 Short circuit ratio, 555 Signal, defined, 587-588 Simplifying assumptions, for hydro transfer functions, 506 Silverstat regulator, 236-237 Simulation methods, 10 Small disturbances: defined, 53 response, 53-80. See also Linear analysis Small impacts, response, 54 Specific inertia of steam turbine generator, 451 Speed droop governor, 413 block diagram, 416 eigenvalues, 416 floating lever, 419 root locus, 417 transient response, 416
Index Speed reference, governor, 408 Speed regulation, 19, 49, 57-58, 563-564. See also Droop Speed relay, governor, 402 sensing, 402 Speed voltage, synchronous machine, 90 Stability: asymptotic, 5 defined, 5, 13,588 dynamic, 6, 53, 310-311, 321-327 effect of excitation, 304, 309-366 effect of inertia constant, 317 excitation system, 588 first swing, 35-37, 46, 315, 320 limit, 33, 588 power systems, 3-12 primitive definition, 5 problem, statement of, 4 simulation in nine-bus system, 353 steady-state, 6, 24, 309 synchronous machines, 6 transient, 6,46, 309-310, 315-321. See also Transient stability Stabilizer, 327, 338, 584. See also Power system stabilizer Stabilizing signal, supplementary, for excitation system, 338-344 State-space equations: current form, synchronous machine, 91-92, 107, 368 excitation system, 285-288, 296-297 flux-linkage form: linear, 217-222 loaded machine, 118-119 neglecting saturation, 111-112 synchronous machine, 109-112 linear current form, 209-113 loaded machine, 117-119 simplified linear machine, 231-232 synchronous machine, 83,91 total system, 390-391 Stator equivalent, nus pu quantities, 129, 136, 139, 151-152,154,369,379 Steady state, 4, 588 Steady-state equations, synchronous machine, 150-153, 157-165 Steady-state stability 6, 24, 309 Steam plant control functions, 446 Steam power plant model, 435 fueled by fossil fuels, 436 fueled by nuclear energy, 436 Steam turbine, 430, 437 Steam turbine control operations, 444 control of , 444 protection of, 444 Steam turbine generator controls, 445 Steam turbine power generation, in combined cycle plants, 525 Steam valve-steam flow, 618 Steam volume, 618 Stevenson, W. D., 14
Index Subtransient: effects, 9 EMF, 132 flux linkage, 132, 134 inductances, synchronousmachine, 123-124, 135 Summation, in control systems, 590 Summer,analog computercomponent,400-40 I Summingbeam, in speed droop governors,413 compensator, 423 Surge tank, 489,494, 495 Swing curve: defined,41 nine-bussystem,44-45 Swingequation, 13-15, 46, 79 approximate, in pu power, 16 classicaln-machinesystem, 37 defined, 13-14 most useful form, 16 normalizedform, 103-5, III simple nonlinear form, 29 Synchronism, loss, 4, 9 Synchronizing power coefficient (Ps), 24, 59-60, 71, 224,227,230 Synchronous machine: analog simulation, 170-184 block diagram,47-48,57-58,67-69,231,340 classicalmodel, 22-26, 55-56 constantfield flux-linkage model, 142-143 digital simulation, 184-206 E' model, 127-132 E" model, 132-138 equivalenttee circuit, 107-109 flux-linkage equations, 85-88 governor,68-69 inductance, 86-87, 108, 111, 122-26, 143 linear models, 56-57, 60, 208-232, 322 linear,regulated, 327-333 linear,unregulated,55-56 load equations, 114--122 local load, 154-157 normalization equations,92-96, 99-103, 545-554 one-axismodel, 14 1-42,354 operationalinductance, 144 parameters, from manufacturers' data, 166-69, 551-552 phasor diagram, 152-165 regulated,66-69, 329-331, 334 saturation,20, 113-14,355,556-558 simplifiedmodel, 56-57, 127-143, 222-228 simulation, 150-207 solid rotor dynamic models, 14}--144
speed voltage, 90 stability,6 state-spaceequations, 83. 91-92, 107, 109-112,368 steady-stateequations, 150-153, 157-165 subtransientinductance, 123-124, 134 time constants, 125-126, 143 two-axismodel, 138-142 typical parameters, 126, 552-553 unregulated,55
657
unsaturatedflux-linkage model, 111-113 voltage equations, 88-91, 110-111 System: continuouslyacting, proportional,250, 271, 584 control, 581 noncontinuously acting, 584 Systemdata, tabulationof typical values, 566-580 Ted> defined, 104 derived from field energy, 106 Tesla, Nikola, 3 Thermal generation,435 Theta (9), defined, 14, 85 Theveninequivalent,77 Thomas, C. H., 83 Throttle valuve, in steam turbine systems, 439 Thyristor,excitation system, 239, 241-244, 266 Tie-line oscillations,7, 55 Time constant, reheater,450 Time constants, synchronousmachine derived, 125, 143 table, 126 Tirrell regulator,250 Torque: accelerating, 13, 14 asynchronous, 21 damping, (D(,), 21, 35, 46,106,326-327,339, 558-560 dc braking, 21 electromagneticor electrical, 13, 20, 105-107, 111, 326 mechanical, 13, 16,46 regulated, 18-20 unregulated, 17-18 negative sequencebraking, 21 normalizationequations, 103-105 per unit, 103-107 synchronous, 21,326 Torque angle, 6 defined, 14,85 effect of excitation, 235 Transient: defined, 588 effects, 9 EMF's, 138 flux linkage, 138 inductance,synchronousmachine, 123-124 reactance,22 Transducers,605 Transfer functions, of steam turbine control, 446 Transient stability: defined,6,309 digital simulation,353-363 effect of excitation,315-321 effect of faults, 316-317, 355-360 first swing, 46,315-316,319-320 steps in problem solution,41-45 Transmissionline equations, 498 Transmissionlines, typical data, 4564-565
658 Trigonometric identities, for three-phase systems, 398-399 Turbine blading, steam, 437 impulse blading, 437, 438 reaction blading, 437, 438 Turbine efficiency, hydro systems, 492 Turbine-following control mode, 432 Turbine mechanical damping, in torque equations, 411 Turbine torque-speed characteristics, 19-20 Two-port network, 27 Typical constants, of steam turbine systems, 451 Units: English, 15 inertia constant, 15-16 MKS, 15 Vapor power cycle, 435 Venikov, V. A., 56-57 Voltage equations, synchronous machine, 88-91, 110-111 Voltage reference: nonlinear bridge circuit, 273-274 transfer function, 272-74 Voltage regulator, 236, 250-254, 560. See also Amplifier backlash, 238, 250-251 boost-buck, 251-252, 256, 262-263, 272 deadhand, 250-251,268,311 direct-acting, 250 electromechanical, 250-251, 268-271, 305 electronic, 251
Index indirect-acting, 250 linear synchronous machine, 66-67 magnetic amplifier, 239, 252-254 models of physical systems, 559-562 rotating amplifier, 25 1-52, 272 solid-state, 254 Voltage response ratio, 244-246. See also Response ratio Water hammer, 484, 491, 494 formula 497 Water starting time. 498, 509 Wave equations for a hydraulic conduit, 631-639 Wave velocity, in hydro penstocks, 493 Western Systems Coordinating Council (WSCC), 560 Westinghouse Electric Corp., 237, 239-240, 242, 252, 269,291-292,297,299-300,304,347, 349-350,561-563 Wicket gate, 487-488 WR, rotor, 15 xd,defined, 151, 166 xd,defined, 166 x:; , defined, 166 xp,defined, 151, 166 x q , defined, 166 xd,defined, 151, 166 x:; , defined, 166 Xo, defined, 166 X2, defined, 166 Young, C. C., 127, 131, 140,316
appendix
c
Normalization
There are many ways that equations can be normalized, and no one system is clearly superior to the others [1,2,3). For the study of system dynamic performance it is important to choose a normalization scheme that provides a convenient simulation of the equations. At the same time it is also important to consider the traditions that have been established over the years [1,2) and either comply wholly or provide a clear transition to a new system. Having carefully considered a number of normalization schemes for synchronous machines and weighed the merits of each, the authors have adopted the following guidelines against which any normalization system should be measured. I. The system voltage equations must be exactly the same whether the equations are in pu or MKS units. This means that the equations are symbolically always the same and
no normalization constants are required in the pu equations.
2. The system power equation must be exactly the same whether the equation is in pu or MKS units. This means that power is invariant in undergoing the normalization.
Thus both before and after normalization we may write
(C. I )
and k is the same both before and after normalization. 3. All mutual inductances must be capable of representation as tee circuits after normalization. This requirement is included to simplify the simulation of the pu equations.
4. The major pu impedances traditionally provided by the manufacturers must be maintained in the adopted system for the convenience of the users. Other pu imped-
ances must be related to and easily derived from the data supplied by the manufacturer.
The normalization scheme used by U.S. manufacturers does not satisfy requirement 2. The manufacturers use the original Park's transformation, as given by (4.22), which is different from the transformation used in this book, as given by (4.5). However, the pu system is to be developed so that the same pu stator and rotor impedance values are obtained. C.l
Normalization of Mutually Coupled Coils
Consider the ideal transformer shown in Figure C.I. First we write the equations in MKS quantities, i.e., volts, amperes, ohms, and henrys. 545
Appendix C
546
+
Ideal
Fig. C. I.
Schematic diagram of an ideal transformer.
+ L" dt + L'2 dt
III
=
V2
di, V
di,
R .
VI
di, di, V R 2 i 2 + L 22 dt + L 21 dt
(C.2)
where, in terms of the mutual permeance (Pm and the coil turns N, L i k = {P,n Nj N, for i, k = 1,2. Now choose base values for voltage, current, and time in each circuit, i.e., For circuit I: V'Bl,oIIB For circuit 2: V2B128 t28 Then since any quantity is the product of its per unit and base quantities, we have, using the subscript u to clearly distinguish pu quantities,
V IBV l u
R 11O'lu I . +
f/
R I ·l
Y2B V2u =
L'I/IB di.; --I'B dt.;
L'2/20 di2u V +t 2B
di-,
L 22 / 2B di2u L 2 , / IB dl.; V +-- + --
2 2B 2u
dt 2u
12B
1'8
di.;
(C.3)
Dividing each equation by its base voltage, we have the pu (normalized) voltage equations Vlu
V2u
R,
.
R2
V 2B / /2B
;211
+ ---+ VIBIIB du;
L 12 / 2B
di 2u
V'Bt 2B
dt 2u
+
L 21 / IB
LIlI IB
'Iu
VIR/liB
+
di,u
L 22 / 2B di 2u
V2BtiB
V2Bt2B dt211
di l u
(C.4)
dtt;
We can define Rill
=
Lilli =
R,
RI
V IB / /IB
RIB
L II V'8 t IB / I IB
R 2u
L II LIB
R2 V28 / /28
L22u =
Now examine the mutual inductance coefficients. quire that LI212B/VIBI2B =
and since L 12
R2
L22
R 28
V28 t2B /1 28
L22
(C.S)
L 2B
To preserve reciprocity, we re-
L21/IB/V2BlIB
L 21 H, we compute
or (C.6)
Appendix C
547
The ideal transformer is also characterized as having the following constraints on primary and secondary quantities:
n = i2 /i, = where n
=
(C.7)
VI/V2
N , / N 2 • Rewriting in terms of base and pu values, we have n
=
12Bi2u/118ilu = V1BVlu/V2BV2u
Thus the pu turns ratio nu must be n, = i 2u/i lu
=
n/'B/128
=
and base quantities are often chosen to make nu I'B/12B
(C.8)
Vlu/ V2u = nV28/VIB
=
I. From (C.8) we compute
= V2B/V1B
or (C.9)
Combining with (C.6), it is apparent that we must have tlB
=
t 2B
~
(C.IO)
tB
and the mutual inductance terms of the voltage equation (C.4) become (C. I I)
Then the voltage equation is exactly the same in pu as in volts, and the first requirement is satisfied. Furthermore, if this identical relationship exists between currents and voltages, the power is also invariant and the second requirement is also met. C.2
Equal Mutual Flux Linkages
To adapt the voltage equations to a pu tee circuit, we divide the coil inductances into a leakage and a magnetizing inductance: i.e., (C.12) From the flux linkage equations we write (in MKS units) (C.13)
Injecting a base current in circuit I with circuit 2 open, i.e., with i, gives the following mutual flux linkages
liB
and ;2 = 0, (C.14)
In pu these flux linkages are Amlu
=
Am2u
=
Ami/AlB Am 2 / A2B
=
LmIIIB/LIBIIB
= L21/1B/L2B/2B
= Lml/L1B
(C.15) (C.16)
Equal pu mutual flux linkages require that
Am l u
=
Am2u
(C.17)
Appendix C
548 or
(C.18) Following a similar procedure, we can show that injecting a base current in circuit 2 with circuit 1 open (i.e., with ;2 = /28 and i, = 0) gives the following pu flux linkages:
x., =
(C.19)
L 12/ 28 / L I8/18
Again equal pu flux linkages give L m2/ L 28
L m2u
=
L12/28/LIB/IB
(C.20) (C.21)
and from (C.20) and (C.2) L m2/L28 = L m2u = L12/18/L28/28
(C.22)
Comparing (C.18) and (C.22), (C.23)
Now using (C.) 2), (C.20), (C.22), and (C.23) in the voltage equation (C.4),
= R. u; I u + {,. I. u + L mu (i I u + /2 u ) V2u = R 2ui 2u + {2 i2u + Lmu(i,u + 12u) VI U
(C.24)
which is represented schematically by the tee circuit shown in Figure C.2. Thus the third requirement is satisfied.
Fig. C.2.
Tee circuit representation of a transformer.
An interesting point to be made here is that the requirement for equal pu mutual flux linkages is the same as equal base M M F's, S8(L ml/L I8)
=
SB(L m2/L2B)
or (Lml/LI8)(/r8LIB) = (Lm2/L28)(/~8L28) Lml/iB
or in terms of the mutual permeance (Pm
=
Lm2/~8
(e.25)
=
o; Ni/iB
(C.26)
(Pm
Nt IrB
or (C.2?) or in terms of MMF (e.28)
Appendix C C.2.1
549
Summary
The first three normalization specifications require that I. All circuits must have the same VA base (C.9). 2. All circuits must have the same time base (C.6), (C.9), and (C.IO). 3. The requirement of a common pu tee circuit means equal pu magnetizing inductance in all circuits (C.23). This requires equal pu mutual flux linkages (C. I?), which in turn requires that the base MMF be the same in all circuits (C.28). C.3
Comparison with Manufacturers' Impedances
We now select the base stator and rotor quantities to satisfy the fourth requirement, namely, to give the same pu impedances as those supplied by the manufacturers. The choice of the stator base voltage VI B and the stator base current liB determines the base stator impedance. Because of a certain awkwardness in the original Park's transformation resulting from the fact that the transformation is not power invariant, a system of stator base quantities is used by U.S. manufacturers that facilitates the choice of rotor base quantities. For this reason it is customary to use a stator base voltage equal to the peak line-to-neutral voltage and a stator base current equal to the peak line current. Such a choice, along with the requirement of equal base ampere turns (or equal pu mutuals), leads to a rotor VA base equal to the threephase stator VA base. Since the transformation used in this book is power invariant, the awkwardness referred to above is not encountered. A variety of possible stator base quantities can be chosen to satisfy the condition of having the same pu stator impedances as supplied by the manufacturers. For example, among the possible choices for the stator base: peak line-to-neutral voltage and peak line current (same as the manufacturers), rms line-to-neutral voltage and rms line current, or rms line voltage and V3 times rms line current. Note that in all these choices the base stator impedance is the same. However, the other three requirements stated in the previous sections may not be satisfied. To illustrate, it would appear that adoption of stator base quantities of rated rms
line voltage and V3 times line current would be attractive. The factor of V3 appearing in the d and q axis equations of Chapter 4 would be eliminated. Careful examination, however, would reveal that the requirement of having the same identical equation hold for the M KS and the pu systems would be violated. For example, if the phase voltage Va = V2v cos (wRl + ex), the d and q axis voltages are obtained by a relation similar to that of (4.146)
u,
=
-
V3Vsin (0 - a)
u, = V3Vcos(o - ex) V
where V = rms voltage to neutral. Choosing V's = Vdu =
vqu
-
(C.29)
v1Vl N (rated), we get
V3Vsin(o - a) . V3 = -(V/VLN)sm(~ - a) pu 3 VLN
= (V/ Vl N )
cos (0 - ex) pu
(C.30)
Note that (C.29) and (C.30) are not identical, and hence this choice of stator base quantities does not meet requirement number I.
Appendix C
550
In this book the stator base quantities selected to meet the requirements stated above are VI B
rated per phase voltarnpere, V A rated rms voltage to neutral, V
liB
rated rms line current, A
SIB
tlB
1/WR, S
(C.31)
The rotor base quantities are selected to meet the conditions of equal and F0 (or Am). Equal VA base gives
S8, 18,
(C.32)
(The subscript 2 is used to indicate any rotor circuit. The same derivation applies to a field circuit or to an amortisseur circuit.) Equal mutual flux linkages require that the mutual flux linkage in the d axis stator produced by a base stator current would be the same as the d axis stator flux linkage produced by a d axis rotor base current. Thus in M KS units, or (C.33)
where z, = kMF/L m l • From (C.32) and (C.33) we obtain for the rotor circuit base voltage V20 = Vlo/ I B /1 28 = i, Vl8
(C.34)
From (C.33) and (C.34) for the rotor resistance base R 2B
V28 / 128
=
=
k }( VI 8 / 118 )
k} R, B
=
The inductance base for the rotor circuit is then given bY( 1 ) L28
= V28tB/12B
= (kMF/Lml)2(VIB/IIB) -
WR
n
= Jcj.L 1B
(C.35)
(C.36)
The base for the mutual inductance is obtained from (C. I I ) and (C.33)
LI 2B
_
-
VIB
f ~\
VIBtB_~_ f2B
-
----
--- -
-,
(Lml/kMF)IIB
-
(C.37)
kFLIB
The pu d axis mutual inductance is then given by
kM F = kM F u L I 28
=
kM F
(kMF/Lml)LtB
= Lm l = LIB
l.Jmlu
(C.38)
Thus the value of the pu d axis mutual inductance of any rotor circuit is the same as the pu magnetizing inductance of the stator.
kM Fu = kM ou
=
M Ru = L mlu
(C.39)
A comparison between the pu system derived in this book and that used by u.s. manufacturers is given in the Table C.I. Note that the base inductances and resistances are the same in both systems.
Appendix C
551
Table c.i. Comparison of BaseQuantities Per unit system used In this book
Quantity /system
By U.S. manufacturers"
VIVl N VII L
VIB
liB
VlNlwRI L (31 2 ) VIBI IB = 3 VLNIL 0(L m 1IM F )I L (3IVI)(M F / Lml ) ~L.N (3/2)(M F 1L ml )2( VLN1I L) (3/2)(M FI L ml )2LIB
LIB SIB = S2B
l2B
V2B
R 2B
L 28 L I2B
(MF/Lml)LI8
Per unit mutual inductances 2 2 V qu + Vdu
M Fu = L mu = L A Du
3 V~
·See reference [4J.
C.4
Complete Data for Typical Machine
To complement the discussion on normalization given in this appendix, we provide a consistent set of data for a typical synchronous generator. Starting with the pu impedances supplied by the manufacturer, the base quantities are derived and all the impedance values are calculated. The machine used for this data is the 160-MVA, two-pole machine that is used in many of the text examples. The method used is that of Section 5.8 of the text. The data given and results computed are the same as in Example 5.5. Computations here are carried to about eight significant figures using a pocket "slide rule" calculator. The following data is provided by the manufacturer (this is actual data on an actual machine with data from the manufacturers bid or "guaranteed' data). Ratings: 160 MVA
0.85 PF
136MW
15 kV
(C.40)
Unsaturated reactances in pu: Xd =
x'q
1.70
=
0.380
x{ = 0.150 x:J = 0.185
x, = 1.64
Xd
=
0.245
x~' X2 Xo
= 0.185 = 0.185 = 0.100
(C.41)
Time constants in seconds: TdO
rd
= 5.9
=
0.023
r, = 0.24
345 V
lr
T~~ = 0.075
(C.42)
Excitation at rated load: VF =
= 926 A
(C.43)
Resistances in ohms at 25°C:
'a
=
0.001113
'F
= 0.2687
(C.44)
Computations are given in Example 5.5. One problem not mentioned there is that of finding the correct value of field resistance to use in the generator simulation. There
552
Appendix C
are three possibilities: I. Compute from (C.43), at operating temperature,
'F 2. Compute from (C.44)
'F =
at
= 345/926 = 0.37257 n
(C.45)
an assumed operating temperature of 125°C:
0.2687[(234.5 + 125.0)/(234.5
+. 25.0»)
=
(C.46)
0.372245 fl
3. Compute from (5.59), using L F from Table C.3
'F = LFlido = 2.189475/5.9 = 0.371097 g The value computed from Lf'/';o must be used if the correct time constant is to result. Working backward to compute the corresponding operating temperature, we have 0.2687[(234.5 + (J)/(234.5 + 25)]
=
(C.48)
0.371097
or the operating temperature is (J = 123.8C, which is a reasonable result. The base quantities for all circuits are given in Table C.2. Stator base values are derived from nameplate data for voltarnperes, voltage, and frequency. The method of relating stator to field base quantities through the constant k F is shown in Example 4.1 where we compute k;
=
kMF/L md = 109.0102349 mH/5.781800664 mH
=
18.85402857 (C.49)
Note that a key element in determining the factor k», and hence all the rotor base quantities, is the value of M F (in H). This is obtained from the air gap line of the magnetization curve provided by the manufacturer. Unfortunately, no such data is given for any of the amortisseur circuits. Thus, while the pu values of the various amortisseur elements can be determined, their corresponding MKS data are not known. Using the base values from Table C.2 and the pu values from Example 5.5, we may construct Table C.3 of d axis parameters and Table C.4 of q axis parameters. The given values are easily identified since they are written to three decimals. Table C.2. Circuit
Base quantity
Stator
So
Vo
10
IB
Ro Ao Field
LB
SFO VFO I Fs I FO R FB
AF B
L FB M FO
Base Values in MKS Units
Formula S03/ 3
VLl / v'3 1/21r60
Ss/Vo VB/I B Vo/s
Ao/Is So
So/IFs 10
10/k F
VFs/I FB VFBt O AFo/IFB
VLBL FB
Numerical value
53.333 333 333 8.660 254 036 2.652 582 384 6158.402 872 1.406 250 22.972 0373 3.730 193 98 53.333 333 333 163 280.677 2.652 582 384 326.635 915 499.885 8653 433.115 4475 1.325 988 441 0.070 329 184
Units
MVA/phase kVLN ms A
n
Wb
mH
MVA/phase V
ms A
n
Wb H H
Appendix C Table C.3.
Direct Axis Parameters in pu and MKS pu value
Symbol
Ld Ld L'd Lmd
1.700 0.245 0.185 1.550 0.150 1.651 1.550 0.101 1.605 1.550 0.055 1.265 1.550 1.265 1.550 1.550 0.028 0.791 1.096
{d
LF
LmF {F
Lo L mD
{D
MF kM F
MD
kM D
MR
L MD 'a 25°C 'a 125°C 'F 25°C 'F Hot
0.742 13.099 90.477 2224.247 320.442 11 .482 8.670
'0
To TdO Td TdO Td
Table C.4. Symbol
Lq
L'q L"q -:
.f q
LQ Lm Q
{Q
MQ kM Q
L MQ
'a 25°C 'a I25·
oC
'Q
T~O T~'o T~'
553
MKS value
202 749 202 749 416 667 416 667 5697 5697 378 3784 607 397 X 10- 3 463 455 X 10- 3 364 135 868 599 450 945 795
295 x 10- 3 90 X 10- 3 44 7 69 726
Units
6.341 329 761
mH
5.781 0.559 2.189 2.055 0.134
mH mH
800 664 529 097 475 759 282 084 193 675
H H H
0.089 006 484 O. 109 010 235
H H
1.113 1.541 90 I 734 0.2687 (not used) 0.371 097 586
roO mO
n n
0.24 5.90 0.85 0.030 459 0.023
s s s s s
Quadrature Axis Parameters in pu and M KS pu value
1.640 0.380 (not 0.185 1.490 0.150 1.525 808 1.490 0.035 808 1.216 579 1.490 0.028 357 0.791 607 1.096 463 0.053 955 203.575 204 28.274 333 3.189 482
used)
581
MKS value
Units
6.117 518 122
mH
5.557 989 025 0.559 529 097
mH mH
1.113 1.541 90 I 734
mn mn
0.54 0.075 8.460 365 85
s s ms
581 905 4715 397 X 10- 3 455 X 10- 3 165 89 785
554
Appendix C
References I. Rankin, A. W. Per unit impedances of synchronous machines. AlEE Trans. 64:509 2. Lewis, W. A. A basic analysis of synchronous machines, Pt. I. AlEE Trails. 77:436 3. Harris, M. R" Lawrenson, P. J.. and Stephenson. J. M. Per Un;1 Svstems: With Electrical Machines. lEE Monogr. Ser. 4. Cambridge Univ. Press, 1970. 4. General Electric Co. Power system stability. Electric Utility Engineering Seminar. nous Mach ines. Schenectady. N. Y.. 1973.
X41. IlJ45. 56.
19~X.
Special Reference 10
Section on Synchro-
