UNDERSTANDING POWER
QUALITY PROBLEMS
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Technical Reviewers
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Damir Novosel, ABB Power T&D Company, Inc., Raleigh, NC
Roger C. Dugan, Electrotck Concepts, Inc., Knoxville, TN
Mohamed E. El-Hawary, Dalhousie University, Halifax, Nova Scotia, Canada
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IEEE PRESS SERIES ON POWER ENGINEERING
P. M. Anderson, Series Editor
Power Math Associates, Inc.
Series Editorial Advisory Committee
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Keith B. Stump
Siemens Power Transmission and
Distribution
UNDERSTANDING
POWER QUALITY
PROBLEMS
Voltage Sags
and Interruptions
Math H. J. Bollen
Chalmers University of Technology
Gothenburg, Sweden
IEEE Industry Applications Society, Sponsor
IEEE Power Electronics Society, Sponsor
IEEE Power Engineering Society, Sponsor
IEEE.
PRESS
~II
SERIES
POWER
ENGINEERING
ON
P. M. Anderson, Series Editor
+IEEE
The Institute of Electrical and Electronics Engineers, lnc., NewYork
ffiWILEY-
~INTERSCIENCE
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Library of Congress Cataloging-in-Publication Data
Bollen, Math H. J., 1960Understanding power quality problems: voltage sags and interruptions
Math H. J. Bollen.
p. em. - (IEEE Press series on power engineering)
Includes bibliographical references and index.
IBSN 0..7803-4713-7
l. Electric power system stability. 2. Electric power failures.
3. Brownouts. 4. Electric power systems-Quality control.
I. Title. II. Series.
IN PROCESS
621.319-dc21
99-23546
CIP
The master said, to learn and at due times to repeat what one has learnt, is
that not after all a pleasure?
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BOOKS IN THE IEEE PRESS SERIES ON POWER ENGINEERING
ELECTRIC POWER APPLICATIONS OF FUZZY SYSTEMS
Edited by Mohamed E. El-Hawary, Dalhousie University
1998 Hardcover
384 pp
IEEE Order No. PC5666
ISBN 0-7803-1197-3
RATING Of' ELECTRIC POWER CABLES: Ampacity Computations/or Transmission,
Distribution, and Industrial Applications
George J. Anders, Ontario Hydro Technologies
1997 Hardcover
464 pp
IEEE Order No. PC5647
ISBN 0-7803-1177-9
ANALYSIS OF FAULTED POWER SYSTEMS, Revised Printing
P. M. Anderson, Power Math Associates, Inc.
1995 Hardcover
536 pp
IEEE Order No. PC5616
ISBN 0-7803-1145-0
ELECTRIC POWER SYSTEMS: Design and Analysis, Revised Printing
Mohamed E. El-Hawary, Dalhousie University
1995 Hardcover
808 pp
IEEE Order No. PC5606
ISBN 0-7803-1140-X
POWER SYSTEM STABILITY, Volumes I, II, III
An IEEE Press Classic Reissue Set
Edward Wilson Kimbark, Iowa State University
1995 Softcover
1008 pp
IEEE Order No. PP5600
ISBN 0-7803-1135-3
ANALYSIS OF ELECTRIC MACHINERY
Paul C. Krause and Oleg Wasynczuk, Purdue University
Scott D. Sudhoff, University of Missouri at Rolla
1994 Hardcover
480 pp
IEEE Order No. PC3789
ISBN 0-7803-1029-2
SUBSYNCHRONOUS RESONANCE IN POWER SYSTEMS
P. M. Anderson, Power Math Associates, Inc.
B. L. Agrawal, Arizona Public Service Company
J. E. Van Ness, Northwestern University
1990 Softcover
282 pp
IEEE Order No. PP2477
ISBN 0-7803-5350-1
POWER SYSTEM PROTECTION
P. M. Anderson, Power Math Associates, Inc.
1999 Hardcover
1,344 pp
IEEE Order No. PC5389
ISBN 0-7803-3427-2
POWER AND COMMUNICATION CABLES: Theory and Applications
Edited by R. Bartnikas and K. D. Srivastava
2000
Hardcover
896 pp
IEEE Order No. PC5665
ISBN 0-7803-1196-5
Contents
PREFACE
xiii
FTP SITE INFORMATION xv
ACKNOWLEDGMENTS xvii
CHAPTER 1 Overvlew of Power Quality and Power Quality
Standards 1
1.1 Interest in Power Quality 2
1.2 Power Quality, Voltage Quality 4
1.3 Overview of Power Quality Phenomena 6
1.3.1 Voltage and Current Variations 6
1.3.2 Events 14
1.3.3 Overview of Voltage Magnitude Events 19
1.4 Power Quality and EMC Standards 22
1.4.1 Purpose of Standardization 22
1.4.2 The tsc Electromagnetic Compatibility Standards 24
1.4.3 The European Voltage Characteristics Standard 29
CHAPTER 2 Long Interruptions and Reliability Evaluation 35
2.1 Introduction 35
2.1.1
2.1.2
2.1.3
2.1.4
Interruptions 35
Reliability Evaluation of Power Systems 35
Terminology 36
Causes of Long Interruptions 36
2.2 Observation of System Performance 37
2.2.1 Basic Indices 37
2.2.2 Distribution of the Duration of an Interruption 40
2.2.3 Regional Variations 42
vii
viii
Con ten ts
2.2.4 Origin of Interruptions 43
2.2.5 More Information 46
2.3 Standards and Regulations 48
2.3.1 Limits for the Interruption Frequency 48
2.3.2 Limits for the Interruption Duration 48
2.4 Overview of Reliability Evaluation 50
2.4.1
2.4.2
2.4.3
2.4.4
Generation Reliability 51
Transmission Reliability 53
Distribution Reliability 56
Industrial Power Systems 58
2.5 Basic Reliability Evaluation Techniques 62
2.5. J
2.5.2
2.5.3
2.5.4
2.5.5
2.5.6
Basic Concepts of Reliability Evaluation Techniques 62
Network Approach 69
State-Based and Event-Based Approaches 77
Markov Models 80
Monte Carlo Simulation 89
Aging of Components 98
2.6 Costs of Interruptions 101
2.7 Comparison of Observation and Reliability Evaluation 106
2.8 Example Calculations 107
2.8.1
2.8.2
2.8.3
2.8.4
A Primary Selective Supply 107
Adverse Weather 108
Parallel Components 110
Two-Component Model with Aging and Maintenance III
CHAPTER 3 Short Interruptions
115
3.1 Introduction 115
3.2 Terminology 115
3.3 Origin of Short Interruptions 116
3.3.1
3.3.2
3.3.3
3.3.4
Basic Principle 116
Fuse Saving 117
Voltage Magnitude Events due to Reclosing 118
Voltage During the Interruption 119
3.4 Monitoring of Short Interruptions 121
3.4.1 Example of Survey Results 121
3.4.2 Difference between Medium- and Low-Voltage Systems 123
3.4.3 Multiple Events 124
3.5 Influence on Equipment 125
3.5.1
3.5.2
3.5.3
3.5.4
Induction Motors 126
Synchronous Motors 126
Adjustable-Speed Drives 126
Electronic Equipment 127
3.6 Single-Phase Tripping 127
3.6.1 Voltage-During-Fault Period 127
3.6.2 Voltage-Post-Fault Period 129
3.6.3 Current-During-Fault Period 134
3.7 Stochastic Prediction of Short Interruptions 136
Contents
ix
CHAPTER 4 Voltage Sags-Characterization 139
4.1 Introduction 139
4.2 Voltage Sag Magnitude 140
4.2.1 Monitoring 140
4.2.2 Theoretical Calculations 147
4.2.3 Example of Calculation of Sag Magnitude 153
4.2.4 Sag Magnitude in Non-Radial Systems 156
4.2.5 Voltage Calculations in Meshed Systems 166
4.3 Voltage Sag Duration 168
4.3.1 Fault-Clearing Time 168
4.3.2 Magnitude-Duration Plots 169
4.3.3 Measurement of Sag Duration 170
4.4 Three-Phase Unbalance 174
4.4.1 Single-Phase Faults 174
4.4.2 Phase-to-Phase Faults 182
4.4.3 Two-Phase-to-Ground Faults 184
4.4.4 Seven Types of Three-Phase Unbalanced Sags 187
4.5 Phase-Angle Jumps 198
4.5.1 Monitoring 199
4.5.2 Theoretical Calculations 201
4.6 Magnitude and Phase-Angle Jumps for Three-Phase Unbalanced
Sags 206
4.6.1 Definition of Magnitude and Phase-Angle Jump 206
4.6.2 Phase-to-Phase Faults 209
4.6.3 Single-Phase Faults 216
4.6.4 Two-Phase-to-Ground Faults 222
4.6.5 High-Impedance Faults 227
4.6.6 Meshed Systems 230
4.7 Other Characteristics of Voltage Sags 231
4.7.1 Point-on-Wave Characteristics 231
4.7.2 The Missing Voltage 234
4.8 Load Influence on Voltage Sags 238
4.8.1 Induction Motors and Three-Phase Faults 238
4.8.2 Induction Motors and Unbalanced Faults 24 t
4.8.3 Power Electronics Load 248
4.9 Sags due to Starting of Induction Motors 248
CHAPTER S Voltage Sags-Equipment Behavior 253
5.1 Introduction 253
5.1.1 Voltage Tolerance and Voltage-Tolerance Curves 253
5.1.2 Voltage-Tolerance Tests 255
5.2 Computers and Consumer Electronics 256
5.2.1 Typical Configuration of Power Supply 257
5.2.2 Estimation of Computer Voltage Tolerance 257
5.2.3 Measurements of PC Voltage Tolerance 261
5.2.4 Voltage-Tolerance Requirements: CBEMA and ITIC 263
5.2.5 Process Control Equipment 264
5.3 Adjustable-Speed AC Drives 265
5.3.1 Operation of AC Drives 266
5.3.2 Results of Drive Testing 267
5.3.3 Balanced Sags 272
x
Con~nh
5.3.4
5.3.5
5.3.6
5.3.7
5.3.8
5.3.9
DC Voltage for Three-Phase Unbalanced Sags 274
Current Unbalance 285
Unbalanced Motor Voltages 289
Motor Deacceleration 292
Automatic Restart 296
Overview of Mitigation Methods for AC Drives 298
5.4 Adjustable-Speed DC Drives 300
5.4.1
5.4.2
5.4.3
5.4.4
5.4.5
5.4.6
Operation of DC Drives 300
Balanced Sags 303
Unbalanced Sags 308
Phase-Angle Jumps 312
Commutation Failures 315
Overview of Mitigation Methods for DC Drives 317
5.5 Other Sensitive Load 318
5.5.1
5.5.2
5.5.3
5.5.4
Directly Fed Induction Motors 318
Directly Fed Synchronous Motors 319
Contactors 321
Lighting 322
CHAPTER 6 Voltage Sags-Stochastic Assessment 325
6.1 Compatibility between Equipment and Supply 325
6.2 Presentation of Results: Voltage Sag Coordination Chart 328
6.2.1
6.2.2
6.2.3
6.2.4
6.2.5
6.2.6
6.2.7
The Scatter Diagram 328
The Sag Density Table 330
The Cumulative Table 331
The Voltage Sag Coordination Chart" 332
Example of the Use of the Voltage Sag Coordination Chart 335
Non-Rectangular Sags 336
Other Sag Characteristics 338
6.3 Power Quality Monitoring 342
6.3.,1 Power Quality Surveys 342
6.3.2 Individual Sites 357
6.4 The Method of Fault Positions 359
6.4.1
6.4.2
6.4.3
6.4.4
Stochastic Prediction Methods 359
Basics of the Method of Fault Positions 360
Choosing the Fault Positions 362
An Example of the Method of Fault Positions 366
6.5 The Method of Critical Distances 373
6.5.1
6.5.2
6.5.3
6.5.4
6.5.5
6.5.6
6.5.7
6.5.8
6.5.9
Basic Theory 373
Example-Three-Phase Faults 374
Basic Theory: More Accurate Expressions 375
An Intermediate Expression 376
Three-Phase Unbalance 378
Generator Stations 384
Phase-Angle Jumps 384
Parallel Feeders 385
Comparison with the Method of Fault Positions 387
Contents
xi
CHAPTER 7 Mitigation of Interruptions and Voltage Sags
389
7.1 Overview of Mitigation Methods 389
7.1.1
7.1.2
7.1.3
7.1.4
7.1.5
7.1.6
7.1.7
From Fault to Trip 389
Reducing the Number of Faults 390
Reducing the Fault-Clearing Time 391
Changing the Power System 393
Installing Mitigation Equipment 394
Improving Equipment Immunity 395
Different Events and Mitigation Methods 395
7.2 Power System Design-Redundancy Through Switching 397
7.2.1
7.2.2
7.2.3
7.2.4
Types of Redundancy 397
Automatic Reclosing 398
Normally Open Points 398
Load Transfer 400
7.3 Power System Design-Redundancy through Parallel
Operation 405
7.3.1 Parallel and Loop Systems 405
7.3.2 Spot Networks 409
7.3.3 Power-System Design-On-site Generation 415
7.4 The System-Equipment Interface 419
7.4.1
7.4.2
7.4.3
7.4.4
7.4.5
7.4.6
7.4.7
7.4.8
Voltage-Source Converter 419
Series Voltage Controllers-DVR 420
Shunt Voltage Controllers-StatCom 430
Combined Shunt and Series Controllers 435
Backup Power Source-SMES, BESS 438
Cascade Connected Voltage Controllers-UPS 439
Other Solutions 442
Energy Storage 446
CHAPTER 8 Summary and Conclusions 453
8.1 Power Quality 453
8.1.1 The Future of Power Quality 454
8.1.2 Education 454
8.1.3 Measurement Data 454
8.2 Standardization 455
8.2.1 Future Developments 455
8.2.2 Bilateral Contracts 456
8.3 Interruptions 456
8.3.1 Publication of Interruption Data 456
8.4 Reliability 457
8.4.1 Verification 457
8.4.2 Theoretical Developments 457
8.5 Characteristics of Voltage Sags 458
8.5.1 Definition and Implementation of Sag Characteristics 458
8.5.2 Load Influence 458
8.6 Equipment Behavior due to Voltage Sags 459
8.6.1 Equipment Testing 459
8.6.2 Improvement of Equipment 460
8.7 Stochastic Assessment of Voltage Sags 460
8.7.1 Other Sag Characteristics 460
8.7.2 Stochastic Prediction Techniques 460
xii
Contents
8.7.3 Power Quality Surveys 461
8.7.4 Monitoring or Prediction? 461
8.8 Mitigation Methods 462
8.9 Final Remarks 462
BIBLIOGRAPHY
465
APPENDIX A Overview of EMC Standards 477
APPENDIX B IEEE Standards on Power Quality
481
APPENDIX C Power Quality Definitions and Terminology
APPENDIX D List of Figures
APPENDIX E List of Tables
INDEX
529
ABOUT THE AUTHOR
543
507
525
485
Preface
The aims of the electric power system can be summarized as "to transport electrical
energy from the generator units to the terminals of electrical equipment" and "to
maintain the voltage at the equipment terminals within certain limits." For decades
research and education have been concentrated on the first aim. Reliability and quality
of supply were rarely an issue, the argument being that the reliability was sooner too
high than too low. A change in attitude came about probably sometime in the early
1980s. Starting in industrial and commercial power systems and spreading to the public
supply, the power quality virus appeared. It became clear that equipment regularly
experienced spurious trips due to voltage disturbances, but also that equipment was
responsible for many voltage and current disturbances. A more customer-friendly definition of reliability was that the power supply turned out to be much less reliable than
always thought. Although the hectic years of power quality pioneering appear to be
over, the subject continues to attract lots of attention. This is certain to continue into
the future, as customers' demands have become an important issue in the deregulation
of the electricity industry.
This book concentrates on the power quality phenomena that primarily affect the
customer: interruptions and voltage sags. During an interruption the voltage is completely zero, which is probably the worst quality of supply one can consider. During a
voltage sag the voltage is not zero, but is still significantly less than during normal
operation. Voltage sags and interruptions account for the vast majority of unwanted
equipment trips.
The material contained in the forthcoming chapters was developed by the author
during a to-year period at four different universities: Eindhoven, Curacao, Manchester,
and Gothenburg. I Large parts of the material were originally used for postgraduate and
industrial lectures both "at home" and in various places around the world. The material
will certainly be used again for this purpose (by the author and hopefully also by
others).
'Eindhoven University of Technology, University of the Netherlands Antilles, University of
Manchester Institute of Science and Technology, and Chalmers University of Technology, respectively.
xiii
xiv
Preface
Chapter 1 of this book gives an introduction to the subject. After a systematic
overview of power quality, the term "voltage magnitude event" is introduced. Both
voltage sags and interruptions are examples of voltage magnitude events. The second
part of Chapter 1 discusses power quality standards, with emphasis on the IEC
standards on electromagnetic compatibility and the European voltage characteristics
standard (EN 50160).
In Chapter 2 the most severe power quality event is discussed: the (long) interruption. Various ways are presented of showing the results of monitoring the number of
interruptions. A large part of Chapter 2 is dedicated to the stochastic prediction of long
interruptions-v-an area better known as "reliability evaluation." Many of the techniques described here can be applied equally well to the stochastic prediction of other
power quality events.
Chapter 3 discusses short interruptions-interruptions terminated by an automatic restoration of the supply. Origin, monitoring, mitigation, effect on equipment,
and stochastic prediction are all treated in this chapter.
Chapter 4 is the first of three chapters on voltage sags. It treats voltage sags in a
descriptive way: how they can be characterized and how the characteristics may be
obtained through measurements and calculations. Emphasis in this chapter is on magnitude and phase-angle jump of sags, as experienced by single-phase equipment and as
experienced by three-phase equipment.
Chapter 5 discusses the effect of voltage sags on equipment. The main types of
sensitive equipment are discussed in detail: single-phase rectifiers (computers, processcontrol equipment, consumer electronics), three-phase ac adjustable-speed drives, and
de drives. Some other types of equipment are briefly discussed. The sag characteristics
introduced in Chapter 4 are used to describe equipment behavior in Chapter 5.
In Chapter 6 the theory developed in Chapters 4 and 5 is combined with statistical
and stochastical methods as described in Chapter 2. Chapter 6 starts with ways of
presenting the voltage-sag performance of the supply and comparing it with equipment
performance. The chapter continues with two ways of obtaining information about the
supply performance: power-quality monitoring and stochastic prediction. Both are
discussed in detail.
Chapter 7, the last main chapter of this book, gives an overview of methods for
mitigation of voltage sags and interruptions. Two methods are discussed in detail:
power system design and power-electronic controllers at the equipment-system interface. The chapter concludes with a comparison of the various energy-storage techniques
available.
In Chapter 8 the author summarizes the conclusions from the previous chapters
and gives some of his expectations and hopes for the future. The book concludes with
three appendixes: Appendix A and Appendix B give a list of EMC and power quality
standards published by the IEC and the IEEE, respectively. Appendix C contains
definitions for the terminology used in this book as well as definitions from various
standard documents.
Math H. J. Bollen
Gothenburg, Sweden
FTP Site Information
Along with the publication of this book, an FTP site has been created containing
MATLAB® files for many figures in this book. The FTP site can be reached at
ftp.ieee.orgjupload/press/bollen.
xv
Acknowledgments
A book is rarely the product of only one person, and this book is absolutely no exception. Various people contributed to the final product, but first of all I would like to
thank my wife, Irene Gu, for encouraging me to start writing and for filling up my tea
cup every time I had another one of those "occasional but all too frequent crises."
For the knowledge described in this book lowe a lot to my teachers, my colleagues, and my students in Eindhoven, Curacao, Manchester, and Gothenburg and to my
colleagues and friends all over the world. A small number of them need to be especially
mentioned: Matthijs Weenink, Wit van den Heuvel, and Wim Kersten for teaching me
the profession; the two Larry's (Conrad and Morgan) for providing me with a continuous stream of information on power quality; Wang Ping, Stefan Johansson, and the
anonymous reviewers for proofreading the manuscript. A final thank you goes to
everybody who provided data, figures, and permission to reproduce material from
other sources.
Math H. J. Bollen
Gothenburg, Sweden
xvii
Voor mijn ouders
Overview of Power Qual ity
and Power Qual ity Standards
Everybody does not agree with the use of the term
powerquality, but they do agree
t hat
it has becomeaveryimportantaspect of power delivery especially in the second half of
the 1990s.There is a lotof disagreementa boutwhat power quality actually incorporates; it looks as if everyone has her or his own
interpretation.In this chaptervarious
ideas will be summarized to clear up some of the confusion. However,author
the
himself is part of the power quality world; thuspart of the confusion. After reading
this book the reader might want to go to the library and form his own picture. The
number of books onpower quality is still rather limited. The book "Electric Power
SystemsQuality" by Dugan et al. [75] gives a useful overviewof the various power
quality phenomenaand the recent developments in this field. There are two more books
with the term power quality in the title:
"Electric Power QualityControl Techniques"
[76] and "Electric PowerQuality" [77]. But despite the general title, reference
[76]
mainly concentrateson transientovervoltage and[77] mainly on harmonicdistortion.
But both books docontainsomeintroductorychapters on power quality. Also many
recent books on electric power systems
containone or more general
chapterson power
quality, for example,[114], [115], and [116]. Information on power qualitycannotbe
found only in books; a large
numberof papers have been written on the subject; overview papers as well as technical papers
aboutsmall detailsof power quality. The main
journals to look for technical papers are the IEEE
Transactionson Industry
Applications, the IEEE Transactionson Power Delivery andlEE ProceedingsGeneration,Transmission,Distribution. Other technicaljournals in the power engineering field alsocontainpapers of relevance. A
journal specially dedicated to power
quality is Power Quality Assurance. Overview articles can be found in many different
journals;two early ones are[104] and [105].
Various sources use the term
"power quality" with different meanings.Other
sources use similar but slightly different terminology like
"quality of power supply"
or "voltage quality." What all these terms have in common that
is they treat the
interaction between the utility and the customer, or in technical terms between the
power system and the load.
Treatmentof this interaction is in itself not new. The
aim of the power system has always been to supply electrical energy to the customers.
1
2
ChapterI •
Overview of PowerQuality and PowerQuality Standards
What is new is theemphasisthat is placedon this interaction,and the treatmentof it as
a separateareaof power engineering.In Section 1.2 the various termsand interpretations will be discussedin moredetail. From the discussionwe will concludethat "power
quality" is still the most suitableterm. The various power quality phenomenawill be
discussedandgroupedin Section1.3. Electromagneticcompatibility and powerquality
standardswill be treatedin detail in Section 1.4. But first Section 1.1 will give some
explanationsfor the increasedinterestin power quality.
1.1 INTEREST IN POWER QUALITY
The fact that powerquality hasbecomean issuerecently,doesnot meanthat it was not
important in the past. Utilities all over the world have for decadesworked on the
improvementof what is now known as power quality. And actually, even the term
has been in use for arather long time already. The oldest mentioning of the term
"power quality" known to the author was in a paper published in 1968 [95]. The
paper detailed a study by the U.S. Navy after specificationsfor the power required
by electronicequipment.That papergives a remarkablygood overview of the power
quality field, including the useof monitoringequipmentandeven thesuggesteduseof a
static transferswitch. Severalpublicationsappearedsoon after, which used theterm
power quality in relation to airborne power systems[96], [97], [98]. Already in 1970
"high powerquality" is beingmentionedas oneof the aimsof industrial powersystem
design,togetherwith "safety," "reliable service,"and "low initial and operatingcosts"
[99]. At about the sametime the term "voltage quality" was used in theScandinavian
countries[100], [101] and in the Soviet Union [102], mainly with referenceto slow
variationsin the voltage magnitude.
The recent increasedinterestin power quality can be explainedin a numberof
ways. The main explanationsgiven aresummarizedbelow. Of courseit is hard to say
which of these came first; some explanationsfor the interestin power quality given
o f the increasedinterestin power
below.. will by othersbe classified asconsequences
quality. To showthe increasedintereston powerquality a comparisonwasmadefor the
numberof publicationsin the INSPECdatabase[118] using theterms"voltagequality"
or "power quality." For the period 1969-1984the INSPEC databasecontains 91
records containing the term "power quality" and 64 containing the term "voltage
quality." The period 1985-1996resulted in 2051 and 210 records, respectively.We
see thus a large increasein number of publicationson this subjectsand also a shift
away from the term "voltage quality" toward the term "power quality."
• Equipment has become more sensitive to voltage
disturbances.
Electronic and power electronicequipmenthas especiallybecomemuch
more sensitivethan its counterparts10 or 20 years ago.T he paperoften cited
as having introduced the term power quality (by Thomas Key in 1978 [I])
treatedthis increasedsensitivity to voltage disturbances.N ot only has equipment becomemore sensitive,companieshave alsobecomemore sensitiveto
loss of productiontime due to their reducedprofit margins.On the domestic
market, electricity is more and more considereda basic right, which should
is that an interruptionof the supply
simply alwaysbe present.Theconsequence
will muchmorethan beforelead tocomplaints,even if thereare nodamagesor
costsrelatedto it. An importantpapertriggering the interestin powerquality
appearedin the journal BusinessWeek in 1991 [103].The article cited Jane
Section 1.1 • Interestin Power Quality
3
Clemmensenof EPRI as estimating that "power-relatedproblems cost U.S.
companies$26 billion a year in lost time and revenue."This value has been
cited overandoveragain eventhoughit was mostlikely only a roughestimate.
• Equipment causes voltage disturbances.
Tripping of equipmentdue to disturbancesin the supply voltageis often
describedby customersas "bad power quality." Utilities on the other side,
often view disturbancesdue to end-userequipmentas themain power quality
problem.Modern(power) electronicequipmentis not only sensitive tovoltage
disturbances,it also causesdisturbancesfor othercustomers.The increaseduse
of converter-drivenequipment(from consumerelectronicsand computers,up
to adjustable-speed
drives) has led to a large
g rowth of voltagedisturbances,
althoughfortunatelynot yet to a level wheree quipmentbecomes sensitive. The
main issue here is thenonsinusoidalcurrent of rectifiers and inverters. The
input current not only contains a power frequency component(50 Hz or
60 Hz) but also so-calledharmoniccomponentswith frequenciesequal to a
multiple of the power frequency. Theharmonicdistortion of the currentleads
to harmoniccomponentsin the supply voltage. Equipmenthas alreadyproduced harmonicdistortion for a numberof decades. But only recently has the
amountof load fed via powerelectronicconvertersincreased enormously:
not
only large adjustable-speed
drives but also smallconsumerelectronicsequipment. The latter cause a largepart of the harmonicvoltage distortion: each
individual device does notgeneratemuch harmoniccurrentsbut all of them
togethercause a serious
d istortion of the supply voltage.
• A growing need forstandardizationand performancecriteria.
The consumerof electrical energy used to be viewed by
most utitilies
simply as a"load." Interruptionsand other voltage disturbanceswere part
of the deal, and the utility decided
w hat was reasonable.Any customerwho
was not satisfied with the offered
reliability and quality had to pay theutility
for improving the supply.
Todaythe utilities have totreat the consumersas"customers."Even if the
n umberof voltagedisturbances,it does have
utility does not need to reduce the
to quantify them one 'way or theother. Electricity is viewed as aproductwith
certain characteristics,which have to bemeasured,predicted, guaranteed,
improved, etc. This is further triggered by the drive towards privatization
and deregulationof the electricity industry.
Opencompetitioncan make the situationeven more complicated.In the
past a consumerwould have acontract with the local supplier who would
deliver the electrical energyw ith a given reliability and quality. Nowadays
the customercan buy electrical energysomewhere,the transport capacity
somewhereelse and pay the local utility, for the actual connectionto the
system. It is nolongerclear who isresponsiblefor reliability andpowerquality.
As long as thecustomerstill has aconnectionagreementwith the local utility,
one canarguethat the latter is responsiblefor the actualdelivery and thus for
reliability andquality. But what aboutvoltagesags due totransmissionsystem
faults? In some cases the
consumeronly has acontractwith a supplier who
only generatesthe electricityand subcontractstransportand distribution. One
could statethat any responsibilityshould be defined bycontract,so that the
generationcompany with which the customerhas a contractualagreement
would be responsiblefor reliability and quality. The responsibility of the
4
Chapter1 • Overview of PowerQuality and PowerQuality Standards
local distributionwould only betowardsthe generationcompanieswith whom
they have acontractto deliver to givencustomers.No matter what the legal
constructionis, reliability and quality will need to be well defined.
• Utilities want to deliver a good product.
Somethingthatis oftenforgottenin the heatof the discussion isthatmany
power quality developmentsare driven by the utilities.M ost utilities simply
want to deliver a goodproduct, and have beencommittedto that for many
decades.Designinga system with a high reliabilityof supply, for a limited cost,
is a technicalchallengewhich appealedto many in thepower industry, and
hopefully still does in the future.
• The power supply has become too good.
Part of the interestin phenomenalike voltage sagsand harmonicdistortion is due to the highquality of the supply voltage. Long interruptionshave
become rare inmost industrializedcountries(Europe, North America, East
Asia), and theconsumerhas, wrongly,gottenthe impressionthat electricity is
somethingthat is alwaysavailableandalwaysof high quality, or at least something that shouldalways be. The factthat there are someimperfectionsin the
supplywhich are veryhard or evenimpossibleto eliminateis easilyforgotten.
In countrieswhere theelectricity supply has a highunavailability, like 2 hours
in
per day, power quality does not appearto be such a big issue as countries
with availabilitieswell over 99.9°~.
• The power quality can be measured.
The availability of electronicdevices tomeasureandshow waveformshas
certainly contributedto the interestin power quality. Harmoniccurrentsand
voltage sags were simplyhard to measureon a large scale in the past.
Measurementswere restrictedto rms voltage, frequency,a nd long interruptions; phenomenawhich are nowconsideredpart of power quality, but were
simply part of power systemoperationin the past.
1.2 POWER QUALITY, VOLTAQE QUALITY
Therehave been(andwill be) a lot of argumentsaboutwhich term to use for theu tilitycustomer (system-load) interactions. Most people use the term"power quality"
although this term is still prone to criticism. The main objection againstthe useof
the term isthat one cannottalk about the quality of a physicalquantity like power.
Despitethe objectionswe will use the term powerquality here, eventhoughit does not
give aperfectdescriptionof the phenomenon.But it has become a widely used term and
it is the best termavailableat themoment.Within the IEEE, the termpowerquality has
gained some officialstatus already, e.g., through the name of see22 (Standards
CoordinatingCommittee):"PowerQuality" [140]. But theinternationalstandardssetting organizationin electrical engineering(the lEe) does not yet usethe term power
quality in any of its standarddocuments.Instead it uses the termelectromagnetic
compatibility, which is not the same aspower quality but there is astrong overlap
between the two terms. Below, numberof
a
different terms will be discussed. As each
term has itslimitations the author feels that power quality remainsthe more general
term which covers all theotherterms. But, beforethat, it is worth to give the following
IEEE and lEe definitions.
Section 1.2 • PowerQuality, Voltage Quality
5
The definition of power quality given in theIEEE dictionary [119] originatesin
IEEE Std 1100(betterknown as theEmeraldBook) [78]: Powerquality is theconceptof
poweringandgroundingsensitiveequipmentin a matter that issuitableto theoperationof
thatequipment.Despitethis definition the term powerquality is clearly used in a more
general waywithin the IEEE: e.g., SCC 22 also covers
standardson harmonicpollution
caused byloads.
The following definition is given in IEC 61000-1-1:Electromagneticcompatibility
is the abilityof an equipmentor system to function
satisfactorilyin its electromagnetic
environmentwithoutintroducing intolerable electromagneticdisturbancesto anythingin
that environment[79].
Recentlythe lEe has alsostarteda project group on power quality [106] which
should initially result in a standardon measurementof power quality. The following
definition of powerquality was adoptedfor describingthe scopeof the project group:
Setofparametersdefining thepropertiesof thepowersupply asdeliveredto the user in
normaloperatingconditionsin termsofcontinuityofsupplyandcharacteristicsofvoltage
(symmetry,frequency,magnitude,waveform).
Obviously,this definition will not stopthe discussionaboutwhat powerquality is.
The author'simpressionis that it will only increase theconfusion,e.g., becausepower
quality is now suddenlylimited to "normal operatingconditions."
From the many publications on this subject and the various terms used, the
following terminology has beenextracted.The readershould realize that there is no
generalconsensuson the useof these terms.
• Voltage quality (the FrenchQualite de latension)is concernedwith deviations
of the voltagefrom the ideal. The idealvoltageis a single-frequencysine wave
of constantfrequencyand constantmagnitude.The limitation of this term is
that it only covers technical aspects, andthat even within those technical
aspectsit neglects thecurrentdistortions.The termvoltagequality is regularly
used, especially inEuropeanpublications.It can beinterpretedas thequality of
the productdelivered by the utility to thecustomers.
• A complementarydefinition would becurrentquality. Currentquality is concernedwith deviationsof the currentfrom the ideal. The idealcurrentis again
a single-frequencysine waveof constantfrequency and magnitude.An additional requirementis that this sine wave is inphasewith the supply voltage.
Thus where voltage quality has to do with what the utility delivers to the
consumer,current quality is concernedwith what the consumertakes from
the utility. Of coursevoltage and current are strongly related and if either
voltageor currentdeviates from the ideal it is
h ard for the other to be ideal.
• Power quality is thecombinationof voltagequality and currentquality. Thus
powerquality is concernedwith deviationsof voltageand/orcurrentfrom the
ideal. Note that powerquality hasnothingto do with deviationsof the product
of voltageand current (the power) from any ideal shape.
• Quality of supplyor quality of powersupply includes atechnicalpart (voltage
quality above)plus a nontechnicalpart sometimesreferredto as "quality of
service."The lattercovers theinteractionbetween thecustomerand the utility,
e.g., the speed with which the
utility reacts tocomplaints,or the transparency
of the tariff structure.This could be a usefuldefinition as long as one does not
want to include the customer'sresponsibilities.The word "supply" clearly
excludes activeinvolvementof the customer.
6
ChapterI • Overview of PowerQuality and PowerQuality Standards
• Quality of consumption would be the
complementaryterm of quality of supply.
This would containthe currentquality plus, e.g., howaccuratethe customeris
in paying the electricity bill.
• In the lEe standardsthe term electromagnetic compatibility
(EMC) is used.
Electromagneticcompatibility has to do with mutual interaction between
equipmentand with interaction betweenequipmentand supply.Within electromagneticcompatibility, two importantterms are used: the "emission" is the
electromagneticpollution producedby a device; the"immunity" is the device's
ability to withstandelectromagneticpollution. Emission is related to the term
currentquality, immunity to the term voltage quality. Based on this term, a
growing setof standardsis being developedby the lEe. The variousaspectsof
electromagneticcompatibility and EMC standardswill be discussed in Section
1.4.2.
1.3 OVERVIEW OF POWER QUALITY PHENOMENA
We saw in theprevioussectionthat power quality isconcernedwith deviationsof the
voltage from its ideal waveform (voltage
quality) and deviationsof the currentfrom its
ideal waveform(currentquality). Such adeviationis called a"power quality phenomenon"or a "powerquality disturbance."Powerquality phenomenacan be divided into
two types, which need to be
treatedin a different way.
factor) is never
• A characteristicof voltage orcurrent(e.g., frequency or power
exactly equal to itsnominal or desired value. The small
deviationsfrom the
nominal or desired value are called
"voltage variations" or "current variations." A property of any variation is that it has a value at anymomentin
time: e.g., the frequency is never exactly equal to 50 Hz or 60 Hz; the power
factor is never exactly unity.Monitoring of a variation thus has totake place
continuously.
• Occasionallythe voltage orcurrent deviates significantly from itsnormal or
ideal waveshape. These
suddendeviationsare called"events."Examples are a
suddendrop to zero of the voltage due to the
operationof a circuit breaker(a
voltage event), and a heavily
distortedovercurrentdue to switching of a nonloadedtransformer(a currentevent).Monitoring of events takes place by using
a triggering mechanismwhere recordingof voltage and/or current startsthe
momenta thresholdis exceeded.
The classification of aphenomenonin one of these two types isn ot always unique. It
may dependon the kind of problemdue to thephenomenon.
1.3.1 Voltage and Current Variations
Voltage andcurrentvariationsare relatively smalldeviationsof voltage orcurrent
characteristicsa roundtheir nominalor ideal values. The two basic examples are voltage
magnitudeand frequency. On average, voltage
magnitudeand voltage frequency are
equal to theirnominal value, but they are never exactly equal. To describe the deviations in a statisticalway, the probability density or probability distribution function
should be used. Figure1.1 shows a fictitiousvariation of the voltagemagnitudeas a
function of time. This figure is the result
of a so-calledMonte Carlo simulation(see
7
Section 1.3 • Overviewof Power QualityPhenomena
240,.----.---...,----.-~---,---,
220' -0
Figure 1.1 Simulatedvoltage magnitudeas a
function of time.
-
..L---
5
-
-L..-
-
--'--
-
--'-
10
15
Time in hours
-
-'
20
Section2.5.5) .The underlyingdistribution was anormal distribution with an expected
value of 230 V and a standarddeviation of 11.9 V. A setof independents amplesfrom
this distribution is filtered by alow-passfilter to preventtoo large short-timechanges.
The probability density function of the voltage magnitudeis shown in Fig. 1.2. The
probability densityfunction gives theprobability that the voltagemagnitudeis within a
certainrange.Of interestis mainly the probability that the voltagemagnitudeis below
or above a certain value. The probability distribution function (the integral of the
density function) gives that information directly. The probability distribution function
for this fictitious variation is shown in Fig . 1.3. Both the probability density function
and the probability distribution function will be defined more accuratelyin Section
2.5.1.
An overviewof voltageandcurrentvariationsis given below. This list is certainly
not complete,it merely aims at giving someexample. There is an enormousrangein
end-userequipment.many with special requirementsand special problems. In the
power quality field new typesof variationsand eventsappearregularly. The following
list usesneither the terms used by thelEe nor the terms recommendedby the IEEE.
. Also is there still
Terms commonly used donot always fully describea phenomenon
0.12,.--------,----- ,-
-
-----.-- ---,
0.1
.~ 0.08
.g
g
0.06
~
or>
£ 0.04
0.02
o
~
Figure 1.2 Probabilitydensityfunct ion of the
voltage magnitudein Fig . 1.1.
220
___'
225
__L
230
Voltage in volts
_L
235
__'
240
8
Chapter I • Overview of PowerQuality and PowerQuality Standards
0.8
5
I:a
U')
0.6
.~
] 0.4
.s
£
0.2
o
...-:=="--_ _
...
220
225
-..1-
230
Voltagein volts
--'-
235
---'
240
Figure 1.3 Probability distribution function
of the voltage magnitude in Fig.
1.1.
some inconsistencybetweendifferent documentsabout which terms should be used.
The termsused in the list below,a ndin a similar list in Section1.3.2arenot meantas an
alternativefor the lEe or IEEE definitions, but simply an attemptto somewhatclarify
the situation.The readeris advisedto continueusing officially recognizedterms,where
feasible.
1. Voltage magnitudevariation. Increaseand decreaseof the voltage magnitude,
e.g., due to
• variation of the total load of a distribution systemor part of it;
• actionsof transformertap-changers;
• switching of capacitorbanksor reactors.
Transformertap-changera ctionsand switching of capacitorbankscan normally
be traced back to load variations as well. Thus the voltage magnitudevariationsare
mainly due to load variations, which follow a daily pattern. The influence of tapchangersand capacitorbanks makes that the daily pattern is not always presentin
the voltage magnitudepattern.
The lEe uses theterm "voltage variation" insteadof "voltage magnitudevariation." The IEEE does not appearto give a nameto this phenomenon.Very fast variation of the voltagemagnitudeis referred to as voltagefluctuation.
2. Voltage frequencyvariation. Like the magnitude,also the frequency of the
supplyvoltageis not constant.Voltagefrequencyvariationis due tounbalancebetween
load and generation.The term "frequency deviation" is also used.Short-duration
frequency transientsdue to short circuits and failure of generatorstationsare often
also included in voltagefrequencyvariations,althoughthey would betterbe described
as events.
The lEe uses theterm "power frequency variation"; the IEEE uses theterm
"frequencyvariation."
3. Currentmagnitudevariation. On the load side, thecurrentis normally also not
constantin magnitude.The variationin voltagemagnitudeis mainly due tovariationin
current magnitude.The variation in currentmagnitudeplays animportantrole in the
design of power distribution systems.The systemhas to bedesignedfor the maximum
Section 1.3 • Overviewof PowerQuality Phenomena
9
current,where the revenueo f the utility is mainly based onaveragecurrent.The more
constantthe current,the cheaperthe system per delivered energy unit.
Neither lEe nor IEEE give a name for thisphenomenon.
4. Currentphasevariation.Ideally, voltageand currentwaveformsare in phase. In
thatcase thepowerfactor of the loadequalsunity, and the reactivepowerconsumption
is zero.Thatsituationenablesthe most efficientt ransportof (active) powerandthusthe
cheapestd istribution system.
Neither lEe nor IEEE give a name for thispowerquality phenomenon,a lthough
the terms"power factor" and "reactivepower" describe itequally well.
5. Voltage andcurrent unbalance.Unbalance,or three-phaseunbalance,is the
phenomenonin a three-phasesystem, in which the nils values
of the voltagesor the
phase anglesbetweenconsecutivephasesare not equal. The severityof the voltage
unbalancein a three-phasesystem can be expressed innumberof
a
ways, e.g.,
• the ratio of the negative-sequence
and thepositive-sequencevoltage component;
and the lowestvoltage magni• the ratio of the difference between the highest
tude, and the averageof the threevoltagemagnitudes;and
• the difference betweenthe largest and the smallestphasedifference between
consecutivephases.
Thesethree severity indicatorscan bereferred to as "negative-sequence
u nbalance,"
"magnitudeunbalance,"and "phaseunbalance,"respectively.
The primary source of voltage unbalanceis unbalancedload (thus current
unbalance).T his can be due to anunevenspreadof (single-phase)low-voltagecustomers over thethreephases,b ut morecommonlyunbalanceis due to a largesingle-phase
load. Examplesof the latter can befound among railway traction suppliesand arc
furnaces. Three-phasevoltage unbalancecan also be the resulto f capacitor bank
anomalies,such as a blown fuse in one
phaseof a three-phasebank.
Voltageunbalanceis mainly of concernfor three-phaseloads.Unbalanceleads to
additionalheatproductionin the winding of inductionandsynchronousmachines;this
reduces the efficiency
a nd requiresderatingof the machine.A three-phasediode rectifier will experience a largecurrent unbalancedue to a smallvoltage unbalance.The
largestcurrentis in the phase with the highest voltage, thus the load hastendencyto
the
mitigate the voltageunbalance.
The IEEE mainly recommendsthe term "voltage unbalance"although some
standards(notably IEEE Std. 1159) use the term
"voltage imbalance."
6. Voltage fluctuation.If the voltage magnitudevaries, thepower flow to equipment will normally also vary. If thevariationsare largeenoughor in a certaincritical
frequencyrange, theperformanceof equipmentcan be affected. Cases in which
voltage
variation affects load behavior are rare, with theexception of lighting load. If the
illumination of a lamp varies withfrequenciesbetweenabout 1 Hz and 10 Hz, our
eyes are very sensitive to andabovea
it
certainmagnitudethe resultinglight flicker can
become rather disturbing. It is this sensitivity of the human eye which explains the
interestin this phenomenon.The fastvariation in voltagemagnitudeis called "voltage
fluctuation," the visualphenomenonas perceived byour brain is called "light flicker."
The term"voltageflicker" is confusingbut sometimesused as ashorteningfor "voltage
fluctuation leadingto light flicker."
10
Chapter1 • Overview of PowerQuality and PowerQuality Standards
To quantify voltagefluctuation and light flicker, aquantity called "flicker intensity" has beenintroduced[81]. Its value is an objectivemeasureof the severityof the
light flicker due to acertainvoltage'fluctuation.The flicker intensitycan betreatedas a
variation,just like voltagemagnitudevariation. It can beplottedas afunction of time,
and probability densityand distribution functionscan beobtained.Many publications
discussvoltage fluctuation and light flicker. Good overviews can befound in, among
others,[141] and [142].
The terms "voltage fluctuation" and "light flicker" are used byboth lEe and
IEEE.
7. Harmonic voltage distortion. The voltage waveform is never exactly a singlefrequency sine wave. Thisphenomenonis called "harmonic voltage distortion" or
simply "voltage distortion." When we assumea waveform to be periodic, it can be
describedas a sumof sine waves withfrequenciesbeing multiples of the fundamental
frequency.The nonfundamentalc omponentsare called"harmonicdistortion."
Thereare threecontributionsto the harmonicvoltagedistortion:
1. The voltage generatedby a synchronousmachineis not exactly sinusoidal
due to smalldeviationsfrom the idealshapeof the machine.This is a small
contribution; assumingthe generatedvoltageto be sinusoidalis a verygood
approximation.
2. The power system transporting the electrical energy from thegenerator
stations to the loads is not completely linear, although the deviation is
small. Somecomponentsin the systemdraw a nonsinusoidalc urrent, even
for a sinusoidal voltage. The classicalexample is the power transformer,
where thenonlinearity is due to saturationof the magneticflux in the iron
core of the transformer.A more recentexampleof a nonlinearpowersystem
componentis the HVDe link. The transformationfrom ac to dcand back
takesplace by usingpower-electronicscomponentswhich only conductduring part of a cycle.
The amount of harmonicdistortion originating in the power system is
normally small. Theincreasinguseof powerelectronicsfor control of power
flow and voltage(flexible ac transmissionsystems orFACTS) carriesthe risk
of increasingthe amount of harmonic distortion originating in the power
system. The same
technologyalso offers thepossibility of removinga large
part of the harmonicdistortion originatingelsewhere in the system or in the
load.
3. The main contribution to harmonicvoltage distortion is due to nonlinear
load. A growing part of the load is fed throughpower-electronicsconverters
drawing a nonsinusoidalcurrent. The harmoniccurrent componentscause
harmonic voltage components,and thus a nonsinusoidalvoltage, in the
system.
Two examplesof distored voltage are shown in Figs. 1.4and 1.5. The voltage
shownin Fig. 1.4containsmainly harmoniccomponentsof lower order(5,7,11,and 13
in this case). Thevoltageshownin Fig. 1.5containsmainly higher-frequencyharmonic
components.
Harmonicvoltagesand currentcan causea whole rangeof problems,with additional lossesand heating the main problem. The harmonicvoltage distortion is normally limited to a fewpercent(i.e., themagnitudeof the harmonicvoltagecomponents
Section 1.3 •
11
Overview of PowerQuality Phenomena
400
300
200
rl
100
($
>
.5
0
0
~
-100
co
S
-200
-300
-400
0
Figure 1.4 Exampleof distortedvoltage,with
mainly lower-orderharmoniccomponents
5
10
15
20
15
20
Time in milliseconds
[211].
400
300
200
~
0
>
.S
0
100
0
r -100
~
-200
-300
Figure 1.5 Exampleof distortedvoltage,with
higher-orderharmoniccomponents[211].
-400
0
5
10
Time in milliseconds
is up to a fewpercentof the magnitudeof the fundamentalvoltage) in which case
equipmentfunctionsasnormal.Occasionallylarge harmonicvoltage distortion occurs,
problem in
which can lead tomalfunction of equipment.This can especially be a big
industrialpower systems, where there is a large
concentrationof distortingload as well
as sensitive load.Harmonicdistortionof voltage andcurrentis the subject ofhundreds
of papersas well as anumberof books[77], [194], [195].
The term "harmonicdistortion" is very commonly used, and"distortion" is an
lEe term referring to loadstaking harmoniccurrentcomponents.Also within theIEEE
the term "distortion" is used to refer toharmonicdistortion; e.g., "distortion factor"
and "voltage distortion."
8. Harmonic current distortion. The complementaryphenomenonof harmonic
voltage distortion is harmoniccurrent distortion. The first is a voltagequality phenomenon,the latter a currentquality phenomenon.As harmonicvoltage distortion is
mainly due to nonsinusoidalload currents,harmonic voltage andcurrent distortion
are strongly linked. Harmonic current distortion requires over-rating of series components like transformersand cables. As the series resistance increases with frequency, adistorted current will cause more losses
t han a sinusoidalcurrent of the
same rms value.
12
Chapter I • Overview of Power Quality and Power Quality Standards
150
100
en
e SO
~
cd
.5
0
=
~ -so
U
-100
-15°0
5
10
15
Time inmilliseconds
20
Figure 1.6 Exampleof distortedcurrent,
leadingto the voltagedistortionshownin Fig.
1.4 [211).
Two examplesof harmoniccurrentdistortionare shown in Figs. 1.6
and 1.7.Both
currents are drawn by an adjustable-speeddrive. The current shown in Fig. 1.6 is
typical for modernac adjustable-speed
drives. Theharmonicspectrumof the current
containsmainly 5th, 7th,11th, and 13thharmoniccomponents.T he currentin Fig. 1.7
is lesscommon.The high-frequencyripple is due to the switching
frequencyof the dc/ac
inverter. As shown in Fig. 1.5 thishigh-frequencycurrent ripple causes a highfrequency ripple in thevoltageas well.
9. Interharmonicvoltage andcurrentcomponents. Some
e quipmentproducescurrent componentswith a frequency which is not an
integermultiple of the fundamental
frequency. Examples are
cycloconvertersand some typeso f heatingcontrollers.These
" interharmoniccomponents."T heir magcomponentsof the currentare referred to as
nitudeis normallysmallenoughnot to cause anyproblem,but sometimesthey can excite
unexpectedresonancesbetweentransformerinductancesand capacitorbanks. More
fundamental
dangerousarecurrentandvoltagecomponentswith a frequency below the
frequency, referred to as
"sub-harmonicdistortion." Sub-harmoniccurrentscan lead to
saturationof transformersand damageto synchronousgeneratorsand turbines.
Anothersourceof interharmonicdistortionare arc furnaces.Strictly speakingarc
furnaces do notproduce any interharmonicvoltage or current components,but a
50
-50
L - . - ._ _
- . . . J ' - -_
o
5
_
----JL..--_ _
__
__J
- - - - J ~
10
Time inmilliseconds
15
20
Figure 1.7 Exampleof distortedcurrent,
leadingto the voltagedistortionshownin Fig.
1.5 [211].
13
Section 1.3 • Overviewof PowerQuality Phenomena
numberof (integer) harmonicsplus acontinuous(voltage andcurrent)spectrum.Due
to resonances in the power system some
of the frequencies in thisspectrumare amplified. The amplified frequencycomponentsare normally referred to asinterharmonics
due to the arc furnace. These voltage
interharmonicshave recently become
o f special
interest as they are responsible for serious light flicker
problems.
A special case ofsub-harmoniccurrentsare those due to oscillations in the
earthmagnetic field following a solar flare. These so-called
geomagneticallyinducedcurrents
have periodsaroundfive minutes and the resulting
transformersaturationhas led to
large-scaleblackouts[143].
10. Periodicvoltagenotching. In three-phaserectifiers thecommutationfrom one
diode or thyristor to the other creates ashort-circuitwith a duration lessthan 1 ms,
which results in areductionin the supply voltage. Thisphenomenonis called"voltage
notching" or simply "notching." Notching mainly results inhigh-order harmonics,
of characwhich are often notconsideredin power engineering. A more suitable way
terizationis throughthe depthand durationof the notchin combinationwith the point
on the sine wave at which the
notchingcommences.
An exampleof voltagenotchingis shown in Fig. 1.8. This voltage wave shape was
caused by anadjustable-speed
drive in which a largereactancewas used to keep the de
currentconstant.
The IEEE uses the term"notch" or "line voltagenotch" in a more general way:
any reductionof the voltage lasting less than
half a cycle.
11. Mainssignalingvoltage.High-frequencysignals aresuperimposedon the supply voltage for thepurposeof transmissionof information in the public distribution
system and tocustomer'spremises.Threetypes of signal arementionedin the European
voltagecharacteristicsstandards[80]:
• Ripple controlsignals: sinusoidal signals between 110 and 3000 Hz. These
signals are, from avoltage-quality point-of-view, similar to harmonic and
interharmonicvoltage components.
• Power-line-carriersignals: sinusoidal signals between 3 and 148.5 kHz. These
signals can be described
both as high-frequencyvoltage noise (see below) and
as high-order(inter)harmonics.
• Mains markingsignals: superimposedshort time alterations (transients)at
selectedpoints of the voltage waveform.
400r---------,-----,------.--------,
300
200
ZJ
~
.5
j
~
100
0
-100
-200
-300
-400
0
Figure 1.8 Example of voltage
notching[211].
5
10
Timeinmilliseconds
15
20
14
ChapterI •
Overview of PowerQuality and PowerQuality Standards
Mains signalingvoltagecan interferewith equipmentusingsimilar frequencies for some
audiblenoise
internalpurpose.The voltages,a nd the associatedcurrents,can also cause
and signals ontelephonelines.
The other way around,harmonicand interharmonicvoltagesmay beinterpreted
by equipmentas beingsignalingvoltages,leadingto wrong functioning of equipment.
12.High-frequencyvoltage noise. Thesupply voltagecontainscomponentswhich
are not periodicat all. These can be called
"noise," althoughfrom the consumerpoint
of view, all above-mentionedvoltagecomponentsare in effect noise. Arcfurnacesare
an important sourceof noise. But also thecombinationof many different nonlinear
loadscan lead tovoltagenoise [196]. Noise can be
presentbetween thephaseconductors (differential mode noise) or cause anequal voltage in all conductors(commonmode noise).Distinguishingthe noise fromothercomponentsis not always simple,but
actually not really needed. Ananalysisis needed only in cases where the noise leads to
some problem with power system orend-userequipment.The characteristicsof the
problemwill dictatehow to measureand describethe noise.
A whole rangeof voltageand currentvariationshas beenintroduced.The reader
will have noticedthat the distinction between thevariousphenomenais not very sharp,
e.g., voltagefluctuation andvoltagevariation show a clearoverlap.One of the tasksof
future standardizationwork is to developa consistenta ndcompleteclassificationof the
variousphenomena.This might look an academictask, as it doesnot directly solve any
equipmentor systemproblems.But when quantifying the powerquality, the classification becomeslessacademic.A good classificationalso leads to abetterunderstanding
of the various phenomena.
1.3.2 Events
Eventsare phenomenawhich only happenevery once in a while. Aninterruption
of the supply voltage is the best-knownexample.This can intheory be viewed as an
extremevoltagemagnitudevariation (magnitudeequalto zero),andcan beincludedin
the probability distribution function of the voltagemagnitude.But this would not give
much usefulinformation; it would in fact give theunavailability of the supply voltage,
assumingthe resolution of the curve was highenough. Instead,events can best be
describedthrough the time between events, and the
characteristicsof the events;both
in a stochasticsense.Interruptionswill be discussed in sufficientdetail in Chapters2
and 3 and voltagesags inChapters4, 5, and6. Transientovervoltagewill be used as an
examplehere. A transientovervoltagerecording is shown in Fig. 1.9: the (absolute
value of the) voltagerises toabout180% of its normalmaximumfor a few milliseconds.
The smoothsinusoidalcurve is acontinuationof the pre-eventfundamentalvoltage.
A transientovervoltagecan becharacterizedin manydifferent ways; threeoftenusedcharacteristicsare:
1. Magnitude: the magnitudeis either the maximum voltage or the maximum
voltagedeviation from the normal sine wave.
2. Duration: the durationis harderto define, as itoften takes a long time before
the voltage has completelyrecovered.Possibledefinitions are:
• the time in which thevoltagehas recoveredto within 10% of the magnitude of the transientovervoltage;
• the time-constantof the averagedecay of the voltage;
• the ratio of the Vt-integral defined below and themagnitudeof the transient overvoltage.
15
Section 1.3 • Overviewof Power Quality Phenomena
1.5,----~--~-- -~-~--~-___,
0.5
5-
.5
~
~ - 0.5
~
-1
Figure 1.9 Example oftransientovervoltage
event: phase
-to-groundvoltage due to fault
clearing in one of theother phases.( Data
obtained from (16].)
- 1.5
I
,
I
60
,
20
30
40
Time in milliseconds
3. Vt-integral : theVt-integral is defined as
V, =
iT
(l.l)
V(t)dt
where t = 0 is thestart of the event, and an
a ppropriatevalue is chosen forT,
e.g., the time in which the voltage has recovered to within 10% of the magnitude of the transientovervoltage. Again the voltageV(t) can be measured
either from zero or as the
deviation from the normal sine wave.
Figure 1.10 gives thenumberof transientovervoltageevents per year, as
obtained
for the average low-voltage site in
Norway [67]. The distribution function for the time
140
120
100
~
....0~
~
80
60
1.0-1.5
1.5-2.0
40
~~
2.0- 3.0
'-$'
'b"
20
3.0-5.0
.~
~
~'I>
0
5.0-10.0
Figure 1.10Numberof transient overvoltage events per year, as a function of
magnitude and voltage integral.
(Data obtained from [67].)
16
Chapter I • Overview of Powe r Qua lity and Power
Quality Standards
1.2r--
-
-
- - --
- - - --
-
-
-
-,
t:
o
.~
E 0.8t--- -en
~
0.6
:E
0.4
..
.0
J:
0.2
o
1.0-1.5
1.5-2.0 2.0-3.0 3.0-5.0
Magnitude range in pu
5.0-10.0
Figure 1.11Probability distribution function
of the magnitude oftransient overvoltage
events, accord ing to Fig. 1.10.
between events has not been determ ined, but onlynumberof
the
events per year with
of
different characteristics. Notethat the average time between events is the reciprocal
the number of events per year. This is the
normal situation; the actual distribution
function is rarelydetermined in powerquality or reliability surveys[107].
Figures 1.11 through 1.14 givestatistical informationaboutthe characteristicsof
the events. Figure 1.11 gives the
probability distribution function of the magnitude of
the event. We see
t hat almost 80% of the events have a
magnitudelessthan 1.5 pu .
Figure 1.12 gives thecorrespond
ing densityfunction. By using alogarithmic scale the
is
visible. Figure 1.13 gives the
numberof events in the high-magn itude rangebetter
probability distribution function of the Vt-integral; Fig. 1.14 theprobability density
function.
1.2r--
- - --
-
-
-
-
-
-
-
-
---,
o
.u;
t:
~
0.1
g
~ 0.01
2
0..
.0
0.001
1.0-1.5
1.5-2.0 2.0- 3.0 3.0-5.0 5.0-10.0
Magnitude range in pu
Figure 1.12Probability density funct ionof
the magn itudeof transient overvoltage events ,
acco rding to Fig. 1.10.
An overview of various types of powerquality events is given below. Power
quality events are thephenomen
a which can lead totripping of equipment, to interrupt ion of the productionor of plant operation , or endangerpower systemoperation.
The treatmentof these in astochasticway is an extensionof the power system reliability
field as will be discussed inC hapter2. A special classof events, the so-called
"voltage
magnitudeevents," will betreatedin more detail in Section 1.3.3. Voltage
magnitude
events are the events which are the main
concernfor equipment,and they are the main
subject for the resto f this book .
Note that below only " voltage events" are discussed, as these canconcernto
be of
of "currentevents" could be added , with their
end-user equipment. But similarly a list
possible effects on power system
equipment. Most powerquality monitors in use,
continuously monitor the voltage and record an event when the voltage exceeds certain
17
Section 1.3 • Overviewof PowerQuality Phenomena
1.2.-- --
;".s
! 0.8+--
-
-
-
- - -- - - - - --
-
- - --
---,
'"
~
0.6
~ 0.4+--
£
- -- - - --
0.2
o
Figure 1.13Probabilitydistribution function
of the Vt-integral oftransientovervoltage
events.accordingto Fig. 1.10.
0-0.005
0.8 . - - --
0.005-0.01 0.01-0.1
Vt-integral range
-
0.1-1
- - -- -- -- --
----,
.~ 0.6+ - -- - - - -- -
~
~ 0.4+---- - - -- -
J
..: 0.2
Figure 1.14Probability density functionof
the Vt-integralof transientovervoltage
events,accordingto Fig. 1.10.
o
0.005-0.01 0.01-0.1
Vt-integral range
0.1-1
thresholds,typically voltagemagnitudethresholds. Although the currentsare often also
recorded they do notnormally trigger therecording. Thus anovercurrentwithout an
over- or undervoltagewill not be recorded. Of course there are no technical
limitations
in usingcurrentsignals to trigger therecordingprocess. In fact mostmonitorshave the
option of triggering oncurrentas well.
I. Interruptions. A "voltageinterruption"[IEEE Std.I159], "supply interruption"
[EN 50160],or just "interruption" [IEEE Std.1250] is a condition in which the voltage
at the supplyterminalsis close to zero. Close to zero is by the IEC defined"lower
as
than I% of the declaredvoltage" and by the IEEE as"lower than 10%" [IEEE Std.
II 59].
Voltage interruptionsare normally initiated by faults whichsubsequentlytrigger
protection measures .O ther causesof voltage interruption are protection operation
when there is no fault present (a so-called
protection maltrip), broken conductors
not triggering protective measures, andoperatorintervention. A further distinction
can be made between
pre-arrangedand accidentalinterruptions. The former allow
the end user to takeprecautionarymeasures to reduce the impact. All
pre-arranged
interruptionsare of course caused by
operatoraction.
Interruptionscan also be subdivided based on their
duration, thus based on the
way of restoring the supply:
• automaticswitching;
• manualswitching;
• repair or replacementof the faultedcomponent.
Cha pter I • Overviewof PowerQuality and Power QualityStandards
18
Various terminologies are in use to distinguish between these. The IEC uses the
term long interruptionsfor interruptions longer than 3 minutes and the term
s hort
interruptions for interruptions lasting up to 3 minutes. Within the IEEE the terms
momentary,temporary,and sustained are used, but different documents give different
duration values. The various definitions will be discussedChapter3.
in
2. Undervoltages.Undervoltages of variousduration are known under different
names.Short-durationundervoltagesare called"voltage sags" or"voltagedips." The
latter term is preferred by thelEe. Within the IEEE and in manyjournal and conference papers on power qua lity, the term voltage sag is used.
Long-durationundervoltage is normall y simply referred to as " undervoltage."
A voltage sag is areductionin the supply voltagemagnitudefollowed by a voltage
recovery after ashort period of time. When a voltage
magnitudereduct ion of finite
duration can actually be called a voltage sag (or voltage dip in the IEC terminology)
remains apoint of debate, even though the official definitions are cleara bout it.
Accord ing to the IEC, a supply voltage dip is a sudden reduction in the supply voltage
to a value between 90% and I % of the declared voltage, followed by a recovery
between 10ms and I minuteater.
l For the IEEE a voltagedrop is only a sag if the
during -sag voltage is between 10% and 90% of the nominal voltage.
Voltage sags are mostly caused short-circuitfaults
by
in the system and by
starting of large motors. Voltage sags will be discussed in detail Chapters4,
in
5, and 6.
3. Voltage magnitude steps. Load switching,
transformer tap-changers,and
switching actions in the system (e.g.,
capacitorbanks) can lead to a sudden change in
the voltage magnitude. Such a voltagemagnitude step is called a " rapid voltage
change" [EN 50160] or "voltagechange" [IEEE Std.1l59] . Normally both voltage
before and after the step are in the
normal operatingrange (typically 90% to 110%
of the nominal voltage).
An example of voltagemagnitudesteps is shown in Fig. 1.15. The figure shows a
2.5hour recording of the voltage in a 10kVistribution
d
system. The steps in the voltage
magnitudeare due to theoperationof transformer tap-changersat various voltage
levels.
4. Overvoltages. Just like with
undervoltage, overvoltage events are given different
names based on their
duration. Overvoltages of veryshort duration, and high magnitude, are called " transient
overvoltages
," "voltage spikes," or sometimes "voltage
surges." The atter
l
term is ratherconfusingas it is sometimes used to refer to overvoltages with adurationbetweenabout 1 cycle and I minute . Thelatter event is more
correctly called"voltage swell" or "temporarypower frequency overvoltage ." Longer
1.05
1.04
:l
0.
1.03
.S 1.02
.,
OIl
~ 1.01
~
0.99
0.98
5:00:00
5:30:00 6:00:00 6:30:00 7:00:00
Clock time (HH:MM:SS)
Figure 1.15 Example of voltage
magnitude
steps due to tran sformetap-changer
r
7:30:00
operation, recorded in a10kV distribution
system insouthernSweden.
Section 1.3 • Overviewof PowerQuality Phenomena
19
duration overvoltagesare simplyreferredto as "overvoltages."Long and short overvoltagesoriginatefrom, amongothers,lightning strokes,switchingoperations,s udden
load reduction,single-phaseshort-circuits,and nonlinearities.
A resonancebetween thenonlinearmagnetizingreactanceof a transformeranda
capacitance(either in the form of a capacitorbank or the capacitanceof an underground cable) can lead to a large
overvoltageof long duration. This phenomenonis
called ferroresonance,a nd it can lead to seriousdamageto power systemequipment
[144].
5. Fast voltage events. Voltage events with a very
short duration, typically one
as
cycle of the power system frequency or less, are referred to"transients,""transient
(over)voltages,""voltagetransients,"or "wave shapefaults." The termtransientis not
fully correct, as it should only be used for thetransition between twosteadystates.
Events due toswitchingactionscould underthat definition be calledtransients;events
due tolightning strokescould not be calledtransientsunderthat definition. But due to
the similarity in time scaleboth are referredto asvoltagetransients.Even veryshortdurationvoltagesags (e.g., due to fuse
clearing)are referred to as
voltagetransients,or
also "notches."
Fastvoltageevents can be dividedinto impulsive transients(mainly due to lightning) and oscillatory transients(mainly due to switching actions).
6. Phase-anglejumps andthree-phaseunbalance. We will see inC hapter4 that a
voltage sag is often associatedwith a phase-anglejump and some three-phase
unbalance.An interestingthought is whetheror not a jump in phase-anglewithout a
drop in voltage magnitudeshould be called avoltagesag. Such an event
could occur
s hortwhen oneof two parallel feeders istakenout of operation.The same holds for a
duration, three-phaseunbalancewithout changein magnitude,thus where only the
phase-angleof the threevoltages changes.
To get acompletepicture,also short-durationphase-angle
j umpsandshort-duration unbalancesshouldbeconsideredas eventsbelongingto the familyof powerquality
phenomena.
1.3.3 Overview of Voltage Magnitude Events
As mentionedin the previoussection,the majority of eventscurrently of interest
are associatedwith eithera reductionor an increasein the voltagemagnitude.We will
refer to these as"voltage magnitudeevents."
A voltage magnitudeevent is a (significant)deviation from the normal voltage
magnitudefor a limited duration.The magnitudecan befound by taking the rmsof the
voltageover a multiple of one half-cycleof the power-systemfrequency.
(1.2)
where V(t) is the voltageas afunction of time, sampledat equidistantpoints t = k Si.
The rms value istaken over a period N ~t, referred to as the "window length."
Alternatively, the magnitudecan bedeterminedfrom the peak voltage or from the
fundamental-frequency
c omponento f the voltage.Most powerquality monitorsdetermine the rmsvoltage once every cycle or once every few cycles. The
momentthe rms
voltagedeviates morethan a pre-setthresholdfrom its nominal value, thevoltageas a
function of time is recorded(the rmsvoltage,the sampledtime-domaindata,or both).
20
Chapter 1 • Overview of Power Quality and Power Quality Standards
Most events show aratherconstantrms voltagefor a certaindurationafter which the
rms voltage returns to a more or less normal value. This isunderstandableif one
realizesthat events are due tochangesin the system followed by therestorationof
the original systemaftera certaintime. Before,during, andafter the event, the system is
more or less in asteadystate.Thusthe event can be
characterizedthroughoneduration
and onemagnitude.We will see inChapter4 that it is not always possible touniquely
determinemagnitudeand duration of a voltage magnitudeevent. For now we will
assumethat this is possible, and define the
magnitudeof the event as theremaining
rms voltage during the event: if the rms voltaged uring the event is 170V in a 230 V
system, themagnitudeof the event is~~g = 73.9%.
Knowing the magnitudeand duration of an event, it can berepresentedas one
point in the magnitude-durationplane. All eventsrecordedby a monitor over acertain
periodcan berepresentedas ascatterof points.Different underlyingcauses may lead to
events indifferent parts of the plane. Themagnitude-durationplot will come back
several times in theforthcoming chapters.Various standardsgive different names to
events indifferent parts of the plane. Astraightforwardclassificationis given in Fig.
1.16. The voltagemagnitudeis split into three regions:
• interruption: the voltagemagnitudeis zero,
• undervoltage:the voltagemagnitudeis below its nominal value, and
• overvoltage:the voltagemagnitudeis aboveits nominal value.
In duration,a distinction is made between:
•
•
•
•
very short, correspondingto transientand self-restoringevents;
short, correspondingto automaticrestorationof the pre-eventsituation;
long, correspondingto manualrestorationof the pre-eventsituation;
very long, correspondingto repair or replacemento f faulted components.
Very
short
overvoltage
Shortovervoltage
Longovervoltage
Very
long
overvoltage
110%
Normaloperatingvoltage
Very
short
undervoltage
Shortundervoltage
Longundervoltage
Very
long
undervoltage
Veryshort int.
Shortinterruption
Longinterruption
Verylong int.
}-10%
1-3 cycles
}-3min
Event duration
1-3 hours
Figure 1.16 Suggested classification of voltage
magnitudeevents.
21
Section 1.3 • Overviewof Power QualityPhenomena
The various bordersin Fig. 1.16 aresomewhatarbitrary; some of the indicated
lEe and IEEE
values (1-3minutes,1-10%,900/0,and 110% ) are those used in existing
standards.F or monitoringpurposes,strict thresholdsare needed todistinguishbetween
the different events. An example is the
thresholddividing betweeninterruptionsand
undervoltages.This one is placed(somewhatarbitrarily) at 1% of nominalaccordingto
% accordingto the.IEEE (see below). Anyothersmall value would be
the IEC and at 10
equally defendable.
The classificationin Fig. 1.16 is only aimed atexplainingthe different types of
events: the termsmentionedin the figures are not all used in practice. Both
lEe and
IEEE give different namesto events in someof the regionsof the magnitude-duration
plane. The IECdefinitionsaresummarizedin Fig. 1.17 and theIEEE definitionsin Fig.
1.18. Thersc definitionswereobtainedfrom CENELECdocumentEN 50160 [80], the
IEEE definitions from IEEE Std.1159-1995.
The methodof classifying eventsthrough one magnitudeand oneduration has
information and knowledge
been shown to be very useful and has resulted in aoflot
about power quality. But the method also has itslimitations, which is important to
Four points should be especially kept in mind.
realize when using this classification.
1. ,Theduring-eventrms voltage isnot alwaysconstant,leading toambiguities
ambiguitiesin
in defining the magnitudeof the event. It may also lead to
defining thedurationof the event.
2. Fastevents (one cycle or less duration)cannotbe
in
characterized,resulting
in unrealistic values for magnitudeand duration or in thesedisturbances
simply being neglected.
3. Repetitive events can giveerroneousresults: theyeither lead to an overof events is
estimationof the numberof events (when each event in a row
countedas aseparateevent), or anunder-estimationof the severityof the
events (when a rowof identical events iscountedas one event).
0
=00
oS
.~]
Temporaryovervoltage
(1)Overvoltage
f-f>
0
110%
Normaloperatingvoltage
?
(supply)Voltagedip
(1)Overvoltage
1%
Shortinterruption
I
0.5 cycle
I,
Longinterruption
3 min
1 min
Eventduration
EN
Figure1.17 Definitions of voltage magnitude events as used in 50160.
22
Chapter1 • Overview of PowerQuality and PowerQuality Standards
=
Q)
';;
110%
Swell
~
Overvoltage
Normaloperatingvoltage
c:
Q)
'r;)
Voltagesag
I
Undervoltage
~
0
Z
100/0
Momentary
0.5 cycle
I
Temporary
3 sec
Sustainedinterruption
1 min
Eventduration
Figure 1.18Definitions of voltage magnitudeevents as used in
I EEE Std.11591995,
4. Equipmentis sometimessensitive toothercharacteristicsthanjust magnitude
and duration.
We will come back to these
problemsin more detail inChapters3 and 4.
Similar classificationscan beproposedfor voltagefrequency events, for voltage
phase-angleevents, forthree-phasevoltage unbalanceevents, etc. But because
most
equipmentproblems are due to an increase or decrease in voltage
magnitude, the
emphasisis on voltagemagnitudeevents.
1.4 POWER QUALITY AND EMC STANDARDS
1.4.1 Purpose of Standardization
Standardsthat define the quality of the supply have beenpresentfor decades
already. Almost any country has standardsdefining the margins in which frequency
and voltage are allowed to vary.
Other standardslimit harmoniccurrent and voltage
distortion, voltage fluctuations, and duration of an interruption.Thereare three reasons fordevelopingpower quality standards.
l. Defining the nominalenvironment.A hypotheticalexampleof such astandard
is: "The voltage shall he sinusoidal with a .frequency
of 50 Hz and an rms
voltageof 230 V." Such astandardis not very practical as it is technically
impossible to keep voltage magnitude and frequency exactlyconstant.
Therefore,existing standardsuse terms like"nominal voltage" or "declared
voltage" in this context.A more practicalversion of the abovestandardtext
would read as: "Thenominalfrequencyshall be 50 Hz and the nominal voltage
shall be 230V," which comes close to the
wording in EuropeanstandardEN
50160[80].
Section 1.4 • PowerQuality and EMC Standards
23
Defining nominal voltage and frequencydoesnot say anythingabout the
actualenvironment.To do this thedeviationsfrom the nominal values have
to be known. Most countrieshave astandardgiving the allowed variation in
the rms voltage, a typical rangebeing betweenfrom 900/0 to 110°A».
2. Defining the terminology. Even if a standard-settingbody does not want to
impose any requirementson equipment or supply, it might still want to
publish power quality standards.A good example is IEEE Std.1346[22]
which recommendsa methodfor exchanginginformationbetweenequipment
manufacturers,utilities, and customers.T he standarddoes not give any suggestionsabout what is consideredacceptable.
This group of standardsaims at giving exact definitions of the various
phenomena,how their characteristicsshould be measured,a nd how equipment should be testedfor its immunity. The aim of this is to enablecommunication betweenthe various partnersin the power quality field. It ensures,
e.g., that the resultsof two power quality monitors can be easilycompared
and that equipmentimmunity can becomparedwith the descriptionof the
environment.Hypotheticalexamplesare: "A short interruption is a situation
J% ofthe nominalrms voltageforless than
in which the rms voltage is less than
3 minutes."and"The durationof a voltage dip is the time during
'which the rms
voltage is less than 90%of the nominalrms voltage. The durationof a voltage
dip shall beexpressedin seconds. The rms voltage shall determinedevery
be
half-cycle," Both IEEE Std. 1159and EN 50160 give these kindo f definitions,
hopefully merginginto a future lEe standard.
3. Limit the number of powerquality problems. Limiting the numberof power
quality problemsis the final aim of all the work on power quality. Power
quality problemscan be mitigated by limiting the amountof voltagedisturbancescausedby equipment,by improving the performanceof the supply,
and by making equipmentlesssensitiveto voltage disturbances.All mitigation methodsrequiretechnicalsolutionswhich can be implementedindependently of any standardization.But proper standardizationwill provide
important incentives for the implementation of the technical solutions.
Proper standardizationwill also solve the problem of responsibility for
power quality disturbances.H ypotheticalexamplesare:
The current taken by a load exceeding
4 k VA shallnot containmore than
J% ofany
evenharmonic.The harmoniccontents shall be measuredas a l-second
average.and
Equipment shall be immune to voltage
variations between85% and 110%of the
terminals,
nominal voltage. This shall be tested by supplying at the equipment
sinusoidalvoltageswith magnitudesof 85.% and J/0% for a duration of 1 hour.
If the pieceofequipment has more than one
distinctiveload state,it shall be tested
for each load state separately, or for what are conceivedthe most sensitive stales.
In this field both IEC and IEEE lack a.good set of standardson power
quality. The lEe has set up a wholeframework on electromagneticcompatibility which alreadyincludessome power quality standards.T he best example is the harmonic standardIEC-61000-2-3 which limits the amount of
harmonic current produced by low-power equipment. The IEEE has a
good recommendedpractice for the limitation of harmonic distortion:
IEEE 519 [82] which gives limitsboth for the harmoniccurrentstaken by
the customerand for the voltagesdeliveredby the utility.
24
ChapterI • Overview of PowerQuality and PowerQuality Standards
1.4.2 The IEC Electromagnetic Compatibility Standards
Within the International ElectrotechnicalCommittee (IEC) a comprehensive
framework of standardson electromagneticcompatibility is under development.
Electromagneticcompatibility (EMC) is defined as:the ability of a device,equipment
or systemto function satisfactorilyin its electromagneticenvironment without introducing
intolerable electromagneticdisturbances toanything in that environment[79].
Thereare two aspects to EMC:
(1) a piece ofequipmentshould be able tooperate
normally in its environment,and (2) itshouldnot pollutethe environmenttoo much. In
EMC terms:immunity and emission. There are
standardsfor both aspects.Agreement
on immunity is at first a matter of agreementbetween themanufacturerand the customer. But the IEC sets
minimum requirementsin immunity standards.The third term
of importanceis "electromagneticenvironment,"which gives the levelof disturbance
againstwhich theequipmentshouldbe immune. Within theEMC standards,a distinction is made betweenradiated disturbancesand conducteddisturbances.Radiated
disturbancesare emitted (transmitted)by one device and received by
anotherwithout
the need for anyconduction. Conducteddisturbancesneed aconductorto transfer
from one device toanother. Theseconducteddisturbancesare within the scopeof
power quality; radiated disturbances(although very important) are outside of the
normal realm of power system engineering or power quality.
A schematicoverview of theEMC terminologyis given in Fig. 1.19. We see
that
of conducteddisturbancesand radiateddisturthe emission of a device may consist
bances.Radiateddisturbancescan reachanotherdevice via any medium.Normally,
radiateddisturbancesonly influenceanotherdevice when it is physically close to the
emitting device.Conducteddisturbancesreach anotherdevice via an electrically conductingmedium, typically thepowersystem. The device being influenced no longer has
to be physically close as the power system is a very good medium for
conductionof
the
is a device which is
many typesof disturbances.Of course also here the rule that
electrically closer(thereis lessimpedancebetween them) is more likely to be influenced.
A device connectedto the power system is exposed to an electrical
environmentnot
only due to thecombinedemissionof all otherdevicesconnectedto the system but also
due to all kinds of events in the power system (like switching actions,
short-circuitfaults,
and lightning strokes). Theimmunity of the deviceshouldbe assessed with reference to
this electromagneticenvironment.A special typeof disturbances,not shown in the
Powersystem
Events
Conducted
disturbances
Figure 1.19Overviewof EMC terminology.
Section 1.4 • PowerQuality and EMC Standards
25
figure, are radiateddisturbanceswhich induce conducteddisturbancesin the power
system.
Immunity Requirements. Immunity standardsdefine theminimum level of electromagneticdisturbancethat a pieceof equipmentshall be able towithstand. Before
being able todeterminethe immunity of a device, aperformancecriterion must be
defined. In other words, it should be agreedupon what kind of behavior will be
called a failure. Inpracticeit will often be clear when a device
performssatisfactorily
and when not, but when testingequipment the distinction may becomeblurred.
It will all dependon the applicationwhetheror not a certain equipmentbehavioris
acceptable.
The basicimmunity standard[IEC-61000-4-1] gives four classes
of equipment
performance:
• Normal performancewithin the specification limits.
• Temporarydegradationor lossof function which is self-recoverable.
• Temporarydegradationor loss of function which requiresoperatorintervention or system reset.
• Degradationor loss of function which is not recoverabledue to damageof
equipment,componentsor software,or lossof data.
These classes are general as descriptionshouldbe
the
applicableto all kinds of equipment. Thisclassificationis further defined in thevariousequipmentstandards.
Emission Standards. Emission standardsdefine themaximum amount of electromagneticdisturbancethat a pieceof equipmentis allowed to produce.Within the
existing lEe standards,emission limits exist forharmonic currents[lEe 61000-3-2
and 61000-3-6], and for voltage
fluctuations[lEe 61000-3-3, 61000-3-5, and 61000-37]. Most power quality phenomenaare not due to equipmentemission but due to
EMC standardsonly apply
operationalactionsor faults in the power system. As the
to equipment, there are no"emission limits" for the power system. Events like
voltage sagsand interruptions are consideredas a "fact-of-life." These events do,
however,contributeto the electromagneticenvironment.
The Electromagnetic Environment.To give quantitativelevels for theimmunity
of equipment,the electromagneticenvironmentshould be known. Theelectromagnetic environmentfor disturbancesoriginating in or conductedthrough the power
system, isequivalentto the voltagequality as defined before. ThelEC electromagnetic compatibility standardsdefine the voltagequality in three ways:
I. Compatibility levels are reference values for
coordinatingemissionandimmunity requirementsof equipment.For a givendisturbance,the compatibility
level is in between the emission level (or the
environment)and the immunity
level. As both emission andimmunity are stochasticquantities,electromagnetic compatibility can never be completelyguaranteed.The compatibility
level is chosen such
t hatcompatibilityis achieved for mostequipmentm ostof
the time: typically 95% of equipmentfor 950/0 of "the time. It isnot always
possible to influenceboth emission and immunity: three cases can be distinguished:
26
ChapterI • Overview of PowerQuality and PowerQuality Standards
• Both emission andimmunitycan be affected.The compatibility level can in
of
principle be freely chosen. But a high level will lead to high costs
equipmentimmunity and a low level to high costs for limiting the emission. Thecompatibility level shouldthereforebe chosen suchthat the sum
of both costs isminimal. An exampleof a disturbancewhere both emission and immunity can be affected ish armonicdistortion. A very good
example of this process is described IEEE
in Std.519 [82].
• The emission level cannot be affected.
The compatibility level should be
chosen suchthat it exceeds theenvironmentfor most equipmentmost of
the time. An exampleof a disturbancewhere the emission level
cannotbe
of occurrencedependson the
affected are voltage sags: their frequency
fault frequency and on the power system,
both of which cannotbe affected
by theequipmentmanufacturer.N ote that the EMC standardsonly apply
of
to equipmentmanufacturers.We will later come back to the choice
compatibility levels for these kindof disturbances.
• The immunity level cannot be affected.The compatibility level should be
chosen suchthat it is less than the immunity level for most equipment
most of the time. An exampleof a disturbancewhere theimmunity level
cannotbe affected is voltagefluctuation leading to light flicker.
2. Voltagecharacteristicsare quasi-guaranteed
limits for someparameters,covering any location. Again the voltagecharacteristicsare based on a95%
value, but now only in time. They hold at anylocation, and are thus an
important parameterfor the customer.Voltage characteristicsare a wayof
describingelectricity as aproduct. Within Europe the EN 50160standard
defines someof the voltagecharacteristics.This standardwill be discussed in
detail in Section 1.4.3.
3. Planning levels are specified by the supply utility and canconsideredas
be
internal quality objectivesof the utility.
These ideas were
originally developed fordisturbancesgeneratedby equipment,for
which other equipmentcould be sensitive: mainlyradio frequency interference.These
ideas have been
extendedtowardsvariationslike harmonicdistortion or voltage fluctuations.The concepthas not yet beenappliedsuccessfullytowardsevents like voltage
sags orinterruptions.
EMC and Variations. Variations can be stochasticallydescribed through a
probability distribution function, as shownin Fig. 1.20. The curve gives the
probabilcompatibility level
ity that the disturbancelevel will not exceed the given value. The
can, accordingto the recommendationsin the IEC standards,be chosen at the95%
percentile, asindicated in Fig. 1.20. The curve can hold for one site or for a large
number of sites. When the curverepresentsa large number of sites it is important
of the sites (typically950/0 of
that it gives thedisturbancelevel not exceeded for most
the sites).Consideras an examplethat the compatibility level of total harmonicdistortion (THO) is 0.08. Supposethe THO is measuredat 100 sitesduring 1000 10minute intervals. A compatibility level of 0.08 impliesthat at 95 sites(out of 100) at
least 950THD samples (outof 1000) have a valueof 0.08 or less.
In case a higher reliability isrequiredfor the successfuloperationof a device, a
higher levelthan 950/0 should be chosen, e.g.,99.9%.
27
Section 1.4 • PowerQuality and EMC Standards
u
~
u
-; 0.75
.S
~
u
~
g
0.5
~
o
g 0.25
i
.J:J
Figure 1.20 Probability distribution function e
~
for a variation, with the compatibilitylevel
indicated.
O~~-------------------'
Disturbancelevel inarbitraryunits
EMC and Events. The EMC framework has not been developed for events
For important power quality
and its application to them has not been defined yet.
phenomenalike voltage sagsand interruptions,the EMC standardscan thusnot be
part why the EMC standardsare not (yet) wellknown
used. This explains for a large
in the powerquality field. Still an attemptshould be made atapplying the concepts
of electromagneticcompatibility to events.
Events onlyhappenoccasionallyand are not present allof the time;applying a
95°~ criterion is thereforeno longer possible. Animmunity to 95% of voltage sags
would dependon the wayof countingthe sags.C ountingall sags below 200 V (in a 230
V supply) would give a much higher
numberthancountingall sagsbelow 150 V. The
immunity requirementin the latter case would be muchstricter than in the former.
In some powerquality monitoringsurveys a95% criterionin space is applied. The
of disturbance(numberof events)
electromagneticenvironmentis defined as the level
of the environmentin itself does
not exceeded for950/0 of the sites. But the knowledge
not sayanythingaboutequipmentimmunity requirements.The immunity requirement
should be based on theminimum time between events exceeding the
immunity level.
Figure 1.21 shows the time between events exceeding
certain
a disturbancelevel as a
function of the disturbancelevel (the severity of the event). The more severe the event
equipment
the more the time between events (the lower the event frequency). A of
piece
or an industrial process to which theequipmentbelongs will have acertain reliability
requirement,i.e., a certain minimum time between events leading to
tripping of the
equipmentor interruption of the process. By using the curve in Fig. 1.21 this can be
translatedinto an immunity requirement.As we will see later, theactual situation is
more complicated:the severityof an event is amultidimensionalquantity as at least
magnitudeand durationplaya role.
A possiblecompatibilitylevel would be the levelnot exceeded morethanten times
a year by95% of the customers.This can be done for anydimensionof the event,
leading to amultidimensionalcompatibility level. Thisconcepthas beenappliedto the
results of the Norwegianpower quality survey [67]. The frequency oftransientover950/0 site, is shown in Fig. 1.22. The
95% site is chosen such
t hat
voltage events, for the
95% of the sites have less
transientovervoltageevents per yearthanthis site.From Fig.
1.22 we can see
t hat reasonablecompatibility levels are:
• 2.5 pu for themagnitudeof the transients.
• 0.3 Vs for theVt-integral,
28
Chapt er I • Overview of Power Quality and Power Quality
Standards
7
6
Desired reliability
a:;
;;.
.!!
.,
5
-5
OJ)
~.,
.,
.,o><
4
ZJ
.,<::
.,;;.
.,<::
.,
~
.,
.,
a
3
.0
2
f::::
Disturbance level in arbitrary units
Figure 1.21 Time between events as a
function of the disturbancelevel.
500
400
~.,;;.
""'d0
Z
300
1.0-1.5
200
100
2.0-3.0
3.0-5.0
0
5.0-10.0
Voltage-integral in Vs
~
. ~~
't>~"
;s.'<S'
~~"<J
1-10
Figure 1.22 Ma ximumnumberof transientovervoltage events for 95% of the lowvoltagecustomers in Norw ay.(Data obtained from [67].)
29
Section 1.4 • PowerQuality and EMC Standards
As a next step, these
levels could be used as a basis for equipment immunity requirements. This concept could be worked out further by giving compatibility
levels for 10
events and 1 event per year. Compatibility
levelsfor 1event per year cannot be obtained
from Fig. 1.22 because of the short monitoring period (about one year).
1.4.3 The European Voltage Characteristics Standard
Europeanstandard50160 [80] describeselectricity as a product, including its
shortcomings.I~ gives the main characteristics of the voltage at the customer's supply
terminals in public low-voltage and medium-voltage networks under normal operating
conditions.
Some disturbances are just mentioned, for others a wide range of typical values
are given, and for some disturbances actual voltage characteristics are given.
Voltage Variations. Standard EN50160 gives limits for some variations. For
each of these variations the value is given which shall notexceededfor
be
95% of
the time. The measurement should be performed with a certain averaging window.
The length of this window is 10 minutes for most variations; thus very short time
scales are not considered in the standard. The following limits forlow-voltage
the
supply are given in the document:
• Voltagemagnitude:950/0 of the 10-minute averages during one week shall be
within ± 10% of the nominal voltage of 230V.
• Harmonicdistortion: For harmonic voltage components up to order 25, values
are given which shall not be
exceededduring 95% of the 10-minute averages
obtained in oneweek. The total harmonic distortion shall not exceed
8%
during 95% of the week. The limits have been reproduced in Table 1.1.
Theselevels appear to originate from a study after harmonic distortion performed by a CIGRE working group
[83], although thestandarddocument does
not refer to that study. Inreference[83] two values are given for the harmonic
voltage distortion:
-
low value: the value likely to be found in the vicinity of large disturbing
loads and associated with a low probability of causing disturbing
effects;
highvalue: value rarely found in the network and with a higher probability
of causing disturbingeffects.
-
TABLE 1.1 HarmonicVoltage Limits According to EN 50160
Order
3
5
7
9
II
13
RelativeVoltage
5
6%
5%
1.5%
3.5%
3%
%
Order
Relative Voltage
15
17
19
21
0.5%
20/0
1.5%
23
25
1.5%
1.5%
0.50/0
30
ChapterI • Overview of Power Quality and PowerQuality Standards
TABLE 1.2 Harmonic VoltageLevels in Europe[83J
Order
3
5
7
9
II
13
Low
1.5°~
4%)
4%
0.80/0
2.5%
2%
High
Order
2.5% .
6%
15
17
1%
5°AJ
19
O.8°.!cJ
1.5%
3.50/0
3%
21
23
25
Low
High
~O.3°~
2%
1.5°.!cJ
~O.30/0
0.80/0
0.8%
1.5%
1.5°AJ
The valuesfound by the CIGRE working group havebeen summarizedin Table
1.2. The valuesused inEN 50160are obviously the valuesrarely exceededanywherein
Europe.This is exactly what is implementedby the term "voltage characteristics."
• Voltage fluctuation: 95% of the 2-hour long-term flicker severity values
obtainedduring oneweek shall not exceed1. The flicker severityis an objective
measureof the severity of light flicker due to voltagefluctuations (81].
• Voltageunbalance:the ratio of negative-and positive-sequence
v oltageshall be
obtainedas 10minute averages,95% of thoseshall not exceed2% during one
week.
• Frequency:95% of the 10secondaveragesshall not be outsidethe range49.5 ..
50.5 Hz.
• Signaling voltages: 99% of the 3- secondaveragesduring one day shall not
exceed9% for frequenciesup to 500 Hz,50/0 for frequenciesbetween1 and 10
kHz, and a thresholddecayingto 1% for higher frequencies.
Events. Standard EN 50160 does not give any voltage characteristicsfor
events. Most event-typephenomenaare only mentioned,but for some an indicative
value of the event frequencyis given. For completenessa list of eventsmentionedin
EN 50160 isreproducedbelow:
• Voltage magnitudesteps: thesenormally do not exceed ±5°AJ of the nominal
voltage, but changesup to ±100/o can occur a numberof times per day.
• Voltagesags:frequencyof occurrenceis betweena few tensand one thousand
events per year. Duration is mostly less than 1 second,and voltage drops
rarely below 40%. At some places sags due to load switching occur very
frequently.
• Short interruptionsoccur betweena few tensand severalhundredstimes per
year. The durationis in about 70% of the cases less
t han 1 second.
• Long interruptionsof the supply voltage:t heir frequencymay be lessthan 10 or
up to 50per year.
• Voltage swells (short overvoltagesin Fig. 1.16) occur under certain circumstances.Overvoltagesdue to short-circuit faults elsewherein the systemwill
generallynot exceed 1.5 kV rms in a 230 V
system.
• Transientovervoltagewill generallynot exceed 6 kVpeak in a 230 V system.
31
Section 1.4 • PowerQuality and EMC Standards
The 95% Limits. One of the recurring criticisms on the EN 50160standardis
that it only gives limits for 95% of the time. Nothing is said about the remaining
5% of the time. Looking at the voltage magnitudeas an example:95% of the time
the voltage is between207V and 253V (10% variation around the nominal voltage
of 230V), but during the remaining 5% of the time the voltage could be zero, or
10000 V, and the voltagewould still conform with the voltagecharacteristics.
The voltage magnitude(rms value) is obtainedevery 10 minutes-thatgives a
total of 7 x 24 x 6 = 1008 samplesper week; all but 50 of thosesamplesshould be in
the givenrange.If we only considernormal operation(as isstatedin the document)it
would be very unlikely that these are far away from the ±lOOiO band. Understanding
this requiressomeknowledgeof stochastictheory. In normal operation,the voltageat
the customeris determinedby a seriesof voltagedropsin the system.All of thoseareof
a stochasticcharacter.According to stochastictheory, a variablewhich is the sum of a
sufficient numberof stochasticvariables,can be describedby a normal distribution.
The normal distribution is one of the basicdistributionsin stochastictheory: its probability densityfunction is
1
(V-Il)2
f(v) = --e-J;2
(1.3)
.J2ira
where v is the value of the stochasticvariable, It its expectedvalue, and (1 its standard
deviation. The well-known bell-shape of this function is shown in Fig. 1.23 for
It = 230V and (1 = 11.7V.
Thereis no analyticalexpressionfor the probability distribution function, but it
can be expressedin the so-callederror function <1>:
F(v) =
[f(t/J)dt/J = <I>[V :
/l]
(1.4)
The voltage characteristicsstandardgives the expectedvalue (230V) and the 950/0
interval (207 .. 253 V).Assumingthat the voltageis normally distributedwe cancalculate the standarddeviationwhich resultsin the given 95% confidenceinterval. As 95%
of the voltagesamplesare between207 and 253 V, 97.50/0 is below 253 V, thus:
<1>[253V ~ 230V]
= 0.975
(1.5)
3.5,.--------.----,----.----.:.--.,....----,
5e
~
3
2.5
.53
.~
a
2
-8
g
1.5
~e
~
0.5
O'---.:=-----L--------J~_---I~_----I--=----'
Figure 1.23 Probability density function of
the normal distribution.
180
200
220
240
Voltagein volts
260
280
32
Chapter 1 • Overview of Power Quality and Power Quality Standards
From a table of theerror function, which can befound in almostany book on statistics
or stochastictheory, we find that <1>(1.96) = 0.975 which givesa> 11.7V. Knowing
expectedvalue and standarddeviation of the normal distribution, the wholedistribution is known. It is thus no longerdifficult to calculatethe probability that the voltage
deviatesmore than 10% from its nominalvalue. The resultsof this calculationare given
in Table 1.3. The firstcolumngives theprobability that the voltageis within the voltage
range in thesecond,third, and fourth columns.The voltagerange is given instandard
deviations,in volts and as a percentageof the nominal voltage. The voltage is thus
between 200 and 260 V for990/0 of the time. The lastc olumn indicateshow often the
voltageis outsideof the range,assumingall samplesto bestochasticallyindependent.In
reality there isstrong correlation between thesampleswhich makesthat large deviations become even more unlikely.
Further, there arevoltage regulation mechanisms
(capacitor banks, transformertap-changers)which become active when the
voltage
deviatestoo much from itsnominal value. Finally, one should realize that the 95%
value given in thestandarddoes not hold for the averagecustomerbut for the worstservedcustomer.All this leads to theconclusionthat voltagemagnitudevariationsof
much morethan 10% are extremelyunlikely.
From this reasoningone should absolutely not draw the conclusion that the
voltage magnitudewill never be lowerthan a value like 80%. The mainassumption
used isthat the voltage variations are due to the sumof a numberof small voltage
drops. During, e.g., a voltage sag, this nolonger holds. This brings us back to the
principal differencebetween"events" and "variations": for variationsthe normal distribution can be used; for events it is the time
betweenevents which isof main importance. Theprobabilitiesin Table 1.3 thus only hold for voltagemagnitudevariations;
absolutelynothing is said yetabout voltagemagnitudeevents.
Scope and Limitations. StandardEN 50160containssome well-defined limits
and measurementprotocols,but it falls short of putting responsibilitywith any party.
This is of courseunderstandablewhen one realizesthat the documentdescribes the
"voltage characteristics"which is the electromagneticenvironmentas it is now, not
as it should be, and not even as it will be infuture. Of coursethe underlying thought
is that the situation will not become worse andthat it is up to the utilities toensure
this.
When interpretingthis standardit is also very important to realize that it only
appliesunder"normal operatingconditions."The documentspecifies a listo f situations
to which the limits do not apply.T his list includes "operationafter a fault," but also
"industrial actions" and such vague terms as
"force majeure" and "power shortages
due to externalevents."This list removes a loto f the potential value from the document. A descriptionof the electromagneticenvironmentshould include all eventsand
TABLE 1.3 Probability of Voltage Exceeding Certain Levels
Probability
95%
99%
99.9%
99.99%
99.999%
99.9999%
Frequency
Voltage Range
± 1.960'
± 2.580'
/l ± 3.290'
/1. ± 3.900'
J.,l ± 4.420'
J.,l ± 4.890'
u
J-L
207-253 V
200-260 V
193-268 V
184-276 V
178-282 V
173-287 V
±IO%
±13%
±17%
±200/o
±23%
±25%
50 per week
10 per week
I per week
5 per year
t per 2 years
1 per 20 years
Section 1.4 • PowerQuality and EMC Standards
33
variationsto which acustomeris exposed, notjust those which occurduring "normal
operating conditions." A voltage sagduring a severe lightningstorm (exceptional
weather) is equallydamagingas a sagduring a sunnyafternoonin May.
Looking at the documentin a more positive light, one can say
that it only gives
limits for what we have called"variations";voltage quality"events"are not covered by
the document.
What Next? Despite all itsshortcomings,EN 50160 is a very gooddocument.
It is probably the bestthat could be achievedunder the circumstances.One should
realize that it is the first time that the electromagneticenvironment has been
described in such detail in an official
document.Although limits are only given for
some of the phenomena,and although the standardonly applies during normal
operation,and althoughabsolutely noguaranteesare given, at least a first step is set.
Based on thisstandardone can see a
numberof developments:
• Utilities all over Europehave startedto characterizetheir voltagequality by
using themeasurements
as defined in EN 50160; thus 10-minute averages are
the
etc. The
takenof the rms voltage, 10-minute averages of harmonicvoltages,
characterizethe
values not exceeded
during 95% of the time are then used to
local voltage quality. Aproblemis that some utilities thencomparethe results
with the EN 5160 limits and state
t hat their voltagequality confirms with the
Europeanstandards.Understandingthe conceptof voltagecharacteristics,it is
TABLE 1.4 Voltage Characteristics as Published by Goteborg Energi
Basic Level
Phenomenon
Voltage Variations
Magnitude variations
Harmonic voltages
Voltage fluctuations
Voltage unbalance
Frequency
Voltage shall be between 207 and 244 V
Up to 4% for odd harmonic distortion
Up to l°,.{, for even harmonic distortion
Up to 60/0 THO
Up to 0.30/0 for interharmonic voltages
Not exceedingthe flicker curve
Up to 20/0
In between 49.5 and 50.5 Hz
Voltage Events
Magnitude steps
Voltage sags
Short interruptions
Long interruptions
accidental
planned
Transients
Frequent events shall be less than
3°.!cl in magnitude
No limits
No limits
On average less than one in three years
On average shorter than 20 minutes
Individual interruptions shorter than 8 hours
On average less than one 18
in years
On average shorter than 90 minutes
Individual interruptions shorter than 8 hours
The utility tries to minimize size and frequency of
transients whichinfluencecustomers
34
ChapterI •
Overview of PowerQuality and PowerQuality Standards
no surprisethat the local voltagequality is betterthan the limits given in the
standard.This result should thus absolutelynot be used by a utility to show
that their supply is goodenough.The statement" our supply confirms with EN
50160" isnonsense,as thestandarddoesnot give requirementsfor the supply,
but only existingcharacteristicsof the worst supply in Europe.
• Some utilities have come up with their own voltage
characteristicsdocument,
which is of coursebetter than the one described in the
s tandard.The local
utility in Gothenburg,Sweden hasdistributeda flyer with the limits given in
Table 1.4. The term"voltage characteristics"is actually not used in the flyer;
insteadthe term "basic level" is used [108].
• Measurementsare beingperformed all over Europe to obtain information
about other power quality phenomena.For voltage sags,interruptions,and
transient voltages no limits are given in the existing
document. A voltage
characteristicfor voltagesags,and for other events, ishard to give asalready
mentionedbefore. An alternativeis to give themaximum numberof events
below a certain severity, for 95°A, of the customers.Figure 1.22 gives this
voltage characteristicfor transient overvoltage, as obtained through the
NorwegianPowerQuality survey [67]. Such a choiceof voltage characteristic
would be inagreementwith the useof this same950/0 level for thedefinition of
the compatibility level.
Long Interruptions and
Reliability Evaluation
2.1 INTRODUCTION
2.1.1 Interruptions
A long interruption is a power quality event during which the voltage at a customerconnectionor at theequipmentterminalsdropsto zero and does not come back
automatically.Long interruptionsare one of the oldestand most severepower quality
concerns.The official IEC definition mentionsthree minutesas theminimum duration
of a long interruption. An interruption with a duration of less than three minutes
shouldbe called a"shortinterruption."Within the IEEE standardsthe term"sustained
interruption" is used forinterruptionslastinglongerthan 3 seconds[IEEE Std. 1159] or
longer than2 minutes[IEEE Std. 1250]. In thischapterthe term"long interruption"will
be used as aninterruption which is terminatedthroughmanualaction, thus not automatic. An interruptionterminatedthroughautomaticreclosureor switching, is called a
"short interruption" and will be treatedin detail in Chapter3.
2.1.2 Reliability Evaluation of Power Systems
An area of researchcalled "power system reliability" has developed,in which
numberand duration of long interruptionsare stochasticallypredicted.This areahas
long beenconfined to universities and to industrial power systems,but the recent
interestin power quality in all its aspects hascausedincreasedactivities in reliability
both at universities and in utilities. Anadditional reasonfor the increasedinterestin
reliability is the availability of cheapfast computers.In the past reliability evaluation
studiesof realistic power systemsrequiredlarge computers,gross simplifications,and
long calculationtimes. Many ideasproposedin the pastcan only now beimplemented.
Someof the basicsbehind reliability evaluationof power systems will be discussed in
Sections 2.4 and 2.5; some
exampleswill be presentedin Section 2.8.
35
36
Chapter2 • Long Interruptionsand Reliability Evaluation
2.1.3 Terminology
In this chapterthree terms willa ppearregularly: failure,outage,and interruption.
In daily life their meaningsare interchangeable,but in the reliability evaluation of
power systems, there are clear and
importantdifferences.
• Failure. The term failure is used in the general
meaningof the term: a device or
system which doesn ot operateasintended.Thuswe can talkabouta failure of
the protectionto clear a fault, but also of the failure oftransformer,and
a
even
about the failure of the public supply.
• Outage. An outageis the removalof a primary componentfrom the system,
e.g., atransformeroutageor the outageof a generatorstation.A failure does
not necessarily have to lead to an
outage,e.g., the failure of the forced
cooling
of a transformer.And the other way around,an outageis not always due to a
failure. A distinction is thereforemade between"forced outages"and "scheduled outages."The former are directly due to failures, the
l atter are due to
operatorintervention.Scheduledoutagesare typically to allow forpreventive
maintenance,but also theaforementionedfailure of the forced cooling of a
transformercould initiate the schedulingof a transformeroutage.
• Interruption. The term interruption has already been used before. It is the
situation in which a customeris no longer supplied with electricity due to
one or moreoutagesin the supply. In reliabilityevaluationthe terminterruption is used as theconsequenceof an outage(or a number of overlapping
outages),which is in most cases the same as the
definition used in the power
quality field (a zero-voltagesituation).
2.1.4 Causes of Long Interruptions
Long interruptionsare always due tocomponentoutages.Componentoutages
are due to threedifferent causes:
I. A fault occurs in thepower system which leads to an
intervention by the
power systemprotection. If the fault occurs in apart of the systemw.hich is
not redundantor of which the redundantpart is out of operationthe intervention by the protectionleads to aninterruptionfor a numberof customers
or piecesof equipment.The fault is typically ashort-circuitfault, but situations like overloadingof transformersor underfrequencymay also lead to
long interruptions.Although the results can be very
disturbingto the affected
customers,this is acorrectinterventionof the protection.Would the protection not intervene,the fault would most likely lead to an
i nterruption for a
damageto the electrical
much largergroupof customers,as well as to serious
equipment.
As distribution systems are oftenoperatedradially (i.e., without redundancy) andtransmissionsystems meshed (with
redundancy),faults in transof the supply,
mission systems do not have much influence on the reliability
but faults indistribution systems do.
2. A protectionrelay intervenesincorrectly, thus causinga componentoutage,
which might again lead to a long
interruption. If the incorrect tripping (or
maltrip) occurs in apart of the systemwithout redundancy,it will always lead
Section 2.2 • Observationof SystemPerformance
37
to an interruption. If it occurs in apart of the system withredundancythe
situationis different. For a completelyrandommaltrip, the chancethat the
redundantcomponentis out of operationis rather small. Randommaltrips
are thus not a serious reliability concern redundantsystems.
in
However
malt rips are often not fullyrandom, but more likely when the system is
protection: a correct
faulted. In that case there will be two trips by the
interventionand anincorrectone. Themaltrip trips the redundantcomponent just at themomentthat redundanceis needed.Fault-relatedmaltripsare
a seriousconcernin redundantsystems.
3. Operatoractionscause acomponentoutage which can also lead to a long
interruption.Some actionsshouldbetreatedas abackupto the power system
protection,either correct or incorrect. But an operatorcan also decide to
switch off certain parts of the system for preventive
maintenance.This is a
very normalactionand normally not of any concern tocustomers.There is in
most cases at least some level redundancyavailable
of
sothat the maintenance does not lead to an
interruptionfor any of the customers.In some lowvoltage networksthere is noredundancypresent at all, which impliesthat
preventive maintenanceand repair or changes in the system can only be
performedwhen the supply to apart of the customersis interrupted.These
interruptions are called "scheduledinterruptions" or "planned interruptions." The customercan take someprecautionsthat make the consequences
of the interruptionlessthan for a nonscheduledinterruption.This of course
assumesthat the utility informs thecustomerwell in advance,which is unfortunatelynot always the case.
2.2 OBSI!RVATION OF SYSTEM PERFORMANCE
Long interruptions have long beenconsideredas somethingworth preventing: the
numberand duration of long interruptionswas viewed as themeasureof how good
the supply was.T oday we would call it a powerquality indicator Of, in lEe terms, a
voltage characteristic.
Many utilities have recordsof numberand durationof interruptions,but mostly
for internal use. Theamount of publishedmaterial is relatively low. That not only
makes ithard to getinformationaboutsupply performancefor educationand research
purposes,but even forcustomersit is often hard to find out what the reliabilityof the
supply is. The former is
j ust an inconvenience, the
latteris a serious concern. A positive
exception to this is theprivatizedelectricity industryin the United Kingdom. The data
presented in theremainderof this section has mainly been
obtainedfrom the reports
published by the British Officeof Electricity Regulation(OFFER) [109]. Some additional information has beenobtainedfor The Netherlands[110], [111].
2.2.1 Basic Indices
As alreadymentionedin Section 1.3.2 the mainstochasticcharacteristicof any
voltagemagnitudeevent is the time between events, or (which is in effect the same) the
numberof events per year. The
latter is indeed oneof the maincharacteristicscollected
for long interruptions.Figure 2.1 shows the average
numberof interruptionsper customer for six consecutiveone-yearperiods. When the U.K. electricityindustry was
privatized in December1990 there was a serious concern
that the reliability of the
Chapter 2 • Long Interruptions and Reliability Evaluation
38
....
E 1.2-,---- --
-
-
-
- - -- - - - ----,
o
'@
o
l:;
~ 0.8
c:
1
o
06
.
.5 0.4
'-
~ 0.2
OJ
§
Z
0
90/9 1
91/92
92/93 93/94
Monitoring period
94/95
95/96
Figure 2.1 Numberof interruptionsper
customer.average forGreat Britain. (Data
obtainedfrom (1091.)
supply woulddeteriorate. Figure 2.1 clearly shows
t hat this has not been the case; the
numberof supply interruptionshas stayedremarkablyconstant.
Individual interruptions arecharacterizedthrough their duration,i.e., the time it
takes until the supply isrestored. Often the averaged uration of an interruptionis not
published but instead the total
durationof all interruptionsduring one year is provided.
This value is referred to as the
"minuteslost perconnectedcustomer"or more correctly
as theunavailability of the supply. Thedata for Great Britain (Wales,Scotland,and
England)is shown in Fig. 2.2. We again see
that the reliability of the supply remained
constant,with the exception of the year 1990/91,
during which severe blizzards made it
impossible to restore the supply within a few
hours. The numberof interruptionsdue to
this severeweatherwas relatively small. as can be
concludedfrom Fig. 2.1, but its
duration had a seriousimpact on the unavailability of the supply .
that
The collectionof this datais less trivial thanit may look . One should realize
most utilities do notautomatically become awarethat the supply to one or more
customersis interrupted. It is typically the customersthat report an interruption to
the utility . The startingmomentof an interruption,and thus theduration,is therefore
not always easy todetermine.The total numberof long interruptionsin the service
territory of a utility can beobtainedsimply by counting them , as eachinterruption
requires anoperatoraction for the supply to be restored. The
numberof customers
affected by aninterruption requires a studyof customerrecords which is often time
consuming. Some utilitiesjust assume a fixedamountof customersconnectedto each
feeder, whileother utilities link the interruption records with theircustomerdatabase
.
250-,----
- -- -- -- -- - --
-
-
--,
~
" 200
~
:.§.
:€
{j
150
100
=a
g 50
;:J
o
90/9 1
91/92
92/93 93/94
Monitoring period
94/95
95/96
Figure 2.2 Unavailability of the supply.
average forGreat Britain. (Data obtained
from [109].)
39
Section 2.2 • Observationof SystemPerformance
The calculationof the indices from the collected
d atacould proceedas follows.
Considera utility serving N,o, customers.During the reporting period (typically one
year) a total of K outagesin the system lead to aninterruption for one or more
customers.Interruption i affects N, customersand has aduration of D; minutes. The
averagenumberof interruptionsper customerper yearI is given by
(2.1)
The underlying assumptionoften used in theinterpretationof this data is that the
system average over 1 year, equals customeraverage
the
overmany years. Thus I
would also be the expected
numberof interruptionsper year for eachcustomer.But
variations in customerdensity, system design and
operation, and weather patterns,
make that not all customersare equal from a reliabilitypoint of view.
The averageunavailabilityper customerq, in minutesper year, may be
calculated
as
K
LN;D;
-
;=1
q=---
»:
(2.2)
The averagedurationof an interruption D is
(2.3)
This value isredundant,as it may becalculatedfrom (2.1) and (2.2) by using the
following relation:
-
q
D==
A
(2.4)
Utilities often publish two of these three values,
X, q, D.
Note that (2.3) gives theaverageduration of an interruption from a customer
perspective.From a utility perspectiveanothervalue is alsoof interest: the average
duration per interruption, Dint, calculatedas
(2.5)
This value givesinformation abouthow fast a utility is able torestorean interruption.
The outcomeof (2.4) and (2.5) iscertainly not the same.I nterruptionsserving more
customers,originating at higher voltage levels, tend to haveshorterduration.
a
Thus
the averagedurationper customeris likely to beshorterthan the averagedurationper
interruption.Which valueshould be used is open for discussion.
40
Chapter2 • Long Interruptionsand Reliability Evaluation
2.2.2 Distribution of the Duration of an Interruption
We will later seethat the costsof an interruption increasenonlinearly with the
durationof the interruption.The averagedurationof an interruptionwill thusnot give
the average cost. Tocalculate the latter, information about the distribution of the
durationshouldbe available.The U.K. utilities publish information aboutthe percentage of interruptions restored within 3 hours and the percentageof interruptions
restoredwithin 24 hours. This is part of the so-called"overall standardsof service"
which we will discuss inSection 2.3. The assumptionmade in almost all reliability
evaluationstudiesis that the componentoutageduration as well as the supplyinterruption durationareexponentiallydistributed.The exponentialdistribution,also called
"negative-exponentialdistribution," is the basicdistribution of most reliability evaluation techniques,as we will see in Section 2.5. The
probability distribution function of
the exponentialdistribution can be expressed as
F(t) = I - e-t
(2.6)
where T is the expected value
o f the stochasticvariable,which will be estimatedby the
averageduration. Knowing the averageduration, e.g., from Table 2.2 and Table 2.3,
the percentageof interruptionsrestoredwithin a time t} may bedeterminedas
(2.7)
Table2.1 gives thepercentageof interruptionsrestoredwithin 3 hoursfor a numberof
British distributioncompanies.The values in thecolumnslabeled"practice" have been
obtainedfrom [109], the values in thecolumnslabeled "theory" have beenobtained
from (2.7) by using theaverageduration of supply interruptionsfor the same year.
Using the averagedurationandassumingan exponentialdistributionwill overestimate
the impact of interruptions:the numberof interruptionslonger than 3 hoursis significantly lessthanwould be expected from the
measuredaverage. This is clearly a case for
more detailedreportingof the distributionof the durationof both componentoutages
and supplyinterruptions.It also calls forincluding nonexponentialdistributionsin the
reliability evaluation.
Figure 2.3 shows theprobability density function of the durationof all interrupt hat the
tions obtained for The Netherlandsbetween 1991 and 1994 [112]. We see
majority of interruptions has a duration between 30minutes and 2 hours, with a
TABLE 2.1 Distribution of Interruption Duration, 1996/97 Values for Various British
Utilities: Theory and Practice
Supply Not RestoredWithin 3 Hours
Company
AverageDuration in
Hours
A
2.38
B
C
1.38
D
E
1.45
1.63
F
G
1.62
2.27
1.38
H
1.42
Source: Data obtainedfrom [109].
Theory
Practice
28.4%
11.4%
12.1o~
12.6%
26.7%
19.3°AJ
9.8°AJ
7.3°AJ
7.0%
11.5%
8.6°AJ
13.4°AJ
11.4%
7.1%
15.90/0
15.7°~
41
Section2.2 • Observationof SystemPerformance
TABLE 2.2 Numberof Interruptions perCustomerper Year X for Some British Utilities
Distribution
Company
A
B
C
D
E
F
G
H
ReportingYear
90/91
91/92
92/93
93/94
0.41
0.58
1.70
0.76
2.85
1.46
0.82
1.69
0.47
0.62
1.11
0.68
2.29
1.29
0.74
0.82
0.38
0.57
1.29
0.96
1.95
1.18
0.86
0.75
0.37
0.56
1.25
0.59
2.14
1.19
0.89
0.92
94/95
0.40
0.70
1.21
0.65
2.20
1.24
0.70
0.96
95/96
0.33
0.61
1.39
0.85
2.23
1.16
0.62
0.97
Source: Data obtainedfrom (109).
TABLE 2.3
Distribution
Company
A
R
C
D
E
F
G
H
SupplyUnavailabilit~
q for Some
British Ut ilities
Repor ting Year
90/91
91/92
92/93
93/94
94/95
95/96
51
88
398
76
325
185
185
1004
67
75
118
65
212
176
108
87
53
52
69
144
63
200
167
121
97
58
70
128
94
212
133
102
105
54
67
151
85
233
111
88
95
77
122
91
212
184
129
87
Source:Data obtainedfrom (109).
6
Figure 2.3 Distribution ofd urationof
interruption, The Netherlands , 1991- 1994.
(Reproducedfrom Hen drik Boers and
Frenken(112).)
50
100
150
200
250
Duration of interruption in minutes
300
long tail up to 5 hours What
.
is a moreimportantconclusionis that the distribution is
absolutely notexponential.(The density functionof the exponentialdistribution has its
maximum for zeroduration and continues to decay afterthat.) To estimate the
expected costsof interruption it is important to take thisdistribution into account.
However, most studies still assume exponentialdistribution.
an
42
Chapter2 • Long Interruptionsand ReliabilityEvaluation
2.2.3 Regional Variations
Both Fig. 2.1 and Fig. 2.2 give the average
supply reliability for the whole of
GreatBritain. An old questionis, how useful is thisdatafor an individual customer.No
informationaboutindividual customersis available,but separatedataare availablefor
each of the 12d istributioncompanies[109]. Someof this datais shownin Table2.2 and
Table 2.3. In Great Britain the distribution companiesoperatethe voltage levels of
132 kV and lower. As will beshownin Table2.4 their systems are
responsiblefor 97°~
of the numberof interruptions,as well as for97% of the unavailability. The comparison between thedifferent utilities can giveinformationabouthow differences in system
design and operation influence the supply performance.Apart from the adverseweatheryear 90/91 thenumber of interruptionsand the supply unavailability have
remained remarkably constant. An accurate stochastic prediction method should
thus be capable of reproducing these numbers, an interesting challenge. We will
come back to thecomparisonbetweenobservationand predictionin Section2.7.
TABLE 2.4 Contributionsto the Supply Performance in Great Britain, 1995/96
Number ofInterruptions Unavailability per Customer
per Customer
per Year
Total
Low voltage (240/415 V)
6.6 and 11 kV
33 kV
132 kV
Other
Scheduled
1.03
0.06
0.63
0.13
0.06
0.03
0.12
158 min
22 min
81 min
12 min
7 min
4 min
32 min
Average
Duration of an
Interruption
150min
140/0
52%
8%
4%
3%
20%
370 min
130min
90 min
120min
130 min
270 min
Source: Data obtainedfrom [109].
From Table 2.2 and Table 2.3 we can also see
t hat companiesC, E, and H
suffered most from the severe
weatherin 90/91. It is possible tocalculatethe average
duration of an interruptionfor eachof the distribution companies,by using (2.4).For
companyH we obtainfor the year 90/91:D = ll~: = 594minutes,almost 10hours.For
the year 91/92 theaverageduration of an interruption was only 106 minutesfor the
samecompany.
An evenfurther subdivisionhas been made in
[109]: for each so-called"operation
unit" within the utility values are given forn umberof interruptionsand unavailability.
Based on thisdata a probability density function has beenobtainedfor the unavailability of operationunits. The results areshown in Fig. 2.4 and Fig. 2.5. Thelatter
figure includes the units with the highest
unavailability. We seethat 50% of the units
have anunavailability between50 and 100minutesper year.The 950/0 percentileof the
distribution is at 350 minutes. It is obvious from this graph that the averageunavailability doesnot give anyinformation aboutthe unavailabilitywhich can be expected by
a specificcustomer.One shouldnote that this is not thedistribution for the customers,
as not all operationunits have the samenumberof customersand not all customers
within one operationunit have the sameunavailability. Getting such agraph for all
customerswould require a much more intensivedata collection effort than currently
being done.
43
Section 2.2 • Observationof SystemPerformance
10
.§tJ
8
C+-c
6
.8
4
0
~
2
O~
0
Figure 2.4 Probability density function for
the averageunavailability in Great Britain.
(Data obtainedfrom [109].)
0
f")
I
N
0
tn
;
I
0
'"
1
\0
0
0\
...!.
00
-,
§
~
0
~
~
~
- -0
V)
,
~
0
'"
I
§
Interrupted minutes
~
I
00
2
0
(5
M
N
~
N
f")
~
0
V)
N
I
~
N
10......--...---------------------,
9
tJ 8
.~
~o
.8
7
6
5
§4
Z 3
2
1
Figure 2.SExtensionof Fig. 2.4 toward
higher values.
2.2.4 Origin of Interruptions
The data on numberand duration of interruptionsmight be veryinterestingby
itself, especially forcustomers,but they donot directly lead to anyunderstandingof the
causes ofinterruptions.For that purpose,additionaldatacollectionis required.A first
step is toobtaindataon the voltage level at which the
outageoccurredwhich led to the
interruption. Table 2.4 gives thisdata for Great Britain over the year 1995/96. The
values for other years are very similar. We see
that the major contribution to the
number of interruptions,as well as to theunavailability, comes from the medium
voltage network (6.6 and 11kV). Anexplanationfor this is not too difficult to give.
Thesenetworkshave noredundancyso thata componentoutageimmediately leads to a
supply interruption. The 33 kV network is partly operatedas a loop, hence its lower
contribution. The low voltage network is also operatedradially, thus without any
redundancy,still its contributionis rathersmall. This is because a low voltage
customer
of medium voltage feederthan of low voltage
is exposed to much more (kilo)meters
feeder. Thus, there will thus be much more outages affectingcustomerat
the
medium
voltagethanat low voltage. Anadditionalfactor is that a largerpart of the low voltage
networkis underground,which accountsfor a lower failure rate. Thedatain Table 2.4
that an
are showngraphically in Fig. 2.6 and Fig. 2.7. These figures again confirm
increased reliability of the supply can only be achieved
throughinvestmentat distribution level. An importantconclusionfrom Table 2.4, Fig. 2.6, and Fig. 2.7 that
is the
longestinterruptionsare due to scheduled outages and outages at low voltage level. But
44
Ch apt er 2 • Long Interruptions andReliability Evaluation
Other
3%
33 kV
12%
Figure 2.6 Contributionsto the numberof
supplyinterruptionsin Great Britain . (Data
obtainedfrom [109].)
Other
3%
132 kV
4%
Figure 2.7 Contributionsto the
unavailability of the supply inGreat Britain.
(Data obtainedfrom [109].)
as theseoccur less oftenthan interruptionsdue tooutagesat medium voltage level, the
latter make the largestcontribution to the unavailability of the supply .
Surveys inother countriesconfirm that the majority of interruptionsis due to
outagesat medium voltage level. Table 2.5 gives
interruption data obtainedin The
Netherlandsover the period 1991 through 1995 [110]. ("High voltage" is typically
150kV and 380kV, "medium voltage" 10 kV, and "low voltage" 400 V.) Here we see
the somewhatremarkablephenomenonthat about one third of the interruptionsfor
urban customersare due tooutagesin high voltage networks. This is due to the large
consumerdensity in the cities, and due to the fact
that all low voltage and medium
voltage distribution is underground. The numberof outages in medium voltage networks is thereforesimply very low. The high voltagenetworksare mainly overhead,
which makes themcomparableto the U.K. situation. We see 6interruptionsper 100
customersin The Netherlandsand 9 per 100customersin the U.K. ("132 kV" and
"others"), indeed a similar number. Like in the U.K ., the
unavailability of the power
supply in TheNetherlandsis mainly due to the medium voltage
distribution network.
Figure 2.8 shows thecontributionsof the three voltage levels to the
interruption
frequency, between 1976 and 1995, for the average low voltage
customerin The
Netherlands.The contribution of the low voltage and medium voltage systems to the
interruptionfrequency isratherconstant.The contributionof the high voltage network
45
Section 2.2 • Observationof SystemPerformance
TABLE 2.5 Supply Performancein The Netherlands,1991-1995
Urban Customers
High Voltage
Medium Voltage
Numberof interruptions 0.06/year 29%
2 minutes 15%
Unavailability
26 minutes
Interruptionduration
0.12/year 58%
9.5 minutes 73%
75 minutes
Low Voltage
Total
O.OI/year 50/0
1.5 minutes 12%
198 minutes
0.21/year
13 minutes
62 minutes
All Customers
High Voltage
Medium Voltage
Low Voltage
Total
Numberof interruptions 0.06/year 22%
Unavailability
2 minutes t 1%
Interruptionduration
26 minutes
0.20/year 740/0
15 minutes 79%
75 minutes
40/0
O.OI/year
2 minutes 110/0
199 minutes
0.27/year
19 minutes
70 minutes
Source: Data obtainedfrom [110].
0.4
i' 0.35
t)
>-
!,
0.3
~ 0.25
6
t
0.2
¢:l
a
r
.:;: 0.15
Figure 2.8 Numberof interruptionsper year
for the averagelow voltagecustomerin The
Netherlands,1976-1995,with contributions
from low voltage(x), mediumvoltage(0), and
high voltage( +) systems.(Reproducedfrom
van Kruining et al. [110].)
..=
0.1
0.05
Ol..------J.------L.----....L.---~
80
85
90
95
Year
varies much more. In some years (1985, 1991)contribution
its
is negligible, while in
other years (1990) they make up
half of the numberof interruptions.This large variation is partly of a stochasticnature(the numberof outagesof high voltagecomponents
leading to aninterruptionis very small)but also due toweathervariationshaving more
influence on the (mainlyoverhead)high voltagenetwork than on the (mainlyunderground)mediumvoltage and lowvoltagenetworks.
Figure2.9 shows theprobabilitydensityfunction for the durationof interruptions
originatingat different voltage levels in The
Netherlands[Ill]. For interruptionsdue to
high voltage componentoutages,the majority of durations is short: about 75% is
shorter than 30 minutes. Outagesin the medium voltageand low voltage networks
(typically 10kV and 400 V, respectively, in The
Netherlands)lead to longer interruptions. For medium voltage onlyabout 15% of the interruptionsis shorter than 30
about 5%. This has to do with the
minutes, for low voltage this value is even lower:
methodsused forrestorationof the supply.Outagesin the high voltage networksare
normally restoredvia operatorinterventionfrom a centralcontrol room. In medium
voltage and low voltage networksthere is no suchcontrol room and both fault locaFrom the density
lization and restorationof the supply has to take place locally.
functions in Fig. 2.9 it is clearthat 30 minutes is about the minimum time needed
46
Chapter2 • Long Interruptionsand Reliability Evaluation
High voltage
60
%
50
40
30
20
10
O'---.£""""",L-L-
0-1/4
114-112
1/2-1
1-2
2-4
Duration in hours
4-8
8-16
16-32
4-8
8-16
16-32
4-8
8-16
16-32
Medium voltage
40
% 35
30
25
20
15
10
5
O'--'=L-L-
0- 1/4
1/4-1/2
1/2-1
1-2
2-4
Duration in hours
Low voltage
30
%
25
20
15
10
5
o'--'"'-=L-.L._
0-1/4
1/4·112 112-1
1-2
2-4
Duration in hours
Figure 2.9 Probability den sity function for duration of interruptions,originating at
three voltage levels in The Netherlandspower systems. (Reproduced
from Waumans[III].)
for this. Almost 100% of medium and low voltage
networksin The Netherlandsare
underground. Restorationof the supply takes place
normally via switching in radially
operatedloops .
2.2.5 More Information
From recording interruption events, much moreinformation can beobtained
than just averageduration and frequency . We already saw origin of the
interruption
and theprobability distribution of the durationas examples ofadditionalinformation.
The amountof information that can beobtaineddepends on how detailed the record of
the interruptionis. There are twoapplicationsfor the recordedinformation, each with
their own requirements
:
This mainly requiresinformation about the
I. Increase the quality of supply.
origin of interruptions and the way in which the supply is restored.
For
47
Section 2.2 • Observationof SystemPerformance
example,the knowledgethat most interruptionsoriginateat mediumvoltage
level teaches usthat most gain can beobtainedby improvementsthere. But
supposethat for a certaincustomerinterruptioncostsare small forinterrupequipmentis supplied
tion durationsup to 2 hours, e.g., because essential
through a battery backup (an uninterruptablepower supply or UPS). By
using Fig. 2.9 it is shownthat improvementsin the low voltage network
become moreappropriate.To make such a decision it is
obvious that more
data is neededthanjust interruption frequencyand unavailability.
2. Serve as inputdatafor reliability evaluation studies. Thisrequiresa lot more
data, not just about interruptions but also about outagesnot leading to
interruptions.Most utilities and industriesdo keepinformation about outage frequenciesand durationsof components,but not much of it is openly
available. Some large surveys have been
performed to obtain outage frequencies: e.g., by theIEEE Industry Applications Society for industrial
power systems [21], and byCIGRE for componentsof high voltage networks [197]. What is clearly still missing aredata on failure of the power
system protection, and probability distributions for time betweenoutages
and time to restorethe component.Especially thelatter could become very
important in future reliability evaluationstudies, as theinterruption costs,
and thus theinterruption duration, becomes the desired
o utput. A detailed
literature survey performedby the author in the early 90s resulted in suggestionsfor expectedcomponentlifetimes [107]. The resultsof that study
are reproducedin Table 2.6.
TABLE 2.6 SuggestedValues for Number of Component Outages
and Failures
Component Type
Number of Outages per Number of Outages per
1000Components per Year Component per Year
MV IL V transformers
MV /MV transformers
HV jMV transformers
MV and LV circuit breakers
Disconnect switches
Electromagnetic relays
Electronic relays (single function)
Electronic relay systems
Fuses
Voltage and current transformers
Standby generators
failure to start
Continuous generators
UPS inverter
UPS rectifier
Underground ·cable (1000 meters)
Cable terminations
Cable joints
Busses(one section)
Large motors
Source: [107].
Failure
Probability
1-2
10-12
14-25
0.2-1
1-4
1-4
5-10
3D-100
0.2-1
0.3-0.5
20-75
0.5-20/0
0.3-1
0.5-2
30-JOO
13-25
0.3-1
0.5-2
0.5-2
30-70
48
Chapter2 •
Long Interruptionsand Reliability Evaluation
2.3 STANDARDS AND REGULATIONS
2.3.1 Limits for the Interruption Frequency
Long interruptions are by far themost severepower quality event; thus any
documentdefining or guaranteeingthe quality of supply should contain limits on
frequencyand durationof interruptions.The internationalstandardson powerquality
do not yet give anylimitations for interruption frequencyor duration. The European
voltage quality standardEN 50160 (see Section 1.4.3) comes closeststating
by
that
"under normaloperatingconditionstheannualfrequencyof voltageinterruptionslonger
thanthreeminutesmay be less than
10 or up to 50dependingon the area."The document
also statesthat Hit is notpossibleto indicate typicalvaluesfor the annualfrequency and
durations 0.[longinterruptions."
Many customerswant more accurate limits for the interruption frequency.
Therefore, some utilities offer their customersspecial guarantees,sometimescalled
"power quality contracts."The utility guaranteesthe customerthat there will be no
more than a certain number of interruptionsper year. If thismaximum number of
interruptionsis exceeded in a given year, the utility will paycertainamountof
a
money
per interruptionto the customer.This can be a fixedamountper interruption,defined
in the contract,or the actualcosts and lossesof the customerdue to theinterruption.
Some utilities offervariouslevelsof quality, with differentcosts. Thenumberof options
is almost unlimited: customerwillingness to payextra for higher reliability and utility
creativity are the maininfluencingfactors at the moment.Technicalconsiderationsdo
not appearto play any role insettinglevels for themaximumnumberof interruptions
or the costsof the various options. For a customerto make adecisionaboutthe best
option, datashouldbe available,not only aboutthe averageinterruptionfrequencybut
also on theprobability distribution of the numberof interruptionsper year.
Contractualagreementsaboutthe voltagequality are mainly aimed atindustrial
customers.But also fordomesticcustomers,utilities offer compensation.Utilities in the
U.K. have to offer a fixedamount to each customerinterruptedfor longer than 24
hours. In The Netherlandsa court has ruled that utilities have to compensatethe
customersfor all interruption costs, unless theutility can provethat they are not to
blame for theinterruption.Also in Sweden some utilities offer
customerscompensation
for an interruption.
2.3.2 Limits for the Interruption Duration
The inconvenienceof an interruptionincreasesvery fast when itsdurationexceeds
a few hours.This holds especially fordomesticcustomers.T hereforeit makessense to
not reduce thenumber of interruptions (which might be very expensive)b ut their
duration. Limiting the durationof interruptionsis a basicphilosophyin power system
design andoperationin almostany country.In the U.K., as anexample,the durationof
interruptionsis limited in three ways:
1. The Officeof Electricity Regulation(OFFER)setstargetsfor the percentage
of interruptionslasting longer than 3 hoursand for the percentageof interruptionslasting longer than 24 hours.These areso-called"overall standards
of service" [109].
49
Section 2.3 • Standardsand Regulations
2. Thedistributioncompanypays all customers whose supplyinterruptedfor
is
longer than 24 hours. This is a so-called
"guaranteedstandardof service"
[109].
3. The design of the systems is such
that a supplyinterruption is likely to be
restored within a certain time.
The OFFERregulationscontain,for eachdistributioncompany,a target for the
percentage ofinterruptionsthat is restored within 3 hours, and targetfor
a
the percentagerestoredwithin 24 hours. At the end of each year the
distributioncompaniesreport
togetherwith the actual achievement.
back to OFFER, which publishes the targets
Table 2.7 shows targets and achievement over 1996/97 for some
of the utilities. We
990/0, and
seethat most utilities meet their targets. All targets for 24 hours are at least
the 3-hour targets are no lower
than 800/0.
The maximumdurationof interruptionis also animportantpart of the designof
systems. As we will see in
Chapter7 the concept of" redundancy"plays a very important role in that. To achieve acertain reliability of supply, the power system should
contain a certain amount of redundancy.A common rule in the design of public
systems isthat the larger the number of customers
that would be affected by the outage
of a component,the more redundancythere should be present and the faster this
redundancyshould be available. Table 2.8 summarizes the way this is implemented
part of a so-called engineering
recommendain the U.K. [119]. These rules used to be
tion, and it has been in use in the U.K. for many years. When the utilities were
privatized thisrecommendationbecamepart of the license agreement.
Dependingon
the load size, maximumdurationsof interruptionare given. The larger the
a mountof
TABLE 2.7 Performance of U.K. Utilities over1996/97
24 hours
3 hours
A
B
C
D
E
F
G
H
Target
Achieved
Target
Achieved
80°A,
85%
950/0
93%
80%
80%
85%
850/0
80.7°A,
90.2%
92.70/0
93.0%
1000/0
99%
1000/0
100%
99%
99%
99%
99%
100%
100%
99.9%
100%
100%
100%
99.3%
100%
88.50/0
91.4%
86.6%
92.9%
Source: Data obtained from[109].
TABLE 2.8 DesignRecommendations
for the U.K. Supply System
Amount of Load Restored
Load Size
Immediately
Within 15 Min
Within 3 Hours
0-1 MW
1-12MW
12-60 MW
60-300 MW
Load - 20 MW
Load - 12 MW or 2/3 load
Total load
Source: U.K. EngineeringRecommendation P2/5
[119].
Load - I MW
Tota11oad
In Repair Time
Total load
Total load
50
Chapter 2 • Long Interruptions and Reliability Evaluation
load affected, the faster the
restorationof the supply. In termsof power systemoperation and design, thisrequires parallel supply for loads above 60 MW,automaticor
remotemanual transferfor loads above 12MW, and local
manualtransferfor loads
above 1 MW. The relation between reliability andpower system design is discussed in
detail in Chapter7.
2.4 OVERVIEW OF RELIABILITY EVALUATION
A number of books and hundredsof papers have beenwritten on power system
reliability. The most well-known books are those by Billintonand Allan [84], [85],
[86], but also thebook by Endreyni [87] and the IEEE Gold Book [21] treat this
subject inconsiderabledetail. Thelatter publicationdoesnot give detailedtheoretical
considerations,but a useful seto f basic calculations.It also gives a seto f component
outage rates, which issomewhatmissing in the other books. Interesting books on
power system reliability have also been
written in the German language:[88], [89],
and probably in other languagesas well. An overview of publications on power
system reliability in theinternational refereed literature, is published about once
every five years in theIEEE Transactionson Power Systems [90], [91], [92].
Other
national and
sourcesof information are reportson power system reliability issued by
international organizations[93], [94]. Also more and morebooks on power system
analysis, design, or operation contain chapterson power system reliability. In the
thoughtswill be presented
remainderof this section,and in Section 2.5, some general
about reliability evaluationof power systems.For more details, thereaderis referred
to the literature.
The power system is often divided into three
functional parts,each with its own
specific design andoperationproblemsand solutions:
• generation
• transport(transmission)
• distribution
In the reliability analysis asimilar distinction is made between three so-called
hierarchicallevelsof reliability:
• level I: generation
• level II: generationand transport
• level III: generation,transport,and distribution
Virtually all books and paperson reliability use thisclassification,either implicitly or
explicitly, but nor everybodyactually uses the term"hierarchicallevels." This being a
various techniques.
useful educationalconcept,it is used in this section to discuss the
The conceptof hierarchicallevels remains anapproximation,as most classifications.
The reliability of a generationstation dependsin part on the auxiliary supply, which
must be treatedas adistribution system, thus level III. Also, asubstantialpart of the
generationhas becomeembeddedin the distributionsystem, in somecountrieswell over
100AJ [120]. The amountof embeddedgenerationis likely to grow further, with more
industrial combinedheat and power(CHP), a growth in the useof small-scale renewable energy and possibly so-called
micro-CHPsappearingwith domesticcustomers.
Section 2.4 • Overviewof Reliability Evaluation
51
Anotherdisadvantageof this conceptof hierarchicallevels isthat it is developed
for the largepublic supplysystem inindustrializedcountries.For developingcountries,
for small insularsystems,andfor industrial power systems,different thoughtsmight be
needed. At the end
o f this section anequivalentof hierarchicallevels for largeindustrial
power systems will beproposed.
Despitethe shortcomingsof the classificationin hierarchicallevels, it still gives a
good insight into the subject. Newdevelopmentsare most likely to appearat those
places where theclassificationno longer holds, but to understandthosethe classification should be understoodfirst.
2.4.1 Generation Reliability
As we saw from theobservationresults presentedin Section 2.2, outagesof
generatorshave no influencewhatsoeveron the interruption frequency nor on the
supply availability experiencedby a customer.Thus, for a customer,level I reliability
studies donot appearvery important.This conclusionis correctfor an existing, wellplanned,and well-operatedpower system. But in theplanning stage, level Istudies
are extremelyimportant. In modern power systems,generationof power takes place
at- the highestvoltagelevel; thus a lack of generationbecomesimmediatelya national
or even internationalproblem. Such asituation should be avoided as much as possible. Because asuitable reserve in generationcapacity has beenplanned and is
availableduring operation,the customerdoes not have toworry about lack of generation anymore.
AnnualPeak Load. The rule that the total generationcapacityin a power system should exceed theannualpeak load is likely to be themost important planning
criterion in power systems. Planning and building of large power stations take
between 5and 10 years, thus decisions
about these have to bem adeseveral years in
advance.The most basic level Ireliability study is to calculatethe probability that
the availablegenerationcapacityis lessthan the annual peak load in a certain year
(e.g., 7 yearsaheadof the decision date). The
i nput data for such astudy consistsof
the expectedannualpeakload, the capacityof eachgeneratorunit, and its forced unavailability. The forced unavailability is the fraction of time during which a unit is
not availabledue to forcedoutages,Le., during which it is in repair. The assumption
to be made is that the probability that the unit is not available during the annual
peak isequal to the forcedunavailability. This gives us sufficienti nformation to calculate the probability that the available capacityis lessthan the annual peak load.
This probability is called the"loss of load expectation"(LOLE) of the annual peak.
Note that scheduledoutagesare not consideredin peak load studies. It isassumed
that preventivemaintenancewill not be scheduledduring the period of the year in
which the peakload can be expected.
Preventive Maintenance. Preventivemaintenanceof generatorscontributessignificantly to their unavailability. The unavailability consistsof two terms: theabovementioned"forced unavailability" and the "scheduledunavailability." The latter is
the fraction of time during which a unit is not available due to scheduledoutages
(Le., maintenance).The scheduledunavailability of a unit may exceed its forced
unavailability. The scheduledunavailability should not be treated as a probability,
like the forced unavailability. Generatormaintenancecan beplannedseveralmonths
or even more than a year ahead.The maintenanceplanning will be such that the
52
Chapter 2 • Long Interruptions and Reliability Evaluation
supply of the daily peak load willn ot be endangered.Typically, maintenanceis
scheduled away from the
a nnualpeak: if theannualpeak occurs in winter,g enerator
maintenanceis done insummerand the other way around. In tropical areas, where
the temperatureand thus the load do not vary much
during the year, this kindof
scheduling of maintenanceis not possible. Theconsequenceis that a higher LOLE
needs to be accepted
part of the time, or that additional units are needed. The problem can be especially
stringentin small systems(insular or isolated systems) where
the unit size is a largefraction of the total load.
A way of including preventivemaintenancein the level Ievaluationis to split the
For each period aLOLE is calculatedfor the peak
year into periods of, e.g., 1 week.
load overthat period. Thegenerationcapacityfor eachperiod excludes the unitsthat
are in maintenance.Such a study is typicallyperformed as an aid inmaintenance
scheduling.
The maintenancefrequency (i.e., how oftenmaintenanceis performed)is normally assumed given in level I studies. When
varying the maintenancefrequency it is
very importantto realizethatthis will influence thecomponentfailure rate. Anaccurate
model requires knowledge
a bout the aging of thecomponentand the influence which
of reliability evaluationwhich is
preventivemaintenancehas on this. This is an aspect
seldomconsideredin power systems. We will come back component
to
aging in Section
2.5.6.
Load-Duration Curve. The loss-of-loadexpectation(LOLE) quantifiesthe risk
that the generatorcapacityis not sufficient to supply the(annual)peak load. It does
not quantify the unavailability of the supply due to insufficientgenerationcapacity.
To obtain the level I contribution to the unavailability, a more detailed study is
required. Not only the unavailability of each generatorunit needs to beknown, but
also its outagefrequency and therepair time distribution. The load variation with
time and scheduledmaintenancehave to betakeninto accounthere as well. A simple
method is to use theload-durationcurve, approximatethis through a number of
steps, andcalculatea LOLE for each load level. Theapplicationof such calculations
is ratherlimited as they are toocomplicatedto be of use inplanningstudies, but the
influence on thecustomeris too small to be of anyimportancethere. Exceptions are
power systems inunderdevelopedor very fast developingcountries, where lack of
generationcan seriouslycontributeto the supplyunavailability.
Derated States. The simplestLOLE calculationsassume twostatesfor a generator unit: available andoutage(unavailable).In reality this is a gross oversimplification, especially for the larger units. It is very
common that due to an auxiliary
failure the unit will reach a so-called
" deratedstate"in which it is only able to generate part of its maximum capacity.An example is the failureof one of theburnersthis limits the combustioncapacityand thus the powercapacity.Consideringsuch a
failure as acompleteoutageof the unit underestimatesthe level I reliability. In the
planning process this leads to an
overestimationof the number of units that have
to be built. As costsreduction becameimportant several years ago, the
interest in
derated state models increased. Anadditional factor explaining the use of more
detailed models is again the
availability of faster computersenablingthe implementation of these more detailed models.
Operating Reserve. Reliability studies are typicallyperformed for planning
purposes,where questionslike "how many generatingcapacity should be available
Section 2.4 • Overview ofReliability Evaluation
53
ten years fromnow" are addressed.In that case it is assumedthat all generating
plants and linesthat are not in repair or in maintenanceare availablefor generation
and transport.For operationalreserve studies the
situation is different: one needs to
take into account only those plants that are actually running or which can be
brought online at short notice and assess the riskthat the total load cannot be
supplied within the next few hours.
2.4.2 Transmission Reliability
Level II reliability concernsthe availability of power at so-called bulk supply
points: typically transmissionsubstationswhere power istransformeddown to distribution voltage. Thepower not only has to begeneratedbut also transportedto the
taken into account.
customers.The availability of sufficient lines or cables has to be
Level II reliability studies are much more difficult
thanlevel I studies, and are still
under
considerabledevelopment.Some of the difficulties and suggested
solutions are discussed in theremainingpartsof this section.
Overloadingof Lines. Due to theoutageof a transmissionline the flow of active and reactive powert hrough the transmissionsystem changes. This can lead to
overloadingof other lines. Thestandardexample is theoverloadingof a parallel line.
Normally parallel lines will be rated such that the outageof one of them will not
lead to overloadingof the other. Thus two lines feeding a 200 MVA loadshouldeach
be able totransport200 MVA. This so-called(n - 1) criterion has been animportant
part in the design of transmissionsystems: a systemconsisting of n components
should be able tooperatewith any combinationof (n - 1) components,thus for any
single-componentoutage. In important parts of the system, more strictcriteria are
used:(n - 2), (n - 3), etc.
Large transmissionnetworkshave become so complex
that it is hardto realize the
actual loction of the parallel paths.In systemsthat are meshed across several voltage
levels,overloadingdue to anoutageis a serious risk as some recent
interruptionsand
"almost-interruptions" have taught us. The risk has been
increasedby the growing
transportof power over large distances.
For level II studies in large systems, a load flow
calculationhas to beperformed
for each outage.Thesecalculationsmake level II studies very timeconsuming.The
processingof overloadeventsdependson the policy used by the utility to rectify the
overload.Typically two models for this are used in reliability studies.
I. The overloadleads to anoutageof the overloadedcomponent,eitherimmediately or after a certaindelay which coulddependon the amountof overload. As this secondoutagecan lead tofurther overloadsa cascadeeffect may
occur.
2. The overloadis assumed to be
alleviatedthrough the sheddingof load.
Reliability of the Protection. The power systemprotection'saim is to remove
faulted componentsfrom the system so as to limit the
damageas much as possible.
Failure of the protection to remove thefaulted componentcan lead to significantly
more damage,including an interruption for customerswhich would normally not be
interrupted.It will be clear that the reliability of the protection is an important part
of the reliability of the supply. Protectionfailure is alreadymentionedas oneof the
54
Chapter2 •
Long Interruptionsand Reliability Evaluation
underlying causes ofcomponentoutages.The power system protection can fail in
several ways.
1. The protectionfails to operatewhen required.In that case thebackupprotection will operateand clear the fault. Thisbackupprotectionoften clears
more than only the faulted componentmaking the impact on the system
much bigger. As thetransmissionsystemoften has only singleredundancy,
such aprotectionfailure canpotentiallyeliminatethe redundancyandlead to
an unnecessaryinterruption.
2. The protectionoperateswhen not supposedto. If this happensindependently
of anotherevent it will simply lead to anoutageof the protectedcomponent.
The redundancyin transmissionsystems makesthat thesemaltrips do not
have a biginfluenceon the reliability of the supply.
3. The powersystemprotectionshows amaltrip when anotherrelay issupposed
to operate.This leads to the loss
o f two componentsat the same time. The
large currentsflowing throughthe systemduring a shortcircuit makethis an
event which has to be
consideredin the calculations.Accuratemodelsfor it
have not beendevelopedyet. The main problem is that each fault can in
theory lead to a malt ripof any of the other relays in thepower system.
4. The power system protection shows amaltrip due to anotherevent in the
system, e.g., aswitchingaction. Although the event itself doesn ot lead to any
required protectionintervention,it can still potentially eliminate the redundancy. Thereasonis that several relays willexperiencea similar disturbance
and thus all might show amaltrip at the samemoment.
The reliability of powersystemprotectionis often split into two aspects," dependability" and "security." The dependabilityis the degreeof certaintythat the protection
will operatecorrectly (point 1 above); thesecurity is the degreeof certainty that the
relay will not operateincorrectly. As shown above this neglects thedifferent aspects
within the "security.'
Dynamic SystemBehavior. Most componentoutagesare due toshort-circuit
faults. Occurrenceand clearing of a fault lead todynamic oscillationsin the system.
These can lead tooverloading or tripping of components.This so-called security
aspectof level II reliability is often not taken into account. To include it, detailed
dynamic models of the system are needed. In the
reliability literature a distinction is
made between adequacy(static evaluation) and security (dynamic evaluation). The
adequacypart is taken care of by most evaluation techniques,but security is often
forgotten. In a well-designedtransmissionsystem ashort circuit should not lead to
loss of any generator,or overloadingof any component.But one canthink of several
situationsin which the dynamic system behaviorcan have asignificant influence on
the level II reliability.
• The system might be secure for each
short circuit in an otherwiseundisturbed
system, but not for short circuits in a system in whichalreadyone or more
componentsare out of operation.Both thestatesbefore andafter the fault (i.e.,
after removal of the faulted component)might be healthy, but the transition
between the twomight not be healthydue to largedynamic oscillations.The
Section2.4 • Overview of Reliability Evaluation
55
system couldappearto have double or triple redundancywhere in reality it
only has singleredundancy.
b ackupprotection;
• Failure of the protectioncan lead to fault clearing by the
this leads to a longerfault-clearing time and thus to more adverse
dynamic
effects. The system might be
stablewhen the fault is cleared by the
primary
protectionbut not when the fault is cleared by the
secondaryprotection.
• In small power systems with two centers generation,a
of
fault close to a
generator might lead to somegeneratorsaccelerating,while others slow
down. The difference between their
rotor increases very fast, leading to large
instabilities.This phenomenonis especially severe for systems withtransmisa
sion grid at voltagesof 10 to 30 kV with mainlyundergroundcables [113]. A
reliability evaluationstudy for such a system
should not just considercable
outagesbut also theunderlyingshort-circuitfaults.
• In industrial power systems themaximum motor load connectedto a bus is
limited to a certain fraction of the short-circuit level of the bus. Theactual
motor load is oftenratherclose to this limit. If in the courseof time theamount
of motor load grows, some faults can lead loss-of-synchronism
to
of synchronous motorsor to stalling of induction motors.
Common-ModeOutages. The componentsin a level II study are often considered independent,i.e., the outageof one componentdoesnot dependon the stateof
the others. But sometimestwo or more componentoutagesoccur at the same time.
of a tower carrying two circuits and excavation
Classical examples are the collapse
of the other aspectsof level II relialeading todamageof two parallel cables. Several
bility studies (failure of theprotection, overloadingof a parallel line) are sometimes
also consideredcommon-modefailures. For example, a malt ripof a relay during a
fault on the parallel line will lead to anoutageof both lines. By modeling this as a
common-modefailure, no detailedprotectionmodel is needed.
Weather-RelatedOutages. The outagerate is in most studies considered- constant, but in reality this is not the case.
Many outagesare weatherrelated (lightning,
storm, snow) and thusstrongly time dependent.For nonredundantsystems this does
not matterat all, but for parallel systems it will significantly increase the
interruption
rate, evenif the averagecomponentoutage rate stays the same. Some
numerical
examplesof this effect are given in Section 2.8.
The IEEE standardfor collecting outagedata [198] recommendsto distinguish
between three levels of
outagerate:
• normal weather
• adverseweather
• major storm disaster
The contribution of adverseweatheron the outageof transmissionand distribution
systemcomponents,for a U.K. utility, is shown in Table 2.9[199]. Different utilities
will have different contributing phenomena,especially when they are indifferent
climates (snowstorms are more likely in Scotland than in Texas), but the general
impressionis that adverseweatherrelated outagesare the biggestc ontribution to the
outagerate.
56
Chapter2 • Long Interruptionsand Reliability Evaluation
TABLE 2.9 VariousContributionsto theOutageRateof Transmissionand Distribution
Componerits
Causeof Outage
TransmissionSystem
Distribution System
9%
52%
32%
50/0
2%
12%
11 %
7%
39%
21°tla
8%
Lightning strikes
Snowlice on lines
High winds
Plant failures
Line interference
Animal/bird strikes
Adjacent loads
2°tla
Source: Data obtainedfrom [199].
2.4.3 Distribution Reliability
Most publishedwork on power system reliability concernsthe generationand
transmissionsystems,what has been called level and
I level II before. Level III (distribution) reliability studiesare rather rare, although this is changingin the last few
years. The lackof interest in distribution reliability is clearly not due to the high
reliability of the distribution system. In fact,both interruptionfrequencyand unavailability are mainlydeterminedby events atdistribution level, both mediumvoltage and
low voltage. A numberof reasonscan be given for the lacko f interestin distribution
system reliability:
• The interestin distributionsystem research is in general
(much)lower thanthat
in transmissionand generation.
• Reliability of power transmissionand generationis of national interest,and
thus requires more effort. An
i nterruptionoriginatingat thetransmissionlevel
will affect a largepart of the system,and is thus more likely to lead to newspaperheadlines.
than in distribution systems
• Investmentsin transmissionsystems are easier
because there are much
moreof the latter. This meansthat a reliability analysis
of variousdistribution alternativesis not attractive.
• Reliability studies in distribution systems are relatively simple, which
make
them lessattractiveto the academicworld.
• A reliability analysis would only beof interestto the customerif it would give
an absolutevalue of the interruption frequency oravailability. A widely held
belief used to bethat the results of reliabilitystudiescan only be used in a
relative sense (i.e., to
c omparealternatives);such astudywould thereforebe of
no use to thecustomer.
But, as already said, the interest in distribution system reliability is growing,
probably due to the increasingattention for the customers'interests. Distribution
system reliability has its ownproblemsand solutions,some of which we will discuss
below.
Radial Systems. Distribution systems are mostoften radially operated.The
consequenceo f this is that each componentoutagewill lead to a supply interruption.
To obtain the interruption frequency one only needs to sum the
outagerates of all
Section 2.4 • Overviewof Reliability Evaluation
57
componentsbetween the' bulk supply point and the customer. Occasionally,
parts of
the system areoperatedin parallel or meshed. As this concerns small
parts of the
system, themathematicaldifficulties for calculatingthe interruptionfrequency remain
limited.
Duration of an Interruption. The main problem indistribution system reliability concerns theduration of the interruption. As we will see later, the costs of interruption increasesnonlinearly with its duration. The probability distribution function
of the interruption duration is of great influence on the expected costs. Itfurther
is
importantto realizethat the restorationtime depends on the
position in the network.
The averageinterruption duration, and thus theinterruption costs, cantherefore
vary significantly throughoutthe network. Theduration of an interruption consists
of a numberof terms, each of which has a stochastic
character.A list of contributing
terms is given, e.g., in[121] and [122]; the most relevant ones are
•
•
•
•
receive alarm,contactor travel to affectedsubstation;
find fault location or faulted section;
perform required switching actions;
restore supply.
A well-known law in stochastictheory isthat the sum of a sufficient number
of
stochasticterms has anormal distribution. Thus thedistribution of the interruption
durationbecause of itsstochasticnatureis likely to benormal and notexponentialas
assumed in mostcalculationmethods. This could give unrealistic values for the interruption costs.
The Availabilityof the Alternative Supply. The list of terms given above, contributing to the duration of an interruption, assumesthat the alternativesupply is
available. Thus, themoment the fault is located (or the faulted section is identified)
the supply can be restored. But this is not always the case, as
alternativesupply
the
can also beinterrupted,or the alternativesupply is only able to take over
part of the
load. In that case the supply can only be completely restored after
repair or replacement of the faultedcomponent.When the supply can be restored by switching, the
customerexperiences a"long interruption." When the supply can only be restored
through repair/replacement,the customerexperiences a"very long interruption" as
defined in Section 1.3.3. The frequency of very long
interruptionswill be rathersmall
in most distribution systems (with the exception
o f remote rural networks), but the
interruption costs may become very large, which makesimportant
it
that they become an essential
part of the reliability evaluationresults.Another reason forputting
special emphasis on very long
interruptionsmay bethat the utility has to publish the
number of interruptionsnot restored within acertain time, or has to pay damages
for these"very long interruptions."
To get exact detailsof the distribution of the duration of interruptions,complicatedstochasticmodelsof the system are needed. But a two-step
approachcan be used
if one is only interested in the frequency of very long
interruptions. For very long
interruptions,the time-scale of interest is longer
thanthe time needed for the
alternative
supply to be made available.
For the assessment
of the numberof very long interruptions the switches used to restore the supply can be considered in a closed
position
already. Toevaluatethe reliability of the resulting system, techniques developed for
58
Chapter2 • Long Interruptionsand Reliability Evaluation
transmissionsystems may be used. The models
requiredfor this are muchmore complicated than for predictingthe total interruption frequency.
Some of the before-mentionedaspectsof transmissionsystem reliability (common-modefailures, adverseweather,overloading)have to beincorporatedin a level
III study if the number of very long interruptionsand/or the interruption duration
distribution are of interest.
Adverse Weather. Adverse weather not only influences thenumber of very
long interruptions(by increasingthe probability that both a feeder and itsbackup
are not available) but it also makesrepair much more difficult. Blizzardsand heavy
thunderstormscause asubstantialfraction of outages.During the storm, repair is
very difficult, if not impossible,and after the storm the largenumberof outagescan
make this processmore difficult given that repair crews have tohandle the outages
one after the other. Such aspectsof the reliability of the supply are extremely difficult
to take into accountin a stochasticmodel. As alreadymentionedbefore, oneof the
problemsis the lack of data, but certainly not the only one. But despite the
mathematical difficulties, more datacollection must be encouraged.Also, the collecteddata
should be madeavailablefor a wider public.
Embedded Generation.The presenceof embeddedgenerationsomewhatcomplicates thereliability calculations.But the amountof embeddedgenerationis seldom
large enoughto have asignificant influence on thereliability of the supply. Industrial
power systems are anexception because in such cases
embeddedgenerationcan be
used toobtain a very high levelo f reliability.
Embeddedgenerationin public distribution systems consist mainlyo f wind turof the distribution system is suchthat the
binesand CHP units. In all cases the design
outageof one generatorunit will not lead to anoverload,and thusnot to an interruption of the supply for any of the customers,Thereforethe presenceof the embedded
generationdoesnot influence theinterruptionfrequency. Anexceptionare those cases
where outageof a generatorleads to aninterruption indirectly, e.g., when theheat
productionof a CHP unit is essentialfor an industrialprocess, or when caontractwith
the utilities requiresload sheddingupon a generatoroutage.
The presenceof embeddedgenerationcan have some influence on the
availability
of the alternativesupply, and thus on the frequency
o f very long interruptions.The
connectedto the
interruptionwill normally lead to the lossof all embedded generation
affected feeder.Thus the alternativesupply also has tosupply this additional load.
Further, embeddedgenerationconnectedto the alternativefeeder can havetripped
on the voltage sag due to the fault which led to the
interruption. The speed with
which this generationbecomesavailableagain will influence theprobability that the
alternativesupply is able to take over allload from the affected feeder.
2.4.4 Industrial Power Systems
Large industrial and commercialusers own andoperatetheir own medium voltage distribution system. Thelargestusers even own ando peratea high voltage network. The point-of-connectionto the public supply is somewhereat distribution or
transmissionlevel: thecustomeris responsiblefor the further distributionto thevarious
structureis
pointsof utilization. In these so-calledindustrialpowersystems the general
often somewhatdifferent than in public systems. Also there is no need for
separate
studies atseparatehierarchicallevels; all that mattersis the continuity (or whatever
59
Section2.4 • Overview of Reliability Evaluation
word one likes to use)o f the supply to theequipmentessential for theproduction
process. A possible listof questionsthat need to be addressed for a reliability study
in an industrial power system is given below. We will only discuss
interruption frecomparedto the
quency below.Restorationof the supply will often take place faster
time it takes torestartthe productionprocess. Of course this is not always the case, and
for someindustrial systems, thequestionsneed to be modified. The list below should
not be blindly followed, but be used as a basis for a specific study.
starting
Each of the questionsgives feedback on the design of the system. The
point may be the existing system, or detailed design based on past experience. The
whole "design process"is shown in Fig. 2.10. The term"layer" has been used here
to distinguish from the "hierarchical levels" used for the reliability analysis
o f the
public supply, but in factboth terms denoteexactly the same.
I. How often will the availablegenerationnot be enough to~upply the load?
This layer correspondsto hierarchicallevel I in the public supply, for
the
which a largenumberof tools are available. Some aspects of calculations
are already mentionedin Section 2.4.1. A few pointsof special interest to
industrial systems need to be
mentioned.
• Maintenanceon generatorunits can play a veryimportant role in industrial systems. The load does not show much
variation through the year,
thus maintenancecannotbe scheduledduring a period of low load. This
means that the generation capacity will influence the scheduling of
Changegeneration
Changetransportsystem
Changestabilityaspects
Changedistributionsystem
Changeequipment
immunity
Changeequipmentreliability
andredundancy
Figure 2.10 Reliability layers in industrial
power systems and their role in system design.
60
Chapter2 • Long Interruptionsand Reliability Evaluation
•
•
•
•
maintenance.The lower the reserve
(differencebetween loadandcapacity)
the less likelythat maintenancecan beperformed.
The influenceof maintenanceon aging can only be assessed
ratherqualitatively as accuratemodels are still lacking.T hereforea constantfailure
rate will often be used. Inthat case oneshouldrealizethat the calculation
resultscannotbe used tooptimize the maintenancefrequency.
Powergenerationunits may be linked, e.g.,through the useof a common
steam circuit. This needs to be
takeninto accountin the reliability studies
as it might increasethe probability that two or more·units have anoutage
at the same time.
During capacity shortagesor when thecapacity margin is Iowa load
sheddingpolicy is often in place. This needs to be
i ncorporatedin the
reliability calculations.
When the plant is connectedto the public supply (which is mostly the
case), itsreliability needs to beconsidered.When the plant is fed via
multiple infeeds,common-modefailures need to beconsidered.
2. How often will a situation occur that the generationis available butthat it
cannotbe transportedto the load?
This layer correspondsto hierarchicallevel II in the public supply. The
variousconsiderationsare very similar,but with somedifferencein emphasis.
• Componentloading is higher in industrial systems,and more constant.
Thereforeassessmentof overloadsdue to outagesbecomesmore important, but load variation often does not need to beconsidered.
• Distancesbetweensubstationsare much smaller,which makessubstation
failures toplayalargerrole (relatively speaking)thanin the public supply.
3. How often willtransientinstability lead to a plant trip?
This is arathernewsubject,correspondingsomewhatto the securitypart
of hierarchicallevel II. In industrial systems, with largemotor load, on-site
generation,and short distancesbetween them,t ransientstability aspectscan
play a very important role. What is needed first is aprediction of the frequencyof variousshort-circuitevents,and next an assessment
of the effects
of each event on the system
stability. The event frequenciesfollow from
earlier reliability calculations.Assessing the effects
o f the event requiresa
detailed model and can become a severe
strain on the computer power.
Performinga detailedtransientstability calculationfor a large system is no
longer too difficult with modern computerspeed and memory, but for a
reliability study such acalculationneeds to beperformedfor many possible
systemstates(preferablyfor all possiblestates).Even amedium-sizedsystem
may requirethousandsof transientstability calculations,which still places a
severestrain on the computerpower. Two optionsare availableto limit the
calculationtime.
• Use a simplecriterion to assess the system
stability, e.g., theratio between
fault levelandmotor load, or the differences inrotor angles at themoment
of fault clearing.For the latter, simple models can be used, e.g., change
the
in active power between thepre-eventand during-event steady states.
Apply this simple criterion to all (or at leastmany) system states.The
criteria might appeargross simplifications, which would never be accep-
Section 2.4 • Overview of Reliability Evaluation
61
table for aconventionaltransient-stabilitycalculation.But all we need to
know here is the sum
of the frequenciesof all events leading to an
unstable
situation.
• Use adetailedsystem model, but limit thenumberof events to bestudied.
A first pruningis to look only at those events which involveshortcircuit
a
and for whichboth the initial steadystateand theresultingsteadystateare
stable. A secondpruning is to stop looking for stateswith more components out, when astate has unstableevents associatedwith it. As an
example, if a fault leads totransientinstability when two of six generators
are out of operation,there is no need to study a fault when three generators are out.
One shouldnote that it is not theactualinstability limit which matters,
but whethergeneratorsor motorswill be tripped by their protection(undervoltage, overcurrent,reverse-power,under-or overfrequency).This can happen in systems which are in
principlestill stable.Thusa detailedmodel would
also requiresufficient detailsof the protectionpresentin the system.
4. How often will the distribution system fail to
transportpower to the plant?
Layer 4of industrial powersystem reliabilitycorrespondsto level III in
the public supply. We can thus apply similar
techniques,with the difference
that the duration of an interruption is often not so important in industrial
of the interruption duration which makes
systems. As it is the assessment
reliability analysis indistribution systemscomplicated,the calculationsin
industrial distribution systems will be simplerthan in public systems.
The distribution systemstartsat the transportsystemstudiedin layer 2
and layer 3, and ends at the
equipmentterminals.The various distribution
systems arenormally consideredindependentof each other. An industrial
of equipmentwith
distributionsystem can be extremely complex: many pieces
many levelsof redundancyand importance.Some kindof pruning needs to
be made before a study' can startedwith
be
a reasonablechanceof success. A
first pruningis to only considerthe supply toequipmentwhich is essential for
the operationof the plant.
A decision to be madebeforehandis where thetransmissionsystemstops
and thedistribution system begins. The answer to this will again
dependon
the detailsof the study. For smaller systems itmight be appropriateto not
make anydistinction betweentransmissionand distribution, while for large
systems eachplant is consideredas aseparatedistribution system.
5. How often will the plantoperationbe interrupteddue to insufficient voltage or
currentquality?
In this layer allother power quality phenomena(i.e., apart from interI through 4) have to be assessed.
ruptions which were discussed in layers
Examplesof voltage quality events to bestudiedare:
• Transientovervoltages.
• Voltage sagsand swells.
• Notching and burstsof harmonicdistortion.
• High-frequencyconducteddisturbances.
To studyall these in as muchdepthas for the longinterruptionswould lead
of
to extremely long studieswithout much hope of useful results. The level
detail againdependson the system. Anappropriatechoice is to only look at
62
Chapter2 •
Long Interruptionsand Reliability Evaluation
first or secondorderevents (firstordereventsare shortcircuits in the normal
system,secondorder eventsare short circuits when alreadyone other componentis out of operation).
Thesekind of studiesare extremelyrare,andwhere they aredonedo not
contain much quantitative details. Still, even the decision to not study a
certain type of event in detail becauseit is not likely to be of influence is
alreadymuch better than simply forgetting aboutit.
To actually determinethe number of equipmenttrips is not possible
without a detailed knowledgeof equipmentimmunity. In the designphase
of the system, thisinformationis simply not available.It will then be easierto
determinethe electromagneticenvironmentwhich the equipmentwill experience and to proposeimmunity requirementsfor the equipmentto be used.
Here it becomesimportant to distinguish between(voltage) variations and
(voltage)events,as describedin Section 1.3.
Currentquality eventswill not directly lead to tripping of the plant, but
utility requirementsmight force a plant shutdown,e.g., when theharmonic
currentdistortionexceeds acertainlevel. If such ashutdownwill have severe
consequences,
it needs to beconsideredin the reliability study.
6. How often will theplant operationbe interrupteddue to the failure ofessential
equipment?
Equipmentfailure is normally hot consideredas part of supply reliability, but in an industrial system it isequally important. There is no need to
build a very reliable power system if theplant will stop twice a week due to
equipmentproblems.Industrial customersoften use theterm interruptionin
a more general meaning than the utility. The descriptive terms "voltage
interruption" and "interruption of plant operation" indicate the difference
in interpretationrather well.
Detailed knowledgeof the plant processis needed toperform a study
like this. Like in severalof the stepsbefore, some seriouspruning will be
neededto make the study feasible. It might even bethat only a qualitative
assessment
is feasible.
Note that there is someoverlapwith layer 4 (distribution systems)and
layer 5 (equipmenttrips due to voltagequality events).
Additional aspectsto be consider~d are:
• redundancyof equipment,e.g., thefunction of a motor being taken over
by anotherone;
• "linkage betweenplantson the productionside, e.g., thesteamproduction
by one plant which is neededto operateanotherplant.
2.5 BASIC RELIABILITY EVALUATION TECHNIQUES
2.5.1 Basic Concepts of Reliability Evaluation Techniques
Stochastic Components.For a reliability evaluation study, the power supply
system is splitinto stochasticcomponents.The choice of componentsis rather arbitrary: the whole transmissionsystem might be one component,but a single relay
could be severalcomponents.Each componentcan be in at least two states:healthy
and nonhealthy,the latter often referredto as theoutagestate.For a two-statecomponent,two eventscan occur: thetransition from the healthyto the nonhealthystate,
Section 2.5 •
63
BasicReliability EvaluationTechniques
an outageor failure event; and the reverse
transition(i.e., from thenonhealthyto the
healthystate), therepair or restore event.
The systemstateis a combinationof all event states; if thestateof one of the
componentschanges,the systemstatechanges. The system
s tatefor a system withN
componentscan bethoughtof as a vectorof rank N. The valueof each element is the
state of the correspondingcomponent.An event is atransition between two system
states, due to the
changein stateof one or morecomponents.
EXAMPLE Consider,as an example, the system in Fig. 2.11:generatorwith
a
generatortransformer,feeding into a large system via two
parallel transmissionlines and atransformer. We areinterestedin the reliability of the supplyinto the large system, thus, at
point C
in the figure.
Ll
Figure 2.11 Power systemexample,for choice
of stochastic components.
A
L2
A possiblesubdivisioninto stochasticcomponentsis as follows:
1. generatorplus generatortransformerTl
2. substationA
3. line Ll
4. line L2
5. substationB
6. transformerT2
In case adetailedstudy is neededof the generatorplus thegeneratortransformer,component1
may besubdividedinto stochasticcomponentsas follows:
1.
2.
3.
4.
the mechanicalside of the generator,including the fuel availability
the electrical side of the
g enerator,including its protection
the generatorcircuit breaker
the auxiliary supply
5. the generatortransformer
6. the protectionof the generatortransformer
The Interruption Criterion. For each systemstateor for each event, an"interruption criterion" is used todetermineif this state or eventshould be countedas an
interruption or not. In most studies theinterruptioncriterion is rather trivial, but for
more detailed studies, especially for
Monte Carlo simulation, the definition of the interruption criterion becomes animportant part of the modeling effort. It is recommended to spend at least some time on defining interruption
the
criterion for a
of interruption criteria are given
reliability evaluationstudy. Some simple examples
below. Note that these arejust examples, andcertainly not the only possibilities.
64
Chapter2 •
Long Interruptionsand Reliability Evaluation
• In a level I studya stateis an interruptionstateif the generatorc apacityis less
thanthe load demand.Note that thereis only oneinterruptioncriterion for the
whole system.Eachcustomeris equal at this level.
• In a level II study a state is an interruption state for a given transmission
substationif the maximum power that can be transportedto this substation
is less than the demand. For level II studies, each substationhas its own
interruptioncriterion, thus its own reliability.
• In a level II security study an event is an interruption event if the transient
phenomenondue to theevent leads totripping of generatorsand/orload.
• In a level III study a stateis an interruptionstatefor a given customerif the
voltageat the customerterminalsis zero.
• In a level III power quality study an eventis an interruptioneventfor a given
device if it leads to thevoltage at the device terminals to exceed certain
magnitudesand durations.
The GeneralComponentModel. Two quantitiesare normally used to describe
the behaviorof a stochasticcomponent:the failure rate and the (expectedor average)
repair time. The meaningof the term "expectedrepair time" is obvious: the expected
value of the time the componentresidesin the nonhealthystate. The failure rate A
gives the averageprobability that the componentwill fail in the next small period of
time:
. Pr(failure in period 6.t)
A = I1m - - - - - - - - 6t.....0
8.1
(2.8)
For componentsrepresentingprimary partsof the power system,which are the majority of the componentsin moststudies,the term outageratemight be used.Herewe shall
use thegeneralterm failure rate.
The definition of failure rate in (2.8) is rathermathematical.It will becomeof use
below. A more practical way of defining the failure rate is through the number of
failures in a population.Considera populationof N similar components(e.g.,distribution transformers).During a period n, this populationshows K componentfailures.
The failure rate may be determinedas
K
A=nN
(2.9)
The two definitions of failure rate are equivalentundera numberof assumptions.T he
most importantof which is that the componentis repaired(within a short time) after
every failure. The definition according to (2.9) is used toobtain failure rates from
observedfailures.
Someother quantitieswhich are in use will bedescribedbelow.
• The expectedtime to failure T is the reciprocalof the failure rate:
1
T=-A
(2.10)
• The repair rate {t is the reciprocalof the expectedtime to repair R:
1
{t=-
R
(2.11)
65
Section 2.5 • Basic Reliability Evaluation Techniques
Note that expectedtime to failure can be defined in asimilar way as the
expectedrepair time, and the repair rate similarly as thefailure rate according
to (2.8).
• The availability of the componentis the probability to find the componentin
the healthystate:
T
p=--
(2.12)
R+T
• The unavailability is the probability that the componentis in the nonhealthy
state:
R
(2.13)
Q=R+T
• The expectedtime betweenfailures (ETBF) is the sum of the expectedtime to
failure (ETTF) and the expectedrepair time. As the repair time is normally
muchsmallerthanthe time tofailure, ETBF and ETTF are aboutequalandas
a consequenceoften mixed up. From a mathematicalp oint of view, this is a
seriousmistake,but in engineeringthese kindof errors are commonand not
consideredvery seriously.
EXAMPLE A distribution company operates 7500 distribution transformers. Over a
period of 10 years, 140 of these transformers fail for various reasons. A small fraction of them
can be repaired, but most failures require replacement with a spare transformer. Records have
been kept of the repair or replacement time needed. Adding all these for the 140 failures gives
areoba total of 7360 hours. From these observation data, the values of the above parameters
tained:
140
_I
(2.14)
A = 10 x 7500= 0.0019yr
1
T
7360
R = 140 = 52.6h
Jl
(2.15)
= 0.0019= 530yr
= 0.006yr
(2.16)
= ~R = 167yr-1
(2.17)
530
p = 0.006+ 530= 0.999989
0.006
(2.18)
.
Q = 0.006+ 530= 0.000011= 6mtn/yr
(2.19)
This can be interpreted in normal words,follows:
as
• Each transformer has a probability of 0.0019 to fail in the coming year. In the whole
population, 14 transformers are expected to fail.
• After such a failure, the repair or replacement of the transformer is expected to last 52.6
hours.
• Each transformer will be out of operation, on average, 6 minutes per year.
66
Chapter2 • Long Interruptionsand Reliability Evaluation
d atato predictfuture behavior.This is the basis for all
Note that we have usedpast-performance
reliability analysis: theassumptionthat the averageperformancein the past gives theexpected
behaviorfor the future.
The Detailed Component Model.Describinga stochasticcom-ponentby means
of two quantities (e.g., failure rate and repair time) is a gross simplification of the
actual situation. Still this model is usedin more than 95% of all reliability evaluation
studies. To understandthe reasonsfor this, we first need to introduce the general
componentmodel. The componentis again assumedto be in one of two states.The
theory can be extendedto multi-statemodels, but describingthosewould not lead to
better understanding.For each of the two statesa probability distribution is defined
for the time the componentstaysin that state.Thereis thus one probability distribution function for the repair time (the time in the nonhealthystate) and one for the
lifetime (the time in the healthystate). Let T be the lifetime (expectedtime to failure)
of the component.The probability distribution function of the lifetime F(t) is the
probability that the componentfails beforeit reachesan age t:
F(t)
= Pr(T s t)
(2.20)
The probability density function is the derivative of the probability distribution
function:
f(t)
= dF = lim
dt
Pr(t < T :::: t + M)
L\t~O
~t
(2.21)
The probability density function I(t) is a measurefor the probability that the component will fail aroundan age t:
l(t)6.1 ~ Pr(1 < T
s 1 + ~t)
(2.22)
The failure rate A( I) is definedas theprobability that the componentfails soonafter the
age 1 assumingthat it has not failed before age t:
A()
t
. Pr(T~t+~tIT>t)
Iim - - - - - - - = L\t~O
~I
(2.23)
The failure rate can be calculatedfrom the probability density function I(t) and the
probability distribution function F( t):
A(t) = f(t)
1 - £(t)
(2.24)
We will discussthe failure rate and its relation to componentaging in more detail in
Section2.5.6.
Similar definitionscanbe given for therepairtime, resultinge.g., in the repairrate
/1(t), a probability density function g(t) and a probability distribution function G(t).
The Weibull Distribution. A distribution often used in stochastictheory is the
so-called Wei bull distribution. The probability distribution function for a Weibull
distributedvariable T is
F(t)
=I -
ex p {
-(~r}
(2.25)
67
Section 2.5 • Basic Reliability Evaluation Techniques
For m = 1 weobtainthe exponentialdistributiondiscussed before. We refer m
to as the
shapefactor and to () as thecharacteristictime of the Weibull distribution. The probability density function .(t) is
nl
f(t)
= m t om
-
1
{
exp
-(0)t nil
(2.26)
The failure rateA(t) for a Weibull distribution is obtainedfrom (2.24):
r:'
A(t)=m-
om
(2.27)
m > 1 and decreases for
m < 1. From a
We seethat the failure rate increases for
relatively simpleexpressionit is possible togeneratea whole rangeof lifetime distributions.
%
The ExponentialDistribution-Lifetime. As alreadystatedbefore, over 95
of
reliability evaluation studies use the simple model with a single failure rate and a
single repair rate. The underlying assumptionis that both repair time and lifetime
are exponentially distributed. The exponential distribution (also called "negative
exponentialdistribution") is defined through the following probability distribution
function:
F(t) = 1 -
e-'At
(2.28)
From the aboveequationsit follows easilythat A in (2.28) is the failure rateaccording
to (2.24).Thus,the negativeexponentialdistributionhas aconstantfailure (repair)rate
and the generalcomponentmodel can be used.
Thereare anumberof reasonswhy this
distribution is almostthe only one used:
• Using nonexponentialdistributions makes that most reliability evaluation
techniquescurrently available can no longer be used.For many years the
choice was between using the
exponentialdistribution or not doing any reliability evaluationat all.
nonexponentialdis• Even the smallnumberof studies which are able to use
tributions(the so-calledMonte Carlo simulationswhich we will discuss below)
often still useexponentialdistributions,becauseof the lackof data.Collection
schemesof componentfailure data normally only provide failure rates and
averagerepair times.
• The lackof experience withnonexponentialdistributionsmakesthat the results
of such a study areratherhard to interpret.
of componentswith different ages
• In an actualpower system there is a mixture
for three reasons:preventive maintenanceis performed on componentsat
different times; componentsare replacedafter failure; and the system is not
built at once but has grown over time. The
mixture of ages makesthat the
systembehavior,being a kindof averageof the componentbehavior,can be
described byassumingall componentsto have aconstantfailure rate.
• Most componentsin use are in their so-called
"useful operatingtime": they
have passed the wear-in time, and have not yet reached theoftime
serious
wear-out.This is based on the
assumptionthat the failure rateof a component
versus time can be
describedthrough a "bathtubcurve." During most of the
Chapter 2 • Long Interruptions and Reliability Evaluation
68
operatingtime of a component,it resides in the flatpart of the bathtubcurve
where the failure rate is
c onstant.
The ExponentialDistribution-Repair Time. For repair time distributions, the
above reasoningsdo not hold. Wealreadysaw in Table 2.1that the duration of an
interruption is nonexponentiallydistributed. If we assume theinterruption duration
to be Weibull distributed,the shapefactor in (2.25) can becalculatedfrom the available data:
In( -In(Fr3 »
m=
(2.29)
In(~)
with Fr3 the fraction of interruptionsnot restoredwithin three hours and () the characteristicrepairtime. If we take theaveragerepairtime as thecharacteristicrepairtime,
we only makea smallerror as long asm > 1. Including the effectof the shapefactor on
the averagerepair time would make thecalculation too complicated.The resulting
shapefactors for the interruption durationsare given inTable 2.10. We find shape
factors somewhatin excessof unity.
The IEEE Gold Book [21] gives,amongothers,repair times for large electrical
motors in an industrial environment.As both the average and the
median value are
given, it is again possible to
estimatethe shapefactor assuminga Weibull distribution.
In mostcases themedianvalue ismuchlarger thanthe average, whichindicatesa shape
factor less than one. An alternativeexplanationis the combinationof two Weibull
distributions,both with shapefactor greaterthan one, but with significantlydifferent
characteristicor averagerepair times.
More theoreticalmodelingand observationwork is needed tovalidatethe useof
the exponential distribution in power system reliability evaluation. Based on the
evidencepresented,the following preliminary conclusionscan bedrawn:
• The exponentialmodel appearsan acceptableapproximationfor lifetime disof preventivemaintributions,with the exceptionof studies in which the effect
tenanceis evaluated.
• The exponentialmodel isnot correctfor the repair time.
A short discussionon componentaging will be given in Section 2.5.6.
TABLE 2.10 ShapeFactor for Weibull Distribution of Interruption
Duration
()
2.38
1.38
1.42
1.45
1.63
1.62
2.27
1.38
Fr3
ShapeFactor
0.193
0.098
0.073
0.070
0.115
0.086
0.134
0.071
2.15
1.09
1.29
1.35
1.27
1.46
2.50
1.25
69
Section 2.5 • BasicReliability EvaluationTechniques
2.5.2 Network Approach
Whenusing the so-callednetworkapproach,the system ismodeledas a"stochastic network." The stochasticbehaviorof the system isrepresentedgraphicallyby means
of a numberof network blocks, connectedin parallelor in series.Eachblock refers to a
stochasticcomponentin the system. The model is such
that the system ishealthy(i.e.,
the supply is available)as long as there is p
aath through the network. This graphical
charactero f the methodmakesit very suitableto get an overviewof the reliability of the
system. An additional advantageof the network approachis the similarity with the
electrical network. Electrically parallel componentsare often modeled as a parallel
connectionin the stochasticnetwork. An electrical seriesconnectionin most cases
results in astochasticseriesconnection.
When the reliability is quantified by using a stochasticnetwork, a number of
mathematicalapproximationsare needed. Thecalculationsassumethat the repair
time and the lifetime areexponentiallydistributedfor all components.
Each block(network element)is characterizedthrough an outagerate A and an
expectedrepair time r. For each element we
further define the"availability" P and the
"unavailability" Q.
P = I - Ar
(2.30)
Q=Ar
(2.31)
Sometimesa different form of these expressionsis used: theoutagerate is given in
failures per year,a ndthe repairtime in hours,leading to the following(mathematically
not fully correct,but very handy)expressionsfor availability and unavailability:
Ar
P = 1 - 8760
(2.32)
Ar
Q = 8760
(2.33)
EXAMPLE Considerthe supply system in Fig. 2.12. A possible
stochasticnetwork
for this system is shown in Fig. 2.13 where the
numbers refer to the following typesof
failure:
Public
supply
Figure 2.12Single-linediagram of a supply
system.
On-sitegeneration
70
Chapter2 • Long Interruptionsand Reliability Evaluation
Figure 2.13Stochasticnetwork
representationof the systemshownin Fig.
2.12.
1.
2.
3.
4.
5.
6.
7.
outage of the public supply
outage of agenerator
bus outage
transformeroutage
circuit breakerfailure (maltrip or short circuit)
circuit breakerfailure (maltrip)
circuit breakerfailure (short circuit)
All componentsin the network in Fig. 2.13 are stochastically
independent,so that simple mathematics can be applied. Note that the capacity of one
generator(5 MW) is not enoughto supply
the load (7 MW). To supply the load the public supply needsbetopresent, or both on-site
generatorsneed to be inoperation.In the networkdiagramthis is shown as the"public supply"
in parallel with both "on-site generators"in series. Also note the difference between a circuit
breakermaltrip and a short circuit in the breaker. In the
lattercase theprotectionon both sides of
the breakerwill trip leading to the loss of twoprimary componentsat the same time.
Various methodsare availableto calculateinterruption rate and expectedinterruption durationfrom componentfailure rateandrepairtime; all thesemethodsreplace
the wholenetwork by one equivalentcomponent.
An obvious methodfor network reductionis to find seriesand parallel components. A parallelconnectionrepresentsredundantcomponents,where thesupplyis not
interrupteduntil all of them are in theoutagestate. A seriesconnectionrepresentsthe
situation where eachcomponentoutageleads to aninterruption of the supply. The
correspondencewith electrical seriesand parallel connectionsis clear but not one-toone. Consideras an example twotransformersin parallel. If one of them fails theother
one can take over the supply. This is clearlystochasticparallelconnection.But
a
if the
total load is much morethan the maximumloading of one transformer,the other one
Section 2.5 •
71
BasicReliability EvaluationTechniques
will also soon fail or be tripped by its overload protection. In that case astochastic
seriesconnectionis a betterrepresentation.
Stochastic Series Connections.Considerthe seriesconnectionof two stochastic
componentswith outagerates AI and A2 and repair time r and '2, as shown in Fig.
2.14. Wewant to derive expressionsfor outagerate As and repair time r s of the series
connection,so that the seriesconnectioncan be replaced by one
equivalentcomponent.
Al
rl
-<.
As
rs
A2
r2
Figure 2.14Stochasticseriesconnection.
A seriesconnectionfails wheneitherof the componentsfails. The outagerate for
the seriesconnectionis thus the sumof the outagerates of the components:
As = Al
+ A2
(2.34)
The seriesconnectionis not availablewhen oneof the componentsis not available,
giving for the unavailability of the seriesconnection:
(2.35)
Using thedefinition of unavailability(2.31) gives anexpressionfor the equivalentrepair
time of the seriesconnection:
Air. + A2r2
r -----
Al
S -
+ A2
(2.36)
For n componentsin series, the followingexpressionscan be derived:
n
As=
LA;
(2.37)
;=1
r.s
= L"'IA'"
'=; I '
(2.38)
LJ=I AJ
In deriving the expressionsfor equivalentoutage rate and repair time a number of
assumptionshave been made, all
coming back to the system being
availablemost of
the time, thusAr « 1. Exact expressionswill be derived inSection2.5.3.
StochasticParallel Connections. A parallel connectionof two stochasticcomponentsis shownin Fig. 2.15.
A parallel connectionfails when one of thecomponentsis not availableand the
other one fails: thus when 1 isunavailableand 2 fails or when 2 isunavailableand 1
fails. The outagerate of the parallel connectionis
72
Chapter2 • Long Interruptionsand Reliability Evaluation
Figure 2.15Stochasticparallel connection.
Ap
= QI A2 + Q2 AI
= AI A2(' 1+ '2)
(2.39)
The parallel connectionis not availablewhen both componentsare not available. The
unavailability of the parallel connectionis
o, =
QI X
Q2
(2.40)
The repair time of the parallel connectionis obtainedfrom (2.39)and (2.40):
'p
=-'1'2
-
(2.41)
'I +'2
The equationscan beextendedto a system with threecomponentsin parallel by considering it as theparallel connectionof one componentand the equivalent of the
parallel connectionof the two other components.This results in the following expressions for outagerate and repair time:
(2.42)
1
1
1
I
'p
'1
'2
'3
-=-+-+-
(2.43)
The same process can be
repeatedseveral times, resulting in the following general
expressionsfor a systemconsistingof n componentsin parallel:
n
n
1
-. = Il
Aj'j L -:
;=1
j=1 ,}
(2.44)
(2.45)
Minimum Cut-Sets. A secondmethodfor analysisof stochasticmethodsis the
so-called "minimum-cut-setmethod." A "cut-set" is a combination of components
whose combinedoutagewould lead to aninterruption. In the stochasticnetwork in
Fig. 2.16 thecombinations{I, 2, 3} and {4,5} are examplesof cut-sets. Acut-setis a
"minimum cut-set" if the removal of anyoneof the componentsfrom the cut-set
would make it no longer a cut-set. Inother words, if the repair of anyonecomponent would restore the supply. In Fig. 2.16 thecut-set {I, 2, 3} is not a minimum
Section 2.5 •
73
BasicReliability EvaluationTechniques
5
Figure 2.16 Example of stochastic network,
for explaining the minimum cut-set method.
cut-set becauserepair of component3 does not restore the supply, even
though
repair of component 1 or component 2 does. The cut-set {4,
5} is a minimum
cut-set becauseboth repair of component4 and repairof component5 restore the
supply. For each network there are a limitednumberof minimum cut-sets.Finding
all minimum cut-sets is the first step of the
minimum-cut-setmethod.
The network in Fig. 2.16 has the following minimum cut-sets:
{1,2}
{4,5}
{1,3,4}
{2, 3, 5}
The supply isinterruptedwhen anycombinationof thesecomponentsis not available.
The systembehaviorcan thus also be described as a series
connectionof four parallel
connections,representingthe four minimum cut-sets. This is shown for this example in
Fig. 2.17'. After having found theminimum cuts-sets, thecalculationproceedsstraightforward: outagerates andrepairtimes aredetermined,first for the parallelconnections,
next for the resulting series
connection.The latter gives the interruption rate and
expectedinterruptionduration for the supply.
2
Figure 2.17 Alternative drawing of the
network in Fig. 2.16: series connection of
parallel connections.
EXAMPLE
Considerthe following outageratesand repair times for thenetwork ele-
mentsin Fig. 2.16:
AI = 1
'1 = 0.2
A2 2
'2
0.1
A] = 0.5 '3 = 0.1
A4 = 0.8 r 4 = 0.15
As = 1.5 's = 0.2
=
=
At') = 0.2
A2'2 = 0.2
A3'3 = 0.05
A4'4 = 0.12
AS'S= 0.3
Equations(2.44) and (2.45) giveequivalentfailure rate and repair time for the parallel connections representingthe four cut-sets.
74
Chapter2 • Long Interruptionsand Reliability Evaluation
(2.46)
'cl
1)-1= 0.067
= ( -1 + '1
'c2
'('3
'('4
'2
I)-I =
= ( -I + -1 + -1)-1 =
1
= ( -+'4
'5
'1
'3
1
1
'2
'3
0.086
(2.47)
0.046
'4
1)-1= 0.04
= ( -+-+'5
The failure rate A and repair time r of the whole system can be
calculatedby consideringit as a
seriesconnectionof the four cut-sets:
(2.48)
r = Ad'cl
+ Ac2' c2 + Ad',,3 + Ac4' c4 = 0.072
+ Ac2 + Ac3 + A('4
(2.49)
Ad
A second example of the use
of the network approachis shown in Fig. 2.18 and Fig.
2.19. The first figure showsp art of a subtransmissionsystem. Thetransmissiongrid is
assumed to be fully reliable. Also
substationsA, B, and C areassumednot to fail. The
load of interestis connectedto substationD. The networkrepresentationfor the system
in Fig. 2.18 is shown in Fig. 2.19.
Component8 representsoutagesin the local substation(D) which lead to aninterruptionfor the loadof interest.This network can no
longer be reducedthrough series andparallel connections,but the minimum cut-set
methodcan still be used.
c
6
D
7
8
Figure 2.18 Exampleof public supply, with
single redundancy.
Section 2.5 •
75
BasicReliability EvaluationTechniques
6
Figure 2.19 Network representationof the
supply in Fig. 2.18.
The following minimum cut-sets can be
found for this network:
{8}
{6,7}
{I,2,4}
{I,2,5}
{I, 3, 7}
{2, 3,4, 6}
{2, 3, 5,6}
These minimum cut-sets are shown in Fig. 2.20 from where the term
..set
cutbecomes
clear. A cut-set cuts allpathsbetween the source and the load.minimum
A
cut-setcan
be described as "ashortestcut."
1----------
Figure 2.20Network representationof the
supply in Fig. 2.18, with minimum cut-sets
indicatedas dottedlines.
A third example is shown in Fig. 2.21. This supply system
containsa substation
with a third bus (4), aconfiguration used in industrial systems toprevent a circuit
breakerfailure from leading to loss of the whole
substation.The variouscomponents
have beennumberedin the figure.Translatingthis to anetworkdiagramis not obvious,
as component3 is in series with 1, 4, and 6, but 1 and 4 are in parallel. A possible
solutionis shown in Fig. 2.22.Components3 and 5,representingbus outages,are now
placed in a triangle with themselves. The
network might seemsomewhatartificial, the
list of minimum cut..sets can beobtainedin a normal way, resulting in
76
Chapter2 • Long Interruptionsand Reliability Evaluation
{8}
{1,2}
{I, 5}
{2,3}
{3,5}
{3, 7}
{5,6}
{6,7}
{I, 4, 7}
{2, 4, 6}
The advantageof the networkapproachis thatit gives a fastunderstandingof the
reliability of the system. It also enables reliability
calculationsin large systems and
provides,through minimum cut-set techniques, an insight into the weak
points of the
supply system.Drawing the stochasticnetwork is a useful exercise in itself, often more
usefulthan the actualresults. Thedisadvantageis that approximationerrorsare made
in each step of thecalculationprocess. This could lead to serious
errorsin the results,
2
3
5
6
7
8
Figure 2.21 Industrial system withthree-bus
substation.
2
3
5
6
7
8
Figure 2.22 Network representationof the
system in Fig. 2.21.
Section 2.5 • BasicReliability EvaluationTechniques
77
especially for large systems. The
errorsare due to theassumptionsmade when replacing
series and parallelconnectionsby one element. Theassumptionsmade arethat the
t hat the elements are
stochasticallyindepenunavailability of the element is small and
dent. Thelatter assumptionis no longer fully correct when the seriesconnectionof
minimum cut-setsis replaced by one element. As the same
network componentcan
appearin more than one minimum cut-set,the minimum cut-setswill becomestochastically dependent.
2.5.3 State-Based and Event-Based Approaches
In the state-baseda pproachthe systembehavioris describedvia states andtransitions between states. sAtateis eitherhealthyor nonhealthy.A healthystateis a statein
which the supply is available,a nonhealthystate one in which the system is not available. Theprobability of all the nonhealthystatesis calculatedand added. This sum is
the probability that the supply isnot available. In addition to probability it is also
possible tocalculateother parameters,like the expectednumberof interruptionsper
year, or the average
d urationof an interruption.
of events.
In the event-basedapproachthe systembehavioris described by means
For each event theconsequencesfor the supply aredetermined.In case analytical
of states and
techniquesare used the system is often still modeled ascollection
a
transitions. But now the transitions are either healthy or nonhealthy.A transition
between twohealthystatesis NOT necessarily healthy.
A Four-State Component Model.The basiccomponentmodel for astate-based
approachconsistsof two states: { inoperation}; and { not in operation}, often shortened to { in } and { out }. A more detailedmodel is shown in Fig. 2.23. This model
consists of four states: {healthy}, { faulted }, { out of operationfor repair }, and
{ out of operationfor maintenance}. We can see from the figure
t hat the component
cannot fail while in maintenance,but that maintenancecan start while the compothat a faulted componentwill first be repairedbefore it
nent is in repair. We also see
becomes"healthy" again. The faultedstaterepresentsa short-circuit fault, the duration of which is much smallerthan of the other states.Thereforethis state is often
combinedwith the repair state. But in studiesof power systemprotection,the faulted
stateplays an essential role.
Figure 2.23Four-statecomponentmodel.
78
Chapter 2 • Long Interruptions and Reliability Evaluation
A Protective Relay. An example of astate model for a protective relay is
shown in Fig. 2.24. We see the same healthy, repair, maintenance
and
statesas in
Fig. 2.23, but now thecomponentcan fail in threedifferent ways. A maltrip leads directly to an outageof the componentto be protected,after which the relay needs to
"dormantfail to trip") meansthat the relay
be repaired. A hidden failure (also called
will no longer trip when it needs to. This failure will only reveal itself when the relay
needs to trip, thus when there isshort
a
circuit in the componentto be protected.A
potentialmaltrip is a situationwhere the relay will send an
incorrecttrip signal under
certain systemconditions.Maintenancecan bring the relay from the"hidden failure"
or "potential maltrip" states back to the"healthy" state.
Figure 2.24 Model for protective
relay,consistingof one healthy and
six nonhealthy states.
An Industrial Supply. Considerthe system shown in Fig. 2.25. The
industrial
load is fed via threeoverheadlines from two generatorunits plus the public supply.
The rating of the componentsis such that one line is sufficient to supply the whole
load; also onegeneratoror the public supply are sufficient. We
further assumethat a
failure of a line and a failure of the public supply are
associatedwith a short circuit,
but that a generatorfailure only involves thetripping of the unit.
It is assumedthat eachcomponentcan be in one of two states. Only failures
of the
public supply, the on-sitegeneratorsand theoverheadlines, areconsidered.This results
in the system states as shown in Fig. 2.26. The system consistscomponents,
of 6
each
with two states. Thenumberof system states is
thereforeequal to 26 = 64, but only 23
states are shown in Fig. 2.26. By assuming
that the three lines are identical, and the two
Section 2.5 •
79
BasicReliability EvaluationTechniques
On-sitegeneration
Public
supply
Figure 2.25 Example of industrial supply
with double redundancy.
Industrial
load
on-site generatorsalso, states can be
aggregated.For example, state 2 {I line out}
representsthree basicstates{line lout, line 2 out, line 3 out}; state 5 {2 lines out}
also representsthree basic states: {line 1 and line 2 out}, {line 1 and line 3 out}, {line 2
and line 3 out}. Thestateshown on top is the one with all
componentsin operation.
From this statethe system can reach three
other states:
• One line outof operation.
• One generatorout of operation.
• The public supplyout of operation.
an
An interruptionof the supply can be due to the system being inunhealthystate
(e.g., three lines out), but also due to an
unhealthytransition between twohealthy
states. Astate-basedstudy would onlyconsiderthe states,n ot the transitionsbetween
states. To includeinterruptionsdue to unhealthytransitions,an event-basedapproach
is more suitable.
In this system it can beassumedthat only short-circuitfaults lead tounhealthy
o f the public supply. Thesepotentially
transitions,thus only line failures and failures
unhealthytransitionsare indicated by an arrow in Fig. 2.26. From the state{2 lines
out}, again, threetransitionsare possible:
• The failure of the last remainingline will anyway result in aninterruption as
the final stateis an unhealthyone. Thistransitiondoesnot need to befurther
studied.
generatorout}
• The failure of a generatorleads to the state {2 lines and one
which is ahealthystate. Thetransitionis not associatedwith a shortcircuit and
does not requirefurther study.
• The failureof the public supply isassociatedwith a shortcircuit and it leads to
a healthystate. Thistransition requiresfurther study.
80
Chapter 2 • Long Interruptions and
Reliability Evaluation
,,
,
,
,,
\
\
,
\
,,
,
I
8
\
"~
~
,,
-,
...
I
...
,
,
/
,
I
...
,
I
,
'",
"
I
I
I
/
/
"
\\,
...
,
,
,
\\,',
"" \
Figure 2.26 Statesand transitionsfor the systemshownin Fig. 2.25.The solid
lines indicate transitionsbetweenhealthystates,the dotted lines
indicate transitionsbetweena healthystateand anonhealthystate,
the arrowsindicatetransitionsassociatedwith a short-circuitevent.
2.5.4 Markov Models
Markov models are amathematicalway of calculating state probabilities and
event frequencies instochasticmodels. In Markov-model calculations all lifetimes
and repair times are assumed
exponentiallydistributed.A Markov model consistsof
a numberof states, withtransitionsin between them; several examples will be given
below.
One-Component Two-State Model.
The simplest Markov model is shown in
Fig. 2.27: atwo-statemodel of one component.In state 1 the componentis healthy,
81
Section2.5 • Basic Reliability EvaluationTechniques
Figure 2.27 Two-state Markov model.
in state2 it is nonhealthy.The transition rates areA and J-L, as indicated.This model
will be used tointroducesomeof the basic concepts and
calculationtechniques.
To derive the expressions for the
state probabilities, one should consider an
infinite number of stochasticallyidentical systems. At a timet a fraction PI of the
systems is instate 1 and afraction P2 in state 2, withPI + P2 = 1. In mathematical
terms: theprobability of finding the system in state 1 is equal PI'
to The transitionrate
from state 1 tostate2 is A. Thus in a veryshortperiod t1t a fraction At1t of the systems
J-Lt1t of the systems in state 2
in state 1transitsto state 2. In the mean time fraction
a
transitsto state I. The fraction of systems instate1 at time t + t1t becomes
(2.50)
A similar expression isobtainedfor the probabilityto find the system in state 2.
Making
the transitionfor !:!t ~ 0 gives the followingdifferential equationsfor the stateprobabilities:
dpi
-dt = -API + J-LP2
+'
We seethat'
dP2
dt
- = JlP2 -API
= 0, which isunderstandableif one realizesthat
PI + P2 = 1
(2.51)
(2.52)
(2.53)
i.e., the sumof state probabilitiesequalscertainty.To calculatethe stateprobabilities
only oneof the expressions (2.51) and (2.52) is needed,
togetherwith (2.53).
From (2.51) and (2.53) we can solve the
probability that the system is instate1,
thus that the componentis healthy. It is assumed
t hat the componentis healthyfor t =
o which correspondsto PI (0) = 1.
P (t)
1
= _Jl_ + _A_e-t(A+Jl)
A+J-L
(2.54)
A+1l-
We seethat the probability reaches aconstantvalue after anexponentiallydecaying
transientwith a timeconstantA~ • For almost any engineering system we may assume
that repairis much faster thanf:ilure, thus A «/1.. When we also realizethat is the
averagerepair time, we seethat the probability reaches aconstantvalue within a time
of interestis normally much largerthan
scale equal to the repair time. The time period
the repair time (years versus hours) so
that we can considerthe system states and
transition frequenciesconstant.This holds not only for atwo-componentmodel but
for every Markov model in which repair takes place much faster
than failure.
k
82
Chapter2 • Long Interruptionsand Reliability Evaluation
Steady-StateCalculation. As the transition between the initialcondition and
the steady-stateprobabilities can be neglected, we can directly
calculatethe steadystate probabilities. In steady state, the state probabilities are constantas a function
of time; thus,
dpi =0
dt
(2.55)
The equationswhich describe thestateprobabilitiesbecomealgebraicequations,which
can be easily solved.For the two-statemodel weobtain
o = -API + ttP2
o = API - IlP2
PI
+ P2 =
(2.56)
I
One of the equationsin this set isredundant,so that only oneof the first twoequations
is needed.From this one and thethird equation,the steady-stateprobability becomes
PI
= A +tt JL
(2.57)
P2
=-A-
(2.58)
A+1l
Operating Reserve. We mentionedbefore that we can neglect thetransition to
the steadystateand thuscalculatesteady-stateprobabilitiesdirectly. Two exceptions
to this rule must be mentioned:one for veryshort time scales,and one with a very
long repair time. When looking at a veryshort time scale theexponentiallydecaying
componentof the stateprobability can nolonger be neglected. Veryshort time scales
that a componentis in
are of interestin operatingreserve studies, where one knows
operation,and wants to know the probability that it fails within a time flt. For a
two-statemodel we derived before:
P2(flt) = I - Pl(flt) = _A
A+Jl
Assumingthat 6.t
A_e-~t(A+tL)
A+Jl
(2.59)
« h« *we obtain the following approximatedexpression:
A
P2(6.t) ~ -Jl6.t = A~t
(2.60)
JL
Note that the same result is
obtainedif we assumethat the componentmay fail but that
it is not repairedwithin the period flt.
Hidden Failures in a Protective Relay.A second example in which the exponentially decaying termcannotbe neglected is aprotectiverelay with hidden failures.
Hidden failures of protective relays havealready been discussed in Section. 2.4.2. If
that repair takes placeinstantawe neglect allother failures of the relay, and assume
neously when thehidden failure is detected,we obtain the statemodel shown in Fig.
2.28. In state 1 the relay is
healthy and a fault in the primary componentto be protected is cleared as
intended.If the relay is in state2, the fault will not be cleared by
this relay, but instead some backup protection needs to take over. The third
state
shown in Fig. 2.28 is therepair state. The failure rateA2 is the fault frequency in the
primary component. We will initially assume that no preventive maintenanceis
performedon the relay.
Section2.5 •
83
BasicReliability EvaluationTechniques
Figure 2.28 Model for relay withhidden
failure (left); the relay ishealthyin stateI and
containsa hiddenfailure in state2. The figure
on the right gives thetwo-statemodel which is
obtainedby neglectingthe repair time 11.
From the three-statemodel in Fig. 2.28 weobtain the following setof equations
for the stateprobabilities:
(2.61)
From this it is possible toobtain expressionsfor the stateprobabilitiesPI, P2, and P3
and for thetransition frequenciesAIPI, A2P2, and J-LP3'
Neglectingthe transientto steadystategives the followingequationsfor the state
probabilitiesin steadystate
AIPI = I-tP3
= AIPI
IlP3 = A2P2
PI + P2 + P3 = 1
A2P2
(2.62)
(2.63)
EliminatingPI andP3 from the first threeexpressionsand substitutingthis in thefourth
one results in
(2.64)
The frequencyof fail-to-trip events insteadystateis
(2.65)
If we assumethat repair (the transitionfrom state3 to state1) takes placemuch faster
than detectionof the hidden fault (from state2 to state3), we can neglects tate3 and
obtain the two-statesystemshownon the right of Fig. 2.28. This model results in the
following equations:
dpi
dt =
PI
-AIPI
+ P2 = 1
+ A2P2
(2.66)
(2.67)
84
Chapter2 •
Long Interruptionsand Reliability Evaluation
which correspondsto the equationsfor the two-statesingle-componentm odel in Fig.
2.27 and (2.51) through (2.53). The resultingprobability of being in the hidden-failure
stateis
P2(t) =
AI
Al +A2
[I _e- /()..I+A2>]
(2.68)
The fail-to-trip frequencyis equalto A2P2 and reachesits steady-statevalue with a time
constant A
This holds if we assumethat hidden failures only reveal themselves
during a f~urt in the primary component.In case maintenanceis performed with a
frequencyA3 the transition rate from state2 to state1 is A2 + A3' The probability that
the relay is instate 2 becomes
LA,'
P2(t) =
[I _e-IO'I+A2+A,l>]
AI
Al
+ A2 + A3
(2.69)
Maintenancereduces the time constant with which the steady-stateprobability is
reached,and (more importantly) it reducesthe steady-stateprobability. The number
of fail-to-trip eventsper year nm l remain equal to A2P2, thus given by the following
expression:
n
(1) =
mt v.
AI A2
Al
+ A2 + A3
[I' _e
3>]
-t(AI +A2+ A
(2.70)
We seethat for maintenanceto be effective, the maintenancefrequency needsto be
higher than the sum of the fault frequencyin the primary componentand the hiddenfailure rate of the relay
(2.71)
Two-Component Model. Considera system that consistsof two components:
component1 and component2, with failure rates At and A2, and repair rates J,Lt and
~2' If we model eachcomponentthrough two states,this systemhasfour states:
•
•
•
•
State 1 with
State 2 with
State 3 with
State4 with
both componentsin operation.
only component2 in operation.
only component1 in operation.
none of the componentsin operation.
The resultingstatemodelis shownin Fig. 2.29.The equationsfor the stateprobabilities
are
dpi
dt = -(AI + A2)PI + JLIP2 + JL2P3
dP2
(2.72)
dt = AIPI -
(JLI
+ A2)P2 + JL2P4
(2.73)
dP3
dt
= A2PI -
(J.t2 + At)P3 + J.tIP4
(2.74)
dP4
dt
=
A2P2
PI + P2 + P3 + P4 = 1
+ AIP3 -
(J.tl + J,L2)P4
(2.75)
(2.76)
Section 2.5 •
8S
BasicReliability EvaluationTechniques
Figure 2.29Two-component,two-state
Markovmodel.
Thesecan be solvedagain like for the previousexamples,but there is an alternative
solution method.We haveassumedthat the two componentsare stochasticallyindependent.This assumptionhas not beenmadeexplicitly but by making the failure and
repair ratesof the componentsindependento f the stateof the othercomponent.If the
componentsare stochasticallydependent,the transition rate from state 1 to state2 is
not the same as the one from
s tate3 to state4 (both representfailure of component1),
etc. For stochasticallyindependentc omponentswe can multiply the componentstate
probabilitiesto get the systemstateprobabilities.Thus, with Pidown and Piup the probabilities that componenti is in the "up" and in the "down" state, respectively, we
obtain for the stateprobabilities
= Plup X P2up
P2 = PIc/own X P2up
P3 = Plup X P2down
P4 = Pldown X P2down
(2.77)
PI
(2.78)
(2.79)
(2~80)
Theseequationshold for eachmomentin time, thusfor the transientto steadystate,as
well as for thesteadystate.Using the expressionsfor the stateprobabilitiesin the onecomponentmodel the steady-stateprobabilitiesin the two-componentmodel become
/-LI/-L2
PI
= 0"1 + ILI)p + 1L2)
o2
(2.81)
AI/-L2
P2
= pol + IL] )().o2 + IL2)
P3
= po] + ILI)P'2 + 1L2)
(2.82)
/-L IA2
AI A2
P4 = (>"] + ILI)O'2
+ 1L2)
(2.83)
(2.84)
Series and ParallelConnections. We can use theseresults to obtain exact
expressionsfor the failure rate and repair time of series and parallel connections,
approximationsfor which were given inSection 2.5.2. For a seriesconnection of
components1 and 2, state 1 is the healthy state. System failure is a transition from
state 1 to state2, or from state 1 to state3. The systemfailure rate As is the sumof
these twotransition rates:
A -
A
s - PI I
A _
+ PI
2-
/-L1/-L2(AI
+ A2)
p.] + ILI)P'2 + IL2)
(2.85)
86
Chapter2 •
Long Interruptionsand Reliability Evaluation
The systemis unavailablewhen it is not in state 1. The systemrepair time's is found
from the unavailability Qs:
(2.86)
As's= Qs = I - PI
The averagerepair time for the seriesconnectionis
AI112 + A2111 + AIA2
111112(AI + A2)
(2.87)
,~=-------
.
In a similar way expressionscan be derived for the parallel connection.For a parallel
connection,states1, 2, and 3 arehealthy,and systemfailure is a transitionfrom state2
to state4 or from state3 to state4. The resulting expressionsfor failure rate Ap and
repair time 'p are
(2.88)
(2.89)
Exact Solutionof Large Markov Models. For a systemwith a large numberof
states, the underlying equationscan be derived in the same way as shown in the
above example. The set of differential equationscan be written in the following
matrix form:
-dP
= AP(t)
dt
(2.90)
with A the matrix of statetransitionsand P the vector of stateprobabilities. For the
Markov model in Fig. 2.29we get
(2.91)
and
o
112
A=
JLl
(2.92)
-J-l1 - J-l2
The off-diagonal element Aij is the transition rate from statej to state i. The diagonal
elementA ii is minus the sum of all transition ratesaway from state i:
(2.93)
Aij=\i
A ii
= - LAij
(2.94)
j
Togetherwith an initial condition for the stateprobability vector
75(0) =
Po
(2.95)
we obtain the following solution for this initial value problem:
P(I) = Sexp[-Al]S-IPo
(2.96)
Section 2.5 •
87
BasicReliability EvaluationTechniques
with S the matrix of eigenvectorsof A and A the diagonalmatrix of eigenvaluesof A.
BecauseA is a singularmatrix (the sumof all transitionsis zero) oneof the eigenvalues
is zero. That leads to aconstantterm in the solution
P(t) =
v: + LPie-~
(2.97)
;>1
In most cases wecan neglect thetransientsand are only interestedin the steady-state
solution
Note that the steady-statesolutionis independento f the initial values.The
steady-statesolution can be obtaineddirectly from the transition rates by setting the
time derivativesto zero:
r;
(2.98)
(2.99)
Approximate Solutionof Large Markov Models. The main problem with the
exact solution of large systemsis that all stateprobabilities have to becalculatedat
the same time, even those with a very low probability. For an N-state model, an
N x N matrix has to beinverted to find the steady-stateprobabilities.Assumingthat
all componentshave two states(up and down) an It-componentsystem requires 2n
states.Thus, a IO-componentsystemalready requires 1000 states,and a 150-component model requires the inversion of a matrix of size 1045• In other words, this
methodhas seriouslimitations. We might be able tosomewhatreducethe numberof
states,but exact solutionsfor systemswith more than 10 componentsare in practice
not possibleto obtain. To overcomethese limitations, one can use anapproximated
method, which gives recursive expressionsfor the state probabilities [145]. The
assumptionsmadeare as follows:
• The statewith all componentsin operationhasa probability equal to one.
• The repair rate of a componentis much larger than its failure rate.
• The probability of a statewith k componentsout of operationis much lower
than the probability of a correspondingstatewith (k - 1) componentsout of
operation.
All these assumptionscan be brought back to one basicassumption:the components
are repaired much faster than they fail. This is a reasonableassumptionfor most
engineeringsystems.An exceptionare theso-called"hiddenfailures" discussedbefore.
For hidden failures the model requiressomeadjustments.
Consideragain the statemodel for an industrial supply, as shown in Fig. 2.26.
Partof this figure has beenreproducedin Fig. 2.30.Here A and JL arefailure and repair
rates,respectively.The index 1 refers to lines, the index 2 to
generators,a nd the index 3
to the public supply.
The exactexpressionsfor the stateprobabilitiesof states1 through4 are
(2.100)
88
Chapter2 •
Long Interruptionsand Reliability Evaluation
Figure 2.30 Part of a multistateMarkov
model. (Reproducedfrom Fig. 2.26.)
+ A3 + J-tl )P2 = 3AtPI + 2J-tIPs + J-t2P6 + J-t3P7
(3AI + A2 + A3 + /l2)P3 = 2A2PI + J-ttP6 + 2J-t2PS+ J-t3P9
(3AI + 2A2 + J-t3)P4 = A3PI + /lIP7 + J-t2P9
(2AI + 2A2
(2.101)
(2.102)
(2.103)
The approximatedmethod starts with assumingthat the system is almost certainly
healthy, thus
PI
=1
(2.104)
According to the third assumption,we neglectthe termswith Ps, P6, P7,pg, and P9 on
the right-hand side of (2.101) through (2.103). That gives the following equationsfor
the states2 through 4:
(2AI + 2A2
+ A3 + J-tl)P2 = 3AIPI
(3AI + A2 + A3 + J-t2)P3 = 2A2Pt
(3A)
+ 2A2 + J-t3)P4 = A3PI
(2.105)
(2.106)
(2.107)
As PI is known we obtain the stateprobabilitiesof thesethreestateswithout having to
know the other stateprobabilities:
3AI
P2=------2A) + 2A2 + A3 + J-tl
2A2
P3=------3AI + A2 + A3 + J-t2
A3
P4 =
3Al + 2A2 + J-t3
(2.108).
(2.109)
(2.110)
A correctioncan be madeby recalculatingthe probability PI from
PI = 1-
LP;
(2.111)
;>1
The samemethod can be applied to states5 through 15, each time resulting in an
equationin which only one stateprobability is unknown. Insteadof having to solve
all stateprobabilitiesat the sametime, this procedureallows solving stateprobabilities
sequentially. For very large systems,not all statesare of equal interest, which can
89
Section 2.5 • Basic Reliability Evaluation Techniques
further reduce thecomputationalrequirements.The recursiveprocedurecan, e.g., be
stoppedwhen the stateprobability drops below acertainvalue.
2.5.5 Monte Carlo Simulation
Basic Principles. In all precedingexamples, theunknown quantitieswere actually calculated. We saw several timesthat approximationsand assumptionswere
needed toobtain a solution. In a Monte Carlo simulation, or simply simulation,
theseassumptionsand approximationsare no longer needed. The
Monte Carlo simulation methoddoes not solve theequationsdescribing the model;insteadthe stochastic behaviorof the model issimulatedand observed.
The behavior of the system(stochasticprocess isactually a better term) is
observed many times or for a long
period of time. The averageobservationis used
as anestimatefor the expectedbehaviorof the system.
The basisof each Monte Carlo simulation involves using a so-calledrandomnumber generator.The random-numbergeneratoris needed to bring thestochastic
element in thecalculations.One could use a physical
random-numbergeneratorlike
a dice or a coin, but anumericalrandom-numbergeneratoris more suitablefor computer-basedcalculations.
A coin can be used to model statewith
a
a probability of 50% • Consideras an
example athree-component
system with500/0 availability for eachcomponent.The coin
is used togeneratecomponentstates, with the second
columnin Table2.11 the resulting
sequence. Thisrepresentsthe stateof one of thecomponentsover 24consecutiveI-hour
periods. The same is done for
component2 and component3, resultingin columns3
TABLE 2.11 MonteCarloSimulationwith 50% Probabilities
Hour
I
2
3
4
5
6
7
8
9
10
II
12
13
14
15
16
17
18
19
20
21
22
23
24
Component1
Component2
Component3
SystemI
System2
System3
up
up
down
down
up
down
up
up
down
down
up
up
up
down
down
up
up
down
up
up
down
up
down
down
up
down
down
up
down
down
up
down
up
up
up
down
up
up
down
up
down
up
down
up
up
up
down
up
down
up
up
down
up
down
up
up
down
down
down
up
down
down
up
down
up
up
down
down
down
up
up
down
up
up
down
down
up
down
up
up
down
down
up
up
up
down
down
up
up
up
down
up
down
up
down
down
up
up
up
down
down
down
up
up
up
down
down
down
up
up
down
down
down
up
up
up
up
up
up
down
up
up
down
down
up
down
up
down
down
down
dow
down
down
down
down
down
down
down
down
up
down
up
down
down
90
Chapter2 •
Long Interruptionsand Reliability Evaluation
and 4, respectively. The
column labeled"systemI" gives thestateof a system which is
available if at least twocomponentsare available.
One can make thisM onte Carlo simulationas complicatedas one wants. In the
column labeled"system2" the system is down if less than two
componentsare available
for two consecutivel-hour periods, and if the system is down it remains down for at
least 3 hours.For "system3" the system needs three
componentsto beavailablein the
hourly periods 8through 18, but only two for theother periods . As a second example
consider threecomponentswhose lifetime is uniformlydistributed between 0 and 6
years. To generate the lifetime of these
componentswe can use a dice. By using this
we simulatethe behaviorof this three-componentsystemduring 10 years. In Fig. 2.31
three possibleoutcomesof this "experiment" are shown . Each possible
outcomeis
called a "sequence."During sequence 1, the firstcomponentfails after 3 years and
again after 6 years; the second
componentfails after 2, 6, 7, 9, and 10 years, etc.
Sequence I
3
21
6
I
~~
I
4
I
I
4
Sequence 2
o--L-o
6
G>>-+-~-~e
6
o~
cr--1--o--i--o--,,-6----,,.--Sequence 3
3
4
4
2
6
010
..
4
5
0
3
--0
I
5
;
10 years
- 0
•
Figure 2.31Threesequencesof a Monte
Carlo simulation.The circles indicate failures
followed by repair ; the numbers in between
indicate times-to-failure.
At time zero all threecomponentsstart their first lifetime. Upon failure they are
repaired and a new lifetime determined.This
is
process isrepeateduntil t = 10 years is
reached .F rom the outcomeof this stochasticexperiment,many different outputparameters can be chosen, for example,
• Total numberof componentfailures in a IO-year period . In this case the values
11, 7, and 8 are found .
• Total numberof events with two or morecomponentfailures in the same year,
with values 3, 2, and 1 being found .
• Probability distribution function of the componentlifetime.
Numerical Random-Number Generators.
In practice one never uses physical
random-number generatorslike dice or coins. The reason is
that it is difficult to
actually use them in acomputer program and hand-calculationsof Monte Carlo
simulationsare verycomplicated,as will be clearafter the preceding examples.
Section 2.5 •
91
BasicReliability EvaluationTechniques
A numerical random-numbergeneratorcreates a rowof integers of pseudorandom nature. The row is not really random as a numerical algorithm is used to
calculateit-therefore,the term"pseudo-randomnumbergenerator."Most computer
simulationsuserandom-numbergeneratorsof the following form:
U;+1
= (aU;)modN
(2.112)
where a and N have to be chosen. The
output of this is a rowof integers with values
between 1 and(N - 1).
EXAMPLE
Considerthe values N = 11 and a = 7. That gives the following row of
integers:
1,7,5,2,3,10,4,6,9,8,1,7,5,2,3,10,4,6,9,8,1,
etc.
The row repeatsitself after 10elements,which isunderstandable
if one realizesthat thereare only
10 possibleoutcomesof (2.112). A cyclelength 10 (in general(N - 1) ) is the longestpossible
value. To showthat shortercycle lengthsare also possible,considerthe random-numbergeneratorwith N = II and a = 5 which has twopossiblerows, eachof cycle length 5:
1,5,3,4,9,1
2,10,6,8,7,2
The random-numbergeneratorsin use in Monte Carlo simulationshave much longer
cycles lengths, andtherefore much higher valuesof N. A popular value is N =
231 - 1 = 2 147483647.Most valuesof a give a cycle length less
t han N - 1. A value
of a which gives themaximumcycle length isa = 950 706376.Startingfrom U = 1 we
get the following rowof integers:
1, 950706376,129027 171,I 782259899,365181143,1966843080,etc.
The resulting integer is often divided by
N to get arandomnumberbetween 0 and 1,
which leads to a slightlydifferent version of (2.115):
u _ aNU;modN
;+1-
N
(2.113)
The resultof (2.113) is arandomdraw from the uniform distribution on the interval
(0,1). Neither zero nor one can beobtainedthrough this method, which is often an
advantageas it preventsdividing by zero in further processingof the result. This
standarduniform distribution is the basis for allM onte Carlo simulations.
For N = 11 and a = 7, (2.113) results in thefollowing row of samples:
0.09,0.63,0.45, 0.18, 0.27, 0.91, 0.36,
0.55,0.82,0.73, 0.09, etc.
EXAMPLE
Simulating a Probability-RandomMonte Carlo Simulation. Two types of
Monte Carlo simulation can bedistinguished:random simulation and sequentialsimulation. An example of random simulation is the simulation shown in Table 2.11.
In a random Monte Carlo simulation each componenthas aprobability of being in
a certainstate. Thesimulation generatescombinationsof componentstates.For each
resulting combination the systemstate (healthy or nonhealthy) is evaluated.This
whole process isrepeateduntil a certainaccuracyis obtained.
92
Chapter 2 • Long Interruptions and Reliability Evaluation
The basisof a randomMonte Carlo simulationis the probability: an eventtakes
place with acertain probability, a quantity has acertain value with a certain probability, or a componentis in a certainstatewith a certainprobability. A probability is
simulated by drawing a value from the standard uniform distribution introduced
before. Let p be the probability that the componentis in state 8 1; otherwise, the
componentwill be in state 8 2, then the Monte Carlo simulation proceedsas follows:
• Draw a value U from the standarduniform distribution.
• If U :s p the componentis in stateSI.
• If U > P the componentis in state S2.
Note that for U = p the componentstateis actually not defined. In this examplethis
situationis attributedto stateSI but it could equally have beenattributedto state S2.
This ambiguity has to do with our discretizationof the uniform distribution. For a
continuousdistribution the probability that U =p is zero. For a random-numbergenerator with a cycle length of 231 - 1 this probability (5 x 10-1°) is small enoughto
neglect in allpracticalcases.
Simulating a Time Distribution. The basis of a sequentialsimulation is the
time distribution. Thereforewe need amethod of obtaining other distributions than
just the standarduniform distribution, Le., the uniform distribution on the interval
(0,1).
The uniform distribution on an interval (T1, T 2) is obtainedfrom a sampleof the
standarduniform distribution U as follows:
(2.114)
where X is a sample from the uniform distribution on the interval (Tt , T 2) . More
general:a stochasticvariable S with a distribution function F(s) is obtainedfrom
(2.115)
where U is a stochasticvariablewith a standarduniform distribution.To provethis, we
look at the probability distribution function of the stochasticvariable S accordingto
(2.115), thus at the probability that S is lessthan a certainvalue s.
Pr{S < s} = Pr{F-1(U) < s}
(2.116)
As F is a non-decreasingfunction, we can write this as
Pr{S < s}
= Pr{ U < F(s)}
(2.117)
The stochasticvariable U has astandarduniform distribution; thus,
Pr{ U < x} = x, for 0 < x < 1
(2.118)
As 0 < F(s) < 1 we get theintendedexpression,which proves that S is distributed
accordingto F(s).
Pr{S < s} = F(s)
(2.119)
93
Section 2.5 • Basic Reliability Evaluation Techniques
Consideras anexample,the Weibull distribution introducedby (2.25).From (2.115) it
follows that a sample W from the Weibull distribution with characteristictime 0 and
shapefactor m is obtainedfrom a sampleU from the standarduniform distribution by
W = Oy!-ln(l - U)
(2.120)
For m = 1 we obtain the exponentialdistribution as a special case
o f the Weibull
distribution. A sample E from the exponentialdistribution with expectedtime 0 is
obtainedby
E
= -Oln(l -
U)
(2.121)
Sequential Monte Carlo Simulation.The examples in Fig. 2.31 show sequena
tial simulation. In a sequentialMonte Carlo simulation, the whole timebehaviorof
a system issimulated,with failure and repair of componentsthe main subject in a
reliability study. But alsoother events, like loadswitching and weatherchanges,can
be part of the simulation. This kind of simulation offers the most opportunitiesof
obtainingoutput, but it also requiresthe most programmingand computingefforts.
"Thedetailsof a sequentialM onte Carlosimulationvary widely anddependon the
particular application,the kind of programminglanguageavailable,and on personal
tasteof the programdevelopers.Below, a possiblestructureis given which was used
successfully by theauthorfor evaluatingthe reliability of industrialpowersystems[61],
[62], [63]. Only onesequenceof a given length is described here. This sequence
shouldbe
repeateda largenumberof times to getstatisticallyrelevantresults.
I. Set up an initial event list. At thestart of each sequence, times for the first
event aredrawn for eachcomponent.The first event is typically a failure or
start of maintenance.These events aresorted on time of occurrenceand
placed in a so-called"event list." Part of an event list would typicallylook
as follows:
0.15 years
component2
failure
1.74 years
component5
maintenance
3.26years
component1
hidden failure
4.91 years
component5
failure
5.67 years
component2
maintenance
6.21 years
componentI
maltrip
This event listshouldbe interpretedas follows: at t = 0.15 years,c omponent
2 will fail; at t = 1.74years,maintenanceon component5 is planned,etc. Not
all events in the list willactuallyoccur. We will see belowthat events may be
removedfrom the' event listand that events may beinserted.Furtheron in
the simulationof this sequence, it will always be the event top
on of the list
which will be processed,a fter which the event list will beupdated.When the
event list isempty the simulationof this sequence is over.
2. Processthe event on top of the event list.
Processingof the event on topof the
eventlist (thusthe next event tohappenin the system) is themain part of the
simulation, which will take up most time in programmingand deciding
about. This is where thestochasticmodel of the power systemand its componentsis implemented.The processingof an event typicallyconsistsof
making changesin the event list and making changesin the electrical
model of thepowersystem.Changesin the powersystem can be the
removal
94
Chapter 2 • Long Interruptions and Reliability Evaluation
of a component(e.g., due to theinterventionof the protection)or the insertion of a branch(e.g., repair of a componentor due to ashort-circuitfault).
To assess the effect
of the eventon the load, eitherthe newsteadystateor the
electrical transientdue to theevent need to beevaluated.The interruption
criterion needs to beappliedto decideif this eventleads to aninterruptionor
not. The changesin the eventlist will be discussedbelow for different events.
(a) Short-circuitevent. The next event after a short-circuit event will be an
intervention of the protection. Some rules areneededto decide which
relays will intervene: the relay or relays which need to
clear this fault;
thosewhich incorrectly intervene;and thosewhich take over the protection in case one orm oreof the primary relays fails totrip. For eachrelay
a time until tripping needs to bedetermined.Tripping of the fault normally takesplace very soon after the short-circuit event. Thereforeone
can decide totreat fault initiation (short-circuitevent)and fault clearing
(protectioninterventionevent)as one event.H ere they areconsideredas
two events.
(b) Protectionintervention event. During the processingof this event one
needs todistinguishbetweenthe last relay totrip and all the other protection interventionevents.After the last relay hastripped the repair of
the faulted componentc an startand also theswitching neededto restore
the nonfaultedcomponentstripped by the protection. For the Monte
Carlo simulation this meansthat times to repair and times to switching
need to bedetermined.Alternatively one can determineall these times
when processingthe short-circuitevent.
(c) Repairevent. When a componentis repaired,it can fail again.Therefore
a time to failure needsto be determinedfor all its failure modes:short
circuit, maltrip, hiddenfailure, etc. Different failure modeswill typically
have different lifetime distributions.
(d) Maltrip event. A maltrip eventis associatedwith the power systemprotection, either with a circuit breakeror with a protectiverelay. The next
eventsto bedeterminedarerepairof failed componenta ndrestorationof
the primary componenttripped.
(e) Hidden failure event. Ahidden failure eventwill not reveal itself immediately. Thereforeit will only changethe way the relay will infuture react
to a short-circuitevent. Only when ahidden failure reveals itself,either
due to ashort circuit or due tomaintenance,will the repair start.
(f) Start of maintenanceevent. Start of maintenancewill require the scheduling of an end of maintenanceevent. For an accuratemaintenance
model, one needs tointroducean additional event called "maintenance
attempt." Maintenanceattemptsare scheduledand either immediately
lead to astart of maintenanceevent or to a new maintenanceattempt
event. Somerules areneededto decideif the systemstateis suitablefor
maintenanceto beperformed.The rules will dependamongotherson the
companyrules for performingmaintenance.S omeexamplesare
• Maintenancecannotbe performedat more than one componentat the
same time, e.g.,becausethere is only one maintenancecrew available.
• Maintenancewill not be performedif it leadsto an interruptionof the
supply for any of the loads.
95
Section 2.5 • Basic Reliability Evaluation Techniques
• Maintenancewill not be performedwhen aparallel or redundantcomponentis out of operation.
When processingthe start of maintenanceevent, the time for an endof
maintenanceevent needs to bedetermined.
(g) End ofmaintenanceevent. When the maintenanceis finished a newmaintenanceattempt or start of maintenanceevent needs to bedetermined.
Also some future fail events will be influenced by the maintenance.
Typically the componentis assumedto be"as-good-as-new"a fter maintenance.In that case allfuture fail eventsare removedfrom the eventlist
and new ones aredrawn from appropriatedistribution functions.
Some additional rules might be neededto control the processingof events.
One might, for instance,decidethat a componentc annotfail while it is out of
operation(for any reason).One can makea checkduring a failure event to
see if a componentis in operation and simply draw a new failure event
without any additional processingif the componentis not in operation.
One can also decide to shift all
failure events belonging to a component
further into the future with a time equal to the time during which it is out
of operation.
3. Update the event list. All new events whichoccur before the end of the
sequenceare placed in the event list; the eventjust processedis removed;
the eventlist is sortedagain;after which the eventthat appearson top of the
event list is processed.
Errors in the Monte Carlo Simulation. An exampleof the result of a Monte
Carlo simulation is shown in Fig. 2.32. The figure has beenobtainedby taking samples from the uniform distribution on the interval (0,1), followed by calculating the
averageover all the proceedingsamples.For an increasingnumber of samples,the
averagevalue approaches0.5. As wecan see from the figure, theerror is still rather
large after 100 samples.
Figure2.33 gives thebehaviorfor a much largernumberof samples.A fter 10000
samples,the error has becomelessthan 1%, but is still not zero. An importantproperty
of the Monte Carlo simulation is that the error approacheszero, but never becomes
zero. Figure 2.33 also showsanotherpropertyof the Monte Carlo simulation: the fact
0.4
~
Q)
~ 0.3
I
0.2
Figure 2.32 Outcomeof a Monte Carlo
simulation.
20
40
60
Samplenumber
80
100
96
Chapter2 • Long Interruptionsand Reliability Evaluation
0.55 .------~--~--~--~--_,
2000
4000
6000
Samp le number
8000
that each simulation may give a different result. The figure gives the result of 10
simulations, each using adifferent starting value of the random-numbergenerator.
Note that exactly the sameresultsare obtainedif the samestarting value is used for
the random-numberg enerator.
The error in the result of a Monte Carlo simulationcan be estimatedby usingthe
so-calledcentral-limit theorem.This theoremstatesthat the sumof a large numberof
stochasticvariableshas anormal distribution. Supposethat eachsequenceof a simulation gives a value Xi for a certain stochasticvariable X. This value can be the total
number of interruptionsduring 20 years, but also the fraction of interruptionswith
durationsbetween1 and 3 hours.What we areinterestedin is the expectedvalueof such
a variable.To estimatethe expectedvaluewe usethe averagevalue, which is astandard
procedurein statistics. Let X be the averageof N samplesof Xi:
(2.122)
For sufficiently large N, X is normally distributedwith expectedvalue ux andstandard
deviation aA" where Ilx and ax areexpectedvalue and standarddeviationof Xi' Thus,
_
'iN
X is an estimatefor Ilx (the expectedvalue of X) . The error in the estimateis proportional to the standarddeviation. Note that obtainingthe valueof ux is the aim of the
simulation.
The Stopping Criterion. The fact that the error in a Monte Carlo simulation
will never becomezero meansthat we have toaccepta certain uncertaintyin the result. This issometimesmentionedas a disadvantageof the Monte Carlo simulation,
but also analytical calculationsare uncertain, due to theassumptionsand approximations madein the model. Where the error in an analytical calculationis often impossible to estimate(unless a better model is used), theuncertainty in the result of a
Monte Carlo simulation can be estimated.The outcomeof any Monte Carlo simulation will be a stochasticquantity with a normal distribution. For the normal distribution we know that 95% of all values are within two standard deviations of the
expectedvalue. We saw above that the standarddeviation after N samplesis equal
to ~. The 95% confidenceinterval of the estimateis thus,
Section 2.5 •
97
BasicReliability EvaluationTechniques
-
ax
-
ax
(2.123)
X-2-</lx<X+2-
./N
./N
The standarddeviation of the stochastic
quantity X, ax , can beestimatedthroughthe
following expression:
I ?=xl- [1
ax ~
N_ I
N
N
]2
(2.124)
N?= Xi
1=1
1=1
At regular momentsduring the simulation, e.g., after every 100 sequences, error
the in
the estimates may be
calculatedand comparedwith the required accuracy. When the
required accuracy is reached the
simulationcan bestopped. Note that to determine the
of the sum of theXi values but also of the sum
error, one needs not only keep a record
of their squares .
Convergence Tests.Because of the slow convergence process Monte
of a Carlo
simulationit is hard to recognize a case in which the average no longer converges to
the expected value. Such situation
a
arises, e.g., when the
random-numbergenerator
has ashort cycle length.
Consider again (2.123), which shows that error
the (X - u.x ) decreases as
-:fN. One
can conclude from this that the function
(2.125)
neither converges nor diverges. The convergence
parameterC has been plotted in Fig.
2.34 for 10simulationsof 10000 samples each. The underlying
simulationis the same as
in Figs. 2.32 and 2.33. We see
that the plotted quantity remains within a bandaround
/lx .
zero, thusthat the averageX indeed converges to the expected value
In Fig. 2.35 the same convergence
parameteris plotted for a simulation which
does not converge. The divergence is clearly visible.
(From sample 2000onward, the
random-numbergeneratorwas given a cycle length of 1000 samples.)
!l
0.5
I
.,
~
~
~
U - 0.5
Figure 2.34 Convergence
p arameterfor 10
identical Monte Carlo simulations.
2000
4000
6000
Sample number
8000
10000
98
Chapter2 •
~
I
Long Interruptionsand Reliability Evaluation
0.5
os
0..
g
"
"e!'
"c>
o
U -0.5 ,
2000
8000
10000
Figure 2.35 Con vergence
parameterfor a
non-con vergence case.
2.5.6 Aging of Components
In most studies it is assumed
that both failure rate and repair rate are
constant.
of dataand a lack ofevaluationtechniques. At the
The underlyingreasons are a lack
moment, only the Monte Carlo simulationis capableof incorporatingnonexponential
nonexponendistributionsfor nontrivial systems. But despite the lack applicationof
of
tiallifetime distributions, it is still worthwhile to have a closer look at the
variousaging
phenomena.Nonexponential repair time distributions are easier to understand,
althoughequally difficult to incorporatein the reliability evaluation.
Two Typesof Aging. Aging is used in daily life as thephenomenonthat the
failure rate of a componentincreases with its age. Here it will be used in a slightly
more general sense: aging is the
phenomenonthat the failure rateof a componentis
dependenton:
• the actual age of thecomponent.
• the time since the last repair or
maintenance
.
To quantify the dependenceof the failure rate on the age
of the component,the
so-called"bathtubcurve" is often used. Acommonway of drawingthe bathtubcurve is
T) is called the wear-in period,
after T 2
shown in Fig. 2.36. The period between 0 and
the wear-out period, and betweenT) and T 2 the useful life or the periodof random
failures. One should realize
that the bathtubcurve is only a stylized version of
whatcan
be a rathercomplicatedfunction of time. The actual failure rate as a functionof time
can beof completely different shapealthoughit
,
is likely to containat least an initial
wear-in period and an overall
increasingfailure rate for oldercomponents[146].
This aging effect can be included in the reliability
evaluationmodels, byrepeating
the calculationsfor different componentage.For each age one assumes
that all failure
rates areconstant.From the expressionsobtainedby usingMarkov models in Section
2.5.4, we knowthat the timeconstantwith which the system reacts to changes is of the
order of the repair times. For such
s hort time scales we can safely assume the failure
rate to be constant.That way one can assess the aging
of the system, e.g., the interruption frequency as afunction of time. When performing such a study one should
99
Section 2.5 • Basic Reliab ilityEvaluationTechniques
Figure 2.36Bathtubcurve :component
failure rate versus age.
o
Component age
realize that also the repair time and the
durationof maintenanceare likely to increase
when thecomponentgrows older. The second type of aging, the fact
that the failure
rate depends on the time elapsed since the last repair
maintenance,
or
is more difficult to
consider in a reliabilityevaluationstudy. Here it is essential
that nonexponentialdistributions are used for thecomponentlifetimes. Techniques like
M arkov modeling and
network representationscan no longer be used. For smaller systems one might use
[123], [215]; for larger systems
highly mathematicaltechniques like renewal theory
only Monte Carlo simulation remains as a practical tool.
As an exampleof the second type of aging, assume that the failure rate only
depends on the time until
maintenanceand that maintenanceis performed at regular
intervals. The failure rate as a function of time is as shown in Fig. 2.37: the failure rate
increases untilmaintenanceis performedon thecomponent,at whichinstantthe failure
rate drops to its initial value again. The
dotted line in Fig. 2.37 represents a kind of
average failure rate .
i
Figure 2.37 Failure rate versus time for
regular maintenanceintervals.
Time -
In Fig. 2.38 the failure rates of two
componentsare plotted (the dashed and the
It is assumed here
dottedline), plus the average of the two failure rates (the solid line).
that maintenanceon the secondcomponenttakes place in between two
maintenance
instantsfor the first component.We seethat the average of the two failure rates varies
less than each of the failure rates. It is easy to imagine
that the failure rate of a large
number ofcomponentsbecomesconstantwhen maintenanceon them isperformedat
different times.
In reality the failure rate not only depends on the time elapsed since the last
maintenancebut also on the time elapsed since the last
maintenanceor repair.
tOO
Chapter2 • Long Interruptionsand Reliability Evaluation
t
• Avejrage
..
••
. :
,','
0.
.
COlmpo~ent I
..
.'
..0:
",1:
""
•
A
, ,'t
o·
.
0
ee
Time----+
Figure 2.38 Failure rate versus time for two
components.
Similar reasoningsas given formaintenancecan be used for failure, with the difference
that the failure instantsare lessregularly positionedthan maintenanceinstants.
As-Good-As-Newor As-Bad-As-Old. In Fig. 2.37 and Fig. 2.38 it was assumed
that the failure ratedropped to its original value after maintenance.This model is
called maintenance(or repair) "as-good-as-new."The oppositemodel is called maintenance(or repair) "as-bad-as-old."In the latter case themaintenanceor repair has
no influence on the failure rate; thus the failure rate
just after maintenanceis the
same asjust before. The two models are
shownin Fig. 2.39.For repair as-bad-as-old
the failure ratedependson the ageof the component,for repair as-good-as-newit
dependson the time since last repair.
The actualfailure rate isnormally a combinationbetweenas-good-as-new
and asbad-as-old.This can bemodeledas the sumof two failure rate, thus twocomponentsin
series: one beingrepairedas-good-as-newand theother being repairedas-bad-as-old.
The latterone will lead to an average increase in failure rate which leads towear-out
the
phase in thebathtubcurve.
i
t=O
As-bad-as-old
As-good-as-new
i
Age of thecomponent-e-->
Repairor
maintenance
Figure 2.39 Repair as-good-as-new and asbad-as-old.
Failure Rate Increase due toMaintenance. Somethingthat should also be considered in reliabilityevaluationis that maintenanceand repair can lead to anactual
increase in failure rate. The
s tandardexampleis the screwdriverleft inside the switchgear. But alsomore subtle effects are possible. In m
a aintenanceoptimization study
Section 2.6 • Costsof Interruptions
101
one has to take this intoaccountone way or the other. Alsod uring maintenancethe
chance of anoutage of anothercomponentis increased: itsloading is higher and
there is activity in theneighborhoodwith the associatedrisk of errors.
Many aspects of aging are extremely difficult quantify,
to
but shouldat least be
consideredin a qualitative way in reliability evaluationstudies. A serious difficulty in
includingcomponentaging is the lackof availabledata:not just componentfailure data
is needed, but alsorepair and maintenancerecordsof all the components.
Aging Data. Information on aging of power systemcomponentsis hard to
publications
find. A few examplesof good data are given below. There are more
addressingthis problem [107], but the total amount of data is not enough to
include aging with sufficient confidence into the reliability
evaluation.
• A number of Dutch utilities published "expert opinions" on the ageof a
componentat which the failure rate significantlys tarts to increase [124]. A
group of expertswas asked to give their
estimationof this age forcomponents
operatedunder "good circumstances,""averagecircumstances,"and "bad
circumstances.
"
[125]. One of the
• Bathtubcurves for transformersare presented in reference
conclusionsis that newer generationsof transformershave not only a lower
overall failure rate but also a longer useful life. The useful life is the period
during which the failure rate is more or less
constant.Newer productiontechwearniques have however not been able to significantly reducenumberof
the
in failures.
• Another interestingstudy is publishedin reference[126]. By using purchasing
records anassessment
is madeof the age at whichtransformersfatally fail, i.e.,
a failure severeenoughfor them to bescrapped.It turnedout that the failure
After that,
rate stayedconstant,at about0.01 per year, for the first 12 years.
the failure rate increased until 1 per year at an age of 29 years.
• A bathtubcurve for circuit breakersis presentedin [127], based on the observation of a largenumberof breakers.The failure rate decreases from 0.2 for
age zerothrough 0.05 for 8 years after which it rises to 0.15 for 10-year-old
breakers.
• In reference [128] the failbehaviorof circuit breakersis studiedby dividing the
causeof failures into three categories:
- initial failures.
- randomfailures.
- wear-outfailures.
By plotting the failure rate as afunction of age for eachcategory,it is shown
that the failure rateof random
that the failure rateof initial failures decreases,
failures staysconstant,and that the failure rateof wear-outfailures increases
with time.
2.8 COSTS OF INT.RRUPTIONS
To considerinterruptionsof the supply in the design andoperationof power systems,
the inconveniencedue tointerruptionsneeds to bequantifiedone way or theother.The
term inconvenienceis rather vague andbroad. Any seriousquantification requires a
102
Chapter2 • Long Interruptionsand Reliability Evaluation
.....Reliability costs
- - . Buildingcosts
- Totalcosts
Reliability
Figure 2.40 Costs versus reliability: costs of
building and operation(dashed curve), costs
of supply interruptions(dottedcurve), and
total costs (solid curve).
translationof all inconvenienceinto amountof money. In theremainderof this section
we will considercostsof interruptionsin dollars, but any othercurrencycan be usedof
course.
Many publications on costs of interruption show a graph with costs against
reliability. Such a curve isreproducedin Fig. 2.40. The ideabehind this curve isthat
a more reliable system is more expensive to build
and operate,but the costsof intertotal costs will
ruption (either over the lifetimeof the system, or per year) are less. The
show aminimum, which correspondsto the optimal reliability. Even if we assumethat
both cost functions can bedeterminedexactly, the curve still has some
seriouslimitations. Figure 2.40 should only be used as a
qualitativedemonstrationof the trade-off
between costs and reliability.
• Additional investmentdoesnot always give a more reliable system: an increase
in the numberof componentscould even decrease the reliabiity.
• Reliability is not a single-dimensionalquantity. Both interruption frequency
and duration of interruption influence theinterruptioncosts.
designercan choose
• Thereis no sliding scaleof reliability and costs. The system
between a limitednumber of design options; sometimesthere arejust two
advantages
options available. The choice becomes simply comparisonof
a
and disadvantageso f the two options.
• The two cost termscannotsimply beadded.One term (building and operational costs) has a smalluncertainty,the other term (interruptioncosts) has a
large uncertaintydue to theuncertaintyin the actualnumberand durationof
interruptions. A more detailed risk analysis is neededthan just adding the
expected, costs.
The cost of aninterruption consistsof a number of terms. Each term has its own
difficulty in being assessed. Again simply
adding the terms toobtain the total costs
of an interruptionis not the right way,but due to lackof alternativesit is often the only
feasibleoption.
attributableto the inter1. Direct costs.These are the costs which are directly
ruption. The standardexample fordomesticcustomersis the lossof food in
the refrigerator. For industrial customersthe direct costs consist,among
others, of lost raw material, lost production, and salary costs during the
non-productiveperiod. For commercialcustomersthe direct costs are the
Section 2.6 • Costsof Interruptions
103
loss of profit and the salary costs
during the non-productiveperiod. When
assessing the direct costs one has towatchful
be
of double-counting.One
shouldat first subtractthe savings made
duringthe interruption.The obvious
savings are in the electricity costs, but for
industrial processes there is also a
saving in useof raw material.An example ofdouble-countingis addingthe
of the productalreadyincludes the
lost salesandthe salary costs (as the price
salarycosts). Also to besubtractedfrom the costsof interruptionis the lost
productionwhich can be recovered later. Some
plants only run part of the
time. Extra salaryduring overtime needed to recover lost
productionshould
be addedto the direct costs.
2. Indirect costs. Theindirect costs are muchharderto evaluate,and in many
casesnot simply to express inamountof money. Acompanycan losefuture
orderswhen aninterruptionleads to delay in delivering paroduct.A domestic customercan decide to take an
insuranceagainstlossof freezercontents.
A commercialcustomermight install a battery backup. A large industrial
customercould even decide to move plant
a
to an area with less supply
interruptions.The main problem with this cost term isthat it cannot be
attributed to a singleinterruption, but to the (real or perceived)
quality of
supply as a whole.
3. Non-material inconvenience. Someinconveniencecannot be expressed in
2
can be a serious
money. Not being able to listen to the radio for hours
inconvenience,but the actual costs are zero. Inindustrial and commercial
environments,the non-materialinconvenience can also be big
without contributing to the director indirect costs. A wayof quantifyingthese costs is to
look at theamountof money acustomeris willing to pay for not having this
interruption.
To evaluatethe costs of supplyinterruptions,different methodshave beenproposed.
For large industrial and commercial customersan inventory of all the direct and
indirect costs can be made, and this can then be used in the system design and
operation. Even for small customerssuch a study could be made, e.g., to decide
about the purchaseof equipmentto mitigate interruptions. However, for small and
domestic customersit is often the non-material inconveniencewhich has a larger
influence on the decisionthan the direct orindirect costs. For a group of customers,
such an individual assessment is nolonger possible. The only generally accepted
method is the large surveyamong customers.Customersget asked anumber of
questions. Based on the answers the average costs
of interruption are estimated.
These results are typically the ones used by utilities in decision
making. When comparing the resultsof different surveys, it isimportant to realizethat they not all ask
the samequestions.Some surveys ask a very specific
question:"What are the costsof
an interruption of 2 hours on a Monday afternoonin January?"Other surveys use
more indirect questioning:"What is a reasonablecompensationfor an interruption"
i nterruption frequency from 4
or "What would you be willing to pay to reduce the
per year to 3 per year?"Different questionsobviously lead to different estimatesfor
the costsof interruption.
To quantify the costs of aninterruption,again differentmethodsare in use. Some
values can be easily
calculatedinto eachother, with some values acertainamountof
care is needed. Worse that
is it is not always clear from thecontext which methodis
actually used.
104
Chapter 2 • Long Interruptions and Reliability Evaluation
• Costsper interruption. For an individual customerthe costsof an interruption
of duration d can beexpressedin dollars.Thereis no confusionpossibleabout
this. For simplicity, we neglectthe fact that the costsnot only dependon the
duration but on many other factors as well. The costsper interruptioncan be
determinedthrough an inventory of all direct and indirect costs.
• Costsper interruptedkW. Let C;(d) be thecostsof an interruptionof duration
d for customeri, and L; the load of this customerwhen therewould not have
been aninterruption.The costsper interruptedkW are defined as
C;(d)
(2.126)
L;
and are expressedin $jkW. For a group of customersexperiencingthe same
interruption,the costsper interruptedkW are defined as theratio of the total
costsof the interruptionand the total load in casetherewould not havebeen
an interruption:
(2.127)
• Costsper kWh not delivered. Inmany studiesthe assumptionis madethat the
costof an interruptionis proportionalto the durationof the interruption.The
cost per kWh not deliveredis defined as
C;(d)
st;
(2.128)
andis constantunderthe assumption.T hecostper kWh is expressedin S/kWh.
For a group of customersthe cost per kWh not deliveredis defined as
L; C;(d)
dL;L;
(2.129)
Someutilities obtain an averagecost per kWh not deliveredfor all their customers.This value isassumedconstantand used as areferencevaluein system
operationand design. The term "value of lost load" is sometimesusedfor the
cost per kWh not deliveredaveragedover all customers.
• Costsof interruption rated to the peak load. A problem in surveysis that the
actual load of individual customersin case there would not have been an
interruption is often not known. One should realize that surveys consider
hypotheticalinterruptions,rarely actual ones. For industrial and commercial
customersthe peak load is much easierto obtain, as it is typically part of the
supplycontract.Thecostof an interruptioncanbe divided by the peakload, to
get a value in$jkW. Somecare is neededwhen interpretingthis value, as it is
not the same as the cost per kW interrupted (also in $/kW). For planning
purposesthe cost of interruption rated to the peak load can still be a useful
value. The design of a systemis basedfor a large part on peak load, so that
rating the cost to the peak load gives adirect link with the design.
• Costsper interruptionratedto the annualconsumption.For domesticcustomers
it is easierto obtain the annualconsumptionthan the peak load. Rating the
lOS
Section 2.6 • Costs ofInterruptions
cost of an interruption to the annual
consumptiongives a value in $/kWh.
Note that this has no relation to the costs per kWh not delivered.
of
are given in
Someof the results of a Swedish survey after costsinterruptions[200]
1993 and
Figs. 2.41 and 2.42. The survey was
conductedamong 4000 customers in
resulted in interruption costs per kW of peak load for interruption
duration of 2
2.41 gives the costs for a forced interminutes, 1hour, 4 hours , and 8 hours . Figure
120
0
•
•
0
~
o
2 min
I hour
4 hours 8 hours
60
1
40
-
r--"
20
f - - - - """
f---
o
W _ .,....
-- J
Domestic
Agriculture
Trade and
services
-
f---
lJ
Small
industry
-
~
Textile
industry
,', '
~
-
--'--'='
Chemical
industry
Food
industry
Figure 2.41 Int erru ption costs in S/kW for different customers, for forced
interruptions . Results from a Swedish stud y 1993
in [200).
120
0 2 min
•
•
I hour
0 4hours 8 hours
~ 60
8
§
}40
r-r-
20
o
f-------
---
Domestic
..r
Agriculture
•
Trade and
services
--f
Small
industry
~ J
Textile
industry
Chemical
industry
Figure 2.42 In terruption costs in S/kW for different customers. forscheduled
interruptions. Res ults from a Swedish
study in 1993[200).
...
Food
industry
106
Chapter2 •
Long Interruptionsand Reliability Evaluation
ruption, i.e., in case thecustomerreceives nopre-warningof the interruption. Figure
2.42 relates to scheduled
i nterruptions where the customer receives sufficient prewarning. An exchange rateof 7.32 Swedishcrowns per U.S. dollar has been used
and an inflation rate of 2.5% per year, toobtain the costs in 1998 dollars.
The valuesindicated are averages over n
aumber of customers.Surveys have
shown that the range betweendifferent customersis very large, even within one type
of industry. Rangesof interruptioncost within one typeof industry are given bySkof
[147]. For a I-minute interruption the cost for automobilefactories varies between
0.001$/kW and 6$/kW. For a l-hour interruption the range is from 0.3 to40$/kW.
of
Thus, an industry averageshould be treated with care when assessing the cost
interruption for a specificindustrial customer.Where possible, it isrecommendedto
otherpublicausecustomer-specificdatainsteadof industrynationalaverages. Several
a nd resultsof otherways toestimatethe interruptioncosts; an
tions give survey results
admittedlyincompletelist is [21], (129], [130], [131], [132], [216].
2.7 COMPARISON OF OBSERVATION AND RELIABILITY EVALUATION
Despite all the reliability analysis toolsavailable, simple past-performancerecords
remain the main sourceof information on systemperformance.This does not imply
that reliability analysis has no value. To the
contrary, analysistechniquescan obtain
results much fasterandwith a higher degreeo f accuracythanpastperformancerecords.
This holds especially forindividual sites. For the evaluationof operationalreserve,
past-performanceis simply not available.Stochasticprediction techniquesare the
only option here. However,comparisonbetweenstochasticprediction techniquesand
pastperformancemeasuresis a highlyundervaluedarea. Very little work has been
done
on this often with thejustification that it is not possible.
Some kind of verification of stochasticprediction techniquesremains needed,
especially asmany engineersremain, rightly or wrongly, skepticalabout the outcome
of reliability evaluations.The emphasison past-performance
recordsis, in the author's
view, also determinedby the skepticism toward stochasticprediction techniques. A
number of ways of comparingobservationsand the resultsof reliability evaluation
are given in the following list:
• Apply stochasticprediction techniquesto a systemthat has not changedtoo
much over alongerperiod,andfor which dataare availableon thenumberanddurationsof supply interruptionsover this period. As the transmissionnetworks in most industrializedcountrieshave remainedmore or less the same
over the last 10 years or so, such a verification
techniquecould be used here.
• Use a largenumberof observationpoints,e.g., allurbandistribution networks
within one utility. Somefurther selection might be needed to get a homogeneous group of systems. Applystochasticprediction techniquesto a typical
observationresultsof
configurationand comparethe results with the average
all existing networks.This verification techniqueis suitablefor level III (distribution) reliability studies.
• Use acommondataset.Choosea system for whichinterruptiondataas well as
componentfailure dataare known over a numberof years. Use the observed
failure rates asinput for the stochasticprediction, thus eliminating the data
uncertainty.Any differences between observed
and predictednumberof interruptionscan becontributedto model limitations.
107
Section 2.8 • ExampleCalculations
• Perform detailed analysis of the underlying events of interruptions. Assess
whetherthese events orcombinationsof events arepart of the stochasticprediction model. Thistechniquemight be somewhattrivial for distribution systems, but itappearsespecially useful fortransmissionand generationsystems
where onlymultiple events lead tointerruptions.
2.8 EXAMPLE CALCULATIONS
2.8.1 A Primary Selective Supply
Consideran industrial customerwith a so-calledprimary selective supply, as
shown in Fig. 2.43.P rimary selectivesuppliesand other ways of improving the reliability are discussed in detail in
C hapter7.
'A,r
At,r,
Figure2.43 Example ofreliability
calculation:primaryselectivesupply.
For this example we use the following
c omponentdata:
• A = 5 year-I, failure rate of each of the two public supplies.
• r = 0.00025 years= 2 hoursand 11minutes,averagerepair time of the public
supply.
• At = 0.02year-I, transformerfailure rate.
• r t = 0.0114 year = 100 hours,transformerrepair time.
• Ps = 30/0, transferswitch failure probability.
The frequencyof interruptionsdue to overlappingoutagesis obtainedfrom the equation for the failure rateof two parallel components(2.39):
Ap
= 2rA2 = 2 x 0.00025X
52
= 0.0125interruptionsper year
(2.130)
The averageduration of an interruption is the equivalentrepair time of the parallel
connectionas obtainedfrom (2.41):
rp =
r
2" = 0.000125 years= 1.1hours
(2.131)
In otherwords, the secondof two overlappingoutagesstartson averagein the middle
of the first outage.From the interruption rate and theinterruption duration, we can
obtain the unavailability due to overlappingoutages:
Qp = Aprp = 1.56 x 10-6 = 0.014hoursper year
(2.132)
108
Chapter2 •
Long Interruptionsand Reliability Evaluation
In a primary selective supply, atransformeroutagecan also lead to aninterruption. The transformeroutagerate (0.02year-I) is of the sameorderof magnitudeas the
outagerate due to overlappingoutagesin the supply. Theduration of transformer
outagesis much longer. Theunavailability due to transformeroutagesis
Q, = A,r, = 2.28 x 10-4
= 2 hoursper year
(2.133)
When very longinterruptionsare aconcern,a secondtransformershould be placed in
parallelwith the existing one and the switching
shouldbe performedon secondaryside.
This leads to the so-called
secondaryselective supply. Theinterruptionfrequency due to
overlappingtransformeroutagesis very small:
A,p
= 2r tA; = 9.1 x
10-6 interruptionsper year
(2.134)
Apart from overlappingsupply outagesand transformerfailures, interruptionscan be
due to a failureof both supplies at the same time and due to a failure
of the transfer
switch. Failure of both supplies at the same time mainly
is
due to outagesat a higher
voltage level, either medium voltage distribution or transmission,dependingon the
supply configuration.Interruption rates associatedwith this vary significantly, with a
typical range between 0.05 and 0.5
interruptionsper year. Aseparatestudy is needed
for eachsupply configuration,or alternativelyinformation needs to beobtainedfrom
the utility. The probability that the transferswitch fails was given as
P.f = 3%, which
meansthat the switch willnot transferthe loadcorrectlyin 3% of the cases for which it
is supposedto do so. The frequency
o f cases in which thetransferswitch issupposedto
transfer the load is equal to the outagerate of one of the supplies. Theinterruption
frequency due totransferswitch failure is thus,
Ps x As = 0.15 per year
(2.135)
We seethat the transferswitch is apparentlya weak part in the supply. Toobtain a
reliable supplyit is thus essential tochoosea reliabletransferswitch. Alsomaintenance
on the transferswitch plays animportantrole.
2.8.2 Adverse Weather
Consideragainthe primary selective supply in Fig. 2.43. We
considerthe factthat
the failure rate is not constantduring the year.Most overheadline outagesare due to
adverseweatherlike snow, storm, or lightning. Overheadline outagesare much more
likely during adverseweatherthan during normal weather.The failure rate as a function of time will look like in Fig. 2.44: the failure rate is low
mostof the year,but high
during a numberof short periodsof adverseweather.
The adverseweatherperiodsare not fixed but stochasticin time as well. AMonte
Carlo simulationwould be anappropriatetool, if sufficient dataandmodel detailswere
available.To enablea simplified analysis, weconsidera two-statemodel, asshownin
Fig. 2.45. The failurerate during adverseweatheris Al and during normal weatherA2'
The adverseweatheris presentduring a fraction T} of the time and normal weather
during a fraction T2• The average failure rate
A is obtainedfrom
A = Al T I
+ A2T2
(2.136)
For both statesan interruption frequency can bedetermined,after which the annual
of these two.Supposeas anexamplethat 75% of
interruptionfrequency is the average
supplyoutagesare due to adverse
weatherwhich takesplaceduring 100hoursper year.
The failure ratesduring adverseandnormalweatherare, respectively:Al = 329 per year
109
Section 2.8 • ExampleCalculations
~ Adverseweather
!
Figure 2.44 Failure rate as a function of
time-normaland adverse weather.
Normalweather
Adverse
weather
Averagefailure rate
Normalweather
A21---------------'
Figure 2.45Two-statemodel with normal
and adverse weather.
1 year
andx, = 1.25per year. The averagefailure rateis the sameas in thepreviousexample:
A = 5 per year. The repair time is also likely to beaffectedby the adverseweather.We
usethe following repairtimes: '1
2.59 hour (during adverseweather)and '2 = 1 hour
(during normal weather)leadingto the sameaveragerepair time as before (r=2 hours
11 min).
=
_
At T
,]
+ A2 T2'2
, = -t- - - - A]T] +A2T2
(2.137)
The normal weatherinterruption rate is found by using the sameexpressionfor the
parallelconnectionas before,with the exceptionthat failure rateand repairtime during
normal weatherare used insteadof the averagevalues.
Ap2 == 2'2A~
= 0.0003566per year
(2.138)
Normal weatheris presentduring a fraction T2 = ~~~~ of the year, which gives for the
expectednumberof interruptionsper year due to normal weather:
T2A p2 == 0.0003525interruptionsper year
(2.139)
The adverseweatherinterruption rate is
ApI
= 2,)AI = 64 per year
(2.140)
110
Chapter2 •
Long Interruptionsand Reliability Evaluation
This is a very high value, but normal weatheris only presentduring a fraction T I =
8170~O = 0.0114of the year. The contribution of adverseweatherto the annualinterruption frequencyis
TIApl
= 0.73 interruptionsper year
(2.141)
The annualinterruptionfrequencyis thereforevery much affectedby adverseweather.
Note the large differencewith the interruption frequencyfound before by assuminga
constantfailure rate (0.0125 per year). It is clear that the influenceof adverseweather
cannotbe neglectedin reliability evaluationstudieswith parallelconnections.F or series
connectionsthe interruption rate is the sum of the componentfailure rates and the
averageinterruption rate is the sum of the averagecomponentfailure rates. Only for
parallel connectionsdo we need to explicitly consideradverseweather.
2.8.3 Parallel Components
Considera systemconsistingof n identical componentsin parallel. Eachcomponent has an outagerate A and an averagerepair time r. The interruption rate of the
systemcan be calculatedfrom expression(2.44), resulting in
(Ar)"
Al = n -
(2.142)
r
Apart from interruptionsdue to theseoverlappingoutages,the systemcan be interruptedwhen a failure in onecomponentl eadsto the outageof all components.O necan
think of failure of the protection, tripping of equipmenton the voltage sag or on
anothertransient, or transient instability. Supposethat there is a probability ex that
the underlyingfailure of a componentoutageleadsto a systeminterruption.For an ncomponentsystemwith a componentoutagerate A, this gives an additional interruption frequencyof
A2 = an):
(2.143)
The total numberof interruptionsis given by the following expression:
Ato l = A) +)...2 = an):
(Ar)n
+ n -r -
(2.144)
For mostcomponents>..r « ] so that the secondterm reducesvery fast for increasingn,
while the first term increaseslinearly with the numberof parallelcomponents.T he first
term will rather quickly start to dominateafter which an increaseof the number of
parallel componentsonly decreasesthe reliability. Assume the following component
data: A 1 per year, r = 0.001 year,ex = 10/0. The resultinginterruptionratesaregiven
in Table2.12. Wenotethe somewhats urprisingresult that threecomponentsin parallel
is less reliable than two componentsin parallel.
=
TABLE 2.12
Rate
Influenceof Numberof ParallelComponentson Interruption
n
Individual
Overlapping Outages
Total Interruption Frequency
1
2
3
0.0 I per year
0.02 per year
0.03 per year
I per year
2 x 10-3 per year
3 x 10-6 per year
1.0I per year
0.022 per year
0.030 per year
111
Section 2.8 • ExampleCalculations
To justify a three-component
model,the interruptionfrequencyfor n = 3 needsto
be lessthan for n = 2, thus
(Ar)3
(Ar)2
3aA+ 3 - - < 2aA+ 2 - r
r
(2.145)
resulting in the following upper bound for the probability that a componentoutage
leadsdirectly to a systemoutage
a < 2AY - 3(Ar)2
(2.146)
For the previousexamplethis givesa < 0.002. Thus,a three-componentsystemis only
justified if the protectionof the componentis very reliable,the risk of transientinstability is low, etc.
2.8.4 Two-Component Model with Aging and Maintenance
To assess the effect
of aging and maintenanceon a parallel connection,we consider two componentswith a time-dependento utagerate:
(2.147)
with t the time since lastm aintenance.F or maintenanceperformedevery 4 years,the
averageoutagerate is
-=4I[
A
0
A(t)dt
= 0.16 outagesper year
(2.148)
We will calculatethe interruption frequency of the parallel connectionof these two
components.We assumethat both repair time r and maintenanceduration m are on
average100 hours. For eachof the modelsto be discussedwe will calculateboth the
interruption rate due to overlappingoutages(AQo ) , and the interruption rate due to
outageduring maintenance(Aom)'
Average FailureRate-OverlappingOutages. Using the averagefailure rate for
the two components,we can calculatethe interruption rate of the parallel connection
due to overlappingoutages:
2
Aoo = X 2r = 5.84 x 10-4 interruptionsper year
(2.149)
The expectednumber of interruptions due to overlapping outagesduring a 4-year
period is equal to 2.34 x 10-3 •
AverageFailure Rate-OutageDuring Maintenance. When one componentis
being maintained, an outage of the other componentwill lead to an interruption.
One of the two componentsis in scheduledoutageduring a period 2m every 4 years.
An outage during this period leads to an interruption. The expected number of
outagesduring maintenanceduring a 4-year period is thus,
4Aom
= Zm):- = 3.65x
10-3 outagesper4 years
(2.150)
Maintenance Every FourYears-OverlappingOutages. When the failure rate
of the componentsis time dependent,it is still possibleto determinethe interruption
112
Chapter2 • Long Interruptionsand Reliability Evaluation
rate due to overlappingoutages.The only difference with the previous case isthat
the outageratesare timedependenta nd thereforethe interruption rate as well:
Aoo( l )
= A(I)22r = 2.28 x
10- 6 16 interruptionsper year
(2.151)
3
The averageinterruption frequency is 1.334 x 10- interruptions per year, and the
maximuminterruptionfrequencyGustbeforemaintenance)is 9.34 x 10- 3 interruptions
per year.The expectednumberof interruptionsdue to overlappingoutages,during a
4-yearperiod, is equal to 5.34 x 10-3 •
Maintenance Every FourYears-OutageDuring Maintenance. Normally maintenancewill not be performed on both componentsat the same timebecausethat
would lead to an interruption. Maintenanceis performed first on one component
and then on theother. During maintenanceon the secondcomponentthe first one is
as-good-as-new,has a failure rate close to zero,and the risk of an outagecan be
neglected.The situation is completely different for maintenanceon the first component, becausethe other componenthas its highest failure rate. The probability that
the secondcomponentwill fail while the first one is beingmaintainedis
4A om
= mA(4) = 8170~0 x 0.64 = 7.31 x 10-3 interruptionsper maintenanceinterval
(2.152)
Maintenance Every TwoYears-OverlappingOutages. Above it was assumed
that maintenanceon the two componentsis done immediately after each other. An
alternative is to spread the maintenanceover time; that is, by performing maintenanceevery 2 years and each time only on one component.Supposethat maintenance has been performed on component 1 at t = 0 and on component 2 at
t = -2. The componentfailure ratesbecome
= 0.0113
(2.153)
A2(1) = 0.01(1+ 2)3
(2.154)
A) (t)
The interruption rate due to overlappingoutagesis
Aoo(/) = A)(t)A2(t)2r = 2.28 x 10-6t3(t + 2)3 interruptionsper year
(2.155)
Note that this expressionis valid betweent = 0 and t = 2 after which component1 and
component2 switch roles. Theaverageinterruption rate over this 2-yearperiod is
-. = ~
1
2
Ap(t)dt= 2.18 x 10- interruptionsper year
4
(2.156)
The expectednumber of interruptionsdue to overlappingoutages,during a 4-year
period, is equal to 0.87 x 10-3.
Maintenance Every TwoYears-OutageDuring Maintenance. Failure during
maintenancecan happenfor eachof the two components.W hen maintenanceis performed on one component,the other componenthas an"age" of 2 years; thus, its
failure rate is 0.08 per year.The expectednumberof outagesof the parallel component during maintenanceon the other componentis equal to
mA(2)
= 8170600 x 0.08 = 0.913 x 10-3 interruptionsper maintenance
(2.157)
113
Section 2.8 • ExampleCalculations
Such asituation occurs twiceduring a 4-year period, sothat the expectednumberof
interruptionsdue to outageduring maintenance,over a 4-yearperiod, is 1.83 x 10-3.
Overview. The results of thevarious models aresummarizedin Table 2.13.
We seethat the aging/maintenancemodel influences theinterruption frequency over
almost afactor of 10. Also notethat the numberof interruptionsdue to outages during maintenanceis, for eachof the models, higherthan the numberof interruptions
due to overlappingoutages.Further optimization studies would be needed to assess
if the total interruption rate can bebrought down. An obvious choice is to reduce
the duration of maintenance,as the number of interruptionsdue to outagesduring
maintenanceis directly proportional to the duration of maintenance.One should
take a certain care with that, as thequality of the maintenancemight also become
less. In the abovecalculationsit has been assumed
t hat the outagerate is brought
back to zero aftermaintenance,and that the outagerate of the parallelc omponentis
not increasedduring the maintenance.
Without any optimizationstudy, it is obvious, however,
that maintenanceshould
be scheduled as much as possible
during periodswith low interruptioncosts.
TABLE2.13 Influenceof Aging and MaintenanceModel on Interruption
Rate
Interruptionsdue to
overlappingoutages
Interruptionsdue to failure
during maintenance
ConstantFailure Rate
MaintenanceEvery
4 Years
MaintenanceEvery
2 Years
2.34 x 10-3 per 4 years
5.34 x 10-3 per 4 years
0.85 x 10-3 per 4 years
3.65 x 10-3 per 4 years
7.31 x 10-3 per 4 years
1.83X 1-0-3 per 4 years
Short Interruptions
3.1 INTRODUCTION
A short interruptionhas the same causes as a long
interruption: fault clearing by the
protection,incorrectprotectionintervention,etc. When thesupplyis restoredautomatically, the resulting event is called short
a
interruption. Long interruptionsand very
long interruptionsresult when the supply is
restoredmanually.Automatic restoration
can take place by reclosing the circuit
breakerwhich cleared the fault or by switching to
a healthy supply. The former takes place mainlyoverhead
in
distributionnetworks,the
latter is a typical solution in industrial systems.
Shortinterruptionsin the public supply are due to
a ttemptsby the utility to limit
the duration of interruptions.We sawalreadyin Section 2.3that the duration of an
interruptionis an importantaspectof distribution and transmissionsystem design.By
using automaticreclosing theduration of an interruption can bebrought back from
typically about 1 hour, to typically lessthan 1 minute. For many yearsinterruptions
shorterthan severalminuteswerenot consideredas a causeof concernto most customers. Recently this has changed: more and more
equipmentis sensitive to veryshort
duration events, and more and more
customers(domestic as well asindustrial) view
short interruptionsas a seriousimperfectionof the supply. This ispart of the trends
mentionedin Section 1.1 for the increased
interestin power quality in general.Short
interruptionsalso occur inindustrial power systems due to the
operationof automatic
transferswitches. We discuss this in
Chapter7.
3.2 TERMINOLOGY
There is some serious
confusionaboutterminologyon interruptionsof different duration. Terms likeshort interruptions,momentaryinterruptions, temporary interruptions, instantaneousinterruptions, and transient outagesare all used with more or
less the same meaning. The
definition of short interruptionsused for thischapteris
not based onduration but on themethodof restoringthe supply. Thischapter(short
115
116
Chapter3 • Short Interruptions
interruptions)discussesautomatic restoration,where Chapter2 (long interruptions)
discussesmanualrestoration.
Below, an overview is givenof the various terms and definitions used in the
EuropeanstandardEN 50160and in three IEEE standards.The definitions used in
EN 50160 areidentical to the IEC definitions.
• EN 50160
- Long interruption: longer than three minutes.
- Short interruption: up to threeminutes.
• IEEE Std.1159-1995
This standardis consideredby many as providing the basic power quality
definitions. It distinguishesbetween momentary, sustained,and temporary
interruptions. Note.the overlap between sustainedand temporary interruptions.
- Momentaryinterruption: between0.5 cyclesand 3 seconds.
- Sustainedinterruption: longer than 3 seconds.
- Temporaryinterruption: between3 secondsand I minute.
• IEEE Std.1250-1995
This standardwas publishedat about the same time asIEEE Std.1159-1995,
but it usessomewhatdifferent definitions. The differenceis especiallystriking
for interruptions.
- Instantaneousinterruption: between0.5 and 30 cycles(half a second).
- Momentaryinterruption: between30 cyclesand 2 seconds.
- Temporaryinterruption: between2 secondsto 2 minutes.
- Sustainedinterruption: longer than 2 minutes.
• IEEE Std.859-1987
This somewhatolder standarddocumentgives definitions for terms relatedto
power system reliability. A distinction is made between different types of
outagesbased on theduration of the outage. This standarddoes not give
specific timerangesbut uses therestorationmethodto distinguishthe different
types. Although outages and interruptions are different phenomena(see
Section2.1.3) they arerelatedclosely enoughto comparethe terminology.
- Transientoutagesare restoredautomatically.
- Temporaryoutagesare restoredby manualswitching.
- Permanentoutagesare restoredthrough repair or replacement.
3.3 ORIGIN OF SHORT INTERRUPTIONS
3.3.1 Basic Principle
Figure 3.1 shows anexampleof an overheaddistribution network. Each feeder
consistsof a main feederand a numberof lateral conductors.Most faults onoverhead
lines aretransient:they requireoperationof the protection,but do not causepermanent
damageto the system. Atypical causeof a transientfault is a lightning stroke to an
117
Section 3.3 • Originof Short Interruptions
overheadline. The lightning stroke injects a very highcurrentinto the line causinga
very fast rising voltage. The lightning current varies between 2and 200 kA in peak
value. Thetypical lightning currenthas apeakvalue of [peak = 20 kA which isreached
within IlJ,s after its initiation. If the wave impedanceZ",ave of the line is 2000, the
voltage cantheoreticallyreach a valueof
Vpeak
Z"'ave
= -2-Ipeak
= 1000 x 20 kA = 2 MV
(3.1)
The voltagewill never reach such a value in reality (with the possible
exceptionof
transmissionsystems withoperatingvoltagesof 400 kV or higher), because flashover
a
to groundor betweentwo phaseswill resultlong before thevoltagereachessuch a high
value. The result is anarcing fault betweenone phaseand ground or between two or
morephaseswith or without ground.Soonafter the protectionremoves thefaultedline
from the system, the arc
disappears.A utomatic reclosingwill restorethe supplywithout
any permanentd amageto the system.
Also, smallerobjectscausinga temporarypath to groundwill only cause atransient shortcircuit. The object(e.g., a smallb ranchfallen from a tree) willeitherdrop to
the ground or evaporatedue to the highcurrentduring the fault, leaving only an arc
which disappearsagainsoonafter the protectionintervenes.The durationof an interruption due to a transient fault can thus be enormouslyreduced by automatically
restoringthe supply after an interruption. In caseof a fault somewhereon the feeder,
the circuit breakeropensinstantaneouslyand closesagainafter a "reclosinginterval"
or "dead time" ranging from lessthan one second up to several
minutes.There is of
coursea risk that the fault wasnot a transientone but permanent.In that case the
protectionwill againnotice a largeovercurrentafter reclosureleadingto a secondtrip
signal.Often the reclosergives the fault a second
chanceat extinguishing,by meansof a
longer tripping time and/ora longer reclosinginterval.
3.3.2 Fus.Saving
A practiceassociatedwith reclosing and short interruptionsis "fuse saving." In
Fig. 3.1 thelateralsaway from themain feeder areprotectedby meansof expulsion
fuses. These are slow fuses which will
not trigger when atransientfault is clearedby the
main breaker/recloser.Thus, a transientfault will be clearedby the recloserand the
supply will be automaticallyrestored.
A permanentfault can also beclearedby the main breaker,but thatwould lead to
a long interruptionfor all customersfed from this feeder.Instead,a permanentfault is
/Lateral
Recloser
J
Distribution
substation
Figure 3.1 Overheaddistribution network
with fusesand reclosers.
2
tt8
Chapter3 • Short Interruptions
cleared by anexpulsionfuse. To achieve this, the recloser has two settings:instanan
taneoustrip and a delayed trip. Theprotectioncoordinationshould be suchthat the
instantaneoustrip is faster than the expulsionfuse and the delayed trip slower, for all
possible faultcurrents.
From the abovedescriptionwe canconcludethat the following trade-offhas been
made: ashort interruption for all customers(fed from this feeder)insteadof a long
interruption for some customers.The alternativewould be more longinterruptions;
however, not everyshort interruptionwould become a longinterruption.
3.3.3 Voltage Magnitude Events due to Reclosing
The combination of reclosing and fuse saving, as
decribed above, leads to
different voltage magnitude events for different customers. Figure 3.2 shows the
events due to one reclosing
action as experienced by caustomeron the faulted feeder
(indicatedby "1" in Fig. 3.1) and by acustomeron anotherfeeder fed from the same
substationbus (indicated by "2"). In Fig. 3.2, A is thefault-clearingtime and B the
reclosing interval. The customeron the faulted feeder (solid line) willexperiencea
decreasein voltage during the fault, similar in causeand magnitudeto a voltagesag.
The difference between the two
customersis in the effectof the fault clearing.For the
customeron the nonfaulted feeder, the voltage recovers to its
pre-eventvalue. The
For the customeron the faulted feeder,
customerwill only experience a voltage sag.
the voltage drops to zero.
The customeron a neighboringfeeder(dashedline) will see avoltagesag with a
durationequal to the fault-clearingtime. The momentthe recloseropens,the voltage
recovers. If the fault is stillpresentat the first reclosure, the
c ustomeron thenonfaulted
feeder will experience a second voltage sag.
Customerson thefaulted feeder will experience a secondshort interruption or a long interruption.
Figure 3.3 [11] shows anactual recordingof a short interruption.The top figure
correspondsto the dashedline in Fig. 3.2 (customeron a nonfaulted feeder). The
bottom figure is for a customeron the faulted feeder (solid line in Fig. 3.2). The
fault-clearing time is about two cycles, the dead timea bout two seconds. The first
top figure shows avoltage sag to
reclosureis not successful, the second one is. The
about75% of two-cycleduration,the bottomfigure avoltagereductionto 50% for two
abouttwo seconds.
cycles followed by zero voltage for
1
Voltage sag
f----
r •••••••••••••••••••••••••••••••••••••••••••
1
~
Short
interruption
-------.
B
+----..~
A
Time
Figure 3.2 RMS voltageduring a recJosure
sequence on the faulted feeder (solid line) and
on thenonfaultedfeeder(dashedline). A =
fault-clearingtime; 8 = reclosing interval.
119
Section 3.3 • Originof Short Interruptions
April 29, 1994at 22:14:20PQNodelocaltrigger
1472
PhaseB voltage
RMSvariation
E
120~
i :ft!
I
60 0-
~[C
-0.05-0-'-.1--0......
15 0.2
Time(seconds)
L
'
-
0.25
I
0.3
Duration
0.050 s
Min 65.80
Ave 90.10
Max 100.5
150
lIOO
'~ 50
f
0
~ -50
~
-1000
25 50 75 100 125 150 175 200
Time(milliseconds)
(a) Uplinemonitoringlocation
April 29, 1994at 22:14:20PQNodelocaltrigger
2592
PhaseB voltage
RMSvariation
J
lJ
Jil_.. . ._..,.. . .~ :
~100
120[
o
I
234
5
Duration
4.983 s
Min 2.257
Ave 8.712
Max 100.2
6
Time(seconds)
J_;;
Figure 3.3 Recorded rms voltage during a
short interruption. (Reproduced from Dugan
et al. [II].)
00 0
I
25 50 75 100 125 150 175 200
Time(milliseconds)
(b) Downlinemonitoringlocation
When comparingFig. 3.2 and Fig. 3.3, note
t hat the horizontalaxis of Fig. 3.2 is
not to scale, B is much larger
than A. This is the typicalsituation. The fault-clearing
time (A) is only a few cycles, whereas the reclosing time (B) can be up to several
minutes.
Anotherexample of theinitiation of a shortinterruptionis shown in Fig. 3.4 [3].
We seethat the voltagemagnitudeinitially drops to about 25% of nominal and to
almost zero after three cycles. The spikes in the voltage are due to the arc becoming
instablearoundthe currentzero-crossing.Apparentlythe arc gets more stable
after two
cycles.
3.3.4 Voltage During the Interruption
The momentthe circuit breakerin Fig. 3.1 opens, the feeder and the load fed from
it are no longer supplied. The effect of this normally
is
that the voltagedrops to zero
very fast. There are, however,
situationsin which the voltagedrops to zero relatively
slow, or even remains at nonzerovalue.
a
Thelatter would strictly speakingnot be an
120
Chapter3 • Short Interruptions
150
100
,-...
50
~
0
e
l!
~ -50
-100
-150
0
25
50
75
100
125
Time inmilliseconds
150
175
200
Figure 3.4 Recorded voltage during the initiation of a short interruption.
(Reproduced from IEEE Std.
I I 59 [3].)
interruption,but the origin is similar tothat of an interruptionso that a shortdescription of the phenomenonis appropriatehere.
• Inductionmotor load is able tomaintainsome voltage in the system forshort
a
time. This contribution is typically rather small because themotors have
already been feeding into ashort circuit for a few cycles; thus,p art of the
rotor field of the inductionmotorswill begone already.M ost inductionmotors
will thus only give a small voltagecontributionand only for a few cycles.
• Synchronousmotorsmaintain their field even when the supply voltage disappears. They will be able tom aintainsome system voltage until their load has
come to astandstill, which can take several seconds.
If there is a significant
amountof synchronousmotor load present,its fault contributioncould make
fault extinguishingdifficult. Typically synchronousmotors will be tripped by
their undervoltageprotectionafter about 1 second, after which they no longer
contributeto the feeder voltage.
• Synchronousand inductiongeneratorsconnectedto the feeder (e.g., wind turbines orcombined-heat-and-power
installations)arecapableof maintainingthe
feeder voltage at n
aonzerovalue evenduringa longinterruption.This could be
a potential problem when largeamountsof generationare connectedto the
feeder. This so-calledembedded generation
is often not equippedwith any
voltage or frequencycontrol (relying on the grid tomaintainvoltageand current within limits) sothat an islandingsituationcan occurin which voltage and
frequency deviate significantly from their
nominal values. Especially overvoltage andoverfrequencycan lead to serious damage. preventsuch
To
asituation,
most embeddedgenerationis equippedwith a loss-of-gridprotectionthat disconnectsthe generatorwhen anunusualvoltage or frequency is detected.
All this assumesthat the short-circuitfault is no longerpresenton the feeder. As
long as the fault ispresent,all above-mentionedmachinesfeedinto the fault sothat the
feeder voltage remains low. The
fault-current contribution makes that the arc is less
likely to extinguish, but after extinguishing of the arc there will be a chance
of a
remaining voltage on the feeder.
For interruptionsdue toincorrectprotectioninterventionthere is noshort-circuit
fault presenton the feeder and themachinesconnectedto the feeder may cause a
121
Section 3.4 • Monitoring of Short Interruptions
temporaryor permanentnonzerovoltage. Also thecontribution of induction motors
will be larger.
3.4 MONITORING OF SHORT INTERRUPTIONS
As shortinterruptionsare due toautomaticswitching actions, their recording requires
automaticmonitoring equipment.Unlike long interruptions,a short interruptioncan
occur without anybodynoticing it. That is one of the reasons why utilities do not yet
collect and publishdataon shortinterruptionson a routine basis. One
of the problems
in collecting thisdata on a routine basis isthat some kind ofmonitoring equipment
surveys have been
performedto obtain
needs to be installed on all feeders.numberof
A
statisticalinformation aboutvoltagemagnitudevariationsand events. With those surveys,monitors were installed at anumberof nodes spreadthrough the system. The
surveys will be discussed in more detail Chapter6.
in
As with long interruptions,
interruption frequency andduration of interruption are normally presented as the
outcome of the survey. Again like with long
interruptionsmuch moredata analysis
is possible, e.g.,interruptionfrequency versus time of day or time of year,
distributions
for the time between events,
variation amongcustomers.
3.4.1 Example of Survey Results
Figures 3.5, 3.6, and 3.7 show some results of analysis ofdataobtainedby
the
a
large North American survey [68]. Figure 3.5 gives theinterruption frequency as a
function of theinterruptionduration. Each vertical bar gives the average
numberof
interruptionsper year, with aduration in the given interval. The average
numberof
interruptionshas beenobtainedas follows:
4
j
3.5
>.
3
~
2.5
!
5
~
8
',=
tt=
~
B
~
2
1.5
I
0.5
-
o O-Ie
- -
2-3c
-- .
4-5c
6-IOc 20e-0.5s 1-2s
Durationof interruption
•
5-108
II
30-60s
120s-
Figure 3.5 Interruption frequency (number of interruptions per year) as a function
of interruption duration. (After data obtained from Dorr
[68].)
122
Chapter3 • Short Interruptions
(3.2)
where Nfl') is the numberof events in ranger observedby monitor i during a monitoring
the resulting averageas plotted in Fig. 3.5. We see from Fig. 3.5
interval T;, and
that the typical eventhas aduration between 1and 30 seconds.Eventsshorterthan six
cycles (100 ms) are very unlikely. These
"very short interruptions"are most likely due
to short-circuit faults close to themonitor position. One should realize that in this
survey anevent is recordedas an interruption if the rms voltage somewhereduring
the eventdrops below 100~ of nominal. Note also that the horizontal scale is nonhomogeneous.F rom the data shown in Fig. 3.5 one cancalculate the probability
of all
densityfunction of the interruption duration by dividing each value by the sum
values:
FIr)
N(r)
f(r)
= I:Fl k )
(3.3)
(k)
The probability distribution function of the interruptiondurationcan beobtained
by addingthe valuesof the density function up to acertainduration.
F(t) =
I:!(r)
(3.4)
(")<1
The resultingprobability distribution function is presentedin Fig. 3.6. This curve gives
the fraction of interruptionswith a durationnot exceeding theindicatedvalue. We see
that 10% of interruptionslasts lessthan 20 cycles,and 80% of interruptionslessthan 2
minutes(thus 20% morethan2 minutes).From an equipmentpoint of view the reverse
dataare of more interest,the fraction of interruptions(or the absolutenumber)lasting
longer than a given duration. This will give information about the numberof times a
device will trip or (for a givenmaximum trip frequency)about the immunity requirementsof the device.Figure3.7 plots the numberof interruptionsper yearlastinglonger
than the indicatedvalue. Apart from a small shift (due to the
discretizationof the data)
1.2,..------------------..-,
s=
o
.~
.&J 0.8
'Een
:.a
~0.6
:.0
.se 0.4
c..
0.2
O................
-==~:::...J----'-___L.---L---Jl.._._.L...._.J..._...L._....L_..J......_.J
Ic
3e
5e
JOe
0.5s
2s
Duration
lOs
60s
info
Figure 3.6 Probability distribution function
of interruption duration. (From the data in
Fig. 3.5.)
123
Section 3.4 • Monitoring of Short Interruptions
18,..--------------------,
16
~
r------ __
14
g.~ 12
~ 10
5
.~
j
8
6
4
2
Figure 3.7 Number of interruptions lasting
longer than the indicated value. (From the
data in Fig. 3.5.)
OL--..a...-....&..-_'___....I----£--L..---L.-L----.I~J..__..&.___'___~_.I_.-L..___I
Oc
2c
4c
6c
20c
Is
5s
30s
120s
Durationof interruption
and amultiplication factor equal to thetotal numberof interruptions,the curve is the
complementof the curve in Fig. 3.6. We can conclude from the figure
that equipment
which trips for aninterruption of 20 cycles will trip on average 14 times per year. To
limit the equipmenttrip frequency to four per year, the
equipmentshould be able to
tolerateinterruptionsup to 30 seconds induration.
3.4.2 Difference between Medium- and Low-Yoltage Systems
The numberof short interruptionshas beenobtainedby various power quality
surveys.Comparisonof the numbersobtainedby each survey gives
information about
the average voltage
quality in the variousareas. Acomparisonbetween thenumberof
short interruptionscountedat various places in the system can teach us how the interruptions "propagate"in the system. Such caomparisonis madein Table 3.1 for two
large North American surveys: theEPRI survey and theNPL survey [54]. TheEPRI
surveymonitoredboth distribution substationsand distribution feeders.
From Table 3.1 we seethat the overall trend is for thenumberof shortinterruptions to increase when moving from the source to the load. This
understandable
is
as
there are more possibletripping points the further one movestowards the load.
Especially interruptions lasting several seconds and longer mainly
originate in the
low-voltage system.F or interruptionslessthan one second induration,the frequency
remainsaboutthe same, which makes us
concludethat they probablyoriginatein the
distributionsubstationor even higher up in the system. The large
numberof very short
TABLE 3.1 Interruption Frequency (number of events per year) for Three
Points in the U.S. Distribution System
Duration
Survey
1-6c
6-IOc
lo-20c
EPRI substation
EPRI feeder
NPL low voltage
0.2
1.6
0.2
0.1
0.1
0.3
0.4
0.2
0.8
0.7
0.8
Source: After data obtained from[54].
20-30c
0.6
0.5-1 sec
0.5
0.5
1.2
1-2 sec
2-10 sec> 10 sec
0.9
1.1
1.5
1.1
2.3
3.3
1.3
1.7
4.2
124
Chapter3 • Short Interruptions
TABLE 3.2 Interruption Frequency(per year) forPrimary and Secondary
Systems inCanada
Duration
Survey
CEA primary side
CEA secondaryside
1-6c
6--IOc
10-20c
2Q-30c
0.5-1 sec
1.9
3.7
0.0
0.0
0.1
0.0
0.0
0.0
0.4
0.2
1-2 sec 2-10 sec
0.0
0.5
0.0
0.5
> 10 sec
0.7
2.1
Source: After data obtainedfrom [69].
TABLE 3.3 Interruption Frequency(per year) forDistribution and
Low-voltageSystems inNorway
Duration
Survey
0.01-0.1 sec
0.1-0.5sec
0.5-1.0sec
EFI distribution
EFI low-voltage
1.5
1.1
0.0
0.7
0.0
0.0
1-3 sec
0.0
0.7
3-20 sec
> 20 sec
0.5
0.9
5.2
5.9
Source: After data obtainedfrom [67].
interruptions(lessthan six cycles) ondistribution feeders ishard to explain, especially
as they donot show up in the low-voltagedata.
Similar conclusionscan bedrawn from the CEA survey [69] and from the
E FI
survey [67], some results of which are shown in Tables 3.2 and 3.3.
againsee
We
a larger
number of interruptions,mainly of 1 second and longer, forlow-voltage than for
medium-voltagesystems. Both theCanadian(CEA) and theNorwegian (EFI) data
show a considerablenumber of very short interruptions,for which no explanation
has been found yet.
3.4.3 Multiple Events
A direct consequenceo f reclosingactionsis that a customermay experience two
or more events within as hortinterval. When theshort-circuitfault is still presentupon
the first reclosure, thecustomersfed from the faulted feeder will experience a second
event. This isanother short interruption if a second attempt at reclosing is made.
Otherwisethe second event will be a long
interruption. A customerfed from a nonfaulted feeder experiences two
voltagesags in ashort period of time.
For a few years a discussion has been goingaboutwhetherto
on
countthis as one
event or as multiple events
[20]. The most recentpublications of North American
surveysconsidera l-minute or 5-minute window. If two or more events take place
within such a window, they are
c ountedas one event. The severity
of the multiple event
(i.e., magnitudeand duration)is the severity of the most severe single event within the
window. Some examples of the
working of a "five-minute filter" are shown in Fig. 3.8.
Using such a"filter" is suitablefor assessment of the
numberof equipmenttrips,
as theequipmentwill trip on the most severe event or not at all. The
cumulativeeffect
of the events is neglected, but the general
impressionis that this effect is small.T his has
however not been confirmed hy
measurements
yet. In some cases it could still be needed
to know thetotal event frequency, thus
countingall events even if they come very close.
Two possibleapplicationsare: (I) componentswhich show acceleratedaging due to
shortundervoltageevents; and (2)equipmentwhich only tripsduring a certainfraction
125
Section 3.5 • Influence onEquipment
Time
i
Q
~
Time
Time
Go)
C)O
~
Q
Figure3.8 Effect of a"five-minute filter" on
the voltage magnitude events. The
figures on
the left show the recorded rms
voltages;the
figures on the right show the equivalent event
after thefilter.
~
Time
Time
TABLE3.4 Number of Singleand Multiple Interruptions per Year, NPL
Low-Voltage Survey
Duration
Survey
1-6c
6-IOc
1(}-20c
2(}-30c
0.5-1 sec
No filter
0.3
5-min filter
0.2
Percent reduction 33°A.
0.3
0.3
0.8
0.7
12%
0.9
0.8
11%
1.4
1.2
14%
1-2 sec 2-10 sec
1.9
1.5
21%
4.2
3.3
21%
> 10 sec
5.7
4.2
26%
Source: After data obtained from[54].
of its load cycle. In thelattercase theequipmenthas aprobability to trip duringeachof
the three events, and the
total probability is of course largerthanthe probability to trip
during the most severe event only.
The NPL low-voltage datafor short interruptionshave beenpresentedwith and
without the above-mentionedfilter in Table 3.4[54]. The three rows give, from top to
bottom: the numberof shortinterruptionswhen each event is
countedas one event no
matter how close it is toanotherevent; thenumberof events when multiple events
within a 5-minute interval arecountedas one event; the
reductionin numberof events
due to theapplicationof this filter.
3.5 INFLUENCE ON EQUIPMENT
During a shortinterruptionthe voltage is zero; thus, there is no supply of power at all
to the equipment.The temporaryconsequences are
that there is no light,that motors
126
Chapter 3 • Short Interruptions
slow down, that screensturn blank, etc. All this only lasts for a few seconds, but the
of contents
consequences
can last much longer:disruptionof productionprocesses, loss
of computermemory,evacuationof buildings due to fire alarms going off, and sometimes damagewhen the voltage comes back
(uncontrolledstarting).
For most sensitiveequipment,there is no strictborderbetween a voltage sag and
an interruption:an interruptioncan be seen as a severe sag, i.e. one with remaining
zero
voltage. The effecto f voltage sags onequipmentis discussed in detail inChapter5.
Many of the conclusionsin thatchapteralso hold forshortinterruptions.In this section
of the load behaviorare pointedout.
only some general aspects
3.5.1 Induction Motors
The effectof a zero voltage on aninduction motor is simple: themotor slows
down. Themechanicaltime constantof an inductionmotor plus its load is in the range
of 1 to 10 seconds. With dead times of several seconds,motor
the has not yet come to a
standstillbut is likely to have slowed down significantly. This
reductionin speedof the
motorsmight disrupt the industrial process so muchthat the processcontrol trips it.
The motor can re-acceleratewhen the voltage comes back, if the system
strong
is
enough.For public distribution systemsre-accelerationis seldom aproblem.
Also the settingof the undervoltageprotectionshouldbe suchthat it does not trip
before the voltage comes back. This calls forcoordinationbetween
a
theundervoltage
setting of themotor protectionand the reclosureinterval setting on the utility feeder.
Induction motors fed via contactorsare disconnectedautomaticallyas the conof the load.
tactordropsout. Without countermeasures
this would always lead to loss
In someindustrial processes the
induction motorsare automaticallyreconnectedwhen
the voltage comes back:
either instantaneouslyor staged (the mostimportant motors
first, the rest later).
3.5.2 Synchronous Motors
Synchronousmotors can normally not restarton full load. They aretherefore
equippedwith undervoltageprotectionto preventstallingwhen the voltage comes back.
For synchronousmotors the delay timeof the undervoltageprotectionshould be less
than the reclosing interval. Especially for very fast reclosure this can problem.We
be a
see here asituationwhere aninterruptioncauses a more serious
threatto the synchronousmotorsthe faster the voltage comes back. With most
otherload thesituationis the
other way around: the shorterthe interruption,the less severe it is to the load.
3.5.3 Adjustable-Speed Drives
Adjustable-speeddrives are very sensitive to
s hort interruptions,and to voltage
sags as we will see in
C hapter5. They normally trip well within I second, sometimes
even within one cycle; thus even the
shortestinterruptionwill cause a lossof the load.
Some of the moremoderndrives are able toautomaticallyreconnectthe momentthe
voltage comes back. But being
disconnectedfrom the supply for several seconds will
often havedisruptedthe processbehindthe drive so muchthat reconnectiondoes not
make much sense
anymore.
127
Section 3.6 • Single-Phase
Tripping
3.5.4 Electronic Equipment
Without countermeasures
electronics devices will trip wellwithin the reclosing
interval. This leads to theinfamous"blinking-clock syndrome":clocks of video recorders, microwave ovens, and
electronicalarmsstart blinking when the supply is interrupted; and they keep on
blinking until manuallyreset. An easysolution is to install a
small rechargeablebattery inside of the equipment,to power the internal memory
during the interruption.
problem. But
Computersand processcontrol equipmenthave basically the same
they require more than a simplebattery. An uninterruptiblepower supply (UPS) is a
much-usedsolution.
3.8 SINGLE-PHASE TRIPPING
Single-phasetripping is used intransmissionsystems tomaintainsynchronicitybetween
both sidesof a line. Single-phasetripping is rarely used indistribution or low-voltage
systems.Not only will it requiremore expensiveequipment,but it will also reduce the
chanceof a successful reclosure. The fault
currentcontinuesto flow via the nonfaulted
phases. This reduces the
chancethat the fault will extinguishand thus increases the
numberof reclosureattemptsand thenumberof long interruptions.But if the reclosure
is successful,single-phasetripping has clearadvantagesover three-phasetripping and
thereforejustifies being discussed here. We will have a look at the voltages experienced
by the customerduring single-phasetripping. A distinction is made between two distinctly different situations,both assuminga single-phase-to-ground
fault followed by
tripping of the faulted phase.
ground (the fault) is
• The low-impedancepath between the faulted phase and
still presentso that the voltage in the faulted phase remains zero or close to
zero. We will call this the"during-fault period."
• The fault hasextinguished,the short circuit has now become an
o pencircuit
because thebreakerin that phase is still open. This we will call the
" post-fault
period."
3.8.1 Voltage-During-Pault Period
The phase-to-neutralvoltages in theduring-fault period are, with a the faulted
phase:
Va =0
Vb
= (-~-~jJ3)E
(3.5)
V(' = (-~+~jJ3)E
with E the magnitudeof the pre-eventvoltage. It has been assumed here
that the preevent voltages form a
balancedthree-phaseset,andthat the voltage in thefaulted phase
is exactly equal to zero. We will in most of the remainderof this book use per unit
voltages, with thepre-eventvoltagemagnitudeas base. Inthat case we getE = 1 and
(3.5) becomes
128
Chapter3 • Short Interruptions
VlI=O
Vb =V =
c
~ - ~jvS
(3.6)
_!+!J·vS
2 2
Figure 3.9 shows thephase-to-neutralvoltages as aphasordiagram.In this and subsequentphasordiagramsthe during-eventvoltage isindicatedvia solid lines, the preevent voltage (i.e., thebalancedthree-phasevoltage) viadottedlines, if different from
the during-eventvoltage. If single-phasetripping would take place in alow-voltage
network, the voltages in Fig. 3.9 would be the voltages experienced bycustomers.
the
Only one outof three customerswould experience aninterruption. The otherswould
not noticeanything. Single-phasetripping would thus reduce then umberof interruption eventsby a factor of three.
Va
........................•
Figure 3.9 Phase-to-neutralvoltages for
single-phase tripping.
For tripping taking place onmedium-voltagefeeders, thephase-to-phase
voltages
are of more importance.Large equipmentfed at medium-voltagelevel is in most cases
connectedin delta; small single-phase
equipmenttends to beconnectedbetween a phase
and neutral but at a lower voltage level fed via delta-starconnectedtransformer.In
a
both cases theequipmentexperiences the pu value
of the phase-to-phase
voltage at the
medium-voltagelevel.
The phase-to-phasevoltages in pu areobtained from the phase-to-neutral
voltages as follows:
(3.7)
The factor .J3 is needed because 1 pu
of the line(phase-to-phase)
voltage is.J3 times as
big as I pu of the phase(phase-to-neutral)
voltage. Themultiplication withj results in a
rotationover 90° suchthat the axisof symmetryof the disturbanceremainsalongphase
a and along the real axis. The
transformationin (3.7) will be the basisof a detailed
analysisof unbalancedvoltage sags in theforthcomingchapters.When we leave away
the prime " weobtain the following expressions for the voltages due to single-phase
tripping at the terminalsof delta-connectedequipment:
129
Section 3.6 • Single-PhaseTripping
~
\ .•..
~~:
A
Vb /
Figure 3.10Phase-to-phase
voltages for
single-phase tripping.
/ .../.
,l
Va = 1
Vb =
_!_!jJ3
Vc =
-~+~jJ3
2
6
(3.8)
Figure 3.10 again shows the voltages at the
equipment terminals in phasordiagramform. Using the definitions given in the
variousstandardsthis shouldnot be
called ashort interruption but a voltage sag. It would again bring up the discussion
betweenconsequence-based
terminologyand cause-based
terminology.In the first case
this event would have to be called a voltage sag, in latter
the case it would be ashort
interruption. But no matterwhich name is given to the event, it is clearly less severe
than the effect ofthree-phasetripping, when all three phase voltages go down to zero.
An exception to this might have to be made for
inductionmotors.The voltagesduring
single-phasetripping contain a large negative sequence voltage
component(0.33 pu)
which may lead tooverheatingof induction motors. With a negative sequence impedance 5through 10 times as small as the positive sequence
impedance,the negative
sequencecurrent would become 170through 330% of the rated (positive sequence)
current.It is unlikely that inductionmotor load is able towithstandsuch anunbalance
for longer than several seconds.
Low-voltage customersalso experience the voltages in Fig. 3.10.
None of the
customersexperiences a zero voltage, but
two-thirds of the customersexperience an
event with aduring-eventvoltage of 580/0 magnitudewith a change in voltage phaseangle of 30°.
3.8.2 Voltage-Poet-Pault
Period
When the fault extinguishes, the
situation in the faulted phase changes from a
short circuit to an open circuit. In many cases a change in voltage occurs, thus the
resulting voltage is no longer equal to zero. The voltage in the faulted
phasedependson
considerthe coupling
the typeof load connected. Tocalculatethis voltage we need to
between the phases or use the
theory of symmetricalcomponents.The latter, which is
normally used for the analysis of
nonsymmetricalfaults, isdescribedin detail in many
reference books. A good and detailed
descriptionof the useof symmetricalcomponents
for the analysis ofnonsymmetricalfaults is, e.g., given in reference [24], and is not
repeatedhere.
To analyze an open circuit, the system has to be modeled as seen from the opencircuit point. This results in three
equivalentcircuits: for the positive sequence, for the
130
Chapter3 • Short Interruptions
~V:J
s,
c~V2:J
[91V0:J
Figure 3.11 Sequencenetworksfor the
analysisof single-phaseopen-circuitfaults:
positive sequence(top), negativesequence
(center),and zerosequence(bottom).
negativesequence,a nd for the zero sequence.T hesethree networksare shown in Fig.
3.11: ZSb ZS2' and Zso are positive, negative, and zero-sequenceimpedanceof the
source; ZL), 2 L2 , and ZLO are positive, negative,and zero-sequenceimpedanceof the
load; 6 V1 , 6 V2 , and 6. Vo are positive, negative,and zero-sequence
v oltagedrop' at the
s ourcevoltage. Negativeand zeroopen-circuitpoint; and E 1 is the positive-sequence
sequencesourcevoltagesare assumedzero, and the load is assumednot to containany
sources.Below we again assumeE) = 1.
Sequencevoltagesand currentsat the open-circuit point can be calculatedfor
different types of open-circuit faults, by connectingthe three sequencenetworks in
different ways. For a single-phaseopen circuit, the voltagedifferencein the two nonfaulted phasesis zero and the current in the faulted phaseis zero:
6. Vb
=0
(3.9)
6. Vi' = 0
III =0
where a is the faulted (open-circuited)phase.Transformingtheseequationsto symmetrical componentsgives thefollowing set of equations:
II
+ 12 + /0 = 0
= 6.V2
6. VI = 6. Vo
(3.10)
6. VI
Theseexpressionscorrespondto a connectionof the sequencenetworks,as shown in
v oltagedrop at the open-circuitpoint
Fig. 3.12. From Fig. 3.12 thepositive-sequence
can bewritten as
1
6. VI
= 6.V 2 = 6.Vo = 1 + 2 Ll +ZS) + Z LI
ZLO
and the voltagedrop in the faulted phaseis
+ Zso
ZL2
+2
SI
+ ZS2
(3.11)
131
Section 3.6 • Single-PhaseTripping
Figure 3.12 Connectionof the sequence
networksin Fig. 3.11 for asingle-phaseopen
circuit.
~ Va
3
= ~ VI + ~ V2 + ~ Vo = 1 + Z Ll + ZSI + ZLl + ZSJ
ZLO
+ ZSO
ZL2
(3.12)
+ ZS2
Normally the load impedance dominates over the source impedance (ZLi»
ZSi' i = 0, 1, 2) sothat we can write with goodapproximation:
~Va =
Z
3
(3.13)
Z
1+~+~
ZLO
ZL2
The voltage at the load side
of the open phase is
V -1-
a-I
3
2 Ll
+-+ZLO
ZL2
which can bewritten as an expressionusing admittancesby introducing
YL2 = -Zl,
and YLO = -zl,
resultingin
L2
LO
Va = I -
1
(3.14)
ZLI
hI
3(YL 1 + YL2 + YLO)
YLI
= -Zl,
LI
(3.15)
From (3.15) the voltage experienced
by the load during the interruptioncan be found
for different types of load. As can be seen it is the
ratio between the sequence impeimpedancedoes have a
dancesof the load whichdeterminesthe voltage. The source
small influence as the load
c urrent will give a voltage drop between the load and the
open-circuitpoint. This influence was neglected when going from
(3.12) to (3.13).
3.6.2.1 Star-connected Static Load.For star-connectedstatic load, the three
sequenceimpedancesare equal: YLI = YL2 = YLO, (3.15) gives
(3.16)
In other words, this typeof load does not affect the voltage in the
openphase. Singlephase,low-voltageload cannormally be representedin this way.
3.6.2.2 Delta-connectedStatic Load. Delta-connectedstatic load is found in
medium-voltagepublic distribution networks. The delta-starconnectedtransformer
feeding thelow-voltagecustomerscan beconsidereda delta-connectedstatic load, as
long as mainly single-phaseload is present. For this kind of load, positive and
negative sequence
impedancesare equal and the zero-sequenceimpedanceis infinite
132
Chapter3 • Short Interruptions
va . --
.....••••••••••·•••••
Figure 3.13Phase-to-groundvoltagesduring
single-phase reclosure with
delta-connected
load .
..
..
:
Figure 3.14Phase-to-phase
voltages during
single-phase reclosure with
delta-connected
load.
because of the lack of any
r eturn path; in admittanceterms, YLI
resulting in
Va
1
= --2
= YL2
and
YLO
= 0,
(3.17)
In high-impedancegrounded or isolated-neutralsystems, the zero-sequence source
impedanceis very large or even infinite.F rom the aboveequationsit is easy to prove
thatthe resultingvoltagein the open phase is again equal to The phase voltages and
the line voltages fordelta-connectedstatic load are shown in Fig. 3.13 and Fig. 3.14,
respectively.
-!.
3.6.2.3 Motor Load. For motor load, a typical load inindustrial systems and
in some public systems, the
zero-sequenceimpedanceis again infinite, and the negative sequenceimpedanceis smaller than the positive-sequence
impedance: YL2 > YLI
and YLO = o. The resulting expression for the open-phase voltage is, with
YL2 = YYLI
y-2
V =-a
y+ I
(3.18)
-!,
For y = 1, whichcorrespondsto staticdelta-connectedload, we againobtain Va =
for y = 2 we obtain Va = O. A typical rangeof the-ratiobetween positive and negative
sequenceimpedanceis: y = 3··· 10 resulting in Va = 0.25···0.73. When theinduction
motors slow down, the negative sequence
impedancestays about the same while the
positive sequence
impedancebecomes smaller, until they are equal when motor
the has
come to astandstill.From equation(3.18) we canconcludethat the open-phasevoltage
decays wheny gets smaller, thus when the
motorsslow down. Theopen-phasevoltage
0AJ and 700/0 of the pre-faultvoltage,
for a system withmotor load is initially between 50
133
Section 3.6 • Single-Phase
Tripping
decaying to -50% of pre-fault voltage (i.e., 500/0 of magnitude,but with opposite
phase).
From the above examples, we can
concludethat the voltage in the open phase
V
varies between-0.50 and + 0.75times thepre-faultvoltage. When we use the symbol
to indicatethis voltage, we get the followingphasorexpression for the voltages in the
three phases:
Va = V
Vb
= _!_!jY'3
Vc
= _!+!jY'3
2 2
2
(3.19)
2
Using the transformationas defined by (3.7), we get for the line voltages (i.e., the
voltages experienced by delta-connected
a
load)
(3.20)
We seethat a delta-connectedload experiences a voltage
drop in two phases, but this
experiencedby a starvoltagedrop is smallerthanthe voltagedrop in the open phase as
connectedload. Also the load is less influenced by
single-phasetripping than by threephasetripping.
3.6.2.4 Transfer to Lower Voltage Levels.
Transfer to lower voltage levels
often takes placethrough delta-starconnectedtransformers.The first transformer
simply changes lineinto phase voltages,resulting in expression(3.20) but for the
phase voltagesinsteadof for the line voltages.
To obtain the line voltagesafter a delta-starconnectedtransformer,or the phase
voltages after two such transformers,the transformation(3.7) has to beapplied a
second time, to (3.20),
resultingin
I
2
Va =-+-V
3 3
2 ) --jY'3
I
Vb = - -1(1-+-V
2 3
3
(3.21)
2
1 (1-+-V
2 ) +-jY'3
1.
V.=-c
2 3 3
2
The resulting voltages fordifferent types of load are summarizedin Table 3.5. The
transferof this kind of voltage events to lower voltage levels is discussed
muchmore
in
detail in Section 4.4.Therewe will denotethe voltage events in
(3.19), (3.20),and (3.21)
as sags of type B withmagnitudeV, of type C with magnitude + ~ V, and of type D
with magnitude!+ ~ V, respectively.
t
Chapter3 • Short Interruptions
134
TABLE 3.5
Load
Voltages Due to Single-Phase
Tripping, for Various Types of
Star-connected
Load
Induction Motor Load
Delta-connected
Load
Initial
Motor Slowed Down
Voltage in the Open Phase
Va=-0.5
Va =0.75
Voltages After theFirst Dy-transformer
Va=O
Phasors
Magnitudes
Va = J
Va = J
Vh = -!-!j~
v, =-!
Vc = -!+~jJ3
V(.=
100%, 57.7%,57.7%
-!
100%, 50%,50%
Va = 0.25
Va = J
Va = I
-! - f2jJ)
Vr = - ! + fijv'3
Vh =
-1- !.iv'3
Vc = -! + iJv'3
Vh =
100%,87.80/0,87.80/0100%, 66.1%, 66.1%
Voltages After the Second
Dy-transformer
=!
Va=0
= -!-!jJ)
Vh = -!jJ)
Vh = -fi -
Vr =-
Vr =
Va
Phasors
Vh
VC = -~+!j~
Magnitudes
33.3%,88.20/0,88.2%
!jJ3
Va =~
Va =!
!jJ3
-fi + !Jv'3
Vb = -!-!JJ)
Vr =
-! + !jv'3
%
50%, 90.1%, 90.1%
0, 86.6%,86.60/0 83.3%, 96.1%, 96.1
3.8.3 Current-During-Fault Period
As we have seen in the previous section, the voltage in the faulted phase
duringthe
post-fault period is not necessarily zero. Anonzerovoltage after fault extinguishing
implies a nonzerocurrent while the fault is present. This makes fault
extinguishing
more difficult.
To calculate the fault current after single-phasetripping but before the fault
extinguishes, weconsider the circuit in Fig. 3.15. Source and load
impedancesare
indicatedby the same symbols as before. Voltages and
currentsat the system side of
the openpoint are indicatedas Va' Vb, etc., and at the load side as
V~, V;" etc.
The electricalbehavior of this system can be described
through 12 equations,
three equationsdescribing the source (with again
£] = 1):
l-ZSlI] = V]
-Zs212 = V2
(3.22)
-ZsoIo = Vo
three equationsdescribing the load:
r; = ZLll{
V~
= ZL2I~
Vo = ZLolo
(3.23)
135
Section 3.6 • Single-Phase
Tripping
ZS2
Zso
Figure 3.15 Single-phase
tripping with the
short circuit still present.
threevoltageequationsat the open point:
V~ =0
v; = Vb
V; = Ve
(3.24)
and threecurrentequationsat the open point:
=0
fb =Ib
fa
(3.25)
t, = l~
If we neglect thesourceimpedances,the voltagesat the systemside of the open point
are equalto the sourcevoltages:
VI
=1
(3.26)
V2 =0
Vo =0
From (3.24) relationscanbe obtainedbetweenthe componentvoltageson both sidesof
the openpoint:
I
VI =
I
V2 =
2
3" VI
I
1
-"3 V2 - "3 Vo
1
2
1
1
2
-"3 VI +"3 V2 - "3 Vo
I
1
Vo = -"3 VI
(3.27)
-"3 V2 + "3 Vo
With (3.26), thecomponentvoltagesat the load side of the open point can be found.
Togetherwith (3.23) and I~ = I~ + 11 + 12we obtain an expressionfor the fault current
after single-phasetripping:
,
2
1
1
I a =- - - - - - - 3ZL1 3ZL2 3ZLO
(3.28)
We seethat the currentdependson the load impedancesin positive, negative,and zero
sequence.As these impedancesare significantly larger than the source impedances
(typically a factor of 10 to 20) thecurrent becomesmuch smaller than the original
fault current.This certainlyhelps theextinguishingof the fault, but still the fault is most
likely to extinguishwhen thecurrentis close tozero,thuswhen: 2YLt ~ YL2 + YLO with
YL l = -Zl,
etc. Not surprisingly this is also thecondition for which the voltage after
LO
fault extinguishingis zero, accordingto (3.15).
136
Chapter3 • Short Interruptions
3.7 STOCHASTIC PREDICTION OF SHORT INTERRUPTIONS
To stochasticallypredict the number of short interruptions experiencedby a
customerfed from acertain feeder, the followinginput data is required:
• Failure rate per km of feeder,different valuesmight be used for the mainand
for the lateral conductors.
• Length of the main feederand of the lateral conductors.
• Successrateof reclosure,if multiple reclosureattemptsare used: success
rateof
the first reclosure,of the secondreclosure,etc.
• Position of reclosingbreakersand fuses.
We will explainthe varioussteps in astochasticpredictionby using the system
shownin
Fig. 3.16.Note that this is ahypotheticalsystem.Stochasticpredictionstudiesin larger,
albeit still hypothetical,systems have been
performedby Warren[139]. The following
datais assumedfor the system in Fig. 3.16:
• The failure rate of the main feeder is:0.1 faults per year per kmof feeder.
• The failure rate of the lateral conductorsis: 0.25 faults per year per kmof
feeder.
• The success rate
o f the first reclosureis 75%; thus, in25% of the cases asecond
trip and reclosureare needed.
• The success rate
o f the secondattemptis 100/0 of the numberof faults. Thus,
for 15% of the faults thesecondattemptdoesnot clear the fault.Thosefaults
are "permanentfaults" leadingto a long interruption.
The reclosingprocedureused is as follows:
I. The circuit breakeropensinstantaneouslyon theovercurrentdue to the fault.
2. The circuit breakerremainsopen for a short time (1 sec);75% of the faults
clearsin this period.
3. The circuit breakercloses. If the fault is stillpresentthe breakeragainopens
instantaneouslyon overcurrent.This is requiredin 25% of the cases.
4. Thecircuit breakernow leaves alongerdeadtime (5 sec).A nother 10% of the
faults clear in this period.
Lateral0: 3 km
Lateral C: 7 km
l----
]] km of main feeder
Recloser
I
Lateral B: 4 km
• --Fuses
LateralA: 8 km
Figure 3.16 Example of overhead
distribution feeder, for stochastic prediction
study.
137
Section 3.7 • Stochastic Prediction of Short
Interruptions
5. The circuit breakercloses for asecondtime. If the fault is still presentthe
breakerremainscloseduntil the fuseprotectingthe lateralconductorhashad
time to blow.
6. If the fault is still present(i.e., if the current magnitudestill exceeds its
threshold)after the time needed for the fuse to
clear the fault, thebreaker
opensfor a third time and now remainsopen. Furtherreclosurehas to take
place manuallyand the whole feeder willexperiencea long interruption.
The total numberof faults on the feeder is
11 km x 0.1faults/kmyear + 22 km x 0.25faults/kmyear = 6.6faults/year
(3.29)
Each fault will lead to a voltage magnitudeevent. There are four different events
possible:
• a short interruptionof 1 secondduration.
• two short interruptions; one of 1 second duration and one of 5 seconds
duration.
• two short interruptionsfollowed by avoltagesag.
• two short interruptionsfollowed by a longinterruption.
Due to short-circuitfaults on this feeder, 6.6 events per
year occur, of which
• 750/0 = 5.0 per year needone trip, leading to one short interruption for all
customers.
leadingto two short interruptionsfor all
• 100/0 = 0.7 per year need two trips,
customers.
• 15% = 1.0 per year arepermanent,leadingto two shortinterruptionsfollowed
by a voltagesag or followed by a longinterruption.
The numberof shortinterruptionsis equalfor everycustomerconnectedto this feeder:
5.0/yearof 1 secondduration.
0.7/yearof 1+ 5 secondsduration.
The numberof long interruptionsdependson the position at the feeder. Apermanent
fault on the main feeder leads to a longinterruption for all customers.A permanent
fault on oneof the lateralsleads to a longinterruptiononly for customersfed from this
lateral. The numberof permanentfaults is, for the different partsof the feeder:
•
•
•
•
•
lateral A: 8 km x 0.25faults/kmyear x 0.15= 0.3faultsperyear
lateral B: 4 km x 0.25faults/kmyear x 0.15 0.15faultsper year
lateral C: 7 km x 0.25faults/kmyear x 0.15= 0.26faultsper year
lateral D: 3 km x 0.25faults/kmyear x 0.15= 0.11faultsper year
main: 11 km x 0.1faults/kmyear x 0.15= 0.17faultsper year
=
The number of long interruptionsexperiencedby customersconnectedto different
partsof the feeder, is
138
Chapter3 • Short Interruptions
•
•
•
•
•
main: 0.17/year
lateral A: 0.17 + 0.3 = 0.47/year
lateral B: 0.17 + 0.15 = 0.32/year
lateral C: 0.17 + 0.26 = 0.43/year
lateral D: 0.17 + 0.11 = 0.28/year
Gettingrid of the reclosureschemeand letting a fuseclearall faults on the lateral
conductorswould lead to long interruptionsonly.
•
•
•
•
•
main: Lljyear
lateral A: 3.1/year
lateral B: 2.I/year
lateral C: 2.9/year
lateral D: 1.9/year
Table 3.6 comparesthe numberof long and short interruptionsfor systemswith
and without a reclosurescheme.For equipmentor production processessensitiveto
long interruptionsonly, the systemwith a reclosureschemeis clearly preferable.It leads
to a reduction of the number of long interruptions by 85%. But when equipment/
productionprocessis sensitiveto short and to long interruptions,it is betterto abolish
the reclosure schemeand trip permanentlyon every fault. That would reduce the
number of equipmenttrips by a factor between2 and 5, dependingon the position
of the load on the feeder. Inreality this decision is not that easy to make, as some
customersprefermoreshortinterruptionsabovea few long ones, while forothersonly
the numberof interruptionsmatters.The first group is mainly the domesticcustomers,
the secondone theindustrialcustomers.A financial assessment
will almostalwaysbe in
the favor of the industrials.An assessment
on numbersof customersor on kWh will be
in favor of the domesticcustomers.
TABLE 3.6 Numberof Short and Long Interruptionsper Year on an
OverheadDistribution Feeder, With andWithout Automatic Reclosure
Long InterruptionsOnly
With
Reclosure
Main feeder
Lateral A
Lateral B
Lateral C
Lateral 0
0.2
0.5
0.3
0.4
0.3
Without
Reclosure
1.1
3.1
2.1
2.9
1.9
All Interruptions
With
Reclosure
Without
Reclosure
6.6
6.6
6.6
6.6
6.6
3.1
2.1
2.9
1.9
1.1
Voltage SagsCharacterization
4.1 INTRODUCTION
Voltage sags areshort duration reductionsin rms voltage, caused by
short circuits,
overloads, andstartingof largemotors.The interestin voltage sags is mainly due to the
problems they cause on several typesequipment:adjustable-speed
of
drives, processcontrol equipment,and computersare notoriousfor their sensitivity. Some pieces of
equipmenttrip when the rms voltagedrops below 900/0 for longer than one or two
cycles. In this and the two following
chapters,it will become clearthat such a piece of
equipmentwill trip tens of times a year.I f this is theprocess-controlequipmentof a
papermill, one can imaginethat the damagedue to voltage sags can be
enormous.Of
course a voltage sag is not damagingto
as
industryas a (long orshort)interruption.But
as there are far more voltage sags
thaninterruptionsthe total damagedue to sags is still
larger. Short interruptionsand most longinterruptionsoriginatein the localdistribution network. However, voltage sags at
equipmentterminalscan be due toshort-circuit
faults hundredsof kilometers away in thetransmissionsystem. A voltage sag is thus
much more of a"global" problem than an interruption. Reducing the number of
interruptionstypically requiresimprovementson one feeder.Reducingthe numberof
voltage sags requires
improvementson several feeders, and often eventransmission
at
lines far away.
An exampleof a voltage sag due to short-circuitfault
a
is shown in Fig. 4.1. We
seethat the voltageamplitudedropsto a valueof about20% of the pre-eventvoltage
for abouttwo cycles. After these two cycles the voltage comes back
aboutthe
to
pre-sag
voltage. Thismagnitudeand duration are the maincharacteristicsof a voltage sag.
Both will be discussed in more detail in the
forthcomingsections. We can also conclude
from Fig. 4.1that magnitudeand durationdo not completelycharacterizethe sag. The
during-sagvoltage containsa rather large amount of higher frequencycomponents.
Also the voltage shows a small
overshootimmediatelyafter the sag.
directedto voltage sags due to
shortMost of the currentinterestin voltage sags is
of equipment
circuit faults. These voltage sags are the ones which cause
majority
the
trips. But also thestartingof inductionmotorsleads to voltage sags.
Figure4.2 gives an
139
140
Chapter4 • VoltageSags-Characterization
--~--~-~--~-·- - ·r ·- ·- -· · _ · - · --,
o
2
3
4
Time in cycles
5
6
Figure 4.1 A voltage sag due to a
shortcircuit fault-voltagein one phase in time
domain. (Data obtainedfrom [16].)
Phase A voltage
106
..
:
:
104 ..
---_ .
5 102
~
t
I- . .. . .
,
I..
............-...................1"....................-..........-......
'1'..........
.... Min:
Max: 93.897
101.46 .....
,
... ........ ... ... . .+...........- . . ... . . ..
1............... Avg: 95.8598.....
5100 I- . . .
._----------_._-----------------:-----_._-------..--------·---------------1-·--------------···_--··-----------------
;'
.,
-
98 I- ... .
..····..··..·..··..··..·········1·........···············..··....··.......j...............................
CI)
96 I- . . .. .
.. .. .... .........."":;;';;;;-
e
~
_ _ _ _0 - • • • •
...
-------
- - --- ----~ ._ ._.- -- - -- --_ ._ -- - _.
__
._-_._-.-_.-.-----
...............j.........................................j.........................................
94 I- ..... ~
50
100
150
Time-cycles
Figure 4.2 A voltage sag due toinduction motor starting.(Data obtainedfrom
ElectrotekConcepts[l9J.)
example of such a voltage sag [19]
. Comparingthis figure with Fig. 4.1 shows that no
longer theactualvoltage as afunction of time is given but the rms voltage versus time.
The rms voltage is typicallycalculatedevery cycle or half-cycleof the power system
frequency. Voltage sags due induction
to
motor startinglast longerthan those due to
short circuits. Typicaldurationsare seconds to tens of seconds. The
remainderof this
chapterwill concentrateon voltage sags due to
shortcircuits. Voltage sags due to
motor
startingwill be discussed inshort in Section 4.9.
4.2 VOLTAGE SAG MAGNITUDE
4.2.1 Monitoring
The magnitudeof a voltage sag can be
determinedin a numberof ways. Most
existing monitors obtain the sagmagnitudefrom the rms voltages
. But this situation
might well change in the future. There are several
alternativeways of quantifying the
voltage level. Two obvious examples are the
magnitudeof the fundamental(power
frequency)componentof the voltage and the peak voltage over each cycle or halfcycle. As long as the voltage is
sinusoidal,it does not matter whether rms voltage,
141
Section 4.2 • Voltage SagMagnitude
fundamentalvoltage, or peak voltage is used to obtain the magnitude
sag
. But especially during a voltage sag this is often not the case
.
4.2.1.1 Rms Voltage. As voltage sags are initially recorded as sampled
points
in time, the rms voltage will have to be calculated from the sampled
time-domain
voltages. This is done by using the following
equation:
1
-Lv?
N
N
;=1
(4.1)
I
where N is the numberof samples per cycle and
V; are the sampled voltages in time
domain.
The algorithm described by (4.1) has been applied to the sag shown in
. 4.1.
Fig
The results are shown in Fig..34and in Fig. 4.4. In Fig. 4.3 the rms voltage has been
calculated over a window of one cycle, which was 256 samples for the record
ing used.
Each point in Fig. 4.3 is the rms voltage over the preceeding 256 points (the first 255
rms values have been made equal to the value for sample: 256)
1.2,--~--,---
5..
0.8
.S
~
0.6
S
~
0.4
0.2
Figure 4.3 One-cycle rms voltagefor the
voltage sagshownin Fig. 4.1.
2
3
4
Time in cycles
5
1.2,--~--.,.---
5..
0.8
.S
~
~
0.6
~
.,. 0.4 '
,
Figure4.4 Half-cycle rms voltagefor the
voltage sagshownin Fig. 4.1.
2
.
3
4
Time in cycles
5
6
Chapter4 • VoltageSags-Characterization
142
i=k
Vrmik) =
L
N
1?;
(4.2)
i=k-N+t
with N = 256. We see that the rms voltage does not immediately
drop to a lower value
but takes one cycle for the
transition.We also seethat the rms value during the sag is
not completelyconstantand that the voltage does not immediately recover after the
fault. A surprisingobservationis that the rms voltage immediately after the fault is only
about90% of the pre-sag voltage. We will come back to this
phenomenonin Section
4.9. From Fig. 4.1 one can see that the voltage in time domain shows a small overvoltage instead. In Fig. 4.4 the rms voltage has been calculated over the preceeding 128
points, N = 128 in (4.2). Thetransition now takes place in one half-cycle. sAhorter
window than one half-cycle is not useful. The window length has to be an integer
multiple of one half-cycle. Anyother window length will produce an oscillation in
the result with a frequency equal to twice the
fundamentalfrequency.For both figures
the rms voltage has been calculated after each sample. In power quality
monitors,this
calculationis typically made once a cycle:
i=kN
VrmikN) =
L
N
v~
(4.3)
i=<k-l)N+l
It is thus very likely that themonitor will give one value with anintermediatemagnitude before its rms voltage value settles down. We will come back to this when discussing sagduration.
4.2.1.2 Fundamental Voltage Component.
Using the fundamentalcomponent
of the voltage has the
advantagethat the phase-angle
j ump can be determined in the
same way. The phase-angle
jump will be discussed in detail in Section 4.5. The fundamentalvoltagecomponentas a function of time may be calculated as
~lund(t) = -T2 j l
v(r)t!Wotdr
(4.4)
i-r
whereWo = 2; and T one cycle of thefundamentalfrequency. Notethat this results in a
complex voltage as a function of time. The absolute value of this complex voltage is the
voltagemagnitudeas a functionof time; its argumentcan be used toobtain the phaseangle jump. In a similar way we can
obtain magnitude and phase angle ofharmonic
a
voltagecomponentas a function of time. This so-called
"time-frequencyanalysis" is a
well-developed area within digital signal processing with a large
applicationpotentialin
power engineering.
The fundamentalcomponenthas beenobtainedfor the voltage sag shown in Fig.
4.1. The absolute value of the
fundamentalcomponentis shown in Fig. 4.5. Each point
represents themagnitudeof the (complex)fundamentalcomponentof the previous
cycle (256 points). Thefundamentalcomponentof the voltage has been
obtained
through a fast-Fourier transform (fft) algorithm [148]. A comparisonwith Fig. 4.3
shows that the behavior of the
fundamentalcomponentis very similar to the behavior
of the rms voltage.
The rms voltage has the
advantagethat it can be applied easily to a half-cycle
window. Obtainingthe fundamentalvoltage from a half-cycle window is more complicated. A possible solution is to take a half-cycle window and to calculate the second
half-cycle by using
143
Section 4.2 • Voltage SagMagnitude
cos(wt
I
,
,
3
4
Time in cycles
2
Figure4.5 Magnitudeof the fundamental
componentof the voltage sag in ig.
F 4.1.
6
5
+ rP + 1l') =- cos(wt + rP)
(4.5)
Let Vi, i = 1. . . ~ be the samplesvoltagesover a half-cycle window. Thefundamental
voltage isobtainedby taking the Fourier transformof the following series:
VI ... v~, -VI' .. -
(4.6)
v~
This algorithm has beenapplied to the voltage sagshownin Fig. 4.1, resultingin Fig.
4.6. The transition from pre-fault to during-voltageis clearly fasterthan in Fig. 4.5.
Note that this method assumesthat there is no de voltagecomponentpresent.The
presence of a de
voltagecomponentwi11lead to anerror in the fundamentalvoltage.
An alternativemethod of obtaining the fundamentalvoltage componentis discussed in Section 4.5.
4.2.1.3 Peak Voltage.The peakvoltage as afunction of time can beobtained
by using the following expression:
I
Vpeak = 0 <max
r < T v(t - r)
6.
(
1lc:
8. 0.8
E
o
<.>
~ 0.6
~
::E
0.2
0'
==l
J
0.4
.~
Figure 4.6 Magnitudeof the fundamental
componentof the voltage sag in Fig
. 4.1,
obtainedby using a half-cycle window.
(4.7)
I
.S
.E
....o
]'"
I
..._.~ _ _~~I
2
345
Time in cycles
6
144
Chapter 4 • Voltage
Sags-Characterization
1.2 I,---~--~-~--~-~~--,
50
0.8
.5
~
~
~
0.6
L
0.4
0.2
234
5
6
Time in cycles
Figure4.7 Half-cycle peak voltage for the
voltage sag shown
i n Fig. 4.1.
with v(t) the sampled voltage waveform and
T an integer multiple of one half-cycle. In
Fig. 4.7, for each sample the
maximum of the absolutevalue of the voltage over the
preceding half-cycle has been
calculated.We seethat this peak voltage shows sharp
a
drop and asharprise, althoughwe will see laterthat they do not coincide with commencement and clearing of the sag.
Contraryto the rms voltage, the peak voltage shows
an overshootimmediately after the sag
, which correspondsto the overvoltage in time
domain. The two methods are
comparedin Fig. 4.8. We seethat the peak voltage tends
to be higher most of the time with the exception of the end of the deep
part of the sag.
:::l
0.
0.8
.5
~
s
~
0.6
0.4
,,
,
,,
,
,,
,
,,
,
,,
0.2
2
3
Time in cycles
4
5
6
Figure4.8 Comparisonbetweenhalf-cycle
peak (solid line) andhalf-cycle rms voltage
(dashed line) for the voltage sag shown in
Fig. 4.1.
4.2.1.4 A One-Cycle Voltage Sag
. Another example of a voltage sag is shown
in Fig. 4.9; contrary to Fig. 4.1, all three phase voltages are shown. The voltage is
low in one phase forabout one cycle and recovers
rather fast after that. Theother
two phases show some
transientphenomenon,but no clear sag or swell. The
latter is
also evident from Fig
. 4.10 which gives the half-cycle rms value for the sag shown in
Fig. 4.9. We see in thelatter figure that the voltage in the twonon-faultedphases
shows a small swell
. Due to theshort duration of the sag the rms voltage curve does
not have a specific flat part. This makes the
determinationof the sagmagnitude
rather arbitrary. If the monitor takes one sample every half-cycle the resulting sag
145
Section 4.2 • Voltage SagMagnitude
al ~
f-:~
~
0
I
2
3
456
al0 ~
~- I l
' , ~
0123456
c:
.;;
OIl
I
'
~I VVV\IVYJ
';;
OIl
0
19 - )
Figure 4.9 Time-domainplot of a one-cycle
sag, plots of the three phase voltages
. (Data
obtainedfrom [16J.)
~
0)
23456
Time in cycles
io:~:
.:I
1:l l
4
5
3
4
Time in cycles
5
~
00
~
I
0
:
I
2
3
4
5
~
o
2
o
2
3
ko:I======
~~-~'-~,~~,
Figure 4.10Half-cycle rms voltages for the
voltage sag shown in Fig
. 4.9.
6
-'I
6
6
magnitudecan be anywhere between 26% and 70% depending onmoment
the
at
which the sample is taken
. In case a one-cycle window is used to calculate the rms
voltage, thesituation becomes worse
.
The two alternativemethods forobtaining the sagmagnitudeversus time have
also been applied to phase b of the event in Fig.
.9. The
4
half-cycle peak voltage is
shown in Fig. 4.11, the half-cycle
fundamentalvoltage componentin Fig. 4.12. The
shape of thelatter is similar to the shape of the half-cycle rms. The half-cycle peak
voltage again shows a much
sharpertransition than theother two methods.
4.2.1.5 Obtaining One Sag Magnitude.Until now, we havecalculatedthe sag
magnitudeas a function of time: either as the rms voltage
, as the peak voltage, or as
the fundamentalvoltage componentobtainedover acertain window. There are various ways of obtaining one value for the sagmagnitudefrom the magnitudeas a
function of time. Most monitors take the lowest value.Thinking about equipment
sensitivity, this correspondsto the assumptionthat the equipment trips instantaneously when the voltage drops below a certain value. As most sags have
rather
a
constantrms valueduring the deeppart of the sag, using the lowest value
appears
an acceptableassumption.
146
Chapter4 • VoltageSags-Characterization
I.2 f
:>
0..
0.8
.5
1iI> 0.6
S
~
0.4
0.2
2
3
4
5
6
Time in cycles
a
Figure 4.11 Half-cyclepeak voltage for phase
b of the sag shown in Fig
. 4.9.
I [_ ~ -- '
!
.5
C
~
8. 0.8
E
o
o
'3
e
0.6
E
.jg
~ 0.4
e-
o
]" 0.2
.~
~
0
~~_~
L
:
.
_ _
~
_
_
~
_ _
234
Time in cycles
~_--'
5
6
Figure 4.12 Half-cyclefundamentalvoltage
F 4.9.
for phase bof the sag shown in ig.
So far there israther generalagreement,both about using the rms value,a nd
about taking the lowest rms value todeterminethe sagmagnitude.But when the sag
magnitudeneeds to bequantified in a number,the agreementis no longer there. One
common practiceis to characterizethe sagthrough the remainingvoltageduring the
sag.This is then given as ap ercentageof the nominal voltage. Thus, a 70% sag in a 120
volt systemmeansthat the voltagedroppedto 84 V. This methodof characterizingthe
sag isrecommendedin a numberofIEEE standards(493-1998,1159-1995,1346-1998)
.
The confusionwith this terminologyis clear. One could be tricked into thinking that a
70% sag refers to a
d rop of 70%, thus a remainingvoltageof 30%. The recommendation is thereforeto use thephrase" a sagdown to 70%" [3]. The lEC has solved this
ambiguity by characterizingthe sagthroughthe actualdrop in the rmsvoltage[4]. This
has somewhatbecomecommon practicein Europe. Characterizinga sag through its
drop in voltagedoes not solve all problemshowever,becausethe nextquestionwill be:
What is the referencevoltage? There are argumentsin favor of using the pre-fault
voltage and there are arguments in favor of using the nominal voltage. The
International Union of Producers and Distributors of Electrical Energy (Union
International des Producteurs et Distributeurs d'Energie Electrique, UNIPEDE)
147
Section 4.2 • Voltage Sag
Magnitude
recommendsto use thenominal voltageas a reference(5]. As severaldefinitions are in
use, it isimportantto clearly define the way in which the sag
magnitudeis defined. In
this book sag magnitudeis defined as theremainingvoltageduring the event.
Using the remainingvoltageas the sagmagnitude,leads to someobviousconfusions. Themain sourceof confusionis thata largersagmagnitudeindicatesa less severe
of terms
event. In fact, a sagm agnitudeof 100% correspondsto no sag at all. The use
like "large sag" and "small sag" would be extremelyconfusing. Insteadwe will talk
abouta "deepsag" and a "shallow sag." A deep sag is a sag with a low
magnitude;a
shallow sag has a largemagnitude.When referring to equipmentbehaviorwe will also
use theterms "severesag" and "mild sag." As far as magnitudeis concerned,these
terms correspondto "deepsag" and "shallow sag," respectively.
4.2.2 TheoreticalC alculations
Considerthe power systemshownin Fig. 4.13, where thenumbers(1 through5)
indicate fault positionsand the letters (A through D) loads. A fault in the transmission network, fault position 1, will causea serioussag for both substationsbordering
the faulted line. This sag isthen transferreddown to all customersfed from these two
substations.As there is normally no generationconnectedat lower voltage levels,
thereis nothing to keep up the voltage. The result that
is a deep sag isexperiencedby
all customersA, B, C, and D. The sagexperiencedby A is likely to be somewhatless
deep, as thegeneratorsconnectedto that substationwill keep up thevoltage. A fault
at position 2 will not causemuch voltagedrop for customerA. The impedanceof the
transformersbetween the transmissionand the sub-transmissionsystem are large
enoughto considerablylimit the voltagedrop at high-voltageside of the transformer.
The sagexperiencedby customerA is further mitigated by the generatorsfeeding in
to its local transmissionsubstation.The fault at position 2 will, however,causea deep
sag at both subtransmissionsubstationsand thus for all customersfed from here (B,
C, and D).
3
Figure4.13 Distribution network with load
positionsand fault positions.
Chapter4 • VoltageSags-Characterization
148
A fault at position 3 will cause a very deep sag for customer D, followed by a
short or longinterruption when theprotectionclears the fault.CustomerC will only
experience a deep sag. If fast reclosure is used indistribution
the
system, customer C
will experience two or more sags shortly after each other forpermanentfault.
a
Customer B will only experience a shallow sag due to the fault at position 3, again
due to thetransformerimpedance.CustomerA will probably not notice anything from
this fault. Finally, fault 4 will cause a deep sag for customer C and a shallow one for
customer D.For fault 5 the result is just the other way around: a deep sag for customer
D and a shallow one for customer C. Customers A and B will not be influenced at all by
faults 4 and 5.
To quantify sag magnitude in radial systems, the voltage divider model, shown in
Fig. 4.14, can be used. This might
appeara rather simplified model, especially for
transmission systems. But as we will see in the course of this and further chapters, it
has turned out to be a rather useful model to predict some of the properties of sags. In
Fig. 4.14 we see two impedances: Zs is the source impedance point-of-common
at the
point-of-commoncoupling and the
coupling; and ZF is the impedance between the
fault. The point-of-commoncoupling is the point from which both the fault and the
load are fed. In other words: it is the place where the load
currentbranches off from the
fault current. We will often abbreviate"point-of-commoncoupling" as pee, In the
voltage divider model, the load current before as well as during the fault is neglected.
There is thus no voltage drop between the load and the pee. The voltage at the pee, and
thus the voltage at the equipment terminals, can be found from
ZF
s+ Z F E
v.rag=Z
(4.8)
In the remainder of this chapter, we will assume that the pre-event voltage is exactly 1
pu, thus E = 1. This results in the following expression for the sag magnitude
v =
sag
ZF
ZS+ZF
(4.9)
Any fault impedance should be included in the feeder impedance
ZF' We see from (4.9)
that the sag becomes deeper for faults electrically closer to the customer ZF
(when
becomes smaller), and for systems with a smaller fault level (when
Zs becomes larger).
Note that a single-phase model has been used here, whereas in reality the system is
three-phase. That means that this
equationstrictly speaking only holds for three-phase
faults. How the voltage divider model can be used for single-phase and phase-to-phase
faults is discussed in Section 4.4.
Equation(4.9) can be used to calculate the sag magnitude as a function of the
ZF = Z x E, with z the impedance of
distance to the fault. Therefore we have to write
£ the distance between the fault and the pee, leading to
the feeder per unit length and
E
Fault
Load
pee
Figure.4.14Voltage divider model for a
voltage sag.
149
Section 4.2 • Voltage Sag Magnitude
v _
sag -
z£
Zs + z£
(4.10)
The sagmagnitudeas afunction of the distanceto the fault has been
calculatedfor a
typical 11kV overheadline, resulting in Fig. 4.15. For the calculationsa 150mnr'
overheadline was used and fault levels
of 750 MVA, 200 MVA, and 75 MVA. The
fault level is used tocalculatethe sourceimpedanceat the pee, the feeder
impedanceto
calculatethe impedancebetween the pee and the fault. It was assumed
that the source
impedanceis purely reactive, thusZs =jO.161 n for the 750 MVA source. The impedance of the 150mrrr'overheadline is 0.117+ jO.315 Q per km [10].
As expected, the sag
magnitudeincreases (i.e., the sag becomes less severe) for
t hat faults at
increasingdistanceto the fault and forincreasingfault level. We also see
tensof kilometersdistancemay still cause a severe sag.
1
0.8
:s
e,
.5
-8
a
75MVA
I
0.6
.~
e 0.4
~
fI)
0.2
Figure 4.15 Sag magnitude as a function of
the distance to the fault, for faults on an
11 kV, 150 mnr' overhead line.
10
20
30
40
Distanceto the fault in kilometers
50
4.2.2.1 Influence of Cross Section. Overheadlines of different cross section
have different impedance,and lines and cables also have
different impedance.It is
thus to be expectedthat the cross sectionof the line or cable influences the sag magnitude as well. To show this influence, Fig. 4.16 plots the magnitudeat
sag
the pee
0.8
6-
.5
] 0.6
)9---T~
300
1/
.~
e0.4
f
fI)
0.2
Figure 4.16 Sag magnitude versus distance,
for 11 kV overhead lines with different cross
sections.
5
10
15
20
Distanceto the fault in kilometers
25
150
Chapter 4 • VoltageSags-Characterization
50
0.8
150
8.5
300
~ 0.6
a
.~
e 0.4
~
en
0.2
5
10
15
20
Distance to the fault in kilometers
25
Figure4.17 Sagmagnitudeversusdistance,
for II kV undergroundcableswith different
cross sections.
as a function of the distancebetweenthe fault and the pee, for 11 kVoverheadlines
with threedifferent cross sections:50, 150,and 300 mm''. A sourceimpedanceof 200
MV A has been used. Thesmaller the crosssection, the higher the impedanceof the
feeder and thus the lower thevoltagedrop. For overheadlines, the influence israther
small as thereactancedominatesthe impedance.For undergroundcables, the influence ismuch bigger asshown in Fig. 4.17, again for cross sectionsof 50, 150,and
300 mrrr'. The inductanceof cablesis significantly smallerthan for overheadlines, so
that the resistancehas more influence on theimpedanceand thus on the sagmagnitude. Theimpedancevalues used toobtain Fig. 4.16 and Fig. 4.17 are given inTable
4.1. All impedancesare for an II kV voltage level.
TABLE 4.1 Line and CableImpedancesfor 11 kV FeedersUsed in Figs.
4.16 and 4.17
Impedance
CrossSection
2
50 mm
150mrrr'
300 mm2
OverheadLine
Cable
0.363+ jO.351 Q
0.117 + jO.315Q
0.061+ jO.298Q
0.492 + jO.116Q
0.159+jO.097Q
0.079 +jO.087 Q
Source: Data obtained from [10].
4.2.2.2 Faults behind Transformers.The impedancebetweenthe fault and the
pee in Fig. 4.14not only consistsof lines or cablesbut also of power transformers.
As transformershave arather large impedance,amongothersto limit the fault level
on the low-voltage side, thepresenceof a transformerbetweenthe fault and the pee
will lead to relatively shallow sags.
To show the influenceof transformerson the sagmagnitude,considerthe situation shownin Fig. 4.18: a 132/33kV transformeris fed from thesamebus as a132kV
line. A 33 kV line is fed from thelow-voltageside of the transformer.Fault levels are
3000 MV A at the 132 kV bus,a nd 900 MV A at the 33 kV bus. Inimpedanceterms,the
source impedanceat the 132 kV bus is5.81 0, and the transformerimpedanceis
13.550, both referred to the 132kV voltage level. The sensitiveload for which we
lSI
Section 4.2 • Voltage Sag
Magnitude
pee
132kV
132 kV line
Load
Figure 4.18Powersystem with faults at two
voltage levels.
33 kV line
want to calculatethe sagmagnitudeis fed from the 132kV bus viaanother132/33 kV
Zs = 5.810, ZF = 13.550+ z x {" z is the
transformer.We can again use (4.9), where
transfeederimpedanceper unit length,and {, the distancebetween the fault and the
former's secondaryside terminals. The feeder
impedancemust also be referred to the
k{ )2x 0.3Qjkm when the feederimpedanceis 0.3Qjkm at 33 kV.
132kV level: z=
(upper
The resultsof the calculationsare shown in Fig. 4.19 for faults on the 33 kV line
curve) and for faults on the 132kV line (lower curve). We see
that sags due to33kV
Not only does the 33 kV curves tart
faults are less severe
thansags due to 132kV faults.
off at a higher level (due to the
t ransformerimpedance),it also rises much faster. The
latter is due to the factt hat the feederimpedanceseen from the 132kV level is (132/3
3)2 = 16 times as high asthat seen from the 33 kV level.
(lilk
Faultsat 33 kV
0.8
Faultsat 132kV
0.2
I.......--_ _L . . - - _ - - J I - - -
Figure 4.19 Comparisonof sag magnitude
for 132 kV and 33 kV faults.
--J-_
__._! __ ..•.•..._ . . . • ._..
20
40
60
80
Distanceto thefault in kilometers
100
4.2.2.3 Fault Levels. Often the sourceimpedanceat a certain bus isnot immediately available, but insteadthe fault level is. One canof coursetranslatethe fault
level into a sourceimpedanceand use (4.9) tocalculatethe sagmagnitude.But one
may calculatethe sagmagnitudedirectly if the fault levelsboth at the peeand at the
fault position are known. LetSFLT be the fault level at the faultposition and Spec at
the point-of-commoncoupling. For a rated voltage Vn the relationsbetween fault level and sourceimpedanceare as follows:
(4.11)
152
Chapter4 • VoltageSags-Characterization
V,;
(4.12)
SPCC=-
Zs
With (4.9) the voltage at the pee can be
written as
Vsag -- I _
SFLT
(4.13)
Spec
We use(4.13)to calculatethe magnitudeof sags behindtransformers.For this we use
typical fault levels in theU.K. power system[13]:
400 V
11 kV
33 kV
132 kV
400 kV
20 MVA
200 MVA
900 MVA
3000 MVA
17000MVA
Considera fault at a typical11 kV bus, i.e., with a fault levelof 200 MVA. The voltage
sag at thehigh-voltageside of the 33/11kV transformeris from (4.13)
v,wg = 1 -
200 MVA
0
900 MVA = 78Yo
In a similar way the wholeof Table 4.2 has been filled. The zeros in this table
indicatethat the fault is at the same or at a higher voltage level. The voltage
dropsto a
low value in such a case. We can see from
Table 4.2 that sags are significantlydamped
when theypropagateupwardsin the power system. In a sagstudy we typically only
of
have to take faults one voltage level down
into account.And even those are seldom
seriousconcern.An exceptionherecould be sags due to faults at
33kV with a pee at
132kV. They could lead to sags down to
70o~.
TABLE 4.2 Upward Propagationof Sags
Point-of-CommonCoupling
Fault Point
II kV
33 kV
132 kV
400 kV
400 V
II kV
33 kV
132 kV
900~
98°~
0
0
0
78%
0
0
99%
93%
70%
0
100%
990/0
950/0
82%
4.2.2.4 Critical Distance. Equation (4.10) gives thevoltage magnitude as a
function of the distanceto the fault. From this equationwe can obtain the distance
at which a fault will lead to a sagof a certain magnitude.If we assumeequal X/R
ratio of sourceand feeder, weobtain
(4.14)
We refer to thisdistanceas the criticaldistancefor a voltage V. Supposethat a pieceof
equipmenttrips when thevoltagedropsbelow acertainlevel (the critical voltage). The
153
Section 4.2 • Voltage Sag
M agnitude
definition of critical distance is such
that each fault within the criticaldistancewill cause
the equipmentto trip . This concept will be used in Section 6.5 to estimate the expected
numberof equipmenttrips .
If we assumefurther that the numberof faults isproportional to the line length
within the critical distance, we would expect
that the numberof sags below a level V is
conclusion.
proportionalto V/( I - V) . Another assumptionis needed to arrive at this
Every feeder connected to every pee needs to be infinitely long
ithout
w any branching
off. Of course this is not the case in reality . Still this
equation has beencomparedwith a
number of large power quality surveys. The results are shown in Fig. 4.20. Power
quality survey results in the Un ited States [IIJ, [l2J, in the
U.K. [l3J and in Norway
[16J are indicated as dots, the
theoretical curve is shown as a solid line. The rresponco
dence is good, despite the obviously serious
approximationsmade.
Even though (4.14) only holds for rad ial systems, it gives a generally usable
relation between thenumberof voltage sags and the voltage. The expression clearly
measurements
.
showsthat the majority of sags are shallow, a fact confirmed by most
-._ - ---_._ - --
. USA [II]
• USA [12]
• UK [13]
x Norway [16]
- Theory
Figure 4.20 Numberof sags versus
magnitude :theoretical results (solid line)
versus monitoring results (dots) .
o
20
40
60
80
100
Sag magnitude in percent
4.2.3 Example of Calculation of Sag Magnitude
We will apply the theoreticalconceptsdeveloped in the previous sections to the
supply shown schematically in Fig. 4.21. This same example will be used again in
forthcoming parts of this book. The supply shown in Fig. 4.21 is the existing supply
to an indust rialcustomersomewhere in the No rth of
England[15J. The sensitive load
consists of several large ac and de adjustable-speed drives. The de drives are fed via
dedicated transformersat 420 V, the moremodernac drives at 660 V.Most of the data
used for the various
calculationsbelow have been obta ined from the local utility. Where
"as typical as posno data was available,d ata have been used which was considered
sible." Like often in these kind of studies, the collection of datarequires
the
at least as
much effort as theactualcalculations. In the restof this book it will always be assumed
that all the requireddata is readily available.
The first step in a sag analysis is to recognize the possible
pee's, For any fault on
one of the II kV feeders, the faultcurrent will flow through the STU-II bus, but
not further towards the ·load . TheS TU-II bus is thus the pee for all faults within the
the
II kV network. In the same way, the ROS-33 bus is the pee for faults on of
any
33 kV feeders. Theother possible pee's are PAD
-I32 and PAD-400. To calculatethe
sag magnitudewe need the sou rce
impedanceand the feeder impedance. The source
154
Chapter 4 • VoltageSags-Characterization
Slines
8 lines
P
---.
J\O-400
--ill
r - - -_ _
EGG-400
3 feeders
Figure 4.21 Example of power supply be
to
used for voltage sag
calculations.
impedanceis given in Table4.3, the feederimpedancein Table4.4. All impedancesare
given for a 100 MVA base.Finally, Table 4.5 gives thetransformerconnectionand
neutral grounding.This information is needed inlater sections,when unbalancedsags
are discussed.
For now weignore the fact that the impedancesare complexand use theabsolute
values for our calculations.We will come back to thecomplex impedancesin Section
4.5 whenphase-anglejumps are discussed.F or faults at II kV we obtain for the impedances:z = 27.75% per km and Zs = 66.08%.The critical distancecan becalculated
from Lcril = 2.381 x I~V'
Calculationsfor the critical distancesat 33 kV and 132kV proceedin exactly the
same way as for the11 kV system.The resultsof thesecalculationsare shownin Table
4.6. We seethat thereare twocolumnsfor the 400 kV system inTable4.3 and in Table
4.6. This has to do with the fact
that thereare twopossiblesourcesfor the short-circuit
power. If the fault issomewherebetweenPAD-400 and PEN-400the fault currentwill
be delivered from thedirectionof EGG-400.Thus,for such a fault, theimpedanceZs is
the sourceimpedanceas seen in thedirectionof EGG-400.The critical distancesresulting from this sourceimpedanceare shownin Table4.6 in thecolumn labeled "toward
PEN-400." Note that for this the sourceimpedancein the direction of EGG-400has
been used.F or faults in thedirection of EGG-400,the sourceimpedancein the direc..
tion of PEN-400has been used.
Thoseresults areshownin the columnlabeled"toward
EGG-400."
WheninterpretingTable4.6 oneshouldrealizethat these values hold for raadial
system with infinitely long lineswithout any sidebranches.In reality all feeders have a
finite length. In this system themaximum distancefrom the pee for afault at 11 kV is
5 km. The distanceto the fault can thus not be more than 5 km and the magnitudeof
the most shallow sag due to afault at 11 kV is
ZF
V:vag
5 x 0.2727
°
= Zs + ZF = 5 x 0.2727+ 0.6608= 67 Yo
(4.15)
Figure4.22plots sagmagnitudeversusdistancefor faults at all thevoltagelevels in Fig.
4.21. Thehorizontalscale isdeterminedby the maximumlength of the feeders att hat
155
Section 4.2 • Voltage Sag Magnitude
TABLE 4.3 Source Impedance for the Supply Shown in Fig. 4.21, at a 100
MVA Base
Zero Sequence
II kV
33 kV
132 kV
400 kV
From EGG
From PEN
Positive and Negative
Sequence
787 + j2200/0
2510/0
0.047 + .i2.75%
4.94 + j65.90/0
1.23 + jI8.3°At
0.09 + j2.86%
0.329 + j2.273%
0.653 + j5.124%
0.084 + jl.061 %
0.132 + j1.94%
TABLE 4.4 Feeder Data for the Supply Shown in Fig. 4.21
Positive and Negative Sequence
II kV
33 kV
132 kV
400 kV
9.7 + j26 %/km
1.435 + j3.102°At/km
0.101 + jO.257°At/km
0.001 + jO.018%/km
Zero Sequence
18.4 + jII2°At/km
2.795 + jI5.256%/km
0.23 + ]U.650/0/km
0.007 + ]U.0500/0/km
Max Length
5 km
10 km
2 km
> 1000km
TABLE 4.5 Transformer Connections and Neutral Grounding for the
Supply Shown in Fig.4.21
Voltage Level
Transformer Winding Connection
400 kV
400/132 kV
132/33kV
solidly grounded
solidly grounded
resistance grounded through zigzag transformer
resistance grounded
solidly grounded
YY autotransformer
Star - Delta
Delta - Star
Delta - Star
33/11 kV
II kV/660 V and
11 kV/420 V
Neutral Grounding at LV Side
TABLE 4.6 Critical Distance Calculation for the Network Shown in Fig.
4.21, According to (4.14)
z
Zs
V= 10°At
V = 30%
V = 500/0
V = 70%
V = 90%
II kV
33 kV
132 kV
27.27%
66.08%
0.3 km
1.0 km
2.4 km
5.6 km
21.4 km
3.418°At
18.34%
0.6 km
2.3 km
5.4 km
12.5 km
48.3 km
0.276%
2.8610/0
1.2 km
4.4 km
10.4 km
24.2 km
93.3 km
400 kV Toward 400 kV Toward
PEN-400
EGG-400
0.018%
1.064%
6.6 km
25.3 km
59.1 km
138 km
532 km
0.018%
1.9440/0
12.0 km
46.3 km
J08 km
252 km
972 km
156
Chapter 4 • VoltageSags-Characterization
)
11 kV faults
33 kV faults
,.-----...---,
132kV faults
400 kV faults
I
I:
0.5
00
1
2
o
o
Distanceinkilometers
.
100
--.JI
200
Distanceinkilometers
Figure 4.22 Magnitude versus distance for
faults at various voltage levelsin the supply in
Fig. 4.21.
voltage level.For 400 kV a lengthof 200 km has been taken. The
short length of the
132kV feeders makesthat sags due to faults at 132kV are always very deep.
4.2.4 Sag Magnitude In Non-Radial Systems
In Section4.2.2 we discussed sag
magnitudeversusdistancein radial systems.
Radial systems arecommonin low-voltage and medium-voltagenetworks.At higher
voltage levelsother supply arrangementsare common.Some typical cases will be discussed below. We will alsopresenta general way ofcalculating sag magnitudesin
meshed systems.
4.2.4.1 Local Generators. The connectionof a local generatorto a distribution
network, as shown in Fig.4.23, mitigates voltage sagsof the indicated load in two
different ways. Thegeneratorincreases the fault level at the
distribution bus, which
mitigates voltage sags due to faults on the
distribution feeders. This especially holds
for a weak system.For a strong system, the fault levelcannot be increased much
without the risk of exceeding themaximum-allowableshort-circuit current of the
switchgear.The installation of local generationrequires a largerimpedanceof the
feeding transformer.
Rest of the system
I'\v
Load
Local
generation
Figure 4.23 Connection of a local generator
to a distribution bus.
157
Section 4.2 • Voltage Sag Magnitude
A local generatoralsomitigatessags due to faults in the rest
of the system.D uring
such a fault thegeneratorkeeps up thevoltageat its local bus by feedinginto the fault.
An equivalentcircuit to quantify this effect has beendrawn in Fig. 4.24: Z4 is the
impedanceof the local generatorduring the fault (typically thetransientimpedance);
ZI the sourceimpedanceat the pee;Z2 the impedancebetween the faultand the pce;
and Z3 the impedancebetween thegeneratorbus and the pee.
Note that the conceptof
point-of-commoncoupling strictly speakingno longer holds. Thisconcept,which was
o f fault current. By addinga
introducedfor radial networks,assumes one single flow
generatorclose to the load asecondflow of fault current is introduced.The pee as
indicatedin Fig. 4.24 is thepoint-of-commoncoupling before theintroduction of the
local generator.Without the local generatorthe voltage at theequipmentterminals
would beequalto the voltageatthepee,Whena local generatoris present,the voltage
at theequipmentterminalsduring the sagequalsthe voltageon thegeneratorbus. This
voltage is related to thevoltageat the peeaccordingto the following equation:
(1 -
2
Vvag)
= Z 3+4Z 4 (1 -
Vpcc)
(4.16)
The voltagedrop at the generatorbus isz ~z times thevoltagedrop at the pee,
The voltagedrop becomes smaller forlarger imped~nce to the pee(weakerconnection)
and for smallergenerationimpedance(larger generator).The fault contributionof the
rest of the system at theg eneratorbus isoften mainly determinedby the impedanceof
the feedingtransformer.In that case thereductionin voltage drop is approximately
equal to thegeneratorcontributionto the fault level at thegeneratorbus. Thus, if the
generatordelivers50% of the fault current,a sagdown to 40% at the pee(60% voltage
drop) will be reducedto a sagdown to 700/0 (30% voltage drop) at the equipment
terminals. From (4.16) we can alsoconcludethat there is anon-zerominimum sag
magnitude.Even a fault at the pee will no
longercause a sag
d own to zerovoltagebut a
sagof magnitude
Vmin
=2 3Z3
+2
4
(4.17)
For the above-mentioned
system, where the local
generatoris responsiblefor 50%
of the fault level at thegeneratorbus, the lowest sagm agnitudedue to a fault at a
higher voltagelevel is 50% • During a fault not only local generatorscontributeto the
fault but also induction motors. Using the abovereasoningwe can concludethat the
minimum voltage at the plant bus equalsthe relative fault levelc ontribution of the
induction motors. We will discussinduction motorsin more detail in Section 4.8.
pee--'---.---'-Load
Figure 4.24 Equivalent circuit for system
with local generation.
Fault
158
Chapter4 • Voltage Sags-Characterization
EXAMPLE An exampleof a system withon-site generationis given in Fig. 4.25: the
industrial system is fed from a 66 kV, 1700 MVAsubstationvia two 66/11 kV transformersin
paraJIel. The fault level at the 11 kV bus is 720 MVA, which includes contribution
the
of two
20 MVA on-site generatorswith a transientreactanceof 170/0. The actual industrial load is fed
from the 11 kV bus, for which we willcalculatethe sagmagnitudedue to faults at 66 kV. The
feeder impedanceat 66 kV is 0.3Q/km.
Public supply
66 kV, 1700MVA
---a._..........._....--a_.L--1_1_k_V,_720
Faulted
feeder
MVA
Figure 4.25 Industrial distribution system
with on-site generation.
Industrial load
With referenceto (4.16) and Fig. 4.24, we get the
following impedancevalues for this
system(referred to 66kV):
Z. == 2.56Q
2 2 = 0.3 O/km x
£,
2 3 = 6.42Q
2 4 = 18.SQ
The calculationresults areshownin Fig. 4.26.The bottomcurvegives the sagmagnitudeat
the 11 kV bus for faults at a 66 kV feeder, when the 11generatoris
kV
not in operation.In that
case the sagm agnitudeat 11 kV equalsthe sagmagnitudeat 66 kV becauseall load currentshave
t op curve gives the sagmagnitudeat the 11kV bus withon-site generator
been neglected. The
connected.Due to thegeneratorkeepingup thevoltageat the 11 kV bus, the sag
magnitudenever
drops below 260/0. Thereare two methodsto further improve the supply. One canincreasethe
numberor sizeof the generators,which correspondsto decreasing2 4 in (4.16).Alternatively one
can increase2 3, which leads to a lower fault level at the 11 kV bus.
0.:
~::-er-a--'t~-rs-----r----.---i
.~a 0.6
I
Without generators
"'0'
'1 ~
0.4
~
V}
0.2
oO~--w-
20
30
4'0
Distanceto the faultin kilometers
--.J
50
Figure 4.26 Sagmagnitudeversus distance,
with and without on-site generator.
IS9
Section 4.2 • Voltage Sag Magnitude
EXAMPLE Another exampleof the useof (4.16) is given bymeansof Fig. 4.27. This
figure representshalf of the transmissionsystem part of the examplein Fig. 4.21, containing
the substationsPAD-400 and EGG·400,plus 30 km of overhead400kV line in betweenthem.
The impedanceshave thefollowing values (in % at a 100 MVA base), withE the distancebetween EGG-400and the fault:
Zt
= 1.4%
Z2 = 0.OI8% / k m x £
2 3 = 0.54%
Z4 == 1.940/0
The impedance2 4 representsthe sourcecontributionfrom PEN-400at PAD-400; 2 3 represents
the impedanceof 30 km line (0.018%/km); 2 2 the impedancebetweenEGG-400and the fault,
and Zt the contribution through the non-faultedlines at EGG-400(excluding the contribution
from PAD-400) during the fault. The latter impedanceis likely to be different for faults on
different lines. In this study we assumedit to be simply equal to the contribution of all lines at
EGG-400minusthe line toPAD-400.As thereare atotal of nine linesconnectedto EGG-400the
error madewill not be very big.
Fault
Figure 4.27 Circuit diagramrepresentation
of two transmissionsubstations.The sensitive
load is fed from thesubstationon the left.
Load
For faults to the right of EGG-400we can use (4.16) tocalculatethe voltageat PAD-400,
knowing the voltageat EGG-400.The latter can be obtainedfrom the voltagedivider equation
with the sourceimpedanceformed by the parallel connectionof 2, and 2 3 + Z4' Note that we
still neglect all loadcurrents,so that both sourcevoltagesare equal in magnitudeand in phase
and can bereplacedby one source. For faults betweenPAD-400 and EGG-400'the voltage
divider model will give the required voltage directly. The sourceimpedanceis now formed by
2 4; the feederimpedanceis O.018% / k m x C. with E the distancebetweenPAD-400and the fault.
The resultingsag magnitudeas afunction of the distanceto the fault isshown in Fig. 4.28. For
0.8
5.
.S
]
0.6
'ts 0.4
~
r.n
0.2
Figure 4.28 Sagmagnitudeas afunction of
the distanceto the fault, for transmission
systems.
I
I
,
20
I
40
60
----1.-------':
80
100
Distance to the fault in kilometers
160
Chapter4 • VoltageSags-Characterization
distances up to 30 km the sag
magnitudechanges with distance like in a radial system; for larger
distances themagnitudeincreases faster. Thus, the sag is less severe than for a fault at the same
distance in a radial system.
4.2.4.2 SubtransmissionLoops. At subtransmissionlevel, the networks often
consist of severalloops-atypical example is shown in Fig. 4.29. The
transmission
system isconnectedto the subtransmissionsystem through two or three transforof thesetransformersa numberof submers. From the busses at the low-voltage side
stationsare fed via a loop. Such n
aetwork configurationis also found inindustrial
of two branchesin parallel. The mathepower systems. Often the loop only consists
matical expressionsthat will be derived below can also be used calculatevoltage
to
sags due to faults on
parallel feeders.
Subtransmission
Figure 4.29 Example of subtransmission
loop.
To calculate the sagmagnitudewe need to identify the load bus, the faulted
branch, and the non-faulted branch. Knowing these theequivalent scheme in Fig.
4.30 isobtained,where Zo is the sourceimpedanceat the bus from which the loop is
fed; Zl is the impedanceof the faulted branchof the loop; Z2 is the impedanceof the
non-faulted branch; and p is the position of the fault on the faultedbranch (p = 0
correspondsto a fault at the bus from which the load is fed,
p = 1 correspondsto a
fault at the load bus).
From Fig. 4.30 the voltage at the load bus can calculated,resulting
be
in the
following expression:
v
_
sag -
p(l-p)Zr
ZO(ZI
+ Z2) + pZ t Z 2 + p(l - p)Z?
(4.18)
Fault
pZl
1
(I - p)ZJ
Load
Figure 4.30 Equivalent circuit for
subtransmission loop.
161
Section 4.2 • Voltage Sag Magn
itude
The voltage is zero forp = 0 (fault at the mainsubtransmissionbus) and forp = 1
(fault at the load bus) and has a maximum somewhere in between
.
. 4.31: a 125-km 132kV loop connectEXAMPLE Consider the system shown in Fig
ing a number ofsubstations.Only the substationfeeding the load of interest is shown in the
figure. This substationis located at 25 km from the main
substation. The fault level at the
Qjkm. Faults occur both in the
point-of-supply is 5000 MVA and the feeder impedance 0.3
25 km part and in the 100 km part of the loop
, so that both may form the faulted branch
. For
a fault on the 25 km branch we
substitutein (4.18): Z\ = 25zand Z2 = 100z, with z the feeder
impedance per km. For a fault on the 100 km branch
, we get Z\ 100z andZ2 25z.
=
=
............. ........ ...... ..............
132 kV
5000MVA
J---
-
. .....100km
......... .....:
---,
Load
Figure 4.31Loop systemoperatingat 132kV.
Figure 4.32gives the magnitudes of sags due to faults in the 132subtransmission
kV
loop.
The dashed (top) curve gives the sag
magnitudefor faults on the 100 km branch, the solid
(bottom) curve holds for the 25 km branch. Note that the
horizontal scale correspondsto
25 km for thebottomcurve and to 100 km for the top curve. Figure
.33gives
4
the sag magnitudes
for the 100 km and 25 km feeder as a function of the actual distance between the fault and the
main 132 kV bus.For comparison, the magnitudeis also given for sags due to faults at a radial
feeder from the same main 132kV bus
(dottedcurve).
0.8
So
=
~
0.6
~ 0.4
e
~
,,
en
r
0.2
r
,,
,
'
Figure 4.32 Sagmagnitudesfor faults on a
132kV loop.
00
~--
0.2
0.4
0.6
Fault position
0.8
We see from Fig. 4.32 and Fig. 4.33that each fault on the loop will cause the
voltage todrop below 50%of the nominal voltage. A sag due to a fault on a loop is
always lower than due to a fault on a radial feeder.
Faultsclose to thepoint-of-supply
will lead to a deep sag
. Faultsclose to the load too
. Somewhere in between there is a
162
Chapter4 • VoltageSags-Characterization
5I':
:g
0.6
2
'10.41 ::: .:
ell
. •
C':.'I
: ,
en
.,. ,
02 b~
o0
\1
--2·0 '"----4,.,.0--·-6
~0:---~
8 0---..,1 00
Fault position in kilometers
Figure 4.33 Sagmagnitudeversusdistance,
for faults on loops (solid anddashedlines)
and on a radial feeder
(dotted line).
maximum magnitudeof the voltage sag due to a fault. The longer the line the higher the
maximum . We see from the figure
that this maximumis not necessarily in the middle
of
the branch. The maximum voltage has beencalculatedas a function of the system
parameters
. The results are shown in Fig. 4.34 and in Fig. 4.35.obtain
To thesegraphs
(4.18) has been rewritten as a function
of ZI = and Z2 =
Zt is the relative impedance of the faulted branch and Z2 of the non-faulted bran~h. Figure 4.34 gives the
maximum voltage as afunction of Z2 for variousvalues of Zl and Fig. 4.35 theother
way around. From both figures it follows that the sags become less severe (higher
maximum) when the faultedbranch becomes longer (higherimpedance)and when
the non-faulted branch becomesshorter. This can be explained as follows. A longer
faulted branchmeansthat the fault can befurther away from both busses. Ashorter
non-faultedbranchgivesstrongervoltage supportat the load bus. These relations can
easily beunderstoodby consideringa fault in the middleof the faulted branch.
The rangeof values used forboth ZI and Z2 is between I and 10.For smaller
Larger values do not give realistic
valuesof the sagmagnitudebecomes very small.
that is proportional to the fault level at thepoint-ofsystems. One has to realize
supply. Thus,Z\ and Z2 indicate the variation in fault level for different points in the
system. A valueof 10 impliesthat there is at least faactor of six between the highest and
the lowest fault level.(Note that the twobranchesareoperatedin parallel.)Such a large
¥
z,
¥;
i
2.5
5
7.5
10
Relative impedance of non-faulted branch
Figure 4.34 Mostshallow sag for a fault in a
loop , as afunction of the impedanceof the
non-faulted branchfor various values of the
impedanceof the faulted branch.
163
Section 4.2 • Voltage Sag
Magnitude
Figure 4.35 Most shallow sag for a fault in a
loop, as a function of the impedance of the
faulted branch, for various values of the
impedance of thenon-faultedbranch.
2.5
5
~5
Relative impedance of faulted branch
10
range in fault level isratherunlikely in subtransmissionsystems, as it will lead to large
variationsin voltage due to loadvariations.
The generalconclusionfrom Figs. 4.34 and 4.35 is
t hat faults on a loop lead to
sags with amagnitudewell below 50%,irrespectiveof the voltage levels. Asmentioned
ZI = Z2. For these we can
before a parallel feeder is a special case of a loop: one in which
concludethat the most shallow sag has magnitudebetween
a
20% and 30% for most
systems.
4.2.4.3 Branches from Loops.W hen a load is fed from a loop, like the ones
discussed above, a fault on branch
a
away from that loop will also cause a sag. In
that case it is often possible to model the system as shown in Fig. 4.36. The feeder to
the fault does not necessarily have to be a single feeder, but could,represent
e.g.,
the
effective impedanceof another loop. The equivalentcircuit for the system in Fig.
4.36 is shown in Fig. 4.37: 21 is the source
impedanceat the mainsubtransmission
bus; 22 is the impedancebetweenthat bus and the bus from which the load is fed;
2 3 is the impedance between the bus from which the load is fed and the bus from
which the fault is fed; 24 and 25 are the
impedancesbetween thelatter bus and the
main subtransmissionbus and the fault, respectively. The voltage at the load bus is
found from
Vsag --
~~+~~+~~+~~
2 122 + 2,23 + 2\24 + 2 522 + 2 523 + 2 524 + 2 422 + 2 423
Subtransmission
Figure 4.36 System with b
aranchaway from
a loop.
(4.19)
164
Chapter4 • Voltage Sags-Characterization
Figure4.37 Equivalentcircuit for system
with a branchaway from aloop, as in Fig.
4.36.
Normally closed
Normally open
Fault
Load
Load
Figure4.38 Industrial system withbreakerat
intermediatevoltagelevel closed (left) and
open(right).
The sameexpressioncan be used to assess an
industrialsystem in which bussplitting is
used at anintermediatevoltagelevel. An exampleof the supplyconfigurationin a large
industrial network is shown in Fig. 4.38. In the leftexample,two transformersare
operatedin parallel. Typically both"transformersfeed into a different part of the substation bus, separatedthrougha circuit breaker.This enablesan uninterruptedsupply
after a bus fault. In thenetwork on the right the substationconsistsof two separate
busses,typically with a normally open breakerin between. In case the
b reakerat an
intermediatevoltage level is closed, the sag due to a fault at this
voltage level will be
experiencedfully by the load. In case thebreakeris open the sag will bemitigated
according to (4.19). On the onehand, the source impedancewill be 'Iess when the
breakeris open, leading to a deepersag at theintermediatevoltage level. But on the
other hand, the sag at theload bus will be less deepthan at the faulted intermediate
voltage level.
EXAMPLE Considerthe systemshown in Fig. 4.38 with thefollowing voltagesand
fault levels: 2500MVA at 66 kV, 500MVA at 11 kV (with the breakerclosed),and 50 MVA
at 660 V. When the breakerconnectingthe two 11 kV busses iso pen, the circuit diagram in
Fig. 4.37 can be used tocalculatethe sagmagnitudeat the 660 V bus for a fault at an 11kV
o f various impedancescan be calculated(all
feeder. From the fault levels given, the values
referred to I] kV):
ZI =0.048(2
Z2=4.75Q
Z3 = 4.36Q
2 4 = 0.388(2
Z5 = 0.3 Q/km x
£,
165
Section4.2 • Voltage Sag Magnitude
Normally open
Normally closed
Figure 4.39 Sag magnitude versus distance to
the fault, for an industrial system with and
without bus-splittingapplied to the
II kV bus.
I
2
3
4
Distanceto the faultinkilometers
5
with £, the distancebetweenthe 11 kV busand the fault, and a feederimpedanceof 0.3 Q/km.
When the 11 kV breakeris closed,the systemcan be treatedlike a radial systemwith a source
impedanceequal to Z.
Z4 and a feederimpedanceequalto Z5' A comparisonbetweenthese
two ways of systemoperationis given in Fig. 4.39.Bus-splitting(operatingthe systemwith the
11 kV breaker normally open) clearly limits the influence of 11 kV faults on the load. The
improvementis especiallylarge for nearby faults. For faults further away from the 11 kV substationthe effectbecomessmaller.But industrialmedium-voltagesystemsare seldomlargerthan
a few kilometers.We will come back to this and other ways of mitigating sagsthrough system
design and operationin Chapter7.
+!
4.2.4.4 Parallel Operation across Voltage Levels.
In many countries the subtransmissionsystem is not fed from thetransmissionsystem at onepoint but at a
number of points, resulting in a systemstructuresimilar to the one shown in Fig.
4.40. Thenumberof supply points for the subtransmissionsystem varies from country to country. The 275kV systems in the U.K. are fed like this; also the 130kV sys[23].
tem in Sweden and the 150kV system in Belgium
This typeof configurationcan betreatedlike a loop thatextends over two voltage
levels.For a fault within the loop we can apply (4.18), for a fault on a feeder away from
the loop (4.19) can be used. The
equationsremain the same
independentof the voltage
level at which the fault takes place. The only thing
that changes are theimpedance
values.
Transmission
Figure 4.40 Parallel operation of
transmission and subtransmission
systems.
Subtransmission
166
Chapter4 • VoltageSags-Characterization
4.2.5 Voltage Calculations In Meshed Systems
When the system becomes more
complicatedthan the examples discussed previously, closed expressions for the voltage during the sag get complicatedand
very
matrix calculations have proven to be
unfeasible to handle.F or meshed systems,
very efficient for computer-basedanalysis. Thecalculation of the voltagesduring a
fault is based on two principles from circuit theory:
Thevenin'ssuperpositiontheorem;
and the nodeimpedancematrix. Both are discussed in detail in many books on power
systems. Here we will only give a brief description.
• According to Thevenin'ssuperpositiontheorem voltages and
currentsin the
systemduring a sag are the sum
of two contributions:currentsand voltages
before the event, and
currentsand voltages due to the change in voltage at the
fault position. Currentsand voltages before the fault are due to generators
all
across the system.
Currentsand voltages due to the fault
originateat a voltage
source at the fault position. Allother voltage sources are considered
shortcircuited during the calculationof the latter contribution.
• The node impedance
matrix Z relates node voltages and node
currents:
(4.20)
V=ZI
with V the vector of (complex) node voltages and
I the vectorof (complex)
node currents.The node voltage is the voltage between a node and the reference node (typicallyground). The nodecurrent is equal to the sumof all
currents flowing toward a node. For most nodes the node
c urrent is zero
according to Kirchhoff's current law. The only exception aregenerator
nodes, where the node
current is the currentflowing from the generatorinto
the system.
Considera system withN nodes plus a reference node. The voltages before the
fault are denoted as viO). A short-circuit fault occurs at nodef. According to
Thevenin'ssuperpositiontheorem we can write the voltageduring the fault at any
node k as
(4.21)
where t:. Vk is the change in voltage at node
k due to the fault. Thislatterterm is due to
a voltage source -vjO) at the fault position. To calculate VAk all othervoltage sources
in the system areshort-circuited,so that nodef is the only node with anon-zeronode
current.After using theinformation, (4.20) becomes
l:1 Vk = Zkflf
At the fault position (k
=f) we know that l:1 Vf
= -
(4.22)
vjO)
so that
V(O)
If=_L
Zff
(4.23)
and
(4.24)
167
Section 4.2 • Voltage Sag
M agnitude
The pre-fault voltagesare normally close to unity, so that (4.24) can be approximated
by
(4.25)
The moment the node impedancematrix is known, calculatingsag magnitudes
becomesvery easy.The drawbackwith this methodis that the nodeimpedancematrix
needs to becalculated.This can be done through a recursive procedurewhere the
matrix is updatedfor each new branch added. Alternatively one can first calculate
the nodeadmittancematrix from the branchimpedances.T he nodeimpedancematrix
is the inverseof the nodeadmittancematrix.
EXAMPLE Considerthe circuit diagram shown in Fig. 4.41. Thiscircuit represents
a 275/400 kV system, with nodes 1 and representing400
2
kV substations;nodes 3, 4, and 5
representing275 kV substations;the branchesbetween 1 and 3 and betweenand
2 4 representing transformers(the latter two transformersin parallel). The impedancevalues indicated in
the figure are inpercentat a 100 MVA base.
Figure 4.41 Circuitdiagramrepresentationof
part of a 400/275kV system.
The node admittancematrix can be built easily from thebranch admittancesor impedances. Anoff-diagonalelement Yk1 of the nodeadmittancematrix is equalto minus the admittanceof the branchbetween nodesk and I. The elementis zero ifthereis no branchbetween these
two nodes. Thediagonalelement Ykk equalsthe sumof all admittancesof branchesto node k
including any branchbetweennode k and the reference node.
For the circuit in Fig. 4.41 this
calculationleads to the nodeadmittancematrix
y=
2.5719 -0.9091 -0.6211
0
0
-0.9091 4.5981
-1.25
0
0
-0.6211
2.0497
0
-1.4286
0
-1.25
0
0
2.7206 -1.4706
0
-1.4286 -1.4706 2.8992
0
(4.26)
The nodeimpedancematrix is obtainedby inverting the nodeadmittancematrix
z=
y- I =
0.5453
0.1771
0.3889
0.2548
0.3209
0.1771
0.3344
0.2439
0.3012
0.2730
0.3889
0.2439
1.2534
0.6144
0.9292
0.2548
0.3012
0.6144
0.9225
0.7707
0.3209
0.2730
0.9292
0.7707
1.1937
(4.27)
The voltage at node 5 due to a fault at node 2 is
=
Vs = 1 - Z52 = 1 _ 0.2730 0.1836
Z22
0.3344
(4.28)
Chapter4 • VoltageSags-Characterization
168
TABLE 4.7
Voltage Sagsin the System Shownin Fig. 4.41
Fault at Node
Voltage at Node
I
2
3
4
5
2
0
0.6753
0.2869
0.5327
0.4116
0.4704
0
0.2706
0.0993
0.1837
3
0.6897
0.8054
0
0.5098
0.2586
4
0.7238
0.6735
0.3340
0
0.1646
5
0.7312
0.7713
0.2216
0.3544
0
Table 4.7 gives the voltage at any node due to a fault atothernode.
any
We see, e.g.,
that for node
5 a fault at node 2 is more severe than a fault at node 1. This
understandable
is
as the source at
l.
node 2 isstrongerthan the source at node
4.3 VOLTAQE SAG DURATION
4.3.1 Fault-Clearing Time
We have seen inSection4.2 that the drop in voltageduring a sag is due to a
s hort
circuit being presentin the system. Themomentthe short-circuitfault is clearedby the
protection,the voltagecan return to its original value. Thedurationof a sag ismainly
determinedby the fault-clearingtime, but it may belongerthan the fault-clearingtime.
We will come back to thisfurther on in this section.
Generallyspeakingfaults in transmissionsystems arecleared fasterthanfaults in
distribution systems. Intransmissionsystems thecritical fault-clearing time is rather
small. Thus, fast protectionand fast circuit breakersare essential.Also transmission
and subtransmissionsystems arenormally operatedas a grid,requiring distanceprotection or differential protection,both of which are ratherfast. The principal form of
protectionin distribution systems isovercurrentprotection.This requiresoften some
time-gradingwhich increasesthe fault-clearingtime. An exceptionare systems in which
current-limiting fuses are used.Thesehave theability to clear afault within one halfcycle [6], [7].
An overview of the fault-clearing time of various protectivedevices is given in
reference [8].
•
•
•
•
•
•
•
current-limiting fuses: lessthan one cycle
expulsionfuses: 10-1000 ms
distancerelay with fast breaker:50-100ms
distancerelay in zone 1:100-200ms
distancerelay in zone 2:200-500ms
differential relay: 100-300ms
overcurrentrelay: 200-2000ms
Some typicalfault-clearingtimes atvariousvoltagelevels for a U.S. utility are given in.
reference [9].
Section4.3 I Voltage Sag Duration
Voltage Level
525 kV
345 kV
230 kV
115 kV
69 kV
34.5 kV
12.47 kV
169
Best Case
33 ms
50 ms
50 ms
83 ms
50 ms
100 ms
100 ms
Typical
50 ms
67 ms
83 ms
83 ms
83 ms
2 sec
2 sec
Worse Case
83 ms
100 ms
133 ms
167 ms
167 ms
3 sec
3 sec
From this list it becomesclear that the sag duration will be longer when a sag originates
at a lower voltage level. Many utilities operatetheir distribution feedersin such a way
that most faults are clearedwithin a few cycles. Such a way of operation was discussed
in detail in Chapter 3. But even for those feeders,a certain percentageof faults will lead
to long sags. The difference between the two ways of operation is discussedin more
detail in Section 7.1.3.
4.3.2 Magnitude-Duration Plots
Knowing the magnitude and duration of a voltage sag, it can be presentedby a
point in a magnitude-durationplane. This way of sag characterizationhas been shown
to be extremely useful for various types of studies. We will use it in forthcoming
chaptersto describeboth equipmentand systemperformance.Various types of magnitude-durationplots will be discussedin Section 6.2. The magnitude-durationplot will
also be usedin Chapter 6 to presentthe results of power quality surveys.An exampleof
a magnitude-durationplot is shown in Fig. 4.42. The numbersin Fig. 4.42 refer to the
following sag origins:
1.
2.
3.
4.
5.
6.
Transmissionsystem faults
Remote distribution system faults
Local distribution system faults
Starting of large motors
Short interruptions
Fuses
Consider the general system configuration shown in Fig. 4.43. A short-circuit
fault in the local distribution network will typically lead to a rather deep sag. This is
lOO%
80%
0%
Figure 4.42 Sags of different origin in a
magnitude-duration
plot.
0.1s
,,7---
Is
Duration
170
Chapter4 • VoltageSags-Characterization
Transmission network
Remote distribution
network
Local distribution
network
Figure 4.43 Generalstructureof power
system, withdistribution and transmission
networks.
Load
due to the limitedlength of distribution feeders. When the fault occurs in remote
a
distribution network, the sag will be much moreshallowdue to thetransformerimpeFor a fault in anydistribution network, the sag
dancebetween the fault and the pee.
durationmay be up to a few seconds.
Transmissionsystem faults are typically cleared within 50 to 100rns, thus
leading
to short-durationsags.Current-limitingfuses lead to· sag
d urationsof one cycle or less,
and rather deep sags if the fault is in the local
distribution or low-voltage network.
Faultsin remotenetworks,clearedby current-limitingfuses, lead toshortand shallow
m otor
sags,not indicatedin the figure. Finally the figurecontainsvoltage sags due to
starting,shallowand long duration(see Section 4.9)and shortinterruptions,deep and
long duration(seeChapter3).
4.3.3 Measurement of Sag Duration
Measurementof sag duration is much less trivialthan it might appearfrom the
previoussection. For a sag like in Fig. 4.1 it isobvious that the duration is about 2!
cycles.However, to come up with anautomaticway for a power quality monitor to
obtain the sagduration is no longer straightforward,A commonly useddefinition of
sag duration is the numberof cycles during which the rms voltage is below a given
threshold.This thresholdwill be somewhatdifferent for eachmonitor but typical values
are around 900/0. A power quality monitor will typically calculatethe rms value once
every cycle. This gives an
overestimationof the sagdurationasshownin Fig. 4.44. The
t
t f
Calculated
rmsvalues
X
X
Calculation
interval
,,,
I
I
~
I
Calculation instants
Figure 4.44 Estimationof sag duration by
power quality monitor for a two-cycle sag:
overestimationby one cycle(uppergraph);
correctestimation(lower graph).
171
Section 4.3 • Voltage Sag
D uration
normal situation is shown in theupper figure. The rmscalculation is performedat
regular instants in time and the voltage sag
starts somewhere in between two
of
thoseinstants.As there is nocorrelationbetween thecalculationinstantsand the sag
commencement,this is the most likelysituation.We seethat the rms value is low for
three samples in a row. The sag
durationaccordingto the monitor will be three cycles.
Here it is assumedthat the sag is deepenoughfor the intermediaterms value to be
below the threshold. For shallow sagsboth intermediatevalues might beabove the
thresholdand themonitor will record a one-cycle sag. The
bottom curve of Fig. 4.44
shows the raresituationwhere the sagcommencement
almostcoincides with one of the
instantson which the rms voltage iscalculated.In that case themonitor gives the
correctsagduration.
Calculating the rms voltage once a· cycle, it is obvious
that the resulting sag
durationwill be an integernumberof cycles.For a 2!-cycle sag thecomputedduration
will be either two or three cycles. But even when a sliding window is used
calculate
to
the rms voltage as faunction of time, anerroneoussagdurationmight result. To show
of
this possibleerror for a measuredsag, we haveplottedin Fig. 4.45 the half-cycle rms
the sag shown in Fig. 4.1,
togetherwith the absolutevalueof the measuredvoltage. The
"actual sag duration" obtainedfrom the suddendrop and rise in the voltage is 2.4
d urationwill be an overestimation.A 90%
cycles.For largethresholdsthe recorded sag
thresholdgives a 2.8 cycle sag
d uration,and 80% thresholda 2.5 cyclesduration. For
lower thresholdsthe recorded sagduration is an underestimation:a 60% threshold
gives a 2.1 cycledurationand a400/0 thresholda 2.0 cycleduration.In reality, thresholds this low willnot be used, but the same effect will be
obtainedwhen thedepthof the
sag is varied and the
thresholdis kept constant.The durationof deep sags will be overestimated,and thedurationof shallow onesunderestimated.
As the shortest-durationwindow for calculatingthe sagmagnitudeis one halfcycle, an error up to one half-cycle must be accepted. Several
methodshave been
suggested tomeasuresaginitiation and voltage recovery more
accurately.These methods also give a moreaccuratevalue of sag duration [134], [201], [202]. Using the
fundamentalvoltage componentresults in a similartransition betweenpre-sagand
during-sagvoltage, thus similarerrorsin sagduration. Using the half-cycle peak voltage will give a muchsharpertransition,as long as sag
initiation and voltage recovery
are close to voltagemaximum.Saginitiation and voltage recoverya roundthe voltage
zero-crossingwill give a smoothertransitionand a largeruncertainlyin sagduration.
1.2r - - - - r - - - - , - - - - - - - , - - - - - - - r - - - r - - - - - ,
I
'~I
I
"
,
I'
"
,'~
I
I,
Q..
'
.....
I
' I
'
,
I
II
I
,
II
I
,
I
"
I
I
I
~
::
04
I I
t
•
;:' I
I
"
r,
,f
"""
0.2 L .:
~,
Figure 4.45 Half-cycle rms voltage
together
with absolutevalue of the voltage(dashed
line) of the sag shown inFig. 4.1.
~
I
I
I
I
I
I
:
,I
I
,I
'
"
,I
•
"
"
"
"
,,
"
"
I
"
"
"
"
I
I
"
"
,,
, "I,
",I
"
"
"
"~
,
I
•
I
I
I
"
"
"
'
"
It'
,
I
\,
I
.1
:I~:
II
I
, I'
,,'
" : ,\'l\:II,:
oU
o
i
I'
I
: :: :
,
I
,, '
:':::
1 I
"
I I
," , ,1
I
I
I
,",
'.
/.
I
,
~ 0.6' :: :::
S
r
I
;'~
"
"
"
't
: :
=' 0.8 :: I
s::
",'\
,\
'~.
"
I
I
I
I
I
I
I
"
"
I
~
_---a....'_'-L..---L.~--...L--___L_:..____:._...:.J._l.___U.__---L-__:.J
1
234
Timein cycles
5
6
172
Chapter 4 • VoltageSags-Characterization
The above-mentionederror in sagduration is only significant forshort-duration
sags.For longer sags it does not really
matter. But for longer sags the so-called postfault sag will give a seriousuncertaintyin sagduration. When the fault is cleared the
voltage does not recover immediately. Some
of this effect can be seen in Fig. 4.3 and
Fig. 4.4. The rms voltage after the sag is slightly lower than before the sag. The effect
can be especially severe for sags due to three-phase faults.explanationfor
The
this
effect is as follows[17], [18]. Due to thedrop in voltage during the sag,induction
motors will slow down. The torqueproducedby an induction motor is proportional
to the squareof the voltage, so even rathersmall
a
drop in voltage can already
produce
a large drop intorque and thus in speed. The moment the fault is cleared and the
voltage comes back, the
induction motorsstart to draw a largecurrent:up to 10 times
their nominalcurrent.Immediately after the sag, the air-gap field will have to be built
up again . Inother words, theinduction motor behaves like ashort-circuitedtransformer. After the flux has come back into the air gap, motor
the can start re-accelerating
which also requires aratherlargecurrent. It is this post-faultinrush currentof induction motorswhich leads to an extended sag. The post-fault sag can last several seconds,
much longer than the actual sag.
Such apost-faultsag will causeuncertaintyin the sagdurationas obtainedby a
power quality monitor: different monitors can give different results. This is shown
schematically in Fig. 4.46. Assume
that monitor I has a setting as indicated, and
monitor 2 a slightly higher setting. Bothmonitors will record a sagduration much
longer than the fault-clearing time. The fault-clearing time canestimated
be
from the
duration of the deeppart of the sag. We see
t hat monitor 2 will record a significantly
longer duration than monitor 1.
A measured sag with a long
post-faultcomponentis shown in Fig. 4.47. The three
phases are shown in the same figurebetterindicate
to
thepost-faultvoltage sag. Note
that the sag isunbalancedduring the fault, but balanced after the fault.
The rms voltage versus time for the sag shown in Fig. 4.47
plottedin
is
Fig. 4.48.
We see a largedrop in voltage in two phases and a small one in third
the phase. The
fault-clearing time isabout four cycles; the fault leading to this sag
took place at
132kV, the voltages were measured at II kV. The sag
duration has been determined
as the timeduring which the rms voltage is belowcertainthreshold
a
. Figure 4.49 plots
of the phases only
this durationas a function of thethreshold,for the three phases. One
drops to 88% sothat any thresholdsetting below 88% will give zero sag
durationfor
that phase . The sag
durationobtainedfor the other two phases isa boutfour cycles for
thresholds below 90% , increasing fast for higher threshold settings.
Duration monitor 1
Duratio n monitor 2
Time
Figure 4.46 Error in sag
durationdue to
post-fault sag.
Section 4.3 •
173
Voltage SagDu ration
0.5
o
-0.5
- IL
Figure4.47 Measuredsagwith a clear postfault component(Data obtainedfrom
ScottishPower.)
o
6-
~----:'=-----;';=---'
5
15
10
Timein cycles
0.8
.S
ll>
;>
0.6
~
en
~ 0.4
0.2
5
Figure 4.48 The rms voltagesversustime for
the sagshownin Fig. 4.47.
15
10
Timein cycles
12
10
c:
0
'p
~eo
oS
'"
-e
~
e
.~
8
6
4
\l.l
2
Figure 4.49 Sagdurationversusthreshold
settingfor the threephasesof the sagshown
in Figs. 4.47 and 4.48.
0
0.8
0.85
0.9
Thresholdin pu
0.95
Chapter 4 • VoltageSags-Characterization
174
4.4 THREE-PHASE UNBALANCE
The analysisof sag magnitudepresentedin the previous sectionsconsidersonly one
phase.For example, the voltage divider model in Fig. 4.14 was
introducedfor threevalues. But
phase faults: theimpedancesused inthat figure are thepositive-sequence
most shortcircuits in powersystems are single phase or two phase.that
In case we need
to take all three phases into
accountor use thesymmetricalcomponenttheory. A good
and detaileddescriptionof the useof symmetricalcomponentstheory for the analysis
of non-symmetricalfaults is given in reference [24] and in several
otherbooks on power
of the theory to
system analysis and not
is repeatedhere. We will only use the results
calculatethe voltages in the three phases due to
non-symmetrical
a
s hort circuit.
For non-symmetricalfaults the voltage divider in Fig. 4.14 can still be used but it
has to be split into its threecomponents:a positive-sequencenetwork, a negativesequencenetwork, and azero-sequencenetwork. The threecomponentnetworksare
shown in Fig. 4.50, whereVI, V2, and Vo representpositive-, negative-, and zerosequence voltage, respectively, at the pee;
ZSb ZS2' and Zso are the source
impedance
values andZFt, ZF2, and ZFO the feederimpedancevalues in the threecomponents.The
three componentsof the fault current are denotedby I., 12 , and 10 , The positivesequence source denotedby
is
E. Thereis no source in the negative and zero-sequence
networks. The threecomponentnetworks have to beconnectedinto one equivalent
circuit at the faultposition.The connectionof the componentnetworksdependson the
fault type. For a three-phasefault all threenetworksare shortedat the fault position.
This leads to thestandardvoltage divider model for the positive sequence, and zero
voltage andcurrentfor the negative and zero sequences.
4.4.1 Single-Phase Faults
For a single-phasefault, the threenetworksshown in Fig. 4.50should be connected in series at the fault
position. The resulting circuit for a single-phase fault in
E
Figure 4.50 Positive- (top), negative- (center),
and zero-(bottom)sequence networks for the
voltage divider shown in Fig. 4.14.
175
Section 4.4 • Three-Phase
Unbalance
F~gure 4.51 Equivalent circuit for a singlephase fault.
phasea, isshownin Fig. 4.51.Ifwe againmakeE = 1, like in the single-phasemodelin
Fig. 4.14, thefollowing expressionsare obtainedfor the componentvoltagesat the pee:
VI
=
ZFI
(2F I
+ ZS2 + ZF2 + Zso+ ZFO
+ ZF2 + 2 FO) + (2s1 + ZS2 + 2 so)
(4.29)
(4.30)
(4.31)
The voltagesin the threephasesat the peeduring the fault are obtainedby transforming back from sequencedomain to phasedomain:
= VI + V2 + Vo
2
Vb = a VI + aV2 + Vo
Va
Vc = aVI
(4.32)
+ a2 V2 + Vo
For the faulted phasevoltage Va we get
Va =
ZFI
(2F t
+ Zn + ZFO
+ ZF2 + ZFO) + (ZSI + ZS2 + ZSO)
(4.33)
We can obtain the original voltagedivider equation(4.9) by defining 2 F = 2 F l +
ZF2 + ZFO and Zs = ZSl + ZS2 + Zso.Thus,the voltagedivider modelof Fig. 4.14and
(4.9) still holds for single-phasefaults. The condition thereby is that the resulting
voltage is the voltage in the faulted phase,and that the impedancevalues used are
the sumof the positive-,negative-,andzero-sequence
i mpedances.F rom(4.29) through
(4.32) wecan calculatethe voltagesin the non-faultedphases,which resultsinto the
following expressionsfor the three voltages:
Chapter4 • VoltageSags-Characterization
176
Va = 1 _
Vb
= a2 _
ZSI +ZS2
(2 F1 + 2 F2 + 2 FO)
+ZSO
+ (2S 1+ ZS2 + ZSO)
2
+ aZS2 + Zso
+ ZF2 + 2 FO) + (ZSI + ZS2 + ZSO)
2ZS2
aZSI + a
+ Zso
(2 F1 + ZF2 + ZFO) + (2 S 1+ ZS2 + 2 so)
a ZSI
(4.34)
(ZFl
V
c
=a _
Note that the expressionfor Va has been slightlyrewritten to explicitly obtain the
voltagedrop as aseparateterm.
Thesevoltagesare shownas aphasordiagramin Fig. 4.52. The voltagedrop in
the non-faultedphasesconsistsof three terms:
s ourceimpedance,a long
• a voltagedrop proportionalto the positive-sequence
the direction of the pre-fault voltage.
sourceimpedance,a long
• a voltagedrop proportionalto the negative-sequence
the direction of the pre-fault voltagein the other non-faultedphase.
s ourceimpedance,a long the
• a voltagedrop proportionalto the zero-sequence
direction of the pre-fault voltagein the faulted phase.
- a2ZS2
-aZsl
-zso \..\
\\Vc
\
Figure 4.52 Phaseto-groundvoltagesduring
a single-phasefault.
The voltage between the twonon-faultedphasesis
(4.35)
We seethat the changein this voltage is only due to thedifferencebetweenpositivesourceimpedances.As these two arenormally about
sequenceand negative-sequence
equal, the voltage betweenthe non-faultedphasesis normally not influenced by the
fault. Below we will simplify the expressions(4.34) and (4.35) for two cases:
• Positive-, negative-,and zero-sequence
s ourceimpedancesare equal.
• Positive- and negative-sequence
sourceand feeder impedancesare equal.
177
Section 4.4 • Three-Phase
Unbalance
4.4.1.1Solidly-GroundedSystems. In a solidly-grounded system, the source impedances in the three sequence
componentsare oftenaboutequal. The three voltage
drops in thenon-faultedphases now cancel, resulting in the following voltages during
the fault:
_ _
Va - 1
3(ZFl
ZSl
+ ZF2 + ZFO) + ZSI
2
Vb = a
(4.36)
Vc =a
The voltage in the faulted phase is the same as during a three-phase fault, the voltages
in the non-faultedphase are not affected.
4.4.1.2 Impedance-GroundedSystems. In a resistance or high-impedance
groundedsystem, the zero-sequence source impedance differs significantly from the
positive and negative-sequence source impedances. We can, however, assume
that the
latter two are equal. Also in systems where the source impedance consists for a large
part of line or cable impedances (e.g., in transmission systems) positive- and zero-sequence impedances can be significantly different. The resulting expressions for the
voltages at the pee during a single-phase fault are, when
ZSI = ZS2 and ZFl = ZF2:
Va = 1 _
Vb = a2 _
V
c
=a _
Zso+ 2Zs1
(2Z F1 + 2 FO)
(2ZFJ
+ (2ZS1 + ZSO)
ZSO - 22s 1
+ ZFQ) + (2Zs1 + Zso)
(4.37)
Zso - 2Zs1
(22F1 + ZFO) + (22s 1 + ZSO)
The voltagedrop in .the non-faultedphases onlycontainsa zero-sequence
component
(it is the same inboth phases). We will see later
that the zero-sequence
componentof
the voltage is rarely ofimportancefor the voltage sag as experienced equipment
at
terminals. Sags at the same voltage level asequipment
the
terminals are rare.During the
transfer of the sag down to lower voltage levels, the
transformersnormally block the
zero-sequence
componentof the voltage. Even if the fault occurs at the same voltage
level as the equipment terminals, the
equipmentis normally connected in delta so it will
not notice the zero-sequence
componentof the voltage. Thus the voltage
drop in the
non-faultedphases is not ofimportancefrom an equipmentpoint of view. We can
therefore add a zero-sequence voltage to (4.37) such
that the voltagedrop in the nonfaulted phases disappears. The resulting expressions are
va, -- Va+
Zso - ZSl
_ 1_
(22F 1 + ZFO) + (2Z S1 + 2 so)
(2Z F l
n= Vb + (2Z
ZSO - ZSl
F 1 + 2 FO) + (22s 1 + 2 so)
= a2
3ZS1
+ 2 FO) + (22s 1 + ZSO)
(4 38)
·
ZSO -ZSI
, V
vc=
c+ (2Z + ZFO) + (2Z + Zso) =a
Ft
S1
The expression for the voltage in the faulted phase is somewhat rewritten, to enable a
comparisonwith (4.36):
(4.39)
178
Chapter4 • VoltageSags-Characterization
E
Neutral
point
Figure 4.53Three-phasevoltage divider
model.
The denominatorcontainsan additional term !(Zso - 2 S1) comparedto (4.36). This
can beinterpretedas anadditionalimpedancebetween the pee
a nd the fault. When this
> ZSI, the sag becomes more shallow. In resisimpedanceis positive, thus when Zso
tance and reactance-groundedsystems, Zso» ZSl' so that even a terminal fault,
ZFI + ZF2 + ZFO = 0, will lead to ashallow sag.
Note that in solidly-groundedsystems, thezero-sequence
sourceimpedancemay
be lessthan the positive-sequence
one, Zso < ZSl' so that the additionalimpedanceis
negative.For nearbyfaults, we will thus obtain a negativevoltage
All this might look like a mathematicaltrick to get rid of the voltagedrop in the
non-faultedphases.T hereis, however, some physical significance to this. To show this,
the three-phasevoltage divider is drawn in a commonly used way [24] in Fig. 4.53.
From this model we cancalculatethe phase-to-neutralvoltages at the pee; with
E= 1
the calculationresultsinto
V-I _
3ZS 1
an (2Z F 1 + ZFO) + (2ZS 1 + 2 so)
(4.40)
2
Vbn = a
V;.
Vcn
=a
The correspondence
between (4.40)a nd (4.38) isobvious. The voltages in (4.38)
thus correspondto thephase-to-neutralvoltages.Note that the "neutral" in Fig. 4.53 is
not a physicalneutralbut a kind of mathematicalneutral.In resistance-or high-impedancegroundedsystems the physical
neutral(Le., thestarpoint of the transformer)is a
good approximationof this "mathematicaln eutral."The expressionsderived not only
hold for resistance-grounded
systems, but for each system in which we can assume
positive- andnegative-sequence
impedancesequal.
EXAMPLE Consider again the system shown in Fig. 4.21, and assume that a singlephase fault occurs on one of the 132 kV feeders. The 132 kV system is solidly grounded, therefore the positive- and zero-sequence source impedances are similar. For the feeders, the zerosequence impedance is about twice the positive- and negative- sequence impedance. Positiveand negative-sequence impedance are assumed equal.
ZSI = ZS2 = 0.09+j2.86%
Zso = 0.047+ j2.75°A>
ZFt = ZF2 = 0.101 + jO.257°A>/km
ZFO = 0.23+ jO.65°A>/km
179
Section 4.4 • Three-Phase Unbalance
0.8
Single-phasefault
Three-phasefault
Figure 4.54 Voltage in the faulted phase for
single-phase and three-phase faults on a 132
kV feeder in Fig. 4.21.
10
20
30
40
50
Distanceto the fault inkilometers
By using the above-given equations, the voltages in the three phases have been calculated for
single-phase as well as for three-phase faults. The results for the faulted phase are shown in Fig.
4.54. The difference is mainly due to the difference in feeder impedance. Note that it is assumed
here that the feeders are at least 50km long, where they are in reality only 2 km long. The zerosequence feeder impedance increases faster than the positive-sequenceimpedance, with increasing
distance to the fault. Therefore single-phase faults lead to slightly smaller voltage drops than
three-phase faults.
As we saw from the equations above, it is the average of the three sequence
impedances which determines the voltage drop due to single-phase faults. The voltages in the nonfaulted phases showed only a very small change due to the single-phase fault.
EXAMPLE The voltages due to single-phase faults have been calculated for the II
kV system in Fig.4.21. As this system is resistance grounded, the zero-sequence source impe-
dance is considerably larger than the positive-sequence impedance.
ZSI
= ZS2 = 4.94+ j65.9
%
Zso= 787+ j220%
= 9.7 +j26%/km
ZFI
=
ZFO
= 18.4+ jI12 % / k m
ZF2
Note the large zero-sequence source impedance, especially its resistive part. The voltage in the
faulted phase for three-phase and single-phase faults is shown in Fig. 4.55 as a function of the
distance to the fault. The larger source impedance for single-phase faults more than compensates
the larger feeder impedance, which makes that single-phase faults cause deeper sags than threephase faults.
In a solidly-groundedsystem the voltage in a
non-faultedphase staysabout the
sameduring a single-phasefault. In a resistance-grounded
system the voltage in the
and 4.57. Figure 4.56
non-faultedphases increases. This effect is shown in Figs. 4.56
p ath of the
shows the voltagemagnitudeversusdistanceto the fault and Fig. 4.57 the
voltages in the complex plane. The circles and arrowsindicatethe
the
complex voltages
during normaloperation.The curvesindicate the path of the complex voltages with
varyingdistanceto the fault. Where thefaulted phase shows d
arop in voltage, the nonof
faulted phases show a large increase in voltage, for one phaseincreasing170%
even
the nominal voltage. From Fig. 4.57 we seethat all three voltages are shifted over a
Chapter 4 • VoltageS ags-Characterization
180
0.8
Three-phasefault
[
.S
Single-phasefault
.s 0.6
·1
~ 0.4
f
tI)
0.2
Figure 4.55 Voltage in the faulted phase for
20 single-phaseand three-phase faults on11an
kV feeder in Fig. 4.21.
5
10
15
Distanceto the fault inkilometers
1.8,..-----r------.,..-------r------,
1.6
~ 1.4
.S 1.2
t
Non-faultedphases
E 0.8
«)
I
0.6
Faultedphase
'0
:> 0.4
0.2
0
0
5
10
15
Distanceto the fault in
kilometers
Figure 4.56 Voltage in the faulted and nonfaulted phases for a single-phase fault on an
20 11 kV feeder in Fig. 4.21, as a function of the
distance to the fault.
1.5...---....---........-----.----r----r----r-------.
,
~,.
«)
~
<a
1\ ,
\
\
\
0.5
\
i
~
·st
,
\
\~
0
I
.>
I
,
I
E
I
....-0.5
/
-1 '---___'___ _- ' - - _ - . . I_ _---'-_ _- ' - - ' _ - - - ' _ - - - - J
-1.5
-1
-0.5
0
0.5
Realpart of voltage
Figure 4.57 Complex voltages due to a fault
on an 11 kV feeder in Fig. 4.21.
181
Section 4.4 • Three-PhaseUnbalance
similar distance in the complex plane. The effect of this
commonshift (a zero-sequence
component)is that the phase-to-phase
voltages do not change much.
The phase-to-phase
voltages have been calculated from the complex phase voltagesby using the following expressions:
v _ Va -
Vb
.J3
ab -
Vb - Vc
(4.41)
= .J3
VIn·
_ V - Va
Vca - c.J3
The factor .J3 is needed to ensure
that the pre-fault phase-to-phase
voltages are 1 pu.
The resulting voltagemagnitudesare shown in Fig. 4.58: note the difference in vertical
scalecomparedto the previous figures. We see
that the phase-to-phase
voltages are not
much influencedby single-phase faults. The lowest voltage
magnitudeis 89°/0, the
highest 101°/0.
Figure 4.59comparesphase-to-ground
voltage, according to (4.37), and
phase-toneutralvoltage,accordingto (4.40). We see
t hat the drop in phase-to-neutralvoltage is
1.05r - - - - - , - - - - - - r - - - - - - . - - - - - - - ,
a
.8
QJ
~
.~ 0.95
e
j
~
Figure 4.58Phase-to-phase
voltages due to a
single-phase fault on an II kV feeder in Fig.
4.21, as a function of the distance to the fault.
0.9
0.85
0
5
10
15
20
Distanceto the fault inkilometers
1'--
-
0.8
a
.8
~
0.6
.~
t 0.4
e
/
/
I
I
f
I
,,
I
(/)
,,
0.2
,
,
,
Figure 4.59Phase-to-ground(dashed) and
phase-to-neutral
(solid) voltages due to singlephase faults on an II kV feeder in Fig. 4.21.
I
5
10
15
Distanceto the fault inkilometers
20
182
Chapter4 • VoltageSags-Characterization
very small. As explained before, this is due to the large zero-sequence source impedance. Also notethat the lowestphase-to-neutral
voltage occurs for anon-zerodistance
to the fault.
4.4.2 Phase-to-Phas. Faults
For a phase-to-phase
fault the positive- and negative-sequence
networksare connected in parallel, as shown in Fig. 4.60. The zero-sequence voltages
currentsare
and
zero for aphase-to-phase
fault.
E
Figure 4.60 Equivalent circuit for a phase-tophase fault.
The sequence voltages at the pee are
=E-E
VI
ZSI
(ZSl
V 2-
+ 2 S2) + (2£1 + 2£2)
ZS2
(4.42)
(ZSI + ZS2)+ (Z£I
+ Z£2)
Vo =0
The phase voltages can be found from (4.42) by using (4.32). This results in the following expressions, again with
E = 1:
Va = 1 _
ZSI - ZS2
(ZSl
V
2
b
V
C
=a =a _
+ ZS2)+ (2 F1 + 2£2)
a
2ZS1
- aZS2
(2s1 + ZS2)+ (2F1 + 2 F2)
(4.43)
2ZS2
aZSI - a
(ZSI + ZS2) + (2 F t
+ 2£2)
In thecalculationof the componentvoltages andcurrents,it has been assumed
that the
fault is between the phases
bandc. Thus a is thenon-faultedphase, andbandc are the
183
Section 4.4 • Three-Phase
Unbalance
faulted phases.F rom (4.43) we seethat the voltagedrop in the non-faulted phase
depends onthe difference between the positive and negative-sequence source impedances. As these are
normally equal, the voltage in the
non-faultedphase will not be
influenced by the phase-to-phasefault. Under the assumption, ZSI = ZS2 (4.43)
becomes
Va = 1
Vb
= a2 _
2
(a - a)Zsl
22s 1 + 2ZF1
(4.44)
(a2 - a)Zsl
Vc=a+-----
2Zs 1 +2ZF 1
We seethat the voltagedrop in the faulted phases is equal magnitude
in
2Z z;~z but
opposite in direction. The direction in which the two phase voltages
drop iss~loJg the
Vb - VC •
pre-fault phase-to-phase
voltage between the faulted phases,
From (4.43) we can derive the following expression for the voltage between the
faulted phases
Vb - Vc
=
(ZSI
ZFI + ZF2
(a2 + ZS2) + (ZFI + ZF2)
a)
(4.45)
When we realizethat (a2 - a) is the pre-fault voltage between the two faulted phases,
the resemblance with the single-phase voltage divider of Fig. 4.14 and (4.9) becomes
immediately clear.t he same expressions as for the
three-phasefault can be used, but
for the voltages between the faulted phases; the impedances in the expression are the
sum of positive and negative sequence values.
faults on one of the 33 kV feeders in the system
EXAMPLE Considerphase-to-phase
shown in Fig. 4.21. The impedance values needed to calculate the voltages
during a phase-tophase fault are as follows:
ZSI
ZFl
= ZS2 = 1.23+j18.3%
= ZF2 = 1.435+ j3.l02
%/km
The resulting complex voltages are shown in Fig. 4.61. The circles and the arrows indicate the prefault voltages; the cross indicates the voltages in the faulted phases for a fault at the 33 kV bus.
, ,,
I',' .
u
,
,
0.5
,,
,,
~
~
o
i
,,
,
\------------~~_:.o
,
..
,,
0
.i
I
I
I
~-0.5
I
I
I
I
•
I
I,
1///
-1 "--------'---_ _--'---_ _
-1
-0.5
0
0.5
....L--
Figure 4.61Complex voltages due to a phaseto-phase fault (solid line).
Realpart of voltage
-..J
Chapter4 • VoltageSags-Characterization
184
We see how the voltages in the two faulted phases move
towardeach other. Thedeviationof their
path from astraight line is due to the difference in
X /R ratio between source and feeder impedance. This is a subject to be discussedfurther
in
detail in Section 4.5.
4.4.3 Two-Phase-to-Ground Faults
Single-phaseand phase-to-phase
faults have beendiscussedin the two previous
sections.The only asymmetricalf ault type remainingis the two-phase-to-groundfault.
For a two-phase-to-ground
f ault the threesequencenetworksare connectedin parallel,
as shown in Fig. 4.62. It isagain possibleto calculatecomponentvoltagesand from
thesecalculatevoltagesin the threephasesin the sameway asdonefor the single-phase
and phase-to-phase
faults.
The sequencevoltagesat the pee for afault betweenphasesbandc and ground
are given by thefollowing expressions:
VI
= 1 _ ZSI (Zso + ZFO + ZS2+ ZF2)
D
V = ZS2(ZSO+ ZFO)
2
D·
V
ZSO(ZS2+ ZF2)
o
D
(4.46)
=
with
(4.47)
From (4.46) it is possibleto calculatethe phase-to-groundvoltagesin the threephases
V-I
+
a-
V h-
V _
l'
2
a
(2 S2 - 2 S1)(2so + 2 FO)
D
+
-a+
(aZS2-
~ZSI)ZO
D
+
(2so - 2 SI)(2s 2 + 2 F2)
D
2ZSI)Z2
(ZSO - a
+
D
(4.48)
2ZS2
(a
- aZsl)Zo (Zso - aZSI)Z2
D
+
D
E
Figure 4.62 Equivalent circuit for a twophase-to-groundfault.
18S
Section 4.4 • Three-PhaseUnbalance
There are two effects which causechangein
a
voltage in thenon-faultedphase(Va): the
difference between the positive- and the
negative-sequence
sourceimpedance;and the
difference between the positive- and the zero-sequence source
impedance
. For both
effects the non-faulted phase voltagedrops when the positive-sequenceimpedance
increases. Negative- and
positive-sequenceimpedanceare normally rather close, so
that the second term in (4.48) may be neglected. The
third term, which dependson
the difference between zero- and
positive-sequencesource impedance,could cause a
seriouschangein voltage. As thezero-sequence
sourceimpedanceis often largerthan
the positive-sequenceone, we expect a rise in voltage in the
non-faultedphase. Like
with single-phasefaults we caneliminate this term by consideringphase-to-neutral
voltagesinsteadof phase-to-groundvoltages .
Looking at the voltages in the
faulted phases and realizing
that ZSI is close toZS2
we seethat the second term is a voltage
drop in the directionof the otherfaulted phase;
2
(a - a ) is the pre-fault voltage between the faulted phases
For
. Zso = ZSI the third
term in (4.48) is a voltagedrop towards the non-faultedphase pre-fault voltage, for
Zso « ZSI the third term is adrop along the positive real axis, as shown in Fig. 4.63.
The voltagedrop accordingto A in Fig. 4.63 is the same
d rop as for aphase-to-phase
fault. The ground-connectioncauses anadditional drop in the voltage in the two
t hat all
faulted phases,somewherein betweendirectionsBand C. It is assumed here
impedanceshave the sameX/R ratio.
. · \· · B ~
A~
~
-.
-. B··
.
Figure 4.63 Voltage drops in the faulted
phase during atwo-phase-to-groundfault. A:
second term in (4.48); B: third term for
ZSI = Zso;C: third term for ZSI « Zso.
As said before, positive- andnegative-sequence
impedancesare normally very
close. In that case we can simplify the expressions substituting
by
ZSI = ZS2 and
ZFt ZF2' But when we are onlyinterestedin phase-to-neutralvoltages it is easier
to use thethree-phasevoltage divider modelintroducedin Fig. 4.53 for single-phase
faults . For two-phase-to-groundfaults theequivalentcircuit is redrawnin Fig. 4.64.
Without any further calculation we can see from Fig. 4.64
t hat the phase-toneutralvoltage in thenon-faultedphase is not influenced by the
two-phase-to-ground
fault. The phase-to-neutralvoltage at the faultpoint, VFN, is found from applying
Kirchhoff's current law to the fault point:
=
2
a - V FIV
------~ +
ZSI+ZFt
a - VFN
V
FN
.
=J
I
ZSJ-ZFI 3(Zso-Zsd+
(4.49)
3(ZFO-ZFI)
Solving (4.49) leads to the following expression for the voltage at the fault
point:
V
FN
= _ (Zso + ZFO) -
(ZSI + ZFt)
2(Zso + ZFO) + (ZSI + ZFI)
(4.50)
186
Chapter 4 • VoltageSags-Characterization
E
4----------- VF
-:
Figure 4.64 Three-phasevoltagedivider
model for a two-phase-to-groundfault.
If zero-sequenceand positive-sequenceimpedances are equal, Zso = ZSI and
2 FO = 2 F J, we find that
(4.51)
If the zero-sequence
i mpedancebecomeslarge, like in aresistance-grounded
system,the
fault-point voltageis
r
1
2
VF~ =--
(4.52)
The latter expressioncorrespondsto the expressionobtainedfor phase-to-phase
faults.
This isratherobviousif we realizethat a largezero-sequence
i mpedanceimplies that the
fault currentthrough the earthreturn is very small.Thus, the presenceof a connection
with earthduring the fault does not influence thevoltages.
Pathof Vcn
Pathof Vbn
Figure 4.6~ Phase-to-neutralvoltagesin the
faulted phasesfor a two-phase-to-ground
fault.
187
Section 4.4 • Three-PhaseUnbalance
The intermediatecase, whereZSI < Zso <
somewhere in between these two extremes:
00,
gives a voltage at the faultpoint
1
2
(4.53)
- - < VFN < 0
This voltage and theresultingvoltages at the pee can be
obtainedfrom Fig. 4.65. The
the former for
voltage at the faultp oint is locatedbetween the origin and the
point
equal positive- negative-, and
zero-sequence
impedances,the latter for very large zerosequenceimpedance.The voltage at the pee for a faulted phasesomewhere
is
between
the voltage at the faultp oint and thepre-fault voltage in that phase. This knowledge
o f three-phaseunbalancedsags.For calculating
will later be used for the classification
sagmagnitudesthis constructionis not of practicaluse, as thefault-to-neutralvoltage
VFN dependson the fault position.
-!:
4.4.4 Seven Types of Three-Phase Unbalanced Sags
The voltage sags due to the
various types of faults have been discussed in the
previous sections:three-phasefaults in Section 4.2, single-phase faults in Section 4.4.1,
phase-to-phase
faults in Section 4.4.2, and finally
two-phase-to-ground
faults in Section
4.4.3.For each typeof fault, expressionshave been derived for the voltages at the pee.
But as alreadymentioned,this voltage is not equal to the voltage at the
equipment
terminals. Equipmentis normally connectedat a lower voltage levelthan the level at
which the fault occurs. The voltages at the
equipmentterminals, therefore,not only
but also on the windingconnectionof the transfordependon the voltages at the pee
and the equipmentterminals. The voltages at theequipment
mers between the pee
terminalsfurther dependon the loadconnection.Three-phaseload is normally connected in delta butstar-connectionis also used.Single-phaseload isnormallyconnected
in star (i.e., between onephaseand neutral) but sometimes indelta (between two
phases).Note that we considerhere the voltage sag as experienced at terminals
the
of end-userequipment,not the voltage asmeasuredby monitoring equipment.The
latter is typically locatedat distribution or even attransmissionlevel.
In this section we will derive a classification for
three-phaseunbalancedvoltage
sags, based on the following
assumptions:
• Positive- andnegative-sequence
impedancesare identical.
• The zero-sequence
c omponentof the voltage does notpropagatedown to the
equipmentterminals,so that we can considerphase-to-neutralvoltages.
• Load currents,before, during, and after the fault, can be neglected.
4.4.4.1 Single-Phase Faults.The phase-to-neutralvoltages due to a singlephase-to-groundfault are, underthe assumptionsmentioned,
Va = V
1
I
Vb = ----j~
2 2
1 I
V = --+-J'~
c
2 2
(4.54)
188
Chapter4 • VoltageSags-Characterization
>------.
Va
Figure 4.66Phase-to-neutralvoltages before
(dashed line) and during (solid line) a phaseto-groundfault.
The resultingphasordiagramis shown in Fig. 4.66. If the load is connected in star,
these are the voltages at the
equipmentterminals. If the load is connected in delta, the
equipmentterminal voltages are the
phase-to-phase
voltages. These can be
obtained
from (4.54) by the followingtransformation:
(4.55)
This transformationwill be an important part of the classification. The factor.J3 is
of the pu values, sothat the normal operatingvoltage
aimed at changing the base
remains at 1000/0. The 90° rotation by using a factorj aims at keeping the axis of
symmetry of the sag along the real axis. We will normally omit the primes from
(4.55). Applying transformation(4.55) results in the following expression for the
three-phaseunbalancedvoltage sag experienced bydelta-connected
a
load, due to a
single-phase fault:
(4.56)
The phasordiagramfor the equipmentterminal voltages is shown in Fig. 4.67: two
voltages show adrop in magnitudeand change in phase angle; the third voltage is not
influenced at all.Delta-connectedequipmentexperiences a sag in two phases due to a
single-phase fault.
189
Section 4.4 • Three-PhaseUnbalance
\
\.
\
\ ...\ ..
\
Figure 4.67 Phase-to -phase voltages before
(dashed line) and during (solid line) a phaseto-ground fault.
4.4.4.2 Phase-to-Phase Faults.For a phase-to-phasefault the voltages in the
two faulted phases move
toward each other. The expressions for the
phase-to-neutral
voltagesduring a phase-to-phase
fault read as follows:
Va = I
Vb
= _!_! VjJ3
V
= _!+!
V)'J3
2 2
c
2
2
(4.57)
Like before, (4.55) can be used to calculate the voltages experienced by a phase-tophase connected load, resulting in
Va = V
Vb =
_!2 V - !2jJ3
(4.58)
1
1
Vc = -2 V +-j"J3
2
The correspondingphasordiagramsare shown in Figs. 4.68 and 4.69. Due to a phaseto-phasefault a star-connectedload experiences adrop in two phases, a delta-
)-- - - - - - . va
i/
//
Figure 4.68Phase-to-neutralvoltages before
(dashed line) and during (solid line) a phaseto-phase fault.
,.<
190
Chapter4 • VoltageSags-Characterization
"-\ Vc
..•.\\-,
} - - - - . . .............................•Va
'/ Vb
Figure 4.69Phase-to-phase
voltages before
(dashed line) andduring (solid line) a phaseto-phase fault.
connectedload experiences ad rop in three phases.F or the star-connectedload the
maximum drop is 50%, for V = O. But for thedelta-connectedload one phase could
drop all the way down to zero. Theconclusion that load could therefore best be
connected.in star is wrong, however .M ost sags do notoriginateat the same voltage
level as theequipment terminals. We will see later that the sag at theequipment
terminals could beeither of the two types shown in Figs. 4.68 and 4.69,
depending
on the transformerwinding connections.
4.4.4.3 Transformer Winding Connections.
Transformerscome with manydifferent winding connections, but a classification into only three types is sufficient to
explain the transfer of three-phaseunbalancedsags from one voltage level to another.
I. Transformersthat do notchangeanything to the voltages F
. or this typeof
transformerthe secondary-sidevoltages (in pu) are equal to the
primary-side
voltages (in pu). The only type of
transformerfor which this holds is thestarstar connectedone with both star points grounded.
2. Transformersthat remove the zero-sequence voltage. The voltages on the
secondaryside are equal to the voltages on the
primary side minus the
zero-sequencecomponent.Examplesof this transformertype are the starstar connectedtransformerwith one or both star points not grounded,and
the delta-deltaconnectedtransformer. The delta-zigzag (Dz)transformeralso
fits into this category.
For thesetransformerseach
3. Transformersthat swap line and phase voltages.
secondary-sidevoltage equals the difference between two
primary-sidevoltages. Examples are the
delta-star(Dy) and thestar-delta(Yd) transformeras
well as thestar-zigzag(Yz) transformer.
Within each of these threecategoriesthere will be transformerswith different clock
number(e.g., Yd I and YdII) leading to a different phase shift between
primary- and
secondary-sidevoltages. This difference is not of any
importancefor the voltage sags as
experienced by theequipment.All that mattersis the change between the
pre-fault
voltages and theduring-fault voltages, inmagnitudeand in phase-angle
. The whole
phasordiagram, with pre-fault and during-fault phasors, can berotatedwithout any
influence on theequipment.Such arotation can be seen as a shift in the zero
point on
191
Section 4.4 • Three-PhaseUnbalance
the time axis which of course has no influence on equipment behavior. The three
transformertypes can be defined mathematicallyby meansof the following transformation matrices:
T1 =
T2 =
[1 0
;]
0 1
o 0
-1]
~ [-~
-1
2 -1
-1
2
-1
3
T
=
(4.59)
~[-:
1
0
-1
-i]
(4.60)
(4.61)
Equation(4.59) isstraightforward:matrix T 1 is the unity matrix. Equation(4.60)
c omponento f the voltage.The matrix T2 can be understood
removesthe zero-sequence
easily by realizing that the zero-sequence
v oltageequals!(Va + Vb + Vc) ' Matrix T3 in
(4.61) describesexactly the sametransformationas expression(4.55). The additional
advantageof the 90°rotation is that twice applying matrix T3 gives thesameresultsas
once applying matrix T2• Thus, Tf = T2 ; in engineeringterms: two Dy transformersin
cascadehavethe sameeffect on the voltagesag asone Dd transformer.
4.4.3.4 Transfero f Voltage Sags across Transformers.
The three types of transfaults. To
formers can be applied to the sags due tosingle-phaseand phase-to-phase
get an overview of the resulting sags, thedifferent combinationswill be systematically treatedbelow.
• Single-phasefault, star-connectedload, no transformer.
This casehas been discussedbefore,resultingin (4.54) and Fig. 4.66. We will
1 gives thesameresultsof course.
refer to this sag as sag X Transformertype
1.
• Single-phasefault, delta-connectedload, no transformer.
The voltagesag for this case is given in (4.56)
and shownin Fig. 4.67.This sag
will be referredto as sag X2.
• Single-phasefault, star-connectedload, transformertype 2.
Transformertype 2 removesthe zero-sequence
c omponentof the voltage.The
zero-sequence
c omponentof the phasevoltagesdue to a single-phasefault is
found from (4.54) to beequalto !(V - 1). This gives thefollowing expressions
for the voltages:
1
Va =
2
3+3 V
Vb = -
1.
-61 - -31V - -]v'3
2
1
1
(4.62)
1.
Vc = ----V+-jv'3
6 3
2
This looks like a new typeof sag, but we will seelater that it is identical to the
one experiencedby a delta-connectedload during a phase-to-phase
fault. But
for now it will be referredto as sag X3.
192
Chapter 4 • VoltageSags-Characterization
• Single-phasefault, delta-connected load, transformer type 2.
The phase-to-phasevoltages experienced by delta-connected
a
load do not
contain any zero-sequencecomponent.Thus transformer type 2 does not
have any influence on the sag voltages. The sag is thus still of type X2.
• Single-phasefault, star-connectedload, transformer type 3.
Transformertype 3 changes phase voltages into line voltages. Thus
star-connected load onsecondaryside experiences the same sagdelta-connected
as
load
on primary side. In this casethat is sag X2.
• Single-phasefault, delta-connected load, transformer type 3.
There are now twotransformations:from star- todelta-connectedload, and
from primary to secondarysideof the transformer.Eachof thesetransformations can be describedthrough matrix T3 defined in (4.61). Two of those
T2 • Thus,
transformationsin cascade have the same effecttransformation
as
the sag experienced by this
delta-connectedload is the same as by the starX3~
connectedload behind atransformerof type 2; thus, sag type
• Phase-to-phase
fault, star-connectedload, no transformer.
This case wastreatedbefore resulting in (4.57) and Fig. 4.68. This will
be sag
type X4.
• Phase-to-phase
fault, delta-connected load, no transformer.
The expression for the sag voltages reads as (4.58) and is shown in Fig. 4.69.
This type will be referred to as X5.
• Phase-to-phase
fault, star-connectedload, transformer type 2.
As phase-to-phase
faults do not result in any zero-sequence voltage,
transformer type 2 (which removes the zero-sequence voltage) does not have any effect.
The sag thus remains
of type X4.
• Phase-to-phase
fault, delta-connected load, transformer type 2.
Like before, the sag is still of type X5.
• Phase-to-phase
fault, star-connectedload, transformer type 3.
Star-connectedload on secondaryside of transformertype 2 experiences the
same sag as
delta-connectedload on primary side. This results in type X5.
• Phase-to-phase
fault, delta-connected load, transformer type 3.
This gives again two identicaltransformationsT3 in cascade, resulting in one
transformationT2 • But that one only removes the zero-sequence
component
and has thus no influence on sags duephase-to-phase
to
faults. The result is,
thus, again X4.
The effectof a secondtransformeron sags Xl throughX5 is shown in Table 4.8. These
results can beobtainedby following the samereasoningas above. It becomes clear
that
TABLE 4.8 FurtherPropagationof Sags
TransformerType
Sag Type
2
3
X2
Xl
Xl
X3
X2
X2
X2
X3
X3
X3
X3
X4
X4
X4
X5
X5
X5
X2
X5
X4
193
Section 4.4 • Three-PhaseUnbalance
the numberof combinationsis limited: at mostfive different sag types arepossibledue
faults.
to single-phaseand phase-to-phase
4.4.4.5 The Basic Types
ofSags. We sawthat single-phasefaults lead to three
types of sags,designatedsag Xl , sag X2, and sag X3. Phase-to-phase
faults lead to
sag X4and sag X5. We sawalreadyfrom the phasordiagramsin Figs. 4.67and 4.68
that single-phaseand phase-to-phase
faults lead tosimilar sags.The sag voltagesfor
sag type X2 are
Va = 1
-~(!+!
V)1J3
2
6 3
Vb =
(4.63)
Vc = .i,
2 (~+~
6 3 V)'iJ3
J
For sag type X4 thevoltagesare
Va = 1
Vb
1 1
= ---VjJ3
2 2
V
= _!+!
V)·J3
2 2
c
0
(4.64)
Comparingthese two setso f equationsshowsthat (4.63) can beobtainedby replacing
V in (4.64)by! + j V. Ifwe define themagnitudeof sag X4 asV, then sag X2 is a sag
of
type X4 with magnitude!+ j V.
In the same way we can
c omparesag X3:
I 2
Va =3+3 V
Vb =
V
c
-~ -~ V -~jJ3
6
3
2
(4.65)
= -~-~
V+~joJ3
6 3
2
and sag X5:
Va = V
I
1.
Vb = - - V - - j J 3
2
2
V
c
(4.66)
1
=--21 V +_joJ3
2
t
Again we obtain (4.65) by replacing V in (4.66) by + ~ V. The result isthat only
three types remain:X l , X4, and X5. A fourth type of sag is the sag due to threephasefaults, with all three voltagesdown the sameamount. The resulting classification is shownin Table 4.9 in equationform and in Fig. 4.70 in phasorform. All sags
in Fig. 4.70 have amagnitudeof 500/0. From the discussionabout sags due to singlephaseand phase-to-phasefaults, togetherwith the definition of the four types, the
origin and the propagationof the sags becomess traightforward. The results are
summarizedin Table 4.10 for theorigin of sagsand in Table 4.11 for their propagation to lower voltagelevels. Thesuperscript(") behindthe sag type inTables4.10 and
194
Chapter 4 • Voltage
Sags-Characterization
TABLE 4.9 Four Types of Sagsin EquationForm
Type A
Type 8
Va = V
Vb =
V - !jV J3
Vc = -t V +!jvJ3
Va = V
-!
Vb
Vc
Type C
Type 0
=V
Vb =Vc = -
Va = 1
Vb =
Vc
= -!-!jJ)
= -! +!jJ3
Va
-! -!jV~
= -!+!jvJ3
V -!jJ3
V +!jJ)
TypeB
...............
~ T~C
TypeD
...............
~./
Figure 4.70 Four types of sag inphasordiagramform.
TABLE 4.10 Fault Type, Sag Type, andLoad Connection
Fault Type
Star-connectedLoad
Delta-connectedLoad
Three-phase
Phase-to-phase
Single-phase
sag A
sag C
sag B
sag A
sagD
sag C*
TABLE 4.11 Transformationof Sag Type to Lower Voltage Levels
Transformer
Connection
YNyn
Yy, Dd, Dz
Yd, Dy, Yz
Sag Type A
Sag Type B
Sag Type C
Sag Type D
type A
type A
type A
type B
type D*
type C*
type C
type C
type D
type D
type D
type C
t
4.11 indicatesthat the sagmagnitudeis not equal to V but equal to + ~ V, with V
the voltage in the faulted phase or between the faulted phases in Table 4.10 and the
magnitudeof the sag onprimary side in Table 4.11.N ote that in effect these two
definitions of V are the same.
195
Section 4.4 • Three-PhaseUnbalance
4.4.4.6 Two-Phase-to-Ground Faults.T wo-phase-to-groundfaults can be treated in the same way as
single-phaseand phase-to-phase
faults. We will assumethat
the voltage in the non-faultedphaseis not influenced by the fault. As we have seen
in Section 4.4.3 this correspondsto the situation in which positive-, negative-, and
zero-sequenceimpedancesare equal. This can be seen as anextremecase. A zerosequenceimpedancelarger than the positive-sequence
impedancewill shift the resulting voltagestoward thosefor a phase-to-phase
fault.
The phase-to-groundvoltagesat the pee due to tawo-phase-to-groundfault are
Va = 1
Vb
= _! V _! Vj-IJ
Vc
= -~ V +~ Vj../3
2
(4.67)
2
After a Dy transformeror any other transformerof type 3, thevoltagesare
Va = V
Vb
1
1V 1
= --j../3
- - - - Vj-IJ
3
2
6
V
I.
= + -J../33
c
1
(4.68)
1.
- V +- V)../3
2
6
After two transformersof type 3 or after one transformerof type 2, we get
2
1
Va
=3+3 V
Vb
= - -31 - -61 V -
V
= _!_~
V +! Vj'-IJ
362
c
1
(4.69)
- Vj../3
2
Thesethree sags aredifferent from the four types found earlier. It is not possibleto
translateone into the other. Two-phase-to-groundfaults lead tothree more types of
sags,resulting in a total of seven. Thethree new types areshown in phasor-diagram
form in Fig. 4.71and in equationform in Table4.12. Sags due totwo-phase-to-ground
faults andsags due tophase-to-phase
faults are comparedin Fig. 4.72.For a type C sag
the voltageschangealong the imaginaryaxis only, for type 0 along the real axis only.
TypeF
..............
Figure 4.71Three-phaseunbalancedsags due
to two-phase-to-groundfaults.
196
Chapter4 • VoltageSags-Characterization
TABLE 4.12
Sags Due toTwo-Phase-to-GroundFaults
Type F
Type E
Va = V
Vh = -ijJ3 -
Va = I
Vh =
Vi' =
-! V - ! VjJ3
-! V +! Vjv'3
Vc =
+ijv1 -
V-
Vjv"j
V + Vjv1
Type G
Va = j+i V
Vh = ~V-
v(' =
-
i! Vjv'3
i - ~ V +! VjJ3
D
~
"·""N
c
G
...-.-
............
c
.....~
.
.
.. DF
i"V
Z-J
D
Figure 4.72Comparisonof three-phase
unbalancedsags due totwo-phase-to-ground
faults (F and G) withthree-phaseunbalanced
a nd single-phasesags due tophase-to-phase
to-groundfaults (C and D). The arrows
indicate the direction of changein the three
complexvoltagesfor the different sag types.
For types F and G the voltages
drop along both axis. The resulting voltages at the
equipmentterminalsare lowerduring a two-phase-to-groundfault. An additionaldifference isthat all three voltagesdrop in magnitudefor a type G sag.N otealsothat for a
type D and type F sag the
drop in the worst-affectedphase is the same, whereas for a
type C and a type G sag the
drop in voltage between the two
worst-affectedphases is
the same. Thispropertywill be used when defining themagnitudeof measuredthreephaseunbalancedsags.
Sag types F and G have been derived by assuming
that positive-, negative-, and
zero-sequenceimpedancesare the same. If the zero-sequence
impedanceis larger than
the positive-sequence
impedance,the resulting sag will be somewhere in between type C
and type G, or in between type D and type F.
4.4.4.7 Seven Types
of Three-Phase Unbalanced Sags.Origin of sags and transof three-phaseunbalancedsags
formation to lower voltage levels for all seven types
of the sagtransformationto
are summarizedin Tables 4.13 and 4.14. An example
TABLE 4.13 Origin of Three-PhaseUnbalancedSags
Fault Type
Star-connectedLoad
Delta-connectedLoad
Three-phase
Two-phase-toground
Phase-to-phase
Single-phase
Type A
Type E
Type A
Type F
Type C
Type B
Type D
Type C·
Note: Asterisk defined as inTables4.10 and 4.11.
197
Section 4.4 • Three-PhaseUnbalance
TABLE 4.14 Transformationof Sag Type to Lower Voltage Levels
Transformer
Connection
YNyn
Yy, Dd, Dz
Yd, Dy, Yz
Sag on Primary Side
Type A
Type B
Type C
Type D
Type E
Type F
TypeG
A
A
A
B
D*
C*
C
C
D
D
D
C
E
G
F
F
G
G
G
F
F
lower voltage levels isshown in Fig. 4.73. A fault at 33 kV causes the
voltage at the
peeto drop to 50% of the nominal voltage. For a three-phasefault the situation is
easy: at any leveland for any load connectionthe sag isof type A and with a magnitude of 50%. For a phase-to-phasefault the voltage betweenthe faulted phasesat
the peedrops to 50%. For star-connectedload the resulting sags are typeC, 50% at
33 kV; type D, 50% at 11 kV; and again type C, 500/0 at 660 V. In case thefault is a
single-phaseone, thevoltage in the faulted phasedrops to 50% at the pee,This correspondsto a sagof type B and magnitude50% at 33 kV. After the first Dy transformer the zero-sequencecomponent of the voltages has been removed. Starconnectedload at 11 kV will experiencea sagof type C with a magnitudeof 67%.
Delta-connectedload will experiencea sagof type D with a magnitudeof 670/0. For
load fed at 660 V thesituation is just the other way around: star-connectedload experiencesa sagof type D; delta-connectedload one of type C.
4.4.4.8Overview. In the beginningof this sectionwe assumedthat the zero-sequencecomponentof the voltagesdid not propagatedown to the equipmentterminals. We used thisassumptionto obtain an expressionfor the voltages during a
single-phase-to-groundfault. Under this sameassumptionwe find that three-phase
unbalancedsagsof type B or type E cannotoccur at the equipmentterminals.At the
equipmentterminalswe only find the following five typesof three-phaseunbalanced
sags:
• type A due tothree-phasefaults.
• type C and type D due tosingle-phaseand phase-to-phase
faults.
• type F and type G due totwo-phase-to-groundfaults.
Iph...gnd
B, 50%
Figure 4.73 Example of sag
transformation,
for star-connectedload.
n 67%
2ph
2ph-gnd
3ph
C, 50% E, 50% A, 50%
n 50%
F, 50% At 50%
C, 50%
o, 50%
At 50%
Chapter4 • VoltageSags-Characterization
198
The latter two types can beconsideredas distortedversions of type C and D. Sags
of
type C and D are also
distorted by the presenceof inductionmotor load. The presence
of inductionmotor load makesthat positive- and negative-sequence source impedances
of the effectsof this is that the voltage in the"non-faulted
are no longer equal. One
100%. This has been the basis for a
phase"for a type C sag is no longer equal to
classification andcharacterizationof three-phaseunbalancedsags into three types,
correspondingto our types A, C, and D[203], [204].
4.5 PHASE-ANGLE JUMPS
A shortcircuit in a power system
not only causes adropin voltagemagnitudebut also a
of the voltage. In a 50 Hz or 60 Hz system, voltage is a
change in the phase angle
complex quantity (a phasor)which hasmagnitudeand phase angle. A change in the
system, like ashortcircuit, causes a change in voltage. This
changeis not limited to the
magnitudeof the phasorbut includes a change in phase angle as well. We will refer to
the latter as thephase-anglejump associatedwith the voltage sag. Thephase-angle
jump manifests itself as a shift in zero crossing of the
instantaneousvoltage. Phaseanglejumps are not of concernfor most equipment.But power electronicsconverters
using phase-angleinformation for their firing instantsmay be affected. We will come
back to the effecto f phase-anglejumps on equipmentin Chapter5.
j ump of +45°: theduring-fault
Figure4.74 shows a voltage sag withphase-angle
a
voltage leads thepre-faultvoltage. A sag with aphase-anglejump of -45° is shown in
Fig. 4.75: theduring-fault voltage lags thepre-fault voltage. Both sags have a magnicontinuedas a dashed
tude of 70%. In both figures, thepre-fault voltages have been
curve. Notethat these aresyntheticsags, notmeasurementresults.
The origin of phase-anglejumpswill be explained for athree-phasefault, asthat
enables us to use the single-phase model.
Phase-angle
j umps during three-phasefaults
X /R ratio between the source and the feeder. A second
are due to the difference in
causeof phase-anglejumps is the transformationof sags to lower voltage levels. This
phenomenonhas already been mentionedwhen unbalancedsags were discussed in
Section 4.4.
0.5
-0.5
2
3
Time in cycles
4
5
Figure 4.74 Synthetic sag with a magnitude
of.70°tlo and a phase-angle jump of +45°,
199
Section 4.5 • Phase-Angle
Jumps
0.5
-0.5
-I
L - . . - _ - - - J ' - - _ - . . . . L_ _- - - L ._ _-...L.._ _
--'
o
Figure4.75Syntheticsag with amagnitude
of 700/0 and aphase-anglejump of -45
2
3
4
5
Time in cycles
0
•
4.5.1 Monitoring
To obtainthe phase-anglejump of a measuredsag, thephase-angleof the voltage
during the sagmust be comparedwith the phase-angleof the voltage before the sag.
The phase-angleof the voltagecan be obtainedfrom the voltagezero-crossingor from
the phaseof the fundamentalcomponentof the voltage. The complex fundamental
voltagecan be obtainedby doing a Fourier transformon the signal. This enablesthe
use of Fast-FourierTransform(FFT) algorithms.
To explain an alternativemethod,considerthe following voltage signal:
v(t) = X cos(wot)'- Y sin(wot) = Re{(X + jY)eia>ot}
(4.70)
with Wo the fundamental(angular)frequency.Two new signalsare obtainedfrom this
signal, as follows:
Vd(t) = 2v(t) x cos(Wot)
(4.71)
= 2v(t) x sin(wot)
(4.72)
vq(t)
which we can write as
Vd(t) = X
+ X cos(2wot) + Y sin(2wot)
vq(t) = - y
+
Y cos(2wot)
(4.73)
+ X sin(2wot)
(4.74)
Averaging the two resulting signalsover one half-cycle of the fundamentalfrequency
gives therequiredfundamentalvoltage.
(4.75)
J
Knowing the valuesof X and Y, the sagmagnitudecanbe calculatedas X 2 + y2 and
the phase-anglejump as arctan
This algorithm has beenappliedto the recordedsag in Fig. 4.1.The resultingsag
magnitudeis shown in Fig. 4.76 and the phase-anglejump in Fig. 4.77. The effect of
averagingVd(t) and vq(t) over one full cycleof the fundamentalfrequencyis shown in
Fig. 4.78 for the sagmagnitudeandin Fig. 4.79 for thephase-anglejump. The effect of
a largerwindow is that the transitionis slower,but the overshootin phase-angleis less.
Which window length needs to bechosendependson the application.
t.
Chapter 4 • VoltageSags-Characterization
200
0.8
a
.5
~
0.6·
2
.~
~ 0.4
0.2
234
Timein cycles
5
6
Figure 4.76 Amplitude of the fundamental
voltage versus time for the voltage sag shown
in Fig. 4.I-a half-cycle window has been
used.
20,-----,.------,-----r----,-----r-----,
10
fI)
~
~
0....-----'
-8
.S -10
Q..
§ -20
'--'
.!!
, -30
~
]
-40'
A.4
-50
234
Timein cycles
a
.5
6
0.8
-8
0.6
~
0.4
-I
5
Figure 4.77 Argument of the fundamental
voltage.versustime for the voltage sag shown
in Fig. 4.I-a half-cycle window has been
used.
,
,
,
234
Timein cycles
. - L _.. __ . _ .. _
5
6
Figure 4..78 Amplitude of the fundamental
voltage versus time for the voltage sag shown
in Fig. 4.I-a one-cyclewindow has been
used.
201
Section 4.5 • Phase-Angle
J umps
20..---~--~--,.---.,.-------r-
10
l
f'J
Ol-----..J
-8
.5 -10
Q.
§ -20
."""\
u
l-30
u
=-40
-sof
Figure 4.79 Argument of the fundamental if
voltage versus time for the voltage sag shown -60O'-------'------"----L------"--~
.
2
3
4
5
in Fig. 4.I-a one..cyc1ewindow has been
Timein cycles
used.
,
,
I
6
4.5.2 Theoretical Calculations
4.5.2.1 Origin of Phase-AngleJumps. To understandthe origin of phase-angle
jumps associated with voltage sags, the single-phase voltage divider model of Fig.
4.14 can be used again, with the difference
that Zs and ZF are complexquantities
which we will denote asZs and ZF. Like before, we neglect all loadcurrentsand
point-of-commoncoupling (pee):
assumeE = 1. This gives for the voltage at the
-V
ZF
sag
---r:
ZS+ZF
(4.76)
Let Zs = R s + jXs and ZF = R F + jXF . The argumentof V.mg , thus the phase-angle
jump in the voltage, is given by the following expression:
11t/J = arg(Vsag)=arctan(~:)
- arctan(~:: ~:)
(4.77)
¥,
If ~ =
expression (4.77) is zero and there is no phase-angle jump. The phase-angle
jump will thus be present if theX/R ratios of the source and the feeder are different.
4.5.2.2 Influenceof Source Strength. Consideragain the power system used to
obtain Fig. 4.15. Insteadof the sagmagnitudewe calculatedthe phase-anglejump,
resulting in Fig. 4.80. We again see
that a strongersource makes the sag less severe:
lessdrop in magnitudeas well as a smaller phase-angle
jump. The only exception is
for terminal faults. The phase-angle
jump for zero distance to the fault is independent of the source strength. Note
that this is only of theoreticalvalue as the phaseanglejump for zero distance to the fault, and thus for zero voltage
magnitude,has
no physical meaning.
4.5.2.3 Influenceof Cross Section. Figure 4.81 plots phase-angle
jump versus
of the
distance for 11 kV overhead lines of different cross sections. The resistance
source has been neglected in these calculations:
Rs = O. The correspondingsag
magnitudeswere shown in Fig. 4.16.From the overhead line impedance
data shown
in Table 4.1 we can calculate the
X/R ratio of the feeder impedances: 1.0 for the
202
Chapter 4 • VoltageS ags-Characterization
Or----..----~----:==:::!::::=:==:::c:=====~
-5
g -10
~ -15
75MVA
"'t'
.5
~ -20
.; -25
bb
~ -30
Go)
~
f
-35
-40
-45
0
10
20
30
40
50
Distance to the fault in kilometers
Figure 4.80Phase-anglejump versus
2
11kV
distance, for faults on a 150 mm
overheadfeeder, withdifferent source
strength.
_______
- - -.-: .....
-:.-:.-:.-:~:-.:-.-:-.:-.7.
g-10
t
c:: -20 '
.-
.[
~ -30':'
=
. .
~
,
.
G)
.
~
-40:
-soL , , '
o
5
10
15
20
Distance to the fault in kilometers
25
Figure 4.81Phase-anglejump versus
distance,for overheadlines with cross section
300mm2 (solid line), 150mm2 (dashedline),
and 50mm2 (dottedline).
50 mrrr' line, 2.7 for the 150 mm", and 4.9 for the 300
mm-; the phase-anglejump
decreases for larger
X/R ratio of the feeder.
The results forundergroundcables are shown in Fig. 4.82. Cables with a smaller
cross section have a larger
phase-anglejump for small distances to the fault, but the
phase-anglejump also decays faster for increasing distance. This is due to the (in
absolutevalue) larger impedance per unit length. The
correspondingsag magnitudes
were shown in Fig. 4.17.
Sagmagnitudeand phase-anglejump, i.e., magnitudeand argumentof the complex during-faultvoltage, can beplottedin onediagram.Figure 4.83 shows the voltage
pathsin the complex plane, where the pre-sag voltage is in the direction of the positive
real axis. Thefurther the complex voltage is from +
1 jO, the larger the change in
complex voltage due to the fault. The difference between the pre-sag voltage and the
actual voltage is referred to as the missing voltage. We will come back to the concept of
missing voltage in Section 4.7.2.
Insteadof splitting the disturbanceinto real andimaginary parts one may plot
magnitudeagainst phase-anglejump as done in Fig. 4.84.F rom the figure we can
conclude that the phase-anglejump increases (inabsolutevalue) when thedrop in
voltage increases (thus, when the sag
magnitude decreases). Both an increase in
203
Section 4.5 • Phase-AngleJumps
Or------y---~---.__--__r_--__,
-10
1-20
-8
.6 -30
,/
Q.
' ,
§ -40
.•
.,
.~
I-50
1::1
~
-60 'f
Q..
:
..c:
'
-70
-80
0
5
10
15
20
Distance to the faultin kilometers
25
Figure 4.82Phase-anglejump versus
distance, forundergroundcables with cross
2
(solid line), 150mm2 (dashed
section 300mm
line), and 50mm2 (dottedline).
O-----,..---~----r-----r---___,
,\
I
'\
'\
.s
t
:s
"
/:'
,
,,
"
,
,
-0.1 '. ,,
',
,
\
,
\
\
,,
I'
I
:
.
/
I
] -0.2
c.e..
o
i- 0.3
~
.s
e
t)I)
..... -0.4
-0.50
-70
0.2
"'--OA-
0.6
0.8
Realpartof voltagein pu
Figure 4.83Pathof the voltage in the
complex plane when the distance to the fault
changes, forundergroundcables with cross
2
(solid line), 150mm2 (dashed
section 300mm
line), and 50mm2 (dottedline).
Figure 4.84Magnitudeversus phase-angle
-80 I.-----'--------'----~-------------' jump, for undergroundcables with cross
o
0.2
0.4
0.6
0.8
section 300mm2 (solid line), 150mm2 (dashed
Sagmagnitudein pu
line), and 50mnr' (dotted line).
204
Chapter 4 • VoltageSags-Characterization
phase-angle
j ump and a decreasein magnitudecan bedescribedas amoresevere event.
Knowing that both voltage drop and phase.. angle jump increasewhen thedistanceto
the fault increases,we can conclude that a fault leads to amore severe event the
closer it is to thepoint-of-commoncoupling. We will later seethat this only holds for
three-phasefaults. For single-phaseand phase-to-phase
faults this is not always the
case.
4.5.2.4 Magnitude and Phase-Angle Jump Versus Distance.
To obtain expressions for magnitudeand phase-anglejump as a function of the distanceto the fault
we substituteZF = z£ in (4.76) with z the complex feederimpedanceper unit length,
resultingin
V
z.c
----
.mg -
(4.78)
Zs+z.c
The phase-anglejump is found from
arg(V.mg ) = arg(z.c) - arg(Zs + z£)
(4.79)
The phase-anglejump is thus equal to the angle in thecomplexplanebetweenz£ and
2 s + u: This is shownin Fig. 4.85, where</J is the phase-anglejump anda is the angle
betweensourceimpedanceZs and feeder impedancez.
ex =
arctan(~;) - arctan(~;)
(4.80)
We will refer to a as the"impedanceangle;" it is positive when theX/R-ratio of the
feeder islargerthan that of the source.Note that this is araresituation:the impedance
angle is inmost casesnegative.Using the cosinerule twice in the lowertriangle in Fig.
4.85 gives the twoexpressions
IZs + z.c12 = tz.c,2 + IZsl2 - 2lz.cIlZ l cos(180°+ a)
(4.81)
s
2
12s1
= IZs + zL:1 2 + IzL:1 2 -
212s + zL:llz£1 cos(-t/J)
(4.82)
Substituting(4.81) into (4.82) and some rewriting gives an expressionfor the phaseangle jump as afunction of distance
Ar.)
cos('P
A + cosa
= --;::::======
Jl + A2 + 2Acosa
(4.83)
where A = z£/Zs is a measureof the "electrical" distance to the fault and a the
impedanceangle. Note that it is not so much the differencein X/R ratio which deter-
Figure 4.85Phasordiagram for calculation
of magnitude and phase-angle jump.
205
Section 4.5 • Phase-Angle Jumps
mines the sizeof the phase-anglejump but the actualangle betweensourceand feeder
X s/ Rs = 40 and a feeder withXF / RF 2 gives
impedance.For example, a source with
an impedanceangle of
=
= -25.2°
a = arctan(2)- arctan(40)= 63.4° - 88.6°
(4.84)
=
where a source withX s /Rs 3 and a feeder withXF / RF = 1 gives animpedanceangle
of a = -26.6°. The latter will result in more severephase-anglejumps.
The maximum angulardifference occurs forundergroundcables indistribution
systems.F or a sourceX/R of 10 and a cableX/R of 0.5 weobtainan impedanceangle
of about-60°. In the forthcomingsections the value of-60° is used as theworstcase.
Although this is aratherrare case, it assists in showing the
variousrelationships.Small
positive phase-anglejumps may occur in transmissionsystems whereX/R ratio of
source and feeder impedanceare similar. Impedanceangles exceeding+ 10° are very
unlikely. For mostof the forthcomingstudies we will assumethat the impedanceangle
varies between0 and -60°.
From (4.83) we can concludethat the maximum phase-anglejump occurs for
[, = 0, A = 0 and that it is equalto the impedanceangle a.
The magnitudeof the sag isobtainedfrom (4.79)as
v _
sag-
Iz£1
Iz.c + Zsl
(4.85)
With (4.81)the following expressionfor the sagmagnitudeas afunction of the distance
to the fault isobtained:
V
_ _A_
(1 + A) -;:=====
1 _ 2A(l-COS a)
(4.86)
sag -
(t+A)2
Note that the first factor in the right-handsideof (4.86) gives the sag
magnitudewhen
the difference inX/R ratio is neglected(a = 0). This is the sameexpressionas (4.9) in
Section4.2. The error in makingthis approximationis estimatedby approximatingthe
secondfactor in (4.86) for small valuesof a:
1-
2A(l-cosa)
(l+A)2
~
1
1-
A(1-cosa)~l+
(1+,)2 -
A
(1 + A)
A
2(1-Cosa)~1+(1+')2a
2
(4.87)
I\,
A
The error is proportionalto a2• Thus, for moderatevaluesof a the simpleexpression
without consideringphase-anglejumpscan be used tocalculatethe sagmagnitude.
4.5.2.5RangeofMagnitudeandPhase-AngleJump. The relation between magnitude and phase-anglejump is plotted for four values of the impedanceangle in
Fig. 4.86.Magnitudeand phase-anglejump have beencalculatedby using (4.83)and
(4.86). During a three-phasefault all three phases will experience the same
changein
magnitudeand phase-angle.The relation shown in Fig. 4.86 thus alsoholds for single-phaseequipment.When testingequipmentfor sags due tothree-phasefaults one
of
should considerthat magnitudeand phase-anglejump can reach the whole range
combinationsin Fig. 4.86.
206
Chapter4 • VoltageSags-Characterization
-.---- ---., --7l
10, . . - - - - : : : : - - - - - - r - - -
o
-~ ~
.... ' .' ... _---~--~.~.~;>;
rJ
~ -10
--
-8
.8 -20
Q.,
§ -30
I
'",,"",
l-40
Cl)
Cl)
~
f
-50
-60
0.2
Figure 4.86 Relation between magnitude and
phase-angle jump for three-phase faults:
impedance angles:
-60 (solid curve);-35
(dashed);- I 0 (dotted);+ I0° (dash-dot).
0.4
0.6
0.8
Sagmagnitudein pu
0
0
0
EXAMPLE Magnitude and phase-anglejump have beencalculatedfor sags due to
three-phasefaults at the various voltage levels in the example supply shown in Fig. 4.21. Using
the data in Tables 4.3 and 4.4 we can
calculatethe complex voltage at the pee for any fault in
the system. Theabsolutevalue andargumentof this complex voltage are shown in Fig. 4.87.
The complex voltage has been
calculatedfor distances to the fault less than the maximum feeder lengthindicatedin the lastcolumn of Table 4.4. As the maximum feeder length at 132kV
is only 2 km, the sagmagnitudedue to 132kV faults does not exceed 20%. We that
see distribution system faults givephase-anglejumps up to 200 , with the largest ones due to 33 kV
faults. Transmissionsystemfaults only cause very mild phase-angle
jumps. Thesemagnitudes
and phase-anglejumps hold for single-phase as well as
three-phaseequipment,connected to
any voltage level and irrespective
of the load beingconnectedin star or in delta.
rJ
o --------'- - - - - =:: = = =---":'"--------- -:. ~.=
... _-....-----
-0
j
..., -5
.5
Q.,
~ -10
'",,"",
u
bo
~ -15
Cl)
~
~
-20
0.2
0.4
0.6
0.8
Sagmagnitudein pu
Figure 4.87 Magnitude and phase-anglejump
for three-phase sags in the example supply in
Fig. 4.21-solidline: II kV; dashed line:
33kV; dotted line: 132kV; dash-dot line:
400kV.
4.8 MAGNITUDE AND PHASE-ANGLE JUMPS FOR THREE-PHASE
UNBALANCED SAGS
4.8.1 Definition of Magnitude and Phase-Angle Jump
4.6.1.1 Three Different Magnitudes and Phase-Angle Jumps.
The magnitudeof
a voltage sag wasdefined in Section 4.2 as the rmsvalue of the voltage during the
fault. As long as thevoltage in only one phaseis consideredthis is an implementable
Section 4.6 • Magnitudeand Phase-Angle Jumps for
Three-PhaseUnbalancedSags
207
definition, despite theproblems with actually obtaining the rms value.For threephase unbalancedsags theproblem becomes morecomplicated as there are now
three rms values to choose from. The most
commonly used definition is: The magnitude of a three-phase unbalanced sag is the rms value
of the lowestof the three vol·
tages.Alternativessuggested earlier are to use the average
of the three rms values, or
the lowest valuebut one [205]. Here we willproposea magnitudedefinition based on
the analysisof three-phaseunbalanced.sags.
First we need todistinguish between three different kindso f magnitudeand
phase-anglejump. In all casesmagnitudeand phase-anglejump are absolutevalue
and argument,respectively, of a complex voltage.
• The initial complexvoltage is the voltage at thepoint-of-commoncoupling at
fault the initial complex
the faulted voltage level.For a single-phase-to-ground
voltage is the voltage between the faulted phase groundat
and
the pee,For a
phase-to-phase
fault the initial complex voltage is the voltage between the two
faulted phases.F or a two-phase-to-ground or a three-phasefault it can be
either the voltage in oneof the faulted phases or between two faulted phases
(as long as pu values are used). The initial magnitudeis
sag
the absolutevalue
of the complex initial voltage; the initialphase-anglejump is the argumentof
the complex initial voltage.
• The characteristiccomplexvoltageof a three-phaseunbalancedsag is defined as
interpretationof the
the valueof V in Tables4.9 and 4.12. We will give an easy
characteristiccomplex voltage later on. The
characteristicsagmagnitudeis the
absolutevalue of the characteristiccomplex voltage. Thecharacteristicphaseanglejump is the argumentof the characteristiccomplex voltage. These can be
viewed as generalizeddefinitions of magnitudeand phase-anglejumps for
three-phaseunbalancedsags.
• The complexvoltages at theequipmentterminals are the valuesof Va' Vb, and
Vc in Tables 4.9 and 4.12 and in several of the
equationsaroundthese tables.
The sagmagnitudeand phase-anglejump at the equipment terminals are
absolutevalue and argument, respectively,of the complex voltages at the
equipmentterminals.For single-phaseequipmenttheseare simply sag magnitude and phase-anglejump as previouslydefined forsingle-phasevoltage sags.
4.6.1.2 Obtaining theCharacteristic Magnitude. In Section 4.4 we haveintroduced seven types
o f sagstogetherwith their characteristiccomplex voltage V. For
type D and type F themagnitudeis the rms valueof the lowestof the three voltages.
For type C and type G it is the rms value
of the difference between the two lowest
voltages (in pu).From this we obtain the following method of determiningthe characteristic magnitudeof a three-phasesag from the voltagesmeasuredat the equipment terminals:
• Determinethe rms valuesof the three voltages.
• Determinethe rms values of the three voltage differences.
• The magnitudeof the three-phasesag is the lowesto f these six values.
It is easy to see from the
expressionsgiven earlier,that this will give the valueof IVI as
used for the definitionof the three-phaseunbalancedsags. Anexceptionare sagsof type
m ethodwould still give the
B and type E.For sagsconformingto (4.54) and (4.67) the
208
Chapter4 • VoltageSags-Characterization
exact value for themagnitude.But the difference between zero-sequence and positivethat
sequence source
impedancemakesthat the actualsags can deviate significantly. In
case themethodis likely to give acompletelywrong picture.Anotherproblemis that
for these sags the
magnitudechanges when they
propagateto a lower voltage level. This
makes measurementsat a medium voltage level not suitable forpredicting the sag
magnitudeat the equipmentterminals.This problem can be solvedby removing the
zero-sequencec omponentfrom the voltage andapplying the methodto the remaining
voltages. The complete
procedureproceeds as follows:
• obtain the three voltages as function
a
of time: Va(t), Vb(t),and Vc(t).
• determinethe zero-sequence
voltage:
(4.88)
• determinethe remainingvoltages aftersubtractingthe zero-sequence voltage:
V~(t) = Va(t)- Vo(t)
Vb(t) = Vb(t)- Vo(t)
V;( t) = Vc( t) - Vo(t)
• determinethe rms valuesof the voltagesV~,
• determinethe three voltage differences:
(4.89)
Vb, and V;.
(4.90)
• determinethe rms valuesof the voltagesVab, Vbc' and Vcao
• the magnitudeof the three-phasesag is the lowest of the six rms values.
In case alsophase-anglejump and sag type are needed, it better
is
to use a more
mathematicallycorrect method. A method based onsymmetrical componentshas
recently beenproposedby Zhang[203], [204].
EXAMPLE This procedure has been applied to the voltage sag shown in Fig. 4.1. At
first the rms values have been determined for the three measured
phase-to-groundvoltages, resulting in Fig. 4.88. The rms value has been determined each half-cycle over the preceeding
128 samples (one half-cycle). We see the behavior typical for a single-phase fault on an overhead feeder: a drop in voltage in one phase and a rise in voltage in the two remaining phases.
component,all three voltages show a drop in
After subtractionof the zero-sequence
magnitude (see Fig. 4.89). The
phase-to-groundvoltages minus the zero-sequence are indicated
through solid lines, thephase-to-phase
voltagesthrough dashed lines. The lowest rms value is
reached for aphase-to-groundvoltage, which indicates a sag of type D. This is not surprising as
the original sag was of type B (albeit with a larger than normal zero-sequence component). After
removal of the zero-sequence voltage a sag of type D remains.characteristic
The
magnitudeof
this three-phase unbalanced sag630/0.
is
209
Section 4.6 • Magnitudeand Phase-Angle Jumps for Three-Phase
UnbalancedSags
0.4
0.2
234
Time in cycles
Figure 4.88The nns values of the phase-togroundvoltages for the sag shown in Fig. 4.1.
5
6
0.8
a
,
.S 0.6
o
~ 0.4
Figure 4.89The rms valuesof phase-to-phase
(dashed lines) andphase-to-groundvoltages
after removal of the zero-sequence
component(solid lines) for the sag shown in
Fig. 4.1.
0.2
234
Time in cycles
5
6
4.8.2 Ph••e-to-Ph.s.F .ults
The impact ofphase-to-phase
faults depends on the
transformerwinding connections between the fault and the equipment. As shown in Section 4.4, the result is a sag
either of typeCor of type D. It was shown in Section 4.4.2
that the voltage between the
faulted phases can be
obtainedby using the same voltage divider model as for the threephase sag. The
latter has been.used to
obtain expressions (4.83) and (4.86) for phaseanglejump and magnitudeversus distance. These expressions can thus also be used to
calculate initialmagnitudeand initial phase-angle jump: absolute value and
argument
of the voltage between the faulted phases atpee,
theThe three-phase
unbalancedsags in
Section 4.4 were all derived under the
assumptionthat the initial voltage drops in
magnitudewithout change in phase angle. In case of a phase-angle
jump in the initial
voltage, thecharacteristicvoltage of the three-phaseunbalancedsag at the pee also
becomes complex. The expressions in Tables 4.9 and 4.12 still hold with the exception
characteristic
that the characteristicvoltage V has become a complex number. The
210
Chapter4 • VoltageSags-Characterization
voltage for sag types Cand D does not changewhen they aretransformeddown to
lower voltage levels, sothat the characteristiccomplex voltage remainsequal to the
initial complex voltage.
4.6.2.1 Sagsof Type C. The phasordiagram for a sag of type C is shown in
Fig. 4.90, where <p is the characteristicphase-anglejump and V the characteristic
magnitude.Dependingon the phaseto which it is connected,single-phaseequipment
will experiencea sag withmagnitude Vb and phase-anglejump ~h, a sag withmagnitude Vc and phase-anglejump ~c, or no sag at all.Due to the initial phase-angle
jump <P the voltage magnitudesin the two faulted phasesare no longer equal. Note
that in Fig. 4.90 <P < 0, ~h < 0, and <Pc > O.
From Fig. 4.90expressionscan bederivedfor magnitudeandphase-angle
j ump at
the equipmentterminals.As a first step the sine ruleand the cosinerule are applied to
the two trianglesindicatedin Fig. 4.90 resultingin
vi = !4 + ~4 V2 -
2·
!2·!2 V..[j cos(90° -l/J)
sin(60° + <Ph) sin(90° - ~)
-----=---V v'3
Vb
!
V~c =!4 +~4 V2 sin(60° -
2.!.!
V..[jcos(90° + l/J)
2 2
~c)
sin(90° + ~)
-----=----
! Vv'3
(4.91)
(4.92)
(4.93)
(4.94)
Vc
from which the following desiredexpressionsare obtained:
Va = 1
Vh
Jt
= 4- + -43 V 2 -
:
-1 V Vrx3 sln(f/J)
2
(4.95)
1/2
Figure 4.90Phasordiagram for a sag of type
C with characteristicmagnitudeV and
characteristicphase-anglejump 4>.
Section 4.6 • Magnitudeand Phase-Angle Jumps for Three-Phase Unbalanced Sags
2
3
Distancetothe fault
f
....., rJ
~~
4
-8
5
50
--------------------
0
I _--~-------------~.s -50~~_.__
~
211
- - - I ._ _- - . . . I_ _
----.J
o
I
2
4
3
Distanceto thefault
5
Figure 4.91Magnitude(top) and phase-angle
jump (bottom) for sags of type C due to
phase-to-phase
faults. Dashedline: zero
impedance angle (no
characteristicphaseangle jump). Solid line:- 600 impedance angle
(largecharacteristicphase-angle jump).
tPa =0
tPh = -60 + arCSinGJ3~ COS(tP»)
0
tPc = 60
0
-
(4.96)
arCSinGJ3~ COS(tP»)
Combining(4.95) and (4.96) with (4.83) and (4.86) gives the
magnitudeand phaseanglejump in the three phases asfunction
a
of the distance to the fault. This is done in
- 60°. The horizontalscalecorrespondsto
Fig. 4.91 forimpedanceangles equal to 0 and
A = ~£ as in (4.83). We see
t hat the severity of sags decreases with increasing distance
whenSthereis no characteristicphase-anglejump. The introductionof a characteristic
phase-anglejump creates asymmetry between the faulted phases. We see,
that
e.g.,
the
fault.
voltage in oneof the phases initially decreases with increasing distance to the For
one of the phases the phase-angle
jump drops to zero ratherquickly, whereas for the
other phase thephase-anglejump remains high much longer.
Figure4.92 plotsmagnitudeversus phase-angle
jump for four values of the impedance angle. We can see
that the characteristicphase-anglejump significantly disturbs
the symmetry between the two faulted phases. Also the voltagedrop
can well below
50% , which is not possiblewithout characteristicphase-anglejump.
60
8
:
.s
I:'
," :''. \ ,
40
~
\
\
\
20
\
e,
§
......
0
u
"EO
; -20
M
f
I
,
-40
I
-60
o
0.2
0.4
0.6
0.8
Sagmagnitudein pu
Figure 4.92Magnitudeversus phase-angle
jump for sag type C due tophase-to-phase
- 600 (solid line),
faults for impedance angle
0
0
-40 (dashed), -20 (dotted), 0 (dash-dot).
212
Chapter 4 • Voltage
Sags-Characterization
4.6.2.2 Sagsof Type D. The phasordiagramfor a type D sag is shown in Fig.
4.93, wherel/J is again thecharacteristicphase-anglejump. One phasewill go down
significantly with a phase-anglejump equal to the characteristicvalue. Equipment
connectedto one of the two other phases will see a smalldrop in voltage and a
phase-anglejump of up to 30°. Severecharacteristicphase-anglejumps can even lead
to voltageswells. The twophaseswith the small voltagedrop can experience positive
drop always
as well asnegativephase-anglejumps. The phase with the large voltage
experiences anegativephase-anglejump.
From Fig. 4.93 magnitudeand phase-anglejump in the three phases can be
calculatedfor a sagof type D. Applying the sine rule and the cosine rule to the two
trianglesindicatedin Fig. 4.93 gives the following expressions:
vI = !4 V2 + ~4 -
2 ·! V.!,J3 cos(90°+ lj)
2
2
sin(30° - l/Jb) sin(90°+ f/J)
-~---=----
!V
(4.98)
Vb
V 2 =! V 2 +~ - 2.! V· !,J3cos(90°- lj)
c
(4.97)
4
4
2
2
sin(30°+ tPc) sin(90° -l/J)
----=---!V
Vc
(4.99)
(4.100)
Rewriting theseexpressionsresults in
Va
=V
Vb
= ~+~ V2 +~ V,J3sin(lj)
Vc
= ~ + ~ V2 - ~ VJ3sin(lj)
(4.101)
Figure 4.93Phasordiagram for a sag of type
D, with characteristicmagnitude V and
phase-angle jumpt/J.
213
Section 4.6 • Magnitudeand Phase-Angle Jumps for
Three-PhaseUnbalancedSags
-------------------------------
2
3
5
4
Distanceto thefault
Figure 4.94Magnitude(top) and phase-angle
jump (bottom) for sagsof type D due to
phase-to-phase
faults. Dashedline: zero
impedanceangle. Solid line:impedanceangle
of -60°.
o
5
234
Distanceto the fault
cPa = cP
f!Jb
= 30 arCSin(2~b COS(f!J»)
f!Jc
= -30 + arCSin(2~c COS(f!J»)
0
(4.102)
-
0
Again we can plotmagnitudeand phase-angle
j ump versus distance and
magnitude
versus phase-angle
jump. Figure 4.94 givesmagnitudeand phase-anglejump as a
function of distance for impedance angles equal to zero and
-60 Here we seethat
the voltagedrop in the non-faultedphases israthersmall; the voltagedropsto about
75%. Thecharacteristicphase-anglejump causes anadditionaldrop in voltage at the
equipmentterminals. Magnitudeversus phase-angle
jump is plotted in Fig. 4.95 for
four values of the impedance angle.
0
•
4.6.2.3 Rangeof Magnitude andPhase-AngleJump. As mentioned before,
phase-to-phase
faults lead to sags of type C or of type D.
Combining the range of
magnitudeand phase-angle
j ump due to type C sags (Fig. 4.92) with the range due
60
"
':',
I
\
"
\
\
'.
...........' ,
.... .:...:'.
"
-
-
-
-
-
-
-
-
-
-
-
_. -
-
-
-
-
-
-
-
~ ..-. ,-~. ~ . .:'. .:.:~:.-.: I
/
.'
.:
I~'~'" '"
Figure 4.95Magnitudeversusphase-angle
jump for sag type D due tophase-to-phase
faults: impedanceangle -60° (solid line),
-400 (dashed),-20° (dotted), 0 (dash-dot).
-60
o
0.2
0.4
0.6
0.8
Sagmagnitudein pu
,
.'
I
,I
214
Chapter4 • VoltageSags-Characterization
60
lj
~
~
.9
~
.~
u
40
20
0 .-------~
-;0
; -20
~
f
-40
-60
o
0.2
0.4
0.6
0.8
Sag magnitude in pu
Figure 4.96 Rangeof sags due tophase-tophase faults, as experienced
by single-phase
equipment.
to type D sags (Fig. 4.95) gives the whole range
of sags experienced by single-phase
equipmentduring phase-to-phase
faults. The merger of the twomentionedfigures is
shown in Fig. 4.96, where only the
outercontourof the area isindicated.
Sags due tothree-phasefaults areautomaticallyincluded in Fig. 4.96. A threephase fault gives a sag with the initial
magnitudeand the initialphase-angle
j ump, in all
the three phases. Such a sag also
appearsin one of the phases for a type D sag due to a
phase-to-phase
fault. This is the largetriangulararea in Fig. 4.96. Sags due to singletreated
phase andtwo-phase-to-ground
faults havenot yet been included. These will be
below.
EXAMPLE: PHASE-TO-PHASEFAULTS, THREE-PHASELOAD The magnitude and phase-angle
jump due to phase-to-phase
faults have beencalculatedfor faults in the
example supply in Fig. 4.21. The
calculationshave beenperformedfor two different types of
load:
• three-phaseload connectedin delta at 660 V.
• single-phase loadconnectedin star (phase-to-neutral)at 420 V.
For a three-phaseload, we can use the classification
introducedin Section 4.4 tocharacterizethe
j ump of thesethree-phaseunbalancedsags are the same as
sag. Themagnitudeand phase-angle
those of sags due to
three-phasefaults. The only difference is the type of sag.phase-to-phase
A
transformer
fault at 11 kV will, for delta-connectedload at 11kV, lead to a sag of type D. The Dy
between the fault (at11 kV) and the load (at 660 V) will change this into a type C sag. Thus, the
of type
delta-connectedload at 660 V will, due to aphase-to-phase
fault at 11kV, experience a sag
C. Thecharacteristicmagnitudeand phase-angle
j ump of this three-phaseunbalancedsag will be
j ump of the voltage (in any phase) due tothree-phase
a
equal to themagnitudeand phase-angle
fault at the same position as the
phase-to-phase
fault. Using the same reasoning we find that
phase-to-phase
faults at 33kV lead to type0 sags and faults at132kVand400kV to sags of type
C. The results of thecalculationsare shown in Fig. 4.97:characteristicmagnitudeand phaseanglejump of three-phaseunbalancedsags due tophase-to-phase
faults. Note the similarity with
Fig. 4.87. The curves are at exactly the same position; the only difference
thatis
the ones due to
33 kV faults are of type D and the others are of typeThree-phase
C.
faults at any voltage level will
lead to a sag of type A.
Section 4.6 • Magnitudeand Phase-Angle Jumps for
Three-PhaseUnbalancedSags
215
5r-----r-----r-----r------r-----~__.
~
0
~
-8 -5
.5
Q..
.[ -10
.£
bO
~Go) -15
]
Figure 4.97 Characteristicmagnitudeand
phase-anglejump for sags due tophase-tophase faults in theexamplesupply in Fig.
4.21-solidline: type C sags,d ashedline: type
D sags.
~ -20
0.4
0.2
0.6
0.8
Sagmagnitudein pu
EXAMPLE: PHASE-TO-PHASEFAULTS, SINGLE-PHASELOAD Magnitude and phase-angle
jump at the equipmentterminals due tophase-to-phase
faults have been
calculated for a single-phase load connected
phase-to-neutralat 420 V. The classification of
three-phase sags no longer fully describes the voltage atequipmentterminals.
the
The additional information needed is the phases between which the fault takes place. One can calculate
the voltage sag in one phase for three different faults; but it is easier to calculate the voltages
in the three phases for one fault. These three voltages are the voltages in one phase for the
three different faults. We saw before that we do not need to calculate the whole
transferof the
sag from the faulted voltage level to the load terminals. All we need to do is determine whether
the equipmentterminal voltagecorrespondsto phase-to-phaseor phase-to-neutralvoltage at
the faulted voltage level. In this example, the
equipment terminal voltagecorrespondsto
phase-to-phasevoltages at II kV, 132kV, and 400 kV and tophase-to-neutralvoltages at
33kV.
The resultingmagnitudeand phase-angle
j ump are plotted in Fig. 4.98.Faultsat 11kV,
132kV, and 400 kV cause a three-phase
unbalancedsag of type D forstar-connectedequipment.
For a type D sag one voltage drops to a low value, and the two remaining voltages show a small
drop with a phase-anglejump up to 30°. Note the symmetry in the sags
originating at 400kV,
which is not present in the sags
originating at 11kV and 132kV. This is due to the large initial
60
I
,
f
I
40
12:
Figure 4.98 Magnitudeandphase-angle
j ump
at the equipmentterminalsdue to phase-tophasefaults in thesupply in Fig. 4.21,
experiencedby single-phaseload connected
phase-to-groundat 420V-solid line: 11 kV,
dashedline: 33 kV, dotted line: 132 kV, dashdot line: 400 kV.
\
= •••••••• _
._~_:~ ~~ ~~~
i-20~
b
~
f
-40
I
,
-60
o
0.2
"
"
"
I
0.4
0.6
0.8
Sag magnitude in pu
,//
V
216
Chapter4 • VoltageSags-Characterization
phase-angle jump for the
latter two. Faults at 33 kV cause a sag of type C, with two voltages
down to about 50% and phase-angle jumps up to ±60°.
4.8.3 Single-Phase Faults
For single-phase faults the situation becomes slightly more complicated.
Expressions(4.83) and (4.86) can still be used to calculate magnitude and phaseangle jump of the voltage in the faulted phaseat the pee (Le., theinitial magnitude
and phase-anglejump). Star-connectede quipmentat the samevoltagelevel 'as thefault
would experiencea sag of type B. But as we have seen
before, this is a rather rare
situation.In almostall cases a sag
d ueto a single-phasefault is of type Cor type D. The
characteristicmagnitudeof thesethree-phaseunbalancedsags is nolongerequalto the
initial magnitude.The sameholds for the phase-anglejump.
4.6.3.1 Initial and Characteristic Magnitude.To obtain an expressionfor the
characteristicmagnitudeand phase-anglejump, we need to goback to the type B
sag. Thevoltagesfor a type B sag are
Va = V cos </J + jV sin </J
Vb
= _! - !j.Jj
V
= --+-J'../3
2 2
c
2
2
I
1
(4.103)
with V the initial magnitudeand </J the initial phase-anglejump. When this three-phase
unbalancedsag propagatesto lower voltage levels, the zero-sequencevoltage is lost.
The zero-sequence
c omponentfor (4.103) is
(4.104)
Subtractingthe zero-sequence
v oltagefrom (4.103) gives athree-phaseunbalancedsag
of type D. Characteristicmagnitudeandphase-angle
j ump for a sagof type D areequal
to the absolutevalue and the argumentof the complex voltage in the worst-effected
phase, Va in this case.
(4.105)
Note that this expressioncan also beobtainedby substitutingV = V cos</J + jV sin </J in
(4.62). For three-phaseunbalancedsags due to single-phasefaults the characteristic
magnitudebecomes
Vchar=
IVai =
2 / 2
1
3'1
V +.Vcos</J+4
(4.106)
with V and t/J the initial magnitudeand phase-anglejump, and Va accordingto (4.105).
The characteristicphase-anglejump is
2Vsin<fJ )
tPchar = arg(Va) = arctan( 1 + 2V costP
(4.107)
217
Section 4.6 • Magnitudeand Phase-Angle Jumps for Three-Phase Unbalanced Sags
For small values ofl/J these expressions can approximatedby
be
using
sinl/J~
l/J
cosl/J~ 1
arctantxe)~ xl/J, x < 1
resulting in
,12
Vchar = 3+3 V
(4.108)
,
2V~
l/Jchar = 1 + 2V
(4.109)
Figures 4.99 and 4.100 show the
error made by using theapproximatedexpressions
The calculationshave been
(4.108)and (4.109). Theerror has been defined as-~.
1
performed forimpedanceangles equal to-60°, -40°,ci'itd -20°. Even for a system
with large phase-anglejumps, an impedance angle of
-60°, the errors are not very
big. Only for calculatingthe characteristicphase-anglejump with deep sags mightit
be needed to use the exact expression. One should realize, however,
that the
0.08....----.,.-----r---...,.u 0.07
]
.1 0.06
~
0.05
.j
0.04
.~
(J
j
'"
0.03
- - .... ,
(J
.~ 0.02
~
Figure 4.99Transformationof sags due to
~ 0.01 "
single-phase
faults--errorin approximate
o
expressions for characteristic magnitude.
o
-600 (solid line); -400
Impedance angle:
(dashed);- 20 (dotted).
I
I
<;
•••• •••••••••• .. •••
........... -- =---....
0.4
0.6
0.8
Initial magnitudein pu
L . -_ _
..&..-_~~
...............
1_'_ _- - L .
0.2
0
0.2r----~----r------.,.-----r-----,
~
.~
1 0.15
4)
t
(J
'i
0.1
J
0.05 \ \
(J
Figure 4.100Transformationof sags due to
single-phase
f aults-errorin approximate
expressions for characteristic phase-angle
0
(solid line);
jump. Impedance angle: _60
0
0
-40 (dashed);- 20 (dotted).
...
.s
~
J3
.......:-..-:-.:-.~"':'".:-:."""._-~.:::s.:.=::~....-.-_----1
0
o
0.2
0.4
0.6
Initial magnitudein pu
0.8
218
Chapter4 • VoltageSags-Characterization
or - - - - - - - r - - - r - - r - - - - - - r - - - -.------r-----.
\
\
\
-10
(/)
8
~ -20·
~
= -30
.~
§
:£-40
~
~ -50
f
-60
0.2
Figure 4.101 Relation betweenphase-angle
jump and magnitudeof sags due to singlephase faults:characteristicvalues(dashed
curve) and initial values (solid curve).
0.4
0.6
0.8
Sag magnitude in pu
characteristicphase-anglejump is close to zero for single-phase faults with a small
initial magnitude,as can be seen from (4.107). The
absoluteerror is even for an
impedanceangle of -60 lessthan 1
Figure 4.101comparesinitial magnitudeand phase-angle
j ump with the characbottom (solid) curve
teristic values. Animpedanceangle of -60 has been used. The
also gives therelation betweencharacteristicmagnitudeand phase-anglejump due to
phase-to-phase
and three-phasefaults. Sags due to single-phase faults are clearly less
severe: inmagnitudeas well as inphase-anglejump.
0
0
•
0
4.6.3.2 Sagsof Type C and Type D. Knowing characteristicmagnitudeand
phase-anglejump for the typeC or type D sag it is again possible calculatemagnito
tude and phase-anglejump at the equipmentterminals.This results insimilar curves
as for sags due tophase-to-phase
faults. The main difference is
t hat voltage sags due
to single-phase faults are less severe
than due to phase-to-phase
faults. Figure 4.102
plots magnitudeversusphase-anglejump for sag typeC, for four valuesof the impedance angle. The lowest sag
magnitudeat theequipmentterminals isabout 58°~, the
largestphase-anglejump is 30
0
•
60
rJ 40
~
"'0
.5
20
.[
0
u
bb
fa -20
~
f
-40
-60
o
0.2
0.4
0.6
0.8
Sag magnitude in pu
Figure 4.102 Rangeof sagsexperiencedby
single-phaseequipmentfor sag type C and
single-phasefault, impedanceangle: _60°
(solid line), _40° (dashed),-20 (dotted),
o(dash-dot).
0
219
Section 4.6 • Magnitude and Phase-Angle Jumps for Three-Phase
UnbalancedSags
60 ~
I
40
~
20
.[
0
..2
eo
~
-20
f
-40
~
Figure 4.103 Range of sags experienced
by
single-phaseequipment for sag type D and
-600
single-phasefault, impedance angle:
0
0
(solid line), -40 (dashed),- 20 (dotted),
o(dash-dot).
\ :.\. ,
~
..:-.- -- '-'~'
- --~'~'~'~'~'- -~.:
..-:.;.
. -~.~.~.~.~
----,.~:j. ~..:~>'.
~---
/.;. ...
-60
o
0.2
0.4
0.6
0.8
Sagmagnitudein pu
60
!
,I
\
\
I
,
I
40
,,
....
I
I
~
\
.S 20
~
.--.
,
... '1
,
\
\
0
bb
; -20
~
,
\
u
f
.,
,
z
-,
....
.... -
I
-40
,
I
I
,I
-60 . .
Figure 4.104 Range of sags due to singlephase faults (solid curve) and due to phase-tophase faults (dashed curve).
o
t,..'"
0.2
0.4
0.6
0.8
Sagmagnitudein pu
Figure 4.103 repeats this for type D sags duesingle-phase
to
faults. The lowest sag
magnitudeis 330/0 with a maximumphase-anglejump of 19°. Sags due to type C and
type D are merged into one
p lot in Fig. 4.104 which gives the whole range
of sags
experiencedby single-phaseequipmentdue to single-phase faults. This
rangeis smaller
than the range due tophase-to..phasefaults, indicated by a dashedline in Fig. 4.104.
EXAMPLE: SINGLE-PHASEFAULTS, THREE-PHASELOAD The calculations for phase..to..phase faults shown in the previous section have been repeated for single4.21, the sag magniphase faults.For single-phase faults at the various voltage levels in Fig.
tude, phase-angle
jump, and type have been calculated for delta..connected (three-phase) load
at 660 V.Equations(4.108) and (4.109) have been derived for a system with equal positive, negative and zero-sequence impedance. This is a good
approximationfor the (solidly grounded)
132kV system but not for the(resistance-grounded)
11 kV and 33kV systems. At 400 kV the
source impedance is mainly determined by overhead lines,
that
sothe zero-sequence source impedance is larger than the positive-sequence value. To calculate
characteristic
the
magnitudeof
three-phase unbalanced sags due to single-phase faults, we can first calculate
phase-to-neuthe
tral voltage in the faulted phase according to (4.40).
Characteristicvalues areobtainedfrom
this by applying (4.108) and (4.109). Alternatively we can calculate the complex phase-to-
220
Chapter4 • VoltageSags-Characterization
5..-----.------r----...----.----..-
l
~
o
_------------- _ .
---6
-5
.S
Qc
g-10
.""""
~
; -15
j
~-20~
-25
0
. _,
0.2 _---'-_
0.4
0.6
0.8
Sagmagnitudein pu
-..L..
--L-_ _
.....L---'
Figure 4.105 Characteristic magnitude and
phase-angle
j umpfor sags due to single-phase
faults in the example supply in Fig. 4.21,
experienced by three-phase load-connected
phase-to-phase at
660V-solid line: II kV,
dashed line: 33kV, dotted line: 132kV, dashdot line: 400kV.
2 transformerto these.A type 2 transformerreground voltages at the pee, and apply a type
moves the zero-sequence voltage and results in a three-phase
unbalancedsagof type D. Magnitude and phase-angle
jump of the worst-affected phase are equal to characteristic
the
values.
In other words, thecharacteristiccomplex voltage can be
obtained by subtractingthe zerosequence voltage from the voltage in the faulted phase atpee.
the
The results are shown in
Fig. 4.105. We seethat single-phase faults at11 kV and 33kV
cause only a small
drop in voltage, but amoderatephase-angle jump. This is due to the resistance
groundingapplied at these voltage
levels, Sagsoriginating in the 132kV and 400 kV networks
show a much largerd rop in voltage magnitudebut a smaller phase-angle jump. Note that the
curves for sags due to 400
Vkfaults do notstartat 33°A. voltage as expected for solidly-grounded
systems. The reason that
is the source impedance in PAD-400 mainly consists of overhead lines.
For faults
Therefore the zero-sequence impedance is larger
thanthe positive-sequence impedance.
in the direction of PEN, the source impedances ZSI
are = 0.084+ jl.061, Zso =0.319+ j2.273,
which gives for the initialphase-to-neutralvoltage duringa terminal fault:
Van = 1 - 22
3ZS1
Z
Sl
+
so
•
= 0.2185+JO.0243
(4.110)
The characteristicmagnitudeat a lower voltage level is found from
v.: = H·+~ VanI= 0.519
(4.111)
For single-phase faults in thedirection of EGG we find: Van = 0.3535 - jO.0026 and
Vchar = 0.571. This is amoderateversion of the effect which leads to very shallow sags in
resistance-grounded
systems. Notethat we still assume the system to
be radial, which gives an
erroneousresult for single-phase faults at 400
kV. This explains the difference in resulting voltage
sags for a terminal fault in the two directions. The actual value is somewhere between 0.519 and
0.571. The difference is small enough to be neglected here.
Figure 4.105 does not plot the sag type: faults atkV33lead to a type C sag; faults at 11kV,
132kV, and 400 kV cause a sag
of type D at theequipmentterminals for delta-connected load. At
the equipmentterminals it is not possible to distinguish between a sag due to a single-phase fault
and a sag due to phase-to-phase
a
fault: they bothcause sags
o f type C or type D. Therefore, we
have merged Figs. 4.97 and 4.105 into one figure. The result is displayed in Fig. 4.106, showing
characteristicmagnitudeand phase-angle
j ump of all three-phaseunbalancedsags due to singlephase andphase-to-phase
faults, as experienced bydelta-connected
a
three-phase load at 660 V.
We seethat the equipmentexperiences the whole range of magnitudes and phase-angle jumps.
These have to be considered when specifyingvoltage-tolerance
the
requirements of equipment. To
221
Section 4.6 • Magnitudeand Phase-AngleJumpsfor Three-PhaseUnbalancedSags
~
0
~
-5
tt
\\
_--------- ==
0
.9
c.
§ -10
.~
i
; -15
J
~ -20
Figure4.106Characteristicmagnitudeand
phase-angle jump for three-phase
unbalanced
sags in Fig. 4.21, experienced by three-phase - 25O'------.L---L-----'.
0.4
0.6
0.8
0.2
delta-connectedload-solidline: type C,
Sagmagnitudein pu
dashed line: type D.
J __ - . - -
be able to fully interpret these results, twomore dimensionsare needed. At first, one has to
realize that not all sags areof equalduration. Typically sags due to11 kV and 33 kV faults are
of longer duration than those due to 132kV and 400kV faults. What is also different for
different sags is itslikelihood. Roughly speakingone can say that deepersags are less likely
than shallower sags. We will come back to probabilities in detail in Chapter 6. To include
magnitude,phase-anglejump, duration,and probability in one, two-dimensional,figure is very
difficult if not impossible.
EXAMPLE: SINGLE-PHASE FAULTS, SINGLE-PHASE LOAD The magnitude and phase-anglejump have been calculatedfor voltage sagsdue to single-phasefaults,
experiencedby single-phasestar-connectedload. For this we havecalculatedeither the phaseto-phase voltage, or the phase-to-groundvoltage minus the zero-sequencevoltage, at the
faulted voltage level. For a single-phasefault at 11 kV, star-connectedload at 420 V experiences a sagof type C. The complex voltages at the equipment terminals are equal to the
phase-to-phase
voltagesat the pee,The samecalculationmethodcan be used forsingle-phase
faults at 132 kVand at 400 kV. Single-phasefaults at 33 kV lead to sagsof type D. The complex voltagesat the equipmentterminalscan be calculatedas the phase-to-groundvoltagesat
the pee minus the zero-sequencecomponent.The results of these calculationsare shown in
Fig. 4.107. We seethat the voltage never drops below 500/0, and that the phase-anglejumps
are between-30° and +30°. Faults at 11 kV and 33 kV again only causeshallow sags due to
the system beingresistance-grounded.
Due to a 33 kV fault, the load can even experiencea
small voltage swell. Faults at 400kV are also somewhatdampedbecausethe zero-sequence
source impedanceis about twice the positive-sequencevalue. Therefore, sags due to singlephasefaults are milder than expectedfor a solidly-groundedsystem. In the 132 kV system, the
zero-sequencesource impedanceis even a bit smaller than the positive sequencevalue, thus
V they appearas a typeC in which the drop in phasevoltages
leadingto deep sags. But at 420
is not below 500/0. For this specificsystem,single-phasefaults do not causevery deep sags for
star-connectedload. Note that this is not a generalconclusion.Had the 11 kV/420 V transformer beenof type Dd, the equipmentwould have experiencedvoltagedropsdown to 300/0 (see
Fig. 4.105).
To get acompletepictureof all sagsexperiencedby the single-phaseload, we havemerged
Fig. 4.87 (three-phasefaults), Fig. 4.98 (phase-to-phasefaults), and Fig. 4.107 (single-phase
faults), resulting in Fig. 4.108. Here we see the wholerange of values both in magnitudeand
in phase-anglejump.
222
Chapter4 • VoltageSags-Characterization
60
~
40
Go)
~
~
.S 20
~
0
'~
Go)
~
S -20
I
~
f
Figure 4.107 Magnitudeand phase-angle
jump for sags due tosingle-phasefaults in the
examplesupply in Fig. 4.21, experiencedby
single-phaseload-connectedphase-to-ground
at 420V-solid line: II kV, dashedline:
33 kV, dotted line: 132kV, dash-dotline:
400kV.
-40
-60
0.2
0
0.4
0.6
0.8
Sag magnitude in pu
60
,
I
r
,
I
~
40
.S
20
j
~~
\
(\
\""
'
.~_---- ~~'_-_--~~~ ~ ~------=-,-~~~'~~-~J~--
0 ------
Ii - 20
I
/~ ~ ~
M
f
...... : .......... \'"
-40
~
,, , / /
V
/
I
I
-60
I
o
0.2
0.4
0.6
0.8
Figure 4.108 Magnitudeand phase-angle
jump for all sags in theexamplesupplyin Fig.
4.2), experiencedby single-phaseloadconnectedphase-to-groundat 420 V-solid
line: I) kV, dashedline: 33 kV, dotted line:
132kV, dash-dotline: 400kV.
4.8.4 Two-Phase-to-Oround Faults
The analysisof two-phase-to-groundfaults does not differ from the treatmentof
phase-to-phase
faults. We saw inSection4.4.4 that two-phase-to-groundfaults lead to
three-phaseunbalancedsagsof type E, type F,or type G. Type E is a rare type which
f ault, the type E
we will not discusshere. Like type B for the single-phase-to-ground
containsa zero-sequencec omponentwhich is normally not transferredto the utility
voltage,and neverseen bydelta-connectedequipment.
For type F and type G we can againplot characteristicmagnitudeagainstphaseanglejump. The relation betweenthe characteristicmagnitudeandphase-angle
j ump of
the unbalancedthree-phasesag isidentical to the relation betweenthe initial magnitude
and phase-anglejump, i.e, magnitudeand phase-anglejump of the voltage in the
faulted phasesat the pee.This relation is describedby (4.83) and (4.86) and is shown
in Fig. 4.86.
4.6.4.1 Sagsof Type F. A detailedphasordiagramof a sagof type F is shown
in Fig. 4.109. Like with a type D sag, one phasedrops significantly in magnitude,
and the other two phasesless.The differencewith the type D sag is in thelatter two
Section 4.6 • Magnitudeand Phase-Angle Jumps for
Three-PhaseUnbalancedSags
223
Figure 4.109Phasordiagramfor three-phase
unbalancedsag of type F with characteristic
magnitudeV and characteristicphase-angle
jump t/J.
-!
phases.With a type D sag theydrop from
± !jJ3 to ± !jJ3, but with a type F
sag theydrop significantly more: to ±!jJ3. The lowest magnitudefor a type D sag
is 86.60/0, whereasit is 57.7% for a type F sag.
In the upper triangle indicatedin Fig. 4.109 wecan again apply the cosineand
sine rule toobtain magnitudeand phase-anglejump at the equipmentterminals.Note
that in Fig. 4.109, rP < 0, rPb > 0, and rPc < O. The cosinerule gives
(4.112)
which resultsin an expressionfor the voltagemagnitude Vc:
(4.113)
The sine rule in thesametriangle gives
sin(30° + rPc)
! vJ3
sin(120° - rP)
=-----
(4.114)
Vc
The phase-anglejump rPc follows as
0
f/Jc = -30
+ arcsin{V~Sin(120° - f/J)}
(4.115)
The same rules can be applied to the lower triangle, which leads to the following
expressionsfor magnitude Vb and phase-anglejump rPb:
(4.116)
224
Chapter 4 • VoltageSags-Characterization
60
l
~
40
.S
20
~
....,
0
---------'::
u
tih
fa -20 .
~
~
f
-40
-60
o
0.2
0.4
0.6
0.8
Sag magnitude in pu
Figure 4.110Magnitudeand phase-angle
jump at theequipmentterminals for a type F
sag, due to atwo-phase-to-groundfault. The
curves are given for an impedance angle of 0
(dashed line) and_600 (solid line).
(4.117)
From theseequationswe can againcalculatemagnitudeand phase-anglejump at
theequipmentterminals,e.g., as afunction of thedistanceto the fault.Figure4.110 plots
magnitudeversusphase-angle
j ump for a type F sag due to two-phase-to-ground
a
fault.
We seethat one phase behaves again like the sag due three-phase
to a
fault. The other
two phase aresomewhatlike the two phases with a shallow sag in the type D sag shown
in Fig. 4.95. The difference is
thatfor a type F sag the voltages show a significantly larger
drop. Themaximumphase-anglejump for these two phases is again 30°.
4.6.4.2 Sagsof Type G. A detailedphasordiagramfor a type G sag is shown
in Fig. 4.111. The complex voltage in phasedrops
a
to a valueof ~ (no drop for a
bandc drop to a valueof
for
sag of type C); the complex voltages in phase
type C).
.
-! (-!
Figure 4.111 Detailedphasordiagram for
three-phaseunbalancedsag of type G with
characteristicmagnitudeV and characteristic
phase-anglejump l/J.
225
Section 4.6 • Magnitudeand Phase-Angle Jumps for Three-Phase
UnbalancedSags
The cosine ruleand the sine ruleapplied to the triangle on the right give the
following expressions:
2
4
a
9
V = -
12
+ -9 V
2 V
0
- 2 x - x - cos( 180 + cP)
(4.118)
= sin(-4>0)
(4.119)
3
sin(180°+ 4»
3
!V
Va
This leadsagainto expressionsfor magnitudeand phase-anglejump at the equipment
terminals.
(4.120)
4>0 = arcsin(3~0 sin4>)
(4.121)
Repeatingthe calculationsfor the other trianglesgives expressionsfor magnitudeand
phase-anglejump in the other two phases.Note the angle 1010 and the factor!../7.
These originate from the triangle formed by the complex numbers 0,
and
-!,
-!±!jv'3.
(4.122)
(4.123)
Vc =
~J 1 + 7V2 -
2V.J7cos(lOI° + 4»
4>c = 60° - arcsinG.J7~ sin(lOlo + 4»)
(4.124)
(4.125)
The resultsfor type G sags areshownin Fig. 4.112. We seethat the type G sag is
somewhatsimilar to the type C sag, as
s hown in Fig. 4.92. Unlike the phase-to-phase
60
\
\
\
\
\
\
/
/
I
Figure4.112Magnitudeand phase-angle
jump at theequipmentterminals for a type G
sag, due to atwo-phase-to-groundfault. The
curves are given for an impedance angle of 0
(dashed line) and-600 (solid line).
I
I
I
,
0.2
,
226
Chapter4 • VoltageSags-Characterization
fault, two-phase-to-groundfaults cause two voltages to
drop to 33% instead of50%.
For faults somedistanceaway from the pee the voltage
magnitudecan even become a
bit less than 33% due to the initial phase-anglejump. Another difference with the
phase-to-phasefault is that all three phasesdrop in magnitude.The third phase,
which is not influenced at all by aphase-to-phase
fault, may drop to 67% during a
two-phase-to-groundfault.
4.6.4.3 Rangeof Magnitude and Phase-Angle Jump.
Merging Fig. 4.110 and
Fig. 4.112 gives the whole range
of magnitudesand phase-anglejumps experienced
by a single-phaseload due totwo-phase-to-groundfaults. In Fig. 4.113 the area due
to two-phase-to-groundfaults (solid curve) iscomparedwith the area due to phaseto-phasefaults (dashedcurve). We seethat there arecertaincombinationsof magnitude andphase-anglejump which can occur due tophase-to-phase
faults but not due
to two-phase-to-groundfaults, but also theother way around. These curves have
beenobtainedunder the assumptionthat zero-sequence and positive-sequence impedances are equal.For a zero-sequenceimpedancelarger than the" positive-sequence
sourceimpedance,the resulting sags due totwo-phase-to-groundfaults are closer toward sags due tophase-to-phasefaults. The results arethat even a larger rangeof
magnitudeand phase-anglejumps can be expected. An increasing zero-sequence impedance will meanthat the area enclosed by the solid curve in Fig. 4.113 will shift
toward the area enclosed by the
dashedcurve. Thelatter is reached for an infinite
zero-sequence
impedancevalue.
60
,
... 1
/
\
~
~
008
.5
\
40
\
\
,
- ....
\
20
\
\
~ Ot----~----·,
,
'''''''''\
«>
bo
; -20
o
]a.- -40
I
I
-60
0.2
0.4
0.6
0.8
Sag magnitude in pu
Figure 4.113Range ofmagnitudeand phaseanglejump at theequipmentterminals due to
phase-to-phase
(dashed curve) and twophase-to-groundfaults (solid curve).
EXAMPLE: TWO-PHASE-TO-GROUNDFAULTS,SINGLE-PHASELOAD
For the same example system as used before (Fig. 4.21) the complex voltages at the equipment
terminals due totwo-phase-to-groundfaults have been calculated.
Characteristicmagnitude
and phase-angle
j ump due to atwo-phase-to-groundfault are the same as due to a phase-tophase fault.For three-phasedelta-connectedequipmentwe can directly use the results obtained
for phase-to-phase
faults in Fig. 4.97.For two-phase-to-groundfaults, the solid lines refer to
sags of type G, the dashed lines to sags of type F.
two-phase-to-ground
A
fault at 1I kV leads
to a sagof type F for delta-connectedload, according to Table 4.13. The Dy
IlkV/660 V
transformerchanges this into a sag
of type G, according to Table 4.14.
Two-phase-to-ground
faults at 33 kV lead to sags of type F, and faults at 132kV and 400kV to type G.
Section 4.6 • Magnitudeand Phase-Angle Jumps for Three-Phase
UnbalancedSags
227
60
I
I
I
I
-60
I
o
0.2
0.4
0.6
0.8
Sagmagnitudein pu
Figure 4.114 Magnitude and phase-angle
jump at the equipment terminals due to twophase-to-ground faults in Fig. 4.21,
experiencedby single-phase load-connected
t 1 kV,
phase-to-ground at 420 V - solid line:
dashed line: 33 kV, dotted line: 132 kV, dashdot line: 400 kV.
For star-connectedsingle-phase load, thesituation is completely different. The zerosequence source and feeder impedances influence the voltages
during a two-phase-to-ground
fault, but notduring a phase-to-phase
fault. The voltage sags experienced by single-phase equipment are shown in Fig. 4.114.
Faultsat 11kV, 132kV, and 400kV cause sags of type G, in which
II kV the zero-sequence
one phase shows a deep sag and the otherphases
two
a shallow sag. At
source impedance is much larger than the positive-sequence one, due to the resistance
grounding
of this voltage level. The resulting sag is very close to the type D sags duephase-to-phase
to a
fault. The large zero-sequence impedance makes that the ground
connectionof a two-phase-togroundfault does notcarrymuch current. The voltage
magnitudein the two phases with shallow
sags is thus only down toa bout 900/0. For faults at 132kV, which is solidlygrounded,these
voltages are down toabout 55°~. The 400kV system is also solidlygrounded,but the line
impedancedominatesthe source impedance, making that the zero-sequence impedance is more
than twice as large as the positive-sequence impedance. In the phase with the largestdrop,
voltage
the voltagemagnitudeis aboutthe same for the three voltage levels.
Faultsat 33 kV will cause a
type G sag. As the system is resistance
groundedthis sag is very close to a type C sag due to a
phase-to-phase
fault.
4.8.5 High-Impedance Faults
In all the previouscalculationsin this chapter,we have assumed the fault impedance to be zero. The
a rgumentationfor this wasthat the fault impedancecould be
incorporatedin the feederimpedance,ZF in (4.9). Thisargumentstill holds as long as
the magnitudeof the sag isconcerned,but the phase-anglejump can be significantly
affected. We will first addressthree-phasefaults and after that single-phasefaults.
High-impedancefaults are more likely forsingle-phase-to-groundfaults than for
three-phasefaults.
4.6.5.1 Three-Phase Faults.Consider again the basic voltage divider expresRtit explicitly included:
sion (4.9), but this time with the fault resistance
V
_
sag -
ZF+Rfll
Z s + Z F + Rfll
(4.126)
In many cases the source
impedanceand the feederimpedanceare largely reactive,
whereas the faultimpedanceis mainly resistive. The angle between source
impedance
228
Chapter 4 • VoltageSags-Characterization
and feeder plus fault impedance gets close to 90°, which can lead to very large phaseangle jumps.
The fault resistance only noticeably affects the voltage ifF I12
« Rfll' thus for
faults close to thepoint-of-commoncoupling with the load.For zero distance to the
Zs =}Xs):
fault we get for the complex voltage (with
V
-
sag -
~t
}Xs + Rflt
(4.127)
The fault resistance is normally not more
than a fraction of the source reactance, in
which case the sag
magnitudeis the ratio of the fault and the source impedances with a
phase-anglejump equal to almost 90°.
To quantify the influence of the fault resistance, the complex voltage
during the
sag was calculated as a function
of the distance to the fault for three-phase faults at
11 kV in Fig. 4.21.The calculationshave beenperformedfor a zero fault resistance and
10%,200/0, and 300/0 of the (absolutevalueof the) source
for fault resistances equal to
impedance. The sag
magnitude(the absolutevalue of the complex voltage) plottedin
is
Fig. 4.115as a function of the distance to the fault. As expected the influence on the sag
magnitudeis limited to small distances to the fault. The fault resistance increases the
impedance between the pee and the fault, and thus reduces the voltage
drop at the pee.
The phase-anglejump is much more influenced, as shown in Fig.
4.116. The
For increasing fault resistance the maximum
phase-anglejump reaches values up to 80°.
phase-anglejump does not reduce much.
4.6.5.2 Single-Phase Faults.To assess the effect
of high-impedance singleof
phase faults on the voltage at the
equipmentterminals, we use the classification
three-phaseunbalancedsags again. At first we consider solidly-groundedsystem,
a
for which we can 'assumethat the two non-faulted phase voltages remain at their
pre-fault values. Inother words, we have a clean type B sag. The voltage in the
faulted phase is influenced by the fault resistance as shown in 4.115
Figs. and 4.116.
At the equipmentterminals the sag will be of type C or D.
Magnitude and phaseanglejump at theequipmentterminals are shown in Fig.
4.117for a type C sag and
in Fig. 4.118for a type D sag.In' Fig. 4.117we see how an increasing fault resistance
increases theunbalancebetween the two affected phases.
Although the characteristic
0.8
~
.5
~
0.6
.E
ie 0.4
ee
~
00
1
2
3
4
Distanceto the fault inkilometers
5
Figure 4.115 Sag magnitude versus distance
for three-phase faults with fault
resistances
equal to zero (solid line),100/0 (dashed line),
20°,lc, (dash-dot line), and30% (dotted line)of
the source impedance.
Section 4.6 • Magnitudeand Phase-Angle Jumps for
Three-PhaseUnbalancedSags
229
o
8-10
-8~ -20
-
.8 -30
I
e,
,
,
§ -40
.
•
I
;
I
,
,
I
M -60
-70
-80
anglejump for three-phase faults with fault
resistances equal to zero (solid line),
to°A.
(dashed line),200/0 (dash-dotline), and 30%
(dotted line) of the source impedance.
I
I
u
~-50
f
I
,
,
."""'\
Figure4.116Sagmagnitudeversus phase-
I
-9°0
I
I
I
I
0.2
0.4
0.6
Sag magnitude in pu
0.8
magnitudeincreases due to the fault resistance, one
of the phasesactually drops in
voltage. The characteristicmagnitude is the difference between the two affected
that the phase-anglejump at the equipmenttermphases in the figure. We also see
inals only slightly exceeds 30°, despite the very large initial
phase-anglejump. The
-31.9°.
largest phase-anglejump occurs for a30% fault resistance at zero distance:
In Fig. 4.118 we seethat for a type D sag, the fault resistance increases the phaseangle jump in the phasewith the large voltagedrop, and that it raises oneof the
300/0 cause a small
other two voltages and reduces the
other. Fault resistances above
swell in oneof the phases.
For Figs. 4.117 and 4.118, the 11 kV system was assumed to
solidlygrounded.
be
Therefore, the zero-sequence source
impedancewas made equal to the positivesequence value. In reality this system is resistive
grounded:positive- andzero-sequence
source impedanceare significantly different. Thephase-to-neutralvoltage is much
lower in this case. Tocalculatethe phase-to-neutralvoltage a slightly revised version
of (4.38) has been used:
32s1
V-I _
an -
22F1 + ZFO
(4.128)
+ 2Z S1 + ZSO + 3R.Jzt
30
,
,, , ,
\
en
Q)
tb
20
.5
10
~
\
, , '\ ,
"
'"
......
'..<:~':..,
... "":~ ..
~
e
::s
.'""'\
0
Q)
bi>
; -10
Figure 4.117Magnitudeversusphase-angle
jump at theequipmentterminals for singlephase faults in a solidlygroundedsystem, sag
type C; fault resistances equal to zero (solid
line), 10% (dashed line),20% (dash-dotline),
and 300/0 (dottedline) of the source
impedance.
"/~~.
/1,
I
~
f
,1'-
" 1,-
-20
III,"
I':
I
,
"
-30
I
0
0.2
I
.' :
:'
:
0.4
0.6
0.8
Sag magnitudein
pu
230
Chapter4 • VoltageSags-Characterization
30
(I)
u
~
20
\
'\
u
"T;:)
.S
\
",\\
10 .
',\\
,
.. ~,~
Figure 4.118 Magnitude versus phase-angle
jump at the equipment terminals for single..
phase faults in a solidly grounded system, sag
type D, fault resistances equal to zero (solid
line), 10% (dashed line),20% (dash-dot line),
and 30% (dotted line) of the source
impedance.
-30
o
0.2
0.4
0.6
0.8
Sag magnitudein pu
5r----,-----.....--------.--------.
.. :--.~~~'?o'
" ..
/
,.
~.
/~:"
,1./, :
1/'
i, : :'"
\
\
\
\
\
"
"
"...' .
-10 '------'--------'-------'-------'
0.95
t
1.1
0.9
1.05
Sag magnitudei
....pu
Figure 4.119 Magnitude versus phase-angle
jumps at the equipment terminals for single..
phase faults in a resistance-grounded system,
sag type D; fault resistances equal to zero
(solid line), 50% (dashed line),100% (dashdot line), and150°A. (dotted line) of the source
impedance.
The influence of the fault resistance is small in this case, as can be seen in Fig. 4.119.
j ump at theequipmentterminalsare plottedfor a type
The magnitudeand phase-angle
D sag. Due to the small fault
currentsarc resistances can reach much higher values in a
resistance-grounded
systemthanin a solidly-groundedsystem. In thecalculationsleading to Fig. 4.119 fault resistances
equal to 50%, 1000/0, and 1500/0 of the positivesequence source
impedanceWere used. The main effectof large fault resistances is
that the sag becomes less severemagnitudeand
in
in phase-anglejump.
4.8.8 Meshed Systems
All calculationsin Sections 4.4and 4.5 were based on the
assumptionthat the
system is radial; thusthat we canuniquelyidentify a point-of-commoncoupling(pee), a
sourceimpedanceZs, and a feederimpedanceZF, as were shown in Fig. 4.14.
From
Fig. 4.14 weobtainedthe basicvoltagedivider equationfor the complex sag voltage:
V-I _
sag-
Zs
ZS+ZF
(4.129)
231
Section 4.7 • OtherCharacteristicsof Voltage Sags
In case the system loaded,we
is
can useThevenin'ssuperpositiontheoremwhich states
that the voltageduring the fault equals the voltage before the fault plus the change in
voltage due to the fault:
z,
(0)
V.vag = Vpee - Z
s+
Z
V(O)
F
f
(4.130)
with V~~e the pre-fault voltage at the pee andV}O) the pre-fault voltage at the fault
position. Notethat the source impedance
Zs includes the effecto f loads elsewhere in
the system.
For a meshed system we need
matrix methods to calculate voltage
during the
fault, asintroducedin Section 4.2.5. Weobtainedthe following expression (4.24) for
the voltageV k at node k due to a fault at node
f:
V
k
= V~O)
_
Zkf V(O)
Zff f
(4.131)
with ViOl the voltage at nodek before the fault andvjO) the voltage at the fault position
Comparingthis
before the fault, andZij element ij of the node impedance matrix.
equationwith (4.129) we seethat they have the same
structure.The voltage divider
model can be used for meshed systems, when the following source and feeder impedances are used:
z, = Zk/
ZF
= Zff -
(4.132)
Zk/
(4.133)
The main difference isthat both Zs and ZF are dependenton the fault location.
Equivalentsource and feeder impedances canobtainedfor
be
positive-, negative-, and
zero-sequence
networks,and all the previously discussed analysis can still be applied.
4.7 OTHER CHARACTERISTICS OF VOLTAGE SAGS
4.7.1 Point-on-Wave Characteristics
The voltage sagcharacteristicsdiscussedhitherto (magnitude, phase-angle
jump,
three-phaseunbalance)are all related to thefundamental-frequency
componento f the
voltage. They require the
calculationof the rms value of the voltage or the complex
voltage over aperiod of one half-cycle or longer. We saw earlier how this leads to an
uncertaintyin the calculationof sagduration.To obtaina moreaccuratevalue for the
sagdurationone needs to be able to
determine"start" and "ending" of the sag with a
higher precision.For this one needs to find the so-called
"point-on-waveof sag initiation" and the "point-on-wave of voltage recovery" [38], [134]. Both require more
advanced analysis techniques, which are still under development. We will see in the
next chapterthat the point-on-wavecharacteristicsalso affect the behavior of some
equipment.
4.7.1.1 Point-on-Waveo f Sag Initiation. The point-on-waveof saginitiation is
the phase angle
o f the fundamentalvoltage wave at which the voltage sag starts. This
anglecorrespondsto the angle at which theshort-circuitfault occurs. As most faults
are associated with a flashover, they are more likely to occur near voltage maximum
than near voltage zero. In the sag shown in Fig. 4.1 point-on-waveof
the
sag initiation is close to voltage maximum. In Fig. 4.9 sag
initiation takes placeabout 35°
232
Chapter4 • VoltageSags-Characterization
after voltage maximum, at least in the phase with the largest voltage
drop. In other
phases the event
startsat anotherangle comparedto the fundamentalvoltagein that
phase.
When quantifying the point-on-wavea referencepoint is needed. Theupward
zero crossingof the fundamentalvoltage is anobvious choice. One is likely to use
the last upward zero crossing of the pre-event voltage as reference, as this closely
resembles thefundamentalvoltage. The sag shown in Fig. 4.1 partly
is
repeatedin
Fig. 4.120: one cycle (1/60
o f a second)startingat the lastupwardzero crossing before
sag initiation. We seethat the point-on-waveof saginitiation is about 275°. A closer
look at the data learns that this point is between 276° and 280°. The slope at the
beginningof the sagactually takes 4°, orabout 185 j.LS. This is probably due to the
low-passcharacterof the measurementcircuit.
was in
Figure 4.12I plots all three phases of the sag for which one phase plotted
Fig. 4.120.For each phase, the zero
point of the horizontalaxis is the lastu pwardzero
crossingbefore thestart of the event inthat phase. We see
t hat the point-on-waveis
different in the three phases. This obviousif
is
one realizesthat the eventstartsat the
samemomentin time in the three phases. As the voltage zero crossings are 120° shifted,
2
--r-·_···~----·r--·---r----'-----r-1
1.5
0.5
;
~
0
F--------~-----ft--~--t
-0.5
-1
-1.5
- 2 '----_-'---
o
i_:
o
'-:
o
50
- A . - - _ - - ' - - _ - - - ' - _ - - ' -_ _ -L..J
100 150 200 250
300 350
Angle of voltage wave in degrees
50
100
150
50
100
150
~~::1
200
250
200
250
300
350
/1
300
350
i_:P=~
o
50
Figure 4.120Enlargemento f the sag shown
in Fig. 4.1indicatingthe point-on-waveof sag
initiation.
100 150 200 250 300
Angle of voltage wave in degrees
Figure 4.121 Eventinitiation in the three
350 phases,comparedto the lastupward voltage
zero crossing.
233
Section 4.7 • Other Characteristicsof Voltage Sags
the point-on-wavevalues differ by 120
°. In casephase-to-phase
voltages are used, the
resultingvalues are again different. When
quantifying point-on-wave it is essential to
clearly define the reference
.
4.7.1.2 Point-on-Waveof Voltage Recovery. The point-on-waveof voltage recovery is the phase angle of the
fundamentalvoltage wave at which the main recovery takes place. We saw before
that most existing powerquality monitors look for
the point at which the voltage recovers to 90% or 95% of nominal
the
voltage. Note
that there is in many cases no link between these two points
. Consideras an example
of this section takes
again the sag shown inig.
F 4.1. Voltage recovery in the meaning
placeabout 2.5 cycles after sag
initiation, even though the voltage does not fully- re
cover for at leastanothertwo cycles, as can be seen in Fig. 4.3.
Voltage recoverycorrespondsto fault clearing, which takes place currentzero
at
crossing. Because the power system is mainly
inductive,current zero crossing corresponds to voltage maximum
. Thus we expect points-on-wave of voltage recovery to be
around90° and 270°. This assumes that we use the pre-event
fundamentalvoltage as
reference,not the during-event voltage
. It is the pre-event voltage which drives the fault
currentand which is thus 90
° shiftedcomparedto the faultcurrent.The recovery of the
sag in Fig. 4.120 is shown in Fig. 4.122. The recovery is, at least in this,case
slower than
the saginitiation. The shape of the voltage recovery
correspondsto the so-called
" transient recovery voltage" well-known in
circuit-breakertesting. The smoothsinusoidal curve in Fig. 4.122 is the continuationof the pre-eventfundamentalvoltage.
Considering thestart of the recovery
, we find a point-on-waveof 52°. If we further
assume this to be the moment of fault-clearing taking place
currentzero,
at
we seethat
the currentlags the voltage by 52
°, which gives anX/R ratio at the fault position equal
to tan-I(52 °) = 1.3.
For a two-phase-to-ground
or three-phasefault, fault clearing does not take place
in all three phases at the same time
. This could make adeterminationof the point-onwave of voltage recovery difficult. Anunambiguousdefinition of the referencepoint
and phase is needed to apply this
conceptto three-phaseunbalancedsags.
1.5
0.5
j
s
0
- 0.5
-I
Figure 4.122 Enlargement of Fig
. 4.1
showing thepoint-on-waveof voltage
recovery. The smoothcurve is the
continuationof the pre-sagfundamental
voltage.
- 1.5
o
50
100
150 200
250
Time in degrees
300
350
234
Chapter4 • VoltageSags-Characterization
4.7.2 The MI••ing Voltage
The missing voltage is
a nothervoltage sagcharacteristicwhich has beenproposed
recently [134]. The missing voltage is a way
o f describing the change in
momentary
voltage experienced by the
equipment.The conceptbecameimportantwith the dimensioning of series-connected
voltage-sourceconvertersto compensatefor the voltage
drop due to the fault. We will see inChapter7 that the voltage injected by the series
compensatoris equal to the missing voltage: the difference between the voltage as it
would have beenwithout the sag, and theactual voltage during the sag.
4.7.2.1 The Complex Missing Voltage.
One can thinkof the missing voltage as
a complex voltage (aphasor),being the difference in the complex plane between the
pre-event voltage and the voltage
during the sag. Theabsolutevalue of this complex
missing voltage can be directly read from a plot like shown in Fig. 4.83. In Fig. 4.83
the missing voltage is the
distancebetween the complex voltage
during the sag (which
top-right corner of the diagram (the point
is on one of the three curves) and the
I + jO).
EXAMPLE Consider a sag on a 50 mrn?
undergroundcable, like in Fig. 4.83, with a
sag magnitudeof 600~. If the pre-event voltage was
100%, the drop in rms value of the vola
tage is40°A.. Having no furtherinformation one would be tempted to say thatcompensator
should inject a voltage with an rms value equal to 40%
of nominal.
Looking in the complex plane, we see
that a magnitudeof 60% correspondsto a complex
voltageV = 0.45 - jO.39. The missing voltage is the difference between the pre-fault voltage and
the voltage during the sag, thus 117- = 0.55+ jO.39. The absolute value
o f the missing voltage is
67% in this example.Comparethis with the 40% drop in rms voltage.
The complex missing voltage can also calculatedfrom
be
the magnitude V and the
phase-anglejump l/J of the sag. The complex voltage
during the sag is
V
= V cos q,+ jV sin q,
(4.134)
The missing voltage is simply
1- V= 1- Vcosq,-jVsinq,
(4.135)
=JI -
(4.136)
with as absolutevalue
Vmiss= 11 - VI
V2- 2 V cosl/J
When we neglect the
phase-anglejump, thus assumethat V = V, the missing voltage is
simply Vmiss = 1 - V. We can assess the
errormade by writing 1 - V = JI + V 2 - 2V.
Comparingthis with (4.136) gives for the difference between the exact andapproxthe
imate expression for the missing voltage:
2
Vmis,f -
-2
V miss = 2V(1 -
cosq,)
(4.137)
4.7.2.2 The Missing Voltage in Time Domain.
The conceptof missing voltage
can become much more useful by
extending it to time domain. A very first step
would be to look at the difference between the
fundamentalpre-event voltage and
the fundamentalduring-eventvoltage. Butthat would not give any extra information
comparedto the complex missing voltage.
235
Section 4.7 • Other Characteristicsof Voltage Sags
2 .-----.,..---.,.------r----,------,-----,
i~
0
-1
234
Timein cycles
5
6
2.---r---,-----.-----r-----r-----.
u
01)
Figure 4.123·T ime-domainvoltage
measurementtogetherwith pre-event
fundamentalvoltage(top curve) andthe timedomainmissing voltagebeing thedifference
of those two(bottom curve).
~
~
0 ..............."'--'~
-1
-2
0
234
Time in cycles
5
6
In the top part of Fig. 4.123 the sag from Fig. 4.1 has been
plotted again.
Togetherwith the actualtime-domainvoltage wave, thefundamentalpre-event voltage
has beenplotted.The latter is obtainedby applyinga fast-Fourier-transformalgorithm
to the first cycle of the voltage wave form.
From the complex coefficient for the fundamental term in theFourierseries Ct , the (time-domain)fundamentalcomponentof the
voltage can becalculated:
(4.138)
This fundamentalc omponentof the pre-eventvoltage (pre-eventfundamentalvoltage,
for short) is the smoothsinusoidalcurve in the toppart of Fig. 4.123.
The missing voltage is
calculatedas the difference between the
actualvoltage and
the pre-eventfundamentalvoltage:
(4.139)
This missing voltage isplottedin the bottompart of Fig. 4.123. Before theinitiation of
the sag 'there isobviously no fundamentalcomponentpresent; during the sag the
fundamentalcomponentof the missing voltage is large;
after the principal sag (after
fault clearing) a smallfundamentalcomponentremains. The reason for this becomes
clear from theuppercurve: the voltage does not immediately fully recover to its preevent value.
Figure4.124repeatsthis for the voltage in oneof the non-faultedphases, for the
same event as in Fig. 4.123 and Fig. 4.1. In the top curve wethat
seethe during-event
voltage has alarger rms valuethan the pre-eventvoltage. In termsof rms voltages, we
would call this an increase in voltage: a voltage swell. looking
But
at the missing voltage
it is not possible to saywhetherthe underlyingevent is a swell or a sag. This might be
shouldrealizethat this
seen as adisadvantageof the missing voltage concept. But one
conceptis not meant to replace theother ways of characterizingthe sag;instead,it
should giveadditionalinformation.
Finally, Fig. 4.125 plots the missing voltage in all three phases. As expected for a
single-phase-to-ground
fault, the missing voltage in the two
non-faultedphases is the
same and in phase with the missing voltage in the faulted phase. After the fault the
missing voltages in the three phases form a positive sequence set. This
probablydue
is
to the re-accelerationof induction motorsfed from the supply.
236
Chapter4 • VoltageSags-Characterization
t:~
- 20
1
2
3
4
Time in cycles
5
6
f_: ~
1
-2 0
2
3
4
Time in cycles
5
6
Figure 4.124 Measured voltage with preevent fundamentalvoltage (top curve) and
missing voltage(boltom curve) during a
voltage swell event.
~.:~
-2 0
I
-2 0
1
- 20
I
2
3
4
5
6
~:~
2
3
4
5
6
~.:~
2
3
4
Time in cycles
5
6
Figure 4.125 Missing voltage for the three
phasesof a sag due to a single-phase fault .
In Figs. 4.124 and 4.125 we used the
fundamentalpre-event voltage as a reference
to obtain the missing voltage. The
conceptof missing voltage has been
introducedto
quantify the deviationof the voltage from its ideal value. In
otherwords: we have used
the fundamentalpre-event voltage as the ideal voltage. This could become
point
a of
discussion, as there are at least three
alternatives:
• Use the full pre-event waveform, including the
harmonicdistortion, as a reference. One can either take the last cycle before the event or the average over a
numberof cycles. Thelatter option is limited in its applicationbecause there
are normally not more than one or two pre-event cycles available.
• Use thefundamentalcomponentof the pre-event waveform as a reference. One
can again choose between the
fundamentalobtainedfrom the last cycle before
the event (as was done in Fig. 4.124 and Fig. 4.125)
obtain
or the fundamental
from a numberof pre-event cycles.
• Use as a reference, sinusoidalwaveform
a
with the sameamplitudeand rms
value as the system nominal voltage and the same phase angle as the fundamental pre-event waveform. The difference between the last two alternatives is
237
Section 4.7 • OtherCharacteristicsof Voltage Sags
the same as the discussion between defining the voltage
drop with reference to
the pre-event rms voltage or with reference to nominal
the
rms voltage. Both
methodshave theiradvantagesand can thus be used. But it important
is
to
alwaysindicatewhich methodis used.
4.7.2.3 Distributionofthe Missing Voltage. An alternativeand potentially very
useful wayof presentingthe missing voltage isthrough the amountof time that the
missing voltage, inabsolutevalue, exceeds given values; other
in
words, theamount
of time during which the deviation from the ideal voltage waveform is larger
than a
given value.
In the top curveof Fig. 4.126 the missing voltage from Fig. 4.123 is shown again.
But this time theabsolutevalue isplotted,insteadof the actualwaveform. We see, e.g.,
that this absolutevalue exceeds the value of 0.5,total
a of six timesduring the event.
The cumulativedurationof these six periods is 1.75 cycles. The
cumulativetime during
which the missing voltage in
absolutevalue exceeds a given level can determinedfor
be
each level. The result
of this calculationis shown in thebottompart of Fig. 4.126. This
curve can be read as follows: the missing voltage is never larger
than 1.53, isduring 1
cycle larger than 0.98, during 1.75 cycle largerthan 0.5, during two cycles largerthan
0.32, etc. The long tail in Fig. 4.126 is due to the
post-faultvoltage sag as well as to the
non-zeropre-event missing voltage. The
latter contributioncan be removed by
either
using the full pre-event waveshape as a reference
calculatethe
to
missing voltage, or by
only consideringthe missing voltage samples from the
instantof sag-initiationonward.
Throughthe sameprocedure,distributionsof the missing voltage can be
obtained
for the other two phases, resulting in the curves shown in Fig. 4.127. The missing
voltage in the faulted phase (solid curve) naturally
is
larger than in the non-faulted
phases. But still, the missing voltage in the
non-faultedphases is significant:during
about1 cycle it exceeds a value of 0.4. We also see a small difference in missing voltage
between the twonon-faultedphases: the value in phase bsomewhathigher
is
than in
phase c.
of definThe missing voltagedistributioncurve can be used as a generalized way
ing the eventduration.The larger thedeviationfrom the ideal voltage one considers,
the shorter the "cumulative duration" of the event. Thecumulative duration of a
2r----..---r------r----~--,-------,
II)
11.5
o
>
.Ef
1
.~ 0.5
~
°0
234
5
6
Timein cycles
2 r-----r---..----r---~--
Figure 4.126 Absolute value of the missing
voltage (top curve) and the
distributionof the
missing voltage(bottom curve) for the sag
shown in Fig. 4.1.
234
Cumulativetimein cycles
5
6
238
Chapter4 • VoltageSags-Characterization
Cl
r------r-----r---.------.---·-..----l
1.5
.2
:s
~
1
fI'.I
~
;
~
L
.S 0.5 '- .. "" '-_'- fI'.I
fI'.I
-
-
-
_--
~
_ '_',-,
.. ....,
~,
.... ':..-...-_...:: :. -- ---:= "::. ----- = .... --
0'
,
o
0.5
,
I
_L-_>______---'
1
1.5
2
Cumulative time in cycles
2.5
3
Figure 4.127 Missing voltaged istribution for
phase a (solid curve), phase(dashedcurve),
b
and phase c(dash-dotcurve).
voltage sag for a givendeviation would be defined as the
t otal amountof time during
which the voltage deviates more
thanthe given value from the ideal
voltagewaveshape.
4.8 LOAD INFLUENCE ON VOLTAGE SAGS
In the calculationof sagmagnitudefor varioussystemconfigurations,in the classification of three-phasesags and in mosto f the examples, we have
assumedthat the load
currentsare zero. In this section we will discuss some
situationsin which the load
currentscan have a significant influence on the voltages
during a fault. The main
load having influence on the voltage
during and after a sag isformed by induction
and synchronousmotors as they have the largest
currentsduring and after a shortcircuit fault. But we will also briefly discuss single-phase and
three-phaserectifiers as
they are a largefraction of the load at manylocations.
4.8.1 Induction Motors and Three-Phase Faults
During a three-phasefault the voltages at the
m otor terminalsdrop in magnitude.
o f this drop are twofold:
The consequences
• The magneticflux in the air gap is no longer inbalancewith the statorvoltage.
During this decay
The flux decays with a timec onstantof up to several cycles.
the induction motor contributesto the fault andsomewhatkeeps up the voltage at themotor terminals.
• The decay in voltage causesdrop
a in electrical torque: the electricaltorqueis
proportional to the squareof the rms valueof the voltage. Themechanical
torque in the mean time remains largely
unchanged.The result isthat the
motor slows down. While themotor slows down it will take alarger current
with a smaller power factor. This could bring down the
voltageeven more.For
small voltagedrops, a new steadystatecould be reached at a lower speed,
dependingon the speed-torquebehavior of the mechanicalload. For deep
sags themotor will continueto slow down until it reachesstandstill, or until
the voltage recovers, whichever comes first. The
mechanicaltime constantof
electricalmotorsis of the orderof one secondand more. Thereforethe motor
will normally not have reached zero speed upon
yet voltagerecovery.
Section 4.8 • LoadInfluenceon Voltage Sags
239
The momentthe voltage recovers the
oppositephenomenaoccur. The flux in the
air gap will build up again. This causes a large
inrush current, which slows down the
voltage recovery. Afterthat, the motor will re-accelerateuntil it reaches its pre-event
speed.During the re-accelerationthe motor againtakes a largerc urrentwith a smaller
power factor, which causes post-faultvoltage
a
sag sometimes lasting for several seconds.
The contribution of the induction motor load to the fault can be modeled as a
voltage sourcebehindreactance.The voltagesource has a value
o f about 1 pu at fault
initiation and decays with thesubtransienttime-constant(between0.5 and 2 cycles).
The reactanceis the leakagereactanceof the motor, which is between100/0 and 20% on
the motor base.Note that this is not the leakagereactancewhich determinesthe starting current, but the leakagereactanceat nominal speed.For double-cageinduction
machines these two can be significantly
different.
EXAMPLE Considera bolted fault at primary side of a 33/11 kV transformerin the
supply shown in Fig. 4.21. Thetotal induction motor load connectedto the 11 kV bus is50/0
of the fault level. Theinduction motors have a leakagereactanceof 10% on the motor base.
We are interestedin the voltage at secondaryside of the transformer.Consideronly the reactive part of the impedances.
33 kV and II kV fault levels:ZT
The transformerimpedanceis the difference between the
= 47.60/0 at a 100 MVA base. The fault level at
II kV is 152 MVA, thus the total motor load is
(5% of this): 7.6 MVA. The leakagereactanceof the motorsis 100/0 at a 7.6 MVA base, which is
Z M = 132%at a 100MVA base. Thevoltageon secondarysideof the transformeris found from
the voltage dividerequation:
V/oad = Z ZT
= 27%
T+ Z M
(4.140)
To assess the increase motor
in
current after the fault, we use thecommon
equivalentcircuit for the induction motor, consistingof the seriesconnectionof the
statorresistanceRs, the leakagereactanceXL and theslip-dependentr otor resistance
~, with s the motor slip. The motor impedanceis
ZM
= s,+jXL + RR
s
(4.141)
The changeof motor impedancewith slip has beencalculatedfor four induction
motorsof four different sizes.Motor parametershave beenobtainedfrom [135], [136],
and themotorimpedancehas beencalculatedby using(4.141). The results are shown in
Fig. 4.128. For each motor, the impedanceat nominal slip is set at I pu, and the
absolutevalue of the impedanceis plotted betweennominal slip and 25% slip. We
m otor
see for eachmotor a decrease inmotor impedance,and thus an increase in
current,by a factor of aboutfive. The decrease inimpedanceis much faster for large
machinesthan for smaller ones.
If we assume the voltage to recover to 1immediatelyupon
pu
fault clearing, the
currenttakenby themotoris the inverseof the impedance(bothequal to 1pu innormal
operation).The path of the currentin the complex plane is shown in Fig. 4.129. The
pathis given for an increase in slip from its
nominalvalue to 250/0. The positive real axis
is in the direction of the motor terminal voltage. For small motors we seepredominantly an increase in resistive
c urrent, for large motors the main increase is in the
inductive part of the current. When the slip increases
further, even the resistivepart
240
Chapter4 • VoltageSags-Characterization
I:
8
-ae
0.8
Jg
,:
:\
, "
I
\
\
\
\
\
\
",
\
\
\
\
I
\
\
0.6
~
\
~ 0.4
'i
~
,
\
\
\
0.2
Figure 4.118Induction motor impedance
0.05
0.1
0.15
0.2
0.25
Motor slip
versusslip; the impedanceat nominal slip is 1
pu; 3 hp 220 V (solid line), 50 hp 460 V
(dashedline), 250 hp 2300 V(dotted line),
1500 hp 2300 V(dash-dotline).
o.-------,.------r----~----..
-I
'"
,
""
"
"
\
"
"
\
\
"
:
\
,
I
,
,
I
-5
2
Resistivemotor current
3
4
Figure 4.129Changein induction motor
currentwith increasingslip; the currentat
nominalslip is 1 pu; 3 hp 220 V (solid line), 50
hp 460 V(dashedline), 250 hp 2300 V(dotted
line), 1500 hp 2300 V(dashedline).
of the currentstartsto decrease. The power factor of the
currentdecreases significantly,
especially for largemotors.
The influenceof large induction motorson voltage sags is described in detail by
Yalcinkaya [136]. Fig. 4.130 shows the voltage sags (top curve) and motor
the
slip
(bottomcurve) due to athree-phasefault in an industrialsystem with a largeinduction
motor load. Without induction motor load, the voltage would have been zero during
the sag and 1 pu after the sag. The voltage
plottedin Fig. 4.130 is theabsolutevalue of a
time-dependentphasor,used in atransient-stabilityprogram.The effectof the induction motor load is that the voltageduring the fault is increased, and after the fault
decreased. The slip
o f all motors increases fast during the sag, and even continues to
increase a bitafter fault clearing.
The voltage after fault clearing, the so-called
post-faultsag, shows anadditional
decreaseabout 200 ms after fault clearing. Thiscorrespondsto the momentthe motor
starts to re-accelerate and draws larger currents. The low voltage immediately after
fault clearing is due to the large
currentneeded to rebuild the air gap flux.
During the fault theinduction motors significantly keep up the voltage. Even
toward the endof the sag the voltage at the
motor busses is still above100/0 of its
pre-event value.
Section 4.8 •
241
Load Influence on Voltage Sags
1.0
0.9
0.8
::l
0.
0.7
.S 0.6
~ 0.5
~
'0 0.4
::> 0.3
0.2
0.1
, 1 , ,, , 1,,,,1,,,,1
0.0 +-r-..,...,...-.-+-..,...,...- r-rr-+-r--,--,-r+-,--,-,--,-h-r-rr-T+-r-rr--r-r-rl
0.5
1.0
1.5
2.0
2.5
3.0
Time in seconds
3.5
3.0
..:
.
C 2.5
~
8. 2.0
.S
.9- 1.5
U;
;:§
1.0
0.5
Figure 4.130 Voltage sag
( top) and induction
motor slip (bottom) for three busses in an
industrial power system.(Reproducedfrom
Yalcinkaya (136).)
0.5
1.0
1.5
2.0
Time in seconds
2.5
3.0
One should realizethat this is a somewhat exceptional case, as motor
the load
connected to the system is very large. Similar but less severe effects have been noticed in
othersystems.A notherphenomenonwhich contributesto the post-fault voltage sag is
that the fault occurs in one of two parallel transformers. The
protectionremoves the
faulted transformer,so that only onetransformeris available for the supply after fault
clearing. Thepost-fault fault level is thus significantly less than its pre-fault value. A
similar effect occurs for a fault in one of two parallel feeders. The post-fault sag,
described here for three-phase faults, has also been observed after single-phase faults.
4.8.2 Induction Motors and Unbalanced Faults
The behaviorof an induction motor during an unbalanced fault is
rathercomplicated . Only a network analysis
programsimulating a large parto f the system can
p laya
give an accuratepicture of thequantitativeeffects. The following phenomena
part in the interactionbetween system and induction
motor during unbalanced faults.
• During the first one or two cycles after fault initiation the
induction motor
contributesto the fault. This causes an increase in positive-sequence voltage.
Negative- and zero-sequence voltage are not influenced.
• The induction motor slows down, causing a decrease in positive-sequence
impedance. This decrease in impedance causes an increase in current and
thus adrop in positive-sequence voltage.
242
Chapter 4 • VoltageSags-Char
acterization
• The negative-sequence impedance of motor
the is low, typically 10-20%of the
nominal positive-sequence impedance . The negative-sequence voltage due to
the fault will thus be significantlydampedat the motor terminals. The negative-sequence impedance independentof
is
the slip. The negat ive-sequence
voltage will thus remain constant
during the event.
• The induct ion motor does not take any zero-sequence
current. The zerosequence voltage will thus not be influenced by induction
the
motor.
4.8.2.1 Simulation Example.Simulationsof the influence ofinduction motor
loads on unbalancedsags are shown in[136], [137]. Some of those results are reproduced here. The systemstudied was a radial one with large induction
m otor
load connectedto each of the low-voltage busses.
Motor sizes andtransformerimpedances were chosen such
that for each bus the fault level
contribution from the
source was 15 times the total
motor load fed from the bus. Voltages and
currents
in the system werecalculated by using the transient analysis packageEMTP. All
transformers in the system wereconnected star-star with both neutral points
earthed .Although this is not a verycommon arrangement
, it helped in understand
ing the phenomena. The voltages at the terminals of one of the motors are shown
of type
in Fig. 4.131.Without induction motor influence we would have seen a sag
B of zero magnitude: zero voltage in phase a, and no change in the voltage in
- _.~--~--~--~----,
.,
_$
~
3000
2000
1000
'"
0
~ - 1000
..d
p... - 2000
- 3000
111111111/\/\/\
11v v
v v v v v v v v v
o'----o.~I---O.~2---0.3--~---'
0.4
0.5
3000
E 2000
'0
>
1000
0
~ -1000
..d
e, -2000
-3000
.0
oL----lL.:...:--:----::'-:----:--:-0.1
0.2
0.3
0.\
--::''-:'''''- - : '
0.4
0.5
0.4
Section 4.8 •
243
Load Influence on Voltage Sags
phase b and phasec. Instead we see a smallnon-zero voltage in phase a and in
the two non-faultedphasesan initial increasefollowed by a slow decay. After fault
clearing the system becomes balanced again, and the three phase voltages thus
equal in amplitude. The motor re-accelerationcausesa post-fault sag of about 100
ms duration.
The non-zerovoltagein the faulted phaseis due to thedrop in negative-sequence
voltage. We saw in (4.32) and (4.34) that the voltage in the faulted phaseduring a
single-phasefault is given as
(4.142)
Theeffect of the inductionmotor is that V2 dropsin absolutevalue,causingan increase
in voltagein the faulted phase.
During the sag, thepositive-sequence
v oltagealso drops,which showsup as the
slow but steadydecreasein voltagein all phases.
The non-faultedphasesshow an initial increasein voltage. The explanationfor
this is as follows.The voltagein the non-faultedphasesduring a single-phasefault is
madeup of a positive-sequence,
a negative-sequence,
and a zero-sequence
c omponent.
For phasec this summationin the complex planeis for the systemwithout induction
motor load.
Vc
2
= Vel + VcO + Vc2 = -a
3
1
3
1
3
- - -cl
=a
(4.143)
Due to the induction motor load, the positive-sequencev oltage will not immediately
dropfrom 1 pu to 0.67pu. The negative-sequence
voltagewill jump from zeroto its new
value immediately. The consequenceis that the resulting voltage amplitude slightly
exceeds itspre-fault value. After a few cycles theinduction motor no longer keeps up
the positive-sequencevoltage. The voltage in the non-faultedphasesdrops below its
pre-eventvalue due to negative-and positive-sequencevoltagesbeing less than 33%
and 67%, respectively.
The currentstaken by the induction motorsare shown in Figs. 4.132and 4.133.
Figure 4.132showsthe motor currentsfor a motor with a small decreasein speed.The
slip of this motor increasesfrom 2% to 6% during the sag.The motor shownin Fig.
4.133 experienceda much largerdecreasein speed: its slipincreasedfrom 3% to 19°A>.
This behavioris difficult to explain without consideringsymmetricalcomponents.But
generallywe canobservethat the currentincreasesinitially in the faulted phase,rises to
a higher value in one of the non-faultedphases,and initially drops in the other nonfaulted phase.The current in the secondnon-faultedphaserises again after a certain
time, determinedby the slowing down of the motor.
For the motor shownin Figs. 4.131 and 4.132the componentvoltagesand currentshavebeen plotted in Figs. 4.134and 4.135. From Fig. 4.134we seethat negative
and zero-sequencevoltage remain constant during the sag, but that the positivesequencevoltageshowsa steadydecay,due to the decreasein positive-sequence
impedancewhen the motor slows down. Figure 4.135clearly showsthe increasein positivesequencecurrentwhen themotor slows down. The zero-sequence
c urrentis zero as the
motor windings are connectedin delta. From Figs. 4.134and 4.135the positive- and
negative-sequence
i mpedanceof the motor load can be calculated,simply through
dividing voltage by current. The resultsare shown in Fig. 4.136,where we seeagain
that the negative-sequence
impedanceremainsconstant,whereasthe positive-sequence
impedancedrops.When the motor reachesstandstill,it is no longera dynamicelement,
and positive- and negative-sequence
impedancebecomeequal.
244
Chapter4 • VoltageSags-Characterization
150
J I~~
tlS
M
~
0
-50
i- IOO
-150 --------'~----'''--_.-'--0.1
o
0.2
0.3
'---_--J
0.4
0.5
150
=
~
~~ 500 ~ "11""
~ ~ "JII'1,HflJIJlI1IJlIIlI
~
100
-a
-50
j~A~~~~1
, ~ ~ ~ ~ V~ ij.
~ V~
~-100
-150 ~--"--o
0.1
0.2
0.3
0.4
0.2
0.3
0.4
0.1
,,--_ _a . - - _ - - J
0.5
Time in seconds
4.8.2.2 Monitoring Example. An exampleof a three-phaseunbalancedsag was
shown in Fig. 4.48. The severe
post-faultsag indicatesthe presenceof induction motor load. For each of the three sampledwaveforms,the complex voltage as a function of time wasdeterminedby using themethoddescribedin Section 4.5.From the
three complex voltages, positive-, negative- and
zero-sequencevoltages have been calculated. Their absolutevalues areplotted in Fig. 4.137 as afunction of time. The
zero-sequencecomponentis very small. The negative-sequence
c omponentis zero
when the fault is notpresentand non-zerobut constantduring the fault. The positive-sequence voltage is I pu before the fault, shows a slow decay
during the fault,
and a slow increase
after the fault. This is exactly incorrespondence
with the abovedescribedtheory and simulation results.
4.8.2.3 Simplified Analysis. From the simulation and monitoring results we
can extractthree stages in the voltage sag:
• The inductionmotor feeds into the fault, raising the
positive-sequence
voltage.
• The positive-sequence
voltage is the same as it would have been
without the
induction motor load.
• The induction motor has sloweddown, drawing additional positive-sequence
current,thus causingthe positive-sequence
voltageto drop.
245
Section 4.8 • LoadInfluenceon Voltage Sags
4000
=
g 3000
='
2000
1000
~
]-10~
': -2000
~
:E - 3000
-4000
"""'--_ ____'__ _- I
L -_ _--'--_ _- - ' "
o
0.2
0.1
0.3
0.4
0.5
4000
= 3000
~
2000
1000
.rJ
.i -10000 II \II H\1 UIII 1111II' 1111" 11111HI H1I
c: ~2000
GJ
~ -3000
- 4000
L . . -_ _
..o.--_ _
o
-'--_~__'__ ____'___ ___I
0.2
0.1
0.3
0.4
0.5
4000
=
3000
~ 2000
~
1000
M 0
-1000
':' - 2000
~ -3000
-4000
..d
Figure 4.133Induction motor currents
during and after a single-line-to-groundfault
in the supply. Thismotor showeda large
decrease in speed.
(Reproducedfrom
Yalcinkaya[136].)
~
~
L . -_ _
- ' - -_ _
---"
o
0.1
0.2
0.3
Timein seconds
0.5
0.4
80 , . . . - - - - - - - - - - - - - - - - - - -
ijo
Positive-sequence
voltage
60
~
e,
.5 40
i
~ 20
Figure 4.134Symmetricalcomponentsfor
the voltagesshown in Fig. 4.131.
(Reproducedfrom Yalcinkaya[136].)
Zero-sequence
voltage
----------------Negative-sequence
voltage
....................... -
Ot------+-----+-----+-----&-....J
100
50
150
200
250
Timeinmilliseconds
fj
~
170·.,..------------------.
Negative-sequence
current
J50
&J30
.S
1: 110
8t:
Figure 4.135 Symmetricalcomponentsfor
the currentsshown in Fig. 4.132.
(Reproducedfrom Yalcinkaya[136].)
.
.,.,..".--
____ -- -'
.,.""...--
-_.....-.----
Positive-sequence
current
90
70 .....-_+-_--.-._-+-_ _--+---+---o.....--._~
90
110
130
150
170
190
Timeinmilliseconds
210
230
250
246
Chapter 4 • VoltageSags-Characterization
6 80
()
[ 60
c=
.;; 40
s
i 20
~
~gativ~s~~n~m..£e~a~e_
Figure 4.136 Positive- andnegative-sequence
230 250 impedance for an induction motor during a
sag. (Reproduced from Yalcinkaya
[136].)
O~---i---+--+--+--+---+--+--+----'
90
110
130 150 170 190 210
Timeinmilliseconds
:::s
a.
.;; 0.8
=
J
~
0.6
5
i= 0.4
o
o
0.2
5
15
10
Timein cycles
Figure 4.137 Positive-, negative- and zero..
sequence voltages for the three-phase
unbalanced sag shown in Fig. 4.47.
The negative-sequence
v oltage is constantduring the fault, but lower than without
induction motor load. To quantify the effect of induction motors, we use atwo-step
v oltage
calculation procedure.At first we calculate positive- and negative-sequence
(V~no), V~no» for the no-load case. As we sawbefore this will lead to voltage sags of
type C or type D with different characteristicmagnitude.We assumeda' zero characjump. As a secondstep the influence of the induction motor is
teristic phase-angle.
incorporated.For this we model the supply as a sourcegeneratinga type C or type
D sag, with a finite sourceimpedance.Note that this is a three-phaseTheveninsource
representationof the supplyduring the fault. Theeffect of the inductionmotor load is a
difference betweenthe sourcevoltagesand the voltagesat the motor terminals, for
positive as well as for negative-se~uence components.T he voltageat the motor terminals are denotedas V}/oaa) and V 2/oad). For the three above-mentioned"stages"these
relationsare assumedto be as follows:
1. The drop in positive-sequencevoltage is reduced by 15%, the negativesequencevoltagedrops by 300/0.
V~/oad) = 0.15 + 0.85V}no)
V~/oad)
= O.7 V~no)
2. The negative-sequence
voltagedrops by 30%.
_ V(no)
V(/oad)
I
1
V~load) = 0.7 vjno)
Section 4.8 •
247
Load Influence on Voltage Sags
3. The positive-sequence voltage
drops by
dropsby 300/0.
100~,
the negative-sequence voltage
V~load) = 0.9V~no)
V~load) =
O.7 V~no)
The voltages at themotor terminals are calculated from the positive- and negativeV~load) and V~load). The resulting phase voltages for the three stages
sequence voltages
For sag type C the voltages are shown for one
of
are shown in Figs. 4.138 and 4.139.
the phases with a deep sag, and for the phase with a shallow sag. The more the
motorsslow down, the more the voltage in this phase drops. The voltage in the worstaffected phase is initially somewhat higher due to induction
the
motor influence, but
dropswhen themotor slows down and the positive-sequence voltage
dropsin value as
well. For type D we seethat the voltage in the least-affected phases
dropsduring all
stages of the sag. The voltage in the worst-affected phase increases initially
but
decreases later.
Figure 4.138 Voltages at the equipment
terminals, for three stages of induction
motor
influence for type C sags. The solid lines are
without induction motor influence, the
dashed lines with.
~ o.~!~~;~~-~~---~~---------~-----I
~Q~
~
Figure 4.139 Voltages at the equipment
terminals, for three stages of induction
motor
influence for type D sags. The solid lines are
without induction motor influence, the
dashed tines with.
j
i
tOt - . .
1
~=I
O.5~_
00
0.2
0.4
0.6
Characteristicmagnitude
0.8
I
248
Chapter4 • VoltageSags-Characterization
From the curves in Figs. 4.138
a nd 4.139 we can see the following two
patterns:
• The lowestvoltageincreases, the highest
voltage.decreases,thus the unbalance
becomes less. This is
understandableif we realize that the negative-sequence
voltagedropssignificantly.
• For longer sags all voltagesdrop. This is due to thedrop in positive-sequence
voltage.
4.8.3 Power Electronics Load
In systems with a largefraction of the loadformed by single-phaseor three-phase
rectifiers, these can also influence the
voltageduring and after the voltagesag. Below
somequalitativeaspectsof the effectof rectifiers on thevoltagewill be discussedbriefly.
Different aspects willdominatein different systems. Thebehaviorof powerelectronics
equipmentduring voltage sags is discussed in
detail in Chapter5.
• Especially for longer and deepersags, a largepart of the electronicsload will
trip. This will reduce theload currentand thus increasethe voltage,during as
well as after the sag.
• Equipment that does not trip will initially take a smaller current from the
supply or even nocurrentat all because the de bus
voltage is larger than the
capacitorhasdischarged
peakof the ac voltage.Within a few cycles the de bus
sufficiently for the rectifier tostartconductingagain. Normally the total power
taken by the load remainsconstantso that the accurrentwill be higher. This
currenthas a highharmoniccontentsso that the harmonicvoltagedistortion
during the sag will increase.
• Upon voltage recovery, the dc busc apacitorswill take a large current pulse
from the supply. This canpostponethe voltage recovery by up to one cycle.
• For three-phaserectifiers, under unbalancedsags, thelargest current flows
between the twophaseswith the largestvoltage difference. The effect isthat
the voltagein thesephasesdropsand increasesin the other phase. The threephaserectifier thusreduces theunbalancebetween thephases.In this sense they
behavesimilar to induction motor load. For unbalancedsags thecurrent to
three-phaserectifiers containsso-called non-characteristicharmonics,noticeably a third harmoniccurrent, so that the voltage during the sagcontainsa
third harmoniccomponenthigher than normal.
• Three-phasecontrolled rectifiers will experiencea longer commutationperiod
because thesourcevoltage is lower during the sag. This leads to
m ore severe
commutationtransients(notches)during the sag.Again this assumesthat the
equipmentwill not trip.
4.9 SAGS DUE TO STARTING OF INDUCTION MOTORS
In the previoussectionsof this chapter,we have discussedvoltage sags due toshortcircuit faults. Thesevoltagesags are the main cause
of equipmentfailure and malfunction, and oneof the main reasonsfor powerquality to become an issue
during the last
decade.Anotherimportantcauseof voltagesags, one which has
actuallybeenof much
more concernto designersof industrialpowersystems in thepast,is the startingof large
249
Section 4.9 • Sags due to
Startingof Induction Motors
inductionmotors. Also the switching on
o f otherloads will cause a voltage sag, just like
the switching offof a capacitorbank. But in thoselatter cases thedrop in voltage is
rather small, and the voltage onlyd rops but does not recover.Thereforethe term
"voltagemagnitudestep" would be moreaccurate.
During start-upan induction motor takes a largercurrentthan normal, typically
five to six times as large. This
currentremainshigh until themotor reaches its nominal
speed, typically between several seconds and one minute.drop
Thein voltage depends
Zs
strongly on the system
parameters.Considerthe system shown in Fig. 4.140, where
is the sourceimpedanceand ZM the motor impedanceduring run-up.
Figure 4.140 Equivalent circuit for voltage
sag due to inductionmotor starting.
The voltage experienced by a load fed from the same bus as
motor
the is found
from the voltage dividerequation:
v
_
.wg -
ZM
ZS+ZM
(4.144)
Like with most previouscalculations,a source voltage of 1 pu has been assumed. When
a motor of rated powerSmotor is fed from a source withshort-circuitpower Ssourc:e,we
can write for the source impedance:
vn_
Zs = __
2
(4.145)
Ssource
and for themotor impedanceduring starting
_ Vn2
ZM---
(4.146)
fJSmotor
with fJ the ratio between the
startingcurrentand the nominalcurrent.
Equation(4.144) can now bewritten as
v _
sag -
S.fOurc:e
S.'iOurc:e
+ /3Smotor
(4.147)
Of course one needs to realize
that this is only anapproximation.The value can be used
to estimate the sag due inductionmotorstarting,but
to
for anaccurateresult one needs
a power system analysis package. The
latter will also enable the user to
incorporatethe
effect of othermotorsduring startingof the concernedmotor. The drop in voltage at
the other motor's terminals will slow them down and cause an
additional increase in
load currentand thus anadditionaldrop in voltage.
250
Chapter4 • VoltageSags-Characterization
EXAMPLE Supposethat a 5 MVA motor is startedfrom a 100 MVA, 11 kV supply.
The startingcurrent is six times thenominal current. This is a ratherlarge motor for a supply
of this strength,as we will see soon. The voltage at the
motor terminals during motor starting
can beestimatedas
_
100MVA
_
°
Vrag - 100MVA + 6 x 5 MVA - 77Yo
(4.148)
In case the voltageduring motor starting is too low for equipmentconnectedto the
same bus, one can decide to usededicatedtransformer.This
a
leads to thenetwork
shown in Fig. 4.141.
Let again Zs be the sourceimpedanceat the pee,ZM the motorimpedanceduring
fun-up, and ZT the transformerimpedance.The magnitudeof the voltage sag experiencedby the sensitive load is
v _
sag -
2 T+ZM
Zs + ZT + 2 M
(4.149)
Introducing,like before, theshort...circuit power of the sourceS.'iource,the rated power of
the motor Smolor and assumingthat the transformerhas the same rated power of the
motor and animpedancef, we get from (4.149):
v
(1 + 6€)Ssource
_
sag -
(1
+ 6f)Ssource+ 6Smotor
(4.150)
Figure 4.141 Induction motor starting with
dedicatedtransformerfor the sensitive load.
EXAMPLE Considera dedicatedsupply for themotor in the previous example. The
motor is fed through a 5 MVA, 5% 33/11 kV transformerfrom a 300 MVA, 33 kV supply.
Note that the fault currentat the 33 kV bus is identical to the fault
currentat the 11 kV in the
previous example. That gives the following parameter values: Ssource= 300 MVA,
Sma tor= 5 MVA, and € = 0.05, giving, from (4.150), a sag
magnitudeof 930/0. Most loads will
be able towithstand such a voltagereduction. Note that the reduction in sag magnitudeis
mainly due to the increased fault level at the pee, not so much due totransformerimpethe
dance. Neglecting the
transformerimpedance(€ = 0 in (4.150»)gives Vsag = 91 % •
The duration of the voltage sag due to
m otor startingdepends on anumberof
motor parameters,of which the motor inertia is the main one. Whendeterminingthe
fun-up time, it is alsoimportantto determinethe sagmagnitudeat themotor terminals.
251
Section 4.9 • Sags due to
Startingof Induction Motors
The torqueproducedby themotor is proportionalto thesquareof the terminalvoltage.
That makes that a sag down to90% causes adrop in torque down to 81%. It is the
difference betweenmechanicalload torque and electricaltorquewhich determinesthe
accelerationof the motor,andthus therun-uptime. Assumethat the mechanicalt orque
is half the electricaltorqueduring most of the run-upif the terminalvoltageis nominal.
This assumptionis based on the general design
criterion that the pull-out torqueof an
induction motor is about twice the torque at nominal operation.When the voltage
drops to 90% of nominal the electrical torque drops to 81 % of nominal which is
162% of the mechanicaltorque. The acceleratingtorque, the difference between electrical and mechanicalt orquedropsfrom 100 % to 62%, a drop of 38%.
EXAMPLE Consideragain the 5 MVA induction motor startedfrom a 100 MVA 11
kV supply. The voltage at the motor terminalsduring run-up drops to 770/0 as we saw before.
The electrical torque drops to 590/0 of nominal which is 118% of the mechanicaltorque. The
acceleratingtorquethus dropsfrom 1000/0 to only 18%, and therun-up time will increaseby a
factor of 6.
A dedicatedtransformeralone cannot solve this problem, as the voltage at the motor
terminalsremainslow. What is needed here is strongersupply.
a
To limit thevoltagedrop at the
motor terminalsto Vmin' the sourcestrength,from (4.147), needs to be
Ssourc(! =
6Smotor
V .
1-
(4.151)
mm
A 5 MVA motor, with a minimum-permissablevoltageof 85% during starting,needs asource
strengthof at least 6x~~5VA = 200 MVA. To keep thevoltage above 90%, the sourcestrength
needs to be 300 MV A.
From these examplesit will be clear that large voltagedrops are not only a
problem for sensitive load, but that they also lead tounacceptablylong run-up
times. The situation becomes even worse if more
motors are connectedto the same
bus, as they willf urther pull down the voltage. Voltaged ropsdue to induction motor
startingare seldom deeper
than 85%.
Voltage SagsEquipment Behavior
In this chapterwe will study theimpact of voltage sags on electrical
equipment.After
the introductionof some generalterminology,we will discuss three types of
equipment
which are perceived as most sensitive to voltage sags.
1. Computers,consumerelectronics, andprocess-controlequipmentwhich will
be modeled as a single-phase diode rectifier.
Undervoltageat the dc bus is the
main cause of tripping.
2. Adjustable-speedac drives which arenormally fed through a three-phase
rectifier. Apart from the undervoltageat the de bus,current unbalance,de
voltage ripple, andmotor speed are discussed.
3. Adjustable-speedde drives which are fedthrough a three-phasecontrolled
rectifier. The firing-anglecontrolwill causeadditionalproblemsdue to phaseangle jumps. Also the effect of the
separatesupply to the field winding is
discussed.
This chaptercloses with a brief discussion
of otherequipmentsensitive to voltage sags:
induction and synchronousmotors,contactors,and lighting.
5.1 INTRODUCTION
5.1.1 Voltage Tolerance and Voltage-Tolerance Curves
Generally speaking electrical
equipmentoperatesbest when the rms voltage is
constantand equal to the nominal value. In case the voltage is zero certainperiod
for a
of time, it will simply stop operatingcompletely. No piece of electrical
equipmentcan
operateindefinitelywithout electricity. Someequipmentwill stop within one second like
most desktopcomputers.Other equipmentcan withstanda supplyinterruptionmuch
longer; like a lap-top computerwhich is designed towithstand (intentional) power
interruptions.But even alap-top computer'sbatteryonly containsenoughenergy for
253
254
Chapter5 • VoltageSags-EquipmentBehavior
typically a few hours. For eachpiece of equipmentit is possibleto determinehow long
it will continueto operateafter the supply becomesinterrupted.A rather simple test
would give the answer.The sametestcan be donefor a voltageof 10% (of nominal),for
a voltageof 20% , etc. If the voltagebecomeshigh enough,the equipmentwill be ableto
operateon it indefinitely. Connectingthe points obtained by performing these tests
results in the so-called"voltage-tolerancecurve." An exampleof a voltage-tolerance
curveis shownin Fig. 5.1. In this caseinformationis providedfor the voltagetolerance
of power stationsconnectedto the Nordic transmissionsystem[149]. The auxiliary
supply should be able to toleratea voltage drop down to 25% for 250 ms. It should
be able to operateon a voltage of 95% of nominal. No requirementsare given for
voltagesbelow 250/0 of nominal as thesearc very unlikely for the infeed to the auxiliary
supplyof a powerstation.Onemay claim that this is not a voltage-tolerancecurve, but
a requirementfor the voltage tolerance.One could refer to this as avoltage-tolerance
requirementand to the result of equipmenttests as avoltage-toleranceperformance.
We will refer to both the measuredcurve, as well as to therequirement,as avoltagetolerancecurve. It will be clear from the context whether one refers to thevoltagetolerancerequirementor the voltage-toleranceperformance.
The concept of voltage-tolerancecurve for sensitive electronic equipmentwas
introduced in 1978 by Thomas Key [1]. When studying the reliability of the power
supplyto military installations,he realizedthat voltagesagsand their resultingtripping
of mainframecomputerscould be a greaterthreat to national security than complete
interruptionsof the supply. He thereforecontactedsomemanufacturersfor their design
criteria and performedsometestshimself. The resultingvoltage-tolerancec urvebecame
known as the"CBEMA curve" severalyearslater. We will comeback to the CBEMA
curve when discussingcomputing equipment further on. Note that curves plotting
minimum voltageagainstmaximumdurationhavebeenused forsynchronousmachines
for many years already, but not for electronicequipment.We will come back to the
voltage toleranceof synchronousmachinesin Section5.5.
The voltage-tolerancecurveis also an importantpart of IEEE standard1346 [22].
This standardrecommendsa method of comparingequipmentperformancewith the
supply powerquality. The voltage-tolerancecurve is the recommendedway of presenting the equipmentperformance.T he conceptof "voltagesag coordinationchart" [20],
which is at the heartof IEEE standard1346, will be presentedin detail in Section6.2.
While describingequipmentbehaviorthroughthe voltage-tolerancec urve,a number of assumptionsare made. The basic assumptionis that a sag can be uniquely
characterizedthrough its magnitudeand duration. We already saw in the previous
100%
.
95%
j
I
t
:
25% f . - - - - - - - < '
I
0% "--_ _---'Oms
250ms
.....t.--
750ms
Duration
_
Figure 5.1 Voltage-tolerancerequirementfor
powerstations.(Data obtainedfrom [149].)
255
Section 5.1 • Introduction
chapterthat this is only anapproximation.From an equipmentpoint of view the basic
assumptionbehind thevoltage-tolerancecurve is: if two sags have the same
magnitude
and duration then they will both lead to tripping
o f the equipmentor both not lead to
chapter,the definitions of
tripping of the equipment.As we have seen in the previous
magnitudeand durationof a sagcurrently in use are far fromunique. Further,phaseanglejumpsand three-phasevoltageunbalancecan significantly influence thebehavior
of equipment.The two-dimensionalvoltage-tolerancecurve clearly has itslimitations,
especially forthree-phaseequipment.We will present someextensionsto the conceptin
the nextchapter.
An overviewof the voltagetoleranceof currentlyavailableequipmentis presented
in Table 5.1. The range in voltagetoleranceis partly due to the difference between
equipment,partly due to theuncertaintiesmentionedbefore.With thesedata,as well as
with the voltage-tolerancedatapresentedin the rest of thischapter,one shouldrealize
that the valuesnot necessarily apply to a specific piece
of equipment.As an example,
Table 5.1 gives formotor startersa voltagetolerancebetween 20 ms,60% and 80 ms,
40%. Using this range to design an
installation could berather unreliable; using the
averagevalue even more. These values are only
meantto give thereaderan impression
of the sensitivity of equipmentto voltage sags, not to serve asdatabase
a
for those
determinethe voltage
designinginstallations.For the time being it is still necessary to
toleranceof each criticalpart of an installationor to subject the wholeinstallationto a
test. In future, voltage-tolerancerequirementsmight make thejob easier.
The values in Table 5.1
shouldbe read as follows. A voltage
toleranceof a rns, bOlo
implies that the equipmentcan toleratea zero voltage ofa ms and a voltageof b% of
nominalindefinitely. Any sag longerthan a ms and deeperthan bOlo will lead to tripping
or malfunction of the equipment.In other words: the equipmentvoltage-tolerance
curve isrectangularwith a "knee" at a ms, bt/«.
TABLE S.1 Voltage-Tolerance Ranges of Various Equipment Presently in Use
Voltage Tolerance
Equipment
Upper Range
Average
Lower Range
PLC
PLC input card
5 h.p. ac drive
ac control relay
Motor starter
Personal computer
20 ms,75%
20 ms,80%
30 ms,800/0
10 ms,75%
20 ms,600/0
30 ms,800/0
260 ms,60°A»
40 ms,55°A»
50 ms,75%
20 ms,65%
50 ms,50°A»
50 ms,60%
620 ms,450/0
40 ms, 30%
80 ms,600/0
30 ms, 60%
80 ms,400/0
70 ms,500/0
Source: As given data obtained from IEEE Std.1346
[22]. This data should not be used as a basis for design of
installations.
5.1.2 Voltage-Tolerance Tests
The only standardthat currently describes how toobtain voltage toleranceof
equipmentis lEe 61000-4-11[25]. This standard,however, doesnot mentionthe term
voltage-tolerancecurve. Insteadit defines anumberof preferredmagnitudesanddurastandarduses the
tions of sags for which theequipmenthas to be tested. (Note: The
term "test levels," which refers to theremainingvoltageduring the sag.) Theequipment
doesnot need to be tested for all these values, but onemore
or of the magnitudesand
256
Chapter5 • VoltageSags-EquipmentBehavior
TABLE S.2 PreferredMagnitudesand Duration for EquipmentImmunity
TestingAccording to IEC-61000-4-11 [25]
Duration in Cyclesof 50 Hz
Magnitude
0.5
5
10
25
50
durationsmay be chosen. The
preferredcombinationsof magnitudeand durationare
o f the matrix shown in Table 5.2.
the (empty) elements
The standardin its currentform does not set any
voltage-tolerancerequirements.
It only defines the way in which the voltage toleranceequipmentshall
of
beobtained.
An informative appendixto the standardmentionstwo examplesof test setups:
• Use atransformerwith two outputvoltages. Make oneo utputvoltage equal to
1000/0 and theother to the requiredduring-sagmagnitudevalue. Switch very
fast between the twooutputs,e.g., by usingthyristor switches.
• Generatethe sag by using a waveform
generatorin cascade with a power
amplifier.
The IEEE standard1346 [22] refers tolEe 61000-4-11for obtainingthe equipment voltagetolerance,and specificallymentionsthe switching between two supply
voltages as a way ofgeneratingsags. Bothmethodsare only aimed at testing one
piece ofequipmentat a time. To make a whole
installationexperience acertainvoltage
sag, each piece needs to be tested
hoping that their interconnectiondoes not cause any
unexpecteddeteriorationin performance.A methodfor testing a wholeinstallationis
presentedin [56]. A three-phasedieselgeneratoris used to power the
installationunder
test. A voltage sag is made by reducing the field voltage.
It takesabouttwo cycles for
the ac voltage to settle down after a sudden change in field voltage,
thatso
this method
can only be used for sags
of five cycles and longer.
5.2 COMPUTERS AND CONSUMER ELECTRONICS
The power supply of acomputer, and of most consumerelectronics equipment
regulator
normally consists of a diode rectifier along with an electronic voltage
(de/deconverter).The power supplyof all these low-power electronic devices is similar
and so is their sensitivity to voltage sags.
What is different are the consequences
of a
sag-inducedtrip. A television will show a black screen for up to a few seconds; a
compactdisc player will reset itself andstart from the beginningof the disc, orjust
wait for a newcommand.Televisions and video recorders
normallyhave a smallbattery
to maintain power to thememory containingthe channel settings. This is to prevent
loss of memory when the television is moved or unplugged for some reason. If this
batteryno longercontainsenoughenergy, a sag orinterruptioncould lead to the loss of
these settings. The same could
happento the settingsof a microwave oven, which is
often not equippedwith a battery.
The process-controlcomputer of a chemical plant is rather similar in power
supply to anydesktopcomputer.Thus, they willboth trip on voltage sags and inter-
257
Section 5.2 • Computersand ConsumerElectronics
ruptions,within one second. But the
desktopcomputer'strip might lead to the loss
of 1
hourof work (typically less), where the
process-controlcomputer'strip easily leads to a
restartingprocedureof 48 hours plus sometimes a very
dangeroussituation.It is clear
thatthe first is merely an inconvenience, whereaslattershould
the
be avoided at any cost.
5.2.1 Typical Configuration of Power Supply
A simplified configurationof the power supply to a
computeris shown in Fig. 5.2.
The capacitorconnected to thenon-regulatedde bus reduces the voltage ripple at the
input of the voltageregulator. The voltageregulator converts thenon-regulatedde
voltage of a fewhundredvolts into a regulated de voltage of the
order of 10 V. If
the ac voltage drops, the voltage on the de side of the rectifiernon-regulated
(the
de
voltage) drops. The voltage
regulatoris able to keep itsoutputvoltageconstantover a
certainrange ofinput voltage. If the voltage at the de bus becomes too low the regulated dc voltage will alsostart to drop and ultimatelyerrors will occur in the digital
electronics. Somecomputersdetect anundervoltageat theinput of the controller and
give a signal for a"controlled" shutdownof the computer,e.g., byparking the hard
drive. Thosecomputerswill trip earlier but in a morecontrolledway.
Nonregulateddc voltage
Regulated
de voltage
1
230 Vac
Voltage
controller
Figure 5.2Computerpower supply.
5.2.2 Estimation of Computer Voltage Tolerance
5.2.2.1DC Bus Voltages. As shown in Fig. 5.2, a single-phase rectifier consists
of four diodes and acapacitor.Twice every cycle thecapacitoris charged to the amplitude of the supply voltage. In between the
chargingpulses thecapacitordischarges
via the load. The diodes only
conduct when the supply voltage exceeds the de voltage. When the supply voltage
drops the diodes no longerconductand thecapacitor
continuesto discharge until the de voltage reaches the reduced supply voltage again.
In normal operation the capacitoris charged during two small periods each cycle,
and dischargesduring the rest of the cycle. In steady state, the
amount of charging
and discharging of thecapacitorare equal.
To study the effect of voltage sags on the voltage at (non-regulated)
the
de bus,
the power supply has been modeled as follows:
• The diodesconductwhen theabsolutevalue of the supply voltage is larger
than
the de bus voltage. While the diodes
conduct,the de bus voltage is equal to the
supply voltage.
• The supply voltage is a 1pu sinewave before the event and
constant-amplia
tude sinewaveduring the eventbut with an amplitude less than 1pu. The
258
Chapter5 • VoltageSags-EquipmentBehavior
voltage only shows a -drop in magnitude,no phase-anglejump. The supply
voltage is not affected by the load current.
• While the diodes do not conduct, the capacitoris dischargedby the voltage
regulator.The power taken by the voltageregulatoris constantand independent of the dc busvoltage.
This model has been used tocalculatethe dc busvoltagesbefore,during, and after a
voltagesag with amagnitudeof 50% (without phase-anglejump). The result is shown
in Fig. 5.3.As a reference,the absolutevalue of the ac voltage hasbeen plotted as a
dashedline.
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10
Figure 5.3 Effect of a voltage sag on de bus
voltage for a single-phase rectifier: absolute
value of the ac voltage (dashed line) and de
bus voltage (solid line).
Due to the voltage drop, the maximum ac voltage becomesless than the de
voltage. Theresultingdischargingof the capacitorcontinuesuntil the capacitorvoltage
drops below the maximum of the ac voltage. After that, a new equilibrium will be
reached.Because aconstantpower load has beenassumedthe capacitordischarges
fasterwhen the de busv oltageis lower. This explainsthe largerdc voltageripple during
the sag.
It is importantto realize that the dischargingof the capacitoris only determined
by the load connectedto the de bus,n ot by the acvoltage.Thus all sagswill causethe
same initial decay in devoltage. But the duration of the decay is determinedby the
magnitudeof the sag.The deeperthe sag thelonger it takesbeforethe capacitorhas
dischargedenoughto enablechargingfrom the supply. In Fig. 5.4 the sags in ac
andde
voltage are plotted for voltagesagsof different magnitude.The top curveshave been
calculatedfor a sag in acvoltagedown to 50%, the bottomonesfor a sag in acvoltage
down to 70% • The dottedlines give the rmsvoltageat ac side(thesag in acvoltage).We
seethat the initial decay in de busvoltageis the samefor both sags.
5.2.2.2 Decayof the DC Bus Voltage. Within a certain rangeof the input voltage, thevoltage regulator will keep its output voltage constant,independento f the
input voltage. Thus, the output power of the voltage regulatoris independento f the
input voltage. If we assumethe regulator to be lossless theinput power is independent of the devoltage. Thus, the load connectedto the de buscan be consideredas a
constantpower load.
259
Section 5.2 • Computersand ConsumerElectronics
u
EO.5
~
0
2
4
6
Time in cycles
8
10
2
4
6
Time in cycles
8
10
.i~ 0.5
Figure 5.4 Voltage sag at ac side
(dashedline)
and at the de bus (solid line) for a sag down to
50% (top) and for a sag
d own to 70%
(bottom).
0
As long as theabsolutevalueof the ac voltage is less
than the de bus voltage, all
electrical energy for the load comes from the energy stored incapacitor.Assume
the
that the capacitor has capacitanceC. The energy a timet after saginitiation is
C{ V(t)}2, with V(t) the de bus voltage. This energy is equal to the energy at sag
initiation minus the energy consumed by the load:
!
1
2
-CV2=1-CVo - Pt
2
2
(5.1)
where Vo is the de bus voltage at sag
initiation and P the loading of the de bus.
Expression (5.1) holds as long as the de bus voltage is higher
than the absolute value
of the ac voltage, thus
during the initial decay period in Figs. 5.3 and 5.4. Solving (5.1)
gives an expression for the voltage
during this initial decay period:
(5.2)
During normal operation,before the sag, the
variation in de bus voltage is small, so
that we can linearize (5.2)around V = Vo, resulting in
(5.3)
wheret is the time elapsed since the last recharge ofcapacitor.The
the
voltage ripple is
defined as the difference between the maximum and the minimum value of the de bus
voltage. The maximum is reached for
t = 0, the minimum fort = f, with T one cycle of
the fundamentalfrequency. The resulting expression for the voltage ripple is
PT
E=
2V2C
(5.4)
o
The voltage ripple is often used as a design
criterion for single-phase diode rectifiers.
Inserting the expression for the de voltage ripple (5.4) in (5.2) gives an expression for the dc voltageduring the discharge period, thus
during the initial cycles of a
voltage sag:
260
Chapter5 • VoltageSags-EquipmentBehavior
(5.5)
where f is the numberof cycleselapsedsince saginitiation. The larger the dc voltage
ripple in normal operation,the faster the devoltagedropsduring a sag.
5.2.2.3 VoltageTolerance. Tripping of a computerduring a voltage sag is attributed to the de busvoltage dropping below the minimum input voltage for which
the voltagecontroller can operatecorrectly. We will refer to this voltageas Vmin. We
will further assumethat in normal operation, before the sag,both ac and de bus
voltage areequal to 1 pu.
A sag with amagnitudeV will result in a newsteady-statede voltage which is also
equal to V, if we neglect the dc voltage ripple.From this we canconcludethat the
d ropsbelow
computerwill not trip for V > Vmin• For V < Vmin' the dc bus voltage only
Vmin if the sag duration exceeds acertain value lmax. The time tmax it takes for the
voltage to reach a levelVmin can befound by solving t from (5.5) with Vo = I:
I - V;';n T
tmax = - - -
(5.6)
4E
When theminimum de bus voltage isknown, (5.6) can be used to
calculatehow long it
will takebefore tripping. Or in otherwords: what is themaximumsagdurationthat the
equipmentcan tolerate. The dc busvoltage at which the equipmentactually trips
dependson the designof the voltage controller: varying between 50% and 90% de
of voltage
voltage, sometimeswith additional time delay.Table 5.3 gives some values
tolerance,calculatedby using (5.6).
Thus, if a computertrips at 50% de bus voltage, and as the
normal operationde
voltage ripple is50/0, a sagof lessthan four cycles indurationwill not cause amaltrip.
Any sag below50°A, for more than four cycles will trip thecomputer.A voltageabove
50% can bewithstood permanentlyby this computer.This results inwhat is called a
"rectangularvoltage-tolerancec urve," as shown in Fig. 5.5. Each voltage
regulatorwill
have anon-zerominimum operatingvoltage. The row for zerominimum de bus voltage
is only insertedas a reference. We can see from
Table5.3 that the performancedoes not
improve much by reducing the minimum operatingvoltage of the voltagecontroller
beyond50%. When the dcvoltagehas droppedto 50°A" the capacitorhas alreadylost
75°A, of its energy.
TABLE 5.3 Voltage Tolerance of Computers and Consumer
Electronics
Equipment:Maximum-AllowableDuration of a Voltage Sag for a Given
Minimum Value of the DC Bus Voltage, for Two Values of the DC Voltage
Ripple
Maximum Sag Duration
Minimum de Bus Voltage
0
50%
70%
900/0
5°AJ ripple
5 cycles
4 cycles
2.5 cycles
I cycle
I % ripple
25 cycles
19cycles
13 cycles
5 cycles
261
Section 5.2 • Computersand ConsumerElectronics
100%
~
Vmin
- --.-.---..--.---------..-..-- -.-- --._-_.. -------
-
-
-
Minimum steady-statevoltage
.~
~
Maximumduration
,/ of zerovoltage
Figure 5.5 Voltage-tolerancecurve of a
computer:an exampleof a rectangular
voltage-tolerancecurve.
Duration
5.2.3 Measurements of PC Voltage Tolerance
The voltage tolerance of personal
computershas been measured bynumberof
a
authors[28], [29], [41], [49],[50]. The voltage-tolerancecurves they present are in the
same range as found from the simplified model presented in the previous section. Figure
5.6 showsmeasuredvoltages andcurrentsfor a personalcomputer.The applied voltage
sag was oneof the most severe the
computercould tolerate.
In Fig. 5.6 we see the de bus voltage
startingto drop the momentthe ac voltage
drops.During the decay in de bus voltage, the input
currentto the rectifier is very small.
The output of the voltagecontroller remainsconstantat first. But when the de bus
voltage hasdroppedbelow acertainvalue, the de voltage
regulatorno longeroperates
properly and itsoutput also startsto drop. In this case a new steady state is reached
where the regulated de voltageapparentlystill
is
sufficient for the digital electronics to
operatecorrectly. During the new steady state, the
input current is no longer zero.
Upon ac voltage recovery, the de bus voltage also recovers quickly. This is associated
Slightde offsetrelated
to instrumentation
Regulated
de voltage
(l V/div)
Figure 5.6 Regulatedand non-regulatedde
voltages for a personalcomputer,during a
200 ms sag down to500/0: (top-to-bottom)ac
voltages; accurrent; regulatedde voltage;
non-regulatedde voltage.(Reproducedfrom
EPRI PowerQuality Database[28].)
Unregulated
de voltage
(100V/div)
Time(SO milliseeonds/div)
262
Chapter5 • Voltage Sags--·EquipmentBehavior
IOO,------r----.-----r-------,
80
20
.5
10
15
20
Duration in cycles
Figure 5.7 Voltage-tolerancecurves for
personalcomputers.(Data obtainedfrom
EPRI PowerQuality Database[29J.)
with a very largecurrentpeak chargingthe dc buscapacitor.This currentcould cause
an equipmenttrip or even a longinterruption if fast-acting overcurrentprotection
devices are used.
The voltage-tolerancecurvesobtainedfrom various tests are shown in Fig. 5.7
and Fig. 5.8. Figure 5.7 shows the result of a U.S. study
[29]. For each personal
computer, the tolerance for zero voltage was
determined, as well as the lowest
steady-statevoltage for which thecomputerwould operateindefinitely. For one computerthe tolerance for800/0 voltage wasdetermined;all othercomputerscould tolerate
this voltage indefinitely. We see
t hat there is a large range in voltage
tolerancefor
o f the computerdid not have any influence.
different computers.The age or the price
The experiments wererepeatedfor various operating states of thecomputer: idle;
calculating; reading; or writing. Itturned out that the operatingstate did not have
any significant influence on the voltage tolerance or on the power
consumption.
Figure 5.7 confirms that the voltage-tolerancecurve has analmostrectangularshape.
Figure 5.8 showsvoltage-tolerancecurves forpersonalcomputersobtainedfrom
a Japanesestudy [49], in the sameformat and scale as the
Americanmeasurements
in
Fig. 5.7. The general shape
o f the curves is identical, but the curves in Fig. 5.7 indicate
less sensitivecomputersthan the ones in Fig.5.8.
100..----,------r-----.-----,
80
20
100
200
300
Duration in milliseconds
400
Figure 5.8 Voltage-tolerancecurves for
personalc omputers-Japanese
tests.(Data
obtainedfrom [49J.)
263
Section 5.2 • Computersand ConsumerElectronics
Summarizingwe can saythat the voltagetoleranceof personalcomputersvaries
over a rather wide range:30-170ms, 50-70% being the rangecontaininghalf of the
88% and 210 ms, 30%.
models. The extreme values found are 8 ms,
5.2.4 Voltage-Tolerance Requirements. CBEMA and ITIC
As mentionedbefore, the firstmodern'voltage-tolerance curve was
introducedfor
mainframecomputers[1]. This curve is shown as a solid line in Fig. 5.9. We see
that its
5.5,5.7,and 5.8.
shape doesn ot correspondwith the shapeof the curves shown in Figs.
This can beunderstoodif one realizesthat these figures give thevoltage-tolerance
performancefor one pieceof equipmentat a time, whereas Fig. 5.9 isvoltage-tolera
ance requirementfor a whole range ofequipment.The requirementfor the voltagetolerance curves ofequipmentis that they should all be above thevoltage-tolerance
requirementin Fig. 5.9. The curve shown in Fig. 5.9 became well-known when the
ComputerBusinessEquipmentManufacturersAssociation(CBEMA) startedto use
the curve as arecommendationfor its members. The curve was
subsequentlytakenup
in an IEEE standard[26] and became a kindo f reference forequipmentvoltage tolerof voltage sags. Anumberof softwarepackagesfor analyzance as well as for severity
ing power quality data plot magnitudeand duration of the sagsagainstthe CBEMA
curve. The CBEMA curve alsocontains a voltage-tolerancepart for overvoltages,
which is not reproducedin Fig. 5.9. Recently a "revisedCBEMA curve" has been
adoptedby the InformationTechnologyIndustryCouncil (ITIC), which is the successor of CBEMA. The new curve isthereforereferred to as theITIC curve; it is shown as
a dashedline in Fig. 5.9.
The ITIC curve givessomewhatstrongerrequirementsthan the CBEMA curve.
This is because power
quality monitoringhas shownthat there are analarmingnumber
of sagsjust below theCBEMA curve [54].
100 . . . . - - - - - - - - - - - - - - - - - - - - - - - - CBEMA
80
---
...
,
+--------.---------~
I
--.------~
ITIC
20
O-----._-..l.--------"'--------L.-------J
0.1
10
100
Durationin (60 Hz) cycles
Figure5.9 Voltage-tolerance requirements for computing equipment:
CDEMA
curve (solid line) and ITIC curve (dashed line).
1000
264
Chapter 5 • VoltageSags-EquipmentBehavior
5.2.5 Process Control Equipment
Processcontrol equipmentis often extremelysensitiveto voltagesags;equipment
has beenreportedto trip when the voltagedropsbelow 800/0 for a few cycles [31], [37],
[39], [41]. The consequences
o f the tripping of processcontrol equipmentcan be enormous. For example,the tripping of a small relay can causethe shutdownof a large
chemical plant, leading to perhaps$IOO~OOO in lost production.Fortunatelyall this is
low-powerequipmentwhich can be fedfrom a UPS, or for which the voltagetolerance
can be improved easily by addingextra capacitors,or somebackupbattery.
Tests of the voltage toleranceof programmablelogic controllers (PLC's) have
been performedin the sameway as the PC testsdescribedbefore [39]. The resulting
voltage-tolerancecurvesfor somecontrollersare shown in Fig. 5.10. It clearly shows
that this equipmentis extremelysensitiveto voltagesags. Asmost sagsare between4
and 10 cycles in duration, we can reasonablyassumethat a PLC trips for each sag
below a given threshold,varying between85% and 35%.
Even more worrying is that some controllers may send out incorrect control
signalsbefore actually tripping. This has to do with the different voltage toleranceof
the various parts of the controller. The incorrect signals could lead to dangerous
processmalfunctions.
Additional voltage-tolerancecurvesfor processcontrol equipment,obtainedfrom
anotherstudy [41], are shown in Fig. 5.11. The numberswith the curvesrefer to the
following devices:
1. Fairly commonprocesscontrollerused for processheatingapplicationssuch
as controlling water temperature.
2. More complicated processcontroller which can be used toprovide many
control strategiessuch as pressure/temperature
compensationof flow.
3. Processlogic controller.
4. Processlogic controller, newer and more advancedversion of 3.
5. AC control relay, usedto power importantequipment.
6. AC control relay, used topower important equipment;samemanufacturer
as 5.
7. AC control relay usedto power motors; motor contactor.
100
80
/
5e
I
8. 60
I
.5
~
~
/
40
~
20
---------
:/
Figure 5.10Voltage-tolerancecurves for
5
10
Duration in cycles
15
20
programmablelogic controllers(PLCs).
(Data obtainedfrom [39].)
265
Section 5.3 • Adjustable-SpeedAC Drives
100.------r-----,..-----r--------,
80
6
20
3
Figure 5.11 Voltage-tolerancecurvesfor
variousprocesscontrol equipment(41].
5
10
Duration in cycles
15
20
This study confirmsthat processcontrol equipmentis extremely sensitive to voltage
disturbances,but alsothat it is possible to buildequipmentcapableof toleratinglong
and deep sags. The fact
that someequipmentalready trips for half-a-cycle sags suggests
a serious sensitivity to voltage
transientsas well. The main steps taken to prevent
control equiptripping of processcontrol equipmentis to power all essential process
ment via a UPS or to ensure anotherway
in
that the equipmentcan withstandat least
short and shallow sags. Devices 2 and 3 in Fig.
5.11 show that it is possible to make
processcontrolequipmentresilient to voltage sags. But even here the costs of installing
a UPS will in almost all cases
be justified.
Here are someotherinterestingobservationsfrom Fig. 5.11:
• Device 2 is the more complicated version of device 1. Despite the higher complexity, device 2 is clearly less sensitive to voltage sags than device 1.
• Device 4 is a newer and more advanced version of device 3. Note
enormous
the
deteriorationin voltage tolerance.
• Devices 5 and 6 come from the same
manufacturer,but show completely
different voltage tolerances.
5.3 ADJUSTABLE-SPEED AC DRIVES
Many adjustable-speed
drives are equally sensitive to voltage sags as process
control
equipmentdiscussed in the previous section.
Tripping of adjustable-speed drives can
occur due to several
phenomena:
• The drivecontroller or protectionwill detect the sudden change operating
in
conditionsand trip the drive to prevent damage to the power electronic components.
• The drop in de bus voltage which results from the sag will cause
maloperation
or tripping of the drivecontroller or of the PWM inverter.
• The increased ac
currentsduring the sag or the post-sag
overcurrentscharging
the decapacitorwill cause anovercurrenttrip or blowing of fusesprotecting
the power electronics
components.
266
Chapter5 • VoltageSags-EquipmentBehavior
• The process driven by the
motor will not be able totoleratethe drop in speed
or the torquevariationsdue to the sag.
After a trip some drivesrestartimmediatelywhen the voltage comes back; some
restart
after a certaindelay time andothersonly after a manualrestart.The variousautomatic
restartoptionsare only relevantwhen the processtoleratesa certainlevel of speedand
torquevariations.In the restof this section we will first look at the results
of equipment
testing. This will give animpressionof the voltagetoleranceof drives. The effecto f the
of equipmenttripping, will be disvoltage sag on the de bus voltage, the main cause
cussed next.Requirementsfor the sizeof the de buscapacitorwill be formulated.The
currentand on themotor terminalvoltagewill also be
effect of the voltage sag on the ac
of automaticrestart. Finally, a short overview of
discussed, as well as some aspects
mitigation methodswill be given.
5.3.1 Operation of AC Drives
Adjustable-speeddrives (ASD's) are fedeither through a three-phasediode rectifier, or througha three-phasecontrolledrectifier. Generallyspeaking,the first type is
We will discuss
found in ac motor drives, the second in de drives and in large ac drives.
small andmediumsize ac drives fedthrougha three-phasediode rectifier in this section,
and de drives fedthroughcontrolled rectifiers in the next section.
The configurationof mostac drives is as shown in Fig. 5.12. The three ac voltages
are fed to athree-phasedioderectifier. Theoutputvoltageof the rectifier issmoothened
by meansof a capacitorconnectedto the de bus. Theinductancepresentin some drives
aims atsmootheningthe dc linkcurrentand soreducingthe harmonicdistortionin the
current taken from the supply.
The devoltageis inverted to an ac voltageof variablefrequencyand magnitude,
by meansof a so-calledvoltage-sourceconverter(VSC). The most commonly used
method for this is pulse-width modulation (PWM). Pulse-width modulation will be
discussed briefly when we' describe the effect
of voltage sags on them otor terminal
voltages.
The motor speed iscontrolledthroughthe magnitudeand frequencyof the output
voltage of the VSC. For ac motors, the rotational speed ismainly determinedby the
frequency of thestator voltages.Thus, by changingthe frequency an easy
methodof
speed control is obtained. The frequency andmagnitudeof the stator voltage are
plotted in Fig. 5.13 as afunction of the rotor speed.For speeds up to thenominal
speed,both frequency andmagnitudeare proportional to the rotational speed. The
50 Hzr-------..
ac
ac
Variable
frequency
de link
dc
dc
ac
Controlsystem
'---
-.J
Figure 5.12Typical ac drive configuration.
267
Section 5.3 • Adjustable-SpeedAC Drives
nom
Rotational speed
. ,-- - -- - - -
... .. .. ._. .
nom
Figure 5.13 Voltage and frequency as a
function of speed for an acadjustable-speed
drive.
nom
Rotational speed
maximum torque of an induction motor is proportional to the squareof the voltage
magnitudeand inverselyproportionalto the squareof the frequency [53], [206]
:
r.:
V2
~ /2
(5.7)
By increasingboth voltage magnitudeand frequency, themaximum torque remains
constant.It is not possible to increase the voltage
magnitudeabove itsnominal value.
Furtherincrease in speed will lead to a fast
drop in maximum torque.
5.3.2 Results of Drive Testing
The performanceof a numberof adjustable-speed
drives inrelationto voltage sag
monitoring in an industrial plant is presentedin Fig. 5.14 [40]: the circlesindicate
magnitudeand duration of voltage sags for which the drives trip
; for the voltage
sags indicated by the crosses, the drives did not trip. Wethat
seethe drives used in
this plant were very sensitive to sags. The voltage
toleranceof these drives is 80%of
voltage for less than six cycles
. The exactduration for which the drivestripped could
not bedeterminedas theresolution of the monitors was only six cycles. Similar high
reported in other
sensitivitiesof adjustable-speeddrives to voltage sags have been
studies [2],[35], [42], [48]. Using thesedataas typical foradjustable-speed
drives carries
a certain risk. If the drives had not been sensitive to ,sags
the study would never have
beenperformed. This warning holds for manypublicationsthat mention a high sensitivity of equipmentto sags. It would thus be very well possible
t hat a largefraction of
the adjustable-speeddrives are not sensitive to sags at all. To
determinethe performance of typical drives
, one needs to apply tests randomlyselected
to
drives.
drives, selected atrandom
Studies after the voltage
toleranceof adjustable-speed
arepresentedin [32],[47]. In oneof the studies [47] tests were
performedfor 20 h.p. and
3 h.p.drives, from several different
manufacturers.Eachmanufacturerprovideda 20 h.p.
and a 3 h.p.drive. Each drive was tested for the following three voltage
magnitudeevents:
Chapter 5 • VoltageSags-EquipmentBehavior
o
20
40
60
Duration in cycles
100
80
Figure 5.14 Voltage sags which led to drive
tripping (0) and voltage sags which did not
lead to drivetripping (x). (Data obtained
from Sarmiento[40].)
• zero voltage for 33 ms.
• 500/0 voltage for 100ms.
• 700/0 voltage for 1 sec.
The driveperformanceduring the event was classified based on the three types
of speed
curves shown in Fig.5.15;
• I: The speed of themotor shows a decrease followed by a recovery.
• II: The speed of themotor reduces to zero after which the drive
restartsautomatically and accelerates the
motor load back to nominal speed.
• III: The motorspeed becomes zero, and the drive is unable
restartthe
to
motor.
The test results are
summarizedin Tables 5.4 and5.5. Eachof the columns in the
tables gives thenumberof drives with the indicatedperformance.For a 500/0, lOOms
sag, fourof the 20 h.p. drives showed performance
a
accordingto curve II in Fig. 5.15
and sevenof the drivesaccordingto curve III. Table 5.4 gives the results for drives at
full load; a distinctionis made between 3 h.p. and 20 h.p. drives.
Table5.5comparesthe
drive behaviorat full load with the drivebehaviorat half-load. These results include
20 h.p. as well as 3 h.p. drives.
Nominal speed
1············.···········.······..··-.-.·.··.···.·····...
I
I
II
II
I
I
I
I!
Stand-I
still i
I
III
···············t···········t··············.L.---......L---------
..--.'
Sag duration
Time
Figure 5.15 Three types
of motor speed
behaviorfor an adjustable-speed
drive due to
a sag.
269
Section 5.3 • Adjustable-SpeedAC Drives
TABLE 5.4 Resultsof Voltage-ToleranceTestingof Adjustable-Speed
Drives: Numberof Drives with the IndicatedPerformance.I: Only Drop in
Speed; II:Automatic Restart;III: Manual Restart
Drive Performance
Applied Sag
00/0 33 ms
50% 100 ms
70% 1000 ms
3 h.p. drives
20 h.p. drives
I
4
II
2
4
5
III
5
7
6
I
12
3
1
II
III
5
4
4
7
Source: Data obtainedfrom [47].
TABLE 5.5 Influenceof Loading on Drive Voltage Tolerance:Numberof
Driveswith the IndicatedPerformance.I: Only Drop in Speed; II:Automatic
Restart;III: Manual Restart
Drive Performance
Applied Sag
I
0% 33 ms
50% 100ms
700/0 1000 ms
Half-Load
Full Load
7
2
1
II
I
4
5
III
2
4
4
I
8
3
1
II
I
4
III
4
5
I
3
Source: Data obtainedfrom [47].
From the results in Tables 5.4 and 5.5 one can
draw the following conclusions:
• 3 h.p. drives are less sensitive
than20 h.p. drives. This does not necessarily hold
in all cases,a lthougha comparisonof 3 h.p. versus 20 h.p. drives for the same
manufacturer,the same voltage sag, and the same drive
loadinggives in 25of
the cases abetterperformancefor the 3 h.p. drive; in 20 cases the
performance
is the same (i.e., in the same class
accordingto the classification above); and
only in three cases does the 20 h.p. drive
perform better.
• Thereis no significant difference between the full load and the
half-load voltage tolerance.F or some loads theperformanceimproves, forothersit deteDoing the same
riorates,but for mostit doesnot appearto have any influence.
comparisonas before shows
t hat in two casesperformanceis betterat full load,
in four cases it isbetterat half-load,and in 24 cases the
performancefalls in the
sameperformanceclass.For drives falling inperformanceclass I it may bethat
at full load thedrop in speed is more severe
thanat half-load,but the study did
not report this amountof detail.
• Very shortinterruptions(0%, 33 ms) can behandledby all 3 h.p. drives and by
a largepart of the 20 h.p. drives.
of 100 ms and longer,
• Adjustable-speeddrives have severe difficulties with sags
especially as one
considersthateven response I could mean a serious
disruption
of sensitivemechanicalprocesses.
270
Chapter5 • VoltageSags-EquipmentBehavior
• The tests confirmthat adjustable-speed
drives are very sensitive to sags; however, the extreme sensitivity (85%, 8 ms)
mentionedby some isnot found in
this test.
The resultsof a similar set of tests arereportedin [32]: two different voltage sags were
applied to 17 drives:
• voltagedown to 50% of nominal for 100ms (6 cycles);
• voltage down to70% of nominal for 167ms (10 cycles).
Their results are shown in
Table 5.6. The classification used is fairly similar to the one
used inTables 5.4 and 5.5, with the exception
t hat a class"drive kept motor speed
constant" is included. This driveperformanceis indicated as class 0 inTable 5.6.
Responseclasses I, II, and IIIcorrespondto the ones used before.
From these studies, it is possible to
obtain a kind of "averagevoltage-tolerance
curve" for adjustable-speed
drives. The resulting curve is shown in Fig. 5.16, with the
measurementpointsindicatedas circles.Toleranceis defined here asperformance0 or
I. Note that the actual drives show a largespreadin voltage tolerance: some drives
o f the drivestoleratedall sags. It
could not tolerateany of theappliedsags, where one
has further beenassumedthat the drives couldoperateindefinitely on 85% voltage.
Conrad et al. [48] obtained voltage tolerancedata for adjustable-speeddrives
througha survey of drivemanufacturers.The voltagetolerancestatedby the manufacturers is shown in Fig. 5.17. The circles indicate
manufacturerswhich gaveminimum
voltage as well asmaximumsagduration.The othermanufacturers,indicatedby triangles in Fig. 5.17, only gave a value for the
maximumsagduration.Note that 10 out of 13
d uration.
manufacturersindicatethat their drives trip for sagso f three cycles or less in
TABLE 5.6
Resultsof Voltage-ToleranceTestson Adjustable-SpeedDrives
Responseof the Drive
Sag Applied
50% 100 ms
70% 170ms
o
II
III
2
9
5
5
II
Source: Data obtainedfrom [32].
100%
.............................
85%
~
a
70%
(l;S
50%
.~
~
.........................
/
33 ms 100 ms 170 ms
Duration
1000ms
Figure 5.16 Averagevoltage-tolerancecurve
for adjustable-speed
drives. Note the nonlinear horizontalscale.
Section 5.3 • Adjustable-Speed AC Drives
100
80
u
00
~>
§
.5
.s
~
271
. I
I.M..
Voltage not stated
-
•
•
•
•
60
40 '-
20 -
0
0
I
I
I
10
20
30
Maximum duration in cycles
Figure
s.t 7 Adjustable-speeddrive voltage tolerance,accordingto the drive
manufacturer.• = Magnitudeand duration; A = durationonly. (Data
obtainedfrom [48].)
5.3.2.1 Acceptance Criterion.When testing anadjustable-speed
drive, without
detailed knowledge of the load driven by the drive, a well-defined
criterion is needed
to distinguish successful from unsuccessful behavior. lEe
The standard61800-3 [52]
gives criteria to assess theperformanceof adjustable-speeddrives for EMC testing.
Thesecriteria are given in Table 5.7; they should also be used for voltage sag testing
of adjustable-speeddrives. The IEC performancecriteria can be summarizedas
follows:
• A: the drive operatesas intended;
• B: the drive temporarily operatesoutsideof its intendedoperatingrange but
recoversautomatically;
• C: the drive shuts down safely.
TABLE 5.7 AcceptanceCriteria for Drives According to IEC 61800-3 [52]
AcceptanceCriterion
A
Specific performance
Torque-generating
behavior
Operationsof power
electronicsand driving
circuits
Information processing
and sensingfunctions
Operationof display and
control panel
No changewithin the
specified tolerance
Torque within tolerances
No maloperationof a
power semiconductor
Undisturbedcommunication and data
exchange
No changeof visible
display information
B
Noticeablechanges,selfrecoverable
Temporarydeviation
outsideof tolerances
Temporarymaloperation
which cannotcause
shutdown
Temporarydisturbed
communication
C
Shutdown,big changes,n ot
self-recoverable
Loss of torque
Shutdown,triggering of
protection
Errors in communication,
loss of dataand
information
Visible temporarychanges Shutdown,obviously wrong
of information
display information
272
Chapter 5 • VoltageSags-EquipmentBehavior
5.3.3 Balanced Sags
Many trips of ac drives are due to a low voltage at the de bus. The trip or
maloperationcan be due to thecontroller or PWM inverter not operatingproperly
when the voltage gets too low. But it can also be due tointerventionof
the
undervoltage
protectionconnectedto the dc bus. ·Most likely, the
protectionwill intervene before any
equipmentmalfunction occurs.
The de bus voltage is
normallyobtainedfrom the three ac voltages
througha diode
rectifier. When the voltage at ac side
drops, the rectifier will stop conductingand the
PWM inverterwill be powered from thecapacitorconnectedto the de bus. This capacitor has only limited energyc ontent(relative to the powerc onsumptionof the motor)
and will not be able to supply the load much longer
than a few cycles. Animproved
drives can be achieved by lowering the setting
of
voltage toleranceof adjustable-speed
the undervoltageprotectionof the de bus. One
shouldtherebyalways keep in mindthat
the protectionshould trip before anymalfunction occurs and beforecomponentsare
damaged.N ot only is theundervoltagea potentialsourceof damagebut also the overcurrentwhen the ac voltage recovers. If the drivenot
is equippedwith additionalovercurrent protection, the de bus undervoltage should also protect against these
overcurrents.Many drives areequippedwith fuses in series with the diodes,
against
large overcurrents.Theseshouldnot be used toprotectagainstthe overcurrentafter a
sag.Havingto replace the fuses
aftera voltage sag only causes
additionalinconvenience.
5.3.3.1 Decayof the DC BusVoltage. The de bus voltage for anadjustableof a
speed driveduring a sag in three phases behaves the same as the de bus voltage
personalcomputer, as discussed in Section 5.2.
When we consider a drive with a
motor load P, a nominal de bus voltageVo, and capacitanceC connectedto the de
bus, we can use (5.2) to
calculatethe initial decayof the de bus voltaged uring the sag:
V(t)
=
J 2;
V6 -
t
(5.8)
It has been assumed
that the de bus voltage at sag
initiation equals thenominalvoltage.
We further assumed aconstantpower load. For the standardPWM invertersthis is
probablynot the case. But one can
translatethe constant-powerassumptioninto the
assumptionthat the load on ac side of the inverter, i.e., the motor,
ac
does not notice
anythingfrom the sag. Thus, the
o utputpower of the inverteris independento f the dc
bus voltage. If we neglect the increaseinverter
in
loss for lower de bus voltage (due to
the highercurrents)we arrive at theconstant-powerassumption.The constant-power
assumptionthus correspondsto assumingan ideal inverter: nodrop in voltage at the
motor terminals, and no increase in losses
during the sag.
5.3.3.2 VoltageTolerance. The adjustable-speed
drive will trip either due to an
active interventionby the undervoltageprotection(which is the mostcommonsituation), or by a maloperationof the inverter or the controller. In both cases the trip
will occur when the de busvoltage reaches acertain value Vmin. As long as the ac
voltage does notd rop below this value, the drive will not trip.For sags below this
value, (5.8) can be used to
calculatethe time it takes for the de bus voltage to reach
the value Vmin:
(5.9)
273
Section 5.3 • Adjustable-Speed AC Drives
EXAMPLE 'Consider the example discussed[42]:
in a drive with nominalde bus voltage Vo = 620V and de buscapacitanceC = 4400j.tF powers an acmotor taking an active
power P = 86 kW. The drive trips when thede bus voltagedrops below Vmin = 560V. The
time-to-trip obtainedfrom (5.9) is
4400j.tF
(
2
2)
t = 2 x 86kW x (620V) .- (560 V) = 1.81ms
(5.10)
The minimum ac bus voltage for which the drive will not trip is 560/620
= 90%. This drive will
900/0.
thus trip within 2 ms when the ac bus voltage drops below
Supposethat it would be possible to reduce the setting of the
undervoltageprotectionof
the de bus, to 310 V(50°tlc»). That would enormouslyreduce thenumberof spurioustrips of the
drive, because thenumberof sags below500/0 is only a small fraction of thenumberof sags
below900/0. But the time-to-trip for sags below50% remains very short. Filling inVmin = 310V
in (5.9) givest = 7.38 ms. In fact, bysubstituting Vmin = 0 we can seethat the capacitanceis
completely empty 9.83 ms after sag
initiation, assumingthat the load power remains
constant.
We can concludethat no matter how good the inverter, the drive will trip for any voltage
interruption longer than 10 ms.
The amountof capacitanceconnectedto the dc bus of anadjustable-speed
drive
can be expressed in
I-tF/kW. If we express the de bus voltage in kV and the time in ms,
(5.9) can bewritten as
O.5(~)(V6 -
t=
V;'in)
(5.11)
with (C/P) in JLF/kW. With (C/P) in JLF/h.p. (5.11) becomes
t
= O.67(~)(V6 -
V;'in)
(5.12)
The amountof capacitanceconnectedto the de busof modernadjustable-speed
[138]. Figure 5.18 plots therelation between the
drives is between 75 and 360 JLF/kW
undervoltagesetting for the de bus (vertical) and the
time-to-trip (horizontalscale), for
capacitanceand motor sizeaccordingto (5.11).
three valuesof the ratio between de bus
The voltagetoleranceof the drive, for balancedsags, can beobtainedas follows:
100 ~ ...
ij
[
80
.5
.tg
60
e
40
.~
-.
,,
\
\
~ 20
Figure5.18 Voltage tolerance of adjustablesizes.
speed drives for different capacitor
Solid line: 75J.LF/kW; dashedline: 165 I-tF/
kW; dotted line: 360J.LF/kW.
\
\
\
\
\
\
\
,
\
\
\
\
\
\
\
,
,
\
\
\
20
40
60
Maximum timeinmilliseconds
80
274
Chapter5 • Voltage Sags-EquipmentBehavior
• The setting of the de busundervoltageprotection determinesthe minimum
voltage for which the drive is able to operate.
• From the appropriatecurve, determinedby the capacitorsize, themaximum
sag duration is found.
We seethat even for very lowvaluesof the settingof the de busundervoltage,the drive
will trip within a few cycles.
5.3.3.3 Capacitor Size. It is obvious from the aboveexamplesthat the amount
of capacitanceconnectedto the de busof an adjustable-speed
drive, is not enoughto
offer any seriousimmunity againstvoltage sags. The immunity can be improved by
adding more capacitanceto the de bus.To calculate the amount of capacitance
neededfor a given voltage tolerance,we go back to (5.8) and assumeV(t max) = Vmin,
leadingto
C-
2Ptmax
2
Vo2 - Vmin
(5.13)
This expressiongives the amount of dc bus capacitanceneededto obtain a voltage
toleranceof Vmin, tmax (Le., thedrive trips when the voltagedropsbelow Vmin for longer
than tmax) .
EXAMPLE Considerthe same drive as in the
previousexampleWe want the drive to
be able totoleratesags withdurationsup to 500 ms. Theundervoltagesetting remainsat 560
V (90% of nominal). The capacitanceneeded to achieve this is
o btained from (5.13) with
tmax = 500msand Vmilf = 560V:
c=
286kW x 500ms = t.12F
(620 V)2 - (560 V)2
(5.14)
This exampleis used in [42] tocomparedifferent ways of improving the drive's voltage
tolerance,including the costsof the variousoptions.The total costsof 1.12 F capacitance,with
about$200,000 and to place these
capacitorswould
enclosures,fuses, bars, and fans, would be
2
require a space 2.5 x 18 m
and 60 em high. Abattery backupwould cost "only" $15,000 and
requirea spaceof 2.5 x 4 x 0.6 rrr'. Howeverthe batteryblock would requiremore maintenance
than the capacitors.
Assumethat an undervoltageprotectionsetting of 310 V (50%) is feasible, andthat the
drive shouldbe able totoleratevoltagesags up to 200 ms in
d uration.Equation(5.13) can again
be used to give therequiredcapacitance,which is 119 mF.
This is only one-tenthof the required capacitancefor the original inverter. The costs of
installing capacitancewould still be higher than for the batteryblock but the lowermaintenance
requirementsof the capacitorsmight well tip the balancetoward them. Making an inverter that
can operatefor even lowervoltageswould not gain much ridethroughtime or savecapacitors.
This is because the
s toredenergy in acapacitoris proportionalto the squareof the voltage. It
would, however, increase the
current through the inverter significantly. Bringing theminimum
operatingvoltage down to 25% would doublethe requiredcurrentrating of the inverter but still
require95 mF of capacitance;a reductionof only 20%.
5.3.4 DC Voltage for Three-Phase Unbalanced Sags
In normal operation,the debus voltageis somewhatsmoothenedby the capacitanceconnectedto the dc bus.T he largerthe capacitance,the smallerthe voltageripple.
Section 5.3 •
275
Adjustable-SpeedAC Drives
I "", :----,~--"o~-""""~-r"__~---r<:------,,
,,
0.98 "
:
,
,I
,
g, 0.96
.8
*'
0.94
]
0.92
~
,,
,
,, ,,
""
\ :
", 'I
I
\
, I
. "
':
,,
,,
,
\
I
I
'
'
,
I
, '
, '
,, ,'
,,
,,
I
,
I
,
,I
,,
I
I
"
"i
g 0.90
Figure 5.19 DC bus voltage behind a threephase rectifier during normal
operation,for
largecapacitor(solid line), smallcapacitor
(dashed line)
, and nocapacitorconnected to
the dc bus (dotted line)
.
0.88
0.2
0.8
Where with a single-phase rectifier the
capacitoris only charged twice a cycle
, it is
chargedsix times every cycle for athree-phaserectifier. Figure 5.19 shows the de bus
voltage behind athree-phaserectifier, for variouscapacitorsize. The load fed from the
de bus was assumed to of
bethe constant-powertype. The size of thecapacitanceswas
chosen as follows: for the large
capacitanceand a de bus voltage
o f 100%, the initial
rate of decayof the voltage is 10% per cycle when the ac side voltage drops; for the
small capacitancethe initial rate of decay is 75% per cycle
. We will relate this to the
drive parametersfurther on.
We saw in Section 4.4 that the most
commonsags experienced by three-phase
a
load are type A, type C, and type . DFor a type A sag all three phases
drop in
magnitudethe sameamount.All six voltage pulses in Fig. 5.19 willdrop in magnitude
and the load will empty thecapacitorconnectedto the de bus, until the de bus voltage
drops below the peak of the ac voltage again
. The voltagetolerancefor this case has
been discussed in the previous section
.
5.3.4.1 Sagsof Type C. For a three-phase
.unbalancedsag of type C or type
D, different phases have different voltage drops. Some phase voltages also show a
jump in phase angle
. The behavior of the dc bus voltage
, and thusof the drive, is
completely different than for a balancedvoltage sag
. The upper plot in Fig. 5.20
shows the voltages at the drive
terminalsfor a sagof type C. Note that these are the
line-to-line voltages, as the drive is
connectedin delta. We see how the voltage
drops
in two phases, while the sine waves move toward each other. The third phase does
not drop in magnitude.A sag with acharacteristicmagnitudeof 50% and zero characteristic phase-anglejump is shown. The voltagemagnitudesat the driveterminals
are 66.1% (in two phases) and 100% in the third phase;
phase-anglejumps are
-19.1°, +19.1°, and zero.
The effect of thisthree-phaseunbalancedsag on the de bus voltage is shown in the
lower plot of Fig. 5.20.The capacitorsizes used are the same as in Fig. 5.19. Wethat
see
even for the smallcapacitance
, the de bus voltage does not
drop below 70%. For the
large capacitance,the dc bus voltagehardly deviates from itsnormal operatingvalue.
In the lattercase, the drive will never tripduring a sag of type C, nomatterhow low the
characteristicmagnitudeof the sag. As one phase remains at its pre-event ,value
the
three-phaserectifier simply operatesas a single-phase rectifier
during the voltage sag
.
The drop in de bus voltage (actually
: the increase in voltage ripple) is only
moderate.
276
Chapter 5 • VoltageS ags-EquipmentBehavior
fO:~
U-0.5
«
- I
j
o
>
gj
o
0.5
I
1.5
.~ ,
..
2.5
--: -', -: ',
I
. ', '
0.8
2
. .'
· \
I
'. ' ,
.
- {
"
..
.. ~'
..~'
;
\
". ', :
.'
: ', :
;
3
",'
.o 0.6
U
Cl
0.5
1.5
2
2.5
3
Time in cycles
Figure 5.20 Voltage during a three-phase
unbalanced sag of type C: ac side voltage
(top) and dc side voltages
(bottom) for large
capacitor (solid line), smallcapacitor(dashed
line), and nocapacitorconnected to the dc
bus (dotted line).
The initial behaviorremains identical to the one discussed before for the balanced
t hat the de bus voltage recovers
sag (due to athree-phasefault). The main difference is
after one half-cycle. This is due to the one phase
that remains atnominal voltage for a
sag of typeC.
5.3.4.2 Sagsof Type D. The voltages on ac side and de side
of the rectifier are
shown in Fig. 5.21 for athree-phaseunbalancedsag of type D with characteristic
magnitude50% and nocharacteristicphase-anglejump. The magnitudeof the voltages at the driveterminals is 50%, 90.14%, and 90.14%, with phase-anglejumps
zero, -13.9° and +13 .9°.
For a sag of type D, all three phases
drop in voltage , thus there is no longer one
phase which can keep up the de bus voltage.
Fortunatelythe drop in voltage is moderate for twoof the three phases. Even for a terminal fault, where the voltage in one
phase drops to zero, the voltage in theother two phases does notd rop below
= 86%. The top curve in Fig. 5.21 shows how one phase
drops significantly in
voltage. Theother two phasesdrop less in voltagemagnitudeand theirmaximamove
away from each other. In the
b ottomcurve of Fig. 5.21 the effecto f this on the de bus
4.j3
~ 0.5
~
"0
>
gj
.0
u -0.5
-e
~
"0
>
.
' 1 '1 :
\" .' \
.
0.8
..
I
:' , ,'
",'
]'" 0.6
- ., '\~-..ron--_J'"'...--....j
..
,
: '...
;
. ',I
,
.' , ,
,'
"
U
Cl
0.5
1.5
Time in cycles
2
2.5
3
Figure 5.21 Voltage during a three-phase
unbalanced sag of type D: ac side voltage
(top) and dc side voltages (bottom) for large
capacitor(solid line), smallcapacitor(dashed
line), and nocapacitorconnected to the dc
bus (dotted line).
277
Section 5.3 • Adjustable-SpeedAC Drives
voltage is shown.F or not too small valuesof the dc buscapacitance,the dc bus voltage
of the voltage in the two phases with the
reaches a value slightly below the peak value
moderatedrop. Again the effect of the sag on the de bus voltage,
andthus on themotor
speed andtorque,is much lessthan for a balancedsag.
5.3.4.3 Phase-Angle Jumps.In Figs. 5.20 and 5.21 it isassumedthat the characteristic phase-anglejump is zero. This makesthat two of the phasevoltages have
the same peak value: the highest phases for a sag of type D (Fig . 5.21); the lowest
phases for a sagof type C (Fig . 5.20). A non -zerocharacteristicphase-angle jump
makesthat one of these .two voltages gets lower, and other
the higher. The effecto f
this is shown in Fig. 5.22 for athree-phaseunbalancedsag of type D, with acharacteristic magnitude of 50%. All phase-anglejumps are assumed negative ; positive
phase-anglejumps would give exactly the same effect. When there is capacitance
no
connectedto the de bus(dotted line) the minimum de bus voltage isdeterminedby
of the phase-anglejump is that the minimum
the lowest ac side voltage. The effect
de busvoltage gets lower. But for a drive with a large
capacitanceconnectedto the
de bus, it is the highest peak voltage which
determinesthe de bus voltage.F or such
a drive, the de bus voltage will increase for
increasing phase-anglejump. For a
than during normal
phase-anglejump of -300 the de bus voltage is even higher
operation. Note that a -300 phase-anglejump is an extremesituation for a sag
with a characteristicmagnitudeof 50%.
~
I
,
~
'0 0.8
:-
]
0.6
Q
0.4
0
o
Figure 5.22 DCbus voltageduring a threephase unbalanced sag of type
D, with
characterist
ic magnitude50% and
characteristicphase-anglejump zero (top
left), 10' (top right), 20' (bottomleft), and 30·
(bottom right). Solid line: largecapacitance
;
dashed line: smallcapacitance
; dotted line: no
capacitanceconnected to the de bus.
"
~
:-
Q
,~
- .
,
"
'.
,J
-
' .'"
1
1
1
,I
,,
1
1
,
,.
., ,
I
0.6
0.4
0
, .,
,,
-,
I
,
,
0.5
1
,
,
"
"
0.4
0
0.5
.
.
1
1
0.8
0.6
I
'0 0.8
..5"'
o
1
1
i
0.8 •
"
0.6
-,
1
1
' ,I
0.4
0.5
Time in cycles
0
0.5
Time in cycles
For three-phaseunbalancedsagsof type C, the de busvoltageis determinedby
not drop in magnitude. The phase-anglejump has
the voltage in the phase which does
no influence on this value: it simply remains at 100%.
Thusfor sagsof type C the de bus
voltage is not influenced by the
phase-anglejump, assumingthe capacitanceconnected
to the de bus is largeenough.
5.3.4.4 EffectofCapacitor Size and Sag Magnitude.Some of the effectsof the
size of the de buscapacitanceon the de bus voltageduring unbalancedsags are
summarizedin Figs. 5.23through 5.30. In all the figures, thehorizontal axis gives
the characteristicmagnitudeof the sag, the solid linecorrespondsto a largecapacitanceconnectedto the de bus, thedashedline holds for smallcapacitance,the dotted
278
Chapter5 • VoltageSags-EquipmentBehavior
~ 0.8
.5
~
S
~ 0.6
]
.g 0.4
.1
~ 0.2
0.2
0.4
0.6
0.8
Characteristic magnitude in pu
Figure 5.23 Minimum de bus voltage as a
function of the characteristicmagnitudeof
three-phaseunbalancedsags of type C. Solid
line: largecapacitance;dashed line: small
capacitance;dotted line: no capacitance
connected to the de bus.
line for no capacitanceat all. Figures 5.23 through 5.26 are for three-phaseunbalancedsags of type C. Figures 5.27 through 5.30 are thecorrespondingfigures
for type D.
Figure 5.23 shows the influence on the
minimum de bus voltage. The de bus
undervoltageprotection normally uses this value as a trip
criterion. There is thus a
direct relation between theminimum dc bus voltageand the voltagetoleranceof the
t hat the presenceof sufficient capacitancemakesthat the
drive. We see from the figure
dc busvoltageneverdropsbelow acertainvalue, nomatterhow deep the sag at ae side
normal
is. This is obviously due to the onephaseof the ac voltage which stays at its
value. For a largecapacitance,the drop in de busvoltageis very small. The smaller the
capacitance,the more thedrop in de bus voltage.
Figure 5.24 shows the influenceof sag magnitude and capacitor size on the
voltage ripple at the de bus. The
largerthe capacitanceand the larger thecharacteristic
magnitude,the smaller thevoltage ripple. Again a largecapacitancemitigates the
voltage disturbanceat the de bus. Some drives use the voltage ripple to detect malfunctioning of the rectifier. This ismore used in controlled rectifiers where a large
voltage ripplecould indicate an error in one of the firing circuits. The figure is some-
I00 ~---r------r----'--r-------r-----.,
--0.2
0.4
0.6
0.8
Characteristic magnitude in pu
Figure 5.24 Voltage ripple at the de bus as a
function of thecharacteristicmagnitudeof
three-phaseunbalancedsags of type C. Solid
line: largecapacitance;dashed line: small
capacitance;dotted line: no capacitance
connected to the de bus.
279
Section 5.3 • Adjustable-SpeedAC Drives
_ _ _ _ _
- -. -:-. '7'.":'.~ .-:'."": ..
[ 0.8
.S
i
($
0.6
;>
j
~
Figure 5.25 Average de bus voltage as a
function of the characteristic magnitude of
three-phase unbalanced sagstypeC.
of
Solid
line: large capacitance; dashed line: small
capacitance; dotted line: no capacitance
connected to the dc bus.
0.4
~u
.( 0.2
0.2
0.4
0.6
0.8
Characteristicmagnitudein pu
what misleadingin this sense, as a large
capacitancewould also make it more difficult to
detectunbalancesin the rectifier (likeerrorsin the thyristor firing). In thatcase,eithera
more sensitive'setting of the voltage rippledetectionshould be used (which would
overrule.the gain in voltagetolerance)or the rectifiercurrentsshould be used as a
detectioncriterion (which might introducemore sensitivity tounbalancedsags).
The average de bus voltageshownin
is
Fig. 5.25, the rms value in Fig. 5.26. These
that the drop
determinehow themotordriven by the drive slows down in speed. We see
in average or rmsvoltageis not asdramaticas thedrop in minimum voltage: although
of the capacitance,the less thedrop in speed. Especially for
also here, the larger the size
Of course
longer voltage sags, or
low-inertia loads, this could be a decisive difference.
one needs to assume
that the inverteris able tooperateduring the voltage sag.That is
more likely for largecapacitance,where the dc bus voltage remains high,
thanfor small
capacitance,where the de bus voltage
drops to a low value twice a cycle.
The results for athree-phaseunbalancedsag of type D are shown in Figs. 5.27
through5.30. We saw in Fig. 5.21
t hat for large capacitance,the new steadystatedoes
not settle inimmediately.All values for the type D sag have been
calculatedfor the
of type D is shown in
third cycleduring the sag. Theminimum de bus voltage for a sag
-----------------------------~~~~~~~.
[ 0.8
.S
&>0
~ 0.6
;>
~
..0
.g 0.4
t+-
o
tI.)
Figure 5.26 The rms of the dc bus voltage as
~ 0.2
a function of the characteristic magnitude of
three-phase unbalanced sags of type C. Solid
line: large capacitance; dashed line: small
capacitance; dotted line: no capacitance
connected to the de bus.
0.2
0.4
0.6
0.8
Characteristicmagnitudein pu
280
Chapter5 •
::l
0.
.S
Voltage Sags-EquipmentBehavior
0.8
"
OIl
.f!0
>
0.6
ee
::l
or>
o
-e 0.4
E
::l
E
'2
~ 0.2
0
0
0.2
0.4
0.6
0.8
Characteristic magnitude in pu
Figure5.27 Minimum de bus voltage as a
function of thecharacteristicmagnitude of
three-phase unbalanced sags of type
. Solid
D
line: largecapacitance
; dashed line: small
capacitance
; dotted line: nocapacitance
connected to the de bus
.
100
;:: 80
....
"~
"0.
.S
"0.
60
Q.
' 1:
.s"
OIl
40
0
>
o
0
20
00
~. ~..
0.2
0.4
0.6
0.8
Characteristic magnitude in pu
Figure5.28 Voltage ripple at the de bus as a
function of the characteristicmagnitudeof
three-phase unbalanced sags of type D. Solid
line: large capacitance; dashed line
: small
capacitance; dotted line
: no capacitance
connected to the dc bus.
5. 0.8
.S
~
~ 0.6
:g
or>
.g 0.4
t
~ 0.2
0.2
0.4
0.6
0.8
Characteristic magnitude in pu
Figure 5.29 Average de bus voltage as a
function of thecharacteristicmagnitudeof
three-phaseunbalancedsags of type D. Solid
line: large capacitance; dashed line
: small
capacitance;dotted line: no capacitance
connected to the de bus
.
28t
Section 5.3 • Adjustable-SpeedAC Drives
a0.8
.s
~
~
0.6
.
>
]
~
0.4
C+-t
o
rJ)
Figure 5.30 The rms of the de bus voltage as~ 0.2
a function of the characteristicmagnitudeof
three-phaseunbalancedsags of type D. Solid
line: large capacitance;d ashedline: small
00
capacitance;d otted line: no capacitance
connectedto the de bus.
0.2
0.4
0.6
0.8
Characteristicmagnitudein pu
Fig. 5.27. Comparisonwith Fig. 5.23 for type C revealsthat for a type D sag the
minimum de busvoltagecontinuesto drop with lower characteristicmagnitude,even
with large capacitorsize. But againan increasein capacitancecan significantly reduce
the voltage drop at the de bus.For the drive with the largecapacitancethe de bus
voltagedoes not drop below 80% , even for thedeepestunbalancedsag.
Figure 5.28 plots' thevoltageripple for type D sags, whichshowsa similar behavior as for type C sags.T he voltage ripple is calculatedas the peak-to-peakripple
related to the normal value. Therefore,the voltage ripple for the drive without capacitancedoes not reach 1000/0 for a sagof zero characteristicmagnitude.
In Figs. 5.29and 5.30, showingaverageand rms valueof the de busvoltage,we
seesimilar values as for sags
o f type C. Again the differenceis that the de busvoltage
continuesto drop for decreasingcharacteristicmagnitude.Deep sags of type D will
causemore drop in motor speed than sags of the samemagnitudeof type C. For
shallow sags the effect on the
m otor speed will beaboutthe same.
5.3.4.5 Sizeof the DC BusCapacitance. In the previous figures, the de bus
voltagewas calculatedfor threevalues of the sizeof the capacitanceconnectedto the
dc bus. Thosewere referred to as "large capacitance,""small capacitance,"and "no
capacitance."Large and small werequantified through the initial decayof the de bus
voltage: 10% per cycle for the largecapacitance,75% per cycle for the smallcapacitance.Here we will quantify the amountof tLF to which this corresponds.
The de busvoltage V(t) during the sag isgovernedby the lawof conservationof
energy: the electricload P is equalto thechangein energystoredin the de buscapacitor
C. In equationform this readsas
2
!!-{!CV
} =p
dt 2
(5.15)
Let Vo be the de busvoltageat saginitiation. This gives at saginitiation
dV
CVo-=P
dt
(5.16)
282
Chapter5 • Voltage Sags-EquipmentBehavior
from which the initial rateof decay of the dc busvoltagecan becalculated:
dV
P
d(= CVo
(5.17)
From (5.16) we can derive an
expressionfor the capacitorsizeneededto get acertain
initial rate of decay of de bus voltage:
p
C=--cw
V
(5.18)
oClt
EXAMPLE For the same driveparametersas before (620 V, 86 kW) we can use
(5.18) to calculate the required size of the
capacitance.As a first step we have totranslatepercent per cycle into volts per second:
75% per cycle = 27,900Vis
100/0 per cycle = 3730Vis
To obtain a rate of decay of750/0 per cycle, we need caapacitanceof
86kW
C = 620 V x 27,900V/s = 4970JlF
(5.19)
or 57.8 /-LF/kW. Similarly we find that 37.3 mF or 433/-LF/kW correspondsto 10% per cycle.
These values need to be
comparedto the amountof capacitancepresentin moderndrives, which
JlF/kW, accordingto [138]. We seethat the "largecapacitance"curves are
is between 75 and 360
feasible withmodernadjustable-speed
drives.
5.3.4.6 Load Influence. The main load influence on voltage sags is the reduction in negative-sequencevoltage due to induction motor load, as explained in
Section 4.8. To see
w hat the effect is onadjustable-speed
drives, wereproducedtype
C and type 0 sags withreducednegative-sequence
voltageand calculatedde bus voltage behind a non-controlledrectifier. The three-phaseunbalancedsags with reduced
negative-sequence
voltage were calculatedin the same way as for Figs. 4.138
and
4.139. Theanalysiswas performedfor a three-phaseunbalancedsag with acharacteristic magnitudeof 50% and zero phase-anglejump. The voltagesat the equipment
terminalsare for a 50% sag of type C:
Va = 1
Vb =
Vc
_!2 - !j./3
4
(5.20)
= -~+~j./3
and for a sagof type 0:
(5.21)
283
Section 5.3 • Adjustable-SpeedAC Drives
Splitting the phasevoltages in sequence
componentsgives
(5.22)
for a sagof type C, and
3
VI =-
4
1
4
(5.23)
V2 =--
for a sagof type D. A "distortedtype C" sag iscreatedby keeping the positive-sequence
voltage constant,while reducingthe negative-sequence
voltage. This is to simulate the
drops
effect of induction motor load. If we assumethat the negative-sequence voltage
by a factor of {J, thus from V2 to (1 - {J) V2, we obtain the phase voltages from
Va = VI +(I-fJ)V2
+ a2( 1 - fJ)V2
VI + a(l - {J)V2
Vb = VI
V(. =
(5.24)
-!
where a =
+ !j,J3. The resulting phase voltages are usedcalculatethe
to
de bus
voltagesduring the sag, in the same way as for the
"nondistorted"sag. The results are
shown in Figs. 5.31through 5.34. Figure 5.31 plots the average de bus voltage as a
voltage. Notethat a drop of 50o~ in negativefunction of the drop in negative-sequence
sequencevoltagerequiresa very largeinductionmotorload. We see from Fig. 5.31 that
the motor load drops the minimum dc bus voltage in case capacitoris
a
used.For a
drive without de buscapacitor,the minimum de bus voltage increases. The
drop in
negative-sequence
voltagemakesthat the three voltages get closer magnitude,so
in
that
the effect of acapacitorbecomes less. The same effect is seen in Fig. 5.33 for type D
sags. Figs. 5.32 and 5.34 show
that also the average de bus voltage
dropsfor increasing
motor load.
[ 0.8
.S
~g 0.6
j
.g 0.4
.1
~ 0.2
Figure 5.31 Induction motor influence on
minimum de bus voltage for sags of type C.
Solid line: large capacitor; dashed line: small
capacitor; dotted line: no capacitor connected
to the de bus.
0.1
0.2
0.3
0.4
Drop innegative-sequence
voltage
0.5
284
Chapter5 • VoltageSags-EquipmentBehavior
~ 0.8
.S
.
Go)
f 0.6
-0
>
:g
.,D
~
0.4
Go)
<G0.2
0.1
0.2
0.3
0.4
Drop in negative-sequence
voltage
0.5
Figure 5.32 Induction motor influenceon
averagede busvoltage for sagsof type C.
Solid line: largecapacitor;dashedline: small
capacitor;dottedline: no capacitorconnected
to the de bus.
&e 0.8
.5
i
]
0.6
]
~ 0.4
§
:~~
0.2
0.1
0.2
0.3
0.4
0.5
Drop in negative-sequence
voltage
a 0.8
------------------_
Figure 5.33 Induction motor influenceon
minimum de busvoltage for sagsof type D.
Solid line: largecapacitor;dashedline: small
capacitor;dottedline: no capacitorconnected
to the de bus.
--.
.53
.
;
-0 0.6
>
]
~
0.4
-<
0.2
t
0.1
0.2
0.3
0.4
Drop in negative-sequence
voltage
0.5
Figure 5.34 Induction motor influenceon
averagede busvoltagefor sagsof type D.
Solid line: largecapacitor;dashedline: small
capacitor;dottedline: no capacitorconnected
to the de bus.
285
Section 5.3 • Adjustable-SpeedAC Drives
5.3.4.7 Powering the Controllers.In older drives thecontrol electronics for the
PWM inverter was powered from the supply. This made the drive very sensitive to
disturbancesin the supply. Inmoderndrives thecontrol electronics is powered from
the de bus which can be more
constantdue to the presence of
capacitors.But even
here the samereasoningcan be used as for process
control equipment.Controllers
are essentiallylow-power equipmentwhich only require a smalla mount of stored
energy to ridethrough sags. The design of the power supply to the drive
controller
should be suchthat the controller stays active at least as long as the power electronics or themotor do not require apermanenttrip. It should not bethat the controller becomes the weak
p art of the drive. Figure 5.35 shows the typical
configuration
for powering the controller. The capacitanceconnected to the de bus between the
rectifier and theinverter is normally not big enoughto supply themotor load and
the controller during a balancedsag longerthan a few cycles. The power supply to
the controller can beguaranteedin a numberof ways:
• By inhibiting firing of the inverter sothat the motor no longer discharges the de
buscapacitance.The power taken by the
controlleris so much smallerthanthe
motor load, that the capacitorcan easily power thecontroller even for long
voltage sags. When the supply voltage recovers,controller
the
can automatically restartthe load.
• Additional capacitancecan be installed on low-voltage side of the de-de
switched mode power supply between the dc bus and control
the
circuitry.
As this capacitanceonly needs to power thecontroller, a relatively small
amountof capacitanceis needed. Also abatteryblock would do the job.
• Some drives use the
rotationalenergy from themotor load to power the controllers during a voltage sag orshortinterruption.This causes small
additional
drop in motor speed, smallenoughto be negligible. A special
control technique
for the inverter is needed, as well as methodto
a
detect the sag[33].
Diode
rectifier
PWM
inverter
ac motor
Figure 5.35 Configurationof the power
supply to the control circuitry in an
adjustable-speed drive.
5.3.5 Current Unbalance
5.3.5.1 Simulations. Unbalanceof the ac voltages not only causes an increased
ripple in the de voltage but also a large
unbalancein ac currents.The unbalancein
currentdependson the typeof sag.Considerfirst a sagof type D, where one voltage
is much lower than the other two. The upperplot in Fig. 5.36 shows the ac side voltages (inabsolutevalue) comparedwith the de bus voltage (solid line near the top)
during one cycle, for a sag of type D with
characteristicmagnitudeequal to 50% •
Here it is assumedthat the de bus voltage does
not change at allduring the sag. The
286
Chapter5 • Voltage Sags
-EquipmentBehavior
fo:o/>:'
-'> ;::>~,- >: :Jj
oL~'.:-><: . . . . :
_~l o~
V
I
' < ;
J_~1
O
J}01
_
.
o
0,'
0:6
A
0:'
M :
0: ;
0.'
0,6
_
0.2
.
.
0.4
0.6
Time in cycles
0,'
0,'
J~
0.8
I
'I Figure 5.36 ACsideline voltages (top) and
currents(phase a, b, and c from top to
I bottom) for a three-phase unbalanced sag of
type D.
rectifier only deliverscurrent when the ac voltage (inabsolutevalue) is largerthan
the dc voltage . We have assumed
that this current is proportional to the difference
between theabsolutevalue of the ac voltage and the de voltage . This results in the
line currentsas shown in the three
remainingplots in Fig. 5.36.
The three voltages in the top plot of Fig. 5.36 are the voltage difference between
phase a and phase (dashed),between
b
phase b and phase(dash-dot),and
c
between
phase c and phase (dotted).The
a
first pulse occurs when the voltage between a and c
exceeds the de voltage
( around t = 0.2 cycle). This results in acurrent pulse in the
phases a and c.
Around t = 0.3 cycle the voltage between
bandc exceeds the dc voltage
leading to acurrent pulse in the phases
b andc. The patternrepeatsitself around t =
0.7 cycle andt = 0.8 cycle. Thecurrentsflow in oppositedirection because the ac voltages areopposite now. Whereasat t = 0.2 cycle the voltage between c and a was
negative resulting in acurrent from a to c, the voltage is positive now
resulting in a
currentfrom c to a. The voltage between a and b has
droppedso muchthat there are no
currentpulses between a and b. This results in two missing pulses per cycle for phase a
as well as for phase b.
Whereas innormal operationthe capacitoris charged 6 times per cycle, this now
only takes place four times per cycle. These four pulses must
carry the sameamountof
charge as the original six pulses. The
consequenceis that the pulses will be up to 50%
higher in magnitude.
For a type C sag thesituationis even worse, as shown in the top plot
of Fig. 5.37.
One line voltage is much higher
than the other two, so that only this voltage leads to
current pulses. The resultingcurrent pulses in the three phases are shown in the three
bottom plots of Fig. 5.37.
Due to a sag of type C the
numberof currentpulses is reduced from 6 per cycle to
2 per cycle, leading to up to 200%
overcurrent.Note that a large overcurrentwould
alreadyarise for a shallow sag. The
momentone or two voltagesdrop below the de bus
voltage, pulses will be missing and the
remainingcurrentpulses will have to be higher to
compensatefor this.
5.3.5.2 Measurements.Figures 5.38, 5.39, and 5.40 show
measurementso f the
input currents of an adjustable-speeddrive [27], [30]. Figure 5.38 shows theinput
287
Section 5.3 • Adjustable-Speed AC Dr ives
Figure 5.37 AC side volta ge (top ) and
cur rents (ph ase a, b, and c from top to
bottom) for a three-phaseu nbalancedsag of
type C.
300
200
100
.5
o
~
3 - 100
./
l
\
I
\
u
-200
-300
o
0.01
300,--- , --
0.02 0.03 0.04
Time in seconds
....,.----,---r-
0.05
0.06
---,,---
,---,
200 l--tHr-+tHl--1---It-Ir--+---+Ht---l
~
~
.5
1001-t-ft-t---HUHH---ttH+---'I-Ht+----i
0 H-l...--li-'r-,.....--lo+--t""'4--l-o,--+1p.o1--jloo~
~
3 -100 H---t-\-Itti---t1tt-t--HH--+-IHl
o
- 200 JV---t-+HF-t----ftt+-t--\tPJ-Figure 5.38 Input cur rent for an ac drive in
normal operation. (Reproduced from
Mans oo r (27).)
- 300 " -_
o
+-ffi
..L-_-'-_ - - ' -_ - - "_ _" - _-'-----'
0.01
0.02 0.03 0.04
Time in seconds
0.05
0.06
currents for the drive under normal operating conditions. Only two currents ar e
shown , the th ird one issimilar to one of the other two. The drive is connectedin
of four pulses in
delt a, so that each current pulse shows up in two phases. total
A
each of the threephasesimplies 6 pulsesper cycle chargingthe capacitor. Therewas a
small unbalancein the supply voltage leading to the difference between thecurrent
pulses. We see
t hat the magnitudeof the currentpulses is between 200 and 250 A.
288
Chapter5 • VoltageSags-EquipmentBehavior
400
300
'"
200
~
100
~
o
.5
.\
s5 - 100
o
-200
- 3000
0.01
0.02 0.03 0.04
Time in seconds
400
i
n
N
n
ru
300
200
0.05
0.06
t\
/\
100
o
.5
~ - 100
8
-200
-300
-400
0
~
~
0.01
lJ\
~
~I
~
0.02 0.03 0.04
Time in seconds
\~
~
0.05
Figure 5.39Input currentfor an ac drive with
voltage unbalance(Reproduced
.
from
0.06
Mansoor[27].)
Figure 5.40Input current for an ac drive during a single-phase fault.
(Reproduced
from Man soor [27).)
289
Section 5.3 • Adjustable-SpeedAC Drives
Figure 5.39 showsthe samecurrents,for an unbalancein the supply voltage. The
highestvoltagemagnitudewas 3.6% higher than the lowest one. This smallunbalance
alreadyleads to two missing pulses
both relatedto the same linevoltage.Thereare now
only four pulsesleft, with a magnitudebetween300 and 350 A, confirming the 500/0
overcurrentpredictedabove.
Figure 5.40 shows the rectifierinput currentfor a single-phasesag at the rectifier
terminals. A measuredsag is reproducedby means of three power amplifiers. As
explainedin Section4.4.4, asingle-phasefault will cause a type D sag on the
terminals
of delta-connectedload. The two remainingpulses per cyclea ndthe peakcurrentof 500
to 600 A confirm the 200% overcurrentpredictedabove.
5.3.6 Unbalanced Motor Voltages
The de busvoltageis convertedinto an acvoltageof the requiredmagnitudeand
frequency, by using a voltage-sourceconverter (VSC) with pulse-width modulation.
The principle of PWM can beexplainedthrough Fig. 5.41. A carrier signal Vcr with
. a frequency of typically a few hundred Hertz, is generatedand comparedwith the
referencesignal Vrej (dashedcurve in the upper figure). The referencesignal is the
required motor terminal voltage, with a certain magnitude, frequency, and phase
angle. If the referencesignal is largerthan the carriersignal, theoutput of the inverter
is equal to the positive input signal V+ and the other way around:
= V+,
Vout
Vout = V_,
V ref
> Vcr
(5.25)
Vr~f < Vcr
The resulting output voltage Vout is shown in the lower plot of- Fig. 5.41. It can be
shown that the output voltage consistsof a fundamentalfrequency sine wave plus
harmonicsof the switching frequency[43]. The latter can beremovedby a low-pass
filter after which the requiredsinusoidalvoltageremains.If the de busvoltagevaries,
both the positive and the negativeoutput voltage V+ and V_will changeproportionally. These variations will thus appearas an amplitude modulation of the output
voltage. Let the requiredmotor voltagesbe
::s
.e
I
0.5
~
0
S
0-0.5
::>
-I
o~----::-.L..:-----:-~--~-_.L.--_--J
0.6
1 r~
.9 0.5
0.8
r--
i
Figure5.41 Principle of pulse-width
modulation:carrier signal with reference
signal (dashed)in the top figure; the pulsewidth modulatedsignal in the bottom figure.
0
0-0.5
::>
.....-.
-1
o
'--
0.2
~
0.4
0.6
Timein cycles
'----
0.8
Chapter 5 • VoltageSags-EquipmentBehavior
290
=
Va Vm cos(2rrfmt)
Vb = Vm cos(2rr.fmt- 120°)
(5.26)
Vc = Vm cos(2rrfmt+ 120°)
We assumethat the high-frequencyharmonicsdue to the PWM switching are all
removedby the low-passfilter, but that the variationsin dc busvoltagearenot removed
by the filter. The motor voltagesfor a de bus voltage Vdc(t) are the product of the
requiredvoltage and the p.u. dc bus voltage:
Va = Vdc(t) X Vmcos(2rrfmt)
Vb = Vdc(t) X Vmcos(2rrfmt- 120°)
Vc = Vdc(t) X Vmcos(2rrfmt+ 120°)
(5.27)
Normally the motor frequencywill not be equalto the systemfrequency,thusthe ripple
in the de voltage is not synchronizedwith the motor voltages.This may lead to unbalancesand interharmonicsin the motor voltages.
The motor terminal voltageshave beencalculatedfor sagsof type C and 0, for
various characteristicmagnitudesand motor frequencies.A small capacitorwas connectedto the de bus. Figure 5.42showsthe resultsfor a 500/0 sag of type C (see Fig.
5.20) anda motor frequencyequalto the fundamentalf requency.We seethat the motor
terminal voltagesare seriouslydistortedby the ripple in the de busvoltage.One phase
dropsto 75% while anotherremainsat 100%. The de busvoltageis shownas adashed
line in the figure. Figure 5.43 showsthe result for a 50% sag of type 0 and a motor
frequencyof 50 Hz. The effect is similar but lessseverethan for the type C sag.
Figure 5.44 plots the three motor terminal voltages for a motor frequency of
40 Hz and a supply frequency of 50 Hz. The motor frequency is now no longer an
integer fraction of twice the power systemfrequency (the de ripple frequency). But
two periods of the motor frequency (50 ms) correspondto five half-cycles of the
power system frequency. The motor terminal voltage is thus periodic with a period
of 50 ms. This subharmonicis clearly visible in Fig. 5.44.
Figure 5.45 shows the unbalanceof the voltages at the motor terminals, as a
function of the motor speed.The unbalanceis indicatedby showingboth the positive
and the negative-sequence
componento f the voltages.The largerthe negative-sequence
component,the larger the unbalance.We seethat the unbalanceis largest for motor
o
234
Time in cycles
5
6
Figure 5.42 Motor terminal voltagedue to a
three-phaseunbalancedsag of type C with a
characteristicmagnitudeof 50%, for a motor
frequencyof 50 Hz. The de busvoltageis
shownas adashedcurve for reference.
291
Section 5.3 • Adjustable-Speed AC Drives
j
0.5
'0
>
]
.~
0
B
~ -0.5
~
Figure 5.43 Motor terminal voltage due to a
three-phase unbalanced sag of type D with a
characteristic magnitude of
500/0, for a motor
frequency of 50 Hz. The de bus voltage is
shown as a dashed curve for
reference.
o
~-:
~ -:
j-:
Figure 5.44 Motor terminal voltages due to a
three-phase unbalanced sag of type C with a
characteristic magnitude of 50%, for a motor
speed of 40 Hz.
5
234
Timein cycles
6
o
2
4
6
8
10
o
2
4
6
8
10
o
2
4
6
8
10
Time in cycles
0.9 ...------r----~------..---------.
0.8
::s 0.7
Q..
.s 0.6
.t
~
0.5
H0.4
g.
0.3
Figure 5.45 Positive- (solid) and negative- rI} 0.2
sequence component (dashed) of the motor
0.1
terminal voltages as a function of the motor
,,'--- .....
speed. A sag of type C with a characteristic
°O~---.....::a....:-.;:l-----"""'O---~-~--_--J-_-----J
50
100
150
200
magnitude of500/0 was applied at the supply
Motorfrequencyin Hz
terminals of the adjustable-speed drive.
292
Chapter5 • VoltageSags-EquipmentBehavior
TABLE 5.8 Motor Terminal and DC Bus Voltagesfor AC Drives Due to a
50% Type C Sag
Positive-sequence
voltage
min
max
Small capacitance
Large capacitance
88.88%
98.250/0
Negative-sequence
voltage
83.44%
96.91%
max
5.56%
0.81 %
de busvoltage
avg.
rms
87.38%
97.83%
87.80%
97.84%
speedsaround50 Hz. For low,speed theunbalanceis very small.Note that the voltage
contains25% of negativeat the supply terminals of the drive (i.e., the type C sag)
sequence and75% of positive-sequence
voltage. Even for a small de bus
capacitorthe
unbalanceat the motor terminalsis significantly lessthan at the supply terminals.
The resultsof the calculationsare summarizedin Table 5.8. Maximum andminimum positive andnegative-sequence
voltageshave beenobtainedas in Fig. 5.45. (The
lowestnegative-sequence
voltagewas lessthan0.01% in bothcases.) The average de bus
For a
voltage wasobtainedas in Fig. 5.25; the rmso f the de bus voltage as in Fig. 5.26.
large dc buscapacitor,the ripple in the de busvoltagebecomes very small, so
that the
motorterminalvoltagesremainbalanced,no matterhow big theunbalancein the supply.
5.3. 7 Motor Deacceleratlon
Most ac adjustable-speed
drives trip on one of thecharacteristicsdiscussed before.
After the tripping of the drive, theinduction motor will simply continueto slow down
until its speed getso ut of the rangeacceptablefor the process. In case the electrical
part
of the drive is able towithstandthe sag, thedrop in systemvoltagewill cause adrop in
voltage at themotor terminals. We will estimatethe motor speed forbalancedand
unbalancedsags. We will use a simplifiedm otor model: the electricalt orqueis proportional to thesquareof the voltage,but independento f the motor speed; themechanical
torque is constant.
5.3.7.1 Balanced Sags.For balancedsags all threephasevoltages drop the
sameamount. We assumethat the voltagesat the motor terminals are equal to the
supply voltages (in p.u.),thus that the sag at themotor terminalsis exactly the same
as the sag at the rectifier
terminals. The de buscapacitorwill somewhatdelay the
drop in voltage at the de bus andthus at the motor terminals; but we sawthat this
effect is relatively small. Thevoltage drop at the motor terminals causes adrop in
torque and thus adrop in speed. Thisdrop in speed candisrupt the production
processrequiring an intervention by the processcontrol. The speed of amotor is
governed by the energy balance:
d
dt
(12: J w2) =
w(Tel
-
Tm£'ch)
(5.28)
where J is the mechanicalmoment of the motor plus the mechanicalload, «o is the
motor speed (inradiansper second),Tel is the electricaltorquesuppliedto the motor,
and Tmech is themechanicall oad torque.The electricaltorque Tel is proportionalto the
squareof the voltage. Weassumethat the motor is runningat steadystatefor a voltage
of I pu, sothat
293
Section 5.3 • Adjustable-SpeedAC Drives
= V 2 Tmech
Tel
(5.29)
For V = 1 electricaland mechanicalt orqueare equal.The resultingexpressionfor the
drop in motor speedis
d to
dt
2
(V -
=
I) Tmech
J
(5.30)
Introduce the inertia constant H of the motor-load combinationas the ratio of the
kinetic energyand the mechanicaloutput power:
H=
IJw2
2
(5.31)
0
lOo T,nech
with lOo the angularfrequencyat nominal speed;and the slip:
lOo - w
s=--lOo
(5.32)
Combining(5.31) and (5.32) with (5.30) gives anexpressionfor the rate of changeof
motor slip during a voltagesag (for w ~ wo):
ds I - V 2
dt = ---:uI
(5.33)
Thus for a sagof duration ~t and magnitude V the increasein slip is
tls
ds
1 - V2
= -tlt
= -2H
-tlt
dt
(5.34)
The largerthe inertia constantH, the less theincreasein slip. For processessensitive to
speedvariations,the voltage tolerancecan be improved by addinginertia to the load.
Figure 5.46 showsthe increasein slip as afunction of the sagmagnitudeandduration,
for an inertia constantH = 0.96 sec.N ote that an increasein slip correspondsto a drop
in speed.The increasein slip is given for four different sagdurations,correspondingto
2.5,5,7.5,and 10 cycles in a50Hz system.As expectedthe speed willdrop more for
v oltage(PWM disabled)the drop in speed
deeperandfor longersags. But even for zero
is only a fewpercentduring the sag.
If the maximum-allowableslip increase(slip tolerance)is equal to tlsmClx , the
minimum-allowablesag magnitude Vmin for a sagduration T is found from
O.I.------r----~--~----..-------..
0.08
~
fI.)
~
0.06
S
.S
Q,)
~
j
0.04
0.02
Figure 5.46Increasein motor slip as a
function of the sagmagnitudefor different
sagduration: 50ms(solid curve), lOOms
(dashed),150ms(dash-dot),200 ms(dotted).
"
......
...... "
0.2
0.4
0.6
Sag magnitude in pu
0.8
294
Chapter5 • VoltageSags-EquipmentBehavior
vmin. --
J
I - 2H f).smax
T
(5.35)
A zero voltage, Vmin = 0, can be tolerated for a duration 2H f:1s max' The resulting
voltage-tolerancecurves have beenplotted in Fig. 5.47 for H = 0.96 secand various
valuesof the slip tolerancef:1s max' Theseare thevoltage-tolerancecurvesfor an adjustable-speeddrive wherethe drop in speedof the mechanicall oad is the limiting factor.
Note that some of the earlier quoted tolerancesof adjustable-speeddrives are
even abovethe 1% or 2% curves.This is mainly due to thesensitivity of the powerelectronicspart of the drive. Note also that it has beenassumedherethat the drive stays
on-line. Temporary tripping of the drive correspondsto zero voltage at the drive
terminals.This will obviously lead to alarger drop in speed.
5.3.7.2 Unbalanced Sags.The curves in Figs. 5.46and 5.47 have been calculated assumingthat the voltagesat the motor terminalsform a balancedthree-phase
set. For a balancedsag this will obviously be the case. But as we have seen in the
previous section, for an unbalancedsag themotor terminal voltagesare also rather
balanced.The larger the de buscapacitance,the more balancedthe motor terminal
voltages. The above calculations of the motor slip are still applicable. When the
motor terminal voltage show a serious unbalance, the positive-sequencevoltage
should be used.
The effect of three-phaseunbalancedsags on themotor speed has been
calculated
underthe assumptionthat the positive-sequence
voltageat themotor terminalsis equal
to the rmsvoltageat the de bus.T his is somewhatan approximation,but we haveseen
that the motor terminalvoltageis only slightly unbalancedeven for a largeunbalancein
the supplyvoltage.This holdsespeciallyfor a drive with a largede buscapacitance.The
de bus rmsvoltageshave been calculatedin the sameway as for Figs. 5.26and 5.30.
Thesewere used tocalculatethe drop in motor speedaccordingto (5.34) and voltagetolerancecurveswere obtained,as in Fig. 5.47.T he resultsfor type C sags areshownin
Figs. 5.48, 5.49,and 5.50. Figures 5.48 and 5.49 presentvoltage-tolerancecurvesfor
different values of the maximum drop in speed which theload can tolerate, for no
capacitanceand for a small capacitance,respectively,presentat the de bus.Even the
small capacitorclearly improvesthe drive's voltage tolerance.Below a certaincharacteristic magnitudeof the sag, the rms value
o f the de busvoltageremainsconstant.This
100
90
1%
=80
5%
G,)
t
70
]0%
0-
.5 60
G,)
]
50
.~ 40
~
30
«I
C/.)
20
10
200
400
600
800
Sag duration in milliseconds
1000
Figure 5.47 Voltage-tolerancecurvesfor
adjustable-speed
drives, for three-phase
balancedsags, fordifferent valuesof the slip
tolerance.
295
Section 5.3 • Adjustable-SpeedAC Drives
100r----r------r-====::::======::::::::~
90
10/0
... 80
2%
[ 70
5%
.S 60
u
]
50
10%
.~ 40
; 30
~
20
fIl
200/0
10
Figure 5.48 Voltage-tolerancecurves for sag
type C, nocapacitanceconnectedto the de
bus, for different values of the slip tolerance.
200
400
600
800
1000
800
1000
Sag duration inmilliseconds
... 80
5
e
&
.5 60
i.~ 40
e
~
fIl
1%
5%
2%
20
Figure 5.49 Voltage-tolerancecurvesfor sag
type C, smallcapacitanceconnectedto ~he de
bus, for different values of the slip tolerance.
200
400
600
Sag duration inmilliseconds
100----r----.,.------r----=~======l
- - -- --
.;
.:--
,',
:,
:
Figure 5.50 Voltage-tolerancecurves for sag
type C, large (solid line), small(dashed),and
no (dotted)capacitanceconnectedto the de
bus.
,
I
200
400
600
Sag duration inmilliseconds
800
1000
296
Chapter5 • VoltageSags-EquipmentBehavior
shows up as a vertical line in Fig. 5.49.
Figure 5.50 comparesdrives with large, small,
and no de buscapacitancefor a load with a slip toleranceof 1%. The capacitorsize has
a very significant influence' on the drive
performance.
The largeimprovementin drive performancewith capacitorsize for type C sags is
obviouslyrelatedto the onephaseof the acsupplywhich doesnot drop in voltage.For
a largecapacitance,this phasekeeps up thesupply voltage as if almost nothing hapsmaller,as even theleast-affectedphasesdrop in
pened.For type D sags, this effect is
o f the capacitorsize on thevoltage
voltage magnitude.Figure 5.51 shows the influence
tolerancefor type D sags.T he threecurveson the left are for a sliptoleranceof 1%, the
ones on the right for10% slip tolerance.The improvementfor the I % casemight look
marginal,but one shouldrealizethat the majority of deepvoltagesags have aduration
around100 ms. The largecapacitanceincreasesthe voltagetolerancefrom 50 to 95 ms
for a 50% sag magnitude. This could imply a serious reduction in the number of
equipmenttrips.
From Figs. 5.48through5.51 it becomesclear that the effectof unbalancedsags
to
by using a large
on themotor speed is small. The best way preventspeedvariationsis
de buscapacitorand by keeping the drive online. The small speedvariations which
would result may becompensatedby a control systemin case theycannotbe tolerated
by the load.
100
.;
+J
e
Q)
,~
80
1%
~
8.
.5 60
/'
.sa
/
,,
.~ 40
eu
/
/
I
e
I
:;
~
I
,"
,
,
en 20 ::
:,
:,
o
o
10%
:''I
:1
,
I
200
400
I
600
Sag duration in milliseconds
I
800
1000
Figure 5.51 Voltage-tolerancecurves for sag
type D, for two valuesof the slip tolerance,
large (solid line), small(dashed),and no
(dotted)capacitanceconnectedto the de bus.
5.3.8 Automatic Restart
As we saw before many drives
trip on undervoltage,for a sagof only a few cycles.
This tripping of the drive doeshowevernot always imply aprocessinterruption.What
happensafter the tripping dependson how themotor reactswhen thevoltage comes
back. A good overviewof options is given in [51], which served as a basis for the list
below.
• Some drives simply tripandwait for a manualrestart.This will certainlylead to
a processinterruption.A drive which doesnot automaticallyrecoveraftera trip
looks like aratherbad choice.Howeverthereare cases in which this is the best
option. On onehandthere areprocesseswhich arenot very sensitive to a drive
outage.The standardexampleis a drive used forair-conditioning.An interruption of the air flow for a fewminutesis seldomany concern.On theothersideof
Section 5.3 • Adjustable-SpeedAC Drives
•
•
•
•
•
•
297
the spectrumone finds processes which are extremely sensitive to speed variation. If a very small speedvariation alreadyseverelydisruptsthe process, it is
best tonot restartthe drive. Restartingthe drivecertainlyleads to a speed and
torquetransient,which could makethe situationworse. Safetyconsiderations
could dictatethat a total stoppageis preferableabove anautomaticrestart.
Some drives wait a few
minutesbeforethe automaticrestart.This ensuresthat
the motor load has come to acompletestop. Thecontrol system simplystarts
the motor in the same way it would do for an ormal start. With a delayed
automaticrestart,safety measureshave to betakento ensurethat nobodycan
be injured by the restartof the motor.
The control system of the drive canapply electrical ormechanicalbraking to
bring the load to a forcedstop, after which a normal restart takes place.
Without specialcontrol measures,it is very hard to restartthe drive successfully before it has come to a
standstill.Thusforced brakingcan reduce the time
to recovery.The requirementis that the process driven by the drive is able to
toleratethe variationsin speed andtorquedue to braking and reacceleration.
Most drives are able tostart under full load, which also impliesthat they
should be able to pick up thealready spinning load. The dangerof already
spinningload is that it might still containsomeair-gapflux causingan opencircuit voltageon themotor terminals.Whenthe drive isrestartedwithout any
synchronizationsevere electricaltransientsare likely to occur due to the residual flux. The solutionis to delay therestartfor aboutone second to allow this
residual flux to decay. Thisoption will imply that the motor load will be
without poweringfor one or two seconds. In this
time the motor speed decays
to a typical valueof 50% of the nominalspeed,dependingon the intertia of the
load. Also at themomentof restartthe inverterfrequency will not beequalto
the motor speed, themechanicaltransientthis causes might not be
toleratedby
the process.
A speedidentification techniquecan be used toensurethat the inverter picks
up the load at the right speed. This reduces mechanical
the
transienton restarts
and makesthe motor recoverfaster. Thespeed-identificationprocessshouldbe
enablea fastrestartof
able todeterminethe motor speed within a few cycles to
the drive.
To seriouslylimit the drop in speed and the time to recovery, the drive needs to
restartvery soonafter the voltagerecovers.For this theinvertershouldbe able
to resynchronize.on the residual stator voltages. This requiresextra voltage
sensors,thus increasingthe priceof the drive.
Insteadof resynchronizingthe drive after the sag, it is possible tomaintain
synchronizationbetween inverter and motor during the sag. This requires a
more complicatedmeasurementand control mechanism.
Figures5.52and 5.53showthe responseof a drive with automaticrestart.In Fig.
5.52 the driverestartssynchronouslywhich leads to adrop in speed well within 10%.
The motor currentdropsto zeroduring the sag. Thisindicatesthat the operationof the
inverter was disabled(by inhibiting the firing of the inverter transistors).The moment
the voltagerecovered,inverteroperationwasenabledleading to the large peak in
motor
current.As the air-gapfield in the motor is low and not synchronizedwith the inverter
voltage, it takesanother hundred milliseconds before themotor is actually able to
298
Chapter 5 • VoltageS ags-Equ
ipment Behavior
Motor speed
(445 rpm/div)
•
.
• • •
,
,
. 0-
0 _
. 1.
,
,
,
. . , . , ,
._----1-------[-------[------r------1-------1-------[-------[-------r-----·..··-j-·.... -l..·..
t···.. ··r....··'j'·..··
r·. ·r·..·)'· · . l. . ·.
.---- . ~ -----_. ~ --_..--r---_•. -l--_ . - - - ~ - - - __
A -
!
1
,
-
-
-
_ .
•
-
_ .
- --
- ~ - ---- -
1
Motor current
(20 A/div)
1
,
- :- -
:
1
,
- - - - -~--
:
I
__ A -
-
;
-
-- -- -
j
1
!
-
-
-~
~-
---
1
,
~- -_.
-
-
-
-
-
-
-~
j
1
I
-
!
-_or -------r ---_.-
,
-
:
-- -
,
- - -:- .
-
-
-
-
-7 --- ---
!
:
Figure 5.52 Drive response with
synchronous
restart.(Reproducedfrom Mansoor[32].)
Time (30 cycles or 0.5 seconds/div)
Ai
Motor speed
(445 rpm/div)
orpm
'------'-I--'----J_--'-~..i......----'
,
. . ..
:
Motor current
(20 A/div)
_
_ ' _ _l.._----'-_.J
4•
•I
!
:
.
, ..
. _. . .
!
• .
. . .....
_
:
.,
--- ~-- - - - -- i - ---· _ · . - - - - - - -~--- _ · - -:.. - ---- ·
!
!
!
:
!
:
!
:
!
!
Figure 5.53 Drive response with nonsynchronousrestart.(Reproducedfrom
Mansoor[32].)
reaccelerate. If the process driven by the
motor is able towithstandthe variation in
speed or torque, this is a successful throughfrom
ride
the process point of view. In Fig.
5.53 we see whathappensduring non-synchronousrestart. It now takesabout one
second before the inverter is enabled, and
another 500 ms for themotor to start
reaccelerating. By tha t time the
m otor speed hasdropped to almost zero. If the
motor is used to power any kind ofproduction process this would almost certainly
not be acceptable . However, if the
motor is used forair-conditioningthe temporary
drop in speed would not be of any concern .
5.3.9 Overview of Mitigation Methods for AC Drives
5.3.9.1 Automatic Restart.The most commonly used mitigation method is to
disable theoperationof the inverter, so that themotor no longer loads the drive.
Section 5.3 • Adjustable-SpeedAC Drives
299
This prevents damagedue to overcurrents,overvoltages, andtorque oscillations.
After the voltage recovers the drive automaticallyrestarted.The
is
disadvantageof
this method is that the motor load slows down morethan needed. When synchronous restartis used thedrop in speed can be
somewhatlimited, but non-synchronous
restartleads to very largedrops in speed or evenstandstill of the motor. An important requirementfor this type of drive is that the controller remain online.Powering
of the controllersduring the sag can be from the dc bus
capacitoror from separate
of the mechanicapacitorsor batteries.Alternatively, one can use the kinetic energy
cal load to power the de bus
capacitorduring a sag orinterruption[33], [35], [150].
5.3.9.2 Installing Additional Energy Storage.The voltage-toleranceproblem of
drives is ultimately an energy problem. In manyapplicationsthe motor will slow
down too much tomaintain the process. This can be solved addingadditionalcaby
pacitorsor a battery block to the de bus. Also the
installation of a motor generator
set feeding into the de bus will give the required energy. A large
amount·of stored
energy is needed to ensure tolerance against
three-phasesags andshort interruptions.
For sags due to single-phase and
phase-to-phase
faults, which are the mostcommon
of the
ones, only a limitedamount of storedenergy is needed as at least one phase
of improvsupply voltage remains at a high value. This
appearsto be the easiest way
ing the voltagetolerancefor the majority of sags.
5.3.9.3 Improving the Rectifier.The useof a diode rectifier ischeapbut makes
control of the de bus voltage difficult. Themoment the ac voltagemaximum drops
below the de bus voltage, the rectifier stops supplying energy andmotor
the is powered from thecapacitor.Using acontrolled rectifier consistingof thyristors,like used
in de drives, gives some
control of the dc bus voltage. When the ac bus voltage
drops
the firing angle of thethyristors can be decreased to
maintain the de bus voltage.
For unbalancedsags different firing angles are needed for the three phases which
could make thecontrol rather complicated.Additional disadvantagesare that the
control system takes a few cycles to react and
that the firing-anglecontrol makes the
drive sensitive tophase-anglejumps.
Anotheroption is to use someadditionalpower electronics todraw more current
from the supplyduring the sag. A kind of power electronic
currentsource isinstalled
between the diode rectifier and the dc bus
capacitor.This currentcan becontrolledin
such a waythat it keeps the voltage at the de bus
constantduring a voltage sag [150],
[151].
By using a rectifier consisting of
self-commutatingdevices (e.g.,IGBTs), complete
control of the dc voltage is possible.
Algorithms have beenproposedto keep the de
voltage constantfor any unbalance,drop, or change in phase angle in the ac voltages
[44], [45],[46]. An additionaladvantageis that theseIGBT inverters enable a
sinusoidal
input current,solving a lot of theharmonicproblems caused by
adjustable-speed
drives.
The main limitation of all thesemethodsis that they have aminimum operating
voltage and willcertainly not operatefor an interruption.
5.3.9.4 Improving the Inverter. Instead ofcontrolling the de bus voltage, it is
also possible tocontrol the motor terminal voltage.Normally the speedcontroller assumes aconstantde bus voltage and calculates the switching
instantsof the inverter
from this. We saw earlierthat the effect of this isthat the de bus voltage is
amplitude
modulatedon the desiredmotor terminal voltages. This effect can be
compensated
300
Chapter5 • VoltageSags-EquipmentBehavior
by consideringthe dc busvoltage in the algorithms used to calculatethe switching
instants.For this (5.25)should be revised as follows,w ith Vdc the de busvoltage:
Vout = V+,
Vre;f
-V >
V
er
de
(5.36)
Vref
V
- < cr
Vde
This in effectincreasesthe referencevoltagewhen the de busvoltagedrops(insteadof
pulse-widthmodulationthis resultsin a kind of "pulse-areamodulation"). The drawback of this method is that it will result in additional harmonicdistortion, especially
when the drive isoperatedclose tonominal speed.Again this methodhas aminimum
voltage below which it will no longer work properly.
5.4 ADJUSTABLE-SPEED DC DRIVES
DC drives havetraditionally been much better suited for adjustable-speedo peration
than ac drives.The speedof ac motors is, in first approximation,proportionalto the
frequencyof the voltage. The speedof dc motors is proportionalto the magnitudeof
the voltage. Voltage magnitudeis much easierto vary than frequency.Only with the
introductionof power transistorshavevariable-frequencyinvertersand thus ac adjustable-speeddrives becomefeasible. In thissectionwe will discuss someaspectsof the
behaviorof dc drives during voltage sags.Modern de drives come in many different
configurations,with different protectionandcontrol strategies.A discussionof all these
is well beyondthe scopeof this book. The behaviordescribedbelow doesnot coverall
types of de drivesand should be viewed as anexampleof the kind of phenomenathat
occur when avoltagesag appearsat the terminalsof a de drive.
5.4.1 Operation of DC Drives
5.4.1.1 Configuration. A typical configurationof a de drive ispresentedin Fig.
5.54. The armaturewinding, which usesmost of the power, is fed via a three-phase
controlled rectifier. The armaturevoltage is controlled through the firing angle of the
thyristors. The more the delay in firing angle, thelower the armaturevoltage. There
is normally no capacitorconnectedto the de bus.The torque produced by the de
motor is determinedby the armaturecurrent, which shows almost no ripple due to
Firing
angle
,--_--J<.---.,.
ae
-----------,
Armature
Control
system
de
Figure 5.54 Modern de drive with separately
excited armatureand field winding.
301
Section 5.4 • Adjustable-SpeedDC Drives
the largeinductanceof the armaturewinding. The field winding takes only a small
amountof power; thus a single-phase rectifier is sufficient. The field winding is powered from oneof the phase-to-phase
voltagesof the supply. In case field-weakening
is used to extend the speed range
of the dc motor, a controlled single-phase rectifier
is needed. Otherwise a simple diode rectifier is sufficient. To limit the field
current,a
resistance is placed in series with the field winding. The resulting field circuit is therefore mainly resistive, sothat voltage fluctuations result in current fluctuations and
thus in torque fluctuations. A capacitor is used to limit the voltage (andtorque)
ripple. To limit thesetorque fluctuations a capacitor is used like the one used to
limit the voltage ripple in single-phase rectifiers.
5.4.1.2 DC Motor Speed Control.The standardequivalent circuit for a dc
motor is shown in Fig. 5.55. This circuit can only be used for
normal operation,
because it only considers the componentof
de
voltages andcurrents.A model including the inductanceof the windings will be discussed
further on.
The voltage Vf over the field winding causes current
a
If accordingto
(5.37)
where Rt is the resistance in the field circuit (the resistance of the winding plus any
external series resistance). This field
currentcreates theair-gapfield
(5.38)
which rotateswith a speedWm thus inducing a voltage F., the so-called
"back-EMF" in
the armaturewinding:
E
= kwmIf
(5.39)
This induced voltage limits the
a rmaturecurrent fa:
Va
= E+Rafa
(5.40)
where Va is the voltage over the
a rmaturewinding andRa the resistanceo f the armature
winding. Field currentand armaturecurrenttogetherproducea torque
(5.41)
which accelerates the
m otor up to the speed at which
m otor torque and load torque
balance.
The designof the motor is typically suchthat the armatureresistance is low and
the field resistance relatively high. Neglecting the
armatureresistance gives the following expression for thearmaturevoltage:
(5.42)
Figure 5.55Equivalent scheme for dc
m otor
during normaloperation.
302
Chapter5 • VoltageSags-EquipmentBehavior
Rewriting this, and using field voltage as an independentvariable, gives the basic
expressionfor the speedcontrol of dc motors:
(5.43)
The speedof a dc motor is increasedby increasingthe armaturevoltageor by decreasing the field voltage. Speedcontrol of a de drive takesplace in two ranges:
1. Armature voltage control range. The field voltage is kept at its maximum
value and the speedis controlled by the armaturevoltage. This is the preferred range. The field current is high, thus the armaturecurrent has its
minimum value for a given torque. This limits the armaturelossesand the
wear on the brushes.
2. Field weakeningrange. Above a certain value the armaturevoltage can no
longer be increased.It is kept constantand the speed isfurther increasedby
reducing the field voltage. As there is a maximum value for the armature
current, the maximum torque decreaseswith increasingspeed.
5.4.1.3 Firing-Angle Control. The de componentof the output voltage of a
thyristor rectifier is varied by meansof firing-angle control. The firing angle determines during which part of the cycle the rectifier conducts,and thus the averageoutput voltage. The output voltage of a non-controlledthree-phaserectifier was shown
•
in Fig. 5.19 in Section5.3. A diode startsconductingthe momentits forward voltage
becomespositive; a thyristor conductsonly when the forward voltage is positive and
a pulse isapplied to its gate. By firing the thyristor at the instant a diode would start
conducting,the output voltage of a controlled rectifier is the sameas that of a noncontrolled one. This is called free-firing. The firing angle of a thyristor is the delay
comparedto the free-firing point. Figure 5.56 shows the output voltage of a threephasethyristor rectifier with a firing angle of 50°. For a controlled rectifier the de
bus voltage still consistsof six pulsesbut shifted comparedto the output voltage of
a non-controlledrectifier. As the conductionperiod is shifted away from the voltage
maximum, the averagevoltage becomeslower.
0.8
a
.5
~0.6
~
]
0.4
U
c
100
150
200
250
Time in degrees
300
Figure 5.56 Output voltageof controlled
rectifier with a firing angleof 50°. No
capacitanceis connectedto the de bus. Note
350 the differencein vertical scalecomparedto
Fig. 5.19.
Section 5.4 • Adjustable-SpeedDC Drives
303
A firing angle a delaysconductionover a period 2Jr x T, with T one cycleof the
fundamentalfrequency.The averageoutputvoltage(i.e., the dccomponent)for a firing
angle a is
(5.44)
with Vmax the outputvoltageof a non-controlledrectifier. The voltagealso containsan
alternatingcomponent,with' a frequency of six times the power system frequency:
300 Hz in a 50 Hzsystem;360 Hz in a 60 Hz system.This voltagecomponentwill not
lead tolargefluctuationsin the currentand in torquedue to the largeinductanceof the
armature'winding.
The firing of the thyristorstakesplaceat acertainpoint of the supplyvoltagesine
wave. For this the control systemneedsinformation about the supply voltage. There
are different methodsof obtainingthe correctfiring instant:
I. The thyristorsare fired with acertaindelay comparedto the zero-crossingof
the actualsupply voltage. In normal operationthe threevoltagesare shifted
1200 comparedto eachother. Therefore,the zero-crossingof one voltageis
used as areferenceand all firing instantsare obtainedfrom this reference
point. This method of control is extremely sensitive to distortion of the
supply voltage.Any changein zero-crossingwould lead to achangein firing
angle and thus to a changein armaturevoltage. The problem is especially
seriousas thyristor rectifiers are themain sourceof notching, creatinglarge
distortion of the supplyvoltagesine wave [53], [55]. Onecould end up with a
situation where the drive isnot immune to its own emission.
2. The output voltage of a phase-lockedloop (PLL) is used as areference.A
phase-lockedloop generatesan output signal exactly in phasewith the fundamentalcomponentof the input signal. The referencesignal is no longer
sensitiveto short-time variationsin the supply voltage. This slow response
will turn out to be aseriouspotentialproblemduring voltage. sagsassociated
with phase-anglejumps.
3. A more sophisticatedsolution is to analyzethe voltage in the so-calledsynchronouslyrotating dq-frame. In the forwardly rotating frame the voltage
consists of a dc componentproportional to the positive-sequencesupply
voltageanda componentw ith twice the fundamentalfrequencyproportional
to the negative-sequence
supplyvoltage.In the backwardlyrotatingframethe
dc componentis proportionalto the negative-sequence
voltage.Using a lowpassfilter will give complexpositive and negative-sequence
voltageand thus
all required information about the systemvoltages.The choice of the lowpass filter's cut-off frequency is again a compromisebetween speed and
sensitivity to disturbances[152], [153].
5.4.2 Balanced Sags
A balancedvoltage sag leads to arathercomplicated.t ransientin the demotor,
with a new steadystateat the samespeed as theoriginal one. The new steadystatewill,
however,rarely be reached.Most existing drives will trip long before, mainly through
the interventionof somekind of protectionin the powerelectronicconverters.But even
if the drive doesnot trip, the voltagesag will typically be over well within one second.
The new steadystatewill only be reachedfor long shallow sags.
304
Chapter5 • VoltageSags-EquipmentBehavior
According to (5.43), themotor speed isproportional to the ratio of armature
voltage and field voltage. The voltage sag in all three phases makes
that armatureand
field voltagedrop the sameamount;the speedshouldthus remain the same. The model
behind (5.43), however, neglects thetransient effects, which are mainly due to the
inductanceof the motor winding and theinertia of the load. A model of the dc
motor, which is valid for transientsas well, is shown in Fig. 5.57, where
La and Lf
are theinductanceof armatureand field winding, respectively.
5.4.2.1 Theoretical Analysis.The qualitative behavior of the motor can be
summarizedas follows, where it is assumed
t hat neither thecontrol system nor the
protectionintervenes.
of the field-winding rectifier
• Becauseof the voltage sag, the voltage on ac side
will drop. This will lead to a decay in fieldcurrent. The speedof decay is
determinedby the amountof energystoredin the inductanceand in thecapacitance. Typically thecapacitorwill give the dominanttime constantso that the
decay in fieldcurrentcan be expressed as follows:
(5.45)
where If o is the initial currentand r is the timeconstantof the decay in field
current.The fieldcurrentwill not decay to zero, as suggested by (5.45),
but the
decay will stop the momentthe field voltage reaches the ac
voltageamplitude
again. For a voltagedrop of 20% the fieldcurrentwill also drop 20%. This is a
similar situationas discussed in Section 5.2. The
only difference isthat the load
is a constantimpedanceinsteadof constantpower. For small dc voltage ripple
it may take 10 cycles or more for the
capacitorvoltage, and thus for the field
current, to decay.Note that the ripple in the fieldcurrent directly translates
into a torqueripple. As thelatter is often not acceptable,a largecapacitanceis
generally used. Some drives useconstant-voltage
a
transformerto supply the
field windings. The effect is againthat the fieldcurrentdropsslowly.
• The voltage sag leads to a direct
drop in armaturevoltage, which leads to a
decay inarmaturecurrent.The decay issomewhatdifferent from the decay in
field current. The armaturecurrent is driven by the difference between the
armaturevoltage and theinducedback-EMF. As this difference isnormally
only a few percent,the changein armaturecurrent can be very large. The
current quickly becomes zero, but not negative because the rectifier blocks
that. From Fig. 5.57 weobtain the following differential equation for the
armaturecurrent I a :
(5.46)
Figure 5.57 Equivalentcircuit for a dc motor
during transients.
305
Section 5.4 • Adjustable-Speed DC Drives
The solution, with /0 the armaturecurrentat time zero, is
I
a=
E (l Va - E)
n, + 0 - n, e
Va -
_L
(5.47)
f
1-.
where Vais the armaturevoltageduring the sag, andT = As we saw before,
the field current remains close to itspre-eventvalue for aDt least a few cycles.
Because themotor speed doesnot immediatelydrop, the back-emfE remains
of a drop in armaturevoltage is thusthatthe currentdrops
the same. The effect
toward a large negative value
(Va - E)I Ra.
We will estimatehow fast thearmaturecurrentreaches zero by
a pproximating
(5.47) for t « T. Using e-f ~ 1 - ~ gives
t, ~ 10 -
E-V
L
at
(5.48)
a
The pre-sagsteady-statecurrent /0 may beobtainedfrom
l-E
/0=--
(5.49)
Ra
where thesteady-statearmaturevoltage is chosenequal to 1pu. The time for
the currentto reach zero is, in cycles
of the fundamentalfrequency:
1
t
(X
a)
1-
E
= 21l' Ra 1 - V
(5.50)
whereX a is thearmaturereactanceat thefundamentalfrequency.For X a/ R a =
31.4 and 1 - E = 0.05 we obtain
t
I
= 10.25
_ V (eye es)
(5.51 )
For a sag down to75% the currentdropsto zero in one cycle; for a90% sag it
takes 2.5 cycles which is still very fast.T hus for the majority of sags thearmature currentand thetorquewill drop to zero within a few cycles.
• The drop in armatureand in field current leads to adrop in torque which
causes adrop in speed. Thedrop in speed and the
d rop in field currentcause a
reductionin back-EMF.
II Sooneror later theback-EMFwill become smallerthan the armaturevoltage,
reversing thedrop in armaturecurrent. Because speed as well as field
current
havedroppedthe newarmaturecurrentis higher than the pre-eventvalue.
• The more the speed
drops,the more theback-EMFdrops,the more thearmature current increases, the more the
torque increases. Inother words, the dc
motor has abuilt-in speedcontrol mechanismvia the back-EMF.
• The torquebecomes higherthan the load torque and the load reaccelerates.
• The load stabilizes at theoriginal speed andtorque, but for a lower field
current and a higherarmaturecurrent. The drop in field current equals the
drop in voltage; thearmaturecurrent increases asm uch as the fieldcurrent
drops, because their
product(the torque) remainsconstant.
306
Chapter5 • Voltage Sags
-EquipmentBehavior
5.4.2.2 Simulationof Balanced Sags. Some simulations have beenperformed
to quantify the behavior described above. The results are shown in Figs. 5.58
through 5.61. The simulated drive was configured as shown in Fig. 5.54, with a
three-phaserectifier to power the armaturewinding and a single-phaserectifier for
the field winding. The drive was operating at nominal speed ,thus with zero firing
angle for the rectifiers. In thissystemthe time constantwas 100 ms,both for the armaturewinding and for the field wind ing . Asupply voltage of 660 V was used result.
moment of
ing in a pre-sagmotor power of 10 kW and a speed of 500 rpm The
inertia of the load driven by themotor was 3.65 kgm/s" ,T he load torque was proportional to the speed.The simulations were performed by solving the differential
equations with a step-by-stepapproximation[154]. The voltage dropped to 80% in
all three phasesduring 500 ms (30 cycles).T he plots show two cyclespre-sag,30 cycles during-sag,and 88 cyclespost-sag.
The armaturecurrentis shownin Fig . 5.58.The armaturecurrentdropsto zero in
the
a veryshorttime due to thephenomenondescribedbefore. As a directconsequence
torque becomes zero also , as
shown in Fig . 5.60. This inturn leads to a fastd rop in
speed, asshown in Fig. 5.61. After a few cycles the fieldcurrent (Fig . 5.59) and the
speed havedroppedsufficiently for the back-EMF to becomelower that the armature
2.5
50 2
.5
~
::l
o
1.5
e
a
!
0.5
0.5
1.5
2
Figure 5.S8 DCmotor armaturecurrent
dur ing balanced sag.
~
t:: 0.6
::l
o
.",
~
0.4
0.2
0.5
I
-~~2
Time in seconds
1.5
Figure 5.59 DCmotor field current during
balanced sag.
307
Section 5.4 • Adjustable-SpeedDC Drives
2.5
0.5
0.5
Figure 5.60 Torqueproducedby de motor
during balancedsag.
1
Time in seconds
2
1.5
1.15
1.1
a
.S 1.05
1
....
~ 0.95
~
0.9
0.85
Figure 5.61 Speedof de motor during
balancedsag.
0.8
0
0.5
1
Time in seconds
1.5
2
voltage. From this momenton thearmaturecurrentand thetorquerecover and a few
hundredmillisecondslater even exceed their pre-sag value. The result
that
is the motor
picks up speed again.
Upon voltage recovery,a round t = 0.5 in the figures, the opposite effect occurs.
The armaturevoltage becomes much larger than the
back-EMF leading to a large
overcurrent,a large torque, and even a significant overspeed. The post-sag
transient
is overafteraboutone second. Notethat the simulated behavior was due to a sag down
to 80% , a rather shallow sag. Due to the fast
drop in armaturecurrent even such a
shallow sag willalreadylead to a serioustransientin torqueand speed.
5.4.2.3 Interventionby the Control System. The control system of a de drive
can control a numberof parameters:a rmaturevoltage, armaturecurrent, torque, or
speed. In case the
control system is able to keep
armatureand field voltageconstant,
the drive will not experience the sag. However, the
control system will typically take
a few cycles to react, so
t hat the motor will still experience the fastd rop in armature
current.The useof such acontrol system may also lead to an even more severe transient at voltage recovery. The
a rmaturevoltage will suddenly become much higher
than the back-emfleading to a very fast rise inarmaturecurrent, torque, and speed.
308
Chapter5 • VoltageSags-EquipmentBehavior
If the motor aims at keeping themotor speedconstant,the drop in speed (as shown
in Fig. 5.61) will be counteractedthrough a decrease in firing angle of the
thyristor
rectifier. For a deep sag the firing angle will quickly reach its
minimum value.
Further compensationof the drop in armaturevoltage would requirecontrol of the
field voltage. But as we saw above, the field voltage is kept
intentionally constantso
that control is difficult.
5.4.2.4 Intervention by the Protection.The typical reason for thetripping of a
dc drive during a voltagesag isthat one of the settingsof the protectionis exceeded.
As shown in Figs. 5.58through 5.61, voltage,current,speed, andtorqueexperience a
large transient.The protectioncould trip on anyof theseparameters,but more often
than not, the protectionsimply trips on de busundervoltage.
DC drives areoften used for processes in which very precise speedpositioning
and
are required,e.g., in robotics. Even smalldeviationsin speedcannotbe toleratedin
such a case. We saw
beforethat the motor torquedropsvery fast, even for shallow sags,
than for an ac drive. A shallow sag
so that the drop in speed will become more severe
will alreadyhave the same effect on a de drive as a zero voltage on an ac drive:
bothin
cases thetorqueproducedby the motor dropsto zero.
5.4.3 Unbalanced Sags
One of the effectsof unbalancedsags on dc drives isthat armatureand field
voltage do not drop the sameamount. The armaturevoltage is obtained from a
three-phaserectifier, the field voltage from a single-phase rectifier.
During an unbalanced sag, thesingle-phaserectifier is likely to give adifferent outputvoltage than the
three-phaserectifier. If the field voltagedropsmore than the armaturevoltage, the new
steady-statespeedcould be higher than the original speed. However, initiallyboth
armature and field current decrease, leading to a decrease torque
in
and thus in
speed. The slowest speed recovery takes place when thevoltageremainsconstant.
field
The back-EMFonly startsto drop when themotor slows down. Thearmaturecurrent
will remain zerolonger when the field voltage stays
constant.
• If the field voltage drops more than the armaturevoltage, theback-emfwill
in
quickly be lessthan the armaturevoltage, leading to an increase armature
current. Also the newsteady-statespeed is higherthan the pre-eventspeed.
Overcurrentin the armaturewinding and overspeed are the
main risk.
• If the field voltagedropslessthan the armaturevoltage, thearmaturecurrent's
decay will only be limited bythe drop in motor speed. It will take a long time
steady-statespeed is lowert han
before themotor torquerecovers. As the new
the pre-eventspeed,underspeedbecomes the main risk.
Simulationshave beenperformedfor the same driveconfigurationas before. But
insteadof a balancedsag, anumberof unbalancedsags were applied to the drive. The
results of two sagso f type D and one sag of type C are shown here. All three sags had a
durationof 10 cycles, acharacteristicmagnitudeof 50%, and zerocharacteristicphaseanglejump. Note that in this case the sag type refers to the
line-to-linevoltages, not the
the line-to-neutralvoltage. The rectifier isdelta-connected;thus the line-to-line voltages
more directly influence the drivebehavior.
309
Section 5.4 • Adjustable-SpeedDC Drives
• SAGI: a sag of type 0 with the large
voltagedrop in the phasefrom which the
field winding is powered.The field voltagethus drops to 50%. The results for
sag I are shown in Figs. 5.62
through 5.65.
• SAGII: a sagof type 0 with a small voltagedrop in the phasefrom which the
field winding is powered,making the field voltage drop to about 90%. The
results for sag11 are shownin Figs. 5.66through 5.69.
• SAGIll: a sagof type C with the field windingpoweredfrom the phasewithout
The resultsfor sag III
voltage drop. The field voltage thus remains at 100%.
are similar to those for sag11 and thereforenot reproducedin detail.
All plots show two cycles before the sag, 10 cycles
during the sag, and 48 cycles
that a deep sag in the field
voltage(sag I) causes
afterthe sag.F romthe figures we can see
a highovershootin the armaturecurrent(Fig . 5.63), in thetorque(Fig. 5.64),andin the
speed (Fig. 5.65).For a shallowsag in the fieldvoltage(sag11) the armaturecurrentand
torqueare zero for a long time, but with smallerovershoot(Figs
a
. 5.67 and 5.68); the
speed shows a large
drop but only a smallovershoot(Fig . 5.69).Note the ripple in the
armaturecurrentduring the sag. Theunbalancein the acvoltageleads to a muchlarger
,
Figure 5.62 Fieldcurrentfor sag type D, with
largedrop in field voltage.
Figure 5.63Armaturecurrentfor sag type D,
with large drop in field voltage.
0.2
0.4
0.6
Time in seconds
0.8
0.8
310
Chapter 5 • Voltage Sags
-Equipment Behavior
5
4
;>
"'-
.S
.,
eB
3
...
B 2
0
::E
0.8
Figure 5.64Motor torque for sag type D,
with large drop in field voltage.
1.3 ~-- ---,---,--~--~---,
1.25
1.2
5.
.S 1.15
1l
~ 1.1
~
1.05
::E
0.95
0.2
0.4
0.6
0.8
Figure 5.65Motor speed for sag type D, with
large drop in field voltage.
0.8
Figure 5.66 Fieldcurrentfor sag type D, with
smal1 drop in field voltage.
Time in seconds
;>
c,
.S
0.8
C
50.6
o
'"
"0
u:
0.4
0.2
0.2
0.4
0.6
Time in seco nds
311
Section 5.4 • Adjustable-SpeedDC Drives
5 c----~----.---_--~-----,
4
0.4
0.6
Time in seconds
Figure 5.67 Armaturecurrentfor sag type D,
with small drop in field voltage .
0.8
5
4
::l
0.
.5
3
<Ll
::l
go
B
....
2
~
~
0.4
0.6
Time in seconds
Figure 5.68 Motor torque for sag type D,
with small drop in field voltage .
0.8
1.15
1.1
5.
.5 1.05
1
J ....
~ 0.95
~
0.9
0.85
Figure 5.69 Motor speed for sag type D, with
small drop in field voltage .
0.2
0.4
0.6
Time in seconds
0.8
312
Chapter5 • VoltageSags-EquipmentBehavior
ripple in armaturevoltage than during normal operation.This ripple disappearsupon
voltagerecovery and is alsonot presentduring a balancedsag (Fig. 5.58).
The maximumand minimum values forcurrent,torque, and speed are shown in
Table 5.9. All values are given as percentage
a
of the averagepre-eventvalue.Tripping
of the drive can be due to
undervoltageor overcurrent.The undervoltageis similar for
the three sags; thus sag I is the most severe one for the electrical
part of the drive
becauseof the largearmaturecurrent. The mechanicalprocess can, however, get disrupted due to torque variations and variations in speed.For a process sensitive to
underspeed,sags II and III aremostsevere; for a process sensitivetorquevariations,
to
sag I is themostsevere one. The main
conclusionis that unbalancedsagsrequiretesting
for all phases; it ishardto predictbeforehandwhich sag will be most severe to the drive.
TABLE 5.9 DC Drive PerformanceDuring UnbalancedSags inDifferent
Phases
Field Current
Sag
I
II
III
Type
FieldVoltage
50%
90%
100%
D
D
C
ArmatureCurrent
min
max
min
59%
900AJ
100%
100%
100%
100%
0
0
0
max
460%
264%
229%
Motor Torque
min
0
0
0
max
367%
256%
229%
Motor Speed
min
max
93%
85%
85%
124%
107%
114%
5.4.4 Phase-Angle Jumps
Phase-anglejumps affect the angle at which the
thyristors are fired. The firing
instantis normally determinedfrom the phase-lockedloop (PLL) output, which takes
at least several cycles to react to the
phase-anglejump.
A calculatedstepresponseof a conventionaldigital phase-lockedloop to a phaseangle jump is shown by Wang [57]. His results arereproducedin Fig. 5.70, where we
can seethat it takesabout400 ms for thePLL to recover. Theerror gets smallerthan
10% after about250 ms, which is still longerthan the durationof most sags. Thus for
our initial analysiswe can assumethat the firing instantsremain fixed to thepre-event
voltage zero-crossings.W ith additional measures it is possible to
m ake PLLs which
respondfaster tophase-anglejumps, but those will be more sensitive to
harmonicsand
other high-frequencydisturbances.
We can reasonablyassumethat the phase-locked-loopoutput does not change
during the sag.The effect of the phase-anglejump is that the actual voltage is shifted
0.....--....----------------.
-0.2
-0.4
-0.6
-0.8
-1
-1.2
......-------I
~
o
0.1
0.2
0.3 0.4 0.5
Time (sec)
0.6
0.7
0.8
Figure 5.70 Stepresponseof a conventional
digital phase-lockedloop. (Reproducedfrom
Wang[57].)
313
Section 5.4 • Adjustable-SpeedDC Drives
Firing
I
I
I
::s 0.8
PLLoutput
0..
,/
.S
'" ,Supplyvoltage
,
I
I
~0.6
\
I
\
I
S
15
\
I
I
;> 0.4
\
I
\
I
\
t
\
I
\
I
0.2
,
\
I
\
I
\
I
\
I
\
I
\
I
o
Figure 5.71 Influenceofphase-lockedloop
on firing angle.
\
50
250
100
150
Timeindegrees
200
Actual firing
Intendedfiring
::s 0.8
e,
.S
~
~
0.6
;> 0.4
0.2
Figure 5.72Influenceof phase-locked loop
on firing angle: with actual voltage as a
reference.
0"----.A---a..---..4.-~-~-..L-----'--J
o
50
100
Timeindegrees
150
200
comparedto the reference voltage. Because of this thyristors
the
are fired at a wrong
point of the supply-voltagesine wave. This is shown in Fig. 5.71 for a negative phaseof
anglejump. The during-sagvoltage lags the pre-sag voltage; thus the zero-crossing
the actualsupply voltage comes later
than the zero crossingo f the PLL output. In Fig.
o f the actualvoltage is used as a reference: due to the negative phase5.72 the sine wave
anglejump t!¢, the thyristorsare fired at an anglet!¢ earlier than intended.
5.4.4.1 Balanced Sags.For balanced sags the phase-angle
jump is equal in the
three phases; thus the shift in firing angle is the same for all three voltages.
If the
shift is lessthan the intendedfiring-angle delay, theoutput voltage of the rectifier
will be higher than it would be without phase-anglejump. This assumesthat the
phase-anglejump is negative, which isnormally the case. A negativephase-angle
jump will thus somewhatcompensatethe drop in voltage due to the sag.
For a positive phase-anglejump the output voltage would be reduced and the
phase-anglejump
would aggravatethe effects of the sag.
For a firing angle equal toa the pre-sagarmaturevoltage equals
Va = cos(a)
(5.52)
314
Chapter5 • VoltageSags-EquipmentBehavior
120,------r-110
=
~ 100
8-
.5
~
70 degrees
90
S
~
80
:g
~ 70
o
60
30 degrees
5
10
15
20
Phase-anglejump in degrees
25
Figure5.73 Influenceof phase-anglejump on
30 the armaturevoltage,for different firing
angles.
The voltage is rated to thearmaturevoltage for zero firing angle.For a sag with
magnitude V (in pu) and phase-anglejump !:14>, the during-eventarmaturevoltageis
V~
= V x cos(a -
/j.l/J)
(5.53)
The phase-anglejump is assumednegative, /j.(j> is its absolutevalue. The ratio between
V~ and Va is the relativemagnitudeof the sag in thearmaturevoltage.This isplottedin
Fig. 5.73 for firing-angle delayso f 30°, 50°, and 70°. Aduring-eventmagnitude V of
500~ has beenassumed,and the phase-anglejump is varied between zero
and 30°.
According to Fig. 4.86 this is the range one can expect for50%
a sag. For large
firing-angle delays thearmaturevoltageis low; thusa jump in phase-anglecan increase
the voltagesignificantly. For a 70° firing-angle delayandphase-anglejumpsof 20° and
higher theduring-eventvoltageis evenhigher than the pre-eventvoltage.Whetherthis
actually makesthe sag less severe
dependson the behaviorof the fieldvoltage.When a
diode rectifier is used topowerthe field winding, the fieldvoltagewill not be influenced
by the phase-anglejump. The consequenceof the phase-anglejump is that the field
voltagedropsmorethanthe armaturevoltage,similar to sag Idiscussedin the previous
section. This can lead to large
overcurrentsin the armaturewinding and to overspeed.
of missing pulses which would make
When a controlled rectifier is used there is a risk
the field voltagemuch lowerthan the armaturevoltage.
If the shift is largerthan the intendedfiring-angle delay, theactualfiring will take
place before the free-firingpoint. As the forward voltage over the thyristors is still
negative it will not commenceconducting.How seriousthis effect isdependson the
o f a shortpulse will makethe drivemoresensitive.
durationof the firing pulse. The use
Note that eitherthe armatureor the field rectifier isoperatedat its maximumvoltageso
that at least one of them always will be
proneto missing pulses.
5.4.4.2 UnbalancedSags. For unbalancedsags thesituation becomes rather
complicated. In most cases thedifferent phases showpositive as well as negative
phase-anglejumps. Thus for some phasesthe phase-anglejump can be animprovement, for othersnot. Somephasesmight miss their firing pulses,o thersnot. The armature winding might be influenceddifferently from the field current as wealready
saw before.
315
Section 5.4 • Adjustable-SpeedDC Drives
1.1r-------.---~----
& 0.9
.~ 0.8
co
11o
0.7
>
] 0.6
~
0.5
0.4
0.5
Figure 5.74 DCvoltage for sag type D, with
rectifier operatingat 10° firing angle.
1
1.5
2
Time in cycles
1.1....----..,-----r------r------,
=' 0.9
Q.
.9
08
4)
•
co
~
>
0.7
.8 0.6
g 0.5
0.4
Figure 5.75 DCvoltage for sag typeC, with
rectifier operatingat 10° firing angle.
0.5
1
1.5
2
Time in cycles
Figures5.74 and 5.75 show the dc bus
voltagebeforeandduring a voltage sag, in
of a
case the rectifier isoperatedat a firing angleof 10°. Figure 5.74 shows the effect
m agnitudethe maxtype D sagof 50% magnitude.As all three voltages go down in
imum de voltage alsodrops. The two voltage pulses belonging to the least-affected
phases come very close
after each other. In the phasor diagram they move away
from eachother, so that the voltagemaxima of the rectified voltage come closer. The
consequence
is that the commutationbetween these two phases takes place natural
at a
commutationpoint. The firing of the thyristor has taken place already before that
momentin time. Thereis thus a risk for a missing pulse which would even more
distort
the de bus voltage.Figure 5.75 shows the effectof a type C sagof 50% magnitude.
5.4.5 Commutation Failures
The momenta thyristor is fired andforwardly biased, itstartsconducting.But the
of the
currentthroughthe conductordoesnot immediatelyreach its full value because
inductive nature of the source.Considerthe situation shown in Fig. 5.76, where the
316
Chapter 5 • VoltageSags-EquipmentBehavior
L
+
+
Figure 5.76 Origin ofcommutationdelay.
currentcommutatesfrom phase1 to phase2. The driving voltagesin these twophases
are shifted by 1200 :
(5.54)
(5.55)
At time zero the two driving voltagesare the same,thus the line-to-line voltageis zero,
which correspondsto the free-firing point. For a firing-delay anglea, thyristor 2 is fired
at lJJot = a. This is the moment the current through thyristor I startsto rise and the
currentthrough thyristor 2 startsto decay.The changein currentis describedthrough
the following differential equation (note that both thyristors conduct, thus the two
phasesare shorted):
Vt(t) - L
di,
di 2
di + L di =
(5.56)
V2(t)
with L the sourceinductance.We can assumethe armaturecurrent Ide to be constant;
thus the changesin i} and i 2 compensateeachother:
di 1 + di2
dt
dt
=0
(5.57)
after which i 2 can be obtainedfrom the differential equation:
di2
di=
J3v sin(wot)
(5.58)
2L
with the following solution:
;2(t) =
~~ [cos(a) -
cos(eoot)],
a
t>-
Wo
(5.59)
Commutation is complete and thyristor 1 ceases to conduct when i2(t) = Ide.
Commutationtakeslonger for smaller valuesof V, thus during voltage sags,and for
a firing-delay anglea closerto 1800 , thus for the drive beingin regenerativemode.The
maximumcurrent the supply voltage is able to cummutateis found from (5.59) as
J3v
I max = 2eoo (l
L
+ cosa)
(5.60)
Section 5.4 • Adjustable-SpeedDC Drives
317
If this is lessthan the actual armaturecurrent, a commutationfailure occurs:both
thyristors will continueto conduct,leading to aphase-to-phase
fault. This will cause
blowing of fuses ordamageof the thyristors.The risk ofcommutationfailure isfurther
increased by the increased
armaturecurrentduring and after the sag.
j ump reduces theactualfiring angle, thus lowering the risk
A negativephase-angle
of commutationfailure. A positive phase-anglejump makes acommutationfailure
more likely. Unbalancedfaults cause acombinationof positive and negative phaseanglejumps, thus increasingthe risk in at least one phase.
5.4.8 Overview of Mitigation Methods for DC Drives
Making de drivestolerant againstvoltage sags is more
complicatedthan for ac
drives. Threepotentialsolutions,to be discussed below, are
addingcapacitanceto the
armaturewinding, improvedcontrol system, andself-commutatingrectifiers.
5.4.6.1ArmatureCapacitance. Installing capacitanceto the armaturewinding,
on dc side of thethree-phaserectifier, makesthat the armaturevoltage no longer
drops instantaneouslyupon sag initiation. Insteadthe armaturevoltage decays in a
similar way to the field voltage. Toobtain a large timeconstantfor the decay of the
armaturevoltage requires a large
c apacitorfor the armaturewinding. Note that the
power taken by thearmaturewinding is much largerthan the power taken by the
field winding. For three-phaseunbalancedsags it may be sufficient to keep up the
voltage during one half-cycle.
Keeping up thearmaturevoltage will still not solve theproblemof missing pulses
due to phase-anglejumps and commutationfailures. Another disadvantageof any
amountof armaturecapacitanceis that it makes the drive react slower to the
control
system.Changesin motor speed areobtainedthrough changes in firing angle. The
armaturecapacitanceslows down the response of the
armaturecurrent and torque
on a change in firing angle. When the drive
applicationrequires fast changes torque
in
and speed, thea rmaturecapacitanceshould be small.
5.4.6.2 Improved ControlSystem. Any control system for a de driveultimately
controls the firing angleof a controlled rectifier. This may be thearmaturerectifier,
the field rectifier, orboth. Due to thenatureof a thyristor rectifier it is unlikely that
the control system will have anopen-loop time constantless than two cycles. We
saw beforethat the drop in armaturecurrent and torque takes place much faster
than this. It is thus not possible toprevent the transient in armaturecurrent and
torque.
Two straightforwardquantitiesto becontrolledare armaturevoltage andmotor
speed.Controlling the armaturevoltage enables the use
of a simplecontroller with a
small open-looptime constant.For the controller to work, sufficientmargin must be
available in the rectifier to bring the
a rmaturevoltage back to1000/0. If sags down to
50% magnitudehave to bemitigated, the normal operatingvoltage on de sideof the
rectifier shouldnot exceed50°A, of maximum. The result is
t hat only half of the control
rangeof the rectifier can be used for speed
control. The otherhalf is needed for voltagesagmitigation.
Speedcontrol is thecommonly-usedmethodof control for de drives. The voltage
sag will cause adrop in speed. The speed
controllerdetects this and reduces the firing
angle tocompensate.If the firing angle is zero thecontrollercan no longer increase the
318
Chapter5 • Voltage Sags-EquipmentBehavior
speed. Speed
control will not mitigate thetransientsin torque and current but it may
reduce the variations in speed.
A disadvantageof both control techniques isthat they will lead to a severe
transient inarmaturecurrent and torque upon
voltage recovery.
5.4.6.3 Improved Rectifiers. The control of the drive may be significantly imenable control of the
proved by using a self-commutating rectifier. These rectifiers
output voltage on a sub-cycle timescale. This will
preverit the drop in armature
voltage and thus the severe drop in torque. Using
advancedcontrol techniquesit
may also be possible to install
additional enery storagewhich is only madeavailable
during a reduction in the supply voltage.
By using self-commutating rectifiers it may also possibleto
be
usea sophisticated
control systemthat detects and mitigates
phase-anglejumps. With such a control
system, the reference signal should no longerobtainedfrom
be
a phase-lockedloop
but from the measured supply voltage
through a suitabledigital filter.
5.4.6.4 Other Solutions. Other solutionsinclude a more critical setting of the
undervoltageand overcurrentprotection; the useof componentswith higher overcurrent tolerance; and disabling the firing of the
t hyristors to prevent tripping on
overcurrent.All these solutions are only feasible when
the load can tolerate rather
large variationsin speed.
5.5 OTHER SENSITIVE LOAD
5.5.1 Directly Fed Induction Motors
Despite the growth in the number of
adjustable-speeddrives, the majority of
induction motors are still directly fed; i.e., the
m otor terminals are connectedto the
constantfrequency,constantvoltage, supply. It will beclear that speedcontrol of the
motor is not possible. Directly fedinduction motors are rather insensitiveto voltage
sags,althoughproblems could occur when too
manymotorsare fedfrom the samebus.
The drop in terminal voltage will cause d
a rop in torquefor an induction motor.
Due to this drop in torquethe motor will slow down until it reachesa new operating
point. If the terminal voltage drops too much the
load torque will be higher than the
pull-out torque and themotor will continue to slow down. An induction motor is
typically operatedat half its pull-out torque. As thepull-out torque is proportional
to the square of the voltage, a voltage
drop to 70% or less will not lead to a new stable
operatingpoint for the induction motor. The
d rop in speed isseldoma seriousconcern
for directly fed induction motors. These kind motorsare
of
usedfor processesthat are
not very sensitive to speed variations; and variation
the
in speedis seldommore than
10% • The effect of voltage sags on
induction motors has already been discussedin
Section 5.3 under the assumption that both
motor and load torqueremainconstant.In
motor
most practical cases the load torque decreases and
the torqueincreaseswhen the
motor slows down. The actual drop in speed will
thus be lessthan indicated.
Although the inductionmotor is normally ratherinsensitiveto voltagesags,there
are a few phenomena
t hat could lead to process
i nterruption due to a sag.
• Deep sags lead to severe torque
oscillationsat sagcommencementand when
the voltage recovers. These could leaddamageto
to
the motor and to process
319
Section 5.5 • Other Sensitive Load
•
•
•
•
interruptions.The recoverytorquebecomes more severe when the internal flux
is out of phasewith the supply voltage, thus when the sag is associated with a
phase-anglejump.
At sagcommencementthe magneticfield will be driven out of the airgap. The
associatedtransientcauses anadditionaldrop in speed for deep sags.
During
this period the motor contributesto the short-circuit current and somewhat
mitigatesthe sag. This effect has been discussed in Section 4.8.
When the voltage recovers, the
airgapfield has to be built up again. In weaker
systems this can last up to 100ms,
during which the motor continuesto slow
down. This could become a
problem in systems where them otor load has
grown over the years. Where in the past a voltage sag would notproblem,
be a
now "suddenly"the process can no longer
withstandthe speeddrop due to a
sag. As deep sags are rare it can take a long time before such
problem
a is
discovered.
Whenthe voltage recovers, the
motor takes a high inrushcurrent:first to build
up the airgap field (the electricalinrush), next to reaccelerate the
motor (the
mechanicalinrush). This inrush can cause apost-faultsag with adurationof
one second or more, and lead tripping
to
of undervoltageand overcurrent
relays. Again thisproblem is more severe for a weak supply, and can thus
become aproblemwhen theamountof motor load increases.
For unbalancedsags themotor is subjected to a positive sequence as well as to
a negative-sequence
voltage at the terminals. The negative-sequence voltage
causes atorqueripple and a large negative-sequence
current.
5.5.2 Directly Fed Synchronous Motors
A synchronousmotor has similar problemswith voltage sags as an
induction
motor: overcurrents,torque oscillations, and drop in speed. But asynchronous
motor can actuallylosesynchronismwith the supply. Aninductionmotor is very likely
able to reaccelerateagain after the fault: it might take too long for the process, the
currentmight be too high for themotor (or its protection),or the supply might be too
weak, but at least it is intheorypossible. When asynchronousmotorloses synchronism
it has to bestoppedand the load has to be removed before it canbroughtback
be
to
nominal speed again.
The lossof synchronismof a synchronousmotor is ruled by theequationfor the
transportof power P from the supply to the motor:
p
= V.vupEsin </J
X
(5.61)
with v'vup the supply voltage,E the back-EMFin the motor, </J the angle between the
back-EMF and the supply voltage, andX the reactance between the supply and the
reactanceof the motor and the source
synchronousmotor (this includes the leakage
reactanceof the supply). Thisrelationis shown in Fig. 5.77.For a givenmotor load the
operatingpoint will be such that the powertaken by the load equals the power transported to the motor. This point is indicatedin Fig. 5.77 as"normal operatingpoint."
When the voltagedrops,e.g.,duringa sag, the powertransportedto themotor becomes
smaller than the power taken by the load. As a result the
m otor slows down, which
meansthat the angle</J increases. The angle will settle down at a new
operatingpoint,
Chapter5 • VoltageSags-EquipmentBehavior
320
Pre-sag power
0.8
Normal
::s
0..
.8
~
~
0
During-sag
power
operating
point
0.6
Operatingpoint
with reduced
voltage
~
0.4
0.2
0
50
100
150
Rotor angle in degrees
0
200
Figure5.77 Powertransferto a synchronous
motor as a function of the rotor angle.
indicated by "operatingpoint with reducedvoltage," where again the power to the
motor and the powertaken by the load are in balance.
It follows from Fig. 5.77that for deep sags there is no longer a stable
operating
point. In that case therotor angle will continueto increase until the supply voltage
the loses synchronism.Looking at
recovers.If the angle has increased too much motor
Fig. 5.78 we see twooperatingpoints: the normaloperatingpoint, labeled as"stable"
and a secondpoint labeled as"instable."In the latterpoint, bothpower flows are again
equal so themotor would be able tooperateat constantspeed. But any small
deviation
will make that the motor drifts away from thisoperatingpoint: either to the left (when
it will end up in the stableoperatingpoint) or to the right (when it will lose synchronism). The motor losessynchronismthe moment its rotor angle exceeds this instable
operatingpoint.
There is a second curve
plotted in Fig. 5.78, which indicates the power
transfer
during the sag. In this case there is no stable
operatingpoint during the sag and the
motor will continueto slow down until the voltage recovers. At
thatmomentthemotor
Operatingangle
I
I
Critical angle
I
I
I
I
I
I
I
0.8
I
::s
I
I
0..
c::
'ii
0.6
~
Q.c
0.4
0.2
0
0
50
100
150
Rotor angle in degrees
200
Figure5.78 Powertransferin normal
situationand for a deep sag.
321
Section 5.5 • Other Sensitive Load
will start to accelerate again but as it still rotates slower than the airgap field (thus
slower than the frequency of the supply voltage) rotor
its angle will continue to
increase. The maximumrotor angle is reached the
moment the motor speed comes
back to nominal. As long as this angle is smaller than the angle for the instable
operatingpoint, themotor does not lose synchronism. The figure shows the maximum
angle at the end of the sag which does not lead to an instable
situation; this angle is
indicated as"critical angle." According to the so-called
"equal-area-criterion"the two
[207].
shadedpartsin the figure are equal in area
The highest possiblesteady-staterotor angle equals 90
°-this occurs when the
motor load equals the maximum power which can transportedto
be
the motor. If the
motor load is onlyhalf this maximum value, a drop in voltage to 50% will bring the
operatingpoint back to the top of the sine wave again. This 50% is, however,
not the
deepest sag the
m otor can withstand for a long time. The
drop in voltage causes the
motor to slow down, thus when the
r otor angle reaches 90° it does not stop but will
continue to increase until the voltage recovers. The deepest
long-durationsag can be
found from Fig. 5.79. Again theequal-areacriteria tells usthat the two shadedparts
have the same area .
Operating angle
I
I
I
I
1
I
I
0.8
I
:s
I
I
,
0.
<:
't
0.6
~
0
I:l-o
0.4
0.2
Figure 5.79 Powertransferin normal
situationand for the deepest
long-duration
sag.
0
0
50
100
150
Rotor angle in degrees
200
5.5.3 Contaetora
Contactorsare a very common way of connecting
motor load to the supply. The
supply voltage is used to power an
electromagnetwhich keeps thecontactin place.
When the supply voltage fails the
contactopens, preventing the
m otor from suddenly
restartingwhen the supply voltage comes back. This works fine for long
interruptions
where the unexpected
starting of motors can be verydangerous.But contactorsalso
drop out for voltage sags and
short interruptionswhere such a behavior is not always
acceptable. Test results for
contactorsare presented in[34]. The measuredvoltage
tolerance curve for a
contactoris shown in Fig. 5.80. We see that the
contactortolerates
any voltage sag down to
a bout70%. When the sag
magnitudeis below 70% for longer
than a few cycles, thecontactordrops out. We also see the
remarkableeffect that the
voltage tolerance becomes better for deeper sags: a zero voltage toleratedfor
can be
3.5
cycles but a 50% voltage only for one cycle. This effectprobablydue
is
to the experimental setup. Sags were generated by switching between
normal
a supply and the out-
322
Chapter5 • VoltageSags-EquipmentBehavior
0.8
a
]
.8 0.6
.~
«S
0.4
~
0.2
246
Duration in cycles
8
Figure 5.80Voltage-tolerancecurve for a
contactor.(Data obtainedfrom [34].)
put of a variable-outputtransformer.It is not the voltagebut the currentthrough the
coil that causes the force keeping the
contactorclosed. Themomentthe currentdrops
below acertainvalue thecontactorwill startto drop out. For lower voltages thecurrent
path through the transformeris smaller, thus there is less
resistanceto damp the
current. As the current dampsmore slowly for smaller voltages, the
c ontactorwill
not drop out as fast as for medium voltages. This shows
that for contactorsthe supply
characteristicscan significantly influence the voltage
tolerance.
The factthat it is the currentand not the voltagethatdeterminesthe droppingout
of the contactorfollows also from thedependenceof the voltagetoleranceon thepointon-wave of sag commencement.The contactorof Fig. 5.80 toleratesa 3.4 cycle sag
startingat voltage zero, but only a 0.5 cycle sag
startingat voltagemaximum. As the
contactorcoil is mainly inductivethe currenthas amaximumat voltagezero andis zero
at voltage maximum.
The influenceof the point-on-waveof sagcommencementhas beenfurther studied by Turner and Collins [38],reporting a voltage toleranceof 30 ms for sag comof the voltage zero crossing, reducing to less
than 8 ms for sags
mencements within 30°
commencingat voltagemaximum.
Note that all this refers to so-called ac
contactors.An alternativeis to use de
contactorswhich are fed from aseparatedc system with their ownbattery backup.
Thesecontactorsdo normally not drop out during voltage sags.However,they require
a separatede system and analternativeprotection againstunexpectedrestart of the
motor.
5.5.4 Lighting
Most lampsjust flicker when a voltage dip occurs.
Somebodyusing the lamp will
probably notice it, but it may not .beconsideredas somethingserious. It isdifferent
when the lamp completely extinguishes and takes several
minutesto recover. In industrial environments,in places where a large
numberof people aregathered,or with street
lighting, this can lead todangeroussituations.
Dorr et a1. [36] havestudiedthe voltagetoleranceof high-pressuresodiumlamps.
Voltage sags can extinguish the lamp, which must cool down for one to several minutes
beforerestarting.The voltage-tolerancecurves for three lamps are shown in Fig. 5.81.
For voltages below50% the lampsalreadyextinguishfor a sagof lessthan two cycles.
323
Section 5.5 • Other Sensitive Load
0.8
[
.s 0.6
~
.~ 0.4
~
0.2
Figure 5.81 Voltage toleranceof highpressuresodium lamps. (Data obtainedfrom
Dorr et al. [36].)
5
10
Duration in cycles
15
20
The lampstook aboutoneminuteto restrike, andanotherthree minutes before the full
light intensity was reached again. The voltage tolerance
of the lamp isfurther dependent on the age. When lamps age they need a larger voltage to operate; they will thus
extinguishalreadyfor a lower drop in voltage. The minimum voltage for longer sags
varied from 450/0 for new lamps to850/0 for lamps at the end of their useful life.
Voltage SagsStochastic Assessment
In this chapterwe discussmethodsto describe, measure, and
predictthe severityof the
voltagesag problem: how many times per year will the
equipmenttrip. Thereare two
methodsavailablethat quantify the severityof the problem: powerquality monitoring
and stochasticprediction. Power quality monitoring gives mainly information about
commonevents.For lesscommoneventsstochasticpredictionis more suitable. In this
chapterboth are discussed in detail.
After explainingthe need forstochasticassessment, the
variousways of presenting the voltagesagperformanceof the supply are discussed. The
chaptercontinueswith
some aspectsof voltage sagmonitoring, including the resultsof a number of large
surveys.Finally, two methodsfor stochasticprediction of voltage sags are discussed,
togetherwith a few examples. The
methodof fault positionsis suitablefor implementation in computersoftwareand is thepreferredtool for studies in meshed
transmission
systems.For radial distribution systems andhandcalculations,the methodof critical
distancesis more suitable.
8.1 COMPATIBILITY BETWEBN EQUIPMENT AND SUPPLY
Stochasticassessment
of voltage sags is needed to find out
whethera pieceof equipment
is compatiblewith the supply. A studyof the worst-casescenariois not feasible as the
of
.worst-casevoltage disturbanceis a very longinterruption. In some cases, a kind
"likely-worst-case-scenario"is chosen, e.g., a fault close to the
equipmentterminals,
clearedby the primary protection,not leading to aninterruption.But that will not give
any informationaboutthe likelihood of an equipmenttrip. To obtaininformation like
that, a "stochasticcompatibility assessment"
is required. Such a study typically consists
of three steps:
1. Obtain system performance.Information must beobtained on the system
performancefor the specific supply point: the expected
number of voltage
sags with different characteristics.There arevarious ways to obtain this
325
326
Chapter6 • VoltageSags-Stochastic
Assessment
information:contactingthe utility, monitoringthe supplyfor severalmonths
or years, or doing astochasticpredictionstudy. Both voltagesagmonitoring
and stochasticprediction are discussed in detail in this
chapter.Note that
contactingthe utility only shifts theproblem, as also the utility needs to
perform either monitoring or a stochasticpredictionstudy.
2. Obtain equipment voltage tolerance.
Information has to beobtainedon the
behaviorof the pieceof equipmentfor variousvoltagesags. Thisinformation
can beobtainedfrom theequipmentmanufacturer,by doing equipmenttests,
or simply by taking typical values for thevoltagetolerance.This part of the
compatibility assessment
is discussed in detail inC hapter5.
3. Determine expected impact. If the two types
of information are availablein
an appropriateformat, it is possible toestimatehow often the pieceof equipmentis expected to trip per year, and
what the (e.g., financial)impactof that
will be. Based on theoutcomeof this study onecandecide toopt for a better
supply, for better equipmentor to remain satisfied with thesituation. An
essentialcondition for this step isthat systemperformanceand equipment
voltage toleranceare presentedin a suitableformat. Some possibleformats
are discussed in Section 6.2.
is given, based on Fig. 6.1.
An exampleof a stochasticcompatibility assessment
The aim of the study is to comparetwo supply alternativesand two equipmenttolerances. The twosupply alternativesare indicated in Fig. 6.1 through the expected
numberof sags as afunction of the sag severity:supply I is indicatedthrougha solid
line; supply II through a dashedline. We further assume the following costs to be
associatedwith the two supply alternativesand the two devices (inarbitrary units):
supply
supply
device
device
I
II
A
B
200 units/year
500 units/year
100units/year
200 units/year
We also assumethat the costsof an equipmenttrip are
to units.
160
140
ft 120
~
8. 100
fI)
bO
~
fI)
~
...
80
0
U
-a
i
60
\
\
\
\
\
40
\
,,
,
I
20
- - __: _-_-__
-_-_-_-__
-_-_-_-__
-_-_- J
o '-----'---"---'------'----'--~-~-.-j
10
20
30
40
50
60
Severityof thesag
70
80
Figure 6.1 Comparisonof two supply
alternatives(solid curve: supply I, dashed
curve: supplyII) and twoequipment
tolerances (solid vertical line: device
A,
dashedline: device B).
327
Section 6.1 • Compatibility BetweenEquipmentand Supply
From Fig. 6.1, one canreadthe numberof spurioustrips per year,for eachof the
four designoptions, at the intersectionbetweenthe supply curve and the device (vertical) line. For device AandsupplyI we find 72.6spuriousequipmenttrips peryear,etc.
The resultsare shown in Table 6.1.
TABLE 6.1 Numberof SpuriousTrips per Year forFour
Design Alternatives
Device A
Device B
Supply I
Supply II
72.6
14.6
29.1
7.9
Knowing the numberof trips per year, theannualcostsof eachof the four design
options,andthe costsper spurioustrip, it is easy tocalculatethe total annualcosts.For
the combinationof device A and supply I thesecostsare
72.6 x 10 + 100+ 200 = 1026units/year
The resultsfor the four designoptionsareshownin Table6.2. From this tableit follows
that the combinationof supply I and device B has thelowest annualcosts.
TABLE 6.2 Total Costs per Year forF our Design
Alternatives
Device A
Device B
Supply I
Supply II
1026
546
891
779
Note the stochasticcharacterof the assessment.
An expectedvalue (the expected
numberof equipmenttrips per year multiplied by the cost of one equipmenttrip) is
addedto a deterministicvalue (the annualcost of supply and device). Assumethat the
voltagetolerancefor a device is thesameunderall circumstances;the voltagetolerance
is thusa deterministicquantity. But the numberof sags willvary from yearto year. We
further assumethe occurrenceof a sag to beindependento f the occurrenceof other
sags. Inthat case thenumberof sags inany given year follows a Poissondistribution.
Let N be thenumberof sags inany given year and JL the expectednumberof sags (as
indicatedin Table 6.1). The probability that N = n for a Poissondistribution is found
from
J1,n
Pr{N
= n} = e-/Ln!
(6.1)
For the four design alternativesin Table 6.1 this distribution has been plotted in Fig.
6.2. It follows from the figure, for example,that the number of trips of design BII
(supply II in combinationwith device B)varies between2 and 18, and for design BI
between7 and26. It is thusnot surethat in a given year,designBII gives lesstrips than
design BI.
From the probability density function for the number of trips (Fig. 6.2) the
probability densityfunction for the total costsper year can be calculated,resultingin
328
Chapter6 • VoltageSags-Stochastic
Assessment
0.15
BII
0.1
g
~
.,J:)
e
~
AI
0.05
20
40
60
80
Numberof sags in a given year
0.15
g
100
Figure6.2 Probabilitydensityfunction of the
numberof sags per year for four design
alternatives.
"BII
0.1
~
£
0.05
400
600
800
1000
Total costs in a given year
1200
Figure6.3 Probabilitydensityfunction of the
costsper year forfour design alternatives.
Fig. 6.3. This figure showsthat design BI is clearlybetterthan any of the otherdesign
options.
6.2 PRESENTATION OF RESULTS: VOLTAGE SAG COORDINATION CHART
In this section we discuss numberof
a
ways to presentthe supplyperformance.The
discussionconcentrateson the presentationof results obtainedfrom power quality
monitoring.The sametechniquecan beappliedto the resultsof a stochasticassessment
study.
8.2.1 The Scatter Diagram
Every power quality monitor will at least givemagnitudeand duration as an
output for a sag. When the supply monitoredfor
is
a certainperiod of time, anumber
of sags will berecorded.Each sag can be
characterizedby a magnitudeand aduration
and be plotted as one point in the magnitude-durationplane. An example of the
resulting scatterdiagramis shown in Fig. 6.4. Thescatterdiagramis obtainedfrom
329
Section 6.2 • Presentationof Results: Voltage Sag
CoordinationChart
1---------------------,
0.9
0.8
••
!
..
r,
aO.7 •
.~ 0.6
~ 0.5
.~ 0.4
~ 0.3
0.2
0.1
Figure 6.4 Seatterdiagramobtainedby one
year of monitoring at an industrial site.
°0
5
10
15
20 2S 30
Duration in cycles
35
40
45
Voltage swells
Lower thresholdfor swells
Upper threshold for sags
Sags due to motor starting
Voltage sags due
to short circuits
Figure 6.5 Scatterdiagramas obtainedfrom
a large power quality survey.
Short interru tions
Duration
one yearof monitoringat anindustrialsite [155]. For a large powerquality survey, the
of the resulting
scatterdiagramsof all the sites can be combined. A stylized version
scatterdiagram is shown in Fig. 6.5. In this figure not only voltage sags, but also
interruptionsand voltage swells are
indicated.
In Fig. 6.5 we see anumberof heavily populatedregions:
• Voltage sags due tos hort circuits, with durationsup to a fewhundredmillisecondsand magnitudesfrom 50% upwards.Deeper and longer sags are present but rare.
• Voltage sags due to
m otorstarting,with durationsof a few seconds and longer,
and magnitudesfrom 800~ upwards.
• Short interruptionsdue to fast reclosing, with voltage
magnitudezero and
durationsfrom about 10 cyclesonward.
• Voltage swells with similardurationsas sags due to
s hortcircuits, but magnitudes up to1200/0.
Next to these densely
populatedareas there are
scattered,long, deep sags, likely due to
the errorsmade inrecordingdurationof sags with a long,post-faultsag. These long,
deep sagsconsistof a short,deep sag followed by a long, shallow sag. This points to one
330
Chapter6 • VoltageSags-Stochastic
Assessment
of the shortcomingsof the commonly used method of sag
characterization:the lowest
rms value as sag
magnitudeand thenumberof cycles below thethresholdas the sag
duration.
No reliableinformationhas been published
a boutthe numberof sags with a large
non-rectangularpart. It is mentionedin [156] that about 100/0 of sags in the U.S.
distribution systems arenon-rectangular.Another indication that this effect is not
very severe is the factthat the duration of most sagscorrespondsto typical faultclearing times in the system.
8.2.2 The Sag Density Table
The scatterdiagramis very useful to give aqualitativeimpressionof the supply
performance,but for aquantitativeassessment
otherways ofpresentationare needed.
A straightforwardway of quantifying the number of sags isthrough a table with
magnitudeand duration ranges. This is done in Table 6.3 for
data obtainedfrom a
large powerquality survey[20]. Each element in the table gives the
numberof events
with magnitudeand duration within a certain range; e.g.,
magnitudebetween 40 and
50% and durationbetween400 and 600 ms. Each element gives the density
of sags in
that magnitudeand durationrange; hence the term
"sagdensitytable" or "sagdensity
function." A combinationof magnituderange anddurationrange is called a"magnitude-durationbin."
The sag density function is typically presented as a bar chart. This is done in Fig.
6.6 for the data shown in Table 6.1. The length of each barproportional
is
to the
numberof sags in thecorrespondingrange. From the barchart it is easier to get an
impressionof the distribution of the sagcharacteristics,but for numerical values the
6.6 that the majority of sags has a
table is more useful. In this case we see from Fig.
magnitudeabove800/0 and adurationless than200ms. There is also caoncentrationof
short interruptionswith durationsof 800 ms and over.
duration ranges. In
In Fig. 6.6 all magnituderanges areof equal size, so are all
most cases the ranges will be of different size. There are moreofsags
shortdurationand
high magnitudethan sags elsewhere in the
magnitude-durationplane. Therefore,the
resolution is chosen higher forshorter duration sags and for shallow sags. Several
examples of the density function in
bar-chartform are shown in Section 6.3.
TABLE 6.3 Exampleof SagDensity Table: Numberof Sags per Year
Magnitude
0-200 ms
200-400ms
400-600ms
600-800ms
> 800 ms
80-90%
70-80°./c,
60-70%
50-600/0
40-50%
30-40%
20-30%
10-20°./c,
0-10%
18.0
7.7
3.9
2.3
l,4
1.0
0.4
0.4
1.0
2.8
0.7
0.6
0.4
0.2
0.2
0.1
0.1
0.3
1.2
0.4
0.2
0.1
0.1
0.1
0.1
0.1
0.1
0.5
0.2
0.1
0.1
0.1
0.0
0.0
0.0
0.0
2.1
0.5
0.2
0.1
0.1
0.1
0.0
0.1
2.1
Source: Data obtainedfrom [20].
Section 6.2 • Presentationof Results : Voltage Sag
C oordinationChart
331
18
16
14
~
..,...c,
;>..
12
'" 10
bIl
....1J!
0
..,...
8
§
6
.c
Z
4
2
0
> 0.8s
Figure 6.6 Two-dimensional bar chart of the sag density function shown in Table
6.3.
8.2.3 The Cumulative Table
Of interest to thecustomeris not so much thenumberof voltage sags in a given
magnitudeand duration range, but the number of times
that a certainpiece of equipnumberof sags worse
ment will trip due to a sag.It therefore makes sense to show the
than a givenmagnitudeand duration. For this a so-called"cumulative sag table" is
calculated. ElementM D of the cumulativesag table is defined as follows:
(6.2)
withfmd elementmd of the density table : the
numberof sags in thedurationranged and
the magnituderangem; and with FMD elementMD of the cumulativetable: thenumber
of sags withdurationlonger thanD and magnitudelessthan M. Durationsare summed
from the value upward because a longer sag is more severe;
magnitudesare summed
from the value down to zero because a lower
magnitudeindicates a more severe sag.
This is a direct consequence of the definition
of sagmagnitude,where a higher magnitude indicates a less severe event.
The cumulative tableobtainedfrom the density table inTable 6.3 is shown in
Table 6.4. The table shows, e.g.,
that the rms voltage drops below 60% for longer
than
200 ms, on average 4.5 times per year. If the
equipmentcan only tolerate a sag
332
Chapter6 • VoltageSags-Stochastic
Assessment
TABLE 6.4
Example ofCumulativeSagTable, Numberof Sags per Year
Magnitude
0
90%
80%
70%
60%
50%
40%
30%
20%
10%
49.9
25.4
15.8
10.9
8.0
6.2
4.9
4.2
3.5
200ms
13.9
7.4
5.5
4.5
3.8
3.4
3.1
2.8
2.5
400 ms
600 ms
8.4
4.7
3.6
3.1
2.9
2.7
2.6
2.4
2.2
6.1
3.6
2.9
2.6
2.5
2.3
2.3
2.2
2.1
800 ms
5.2
3.1
2.6
2.4
2.3
2.3
2.2
2.2
2.1
Source: Data obtainedfrom Table 6.3.
below 60% for 200 ms, it will trip on average 4.5 times per year.
From such a table the
number ofequipmenttrips per year can be
obtainedalmost directly.
6.2.4 The Voltage Sag Coordination Chart
Table 6.4 is shown as a bar
chart in Fig. 6.7. The values in the cumulative table
belong to acontinuousmonotonefunction: the values increase toward the left-rear
cornerin Fig. 6.7. The values shown Table6.4
in
can thus be seen as a two-dimensional
function of numberof sags versusmagnitudeand duration. Mathematicallyspeaking,
50
45
40
~ 35
&30
~
25
~
20
~
15
'"
'o
~fJ.ril~~~~~
90%
~~
80%
70%
60%
.0
10
50%
40%
30%
. ,&0(,
<$''bo~"
llc
e.,'bo
5
o
Figure 6.7 Barchart of the cumulativevoltage sag table shown in
Table 6.4.
333
Section 6.2 • Presentationof Results: Voltage Sag
CoordinationChart
25 ~~-----l~"-'£'-+-------:~~-----t-------;- 80%
J-,C--~rJ----+---7"G.-_---+-----+-------t-70%
a--.,t;-----~------+-----+-------t-60%
4)
]
J----~t.--_+_------+-----+_----___t_ 50% .~
8
~-~---+-------+-----+-------t-40%
l
5 sags/year
I - - - - - - - + - - - - - - - - + - - - - - - f - - - - - - - - t - 20%
1--------+-------+------+------.....-,- 10%
0.6 s
0.8 s
Os
0.2 s
0.4 s
Sag duration
6.4.
Figure 6.8 Contourchart of the cumulativesag function, based on Table
this function is defined for the whole
magnitude-durationplane. Whenobtainedfrom
power quality monitoring the function is not continuous.Stochasticpredictiontechniques will normally also not lead to acontinuousfunction. Whether the function is
continuousor not, acommonway of presentinga two-dimensionalfunction isthrough
a contour chart. This was done byConrad for the two-dimensionalcumulative sag
function, resulting in Fig. 6.8[20].
The contourchartis recommendedas a"voltagesagcoordinationchart" in IEEE
Standard493 [21] and in IEEE Standard1346[22]. In a voltage sagcoordinationchart
thecontourchartof the supply iscombinedwith the equipmentvoltage-tolerancecurve
reproduced
to estimate thenumberof times theequipmentwill trip. Figure 6.8 has been
in Fig. 6.9 including twoequipmentvoltage-tolerancecurves. Both curves are rectangular; i.e., theequipmenttrips when the voltage drops below certain
a
voltage for
longer than a givenduration. Device A trips when the voltagedrops below 65% of
number
nominal for longer than200 ms. According to the definition given before, the
of voltage sags below65% for longer than 200 ms is equal to the element of the
cumulative table for 65%, 200 ms. The values in the cumulative sag table are the
underlying function of the contour chart in Figs. 6.8 and 6.9. In short, the number
of spurioustrips is equal to thefunction value at the kneeof the voltage-tolerance
curve,indicatedas a circle in Fig. 6.9.F or device A thispoint is located exactly on the
For device B, the
five sags per year
contour.Thus, device A will trip five times per year.
knee is located between the
15 and 20 sags per yearcontours.Now we use the knowledgethat the underlyingfunction is continuousand monotone.The numberof trips will
thus be between15 and 20 per year; usinginterpolationgives anestimatedvalue of 16
trips per year.
For a non-rectangularequipmentvoltage-tolerancecurve, as shown in Fig. 6.10,
the procedurebecomessomewhatmore complicated.Considerthis device as consisting
of two components,each with arectangular.voltage-tolerancecurve.
• ComponentA trips when the voltagedropsbelow 50% for longerthan 100ms;
accordingto the contourchart this happenssix times per year.
334
Chapter6 • VoltageSags-Stochastic
Assessment
17"~"7""""':::r-::;lI..-,.-,..,r----~-~-------r------__
90%
..,.llIIIIIIf----..,......... DeviceB .....-----_r80%
25 r-:7'--....
-~--
20
t7----t'7l'----tr-.--.."e-----+------4-------I-70%
15
DeviceA
10
60% ~
t----t----:r---tr-.--------+------+-------4-50% .~
8
~
t--""7'"t----t-------+-------+-------I-40% U)
5 t-----t-----Ir.--------+------+------4- 30%
t----t----tr-.--------+------+-------I- 20%
t-----t----1I----------+------f-------+. 100/0
0.2 s
0.68
0.48
08
0.88
Sag duration
Figure 6.9 Voltage sag coordinationchart, reproducedfrom Fig. 6.8, with two
equipmentvoltage-tolerancecurves.
~.....,.._~7"_::l~--,.,r-----~---y------~-----~
90%
B
........,.:....----~..-------+------+------~60%
-8
a
r---:-i==:::;~~~-------t------;-------;- 50% .~
10
40%
J---....,.r..t-----4I---------f.-------t------_+_
51o------II-------4I---------f.-------t--------t-
e
tf
en
30%
J------tl------II---------+-------+--------t- 20%
t------tI...------I'-------4-------+----------- 10%
0.28
0.6s
0.4 s
0.88
Os
Sag duration
Figure 6.10 Voltage sag coordinationchart, reproducedfrom Fig. 6.8, with nonrectangularequipmentvoltage-tolerancecurve.
• ComponentB trips when the voltagedropsbelow 85% for longerthan200 ms,
which happens12 times per year.
Adding these twonumbers(6 + 12 = 18) would count double those voltage sags for
which both componentstrip. Both componentstrip when thevoltagedropsbelow 50%
for longer than 200 ms;aboutfour times per year. Thiscorrespondsto point C in the
chart. The numberof equipmenttrips is thus equal to
FA
+ En -
Fe = 6 + 12- 4
= 14
(6.3)
Section 6.2 • Presentationof Results : Voltage SagC oordination Chart
335
Note that assuming arectangularequipmentvoltage-tolerancecurve (100 rns, 85%)
would have resulted in the incorrect value of 20 trips per year.
By using thisprocedure,the voltage sagcoordinationchartprovidesfor a simple
and straightforwardmethodto predict thenumberof equipmenttrips.
8.2.5 Example of the Use of the Voltage Sag Coordination Chart
The dataobtainedfrom a large survey [68] has been usedplot
to the sag density
quality of the voltage at the
bar chart shown in Fig. 6.11. The survey measured the
terminals of low-voltageequipment(at the wall outlet) at many sites across the
United
States andCanada. Figure 6.11 can thus be
interpretedas the average voltage
quality
experienced by low-voltage
equipment.
From Fig. 6.11, a voltage sag
coordinationcharthas beenobtained,shown in Fig.
indicatedby the points A, B, C, and D.
6.12. Four equipmentvoltage tolerances are
The meaningof these will be explained next.
Supposethat a computermanufacturerconsiders differentoptionsfor the power
supply of personalcomputers. The choice is between two
different de/deconverters,
with minimum operatingvoltages of 100V and 78 V, and between two
capacitorsizes,
leading to 5% and 1% de voltage ripple. Using (5.6) we can
calculate the voltage
toleranceof the four designoptions. For a minimum operatingvoltage of 100V and
a de voltage rippleof 5% we find a voltage tolerance of 84% (100 V) and 1.5 cycles, etc.
The results are shown in column 4 of Table 6.5. The voltage
tolerancefor the four
options(A , B, C, and D) is indicated by the four dots in Fig. 6.12.
From this voltage sag
70
60
[
.,...
50
Co
'"
40
.,...o
30
OIl
~
e-
.r>
e
z'"
20
10
6-10 c 20 c0.5s
Sag duration
Figure 6.11 Sag density for the average low-voltage supply in the United
Statesand
Canada. (Data obtainedfrom Dorr [681.)
336
Chapter6 • VoltageSags-Stochastic
Assessment
TABLE 6.5 Comparisonof Four Design Optionsfor the PowerSupplyof a
PersonalC omputer
Option
Minimum Operating
Voltage
A
de Ripple
5%
1%
IOOV
IOOV
78 V
78 V
B
C
D
Voltage Tolerance
EstimatedTrip
Frequency
84%, 1.5 cycles
84°tlo, 8 cycles
65%, 3 cycles
650/0, 15 cycles
5°tlo
10/0
10 sagsperyear
A
~ t:::::::;
--
r-'WB
V..-- -::::: ~~ ~
60
V
90
......... ::--
10-
40
30
/---
~
--
-------/
-
~
~
~
.--/
---
~
~
I-'
.."I
~
II
J
I..- /
-~
f.--
104V
lOOV
>
.8
90V
]
78V
/D
:l
(5
96V
84V
J
C~
20 -
)
l-/V
)
IOO/year
50/year
25/year
20/year
'f
f
(/)
,
60V
10V
1 c 2 c 3 c 4 c 5 c 6 c 10 c 20 c 0.5
sis 2 s 5 s lOs 30 s 60 s 120 s
Sag duration in cycles (c) and seconds (s)
Figure 6.12 Voltage sag
c oordinationchart for the averagelow-voltagesupplyin
the United Statesand Canada.(Obtainedfrom the sagdensitychart
in Fig. 6.11.)
coordination chart the trip frequency can easily be
estimated,resulting in the last
column of Table 6.5.
8.2.8 Non-Rectangular Sags
Characterizingvoltage sagsthrough their magnitudeand duration assumes a
static load, a static system, and no changes in the fault. In reality
both the load and
the system are dynamic and the fault can develop, e.g., from a single-phase to.. a three
phase fault.Simulationsand measurements
have shownthat inductionmotor load can
lead to longpost-faultvoltage sags. A few examples
of non-rectangularvoltage sags
were shown inChapter4: Figs. 4.47,4.48,and 4.130.
There are two ways
o f presentingnon-rectangularsags intwo-dimensionalcharts
like Figs. 6.8 and 6.12.
1. Define themagnitudeas the minimum rms voltage
during the disturbance
and thedurationas the timeduring which the rms voltage is below a thresh
..
old, typically 90% of nominal voltage. Thismethodis used in most power
quality monitors. The consequenceof this is that non-rectangularsags are
characterizedas more severe
t han they actually are. Alternativesare to use
the average or the rms of the one-cycle rms values latter
(the is a measure of
the energy remainingduring the sag).
337
Section 6.2 • Presentationof Results: Voltage Sag
C oordinationChart
2. Characterizethe voltagequality by the numberof times the voltage drops
below a given value for longer
thana given time. This again results ingraph
a
like Fig. 6.8, but nowwithout the need tocharacterizesags individually. Such
a methodwas firstproposedin [17] and used in [18],andbecamepartof IEEE
Std. 493 [21]. A similarmethodis proposedin [156] for inclusion incontracts
between utility andcustomers.The argumentfor the latter proposalbeing
that utilities shouldnot be overlypunishedfor non-rectangularsags.
To explain the secondmethod,the cumulativetable will be introducedin a different
way. We define each element ascountercountingthe
a
numberof sags worsethan the
magnitudeand duration belonging to this element. Each sag
that occurs increases the
value ofpartof the elements by one. The elements whose value is increased are those for
which the sag is more severe
thanthe element. Inotherwords, those elements less severe
than the sag; in the table, the elements above the sag. This is shown in Fig. 6.13 for a
rectangularsag.
o f points correspondingto the cumulativesag
Figure 6.14 again shows the grid
function. But this time anon-rectangularsag is shown. Theprocedureis exactly the
"The function valueshouldbe increasedby one for all points above the
same as before:
sag."
®
®
®
X
X
X
X
®
®
®
X
X
X
X
Q9
®
®
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Figure 6.13 Updateof cumulativetable for
rectangularsag.
Figure 6.14 Updateof cumulativetable for
non-rectangularsag.
Duration
®
®
®
®
®
Ix
X
®
®
®
X
X
X
X
®
®
®
X
X
X
X
®
®
X
X
X
X
X
®
®
X
X
X
X
X
X
X
X
X
X
X
X
Duration
338
Chapter6 • VoltageSags-Stochastic
Assessment
Using this methodit is possible toquantify the quality of the supply including
non-rectangularsags. But thismethodcannotbe used tocharacterizeindividual sags.
Note that this is oftennot a seriousconcernwhen one isinterestedin merelyquantifying the supply performance.
Some sags will still escape
quantification,as shown in Fig. 6.15. A possible choice
here is tomeasurethe time the sag is in each
magnituderange in the table, and then
increase thepoints to the left of the table inthat magnituderange. This would lead to
an equivalentsag asindicatedin Fig. 6.15. Themethodproposedin [156] treatsthese
"very non-rectangularsags" in a similar way. To understandthe limitation of the
method in Figs. 6.13, 6.14, and 6.15 the term
" rectangularvoltage-tolerancecurve"
is introduced. A piece of equipmenthas a rectangularvoltage-tolerancecurve if its
tripping is determinedby one magnitude and one duration. Thus, the equipment
trips when thevoltage drops below a certain magnitudefor longer than a certain
duration. The actual shapeof the rms voltage versus time has no influence on the
equipmentbehavior. Examplesof such equipmentare undervoltagerelays (e.g., used
to protect induction motors) and mostnon-controlledrectifiers. Alsocomputersand
otherconsumerelectronicsequipmentfit in this category. Manyadjustable-speed
drives
trip due to anundervoltage-timerelay at the dc bus or on the ac terminals. Also those
can beconsideredas having a rectangularvoltage-tolerancecurve.
For equipmentwith a rectangularvoltage-tolerancecurve this method directly
gives the expectednumber of spurious trips. For non-rectangularvoltage-tolerance
curves themethod no longer works.That might appeara seriousdisadvantageuntil
one realizesthat a non-rectangularvoltage-tolerancecurve will normally be obtained
for rectangularsags.Applying it directly to non-rectangularsags isproneto uncertainties anyway, nomatter which definition of magnitudeand duration is used. When
assessing the influence
of non-rectangularsags on a piece ofequipmentit is recommended to use a
r ectangularapproximationof the voltage-tolerancecurve unless more
detailedinformation on its behaviorundernon-rectangularsags is available.
Q9
@
@
@
@
@I
X
Q9
@
Q9
@
X
X
X
Q9
®
@
®
X
X
X
@
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
@
X
-
Duration
Figure 6.15 Problemsin updatingthe
cumulativetable for a verynon-rectangular
sag.
8.2.7 Other Sag Characteristics
In the previouspart of this section, we onlyconsideredmagnitudeand durationof
o ther
the sags. We sawbefore that the equipmentbehaviormay also be affected by
characteristics:phase-anglejump, three-phaseunbalance,point-on-waveof sag initiation. Below, somesuggestionsare given for thepresentationof the results when these
339
Section 6.2 • Presentationof Results: Voltage Sag
C oordinationChart
additional characteristics need to be
incorporated.Note that, unlike magnitude and
duration, no monitoring data are available on phase-angle jump, three-phase unbalance, and point-on-wave of sag initiation. This makes that some of the suggestions
remainrather theoretical, without the chance to apply them to actual
data.
6.2.7.1 Three-Phase
Unbalance. We saw in Section 4.4 that three-phase unbalanced sags come in a number of types. The fundamental types were referred to as
A, C, and D. The concept of voltage sag
coordinationchart can be extended to
three-phase unbalance by creating one chart for each type, as shown in Fig. 6.16. A
contour chart is created for the number of sags more severe than a given magnitude
and duration, for each type. Also the equipment voltage-tolerance curve is obtained
for each type. In exactly the same way as before, the number of equipment trips can
be found for each type
; in this example:N A , Nc, and ND' The total number of equipment trips N is the sum of these three values:
(6.4)
The method can be extended toward other types. The main problem remains to obtain
the type of sag frommonitoringdata. A technique for this has been proposed
in
[203],
[204] which requires the sampled waveforms.
6.2.7.2 Phase-Angle Jumps.Including phase-angle jumps in the compatibility
assessment for single-phase equipment creates a three-dimensional problem. The
three dimensions are magnitude,
duration, and phase
-anglejump. Next to this there
are twoadditionalcomplications:
Type A
Duration
_..
Tn'~~
Duration
._. _....!~e _~. "_ .
Figure 6.16 Useof the voltagesag
coordinationchart when three-phase
unbalanceneeds to beconsidered.
Duration
.
.__..._..
340
Chapter6 • Voltage Sags
-StochasticAssessment
• Phase-anglejumps can beboth positive and negative, with the
majority of
values likely to be foundaroundzero phase-angle
j ump. Using a cumulative
function requires thesplitting up of the three-dimensionalspace in two halfspaces: one for positive
phase-anglejump, one for negativephase-anglejump.
Note that equipmentbehaviormay be completelydifferent for positive and for
negative phase-angle
jump.
• An increasingphase-anglejump (in absolutevalue) not necessarily leads to a
more severe event for the
equipment.With both magnitudeand durationit was
possible to indicate adirection in which the event becomes more severe
(decreasingmagnitudeand increasingduration). For phase-anglejumps this
is not possible.
Especially thelatter complicationmakes athree-dimensionalversion of the voltage sag
coordinationchartnot feasible. A possible
solutionis to split thephase-angle
j ump axis
in a numberof ranges, e.g.,[-60°, - 30°], [-30°, - 10°], [_10°, + 10°], [+10°, + 30°],
[+30 °, + 60°]. For each range then umberof equipmenttrips isdeterminedlike before.
The total numberof equipmenttrips is the sum of the values
obtainedfor each rangeof
phase-angle
j ump. A plot of magnitudeversusphase-angle
j ump for single-phaseequipment was shown in Fig. 4.108.
Splitting the phase-anglejump axis in a number of
ranges showsthat not all charts will contain the whole rangeof magnitudevalues.
Only in the rangearound zero phase-anglejump do we expectmagnitudevalues
between zero and100%. The range[+30 °, + 60°] may only contain magnitudevalues
around 50% of nominal. An alternativeis to split the duration axis in a numberof
ranges. In astochasticprediction study this could correspondto the typical faultclearing time in differentparts of the system, e.g., atdifferent voltage levels.For
eachduration range, a plot ofmagnitudeversusphase-anglejump results, similar to
the oneplotted in Fig. 4.108. Within this plot, anequipmentvoltage-tolerancecurve
can be drawn . Ahypotheticalexample is shown in Fig. 6.17.
Note that this curve has a
different shape than the
voltage-tolerancecurve in themagnitude-durationplane.Note
further that it is no longer possible to use cumulativefunction
a
for the number of
events like in the voltage sag
coordinationchart. Insteada density function must be
used, and thenumberof eventsoutsideof the voltage-tolerancecurve added.
For three-phaseequipmentthe problem becomes slightly less
complicated.Using
characteristicmagnitudeand phase-anglejump results in negative phase -angle
jump
values only. But a larger (negative)
phase-anglejump could still be a less severe event
for the equipment.Presentingequipmentand supplyperformancestill requiressplitting
up thephase-anglejump axis or thedurationaxis.
Trip
No trip
0.
.[
ll)
1ib 0° t - - - - - --+--
~
ll)
- - - - <:f)--Magnitude
;{l
..c
c..
Figure 6.17Hypotheticalexample of the
voltage-tolerance curve for
magnitudeagainst
phase-anglejump. The sagdurationis
consideredconstant.
341
Section 6.2 • Presentationof Results: Voltage Sag
C oordinationChart
6.2.7.3 Point-on-Wave. Point-on-wavecharacteristicsmay be easier to include
in the compatibility assessmentthan phase-anglejumps, because thepoint-on-wave
of sag initiation is likely to be independento f the other characteristics.F or here we
will assumethat this is the case. Analysis
o f monitoring data is needed to check this
assumption.
As the point-on-waveof saginitiation is independentof the sagmagnitudeand
duration, there is no need for athree-dimensionaltreatment.Next to the standard
contourchart of magnitudeversusduration,a one-dimensionalp lot is needed for the
point-on-wave.A hypotheticalexample is shown in Fig. 6.18.
Note that only values
0
are shown;o thervalues can betranslatedinto a value in this range.
between zero and 90
For a number of values avoltage-tolerancecurve needs to beobtained and
plotted in the standardvoltage sagcoordinationchart; see Fig. 6.19. The resulting
number of equipmenttrips N; from each voltage-tolerancecurve is weightedby the
fraction of sags~; with a point-on-wavevalue equal toi, and addedto get thetotal
numberof equipmenttrips N:
(6.5)
In the example shown in Figs. 6.18 and 6.19, this
total numberof equipmenttrips is
obtainedfrom
N
= ~oNo + ~30N30 + ~6oN60 + ~90N90
Figure 6.18Hypotheticalexample of the
fraction of sags with a given
point-on-wave
value.
30°
60°
90°
Point-on-wave
0°
No ~ N
30
30°
N60
60°
N90
Figure 6.19Hypotheticalexample of the
voltage-tolerancecurves for differentpointon-wave of saginitiation.
(6.6)
Duration
90°
342
Chapter6 • VoltageSags-Stochastic
Assessment
6.3 POWER QUALITY MONITORING
A common way of obtaining an estimatefor the performanceof the supply is by
recording the disturbanceevents. For interruptionsof the supply this can be done
manuallyas describedin Chapter2. For voltage sags andother short-durationevents
an automaticrecording method is needed. A so-called power
quality monitor is an
appropriatetool for that, although modern protective relays canperform the same
function. Powerquality monitors come in various types and for a range of prices. A
further discussionabout them isbeyondthe scope of this book.
For each event themonitor recordsa magnitudeand aduration plus possibly a
few othercharacteristicsand often also acertainnumberof samplesof raw data: time
domainas well as rms values. This could result in enormousamountof
an
data,but in
the end onlymagnitudeand durationof individual events are used for
quantifying the
performanceof the supply.
Two typesof power quality monitoring need to bedistinguished:
• monitoringthe supplyat a (large)numberof positionsat the same time, aimed
at estimatingan "averagepower quality": a so-called powerquality survey.
• monitoringthe supplyat one site, aimed at
estimatingthe powerquality at that
specific site.
Both will be discussed inmore detail below.
8.3.1 Power Qualltv Survey.
Large power quality surveys have been
performedin severalcountries.Typically
ten to ahundredmonitorsare installedat one or two voltage levels spread over a whole
country or the serviceterritory of a utility. Because not allsubstationsand feeders can
bemonitored,a selection has to be made. The selection
shouldbe suchthat the average
power quality, as measured,is also representativefor the substationsand feedersnot
monitored.Making such a fullyrepresentativechoice is very difficult ifnot impossible.
different from a sag
Sites come indifferent types,but it is hard to decide which sites are
viewpoint without first doing the survey. Afurther analysis ofdata from the current
generationof surveys will teach us more about the differences between sites. This
knowledge can be used for
choosingsites infuture surveys.
Some aspectsof power quality surveys and the way in which the
data can be
processed,are discussed below by using
datafrom four surveys:
• The CEA survey.A three-yearsurvey performedby the CanadianElectrical
Association (CEA). A total of 550 sites wasmonitored for 25 days each.
Residential,commercial, and industrial sites weremonitored at their 120V
or 347 V serviceentrancepanels.Approximately10% of the sites had metering
on primary side of the servicetransformerto provide an indication of the
power quality characteristicsof the utility's distribution system[54], [65], [66].
• The NPL survey. A five-year surveyperformedby NationalPowerLaboratory
(NPL). At 130 siteswithin the continentalUS and Canada,single-phase lineto-neutral data were connectedat the standardwall receptacle. The survey
resulted in atotal of 1200monitor monthsof data[54], [68], [69].
• The EPRIsurvey. A survey performedby the Electric Power Research
Institute
(EPRI) between June 1993 and September1995. Monitoring took place in
343
Section 6.3 • PowerQuality Monitoring
distribution substationsand on distribution feeders at voltages from 4.16 to
34.5 kV. Monitoring at 277 sites resulted in
5691 monitor monthsof data. In
mostcases threemonitorswere installed for each
randomlyselected feeder: one
at thesubstationand two atrandomlyselected places along the feeder
[54], [70].
• The EFI survey. The Norwegian Electric Power ResearchInstitute (EFI,
recently renamed"SINTEF Energy Research")has measuredvoltage sags
and other voltage disturbancesat over 400 sites in Norway. The
majority
(379) of the sites were at low-voltage
(230 and 400V), 39 of them were at
[67].
distribution voltages, and the rest at various voltage levels
The resultsof these surveys will be presented and discussed in the following
paperscited.
paragraphs.For more details about the surveys refer to the various
These are by far the only surveys,
but they were the ones for which detailed results
were available. With the exceptionof the EFI survey all the resultspresentedbelow
werepublishedin the internationalliterature.Especially the paper by
Dorr [54] contains
very usefulinformation. The amountof results published, even in
reports,is still very
limited. There must still be gigabytesof very interestingmonitoring data stored at
utilities all over the world, waiting to be processed.numberof
A
observationscan be
made from thevarious surveys, someof which are mentionedbelow. To explain or
check all this,further analysis of thedatais needed.
6.3.1.1 MagnitudeVersus Duration: CEA Survey. The cumulative number of
6.7 for
sags per year, as
o btainedfrom the CEA survey is shown in Tables 6.6 and
primary as well assecondaryside of the servicetransformer.Bar charts of the sag
density function are shown in Figs. 6.20 and 6.22. A voltage sag
coordinationchart
for the secondaryside datais shown in Fig. 6.21.
TABLE 6.6 CumulativeVoltage Sag Table for CEA SecondarySide Data: Numberof Sags perYear
Duration
Magnitude
I cycle
6 cycles
10 cycles
20 cycles
0.5 sec
1 sec
90%
80%
70%
500/0
10%
98.0
19.2
14.4
10.5
6.5
84.0
9.2
5.7
3.5
2.8
84.0
9.2
5.7
3.5
2.8
67.3
5.5
4.4
3.2
2.8
63.8
5.0
4.2
3.2
2.8
35.8
3.2
3.1
2.8
2.6
2 sec
6.6
2.3
2.3
2.2
2.1
Source: Data obtainedfrom Dorr et al. [54].
TABLE 6.7 CumulativeVoltage Sag Table for CEA Primary Side Data: Numberof Sags perYear
Duration
Magnitude
I cycle
6 cycles
10 cycles
20 cycles
90%
80%
20.3
12.0
9.4
4.8
3.1
11.2
5.8
3.6
1.2
1.2
10.8
5.4
3.3
1.2
1.2
5.5
3.2
2.0
1.1
1.1
700~
500/0
10%
Source: Data obtainedfrom Dorr et al. [54].
0.5 sec
5.2
3.1
1.9
1.1
1.1
I sec
1.9
0.9
0.7
0.7
0.7
2 sec
1.3
0.7
0.7
0.7
0.7
344
Chapter6 • Voltage Sags
-StochasticAssessment
30.0
25.0
:a
...;".,
"e,
20.0
.....0~
15.0
'"
OJ)
...
'"
1
10.0
Z
5.0
.,J§'
10-50% ~'Ir~
50-70%
0-10%
Duration in seconds
Figure 6.20 Sagdensityfunction for CEA secondaryside data,correspondingto
Table 6.6.
80
---
::::--:::
~
::::::::::
I-----
50
20 10 sags/year
/'i/ Wi
80%
/
/
90%
70%
t
~
50%
17 ms
lOOms
167 ms
333 ms
0.5 s
Duration
I s
2s
10%
10 s
Figure 6.21 Voltage sag coordinationchart for CEA secondaryside data,
correspondingto Table 6.6.
We seethat the numberof sags onsecondaryside is significantly highert han the
numberof sags onprimary side.Partof the secondaryside sagsoriginatesat secondary
side, i.e., within thecustomerpremises. The largenumber of long shallow sags at
secondaryside can be explained as
motor starting on secondaryside. As we saw in
Section 4.9, these sags are not
noticeable(i.e., magnitudeabove90%) on primary side
of the transformer.
Section 6.3 •
PowerQuality Monitoring
345
30
25
5
o
Duration in seconds
Figure 6.22 Sag dens ity of
primary side CEA data,correspondingto Table 6.7.
Anotherinterestingobservationis the largenumberof deepshortsags (0-100 ms,
0-50%). The numberis less onsecondaryside, but still significant. Acomparisonwith
othersurveys showsthat this is a typical feature of the
C EA survey.Furtheranalysisof
the data is needed to explain this.
With any interpretationof the CEA primary side data one should also consider
the uncertaintyin the results. Asmentionedabove, about 10% of the 550 sites was
located onprimary side of adistribution transformer. As each site wasmonitoredfor
only 25 days, this resulted in only 3.7
monitoring-yearsof data.The uncertaintyin sag
of two for each of the bins in the sag density table . In the
frequency is at least a factor
CEA secondarysidedatathe uncertaintyis smaller as theamountof datais equivalent
to 38 monitor years.
6.3.1.2 MagnitudeVersus Duration: NPL Survey. The number of sags per
year, asobtainedfrom the NPL survey, is shown incumulative form in Tables 6.8
and 6.9. Table 6.8 shows the original
data, where eachindividual event iscounted,
even if they are due to the same reclosure cycle. In Table 6.9
5-minute
a
filter is applied: all events within 5 minutes are
countedas one event: the one with the worst
magnitudebeing the onecounted.The sag densities are shown in Figs. 6.23 and 6.24
without and with filter, respectively. A voltage sag
coordinationchart for the filtered
data is shown in Fig. 6.25.
ComparingFigs. 6.23 and 6.24, we see
that there is somereductionin the number
of shortinterruptions(voltage below 10%) as
alreadydiscussed inChapter3. The most
serious reduction is the number of long, shallow sags, the ones
attributed to load
switching. Apparentlyload switching sags come in clusters , with on average
about 15
events within 5 minutes. This clearly
distortsthe quality of supply picture asdrawn by
346
Chapter 6 •
Voltage Sags-Stochast
ic Assessment
TABLE 6.8 Cumul ative Voltage Sag Table for
NPL Data Without Filter:
Numberof Sags per Yea r
Duration
Magnitude
1 cycle
6 cycles
351.0
59.5
31.4
20.9
15.5
259.8
32.3
23.2
18.3
15.2
87%
80%
70%
50%
10%
10 cycles 20 cycles
0.5 sec
157.9
19.0
17.1
15.4
14.1
134.0
16.2
15.2
14.1
13.2
211.9
23.7
19.4
16.8
14.9
I sec
2 sec
10 sec
108.2
13.1
12.7
12.2
11.8
90.3
10.4
10.3
10.2
9.9
13.7
5.8
5.8
5.8
5.7
Source :Data obtained from Dorr et al. [54).
TABLE 6.9 CumulativeVoltage SagTable for NPL Data with 5-minute
Filter : Numberof Sags per Year
Duration
Magnitude
I cycle
6 cycles
126.4
44.8
23.1
15.9
12.2
56.8
23.7
17.3
14.1
12.0
87%
80%
70%
50%
10%
10 cycles
36.4
17.0
14.5
12.9
11.7
20 cycles 0.5 sec
27.0
13.9
12.8
11.8
11.0
23.0
12.2
11.5
10.6
10.2
2 sec
I sec
18.1
10.0
9.7
9.4
9.0
14.5
8.0
7.9
7.8
7.5
Source:Data obtained from Dorr et al. [54).
80
70
...
"'"
...>-
60
'0."
50
....0~
40
'"
l>
e
30
:s
Z
20
10
50-70% ..,s>"O'lJ
10-50%
0-10%
o f NPL data,no filter, corresponding to Table 6.8.
FIgure 6.23 Sag density
~'!1q
10 sec
5.2
4.3
4.3
4.3
4.2
Secti on 6.3 • PowerQuality Monitoring
347
80
70
Ii!
.,
>.
.,...
Co
.,
60
50
bO
.,
'"
...
'0
~
~
Z
40
30
20
10
o f NPL data, 5-minute filter, correspondingto Table 6.9.
Figure 6.24 Sag dens ity
20
10
sags/year
F-."""""'=-r"""t--,,...,:==-t----j----+-----ji"""""---t----'-----'---j 80%
f--.,-:==-t--- - f - - - - + -- -+----f-1f-- - - + - - - - j 70%
1
~
::8
f - - - - f - - - - f - - - - +---+--+--1f----+- - - - j 50%
L-_
17 ms
_
--!
100 ms
--'-
167 ms
-1-
...e..-.'--_
333 ms
0.5 s
Duration
_
!--_ _-+
1s
2s
-' 10%
10 s
Figure 6.25 NPL data: voltage sagcoordination chart, 5-minute filter,
corresponding to Table 6.9.
the survey.F urtherinvestigation of the datais needed to find out whether most
starting
events areclusteredor whetherit is all due to a smallnumberof sites. Acomparison
between theNPL dataand the CEAdatashows a much larger
numberof events for the
former . The most likelyexplanationis the much lower lightning activity inCanadaas
comparedto the United States .
348
Chapter6 • VoltageSags-Stochastic
Assessmen1
6.3.1.3 MagnitudeVersusDuration: EPRI Survey. The cumulative number of
and 6.11.
sags per year, as
o btainedfrom the EPRI survey, is shown in Tables 6.10
Table 6.11 gives the results forsubstations,while Table 6.10 isobtainedfrom measurementsalong feeders.For both tables a5-minutefilter was applied. The sag density function is shown in Figs. 6.26 and 6.28.Figures 6.27 and 6.29 give the
correspondingvoltagesagcoordinationcharts.
The differences between the feeder
dataand thesubstationdataare small: in total
only seven events per year, whichabout
is
10% (this is the value in theupper-leftcorner
magnitude-duraof the tables).The seven-eventdifference is found in two areas in the
tion plane:
• Eventsup to 10 cycles withmagnitudesbelow 700/0. Here we find 13.6 events
for the feeders,b ut only 8.3 for thesubstation.
substation,5.1 for the
• Interruptionsof 1 second and longer: 3.4 events for the
feeder.
Where thetotal numberof events isremarkablysimilar, the relative difference in the
numberof severe events is significant.
Table6.12comparesthe numberof events below
certain voltage levels, including events recordedat low voltage (NPL survey). Only
events with aduration lessthan 20 cycles(about 300ms)are 'included in thecomparison: i.e.mainly events due toshort circuits. Looking at Table 6.12 we see moreinter..
ruptions and deep sags on the feeder comparedto
as
the substation.The increased
TABLE 6.10 CumulativeVoltage Sag Table for EPRI FeederData
with 5-minute Filter: Numberof Sags perYear
Duration
Magnitude
I cycle
6 cycles
10 cycles
90%
80%
70%
50%
10%
77.7
36.3
23.9
14.6
8.1
31.2
17.4
13.1
9.5
6.5
19.7
12.4
10.3
8.4
6.4
20 cycles
13.5
9.3
8.3
7.5
6.2
I sec
2 sec
10 sec
7.4
6.4
6.2
5.9
5.1
5.4
4.9
4.8
4.6
4.0
1.8
1.7
1.7
1.7
1.7
0.5 sec
I sec
2 sec
10 sec
8.6
5.6
4.9
4.4
3.9
5.4
4.3
4.0
3.8
3.4
3.7
3.2
3.0
2.9
2.5
1.5
1.4
1.4
1.4
1.4
0.5 sec
10.7
7.9
7.2
6.6
5.6
Source: Data obtainedfrom Dorr et at. [54].
TABLE 6.11 CumulativeVoltage Sag Table for EPRI SubstationData
with 5-minute Filter: Numberof Sagsper Year
Duration
Magnitude
90%
80%
70%
50%
100/0
I cycle
6 cycles
10 cycles
70.8
29.1
16.1
7.9
5.4
28.1
14.7
9.8
6.6
5.2
17.4
10.1
7.8
6.1
5.1
Source: Data obtainedfrom Dorr et al. [54].
20 cycles
11.4
7.1
6.0
5.3
4.7
349
Section 6.3 • Power QualityMonitoring
30
25
:a
...>.
0.
.,
bll
.,
....0
...
.&J
§
Q)
Q)
20
~
15
Q)
10
Z
5
0
Figure 6.26 EPRI feeder data : sag density function , correspond ing to Table 6.10.
50
rrT"rrrTrTTTr---r-
20
10
...,,-- - , - , - -- - - ,r-r- -
5 sags/year
- ,--....::....--n------,- 90%
~"....r£"....r£'_A----r'=-+--T+------1I-----+----(t-------j -
80%
f--+--A---+~--+-----I----t-----j'+---+
70% . ~
]
~
:::E
~--_A---+---+-----I----t--+-+---+ 50%
L -_ _
17 ms
~
100 ms
......L
167ms
-l--_
_
333 ms
----'
0.5 s
-+-L__ _--'-_ _-----l 10%
2s
1s
10 s
Duration
Figure 6.27 EPRI feeder da ta: voltage sag coordination chart
correspond
,
ing to
Table 6.10.
numberof interruptionsis understandable
: someinterruptionsonly affect part of the
feeder; the closer to the
equipment,the higher thenumber of interruptionssimply
because the path
t hat can be interrupted is longer.
For the increase in thenumberof
deepshort sags there is no ready
explanation.Three possibleexplanations,which will
probablyall somewhatcontribute, but for which more investigations are needed to give
a definiteexplanationare:
Chapter6 • Voltage Sags- Stochastic Assessment
350
30
25
5
50-70%
o
10-50%
.J'¢)
~i'
0-10%
Figure 6.28 EPRI substation data : sag density function
correspond
,
ing to Table
6.1 1.
50
20
10
5 sags/year
r-r-r=-.l'~---+'~--+----+---r""---+----+----\·
80%
.g
a
h<:=:::..--.....,f=--- - + -- - + - - - - I - + - - - - + -- - - + - - - - \·70% '§,
os
~
1----- - - + - - - + - - - + - - + -- + - -- - + - - - - + - - - - \·50%
10%
10 s
' - - - - -- - ' - -- ---+--""'' - - - - ' - - - - ' - - - - - - ' -- - - - ' - - - --'-.
17 ms
100ms
167 ms
333 ms
0.5 s
Duration
1s
2s
Figure 6.29 EPRI substation data : voltage sag coord ination
chart,corr esponding
to Table 6.11.
• Reclosing actions on the feeder beyond the point where the monitor is connected. Themonitor on the feeder will record a deeper sag than the one in the
substation. This would explain the deep
short sags. As thedistribution transformer is often Dy-connected, deep sags due to single-phase faults will not
transfer fully to low voltage. This explains the smaller
numberof deep short
sags measured at low voltage (NPL survey).
351
Section 6.3 • PowerQuality Monitoring
TABLE 6.12 Numberof Events with aDurationLess than20 Cycles: NPL Survey (LV) andEPRI Survey (Feeder,
Substation)
Events per Year
Distribution
Voltage Range
80-900AJ
70-800/0
50-70%
10-50%
0-10%
LV
68.5
20.6
6.2
2.9
1.1
Feeder
Substation
37.2
11.4
8.5
5.8
1.9
37.4
12.0
7.5
1.9
0.7
Source: Data obtainedfrom Dorr et al. [54].
• The normal operatingvoltage at the feeder is lower. As the sag
magnitudeis
given as apercentageof the nominalvoltage, the sag willappeardeeper at the
feederthan at thesubstation.Giving the sagmagnitudeas apercentageof the
pre-event voltage wouldcompensatethis effect. This may explain the increase
in the numberof shallow sagsalong the feeder.
• Induction motor influence.Induction motorsslow down more for deeper sags
and thus reduce the positive sequence voltage.
reductionin
A
positive sequence
voltage would imply areduction(also) in the lowest phase voltage and thus a
reductionin sagmagnitude.
Comparinglow voltage andmedium voltage data we seethat the numberof shallow
sags is much higher at low voltage
thanat mediumvoltage, whereas the
numberof deep
sags is smaller at low voltage.
6.3.1.4 MagnitudeVersus Duration: EFI Survey. The cumulative voltage sag
tables, asobtained by the EFI survey, are shown in Tables 6.13
through 6.16. The
sag densityfunctions are presentedin Figs. 6.30through 6.33. Table 6.13 and Fig.
6.30 give the average results for the
low-voltage sites, Table 6.14 and Fig. 6.31 refer
to the distribution sites.
We seethat the averagedistribution site experiencessomewhatlesslonger-duration events but clearly more
s hort-durationevents. The increase numberof
in
interruptions for lower voltage levels isconsistentwith the findings of U.S. surveys. To
TABLE 6.13 CumulativeVoltage Sag Table for EFID ata, All Low-Voltage
Networks: Numberof Sags per Year
Duration (sec)
Magnitude
90%
700/0
40%
1%
0.01
0.1
0.5
1.0
3.0
20.0
74.7
26.3
16.6
9.3
36.5
11.9
9.8
8.2
18.5
8.2
7.5
7.5
12.1
7.5
7.5
7.5
8.6
6.8
6.8
6.8
5.9
5.9
5.9
Source: Data obtainedfrom Seljeseth[67].
6.8
352
Chapter6 • VoltageSags-Stochastic
Assessment
TABLE 6.14 CumulativeVoltage Sag Table for EFI Data, All Distribution
Networks: Numberof Sags perYear
Duration (sec)
Magnitude
90%
70%
40%
1%
0.01
0.1
0.5
1.0
3.0
20.0
112.2
40.5
15.2
7.2
39.2
16.9
7.6
5.7
15.5
11.4
6.8
5.7
7.9
6.6
6.0
5.7
6.0
6.0
5.7
5.7
5.2
5.2
5.2
5.2
20.0
Source: Data obtainedfrom Seljeseth [67].
TABLE 6.15 CumulativeVoltage Sag Table for EFI Data, 950/0 Percentile
for Low-Voltage Networks: Numberof Sags perYear
Duration (sec)
Magnitude
0.01
0.1
0.5
1.0
3.0
90%
70%
40%
10/0
315
120
128
39
25
11
47
II
11
11
20
11
11
11
11
11
11
11
11
11
11
66
25
II
Source: Data obtainedfrom Seljeseth [67).
TABLE 6.16 CumulativeVoltage Sag Table for EFI Data, 95% Percentile
for ·Distribution Networks: Numberof Sagsper Year
Duratjo~ (sec)
Magnitude
0.01
0.1
90%
70%
40%
1%
388
130
45
18
159
53
21
12
0.5
1.0
3.0
20.0
57
22
12
12
20
12
12
12
12
12
12
12
12
12
12
12
Source:Data obtainedfrom Seljeseth [67].
understandall effects, one needs to
understandthe propagationof sags to lower voltage
levels, for which thestudy of more individual events is needed.
Tables 6.15 and 6.16 give the950/0 percentile of the sagdistribution over the
various sites. A stochasticdistribution function was createdfor the total numberof
sagsmeasuredat one single site. The
95% percentileof this distributionwas chosen as a
5%the sites.
reference site. Then umberof sags at this site is thus exceeded by only of
The 95% value was suggested in
Chapter1 as a way ofcharacterizingthe electromagnetic environment(the term used by thel Ee for the quality of the supply). Thus, we
could say that Table 6.15 characterizesthe electromagneticenvironment for the
Norwegianlow-voltagecustomer.
6.3.1.5 Variation in Time-LightningStrokes. A large fraction of the voltage
sags is due tolightning strokeson overheadlines. Two phenomenaplay a role here:
short circuits due to lightning strokesand triggering ofspark gaps due to lightning-
Section 6.3 • PowerQuality Monitoring
353
50
45
40
.,til
..
.,
0.
>.
35
., 30
OIl
., 25
....'"
.,
..
20
§
15
0
~
Z
70-90%
10
40-70%
5
~q
1-40%
0
~
,s.'/!!
's
~
e,'bo"Jo
Sag duration in seconds
Figure 6.30 Sagdensity for EFI low-voltagenetworks,correspondingto Table
6.13.
50
45
40
..~
>.
.,0.
.,
OIl
.,
....0'"
...,
35
30
25
20
~
§
Z
15
70-90%
10
40-70%
5
1-40%
0
$'
e,'bo"Jo
Sag duration in seconds
Figure 6.31 Sagdensity for EFI distribution networks.correspondingto Table
6.14.
~
.,s.'/!!
~q
Chapter6 • Voltage Sags- Stochastic Assessment
354
160
140
:....
;...
.0....
..
120
100
VI
bO
....'0"
VI
.....
.D
80
60
E
::l
Z
70-90%
40
~
40-70%
20
~'tS
~~
<$'
's
1-40%
0
~"'~
Sag duration in seconds
20-180
Figure6.32 Sag density for 95% percentile of EF I low-voltage networks,
correspondingto Table6.15.
160
140
:.
...
.....
;...
120
100
0..
VI
bO
....'"0
80
~
60
Z
40
VI
..
§
70-90%
40-70% ~
20
.s>
1-40%
0
~~
<$'
~"'~
Sag duration in seconds
20-180
Figure 6.33 Sag density for 95% percentile of EFI distrib ution networks,
corresponding to Table 6.16.
355
Section 6.3 • Power Quality Mon
itoring
induced overvoltages. The effect of a lightning stroke is to induce a large overvoltage
on the line. If this voltage exceeds the
insulation withstand level it results in a short
circuit, otherwise the voltage peak will
start to propagatethrough the system. If the
peak voltage is not high enough to cause a flashover on the line, it might still trigger
a spark gap or a (ZnO)
varistor. A sparkgap mitigates the overvoltage by creating a
temporaryshort circuit, which in its turn causes a.sag of one or two cycles. A varistor will only cap the overvoltage. Aconclusionfrom one of the first power quality
surveys[72] was that the number of voltagetransientsdid not increase in areas with
more lightning; instead the number of voltage sags increased.
For a few sites in the EPRI survey, the sag frequency comparedwith
was
the
lightning flash density[70]. This comparisonshowedthat the correlationbetween sags
and lightning was much stronger than expected.
Plotting the sag frequency against the
2
flash density (numbero f lightning flashes per km
per year) for five sites resulted in
of
almost a straight line. This justifies the conclusion
that lightning is the main cause
voltage sags in U.S.distribution systems.
As sags are correlated with lightning and lightning activity varies with time, we
expect the number of sags to vary with time. This is shown in Fig. 6.34 for the NPL
survey[68]. The sag frequency is at its maximum in summer, when also the lightning
activity is highest. This effect has been confirmedothercountries.
in
Also the distribution of sagsthroughthe day follows the lightning activity, with its peak in the evening.
18
16
14
E 12
'"
>
....'"0
fl'"
s::
'"g
e,
'"
10
OJ)
8
6
-
-
-
.-
:?i;;~
f ";).
4
2
-
~
I!
..,
h ~
:~~
0
~'.!:.-"
Jan
'-
ff41
.~~
,....--
f--
1-
:f\,'-!$1.
Feb March April May June July Aug
Month of the year
Sept
Oct
Nov
Dec
Figure 6.34 Variation of voltage sag frequency
through the year .(Data obtainedfrom Dorr [68J.)
6.3.1.6 Correcting for Short Monitoring Periods.The variation of the sag frequency through the year indicates
that the monitoring period should be at least I
year to get a good impression of the power quality at a certain site. As weather activity varies from year to year, it is even neededmonitor
to
several years. In case a limof the
ited monitoring period is used, it is still possible to get a rough estimate
[49]. To do this, faultdata are needed
average number of sags over a longer period
of time.
over themonitoring period as well as over a longer period
356
Chapter6 • VoltageSags-Stochastic
Assessment
The basicassumptionbehindthe correctionmethodis that voltagesags are due to
short circuits: thus that the numberof sags isproportional to the numberof shortcircuit faults. In equationform this readsas
N sags Njaults
N
sags= ~
faults
(6.7)
where N.r;ag.'l and Nfaults are thenumberof sagsand faults, respectively,recordedduring
the monitoring period, and Nsag.'l and Njaults the (average)number during a longer
period of time. The numberof sags over alonger period of time can thus beobtained
from
Njaults
N sags -- N sags x
N- -
(6.8)
faults
Ideally, one would like toknow the numberof faults in theareaof the system in which
the sagsoriginate. Often this information is not available: one is likely to only have
fault data over the whole servicearea of the utility. This method also neglects the
above-mentionedshort-durationsags due totriggering of overvoltagedevices and
sags due totransientfaults which are not recorded.
The correction method can beimproved if the sags can betraced back to the
voltage levels at which theyoriginated:
N
sags=
L[
I
Fli)]
faults
sags X N(')
N(i)
(6.9)
faults
with N.~2gs the numberof sagsduring the monitoring period originatingat voltage level
i, etc. In most cases it will
not bepossibleto traceback all sags. Only for a small
number
of sites thismethodmight be suitable.It has been used in [49] to
q uantify the average
supply performancein Japan.
6.3.1.7 Variation in Space. The basic assumptionof a large power quality survey is that the averagepower quality, over a number of sites, givesinformation
aboutthe power quality for each individual site. Thus, if the conclusionof the survey
is that there are onaverage25 sagswithin a certain magnitudeand duration range,
this number should at least be anindication of the numberof sags at anindividual
site, in an individual year. Obtaining information about the differences between different sites is difficult;partly becausemainly the averageresults have been
published;
partly because differences
betweensites arenot always statistically significant after a
short monitoring period.
Someindication of the differencebetween sites iso btainedfrom the EFI survey.
95% site and the averageof all sites is very large, as can be
The difference between the
seen bycomparingTables6.13 and 6.15. At least5% of the sites haveaboutfour times
as many sags as theaverageof all sites. For those sites theaveragevalues donot give
much usefulinformation. The problem is that without a prior study it is difficult to
know whetherthe averagedataappliesto a certainsite. Furthersplitting up thedataset
in different types of sites, e.g., systems with
mainly overheadlines and systems with
mainly undergroundcables, canreducethe spreadamongthe sites within onegroup.
But reducing thedataset will' also increasethe statisticalerror in the estimates.
Information on the spreadin power quality amongdifferent sites is also given in
[72]. Sags and someo ther voltage disturbanceswere measuredat 24 sites from May
1977through September1979,leading to a total of 270 monitor-monthsof data. The
357
Section 6.3 • PowerQuality Monitoring
TABLE 6.17 Distribution Over the Sitesof the Numberof Sagsand
Interruptions
Maximum Numberof Sags LongerThan the IndicatedDuration
Number of Sites
10%
250/0
50%
75%
900/0
I cycle
100ms
200 ms
0.5 sec
I sec
II
6
9
3
5
13
19
26
8
2
3
5
12
17
8
0
2
3
5
12
8
17
25
36
51
Source: Data obtainedfrom [72].
total amountof dataof this survey is not very large, but the
monitor period at each site
is long enoughto make some comparisonbetween the different sites. Some of the
results are shown inT able 6.17. This table gives, for various minimum
durations,the
maximum number of sags andinterruptionsfor a certain percentageof sites. As an
example:25°~ of the sites has fewerthan five events per year longer
than 200 milli11 and 51 events per year longer
than one
seconds. Also:80% of the sites has between
cycle in duration,the remaining20% of sites are outsideof that range.For about half
of the sites themedianvalue is areasonableindicatorof the numberof sagsthat can be
500/0
expected. Asalreadymentionedbefore, it ishard to know if a site belongs to the
average sites or not,without monitoring the supply.
8.3.2 IndividualSites
Monitoring is not only usedfor large power quality surveys, it is also used for
assessing thepower quality of individual sites.For harmonicsand voltagetransients,
reliable results can be
obtainedin a relativelyshortperiod of time. Someinterestingsite
surveys inCanadianrural industry have beenperformedby Koval [58]. One of the
conclusionsof his studieswas that a monitoring period of two weeks gives a good
that this
impressionof the power quality at a site[59]. Again it needs. to be stressed
holds only for relativelyfrequentevents like voltagetransientsand motor startingsags
and for phenomenalike harmonicsand voltagefluctuation. Voltage sags andinterruptions of interest for compatibility assessment have occurrence frequencies
of once a
month or less.Much longer monitoring periods are needed for those events.
6.3.2.1 The Required Monitoring Period.To estimate how long the
monitoring
period needs to be, we assume
that the time-between-events exponentiallydistribuis
ted. This meansthat the probability of observing an event, in let's say the next minute, is independentof the time elapsed since the last event.
Thus, events occur
completelyindependentfrom each other.Under that condition the numberof events
capturedwithin a certain period is a stochasticvariable with a so-called Poisson distribution.
numberof
Let Jl be the expectednumberof events per year, then the observed
eventsK, over amonitoringperiod of n years is a discrete
stochasticvariablewith the
following distribution:
(6.10)
358
Chapter6 • VoltageSags-Stochastic
Assessment
This Poissondistribution has anexpectedvalue nil anda standarddeviation ..jifii. The
result of monitoringis an estimateof the expectednumberof events per year,obtained
as follows:
K
(6.11)
Ilest =-
n
This estimatehas an expectedvalue JL (it is a true estimate)and a standarddeviation
~. For a largeenoughvalue of nil (i.e., for a sufficientnumberof observedevents)the
Poissondistributioncan be approximatedby a normaldistributionwith expectedvalue
J-L and standarddeviation ~. For a normal distribution with expectedvalue J-L and
standarddeviation (J the so-called95% confidenceinterval is betweenIl - 1.96(1 and
JL + 1.96(1,with (1 the standarddeviation.The relativeerror in the estimateof JL after n
samplesis thus,
1.96(1
1.96
2
(6.12)
-,;- = ..jifii ~ ,IN
with N = nil the expectednumberof events inn years, i.e., in the wholeobservation
period. To limit the relativeerror to E the monitoringperiod n shouldfulfill the following inequality:
2
(6.13)
--<E
~
or
4
(6.14)
n > -2
J-LE
For an eventwith a frequencyof JL times per year, themonitoringperiod shouldbe at
least ~
yearsto obtain an accuracyE.
/-U
Table 6.18 gives theminimum monitoring period for various event frequencies
and accuracies.N ote that sag frequenciesare ultimately used topredictequipmenttrip
frequencies.It showsthat site monitoringcan only giveaccurateresultsfor very sensitive equipment(high frequency of tripping events).When equipmentbecomesmore
compatiblewith the supply (and thus trips lessoften) site monitoringcan no longer be
used topredict the numberof trips.
As mentionedbefore, the approximationof a Poissondistribution by a normal
distributionholdsfor a sampleof large size.N othingwas saidaboutwhat this large size
is. A more accurateexpressionfor the uncertaintyis obtainedby using theso-called
Student'st-distribution. Using this distributiongivesanotherfactor in (6.12) insteadof
1.96.The deviationis small: for 10eventswe find afactor of 2.228, which is anincrease
of 14%; for five eventsthe value is 2.571.F or 16 events(50 % accuracyaccordingto the
TABLE 6.18 Minimum Monitoring Period Needed toObtain a Given
Accuracy
Event Frequency
50°At Accuracy
10% Accuracy
2% Accuracy
I per day
I per week
I per month
1 per year
2 weeks
4 months
I year
16 years
I year
7 years
30 years
400 years
25 years
200 years
800 years
10,000 years
Section 6.4 • TheMethod of Fault Positions
359
approximation)the Student'st-distributiongives anaccuracyof 53%. The effecto f this
on Table 6.18 is small.
6.3.2.2 More Uncertainties. The abovereasoningassumes astationarysystem
with exponentiallydistributed times between events, thus where events
appearcompletely at random. For a stationary system it is possible toobtain the event frequency with anyrequiredaccuracy byapplying a long-enoughmonitoring period. In
that monitoring results
the actual situation there are two more effects which make
have a limited predictive value:
lightning, heavy wind,
• A large fraction of voltage sags is due to bad weather:
snow, etc. The sag frequency thereforenot
is
at all constantbut follows the
annual weatherpatterns.But the amountof weatheractivity also varies significantly from year to year. Due to the
relation betweenvoltage sags and
adverseweather,the sags come in clusters. To getcertain
a
accuracyin the
estimate,one needs to observe more
than a minimum numberof clusters. It is
obviousthat this will increase therequiredmonitoring period. To get a longterm average a long
monitoringperiodis needed. Acorrectionmadeaccording
to (6.8) might increase the accuracy.
• Power systems themselves are not static but change
continuouslyfrom year to
year. This especially holds for
distribution networks. The numberof feeders
connectedto a substation·can change; ora notherprotectiverelay is used. Also
componentfailure rates can change, e.g., due to aging;
increasedloading of
components;different maintenancepolicies; or because the
amountof squirrels
in the areasuddenlydecreases.
Despite thesedisadvantages,site monitoring can be very helpful in finding and
hard to predict. In
solving power quality problems,as some things are simply very
addition, stochasticassessment requires certain
a
level of understandingof voltage
disturbancesand their origin. Thisunderstandingcan only be achievedthroughmonitoring.
8.4 THE METHOD OF FAULT POSITIONS
8.4.1 Stochastic Prediction Methods
The great advantageof stochasticprediction as comparedto monitoring is that
the required accuracyis obtained right away. With stochasticprediction it is even
possible to assess the power
quality of a systemthat does not yet exist; something
which is impossible to achieve by power
quality monitoring.
Stochasticprediction methodsuse modeling techniquesto determineexpected
value, standarddeviation, etc., of a stochasticvariable. With' stochasticpredictions
one should not think of a prediction like a voltage sag down to35% will occur at
7:30 in the evening on July21. Instead,the kind of predictionsare more like in July
one canexpect10 sags below 70%,halfofwhich areexpectedto occur between5 and 9 in
the evening.
Stochasticpredictionmethodshave been used for many yearspredictfrequency
to
and duration of long interruptionsas discussed in detail in
C hapter2. For shorter
duration events, the useof stochasticprediction techniquesis still very uncommon.
360
Chapter6 • VoltageSags-Stochastic
Assessment
Those events tend to have a higher
occurrencefrequency, making monitoring more
feasible. Also the required electrical models have a higher
complexity than for long
interruptions.A final explanationis that power quality is still very much anindustrydriven area, whereas reliability
evaluationis much more auniversity-drivensubject.
Stochasticpredictionmethodsare asaccurateas the model used and as
accurate
as thedataused. The accuracy
of the models can be influenced; the accuracy
of the data
is often outsideour control. Any stochasticpredictionstudy in power systems requires
two kinds of data:power systemdata and componentreliability data. The main data
concern is thelatter one. Componentreliability data can only beobtainedthrough
observing thebehavior of the component.From a stochasticpoint of view this is
identical to the powerquality monitoring of one individual site we discussed earlier.
Componentreliability data has therefore the sameuncertaintiesas the outcomeof
power quality monitoring. One could now betempted to draw the conclusion that
we did not gain anythingby usingstochasticprediction.This conclusionis fortunately
not correct. Many utilities have records ofc omponentfailures over several decades.
Componentsdo not need to beconsideredseparatelybut can begroupedinto "stochastically identical" types: like alldistribution transformers.This enormouslyreduces
the error in the componentfailure rate.
Someproblemsremain of course:maintenancemethodschange; the failure rate
of
new componentsis hard to assess;c omponentloading patternscan change; even
weatherpatternsare prone to change. The same
uncertaintiesare presentwith power
quality monitoring, but with stochasticassessment one is able somewhatassess
to
the
influence of theseuncertainties.
8.4.2 Basics of the Method of Fault Positions
The method of fault positions is a straightforward method to determine the
expectednumberof sags. It wasproposedindependentlyby a numberof authorsbut
probablyfirst used byConrad[48] whose work has become
part of IEEE Std-493 [8],
[21]. The methodis also used byEdF (Electricite deFrance)to estimatethe numberof
sags due to faults in their
distribution systems [60]. Themethodof fault positionswas
combinedwith Monte Carlo simulationby the authorin [61], [63], extendedwith nonrectangularsags due tomotor re-accelerationin [18], [62] and extendedwith generator
outagesin [64]. At least onecommercialsoftwarepackageis availableusing themethod
of fault positions. Morepackageswill almostcertainlyfollow as themethodis computationally very simple,althoughit often requires excessive
calculationtime. The accuracy of the results can be increased increasingthe
by
numberof fault positions.Nonrectangularsags can betaken into account by using dynamic generatorand load
models; phase-angle
j umps by working with complex impedancesand voltages; threephaseunbalanceby including single-phase andphase-to-phase
faults.
6.4.2.1 Outlineof the Method. The method of fault positions proceeds, schematically, as follows:
• Determinethe area of the system in which
short circuits will be considered.
• Split this area into smallparts. Short circuits within one part should lead to
voltage sags with similarcharacteristics.Each smallpart is representedby one
fault position in an electriccircuit model of thepower system.
361
Section 6.4 • TheMethod of Fault Positions
• For each faultposition, the short-circuitfrequency isdetermined.The shortcircuit frequency is thenumberof short-circuitfaults per year in the small
part
of the systemrepresentedby a fault position.
• By usingthe electric circuit modelo f the power system the sag
characteristics
are calculatedfor each faultposition. Any power system model and any calculationmethodcan be used. The choice will depend on the availability of tools
and on thecharacteristicswhich need to be calculated.
• The results from the two previous steps (sag
characteristicsand frequency of
occurrence) arecombinedto obtainstochasticalinformationaboutthe number
of sags withcharacteristicswithin certainranges.
6.4.2.2 Hypothetical Example.Considera lOOkm line as shown in Fig. 6.35.
Short circuits in this part of the system arerepresentedthrougheight fault positions.
The choiceof the fault positionsdependson the sagcharacteristicswhich are of interest. In this example we
considermagnitudeand duration. Fault position I (representingbusbarfaults in the localsubstation)and fault position 2 (faults close to the
local substation)will result in the same sag
magnitude.But the fault-clearing time is
different, thereforetwo fault positions have been chosen. The fault positions along
the line (2, 3, 4, and 5) have similar
fault-clearingtime but different sag magnitude.
Fault positions6, 7, and 8 result in the same sag
magnitudebut different duration.
For each faultpositiona frequency, amagnitude,and adurationare determined,
as shown inTable6.19. Failure ratesof eight faults per 100kmof line per year and 10
faults per 100substationsper year have been used. It should be realized that
herenot all
fault positions along the linerepresentan equal fraction of the line: e.g., position 5
represents 25 km (between
5/8th and 7/8th of the line) but position 6 only 12.5km
18th and 1).
(between 7
The resulting sags (1
through8 in Table 6.19) are placed in bins or immediately in
a cumulative form.Table6.20 shows how the various sags fit in the bins. Filling in the
frequencies (failure rates) leads to Table 6.21 andcumulativeequivalentshown
its
in
Table 6.22.Alternatively it is possible toupdatethe cumulative table after each fault
8
3
4
5
l
6 .-..---
Figure 6.35 Part of power system with fault
positions.
Load
TABLE 6.19 Fault Positions with ResuJtingSag Magnitude and Duration
Fault Position
I
2
3
4
5
6
7
8
Busbar fault in local substation
Fault on a line close to local substation
Fault at 25%. of the line
Fault at 50% of the line
Fault at 75% of the line
Fault at 1000/0 of local line
Fault at 0% of remote line
Busbar fault in remote substation
Frequency
Magnitude
Duration
O.ljyr
4jyr
2/yr
2/yr
2/yr
l/yr
2/yr
O.l/yr
%
0
0%
320/0
180 ms
80 ms
49%
57%
105 ms
110ms
250 fiS
64%
64%
64%
90 ms
90 ms
180ms
362
Chapter6 • VoltageSags-Stochastic
Assessment
TABLE 6.20 Fault Positions Sorted for Magnitude and
Duration Bins
60-80%
40-60%
20-40°A>
0-200/0
0-100 ms
100-200 ms
200-300ms
7
8
4 and 5
6
3
2
TABLE 6.21 Table with Event Frequencies for Example of Method
of Fault Positions
60-80%
40-60%
20-40%
0-20%
0-100 IDS
100-200 ms
2.0
0.1
4.0
2.0
4.0
200-300IDS
1.0
0.1
TABLE 6.22 Cumulative Table for Example of Method of Fault Positions
800/0
600/0
40%
20o~
oIDS
100 ms
200 ms
13.2
10.1
6.1
4.1
5.2
4.1
0.1
0.1
1.0
0.0
0.0
0.0
position. As we have seen inSection6.2 this is neededanywaywhen non-rectangular
sags areconsidered.Pleasenote that this is acompletelyfictitious example.No calculaobtain the magnitudeand durationsin Table 6.19.
tion at all has been used to
6.4.3 Choosing the Fault Positions
The first step inapplying the methodof fault positionsis the choiceof the actual
fault positions. It will be obvious that to obtain more accurateresults, more fault
positions are needed.But a random choice of new fault positions will probably not
increasethe accuracy,only increasethe computationaleffort.
Threedecisionshave to bemadewhen choosingfault positions:
applying
1. In whichpart of the power system do faults need to be applied? Only
faults to one feeder iscertainly not enough; applyingfaults to all feeders in
the wholecountryis certainlytoo much. Some kindof compromiseis needed.
This questionneeds to beaddressedfor each voltagelevel.
2. How muchdistancebetween fault positions is needed? Do we only need fault
positionsin the substationsor also eachkilometeralongthe lines?Again this
questionneeds to beaddressedfor eachvoltagelevel.
For each fault position, different events
3. Which events need to be considered?
can beconsidered.One can decide to onlystudy three-phasefaults, only
363
Section 6.4 • TheMethod of Fault Positions
single-phasefaults, or all types of faults. One can considerdifferent fault
impedances,d ifferent fault-clearingtimes, or different schedulingof generators, eachwith its own frequencyof occurrenceand resultingsag characteristics.
Below are somesuggestionsfor the choice of the fault positions. A numberof those
suggestionsare borrowed from the method of critical distancesto be discussedin
Section 6.5. In this section only the results will be used; for more theoretical background one is advisedto read Section6.5 first.
The main criterion in choosingfault positionsis: a fault position should represent
This criterion has been
short-circuit faults leading to sags with similar characteristics.
applied in choosingthe fault positionsin Fig. 6.35 and Table 6.19.
6.4.3.1 DistancebetweenFault Positions. To understandhow the distancebetween fault positions influencesthe result, considerthe sagmagnitudeas a function
of the distancebetweenthe fault and the substationfrom which the load is fed. The
sag magnitudeis plotted in Fig. 6.36. The shapeof the curve can be obtainedfrom
the equationsin Section 6.5. By choosing one fault position to representa certain
rangeof possiblefaults, we make the sag magnitudefor the whole rangeequal to the
sag magnitudefor that one position. The approximatedmagnitudeversusdistanceis
shownin Fig. 6.37. We seethat the error is largestwhen theexactcurve is at its steepest, which is close to the load. Here we would need ahigher density of fault positions. For more remote faults, the curve becomesmore flat, and the error smaller.
Furtheraway from the load, a lower density of fault positionswould be acceptable.
To quantify this, considera radial systemasshownin Fig. 6.38. Aload is fed from
a substationwith a nominal (phase-to-phase)voltage V nom. The fault current for a
terminal fault on the indicatedfeederis [fault, thus the sourceimpedanceis
Z s=
Vnom
(6.15)
v'3 x [fault
0.8
.e~ 0.6
Q
~
c=
8
fO.4
0
·3en
r/)
J:J
~
]
0.2
.s
0
0
0.25
0.5
0.75
1
1.25
Distanceto the fault
1.5
Figure 6.36 Voltage as afunction of the distanceto the fault.
1.75
2
364
Chapter6 • Voltage Sags-Stochastic
Assessment
\
0.8
~
lO.6
~
: 0.4
en
/'
../
0.2
..... ....
~Approximated voltage
........~ Actualvoltage
O...----I---+----t--~~---I----+-----I~---I
o
0.25
0.5
0.75
1
1.25
Distanceto the fault
1.5
1.75
2
Figure 6.37Approximatedvoltage as a functionof the distanceto the fault.
Source
Feeder
Load
Figure 6.38 Faults ina radial system.
The feederhasan impedancez per unit length and the distancebetweenthe substation
and the fault is x, leading to a feederimpedanceof ZF = zx. The voltage at the substationduring the fault (as afraction of the pre-fault voltage)is found from
V
sag -
ZF
_
ZS+ZF -
xz
~+xz
(6 16)
·
.../31/ou11
For a given sag magnitude Vsag, we can calculatethe distanceto the fault:
x
=
Vnom
./3Z[/ault
Vsag
X ------~
1 - Vsag
(6.17)
Note that someapproximationsare madehere, which will be discussedin Section6.5.
Consideras an example a 34.5 kV system with 10kA
availablefault currentand a
feederimpedanceof 0.3 O/km. This gives the following distances to the fault:
• Vsag = 10%: x = 750m
• Vsag = 20%: x = 1650m
• v,rag = 50%: x = 6.5 km
•
Vsag=700;O:x=15km
• Vsag = 80%: x = 27 km
• Vrag = 90%: x = 60km
Section 6.4 • TheMethod of Fault Positions
365
If we want to distinguish between a sag down 10%
to and one down to 20%, we need
fault positionsat least every kilometer. But if the
bordersof the bins in the sag density
table are at500~, 70%, 80%, and900~, fault positionsevery' 5 km are sufficient.Note
also that the required distance between fault
positionsincreases very fast when moving
away from the load position. Thus, the required density
of fault positionsdecreases fast
for increasing distance to the fault.
Equation (6.17) gives anindication of the distancebetween faultpositions for
linesoriginatingin thesubstationfrom which the load is fed.For otherlines, one or two
fault positions per line is normally enough, if thesubstationsare not too close. A
possible strategy is to first calculate the resulting magnitudefor
sag
faults in the substation and to insert fault positions in between when the
resulting sagmagnitudefor
two neighboringsubstationsdiffers too much.
Choosingtwo fault positions per line instead
o f one couldactually speed up the
calculationsif the fault positions are chosen at the
beginningand end of the line. This
way, all tinesoriginating from the samesubstationneed only one voltagecalculation.
The situationbecomes morecomplicatedwhen networksare meshed across voltage levels, like thetransmissionvoltage levels in theUnited Statesand in severalo ther
countries.Considera system like in Fig. 6.39. A safe
strategyis to use multiple fault
positions on the indicated lines and only one or two fault
positionson the other lines,
including 138kV, 230kV, and 345kV. Due to the
multiple pathsfor the fault current
not
and the relatively largetransformerimpedances, faults at 138kV and higher will
cause very deep sags; and the precise fault
positionwill not have much influence on the
sagmagnitude.For 230kV and 345kV, one faultposition per substationis probably
still too much. The main problem is
that no definite rules can be given for the required
numberof fault positions. In case
computationtime is noconcern,and the selectionof
fault positionsis automatic,one might simply choose 10 or even more fault
positions
for each line.
In the above, only the sag
magnitudehas been used to
determinethe numberof
fault positions.Apart from the sagmagnitude,the sagduration will also have to be
considered. The sag
d uration depends on theprotectionused for thevarious feeders
and substationcomponents.It is especiallyimportant to considerparts of the system
and thus to a longer sagduration.
where faults lead to longer fault-clearing time
Possible examples are busbars
protectedby the backup protection of the infeeding
lines; faults toward the remote endof a transmissionline cleared by thedistance
protectionin its zone 2.
345kV
Figure 6.39Network meshed across voltage
levels, with suggested fault positions.
366
Chapter6 • Voltage Sags-Stochastic
Assessment
6.4.3.2 Extentof the Fault Positions. In the precedingsection, the requirements for thedistancebetween faultpositions were discussed. The resulting recommendationwas to use one or two faultpositionsper line for all but those lines which
are directly feeding the load.T he next question that comes up is: How far do we
have to go with this? Is it, e.g., needed considera
to
345 kV substationat 1000km
away?Probablynot, but howaboutone at 200 km?There are two possible ways forward, both of which arenot really satisfying:
1. Use (6.17) toestimateat which distancea fault would lead to a sag down to
90%, or anyothervalue for the"mostshallow sagof interest."For transmission voltages this will give very large values (600 km for a 345 kV system with
10 kA availablefault current),which areprobablymuch higherthanactually
needed.
2. Startwith fault positionsin a restrictedarea, and look at the sag
magnitudes
for faults at theborderof this area.If thesemagnitudesare below900/0, the
area needs to be
extended.If the system isavailablein the right format for a
suitablepower systemanalysispackage,this might still be the fastestmethod.
6.4.3.3 Failure of the Protection. Failure of the protection is of concern for
voltage sagcalculationsbecause it leads to laonger fault-clearing time, and thus a
longer sagduration. This longer sagduration, often significantly longer,could be important for the compatibility assessment.
The equipmentmight toleratethe sag when
the primary protection clears the fault, but not when the
backup protection has to
take over.
To include failureof the protection,two events have to be
consideredfor each
fault position: onerepresentingclearing by the primary protection, the other fault
clearing by thebackup.The two events will typically be givendifferent fault frequencies.Alternatively one can use a fixed failure rate
of the protectionand a fixed faultclearing time forboth the primary and the backupprotection.In that case the resulting
magnitudedistribution only needs to be shiftedtoward the relevantduration.
6.4.3.4 Multiple Events. The method of fault positions in its basic form only
considersshort-circuit faults in an otherwisenormal system. Multiple events like a
fault during the failure of anearbypower stationare normally not considered.To include these, faultcalculationsneed to beperformedfor the system with the power
station out of operation.The choice of fault positions becomes even more complicated now. Only those faults need to be
consideredfor which the outage of the
power station influences the sag.When an automaticmethod is used, it isprobably
simplest toconsiderall situations.The beststrategyappearsagain to' start with generator stationsnear the load,and move further away from the load until there is no
longer any significant influence on the sag
magnitude.Significant influence should be
defined as likely to affectbehaviorof equipment.
8.4.4 An Example of the Method of Fault Positions
In this section we discuss an
exampleof the useof the methodof fault positions.A
small system is used for this: the
reasonbeing that the data was readilyavailableand
that the data processingwas limited sothat various options could be studied in a
367
Section 6.4 • TheMethod of Fault Positions
relatively short time. A study in a U.S. transmissionsystem isdescribedin [8], and a
study in a large Europeantransmissionsystemin [71], [74].
6.4.4.1 The Reliability Test
System. The reliability test system(RTS) was proposedby the IEEE subcommitteeon the applicationof probability methodsto compare stochasticassessmenttechniquesfor generationand transmissionsystems [73].
The RTS has been used by
Qader[64], [71] to demonstratethe method of fault positions. The reliability test system consists of 24 bussesconnectedby 38 lines and
cables, as shown in Fig. 6.40. Ten generatorsand one synchronouscondenserare
connectedat 138kV and at 230kV.
6.4.4.2 Voltages Due to One Fault.F igure 6.41 shows the effectof a fault halfway between busses 2and 4 on the voltages throughout the system. Only bus 4
BUS 22
230kV
BUSt3
Trans. 4
BUS 10""'''''''''
138kV
BUS 4
BUS 8
BUSS
BUst
BUS2
Figure 6.40 Reliability test system.( Reproducedfrom Qader[71].)
368
Chapter6 • VoltageSags-Stochastic
Assessment
Figure 6.41 Voltage sags at
different busses due to a fault halfway between bus 2
and bus 4 in Fig. 6.40.(Reproducedfrom Qader [7IJ.)
shows avoltage drop below 50%, but the voltagedrops below 900/0 in a large part
of the 138kV system.Note that the voltagedrops to 280/0 at bus 4, but only to58%
at bus 2, while theshort-circuit fault is exactly in the middleof the line between bus
generatorsat bus I and bus 2 keeping up
2 and bus 4. This difference is due to the
the voltage. Bus 4 is far away from any
generatorstation, thus the voltagedrops to
a much lower value. The dense
c oncentrationof generatorstationskeeps up the voltage in most of the 230 kV system, thuspreventingmore serious voltagedrops. Also,
the relatively hightransformerimpedancemakesthat the voltagedrops at 230 kV level are small. This figure shows some well-known and trivial facts which are still
worth repeatinghere:
• The voltagedrop is highestnearthe fault positionand decreases when moving
further away from the fault.
• The voltagedrop diminishesquickly when movingtoward a generatorstation.
Section 6.4 • The Method ofFault Positions
369
• The voltage drop diminishes when moving acrosstransformertoward
a
a
higher voltage level. This assumes
that more generation is connected to higher
voltagelevels.The high-voltage side of the
transformeris closer to the source,
so that the voltage drops less in magnitude.
6.4.4.3 Exposed Area.In Fig. 6.41 the fault position was fixed and voltage
sags were calculated for all busses. Figure 6.42 gives the reversed
situation: the voltage magnitudeis calculated for one bus but for many fault positions. In this case,
the sagmagnitudeat bus 4 is calculated. Positions leading to equal magnitudes
sag
at bus 4 are connected
through "contour lines" in Fig. 6.42.Contourlines have been
plotted for sag magnitudes of30% , 50% , 60%, 70% , and 80% • The area in which
faults lead to a sag below a certain voltage is called the "exposed
area."The term exposed area was originally linked to
equipmentbehavior. Suppose
t hat the equipment
Figure 6.41 Exposed area contours for bus
4. (Reproduced from Qader
[71].)
370
Chapter6 • Voltage Sags-Stochastic
Assessment
trips when the voltagedrops below 600/0. In that case theequipmentis "exposed"to
all faults within the 60% contour in the figure; hence the term exposed .area. As
faults can only occur onprimary components(lines, cables,transformers,busses,
etc.), the exposed area is strictly
speakingnot an area, but acollection of points (the
substations)and curves (the lines and cables). But
drawing a closedcontourhelps to
visualize the concept.Knowing which primary componentsare within the exposed
area can be morevaluable information than the actual number of sags. Suppose
there is anoverheadline across amountainprone to adverseweather,within the exposed area. Then it might be
worth to consideradditional protection measuresfor
this line, or to change the system
structureso that this line no longer falls within the
exposed area, or to
improve equipmentimmunity so that the exposedareano longer
con tains this line.
From Fig. 6.42 andother exposed areacontours,the following conclusionsare
drawn:
• The exposed area extends
further toward large concentrationsof generation,
than toward partsof the systemwithout generation.
• The shape of the exposed area
contour near transformerstationsdependson
the amount of generationpresent on theother side of the transformer.The
exposed area typically extends far into
higher-voltagenetworksbut rarely into
lower-voltagenetworks.If the fault takes place in a
lower-voltagenetwork the
voltage drop over thetransformerimpedancewill be large. This assumes
t hat
the maingenerationis at a higher 'voltage level
than the fault. Consideringthe
simple network structuresin Chapter4 explains thisbehavior.
6.4.4.4 Sag Frequency.Thesecalculationscan beperformedfor all busses, resulting in a setof exposed areacontoursfor each bus.Plotting them in one figure
would not result in somethingeasily interpretable.Instead Fig. 6.43 gives the expectednumberof sags to a. voltage below
80% for each bus. The average
numberof
sags per bus is 6.85 per year; the
various percentilesare given inTable 6.23. We see
that 80% of the busses has a sag frequency within
30% of the average sag frequency
for all busses. Notethat we assumedthe same fault rate (in faults per km per year).
for all lines. In reality some lines are more
prone to faults than others, which can
give larger variationsin the sag frequency.
It is difficult to draw generalconclusionsaboutthe sag frequency, because each
system is different.From this and otherstudies, however, one
might, draw the conclusion that sag frequencies are lower
towards large concentrationsof generationand
higher further away from thegeneratorstations.
TABLE 6.23 Percentiles of the Sag Frequency Distribution Over the
Busses in the Reliability Test System
Percentile
90%
75%
50%
25%
10%
Sag Frequency
4.7 per
5.2 per
6.8 per
8.2 per
9.0 per
year
year
year
year
year
Percent of Average
700/0
75%
100%
120%
130%
371
Section 6.4 • TheMethod of Fault Positions
8.58
138kV
6.81
7.14
4.72
Figure 6.43 Voltage sag frequencyfor all busses in the RTS:numberof sags
below 800/0. (Reproducedfrom Qader[71].)
6.4.4.5 Generator Scheduling.In the precedingstudy it was assumedthat all
generatorswere in operation.In reality this is an unlikelysituation.We sawthat generator stations have a significant influence on the voltages in the system
during a
fault, and on the sag frequency. To
quantify this influence, thecalculationsin the reliability test system have been
repeatedfor the situation in which all 138kV substations are out of operation. The resulting sag frequency isshown in Fig. 6.44.
Comparingthis figure with Fig. 6.43 showsthat the sagfrequencyis increased at all
busses but most significantly at the 138kV busses. The sag frequency is very similar
nearby in
for all 138kV busses. The reason that
is faults in the 138kV system, and·
the 230kV system, makethat the voltage drops below 800/0 for all 138kV busses. If
the sag frequency is defined as the
number of sags below65% the differences between the 138kV busses become larger, Table
see 6.24.
As a next step it has been assumed
that the three 138kVgeneratorsare each out
of operationduring four months of the year, andthat there is nooverlap in these
periods; thus there are always two 138 kV
generatorsin operation.For each of these
periods (i.e., for eachc ombinationof one generatorout and two in operation)the sag
frequency has been
calculatedin exactly the same way as before. The results for the
372
Chapter6 • VoltageSags-Stochastic
Assessment
12.18
138kV
12.18
12.18
12.18
Figure 6.44 Voltage sag frequency(numberof sags per year) for all busses in the
reliability test system when the 138 kV
g eneratorsare out of operation.
(Reproducedfrom Qader[71].)
TABLE 6.24 Influenceof GeneratorSchedulingon the SagFrequencyin the Reliability Test System,Numberof
Sags perYear below 65%
138 kV Bus
Generator
Scheduling
Generatorlout
Generator2 out
Generator7 out
Average
All generatorsin
All generatorsout
2
2.80
2.43
1.54
2.26
1.34
7.37
2.77
2.79
1.40
2.32
1.40
7.37
3
3.24
3.06
3.06
3.12
2.85
6.73
4
3.65
3.77
2.81
3.41
2.19
7.43
5
3.42
3.44
3.20
3.35
2.16
7.06
6
3.16
3.18
3.18
3.17
2.60
5.19
7
0.80
0.80
4.42
2.01
0.80
6.66
8
9
10
1.47
1.49
4.42
2.46
1.34
6.66
2.65
2.64
3.11
2.80
2.59
5.88
3.38
3.40
3.44
3.41
2.81
5.96
373
Section 6.5 • TheMethod of Critical Distances
138kV busses are shown in Table 6.24. The table shows
numberof
the
sags below65%
for all 138kV substations,for a number of generatorscheduling options. The sag
frequency for the three4-monthperiods mentioned, is given in the rows labeled "generator lout," "generator2 out," and "generator7 out." The numberof sags per year
has beencalculatedas the averageo f these three sag frequencies, and included in the
the
row labeled"average."For reference the sag frequency is also given for situation
when all generatorsare in operation("all generatorsin") and when all three 138kV
generatorsare out of operation("all generatorsout").
8.5 THE METHOD OF CRITICAL DISTANCES
The methodof critical distances does
not calculate the voltage at a given fault
position,
but the fault position for a given voltage. By using some simple expressions, it is
possible to findout where in thenetwork a fault would lead to a voltage sag down
to a givenmagnitudevalue. Each fault closer to the load will cause a deeper sag. The
numberof sagsmore severethan this magnitudeis the numberof short-circuitfaults
closer to the loadthan the indicated positions.
We first describe the basic theory and give the outline
of the method. A simple
exampledemonstrateshow to apply the method. In the
derivationof the basic expression, anumberof approximationshave been made. More exact expressions and expresof the method are
sions for non-radial systems are derived next. Finally the results
comparedwith the resultsof the methodof fault positions.
8.5.1 Basic Theory
The method of critical distances is based on the voltage divider model for the
voltage sag, asintroducedin Fig. 4.14. Neglecting loadcurrentsand assuming the preevent voltage to be one, we
obtainedfor the voltage at thepoint-of-commoncoupling
(pee)during the fault:
ZF
Vsag
= ZF + Zs
(6.18)
where ZF is the impedancebetween the pee and the fault, and
Zs the source impedance
at the pee. LetZF = z£, with z the feeder impedance per unit length and
£, the distance
between the peeand the fault. This results in the following expression for the sag
magnitude:
V:,ag =
z£~ Zs
(6.19)
The "critical distance"is introducedas follows: themagnitudeat the peedropsbelow a
critical voltage V whenever a fault occurs within the critical distance from the pee. An
expression for the critical distance
£'crit is easily beobtainedfrom (6.19):
Zs
LCrit
V
=---; x 1 _ V
(6.20)
Here it isassumedthat both source and feeder impedance are purely reactive
rather
(a
commonassumptionin power system analysis), or more general:
that the angle in the
complex plane between these two impedances is zero.
For three-phase
Strictly speaking(6.20) only holds for a single-phase system.
faults in a three-phasesystem, the expressions are valid if for
Zs and z the positive-
374
Chapter6 • VoltageSags-Stochastic
Assessment
sequenceimpedancesare used.For single-phasefaults the sum of positive-, negative- ,
and zero-sequenceimpedancesshould be used; forphase-to-phasefaults the sumof
positive and negativesequence.The voltage in the expressionsaboveis the phase-toneutral voltage in the faulted phasein case of a single-phasefault and thevoltage
between thefaulted phasesin case of a phase-to-phase
fault. We will come back to
single-phasefaults and phase-to-phase
faults below.
Equation(6.20) can be used to
e stimatethe exposed area at every
voltagelevel in
the supply to a sensitiveload. The exposedareacontainsall fault positionsthat lead to
a voltagesagcausinga spuriousequipmenttrip . The expectednumberof spurioustrips
is found by simply addingthe failure ratesof all equipmentwithin the exposed area.
Transformerimpedancesare a largepart of the sourceimpedanceat any point in
the system .Therefore,faults on thesecondaryside do not cause a deep sag on the
primary side. To estimatethe numberof sags below acertainmagnitudeit is sufficient
to add alllengthsof lines andcableswithin the critical distancefrom the pee. Thetotal
length of lines and cableswithin the exposedareais called the"exposedlength." The
resultingexposedlengthhas to bemultiplied by the failure rate peru nit lengthto obtain
the numberof sags per year.
8.5.2 Example-Three-Phase Faults
Considerthe II kV network in Fig. 6.45.The fault level at themain 11 kV bus is
151 MVA (sourceimpedance0.663 pu on a 100 MVA base), the feeder
impedanceis
0.336 Q/km (0.278pu/km on the 100 MV A base).
The critical distancefor different critical voltages,calculatedfrom (6.20), is given
in Table 6.25.The next-to-lastcolumn (labeled"exposedlength") gives thetotal feeder
length within theexposedarea.Figure 6.45 gives thecontoursof the exposed area for
variouscritical voltages. Eachfault betweenthe main II kV bus (the pee) and the 50%
contourwill lead to a voltagesag at the pee with magnitudebelow
a
50%. All pointson
the 50%contourare at adistanceof 2.4 km (seeTable6.25)of the main II kV bus. The
last column in Table 6.25 gives theexpectednumberof equipmenttrips per year. A
value of 0.645 faults per km per year has been used .
II kV. 15 1 MVA
- - ---- - - - -- - 80%
-
.-..
__------- 90%
Figure 6.45 An II kV network used as an
example for the method of critical distances.
375
Section 6.5 • The Methodof Critical Distances
TABLE 6.25 Results of Method of Critical Distances, Three-Phase Faults
Critical Voltage
Critical Distance
Exposed Length
90%
80%
21.4 km
9.6 km
5.6 km
3.6 km
2.4 km
1.6km
1.0 km
0.6 km
0.3 km
24.0 km
21.6 km
16.8 km
12.2 km
8.6 km
5.4 km
3.0 km
1.8km
0.9 km
700~
60%
50%
40%
300/0
200/0
10%
Number of Trips per Year
15.5
13.9
10.8
7.9
5.5
3.5
1.9
1.1
0.6
8.5.3 Basic Theory: More Accurate Expressions
To obtain a more accurateexpression,we have toconsiderthat both the feeder
and the sourceimpedanceare complex.The basicexpressionis againobtainedfrom the
voltagedivider shownin Fig. 4.14,but with complexvoltageand impedances:
v=
ZF
ZS+ZF
(6.21)
where Zs = Rs + jXs is the sourceimpedanceat the pee,ZF = (r + jx)£' is the impedancebetweenthe fault and the pee,.c is the distancebetweenthe fault and the pee,
z = r + jx is the feeder impedanceper unit length. The load currents have been
neglected; thepre-fault voltageat the peeequalsthe sourcevoltageequals 1000/0.
In Section4.5 expressionshave beenderivedfor the magnitudeV and the phaseanglejump as afunction of the distancebetweenthe peeand the fault. Equation(4.87)
for the magnitudeof the voltagereadsas follows:
v = -1-~-A --;::;::===:::::::::::====
i 2A(l-COSa)
-
(6.22)
(1+Ai
with
A = ZF =
Zs
Z X £,
Zs
(6.23)
a the angle in thecomplex plane betweensourceand feeder impedance,the so-called
impedanceangle:
a
= arctan(~~) - arctan(~)
(6.24)
and Zs = IRs + jXsl, Z = Ir + jxl, V = IVI, etc.
To obtainan expressionfor the critical distance,A needs to besolvedfrom (6.22)
for known V. Therefore,this equationis rewritten into the second-orderp olynomial
equation
(6.25)
376
Chapter6 • VoltageSags-Stochastic
Assessment
The positivesolution of this equationcan bewritten as
(6.26)
Togetherwith (6.23) the desiredexpressionfor the critical distanceis obtained:
2
c . _Zs x_v_[vcosa+JI-V2 sin a ]
crtt Z
1- V
V+ I
(6.27)
The first part of (6.27)
(6.28)
is the expression for the critical
distanceobtained(6.20).For most applications(6.20) is
sufficient, especially as the
d ata are not alwaysavailable to calculatethe impedance
angle. To assess the
error made by using theapproximatedexpression the critical
distance has been
calculatedfor different valuesof a.
Figure6.46 gives the critical length as function
a
of the critical voltage for 11kV
overheadlines. A sourceimpedanceof 0.663 pu and a feeder
impedanceof 0.278pu/km
have been used.
Note that these are the same values as used in the previous example
(Fig. 6.45). We seethat the error only becomes significant for large
impedanceangles
(more than 30°). In that case moreaccurateexpressionsshould be used. In the next
section a simple butaccurateapproximationfor the critical distanceis derived.
25r - - - - - - - r - - - - - - , . - - - - - , - - - - - - - , . - - - ,
0.2
0.4
0.6
Critical voltage in pu
0.8
Figure 6.46 Critical distanceas afunction of
the critical voltagefor impedanceangle 00
(solid line), -300 (dashedline), -600 (dashdot line).
8.5.4 An Intermediate Expression
In the previous sections an exact and approximateexpression
an
for the critical
distance have been derived: (6.27) and (6.20), respectively. The difference between these
two expressions is the
factor betweensquarebracketsin the right-handside of (6.27):
k= Vcoscx+Jl- V2 sin 2 cx
1+ V
(6.29)
377
Section 6.5 • TheMethod of Critical Distances
50 r - - - - - . , . . - - - - - . . , . - - - - - . - - - - - , - - - - ,
40
d
~ 30
&
.5
~ 20
Jj
~.
/'
10
Figure 6.47 Error madein the simplified
expressionof critical distance;impedance
angle: -200 (solid line), -400 (dashedline),
and -600 (dash-dotline).
0.2
0.4
0.6
0.8
Critical voltage in pu
The more thisfactor deviates from one, the larger the
errormadeby using the simplified
expression (6.20). This
e rror has beencalculatedas (1 - k) * 100% and plotted in Fig.
6.47 for three valueso f the impedanceangle. The simplified expression (6.20) overestimates the criticaldistance(and thus thenumberof sags) as is also shown in Fig. 6.46.
The error is, however, small in most cases, with the exception of systems with large
impedance angles like
undergroundcables indistributionsystems. A first-order correction to the simplified expression (6.20) can obtainedby
be
approximating(6.29)around
V=O:
(6.30)
k ~ 1 - V(l - cosa)
(6.31)
The error made by usingapproximation(6.31) is shown in Fig. 6.48 for different
impedanceangles. Theerror made never exceeds a few percent.
An importantconclusionfrom Fig. 6.48 isthat the following expression gives the
critical distance in systems with a large
impedanceangle:
L,crit
z,
=--;- x
V
I _ V (I - V(l - cosa)}
0
(6.32)
-.......::---I
-0.5
l
,
,
-1
\
5 -15
U
.
[
.S
...
-2
,
,
,
,
,
,
\
,,
\
~ -2.5
,
,
,
\
~
\
,
I
-3
\
I
\
,
I
Figure 6.48 Error madeby usinga first-order
approximationfor the critical distance;
impedanceangle: - 20° (solid line), -400
(dashedline), and -600 (dash-dotline).
-3.5
I
/
-4
0
0.2
0.4
0.6
Critical voltage in pu
0.8
378
Chapter6 • VoltageSags-Stochastic
Assessment
6.5.5 Three-Phase Unbalance
The abovereasoningapplies to three-phasefaults only. For unbalancedfaults
(single-phase,phase-to-phase)the method needsadjustment.Most of the discussion
below follows directly from thetreatmentof three-phaseunbalancedsags in Section 4.4.
6.5.5.1 Phase-to-PhaseFaults. Phase-to-phase
faults lead to sagso f type C or
type D, with a characteristicmagnitudeequal to the initial(phase-to-phase)
voltage
at the point-of-commoncoupling. Themethodof critical distances applies to the voltage at the pcc and can thus be used
without modification for phase-to-phase
faults.
The impedancevalues to be used are the average of positive- and negative-sequence
values. As these are
normally about equal, the positive-sequence
impedancecan be
usedjust like for three-phasefaults. In termsof characteristicmagnitude:the critical
distancefor phase-to-phase
faults equals the critical distance for
three-phasefaults.
In case the voltage at the
equipmentterminals is of interest (e.g., for single-phase
equipment),the strategyis to translatethis voltage back tocharacteristicmagnitude
and apply the equationsfor the critical distance to thecharacteristicmagnitude.Of
importancehere is todeterminewhether a fault at acertainvoltage level leads to a type
C or type D sag.
Supposethat the fault leads to a type C sag. In
that case of the single-phase
equipmentwill not see any sag at all, where
j will see a sag between
50% and 100%. Let
Veq be the sagmagnitudeat theequipmentterminals andVchar the characteristicmagnitude of the three-phaseunbalancedsag. These twomagnitudesrelate accordingto
t
Veq =
~ j I + 3V;har
(6.33)
This expressionis obtainedfrom Fig. 4.90 when neglecting the
characteristicphaser ather
anglejump (l/J = 0). Including phase-anglejumpsis possible, but would result in
complicatedexpressions.
The characteristicmagnitudecan beobtainedfrom the magnitudeat the equipment terminalsby using
Vchar =
1,
J~ V;q - ~
(6.34)
1
For Veq < there are no sags.
For < Veq < 1, (6.20) can be used to calculate the
critical distance,with V = Vchar• The resulting sag frequency should be multiplied~by
to accountfor the fact that one in three faults does not lead to a sag at equipment
the
a agnitudeof Vchar
terminals. For a type D sagof magnitude Vcha" one phase has m
also. The expression for the critical distance can be applied directly, but the resulting
sag frequency needs to be multiplied
by!. The two other phasesdrop to
Veq = ~
j n: + 3
(6.35)
For Veq < !"f3 this gives nocontribution.For!"f3 < Veq < 1, the critical distance can
be calculatedby using
(6.36)
and the resulting sag frequency should be multiplied j.byNote that the two sag
frequencies for the type D sag
should be added.
379
Section 6.5 • TheMethod of Critical Distances
6.5.5.2 Example: Phase-to-Phase Faults.
Consider the same system as in the
examplefor three-phasefaults. We areinterestedin the numberof spurioustrips for
phase-to-phase(delta) connectedsingle-phaseload at 660V. A Dy-connectedllkV/
660V transformeris used.The sag type at theequipmentterminals is determinedas
follows:
• The phase-to-phase
fault leads to athree-phaseunbalancedsag of type C for
star-connectedload at 11 kV.
.
• For delta-connectedload at 11 kV the sag isof type D.
• For delta-connectedload at 660V it is of type C.
Thecalculationof the trip frequencyas afunction of the equipmentvoltagetoleranceis
summarizedin Table 6.26. It proceedsas follows:
• For a given critical voltage at the equipmentterminals Veq, the critical characteristicmagnitude Vchar is calculatedby using
(6.37)
The resultis shownin the secondcolumnof Table6.26. For Veq < 0.5 thevalue
underthe squareroot is negative, whichmeansthat even for aterminal fault
(distancezero), thevoltageat the equipmentterminalsis higher than the critical voltage. The contribution to the exposedlength is thus zero, hence the
zeros in the first few rowsof the table.
• From the critical characteristicmagnitude,the critical distanceis calculatedin
the standardway, by using
z,
v-;
(6.38)
Vcru=-x--z
1 - Vchar
with Zs = 0.661 pu and z= 0.278pu/km, The resultingcritical distanceis given
in the third column of Table 6.26.
Faults,Type C Sags
TABLE 6.26 Method of Critical Distances-Phase-to-Phase
Sag Magnitudeat
EquipmentTerminals
Characteristic
Magnitude
Critical Distance
(km)
0
0
0.1
0.2
0
0
0
0
0
0
0
1.5
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0
0
0
0.38
0.57
0.72
0.86
3.2
6.1
14.7
ExposedLength
(km)
Trip Frequency
(per year)
0
0
0
0
0
0
0
0
5.0
0
0
0
0
11.4
18.2
24
2.1
4.9
7.8
10.3
380
Chapter6 • VoltageSags-Stochastic
Assessment
• From the critical distance, the exposed lengthcalculatedfor
is
the 11kV distribution system in Fig. 6.45. The
methodused for this is the same as shown in
Fig. 6.45 for three-phasefaults.
• Knowing the exposed length it is possible calculatethe
to
trip frequency. Here
it is assumedthat the numberof phase-to-phasefaults is equal to thenumberof
three-phase faults: 0.645 per km per year. This is not a realistic
assumption,but
it enables an easier
comparisonof the influenceof the different typesof fault.
Because the voltage is only down on two phases for a type C sag, this fault
frequency has to be multiplied by
j to get the trip frequency. Thelatteris given
in the last rowof the table.
Consider, as a second example,
that the low-voltage load isconnectedin star (thus
phase-to-neutralsingle-phase load). The
three-phaseunbalancedsag will beof type D,
with one deep sag and two shallow sags atequipmentterminals.
the
Acalculationof the
trip frequency using themethodof critical distancesis summarizedin Table6.27. Only
calculationfor
critical voltages between80% and 960/0 are shown in the table. The
other voltage values proceeds in a similar way.
• Like for delta-connected
load, thecalculationstartswith the choiceof a critical
voltage at theequipmentterminals. Next,separatecalculationsare needed for
the deep sag and for the shallow sag.
• The calculationsfor the deep sag (labeled
"lowest voltage" in Table 6.27) are
almost identical to thecalculationsfor a three-phasefault. The magnitudeof
the deep sag at the
equipmentterminals is equal to thecharacteristicmagnitude, sothat the standardequationfor the critical distancecan be used. The
only difference isthat the fault frequency needs to be divided by three to
accommodatefor the fact that only one in three voltages shows a deep sag.
Thus, from the viewpointo f single-phaseequipment:only one in three faults
leads to a deep sag. Critical distance, exposed length, and trip frequency for the
of4Table 6.27.Note that the exposed
deep sag are given in columns 2, 3, and
length and the trip frequency no longer increase for critical voltages above
84%. This is because the exposed area
alreadyincludes the whole lengthof
the 11 kV feeders.
TABLE 6.27 Method of Critical Distances-Phase-to-Phase
Faults,Type D Sags
Lowest Voltage
Magnitude
Equipment
Terminals
(pu)
Critical
Distance
(km)
0.80
0.82
0.84
0.86
0.88
0.90
0.92
0.94
0.96
9.5
10.9
12.5
14.7
17.5
21.5
27.4
37.4
57.2
HighestVoltage
Exposed
Trip
Characteristic Critical
Length Frequency Magnitude
Distance
(pu)
(km)
(per year)
(km)
21.5
22.9
24
24
24
24
24
24
24
4.6
4.9
5.2
5.2
5.2
5.2
5.2
5.2
5.2
0
0
0
0
0.31
0.49
0.62
0.73
0.83
0
0
0
0
1.1
2.3
3.9
6.4
11.6
Trip
Total Trip
Exposed
Length FrequencyFrequency
(km)
(per year) (per year)
0
0
0
0
3.4
8.2
12.8
18.4
23.6
0
0
0
0
1.5
3.5
5.5
7.9
10.1
4.6
4.9
5.2
5.2
6.7
8.7
10.7
13.1
15.3
381
Section 6.5 • TheMethod of Critical Distances
• The calculationsfor the shallow sag proceed fairly similar to the calculations
for thedelta-connected
load. As a first step the critical voltage at the equipment
terminals istranslatedinto a criticalcharacteristicmagnitude,using the following expression:
(6.39)
resulting in the values incolumn 5. For Veq < 0.866 thecharacteristicmagnitude is set to zero. The shallow sag at equipmentterminals
the
never becomes
lower than this value.Calculationof critical distance, exposed length, and trip
For the trip frequency, the fault frequency
frequency proceeds like before.
needs to be multiplied bybecause only two of the three phases show a shallow
sag. The results for the shallow sag are summarized in columns
through8.
5
• Finally the total trip frequency is the sum
o f the trip frequency due to deep sags
and the trip frequency due to shallow sags. The total trip frequency is given in
the last column.
1
6.5.5.3 Single-PhaseFaults-Solidly Grounded Systems.Single-phase faults
lead to sagsof type B, C, or D' at theequipmentterminals. Thetranslationfrom
equipmentterminal voltages to the voltage to be used in the expressions for the critiof sag.
cal distance depends on the type
A type B sag only occurs in case of
equipmentconnected in star and the singlephase fault at the same voltage level as equipment(or
the
at a higher level with only
YnYn transformersbetween the fault and the equipment).
For a type B sag the terminal
voltage can be directly used in the expressions for the critical distance. As only one
by! for single.phase drops in voltage, the resulting sag frequency should be multiplied
phase equipment.F or the impedances the sum of positive-, negative-, and zerosequence values should be used.
Sags of type C or type D occur in all
other cases.For these thecharacteristic
magnitude deviates from the initial voltage (the voltage in the faulted phase at the pee).
For solidly groundeddistribution systems (where positive- and zero-sequence source
impedances are equal), the following relation between
characteristicmagnitude Vchar
and initial magnitudeVinit has been derived (4.109):
Vchar
1
2
= 3" + 3v.;
(6.40)
Knowing the characteristic magnitude of the three-phase unbalanced sag, and
Vchar < 1, the initial voltage isobtainedfrom
!<
3
V init
1
= 2 V char - 2
(6.41)
The characteristicmagnitudeneeds to betranslatedto an initial magnitude,by using
(6.41). In case themagnitudeat the equipmentterminals is ofimportance,a second
translationhas to be made: from
magnitudeat theequipmentterminals tocharacteristic
magnitude. Thistranslationproceeds in exactly the same way as phase-to-phase
for
faults.
6.5.5.4 Example: Single-Phase Faults in a Solidly
GroundedSystem. When considering single-phase faults, we need to include the zero-sequence impedance of
that
source and feeder.For a solidly groundeddistribution system we can assume
382
Chapter6 • VoltageSags-Stochastic
Assessment
positive- and zero-sequencesourceimpedanceare equal. But this cannotbe assumed
for the feeder impedances.From Table 4.4 we get 1.135pu/km for the zero-sequence
feeder impedance,and 0.278pu/km for the positive-sequenceimpedance.In the calculations we use the sum of positive-, negative-,and zero-sequenceimpedanceleading to Zs = 1.989pufor the sourceand z = 1.691pu/km for the feeder.
The calculationof the critical distancefor single-phase'faults from a given critical
characteristicmagnitudeis summarizedin Table 6.28.
• The first step is the translation from the characteristicvoltage to the initial
voltage,for which expression(6.41) is used.The characteristicmagnitudecannot be lessthan0.33 pu, hencethe zerosin the tablesfor lower valuesthanthis.
• From the critical initial voltage,the critical distancecan be calculatedby using
the standardexpression
r
J-crit
Zs
= -
z
X
Vinit
1 - Vinit
(6.42)
with Zs = 1.989puand z = 1.691pu/km,
• From the critical distance,the exposedlength and the trip frequencycan be
calculatedlike before. For single-phasefaults againa fault frequencyof 0.645
faults per km per year has beenused.
TABLE 6.28 Method of Critical Distances-Single-Phase
Faults,Solidly Grounded
System
Characteristic
Magnitude(pu)
o
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Initial Magnitude
(pu)
Critical Distance
(km)
o
o
o
o
o
o
o
o
0.10
0.25
0.40
0.55
0.70
0.85
0.1
0.4
0.8
1.4
2.7
6.6
ExposedLength
(km)
o
Trip Frequency
(per year)
o
o
o
o
o
o
o
0.3
1.2
2.4
4.6
9.8
0.2
0.8
1.5
18.6
12.0
3.0
6.3
6.5.5.5 Single-Phase
Faults-GeneralSolutions. In resistance-groundedd istribution systems,the assumptionthat positive- and zero-sequence
i mpedanceare equal
no longer holds. The assumptionis also not valid when line impedancesare a large
part of the sourceimpedance.This is the casein the 400kV supply in Fig. 4.21, as
was shown in Fig. 4.105. To obtain a more general expressionfor the critical distance, we can use the phase-to-neutralv oltage in the faulted phase according to
(4.40):
V-I an -
(2Z F t
3Zs 1
+ Z£o) + (2ZS1 + Zso)
(6.43)
The phase-to-neutral
voltagesin the non-faultedphasesarenot affectedby single-phase
faults. We canthus treatthe phase-to-neutralvoltagesthe sameas the phase-to-ground
383
Section 6.5 • TheMethod of Critical Distances
voltages in a solidlygroundedsystem. Thecharacteristicmagnitudeis related to the
(initial) phase-to-neutralvoltage as follows:
I
2
Vchar = 3" + '3 Van
(6.44)
With this knowledgeit is possible totranslatesagmagnitudesat the equipmentterminals to characteristicmagnitudesand to phase-to-neutralvoltages. It is possible to
translatephase-to-neutralvoltages tophase-to-groundvoltages, but one canalternatively derive anexpressionfor the critical distancefor phase-to-neutralvoltages. For
this weintroducepositive- and zero-sequence feeder
impedanceper unit length,Zl and
zo, respectively, and thedistanceto the fault L. Expression (6.43)changesinto
32s1
V - 1an -
(2z)
+ zo)£ + (2ZS1 + Zso)
(6.45)
The distanceto the fault £erit can beobtainedfor a given (critical) phase-to-neutral
voltage Van:
Lcrit
= (ZSI -
Zso) + Van(2ZS1 + Zso)
(2z1 + zo)(1 - Van)
(6.46)
For ZSI = Zso weobtainthe expression used for solidly
groundeddistributionsystems.
Note that normally ZSI < Zso so that the critical distancecan become negative for
small valuesof Van' Even for aterminal fault the phase-to-neutralvoltageis not zero.
Any critical voltagelessthan this minimum value will give a negative criticaldistance.
This has no physicalmeaning,and for calculating the exposed length (and sag frequency) a criticaldistanceof zero should be used.Alternatively one cancalculatethe
critical distancedirectly from thecharacteristicmagnitude.For this we useVchar = VI
+ V2 togetherwith (4.29) and (4.30) which give the positive- andnegative-sequence
voltages at thepeedue to a single-phase fault. Using the same
notationas before, we
get the followingexpressionfor the characteristicmagnitudeas a function of the distance to the fault:
v
_ Z£+ZSO
z£ + Zs
char -
(6.47)
with Zs = 2Zs1+ Zso and z = 2z1+ Z00 Solving the criticaldistancegives
z,
Vchar
£crit=-x--Z
1 - Vehar
Zso
z(l - Vchar)
(6.48)
6.5.5.6 Example: Single-Phase Faults in Resistance-Grounded
System. In a resistance-grounded
system we can no longer assume
that positive- andzero-sequence
sourceimsourceimpedanceare equal.From Table 4.3 we get for thezero-sequence
pedancea value of Zso = 8.172 pu. Thecalculationresults aresummarizedin Table
6.29. The results are only shown for critical voltages between 86% and 98%.
For
smaller valuesof the critical voltage, the trip frequency is zero.
Single-phasefaults in
resistance-grounded
systems typically lead to very shallow sags. The critical
distance
is calculateddirectly from the critical characteristicmagnitudeby using (6.48) with
Zs = 9.494pu, Zso = 8.172pu, and z= 1.691 pu/krn, Calculation of exposed area
and trip frequencyproceedslike before.
384
Chapter6 • VoltageSags-Stochastic
Assessment
TABLE 6.29 Method of Critical Distances-Single-Phase
Faults, Resistance-Grounded
System
CharacteristicMagnitude
(pu)
Critical Distance
(km)
ExposedLength
(km)
Trip Frequency
(per year)
0.86
0.88
0.90
0.92
0.94
0.96
0.98
0
0.9
0
2.7
7.8
13.3
19.4
0
1.7
2.2
4.2
7.4
13.9
24
33.5
24
5.0
8.9
12.5
15.5
15.5
8.5.8 Generator Stations
In Section 4.2.4 expression (4.16) was derived
describingthe effectof a generator
on the sagmagnitude.The equivalentcircuit used toobtain this is shown in Fig.4.24.
The expression has the following form:
(1 - Vsag)
=2
24
3
+
(6.49)
Z (1 - Vpcc)
4
To obtain the voltage at the pee we have to realize
that all load currentshave been
neglected here.There are no pre-fault power flows, andboth generatorsin Fig. 4.24
have exactly the same
outputvoltage, sothat they canbe replaced by onesourcein the
equivalent scheme. The following expression for the voltage obtained
is
from this
scheme:
2
V
pee
2
= Z3 + ZIII(2
3 + Z4)
(6.50)
where ZAI/ZB = f~l is the parallelconnectionof ZA and ZB' Combining(6.49) and
(6.50) gives thefoll~wi~g expression for theduring-sagvoltageexperiencedby the load
v -
1-
sag -
Z1Z 4
2 2(Z I + 2 3 + 2 4) + ZI(Z3
+ 2 4)
(6.51)
To obtain an expression for the critical
distance,we substitute2 2 = Z X L. The critical
distanceis obtainedby solving v,rag = Vcrit toward £. The resultingexpressionis
Lail
=21 {
Z
24
2 1+ 2 3 + 24
X
Vcrit
1 - Vcrit
_
23
}
2 1+ 23 + 24
(6.52)
The critical distancein (6.52)is not thedistancebetween the faulta nd the load, but the
distancebetween the fault and the main supply
point.
8.5.7 Phase-Angle Jumps
As we have seen inC hapter5, someequipmentis sensitive to thejump in phase
angle between the pre-sag voltage and during-sagvoltage.
the
Inthat case it is reasonable to find an expression for the critical
distanceas afunction of the "critical phaseanglejump." In otherwords, at whichdistancedoes a fault lead to a sag with a phaseangle jump equal to a given value? Too btain such an expression we
start with the
385
Section 6.5 • TheMethod of Critical Distances
expressionfor the phase-anglejump as afunction of the distanceto the fault: (4.84) in
Section 4.5.
).. + coso
cos</J = --;=======
+)..2 + 2Acosa
Jl
(6.53)
where a is the angle in thecomplexplanebetweenthe feederandthe sourceimpedance
and A the ratio betweentheir absolutevalues:
ZL
A=-
(6.54)
Zs
To obtain an expressionfor the critical distance,we need to solve xfrom (6.53) for
given phase-anglejump f/J. Taking the squareof both sides of (6.53) and using sin2 =
1 - cos2 gives thefollowing second-orderalgebraicequationfor A:
2
+ 2Acosa+ 1 = -sin2a-
2
A
sin
f/J
(6.55)
This can be solved by using the
standardexpressionfor the roots of a second-order
polynomial, or by further rewriting the expression.In any way it will lead to the
following (positive) root:
sin a
A.=---cosa
tanf/J
(6.56)
Combining(6.56) with (6.54) gives thefollowing expressionfor the critical distancefor
a critical phase-anglejump cP:
Leril
= -z, {Sina
-----:i: - cosa }
z tan 'P
(6.57)
8.5.8 Parallel Feeder.
Voltage sags onparallel feedersand other loops have beendiscussedin Section
4.2.4. Therewe sawthat most faults on parallel feederstoward the load, lead to deep
sags. It is anacceptableapproximationto makethe sagmagnitudezerofor all faults on
the parallel feeders. In caseof long feeders(feederimpedancemore than two or three
times the sourceimpedance)some additional calculation is needed.It is possibleto
derive anexpressionfor the critical distancefor parallel feeders from (4.18)but that
expressionwould be too complicatedto beof any use.Insteada simplified calculationis
proposed.
The voltage profile along the feedercan be approximatedas a (second-order)
parabola:
v.rag ~
4Vmaxp(1 - p)
(6.58)
with p indicating the position of the fault along the feeder, 0~ p :5 1, and Vmax the
maximumsagvoltagedue to afault anywhereon the feeder.T hereis no simpleexpression for Vmax; it needs to beobtainedgraphicallyfrom Fig. 4.34 or Fig. 4.35.W hen the
maximumvalue isknown, the "critical fraction" is readily obtained:
Peril ~
I -
I _ Veril
Vmax'
(6.59)
1400
11000
!2000
j 1500
x....-=~x-~
100
XC==40
60
8'0
800
0
0
200
400
40
60
80
Sagmagnitudein percent
20
" " ;,;z.
-'II'-Z-;r
100
i
00
&1 100
.!400
]
300
8. 200
40
60
80
40
60
I
80
I
40
60
80
Sagmagnitudein percent
20
Sagmagnitudein percent
20
i
,~,
Sagmagnitudein percent
20
o
x
100
100
I
100
'
s
1600
1400
2500
1500
~
~
e
CIJ
00
500
-g 1000
.£
-=i
.S 2000
]
°0
200
&400
~
~600
fa
.S 1200
1000
]
800
~
40
60
40
60
80
Sagmagnitudein percent
20
,",z-r-*""I
~/z
1
80
Sagmagnitudein percent
20
*_*__X#..
J
JX
~x
100
100
Figure 6.49 Exposedlength for nine 400 kV substations:c omparisonbetweenthe methodof fault positions(crosses)and the methodof critical distances(diamonds).
!
~
"d
]600
fo
600
500
.5
oI
o
800
700
~
ttl
- 600
~
~ 400
Q.
~ 200
a
z
00
1
1400
= 1200
~ 1000
io 800
]
0
0
500
1000
B 1200
Sagmagnitudein percent
20
h
h
r:
h
*
2
~
~.
&
"'0
2000
] 1500
t
3000
~ 2500
.s 1000
~ 500
o0
o
J
~x..~-;...x
.
20
40
60
80
Sagmagnitudein percent
_.
3000~--------------.
200
00
400
600
~ 2500
~
&
~
"'0
= 1200
:B 1000
j 800
]
387
Section 6.5 • TheMethod of Critical Distances
The contribution of the feeder to the exposed length equals the critical fraction times
the feeder length.F or Veri' > Vmax the whole feedercontributesto the exposed length.
8.5.9 Comparison with the Method of Fault Positions
The transmissionsystem study performed by
Qader [71], [74] resulted in the
number of sags as afunction of magnitudefor all substationsin the U.K. 400-kV
transmissionsystem. Themethod of fault positions was used for this study.
For a
comparedwith the resultsobtainedby
numberof substationsthose results have been
using themethodof critical distances. The critical distance was calculated as a function
of the sagmagnitudeV by using theapproximatedexpression
z,
V
£crit = ~ 1 _ V
(6.60)
where Zs is the sourceimpedanceand z the feeder impedance per unit length. All the
lines originating at the substationare assumed infinitely long; the exposed length is
simply the criticaldistancetimes thenumberof lines.
The sourceimpedanceZs is calculated by assuming
t hat all lines contribute
equally to theshort-circuitcurrent for a busbarfault. During a fault on oneof these
lines, only (N - 1) out of N lines contribute to the short-circuit current. Thus, the
sourceimpedancein p.u. equals
z, = -.!!.-.- Sbase
N - I
(6.61)
Sjault
with N the numberof linesoriginatingat thesubstation,Sba.vethe base power, and
S/auft
the short-circuitpower for asubstationfault. The exposed length is found from
~
2
r
r
'-exp = N x '-erit
= NN_
I
--z 1 _V V
Slaul,
(6 2)
.6
The exposed length for the nine
substationsis shown in Fig. 6.49, where the crosses
indicate the resultsof the method of fault positions. There are obviously differences
of the twomethods,with the method of fault positions viewed as the
between the results
most accurateone. But for themethodof fault positions a largepart of the national
grid needs to be modeled. All the
data needed for the method of critical distances is,
from equation(6.62):
• numberof lines originating from the substation;
• fault level of the substation;and
• feederimpedanceper unit length.
All this datacan beobtainedwithout much difficulty.
Another interestingobservationfrom (6.62) concerns thevariation in sag frequency among different substations.The main variation can be brought back to
fault level,numberof lines originating at thesubstation,and fault frequency.
Mitigation of
Interruptions and
Voltage Sags
This chaptergives an overviewof methodsto mitigate voltagesags andinterruptions.
o f the variousforms of mitigation, we concentrateon power
After a general discussion
system design and on
mitigation equipmentto be installed between thepower system
and the sensitiveequipment.Especially thelatter is underfast developmentsince a few
years. Anattemptis made to give aneutraloverview of the variousoptions,knowing
that new developmentsare veryhard to predict. Powersystem design is a
m ore traditional area,althoughnew developmentsin power electronicsare also expected to have
an impact here.
7.1 OVERVIEW OF MITIGATION METHODS
7.1.1 From Fault to Trip
In the previous chapterswe discussed voltagemagnitudeevents (voltage sags,
shortinterruptions,and longinterruptions)in considerabledetail: their origin,methods
of characterization,monitoring and prediction,and their effects onequipment.In this
chapterwe look at existing and future ways
of mitigating voltagemagnitudeevents. To
understandthe variouswaysof mitigation, the mechanismleading to anequipmenttrip
needs to beunderstood.Figure 7.1 shows how ashort circuit leads to anequipment
trip. The equipmenttrip is what makes the event problem;if
a
there were noequipment
trips, there would not be any voltage
quality problem. The underlying event of the
equipmenttrip is a short-circuit fault: a low-impedanceconnectionbetween two or
more phases, or between one or more phases ground.
and
At the fault position the
voltage drops to a low value. The effecto f the short circuit at other positionsin the
system is an event of caertain magnitudeand duration at the interface between the
equipmentand the power system. The
short-circuitfault will always cause avoltagesag
for somecustomers.If the fault takes place in a
radial part of the system, theprotection
intervention clearing the fault will also lead to an
interruption. If there is sufficient
redundancypresent, theshort circuit will only lead to a voltage sag. If theresulting
event exceeds caertain severity, it will cause anequipmenttrip. Admittedly, not only
389
390
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Reduce number
of faults
Improve system
design
Mitigate
disturbance
Improve
equipment
Figure 7.1 The voltagequality problemand
ways of mitigation.
shortcircuits lead toequipmenttrips, but also events likecapacitorswitching or voltage
sags due tomotor starting. But the largemajority of equipmenttrips will be due to
short-circuit faults. Most of the reasoningto follow also applies to anyother event
potentially leading to anequipmenttrip.
Figure 7.1 enables us todistinguishbetween thevariousmitigation methods:
• reducing thenumberof short-circuitfaults.
• reducing thefault-clearingtime.
• changingthe system suchthat short-circuitfaults result in less severe events at
the equipmentterminalsor at thecustomerinterface.
• connectingmitigation equipment between the sensitiveequipmentand the
supply.
• improving the immunity of the equipment.
These four types ofmitigation are discussed briefly next. Power system design and
mitigation equipmentat the system-equipmentinterfaceare discussed in detail in the
remainderof this chapter.Power engineers have always usedcombinationof
a
these
mitigation methodsto ensurea reliableoperationof equipment.Classically the emphasis has been on reducing the
number of interruptions, while recently emphasishas
shifted toward mitigating voltagesags.
7.1.2 Reducing the Number of Faults
Reducingthe numberof short-circuitfaults in a systemnot only reduces the sag
frequency but also the frequency
of sustainedinterruptions.This is thus a very effective
way of improvingthe quality of supplyand manycustomerssuggest this as the obvious
solution when a voltage sag or
short interruptionproblemoccurs. Unfortunately,the
solution is rarely that simple. A short circuit not only leads to a voltage sag or interruption at thecustomerinterfacebut may also causedamageto utility equipmentand
plant. Thereforemost utilities will alreadyhave reduced the fault frequency as far as
economically feasible. Inindividual cases therecould still be room for improvement,
e.g., when themajority of trips is due to faults on one or two
distribution lines. Some
examples of faultmitigation are:
Section 7.1 • Overview of Mitigation Methods
391
• Replaceoverheadlines byundergroundcables. A largefraction of short-circuit
faults is due to adverseweatheror other external influences.U nderground
cables are much less affected externalphenomena
by
(with the obvious exception of excavation).The fault rate on anundergroundcable is anorder of
magnitudelessthan for an overheadline. The effect is a bigreductionin the
number of voltage sags andinterruptions. A disadvantageof underground
cables isthat the repair time is much longer.
• Use covered wires foroverheadline. A recentdevelopmentis the construction
of overheadlines with insulatedwires. Normally the wires of anoverheadline
are bareconductors.With covered wires, theconductorsare covered with a
thin layer of insulatingmaterial.Even thoughthe layer is not a fullinsulation,
it has proven to be efficient inreducingthe fault rate ofoverheadlines [208],
[212]. Also other types ofconductorsmay reduce the fault rate
[213].
• Implementa strict policyof tree trimming.Contactbetween treebranchesand
wires can be animportantcauseof short-circuitfaults, especiallyduring heavy
loadingof the line. Due to theheatingof the wires their sag increases, making
contactwith trees more likely.N ote that this is also the timeduring which the
consequences
of a short circuit are most severe.
• Install additional shielding wires.Installation of one or two shielding wires
reduces the riskof a fault due to lightning. The shielding wires are located
such that severelightning strokesare most likely to hit a shielding wire. A
lightning stroke to a shielding wire isnormally conductedto earththrougha
tower.
• Increasethe insulation level. This generally reduces the risk short-circuit
of
faults. Note that many short circuits are due toovervoltagesor due to a
deteriorationof the insulation.
• Increasemaintenanceand inspectionfrequencies. This again generally reduces
the risk of faults. If themajority of faults are due to adverse weather, as is often
the case, the effect
of increasedmaintenanceand inspectionis limited.
that these measures may be very expensive that
and
One has to keep in mind, however,
its costs have to be weighted
againstthe consequences
of the equipmenttrips.
7.1.3 Reducing the Fault-Clearing Time
Reducingthe fault-clearingtime does not reduce the
numberof events but only
their severity. It does not doanything to reduce thenumberor duration of interruptions. Thedurationof an interruptionis determinedby the speed with which the supply
is restored.Fasterfault-clearingdoes also not affect the
numberof voltage sags but it
can significantly limit the sagduration.
The ultimate reductionin fault-clearingtime is achieved by using
current-limiting
fuses [6],[7]. Current-limitingfuses are able to clear a fault within one half-cycle,that
so
the durationof a voltage sag will rarely exceed one cycle.
If we further realizethat fuses
have an extremely small
chanceof fail-to-trip, we have what looks like theultimate
solution. The recentlyintroducedstatic circuit breaker[171], [175] also gives a faultclearing time within one half-cycle; but it is obviously much more expensive
than a
current-limiting fuse. No information is availableaboutthe probability of fail-to-trip.
Additionally several types offault-currentlimiters have beenproposedwhich not so
392
Chapter7 • Mitigation of Interruptionsand Voltage Sags
much clear the fault, but significantly reduce the
fault-currentmagnitudewithin one or
two cycles.
One importantrestrictionof all these devices is
t hat they can only be used for lowand medium-voltagesystems. Themaximumoperatingvoltage is a few tenso f kilovolts.
Staticcircuit breakersshow thepotentialto be able tooperateat higher voltage levels in
the future.
breakerbut also
But thefault-clearingtime isnot only the time needed to open the
the time needed for the
p rotectionto make a decision. Here we need considertwo
to
significantly different types of distribution networks,both shown in Fig. 7.2.
The topdrawingin Fig. 7.2 shows a system with one circuit
breakerprotectingthe
whole feeder.The protectionrelay with thebreakerhas acertaincurrentsetting. This
not exceeded for
setting is suchthat it will be exceeded for any fault on the feeder, but
nor for any loadingsituation.The momentthe current
any fault elsewhere in the system
value exceeds thesetting (thus for any fault on the feeder) the relay
instantaneously
gives a trip signal to thebreaker. Upon receptionof this signal, thebreakeropens
within a few cycles. Typicalfault-clearingtimes in these systems are
around 100 milliseconds. To limit thenumberof long interruptionsfor the customers,reclosing is used
in combination with (slow) expulsion fuses in the laterals or incombination with
interruptorsalong the feeder. This typeof protectionis commonly used inoverhead
systems.Reducingthe fault-clearing time mainly requires a fasterbreaker.The static
circuit breakeror severalof the othercurrentlimiters would be goodoptionsfor these
systems. Acurrent-limitingfuse to protectthe whole feeder is notsuitableas it makes
fast reclosingmore complicated.Current-limiting fuses can also not be used for the
protection of the laterals because they wouldstart arcing before the mainbreaker
opens. Using a fasterclearing with the main breakerenables fasterclearing in the
lateralsas well.
The network in the bottom drawing of Fig. 7.2 consistsof a numberof distribution substationsin cascade. To achieve selectivity,
time-grading of the overcurrent
relays is used. The relays
furthest away from the source tripinstantaneouslyon overcurrent.When moving closer to the source, the
tripping delay increases each time with
typically 500 ms. In theexamplein Fig. 7.2 the delay times would be 1000ms, 500 ms,
and zero(from left to right). Close to the source,
fault-clearing times can be up to
several seconds. These kind
of systems are typically used underground
in
networksand
in industrial distribution systems.
pr~
. .
overcient
Figure 7.2Distribution system with one
circuit breakerprotectingthe whole feeder
(top) and with anumberof substations
(bottom).
Section 7.1 • Overviewof Mitigation Methods
393
The fault-clearingtime can be reduced by using inverse-time
overcurrentrelays.
For inverse-timeovercurrentrelays, the delay time decreases for
increasingfault current. But even with these schemes,
fault-clearingtimes above one second are possible.
The varioustechniquesfor reducingthe fault-clearingtime without loosing selectivity
are discussed invariouspublicationson power systemprotection,e.g., [176] and[10].
To achieve a seriousreduction in fault-clearing time one needs to reduce the
grading margin, therebyallowing a certain loss of selectivity. The relay setting rules
described in mostpublicationsare based onpreventingincorrecttrips. Futureprotection settings need to be based onmaximumfault-clearingtime.
a
A methodof translating a voltage-tolerancecurve into atime-currentcurve is described in[167]. The latter
curve can be used in
c ombinationwith relay curves toobtain the varioussettings. The
opening timeof the downstreambreakeris an importantterm in theexpressionfor the
gradingmargin. By using fasterbreakers,or evenstatic circuit breakers,the grading
margin can be significantly reduced, thus leading to a significant
reduction in faultclearing time. Theimpactof staticcircuit breakersmight be bigger in these systems
than
in the ones with onebreakerprotectingthe whole feeder.
In transmissionsystems thefault-clearingtime is often determinedby transientstability constraints.These constraintsare much more strictt han the thermal constraintsin distribution systems,requiring shorterfault-clearingtimes, rarely exceeding
200ms. This also makes
t hat further reductionof the fault-clearingtime becomes much
more difficult. Someremainingoptions for the reductionof the fault-clearingtime in
transmissionsystems are:
• In some cases faster circuit
breakerscould beof help. This againnot only limits
the fault-clearingtime directly but it also limits thegradingmarginfor distance
protection. One should realize howeverthat faster circuit breakerscould be
very expensive.
• A certain reductionin grading margin is probably possible. This willnot so
much reduce thefault-clearingtime in normal situations,but in case the proreducingthe grading
tection failsand a backuprelay has to intervene. When
margin oneshouldrealizethat lossof selectivity isunacceptablein most transmission systems as it leads to the loss of twomorecomponents
or
at the same
time.
• Fasterbackupprotectionis one of the few effective meanso f reducing faultclearing time intransmissionsystems. Possible
optionsare to useintertripping
for distanceprotection,and breaker-failureprotection.
7.1.4 Changing the Power System
By implementingchangesin the supply system, the severity
of the event can be
reduced. Here again the costs can become very high, especially
transmissionand
for
subtransmissionvoltage levels. The mainmitigation methodagainstinterruptionsis the
installationof redundantcomponents.
Some examples of
mitigation methodsespecially directedtowardvoltage sags are:
• Install a generatornearthe sensitive load. The
generatorswill keep the voltage
up during a sag due to aremotefault. The reductionin voltagedrop is equal to
the percentagecontributionof the generatorstationto the faultcurrent.In case
394
Chapter7 • Mitigation of Interruptionsand Voltage Sags
a combined-heat-and-power
station is planned, it is worth to consider the
position of its electricalconnectionto the supply.
• Split busses orsubstationsin the supplypathto limit the numberof feeders in
the exposed area.
• Install current-limiting coils at strategicplaces in the system to increase the
"electricaldistance"to the fault. Oneshouldrealizethat this canmakethe sag
worse for other customers.
• Feed the bus with the sensitive
equipmentfrom two or more substations.A
voltage sag in onesubstationwill be mitigated by the infeed from theother
substations.The moreindependentthe substationsare themorethe mitigation
effect. The bestmitigation effect is by feeding from twodifferent transmission
substations.Introducing the second infeed increases the
numberof sags, but
reduces their severity.
The numberof short interruptionscan bepreventedby connectinglesscustomersto
one recloser (thus, byinstalling more reclosers), or bygetting rid of the reclosure
schemealtogether.Short as well as longinterruptions are considerablyreduced in
frequency byinstalling additional redundancyin the system. The costs for this are
only justified for large industrial and commercialcustomers.Intermediatesolutions
reduce theduration of (long) interruptionsby having a levelo f redundancyavailable
within a certain time. The relations betweenpower system design,interruptions,and
voltage sags are discussed in detail in Sections 7.2 and 7.3.former
The mainly considers
methodsof reducing thedurationof an interruption,where thelatterdiscussesrelations
between sag frequency and system design.
7.1.5 Installing Mitigation Equipment
The mostcommonlyappliedmethodof mitigation is theinstallationof additional
equipmentat the system-equipmentinterface. Recentdevelopmentspoint toward a
continuedinterest in this wayof mitigation. The popularity of mitigation equipment
is explained by it being the only place where the
customerhascontrolover thesituation.
Both changes in the supply as well improvementof
as
the equipmentare often completely outsideof the control of the end-user.
Some examples ofmitigation equipmentare:
• Uninterruptiblepower supplies (UPSs) are extremely
popularfor computers:
personalcomputers,central servers, andprocess-controlequipment.For the
latter equipmentthe costs of a UPS are negligible
comparedto the total costs.
• Motor-generatorsets are oftendepictedas noisyand as needingmuch maintenance. But inindustrial environmentsnoisy equipmentand maintenanceon
rotating machines arerathernormal. Large batteryblocks alsorequiremaintenance, expertise on which is much less available.
• Voltage sourceconverters (VSCs) generatea sinusoidal voltage with the
required magnitudeand phase, by switching a de voltage inparticularway
a
over the three phases. This voltage source can be used
mitigatevoltage
to
sags
and interruptions.
Mitigation equipmentis discussed in detail in Section 7.4.
Section 7.1 • Overviewof Mitigation Methods
395
7.1.8 Improving Equipment Immunity
Improvementof equipmentimmunity is probably the most effectivesolution
against equipmenttrips due to voltage sags. But it is often
not suitable as a shorttime solution. A customeroften only findsout about equipmentimmunity after the
equipmenthas been installed.
For consumerelectronics it is veryhardfor a customerto
find out about immunity of the equipmentas he is not in directc ontact with the
manufacturer.Even mostadjustable-speed
drives have become
off-the-shelfequipment
where thecustomerhas no influence on the specifications. Only large
industrial equipment is custom-madefor a certain application, which enables theincorporationof
voltage-tolerancerequirements.
Severalimprovementoptions have been discussed in detail Chapter5.
in
Some
specificsolutionstoward improved equipmentare:
• The immunity of consumerelectronics,computers,and controlequipment(i.e.,
single-phase low-powerequipment)can be significantlyimproved by connecting more capacitanceto the internal de bus. This will increase the maximum
sagdurationwhich can betolerated.
• Single-phase low-powerequipmentcan also beimproved by using a more
sophisticatedde/de converter: one which is able to
operate over a wider
range ofinput voltages. This will reduce the
minimum voltage for which the
equipmentis able tooperateproperly.
• The main source ofconcernare adjustable-speed
drives. We sawthat ac drives
can be made totoleratesags due to single-phase and
phase-to-phase
faults by
adding capacitanceto the de bus. To achieve
toleranceagainst sags due to
three-phasefaults, seriousimprovementsin the inverteror rectifier are needed.
drives is very difficult because
• Improving the immunity of de adjustable-speed
the armaturecurrent, and thus thetorque, drops very fast. Themitigation
methodwill be very muchdependenton restrictionsimposed by theapplication
of the drive.
• Apart from improving (power) electronicequipmentlike drives and processcontrol computersa thorough inspection of theimmunity of all contactors,
relays, sensors, etc., can also significantly
improve the processridethrough.
• When newequipmentis installed, information about its immunity should be
obtained from the manufacturer beforehand. Where possible,immunity
requirementsshould be included in theequipmentspecification.
For short interruptions,equipmentimmunity is very hard to achieve; for long interruptions it is impossible to achieve. The
equipmentshould in so far be immune to
interruptions, that no damageis caused and nodangeroussituation arises. This is
especiallyimportantwhen consideringa completeinstallation.
7.1.7 Different Events and Mitigation Methods
Figure 7.3 shows themagnitudeand duration of voltage sags andinterruptions
For differenteventsdifferent mitigation strategies
resulting from various system events.
apply.
396
Chapter7 • Mitigation of Interruptionsand Voltage Sags
100%
800/0
]
.~
~
~
50%
Local
MVnetworks
Interruptions
0% - - - - - -....- - - - - -.....- - - - - - - - - - - - - 0.1 s
1s
Duration
Figure 7.3 Overviewof sags andinterruptions.
• Sags due toshort-circuitfaults in thetransmissionand subtransmissionsystem
are characterizedby a short duration, typically up to lOOms. These sags are
very hard to mitigate at the source and also
improvementsin the system are
seldom feasible. The only way
of mitigating these sags is by
improvementof the
equipmentor, where thisturnsout to be unfeasible,installingmitigation equipment. For low-powerequipmenta UPS is astraightforwardsolution; for highpower equipmentand for completeinstallationsseveralcompetingtools are
emerging.
• As we saw in Section 7.1.3, the
duration of sags due todistribution system
faults dependson the typeof protectionused,rangingfrom lessthana cycle for
current-limiting fuses up to several seconds for
overcurrentrelays in underground or industrial distribution systems. The long sag
duration makesthat
equipmentcan also trip due to faults on
distribution feeders fed fromanother
HV/MV substation.For deep long-duration sags,equipmentimprovement
becomes more difficulta nd systemimprovementeasier. Thelatter could well
become thepreferredsolution, althougha critical assessment of the various
options is certainly needed.Reducingthe fault-clearing time and alternative
designconfigurationsshould be considered.
• Sags due to faults in
remotedistributionsystems and sags due motorstarting
to
should not lead to equipmenttripping for sags down to85%. If there are
problemsthe equipmentneeds to be improved. Ifequipmenttrips occur for
long-durationsags in the70%-80% magnituderange, improvementsin the
system have to be
consideredas anoption.
improving the equipmentimmu• For interruptions,especially the longer ones,
nity is no longer feasible. System
improvementsor a UPS incombinationwith
an emergencygeneratorare possiblesolutions here. Somealternativesare
presentedin Sections 7.2 and 7.3.
Section 7.2 • Power System
D esign-Redundancy
ThroughSwitching
397
7.2 POWER SYSTEM DESIGN-REDUNDANCY THROUGH SWITCHING
This and the next section discuss some
of the relationsbetweenstructureandoperation
of power systemsand the numberof voltage sags andinterruptions.The reductionof
interruptionfrequency is animportantpart of distributionsystem design and as such it
is treatedin detail in a numberof books and in many papers.Often cited books on
distribution system design are"Electricity Distribution Network Design" by Lakervi
and Holmes [114] and "Electric Power Distribution SystemEngineering" by Gonen
[164]. Other publicationstreatingthis subject inpart are [23], [115], [116], [165], [209],
[214]. Many case studies have
appearedover the years in conferences
and transactions
of the IEEE Industry ApplicationsSociety and to a lesser degree in the
publicationsof
the PowerEngineeringSociety andof the Institute of Electrical Engineers.
7.2.1 Types of Redundancy
The structureof the distribution system has a big influence on the
numberand
durationof the interruptionsexperienced by thecustomer.The influence of thetransmission system ismuch smaller becauseof the high redundancyused. Interruptions
originating in the distribution system affect lesscustomersat a time, but any given
customerhas a muchhigherchanceof experiencing adistribution-originatedinterrupone. The largeimpact of interruptionsoriginating
tion than a transmission-originated
in the transmissionsystemmakesthat they shouldbe avoided atalmostany cost. Hence
the high reliability of transmissionsystems.
Number and duration of interruptionsis determinedby the amount of redundancy presentand the speed with which the
redundancycan be made available.
Table
of redundancyand thecorrespondingdurationof the interruption.
7.1 gives some types
Whetherthe supply to a certainload is redundantdepends on the time scale at which
one islooking. In otherwords, on themaximuminterruptiondurationwhich the load
cantolerate.
When apowersystemcomponent,e.g., atransformer,fails it needs to berepaired
or its function takenover byanothercomponentbefore the supply can be
restored.In
casethereis no redundanttransformeravailable, the faultedtransformerneeds to be
repairedor a spareone has to beb roughtin. The repairor replacementprocess can take
severalhours or, especially withpower transformers,even days up to weeks.
Repair
times of up to onemonth have beenreported.
TABLE 7.1 Various Types of Redundancyin Power System Design
Duration of Interruption
No redundancy
Redundancythroughswitching
- Local manualswitching
- Remotemanualswitching
- Automatic switching
- Solid stateswitching
Redundancythrough parallel
operation
Typical Applications
Hours throughdays
Low voltage in rural areas
1 hour and more
5 to 20minutes
Low voltage anddistribution
Industrial systems,
future public distribution
Industrial systems
Futureindustrial systems
Transmissionsystems,
industrial systems
I to 60 seconds
I cycle and less
Voltage sag only
398
Chapter7 • Mitigation of Interruptionsand Voltage Sags
In most cases the supply is
n ot restored through repair or replacementbut by
switching from the faulted supply to backupsupply.
a
The speed with which this takes
place dependson the type of switching used. The various types will be discussed in
detail in theremainderof this section.
A smoothtransitionwithout any interruptiontakes place when two
c omponents
are operatedin parallel. This will however notmitigate the voltage sag due to the fault
which often precedes the
interruption.Various optionsand their effect on voltage sags
are discussed in Section 7.3.
7.2.2 Automatic Recloslng
Automatic reclosing was discussed in detail Chapter3.
in
Automatic reclosing
after a short-circuitfault reduces thenumberof long interruptionsby changingthem
into short interruptions.Permanentfaults still lead to longinterruptions,but on overhead distribution lines this is lessthan 25% of the total numberof interruptions.We
saw in Chapter3 that the disadvantageof the commonly usedmethod of automatic
reclosing isthat more customersare affected by a fault. A long
i nterruptionfor part of
a feeder ischangedinto a shortinterruptionfor the whole feeder. This is
not inherentto
automatic reclosing, but to themethod of fuse saving used. If all fuses would be
replaced by reclosers, the
numberof shortinterruptionswould be significantly reduced.
A customerwould only experience ashort interruption for what would have been a
long interruptionwithout reclosing. This wouldof course make the supply more expensive, which is not alwaysacceptablefor remote rural areas.
7.2.3 Normally Open Points
The simplest radial system possible is shown in Fig. 7.4:
number
a
of feeders
originate from a distribution substation.When a fault occurs on one
o f the feeders,
the fuse will clear it, leading to an
interruptionfor all customersfed from this feeder.
The supply can only berestoredafter the faultedcomponenthas ·beenrepaired or
replaced. Such systems can be found rural
in low-voltage anddistribution systems
with overheadfeeders.Protection is through fuses in thelow-voltage substations.
Repair of a faulted feeder (orreplacemento f a blown fuse) can take several
hours,
repair or replacemento f a transformerseveral days. As the feeders are
overheadthey
are prone to weather influences;stormsare especiallynotoriousfor it can take days
before all feeders have been repaired.
A commonlyusedmethodto reduce thedurationof an interruptionis to install a
normally open switch, often called"tie switch." An example is shown in Fig. 7.5.
Lateral
Figure 7.4 Power systemwithout
redundancy.
399
Section 7.2 • Power System
D esign-Redundancy
ThroughSwitching
33/11 kV
n/o switch
----:
~
ntc¥nto
0/0
11kvt400~
Figure 7.5 Distribution system with
redundancy through manual switching.
The system is stilloperatedradially; this prevents the fault level from getting too
high and enables the use
of (cheap)overcurrentprotection.If a fault occurs it is cleared
by a circuitbreakerin the substation.The faulted section is removed, the
normallyopen
switch is closed, and the supply can restored.The
be
varioussteps in therestorationof
the supply are shown in Fig. 7.6.
(a) Normaloperation
I
Nonnallyopen
point
T$ $ $/' $ $
(b) Fault clearing
(c) Interruption
---r-
Interruptionfor
these customers
____T
(d) Isolatingthe fault
---r-
n---~$ $
(e) Restoringthe supply
Figure 7.6 Restoration procedure in a
distribution system with normally open
points. (a) Normal operation, (b) fault
clearing, (c) interruption, (d) isolating the
fault, (e) restoring the supply.
400
Chapter7 • Mitigation of Interruptionsand Voltage Sags
In normal operation(a) the feeder isoperatedradially. A normally open switch is
located between this feeder and
anotherfeeder,preferablyfed from anothersubstation.
When a fault occurs (b) the
breakerprotectingthe feeder opens leading to an
interruption for all customersfed from this feeder (c).After the fault is located, it is isolated
from the healthypartsof the feeder (d) and the supply to these healthy
partsis restored
by closing the circuitbreakerand thenormally open switch (e).Repairof the feeder
only startsafter the supply has been restored.
.
This procedurelimits the durationof an interruptionto typically one or two hours
in case the switching is done locally (i.e.,
somebodyhas to go to the switches to open or
close them). If faultlocation and switching is done remotely (e.g., in a regional
control
center) the supply can be
restoredin several minutes.Locatingthe fault may take longer
than theactualswitching. Especially in case
of protectionor signaling failure, locating
the fault can take a long time.
Various techniques are in use for identifying the faulted
section of the feeder. More precise fault
location,needed for repair, can be done afterwards.
The type of operationshown in Figs. 7.5 and 7.6 is very commonly used in
undergroundlow-voltage and medium-voltagedistribution systems. Therepair of
undergroundcables can take several days that
so system operationlike in Fig. 7.4
becomes totallyunacceptable.Similar restorationtechniquesare in use for mediumvoltage overheaddistribution, especially in the moreurbanpartsof the network. The
high costs for signalingequipmentand communicationlinks make remote switching
only suitable for higher voltages
andin industrialdistributionsystems. Whencustomer
demands forshorterdurationsof interruptionscontinueto increase, remote signaling
and switching will find its way into publicdistribution systems as well.
The additional costs for the system in Fig. 7.5 are not only switching, signaling
and communicationequipment.The feeder has to be
dimensionedsuch that it can
handle the extra load. Also the voltage
drop over the, nowpotentially twice as long,
feeder should not exceed the margins.
Roughlyspeakingthe feeder can only feed
half as
much load. This will increase the
n umberof substationsand thus increase the costs.
7.2.4 Load Transfer
A commonly used and very effective way
of mitigating interruptionsis transferring the load from theinterruptedsupply to ahealthysupply. Load transferdoes not
affect thenumberof interruptions,but it can significantly reduce thed uration of an
interruption. Load transfercan bedone automaticallyor manually; automatictransfer is faster and therefore more effective in reducing interruption
the
duration. An
example of manual switching was discussed before. Here we will
concentrateon
automatictransfer of load, although the proposedschemes are equally suitable for
manualtransfer.
7.2.4.1 Maximum Transfer Time. An importantcriterion in the designof any
transfer scheme is themaximum interruption duration that can betolerated by the
equipment.The transfershould take place within this time, otherwise the load would
trip anyway. In anindustrialenvironmentthe rule for themaximum transfertime is
relatively simple: theshort interruptionof the voltageshould not lead to aninterruption of plant operation.An example is apapermill, where the interruption should
not lead totripping of the papermachine. Below acertain interruptionduration the
machine will not trip, for interruptionslasting longer it will trip. The choice is not
always that straightforward,e.g., with lighting of public buildings. A general rule is
Section 7.2 • Power System
D esign-Redundancy
ThroughSwitching
401
that one should in all cases choose taransfer time such that the transfer does not
unacceptableis
lead to unacceptableconsequences.W hat should be considered as
simply part of the decision process. In practice the load
of a power system is not
constant,and decisionsabouttransfertime may have to be revised several years later
because more sensitive
equipmentis being used, as, e.g., described[163].
in
7.2.4.2 Mechanical LoadTransfer. Most transfer schemes use a mechanical
switch or circuit breakerto transferfrom one supplypoint to another.A typical configuration as used inindustrial distribution systems is shown in Fig. 7.7. Two transformers eachsupply part of the load. If oneof them fails, thenormally open switch
is closed and thetotal load is fed from onetransformer.Each transformershould be
able to supply thetotal load or a load shedding scheme
should be in place. When a
short circuit occurs close to thetransfer switch, it is essentialt hat the load is not
transferred before the fault has been cleared: a so-called
"break-before-make"
scheme. A"make-before-break"scheme would spread the fault to the healthy supply
leading to possibleintervention by the protectionin both feeders. In case one transformer is taken out of operation for maintenance,a (manual) make-before-break
scheme can be used. This reduces the risk of a interruptiondue
long
to failure of the
transfer switch. During the parallel operation,a short circuit could lead to serious
switchgeardamage.
The advantagesof this schemecomparedto parallel operationare that the protection is simpler andthat the fault currentis lower. As long as the load can
toleratethe
shortinterruptionduring load transfer,the reliability of the supply is similar tothat of
parallel operation.As we saw in Section 2.8, load
interruptionsfor a transferscheme
are mainly due to failureof the transferswitch and due to any kind of
common-mode
effect in the two supplies. In an
industrial environment,maintenanceand excavation
activities could seriously effect the supply reliability.
66 kV substation
Figure 7.7 Industrial power system with
redundancythroughautomaticswitching.
Variousindustrialload
7.2.4.3 Transfer of Motor Load. A problem with automaticswitching is the
presenceof large numbersof induction motors in most industrial systems. When the
supply is interrupted,the remaining airgap flux generates a voltage over the
motor
terminals. This voltage decays inmagnitudeand in frequency. The switching has to
take placeeither very fast (before themotor voltage has shifted much in phase comparedto the system voltage) or very slow
(after the motor voltage has become zero).
As the first option is expensive, the second onenormally
is
used.
402
Chapter7 • Mitigation of Interruptionsand Voltage Sags
The airgap field in a induction motor decayswith a certaintime constantwhich
varies from less than one cycle for small motors up to about 100 ms for large motors.
The time constantwith which the motor slows down is much larger: typically between
one and five seconds.
The momentthe motor is reconnected,the sourcevoltagewill normally not be in
phasewith the motor voltage. In case they are in opposite phasea large current will
flow. This currentcan be morethantwice thestartingcurrentof the motor. It caneasily
damagethe motor or lead to tripping of the motor by the overcurrentprotection.
The inducedvoltage has the following form:
E = isinro!
(7.1)
with co the angularspeedof the motor, which decaysexponentially:
(J) =
Wo(1 - e-f.;)
(7.2)
and E dependenton the frequency and the exponentially decaying rotor current.
Assume for simplicity that the magnitudeof the induced voltage remains constant
and considera linear decayin motor speed:
(J) ~ (J)O(1 -
L)
(7.3)
This gives for the voltage at the motor terminals:
E(t) =
Sin(Wo(1 - L)r) = sin(Wot _ ~~2)
(7.4)
The secondterm underthe sinefunction is the phasedifferencebetweenthe supplyand
the induced voltage. As long as this phasedifference is less than 60°, the voltage
difference betweenthe sourceand the motor is less than 1 pu. A phasedifference of
60° (1) is reachedfor
~
t=y6KJ
(7.5)
For a mechanicaltime constantT:m = 1 secand a frequencyof 10 = 50 Hz an angular
differenceof 60° is reachedafter 58 ms. In thecalculationit is assumedthat the motor
has not slowed down during the fault. If this is also considered,the value of 60° is
reachedfaster.Only very fast transferschemesareableto switch within this shorttime.
A secondchanceat closing the transferswitch is when the angulardifferenceis about
360° (i.e., sourceand motor are in phaseagain). This takesplacefor
&
t=Yh
(7.6)
which is 140IDS in the aboveexample.Theseso-calledsynchronoustransferschemesare
very expensiveand may still leadtotransfertimes above 100 ms. In most cases asynchronoustransfer is used where the transfer switch is only closed after the induced
voltagehassufficiently decayed,leadingto transfertimes aroundonesecondor longer.
For synchronousmachinesthe airgapfield decayswith the sametime 'constantas
the motor speed,so that the terminal voltagemay be presentfor severalseconds.In a
systemwith a large fraction of synchronousmotor load, synchronoustransferbecomes
403
Section 7.2 • Power System
D esign-Redundancy
ThroughSwitching
more attractive. Note that asynchronoustransferwill always lead to lossof the synchronousmotor load.
7.2.4.4 Primary and Secondary Selective Supplies.
Figures 7.8 and 7.9 show
two ways of providing a medium-voltagecustomerwith a reliable supply. In a primary selective system (Fig. 7.8) the
transfer takes place on theprimary side of the
but there is a
transformer.A secondaryselective system (Fig. 7.9) is more expensive
much reduced chance of very long
interruptionsdue to transformerfailure. A numerical analysisof such atransferscheme is given in Section 2.8.
The actual transferis identical to thetransferin the industrial supply shown in
Fig. 7.7: the load istransferredfrom the faulted to thehealthyfeeder as soon as possible
after fault clearing. With aprimary selective supply amake-before-break
scheme would
directly connecttwo feeders. It is unlikelythat the utility allows this. Thetransfertakes
place behind atransformerwith the secondaryselective supply. The possible consequencesof a make-before-breakscheme are less severe for the utility.
With the design ofprimary and secondaryselective supplies, it is again very
important to determinethe tolerance of the load to
s hort interruptions.The choice
for a certaintype of transferscheme should depend on this tolerance.
Medium-voltage
substation1
Medium-voltage
substation2
. -Automatic
transfer
switch
Industrial
customer
Figure 7.8 Primary selective supply.
Medium-voltage
substation1
Figure 7.9 Secondaryselective supply.
Medium-voltage
substation2
404
Chapter7 •
Mitigation of Interruptionsand Voltage Sags
7.2.4.5 Static Transfer Switches.Static transfer switches have been used
already for several years inlow-voltage applications,e.g., in uninterruptablepower
supplies to be discussed in Section 7.4.
Currently, static transfer switches are also
available for medium voltages [166], [171], [173]. Astatic transfer switch consistsof
two pairs of anti-parallelthyristors as shown in Fig. 7.10.During normal operation,
thyristor pair I is continuouslyfired, and thus conductingthe load current. Thyristor
pair II is not fired. In termsof switches,thyristor pair I behaves like a closed switch,
pair II like an open switch.
When adisturbanceis detectedon thenormalsupply,the firing of thyristor pair I
is disabled and the firingo f thyristor pair II enabled.The effectof this is that the load
currentcommutatesto the backupsupply within half a cycleof detectingthe disturbance.Actual transfertimes are lessthan 4ms [166]. The three small figures show the
voltages in a stylized way. In reality
voltagesare sinusoidal,but the principle remains
interruptionat time I.
the same.P oint A experiences adrop in voltage due to a sag or
This drop in voltage is alsoexperiencedby the load at point C. We assumethat the
backupsupply does not experience this. At time 2, the
disturbanceis detected,the firing
of thyristor pair I is disabled,andthe firing of thyristorpair II enabled.At that moment
the commutationof the current from the normal supply to the backupsupply starts.
During commutationthe voltage atpoints A, B, and C is equal asboth thyristor pairs
are conducting.This voltage issomewherein between the twosupplyvoltages. At time
3 the commutationis complete(the thyristor currentin pair I extinguisheson the first
zero crossingafter the firing beingdisabled)and thevoltageat BandC comes back to
its normal value. Note that the current through the thyristors never exceeds the load
current,also not for a fault close to thestatic switch.
A static transfer switch can be used in any
o f the transfer schemes discussed
before: industrial distribution, primary selective,secondaryselective. The speed with
which the transfer takes place makes .the
distinction betweensynchronizedand nonsynchronizedtransferno longer relevant. Load transfer by a static transferswitch is
always synchronized.
To ensure very fasttransfer,any voltage sag orinterruptionin the normal supply
shouldbedetectedvery fast. Thecommutationof the currentfrom onethyristorpair to
the other takes lessthan half a cycle sothat we need adisturbancedetectionwhich is
equally fast. Static transfer schemes can use the missing
voltage or a half-cycle rms
value to detect a sag or
interruption. For the missingvoltage detectionscheme, the
Backup
supply
Normal
supply
II
~'----Ct---+---fc~
1
bL=
123
B
Dc
23
Figure 7.10 Constructionand principle of
operationof a static transferswitch.
Section 7.3 • Power System
D esign-Redundancy
ThroughParallelOperation
405
actual voltage iscomparedon a sample-by-samplebasis with theoutput voltage of a
phase-locked-loop(PLL). When the deviation becomes too large for too long, the
transferis initiated. With the rms scheme,transferis initiated when the rms voltage
drops below acertain threshold. The latter scheme is slower as it will lead to an
additional half-cycle delay, but it has a smaller chanceincorrect
of
transfer.
A transferscheme using a
static transferswitch enables thed urationof a voltage
sag to be limited tohalf a cycle by switching to thebackupsupply when a sag occurs in
the normalsupply. For sensitive load, astatictransferswitch might bepreferableabove
parallel operation.Voltage sagsoriginating in the transmissionsystemcannotbe mitigated by such atransfer scheme as the voltage sag is likely to be presentboth
in
supplies; but for sags
originating in the distribution system the statictransferscheme
is very effective. The mainlimitations are theunknownreliability of the transferswitch
and the degree in which the two sources independent.
are
The notch due to loadtransfercould be aconcern,especially for the load on the
healthy feeder. Whencomparingstatic transfer with parallel operation,a notch of
millisecond duration replaces the voltage sags
of several cyclesduration. When comparingwith the mechanicaltransferscheme, thenotchin the backupsupply constitutes
a deteriorationof the voltagequality, albeit not a severedeterioration.Some utilities do
not allow parallel operation of feeders, requiring a so-called"break-before-make"
transfer scheme. Thestatic transfer switch as described here is essentially"makea
before-break"scheme. It isimpossibleto predicthow strict utilities will apply this rule
on a sub-cycle timescale. As an
alternativeone could enable firing ofthyristor pair II
only after the current through pair I has extinguished. Such b
areak-before-make
scheme willobviously make the transferslower and couldactually make the voltage
transientin the healthy supplymore severe.
A final potential problem with static transferis that the normal supply and the
healthy supply are not exactly in phase. The
phase-angledifference could lead to a small
0
phase-anglejump at the loadterminals.Values up to 6
have beenreported.As long as
there are nostandardson equipmenttoleranceto phase-angle
jumps,it is hardto assess
the impact of this. The successful use
of medium-voltagestatic transferswitches on a
numberof sitesindicatesthat the equipmentis able totoleratethe transient.
7.3 POWI!R SYSTEM DI!SIGN-REDUNDANCY THROUGH
PARALLI!L OPERATION
7.3.1 Parallel and Loop Systems
Figure 7.11 shows a publicdistribution network with a higher nominal voltage
than the one in Fig. 7.5. It serves more
customersso it is worth to invest more in
reliability. Partof the system is stillo peratedin a radial way withnormally open points.
These are serving less densely
populatedareas, and areas with less
industrial activity.
The majority of the 33 kV system isoperatedwith parallel feeders. Bothpathscarry
part of the load. If onepathfails, the otherpathtakes over the supply
instantaneously.
Also the 33/1I kV transformerand the 33 kV substationbus areoperatedin parallel.
The rating of eachcomponentis such that the load can be fully supplied if one componentfails.
of parallel operation:two feeders in parallel and a
We see in Fig. 7.11 two types
loop system. Inboth cases there is single
redundancy.The loop system is significantly
of transformerconnections.But the voltagecontrol of loop
cheaper, especially in case
systems is more difficult, and the
various loads are moreprone to disturbing each
406
Chapter7 • Mitigation of Interruptionsand Voltage Sags
33 kV
loop
6.6kV
llkV
Another33kV
network
~----t
n/o
Figure 7.11Distribution network with
redundancythroughparallel operation.
other'ssupply. Loop systems arethereforelesspopularin industrial systems,although
somesmallerloops (three or four busses) are used to limit numberof
the
transformers.
7.3.1.1 Design Criteria for Parallel and Loop Systems.The design of parallel
(n - 1) criterion, which statesthat the
and loop systems is based on the so-called
system consisting of n componentsshould be able to operate with only (n - 1)
componentsin operation, thus with onecomponentout of operation. This should
hold for anyonecomponentout of operation. The (n - 1) criterion is very commonly used inpower system design. It enables a high reliability
without the need
for stochasticassessment. In some cases (large
transmissionsystems,generatorschea
duling), (n - 2) or (n - 3) criteria are used. As we saw in Section 2.8,thorough
trustfully use
assessmento f all "common-modefailures" is needed before one can
such ahigh-redundancydesign criterion.
Here we will concentrateon the (n - 1) criterion, also referred to as "single
redundancy."This criterion is very commonly used in the designof industrial medium-voltagedistribution as well as in publicsubtransmissionsystems. The main design
rule is that no single eventshould lead to aninterruption of the supply to any of the
customers.In an industrial environmentthe wording is somewhatdifferent: no single
eventshouldlead to aproductionstop for any of the plants. How these basic rules are
further developeddependson the kind of system. A list of things that have to be
consideredis given.
1. The obviousfirst rule is that no componentoutageshouldlead to an interruption. Thereshouldthus be analternatepath for the power flowthrough
any component.
2. Not only shouldthere be analternatepathfor the power flow, thisalternate
path shouldalso not lead to anoverloadsituation. In the public supply the
Section7.3 • Power SystemDesign-Redundancy
ThroughParallel Operation
3.
4.
5.
6.
407
load demandvaries significantlyduring the day. Acertainamountof overload can betoleratedfor a few hours. Inindustrial systems the load is typically more constant,so that any overloadwould bepermanent.However in
industrialsystems it is often easier to reduce the load on a time scale
hours
of
or to start on-site generation.
The power systemprotectionshouldbe able to clear any faultwithout causing an interruptionfor any of the customers. This
requiresmore complicated
protectionsystemsthan for radial-operatednetworks. Theseprotectionsystems require additional voltage transformersand/or communicationlinks.
Also thenumberof circuit breakersincreases: two circuitbreakersare needed
for eachconnectionbetween twosubstationsin a looped orparallel system.
Voltage fluctuationsdue to rapid loadfluctuations and voltage sags due to
motor startingshouldbe within limits for anyonecomponentout of operation. This translatesinto a minimum fault level for any load bus. The switchgearratingdictatesa maximumfault level for the system with allcomponents
in operation.The optimal use of this margin betweenmaximum and minimum fault levels is oneof the main challenges in the design industrial
of
medium-voltagedistribution systems.
The electromechanicaltransientdue to ashort circuit in the system with all
componentsin operationshould not lead to loss of any load. In
industrial
systems with a large
fraction of inductionmotor load, it must beensuredthat
thesemotorsare able to re-acellerate after the fault.
The voltage sag due to any fault in the system
shouldnot lead to tripping of
essentialload with any of the customers.
From this list it becomesobviousthat the designof a parallel or loop system
could be a
serious challenge. But the reliability
demandsof largeindustrialplantsare suchthat no
radial system could deliver this. The increased reliability is more
than worth the higher
installationcosts and costso f operation.
7.3.1.2 Voltage Sags in Parallel and Loop Systems.
Considerthe system shown
in Fig. 7.12: three supplyalternativesfor an industrial plant. In theradial system on
the left, theplant is fed through a 25 km overheadline; two more overheadlines originate from the samesubstation,each with a lengthof 100km. In thecenterfigure
the plant is fed from a loop bymaking a connectionto the nearestfeeder. In the
third alternativeon the right aseparateoverheadline has beenconstructedin parallel with the existing 25 km line. Themagnitudeof voltage sags due to faults in this
system is shown in Fig. 7.13. The
calculationsneeded toobtain this figure are discussed in Section 4.2.4. We will use Fig. 7.13 to assessnumberof
the
voltage sags experiencedby the plant for the three designalternatives.
For the radial system, theplant will experienceinterruptionsdue to faults on
kmline. The relation
25 km of overheadline, and voltage sags due to faults on 200 of
between sagmagnitudeand distanceto the fault isaccordingto the dottedline in Fig.
7.13. Improving the voltage toleranceof the equipmentwill significantly reduce the
exposed length. The exposed length for
radial operation is given in Table 7.2 for
different equipmentvoltage tolerances. By simplyadding the exposed lengths, it is
is
assumedthat the impact of interruptionsand voltage sags is the same, whichnot
always the case. Even if the process trips due to a voltage sag, it mightrequire
still
power from the supply for a safes hutdownof the plant.
408
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Substation
II
Substation
III
II
]
]
an
an
N
Substation
III
II
]
III
an
N
N
X
N
]
.e
B
§
§
0
~
lOOkm
100km
100km
Figure 7.12 Threesupply alternativesfor an industrial plant: radial (left), looped
(center),and parallel (right).
0.8
a
.8
-8
a
.~
0.6
m 0.4
f
,
:
f/}
I
.-
I
........
.,
"
,
....
0.2 :/
:t
:'
\
\
:'
Figure 7.13 Sag magnitudeas a function of
fault position for faults in the systemshown
\
in Fig. 7.12. Solid line:faults on the 25 km
,
branchof a 125 km loop; dashedline: faults
100 on the 100km branchof a 125km loop;
dotted line: faults on aradial feeder.
\
20
40
60
80
Fault positionin kilometers
TABLE 7.2 ExposedLengthfor VariousEquipmentVoltageTolerancesfor Radial
Operationin Fig. 7.12
ExposedLength
VoltageTolerance
Trips on
Trips on
Trips on
Trips on
interruptionsonly
sags below20%
sags below50%
sags below900/0
FeederI
25
25
25
25
km
km
km
km
FeederII
FeederIII
3km
12 km
100 km
3 km
12 km
100 km
Total
25 km
31km
45 km
225 km
The calculationshave beenrepeatedfor loopedoperationas in thecenterdrawing
in Fig. 7.12, resulting in the values shown in Table 7.3. Only equipmentimmuneto
for
all voltage sags will thenumberof equipmenttrips be lessthan for the radial supply.
The exposed length for the
variousequipmentvoltagetolerancesis given in Table
7.4 for parallel operation. For a voltage toleranceof 50% this option is preferable
above loopedoperation.Knowledgeof the various costs involved is needed to decide
if this reductionin trip frequency isworth the investment.
Section 7.3 • Power System
D esign-Redundancy
Through ParallelOperation
409
TABLE 7.3 ExposedLength for Various Equipment Voltage Tolerances for
Looped Operation in Fig. 7.12
Exposed Length
VoltageTolerance
Trips on interruptions only
200/0
Trips on sags below
Trips on sags below
50°A»
Trips on sags below
90°A»
Feeder I
Feeder II
Feeder III
Total
25 km
25 km
25 km
14km
100 km
100 km
3 km
12 km
100 km
42 km
137 km
225 km
TABLE 7.4 Exposed Length for Various Equipment Voltage Tolerances for
Parallel Operation in Fig. 7.12
Exposed Length
VoltageTolerance
Trips on interruptions only
20%
Trips on sags below
50°A»
Trips on sags below
90%
Trips on sags below
Feeder I
50 km
50 km
50 km
Feeder II
3 km
12 km
100 km
Feeder III
Total
3 km
12 km
100 km
56 km
74 km
250 km
7.3.2 Spot Networks
The basiccharacteristicof a spot network isthat a bus is fed from two or more
different busses at a higher voltage level. In the previous section we looked at parallel
and loop systems
originatingat the same bus or at two busses
connectedby a normally
closed breaker. When a bus is fed from two different busses, the same design
problems
- 1) criterion remains the
have to be solved as for parallel and loop systems. (n
The
underlying rule. Themagnitudeof voltage sags is significantly lower for spot networks,
comparedto parallel networks. Also thenumber of interruptionswill be somewhat
lower, but that difference will not be significant as the
numberis already low.
7.3.2.1 Magnitudeof Voltage Sags. Considerthe system in Fig. 7.14: the busbar with the sensitive load is fed from two different
busbarsat a higher voltage level,
ZSI and ZS2 are source impedances at the higher voltage level,
Ztt and Zt2 are transformer impedances, z is the feeder impedance per unit length, {, the distance between
bus I and the fault. The two busses can be in the same
substationor in two different
substations.The reliability in thelatter case is likely to besomewhathigher, although
it is hard to exactlyquantify this difference.
Consider a fault on a feeder
originatingfrom bus I at a distance£, from the bus.
The magnitudeof the voltage at bus I is found from the
voltage-dividerequation
(7.7)
where we neglect the effect of the second source on the voltage at bus I. This is a
reasonableassumptionas the impedanceof the two transformersin series will be
much higher than the source
impedanceat bus I. If we assume the two sources to be
410
Chapter7 • Mitigation of Interruptionsand Voltage Sags
ZSl
BusI-..........- . - -
-
..........--BusIl
Figure 7.14 Busbarfed from two different
busbarsat a higher voltage level.
Fault
Sensitive load
completelyindependent,so that the sourcevoltageat bus II doesnot drop due to the
fault, the voltageat the load bus isfound from
v.wg = VI + Z
II
+
~tl
12
+
Z
(1 - VI)
(7.8)
SI
We simplify the expressionssomewhatto be betterable to assess the effect
of the double
supply. Assumethat z == ZSI, which is alwayspossibleby choosingthe properdistance
Z,t and ZS2 « 2 ,2, The voltageat the
unit. Assumealso that Z,1 = Zt2 and that ZSl
load bus is, undertheseassumptions:
«
t:
V
sag -
+12
.c + 1
(7.9)
and at bus I:
c
VI
= £+ 1
(7.10)
For a radially operatedsystem,without a connectionto bus II the voltageat the
load bus is equal to the voltage at bus I, given by (7.10).Figure 7.15 comparesthe
voltage magnitudeat the load bus for the two designalternatives.It is immediately
obvious that the secondinfeed significantly reducesthe voltagedrop. The deepestsag
will have a magnitudeof 50% of nominal. Here it is assumedthat the secondtransformer has the sameimpedanceas the first one. Inpractice this translatesto them
having the samerating. If the secondtransformerhas asmaller rating, its impedance
will typically be higher and the voltagesag will bedeeper.
From the expressionsfor the voltageversusdistance,we can obtain expressions
for the critical distance,like in Section6.5. For the radial systemwe obtain the same
expressionas before:
(7.11)
For the systemwith doubleinfeed, we obtain
V-!
Lcrit
= 1 _ ~, V ~ 0.5
(7.12)
411
Section 7.3 • Power System
D esign-Redundancy
Through ParallelOperation
0.8
a
/
.S
~ 0.6 "
.a
"
'
.~
8 0.4
~
r:J)
0.2
·Figure 7.15 Sagmagnitudeas adistanceto
the fault, without (solid line) and with (dashed
line) a connectionto a secondsubstationat a
higher voltage level.
2
10
4
6
8
Distance to fault (arbitr. units)
10,..-----y------r-----r-----,..-..,..,....----,
I
I
,
,
I
I
I
,
I
I,,
, ,
i
,, ,,'
, ,
, ,
,,
,
I
I
I
I
I
.'
.'
I
I
I
I
Figure 7.16 Exposed length for radial supply
(solid line) and for aconnectionto a second
substationat a higher voltage level: same
numberof feeders from bothsubstations
(dashed line); twice as many feeders from the
secondsubstation(dash-dotline).
, ,
I
"
,II / '
". ,,"
"."" .,
0.2
0.4
".:'" "
0.6
0.8
Sag magnitude in pu
and L,crit = 0 for V < 0.5. From the critical distancethe exposed length can be calculated, resultingin Fig. 7.16. Themain featureis that the exposed length is zero in case
i mportant
the equipmentcan toleratea sag down to50% of nominal. This could be an
pieceof informationin decidingaboutthe voltage-tolerancerequirementsfor the load.
For higher critical voltages(more sensitiveequipment)the exposed lengthdependson
the numberof feedersoriginatingfrom the two busses. Let INbe thenumberof feeders
fed from busI and N 2 the numberof feeders fed from bus II. The
total exposed length
for the load fed from both feeders is found from
(7.13)
for the spot network and
(7.14)
double infeed is
for the radial system. In case N
I = N 2, the exposed length for the
always lessthan for single infeed. WhenN2 > N, the double-infeedoption becomes
lessattractivewhen theequipmentbecomes too sensitive. In the example shown by a
412
Chapter7 • Mitigation of Interruptionsand Voltage Sags
dash-dottedline in Fig. 7.16, N2 = 2N}, the cross-overpoint is at 75% remaining
voltage.
It is important to realize that the second bus does not have to beanother
at
substation.By operatinga substationwith two bussesconnectedby a normally open
breaker,the same effect is achieved. Suchconfigurationmight
a
not be feasible in the
public supply as it reduces the reliability for
customersfed from aradial feeder. But for
industrial distribution systems it is an easy
methodof reducing the sagmagnitude.
7.3.2.2 Public Low- Voltage
Systems. An example of a low-voltagespotnetwork
is shown in Fig. 7.17. A low-voltage bus is fed by two or more feeders
originating
from different substationsor from busses notoperatedin parallel. Theprotectionof
the feeders takes place by
overcurrentprotectionin the medium-voltagesubstations
and by a sensitive reverse-power relay (the
"network protector") at the low-voltage
bus. In public systems it is not always possible to supply from
different substations.
This will still lead to a low numberof interruptions,but the numberof voltage sags
will not be reduced, and will even be
somewhatincreased due to faults on the
parallel
feeders.
The system shown in Fig. 7.18 is also referred to as
spotnetwork;otherscall
a
it a
distributedgrid network, or simply asecondarynetwork. Suchnetworksare common
in the downtown areas of large cities (NewYork, Chicago, London, Berlin).
Distributedlow-voltagenetworkswith an operatingvoltage of 120 V typically use no
protection against low-voltage faults. The faultcurrent is so high that every short
circuit will burn itself free in a short time. For voltage levels of 200 V and higher,
expulsionfuses orcurrent-limitingfuses are used. A
networkprotectoris againinstalled
on secondaryside of everytransformerto preventbackfeed from the low-voltage network into medium-voltagefaults. Thesedistributedlow-voltage networksoffer a high
reliability. Outageson any of thedistribution feeders willnot be noticed by thecustomers. For the mitigation of sags it is essentialthat the feedersoriginate in different
substations,otherwise thenumberof sags will even be increased. Any fault in the lowvoltage network will cause a sag for allcustomerssuppliedfrom this network.The use
of current-limitingfuses will significantly reduce the sag
duration,so that these sags are
not of much concern.
T
Oifferent MV
substations
SecondaryLVfeeders
Figure 7.17Low-voltage spot network.
Section 7.3 • Power System
D esign-Redundancy
ThroughParallelOperation
413
Substation2
Substation 1
MVILV
transformers
Low-voltage
network
Substation 3
Figure 7.18 Low-voltagedistributedgrid.
A comparisonof different designoptionsfor the public supply is given in[165].
Both stochastic predictiontechniques and site
monitoringwere used in thecomparison.
Spot networks turned out to have much less
interruptionsthan any other network
configuration.Looking at the sag frequency,
undergroundnetworksperformedbetter
of the numberof sags. The supply
thanoverheadnetworks, experiencing only one third
configurationhad onlyminor effect on the sag frequency.
7.3.2.3 Industrial Medium-VoltageSystems. In industrial systems spot networks are in use at almost any voltage level; the feeders are typically
protectedby
using differential protection. A configuration with three voltage levels is shown in
Fig. 7.19.
At each voltage level, a bus is fed from two different busses at a higher voltage
level. These two busses might well be in the same
substation,as long as they are not
operatedin parallel. The effect of this supply
configurationhas been discussed in Figs.
4.37, 4.38, and 4.39 in Section 4.2.4. By opening breakerin
the
the substationat an
intermediatevoltage level, thuschangingfrom parallel operationto a spot supply, the
lnfeedfrom transmission network
--.-.........---.......- Medium-voltageload
Figure 7.19 Industrial spot network.
Low-voltage load
414
Chapter7 • Mitigation of Interruptionsand Voltage Sags
magnitudeof deep sags is significantly reduced (Fig. 4.39). The effect on shallow sags is
more limited.
7.3.2.4 Transmission Systems. Another example of a spot network is the
275 kV system in the UK. These systems form the
subtransmissionnetwork around
the big cities. Each 275 kV system
consistsof about 10 busses in a loop-likestructure,
fed at three to five places from the 400 kV
national grid. The structureof the grid
aroundManchesteris shown in Fig. 7.20: thick linesindicate400kV substationsand
lines, and thin lines 275 kV.
Similar configurationsare used inother Europeancountries,e.g., 150kV and
400 kV in Italy and Belgium, 150kV and 380 kV in
p arts of The Netherlands,130kV
and 400 kV in Sweden[23]. The number of supply points for the subtransmission
systems varies from twothrough ten. In theUnited States this type ofconfiguration
is in use across all voltage levels, down to 69 kV, as shown in Fig. 6.39.
The effect of supplyconfigurationsas shown in Fig. 7.20 isthat faults in the
400 kV grid only cause shallow sags at the 275 kV
substations.If we neglect the
275 kV line impedancescomparedto the transformerimpedances,the voltage in the
275 kV system is the average of the voltages at the 400 kV sides of
transformers.
the
A
fault close to one of thesubstationswill drop the voltage to a low value at this substation,but othersubstationswill be less affected. With ninetransformers,the shallow
sags willdominate.The effect of this"averaging" is that the customerexperiences less
deep but more shallow sags. To
illustratethis effect, we againconsiderthe transmission
system shown earlier in Fig. 4.27. The
distance between thesubstationshas been
increased to 100km, allother parameterswere kept the same.Figure 7.21 plots the
sagmagnitudeas afunction of the fault position; position 0 is a fault insubstation1,
position 100 (km) a fault insubstation2. Considernext asubtransmissionsystem fed
from substation1 and substation2. The voltage in thesubtransmissionsystem is
approximatedby the averagevoltagein the two transmissionsubstations;this voltage
is indicated by the dotted line in Fig. 7.21. Due to thelooped operationacross the
voltage levels, the deepest sags become
shallower,and someof the shallow sags deeper.
The disadvantageof the way of operationlike in Fig. 7.20 isthat faults in the
275 kV networks lead to deep sags. The
interconnectedoperation makes that the
exposed areacontainsmore lengthof lines than in caseof radial operation.If these
Figure 7.20 Spot network at subtransmission
level: 400 kV (thick lines) and 275 kV (thin
lines) system in theNorth of England.(Data
obtainedfrom [177].)
415
Section 7.3 • Power System
D esign-Redundancy
ThroughParallelOperation
::I
'-
Qc
.S
-8
.S
t
0.6
",,
,,
/'
"
"
0.4
""
",,
~
t:I}
Figure 7.21Sagmagnitudein transmission
and subtransmissionsystems. Solid line:
transmissionsubstationI, dashed line:
transmissionsubstation2, dottedline:
subtransmission.
,
/
I
,,
I
I
,,
,,
,, ,
,, ,
0.2
"
'"
""
,,
I
I
I
J
.PI00
-50
0
"
50
100
Faultposition
150
200
loops cross several voltage levels, like in the
United States, the net effect is likely to be a
reductionin sag frequency.
7.3.3 Power System Deslgn-on-slte Generation
7.3.3.1 Reasons for Installing a Generator.
Local generatorsare used for two
distinctly different reasons:
1. Generatingelectricity locally can becheaperthan buying it from the utility.
This holds especially forcombined-heat-and-power
(CHP) where the waste
heat from the electricitygenerationis used in theindustrialprocess. Thetotal
than in conventionalgenefficiencyof the process is typically much higher
eratorstations.
2. Havingan on-sitegeneratoravailableincreases the reliabilityo f the supply as
it can serve as abackupin case the supply is
interrupted.Some large industrial plants have the ability tooperatecompletely in island mode. Also
hospitals,schools,governmentoffices, etc., often have satandbygenerator
to take over the supply when the public supplyinterrupted.
is
Here we onlyconsiderthe secondsituation, which might be anadditional advantage
next to theeconomicand environmentalbenefits of on-sitegeneration.We first assess
the effectof the generatoron the availability. Supposethat the public supply has an
availability of 98%. This might soundhigh, but anunavailability of 2°~ implies that
there is no supply for 175
h ourseach year, or on average 29
minutesper day, or 40 4hour interruptionsper year. Inother words, 980/0 availability is for many industrial
customersunacceptablylow. We assumethat an on-sitegeneratoris installed which can
take over all essential load.
Supposethat the on-sitegeneratorhas anavailability of
900/0. The supply isguaranteedas long aseitherthe public supply or thegeneratorare
available. Themethodsintroducedin Chapter2 can be used tocalculatethe reliability
of the overall system. The resulting
availability is 99.8%, or an unavailability of 18
hoursper year, four to five4-hourinterruptionsper year. In case faurther increase in
reliability is needed, one can
considerto install two or even threegeneratorunits. Each
of these is assumed to be able to supply all the essential load. Withgenerators
two
we
416
Chapter7 • Mitigation of Interruptionsand Voltage Sags
obtain an unavailability of 2 hours per year; with three, the unavailability is only 10
minutesperyear,neglectingall common-modeeffects. As we saw inChapter2 the latter
assumptionis no longer valid for highly reliable systems.Any attempt to further
increasethe reliability by adding more generatorunits is unlikely to be successful.
Emergencyor standbygeneratorsare often startedwhen an interruptionof the public
supply occurs. Instead of calculating unavailabilitiesit is more suitable to calculate
interruption frequencies.Supposethat the public supply is interrupted40 times per
year. The failure to startof an emergencygeneratoris typically somewherebetween10/0
and 5%. A valueof 5% will reducethe numberof interruptionsfrom 40 peryearto two
per year. This assumesthat the generatoris alwaysavailable.In reality one hasto add
anotherfew percentunavailability due to maintenanceand repair. The resultinginterruption frequencywill be aroundfive per year. Again an industrial useris likely to opt
for two units, which brings the interruptionfrequencydown to lessthan one per year.
7.3.3.2 Voltage Sag Mitigating Effects.We saw inSection4.2.4 and in Section
6.4 that a generatormitigatessags near its terminals. To mitigate sags thegenerator
has to be on-line; an off-line generatorwill not mitigate any voltage sags.The effect
of a generatoron the sag magnitudewas quantified in Fig. 4.26 and in (4.16). The
latter equationis reproducedhere:
(1 - Vsag)= Z Z4 (1 - Vpcc)
3+ Z 4
(7.15)
with Z3 the impedancebetween the generator/loadbus and the pee (typically the
impedanceof a distribution transformer)and 2 4 the (transient)impedanceof the generator. If we further assumethat Vpcc = .c~1' with.Z the distanceto the fault, and
introduce ~ =~, we get the following expressionfor the sagmagnitudeat the load
bus as a functio~ of the distanceto the fault:
V
=1
sag
1_ _
(1 + ~)(1 + £)
(7.16)
This expressionhasbeenused toobtain the curvesin Fig. 7.22: the sagmagnitudeas a
function of distanceis shownfor different valuesof the impedanceratio ~. A value ~ =
o correspondsto no generator;increasingt; correspondsto increasinggeneratorsizeor
increasingtransformerimpedance.C onsidera typical transformerimpedanceof 50/0 of
0.8
6-
.5
~
a
0.6
.~
.
,I
~ 0.4 i,'
"
8
~
",',
C/)
,
0.2
Figure 7.22Sagmagnitudeversusdistance
2
4
6
8
Distance to the fault (arbitr. units)
10
for different generatorsizes. Theratio
betweentransformerand generator
impedanceused was 0 (solid line), 0.2
(dashed
line), 0.4(dash-dotline), and 0.8(dottedline).
417
Section 7.3 • Power System
D esign-Redundancy
Through ParallelOperation
its rated power, and a typical generatortransientimpedanceof 18%. For equal generatorandtransformerrating, we find t; = 0.28; ~ = 0.8 correspondsto a generatorsize
about three times the transformerrating, thus also about three times the sizeof the
load. We sawbeforethat generatorcapacityof more than threetimes theload doesnot
have any improving effect on the reliability. It is thus unlikely that the generator
capacity is more than three times the load. Anexception are someCHP schemes
where theindustry sellsconsiderableamountsof energy to theutility.
We see in Fig. 7.22 how the
g eneratormitigatesthe voltage sag. The larger the
generator,the more the reduction in voltage drop. From the expressionfor the sag
magnitudeas afunction of distance,one canagainderive anexpressionfor the critical
distance:
1
Lcrtl
= (1 + ~)(1 _
(7.17)
V) - 1
This expressionhas been used to
calculatethe critical distancefor different generator
sizes,resultingin Fig. 7.23. The curves are simply the inverse
of the curvesin Fig. 7.22.
We see areductionin critical distancefor each valueof the sagmagnitude.Note that
the installationof an on-sitegeneratordoesnot introduceany additionalsags (with the
exceptionof sags due to faults in ornear the generator,but thoseare rare). The sag
frequencyfor the different alternativescan thus be comparedby comparingthe critical
distances.
A betterpictureof the reductionin sagfrequencycan beobtainedfrom Fig. 7.24.
The various curves show the percentagereduction in sag frequency betweenthe site
without generatorand the site with agenerator.Again three generatorsizes have been
compared.For small sagmagnitudesthe reductionin sagfrequencyis 100%; thereare
no sags left with thesemagnitudes.For higher magnitudesthe relative reduction
becomesless. Thismitigation methodworks best forequipmentwhich alreadyhas a
certainlevel of immunity againstsags.
10r----...----.------y-----,-----rr-..---,
Figure 7.23 Critical distance versus
magnitude for different generator
sizes.The
ratio between transformer and generator
impedance used was 0 (solid line), 0.2 (dashed
line),0.4 (dash-dot line), and 0.8 (dotted line).
0.2
0.4
0.6
Sagmagnitudein pu
0.8
7.3.3.3 Island Operation. On-site generatorsare fairly commonin large industrial and commercialsystems. Theon-site generationis operatedin parallel with the
public supply. When the public supply fails, the on-site generatorgoes into island
operation.This "island" can consistof the whole load or part of the load. The latter
situation is shown in Fig. 7.25.The island systemshould be mademore reliable than
418
Chapter7 • Mitigation of Interruptionsand Voltage Sags
5 100
[
-'-'-'-,-"-'-';"
\
t!=
.
\
\
\
\
\
.5 80
i~
,
\
\
\
\
\
60
" "'-.
~
.8 40
.s=
.g
20
Figure 7.24Reductionin sag frequencydue
~
0.2
0.4
0.6
Sag magnitude inpu
Infeed from
0.8
to the installationof an on-sitegenerator.The
ratio betweentransformerand generator
impedanceused was 0.2(dashedline), 0.4
(dash-dotline), and 0.8 (dotted line).
On-site
publicsupply
generation
Radial
network
Island system
(meshed)
n/o
Nonessential load
Essential load
Figure 7.25Industrial power system with
islandingoption.
the rest of theindustrial distribution system (e.g., by using a meshed
network and
differential protection).The island system also serves asbackupfor
a
the restof the
industrial distribution system. A big problem in large industrial systems isthat
islanding cannot be tested. One has to wait for an
interruption to occur to seeif it
works.
7.3.3.4 Emergency and
StandbyGeneration. Emergencyand standbygenerators
are typically started the moment an interruption is detected. They come online
between one second and one
minute after the start of the interruption. Note that
there is no technical difference between emergency
generationand standby generation. The term "emergencygeneration"is used when there is a legal
obligation to
have ageneratoravailable; in allother cases the term"standbygeneration"is used
[26]. When installing standbygenerationto improve voltage quality it is important
that essentialequipmentcan toleratethe short interruption due to thetransferto the
419
Section 7.4 • TheSystem-EquipmentInterface
standby generation.Standbygenerationis often used incombination with a small
amountof energystoragesupplying the essential load
during the first few seconds of
an interruption.
7.4 THE SYSTEM-EQUIPMENT INTERFACE
The interface between the system and equipmentis
the
the mostcommon place to
mitigate sags andinterruptions.Most of the mitigation techniques are based on the
injection of active power, thuscompensatingthe lossof active power supplied by the
system. Allmoderntechniques are based on power electronic devices, with the voltagesourceconverterbeing the main building block. Next we discuss the
various existing
and emerging technologies, withemphasis on the voltage-source converter.
Terminology is still very confusingin this area, terms like"compensators,""conditioners," "controllers," and "active filters" are in use, all referring to similar kind
of
devices. In theremainderof this section, the term"controller" will be used, with
reference toother terms in general use.
7.4.1 Voltage-Source Converter
Most modernvoltage-sagmitigation methodsat thesystem-equipment
interface
contain a so-calledvoltage-sourceconverter.A voltage-sourceconverteris _a power
electronic device which cangeneratea sinusoidalvoltage at any required frequency,
magnitude,and-phase angle. We
alreadysaw thevoltage-sourceconverteras an important part of ac adjustable-speed
drives. In voltage-sagmitigation it is used totemporarily replace the supply voltage or to
generatethe part of the supply voltage which is
missing.
The principle of thevoltage-sourceconverteris shown in Fig. 7.26. Athree-phase
converterswith a commondc
voltage-sourceconverterconsistsof three single-phase
voltage. By switching the power electronic devices onoff
orwith a certainpatternan ac
voltage isobtained.One can use a simple
squarewave or apulse-width modulated
pattern. The latter gives lessharmonicsbut somewhathigher losses. Details of the
/
II
Commondc
bus with capacitor
or battery block
Self-commutating
device (GTO/IGBT)
...----+-----0 Three-phase
ac output
Controllergenerating
required switching pattern
Figure 7.26 Three-phasevoltage-sourceconverter.
420
Chapter7 • Mitigation of Interruptionsand Voltage Sags
operationand control of the voltage-sourceconvertercan be found in most books on
power electronics, e.g., [53],
[55].
In circuit-theorymodels,. thevoltage-sourceconvertercan simply be modeled as
an ideal voltage source.
To assess the effect
of this on voltages andcurrents,no knowledge is neededa bout the powerelectronicdevices and thecontrol algorithms. In the
forthcomingsections thevoltage-sourceconverteris modeled as an ideal voltage source
to analyze themitigation effect of variousconfigurations.
The samevoltage-sourceconvertertechnologyis also used for so-called
"Flexible
AC TransmissionSystems"or FACTS [180], [181] and for mitigation of harmonic
distortion [179], [182], [183] and voltagefluctuations [170], [178]. In this chapterwe
will only discuss their use formitigating voltage sags andinterruptions.The whole set
of power electronicsolutions to power quality problems, including static transfer
switches, activeharmonicfilters, and voltage control, is often referred to as"custom
power" [184], [191].
7.4.2 Series Voltage Controllers-DVR
7.4.2.1 Basic Principle. The series voltagecontroller consists of a voltagesourceconverterin series with the supply voltage, as shown in Fig. 7.27. The voltage
at the load terminalsequalsthe sumof the supply voltage and the
output voltage of
the controller:
(7.18)
A convertertransformeris used toconnectthe output of the voltage-sourceconverter
to the system. A relatively small
capacitoris presenton de sideof the converter. The
voltage over thiscapacitoris kept constant,by exchangingenergy with the energy
storage reservoir. The
requiredoutputvoltage isobtainedby using a pulse-width modulation switchingpattern.As thecontrollerwill have to supply active as well as reactive
power, some kindof energy storageis needed. The termDynamic Voltage Restorer
(DVR) is commonlyusedinsteadof series voltagecontroller [184], [185]. In the DVRs
that are currently commercially available large capacitorsare used as a source of
energy.Other potential sources are being considered:
battery banks, superconducting
coils, flywheels. We will for now assumethat there is some kind of energy storage
available. Thevariousstorageoptionswill be discussed later.
Supply
voltage
+
Injected
voltage
Load
voltage
dcbus
Energy
storage
Figure 7.27 Series voltagecontroller.
421
Section 7.4 • TheSystem-EquipmentInterface
converter
The amountof energystoragedepends on the power delivered by the
and on themaximumdurationof a sag. Thecontrolleris typically designed for a certain
maximum sagdurationand acertainminimum sag voltage. Some
practicalaspects of a
series voltagecontroller are discussed in[174].
7.4.2.2 Active Power Injection. To assess the storage
requirementswe calculate
the active power deliveredby the controller, using the notation in Fig. 7.28. We
assumethat the voltage at the load terminals is 1pu along the positive real axis:
V/oad = 1 + OJ
(7.19)
The loadcurrentis 1pu in magnitude,with a lagging power factor cos
ljJ:
[load
= cosljJ - jsinljJ
(7.20)
of the controllerhas amagnitudeV and phase-angle
The voltage sag at the system side
jump y,:
Vsag = V cos1/1 + jV sin y,
(7.21)
The complex powertaken by the load is found from
P10ad + jQload = V load7;oad = cosljJ + j sinl/J
(7.22)
The complexpower takenfrom the system is
PsyS+ jQsys = Vsagl ;oad
= V cos(l/J+ y,) + jV sin(ljJ + y,)
(7.23)
The active powerthat needs to begeneratedby thecontrolleris the difference between
the activepower takenfrom the system and the active
part of the load:
P eonl
= p/oad -
(7.24)
P syS
This can bewritten as
P COnl
= [ 1-
V cos(ljJ + 1/1)]
cosf/>
X Plood
(7.25)
For zero phase
...anglejump we obtain the following simple expression for the activepower requirementof the controller:
Peon'
= [1 -
V]P/oad
(7.26)
The active powerrequirementis linearly proportionalto the drop in voltage. When
phase-anglejumps are consideredthe relation is no longer linear and becomes dependent on the powerfactor also. To assess the effect of phase-angle
jump and power
factor, we have used the
relations between sagmagnitudeand phase-angle
j ump as
derived in Chapter4. The active powerrequirementfor different power factor and
Figure 7.28 Circuitdiagramwith power
system, series
controller, and load.
422
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Alpha=O
Alpha =- 20 degrees
Alpha = -40 degrees
Alpha = - 60 degrees
0.5
00
0.5
1
00
0.5
1
Sag magnitude in pu
Sag magnitude in pu
Figure 7.29 Active powerrequirementfor a
series voltagecontroller, for different
impedanceangles(a=O, -20°, -40°, -60°)
and different lagging power factors: 1.0 (solid
lines), 0.9(dashedlines), 0.8(dash-dotlines),
0.7 (dotted lines).
different phase-anglejump is shown in Fig. 7.29. Sag
magnitudeand phase-anglejump
have beencalculatedas a function of the distanceto the fault by using expressions
j ump werecalculatedfor different values
(4.84) and (4.87).Magnitudeand phase-angle
of the impedanceangle and next filled in in (7.25) toobtain the active power requirement. The latter is plotted in Fig. 7.29 as afunction of the sagmagnitude V.
As shown in (7.26), thepower factor of the load does not influence the active
power requirementsfor sagswithout phase-anglejumps (upperleft). For unity power
requirement.This is
factor, the phase-anglejump somewhatinfluences the active power
mainly due to the voltage over the
controller no longer being equal toI-V. For
decreasingpowerfactor and increasingphase-angle
j ump, the active powerrequirement
becomes less. One
shouldnot concludefrom this that a low powerfactor is preferable.
The lower thepower factor, the larger the loadcurrentfor the sameamountof active
power, thus the higher therequiredrating of the converter.
The reductionin active powerrequirementwith increasing (negative)phase-angle
jump is explainedin Fig. 7.30. Due to thephase-angle
j ump the voltage at system side
of
the controllers becomes more in phase with the load
current. The amount of active
Sag without
phase-angle jump
....
Load
voltage
,
.. ..
Sag with
phase-angle jump
Lagging load
current
Figure 7.30 Phasordiagramfor a series
voltage controller. Dashedline: with negative
phase-anglejump. Solid line: without phaseanglejump.
423
Section 7.4 • TheSystem-EquipmentInterface
Alpha = - 20 degrees
Alpha=O
I
~ 0.5
.s> . 0
l.--
o
o
o
L.-
--J
0.5
1
Alpha = -40 degrees
a
.. 0.5
-" ~.,,:<~.:,:~,~ . .
J
" .:-~~~~:-..
0.5
!
o
L--
o
--J
0.5
Sagmagnitudein pu
1
1
Alpha = - 60 degrees
~
Figure 7.31 Active power requirementfor a
seriesvoltagecontroller, for different
impedanceangles(a=O, -20°, -40°, -60°)
and different leadingpowerfactors: 1.0(solid
lines), 0.9(dashedlines), 0.8(dash-dotlines),
0.7 (dotted lines).
--J
0.5
,".v v,
,,,,
'~\,
o
'---
o
-..J
0.5
1
Sagmagnitudein pu
power taken from the supply thus increases and the active
power requirementof the
controlleris reduced. This holds for a negative
phase-anglejump and a lagging power
phase-anglejump increases the active
factor. For a leading power factor, a negative
power requirements,as shown in Fig. 7.31.
7.4.2.3 Three-Phase Series Voltage Controllers.
The seriescontrollerscurrently
commercially available consist of three single-phase
converterswith a commonde capacitor and storagereservoir. The power taken from the
storagereservoir is the sum
of the power in the three phases.
For eachof the phases, (7.25) can be used to calcuto a
late the active power.For a three-phase balanced sag (Le., a sag due three-phase
requirementis
fault) the sameamountof power is injected in each phase. The power
multiplied by three. But also the active power
taken by the load is three times as
large, sothat (7.25) still holds, with the differencethat Pload is the total load in the
three phases.
To considerthe powerrequirementsfor three-phaseunbalancedsags, we write
(7.25) in a somewhatdifferent form. Let the (complex)remaining voltage (the sag
magnitude)be V, so that the voltage injected by thecontroller is I - V. The load
currentis e-jt/J, which gives for the complex power delivered by the
controller:
(7.27)
Considera three-phaseunbalancedsag of type C: two phases down in voltage; one
phase not affected. To calculate the injected power in phase apply
b, we the same lineof
thoughtas leading to (7.27). The load voltage in phase b is
-
Vload
I r:;
= - -2I - -J'v
2 3
(7.28)
The complex voltageduringthe sag is
1 1r:
Vsag=-"2-2jVeharv3
(7.29)
with Vellar the complexcharacteristicvoltage of the sag. The voltage injected by the
controller is the difference between the load voltage and the sag voltage:
(7.30)
424
Chapter7 • Mitigation of Interruptionsand Voltage Sags
0
The loadcurrentin phase b isshifted over 120 comparedto the currentin phase a:
i.: = e-j ¢(-~ - ~jJ3)
(7.31)
The complex injected power in
phaseb is
(7.32)
For phase c we find
(7.33)
(7.34)
(7.35)
(7.36)
(7.37)
Adding the complex powers in
phaseb and phasec gives thetotal injected power (the
voltage in phase a is
n ot affected by the sag):
-
s.; -_32(1 -
if/>
Vchar)e
(7.38)
This is identical to (7.27), except for the
factor j, Repeatingthe calculationsfor a threephaseunbalancedsagof type D, gives exactly the same injected power as for a type C
sag. For the analysisof three-phaseunbalancedsags we have neglected the zerosequencecomponent.This is an acceptableapproximationat the terminals of enduserequipment,but not always inmedium-voltagedistribution, where DVRs are currently being installed.Adding a zero-sequence
voltageto all three-phasevoltages in the
above reasoningwill lead to an additional term in thecomplexpower expressions for
the three phases. These
additionalterms add to zero, so
t hat the zero-sequence voltage
does not affect thetotal active power demandof the seriescontroller.
The power injectedduring a three-phasesag is three times the power injected in
one phase. Bycomparing(7.38) with (7.27) we canconcludethat the power injected
during a sagof type C or type D ishalf the powerinjected during a balancedsag with
the samecharacteristicmagnitude,phase-anglejump, and duration.
7.4.2.4 Single-Phase Series Voltage Controllers.
For single-phasecontrollers,
the actual voltage in one phase (the
voltage at the equipmentterminalsin the terminology from Chapter4) determinesthe amountof active power which needs to be injected. This is not onlydeterminedby the characteristicmagnitudebut also by the
type of sag and the phase to which the
controller is connected.
of the
What mattersto a single-phasecontroller are the injected powers in each
three phases, i.e. the real
part of Sb in (7.32) and of Sc in (7.37). Thesecalculations
have beenperformedfor three-phaseunbalancedsagsof type C and type D, resulting
in Figs. 7.32 and 7.33, respectively.
For each sag type only two phases have been
plotted: the two phases with the deep sag for type C, and the two phases with the
425
Section 7.4 • TheSystem-EquipmentInterface
shallow sag for type D. Thethird phase for a type C sag does not require any injected
power; the activepower requirementsfor the third phase of a type 0 sag are identical
to (7.25). Both in Fig. 7.32and in Fig. 7.33 the injected power has been
plotted for
two valuesof the impedanceangle (0 and 30°) and four valuesof the power factorof
the loadcurrent (1.0,0.9,0.8,0.7).We can conclude from the figures
that the power
factor has significant influence on the power injection. The
characteristicphase-angle
not change the
jump makes that the two phases behave slightly differently, but does
overall picture.
For a single-phasecontroller,the characteristicvoltage does not have much practical meaning.Thereforethe activepowerrequirementshave beenplottedin a different
way in Figs. ·7.34and 7.35. Thehorizontal axis is theabsolutevalue of the complex
voltage during the sag; inother words, the sagmagnitudeat theequipmentterminals.
The different curves in eachsubplot give the relation between sagmagnitudeand
injected power for each of the phasesof a type C or type Dthree-phaseunbalanced
sag. This leads to m
a aximumof five curves, two from a type C sag, three from a type D
sag. We seethat there is no generalrelation between the injected power and the sag
Alpha =- 30 degrees
Alpha = 0
t
l
1
~ 0.5 ..__"," _, ,
~.:~:.~~::~.~.~ ..~.:-:.:~..
0.5 , .. ...
~
o
o
~&t
0.5
................
o
o
. . . "," -·w.
~~..• ~ ...
...
~ 0.5
' ~.~::~,...
j
'~'::
0"'---
0'---
o
o
---'
0.5
1
Characteristicmagnitude
~
.
--
~:~ ~~::':·?~~~2~.~.~. ~~.,.
".
.
~ 0.2
~
0
S-O.2
o
.....-J
~
o
0.5
~ 0.6
Figure 7.33Active power requirementsfor a
single-phaseseriesvoltagecontroller, for two
phasesof a type D unbalancedsag, for
impedanceangle zero (left) and -300 (right).
Powerfactor 1.0(solid lines), 0.9(dashed),0.8
(dash-dot),0.7 (dotted).
a 0.4
~ 0.2
j -o.~ ~~~~~~.:.:.~~~~~.~~c~.,,~',....
o
--'
1
Alpha = - 30 degrees
.
0.4 '- ..--
0.5
Characteristicmagnitude
Alpha=O
t 06
8.
0.5
1',~~>~....
Figure 7.32Active power requirementsfor a
single-phaseseriesvoltagecontroller, for two
phasesof a type C unbalancedsag, for
impedanceanglezero (left) and -300 (right).
Powerfactor 1.0(solid lines), 0.9 (dashed),0.8
(dash-dot),0.7 (dotted).
"
--.J
~
0.5
I
Characteristicmagnitude
~
-0.2
'--
o
-.1
0.5
0.6
0.4
0.2
. 0 ..
-0.2 ...:. :..~..~ ..-:-..:-:.::-....
o
0.5
I
Characteristicmagnitude
426
Chapter7 •
pf= 0.9
pf= 1.0
~
Q>
~
Mitigation of Interruptionsand Voltage Sags
I
&
t 0.5
0.5
~
Q>
>
~
0
0
0
0.5
pf= 0.8
0
0.5
pf= 0.7
~
QJ
R
t 0.5
0.5
0
0
ti
.s>
0
0.5
Sag magnitude
0
pf= 1.0
~
Go)
~
0.5
Sag magnitude
Figure 7.34Active power requirementsfor a
single-phase series voltage
controller as a
function of the sagmagnitude-forzero
impedanceangle and four values
of the power
factor of the loadcurrent.
pf= 0.9
1
a
t 0.5
0.5
J3 0
0
ii>
0
0.5
pf= 0.8
0
0.5
pf= 0.7
~
~
0
c,
t 0.5
0.5
~
Go)
>
.s
0
0
0
0.5
Sag magnitude
0
0.5
Sag magnitude
Figure 7.35Active power requirementsfor a
single-phase series voltage
controller as a
function of the sagmagnitude-foran
impedanceangle equal to - 30° and four
valuesof the power factorof the loadcurrent.
magnitude,especially for small values
o f the power factor. Note also
thatfor low power
factor, a zero-magnitudesag is not the one with the highest active power requirements.
Figures 7.34 and 7.35 have been
reproducedin Figs. 7.36 and 7.37 with yet
anotherhorizontalaxis. The active powerrequirementshave beenplottedas a function
of the absolutevalueof the complex missing voltage (see Section 4.7.1). We seethat
also
the missingvoltage does not uniquely determinethe injected power. The load power
factor and, to a lesserextent,the characteristicphase-anglejump influence the injected
poweras welland shouldthus beconsideredin dimensioningthe energystorageof the
controller.
7.4.2.5 Effect of the Voltage Rating. The voltage ratingof the voltage-source
converter directly determinesthe maximum voltage (magnitude)which can be injected. This inturn determinesagainstwhich sags the load is
protected.In the above
calculations,it was assumedthat the load voltage would remain exactly at its pred rop and some phase-angle
event value. This isnot strictly necessary: small voltage
jump can betoleratedby the load. Figure 7.38 shows how theprotectedarea of the
complex (voltage) plane can beobtained for a given voltage rating. The voltage
427
Section 7.4 • TheSystem-EquipmentInterface
pf= 0.9
pf= 1.0
...
u
~
8-
1
~
0.5
0.5
~
0
0
t:u
0
t>
0.5
pf= 0.8
0
1
~
... 0.5
0.5
0
0
Figure 7.36 Active power requirements for a ~
u
single-phase series voltage
controller as a
,....~
function of the missingvoltage-forzero
impedance angle and four values of the power
factor of the loadcurrent.
0
0.5
Missing voltage
0
~
0.5
Missing voltage
pf= 0.9
pf= 1.0
t)
0.5
pf= 0.7
I
8-
0.5
b 0.5
\3
.s
0
0
0
0.5
pf= 0.8
0
0.5
pf= 0.7
...
l
Figure 7.37 Active power requirements for a
single-phase series voltage
controlleras a
function of the missingvoltage-foran
-300 and four
impedance angle equal to
values of the power factor of the load
current.
0.5
t) 0.5
i
~
0
0
0
0.5
Missing voltage
0
0.5
Missing voltage
rating of the voltage-sourceconverter is translatedto the same base as the load
voltage. Theactual rating dependson the turns ratioof the convertertransformer.
The voltage tolerance, as indicated in the figure, gives the lowest
voltagemagnitude and the largest
phase-angle
j ump for which the load canoperatenormally. The sag
voltage shouldnot deviate morethan the maximum injectable voltage (Le., the
voltage
dashedcurve,
rating of the converter)from the voltage tolerance. This leads to the
which givesmagnitudeand phase-angle
j ump of the worst sagsthatcan bemitigatedby
the controller;i.e., the voltage tolerance of the
combinationof load andcontroller.The
possible range of sags indicated
is
by a thick solid line. The rangeo f sags caneither be
the range for a varietyo f supplies, like in Fig. 4.96, or for a specific supply, like in Fig.
4.108. It. is very well possible to cover the whole range of possible sags
choosinga
by
large enoughvoltage rating. However, the
numberof sags decreases for lower
magnitudes, and the costs of the
controller increase with increasing voltage
rating. Therefore
the seriescontrollerscurrentlyin use have a minimum voltage
of typically 50%, so that
sags with amagnitudebelow 50% of nominalare notprotected.With reducingcostsof
'power electronics, it is very well possible
t hat future controllerswill cover the whole
range of possible sags.
428
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Voltagetolerance
~
t
Range of
possible sags
Figure 7.38 Part of the complex (voltage)
plane protectedby a series voltagecontroller
with the indicatedvoltage rating.
7.4.2.6 Effectofthe Storage Capacity. The voltage rating of the controller determineswhich range of magnitudeand phase-anglejump of sags can be mitigated.
For a given magnitudeand phase-anglejump the active power requirementis found
from (7.25). The active power requirementand the amount of energystoragedetermine the longestsag durationwhich can be mitigated.
During the designof a seriescontroller, a sagmagnitudeand a sagdurationare
chosen.The sagmagnitudegives thevoltagerating, the sagdurationgives therequired
storagecapacity.Togetherthey determinethe "designpoint" in Fig. 7.39.The voltage
toleranceof the load without controller is shownas adashedline (in this examplethe
voltagetoleranceof the load is 200 ms, 90% ) . The influenceof the phase-anglejump is
neglectedhere.(Including the phase-angle
j ump would give arangeof voltage-tolerance
curves, both with and without the controller.) Any sag with amagnitudeabove the
design magnitudeand with a duration lessthan the designduration,will be mitigated
by the controller: i.e., the resulting load voltage will be above the voltage-tolerance
curveof the load. Sagslonger than the designdurationareonly toleratedif they do not
depletethe storagecapacity.Neglectingthe phase-angle
j ump, we can use (7.26) for the
injected power:
Peont
= (1 -
V)P1oad
(7.39)
The energyneededto ride through a sagof magnitude V and duration T is
£ = (1 - V)TPload
(7.40)
--------------------~-----------------;
0.8
:::s
Q..
.S 0.6
]
.~ 0.4
Design point
~
0.2
2
4
6
Duration in seconds
8
10
Figure 7.39 Voltage-tolerancecurve without
(dashed line) and with (solid line) series
voltage controller. The designpoint gives the
lowest magnitudeand the longestd uration
which theload-controllercombinationis able
to tolerate.
429
Section 7.4 • TheSystem-EquipmentInterface
Let (To, Vo) be the designpoint. The availableenergystorageis
[avail
= (1 -
VO)TOPload
(7.41)
The minimum sagmagnitudeVmin for a duration T is found from
[avail
= (1 -
Vmin)TPload
(7.42)
This gives the following expression for the
voltage-tolerancecurve:
V min
= 1-
(1 -
To
VO)T
(7.43)
This is is shown in Fig. 7.39 as the curve from the design
point toward the right and
upward. Thevoltage-tolerancecurve of the load withcontroller gets its final shape by
realizingthat any sagtoleratedwithout controllercan also betoleratedwith controller.
The area between the curves is the gain in voltage
tolerancedue to thecontroller. To
assess thereductionin numberof trips, a sag densityc hart is needed.
7.4.2.7 Interruptions. A series voltage controller does not function during an
interruption. It needs a closedpath for the load current,which is not always present
during an interruption. If there is loadpresentupstreamof the controller and downstream of the circuit breaker causing the interruption, this load will form a path
throughwhich theconvertercurrentcan close, as shown Fig. 7.40.
The seriescontroller will aim to keep the voltageVI and thus thecurrent /load
constant.The effect isthat the current[load is forced into theupstreamload impedance
Z2 leading to a voltageV2 = Z2//oad on system sideof the controller, but in opposite
phasecomparedto VI' Using VI = Zt[/oad we get
V2
Z2
=-ZI
V.
(7.44)
with ZI the impedanceof the load to beprotectedby the controller. If the upstream
load is smallerthan the protectedload, 2 2 > Z 1, this could lead todangerousovervoltages. With the existing devices this effect is limited in two ways:
• The voltage difference over the
controller is V t + V2 which is significantly
larger than 1pu if.Z 2 > Zt. For a controller with a maximumoutputvoltage
of 0.5 pu (a typical value) the resulting voltage over the
upstreamload can
never be morethan 0.5 pu.
Circuit breaker
causing the
interruption
~
----/--r--f
Upstream----...-
load
Figure 7.40 Series voltage
c ontroller with
upstreamload during an interruption.
Series
controller
Loadprotected
Jontroner
430
Chapter7 • Mitigation of Interruptionsand Voltage Sags
• The energy reservoir is limited, so
that this overvoltage willdisappearwithin a
few seconds.Note that both the protectedload and theupstreamload will
deplete the energy reservoir.
This could, however, become problem
a
in the future when therating of voltage controllers increases,both in injected voltage and instoredenergy. The effect of the sudden
inversion of the voltage on theupstreamload should be studied as well.
7.4.3 Shunt Voltage Controllers-StatCom
A shunt-connected
voltagecontrolleris normally not used for voltage sag mitigation but for limiting reactive powerfluctuationsor harmoniccurrentstakenby the load.
Such acontroller is commonly referred to as a"Static Compensator"or "StatCom."
Alternativeterms in use are"AdvancedStaticVar Compensator"(ASVC) and "Static
Condensor"(StatCon).A StatComdoesnot containany active powerstorageand thus
only injects or draws reactive power. Limited voltage sag
mitigation is possible with the
injection of reactive power only [57], [157], [210],
but active power is needed both
if
constant.
magnitudeand phase angleof the pre-eventvoltage need to be kept
The principle of a shunt voltage controller is shown in Fig. 7.41. The actual
controller has the sameconfigurationas the seriescontroller. But instead of injecting
the voltage difference between the load and the system,
current
a is injected which
pushes up thevoltageat the loadterminals,in a similar way to the sagmitigation by
a generatordiscussed in Section 7.2.
The circuit diagramused to analyze the
controller'soperationis shown in Fig.
7.42. The load voltaged uring the sag can be seen as the
superpositionof the voltage due
to the system and the
voltagechangedue to thecontroller.The former is the voltage as
it would have beenwithout a controller present,the latter is the change due to the
injectedcurrent.
Assumethat the voltagewithout controller is
V.sag
= V cos1/1 + jV sin 1/1
(7.45)
The load voltage is again equal to 1pu:
V/oad = 1 + OJ
Distribution
substation
Transmission
system
Supply transformer
t----~
Load
Shunt voltage
controller
Figure 7.41 Shuntvoltagecontroller.
(7.46)
431
Section 7.4 • TheSystem-EquipmentInterface
Figure 7.42Circuit diagramwith power
system, series
c ontroller,andload. Full circuit
(top), voltageswithout controller(center),
effect of thecontroller (bottom).
The requiredchangein voltagedue to the injectedcurrentis the differencebetweenthe
load voltage and the sag voltage:
~V
=1-
V cos 1/1 - jV sin 1/1
(7.47)
This changein voltagemust be obtainedby injecting a currentequal to
leont
= P - jQ
(7.48)
with P the active powerand Q the reactivepowerinjectedby the controller. The active
power will deterrninethe requirementsfor energystorage.Let the impedanceseen by
the shuntcontroller(sourceimpedancein parallelwith the load impedance)be equalto
Z=R+jX
(7.49)
The effect of the injected currentis a changein voltageaccordingto
~ V = leontZ = (R
+ jX)(P - jQ)
(7.50)
The requiredvoltageincrease(7.47) and the achievedincrease(7.50) haveto be equal.
This gives the following expressionfor the injectedcomplex power:
p _ 0Q = I - V cos"" - jV sin ""
}
R+jX
(7.51)
Splitting the complexpowerin a realandan imaginarypart, givesexpressionsfor active
and reactivepower:
P = R(l - V cos 1/1) - VX sin 1/1
R2 + X 2
Q
= RV sin 1/1 + X(l
- V cos 1/1)
2+X2
R
(7.52)
(7.53)
The main limitation of the shuntcontroller is that the sourceimpedancebecomesvery
small for faults at the samevoltagelevel close to theload. Mitigating suchsagsthrough
a shuntcontroller is impractical as it would require very large currents.We therefore
432
Chapter7 • Mitigation of Interruptionsand Voltage Sags
only consider faultsupstreamof the supplytransformer.The minimum value of the
sourceimpedanceis the transformerimpedance. One can think of this
configurationas
a dedicatedsupply to a sensitive load (e.g., an
automobileplant), where the task of the
controller is to mitigate sagsoriginating upstreamof the transformer.
The resultsof somecalculationsfor this configurationare shown in Figs. 7.43 and
impedance(transformerimpedance) have
7.44. Four different values for the source
For the load impedance a value of 1pu
been used: 0.1, 0.05, 0.033, and 0.025 pu.
resistive has been chosen.
For a 0.05 pu source impedance, the fault level is 20 times
the load power.Fault levels of 10 to 40 times the load are typical distribution
in
systems.
controllerto mainFigure 7.43 shows the
a mountof active power injected by the
tain the voltage at its pre-event value. We see
that for zero impedance angle the active
power requirementis independentof the source impedance. This does not hold in
general, but only for this specific case with a pure reactance in parallel with a pure
resistance.F or increasingimpedanceangle we see an increase in active power, especially
for smaller valuesof the source impedance. The reactive power shown in Fig. 7.44 is
ratherindependento f the impedanceangle. The reactive power requirements decrease
significantly with increasing source impedance. As the (reactive) source impedance
Alpha = 0
Alpha = - 20 degrees
6r---------,
5.S
t
4
~ 0.5
.. '
Q)
.~
<
00
0.5
I
Alpha =-40 degrees
6-
8,..-----:-:-:------,
.:
6
~
Q.,
~
.~
<
'
4
...
o'.,
.:'<": ~
: /
.
.. ,-° '.
10
0
"
,
,
2..{:""
,
Alpha = -60 degrees
15r - - - - - - - - - - ,
.. ', ,".....:,".
, \"'.
0.5
I
Sag magnitude in pu
Alpha = 0
40r-:-·..- - - - - - - - - ,
.: 30, ,
.
~
~
8. 20
.~ 10
",
.
"""
", .....
' ".
8
~ 00
"
",
..
5 :.~.~:~ , ~ .,
00
40
".
- - - _....... '-0.
" ..,\.'~'"
~"
'\
00
- -'- ,,0.
.... ".-'
o"
."
'~
,
0.5
1
Sag magnitude in pu
Alpha = - 20 degrees
.
30,.
'.
20
10
0.5
1
::s
Alpha = -40 degrees
Q., 40rr-·.-.- -......----..,
.S
".
0.5
1
Alpha = - 60 degrees
40
.
l) 30 ....
30 -.-.
~
&20
-0 •
.~ 10
00
"
" .....
10
~
~
00
20
......
0.5
1
Sag magnitude in pu
Figure 7.43 Activepowerinjected by a shunt
voltage controller, for different impedance
angles(0, -20° -40°, -60°) and different
sourceimpedances:0.1pu (solid line),0.05pu
(dashedline), 0.033pu (dash-dotline),
0.025pu (dottedline).
00
0.5
1
Sag magnitude in pu
Figure 7.44 Reactivepower injected by a
shuntvoltagecontroller, for different
impedanceangles(0, -20°, -40°, -60°) and
different sourceimpedances:0.1pu (solid
line), 0.05pu (dashedline), 0.033pu (dash-dot
line), 0.025pu (dotted line).
433
Section 7.4 • TheSystem-EquipmentInterface
increases, less injected
current is needed to get the same change in voltage. Note the
difference in vertical scale between Figs 7.43 and 7.44. The reactive power exceeds the
active power injected in all shown
situations.
The current rating of the controller is determinedby both active and reactive
power. From (7.52) and (7.53) we find for the
absolutevalue of the injected current:
1 - 2V cos1/1 + V2
R2+X2
I cont =
(7.54)
1/1) increases
We seethat an increasing phase-angle
jump (increasing1/1, decreasing cos
thecurrentmagnitude. Thecurrentmagnitudeis plottedin Fig. 7.45 in the same
format
as the active power in Fig. 7.43 and the reactive power in Fig. 7.44.
ComparingFig. 7.45 with Fig. 7.44 showsthat the currentmagnitudeis mainly
determinedby the reactive power. Like the reactive power, the
current magnitudeis
only marginally affected by the phase-angle
jump.
The large increase in active power injected with increasing phase-angle
jump is
explained in Fig. 7.46. The injected voltage is the
requiredvoltage rise at the load due to
Alpha =- 20 degrees
Alpha=O
40 '.
30
a
6 20
.S
u~
.
.." ....
..'.
30 ..
20
10
:s
~
.S
. 00
O.S
1
Alpha = -40 degrees
40 ....
30..
"
.'.
".
..
0.5
1
Alpha = - 60 degrees
40··..
.
5 20
o~
......
10
00
Figure 7.45 Magnitudeof the currentinjected
by a shuntvoltagecontroller, for different
impedanceangles (0, -200 , -400 , -60°) and
different sourceimpedances:0.1 pu (solid
line), 0.05 pu(dashedline), 0.033 pu(dash-dot
line), 0.025 pu(dotted line).
40·...
......
30 ' ,
....
20
'eo
10
'"
....
10
00
0.5
1
00
0.5
1
Sagmagnitudein pu
Sagmagnitudein pu
Source
impedance
,"Injected
.
Normaloperating
....,..
voltage
\
\
,,
\
-------
,,
,
\
\
\
\
\
\
,
\
\
Figure 7.46 Phasordiagramfor shuntvoltage
controller. Solid lines: without phase-angle
jump. Dashedlines: with phase-anglejump.
,,
~
Injected
current
voltage
Sag
voltage
----a.,.
434
Chapter7 • Mitigation of Interruptionsand Voltage Sags
the injection of a currentinto the sourceimpedance.T his injectedvoltageis the differencebetweenthe normal operatingvoltageand the sag voltageas it would be without
controller. The injectedcurrentis the injectedvoltagedivided by the sourceimpedance.
In phasorterms: theargument(angle,direction)of the injectedcurrentis the argument
of the injected voltageminus the argumentof the sourceimpedance.T he sourceimpedance is normally mainly reactive. In case of a sag without phase-anglejump, the
injected current is also mainly reactive. A phase-anglejump causesa rotation of the
injectedvoltageasindicatedin the figure. This leadsto a rotationof the injectedcurrent
away from the imaginaryaxis. From the figure it becomesobviousthat this will quickly
causea seriousincreasein the active part of the current (i.e., the projection of the
current on the load voltage). The changein the reactive part of the current is small,
so is thechangein currentmagnitude.
7.4.3.1 Disadvantagesof the Shunt Controller. It is clear from the above
reasoningthat the main disadvantageof the shuntcontroller is its high active power
demand. In case of a large load with a dedicatedsupply from 'a transmissionnetwork, a shunt controller might be feasible. Voltage sags in transmissionnetworks
show smallerphase-anglejumps, and the transformerlossesare very small. The latter
have not been taken into considerationin the above calculations,as they are rarely
more than a few percentof the load. If the load is suppliedthrough an underground
cable network, these lossescould dominatethe active power requirementof the controller. Another disadvantageof the shunt controller is that it not only increasesthe
voltage for the local load but for all load in the system.Again for a load with a dedicatedsupply through a large transformer,this effect issmall, but for a load fed from
a distribution feeder with many other customersit is not feasible to install a shunt
controller. In caseof a load fed from a distribution feeder, the controller will not be
able to mitigate sags originating at distribution level. The sourceimpedanceduring
the sag willsimply be too small to enableany seriousincreasein voltage.
The behaviorof the shuntvoltage controller during an interruption dependson
the amountof load involved in the interruption. When the supply is interrupted,the
injectedcurrentclosesthrough the load, and the (active and reactive)power demands
are formed by the total load involved in the interruption.If this is only the load to be
protected,the controllerwill haveno problemproviding this power. If a lot more load
is interruptedthe controllerwill probablyreachits currentlimits or its energyreservoir
will be depletedvery fast.
If the controlleris able to maintainthe load during the interruption,synchronization problemscan occur when the voltage comesback. If the supply voltage differs
significantly in phasewith the voltage generatedby the controller, large currentswill
start to flow leadingto relay tripping and/orequipmentdamage.A phasedifferenceof
600 gives an rmsvoltageof 1pu over the terminalsof the recloser.A phasedifferenceof
1800 gives 2 puover the terminals.Considerthat the nominalsystemfrequencyis 60 Hz
and that the voltage comes back after 3 seconds.If we want to limit the angular
differenceto 300 , the relative error in frequencyshould not be more than:
30°
= 5 X 10-4
3 s x 60cyclesjsx 360 0 jcycle
(7.55)
From this it follows that the frequencyneedsto be between59.97 and 60.03 Hz. To
operatethe voltage-sourceconverterwithin this frequencyrangeis not a problem:modern clocksachieveaccuracieswhich areseveralordersof magnitudebetterthanthis. But
the systemfrequencycaneasilydeviatemore than0.03 Hz from its nominal value.
435
Section 7.4 • TheSystem-EquipmentInterface
The mainadvantageof a shuntcontrolleris thatit can also be used to improve the
currentquality of the load. By injecting reactive power, the power factor can be kept at
unity or voltagefluctuationsdue tocurrentfluctuations(the flickerproblem)can be kept
to a minimum. Theshuntcontrollercan also be used to
a bsorbthe harmoniccurrents
generatedby the load. In case such controller
a
is present, it isworth considering the
installation of some energystorageto mitigate voltage sags. It will be clear from the
previouschaptersthat a stochasticassessmentof the variousoptionsis needed.
7.4.4 Combined Shunt and Serle. Controller.
The seriescontroller, as discussed before, uses an energy storage reservoir to
t hat the seriescontroller cannot
power part of the load during a voltage sag. We saw
mitigate any interruptions,and that it is normally not designed to mitigate very deep
'sags (much below50% of remaining voltage). There is thusnormally some voltage
remainingin the powersystem. This voltage can be usedextractthe
to
required energy
from the system.A series-connected
converterinjects the missing voltage, and a shuntconnectedconvertertakes acurrent from the supply. The powertaken by the shunt
controllermust be equal to the power injected by the series
controller.The principle is
shown in Fig.7.47.Series- andshunt-connected
convertershave acommonde bus. The
change instoredenergy in thecapacitoris determinedby the difference between the
power injected by the series
converterand the powertakenfrom the supplyby the shunt
converter.Ensuringthat both are equal minimizes the size
of the capacitance.
Iseries
~ag
-----.
~
load
Load
System
o
00
Figure 7.47 Shunt-series-connected
voltage
controller: theshunt-connectedconverteris
placed on system side of the series
controller.
>
7.4.4.1 Current Rating. The active powertaken from the supply by the shuntconnectedconverteris
(7.56)
We assumethat the shunt-connectedconvertertakes acurrent from the supply with
magnitude[shunt and in phase with the system voltage
IShunt
= [shunt COS t/J +Jrtthunt sin t/J
(7.57)
where 1/1 is thephase-angle
j ump of the sag.Taking the currentin phase with the system
voltage minimizes thecurrent amplitude for the sameamount of active power. The
active powertakenfrom the supply is
Pshunt = VIshunt
(7.58)
436
Chapter7 • Mitigation of Interruptionsand Voltage Sags
with V the sagmagnitude.The active power injected by the series
controller was
calculated before, (7.25):
Pseries =
[ 1-
V cos(¢+ 1/1)]
cos ¢
Pload
(7.59)
The powertakenby the shunt-connected
converterPshunt should be equal to the power
injected by the series-connected
converterP.reries' This gives the following expression for
the magnitudeof the shuntcurrent:
1 cos(¢+ 1/1)]
cos ¢
Plood
[V-
I ,rhunt =
(7.60)
The resultsof this equationare shown in Fig. 7.48 in the same format and with the same
parametervalues as before (e.g., Fig. 7.29). The magnitude ofshunt
the current has
beenplottedfor values up to 4 pu, i.e. four times the active
partof the loadcurrent.The
influence ofphase-angle
j ump and power factor is similar to their influence on the active
power as shown in Fig. 7.29. But the overriding influence onshuntcurrentis
the
the sag
magnitude.The less voltage remains in the system, the more
currentis needed to get the
sameamount of power. As the power requirement increases with decreasing system
voltage, the fast increase current
in
for decreasing voltage is
understandable
.
Alpha = 0
Alpha = - 20 degrees
4,----;--
-
--='---,
3
2
00
0.5
I
Alpha = - 40 degrees
4 .-.:..r--
---='---,
4
3
\,
,
•
'\,
~
...
:::: .'::.. ....-.;:
0.5
Sag magnitude in pu
0.5
1
Alpha = - 60 degrees
I
I,
\
.~\
-v
Figure 7.48Shuntcurrent for a shunt-series
voltage controller, for different impedance
angles(0, _20°, _40°, _60°)and different
" .c- ~.~ ":.."'leading powerfactors: 1.0 (solid lines), 0.9
00
0.5
1 (dashed lines), 0.8
( dash-dotlines), 0.7(dotted
Sag magnitude in pu
lines).
2
.~\
00
00
..
~
.\ ,
7.4.4.2 Shunt Converter on Load Side.
Figure 7.49 again shows shunt
a
-series
controller. The difference with Fig. 7.47 is
t hat the shuntcurrentis taken off the load
voltage.
To assess the effect of this, we again calculate the requirements for the shunt and
seriescurrents.We use the same
n otationas before:
V load
[load
Vsag
=
= 1 + OJ
COS¢-
jsin¢
= V cos1/1 + jV sin 1/1
(7.61)
(7.62)
(7.63)
437
Section 7.4 • TheSystem-EquipmentInterface
~oad
~ag
Load
System
Figure 7.49Shunt-seriesconnectedvoltage
controller; the shunt-connectedconverteris
placed on load sideo f the seriescontroller.
We assumethat the shuntcurrentis taken at a lagging power
factor COs~:
I.vlzunt =
I cos~ - jI sin ~
(7.64)
The total currenttaken off the supply,throughthe series-connected
converter,is
[series
= IShunt + [load = cosl/J + I cos~ -
j sin l/J- jI sin ~
(7.65)
The active power taken off the supply should be equal to the power
takenby the load.
The power injected by the series
converteris taken off again by theshuntconverter.As
there is no active power storage, the total active power still has to come off the supply.
This gives the following expression:
(7.66)
From this the following expression for the
s huntcurrentcan beobtained:
I = cosl/J - V cos(l/J + 1/1)
V cos(1/1 + ~)
(7.67)
To minimize theshuntcurrent,the angle~ is taken suchthat 1/1 + ~ = 0; thus theshunt
If we further rate theshuntcurrentto the
currentis in phase with the supply voltage.
active part of the loadcurrent,we obtain
I
= -!. _cos(1/1 + e/»
V
cose/>
(7.68)
which is exactly the same
currentas for a system-side shunt.
7.4.4.3 Single-PhaseController. For a single-phasecontroller, we have again
calculatedthe invertercurrentas a functionof the sagmagnitudein a similar way as
for Figs. 7.34 and 7.35. The results are shown in Figs. 7.50
and 7.51 for different
power factor of the load current. Fig. 7.50 is for sagswithout phase-anglejumps
(zero impedance angle), Fig. 7.51 for sags with a serious
phase-anglejump (an impedance angle equal to
-30°). The overallbehavioris dominatedby the fast increase in
current for deep sags. But for small power factor, especially, phase-angle
the
jump
also plays animportantrole.
7.4.4.4 Advantagesand Disadvantages. The main advantageof the shunt-series
controller is that it does not require any energy storage. It can be designed to mitigate any sag above caertain magnitude,independentof its duration. This could
result in a relatively cheap device, able to compete with the UPS (see below) for the
438
Chapter7 • Mitigation of Interruptionsand Voltage Sags
4
pf= 1.0
4
=
~ 3
:s
3
2
2
(J
~
~
pf= 0.9
~
.s
00
c:
~
(J
~
u
4
0.5
pC= 0.8
0
0
4
3
3
2
2
0.5
pC= 0.7
t:
u
>
.s
0
0
0.5
0
0
Sag magnitude
4
pf= 1.0
=
~ 3
(J
~
i>
4
0.5
Sag magnitude
Figure 7.50 Shuntcurrentfor a single-phase
shunt-seriesv oltagecontrolleras afunction of
the sagmagnitude,for zero impedanceangle
and four valuesof the power factor of the
load current.
pC= 0.9
3
2
2
.s
00
4
d
~
(J
~
u
t:
0.5
pC= 0.8
0
0
4
3
3
2
2
0.5
pf= 0.7
u
]
°0
0.5
Sag magnitude
0
0
0.5
Sag magnitude
Figure7.51 Shuntcurrentfor a single-phase
shunt-seriesv oltagecontrolleras afunction of
the sagmagnitude,for impedanceangle - 30°
and four values of the power factor of the
load current.
protectionof low-power, low-voltageequipment.The shuntconverterof a shunt-series controller can also be used to mitigate
current quality problems, as mentioned
o f the shuntcontroller.
above with the discussion
The main disadvantageof the shunt-seriescontroller is the largecurrent rating
required to mitigate deep sags.
For low-power, low-voltageequipmentthis will not be a
serious concern,b ut it might limit the number of large power andmedium-voltage
applications.
7.4.5 Backup Power Source-SMES, BESS
One of the maindisadvantagesof a seriescontroller is that it cannot operate
during an interruption. A shunt controller operatesduring an interruption, but its
storage requirementsare much higher. We saw
that the shunt-connectedcontroller
operatesperfectly when only thecontroller and the protectedload are interrupted.
The controller is in that case only feeding theprotectedload. This principle can be
used by creating the rightinterruption. This results in theshunt-connectedbackup
power source as shown in Fig. 7.52. The
configuration is very similar to the shunt
439
Section 7.4 • TheSystem-EquipmentInterface
------t
System
Statict--_.._-------switch
Load
Energy
storage
reservoir
u
00
>
Figure 7.52 Shunt-connected backup power
source.
_ _~ Static1 - - - . . . . , . - - - - - - ' \
System switch
1
Load
Static
switch
2
Figure 7.53 Series-connected backup power
source.
Energy
storage
reservoir
controller. The difference is the static switch which is present between the system and
the load bus. Themomentthe system voltagedropsbelow a pre-set rms value, the static
switch opens and the load is supplied from the energy storage reservoir
through the
voltage-sourceconverter.Various formsof energy storage have been
proposed.A socalled superconductingmagneticenergy storage (SMES) stores electrical energy in a
superconducting coil [57], [158], [159], [160], [161], [162]. A BESSor battery energy
storagesystemuses a largebatterybankto store the energy
[186], [187],[188]. For small
devices the energy
storageis not a problem,but using a SMES, BESS, or any
otherway
of storage at medium voltage will
put severestrainson the storage. A backup power
source is only feasible if it can ride
t hrough a considerablefraction of short interruptions. Looking at some statistics forshort interruptions,Figs. 3.5, 3.6, and 3.7, shows
that the amountof storageshouldbe able to supply the load for
10 to 60 seconds. Less
storage would not give any serious
improvementin the voltage tolerance
comparedto
the seriescontroller.
All backuppower sources suggested in the
literatureuse ashuntconnection,but it
aseriesconnectionas in Fig. 7.53. This device could
operateas a
is also feasible to use
seriescontrollerfor sags and as baackuppower source forinterruptions.The moment a
deep sag is detected, static switch 1 opens and static switch 2 closes.
7.4.8 Cascade Connected Voltage Controllers-UPS
The main device used to
mitigatevoltage sags and
interruptionsat the interface is
the so-calleduninterruptablepowersupply(UPS).The popularityof the UPS isbasedon
its low costs and easy use.
For an office worker the UPS isjust anotherpiece of
440
Chapter7 • Mitigation of Interruptionsand Voltage Sags
equipmentbetween the wallo utlet and acomputer.All that is needed is to replace the
batteriesevery few years, and as long as one does
not power the kettle and the microwave from the same UPS, virtually
a
problem-freesupply iscreated.
7.4.6.1 Operationof a ups. The UPS isneither a shunt nor a series device,
but what could bedescribedas acascadeconnectedcontroller. The basic configuration of a typical UPS isshown in Fig. 7.54. Itsoperationis somewhatsimilar to the
drive (compareFig. 5.12): a diode rectifier
converterpart of an ac adjustable-speed
followed by an inverter. The main difference is the energystorageconnectedto the
de bus of a UPS. In allcurrently commerciallyavailable UPSs the energy
storageis
in the form of a battery block. Other forms of energystoragemight become more
suitablein the future.
During normal operation,the UPS takes its power from the supply, rectifies the
ac voltage to dc andinverts it again to ac with the same frequency and rms value. The
designof the UPS is suchthat the de voltageduring normal operationis slightly above
the batteryvoltageso that the batteryblock remainsin standbymode. All power comes
from the source. The onlyp urposeof the batteryblock in normal operationis to keep
the de busvoltageconstant.The load ispoweredthroughthe inverterwhich generates a
sinusoidalvoltagetypically by using aPWM switching pattern.To preventload interruptions due to inverter failure, a static transfer switch is used. In case the inverter
output drops below acertainthresholdthe load is switched back to the supply.
During a voltage sag orinterruptionthe batteryblock maintainsthe voltage at the
de bus for severalminutesor evenhours,dependingon the batterysize. The load will
thus tolerate any voltage sag ors hort interruption without problem. For long interruptions, the UPSenablesa controlledshutdown,or the start of a backupgenerator.
Bypass
de
ac
System
de
Energy
storage
Figure 7.54 Typicalconfigurationof an
uninterruptablepower supply (UPS).
7.4.6.2 Advantages and Disadvantages.
The advantageof the UPS is its simple
operationand control. The power electronic componentsfor low-voltage UPSs are
readily availableand the costsof a UPS arecurrently not more than the costsof'.a
personalcomputer.It is probably not worth installing a UPS for eachpersonalcomputer in an office (making regular backupswould be moresuitable), but when a
computer(or any other low-power device) is an essentialpart of a production process the costsof the UPS are negligible. As the UPS will
mitigate all voltage sags
and short interruptionsa stochasticassessment
is not even needed.
The main disadvantageof the UPS is thenormal-operatingloss because of the
two additional conversions,and the useof batteries.Contrary to general belief, batteries do needmaintenance.They should be regularly tested to ensure
that they will
operatein caseof an interruption; also they should not be exposed to high or low
441
Section 7.4 • TheSystem-EquipmentInterface
temperaturesand sufficient cooling should be installed preventoverheating.All
to
this
is not so much aconcernfor the small UPSs used in an office
environment,but for large
installationsthe maintenancecosts of a UPSinstallationcould becomeratherhigh.
7.4.6.3 Alternatives. As a long-term solution to mitigate voltage sagsand
interruptions,the UPS is not the mostappropriateone. The twoadditional conversions are not really needed, as can be seen in Fig. 7.55. Thedrawing
top shows the
normal configuration:the ac voltage isconvertedinto de and back to ac by the UPS.
In the computer the ac voltage is againconvertedinto de and nextconverted to
the utilization voltage for the digital electronics. This scheme represents
almost any
modernconsumerelectronics device.
Alternatively, one can directlyconnectthe batteryblock to the de bus inside the
computer.In fact a laptop computergets its power in such a way. Some
mitigation
methodsfor ac adjustable-speed
drives also use a direct infeed into the dc bus.
From an
engineeringviewpoint this is a more elegant
solution than using a UPS, but the user
doesnot always have the technical knowledge to do this.
solutionlike
A
this can only be
initiated by the equipmentmanufacturers.
One can extend this idea
further, ending up with a denetwork for an office
building providing backuppower to all sensitiveequipment.By connectingan array
of solar cells to this denetwork the situation could arise where the utilitysupply
becomes thebackupfor the internal de network.
UPS
.- -. ---- -----Computer
---------.. -.... ---.. ----.
f
f
t-----:--t
_
..
i
Digital
electronics :
-
-
. _ .. -
__ -
-
I
f
I
_ _ eI
Computer
Digital
electronics
I
I
I
I
,.
-.-- ---
_--------._.
Figure 7.55 Powerconversionsfor a UPS poweringa computer,and for an
alternativesolution.
7.4.6.4 UPS and Backup
Generators. Figure 7.56 shows a power system where
both UPSs and backup generationare used to mitigate voltage sags and
interruptions. The UPS is used to
protect sensitive essential load against voltage sags and
short interruptions.But especially for large loads, it is not feasible to have more
than
a few minutes energy supply stored in the batteries. In case
of an interruption, the
so-called "islanding switch" opens, disconnectingthe sensitive load from theutility
system. During the interruption the sensitive load is completelypowered from a
backup generator.This generatorcan be eitherrunning in parallel with the utility
442
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Utility
infeed
Islanding
switch
Nonessential
load
Nonsensitive
essentialload
Figure 7.56 UPS combined with backup
Sensitive
generation to mitigate voltage sags, short and
essential load long interruptions.
supply, or bestartedthe momentan interruption is detected. All essential load is fed
from the backup generator,where only the essential load which is sensitive to sags
and short interruptionsneeds to be powered from the UPS.
Decreasingthe time to
switch over to islandoperation decreases the energy
storage requirementsin the
ups. The energystoragerequirementis proportional to the switch-over time. The
UPS only needs to supply the load which
cannottoleratethe interruptiondue to the
switch-overto islanding operation.The faster the switch-over, the less load needs to
be powered from the UPS.
An interestingexample of the use
o f UPSs incombinationwith on-sitegenerators
to achieve a high reliability is discussed [172].
in
7.4.7 Other Solutions
Somemitigation equipmentis not based on thevoltage-sourceconverter;a few
examples are discussed below.
Motor-generatorsets andferroresonanttransformers
have beenaround for many years to mitigate voltage sags;
electronic tap changers
form an interestingnew technique.
7.4.7.1 Motor-Generator Sets. A motor-generatorset is an oldsolution against
voltage sags,making use of the energy stored in a flywheel. The basic principle is
shown in Fig. 7.57: a(synchronousor induction) motor and asynchronousgenerator
are connectedto a common axis together with a large flywheel. When the power
supply to themotor is interrupted,the flywheel makes that the systemcontinuesto
rotate and thuscontinuesto supply the load. These kind
o f systems are still in use
(and new ones are still being installed) industrial
in
installations. The ridethrough
time of several seconds enables
transferschemes withmechanicalswitches. The noise
of a motor-generatorset and themaintenancerequirementsof the rotating machines
are not a concernin most industrial environments.They do however makemotorgeneratorsetsunsuitablefor an officeenvironment.
In the configurationshown in Fig. 7.57, thenormaloperationlosses are very high
which makes this an expensive
solution. A numberof alternativeshave beenproposed
to limit the losses. Oneoption is to have themotor-generatorset operatingin no-load
when the supply voltage is within its
normalrange. Themomenta sag orinterruptionis
detected, a (static) switch is opened and generator
the
takes over the supply. A possible
configurationis shown in Fig. 7.58.
In normal operationthe synchronousmachineoperatesas asynchronouscondensorwhich can, e.g., be used for reactive power
compensationor for voltagecontrol.
When the supply isinterruptedthe static switch opens and the
synchronousmachine
443
Section 7.4 • The System
-EquipmentInterface
r-r-
Flywheel-
=
Power
system
Generator ~
Motor
I--
Sensitive
load
-
Figure 7.57 Principle of motor-generatorset.
Static
switch
Power
system- - - - - I
1-----,.-
-
-
-
-
-
- Load
Synchronous
machine
Flywheel
Figure 7.58 Configurationof ofT-line UPS
with diesel enginebackup.
Diesel
engine
starts tooperateas asynchronousgenerator,injecting both active and reactive power .
This will provide power for one or two seconds. By using a large reactance between the
load and the power system,certain
a
level of voltage-sagmitigation is achieved. The
effect is the same as for an on-site
generator.By opening the static switch on an
undervoltageit is even possible tooperate the synchronousmachine as abackup
power sourceduring sags as well. While the flywheel provides
backuppower, the diesel
engine isstarted.
More recentimprovementsare the use ofwritten-polemotorsand thecombination of a motor-generatorset with power electronics. Awritten-pole motor is an ac
motor in which the magnetic pole pairs are not
obtainedfrom windings but instead are
magnetically written on therotor [193]. This enables aconstantoutputfrequency of the
generator,independentof the rotationalspeed. The mainadvantagefor use in amotorgeneratorset isthat the generatorcan be used over a much larger range of speed, so
that more energy can be
extractedfrom the flywheel.
A combinationof the motor-generatorset with power electronicconvertersis
shown in Fig. 7.59. Themotor is no longer directlyconnectedto the power system,
but through an adjustable-speeddrive. This enablesstarting of the flywheelwithout
causing voltage sags in the system, overspeed of the flywheel increasing
ridethrough
the
time, and lossreduction while the set is instandby.The output of the generatoris
rectified to a constantde voltage which can be utilized
through a series- or shuntconnected voltage-source
converteror directly fed into the de buso f an adjustablespeed drive. The ac/dc
converterenables theextractionof power from the flywheel over
a much larger range of speed.
Supposethat a normalmotor-generatorset gives anacceptableoutputvoltage for
a frequency down to 45 Hz (in a 50 Hz system). A frequency of 45 Hz is reached when
the speed has
droppedto 90%. Theamountof energy in the flywheel is still 81% of the
energy at maximum speed. This implies
that only 19% of thestoredenergy is used.
444
Chapter7 • Mitigation of Interruptionsand Voltage Sags
Adjustable-speed
drive
ac motor
Power
system
Figure 7.59 Powerelectronicconvertersin
combination with a motor-generatorset.
Supposethat we cangeneratea constantde voltage for a speed down to 50% , by using
an ac/dcconverter.The energythat can beextractedis 75% of the total energy, an
increase by a factoro f four. The ridethroughtime is thus also increased by a factor of
four-for example, from 5 to 20 seconds. The
ridethroughcan befurther increased by
running the acmotor above nominal speed. Byacceleratingthe flywheel slowly, the
m otor can be kept small. As the kinetic energyproportional
is
mechanical load on the
to the square of the speed,rathersmall
a
increase in speed can
alreadygive a serious
increase inridethroughtime.Supposean overspeedof 20%. which increases the energy
in the flywheel to 144% of theoriginal maximum. The extraction of energy from the
flywheel stops when 25% of the original maximum remains, that
so the amount of
energyextractedfrom the flywheel is 119%: afactor of six more thanwith the original
setup . The resultingridethroughtime is 30 seconds .
7.4.7.2 Electronic TapChangers. Electronic tap changers use fast static
switches to change the
t ransformation rat io of a transformer. Th is can either be a
distr ibution transformeror a dedicatedtransformerfor a sensitive load. The principle
of its operation is shown in Fig. 7.60, in this case with three static switches. The
number of turns of the fourparts of the secondarywinding are (top tobottom):
100%, 40% , 20%, and 10% of the
nominal turns ratio . Byopeningor closing these
three switchestransformationratios between 100% and 170% can be achieved, with
10% steps. If all three switches are closed, the
turns ratio is 100%; with switch 1
closed and 2 and 3 open it is 130% , etc. By using this electronicchanger
tap , the
o f nominal for input voltages down to
output voltage is between 95% and 105%
currently available as
56% of nominal. Transformerswith electronictap changers are
.....
Power
system
-
>>-
Load
.....
>>>>>-
>-
,'1
:'2
1'3
Static
switehe
Figure 7.60 Basic principleof the
constructionof an electron ic tap changer.
445
Section 7.4 • TheSystem-Equ
ipment Interface
an additional series
componentbetween the source and the load. In future it may be
feasible to install electronic tap changers on distribution transformers and save the
additional component.
7.4.7.3 FerroresonantTransformers. A ferroresonanttransformer,also known
as aconstant-voltagetransformer,is mainly designed to maintain constantvoltage
a
on its output over a range of input voltage. The basic
constructionof a ferroresonant
transformeris shown in Fig. 7.61. The third winding of a three-winding
transformer
is connected to a large capacitor.
Without this capacitor,the device operates as a
normal transformer.The effect of thecapacitoris explainedthrough Fig. 7.62. The
solid line is the relation between voltage and current for the nonlinear inductance.
The dashed line holds for the capacitor. The place where the curves cross is the operating point. Note that these curves give the voltage and current magnitude for one
frequency, in this case the power system frequency as that is the frequency exciting
the system. Thisoperatingpoint is independent of the supply voltage, thus the flux
through the iron core is independent of the supply voltage (assuming that the ferroresonantwinding has a smaller leakage than the input winding). The
output voltage
is related to this flux, thus also independent of the input voltage.
The energy stored in the ferroresonant winding is able to provide some ridethrough during voltage dips. A disadvantage of a ferroresonant transformer is its
dependence on load changes. The inrush current of the load can lead to a collapse of
the flux and a long undervoltage . A modern version of the ferroresonant
transformer
uses power electronic converters to keep the load current at unity power factor, thus
optimizing theoperation of the transformer.
power~ ~sensitive
0----3
system
tl
LJ
Figure 7.61 Basic principle of the
construction of a ferroresonant transformer .
Figure 7.62 Voltage versus current diagram
for a saturableinductor (solid line) and for a
capacitor(dashed line).
~Ioad
.:
Current
Ferroresonant
winding
446
Chapter7 • Mitigation of Interruptionsand Voltage Sags
7.4.8 Energy Storage
Severalof the controllersdiscussedabove,needenergystorageto mitigate a sag.
All of them needenergystorageto mitigatean interruption.Herewe comparedifferent
types of energystoragewhich arecurrently being usedandconsidered.T he comparison
is basedon three different time scales,relatedto threedifferent controllers.
• A seriesvoltagecontrolleris only able tomitigatevoltagesags. Atypical design
value is 50%, 1 second;i.e., the controller is able to deliver 50% of nominal
voltagefor 1 second.In termsof energy-storagerequirementsthis corresponds
to full load for 500 ms.
• A (shunt-connected)b ackup power sourceis also able tomitigate interruptions. To be able toimprove the voltage tolerancesignificantly a ridethrough
between10 and 60 secondsis needed.We considerthe requirement:full load
for 30 seconds.
• To achievevery high reliability, sensitiveload is typically poweredvia a UPS
which can supply the load for 10 to 60 minutes. During this period, backup
generatorscome on line to take over the supply. The third energy-storage
requirementwill be full load for 30 minutes.
7.4.8.1 DC Storage Capacitors.Capacitorsare mainly used to generatereactive power on an ac system. But in a de systemthey can be used togenerateactive
power. The amountof energystoredin a capacitanceC with a voltage V is
(7.69)
The voltagedecreaseswhen theenergyis extractedfrom the capacitor.Capacitorscan
thus not be used tosupplyelectric power to a constant-voltagede bus, asneededfor a
voltage-sourceconverter.A second(de/de)'converteris neededbetweenthe capacitors
and the constant-voltagebus, asshown in Fig. 7.63. Alternatively, the control algorithm of the voltage-sourceconvertercan be adjustedto variablede voltage.
In either case,there will be a minimum voltage below which the converteris no
longerableto operate.It is thusnot possibleto extractall energyfrom the capacitors.I f
the converteroperatesdown to 50% of the maximumvoltage,75% of the energycanbe
extracted.A converteroperatingdown to 25% canextract 940/0 of the energy.
Considera medium-voltagecontroller using 4200 V, 1500J.LF storagecapacitors.
The amountof energystoredin one capacitoris
(7.70)
PWM voltage-source
converter
Storage
capacitors
\
de
de
Variable
de voltage
O__
_ _....J
ac
Power
system
interface
Figure 7.63 Energyextractionfrom de
storagecapacitors.
Section 7.4 • TheSystem-EquipmentInterface
447
Supposethat the converteris able tooperatedown to 50% of voltage. Eachcapacitor
unit is able tosupply: 0.75 x 13kJ= 9.75 kJ.
For a 500 msridethrough,eachunit cansupply 19.5 kW of load. A small mediumvoltage load of 500 kW requires26 capacitorunits; a largemedium-voltageload of
10MWover 1000 units. For a 30secondridethrougheach unit can only power 325W
of load, alreadyrequiring 1500units for a small medium-voltageload. Thus de capacitors are feasible for series
controllerswith ridethroughup to about 1 second,but not
for backupvoltagesourcesrequiring ridethroughof 30 secondsand more.
Various energy storageoptionsfor adjustable-speed
drives arecomparedin [42].
A price of $35 is given for a 4700
JtF, 325 V capacitor.The amountof energystoredin
one suchcapacitoris 250J,of which 188J(75%) can be used,enoughto powera 375W
load for 500 ms or a 6.25 W load for 30
seconds.
To power a small low-voltageload of 1000Wduring 500msrequiresthreecapacitorscosting$105; topowerit for half a minuterequires160capacitors,c osting$5600.
For a completelow-voltageinstallationof 200 kW we need 534
c apacitors($18,700) for
500msridethroughand 32,000capacitors($1,120,000) for 30seconds.T he conclusion
is the same as before:
capacitorstorageis suitablefor 1 secondridethroughbut not for
1 minute ridethrough.
7.4.8.2 Batteries. Batteriesare a verycommonly used method of storing electric energy. They are used in the vast
majority of UPSs sold,not only in the small
one used topower a single PCbut also in larger ones whichcan power a complete
installation. Batteries provide a constantvoltage so that they can be directly connected to thevoltage-sourceconverter. A 5 MVA, 2.5 MWh battery energy storage
system (BESS) has been
installed to power critical equipment in a large chemical
facility [188]. The amountof storedenergy in this system is GJ,
9 much more than in
any of the aboveexamples.An even larger installation has beeninstalled in California in 1988 for load-levelingpurposes[186]. This BESS is able tosupply 10MW during 4 hours, correspondingto 144GJ of stored energy. This installation covers an
areaof 4200 m2 for the batteriesonly.
Looking at smallersizes,considera car batterywith a storagecapacityof 1MJ
(12 V,'23 Ah) costingabout$50. This simplebatterycontainsenoughenergyto powera
2 MW load during 500 ms, a 33 kWload during 30 seconds,or a 550 Wload during 30
minutes. One car battery containsthe sameamountof energyas 77medium-voltage
storagecapacitors.
The limitation with a batteryis not so muchthe amountof energystoredin it, but
the speed with which this energy can madeavailable.Emptyingour
be
car batteryin 30
secondsrequiresa currentof 2760 A. The batterywill never beableto supplythis. If we
considera maximumcurrentof 200 A, the maximumload which can be suppliedfrom
one battery is 2400W. The battery can power this load for 7 minutes,which can be
consideredas theoptimumridethroughtime for this battery.This fits well in equipment
to mitigate interruptionsfor the time until on-sitegenerationbecomesavailable.
The numberof batteriesneededand the costsof these, are given inT able 7.5 for
the load sizesa nd ridethroughtimes given before. Only fors hortridethroughtimes will
capacitorsbe able tocompetewith batteries.
Batterieshave anumber of disadvantagesc omparedto capacitors,which may
compensatethe higher costs of the latter. The commonly used lead-acid battery (on
which this calculationis based),containsenvironmentally unfriendlymaterials,has a
limited lifetime (in numberof rechargingcycles),and requiresregularmaintenanceto
ensurea high reliability. The newer typesof batteries,which arebeing developedfor use
448
Chapter7 • Mitigation of Interruptionsand Voltage Sags
TABLE 7.5 Numberof Batteries (inbrackets)and CostsNeeded toPowerSeveral
Load Sizes for SeveralRidethroughTimes
500 ms
30 sec
30 min
I kW
200 kW
500 kW
(I) S50
(I) S50
(2) stoo
(84) S4200
(84) S4200
(364) SI8,000
(209) SIo,oOO
(209) $10,000
(910) $46,000
IOMW
(4167) $210,000
(4167) $210,000
(18182) S910,000
in electrical vehicles, do not have these
disadvantagesbut they obviously have higher
costs.
7.4.8.3 Supercapacitors. Supercapacitors(or double-layercapacitors)are propagated as a future solution for energy storage to improve
equipmentvoltage tolerance. They have energy densities
comparableto batteries, but much longer lifetime
and much lessmaintenancerequirements. Theirdisadvantageis that they are only
of 3.3 F, 5.5 V ismentionedin [189].
available for voltages of a few volts. A value
The amountof stored energy is5 0J, only 1/5th of the4700J.l,F, 325V capacitor. Like
with a battery, there is a limit to the speed with which energy be
canextracted from
a supercapacitor.F or the supercapacitorscurrently in operation, the discharge time
cannotbe less thanabout 1 minute . This makes them somewhat faster than batteries
but still much slower thancapacitors. The development ofsupercapacitorsis mainly
o f electric vehicles, where the
a mountof storedenergy is
driven by the requirements
of more importancethan the speed with which it can be extracted .
7.4.8.4 Flywheels. An alternative which is currently being investigated is the
storage of energy in fast-spinning
flywheels. The classicalmotor-generatorset, discussed before, already uses this principle, but the modern equivalent
rotates at a
much higher speed. By using magnetic bearings and vacuum sealing of
rotating
the
[192], values up to 90,000 rpm
parts , very highrotational speeds can be achieved
have been reported [l90J. A possible
configuration is shown in Fig. 7.64. The flywheel isbroughtup to speed by an ac
adjustable-speed
drive. This drive also ensures
that rotational speed of theflywheel remains within a certain range
during standby
operation.During a voltage sag or an
interruptionthe brushless de
generatorextracts
From the
power
system
Brushless de generator
~
To the
power
''' _ ~~~ ~~ , - - system
Inertia
Figure 7.64 Configurationof a flywheel energystoragesystem and itsinterfaceto
the power system.
449
Section 7.4 • TheSystem-EquipmentInterface
energy from the flywheela nd suppliesthis to the power system via ade/deconverter
and a voltage-source(dc/ac) converter.
Considera solid cylindrical pieceof materialwith a length of 50 em and a radius
of 25 em. Theinertia of this pieceof material,for rotationalongthe axisof the cylinder,
is
J
= ~mR2
(7.71)
with m the massand R the radiusof the cylinder. With a specific massof 2500 kg/m" we
find for the mass:
m=n
X
0.252 x 0.50 x 2500 = 245 kg
and for the inertia:
J
= 2:1 x 245 x 0.252 = 7.7kgm2
The kinetic energyof an intertia J rotating with an angularvelocity (J) is
£ = !J(J)2
2
(7.72)
If we rotate our cylinder at the "moderate" speed of 3000 rpm (w =
21r X 3~ = 314radjs,the amountof kinetic energystoredin the rotating cylinder is
1
£ =2 x 7.7 x 3142 = 380kJ
This energycannotbe extractedcompletely,as the energyconversionbecomes inefficient below acertain speed. Supposethis to be 50% of the maximum speed. The
amount of useful energy is again 750/0 of total energy, in this case
0.75 x 380kJ = 285kJ. This flywheel is thus able to power a 570kW load for 500ms,
a 9.5kW load for 30 seconds,o r a 160W load for 30 minutes.
Increasingthe rotational speed to 25,000 rpm by using the newest
technologies,
increasesthe amountof storedenergyto
1
£ = 2 x 7.7 X 26182 = 26 MJ
The useful energy of 0.75 x 26MJ is enoughto power a 40MW load for 500ms, a
650kW load for 30 seconds,or an II kW load for 30 minutes.
7.4.8.5 Superconducting Coils.It is well known that an inductor L, carrying a
current i, containsan amountof energyin its magneticfield equal to
(7.73)
This would makean inductor an alternativeform of energystorage,next to thecapacitor. The reasonthat inductorstorageis not commonlyused isthat the currentcauses
high losses in the wirem akingup the inductor.The losses due to current
a
i are equalto
(7.74)
with R the total seriesresistance.S upposethat we can achieve anX jR ratio of 100 for
the inductor. In that case we find for the losses:
450
Chapter7 • Mitigation of Interruptionsand Voltage Sags
(7.75)
To compensatefor the resistivelosses,the energycontentsin the coil has to be
suppliedthree times a second.
A solutionsuggestedseveralyearsago is tostorethe energyin a superconducting
coil. The resistanceof a superconductoris (exactly) zero sothat the current will flow
forever without any reductionin magnitude.A possibleconfigurationfor sucha superconducting magnetic energy storage
(SMES)is shownin Fig. 7.65.The variablecurrent
through the superconductingcoil is convertedto a constantvoltage. The constantvoltage de bus isconnectedto the (ac) power system by meansof a voltage-source
converter.The coil currentclosesthroughthe de/deconverterwhich causesa small loss.
The configurationofSMESdevices isdiscussedin moredetail in [57], [158], [160], [162],
[169].
Refrigerator
Constant-voltage
de bus
Superconducting
coil
t
Power
system
interface
Figure 7.65 Energystoragein a
superconductingcoil and interfacewith the
power system.
Oneapplication[158] uses a 1000 Acurrentthrougha 1.8 H inductor. The energy
storedin the magneticfield is
1
£ == 2" x 1.8 X 10002 = 900kJ
(7.76)
Assumethat the de/deconverteroperatesfor currentsdown to 50% of the maximum
current. The usable energy is in this case 0.75x 900kJ = 675kJ. This is enough to
power a 1.35MW load for 500 ms, a 22.5 kWload for 30 seconds,or a 375 W load
for 30 minutes. The device describedin [158] operatesas a shunt-connectedbackup
power source;it is used tomitigatevoltagesagsand shortinterruptionswith durations
up to a fewseconds.
Commercialapplicationsof SMES devices arereportedfor storedenergyup to
2.4 MJ and power ratings up' to 4 MV A. The devicescurrently in operationuse lowtemperaturesuperconductors
w ith liquid helium as acoolingmedium.A demonstration
SMES using high-temperaturesuperconductorshas beenbuilt which is able tostore
8 kJ of energy.This is still two ordersor magnitudeaway from the devices using lowtemperaturesuperconductors,but the manufacturerexpectsto build 100 kJ devices in
the nearfuture. A study after the costsof SMES devices nowand in 10 years'time, is
describedby Schoenunget al. [168]. For example, a 3MW, 3 MJ unit would cost
$2,200,000now, but "only" $465,000in 10 years' time. The main cost reduction is
basedon the so-calledlearningcurve due to the productionof about 300 units in 10
years.By using the datain [168] the costshavebeenplotted as afunction of the stored
energy,resultingin Fig. 7.66.
In Table7.6 thecostsof energystoragein a SMESare comparedwith the costsof
batteriesand capacitors.The costs of the power electronicconvertershave not been
451
Section 7.4 • TheSystem-EquipmentInterface
5-------------------,
• Costs now
in 10 years time
o Costs
4
~
3
~
2
.8
o
..
Figure 7.66 Costs ofsuperconducting
magnetic energystorage(SMES) including
the power system interface, asfunction
a
of
the amountof stored energy.( Data obtained
from [168].)
TABLE 7.6
n
0
o
0
0
0 0
o
..
o 0
o
o
o
00 0
0
00
o~_w.......:==----+----+-----+-----f
10
0.1
100
1000
Stored energy in MJ
CostsComparisonof SMES, BESS andCapacitors
Costs of EnergyStorage
RidethroughTime
Power
300 kW
3MW
I sec
60 sec
I sec
60 sec
BESS
SMES
$6300
$6300
$63,000
$63,000
$183,000
$389,000
$411,000
$1,064,000
Capacitors
$56,000
$3,350,000
$558,000
$33,500,000
included, as these are similar for all energy storage methods. The costsbattery
of a
energy storage system (BESS) is based on the same
batteriesas used before: 1MJo f
storage, 2400Wof power for $50. The costs of
capacitorstorage is based on 188
J of
storage for $35 as used before.
Additional costs ofconstruction,wiring, protection,
cooling, etc., have not been included for the
capacitorsor for the batteries.
We see that, withcurrent prices, battery storage remains by far the
cheapest
solution, even if we consider a factor
of two to three for additional costs. But the
lifetime of a battery is limited in number of discharge cycles, andbatteriescontain
environmentallyunfriendly products.When the costsof SMES devices go down and
the costs of batteries go up in the future, the former will become a more
attractive
option for high-power short-time ridethrough.For short-time ridethroughcapacitor
storage is still moreattractive,especially if one realizes
that we used low-voltage capacitors where medium-voltage
capacitorsare likely to form acheaperoption.
Note that the amountof energystoredin an SMES is similar to theamountof
energy stored in abattery.The main difference isthat the energy in asuperconducting
coil can be made available much faster. The units
currently in operationare able to
extract 1MJ of energy from the coil in 1 second. The
limitation in energyextractionis
the voltage over aninductor when thecurrentchanges:
di
V;nd
dc
= L Cit
(7.77)
The energyextraction p/oad is related to the change in
currentaccordingto
~
H3
Li
c}
= P10ad
(7.78)
452
Chapter7 • Mitigation of Interruptionsand Voltage Sags
which gives for thevoltage over the inductor:
. V ind -
P/oad
.
'de
(7.79)
With constantenergy extraction (constant p/oad) , the induced voltage increaseswith
decreasingcurrent. For a 500 kW load and a minimum current of 500 A, the voltage
over the coil is
500kW
V;nd
= 500A
= lOOOV
(7.80)
For a 3MW unit we get V;nd = 6 kV. The de/de converter should be able to
operatewith this voltageover its input terminals.
Summary and
Conclusions
This chaptersummarizesthe conclusionsfrom the previouschapters.Next to that some
thoughtsare givenconcerningthe future of this area of power engineering. Just like in
the rest of thebook, the emphasisis on voltage sags and
interruptions.
8.1 POWER QUALITY
In Chapter I the term "power quality" and several related terms are defined. Power
quality is shown to consistof two parts: "voltage quality" and "currentquality." The
voltagequality describes the way in which the power supply affects
equipment;as such
it is part of the quality of supply. Current quality describes the way in which the
equipmentaffects the power system and part
is of the so-called"quality of consumption." The termelectromagneticcompatibility (EMC) has a largeoverlapwith "power
quality" and the terms can often be used as synonyms.
An overview is givenof the various types of powerquality disturbances.An
important distinction is made between"variations" and "events." Variations are a
continuous phenomenon, e.g., the variation of the power system frequency.
Measuringvoltage andcurrentvariationsrequirescontinuousrecordingof their values.
Events only occur occasionally: voltage sags and
interruptionsare typical examples.
Measuringvoltage andcurrentevents requires a triggering process: e.g.,ems
the voltage
becoming lessthan a pre-definedthreshold.These two typesof power quality disturbances also requiredifferent analysis methods: average andstandarddeviation for
variations;frequencyof occurrencefor events.
The main subjecto f this bookis formed by voltage sags and
interruptions:the two
mostimportantexamples from a familyo f voltage events known as
"voltagemagnitude
events." Voltage magnitudeevents aredeviationsfrom the normal magnitude(ems
value) of the voltage with arather well-definedstarting and end time. Themajority
of these events can be
characterizedby one magnitudeand oneduration. Different
initiating events and differentrestorationprocesses lead to
different rangesof magnitude andduration.Based on these ranges, a classification
of voltagemagnitudeevents is
proposed.
453
454
Chapter8 • Summaryand Conclusions
8.1.1 The Future of Power Quality
There is one questionthat always comesup when thinking about the future of
powerquality: "Will the powerquality problemstill be amongus in 10 years time?" It
may well bethat equipmentwill be improvedin such a waythat it no longeris sensitive
to the majority of voltagedisturbancesand that it no longer producesseriouscurrent
disturbances.In other words, equipmentwill have becomefully compatiblewith the
power supply. At the moment, however, there is no indication that this will happen
soon.Equipmentappearsto be assensitiveandpolluting as ever. Abrowsethroughthe
advertisementsin power-qualityorientedjournalsshows that the emphasisis on mitigation equipment (surge suppressors,UPSs, custom power) and on power-quality
measuremente quipment.Advertisementsin which equipmentwith improved voltage
toleranceis offered are extremelyrare.
The main drive for improved equipmentis likely to come from standards,in
particular the IEC standardson electromagneticcompatibility. When the standards
on harmonic currents produced by end-userequipment (lEe 61000-3-2 and -3-4)
becomewidely accepted,the harmonic distortion problem may be the first one to
move to thebackground.
Voltage quality eventslike voltage sags will take even longer to becomepart of
equipmentstandards.A t leastvoltagesags arereasonablyunderstoodnowadays(read
Chapters4, 5, and 6). Higher frequencyphenomenalike switching transientsare less
well understood,more difficult to model, and their statistics probably show more
variations among different customers.Still they causeequipmentproblems. Highfrequencydisturbancesmay well becomethe next bigpower-qualityissue.
8.1.2 Education
An importantaspectof powerquality is education:educationof those who come
in touch with power quality problemsas well as newgenerationsof engineers.Power
quality may bring power engineeringeducationcloser to the actual aim of power
engineering:generatingelectrical energyand delivering it to electricalend-userequipment. Educatinga newgenerationof engineersis obviouslya taskfor universities. And
with engineersI am not only referring to power engineers.Every studentin electrical,
electronic,and mechanicalengineeringshould know aboutpotential problemsdue to
the connectionof equipmentto the powersupply.Note that these are thepersonsto use
electricalequipmentandto designfuture equipment.When they areawareof potential
compatibility problems,they are more likely to comeup with equipmentthat is compatible with the supply.
Postgraduateeducationis importantand not necessarilya task for a university.
Severalcompaniesoffer good power-qualitycoursesthat enablepeoplein industry to
solve the problems they encounter.However, universities are better suited to give
theoreticalbackgroundsneededto solve future problems,next to providing an understandingof existing problems.
8.1.3 Measurement Data
From the beginning,power quality has been anareavery much based on measurementsand observations.T he standardtools in use atuniversities,simulationsand
theoreticalanalysis;are much less used in thepowerquality work. In fact, theamount
of universityresearchon powerquality is still very limited. This will certainlychangein
Section 8.2 • Standardization
4SS
the near future; powerquality will not only find its way into educationbut also into
university research.There is a serious risk herethat a gap will develop between the
heavily measurement-based
power-quality practice and the very much theory- and
simulation-baseduniversity research. Such situation
a
may be preventedif utilities
education.A
make much moreof their data available for university research and
very good example is set by
IEEE Project group 1159.2. At their Website (accessible
through www.standards.ieee.org)
a number of voltage recordings are available for
downloading.I would like to see much more utilities making
d ata available in this
way: not only theactual voltage andcurrent recordings but also some basic
data
aboutthe kind of event and the kind of power system involved.
8.2 STANDARDIZATION
In the secondpart of Chapter1, power qualitystandardsare discussed. The IEC set
of
standardson electromagneticcompatibility offers the opportunity to seriously solve
several powerquality problems.The standardsdescribe variouspower-qualitydisturbances, define testingtechniquesand give requirementsfor equipmentand system
performance.A large number of standardsis still under developmentand even
more arerequired to fully standardizeequipmentas far as power quality and EMC
is concerned.In Chapter1 some suggestions are given for the extension of the concept
"compatibility level" from variationsto events.
The Europeanvoltage characteristicsstandard,EN 50160, is described in detail.
The standardgives a gooddescriptionof the voltage quality forvoltage.variations,but
is ratherweak for voltage events.
8.2.1 Future Developments
Developmentsin this area will unfortunatelytake a long time, sothat power
quality problemswill be aroundfor at least several more years. This is simply
inherent
to the standard-settingprocess.During my work on someIEEE standards,it became
of
clear that one can only take one step at a time. The first step, making people aware
power quality problems,has beentakenboth within the IEC and within the IEEE. The
recently publishedIEEE standardon compatibility between electronic process equipment and the power system
( IEEE Std. 1346-1998) may be the first of a long series of
IEEE standardson this subject. AlsoChapter9 of the 1997 editionof the IEEE Gold
Book (IEEE Std.493-1997) willcontributeto the power-qualityawareness. It is interesting to notice that both documentswere already being used and referred to several
years before theyactually became accepted as
standarddocuments.The same has
happenedwith several IECstandards,noticeablythe one limiting theharmoniccurrent
distortion by low-powerequipment(IEC 61000-3-2). BothIEEE and lEe shouldmake
their draft documentsavailableto a much wider audience. This will not only widen the
discussion but also speed up the
acceptanceprocess of thestandard.
The EuropeanvoltagecharacteristicsstandardEN 50160 is one of the first documents quantifying the voltage quality experienced by customers. Despite allshortits
comings, thepublicationof this standardhas triggered morecoordinatedmeasurement
campaignsthan before. Thefuture will bring the publicationof local equivalentsof EN
50160.
456
Chapter8 • Summaryand Conclusions
8.2.2 Bilateral Contracts
An arearelated to powerquality standards,but likely showingmuchfaster development,is formed by thebilateral contractsbetween utilitiesand customers.Several
examples arealreadyin place where the utility pays
compensationto its customerswhen
the quality of supply drops below a certainlevel. The typicalcontractdefines a maximum-acceptablenumberper year for each event type, e.g., two long
interruptions,five
shortinterruptions.When thisnumberis exceeded within acertainyear, the utility pays
a predefinedamountof compensationfor eachadditional event. The initialcontracts
only containedinterruptions,but voltage sags have been
implementedin a numberof
contractsas well. When setting up these
contracts,a precise definitionof the various
events is essential. Next to these
bilateral contracts,utilities are likely to come up with
generalcompensationschemes forcustomerswith a bad voltagequality. When utilities
refuse to take these steps they may be forced into worse
constructionsby political and
legal developmentsoutsideof their control.
The conceptof bilateralcontractsis likely to beextendedto the interfacebetween
transmissionand distribution systems. At thisinterfacevoltage quality becomes even
more two-directionalthan at the utility-eustomerinterface. Voltagedisturbancesmay
originatein either system.
8.3 INTERRUPTIONS
A long interruption is an interruption of the power supply followed by amanual
restoration.When the supply isrestoredautomaticallythe result is ashortinterruption.
Long interruptionsare discussed inC hapter2, short interruptionsin Chapter3. Long
interruptionsare by far themostserious voltagequality disturbance.M ost utilities keep
a record of frequencyand durationof long interruptions.Unfortunatelymuch of this
very usefuldata is not generally accessible. A positive
exceptionto this is theUnited
Kingdom where utilities are obliged topublish data on the supply performance.
Currently this only includesinterruption data but it is likely to beextendedto other
types of events.
Short interruptionsare shown to be due to combinationof
a
automaticreclosing
and a system design aimed at limiting the
numberof reclosers.Automatic reclosing
a itigation
makesthat a longinterruptionbecomes ashortinterruptionand is as such m
method.But limiting the numberof reclosers makesthat customersexperience ashort
interruptionthat otherwisewould have experienced a voltage sag.
Removingthe whole
the supply for somecustomersbut an improvereclosure scheme is deteriorationof
a
ment for others.
A detailed analysis ispresentedof voltages andcurrentsassociatedwith singlephasetripping. It is shown that single-phasetripping leads to less severe voltage events
at theequipmentterminals,but it may also lead to a higher
percentageof second trips.
A numberof pilot schemesshouldbe set up wheresingle-phasetripping is used for the
first attemptand three-phasetripping for the secondattempt.
8.3.1 Publication of Interruption Data
In the future more utilities will publish interruptionfrequency andsupply availability. For customersto be able to assess the
compatibility betweenequipmentand
supply, it is essentialthat utilities publishthe supplyperformance.As interruptiondata
Section 8.4 • Reliability
457
are already available, this will be the first to be published. A likely
developmentis that
utilities publish more than just frequency and availability over the whole
country.
Details like "worst-servedcustomers,"regional variations, and distribution of the
interruptiondurationwill give more insight into thequality of supply experienced by
a
individual customers.Publicationof more statistics will inevitably lead to comparison
between different utilities and regions. To
obtain a fair comparison,many years of
observationmay be needed. Alternatively satandardizedreliability evaluation tool
can be used to predict the supply
performance.As most interruptionsoriginate in
the distribution system, relatively simple techniques may be sufficient.
The increase inobservationdata will probably not include data on short interruptions, at least not initially.Getting data on short interruptionsfor all customers
requires an extensive
monitoring effort. For short interruptions,prediction methods
of getting datafor all customers. These
predictionmethmay be the only suitable way
ods may be"calibrated"throughmonitoring at a limitednumberof sites.
8.4 RBLIABILITY
The secondpart of Chapter2 summarizes the various aspects of power system reliability and the stochastic analysis techniques
currently in use: network modeling,
Markov models, andMonte Carlo simulation. Various examples are given for each
of these techniques. Different aspects are given for the reliability analysis
of generation,
transmission,and distribution systems (the three so-called
"hierarchicallevels"). For
the industrial power supply a systematic
methodologyis given that can be used to
obtain the reliability of the supply. Thismethodologyconsistsof six layers, partly
correspondingto the hierarchical levels but also including power
quality and equipment
failure.
8.4.1 V.rlflcatlon
Power system reliability has two distinctly different faces: the observed reliability
of numberand
and the predicted reliability. Observed reliability, i.e., keeping records
durationof interruptions,is the domainof the utilities; predictedreliability, i.e., reliability evaluation,is thedomainof universities;without much overlapbetween these two
sides. A comparisonbetween observed and predicted reliability is needed to move
forward in reliability evaluation.For this, utilities should provide thedata and universities the analysis and
prediction techniques. Only such caomparisonwill give a
clear answera boutthe accuracyof the variousstochasticpredictiontechniques. Such a
of stochasticpredictiontechniques and
comparisonwill also lead to a wider acceptance
to a wider useof them within the utilities.
8.4.2 Theoretical Developments
Potentialdevelopments on the
theoreticalside are the inclusionof nonexponential repair-time distributions and of common-modeeffects. In both cases thedata
requirements are high. This again calls for a closer
cooperationbetween utilities and
universities. Muchof the theoretical work on power system reliability has been directed toward transmission systems. In the near future,
distribution networks will
become much more a focus of the research. The main
theoretical bottleneck is
again thedistribution of the interruptionduration. By using theexponentialdistribu-
458
Chapter8 • Summary andConclusions
tion erroneousresults areobtained,especially for thenumberof very long interruptions.
8.5 CHARACTERISTICS OF VOLTAGE SAGS
In Chapter4 the various characteristicsof voltagesags are discussed. After the more
"classical"characteristics,m agnitudeand duration, two newercharacteristics,phaseanglejump and three-phaseunbalance,are treatedin considerabledetail. Techniques
are presentedto calculatethese sagcharacteristicsfor a given fault and loadposition
and fault type. Thetechniquesare applied to an example supplyconsistingof several
voltage levels.
Chapter4.
Phase-angle
j ump andthree-phaseunbalanceare discussed in detail in
Especially three-phaseunbalanceis an important characteristic.The currently used
definition of sagmagnitudeis not suitablefor three-phaseequipment.The definition
of sagmagnitudeis generalized forthree-phaseunbalancedsags leading to a classificaof which two types (C and D)
tion of three-phaseunbalancedsags into seven types,
cover the majority of sags. A three-phaseunbalancedsag is quantified through a
characteristiccomplex voltage which isindependento f voltage level or loadconnection.
Magnitudeand phase-anglejump are absolutevalue andargument,respectively,of the
characteristiccomplex voltage. The possible rangemagnitudeand
in
phase-anglejump
is calculated,for single-phaseas well as forthree-phaseequipment,for the example
supply as well as in general.
Chapter4 concludeswith a treatmentof two additionalsagcharacteristics,pointon-wave and missing voltage, discussionaboutload
a
influence on voltage sags, and a
brief treatmentof voltagesags due toinduction motor starting.
8.5.1 Definition and Implementation of Sag Characteristics
The variouscharacteristicsdiscussed here
a nd othersrecentlyintroduced,need to
be applied tomeasuredvoltage sags. This will giveinformation about their statistics
be these
and about the rangeof values that can be expected. The next step will that
additional characteristicsare implementedin commercially available power quality
monitors. Before that stage isreached,it is essentialthat all sag characteristicsare
uniquely defined. This willprevent confusion due to different manufacturersusing
different definitions. Missingvoltagemay become acompromisebetween the different
magnitudedefinitions used onboth sides of theAtlantic (voltagedrop versusremaining
voltage). Thedisadvantageof using missing voltage is
t hat the majority of single-phase
equipmentis affected by theremainingvoltage, not by the missing voltage. The application of point-on-wavecharacteristicsmay be limitedto.a small group of equipment.
But in any case, all these
characteristicsdescribepart of the quality of supply and
statisticalinformationaboutthem shouldbe partof the outcomeof voltage sag surveys.
8.5.2 Load Influence
An areathat has beensomewhatforgottenin the variousvoltage sag studies is the
effect of load on the voltage sag
characteristics.A qualitativestudyof the effect of large
inductionmotorsis describedin Chapter4. For a quantitativestudy of all types of load,
a detailed analysisof measuredvoltage sags is needed. Such a study
should include
large and smallm otor and electronicload as well asembeddedgeneration.The effect of
the loaddetermineshow the sagcharacteristicschangewhen a voltage sag
propagates
Section 8.6 • EquipmentBehaviordue to Voltage Sags
459
from high voltageto low voltage.Observationshaveshownthat a sag with amagnitude
(remainingvoltage) of 40% at 132 kV is seen as a sag with magnitudeof
a
60% at
400V.
8.8 EQUIPMENT BEHAVIOR DUE TO VOLTAGE SAGS
In Chapter5 the effectof voltagesags onequipmentis discussed. Theemphasisis on
single-phase rectifiers (computers, consumer electronics, process controllers), ac
adjustable-speeddrives, and de adjustable-speeddrives. Single-phaserectifiers are
affected bymagnitudeand duration of the voltage sag. They trip when thevoltage
drops below acertainmagnitudefor longer than a certainduration(resultingin a socalled "rectangularvoltage-tolerancecurve"). The voltage toleranceof the equipment
can easily beimprovedby addingadditionalcapacitanceto the internaldc bus. Using a
elegantbut also
voltage regulatorthat can operatedown to a lower voltage is a more
more difficult solution.
For three-phaserectifiers, as used in ac
adjustable-speed
drives, it is mainly the
characteristicmagnitudeand the sag typethat affect the de busvoltage and thus the
drive behavior.The amountof capacitancecurrentlyin use in ac drives is too small for
the sagdurationto haveany influence.Making the drivetolerantagainstbalancedsags
requiresseriousimprovementsin the designof the PWM inverter. For balancedsags
the dc-busvoltagedropsto a lower value(equalto the sagmagnitude,in pu) within one
o f the dc-buscapacitanceis very
or two cycles.For three-phaseu nbalancedsags the size
important. If the capacitanceis large enough(in the upper range of the amount of
capacitancecurrentlyin use) thedc-busvoltagewill not drop below 80% for any threephaseunbalancedsag. If the drive is able to stay
on-line, the effectof the sag on the
load will be very small.
DC adjustable-speed
drives areshown to be very sensitive tovoltage sags. The
armaturecurrent and the torque drop to zero almost immediately,even for arather
shallow sag. As de drives are typically used for
speed-sensitiveprocesses, thed rop in
speedassociatedwith the zerotorquewill easily lead to adisruption of the process.
8.8.1 Equipment Testing
An importantfuture step is thedevelopmento f a testingprotocol for equipment.
This will enablethe customerto comparethe voltagetoleranceof different devices.For
single-phaseequipmentit is probably sufficient to test fordifferent magnitudeand
duration. Possibleexceptionsare contactors(affected by point-on-wave)and equipment with controlled rectifiers (affectedby phase-anglejump).
Testingof three-phaseequipmentwill be much more complicated:even for noncontrolledrectifiers, thecharacteristicphase-anglejump affects thedc-busvoltage. For
de drives the threephasesare nolongerequivalentso that the numberof testsrequired
increasesby a factor of three. Three-phaseequipmentneeds to be tested for several
types of three-phaseunbalancedsags and for a range of magnitude,duration, and
phase-anglejump. Further analysisof monitoring results is needed too btain realistic
values for therangeof characteristicsto be included in the tests.
Another problem that needs to be solved is the
definition of the testcriterion.
Whethera certainreactionis acceptabledependsto a largeextendon the processdriven
by the drive. A possiblesolution is to give the variation in speedand torque as a
of the effectof the
function of the sagcharacteristics.This will enablean assessment
sag on theprocesswhen using acertaindrive.
460
Chapter8 • Summary and Conclusions
8.6.2 Improvement of Equipment
Improvementof equipmentoffers the only long-termsolutionto the power quality problem. As shown inChapter5, the effect of the sag can be mitigated for many
devices by installingadditionalcapacitance.There are somedrawbackswith this, the
first being the additional costs. A risk of additional capacitanceis that the inrush
of fuses
current on voltage recovery becomes more severe. This may lead to blowing
or to damageon power electroniccomponents.
Installing additionalcapacitancehas its limits. It is not feasible for making drives
toleratebalancedsags and it is in most cases not feasible at all for de drives. More
advanced rectifiers, inverters, and
control algorithmsare needed to achieve this. There
is not yet a drivetoward improvedequipmentbut somewhere in the (hopefully not too
remote) future this willhappen.Possible driving forces are standardized
a
testing protocol; equipmentimmunity requirementsaspartof the EMC standards;and, of course,
a demandfor improvedequipmentfrom the sideof the customer.
8.7 STOCHASTIC ASSESSMENT OF VOLTAGE SAGS
Chapter6 discusses the
stochasticand statisticaltreatmentof the compatibility between
equipmentand supply.Dataaboutthe performanceof the supply can be
obtainedfrom
power quality monitoringand fromstochasticpredictionstudies.Monitoring may give
a more accuratepicture of the kind of disturbancesto be expected, butstochastic
predictionwill give results in a muchshortertime.
Different methodsare discussed to present the results of
stochasticassessment
a
study (either powerquality monitoring or stochastic prediction). The so-called
"voltage-sagcoordinationchart" is shown to be a useful
instrumentfor the compatibility assessment. The results
of a numberof largepower-qualitysurveys are presented
and compared.One of the conclusionsis that a further treatmentis needed of the
propagationof voltage sags from the faultposition to lower voltage levels. The
above-mentionedeffect of load on the sag
characteristicswill play an important role
in such studies.
Two methodsare presentedfor the stochasticprediction of voltage sags: the
method of fault positionsand themethod of critical distances. Themethod of fault
positionsis suited for computerizedcalculationsin large meshed(transmission)systems. Themethodof critical distancesis suitable for simplehandcalculationsand for
calculationsin radial (distribution) systems.
8.7.1 Other Sag Characteristics
All the techniques discussed Chapter6
in
concentrateon magnitudeand duration
of voltage sags. To cover a wider range equipment,new
of
techniques have to be
developed for theother sagcharacteristics:phase-angle jump;three-phaseunbalance;
point-on-wave.Some suggestions are given in the text.problemwith
A
theseadditional
sagcharacteristicsis that the equipment'sreaction to them is not known, not even in a
qualitativeway.
8.7.2 Stochastic Prediction Techniques
Stochasticprediction techniques willcontinue to be further developed: both
detailedcomputerizedtechniques using the
methodof fault positions as well as simpli-
Section 8.7 • StochasticAssessment of Voltage Sags
461
tied methodslike the methodof critical distances. These
developmentswill reduce the
gap between powerquality and reliability evaluation.In fact, stochasticprediction of
voltage sags may be
consideredaspart of the reliability evaluationof the power supply.
Stochasticpredictionof voltage sags based on the
methodof fault positionsis likely to
become astandardpart of power-system analysis software, next to load flow,
shortcircuit currentcalculations,transientstability, etc.Calculatingthe expectednumberof
voltage sags may become common
as
as calculatingthe short-circuit current or the
normal operatingvoltage.
It is likely that the first commercially availableprogramswill only give results for
magnitudeand duration.But soon morecharacteristicsmay becomepart of the calculation results:three-phaseunbalancebeing the most essential one.
The methodof critical distances willcontinueto playarole. It may becomepart
of the stochasticprediction software, e.g., to estimate the extent of and
distance
between the fault positions. The
method of critical distances remains much more
powerful than the methodof fault positionsfor fast "back-of-the-envelope"calculations. An exampleof the latter is the simple expression derived in the last section of
Chapter6. This expression estimates the
number of sags due to faults in a meshed
transmissionsystem. Thedrawbackwith this expression isthat there is(not yet) any
theoretical basis for it. Further studies andcomparisonsmay teach usabout this
expression'saccuracy level.
8.7.3 Power Quality Survey.
Power quality surveys will alsocontinueto be performed.In fact quite a large
numberof them is going on at the moment, even
thoughthe statisticsare not actually
being collected in all cases. The
numberof publicationsof survey results will however
become less, as they are likely to show
"more of the same."This is anunfortunatebut
understandable
development. There is a small hope however
that the datawill be made
available for further research, e.g., resulting in statistics for
three-phaseunbalance,
phase-anglejump, point-on-wave, and any other possible sagcharacteristic.Such
data provide very useful results needed to assess voltage-tolerance
the
requirements
of equipment.
in
reports, is still very
The amount of survey results published, even internal
limited. There must be gigabytes
of very interestingmonitoringdatastoredat utilities
all over the world, waiting to be processed. Only ten years ago it was very difficult to get
power systemmeasurementsfor research purposes. Soon the
situation may be that
there is a surplusof data for which there are no directapplications.This should of
course not stop any utility from installing
monitors.The only way of getting an accurate picture of the quality of supply at any given location (i.e., not only sags and
interruptionsbut the wholespectrumof disturbances)is still by meansof measuring.
8.7.4 Monitoring or Prediction?
Both monitoring and stochasticpredictionare mentionedas a wayof obtaining
information aboutthe supplyperformance.Monitoring is still the methodmost commonly used: it gives not onlyinformation on voltage sags but also on
other voltage
events andvariations.Much of thisinformationis still very hardto obtainby stochastic
predictiontechniquesexist andobtainprediction.For voltage sags, however, powerful
ing accurateresults through monitoring may take many years.For individual sites
stochasticprediction is most suitable; toobtain the average powerquality over a
462
Chapter8 • Summary andConclusions
large area (e.g., a whole
country) monitoring is more suitable. Bycomparingmonitoring and prediction results thetrust in prediction techniquesis likely to grow, and the
comparisoncan be used tofurther develop thepredictiontechniques.
8.8 MITIGATION METHODS
In Chapter7 various methodsfor the mitigation of voltage sags andinterruptionsare
discussed. This is theultimate aim of any powerquality investigation: to solve the
problem. Thechapterstartswith an overviewof mitigation methods.Each methodis
briefly discussed: reducing the
numberof faults; reducing thefault-clearingtime; changing the power system;installing mitigation equipment; and improving equipment
immunity. For different types of events,different mitigation methodsare most suitable:
improving the equipmentfor short-durationevents, improving the system for longdurationevents.
Power system design and
mitigation equipmentare discussed in more detail. The
two improvementmethodsin power system design are
parallel operationof components and switching to an
alternativesupply. Until a few years ago, thelatter would
only besuitableas amitigation methodagainstlong interruptions.For sagmitigation
only certaintypesof parallel operationwere suitable. Theintroductionof the mediumvoltagestaticswitch makes it possible to
mitigatevoltage sags by very quickly switching
to a healthysupply. This may make radial
operationa more reliable supplyalternative
than parallel operation.
Several types ofmitigation equipmentare discussed in
C hapter7. Theemphasisis
on shunt and seriescontrollers based onpower-electronicvoltage-sourceconverters.
Throughtheseconvertersit is possible tocompensatefor the drop in system voltage or
even totemporarilytake over thesupplycompletely.For not too deep voltage sags it is
possible tocompensatethe drop in voltage magnitudeby injecting reactivepoweronly,
latter calls
but for a full compensationboth reactive and active power are needed. The
for a certain amountof energystorage.A numberof energystorageoptions are discussed in the last section Chapter7:
of
both classical ones(batteries,capacitors)as well
as some of the more recently
introducedones (superconductingcoils, high-speed flywheels,supercapacitors).
A comparisonof the variousoptionsshowsthat batteriesand
capacitorsremain themost-suitable options:capacitorsfor ridethroughtimes around
one second;batteriesfor ridethroughtimes of 10 minutesand longer.
The mostcommonly used method remains theinstallation of mitigating equipment at theutility-customerinterfaceor at theequipmentterminals.The uninterruptible power supply has become standardpieceof
a
equipmentin many installations.
This simply takes away lots
o f worries aboutthe quality of the supply. It is also in many
cases the only possible
solution: manycustomersdo not have the possibility toopt for
improved equipmentor for an improved power supply. A recentdevelopmentis the
installationof large mitigation equipmentat theutility-eustomerinterfaceprotectinga
whole plant againstsupply disturbances.This may be thecheapestshort-termsolution,
but it should not be used as an excusestop
to the installationand developmentof lesssensitiveequipment.
8.9 FINAL REMARKS
Powerquality is an areaof power engineeringthat did not exist only 10 years ago.
Powerquality and reliability have formanyyears beenpart of power system design and
operation,but they were rarelyconsideredas a separatearea. Being a new area, the
Section 8.9 • FinalRemarks
463
developments in power quality are fast and difficult to predict. A new device may be
inventedtomorrow solving all voltage sagproblems.
A more likely developmentis that sensitiveequipmentwill stay amongus for a
long time to come.Certainlyshort and longinterruptionswill remain a problem.The
power quality area willfurther expandand likely develop into two new areas: a nontechnical area covering"customer-utility interactions"and a technical onethat will
merge withelectromagneticcompatibility("equipment-system
interactions").An additional spin-off of the developmentsin power quality will be that power system education and research will be much more
measurementbasedthan in the past.
Regardlessof what the future will bring, powerquality in all its varieties will offer
utilities, equipmentmanufacturers,customers,and universities a very
interestingfield of
study, on which lots ofcooperationis needed and possible.
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Computation Conference, August 1996, Dresden, Germany, pp. 1028-1034.
[19] This figure was obtained from the Power Quality monitoring demonstration at the
Electrotek Concepts Website. http://www.electrotek.com.
(20] L. E. Conrad and M. H. J. Bollen, Voltage sag coordination for rel