01 Elements of Power System Analysis Stevenson 1

power
system
analysis
Fourth Edition
William D.
Stevenson, Jr.
'
ELEMENTS OF POWER SYSTEM ANALYSIS
McGraw-Hill Series in Electrical Engineering
Consulting Editor
Stephen W. Director, Camegie-Mellon University
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ELEMENTS
OF POWER
SYSTEM ANALYSIS
Fourth Edition
William D. Stevenson, Jr.

Professor of Electrical Engineering, Emeritus
North Carolina State University
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HDHD 99876543
Library of Congress Cataloging in Publication Data
Stevenson, William D.
Elements of power system analysis.
(McGraw-Hill series in electrical engineering.

Power and energy)
Includes index.
1. Electric power distribution. 2. Electric
power systems. I. Title. II. Series.
TK3001.S85 1982 621.319 81-3741
ISBN 0-07-061278-1 (Text) AACR2
ISBN 0-07-061279-X (Solutions manual)
CONTENTS
Preface xi
Chapter 1 General Background 1
l.i The Growth of Electric Power Systems 1

1.2 Energy Production 3
1.3 Transmission and Distribution 4
1.4 Load Studies 5
1.5 Economic Load Dispatch 6
1.6 Fault Calculations 6
1.7 System Protection 7
1.8 Stability Studies 8
1.9 The Power-System Engineer 9
1.10 Additional Reading 9
Chapter 2 Basic Concepts 10
2.1 Introduction 10
2.2 Single-Subscript Notation 11
2.3 Double-Subscript Notation 12
2.4 Power in Single-Phase AC Circuits 13
2.5 Complex Power 18
2.6 The Power Triangle 18
2.7 Direction of Power Flow 19
2.8 Voltage and Current in Balanced Three-Phase Circuits 22
2.9 Power in Balanced Three-Phase Circuits 28
2.10 Per-Unit Quantities 30
2.11 Changing the Base of Per-Unit Quantities 33
Problems 34
Chapter 3 Series Impedance of Transmission Lines 37
3.1 Types of Conductors 38

3.2 Resistance 40
3.3 Tabulated Resistance Values 42
3.4 Definition of Inductance 43
VI CONTENTS
3.5 Inductance of a Conductor Due to Internal Flux 44

3.6 Flux Linkages between Two Points External to
an Isolated Conductor 47
3.7 Inductance of a Single-Phase Two-Wire Line 48
3.8 Flux Linkages of One Conductor in a Group 50
3.9 Inductance of Composite-Conductor Lines 52
3.10 The Use of Tables 56
3.11 Inductance of Three-Phase Lines with Equilateral Spacing 58
3.12 Inductance of Three-Phase Lines with Unsymmetrical
Spacing 58
3.13 Bundled Conductors 61
3.14 Parallel-Circuit Three-Phase Lines 63
3.15 Summary of Inductance Calculations for Three-Phase Lines 64
Problems 65
Chapter 4 Capacitance of Transmission Lines 68
4.1 Electric Field of a Long Straight Conductor 69

4.2 The Potential Difference between Two Points
Due to a Charge 70
4.3 Capacitance of a Two-Wire Line 71
4.4 Capacitance of a Three-Phase Line with Equilateral Spacing 76
4.5 Capacitance of a Three-Phase Line with Unsymmetrical
Spacing 78
4.6 Effect of Earth on the Capacitance of Three-Phase
Transmission Lines 81
4.7 Bundled Conductors 83
4.8 Parallel-Circuit Three-Phase Lines 85
4.9 Summary 86
Problems 87
Chapter 5 Current and Voltage Relations on a

Transmission Line 88
5.1 Representation of Lines 90
5.2 The Short Transmission Line 91
5.3 The Medium-Length Line 93
5.4 The Long Transmission Line: Solution of the Differential
Equations 94
5.5 The Long Transmission Line: Interpretation of the
Equations 97
5.6 The Long Transmission Line: Hyperbolic Form of the
Equations 99
5.7 The Equivalent Circuit of a Long Line 104
5.8 Power Flow through a Transmission Line 107
5.9 Reactive Compensation of Transmission Lines 110
5.10 Transmission Line Transients 113
5.11 Transient Analysis: Traveling Waves 114
5.12 Transient Analysis: Reflections 117
5.13 Direct-Current Transmission 122
5.14 Summary 123
Problems 124
CONTENTS vii
Chapter 6 System Modeling 127
6.1 Construction of the Synchronous Machine 128

6.2 Armature Reaction in a Synchronous Machine 131
6.3 The Circuit Model of a Synchronous Machine 133
6.4 The Effect of Synchronous-Machine Excitation 137
6.5 The Ideal Transformer 138
6.6 The Equivalent Circuit of a Practical Transformer 142
6.7 The Autotransformer 145
6.8 Per-Unit Impedances in Single-Phase Transformer Circuits 146
6.9 Three-Phase Transformers 149
6.10 Per-Unit Impedances of Three-Winding Transformers 153
6.11 The One-Line Diagram 155
6.12 Impedance and Reactance Diagrams 157
6.13 The Advantages of Per-Unit Computations 161
6.14 Summary 162
Problems 163
Chapter 7 Network Calculations 166
7.1 Equivalence of Sources 166

7.2 Node Equations 168
7.3 Matrix Partitioning 172
7.4 Node Elimination by Matrix Algebra 174
7.5 The Bus Admittance and Impedance Matrices 178
7.6 Modification of an Existing Bus Impedance Matrix 182
7.7 Direct Determination of a Bus Impedance Matrix 187
7.8 Summary 190
Problems 191
Chapter 8 Load-Flow Solutions and Control 193
8.1 Data for Load-Flow Studies 193

8.2 The Gauss-Seidel Method 194
8.3 The Newton-Raphson Method 196
8.4 Digital-Computer Studies of Load Flow 203
8.5 Information Obtained in a Load-Flow Study 204
8.6 Numerical Results 206
8.7 Control of Power into a Network 206
8.8 The Specification of Bus Voltages 209
8.9 Capacitor Banks 212
8.10 Control by Transformers 214
8.11 Summary 223
Problems 224
Chapter 9 Economic Operation of Power Systems 227

9.1 Distribution of Load between Units within a Plant 228
9.2 Transmission Loss as a Function of Plant Generation 234
9.3 Distribution of Load between Plants 238
9.4 A Method of Computing Penalty Factors and Loss
Coefficients 242
9.5 Automatic Generation Control 243
Problems 246
viii CONTENTS
Chapter 10 Symmetrical Three-Phase Faults *..248
10.1 Transients in RL Series Circuits 248

10.2 Short-Circuit Currents and the Reactances of Synchronous
Machines 251
10.3 Internal Voltages of Loaded Machines under Transient
Conditions 255
10.4 The Bus Impedance Matrix in Fault Calculations 261
10.5 A Bus Impedance Matrix Equivalent Network 264
10.6 The Selection of Circuit Breakers 267
Problems 272
Chapter 11 Symmetrical Components 275

11.1 Synthesis of Unsymmetrical Phasors from Their Symmetrical
Components 275
11.2 Operators 277
11.3 The Symmetrical Components of Unsymmetrical Phasors 278
11.4 Phase Shift of Symmetrical Components in Y-A
Transformer Banks 281
11.5 Power in Terms of Symmetrical Components 288
11.6 Unsymmetrical Series Impedances 289
11.7 Sequence Impedances and Sequence Networks 291
11.8 Sequence Networks of Unloaded Generators 291
11.9 Sequence Impedances of Circuit Elements 294
11.10 Positive- and Negative-Sequence Networks 295
11.11 Zero-Sequence Networks 296
11.12 Conclusions 302
Problems 303
Chapter 12 Unsymmetrical Faults 305
12.1 Single Line-to-Ground Fault on an Unloaded Generator 306

12.2 Line-to-Line Fault on an Unloaded Generator 309
12.3 Double Line-to-Ground Fault on an Unloaded Generator 312
12.4 Unsymmetrical Faults on Power Systems 316
12.5 Single Line-to-Ground Fault on a Power System 318
12.6 Line-to-Line Fault on a Power System 319
12.7 Double Line-to-Ground Fault on a Power System 319
12.8 Interpretation of the Interconnected Sequence Networks 320
12.9 Analysis of Unsymmetrical Faults Using the Bus
Impedance Matrix 328
12.10 Faults through Impedance 331
12.11 Computer Calculations of Fault Currents 334
Problems 334
Chapter 13 System Protection 337

13.1 Attributes of Protection Systems 338
13.2 Zones of Protection 340
13.3 Transducers 342
13.4 Logical Design of Relays 345

CONTENTS IX
13.5 Primary and Backup Protection 352

13.6 Transmission Line Protection 354
13.7 Protection of Power Transformers 366
13.8 Relay Hardware 370
13.9 Summary 370
Problems 371
Chapter 14 Power System Stability 373

14.1 The Stability Problem 373
14.2 Rotor Dynamics and the Swing Equation 375
14.3 Further Considerations of the Swing Equation 379
14.4 The Power-Angle Equation 383
14.5 Synchronizing Power Coefficients 390
14.6 Equal-Area Criterion of Stability 393
14.7 Further Applications of the Equal-Area Criterion 399
14.8 Multimachine Stability Studies: Classical Representation 401
14.9 Step-by-Step Solution of the Swing Curve 409
14.10 Digital-Computer Programs for Transient Stability Studies 416
14.11 Factors Affecting Transient Stability 418
Problems 420
Appendix 423
Index 431
V*
* ■*
v
PREFACE
Each revision of this book has embodied many changes, this one more so than
usual. Over the years, however, the objective has remained the same. The
approach has always been to develop the thinking process of the student in
reaching a sound understanding of a broad range of topics in the power-system
area of electrical engineering. At the same time, another goal has been to promote
the student’s interest in learning more about the electric-power industry. The
objective is not great depth, but the presentation is thorough enough to give the
student the basic theory at a level that can be understood by the undergraduate.
With this beginning, the student will have the foundation to continue his educa¬
tion while at work in the field or in graduate school. Footnotes throughout the
book suggest sources of further information on most of the topics presented.
As in preparing previous revisions, I sent a questionnaire to a number of
faculty members across the country, and I appreciate greatly the prompt and in
many cases, the detailed responses to specific questions as well as the valuable
additional comments. The most popular suggestion was for a chapter on system
protection, and accordingly that subject has been added to the other four main
topics of load flow, economic dispatch, fault calculations, and system stability.
Perhaps surprisingly there were many requests to retain the material on trans¬
mission-line parameters. The per-unit system is first introduced in Chapter 2 and
developed gradually to allow the student to become accustomed to scaled, or
normalized, quantities. The need to review steady-state alternating-current circuits
still exists and so the chapter on basic concepts is not altered. Direct formulation
of the bus impedance matrix has been added. The Newton-Raphson method for
load-flow calculations has been developed more fully. Increased attention has
been given to developing the equivalent circuits of transformers and synchronous
machines to help the many students who study power systems before they have
had a course in machinery. Equations for transients on a lossless line have been
developed to lead into the discussion of surge arresters. Other topics discussed
briefly are direct-current transmission, reactive compensation, and underground
cables. The subject of automatic load dispatch has been expanded.
xii PREFACE
I am particularly fortunate to have had two principal contributors to this

edition. Arun G. Phadke, Consulting Engineer, American Electric Power Service
Corporation, is the author of the new chapter on system protection. John J.
Grainger, my colleague at N. C. State University, has completely rewritten the
chapter on power-system stability. To both of these people who have added so
much to this edition, I extend my sincere thanks. W. H. Kersting of New Mexico
State University contributed to the section on reactive compensation. He as well
as J. M. Feldman of Northeastern University, G. T. Heydt of Purdue University,
and H. V. Poe of Clemson University provided new problems for a number of
chapters. To all of them, I am extremely grateful.
I must thank three people who have always willingly given their time when
I needed advice about this revision. Homer E. Brown with his long experience in
power-system work as well as his secondary career in teaching has been a great
help to me. A. J. Goetze who has often taught the course based on this book
here at North Carolina State University has always been ready with suggestions
when asked, and finally John Grainger should be mentioned again because he has
provided me with much up-to-date information and counseling.
My thanks go also to the companies which have been so willing to provide
information, photographs and even the review of some of the new descriptive
material. These companies are: Carolina Power and Light Company, Duke
Power Company, General Electric Company, Leeds and Northup Company,
Utility Power Corporation, Virginia Electric and Power Company, and Westing-
house Electric Corporation.
As always, I have profited by the letters from users of past editions. I hope
this correspondence will continue.
William D. Stevenson, Jr.

CHAPTER
ONE
GENERAL BACKGROUND
Development of sources of energy to accomplish useful work is the key to the

industrial progress which is essential to the continual improvement in the stan¬
dard of living of people everywhere. To discover new sources of energy, to obtain
an essentially inexhaustible supply of energy for the future, to make energy
available wherever needed, and to convert energy from one form to another
and use it without creating the pollution which will destroy our biosphere are
among the greatest challenges facing this world today. The electric power system
is one of the tools for converting and transporting energy which is playing an
important role in meeting this challenge. The industry, by some standards, is the
largest in the world. Highly trained engineers are needed to develop and imple¬
ment the advances of science to solve the problems of the electric power industry
and to assure a very high degree of system reliability along with the utmost
regard for the protection of our ecology.
An electric power system consists of three principal divisions: the generating
stations, the transmission lines, and the distribution systems. Utilization of the
energy delivered to the customers of the operating companies is not within the
responsibilities of the utility companies and will not be considered in this book.
Transmission lines are the connecting links between the generating stations and
the distribution systems and lead to other power systems over interconnections.
A distribution system connects all the individual loads to the transmission lines
at substations which perform voltage transformation and switching functions.
The objective of this book is to present methods of analysis, and we shall
devote most of our attention to transmission lines and system operation. We
shall not be concerned with distribution systems or any aspects of power plants
other than the electrical characteristics of generators.
1.1 THE GROWTH OF ELECTRIC POWER SYSTEMS
The development of ac systems began in the United States in 1885, when George
Westinghouse bought the American patents covering the ac transmission system
developed by L. Gaulard and J. D. Gibbs of Paris. William Stanley, an early
l
2 ELEMENTS OF POWER SYSTEM ANALYSIS
associate of Westinghouse, tested transformers in his laboratory in Gj.eat

Barrington, Massachusetts. There, in the winter of 1885-1886, Stanley installed
the first experimental ac distribution system which supplied 150 lamps in the
town. The first ac transmission line in the United States was put into operation
in 1890 to carry electric energy generated by water power a distance of 13 mi
from Willamette Falls to Portland, Oregon.
The first transmission lines were single-phase, and the energy was usually
consumed for lighting only. Even the first motors were single-phase, but on May
16, 1888, Nikola Tesla presented a paper describing two-phase induction and
synchronous motors. The advantages of polyphase motors were apparent im¬
mediately, and a two-phase ac distribution system was demonstrated to the
public at the Columbian Exposition in Chicago in 1893. Thereafter, the trans¬
mission of electric energy by alternating current, especially three-phase alterna¬
ting current, gradually replaced dc systems. In January 1894, there were five
polyphase generating plants in the United States, of which one was two-phase
and the others three-phase. Transmission of electric energy in the United States
is almost entirely by means of alternating current. One reason for the early
acceptance of ac systems was the transformer, which makes possible the trans¬
mission of electric energy at a voltage higher than the voltage of generation or
utilization with the advantage of greater transmission capability.
In a dc transmission system ac generators supply the dc line through a
transformer and electronic rectifier. An electronic inverter changes the direct
current to alternating current at the end of the line so that the voltage can be
reduced by a transformer. By providing both rectification and inversion at each
end of the line power can be transferred in either direction. Economic studies
have shown that dc overhead transmission is not economical in the United
States for distances of less than 350 mi. In Europe, where transmission lines are
generally much longer than in the United States, dc transmission lines are in
operation in several locations for both underground and overhead installations.
In California large amounts of hydro power are transferred from the Pacific
Northwest to the southern part of California over 500-kV ac lines along the
coast and farther inland through Nevada by direct current at.800 kV line to line.
Statistics reported from 1920 until early in the 1970-1980 decade showed an
almost constant rate of increase of both installed generating capacity and annual
energy production which amounted quite closely to doubling each of these
values every 10 years. Growth then became more erratic and unpredictable, but
in general somewhat slower.
In the early days of ac power transmission in the United States, the opera¬
ting voltage increased rapidly. In 1890 the Willamette-Portland line was operated
at 3300 V. In 1907 a line was operating at 100 kV. Voltages rose to 150 kV in
1913, 220 kV in 1923, 244 kV in 1926, and 287 kV on the line from Hoover Dam
to Los Angeles, which began service in 1936. In 1953 came the first 345-kV line.
The first 500-kV line was operating in 1965. Four years later in 1969 the first
765-kV line was placed in operation.
Until 1917, electric systems were usually operated as individual units be¬
cause they started as isolated systems and spread out only gradually to cover the
GENERAL BACKGROUND 3
whole country. The demand for large blocks of power and increased reliability
suggested the interconnection of neighboring systems. Interconnection is advan¬
tageous economically because fewer machines are required as a reserve for oper¬
ation at peak loads (reserve capacity) and fewer machines running without load
are required to take care of sudden, unexpected jumps in load (spinning reserve).
The reduction in machines is possible because one company can usually call on
neighboring companies for additional power. Interconnection also allows a com¬
pany to take advantage of the most economical sources of power, and a com¬
pany may find it cheaper to buy some power than to use only its own generation
during some periods. Interconnection has increased to the point where power is
exchanged between the systems of different companies as a matter of routine.
The continued service of systems depending on water power for a large part of
their generation is possible in times of unusual and extreme water shortage only
because of the power obtained from other systems through interconnections.
Interconnection of systems brought many new problems, most of which have
been solved satisfactorily. Interconnection increases the amount of current which
flows when a short circuit occurs on a system and requires the installation of
breakers able to interrupt a larger current. The disturbance caused by a short
circuit on one system may spread to interconnected systems unless proper relays
and circuit breakers are provided at the point of interconnection. Not only must
the interconnected systems have the same nominal frequency, but also the syn¬
chronous generators of one system must remain in step with the synchronous
generators of all the interconnected systems.
Planning the operation, improvement, and expansion of a power system
requires load studies, fault calculations, the design of means of protecting the
system against lightning and switching surges and against short circuits, and
studies of the stability of the system. An important problem in efficient system
operation is that of determining how the total generation required at any time
should be distributed among the various plants and among the units within each
plant. In this chapter we shall consider the general nature of these types of problems
after a brief discussion of energy production and of transmission and distribution.
We shall see the great contributions made by computers to the planning and
operation of power systems.
1.2 ENERGY PRODUCTION
Most of the electric power in the United States is generated in steam-turbine

plants. Water power accounts for less than 20% of the total and that percentage
will drop because most of the available sources of water power have been
developed. Gas turbines are used to a minor extent for short periods when a
system is carrying peak load.
Coal is the most widely used fuel for the steam plants. Nuclear plants fueled
by uranium account for a continually increasing share of the load, but their
construction is slow and uncertain because of the difficulty of raising capital to
meet the sharply rising costs of construction, constantly increasing safety require-
merits which cause redesign, public opposition to the operation of the plants,
and delays in licensing.
Many plants converted to oil between 1970 and 1972, but in the face of the
continuous escalation in the price of oil and the necessity of reducing depen¬
dence on foreign oil reconversion from oil to coal has taken place wherever
possible.
The supply of uranium is limited, but the fast breeder reactors now pro¬
hibited in the United States, have greatly extended the total energy available
from uranium in Europe. Nuclear fusion is the great hope for the future, but a
controllable fusion process on a commercial scale is not expected to become
feasible until well after the year 2000, if ever. That year, however, is now the
target date for demonstrating the first pilot model of a controlled fusion reactor.
As this comes to pass, electric power systems must continue to grow and take
over the direct fuel applications. For instance, the electric car will probably be
widely used in order to reserve the fossil fuels (including petroleum and gas
synthesized from coal) for aircraft and long-distance trucking.
There is some use of geothermal energy in the form of live steam issuing
from the ground in the United States and foreign countries. Solar energy, now
chiefly in the form of direct heating of water for residential use, should even¬
tually become practical through research on photovoltaic cells which convert
sunlight to electricity directly. Great progress has been made in increasing the
efficiency and reducing the cost of these cells, but the distance still to go is
extremely large. Windmills driving generators are operating in a number of
locations to provide small amounts of power to power systems. Efforts to extract
power from the changing tides and from waves are under way. An indirect form
of solar energy is alcohol grown from grain and mixed with gasoline to make an
acceptable fuel for automobiles. Synthetic gas from garbage and sewage is
another indirect form of solar energy.
Finally, in producing energy by any means, protection of our environment is
extremely important. Atmospheric pollution is all too apparent to residents of
industrialized countries. Thermal pollution is less obvious, but cooling water for
nuclear reactors is very important and adds greatly to construction costs. Too
great a rise in the temperature of rivers is harmful to fish, and artificial lakes for
cooling water often use up too much productive land. Here cooling towers,
though expensive, seem to be the answer for cooling at nuclear plants.
1.3 TRANSMISSION AND DISTRIBUTION
The voltage of large generators usually is in the range of 13.8 kV to 24 kV. Large
modern generators, however, are built for voltages ranging from 18 to 24 kV. No
standard for generator voltages has been adopted.
Generator voltage is stepped up to transmission levels in the range of 115 to
765 kV. The standard high voltages (HV) are 115, 138, and 230 kV. Extra-high
voltages (EHV) are 345, 500, and 765 kV. Research is being conducted on lines
in the ultra-high-voltage (UHV) levels of 1000 to 1500 kV. The advantage of
higher levels of transmission-line voltage is apparent when consideration is given
to the transmission capability in megavoltamperes (MVA) of a line. Roughly, the

capability of lines of the same length varies at a rate somewhat greater than the
square of the voltage. No definite capability can be specified for a line of any
given voltage, however, because capability is dependent on the thermal limits of
the conductor, allowable voltage drop, reliability, and requirements for main¬
taining synchronism between the machines of the system, which is known as
stability. Most of these factors are dependent on line length.
Undergound transmission cables for a particular voltage seem to be
developed about 10 years after operation is initiated at that voltage on open-wire
lines. Underground transmission is negligible in terms of mileage but is increas¬
ing significantly. It is mostly confined to heavily populated urban areas, or used
under wide bodies of water.
The first step-down of voltage from transmission levels is at the bulk-power
substation, where the reduction is to a range of 34.5 to 138 kV, depending, of
course, upon transmission-line voltage. Some industrial customers may be
supplied at these voltage levels. The next step-down in voltage is at the distribu¬
tion substation, where the voltage on lines leaving the substation ranges from 4
to 34.5 kV and is commonly between 11 and 15 kV. This is the primary distribu¬
tion system. A very popular voltage at this level is 12,470 V line to line, which
means 7200 V from line to ground, or neutral. This voltage is usually described
as 12,470Y/7200 V. A lower primary-system voltage which is less widely used is
4160Y/2400 V. Most industrial loads are fed from the primary system, which
also supplies the distribution transformers providing secondary voltages over
single-phase three-wire circuits for residential use. Here the voltage is 240 V
between two wires and 120 V between each of these and the third wire, which is
grounded. Other secondary circuits are three-phase four-wire systems rated
208Y/120 V, or 480Y/277 V.
1.4 LOAD STUDIES
A load study is the determination of the voltage, current, power, and power
factor or reactive power at various points in an electric network under existing
or contemplated conditions of normal operation. Load studies are essential in
planning the future development of the system because satisfactory operation of
the system depends on knowing the effects of interconnections with other power
systems, of new loads, new generating stations, and new transmission lines before
they are installed.
Before the development of large digital computers load-flow studies were
made on ac calculating boards, which provided small-scale single-phase replicas
of actual systems by interconnecting circuit elements and voltage sources. Setting
up the connections, making adjustments, and reading the data was tedious
and time-consuming. Digital computers now provide the solutions of load-flow
studies on complex systems. For instance, the computer program may handle
more than 1500 buses, 2500 lines, 500 transformers with tap changing under load,
and 25 phase-shifting transformers. Complete results are printed quickly and
economically.
System planners are interested in studying a power system as it will exist 10

or 20 years in the future. More than 10 years elapse between initiating the plans
for a new nuclear plant and bringing it on the line. A power company must
know far in advance the problems associated with the location of the plant and
the best arrangement of lines to transmit the power to load centers which do not
exist when the planning must be done.
We shall see in Chap. 8 how load-flow studies are made on the computer.
Figure 8.2 shows the computer printout of the load flow of a small system which
we shall be studying.
1.5 ECONOMIC LOAD DISPATCH
The power industry may seem to lack competition. This idea arises because each
power company operates in a geographic area not served by other companies.
Competition is present, however, in attracting new industries to an area. Favor¬
able electric rates are a compelling factor in the location of an industry, although
this factor is much less important in times when costs are rising rapidly and rates
charged for power are uncertain than in periods of stable economic conditions.
Regulation of rates by state utility commissions, however, places constant pres¬
sure on companies to achieve maximum economy and earn a reasonable profit
in the face of advancing costs of production.
Economic dispatch is the name given to the process of apportioning the total
load on a system between the various generating plants to achieve the greatest
economy of operation. We shall see that all the plants on a system are controlled
continuously by a computer as load changes occur so that generation is al¬
located for the most economical operation.
1.6 FAULT CALCULATIONS
A fault in a circuit is any failure which interferes with the normal flow of current.
Most faults on transmission lines of 115 kV and higher are caused by lightning,
which results in the flashover of insulators. The high voltage between a conduc¬
tor and the grounded supporting tower causes ionization, which provides a path
to ground for the charge induced by the lightning stroke. Once the ionized path
to ground is established, the resultant low impedance to ground allows the flow
of power current from the conductor to ground and through the ground to the
grounded neutral of a transformer or generator, thus completing the circuit.
Line-to-line faults not involving ground are less common. Opening circuit
breakers to isolate the faulted portion of the line from the rest of the system
interrupts the flow of current in the ionized path and allows deionization to take
place. After an interval of about 20 cycles to allow deionization, breakers can
usually be reclosed without reestablishing the arc. Experience in the operation of
transmission lines has shown that ultra-high-speed reclosing breakers suc¬
cessfully reclose after most faults. Of those cases where reclosure is not success-
ful, an appreciable number are caused by permanent faults where reclosure

would be impossible regardless of the interval between opening and reclosing.
Permanent faults are caused by lines being on the ground, by insulator strings
breaking because of ice loads, by permanent damage to towers, and by lightning-
arrester failures. Experience has shown that between 70 and 80% of
transmission-line faults are single line-to-ground faults, which arise from the
flashover of only one line to the tower and ground. The smallest number of
faults, roughly 5%, involve all three phases and are called three-phase faults.
Other types of transmission-line faults are line-to-line faults, which do not in¬
volve ground, and double line-to-ground faults. All the above faults except the
three-phase type are unsymmetrical and cause an imbalance between the phases.
The current which flows in different parts of a power system immediately
after the occurrence of a fault differs from that flowing a few cycles later just
before circuit breakers are called upon to open the line on both sides of the fault,
and both these currents differ widely from the current which would flow under
steady-state conditions if the fault were not isolated from the rest of the system
by the operation of circuit breakers. Two of the factors upon which the proper
selection of circuit breakers depends are the current flowing immediately after
the fault occurs and the current which the breaker must interrupt. Fault calcula¬
tions consist in determining these currents for various types of faults at various
locations in the system. The data obtained from fault calculations also serve to
determine the settings of relays which control the circuit breakers.
Analysis by symmetrical components is a powerful tool which we shall study
later and which makes the calculation of unsymmetrical faults almost as easy as
the calculation of three-phase faults. Again it is the digital computer which is
invaluable in making fault calculations. We shall examine the fundamental oper¬
ations called for by the computer programs.
1.7 SYSTEM PROTECTION
Faults can be very destructive to power systems. A great deal of study, develop¬
ment of devices, and design of protection schemes have resulted in continual
improvement in the prevention of damage to transmission lines and equipment
and interruptions in generation following the occurrence of a fault.
We shall be discussing the problem of transients on a transmission line for a
very simplified case. This study will lead us to a discussion of how surge arresters
protect apparatus such as transformers at plant buses and substations against
the very high voltage surges caused by lightning and, in the case of EHV and
UHV lines, by switching.
Faults caused by surges are usually of such short duration that any circuit
breakers which may open will reclose automatically after a few cycles to restore
normal operation. If arresters are not involved or faults are permanent the
faulted sections of the system must be isolated to maintain normal operation of
the rest of the system.
Operation of circuit breakers is controlled by relays which sense the faulL In

the application of relays zones of protection are specified to define the parts of
the system for which various relays are responsible. One relay will also back up
another relay in an adjacent zone or zones where the fault occurs in case the
relay in the adjacent zone fails to respond. In Chap. 13 we shall discuss the
characteristics of the basic types of relays and look at some numerical examples
of relay applications and coordination.
1.8 STABILITY STUDIES
The current which flows in an ac generator or synchronous motor depends on

the magnitude of its generated (or internal) voltage, on the phase angle of its
internal voltage with respect to the phase angle of the internal voltage of every
other machine in the system, and on the characteristics of the network and loads.
For example, two ac generators operating in parallel but without any external
circuit connections other than the paralleling circuit will carry no current if their
internal voltages are equal in magnitude and in phase. If their internal voltages
are equal in magnitude but different in phase, the voltage of one subtracted from
the voltage of the other will not be zero, and a current will flow, as determined
by the difference in voltages and the impedance of the circuit. One machine will
supply power to the other, which will run as a motor rather than a generator.
The phase angles of the internal voltages depend upon the relative positions
of the rotors of the machines. If synchronism were not maintained between the
generators of a power system, the phase angles of their internal voltages would
be changing constantly with respect to each other and satisfactory operation
would be impossible.
The phase angles of the internal voltages of synchronous machines remain
constant only as long as the speeds of the various machines remain constant at
the speed which corresponds to the frequency of the reference phasor. When the
load on any one generator or on the system as a whole changes, the current in
the generator or throughout the system changes. If the change in current does
not result in a change in magnitude of the internal voltages of the machines, the
phase angles of the internal voltages must change. Thus, momentary changes in
speed are necessary to obtain adjustment of the phase angles of the voltages with
respect to each other, since the phase angles are determined by the relative
positions of the rotors. When the machines have adjusted themselves to the new
phase angles, or when some disturbance causing a momentary change in speed
has been removed, the machines must operate again at synchronous speed. If
any machine does not remain in synchronism with the rest of the system, large
circulating currents result; in a properly designed system, the operation of relays
and circuit breakers removes the machine from the system. The problem of
stability is the problem of maintaining the synchronous operation of the genera¬
tors and motors of the system.
Stability studies are classified according to whether they involve steady-state
or transient conditions. There is a definite limit to the amount of power an ac
generator is capable of delivering and to the load which a synchronous motor
can carry. Instability results from attempting to increase the mechanical input to
a generator or the mechanical load on a motor beyond this definite amount of
power, called the stability limit. A limiting value of power is reached even if the
change is made gradually. Disturbances on a system, caused by suddenly applied
loads, by the occurrence of faults, by the loss of excitation in the field of a
generator, and by switching, may cause loss of synchronism even when the
change in the system caused by the disturbance would not exceed the stability
limit if the change were made gradually. The limiting value of power is called the
transient stability limit or the steady-state stability limit, according to whether
the point of instability is reached by a sudden or a gradual change in conditions
of the system.
Fortunately, engineers have found methods of improving stability and of
predicting the limits of stable operation under both steady-state and transient
conditions. The stability studies we shall investigate for a two-machine system
are less complex than studies of multimachine systems, but many of the methods
of improving stability can be seen by the analysis of a two-machine system.
Digital computers are used to advantage in predicting the stability limits of a
complex system.
1.9 THE POWER-SYSTEM ENGINEER

This chapter has attempted to sketch some of the history of the basic develop¬
ments of electric power systems and to describe some of the analytic studies
important in planning the operation, improvement, and expansion of a modern
power system. The power-system engineer should know the methods of making
load studies, fault analyses, and stability studies and the principles of economic
dispatch because such studies affect the design and operation of the system and
the selection of apparatus for its control. Before we can consider these problems
in more detail, we must study some fundamental concepts relating to power
systems in order to understand how these fundamental concepts affect the larger
problems.
1.10 ADDITIONAL READING

Footnotes throughout the book provide sources of further information about
many of the topics which we will be discussing. The reader is also referred to
the books listed below which treat most of the same subjects as this text
although some include other topics or the same ones in greater depth.
Elgerd, O. I., “Electric Energy Systems Theory: An Introduction,” 2d ed.,
McGraw-Hill Book Company, New York, 1982.
Gross, C. A., “Power System Analysis,” John Wiley & Sons, New York, 1979.
Neuenswander, J. R., “Modern Power Systems,” Intext Educational Publishers,
New York, 1971.
Weedy, B. M., “Electric Power Systems,” 3d ed., John Wiley & Sons Ltd.,
London, 1979.
CHAPTER
TWO
BASIC CONCEPTS
The power-system engineer is just as concerned with the normal operation of the
system as he is with the abnormal conditions which may occur. Therefore, he
must be very familiar with steady-state ac circuits, particularly three-phase cir¬
cuits. It is the purpose of this chapter to review a few of the fundamental ideas of
such circuits, establish the notation which will be used throughout the book, and
introduce the expression of values of voltage, current, impedance, and power in
per unit.
2.1 INTRODUCTION
The waveform of voltage at the buses of a power system can be assumed to be

purely sinusoidal and of constant frequency. In developing most of the theory in
this book we shall be concerned with the phasor representations of sinusoidal
voltages and currents and shall use the capital letters V and I to indicate these
phasors (with appropriate subscripts where necessary). Vertical bars enclosing V
and I, that is, | V | and | /1, will designate the magnitude of the phasors. Lower¬
case letters will indicate instantaneous values. Where a generated voltage (elec¬
tromotive force) is specified, the letter E rather than V will be used for voltage to
emphasize the fact that an emf rather than a general potential difference between
two points is being considered.
If a voltage and a current are expressed as functions of time, such as
v — 141.4 cos (cor + 30°)
and
i = 7.07 cos cot
10
BASIC CONCEPTS 11
their maximum values are obviously Fmax = 141.4 V and 7max = 7.07 A, respec¬
tively. Vertical bars are not needed when the subscript max with V and / is used
to indicate maximum value. The term magnitude refers to root-mean-square or
(rms) values, which equal the maximum values divided by yj2. Thus, for the
above expressions for v and i,
| V\ = 100 V and |/| = 5 A
These are the values read by the ordinary types of voltmeters and ammeters.
Another name for the rms value is the effective value. The average power ex¬
pended in a resistor is \I\2 R-
To express these quantities as phasors a reference must be chosen. If the
current is the reference phasor
/ = 5/p = 5+j0 A
the voltage which leads the reference phasor by 30° is
V = 100/30° = 86.6 + j50 V
Of course, we might not choose as the reference phasor either the voltage or
current whose instantaneous expressions are v and i, in which case their phasor
expressions would involve other angles.
In circuit diagrams it is often most convenient to use polarity marks in the
form of plus and minus signs to indicate the terminal assumed positive when
specifying voltage. An arrow on the diagram specifies the direction assumed
positive for the flow of current. In the single-phase equivalent of a three-phase
circuit single-subscript notation is usually sufficient, but double-subscript nota¬
tion is usually simpler when dealing with all three phases.
2.2 SINGLE-SUBSCRIPT NOTATION
Figure 2.1 shows an ac circuit with an emf represented by a circle. The emf is Eg,
and the voltage between nodes a and o is identified as Vt. The current in the
circuit is IL and the voltage across ZL is VL. To specify these voltages as phasors,
however, the + and - markings, called polarity marks, on the diagram and an
arrow for current direction are necessary.
In an ac circuit the terminal marked + is positive with respect to the
terminal marked - for half a cycle of voltage and is negative with respect to the
\J
V#.
\p2_
Q.
Figure 2.1 An ac circuit with emf Eg and load impe¬

dance ZL.
other terminal during the next half cycle. We mark the terminals to enable «us to
say that the voltage between the terminals is positive at any instant when the
terminal marked plus is actually at a higher potential than the terminal marked
minus. For instance, in Fig. 2.1 the instantaneous voltage v, is positive when the
terminal marked plus is actually at a higher potential than the terminal marked
with a negative sign. During the next half cycle the positively marked terminal is
actually negative, and vt is negative. Some authors use an arrow but must specify
whether the arrow points toward the terminal which would be labeled plus or
toward the terminal which would be labeled minus in the convention described
above.
The current arrow performs a similar function. The subscript, in this case L,
is not necessary unless other currents are present. Obviously the actual direction
of current flow in an ac circuit reverses each half cycle. The arrow points in the
direction which is to be called positive for current. When the current is actually
flowing in the direction opposite that of the arrow, the current is negative. The
phasor current is
(2.1)
and
X=£ ,-hZ, (2.2)

Since certain nodes in the circuit have been assigned letters, the voltages
may be designated by the single-letter subscripts identifying the node whose
voltages are expressed with respect to a reference node. In Fig. 2.1 the instanta¬
neous voltage va and the phasor voltage Va express the voltage of node a with
respect to the reference node o, and va is positive when a is at a higher potential
than o. Thus
v„ = v, vb = vL
V = V, K = V,
2.3 DOUBLE-SUBSCRIPT NOTATION
The use of polarity marks for voltages and direction arrows for currents can be
avoided by double-subscript notation. Understanding of three-phase circuits is
considerably clarified by adopting a system of double subscripts. The convention
to be followed is quite simple.
In denoting a current the order of the subscripts assigned to the symbol for
current defines the direction of flow of current when the current is considered to
be positive. In Fig. 2.1 the arrow pointing from a to b defines the positive
direction for the current IL associated with the arrow. The instantaneous current
iL is positive when the current is actually in the direction from a to b, and in
double-subscript notation this current is iab. The current iab is equal to - iba.
BASIC CONCEPTS 13
In double-subscript notation the letter subscripts on a voltage indicate the

nodes of the circuit between which the voltage exists. We shall follow the con¬
vention which says that the first subscript denotes the voltage of that node with
respect to the node identified by the second subscript. This means that the
instantaneous voltage vab across ZA of the circuit of Fig. 2.1 is the voltage of
node a with respect to node b and that vab is positive during that half cycle when
a is at a higher potential than b. The corresponding phasor voltage is Vab, and
Vab = IabZA (2.3)
where ZA is the complex impedance through which Iab flows between nodes a
and b, which may also be called Zab.
Reversing the order of the subscripts of either a current or voltage gives a
current or voltage 180° out of phase with the original; that is,
v.„ = vb„imL = - vbo

The relation of single- and double-subscript notation for the circuit of
Fig. 2.1 is summarized as follows:
V
vt = V
va = V
v ao V,
rL = K
rb = K
v bo
IL Iab
In writing Kirchhoff’s voltage law the order of the subscripts is the order of
tracing a closed path around the circuit. For Fig. 2.1,
K. + K, + K, = 0 (2.4)
Nodes n and o are the same in this circuit, and n has been introduced to identify
the path more precisely. Replacing Voa by - Vao and noting that Vab = IabZA
yields
— K o + KbZA + Kn — 0 (2-5)
and so
U= (2.6)
2.4 POWER IN SINGLE-PHASE AC CIRCUITS
Although the fundamental theory of the transmission of energy describes the

travel of energy in terms of the interaction of electric and magnetic fields, the
power-system engineer is almost always more concerned with describing the rate
of change of energy with respect to time (which is the definition of power) in
terms of voltage and current. The unit of power is a watt. The power in watts
being absorbed by a load at any instant is the product of the instantaneous
voltage drop across the load in volts and the instantaneous current into the load
in amperes. If the terminals of the load are designated a and n, and if the voltage
Figure 2.2 Current, voltage, and power plotted versus time.
and current are expressed by
Van = Knax cos cot and ian = 7max cos (cat - 9)
the instantaneous power is
P = cos lot cos (tot - 0) (2.7)
The angle 6 in these equations is positive for current lagging the voltage and
negative for leading current. A positive value of p expresses the rate at which
energy is being absorbed by the part of the system between the points a and n.
The instantaneous power is obviously positive when both van and ian are positive
but will become negative when van and ian are opposite in sign. Figure 2.2
illustrates this point. Positive power calculated as van ian results when current is
flowing in the direction of a voltage drop and is the rate of transfer of energy to
the load. Conversely, negative power calculated as van ian results when current is
flowing in the direction of a voltage rise and means energy is being transferred
from the load into the system to which the load is connected. If van and ian are in
phase, as they are in a purely resistive load, the instantaneous power will never
become negative. If the current and voltage are out of phase by 90°, as in a
purely inductive or purely capacitive ideal circuit element, the instantaneous
power will have equal positive and negative half cycles and its average value will
be zero.
By using trigonometric identities the expression of Eq. (2.7) is reduced to
where Fmax/max/2 may be replaced by the product of the rms voltage and current
I Km | • \lan\ Ot\V\ ' |/|.
Another way of looking at the expression for instantaneous power is to
consider the component of the current in phase with van and the component 90°
out of phase with van. Figure 2.3a shows a parallel circuit for which Fig. 23b is
the phasor diagram. The component of ian in phase with van is iR, and from
Fig. 23b, 17*1 = \Ian \ cos 9. If the maximum value of ian is 7max, the maximum
value of iR is 7max cos 9. The instantaneous current iR must be in phase with van.
BASIC CONCEPTS 15
ao
Figure 2.3 Parallel RL circuit and the no

corresponding phasor diagram.
For van = Fmax cos cot,
iR = /max cos d cos col (2.9)

max iR
Similarly the component of ian lagging van by 90° is ix, whose maximum value is
/max sin 9. Since ix must lag van by 90°,
ix = /max sin 9 sin cot (2.10)

max ix
Then
^ an Ir fnlax f n cos 9 cos2 cot
V I max
r max1
cos 0(1+ cos 2cot) (2.11)
which is the instantaneous power in the resistance and is the first term in
Eq. (2.8). Figure 2.4 shows vaniR plotted versus t.
Similarly,
vanh = Fmax/max sin 9 sin cot cos cot
V 1
=---sin 0 sin 2ojt (2.12)
which is the instantaneous power in the inductance and is the second term in
Eq. (2.8). Figure 2.5 shows van, ix, and their product plotted versus t.
Figure 2.4 Voltage, current in phase with the voltage, and the resulting power plotted versus time.
Figure 2.5 Voltage, current lagging the voltage by 90°, and the resulting power plotted versus time.
Examination of Eq. (2.8) shows that the first term, the term which contains
cos 9, is always positive and has an average value of
P = Fmax2/max cos 9 (2.13)
or, when rms values of voltage and current are substituted,
P = | V \ • |/| cos 9 (2.14)
P is the quantity to which the word power refers when not modified by an
adjective identifying it otherwise. P, the average power, is also called the real
power. The fundamental unit for both instantaneous and average power is the
watt, but a watt is such a small unit in relation to power-system quantities that P
is usually measured in kilowatts or megawatts.
The cosine of the phase angle 9 between the voltage and the current is called
the power factor. An inductive circuit is said to have a lagging power factor, and
a capacitive circuit is said to have a leading power factor. In other words, the
terms lagging power factor and leading power factor indicate, respectively,
whether the current is lagging or leading the applied voltage.
The second term of Eq. (2.8), the term containing sin 9, is alternately posi¬
tive and negative and has an average value of zero. This component of the
instantaneous power p is called the instantaneous reactive power and expresses
the flow of energy alternately toward the load and away from the load. The
maximum value of this pulsating power, designated Q, is called reactive power or
reactive voltamperes and is very useful in describing the operation of a power
system, as will become increasingly evident in further discussion. The reactive
power is
Q = Kmax2/max sin 9 (2.15)
or
Q = |V|-|/| sin 6 (2.16)

BASIC CONCEPTS 17
The square root of the sum of the squares of P and Q is equal to the product
of | V | and | /1, for
yP2 + Q2 = ^(\V\• |/| cos 9)2 + (\V\- |/| sin 9f = |K| • |/| (2.17)
Of course P and Q have the same dimensional units, but it is usual to designate
the units for Q as vars (for voltamperes reactive). The more practical units for Q
are kilovars or megavars.
In a simple series circuit where Z is equal to P + jX, we can substitute
| / | • | Z | for | V | in Eqs. (2.14) and (2.16) to obtain
P=\I \2 ■ |Z| cos 9 (2.18)
Q=\i \2 • | Z| sin 9 (2.19)

Then recognizing that R = \ Z\ cos 9 and X = | Z | sin 9, we find
P = \I\2R and Q = \I\2X (2.20)

as expected.
Equations (2.14) and (2.16) provide another method of computing the power
factor since we see that Q/P = tan 9. The power factor is therefore
cos 9 = cos tan 1

Q
p
or from Eqs. (2.14) and (2.17)
cos 6 =
P2 + Q<
If the instantaneous power expressed by Eq. (2.8) is the power in a pre¬
dominantly capacitive circuit with the same impressed voltage, 9 would be nega¬
tive, making sin 9 and Q negative. If capacitive and inductive circuits are in
parallel, the instantaneous reactive power for the RL circuit would be 180° out of
phase with the instantaneous reactive power of the RC circuit. The net reactive
power is the difference between Q for the RL circuit and Q for the RC circuit. A
positive value is assigned to Q for an inductive load and a negative sign to Q for
a capacitive load.
Power-system engineers usually think of a capacitor as a generator of posi¬
tive reactive power rather than a load requiring negative reactive power. This
concept is very logical, for a capacitor drawing negative Q in parallel with an
inductive load reduces the Q which would otherwise have to be supplied by the
system to the inductive load. In other words, the capacitor supplies the Q
required by the inductive load. This is the same as considering a capacitor as a
device that delivers a lagging current rather than as a device which draws a
leading current, as shown in Fig. 2.6. An adjustable capacitor in parallel with an
inductive load, for instance, can be adjusted so that the leading current to the
capacitor is exactly equal in magnitude to the component of current in the
Figure 2.6 Capacitor considered (a) as a passive circuit

element drawing leading current and (b) as a generator I leads V by 90° I lags V by 90°
ru3^
supplying lagging current.
inductive load which is lagging the voltage by 90°. Thus, the resultant current is
in phase with the voltage. The inductive circuit still requires positive reactive
power, but the net reactive power is zero. It is for this reason that the power-
system engineer finds it convenient to consider the capacitor to be supplying this
reactive power to the inductive load. When the words positive and negative are
not used, positive reactive power is assumed.
2.5 COMPLEX POWER
If the phasor expressions for voltage and current are known, the calculation of
real and reactive power is accomplished conveniently in complex form. If the
voltage across and the current into a certain load or part of a circuit are ex¬
pressed by V — \ V\[cl_ and / = 17 the product of voltage times the conju¬
gate of the current is
VI* = V/a x 1/-P= \V\ ■ |/|/«- p (2.21)
This quantity, called the complex power, is usually designated by S. In rectangu¬

lar form
S — | V | • | / | cos (a — j8) + j | V | • 111 sin (a — ft) (2.22)
Since a — /?, the phase angle between voltage and current, is 6 in the previous
equations,
S = P +jQ (2.23)
Reactive power Q will be positive when the phase angle a — between voltage
and current is positive, that is, when a > /I, which means that current is lagging
the voltage. Conversely, Q will be negative for /? > a, which indicates current
leading the voltage. This agrees with the selection of a positive sign for the
reactive power of an inductive circuit and a negative sign for the reactive power
of a capacitive circuit. To obtain the proper sign for Q, it is necessary to calcu¬
late S as VI*, rather than V*I, which would reverse the sign for Q.
2.6 THE POWER TRIANGLE
Equation (2.23) suggests a graphical method of obtaining the overall P, Q, and

phase angle for several loads in parallel since cos 9 is P/1S \. A power triangle
can be drawn for an inductive load, as shown in Fig. 2.7. For several loads in
BASIC CONCEPTS 19
Figure 2.7 Power triangle for an inductive load. P
parallel, the total P will be the sum of the average powers of the individual loads,
which should be plotted along the horizontal axis for a graphical analysis. For
an inductive load, Q will be drawn vertically upward since it is positive. A
capacitive load will have negative reactive power, and Q will be vertically down¬
ward. Figure 2.8 illustrates the power triangle composed of Pl5 Qu and Sj for a
lagging load having a phase angle 9X combined with the power triangle
composed of P2, Q2, and S2, which is for a capacitive load with a negative 02.
These two loads in parallel result in the triangle having sides Px + P2, Qi + Q2
and hypotenuse SR. The phase angle between voltage and current supplied to
the combined loads is 0R.
2.7 DIRECTION OF POWER FLOW
The relation between P, Q, and bus voltage V, or generated voltage E, with

respect to the signs of P and Q is important when the flow of power in a system
is considered. The question involves the direction of flow of power, that is,
whether power is being generated or absorbed when a voltage and current are
specified.
The question of delivering power to a circuit or absorbing power from a
circuit is rather obvious for a dc system. Consider the current and voltage
relationship shown in Fig. 2.9, where dc current / is flowing through a battery. If
/ = 10 A and E = 100 V, the battery is being charged (absorbing energy) at the
rate of 1000 W. On the other hand, with the arrow still in the direction shown,
the current might be I = —10 A. Then the conventional direction of current is
p.
Figure 2.8 Power triangle for

Pi
combined loads. Note that Q2 is
negative. Pi+ P2 - PR
I E
Figure 2.9 A dc representation of charging a battery if E and / are
both positive or both negative.
opposite to the direction of the arrow, the battery is discharging (delivering

energy), and the product of E and / is —1000 W. By drawing Fig. 2.9 with /
flowing through the battery from the positive to the negative terminal, charging
of the battery seems to be indicated, but this is the case only if E and / are
positive so that the power calculated as the product of E and I is positive. With
this relationship between E and / the positive sign for power is obviously as¬
signed to charging the battery.
If the direction of the arrow for / in Fig. 2.9 had been reversed, discharging
of the battery would be indicated by a positive sign for I and for power. Thus the
circuit diagram determines whether a positive sign for power is associated with
charging or discharging the battery. This explanation seems unnecessary, but it
provides the background for interpreting the ac circuit relationships.
For an ac system Fig. 2.10 shows an ideal voltage source (constant magni¬
tude, constant frequency, zero impedance) with polarity marks which, as usual,
indicate the terminal which is positive during the half cycle of positive instanta¬
neous voltage. Of course, the positively marked terminal is actually the negative
terminal during the negative half cycle of the instantaneous voltage. Similarly the
arrow indicates the direction of current during the positive half cycle of current.
In Fig. 2.10a a generator is expected since the current is positive when
flowing away from the positively marked terminal. However, the positively
marked terminal may be negative when the current is flowing away from it. The
approach to understanding the problem is to resolve the phasor I into a com¬
ponent along the axis of the phasor E and a component 90° out of phase with E.
The product of | E | and the magnitude of the component of I along the E axis is
P. The product of | E | and the magnitude of the component of / which is 90°
out of phase with E is Q. If the component of I along the axis of E is in phase
with E, the power is generated power which is being delivered to the system, for
this component of current is always flowing away from the positively marked
terminal when that terminal is actually positive (and toward that terminal when
the terminal is negative). P, the real part of El*, is positive.
If the component of current along the axis of E is negative (180° out of phase
with £), power is being absorbed and the situation is that of a motor. P, the real
part of El*, would be negative.
The voltage and current relationship might be as shown in Fig. 2.10b, and a
motor would be expected. However, an average power absorbed would occur
only if the component of the phasor / along the axis of the phasor E was found
I I
Figure 2.10 An ac circuit represen¬
tation of an emf and current to
illustrate polarity marks.
O (a)
O
(6)
BASIC CONCEPTS 21
I ►
- I
--
Figure 2.11 Alternating emf applied (a) to a purely induc¬

tive element and (b) to a purely capacitive element. (a) (6)
Table 2.1
Circuit diagram Calculated from El*
If P is +, emf supplies power

If P is —, emf absorbs power
If Q is +, emf supplies reactive power (/ lags E)
Generator action assumed If Q is —, emf absorbs reactive power (/ leads E)
If P is + , emf absorbs power

If P is —, emf supplies power
If Q is +, emf absorbs reactive power (/ lags E)
Motor action assumed If Q is emf supplies reactive power (/ leads E)
to be in phase rather than 180° out of phase with E, so that this component of
current would be always in the direction of the drop in potential. In this case P,
the real part of El* would be positive. Negative P here would indicate generated
power.
To consider the sign of Q, Fig. 2.11 is helpful. In Fig. 2.11a positive reactive
power equal to \I\2X is supplied to the inductance since inductance draws
positive Q. Then / lags E by 90°, and Q, the imaginary part of El*, is positive. In
Fig. 2.11h negative Q must be supplied to the capacitance of the circuit, or the
source with emf E is receiving positive Q from the capacitor. 1 leads E by 90°.
If the direction of the arrow in Fig. 2.11a is reversed, / will lead E by 90° and
the imaginary part of El* would be negative. The inductance could be viewed as
supplying negative Q rather than absorbing positive Q. Table 2.1 summarizes
these relationships.
Example 2.1 Two ideal voltage sources designated as machines 1 and 2 are
connected as shown in Fig. 2.12. If Ex = 100/0° V, E2 = 100/30° V, and
Z = 0 + j5 fi, determine (a) whether each machine is generating or consum¬
ing power and the amount, (b) whether each machine is receiving or supply¬
ing reactive power and the amount, and (c) the P and Q absorbed by the
impedance.
Figure 2.12 Ideal voltage sources connected through

impedance Z.
Solution
- E2 100 + y'O - (86.6 + ;50)
1 ~ Z ~ 75
= l3A~J— = -10 -7I68 = 10.35/195°
ElI* = 100(-10 + 72.68) = -1000 + 7'268

E2I* = (86.6 + 7'50)( —10 + 72.68)
= -866 + j232 -y500 - 134 = -1000 -7268
\I\2X = 10.352 x 5 = 536 var

Machine 1 may be expected to be a generator because of the current
direction and polarity markings. Since P is negative and Q is positive, the
machine consumes energy at the rate of 1000 W and supplies reactive power
of 268 var. The machine is actually a motor.
Machine 2, expected to be a motor, has negative P and negative Q.
Therefore, this machine generates energy at the rate of 1000 W and supplies
reactive power of 268 var. The machine is actually a generator.
Note that the supplied reactive power of 268 + 268 is equal to 536 var,
which is required by the inductive reactance of 5 Q. Since the impedance is
purely reactive, no P is consumed by the impedance, and all the P generated
by machine 2 is transferred to machine 1.
2.8 VOLTAGE AND CURRENT

IN BALANCED THREE-PHASE CIRCUITS
Electric power systems are supplied by three-phase generators. Usually the gen¬
erators are supplying balanced three-phase loads, which means loads with identi¬
cal impedances in all three phases. Lighting loads and small motors are, of
course, single-phase, but distribution systems are designed so that overall the
phases are essentially balanced. Figure 2.13 shows a Y-connected generator with
neutral marked o supplying a balanced-Y load with neutral marked n. In discus¬
sing this circuit we shall assume that the impedances of the connections between
the terminals of the generator and the load, as well as the impedance of the
direct connection between o and n, are negligible.
The equivalent circuit of the three-phase generator consists of an emf in each
of the three phases, as indicated by circles on the diagram. Each emf is in series
BASIC CONCEPTS 23
Figure 2.13 Circuit diagram of a Y-connected generator connected to a balanced-Y load.
with a resistance and inductive reactance composing the impedance Zg. Points
a', b', and c' are fictitious since the generated emf cannot be separated from the
impedance of each phase. The terminals of the machine are the points a, b, and c.
Some attention will be given to this equivalent circuit in a later chapter. In the
generator the emfs Ea o, Ebo, Ec o are equal in magnitude and displaced from
each other 120° in phase. If the magnitude of each is 100 V with Ea.0 as reference,
Ea o = 100/0! V Ebo = 100/240° V Ec o = 100/l20° V

provided the phase sequence is abc, which means that Ea o leads Eb.0 by 120° and
Eb 0 in turn leads Ec 0 by 120°. The circuit diagram gives no indication of phase
sequence, but Fig. 2.14 shows these emfs with phase sequence abc.
At the generator terminals (and at the load in this case) the terminal vol¬
tages to neutral are
Figure 2.14 Phasor diagram of the emfs of the circuit shown in Fig. 2.13.
Since o and n are at the same potential, Vao, Vbo, and Vc0 are equal to Va^%, Vbn,
and Vcn, respectively, and the line currents (which are also the phase currents for
a Y connection) are
Eao Van
Kn =
Zg + ZR ~zR
Eb0
13
(2.25)
1
hn =
Zg + ZR
Ec'o Kn
Kn =
Zg + ZR ZR
Since Ea.0, Eb.0, and Ec 0 are equal in magnitude and 120° apart in phase and
the impedances seen by each of these emfs are identical, the currents will also be
equal in magnitude and displaced 120° from each other in phase. The same must
also be true of Van, Vbn, and Vcn. In this case we describe the voltages and
currents as balanced. Figure 2.15a shows three line currents of a balanced
system. In Fig. 2.15h the sum of these currents is shown to be a closed triangle. It
is obvious that their sum is zero. Therefore, /„ in the connection shown in
Fig. 2.13 between the neutrals of the generator and load must be zero. Then the
connection between n and o may have any impedance, or even be open, and n
and o will remain at the same potential.
If the load is not balanced, the sum of the currents will not be zero and a
current will flow between o and n. For the unbalanced condition, in the absence
of a connection of zero impedance, o and n will not be at the same potential.
The line-to-line voltages are K„, Vbc, and Vca. Tracing a path from a to b
through n in the circuit of Fig. 2.13 yields
Vab = K„ + Vnb = van - vbn (2.26)
Although Ea o and Van are not in phase, we could decide to use Van rather than
Ea.0 as reference in defining the voltages. Then Fig. 2.16a is the phasor diagram
of voltages to neutral, and Fig. 2.16b shows how Vab is found. The magnitude of
Kh is
\Vab\ =2\Van\ cos 30°
= v/3K.| (2.27)
Figure 2.15 Phasor diagram of currents in a balanced

three-phase load: (a) phasors drawn from a common
point; (b) addition of the phasors forming a closed triangle.
BASIC CONCEPTS 25
Figure 2.16 Voltages in a balanced three-

phase circuit: (a) voltages to neutral;
(b) relation between a line voltage and
voltages to neutral. (.b)
As a phasor, leads Van by 30°, and so
Kt = V5 K,M (2.28)
The other line-to-line voltages are found in a similar manner, and Fig. 2.17
shows all the line-to-line and line-to-neutral voltages. The fact that the magni¬
tude of line-to-line voltages of a balanced three-phase circuit is always equal to
x/3 times the magnitude of the line-to-neutral voltages is very important.
Figure 2.18 is another way of displaying the line-to-line and line-to-neutral
voltages. The line-to-line voltage phasors are drawn to form a closed triangle
oriented to agree with the chosen reference, in this case Van. The vertices of the
triangle are labeled so that each phasor begins and ends at the vertices corre¬
sponding to the order of the subscripts of that phasor voltage. Line-to-neutral
V
Y ca V,
ao
Figure 2.17 Phasor diagram of voltages in a balanced

three-phase circuit.
Figure 2.18 Alternative method of drawing the phasors of Fig. 2.17. c

Figure 2.19 Phasor diagram of voltages for Example 2.2. c
voltage phasors are drawn to the center of the triangle. Once this phasor dia¬
gram is understood, it will be found to be the simplest way to determine the
various voltages.
The order in which the vertices a, b, and c of the triangle follow each other
when the triangle is rotated counterclockwise about n indicates the phase
sequence. We shall see later an example of the importance of phase sequence
when we discuss symmetrical components as a means of analyzing unbalanced
faults on power systems.
A separate current diagram can be drawn to relate each current properly
with respect to its phase voltage.
Example 2.2 In a balanced three-phase circuit the voltage is 173.2/02 V.

Determine all the voltages and the currents in a Y-connected load having
ZL = 10/20° Q. Assume that the phase sequence is abc.
Solution The phasor diagram of voltages is drawn as shown in Fig. 2.19,

from which it is determined that
Kt, = 173.2/02 V Van = 100/-300 V

Vbc = 173.2/240° V Vbn = 100/210° V
Vca = 173.2/120° V Vcn = 100/90° V
Each current lags the voltage across its load impedance by 20°, and each
current magnitude is 10 A. Figure 2.20 is the phasor diagram of the currents
Ian = 10/-50° A Ibn = 10/1902 A Icn = 10/70! A
Balanced loads are often connected in A, as shown in Fig. 2.21. Here it is left to
the reader to show that the magnitude of a line current such as Ia is equal to
y/3 times the magnitude of a phase current such as Iab and that Ia lags Iab by
30° when the phase sequence is abc.
When solving balanced three-phase circuits it is never necessary to work with
the entire three-phase circuit diagram of Fig. 2.13. To solve the circuit a neutral
connection of zero impedance is assumed to be present and to carry the sum of the
three phase currents, which is zero, however, for balanced conditions. The circuit
is solved by applying Kirchhoff’s voltage law around a closed path which includes
one phase and neutral. Such a closed path is shown in Fig. 2.22. This circuit is
BASIC CONCEPTS 27
Figure 2.20 Phasor diagram of currents for

Example 2.2.
the single-phase equivalent of the circuit of Fig. 2.13. Calculations made for this
path are extended to the whole three-phase circuit by recalling that the currents in
the other two phases are equal in magnitude to the current of the phase calculated
and are displaced 120 and 240° in phase. It is immaterial whether the balanced
load, specified by its line-to-line voltage, total power, and power factor, is A- or
Y-connected since the A can always be replaced for purposes of calculaton by its
equivalent Y. The impedance of each phase of the equivalent Y will be one-third
the impedance of each phase of the A which it replaces.
Ia
Figure 2.21 Circuit diagram of A-connected three-phase

load.
Figure 2.22 One phase of the circuit of Fig. 2.13. n

127/- 30° A
__ V*
Figure 2.23 Circuit diagram with values for

Example 2.3.
Example 2.3 The terminal voltage of a Y-connected load consisting of three

equal impedances of 20/30° Q is 4.4 kV line to line. The impedance in each
of the three lines connecting the load to a bus at a substation is
ZL = 1.4/752 ft. Find the line-to-line voltage at the substation bus.
Solution The magnitude of the voltage to neutral at the load is

4400/^3 = 2540 V. If Van, the voltage across the load, is chosen as reference,
Vm = 2540/W V and l„ = = 127.0 ^30“ A
The line-to-neutral voltage at the substation is
Van + IanZL = 2540/02 + 127/-300 x 1.4/75°

= 2540/0° + 177.8/45°
= 2666 + j125.7 = 2670/2.70° V
and the magnitude of the voltage at the substation bus is
^3 x 2.67 = 4.62 kV
Figure 2.23 shows the circuit and quantities involved.
2.9 POWER IN BALANCED THREE-PHASE CIRCUITS
The total power delivered by a three-phase generator or absorbed by a three-

phase load is found simply by adding the power in each of the three phases. In a
balanced circuit this is the same as multiplying the power in any one phase by 3,
since the power is the same in all phases.
If the magnitude of the voltages to neutral Vp for a Y-connected load is
K= \vm\ = Hi. I = IK, I (2.29)
and if the magnitude of the phase current Ip for a Y-connected load is
/,= iu = uj = k,i (2.30)

BASIC CONCEPTS 29
the total three-phase power is
^ = 3Fp/p cos 6P (2.31)
where 9p is the angle by which phase current lags the phase voltage, that is, the
angle of the impedance in each phase. If VL and IL are the magnitudes of
line-to-line voltage and line current, respectively.
Vp - ^3 and h ~ h (2.32)
and substituting in Eq. (2.31) yields
P= sJ3 VJl cos 9p (2.33)

The total vars are
Q = WpIp sin 9P (2.34)
Q = s/3 VJl sin 9P (2.35)
and the voltamperes of the load are
|s| =yp2+e2 = jivLiL (2.36)
Equations (2.33), (2.35), and (2.36) are the usual ones for calculating P, Q,
and | S | in balanced three-phase networks since the quantities usually known
are line-to-line voltage, line current, and the power factor, cos 9P. In speaking of
a three-phase system, balanced conditions are assumed unless described other¬
wise; and the terms voltage, current, and power, unless identified otherwise, are
understood to mean line-to-line voltage, line current, and total power of all three
phases.
If the load is connected A, the voltage across each impedance is the line-to-
line voltage and the current through each impedance is the magnitude of the line
current divided by ^/3, or
The total three-phase power is
P = 3Fp/p cos 9p (2.38)
and substituting in this equation the values of Vp and Ip in Eq. (2.37) gives
P — >/3 VLIL cos 9P (2.39)
which is identical to Eq. (2.33). It follows that Eqs. (2.35) and (2.36) are also
valid regardless of whether a particular load is connected A or Y.
2.10 PER-UNIT QUANTITIES
Power transmission lines are operated at voltage levels where the kilovolt is the
most convenient unit to express voltage. Because of the large amount of power
transmitted kilowatts or megawatts and kilovoltamperes or megavoltamperes
are the common terms. However, these quantities as well as amperes and ohms
are often expressed as a percent or per unit of a base or reference value specified
for each. For instance, if a base voltage of 120 kV is chosen, voltages of 108, 120,
and 126 kV become 0.90, 1.00, and 1.05 per unit, or 90, 100, and 105%, respec¬
tively. The per-unit value of any quantity is defined as the ratio of the quantity
to its base value expressed as a decimal. The ratio in percent is 100 times the
value in per unit. Both the percent and per-unit methods of calculation are
simpler than the use of actual amperes, ohms, and volts. The per-unit method
has an advantage over the percent method because the product of two quantities
expressed in per unit is expressed in per unit itself, but the product of two
quantities expressed in percent must be divided by 100 to obtain the result in
percent.
Voltage, current, kilovoltamperes, and impedance are so related that selec¬
tion of base values for any two of them determines the base values of the
remaining two. If we specify the base values of current and voltage, base im¬
pedance and base kilovoltamperes can be determined. The base impedance is
that impedance which will have a voltage drop across it equal to the base
voltage when the current flowing in the impedance is equal to the base value of
the current. The base kilovoltamperes in single-phase systems is the product of
base voltage in kilovolts and base current in amperes. Usually base megavolt¬
amperes and base voltage in kilovolts are the quantities selected to specify the
base. For single-phase systems, or three-phase systems where the term current
refers to line current, where the term voltage refers to voltage to neutral, and
where the term kilovoltamperes refers to kilovoltamperes per phase, the follow¬
ing formulas relate the various quantities:
base kVA1(x
Base current, A = --—- (2.40)
base voltage, k\LN '
Base impedance - ^ Vol,age’ V“ (2.41)

base current, A
Base impedance - <baSe vol,age' kV“>2 x 1000 (2.42)

base kVA^
Base impedance-(baSeuVOltage'kV“)2 (2.43)

base MVA10
Base power, kW^ = base kVA^ (2.44)

Base power, MW1(jl = base MVA10 (2.45)
Per-unit impedance actual impedance, ft
(2.46)
of a circuit element base impedance, ft
BASIC CONCEPTS 31
In these equations the subscripts 10 and LN denote “per phase” and “line-to-
neutral,” respectively, where the equations apply to three-phase circuits. If the
equations are used for a single-phase circuit, k\LN means the voltage across the
single-phase line, or line-to-ground voltage if one side is grounded.
Since three-phase circuits are solved as a single line with a neutral return,
the bases for quantities in the impedance diagram are kilovoltamperes per phase
and kilovolts from line to neutral. Data are usually given as total three-phase
kilovoltamperes or megavoltamperes and line-to-line kilovolts. Because of this
custom of specifying line-to-line voltage and total kilovoltamperes or megavolt¬
amperes, confusion may arise regarding the relation between the per-unit value
of line voltage and the per-unit value of phase voltage. Although a line voltage
may be specified as base, the voltage in the single-phase circuit required for the
solution is still the voltage to neutral. The base voltage to neutral is the base
voltage from line to line divided by y/3. Since this is also the ratio between
line-to-line and line-to-neutral voltages of a balanced three-phase system, the
per-unit value of a line-to-neutral voltage on the line-to-neutral voltage base is
equal to the per-unit value of the line-to-line voltage at the same point on the
line-to-line voltage base if the system is balanced. Similarly, the three-phase kilo¬
voltamperes is three times the kilovoltamperes per phase, and the three-phase
kilovoltamperes base is three times the base kilovoltamperes per phase. There¬
fore, the per-unit value of the three-phase kilovoltamperes on the three-phase kilo¬
voltampere base is identical to the per-unit value of the kilovoltamperes per phase
on the kilovoltampere-per-phase base.
A numerical example may serve to clarify the relationships discussed. For
instance, if
Base kVA3^ - 30,000 kVA
and
Base kVLL = 120 kV
where the subscripts 30 and LL mean “three-phase” and “line-to-line,”

respectively,
30,000
Base kVA^ = = 10,000 kVA
3
and
120
Base kVLAf = = 69.2 kV
75
For an actual line-to-line voltage of 108 kV, the line-to-neutral voltage is
108/^/i = 62.3 kV, and
„ 108 62.3
Per-unit voltage = — = 0.90
692
For total three-phase power of 18,000 kW, the power per phase is 6000 kW, and
18,000 6000
Per-unit power =—=^=0.6
Of course, megawatt and megavoltampere values may be substituted for kilowatt

and kilovoltampere values throughout the above discussion. Unless otherwise
specified, a given value of base voltage in a three-phase system is a line-to-line
voltage, and a given value of base kilovoltamperes or base megavoltamperes is
the total three-phase base.
Base impedance and base current can be computed directly from three-phase
values of base kilovolts and base kilovoltamperes. If we interpret base kilovolt¬
amperes and base voltage in kilovolts to mean base kilovoltamperes for the total of
the three phases and base voltage from line to line, we find
base kVA3^
Base current, A = (2.47)
y/3 x base voltage, kVtL
and from Eq. (2.42)
(base voltage, WLL/y/?>)2 x 1000

Base impedance = (2.48)
base kVA30/3
(base voltage, kVLi)2 x 1000

Base impedance = (2.49)
base kVA3(#)
Base impedance = (baSr llZ

0 MVA3(>
base \kVLL>2 (2.50)
Except for the subscripts, Eqs. (2.42) and (2.43) are identical to Eqs. (2.49) and
(2.50), respectively. Subscripts have been used in expressing these relations in
order to emphasize the distinction between working with three-phase quantities
and quantities per phase. We shall use these equations without the subscripts,
but we must (1) use line-to-line kilovolts with three-phase kilovoltamperes or
megavoltamperes and (2) use line-to-neutral kilovolts with kilovoltamperes or
megavoltamperes per phase. Equation (2.40) determines the base current for
single-phase systems or for three-phase systems where the bases are specified in
kilovoltamperes per phase and kilovolts to neutral. Equation (2.47) determines
the base current for three-phase systems where the bases are specified in total
kilovoltamperes for the three phases and in kilovolts from line to line.
Example 2.4 Find the solution of Example 2.3 by working in per unit on a
base of 4.4 kV, 127 A so that both voltage and current magnitudes will be
1.0 per unit. Current rather than kilovoltamperes is specified here since the
latter quantity does not enter the problem.
BASIC CONCEPTS 33
Solution Base impedance is f \rfr

^ I' iA f '
^
127
= 20.on
and therefore the magnitude of the load impedance is also 1.0 per unit. The
line impedance is
1.4/75°
= 0.07/75° per unit
20
Van = 1.0/0° + l.o/- 30° X 0.07/75°
= 1.0/0° + 0.07/45°
= 1.0495 + y'0.0495 = 1.051/2.70° per unit
4400
VLN = 1.051 x —j=- = 2670 V, or 2.67 kV
VLL = 1.051 x 4.4 = 4.62 kV
When the problems to be solved are more complex and particularly when
transformers are involved the advantages of calculations in per unit will be
more apparent.
2.11 CHANGING THE BASE OF PER-UNIT QUANTITIES
Sometimes the per-unit impedance of a component of a system is expressed on a

base other than the one selected as base for the part of the system in which the
component is located. Since all impedances in any one part of a system must be
expressed on the same impedance base when making computations, it is neces¬
sary to have a means of converting per-unit impedances from one base to
another. Substituting the expression for base impedance given by Eq. (2.42) or
(2.49) for base impedance in Eq. (2.46) gives
Per-unit impedance of a circuit element
(actual impedance, Q) x (base kVA)

(2.51)
(base voltage, kV)2 x 1,000
which shows that per-unit impedance is directly proportional to base kilovolt¬

amperes and inversely proportional to the square of the base voltage. Therefore,
to change from per-unit impedance on a given base to per-unit impedanceôn a

new base, the following equation applies:
fbase kV given \ / base kVAnew \ (2 52)
Per-unit Znew = per-unit Z
g,ven base kV^new/ \base kVAgiven/
This equation has nothing to do with transferring the ohmic value of impedance
from one side of the transformer to another. The great value of the equation is in
changing the per-unit impedance given on a particular base to a new base.
Rather than using Eq. 2.52, however, the change in base may also be accom¬
plished by converting the per-unit value on the given base to ohms and dividing
by the new base impedance.
Example 2.5 The reactance of a generator designated X" is given as 0.25 per
unit based on the generator’s nameplate rating of 18 kV, 500 MVA. The
base for calculations is 20 kV, 100 MVA. Find X" on the new base.
Solution By Eq. (2.52)

(18\2/100\
20/ (500) =°.0405 Per uni,
or by converting the given value to ohms and dividing by the new base
impedance
0.25(182/500)
X" = 0.0405 per unit
202/100
Resistance and reactance of a device in percent or per unit are usually

available from the manufacturer. The base is understood to be the rated kilo¬
voltamperes and kilovolts of the device. Tables A.4 and A.5 in the Appendix list
some representative values of reactance for generators and transformers. We
shall discuss per-unit quantities further in Chap. 6 in connection with our study
of transformers.
PROBLEMS
2.1 If v = 141.4 sin (cot + 30°) V and i = 11.31 cos (cof — 30°) A, find for each (a) the maximum
value, (b) the rms value, and (c) the phasor expression in polar and rectangular form if voltage is the
reference. Is the circuit inductive or capacitive?
2.2 If the circuit of Prob. 2.1 consists of a purely resistive and a purely reactive element, find R and
X, (a) if the elements are in series and (b) if the elements are in parallel.
23 In a single-phase circuit Va = 120/45° V and Vb = 100 /-15° V with respect to a reference node
o. Find Vba in polar form.
2.4 A single-phase ac voltage of 240 V is applied to a series circuit whose impedance is 10 /60° Q.
Find R, X, P, Q, and the power factor of the circuit.
BASIC CONCEPTS 35
connected in parallel with the circuit of Prob. 2.4 and if this capacitor supplies
and Q supplied by the 240-V source, and find the resultant power factor,
inductive load draws 10 MW at 0.6 power factor lagging. Draw the power
ine the reactive power of a capacitor to be connected in parallel with the load to
or to 0.85.
induction motor is operating at a very light load during a large part of every day
jm the supply. A device is proposed to “ increase the efficiency ” of the motor,
'ation the device is placed in parallel with the unloaded motor and the current
iply drops to 8 A. When two of the devices are placed in parallel the current drops
le device will cause this drop in current? Discuss the advantages of the device. Is
e motor increased by the device? (Recall that an induction motor draws lagging
e between machines 1 and 2 of Example 2.1 is Z = 0 — ;5 Cl determine (a) whether

nerating or consuming power, (b) whether each machine is receiving or supplying
wer and the amount, and (c) the value of P and Q absorbed by the impedance.
1.8 if Z = 5 + jO n.
ource Ean = —120/210° V and the current through the source is given by
nd the values of P and Q and state whether the source is delivering or receiving
»le 2.1 if £j = 100/0° V and E2 = 120/30° V. Compare the results with Example
: conclusions about the effect of variation of the magnitude of E2 in this circuit.
;al impedances of 10 !—15° Q are Y-connected to balanced three-phase line vol-
>ecify all the line and phase voltages and the currents as phasors in polar form
ce for a phase sequence of abc.
d three-phase system the Y-connected impedances are 10/30° Cl. If Vbc = 416/^0° V,
r form.
Is of a three-phase supply are labeled a, b, and c. Between any pair a voltmeter
A resistor of 100 D and a capacitor of 100 Cl at the frequency of the supply are
connecteu m senes from a to b with the resistor connected to a. The point of connection of the
elements to each other is labeled n. Determine graphically the voltmeter reading between c and n if
phase sequence is abc and if phase sequence is acb.
2.15 Determine the current drawn from a three-phase 440-V line by a three-phase 15-hp motor
operating at full load, 90% efficiency, and 80% power factor lagging. Find the values of P and Q
drawn from the line.
2.16 If the impedance of each of the three lines connecting the motor of Prob. 2.15 to a bus is
0.3 + j'1.0 D, find the line-to-line voltage at the bus which supplies 440 V at the motor.
2.17 A balanced-A load consisting of pure resistances of 15 D per phase is in parallel with a
balanced-Y load having phase impedances of 8 + j6 D. Identical impedances of 2 + j5 D are in each
of the three lines connecting the combined loads to a 110-V three-phase supply. Find the current
drawn from the supply and the line voltage at the combined loads.
2.18 A three-phase load draws 250 kW at a power factor of 0.707 lagging from a 440-V line. In
parallel with this load is a three-phase capacitor bank which draws 60 kVA. Find the total current
and resultant power factor.
2.19 A three-phase motor draws 20 kVA at 0.707 power factor lagging from a 220-V source. Deter¬
mine the kilovoltampere rating of capacitors to make the combined power factor 0.90 lagging, and
determine the line current before and after the capacitors are added.
2.20 A coal mining “drag line” machine in an open-pit mine consumes 0.92 MVA at 0.8 power
factor lagging when it digs coal, and it generates (delivers to the electric system) 0.10 MVA at 0.5 power
factor leading when the loaded shovel swings away from the pit wall. At the end of the “dig” period,
the change in supply current magnitude can cause tripping of a protective relay which is constructed
of solid-state circuitry. Therefore it is desired to minimize the change in current magnitude. Consider
the placement of capacitors at the machine terminals and find the amount of capacitive correction
(in kvar) to eliminate the change in steady-state current magnitude. The machine is energized from a
36.5 kV, three-phase supply. Start the solution by letting Q be the total three-phase megavars of the
capacitors connected across the machine terminals, and write an expression for the magnitude of the
line current drawn by the machine in terms of Q for both the digging and generating operations.
2.21 A generator (which may be represented by an emf in series with an inductive reactance) is rated
500 MVA, 22 kV. Its Y-connected windings have a reactance of 1.1 per unit. Find the ohmic value of
the reactance of the windings.
2.22 The generator of Prob. 2.21 is in a circuit for which the bases are specified as 100 MVA, 20 kV.
Starting with the per-unit value given in Prob. 2.21, find the per-unit value of reactance of the
generator windings on the specified base.
2.23 Draw the single-phase equivalent circuit for the motor (an emf in series with inductive reactance
labeled Zm) and its connection to the voltage supply described in Probs. 2.15 and 2.16. Show on the
diagram the per-unit values of the line impedance and the voltage at the motor terminals on a base of
20 kVA, 440 V. Then using per-unit values find the supply voltage in per unit and convert the per-unit
value of the supply voltage to volts.
CHAPTER
THREE
SERIES IMPEDANCE OF
TRANSMISSION LINES
An electric transmission line has four parameters which affect its ability to fulfill
its function as part of a power system: resistance, inductance, capacitance, and
conductance. In this chapter we discuss the first two of these parameters, and we
shall consider capacitance in the next chapter.
Conductance between conductors or between conductors and the ground
accounts for the leakage current at the insulators of overhead lines and through
the insulation of cables. Since leakage at insulators of overhead lines is negli¬
gible, the conductance between conductors of an overhead line is assumed to be
zero.
When current flows in an electric circuit, we explain some of the properties
of the circuit by the magnetic and electric fields present. Figure 3.1 shows a
single-phase line and its associated magnetic and electric fields. The lines of
magnetic flux form closed loops linking the circuit, and the lines of electric flux
originate on the positive charges on one conductor and terminate on the nega¬
tive charges on the other conductor. Variation of the current in the conductors
causes a change in the number of lines of magnetic flux linking the circuit. Any
change in the flux linking a circuit induces a voltage in the circuit which is
proportional to the rate of change of flux. Inductance is the property of the
circuit that relates the voltage induced by changing flux to the rate of change of
current.
Capacitance exists between the conductors and is the charge on the conduc¬
tors per unit of potential difference between them.
37
The resistance and inductance uniformly distributed along the line form the
series impedance. The conductance and capacitance existing between conductors
of a single-phase line or from a conductor to neutral of a three-phase line form
the shunt admittance. Although the resistance, inductance, and capacitance are
distributed, the equivalent circuit of a line is made up of lumped parameters, as
we shall see when we discuss them.
3.1 TYPES OF CONDUCTORS
In the early days of the transmission of electric power, conductors were usually
copper, but aluminum conductors have completely replaced copper because of
the much lower cost and lighter weight of an aluminum conductor compared
with a copper conductor of the same resistance. The fact that an aluminum
conductor has a larger diameter than a copper conductor of the same resistance
is also an advantage. With a larger diameter the lines of electric flux originating
on the conductor will be farther apart at the conductor surface for the same
voltage. This means a lower voltage gradient at the conductor surface and less
tendency to ionize the air around the conductor. Ionization produces the
undesirable effect called corona.
Symbols identifying different types of aluminum conductors are as follows:
AAC all-aluminum conductors
AAAC all-aluminum-alloy conductors
ACSR aluminum conductor, steel-reinforced
ACAR aluminum conductor, alloy-reinforced
Aluminum-alloy conductors have higher tensile strength than the ordinary

electrical-conductor grade of aluminum. ACSR consists of a central core of steel
strands surrounded by layers of aluminum strands. ACAR has a central core of
higher-strength aluminum surrounded by layers of electrical-conductor-grade
aluminum.
SERIES IMPEDANCE OF TRANSMISSION LINES 39
Alternate layers of wire of a stranded conductor are spiraled in opposite

directions to prevent unwinding and make the outer radius of one layer coincide
with the inner radius of the next. Stranding provides flexibility for a large cross-
sectional area. The number of strands depends on the number of layers and on
whether all the strands are the same diameter. The total number of strands in
concentrically stranded cables, where the total annular space is filled with
strands of uniform diameter, is 7, 19, 37, 61, 91, or more.
Figure 3.2 shows the cross section of a typical steel-reinforced aluminum
cable (ACSR). The conductor shown has 7 steel strands forming a central core,
around which are two layers of aluminum strands. There are 24 aluminum strands
in the two outer layers. The conductor stranding is specified as 24 Al/7 St, or
simply 24/7. Various tensile strengths, current capacities, and conductor sizes are
obtained by using different combinations of steel and aluminum.
Appendix Table A.l gives some electrical characteristics of ACSR. Code
names, uniform throughout the aluminum industry, have been assigned to each
conductor for easy reference.
A type of conductor known as expanded ACSR has a filler such as paper
separating the inner steel strands from the outer aluminum strands. The paper
gives a larger diameter (and hence, lower corona) for a given conductivity and
tensile strength. Expanded ACSR is used for some extra-high-voltage (EHV)
lines.
Cables for underground transmission are usually made with stranded copper
conductors rather than aluminum. The conductors are insulated with oil-
impregnated paper. Up to voltages of 46 kV the cables are of the solid type
which means that the only insulating oil in the cable is that which is im¬
pregnated during manufacture. The voltage rating of this type of cable is limited
by the tendency of voids to develop between the layers of insulation. Voids cause
early breakdown of the insulation. A lead sheath surrounds the cable which may
consist of a single conductor or three conductors.
At voltages from 46 to 345 kV low-pressure oil-filled cables are available.
Oil reservoirs at intervals along the length of the cable supply the oil to ducts in
the center of single-conductor cables or to the spaces between the insulated
conductors of the three-phase type. These conductors are also enclosed in a lead
sheath.
Figure 3.2 Cross section of a steel-reinforced conductor, 7 steel

strands, 24 aluminum strands.
High-pressure pipe-type cables are the most widely used cables for under¬
ground transmission at voltages from 69 to 550 kV. The paper-insulated cables
he in a steel pipe of diameter somewhat larger than necessary to contain the
insulated conductors which lie together along the bottom of the pipe.
Gas-insulated cables are available at voltages up to 138 kV. Research is
constantly conducted on other types of cables especially for voltage levels of 765
and 1100 kV. Manufacturers furnish details of construction for the various
cables.
In this book we shall devote our attention to overhead lines almost exclu¬
sively since underground transmission is usually restricted to large cities or
transmission under wide rivers, lakes, and bays. Underground lines cost at least
eight times as much as overhead lines and 20 times as much at the highest
voltage.
3.2 RESISTANCE
The resistance of transmission-line conductors is the most important cause of

power loss in a transmission line. The term resistance, unless specifically
qualified, means effective resistance. The effective resistance of a conductor is
power loss in conductor
R— | j |2 ^ (3*1)
where the power is in watts and / is the rms current in the conductor in amperes.
The effective resistance is equal to the dc resistance of the conductor only if the
distribution of current throughout the conductor is uniform. We shall discuss
nonuniformity of current distribution briefly after reviewing some fundamental
concepts of dc resistance.
Direct-current resistance is given by the formula
Ro = j n (3.2)
where p = resistivity of conductor
/ = length
A = cross-sectional area
Any consistent set of units may be used. In power work in the United States, l is
usually given in feet, A in circular mils (cmil), and p in ohm-circular mils per
foot, sometimes called ohms per circular mil-foot. In SI units / is in meters, A in
square meters, and p in ohm-meters.f
A circular mil is the area of a circle having a diameter of 1 mil. A mil is
equal to 10"3 in. The cross-sectional area of a solid cylindrical conductor in
circular mils is equal to the square of the diameter of the conductor expressed in
mils. The number of circular mils multiplied by n/4 equals the number of square
t SI is the official designation for the International System of Units.
Figure 3.3 Resistance of a metallic conductor as a func¬

tion of temperature.
mils. Since manufacturers in the United States identify conductors by their cross-
sectional area in circular mils we must use this unit occasionally. The area
in square millimeters equals the area in circular mils multiplied by 5.067 x 10“ 4
The international standard of conductivity is that of annealed copper. Com¬
mercial hard-drawn copper wire has 97.3% and aluminum 61% of the conduc¬
tivity of standard annealed copper. At 20°C for hard-drawn copper p is
1.77 x 10“8Qm (10.66 Q-cmil/ft). For aluminum at 20°C p is 2.83 x
r8H m (17.00 Q-cmil/ft).
The dc resistance of stranded conductors is greater than the value computed
by Eq. (3.2) because spiraling of the strands makes them longer than the conduc¬
tor itself. For each mile of conductor the current in all strands except the one in
the center flows in more than a mile of wire. The increased resistance due to
spiraling is estimated as 1% for three-strand conductors and 2% for concen¬
trically stranded conductors.
The variation of resistance of metallic conductors with temperature is prac¬
tically linear over the normal range of operation. If temperature is plotted on the
vertical axis and resistance on the horizontal axis, as in Fig. 3.3, extension of the
straight-line portion of the graph provides a convenient method of correcting
resistance for changes in temperature. The point of intersection of the extended
line with the temperature axis at zero resistance is a constant of the material.
From the geometry of Fig. 3.3
R2_T+t2
(3.3)
Ri T+ u
where 1% and R2 are the resistances of the conductor at temperatures and t2,
respectively, in degrees Celsius and T is the constant determined from the graph.
Values of the constant T are as follows:
1234.5 for annealed copper of 100% conductivity
T— 241 for hard-drawn copper of 97.3% conductivity
I 228 for hard-drawn aluminum of 61% conductivity
Uniform distribution of current throughout the cross section of a conductor

exists only for direct current. As the frequency of alternating current increases,
the nonuniformity of distribution becomes more pronounced. An increase in
frequency causes nonuniform current density. This phenomenon is called skin
effect. In a circular conductor the current density usually increases from the
interior toward the surface. For conductors of sufficiently large radius, however,
a current density oscillatory with respect to radial distance from the center may
result.
As we shall see when discussing inductance, some lines of magnetic flux exist
inside a conductor. Filaments on the surface of a conductor are not linked by
internal flux, and the flux linking a filament near the surface is less than the flux
linking a filament in the interior. The alternating flux induces higher voltages
acting on the interior filaments than are induced on filaments near the surface of
the conductor. By Lenz’s law the induced voltage opposes the change of current
producing it, and the higher induced voltages acting on the inner filaments cause
the higher current density in filaments nearer the surface and therefore higher
effective resistance. Even at power frequencies skin effect is a significant factor in
large conductors.
3.3 TABULATED RESISTANCE VALUES
The dc resistance of various types of conductors is easily found by Eq. (3.2), and
the increased resistance due to spiraling can be estimated. Temperature correc¬
tions are determined by Eq. (3.3). The increase in resistance caused by skin effect
can be calculated for round wires and tubes of solid material, and curves of R/R0
are available for these simple conductors.! This information is not necessary,
however, since manufacturers supply tables of electrical characteristics of their
conductors. Table A.l is an example of some of the data available.
Example 3.1 Tables of electrical characteristics of all-aluminum Marigold

stranded conductor list a dc resistance of 0.01558 Q per 1000 ft at 20°C and
an ac resistance of 0.0956 fi/mi at 50°C. The conductor has 61 strands and
its size is 1,113,000 cmil. Verify the dc resistance and find the ratio of ac to
dc resistance.
Solution At 20°C from Eq. (3.2) with an increase of 2% for spiraling
„ 17.0 x 1000
Ro = TT73-x 102 = 0 01558 Q per 1000 ft
1,113 x 10J
t See The Aluminum Association, “Aluminum Electrical Conductor Handbook ” New York
1971.
At a temperature of 50°C from Eq. (3.3)
228 4- 50
R0 = 0.01558 ——— = 0.01746 Q per 1000 ft
228 + 20
R 0.0956
R0 0.01746 x 5.280
Skin effect causes a 3.7% increase in resistance.
3.4 DEFINITION OF INDUCTANCE
Two fundamental equations serve to explain and define inductance. The first
equation relates induced voltage to the rate of change of flux linking a circuit.
The induced voltage is
where e is the induced voltage in volts and x is the number of flux linkages of the
circuit in weber-turns (Wbt). The number of weber-turns is the product of each
weber of flux and the number of turns of the circuit linked. For the two-wire line
of Fig. 3.1 each line of flux external to the conductors links the circuit only once.
If we had been considering a coil instead of the circuit of Fig. 3.1, most of the
lines of flux produced would have linked more than one turn of the coil. If some
of the flux links less than all the turns of a coil, the total flux linkages are
reduced. In terms of lines of flux, each line is multiplied by the number of turns it
links, and these products are added to obtain the total flux linkages.
When the current in a circuit is changing, its associated magnetic field
(which is described by the flux linkages) must be changing. If constant permeabi¬
lity is assumed for the medium in which the magnetic field is set up, the number
of flux linkages is directly proportional to the current, and therefore the induced
voltage is proportional to the rate of change of current. Thus our second fun¬
damental equation is
(3.5)
where L = constant of proportionality

L = inductance of circuit, H
e = induced voltage, V
di/dt = rate of change of current, A/s
Equation (3.5) may be used where the permeability is not constant, but in such a
case the inductance is not a constant.
When Eqs. (3.4) and (3.5) are solved for L, the result is
(3.6)
If the flux linkages of the circuit vary linearly with current, which means that the
magnetic circuit has a constant permeability,
L =- H (3.7)
i
from which arises the definition of the self-inductance of an electric circuit as the
flux linkages of the circuit per unit of current. In terms of inductance the flux
linkages are
t = Li Wbt (3.8)
In Eq. (3.8), since i is instantaneous current, t represents instantaneous flux

linkages. For sinusoidal alternating current, flux linkages are sinusoidal. Where
ip is the phasor expression for the flux linkages,
ip = LI Wbt (3.9)
Since tp and / are in phase, L is real, as is consistent with Eqs. (3.7) and (3.8).
The phasor voltage drop due to the flux linkages is
V = jcoLI V (3.10)
V=jco\p V (3.11)
Mutual inductance between two circuits is defined as the flux linkages of one
circuit due to the current in the second circuit per ampere of current in the
second circuit. If the current I2 produces ipl2 flux linkages with circuit 1, the
mutual inductance is
The phasor voltage drop in circuit 1 caused by the flux linkages of circuit 2 is
Vl=j(oM12I2=ja)\p12 V
Mutual inductance is important in considering the influence of power lines

on telephone lines and the coupling between parallel power lines.
3.5 INDUCTANCE OF A CONDUCTOR

DUE TO INTERNAL FLUX
Only flux lines external to the conductors are shown in Fig. 3.1. Some of the
magnetic field, however, exists inside the conductors, as we mentioned when
considering skin effect. The changing lines of flux inside the conductors also
contribute to the induced voltage of the circuit and therefore to the inductance.
The correct value of inductance due to internal flux can be computed as the ratio
of flux linkages to current by taking into account the fact that each line of
internal flux links only a fraction of the total current.
To obtain an accurate value for the inductance of a transmission line, it is

necessary to consider the flux inside each conductor as well as the external flux.
Let us consider the long cylindrical conductor whose cross section is shown in
Fig. 3.4. We shall assume that the return path for the current in this conductor is
so far away that it does not appreciably affect the magnetic field of the conduc¬
tor shown. Then the lines of flux are concentric with the conductor.
The magnetomotive force (mmf) in ampere-turns around any closed path is
equal to the current in amperes enclosed by the path. The mmf is also equal to
the integral of the tangential component of the magnetic field intensity around
the path. Thus
mmf = <j> H ds — I At (3-12)
where H = magnetic field intensity, At/m

s = distance along path, m
/ = current, A, enclosed!
The dot between H and ds indicates that the value of H is the component of the
field intensity tangent to ds.
Let the field intensity at a distance x meters from the center of the conductor
be designated Hx. Since the field is symmetrical, Hx is constant at all points
equidistant from the center of the conductor. If the integration indicated in
Eq. (3.12) is performed around a circular path concentric with the conductor at
x meters from the center, Hx is constant over the path and tangent to it. Equa¬
tion (3.12) becomes
j> Hx ds = Ix (3.13)
2nxHx — Ix (3.14)
t Our work here applies equally to alternating and direct current. As shown, H and / are phasors
and represent sinusoidally alternating quantities. For simplicity the current / could be interpreted as
a direct current and H as a real number.
where Ix is the current enclosed. Then, assuming uniform current density,
(3.15)
where / is the total current in the conductor. Then substituting Eq. (3.15) in
Eq. (3.14) and solving for Hx, we obtain
».-2At/m (316)
The flux density x meters from the center of the conductor is
B, = pH, = ^ Wb/m2 (3.17)
where /r is the permeability of the conductor.!

In the tubular element of thickness dx, the flux dtp is Bx times the cross-
sectional area of the element normal to the flux lines, the area being dx times the
axial length. The flux per meter of length is
d(j)=JpLdx Wb/m (3.18)
The flux linkages dip per meter of length, which are caused by the flux in the
tubular element, are the product of the flux per meter of length and the fraction
of the current linked. Thus
dip = ——j d(p = dx Wbt/m (3-19)

nr 2nr*
Integrating from the center of the conductor to its outside edge to find ipint, the
total flux linkages inside the conductor, we obtain
K, = j r IIIX3
r, 2nr4
dx
•Aim = ^ Wbt/m (3.20)
For a relative permeability of 1, p. = An x 10 7 H/m, and
•Aim x 10"7 Wbt/m (3.21)
Iin, = ^xl0-7 H/m (3.22)
t In SI units the permeability of free space is n0 = 47t x 10 7 H/m, and the relative permeability
is fir = n/n0.
We have computed the inductance per unit length (henrys per meter) of a
round conductor attributed only to the flux inside the conductor. Hereafter, for
convenience, we shall refer to inductance per unit length simply as inductance, but
we must be careful to use the correct dimensional units.
The validity of computing the internal inductance of a solid round wire by
the method of partial flux linkages can be demonstrated by deriving the internal
inductance in an entirely different manner. Equating energy stored in the mag¬
netic field within the conductor per unit length at any instant to Lin( i2/2 and
solving for Lint will yield Eq. (3.22).
3.6 FLUX LINKAGES BETWEEN TWO POINTS

EXTERNAL TO AN ISOLATED CONDUCTOR
As a step in computing inductance due to flux external to a conductor, let us

derive an expression for the flux linkages of an isolated conductor due only to
that portion of the external flux which lies between two points distant Dr and D2
meters from the center of the conductor. In Fig. 3.5, Pt and P2 are two such
points. The conductor carries a current of / A. Since the flux paths are concen¬
tric circles around the conductor, all the flux between Ej and P2 lies within the
concentric cylindrical surfaces (indicated by solid circular lines) which pass
through Pj and P2. At the tubular element which is x meters from the center of
the conductor the field intensity is Hx. The mmf around the element is
2nxHx = / (3.23)
Solving for Hx and multiplying by p yields the flux density Bx in the element, so
that
, _ P1 Wb/m2 (3.24)
2nx
Figure 3.5 A conductor and external points Pl and P2.

The flux d(j) in the tubular element of thickness dx is
d(b - dx Wb/m (3.25)

2nx
The flux linkages dip per meter are numerically equal to the flux dcp, since flux
external to the conductor links all the current in the conductor once and only
once. The total flux linkages between P1 and P2 are obtained by integrating dip
from x = Dy to x = D2. We obtain
*Al2 —
. 1)2
pi ,
-— dx — — In —
pi , D2
Wbt/m (3.26)
Di
27TX 271 Dj
or, for a relative permeability of 1,
ip12 — 2 x 10 7/ In ^ Wbt/m (3.27)
The inductance due only to the flux included between Py and P2 is
Ll2 = 2 x 10 7 In H/m (3.28)

Di
3.7 INDUCTANCE OF A SINGLE-PHASE TWO-WIRE LINE
Before proceeding to the more general case of multiconductor lines and three-
phase lines, let us consider a simple two-wire line composed of solid round
conductors. Figure 3.6 shows a circuit having two conductors of radii rx and r2.
One conductor is the return circuit for the other. First consider only the flux
linkages of the circuit caused by the current in conductor 1. A line of flux set up
Figure 3.6 Conductors of different radii and

the magnetic field due to current in conduc¬
tor 1 only.
by current in conductor 1 at a distance equal to or greater than D + r2 from the

center of conductor 1 does not link the circuit and cannot induce a voltage in the
circuit. Stated in another manner, such a line of flux links a net current of zero,
since the current in conductor 2 is equal in value and opposite in direction to
the current in conductor 1. The fraction of the total current linked by a line of
flux external to conductor 1 at a distance equal to or less than D — r2 is 1.
Between D — r2 and D + r2 (that is, over the surface of conductor 2), the fraction
of the total current in the circuit linked by a line of flux set up by current in
conductor 1 varies from 1 to 0. Therefore, it is logical to simplify the problem,
when D is much greater than rj and r2 and the flux density through the conduc¬
tor is nearly uniform, by assuming that all the external flux set up by current in
conductor 1 extending to the center of conductor 2 links all the current I and
that flux beyond the center of conductor 2 links none of the current. In fact, it
can be shown that calculations made on this assumption are correct even when
D is small.
The inductance of the circuit due to current in conductor 1 is determined by
Eq. (3.28), with the distance D between conductors 1 and 2 substituted for D2
and the radius rx of conductor 1 substituted for Dv For external flux only
L\'exl = 2 x 10-7 In — H/m (3.29)

ri
For internal flux only
x 10-’ H/m (3.30)
The total inductance of the circuit due to the current in conductor 1 only is
Lx = + 2 In x 10"7 H/m (3.31)
The expression for inductance may be put in a more concise form by factor¬
ing Eq. (3.31) and by noting that In e1/4 = 1/4, whence
L, = 2x 10-7|lne1/4 + ln (3.32)
Upon combining terms, we obtain
L, = 2 x 10-’ to -4* (3-33)

ri£
If we substitute r\ for rx e_1/4,
I, =2 x 10-7 In ^ H/m (3.34)

U
The radius r\ is that of a fictitious conductor assumed to have no internal flux
but with the same inductance as the actual conductor of radius rv The quantity
e~ 1/4 is equal to 0.7788. Equation (3.34) gives the same value for inductance as
Eq. (3.31). The difference is that Eq. (3.34) omits the term to account for internal
flux but compensates for it by using an adjusted value for the radius of the
conductor. We should note carefully that Eq. (3.31) was derived for a solid
round conductor and that Eq. (3.34) was found by algebraic manipulation of
Eq. (3.31). Therefore, the multiplying factor of 0.7788 to adjust the radius in
order to account for internal flux applies only to solid round conductors. We
shall consider other conductors later.
Since the current in conductor 2 flows in the direction opposite to that in
conductor 1 (or is 180° out of phase with it), the flux linkages produced by
current in conductor 2 considered alone are in the same direction through the
circuit as those produced by current in conductor 1. The resulting flux for the
two conductors is determined by the sum of the mmfs of both conductors. For
constant permeability, however, the flux linkages (and likewise the inductances)
of the two conductors considered separately may be added.
By comparison with Eq. (3.34) the inductance due to current in conductor 2 is
L2 = 2 x 10 7 In ~ H/m (3.35)
r2
and for the complete circuit
L = L1 + L2=4-x 10"7 In ^— H/m (3.36)

V r\ r'2
If rj = r'2 = r', the total inductance reduces to
D
L = 4 x 1(T7 In H/m (3.37)
7
Equation (3.37) is the inductance of the two-wire line taking into account the
flux linkages caused by current in both conductors, one of which is the return
path for current in the other. This value of inductance is sometimes called the
inductance per loop meter or per loop mile to distinguish it from that compon¬
ent of the inductance of the circuit attributed to the current in one conductor
only. The latter, as given by Eq. (3.34), is one-half the total inductance of a
single-phase line and is called the inductance per conductor.
3.8 FLUX LINKAGES OF ONE CONDUCTOR IN A GROUP
A more general problem than that of the two-wire line is presented by one
conductor in a group of conductors where the sum of the currents in all the
conductors is zero. Such a group of conductors is shown in Fig. 3.7. Conductors
1, 2, 3, ..., n carry the phasor currents Iu I2,J3, /„. The distances of these
conductors from a remote point P are indicated on the figure as D1P, D2P, D3P,
..., DnP. Let us determine iJ/1PU the flux linkages of conductor 1 due to l[
including internal flux linkages but excluding all the flux beyond the point P.
Figure 3.7 Cross-sectional view of a group of n conductors carrying currents whose sum is zero.
Point P is remote from the conductors.
By Eqs. (3.21) and (3.27),
h
*i«-Ij + 2/, In ^) 10-7 (3.38)
IP
ij/iPi = 2 x 10 7/j In Wbt/m (3.39)
V1
The flux linkages tj/1P2 with conductor 1 due to I2, but excluding flux beyond
point P, is equal to the flux produced by I2 between the point P and conductor 1
(that is, within the limiting distances D2P and Dl2 from conductor 2), and so
D 2P
•Aip2 = 2 x 10 12 In (3.40)
D i:
The flux linkages iJ/1P with conductor 1 due to all the conductors in the group, but
excluding flux beyond point P, is
•Aip = 2 x 10 7(/x In —^ + I2 In —— + /3 In —— + ••• + /„ In ~r—j (3.41)

’ rl ^12 ^13
which becomes, by expanding the logarithmic terms and regrouping,
•Aip = 2 x 10“7//x In — 4 12 In —-1- /3 In —-f •■• + /„ In ——

-
\ rl ^12 ”l3 Uln
+ 1 j In DiP 4" I2 In D2P 4- /3 In D2P + ■••-(-/„ In Dnp| (3.42)
Since the sum of all the currents in the group is zero,
11 + f2 + /3 + ,,,4-/„ = 0
and, solving for /„, we obtain
I„ = —(h + h + h + ••• 4- 1) (3.43)

Substituting Eq. (3.43) in the second term containing /„ in Eq. (3.42) and recom¬
bining some logarithmic terms, we have
tplP = 2 x 10“7(/t In — -h /2 In —— + /3 In —— + ••• + /, In

12 Du Du
+ /, In ^ + I2 In + h In ^ + ' ' + to ) (3.44)

AiR AnP
Now letting the point P move infinitely far away so that the set of terms contain¬
ing logarithms of ratios of distances from P becomes infinitesimal, since the
ratios of the distances approach 1, we obtain
ij/1 = 2 x 10’ h In — + 12 In —-1-13 In —-f ••• + /„ In j Wbt/m

ri Ai: Ai
(3.45)
By letting point P move infinitely far away we have included all the flux linkages
of conductor 1 in our derivation. Therefore, Eq. (3.45) expresses all the flux
linkages of conductor 1 in a group of conductors, provided the sum of all the
currents is zero. If the currents are alternating, they must be expressed as instan¬
taneous currents to obtain instantaneous flux linkages or as complex rms values
to obtain the rms value of flux linkages as a complex number.
3.9 INDUCTANCE OF COMPOSITE-CONDUCTOR LINES
Stranded conductors come under the general classification of composite conduc¬

tors, which means conductors composed of two or more elements or strands
electrically in parallel. We are now ready to study the inductance of a transmis¬
sion line composed of composite conductors, but we shall limit ourselves to the
case where all the strands are identical and share the current equally. The
method can be expanded to apply to all types of conductors containing strands
of different sizes and conductivities, but this will not be done here since
the values of internal inductance of specific conductors are generally available
from the various manufacturers and can be found in handbooks. The method to
be developed indicates the approach to the more complicated problems of non-
homogeneous conductors and unequal division of current between strands. The
method is applicable to the determination of inductance of lines consisting of
circuits electrically in parallel since two conductors in parallel can be treated as
strands of a single composite conductor.
Figure 3.8 shows a single-phase line composed of two conductors. In order
to be more general, each conductor forming one side of the line is shown as an
arbitrary arrangement of an indefinite number of conductors. The only restric¬
tions are that the parallel filaments are cylindrical and share the current equally.
o
.o
co
o
°'0 „o
o
Figure 3.8 Single-phase line consisting of two com¬
posite conductors.
V
Cond. X
^- v-'
Cond. y
Conductor X is composed of n identical, parallel filaments, each of which carries

the current I/n. Conductor Y, which is the return circuit for the current in
conductor X, is composed of m identical, parallel filaments, each of which carries
the current —I/m. Distances between the elements will be designated by the
letter D with appropriate subscripts. Applying Eq. (3.45) to filament a of conduc¬
tor X, we obtain for flux linkages of filament a
•Aa = 2 x 10 7 - (In \ + In ~ + In + • ■ • + In
n\ ra Dab Dac DJ
— 2 x 10 7 — (In —b In —-—b In ~—b ■ • ■ + In (3.46)

m\ Daa Dab Dac Dn
from which
i m-7r i \fDaa Dab'Dac‘ Dam \\tU+
\i)n — 2 x 10 1 In —, Wbt/m (3.47)
Dividing Eq. (3.47) by the current I/n, we find that the inductance of filament a is
L=t± = 2n x 10-7 In ^Daa’Dab' °ac'" H/m (3.48)

" I/n ' !jr'aDabDac Dn
Similarly, the inductance of filament b is
\/Dba'Dbb'Dbc' ' ' ' Dbm

Lh =— = 2n x 10" 7 In H/m (3.49)
I/n ZjDbar'bDbc---Dbn
The average inductance of the filaments of conductor X is
La + Lb + Lc + ■ ■ ■ + Ln
(3.50)
D av
Conductor X is composed of n filaments electrically in parallel. If all the

filaments had the same inductance, the inductance of the conductor would be 1 /n
times the inductance of one filament. Here all the filaments have different induc¬
tances, but the inductance of all of them in parallel is 1 /n times the average
inductance. Thus the inductance of conductor X is
La + Lb + Lc + • • • + L
(3.51)
Lx ~
Substituting the logarithmic expression for inductance of each filament in

Eq. (3.51) and combining terms, we obtain v,
Lx = 2 x 1(T7
x >/(A»a''Dgb'Dgc' ' ’ ' Dam){Dba'Dbb'Dbe' ' ' ' Abm) ' ' ' (Dna‘Dnb'Dnc' Dnm)
y(DaaDabDac • • • Dan)(DbaDbbDbc • • • Dbn) • • • {DnaDnbDnc ■ ■ • Z>J

H/m (3.52)
where r^, rj,, and r^, have been replaced by Daa, Dbb, and Dnn, respectively, to
make the expression appear more symmetrical.
Note that the numerator of the argument of the logarithm in Eq. (3.52) is the
mnth root of mn terms, which are the products of the distances from all the n
filaments of conductor X to all the m filaments of conductor Y. For each
filament in conductor X, there are m distances to filaments in conductor Y, and
there are n filaments in conductor X. The product of m distances for each of n
filaments results in mn terms. The mnth root of the product of the mn distances is
called the geometric mean distance between conductor X and conductor Y. It is
abbreviated Dm or GMD and is also called the mutual GMD between the two
conductors.
The denominator of the argument of the logarithm in Eq. (3.52) is the n2
root of n2 terms. There are n filaments, and for each filament there are n terms
consisting of r' for that filament times the distances from that filament to every
other filament in conductor X. Thus we account for n2 terms. Sometimes r'a is
called the distance from filament a to itself, especially when it is designated as
Daa. With this in mind the terms under the radical in the denominator may be
described as the product of the distances from every filament in the conductor to
itself and to every other filament. The n2 root of these terms is called the self
GMD of conductor X, and the r' of a separate filament is called the self GMD of
the filament. Self GMD is also called geometric mean radius, or GMR. The
correct mathematical expression is self GMD, but common practice has made
GMR more prevalent. We shall use GMR in order to conform to this practice
and identify it by Ds.
In terms of Dm and Ds, Eq. (3.52) becomes
\
Lx = 2 x FT7 In ^ H/m (3.53)
If we compare Eq. (3.53) with Eq. (3.34), the similarity between them is
apparent. The equation for the inductance of one conductor of a composite-
conductor line is obtained by substituting in Eq. (3.34) the GMD between con¬
ductors of the composite-conductor line for the distance between the solid
conductors of the single-conductor line and by substituting the GMR of the
composite conductor for the GMR (r') of the single conductor. Equation (3.53)
gives the inductance of one conductor of a single-phase line. The conductor is
composed of all the strands which are electrically in parallel. The inductance is
the total number of flux linkages of the composite conductor per unit of line
current. Equation (3.34) gives the inductance of one conductor of a single-phase
line for the special case where the conductor is a solid round wire.
The inductance of conductor Y is determined in a similar manner, and the

inductance of the line is
■L — -(- Ly
Example 3.2 One circuit of a single-phase transmission line is composed of

three solid 0.25-cm-radius wires. The return circuit is composed of two
0.5-cm-radius wires. The arrangement of conductors is shown in Fig. 3.9.
Find the inductance due to the current in each side of the line and the
inductance of the complete line in henrys per meter (and in millihenrys per
mile).
Solution Find the GMD between sides X and Y:
Dm — Dad Dae Dbd Dbe Dcd Dce
Dad = Dbe = 9 m
Dae = Dbd = Dce = yj62 + 92 = y 177
Dcd - J92 + 122 = 15 m

Dm - ^92 x 15 x 1773/2 = 10.743 m
Then find the GMR for side X
Ds — y Daa Dab Dac Dba Dbb Dbc Dca Dch Dcc

= y(0.25 x 0.7788 x 10“2)3 x 64 x 122 = 0.481 m
o.
9m
o a
6m
6m
V
o C
Figure 3.9 Arrangement of conductors for
Example 3.2. Side X Side Y
and for side Y
Ds = ^(0.5 x 0.7788 x 10-2)2 x 62 = 0.153 m
Lx - 2 x 1(T7 In = 6.212 x 1CT7 H/m

x 0.481
LY = 2 x 10-7 In = 8.503 x 10" 7 H/m
L = LX + LY= 14.715 x 10"7 H/m
(L = 14.715 x 10“7 x 1609 x 103 = 2.37 mH/mi)
If a single-phase line consists of two stranded cables it is seldom necessary

to calculate the GMD between strands of the two sides, for the GMD would be
almost equal to the distance between centers of the cables. The calculation of
mutual GMD is important only where the various strands (or conductors)
electrically in parallel are separated from each other by distances more nearly
approaching the distance between the two sides of the circuit. For instance, in
Example 3.2 the conductors in parallel on one side of the line are separated by
6 m, and the distance between the two sides of the line is 9 m. Here the calculation
of mutual GMD is important. For stranded conductors the distance between sides
of a line composed of one conductor per side is usually so great that the mutual
GMD can be taken as equal to the center-to-center distance with negligible error.
If the effect of the steel core of ACSR is neglected in calculating inductance,
a high degree of accuracy results, provided the aluminum strands are in an even
number of layers. The effect of the core is more apparent for an odd number of
layers of aluminum strands, but the accuracy is good when the calculations are
based on the aluminum strands alone.
3.10 THE USE OF TABLES
Tables listing values of GMR are generally available for standard conductors
and provide other information for calculating inductive reactance as well as
shunt capacitive reactance and resistance. Since industry in the United States
continues to use units of inches, feet, and miles so do these tables. Therefore
some of our examples will use feet and miles, but others will use meters and
kilometers.
Inductive reactance rather than inductance is usually desired. The induc¬
tive reactance of one conductor of a single-phase two-conductor line is
XL = 2nfL = 2nfx2x 10"7 In ~

Ds
= 4nf x 10"7 In ~ Q/m (3.54)

Us
or -2.022 x l(T3/ln ~ Q/mi (3.55)

U*
where Dm is the distance between conductors. Both Dm and Ds must be in the

same units, usually either meters or feet. The GMR found in tables is an equiva¬
lent Ds which accounts for skin effect where it is appreciable enough to affect
inductance. Of course skin effect is greater at higher frequencies for a conductor
of a given diameter. Values of Ds listed in Table A.l are for a frequency of 60 Hz.
Some tables give values of inductive reactance in addition to GMR. One
method is to expand the logarithmic term of Eq. (3.55) as follows:
XL = 2.022 x 10~3/ln ~ + 2.022 x HT3/lnDm Q/mi (3.56)

Us
If both Ds and Dm are in feet, the first term in Eq. (3.56) is the inductive reactance
of one conductor of a two-conductor line having a distance of 1 ft between
conductors, as may be seen by comparing Eq. (3.56) with Eq. (3.55). Therefore,
the first term of Eq. (3.56) is called the inductive reactance at 1 ft spacing Xa. It
depends upon the GMR of the conductor and the frequency. The second term of
Eq. (3.56) is called the inductive reactance spacing factor Xd. This second term is
independent of the type of conductor and depends on frequency and spacing
only. The spacing factor is equal to zero when Dm is 1 ft. If Dm is less than 1 ft,
the spacing factor is negative. The procedure for computing inductive reactance
is to look up the inductive reactance at 1-ft spacing for the conductor under
consideration and to add to this value the value of the inductive reactance
spacing factor, both at the desired line frequency. Table A.l includes values of
inductive reactance at 1-ft spacing, and Table A.2 lists values of the inductive
reactance spacing factor.
Example 33 Find the inductive reactance per mile of a single-phase line

operating at 60 Hz. The conductor is Partridge, and spacing is 20 ft between
centers.
Solution For this conductor, Table A.l lists Ds = 0.0217 ft. From Eq.
(3.55), for one conductor,
20
= 2.022 x lO"3 x 60 In
= 0.828 Q/mi
The above calculation is used if only Ds is known. Table A.l, however,

lists inductive reactance at 1-ft spacing Xa = 0.465 Q/mi. From Table A.2
the inductive reactance spacing factor is Xd = 0.3635 Q/mi, and so the
inductive reactance of one conductor is
0.465 + 0.3635 = 0.8285 Q/mi
Since the conductors composing the two sides of the line are identical,
the inductive reactance of the line is
XL = 2 x 0.8285 = 1.657 Q/mi

Figure 3.10 Cross-sectional view of the equilaterally

spaced conductors of a three-phase line.
3.11 INDUCTANCE OF THREE-PHASE LINES

WITH EQUILATERAL SPACING
So far in our discussion we have considered only single-phase lines. The equa¬
tions we have developed are quite easily adapted, however, to the calculation of
the inductance of three-phase lines. Figure 3.10 shows the conductors of a three-
phase line spaced at the corners of an equilateral triangle. If we assume that
there is no neutral wire, or if we assume balanced three-phase phasor currents,
la + ib + jc = o. Equation (3.45) determines the flux linkages of conductor a:
ipa = 2 x l(T7(/a In ^ + Ib In ^ + Ic In Wbt/m (3.57)
Since Ia = -(/„ + Ic), Eq. (3.57) becomes
Ih = 2 X KT7(/„ In 7- - /. In 7) = 2 x lO’7;. In ^ Wbt/m (3.58)
and
La = 2 x 10“7 In ~ H/m , (3.59)
Equation (3.59) is the same in form as Eq. (3.34) for a single-phase line except
that Ds replaces r'. Because of symmetry, the inductances of conductors b and c
are the same as the inductance of conductor a. Since each phase consists of only
one conductor, Eq. (3.59) gives the inductance per phase of the three-phase line.
3.12 INDUCTANCE OF THREE-PHASE LINES

WITH UNSYMMETRICAL SPACING
When the conductors of a three-phase line are not spaced equilaterally, the
problem of finding the inductance becomes more difficult. Then the flux linkages
and inductance of each phase are not the same. A different inductance in each
Figure 3.11 Transposition cycle.
phase results in an unbalanced circuit. Balance of the three phases can be

restored by exchanging the positions of the conductors at regular intervals along
the line so that each conductor occupies the original position of every other
conductor over an equal distance. Such an exchange of conductor positions is
called transposition. A complete transposition cycle is shown in Fig. 3.11. The
phase conductors are designated a, b, and c, and the positions occupied are
numbered 1, 2, and 3. Transposition results in each conductor having the same
average inductance over the whole cycle.
Modern power lines are usually not transposed at regular intervals, although
an interchange in the positions of the conductors may be made at switching
stations in order to balance the inductance of the phases more closely. For¬
tunately, the dissymmetry between the phases of an untransposed line is small
and is neglected in most calculations of inductance. If the dissymmetry is neg¬
lected, the inductance of the untransposed line is taken as equal to the average
value of the inductive reactance of one phase of the same line correctly
transposed. The derivations to follow are for transposed lines.
To find the average inductance of one conductor of a transposed line, the
flux linkages of a conductor are found for each position it occupies in the
transposition cycle, and the average flux linkages are determined. Let us apply
Eq. (3.45) to conductor a of Fig. 3.11 to find the phasor expression for the flux
linkages of a in position 1 when b is in position 2 and c is in position 3, as
follows:
>j/al = 2 x In 77 + A In yr— + /c In Wbt/m (3.60)

\ L>s U\2 U3l!
With a in position 2, b in position 3, and c in position 1,
1/^2 = 2 x 10-7(/a In + /fc In —^-1- Ic In ) Wbt/m (3.61)

\ Vs U23 U12/
and, with a in position 3, b in position 1, and c in position 2,
^, = 2xl0-’(/.lnT+/tlnA.+/tto
+ Ic In A.) Wbt/m (3.62)
The average value of the flux linkages of a is
•Aal + *Aa2 + *Aa3

<K =
2 x 10
(3Ia ln Ds + h ln Dl2D23D31 + Ic ^ Dl2D23D3 J (3 63)
With the restriction that Ia = —(Ib + Ic),
2 x 10- 1 1
•Aa =
3;“ln^-'“ln Dl2D23D33
'a “*
2 x 10-7/. In ■f’hlJhl1’!! Wbt/m (3.64)
and the average inductance per phase is
Lfl = 2 x 10“7 ln H/m (3.65)
where
êq — JD12D23 D3 i (3.66)
and Ds is the GMR of the conductor. Deq, the geometric mean of the three
distances of the unsymmetrical line, is the equivalent equilateral spacing, as may
be seen by a comparison of Eq. (3.65) with Eq. (3.59). We should note the
similarity between all the equations for the inductance of a conductor. If the
inductance is in henrys per meter, the factor 2 x 10“7 appears in all the equa¬
tions, and the denominator of the logarithmic term is always the GMR of
the conductor. The numerator is the distance between wires of a two-wire line,
\
the mutual GMD between sides of a composite-conductor single-phase line,
the distance between conductors of an equilaterally spaced line, or the equivalent
equilateral spacing of an unsymmetrical line.
Example 3.4 A single-circuit three-phase line operated at 60 Hz is arranged

as shown in Fig. 3.12. The conductors are ACSR Drake. Find the inductive
reactance per mile per phase.
'Jo-
Figure 3.12 Arrangement of conductors for Example 3.4 38'

Solution From Table A.l,
Ds = 0.0373 ft Deq = ^20 x 20 x 38 = 24.8 ft
24 8
L = 2 x 10 7 In - 13.00 x 10~7 H/m
Xl = 2n:60 x 1609 x 13.00 x 10~7 = 0.788 Q/mi per phase
Equation (3..55) may be used also, or, from Tables A.l and A.2,
= 0.399
and for 24.8 ft,
Xd = 0.389
XL — 0.399 4- 0.389 = 0.788 Q/mi per phase
3.13 BUNDLED CONDUCTORS
At extra-high voltages (EHV), that is, voltages above 230 kV, corona with its
resultant power loss and particularly its interference with communications is
excessive if the circuit has only one conductor per phase. The high-voltage
gradient at the conductor in the EHV range is reduced considerably by having
two or more conductors per phase in close proximity compared with the spacing
between phases. Such a line is said to be composed of bundled conductors. The
bundle consists of two, three, or four conductors. The three-conductor bundle
usually has the conductors at the vertices of an equilateral triangle, and the
four-conductor bundle usually has its conductors at the corners of a square.
Figure 3.13 shows these arrangements. The current will not divide exactly
between the conductors of the bundle unless there is a transposition of the con¬
ductors within the bundle, but the difference is of no practical importance, and
the GMD method is accurate for calculations.
Reduced reactance is the other equally important advantage of bundling.
Increasing the number of conductors in a bundle reduces the effects of corona
and reduces the reactance. The reduction of reactance results from the increased
GMR of the bundle. The calculation of GMR is, of course, exactly the same as
that of a stranded conductor. Each conductor of a two-conductor bundle, for
instance, is treated as one strand of a two-strand conductor. If we let Dbs indicate
the GMR of a bundled conductor and Ds the GMR of the individual conductors
composing the bundle, we find, referring to Fig. 3.13,
—Q
©—r-©
Figure 3.13 Bundle arrangements.
For a two-strand bundle *•.
Db = J(DS x d)2 = jDsxd (3.67)
For a three-strand bundle
Dbs = jf(Ds xdxd)3 = ^Ds x d2 (3.68)
For a four-strand bundle
Db = \f/(Ds xdxdxdx 21/2)4 = 1.09^DS x d3 (3.69)
In computing inductance using Eq. (3.65), £>* of the bundle replaces Ds of a

single conductor. To compute Z)eq, the distance from the center of one bundle to
the center of another bundle is sufficiently accurate for Dab, Dbc, and Dca.
Obtaining the actual GMD between conductors of one bundle and those of
another would be almost indistinguishable from the center-to-center distances
for the usual spacing.
Example 3.5 Each conductor of the bundled-conductor line shown in

Fig. 3.14 is ACSR, 1,272,000-cmil Pheasant. Find the inductive reactance in
ohms per km (and per mile) per phase for d = 45 cm. Also find the per-unit
series reactance of the line if its length is 160 km and base is 100 MVA,
345 kV.
Solution From Table A.l Ds = 0.0466 ft, and we multiply feet by 0.3048 to
convert to meters.
Db = ^0.0466 x 0.3048 x 0.45 = 0.080 m
Deq = ^8 x 8 x 16 = 10.08 m
XL = 2;r60 x 2 x 10~7 x 103 In

0.08
= 0.365 Q/km per phase
(0.365 x 1.609 = 0.587 fi/mi per phase)

(345)2
Base Z = - ; = 1190 Q
100
„ 0.365 x 160
X =-- = 0.049 per unit
HH
bO Ob’
Figure 3.14 Spacing of
8m <- 8m
conductors of a bundled-
conductor line. d = 45 elm
°G-18'-G~—r
10'
°i-2r-
10'
Figure 3.15 Typical arrangement of conductors of

a parallel-circuit three-phase line. CG-18'-o—L
3.14 PARALLEL-CIRCUIT THREE-PHASE LINES
Two three-phase circuits that are identical in construction and electrically in

parallel have the same inductive reactance. The inductive reactance of the single
equivalent circuit, however, is half that of each of the individual circuits con¬
sidered alone only if they are so widely separated that there is negligible mutual
inductance between them. If the two circuits are on the same tower, the method
of GMD can be used to find the inductance per phase by considering all the
conductors of any particular phase to be strands of one composite conductor.
Figure 3.15 shows a typical arrangement of a parallel-circuit three-phase
line. Although the line will probably not be transposed, we obtain a practical
value for inductance and the calculations are simplified if transposition is
assumed. Conductors a and a' are in parallel to compose phase a. Phases b and c
are similar. We assume that a and a' take the positions of b and b' and then of c
and c' as those conductors are rotated similarly in the transposition cycle.
To calculate Deq the GMD method requires that we use Dpb, D£c, and Dpa,
where the superscript indicates that these quantities are themselves GMD values
and where Dpb means the GMD between the conductors of phase a and those of
phase b.
The Ds of Eq. (3.65) is replaced by Dp, which is the geometric mean of the
GMR values of the two conductors occupying first the positions of a and a', then
the positions of b and b’, and finally the positions of c and c'. Following each
step of Example 3.6 is possibly the best means of understanding the procedure.
Example 3.6 A three-phase double-circuit line is composed of 300,000-cmil

26/7 ACSR Ostrich conductors arranged as shown in Fig. 3.15. Find the
60-Hz inductive reactance in ohms per mile per phase.
Solution From Table A.l for Ostrich
Ds = 0.0229 ft
Distance a to b: Original position = ^/iO2 + 1.52 = 10.1 ft
Distance a to b': Original position = ^/lO2 + 19.52 = 21.9 ft
The GMDs between phases are
Dpab = Dl = ^/(10.1 x 21.9)2 = 14.88 ft
Dpca = ^(20 x 18)2 = 18.97 ft
Deq = ^14.88 x 14.88 x 18.97 = 16.1 ft
The GMR for the parallel-circuit line is found after first obtaining the GMR
values for the three positions. The actual distance from a to a' is
Ô2 4- 182 = 26.9 ft. Then GMR of each phase is
In position a-ayj26.9 x 0.0229 = 0.785 ft
In position b-b': ^/21 x 0.0229 = 0.693 ft
In position c-c'\ ^26.9 x 0.0229 = 0.785 ft
Therefore
Dp = ^0.785 x 0.693 x 0.785 = 0.753 ft
L = 2 x 10“7 In }^ = 6.13 x 10“7 H/m per phase

0.753 ' F y
XL = 27r60 x 1609 x 6.13 x 10 “7 = 0.372 Q/mi per phase
3.15 SUMMARY OF INDUCTANCE CALCULATIONS

FOR THREE-PHASE LINES \
Although computer programs are usually available or written rather easily for
calculating inductance of all kinds of lines, some understanding of the develop¬
ment of the equations used is rewarding from the standpoint of appreciating the
effect of variables in designing a line. However, tables like A.l and A.2 make the
calculations quite simple except for parallel-circuit lines. Table A.l also lists
resistance.
The important equation for inductance per phase of single-circuit three-
phase lines is given here for convenience
1 = 2 x 10“7 In H/m per phase (3.70)
Inductive reactance in ohms per kilometer at 60 Hz is found by multiplying

inductance in henrys per meter by 27i60 x 1000:
XL = 0.0754 x In Q/km per phase (3.71)
or
XL = 0.1213 In —^ Q/mi per phase (3.72)
Both Deq and Ds must be in the same units, usually feet. If the line has one
conductor per phase, Ds is found directly from tables. For bundled conductors
Dbs, as defined in Sec. 3.13, is substituted for Ds. For both single- and bundled-
conductor lines
Deq=^DahDbcDca (3.73)
For bundled-conductor lines Dab, Dbc, and Dca are distances between the centers
of the bundles of phases a, b, and c.
For lines with one conductor per phase it is convenient to determine XL by
adding Xa for the conductor as found in tables like A.l to Xd as found in Table
A.2 corresponding to Deq.
Inductance and inductive reactance of parallel-circuit lines are calculated by
following the procedure of Example 3.6.
PROBLEMS
3.1 The all-aluminum conductor identified by the code word Bluebell is composed of 37 strands of
diameter 0.1672 in. Tables of characteristics of all-aluminum conductors list an area of 1,033,500 cmil
for this conductor. Are these values consistent with each other? Find the area in square millimeters.
3.2 Determine the dc resistance in ohms per km of Bluebell at 20°C by Eq. (3.2) and the information
in Prob. 3.1, and check the result against the value listed in tables of 0.01678 D per 1000 ft. Compute
the dc resistance in ohms per kilometer at 50°C and compare the result with the ac 60-Hz resistance
of 0.1024 Q/mi listed in tables for this conductor at 50°C. Explain any difference in values.
33 An all-aluminum conductor is composed of 37 strands each having a diameter of 0.333 cm.
Compute the dc resistance in ohms per kilometer at 75°C.
3.4 A single-phase 60-Hz power line is supported on a horizontal crossarm. Spacing between con¬
ductors is 2.5 m. A telephone line is supported on a horizontal crossarm 1.8 m directly below the
power line with a spacing of 1.0 m between the centers of its conductors. Find the mutual inductance
between the power and telephone circuits and the 60-Hz voltage per kilometer induced in the
telephone line if the current in the power line is 150 A.
3.5 If the power and telephone lines described in Prob. 3.4 are in the same horizontal plane and the
distance between the nearest conductors of the two lines is 18 m, find the mutual inductance between
the circuits and the voltage per mile induced in the telephone line for 150 A in the power line.
3.6 The conductor of a single-phase 60-Hz line is a solid round aluminum wire having a diameter of
0.412 cm. The conductor spacing is 3 m. Determine the inductance of the line in millihenrys per mile.
How much of the inductance is due to internal flux linkages? Assume skin effect is negligible.
3.7 Find the GMR of a three-strand conductor in terms of r of an individual strand.
Figure 3.16 Cross-sectional view of un¬

conventional conductors for Prob. 3.8. (a)
3.8 Find the GMR of each of the unconventional conductors shown in Fig. 3.16 in terms of the
radius r of an individual strand.
3.9 The distance between conductors of a single-phase line is 10 ft. Each conductor is composed of
seven equal strands. The diameter of each strand is 0.1 in. Show that Ds for the conductor is 2.177
times the radius of each strand. Find the inductance of the line in millihenrys per mile.
3.10 Find the inductive reactance of ACSR Rail in ohms per kilometer at 1-m spacing.
3.11 Which conductor listed in Table A.l has an inductive reactance at 7-ft spacing of 0.651 Q/mi?
3.12 A three-phase line is designed with equilateral spacing of 16 ft. It is decided to build the line
with horizontal spacing (D13 = 2Dl2 = 2D23). The conductors are transposed. What should be the
spacing between adjacent conductors in order to obtain the same inductance as in the original
design ?
3.13 A three-phase 60-Hz transmission line has its conductors arranged in a triangular formation so
that two of the distances between conductors are 25 ft and the third distance is 42 ft. The conductors
are ACSR Osprey. Determine the inductance and inductive reactance per phase per mile.
3.14 A three-phase 60-Hz line has flat horizontal spacing. The conductors have a GMR of 0.0133 m
with 10 m between adjacent conductors. Determine the inductive reactance per phase in ohms per
kilometer. What is the name of this conductor?
3.15 For short transmission lines if resistance is neglected the maximum power which can be trans¬
mitted per phase is equal to
where Vs and VR are the line-to-neutral voltages at the sending and receiving ends of the line and X is
the inductive reactance of the line. This relationship will become apparent in the study of Chap. 5. If
the magnitudes of Vs and VR are held constant and if the cost of a conductor is proportional to its
cross-sectional area, find the conductor in Table A.l which has the maximum power-handling
capacity per cost of conductor.
3.16 A three-phase underground distribution line is operated at 23 kV.' The three conductors are
insulated with 0.5 cm solid black polyethylene insulation laid flat and side by side directly in a dirt trench.
The conductor is circular in cross section and has 33 strands of aluminum. The diameter of the
conductor is 1.46 cm. The manufacturer gives the GMR as 0.561 cm and the cross section of the
conductor as 1.267 cm2. The thermal rating of the line buried in normal soil whose maximum
temperature is 30°C is 350 A. Find the dc and ac resistance at 50°C and the inductive reactance in
ohms per kilometer. To decide whether to consider skin effect in calculating resistance determine the
percent skin effect at 50°C in the ACSR conductor of size nearest that of the underground conductor.
Note that the series impedance of the distribution line is dominated by R rather than XL because of
the very low inductance due to the close spacing of the conductors.
3.17 The single-phase power line of Prob. 3.4 is replaced by a three-phase line on a horizontal
crossarm in the same position as that of the original single-phase line. Spacing of the conductors of
the power line is D13 = 2Dl2 = 2D23, and equivalent equilateral spacing is 3 m. The telephone line
remains in the position described in Prob. 3.4. If the current in the power line is 150 A, find the
voltage per kilometer induced in the telephone line. Discuss the phase relation of the induced voltage
with respect to the power-line current.
3.18 A 60-Hz three-phase line composed of one ACSR Bluejay conductor per phase has flat horizon¬
tal spacing of 11 m between adjacent conductors. Compare the inductive reactance in ohms per
kilometer per phase of this line with that of a line using a two-conductor bundle of ACSR 26/7
conductors having the same total cross-cectional area of aluminum as the single-conductor line and
11 m spacing measured from the center of the bundles. The spacing between conductors in the
bundle is 40 cm.
3.19 Calculate the inductive reactance in ohms per kilometer of a bundled 60-Hz three-phase line
having three ACSR Rail conductors per bundle with 45 cm between conductors of the bundle. The
spacing between bundle centers is 9, 9, and 18 m.
3.20 Six conductors of ACSR Drake constitute a 60-Hz double-circuit three-phase line arranged as
shown in Fig. 3.15. The vertical spacing, however, is 14 ft; the longer horizontal distance is 32 ft; and
the shorter horizontal distances are 25 ft. Find the inductance per phase per mile and the inductive
reactance in ohms per mile.
CHAPTER
FOUR
( \['\( [ [ \\( I OF TRANSMISSION LINES
As we discussed briefly at the beginning of Chap. 3, the shunt admittance of a

transmission line consists of conductance and capacitive reactance. We have also
mentioned that conductance is usually neglected because its contribution to
shunt admittance is very small. For this reason this chapter has been given the
title of capacitance rather than shunt admittance.
Another reason for neglecting conductance is that there is no good way of
taking it into account because it is quite variable. Leakage at insulators, the
principal source of conductance, changes appreciably with atmospheric condi¬
tions and with the conducting properties of dirt that collects on the insulators.
Corona, which results in leakage between lines, is also quite variable with atmos¬
pheric conditions. It is fortunate that the effect of conductance is such a neglig¬
ible component of shunt admittance.
Capacitance of a transmission line is the result of the potential difference
between the conductors; it causes them to be charged in the same manner as the
plates of a capacitor when there is a potential difference between them. The
capacitance between conductors is the charge per unit of potential difference.
Capacitance between parallel conductors is a constant depending on the size and
spacing of the conductors. For power lines less than about 80 km (50 mi) long,
the effect of capacitance is slight and is usually neglected. For longer lines of
higher voltage, capacitance becomes increasingly important.
An alternating voltage impressed on a transmission line causes the charge on
the conductors at any point to increase and decrease with the increase and
decrease of the instantaneous value of the voltage between conductors at the
point. The flow of charge is current, and the current caused by the alternate
charging and discharging of a line due to an alternating voltage is called the
68
CAPACITANCE OF TRANSMISSION LINES 69
charging current of the line. Charging current flows in a transmission line even
when it is open-circuited. It affects the voltage drop along the line as well as
the efficiency and power factor of the line and the stability of the system of which
the line is a part.
4.1 ELECTRIC FIELD OF A

LONG STRAIGHT CONDUCTOR
Just as the magnetic field is important in considering inductance, so the electric

field is important in studying capacitance. In the preceding chapter we discussed
both the magnetic and electric fields of a two-wire line. Lines of electric flux
originate on the positive charges of one conductor and terminate on the negative
charges of the other conductor. The total electric flux emanating from a conduc¬
tor is numerically equal to the number of coulombs of charge on the conductor.
Electric flux density is the electric flux per square meter and is measured in
coulombs per square meter.
If a long straight cylindrical conductor lies in a uniform medium such as air,
has a uniform charge throughout its length, and is isolated from other charges so
that the charge is uniformly distributed around its periphery, the flux is radial.
All points equidistant from such a conductor are points of equipotential and
have the same electric flux density. Figure 4.1 shows such an isolated conductor
carrying a uniformly distributed charge. The electric flux density at .x meters
from the conductor can be computed by imagining a cylindrical surface concen¬
tric with the conductor and x meters in radius. Since all parts of the surface are
equidistant from the conductor, which has a uniformly distributed charge, the
cylindrical surface is a surface of equipotential and the electric flux density on
Figure 4.1 Lines of electric flux originating on the

positive charges uniformly distributed over the
surface of an isolated cylindrical conductor.
the surface is equal to the flux leaving the conductor per meter of length divided
by the area of the surface in an axial length of 1 m. The electric flux density is
D=^~ C/m2 (4.1)

2nx
where q is the charge on the conductor in coulombs per meter of length and x is
the distance in meters from the conductor to the point where the electric flux
density is computed. The electric field intensity, or the negative of the potential
gradient, is equal to the electric flux density divided by the permittivity! of the
medium. Therefore, the electric field intensity is
(4.2)
4.2 THE POTENTIAL DIFFERENCE BETWEEN

TWO POINTS DUE TO A CHARGE
The potential difference between two points in volts is numerically equal to the
work in joules per coulomb necessary to move a coulomb of charge between the
two points. Electric field intensity is a measure of the force on a charge in
the field. Electric field intensity in volts per meter is equal to the force in newtons
per coulomb on a coulomb of charge at the point considered. Between two
points the line integral of the force in newtons acting on a coulomb of positive
charge is the work done in moving the charge from the point of lower potential
to the point of higher potential and is numerically equal to the potential differ¬
ence between the two points.
Consider a long straight wire carrying a positive charge of q C/m, as shown
in Fig. 4.2. Points Pi and P2 are located at distances and D2 meters from the
center of the wire. The positive charge on the wire will exert a repelling force on
a positive charge placed in the field. For this reason and because D2 in this case
is greater than Du work must be done on a positive charge to move it from P2
to Pl5 and Pi is at a higher potential than P2. The difference in potential is the
amount of work done per coulomb of charge moved. On the other hand, if the
coulomb moves from Pi to P2, it expends energy, and the amount of work, or
energy, in newton-meters is the voltage drop from Px to P2. The potential
difference is independent of the path followed. The simplest way to compute the
voltage drop between the two points is to compute the voltage between the
t In SI units the permittivity of free space k0 is 8.85 x 10“12 F/m. Relative permittivity kr is the
ratio of the actual permittivity k of a material to the permittivity of free space. Thus, kr = k/k0. For
dry air kr is 1.00054 and is assumed equal to 1.0 in calculations for overhead lines.
distributed positive charge. /
equipotential surfaces passing through Pt and P2 by integrating the field inten¬

sity over a radial path between the equipotential surfaces. Thus the instanta¬
neous voltage drop between Px and P2 is
. 02 . D2
vl2 = I $ dx = I dx =
q , °2
(4.3)
•D1 jD1
2nkx Ink Dj
where q is the instantaneous charge on the wire in coulombs per meter of length.
Note that the voltage drop between two points, as given by Eq. (4.3), may be
positive or negative depending on whether the charge causing the potential
difference is positive or negative and on whether the voltage drop is computed
from a point near the conductor to a point farther away, or vice versa. The sign
of q may be either positive or negative, and the logarithmic term is either positive
or negative depending on whether D2 is greater or less than Dt.
4.3 CAPACITANCE OF A TWO-WIRE LINE
Capacitance between the two conductors of a two-wire line was defined as the
charge on the conductors per unit of potential difference between them. In the
form of an equation, capacitance per unit length of the line is
C=- F/m (4.4)

v
where q is the charge on the line in coulombs per meter and v is the potential
difference between the conductors in volts. Hereafter, for convenience, we shall
refer to capacitance per unit length as capacitance and indicate the correct
dimensions for equations derived. The capacitance between two conductors can
be found by substituting in Eq. (4.4) the expression for v in terms of q from
r.
D
Figure 4.3 Cross section of a parallel-wire line.
Eq. (4.3). The voltage vab between the two conductors of the two-wire line shown
in Fig. 4.3 can be found by determining the potential difference between the two
conductors of the line, first by computing the voltage drop due to the charge qa
on conductor a and then by computing the voltage drop due to the charge qb on
conductor b. By the principle of superposition the voltage drop from conductor
a to conductor b due to the charges on both conductors is the sum of the voltage
drops caused by each charge alone.
Consider the charge qa on conductor a, and assume that conductor b is
uncharged and merely an equipotential surface in the electric field created by the
charge on a. The equipotential surface of conductor b and the equipotential
surfaces due to the charge on a are shown in Fig. 4.4. The distortion of the
equipotential surfaces near conductor b is caused by the fact that conductor b is
also an equipotential surface. Equation (4.3) was derived by assuming all the
equipotential surfaces due to a uniform charge on a round conductor to be
cylindrical and concentric with the conductor. Such is actually true for the case
under discussion except in the region near conductor b. The potential of conduc¬
tor b is that of the equipotential surface intersecting b. Therefore, in determining
vab a path may be followed from conductor a through a region of undistorted
equipotential surfaces to the equipotential surface intersecting conductor b.
Then, moving along the equipotential surface to b gives no further change in
voltage. This path of integration is indicated in Fig. 4.4 together with the direct
path. Of course, the potential difference is the same regardless of the path over
which the integration of the field intensity is taken. By following the path
through the undistorted region, we see that the distances corresponding to D2
and Dj of Eq. (4.3) are D and ra, respectively, in determining vab due to qa.
Similarly, in determining vab due to qb, the distances corresponding to D2 and
of Eq. (4.3) are rb and D, respectively. Converting to phasor notation (qa and qb
become complex numbers), we obtain
(4.5)
due to qa due to qt>
and, since qa— —qb for a two-wire line.
(4.6)
Figure 4.4 Equipotential surfaces of a portion of the electric

field caused by a charged conductor a (not shown). Conductor
b causes the equipotential surfaces to become distorted.
Arrows indicate optional paths of integration between a point
on the equipotential surface of conductor b and the conductor
a, whose charge qa creates the equipotential surfaces shown.
or, by combining the logarithmic terms,
K& = In — (4.7)
2nk rna r,
'b
The capacitance between conductors is
_ 'la _ Ink
^ ab T/
Kb
( t\2
^ (D2/rar„)
F/m (4.8)
If ra = rb =
nk
F/m (4.9)
ab~ In (D/r)
Equation (4.9) gives the capacitance between the conductors of a two-wire

line. Sometimes it is desirable to know the capacitance between one of the
conductors and a neutral point between them. For instance, if the line is supplied
by a transformer having a grounded center tap, the potential difference between
each conductor and the ground is half the potential difference between the two
conductors and the capacitance to ground, or capacitance to neutral, is the charge
on a conductor per unit of potential difference between the conductor and
ground. Thus, the capacitance to neutral for the two-wire line is twice the line-to-
line capacitance (capacitance between conductors). If the line-to-line capacitance
is considered to be composed of two equal capacitances in series, the voltage
across the line divides equally between them and their junction is at the ground
potential. Thus, the capacitance to neutral is that of one of the two equal series
o —k-
Cab
(a) Representation of line-to-line capacitance (b) Representation of line-to-neutral capacitance
Figure 4.5 Relationship between the concepts of line-to-line capacitance and line-to-neutral
capacitance.
capacitances, or twice the line-to-line capacitance. Therefore,
C„ = Can = Cbn = ln" F/m to neutral (4.10)
The concept of capacitance to neutral is illustrated in Fig. 4.5.

Equation (4.10) corresponds to Eq. (3.34) for inductance. One difference
between the equations for capacitance and inductance should be noted carefully.
The radius in the equation for capacitance is the actual outside radius of the
conductor and not the GMR of the conductor, as in the inductance formula.
Equation (4.3), from which Eqs. (4.5) to (4.10) were derived, is based on the
assumption of uniform charge distribution over the surface of the conductor.
When other charges are present, the distribution of charge on the surface of the
conductor is not uniform and the equations derived from Eq. (4.3) are not
strictly correct. The nonuniformity of charge distribution, however, can be neg¬
lected entirely in overhead lines since the error in Eq. (4.10) is only 0.01% even
for such a close spacing as that where the ratio D/r = 50.
A question arises about the value to be used in the denominator of the
argument of the logarithm in Eq. (4.10) when the conductor is a stranded cable,
since the equation was derived for a solid round conductor. Since electric flux is
perpendicular to the surface of a perfect conductor, the electric field at the
surface of a stranded conductor is not the same as the field at the surface of a
\
cylindrical conductor. Therefore, the capacitance calculated for a stranded con¬

ductor by substituting the outside radius of the conductor for r in Eq. (4.10) will
be slightly in error because of the difference between the field in the neighbor¬
hood of such a conductor and the field near a solid conductor for which
Eq. (4.10) was derived. The error is very small, however, since only the field very
close to the surface of the conductor is affected. The outside radius of the
stranded conductor is used in calculating the capacitance.
After the capacitance to neutral has been found, the capacitive reactance
existing between one conductor and neutral for relative permittivity kr = 1 is
found by using the expression for C given in Eq. (4.10) to yield
1 2.862
xc = x 109 In — Q ■ m to neutral (4.11)
2njT ~J~ r
Since C in Eq. (4.11) is in farads per meter, the proper units for Xc must be
ohm-meters. We should also note that Eq. (4.11) expresses the reactance from
line to neutral for 1 m of line. Since capacitive reactance is in parallel along the
line, Xc in ohm-meters must be divided by the length of the line in meters to
obtain the capacitive reactance in ohms to neutral for the entire length of the
line.
When Eq. (4.11) is divided by 1609 to convert to ohm-miles, we obtain
1.779 D
Xc = x 106 In Q - mi to neutral (412)
Table A.l lists the outside diameters of the most widely used sizes of ACSR.
If D and r in Eq. (4.12) are in feet, capacitive reactance at 1-ft spacing X'a is the
first term and capacitive reactance spacing factor X’d is the second term when the
equation is expanded as follows:
1 779 1 1 779
Xc = —- x 106 In - + —— x 106 In D Q mi to neutral (4.13)
/ r f
Table A.l includes values of X'a for common sizes of ACSR, and similar tables
are readily available for other types and sizes of conductors. Table A.3 lists
values of X'd.
Example 4.1 Find the capacitive susceptance per mile of a single-phase line
operating at 60 Hz. The conductor is Partridge, and spacing is 20 ft between
centers.
Solution For this conductor Table A.l lists an outside diameter of

0.642 in, and so
0.642
r = 0.0268 ft
2 x 12
and from Eq. (4.12)

1 77Q 20
Xr = —- x 106 In - = 0.1961 x 106 Q mi to neutral
c 60 0.0268
Br = — = 5.10 x 10“6 U/mi to neutral

C
or in terms of capacitive reactance at 1-ft spacing and capacitive reactance
spacing factor from Tables A.l and A.3
A; = 0.1074 Mfi-mi
X'd = 0.0889 Mfi-mi
Xc = 0.1074 + 0.0889 = 0.1963 MO-mi per conductor

Line-to-line capacitive reactance and susceptance are
Ic = 2 x 0.1963 x 106 = 0.3926 x 106 Q • mi
Bc =— =2.55 x 10“6 U/mi
4.4 CAPACITANCE OF A THREE-PHASE LINE

WITH EQUILATERAL SPACING
The three identical conductors of radius r of a three-phase line with equilateral

spacing are shown in Fig. 4.6. Equation (4.5) expresses the voltage between two
conductors due to the charges on each one if the charge distribution on the
conductors can be assumed to be uniform. Thus the voltage Vab of the three-
phase line due only to the charges on conductors a and b is
V (4.14)
due to qa and qb.
Equation (4.3) enables us to include the effect of qc since uniform charge distri¬
bution over the surface of a conductor is equivalent to a concentrated charge at
the center of the conductor. Therefore, due only to the charge qc,
which is zero since qc is equidistant from a and b. However to show that we are
considering all three charges we can write
(4.15)
Figure 4.6 Cross section of a three-phase line with

equilateral spacing. a D c
Figure 4.7 Phasor diagram of the balanced voltages of a

three-phase line.
Similarly
1 / , D , D r\
V = (la In - + qb In - + qc In - (4.16)
* 2nk
Adding Eqs. (4.15) and (4.16) gives
D r
V A- V
r ab ' r ac
— 2qa In - + (qb + qc) In - (417)
2nk
In deriving these equations we have assumed that ground is far enough away to
have negligible effect. Since the voltages are assumed to be sinusoidal and ex¬
pressed as phasors, the charges are sinusoidal and expressed as phasors. If there
are no other charges in the vicinity, the sum of the charges on the three conduc¬
tors is zero and we can substitute — qa in Eq. (4.17) for qb + qc and obtain
3 <7
Kb + Kr — •in® (418)
2nk
Figure 4.7 is the phasor diagram of voltages. From this figure we obtain the
following relations between the line voltages Vab and Vac and the voltage Van from
line a to the neutral of the three-phase circuit:
Kt = s/3U0.866 + j0.5) (419)
Vu = - V„ = V3Vm(0.866 - j0.5) (420)
Adding Eqs. (4.19) and (4.20) gives
V.
r ab +
' V
r ac = 3Van
*■' r
(421)
Substituting 3Van for Vab + Vac in Eq. (4.18), we obtain
D
V (422)
r
Since capacitance to neutral is the ratio of the charge on a conductor <to the
voltage between that conductor and neutral,
c-=|; =taW) F/mtoneu,ral <423>

Comparison of Eqs. (4.23) and (4.10) shows that the two are identical. These
equations express the capacitance to neutral for single-phase and equilaterally
spaced three-phase lines, respectively. We saw in Chap. 3 tTiat the equations for
inductance per conductor were the same for single-phase and equilaterally
spaced three-phase lines.
The term charging current is applied to the current associated with the
capacitance of a line. For a single-phase circuit, the charging current is the
product of the line-to-line voltage and the line-to-line susceptance, or, as a
phasor,
IcH=joCabVab (4.24)
For a three-phase line, the charging current is found by multiplying the voltage
to neutral by the capacitive susceptance to neutral. This gives the charging
current per phase and is in accord with the calculation of balanced three-phase
circuits on the basis of a single phase with neutral return. The phasor charging
current in phase a is
/chg =ju>CnVan A/mi (4.25)
Since the rms voltage varies along the line, the charging current is not the same
everywhere. Often the voltage used to obtain a value for charging current is the
normal voltage for which the line is designed, such as 220 or 500 kV, which is
probably not the actual voltage at either a generating station or a load.
4.5 CAPACITANCE OF A THREE-PHASE LINE

WITH UNSYMMETRICAL SPACING
When the conductors of a three-phase line are not equilaterally spaced, the
problem of calculating capacitance becomes more difficult. In the usual un¬
transposed line the capacitances of each phase to neutral are unequal. In a
transposed line the average capacitance to neutral of any phase for the complete
transposition cycle is the same as the average capacitance to neutral of any other
phase, since each phase conductor occupies the same position as every other
phase conductor over an equal distance along the transposition cycle. The
dissymmetry of the untransposed line is slight for the usual configuration, and
capacitance calculations are carried out as though all lines were transposed.
For the line shown in Fig. 4.8 three equations are found for Vab for the three
different parts of the transposition cycle. With phase a in position 1, b in
position 2, and c in position 3,
K„ = 2nk 1 ^12 ,
qa In- + qb In
r
r
D i;
+ qc In
^23)
dJ
V (4.26)
Figure 4.8 Cross section of a three-phase line with

unsymmetrical spacing. 3
With a in position 2, b in position 3, and c in position 1,
V (4.27)
and, with a in position 3, b in position 1, and c in position 2.
V (4.28)
Equations (4.26) to (4.28) are similar to Eqs. (3.60) to (3.62) for the flux
linkages of one conductor of a transposed line. However, in the equations for
flux linkages we noted that the current in any phase was the same in every part
of the transposition cycle. In Eqs. (4.26) to (4.28), if we disregard the voltage
drop along the line, the voltage to neutral of a phase in one part of a transposi¬
tion cycle is equal to the voltage to neutral of that phase in any part of the cycle.
Hence, the voltage between any two conductors is the same in all parts of the
transposition cycle. It follows that the charge on a conductor must be different
when the position of the conductor changes with respect to other conductors. A
treatment of Eqs. (4.26) to (4.28) analogous to that of Eqs. (3.60) to (3.62) is not
rigorous.
The rigorous solution for capacitance is too involved to be practical except
perhaps for flat spacing with equal distances between adjacent conductors. With
the usual spacings and conductors, sufficient accuracy is obtained by assuming
that the charge per unit length on a conductor is the same in every part of the
transposition cycle. When the above assumption is made with regard to charge,
the voltage between a pair of conductors is different for each part of the transpo¬
sition cycle. Then an average value of voltage between the conductors can be
found, and the capacitance calculated from the average voltage. We obtain the
average voltage by adding Eqs. (4.26), (4.27), and (4.28) and dividing the result
by 3. The average voltage between conductors a and b, based on the assumption
of the same charge on a conductor regardless of its position in the transposition
cycle, is
D 12^23^31
Dl2D23 d31
V (4.29)
r
where
Dn=^Dl2D13D>t (4.30)
Similarly, the average voltage drop from conductor a to conductor c is
v (4'31>
Applying Eq. (4.21) to find the voltage to neutral, we have
= + V (4.32)
Since qa + qb + qc = 0 in a balanced three-phase circuit,
3V
** r an
v (4.33)
and
Qa_ 2nk
F/m to neutral (4.34)
Vm ln iD'Jr)
Equation (4.34) for capacitance to neutral of a transposed three-phase line corre¬
sponds to Eq. (3.65) for the inductance per phase of a similar line. In finding
capacitive reactance to neutral corresponding to C„ the reactance can be split
into components of capacitive reactance to neutral at 1-ft spacing X’a and capaci¬
tive reactance spacing factor X'd as defined by Eq. (4.13).
Example 4.2 Find the capacitance and the capacitive reactance for 1 mi of
the line described in Example 3.4. If the length of the line is 175 mi and the
normal operating voltage is 220 kV, find capacitive reactance to neutral for
the entire length of the line, the charging current per mile, and the total
charging megavolt-amperes.
Solution
1.108
r = 0.0462 ft
2 x 12
Deq 24.8 ft
2n x 8.85 x 10 12
Cn = 8.8466 x 10“12 F/m
ln (24.8/0.0462)
1012
Xc = 0.1864 x 106 Q- mi
2;r x 60 x 8.8466 x 1609
or from tables
X'a = 0 0912 x 106 X'd = 0.0953 x 106
Xc = (0.0912 + 0.0953) x 106 = 0.1865 x 106 Q - mi to neutral

For a length of 175 mi
0.1865 x 106
Capacitive reactance = = 1066 Q to neutral
175
220,000
/chg = 27r60 x 8.8466 x 10“12 x 1609 = 0.681 A/mi
73
or 0.681 x 175 = 119 A for the line. Reactive power is Q = ^3 x 220 x
119 x 10 3 = 45.3 Mvar. This amount of reactive power absorbed by the
distributed capacitance is negative in keeping with the convention discussed
in Chap. 2. In other words, positive reactive power is being generated by the
distributed capacitance of the line.
4.6 EFFECT OF EARTH ON THE CAPACITANCE OF

THREE-PHASE TRANSMISSION LINES
Earth affects the capacitance of a transmission line because its presence alters the
electric field of the line. If we assume that the earth is a perfect conductor in the
form of a horizontal plane of infinite extent, we realize that the electric field of
charged conductors above the earth is not the same as it would be if the equi-
potential surface of the earth were not present. The electric field of the charged
conductors is forced to conform to the presence of the earth’s surface. The
assumption of a flat, equipotential surface is, of course, limited by the irregular¬
ity of terrain and the type of surface of the earth. The assumption enables us,
however, to understand the effect of a conducting earth on capacitance
calculations.
Consider a circuit consisting of a single overhead conductor with a return
path through the earth. In charging the conductor, charges come from the earth
to reside on the conductor, and a potential difference exists between the conduc¬
tor and earth. The earth has a charge equal in magnitude to that on the conduc¬
tor but of opposite sign. Electric flux from the charges on the conductor to the
charges on the earth is perpendicular to the earth’s equipotential surface, since
the surface is assumed to be a perfect conductor. Let us imagine a fictitious
conductor of the same size and shape as the overhead conductor lying directly
below the original conductor at a distance equal to twice the distance of the
conductor above the plane of the ground. The fictitious conductor is below
the surface of the earth by a distance equal to the distance of the overhead
conductor above the earth. If the earth is removed and a charge equal and
opposite to that on the overhead conductor is assumed on the fictitious conduc¬
tor, the plane midway between the original conductor and the fictitious con¬
ductor is an equipotential surface and occupies the same position as the
equipotential surface of the earth. The electric flux between the overhead con¬
ductor and this equipotential surface is the same as that which existed between
the conductor and the earth. Thus, for purposes of calculation of capacitance,
the earth may be replaced by a fictitious charged conductor below the surface of
the earth by a distance equal to that of the overhead conductor above the earth.
Such a conductor has a charge equal in magnitude and opposite in sign to that of
the original conductor and is called the image conductor.
The method of calculating capacitance by replacing the earth by the image
of an overhead conductor can be extended to more than one conductor. If we
locate an image conductor for each overhead conductor, the flux between the
original conductors and their images is perpendicular to the plane which re¬
places the earth, and that plane is an equipotential surface. The flux above the
plane is the same as it is when the earth is present instead of the image
conductors.
To apply the method of images to the calculation of capacitance for a
three-phase line, refer to Fig. 4.9. We shall assume that the line is transposed and
Figure 4.9 Three-phase line and

its image. 2
that conductors a, b, and c carry the charges qa, qb, and qc and occupy positions
1, 2, and 3, respectively, in the first part of the transposition cycle. The plane of
the earth is shown, and below it are the conductors with the image charges —qa,
— qb, and —qc. Equations for the three parts of the transposition cycle can be
written for the voltage drop from conductor a to conductor b as determined by
the three charged conductors and their images. With conductor a in position 1, b
in position 2, and c in position 3,
1 H12)
Vat qfl|ln - In + qb In
2nk HJ
D23
+ <Jc (4.35)
d3 H31J
Similar equations for Vab are written for the other parts of the transposition
cycle. Accepting the approximately correct assumption of constant charge per
unit length of each conductor throughout the transposition cycle allows us to
obtain an average value of the phasor Vat . The equation for the average value of
the phasor Vac is found in a similar manner, and 3Van is obtained by adding the
average values of Kt and Vac. Knowing that the sum of the charges is zero, we
then find
2nk
In (£>e,A)-ln
Comparison of Eqs. (4.34) and (4.36) shows that the effect of the earth is
to increase the capacitance of a line. To account for the earth the denominator
of Eq. (4.34) must have subtracted from it the term
log
If the conductors are high above ground compared with the distances between
them, the diagonal distances in the numerator of the correction term are nearly
equal to the vertical distances in the denominator and the term is very small.
This is the usual case, and the effect of ground is generally neglected for three-
phase lines except for calculations by symmetrical components when the sum of
the three line currents is not zero.
4.7 BUNDLED CONDUCTORS
Figure 4.10 shows a bundled-conductor line for which we can write an equation
for the voltage from conductor a to conductor b as we did in deriving Eq. (4.26)
except that now we must consider the charges on all six individual conductors.
The conductors of any one bundle are in parallel, and we can assume the charge
per bundle divides equally between the conductors of the bundle since the separ¬
ation between bundles is usually more than 15 times the spacing between the
conductors of the bundle. Also, since D12 is much greater than d, we can use Dl2
£>31
Figure 4.10 Cross section of a ■* Diz ^23

bundled-conductor three-phase a o q a' b O O b c o O c'
line. b-«M
in place of the distances D12 - d and D12 + d and make ather similar substitu¬
tions of bundle separation distances instead of using the more exact expressions
that occur in finding Vab. The difference due to this approximation cannot be
detected in the final result for usual spacings even when the calculation is carried
to five or six significant figures.
If the charge on phase a is qa, conductors a and a' each have the charge
qa/2; similar division of charge is assumed for phases b and c. Then
Kt = 2nk ||ln ^13 +ln 0,2
D 23
+ In + In (4.37)
+ t(ln^ D + "(,n§77 D 31
b'
The letters under each logarithmic term indicate the conductor whose charge is
accounted for by that therm. Combining terms gives
^12 , , s/rd , D23

Kb = In qb In —- + qc In (4.38)
2nk sj~rd D12 Dilt
Equation (4.38) is the same as Eq. (4.26) except the /rd has replaced r. It
therefore follows that if we considered the line to be transposed, we would find
C"= , /n . r-jv F/m to neutral (4.39)

In (êq I/rd)
The y/rd is the same as Dbs for a two-conductor bundle except that r has replaced
Ds. This leads us to the very important conclusion that a modified GMD
method applies to the calculation of capacitance of a bundled-conductor three-
phase line having two conductors per bundle. The modification is that we are
using outside radius in place of the GMR of a single conductor.
It is logical to conclude that the modified GMD method applies to other
bundling configurations. If we let DbsC stand for the modified GMR to be used in
capacitance calculations to distinguish it from Db used in inductance calcula¬
tions, we have
2nk
In (DJDb/)
Then for a two-strand bundle
DbsC = </(r x d)2 = Jrd (4.41)

for a three-strand bundle
DbsC = \/(r x d x df = *Jrd2 (4.42)

and for a four-strand bundle
DbsC = '*/(r xdxdxdx 21/2)4 = 1.09 ijrd2 (4.43)
Example 43 Find the capacitive reactance to neutral of the line described in

Example 3.5 in ohm-kilometers (and in ohm-miles) per phase.
Solution Computed from the diameter given in Table A.l
1.382 x 0.3048
= 0.01755 m
2 x 12
DbsC = yo.01755 x 0.45 = 0.0889 m
£>eq = */8 x 8 x 16 = 10.08 m
„ 2n x 8.85 x 10~12
Cm = . _ = 11.754 x 10-12 F/m
In (10.08/0.0889)
1012 x KT3
Xr = = 0.2257 x 106 Q - km per phase to neutral
2tt60 x 11.754
/v _ 0.2257
2257 x 106
= 0.1403 x 106 Q-mi per phase to neutral
\Xc ~ 1.609
4.8 PARALLEL-CIRCUIT THREE-PHASE LINES
Throughout our discussion of capacitance we have noted the similarity of the

equations for inductance and capacitance. A modified GMD method has been
found to apply in finding capacitance of bundled-conductor lines. We could
show that this method is equally good for transposed three-phase lines with
equilateral spacing (conductors at the vertices of a hexagon) and for flat vertical
spacing (conductors of the three phases of each circuit lying in the same vertical
plane). It is reasonable to assume that the modified GMD method can be used
for arrangements intermediate between equilateral and flat-vertical spacing
Even though transpositions are not made, the method is generally used. An example
should be sufficient to illustrate the method.
Example 4.4 Find the 60-Hz capacitive susceptance to neutral per mile per
phase of the double-circuit line described in Example 3.6
Solution From Example 3.6, Deq = 16.1 ft.

The calculation of Dpc is the same as that of Dp in Example 3.6 except
that the outside radius of the Ostrich conductor is used instead of its GMR.
The outside diameter of 26/17, ACSR Ostrich is 0.680 in.
r= = 0.0283 ft
2 x 12
DpsC = (^26.9 x 0.0283^/21 x 0.0283^26.9 x 0.0283)1/3
= ^0.0283 (26.9 x 21 x 26.9)1/6 = 0.837 ft
2n x 8.85 x 10"12
= 18.807 x 10"12 F/m
C" In (16.1/0.837)
Bc = 2n x 60 x 18.807 x 1609 = 11.41 x 10 6 U/mi per phase to neutral
4.9 SUMMARY
The similarity between inductance and capacitance calculations has been em¬
phasized throughout our discussions. As in inductance calculations, computer
programs are recommended if a large number of calculations of capacitance are
required. Tables like A.l and A.3 make the calculations quite simple, however,
except for parallel-circuit lines.
The important equation for capacitance to neutral for a single-circuit, three-
phase line is
2nk
cn = - n /n F/m to neutral (4.44)
In Deq/DsC
DsC is the outside radius r of the conductor for a line consisting of one conductor
per phase. For overhead lines k is 8.85 x 10"12 since kr for air is 1.0. Capacitive
reactance in ohm-meters is l/2nfC where C is in farads/meter. So at 60 Hz
\
Xc = 4.77 x 104 In ^ fi-km to neutral (4-45)

Osc
or upon dividing by 1.609 km/mi
Xc = 2.965 x 104 In Q mi to neutral (4.46)

Dsc
Values for capacitive susceptance in mhos/kilometer (siemens/kilometer) and
mhos/mile (siemens/mile) are the reciprocals of Eqs. (4.45) and (4.46) respectively.
Both Deq and DsC must be in the same units, usually feet. For bundled
conductors DbsC, as defined in Sec. 4.7, is substituted for DsC. For both single-
and bundled-conductor lines
Deq — y/DabDbcDca
For bundled-conductor lines Dab, Dbc, and Dca are distances between the centers
of the bundles of phases a, b, and c.
For lines with one conductor per phase it is convenient to determine Xc by
adding X'a for the conductor as found in tables like A.l to X'd as found in Table
A.3 corresponding to Deq.
Capacitance and capacitive reactance of parallel-circuit lines are found by
following the procedure of Example 4.4.
PROBLEMS
4.1 A three-phase transmission line has flat horizontal spacing with 2 m between adjacent conduc¬
tors. At a certain instant the charge on one of the outside conductors is 60 pC/km, and the charge on
the center conductor and on the other outside conductor is —30 /rC/km. The radius of each conduc¬
tor is 0.8 cm. Neglect the effect of ground, and find the voltage drop between the two identically
charged conductors at the instant specified.
4.2 The 60-Hz capacitive reactance to neutral of a solid conductor, which is one conductor of a
three-phase line with an equivalent equilateral spacing of 5 ft, is 196.1 k 12-mi. What value of reac¬
tance would be specified in a table listing the capacitive reactance in ohm-miles to neutral of the
conductor at 1-ft spacing for 25 Hz? What is the cross-sectional area of the conductor in circular
mils?
4.3 Derive an equation for the capacitance to neutral in farads per meter of a single-phase line,
taking into account the effect of ground. Use the same nomenclature as in the equation derived for
the capacitance of a three-phase line where the effect of ground is represented by image charges.
4.4 Calculate the capacitance to neutral in farads per meter of a single-phase line composed of two
single-strand conductors each having a diameter of 0.229 in. The conductors are 10 ft apart and 25 ft
above ground. Compare the values obtained by Eq. (4.10) and by the equation derived in Prob. 4.3.
4.5 A three-phase 60-Hz transmission line has its conductors arranged in a triangular formation so
that two of the distances between conductors are 25 ft and the third is 42 ft. The conductors are
ACSR Osprey. Determine the capacitance to neutral in microfarads per mile and the capacitive
reactance to neutral in ohm-miles. If the line is 150 mi long, find the capacitance to neutral and
capacitive reactance of the line.
4.6 A three-phase 60-Hz line has flat horizontal spacing. The conductors have an outside diameter of
3.28 cm with 12 m between conductors. Determine the capacitive reactance to neutral in ohm-
meters and the capacitive reactance of the line in ohms if its length is 125 mi.
4.7 A 60-Hz three-phase line composed of one ACSR Bluejay conductor per phase has flat horizon¬
tal spacing of 11 m between adjacent conductors. Compare the capacitive reactance in ohm-
kilometers per phase of this line with that of a line using a two-conductor bundle of ACSR 26/7
conductors having the same total cross-sectional area of aluminum as the single-conductor line and
the 11-m spacing measured between bundles. The spacing between conductors in the bundle is
40 cm.
4.8 Calculate the capacitive reactance in ohm-kilometers of a bundled 60-Hz three-phase line having
three ACSR Rail conductors per bundle with 45 cm between conductors of the bundle. The spacing
between bundle centers is 9, 9, and 18 m.
4.9 Six conductors of ACSR Drake constitute a 60-Hz double-circuit three-phase line arranged as
shown in Fig. 3.15. The vertical spacing, however, is 14 ft; the longer horizontal distance is 32 ft; and
the shorter horizontal distances are 25 ft. Find the capacitive reactance to neutral in ohm-miles and
the charging current per mile per phase and per conductor at 138 kV.
CHAPTER
FIVE
CURRENT AND VOLTAGE RELATIONS
ON A TRANSMISSION LINE
We have examined the parameters of a transmission line and are ready to

consider the line as an element of a power system. Figure 5.1 shows a 500-kV
line having bundled conductors. In overhead lines the conductors are suspended
from the tower and insulated from it and from each other by insulators, the
number of which is determined by the voltage of the line. Each insulator string
in Fig. 5.1 has 22 insulators. The two shorter arms above the phase conductors
support wires usually made of steel. These wires being of much smaller diameter
than the phase conductors are not visible in the picture, but they are electrically
connected to the tower and are therefore at ground potential. These wires are
referred to as ground wires and shield the phase conductors from lightning
strokes.
A very important problem in the design and operation of a power system is
the maintenance of the voltage within specified limits at various points in the
system. In this chapter we shall develop formulas by which we can calculate the
voltage, current, and power at any point on a transmission line provided we
know these values at one point, usually at one end of the line. The chapter also
provides an introduction to the study of transients on lossless lines in order to
indicate how problems arise due to surges caused by lightning and switching.
The purpose of this chapter, however, is not merely to develop the pertinent
equations; it also provides an opportunity to understand the effects of the par¬
ameters of the line on bus voltages and the flow of power. In this way we can see
the importance of the design of the line and better understand the discussions to
come in later chapters.
88
CURRENT AND VOLTAGE RELATIONS ON A TRANSMISSION LINE 89
Figure 5.1 A 500-kV transmission line. Conductors are 76/19 ACSR with aluminum cross section of
2,515,000 cmil. Spacing between phases is 30 ft 3 in and the two conductors per bundle are 18 in
apart. (Courtesy Carolina Power and Light Company.)
In the modern power system data from all over the system are being fed
continuously into on-line computers for control purposes and for information.
Load-flow studies performed by a computer readily supply answers to questions
concerning the effect of switching lines into and out of the system or of changes
in line parameters. Equations derived in this chapter remain important, however,
in developing an overall understanding of what is occurring on a system and in
calculating efficiency of transmission, losses, and limits of power flow over a line
for both steady-state and transient conditions.
5.1 REPRESENTATION OF LINES
The general equations relating voltage and current on a transmission line recog¬
nize the fact that all four of the parameters of a transmission line discussed in the
two preceding chapters are uniformly distributed along the line. We shall derive
these general equations later but first we shall use lumped parameters which give
good accuracy for short lines and for lines of medium length. If an overhead line
is classified as short, shunt capacitance is so small that it can be omitted entirely
with little loss of accuracy, and we need to consider only the series resistance R
and the series inductance L for the total length of the line. Figure 5.2 shows a
Y-connected generator supplying a balanced-Y load through a short transmis¬
sion line. R and L are shown as concentrated, or lumped, parameters. It makes
no difference, as far as measurements at the ends of the line are concerned,
whether the parameters are lumped or uniformly distributed if the shunt admit¬
tance is neglected since the current is the same throughout the line in that case.
The generator is represented by an impedance connected in series with the
generated emf of each phase.
A medium-length line can be represented sufficiently well by R and L as
lumped parameters, as shown in Fig. 5.3, with half the capacitance to neutral of
the line lumped at each end of the equivalent circuit. Shunt conductance G, as
mentioned previously, is usually neglected in overhead power transmission lines
when calculating voltage and current.
Insofar as the handling of capacitance is concerned, open-wire 60-Hz lines
R L
Figure 5.2 Generator supplying a balanced-Y load through a transmission line where the resistance
R and inductance L are values for the entire length of the line. Line capacitance is omitted.
Figure 5.3 Single-phase equivalent of the circuit of Fig. 5.2 with the addition of capacitance
to neutral for the entire length of the line divided between the two ends of the line.
less than about 80 km (50 mi) long are short lines. Medium-length lines are
roughly between 80 km (50 mi) and 240 km (150 mi) long. Lines longer than
240 km (150 mi) require calculations in terms of distributed constants if a high
degree of accuracy is required, although for some purposes a lumped-parameter
representation can be used for lines up to 320 km (200 mi) long.
Normally, transmission lines are operated with balanced three-phase loads.
Although the lines are not spaced equilaterally and not transposed, the resulting
dissymmetry is slight and the phases are considered to be balanced.
In order to distinguish between the total series impedance of a line and the
series impedance per unit length, the following nomenclature is adopted:
z = series impedance per unit length per phase

y = shunt admittance per unit length per phase to neutral
/ = length of line
Z = zl = total series impedance per phase
Y = yl = total shunt admittance per phase to neutral
5.2 THE SHORT TRANSMISSION LINE
The equivalent circuit of a short transmission line is shown in Fig. 5.4, where Is
and IR are the sending- and receiving-end currents and Vs ahd VR are the
sending- and receiving-end line-to-neutral voltages.
Figure 5.4 Equivalent circuit of a short transmission line where the resistance R and inductance L
are values for the entire length of the line.
The circuit is solved as a simple series ac circuit. Since there are no-shunt
arms, the current is the same at the sending and receiving ends of the line and
IS = IR (5-1)
The voltage at the sending end is
l's=Vr + IrZ (5.2)
where Z is zl, the total series impedance of the line.

The effect of the variation of the power factor of the load on the voltage
regulation of a line is most easily understood for the short line and therefore will
be considered at this time. Voltage regulation of a transmission line is the rise in
voltage at the receiving end, expressed in percent of full-load voltage when full
load at a specified power factor is removed while the sending-end voltage is held
constant. In the form of an equation
Percent regulation - ^ ^ FL^ x 100 (5.3)

I 'R, FL I
where | VR NL | is the magnitude of receiving-end voltage at no load and | FR FL |
is the magnitude of receiving-end voltage at full load with | Vs | constant. After
the load on a short transmission line, represented by the circuit of Fig. 5.4, is
removed, the voltage at the receiving end is equal to the voltage at the sending
end. In Fig. 5.4, with the load connected, the receiving-end voltage is designated
by VR, and | VR | = | VR FL |. The sending-end voltage is Vs, and | Vs \ = | VRy NL |.
The phasor diagrams of Fig. 5.5 are drawn for the same magnitudes of receiving-
end voltage and current and show that a larger value of sending-end voltage is
required to maintain a given receiving-end voltage when the receiving-end cur¬
rent is lagging the voltage than when the same current and voltage are in phase.
A still smaller sending-end voltage is required to maintain the given receiving-
end voltage when the receiving-end current leads the voltage. The voltage drop
is the same in the series impedance of the line in all cases, but because of the
different power factors the voltage drop is added to the receiving-end voltage at
a different angle in each case. The regulation is greatest for lagging power factors
and least, or even negative, for leading power factors. The inductive reactance of
a transmission line is larger than the resistance, and the principle of regulation
illustrated in Fig. 5.5 is true for any load supplied by a predominantly inductive
Figure 5.5 Phasor diagrams of a short transmission line. All diagrams are drawn for the same
magnitudes of and IR.
circuit. The magnitudes of the voltage drops IR R and IR XL for a short line have
been exaggerated with respect to VR in drawing the phasor diagrams in order to
illustrate the point more clearly. The relation between power factor and regula¬
tion for longer lines is similar to that for short lines but is not visualized so
easily.
53 THE MEDIUM-LENGTH LINE
The shunt admittance, usually pure capacitance, is included in the calculations

for a line of medium length. If the total shunt admittance of the line is divided
into two equal parts placed at the sending and receiving ends of the line, the
circuit is called a nominal n. We shall refer to Fig. 5.6 to derive equations. To
obtain an expression for Vs we note that the current in the capacitance at the
receiving end is VR Y/2 and the current in the series arm is IR + VR Y/2. Then
VS=(v«j +I„)z+VR (5.4)
Ks=(f +lk + Z,R (5.5)
To derive Is we note that the current in the shunt capacitance at the sending end
is Vs Y/2, which added to the current in the series arm gives
h=Vs\ +^1 +/„ (5.6)
Substituting Vs, as given by Eq. (5.5), in Eq. (5.6) gives
Wlr(1+f) + (f + i),. (5,)
Corresponding equations can be derived for the nominal T which has all of the
shunt admittance of the line lumped in the shunt arm of the T and the series
impedance divided equally between the two series arms.
Equations (5.5) and (5.7) may be expressed in the general form
Vs=AVr + B1r (5.8)
h = CVR + DI„ (5.9)
where
(5.10)
These ABCD constants are sometimes called the generalized circuit constants of
the transmission line. In general they are complex numbers. A and D are dimen¬
sionless and equal each other if the line is the same when viewed from either end.
The dimensions of B and C are ohms and mhos, respectively. The constants
apply to linear, passive, and bilateral four-terminal networks having two pairs of
terminals.
A physical meaning is easily assigned to the constants. By letting IR be zero
in Eq. (5.8) we see that A is the ratio VS/VR at no load. Similarly, B is the ratio
VS/IR when the receiving end is short-circuited. The constant A is useful in
computing regulation. If VR FL is the receiving-end voltage at full load for a
sending-end voltage of Vs, Eq. (5.3) becomes
Percent regulation = -1 AfLl x 100 (5-11)

I ’R, Fl|
ABCD constants are not widely used. They are introduced here because they
simplify working with the equations. Appendix Table A.6 lists ABCD constants
for various networks and combinations of networks.
5.4 THE LONG TRANSMISSION LINE:

SOLUTION OF THE DIFFERENTIAL EQUATIONS
The exact solution of any transmission line and the one required for a high
degree of accuracy in calculating 60-Hz lines more than approximately 150 mi
long must consider the fact that the parameters of the lines are not lumped but
are distributed uniformly throughout the length of the line.
Figure 5.7 shows one phase and the neutral connection of a three-phase line.
Lumped parameters are not shown because we are ready to consider the solu¬
tion of the line with the impedance and admittance uniformly distributed. The
same diagram also represents a single-phase line if the series impedance of the
line is the loop series impedance of the single-phase line instead of the series
impedance per phase of the three-phase line and if the shunt admittance is
the line-to-line shunt admittance of the single-phase line instead of the shunt
admittance to neutral of the three-phase line.
Figure 5.7 Schematic diagram of a transmission line showing one phase and the neutral return.
Nomenclature for the line and the elemental length are indicated.
Let us consider a very small element in the line and calculate the difference
in voltage and the difference in current between the ends of the element. We shall
let x be the distance measured from the receiving end of the line to the small
element of line, and we shall let the length of the element be Ax. Then z Ax is the
series impedance of the elemental length of the line, and y Ax is its shunt admit¬
tance. The voltage to neutral at the end of the element toward the load is V, and
V is the complex expression of the rms voltage, whose magnitude and phase vary
with distance along the line. The voltage at the end of the element toward the
generator is V + AV. The rise in voltage over the elemental length of line in the
direction of increasing x is AV, which is the voltage at the end toward
the generator minus the voltage at the end toward the load. The rise in voltage
in the direction of increasing x is also the product of the current in the element
flowing opposite to the direction of increasing x and the impedance of
the element, or Iz Ax. Thus
AV = Iz Ax (5.12)
or
(5.13)
and as Ax ^ 0 the limit of the above ratio becomes
dV
~ = Iz 1/ (5.14)
Similarly, the current flowing out of the element toward the load is /. The
magnitude and phase of the current / vary with distance along the line because
of the distributed shunt admittance along the line. The current flowing into the
element from the generator is / + A/. The current entering the element from
the generator end is higher than the current flowing away from the element in
the direction of the load by the amount A/. This difference in current is the
current Vy Ax flowing in the shunt admittance of the element. Thus
A / = Vy Ax
and pursuing steps similar to those of Eqs. (5.12) and (5.13) we obtain
dl
= Vy (5.15)
dx
Let us differentiate Eqs. (5.14) and (5.15) with respect to x, and obtain
d2V dl
= z- (5.16)
dx2 dx
and
d2I dV
= y~i7.
(5.17)
dx2 dx
If we substitute the values of dl/dx and dV/dx from Eqs. (5.15) and (5.14) in
Eqs. (5.16) and (5.17), respectively, we obtain
d2V
= yzV (5.18)
dx2
and
(5.19)
Now we have an equation (5.18) in which the only variables are V and x, and
another equation (5.19) in which the only variables are I and x. The solutions of
Eqs. (5.18) and (5.19) for V and /, respectively, must be expressions which when
differentiated twice with respect to x yield the original expression times the
constant yz. For instance, the solution for V when differentiated twice with
respect to x must yield yzV. This suggests an exponential form of solution.
Assume that the solution of Eq. (5.18) is+
V = Ax exp (yfyz x) 4- A2 exp (-sfyz x) (5.20)
Taking the second derivative of V with respect to x in Eq. (5.20) yields
d2V
= yz[Ax exp {JJz x) + A2 exp (-y/yz x)] (5.21)
dx‘
which is yz times the assumed solution for V. Therefore, Eq. (5.20) is the solution
of Eq. (5.18). When we substitute in Eq. (5.14) the value for V given by
Eq. (5.20), we obtain,
^ _ eX^ •*) yj~z[y^2 eX^ ^ — *) (5.22)
The constants At and A2 can be evaluated by using the conditions at the
t The term exp (s/yz x) in Eq. (5.20) and similar equatons is equivalent to e raised to the power
receiving end of the line, namely when x = 0, V = VR and I = IR. Substitution of

these values in Eqs. (5.20) and (5.22) yields
VR — Ax + A 2 and /» = (î “ Z2)

R~^[y
Substituting Zc = yfzjy and solving for Ax give
VR + IRZC VR~IrZc
Ax = and A2 —
Then, substituting the values found for Ax and A2 in Eqs. (5.20) and (5.22) and
letting y = y/yz, we obtain
Vr + IrZc ^yx
VR-1RZC yx
V = + (5.23)
Vr/Zc + IR £yx _
VR/ZC - 7# yx
/ - -
(5.24)
where Zc = xfzjy and is called the characteristic impedance of the line, and
y = sfyz and is called the propagation constant.
Equations (5.23) and (5.24) give the rms values of V and / and their phase
angles at any specified point along the line in terms of the distance x from the
receiving end to the specified point, provided VR, IR, and the parameters of the
line are known.

INTERPRETATION OF THE EQUATIONS
Both y and Zc are complex quantities. The real part of the propagation constant
y is called the attenuation constant a and is measured in nepers per unit length.
The quadrature part of y is called the phase constant /? and is measured in
radians per unit length. Thus
y = & + jfi (5.25)
and Eqs. (5.23) and (5.24) become
y _ Vr + IR Zc E0LXEjpx _|_ Vr !r Zc e~axE-j(ix

(5.26)
2 2
and
Vr/Zc + IR VrIZc ~ IR e-ax£- jfix
l = £«- (5.27)
The properties of exx and ejpx help to explain the variation of the phasor values
of voltage and current as a function of distance along the line. The term eax
changes in magnitude as x changes, but eJfix, which is identical to cos /?x +
j sin fix, always has a magnitude of 1 and causes a shift in phase of rad per unit
length of line.
The first term in Eq. (5.26), [(VR + IR Zc)/2]eaV/i*, increases in magnitude
and advances in phase as distance from the receiving end increases. Conversely,
as progress along the line from the sending end toward the receiving end is
considered, the term diminishes in magnitude and is retarded in phase. This is
the characteristic of a traveling wave and is similar to the behavior of a wave in
water, which varies in magnitude with time at any point, whereas its phase is
retarded and its maximum value diminishes with distance from the origin. The
variation in instantaneous value is not expressed in the term but is understood
since 1^ and IR are phasors. The first term in Eq. (5.26) is called the incident
voltage.
The second term in Eq. (5.26), [(F* — IR Zc)/2]e~xxe~jPx, diminishes in mag¬
nitude and is retarded in phase from the receiving end toward the sending end. It
is called the reflected voltage. At any point along the line the voltage is the sum
of the component incident and reflected voltages at that point.
Since the equation for current is similar to the equation for voltage, the
current may be considered to be composed of incident and reflected currents.
If a line is terminated in its characteristic impedance Zc, the receiving-end
voltage VR is equal to IR Zc and there is no reflected wave of either voltage or
current, as may be seen by substituting IRZC for VR in Eqs. (5.26) and (5.27). A
line terminated in its characteristic impedance is called a flat line or an infinite
line. The latter term arises from the fact that a line of infinite length cannot have
a reflected wave. Usually power lines are not terminated in their characteristic
impedance, but communication lines are frequently so terminated in order to
eliminate the reflected wave. A typical value of Zc is 400 Q for a single-circuit
overhead line and 200 fl for two circuits in parallel. The phase angle of Zc is
usually between 0 and —15°. Bundled-conductor lines have lower values of Zc
since such lines have lower L and higher C than lines with a single conductor per
phase.
In power-system work, characteristic impedance is sometimes called surge
impedance. The term surge impedance, however, is usually reserved for the
special case of a lossless line. If a line is lossless, its resistance and conductance
are zero and the characteristic impedance reduces to yjL/C, a pure resistance.
When dealing with high frequencies or with surges due to lightning, losses are
often neglected and the surge impedance becomes important. Surge-impedance
loading (SIL) of a line is the power delivered by a line to a purely resistive load
equal to its surge impedance. When so loaded, the line supplies a current of
where \VL\ is the line-to-line voltage at the load. Since the load is pure
resistance,
W
or, with | VL | in kilovolts,
MW (5.28)
Power-system engineers sometimes find it convenient to express the power

transmitted by a line in terms of per unit of SIL, that is, as the ratio of the power
transmitted to the surge-impedance loading. For instance, the permissible load¬
ing of a transmission line may be expressed as a fraction of its SIL, and SIL
provides a comparison of load-carrying capabilities of lines.t
A wavelength X is the distance along a line between two points of a wave
which differ in phase by 360°, or In rad. If /? is the phase shift in radians per
mile, the wavelength in miles is
2 = % (5.29)
At a frequency of 60 Hz, a wavelength is approximately 3000 mi. The velocity of

propagation of a wave in miles per second is the product of the wavelength in
miles and the frequency in hertz, or
Velocity = fX (5.30)
If there is no load on a line, IR is equal to zero, and, as determined by

Eqs. (5.26) and (5.27), the incident and reflected voltages are equal in magnitude
and in phase at the receiving end. In this case the incident and reflected currents
are equal in magnitude but 180° out of phase at the receiving end. Thus, the
incident and reflected currents cancel each other at the receiving end of an open
line but not at any other point unless the line is entirely lossless so that the
attenuation a is zero.

HYPERBOLIC FORM OF THE EQUATIONS
The incident and reflected waves of voltage are seldom found when calculating
the voltage of a power line. The reason for discussing the voltage and current of
a line in terms of the incident and reflected components is that such an analysis
is helpful in obtaining a fuller understanding of some of the phenomena of
transmission lines. A more convenient form of the equations for computing
current and voltage of a power line is found by introducing hyperbolic functions.
Hyperbolic functions are defined in exponential form as follows:
sinh 6 = £ £ (5.31)
2
4. E~e
cosh 6 = ——— (5.32)
t See R. D. Dunlop, R. Gutman, and P. P. Marchenko, “Analytical Development of Loadability

Characteristics for EHV and UHV Transmission Lines,” IEEE Trans. PAS, vol. 98, no. 2, 1979,
pp. 606-617.
By rearranging Eqs. (5.23) and (5.24) and substituting hyperbolic functions for
the exponential terms, a new set of equations is found. The new equations, giving
voltage and current anywhere along the line, are
V =VR cosh yx + IRZC sinh yx (5.33)
and
I = IR cosh yx + ~ sinh (5.34)

Zc
Letting x — l to obtain the voltage and current at the sending end, we have
Vs = VR cosh yl + IRZC sinh yl (5.35)
and
y
Is = IR cosh yl + -y sinh yl (5.36)
Zc
From examination of these equations we see that the generalized circuit con¬
stants for a long line are
sinh yl
A = cosh yl C = (5.37)
B = Zc sinh yl D = cosh yl
Equations (5.35) and (5.36) can be solved for VR and IR in terms of Fs and Is
to give
VR = Vs cosh yl — ISZC sinh yl (5.38)
and
y
IR = Is cosh yl — ~ sinh yl (5 39)
\
For balanced three-phase lines the current in the above equations is the line
current, and the voltage is the line-to-neutral voltage, that is, the line voltage
divided by ^/3. In order to solve the equations, the hyperbolic functions must be
evaluated. Since yl is usually complex, the hyperbolic functions are also complex
and cannot be found directly from ordinary tables or electronic calculators.
Before the widespread use of the digital computer, various charts, some of them
especially adapted to the values usually encountered in transmission-line calcula¬
tions, were frequently used to evaluate hyperbolic functions of complex argu¬
ments. Now the digital computer provides the usual means of incorporating such
functions into our calculations.
For solving an occasional problem without resorting to a computer or
charts there are several choices. The following equations give the expansions of
hyperbolic sines and cosines of complex arguments in terms of circular and
hyperbolic functions of real arguments:
cosh (a/ + jfil) = cosh a/ cos pi + j sinh a/ sin /1/ (5.40)
sinh (al + jpi) = sinh a/ cos /?/ + j cosh a/ sin pi (5-41)
Equations (5.40) and (5.41) make possible the computation of hyperbolic func¬
tions of complex arguments. The correct mathematical unit for pi is the radian,
and the radian is the unit found for (11 by computing the quadrature component
of yl. Equations (5.40) and (5.41) can be verified by substituting in them the
exponential forms of the hyperbolic functions and the similar exponential forms
of the circular functions.
Another convenient method of evaluating a hyperbolic function is to expand
it in a power series. Expansion by Maclaurin’s series yields
q2 q4 q6
cosh 0 = 1 + — + — + — + "• (5.42)
and
q3 q5 al
sinh0 = 9 + - +- + - +- (5.43)
The series converge rapidly for the values of yl usually found for power lines, and
sufficient accuracy may be found by evaluating only the first few terms.
A third method of evaluating complex hyperbolic functions is suggested by
Eqs. (5.31) and (5.32). Substituting a + jfi for 6, we obtain
£ag/p + g “e
cosh (a: + jfi) - = |(ea /£_ + e~a (5.44)
and
sinh (a + j(3) - --- = k2{ea /£ - e~a /-jS) (5.45)
Example 5.1 A single-circuit 60-Hz transmission line is 370 km (230 mi)

long. The conductors are Rook with flat horizontal spacing and 7.25 m
(23.8 ft) between conductors. The load on the line is 125 MW at 215 kV
with 100% power factor. Find the voltage current, and power at the sending
end and the voltage regulation of the line. Also determine the wavelength and
velocity of propagation of the line.
Solution Feet and miles rather than meters and kilometers are chosen for
the calculations in order to use Tables A.1-A.3.
Deq = ^23.8 x 23.8 x 47.6 £ 30.0 ft

and from the tables for Rook

X Xo_ .
z = 0.1603 + 7(0.415 + 0.4127) = 0.8431/79.04° fi/mi
y = /[l/(0.0950 + 0.1009)] x 10“ 6 = 5.105 x KT6/90° U/mi
/79.04° + 90°
yl = l = 230^0.8431 x 5.105 x 10'
= 0.4772/84.52° = 0.0456 + /0.4750
0.8431 /79.04° - 90
= 406.4/-5.48° Q
z* = y 5.105 x 10 -6
215,000
VR =- =— = 124,130/(F V to neutral
1
V3
125,000,000
ID = = 335.7/0_° A
R ■sfe x 215,000
From Eqs. (5.40) and (5.41)
cosh yl = cosh 0.0456 cos 0.475 + j sinh 0.0456 sin 0.475+

= 1.0010 x 0.8893 + 70.0456 x 0.4573
= 0.8902 + 70.0209 = 0.8904/1,34°
sinh yl = sinh 0.0456 cos 0.475 + j cosh 0.0456 sin 0.475
= 0.0456 x 0.8893 + 71.0010 x 0.4573
= 0.0405 + 70.4578 = 0.4596/84.94°
Then from Eq. (5.35)

V^ ceo k *7 )- fv +k c, Aam- k VI ^
Vs = 124,130 X 0.8904/1.34° + 335.7 x 406.4 /-5.48°'x 0.4596/84,94°
= 110,495 + 72,585 + 11,480 + 761,642
= 137,851/27.77 V
and from Eq. (5.36)

In. X (L 124 130 if L
Is = 335.7 x 0.8904/1,34° + --:^ . x 0.4596/84.94°
406.4/-5.48 -
= 298.83 + 76.99 - 1.03 + 7140.33
= 332.27/26.33°
t 0.475 rad = 27.2°

At the sending end
Line voltage = ^3 x 137.85 = 238.8 kV
Line current = 332.3 A
Power factor = cos (27.78° - 26.33°) = 0.9997 s 1.0
Power = ^3 x 238.8 x 332.3 x 1.0 = 137,440 kW

From Eq. (5.35) we see that at no load (IR = 0)
Vs
VR =
cosh yl
So the voltage regulation is
137.85/0.8904 - 124.13
100 = 24.7%
124.13
The wavelength and velocity of propagation are computed as follows:
0.4750
= 0.002065 rad/mi
230
. 2n 2n
A = 7=OS02065 = 3043 ™
Velocity =fX = 60 x 3043 = 182,580 mi/s
We note particularly in this example that in the equations for Vs and Is

the value of voltage must be expressed in volts and must be the line-to-
neutral voltage.
Example 5.2 Solve for the sending-end voltage and current found in Exam¬
ple 5.1 using per-unit calculations.
Solution We choose a base of 125 MV A, 215 kV to achieve the simplest

per-unit values and compute base impedance and base current as follows:
(215)2
Base impedance = = 370 Q
125
125,000
Base current = = 335.7 A
s/3 x 215
So
406.4/-5.48c
Zc = 1.098 /-5.48° per unit
370
215 _ 215/^3
1.0 per unit
215 “215/71
For use in Eq. (5.35) we chose VR as the reference voltage. So
VR = 1.0/0° per unit (as a line-to-neutral voltage)
and since the load is at unity power factor
337.5/0°
IR — = l.o/cr
337.5
If the power factor had been less than 100%, IR would have been greater
than 1.0 and would have been at an angle determined by the power factor.
By Eq. (5.35)
Vs = 1.0 x 0.8904 + 1.0 x 1.098 /-5.48° x 0.4596/84.94°
= 0.8902 + ;'0.0208 + 0.0923 + y'0.4961
- 1.1102/27.75° per unit
and by Eq. (5.36)
Is = 1.0 x 0.8904/1.34° + { x 0 4596/84.94^
= 0.8902 + yO.0208 - 0.0031 + jO.4186
= 0.990/26.35° per unit
At the sending end

Line voltage = 1.1102 x 215 = 238.7 V
Line current = 0.990 x 335.7 — 332.3 A
Note that we multiply line-to-line voltage base by the per-unit magnitude of

the voltage to find the line-to-line voltage magnitude. We could have mul¬
tiplied line-to-neutral voltage base by the per-unit voltage to find line-to-
neutral voltage magnitude. The factor ^/3 does not enter the calculations
after we have expressed all quantities in per unit.
5.7 THE EQUIVALENT CIRCUIT OF A LONG LINE
The nominal-T and nominal-7i circuits do not represent a transmission line

exactly because they do not account for the parameters of the line being uni¬
formly distributed. The discrepancy between the nominal T and n and the actual
line becomes larger as the length of line increases. It is possible, however, to find
the equivalent circuit of a long transmission line and to represent the line ac¬
curately, insofar as measurements at the ends of the line are concerned, by a
network of lumped parameters. Let us assume that a n circuit similar to that of
Fig. 5.6 is the equivalent circuit of a long line, but let us call the series arm of our
equivalent-;! circuit Z' and the shunt arms Y'/2 to distinguish them from the
arms of the nominal-^: circuit. Equation (5.5) gives the sending-end voltage of a
symmetrical-^ circuit in terms of its series and shunt arms and the voltage and
current at the receiving end. By substituting Z' and Y'/2 for Z and Y/2 in
Eq. (5.5), we obtain the sending-end voltage of our equivalent circuit in terms of
its series and shunt arms and the voltage and current at the receiving end:
IZ'Y' \
Vs=[~2T + 1)F« + Z7« (5-46)
For our circuit to be equivalent to the long transmission line, the coefficients of
VR and IR in Eq. (5.46) must be identical, respectively, to the coefficients of VR
and IR in Eq. (5.35). Equating the coefficients of IR in the two equations yields
Z' = Zc sinh yl (5.47)
sinh yl
Z' = sinh yl = zl
/
sinh yl
z' = z (5.48)
yl
where Z is equal to zl, the total series impedance of the line. The term (sinh yl)/yl
is the factor by which the series impedance of the nominal n must be multiplied
to convert the nominal n to the equivalent n. For small values of yl, sinh yl and
yl are almost identical, and this fact shows that the nominal n represents the
medium-length transmission line quite accurately, insofar as the series arm is
concerned.
To investigate the shunt arms of the equivalent-^: circuit, we equate the
coefficients of VR in Eqs. (5.35) and (5.46) and obtain
+ 1 = cosh yl (5.49)
Substituting Zc sinh yl for Z' gives
Y’ZC sinh yl
+ 1 = cosh yl (5.50)
Y~
and
Y' 1 cosh yl — 1
(5.51)
2 Zc sinh yl
Another form of the expression for the shunt admittance of the equivalent circuit
can be found by substituting in Eq. (5.51) the identity
, yl cosh yl - 1
(5.52)
,a"h 2 = sinh yl
The identity can be verified by substituting the exponential forms of Eqs. (5.31)
Z' = Zc sinh 71 =
Y_'
2
Y tanh 11/2
Figure 5.8 Equivalent-^ circuit of
a transmission line. I 2 11/2
and (5.32) for the hyperbolic functions and by recalling that tanh 9 =
sinh 0/cosh 9. Now
(5.53)
r Y tanh (yl/2)
(5.54)
~2~2 yl/2.
where Y is equal to yl, the total shunt admittance of the line. Equation (5.54)
shows the correction factor used to convert the admittance of the shunt arms of
the nominal n to that of the equivalent n. Since tanh {yl/2) and yl/2 are very
nearly equal for small values of yl, the nominal n represents the medium-length
transmission line quite accurately, for we have seen previously that the correc¬
tion factor for the series arm is negligible for medium-length lines. The
equivalent-7r circuit is shown in Fig. 5.8. An equivalent-T circuit can also be
found for a transmission line.
Example 5.3 Find the equivalent-7r circuit for the line described in Exam¬
ple 5.1 and compare it with the nominal n.
Solution Since sinh yl and cosh yl are already known from Example 5.1,
Eqs. (5.47) and (5.51) will be used.
Z' = 406.4 /—5.48° x 0.4596 784,94° = 186.78 /79.46° H in series arm
r _ 0.8902 + yQ.0208 - 1 _ 0.1118 /169.27°

2 ~~ 186.78 /79.46° ~~ 186.78 /79.460
= 0.000599 /89.81° ^ in each shunt arm
The nominal-7r circuit has a series impedance of
Z = 230 x 0.8431 /79.04° = 193.9 /79.04°
and equal shunt arms of

Y 5.105 x 10-6 /90°
x 230 = 0.000587 /90! O
2 2
For this line the impedance of the series arm of the nominal n exceeds
that of the equivalent n by 3.8%. The conductance of the shunt arms of the
nominal n is 2.0% less than that of the equivalent n. So we conclude that the
nominal n may represent long lines sufficiently well if a high degree of
accuracy is not required.
5.8 POWER FLOW THROUGH A TRANSMISSION LINE
Although power flow at any point along a transmission line can always be found
if the voltage, current, and power factor are known or can be calculated, very
interesting equations for power can be derived in terms of A BCD constants. Of
course, the equations apply to any two-terminal-pair network. Repeating
Eq. (5.8) and solving for the receiving-end current IR yields
Is — AFfl + BIr (5.55)

_Vs~AVr
(5.56)
R B
Letting
A = MlZs. b=\b\h
Vr=\vr l/r vs= | vs | [d

we obtain
(5.57)
,R |B| |B| te—L
Then the complex power VR /£ at the receiving end is
p +i0 - lK>NK«l
Pr+iQr |B|
,»
IS_S
* \a\-\vr\2
|fl| l»_* (5.58)
and real and reactive power at the receiving end are
Pr= COS (/J - i) - cos (^ - a) (5.59)
Qr * |'J| ' sm(p S) 1 sin (ft a) (5.60)
Noting that the expression for complex power PR + jQR is shown by

Eq. (5.58) to be the resultant of combining two phasors expressed in polar form,
we can plot these phasors in the complex plane whose horizontal and vertical
coordinates are in power units (watts and vars). Figure 5.9 shows the two com¬
plex quantities and their difference as expressed by Eq. (5.58). Figure 5.10 shows
the same phasors with the origin of the coordinate axes shifted. This figure is a
power diagram with the resultant whose magnitude is |PR + jQR |, or | FR |- |/R |,
V3T -%.*
Figure 5.9 Phasors of Eq. 5.58 plotted in the Figure 5.10 Power diagram obtained by
complex plane, with magnitudes and angles shifting the origin of the coordinate axes of
as indicated. Fig. 5.9.
at an angle 0R with the horizontal axis. As expected, the real and imaginary
components of | PR + jQR | are
Pr=\Vr\-\Ir\cosOr (5.61)
and
Qr = |^rM/k| sin (5.62)

where 0R is the phase angle by which VR leads IR, as discussed in Chap. 2. The
sign of Q is consistent with the convention which assigns positive values to Q
when current is lagging the voltage.
Now let us determine some points on the power diagram of Fig. 5.10 for
various loads with fixed values of | Vs | and | VR |. First, we notice that the
position of point n is not dependent on the current IR and will not change so
long as | VR | is constant. We note further that the distance from point n to
point k is constant for fixed values of | Fs | and 11^). Therefore, as the distance
0 to k changes with changing load, the point k, since it must remain at a constant
distance from the fixed point n, is constrained to move in a circle whose center
is at n. Any change in PR will require a change in QR to keep k on the circle.
If a different value of | | is held constant for the same value of | VR |, the
location of point n is unchanged but a new circle of radius nk is found.
In Eqs. (5.50) to (5.62) |FS| and |I^| are line-to-neutral voltages and
coordinates in Fig. 5.10 are watts and vars per phase. However, if 11^| and | VR \
are line-to-line voltages each distance in Fig. 5.10 is increased by a factor of 3
and the coordinates on the diagram are total three-phase watts and vars. If the
voltages are kilovolts the coordinates are megawatts and megavars.
If the receiving-end voltage is held constant and receiving-end circles are

drawn for different values of sending-end voltage, the resulting circles are con¬
centric because the location of the center of the receiving-end power circles is
independent of the sending-end voltage. A family of receiving-end circles is
shown in Fig. 5.11 for a constant receiving-end voltage. The load line marked on
Fig. 5.11 is convenient if the load changes in magnitude while its power factor
remains constant. The angle between the load line through the origin and the
horizontal axis is the angle whose cosine is the power factor of the load. The
load line of Fig. 5.11 is drawn for lagging loads since all the points on the line
are in the first quadrant and have positive vars.
Since the advent of digital computers circle diagrams have been of little
practical use. They have been introduced here to illustrate some concepts of
transmission-line operation. For instance, examination of Fig. 5.10 shows that
there is a limit to the power that can be transmitted to the receiving end of the
line for specified magnitudes of sending- and receiving-end voltages. An increase
in power delivered means that the point k will move along the circle until the
angle P — S is zero; that is, more power will be delivered until d equals ft.
Further increases in 5 result in less power received. The maximum power is
H-l^l cos (p — a) (5.63)

PR, max
IBI
The load must draw a large leading current to achieve the condition of maxi¬
mum power received.
In Fig. 5.11 the length of the vertical line from point a at the intersection of
the load line and the | VS4. | circle to point b on the | Vs3 \ circle is the amount of
Figure 5.11 Receiving-end power circles for vari¬

ous values of | Ks | and a constant 1|. IVy constant
negative reactive power that must be drawn by capacitors added in parallelwith

the load to maintain constant | VR | when the sending-end voltage is reduced
from | VS41 to | FS31. Addition of slightly more capacitive kilovars will result in a
combined load of unity power factor and a further reduction in | Vs | for the
same | VR |. Of course, this analysis means that a constant | Vs | would result in
higher | VR | capacitors are added in parallel with the lagging load.
One of many results made apparent by studying the circle diagram of
Fig. 5.11 is the variation of sending-end voltage to maintain constant receiving-
end voltage for different values of real and reactive power received. For instance,
for a constant dR at the load the coordinates of the intersection of the load line
with a circle of constant sending-end voltage are the P and Q of the load for that
value of | Vs | and the | VR | for which the diagram is drawn.
5.9 REACTIVE COMPENSATION OF

TRANSMISSION LINES
The performance of transmission lines, especially those of medium length and

longer, can be improved by reactive compensation of a series or parallel type.
Series compensation consists of a capacitor bank placed in series with each
phase conductor of the line. Shunt compensation refers to the placement of
inductors from each line to neutral to reduce partially or completely the shunt
susceptance of a high-voltage line, which is particularly important at light loads
when the voltage at the receiving end may otherwise become very high.
Series compensation reduces the series impedance of the line, which is the
principal cause of voltage drop and the most important factor in determining the
maximum power which the line can transmit. In order to understand the effect of
series impedance Z on maximum power transmission we examine Eq. (5.63) and
see that maximum power transmitted is dependent upon the reciprocal of the
generalized circuit constant B which for the nominal n equals Z and for the
equivalent n equals Z (sinh yl)/yl. Because the A, C, and D constants are func¬
tions of Z they will also change in value, but these changes will be small in
comparison to the change in B.
The desired reactance of the capacitor bank can be determined by compen¬
sating for a specific amount of the total inductive reactance of the line. This leads
to the term “compensation factor” which is defined by XC/XL where Xc is the
capacitive reactance of the series capacitor bank per phase and XL is the total
inductive reactance of the line per phase.
When the nominal-7r circuit is used to represent the line and capacitor bank,
the physical location of the capacitor bank along the line is not taken into
account. If only the sending- and receiving-end conditions of the line are of
interest, this will not create any significant error. However, when the operating
conditions along the line are of interest, the physical location of the capacitor
bank must be taken into account. This can be accomplished most easily by
determining the ABCD constants of the portions of line on each side of the
capacitor bank and representing the capacitor bank by its ABCD constants. The
equivalent constants of the series combination of line-capacitor-line can then be

determined by applying the equations found in Table A.6 in the Appendix.
In the southwestern part of the United States series compensation is
especially important because large generating plants are located hundreds of
miles from load centers and large amounts of power must be transmitted over
long distances. The lower voltage drop in the line with series compensation is an
additional advantage. Series capacitors are also useful in balancing the voltage
drop of two parallel lines.
Example 5.4 In order to show the relative changes in the B constant with
respect to the change of the A, C, and D constants of a line when series
compensation is applied, find the constants for the line of Example 5.1 un¬
compensated and for a series compensation factor of 70%.
Solution The equivalent-^ circuit and quantities found in Examples 5.1

and 5.3 can be used with Eqs. (5.37) to find, for the uncompensated line
A = D = cosh yl = 0.8904^.34°
B = Z' = 186.78/79.46° Q
^ sinh yl 0.4596/^4.94°
Zc ~ 406.4/-5.48°
= 0.001131/90.42° U
The series compensation alters only the series arm of the equivalent-?:
circuit. The new series arm impedance is also the generalized constant B. So
B = 186.78/79.46° - j'0.7 x 230(0.415 + 0.4127)

- 34.17 + j50.38 = 60.88/55.85° Q
and by Eqs. (5.10)
A = 60.88/55.85° x 0.000599/89.81° + 1 = 0.970/1.24°

C = 2 x 0.000599/89.81° + 60.88/55.85°(O.OOQ599/89.81 )2
= 0.001180/90.41° 13
The example shows that compensation has reduced the constant B to

about one-third of its value for the uncompensated line without affecting the
A and C constants appreciably. Thus, maximum power which can be trans¬
mitted is increased by about 300%.
When a transmission line, with or without series compensation, has the

desired load transmission capability, attention is turned to operation under light
loads or at no load. Charging current is an important factor to be considered
and should not be allowed to exceed the rated full-load current of the line.
Equation (4.25) shows us that the charging current is usually defined as

Bc\V\ if Bc is the total capacitive susceptance of the line and | V | is the
rated voltage to neutral. As noted following Eq. (4.25) this calculation is not
an exact determination of charging current because of the variation of \V\
along the line. If we connect inductors from line to neutral at various points
along the line so that the total inductive susceptance is BL, the charging current
becomes
Lh„ = (Bc-BL)|P|
= Bcln(l-|) (564)
We recognize that the charging current is reduced by the term in parentheses.

The shunt compensation factor is BJBC ■
The other benefit of shunt compensation is the reduction of the
receiving-end voltage of the line which on long high-voltage lines tends to
become too high at no load. In the discussion preceding Eq. (5.11) we noted
that |E}|/|A| equals |Er,nl|- We also have seen that when shunt capacitance
is neglected A equals 1.0. In the medium-length and longer lines, however,
the presence of capacitance reduces A. Thus, the reduction of the shunt
susceptance to a value of (Bc — BL) can limit the rise of the no-load voltage
at the receiving-end of the line if shunt inductors are introduced as load is
removed.
By applying both series and shunt compensation to long transmission
lines it is possible to transmit large amounts of power efficiently and within
the desired voltage constraints. Ideally the series and shunt elements should
be placed at intervals along the line. Series capacitors can be bypassed and
shunt inductors switched off when desirable. As with series compensation
ABCD constants provide a straightforward method of analysis of shunt
compensation.
Example 5.5 Find the voltage regulation of the line of Example 5.1 when a
shunt inductor is connected at the receiving end of the line during no-load
conditions if the reactor compensates for 70% of the total shunt admittance
of the line.
Solution From Example 5.1 the shunt admittance of the line is
y = y‘5.105 x 10“ 6 13/mi

and for the entire line
Bc = J5.105 x 10“ 6 x 230 = 0.001174 U

For 70% compensation
Bl = —jO.l x 0.001174 = 0.000822

We know the ABCD constants of the line from Example 5.4. Table A.6 in the
Appendix tells us that the inductor alone is represented by the generalized

constants
A = D=\ B=0 C = -;0.000822 U
The equation in Table A.6 for combining two networks in series tells us that
for the line and inductor
Aeq = 0.8904/1.34° + 186.78/79.46°(0.000822 /-90°)

= 1.0411/-0,40°
The voltage regulation with the shunt reactor connected at no load

becomes
137.85/1.0411 - 124.13
6.67%
124.13
which is a considerable reduction from the value of 24.7% for the regulation
of the uncompensated line.
5.10 TRANSMISSION LINE TRANSIENTS
The transient overvoltages which occur on a power system are either of external
origin (for example, a lightning discharge) or are generated internally by
switching operations. In general the transients on transmission systems are caused
by any sudden change in the operating condition or configuration of the systems.
Lightning is always a potential hazard to power system equipment, but
switching operations can also cause equipment damage. At voltages up to about
230 kV the insulation level of the lines and equipment is dictated by the need to
protect against lightning. On systems where voltages are above 230 kV but less
than 700 kV switching operations as well as lightning are potentially damaging
to insulation. At voltages above 700 kV switching surges are the main determinant
of the level of insulation.
Underground cables are, of course, immune to direct lightning strokes and
can be protected against transients originating on overhead lines. However, for
economic and technical reasons overhead lines at transmission voltage levels
prevail except under unusual circumstances and for short distances such as
under a river.
Overhead lines can be protected from direct strokes of lightning in most
cases by one or more wires at ground potential strung above the power-line
conductors as mentioned in the description of Fig. 5.1. These protecting wires,
called ground wires, are connected to ground through the transmission towers
supporting the line. The zone of protection is usually considered to be 30° on
each side of vertical beneath a ground wire; that is, the power lines must come
within this 60° sector. The ground wires, rather than the power line, receive the
lightning strokes in most cases.
Figure 5.12 Schematic diagram of an elemental section

of a transmission line showing one phase and neutral
return. Voltage v and current i are functions of both
x and t. The distance x is measured from the sending
end of the line.
Lightning strokes hitting ground wires or power conductors cause an injec¬

tion of current which divides with half the current flowing in one direction and
half in the other. The crest value of current along the struck conductor varies
widely because of the wide variation in the intensity of the strokes. Values of
10,000 A and upward are typical. In the case where a power line receives a direct
stroke the damage to equipment at line terminals is caused by the voltages
between line and ground resulting from the injected charges which travel along
the line as current. These voltages are typically above a million volts. Strokes to
the ground wires can also cause high-voltage surges on the power lines by
electromagnetic induction.
5.11 TRANSIENT ANALYSIS: TRAVELING WAVES
The study of surges on transmission lines regardless of their origin is very

complex and we can consider here only the case of a lossless line.t
A lossless line is a good representation for lines of high frequency where a>L
and coC become very large compared to R and G. For lightning surges on a
power transmission line the study of a lossless line is a simplification that
enables us to understand some of the phenomena without becoming too in¬
volved in complicated theory.
Our approach to the problem is similar to that used earlier for deriving the
steady-state voltage and current relations for the long line with distributed con¬
stants. We shall now measure the distance x along the line from the sending end
(rather than from the receiving end) to the differential element of length Ax
shown in Fig. 5.12. The voltage v and the current i are functions of both x and t
so that we need to use partial derivatives. The series voltage drop along the
elemental length of line is
i(RAx) + (LAx)-^
dt
and we can write
t For further study see A. Greenwood, Electrical Transients in Power Systems, Wiley-Interscience,
New York, 1971.
The negative sign is necessary because v + (dv/dx) Ax must be less than v for
positive values of i and di/dt. Similarly
diA
|gu + C Ax (5.66)
Jt)
We can divide through both Eqs. (5.65) and (5.66) by Ax, and since we are
considering only a lossless line, R and G will equal zero to give
8v _ , di
(5.67)
dx dt
and
— = -r— (5.68)
dx dt
Now we can eliminate i by taking the partial derivative of both terms in
Eq. (5.67) with respect to x and the partial derivative of both terms in Eq. (5.68)
with respect to t. This procedure yields d2i/dxdt in both resulting equations, and
eliminating this second partial derivative of i between the two equations yields
1 d2v d2v
(5.69)
LC dx2 dt2
Equation (5.69) is the so-called traveling-wave equation of a lossless trans¬
mission line. A solution of the equation is a function of (x — vt), and the voltage
is expressed by
V = f(x- Vt) (5.70)
The function is undefined but must be single valued. The constant v must have
the dimensions of meters per second if x is in meters and t is in seconds. We can
verify this solution by substituting this expression for v into Eq. (5.69) to determine
v. First we make the change in variable
u = x — vt (5.71)
and write
v(x, t)=f(u) (5.72)
Then
dv df(u) du
dt du dt
d/(“) (5.73)
du
and
d2v = 2 d2f(u)
(5.74)
dt2 du2
Similarly we obtain
82v d2f(u)
(5.75)
= 8u2
Substituting these second partial derivatives of v into Eq. (5.69) yields
J_ 82f{u) = d2f{u)
(5.76)
LC du2 du2
and we see that Eq. (5.70) is a solution of Eq. (5.69) if
1
v = —,- (5.77)
s/lc
The voltage as expressed by Eq. (5.70) is a wave traveling in the positive x

direction. Figure 5.13 shows a function of (x — vt) which is similar to the shape
of a wave of voltage traveling along a line which has been struck by lightning.
The function is shown for two values of time tt and t2 where t2 > tv An obser¬
ver traveling with the wave and staying at the same point on the wave sees no
change in voltage at that point. To the observer
x — vt = a constant
from which it follows that
dx 1
~r = v = .— m/s
dt yic
for I and C in H/m and F/m, respectively. Thus the voltage wave travels in the
positive x direction with the velocity v.
A function of (x + vt) can also be shown to be a solution of Eq. (5.69) and,
by similar reasoning, can be properly interpreted as a wave traveling in the
negative x direction. The general solution of Eq. (5.69) is thus
v =/l(v - vt) +/2(x + Vt) (5.79)
which is a solution for simultaneous occurrence of forward and backward com¬

ponents on the line. Initial conditions and boundary (terminal) conditions deter¬
mine the particular values for each component.
Figure 5.13 A voltage wave which is a function of (x - vt) is shown for values of t equal to q and t2.
If we express a forward traveling wave, also called an incident wave, as
=fi{x - vt) (5.80)

a wave of current will result from the moving charges and will be expressed by
r=jhcMx-n) <5-8i>
which can be verified by substitution of these values of voltage and current in

Eq. (5.67) and recalling that v is equal to 1 /y/LC.
Similarly for a backward moving wave of voltage where
v~ =fi{x + vt) (5.82)

the corresponding current is
r = -7hcMx+n) (5.83)
From Eqs. (5.80) and (5.81) we note that
(5.84)
and from Eqs. (5.82) and (5.83) that
(5.85)
If we had decided to assume the positive direction of current for i~ to be in the

direction of travel of the backward traveling wave the minus signs would change
to plus signs in Eqs. (5.83) and (5.85). We choose however to keep the positive x
direction as the direction for positive current for both forward and backward
traveling waves.
The ratio of v+ to i+ is called the characteristic impedance Zc of the line. We
have encountered characteristic impedance previously in the steady-state solu¬
tion for the long line where Zc was defined as yfzjy which equals JL/C when R
and G are zero.
5.12 TRANSIENT ANALYSIS: REFLECTIONS
We shall now consider what happens when a voltage is first applied to the
sending end of a transmission line which is terminated in an impedance ZR. For
our very simple treatment we will consider ZR to be a pure resistance. If the
termination is other than a pure resistance we would resort to Laplace trans¬
forms. The transforms of voltage, current, and impedance would be functions of
the Laplace transform variable s.
When the switch is closed applying a voltage to a line a wave of voltage v+
accompanied by a wave of current i+ starts to travel along the line. The radio of
the voltage vR at the end of the line at any time to the current iR at the end of the
line must equal the terminating resistance ZR. Therefore the arrival of v+ and i+
at the receiving end where their values are vR and iR must result in backward
traveling or reflected waves v~ and i~ having values vR and iR at the receiving
end such that
VR _ VR T VR _ 7
— .+ . — ^R
(5.86)
lR lR + lR
where vR and iR are the reflected waves v and i measured at the receiving end.
If we let Zc = yjL/C we find from Eqs. (5.84) and (5.85)
a =$ (5-87)
and
ir —
(5.88)
Then substituting these values of iR and iR in Eq. (5.86) yields
Zr zc
Vr Vr (5.89)
ZR + Zc
The voltage vR at the receiving end is evidently the same function of t as vR (but
with diminished magnitude unless ZR is zero or infinity). The reflection
coefficient pR for voltage at the receiving end of the line is defined as vR/vR so,
for voltage,
Zr Zc
Pr — (5.90)
Zr + Zc
We note from Eqs. (5.87) and (5.88) that
(5.91)
and that therefore the reflection coefficient for current is always the negative of
the reflection coefficient for voltage.
If the line is terminated in its characteristic impedance Zc, we see that the
reflection coefficient for both voltage and current will be zero. There will be no
reflected waves, and the line will behave as though it is infinitely long. Only
when a reflected wave returns to the sending end does the source sense that the
line is not either infinitely long or terminated in Zc.
Termination in a short circuit results in a pR for voltage of — 1. If the
termination is an open circuit, ZR is infinite and pR is found by dividing the
numerator and denominator in Eq. (5.90) by ZR and allowing ZR to approach
infinity to yield pR = 1 in the limit for voltage.
We should note at this point that waves traveling back toward the sending
end will cause new reflections as determined by the reflection coefficient at the
sending end ps. For impedance at the sending end equal to Zs Eq. (5.90)
becomes
zs-zc
Ps = (5.92)
Zs + ze
With sending-end impedances of Zs the value of the initial voltage impressed
across the line will be the source voltage multiplied by ZJ(ZS + Zc). Equation
(5.84) shows that the incident wave of voltage experiences a line impedance of
Zf, and at the instant when the source is connected to the line Zc and Zs in series
act as a voltage divider.
Example 5.6 A dc source of 120 V with negligible resistance is connected

through a switch S to a lossless transmission line having Zc = 30 Q. If the
line is terminated in a resistance of 90 Q, plot vR versus time until t — 5T
where T is the time for a voltage wave to travel the length of the line. The
circuit is shown in Fig. 5.14a.
Solution When S is closed the incident wave of voltage starts to travel

along the line and is expressed by the equation
v = 120C/(vf - x)
where U(vt — x) is the unit step function which equals zero when (vr — x) is
negative and equals unity when (vr — x) is positive. There can be no reflected
wave until the incident wave reaches the end of the line. Impedance to the
incident wave is Zc = 30 Q. Since resistance of the source is zero,
v+ = 120 V
90- 30 1
pR ~ 90 + 30 ” 2
When v+ reaches the end of the line a reflected wave originates of value
iT = (^)120 = 60 V
and so
vR = 120 + 60 = 180 V
When t = 2 T the reflected wave arrives at the sending end where the sending
end reflection coefficient ps is calculated by Eq. (5.92). The line termination
for the reflected wave is Zs, the impedance in series with the source, or zero
in this case. So
0-30
= -1
Ps ~ 0 + 30
and a reflected wave of — 60 V starts toward the receiving end to keep the
sending-end voltage equal to 120 V. This new wave reaches the receiving end
at t = 3T and reflects toward the sending end a wave of
i(_60)= -30 V
and the receiving-end voltage becomes
180 — 60 — 30 = 90 V
An excellent method of keeping track of the various reflections as they

occur is the lattice diagram shown in Fig. 5.14b. Here time is measured
along the vertical axis in intervals of T. On the slant lines are recorded the
values of the incident and reflected waves. In the space between the slant
lines is shown the sum of all the waves above and is the current or voltage
for a point in that area of the chart. For instance, at x equal to three-fourths
of the line length and t = 4.25 T the intersection of the dashed lines through
these points is within the area which indicates the voltage is 90 V.
Figure 5.14c shows the receiving-end voltage plotted against time. The
voltage is approaching its steady-state value of 120 V.
Lattice diagrams for current may also be drawn. We must remember, how¬
ever, that the reflection coefficient for current is always the negative of the
reflection coefficient for voltage.
If the resistance at the end of the line of Example 5.6 is reduced to 10 Cl
as shown in the circuit of Fig. 5.15a, the lattice diagram and plot of voltage are
as shown in Figs. 5.15b and c. The resistance of 10 £2 gives us a negative value
for the reflection coefficient for voltage, which will always occur for resistance
where ZR is less than Zc. As we see by comparing Figs. 5.14 and 5.15 the
negative pR causes the receiving-end voltage to build up gradually to 120 V
while a positive pR causes an initial jump in voltage to a value greater than
that of the voltage originally applied at the sending end.
Reflections do not necessarily arise only at the end of a line. If one line is
joined to a second line of different characteristic impedance, as in the case of an
overhead line connected to an underground cable, a wave incident to the junction
will behave as though the first line is terminated in the Zc of the second line.
However, the part of the incident wave which is not reflected will travel (as a
refracted wave) along the second line at whose termination a reflected wave will
occur. Bifurcations of a line will also cause reflected and refracted waves.
It should now be obvious that a thorough study of transmission-line
transients in general is a complicated problem. We realize, however, that a
voltage surge such as that shown in Fig. 5.13 encountering an impedance at
the end of a lossless line (for instance, at a transformer bus) will cause a
voltage wave of the same shape to travel back toward the source of the surge.
The reflected wave will be reduced in magnitude if the terminal impedance is
other than a short or open circuit, but if ZR is greater than Zc our study has
shown that the peak terminal voltage will be higher than, often close to, double
the peak voltage of the surge.
Figure 5.14 Circuit diagram, lattice dia¬ Figure 5.15 Circuit diagram, lattice dia¬
gram, and plot of voltage versus time for gram, and plot of voltage versus time when
Example 5.6 where the receiving-end resis¬ the receiving-end resistance for Example 5.6
tance is 90 fL is changed to 10 £1
Terminal equipment is protected by surge arresters which are also called

lightning arresters. An ideal arrester connected from the line to a grounded
neutral would (a) become conducting at a design voltage above the arrester
rating, (b) limit the voltage across its terminals to this design value, and
(c) becomes nonconducting again when the line-to-neutral voltage drops below
the design value.
Originally, an arrester was simply an air gap. In this application when the
surge voltage reaches a value for which the gap is designed an arc occurs to
cause an ionized path to ground, essentially a short circuit. However, when
the surge ends the 60-Hz current from the generators still flows through
the arc to ground. The arc has to be extinguished by the opening of circuit
breakers.
Arresters capable of extinguishing 60-Hz current after conducting Surge

current to ground were developed later. These arresters are made using non¬
linear resistors in series with air gaps to which an arc quenching capability
has been added. The nonlinear resistance decreases rapidly as the voltage
across it rises. Typical resistors made of silicon carbide conduct current
proportional to approximately the fourth power of the voltage across the
resistor. When the gaps arc over as a result of a voltage surge, a low-
resistance current path to ground is provided through the nonlinear resistors.
After the surge ends and the voltage across the arrester returns to the
normal line-to-neutral level the resistance is sufficient to limit the arc current
to a value which can be quenched by the series gaps. Quenching is usually
accomplished by cooling and deionizing the arc by elongating it magnetically
between insulating plates.
The most recent development in surge arresters is the use of zinc oxide
in place of silicon carbide. The voltage across the zinc oxide resistor is
extremely constant over a very high range of current, which means that its
resistance at normal line voltage is so high that a series air gap is not
necessary to limit the drain of 60-Hz current at normal voltage.t
5.13 DIRECT-CURRENT TRANSMISSION
The transmission of energy by direct current becomes economical when

compared to ac transmission only when the extra cost of the terminal equipment
required for dc lines is offset by the lower cost of building the lines. Converters
at the two ends of the dc lines operate both as rectifiers to change the generated
alternating to direct current and as inverters for converting direct to alternating
current so that power can flow in either direction.
The year 1954 is generally recognized as the starting date for modern high-
voltage dc transmission when a dc line began service at 100 kV from Vastervik
on the mainland of Sweden to Visby on the island of Gotland, a distance of
100 km (62.5 mi) across the Baltic Sea. Static conversion equipment was in
operation much earlier to transfer energy between systems of 25 and 60 Hz,
essentially a dc transmission line of zero length. In the United States a dc line
operating at 800 kV transfers power generated in the Pacific Northwest to the
southern part of California. As the cost of conversion equipment decreases with
respect to the cost of line construction the economical minimum length of dc
lines also decreases and at this time is about 600 km (375 mi).
Operation of a dc line began in 1977 to transmit power from a mine-mouth
generating plant burning lignite at Center, North Dakota to near Duluth,
t See E. C. Sakshaug, J. S. Kresge, and S. A. Miske, Jr., “A New Concept in Station Arrester
Design,” IEEE Trans. Power Appar. Syst., vol. PAS-96, no. 2, March/April 1977, pp. 647-656.
Minnesota, a distance of 740 km (460 mi). Preliminary studies showed that the
dc line including terminal facilities would cost about 30% less than the compar¬
able ac line and auxiliary equipment. This line operates at + 250 kV (500 kV
line-to-line) and transmits 500 MW.
Direct-current lines usually have one conductor which is at a positive poten¬
tial with respect to ground and a second conductor operating at an equal nega¬
tive potential. Such a line is said to be bipolar. The line could be operated with
one energized conductor with the return path through the earth, which has a
much lower resistance to direct than to alternating current. In this case, or with
a grounded return conductor, the line is said to be monopolar.
In addition to the lower cost of dc transmission over long distances there are
other advantages. Voltage regulation is less of a problem since at zero frequency
a>L is no longer a factor, whereas it is the chief contributor to voltage drop in an
ac line. Another advantage of direct current is the possibility of monopolar
operation in an emergency when one side of a bipolar line becomes grounded.
The fact that underground ac transmission is limited to about 50 km be¬
cause of excessive charging current at longer distances, direct current was
chosen to transfer power under the English Channel between Great Britain and
France. The use of direct current for this installation also avoided the difficulty
of synchronizing the ac systems of the two countries.
No network of dc lines is possible at this time because no circuit breaker is
available for direct current comparable to the highly developed ac breakers. The
ac breaker can extinguish the arc which is formed when the breaker opens
because zero current occurs twice in each cycle. The direction and amount of
power in the dc line is controlled by the converters in which grid-controlled
mercury-arc devices are being displaced by the semiconductor rectifier (SCR). A
rectifier unit will contain perhaps 200 SCRs.
Still another advantage of direct current is the smaller amount of right of
way required. The distance between the two conductors of the North Dakota-
Duluth 500-kV line is 25 feet. The 500-kV ac line shown in Fig. 5.1 has 60.5 feet
between the outside conductors. Another consideration is the peak voltage of
the ac line which is yjl x 500 = 707 kV. So the line requires more insulation
between the tower and conductors as well as greater clearance above the earth.
We conclude that dc transmission has many advantages over alternating
current, but dc transmission remains very limited in usage except for long lines
because there is no dc device which can provide the excellent switching opera¬
tions and protection of the ac circuit breaker. There is also no simple device to
change the voltage level, which the transformer accomplishes for ac systems.
5.14 SUMMARY
The long-line equations given by Eqs. (5.35) and (5.36) are, of course, valid for a
line of any length. The approximations for the short and medium-length lines
make analysis easier in the absence of a computer.
Circle diagrams were introduced because of their instructional value in

showing the maximum power which can be transmitted by a line and also in
showing the effect of power factor of the load or the addition of capacitors.
ABCD constants provide a straightforward means of writing equations in a
more concise form and are very convenient in problems involving network
reduction. Their usefulness is apparent in the discussion of series and shunt
reactive compensation.
The simple discussion of transients although confined to lossless lines and
dc sources should give some idea of the complexity of the study of transients
which arise from lightning and switching in power systems.
PROBLEMS
5.1 An 18-km 60-Hz single circuit three-phase line is composed of Partridge conductors equilaterally
spaced with 1.6 m between centers. The line delivers 2500 kW at 11 kV to a balanced load. What
must be the sending-end voltage when the power factor is (a) 80% lagging, (b) unity, and (c) 90%
leading? Assume a wire temperature of 50°C.
5.2 A 100-mi, three-phase transmission line delivers 55 MVA at 0.8 power factor lagging to the load
at 132 kV. The line is composed of Drake conductors with flat horizontal spacing of 11.9 ft between
adjacent conductors. Determine the sending-end voltage, current, and power. Assume a wire temper¬
ature of 50°C.
5.3 Find the ABCD constants of a n circuit having a 600-D resistor for the shunt branch at the
sending end, a 1-kD resistor for the shunt branch at the receiving end, and an 80-D resistor for the
series branch.
5.4 The ABCD constants of a three-phase transmission line are
A = D - 0.936 + /0.016 = 0.936/0.98°

B = 33.5 + y'138 = 142.0/76.4° D
C = (-5.18 + ;'914) x 10“6U
The load at the receiving end is 50 MW at 220 kV with a power factor of 0.9 lagging. Find the
magnitude of the sending-end voltage and the voltage regulation. Assume the magnitude of the
sending-end voltage remains constant.
5.5 Use per-unit values on a base of 230 kV, 100 MVA to find voltage, current, power, and power
factor at the sending end of a transmission line delivering a load of 60 MW at 230 kV with 0.8
power-factor lagging. The three-phase line is arranged in flat, horizontal spacing with 15 ft between
adjacent Ostrich conductors. Line length is 70 mi. Assume a wire temperature of 50°C. Note that base
admittance must be the reciprocal of base impedance.
5.6 Evaluate cosh 9 and sinh 9 for 9 = 0.5/82°.
5.7 Justify Eq. (5.52) by substituting for the hyperbolic functions the equivalent exponential
expressions.
5.8 A 60-Hz three-phase transmission is 175 mi long. It has a total series impedance of 35 + )140 D
and a shunt admittance of 930 x KT6/901U. It delivers 40 MW at 220 kV, with 90% power factor
lagging. Find the voltage at the sending end by (a) the short-line approximation, (b) the nominal-n
approximation, (c) the long-line equation.
5.9 Determine the equivalent-71 circuit for the line of Prob. 5.8.
5.10 Determine the voltage regulation for the line described in Prob. 5.8. Assume that the sending-
end voltage remains constant.
5.11 A three-phase 60-Hz transmission line is 250 mi long. The voltage at the sending end is 220 kV.
The parameters of the line are R = 0.2 D/mi, X = 0.8 D/mi, and Y = 5.3 Û/mi. Find the sending-
end current when there is no load on the line.
5.12 If the load on the line described in Prob. 5.11 is 80 MW at 220 kV, with unity power factor,
calculate the current, voltage, and power at the sending end. Assume that the sending-end voltage is
held constant, and calculate the voltage regulation of the line for the load specified above.
5.13 Rights of way for transmission circuits are becoming difficult to obtain in urban areas and
existing lines are often upgraded by reconductoring the line with larger conductors or by reinsulating
the line for operation at higher voltage. Thermal considerations and maximum power which the line
can transmit are the important considerations. A 138-kV line is 50 km long and is composed of
Partridge conductors with flat horizontal spacing of 5 m between adjacent conductors. Neglect
resistance and find the percent increase in power which can be transmitted for constant | Ks | and
| VR | while 6 is limited to 45° (a) if the Partridge conductor is replaced by Osprey which has more
than twice the area of aluminum in square millimeters, (f>) if a second Partridge conductor is placed
in a two-conductor bundle 40 cm from the original conductor and a center-to-center distance between
bundles of 5 m, and (c) if the voltage of the original line is raised to 230 kV with increased conductor
spacing of 8 m.
5.14 Construct a receiving-end power-circle diagram similar to Fig. 5.10 for the line of Prob. 5.8.
Locate the point corresponding to the load of Prob. 5.8, and locate the center of circles for various
values of | Fs| if | VR | = 220 kV. Draw the circle passing through the load point. From the measured
radius of the latter circle determine I U I, and compare this value with the values calculated for
Prob. 5.8.
5.15 Use the diagram constructed for Prob. 5.14 to determine the sending-end voltage for various
values of kilovars supplied by synchronous condensers in parallel with the designated load at the
receiving end. Include values of added kilovars to give unity power factor and a power factor of 0.9
leading at the receiving end. Assume that the sending-end voltage is adjusted to maintain 220 kV at
the load.
5.16 A receiving-end power-circle diagram is drawn for a constant receiving-end voltage. For a
certain load at this receiving-end voltage the sending-end voltage is 115 kV. The receiving-end circle
for | Fs| = 115 kV has a radius of 5 in. The horizontal and vertical coordinates of the receiving-end
circles are —0.25 and —4.5 in, respectively. Find the voltage regulation for the load.
5.17 A three-phase transmission line is 300 mi long and serves a load of 400 MVA, 0.8 lagging-power
factor at 345 kV. The A BCD constants of the line are
A = D = 0.8180/1.3°
B = 172.2/84.2° ft
C = 0.001933/90.4° 15
(a) Determine the sending-end line-to-neutral voltage, the sending-end current and the percent
voltage drop at full load.
(b) Determine the receiving-end line-to-neutral voltage at no load, the sending-end current at
no load, and the voltage regulation.
5.18 A series capacitor bank is to be installed at the midpoint of the 300-mi line of Prob. 5.17. The
A BCD constants for 150 mi of line are
A=D = 0.9534/0.3°
B = 90.33Ml° ft
C = 0.001014/90.1° 15
The A BCD constants of the series capacitor bank are
A=D = l/W
B = 146.6/-90° ft
C =0
(a) Determine the equivalent ABCD constants of the series combination of the line-capacitor-
line. (See Table A.6 in the Appendix.)
(b) Rework Prob. 5.17 using these equivalent ABCD constants.
5.19 The shunt admittance of a 300-mi transmission line is
yc = 0 + j'6.87 x 10“ 6 D/mi
Determine the ABCD constants of a shunt reactor that will compensate for 60% of the total shunt
admittance.
5.20 A 250-Mvar, 345-kV shunt reactor whose admittance is 0.0021 /—90Q U is connected to the
receiving end of the 300-mi line of Prob. 5.17 at no load.
(a) Determine the equivalent ABCD constants of the line in series with the shunt reactor. (See
Table A.6 in the Appendix.)
(b) Rework part (b) of Prob. 5.17 using these equivalent ABCD constants and the sending-end
voltage found in Prob. 5.17.
5.21 Draw the lattice diagram for current and plot current versus time at the sending end of the line
of Example 5.6 for the line terminated in (a) an open circuit and (b) a short circuit.
5.22 Plot voltage versus time for the line of Example 5.6 at a point distant from the sending end
equal to one-fourth of the length of the line if the line is terminated in a resistance of 10 Q.
5.23 Solve Example 5.6 if a resistance of 54 D is in series with the source.
5.24 Voltage from a dc source is applied to an overhead transmission line by closing a switch. The
end of the overhead line is connected to an underground cable. Assume both the line and the cable
are lossless and that the initial voltage along the line is v+. If the characteristic impedances of the line
and cable are 400 fi and 50 Q, respectively, and the end of the cable is open-circuited, find in terms of
v+ (a) the voltage at the junction of the line and cable immediately after the arrival of the incident
wave and (b) the voltage at the open end of the cable immediately after arrival of the first voltage
wave.
CHAPTER
SIX
SYSTEM MODELING
At this point in our discussion of power systems we have completed develop¬

ment of the circuit model of a transmission line and have looked at some
calculations of voltage, current, and power on the line. In this chapter we shall
develop circuit models for the synchronous machine and for the power
transformer.
The synchronous machine as an ac generator driven by a turbine is the
device which converts mechanical energy to electrical energy. Conversely as a
motor the machine converts electrical energy to mechanical energy. We are
chiefly concerned with the synchronous generator, but we shall give some con¬
sideration to the synchronous motor. The induction motor is more widely used
than a synchronous motor in industry, but its treatment is beyond the scope of
this book. We cannot treat the synchronous machine fully, but there are many
books on the subject of ac machinery which provide quite adequate analysis of
generators, motors, and transformers.t Our treatment of the synchronous ma¬
chine will enable us to have confidence in the equivalent circuit sufficient for
understanding our further study of the role of the generator in power system
analysis and will constitute a review for those who have devoted previous study
to the subject.
t For a much more detailed discussion of synchronous machines and transformers, consult any of
the texts on electric machinery such as A. E. Fitzgerald, C. Kingsley, Jr., and A. Kusko, Electric
Machinery, 3d ed., McGraw-Hill Book Company, New York, 1971, or L. W. Matsch, Electromag¬
netic and Electromechanical Machines, 2d ed., IEP, New York, 1977.
127
The transformer provides the link between the generator and the transmis¬
sion line and between the transmission line and the distribution system which
delivers energy through still other transformers to the loads on the system. As
with the synchronous machine our treatment of the transformer will not be
comprehensive but will provide us with the circuit model suitable to our further
study.
Then we shall see how a one-line diagram describes the combination of the
component models to form a complete system, and we shall devote some further
study to the application of per-unit quantities to calculations for the system.
6.1 CONSTRUCTION OF THE SYNCHRONOUS MACHINE
The two principal parts of a synchronous machine are ferromagnetic structures.

The stationary part which is essentially a hollow cylinder is called the stator or
armature and has longitudinal slots in which are coils of the armature winding.
This winding carries the current supplied to an electrical load or system by a
generator or the current received from an ac supply by a motor. The rotor is the
part of the machine which is mounted on the shaft and rotates inside the hollow
stator. The winding on the rotor is called the field winding and is supplied with
dc current. The very high magnetomotive force (mmf) produced by this current
in the field winding combines with the mmf produced by current in the armature
winding. The resultant flux across the air gap between the stator and rotor
generates voltage in the coils of the armature winding and provides the electro¬
magnetic torque between the stator and rotor. Figure 6.1 shows the threading of
a four-pole cylindrical rotor into the stator of a 1525-MVA generator.
The dc current is supplied to the field winding by an exciter which may be a
generator mounted on the shaft or a separate dc source connected to the field
winding through brushes bearing on slip rings. Large ac generators usually have
exciters consisting of an ac source with solid-state rectifiers.
If the machine is a generator the shaft is driven by a prime mover which is
usually a steam or hydraulic turbine. The electro-magnetic torque developed in
the generator when it delivers power opposes the torque of the prime mover. The
difference between these two torques is due to losses in the iron core and friction.
In a motor the electromagnetic torque developed in the machine except for core
and friction losses is delivered to the shaft which drives the mechanical load.
Figure 6.2 shows a very elementary three-phase generator. The field winding
is merely indicated by a coil. The generator in this figure is called a nonsalient
pole machine because it has a cylindrical rotor. The rotor of Fig. 6.1 is the
cylindrical type. In the actual machine the winding has a large number of turns
distributed in slots around the circumference of the rotor. The strong magnetic
field produced links the stator coils to induce voltage in the armature winding as
the shaft is turned by the prime mover.
The stator is shown in cross section in Fig. 6.2. Opposite sides of a coil,
which is almost rectangular, are in slots a and a' 180° apart. Similar coils are in
SYSTEM MODELING 129
Figure 6.1 Photograph showing the threading of a four-pole cylindrical rotor into the stator of a
1525-MVA generator. (Courtesy Utility Power Corporation, Wisconsin.)
Figure 6.2 Elementary three-phase ac gener¬

ator showing end view of the two-pole cylin¬
drical rotor and cross section of the stator.
dc field
winding cdils
Stator
Rotor
Figure 63 Cross section of an elementary

stator and salient-pole rotor.
slots b and b' and slots c and d. Coil sides in slots a, b, and c are 120° apart. The
conductors shown in the slots indicate a coil of only one turn, but such a coil
may have many turns and is usually in series with identical coils in adjacent slots
to form a winding having ends designated a and a'. Windings designated b and b'
and c and c' are the same as the a-a! winding except for their location around the
armature.
Figure 6.3 shows a salient-pole machine which has four poles. Opposite sides
of an armature coil are 90° apart. So there are two coils for each phase. Coil
sides a, b, and c of adjacent coils are 60° apart. The two coils of each phase may
be connected in series or in parallel.
Although not shown in Fig. 6.3 salient-pole machines usually have damper
windings which consist of short-circuited copper bars through the pole face
similar to part of a “ squirrel cage ” winding of an induction motor. The purpose
of the damper winding is to reduce the mechanical oscillations of the rotor about
synchronous speed which is determined, as we shall soon see, by the number of
poles of the machine and the frequency of the system to which the machine is
connected.
In the two-pole machine one cycle of voltage is generated for each revolu¬
tion of the two-pole rotor. In the four-pole machine two cycles are generated in
each coil per revolution. Since the number of cycles per revolution equals the
number of pairs of poles the frequency of the generated voltage is
P N
/=2fio HZ <61>
where P is the number of poles and N is the rotor speed in revolutions per
minute.
Since one cycle of voltage (360° of the voltage wave) is generated every time
a pair of poles passes a coil we must distinguish between electrical degrees used
SYSTEM MODELING 131
to express voltage and current and mechanical degrees to express position of the
rotor. In a two-pole machine electrical and mechanical degrees are equal. In a
four-pole machine two cycles, or 720 electrical degrees, are produced per revolu¬
tion of 360 mechanical degrees. The number of electrical degrees equals P/2 times
the number of mechanical degrees in any machine.
By proper design of the rotor and by proper distribution of the stator
windings around the armature very pure sinusoidal voltages will be generated.
These voltages are called no-load generated voltages or simply generated voltages.
For our analysis in later sections we shall consider the mmf produced by the
rotor to be sinusoidally distributed.
If the ends of the windings are connected to each other and the junction is
designated o, the generated voltages (labeled Ea0, Ebo, and Eco to agree with the
notation adopted in Chap. 2) are displaced 120 electrical degrees from each
other.
6.2 ARMATURE REACTION IN A SYNCHRONOUS MACHINE
If a balanced three-phase load is connected to a three-phase generator, balanced

three-phase currents will flow in the phases of the armature winding. These
armature currents will create additional mmfs which we need to investigate. We
shall look at the mmf produced by each of the three windings which are spaced
120° apart around a two-pole machine as in Fig. 6.2. We shall also specify a
cylindrical-rotor machine, again as in Fig. 6.2, so that all flux paths across the
air gap of the machine will have essentially the same reluctance.
Since we may choose t to be zero when ia has its maximum value the
balanced three-phase currents in each phase may be expressed by the equations
ia = Im cos cot (6.2)
ib = Im cos (cot - 120°) (6.3)
ic = Im cos (cot — 240°) (6.4)
where co is in electrical degrees per second. For our two-pole machine to is also
the angular velocity of the rotor in mechanical degrees per second. We shall
assume that the positive direction chosen for current is toward the observer in
conductor a in Fig. 6.2 and make similar assumptions for currents in conductors
b and c.
Positions around the armature will be identified by the angular displacement
9d measured from the axis of coil a as in Fig. 6.2. The subscript d is used to
distinguish the displacement angle 6d from the angle 9 which we are accustomed
to use to express the time phase between a voltage and a current. We recognize
that the mmf around the armature produced by the armature current in any
phase must be a function of both displacement 9d and time t. We shall designate
the mmf due to the current in phase a as &a(9d, t). Since this mmf is produced
by ia it must be a sinusoidal function of time in phase with ia.
Figure 6.4 Distribution around the armature of mmf produced by the current in phase a of the
generator of Fig. 6.2 for various values of cot when ia is expressed by Eq. (6.2).
The distribution of mmf around the periphery of the armature due to arma¬
ture current is not sinusoidal. In a practical machine each coil occupies a
number of slots and the distribution of mmf is nearly triangular, but in the
analysis we are about to make it is customary to consider only the fundamental
harmonic of the triangular wave, and this we shall do.
For the assumption of sinusoidal distribution of mmf the solid line in
Fig. 6.4 shows the mmf of phase a around the armature as a function of 9d at
t — 0; that is when ia has its maximum value. At this time the distribution of a
around the armature is expressed by
0) = cos 6d (6.5)
where 3Fm is the maximum value of !Fa.

For some values of t other than zero the dashed lines of Fig. 6.4 show the
distribution of &a around the armature. We note that the maximum value of F a
is 0.5 2F m when cot is 60° because at that instant of time ia equals 0.5 Im from
Eq. (6.2). At the fixed position 9d = 45°, J5"a is varying sinusoidally between
values indicated by points a and b on Fig. 6.4. At 9d = 90° &a is always zero. So
to account for time variation as well as spatial variation of 3F a we have, since .F a
is in time phase with ia,
Fa(9d, t) = Fm x cos 9^ x cos cot (6.6)

to account for to account for
position on time variation
the armature
Similarly for phases b and c
t) = Pm cos (ed - 120”) cos [tot - 120°) (6.7)

SYSTEM MODELING 133
and
(Od, t) = &m cos (9d - 240°) cos (cot - 240°) (6.8)
The sum of these three mmfs is the resultant mmf 3F ar which is called the mmf of
armature reaction. By using the trigonometric identity
cos a cos p = { cos (a - /?) + \ cos (a + f}) (6.9)
and by recognizing that the sum of three sinusoidal terms of equal magnitude
displaced in phase by 120° and 240° is zero we find
& or = & a + &b + & c = \&m COS (9d - (Ot) (6.10)

Equation (6.10) describes a wave of mmf traveling around the armature in
the direction of increasing 9d. To an observer moving with any point on the
wave the mmf is constant and
6d — cot = a constant (6.11)
from which we obtain
(6.12)
Equation (6.12) tells us that the mmf of armature reaction is rotating around
the armature at the angular velocity a> equal to the angular velocity of the rotor.
Therefore this mmf is stationary with respect to the mmf produced by the dc
winding of the rotor. The net flux across the air gap between the stator and rotor
is produced by the resultant of these two mmfs.
If we neglect saturation and recall that we are considering a cylindrical-rotor
machine, we can consider separately the fluxes produced by each of these mmfs
and speak of <j)f produced by the dc current in the rotor and (j)ar produced by
armature reaction. When a flux having sinusoidal distribution is rotating around
the armature the maximum flux linkages with a coil occur when the direction of
the mmf causing the flux coincides with the axis of the coil, but the rate of
change of flux linkages is then zero. Likewise when the rotor has turned through
90° the flux linkages become zero, but their rate of change is a maximum. Thus
the voltage induced in the coil is 90° out of phase with the flux linkages. By
applying Lenz’s law and taking into account the assumed positive directions of
current in the coil and of the mmf in Fig. 6.2 we could show that the induced
voltage is lagging rather than leading the mmf.
6.3 THE CIRCUIT MODEL OF A SYNCHRONOUS MACHINE
To draw the phasor diagram for phase a we visualize the separate components of
the flux at the axis of coil a; that is where 9d equals zero. The rotor flux (pf is the
only one to be considered when the armature current is zero. This flux (j)f
generates the no-load voltage Eao which we shall designate here as Ef. The flux
Figure 6.5 Phasor diagram showing time relation

of components of flux at the axis of coil a (9d = 0)
and the voltages and currents of phase a of the
generator of Fig. 6.2. Similar diagrams may be drawn
for phases b and c and apply to all cylindrical-
rotor generators.
<par due to the armature-reaction mmf .9' aT is in phase with the current ia
(at 6d = 0) as may be seen by comparing Eqs. (6.2) and (6.10) and recognizing
that cos cot = cos ( — cot). The sum of cpf and 0ar, saturation neglected, is cf)r, the
resultant flux which generates the voltage Er in the coil windings composing
phase a. The phasor diagram for phase a is shown in Fig. 6.5. Voltages Ef and
Ear lag by 90° the fluxes 0f and cpar which generate them. The resultant flux 0r
is the flux across the air gap of the machine and generates Er in the stator.
Similar diagrams can be drawn for each phase.
In Fig. 6.5 we note particularly that Ear is lagging Ia by 90°. The magnitude
of Ear is determined by <par which in turn is proportional to |/a| since it is the
result of armature current. So we can specify an inductive reactance Xar such that
Ear — ~jIaXar (6.13)

Equation (6.13) defines Em so that it has the proper phase angle with respect to
Ia. Then the voltage generated in phase a by the air-gap flux is Er where
Er = Ef + E^ = Ej- — jlaX m (6.14)
The voltage generated in each phase by the resultant flux exceeds the ter¬
minal voltage Vt of the phase only by the voltage drop due to the armature
current times the leakage reactance Xt of the winding if resistance is neglected. If
the terminal voltage is Vt,
Vt = Er-jIaXl (6.15)
The product IaXt accounts for the voltage drop caused by that portion of the
flux (produced by armature current) which does not cross the air gap of the
machine. So from Eqs. (6.14) and (6.15)
K= Ef ~ £aXar ~ £(6.16)
generated due to armature due to armature
at no load reaction leakage reactance
or
Vi = Ef — jIaX. (6.17)
SYSTEM MODELING 135
^T
r-^rOTT- -TttftP—W^—
6 Vi
Figure 6.6 Equivalent circuit of an ac generator.
where Xs, called the synchronous reactance, is equal to Xar + Xt. If the resist¬
ance of the armature Ra is to be considered, Eq. (6.17) becomes
V, = E,-I.(R.+jX.) (6.18)
Ra is usually so much smaller than Xs that its omission is not of great con¬
sequence here, where we are most interested in a qualitative approach.
Now we have arrived at a relationship which allows us to represent the
generator by the simple but very useful equivalent circuit shown in Fig. 6.6
which corresponds to Eq. (6.18).
Examination of Fig. 6.5 reveals a very important point about a synchronous
generator. When the current Ia lags the no-load generated voltage Ef by 90°, (f)ar
subtracts directly from <j>f, and <pr is greatly reduced. Conversely, armature
current leading the no-load voltage by 90° causes (f)ar to add directly to <j)f, and
4>r is greatly increased. The relationship between Ef, Ear, and E, for these two
cases is shown in Fig. 6.7. If a highly inductive load is applied to a generator the
terminal voltage will be considerably below the no-load terminal voltage. On the
other hand a capacitive load will cause the terminal voltage to rise considerably
above its no-load value. These results help to verify our phasor diagram and
equivalent circuit.
In developing this theory we restricted ourselves to considering a two-pole
machine. The theory applies equally well to multiple-pole machines but is a little
more complicated to develop because of keeping in mind the differences between
electrical and mechanical degrees.
The principles we have discussed could be extended to a synchronous motor.
Er Ef
Em Ef Er
(a) (b)
Figure 6.7 Phasor diagrams showing the relation between Ef and E„ when current delivered by a
generator is (a) lagging Ef by 90° and (b) leading Ef by 90°.
E,m
Figure 6.8 Circuit diagram for a generator and motor. /„ is

the current delivered by the generator and received by the
Generator Motor
motor.
The equivalent circuit for the motor is identical with that of the generator with
the direction shown for Ia reversed. The generated voltages of the generator and
motor are often identified by single-subscript notation as Eg and Em, respec¬
tively, instead of Ef, especially when they are in the same circuit as in Fig. 6.8,
for which the equations are.
(6.19)
and
(6.20)
Synchronous reactances of the generator and motor are Xg and Xm, respectively,
and armature resistance is neglected.
When we study faults on a synchronous machine in Chap. 10 we shall see
that the current flowing immediately after the occurrence of a fault differs from
the steady-state value. Instead of synchronous reactance Xs, we use subtransient
reactance X" or transient reactance X' in modeling the synchronous machine for
fault calculations. These reactances will be used in some problems before our
further study of them in Chap. 10.
If we had considered salient-pole machines we would have had to account
for the difference between the flux path directly into the pole face (called the
direct axis) and the path between poles (called the quadrature axis). To do so the
armature current is divided into two components. One component is 90° out of
phase with the no-load generated voltage Ef, and the other is in phase with Ef.
The first component produces the mmf whose flux causes a voltage drop ac¬
counted for by the product of the current and the so-called direct-axis syn¬
chronous reactance Xd. The other component is in phase with Ef and produces
the mmf and flux which causes a voltage drop accounted for by the product of
this component of current and the quadrature-axis synchronous reactance Xq.
Table A.4 in the Appendix lists per-unit values of various reactances for
synchronous machines. As the table shows, values of Xd and Xq in cylindrical-
rotor machines are essentially equal. For this reason we did not need to consider
Xd and Xq separately in our discussion of armature reaction but simply called
the synchronous reactance Xs. To simplify our work we shall continue to
assume all synchronous machines to have cylindrical rotors. Two-reaction
theory discussing direct- and quadrature-axis reactances can be found in most
textbooks on ac machinery. Table A.4 also gives values of X'd and Xd.
SYSTEM MODELING 137
6.4 THE EFFECT OF

SYNCHRONOUS-MACHINE EXCITATION
Changing the excitation of synchronous machines is an important factor in

controlling the flow of reactive power. First we shall consider a generator con¬
nected at its terminals to a very large power system, a system so large that the
voltage Vt at the terminals of the generator will not be altered by any changes in
the excitation of the generator. The bus to which the generator is connected
is sometimes called an infinite bus, which means that its voltage will remain
constant and no frequency change will occur regardless of changes made in
power input or field excitation of the synchronous machine connected to it. If we
decide to maintain a certain power input from the generator to the system,
| Vt | • | Ia | cos 9 will remain constant as we vary the dc field excitation to vary
| Eg |. Then, for a high and a low value of \Eg\ the phasor diagrams of the
generator are given by Fig. 6.9. The angle S is called the torque angle or power
angle of the machine. Normal excitation is defined as the excitation when
| Eg \ cos <3= Vt (6.21)

For the condition of Fig. 6.9a the generator is overexcited and supplies lagging
current to the system. The machine can also be considered to be drawing leading
current from the system. Like a capacitor, it supplies reactive power to the system.
Figure 6.9b is for an underexcited generator supplying leading current to the
system, or it may be considered to be drawing lagging current from the system.
The underexcited generator draws reactive power from the system. This action
can be explained by the mmf of armature reaction. For instance, when the
generator is overexcited, it must deliver lagging current since lagging current
produces an opposing mmf to reduce the overexcitation.
We note that Eg leads Vt in Fig. 6.9, which is always true for a generator and
is necessary to satisfy Eq. (6.19).
Figure 6.10 shows overexcited and underexcited synchronous motors draw¬
ing the same power at the same terminal voltage. The overexcited motor draws
leading current and acts like a capacitive circuit when viewed from the network
to which it supplies reactive power. The underexcited motor draws lagging
current, absorbs reactive power, and is acting like an inductive circuit when
(a) Over-excited generator (b) Under-excited generator
Figure 6.9 Phasor diagrams of (a) overexcited and (b) underexcited generator. Ia is current delivered
by the generator.
Figure 6.10 Phasor diagrams of (a) overexcited and (b) underexcited motor. Ia is current drawn by
the motor.
viewed from the network. We see from Fig. 6.10 that Em lags V, in order to
satisfy Eq. (6.20), and this is always true for a synchronous motor. Briefly,
Figs. 6.9 and 6.10 show us that overexcited generators and motors supply reac¬
tive power to the system and underexcited generators and motors absorb reactive
power from the system.
6.5 THE IDEAL TRANSFORMER
We now have models for transmission lines and synchronous machines and are
ready to consider transformers which consist of two or more coils placed so that
they are linked by the same flux. In a power transformer the coils are placed on
an iron core in order to confine the flux so that almost all of the flux linking any
one coil links all the others. Several coils may be connected in series or parallel
to form one winding, the coils of which may be stacked on the core alternately
with those of the other winding or windings.
Figure 6.11 is the photograph of a three-phase transformer which raises the
voltage of a generator in a nuclear power station to the transmission-line
voltage. The transformer is rated 750 MVA 525/22.8 kV.
Figure 6.12 shows how two windings may be placed on an iron core to form
a single-phase transformer of the so-called shell type. The number of turns in a
winding may be several hundred up to several thousand.
We shall begin our analysis by assuming that the flux varies sinusoidally in
the core and that the transformer is ideal, which means that the permeability p of
the core is infinite and the resistance of the windings is zero. With infinite
permeability of the core all of the flux is confined to the core and therefore links
all of the turns of both windings. The voltage e induced in each winding by the
changing flux is also the terminal voltage v of the winding since the winding
resistance is zero.
We can see from the relationship of the windings shown in Fig. 6.12 that
voltages et and e2 induced by the changing flux are in phase when they are
defined by the -I- and — polarity marks indicated. Then by Faraday’s law
(6.22)
= =
SYSTEM MODELING 139
Figure 6.11 Photograph of a three-phase transformer rated 750 MV A 525/22.8 kV. (Courtesy Duke
Power Company.)
and
v2 = e2 = N2^ (6.23)
where 0 is the instantaneous value of the flux and Nt and N2 are the number of
turns on windings 1 and 2, as shown in Fig. 6.12. Since we have assumed sinu¬
soidal variation of the flux we can convert to phasor form after dividing Eq. (6.22)
Figure 6.12 Two-winding transformer.

h
+ ' +
Vl V2
Figure 6.13 Schematic representation of a two-winding
transformer. r
by Eq. (6.23) to yield
Usually we do not know the direction in which the coils of a transformer are
wound. One device to provide winding information is to place a dot at the end
of each winding such that all dotted ends of windings are positive at the same
time; that is, voltage drops from dotted to unmarked terminals of all windings
are in phase. Dots are shown on the two-winding transformer in Fig. 6.12
according to this convention. We also note that the same result is achieved by
placing the dots so that current flowing from the dotted terminal to the un¬
marked terminal of each winding produces a magnetomotive force acting in the
same direction in the magnetic circuit. Figure 6.13 is a schematic representation
of a transformer and provides the same information about the transformer as is
provided by Fig. 6.12.
To find the relation between the currents and i2 in the windings we apply
Ampere’s law which states that the magnetomotive force around a closed path is
jHds = i (6.25)
where i is the current enclosed by the line integral of the field intensity H around
the path. In applying the law around each of the closed paths of flux shown by
dotted lines in Fig. 6.12, it is enclosed Nt times and the current i2 is enclosed N2
times. However, IVji! and lV2i2 produce magnetomotive' forces in opposite
directions and
■ds = Nr ij — N2i2 (6.26)
The minus sign would change to plus if we had chosen the opposite direction for
the current i2. The integral of the field intensity H around the closed path is zero
when permeability is infinite. So upon converting to phasor form we have
N1Il - N2I2 = 0 (6.27)

So
h=N2
(6.28)
I2 Nt
SYSTEM MODELING 141
and /j and I2 are in phase. Note then that Ix and /2 are in phase if we choose
the current to be positive when entering the dotted terminal of one winding and
leaving the dotted terminal of the other. If the direction chosen for either cur¬
rent is reversed they are 180° out of phase.
From Eq. (6.28)
(6.29)
and in the ideal transformer I x must be zero if 12 is zero.

The winding across which an impedance or other load may be connected is
called the secondary winding and any circuit elements connected to this winding
are said to be on the secondary side of the transformer. Similarly the winding
which is toward the source of energy is called the primary winding on the
primary side. In the power system energy often will flow in either direction
through a transformer and the designation of primary and secondary loses its
meaning. The terms are in general use, however, and we shall use them wherever
they do not cause confusion.
If an impedance Z2 is connected across winding 2 of the circuit of Figs. 6.12
or 6.13
= r
12
(«-3<>)
but upon substituting for V2 and /2 values determined from Eqs. (6.24) and
(6.28)
(N2/NX)VX
2 (N1/N2)I1
and the impedance as measured across the primary winding is
Thus the impedance connected to the secondary side is referred to the primary
side by multiplying the impedance on the secondary side of the transformer by
the square of the ratio of primary to secondary voltage.
We should note also that Vx I x and V212 are equal as shown by the following
equation which again makes use of Eqs. (6.24) and (6.28)
(6.33)
and similarly
VXI* = V2I* (6.34)
so the voltamperes and complex power input to the primary winding equal the
output of these same quantities from the secondary winding since we are con¬
sidering an ideal transformer.
Example 6.1 If Nt = 2000 and N2 = 500 in the circuit of Fig. 6.13 and if
y = 1200/0° V and /x = 5 /—30° A with an impedance Z2 connected across
winding 2, find V2, I2, Z2 and the impedance Z'2 which is defined as the
value of Z2 referred to the primary side of the transformer.
Solution
500
Vo = (1200/tT) = 300/(F V
2000
2000
Io = (5 /— 30°) = 20 /— 30° A
500
300/0^
Z, = = 15/30° fi
20/-30
Zi = (15^0!)(^)'2 = 240/30! O
or
6.6 THE EQUIVALENT CIRCUIT OF

A PRACTICAL TRANSFORMER
The ideal transformer is a first step in studying a practical transformer where

(1) permeability is not infinite, (2) winding resistance is present, (3) losses occur
in the iron core due to the cyclic changing of direction of the flux, and (4) not
all the flux linking any one winding links the other windings.
When a sinusoidal voltage is applied to a transformer winding on an iron
core with the secondary winding open a small current will flow in the primary
such that in a well-designed transformer the maximum flux density Bm occurs at
the knee of the B-H, or saturation curve of the transformer. This current is called
the magnetizing current. Losses in the iron occur due, first, to the fact that the
cyclic changes of the direction of the flux in the iron require energy which is
dissipated as heat and called hysteresis loss. The second loss is due to the fact
that circulating currents are induced in the iron due to the changing flux, and
these currents produce an \I\2R loss in the iron called eddy-current loss. Hyster¬
esis loss is reduced by the use of certain high grades of alloy steel for the core.
Eddy-current loss is reduced by building up the core with laminated sheets of
steel. With the secondary open the transformer primary circuit is simply one of
very high inductance due to the iron core. The current lags the applied voltage
by slightly less than 90°, and the component of current in phase with, the voltage
accounts for the energy loss in the core. In the equivalent circuit-magnetizing
SYSTEM MODELING 143
Figure 6.14 Transformer equivalent circuit using the ideal transformer concept.
current lE is taken into account by an inductive susceptance BL in parallel with a

conductance G.
In the practical two-winding transformer some of the flux linking the pri¬
mary winding does not link the secondary. This flux is proportional to the
primary current and causes a voltage drop that is accounted for by an inductive
reactance xl5 called leakage reactance, which is added in series with the primary
winding of the ideal transformer. Similar leakage reactance x2 must be added to
the secondary winding to account for the voltage due to the flux linking the
secondary but not the primary. When we also account for the resistances and
r2 of the windings we have the transformer model shown in Fig. 6.14. In this
model the ideal transformer is the link between the circuit parameters ru xu G,
and Bl added to the primary side of the transformer and r2 and x2 added to the
secondary side.
The ideal transformer may be omitted in the equivalent circuit if we refer all
quantities to either the high- or the low-voltage side of the transformer. For
instance, if we refer all voltages, currents, and impedances of the circuit of
Fig. 6.14 to the primary circuit of the transformer having Nx turns, and for
simplicity let a = N1/N2, we have the circuit of Fig. 6.15. Very often we neglect
magnetizing current because it is so small compared to the usual load currents.
To further simplify the circuit we let
R1=r1 + a2r2 (6.35)
and
Xt = Xj -I- a2x2 (6.36)
rrW-^TOff
Figure 6.15 Transformer equivalent

circuit with path for magnetizing current.
R1 Xi
~t+—W—p
h
Vi aV2
Figure 6.16 Transformer equivalent circuit with magnetizing
current neglected. -—-^
to obtain the equivalent circuit of Fig. 6.16. All impedances .and voltages in the
part of the circuit connected to the secondary terminals must now be referred to
the primary side.
Example 6.2 A single-phase transformer has 2000 turns on the primary

winding and 500 turns on the secondary. Winding resistances are r1 = 2.0 Q
and r2 = 0.125 Q. Leakage reactances are = 8.0 Q and x2 = 0.50 Q. The
resistance load Z2 is 12 Q. If applied voltage at the terminals of the pri¬
mary winding is 1200 V, find V2 and the voltage regulation. Neglect mag¬
netizing current.
Solution
Nt 2000
= 4
N2 500
R1 = 2 + 0.125(4)2 = 4.0 Q
Xy = 8 + 0.5(4)2 = 16 Q
Z'2 = 12 x (4)2 = 192 Q
The equivalent circuit is shown in Fig. 6.17
1200
/i = = 6.10 7-4.67° A
192 + 4 + j 16
aV2 = 6.10/-4.67° x 192 = 1171.6/-4.67° V

\
1171.6/-4.67°
V-> = = 292.9 7-4.67° V
1200/4 — 292 9
Voltage regulation =-^929—~ = 00242 or 2-42%
Although magnetizing current may be neglected as in Example 6.2 for most
4 Q j16 Q
1200 V 192 Q
Figure 6.17 Circuit for Example 6.2.

SYSTEM MODELING 145
power-system calculations, G and BL can be calculated for the equivalent circuit

by an open-circuit test. Rated voltage is applied to the primary winding with the
other windings open. The impedance measured across the terminals of this wind¬
ing whose resistance and leakage reactance are rx and xx is
(6.37)
and since rx and xx are very small compared to the measured impedance, G and
Bl can be determined in this manner.
The parameters R and X of the two-winding transformer are determined by
the short-circuit test where impedance is measured across the terminals of one
winding when the other winding is short-circuited. Just enough voltage is
applied to circulate rated current. Since only a small voltage is required the
magnetizing current is insignificant and the measured impedance is essentially
equivalent to R + jX.
6.7 THE AUTOTRANSFORMER
An autotransformer differs from the ordinary transformer in that the windings of

the autotransformer are electrically connected as well as being coupled by a
mutual flux. We shall examine the autotransformer by connecting electrically the
windings of an ideal transformer. Figure 6.18a is a schematic diagram of an ideal
transformer, and Fig. 6.186 shows how the windings are connected electrically to
form an autotransformer. Here the windings are shown so their voltages are
additive although they could have been connected to oppose each other. The
great disadvantage of the autotransformer is that electrical isolation is lost, but
the following example will demonstrate the increase in power rating obtained.
Example 6.3 A 30 kVA, single-phase transformer rated 240/120 V is con¬

nected as an autotransformer as shown in Fig. 6.186. Rated voltage is
applied to the low-tension winding of the transformer. Consider the transfor¬
mer to be ideal and the load to be such that rated currents |/x | and \I2\
flow in the windings. Determine | V21 and the kilovoltampere rating of the
autotransformer.
Figure 6.18 Schematic diagram

of an ideal transformer con¬
nected (a) in the usual manner
and (b) as an autotransformer. (a) (b)
Solution
_ 30,000
= 250 A
120
_ 30,000
= 125 A
240
| K21 = 240 + 120 = 360 V
The directions chosen for positive current in defining 7X and /2 in rela¬

tion to the dotted terminals show that these currents are in phase. So the
input current is
\L\/&= \h\/l
|/,„| = 250 + 125 = 375 A
Input kVA is
375 x 120 x 10~3 = 45 kVA
Output kVA is
125 x 360 x 10-3 = 45 kVA
From this example we see that the autotransformer has given us a larger
voltage ratio than the ordinary transformer and transmitted more kilovolt¬
amperes between the two sides of the transformer. So an autotransformer provides
a higher rating for the same cost. It also operates more efficiently since the losses
remain the same as for the ordinary connection. However, the loss of electrical
isolation between the high- and low-voltage sides of the autotransformer is
usually the decisive factor in favor of the ordinary connection in most applica¬
tions. In power systems three-phase autotransformers are frequently used to
make small adjustments of bus voltages.
6.8 PER-UNIT IMPEDANCES IN SINGLE-PHASE

TRANSFORMER CIRCUITS
The ohmic values of resistance and leakage reactance of a transformer depend

on whether they are measured on the high- or low-tension side of the transfor¬
mer. If they are expressed in per unit, the base kilovoltamperes is understood to
be the kilovoltampere rating of the transformer. The base voltage is understood
to be the voltage rating of the low-tension winding if the ohmic values of resist¬
ance and leakage reactance are referred to the low-tension side of the transfor¬
mer and to be the voltage rating of the high-tension winding if they are referred
to the high-tension side of the transformer. The per-unit impedance of a trans¬
former is the same regardless of whether it is determined from ohmic values
SYSTEM MODELING 147
referred to the high-tension or low-tension sides of the transformers, as shown by

the following example.
Example 6.4 A single-phase transformer is rated 110/440 V, 2.5 kVA. Leak¬

age reactance measured from the low-tension side is 0.06 Q. Determine
leakage reactance in per unit.
Solution
t * • u , 0.1102 x 1000
Low-tension base impedance =-—- = 4.84 Q
2.5
In per unit
0.06
X = —— = 0.0124 per unit
4.84
If leakage reactance had been measured on the high-tension side, the value
would be
440
X = 0.06 = 0.96 Q
TlO
0.4402 x 1000
High-tension base impedance = —-—- = 77.5 Q
2.5
In per unit
0.96
X - = 0.0124 per unit
A great advantage in making per-unit computations is realized by the proper

selection of different bases for circuits connected to each other through a trans¬
former. To achieve the advantage in a single-phase system, the voltage bases for
the circuit connected through the transformer must have the same ratio as the
turns ratio of the transformer windings. With such a selection of voltage bases and
the same kilovoltampere base, the per-unit value of an impedance will be the same
when it is expressed on the base selected for its own side of the transformer as when
it is referred to the other side of the transformer and expressed on the base of
that side.
So the transformer is represented completely by its impedance (R -l- jX) in per
unit when magnetizing current is neglected. No per-unit voltage transformation
occurs when this system is used, and the current will also have the same per-unit
value on both sides of the transformer if magnetizing current is neglected.
Example 6.5 Three parts of a single-phase electric system are designated A,

B, and C and are connected to each other through transformers, as shown in
Fig. 6.19. The transformers are rated as follows:
A-B 10,000 kVA, 138/13.8 kV, leakage reactance 10%
B-C 10,000 kVA, 138/69 kV, leakage reactance 8%
If the base in circuit B is chosen as 10,000 kVA, 138 kV, find the per»unit
impedance of the 300-D resistive load in circuit C referred to circuits C, B,
and A. Draw the impedance diagram neglecting magnetizing current, trans¬
former resistances, and line impedances. Determine the voltage regulation if
the voltage at the load is 66 kV with the assumption that the voltage input
to circuit A remains constant.
Solution
Base voltage for circuit A — 0.1 x 138 = 13.8 kV

Base voltage for circuit C = 0.5 x 138 = 69 kV
692 x 1000
Base impedance of circuit C = = 476 Q
10,000
300
Per-unit impedance of load in circuit C = = 0.63 per unit
Because the selection of base in various parts of the system was

determined by the turns ratio of the transformers, the per-unit impedance of
the load referred to any part of the system will be the same. This is verified
as follows:
„ , „ . . 1382 x 1000 _
Base impedance of circuit B = — — = 1900 Q
10,000
Impedance of load referred to circuit B — 300 x 22 = 1200 Q

1200
Pre-unit impedance of load referred to B = —— = 0.63 per unit
iyoo
„ . f . . , 13.82 x 1000 ^
Base impedance of circuit A =-—— = 19 D
F 10,000
Impedance of load referred to A = 300 x 22 x 0.12 = 12 Q
12
Per-unit impedance of load referred to A = — = 0.63 per unit
1-10
A-B

SYSTEM MODELING 149
y'0.1 j 0.08
-npsw'—nnnnp-
;0.63+y'0
Figure 6.20 Impedance diagram for Example 6.5. Im¬
pedances are marked in per unit.
Figure 6.20 is the required impedance diagram with impedances marked

in per unit.
The calculation of regulation proceeds as follows:
Voltage at load = — = 0.957 4- ;0 per unit
j . 0.957 +j0 n
Load current = —--— = 1.52 + ;0 per unit
0.63 4- jO
Voltage input = (1.52 4- ;0)(;'0.10 -I- ;0.08) + 0.957
= 0.957 4- y'0.274 — 0.995 per unit
Voltage input = voltage at load with load removed
Therefore
0.995 - 0.957
Regulation =--x 100 = 3.97 %
Because of the advantage previously pointed out, the principle followed in

the above example in selecting the base for various parts of the system is always
followed in making computations by per unit or percent. The kilovoltampere
base should be the same in all parts of the system, and the selection of the base
kilovolts in one part of the system determines the base kilovolts to be assigned,
according to the turns ratios of the transformers, to the other parts of the
system. Following this principle of assigning base kilovolts allows us to combine
on one impedance diagram the per-unit impedances determined in different parts
of the system.
6.9 THREE-PHASE TRANSFORMERS
Three identical single-phase transformers may be connected so that the three

windings of one voltage rating are A-connected and the three windings of the
other voltage rating are Y-connected to form a three-phase transformer. Such a
transformer is said to be connected Y-A or A-Y. The other possible connections
are Y- Y and A-A. If the three single-phase transformers each have three windings
(a primary, secondary, and tertiary), two sets might be connected in Y and one
in A or two could be A-connected with one Y-connected. Rather than use three
identical single-phase transformers, a more usual unit is a three-phase trans¬

former where all three phases are on one iron structure. The theory is the same for
a three-phase transformer as for a three-phase bank of single-phase transformers.
Let us consider a numerical example of a Y-Y transformer composed of
three single-phase transformers each rated 25 MVA, 38.1/3.81 kV. The rating as
a three-phase transformer is, therefore, 75 MVA, 66/6.6 kV x 38.1 = 66).
Figure 6.21 shows the transformer with a balanced resistive load of 0.6 Q per
phase on the low-tension side. Windings of the primary and secondary drawn in
parallel directions are on the same single-phase transformer. Since the circuit is
balanced and we are assuming balanced three-phase voltages the neutral of the
load and the neutral of the low-voltage winding are at the same potential.
Therefore each 0.6-Q resistor is considered as directly connected across a
3.81-kV winding whether the neutrals are connected or not. On the high-voltage
side the impedance measured from line to neutral is
Figure 6.22a shows the same three transformers connected Y-A to the same
resistive load of 0.6 Q per phase. So far as the voltage magnitude at the low-
tension terminals is concerned the Y-A transformer can be replaced by a Y-Y
transformer bank having a turns ratio for each individual transformer (or for
each pair of phase windings of a three-phase transformer) of 38.1/2.2 kV as
shown in Fig. 6.22b. The transformers of Fig. 6.22a and b are equivalent if we are
not concerned with phase shift. As we shall see in Chap. 11, there is a phase shift
of the voltages between sides of the Y-A transformer that need not be considered
here. Figure 6.22b shows us that, viewed from the high-tension side of the trans¬
former, the resistance of each phase of the load is
0.6 = 180 Q
Here the multiplying factor is the square of the ratio of line-to-line voltages and
66'kN
Figure 6.21 Y-Y transformer rated 66/6.6 kV.

SYSTEM MODELING 151
(b)
Figure 6.22 Transformer of Fig. 6.21 (a) connected Y-A and (b) replaced by a Y-Y transformer with
line-to-line voltage ratio equal to that of the Y-A.
not the square of the turns ratios of the individual windings of the Y-A
transformer.
This discussion leads to the conclusion that to transfer the ohmic value of
impedance from the voltage level on one side of a three-phase transformer to the
voltage level on the other, the multiplying factor is the square of the ratio of
line-to-line voltages regardless of whether the transformer connection is Y-Y or
Y-A. Therefore in per-unit calculations involving transformers in three-phase
circuits we follow the same principle developed for single-phase circuits and
require the base voltages on the two sides of the transformer to have the same
ratio as the rated line-to-line voltages on the two sides of the transformer. The
kVA base is the same on each side.
Example 6.6 The three transformers rated 25 MVA, 38.1/3.81 kV are con¬
nected Y-A as shown in Fig. 6.22a with the balanced load of three 0.6-fi,
Y-connected resistors. Choose a base of 75 MVA, 66 kV for the high-tension
side of the transformer and specify the base for the low-tension side. Deter¬
mine the per-unit resistance of the load on the base for the low-tension side.
Then determine the load resistance RL referred to the high-tension side and
the per-unit value of this resistance on the chosen base.
Solution The rating of the transformer as a three-phase bank is 75 MV A,

66Y/3.81A kV. So base for the low-tension side is 75 MVA, 3.81 kV.
Base impedance on the low-tension side is
(3.81)2
0.1935
75
and on the low-tension side
0.6
rl = = 3.10 per unit
0.1935
Base impedance on the high-tension side is

(66)2
= 58.1 Q
75
and we have seen that the resistance per phase referred to the high-tension
side is 180 Q. So
n =j«0 = 3.10 per unit
L 58.1
The resistance R and leakage reactance X of a three-phase transformer

are measured by the short-circuit test as discussed for single-phase transformers.
In a three-phase equivalent circuit R and X are connected in each line to an
ideal three-phase transformer. Since R and A" will have the same per-unit value
whether on the low-tension or the high-tension side of the transformer, the
single-phase equivalent circuit will account for the transformer by the impedance
R + jX in per unit without the ideal transformer if all quantities in the circuit
are in per unit with the proper selection of base.
Table A. 5 in the Appendix lists typical values of transformer impedances
which are essentially equal to the leakage reactance since the resistance is
usually less than 0.01 per unit.
Example 6.7 A three-phase transformer is rated 400 MVA, 220Y/22A kV.

The short-circuit impedance measured on the low-voltage side of the trans¬
former is 0.121 £2 and because of the low resistance this value may be con¬
sidered equal to the leakage reactance. Determine the per-unit reactance of
the transformer and the value to be used to represent this transformer in a
system whose base on the high-tension side of the transformer is 100 MVA,
230 kV.
Solution On its own base the transformer reactance is
0.121
= 0.10 per unit
(22)2/400
On the chosen base the reactance becomes

2
100
0.1 = 0.0228 per unit
400
SYSTEM MODELING 153
6.10 PER-UNIT IMPEDANCES OF

THREE-WINDING TRANSFORMERS
Both the primary and secondary windings of a two-winding transformer have

the same kilovoltampere rating, but all three windings of a three-winding trans¬
former may have different kilovoltampere ratings. The impedance of each wind¬
ing of a three-winding transformer may be given in percent or per unit based on
the rating of its own winding, or tests may be made to determine the im¬
pedances. In any case, however, all the per-unit impedances in the impedance
diagram must be expressed on the same kilovoltamepre base.
Three impedances may be measured by the standard short-circuit test, as
follows:
Zps = leakage impedance measured in primary with secondary short-circuited
and tertiary open
Zpt — leakage impedance measured in primary with tertiary short-circuited and
secondary open
Zst = leakage impedance measured in secondary with tertiary short-circuited
and primary open
If the three impedances measured in ohms are referred to the voltage of one of
the windings, the impedances of each separate winding referred to that same
winding are related to the measured impedances so referred as follows:
Zps = Zp + Zs
Zpt — Zp + Z( (6.38)
zst = zs + z,
where Zp, Zs, and Z, are the impedances of the primary, secondary, and tertiary
windings referred to the primary circuit if Zps, Zpt, and Zst are the measured
impedances referred to the primary circuit. Solving Eqs. (6.38) simultaneously
yields
(6.39)
The impedances of the three windings are connected in star to represent the
single-phase equivalent circuit of the three-winding transformer with magnetiz¬
ing current neglected, as shown in Fig. 6.23. The common point is fictitious and
unrelated to the neutral of the system. The points p, s, and t are connected to the
parts of the impedance diagrams representing the parts of the system connected
to the primary, secondary, and tertiary windings of the transformer. Since the
ohmic values of the impedances must be referred to the same voltage, it follows
that conversion to per-unit impedance requires the same kilovoltampere base for
all three circuits and requires voltage bases in the three circuits that are in the
same ratio as the rated line-to-line voltages of the three circuits of the
transformer.
Figure 6.23 The equivalent circuit of a three-winding transformer.

Points p, s, and t link the circuit of the transformer to the appro¬
priate equivalent circuits representing parts of the system con¬
nected to the primary, secondary, and tertiary windings.
Example 6.8 The three-phase ratings of a three-winding transformer are:

Primary Y-connected, 66 kV, 15 MVA
Secondary Y-connected, 13.2 kV, 10.0 MVA
Tertiary A-connected, 2.3 kV, 5 MVA
Neglecting resistance, the leakage impedances are
Zps = 7% on 15-MVA 66-kV base
Zpt = 9% on 15-MVA 66-kV base
ZSI = 8% on 10.0-MVA 13.2-kV base
Find the per-unit impedances of the star-connected equivalent circuit for a

base of 15 MVA, 66 kV in the primary circuit.
Solution With a base of 15 MVA, 66 kV in the primary circuit, the proper

bases for the per-unit impedances of the equivalent circuit are 15 MVA,
66 kV for primary-circuit quantities, 15 MVA, 13.2 kV for secondary-circuit
quantities, and 15 MVA, 2.3 kV for tertiary-circuit quantities.
Zps and Zpt were measured in the primary circuit and are therefore
already expressed on the proper base for the equivalent circuit. No change of
voltage base is required for Zst. The required change in base kVA for Zsf is
made as follows:
Zsr = 8% x 15/10 = 12%
In per unit on the specified base
Zp = tC/0.07 + 70.09 — 70.12) = 70.02 per unit
Zs — i(y0.07 + 70.12 — 70.09) = 70.05 per unit
Zt — i(70.09 + 70.12 — 70.07) = 70.07 per unit
Example 6.9 A constant-voltage source (infinite bus) supplies a purely resis¬

tive 5-MW 2.3-kV load and a 7.5-MVA 13.2-kV synchronous motor having a
subtransient reactance of X" = 20%. The source is connected to the primary
of the three-winding transformer described in Example 6.8. The motor and
SYSTEM MODELING 155
>0.05
Figure 6.24 Impedance diagram for Example 6.9.
resistive load are connected to the secondary and tertiary of the transformer.
Draw the impedance diagram of the system and mark the per-unit im¬
pedances for a base of 66 kV, 15 MVA in the primary.
Solution The constant-voltage source can be represented by a generator

having no internal impedance.
The resistance of the load is 1.0 per unit on a base of 5 MYA, 2.3 kV in
the tertiary. Expressed on a 15-MVA 2.3-kV base the load resistance is
15
R — 1.0 x — = 3.0 per unit
Changing the reactance of the motor to a base of 15 MVA, 13.2 kV

yields
X" — 0.20— = 0.40 per unit
Figure 6.24 is the required impedance diagram.
6.11 THE ONE-LINE DIAGRAM
We now have the circuit models for transmission lines, synchronous machines,
and transformers. We shall see next how to portray the assemblage of these
components to model a complete system. Since a balanced three-phase system is
always solved as a single-phase circuit composed of one of the three lines and a
neutral return, it is seldom necessary to show more than one phase and the
neutral return when drawing a diagram of the circuit. Often the diagram is
simplified further by omitting the completed circuit through the neutral and by
indicating the component parts by standard symbols rather than by their equiva¬
lent circuits. Circuit parameters are not shown, and a transmission line is repre¬
sented by a single line between its two ends. Such a simplified diagram of an
electric system is called a one-line diagram. It indicates by a single line and
standard symbols the transmission lines and associated apparatus of an electric
system.
Machine or rotating Power circuit breaker,

armature (basic) oil or other liquid
Air circuit breaker

Two-winding power
transformer Three-phase, three-wire
delta connection
Three-winding power
transformer
Fuse
Three-phase wye,
neutral ungrounded T
Three-phase wye,
Current transformer
neutral grounded
Potential transformer —3 * -3*-
Ammeter and voltmeter
Figure 6.25 Apparatus symbols.
The purpose of the one-line diagram is to supply in concise form the

significant information about the system. The importance of different features of a
system varies with the problem under consideration, and the amount of informa¬
tion included on the diagram depends on the purpose for which the diagram is
intended. For instance, the location of circuit breakers and relays is unimportant
in making a load study. Breakers and relays are not shown if the primary
function of the diagram is to provide information for such a study. On the other
hand, determination of the stability of a system under transient conditions result¬
ing from a fault depends on the speed with which relays and circuit breakers
operate to isolate the faulted part of the system. Therefore, information about
the circuit breakers may be of extreme importance. Sometimes one-line diagrams
include information about the current and potential transformers which connect
the relays to the system or which are installed for metering. The information
found on a one-line diagram must be expected to vary according to the problem
at hand and according to the practice of the particular company preparing the
diagram.
The American National Standards Institute (ANSI) and the Institute of
Electrical and Electronics Engineers have published a set of standard symbols
for electrical diagrams.! Not all authors follow these symbols consistently,
especially in indicating transformers. Figure 6.25 shows a few symbols which are
commonly used. The basic symbol for a machine or rotating armature is a circle,
but so many adaptations of the basic symbol are listed that every piece of
rotating electric machinery in common use can be indicated. For anyone who is
not working constantly with one-line diagrams, it is clearer to indicate a particu¬
lar machine by the basic symbol followed by information on its type and rating.
+ See Graphic Symbols for Electrical and Electronics Diagrams, IEEE Std 315-1975.
SYSTEM MODELING 157
It is important to know the location of points where a system is connected

to ground in order to calculate the amount of current flowing when an unsym-
metrical fault involving ground occurs. The standard symbol to designate a
three-phase Y with the neutral solidly grounded is shown in Fig. 6-25. If a
resistor or reactor is inserted between the neutral of the Y and ground to limit the
flow of current to ground during a fault, the appropriate symbol for resistance or
inductance may be added to the standard symbol for the grounded Y. Most
transformer neutrals in transmission systems are solidly grounded. Generator
neutrals are usually grounded through fairly high resistances and sometimes
through inductance coils.
Figure 6.26 is the one-line diagram of a very simple power system. Two
generators, one grounded through a reactor and one through a resistor, are
connected to a bus and through a step-up transformer to a transmission line.
Another generator, grounded through a reactor, is connected to a bus and
through a transformer to the opposite end of the transmission line. A load is
connected to each bus. On the diagram information about the loads, the ratings
of the generators and transformers, and reactances of the different components
of the circuit is often given.
6.12 IMPEDANCE AND REACTANCE DIAGRAMS
In order to calculate the performance of a system under load conditions or upon

the occurrence of a fault, the one-line diagram is used to draw the single-phase
equivalent circuit of the system. Figure 6.27 combines the equivalent circuits of
Generators Load Transformer T, Transmission line Transformer T2 Load Gen. 3

1 and 2 A B
Figure 6.27 Impedance diagram corresponding to the one-line diagram of Fig. 6.26.
the various components shown in Fig. 6.26 to form the impedance diagram of
the system. If a load study is to be made, the lagging loads A and B are
represented by resistance and inductive reactance in series. The impedance dia¬
gram does not include the current-limiting impedances shown in the one-line
diagram between the neutrals of the generators and ground because no current
flows in the ground under balanced conditions and the neutrals of the generators
are at the potential of the neutral of the system. Since the magnetizing current of
a transformer is usually insignificant compared with the full-load current, the
shunt admittance is usually omitted in the equivalent circuit of the transformer.
As previously mentioned, resistance is often omitted when making fault
calculations even in digital-computer programs. Of course, omission of resis¬
tance introduces some error, but the results may be satisfactory since the induc¬
tive reactance of a system is much larger than its resistance. Resistance and
inductive reactance do not add directly, and impedance is not far different from
the inductive reactance if the resistance is small. Loads which do not involve
rotating machinery have little effect on the total line current during a fault and
are usually omitted. Synchronous motor loads, however, are always included in
making fault calculations since their generated emfs contribute to the short-
circuit current. The diagram should take induction motors into account by a
generated emf in series with an inductive reactance if the diagram is to be used
to determine the current immediately after the occurrence of a fault. Induction
motors are ignored in computing the current a few cycles after the fault occurs
because the current contributed by an induction motor dies out very quickly
after the induction motor is short-circuited.
If we decide to simplify our calculation of fault current by omitting all static
loads, all resistances, the magnetizing current of each transformer, and the capa¬
citance of the transmission line, the impedance diagram reduces to the reactance
diagram of Fig. 6.28. These simplifications apply to fault calculations only and
not to load-flow studies, which are the subject of Chap. 8. If a computer is
available, such simplification is not necessary.
The impedance and reactance diagrams discussed here are sometimes called
the positive-sequence diagrams since they show impedances to balanced currents
in a symmetrical three-phase system. The significance of this designation will
become apparent when Chap. 11 is studied.
If data had been supplied with the one-line diagram we could mark values of
reactance on Fig. 6.28. If ohmic values were to be shown, all would have to be
Figure 6.28 Reactance diagram adapted from Fig. 6.27 by omitting all loads, resistances, and shunt
admittances.
SYSTEM MODELING 159
referred to the same voltage level such as the transmission-line side of the
transformer. As we have concluded, however, when bases are specified properly
for the various parts of a circuit connected by a transformer the per-unit values
of impedances determined in their own part of the system are the same when
viewed from another part. Therefore it is necessary only to compute each im¬
pedance on the base of its own part of the circuit. The great advantage of using
per-unit values is that no computations are necessary to refer an impedance from
one side of a transformer to the other.
The following points should be kept in mind:
1. A base kilovolts and base kilovoltamperes is selected in one part of the

system. The base values for a three-phase system are understood to be line-to-
line kilovolts and three-phase kilovoltamperes or megavoltamperes.
2. For other parts of the system, that is, on other sides of transformers, the base
kilovolts for each part is determined according to the line-to-line voltage
ratios of the transformers. The base kilovoltamperes will be the same in all
parts of the system. It will be helpful to mark the base kilovolts of each part
of the system on the one-line diagram.
3. Impedance information available for three-phase transformers will usually be
in per unit or percent on the base determined by the ratings.
4. For three single-phase transformers connected as a three-phase unit the three-
phase ratings are determined from the single-phase rating of each individual
transformer. Impedance in percent for the three-phase unit is the same as that
for each individual transformer.
5. Per-unit impedance given on a base other than that determined for the part of
the system in which the element is located must be changed to the proper
base by Eq. (2.52).
Example 6.10 A 300 MVA, 20 kV three-phase generator has a subtransient

reactance of 20%. The generator supplies a number of synchronous motors
over a 64-km (40-mi) transmission line having transformers at both ends, as
shown on the one-line diagram of Fig. 6.29. The motors, all rated 13.2 kV,
are represented by just two equivalent motors. The neutral of one motor
is grounded through reactance. The neutral of the second motor M2 is not
connected to ground (an unusual condition). Rated inputs to the motors are
200 MVA and 100 MVA for M, and M2, respectively. For both motors
X" = 20%. The three-phase transformer 7j is rated 350 MVA, 230/20 kV
03.8 kV) Mi
(20 kV)
r. (230 kV)
Figure 6.29 One-line diagram for Example 6.10.

with leakage reactance of 10%. Transformer T2 is composed of three single¬

phase transformers each rated 127/13.2 kV, 100 MVA with leakage reac¬
tance of 10%. Series reactance of the transmission line is 0.5 fi/km. Draw the
reactance diagram with all reactances marked in per unit. Select the genera¬
tor rating as base in the generator circuit.
Solution The three-phase rating of transformer T2 is
3 x 100 = 300 kVA
and its line-to-line voltage ratio is
y/3 x 127/13.2 = 220/13.2 kV

A base of 300 MVA, 20 kV in the generator circuit requires a 300 MVA base
in all parts of the system and the following voltage bases:
In the transmission line: 230 kV (since Tt is rated 230/20 kV)
13.2
In the motor circuit: 230—— = 13.8 kV
These bases are shown in parentheses on the one-line diagram of Fig. 6.29.
The reactances of the transformers converted to the proper base are
300
Transformer 7% X = 0.1 x = 0.0857 per unit
/13.2\2
Transformer T2: X = 0.l(^-^l = 0.0915 per unit
The base impedance of the transmission line is
(230)2
= 176.3 Q
300
and the reactance of the line is
0.5 x 64
= 0.1815 per unit
176.3
300
Reactance of motor =0.2
200 )(§)’-»■ 2745 per unit
'300
Reactance of motor M, = 0.2
100 )(§)■-» 5490 per unit
Figure 6.30 is the required reactance diagram.
Example 6.11 If the motors Mx and M2 of Example 6.10 have inputs of 120
and 60 MW respectively at 13.2 kV, and both operate at unity power factor,
find the voltage at the terminals of the generator.
SYSTEM MODELING 161
Figure 6.30 Reactance diagram for Example 6.10. Reactances are in per unit on the specified base.
Solution Together the motors take 180 MW, or
180
300 = 0 6 per Umt
Therefore with V and / at the motors in per unit
| V | • | /1 =0.6 per unit

and since
V = j.~ = 0.9565/0° per unit

lJ.o
I =
05^5 = °'6273^ unit
At the generator
V = 0.9565 + 0.6273(;0.0915 + yO.1815 + ;0.0857)

= 0.9565 + ;0.2250 = 0.9826/13.2° per unit
The generator terminal voltage is
0.9826 x 20 = 19.65 kV
6.13 THE ADVANTAGES OF PER-UNIT COMPUTATIONS
Making computations for electric systems in terms of per-unit values simplifies

the work greatly. A real appreciation of the value of the per-unit method comes
through experience. Some of the advantages of the method are summarized
briefly below.
1. Manufacturers usually specify the impedance of a piece of apparatus in per¬

cent or per unit on the base of the nameplate rating.
2. The per-unit impedances of machines of the same type and widely different
rating usually lie within a narrow range, although the ohmic values differ
materially for machines of different ratings. For this reason, when the^ im-
pedance is not known definitely, it is generally possible to select from tab¬
ulated average values a per-unit impedance which will be reasonably correct.
Experience in working with per-unit values brings familiarity with the proper
values of per-unit impedance for different types of apparatus.
3. When impedance in ohms is specified in an equivalent circuit, each impedance
must be referred to the same circuit by multiplying it by the square of the
ratio of the rated voltages of the two sides of the transformer connecting
the reference circuit and the circuit containing the impedance. The per-unit
impedance, once it is expressed on the proper base, is the same referred to
either side of any transformer.
4. The way in which transformers are connected in three-phase circuits does not
affect the per-unit impedances of the equivalent circuit, although the transfor¬
mer connection does determine the relation between the voltage bases on the
two sides of the transformer.
6.14 SUMMARY
The introduction in this chapter of the simplified equivalent circuits for the
synchronous generator and transformer is of great importance for the remainder
of our discussion throughout this book.
We have seen that the synchronous generator will deliver an increasing
amount of reactive power to the system to which it is connected as its excitation
is increased. Conversely, as its excitation is reduced it will furnish less reactive
power and when underexcited will draw reactive power from the system. This
analysis was made on the assumption of a generator supplying such a large
system that the terminal voltage remained constant. In Chap. 8, we will extend
this analysis to a generator supplying a system represented by its Thevenin
equivalent.
Per-unit calculations will be used almost continuously throughout the chap¬
ters to follow. We have seen how the transformer is eliminated in the equivalent
circuit by the use of per-unit calculations. It is important to remember that the
square root of three does not enter the detailed per-unit computations because of
the specification of a base line-to-line voltage and base line-to-neutral voltage
related by the square root of three.
The concept of proper selection of base in the various parts of a circuit
linked by transformers and the calculation of parameters in per unit on the base
specified for the part of the circuit in which the parameters exist is fundamental
in building an equivalent circuit from a one-line diagram.
SYSTEM MODELING 163
PROBLEMS
6.1 Show the steps by which the sum of the three mmfs expressed in Eqs. (6.6) to (6.8) can be
equated to the traveling wave of mmf given in Eq. (6.10).
6.2 Determine the highest speed at which two generators mounted on the same shaft can be driven
so that the frequency of one generator is 60 Hz and the frequency of the other is 25 Hz. How many
poles does each machine have?
6.3 The synchronous reactance of a generator is 1.0 per unit, and the leakage reactance of its
armature is 0.1 per unit. The voltage to neutral of phase a at the bus of a large system to which the
generator is connected is 1.0/0° per unit, and the generator is delivering a current Ia equal
to 1.0 /—30° per unit. Neglect resistance of the windings and find (a) the voltage drop°in the
machine due to armature reaction, (b) the no-load voltage to neutral Eg of phase a of the generator,
and (c) the per-unit values of P and Q delivered to the bus.
6.4 Solve parts (b) and (c) of Prob. 6.3 for Ia = 1.0/30 per unit and compare the results of these two
problems.
6.5 For a certain armature current in a synchronous generator the mmf due to the field current is
twice that due to armature reaction. Neglect saturation and find the ratio of the voltage Er generated
by the air-gap flux to the no-load voltage of the generator (a) when armature current Ia is in phase
with £r, (b) when I0 lags Er by 90°, and (c) when /„ leads Er by 90°.
6.6 A single-phase transformer is rated 440/220 V, 5.0 kVA. When the low-tension side is short-
circuited and 35 V is applied to the high-tension side, rated current flows in the windings and the
power input is 100 W. Find the resistance and reactance of the high- and low-tension windings if the
power loss and ratio of reactance to resistance is the same in both windings.
6.7 A single-phase transformer rated 30 kVA, 1200/120 V is connected as an autotransformer to
supply 1320 V from a 1200-V bus.
(a) Draw a diagram of the transformer connections showing the polarity marks on the wind¬
ings and directions chosen as positive for current in each winding so that the currents will be in
phase.
(b) Mark on the diagram the values of rated current in the windings and at the input and
output.
(c) Determine the rated kilovoltamperes of the unit as an autotransformer.
(d) If the efficiency of the transformer connected for 1200/120-V operation at rated load unity
power factor is 97%, determine its efficiency as an autotransformer with rated current in the wind¬
ings and operating at rated voltage to supply a load at unity power factor.
6.8 Solve Prob. 6.7 if the transformer is to supply 1080 V from a 1200-V bus
6.9 A A-connected resistive load of 8000 kW is connected to the low-tension, A-connected side of a
Y-A transformer rated 10,000 kVA, 138/13.8 kV. Find the load resistance in ohms in each phase as
measured from line to neutral on the high-tension side of the transformer. Neglect transformer
impedance and assume rated voltage is applied to the transformer primary.
6.10 Solve Prob. 6.7 if the same resistors are reconnected in Y.
6.11 Three transformers, each rated 5 kVA, 220 V on the secondary side, are connected A-A and
have been supplying a 15 kW purely resistive load at 220 V. A change is made which reduces the
load to 10 kW, still purely resistive. Someone suggests that, with two-thirds of the load, one trans¬
former can be removed and the system can be operated open-A. Balanced three-phase voltages will
still be supplied to the load since two of the line voltages (and thus also the third) will be unchanged.
To investigate further the suggestion
(a) Find each of the line currents (magnitude and angle) with the 10 kW load and the transformer
between a and c removed. (Assume Vab = 220/0° V, sequence a b c.)
(ib) Find the kilovoltamperes supplied by each of the remaining transformers.
(c) What restriction must be placed on the load for open-A operation with these transformers?
(d) Think about why the individual transformer kilovoltampere values include a Q component when
the load is purely resistive.
6.12 A transformer rated 200 MVA, 345Y/20.5A kV connects a load of rated 180 MV A, 22.5 kY*0.8
power factor lag to a transmission line. Determine (a) the rating of each of three single-phase
transformers which when properly connected will be equivalent to the three-phase transformer and
(b) the complex impedance of the load in per unit in the impedance diagram if the base in the
transmission line is 100 MVA, 345 kV.
6.13 A 120 MVA, 19.5 kV generator has Xs = 1.5 per unit and is connected to a transmission line by
a transformer rated 150 MVA, 230Y/18A kV with X = 0.1 per unit. If the base to be used in the
calculations is 100 MVA, 230 kV for the transmission line, find the per-unit values to be used for the
transformer and generator reactances.
6.14 A transformer’s three-phase rating is 5000 kVA, 115/13.2 kV, and its impedance is 0.007 +
y'0.075 per unit. The transformer is connected to a short transmission line whose impedance is
0.02 + y0.10 per unit on a base of 10 MVA, 13.2 kV. The line supplies a three-phase load rated
3400 kW, 13.2 kV, with a lagging power factor of 0.85. If the high-tension voltage remains constant
at 115 kV when the load at the end of the line is disconnected, find the voltage regulation at the load.
Work in per unit and choose as base 10 MVA, 13.2 kV at the load.
6.15 The one-line diagram of an unloaded power system is shown in Fig. 6.31. Reactances of the two
sections of transmission line are shown on the diagram. The generators and transformers are rated as
follows:
Generator 1: 20 MVA, 13.8 kV, X" = 0.20 per unit
Generator 2: 30 MVA, 18 kV, X" = 0.20 per unit
Generator 3: 30 MVA, 20 kV, X" = 0.20 per unit
Transformer T,: 25 MVA, 220Y/13.8A kV, X = 10%
Transformer T2 : Single-phase units each rated 10 MVA, 127/18 kV, X = 10%
Transformer T3 : 35 MVA, 220Y/22Y kV, X = 10%
Draw the impedance diagram with all reactances marked in per unit and with letters to indicate
points corresponding to the one-line diagram. Choose a base of 50 MVA, 13.8 kV in the circuit of
generator 1.
6.16 Draw the impedance diagram for the power system shown in Fig. 6.32. Mark impedances in per
unit. Neglect resistance, and use a base of 50 kVA, 138 kV in the 40-fi line. The ratings of the
generators, motors, and transformers are:
Generator 1: 20 MVA, 18 kV, X" = 20%
Generator 2: 20 MVA, 18 kV, X" = 20%
Synchronous motor 3: 30 MVA, 13.8 kV, X" = 20%
Three-phase Y-Y transformers: 20 MVA, 138Y/20Y kV, X = 10%
Three-phase Y-A transformers: 15 MVA, 138Y/13.8A kV, X = 10%
Figure 6.31 One-line diagram for Prob. 6.15.

SYSTEM MODELING 165
YYi
;'40n
^(Thd- —c}-(5)
_J20
-D[-D- j20n £L-^[-a-
5. 57" B
Figure 632 One-line diagram for Prob. 6.16.
6.17 If the voltage of bus C in Prob. 6.16 is 13.2 kV when the motor draws 24 MW at 0.8 power
factor leading, calculate the voltages of buses A and B. Assume that the two generators divide the
load equally. Give the answer in volts and in per unit on the base selected for Prob. 6.16. Find the
voltages at A and B when the circuit breaker connecting generator 1 to bus A is open while
the motor draws 12 MW at 13.2 kV with 0.8 power factor leading. All other circuit breakers remain
closed.
CHAPTER
SEVEN
NETWORK CALCULATIONS
The continued development of large, high-speed digital computers has brought

about a change in the relative importance of various techniques in the solution
of large networks. Digital-computer solutions depend upon network equations.
So it is important that the power-system engineer understand the formulation
of the equations from which, in obtaining a solution, the program that is
followed by the computer is derived.
This chapter is not meant to be a comprehensive review of network equa¬
tions but will serve to review and expand upon those methods of analysis upon
which programs for computer solutions of power-system problems are very
dependent.
Of particular importance in this chapter is the introduction of bus admit¬
tance and impedance matrices which will prove to be very useful in later work.
7.1 EQUIVALENCE OF SOURCES
A helpful procedure in some problems in network analysis is the substitution of

a source of constant current in parallel with an impedance for a constant emf
and series impedance. The two parts of Fig. 7.1 illustrate the circuits. Both
sources with their associated impedances are connected to a two-terminal
network having an input impedance ZL. The load may be considered a passive
network for the present; that is, any internal emfs in the load network are
assumed to be short-circuited and any current sources opened.
166
NETWORK CALCULATIONS 167
Figure 7.1 Circuits illustrating the

equivalence of sources.
For the circuit having the constant emf Eg and series impedance Zg, the
voltage across the load is
Vl = Eg — IL Zg (7.1)
where IL is the load current. For the circuit having a source of constant current
Is with the shunt impedance Zp, the voltage across the load is
VL = (Is-IL)Zp = IsZp-ILZp (7.2)
The two sources and their associated impedances will be equivalent if the voltage
VL is the same in both circuits. Of course, equal values of VL will mean equal
load currents IL for identical loads.
Comparison of Eqs. (7.1) and (7.2) shows that VL will be identical in both
circuits and therefore that the emf and its series impedance can be interchanged
with the current source and its shunt impedance provided
Eg = ISZP (7.3)
z9 = zp (7.4)
These relations show that a constant-current source and shunt impedance can be
replaced by a constant emf and series impedance if the emf is equal to the
product of the constant current and the shunt impedance and if the series im¬
pedance equals the shunt impedance. Conversely, a constant emf and series
impedance can be replaced by a constant-current source and shunt impedance if
the shunt impedance is identical to the series impedance and if the constant
current is equal to the value of the emf divided by its series impedance.
We have shown the conditions for equivalence of sources connected to a
passive network. By considering the principle of superposition we can show that
the same provisions apply if the output is an active network, that is, if the output
network includes voltage and current sources. To determine the contribution
from the supply if the output network is active, the principle of superposition
calls for shorting emfs in the output network and replacing current sources by
open circuits while impedances remain intact. Thus the output is a passive
network so far as the current component from the interchangeable sources is
concerned. To determine the current components due to the sources in the load
network, the emf of the supply source is shorted in one case and the current
source is opened in the other case. Thus, only Z0 or its equivalent Zp is con¬
nected across the input to the load to determine the effect of sources in the load
network regardless of which type of source is the supply. So, in applying super¬
position, the components contributed by the sources in the load network are
independent of the type of supply so long as the series impedance of the emf
equals the shunt impedance of the constant-current source. Therefore, the same
provisions for equivalence apply whether the load network is active or passive.
7.2 NODE EQUATIONS
The junctions formed when two or more pure elements (R, L, or C, or an ideal
source of voltage or current) are connected to each other at their terminals are
called nodes. Systematic formulation of equations determined at nodes of a
circuit by applying Kirchhoff’s current law is the basis of some excellent com¬
puter solutions of power-system problems. Usually it is convenient to consider
only those nodes to which more than two elements are connected and to call
these junction points major nodes.
In order to examine some features of node equations, we shall begin with the
one-line diagram of a simple system shown in Fig. 7.2. Generators are connected
through transformers to high-tension buses 1 and 3 and supply a synchronous
motor load at bus 2. For purposes of analysis, all machines at any one bus are
treated as a single machine and represented by a single emf and series reactance.
The reactance diagram, with reactances specified in per unit, is shown in Fig. 7.3.
Nodes are indicated by dots, but numbers are assigned only to major nodes. If
the circuit is redrawn with the emfs and the impedances in series connecting
them to the major nodes replaced by the equivalent current sources and shunt
admittances, the result is the circuit of Fig. 7.4. Admittance values in per unit are
shown instead of impedance values.
Figure 7.2 One-line diagram of a simple Figure 7.3 Reactance diagram for the system
system. of Fig. 7.2. Reactance values are in per unit.
Figure 7.4 Circuit of Fig. 7.3 with current sources replacing

the equivalent voltage sources. Values shown are admittances
in per unit.
Single-subscript notation will be used to designate the voltage of each bus

with respect to the neutral taken as the reference node 0. Applying Kirchhoff’s
current law at node 1 with current into the node from the source equated to
current away from the node gives
h = V1Ya + (V\ - V3)Yf + (Vx - V4)Yd (7.5)

and for node 4
0 = (U - V,)Yd + (V4 - V2)Y„ + (V4 - V3)re (7.6)
Rearranging these equations yields
h = Vx{Ya +Yf+ Yd) -V3Yf- V4Yd (7.7)
0=-V1Yd-V2Yh-V3Ye+ V4(Yd +Ye+ Yh) (7.8)
Similar equations can be formed for nodes 2 and 3, and the four equations can
be solved simultaneously for the voltages Vt, V2, V3, and V4. All branch currents
can be found when these voltages are known, and so the required number of
node equations is one less than the number of nodes in the network. A node
equation formed for the reference node would yield no further information. In
other words, the number of independent node equations is one less than the
number of nodes.
We have not written the other two equations because we can already see
how to formulate node equations in standard notation. In both Eqs. (7.7) and
(7.8) it is apparent that the current flowing into the network from current
sources connected to a node is equated to the sum of several products. At any
node one product is the voltage of that node times the sum of all the admittances
which terminate on the node. This product accounts for the current that flows
away from the node if the voltage is zero at each other node. Each other product
equals the negative of the voltage at another node times the admittance con¬
nected directly between the other node and the node at which the equation is
formulated. For instance, at node 1 a product is -V3Yf, which accounts for the
current away from node 1 when all node voltages are zero except that at node 3.
The standard form for the four independent equations in matrix form is
11 Y^ Yl2 Y13 Yl4 Vi

I2 Y21 y22 y23 y24 V2
h y31 y32 y33 f34 V3
_1
-1
>7
>7
>7
>7
U V4
IN
m
The symmetry of the equations in this form makes them easy to remember, and
their extension to any number of nodes is apparent. The order of the Y subscripts
is effect-cause', that is, the first subscript is that of the node at which the current
is being expressed, and the second subscript is that of the voltage causing this
component of current. The Y matrix is designated Ybus and called the bus admit¬
tance matrix.f It is symmetrical around the principal diagonal. The admittances
*ii, y22, y33, and Y44 are called the self-admittances at the nodes, and each
equals the sum of all the admittances terminating on the node identified by the
repeated subscripts. The other admittances are the mutual admittances of the
nodes, and each equals the negative of the sum of all admittances connected
directly between the nodes identified by the double subscripts. For the netwqrk
of Fig. 7.4 the mutual admittance T13 equals —Yf. Some authors call the self-
and mutual admittances of the nodes the driving-point and transfer admittances
of the nodes.
The general expression for the source current toward node k of a network
having N independent nodes, that is, N buses other than the neutral, is
h= i I',/. (7.10)
n = 1
One such equation must be written for each of the N buses at which the voltage
of the network is unknown. If the voltage is fixed at any node, the equation is
not written for that node. For instance, if both the magnitude and angle of the
voltages at two of the high-tension buses of our example are fixed, only two
equations are needed. Node equations would be written for the other two buses,
the only ones at which the voltage would be unknown. A known emf and series
impedance need not be replaced by the equivalent current source if one terminal
of the emf element is connected to the reference node, for then the node which
separates the emf and series impedance is one where voltage is known.
Example 7.1 Write in matrix form the node equations necessary to solve
for the voltages of the numbered buses of Fig. 7.4. The network is equiv¬
alent to that of Fig. 7.3. The emfs shown in Fig. 7.3 are Ea — 1.5/0°.
Eb = 1.5 /—36.87°. and Ec = 1.5/0^, all in per unit.
t Boldface type is used where one letter designates a matrix.

Solution The current sources are
1.5/0°
^1 = I3 — ~Ji25 = ^ ® — 71-20 per unit
r 1.5 7-36.87°
I2 —-JY25- = 1-2 !—126.87° = —0.72 — 70.96 per unit
Self-admittances in per unit are
7U= -75.0 — )4.0 -jO.8 = -79.8

y22 = -75.0 -72.5 -7O.8 = -78.3
^33 = -;4.0 -72.5 -78.0 -7O.8 = -715.3
I44 - —75.0 —75.0 —78.O = — 7I8.O
and the mutual admittances in per unit are
7i2=y2i=0 723 = Y32 = +72.5
>13 = >31 = +74.o y24 = r42 = +75.0

y14 = t41 = +75.0 y34 = y43 - +78.0
The node equations in matrix form are
0 -71.20 -j 9.8 70.0 74.0 75.0 Vi

-0.72 - 70.96 70.0 -78.3 72.5 75.0 V2
0 -71.20 74.0 A5 -715.3 78.0 y3
0 75.0 75.0 78.0 -718.0 K
The square matrix above is recognized as the bus admittance matrix Ybus.
Example 7.2 Solve the node equations of the preceding example to find the
bus voltages by inverting the bus admittance matrix.
Solution Premultiplying both sides of the matrix equation of Example 7.1

by the inverse of the bus admittance matrix (determined by using a standard
program on a digital computer) yields
70.4774 70.3706 70.4020 70.4142 0 -71.20’ 1000' Vt

70.3706 70.4872 70.3922 70.4126 -0.72 -70.96 0 10 0 V2
_1
<N
O
0 0 10
0 0
70.4020 70.3922 70.4558 70.4232 V3

1
70.4142 70.4126 70.4232 70.4733 0 0 0 1 V4

1
The square matrix above obtained by inverting the bus admittance matrix is
called the bus impedance matrix Zbus. Performing the indicated matrix mul¬
tiplication yields
1.4111 -70.2668 Vi
1.3830 - 70.3508 V2
1.4059 -70.2824 y3
1.4009-70.2971 K
and so the node voltages are
Vx = 1.4111 - 70.2668 = 1.436 /-10.71° per unit

V2 = 1.3830 -70.3508 = 1.427 /-14.24° per unit
V3 = 1.4059 — 70.2824 = 1.434 /-11.36° per unit
V4 = 1.4009 -;0.2971 = 1.432 /-11.97° per unit
7.3 MATRIX PARTITIONING
A useful method of matrix manipulation, called partitioning, consists in recogniz¬

ing various parts of a matrix as submatrices which are treated as single elements
in applying the usual rules of multiplication and addition. For instance, assume
a 3 x 3 matrix A, where
an a12 a13
A =
a21 a22 a23 (7.11)
a31 a3 2 a33
The matrix is partitioned into four submatrices by the horizontal and vertical
dashed lines. The matrix may be written
D E
A = (7.12)
F G
where the submatrices are
au a12 a13
D = E =
a21 a22_ a23.
F — [a31 a32] G — 033
To show the steps in matrix multiplication in terms of submatrices let us

assume that A is to be postmultiplied by another matrix B to form the product
C, where
*11
B= *21
(7.13)
*31
With partitioning as indicated,
H
B = (7.14)
J
where the submatrices are
and J = h31
Then the product is
D E H
C = AB = (7.15)
F G J
The submatrices are treated as single elements to obtain
DH + EJ
(7.16)
FH + GJ
The product is finally determined by performing the indicated multiplication and

addition of the submatrices.
If C is composed of the submatrices M and N so that
(7.17)
comparison with Eq. (7.16) shows
M = DH + EJ (7.18)
N = FH + GJ (7.19)
If we wish to find only the submatrix N, partitioning shows that
N — [a3i a32] *., “I- d'l'ib-

33^31
b21
— a31*11 T fl32*21 T a33*31 (7.20)
The matrices to be multiplied must be compatible originally. Each vertical

partitioning line between columns r and r + 1 of the first factor requires a
horizontal partitioning line between rows r and r + 1 of the second factor in
order for the submatrices to be multiplied. Horizontal partitioning lines may be
drawn between any rows of the first factor, and vertical lines between any
columns of the second, or omitted in either or both. An example that applies
matrix partitioning appears at the end of the next section.
7.4 NODE ELIMINATION BY MATRIX ALGEBRA
Nodes may be eliminated by matrix manipulation of the standard node equa¬

tions. However, only those nodes at which current does not enter or leave the
network can be eliminated.
The standard node equations in matrix notation are expressed as
I = Yb„,V (7.21)
where I and V are column matrices and Ybus is a symmetrical square matrix. The
column matrices must be so arranged that elements associated with nodes to be
eliminated are in the lower rows of the matrices. Elements of the square admit¬
tance matrix are located correspondingly. The column matrices are partitioned
so that the elements associated with nodes to be eliminated are separated from
the other elements. The admittance matrix is partitioned so that elements
identified only with nodes to be eliminated are separated from the other ele¬
ments by horizontal and vertical lines. When partitioned according to these
rules, Eq. (7.21) becomes
lA K L
h LT M V*
where I* is the submatrix composed of the currents entering the nodes to be

eliminated and \x is the submatrix composed of the voltages of these nodes. Of
course, every element in lx is zero, for the nodes could not be eliminated other¬
wise. The self- and mutual admittances composing K are those identified only
with nodes to be retained. M is composed of the self- and mutual admittances
identified only with nodes to be eliminated. It is a square matrix whose order is
equal to the number of nodes to be eliminated. L and its transpose LT are
composed of only those mutual admittances common to a node to be retained
and to one to be eliminated.
Performing the multiplication indicated in Eq. (7.22) gives
lA = KV^ + LV* % (7.23)

and
I* = L t\a + MV* (7.24)
Since all elements of I* are zero, subtracting L^V,, from both sides of Eq. (7.24)
and multiplying both sides by the inverse of M (denoted by M-1) yields
-M~1Lt\a = V* (7.25)
This expression for V* substituted in Eq. (7.23) gives
G = KV^ — LM_1LrY^ (7.26)
which is a node equation having the admittance matrix
Ybus = K-LM1LT (7.27)

This admittance matrix enables us to construct the circuit with the unwanted
nodes eliminated, as we shall see in the following example.
Example 7.3 If the generator and transformer at bus 3 are removed from the
circuit of Fig. 7.3, eliminate nodes 3 and 4 by the matrix-algebra procedure
just described, find the equivalent circuit with these nodes eliminated, and
find the complex power transferred into or out of the network at nodes 1
and 2. Also find the voltage at node 1.
Solution The bus admittance matrix of the circuit partitioned for elimina¬
tion of nodes 3 and 4 is
-79.8 0.0 : 74.0 75.0

K L 0.0 -78.3 : 72.5 75.0
17 M 74.0 A.5 : -714.5 78-0
L A0 75.0 : 78.0 -718.0
The inverse of the submatrix in the lower right position is
-718.0 -78.0 70.0914 70.0406

M~1 - 1
-197 -78.0 -714.5 J [70.0406 70.0736
Then
iO
O
—
ON
■'T
O
O
Tj"
O
NO
74.0 75.0 74.0 72.5

i
LM xLr =
[72.5 75.0 J 70.0406 70.0736 (75.0 75.0J
74.9264 74.0736
74.0736 73.4264
0.0
Ybus = K — LM1 - LM_1Lr
-78.3
—74.8736 74.0736
Ybus
y'4.0736 -74.8736
Examination of the matrix shows us that the admittance between the two
remaining buses 1 and 2 is —74.0736, the reciprocal of which is the per-unit
impedance between these buses. The admittance between each of these buses
and the reference bus is
—74.8736 — (—74.0736) = —70.800 per unit
The resulting circuit is shown in Fig. 1.5a. When the current sources are
converted to their equivalent emf sources the circuit, with impedances in per
unit, is that of Fig. 1.5b. Then the current is
1.5/01 - 1.5 7-36.87° 1.5 - 1.2 + 70.9

1 ~ 7(1.25 4- 1.25 + 0.2455) _ 7(2.7455)
= 0.3278 - 70.1093 = 0.3455 A 18.44 per unit

—J4.0736 J 1.25 JO.2455 J1.25

——*
© © © ©
) 1 1-J0.8 —JO. 81
3 <= c>
e/J 1.5/0! Ebf
(a)
1 (b)
Figure 7.5 Circuit of Fig. 7.3 without the source at node 3 (a) with the equivalent current sources
and (b) with the original voltage sources at nodes 1 and 2.
Power out of source a is
1.5/0° x 0.3455/18.44° = 0.492 + jO.164 per unit
And power into source b is
1.5/- 36.87° x 0.3455/18.44° = 0.492 - jO.164 per unit
Note that the reactive voltamperes in the circuit equal
(0.3455)2 x 2.7455 = 0.328 = 0.164 + 0.164
The voltage at node 1 is
1.50 — jl.25(0.3278 - J0.1093) = 1.363 - j0.410 per unit
In the simple circuit of this example node elimination could have been
accomplished by Y-A transformations and by working with series and parallel
combinations of impedances. The matrix partitioning method is a general method
which is thereby more suitable for computer solutions. However, for the elimin¬
ation of a large number of nodes, the matrix M whose inverse must be found
will be large.
Inverting a matrix is avoided by eliminating one node at a time, and the
process is very simple. The node to be eliminated must be the highest numbered
node, and renumbering may be required. The matrix M becomes a single element
and M-1 is the reciprocal of the element. The original admittance matrix
partitioned into submatrices K, L, LT, and M is
K
... y,. ...
*11 *i„
v
1 bus Yki ••• YkJ ••• *kn (7.28)
Yn i ... Ynj •••

*nn
I/r M
the reduced (n — 1) x (n — 1) matrix will be, according to Eq. (7.27),
[Yn 1 (7.29)
and when the indicated manipulation of the matrices is accomplished, the ele¬
ment in row k and column j of the resulting (n — 1) x (n — 1) matrix will be
i<7»)
^nn
Each element in the original matrix K must be modified. When Eq. (7.28) is
compared to Eq. (7.30) we can see how to proceed. We multiply the element in
the last column and the same row as the element being modified by the element
in the last row and the same column as the element being modified. We then
divide this product by Enn and subtract the result from the element being
modified. The following example illustrates the simple procedure.
Example 7.4 Perform the node elimination of Example 7.3 by first removing
node 4 and then by removing node 3.
Solution As in Example 7.3, the original matrix now partitioned for re¬
moval of one node is
' -79.8 0.0 7*4.0 7*5.0"
0.0 --7*8.3 7*2-5 7*5.0
74.0
© -7*14.5 m
75-0 7*5-0 7*8-0 -7*18.0 _
To modify the element j2.5 in row 3, column 2 subtract from it the product
of the elements enclosed by rectangles and divided by the element in the
lower right corner. We find the modified element
78.0 x j5.0
Y32=j 2.5 = 74.7222
-7*18.0
Similarly the new element in row 1, column 1 is
7*5-0 x j5.0
Yn = -7*9.8 - —78.4111
-7*18.0
Other elements are found in the same manner to yield
—7*8.4111 j1.3889 j6.2222

Ybus 71.3889 —76.9111 74.7222
76.2222 74.7222 -710.9444
Reducing the above matrix to remove node 3 yields
-74.8736 74.0736
Y bus
74.0736 -74.8736
which is identical to the matrix found by the matrix-partitioning method

where two nodes were removed at the same time.
7.5 THE BUS ADMITTANCE AND IMPEDANCE MATRICES
In Example 7.2, we inverted the bus admittance matrix Ybus and called the
resultant matrix the bus impedance matrix Zbus. By definition
1
^bus =
- Y~1
* bus
(7.31)
and for a network of three independent nodes
Z11 Z12 Z13

^bus — Z21 Z22 Z23 (7.32)
Z31 Z32 Z33
Since Ybus is symmetrical around the principal diagonal, Zbus must be symmetri¬
cal in the same manner.
The impedance elements of Zbus on the principal diagonal are called driving-
point impedance of the nodes, and the off-diagonal elements are called the transfer
impedances of the nodes.
The bus admittance matrix need not be determined in order to obtain Zbus,
and in another section of this chapter we shall see how Zbus may be formulated
directly.
The bus impedance matrix is important and very useful in making fault
calculations as we shall see later. In order to understand the physical significance
of the various impedances in the matrix we shall compare them with the node
admittances. We can easily do so by looking at the equations at a particular
node. For instance, starting with the node equations expressed as
I = YbusV (7.33)
we have at node 2 of the three independent nodes
h=Y21vl + y22 v2 + y23 v3 (7.34)

® ©
Figure 7.6 Circuit for measuring Y22,

Y12, and Y32.
If V1 and F3 are reduced to zero by shorting nodes I and 3 to the reference node
and current I2 is injected at node 2, the self-admittance at node 2 is
(7.35)
Tl = C3=0
Thus, the self-admittance of a particular node could be measured by shorting all

other nodes to the reference node and then finding the ratio of the current
injected at the node to the voltage resulting at that node. Figure 7.6 illustrates
the method for a three-node reactive network. The result is obviously equivalent
to adding all the admittances directly connected to the node, as has been our
procedure up to now.
Figure 7.6 also serves to illustrate mutual admittance. At node 1 the equa¬
tion obtained by expanding Eq. (7.33) is
Ii ~ Y11V1 + Y12V2 + Yj3 V3 (7.36)
from which we see that
Y12 =
h (7.37)
V,
Thus the mutual admittance is measured by shorting all nodes except node 2 to
the reference node and injecting a current I2 at node 2, as shown in Fig. 7.6.
Then Y12 is the ratio of the negative of the current leaving the network in the
short circuit at node 1 to the voltage V2. The negative of the current leaving
node 1 is used since It is defined as the current entering the network. The
resultant admittance is the negative of the admittance directly connected beween
nodes 1 and 2, as we would expect.
We have made this detailed examination of the node admittances in order to
differentiate them clearly from the impedances of the bus impedance matrix.
We solve Eq. (7.33) by premultiplying both sides of the equation by
Ybus = Zbus to yield
V = Zbus I (7.38)
and we must remember when dealing with Zbus that V and I are column matrices
of the node voltages and the currents entering the nodes from current sources,
respectively. Expanding Eq. (7.38) for a network of three independent tjpdes

yields
V\ — Zii/i + Z12/2 + Z13/3 (7.39)
V2 = Z2Xll + Z22I2 + Z23I3 (7.40)
F3 = Z31 /j + Z32/2 + Z33/3 (7-41)
From Eq. (7.40) we see that the driving-point impedance Z22 is determined
by open-circuiting the current sources at nodes 1 and 3 and injecting the current
12 at node 2. Then
Z22 = £ (7.42)
Figure 7.7 shows the circuit described. Since Z22 is defined by opening the
current sources connected to the other nodes whereas T22 was found with the
other nodes shorted, we must not expect any reciprocal relation between these
two quantities.
The circuit of Fig. 7.7 also enables us to measure some transfer impedances,
for we see from Eq. (7.39) that with current sources I x and /3 open-circuited
/l=/3 = 0
and from Eq. (7.41)
/l=/3 = 0
Thus we can measure the transfer impedances Z12 and Z32 by injecting current
at node 2 and finding the ratios of Vx and V3 to 12 with the sources open at all
nodes except node 2. We note that a mutual admittance is measured with all but
one node short-circuited and that a transfer impedance is measured with all
sources open-circuited except one.
Equation (7.39) tells us that if we inject current into node 1 with current
sources 2 and 3 open, the only impedance through which Ix flows is Zn. Under
© ©
Figure 7.7 Circuit for measuring

^22> ^12> Z32-
the same conditions Eqs. (7.40) and (7.41) show that lx is causing voltages at
buses 2 and 3 expressed by
V2 = hZ21 and V3 = /,Z31 (7.43)
We cannot set up a physically realizable passive circuit with these coupling

impedances, but it is important to realize the implications of the preceding
discussion, for Zbus is sometimes used in load-flow studies and is extremely
valuable in fault calculations, as we shall see later.
Example 7.5 A capacitor having a reactance of 5.0 per unit is connected to

node 4 of the circuit of Examples 7.1 and 7.2. The emf’s Ea, Eb, and Ec
remain the same as in those examples. Find the current drawn by the
capacitor.
Solution The Thevenin equivalent of the circuit behind node 4 has an

emf of
Eth = 1.432/-11,97°
which is the voltage at node 4 before the capacitor is connected and is the
voltage V4 found in Example 7.2.
To find the Thevenin impedance the emfs are short-circuited or the
current sources are open-circuited, and the impedance between node 4 and
the reference node must be determined. From V = ZÎ we have at node 4
V4 — Z41IJ + Z42^2 T Z43/3 + Z44/4
With emfs short-circuited (or with the emfs and their series impedances
replaced by the equivalent current sources and shunt admittances with the
current sources open) no current is entering the circuit from sources at
nodes 1, 2, and 3. The ratio of a voltage applied at node 4 to the current this
voltage will cause to flow in the network is Z44, and this impedance is
known since Zbus was calculated in Example 7.2. By referring to that
example we find
Zth - Z44 =;0.4733
The current drawn by the capacitor is

1 477 /_ 11 Q7°
= 1. j / • = 0.316/78.03° per unit
c 7O.4733-;5.0 L-F
Example 7.6 If a current of -0.316/78.03° per unit is injected into the

network at node 4 of Examples 7.1, 7.2, and 7.5, find the resulting voltages at
nodes 1, 2, 3, and 4.
Solution With original emfs short-circuited, the voltages at the nodes due
only to the injected current will be calculated by making use of the bus
impedance network which was found in Example 7.2. The required^ im¬
pedances are in column 4 of Zbus. From V = ZbusI, the voltages with all
emfs shorted are
V1 = I4Z14 = -0.316/78.03° x 0.4142/90° = 0.1309/-11.97°

V2 = I4Z24 = -0.316/78.03° x 0.4126/90° = 0.1304/- 11.97°
V3 = I4Z34 = -0.316/78.03° x 0.4232/90° = 0.1337/-11.97°
V4 = 14Z44 = -0.316/78.03° x 0.4733/90° = 0.1496/-11.97°
By superposition, the resulting voltages are determined by adding the vol¬
tages caused by the injected current with emfs shorted to the node voltages
found in Example 7.2. The new node voltages are
Fj = 1.436/-10.71° + 0.1309/-11.97° = 1.567/-10.81° per unit
V2 = 1.427/-14.2° + 0.1304/-11.97° = 1.557/-14.04° per unit

V3 = 1.434/-11.4° + 0.1337/-11.97° = 1.568/-11.41° per unit
V4 = 1.432/-11.97° + 0.1496/-11.97° = 1.582/-11.97° per unit
Since the changes in voltages due to the injected current are all at the same
angle and this angle differs little from the angles of the original voltages, an
approximation will give satisfactory answers. The change in voltage magnitude
at a bus is about equal to the product of the magnitude of the per-unit current
and the magnitude of the appropriate impedance. These values added to the
original voltage magnitudes give the magnitudes of the new voltages very closely.
This approximation is valid because the network is purely reactive, but it pro¬
vides a good estimate where the reactance is considerably larger than the resis¬
tance, as is usually the case.
The last two examples illustrate the importance of the bus impedance matrix
and incidentally show how adding a capacitor at a bus will cause a rise in bus
voltages. The assumption that the angles of voltage and current sources remain
constant after connecting capacitors at a bus is not entirely valid if we are
considering operation of a power system. We shall consider capacitors again in
Chap. 8 and see an example using a computer load-flow program to calculate
the effect of capacitors.
7.6 MODIFICATION OF AN EXISTING

BUS IMPEDANCE MATRIX
Since Zbus is such an important tool in power-system analysis we shall now

examine how an existing ZbUS may be modified to add new buses or connect new
lines to established buses. Of course we could create a new Ybus and invert it, but
direct methods of modifying Zbus are available and are very much simpler than a
matrix inversion even for a small number of buses. Also when we know how to
modify Zbus we can see how to build it directly.t
We recognize several types of modifications involving the addition of a
branch having impedance Zb to a network whose original Zbus is known and is
identified as Zorig, annxn matrix.
In our analysis existing buses will be indentified by numbers or the letters h,
i, j, and k. The letter p will designate a new bus to be added to the network to
convert Zorig to an (n + 1) x (n + 1) matrix. Four cases will be considered.
Case 1: Adding Zbfrom a new bus p to the reference bus The addition of the
new bus p connected to the reference bus through Zb without a connection to
any of the buses of the original network cannot alter the original bus voltages
when a current Ip is injected at the new bus. The voltage Vp at the new bus is
equal to Ip Zb. Then
h
h
(744)
Z bus(new)
We note that the column matrix of currents multiplied by the new Zbus will not
alter the voltages of the original network and will result in the correct voltage at
the new bus p.
Case 2: Adding Zbfrom a new bus p to an existing bus k The addition of a

new bus p connected through Zb to an existing bus k with Ip injected at bus p
will cause the current entering the original network at bus k to become the sum
of Ik which is injected at bus k plus the current Ip coming through Zb as shown
in Fig. 7.8.
The current Ip flowing into bus k will increase the original Vk by the voltage
IpZkk; that is
Ho,,+ l,Za (7.45)
and Vp will be larger than the new Vk by the voltage Ip Zb. So
V, = n(„H0 + '„z«» + Lz» (7'46>

and
Vp = I j Zkk + I2Zk2 + --- + /A + Ip(Zkk + Zb) (7.47)
^k(orig)
t See H. E. Brown, Solutions of Large Networks by Matrix Methods, John Wiley & Sons, Inc.,
New York, 1975.
Orig. network
with bus k
and the
reference bus
extracted
Figure 7.8 Addition of new bus p connected

through impedance Zb to existing bus k.
We now see that the new row which must be added to Zorig in order to find Vp is
Zjci zk2 ■■■ Zkn (Zkk + Zb)
Since Zbus must be a square matrix around the principal diagonal we must add a
new column which is the transpose of the new row. The new column accounts
for the increase of all bus voltages due to Ip. The matrix equation is
(7.48)
Z bus(new)
Note that the first n elements of the new row are the elements of row k of Zorig
and the first n elements of the new column are the elements of column k of Znrio.
Case 3: Adding Zbfrom an existing bus k to the reference bus To see how to
alter Z(orig) by connecting an impedance Zb from an existing bus k to the refer¬
ence bus we shall add a new bus p connected through Zb to bus k. Then we
short-circuit bus p to the reference bus by letting Vp equal zero to yield the same
matrix equation as Eq. (7.48) except that Vp is zero. So for the modification we
proceed to create a new row and new column exactly the same as in case 2 but
we then eliminate the (n + 1) row and (n + 1) column, which is possible because
of the zero in the column matrix of voltages. We use the method developed in
Eqs. (7.28) to (7.30) to find each element Zhi in the new matrix where
Zb(n+ l)Z(„+ i)j

■^/ii'(new) Zbi{orig) (7.49)
Zkk + zb
4: Adding Zb between two existing buses j and k To add a branch
Case
impedance Zb between already established buses j and k we examine Fig. 7.9
which shows these buses extracted from the original network. The current Ib is
Orig. network
with buses
j, k and
reference bus
extracted
Figure 7.9 Addition of impedance Zb be¬

tween existing buses j and k. ■±
shown flowing through Zb from bus k to bus j. We now write some equations for
node voltages.
Vi = ZliI1 + ■" + Zjj(lj + Ib) + z i k(h ~ Ib) + "• (7.50)

and upon rearranging
V1 = Z11I1+-- + Zjjlj + ZlkIk + ■■ ' + h(Z\j — Zlk) (7.51)
Similarly
dS
(7.52)
+
' + Zjjlj + zjkik + •• ' + h(Zjj - Zjk)

II
*—»
N
*»<
• -l- ZkjIj + ZkkIk -l- • • ‘ + h(Zkj ~ Zkk) (7.53)

+
II
We need one more equation since Ib is unknown. So we write
Vk ~ Vj = ibzb (7.54)
or
0 = IbZb + Vj - vk (7.55)
and substituting the expressions for Vj and Vk given by Eqs. (7.52) and (7.53) in
Eq. (7.55) we obtain
0 — 7bZb + (Zj! — Zfcl)/ j + + (Zjj — Zkj)Ij + ••■
+ (Zjk — Zkk)Ik + + (Zjj + Zkk — 2 Zjk)Ib (7.56)

Collecting the coefficients of Ib and naming their sum Zbb yields
Zbb = Zb + Zjj + Zkk - 2Zjk (7.57)
By examining Eqs. (7.51) to (7.53) and (7.56) we can write the matrix
equation
N
N
>r
■ •
Vj Zjj Zjk
7
K orig Zkj — zkk
(7.58)
K Z„j — Znk
0 (Zji — zkl) ‘‘‘ (Zkj ~ zkk) Zbb

The new column is column j minus column k of Zorig with Zbb in the (n + 1) row.
The new row is the transpose of the new column.
Eliminating the (n + 1) row and (n + 1) column of the square matrix of
Eq. (7.58) in the same manner as previously we see that each element Zhi in the
new matrix is
7 _ 7_Zh(„ + i)Z(n+ 1)f . .
^hi(new) ^/ii(orig) 7 , 7 . 7 _ 97 \ ' y)
Z/j, T" Z, jj 1 t-'kk jk
We need not consider the case of introducing two new buses connected
by Zb because we could always connect one of these new buses through an
impedance to an existing bus or to the reference bus before adding the second
new bus.
Example 7.7 Modify the bus impedance matrix of Example 7.2 to account
for the connection of a capacitor having a reactance of 5.0 per unit between
bus 4 and the reference bus of the circuit of Fig. 7.4. Then find V4 using the
impedances of the new matrix and the current sources of Example 7.2.
Compare this value of V4 with that found in Example 7.6.
Solution We use Eq. (7.48) and recognize that Zorig is the 4x4 matrix of
Example 7.2, that subscript k = 4, and that Zb = — y'5.0 per unit to find
Vi 70.4142
V2 70.4126
7 70.4232
y3 ong
K 70.4733
0 70.4142 70.4126 7‘0.4232 7’0.4733 -74.5267
The terms in the fifth row and column were obtained by repeating the fourth
row and column of Zorig and noting that
Z55 = Z44 + Zb - j0.4733 -j5.0 = -j4.5267

\
Then eliminating the fifth row and column we obtain for Zbus(new) from
Eq. (7.49)
;0.4142 x ;0.4142
Zn =70.4774 - = 70.5153
—74.5267
70.4733 x ;0.4126
Z24 = ;0.4126 - - j0.4557
-74.5267
and other elements in a similar manner to give
70.5153 70.4084 70.4407 70.4575

7*0.4084 70.5248 70.4308 70.4557
Zb us (new)
70.4407 70.4308 7O.4954 70.4674
70.4575 70.4557 70.4674 70.5228
The column matrix of currents by which the new Zbus is multiplied to

obtain the new bus voltages is the same as in Example 7.2. So
V4 = j0.4575(-;1.2Q) + j0.4557(-0.72 -j0.96) + ;0.4674(-/1.20)
= 1.5474 — jO.3281 = 1.582/-11,97° per unit
as found in Example 7.6.
7.7 DIRECT DETERMINATION OF A BUS

IMPEDANCE MATRIX
We have seen how to determine Zbus by first finding Ybus and inverting it.
However, formulation of ZbUS directly is a straightforward process on the com¬
puter and simpler than inverting Ybus for a large network.
To begin we have a list of the impedances showing the buses to which they
are connected. We start by writing the equation for one bus connected through
an impedance Za to the reference bus as
vl = ilzm
and this can be considered as a matrix equation where each of the three matrices
has one row and one column. Now we might add a new bus connected to the
first bus or to the reference bus. For instance, if the second bus is connected to
the reference but through Zb we have the matrix equation
>1 Za 0 11
V2_ 0 zb h.
and we proceed to modify our matrix by adding other buses following the
procedures described in Sec. 7.6. Usually the buses of a network must be re¬
numbered to agree with the order in which they are to be added to Zbus as it is
built up.
Example 7.8 Determine Zbus for the network shown in Fig. 7.10 where im¬
pedances are shown in per unit. Preserve all three nodes.
70.3
Figure 7.10 Network for Example 7.8. Reference bus

Solution We start by establishing bus 1 with its impedance to the referqpce

bus and write
Vi
We then have the 1 x 1 bus impedance matrix
^bus — 71-2
To establish bus 2 with its impedance to bus 1 we follow Eq. (7.48) to write
j1.2 j1-2
Z* us (new)
71-2 71.4
The term j 1.4 above is the sum of j 1.2 and j0.2. The elements 71.2 in the new
row and column are the repetition of the elements of row 1 and column 1 of
the matrix being modified.
Bus 3 with the impedance connecting it to bus 1 is established by writing
71-2 71-2 yl.2

7*1-2 7*1.4 7*1.2
7*1-2 7*1.2 7*1.5
Since node 1 is the node to which the new node 3 is being connected, the
term j 1.5 above is the sum of Z1X of the matrix being modified and the
impedance Zb of the branch being connected to bus 1 from bus 3. The other
elements of the new row and column are the repetition of the row 1 and
column 1 of the matrix being modified since the new node is being con¬
nected to bus 1.
If we now decide to add the impedance Zb — j 1.5 from node 3 to the
reference bus, we follow Eq. (7.48) to connect a new bus 4 through Zb and
obtain the impedance matrix
7*1-2 7*1-2 7*1-2 ' 7*1-2

7*1-2 7*1-4 7*1-2 7*1-2
7*1-2 7*1-2 7*1-5 71.5
7*1-2 7*1-2 7*1-5 ^73X)
where 73.0 above is the sum of Z33 + Zb. The other elements in the new row
and column are the repetition of row 3 and column 3 of the matrix being
modified since bus 3 is the one which we are connecting to the reference bus
through Zb.
We now eliminate row 4 and column 4. Some of the elements of the new
matrix from Eq. (7.49) are
;1-2x;1.2
Zn=i 1.2 = jOJ2
AO
jl-2 x 71.2
Z22=jlA = j0.92
7 3.0
-7 _-7 _ .-1 O ;1-2x;'1.5 ,Arn

•^23 ~ Z32 — 71-2 2T-TT — 7O.6O
J 3.0
When all the elements are determined we have
;0.72 ;0.72 A60

Z„ us (new) A 72 A92 A 60
7O.6O A 60 A-75
Finally we add the impedance Zb — A15 between buses 2 and 3. If we

let j and k in Eq. (7.58) equal 2 and 3, respectively, we obtain the elements
for row 4 and column 4.
Z14 — Z12 — Z13 =j0.12 — j0.60 — JO. 12
Z24 = Z22 — Z23 = j0.92 — j0.60 = j0.32
Z34 = Z32 - Z33 = ;0.60 -J0.15 = -A15
and from Eq. (7.57)
z44 = Zb + Z22 + Z33 — 2 Z23

= A15 + ;'0.92 + ;0.75 - 2(A60) = j0.62
So we write
A 72 A 72 A60 A12
J0.12 J0.92 ;0.60 70.32
j0.60 A 60 jo.is -A15
J0.12 A32 —A 15 A 62
and from Eq. (7.59) we find
A6968 A-6581 A.6290

Zb us (new) A-6581 70.7548 70.6774
A6290 ;0.6774 A7137
which is the bus impedance matrix to be determined.

The procedure is simple for a computer which first must determine, the
types of modification involved as each impedance is added. However, the
operations must follow a sequence such that we avoid connecting an im¬
pedance between two new buses.
As a matter of interest we can check the impedance values of Zbus by the
network calculations of Sec. 7.5.
Example 7.9 Find Zn of the circuit of Example 7.8 by determining the

impedance measured between node 1 and the reference bus when currents
injected at nodes 2 and 3 are zero.
Solution The equation corresponding to Eq. (7.42) is
We recognize two parallel paths between nodes 1 and 3 of the circuit of

Fig. 7.10 with the resulting impedance of
y'0.3 x y'0.3 5
= y'0.1615
y'0.3 + 70.35
This impedance in series with y'1.5 is in parallel with y'1.2 to yield
yT.2(y'1.5+y'0.1615)
= 70.6968
11 7(1.2+1.5 + 0.1615)
which is identical with the value found in Example 7.8.

Although the network reduction method of Example 7.9 may appear to
be simpler by comparison with other methods of forming Zbus such is not
the case because a different network reduction is required to evaluate each
element of the matrix. In Example 7.9 the network reduction to find Z22, for
instance, is more difficult than that for finding Zn. The digital computer
could make a network reduction by node elimination but would have to
repeat the process for each node.
7.8 SUMMARY
Equivalence of sources and node equations were reviewed briefly in this chapter
to provide the essential background for understanding the bus admittance
matrix which is the basis for most load-flow studies. Matrix partitioning was
reviewed because of its usefulness in node-elimination methods.
The bus impedance matrix is preferred by some engineers for load-flow
studies but finds its greatest value in fault calculations which we shall discuss
later.
The modification of Zbus was discussed to show the simplicity of accounting

for the addition or removal of a transmission line without having to invert Ybus
each time a change is made. Direct formulation of Zbus is a process which can be
programmed in a straightforward manner.
PROBLEMS
7.1 Write the two node equations similar to Eqs. (7.5) and (7.6) required to find the voltages at
nodes 1 and 2 of the circuit of Fig. 7.11 without changing the emf sources to current sources. Then
write the equations in standard form after changing the emf sources to current sources.
7.2 Find the voltages at nodes 1 and 2 of the circuit of Fig. 7.11 by solving the equations determined
in Prob. 7.1.
73 Eliminate nodes 3 and 4 of the network of Fig. 7.12 simultaneously by the method of partitioning
employed in Example 7.3 to find the resulting 2x2 admittance matrix Ybus. Draw the circuit
corresponding to the resulting matrix and show on the circuit the values of the parameters. Solve for
Kj and V2 by matrix inversion.
7.4 Eliminate nodes 3 and 4 of the network of Fig. 7.12 to find the resulting 2x2 admittance matrix
by eliminating node 4 first and then node 3 as in Example 7.4.
7.5 Modify Zbus given in Example 7.2 for the circuit of Fig. 7.4 by adding a new node connected to
bus 4 through an impedance of j 1.2 per unit.
7.6 Modify Zbus given in Example 7.2 by adding a branch having an impedance of y 1.2 per unit
between node 4 and the reference bus of the circuit of Fig. 7.4.
7.7 Determine the impedances in the first row of Zbus for the circuit of Fig. 7.4 with the impedance
connected between bus 3 and the reference bus removed by modifying the Zbus found in Example 7.2.
Then with the current sources connected only at buses 1 and 2 find the voltage at bus 1 and compare
this value with that found in Example 7.3.
7.8 Modify Zbus given in Example 7.2 by removing the impedance connected between nodes 2 and 3
of the network of Fig. 7.4.
7 0.2
Figure 7.11 Circuit for Probs.

7.1 and 7.2. Values shown are
voltages and impedances in per
unit. o
-710
Figure 7.12 Circuit for Probs. 7.3 and 7.4. Values shown are currents and admittances in per unit.
Jl-O
Figure 7.13 Circuit for Prob. 7.9. Values shown are reac¬
tances in per unit. Reference bus
7.9 Find Zbus for the network of Fig. 7.13 by the direct determination process discussed in Sec. 7.7.
7.10 For the reactance network of Fig. 7.14 find (a) Zbus by direct formulation or by inversion of
Ybus, (b) the voltage at each bus, (c) the current drawn by a capacitor having a reactance of 5.0 per
unit connected from bus 3 to neutral, (d) the change in voltage at each bus when the capacitor is
connected at bus 3, and (e) the voltage at each bus after connecting the capacitor. The magnitude
and angle of each of the generated voltages may be assumed to remain constant.
© 70.5 ©
-nror-
70.2
-'’TT57P'— 70.4
70.1 71.0
Figure 7.14 Circuit for Prob. 7.10. ^ 28/0° 72.0 1.20/30°

Voltages and impedances are in
per unit. 1
CHAPTER
EIGHT
LOAD-FLOW SOLUTIONS AND CONTROL
The great importance of load-flow studies in planning the future expansion of

power systems as well as in determining the best operation of existing systems
was discussed in Chap. 1. The principal information obtained from a load-flow
study is the magnitude and phase angle of the voltage at each bus and the real
and reactive power flowing in each line. However, much additional information
of value is provided by the printout of the solution from computer programs
used by the power companies. Most of these features will be brought out in our
discussion of load-flow studies in this chapter which also treats the principles of
the control of load flow.
We shall examine two of the methods upon which solutions to the load-flow
problem are based. The great value of the digital computer in power-system
design and operation will become apparent.
8.1 DATA FOR LOAD-FLOW STUDIES
Either the bus self- and mutual admittances which compose the bus admittance
matrix Ybus or the driving-point and transfer impedances which compose Zbus
may be used in solving the load-flow problem. We shall confine our study to
methods using admittances. The starting point in obtaining the data which must
be furnished to the computer is the one-line diagram of the system. Values of
series impedances and shunt admittances of transmission lines are necessary so
that the computer can determine all the Ybus or Zbus elements. Other essential
information includes transformer ratings and impedances, shunt capacitor rat¬
ings, and transformer tap settings.
Operating conditions must always be selected for each study. At each bus
except one the net real power into the network must be specified. The power
drawn by a load is negative power input to the system. The other power inputs
are from generators and positive or negative power entering over interconnec¬
tions. In addition, at these buses either the net flow of reactive power into the
193
network or the magnitude of the voltage must be specified; that is, at each b^s a
decision is required whether the voltage magnitude or the reactive-power flow is
to be maintained constant. The usual case is to specify reactive power at load
buses and voltage magnitude at generator buses, although sometimes reactive
power is specified for the generators. In digital-computer programs provision is
made for the calculation to consider voltage to be maintained constant at a bus
only so long as the reactive-power generation remains within designated limits.
The one bus at which real-power flow is not specified, called the swing bus, is
usually a bus to which a generator is connected. Obviously, the net power flow
to the system cannot be fixed in advance at every bus because the loss in the
system is not known until the study is complete. The generators at the swing bus
supply the difference between the specified real power into the system at the
other buses and the total system output plus losses. Both voltage magnitude and
angle are specified at the swing bus. Real and reactive power at this bus are
determined by the computer as part of the solution.
8.2 THE GA USS-SEID EL METHOD
The complexity of obtaining a formal solution for load flow in a power system
arises because of the differences in the type of data specified for the different
kinds of buses. Although the formulation of sufficient equations is not difficult,
the closed form of solution is not practical. Digital solutions of the load-flow
problems we shall consider at this time follow an iterative process by assigning
estimated values to the unknown bus voltages and calculating a new value for
each bus voltage from the estimated values at the other buses, the real power
specified, and the specified reactive power or voltage magnitude. A new set of
values for voltage is thus obtained for each bus and used to calculate still
another set of bus voltages. Each calculation of a new set of voltages is called an
iteration. The iterative process is repeated until the changes at each bus are less
than a specified minimum value.
We shall examine first the solution based on expressing the voltage of a bus
as a function of the real and reactive power delivered to a bus from generators or
supplied to the load connected to the bus, the estimated or previously calculated
voltages at the other buses, and the self- and mutual admittances of the nodes.
The derivation of the fundamental equations starts with a node formulation of
the network equations. We shall derive the equations for a four-bus system and
write the general equations later. With the swing bus designated as number 1,
computations start with bus 2. If P2 and Q2 are the scheduled real and reactive
power entering the system at bus 2,
V2I*2 = P2+jQ2 (8.1)

from which I2 is expressed as
r P2 ~ jQ2
(8.2)
LOAD-FLOW SOLUTIONS AND CONTROL 195
and in terms of self- and mutual admittances of the nodes, with generators and
loads omitted since the current into each node is expressed as in Eq. (8.2),
^2 — jQl
Y2lVl + Y22V2 + Y23V3 + Y2AV4 (8.3)
V\
Solving for V2 gives
p2 — iQ2
V2 v* ~(y21v1 + y23v3 + y24v4) (8.4)
Equation (8.4) gives a corrected value for V2 based upon scheduled P2 and Q2
when the values estimated originally are substituted for the voltage expressions
on the right side of the equation. The calculated value for V2 and the estimated
value for V\ will not agree. By substituting the conjugate of the calculated value
of V2 for V\ in Eq. (8.4) to calculate another value for V2, agreement would be
reached to a good degree of accuracy after several iterations and would be the
correct value for V2 with the estimated voltages and without regard to power at
the other buses. This value would not be the solution for V2 for the specific
load-flow conditions, however, because the voltages upon which this calculation
for V2 depends are the estimated values of voltage at the other buses and the
actual voltages are not yet known. Two successive calculations of V2 (the second
being like the first except for the correction of V\) are recommended at each bus
before proceeding to the next one.
As the corrected voltage is found at each bus, it is used in calculating the
corrected voltage at the next. The process is repeated at each bus consecutively
throughout the network (except at the swing bus) to complete the first iteration.
Then the entire process is carried out again and again until the amount of
correction in voltage at every bus is less than some predetermined precision
index.
This process of solving linear algebraic equations is known as the Gauss-
Seidel iterative method. If the same set of voltage values is used throughout a
complete iteration (instead of immediately substituting each new value obtained
to calculate the voltage at the next bus), the process is called the Gauss iterative
method.
Convergence upon an erroneous solution may occur if the original voltages
are widely different from the correct values. Erroneous convergence is usually
avoided if the original values are of reasonable magnitude and do not differ too
widely in phase. Any unwanted solution is usually detected easily by inspection
of the results since the voltages of the system do not normally have a range in
phase wider than 45° and the difference between nearby buses is less than about
10° and often very small.
For a total of N buses the calculated voltage at any bus k where Pk and Qk
are given is
1 (Pk-jQk (8.5)
M v*k
where nfk. The values for the voltages on the right side of the equation are, the
most recently calculated values for the corresponding buses (or the estimated
voltage if no iteration has yet been made at that particular bus).
Experience with the Gauss-Seidel method of solution of power-flow prob¬
lems has shown that an excessive number of iterations are required before the
voltage corrections are within an acceptable precision index if the corrected
voltage at a bus merely replaces the best previous value as the computations
proceed from bus to bus. The number of iterations required is reduced con¬
siderably if the correction in voltage at each bus is multiplied by some constant
that increases the amount of correction to bring the voltage closer to the value it
is approaching. The multipliers that accomplish this improved convergence are
called acceleration factors. The difference between the newly calculated voltage
and the best previous voltage at the bus is multiplied by the appropriate acceler¬
ation factor to obtain a better correction to be added to the previous value. The
acceleration factor for the real component of the correction may differ from that
for the imaginary component. For any system, optimum values for acceleration
factors exist, and poor choice of factors may result in less rapid convergence or
make convergence impossible. An acceleration factor of 1.6 for both the real and
imaginary components is usually a good choice. Studies may be made to deter¬
mine the best choice for a particular system.
At a bus where voltage magnitude rather than reactive power is specified,
the real and imaginary components of the voltage for each iteration are found by
first computing a value for the reactive power. From Eq. (8.5)
n - iQk = (n* k £ n. l) vt
+ (8.6)
where n f k. If we allow n to equal k
Pt-jQt=V!I.YknV„ (8.7)
n=1
e,= -Im [vt £y,„F„! (8.8)
\ n=1 I
where Im means “ imaginary part of.”
Reactive power Qk is evaluated by Eq. (8.8) for the best previous voltage
values at the buses, and this value of Qk is substituted in Eq. (8.5) to find a new
Vk. The components of the new Vk are then multiplied by the ratio of the
specified constant magnitude of Vk to the magnitude of the Vk found by Eq. (8.5).
The result is the corrected complex voltage of the specified magnitude.
8.3 THE NEWTON-RAPHSON METHOD
Taylor’s series expansion for a function of two or more variables is the basis for
the Newton-Raphson method of solving the load-flow problem. Our study of the
method will begin by a discussion of the solution of a problem involving only
two equations and two variables. Then we shall see how to extend the analysis to
the solution of load-flow equations.
Let us consider the equation of a function of two variables xt and x2 equal
to a constant Kx expressed as
(8.9)
II
<N
and a second equation
f2(xi, x2) = K2 (8.10)

where Ky and K2 are constants.
We then estimate the solutions of these equations to be xf] and x(20). The
superscripts indicate that these values are initial estimates. We designate Ax^
and Ax(20) as the values to be added to x(j0) and x20) to yield the correct solutions.
So we can write
Ki =/i(*i, x2) =/1(x(10) + Ax[0), x(20) + Ax(20)) (8.11)
K2 =f2(xi, x2) =/2(x<°> + Ax[°\ x(20) + Ax(20)) (8.12)

Our problem now is to solve for Axf'* and Ax(20), which we shall do by expanding
Eqs. (8.11) and (8.12) in Taylor’s series to give
*1 =/i(*(!0), *(20)) + Ax<0)

8A
+ Ax(,0)
dfi (8.13)
dx2 (0) dxj (0)
k2 -/2W0>, 4”) + A4°> dfi

+ AX(20)
Sf2 + (8.14)
5Xj (0) dx-t (0)
where the partial derivatives of order greater than 1 in the series of terms of the
expansion have not been listed. The term d/j/dxj |(0) indicates that the partial
derivative is evaluated for the values of x^ and x(20). Other such terms are
evaluated similarly.
If we neglect the partial derivatives of order greater than 1 we can rewrite
Eqs. (8.13) and (8.14) in matrix form. We then have
dA dA
K i ~fi{x xf) dx2 dx2 Ax<0)
(8.15)
K2 -fi{x x<0)) df2 8A Ax(20)
QXy dx2
where the square matrix of partial derivatives is called the jacobian J or in this
case J(0) to indicate that the initial estimates x(10) and x(20) have been used to
compute the numerical value of the partial derivatives. We note that/i(x(10), x(20))
is the calculated value of Kt for the estimated values of x{0) and x(20) but this
calculated value of is not the value specified by Eq. (8.9) unless our estimated
values x(i} and x(20) are correct. If we designate as Athe specified value of Kx
minus the calculated value of Kx and define A/C(20) similarly we have
A K[0) Ax*0’
= J<°> (8.16)
A Xl20) Ax(20)
So by finding the inverse of the jacobian we can determine Axf* and A*(20).
However, since we truncated the series expansion, these values added to our
initial guess do not determine for us the correct solution and we must try again
by assuming new estimates and x(2' where
x[l) = x<°> + Ax<0)
x<x> = x<20) + Ax(20)
and repeat the process until the corrections become so small that they satisfy a
chosen precision index.
To apply the Newton-Raphson method to the solution of load-flow equa¬
tions we may choose to express bus voltages and line admittances in polar form
or rectangular form. If we choose polar form and separate Eq. (8.7) into its real
and imaginary components with
X-IKIA and n.-IX.IZ®*.

we have
N
p,-sQk= y. ixxu/«*+*„-«* (8.17)

n= 1
N
Pk- I | X X X. 1 cos (0„+ «„-«,) (8.18)

1
n=
e. = - I IKK X.I sin (0ta + S,~ St) (8.19)

n= 1
As in the Gauss-Seidel method the swing bus is omitted from the iterative
solution to determine voltages since both magnitude and angle of the voltage at
the swing bus are specified. If we postpone consideration of voltage-controlled
buses until later, we specify P and Q at all buses except the swing bus and
estimate voltage magnitude and angle at all buses except the swing bus where
voltage magnitude and angle are specified. The specified constant values of P
and Q correspond to the K constants in Eq. (8.15). The estimated values of
voltage magnitude and angle correspond to the estimated values for X! and x2 in
Eq. (8.15). We use these estimated values to calculate values of Pk and Qk from
Eqs. (8.18) and (8.19) and define
^Pk Pk, spec Pk, calc
^Qk Qk, spec Qk, calc
which correspond to the AK values of Eq. (8.16).

The jacobian consists of the partial derivatives of P and Q with respect to
each of the variables in Eqs. (8.18) and (8.19). The column matrix elements A<5(k0>
and A| Vk\( ’ correspond to Ax'/*' and Ax^01 and are the corrections to be added
to the original estimates 5^°’ and I Ft|(0) to obtain new values for computing
Aand AQ[l\ *
For the sake of simplicity we shall write the matrix equation for a system of
only three buses. If the swing bus is number 1 we start our calculations at bus 2
since voltage magnitude and angle are specified at the swing bus. In matrix form
8P2 dP 2 dP2 8P2

A P2 Ad 2
dS2 dd3 W aim
sp3 sp3 dP3 sp3
A P3 AS3
ds2 dS3 S\Vz\ aim
(8.20)
SQi dQ2 sq2 dQ2
a<22 av2
dS2 ss3 ainl aim
8Q3 sq3 dQ3 8Q3
A£>3
dS2 8S3 aim aim
The superscripts which would indicate the number of the iteration are omitted in
Eq. (8.20) because, of course, they change with each iteration. The elements of
the jacobian are found by taking the partial derivatives of the expressions for Pk
and Qk and substituting therein the voltages assumed for the first iteration or
calculated in the last previous iteration. The jacobian has been partitioned to
emphasize the different general types of partial derivatives appearing in each
submatrix. For instance, from Eq. (8.18) we find
dPk
VkV„Yk„| sin (0t, + <>„ - St) (8.21)
K
where n ^ k and
dP N
-rr‘= I KLUsin (e^ + S.-S,) (8.22)
„=1
n^k
In the above summation n ± k, which is apparent since 3k drops out of Eq. (8.17)
when n = k. Similar general forms of the partial derivatives may be found from
Eqs. (8.18) and (8.19) for calculating the elements in the other submatrices.
Equation (8.20) and similar equations involving more buses are solved by
inverting the jacobian. The values found for A<5* and A | Vk | are added to the
previous values of voltage magnitude and angle to obtain new values for P[]
and g^calc for starting the next iteration. The process is repeated until the
precision index applied to the quantities in either column matrix is satisfied. To
achieve convergence, however, the initial estimates of voltage must be reason¬
able, but this is seldom a problem in power-system work.
Voltage controlled buses are taken into account easily. Since the voltage
magnitude is constant at such a bus we omit in the jacobian the column of
partial differentials with respect to voltage magnitude of the bus. We are not
interested at this point in the value of Q at the bus so we omit the row of partial
differentials of Q for the voltage controlled bus. The value of Q at the bus can be
determined after convergence by Eq. (8.19).t
The Newton-Raphson method, as noted previously, may also be used when
equations are expressed in rectangular form. We chose to develop the equations
in polar form because the jacobian provides interesting information which is
lacking in the rectangular form. For instance, the dependence of Pk on Sk and of
Qk on | Vk\ is seen immediately in the jacobian in polar form. Later in Sec. 8.10
we shall see how voltage regulating transformers in transmission lines principally
affect the transfer of Q in a system while phase-shifting transformers principally
affect the transfer of P.
Example 8.1 Figure 8.1 shows the one-line diagram of a very simple power
system. Generators are connected at buses 1 and 3. Loads are indicated at
buses 2, 4, and 5. Base values for the system are 100 MVA, 138 kV in the
high-tension lines considered here. Table 8.1 gives impedances for the six
lines which are identified by the buses on which they terminate. The charg¬
ing megavars listed in the table account for the distributed capacitance of
the lines and will be neglected in this example but discussed in Sec. 8.4 and
t For a more detailed explanation of the Newton-Raphson method and for excellent numerical
examples carried through to a converging solution for both Gauss-Seidel and Newton-Raphson
methods, see G. W. Stagg and A. H. El-Abiad, Computer Methods in Power System Analysis, chaps. 7
and 8, McGraw-Hill Book Company, New York, 1968.
Figure 8.1 One-line diagram

for Example 8.1.
Table 8.1
Line, Length
bus to R X R X Charging
bus km mi Q n per unit per unit Mvart
1-2 64.4 40 8 32 0.042 0.168 4.1

1-5 48.3 30 6 24 0.031 0.126 3.1
2-3 48.3 30 6 24 0.031 0.126 3.1
3-4 128.7 80 16 64 0.084 0.336 8.2
3-5 80.5 50 10 40 0.053 0.210 5.1
4-5 96.5 60 12 48 0.063 0.252 6.1
+ At 138 kV.
included in the computer runs for the system. Table 8.2 lists values of P, Q,
and V at each bus. Since values of P and Q in Eqs. (8.18) and (8.19) are
positive for real power and inductive reactive voltamperes input to the
network at each bus, net values of P and Q for these equations are negative
at buses 2, 4, and 5. Generated Q is not specified where voltage magnitude is
constant. In the voltage column the values for the load buses are the original
estimates. Listed values of voltage magnitude and angle are to be held
constant at the swing bus, and the listed voltage magnitude is to remain
constant at bus 3. A load-flow study is to be made by the Newton-Raphson
method using the polar form of the equations for P and Q. Determine the
number of rows and columns in the jacobian. Calculate AP(20) and the value
of the second element in the first row of the jacobian using the specified
values or initial estimates of the voltages.
Solution Since the swing bus does not require a row and column of the
jacobian an 8 x 8 matrix would be necessary if P and Q are specified for the
remaining four buses. However, voltage magnitude is specified (held
constant) at bus 3, and the jacobian will be a 7 x 7 matrix.
Table 8.2
Generation Load
Bus P, MW Q, Mvar P, MW Q, Mvar V, per unit Remarks
1 65 30 1.04^1 Swing bus

2 0 0 115 60 1.00/01 Load bus
(inductive)
3 180 70 40 1.024)! Voltage magnitude
constant
4 0 0 70 30 1.00/b: Load bus
(inductive)
5 0 0 85 40 l.ooAr Load bus
(inductive)
In order to calculate P(20) for the estimated-voltage and fixed-voltage

values of Table 8.2 we need only the admittances
1
5.7747/104.04°
0.042 + 70.168
1
7.7067/103.82°
0.031 + 70.126
and Y22, which (since no other admittances terminate on bus 2) is

expressed by
y22= -y21 - Y„ = -\Y2tye21- t23|^23 |
From Eq. (8.18) since Y24 and Y25 are zero and since the initial values
6[°> = S(20> = d\30) = 0
P(20)calc = \V2V1Y21\ cos 021 - I V2 V2Y211 cos e21
- |v2v2y23| cos e23 + \v2v3y23| cos e23
= (1.0 x 1.04 - 1.0 x 1.0)1 y21| cos 021

- (1.0 x 1.0 - 1.0 x 1.02)1 y23| cos 023
= 0.04 X 5.7747 cos 104.04° + 0.02 x 7.7067 cos 103.82°
- -0.0560 - 0.0368 - —0.0928 per unit
Scheduled power into the network at bus 2 is
115
= —1.15 per unit
100
So
AP<20) - -1.15 - (-0.0928) = -1.0572 per unit
To find dP2/dd3 we use Eq. (8.21) and obtain
dP2
= ~\V2V3Y23\ sin (d23 + d3-S2)
85 -.
= —1.0 x 1.02 x 7.067 sin 103.82 = —7.6333 per unit
The Newton-Raphson method is summarized in the following steps:
1. Determine values of Pk calc and Qk caIc flowing into the system at every bus for
the specified or estimated values of voltage magnitudes and angles for the first
iteration or the most recently determined voltages for subsequent iterations.
2. Calculate AP at every bus.
3. Calculate values for the jacobian using estimated or specified values of vol¬
tage magnitude and angle in the equations for partial derivatives determined
by differentiation of Eqs. (8.18) and (8.19).
4. Invert the jacobian and calculate voltage corrections Adk and A | Vk | at every
bus.
5. Calculate new values of Sk and | Vk | by adding ASk and A | Vk | to the previous
values.
6. Return to step 1 and repeat the process using the most recently determined
values of voltage magnitudes and angles until either all values of AP and AQ
or all values of A<5 and A | V | are less than a chosen precision index.
P and Q at the swing bus and Q at voltage controlled buses can be

determined from Eqs. (8.18) and (8.19). Line flow can be determined from the
differences in bus voltages.
The number of iterations required by the Newton-Raphson method using
bus admittances is practically independent of the number of buses. The time for
the Gauss-Seidel method (bus admittances) increases almost directly with the
number of buses. On the other hand, computing the elements of the jacobian is
time-consuming, and the time per iteration is considerably longer for the
Newton-Raphson method. The advantage of shorter computer time for a solu¬
tion of the same accuracy is in favor of the Newton-Raphson method for all but
very small systems.
8.4 DIGITAL-COMPUTER STUDIES OF LOAD FLOW
Power companies use very elaborate programs for making load-flow studies. A
typical program is capable of handling systems of more than 2000 buses, 3000
lines, and 500 transformers. Of course programs can be expanded to even greater
size provided the available computer facilities are sufficiently large.
Data supplied to the computer must include the numerical values given in
Tables 8.1 and 8.2 and an indication of whether a bus is a swing bus, a regulated
bus where voltage magnitude is held constant by generation of reactive power Q,
or a bus with fixed P and Q. Where values are not to be held constant the
quantities given in the tables are interpreted as initial estimates. Limits of P and
Q generation usually must be specified as well as the limits of line kilovolt¬
amperes. Unless otherwise specified, programs usually assume a base of
100 MVA.
Total line charging in megavars specified for each line accounts for shunt
capacitance and equals y/3 times the rated line voltage in kilovolts times Jchg, as
defined by Eqs. (4.24) and (4.25), divided by 103. This equals a>Cn \V\2 where
| V | is the rated line-to-line voltage in kilovolts, and C„ is line-to-neutral capaci¬
tance in farads for the entire length of the line. The program creates a nominal-/!
representation of the line by dividing equally between the two ends of the line
the capacitance computed from the given value of charging megavars. For a long
line, the computer could be programmed to compute the equivalent n for capaci¬
tance distributed evenly along the line.
8.5 INFORMATION OBTAINED IN A LOAD-FLOW STUDY
The information which is obtained from digital solutions of load flow is an

indication of the great contribution digital computers have made to the power-
system engineer’s ability to obtain operating information about systems not yet
built and to analyze the effects of changes on existing systems. The following
discussion is not meant to list all the information obtainable but should provide
some insight into the great importance of digital computers in power-system
engineering.
The printout of results provided by the computer consists of a number of
tabulations. Usually the most important information to be considered first is the
table which lists each bus number and name, bus-voltage magnitude in per unit
and phase angle, generation and load at each bus in megawatts and megavars,
line charging, and megavars of static capacitors or reactors on the bus. Accom¬
panying the bus information is the flow of megawatts and megavars from that
bus over each transmission line connected to the bus. The totals of system
generation and loads are listed in megawatts and megavars. The tabulation
described is shown in Fig. 8.2 for the system of five buses of Example 8.1.
In the operation of power systems any appreciable drop in voltage on the
primary of a transformer caused by a change of load may make it desirable to
change the tap setting on transformers provided with adjustable taps in order to
maintain proper voltage at the load. Where a tap-changing transformer has been
specified to keep the voltage at a bus within designated tolerance limits, the
voltage is examined before convergence is complete. If the voltage is not within
the limits specified, the program causes the computer to perform a new set of
iterations with a one-step change in the appropriate tap setting. The process is
repeated as many times as necessary to cause the solution to conform to the
desired conditions. The tap setting is listed in the tabulated results.
A system may be divided into areas, or one study may include the systems of
several companies each designated as a different area. The computer program
will examine the flow between areas, and deviations from the prescribed flow will
be overcome by causing the appropriate change in generation of a selected
generator in each area. In actual system operation interchange of power between
areas is monitored to determine whether a given area is producing that amount
of power which will result in the desired interchange.
Among other information that may be obtained is a listing of all buses
where the voltage magnitude is above or below 1.05 or 0.95, respectively, or
other limits that may be specified. A list of line loadings in megavoltamperes can
be obtained. The printout will also list the total megawatt (1112 R) and megavar
(1112 X) losses in the system and both P and Q mismatch at each bus. Mismatch
is an indication of the preciseness of the solution and is the difference between P
(and also usually Q) entering and leaving each bus.
a a
• •
• •
• •
• •
• • 1
• 1 •
rvi •• in >0 •• O' o 11
— to in>©co 1 oa 1 =jr\jr-
• Ml 1 *0*0 1 m>o 1 •DOLT* I O' o I —>0r—
o • • *i
o COO • -«© I iSHO >0 I oo — I O' O' CO
o ■ mm 1 rum 1 m—— 1 —— 1 rw—
O
o
X>_J l 1 l
l 1
OIL l 1
UJ Z • 1
X M | I CO 00 • 19^1 O'crm
—O' • O'>01n
— X 1 o- >0 • • r-ru 1
=»in 1 tn^ro- in 3X0
►— (Oui • • •I • «• 1 • •»
1 m»n • —m 1 3 0 3 1 ec — l nj^ru
• rô* 1 1 rvj 1 mm 1 O rum
1
•
1 • 1 I
1 1 • a 1 •a
1 a
1 1 1 •
>0 o_ • •
nj *— • 1
a a
< I a a
cr •
UJ 1
»- • a a
M | a a a a a
I a a ioJ a a oj a xui
m 1 a uj a u_j a a _juj a 1_>_J
l a xz a xa. a ix:z I Q.Z I CKO. *
• a _j— a —«r a a<Mi — ■< C
I a ujcl a 11 a UJQQ. 1 in. 1 CO X o
a a a a
a 'urn a njîn a min a —m o
a cd a a a a
a *-£0 a a
a a
Figure 8.2 Digital-computer solution of load-flow for the system of Example 8.1. Base is 100 MVA.
toco
ao CC
a <1 zz
I uj Q: a • CO
I CC «1 1 c
I N> U1UJ
1 a.r yjto
1<
ao O^T
a —in
a • •
a x 00
a a x
1 a < ac
a a > ao
a a x a rvi
a q a u in J
XX
a _j
a a
a a x
» • 7 • in
ao
• cj- hO
DK
a x
a — o
< a x
z-< ac
a—
— —X a rv
UJ UJ
X X I z
C *3. U«3f
_l OJI a -r
u or a am
CD • a7
C u X
<1 x
3 to
CD
Z
> <1
z _i a
<t d a
Cl l_) I
X _ I
C < l rv/
u u a O'
a
a x a > — G —
o a
a a
•— 1
oc a
co
3
oj a 33
or x
205
To bus 1 To bus 3
Figure 8.3 Flow of P and Q at bus 2 for the

system of Example 8.1. Numbers beside the
arrows show the flow of P and Q in mega¬
watts and megavars. The bus voltage is shown in
per unit.
8.6 NUMERICAL RESULTS
The load-flow study for the system described in Example 8.1 for which Fig. 8.2 is
the printout was run on a program attributed to the Philadelphia Electric Com¬
pany and subsequently modified. Three Newton-Raphson iterations were
required. Similar size studies run using other programs also required three
Newton-Raphson iterations but required 22 Gauss-Seidel iterations with the
same precision index. Figure 8.2 may be examined for more information than
just the tabulated results. For instance, the megawatt loss in any of the lines can
be found by comparing the values of P and Q at the two ends of the line. As an
example we see that 95.68 MW flow from bus 1 into line 1-5 and 92.59 MW
flow into bus 5 from the line. Evidently the \I\2 R loss in the line is 3.09 MW.
On another page of the printout, not reproduced here, \I\2 R losses of the
system are listed as 9.67 MW.
Information provided by Fig. 8.2 can be displayed on a diagram showing the
entire system. Figure 8.3 shows a portion of such a diagram at bus 2.
8.7 CONTROL OF POWER INTO A NETWORK
We studied some characteristics of synchronous machines in Chap. 6. We

developed the principle that an overexcited generator supplies reactive power Q
to a system and an underexcited generator absorbs reactive power Q from a
system. Thus, we understand how the exciter controls the flow of reactive power
between the generator and the system.
Now we turn our attention to real power P. Assume that a generator supply¬
ing a large system is delivering power under stable conditions so that a certain
angle S exists between Vt, the voltage at the system bus, and Eg, the generated
voltage of the machine. If Eg leads V, we have the phasor diagram of Fig. 8.4a
which is identical to Fig. 6.9a. If the power input to the generator is increased by
a larger opening of the valves through which steam (or water) enters a turbine
while |Eg\ is constant, the rotor speed will start to increase and the angle
between Eg and Vt will increase. Increasing 5 results in a larger Ia and lower d, as
Figure 8.4 Phasor diagrams of a generator having constant values of \Eg\ and | V,\ for (a) angle <5
small and (b) angle 6 larger to show the increase in power delivered as 5 increases.
may be seen by comparing Figs. 8.4a and 8.4b. The generator will therefore
deliver more power to the network, and the input from the prime mover will
again equal the output to the network if losses are disregarded. Equilibrium will
be reestablished at the speed corresponding to the frequency of the infinite bus
with a larger 5. Figure 8.4b is drawn for the same dc field excitation and there¬
fore the same \Eg\ as Fig. 8.4a, but the power output equal to | Vt | • | Ia | cos 9
is greater for the condition of Fig. 8.4b, and the increase in d has caused the
generator to deliver the additional power to the network.
The dependence of power on the power angle is also shown by an equation
giving P + jQ supplied by a generator in terms of 5. If
Vt=\Vt\/01 and Eg=\Eg\/s
where Vt and Eg are expressed in volts to neutral or in per unit, then
, \E.\/S- \Vt\ (8.23)
and
(8.24)
Therefore
P+jQ = v,i*
\V,\-)E.\/=£- \Vl
~jX,
(8.25)
(8.26)
and the imaginary part of Eq. (9.13) is
Q= sin (90
w I Eg\ COS d - \ Vt\ (8.27)

x„
When volts rather than per-unit values are substituted for Vt and Eg in
Eqs. (8.26) and (8.27), we must be careful to note that V, and Eg are line-to-
neutral voltages and P and Q will be per-phase quantities. However, line-to-line
voltage values substituted for Vt and Eg will yield total three-phase values for P
and Q. The per-unit P and Q of Eqs. (8.26) and (8.27) are multiplied by base
three-phase megavoltamperes or base megavoltamperes per phase depending on
whether total three-phase power or power per phase is wanted.
Equation (8.26) shows very clearly the dependence of power transferred to
the network on the power angle 5 if \Eg\ and | Vt | are constant. However, if P
and Vt are constant, Eq. (8.26) shows that 5 must decrease if \Eg\ is increased by
increasing the dc field excitation. In Eq. (8.27) with P constant both an increase
in | Eg | and a decrease in d mean that Q will increase if already positive or
decrease in magnitude and perhaps become positive if Q is negative before
increasing the field excitation. This agrees with the conclusions reached in
Sec. 6.4.
Equation (8.26) can be interpreted as the power transferred from one bus in
a network to another bus through a reactance X connecting the two buses. If the
bus voltages are V1 and V2 and d is the angle by which V1 leads V2,
P= sin 5 (8.28)
Similarly from Eq. (8.27), Q received at bus 2 is
Q = î(\Vl\ cos ,5- \V2\) (8.29)
The equations derived in Sec. 5.8 to develop circle diagrams are more general
than Eqs. (8.28) and (8.29) because they account for resistance and capacitance.
Equations (5.59) and (5.60), however, are identical with Eqs. (8.28) and (8.29) if
the only line parameter considered is the inductance.
From Eqs. (8.28) and (8.29) we see that an increase in d causes a larger
change in P than in Q when S is small. This difference is explained when we
recognize that sin <S changes widely but cos <5 changes by only a small amount
with a change in S when S is less than 10 or 15°.
8.8 THE SPECIFICATION OF BUS VOLTAGES
In Chap. 6 and Sec. 8.7 we have been considering the synchronous generator
from the viewpoint of supplying power to an infinite bus. We have examined the
effect of generator excitation and power angle when the terminal voltage of the
generator remains constant. In load-flow studies on a digital computer, however,
we found that it was necessary to specify voltage magnitude or reactive power at
every bus except the swing bus, where voltage was specified by both magnitude
and angle. Although the computer can easily tell us the results over the entire
system of specifying various voltage magnitudes at particular buses, it may be
helpful to look at what happens in a very simple case.
Usually it is at the buses on which there is generation that the voltage
magnitude is specified when a load-flow study is made on a computer. At such
buses real power P supplied by the generator is also specified. The reactive
power Q is then determined by the computer in solving the problem. Therefore,
our purpose at this point is to examine the effect of the magnitude of the
specified bus voltage on the value of Q supplied by the generator to the power
network.
Figure 8.5 shows a generator represented by its equivalent circuit with the
relatively small resistance neglected in order to simplify our analysis. The power
system is represented by its Thevenin equivalent voltage Eth in series with the
Thevenin impedance Xth, where again resistance has been neglected. Any local
load on the bus is included in the Thevenin equivalent. For constant power
delivered by the generator the component of / in phase with Elh must remain
constant. The voltage specified at the bus is | Vt |, and
Vt = Eth+jIXth (8.30)
The phasor diagram for the circuit of Fig. 8.5 is shown in Fig. 8.6 for three
different phase angles between Eth and I. In all three cases, however, the com¬
ponent of / in phase with Eth is constant.
Figure 8.6 shows that larger magnitudes of bus voltage Vt with constant
power input to the bus require a larger \Eg\, and, of course, the larger \Eg\ is
obtained by increasing the excitation of the dc field winding of the generator.
Increasing the bus voltage by increasing | Eg | causes the current to become more
lagging, as we see from Fig. 8.6 and as we expect from our discussion of the
synchronous generator. When we are making a load-flow study, increasing the
voltage specified at a generator bus means that the generator feeding the bus will
xth
Figure 8.5 Generator with internal voltage Eg connected Eth
to a power system represented by its Thevenin

equivalent.
Figure 8.6 Phasor diagrams of a generator supply¬

ing the same power to a system at three different
values of bus voltage V,.
increase its output of reactive power to the bus. From the standpoint of opera¬
tion of the system we are controlling bus voltage and Q generation by adjusting
the generator excitation.
Since we have represented the system by its Thevenin equivalent, we are
assuming that all Eg and Em values in the system remain constant in magnitude
and angle. This assumption is not strictly true under actual operating conditions.
When a change is made in the excitation of one generator, other changes may be
made elsewhere in the system. An example which would require changing Eg of
generators or motors at other buses is the specification to hold voltage constant
at these buses. The computer program takes care of all such conditions that are
imposed. However, our assumption of constant Eg and Em values in the system
except where we are making a change is very useful to illustrate the effect of a
change of voltage magnitude at a particular bus.
Example 8.2 A generator is supplying a large system which can be repre¬

sented by its Thevenin equivalent circuit, consisting of a generator with a
voltage Eth in series with Zth = j0.2 per unit. The voltage at the generator
terminals is Vt — 0.97/0° per unit when delivering a current of 0.8 — j0.2 per
unit. Synchronous reactance of the generator is 1.0 per unit. Find P and Q
into the system at the generator terminals and compute Eg (a) for the condi¬
tions described above and (b) if \ Vt\ =1.0 per unit when the generator is
delivering the same power P to the system. Assume that the system is so
large that Eth is not affected by the change in | Vt \. The bus at the generator
terminals is not an infinite bus, however, because Zth is not zero.
Solution (a) From the generator into the system
P + jQ = 0.97(0.8 + j0.2) = 0.776 + y'0.194 per unit

and
Eg = 0.97 + ;1 (0.8 — j0.2)
= 1.17 + jO.S = 1.42/34.4° per unit
(b) To find P and Q when \Vt\ = 1.0 we must find the phase angle of Vt,
as follows:
Eth = 0.97 —70.2(0.8 — j0.2)

— 0.93 — y'0.16 = 0.9441—9.16° per unit
The phase angle of Vt is determined by finding the angle 3 between V, and Eth
for | Vt\ = 1.0 and P = 0.776. By Eq. (8.28)
1.0 x 0.944
sin 3 = 0.776
0.2
S = 9.46°
(by which Vt leads Eth). Therefore,
Vt = 1.0/-9.76° + 9.46° = 1.0 /-0.3° = 1.0 -;0.005
1.0 -j0.005 - (0.93 -JO. 16)

J0.2
0.07 + ;0.155
= 0.775 -J0.350
j0-2
= 0.850 7-24.3°
and
Eg = 1.0 — y'0.005 + ;1 (0.775 -/0.350)

= 1.350 + jOJIO = 1.55/29.7° per unit
At the generator terminals into the system
P +jQ= 1.0 7-0.3° x 0.850/24.3°
= 0.850/24.0° = 0.776 + ;0.346 per unit
This example verifies our reasoning that specifying a higher terminal

voltage at a system bus to which a generator is connected results in a larger
reactive power supplied to the system by the generator and requires larger
generated voltage obtained by increasing the dc field excitation of the gener¬
ator. In this example Q increased from 0.194 to 0.346 per unit and | Eg\ had
to be increased from 1.42 to 1.55 per unit.
8.9 CAPACITOR BANKS
Another very important method of controlling bus voltage is by shunt capacitor

banks at the buses at both transmission and distribution levels along lines or at
substations and loads. Essentially capacitors are a means of supplying vars at the
point of installation. Capacitor banks may be permanently connected, but as
regulators of voltage they may be switched on and off the system as changes in
load demand. Switching may be manually or automatically controlled either by
time clocks or in response to voltage or reactive-power requirements. When they
are in parallel with a load having a lagging power factor, the capacitors are the
source of some or perhaps all of the reactive power of the load. Thus, capacitors
reduce the line current necessary to supply the load and reduce the voltage drop
in the line as the power factor is improved. Since capacitors lower the reactive
requirement from generators, more real-power output is available. The reader
may wish to review the effect of power factor on voltage regulation by referring
to Fig. 5.5.
In the load-flow computer program, voltage magnitude can be specified only
if there is a source of reactive-power generation. Therefore at load buses at
which there are no generators, capacitor banks must be assumed, and the com¬
puter will specify the value of Q required.
If capacitors are applied at a particular node, the increase in voltage at the
node can be determined by Thevenin’s theorem. Figure 8.7 shows the system
represented by its Thevenin equivalent at the node where capacitors will be
applied by closing the switch. The resistance in the equivalent circuit is indicated
but is always much smaller than the inductive reactance. With the switch open
the node voltage Vt is equal to the Thevenin voltage Eth. When the switch is
closed, the current drawn by the capacitor is
■‘th
Ir = (8.31)
Zth — jX c
The phasor diagram is shown in Fig. 8.8. The increase in Vt caused by adding the
capacitor is very nearly equal to \IC\ Xth if we assurfie that Elh remains un-
Figure 8.7 Circuit showing a capacitor to be Figure 8.8 Phasor diagram of the circuit of
connected through switch S to a system Fig. 8.7 with the capacitor connected. Before
represented by its Thevenin equivalent. connection of the capacitor V, = Eth.
changed since Eth and Vt are identical before adding the capacitor. This phasor
diagram serves to explain the increase in voltage at the bus where the capacitor
is installed. Example 7.6 was introduced as a part of our study of Zbus; and this
example should be reviewed because it shows how the change in voltage magni¬
tude due to the added capacitor can be calculated at all the buses of a system
where there are no regulated buses and loads are represented by impedances.
Here again we have made the assumption that Eg and Em values in the
system remain constant. As described in Sec. 8.8, the assumption is not strictly
true but provides a good estimate of the increase of bus voltages due to adding
capacitors except at buses where voltage is held constant. If the capacitors are
added to a load bus which is remote from any generation, the estimate is quite
good for nearby buses.
The digital load-flow printout of Fig. 8.2 shows a voltage at bus 4 of 0.920
per unit. The same load-flow program can be used to determine the amount of
reactive power which must be supplied by capacitors at this bus to raise the
voltage to any specified value. The procedure is to designate bus 4 as a regulated
bus to be held at the specified voltage and with a generator at the bus to supply
only reactive power. If generator losses are neglected, such a generator is equiva¬
lent to a lossless, unloaded, and overexcited synchronous motor and is known as
a synchronous condenser. The computer determines the necessary amount of
reactive power, which may be supplied to the system by either static capacitors
or a synchronous condenser.
When the voltage of bus 4 is specified to be 0.950 per unit, the required
reactive-power generation is found to be 15.3 kvar. This reactive-power input at
bus 4 also raises the voltage of bus 5 from 0.968 to 0.976 per unit. At bus 2, the
only other unregulated bus, the voltage is unchanged because of its separation
from bus 4 by regulated buses 1 and 3.
The flow of real and reactive power determined by the computer in the lines
connected to bus 4 with and without the added capacitors is shown in Fig. 8.9.
To bus 5 To bus 3 To bus 5 To bus 3
condenser
(a) ib)
Figure 8.9 Flow of P and Q at bus 4 of the system of Example 8.1 (a) as found in the original
load-flow study and (b) with capacitors added at the bus to raise the voltage to 0.950 per unit.
The voltage drops on the lines from buses 3 and 5 to bus 4 are reduced by
supplying reactive power at bus 4 because the reactive power flowing in these
lines is reduced. The increase of voltage at bus 4 obtained by applying capacitors
at the bus results in causing the reactive power reaching bus 4 over the two
transmission lines to be apportioned between the lines to accomplish the
required voltage drop in each.
8.10 CONTROL BY TRANSFORMERS

Transformers provide an additional means of control of the flow of both real
and reactive power. Our usual concept of the function of transformers in a
power system is that of changing from one voltage level to another, as when a
transformer converts the voltage of a generator to the transmission-line voltage.
However, transformers which provide a small adjustment of voltage magnitude,
usually in the range of ± 10%, and others which shift the phase angle of the line
voltages are important components of a power system. Some transformers regu¬
late both the magnitude and phase angle.
Almost all transformers provide taps on windings to adjust the ratio of
transformation by changing taps when the transformer is deenergized. A change
in tap can be made while the transformer is energized, and such transformers are
called load-tap-changing (LTC) transformers or tap-changing-under-load (TCUL)
transformers. The tap changing is automatic and operated by motors which
respond to relays set to hold the voltage at the prescribed level. Special circuits
allow the change to be made without interrupting the current.
A type of transformer designed for small adjustments of voltage rather than
for changing voltage levels is called a regulating transformer. Figure 8.10 shows a
Figure 8.10 Regulating transformer for control of voltage magnitude.

Figure 8.11 Regulating transfor¬

mer for control of phase angle.
Windings drawn parallel to each
other are on the same iron core.
regulating transformer for control of voltage magnitude, and Fig. 8.11 shows a
regulating transformer for phase-angle control. The phasor diagram of Fig. 8.12
helps to explain the shift in phase angle. Each of the three windings to which
taps are made is on the same magnetic core as the phase winding whose voltage
is 90° out of phase with the voltage from neutral to the point connected to the
center of the tapped winding. For instance, the voltage to neutral Van is increased
by a component AV^, which is in phase or 180° out of phase with Vbc. Figure 8.12
shows how the three line voltages are shifted in phase angle with very little
change in magnitude.
The procedure to determine Ybus and Zbus in per unit for a network contain¬
ing a regulating transformer is the same as the procedure to account for any
transformer whose turns ratio is other than the ratio used to select the ratio of
base voltages on the two sides of the transformer. Such a transformer, which we
shall now investigate, is said to have an off-nominal turns ratio.
If we have two buses connected by a transformer, and if the ratio of the
line-to-line voltages of the transformer is the same as the ratio of the base
Figure 8.12 Phasor diagram for the regulating transfor¬ b

mer shown in Fig. 8.11.
Figure 8.13 Transformers with differing turns ratios connected in parallel: (a) the one-line diagram;
(b) the reactance diagram in per unit. The turns ratio 1/a is equal to n/ri.
voltages of the two buses, the equivalent circuit (with the magnetizing current
neglected) is simply the transformer impedance in per unit on the chosen base
connected between the buses. Figure 8.13a is a one-line diagram of two transfor¬
mers in parallel. Let us assume that one of them has the voltage ratio 1/n, which
is also the ratio of base voltages on the two sides of the transformer, and that the
voltage ratio of the other is 1/n'. The equivalent circuit is then that of Fig. 8.13b.
We need the ideal (no impedance) transformer with the ratio 1/a in the per-unit
reactance diagram to take care of the off-nominal turns ratio of the second
transformer because base voltages were determined by the turns ratio of the first
transformer.
If we have a regulating transformer (rather than the LTC, which changes
voltage level as well as providing the tap-changing feature), Fig. 8.13b may be
interpreted as two transmission lines in parallel with a regulating transformer in
one line.
Evidently our problem is to find the node admittances of Fig. 8.14, which is
a more detailed representation of the LTC with a turns ratio of 1/n' or of the
regulating transformer with the transformation ratio 1/a. The admittance Y in
the figure is the reciprocal of the per-unit impedance of the transformer. Since
the admittance Y is shown on the side of the ideal transformer nearest node 1,
the tap-changing side (or the side corresponding to n') is nearest node 2. This
designation is important in using the equations which are to be derived. If we are
considering a transformer with off-nominal turns ratio, a is the ratio n'/n. If we
have a regulating transformer, a may be real or imaginary, such as 1.02 for a 2%
boost in magnitude or ejn/6° for a 3° shift per phase.
transformer )
©
-rAAA/—cnnp
+■ V I
Figure 8.14 More detailed per-unit reactance dia¬

gram of the transformer of Fig. 8.13b, whose turns
ratio is 1/a.
Figure 8.14 has been labeled to show currents Ix and I2 entering the two
nodes, and the voltages are V1 and V2 referred to the reference node. The
complex expression for power into the ideal transformer in the direction from
node 1 is
Si (8.32)
and into the transformer from node 2
S2 - V2I*2 (8.33)
Since we are assuming that we have an ideal transformer with no losses, the
power into the ideal transformer from node 1 must equal the power out of the
transformer from node 2, and so
— f* — _ v r* (8.34)
a h ~ Vlh
and
/j = —a* 12 (8.35)
The current 11 can be expressed by
a 1^
27
*-<
(8.36)
II
or
h = VlY-V2- (8.37)
a
Substituting —a*I2 for 7j and solving for I2 yields
Y Y
h= -Ki— +V2—- (8.38)
a* aa*
Comparing Eqs. (8.37) and (8.38), we have, since aa* = \a\2, the node
admittances
*n = *
(8.39)
Y12 =
The equivalent n corresponding to these values of node admittances can be

found only if a is real, so that *21 = *12- If the transformer is changing magni¬
tude, not phase shifting, the circuit is that of Fig. 8.15. This circuit cannot be
realized if Y has a real component, which would require a negative resistance in
the circuit. The important factor, however, is that we can now account for
magnitude, phase shifting, and off-nominal-turns-ratio transformers in calcula¬
tions to obtain Ybus and Zbus.
Figure 8.15 Circuit having the node admittances of

Eqs. (8.39) providing a is real.
Example 8.3 Two transformers are connected in parallel to supply an im¬

pedance to neutral per phase of 0.8 + j0.6 per unit at a voltage of
V2 = 1.0/0° per unit. Transformer Ta has a voltage ratio equal to the ratio of
the base voltages on the two sides of the transformer. This transformer has
an impedance of j0.\ per unit on the appropriate base. The second transfor¬
mer Tb has a step-up toward the load of 1.05 times that of Ta (secondary
windings on 1.05 tap), and its impedance is jOA per unit on the base of the
circuit on its low-tension side. Figure 8.16 shows the equivalent circuit with
transformer Tb represented by its impedance and an ideal transformer. Find
the complex power transmitted to the load through each transformer.
Solution
1.0
I2 = -0.8 + j0.6
0.8 + j0.6
a = 1.05
To determine the current in each transformer we need to find Vx from the

equation
h = v, r2i + v2y22
where the node admittances are those of the parallel combination of the two
transformers. For transformer Ta alone
*21 =
i'0.1
=;io
y“ “ jSI - -jl°
0.8
Figure 8.16 Circuit for Example 8.3. ;0.6

Values in per unit.
For transformer Tb alone
n, = -^
21 1.05
= 7 9.52
y _ Vft-l
= -)9.07
22 11.0512
For the two transformers in parallel
y2i =;10+;9.52 =;19.52
T22 = -710 -79.07 = -719.07
Then, from the node equation for I2,
-0.8 + 70.6 = F,(719.52) -719.07 x 1.0
V1 = 1.008 + 70.041
- V2 = 0.008 + 70.041
Therefore
Itb = (V1-V2)( -710) = 0.41 -70.08

From Eq. (8.35) the current into bus 2 from the transformer Tb is /Tb/a*,
and from Fig. 8.16 this current is ~(ITa + I2), giving
^ = -l2 ~ lTa = 0.8 - 70.6 - (0.41 -70.08) = 0.39 - 70.52
The complex powers are
STa = F2/*a = 0.41 + y'0.08 per unit
Sjb = V2i^-~\ = 0.39 + 70.52 per unit
An approximate solution to this problem is found by recognizing that

Fig. 8.17 with switch S closed is also an equivalent circuit for the problem if the
voltage AV, which is in the branch of the circuit equivalent to transformer Tb, is
;o.i
Figure 8.17 A circuit equivalent

when switch S is closed to that of
Fig. 8.16.
equal to a - 1 in per unit. In other words, if Ta is providing a voltage ratio 5%

higher than Tb, a equals 1.05 and AT equals 0.05 per unit. To the extent that we
can say that the current set up by AV circulates around the loop indicated by
/circ with switch S open and that with S closed none of that current goes through
the load impedance because it is much larger than the transformer impedance,
we can use the superposition principle. Then
005
/circ = Jo! = ~j025 pCr Umt
With AT short-circuited the current in each path is half of the load current, or
0.4 — y‘0.3. Then superimposing the circulating current gives
ITa = 0.4 — j0.3 - (—y’0.25) = 0.4 -jO.05
ITb = 0.4 — j0.3 + (-j0.25) = 0.4 - j0.55
so that
STa = 0.40 + ;0.05 per unit
and
STb = 0.40 + y'0.55 per unit
These values, although approximate, are so nearly equal to values originally

found that the method is often used because of its simplicity.
This example shows that the transformer with the higher tap setting is
supplying most of the reactive power to the load. The real power is dividing
equally between the transformers. Since both transformers have the same
impedance, they would share both the real and reactive power equally if they
had the same turns ratio. In that case each would be represented by the same
per-unit reactance of ;'0.1 between the two buses and would carry equal current.
When two transformers are in parallel, we can vary the distribution of reactive
power between the transformers by adjusting the voltage-magnitude ratios.
When two paralleled transformers of equal kilovoltamperes do not share the
kilovoltamperes equally because their impedances differ, the kilovoltamperes
may be more nearly equalized by adjustment of the voltage-magnitude ratios
through tap changing.
If a particular transmission line in a system is carrying too small or too large
a reactive power, a regulating transformer to adjust voltage magnitude can be
provided at one end of the line to make the line transmit a larger or smaller
reactive power. We can investigate this by means of the automatic tap-changing
feature in the load-flow program on a digital computer. For instance, we can
raise the voltage at bus 4 of Example 8.1 by inserting a magnitude-regulating
transformer in the line from bus 5 to bus 4 at bus 4, and we tell the computer to
consider this as an LTC with a tap setting to hold the bus voltage at about 0.950
per unit. There is a definite step between tap settings, and the voltage will not
To bus 5
Figure 8.18 Flow of P and Q at bus 4 of

the system of Fig. 8.1 when a regulating
transformer in line 5-4 at bus 4 raises V4 to
0.946 per unit. Load
necessarily be exactly 0.950 per unit. The results achieved at bus 4 are shown in
the one-line diagram of Fig. 8.18. A per-unit reactance of 0.08 was assumed for
the LTC.
When the voltage of bus 4 is raised by the LTC in line 5-4, the voltage drop
over line 3-4 must be less and we expect this to be accomplished by a reduced
flow of reactive power through the line with little change in real power. By
comparing Fig. 8.18 with Fig. 8.9a we see that the Q flowing into bus 4 through
line 3-4 is reduced from 18.90 to 12.26 Mvar without much change in P. To
supply the 30 Mvar required by the load 17.71 Mvar now flows into bus 4
through the line from bus 5 and the LTC. The increased megavars on the line
cause the voltage on the low-tension side of the LTC to be quite low, but the
transformer steps up the voltage to 0.946 per unit at bus 4 by selecting the
proper tap setting.
The voltage at bus 5 dropped from the original value of 0.968 per unit to
0.962 per unit. As a comparison, we noted in Sec. 8.9 that the voltage at bus 5
increased to 0.976 per unit when capacitors were added to bus 4. The reason for
the decrease in the voltage at bus 5 in the present case is that the increased
reactive power supplied to bus 4 from bus 5 caused an increase in the reactive
power which had to be fed to bus 5 from the voltage-regulated buses 1 and 3.
To determine the effect of phase-shifting transformers we need only let a be
complex with a magnitude of unity in Eqs. (8.39).
Example 8.4 Repeat Example 8.3 except that Tb includes both a transformer
having the same turns ratio as Ta and a regulating transformer with a phase
shift of 3° (a = cjnl60 = 1.0/^). The impedance of the two components of Tb
is y'0.1 per unit on the base of Ta.
Solution For transformer Ta alone, as in Example 8.3,
^2i =710 Y2 2=-jl0

and for transformer Tb
l, = ^ = io£r
Y — _ _ _/io
22 “11.0/3! I2 ;
Combining the transformers in parallel gives
Y21 = 10^0! 4- 10^3! = -0.523 + ;20.0
Y22 = -jlO -jlO = -j20
Following the procedure of Example 8.3, we have
-0.8 + j0.6 = Fj( —0.523 + j20) 4- (-/20)(1.0)
-0.8+./20.6 0.418 —y 10.77 4- y'16.0 4- 412.0

Vl ~ -0.523 + j20 “ 400
= 1.03 + ;0.013
V1-V2 = 0.03 4- ;0.013

ITa = (0.03 4 ;0.013)(-;10) = 0.13 -;0.30
^ = 0.8 — j0.6 - (0.13 -;0.30) = 0.67 - ;0.30
STa = 0.13 4- j0.30 per unit

STb = 0.67 + j0.30 per unit
As in Example 8.3, we can obtain an approximate solution of the problem

by inserting a voltage source AV in series with the impedance of transformer
Tb. The proper per-unit voltage is
a - 1 = 1.0/3! ~ 10/0! = (2 sin 1.5°)M.5° = 0.0524/^1.5°

0.0524/91.5°
= 0.262 + ;'0.0069
0.2/90°
ITa = 0.4 — j0.3 - (0.262 + ;0.007) = 0.138 -;0.307

ITb = 0.4 — j0.3 4- (0.262 + jO.OOl) = 0.662 - ;0.293
So
STa = 0.138 + j0.307 per unit

STb = 0.662 4- j0.293 per unit
Again the approximate values are close to the values found previously.
To bus 5
Figure 8.19 Flow of P and Q at bus 4 of the

system of Fig. 8.1 when a regulating transformer
in line 5-4 at bus 4 causes a phase shift of 3°
across its terminals.
The example shows that the phase-shifting transformer is useful to control

the amount of real power flow but has less effect on the flow of reactive power.
Both Examples 8.3 and 8.4 are illustrative of two transmission lines in parallel
with a regulating transformer in one of the lines. For instance, Eqs. (8.39) would
apply to a transmission line having a regulating or off-nominal-turns-ratio trans¬
former at one end and with shunt admittance and the impedance of the transfor¬
mer neglected or included in the series impedance of the line. In that case, Y of
Eqs. (8.39) would be the reciprocal of the series impedance of the line in per unit.
In a load-flow study on the digital computer a transformer at the end of a line
may be taken into account by adding a bus so that the transformer is directly
connected to buses on both sides.
Figure 8.19 shows the flow of real and reactive power and the voltage at bus
4 of the system of Example 8.1 when a phase-shifting transformer is placed in
line 5-4 at bus 4. Input data to the computer specified a shift of 3° across the
transformer. The result was a shift in real power from line 3-4 to line 5-4, which
would be expected from our discussion of transformers or transmission lines in
parallel. In this case the two lines being compared are not in parallel and there is
a significant change (about half as great as the change in P) in reactive power in
the lines into bus 4. This change in Q is consistent with Eq. (8.29) even though
we are not neglecting resistance and is explained by the reduction in <5 between
buses 3 and 4 which increases Q over that line.
8.11 SUMMARY
In addition to discussing how load-flow studies are made on a computer this

chapter has presented some methods of controlling voltage and the flow of
power from the standpoint of understanding how this control is accomplished.
The load-flow study on a computer is the best way to obtain quantitative ans¬
wers for the effect of specific control operations.
The analysis of the effect of excitation of the synchronous generator con¬

nected to a bus having a constant voltage as discussed in Chap. 6 has been
extended to a generator supplying a system represented by its Thevenin
equivalent.
When we looked at the application of capacitors at a load, we saw that the
reactive power furnished by the capacitors caused the voltage at the load to rise.
Since increasing the excitation of the synchronous generator causes input of
reactive power to the system, the effect is the same as the addition of capacitors
and will cause the voltage at the generator bus to rise unless the system is very
large.
Since voltage magnitude and generator real power delivered are usually
specified for a load-flow study, we examined how generator excitation must be
varied to meet the specified bus voltage for constant P from the generator.
Finally we derived expressions for P and Q from the generator in terms of \Vt\,
\Eg\, and power angle S to show the dependence of real power on S.
The results of paralleling two transformers when the voltage-magnitude
ratios were different or when one provided a phase shift were examined. Equa¬
tions (8.39) provide us with the equations for the node admittances of the equi¬
valent circuits of such transformers. Examples were provided to show that LTC
transformers controlling voltage magnitude and regulating transformers of the
magnitude- and phase-shifting types could control the flow of real and reactive
power on transmission lines.
PROBLEMS
8.1 Evaluate AP(°] for continuing Example 8.1.
8.2 Determine the value of the element (dP3/d<54) in the third column and second row of the jacobian
for the first iteration in continuing Example 8.1.
8.3 Evaluate for the first iteration the element in the third column and third row of the jacobian of
Example 8.1.
8.4 Evaluate for the first iteration the element in the sixth column and third row of the jacobian of
Example 8.1.
8.5 Draw a diagram similar to Fig. 8.3 for bus 3 of the system of Example 8.1 from the information
provided by the printout of load flow in Fig. 8.2. What is the apparent megawatt and megavar
mismatch at this bus?
8.6 Reproduce Fig. 8.20 and on it indicate for Example 8.1 the values of
(a) P and Q leaving bus 5 on line 5-4.
(b) Q supplied by the fixed capacitance of the nominal n of line 5-4 at bus 5. (Note that this value of
Q varies as | V512.)
(c) P and Q at both ends of the series part of the nominal n of the line.
© * ©
Figure 8.20 Diagram for

Prob. 8.6.
(d) Q supplied by the fixed capacitance of the nominal n of line 5-4 at bus 4.
(e) P and Q into bus 4 on line 5-4.
8.7 As part of the load-flow solution of Example 8.1 the computer listed a total line loss of 9.67 MW.
How does this compare with the sum of the losses which can be found from the load-flow listings of
each individual line?
8.8 The effect of field excitation discussed in Sec. 6.4 can now be calculated. Consider a generator
having a synchronous reactance of 1.0 per unit and connected to a large system. Resistance may be
neglected. If the bus voltage is 1.0/0° per unit and the generator is supplying to the bus a current of
0.8 per unit at 0.8 power-factor lagging, find the magnitude and angle of the no-load voltage Eg of the
generator and P and Q delivered to the bus. Then find the angle S between Eg and the bus voltage,
the current Ia, and Q delivered to the bus by the generator if the power output of the generator
remains constant but the excitation of the generator is (a) decreased so that | Eg\ is 15% smaller and
(b) increased so that |£j is 15% larger. What is the percent change in Q upon reducing and
increasing |Ej ? Do the results of this problem agree with the conclusions reached in Sec. 6.4?
8.9 A power system to which a generator is to be connected at a certain bus may be represented by
the Thevenin voltage E,h = 0.9/0° per unit in series with Zlh = 0.25/90° per unit. When connected to
the system, Eg of the generator is 1.4/30° per unit. Synchronous reactance of the generator on the
system base is 1.0 per unit, (a) Find the bus voltage V, and P and Q transferred to the system at the
bus; (b) if the bus voltage is to be raised to \V,\ = 1.0 per unit for the same P transferred to
the system, find the value of Eg required and the value of Q transferred to the system at the bus.
Assume all other system emfs are unchanged in magnitude and angle; that is, Elh and Zlh are
constant.
8.10 In Prob. 7.10 voltages at the three buses were calculated before and after connecting a capacitor
from neutral to bus 3. Determine P and Q entering or leaving bus 3 over transmission lines, through
the reactance connected between the bus and neutral, and from the capacitor before and after the
capacitor is connected. Assume generated voltages remain constant in magnitude and angle. Draw
diagrams similar to those of Fig. 8.9 to show the values calculated.
8.11 Figure 8.9 shows that 15.3 Mvar must be supplied by a capacitor bank at bus 4 of the 60-Hz
system of Example 8.1 to raise the bus voltage to 0.950 per unit. If the base voltage is 138 kV, find
the capacitance in each phase if the capacitors are (a) connected in Y and (b) connected in A.
8.12 Two buses a and b are connected to each other through impedances X 2 =0.1 and X 2 = 0.2 per
unit in parallel. Bus b is a load bus supplying a current / = 1.0/— 30° per unit. The per-unit bus
voltage Vb is 1.0/0°. Find P and Q into bus b through each of the parallel branches (a) in the circuit
described, (b) if a regulating transformer is connected at bus b in the line of higher reactance to give a
boost of 3% in voltage magnitude toward the load (a = 1.03), and (c) if the regulating transformer
advances the phase 2° (a = sinl90). Use the circulating-current method for parts (b) and (c), and
assume that Va is adjusted for each part of the problem so that Vb remains constant. Figure 8.21 is the
one-line diagram showing buses a and b of the system with the regulating transformer in place.
Neglect the impedance of the transformer.
8.13 Two reactances A", = 0.08 and X2 = 0.12 per unit are in parallel between two buses a and b in a
power system. If Va = 1.05/10° and Vb = 1.0/0° per unit, what should be the turns ratio of the regulating
transformer to be inserted in series with X2 at bus b so that no vars flow into bus b from the
branch whose reactance is A" j ? Use the circulating-current method, and neglect the reactance of the
regulating transformer. P and Q of the load and Vb remain constant.
© ©
X = >0.1
Figure 8.21 Circuit for Prob. 8.12.

8.14 Two transformers each rated 115Y/13.2A kV operate in parallel to supply a load of 35 MVA,
13.2 kV at 0.8 power-factor lag. Transformer 1 is rated 20 MVA with X = 0.09 per unit, and trans¬
former 2 is rated 15 MVA with X =0.07 per unit. Find the magnitude of the current in per unit
through each transformer, the megavoltampere output of each transformer, and the megavolt-
amperes to which the total load must be limited so that neither transformer is overloaded. If the taps
on transformer 1 are set at 111 kV to give a 3.6% boost in voltage toward the low-tension side of
that transformer compared to transformer 2, which remains on the 115 kV tap, find the megavolt¬
ampere output of each transformer for the original 35 MVA total load and the maximum megavolt-
amperes of the total load which will not overload the transformers. Use a base of 35 MVA, 13.2 kV
on the low-tension side. The circulating-current method is satisfactory for this problem.
8.15 If the impedance of the load on bus b of the circuit described in Prob. 8.12 is 0.866 + y'0.5 per
unit, and if Va is 1.04/0° per unit (the voltage Vb and the load current no longer specified), find Vb for
the conditions described in parts (a), (b), and (c) of Prob. 8.12. Also find P and Q into bus b through
each of the parallel branches in all three cases. Equations (8.39) should be used in this problem, and
the load impedance can be included in Y22 of the node admittance equations of the complete circuit.
CHAPTER
NINE
ECONOMIC OPERATION OF POWER SYSTEMS
An engineer is always concerned with the cost of products and services. For a
power system to return a profit on the capital invested, proper operation is very
important. Rates fixed by regulatory bodies and the importance of conservation
of fuel place extreme pressure on power companies to achieve maximum
efficiency of operation and to improve efficiency continually in order to maintain
a reasonable relation between cost of a kilowatthour to a consumer and the cost
to the company of delivering a kilowatthour in the face of constantly rising
prices for fuel, labor, supplies, and maintenance.
Engineers have been very successful in increasing the efficiency of boilers,
turbines, and generators so continuously that each new unit added to the gener¬
ating plants of a system operates more efficiently than any older unit on the
system. In operating the system for any load condition the contribution from
each plant and from each unit within a plant must be determined so that the cost
of delivered power is a minimum. How the engineer has met and solved this
challenging problem is the subject of this chapter.
An early method of attempting to minimize the cost of delivered power
called for supplying power from only the most efficient plant at light loads. As
load increased, power would be supplied by the most efficient plant until the
point of maximum efficiency of that plant was reached. Then for further increase
in load the next most efficient plant would start to feed power to the system, and
a third plant would not be called upon until the point of maximum efficiency of
the second plant was reached. Even with transmission losses neglected this
method fails to minimize cost.
We shall study first the most economic distribution of the output of a plant
between the generators, or units, within the plant. Since system generation is
often expanded by adding units to existing plants, the various units within a
plant usually have different characteristics. The method that will be developed is
also applicable to economic scheduling of plant outputs for a given loading of
the system without consideration of transmission losses. We shall proceed to
227
develop a method of expressing transmission loss as a function of the outputs of

the various plants. Then we shall determine how the output of each of the plants
of a system is scheduled to achieve minimum cost of power delivered to the load.
9.1 DISTRIBUTION OF LOAD BETWEEN UNITS

WITHIN A PLANT
To determine the economic distribution of load between the various units con¬
sisting of a turbine, generator, and steam supply the variable operating costs of
the unit must be expressed in terms of the power output. Fuel cost is the
principal factor in fossil-fuel plants, and cost of nuclear fuel can also be ex¬
pressed as a function of output. Most of our electric energy will continue to
come from fossil and nuclear fuels for many years until other energy sources are
able to assume some of the task. We shall base our discussion on the economics
of fuel cost with the realization that other costs which are a function of power
output can be included in the expression for fuel cost. A typical input-output
curve which is a plot of fuel input for a fossil-fuel plant in Btu per hour versus
power output of the unit in megawatts is shown in Fig. 9.1. The ordinates of the
curve are converted to dollars per hour by multiplying the fuel input by the cost
of fuel in dollars per million Btu.
If a line is drawn through the origin to any point on the input-output curve,
the reciprocal of the slope can be expressed in megawatts divided by input in
millions of Btu per hour, or the ratio of energy output in megawatthours to fuel
input measured in millions of Btu. This ratio is the fuel efficiency. Maximum
efficiency occurs at that point where the slope of the line from the origin to a
point on the curve is a minimum, that is, at the point where the line is tangent to
ECONOMIC OPERATION OF POWER SYSTEMS 229
the curve. For the unit whose input-output curve is shown in Fig. 9.1, the maxi¬
mum efficiency is at an output of approximately 280 MW, which requires an
input of 2.8 x 109 Btu/h. The fuel requirement is 10.0 x 106 Btu/MWh. By
comparison, when the output of the unit is 100 MW, the fuel requirement is
11.0 x 106 Btu/MWh.
Of course the fuel requirement for a given output is easily converted into
dollars per megawatthour. As we shall see, the criterion for distribution of the
load between any two units is based upon whether increasing the load on one
unit as the load is decreased on the other unit by the same amount results in an
increase or decrease in total cost. Thus, we are concerned with incremental cost,
which is determined by the slopes of the input-output curves of the two units. If
we express the ordinates of the input-output curve in dollars per hour and let
Fn = input to unit n, dollars per hour
Pn = output of unit n, MW
the incR “'ental fuel cost of the unit in dollars per megawatthour is dFn/dPn.
Incre lental fuel cost for a unit for any given power output is the limit of the
ratio of the increase in cost of fuel input in dollars per hour to the corresponding
increase in power output in megawatts as the increase in power output
approaches zero. Approximately, the incremental fuel cost could be obtained by
determining the increased cost of fuel for a definite time interval during which
the power output is increased by a small amount. For instance, the approximate
incremental cost at any particular output is the additional cost in dollars per
hour to increase the output by 1 MW. Actually incremental cost is determined
by measuring the slope of the input-output curve and multiplying by cost per
Btu in the proper units. Since mills (tenths of a cent) per kilowatthour are equal
to dollars per megawatthour, and since a kilowatt is a very small amount of
power in comparison with the usual output of a unit of a steam plant, incremen¬
tal fuel cost may be considered to be the cost of fuel in mills per hour to supply
an additional kilowatt output.
A typical plot of incremental fuel cost versus power output is shown in
Fig. 9.2. This figure is obtained by measuring the slope of the input-output curve
of Fig. 9.1 for various outputs and applying a fuel cost of $1.30 per million Btu.
However, the cost of fuel in terms of Btu is not very predictable, and the reader
should not assume that cost figures throughout this chapter are applicable at
any particular time. Figure 9.2 shows that incremental fuel cost is quite linear
with respect to power output over an appreciable range. In analytical work the
curve is usually approximated by one or two straight lines. The dashed line in
the figure is a good representation of the curve. The equation of the line is
dF
—- = 0.0126 P + 8.9
dPn
so that when the power output is 300 MW, the incremental cost determined by
the linear approximation is $12.68 per megawatthour. This value is the approxi¬
mate additional cost per hour of increasing the output by 1 MW and the saving
Figure 9.2 Incremental fuel cost versus power output for the unit whose input-output curve is shown
in Fig. 9.1.
in cost per hour of reducing the output by 1 MW. The actual incremental cost at
300 MW is $12.50 per megawatthour, but this power output is near the point of
maximum deviation between the actual value and the linear approximation of
incremental cost. For greater accuracy two straight lines may be drawn to repre¬
sent this curve in its upper and lower range.
We now have the background to understand the guiding principle for dis¬
tributing the load among the units within a plant. For instance, suppose that the
total output of a plant is supplied by two units and that the division of load
between these units is such that the incremental fuel cost of one is higher than
that of the other. Now suppose that some of the load is transferred from the unit
with the higher incremental cost to the unit with the lower incremental cost.
Reducing the load on the unit with the higher incremental cost will result in a
greater reduction of cost than the increase in cost for adding that same amount
of load to the unit with the lower incremental cost. The transfer of load from one
to the other can be continued with a reduction in total fuel cost until the
incremental fuel costs of the two units are equal. The same reasoning can be
extended to a plant with more than two units. Thus the criterion for economical
division of load between units within a plant is that all units must operate at the
same incremental fuel cost. If the plant output is to be increased, the incremental
cost at which each unit operates will rise but must remain the same for all.
The criterion which we have developed intuitively can be found mathema¬
tically. For a plant with K units, let
FT = Ft+F2 + --- + FK=YF- <9‘)

n=1
Pr = p> + P2+- + p„= Ip. (9.2)

n=1
where FT is the total fuel cost and PR is the total power received by the plant bus
and transferred to the power system. The fuel costs of the individual units are F1?
•••» Fk wdh corresponding outputs Pu P2, PK. Our objective is to
obtain a minimum FT for a given PR, which requires that the total differential
dFT = 0. Since total fuel cost is dependent on the power output of each unit
dFj
dFT = ^dP1+^~dP2 + dPk = 0 (9.3)
dP i dP, dP>
With total fuel cost FT dependent upon the various unit outputs, the require¬
ment of constant PR means that Eq. (9.2) is a constraint on the minimum value
of Ft. The restriction that PR remain constant requires that dPR = 0, and so
dP i + dP + ■ ■ ■ T dPR — 0
2 (9.4)
Multiplying Eq. (9.4) by X and subtracting the resulting equation from Eq. (9.3)
yields, when terms are collected.
dFT dFT
dP i + dP f ‘ ” T
2 X dPk- = 0 (9.5)
8P~i dP~K
This equation is satisfied if each term is equal to zero. Each partial derivative
becomes a full derivative since only the fuel cost of any one unit will vary if only
the power output of that unit is varied. For example dFT/dPK becomes
dFK/dPK. Equation (9.5) is satisfied if
dF i _ ^ dF 2 _ dFK,
dp[~ ,dF2 ~ ,'",dFK ~
and so all units must operate at the same incremental fuel cost X for minimum
cost in dollars per hour. Thus we have proved mathematically the same criterion
which we reached intuitively. The procedure is known as the method of lagran-
gian multipliers. We shall need this mathematical method when we consider the
effect of transmission losses on the distribution of loads between several plants
to achieve minimum fuel cost for a specified loading of a power system.
When the incremental fuel cost of each of the units in a plant is nearly linear
with respect to power output over a range of operation under consideration,
equations that represent incremental fuel costs as linear functions of power
output will simplify the computations. A schedule for assigning loads to each
unit in a plant can be prepared by assuming various values of X, obtaining the
corresponding outputs of each unit, and adding outputs to find plant load for
each assumed X. A curve of X versus plant load establishes the value of X at
which each unit should operate for a given total plant load. If maximum and
minimum loads are specified for each unit, some units will be unable to operate
at the same incremental fuel cost as the other units and still remain within the
limits specified for light and heavy loads.
Example 9,1 Incremental fuel costs in dollars per megawatthour for a plant
consisting of two units are given by
^ = 0.0080?! + 8.0 ^ = 0.0096? 2 + 6.4

dPt dF 2
Assume that both units are operating at all times, that total load varies from
250 to 1250 MW, and that maximum and minimum loads on each unit are
to be 625 and 100 MW, respectively. Find the incremental fuel cost and the
allocation of load between units for the minimum cost of various total loads.
Solution At light loads unit 1 will have the higher incremental fuel cost
and will operate at its lower limit of 100 MW for which dFj jdPx is $8.8 per
megawatthour. When the output of unit 2 is also 100 MW, dF2/dP2 is $7.36
per megawatthour. Therefore, as plant output increases, the additional load
should come from unit 2 until dF2/dP2 equals $8.8 per megawatthour. Until
that point is reached the incremental fuel cost 2 of the plant is determined by
unit 2 alone. When the plant load is 250 MW, unit 2 will supply 150 MW
with dF2/dP2 equal to $7.84 per megawatthour. When dF2/dP2 equals $8.8
per megawatthour
0.0096P2 + 6.4 = 8.8
2.4
P2 = 250 MW
0.0096
and the total plant output is 350 MW. From this point on the required
output of each unit for economic load distribution is found by assuming
various values of X and calculating each unit’s output and the total plant
output. Results are shown in Table 9.1.
Figure 9.3 shows plant X plotted versus plant output. We note in Table
9.1 that at X = 12.4, unit 2 is operating at its upper limit, and that additional
load must come from unit 1, which then determines the plant X.
If we wish to know the distribution of load between the units for a plant
output of 1000 MW, we could plot the output of each individual unit versus
plant output as shown in Fig. 9.4 from which each unit’s output can be read
for any plant output. A digital computer could easily determine the correct
Table 9.1 Outputs of each unit and

total output for various values of X for
Example 9.1
Unit 1 Unit 2 Plant

Plant X, Pi, Pi, Pi + P2
$/MWh MW MW MW
7.84 100 150 250

8.8 100 250 350
9.6 200 333 533
10.4 300 417 717
11.2 400 500 900
12.0 500 583 1083
12.4 550 625 1175
13.0 625 625 1250
Figure 9.3 Incremental fuel cost

versus plant output with total
plant load economically distri¬
buted between units, as found
in Example 9.1. Plant output, MW
output of each of many units by requiring all unit incremental costs to be

equal for any total plant output. For the two units of the example for a total
output of 1000 MW
pl + p2= 1000
and
0.008P! + 8.0 - 0.0096(1000 - Pt) + 6.4
P1 = 454.55 MW
P2 = 545.45 MW
0 200 400 600 800 1000 1200 1400
Plant output, MW
Figure 9.4 Output of each unit versus plant output for economical operation of the plant of
Example 9.1.
and for each unit X = 11.636. Such accuracy, however, is not necessary-be-
cause of the uncertainty in determining costs and the use of an approximate
equation in this example to express the incremental costs.
The savings effected by economic distribution of load rather than some
arbitrary distribution can be found by integrating the expression for in¬
cremental fuel cost and comparing increases and decreases of cost for the
units as load is shifted from the most economical allocation.
Example 9.2 Determine the saving in fuel cost in dollars per hour for the
economic distribution of a total load of 900 MW between the two units of
the plant described in Example 9.1 compared with equal distribution of the
same total load.
Solution Example 9.1 shows that unit 1 should supply 400 MW and unit 2
should supply 500 MW. If each unit supplies 450 MW, the increase in cost
for unit 1 is
450
0.004P\ + 8 P1 = $570 per hour
400
Similarly, for unit 2

.450 450
(0.0096P2 + 6.4) dP2 0.0048P2 + 6.4P - —$548 per hour
500 500
The negative sign indicates a decrease in cost, as we expect for a decrease in

output. The net increase in cost is $570 — $548 = $22 per hour. The saving
seems small, but this amount saved every hour for a year of continuous
operation would reduce fuel cost by $192,720 for the year.
The saving effected by economic distribution of load justifies devices for

controlling the loading of each unit automatically. We shall consider automatic
control of generation after investigating the coordination of transmission losses
in the economic distribution of load between plants.
9.2 TRANSMISSION LOSS AS A

FUNCTION OF PLANT GENERATION
In determining the economic distribution of load between plants we encounter
the need to consider losses in the transmission lines. Although the incremental
fuel cost at one plant bus may be lower than that of another plant for a given
distribution of load between the plants, the plant with the lower incremental cost
at its bus may be much farther from the load center. The losses in transmission
from the plant having the lower incremental cost may be so great that economy
may dictate lowering the load at the plant with the lower incremental cost and
Plant 1 Plant 2
a a o
©
Figure 9.5 A simple system connecting two gener¬

ating plants to one load. Load
increasing it at the plant with the higher incremental cost. To coordinate trans¬
mission loss in the problem of determining economic loading of plants, we need
to express the total transmission loss of a system as a function of plant loadings.
Determining the transmission loss in a simple system consisting of two
generating plants and one load will help us see the principles involved in express¬
ing loss in terms of power output of the plants.! Figure 9.5 shows such a
system. If Ra, Rb, and Rc are the resistances of lines a, b, and c, respectively, the
total loss for the three-phase transmission system is
Pl = 3 17, |2R. + 3 112\2R„ + 3 17, + I2 |2R, (9.7)
If we assume that !, and l2 are in phase,
|/t +/2| = |/i| + P2I (9.8)
and upon simplification
Pl = 3|/t |2(Kfl + Rc) + 3 x 2\I, | \I2\Rc + 3\I2\2(Rb + Rc) (9.9)
If P1 and P2 are the three-phase power outputs of plants 1 and 2 at power

factors of p/t and pf2 and if and V2 are the bus voltages at the plants,
Pi (9.10)
I'll = v^lFil pf, and /,! =
v/3|ni/>/2
Upon substitution in Eq. (9.9), we obtain
R, B-b + Rc
Pl = Pl , f+2 PtP2 + P\
KrlW ' “ * WIVlWMl) V2\2(pf2)1
= P\Bn + 2P1P2Bi2 + P2 B22 (9.11)
t For a simple discussion of loss formulas in a manner similar to our approach, see D. C. Harker,
A Primer on Loss Formulas, Trans. AIEE, vol. 77, pt. Ill, 1958, pp. 1434-1436.
where
Ra + Rc
5n =
W1?
R,
B12 '
(9.12)
V1\\V2\(pf1)(pf2)
+ Fc
B22 ~
V7\2 (pfif
The terms Blx, B12, and B22 are called loss coefficients or B coefficients. If the
voltages in Eq. (9.12) are line-to-line kilovolts with resistances in ohms, the units
for the loss coefficients are reciprocal megawatts. Then, in Eq. (9.11), with three-
phase powers Px and P2 in megawatts, PL will be in megawatts also. Of course,
the computations may be made in per unit.
For the system for which they are derived and the assumption of I x and 12
in phase, these coefficients yield the exact loss by Eq. (9.11) only for the particu¬
lar values of Px and P2 which result in the voltages and power factors used in
Eqs. (9.12). The B coefficients are constant, as Px and P2 vary, only insofar as
bus voltages at the plants maintain constant magnitude and plant power factors
remain constant. Fortunately, the use of constant values for the loss coefficients
in Eq. (9.11) yields reasonably accurate results when the coefficients are cal¬
culated for some average operating condition and if extremely wide shifts of load
between plants or in total load do not occur. In practice large systems are
economically loaded by calculations based on several sets of loss coefficients
depending on load conditions.t
Example 93 For the system whose one-line diagram is shown in Fig. 9.5,
assume I x = 1.0/0° per unit and I2 = 0.8/0° per unit. If the voltage at bus 3
is V3 — 1.0/0° per unit, find the loss coefficients. Line impedances are 0.04 +
;0.16 per unit, 0.03 + ;'0.12 per unit, and 0.02 + /0.08 per unit for sections a,
b, and c, respectively.
Solution Ordinarily load currents and bus voltages are available from load
studies. For this problem bus voltages can be calculated from the data given:
Vx = 1.0 + (1.0 + y'0)(0.04 + ;'0.16) = 1.04 + y'0.16 per unit

V2 — 1.0 + (0.8 + ;0)(0.03 + yO. 12) = 1.024 + ;0.096 per unit
Since all currents have phase angles of zero, the power factor at each source
node is the cosine of the angle of the voltage at the node, and voltage
t For two of many sources of further information see L. K. Kirchmayer, Economic Operation of
Power Systems, John Wiley and Sons, Inc., New York, 1958; W. S. Meyer and V. D. Albertson,
Improved Loss Formula Computation by Optimally Ordered Elimination Techniques, IEEE
Trans. Power Appar. Syst., vol. 90, 1971, pp. 716-731. Also see the footnotes in Sec. 9.4 for a method
used by a number of companies.
magnitude times power factor equals the real part of the complex expression
for voltage. Therefore
n 0.04 + 0.02
Bi i =-— = 0.0554 per unit
B,2=rd24°x21.04 = 00188 per unit

_ 0.03 + 0.02
= 0.0477 per unit
1.0242
Example 9.4 Calculate the transmission loss for Example 9.3 by the loss
formula of Eq. (9.11), and check the result.
Solution Source powers are ordinarily available from the load study for
the operating condition used to calculate loss coefficients. For this problem
Pj and P2 must be calculated.
Px = Re {(1.0 + y'0)(1.04 + yO. 16)} = 1.04 per unit

P2 = Re {(0.8 + j0)( 1.024 + y'0.096)} = 0.8192 per unit
pL = 1.042 x 0.0554 + 2 x 1.04 x 0.8192 x 0.0188 + 0.81922 x 0.0477
= 0.06 + 0.032 + 0.032 = 0.124 per unit
Adding the loss in each section computed by I2R yields
PL = 1.02 x 0.04 + 1.82 x 0.02 + 0.82 x 0.03

= 0.04 + 0.0648 + 0.0192 = 0.124 per unit
Exact agreement between methods is expected since the loss coefficients were
determined for the condition for which loss was calculated. The amount of
error introduced by using the same loss coefficients for two other operating
conditions may be seen by examining the results shown in Table 9.2.
Table 9.2 Comparison of transmission loss calculated by loss coeffi¬

cients and 72R for data of Example 9.3 with several operating conditions
All quantities are in per unit
Pi. by
loss
coeffi¬
h P2 cients PL byI2R Conditions
1.0 0.8 1.040 0.819 0.124 0.124 Original case

0.5 0.4 0.510 0.405 0.030 0.031 Pj, P2 reduced 50%
0.5 1.3 0.510 1.351 0.128 0.126 0.53 shift Pj to P2
The general form of the loss equation for any number of sources is
m n
<913>
where and indicate independent summations to include all sources. For
instance, for three sources,
PL = p\Bu + P\B21 + P\B33 + 2P1P2B12 + 2P2P3B23 + 2P1P3B13

(9.14)
The matrix form of the transmission-loss equation is
PL = PTBP (9.15)
where for a total of s sources

1_
1
0., ^ ••
Bn B12 B13 • Bu
-H
B2i • BU
<N
B22 B23
p = and B =
Ps Bsl B's2 Bs3 • ' Bss
9.3 DISTRIBUTION OF LOAD BETWEEN PLANTS
The method developed to express transmission loss in terms of plant outputs

enables us to coordinate transmission loss in scheduling the output of each plant
for maximum economy for a given system load. The mathematical treatment is
similar to that of scheduling units within a plant except that we shall now
include transmission loss as an additional constraint.
In the equation
FT=Fl+F2 + ---+FK= Y.F, (9.16)

n=1
Ft is now the total cost of all the fuel for the entire system and is the sum of the
fuel costs of the individual plants Fu F2,..., FK. The total input to the network
from all the plants is
Pt = Pi + Pi + - + Pk- Ip. (9.17)

n=1
where Pu P2, ..., PK are the individual plant inputs to the network. The total
fuel cost of the system is a function of the power inputs. The constraining
relation on the minimum value of FT is
Z,P,-PL-PR= 0 (9.18)
n= 1
where PR is the total power received by the loads on the system and PL is the
transmission loss expressed as a function of the loss coefficients and the power
input to the network from each plant. Since PR is constant, dPR = 0; therefore
K
I dP„ — dPL — 0 (9.19)

n= 1
and since minimum cost means dFT = 0,
K dF
dFr=YJ~~ dPn = 0 (9.20)
n=1 crn
Transmission loss PL is dependent upon plant outputs, and dPL is expressed by

K PP
dpL=
n=1
Z j+dP.
vrn
(9.21)
Substituting dPL from Eq. (9.21) in Eq. (9.19), multiplying by X, and subtracting
the result from Eq. (9.20) yield
dFT .dPL
X\dP = 0 (9.22)
SPn dPn
This equation is satisfied provided that
dFT dJ\
+ A X=0 (9.23)
Wn 8P„
for every value of n. Rearranging Eq. (9.23) and recognizing that changing the
output of only one plant can affect the cost at only that plant, we have
dF„ 1
= X (924)
dPn l - dPL/dPn
or
dF n
(925)
dPnLn~
where Ln is called the penalty factor of plant n and
i,=_I_ (926)
” i - spjdp.
The multiplier X is in dollars per megawatthour when fuel cost is in dollars per
hour and power is in megawatts. The result is analogous to that for scheduling
units within a plant. Minimum fuel cost is obtained when the incremental fuel
cost of each plant multiplied by its penalty factor is the same for all plants in the
system. The products are equal to X, which is called the system X and is approxi¬
mately the cost in dollars per hour to increase the total delivered load by 1 MW.
For a system of three plants, for instance,
dF, r dF 2 r dF3
Lr, = X (9.27)
dP^1 ~ dP2Ll~ dP3
The transmission loss PL is expressed by Eq. (9.13). For K plants partial
differentiation with respect to Pn yields
dP,
X ep.v. / j pmBn
2 y m rt (9.28)
dP„ dP n m=1 n=1 m= 1
The simultaneous equations obtained by writing Eq. (9.24) for each plqqt of
the system can be solved by assuming a value for X. Economical loading for each
plant is found for the assumed X. By solving the equations for several values of X,
data are found for plotting generation at each plant against total generation. If
transmission loss is calculated for each X, plant outputs can be plotted against
total received load. If fixed amounts of power are transferred over tie lines with
other systems or received from hydro stations, the distribution of the remaining
load among the other plants is affected by the changes in transmission loss
brought about by the flow through these additional points of entry to the
system. No new variables are introduced, but additional loss coefficients are
required. For instance, a system having five steam plants, three hydro plants, and
seven interconnections would require a 15 x 15 matrix of loss coefficients, but
the only unknowns to be found for any given X would be the five inputs to the
system from the steam plants.
Example 9.5 A system consists of two plants connected by a transmission

line. The only load is located at plant 2. When 200 MW is transmitted from
plant 1 to plant 2 power loss in the line is 16 MW. Find the required
generation for each plant and the power received by the load when X for the
system is $12.50 per megawatthour. Assume that the incremental fuel costs
can be approximated by the following equations:
= 0.010Pi + 8.5 $/MWh

dl j
^ = 0.015P2 + 9.5 $/MWh

di 2
Solution For a two-plant system
= P\Bn + 2PlP2B12 + P\B22

Since all the load is at plant 2, varying P2 cannot affect PL. Therefore
B22 = 0 and B12 — 0

When Px = 200 MW, PL = 16 MW; so
16 = 2002£11
Bn = 0.0004 MW”1
and
SPl
= 2PlBll + 2P2Bl2 = 0.0008 Px
dP 1
dP2
= 2P 2B22 + 2Pj Bl2 = 0
W,
Penalty factors are
1
Li = and L-, = 1.0
1 - 0.0008Pj
For X = 12.5
0.010^+8.5
1 - O.OOO8P1
P, = 200 MW
0.015P2 + 9.5 = 12.5
P2 = 200 MW
Economic load dispatching therefore requires equal division of load between

the two plants for X = 12.5. The power loss in transmission is
PL = 0.0004 x 2002 = 16 MW
and the delivered load is
PR = p1+ p2- pL = 384 MW
Example 9.6 For the system of Example 9.5 with 384 MW received by the
load, find the savings in dollars per hour obtained by coordinating rather
than neglecting the transmission loss in determining the load of the plants.
Solution If transmission loss is neglected, the incremental fuel costs at the

two plants are equated to give
0.010?! + 8.5 = 0.015P2 + 9.5
The power delivered to the load is
Pi + P2 — 0.0004P? = 384
Solving these two equations for Pl and P2 gives the following values for
plant generation with losses not coordinated:
Pi = 290.7 MW and P2 = 127.1 MW
The load on plant 1 is increased from 200 to 290.7 MW. The increase in
fuel cost is
290.7
0.010
P\ + 8.5Pi
200
= 222.53 + 770.95 = 993.48
The load on plant 2 is decreased from 200 to 127.1 MW. The decrease
(negative increase) in cost for plant 2 is
. 127.1 omsp2 127.1
(0.015P2 + 9.5) dP2 = + 9.5 P2

200 200
= 178.84 + 692.55 = 871.39

The net savings by accounting for transmission loss in schedulings .the

received load of 384 MW is
993.48 - 871.39 - $122.09 per hour
9.4 A METHOD OF COMPUTING PENALTY FACTORS

AND LOSS COEFFICIENTS
Expressing loss in a transmission system as a function of plant outputs in terms

of B coefficients is the most widely used method of calculating dPL/dPn for cost
minimization. The simplicity of the loss equation in terms of B coefficients is the
principal advantage of this method which has resulted in great savings in system
operating costs. The rapid development of digital computers has made other
methods attractive.
One method of evaluating dPL/dPn from load-flow studies and used by a
number of power companies as part of a program to determine B coefficients
will be discussed briefly.! The method depends on the fact that
dj\= £ dP^ddi
(9.29)
dPn ,4 d6jdPn
where 9j is the phase angle of the voltage at node j in a system of N buses. If bus
voltages are assumed constant, it can be shown that in terms of voltage phase
angles,
N
dP_L
= 21 Vk\Gjk sin (9k 0j) (9.30)
d9j k= 1
where Gjk is the real part of Yjk of the bus admittance matrix. The difficulty with
Eq. (9.29) is in expressing d9j/dPn; direct differentiation is impossible because
voltage phase angles cannot be expressed in terms of plant generated powers.
The terms d9j/dPn remain quite constant for changes in load levels and
system generation schedules. Since the terms express a Change in the voltage
phase angles 9j for a change in plant generation Pn with all other plant genera¬
tion remaining constant, these terms can be approximated very closely through
load-flow studies. For some typical load pattern the total received load is in¬
creased by increasing each individual load by the same small amount, say 5%.
The change in total received power plus losses is provided by one plant n while
other plant outputs are held constant. The changes in each voltage phase angle
9j are determined, and the ratios of change in phase angle to change in plant
input to the system A9j/AP„ are found for all values of j for plant rt. The
load-flow program of the digital computer is used and the process is repeated for
t E. F. Hill and W. D. Stevenson, Jr., An Improved Method of Determining Incremental Loss

Factors from Power System Admittances and Voltages, IEEE Trans. Power Appar. Syst.,
vol. PAS-87, June 1968, pp. 1419-1425.
every plant supplying the load change in turn. A set of Ain coefficients is found
where
Ain = J (9.31)
Jn A Pn
Then incremental loss for plant n is given by
dPL " dPL

(9.32)
spn M ddj »
The Ajn values are essentially constant regardless of the various combinations of
generation scheduling and load levels. Thus once a matrix of AJn coefficients has
been determined, an on-line computer monitoring load flow can calculate plant
penalty factors continually by solving Eqs. (9.30) and (9.32).
The usual practice, however, is to use the Ajn coefficients to calculate B
coefficients by the equation.!
1 N N d2P
Bmn ~ 2 ;?! ;?! d0t 86j ^'33^
In determining the second partial derivative of PL with respect to 6t and 9j

Eq. (9.30) is used with the values of 0 found in the load-flow solution of the
typical load pattern used in determining the Ajn coefficients. The on-line com¬
puter will then calculate incremental loss dPL/dP„ from the B coefficients and
control the system for economic load distribution, as described in the following
section.
9.5 AUTOMATIC GENERATION CONTROL
Computer control of the output of each plant and of each unit within a plant is
common practice in power-system operation. By continually monitoring all
plant outputs and the power flowing in interconnections interchange of power
with other systems is controlled. Most control systems are digital or a combina¬
tion of digital and analog. In this section we shall consider one of a variety of
ways in which computer control is accomplished.
In discussing control the term area means that part of an interconnected
system in which one or more companies control their generation to absorb all
their own load changes and maintain a prearranged net interchange of power
with other areas for specified periods. Monitoring the flow of power on the tie
lines between areas determines whether a particular area is absorbing satisfac¬
torily all the load changes within its own boundaries. The function of the com¬
puter is to require the area to absorb its own load changes, to provide the agreed
t See E. F. Hill and W. D. Stevenson, Jr., A New Method of Determining Loss Coefficients, IEEE
Trans. Power Appar. Syst., vol. PAS-87, July 1968, pp. 1548-1552.
Total actual
generation
Figure 9.6 Block diagram to illustrate the operation of a computer controlling a particular area.
net interchange with neighboring areas, to determine the desired generation of

each plant in the area for economic dispatch, and to cause the area to do its
share to maintain the desired frequency of the interconnected system.
The block diagram of Fig. 9.6 indicates the flow of information in a com¬
puter controlling a particular area. The numbers enclosed by circles adjacent to
the diagram identify positions on the diagram to simplify our discussion of the
control operation. The larger circles on the diagram enclosing the symbols x or
£ indicate points of multiplication or algebraic summation of incoming signals.
At position 1 processing of information about power flow on tie lines to
other control areas is indicated. The actual net interchange Pa is the algebraic
sum of the power on the tie lines and is positive when net power is out of the
area. The prearranged net interchange is called the scheduled net interchange Ps.
At position 2 the scheduled net interchange is subtracted from the actual net
interchanged' We shall discuss the condition where both actual and scheduled
net interchange are out of the system and therefore positive.
+ Subtraction of standard or reference value from actual value to obtain the error is the accepted
convention of power-system engineers and is the negative of the definition of control error found in
the literature of control theory.
Position 3 on the diagram indicates the subtraction of the scheduled

frequency fs (for instance 60 Hz) from the actual frequency fa to obtain Af the
frequency deviation. Position 4 on the diagram indicates that the frequency bias
setting Bf, a factor with negative sign, is multiplied by Af to obtain a value of
megawatts called the frequency bias Bf Af
The frequency bias which is positive when the actual frequency is less than
the scheduled frequency is subtracted from Pa - Ps at position 5 to obtain the
area control error (ACE), which may be positive or negative. As an equation
ACE = Pa - Ps — Bf( fa —fs) (9.34)
A negative ACE means that the area is not generating enough power to send the
desired amount out of the area. There is a deficiency in net power output.
Without frequency bias the indicated deficiency would be less because there
would be no positive offset BAf added to Ps (subtracted from Pa) when actual
frequency is less than scheduled frequency and the ACE would be less. The area
would produce sufficient generation to supply its own load and the prearranged
interchange but would not provide the additional output to assist neighboring
interconnected areas to raise the frequency.
Station control error (SCE) is the amount of actual generation of all the area
plants minus the desired generation as indicated at position 6 of the diagram.
This SCE is negative when desired generation is greater than existing generation.
The key to the whole control operation is the comparison of ACE and SCE.
Their difference is an error signal, as indicated at position 7 of the diagram. If
ACE and SCE are negative and equal, the deficiency in the output from the area
equals the excess of the desired generation over the actual generation and no
error signal is produced. However this excess of desired generation will cause a
signal indicated at position 11 to the plants to increase their generation to
reduce the magnitude of the SCE, and the resulting increase in output from the
area will reduce the magnitude of the ACE at the same time.
If ACE is more negative than SCE, there will be an error signal to increase
the X of the area and this increase will in turn cause the desired plant generation
to increase (position 9). Each plant will receive a signal to increase its output as
determined by the principles of economic dispatch.
This discussion has considered specifically only the case of scheduled net
interchange out of the area (positive scheduled net interchange) greater than
actual net interchange with ACE equal to or more negative than SCE. The
reader should be able to extend the discussion to the other possibilities by
referring to Fig. 9.6.
Position 10 on the diagram indicates the computation of penalty factors for
each plant. Here the B coefficients are stored to calculate dPJdP,,. Non-
conforming loads, plants where generation is not to be allowed to vary, and
tie-line loading enter the calculation of penalty factors. The penalty factors are
transmitted to the section (position 9) which calculates the individual plant
generation to provide with economic dispatch the total desired plant generation.
One other point of importance (not indicated on Fig. 9.6) is the offset in
scheduled net interchange of power that varies in proportion to the integral in

cycles between actual and rated (60 Hz) frequency. The offset is in the direction
to help in reducing the integrated difference to zero and thereby keep electric
clocks accurate.
PROBLEMS
9.1 For a certain generating unit in a plant the fuel input in millions of Btu per hour expressed as a
function of power output F in megawatts is
0.000 IF3 + 0.015F2 + 3 .OF + 90
(a) Determine the equation for incremental fuel cost in dollars per megawatthour as a function of
power output in megawatts based on a fuel cost of $1.40 per million Btu.
(b) Find the equation for a good linear approximation of incremental fuel cost as a function of power
output over a range of 20 to 120 MW.
(c) What is the average cost of fuel per megawatthour when the plant is delivering 100 MW?
(d) What is the approximate additional fuel cost per hour to raise the output of the plant from 100
to 101 MW?
9.2 The incremental fuel costs for two units of a plant are
= 0.010F, + 11.0 and —p = 0.012P2 + 8.0

dP1 dP 2
where F is in dollars per hour and P is in megawatts. If both units operate at all times and maximum
and minimum loads on each unit are 625 and 100 MW, plot A in dollars per megawatthour versus
plant output in megawatts for lowest fuel cost as total load varies from 200 to 1250 MW.
9.3 Find the savings in dollars per hour for economical allocation of load between the units of
Prob. 9.2 compared with their sharing the output equally when the total output is 750 MW.
9.4 A plant has two generators supplying the plant bus, and neither is to operate below 100 MW or
above 625 MW. Incremental costs with F, and P2 in megawatts are
dF
= 0.012F1 + 8.0 $/MWh
—- = 0.018F, + 7.0 $/MWh

dP 2
For economic dispatch find the plant A when Ft + P2 equals (a) 200 MW, (b) 500 MW, and
(c) 1150 MW.
9.5 Calculate the power loss in the system of Example 9.3 by the loss coefficients of the example and
by | /121R | for / j = 1.5/01 per unit and I2 = 1.2/01 per unit. Assume V3 = l.O/Ol per unit.
9.6 Find the loss coefficients that will give the true power loss for the system of Example 9.3 for
/ j = 0.8/01 per unit, I2 = 0.8/01 per unit, and V3 = 1.0/01 per unit.
9.7 A power system has only two generating plants, and power is being dispatched economically
with F, = 140 MW and P2 = 250 MW. The loss coefficients are:
Bn =0.10 x 10“2 MW'1
Bl2 = -0.01 x 10“2 MW-1
B22 =0.13 x 10'2 MW'1

To raise the total load on the system by 1 MW will cost an additional $12 per hour. Find (a) the
penalty factor for plant 1, and (b) the additional cost per hour to increase the output of this plant by
1 MW.
9.8 On a system consisting of two generating plants the incremental costs in dollars per megawatt-
hour with Pl and P2 in megawatts are
dF. dF2
= 0.008P. + 8.0 —- =0.012P2 +9.0
dP2 dP2
The system is operating on economic dispatch with Pt = P2 = 500 MW and SPL/dP.2 = 0.2. Find
the penalty factor of plant 1.
9.9 A power system is operating on economic load dispatch with a system 2 of $12.5 per megawatt-
hour. If raising the output of plant 2 by 100 kW (while other outputs are kept constant) results in
increased | /121R | losses of 12 kW for the system, what is the approximate additional cost per hour
if the output of this plant is increased by 1 MW?
9.10 A power system is supplied by only two plants, both of which are operating on economic
dispatch. At the bus of plant 1 the incremental cost is $11.0 per megawatthour, and at plant 2 is $10.0
per megawatthour. Which plant has the higher penalty factor? What is the penalty factor of plant 1 if
the cost per hour of increasing the load on the system by 1 MW is $12.5?
9.11 Calculate the values listed below for the system of Example 9.5 with system X = $13.5 per
megawatthour. Assume fuel costs at no load of $200 and $400 per hour for plants 1 and 2,
respectively.
(a) Pj, P2, and power delivered to the load for economic dispatch with transmission loss
coordinated.
(b) P, and P2 for the value of power delivered to the load found in part (u) but with transmission
loss not coordinated. Transmission loss must be included, however, in determining the total
power input to the system.
(c) Total fuel cost in dollars per hour for parts (a) and (b).
CHAPTER
TEN
SYMMETRICAL THREE-PHASE FAULTS
When a fault occurs in a power network, the current flowing is determined by

the internal emfs of the machines in the network, by their impedances, and by
the impedances in the network between the machines and the fault. The current
flowing in a synchronous machine immediately after the occurrence of a fault,
that flowing a few cycles later, and the sustained, or steady-state, value of the
fault current differ considerably because of the effect of the armature current on
the flux that generates the voltage in the machine. The current changes relatively
slowly from its initial value to its steady-state value. This chapter discusses the
calculation of fault current at different periods and explains the changes in
reactance and internal voltage of a synchronous machine as the current changes
from its initial value upon the occurrence of a fault to its steady-state value.
Description of a computer program for calculating fault currents will be
postponed until we have discussed unsymmetrical faults since programs are not
confined to three-phase faults.!
10.1 TRANSIENTS IN RL SERIES CIRCUITS
The selection of a circuit breaker for a power system depends not only upon the
current the breaker is to carry under normal operating conditions but also upon
the maximum current it may have to carry momentarily and the current it may
have to interrupt at the voltage of the line in which it is placed.
t For a book devoted entirely to fault studies see P. M. Anderson, Analysis of Faulted Power
Systems, Iowa State University Press, Ames, Iowa, 1973.
248
SYMMETRICAL THREE-PHASE FAULTS 249
In order to approach the problem of calculating the initial current when a

synchronous generator is short-circuited, consider what happens when an ac
voltage is applied to a circuit containing constant values of resistance and induc¬
tance. Let the applied voltage be Fmax sin (cot -I- a), where t is zero at the time of
applying the voltage. Then a determines the magnitude of the voltage when the
circuit is closed. If the instantaneous voltage is zero and increasing in a positive
direction when it is applied by closing a switch, a is zero. If the voltage is at its
positive maximum instantaneous value, a is n/2. The differential equation is
Fmax sin (cot + a) = Ri + L ^ (10.1)

at
The solution of this equation is
i= [sin (cot + a — 9) — e~Rl,L sin (a — 0)] (10-2)

I%|
where |Z | is yjR2 + (coL)2 and 6 is tan-1 (coL/R).

The first term of Eq. (10.2) varies sinusoidally with time. The second term is
nonperiodic and decays exponentially with a time constant of L/R. This nonperi¬
odic term is called the dc component of the current. We recognize the sinusoidal
term as the steady-state value of the current in an RL circuit for the given
applied voltage. If the value of the steady-state term is not zero when t = 0, the
dc component appears in the solution in order to satisfy the physical condition
of zero current at the instant of closing the switch. Note that the dc term does
not exist if the circuit is closed at a point on the voltage wave such that
a — 6 = 0 or a — 9 — n. Figure 10.1 shows the variation of current with time
according to Eq. (10.2) when a — 6 = 0. If the switch is closed at a point on the
voltage wave such that a — 9 = ±n/2, the dc component has its maximum initial
value, which is equal to the maximum value of the sinusoidal component. Figure
10.2 shows current versus time when a — 9 = — n/2. The dc component may
have any value from 0 to Fmax/1Z |, depending on the instantaneous value of the
voltage when the circuit is closed and on the power factor of the circuit. At the
instant of applying the voltage, the dc and steady-state components always have
the same magnitude but are opposite in sign in order to express the zero value of
current then existing.
i I
\ Time
Figure 10.1 Current as a function of time in an Figure 10.2 Current as a function of time in an
RL circuit for a - 6 = 0, where 0 = tan_1 RL circuit for a - 6 = - nil, where 6 = tan “1
(coL/R). The voltage is Fmaxsin (cot + a) applied (coL/R). The voltage is Fmax sin (cot + a)
at t = 0. applied at t = 0.
In Secs. 6.2 and 6.3 we discussed the principles of operation of a Synch¬

ronous generator consisting of a rotating magnetic field which generates a vol¬
tage in an armature winding having resistance and reactance. The current
flowing when a generator is short-circuited is similar to that flowing when an
alternating voltage is suddenly applied to a resistance and an inductance in
series. There are important differences, however, because the current in the
armature affects the rotating field.
A good way to analyze the effect of a three-phase short circuit at the ter¬
minals of a previously unloaded alternator is to take an oscillogram of the
current in one of the phases upon the occurrence of such a fault. Since the
voltages generated in the phases of a three-phase machine are displaced 120
electrical degrees from each other, the short circuit occurs at different points on
the voltage wave of each phase. For this reason the unidirectional or dc transient
component of current is different in each phase. If the dc component of current is
eliminated from the current of each phase, the resulting plot of each phase
current versus time is that shown in Fig. 10.3. Comparison of Figs. 10.1 and 10.3
shows the difference between applying a voltage to the ordinary RL circuit and
applying a short circuit to a synchronous machine. There is no dc component in
either of these figures. In a synchronous machine the flux across the air gap of
the machine is much larger at the instant the short circuit occurs than it is a few
cycles later. The reduction of flux is caused by the mmf of the current in the
armature. Our discussion in Sec. 6.2 concerned the effect of armature current,
which is called armature reaction. The equivalent circuit developed in Sec. 6.3
Figure 10.3 Current as a function of time for a synchronous generator short-circuited while running
at no load. The unidirectional transient component of current has been eliminated in redrawing the
oscillogram.
accounts for the reduction of flux due to armature reaction and applies to the
steady-state condition after the dc transient has disappeared and after the ampli¬
tude of the wave shown in Fig. 10.3 has become constant. When a short circuit
occurs at the terminals of a synchronous machine, time is required for
the reduction in flux across the air gap. As the flux diminishes, the armature
current decreases because the voltage generated by the air-gap flux determines
the current which will flow through the resistance and leakage reactance of
the armature winding.
10.2 SHORT-CIRCUIT CURRENTS AND THE

REACTANCES OF SYNCHRONOUS MACHINES!
Certain terms that are valuable in the calculation of short-circuit current in a

power system can be defined from Fig. 10.3. The reactances that we shall discuss
are direct-axis reactances which we mentioned in Sec. 6.3. We recall that direct-
axis reactance is used for computing voltage drops caused by that component of
the armature current which is in quadrature (90° out of phase) with the voltage
generated at no load. Since the resistance in a faulted circuit is small compared
with the inductive reactance, current during a fault is always lagging by a large
angle, and the so-called direct-axis reactance is required. In the discussion to
follow, it should be remembered that the current shown in the oscillogram of
Fig. 10.3 is that which flows in an alternator operating at no load before the
fault occurs.
In Fig. 10.3 the distance oa is the maximum value of the sustained short-
circuit current. This value of current times 0.707 is the rms value 111 of the
sustained, or steady-state, short-circuit current. The no-load voltage of the alter¬
nator | Eg | divided by the steady-state current 11 | is called the synchronous
reactance of the generator or the direct-axis synchronous reactance Xd since the
power factor is low during the short circuit. The comparatively small resistance
of the armature is neglected.
If the envelope of the current wave is extended back to zero time and the
first few cycles where the decrement appears to be very rapid are neglected, the
intercept is the distance ob. The rms value of the current represented by this
intercept, or 0.707 times ob in amperes, is known as the transient current | /' |. A
new machine reactance may now be defined. It is called the transient reactance,
or in this particular case the direct-axis transient reactance X'd and is equal to
| Eg | /1 /' | for an alternator operating at no load before the fault. If the rapid
decrement of the first few cycles is neglected, the point of intersection that the
current envelope makes with the zero axis can be determined more accurately by
plotting on semilogarithmic paper the excess of the current envelope over the
t For a more complete discussion see C. F. Wagner, “ Machine Characteristics,” in Central Station
Engineers of the Westinghouse Electric Corporation, Electrical Transmission and Distribution Refer¬
ence Book, 4th ed., Chap. 6, pp. 145-194, East Pittsburgh, Pa., 1964; and A. E. Fitzgerald, C. Kings¬
ley, and A. Kusko, Electric Machinery, 3d ed., pp. 312-319, 479-492, McGraw-Hill Book Company,
New York, 1971.
Figure 10.4 Excess of the current envelope of

Fig. 10.3 over the sustained maximum current,
plotted on semilogarithmic scales. Time
sustained value represented by oa, as shown in Fig. 10.4. The straight-line por¬
tion of this curve is extended to the zero-time axis, and the intercept is added to
the maximum instantaneous value of the sustained current to obtain the maxi¬
mum instantaneous value of transient current corresponding to ob in Fig. 10.3.
The rms value of the current determined by the intercept of the current
envelope with zero time is called the subtransient current | I" |. In Fig. 10.3 the
subtransient current is 0.707 times the ordinate oc. Subtransient current is often
called the initial symmetrical rms current, which is more descriptive because it
conveys the idea of neglecting the dc component and taking the rms value of the
ac component of current immediately after the occurrence of the fault. Direct-
axis subtransient reactance X'J for an alternator operating at no load before the
occurrence of a three-phase fault at its terminals is | Eg \/ | 7" |.
The currents and reactances discussed above are defined by the following
equations, which apply to an alternator operating at no load before the occur¬
rence of a three-phase fault at its terminals:
oa \E„\
I' xd
(10.3)
ob 1 E,\
\r X’d
(10.4)
r, OC |E,I (10.5)
X*4
where | / | = steady-state current, rms value

11 | = transient current, rms value excluding dc component
11" | — subtransient current, rms value excluding dc component
Xd = direct-axis synchronous reactance
X'd = direct-axis transient reactance
Xd = direct-axis subtransient reactance
| E91 = rms voltage from one terminal to neutral at no load
oa, ob, oc = intercepts shown in Fig. 10.3.
In analytical work the steady-state, transient, and subtransient currents may be

expressed as phasors, usually with Eg as the reference.
The subtransient current | /" | is much larger than the steady-state current
| / | because the decrease in flux across the air gap of the machine caused by the
armature current, as described in Sec. 6.3, cannot take place immediately. So a
larger voltage is induced in the armature windings just after the fault occurs than
exists after steady state is reached. However, we account for the difference in
induced voltage by using different reactances in series with the no-load voltage
Eg to calculate currents for subtransient, transient, and steady-state conditions.
We shall examine the transients for machines carrying a load in the next section.
Equations (10.3) to (10.5) indicate the method of determining fault current in
a generator when its reactances are known. If the generator is unloaded when
the fault occurs, the machine is represented by the no-load voltage to neutral in
series with the proper reactance. The resistance is taken into account if greater
accuracy is desired. If there is impedance external to the generator between its
terminals and the short circuit, the external impedance must be included in the
circuit.
Example 10.1 Two generators are connected in parallel to the low-voltage

side of a three-phase A-Y transformer as shown in Fig. 10.5. Generator 1 is
rated 50,000 kVA, 13.8 kV. Generator 2 is rated 25,000 kVA, 13.8 kV. Each
generator has a subtransient reactance of 25%. The transformer is rated
75,000 kVA, 13.8A/69Y kV, with a reactance of 10%. Before the fault occurs,
the voltage on the high-tension side of the transformer is 66 kV. The trans¬
former is unloaded, and there is no circulating current between the genera¬
tors. Find the subtransient current in each generator when a three-phase
short circuit occurs on the high-tension side of the transformer.
Solution Select as base in the high-tension circuit 69 kV, 75,000 kVA.

Then the base voltage on the low-tension side is 13.8 kV.

Generator 1: v*
75,000
X* 0.25 = 0.375 per unit
50,000
66
E,i 69
= 0.957 per unit
Generator 2:
75,000
x2 0.25
25,000
= 0.750 per unit
66
— - 0.957 per unit
69
Transformer:
X = 0.10 per unit
Figure 10.6 shows the reactance diagram before the fault. A three-phase fault
at P is simulated by closing switch S. The internal voltages of the two
machines may be considered to be in parallel since they must be identical in
magnitude and phase if no circulating current flows between them. The
equivalent parallel subtransient reactance is
0.375 x 0.75
= 0.25 per unit
0.375 + 0.75
Therefore, as a phasor with Eg as reference, the subtransient current in the
short circuit is
0.957
/" —y'2.735 per unit
;0.25 + ;0.10
The voltage on the delta side of the transformer is
(—/2.735)(/0.10) = 0.2735 per unit

and in generators 1 and 2
0.957 - 0.274
n y'0.375
—jl.823 per unit
n 0.957 - 0.274
i2 —y'0.912 per unit
70.75
Figure 10.6 Reactance diagram for

Example 10.1.
To find the current in amperes, the per-unit values are multiplied by the
base current of the circuit:
|/'i'| = 1.823 /Z5,00°

V3 x 13.8
= 5720 A
I/2I = 0.912 -y?5,000 - 2860 A

V3 x 13.8
Although machine reactances are not true constants of the machine and
depend on the degree of saturation of the magnetic circuit, their values usually
lie within certain limits and can be predicted for various types of machines.
Table A.4 gives typical values of machine reactances that are needed in making
fault calculations and in stability studies. In general, subtransient reactances of
generators and motors are used to determine the initial current flowing on the
occurrence of a short circuit. For determining the interrupting capacity of circuit
breakers, except those which open instantaneously, subtransient reactance is
used for generators and transient reactance is used for synchronous motors. In
stability studies where the problem is to determine whether a fault will cause a
machine to lose synchronism with the rest of the system if the fault is removed
after a certain time interval, transient reactances apply.
10.3 INTERNAL VOLTAGES OF LOADED MACHINES

UNDER TRANSIENT CONDITIONS
All the preceding discussion pertains to a synchronous generator that carries no

current at the time a three-phase fault occurs at the terminals of the machine.
Now consider a generator that is loaded when the fault occurs. Figure 10.7a is
the equivalent circuit of a generator that has a balanced three-phase load. Exter¬
nal impedance is shown between the generator terminals and the point P where
the fault occurs. The current flowing before the fault occurs at point P is IL, the
voltage at the fault is Vf, and the terminal voltage of the generator is Vt. As
discussed in Chap. 6, the equivalent circuit of the synchronous generator is its
no-load voltage Eg in series with its synchronous reactance Xs. If a three-phase
fault occurs at P in the system, we see that a short circuit from P to neutral in
the equivalent circuit does not satisfy the conditions for calculating subtransient
current since the reactance of the generator must be X'd if we are calculating
subtransient current I" or X'd if we are calculating transient current /'.
The circuit shown in Fig. 10.76 gives us the desired result. Here a voltage £"
in series with Xd supplies the steady-state current IL when switch S is open and
supplies the current to the short circuit through Xd and Zext when switch S is
closed. If we can determine this current through Xd will be /". With switch S
open we see that
e; = v,+jilxj (10.6)
Figure 10.7 Equivalent circuits for a generator supplying a balanced three-phase load. Application of
a three-phase fault at P is simulated by closing switch S. (a) Usual steady-state generator equivalent
circuit with load. (b) Circuit for calculation of I".
and this equation defines E", which is called the subtransient internal voltage.
Similarly when calculating transient current /' which must be supplied through
the transient reactance X'd the driving voltage is the transient internal voltage E'g,
where
Eg — Vt + jILX'd (10.7)
Voltages Eg and E'g are determined by IL and are both equal to the no-load
voltage Eg only when IL is zero, at which time Eg is equal to Vt.
At this point it is important to note that £" in series with Xd represents the
generator before the fault occurs and immediately after the fault only if the
prefault current in the generator is IL. On the other hand, Eg in series with
the synchronous reactance Xs is the equivalent circuit of the machine under
steady-state conditions for any load. For a different value of IL in the circuit of
Fig. 10.7 Eg would remain the same but a new value of £" would be required.
Synchronous motors have reactances of the same type as generators. When
a motor is short-circuited, it no longer receives electric energy from the power
line, but its field remains energized and the inertia of its rotor and connected
load keeps it rotating for an indefinite period. The internal voltage of a
synchronous motor causes it to contribute current to the system, for it is then
acting like a generator. By comparison with the corresponding formulas for a
generator, the subtransient internal voltage and transient internal voltage for a
synchronous motor are given by
E’m=Vt-jILX: (10.8)
E'm = Vt-jlLX'd (10.9)
Systems that contain generators and motors under load may be solved either
by Thevenin’s theorem or by the use of transient or subtransient internal vol¬
tages, as is illustrated in the following examples.
Example 10.2 A synchronous generator and motor are rated 30,000 kVA,
13.2 kV, and both have subtransient reactances of 20%. The line connecting
them has a reactance of 10% on the base of the machine ratings. The motor
is drawing 20,000 kW at 0.8 power factor leading and a terminal voltage of
12.8 kV when a symmetrical three-phase fault occurs at the motor terminals.
Find the subtransient current in the generator, motor, and fault by using the
internal voltage of the machines.
Solution Choose as base 30,000 kVA, 13.2 kV.

Figure 10.8a shows the equivalent circuit of the system described. We
see that Fig, 10.8a is similar to Fig. 10.7ft and that before the fault, £" and
E" could be replaced by Eg and Em provided we replaced subtransient reac¬
tances by synchronous reactances. However, to find subtransient current we
need the representation of Fig. 10.8a.
If we use the voltage at the fault Vj as the reference pha:
12 8
vf =
- 0.97/0° per unit
13.2
30,000
Base current = = 1312 A
V3 x 13.2
20,000 /36.9°
4 =
0.8 x V3 x 12.8
= 1128 /36.9° A
1128 736.9
0.86 736.9° per unit
1312
= 0.86(0.8 + j0.6) = 0.69 + y'0.52 per unit
For the generator,
Vt = 0.970 + /). 1(0.69 + j0.52) = 0.918 + ;0.069 per unit

Eg = 0.918 -(- 7'0.069 + y'0.2(0.69 -I- ;'0.52) = 0.814 + j0.207 per unit
0.814 + yO.207
/" = = 0.69 - 72.71 per unit
19
70.3
= 1312(0.69 —72.71) = 905 -;3550 A
Figure 10.8 Equivalent circuits for Example 10.2.

For the motor,
Vt= Vf — 0.97/0° per unit
E'm = 0.97 + jO -70.2(0.69 + j0.52) = 0.97 - 7‘0.138 + 0.104
= 1.074 — 7'0.138 per unit
= 1.074 jO. 138 = _ — j531 per unit

m j0.2
= 1312( — 0.69 -75.37) = -905 - 77050 A
In the fault,
1} = /" + /" = 0.69 -7'2.71 - 0.69 -j5.37 = -78.O8 per unit
= -78.O8 x 1312 = -710,600 A
Figure 10.86 shows the paths of /", /", and I"f.
The subtransient current in the fault can be found by Thevenin’s theorem,

which is applicable to linear, bilateral circuits. When constant values are used for
the reactances of synchronous machines, linearity is assumed. When the theorem
is applied to the circuit of Fig. 10.76, the equivalent circuit is a single generator
and a single impedance terminating at the point of application of the fault. The
new generator has an internal voltage equal to Vf, the voltage at the fault point
before the fault occurs. The impedance is that measured at the point of applica¬
tion of the fault looking back into the circuit with all the generated voltages
short-circuited. Subtransient reactances should be used if the initial current is
desired. Figure 10.9 is the Thevenin equivalent of Fig. 10.76. The impedance Zth
is equal to (Zext 4- jXj)ZL/(ZL + Zext + jXj). Upon the occurrence of a three-
phase short circuit at P, simulated by closing S, the subtrapsient current in the
fault is
r = i/
~ z„ + Z», +jXj)
zL(zm+jx:)
(10.10)
Zth
*6 sN
Figure 10.9 Thevenin equivalent of the circuit of Fig. 10.7b.
Example 103 Solve Example 10.2 by the use of Thevenin’s theorem.
Solution
/0.3 x j0.2
= j0.12 per unit
j0.3 + j0.2
Vf = 0.97/0° per unit
In the fault,
„ _ 0-97 + jO
—j8.08 per unit
f jO.12
The above current, found by applying Thevenin’s theorem, is that which

flows out of the circuit at the fault because of the reduction of the voltage to
zero at that point. If this current caused by the fault is divided between the
parallel circuits of the machines inversely as their impedances, the resulting
values are the currents from each machine due only to the change in voltage
at the fault point. To the fault currents thus attributed to the two machines
must be added the current flowing in each before the fault occurred to find
the total current in the machines after the fault. The superposition theorem
supplies the reason for adding the current flowing before the fault to the
current computed by Thevenin’s theorem. Figure 10.10a shows a generator
having a voltage Vf connected at the fault and equal to the voltage at the
fault before the fault occurs. This generator has no effect on the current
flowing before the fault occurs, and the circuit corresponds to that of
Fig. 10.8a. Adding in series with Vf another generator having an emf of
equal magnitude but 180° out of phase with Vf gives the circuit
of Fig. 10.106, which corresponds to that of Fig. 10.86. The principle
of superposition, applied by first shorting £", £", and Vf, gives the currents
found by distributing the fault current between the two generators inversely
as the impedances of their circuits. Then shorting the remaining generator
— Vf with £", £", and Vf in the circuit gives the current flowing before the
fault. Adding the two values of current in each branch gives the current in
Figure 10.10 Circuits illustrating the application of the superposition theorem to determine the
proportion of the fault current in each branch of the system.
the branch after the fault. Applying the above principle to the present
example gives
Fault current from generator = —j8.08 x = — j3.23 per unit

jO.5
i03
Fault current from motor = —j8.08 x —y'4.85 per unit
jO.5
To these currents must be added the prefault current IL to obtain the total
subtransient currents in the machines:
/" = 0.69 + j0.52 - j3.23 = 0.69 — 7*2.71 per unit
J" - -0.69 -;'0.52 -;'4.85 - -0.69 -;5.37 per unit
Note that IL is in the same direction as /" but opposite to /". The per-unit
values found for I"f, /", and /" are the same as in Example 10.2, and so the
ampere values will also be the same.
Usually load current is omitted in determining the current in each line
upon the occurrence of a fault. In the Thevenin method neglect of load
current means that the prefault current in each line is not added to the
component of current flowing toward the fault in the line. The method of
Example 10.2 neglects load current if the subtransient internal voltages of all
machines are assumed equal to the voltage Vf at the fault before the fault
occurs, for such is the case if no current flows anywhere in the network prior
to the fault.
Neglecting load current in Example 10.3 gives
Fault current from generator = 3.23 x 1312 = 4240 A
Fault current from motor = 4.85 x 1312 = 6360 A

\
Current in fault = 8.08 x 1312 = 10,600 A
The current in the fault is the same whether or not load current is con¬
sidered, but the contributions from the lines differ. When load current is
included, we find from Example 10.2
Fault current from generator = 1905 — j3550| = 3660 A
Fault current from motor = | —905 — /70501 = 7200 A
The arithmetic sum of the generator and motor current magnitudes does not
equal the fault current because the currents from the generator and motor
are not in phase.
Ea j 0.2 j 0.1 _
E'c' j0.2 70.1
E'b j'0.2 j'0.1
Figure 10.11 Reactance diagram obtained from Fig. 7.3

by substituting subtransient for synchronous reactances Figure 10.12 Circuit of Fig. 10.11 with
of the machines and subtransient internal voltages for admittances marked in per unit and a
no-load generated voltages. Reactance values are marked three-phase fault on bus 4 of the system
in per unit. simulated by Vf and — Vf in series.
10.4 THE BUS IMPEDANCE MATRIX

IN FAULT CALCULATIONS
Our discussion of fault calculations has been confined to simple circuits, but now
we shall extend our study to general networks. However, let us proceed to the
general equations by starting with a specific network with which we are already
familiar. If we change the reactances in series with the generated voltages of the
circuit shown in Fig. 7.3 to subtransient reactances, and if the generated voltages
become subtransient internal voltages, we have the network shown in Fig. 10.11.
If this network is the single-phase equivalent of a three-phase system and we
choose to study a fault at bus 4, we can follow the procedure of Sec. 10.3 and call
Vf the voltage at bus 4 before the fault occurs.
A three-phase fault at bus 4 is simulated by the network of Fig. 10.12 where
the impedance values of Fig. 10.11 have been changed to admittances. The gen¬
erated voltages Vf and — Vf in series constitute the short circuit. Generated
voltage Vf alone in this branch would cause no current in the branch. With Vf
and — Vf in series the branch is a short circuit, and the branch current is I"f.
Admittances rather than impedances have been marked in per unit on this
diagram. If El, El, and Vf are short-circuited, the voltages and currents are
those due only to — Vf. Then the only current entering a node from a source is
that from — Vf and is —I"f into node 4 (/} from node 4) since there is no current
in this branch until the insertion of - Vf. The node equations in matrix form for
the network with — Vf the only source are
I
1
_J
-1
o
12.33 0.0 4.0 5.0
0 0.0 -10.83 2.5 5.0 vi

=j (10.11)
0 4.0 2.5 -17.83 8.0 Vi
-
i_
_1
-1
5.0 5.0 8.0 -18.0

1
1
i
when the superscript A indicates that the voltages are due only to — Vf. The A
sign is chosen to indicate the change in voltage due to the fault.
By inverting the bus admittance matrix of the network of Fig. 10.12 we
obtain the bus impedance matrix. The bus voltages due to — Vf are given by
1_
0 '
Vj 0
^bus
(10.12)
_1
1
Vi 0
1
and so
(10.13)
and
'14
VA1 — T" 7
i/Z' 14 -
-44
VA2 (10.14)
When the generator voltage — Vf is short-circuited in the network of

Fig. 10.12 and £", E'l, E'l, and Vf are in the circuit, the currents and voltages
everywhere in the network are those existing before the fault. By the principle of
superposition these prefault voltages added to those given by Eqs. (10.14) yield
the voltages existing after the fault occurs. Usually the faulted network is
assumed to have been without loads before the fault. In such a case no current is
flowing before the fault, and all voltages throughout the network are the same
and equal to Vf. This assumption simplifies our work considerably, and apply¬
ing the principle of superposition gives
T" J
V1 = Vf + V'i = Vf 1
V2 = Vf+VA2 = Vf J" 7
24
(10.15)
V3 = Vf+V*= Vf T" 7
i/z' 34
V4=Vf-Vf = 0
These voltages exist when subtransient current flows and ZbUS has been formed
for a network having subtransient values for generator reactances.
In general terms for a fault on bus k, and neglecting prefault currents,
(10.16)
and the postfault voltage at bus n is
K = vf- (10.17)
Using the numerical values of Eq. (10.11), we invert the square matrix Ybus
of that equation and find
0.1488 0.0651 0.0864 0.0978

0.0651 0.1554 0.0799 0.0967
^bus j (10.18)
0.0864 0.0798 0.1341 0.1058
0.0978 0.0967 0.1058 0.1566
Usually Vf is assumed to be 1.0/0° per unit, and with this assumption for our
faulted network
r- 1 = —76.386 per unit

f ;0.1566
j0.0978
Fi = l = 0.3755 per unit
jO. 1566
y'0.0967
K, = l- = 0.3825 per unit
jO. 1566
J0.1058
K = 1 — = 0.3244 per unit
jO. 1566
Currents in any part of the network can be found from the voltages and
impedances. For instance, the fault current in the branch connecting nodes 1 and
3 flowing toward node 3 is
Vt - V3 0.3755 - 0.3244
1l"13 —
j0.25 j0.25
= -y'0.2044 per unit
From the generator connected to node 1 the current is
E"a -V, 1 - 0.3755

° j0.3 70.3
= — 7'2.0817 per unit
Other currents can be found in a similar manner, and voltages and currents
with the fault on any other bus are calculated just as easily from the impedance
matrix.
Equation (10.16) is simply an application of Thevenin’s theorem, and we
recognize that the quantities on the principal diagonal of the bus impedance
matrix are the Thevenin impedances of the network for calculating fault current
at the various buses. Power companies furnish data to a customer who must
determine the fault current to specify circuit breakers for an industrial plant or
distribution system connected to the utility system at any point. Usually the data
supplied lists the short-circuit megavoltamperes, where
Short-circuit MVA = >/3 x (nominal kV) x /sc x 10“3 (10.19)

With resistance and shunt capacitance neglected, the single-phase Thevenin equi¬
valent circuit which represents the system is an emf equal to the nominal line
voltage divided by y/3 in series with an inductive reactance of
(nominal kV/^/3) x 1000
X» = n (10.20)
Solving Eq. (10.19) for Isc and substituting in Eq. (10.20) yield
(nominal kV)2
(10.21)
th short-circuit MVA
If base kilovolts is equal to nominal kilovolts, converting to per unit yields
base MVA
xth = short-circuit
-:-.
MVA
per unit (10.22)
[ base
X* = per unit (10.23)
10.5 A BUS IMPEDANCE MATRIX EQUIVALENT NETWORK

Although we cannot devise a physically realizable circuit employing the im¬
pedances of the bus impedance network, we can draw a circuit with transfer
impedances indicated between branches. Such a diagram will be helpful in under¬
standing the significance of the equations developed in Sec. 10.4.
In Fig. 10.13 brackets have been drawn between branch 4 and the other
three branches of a network which has four nodes in addition to the reference
node.t Associated with these brackets are the symbols Z14, Z24, and Z34, which
t This equivalent network is drawn in the manner adopted in J. R. Neuenswander, Modern

Power Systems, Intext Educational Publishers, New York, 1971, which refers to the bus impedance
matrix equivalent network as the rake equivalent.
Figure 10.13 Bus impedance matrix

equivalent network with four indepen¬
dent nodes. Closing switch S simu¬
lates a fault on node 4. Only the
transfer admittances for node 4 are
shown.
identify transfer impedances of node 4 of the bus impedance matrix. The driving-
point impedances of the bus impedance matrix are Zu, Z22, Z33, and Z44. No
current can flow in any branches when switch S is open. When S is closed,
current flows in the circuit only toward node 4. This current is Vf/Z44, which,
according to Eq. (10.13), is If for a fault on node 4. We shall interpret the
brackets to mean that the current If toward node 4 in the network induces
voltage drops of /}Z14, /}Z24, and /}Z34 in the branches connected to nodes
1, 2, and 3, respectively. These voltage drops are in the direction toward the
respective nodes.
If switch S in the circuit of Fig. 10.13 is open, all nodes will be at the voltage
of Vf, as in Fig. 10.11 if £", ££, and E” equal Vf. If S is closed, examination of
the circuit shows that the voltages at all four nodes with respect to the reference
node 0 will be the values specified by Eqs. (10.15). Therefore, if we interpret the
indicated transfer impedances of this circuit as described above, the circuit is the
equivalent of that shown in Fig. 10.11 with S open and Fig. 10.12 with switch S
closed, where we are still neglecting prefault current.
Of course, we can simulate short circuits at the other buses in a similar
manner and extend this approach to a general network having any number of
nodes. We could indicate the other transfer impedances of the equivalent circuit
by additional brackets and have not done so only because it becomes confusing
to have so many brackets to indicate the transfer impedances. In fact we shall
usually omit the brackets when drawing this network equivalent for the bus
impedance matrix, but we have to realize that the transfer impedances do exist
and must be considered in interpreting the network.
Example 10.4 Determine the bus impedance matrix for the network of
Example 8.1, for which the results of a load-flow study are shown in Fig. 8.2.
Generators at buses 1 and 3 are rated 270 and 225 MV A, respectively. The
generator subtransient reactances plus the reactances of the transformers
connecting them to the buses are each 0.30 per unit on the generator rating
as base. The turns ratios of the transformers are such that the voltage base in
each generator circuit is equal to the voltage rating of the generator. Include
the generator and transformer reactances in the matrix. Find the subtran¬
sient current in a three-phase fault at bus 4 and the current coming to the
faulted bus over each line. Prefault current is to be neglected and all voltages
are assumed to be 1.0 per unit before the fault occurs. System base is
100 MVA. Neglect all resistances.
Solution Converted to the 100-MVA base, the combined generator and

transformer reactances are
100
Generator at bus 1: X = 0.30 x =0.1111 per unit
100
Generator at bus 3: X = 0.30 x — = 0.1333 per unit
Figure 10.14 Admittance

diagram for Example 10.4.
The network with admittances marked in per unit is shown in Fig. 10.14
from which the node admittance matrix is
22.889 5.952 0.0 0.0 7.937

5.952 --13.889 7.937 0.0 0.0
^bus j 0.0 7.937 -23.175 2.976 4.762
0.0 0.0 2.976 -6.944 3.968
7.937 0.0 4.762 3.968 -16.667
i inverted on a digital computer to yield the she

matrix
0.0793 0.0558 0.0382 0.0511 0.0608

0.0558 0.1338 0.0664 0.0630 0.0605
^bus J 0.0382 0.0664 0.0875 0.0720 0.0603
0.0511 0.0630 0.0720 0.2321 0.1002
0.0608 0.0605 0.0603 0.1002 0.1301
Visualizing a network like that of Fig. 10.13 will help

desired currents and voltages.
The subtransient current in a three-phase fault on bus 4 is
1.0
/" - = —;4.308 per unit
y'0.2321
\
At buses 3 and 5 the voltages are
V3 = 1.0 — (— j4.308)(/0.0720) = 0.6898 per unit

V5 = 1.0 - ( —/4.308)(;0.1002) = 0.5683 per unit
Currents to the fault are
From bus 3: 0.6898(-/2.976) = -/2.053
From bus 5: 0.5683(-/'3.968) = -;'2.255
—;4.308 per unit
From the same short-circuit matrix we can find similar information for
faults on any of the other buses.
10.6 THE SELECTION OF CIRCUIT BREAKERS
Much study has been given to circuit-breaker ratings and applications and our
discussion here will give some introduction to the subject. For additional gui¬
dance necessary to specify the breakers the reader should consult the ANSI
publications listed in the footnotes which accompany this section.
The subtransient current to which we have devoted most of our attention to
this point is the initial symmetrical current and does not include the dc compo¬
nent. As we have seen, inclusion of the dc component results in a rms value of
current immediately after the fault, which is higher than the subtransient current.
For oil circuit breakers above 5 kV the subtransient current multiplied by 1.6 is
considered to be the rms value of the current whose disruptive forces the breaker
must withstand during the first half cycle after the fault occurs. This current is
called the momentary current, and for many years circuit breakers were rated in
terms of their momentary current as well as other criteria.!
The interrupting rating of a circuit breaker was specified in kilovoltamperes
or megavoltamperes. The interrupting kilovoltamperes equal ^/3 times the kilo¬
volts of the bus to which the breaker is connected times the current which the
breaker must be capable of interrupting when its contacts part. This current is,
of course, lower than the momentary current and depends on the speed of the
breaker, such as 8, 5, 3, or 1^ cycles, which is a measure of the time from the
occurrence of the fault to the extinction of the arc.
The current which a breaker must interrupt is usually asymmetrical since it
still contains some of the decaying dc component. A schedule of preferred
ratings for ac high-voltage oil circuit breakers specifies the interrupting current
ratings of breakers in terms of the component of the asymmetrical current which
is symmetrical about the zero axis. This current is properly called the required
symmetrical interrupting capability or simply the rated symmetrical short-circuit
current. Often the adjective symmetrical is omitted. Selection of circuit breakers
may also be made on the basis of total current (dc component included)-! We
shall limit our discussion to a brief treatment of the symmetrical basis of breaker
selection.
Breakers are identified by nominal-voltage class, such as 69 kV. Among
other factors specified are rated continuous current, rated maximum voltage,
voltage range factor K, and rated short-circuit current at rated maximum kilo¬
volts. K determines the range of voltage over which rated short-circuit current
times operating voltage is constant. For a 69-kV breaker having a maximum
t See G. N. Lester, “High Voltage Circuit Breaker Standards in the USA: Past, Present, and
Future,” IEEE Trans. Power Appar. Syst., vol. 93, 1974, pp. 590-600.
J See Schedules of Preferred Ratings and Related Required Capabilities for AC High-Voltage
Circuit Breakers Rated on a Symmetrical Current Basis, ANSI C37.06-1971 and Guide for Calculation
of Fault Currents for Application of AC High-Voltage Circuit Breakers Rated on a Total Current
Basis, ANSI C37.5-1979, American National Standards Institute, New York.
rated voltage of 72.5 kV, a voltage range factor K of 1.21, and a continuous
current rating of 1200 A, the rated short-circuit current at the maximum voltage
(symmetrical current which can be interrupted at 72.5 kV) is 19,000 A. This
means that the product 72.5 x 19,000 is the constant value of rated short-circuit
current times operating voltage in the range 72.5 to 60 kV since 72.5/1.21 equals
60. The rated short-circuit current at 60 kV is 19,000 x 1.21, or 23,000 A. At
lower operating voltages this short-circuit current cannot be exceeded. At 69 kV
the rated short-circuit current is
x 19,000 = 20,000 A
Breakers of the 115-kV class and higher have a K of 1.0.

A simplified procedure for calculating the symmetrical short-circuit current,
called the E/X method,t disregards all resistance, all static loads, and all prefault
current. Subtransient reactance is used for generators in the E/X method, and
for synchronous motors the recommended reactance is the X'/, of the motor
times 1.5, which is the approximate value of the transient reactance of the motor.
Induction motors below 50 hp are neglected, and various multiplying factors are
applied to the X'/ of larger induction motors depending on their size. If no
motors are present, symmetrical short-circuit current equals subtransient
current.
The impedance by which the voltage Vf at the fault is divided to find
short-circuit current must be examined when the E/X method is used. In specify¬
ing a breaker for bus k this impedance is Zkk of the bus impedance matrix with
the proper machine reactances since the short-circuit current is expressed by
Eq. (10.16). If the ratio of X/R of this impedance is 15 or less, a breaker of the
correct voltage and kilovoltamperes may be used if its interrupting current
rating is equal to or exceeds the calculated current. If the X/R ratio is unknown,
the calculated current should be no more than 80% of the allowed value for the
breaker at the existing bus voltage. The ANSI application guide specifies a
corrected method to account for ac and dc time constants for the decay of the
current amplitude if the X/R ratio exceeds 15. The corrected method also con¬
siders breaker speed.
This discussion of the selection of circuit breakers is presented not as a study
of breaker applications but to indicate the importance of understanding fault
calculations. The following example should clarify the principle.
Example 10.5 A 25,000-kVA 13.8-kV generator with X"d = 15% is connected

through a transformer to a bus which supplies four identical motors, as
shown in Fig. 10.15. The subtransient reactance X'/ of each motor is 20% on
a base of 5000 kVA, 6.9 kV. The three-phase rating of the transformer is
+ See Application Guide for AC High-Voltage Circuit Breakers Rated on a Symmetrical Current
Basis, ANSI C37.010-1972, American National Standards Institute, New York. This publication is
also IEEE Std 320-1972.
-D-O
-D-O
Gen. -o-0:
Figure 10.15 One-line diagram for Example 10.5. A 'p
25,000 kVA, 13.8/6.9 kV, with a leakage reactance of 10%. The bus voltage
at the motors is 6.9 kV when a three-phase fault occurs at the point P. For
the fault specified, determine (a) the subtransient current in the fault, (b) the
subtransient current in breaker A, and (c) the symmetrical short-circuit
interrupting current (as defined for circuit-breaker applications) in the fault
and in breaker A.
Solution (a) For a base of 25,000 kVA, 13.8 kV in the generator circuit,
the base for the motors is 25,000 kVA, 6.9 kV. The subtransient reactance of
each motor is
25,000
X'i = 0.20—%— = 1.0 per unit
d 5000 F
Figure 10.16 is the diagram with subtransient values of reactance marked.

For a fault at P,
Vf — 1.0 per unit
Zth = ;0.125 per unit
I"f = = —/8.0 per unit

f ;0.125 J F
The base current in the 6.9-kV circuit is
25,000
2090 A
73 x 6.9
8 x 2090 = 16,720 A
Figure 10.16 Reactance diagram for Example 10.5.

(b) Through breaker A comes the contribution from the generator-and

three of the four motors.
The generator contributes a current of
0.25
-j8.0 x = -;'4.0 per unit
Each motor contributes 25% of the remaining fault current, or -jl.O A

per unit.
Through breaker A,
J" = — j'4.0 + 3(— jl.O) = —j7.0 per unit = 7 x 2090 = 14,630 A
(c) To compute the current through breaker A to be interrupted, replace
the subtransient reactance of j 1.0 by the transient reactance of j 1.5 in the
motor circuits of Fig. 10.16. Then
.0.375 x 0.25
Zth = y'0.15 per unit
J 0.375 + 0.25
The generator contributes a current of
1.0 0.375
—;4.0 per unit
j0A5 X 0.625
Each motor contributes a current of
1 10 025 —;'0.67 per unit

4 X j0A5 X 0.625
The symmetrical short-circuit current to be interrupted is
(4.0 + 3 x 0.67) x 2090 = 12,560 A
The usual procedure is to rate all the breakers connected to a bus on the
basis of the current into a fault on the bus. In that case the short-circuit
current interrupting rating of the breakers connected to the 6.9-kV bus must
be at least
4 + 4 x 0.67 = 6.67 per unit
or
6.67 x 2090 = 13,940 A
A 14.4-kV circuit breaker has a rated maximum voltage of 15.5 kV and

a K of 2.67. At 15.5 kV its rated short-circuit interrupting current is 8900 A.
This breaker is rated for a symmetrical short-circuit interrupting current of
2.67 x 8900 = 23,760 A, at a voltage of 15.5/2.67 = 5.8 kV. This current is
the maximum that can be interrupted even though the breaker may be in a
circuit of lower voltage. The short-circuit interrupting current rating at
6.9 kV is
15.5
x 8900 - 20,000 A
6.9
The required capability of 13,940 A is well below 80% of 20,000 A, and the
breaker is suitable with respect to short-circuit current.
The short-circuit current could have been found by using the bus im¬
pedance matrix. For this purpose two nodes have been identified in
Fig. 10.16. Node 1 is the bus on the low-tension side of the transformer, and
node 2 is on the high-tension side. For motor reactance of 1.5 per unit
*ii = -;io + -/12.67

j1-5/4
y12=no
Y22 = ~j 10 -j6.67 = -j 16.67
The node admittance matrix is
-12.67 10.0 ]
10.0 -16.67 J
and its inverse is
0.150 0.090
bus-;[o.090 0.114
Figure 10.17 is the network corresponding to the bus impedance matrix.

Closing Sj with S2 open represents a fault on bus 1.
The symmetrical short-circuit interrupting current in a three-phase fault
at node 1 is
1.0
Isc =j015 = ~/'6'67 per unit
which agrees with our previous calculations. The bus impedance matrix also
gives us the voltage at bus 2 with the fault on bus 1.
V2 = 1.0 - IscZ2l = 1.0 - (-76.67)00.09) - 0.4
and since the admittance between nodes 1 and 2 is —710, the current into the
Figure 10.17 Bus impedance equivalent network for

the bus impedance matrix of Example 10.5.
fault from the transformer is
(0.4 - 0.0)(—/10) = —/4.0 per unit
which also agrees with our previous result.

We also know immediately the short-circuit current in a three-phase
fault at node 2 which, by referring to Fig. 10.17 with open and S2
closed is
;“=j0^ = -/8'77perunit
Even this simple example illustrates the value of the bus impedance
matrix where the effects of a fault at a number of buses are to be studied.
Matrix inversion is not necessary, for, as we have seen in Sec. 7.7, Zbus can
be generated directly by the computer.
PROBLEMS
10.1 A 60-Hz alternating voltage having a rms value of 100 V is applied to a series RL circuit by
closing a switch. The resistance is 15 D, and the inductance is 0.12 H.
(a) Find the value of the dc component of current upon closing the switch if the instantaneous value
of the voltage is 50 V at that time.
(b) What is the instantaneous value of the voltage which will produce the maximum dc component
of current upon closing the switch?
(c) What is the instantaneous value of the voltage which will result in the absence of any dc
component of current upon closing the switch?
(d) If the switch is closed when the instantaneous voltage is zero, find the instantaneous current 0.5,
1.5, and 5.5 cycles later.
10.2 A generator connected through a 5-cycle circuit breaker to a transformer is rated 100 MV A,
18 kV, with reactances of X"d = 19%, X'd = 26%, and Xd = 130%. It is operating at no load and rated
voltage when a three-phase short circuit occurs between the breaker and the transformer. Find
(a) the sustained short-circuit current in the breaker, (b) the initial symmetrical rms current in the
breaker, and (c) the maximum possible dc component of the short-circuit current in the breaker.
10.3 The three-phase transformer connected to the generator described in Prob. 10.2 is rated
100 MVA, 240Y/18A kV, X = 10%. If a three-phase short circuit occurs on the high-tension side of
the transformer at rated voltage and no load, find (a) the initial symmetrical rms current in the
transformer windings on the high-tension side and (b) the initial symmetrical rms current in the line
on the low-tension side.
10.4 A 60-Hz generator is rated 500 MVA, 20 kV, with X"d =0.20 per unit. It supplies a purely
resistive load of 400 MW at 20 kV. The load is connected directly across the terminals of the
generator. If all three phases of the load are short-circuited simultaneously, find the initial symmetri¬
cal rms current in the generator in per unit on a base of 500 MVA, 20 kV.
10.5 A generator is connected through a transformer to a synchronous motor. Reduced to the same
base, the per-unit subtransient reactances of the generator and motor are 0.15 and 0.35, respectively,
and the leakage reactance of the transformer is 0.10 per unit. A three-phase fault occurs at the
terminals of the motor when the terminal voltage of the generator is 0.9 per unit and the output
current of the generator is 1.0 per unit at 0.8 power factor leading. Find the subtransient current in
per unit in the fault, in the generator, and in the motor. Use the terminal voltage of the generator as
the reference phasor, and obtain the solution (a) by computing the voltages behind subtransient
reactance in the generator and motor and (b) by using Thevenin’s theorem.
Figure 10.18 Network for Prob. 10.8.
10.6 Two synchronous motors having subtransient reactances of 0.80 and 0.25 per unit, respectively,
on a base of 480 V, 2000 kVA are connected to a bus. This motor bus is connected by a line having a
reactance of 0.023 fl to a bus of a power system. At the power-system bus the short-circuit megavolt-
amperes of the power system are 9.6 MVA for a nominal voltage of 480 V. When the voltage at the
motor bus is 440 V, neglect load current and find the initial symmetrical rms current in a three-phase
fault at the motor bus.
10.7 The bus impedance matrix of a four-bus network with values in per unit is
0.15 0.08 0.04 0.07

0.08 0.15 0.06 0.09
0.04 0.06 0.13 0.05
0.07 0.09 0.05 0.12
Generators are connected to buses 1 and 2, and their subtransient reactances were included when
finding Zbus. If prefault current is neglected, find the subtransient current in per unit in the fault for a
three-phase fault on bus 4. Assume the voltage at the fault is 1.0 per unit before the fault occurs. Find
also the per-unit current from generator 2 whose subtransient reactance is 0.2 per unit.
10.8 For the network shown in Fig. 10.18 find the subtransient current in per unit from generator 1
and in line 1-2 and the voltages at buses 1 and 3 for a three-phase fault on bus 2. Assume that no
current is flowing prior to the fault and that the prefault voltage at bus 2 is 1.0 per unit. Use the bus
impedance matrix in the calculations.
10.9 If a three-phase fault occurs at bus 1 of the network of Fig. 10.11 when there is no load (all
node voltages 1.0 per unit), find the subtransient current in the fault, the voltages at buses 2, 3, and 4,
and the current from the generator connected to bus 2.
10.10 Find the subtransient current in per unit in a three-phase fault on bus 5 of the network of
Example 8.1. Neglect prefault current, assume all bus voltages are 1.0 per unit before the fault occurs,
and make use of calculations already made in Example 10.4. Find the current in lines 1-5 and 3-5
also.
10.11 A 625-kVA 2.4-kV generator with X"d = 0.20 per unit is connected to a bus through a circuit
breaker as shown in Fig. 10.19. Connected through circuit breakers to the same bus are three
Figure 10.19 One-line diagram for Prob. 10.11.

synchronous motors rated 250 hp, 2.4 kV, 1.0 power factor, 90% efficiency, with X"d = 0.20 per unit.
The motors are operating at full load, unity power factor, and rated voltage, with the load equally
divided between the machines.
(a) Draw the impedance diagram with the impedances marked in per unit on a base of 625 kVA,
2.4 kV.
(b) Find the symmetrical short-circuit current in amperes which must be interrupted by breakers A
and B for a three-phase fault at point P. Simplify the calculations by neglecting the prefault
current.
(c) Repeat part (b) for a three-phase fault at point Q.
(d) Repeat part (b) for a three-phase fault at point R.
10.12 A circuit breaker having a nominal rating of 34.5 kV and a continuous current rating of
1500 A has a voltage range factor K of 1.65. Rated maximum voltage is 38 kV, and. the rated
short-circuit current at that voltage is 22 kA. Find (a) the voltage below which rated short-circuit
current does not increase as operating voltage decreases and the value of that current and (b) rated
short-circuit current at 34.5 kV.
CHAPTER
ELEVEN
SYMMETRICAL COMPONENTS
In 1918 one of the most powerful tools for dealing with unbalanced polyphase
circuits was discussed by C. L. Fortescue at a meeting of the American Institute
of Electrical Engineers.! Since that time the method of symmetrical components
has become of great importance and has been the subject of many articles and
experimental investigations. Unsymmetrical faults on transmission systems,
which may consist of short circuits, impedance between lines, impedance from
one or two lines to ground, or open conductors, are studied by the method of
symmetrical components.
11.1 SYNTHESIS OF UNSYMMETRICAL PHASORS

FROM THEIR SYMMETRICAL COMPONENTS
Fortescue’s work proves that an unbalanced system of n related phasors can be

resolved into n systems of balanced phasors called the symmetrical components of
the original phasors. The n phasors of each set of components are equal in
length, and the angles between adjacent phasors of the set are equal. Although
the method is applicable to any unbalanced polyphase system, we shall confine
our discussion to three-phase systems.
t C. L. Fortescue, “Method of Symmetrical Coordinates Applied to the Solution of Polyphase

Networks,” Trans. AIEE, vol. 37, pp. 1027-1140, 1918.
275
According to Fortescue’s theorem, three unbalanced phasors of a three-

phase system can be resolved into three balanced systems of phasors. The bal¬
anced sets of components are:
1. Positive-sequence components consisting of three phasors equal in magni¬

tude, displaced from each other by 120° in phase, and having the same phase
sequence as the original phasors
2. Negative-sequence components consisting of three phasors equal in magni¬
tude, displaced from each other by 120° in phase, and having the phase
sequence opposite to that of the original phasors
3. Zero-sequence components consisting of three phasors equal in magnitude
and with zero phase displacement from each other
It is customary, when solving a problem by symmetrical components, to

designate the three phases of the system as a, b, and c in such a manner that the
phase sequence of the voltages and currents in the system is abc. Thus the phase
sequence of the positive-sequence components of the unbalanced phasors is abc,
and the phase sequence of the negative-sequence components is acb. If the ori¬
ginal phasors are voltages, they may be designated Va, Vb, and Vc. The three sets
of symmetrical components are designated by the additional subscript 1 for the
positive-sequence components, 2 for the negative-sequence components, and 0
for the zero-sequence components. The positive-sequence components of Va, Vb,
and Vc are Val, Vbl, and Vcl. Similarly, the negative-sequence components are Va2,
Vb2, and Vc2, and the zero-sequence components are Va0, Vb0, and vc0. Figure
11.1 shows three such sets of symmetrical components. Phasors representing
currents will be designated by I with subscripts as for voltages.
Since each of the original unbalanced phasors is the sum of its components,
the original phasors expressed in terms of their components are
Va=Val + Va2+Va0 (11.1)
Vb=Vbl + Vb2+Vb0 (11.2)
K= Ki + Vc2 + Vc0 ' (11-3)
Ki val
Positive-sequence Negative-sequence Zero-sequence

components components components
Figure 11.1 Three sets of balanced phasors which are the symmetrical components of three
unbalanced phasors.
SYMMETRICAL COMPONENTS 277
Figure 11.2 Graphical addition of the components shown

in Fig. 11.1 to obtain three unbalanced phasors.
The synthesis of a set of three unbalanced phasors from the three sets of symme¬
trical components of Fig. 11.1 is shown in Fig. 11.2.
The many advantages of analysis of power systems by the method of symme¬
trical components will become apparent gradually as we apply the method to the
study of unsymmetrical faults on otherwise symmetrical systems. It is sufficient
to say here that the method consists in finding the symmetrical components of
current at the fault. Then the values of current and voltage at various points in
the system can be found. The method is simple and leads to accurate predictions
of system behavior.
11.2 OPERATORS
Because of the phase displacement of the symmetrical components of the vol¬

tages and currents in a three-phase system, it is convenient to have a shorthand
method of indicating the rotation of a phasor through 120°. The result of the
multiplication of two complex numbers is the product of their magnitudes and
the sum of their angles. If the complex number expressing a phasor is multiplied
by a complex number of unit magnitude and angle 6, the resulting complex number
represents a phasor equal to the original phasor displaced by the angle 6.
The complex number of unit magnitude and associated angle 6 is an opera¬
tor that rotates the phasor on which it operates through the angle 6.
We are already familiar with the operator j, which causes rotation through
90°, and the operator -1, which causes rotation through 180°. Two successive
applications of the operator j cause rotation through 90° + 90°, which leads us
to the conclusion that j x j causes rotation through 180°, and thus we recognize
that j2 is equal to -1. Other powers of the operator j are found by similar
analysis.
The letter a is commonly used to designate the operator that causes a
rotation of 120° in the counterclockwise direction. Such an operator is a com¬
plex number of unit magnitude with an angle of 120° and is defined by
a = 1/120° = Uj2nl3 = -0.5 + j0.866

Figure 113 Phasor diagram of the various powers of the

operator a.
If the operator a is applied to a phasor twice in succession, the phasor is rotated

through 240°. Three successive applications of a rotate the phasor through 360°.
Thus,
a2 = 1/240° = -0.5 — j0.866
and
a3 = 1/360° = 1/0° = 1
Figure 11.3 shows phasors representing various powers of a.
11.3 THE SYMMETRICAL COMPONENTS OF

UNSYMMETRICAL PHASORS
We have seen (Fig. 11.2) the synthesis of three unsymmetrical phasors from three
sets of symmetrical phasors. The synthesis was made in accordance with
Eqs. (11.1) to (11.3). Now let us examine these same equations to determine how
to resolve three unsymmetrical phasors into their symmetrical components.
First, we note that the number of unknown quantities can be reduced by
expressing each component of Vb and Vc as the product of some function of the
operator a and a component of Va. Reference to Fig. 11.1 verifies the following
relations:
Vbl = a2Val Vcl = aVal
Vb2 = aK2 Vc2 = a2Vo2 ' (11.4)
Vb0 = Va0 Vc0 = Va0
Repeating Eq. (11.1) and substituting Eqs. (11.4) in Eqs. (11.2) and (11.3) yield
Va=Vai + Va2+Va0 (11.5)
Fb = a2Ffll + aFfl2 + Fo0 (11.6)

K = aKi + a2Va2 + Va0 (11.7)
or in matrix form
1 1 VaO
1 a Kx (11.8)
1 a k2
For convenience we let
1 1 1
A = 1 a2 a (11.9)
1 a a2
Then, as may be verified easily.
1 1
A'1 = 1 a a (11.10)
1 a2 a
and premultiplying both sides of Eq. (11.8) by A-1 yields
Va0 l l l
Ki 1 a a2 K (11.11)
Kz 1 a2 a V,
which shows us how to resolve three unsymmetrical phasors into their symmetri¬
cal components. These relations are so important that we shall write the separate
equations in ordinary fashion. From Eq. (11.11)
Ko = MK+vb+vc) (11.12)
Ki = UK + aVb + a2Vc) (11.13)
va2 =UK + a2Vb + aVc) (11.14)
If required, the components Vb0, Vbl, Vb2, Vc0, Vcl, and Vc2 can be found by
Eqs. (11.4).
Equation (11.12) shows that no zero-sequence components exist if the sum
of the unbalanced phasors is zero. Since the sum of the line-to-line voltage
phasors in a three-phase system is always zero, zero-sequence components are
never present in the line voltages, regardless of the amount of unbalance. The
sum of the three line-to-neutral voltage phasors is not necessarily zero, and
voltages to neutral may contain zero-sequence components.
The preceding equations could have been written for any set of related
phasors, and we might have written them for currents instead of for voltages.
They may be solved either analytically or graphically. Because some of the
preceding equations are so fundamental, they are summarized for currents:
fa Ial T fa2 "F âO (11.15)
h — a2Kl + aKl + I aO (11.16)
K ~ a 1 + aZIa2 + IaO
(11.17)
aO = Wa + lb + Ic)
(11.18)
al = Wa + + a2lc) (11.19)
al = Wa + a2Ib + ale) (11.20)

In a three-phase system the sum of the line currents is equal to the current/„ in
the return path through the neutral. Thus,
/fl + /|, + lt - /, (11.21)

Comparing Eqs. (11.18) and (11-21) gives
K— (11.22)
In the absence of a path through the neutral of a three-phase system, /„ is zero,

and the line currents contain no zero-sequence components. A A-connected load
provides no path to neutral, and the line currents flowing to a A-connected load
can contain no zero-sequence components.
Example 11.1 One conductor of a three-phase line is open. The current

flowing to the A-connected load through line a is 10 A. With the current in
line a as reference and assuming that line c is open, find the symmetrical
components of the line currents.
Solution Figure 11.4 is a diagram of the circuit. The line currents are
Ia = 10/01 A Ib = 10/180° A /c = 0 A
From Eqs. (11.18) to (11.20)
Iao = i(10/01 + 10/180° + 0) = 0
I aX = i(10/01 + 10/180° + 120° + 0)
= 5 -/2.89 = 5.78/-30° A
hi = i(10/01 + 10/180°+ 240° + 0)
= 5 + ;2.89 = 5.78/30! A
From Eqs. (11.4)
Ibl = 5.78 /—150° A Icl = 5.78/90! A
Ib2 = 5.78/150° A Ic2 = 5.78 /—90° A
ho = 0 Ic o = 0
7„ = 10/0° amp
a
b
Z
c
Ic= 0
We note that components Icl and Ic2 have definite values although line c
is open and can carry no net current. As is expected, therefore, the sum of
the components in line c is zero. Of course, the sum of the components in
line a is 10^F A, and the sum of the components in line b is 10/l80° A.
11.4 PHASE SHIFT OF SYMMETRICAL COMPONENTS

IN Y-A TRANSFORMER BANKS
In discussing symmetrical components for three-phase transformers, we need to

examine the standard method of marking transformer terminals. In Sec. 6.5 we
discussed the placing of dots at one end of each winding on the same iron core of
a transformer to indicate that current flowing from the dotted terminal to the
unmarked terminal of each winding produced a magnetomotive force acting in
the same direction in the magnetic circuit. We noted that, if the small effect of
magnetizing current is neglected, two currents Jj and I2 flowing in the only two
windings on a common transformer core would be in phase if we chose the
current to be positive when entering the dotted terminal of one winding and
when leaving the dotted terminal of the other winding.
The standard marking of single-phase, two-winding transformers substitutes
7/j and Xx for the dots on the high- and low-tension windings, respectively. The
other ends of the windings are marked H2 and X2. Figure 11.5 shows both dots
and standard markings, and Ip and Is must be in phase. In Sec. 6.5 we noted that
the dots on the windings of a single-phase transformer indicated that the voltage
drops from dotted to unmarked terminals of the two windings are in phase. So
in the single-phase transformer the terminals and Xt are positive at the same
time with respect to H2 and X2. If the direction of the arrow marked Is in
Fig. 11.5 were reversed while the direction of the arrow marked Ip remained the
same, Is and Ip would be 180° out of phase. Therefore, the primary and secon¬
dary currents are either in phase or 180° out of phase, depending on the direc¬
tion assumed to be positive for the flow of current. Similarly, primary and
secondary voltages may be in phase or 180° out of phase, depending on which
terminal is assumed to be positive for specifying the voltage drop of each.
The high-tension terminals of three-phase transformers are marked Hu H2,
and H3, and the low-tension terminals are marked Xu X2, and X3. In Y-Y or
A-A transformers the markings are such that voltages to neutral from terminals
Figure 11.5 Schematic diagram of single-phase transformer wind¬

ings showing standard markings and the directions assumed posi¬
tive for primary and secondary current.
Positive sequence Negative sequence
(b) Voltage components
Figure 11.6 Wiring diagram and voltage phasors for a three-phase transformer connected Y-A where
the Y side is the high-tension side.
Ht, H2, and H3 are in phase with the voltages to neutral from terminals Xu X2,
and X3, respectively.
Figure 11.6a is the wiring diagram of a Y-A transformer. The high-tension
terminals Ht, H2, and H3 are connected to phases A, B, and C, respectively, and
the phase sequence is ABC. The arrangement and notation of the diagram
conform to a convention that we shall follow in all our computations. Windings
that are drawn in parallel directions are those linked magnetically by being
wound on the same core. When capital letters are assigned to phases on one side
of the transformer, lowercase letters will be assigned to the phases on the other
side. It is customary to use uppercase letters on the high-tension side of the
transformer and lowercase letters on the low-tension side. In Fig. 11.6a winding
AN is the phase on the Y-connected side which is linked magnetically with the
phase winding be on the A-connected side. The location of the dots on
the windings shows that Van is in phase with vbc. We shall examine later the case
where the Y-connected side is the low-tension winding. If H3 is the terminal to
which line A is connected, it is customary to connect phase B to H2 and phase C
to H3.
The American standard for designating terminals H3 and Xt on Y-A trans¬
formers requires that the positive-sequence voltage drop from H3 to neutral lead
the positive-sequence voltage drop from Xt to neutral by 30°, regardless of
whether the Y or the A winding is on the high-tension side. Similarly, the voltage
from H2 to neutral leads the voltage from X2 to neutral by 30°, and the voltage
from H3 to neutral leads the voltage from X3 to neutral by 30°. The phasor
(a) VM leads V61 by 30° (b) VAl leads Val by 30°
Figure 11.7 Labeling of lines connected to a three-phase Y-A transformer.
diagrams for the sequence components of voltage are shown in Fig. 11.6b. We
designate the positive-sequence voltage VAtn as VA1 and other voltages to neutral
in a similar manner and see that VA1 leads Vbl by 30°, which enables us to
determine that the terminal to which phase b is connected should be labeled Xv
Figure 11.7a shows the connections of the phases to the transformer ter¬
minals so that the positive-sequence voltage to neutral VA1 leads the positive-
sequence voltage to neutral Vbl by 30°. It is not necessary, however, to label the
lines attached to the transformer terminals as we have done since no standards
have been adopted for such labeling. Very frequently the designation of lines will
be as shown in Fig. 11.7b. We shall follow the scheme of Fig. 11.7a which con¬
forms to the wiring and phasor diagrams of Fig. 11.6 since such nomenclature is
most convenient for computations. If the scheme of Fig. 11.7b is preferred, it is
necessary only to exchange a for b, b for c, and c for a in the work which follows.
Inspection of the positive- and negative-sequence phasor diagrams of
Fig. 11.6 shows that Val leads VA1 by 90° and that Va2 lags VA2 by 90°. The
diagrams show VA1 and VA2 in phase, which is not necessarily true, but phase
shift between VA1 and VA2 does not alter the 90° relation between VAl and Val or
between VA2 and K 2-
Since the direction specified for IA in Fig. 11.6a is away from the dot in the
transformer winding and the direction of Ibc is also away from the dot in its
winding, these currents are 180° out of phase. Therefore, the phase relation
between the Y and A currents is as shown in Fig. 11.8. We note that 7al leads IA1
Negative-sequence
components
Figure 11.8 Current phasors of a three-phase transformer connected Y-A where the Y side is the
high-tension side.
(a) Wiring diagram
Positive sequence Negative sequence

(b) Voltage components
Figure 11.9 Wiring diagram and voltage phasors for a three-phase transformer connected Y-A where
the A side is the high-tension side.
by 90° and Ia2 lags IA2 by 90°. Summarizing the relations between the symmetri¬
cal components of the line currents on the two sides of the transformer gives
Ki = +jVa i Ki — +j!ai
(11.23)
fa2 = ~jÂ2 Ia2 = ~jÂ2
where each voltage and current is expressed in per unit. Transformer impedance
and magnetizing current are neglected, which explains why the per-unit magni¬
tudes of voltage and current are exactly the same on both sides of the transfor¬
mer (for instance, | Val | equals | V411).
Up to this point our discussion of the Y-A transformer has been confined to
the case where the high-tension windings are connected in Y. Figure 11.9 shows
the A-connected windings on the high-tension side of the transformer. The figure
shows that in order to have the positive-sequence voltage from to neutral
lead the positive-sequence voltage from Xx to neutral by 30 I'bci and Val must
be 180° out of phase, and the currents /BC1 and Ial must be 180° out of phase as
shown in Fig. 11.10. The phasor diagrams for the voltages and currents show
that Eqs. (11.23) are still valid.
We have assumed the power flow from the high-tension to the low-tension
windings by showing IA, JB, and Ic toward the transformer and Ia, Ib, and Ic
away from the transformer. If we were assuming power flow in the opposite
direction, voltage relationships would remain the same but all line currents
would be shown in the opposite direction. However, this would cause no change
Ic 2
Ib2
Positive-sequence Negative-sequence
components components
Figure 11.10 Current phasors of a three-phase transformer connected Y-A where the A side is the
high-tension side.
in the phase angles of primary and secondary line currents with respect to each
other. Therefore Eqs. (11.23) are valid for both voltages and currents regardless
of which windings are the primary.
Example 11.2 Three identical resistors are Y-connected to the low-tension Y

side of a A-Y transformer. The voltages at the resistor load are
| Vab\ = 0.8 per unit \Vbc\ = 1.2 per unit \Vca\ = 1.0 per unit
Assume that the neutral of the load is not connected to the neutral of the
transformer secondary. Find the line voltages and currents in per unit on the
A side of the transformer.
Solution Assuming an angle of 180° for Vca and using the law of cosines to
find the angles of the other line voltages, we have
Vab = 0.8/82.8° per unit

Kc = 1-2 /-41,4° per unit
Vca = 1.0/180° per unit
The symmetrical components of the line voltages are
Vab! = 4(0.8/82.8° + 1.2/120° - 41.4° + 1.0/240° + 180°)

= 4(0.1 + j0.794 + 0.237 +;1.177 + 0.5 + ;'0.866)
= 0.279 -I- jO.946 = 0.985/73.6° per unit (line-line voltage base)
Vab2 = 4(0.8/82.8° + 1.2/240° - 41.4° + 1.0/120° + 180°)

= 4(1.0 + j0.794 - 1.138 -yO.383 + 0.5 -jO.866)
= —0.179 — y'0.152 = 0.235/220.3° per unit (line-line voltage base)
To determine the positive- and negative-sequence voltages to neutral we

need to examine the phase difference between line and phase voltages of
Positive-sequence components Negative-sequence components
Figure 11.11 Positive- and negative-sequence components of line-to-line and line-to-neutral

voltages of a three-phase system.
balanced-Y loads for positive and negative sequences. Consider Fig. 11-11,
where Vabl and Vab2 are taken as reference arbitrarily. The choice of the
reference has no effect on the results. We see that
vml =4^,/-30° (11.24)
and
K. 2 = -4 (11.25)
We find Van as the sum of its components:
Kn ~ Kn 1 + Van2 (11.26)
The other voltages to neutral are found by obtaining their components from
and Vm2 by Eqs. (11.4). If the voltages to neutral are in per unit referred
to the base voltage to neutral and the line voltages are in per unit referred to
the base voltage from line to line, the 1/^/3 factor must be omitted in
Eqs. (11.24) and (11.25). If both voltages are referred to the same base, the
equations are correct as given.
The absence of a neutral connection means that zero-sequence currents
are not present. Therefore, the phase voltages at the load contain positive-
and negative-sequence components only. The phase voltages are found from
Eqs. (11.24) and (11.25) with the factor 1/^/3 omitted, since the line voltages
are expressed in terms of the base voltage from line to line and the phase
voltages are desired in per unit of the base voltage to neutral. Thus
Vanl = 0.985/73.6° - 30°
= 0.985/43.6° per unit (line-neutral voltage base)
Van2 = 0.235/220.3° + 30°

Since each resistor has an impedance of 1.0/0° per unit,
1 di = Yô = 0.985^M! per unit
Ki = = 0-235/250.3° per unit
The direction assumed to be positive for the currents is from the supply
toward the A primary of the transformer and away from the Y side toward
the load.
Multiplying both sides of Eqs. (11.23) by j, we obtain for the high-
tension side of the transformer
VA1 = ~jVal = 0.985 /— 46.4° = 0.680 - jO.713
VA2 = jVa2 = 0.235 /- 19.7° - 0.221 - ;0.079
VA = VAi + Va2 = 0.901 — j'0.792

= 1.20 /—41.3° per unit
VBl = a2VA1 = 0.985/193.6° = -0.958 - j0.232

VB2 = aV42 = 0.235/100.3° - -0.042 + ;0.232
VB=VBl + VB2= -1.0

= 1.0/180° per unit
Vcl = aVA1 = 0.985/73.6° = 0.278 + ;0.944

VC2 = a2VA2 = 0.235/220.3° = -0.179 -;0.152
Vc = VC1 + VC2 = 0.099 + y'0.792
= 0.8/82.9° per unit
VAB =VA-VB = 0.901 — /0.792 + 1.0 = 1.901 -/0.792

= 2.06 /— 22.6° per unit (line-neutral voltage base)
2.06
/— 22.6 = 1.19 /—22.6° per unit (line-line voltage base)
73
FBC = - Kc = -1.0 - 0.099 - jO.792 = -1.099 - /0.792
_ 1.355
= 0.782/215.8° per unit (line-line voltage base)
= 7^
VCA = VC-VA = 0.099 + ;0.792 - 0.901 + /0.792 = -0.802 + jl.584
1.78
/116.9° = 1.028/116.9° per unit (line-line voltage base)
Since the load impedance in each phase is resistance of 1.0/0° per unit,
Ial and Val are found to have identical per-unit values in this problem.
Likewise, Ia2 and K 2 are identical in per unit. Therefore, IA must be identi¬
cal to VA expressed in per unit. Thus
IA = 1.20 /—41,3° per unit
IB— 1.0/180° per unit
Ic = 0.80/82.9° per unit
When problems involving unsymmetrical faults are solved, positive- and

negative-sequence components are found separately and phase shift is taken
into account, if necessary, by applying Eq. (11.23). Digital-computer pro¬
grams can be written to incorporate the effects of phase shift.
11.5 POWER IN TERMS OF SYMMETRICAL COMPONENTS
If the symmetrical components of current and voltage are known, the power
expended in a three-phase circuit can be computed directly from the compo¬
nents. Demonstration of this statement is a good example of the matrix manipula¬
tion of symmetrical components.
The total complex power flowing into a three-phase circuit through three
lines a, b, and c is
s = p +jQ = VaI*a + VbI* + VCI* (11.27)
where Va, Vb, and Vc are voltages to neutral at the terminals and Ia, Ib, and Ic
are the currents flowing into the circuit in the three lines. A neutral connection
may or may not be present. In matrix notation
S = [Ea Vb K] (11.28)
where the conjugate of a matrix is understood to be composed of elements that

are the conjugates of the corresponding elements of the original matrix.
To introduce the symmetrical components of the voltages and currents we
make use of Eqs. (11.8) and (11.9) to obtain
S = [AV]r[AI]* (11.29)
where
va0 I aO
V = Ki and I= Ll (11.30)
va2 h 2.
The reversal rule of matrix algebra states that the transpose of the product of
two matrices is equal to the product of the transposes of the matrices in reverse
order. According to this rule
[AV]r = VrAr (11.31)

and so
S = VrAr[AI]* = VrArA*I* (11.32)
Noting that \T = A and that a and a2 are conjugates, we obtain
1 1 1 1 1 1 I aO
s = [Va0 Val Va2] l a2 a 1 a a2 (11.33)
1 a a2 1 a2 a Kl
or, since ArA* is equal to
1 0 0
0 1 0
0 0 1
I aO
(11.34)
S = 3[Fa0 Val Va2]
Kl
So complex power is
VJ*a + VbI£ + VCI* = 3V0I$ + 3 Vyl\ + 3 V2I*2 (11.35)
which shows how complex power can be computed from the symmetrical com¬
ponents of the voltages and currents of an unbalanced three-phase circuit.
11.6 UNSYMMETRICAL SERIES IMPEDANCES
We shall be concerned particularly with systems that are normally balanced and
become unbalanced only upon the occurrence of an unsymmetrical fault. Let us
look, however, at the equations of a three-phase circuit when the series im¬
pedances are unequal. We shall reach a conclusion that is important in analysis
by symmetrical components. Figure 11.12 shows the unsymmetrical part of a
system with three unequal series impedances Za, Zb, and Zc. If we assume no
la za
a — —wvw-- a'
h
b - —VVNAA-— b'
Figure 11.12 Portion of a three-phase system showing Ic Zc

three unequal series impedances. c — —WWV-- c'
mutual inductance (no coupling) between the three impedances, the voltage
drops across the part of the system shown are given by the matrix equation
Ka' z„ 0 o ■ la
vw — 0 Z„ 0 h (11.36)
_1
N
O
VCC‘ h
u
1
and in terms of the symmetrical components of voltage and current
âa'O 0 0 ' l aO
A 1 = 0 0 A lay
(11.37)
Vaa. 2_ 0 0 lal
where A is the matrix defined by Eq. (11.9). Premultiplying both sides of the
equation by A'1 yields the matrix equation from which we obtain
Kai = + Zb + Zc) + \la2{Za + a2Zb + aZc)

+ iIao{Za + QZb + a2Zc)
Ka'2 = }fai(Za + aZb + a2Zc) + jIa2(Za + Zb + Zc)

(11.38)
+ ilao{Za + a2Zb + aZc)
Ka'o = }lai(Za + a2Zb + aZc) + \la2(Za + aZb + a2Zc)
+ jIao{Za + Zb + Zc)
If the impedances are made equal (that is, if Za = Zb = Zc), Eqs. (11.38)
reduce to
V — 17 (11.39)
y aa' 1 1 a 1 â Kia'2 — al ■Z.
In') âa'O IaO â
Thus, we conclude that the symmetrical components of unbalanced currents

flowing in a balanced-Y load or in balanced series impedances produce voltage
drops of like sequence only, provided no coupling exists between phases. If the
impedances are unequal, however, Eqs. (11.38) show that the voltage drop of any
one sequence is dependent on the currents of all three sequences. If coupling
such as mutual inductance existed between the three impedances of Fig. 11.13,
the square matrix of Eqs. (11.36) and (11.37) would contain off-diagonal ele¬
ments and Eqs. (11.38) would have additional terms.
Although current in any conductor of a three-phase transmission line
induces a voltage in the other phases, the way in which reactance is calculated
eliminates consideration of coupling. The self-inductance calculated on the basis
of complete transposition includes the effect of mutual reactance. The assump¬
tion of transposition yields equal series impedances. Thus the component currents
of any one sequence produce only voltage drops of like sequence in a transmis¬
sion line; that is, positive-sequence currents produce positive-sequence voltage
drops only. Likewise negative-sequence currents produce negative-sequence vol¬
tage drops only, and zero-sequence currents produce zero-sequence voltage
drops only. Equations (11.38) apply to unbalanced-Y loads because points a\ b\
and c may be connected to form a neutral. We could study variations of these

equations for special cases such as single-phase loads where Zb = Zc — 0, but we
shall confine our discussion to systems that are balanced before a fault occurs.
11.7 SEQUENCE IMPEDANCES AND SEQUENCE NETWORKS
In any part of a circuit, the voltage drop caused by current of a certain sequence
depends on the impedance of that part of the circuit to current of that sequence.
The impedance of any section of a balanced network to current of one sequence
may be different from impedance to current of another sequence.
The impedance of a circuit when positive-sequence currents alone are
flowing is called the impedance to positive-sequence current. Similarly, when only
negative-sequence currents are present, the impedance is called the impedance to
negative-sequence current. When only zero-sequence currents are present, the
impedance is called the impedance to zero-sequence current. These names of the
impedances of a circuit to currents of the different sequences are usually
shortened to the less descriptive terms positive-sequence impedance, negative-
sequence impedance, and zero-sequence impedance.
The analysis of an unsymmetrical fault on a symmetrical system consists in
finding the symmetrical components of the unbalanced currents that are flowing.
Since the component currents of one phase sequence cause voltage drops of like
sequence only and are independent of currents of other sequences, in a balanced
system, currents of any one sequence may be considered to flow in an indepen¬
dent network composed of the impedances to the current of that sequence only.
The single-phase equivalent circuit composed of the impedances to current of
any one sequence only is called the sequence network for that particular
sequence. The sequence network includes any generated emfs of like sequence.
Sequence networks carrying the currents Ial, Ia2, and Ia0 are interconnected to
represent various unbalanced fault conditions. Therefore, to calculate the effect
of a fault by the method of symmetrical components, it is essential to determine
the sequence impedances and to combine them to form the sequence networks.
11.8 SEQUENCE NETWORKS OF UNLOADED GENERATORS
An unloaded generator, grounded through a reactor, is shown in Fig. 11.13.

When a fault (not indicated in the figure) occurs at the terminals of the genera¬
tor, currents la, lb, and Ic flow in the lines. If the fault involves ground, the
current flowing into the neutral of the generator is designated /„. One or two of
the line currents may be zero, but the currents can be resolved into their symme¬
trical components regardless of how unbalanced they may be.
Drawing the sequence networks is simple. The generated voltages are of
positive sequence only, since the generator is designed to supply balanced three-
phase voltages. Therefore, the positive-sequence network is composed of an emf
Figure 11.13 Circuit diagram of an unloaded generator

grounded through a reactance. The emfs of each phase
are £„, Eb, and Ec.
in series with the positive-sequence impedance of the generator. The negative-

and zero-sequence networks contain no emfs but include the impedances of the
generator to negative- and zero-sequence currents, respectively. The sequence
components of current are shown in Fig. 11.14. They are flowing through im¬
pedances of their own sequence only, as indicated by the appropriate subscripts
on the impedances shown in the figure. The sequence networks shown in
Fig. 11.14 are the single-phase equivalent circuits of the balanced three-phase
circuits through which the symmetrical components of the unbalanced currents
are considered to flow. The generated emf in the positive-sequence network is
the no-load terminal voltage to neutral, which is also equal to the transient and
subtransient internal voltages since the generator is not loaded. The reactance in
the positive-sequence network is the subtransient, transient, or synchronous
reactance, depending on whether subtransient, transient, or steady-state condi¬
tions are being studied.
The reference bus for the positive- and negative-sequence networks is the
neutral of the generator. So far as positive- and negative-sequence components
are concerned, the neutral of the generator is at ground potential if there is a
connection between neutral and ground having a finite or zero impedance since
the connection will carry no positive- or negative-sequence current.
The current flowing in the impedance Z„ between neutral and ground is
3Ia0. By referring to Fig. 11.14c, we see that the voltage drop of zero sequence
from point a to ground is — 3Ia0 Z„ — Ia0 Zg0, where Zg0 is the zero-sequence
impedance per phase of the generator. The zero-sequence network, which is a
single-phase circuit assumed to carry only the zero-sequence current of one
phase, must therefore have an impedance of 3Z„ + Zg0, as shown in Fig. 11.14/
The total zero-sequence impedance through which Ia0 flows is
Z0 = 3Z„ + Zg0 (11.40)
Usually the components of current and voltage for phase a are found from
equations determined by the sequence networks. The equations for the compo¬
nents of voltage drop from point a of phase a to the reference bus (or ground) are,
Ial
Reference bus
hi
(a) Positive-sequence current paths (b) Positive-sequence network
(c) Negative-sequence current paths (d) Negative-sequence network
Reference bus
13 Zn
"Z0 va0
Li
*a0
(e) Zero-sequence current paths (/) Zero-sequence network
Figure 11.14 Paths for current of each sequence in a generator and the corresponding sequence
networks.
as may be deduced from Fig. 11.14,
Kl _ Eg Igl Zy (11.41)
Va2=-Ia2Z2 (11.42)
N
(11.43)
II
1
o
o
3
where Ea is the positive-sequence no-load voltage to neutral, Zt and Z2 are the

positive- and negative-sequence impedances of the generator, and Z0 is defined
by Eq. (11.40). The above equations, which apply to any generator carrying
unbalanced currents, are the starting points for the derivation of equations for
the components of current for different types of faults. They apply to the case of
a loaded generator under steady-state conditions. When computing transient or
subtransient conditions the equations apply to a loaded generator if E'g or E" is
substituted for Ea.
11.9 SEQUENCE IMPEDANCES OF CIRCUIT ELEMENTS -
The positive- and negative-sequence impedances of linear, symmetrical, static

circuits are identical because the impedance of such circuits is independent of
phase order provided the applied voltages are balanced. The impedance of a
transmission line to zero-sequence currents differs from the impedance to
positive- and negative-sequence currents.
The impedances of rotating machines to currents of the three sequences will
generally be different for each sequence. The mmf produced by negative-
sequence armature current rotates in the direction opposite to that of the rotor
on which is the dc field winding. Unlike the flux produced by positive-sequence
current, which is stationary with respect to the rotor, the flux produced by the
negative-sequence current is sweeping rapidly over the face of the rotor. The
currents induced in the field and damper windings by the rotating armature flux
keep the flux from penetrating the rotor. This condition is similar to the rapidly
changing flux immediately upon the occurrence of a short circuit at the terminals
of a machine. The flux path is the same as that encountered in evaluating
subtransient reactance. So, in a cylindrical-rotor machine subtransient and
negative-sequence reactances are equal. Values given in Table A.4 confirm this
statement.
When only zero-sequence current flows in the armature winding of a three-
phase machine, the current and mmf of one phase are a maximum at the same
time as the current and mmf of each of the other phases. The windings are so
distributed around the circumference of the armature that the point of maximum
mmf produced by one phase is displaced 120 electrical degrees in space from the
point of maximum mmf of each of the other phases. If the mmf produced by the
current of each phase had a perfectly sinusoidal distribution in space, a plot of
mmf around the armature would result in three sinusoidal curves whose sum
would be zero at every point. No flux would be produced across the air gap, and
the only reactance of any phase winding would be that due to leakage and end
turns. In an actual machine, the winding is not distributed to produce perfectly
sinusoidal mmf. The flux resulting from the sum of the mmfs is very small but
makes the zero-sequence reactance somewhat higher than in the ideal case where
there is no air-gap flux due to zero-sequence current.!
In deriving the equations for inductance and capacitance of transposed
transmission lines, we assumed balanced three-phase currents and did not
specify phase order. Therefore, the resulting equations are valid for both
positive- and negative-sequence impedances. When only zero-sequence current
flows in a transmission line, the current in each phase is identical. The current
t The reader who wishes to study machine impedances should consult such books as Central
Station Engineers of Westinghouse Electric Corporation, Electrical Transmission and Distribution
Reference Book, 4th ed., chap. 6, pp. 145-194, East Pittsburgh, Pa., 1964; or A. E. Fitzgerald, C.
Kingsley, Jr., and A. Kusko, Electric Machinery, 3d ed., chaps. 6 and 9, McGraw-Hill Book
Company, New York, 1971.
returns through the ground, through overhead ground wires, or through both.
Because zero-sequence current is identical in each phase conductor (rather than
equal only in magnitude and displaced in phase by 120° from other phase
currents), the magnetic field due to zero-sequence current is very different from
the magnetic field caused by either positive- or negative-sequence current. The
difference in magnetic field results in the zero-sequence inductive reactance of
overhead transmission lines being 2 to 3.5 times as large as the positive-sequence
reactance. The ratio is toward the higher portion of the specified range for
double-circuit lines and lines without ground wires.
A transformer in a three-phase circuit may consist of three individual single¬
phase units, or it may be a three-phase transformer. Although the zero-sequence
series impedances of three-phase units may differ slightly from the positive- and
negative-sequence values, it is customary to assume that series impedances of all
sequences are equal regardless of the type of transformer. Table A.5 lists trans¬
former reactances. Reactance and impedance are almost equal for transformers
of 1000 kVA or larger. For simplicity in our calculations we shall omit shunt
admittance, which accounts for exciting current.
The zero-sequence impedance of balanced Y- and A-connected loads equals
the positive and negative-sequence impedance. The zero-sequence network for
such loads is discussed in Sec. 11.11.
11.10 POSITIVE- AND NEGATIVE-SEQUENCE NETWORKS
The object of obtaining the values of the sequence impedances of a power system
is to enable us to construct the sequence networks for the complete system. The
network of a particular sequence shows all the paths for the flow of current of
that sequence in the system.
We discussed the construction of some rather complex positive-sequence
networks in Chap. 6. The transition from a positive-sequence network to a
negative-sequence network is simple. Three-phase synchronous generators and
motors have internal voltages of positive sequence only, since they are designed
to generate balanced voltages. Since the positive- and negative-sequence im¬
pedances are the same in a static symmetrical system, conversion of a positive-
sequence network to a negative-sequence network is accomplished by changing,
if necessary, only the impedances that represent rotating machinery and by omit¬
ting the emfs. Electromotive forces are omitted on the assumption of balanced
generated voltages and the absence of negative-sequence voltages induced from
outside sources.
Since all the neutral points of a symmetrical three-phase system are at the
same potential when balanced three-phase currents are flowing, all the neutral
points must be at the same potential for either positive- or negative-sequence
currents. Therefore the neutral of a symmetrical three-phase system is the logical
reference potential for specifying positive- and negative-sequence voltage drops
Reference bus
Figure 11.15 Negative-sequence

network for Example 11.3.
and is the reference bus of the positive- and negative-sequence networks. Im¬
pedance connected between the neutral of a machine and ground is not a part of
either the positive- or negative-sequence network because neither positive- nor
negative-sequence current can flow in an impedance so connected.
Negative-sequence networks, like the positive-sequence networks of Chap. 6,
may contain the exact equivalent circuits of parts of the system or may be
simplified by omitting series resistance and shunt admittance.
Example 11.3 Draw the negative-sequence network for the system des¬
cribed in Example 6.10. Assume that the negative-sequence reactance of each
machine is equal to its subtransient reactance. Omit resistance.
Solution Since all the negative-sequence reactances of the system are equal
to the positive-sequence reactances, the negative-sequence network is identi¬
cal to the positive-sequence network of Fig. 6.30 except for the omission of
emfs from the negative-sequence network. The required network is drawn in
Fig. 11.15.
11.11 ZERO-SEQUENCE NETWORKS
A three-phase system operates single phase insofar as the zero-sequence currents

are concerned, for the zero-sequence currents are the same in magnitude and
phase at any point in all the phases of the system. Therefore zero-sequence
currents will flow only if a return path exists through which a completed circuit
is provided. The reference for zero-sequence voltages is the potential of the
ground at the point in the system at which any particular voltage is specified.
Since zero-sequence currents may be flowing in the ground, the ground is not
necessarily at the same potential at all points and the reference bus of the
zero-sequence network does not represent a ground of uniform potential. The
impedance of the ground and ground wires is included in the zero-sequence
impedance of the transmission line, and the return circuit of the zero-sequence
network is a conductor of zero impedance, which is the reference bus of
the system. It is because the impedance of the ground is included in the zero-
sequence impedance that voltages measured to the reference bus of the zero-
sequence network give the correct voltage to ground.
If a circuit is Y-connected, with no connection from the neutral to ground or

to another neutral point in the circuit, the sum of the currents flowing into the
neutral in the three phases is zero. Since currents whose sum is zero have no
zero-sequence components, the impedance to zero-sequence current is infinite
beyond the neutral point; this fact is indicated by an open circuit in the zero-
sequence network between the neutral of the Y-connected circuit and the refer¬
ence bus, as shown in Fig. 11.16a.
If the neutral of a Y-connected circuit is grounded through zero impedance,
a zero-impedance connection is inserted to connect the neutral point and the
reference bus of the zero-sequence network, as shown in Fig. 11.166.
If the impedance Z„ is inserted between the neutral and ground of a Y-
connected circuit, an impedance of 3Z„ must be placed between the neutral and
reference bus of the zero-sequence network, as shown in Fig. 11.16c. As ex¬
plained in Sec. 11.8, the zero-sequence voltage drop caused in the zero-sequence
network by Ia0 flowing through 3Z„ is the same as in the actual system where
Reference bus
■VNAAA-*N
z
(a)
Reference bus
YWW 'N
z
I
(b)
Reference bus
fqQ ^
l-vww
i z
(c)
Figure 11.16 Zero-sequence networks for Y-connected loads.

Reference bus
b'AVJ
Figure 11.17 A-connected load Z
and its zero-sequence network.
3Ia0 flows through Z„. Impedance consisting of a resistor or reactor is usually

connected between the neutral of a generator and ground to limit the zero-
sequence current during a fault. The impedance of such a current-limiting re¬
sistor or reactor is represented in the zero-sequence network in the manner
described.
A A-connected circuit, since it can provide no return path, offers infinite
impedance to zero-sequence line currents. The zero-sequence network is open at
the A-connected circuit. Zero-sequence currents may circulate inside the A cir¬
cuit since the A is a closed series circuit for circulating single-phase currents.
Such currents would have to be produced in the A, however, by induction from
an outside source or by zero-sequence generated voltages. A A circuit and its
zero-sequence network are shown in Fig. 11.17. Even when zero-sequence vol¬
tages are generated in the phases of the A, no zero-sequence voltage exists
between the A terminals, for the rise in voltage in each phase of the generator is
matched by the voltage drop in the zero-sequence impedance of each phase.
The zero-sequence equivalent circuits of three-phase transformers deserve
special attention. The various possible combinations of the primary and second¬
ary windings in Y and A alter the zero-sequence network. Transformer theory
enables us to construct the equivalent circuit for the zero-sequence network. We
remember that no current flows in the primary of a transformer unless current
flows in the secondary, if we neglect the relatively small magnetizing current. We
know, also, that the primary current is determined by the secondary current and
the turns ratio of the windings, again with magnetizing current neglected. These
principles guide us in the analysis of individual cases. Five possible connections
of two-winding transformers will be discussed. These connections are shown in
Fig. 11.18. The arrows on the connection diagrams show the possible paths for
the flow of zero-sequence current. Absence of an arrow indicates that the trans¬
former connection is such that zero-sequence current cannot flow. The zero-
sequence approximately equivalent circuit, with resistance and a path for
magnetizing current omitted, is shown in Fig. 11.18 for each connection. The
letters P and Q identify corresponding points on the connection diagram and
equivalent circuit. The reasoning to justify the equivalent circuit for each connec¬
tion follows.
1: Y-Y Bank, One Neutral Grounded If either one of the neutrals of a

Case
Y-Y bank is ungrounded, zero-sequence current cannot flow in either winding.
The absence of a path through one winding prevents current in the other. An
open circuit exists for zero-sequence current between the two parts of the system
connected by the transformer.
Case 2: Y-Y Bank, Both Neutrals Grounded Where both neutrals of a Y-Y
bank are grounded, a path through the transformer exists for zero-sequence
currents in both windings. Provided the zero-sequence current can follow a
complete circuit outside the transformer on both sides, it can flow in both
windings of the transformer. In the zero-sequence network, points on the two
sides of the transformer are connected by the zero-sequence impedance of the
transformer in the same manner as in the positive- and negative-sequence
networks.
Figure 11.18 Zero-sequence equivalent circuits of three-phase transformer banks, together with dia¬
grams of connections and the symbols for one-line diagrams.
Case 3: Y-A Bank, Grounded Y If the neutral of a Y-A bank is grounded,

zero-sequence currents have a path to ground through the Y because corre¬
sponding induced currents can circulate in the A. The zero-sequence current
circulating in the A to balance the zero-sequence current in the Y cannot flow in
the lines connected to the A. The equivalent circuit must provide for a path from
the line on the Y side through the equivalent resistance and leakage reactance of
the transformer to the reference bus. An open circuit must exist between the line
and the reference bus on the A side. If the connection from neutral to ground
contains an impedance Z„, the zero-sequence equivalent circuit must have an
impedance of 3Z„ in series with the equivalent resistance and leakage reactance
of the transformer to connect the line on the Y side to ground.
Case 4: Y-A Bank, Ungrounded Y An ungrounded Y is a case where the

impedance Z„ between neutral and ground is infinite. The impedance 3Z„ in the
equivalent circuit of case 3 for zero-sequence impedance becomes infinite. Zero-
sequence current cannot flow in the transformer windings.
Case 5: A-A Bank Since a A circuit provides no return path for zero-
sequence current, no zero-sequence current can flow into a A-A bank, although it
can circulate within the A windings.
Zero-sequence equivalent circuits determined for various parts of the system

separately are readily combined to form the complete zero-sequence network.
Figures 11.19 and 11.20 show one-line diagrams of two small power systems and
their corresponding zero-sequence networks simplified by omitting resistances
and shunt admittances.
<y%
oy
Reference bus
-*W'J
Figure 11.19 One-line diagram of a small power system and the corresponding zero-sequence
network.
YIlro^L
Q.
A SCL A
<'1% A rv~|^n> A
JTsJC
o V- £O
w ■
m
<Yi xYY
Figure 11.20 One-line diagram of a small power system and the corresponding zero-sequence
network.
Example 11.4 Draw the zero-sequence network for the system described in
Example 6.10. Assume zero-sequence reactances for the generators and
motors of 0.05 per unit. Current limiting reactors of 0.4 Q each are in the
neutral of the generator and the larger motor. The zero-sequence reactance
of the transmission line is 1.5 Q/km.
Solution The zero-sequence leakage reactance of transformers is equal to

the positive-sequence reactance. So, for the transformers, X0 = 0.0857 per
unit and 0.0915 per unit, as in Example 6.10.
Zero-sequence reactances of the generator and motors are:
Generator: X0 = 0.05 per unit

2
3 00,
Motor 1: X0 = 0.05 _
0 200' o | = 0.0686 per unit
300,f 13.21 2
;
Motor 2: | = 0.1372 per unit

u>
bo
*°=°-05iJ
In the generator circuit
Base Z = = 1.333 Q
300
Reference bus
and in the motor circuit
Base Z = = 0.635 Q
300
In the impedance network for the generator
0.4
3Z„ = 3 -= 0.900 per unit
1.333
and for the motor
3Z„ = 3 U.oij = 1890 Per unit
For the transmission line
x0= {5\i(>Y = 0,5445 per unit

The zero-sequence network is shown in Fig. 11.21.
11.12 CONCLUSIONS
\
Unbalanced voltages and currents can be resolved into their symmetrical com¬
ponents. Problems are solved by treating each set of components separately and
superimposing the results.
In balanced networks having no coupling between phases the currents of
one phase sequence induce voltage drops of like sequence only. Impedances of
circuit elements to currents of different sequences are not necessarily equal.
A knowledge of the positive-sequence network is necessary for load studies
on power systems, for fault calculations, and for stability studies. If the fault
calculations or stability studies involve unsymmetrical faults on otherwise sym¬
metrical systems, the negative- and zero-sequence networks are needed also.
Synthesis of the zero-sequence network requires particular care, because the
zero-sequence network may differ from the others considerably.
PROBLEMS
11.1 Evaluate the following expressions in polar form:

(a) a- 1
(b) 1 — a2 + a
(c) a2 + a +j
(d) ja + a2
11.2 If Vani = 50/0°, Van2 = 20/90°. and Van0 = 10/l80° V. determine analytically the voltages to
neutral Van, Vbn and Vc„, and also show graphically the sum of the given symmetrical components
which determine the line-to-neutral voltages.
113 When a generator has terminal a open and the other two terminals are connected to each other
with a short circuit from this connection to ground, typical values for the symmetrical components of
current in phase a are Ial = 600/-90° A, Ia2 = 250/90° A, and Ia0 = 350/90° A. Find the current
into the ground and the current in each phase of the generator.
11.4 Determine the symmetrical components of the three currents /„ = l()/o°, Ib — 10/230°, and
Ic = 10/l30° A.
11.5 The currents flowing in the lines toward a balanced load connected in A are Ia = 100/0°
/;,= 141.4/225°, and Ic = 100/90° A. Determine a general expression for the relation between the
symmetrical components of the line currents flowing toward a A-connected load and the phase
currents in the load, that is, between Ial and Iabl and between Ia2 and Iab2. Start by drawing phasor
diagrams of the positive- and negative-sequence line and phase currents. Find Iab in amperes from
the symmetrical components of the given line currents.
11.6 The voltages at the terminals of a balanced load consisting of three 10-0 resistors connected in
Y are Vab = 100/0°, Vbc = 80.8 /— 121.44°, and Vca = 90/l30° V. Determine a general expression for
the relation between the symmetrical components of the line and phase voltages, that is, between Vabl
and Vml and between Vab2 and Vm2. Assume that there is no connection to the neutral of the load.
Find the line currents from the symmetrical components of the given line voltages.
11.7 Find the power expended in the three 10-0 resistors of Prob. 1F6 from the symmetrical com¬
ponents of currents and voltages. Check the answer.
11.8 Three single-phase transformers are connected as shown in Fig. 11.22 to form a Y-A transfor¬
mer. The high-tension windings are Y-connected with polarity marks as indicated. Magnetically
coupled windings are drawn in parallel directions. Determine the correct placement of polarity
marks on the low-tension windings. Identify the numbered terminals on the low-tension side (a) with
the letters a, b, and c where lAl leads lal by 30° and (b) with the letters a', b', and c! so that Ia i is 90°
out of phase with IAl.

11.9 Assume that the currents specified in Prob. 11.5 are flowing toward a load from lines connected
to the Y side of a A-Y transformer rated 10 MVA, 13.2A/66Y kV. Determine the currents flowing in
the lines on the A side by converting the symmetrical components of the currents to per unit on the
base of the transformer rating and by shifting the components according to Eq. (11.23). Check the
results by computing the currents in each phase of the A windings in amperes directly from
the currents on the Y side by multiplying by the turns ratio of the windings. Complete the check by
computing the line currents from the phase currents on the A side.
■
11.10 Balanced three-phase voltages of 100 V line-to-line are applied to a Y-connected load consisting
of three resistors. The neutral of the load is not grounded. The resistance in phase a is 10 ft, in phase
b is 20 ft, and in phase c is 30 ft. Select voltage to neutral of the three-phase line as reference and
determine the current in phase a and the voltage Van.
11.11 Draw the negative- and zero-sequence impedance networks for the power system of Prob. 6.15.
Mark the values of all reactances in per unit on a base of 50 MVA, 13.8 kV in the circuit of generator
l. Letter the networks to correspond to the one-line diagram. The neutrals of generators 1 and 3 are
connected to ground through current-limiting reactors having a reactance of 5%, each on the base of
the machine to which it is connected. Each generator has negative- and zero-sequence reactances of
20 and 5%, respectively, on its own rating as base. The zero-sequence reactance of the transmission
line is 210 ft from B to C and 250 ft from C to E.
11.12 Draw the negative- and zero-sequence impedance networks for the power system of Prob. 6.16.
Choose a base of 50 MVA, 138 kV in the 40-ft transmission line, and mark all reactances in per unit.
The negative-sequence reactance of each synchronous machine is equal to its subtransient reactance.
The zero-sequence reactance of each machine is 8% based on its own rating. The neutrals of the
machines are connected to ground through current-limiting reactors having a reactance of 5%, each
on the base of the machine to which it is connected. Assume that the zero-sequence reactances of the
transmission lines are 300% of their positive-sequence reactances.
CHAPTER
TWELVE
UNSYMMETRICAL FAULTS
Most of the faults that occur on power systems are unsymmetrical faults, which
may consist of unsymmetrical short circuits, unsymmetrical faults through im¬
pedances, or open conductors. Unsymmetrical faults occur as single line-to-
ground faults, line-to-line faults, or double line-to-ground faults. The path of the
fault current from line to line or line to ground may or may not contain
impedance. One or two open conductors result in unsymmetrical faults, either
through the breaking of one or two conductors or through the action of fuses
and other devices that may not open the three phases simultaneously.
Since any unsymmetrical fault causes unbalanced currents to flow in the
system, the method of symmetrical components is very useful in an analysis to
determine the currents and voltages in all parts of the system after the occur¬
rence of the fault. First, we shall discuss faults at the terminals of an unloaded
generator. Then we shall consider faults on a power system by applying Theven-
in’s theorem, which allows us to find the current in the fault by replacing the
entire system by a single generator and series impedance. Finally, we shall in¬
vestigate the bus impedance matrix as applied to the analysis of unsymmetrical
faults.
Regardless of the type of fault which occurs at the terminals of a generator
we can apply Eqs. (11.41) to (11.43), derived in Sec. 11.8. In matrix form these
equations become
Ko 'o' Z0 0 0 ' IaO

Ki — Ea — 0 Zi 0 hi (12.1)
va2 0 0 0 z2 Ja2.
305
For each type of fault we shall use Eq. (12.1), together with equations that
describe conditions at the fault, to derive Ial in terms of Ea, Zu Z2, and Z0.
12.1 SINGLE LINE-TO-GROUND FAULT ON AN

UNLOADED GENERATOR
The circuit diagram for a single line-to-ground fault on an unloaded Y-

connected generator with its neutral grounded through a reactance is shown in
Fig. 12.1, where phase a is the one on which the fault occurs. The relations to be
developed for this type of fault will apply only when the fault is on phase a, but
this should cause no difficulty since the phases are labeled arbitrarily and any
phase can be designated as phase a. The conditions at the fault are expressed by
the following equations:
4=0 Ic= 0 Fa = 0
With Ib — 0 and Ic = 0 the symmetrical components of current are given by
Uo 1 1 Ia
Ll 1 a a 0
hi 1 a2 a 0
so that Ia0, Ial, and Ia2 each equal Ia/3 and
Ial Ial IaO (12.2)

Substituting Ial for Ia2 and Ia0 in Eq. (12.1), we obtain
VaO 0 ' Z0 0 0 ' Ial
Kl =
Ea
— 0 Zi 0 Ial (12.3)
Kl 0 0 0 z2 Ial
Performing the indicated matrix multiplication and subtraction yields an

equality of two column matrices. Premultiplying both column matrices by the
row matrix [1 1 1] gives
Ko + Kl + Kl = - Ial Z0 + Ea - Ial Z1 - IaX Z2 (12.4)
Figure 12.1 Circuit diagram for a single line-to-

ground fault on phase a at the terminals of an
unloaded generator whose neutral is grounded
through a reactance.
UNSYMMETRICAL FAULTS 307
Figure 12.2 Connection of the sequence networks of an

unloaded generator for a single line-to-ground fault on
phase a at the terminals of the generator.
Since Va = Va0 + Val + Va2 = 0, we solve Eq. (12.4) for Ial and obtain
Eg
/ a 1 (12.5)
Z\ + Z2 + Z0
Equations (12.2) and (12.5) are the special equations for a single line-to-
ground fault. They are used with Eq. (12.1) and the symmetrical-component
relations to determine all the voltages and currents at the fault. If the three
sequence networks of the generator are connected in series, as shown in
Fig. 12.2, we see that the currents and voltages resulting therefrom satisfy the
equations above, for the three sequence impedances are then in series with the
voltage Ea. With the sequence networks so connected, the voltage across each
sequence network is the symmetrical component of Va of that sequence.
The connection of the sequence networks as shown in Fig. 12.2 is a convenient
means of remembering the equations for the solution of the single line-to-ground
fault, for all the necessary equations can be determined from the sequence-
network connection.
If the neutral of the generator is not grounded, the zero-sequence network is
open-circuited and Z0 is infinite. Since Eq. (12.5) shows that Ial is zero when Z0
is infinite, Ia2 and Ia0 must be zero. Thus no current flows in line a since Ia is the
sum of its components, all of which are zero. The same result can be seen
without the use of symmetrical components since inspection of the circuit shows
that no path exists for the flow of current in the fault unless there is a ground at
the generator neutral.
Example 12.1 A salient-pole generator without dampers is rated 20 MVA,

13.8 kV and has a direct-axis subtransient reactance of 0.25 per unit. The
negative- and zero-sequence reactances are, respectively, 0.35 and 0.10 per
unit. The neutral of the generator is solidly grounded. Determine the sub¬
transient current in the generator and the line-to-line voltages fof sub-
transient conditions when a single line-to-ground fault occurs at the
generator terminals with the generator operating unloaded at rated voltage.
Neglect resistance.
Solution On a base of 20 MV A, 13.8 kV, Ea= 1.0 per unit since the inter¬
nal voltage is equal to the terminal voltage at no load. Then, in per unit,
1-0 +;Q ,'1 A'I

Ial =
7. 4- 7, + 7,
Ia — 3/„i = — 74.29 per unit
„ 20,000
Base current = —7=- = 837 A
^3 x 13.8
Subtransient current in line a is
Ia = — 74.29 x 837 = — y‘3,590 A
The symmetrical components of the voltage from point a to ground are
Ki =Ea- Ial Z! = 1.0 - (—^1.43)00.25)
= 1.0 — 0.357 = 0.643 per unit
Va2 = -Ia2Z2 = — (—yl.43)(j0.35) = —0.50 per unit
Vao = ~7a0Z0 = — (— ;1.43)(70.10) = —0.143 per unit
Line-to-ground voltages are
K = Val + Va2 + Va0 = 0.643 - 0.50 - 0.143 = 0
K = a2Ki + aK2 + Va0
= 0.643( —0.5 -;0.866) - 0.50(—0.5 + yO.866) - 0.143
= -0.322 -70.557 + 0.25 - 70.433 - 0.143
= —0.215 — 70.990 per unit
K = aKi + a2Ki + Ko
= 0.643( —0.5 + 70.866) - 0.50( —0.5 -70.866) - 0.143
= -0.322 + 70.557 + 0.25 + 70.433 - 0.143
= —0.215 + 70.990 per unit
Line-to-line voltages are
Kb = K~K = 0.215 +70.990 = 1.01/77.7° per unit

Kc =K-Vc = 0 -71.980 = 1.980/270° per unit
Ka =K~K = -0.215 +70.990 = 1.01 A02.3° per unit

Figure 12.3 Phasor diagrams of the line voltages

of Example 12.1 before and after the fault. (a) Prefault (6) Postfault
Since the generated voltage-to-neutral Ea was taken as 1.0 per unit, the
above line-to-line voltages are expressed in per unit of the base voltage to
neutral. Expressed in volts, the postfault line voltages are
13.8
Vab = 1.01 x -W77.7° = 8.05/77,7° kV
13.8
Vbc = 1.980 x -W270° = 15.78/270° kV
13.8
Vca = 1.01 x —^/102.3° = 8.05/102.3° kV
Before the fault the line voltages were balanced and equal to 13.8 kV. For
comparison with the line voltages after the fault occurs, the prefault voltages,
with Vm = Ea as reference, are given as
Vab = 13.8/301 kV Vbc = 13.8/270° kV Vca = 13.8 kV
The phasor diagrams of prefault and postfault voltages are shown in
Fig. 12.3.
12.2 LINE-TO-LINE FAULT ON AN

UNLOADED GENERATOR
The circuit diagram for a line-to-line fault on an unloaded, Y-connected genera¬

tor is shown in Fig. 12.4 with the fault on phases b and c. The conditions at the
fault are expressed by the following equations:
K=VC Ia — 0 h = -J,
With Vb = Vc the symmetrical components of voltage are given by
l 1 V„
a a2
a2
a
Figure 12.4 Circuit diagram for a line-to-line fault be¬

tween phases b and c at the terminals of an unloaded
generator whose neutral is grounded through a
reactance.
from which we find
Ki = Vm2 (12.6)
Since Ib = — Ic and Ia = 0, the symmetrical components of current are given by
l aO 1 1
/at a 32
Ia2 a
and therefore
I aO=0 (12.7)
la2 l aX (12.8)
With a connection from the generator neutral to ground, Z0 is finite, and so
Va0 = 0 (12.9)
since Ia0 is zero by Eq. (12.7).
Equation (12.1), with the substitutions allowed by Eqs. (12.6) to (12.9),
becomes
■ 0 ‘ 'O' Z0 0 0 ' 0 '
Ki = Ea — 0 Zi 0 lax (12.10)
Vai 0 0 0 z2 -lax
Performing the indicated matrix operations and premultiplying the resulting

matrix equation by the row matrix [1 1 -1] gives
0 — Ea Ial — Ial Z2 (12.11)

and solving for Ial yields
Iai = (12.12)
Zt + Z2
Equations (12.6) to (12.8) and (12.12) are the special equations for a line-to-
line fault. They are used with Eq. (12.1) and the symmetrical-component rela¬
tions to determine all the voltages and currents at the fault. The special
Figure 12.5 Connection of the sequence networks of

an unloaded generator for a line-to-line fault be¬
tween phases b and c at the terminals of the
generator.
equations indicate how the sequence networks are connected to represent the
fault. Since Z0 does not enter into the equations, the zero-sequence network is
not used. The positive- and negative-sequence networks must be in parallel since
Val — K2 ■ The parallel connection of the positive- and negative-sequence
networks without the zero-sequence network makes Ial = — Ia2, as specified by
Eq. (12.8). The connection of the sequence networks for a line-to-line fault is
shown in Fig. 12.5. The currents and voltages in the sequence networks, when so
connected, satisfy all the equations derived for the line-to-line fault.
Since there is no ground at the fault, there is only one ground in the circuit
(at the generator neutral) and no current can flow in the ground. In the deriva¬
tion of the relations for the line-to-line fault we found that Ia0 = 0. This is
consistent with the fact that no ground current can flow, since the ground
current /„ is equal to 3Ia0. The presence or absence of a grounded neutral at the
generator does not affect the fault current. If the generator neutral is not
grounded, Z0 is infinite and Va0 is indeterminate, but line-to-line voltages may be
found since they contain no zero-sequence components.
Example 12.2 Find the subtransient currents and the line-to-line voltages at
the fault under subtransient conditions when a line-to-line fault occurs at the
terminals of the generator described in Example 12.1. Assume that the gener¬
ator is unloaded and operating at rated terminal voltage when the fault
occurs. Neglect resistance.
Solution
1-0+J0
Ial
—j 1.667 per unit
j0.25 + ;0.35
Ial — Ial = j1.667 per unit
I aO 0
la Ial + Ial + Iao = ~jI-667 + jl.667 = 0
h a2Ia\ + alai + Iao

—/1.667( —0.5 — y‘0.866) + ;1.667(-0.5 + y0.866)
y'0.833 - 1.443 -;0.833 - 1.443 = -2.886 + jO per unit
Ic -Ib = 2.886 -I- jO per unit

As in Example 12.1, base current is 837 A, and so
/. = o
Ib = -2.886 x 837 = 2416/180° A
/c = 2.886 x 837 = 2416/01A
The symmetrical components of the voltage from a to ground are
Val = Va2 = 1 - (—/1.667)(;0.25) = 1 - 0.417 = 0.583 per unit
Ko = 0 (neutral of generator grounded)
Line-to-ground voltages are
va = Val + Va2 + Va0 = 0.583 + 0.583 = 1.166/01 per unit
K — a2Ki + aKi + va0

Vc=Vb = 0.583( —0.5 — y'0.866) + 0.583(-0.5 + j0.866)
= -0.583 per unit
Line-to-line voltages are
vab = Va - Vb = 1.166 + 0.583 = 1.749/T per unit

Vbc =Vb-Vc= -0.583 + 0.583 = 0 per unit
Vca =VC-Va= -0.583 - 1.166 = 1.749/180° per unit
Expressed in volts, the line-to-line voltages are
Vab = 1.749 x^!= 13.94/bl kV

73
Vbc = 0 kV
13.8
Vca - - 1.749 x = 13.94/180° kV
12.3 DOUBLE LINE-TO-GROUND FAULT

ON AN UNLOADED GENERATOR
The circuit diagram for a double line-to-ground fault on an unloaded, Y-

connected generator having a grounded neutral is shown in Fig. 12.6. The
faulted phases are b and c. The conditions at the fault are expressed by the
following equations:
K=o vc = o ia = o
Ia
a
Figure 12.6 Circuit for a double line-to-ground fault

on phases b and c at the terminals of an unloaded
generator whose neutral is grounded through a
reactance.
With Vb = 0 and Vc = 0, the symmetrical components of voltage are given by
Ko 11 1' K
_ 1
Ki 1 a a2 0
_ 3 0
Val l a2 a
Therefore Va0, Val, and Va2 equal Ka/3, and
K i = va2 = va0 (12.13)
Substituting £a - /fllZ, for Val, Va2, and Va0 in Eq. (12.1) and premultiplying
both sides by Z~\ where
1
Zo 0 0 ' -1 0 0
1
0 Zi 0 = 0 0
Zt
1
0 0 z2 0 0
Zi_
give
0 0
Z~o Zo
0 0 0 Ial (12.14)
J_
0 0 0 hi
Z^
Premultiplying both sides of Eq. (12.14) by the row matrix [1 1 1] and recog¬
nizing that Ial -I- Ia2 + Ia0 = Ia = 0, we have
E* (12.15)
Zo
and upon collecting terms we obtain
, /, . z- . Z.) g.(Z2 + Zo) (12.16)

“V z„ zj Z2Z„
Figure 12.7 Connection of the sequence networks of an unloaded generator for a double line-to-
ground fault on phases b and c at the terminals of the generator.
and
Ea{^2 + z0) (12.17)

Z,Z2 + ZjZ0 + Z2Z0 Zl + Z2Z0/(Z2 + Z0)
Equations (12.13) and (12.17) are the special equations for a double line-to-
ground fault. They are used with Eq. (12.1) and the symmetrical-component
relations to determine all the voltages and currents at the fault. Equation (12.13)
indicates that the sequence networks should be connected in parallel, as shown
in Fig. 12.7, since the positive-, negative-, and zero-sequence voltages are equal
at the fault. Examination of Fig. 12.7 shows that all the conditions derived above
for the double line-to-ground fault are satisfied by this connection. The diagram
of network connections shows that the positive-sequence current Ial is
determined by the voltage Ea impressed on Zx in series with the parallel combin¬
ation of Z2 and Z0. The same relation is given by Eq. (12.17).
In the absence of a ground connection at the generator no current can flow
into the ground at the fault. In this case Z0 would be infinite and Ia0 would be
zero. Insofar as current is concerned, the result would be the same as in a
line-to-line fault. Equation (12.17) for a double line-to-ground fault approaches
Eq. (12.12) for a line-to-line fault as Z0 approaches infinity, as may be seen by
dividing the numerator and denominator of the second term in the denominator
of Eq. (12.17) by Z0 and letting Z0 be infinitely large.
Example 12.3 Find the subtransient currents and the line-to-line voltages at
the fault under subtransient conditions when a double line-to-ground fault
occurs at the terminals of the generator described in Example 12.1. Assume
that the generator is unloaded and operating at rated voltage when the fault
occurs. Neglect resistance.
Solution
1.0 +;0
Iai =
Zi + Z2Z0/(Z2 + Z0) j0.25 + (jO.35 x ;0.10)/(/0.35 + y'0.10)
1.0 1.0
= —j3.05 per unit
j0.25 + ;0.0778 j0.3278
Ki = Ea-I .iZ, = l -(-73.05)00.25)
II
II
(N
o
= 1.0 - 0.763 = 0.237 per unit
Va2 0.237
= j0.68 per unit
Z2 ~ j0.35
Ko 0.237
IaO = 7*2.37 per unit
‘ 20 ~ J0.10
K = Kl + I a2 + I aO -7*3.05 + ;0.68 + 7*2.37 = 0
h= a2Ial + ala 2 3” Ko
= (— 0.5 — ;'0.866)( — ;3.05) + (-0.5 + ;0.866)(/0.68) + ;2.37
= 7*1.525 - 2.641 -;0.34 - 0.589 + j2.37
= -3.230 + 7*3.555 = 4.80/132.3° per unit
Ic = alal + a2Ia2 + /a0

= (-0.5 + J0.866)(—7*3.05) + (-0.5 - j0.866)(/0.68) + 7*2.37
= 7*1.525 + 2.641 - j0.34 + 0.589 + 7*2.37 = 3.230 + 7*3.555
= 4.80/47.7° per unit
In = 3Ia0 = 3 x ;2.37 =7*7.11 per unit

jn = Ib + Ic = -3.230 + 7*3.555 4- 3.230 + ;3.555 = 7*7.11 per unit
K = Ki + Va2 + Ko = 3Val = 3 x 0.237 = 0.711 per unit

Vb = Vc = 0
Vab = K - K = 0.711 per unit

vbe = o
Ka = K - K — -0.711 per unit
Expressed in amperes and volts,
/. = 0
lb = 837 x 4.80/132.3° = 4017/132.3° A
Ic = 837 x 4.80/47.7° = 4017/47.7° A
/„ = 837 x 7.11/90° -- 5951/90° A
13.8
Kb = 0.711 X 5.66/01 kV
73
Kc = o
V„ = -0.711 x ~ = 5.66/mi kV
12.4 UNSYMMETR1CAL FAULTS ON POWER SYSTEMS
In the derivation of equations for the symmetrical components of currents and

voltages in a general network during a fault, we shall designate as Ia, Ib, and Ic
the currents flowing out of the original balanced system at the fault from phases
a, b, and c, respectively. We can visualize the currents Ia, Ib, and Ic by referring
to Fig. 12.8, which shows the three lines of the three-phase system at the part of
the network where the fault occurs. The flow of current from each line into the
fault is indicated by arrows shown on the diagram beside hypothetical stubs
connected to each line at the fault location. Appropriate connections of the stubs
represent various types of faults. For instance, connecting stubs b and c produces
a line-to-line fault through zero impedance. The current in stub a is then zero,
and Ib is equal to — Ic.
The line-to-ground voltages at the fault will be designated Va, Vb, and Vc.
Before the fault occurs, the line-to-neutral voltage of phase a at the fault will be
called Vf, which is a positive-sequence voltage since the system is assumed to be
balanced. We met the prefault voltage Vf previously in Chap. 10 in calculations
to determine the currents in a power system when a symmetrical three-phase
fault occurred.
A single-line diagram of a power system containing three synchronous
machines is shown in Fig. 12.9. Such a system is sufficiently general for equa¬
tions derived therefrom to be applicable to any balanced system regardless of the
complexity. Figure 12.9 also shows the sequence networks of the system. The
point where a fault is assumed to occur is marked P on the single-line diagram
and on the sequence networks. As we saw in Chap. 10, the load current flowing
in the positive-sequence network is the same, and the voltages to ground exter¬
nal to the machines are the same, regardless of whether the machines are repre¬
sented by their subtransient internal voltages and their subtransient reactances,
by their transient internal voltages and their transient reactances, or by their no
load voltages and their synchronous reactances.
Since linearity is assumed in drawing the sequence networks, each of the
networks can be replaced by its Thevenin equivalent between the two terminals
composed of its reference bus and the point of application of the fault. The
Thevenin equivalent circuit of each sequence network is shown adjacent to
the diagram of the corresponding network in Fig. 12.9. The internal voltage of
the single generator of the equivalent circuit for the positive-sequence network is
Vf, the prefault voltage to neutral at the point of application of the fault. The
Ml
b
Figure 12.8 Three conductors of a three-phase sys¬
tem. The stubs carrying currents /„, Ib, and Ic may
Ml
C
be interconnected to represent different types of
faults. M
J?*"T (^)—
HF
jT t>
(a) One-line diagram of balanced three-phase system
(e) Thevenin
equivalent of the positive-
sequence network
h2
(c) Negative-sequence network (f) Thevenin
equivalent of the negative-
sequence network
Tao
(of) Zero-sequence network (g) Thevenin

equivalent of the zero-
sequence network
Figure 12.9 One-line diagram of a three-phase system, the three sequence networks of the system,
and the Thevenin equivalent of each network for a fault at P.
impedance Zx of the equivalent circuit is the impedance measured between point

P and the reference bus of the positive-sequence network with all the internal
emfs short-circuited. The value of Zj is dependent on the reactances used in the
network. We recall, for instance, that subtransient reactances of generators and
1.5 times the subtransient reactances of synchronous motors or the transient
reactances of the motors are the values used to calculate the symmetrical current
to be interrupted.
Since no negative- or zero-sequence currents are flowing before the fault
occurs, the prefault voltage between point P and the reference bus is zero in the
negative- and zero-sequence networks. Therefore, no emfs appear in the equiva¬
lent circuits of the negative- and zero-sequence networks. The impedances Z2
and Z0 are measured between point P and the reference bus in their respective
networks and depend on the location of the fault.
Since Ia is the current flowing from the system into the fault, its components
/ai, Ia2, and Ia0 flow out of their respective sequence networks and out of the
equivalent circuits of the networks at P, as shown in Fig. 12.9. The Thevenin
equivalents of the positive-, negative, and zero-sequence networks of the system
are the same as the networks of a single generator. The matrix equations for the
symmetrical components of voltages at the fault, therefore, must be the same as
Eq. (12.1) except that Vf replaces Ea; that is,
va0~ o' 'Zo 0 0 " ho

Ki =
Vf — 0 Zi 0 hi (12.18)
Ki 0 0 0 Z2 Ja2_
Of course, we must evaluate the sequence impedances properly according to

Thevenin’s theorem and realize that the currents are the sequence components in
the hypothetical stubs.
12.5 SINGLE LINE-TO-GROUND FAULT ON

A POWER SYSTEM
For a single line-to-ground fault, the hypothetical stubs on the three lines are
connected as shown in Fig. 12.10. The following relations exist at the fault:
Ib = 0 Ic = 0 Va = 0
These three equations are the same as those which apply to a line-to-ground
fault on a single generator. These equations with Eq. (12.18) and the relations of
symmetrical components must have the same solutions as are found for similar
equations in Sec. 12.1, except that Vf replaces Ea. Thus, for a line-to-ground
fault,
hi = hi = ho (12.19)
and
/a 1 Vf (12.20)
Zi + Z2 + z0
Equations (12.19) and (12.20) indicate that the three sequence networks should
be connected in series through the fault point in order to simulate a single
line-to-ground fault.
c
Figure 12.10 Connection diagram of the hypotheti¬
cal stubs for a single line-to-ground fault.
UNSYMMHTRICAL FAULTS 319
4j

cal stubs for a line-to-line fault. 4
12.6 LINE-TO-LINE FAULT ON A POWER SYSTEM
For a line-to-line fault, the hypothetical stubs on the three lines at the fault are
connected as shown in Fig. 12.11. The following relations exist at the fault:
vb=K ia = o i„=-ic
The above equations are identical in form to those which apply to a line-to-line
fault on an isolated generator. Their solution in the manner of Sec. 12.2, with
Eq. (12.18) replacing Eq. (12.1), yields
Ki = V.2 (1221)
/a 1 Vf (12.22)
Zi + z2
Equations (12.21) and (12.22) indicate that the positive- and negative-sequence
networks should be connected in parallel at the fault point in order to simulate a
line-to-line fault.
12.7 DOUBLE LINE-TO-GROUND FAULT ON A

POWER SYSTEM
For a double line-to-ground fault, the stubs are connected as shown in

Fig. 12.12. The following relations exist at the fault:
vb = K =o
la = 0
b
41
c Is
4J
cal stubs for a double line-to-ground fault.
By comparison with the derivation made in Sec. 12.3,
K i = va2 = K
aO
(12.23)
/a1 Vf (12.24)
Zx + Z2Z0/(Z2 + z0)
Equations (12.23) and (12.24) indicate that the three sequence networks should
be connected in parallel at the fault point in order to simulate a double line-to-
ground fault.
12.8 INTERPRETATION OF THE

INTERCONNECTED SEQUENCE NETWORKS
In the preceding sections we have seen that the sequence networks of a power
system can be so interconnected that solving the resulting network yields the
symmetrical components of current and voltage at the fault. The connections of
the sequence networks to simulate various types of faults, including a symmetri¬
cal three-phase fault, are shown in Fig. 12.13. The sequence networks are in¬
dicated schematically by rectangles enclosing a heavy line to represent the
reference bus of the network and a point marked P to represent the point in the
network where the fault occurs. The positive-sequence network contains emfs
that represent the internal voltages of the machines.
Pos.-seq. net.
Val
Pos.-seq. net. Neg.-seq. net.
Tvo2
_!£L_ _E_
i± ±L
(a) Three-phase fault Tal Ta2
(b) Single line-to-ground fault
Figure 12.13 Connections of the sequence networks to simulate various types of faults. The sequence
networks are indicated by rectangles. The point at which the fault occurs is P.
Reference bus Reference bus
Figure 12.14 Positive-sequence

(a) Positive-sequence network (b) Thevenin
network and its Thevenin
equivalent of the positive
equivalent. sequence network
If the emfs in a positive-sequence network like that shown in Fig. 12.14a are
replaced by short circuits, the impedance between the fault point P and the
reference bus is the positive-sequence impedance Zx in the equations developed
for faults on a power system and is the series impedance of the Thevenin equiva¬
lent of the circuit between P and the reference bus. If the voltage Vf is connected
in series with this modified positive-sequence network, the resulting circuit,
shown in Fig. 12.146, is the Thevenin equivalent of the original positive-sequence
network. The circuits shown in Fig. 12.14 are equivalent only in their effect on
any external connections made between P and the reference bus of the original
networks. We can easily see that no current flows in the branches of the equiva¬
lent circuit in the absence of an external connection, but current will flow in the
branches of the original positive-sequence network if any difference exists in the
phase or magnitude of the two emfs in the network. In Fig. 12.14a the current
flowing in the branches in the absence of an external connection is the prefault
load current.
When the other sequence networks are interconnected with the positive-
sequence network of Fig. 12.14a or its equivalent shown in Fig. 12.146, the cur¬
rent flowing out of the network or its equivalent is Ial and the voltage between P
and the reference bus is Val. With such an external connection, the current in any
branch of the original positive-sequence network is the positive-sequence current
in phase a of that branch during the fault. The prefault component of this
current is included. The current in any branch of the Thevenin equivalent of
Fig. 12.146, however, is only that portion of the actual positive-sequence current
found by apportioning Ial of the fault among the branches according to their
impedances and does not include the prefault component.
An alternate method of studying unsymmetrical faults is by means of the bus
impedance matrix. We shall discuss this method after we look at the following
example to become more familiar with the sequence networks.
Example 12.4 A group of identical synchronous motors is connected

through a transformer to a 4.16-kV bus at a location remote from the
generating plants of a power system. The motors are rated 600 V and oper¬
ate at 89.5 efficiency when carrying full load at unity power factor and rated
voltage. The sum of their output ratings is 4476 kW (6000 hp). The reac¬
tances in per unit of each motor based on its own input kVA rating are
X" = 0.20, X2 = 0.20, and X0 = 0.04 and each is grounded through a<reac-
tance of 0.02 per unit. The motors are connected to the 4.16-kV bus through
a transformer bank composed of three single-phase units, each of which is
rated 2400/600 V, 2500 kVA. The 600-V windings are connected in A to the
motors and the 2400-V windings are connected in Y. The leakage reactance
of each transformer is 10%.
The power system which supplies the 4.16 kV bus is represented by a
Thevenin equivalent generator rated 7500 kVA, 4.16 kV with reactances of
X" = X2 = 0.10 per unit, X0 = 0.05 per unit, and Xn from neutral to ground
equal to 0.05 per unit.
Each of the identical motors is supplying an equal share of a total load
of 3730 kW (5000 hp) and is operating at rated voltage, 85% power-factor
lag, and 88% efficiency when a single line-to-ground fault occurs on the
low-tension side of the transformer bank. Treat the group of motors as a
single equivalent motor. Draw the sequence networks showing values of the
impedances. Determine the subtransient line currents in all parts of the
system with prefault current neglected.
Solution The one-line diagram of the system is shown in Fig. 12.15. The
600-V bus and the 4.16 kV bus are numbered 1 and 2, respectively. Choose
the rating of the equivalent generator as base: 7500 kVA, 4.16 kV at the
system bus.
Since
y/3 x 2400 = 4160 V

and
3 x 2500 - 7500 kVA
the three-phase rating of the transformer is 7500 kVA, 4160Y/600A V. So the

base for the motor circuit is 7500 kVA, 600 V.
The input rating of the single equivalent motor is
6000 x 0.746
= 5000 kVA
0.895
® _
© KD YWl
-O Y®^
,E"r 0“
Equivalent Yêl
jfTP
generator -=r
Figure 12.15 One-line diagram of the system of Y^g_

Example 12.4.
Motors
Figure 12.16 Connection of the sequence

networks of Example 12.4. Subtransient cur¬
rents are marked in per unit for a single line-
to-ground fault at P. Prefault current is
included.
and the reactances of the equivalent motor in percent are the same on the
base of the combined rating as the reactances of the individual motors on
the base of the rating of an individual motor. The reactances of the equiva¬
lent motor on the selected base are
7500
X" = 0.2—— =0.3 per unit
5000 F
7500
X2 = 0.2—— = 0.3 per unit
2 5000 F
7500
X0 = 0.04—— = 0.06 per unit
0 5000 F
The reactance in the zero-sequence network to account for the reactance

between neutral and ground of the equivalent motor is
7500
3X„ = 3 x 0.02= 0.09 per unit
For the equivalent generator the reactance from neutral to ground in the
zero-sequence network is
3X„ = 3 x 0.05 = 0.15 per unit
Figure 12.16 shows the connection of the sequence networks. Reactances are
shown in per unit.
Since the motors are operating at rated voltage equal to the base voltage
of the motor circuit, the prefault voltage of phase a at the fault is
Vf = 1.0 per unit
Base current for the motor circuit is
7,500^000 _ 72i7 A
73 X 600
and the actual motor current is
746 x 5000
= 4798 A
0.88 x ^3 x 600 x 0.85
The per-unit current drawn by the motor through line a before the fault
occurs is
4798
/— cos~1 0.85 = 0.665 7-31.8° - 0.565 -70.350 per unit
72l7
If prefault current is neglected, £" and £" are made equal to 1.0/0°, or
the positive-sequence network is replaced by its Thevenin equivalent circuit
which is shown in Fig. 12.17. The computations follow.
(,/0.1 +,/0.1)(,/Q-3)
= y'0.12 per unit
1 ;(0.1 + 0.1 + 0.3)
(70.1 +70-1X70.3)
= 70.12 per unit
2 7(0.1+0.1+0.3)
Z0 = jO. 15 per unit
1.0 1.0
/ V' -72.564
al Zj + Z2 + Z0 70.12 + 70.12 + 70.15 7039
L2 = Li = -72.564
Lo = Iai = -72.564
Current in the fault = 3Ia0 = 3(-72.564) = —77.692 per unit. The compo¬
nent of Ial flowing toward P from the transformer is
-72.564 x 70.30
= -71.538
70.50
Reference bus
vs
S
y'O.lO § 3; 0.30
■nm^
Figure 12.17 Thevenin equivalent of the positive-sequence j 0.10
network of Example 12.4.
and the component of 7a] flowing from the motor toward P is
-j2.564 x jO.20
-j 1-026
jO.50
Similarly the component of Ia2 from the transformer is -/1.538, and the
component of Ia2 from the motor is —j1.026. All of Ia0 flows toward P from
the motor.
Currents in the lines at the fault are:
To P from the transformer in per unit:
la 11 1 ' ' 0 —j'3.076"

h 1 a2 a —>1.538 — >1.538
Ic 1 a a2 ->1-538 >1.538
To P from the motors in per unit:
la li r -J 2.564 —>4.616
.•00*00
h — 1 a2 a -j 1.026 — 1
Ic 1 a a2 —jf 1.026
1
1
Our method of labeling the lines is such that currents IA1 and IA2 on
the high-tension side of the transformer are related to the currents 7al and
Ia2 on the low-tension side by
I al =jÂl I a2 ~ ~jÂ2 (12.25)

So,
Iai = —7(—jl-538) = —1.538

IA2 == —jl.538) = 1.538
and IA0 = 0
since there are no zero-sequence currents on the high-voltage side of the

transformer. Then
â = Cu + 1a2 — 0
IB1 = a2IAl = (-0.5 — y'0.866)( —1.538) = 0.769 + >1.332
IB2 = aIA2 = (-0.5 + ;0.866)(1.538) = -0.769 + >1.332
Ig — /gi -(- IB2 = 0 + ;2.664 per unit
7C1 = aIAl = (-0.5 + >0.866)(-1.538) = 0.769 ->1.332
/C2 - a2IA2 = (-0.5 — >0.866)(1.538) = -0.769 —>1.332
Ic = Ic 1 + Ic2 — 0 — j2.664 per unit

If voltages throughout the system are to be found, their components at

any point can be calculated from the currents and reactances of the sequence
networks. Components of voltages on the high-voltage side of the transformer
are found first without regard for phase shift. Then the effect of phase shift
must be determined.
By evaluating the base currents on the two sides of the transformer we
can convert the above per-unit currents to amperes. Base current for the
motor circuit was found previously and equals 7217 A. Base current for the
high-voltage circuit is
7,500,000
= 1041 A
•v/3 x 4160
Current in the fault is
7.692 x 7217 = 55,500 A
Currents in the lines between the transformer and the fault are:
In line a: 3.076 x 7217 = 22,200 A
In line b: 1.538 x 7217 = 11,100 A
In line c: 1.538 x 7217 = 11,100 A
Currents in the lines between the motor and the fault are:
In line a: 4.616 x 7217 = 33,300 A
In line b: 1.538 x 7217 = 11,100 A
In line c: 1.538 x 7217 = 11,100 A
Currents in the lines between the 4.16 kV bus and the transformer are:
\
In line A: 0
In line B: 2.664 x 1041 = 2773 A
In line C: 2.664 x 1041 = 2773 A
The currents we have calculated are those which would flow upon the
occurrence of a fault when there is no load on the motors. These currents are
correct only if the motors are drawing no current whatsoever. The statement
of the problem specifies the load conditions at the time of the fault, however,
and the load can be considered. To account for the load, we add the per-unit
current drawn by the motor through line a before the fault occurs to the
component of Ial flowing toward P from the transformer and subtract the
same current from the component of Ial flowing from the motor to P. The
Figure 12.18 Per-unit values of subtransient line currents in all parts of the system of Example 12.4,
prefault current neglected.
new value of positive-sequence current from the transformer to the fault in

phase a is
0.565 -y0.350 - jl.538 = 0.565 - jl.888
and the new value of positive-sequence current from the motor to the fault in
phase a is
-0.565 + y'0.350 -71.026 = -0.565 -j0.676
These values are shown in Fig. 12.16. The remainder of the calculation,
using these new values, proceeds as in the example.
Figure 12.18 gives the per-unit values of subtransient line currents in all
parts of the system when the fault occurs at no load. Figure 12.19 shows the
values for the fault occurring on the system when the load specified in the
example is considered. In a larger system where the fault current is much
higher in comparison with the load current, the effect of neglecting the load
current is less than is indicated by comparing Figs. 12.18 and 12.19. In the
large system, however, the prefault currents determined by a load-flow study
could simply be added to the fault current found with the load neglected.
0.665-y'2.685 -0.586+71.224
Figure 12.19 Per-unit values of subtransient line currents in all parts of the system of Example 12.4,
prefault current considered.
12.9 ANALYSIS OF UNSYMMETRICAL FAULTS

USING THE BUS IMPEDANCE MATRIX
In Chap. 10 we used the bus impedance matrix composed of positive-sequence

impedances to determine currents and voltages upon the occurrence of a three-
phase fault. The method can be extended easily to apply to unsymmetrical faults
Figure 12.20 Connections of the bus impedance equivalent sequence networks of a three-bus system
to simulate various types of faults. Transfer impedances not shown.
by realizing that the negative- and zero-sequence networks can be represented by

bus impedance equivalent networks just as the positive-sequence network was.
Figure 12.20 corresponds to Fig. 12.13 and shows the interconnection of the bus
impedance networks for a three-bus system with the fault on bus 3. The actual
networks have merely been replaced by the bus impedance equivalent network.
The additional subscripts 1, 2, and 0 have been attached to the impedances to
identify the sequence networks to which they belong.
For the single line-to-ground fault on bus 3 examination of Fig. 12.20 shows
Ial = (12.26)
'33 ■1 + Z 33-2 + z 33-0
which should be compared with Eq. (12.20). Obviously Z33_x, Z33_2, and
Z33_0 are equal to the values of Zl5 Z2, and Z0 of Eq. (12.20) if the fault is on
bus 3. The positive-, negative-, and zero-sequence bus impedance matrices enable
us to see immediately the values to be used for Zu Z2, and Z0 in Eqs. (12.20),
(12.22), and (12.24). The transfer admittances (not shown in Fig. 12.20) enable us
to calculate the voltages at the unfaulted buses, from which the currents in the
lines are found.
Example 12.5 Solve for the subtransient current in a single line-to-ground

fault first on bus 1 and then on bus 2 of the network of Example 12.4. Use
the bus impedance matrices. Also find the voltages to neutral at bus 2 with
the fault on bus 1.
Solution We refer to Fig. 12.16 to find the elements of the node admittance
matrices of the three sequence networks, as follows:
y
Ui-i n-2■> = —-|—-
,, . = y,, ai 0_3 = -/'13.3
J
;0.1 jO.
^12-1 = *12-2 = -1 =jio

;'0.i
^22- 1 1 +i = -j20
yO.l jO.l
1
^11-0 = —j'6.67 F12-o = 0
A15
1 1
-j 15.0
Y22-°~]02 + jO.l
-13.3 10.0
^bus— 1 = Y bus - 2 = J
10.0 - 20.0
-6.67 0.0
^bus — 0 =J 0.0 -15.0
Inverting the three matrices above gives the three bus impedance
matrices
0.12 0.06
^bus-l ^bus—2 j
0.06 0.08
0.150 0.0
Zbus-0 ~j
0.0 0.067
The current in the fault on bus 1 is
r 3 x 1.0
= — /7.692 per unit
f ;0.12+;0.12+;'0.15
which agrees with the value found in Example 12.4. If the fault is on bus 2,
3 x 1.0
= — y'13.216 per unit
If y'0.08 + y0.08 + ;0.067
To find the voltages to neutral at bus 2 with the fault at bus 1 we

observe first that
/. i = 1.2 = l.o = = -/2.564
By studying Fig. 12.20 and realizing that transfer impedances are present we
see that at bus 2 with the fault on bus 1 and neglecting phase shift the
sequence components of voltage are
Vai = Vf- IalZ21_x = 1 - (-;2.564)00.06) = 0.8462
Ki =-Ia2Z 2i-2= -(-y2.564)00.06)= -0.1538

and, since Z21_0 is zero,
VaO = 0
Accounting for phase shift we have from Eqs. (11.23)
Vai = -JKi = -70.8462

VA2=jK2 = -70.1538
VA = yAi + yA2 = —7’0.8462 — 7'0.1538 = 0 — jl.OOO per unit

yB = a2yAi + aVA2 -70.4231 - 0.7328 + 7'0.0769 + 0.1332
= —0.600 + 7'0.500 per unit
Vc = aVAi + a2VA2 -7'0.4231 + 0.7328 + 70.0769 - 0.1332

= 0.600 -|- 7O.5OO per unit
All per-unit values are on a line-to-neutral base. Examination of Fig. 12.18

will show why the magnitude VA is 1.0.
Although the method of solution using the bus impedance matrix does not
appear to have any great advantage over the method of Example 12.4 in this
very simple network, it does give us the current for the fault on each of the buses.
For a large network the method is well suited to the digital computer, which can
build the bus impedance matrix directly and add or remove particular lines quite
easily. Thus, with the bus impedance matrix for each sequence network all the
features of digital-computer solutions for symmetrical three-phase faults can be
extended to unsymmetrical faults.
12.10 FAULTS THROUGH IMPEDANCE
All the faults discussed in the preceding sections consisted of direct short circuits
between lines and from one or two lines to ground. Although such direct short
circuits result in the highest value of fault current and are therefore the most
conservative values to use when determining the effects of anticipated faults, the
fault impedance is seldom zero. Most faults are the result of insulator flashovers,
where the impedance between the line and ground depends on the resistance of
the arc, of the tower itself, and of the tower footing if ground wires are not used.
Tower-footing resistances form the major part of the resistance between line and
ground and depend on the soil conditions. The resistance of dry earth is 10 to
100 times the resistance of swampy ground. The effect of impedance in the fault
is found by deriving equations similar to those for faults through zero
impedance. Connections of the hypothetical stubs for faults through impedance
are shown in Fig. 12.21.
A balanced system remains symmetrical after the occurrence of a three-phase
fault having the same impedance between each line and a common point. Only
(a) Three-phase fault (6) Single line-to-ground fault
4| Ia\
h\ M
JZf Zf
(c) Line-to-line fault (d) Double line-to-ground fault
Figure 12.21 Connection diagrams of the hypothetical stubs for various faults through impedance.
(b) Single line-to-ground fault
Figure 12.22 Connections of the sequence networks to simulate various types of faults through
impedance at point P.
positive-sequence currents flow. With the fault impedance Zf equal in all phases,
as shown in Fig. 12.21a, the voltage at the fault is
K = i.zf
and since only positive-sequence currents flow,
Val = IalZf=Vf-IalZ1
and
/a 1 Vf (12.27)
Z\ + zf
The sequence-network connection is shown in Fig. 12.22a.
A formal derivation can be made for the single line-to-ground and double
line-to-ground faults through impedance shown in Fig. 12.21b and d, but we shall
arrive at the correct sequence-network connections by comparison with faults
without impedance. Consider a generator with all terminals open and its neutral
grounded. On such a generator a single or double line-to-ground fault through
Zf is no different with respect to the value of the fault current than the same
type of fault without impedance but with Zf placed in the connection between
the generator neutral and ground. To account for an impedance Z{ in the
neutral of a generator we add 3Zf to the zero-sequence network. Thevenin’s
theorem enables us to apply the same reasoning to these types of faults on a
power system, and so the sequence-network connections for a single line-to-

ground fault and for a double line-to-ground fault are as shown in Fig. 12.22ft
and d. From these figures, for a single line-to-ground fault through Zf
Ial I a2 I aO
/.! = Zj -f Z2 + Z0 + 3Zj
(12.28)
and for a double line-to-ground fault through Zf
Ki = Ka2
/a 1 Yl (12.29)
z, + Z2(Z0 + 3Z/)/(Z2 + Z0 + 3 Zf)
A line-to-line fault through impedance is shown in Fig. 12.21c. The condi¬
tions at the fault are
/fl = 0 Ib=-Ic Vc — Vb — I bZf

Ia, Ib, and Ic bear the same relations to each other as in the line-to-line fault
without impedance. Therefore,
I al ~ I a2
The sequence components of voltage are given by
ya0 l l l
Ki 1 a a2 K (12.30)
va2 1 a2 a K - hZf
or
3Ffll = Va + (a + a2)Vb - a2IbZf (12.31)
3 Va2 = Va + (a + a2)Vb-aIbZf (12.32)

therefore
3(Ffll - Va2) = (a-a2)IbZf=jy/3IbZJ (12.33)
Since Ial = -Ia2,
h = a2lai + ala2 = {a2 - a)Ial = -jy/3Ial (12.34)
and, upon substituting Ib from Eq. (12.34) in Eq. (12.33), we obtain
Ki-Va2 = ImlZf (12.35)
Equation (13.35) requires the insertion of Zf between the fault points in the
positive- and negative-sequence networks to fulfill the required conditions for
the fault. The connections of the sequence networks for a line-to-line fault
through impedance are shown in Fig. 12.22c. Of course the bus impedance
matrix can be used to advantage to find Zl5 Z2, and Z0 of Eqs. (12.27), (12.28),
(12.29), and (12.35).
Faults through impedance are similar to single-phase loads. The impedance

Zf of the single line-to-ground fault is equivalent to connecting a single-phase
load Zy from line a to neutral. The impedance Zy of the line-to-line fault is
equivalent to connecting a single-phase load Zf from line b to line c.
12.11 COMPUTER CALCULATIONS OF FAULT CURRENTS
Modern fault-current programs for the digital computer are usually based on the
bus impedance matrix. Three-phase and single line-to-ground faults are usually
the only types of fault studied. Since circuit-breaker applications are made
according to the symmetrical short-circuit current that must be interrupted, this
current is calculated for the two types of fault. The printout includes the total
fault current and the contributions from each line. The results also list those
quantities when each line connected to the faulted bus is opened in turn while all
others are in operation.
The program uses the data listed for the lines and their impedances as
provided for the load-flow program and includes the appropriate reactance for
each machine in forming the positive- and zero-sequence bus impedance
matrices. As far as impedances are concerned the negative-sequence network is
the same as the positive-sequence network. So, for a single line-to-ground fault
at bus 1, Ial is calculated in per unit as 1.0 divided by the sum of 2Z11_1 and
Zu —o-
The bus voltages are included in the computer printout, if called for, as well
as the current in lines other than those connected to the faulted bus since this
information can easily be found from the bus impedance matrix.
PROBLEMS
12.1 A 60-Hz turbogenerator is rated 500 MVA, 22 kV. It is Y-connected and solidly grounded and
is operating at rated voltage at no load. It is disconnected from the rest of the system. Its reactances
are X" = X2 = 0.15 and X0 = 0.05 per unit. Find the ratio of the subtransient line current for a
single line-to-ground fault to the subtransient line current for a symmetrical three-phase fault.
12.2 Find the ratio of the subtransient line current for a line-to-line fault to the subtransient current
for a symmetrical three-phase fault on the generator of Prob. 12.1.
12.3 Determine the ohms of inductive reactance to be inserted in the neutral connection of the
generator of Prob. 12.1 to limit the subtransient line current for a single line-to-ground fault to that
for a three-phase fault.
12.4 With the inductive reactance found in Prob. 12.3 inserted in the neutral of the generator of
Prob. 12.1, find the ratios of the subtransient line currents for the following faults to the subtransient
line current for a three-phase fault: (a) single line-to-ground fault, (b) line-to-line fault, (c) double
line-to-ground fault.
12.5 How many ohms of resistance in the neutral connection of the generator of Prob. 12.1 would
limit the subtransient line current for a single line-to-ground fault to that for a three-phase fault?
12.6 A generator rated 100 MVA, 20 kV has X" = X2 = 20% and X0 = 5%. Its neutral is grounded
through a reactor of 0.32 fl The generator is operating at rated voltage without load and is discon¬
nected from the system when a single line-to-ground fault occurs at its terminals. Find the sub¬
transient current in the faulted phase.
12.7 A 100-MVA 18-kV turbogenerator having X" = X2 = 20% and X0 = 5% is about to be con¬
nected to a power system. The generator has a current-limiting reactor of 0.162 12 in the neutral.
Before the generator is connected to the system, its voltage is adjusted to 16 kV when a double
line-to-ground fault develops at terminals b and c. Find the initial symmetrical rms current in the
ground and in line b.
12.8 The reactances of a generator rated 100 MVA, 20 kV, are X" = X2 = 20% and X0 = 5%. The
generator is connected to a A-Y transformer rated 100 MVA, 20A-230Y kV, with a reactance of
10%. The neutral of the transformer is solidly grounded. The terminal voltage of the generator is
20 kV when a single line-to-ground fault occurs on the open-circuited, high-tension side of the
transformer. Find the initial symmetrical rms current in all phases of the generator.
12.9 A generator supplies a motor through a Y-A transformer. The generator is connected to the Y
side of the transformer. A fault occurs between the motor terminals and the transformer. The
symmetrical components of the subtransient current in the motor flowing toward the fault are
Iai — ~0-8 — j2.6 per unit, Ia2 = —j2.0 per unit, and Ia0 = —j3.0 per unit. From the transformer
toward the fault Ial = 0.8 -;0.4 per unit, Ia2 = —j 1.0 per unit, and Ia0 = 0. Assume X'[ = X2 for
both the motor and the generator. Describe the type of fault. Find (a) the prefault current, if any, in
line a, (b) the subtransient fault current in per unit, and (c) the subtransient current in each phase of
the generator in per unit.
12.10 Calculate the subtransient currents in all parts of the system of Example 12.4 with prefault
current neglected if the fault on the low-tension side of the transformer is a line-to-line fault.
12.11 Repeat Prob. 12.10 for a double line-to-ground fault.
12.12 The machines connected to the two high-tension buses shown in the one-line diagram of
Fig. 12.23 are each rated 100 MVA, 20 kV with reactances of X" = X2 = 20% and Y0 = 4%. Each
three-phase transformer is rated 100 MVA, 345Y/20A kV, with leakage reactance of 8%. On a base
of 100 MVA, 345 kV the reactances of the transmission line are X2 = X2 = 15% and A"0 = 50%.
Find the 2x2 bus impedance matrix for each of the three sequence networks. If no current is
flowing in the network, find the subtransient current to ground for a double line-to-ground fault on
lines B and C at bus 1. Repeat for a fault at bus 2. When the fault is at bus 2, determine the current in
phase b of machine 2 if the lines are so named that VAl and Val are 90° out of phase. If the phases are
named so that IA1 leads Ial by 30° what letter (a, b, or c) would identify the phase of machine 2
which would carry the current found for phase b above?
12.13 Two generators Gl and G2 are connected through transformers 7j and T2 to a high-tension
bus which supplies a transmission line. The line is open at the far end at which point F a fault occurs.
The prefault voltage at point F is 515 kV. Apparatus ratings and reactances are:
- 1000 MVA, 20 kV, Xs = 100% X" = X2 = 10% Y0 = 5%
G2 - 800 MVA, 22 kV, Xs = 120% X" = X2=15% X0 = 8%
Tt - 1000 MVA, 500Y/20A kV, X = 17.5%
T2 = 800 MVA, 500Y/22Y kV, X = 16.0%
Line - X1 = 15%, X0 = 40% on base of 1500 MVA, 500 kV
© Y C>
The neutral of Gj is grounded through a reactance of 0.04 fl. The neutral of G2 is not grounded.
Neutrals of all transformers are solidly grounded. Work on a base of 1000 MV A, 500 kV in the
transmission line. Neglect prefault current and find subtransient current (a) in phase c of Gl for a
three-phase fault at F, (b) in phase B at F for a line-to-line fault on lines B and C, (c) in phase A at F
for a line-to-ground fault on line A, and (d) in phase c of G2 for a line-to-ground fault on line A.
Assume VAl leads Val by 90° in Tv
12.14 For the network shown in Fig. 10.18, find the subtransient current in per unit (a) in a single
line-to-ground fault on bus 2, and (b) in the faulted phase of line 1-2. Assume no current is flowing
prior to the fault and that the prefault voltage at all buses is 1.0 per unit. Both generators are
Y-connected. Transformers are at the ends of each transmission line in the system and are Y-Y with
grounded neutrals except that the transformers connecting the lines to bus 3 are Y-A with the neutral
of the Y solidly grounded. The A sides of the Y-A transformers are connected to bus 3. All line
reactances shown in Fig. 10.18 between buses include the reactances of the transformers. Zero-
sequence reactance values for these lines including transformers are 2.0 times those shown in
Fig. 10.18. Zero-sequence reactances of generators connected to buses 1 and 3 are 0.04 and 0.08 per
unit, respectively. The neutral of the generator at bus 1 is connected to ground through a reactor of 0.02
per unit, and the neutral of the generator at bus 3 is solidly grounded.
12.15 Find the subtransient current in per unit in a line-to-line fault on bus 2 of the network of
Example 8.1. Neglect resistance and prefault current, assume all bus voltages are 1.0 before the fault
occurs, and make use of calculations already made in Example 10.4. Find the current in lines 1-2 and
3-2 also. Assume that lines 1-2 and 3-2 are connected to bus 2 directly rather than through transfor¬
mers and that the positive- and negative-sequence reactances are identical.
CHAPTER
THIRTEEN
SYSTEM PROTECTION
In connection with our study of transmission-line transients in Chap. 5 we dis¬

cussed briefly the protection of apparatus against surges resulting from light¬
ning and switching. Failure of apparatus due to surges or other causes leads to
faults on a power system. Now that we have studied balanced and unbalanced
faults that can occur on a power system and know how to calculate the currents
and voltages that exist during short circuits we are ready to study protection of a
system by isolation of the faulted portion. Although occurrence of short circuits
is somewhat of a rare event, it is of utmost importance that steps be taken to
remove the short circuits from a power system as quickly as possible. In modern
power systems this short circuit removal process is executed automatically, that
is, without human intervention. The equipment that does this job is known
collectively as the protection system. We shall stress protection of transmission
lines and transformers in order to develop some important principles, but we
shall also discuss a method that is used for the protection of generators, motors,
and buses. An extensive treatment of system protection is beyond the scope of
this book.
Strictly speaking a fault is any abnormal state of the system, so that faults in
general consist of short circuits as well as open circuits. We will limit our
discussion here to faults that are short circuits. Open circuit faults are much
more unusual than short circuits, and often they are transformed into short
circuits by subsequent events. In terms of the seriousness of consequences of a
fault, short circuits are of far greater concern than open circuits, although some
open circuits may present a potential hazard to personnel.
337
If short circuits are allowed to persist on a power system for an extended

period, many or all of the following undesirable effects are likely to occur:'
1. Reduced stability margins for the power system, a subject discussed in

Chap. 14.
2. Damage to the equipment that is in the vicinity of the fault due to heavy
currents, unbalanced currents, or low voltages produced by the short circuit.
3. Explosions which may occur in equipment containing insulating oil during a
short circuit and which may cause fire resulting in a serious hazard to person¬
nel and damage to other equipment.
4. Disruptions in the entire power system service area by a succession of protec¬
tive actions taken by different protection systems, an occurrence known as
cascading.
Which one of these effects will predominate in a given case depends upon the
nature and operating conditions of the power system.
13.1 ATTRIBUTES OF PROTECTION SYSTEMS
Speedy elimination of a fault by the protection system requires correct operation

of a number of subsystems of the protection system. The job of each of these
subsystems can best be understood by describing the events that take place from
the time of occurrence of a fault to its eventual elimination from the power
system. Although complex sequential faults do occur occasionally on a power
system, and some unusual protection-system operations may come into play
from time to time, we will devote most of our attention to the simple case of the
occurrence of a three-phase short circuit on a transmission line and the resulting
operation of the appropriate protection system. Consider the system shown in
Fig. 13.1. Buses 1 and 2 are at the two ends of a transmission line. At each end of
the transmission line, two identical protection systems are shown enclosed by
dotted lines. These constitute the protection system for transmission line 1-2.
The protection system can be subdivided into three subsystems:
1. Circuit breakers (CB, or B)

2. Transducers (T)
3. Relays (R)
Circuit breakers have been mentioned briefly in Chap. 10. Relays are the devices
which sense the fault and cause the circuit-breaker trip circuits to be energized
and the breakers to open their contacts. Transducers provide the input to the
relays. Each of these subsystems will be discussed further as we continue to
develop the material of this chapter.
Usually we will use a double-numbering notation to identify circuit breakers
and relays. Thus the line 1-2 in Fig. 13.1 has a circuit breaker B12 at the bus-1
SYSTEM PROTECTION 339
Figure 13.1 One-line diagram showing two transmission lines and elements of the protection system
for line 1-2.
end of the line and a circuit breaker B21 at the bus-2 end. Relays at these points
are labeled R12 and R21, respectively. A relay R23, although not shown here, is
understood to be associated with B23. Sometimes, however, for the simple
systems we will be examining it is more convenient to refer to circuit breakers by
letters. For example, B12 and B21 could have been labeled A and B without
numerical designation.
Separate breakers may be operated in each phase, or the relays may control
one three-phase breaker which will open all three phases upon operation of any
one of the relays.
When a fault occurs at P on the lines of Fig. 13.1, increased currents flow
from both terminals of the transmission line toward the fault if we assume that
sources of power are available beyond both buses 1 and 2. When this assump¬
tion is not the case, the protection system becomes somewhat simpler. We will
consider the protection of such radial systems later. The increase in current at
the line terminals is accompanied by a reduction in voltages. It should be
realized that the currents and voltages of the transmission line are at kiloampere
and kilovolt levels. These high-level signals are unsuitable for use by the protec¬
tion system. The power line signals are converted to a lower level (tens of
amperes and volts) by the transducers T. Transducers will be discussed at greater
length a little later.
The increase in current and reduction in voltage caused by the fault can be
used to detect that a fault has occurred on the transmission line. Relays are the
logic elements of the protection system. The lower-level signals produced by the
transducers are reasonably faithful reproductions of the actual voltages and
currents of the transmission line. Relays R12 and R21 process these input signals
and make the decision that a fault has in fact occurred on the transmission line
1-2. This decision is reached within a very short time after the occurrence of the
fault, typically 8 to 40 milliseconds depending upon the design of the relays.
The decision by relays R12 and R21 that a fault has occurred on the line
leads to the tripping of their associated circuit breakers B12 and B21. Circuit
breakers are the final link in the fault removal process. They were mentioned
briefly in Chap. 10. When the trip circuit of a circuit breaker is energized by its
relay the contacts of the circuit breaker—which are in series with the transmis¬
sion line—begin to move apart very rapidly. As the current through the breaker
contacts (the fault current) passes through zero, the space between the contacts
becomes a dielectric, and is able to prevent the fault current from flowing again
through the circuit breaker. This leads to the disconnection of the transmission
line from the rest of the system and the elimination of the fault. The entire
process from the time of initiation of the fault to its final clearance takes between
30 and 100 milliseconds depending upon the type of protective system employed.
Certain attributes of a relay are important measures of the quality of its
performance. The sequence of events described above indicates that to do its job
properly a relay must be fast and reliable (that is, dependable and selective.) The
first attribute—speed—is self-explanatory. A relay should make its decision as
quickly as possible, consistent with other requirements placed upon it. The
attribute of dependability means that the relay should operate consistently for all
the faults for which it is designed to operate, and should refrain from operating
for any other system condition. Selectivity of a relay refers to the requirement
that the smallest possible portion of a system should be isolated following a
fault. Selectivity can be illustrated with the help of Fig. 13.1. Consider the relay
R23 connected at terminal 2 of line 2-3. The current and voltage inputs to this
relay will also change due to the fault at P. This effect of the fault at P upon the
relay R23 is often described by saying that the relay R23 also “ sees ” the fault at
P. However, the relay R23 must be selective so that it does not operate for the
fault at P if P is outside the area of responsibility known as reach of this relay. In
general the reliability requirements of a relay are in conflict with its speed
requirement, and a compromise must be made in designing the protection
system so as to obtain a reasonable measure of these attributes.
13.2 ZONES OF PROTECTION
The idea of an area of responsibility of a protection system mentioned above has

been formalized by assigning zones of protection to various protection systems.
The concept of zones helps define the reliability requirements for different
protection systems. We will explain the concept of zones of protection with the
help of Fig. 13.2. In this figure, a portion of a power system consisting of a
generator, two transformers, two transmission lines, and three buses is repre¬
sented by a one-line diagram. The closed dashed lines indicate the five zones of
protection in which this power system is divided. Each zone contains one or
more power system components in addition to two circuit breakers. Each
breaker is included in two neighboring zones of protection. Zone 1, for example,
contains the generator, its associated transformer, and the connecting leads be¬
tween the generator and the transformer. Zone 3 contains a transmission line
only. Note that zones 1 and 5 contain two power-system components each.
The boundary of each zone defines a portion of the power system such that
for a fault anywhere within that zone the protection system responsible for that
zone takes action to isolate everything within that zone from the rest of the
Zone 2 Zone 4
Figure 13.2 Zones of protection indicated by dashed lines enclosing power-system components in
each zone.
system. Since the isolation (or deenergization) under faulted conditions is done
by circuit breakers, it should be clear that at each point where connection is
made between the equipment inside the zone and the rest of the power system, a
circuit breaker should be inserted. In other words, the circuit breakers help
define the boundaries of the zones of protection.
Another important aspect of zones of protection is that the neighboring
zones always overlap. This overlap is necessary, since without it a small part of
the system which falls between the neighboring zones, however small it may be,
would be left without protection. By overlapping neighboring zones no part of
the power system is left without protection, although clearly if a fault should
occur within the overlapped region, a much larger portion of the power system
(that corresponding to both the zones involved in the overlap) will be isolated
and lost from service. To reduce such a possibility to a minimum the region of
overlap is made as small as possible.
Example 13.1 (a) Consider the power system shown in Fig. 13.3a with gen¬
erating sources beyond buses 1, 3, and 4. What are the zones of protection in
which this system should be divided? Which circuit breakers will open for
faults at Pj and P2?
(b) If three circuit breakers are added at the tap point 2, how would the
zones of protection be modified? Which circuit breakers will operate for
faults at Pj and P2 under these conditions?
Solution (a) Using the principles of defining the zones of protection, the
system of Fig. 13.3a can be divided into zones as shown by dashed lines in
that figure. For the fault at Pj breakers A, B, and C will operate. For the
fault at P2 breakers A, B, C, D, and E will operate.
(ib) If three circuit breakers F, G, and H are added at bus 2 as shown in
Fig. 13.3b the zones of protection will be as shown by the dashed lines in
that figure. In this case breakers A and F will operate for the fault at P1;
whereas breakers G, C, D, and E will operate for the fault at P2. Note that in
this case a much smaller portion of the power system is deenergized follow¬
ing the two faults. This improved performance is achieved at the expense of
three additional circuit breakers and the associated protection equipment.
Figure 13.3 One-line diagram for Example 13.1 showing (a) original zones of protection and
(b) modified zones when additional breakers are added at bus 2.
13.3 TRANSDUCERS
Currents and voltages of the protected power equipment are converted by cur¬
rent and voltage transformers to low levels for relay operation. These reduced
levels are necessary for two reasons: (1) the lower level input to the relays
ensures that the physical hardware used to construct the relays will be quite
small and thus less expensive; (2) the personnel who work with the relays will be
working in a safe environment. In principle these transducers are no different
from the power transformers discussed in Chap. 6. However, the use made of
these transformers is rather specialized. For example, it is necessary that a cur¬
rent transformer reproduce in its secondary winding a current which duplicates
the primary current waveform as faithfully as possible. It performs this function
quite well. Similar considerations hold for a voltage transformer. The amount of
power delivered by these transformers is rather modest, since the load connected
to them consists only of relays and meters that may be in use at a given time.
The load on current transformers (CTs) and voltage transformers (VTs) is
commonly known as their burden. The term burden usually describes the im¬
pedance connected to the transformer secondary winding but may specify the
voltamperes delivered to the load. For example, a transformer delivering 5 am¬
peres to a resistive burden of 0.1 ohm may also be said to have a burden of 2.5
voltamperes at 5 amperes.
We will now consider current transformers and voltage transformers
separately.
(1) Current Transformers

There are two types of current transformers found in practice. Certain power
equipment is of the dead-tank type, having a grounded metal tank in which the
power equipment is contained in an insulating medium (usually oil). Examples
are power transformers, reactors, and oil circuit breakers. This type of equip¬
ment has a bushing through which a terminal of the power equipment is brought
out. Current transformers are built within this bushing and are known as bush¬
ing CTs. Where such a dead-tank system is not available, for example at an EHY
switching station where live-tank circuit breakers are in use, free-standing cur¬
rent transformers are used.
The schematic representation for current transformers is shown in Fig. 13.4.
The primary winding of a current transformer usually consists of a single turn,
and is represented in Fig. 13.4 by a straight line marked a and b. This single turn
is obtained by threading the primary conductor through one or more toroidal
steel cores. The secondary windings, the terminals of which are marked as a' and
b' in Fig. 13.4, are multiple-turn windings wound on the toroidal cores. The dots
placed at the terminals a and a' of the current transformer windings have the
same connotation as for a conventional transformer. When the primary current
enters terminal a (the terminal with the dot marking) the current leaving the
dotted terminal a' of the secondary winding is in phase with the primary current
if magnetizing current is neglected.
Current transformers have ratio errors which for some types can be cal¬
culated and for other types must be determined by test. The error can be quite
high if the impedance burden is too large, but with proper selection of the CT
with respect to the burden the error can be maintained at an acceptable value.
Since we are mainly concerned with protection methods, we shall not discuss CT
errors any further, but we must remain aware of them in our consideration of
relays.
Figure 13.4 Schematic representation to show connection

of a current transformer to the line of a power system.
The normal current rating of CT secondaries has been standardized at 5 A,

with a second standard of 1 A being used in Europe and, to a lesser extent, in the
United States. For short periods of time this rating of the CT secondary wind¬
ings can be exceeded without damaging the windings. Currents of more than 10
or 20 times normal are often encountered in CT windings during short circuits
on the power system.
Standard CT current ratios have been established, and some of these are
given in Table 13.1.
(2) Voltage Transformers

Two types of voltage transformers are commonly found in relaying applications.
For certain low-voltage applications (system voltages about 12 kV or lower)
transformers with a primary winding at the system voltage and secondary wind¬
ings at 67 V (representing the system line-to-neutral voltage) and 67 x y/3 =
116 V (representing the system line-to-line voltage) are an industry standard.
This type of voltage transformer is quite similar to a multiwinding power trans¬
former, and becomes expensive at higher system voltages. For voltages at HV
and EHV levels, a capacitance potential-divider circuit is used as shown in
Fig. 13.5. Capacitors Cy and C2 are so adjusted that a voltage of a few kilovolts
is obtained across C2 when terminal A is at system potential. In such a coupling-
capacitor voltage transformer (CVT) the tapped voltage is further reduced to
relaying voltage level by a transformer as shown in Fig. 13.5.
The voltage at point A in the circuit of Fig. 13.5 is essentially that of an
infinite bus so far as the connected capacitors are concerned. The Thevenin
impedance looking toward the system across the terminals of C2 is
l/co(Cj + C2). Adjusting L so that coL equals the Thevenin impedance results in
series resonance, and the output of the CVT is in phase with the line potential
with no phase-angle error introduced in the CVT output. The CVT is a free¬
standing device housed in its own supporting insulator structure, and finds
application in HV and EHV systems. Whenever a power-system component has
a bushing through which a conductor at system voltage passes, such as in a
power transformer or in certain types of circuit breakers, a bushing type of CVT
built inside the bushing can be made available at little additional cost. In such a
CVT, the capacitors Q and C2 are built within the structure of the bushing. In
general, bushing type CVTs are capable of supplying smaller burdens than the
free-standing CVTs.
Table 13.1 Standard CT ratios
Current ratio Current ratio Current ratio
50 : 5 300 : 5 800 : 5
100 : 5 400 : 5 900 : 5
150 : 5 450: 5 1000 : 5
200 : 5 500 : 5 1200 : 5
250 : 5 600 : 5
Figure 13.5 Circuit diagram of a capacitor-coupled voltage transformer

CVT with its tuning inductance L.
Voltage transformers are generally far more accurate than the current trans¬
formers, and their ratio and phase-angle errors are generally neglected. On the
other hand it is often necessary to pay attention to the transient response of the
CVTs under fault conditions, as errors under these conditions are possible.
Transient response of CVTs is beyond the scope of our discussion.
13.4 LOGICAL DESIGN OF RELAYS
The job of a relay is to discriminate between a fault within its zone of protection
and all other system conditions. It must act (energize the trip coil of its asso¬
ciated circuit breakers) dependably for faults within its zones of protection, and
provide security against false tripping for faults outside those zones. A relay is
made secure and dependable by designing into it a logical decision-making
capability such that, based upon the condition of its input signals, it is able to
produce the correct output for every possible state of its input signals. We will
now consider several classes of relays and their logical functional descriptions. In
spite of the great variety of relays found on power systems, a majority of them
fall into five categories. Their logical performance can be defined in terms of the
inputs and outputs of the relay independently of the hardware used in building
the relay. For each type of relay, we will specify conditions on their input signals
(usually voltages and currents) and the corresponding state of the relay output.
The relay output of interest in the present context is the input to the breaker trip
coil. Consequently the output state of the relay will be, with its contacts closed,
called trip, or with its contacts open, called block or block to trip. The five relay
classes to be considered here are:
1. Magnitude relays
2. Directional relays
3. Ratio relays
4. Differential relays
5. Pilot relays
1. Magnitude relays In their most common form these are current magnitude
relays—or overcurrent relays. They respond to the magnitude of their input
current, and operate to trip whenever the current magnitude exceeds a certain
value which is adjustable. If a value | Ip | expressed in terms of the secondary
winding of the CT can be found from system short circuit studies such that for
all faults within the zone of protection of a relay, the fault current magnitude
| If | also expressed in terms of the secondary winding will be greater than \Ip\,
then the following functional description will produce a dependable and secure
relay:
h\ > Up\ triP
(13.1)
\lf\ < \Ip\ block
The inequalities expressed by (13.1) are the logical description of an overcurrent

relay and can be represented graphically by the phasor diagram shown in
Fig. 13.6. The current magnitude \Ip\ is known as the pickup value of the relay.
The fault current phasor If is drawn in the complex plane with an arbitrary
phasor assumed to be the reference. The phase angle of the fault current can lie
anywhere between 0 and 360 degrees, since the reference phasor is arbitrary. A
circle drawn with the origin as its center and the pickup current magnitude | Ip |
as its radius divides the complex phasor plane in two regions labeled trip and
block in Fig. 13.6. Any fault current whose phasor representation lies outside the
circle in the shaded region will cause the relay to trip. Phasors representing fault
currents with magnitude smaller than \lp\ will lie within the circle, and will
result in a block decision by the relay. Diagrams such as these are very useful in
understanding relay characteristics and are used extensively in relaying
literature.
As will be pointed out later, this simplest form of an overcurrent relay is not
found to be sufficiently versatile in many cases. It is necessary to introduce
another parameter—the time it takes the relay to operate after \If\ exceeds
\IP\. One could supplement the conditions (13.1) with the equation
T=<K\lf\ - Kl) if|'/l>lLl (13.2)
Figure 13.6 Graphical representation of the

operating and blocking regions of a time over-
current relay in the complex plane. Plotting of
the phasor current of the relay operating coil
on this diagram will show operation or
blocking and operating time. Time T2 is earlier
than T,.
where T is the relay operating time, and 0 is a function which describes its
dependence upon the fault-current level. This functional dependence can be
illustrated by adding the time circles such as Tj and T2 to the phasor diagram of
Fig. 13.6 as shown. The length of phasor 11f\ then falls on a time line (or
between two time lines) which represents the operating time of the relay for that
fault current. The traditional method of representing the characteristics of a time
overcurrent relay is as shown in Fig. 13.7. The pickup setting \lp\ of a relay is
adjustable through the taps on its input winding. For example, the relay IFC-53
(General Electric Company) whose characteristic curves are shown in Fig. 13.7
is available with tap settings of 1.0, 1.2, 1.5, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 10.0,
12.0 A. The function </> is usually asymptotic to the pickup value and decreases
as some inverse power of the current magnitude for \If\ > \Ip\. The character¬
istic curves are generally presented with multiples of pickup amperes as the
abscissa and operating time as the ordinate. Multiples of pickup amperes means
the ratio of relay current to pickup current. The inverse-time characteristic can
be shifted up or down by an adjustment known as the time-dial setting. In
Fig. 13.7, a time-dial setting of 1/2 produces the fastest operation of the relay,
whereas a setting of 10 produces the slowest operation for a given current.
Although time-dial adjustments are specified as discrete settings, intermediate
values can be obtained by interpolating between the discrete curves.
2. Directional relays In some applications, the zone of a relay includes all of the
power system that is situated in only one direction from the relay location. For
example, consider the relay R21 shown in Fig. 13.8a. This relay is required to
operate for faults to the left of its location, and block for all other conditions.
Since the transmission-line impedance is mostly reactive, the faults to the left of
R21 have currents flowing from bus 2 toward bus 1 which lag the voltage at bus
2 by an angle of about 90 degrees. On the other hand, for faults to the right of
bus 2, the current from bus 2 to bus 1 will lead the voltage at bus 2 by an angle
of about 90 degrees. The operation of the relay is described by dividing the
complex plane of the phasor diagram of Fig. 13.86 such that for all faults pro¬
ducing current phasors lying in the shaded region (when the voltage at bus 2 is
used as a reference) the relay would trip, and for all other faults it would block.
Such a relay is called directional, since it depends for its operation upon the
direction of the current with respect to the voltage. The quantity that provides
the reference phasor is called a polarizing quantity. Thus the directional relay
described above uses a polarizing voltage. Sometimes certain current signals may
also be used as polarizing signals. The relay can be made more selective by
defining a narrower region around the fault-current phasor. In general the oper¬
ating principle of a directional relay can be described by
0min > flop > flmax trip
Amin < flop < flmax block

Operating time in seconds
Time-dial, settings
Figure 13.7 Characteristic curves of type IFC-53 time overcurrent relays (Courtesy General Electric
Company).
Trip i Block
-hi (reverse)
©
-D
R21
(6)
Figure 13.8 Operating principle of a directional relay: (a) one-line diagram to show location and
(b) relay performance characteristic in the complex plane.
where 9op is the phase angle of the operating quantity measured with the polariz¬
ing phasor as the reference, and 0min and 0max are the two angles defining the
boundary of the operating characteristic.
3. Ratio relays Consider the relay R12 shown in Fig. 13.9a. In some applica¬
tions, it is necessary that the relay operate for faults within a certain distance of
its location on any of the lines originating at bus 1. The vicinity is described by
the distance along the lines, or equivalently by the impedance between bus 1 and
the fault location. The zone of protection is thus a region such that the length of
a line originating at bus 1 and having an impedance less than the required
setting |Zr| is included in the zone. This condition can be conveniently ex¬
pressed as a requirement on the ratio of the voltage and current at the location
of R12. Let this ratio (which has the dimensions of an impedance) be
(13.4)
The relay performance can then be specified by:
IZ | < |Zr | trip

(13.5)
|Z| > \Zr\ block
and the relay is called an impedance or distance relay. In the complex impedance
plane, the locus of constant \Zr \ is a circle as shown in Fig. 13.9b. Note that the
impedance Z is defined in Eq. (13.4) as a ratio of the voltage and current at relay
location 1. During normal system conditions this ratio will be a complex number
with some arbitrary phase angle determined by the load power factor. Since the
load current is usually much smaller than the fault current, the ratio Z will have
a large magnitude (and an arbitrary phase angle) during normal system condi¬
tions. Therefore Z plotted in the complex plane under normal system conditions
will lie outside the circle of radius | Zr | and consequently the circuit breaker will
not trip during normal system conditions. Under faulted conditions Z appears to
the relay to be a load whose impedance is that of the line between the relay
location and the fault. The angle associated with Z is 6 or n 4- 9 depending upon
whether the fault is to the right or left of bus 1 in the circuit of Fig. 13.9a.
A simple modification of an impedance relay is often found to be quite
useful. The circle in Fig. 13.9h, which is centered at the origin, can be offset by an
amount Z' producing the characteristic of the offset impedance relay shown in
Fig. 13.9c. The performance of this type of relay is described by
|Z — Z'| < \Zr\ trip

(13.6)
|Z-Z'|>|Zr| block
By selecting \Z'\ to be equal to |Zr|, the relay characteristic can be made to

pass through the origin. This is the case illustrated in Fig. 13.9c, and the charac¬
teristic illustrated here is known as a “ mho ” characteristic. The impedance relay
having a characteristic of Fig. 13.9& is not directional: a fault either to the right
or to the left of the relay location (and having | Z | less than \Zr\) will lead to a
trip decision by the relay. A mho relay having the characteristic of Fig. 13.9c on
the other hand is inherently directional. A fault to the left of the relay at bus 1,
no matter how close to bus 1 it may be, will result in a no-trip decision by the
relay because Z as defined by Eq. (13.4) will lie in the third quadrant. We will
see later that this is a very desirable characteristic in many applications.
4. Differential relays When the entire zone of protection of a relay occupies a

relatively small physical space near the relay, it is possible to employ the prin¬
ciple of current continuity to devise a very simple and effective relaying scheme.
Consider the zone of protection of one phase of a generator winding shown in
Fig. 13.10. Two current transformers having the same turns ratios are placed at
the boundaries of the zone of protection (two for each phase of a three-phase
Figure 13.9 Impedance relay charac¬

teristics showing (a) zone of protection Block
for R12, (b) complex plane in which
measured impedance is plotted for a
nondirectional relay, and (c) for a mho
relay. In both (b) and (c) the impedance
is indicated for a fault to the left
of R12.
Figure 13.10 Wiring diagram for differential protec¬

tion of a generator winding.
unit). Then for normal conditions, as well as for faults outside the zone of
protection
I\ — I2 = 0
Whereas for a fault inside the protected zone
11 12 — If
where If is the fault current as seen from the secondary side of the CTs. It
should be realized that due to the errors of the current transformers, these
equations will not exactly hold in practice. To account for these inaccuracies, a
low value of current | Ip | may be chosen such that
lc-ci<ici
for normal system conditions or for faults external to the zone of protection; and
I'.-cl > lei

for internal faults. The operating principle of the relay can therefore be
defined by
C-dHCl Tip
1C -cl < ICl Mock
If we connect an overcurrent relay of the type described previously so that its
operating coil is coil 3 in the position shown in Fig. 13.10, we see that the
current through the coil is — I2 and the relay will protect the generator
winding by tripping a breaker or breakers according to the principle of differen¬
tial relaying expressed in (13.7). Often the current transformer errors discussed in
Sec. 13.3 increase with increasing values of Ix and 12. For such cases, it is
possible to make the value Ip dependent upon the average of Ix and /2. A relay
can be designed in this fashion such that the operating principle for the relay
becomes
IC-CI >k|(C+d|/2 trip
(13.8)
1C - Cl < k|(C + C)l/2 block
Such a relay is known as a percent differential relay. The current (/x + 12)/,2 is
called the restraining current, and the current (I{ — I2) is the tripping current of
the relay. The relay coils 1 and 2 in Fig. 13.10 have currents It and I2 flowing
through them. If a differential relay is constructed in such a fashion that the
currents through coils 1 and 2 oppose the effect of current through 3, then such a
relay will exhibit the percent differential characteristic expressed by (13.8). The
relative effectiveness of coils 1 and 2 compared to coil 3 is determined by the
constant k of the percent differential relay. In an electromechanical percent
differential relay, coils 1, 2, and 3 are wound on a common magnetic core in such
a direction that currents through 1 and 2 produce a magnetomotive force that is
in opposition to that produced by current in 3. In an electronic relay the desired
characteristic is obtained by amplification factors in the appropriate signal
paths.
A relay of this type can also be used to protect a bus and, of course, a motor.
5. Pilot relays The differential relays discussed above require that the boundary
points of the protected zone be physically close to each other so that the signals
from these boundary points can be connected to the relay. This is possible only
when the zone of protection contains some power equipment of limited size such
as a transformer, a generator, or a bus. When transmission lines are to be
protected by a relay, their terminals may be hundreds of miles apart, and it
becomes impractical to connect the signals from the ends of the transmission line
to one relay. Pilot relaying provides a technique of communicating information
from a remote zone boundary to the relay at each terminal. Although a direct
substitution of pilot communication channels for each differential current signal
wire is not economically or technically feasible, equivalent information trans¬
mission schemes have been devised. The physical medium used for pilot channels
could be conductors of a telephone circuit, high-frequency signals coupled on to
the power transmission line itself (known as the power-line carrier), or micro-
wave channels. Application of pilot relaying will be considered in Sec. 13.6.
13.5 PRIMARY AND BACKUP PROTECTION
It was pointed out at the beginning of this chapter that the protection system
contains many subsystems, and the successful removal of a fault requires that
each subsystem and component function correctly. The protection systems con¬
sidered so far were primarily responsible for the removal of the fault as soon as
possible while deenergizing as little of the system as required. These protection
systems are known as the primary protection systems. However, it is conceivable
that some components or subsystems of the primary protection systems may fail
to operate correctly, and it is the usual practice to allow for a backup protection
system which would take over the job of protection in case the appropriate
primary protection system fails to clear the fault. Consider the power system
shown in Fig. 13.11. For a fault at P, the primary (main) protection system must
Figure 13.11 One-line diagram of

a system having backup protection.
open circuit breakers F and G. One method of backing up the primary protec¬
tion system is to duplicate it entirely (or as much of it as is economically
feasible) so that the failure of one primary system does not prevent the removal
of the fault. Such a backup protection system is known as duplicate primary, for
obvious reasons, and is used on important circuits where the added cost can be
justified. However, there are certain components that are inevitably common to
the first primary and the duplicate primary. (Examples are circuit breakers,
batteries which operate breaker trip coils, CTs and CVTs.) It is therefore pos¬
sible that both primaries may be affected by the failure of one of these common
components, and provision must be made to furnish backup protection from a
remote location where the possibility of a common mode of failure with the
primary protection system is slight. This remote backup function is easily incor¬
porated in the primary protection system at the remote location. For example,
suppose that the primary protection system at bus 1 in Fig. 13.11 has failed to
clear the fault at P. (The bus-5 end is assumed to operate correctly.) Recognizing
this failure, we can arrange the primary protection systems at buses 2, 3, and 4 to
trip circuit breakers A, D, and H, respectively. The protection systems at 2, 3,
and 4, in addition to providing their primary protection for lines 2-1, 3-1, and
4-1, will also provide a remote backup protection for the primary protection
system at bus 1 for line 1-5. The operation of a remote backup system removes a
far greater portion of the power system from service than does the operation of
the primary protection system. In the above example, lines 2-1, 3-1, and 4-1 are
removed in addition to the originally faulted line 1-5 when the fault is removed
by the remote backup protection system. The service to any loads that may be
connected at buses 2, 3, and 4 may be affected, and there will be no service to
bus 1. Secondly, the backup system must allow a sufficient time for the primary
protection system to function normally. A too-hasty operation of the backup
protection may lead to unnecessary removal of the larger portion of the power
system. The backup system is thus made slower-acting by introducing a delay
between the maximum time for fault clearing by the primary system, and the
fastest possible response of the backup system. This delay is called the coordina¬
tion time delay, and it is required to help coordinate the operation of the primary
and the backup protection system.
Returning once again to Fig. 13.11 we note that, upon the failure of the
primary protection at bus 1 associated with line 1-5, a backup system equivalent
to the remote backup system described so far may be designed to trip circuit
breakers B, C, and E all located at bus 1. Such a protection system is known as
the local backup protection, and is generally provided to back up the failure of
the circuit breaker responsible for fault clearing (breaker F in this example). For
this reason, the local backup protection system is also known as breaker-failure
protection. Since the primary protection system and the local breaker-failure
protection system may share certain subsystems—such as station battery—there
are certain modes of failure which are common to the two systems, and some
form of remote backup protection is considered essential to a well-designed
protection system.
13.6 TRANSMISSION LINE PROTECTION
Transmission line protection has a central role in power-system protection be¬

cause transmission lines are vital elements of a network which connects the
generating plants to the load centers. Also because of the long distances
traversed by transmission lines over open countryside, transmission lines are
subject to a majority of the faults occurring on the power system. The simplest
protection system used at the lowest system voltages consists of fuses which act
as relays and circuit breakers combined. We will not consider protection with
fuses and reclosers (which are also used in distribution circuits) in this book.
Instead, we will concentrate on the protection of medium- and high-voltage
transmission lines. The protection system used for medium-voltage transmission
lines is somewhat simpler than that used for FIV and EHV transmission lines
which provide major bulk transmission facilities. Since the consequence of
outage of a high-voltage line is far more serious than that of a distribution or
subtransmission line, the protection of the bulk power transmission line is gen¬
erally more elaborate, with greater redundancy, and is also more expensive.
(a) Protection of Subtransmission Lines

The simplest form of protection system can be devised when the generation-load
system is radial in nature. Consider the power system shown in Fig. 13.12a. The
generator at bus 1 (which may be an equivalent representation of one or more
transformers feeding bus 1 from a higher-voltage supply point) supplies loads at
buses 1, 2, 3, and 4 through three transmission lines. Such a system is known as a
radial system because transmission lines radiate from a generating source to
supply the loads. Since the source of power is only to the left of each of the
transmission lines, it is sufficient to provide only one circuit breaker for each line
at the source end. Clearly for any fault on line 1-2, breaker B12 must be opened.
In this case all the loads at buses 2, 3, and 4 downstream from breaker 1 will be
interrupted.
Overcurrent relays described earlier can be used to protect the transmission
lines of this system. Fault current produced by a fault on any of the lines will
depend upon the fault location, and since the fault path impedance will increase
Figure 13.12 Protection of a radial system:

(a) one-line diagram of the system and (b)
qualitative curve showing fault current | If \
for faults located along the line.
with the distance to the fault from the generator, the fault current will be in¬
versely proportional to this distance. The fault current If as a function of dist¬
ance from bus 1 is shown qualitatively in Fig. 13.12b. Furthermore, the fault
current magnitudes will change depending upon the type of fault and the
amount of generation connected at bus 1. For example, if the generator at bus 1
is an equivalent representation of two parallel transformers, then fault currents
would be lower when one of the transformers is out of service for any reason. In
general, there will be a fault current magnitude curve as shown in Fig. 13.12b for
maximum fault-current levels (obtained when maximum generation is in service
and a three-phase fault is considered), and one for minimum fault levels (ob¬
tained when minimum generation is in service and a line-to-line or a line-to-
ground fault, whether or not through impedance to ground, is considered.) The
time-overcurrent relays discussed earlier can be set to provide primary protec¬
tion for a line, as well as the remote backup for a neighboring line for this
system. Relays at each of the three buses 1, 2, and 3, are provided to protect their
respective lines as primary protection relays, and to provide remote backup
protection to one line downstream from the relay location. Thus the relay at 1,
in addition to providing primary protection for line 1-2, also provides the
remote backup protection for line 2-3. The relay at 3 need provide only primary
protection for line 3-4, since there is no other line to the right of line 3-4. When
the relay at 1 provides backup protection for line 2-3, it must be so adjusted that
it operates with a sufficient time delay (its coordination time delay) such that the
relay at bus 2 will always be expected to operate first for faults on line 2-3. It is
not considered necessary, nor is it practical, to provide backup protection for
any line beyond bus 3 with the relay at bus 1. We will illustrate these concepts
with the following numerical examples.
Example 13.2 A portion of a 13.8-kV radial system is shown in Fig. 13.13.

The system may be operated with only one rather than two source trans¬
formers under certain operating conditions. Assume the high-voltage bus of
J5.0 ®
-^Nbi
o j 9.6
j5-0
Figure 13.13 One-line diagram of the radial system for Examples 13.2 and 13.3. Line and
transformer reactance values are marked in ohms.
the transformer is an infinite bus. The protection system for line-to-line and
three-phase faults is to be designed. Transmission line reactances in ohms
are shown in Fig. 13.13, and the transformer reactances are in ohms referred
to the 13.8-kV side. Neglect resistance and calculate the minimum and maxi¬
mum fault currents for a fault at bus 5.
Solution Maximum fault current will occur for a three-phase fault with
both transformers in service. At bus 5 for this case
13,800/^/3
—/202.75 A
f ~ j(2.5 + 9.6 + 6.4 + 8.0 + 12.8)
Minimum fault current will occur with only one transformer in service
for a line-to-line fault. For a three-phase fault with just one transformer
13,800/^/3
—7190.6 A
f ~ j(5.0 + 9.6 + 6.4 + 8.0 + 12.8)
However, a line-to-line fault would produce a fault current equal to /3/2

times the three-phase fault current. This relationship can be verified by
solving Prob. 12.2. So the minimum fault current for a, fault at bus 5 is
(—;'190.6) = -7165.1 A
2
Similar calculations lead to the maximum and minimum fault currents

shown in Table 13.2.
Table 13.2 Maximum and minimum fault currents for Example

13.3
Fault at bus 1 2 3 4 5
Maximum fault current, A 3187.2 658.5 430.7 300.7 202.7
Minimum fault current, A 1380.0 472.6 328.6 237.9 165.1
As will be explained in Example 13.3, the principle of backup protection

with overcurrent relays for any relay X, backing up the next downstream relay
Y, is that X must pick up
(a) for one third of the minimum current seen by Y and

(b) for the maximum current seen by Y but no sooner than 0.3 s after Y should
have picked up for that current.
Example 133 Select CT ratios, relay tap (pickup) settings, and relay time-
dial settings for the system of Example 13.2. Use at all locations the IFC-53
relay whose characteristic curves are given in Fig. 13.7 and whose tap set¬
tings are listed in Sec. 13.4. Since each line has a breaker at only one end,
simplify the notation by designating the relays at buses 1, 2, 3, and 4 by Rl,
R2, R3, and R4, respectively. The breaker at each bus will open all three
phases when tripped by any of the three associated relays. All three relays at
bus 1, for example, will be designated Rl.
Solution Settings for relay R4: This relay must operate for all currents
above 165.1 A, but for reliability a relay would be selected which will oper¬
ate when current in the line is one third of minimum, or
165.1
/p 55 A
3
For this current a CT ratio of 50/5 (Table 13.1) will result in a relay
current of
L= 5.5 A
So a relay tap setting of 5.0 A is the proper value.

Since this relay is at the end of a radial system, no coordination with
any other relay is necessary. Consequently, the fastest possible operation is
desirable. The time-dial setting is therefore chosen to be 1/2.
Settings for relay R3: This relay must back up relay R4, and therefore it
must pick up reliably for the smallest fault current seen by R4 which is
165.1 A. So, just as for R4, we use a CT ratio of 50/5 and a relay tap of 5 A
for R3 also.
To determine the time-dial setting it is customary to require that the
backup relay (in this case R3) operate at least 0.3 s after the time the relay
being backed up (R4) should have operated. This interval is the coordina¬
tion delay time. As we shall soon see we must provide for R3 a delay of 0.3 s
for the highest fault current seen by R4 (rather than the lowest fault current).
Then R3 will operate no less than 0.3 s after R4 for every possible fault seen
by R4.
The highest fault current seen by R4 is the current for a fault»just

beyond R4 toward bus 5, or 300.7 A according to Table 13.2. The relay
current of both R3 and R4 is then
300.7 x ^ = 30.1 A
and for a relay tap setting of 5 the ratio of relay current to tap setting for
both is 30.1/5 = 6.0. Figure 13.7 tells us the operating time for R4 is 0.135 s
since the time-dial setting is 1/2. So in the event of failure of R4, relay R3
must operate in
0.135 + 0.3 = 0.435 s
and Fig. 13.7 tells us the required time-dial setting for R3 is 2.0.
If we had provided for R3 a delay of 0.3 s for the lowest rather than the
highest fault current seen by R4, we would have determined a time-dial
setting of less than 2.0, which would not have provided a delay of 0.3 s for
the highest current seen by R4.
Setting for relay R2: The smallest fault current for which R2 must pick
up to provide backup for R3 is 237.9 A as given in Table 13.2. We might
choose a CT ratio of 100/5. Then with the required reliability which causes
us to design for pickup at one third of the minimum fault current we com¬
pute a pickup setting of
l x 237.9 x ~ 3.9 A
3 100
and we specify the 4.0 tap.

To find the time-dial setting for R2 we see that the maximum fault
current at bus 3 is 430.7 A. Relay R3 for this current will have a ratio of
relay current to pickup setting of
430.7 x A x i = 8.6
50 5
\
Since the time-dial setting of R3 is 2.0 that relay will operate in 0.31 s as
read from Fig. 13.7. So for proper coordination with R3 relay R2 must
operate in
0.31+0.3 =0.61 s
In backing up R3 relay R2 also sees the fault current of 430.7 A for

which this ratio of relay current to pickup setting is
430.7 = 5.4
We read from Fig. 13.7 a time-dial setting of 2.6.

The relay R1 is set similarly. The final CT ratios, pickup values, and
time-dial settings for all the relays are given in Table 13.3.
Table 13.3 Relay settings for Example 13.3
R1 R2 R3 R4
CT ratio 100: 5 100 : 5 50: 5 50 : 5

Pickup setting, A 5 4 5 5
Time-dial setting 2.9 2.6 2.0 2
It is worth pointing out that there is a slight danger of relay R3 operating

before R4 if the current in line 4-5 happens to be close to the relay pickup value
during a heavy load or light fault condition. Both R3 and R4 see the same current
(since they have the same CT ratios and pickup settings) and there may be just
enough of an error in the CTs or relays that R3 may see this as a trip condition
(current slightly greater than its pickup value) while R4 sees a current slightly
below its pickup value. To avoid possible problems of this nature, the pickup
tap setting of R3 should be set at a value somewhat greater than that for R4.
The system shown in Fig. 13.13 can be protected with time-over-current
relays (which are simple and relatively inexpensive) because it is a radial system.
Consider the system shown in Fig. 13.14a which has multiple sources, and the
system of Fig. 13.14i>, both of which are similar with respect to protection
methods since both are loop systems. In such systems the fault current will flow
from both ends of a transmission line for a fault on the line. Therefore to remove
a faulted line from the system circuit breakers must be provided at both ends of
each line. If, however, every relay responds only to the flow of currents in the
forward direction (toward its zone of protection) as shown by the arrows in
Fig. 13.14 and does nothing for currents in the reverse direction, then the loop
system can be protected much like a radial system. Relays associated with circuit
breakers A, C, E must be coordinated among themselves, and relays F, D, B are
coordinated together. The overcurrent relays are made directional by using an
additional directional relay at each location, and arranging the outputs of the
© ©
_ A B C D E F
O-o- —Q-O
(a)
Figure 13.14 One-line diagrams for loop systems. Heavy arrows beside each circuit breaker show
the direction to the fault for which the relay will respond. Relays having all arrows pointing in the
same direction around the loop coordinate with each other.
directional and overcurrent units in such a manner that a logical “and” operation
between their outputs is performed. Their associated breakers will not operate
unless both relays provide a trip signal.
(b) Protection of HV and EHV transmission lines

In a bulk power network there are no radial or single-loop systems. Many
generating stations and subtransmission feed points are interconnected to form a
network, so that no simple loops can be identified. In such a system it becomes
impossible to coordinate directional overcurrent relays to provide protection for
the transmission lines, since for a given fault location the current seen by the
relay varies over a very wide range depending upon the system operating
conditions.
The impedance relay described earlier provides a method of protecting
transmission lines connected in a network. The relay is made to respond to the
impedance between the relay location and the fault point. This impedance is
proportional to the distance to the fault, hence the name distance relay, and does
not depend upon the fault current levels. Consider the system shown in
Fig. 13.15a to be a portion of a large system. For a fault at I3! the relay R12,
whose forward direction is in the direction from bus 1 to bus 2, is designed to
respond to the positive sequence impedance (or distance) between bus 1 and Pv
Similarly, a relay designated R21 is located at bus 2 with a forward direction
from bus 2 to bus 1.
©'/ Pl'\
-D- B21
hO t>|
B12 P B23 3QO
B32 1
^3 Zone 1 } P:
-OH B24 @
l\
\V _Zone_2_ 4/
o-j
B42'
Zone 3
(a)
(b)
Figure 13.15 Coordination of distance (impedance) relays. The zone of protection shown by the solid
line in (a) is replaced by zones 1 and 2 identified by dashed lines. Zone 3 provides backup protection
for neighboring protection systems. Time delay and operating time is shown in (b) for R12 R23 and
R24.
The impedance relays which respond to line-to-line voltages (such as

K — K) and the difference between line currents (such as Ia — Ib, called delta
currents) are known as phase relays. They detect the positive-sequence impedance
between the fault point and the relay location. Three such relays respond cor¬
rectly to all possible line-to-line faults, double line-to-ground faults, and three-
phase faults. These relays, however, do not respond correctly to line-to-ground
faults. Three additional relays which utilize line-to-neutral voltages Va, Vb, Vc,
line currents Ia, Ib, Ic and the zero-sequence current I0 are provided, and
which detect the positive-sequence impedance between the fault and the relay
location for all line faults involving ground.
The distance relays are made directional by incorporating a directional unit
similar to that in a directional overcurrent relay. There are distance relays which
do not need an added directional unit because they have directionality inherent
in their design. The principal example of such a relay is the mho relay described
earlier. The directionality is necessary so that a relay will respond to the distance
in the forward direction (that is, looking into its zone of protection) and block
its operation for all faults in the reverse direction.
Consider the application of the directional distance relays to the protection
of line 1-2 in Fig. 13.15a. The zone of protection for the relays R12 and R21
which protect this line is shown by a solid line. For the relay R12 the faults at
P2, P3, and P4 all appear to be at the same distance from bus 1; yet faults at P3
and P4 are clearly outside the zone of protection of R12. Consequently if the
distance relay is set to respond to the fault at P2, it will also respond to the
faults at P3 and P4. This is an improper operation of relay R12. To avoid this
basic problem, the zone of the distance relay is modified as shown for R12 by
the dashed lines in Fig. 13.15a. The single zone of protection shown by the solid
line is replaced by two zones: zone 1 and zone 2. Zone 1 extends a shorter
distance than the zone shown by the solid line and is usually about 80% of
the line length. For faults within this zone, the distance relay at bus 1 operates
normally (that is, as quickly as possible). This shortened zone 1 is commonly
known as an underreaching zone. Zone 2, on the other hand, extends beyond
the line terminal well into the lines connected to the remote bus and is said to be
overreaching. The relay R12 responds to a zone-2 fault with a time delay so that
it may coordinate with R23 and R24.
Similar zone 1 and zone 2 settings are available for the relay R21 at bus 2.
For a fault at Pl9 both relays R12 and R21 operate at highest possible speed
since this fault is in zone 1 of both relays. A fault at P2 is in zone 1 of R21 and
consequently breaker B21 will be tripped at high speed. Relay R12 however will
not clear the fault at high speed, since this fault lies in zone 2 of R12. After its
zone 2 time delay has expired, the relay R12 will operate and trip circuit breaker
B12. There is thus a delayed fault clearing from bus 1 while the bus 2 end is
cleared at high speed for faults such as P2.
Now consider a fault at P3. This fault, lying in zone 1 of relay R23 will be
cleared by R23 and circuit breaker B23 at a high speed. Relay R12 will trip
circuit breaker B12 at bus 1 to isolate the fault in the zone-2 time of relay R12 if
B23 has failed to operate. Clearly the zone-2 clearing time must be slower than
the slowest possible zone-1 clearing time of relay R23, so that R12 does not trip
for the fault at P3 prematurely. A similar zone-2 setting also exists at bus 4 for
relay R42 and circuit breaker B42. The fault at P3 lies in zone 2 of relay R42.
The response time of relays R12, R23, and R24 to their respective zone-1 and
zone-2 faults is shown schematically in Fig. 13.156. The abscissa in the response¬
time diagram is the fault location along the appropriate lines, and the ordinate is
the relay operating time. The zone-1 operating time is of the order of 1 cycle,
whereas the zone-2 operating time varies between 15 and 30 cycles.
In most cases, the distance relays are provided with another zone of protec¬
tion known as their zone 3 to provide remote backup for the neighboring lines.
The third zone of a relay must reach beyond the longest line emanating from the
bus at the remote end of its protected line. The remote backup function must
coordinate with the primary protection which it backs up. Thus the third zone of
relay R12 must be coordinated with zone 2 of the relays at bus 2 (R23 and R24).
Notice also that the Fig. 13.156 points to an important principle of relay coor¬
dination. The coordination is by time and also by distance. A faster zone of
protection must reach in distance beyond the reach of its slower backup. Thus
the zone-2 reach of relay R12 is shorter than the zone-1 reach of relay R23 or
R24. Similarly zone-3 reach of relay R12 is shorter than the zone-2 reach of R23
and R24. Unless this coordination by distance was employed, for certain faults,
instead of high-speed clearing an unnecessarily slow backup time clearing would
result. The zone-3 coordinating time is generally of the order of one second. The
three zones of protection of relay R12 and the operating times for the three
zones are shown schematically in Fig. 13.156.
The characteristic of a directional distance relay in the complex R-X plane is
shown in Fig. 13.16a. A straight line called the line impedance locus is shown in
the figure. Along this line the positive-sequence impedances of the protected line
Figure 13.16 Characteristics of (a) directional impedance relay and (b) mho relay for Example 13.4.
as seen by the relay between its location and different points along the protected
line can be plotted. The directional unit of the relay causes separation of the trip
and block regions of the relay characteristic in Fig. 13.16a by a line drawn
perpendicular to the line impedance locus. The impedance measured from bus 1
to buses 2 and 4 along the line impedance locus are indicated by numerals
enclosed in circles. Zone circles whose centers are at the origin of the R-X plane
have radii equal to the magnitudes of the impedances of the protected line seen
by R12 from bus 1 to the end of the zone identified by the numbers of the circles.
Thus, the intersection of a zone circle with the line impedance locus is the line
impedance between the relay and the end of the zone. After a fault occurs the
impedance seen by the relay is very small compared to the load impedance seen
during normal operation. The relays operate when the impedance seen by the
relay lies within a zone circle. For impedances inside the zone-1 circle, operation
occurs in minimum time. Time delays are set for later operation for zone-2 and
then zone-3 faults.
The characteristic of a mho relay is shown in Fig. 13-166 with the centers
of the zone circles lying on the line impedance locus. Note that the radii of the
zone circles are half of the radii of the corresponding zone circles of the direc¬
tional distance relay for the same impedances since the intersection of a zone
circle with the line impedance locus must still be the impedance of the line seen
by the relay between its location and the end of the zone.
The three-step distance relaying scheme described here provides a versatile
protection system for high-voltage transmission lines. With minor modifications
to accommodate the nature of a given system this protection is used almost
universally to protect transmission lines of a modern power network. In many
cases ground faults are provided for by the directional time overcurrent relays of
the type described earlier, while three-phase and line-to-line faults are covered by
distance relays. Occasionally it becomes very difficult to coordinate the third
zone times of a relay with the second zones of the neighboring lines; especially at
EHV levels. In such cases, the remote backup (zone-3) function of the distance
relays may be omitted.
During emergency load-flow conditions when the load is quite high the
impedance seen by the relay is low and must be checked to make sure that it
does not fall within one of the zone circles of the relay characteristic.
Example 13.4 Consider again the portion of a 138-kV transmission system

shown in Fig. 13.15a. Lines 1-2, 2-3, and 2-4 are respectively 64, 64, and
96 km (40, 40, and 60 mi) long. The positive-sequence impedance of the
transmission lines is 0.05 + y'0.5 ohm per kilometer. The maximum load
carried by line 1-2 under emergency conditions is 50 MV A. Design a three-
zone step distance relaying system to the extent of determining for R12 the
zone settings which are the impedance values in terms of CT and CVT
secondary quantities. The zone settings give points on the R-X plane
through which the zone circles of the relay characteristic must pass.
Solution The positive sequence impedances of the three lines are
Line 1-2 3.2 + y'32.0 LI

Line 2-3 3.2 + ;32.0 f2
Line 2-4 4.8 + ;48.0 Q
Since distance relays depend on the ratio of voltage to current, both a

CT and CVT will be needed for each phase. The maximum load current is
50 x 106
209.2 A
^3 x 138 x 103
and we select a CT ratio of 200/5 which will produce about 5 A in the

secondary winding under maximum loading conditions.
System voltage to neutral is
138 x 103
= 79.67 x 103 V
^7T
It was pointed out in Sec. 13.3 that the industry standard for CVT secondary
voltage is 67 V for line-to-neutral voltages. Consequently we select a CVT
ratio of
79.67 x 103
1189.1/1
67
in which case a normal system voltage will produce 67 volts for each phase
on the CVT secondary. Denoting primary and secondary voltages of the
CVT at bus 1 as Vp and the primary current of the CT as Ip, we have for the
impedance measured by the relay
7P/1189.1
= Z,line X 0.0336
V40
Thus the impedances of the three lines as seen by the relay R12 are
approximately
Line 1-2 0.11 + j 1.1 Q, secondary
Line 2-3 0.11 + jl.l Q, secondary
Line 2-4 0.16 +j 1.6 Q, secondary
The maximum load current of 209.2 A appears to the relay, assuming a

power factor of 0.8 lagging, to be
67 .
Z|oad ~ 209.2(5/200) ^°'8 +;0'6)
= 10.2 -(- jl.l Q, secondary
The zone-1 setting of the relay R12 must underreach the line 1-2, so that
the setting should be
0.8 x (0.11 + jl.l) = 0.088 -f j0.88 Q, secondary

The zone-2 setting should reach past terminal 2 of the line 1-2. To allow for
the various possible inaccuracies of the transducer-relay system, zone 2 is
usually set at about 1.2 times the length of the line being protected. Zone 2
for R12 is therefore set at
1.2 x (0.11 -I- jl.l) = 0.13 + yl.32 O, secondary
The zone-3 setting should reach beyond the longest line connected to bus 2.
Thus the zone-3 setting must be
(0.11 + 71.1) + 1.2 x (0.16 + 7*1.6) = 0.302 + ;3.02 Q, secondary
Note that once again a factor of 1.2 is used as a multiplier on the impedance
of the longest line connected to bus 2 to make sure that zone 3 of relay R12
will reach past bus 4 even in the presence of inaccuracies introduced in the
relaying system. A directional impedance relay with the characteristic shown
in Fig. 13.16a can be used.
Both Figs. 13.16a and 13.166 apply to this example, and points 2 and 4
on the line impedance locus correspond to the calculated impedances from
bus 1 to buses 2 and 4. As seen by the relay the load impedance Z,oad is more
than 3 times the line impedance from bus 1 to bus 4 and lies well beyond the
zone-3 circle of both the directional impedance relay and the mho relay.
Consequently there is no danger of tripping the line during any load swings
that may occur on the transmission line. If the maximum load had been too
close to the zone-3 setting of the directional impedance relay, it might have
been necessary to replace it by the mho relay whose zone-3 circle encloses a
smaller area of the R-X plane, as is well apparent in Fig. 13.16.
(c) Line protection with pilot relays

It was pointed out in the previous section that a line (such as 1-2 in Fig. 13.17)
is protected by zone 1 and zone 2 of a distance relay and that the normal reach
for zone 1 is about 80 percent of the line length. This zone is often called the
direct-trip zone or a high-speed zone, and relay operating times of the order of
one cycle are quite common for zone-1 faults. A fault such as that at P2 in
Fig. 13.17 falls in zone 2 of relay R12 and will be cleared by relay R12 in its
zone-2 time. Of course an identical set of relays exists at bus 2, and these relays
will see the fault P2 as being in their zone 1. Consequently the breaker at bus 2
will operate in high-speed relaying time for the fault at P2. Thus for faults in the
middle 60 percent of the line, both ends clear the fault with high speed, whereas
for faults in the 20 percent line length at either end, the nearest end will clear the
fault at high speed while the distant end will clear the fault with a zone-2 time
delay.
G)
Pi
—□“
B12
Figure 13.17 Line to be protected by pilot relaying.
In modern HV and EHV systems, this delayed clearing from the remote end
is often found to be unacceptable because of the complex nature of the modern
interconnected network and the tighter stability margins. It then becomes neces¬
sary to provide high-speed protection for the entire line (instead of the middle 60
percent). High-speed protection for the entire line is provided by pilot relaying of
the type described in Sec. 13.4. For a fault anywhere on the protected line, the
directional relays R12 and R21 see a unique condition: both relays see the fault
current flowing in the forward direction. This information, when communicated
to the remote ends over a pilot channel, confirms that a fault is indeed on the
protected line. Consider faults at P2 and P3 on line 1-2 shown in Fig. 13.15.
Although the relay R12 sees no difference between the two faults, the relay R21
sees P2 as an internal fault and P3 as an external fault (being in the reverse
direction from its zone of protection). Upon receiving this directional informa¬
tion at R12, that relay will be able to block (prevent) tripping for the fault at P3.
(Note that the fault at P3 is in the zone of protection of bus 2 but not of the line
1-2). The fault at P2 is tripped simultaneously at high speed from both ends.
Such a system is called a “directional comparison” pilot scheme. Equivalent
performance can be obtained by comparing the phase angles of the fault currents
seen at the two ends of a line, and exchanging this phase angle information over
the pilot channel. Such a scheme is known as a “ phase comparison ” scheme. A
discussion of the merits and demerits of the two schemes is beyond the scope of
this book, although it may be mentioned that both types of protection schemes
are used in practice.
13.7 PROTECTION OF POWER TRANSFORMERS
As in the case of transmission lines, the type of protection used for power
transformers depends upon their size, voltage rating, and the nature of their
application. For small transformers (smaller than about 2 MVA) protection with
fuses may be adequate; whereas for transformers of greater than 10-MVA capa¬
city differential relays with harmonic restraint may be used-
Consider first the differential protection of a single-phase two-winding
power transformer shown in Fig. 13.18. If the power transformer is carrying load
ii v2
-► • • -
Figure 13.18 Connection diagram for differential

protection of a transformer.
currents /j and /'2 in the primary and secondary windings, we know that with
magnetizing current neglected
(13.9)
12 Nt
where Nt and N2 are the turns of the primary and secondary windings of the
power transformer, respectively. The CT secondary currents are Ix and /2, and
the turns ratio of the CTs on the primary and secondary side of the power
transformer are n1 and n2, respectively (one turn on the primary of a CT means
n turns on the secondary). So
h=— 12 = - (13.10)
»1 n2
To prevent tripping under normal conditions when Eq. (13.9) must be true the
current /j - I2 through the trip coil must be zero; that is Ix must equal I2. So
from Eqs. (13.9) and (13.10) we require that
«I =N2
(13.11)
n2 N,
With an internal fault on the secondary side of the power transformer and with
current I'f in the fault
Ii-I2 = IjL (13-12)

n2
For a fault on the primary side of the transformer, the right-hand side of
Eq. (13.12) would be If the relay is set with a sufficiently small value of
pickup current | Ip | the differential relay would trip for an internal fault, while
blocking for external faults or normal load.
As we discussed in Sec. 8.10, a power transformer is usually equipped with
variable tap settings, which allow its secondary voltage to be adjusted over a
certain range. The adjustments usually vary in small steps over a range ± 10
percent from the nominal turns ratio of Nl/N2. If tap settings result in an
off-normal turns ratio, the relay will see a differential current during normal load
conditions. To avoid improper operation in this case a percent differential relay
must be used.
Three-phase transformers with Y-A windings require further discussion. As
pointed out in Sec. 11.4 the primary and secondary currents of such transformers
differ in magnitude and phase angles during normal operating conditions. The
current transformers must therefore be connected in such a manner that the CT
secondary line currents as seen by the relay are in phase, and the CT ratios must
also be adjusted so that the current magnitudes as seen by the relay are equal
under normal (unfaulted) conditions. The correct phase-angle relationship is
obtained by connecting the CTs on the wye side of the power transformer in
delta, and those on the delta side of the power transformer in wye. In this
manner, the CT connections compensate for the phase shift created by> the
Y-A-connected power transformer. These considerations are illustrated by the
following example.
Example 13.5 A three-phase 345/34.5 kV transformer is rated at 50 MVA

and its short-term emergency rating is 60 MVA. Using standard CT ratios
available, determine the CT ratios, CT connections, and the currents in the
power transformer and the CTs. The 345-kV side is Y-connected, while the
34.5-kV side is A-connected.
Solution The currents on the 345-kV side and 34.5-kV side of the trans¬
former when it is carrying its maximum expected load are
60 x 106 60 x 106
100.4 A and = 1004.1 A
73 x 345 x 103 73 x 34.5 x 103
We will use CT ratio of 1000/5 on the 34.5-kV side. Since the CTs on
this side are connected in wye, the current flowing to the differential relay
from this side will be
1004 x £ 5.0 A
1000
To balance this current, the line currents produced from the A-connected
CTs on the 345-kV side must also be 5.0 A. This will require that each of the
secondary windings of the A-connected CTs have a current of
5.0
2.9 A
75
This current in the CT secondary windings would require CT ratios of
100.4
34.64
2.9
for the CTs on the 345-kV side. The nearest available* standard CT ratio is
200/5. If we use this ratio, the CT secondary currents would be
100.4 2.51 A
and the line currents from the A-connected CTs to the differential relays
would be
2.51 x ^3 = 4.35 A
Clearly this current cannot balance the 5.0 A produced by the 34.5-kV side.
This situation of mismatched currents when standard CT ratios are used
is quite commonly encountered in designing the protection system for Y-A-
connected transformers. A convenient solution is provided by auxiliary cur¬
rent transformers which provide a wide range of turn ratios. These auxiliary
Differential CT Auxiliary CT 1/1.155
Figure 13.19 Wiring diagram showing currents in amperes for the power and relaying circuits
of Example 13.5.
CTs are small and inexpensive devices since their primary and secondary
windings are low-voltage low-current circuits. Using a set of three auxiliary
CTs with turns ratios of
5.0
1.155
435
would produce a balanced set of currents in the differential relay when the
power transformer is carrying normal load. For an assumed power factor 0.8
lagging, the currents in the power transformer and the various CTs are as
shown in Fig. 13.19.
Although the auxiliary transformers can be used as discussed here, in general it

is a good practice to use these only as a last resort. The auxiliary CTs add their
own burden to the main CTs, and also add to the total errors of transformation. A
much more suitable approach is to use tap settings on relay coils themselves, which
serve the same purpose as a variable turns ratio auxiliary CT. In most cases, relay
coil tap settings provide adequate margin to correct for most practical ratio
mismatches.
Notice that in assuming the primary and secondary currents of power trans¬
formers are related by their turns ratio, we have assumed that their magnetizing
currents are negligible. This is a good approximation when the transformer is in
normal operation and the magnetizing current is very small. However, wjien a
transformer is energized, it can draw very heavy magnetizing currents (known as
the magnetizing inrush currents) which decay with time to the very small steady-
state value. If this high inrush current does flow during energization of the trans¬
former, it appears as differential current since it flows in the primary winding only.
It is essential to detect such a condition and stop the differential relay from tripping
the transformer. One of the more common methods used to accomplish this func¬
tion is based upon the fact that the magnetizing inrush current is rich in
harmonics whereas a fault current is a purer fundamental frequency sinusoid. To
take advantage of this fact, in addition to the fundamental frequency restraining
current of (7j + l2)/2, another restraining signal proportional to the harmonic
content of the differential current is generated in the differential relay. By selecting
a proper weighting coefficient for the harmonic component, it is possible to prevent
the differential relay from tripping the transformer during its energization, even
though a substantial tripping current may be produced due to the magnetizing
inrush current.
13.8 RELAY HARDWARE

In Sec. 13.4 we described the logical design of certain types of relays. Most of the
relays discussed so far have been built with electromechanical devices. Some of
the more common types include plunger type relays, balance-beam type relays,
and rotating cup (or disc) relays which are similar to the watt-hour meters found
on most residential circuits. The electromechanical relays have served well since
the earliest days, and are an important part of the present protection system
design practices. These relays are robust, inexpensive, and relatively immune to
the harsh environment in an electric power substation. Their response time is
somewhat slow in terms of modern power system needs, and also their design is
somewhat inflexible in terms of available characteristics, burden capability, and
tap settings.
In the late 1950s relays using solid-state circuitry were introduced. These
relays use analog circuits in conjunction with logic gates to produce the desired
relay characteristic. Solid-state relays are capable of providing characteristics
similar to those of electromechanical relays, and in fact a number of newer
characteristic shapes are available with solid-state relays. Although earlier
models of these relays were prone to frequent component failures under the
harsh operating environment of a substation, newer models do perform very well
and are providing excellent service on HV and EHV power systems. Quite
recently, relays based on microcomputers have been proposed, and are under
active investigation at this time.
13.9 SUMMARY
In this chapter we have concentrated on the discussion of protective relaying
systems for modern high-voltage power networks. We did not study the applica¬
tion of fuses and reclosers, which are more commonly used on distribution
CD
Figure 13.20 One-line diagram for Prob.

13.4.
systems. In many ways, the coordination of these devices is similar to that of

time overcurrent relays discussed in Secs. 13.4 and 13.6. In terms of equipment
protection we discussed transformer protection in Sec. 13.7. Protection of gener¬
ators has been mentioned only in connection with a differential relay which can
also be used to protect bus bars. Protective systems for equipment and lines offer
unique challenges worthy of extensive treatment.!
PROBLEMS
13.1 In the system whose one-line diagram and zones of protection are shown in Fig. 13.36 where is
the fault located if the breakers tripped are (a) G and C, (b) F, G, and H, (c) F, G, H, and B, and
(d) D, C, and F?
13.2 Determine the time-dial setting for R3 of Example 13.3 if the setting had been determined for
the lowest fault current seen by R4? Why is the setting made for the highest rather than the lowest
fault current seen by R4?
13.3 In Example 13.3, assume that a line-to-line fault occurs at the midpoint of line 2-3. Which relay
will operate for this fault? What will be its operating time? Assume that this relay fails to clear the
fault, which relay will operate to clear the fault now? How long will it take?
13.4 An 11-kV radial system is shown in Fig. 13.20. The positive- and zero-sequence impedances of
line 1-2 are 0.8 Q and 2.5 fl, respectively. Impedances of line 2-3 are three times as large. The positive-
and zero-sequence impedances of each of the two transformers are j'2.0 ohms and j 3.5 ohms, respectively.
Under emergency conditions, the system may be operated with one transformer out of service. Deter¬
mine the CT ratios, pickup values, and time-dial settings for IFC-53 relays designed to provide single
line-to-ground fault protection for this system. Assume the high-tension bus is an infinite bus whose
voltage will provide 11 kV at the low-tension bus at no load.
13.5 A portion of a 765-kV network is shown in Fig. 13.21. The positive and zero sequence impedances
of the transmission lines are 0.01 -I- j0.6 ohm and 0.1 + yl .8 ohms per mile respectively. Assume that
the generator impedance isylO.O ohms for positive and negative sequence, andy'20.0 ohms for the zero
sequence.
(a) Relays R12, R23, and R34 use inverse-time directional IFC-53 overcurrent relays for single
line-to-ground fault protection of this system. Determine the fault current needed for setting the pickup
value of R12 ground relay. You may neglect line resistances for this calculation.
t See, for instance, C. R. Mason, The Art and Science of Protective Relaying, John Wiley and
Sons, Inc., New York, 1956, and Westinghouse Electric Corporation, Applied Protective Relaying,
Relay Instruments Division, Newark, N.J., 1976.
©B12 @
B21 B23 B32 B34 B43
o-o- -Q
Figure 13.21 One-line diagram for Prob.
-300 mi ■ -150 mi ■ -100 mi-
13.6.
(b) Select CT and CVT ratios for phase distance relays at bus 1. You may assume that the relay
current coils can carry 10 A continuously, and that the emergency line loading limit is 3000*MVA.
Use standard CT ratios.
(c) Determine and show on the (secondary) R-X diagram the three zones of a directional im¬
pedance relay at bus 1 for phase fault protection.
(d) Also show the location of the equivalent impedance of the emergency load on the R-X
diagram. Do you see any problems with line operating at emergency load? What solution would you
propose ?
13.6 A three-phase fault occurs at the terminals of the delta winding in Fig. 13.19 within the zone of
protection of the differential relay. Assume that the positive-sequence impedance of the transformer
as seen from the 345-kV side is j250 ohms, and that the power system feeding the 345-kV side is of
infinite short-circuit capacity. What are the currents flowing in all the leads shown in Fig. 13.19 in
this case? Neglect prefault current and assume no fault current originates in the low-tension part of
the system.
CHAPTER
FOURTEEN
POWER SYSTEM STABILITY
When ac generators were driven by reciprocating steam engines, one of the

major problems in the operation of machinery was hunting. The periodic varia¬
tions in the torque applied to the generators caused periodic variations in speed.
The resulting periodic variations in voltage and frequency were transmitted to
the motors connected to the system. Oscillations of the motors caused by the
variations in voltage and frequency sometimes caused the motors to lose syn¬
chronism entirely if their natural frequency of oscillation coincided with the
frequency of oscillation caused by the engines driving the generators. Damper
windings were first used to minimize hunting by the damping action of the losses
resulting from the currents induced in the damper windings by any relative
motion between the rotor and the rotating field set up by the armature current.
The use of turbines has reduced the problem of hunting, although it is still
present where the prime mover is a diesel engine. Maintaining synchronism
between the various parts of a power system becomes increasingly difficult,
however, as the systems and interconnections between systems continue to grow.
14.1 THE STABILITY PROBLEM
Power system stability may be defined as that property of the system which
enables the synchronous machines of the system to respond to a disturbance
from a normal operating condition so as to return to a condition where their
operation is again normal. Stability studies are usually classified into three types
depending upon the nature and order of magnitude of the disturbance. These are
transient, dynamic, and steady-state stability studies.
373
Today, transient stability studies constitute the major analytical approach to

the study of power-system electromechanical dynamic behavior. Transient sta¬
bility studies are aimed at determining if the system will remain in synchronism
following major disturbances such as transmission system faults, sudden load
changes, loss of generating units, or line switching. Such studies began more than
50 years ago but were then confined to consideration of the dynamic problems of
not more than two machines. Present-day power systems1 are vast, heavily inter¬
connected systems with many hundreds of machines which can dynamically
interact through the medium of their extra-high voltage and ultra-high voltage
networks. These machines have associated excitation systems and turbine-
governing control systems which, in some but not all cases, must be modeled in
order to properly reflect the correct dynamic response of the power system to
certain system disturbances.
Dynamic and steady-state stability studies are less extensive in scope and
involve one or just a few machines undergoing slow or gradual changes in
operating conditions. Therefore, dynamic and steady-state stability studies con¬
cern the stability of the locus of essentially steady-state operating points of the
system. The distinction made between steady-state and dynamic stability studies
is really artificial since the stability problems are the same in nature; they differ
only in the degree of detail used to model the machines. In dynamic stability
studies, the excitation system and turbine-governing system are represented
along with synchronous machine models which provide for flux-linkage varia¬
tion in the machine air-gap. Steady-state stability problems use a very simple
generator model which treats the generator as a constant voltage source. The
solution technique of steady-state and dynamic stability problems is to examine
the stability of the system under incremental variations about an equilibrium
point. The nonlinear differential and algebraic equations for the system can be
replaced by a set of linear equations which are then solved by methods of linear
analysis to determine whether the machine or machines will remain in synchro¬
nism following small changes from the operating point.
Transient stability studies are much more commonly undertaken thereby
reflecting their greater importance in practice. Such problems involve large dis¬
turbances which do not allow the linearization process to be used and the non¬
linear differential and algebraic equations must be solved by direct methods or by
iterative step-by-step procedures. Transient stability problems can be subdivided
into first-swing and multiswing stability problems. First-swing stability is based
on a reasonably simple generator model without representation of control
systems. Usually the time period under study is the first second following a
system fault. If the machines of the system are found to remain in synchronism
within the first second, the system is said to be stable. Multiswing stability
problems extend over a longer study period and therefore must consider effects
of generator control systems which affect machine performance during the ex¬
tended time period. Machine models of greater sophistication must be repre¬
sented to reflect proper behavior.
In all stability studies, the objective is to determine whether or not the rotors
POWER SYSTEM STABILITY 375
of the machines being perturbed return to constant speed operation. Obviously,

this means that the rotor speeds must depart at least temporarily from syn¬
chronous speed. To facilitate computation, three fundamental assumptions are
made in all stability studies:
1. Only synchronous frequency currents and voltages are considered in the

stator windings and the power system. Consequently, dc offset currents and
harmonic components are neglected.
2. Symmetrical components are used in the representation of unbalanced faults.
3. Generated voltage is considered unaffected by machine speed variations.
These assumptions permit the use of phasor algebra for the transmission
network and solution by load-flow techniques using 60-Hz parameters. Also,
negative- and zero-sequence networks can be incorporated into the positive-
sequence network at the fault point. As we shall see, three-phase balanced faults
are generally considered. However, in some special studies, circuit-breaker clear¬
ing operations may be such that consideration of unbalanced conditions is
unavoidable.t
14.2 ROTOR DYNAMICS AND THE SWING EQUATION
The equation governing the motion of the rotor of a synchronous machine is

based on the elementary principle in dynamics which states that accelerating
torque is the product of the moment of inertia of the rotor times its angular
acceleration. In the MKS system of units, this equation can be written for the
synchronous generator in the form
J~w=T“= T”~ T* N-m (141)
where the symbols have the following meanings:
J the total moment of inertia of the rotor masses, in kg-m2

6m the angular displacement of the rotor with respect to a stationary axis,
in mechanical radians
t time, in seconds
Tm the mechanical or shaft torque supplied by the prime mover less retard¬
ing torque due to rotational losses, in N-m
Te the net electrical or electromagnetic torque, in N-m
Ta the net accelerating torque, in N-m
The mechanical torque Tm and the electrical torque Te are considered positive for
the synchronous generator. This means that Tm is the resultant shaft torque
t For information beyond the scope of this book, see P. M. Anderson and A. A. Fouad, Power
System Control and Stability, The Iowa State University Press, Ames, Iowa, 1977.
Figure 14.1 Representation of a machine rotor comparing direction of rotation and mechanical and
electrical torques for (a) a generator and (b) a motor.
which tends to accelerate the rotor in the positive 6m direction of rotation as

shown in Fig. 14.1a. Under steady-state operation of the generator Tm and Te are
equal and the accelerating torque Ta is zero. In this case there is no acceleration
or deceleration of the rotor masses and the resultant constant speed is the
synchronous speed. The rotating masses which include the rotor of the generator
and the prime mover are said to be in synchronism with the other machines
operating at synchronous speed in the power system. The prime mover may be a
hydro turbine or a steam turbine for which models of different levels of complex¬
ity exist to represent their effect on Tm. In this text Tm is considered constant at
any given operating condition. This assumption is a fair one for generators even
though input from the prime mover is controlled by governors. Governors do
not act until after a change in speed is sensed and so are not considered effective
during the time-period in which rotor dynamics are of interest in our stability
studies here. The electrical torque Te corresponds to the net air-gap power in the
machine and thus accounts for the total output power of the generator plus
\I\2R losses in the armature winding. In the synchronous motor the direction of
power flow is opposite to that in the generator. Accordingly, for a motor both Tm
and Te in Eq. (14.1) are reversed in sign as shown in Fig. 14.lb. Te then corre¬
sponds to the air-gap power supplied by the electrical system to drive the rotor
while Tm represents the counter-torque of the load and rotational losses tending
to retard the rotor.
Since 9m is measured with respect to a stationary reference axis on the stator,
it is an absolute measure of rotor angle. Consequently, it continuously increases
with time even at constant synchronous speed. Since the rotor speed relative to
synchronous speed is of interest it is more convenient to measure the rotor
angular position with respect to a reference axis which rotates at synchronous
speed. Therefore, we define
dm = cosmt + Sm (14.2)
where cosm is the synchronous speed of the machine in mechanical radians per
second and 5m is the angular displacement of the rotor, in mechanical radians,
from the synchronously rotating reference axis. The derivatives of Eq. (14.2) with
respect to time are
djL ,ddm
(143)
dt a,- + ~df
and
d2em = d2Sm
(14.4)
dt2 dt2
Equation (14.3) shows that the rotor angular velocity d6m/dt is constant and
equals the synchronous speed only when ddm/dt is zero. Therefore dSm/dt repre¬
sents the deviation of the rotor speed from synchronism and the units of measure
are mechanical radians per second. Equation (14.4) represents the rotor accelera¬
tion measured in mechanical radians per second-squared.
Substituting Eq. (14.4) into Eq. (14.1), we obtain
jd^?=Ta=Tm-Te N-m (14.5)
It is convenient for notational purposes to introduce
dOm
co m (14.6)
dt
for the angular velocity of the rotor. We recall from elementary dynamics that
power equals torque times angular velocity and so, multiplying Eq. (14.5) by com,
we obtain
Jcom^? = P. = Pm-P, W (14.7)
where Pm is the shaft power input to the machine less rotational losses, Pe is the
electrical power crossing its air-gap and Pa is the accelerating power which
accounts for any unbalance between those two quantities. Usually we will neg¬
lect rotational losses and armature \I\2R losses and think of Pm as power
supplied by the prime mover and Pe as the electrical power output.
The coefficient Jcom is the angular momentum of the rotor; at synchronous
speed cosm, it is denoted by M and called the inertia constant of the machine.
Obviously the units in which M is expressed must correspond to those of J and
(om. A careful check of the units in each term of Eq. (14.7) shows that M is
expressed in joule-seconds per mechanical radian and we write
M^r = P. = Pm-P, w (14.8)

While we have used M in this equation, the coefficient is not a constant in the
strictest sense because com does not equal synchronous speed under all conditions
of operation. However, in practice com does not differ significantly frorq,syn-

chronous speed when the machine is stable and since power is more convenient
in calculations than torque, Eq. (14.8) is preferred. In machine data supplied for
stability studies, another constant related to inertia is very often encountered.
This is the so-called H constant which is defined by
stored kinetic energy in megajoules at synchronous speed

machine rating in MV A
and
iJoiL iMUsm
MJ/MVA (14.9)
H
5mach 5mach
where Smach is the three-phase rating of the machine in MVA. Solving for M in
Eq. (14.9), we obtain
2H
M = Smach MJ/mech rad (14.10)
co«
and substituting this equation in Eq. (14.8), we find
2H d*Sm P. =K^P,
(14.11)
^sm dt ^mach “^mach
This equation leads to a very simple result.

Note that Sm is expressed in mechanical radians in the numerator of
Eq. (14.11) while cosm is expressed in mechanical radians per second in the de¬
nominator. Therefore, we can write the equation in the form
2Hd*d
= p
x a
= xp m - xpe per unit (14.12)
cos dt2
provided both S and cos have consistent units which may be mechanical or
electrical degrees or radians. H and t have consistent units since megajoules per
megavoltampere is in units of time in seconds and Pa, Pm, and Pe must be in per
unit on the same base as H. When the subscript m is associated with co, cos, and 5
it means mechanical units are used; otherwise electrical units are implied. Ac¬
cordingly, cos is the synchronous speed in electrical units. For a system with an
electrical frequency of/hertz, Eq. (14.12) becomes
H^cPS
= P
A a
= APm - per unit (1413)
nf dt2
when <5 is in electrical radians while
H d2S
= P
1 a = 1P m - x
Pe per unit (14.14)
180/ dt2
applies when S is in electrical degrees.

Equation (14.12), called the swing equation of the machine, is the fundamen¬
tal equation which governs the rotational dynamics of the synchronous machine
in stability studies. We note that it is a second-order differential equation which
can be written as the two first-order differential equations
IHdco
— — = pm-Pe per unit (14.15)
cos at
dS
~ = (O-0)s (14.16)
in which a>, cos, and S involve electrical radians or electrical degrees. We will use
the various equivalent forms of the swing equation throughout this chapter to
determine the stability of a machine within a power system. When the swing
equation is solved we obtain the expression for d as a function of time. A graph
of the solution is called the swing curve of the machine and inspection of the
swing curves of all the machines of the system will show whether the machines
remain in synchronism after a disturbance.
14.3 FURTHER CONSIDERATIONS OF THE

SWING EQUATION
The MVA base used in Eq. (14.11) is the machine rating which is introduced by
the definition of H. In a stability study of a power system with many synchro¬
nous machines, only one MVA base common to all parts of the system can be
chosen. Since the right-hand side of the swing equation for each machine must
be expressed in per unit on this common system base, it is clear that H on the
left-hand side of each swing equation must be also consistent with the system
base. This is accomplished by converting the H constant for each machine based
on its own individual rating, to a value determined by the system base, Ssystem.
Equation (14.11), multiplied on each side by the ratio (Smach/Ssystem), leads to the
conversion formula
(14.17)
^system
in which the subscript for each term indicates the corresponding base being used.
In industry studies the system base usually chosen is 100 MVA.
The inertia constant M is rarely used in practice and the forms of the swing
equation involving H are more often encountered. This is because the value of M
varies widely with the size and type of the machine whereas H assumes a much
narrower range of values as shown in Table 14.1. Machine manufacturers also
use the symbol WR2 to specify for the rotating parts of a generating unit (includ¬
ing the prime mover) the weight in pounds multiplied by the square of the radius
of gyration in feet. Hence WR2/32.2 is the moment of inertia of the machine in
slug-feet squared.
Table 14.1 Typical inertia constants of synch¬

ronous machinest
Inertia constant H%
Type of machine MJ/MVA
Turbine generator:
Condensing, 1800 r/min 9-6
3600 r/min 7-4
Noncondensing, 3600 r/min 4-3
Waterwheel generator:
Slow-speed, < 200 r/min 2-3
High-speed, >200 r/min 2-4
Synchronous condenser :§
Large 1.25
Small 1.00
Synchronous motor with load 2.00
varies from 1.0 to 5.0 and
higher for heavy flywheels
t Reprinted by permission of the Westinghouse Electric

Corporation from “Electrical Transmission and Distribu¬
tion Reference Book.”
J Where range is given, the first figure applies to mach¬
ines of smaller megavoltampere rating.
§ Hydrogen-cooled, 25% less.
Example 14.1 Develop a formula to calculate the H constant from WR2 and
evaluate H for a nuclear generating unit rated at 1333 MVA, 1800 r/min
with WR2 = 5,820,000 lb-ft2.
Solution The kinetic energy of rotation in foot-pounds at synchronous

speed is
1 WR2 27i(r/min)
KE = ft-lb
2 32.2 60
Since 550 ft-lb/s equals 746 W, it follows that 1 ft-lb equals 746/550 J. Hence,
converting foot-pounds to megajoules and dividing by the machine rating in
megavoltamperes, we obtain
/ 746 _ J 1 WR2 27i(r/min)

—— x 10-6
\550 2 32.2 60
^mach
which yields upon simplification
h_2.31 x 10 WR2 (r/min)2

c
*Jmach
Inserting the given machine data in this equation, we obtain
^ _ 2.31 x 10“ 10(5.82 x 106)(1800)2

1333
= 3.27 MJ/MVA
Converting H to a 100-MVA system base, we obtain
1333
h = 3 27 x 1UU
= 4156 mJ/mva
In a stability study for a large system with many machines geographically

dispersed over a wide area, it is desirable to minimize the number of swing
equations to be solved. This can be done if the transmission line fault, or other
disturbance on the system, affects the machines within a power plant so that
their rotors swing together. In such cases, the machines within the plant can be
combined into a single equivalent machine just as if their rotors were mechani¬
cally coupled and only one swing equation need be written for them. Consider a
power plant with two generators connected to the same bus which is electrically
remote from the network disturbances. The swing equations on the common
system base are
2Hx d25x .
—l--nr = Pmi-Pei per unit (14.18)
cos at
—-~nr = Pm2~ Pe2 per unit (14.19)

cos dt
Adding the equations together, and denoting <5, and S2 by S since the rotors
swing together, we obtain
— dJi=Pm-Pe Per unit (14.20)

ojs dt
where H = (Hx + H2), Pm = (Pml + Pm2) and Pe = (Pel + Pe2). This single
equation, which is in the form of Eq. (14.12), can be solved to represent the plant
dynamics.
Example 14.2 Two 60-Hz generating units operate in parallel within the
same power plant and have the following ratings:
Unit 1: 500 MVA, 0.85 power factor, 20 kV, 3600 r/min
Hx = 4.8 MJ/MVA
Unit 2: 1333 MVA, 0.9 power factor, 22 kV, 1800 r/min
H2 = 3.27 MJ/MVA
Calculate the equivalent H constant for the two units on a 100-MVA base.
Solution The total kinetic energy of rotation of the two machines is
KE = (4.8 x 500) + (3.27 x 1333) = 6759 MJ
Therefore the H constant for the equivalent machine on 100-MVA base is
H = 67.59 MJ/MVA
and this value can be used in a single swing equation provided the machines
swing together so that their rotor angles are in step at each instant of time.
Machines which swing together are called coherent machines. It is noted that,
when both cos and S are expressed in electrical degrees or radians, the swing
equations for coherent machines can be combined together even though, as in
the example, the rated speeds are different. This fact is often used in stability
studies involving many machines in order to reduce the number of swing equa¬
tions which need to be solved.
For any pair of non-coherent machines in a system, swing equations similar to
Eqs. (14.18) and (14.19) can be written. Dividing each equation by its left-hand-
side coefficient and subtracting the resultant equations, we obtain
d28l d2S2 _ (Os IP ml — Pel P m2 ~ P e2\

dt2 dt2 ~ 2 \ Hx H2 ) {14 21)
Multiplying each side by Hx H2/(H1 + H2) and rearranging, we find that
2I( HxH2 \|î ~h) PmlH2-Pm2Hx PelH2-Pe2H1

(14.22)
coj[Hi + H2) dt2 Hx +H2 Hx + H2
which also may be written more simply in the form of the basic swing equation,
Eq. (14.12), as follows
1H Ahi-P — p el 2 (14.23)
CO, 12 dt2 ~Fm12
Here the relative angle S12 equals 5X — S2 and an equivalent inertia and
weighted input and output powers are defined by
HiH 2
H12 = (14.24)
H i + H2
Pm\H2 ~ Pm2Pd\
Pm 12 — (14.25)
Hx + H2
PelH2-Pe2Hx
Pe 12 — (14.26)
Hx + H2
A noteworthy application of these equations concerns a two-machine system

having only one generator (machine one) and a synchronous motor (machine
two) connected by a network of pure reactances. Whatever change occurs in the

generator output is thus absorbed by the motor and we can write
Under these conditions, Pml2 = Pm, Pe 12 = Pe and Eq. (14.22) reduces to
2H11d^S11 =
cos dt2 m e
which is also the format of Eq. (14.12) which applies for a single machine.
Equation (14.22) demonstrates that stability of a machine within a system is
a relative property associated with its dynamic behavior with respect to the
other machines of the system. The rotor angle of one machine, say <5l5 can be
chosen for comparison with the rotor angle of each other machine, symbolized
by S2. In order to be stable the angular differences between all machines must
decrease after the final switching operation such as the opening of a circuit-
breaker to clear a fault. While we may choose to plot the angle between a
machine’s rotor and a synchronously rotating reference axis, it is the relative
angles between machines which are important. Our discussion above emphasizes
the relative nature of the system stability property and shows that the essential
features of a stability study are revealed by consideration of two-machine prob¬
lems. Such problems are of two types; those having one machine of finite inertia
swinging with respect to an infinite bus and those having two finite-inertia
machines swinging with respect to each other. An infinite bus may be considered
for stability purposes as a bus at which is located a machine of constant internal
voltage, having zero impedance and infinite inertia. The point of connection of a
generator to a large power system may be regarded as such a bus. In all cases the
swing equation assumes the form of Eq. (14.12), each term of which must be
explicitly described before it can be solved. The equation for Pe is essential to
this description and we now proceed to its characterization for a general two-
machine system.
14.4 THE POWER-ANGLE EQUATION
In the swing equation for the generator, the input mechanical power from the
prime mover Pm will be considered constant. As we have mentioned previously,
this is a reasonable assumption because conditions in the electrical network can
be expected to change before the control governor can cause the turbine to react.
Since Pm in Eq. (14.12) is constant, the electrical power output Pe will determine
whether the rotor accelerates, decelerates, or remains at synchronous speed.
When Pe equals Pm the machine operates at steady-state synchronous speed;
when Pe changes from this value the rotor deviates from synchronous speed.
Changes in Pe are determined by conditions on the transmission and distribu¬

tion networks and the loads on the system to which the generator supplies
power. Electrical network disturbances resulting from severe load changes,
network faults, or circuit breaker operations, may cause the generator output Pe
to change rapidly in which case electromechanical transients exist. Our fun¬
damental assumption is that the effect of machine speed variations upon the
generated voltage is negligible so that the manner in which Pe changes is
determined by the load flow equations applicable to the state of the electrical
network and by the model chosen to represent the electrical behavior of the
machine. Each synchronous machine is represented for transient stability studies
by its transient internal voltage E' in series with the transient reactance X'd as
shown in Fig. 14.2a in which Vt is the terminal voltage. This corresponds to the
steady-state representation in which synchronous reactance Xd is in series with
the synchronous internal or no-load voltage E. Armature resistance is negligible
in most cases so that the phasor diagram of Fig. 14.2b applies. Since each ma¬
chine must be considered relative to the system of which it is part, the phasor
angles of the machine quantities are measured with respect to the common
system reference.
Figure 14.3 schematically represents a generator supplying power through a
transmission system to a receiving-end system at bus 2. The rectangle represents
the transmission system of linear passive components such as transformers,
transmission lines, and capacitors, and includes the transient reactance of the
generator. Therefore, the voltage E\ represents the transient internal voltage of
the generator at bus 1. The voltage E'2 at the receiving end is regarded here as
that of an infinite bus or as the transient internal voltage of a synchronous
motor whose transient reactance is included in the network. Later we shall
consider the case of two generators supplying constant-impedance loads within
i
\S>mSLr
(a)
Figure 14.2 Phasor diagram of a synchronous machine

for transient stability studies. (6)
Figure 14.3 Schematic diagram for stability studies. E[

Transient reactances associated with E\ and E\
are included in the transmission network.
the network. The elements of the bus admittance matrix for the network reduced
to two nodes in addition to the reference node is
Y ^11 ^12
K bus (14.28)
^21 ^22
Repeating Eq. (8.7) we have
Pk-jQk=vtYJYknv„ (14.29)
n=1
which upon letting k and N equal 1 and 2, respectively, and substituting E2 for V,
can be written
Pi +JQi = £'i(ni£i)* + E'l(Y12E'2)* (14.30)

or where
E\= |£i|^_i E'2= \E'2\/S_2
^11 = ^11 + 7-®U ^12 — | ^12 | /^12
we obtain
Pi= \E'1\2G11+ \E\ | \E'211 îi | cos (<51 S2 $12) (14.31)
Qi = -\E'1\2B11+ \E'i 11 £'211 Y121 sin («5i — S2 — 012) (14.32)
Similar equations apply at bus 2 by interchanging the subscripts in the two

equations above.
If we let
S = S l — S2
and define a new angle y such that
_ Q n
y-el2 2
we obtain from Eqs. (14.31) and (14.32)
Pi = \E'1\2G11+ |£i||F2||y12|sin {S - y) (14.33)
Qi = — \E'i \2Bu — \E\ | \E'2\ I Y12 \ cos (S-y) (14.34)
Equation (14.33) may be written more simply as
Pe = Pc + Pmax (<5 ~ V) (14.35)

where
Pc= |£'i|2Gii Pm**=\E,i\\E'2\\Yi121 (14.36)
Since Pt represents the electric power output of the generator (armature. Joss
neglected) we have replaced it by Pe in Eq. (14.35) which is often called the
power-angle equation; its graph as a function of <5 is called the power-angle curve.
The parameters Pc, Pmax, and y are constants for a given network configuration
and constant voltage magnitudes |E\ | and \E'2\. When the network is con¬
sidered without resistance all the elements of Ybus are susceptances and so both
Gn and y are zero. The power-angle equation which then applies for the pure
reactance network is simply the familiar equation
Pe = Pmaxsin<5 (14.37)
where Pmax = \E\ \ \E'2\/X and X is the transfer reactance between E\ and E'2.
Example 14.3 The single-line diagram of Fig. 14.4 shows a generator con¬
nected through parallel transmission lines to a large metropolitan system
considered as an infinite bus. The machine is delivering 1.0 per unit power
and both the terminal voltage and the infinite-bus voltage are 1.0 per unit.
Numbers on the diagram indicate the values of the reactances on a common
system base. The transient reactance of the generator is 0.20 per unit as
indicated. Determine the power-angle equation for the system applicable to
the operating conditions.
Solution The reactance diagram for the system is shown in Fig. 14.5a. The
series reactance between the terminal voltage and the infinite bus is
0.4
X = 0.10 + — =0.3 per unit
and therefore the 1.0 per unit power output of the generator is determined
by the power-angle equation
\vt\\v\ . (1.0)(1.0) .
' 1 ' 1 Sin a = ~ 1 sin a = 1.0
A (J.3
where V is the voltage of the infinite bus and a is the angle of the terminal
voltage relative to the infinite bus. Solving for a, we obtain
a = sin"1 0.3 = 17.458°

so that the terminal voltage is
Vt — 1.0/17.458° = 0.954 + y'0.300 per unit
Figure 14.4 One-line diagram for

Examples 14.3 and 14.4. Point P is
O-
at the center of the line.
j‘0-40
1.0/0°
1.0/0°
-j 3.333
-nmm-
-72.5
+
/ 1.05/6. 1.0/0°
-7'5.0 -j 5.0S
(c)
Figure 14.5 Reactance diagram (a) for prefault network for Example 14.3 with impedances in
per unit and (b) and (c) for the faulted network for Example 14.4 with the same impedances con¬
verted to admittances and marked in per unit.
The output current from the generator is now calculated as
1.0/17.458° - 1.0/CT
/ =
j0.3
= 1.0 + ;0.1535 = 1.012/8.729° per unit
and the transient internal voltage is then found to be
E = (0.954 + jO.30) + ;'(0.2)(1.0 +;0.1535)

= 0.923 + ;'0.5 = 1.050/28.44° per unit
The power-angle equation relating the transient internal voltage E and the
infinite bus voltage V is determined by the total series reactance
0.4 ^ r
X = 02 + 0.1+— = 0.5 per unit
Hence, the desired equation is
_ (l-050)(l-0) • ^ = 2.10 sin 3 per unit

0.5
where 3 is the machine rotor angle with respect to the infinite bus.
Figure 14.6 Plot of power-angle curves

found in Examples 14.3 to 14.5.
This power-angle equation is plotted in Fig. 14.6. Note that the mechanical
input power Pm is constant and intersects the sinusoidal power-angle curve at
the operating angle = 28.44°. This is the initial angular position of the genera¬
tor rotor corresponding to the given operating conditions. The swing equation
for the machine may be written
—= 1-0 — 2.10 sin S per unit (14.38)

180/ dt2 F 1 '
where H is in megajoules per megavoltampere, / is the electrical frequency of the

system, and <5 is in electrical degrees. We can easily check the example results
since, under the operating conditions, Pe = 2.10 sin 28.44° =1.0 per unit which
corresponds exactly to the mechanical power input Pm and the acceleration is
zero.
In the next example, we determine the power-angle equation for the same
system with a three-phase fault at P, the midpoint of one of the transmission
lines. Positive acceleration is shown to exist while the fault is on.
Example 14.4 The system of Example 14.3 is operating under the indicated
conditions when a three-phase fault occurs at point P in Fig. 14.4. Deter¬
mine the power-angle equation for the system with the fault on and the
corresponding swing equation. Take H = 5 MJ/MVA.
\
Solution The reactance diagram is shown in Fig. 14.56 with the fault on
the system at point P. Values shown are admittances in per unit. The effect
of the short circuit caused by the fault is clearly shown by redrawing the
reactance diagram as in Fig. 14.5c. As calculated in Example 14.3, transient
internal voltage of the generator remains at E = 1.05/28.44° based on the
assumption of constant flux linkages in the machine. The net transfer admit¬
tance connecting the voltage sources remains to be determined. The buses
are numbered as shown and the Ybus is formed by inspection of Fig. 14.5c as
follows
'-3.333 0 3.333'
Ybus —j 0 -7.50 2.50
. 3.333 2.50 -10.833.
Bus 3 has no external source connection and it may be removed by the node
elimination procedure of Sec. 7.4 to yield the reduced bus admittance matrix
i'll t12 -2.308 0.769

=j
^21 ^22 0.769 -6.923
The magnitude of the transfer admittance is 0.769 and therefore
^max = |E\ |\E'2\| y121 = (1.05)(1.0)(0.769) = 0.808 per unit
The power-angle equation with the fault on the system is therefore
Pe = 0.808 sin S per unit
and the corresponding swing equation is
5 d2S
= 1-0 ~ 0.808 sin <5 per unit (14.39)
For later reference note that, because of its inertia, the rotor cannot change
position instantly upon occurrence of the fault. Therefore, the rotor angle
5 is initially 28.44° as in Example 14.3 and the electrical power output is
Pe = 0.808 sin 28.44° = 0.385. The initial accelerating power is
Pa — 1.0 — 0.385 = 0.615 per unit
and the initial acceleration is positive with the value given by
_ 180/ (g _ 22.14/ elec deg/s2
where / is the system frequency.

The line-relaying schemes will sense the fault on the line and will act to
clear the fault by simultaneous opening of the line-end breakers. When this
occurs another power-angle equation applies since a network change has
occurred.
Example 14.5 The fault on the system of Example 14.4 is cleared by simul¬
taneous opening of the circuit breakers at each end of the affected line.
Determine the power-angle equation and the swing equation for the post¬
fault period.
Solution Inspection of Fig. 14.5a shows that, upon removal of the faulted
line, the net transfer admittance across the system is
1
—j1.429 per unit
yi2 ~ j(0.2 + 0.1 +0.4)
or in the bus admittance matrix
F12 =71.429
Therefore the postfault power-angle equation is
Pe = (1.05)(1.0)(1.429) sin 3 = 1.500 sin 3
and the swing equation is

5 d23
--- 1.0 — 1.500 sin 3
mflt2
The acceleration at the instant of clearing the fault depends upon the angu¬
lar position of the rotor at that time. The power-angle curves for Examples
14.3 to 14.5 are compared in Fig. 14.6.
14.5 SYNCHRONIZING POWER COEFFICIENTS
In Example 14.3 the operating point on the sinusoidal Pe curve of Fig. 14.6 was
found to be at 30 equal to 28.44° where the mechanical power input Pm equals
the electrical power output Pe. In the same figure it is also seen that Pe equals
Pm at 3 = 151.56° and this might appear to be an equally acceptable operating
point. However, this is now shown not to be the case.
A common-sense requirement for an acceptable operating point is that the
generator shall not lose synchronism when small temporary changes occur in the
electrical power output from the machine. To examine this requirement, for fixed
mechanical input power Pm, consider small incremental changes in the operating
point parameters, that is, consider
^ = <50 + <5a Pe = Pe 0 + Pe A (14-40)

where the subscript zero denotes the steady-state operating point values and the
subscript delta identifies the incremental variations from those values. Substitut¬
ing Eqs. (14.40) into Eq. (14.37), we obtain the power-angle equation for the
general two-machine system in the form
Pe0 + PeA = Pmax Sm (<50 + <5A)

\
= Pmax (sin <5o cos 3A + cos <50 sin <5A)
Since 3A is a small incremental displacement from <50,
sin 3A ~ 3a and cos 3A ^ 1 (14.41)
and the previous equation becomes
Pe0 + PeA = Pmax Sm 30 + (Pmax COS (50>5A (14.42)

where strict equality is now used. At the initial operating point <50,
Pm = PeO = Pmax Sm <50 (14.43)

and it therefore follows from the last two equations that
Pm - (PeO + PeA) = ~(Pmax COS 30)3A (14.44)

Substituting the incremental variables of Eq. (14.40) into the basic swing equa¬
tion, Eq. (14.12), we obtain
2H d2(S0 + <5a)
— Pm ~ {Pe0 + PeA) (14.45)
cos dt2
Replacing the right-hand side of this equation by Eq. (14.44) and transposing
terms, we obtain
2H d2i5a ,
+ (pmax cos <50)<SA = 0 (14.46)
since <50 is a constant value. Noting that Pmax cos <50 is the slope of the power-
angle curve at the angle <50, we denote this slope as SP and define it as
dP
SP = -JT = pmax cos S0 (14.47)
dO 6=6o
where SP is called the synchronizing power coefficient. When SP is used in

Eq. (14.46), the swing equation governing the incremental rotor-angle variations
may be rewritten in the form
d23A
(14.48)
dt2 + “ws*=0
This is a linear, second-order differential equation, the solution to which depends
upon the algebraic sign of SP. When SP is positive, the solution SA(t) corresponds
to that of simple harmonic motion; such motion is represented by the oscilla¬
tions of an undamped swinging pendulum.! When SP is negative, the solution
SA(t) increases exponentially without limit. Therefore, in Fig. 14.6 the operating
point <5 = 28.44° is a point of stable equilibrium, in the sense that the rotor angle
swing is bounded following a small perturbation. In the physical situation,
damping will restore the rotor angle to S0 = 28.44° following the temporary
electrical perturbation. On the other hand, the point S = 151.56° is a point of
unstable equilibrium since SP is negative there. So this point is not a valid
operating point.
The changing position of the generator rotor swinging with respect to the
infinite bus may be visualized by an analogy. Consider a pendulum swinging
from a pivot on a stationary frame, as shown in Fig. 14.7a. Points a and c are the
maximum points of the oscillation of the pendulum about the equilibrium point
b. Damping will eventually bring the pendulum to rest at b. Now imagine a disk
rotating in a clockwise direction about the pivot of the pendulum, as shown in
Fig. 14.76, and superimpose the motion of the pendulum on the motion of the
disk. When the pendulum is moving from a to c, the combined angular velocity
t The equation of simple harmonic motion is d2x/dt2 + w2nx = 0 which has the general solution
A cos co„ t + B sin a>n t with constants A and B determined by the initial conditions. The solution
when plotted is an undamped sinusoid of angular frequency co„.
Figure 14.7 Pendulum and rotat¬

ing disk to illustrate a rotor
swinging with respect to an in¬
finite bus.
is slower than that of the disk. When the pendulum is moving from c to a, the
combined angular velocity is faster than that of the disk. At points a and c,
the angular velocity of the pendulum alone is zero and the combined angular
velocity equals that of the disk. If the angular velocity of the disk corresponds to
the synchronous speed of the rotor, and if the motion of the pendulum alone
represents the swinging of the rotor with respect to an infinite bus, the super¬
imposed motion of the pendulum on that of the disk represents the actual
angular motion of the rotor.
From the above discussion, we conclude that the solution to Eq. (14.48)
represents sinusoidal oscillations provided the synchronizing power coefficient
SP is positive. The angular frequency of the undamped oscillations is given by
OJ n elec rad/s (14.49)
which corresponds to a frequency of oscillation given by
1 / cosSP
Hz (14.50)
2n\J ~2H~
Example 14.6 The machine of Example 14.3 is operating at <5 = 28.44° when
it is subjected to a slight temporary electrical-system disturbance. Determine
the frequency and period of oscillation of the machine rotor if the distur¬
bance is removed before the prime mover responds. H = 5 MJ/MVA.
Solution The applicable swing equation is Eq. (14.48) and the synchroniz¬
ing power coefficient at the operating point is
SP — 2.10 cos 28.44°= 1.8466

The angular frequency of oscillation is therefore
1311 x 1,8466
co,
V 2x5
= 8.343 elec rad/s
The corresponding frequency of oscillation is
8.343
= 1.33 Hz
271
and the period of oscillation is
1
T = 0.753 s
fn
This example is an important one from the practical viewpoint since it indicates
the order of magnitude of the frequencies which can be superimposed upon the
nominal 60-Hz frequency in a large power system having many interconnected
machines. As load on the system changes randomly throughout the day, inter¬
machine oscillations involving frequencies of the order of 1 Hz tend to arise but
these are quickly damped out by the various damping influences caused by the
prime mover, the system loads, and the machine itself. It is worthwhile to note
that even if the transmission system in our example contains resistance, nonethe¬
less, the swinging of the rotor is harmonic and undamped. Problem 14.8
examines the effect of resistance on the synchronizing power coefficient and the
frequency of oscillations. In a later section we again discuss the concept of
synchronizing coefficients. In the next section we examine a method of determin¬
ing stability under transient conditions caused by large disturbances.
14.6 EQUAL-AREA CRITERION OF STABILITY
In Sec. 14.4 we developed swing equations which are nonlinear in nature.

Formal solution of such equations cannot be explicitly found. Even in the case of
a single machine swinging with respect to an infinite bus it is very difficult to
obtain literal-form solutions and therefore digital computer methods are
normally used. To examine the stability of a two-machine system without solv¬
ing the swing equation, a direct approach is possible and will now be discussed.
The system shown in Fig. 14.8 is the same as that considered previously
except for the addition of a short transmission line. Initially circuit breaker A is
closed but circuit breaker B at the opposite end of the short line is open.
Therefore, the initial operating conditions of Example 14.3 may be considered
—o
cHb — ©
i—i FT-
"LI 1_r
Figure 14.8 One line-diagram of the system -q£
of Fig. 14.4 with the addition of a short A
transmission line. I Open
Figure 14.9 Power-angle curves for the

generator shown in Fig. 14.8. Areas Ay and
A2 are equal as are areas A3 and A4.
unaltered. At point P close to the bus, a three-phase fault Qccurs and is cleared
by circuit breaker A after a short period of time. Therefore, the effective trans¬
mission system is unaltered except while the fault is on. The short circuit caused
by the fault is effectively at the bus and so the electrical power output from the
generator is zero until the fault is cleared. The physical conditions before, during,
and after the fault can be understood by analyzing the power-angle curves in
Fig. 14.9.
Originally the generator is operating at synchronous speed with a rotor
angle of S0 and the input mechanical power Pm equals the output electrical
power Pe as shown at point a in Fig. 14.9a. When the fault occurs at t = 0, the
electrical power output is suddenly zero while the input mechanical power is
unaltered as shown in Fig. 14.9b. The difference in power must be accounted for
by a rate of change of stored kinetic energy in the rotor masses. This can be
accomplished only by an increase in speed which results from the constant
accelerating power Pm. If we denote the time to clear the fault by tc, then for
time t less than tc the acceleration is constant and is given by
d2S a)s
(14.51)
~dtI = 2HPm
While the fault is on, the velocity increase above synchronous speed is found by
integrating this equation to obtain
dS CO, „ , co„
—- p dt = —- P t (14.52)
dt 02 H m 2H m
A further integration with respect to time yields
(14.53)
s=-w‘ +s°
for the rotor angular position.

Equations (14.52) and (14.53) show that the velocity of the rotor relative to
synchronous speed increases linearly with time while the rotor angle advances
from <50 to the angle at clearing 8C; that is, in Fig. 14.9 the angles 5 goes from b
to c. At the instant of fault clearing, the increase in rotor speed and the angle
separation between the generator and the infinite bus are given, respectively, by
dS = cosPm
(14.54)
dt t=tc 2H ‘
and
m MsPn
4H
te + <50 (14.55)
When the fault is cleared at the angle Sc, the electrical power output abruptly
increases to a value corresponding to point d on the power-angle curve. At d the
electrical power output exceeds the mechanical power input and thus the
accelerating power is negative. As a consequence the rotor slows down as Pe
goes from d to e in Fig. 14.9c. At e the rotor speed is again synchronous
although the rotor angle has advanced to 5X. The angle Sx is determined by the
fact that areas Ax and A2 must be equal, as will be explained later. The acceler¬
ating power at e is still negative (retarding), and so the rotor cannot remain at
synchronous speed but must continue to slow down. The relative velocity is
negative and the rotor angle moves back from Sx at e along the power-angle
curve of Fig. 14.9c to point a at which the rotor speed is less than synchronous.
From a to f the mechanical power exceeds the electrical power and rotor in¬
creases speed again until it reaches synchronism at f Point / is located so that
areas A3 and A4 are equal. In the absence of damping the rotor would continue
to oscillate in the sequence f-a-e, e-a-f etc., with synchronous speed occurring at
e and f.
As noted, we shall soon show that the shaded areas Ax and A2 in Fig. 14.9b
must be equal, and similarly, areas A3 and A4 in Fig. 14.9c must be equal. In a
system where one machine is swinging with respect to an infinite bus we may. use
this principle of equality of areas, called the equal-area criterion, to determine the
stability of the system under transient conditions without solving the swing
equation. Although not applicable to multimachine systems, the method helps in
understanding how certain factors influence the transient stability of any system.
The derivation of the equal-area criterion is made for one machine and an
infinite bus although the considerations in Sec. 14.3 show that the method can
be readily adapted to general two-machine systems. The swing equation for the
machine connected to the bus is
2Md*d
= Pm~ (14.56)
a)s dt2
Define the angular velocity of the rotor relative to synchronous speed by
dd
cor - — = CO ~0)s (14.57)
Differentiating Eq. (14.57) with respect to t and substituting in Eq. (14.56) we

obtain
2 H dcor
= p
1 m - 1p P (14.58)
cos dt
It is clear that when the rotor speed is synchronous, co equals cos and cor is zero.
Multiplying both sides of Eq. (14.58) by cor = dS/dt, we have
. dco. , „ „ ,dS
2cor^ = (Fm-r,)- (14.59)
cos
The left-hand side of the equation can be rewritten to give
IL(M1 = (P _p)^ (14.60)

cos dt { m e} dt
Multiplying by dt and integrating, we obtain
~ K22 - O =
Ws
(
JSi
V™ - p.) ds (14.61)
The subscripts for the cor terms correspond to those for the S limits, that is, the
rotor speed corl corresponds to that at the angle ^ and cor2 corresponds to S2.
Since cor represents the departure of the rotor speed from synchronous speed, we
readily see that if the rotor speed is synchronous at (5X and d2, then, corre¬
spondingly, corl = cor2 = 0. Under this condition Eq. (14.61) becomes
*62
j (Pm-Pe)dS = 0 (14.62)
This equation applies to any two points <5j and <5 2 on the power-angle diagram,
provided they are points at which the rotor speed is synchronous. In Fig. 14.96
two such points are a and e corresponding to <50 and 5X. If we perform the
integration in two steps, we can write
V* - Pe) dS + j Vm - Pe) dS = 0 (14.63)
‘ \pm - Pe) dd = f 6\pe - Pm) dd (14.64)
The left integral applies to the fault period while the right integral corresponds
to the immediate postfault period up to the point of maximum swing Sx. In
Fig. 14.% Pe is zero during the fault. The shaded area At is given by the
left-hand side of Eq. (14.64) and the shaded area ^42 is given by the right-hand
side. So the two areas Ax and A2 are equal.
Since the rotor speed is synchronous at <5^. and also at dy in Fig. 14.9c the
same reasoning as above shows that A3 equals A4. The areas A3 and A4 are
directly proportional to the increase in kinetic energy of the rotor while it is
accelerating, whereas areas A2 and A3 are proportional to the decrease in kinetic
energy of the rotor while it is decelerating. This can be seen by inspection of
both sides of Eq. (14.61). Therefore the equal-area criterion merely states that
whatever kinetic energy is added to the rotor following a fault must be removed
after the fault to restore the rotor to synchronous speed.
The shaded area Ax is dependent upon the time taken to clear the fault. If
there is delay in clearing, the angle Sc is increased; likewise the area Ax increases
and the equal-area criterion requires that area A2 also increase to restore the
rotor to synchronous speed at a larger angle of maximum swing dx. If the delay
in clearing is prolonged so that the rotor angle S swings beyond the angle (5max in
Fig. 14.9 then the rotor speed at that point on the power-angle curve is above
synchronous speed when positive accelerating power is again encountered.
Under the influence of this positive accelerating power the angle <5 will increase
without limit and instability results. Therefore there is a critical angle for clear¬
ing the fault in order to satisfy the requirements of the equal-area criterion for
stability. This angle, called the critical clearing angle dcr, is shown in Fig. 14.10.
The corresponding critical time for removing the fault is called the critical clear¬
ing time tCT.
Figure 14.10 Power-angle curve showing the

critical-clearing angle 5cr. Areas A j and A2 are
equal.
In the particular case of Fig. 14.10, both the critical clearing angle the
critical clearing time can be calculated as follows. The rectangular area is
,6 c
Al = \ Pmd3 = Pm(Scr-d0) (14.65)
J6o
while the area Ar is

,dn
A') — (Pmax sin 3 - Pm) dS
(14.66)
— Pmax(cOS <5cr COS <5max) Fm((5max <5Cr)
Equating the expressions for At and A2, and transposing terms, yields
cos <5cr = (Pm/Pmax){3max ~ d0) + cos <5max (14.67)
We see from the sinusoidal power-angle curve that
<5max = n - S0 elec rad (14.68)
and
Pm = Pmax SHI (14.69)
Substituting for <5max and Pm in Eq. (14.67), simplifying the result and solving for
<5cr, we obtain
Scr = cos 1 [(tt — 2<50) sin (50 — cos <50] (14.70)
for the critical clearing angle. The value for <5cr calculated from this equation,
when substituted for the left-hand side of Eq. (14.55), yields
s _ ^sPm .2 . <;
(14.71)
^cr 4// tcr +
from which is found
(14.72)
for the critical clearing time.
Example 14.7 Calculate the critical clearing angle and the critical clearing
time for the system of Fig. 14.8 when the system is subjected to a three-phase
fault at point P on the short transmission line. The initial conditions are the
same as those in Example 14.3 and H is 5 MJ/MVA.
Solution From Example 14.3 the power-angle equation is
Pe - Pmax sin 5 = 2.10 sin S
the initial rotor angle is
«50 = 28.44° = 0.496 elec rad

and mechanical input power Pm is 1.0 per unit. Therefore, from Eq. (14.70)
we obtain
Scr = cos - '[(it - 2 x 0.496) sin 28.44° - cos 28.44°]
= 81.697° = 1.426 elec rad
for the critical clearing angle. Inserting this value with the other known
quantities in Eq. (14.72) we obtain
/4 x 5(1.426 - 0.496)
tcr ~ V 377 x 1
= 0.222 s
This value is equivalent to a critical clearing time of 13.3 cycles based on a

frequency of 60 Hz.
This example serves to establish the concept of critical clearing time which is
essential to the design of proper relaying schemes for fault clearing. In more
general cases, the critical clearing time cannot be explicitly found without solv¬
ing the swing equations by digital computer simulation.
14.7 FURTHER APPLICATIONS OF THE

EQUAL-AREA CRITERION
Although the equal-area criterion can be applied only for the case of two ma¬
chines or one machine and an infinite bus, it is a very useful means for beginning to
see what happens when a fault occurs. The digital computer is the only practical
way to determine the stability of a large system. However, because the equal-
area criterion is so helpful in understanding transient stability, we shall continue
to examine it briefly before discussing the determination of swing curves and the
digital-computer approach.
When a generator is supplying power to an infinite bus over two parallel
transmission lines, opening one of the lines may cause the generator to lose
synchronism even though the load could be supplied over the remaining line
under steady-state conditions. If a three-phase short circuit occurs on the bus to
which two parallel lines are connected, no power can be transmitted over either
line. This is essentially the case in Example 14.7. However, if the fault is at the
end of one of the lines, opening breakers at both ends of the line will isolate the
fault from the system and allow power to flow through the other parallel line.
When a three-phase fault occurs at some point on a double-circuit line other
than on the paralleling buses or at the extreme ends of the line, there is some
impedance between the paralleling buses and the fault. Therefore, some power is
transmitted while the fault is still on the system. The power-angle equation in
Example 14.4 demonstrates this fact.
Figure 14.11 Equal-area criterion ap¬

plied to fault clearing when power is
transmitted during the fault. Areas
At and A2 are equal.
When power is transmitted during a fault, the equal-area criterion is applied

as shown in Fig. 14.11 which is similar to the power-angle diagram of Fig. 14.6.
Before the fault, Pmax sin 5 is the power which can be transmitted; during the
fault, Pmax sin d is the power which can be transmitted; and r2 Pmax sin S is the
power which can be transmitted after the fault is cleared by switching at
the instant when S = <5tr. Examination of Fig. 14.11 shows that dcr is the critical
clearing angle in this case. By evaluating the areas At and A2 using
the procedural steps of the previous section, we would find
coS(5 = (f>m/f>max)(^max ~ <$o) + r2 COS Smax - Tl COS dp ^
" r2 ~ rl
A literal form solution for the critical clearing time tcr is not possible in this case.
For the system and fault location shown in Fig. 14.8, the values are rx = 0,
r2 — 1 and the equation then reduces to Eq. 14.67.
Regardless of their location, short-circuit faults not involving all three
phases allow the transmission of some power, because they are represented by
connecting some impedance rather than a short circuit between the fault point
and the reference bus in the positive-sequence impedance diagram. The larger the
impedance shunted across the positive-sequence network to represent the fault,
the larger the power transmitted during the fault. The amount of power trans¬
mitted during the fault affects the value of A1 for any given clearing angle. Thus,
smaller values of rx result in greater disturbances to the system, as low rt means
low power transmitted during the fault and larger Ax. In order of increasing
severity (decreasing /q Pmax) the various faults are:
1. single line-to-ground fault

2. line-to-line fault
3. double line-to-ground fault
4. three-phase fault
The single line-to-ground fault occurs most frequently, and the three-phase fault
is least frequent. For complete reliability a system should be designed for tran¬
sient stability for three-phase faults at the worst locations, and this is virtually
the universal practice.
Example 14.8 Determine the critical clearing angle for the three-phase fault
described in Examples 14.4 and 14.5 when the initial system configuration
and prefault operating conditions are as described in Example 14.3.
Solution The power-angle equations obtained in the previous examples are
Before the fault: Pmax sin 3 = 2.100 sin <5
During the fault: Pmax sin d = 0.808 sin 3
After the fault: r2Pmax sin 3 = 1.500 sin 3

Hence
0-808 A_ 1.500 _
r, = 0.385
2.100 r2“zioo ~ 0/714
From Example 14.3, we have
d0 = 28.44° = 0.496 rad
and from Fig. 14.11, we calculate
1 000
<5max = 180° - sin-1 = 138.190° = 2.412 rad
Therefore, inserting numerical values in Eq. (14.73), we obtain
_ (1.0/2.10)(2.412 - 0.496) + 0.714 cos (138.19°) - 0.385 cos (28.44°)

C°S cr_ 0.714-0.385
= 0.127
Hence
3cr = 82.726°
To determine the critical clearing time we must obtain the swing curve of 3
versus t for this example. In Sec. 14.9 we shall discuss one method of com¬
puting such swing curves.
14.8 MULTIMACHINE STABILITY STUDIES:

CLASSICAL REPRESENTATION
The equal area criterion cannot be used directly in systems where three or more
machines are represented. Although the physical phenomena observed in the
two-machine problems basically reflect that of the multimachine case, nonethe¬
less, the complexity of the numerical computations increases with the number of
machines considered in a transient stability study. When a multimachine system
operates under electromechanical transient conditions, intermachine oscillations
occur between the machines through the medium of the transmission system
which connects them. If any one machine could be considered to act alone as the
single oscillating source, it would send into the interconnected system an

electromechanical oscillation determined by its inertia and synchronizing power.
A typical frequency of such oscillation is of the order of 1 to 2 Hz and this is
superimposed upon the nominal 60-Hz frequency of the system. When many
machine rotors are simultaneously undergoing transient oscillation, the swing
curves will reflect the combined presence of many such oscillations. Therefore,
the transmission system frequency is not unduly perturbed from nominal fre¬
quency, and the assumption is made that the 60-Hz network parameters are still
applicable. To ease the complexity of system modeling and, thereby, the compu¬
tational burden, the following additional assumptions are commonly made in
transient stability studies:
(a) The mechanical power input to each machine remains constant during the
entire period of the swing curve computation.
(b) Damping power is negligible.
(c) Each machine may be represented by a constant transient reactance in series
with a constant transient internal voltage.
(d) The mechanical rotor angle of each machine coincides with 3, the electrical
phase angle of the transient internal voltage.
(ie) All loads may be considered as shunt impedances to ground with values
determined by conditions prevailing immediately prior to the transient
conditions.
The system stability model based on these assumptions is called the classical
stability model and studies which use this model are called classical stability
studies. These assumptions, which we shall adopt, are in addition to the fun¬
damental assumptions set forth in Sec. 14.1 for all stability studies. Of course,
detailed computer programs with more sophisticated machine and load models
are available to modify one or more of assumptions (n) to (c). However,
throughout this chapter the classical model is used to study system disturbances
originating from three-phase faults.
As already seen, in any transient stability study, the system conditions before
the fault and the network configuration during and after its occurrence must be
known. Consequently, in the multimachine case, two preliminary steps are
required:
1. The steady-state prefault conditions for the system are calculated using a
production-type load-flow program.
2. The prefault network representation is determined and then modified to
account for the fault and for the postfault conditions.
From the first preliminary step we know the values for power, reactive power,
and voltage at each generator terminal and load bus with all angles measured
with respect to the swing bus. The transient internal voltage of each generator is
then calculated using the equation
E = Vt +jX'dI (14.74)
where Vt is the corresponding terminal voltage and /, the output current. Each
load is converted into a constant admittance to ground at its bus using the
equation
(14.75)
where PL + jQL is the load and \ VL \ is the magnitude of the corresponding bus
voltage. The bus-admittance matrix used for the prefault load-flow calculation is
augmented to include the transient reactance of each generator and the shunt load
admittances as suggested in Fig. 14.12. Note that the injected current is zero at
all buses except the three-generator internal buses. The second preliminary step
determines the modified bus-admittance matrices corresponding to the faulted
and postfault conditions. Since only the generator internal buses have injections,
all other buses can be eliminated to reduce the dimensions of the modified
matrices to correspond to the number of generators. During and after the fault,
the power flow into the network from each generator is calculated by the corre¬
sponding power-angle equations. For example, in Fig. 14.12 the power out of
generator 1 is given by
Pe 1 = |£'1|2G11+ |£j| |£'2| |y12| cos (S12 — dl2)

+ |Ei||E'3||y13| cos (<513-013) (14.76)
where Sl2 equals <5X — S2. Similar equations are written for Pe2 and Pe3 with the
Yij values chosen from the 3 x 3 bus-admittance matrices appropriate to the
fault or postfault condition. The power-angle equations form part of the swing
equations
(14.77)
to represent the motion of each rotor for the fault and postfault periods. The
solutions depend upon the location and duration of the fault, and the Ybus which
Boundary of
/ augmented network
r n
l!
5
Figure 14.12 Augmented network

of a power system.
results when the faulted line is removed. The basic procedures used in digital
computer programs for classical stability studies are revealed in the following
examples.
Example 14.9 A 60-Hz, 230-kV transmission line has two generators and an
infinite bus as shown in Fig. 14.13. The transformer and line data are given
in Table 14.2. A three-phase fault occurs on line 4-5 near bus 4. Using the
prefault load-flow solution shown in Table 14.3 determine the swing equa¬
tion for each machine during the fault period. The generators, with reac¬
tances and H values expressed on a 100-MVA base, are described as follows:
Generator 1 400 MVA, 20 kV, X'd = 0.067 per unit, H = 11.2 MJ/MVA
Generator 2 250 MVA, 18 kV, X'd = 0.10 per unit, H = 8.0 MJ/MVA
Table 14.2 Line and transformer data

for Example 14.9, all values in per unit
on 230-kV, 100-MVA base.
Bus Series Z
to Shunt T
bus R X B
Trans. 1-4 0.022

Trans. 2-5 0.040
Line 3-4 0.007 0.040 0.082
Line 3-5(1) 0.008 0.047 0.098
Line 3-5(2) 0.008 0.047 0.098
Line 4-5 0.018 0.110 0.226
Table 14.3 Bus data and prefault load-flow values

in per unit on 230-KV, 100-MVA base.
Generation Load
Bus Voltage P Q P Q
1 1.030/8.88° 3.500 0.712

2 1.020/6.38° 1.850 0.298
3 1.000/0°
4 1.018/4.68° 1.00 0.44
5 1.011/2.27° 0.50 0.16
Solution To determine the swing equations we need to find transient inter¬

nal voltages. The current into the network at bus 1 based on the data in
Table 14.3 is
{Pi+jQi)* 3.50—,0.712
h = = 3.468 A2.619c
v* 1.030 A 8.88 ‘
and similarly
1.850-,0.298
/, = 1.020 7-6.38°
= 1.837 /-2.771
E\ = 1.030/8.88° + ,0.067 x 3.468 /-2.619 = 1.100/20.82°
E'2 = 1.020/6.38° + ,0.10 x 1.837 A 2.771 = 1.065/16.19°
At the infinite bus
f3 = e3 = 1.000/0.0°
and so
<513 = d1 and <523 — A
The loads at buses 4 and 5 are represented by the admittances calculated by

Eq. (14.75) to yield
1.00-,0.44
= 0.9649 - ,0.4246
"L4 “ (1.018)2
0.50-,0.16
= 0.4892 - ,0.1565
"L5_ (1.011)2
These admittances, together with the transient reactances, are used with the
line and transformer parameters to form for the prefault system the aug¬
mented bus admittance matrix which includes the transient reactances of the
Table 14.4 Elements of prefault bus admittance matrix for Example

14.9, admittances in per unit.
Bus 1 2 3 4 5
1 -711.2360 0.0 0.0 711.2360 0.0
2 0.0 -77.1429 0.0 0.0 77.1429
3 0.0 0.0 11.2841 -4.2450 -7.0392

-765.4731 +724.2571 +741.3550
4 711.2360 0.0 -4.2450 6.6588 • -1.4488

+724.2571 -744.6175 +78.8538
5 0.0 77.1429 -7.0392 -1.4488 8.9772

+741.3550 +78.8538 -757.2972
machines. Therefore, we will now designate as buses 1 and 2 the fictitious

internal nodes between the internal voltages and the transient reactances of
the machines. So, in the matrix, for example:
1
Yu —7*11.236
y'0.067 + y'0.022
1
^34= - = -4.2450 + 724.2571
0.007 + ;0.040
The sum of the admittances connected to nodes 3, 4, and 5 must include the
shunt capacitances of the transmission lines. So
/0.082 /0.226
Y44 = -711.236 + + 4.2450
-724.2571 + --- - „ + 0.9649 -70.4246

J 0.018 + 70.110 % J
= 6.6587 -744.6175
The entire augmented matrix is displayed as Table 14.4.

During the fault, bus 4 must be short circuited to ground. Row and column
4 of Table 14.4 disappear because node 4 is merged with the reference node. The
new row 4 and column 4 (node 5) are eliminated by Eq. (7.30) to reduce the bus-
admittance matrix for the faulted network to that shown in the upper half of
Table 14.5. The faulted-system Ybus shows that bus 1 decouples from the
other buses during the fault and that bus 2 is connected directly to bus 3.
This reflects the physical fact that the short circuit at bus 4 reduces to zero
the power injected into the system from generator 1 and causes generator 2
to deliver its power radially to bus 3. Under fault conditions we find by
using values from Table 14.5 for the per-unit power-angle equations
Table 14.5 Elements of faulted and postfault bus admittance matrices

for Example 14.9, admittances in per unit.
Faulted network
Bus 1 2 3
1 0.0000 -j 11.2360 0.0 + 70.0 0.0 + 70.0

(11.2360/-90°)
2 0.0 + jO.O 0.1362 -76.2737 -0.0681 + 75.1661

(6.2752/-88.7563°) (5.1665/90.7552)
3 0.0 + jO.O -0.681 + 75.1661 5.7986 -735.6299

(5.1665/90.7552°) (36.0987 /-80.7564)
Postfault network
1 0.5005 — J7.7897 0.0 + 70.0 -0.2216 + 77.6291

(7.8058/-86.3237°) (7.6323/91.6638°)
2 0.0 + jO.O 0.1591 — /6.1168 -0.0901 + 76.0975

(6.1189 /—88.5101°) (6.0982/90.8466°)
3 -0.2216 + /7.6291 -0.0901 -76.0975 1.3927 -713.8728

(7.6323/91.6638°) (6.0982/90.8466°) (13.9426/-84.2672)
Pe 1=0
Pel = | E'l |2G22 + | E2 | | £31 | Y23 | cos (<523 - 023)
= (1.065)2(0.1362) + (1.065)(1.0)(5.1665) cos (<52 - 90.755°)
= 0.1545 + 5.5023 sin (d2 - 0.755°)
Therefore, while the fault is on the system, the desired swing equations (values
of Pml and Pm2 from Table 14.3) are
d2Sl 180/ „ , 180/ „

dt2 H,
(Pm 1 f.i)- P. 1
180/
(3.5) elec deg/s2
11.2
d252 180/
(Pm 2 P l-fip
dt2 Hi e2’~ Hi al
p, y
180/
{1.85 [0T545 + 15023 sin (62 0T75T0)]}
8.0
180/
[L6955 15023 sin (<52 - 0755°)] elec deg/s2
8.0 Pm ~ P, y
Example 14.10 The three-phase fault in Example 14.9 is cleared by simul¬

taneously opening the circuit breakers at the ends of the faulted line. Deter¬
mine the swing equations for the postfault period.
Solution When the fault is cleared by removing line 4-5, the prefault Ybus
of Table 14.4 must again be modified. This is accomplished by substituting
zero for Y45 and Y54 and subtracting the series admittance of line 4-5 and
the capacitive susceptance of one-half the line from elements Y44 and Y55 of
Table 14.4. The reduced bus-admittance matrix applicable to the postfault
network is shown in the lower half of Table 14.5 and it is noted that a zero
element appears in the first and second rows. This reflects the fact that,
physically, the generators are not interconnected when line 4-5 is removed.
Accordingly, each generator is connected radially to the infinite bus. There¬
fore, the per-unit power-angle equations for postfault conditions are
Pe 1= \E'1\2G11+ | £'i | | £31 | T131 cos (<513-013)
= (1.100)2(0.5005) + (1.100)(1.0)(7.6323) cos (S1 - 91.664°)
= 0.6056 + 8.3955 sin (<5j - 1.664°)
and
Pel = | E2 |2G22 + IE2 I | E31 I Y231 cos («523 - d23)

= (1.065)2(0.1591) + (1.065)(1.0)(6.0982) cos (<52 - 90.847°)
= 0.1804 + 6.4934 sin (S2 - 0.847°)
For the postfault period the applicable swing equations are
I3'5 “ t0 6056 + 8 3955 sin (<5i - 1-664°)]}
180 f
[2 8944 — 8.3955 sin (<5X — 1.664°)] elec deg/s2
\
and
d2b2 180 f
-gy {1.85 - [0.1804 + 6.4934 sin (S2 - 0.847°)]}
dt2
ISQf
= -j~ [1.6696 - 6.4934 sin (S2 - 0.847°)] elec deg/s2
Each of the power-angle equations obtained in Example 14.9 and in this example
is of the form of Eq. (14.35). The resultant swing equation in each case assumes
the form
d*8 180/
dt2 H
[Pm ~ Pc ~ Pmax ^ (5 ~ y)] (14.78)
where the bracketed right-hand term represents the accelerating power on the
rotor. Accordingly, we may write the equation in the form
d1 25 180/ , , , ,
-^2 = Pa elec deg/s2 (14.79)
where
pa = Pm- Pc- pmax sin (d - y) (14.80)
In the next section we shall discuss how to solve equations of the form of Eq. (14.79)
to obtain 3 as a function of time for specified clearing times.
14.9 STEP-BY-STEP SOLUTION OF THE SWING CURVE
For large systems we depend on the digital computer which determines d versus
t for all the machines in which we are interested; and S may be plotted versus t
for a machine to obtain the swing curve of that machine. The angle <5 is cal¬
culated as a function of time over a period long enough to determine whether <5
will increase without limit or reach a maximum and start to decrease. Although
the latter result usually indicates stability, on an actual system where a number
of variables are taken into account it may be necessary to plot 3 versus t over a
long enough interval to be sure 5 will not increase again without returning to a
low value.
By determining swing curves for various clearing times the length of time
permitted before clearing a fault can be determined. Standard interrupting times
for circuit breakers and their associated relays are commonly 8, 5, 3, or 2 cycles
after a fault occurs, and thus breaker speeds may be specified. Calculations
should be made for a fault in the position which will allow the least transfer of
power from the machine and for the most severe type of fault for which protec¬
tion against loss of stability is justified.
A number of different methods are available for the numerical evaluation of
second-order differential equations in step-by-step computations for small incre¬
ments of the independent variable. The more elaborate methods are practical
only when the computations are performed on a digital computer. The step-by-
step method used for hand calculation is necessarily simpler than some of the
methods recommended for digital computers. In the method for hand calcula¬
tion the change in the angular position of the rotor during a short interval of
time is computed by making the following assumptions:
1. The accelerating power Pa computed at the beginning of an interval is con¬

stant from the middle of the preceding interval to the middle of the interval
considered.
2. The angular velocity is constant throughout any interval at the value
computed for the middle of the interval. Of course, neither of the assump¬
tions is true, since 3 is changing continuously and both Pa and co are functions
Figure 14.14 Actual and assumed values

of Pa, u>r, and 6 as functions of time.
of S. As the time interval is decreased, the computed swing curve approaches

the true curve.
\
Figure 14.14 will help in visualizing the assumptions. The accelerating power
is computed for the points enclosed in circles at the ends of the n — 2, n — 1, and
n intervals, which are the beginnings of the n — 1, n, and n + 1 intervals. The step
curve of Pa in Fig. 14.14 results from the assumption that Pa is constant between
midpoints of the intervals. Similarly, a>r, the excess of the angular velocity o>
over the synchronous angular velocity cos, is shown as a step curve that is
constant throughout the interval at the value computed for the midpoint. Be¬
tween the ordinates n — § and n — \ there is a change of speed caused by the
constant accelerating power. The change in speed is the product of the accelera¬
tion and the time interval, and so
d28 180/
1/2 - n-3/2= ~^2 At = ~~ Pa,n^ At (14.81)
The change in 3 over any interval is the product of a>r for the interval and the
time of the interval. Thus, the change in 3 during the n — 1 interval is
3n _ i 3„-2 — ^t(Or<n-3/2 (14.82)

and during the nth interval
~ <5„-i = Atcor 1/2 (14.83)

Subtracting Eq. (14.82) from Eq. (14.83) and substituting Eq. (14.81) in the re-
suiting equation to eliminate all values of cor yields
A3„ = A<5„_! + kP^î (14.84)

where
(14.85)
Equation (14.84) is the important one for the step-by-step solution of the
swing equation with the necessary assumptions enumerated, for it shows how to
calculate the change in 3 during an interval if the change in 3 for the previous
interval and the accelerating power for the interval in question are known.
Equation (14.84) shows that, subject to the stated assumptions, the change in
torque angle during a given interval is equal to the change in torque angle
during the preceding interval plus the accelerating power at the beginning of the
interval times k. The accelerating power is calculated at the beginning of each
new interval. The solution progresses through enough intervals to obtain points
for plotting the swing curve. Greater accuracy is obtained when the duration of
the intervals is small. An interval of 0.05 s is usually satisfactory.
The occurrence of a fault causes a discontinuity in the accelerating power Pa
which is zero before the fault and a definite amount immediately following the
fault. The discontinuity occurs at the beginning of the interval, when t = 0.
Reference to Fig. 14.14 shows that our method of calculation assumes that the
accelerating power computed at the beginning of an interval is constant from
the middle of the preceding interval to the middle of the interval considered.
When the fault occurs, we have two values cr Pa at the beginning of an interval,
and we must take the average of these two values as our constant accelerating
power. The procedure is illustrated in the following example.
Example 14.11 Prepare a table showing the steps taken to plot the swing
curve for machine 2 for the fault on the 60-Hz system of Examples 14.9 and
14.10. The fault is cleared by simultaneous opening of the circuit breakers at
the ends of the faulted line at 0.225 s.
Solution Without loss of generality, we will consider the detailed computa¬

tions for machine 2. Computations to plot the swing curve for machine 1 are
left to the student. Accordingly we drop the subscript 2 as the indication of
the machine number from all symbols in what follows. All our calculations
are made in per unit on 100 MVA base. For the time interval At = 0.05 s the
parameter k applicable to machine 2 is
180/ 180 x 60
k = (At)2 = x 25 x 10 4 = 3.375 elec deg
H 8.0
When the fault occurs at t = 0 the rotor angle of machine 2 cannot change
instantly. Hence, from Example 14.9,
<50 = 16.19°
and, during the fault,
Pe = 0.1545 + 5.5023 sin (3 - 0.755°)
Therefore, as already seen in Example 14.9
pa = pm- pe= 1.6955 - 5.5023 sin (S - 0.755°)
At the beginning of the first interval there is a discontinuity in the accelera¬

ting power of each machine. Just before the fault occurs, Pa = 0 and just after
the fault occurs
Pa = 1.6955 - 5.5023 sin (16.19° - 0.755°) = 0.231 per unit
The average value of Pa at t = 0 is \ x 0.2310 = 0.1155 per unit. We then

find
kPa = 3.375 x 0.1155 = 0.3898°
and consequently where we now identify the interval by numerical subscripts
A S1 = 0 + 0.3898 = 0.3898°
is the change in rotor angle of machine 2 as time advances over the first
interval from 0 to At. Therefore at the end of the first time interval
= <50 -b A<5j = 16.19 + 0.3898 = 16.5798°

and
5l - y = 16.5798 - 0.755 = 15.8248°
\
We then find at t = Af = 0.05 s
kPa, i = 3.375[(Pm -Pc)-P max sin (6, - y)]
= 3.375[1.6955 - 5.5023 sin (15.8248)] = 0.6583°
and it follows that the increase in rotor angle over the second time interval is
AS2 = A<3t + /cPa,! = 0.3898 + 0.6583 = 1.0481°
Hence, at the end of the second time interval
S2 = + A S2 = 16.5798 + 1.0481 = 17.6279°
The subsequent steps in the computations are shown in Table 14.6. Note
that the postfault equation found in Example 14.10 is needed.
In Table 14.6 the terms Pmax sin (<5 — y), Pa, and bn are values computed
at the time t shown in the first column but A<5„ is the change in rotor angle
during the interval that begins at the time indicated. For example, in the row
for t = 0.10 s the angle 17.6279° is the first value calculated and is found
by adding the change in angle during the preceding time-interval
Table 14.6 Computation of swing curve for machine 2 of Example 14.11 for
clearing at 0.225 s.
k= (180//H)(Ar)2 = 3.375 elec deg. Before clearing Pm — Pc = 1.6955p.u., Pmax = 5.5023p.u., and
y = 0.755°. After clearing these values become 1.6696, 6.4934, and 0.847, respectively.
<5-. - y Pm,x sin (<5„ - y) Pa kP0

t, s elec deg per unit per unit elec deg elec deg elec deg
0- 0.00 16.19
0+ 15.435 1.4644 0.2310 16.19
0 av 0.1155 0.3898 16.19
0.3898
0.05 15.8248 1.5005 0.1950 0.6583 16.5798
1.0481
0.10 16.8729 1.5970 0.0985 0.3323 17.6279
1.3804
0.15 18.2533 1.7234 -0.0279 -0.0942 19.0083
1.2862
0.20 19.5395 1.8403 -0.1448 -0.4886 20.2945
0.7976
0.25 20.2451 2.2470 -0.5774 -1.9487 21.0921
-1.1511
0.30 19.0940 2.1241 -0.4545 -1.534 19.9410
-2.6852
0.35 16.4088 1.8343 -0.1647 -0.5559 17.2558
-3.2410
0.40 13.1678 1.4792 0.1904 0.6425 14.0148
-2.5985
0.45 10.5693 1.1911 0.4785 1.6151 11.4163
-0.9833
0.50 9.5860 1.0813 0.5883 1.9854 10.4330
1.0020
0.55 10.5880 1.1931 0.4765 1.6081 11.4350
2.6101
0.60 13.1981 1.4826 0.1870 0.6312 14.0451
3.2414
0.65 16.4395 1.8376 -0.1680 -0.5672 17.2865
2.6742
0.70 19.1137 2.1262 -0.4566 -1.5411 19.9607
1.1331
0.75 20.2468 2.2471 -0.5775 -1.9492 21.0938
-0.8161
0.80 19.4307 2.1601 -0.4905 —1.6556 20.2777
-2.4716
0.85 17.8061
(0.05 to 0.10 s) to the angle at t = 0.05 s. Next Pmax sin (S - y) is calculated

for S = 17.6279°. Then, Pa = {Pm - Pe) - Pmax sin (<5 - y) and kPa are'cal¬
culated. The value of kPa is 0.3323°, which is added to the angular change of
1.0481° during the preceding interval to find the change of 1.3804° during
the interval beginning at t — 0.10 s. This value added to 17.6279° gives the
value d = 19.0083 at t = 0.15 s. Note that at 0.25 s the value of Pm - Pc has
changed because the fault was cleared at 0.225 s. The angle y has also
changed from 0.755 to 0.847°.
Whenever a fault is cleared, a discontinuity occurs in the accelerating power

Pa. When clearing is at 0.225 s, as is the case for the calculations in Table 14.6, no
special approach is required since our procedure assumes a discontinuity at the
middle of an interval. At the beginning of the interval following clearing the
assumed constant value of Pa is that determined for S at the beginning of the
interval following clearing.
When clearing is at the beginning of an interval such as at three cycles (0.05 s),
two values of accelerating power result from the two expressions (one during the
fault and one after clearing) for the power output of the generator at the beginning
of the interval. For the system of Example 14.11, if the discontinuity occurs at
0.05 s, the average of the two values is assumed as the constant value of Pa from
0.025 to 0.075 s. The procedure is the same as that followed upon occurrence of the
fault at t = 0 as demonstrated in Table 14.6.
In the same manner as followed in preparing Table 14.6, we could determine
d versus t for machine 1 for clearing at 0.225 s and for both machines for clearing at
0.05 s. In the next section, we shall see computer printouts of <5 versus t for both
machines calculated for clearing at 0.05 and 0.225 s. The swing curves are plotted for
the two machines in Fig. 14.15 for clearing at 0.225 s. Evidently, machine 1 is
unstable in this case.
Swing curves for clearing at 0.20 s would show the system to be stable. Since the
output of the unstable machine is zero during the fault the equal-area criterion can
be applied by solving Prob. 14.16 to find the actual critical clearing time which
will be between 0.20 and 0.225 s.
It is noted from the swing curves of Fig. 14.15 that, even though clearing does
not occur until 13.5 cycles after the onset of the fault, the change in the rotor angle of
machine 2 is quite small. Consequently, it is interesting to calculate the approximate
frequency of oscillation of the rotor based on the linearization procedure presented
in Sec. 14.5. The synchronizing power coefficient calculated from the postfault
power-angle equation for machine 2 is given by
SP = ^ =~ [0.1804 + 6.4934 sin (5 - 0.847°)]

do do
= 6.4934 cos (8 - 0.847°)]
We note that for the points calculated for Table 14.6, the angle of S for machine 2
varies between 10.43 and 21.09°. Using either angle makes little difference in the
Figure 14.15 Swing curves for machines 1 and 2 of Examples 14.9 to 14.11 for clearing at 0.225 s.
value found for SP. If we used the average value of 15.76°, we find
SP = 6.274 per-unit power/elec rad
and by Eq. (14.50) the frequency of oscillation is
1 / 377 x 6.274
/„ = 2w 2x8
= 1.935 Hz
and the period of oscillation is
T= = 0.517 s
1.935
Examination of Fig. 14.15 or Table 14.6 confirms the applicability of these

calculations for machine 2 when the fault is on for 0.225 s. For faults of shorter
duration, similar results can be expected since the rotor-angle swing is corre¬
spondingly smaller.
It is possible to calculate the swing curves separately for each machine in the
above examples because of the fault location considered. When other fault locations
are chosen, the intermachine oscillations between the two generators occur because
there is no decoupling of the machines. The swing-curve computations are more
unwieldy. For such cases, manual calculations are time consuming and should be
avoided. Digital computer programs of great versatility are generally available and
should be used.
14.10 DIGITAL-COMPUTER PROGRAMS FOR

TRANSIENT STABILITY STUDIES
Present-day digital computer programs for transient stability studies have

evolved from two basic needs (a) the requirement to study very large inter¬
connected systems with very many machines and (b) the need to represent ma¬
chines and their associated control systems by more detailed models. The classical
machine representation is suitable for many studies. However, more elaborate
models may be required to represent modern turboalternators with dynamic
characteristics determined by the many technological advances in design of
machine and control systems.
The simplest possible synchronous machine model is that used in classical
stability studies. Much more complicated two-axis machine models are available
which provide for direct- and quadrature-axis flux conditions during the sub¬
transient and transient periods following a system disturbance. For example,
unless the machine model explicitly provides for varying flux linkages of the field
winding in the direct axis, it is not possible to represent the action of the
continuously acting automatic voltage regulator and excitation system with
which all modern machines are equipped. Turbine control systems, which auto¬
matically govern the mechanical power input to the generating unit, also have
dynamic response characteristics which can influence rotor dynamics. If these
control schemes are to be represented, the generating unit model must be further
extended. The more complex generator models give rise to a larger number of
differential and algebraic equations for each machine. In large system studies,
many generators are interconnected with widely dispersed load centers by an
extensive transmission system whose performance also must be represented by a
very large number of algebraic equations. Therefore, two sets of equations must
be solved simultaneously for each interval following the occurrence of a system
disturbance. One set consists of the algebraic equations for the steady-state
behavior of the network and its loads and the algebraic equations relating Vt and
E of the synchronous machines. The other set consists of the differential equa¬
tions which describe the dynamic performance of the machines and associated
control systems.
The Newton-Raphson load-flow procedure described in Chap. 8 is probably
the most commonly used solution technique for the network equations. Any one
of several well known step-by-step procedures may be chosen for numerical
integration of the differential equations. The fourth-order Runge-Kutta method
Table 14.7 Computer printout of swing curves for ma¬

chines 1 and 2 of Examples 14.9 to 14.11 for clearing at
0.225 and 0.05 s.
CLEARING AT 0.225 S CLEARING AT 0.05 S
MACH I HACK 2 MACH 1 MACH 2
TIME ANGLE ANGLE TIME ANGLE ANGLE
0.00 20.a 16.2 0.00 20.e 16*2
0.05 25. 1 16.6 0.05 25. 1 16.6
0. 10 37.7 17.6 0.10 32.9 17. 2
0.15 58.7 19.0 0.15 37.3 17.2
0.20 8B • 1 20. 3 0.20 36. 8 16.7
0.25 123.1 20.9 0.25 31 • 7 15.9
O. 30 151.1 19. 9 0. 30 23* 4 15.0
0.35 175.5 17.4 0.35 14.6 14.4
0. 40 205. 1 14.3 0.40 8. 6 14.3
0.45 249.9 11.8 0.45 6. 5 14.7
0.50 319.3 1 0. 7 0.50 10* 1 15.6
0.55 407.0 1 1.4 0.55 17. 7 16.4
0. 60 489.9 13.7 0.60 26.6 17. 1
0.65 566.0 16.8 0.65 34*0 17.2
0.70 656.4 1 9.4 0.70 37. 6 16. e
0. 75 767.7 20. 8 0.75 36.2 16.0
is very often used in production-type transient-stability programs. Other

methods known as the Euler method, the modified-Euler method, the trapezoi¬
dal method, and predictor-corrector methods like the step-by-step method
developed in Sec. 14.9 are alternatives. Each of these methods has advantages
and disadvantages associated with numerical stability, time-step size, computa¬
tional effort per integration step, and accuracy of solutions obtained.!
Table 14.7 shows the computer printout for plotting the swing curves of
machines 1 and 2 of Example 14.11 for clearing at 0.225 and at 0.05 s obtained
by use of a production-type stability program which couples a Newton-Raphson
load-flow program with a fourth-order Runge-Kutta procedure. It is interesting
to compare the closeness of our hand-calculated values in Table 14.6 with those
for machine 2 in Table 14.7 when the fault is cleared at 0.225 s.
t For further information, see G. W. Stagg and A. H. El-Abiad, Computer Methods in Power
System Analysis, chaps. 9 and 10, McGraw-Hill Book Company, New York, 1968.
Our assumption of constant admittances of the loads allowed us to absorb

these admittances into Ybus and avoid load-flow calculations required when the
more accurate solutions using Runge-Kutta calculations are desired. The latter,
being fourth order, require four iterative load-flow computations per time step.
14.11 FACTORS AFFECTING TRANSIENT STABILITY
There are two factors which can act as guideline criteria for the relative stability
of a generating unit within a power system. These are the angular swing of the
machine during and following fault conditions and the critical clearing time. It is
apparent from the equations which we have developed in this chapter that the H
constant and the transient reactance X'd of the generating unit have a direct effect
on both of these criteria.
Inspection of Eqs. (14.84) and (14.85) indicates that the smaller the H
constant, the larger the angular swing during any time interval. On the other
hand, Eq. (14.36) shows that Pmax decreases as the transient reactance of the
machine increases. This is so because the transient reactance forms part of the
overall series reactance of the system which is the reciprocal of the transfer
admittance. Examination of Fig. 14.11 shows that all three power curves are
lowered when Pmax is decreased. Accordingly, for a given shaft power Pm, the
initial rotor angle 50 is increased, <5max is decreased, and a smaller difference
between <50 and <5cr exists for a smaller Pmax. The net result is that a decreased
Pmax constrains a machine to swing through a smaller angle from its original
position before it reaches the critical clearing angle. Thus, any developments
which lower the H constant and increase the transient reactance of a machine
cause the critical clearing time to decrease and lessens the probability of main¬
taining stability under transient conditions. As power systems continually in¬
crease in size, there is a corresponding need for higher-rated generating units.
These larger units have advanced cooling systems which allow higher-rated
capacities without comparable increase in rotor size. As a result, H constants
continue to decrease with potential adverse impact on generating unit stability.
At the same time, this uprating process tends to result in higher transient and
synchronous reactances which makes the job of designing a reliable and stable
system even more challenging.
Fortunately, stability control techniques and transmission system designs
have also been evolving to increase overall system stability. The control schemes
include
Excitation systems
Turbine valve control
Single-pole operation of circuit breakers
Faster fault clearing times
The system design strategies, aimed at lowering system reactance, include
Minimum transformer reactance

Series capacitor compensation of lines
Additional transmission lines
When a fault occurs on a system the voltages at all buses are reduced. At
generator terminals the reduced voltages are sensed by the automatic voltage
regulators which act within the excitation system to restore generator terminal
voltages. The general effect of the excitation system is to reduce the initial rotor
angle swing following the fault. This is accomplished by boosting the voltage
applied to the field winding of the generator through action of the amplifiers in
the forward path of the voltage regulators. The increased air-gap flux exerts a
restraining torque on the rotor which tends to slow down its motion. Modern
excitation systems employing thyristor controls rapidly respond to bus voltage
reduction and can effect from one-half to one-and-one-half cycles gain in critical
clearing times for three-phase faults on the high-side bus of the generator step-up
transformer.
Modern electrohydraulic turbine governing systems have the ability to close
turbine valves to reduce unit acceleration during severe system faults near the
unit. Immediately upon detecting differences between mechanical input and elec¬
trical output, control action initiates the valve closing which reduces the power
input. A gain of one to two cycles in critical clearing time can be achieved.
Reducing the reactance of the system during fault conditions increases
vi Pmax decreasing the acceleration area of Fig. 14.11, and thereby enhances the
possibility of maintaining stability. Since single-phase faults occur more often
than three-phase faults, relaying schemes, allowing independent or selective
circuit-breaker pole operation, can be used to clear the faulted phase while
keeping the unfaulted phases intact. Separate relay systems, trip coils, and oper¬
ating mechanisms can be provided for each pole so that stuck breaker contin¬
gencies following three-phase faults can be mitigated in effect. Independent-pole
operation of critical circuit breakers can extend the critical clearing time by 2 to
5 cycles depending upon whether 1 or 2 poles fail to open under fault conditions.
Such gain in critical clearing time can be important especially if backup clearing
times are a problem for system stability.
Reducing the reactance of a transmission line is another way of raising Pmax.
Compensation for line reactance by series capacitors is often economical for
increasing stability. Increasing the number of parallel lines between two points is
a common means of reducing reactance. When parallel transmission lines are
used instead of a single line, some power is transferred over the remaining line
even during a three-phase fault on one of the lines unless the fault occurs at a
paralleling bus. For other types of faults on one line, more power is transferred
during the fault if there are two lines in parallel than is transferred over a single
faulted line. For more than two lines in parallel the power transferred during the
fault is even greater. Power transferred is subtracted from power input to obtain
accelerating power. Thus increased power transferred during a fault means'lower
accelerating power for the machine and increased chance of stability.
PROBLEMS
14.1 A 60-Hz four-pole turbogenerator rated 500 MVA, 22 kV has an inertia constant of
H = 7.5 MJ/MVA. Find (a) the kinetic energy stored in the rotor at synchronous speed and (b) the
angular acceleration if the electrical power developed is 400 MW when the input less the rotational
losses is 740,000 hp.
14.2 If the acceleration computed for the generator described in Prob. 14.1 is constant for a period of
15 cycles, find the change in S in electrical degrees in that period and the speed in revolutions per
minute at the end of 15 cycles. Assume that the generator is synchronized with a large system and has
no accelerating torque before the 15-cycle period begins.
14.3 The generator of Prob. 14.1 is delivering rated megavolt-amperes at 0.8 power factor lag when a
fault reduces the electric power output by 40%. Determine the accelerating torque in newton-meters at
the time the fault occurs. Neglect losses and assume constant power input to the shaft.
14.4 Determine the WR2 of the generator of Prob. 14.1.
14.5 A generator having H = 6 MJ/MVA is connected to a synchronous motor having// = 4 MJ/MVA
through a network of reactances. The generator is delivering power of 1.0 per unit to the motor when a
fault occurs which reduces the delivered power. At the time when the reduced power delivered is 0.6 per
unit determine the angular acceleration of the generator with respect to the motor.
14.6 A power system is identical to that of Example 14.3 except that the impedance of each of the
parallel transmission lines is j0.5 and the delivered power is 0.8 per unit when both the terminal voltage
of the machine and the voltage of the infinite bus are 1.0 per unit. Determine the power-angle equation
for the system during the specified operating conditions.
14.7 If a three-phase fault occurs on the power system of Prob. 14.6 at a point on one of the transmission
lines at a distance of 30% of the line length away from the sending-end terminal of the line, determine
(a) the power-angle equation during the fault and (b) the swing equation. Assume the system is operating
under the conditions specified in Prob. 14.6 when the fault occurs. Let H = 5.0 MJ/MVA as in
Example 14.4.
14.8 Series resistance in the transmission network results in positive values for Pc and y in Eq. (14.80).
For a given electrical power output, show the effects of resistance on the synchronizing coefficient SP,
the frequency of rotor oscillations, and the damping of these oscillations.
14.9 A generator having H = 6.0 MJ/MVA is delivering power of 1.0 per unit to an infinite bus through
a purely reactive network when the occurrence of a fault reduces the generator output power to zero.
The maximum power that could be delivered is 2.5 per unit. When the fault is cleared the original
network conditions again exist. Determine the critical clearing angle and critical clearing time.
14.10 A 60-Hz generator is supplying 60% of Pmax to an infinite bus through a reactive network. A fault
occurs which increases the reactance of the network between the generator internal voltage and the
infinite bus by 400%. When the fault is cleared the maximum power that can be delivered is 80% of the
original maximum value. Determine the critical clearing angle for the condition described.
14.11 If the generator of Prob. 14.10 has an inertia constant of H = 6 MJ/MVA and Pm (equal to
0.6 Pmax) is 1.0 per-unit power, find the critical clearing time for the condition of Prob. 14.10. Use
At = 0.05 s to plot the necessary swing curve.
14.12 For the system and fault conditions described in Probs. 14.6 and 14.7 determine the power-angle
equation if the fault is cleared by the simultaneous opening of breakers at both ends of the faulted line
at 4.5 cycles after the fault occurs. Then plot the swing curve of the generator through r = 0.25 s.
14.13 Extend Table 14.6 to find <5 at t = 1.00 s.
14.14 Calculate the swing curve for machine 2 of Examples 14.9 to 14.11 for fault clearing at 0.05 s by
the method described in Sec. 14.9. Compare the results with the values obtained by the production-type
program and listed in Table 14.7.
14.15 If the three-phase fault on the system of Example 14.9 occurs on line 4-5 at bus 5 and is cleared by
simultaneous opening of breakers a*, both ends of the line at 4.5 cycles after the fault occurs prepare a
table like that of Table 14.6 to plot the swing curve of machine 2 through t = 0.30 s.
14.16 By applying the equal-area criterion to the swing curves obtained in Examples 14.9 and 14.10 for
machine 1, (a) derive an equation for the critical clearing angle, (b) solve the equation by trial and error
to evaluate <5cr, and (c) use Eq. (14.72) to find the critical clearing time.
V*
'
.
APPENDIX
423
APPENDIX 425
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Table A.l Electrical characteristics of bare aluminum conductors steel-reinforced (ACSR)t
t Data, by permission, from Aluminum Association, "Aluminum Electrical Conductor Handbook,” New York, September 1971.
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426
“Electrical Transmission and Distribution Reference Book,” by permission of the Westinghouse Electric Corporation.
Table A.3 Shunt capacitive-reactance spacing factor Xd at 10 Hzt (megohm-miles per conductor) APPENDIX
427
Table A.4 Typical reactances of three-phase synchronous machines.t
Values are per unit. For each reactance a range of values is listed below the typical value.J
Turbine-generators
2-pole 4-pole Salient-pole generators
Conventional Conductor Conventional Conductor With Without

cooled cooled cooled cooled dampers dampers
1.76 1.95 1.38 1.87 i 1

1.7-1.82 1.72-2.17 1.21-1.55 1.6-2.13 0.6-1.5 0.6-1.5
1.66 1.93 1.35 1.82 0.6 0.6

1.63-1.69 1.71-2.14 1.17-1.52 1.56-2.07 0.4-0.8 0.4-0.8
xd 0.21 0.33 0.26 0.41 0.32 0.32

0.18-0.23 0.264-0.387 0.25-0.27 0.35-0.467 0.25-0.5 0.25-0.5
x: 0.13 0.28 0.19 0.29 0.2 0.30

0.11-0.14 0.23-0.323 0.184-0.197 0.269-0.32 0.13-0.32 0.2-0.5
X2 =x: = X'J 0.2 0.40

0.13-0.32 0.30-0.45
*0§
t Data furnished by Westinghouse Electric Corporation.

'} Reactances of older machines will generally be close to minimum values.
§ X0 varies so critically with armature winding pitch that an average value can hardly be given.
Variation is from 0.1 to 0.7 of X'j.
Table A.5 Typical range of transformer reactancest

Power transformers 25,000 kVA and larger
Nominal system voltage, kV Forced-air-cooled, % Forced-oil-cooled, %
34.5 5-8 9-14

69 6-10 10-16
115 6-11 10-20
138 6-13 10-22
161 6-14 11-25
230 7-16 12-27
345 8-17 13-28
500 10-20 16-34
700 11-21 19-35
t Percent on rated kilovoltampere base. Typical transformers are now designed for the minimum
reactance value shown. Distribution transformers have considerably lower reactance. Resistances of
transformers are usually lower than 1%.
APPENDIX 429
Table A.6 ABCD constants for various networks
z
A
A/W
A = 1
B = Z
C = 0
D = 1
Series impedance
Is Ir
A = 1
B = 0
C = Y
D = 1
Shunt admittance
A = 1 + YZi
B = Z\ A Zi A YZ\Zi
C - Y
D = 1 + YZi
Unsymmetrical T
Is Z IR
A = 1 + YYZ
B = Z
C = Y\ -(- Yi A ZY1Y2
D = 1 A FiZ
Unsymmetrical 7r
/s _ _
-► A = A1A2 A B\Ci
n-
A
B = A1B2 A B1D2
vs A\BxClDl A2 ^2 C2&2
C — A2C1 A C2D1
1
- ♦
D = BiCi + DiDi
Networks in series
A = (A1B2 A AiBi)/(B\ A B2)

£ = BiB2/(Bi A £2)
C = Cx A Ci A (Ai - A2) (£2 - Di)/(Bi A £2)
£ = (£2£>i A B1D1)/(£1 A #2)
Networks in parallel
INDEX
ABCD constants, 94 Bus impedance matrix:

in circle-diagram equations, 108-110 equivalent network: for three-phase faults,
in power-flow equations, 107 264-265
table of, for networks, 429 for unsymmetrical faults, 328-331
Acceleration factors, 196 formation of, 178, 187-190
Admittance: measurement of element values, 180
mutual-, of nodes, 170 modification of, 182-187
as data in load-flow studies, 193, 194 in three-phase fault analysis, 261-263
measurement of, 179 in unsymmetrical fault analysis, 328-331
self-, of nodes, 170
as data in load-flow studies, 193, 194 Capacitance:
measurement of, 179 calculation by method of modified geometric
Admittance matrix (see Bus admittance matrix) mean distance, 84-86
Alternating-current (ac) circuits: definition, 37, 68
balanced three-phase, 22-29 effect of earth on, 81-83
double-subscript notation for, 12-13 effect of nonuniform charge distribution, 74
power in balanced three-phase, 28-29 effect of stranded conductors, 74
power in single-phase, 13-18 to neutral, 73-74
single-subscript notation for, 11-12 of parallel circuit three-phase lines, 85-86
Alternator (see Synchronous machines) summary of calculations, 86-87
American National Standards Institute (ANSI): of three-phase lines: with equilateral spacing,
graphical symbols for diagrams, 156 76-78
guide for circuit-breaker applications, 267 with unsymmetrical spacing, 78-81
Area control error (ACE), 245 transposition to balance, 78-79
Armature reaction, 131-135, 250-251 of two-wire lines, 71-75
Attenuation constant, 97 (See also Reactance)
Automatic generation control, 243-246 Capacitive reactance (see Reactance)
Autotransformers, 145-146 Capacitors:
to control bus voltage, 181-182, 212-214
as generators of reactive power, 17-18
B coefficients, 234-243 for series compensation, 110-111
calculation of, 234-237, 242-243 Characteristic impedance, 97-98, 117
Bundled conductors, 61 termination of line by, 98, 118
capacitive reactance of, 83-85 Charging current, 69, 78
inductive reactance of, 61-62 Circle diagrams of transmission line P and Q,
Bus admittance matrix: 108-110
definition of, 170 Circuit breakers, 3, 6, 289-294, 338-340
measurement of element values, 179 Compensation factor:
Bus impedance matrix: series, 110
definition, 178 shunt, 112
431
432 INDEX
Complex power, 18 Double-subscript notation, 12-13

Computer (see Digital computer) Driving-point impedance of nodes, 178
Conductance, 37, 68 measurement of, 180
Conductors:
table of characteristics of ACSR, 425
Earth, effect on capacitance, 81-83
types, 38-39
Economic dispatch (see Economic load
Control:
distribution)
of P and Q over transmission lines: by
Economic load distribution:
capacitors, 214
between plants: example of, 240-241
by generator adjustment, 137-138,
mathematical development, 238-240
206-208
savings effected by, 241-242
by transformers, 214-223
between units in a plant, 228-234 .
of plant generation, 243-246
example of, 231-233
of power interchange between areas,
intuitive concept, 230
243-246
mathematical development, 230-231
Coordination time delay, 353, 355
savings effected by, 234
Corona, 38, 68
Economic operation of power systems (see
Critical clearing angle, 397-400
Economic load distribution)
Critical clearing time, 397-399
Eddy-current loss, 142
CT (see Current transformers)
Energy sources, 3, 4
Current:
Equal-area criterion, 393-400
direction of, 12
Equivalence of sources, 166-168
equations of: for long transmission lines,
Equivalent circuit:
96-97, 100
for bus impedance matrix, 264-265, 328-331
for medium-length transmission lines,
of long transmission line, 104-106
93-94
of medium-length transmission line, 93
for short transmission lines, 92
of short transmission line, 91
incident wave of, 98, 117, 118, 120
Equivalent circuit:
momentary, in circuit breakers, 267
of synchronous machines, 135-136
reflected wave of, 98, 118, 120
of transformers, 142-145
Current:
Equivalent equilateral spacing Z)eq, 60, 80
subtransient, 252-260
transient: in RL circuits, 249
in synchronous machines, 251-253 Faults:
Current transformers, 343-344 calculations using Zbus, 261-266, 328-331
CT ratios, 344 definition, 6
pickup settings, 347 double line-to-ground, 312-316, 319-320
tap settings, 347 through impedance, 331-334
CVT (see Voltage transformers for system line-to-line, 309-312, 319
protection) single line-to-ground, 306-309, 318
three-phase: on power systems, 261-266
on synchronous machines, 250-255
A-Y transformers: various types: occurrence of, 7
per-unit impedance of, 151 increasing severity of, 400
phase shift in, 281-288 Field intensity:
protection of, 367-369 electric, 70
Digital computer: magnetic, 45-46
for automatic generation control, 243-246 Flux linkages:
in fault calculations, 334 of coils, 43
in load-flow studies, 203-206 internal, 44-47
in stability studies, 416-418 of one conductor in a group, 50-52
Direct current (dc) transmission, 2, 122-123 partial, 44-47
Double line-to-ground faults: Fortescue, C. L., 275
through impedance, 331-334 Fuel cost, 228-234, 238-241
on a power system, 319-320 Frequency bias, 245
on unloaded generators, 312-315 Frequency of oscillation, 392-393, 415
INDEX 433
Gauss-Seidel method for load-flow studies, Inductance:

194-196 mutual, 44
Generalized circuit constants (see ABCD of parallel-circuit three-phase line, 63-64
constants) of single-phase two-wire line, 50
Geometric mean distance (GMD): summary of calculations, 64-65
conductor to conductor, 54 of three-phase line: with equilateral spacing,
method for inductance calculations, 54-65 58
method for capacitance calculations, 83-87 with unsymmetrical spacing, 58-61
mutual, 54 transposition to balance, 59
self (see Geometric mean radius) (see also Reactance)
Geometric mean radius (GMR): Inductive reactance (see Reactance)
of bundled conductors, 60, 61, 83-85 Inertia constant, 377
definition, 54 (See also H constants)
examples of calculation: to compute Infinite bus, 137
inductance, 54-65 in stability studies, 383
modified to compute capacitance, 83-87 Infinite line, 98, 118
table of, for ACSR, 425 Input-output curve, 228
Ground wires, 88, 113 Interchange of power, 244-245
H constants: Kinetic energy, 380

calculation of, 379-382
definition, 378 Lagrangian multipliers, 231
table of, 380 Lightning:
Hyperbolic functions, 99-101 arresters, 121-122
Hysteresis loss, 142 ground wires for protection against, 88, 113
Line-to-line faults (see Faults)
Interconnection of power systems, 3
Load dispatching (see Economic load
as element of automatic generation control,
distribution)
243-245
Load-flow control by transformers, 214-223
Image conductors, 82
Load-flow studies:
Impedance:
data for, 193-194, 203
characteristic (see Characteristic impedance)
on digital computer: data for, 203
diagrams, 157-159
information obtained, 204
driving-point, of nodes, 178
numerical results, 204-206
measurement of, 180
by Gauss-Seidel method, 194-196
measured through transformes, 141
by Newton-Raphson method, 196-200
negative-sequence (see Negative-sequence
Long-line equations:
impedance)
hyperbolic form of, 99-104
positive sequence (see Positive-sequence
interpretation of, 97-99
impedance)
solution of, 94-97
surge, 98
Loss coefficients (see B coefficients)
transfer, of nodes, 178
Losses as function of plant generation, 234-238
measurement of, 180
of transformers, 143-145 Matrix (see Bus admittance matrix, Bus
table, 428 impedance matrix, Matrix partitioning)
zero-sequence (see Zero-sequence Matrix partitioning, 172-173
impedance) node elimination by, 174-178
(See also Reactance) Medium-length line, 93-94
Impedance matrix (see Bus impedance matrix) Mismatch in load-flow studies, 204
Incident voltage and current, 98, 117-120 Mutual-admittance of nodes (see Admittance)
Incremental fuel cost, 228-231
Inductance: Negative-sequence components, 276, 278-281
calculation by method of geometric mean Negative-sequence impedance, 291
distance, 54-65 of circuit elements, 294, 295
definition, 43-44 of synchronous machines, 294
due to internal flux, 44-47 table of, 428
434 INDEX
Negative-sequence networks (see Sequence) Power angle, 137, 206-208

networks) (See also Power angle equations, Power 'angle
Network reduction, 174-178 curves)
Networks: Power angle curves, 386
ABCD constants of, table of, 429 in equal-area analysis, 393-400
(See also Sequence networks) (See also Power angle equations)
Newton-Raphson method for load-flow studies, Power angle equations, 383-390, 403
196-200 definition, 386
Node: examples of, 386-390, 404-408
definition, 168 Power factor, 16, 17
elimination, 174-178 Power triangle, 18-19
major, 168 Precision index, 195, 198, 203
Node equations, 168-172 Propagation constant, 97
for load-flow problems, 194-195 Protection:
backup, 352-354, 357-360
One-line diagrams, 155-157 of buses, 352
of generator and motor, 350-352
Penalty factor, 239 of transformers, 366-370
Per-unit quantities: of transmission lines: HV and EHV, 360-365
advantages of, 159, 161-162 with pilot relays, 365-366
change of base, 33-34 Protection:
definition of, 30 of subtransmission lines, 354-360
Per-unit quantities (See also Protection systems, Relays)
selection of base, 147, 159 Protection systems, 337
for three-phase quantities, 31-33 subsystems of, 338-340
Per-unit reactance: (See also Protection, Relays)
of synchronous machines, table of, 428
of three-winding transformers, 153-155
of transformers, table of, 428
(See also Per-unit quantities) Reactance:
Phase constant, 97 of bundled conductors: capacitive, 83-85
Phase relays, 361 inductive, 61-62
Phase sequence, 23-25 capacitive: at 1—ft spacing, 75
Phase shift in three-phase transformers, table of, for ACSR, 425
281-288 spacing factor, 75
Polarity marks: table of, 427
for transformers, 140, 281-285 direct-axis, 136, 251
for voltages, 11-12 inductive: at 1—ft spacing, 57
Positive-sequence components, 276, 278-281 table of, for AGSR, 425
Positive-sequence impedance, 291 spacing factor, 57
of circuit elements, 294-295 table of, for ACSR, 426
Positive-sequence networks (see Reactance) leakage, of transformers: 143
diagrams, Sequence networks) in synchronous machines, 134
Power: sequence (positive, negative, and zero), 291
average value, 16 of circuit elements, 294-295
complex, 18 of synchronous machines, 294
control of flow by transformers, 214-223 table of, 428
direction of flow, 19-22 subtransient, 252
instantaneous, 14 synchronous, 251
reactive, sign of, 17, 21 transient, 251
in single-phase circuits, 13-18 (See also Capacitance, Inductance)
by symmetrical components, 288-289 Reactance diagrams, 157-159
in three-phase circuits, 22-28 Reactive compensation of transmission lines:
transmitted over transmission line, 107-109 series, 110, 111
(See also Reactive power) shunt, 110, 112
INDEX 435
Reactive power: Stability:

control of flow: types of, 373-374
by machine excitation, 137, 138 (See also Stability studies)
by regulating transformer, 214-223 Stability limit, 9
definition, 16 Stability studies, 8-9
direction of flow, 19-22 assumptions for all, 375
generated by capacitors, 17-18 assumptions for classical machine model, 402
sign of, 17, 21 rotor dynamics, 375-379
of synchronous machines: swing equation derived, 375-379
with overexcitation, 137-138 types of, 373-374
with underexcitation, 137-138 Station control error (SCE), 245
Reflected voltage and current, 98, 117-120 Surge arresters, 121-122
Reflection coefficient, 118 Surge impedance loading, 98-99
Regulation (see Voltage regulation) Swing bus, 194, 198
Relays Swing curves, 379
characteristic curves of IFC-53, 348 computer solution of, 416-418
definition of, 338 step-by-step solution of, 409-416
differential, 350-352 Swing equations, 378-379
directional, 347, 349 (See also Swing curves)
distance, 349-350, 360-365 Symmetrical components:
impedance, 349-350, 360-365 definition of, 275-277
magnitude, 345-347 phase shift of, in Y-A transformers, 281-288
mho, 350, 363 power in terms of, 288-289
overcurrent, 345-347, 357-360 of unsymmetrical phasors, 275-277, 278-281
percent differential, 352 Synchronizing power coefficients, 390-392,
phase, 361 414-415
pilot, 350 Synchronous condenser, 213
ratio, 349, 350 Synchronous machines, 127-138
reach of, 340 armature reaction, 131-135
(See also Protection) construction, 128-131
Resistance, 40-43 equivalent circuit, 133-136
table of, for ACSR, 425 excitation, 128, 137-138
frequency, 130
mmf in, due to armature current, 132-135
phasor diagram for, 133-135
Self-admittance of nodes (see Admittance) power angle, 137, 206-208
Sequence impedances of circuit elements, reactance: leakage, 134
sequence (positive, negative, and zero), 291
294-295
Sequence networks, 291-293, 295-302 subtransient, 252
interconnection for fault calculations, synchronous, 135, 251
316-322, 328-334 table of, 428
Short transmission line, 91-93 transient, 251
Short-circuit megavoltamperes, 263-264 reactive power, dependence on excitation,
Single line-to-ground faults (see Faults) 137-138
Single-line diagrams (see One-line diagrams) speed, 130
Single-phase loads, 334 System protection (see Protection, Protection
Single-subscript notation, 11-12 systems)
Skin effect, 42
Sources: Thevenin’s theorem:
in calculation of three-phase fault currents,
of electric power, 3-4
equivalence of, 166-168 256, 258-260, 263
in representing power systems, 181, 209-211
Stability:
definition of, 373 Three-phase circuits:
factors affecting, 418-420 power in, 28, 29
rotor dynamics for, 375-379 voltage and current in, 22-28
436 INDEX
Three-phase faults, 248-266 Transposition of transmission lines:

Torque angle (see Power angle) to balance capacitance, 78-80 v.”
Transducers, 338, 339 to balance inductance, 59
current transformers (CT), 343-344
voltage transformers, 344-345
Underground transm, ->n, 7
Transfer impedance of nodes, 178
cables of, 39-40
measurement of, 180
Unsymmetrical faults (see Faults)
Transformers:
Unsymmetrical series impedances, 289-291
autotransformers, 145-146
dot convention for, 140
eddy-current loss in, 142 Velocity of propagation, 99, 116
equivalent circuits of, 142-145
Voltage:
for zero-sequence current, 298-302
effect of capacitors on, 110-111, 181-182,
hysteresis loss in, 142 212-214
ideal, 138-142 effect of specification at bus, 209-211
impedance seen by, 141 incident wave of, 98, 117, 118, 120
leakage reactance of, 143 levels of: in generators, 4
load-tap-changing (LTC), 214
in distribution systems, 5
magnetizing current in, 142-145
in transmission lines, 4
off-nominal turns ratio, 215-217
polarity marks for, 11-12
open-circuit test, 145
reflected wave of, 98, 118, 120
phase shift in Y-A, 281-288
subtransient internal, 256
polarity markings for, 140, 281-285 in three-phase circuits, 22-28
protection of, 366-369
transient internal, 256
reactances, table of, 428
Voltage regulation:
regulating, 214-223
definition of, 92
to adjust voltage magnitude, 214-215,
equation using constant A, 94
218-221
reduction by reactive compensation, 112-113
to adjust voltage phase angle, 214-215,
Voltage transformers for system protection,
221-223
344-345
short-circuit test, 145
tap-changing under load (TCUL), 214
three-phase, 149-152 Wavelength, 99
three-winding, 153-155 Waves, incident and reflected, 98, 117-120
Transient analysis of transmission lines:
lattice diagrams, 120 Ybus (see Bus admittance matrix)
Y-A transformers (see A-Y transformers)
reflections, 117-120
reflection coefficient, 118, 120
traveling waves, 114-117 ^bus (see Bus impedance matrix)
incident, 117-120 Zero-sequence components, 276, 278-281
reflected, 118-120 Zero-sequence impedance, 291
velocity of, 116 of circuit elements, 294-295
Transient stability, 373-375 of synchronous machines, 294
factors affecting, 418-419 table of, 428
(See also Stability studies) Zero-sequence networks:
Transients on transmission lines, 113-122 of delta-connected loads, 298
Transmission lines: reference point of, 296
classification of, by length, 90, 91 of three-phase transformers, 298-300
long, 94-107 of unloaded generators, 291-293
medium length, 93-94 of Y-connected loads, 297-298
short, 91-93 Zones of protection, 340-342, 360-365
transients on, 113-122 overreaching, 361
zero-sequence reactance of, 294-295 underreaching, 361

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Power and Energy

Elgerd: Electric Energy Systems Theory: An Introduction

William D. Stevenson, Jr.

ELEMENTS OF POWER SYSTEM ANALYSIS

Library of Congress Cataloging in Publication Data

(McGraw-Hill series in electrical engineering.

Chapter 1 General Background 1

l.i The Growth of Electric Power Systems 1

Chapter 2 Basic Concepts 10

Chapter 3 Series Impedance of Transmission Lines 37

3.1 Types of Conductors 38

3.5 Inductance of a Conductor Due to Internal Flux 44

Chapter 4 Capacitance of Transmission Lines 68

4.1 Electric Field of a Long Straight Conductor 69

Chapter 5 Current and Voltage Relations on a

Chapter 6 System Modeling 127

6.1 Construction of the Synchronous Machine 128

Chapter 7 Network Calculations 166

7.1 Equivalence of Sources 166

Chapter 8 Load-Flow Solutions and Control 193

8.1 Data for Load-Flow Studies 193

Chapter 9 Economic Operation of Power Systems 227

Chapter 10 Symmetrical Three-Phase Faults *..248

10.1 Transients in RL Series Circuits 248

Chapter 11 Symmetrical Components 275

Chapter 12 Unsymmetrical Faults 305

12.1 Single Line-to-Ground Fault on an Unloaded Generator 306

Chapter 13 System Protection 337

13.3 Transducers 342

13.4 Logical Design of Relays 345

13.5 Primary and Backup Protection 352

Chapter 14 Power System Stability 373

I am particularly fortunate to have had two principal contributors to this

William D. Stevenson, Jr.

Development of sources of energy to accomplish useful work is the key to the

1.1 THE GROWTH OF ELECTRIC POWER SYSTEMS

associate of Westinghouse, tested transformers in his laboratory in Gj.eat

1.2 ENERGY PRODUCTION

Most of the electric power in the United States is generated in steam-turbine

1.3 TRANSMISSION AND DISTRIBUTION

to the transmission capability in megavoltamperes (MVA) of a line. Roughly, the

1.4 LOAD STUDIES

System planners are interested in studying a power system as it will exist 10

1.5 ECONOMIC LOAD DISPATCH

1.6 FAULT CALCULATIONS

ful, an appreciable number are caused by permanent faults where reclosure

1.7 SYSTEM PROTECTION

Operation of circuit breakers is controlled by relays which sense the faulL In

1.8 STABILITY STUDIES

The current which flows in an ac generator or synchronous motor depends on

1.9 THE POWER-SYSTEM ENGINEER

1.10 ADDITIONAL READING

The waveform of voltage at the buses of a power system can be assumed to be

i = 7.07 cos cot

| V\ = 100 V and |/| = 5 A

the voltage which leads the reference phasor by 30° is

V = 100/30° = 86.6 + j50 V