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POWER SYSTEM ANALYSIS
Professor,' Department
John J. Grainger
of EleClrica/ and Compl/{er Engineering
.Yonh larolina S{o{e Uniccrsi{y
WUliam
D. Stevenson, Jr.
Lale Professor uf E/ec/m:o/ Engineering
Nonh Camfit/(J .\{(J/c Ulllcersi{y
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POWER SYSTEM ANALYSIS
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Grainger, John J
Power system analysis / John J. Grainger, William D. Stevenson.
p.
cm.
Based on: Elements of power system anal ys i s by William D. Stevenson.
.
,
0-07-061293-5
1. Electric power di stri b ut i on
Includes index.
ISBN
Stevenson, W illi am D.
power system analys i s
I.
TK3001.G73
621.319-dc20
.
1994
2. Electric power systems.
II. Stevenson, William D. Elements of
.
III. Title.
When ordering this title, use ISBN 0-07-113338-0
Printed i1' Singapore
93-39219
To THE MEMORY OF
William D. Stevenson, Jr.
1912-1988
True friend and colleague
ABOUT THE AUTHORS
1" '-
1
John J . Grainger is P rofes sor of E l e ct r i c a l and Computer Engineering a t North
Carolina State Unversity.He is a graduate of the ;'\ ati o n a l University of Ireland
and re c e i ve d his M.S.E.E. and Ph�D. 'degrees at - the' L:"nivcrsity of Wisconsin­
G ra inger is the founding Director of the El e c t ric Power Research
Cente r (It North Carolin,-l State University. a joint uni\'er�ity/industry coopera­
tive research center in electric power systems engineering, He l e ads the
Center's major research programs in transmission and distribution systems
planning, design, (}uto01ation, and c on t rol areas. as well as power system
Madison.
Dr.
dynamics_
Madison, The I llinois Institute of Technology, 't>larquette University, and North
Profes�or
Grainger has
aho
«llight
at
the
of
University
Wisconsin­
ity Supply Board of Ireland; Commonwealth Edison Company, Chicago; \Vis­
consin Electric Po\vcr Company, MilwaUKee; and Carolina Pmver & Light
Company, Raleigh. Dr. Grainger is an active consullant with t h e Pacific Gas and
El e ct ric Company, San Francisco; Southern California Edison Company, Rose­
mead; a n d mimy other power industry organizJtions. His educational ane!
Carolina State University. His industrial experience has been with the Electric­
Am eri c a n Society of Engineering Education, the American Power Conference,
technical
involvements
erRED, and CIGRE.
include
the
IEEE
Povv'cr
Engineering Society,
The
Dr. Grainger is the aut h or of numerous papers in the IEEE Power
Engineering So c i e ty ' s Transactiotls and was recognized by the IEEE Transmis­
s ion and Distribution Committee for the 1985 Prize Paper Award.
In 1984, P ro fessor Grainger was chosen by the Edison Electric Institute
for the EEl Power Engineering Ed u ca tor Award.
William D. Stevenson, Jr. (deceased) was a professor and the As oci at e Head of
the Electrical Engineering De p a rt m e n t of N ort h Carolina State University. A
Fellow of the I n sti tu te of Electrical and Electronics E ng i n ee r s , he worked in
private industry and taught at both Clemson U ni v e rs i ty and P r inc e t on Univer­
s
en?ineering for t h e McGraw-HilI Encyc!opedin of Science and Technology, He
was the recipient of several teaching and professional awards,
sity.
Dr.
Stevenson also served
as a consulting editor in electrical power
CONTEN'fS
Preface
1
Basic Concepts
1.1
1.2
1.3
1.4
1.6
1.5
1.7
1.9
1.8
1.10
1.11
I.J2
1.13
1.14
US
2
Introduction
Single-Subscript NO{dtioll
Do ub le -Sub scr i pt N o ta tio n
Power in Single-Phase AC Circuits
Complex Power
The Power Tr i a n gle
Direction of Pow er flow
Vol tag e anc1 Current in BalanccJ Three-Phase Circuits
Power in Balllnccu Three-Phase Circuits
Per-Unit Qualltities
C ha n ging the B,lse of Per-Unit OU(lntities
Node Equations
The S i ngle - L i n e or One-Line Diagram
Impedance and Reactance Diag r ams
Summary
Proble ms
Transformers
2.1 The Ideal Transformer
2.2 Magnetically Coupled Coils
2.3 The Equivalent Circuit of a Single-Phase Transformer
2.4 Per-Unit Impedances in Single-Phase Transformer Circuits
2.5 Three-Phase Transformers
2.6 Three-Phase Transformers: Phase Shift and Equivalent Circuits
2.7 The Autotransformer
2.8 Per-Unit Impedances of Three-Winding Transformers
2.9 Tap-Changing and Regulating Transformers
2.10 The Advantages of Per-Unit Computations
2.11 Summary
P ro bl ems
1
.-,
.)
4
1
5
0
10
14
1J
24
25
29
30
34
36
17
37
41
41
46
51
56
59
64
71
72
76
80
82
82
XI
xii
3
CONTENTS
The Synchronous Machine
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
3.10
4
5
Description of the Synchronous Machine
Three-Phase Generation
Synchronous Reactance and Equivalent Circuits
Real and Reactive Power Control
Loading Capabi lity Diagram
Thc Two-Axis Machinc Modcl
Voltagc Equations: Sal icnt-Polc Machine
Transient and Subtransicnt Effects
Short-Circuit Currents
Summary
Problcms
Series Impedance of Transmission Lines
4.1
Typ es of Conductors
4.2
Resistance
4.3
Tab u lated Resistance Values
4.4
Inductance of a Conductor Due to Internal Flux
4.5
Flux: Linkages between Two Points External t o an Isolated
Conductor
4.6
Inductance of a Single-Phase Two-Wire Line
Flux Linkages of One Cond u ctor in a Group
4.7
4.8
Inductance of Composite-Conductor Lines
4.9
The Use of Tables
4.10 Inductance of Three-Phase Lines with Equilateral Spacing
4.11 Inductance of Three-Phase Lines with Unsymmetrical Spacing
4.12 Inductance Calculations for Bundled Conductors
4.13 Summary
Problems
136
141
142
143
146
146
149
151
153
155
159
161
161
164
166
167
Capacitance of Transmission Lines
Electric Field of a Lo ng Straight Conductor
5.1
5.2
The Potential Differen ce between Two Points Due to a Charge
5.3
Capacitance of a Two-Wire Line
Capacitance of a Three-Phase Line with Equilateral Spacing
5.4
Capacitance of a Three-Phase Line with Unsymmetrical Spacing
5.5
Effect of Earth on the Capacitance of Three-Phase Transmission
5.6
Lines
170
171
172
5.7
186
,
5.8
5.9
6
87
88
91
100
105
110
117
123
127
132
136
Capacitance Calculations for Bundled Conductors
Parallel-Circuit Three-Phase Lines
Summary
Problems
173
1 77
180
183
188
190
191
Current and Voltage Relations on a Transmission
Line
6. 1
62
.
Representation of Lines
The Short Transmission Line
193
195
196
CONTENTS
200
6.4
6.3
The Medium-Length Line
6.5
The Long Transmission Line: lnterpretation of the Equations
6.6
6.7
6.8
6.9
Equ a t i on s
The Long Transmission Line: Solution of the D i ffere ntial
The Long Transmission Line: Hyperbolic Form of the Equations
207
The Equivalent Circuit of a Long Line
212
Power Flow through a Transmission Line
215
Reactive Compensation of Transmission Lines
218
221
222
6.12
Transient Analysis: Traveling Waves
Transient Analysis: Rct1ections
226
233
231
Direct-Current Transmission
Summary
233
Problems
The Admittance Model and Netw o r k
Node
7.1
Branch and
7.2
Mutually Coupled Branches in Y hu,
An Equivalent AdmittZlncc i'-Jc(wnrk
7.�
7.4
Admittances
Calculations
238
239
245
251
,-­
-))
7.6
Modifie,l(ion o r VioL"
The Network Incidence Matri\ and Y QUI
The Method of Successive Elimination
274
7.0
Triangular F,lctoriz(ltioll
Summar),
280
7.5
7.7
7.8
7.10
Node Elimination
Sparsity
ancl
(Kron
Reduction)
Ncar-Optilll(li Or(lcring
Prohlcms
The Impedance Model and Network Calcu lations
8.2
8.1
8.4
8.3
The Bus Admittance and Impedance Matrices
Thcvcnin's Thcorcm and Zbus
Mpclification of (Ill Existing Zhu:>
8.5
Direct Determination of Zou:>
Calculation of Zous Elements from YbU5
8.7
Mutually Coupled
8.8
Summary
8.6
Power Invariant
Transformations
B r an ch e s
in Zbus
Power-Flow Solutions
9.1 The Power-Flow Problem
9.3
The Gauss-Seidel Method
9.5
Power-Flow
9.2
9.4
9.6
257
263
271
279
280
283
284
287
30 1
294
306
310
316
324
324
Problems
9
205
Transmission-Line Transients
6.13
6.14
8
202
6.1 0
6. 1 1
7
xiii
329
329
335
The Newton-Raphson Method
F
342
347
The Newton-Raphson Power - l o w Solution
356
Regulating Transformers
361
S tu d i e s
in System Design and Operation
xiv
CONTENTS
9.7
9.8
10
11
The Decoupled Power-Flow Method
S ummary
Problems
Symmetrical Faults
10.1
10.2
10.3
10.4
10.5
10.6
13
380
Transients in RL Series Circuits
I n ternal Voltages of Loaded Machines under Fault Conditions
Fault Calculations Using Zbus
Fault Calculations Using ZhU5 Equivalent Circuits
The Selection of Ci rcu it Breakers
Summary
Problems
Symmetrical Components and Sequence Networks
11.1 Synt hesis of U nsymmetrical P hasors from Thei r Symmetrical
Components
11.2 The Symmetrical Components o f U nsymmetrical P h asors
11.3 Symmetrical Y and!:::. Circuits
11.4 Power in Terms of Symmet rical Com ponents
11.5 Sequence Circuits of and tl I mpedances
11.6 Sequence Circui ts of a Symmetrical Transmission Lin e
11.7 Seq uence Circuits of the Synchronous M achine
1 1 8 Sequence Circuits of -tl Transformers
11.9 Unsymmetrical Series Impedances
11.10 Sequence Networks
11.11 Summary
Problems
Y
Y
.
12
368
374
376
Unsymmetrical Faults
The Transmission-Loss Equation
Interpre tation of Transformation C
13.5 Classical Economic Dispatch with Losses
13.6 Automatic Generation Con t rol
13.7 Unit Commitment
13.4
An
402
411
412
416
41 7
418
422
427
429
435
442
449
459
461
467
467
512
523
527
Economic Operation of Power Systems
13.3
]95
470
470
482
488
494
500
12.1 Unsymmetrical Faults on Power Systems
12.2 Single Line-to-Ground Faults
1 2.3 Line-to-Line Faults
12.4 Double Line-to-Ground Fau l ts
12.5 Demonstration Problems
12.6 Open-Conductor Faults
12.7 Summary
Problems
13.1 D istribution of Load between U n i ts within
13.2 Distribution of Load between P l an ts
381
383
390
a
Plant
531
532
540
543
552
555
562
572
13.8
13.9
14
14.2
578
586
Problems
587
591
14.1
Adding and Removing Multiple Lines
14.3
Analysis of Single Contingencies
611
Analysis of Multiple Contingencies
620
14.5
Contingency Analysis by dc Model
626
System Reduction for Contingency and Fault Studies
Summary
636
Problems
636
14.6
14.7
Piecewise Solution of Interconnected Systems
State Estimation of Power Systems
592
601
628
641
15.1
The Method of Least SqU;:HCS
642
1 5.3
Test for Bad Data
655
Power System State Estimation
664
Sumnnry
688
15.2
15.4
1 is
1 5.6
16
Commitment Problem
xv
Summary
Zbus Methods in Contingency Analysis
14.4
15
So lv ing the Unit
CONTENTS
Statistics, Errors. ,wd Estimates
Thc Structure «(Ill! r:orm;ltion of H \
Problems
Power System Stability
16.1
16.2
16.3
16.4
16.5
16.6
16.7
16.8
16.9
Rotor Dynamics ;lnd the Swing Equation
Further Considerations of the S wing Equation
Thc Power-Angle Equ(ltion
Synchronizing rowcl Cocflicicnls
EClual-Area Criterion of S t abili ty
Further Applications of the Equal-Area Criterion
Multimachine Stability Studies: Classical Representatio!1
Step-by-Slep Solution of the Swing CunT
16.10 Computer Programs for Transient Stability Studies
16.11 Factors Affecting Transient S t abi li ty
16.12 Summary
Problems
A.i
A.2
Quantities
Distributed Windings of the Synchronous Machine
P-Transformation of Stator
Appendix B
()77
687
69S
The Stability Problem
Appendix A
650
695
60S
702
707
714
717
724
727
734
741
743
746
745
748
754
763
766
B.l
Sparsity and Near-Optimal Ordering
766
B.2
Sparsity of the Jacobian
771
Index
777
PREFACE
Elemen ts
the long-standing McGraw-Hill textbook by Professor William D.
This book embodies the principles and objectives of
Analysis,
of Power System
Stevenson, Jr., who was for many years my friend and colleague emeritus at
North Carolina State University. Sadly, Professor Stevenson passed away on
great efforts to continue the student -oriented style and format of his own famous
May
1, 1988, shortly after planning this joint venture. In
students for a considerable number of years.
my writing I have made
textbook that has guided the education of numerous power system engineering
The aim here is to instill confidence and understanding of those concepts
of power system analysis that are likely to to
be
encountered in the study and
practice of electric power engineering. The presentation is tutorial with empha­
sis on a thorough understanding of fundamentals and underlying principles. The
approach and level of treatment are directed toward the senior undergraduate
and universities. The coverage, however, is quite comprehensive and spans a
and first-year graduate student of electrical engineering at technical colleges
wide range of topics commonly encountered in electric power system engineer­
ing practice. In this regard, electric utility and other industry-based engineers
will find this textbook of much benefit in their everyday work.
Modern power systems have grown larger and more geographically expan­
sive with many interconnections between neighboring systems. Proper planni ng,
operation, and control of such large-scale systems require advanced computer­
based techniques, many of which are explained in a tutorial manner by means of
numerical examples throughout this book. The senior undergraduate engineer­
ing student about to embark on a career in the electric power industry will most
certainly benefit from the exposure to these techniques, which are presented
here in the detail appropriate to an i ntroductory level. Lik�wise, electri c utility
engineers, even those with a previous course in power system analysis, may find
that the explanations of these commonly used analytic techniques more ade­
quately prepare them to move beyond routine work.
Power System Analysis
can serve as a basis for two semesters of undergrad­
uate study or for first-semester graduate study. The wide range of topics
facilitates versatile selection of chapters and sections for completion in the
semester or quarter time frame. Familiarity with the basic principles of electric
XVII
xviii
PREFACE
circuits, p h asor a lgebra, a n d t h e rudiments of d ifferential equations is assumed.
The reader should a lso h ave some u n d e rstanding of matrix operations and
notation a s they a re used t hroughout the t ext. The coverage includes newer
. /
tOPlCS suc h a s state estima tion a n d unit commitment, as wel l as more detailed
presen tations and newer approaches to traditional s u bj e ct s such as transform­
ers, synchronous machines, and n etwork fau lts. Where appropriate, summary
tables allow quick reference of i mporta n t i d e as. Basic concepts of computer­
based a lgorithms are presented so that students can implement their own
compu ter p rograms.
Chapters 2 and 3 a re devo ted to the t r a n sfo rm e r and sync hrono us ma­
chine, respectively, and should comple ment m a te r i al covered in other electric
circuits and machines courses. Transmiss ion-line p a r a m e t e rs and c alc u l at i on s
are studied in Chapters 4 t h ro u gh 6. N e tw o rk m ode l s based on the adm i t t ance
and impedance representations a re developed i n Ch ap t er s -; and 8, which also
intro duce gaussian elimination, Kron reduction, triangular factorization, a nd the
Zbus building a lgorith m . The power-flow p roblem, sym m e trical com p onents, and
unsymmetrical fau lts are presented in Chapters 9 thro u g h 12: \\h e reas Chap ter
13 p rovides a self-contained d evelopmen t of economic dispatch and the basics
of u n i t commitment. Con t i ngency analysis a n d external equivalents are the
subjects of Chapter 14. Power system state estima t ion is covered in Chapter 15,
whi l e power system stability is introduced i n Cha p t e r 16. Homework problems
and exercises a re provi ded a t the end of each c h apter.
I am most p leased to acknowledge the assistance given to me by a number
of people with whom I h ave been associated within the Department of Electri­
cal a n d Computer Engineering at North Carolina State University. Dr. Stan
S. H. Lee , my colleague a n d friend for m any yea rs, has always willingly g iven his
time a n d effort when I needed help, advice, or suggestions at the various stages
of d evelopment of this textbook. A n u mber of the homework problems a n d
solutions were contributed b y h i m and b y D r. Gamini Wickramasekara, o n e o f
m y former gradu ate s t u d e n t s at No rt h Carolina S t a te University. Dr. Michael 1.
Gorm a n , another of my recent gradu;lle sLucJents, gave ullstintingly or himselr in
d eveloping t he com p u ter-based figures and solu tions fo r many of the n u me r ic a l
examples throughou t the various chapters of the text . Mr. W. Ad r i a n Buie, a
recen t graduate of the Department of E l ectrical and Computer Engineering,
-un dertook the chall enge of committing the e ntire textbook to the computer and
p roduced a truly professional m a n uscript; i n this regard , Mr. Barry W . Tyndall
was also most he l pfu l in t h e early stages of the w r it ing My loyal secretary, Mrs.
Paulette Cannady-Kea, has a lways enth usiastically assisted in t he overall pro­
ject. I am greatly indebted a n d extremely grateful to each and a l l of these
individuals for their generous efforts.
Also within the Department of Electrical and Computer Engineering at
North Carol ina State Univers i ty, the successive le adership of Dr. Larry K.
Monteith (now Chancellor of the University), Dr. Nino A. Mas nari (now
Director of the Engineering R esearch Center for Advanced Electronic M ateri­
als Process i n g), and Dr. Ralph K. Cavin III ( p re s entl y Head of the Depart�ent),
.
with my faculty colleagues, particularly Dr. AJ fred 1.
environment of s u p po r t that I am very ple a sed to record.
xix
PREFACE
provided
an
source of patient understanding and encouragement during the preparation
of
along
Goetze,
The members of my family, especially my wife, Barbara, have been a great
this book. I ask
sincere t hanks.
each
of
them,
and
my frie nd Anne Stevenson, to accept my
would like to thank the following reviewers for
h e l p fu l comments and suggestions: Vernon D. Albertson, U niversity of
Mi n nesota; David R. Brown, University of Texas at Austin; Mehdi
Etezadi-Amoli, Un i ve r s i ty of Nevada. Reno; W. Mack Grady, l)nivcrsity of
McGraw-Hill and
I
many
their
Texas at Austin; Clifford Grigg, Rose-Hulman Institute of Technology; William
H. Kersting, Ne"v Mexico State University;
of
Kenneth KsuempeI, Iowa State
G. Phadke, Virginia Polytechnic Institute and State Uni\'ersily; B. Don Russell,
Texas A & M University; Peter W. Sauer, Uni\'ersit) of Illinois, Urbana­
Champaign; <lnd Ernie L. Stagliano. Jr . Drexel Uni\'ersity.
University;
Mangalore
A.
Pai, Unin'rsiry
Illinois. Urbana-Champaign; Arlin
.
John 1. Grainger
CHAPTER
1
BASIC
CONCEPTS
Normal and abnormal conditions of op e rati o n of the s ys tem are the concern of
the power system engineer who mu s t be very fa mili ar with steady-state ac
circuits, part icu l arly three-phase circuits. The purpos e of th i s chapter is to
review a few of the fund3mental i de Cl s o f s u ch circuits; t o es t a bl i sh the notation
used throughout the book; and to introduce the expression o f va lues of vol t age,
current, imp e da nce a nd power i n per u n i t . Moder n power system ana lysis rel ies
a lm os t exclusively on n o da l n erwork repres e n tatio n wh ich is i n t ro d uce d in the
for m o f the b u s admittance a nd the b u s impedclnce matrices.
,
1.1
INTRODUCTION
The waveform of voltage at the b u s es of a power system ca n be assumed to be
purely sinusoidal and of constant frequency. In developing most of the theory in
this book, we are concerned with the phasor representations of sinusoidal
volt?ges and currents a n d use the capital letters V and I to i ndicate t h ese
phasors (with appropriate subscripts where necessary). Vertical bars e nclosin g V
and I, that is, I VI and III, designate the magnitudes of the phasors. Magnitude s
of complex numbers such as impedance Z and admittance Ya re a lso indic a te d
by vertical bars. Lowercase letters generally indicate in s t antane o u s v a lues.
Where a generated voltage [electromotive force (emf)] is specified, the lett e r E
rather than V is often used for voltage to emphasize the fact that an e m f r a t her
tha n a general p o te n ti a l difference between two points is being considered.
1
2
CHAPTER 1
BASIC CONCEPTS
If a voltage and a current are expressed as fu nctions of time, such as
v = 141 .4 cos ( w t + 30° )
i = 7 .07 cos w t
and
their m ax i m u m values are obviously Vmax = 1 41.4 V and I m = 7.07 A, respec­
t ively. Vertical bars are not needed when t he subscript max with V and I is used
to indicate
value. The term magnitude refe rs to root-mean-squ a re (or
rms) values, which equal the maximum values divided by Ii. Thus, for the
above expressions for v and i
ax
maximum
I VI = 1 00 V
III
and
=
SA
These are the values read b y the ordinary types of voltmeters a n d ammeters.
Another n a me for the rms value is the effective value. The average p ower
expended i n a resistor by a current of magnitude III is 1 I12R.
To exp ress these quanti ties as p hasors, we e mploy Euler's i d e n ti ty SiG =
cos e + j sin e, which gives
cos e
=
Re{8JO} = Re{cos e + j sin e}
(1 . 1 )
where Re m e ans the real part of. We now write
If t h e c u r rent is
the referen ce
I
=
p hasor, we h ave
58 jO°
=
5LQ:
=
�
5 + j0 A
a n d t he voltage which leads the refe rence phasor by 30°
°
V = 100£J30
=
1 00
IS
= 86.6 + j50 V
Of course, we might not c hoose as the refe rence phasor either the vol tage
or the curre n t whose instantaneous expressions are v and i, respectively, i n
w hich case their p hasor expressions would i nvolve other angles.
In circuit d iagrams it is often most convenient to use polarity marks in the
form of p lus and minus signs to i ndicate the terminal assumed positive when
specifying voltage. An arrow on t he diagram specifies the di rection assumed
positive for t h e fl ow of current. In t h e single-phase equivalent of a three-p hase
circuit single-subscript notation is usually sufficient, but double-subscrip t ,nota­
tion is u su a l ly simpler when deal i ng with all three p hases.
1.2
1.2
SINGLE..SUBSCRIPT NOTATION
SINGLE-SUBSCRIPT NOTATION
3
Eg, and the voltage between nodes a and 0 is identified as v,. The current in
the circuit is lL and the voltage across ZL is VL. To specify these voltages as
phasors, however, the + and - markings, called polarity marks, on the diagram
Figure
1.1
shows an ac circuit with an emf represented by a circle. The emf is
and an arrow for current direction are necessary.
terminal marked - for half a cycle of voltage and is negative with respect to the
In an ac circuit the terminal marked + is positive with respect to the
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
is positive when the
terminal marked plus'is actually at a higher potential than the terminal marked
tenninal marked plus is actually at a higher porential th a n the terminal marked
minus. For instance, in Fig. 1.1 the instantaneous voltage
V
,
with a negative sign. During the next half cycle the posi t ive ly marked terminal is
is negative. Some authors use an arrow but must
specify whether the arrow points toward the te rm i na l which would be labeled
plus or toward the terminal which would be labeled m i n us In the convention
actua)ly negative, and
v,
The current arrow p e r form s
similar function. The sub sc r ipt, in this case
L, is not necessary llnles� other currents arc �rcsenl. Obviollsly, 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. \Vhen the
current is actually flowing in the dircciion op posit e to that of the arrow, the
current is negative. The phasor ,'urrenl is
described above.
1
I.
a
vI -vf.
( 1.2)
= -z,.,j .-
( 1.3)
and
Since certain nodes ill t he circuit ha ve been assigned letters, the voltages
voltages are expressed with respect to a reference node.
In Fig. 1.1 the
may be designated by the single-letter subscripts identifying the node whose
instantaneous voltage
a
Va
and the phasor voltage
with respect to the reference node
0,
and
V
a
v"
express the voltage of node
is positive when
a
is at a higher
FIGURE 1.1
An ac circuit with emf E� and load impedance Z[.
4
CHA PTER 1
B AS I C CONCEPTS
potential than o. T hus,
1.3
DOUBLE-SUBSCRIPT NOTATION
T h e use of polarity marks fo r vol tages a n d d i rect ion a r rows for cu rre n ts can be
avoided by double-subscript notation. The u n dersta n d ing of t h ree-phase c i rcuits
is considerably cla ri fied by adop ting a syst e m of double subscripts. The conven­
t ion to be fol lowed i s quite s i mp le.
In denoting a current the order of the s ubscripts assigned to the symbol
for current d efines the d irection of the flow of current when the curre n t is
considered to b e positive. In Fig. 1 . 1 the arrow pointing from a to b defi nes the
positive d i rection for the curren tIL associated with t he arrow. The i ns ta nta­
neous c urre nt i L is positive when the current is actua ll y in the dire ction from a
to b, a n d in double-subscript n otation this current is iab' The cur rent iab is
equal to iba'
In double-subscript notation the letter subs cripts on a vol tage ind icate the
n odes of the circ uit between which the voltage e xists. We shall fol low t he
convention which says that the fi rst s ubsc ri pt d e n otes the voltage of t h a t n ode
with respect to the node identified by the secon d subscript. This m e a ns tha t the
instantaneous voltage Vab across Z A of the circu i t of Fig. 1 . 1 is the voltage of
node a with r espect to node b a n d t h a t vab is positive du ring that half cycle
when a is a t a higher potenti al than b. The cor responding phasor voltage is V,,/J'
which is rel ated to the cu rrent fal) flowi ng from node a to node b by
-
( 1 .4 )
and
where 2/1 is the complex impedance (a l so cal l e d 2(/h) and Y/1 = 1/2/1 is t he
c omplex admittance (also called Y:II).
Reversing the order of the subscripts of either a current or a voltage gives
a cur rent or a voltage 1 800 out of p hase with the original; that is,
vbu
=
Vab c j 1800
=
Vab
/ 1800
=
-
Vab
The relation of single- a n d doub le- subscript notat ion for the ci rcu i t of Fig.
1.1 is summ arized as follows:
1.4
In writing
POWER
1:-" SINGLE,PHASE AC CIRCUITS
Kirchhoff's voltage law, the o rd er of the
of tracing a closed p a th around the circuit. For Fig. 1.1
Nodes n a n d 0 are the same in this c i rcui t
i denti fy the path more precis e l y Re placing
Iob ZA yie ld
Vab
,
,
=
and
VaG
by
11
subscripts is
the order
has been introduced to
-
and noting
Van
PO\VER IN SINGLE-PHASE AC CIRCUITS
Although the
in
terms of
[unciamcntJI theory o[ the transmissicn of
travel of energy
energy ciescIihc� the
the interaction of cleuri...- Hnd nngnctic field". tlie
change of energy with respect to time (v-,:hich is the dcilnilion of /)()l\'t'!') in
power system engineer is usually
of
more
concerned \\ltll llc.'>cribing the r�!IC o[
voltage anel current. The unit of power is a
absorbed
drop
by
a
The pmvcr
H'(I/[.
ill
and il· the vlllL!gc
the 10,1(.1 in volts and the il1\(,llltar1Cnu:-- (linenl inlo the I(}�!d
a n d current are expressed by
and
ion
=
Ci
I,,, ...
,Inel
e qu a
t i on s
fI,
l'l)s( (ut
the instantaneous power is
The angle
in Wd!l" being
term"
lond at any instant is the product of [he inst,1f11<ll1coU\ \Ull,q;e
1f the terminals of the IO,ld arc designated
across
amperes.
that
( 17)
and so
1.4
5
-
0)
( J
.S)
for current lagging the vult<\gc and
leading current. A positive value of fJ expresses the r,lte at which
energy is being absorbed by the part of the s ystem between the p o i nt s {l and II.
The i n s t a n ta n eous power is obviously po s itive when both Vall and illn arc
positive and becomes negative when Vall and i,," are opposite in sign. Figure 1.2
e in these
negative for
is positive
v'lIi
results when current is
flowjng in the direction of a vo l tage drop and is t h e rate of tr a n s fe r of energy to
the load . Conversely, n e gat i v e power c a l cu l a te d as vaniall r e su lt s when curren t is
flowing in the d irection of a voltage rise a n d means energy is being transferred
from the load into the system to which the load is connected. I f Van and ian are
jn phase, as they are in a purely re s i s t ive load, the instantaneous power will
never become negative . If the current and vo ltag e are out of ph as e by 90"', as in
a p u rely inductive or purely capacitive idea l circui t elemen t, the instantaneous
illustrates this point. P o s i tiv e power calculated as
all
6
CHAPTER 1
BASIC CONCEPTS
o --�--��----�--�-+--
FIG U RE 1.2
Current, voltage, and power plotted
versus
time.
power will h ave e qual positive a n d negative half cycles a n d its ave rage value wi l l
a lways be zero.
By using t rigonometric identities the expression of Eq. ( 1.8) is reduced to
p =
Vmax Imax
cos B( 1
2
+
cos 2 w t) +
Vmax J max
2
sin 8 sin 2 w t
( 1 .9)
where Vmaxlmax/2 may be repl ac e d by t h e product of the rms voltage a n d
curre nt, t h a t i s , b y \ Van \ \Ian \ or I V\ \11.
Another way of looking a t the exp ression for i nstantaneous power is to
consider the component of the curre n t in p hase with Van and the component
90° out of phase with Vall. Figure 1 .3(a) s hows a parallel circuit for w hi ch Fig.
1 .3( b ) is the p h asor diagram. The component of ian in phase with Van is iR, a n d
from Fig. 1 .3( b), 11R \ = Ilan \ cos 8 . I f the maxi mu m value of ian i s Imax, the
maximum v a l ue of iR is I max cos 8. The instantaneous current iR must be i n
phase w i t h Vall. For Vall = Vmax cos wt
iR = 1max cos 8 c os
(1 . 1 0)
wt
max In
Si milarly, the component of iall l agging Van by 90°
t+
!-
a
Van
n
ian
-
iR
t
(a)
IR
X
IS
iX with m aximu m value
Van
FIGURE 1 . 3
(b)
Parallel RL c irc u i t and the corresponding phasor diagram.
1 .4
o
POWER IN SINGLE·PHASE AC CIRCUITS
7
- -.-=-----+-1- ---�....,..----""-t
=
0
FIGURE 1 .4
Vol t age, curre o t in ph ase wi t h t h e vo l t a ge,
fm ax. s i n
8. Since i x m u s t la g
by 90° ,
v ll ll
ix
a n d t he re s u l t i n g power p lone d , ersus ti me .
J m:c( sin 8 sin
=
6J (
m ax i "
The n ,
m;!; ".1''' cos
v .
Jr.-
e(
1
+
cos 2 6J [ )
(1.11)
( 1 . 1 2)
whi ch is t h e i n sta n t a n e o u s p ow e r i n t h e r es is t a n c e a n d t he fi r s t t e r m I n
(1 .9). Figu re 1 . 4 s h ows V,"J !? p l o t ted ve r s u s ( .
Eq .
S i m i l arly,
( 1 .13)
which is the instantaneous powe r in t h e i n d u c t a n c e a n d th e second t e r m i n
Eq. (1 .9). Figure 1 .5 shows Van ' ix a n d t h e ir prod uct p l o t t e d versu s t .
Examinat ion of Eq . ( 1 . 9) shows that t he term co n t a i n i n g cos f) IS alw ays
positive and has a n a v er a g e value of
p =
or
Vmax
when rms val u e s of voltage and
P
Imax
cos ()
2
current
=
are substitu ted,
I VI I II cos 8
( 1 . 1 4)
( 1 .15)
8
CHA PT E R 1
BAS I C CONCEPTS
O ----r-----��--�--_7-
FIGURE 1 .5
Vol t a ge, c u r re nt la ggi ng the vo ltage by 90° , a n u t h e re s ul t i ng pow e r plot t e d
ve r s u s t i m e .
P is the q uantity to which the word power refers when not mod ified by an
adj ective identifying it otherwise. P, the average power, is a lso ca lled the real or
active power. The fundamental u ni t for both instanta neous and average powe r
is the w a t t , but a watt is s u c h a s m a l l u n i t i n rel a t ion to power system q u a n t i t i es
that P is usually measured i n kilowatts o r m egawa tts.
The cosine of t he phase a n gle e between the voltage and the curre n t is
called the power factor . An i n d uc tive circuit is said to h ave a lagging power
factor, .and a capacitive circuit is said to h ave a leading power factor. In other
words, the terms lagging power fa clOr and leading power factor i n dicate,
respectively, whether the c urrent is l agging o r leading the app l i ed vol tage.
The second term of Eq. ( 1 .9), the term con t a i n in g sin e, is altern ately
positive a nd negative and has an average value of zero. This component of the
instant aneous power p i s called t he instantaneous reactive power and expresses
the flow of energy alternately toward the load and away from the load. The
maximum value of this pulsating pow e r, d esign a ted Q , is ca l led reactive power
or reactive IJo/lampcres and is very use ful in d escribi n g the operat ion of a power
system, a s b e c o m cs i n crca s i n g l y cv i d c n t i l l ru rl h c r d iscuss i o n . T h e re a c t i v e
power i s
or
Q
Q
�T1aX J
2
max
------
=
sin e
I V I I I l sin e
The square root of the sum of the squares of P a n d
product of I V I and I I I ) fo r
( 1 . 1 6)
( 1 . 1 7)
Q
IS
equal to the
( 1 .18)
Of course, P and Q h ave t h e s a m e d i mensio n a l u n its, b u t i t is us ual to
1 .4
(for vol t a m pe res
u n it s for Q a re k il ovars o r megav a rs .
designate t h e units for Q a s
vars
POWER IN SINGLE·PHASE AC CI RCU ITS
In a simpit: s e i es circ u i t w he e Z i s e q u a l
Il l i Z I for I VI in Eqs. (1 . 1 5) a n d ( 1 . 1 7) to o b t a i n
r
r
re a c t
to R
i ve }
9
The more p ractical
+ )X w e c a n
substitute
( 1 . 1 9)
( 1 -20)
and
Re c o g n i z i n g
that R
=
I Z l co s e
and
X
=
I Z l sin e, we then find
and
() II >
E q u ,\ t i o n s ( 1 . 1 5 ) il n d ( 1 . 1 7 ) r rov i d e
[Jowe r r a c t o r :--. i n c e w e
sec
o r from £q s . 0 . 1 5) and
[ il,L(
c.
[ ;t i l
a n o t h e r m e t hod
( 1 .21 )
of
(J . T h e p owe r L l c l o r
co m p u t i n g t h e
i s t i l e r'do re
( 1 . 1 8)
i s t h e power i n a
b e c om e s
negative, m a k i n g sl n e a n d Q negat ive . I f capacitive a n d inductive c i rc u i ts are in
p a r al i c l , t h e i n s tanta n e o u s r e a c t ive p o w e r for the RL c i rc u i t is 1 800 o u t o f
p h a s e wi t h the i n s t a n t a n e o u s r e a c t i v e p ower of the R C circu i t . T h e net reactive
pO\l, e r i s t h e d i fference b e twe e n Q fo r t he RL ci rcu i t and Q fo r the RC ci rcu i t .
/\ I.� ()s i t ivc va l u e i s a ss i gn ee1 t o Q d ra w n by a n i n d u c t i v e 1 0 ,l cl ;1 n (1 (l n e g a t ive sign
to Q d rawn b y a c a p a c i t i v e l oa o .
P o w e r sys t e m e n g i n e e r s u s u a l l y t h i n k o f a c a p a c i t o r a s a g e ne r a t o r o f
p m i t i ve react ive pow e r ra t h e r t h a n a l o a d requ i r i n g n ega t i v e reactive p o wer.
Th t s con c e p t is ve ry l og ic a l , for a ca p a c i t o r d rawi n g nega rive Q i n para l l e l w i t h
a n ind u c t i ve loa u red u ce s t h e Q w h i c h w o u l d o t h e rw i s e h ave to b e supplied b y
t h e s y s t e m ( 0 t h e i nduct ive l o a d . I n o t h er wo r d s, t h e c a p a c i t o r supplies t h e Q
req u ired by t h e i n d u c t i ve l o a d . Th i s i s t h e s a m e as c o n s i d er i n g a c a p a c i t o r a s a
d ev i ce t h a t d e l ive rs a l a gg i n g cu rrent r a t h e r t h a n a s a device w h i c h d raws a
l e a d i n g c u r re n L as shown i n Fig. 1 . 6 . An a dj u stable capacitor i n p a r a l l el wi t h a n
i n d uct ive l oad, fo r i nstance, can b e a dj u s ted so that t h e l eading c u r re n t to the
c a p a c i t o r i s e x a c t l y e q u a l i n m a g n i t u d e t o t h e co m p o n e n t o f c u r r e n t i n t h e
i n d u c t ive load which is l a gg i n g thc v o l t a g e by 90° . Thus, t h e r es u l t a n t c u r r e n t i s
i n p h ase w i t h t h e v o l t a g e . Th e i n d uc tivc c i rc u i t stil l re q u ires positive reactive
pred o m i na n t l y ca p a ci tive c i r c u i t w i t h t h e s a m e i m p ressed voltage, e
If
the i nstan t a n eous p ower expressed by E q .
( 1 .9)
10
CHA PTE R
+t
I
-----
v
-!
I leads
1
BASIC CONCEPTS
lc
T
+t
I
-
v
-+
I lags
V by 90°
(al
V by
Iv)
Ie
T
90°
Capaci tor co nside red a s : (a) a passive circu i t e le m e n t
draw i ng le a d i n g cu r re n t ; ( b) a g e n erator s u ppl y i ng lag­
g i n g cu r r e n t .
FIG U R E 1 . 6
power, b ut the net reactive power i s zero . I t i s fo r t h i s r e a son t h a t t h e power
syst e m engi n e e r finds i L conve n i e n t t o c o n s i u e r t h e c a p a c i t o r t o be suppl ying
rea ctive power to the i nductive load. W hen t he words positive a n d n egative a re
not used , posi tive react ive powe r is ass u m e d .
1.5
COM PLEX POWER
the ca l c u l ation of
rea l and reactive power is a ccomplished conven i e n t ly in complex form. I f the
vol tage across and the curren t i n to a certain load or p a rt of a circuit a re
expressed by V = I V I
, respectively, the p roduc t of voltage
a nd I = 1 1 1
times the conj ugate of current in pol a r for m is
If the phasor expressions for vol tage and cu r rent are k now n ,
il.
�
VI *
=
I v!c j a
X 1 1 1 £ -jt3
=
I V I I l l c i(a -t3)
=
I VI 1 1 1
/
a
This q u antity, cal led the complex po wer , is us ually designated by
-
f3
S . In
( 1 22 )
.
rectangu­
lar form
5
=
VI *
=
I V l l l l cos ( a
-
f3 ) + j I V l l l l sin ( a
-
(3 )
( 1 23 )
.
Since a f3 , the phase a ngle b etwee n voltage and cu rrent, i s e in the p revious
equ ations,
-
5
=
P -\
jQ
( 1 . 24 )
Reactive power Q will be pos i t ive when t he phase angle a f3 between voltage
a n d cu r rent i s positive, t h at is, w hen a > f3 , wh ich means t h a t current i s lagg i ng
the vol tage. Conversely, Q will b e n egative for f3 > a, which indicates t h a t
c u rrent i s leading the voltage. Th i s a g rees with t h e selection o f a positive sign
for t he re a c t ive powe r o f a n i n d u ct i ve c i rcu i t a n d a n e ga tive si g n for t he r e a c tive
p ower of a capacitive ci rcuit. To obta i n t h e p roper sign for Q, it is necessary t o
calculate 5 a s VI * rather t h a n V * I , w h i c h would reverse t h e sign f o r Q .
-
1.6
__
THE POWER TRIAN GLE
Equ ati on ( 1 .24) suggests a graphical m e thod of obtain ing the overa l l P, Q, and
p hase a ngle for several l oa ds i n p a r a l l e l si nce cos e i s P/ 1 5 1 . A power tri an gle
. c a n be drawn fo r a n i n ductive load, as s hown i n Fig. 1.7. For several loa d s i n
1 .7
s
FIGURE 1.7
Power t ri a ngle for a n induct ive load .
p
"- - .
. ,_ .
----
11
Q
e
"-
D I R ECTION OF POWER FLOW
�-- ---'
p
-, :. -'- --- ----y-------- ..__.-..,---../
1)1
I
j '2
""
F"
FIG U R E 1 . 8
P o w er
10,ld:-"
t r i ,m g l e
Note that
Q2 i s
fo r
com b i n e d
n e ga t i ve ,
p a r <1 ! ! e i t h e t o t a l P wi l l be t h e s u m o f t h e average pow e r s o f t h e individual
l o a d s . which s h o u l d be p l o t t e d a l ong t h e h o rizo n t 8 1 axis fo r a graph ical a n alys i s .
For ,H i i n d u c t i v e l o a d Q w i l l b e d r a w n v e r t i ca l ly u p w a r d s ince i t i s positive . A
c a p a c i t i ve l oa d \vi l l have n e g a t i v e r e a c t ive pow e r, a n d Q wi l l be vert ical ly
downward, F i g u r e 1 . 8 i l l u s t r a t e s t h e powe r t ri a ngle co m p osed o f P I ' Q \ , an d S l
fo r a l a gg : ng power - f a c t o r l o ad h a v i n g a p h a s e ang l e e l c o m b i ned w i t h the
p ow e r t r i a n g l e c o m po s e d of P2 ' Q7" a n d S2 ) which i s fo r a c a p ac i t ive load w i th a
ll c g Zl t i v e e :. , T h e se t w o l oad s i n p ar a l l e l r es u l t in t h e tri angl e h a v i ng s i d es
P 1 'i e-; , Q ! + Q "2 ' � 1 11 <J h y p ( ) t e n u s c S f( . 1 11 g C 1 1 C r a I , I ,)' I< l i s !I0 ! c q u a I t o I S I I +
I S 2 i . T i l c p h a � (: a n g l e b e t w e e n vo l t a ge d n c\ C U f r e nt s u p p l ie d to the co m b i n e d
l o a d s i s Bj( '
D IRECTI ON OF POWE R
1 .7
FLOW
Q, a n d b u s VO Il <lgC V , or ge n e r a t e d vol tage £ , w i t h
;; i g l1 s o f f ' ;t l l li Q i s i m po rt ;t n t w h e n t h e n ow o f pow e r i n a system
is C U I I � i l k n.: J , T h e q u c s t i o n i n vo i v es t h e d i r e c t i o n o f n ow o f powe r , t h a t i s ,
w h e t h e r p o w e r i s be i n g gel / era /ed o r a bsorhed w h e n a vo l t age a n d a c u r re n t a re
spec i fie d .
The q u e s t io n o f d e l ive r i n g p o w e r to a c i rcu i t o r a bs o r bi n g power from a
c i r c u i t i s r a t h e r o b v i o u s fo r a d c sys t e m . Cons i d e r the current a n d v o l ta g e of
Fig. 1 . 9( 0 ) w h e r e de c u rre n t 1 is fl ow i ng t h r o ugh a battery , I f the vol tmeter Vm
a n d t h e a m m e t c r A m bo t h r e a d u p sc a l e t o show E
1 00 V and 1 = 1 0 A , the
b a t t e ry is b e i n g c h a r g e d ( a b so r b i n g e n e rgy) a t the rate given by the p roduct
1 000 \V , On the other h a n d , i f t he a m m e t e r connection s have to be
E1
T h c r e l a t i o n a m o n g I) ,
r e s pe c t
t�)
the
=
=
12
C H A PTE R 1
BASIC CONCEPTS
--....;.�
+
( a)
I
�
--,.
.. I
E
Wattmeter
+
Voltage
coil
E
(6)
Con n ections of: (a) ammeter a n d vol t me t e r t o measure de c u rre n t I a n d volt age E o f a b a t tery ;
( b ) watt m e t e r t o measu re rea l power absorbed by ideal a c volt age sou rce E.
FI G U RE 1 .9
reverse d i n order that i t reads u pscale with t he current arrow still in t h e
d irection s h own, t hen 1 = - 1 0 A and the product EI = - 1 000 W ; that is, the
b a tt e ry i s d i scharging (delivering energy). The same considerations apply to
the ac c i rc u i t relationships.
For a n ac system Fig. 1 .9( b ) shows within t he box a n ideal voltage sou rce
E (con s ta n t magnitude, constant frequency, zero impedance) with polarity
marks t o i n dicate, a s usual, the termi n a l which is positive during the half cycle
of positive instantaneous vol tage. Similarly, the a rrow indicates the d irection of
current 1 i n to the box d uring t he positive half cycle of current. The wattmeter
of Fig. 1 .9( b) has a current coi l and a vol t age coil corresponding, respectively, to
the a mmeter A m and the voltmeter yO) of Fig. 1 . 9( a ). To measure active p ower
the coi l s m ust be correctly conn ected so as to obta i n a n upscale rea ding. By
definition we know that the power absorbed ins i d e the box i s
S
= VI * = P + j Q = I V l l i l cos () + j l V l l i l s i n
e
( 1 .2S )
where 8 is the phase a ngle by w h i ch 1 lags V. Hence, if the watt meter reads
u pscale for t he connections shown i n Fig. 1 .9( b ), P = I V i l l i cos 8 is positive
and real power is being a bsorbed by E. If t he wattmeter tries to deflect
downscale, t he n P = I VI I I l cos 8 is n egative and reversing the connections of
the curre n t coil or the voltage coil, but not b ot h, causes the meter to read
upscale, i n d icat ing t h a t positive power is b e i ng supplied b y E ins i d e t h e box.
Thi s is e quivalent to saying t h a t negative power is being absorbed by E. If the
wattmete r is repl aced by a varmeter , similar considerat ions apply t o t h e sign of
the reactive power Q absorbed or supplied by E . In gene ral, w e can determine
the P and Q absorbed or s upplied by any a c circuit simp ly by regard i n g the
circu i t a s enclosed i n a box with ente ring cur rent 1 and voltage V h aving t he
polarity show n in Tabl e 1 . 1 . Then, the nume rical values of t he real a n d
imagina ry p arts of t h e p roduct S = VI * dete rmi ne the P and Q absorbed
or
,
suppli e d by the e nclosed c i rcuit or network. When current 1 lags voltage V by
TABLE 1 . 1
Dir ection o f P and
"
Q flow where S
ac
�
equi vale n t
+
c ircuit
V
I-
or
-,
P
+
element
jQ
=
P
+ jQ
If P
<
0, c ircu it su pp l i e s real power
If
>
0 , circuit absorbs re l a tive power
<
O. circ u i t supp l ies react i ve power ( I leads
Q
<1 n g l c
=
Exa m p l e
1. 1 .
Two
i d e a l vol t a g e
13
0, circuit absorbs real power
>
0 be t w e e n 0° a n d 90" , we n n d t h tl t
I V I J l sin 0 <I n.: bO l h pos i t iv e, i n d i ca t in g \v a t t s a l1 u
t h e i n d u c t ive c i rc u it i nsi de t he box . When I leads
a n d 90° , P i s s t i l l pos i t ive b u t e a n d Q
I VI
i n d ica t i n g t h at n e gative 'lars are being absorbed
s u p p l i e d by t he c a p ac i tive c i rcu i t insid e t he box.
an
D I RECTI O N OF POWER FLOW
If P
If Q
C i rcuit
-
s
VI *
=
1 .7
sourccs
V)
I l l cos D a n d Q =
vars a r c bei ng absorbed by
V by an a n g l c b e tween 0°
P
=
I
( I lags V )
vi
I I I s i n e are bot h
or positive
ft
d e s ign a t e d
3S
nega t ive,
vars a re b e i n g
mach ines 1 an d 2 are
1 00/300 V,
j5 fl , d e t e rm i n e ( a ) wh e t h e r C el ch m a c h i n e is ge n e ra t i n g or consu m i ng
Z
r e a l pow e r a n d t h e cl mou n t , ( b ) whet h e r each mach i n e i s receiv i n g or su p p ly i ng
re(). c t i v e powc r a n d t h e a m o u n t , and ( d the P a n d Q absorbed by the i m p e d a nce.
co n n e c t e d ,
=
0
+
3S
shown
in
Fig.
1 . 1 0.
If E I
100
=
Solution
1 00
---
_ .
z
1 3.4
-
+
jO
j50
=
=
jS
-
10
-
-
(86.6
jS
j 2 .68 =
+
V,
£'2
=
and
j50)
1 0.35/ 1 95" A
1
--
z
#1
#2
FI GURE 1 . 1 0
I de a l v o l t a ge
i mpeda nce Z.
so u rc e s
c o n n ected
t h ro u gh
14
CHAPTER 1
BASIC CO N C E P TS
The curren t entering box 1 is
51 =
52
£ l ( - 1 )* = PI
= £2 /*
=
+
jQ l
-I
=
and that entering box 2 is f so that
1 00(10
+ j2.68) *
= 1000
P2 + j Q2 = (86.6 + j50) ( - 10 + j2.68)
- j268 VA
=
- 1000
- j268 VA
The reactive power absorbed in the series impedance is
Machine I
ma
y
be expected t o
be
P I is
a ge n e r a t o
d i re c t i o n a n d po l a r i ty m a r k i n g s . H o weve r , s i n ce
t h e m a c h i n e co n s u l ll c s e n c rgy �I l t h c r : l t <.: 0 [' l ( )()() W � l l l li
1.8
r
Q1
bec a u s e o f t h e c u rr e n t
s u ppl ies
pos i t ivc �l n d
i s n e ga t i v e ,
rcact ive power o r
268 v al'. The m ac h i n c i s , l c t U ll l l y � 1
M achine 2, expected to be a motor, h a s n e g a ti v e P 2 and n e g a t i ve Q2 '
Therefore, this machine generates energy at the r a t e o f 1 000 W and supplies
reactive power of 2 6 8 var. The machine is actu a l ly a generator.
Note that the supplied reactive power of 268 + 268 is e qual t o 536 var,
w hich is required by the inductive reactance of 5 n . Since the impedance is purely
reactive, no P i s consumed by the impedance, and al l the w a tts generated by
machine 2 are transferred to machine 1 .
I l l u t or .
VOLTAGE AND CURRENT I N BALANCED
THREE-PHASE CIRC UITS
Electric power systems are sup plied by t hree-ph ase generators. I deally, t h e
'generators are supplying b a l anced three-p h ase loads, w hich m e a n s loads wit h
identical i mpeda nces i n a ll three phases. Lighting loads and sm a l l motors are, of
course, sin gle-phase, but distribu tion sys tems are designed so t h a t overall t h e
p hases a re essentially balanced . Figu re 1 . 1 1 shows a Y-connected generator
with neutral m a rk e d 0 s u p p ly i n g a b a l a n c e d - Y l o a d with neutral m arked 11 . I n
d iscussing this circuit , w e assume t h a t t h e i m p e d a n ces o f t h e con nections
between the termin als of the generator and the load, as wel l as the impe d a n ce
of the d i rect connection between 0 a n d n , are negl igib l e.
The equivalent circu it o f the t h re e- p h ase gen erator co n s i s ts o f a n e m f i n
e a c h o f the three phases, as indicated by circles on the diagram . Each emf i s i n
series wit h a resista nce and inductive reacta nce composing the im ped ance Zd '
Points a', b' , a nd c' are fictitious since t h e generated emf c a n n o t be separated
from t h e i mpedance o f ea ch phase. T h e termin als of the m achine a re the points
a , b , a nd c . Some a ttention is given to this e quivalent circuit in Chap. 3 . I n t h e
g enerator t h e emfs E(lI O J Eb1o ' and £c'o a re equ a l i n magnitude a nd d isplaced
from each other 1 200 in phase. If the m agn i tude of each is 1 0 with Ea1 0 a s
reference,
Ea'o
=
100LQ: V
E/,'o =
100/2400 V
0 V
Ul
VOLTA G E A N D C U R R E NT IN BALANCED TH R E E - P HASE
CI RCU ITS
15
a ,-----------------------____________________________________-,
b
e
c L-------�
--i"-
f Cit
FIG U R E 1 - 1 1
Ci rc u I t d i �l gr a lll
of
�I
!' - co n n <:c t nl
b C J l C ra t o r (,OIl I I C C 1 C U [ 0
a
h,t l a n cc d - Y l o ;, d .
p rov i d ed t h e p h a s e s e q u e n c e i s abc , which m e a n s t h a t Eo'o leads EI/o b y 1 2 00
and £b' o i n t u rn l e a d s L,' n by 1 200 The circu i t d i agram gives no i n d i c a t ion of
phase se que nce , b u t Fi g . 1 . 1 2 s hows these e mfs w i t h p h ase seque nce abc .
A t t h e gene rator te rm inals ( a n d a t the l o a d i n t h is case) t h e t e rmi n a l
vol l ages t o n e u t r a l a r e
_
V -E
ao
vc o
-
-
-
UfO
F-" ( ' U
-I Z
un
-
d
I( 1 / Z
-I e!
( 1 .26)
S i n c e 0 an d / I a rc a t t h e s <.nn e pot e n t i (t i , V;w , ViI () ' a n d V;,o a re e q u a l to
a nci V:'r! ' re spe c t ive ly, a n d t h e l i n e c urre n t s (which a rc a l so t h e p h a s e c ur r e n t s
FI G U R E 1 . 1 2
P h a �or d i agram o f t h e e m fs o f t he c i rc l l i t shown i n Fig . 1 A 1 .
v'w ' Von '
16
CHAPTER 1
BASIC CONCEPTS
FIGURE 1 . 13
P hasor d i a g r a m of c u rr e n t s in a b a l a n c e d t h re e - p h as e loa d :
(al
( a ) phasors d rawn from
(bJ
a
common po i n t ;
( b) a d d i t i o n
of t h e
p h a sors form i n g a closed t r i a n g l e .
v,/Il
Zu
for a Y co n n ection) a re
lUll
�
Zd Z I<
F.J·lJ'u
+
lbll Z" Z "
=
fCI I
Eh, o
+
ZJ + Z 1\
Ec' o
Vb
=
-
II
Za
ZR
( 1 .27)
VO l
S in ce Ea, o ' Eb,o, and Ec'o are e q u a l in ma gn i t u d e and 1 20° a p a rt i n pha se,
and s i nce t he impedances seen by these emfs are i d e n t ic a l , the currents w i l l also
be equal in magni t ude and d i sp laced 1 20° from each o ther in phase. The same
must also be true of Vall ' Vbll , a n d VO l ' In t h i s case we describe the vol t ages and
currents as balance d . Figu re 1 . 1 3( a ) shows t h re e line cu rren ts of a b a la nced
system. In Fig. l . 13 ( b ) these curr e n ts fo rm a closed triangle a n d it is obvious
t h a t t h e i r s u m is ze ro. Therefo re, III m ust b e zero in t h e con nect ion s hown in
Fig. 1 . 1 1 b e tween the neutrals of the gen erator a nd loa d . Then, the con nection
between lZ and 0 may have any i m p e d a n ce , o r eve n bc open, and Il and 0 w i l l
rem a i n a t t h e s a m e p o t e n t i a l . H t h e l o a d i s n o t b a l a n c e d , t h e s u m or t h e
currents w i ll n o t be zero a n d a c u rre n t w ill fl ow b e tween 0 a nd ! 1 . For the
unbalanced condit ion 0 and n w i l l n o t be at the same poten tial u n less t hey a re
con nected by zero imped a nce.
Because of the phase d i spl acement of t h e vol tages and currents in a
b a l anced th ree-phase sys tem, i t is conven i e n t to have a shorth a n d method of
indica ting t h e rot ation of a p h asor t hro u gh 1 20° . The result of the mu l t ip l i c a ­
tion of two complex nu mbers i s the p ro d u c t of t he ir mag n i tu d es a n d t h e su m of
their angles. If the complex n umber exp ressing a p h asor is m u ltipl ied by a
com plex n u mber of u n i t magn i t u d e a n d a n gle e , t he resu lting complex n umber
represents a p h asor equal to the origin a l p hasor d ispl aced b y the a n g l e e. The
complex n u mber of u n it magn i tu d e and a ssoci ated angle e is a n opera tor that
rotates th e p h asor on which it o perates t h ro u g h t he angle e. We a re a lready
familiar with the operator j, which cau ses rota tion through 90° , and the
operator - 1 , w h ich cau ses ro tation t h rou g h 1 800 • Two successive a p p l ications
of t h e o p erator j c au se r ota tion t h ro u g h 9 0 ° + 90° , w h i c h l e a d s U S to t h e
1 .8
V O LTAG E A N D C U R RENT IN BALANCED THREE-PHASE C I R C U ITS
17
conclusion that j x j causes rotation through 1 800 and thus w e recognize t hat
j 2 is equal to
1. O t h e r powers of t h e oper ator j a r e found by s im i l a r analysis.
Th e l e t t er a is commonly used to designate the operator that c auses a
rotation of 1 200 in the counterclockwise direction. Such an operator is a
c o m p l e x number of u nit magnitude with an angle of 1 200 and is d efined by
l
-
a
I f the
=
1/ 1200
1 £ j 2 17 / 3
=
=
+ jO .866
- 0 .5
operator a is appl i e d to a p h a s o r twice in succession,
through 2400 • Th ree successive appl ica tions of a ro t a t e
360° . Th us,
a2
a�
I t is evi d e n t t ha t 1
various
/3
a - I
/150e
powe rs
=
=
1
/ 2400
=
l L36(r'
+a +
a
2
=
=
and fu n c t i o n s of
O.
1 £ j4 rr j J
I
r-; J 2 rr
=
Figure
([ .
� O.5
lLQ':
_.,-
=
1
1 . 1 4 sh ows p h asors represe n t ing
!
I
I
(
\
(1 f
\ d
r i ( ) u , Po\\ u \ ; tn d
fu n e t i o n s
()
is rotated
- j O . 86G
/
FI G U RE 1 . 1 4
P h . � " or r . kl g r:lJ1l
the p h a s o r through
the p h a s o r
r t h e or//..: r;\ I o r
II .
18
CHA PTER I
BAS I C CONCE PTS
Fl G UR E 1 . 1 5
Phasor diagram o f l i n e-to- l i n e vo l t ages i n r e l a t i o n t o l i ne-to- n eu t r a l vol tages i n a b a l anced t h ree­
phase circu i t .
The l ine-to-line voltages in the c i rc u i t of Fig. 1 . 1 1 a re
Tracing a path from a to b through n y i elds
Va b , Vb C '
and Vc a .
( 1 . 28)
Although £a'o and Van of Fig. 1 . 1 1 are not in p hase, we could decide to use �I/l
rather than £a'o as reference i n d e fi n i ng t h e voltages. Then, Fig. 1 . 1 5 shows the
p h asor d iagram of voltages to neutral and how Vab is fou n d . I n terms of
2
o perator a we see that Vbll = a VU II ' and so we have
Figure 1 . 14 shows t hat 1
-
Vab
a 2 = 13
=
�, which means t h a t
= /3- Vall �
13 Vl1 r1 t )JO
°
( 1 .29 )
( 1 .30 )
So, a s a phasor, Va b leads Va ll by 30° and is f3 t i mes l arger in magnitude. The
other l i ne-to-li ne voltages are fou n d in a similar manner. Figure 1 . 1 5 shows a l l
t h e l i ne-to-line voltages i n relation t o t h e l ine-to-neutral voltages. T h e fa ct that
1 ./l
V O LTA G E A N D C U R R ENT IN BALANCED TH R EE-PHAS E C I R CU ITS
19
b
Q T---
FIGURE 1 . 16
A l t erna t ive m e t hod of d rawing t h e p h asors of Fig. 1 . 1 5 .
c
the magnitude of ba la n ced l i n e-to- l in e vol tages of a t h ree-phase circui t is a lw ays
e q u al to !3 t im e s the m agn i tude of t h e l i ne-to-neutral voltages is very impor­
tant.
F i g u re 1 . 1 6 shows another way o f d isplaying the l i ne t o lin e a n d l i n e-to­
n e u t r a l v o l t ag e s The l i n e t o l i ne vUl tage phasors are d rawn to form a close d
t r i a n g l e u r i e n t e d t o ag r e e w i t h t h e c h o s e n refe rence . i n t h i s case �/fl ' T he
v e rt i c e s o f t h e t r i a n g l e ,\ IT l a b e l e d so t h a t e a c h phasor begins and ends a t t h e
v e r t i c e s corre s p o n d i n g t o t h e o r d e r o f t he s u b s c r i p t s o f that phasor voltage.
L i n c - t o - Il e u t r a l vol tage pha s ors are d rawn to the cen ter of the triangle. O nce
t h i s p h a s o r d iagram i s understood, it w i l l be found to be the simplest way t o
d etermine the va rious vol t ages.
T h e o rd e r i n which t he vertices a, b , and c of the t ri angle fol low each
o th e r w h e n the triangle i s ro t a t e d co u nt e rclockwise about n i ndicates the p h ase
s e q u .: n c c . T h e i m po r t anc e o f p h ase sequence becomes clear when w e d is c u s s
t r a ns form e r s and \vh en symmetrical components a r e used to a nalyze u nbalanced
fa u I ts on power systems.
A s e p a r a t e c u rre n t d i a g r a m can b e d rawn to re late each curre n t p roperly
wi t h r e spect t o i t s p ha s e \·ol t a g c .
-
.
1O.L20"
-
E X <1 m p l e 1 . 2 . I n
-
b a la nced t hree-phase c i rcu i t t he vo l t age
a
-
V;/b
is
173.2!!!...
D c t l.:: rm i n e 3 1 1 the vo l t ag e s and t h e cu rre n t s i n a Y-con n c e t e d l o a d h av i n g ZL
D . Assu me t h a t t h e p ha:;c seq u e n c e i s a h c .
Solution . V/ i t h
r�J !)
;1 <; r e f e re n ce , t h e p h a s o r d i a g r a m
in F i g . 1 . 1 7. from w h i ch it i s d e t c r m i n e d t h a t
1 .'
1 73 . 2
v be
l/c o
=
v(lit
1../_ 2.400 V
vh ,,
1 n 'j L
/ 1 2()O V
· .....
VOl
__ _
_
=
of vol t agcs is
I (lo/I
-
=
3 (t
d rawn as s h own
V
�V
100
=
/ 2 1 0° V
�
_
-
-
V.
1 OO
E a c h c u r r e n t l a g s t h e \ o l t age a c r o s s i t s :lo a d i m p e d a n ce by 20° a n d each c u rrent
m a gn i t u d e i s 1 0 A. F i g u r e
1.18
i s the p h a so r d i agram of the currents
J
,I " 1
==
/ 1 90° A
�
10
20
CHAPTER 1
a
BASIC CONCEPTS
b
FIGURE 1 . 1 7
P h a s o r d i agram o f vol t a g es for Exa m p l e 1 .2.
c
FI GURE 1 . 1 8
Phasor d i agram o f c u r r e n t s fo r Exa m p l e 1 .2.
Balance d loads are often connected in 6., as shown in Fig. 1 . 1 9. Here i t i s
left to t h e reader using the prope rties of the operator a t o s how t h a t the
magnitude of a line current such as I II is equal to 13 times the magnitude of
a-phase c urrent lab and that la l ags lab b y 30° when the phase sequence i s
abc. Figure 1 .20 shows t h e current relationships when lab i s chosen a s refer­
ence.
W hen solving balanced three-p hase circuits, i t is not necessary to work
with the e n t i re three-phase ci rcu it d iagram of Fig. 1 . 1 1 . To solve the c ircuit a
neutral con n ection of zero impedance is assumed to be present and to carry the
sum o f the t h ree phase cu rrents, w h ich is zero for bal anced conditions. The
circu i t is solved by applying Kirchhoff's voltage law around a closed path w h ich
includes one p hase and neutral. Such a closed path is shown i n Fig. 1 .2 1 . This
a .-------�
b
C
lb
--,�
[be
-
-
Te
FIG URE 1 . 19
C i r c l l i t d i a ):, r a rn of (:, co n n e c ted t h ree-phase l o a d .
1 .8
vO LTA G E A N D CURR ENT IN BALANCED THREE-PHASE CI RCUITS
21
I f, --
FI G L R E
1 .2 0
P h a s o r d i a g r a m o f t h e l i n e c u r r e n t s i n re l ;] t i o n t o t h e p h a s e c u r re n t s i n a b a l a nc e d .1 - c o n il e c t e d
t h re e - p h a s e l o a d _
of the c i r c u i t of Fig. 1 . 1 1 .
C a l c u l a t io n s m a d e for t h i s p ,l t h a r e e xt ended to t he w h o l e t h ree-phase c i rc u i t by
r e c a. l l i n g t h a t t h e c u r r e n t s in t h e o t h e r two p h a s e s a re equal in magnitude to t h e
c ur r e n t o f t h e p h a s e c a l c u l a t e d a n d a r e d i s p l a c e d 1 200 a n d 2400 i n p h a se . I t is
i m m a t e r i a l w h e t h e r t h e b a l a n c e d l oa d ( s p e c i fi e d by i t s l i n e - t o- l i n e voltage, total
P Q \V -': r, a n d p ow e r fac t o r ) i s 6 - o r Y - c o n n e c t e d s i n c e t h e 6 can a lways be
re p l a c e d fo r p u r p o ses of c a l c u l at i o n b y i t s c q u i v Zl ] c n t Y, a s s h o w n i n Table 1 .2 .
I t i s J. p P 3 r e n t fro m t h e t a b l e t h a t t h e g e n e r,l I exp r e s s i o n for a wye i m pedance
c i rc u i t is t h e single-ph ase o r per-phase equ iL'a len t
1" .,
------...
FI G U RE 1 . 2 1
O n e p h a s e of t h e c i rc u i t o f F i g . 1 . 1 1 .
22
CHAPTER 1
BASIC CONCEPTS
TABLE 1 . 2
y.� an d �.y transformations t
,
t A d m i ttances
A
B
A
,
"
'-
'-
'-
'-
'-
'-
I
,
,
\
\
\
,
'-
'-
Y
/
/
..-
..-
..-
..-
..-
..-
..-
B
'y'
a n d i m pedances w i th the same subscripts are reciprocals o f o n e a n o t h e r .
Zy in terms of the delta impedances Z /).'s is
Zy -
p roduct of a dj acent 2/). 's
sum o f 2/). 's
( 1 .3 1 )
ced 2/).'s) , t h e
So, when all the impe dance s in the !:i are equa l (that is, balan
impe dance o f
impe dance 2 y o f each phase of the equiv alent Y i s one-t hird the
from Z y 'S t o
each p ha se o f the 11 w hich i t r eplac es. Likew ise, in transfonnin g
,
I .H
V O LTA G E A N D C U R IU::NT I N BALAN CED TH R E E-PHASE CI RCU ITS
23
Zt.'s, Table 1 .2 shows that
ZlJ.
s u m of pairw ise products of Zy 'S
=
( 1 .3 2 )
the opposite Zy
Similar stateme nts apply to the a d m i tta nce transformations.
20�n 4.4
E xa m p l e 1 .3 . T h e t e rm i n a l v o l t a ge o f
i m pe d a n ce s o f
is
Y -co n n e ct e d l o ad c o n s i s t i ng of t h ree e q u a l
(\
ZL
k Y l i n e t o l i n e . T h e i m p e d a n c e o f e a ch of t h e t h r e e
l i n e s con n e c t i n g t h e l o () d t o a b u s a t a s u b s t a t i o n i s
l i n e - t o - l i n e v o l t a g e a t t h e s u bs t a t i o n h u s .
2540
Solution . T h e
Y. I f
V:""
m a g n i t u d e o f t h e v o l t ()ge t o n e u t r a l
l.4�fl.
=
a t t he
/J!...
20�
load
t h e v o l t :l g e a c r os s t h e l o a d . i s c h os e n as r de re n c e ,
2540
'UII
==
127.0/
is
-
Find t h e
4400/13 =
30° A
The l i n e - t o - n e u t ra l v o l t ag e at t h e s u b s t a t i o n is
=
2 5 4(J
=
2666
L.Q:
+
+
j 1 25 . 7
=
�
2670/ 2.700
1 7 7 .8
Y
J n el t h e m a g n i t u d e o f t h e v o J t <1 g e a t t he s u b s t a t i o n b u s i s
\;\
x
2 . 67
=
-U<� J.;, V
f i g u r e 1 . 2 2 s h ows t h e p er - p h as e c q \ l i v ; t !L' n t c i rc l I i t , l I l l l t] l I , l IH i t i c s i n v o l v e d .
127
:,
-
. 3Q�
A
-T�
I
2 54 0 L O� V
I
1�
F l G U R E 1 .22
P e r - p h ase e q u iva l e n t c i rcuit for Example 1 .3.
24
1.9
CHAPTER 1
BASIC CON CEPTS
POWER IN BALANCED THREE-PHASE
CIRCUITS
The total power delivered by a three-ph ase generator or absorbed by a t hree­
phase load is found simply by adding the power in each of the t hree p h ases. In a
b alanced circuit this is the same as m u ltip lying the power i n a ny one p h ase by 3
since t h e power is the same i n a l l phases.
If t h e m a gnitude of the voltages to neutra l Vp for a Y-connected load is
and if the
magnitude
of t h e p hase c u r r e n t 1"
for
a Y co n n ec t e d
-
load is
( 1 .33 )
( 1 .34)
t h e total t hree-phase power i s
( 1 .35)
where f)p i s t h e angle by which ph ase cu rren t Ip l a gs the phase /,,' o ltage Vp , t h a t
is, the a ngle of the impedance i n e ach phase. I f I VL I and I lL I are t h e
m agnitudes of l i ne-to- line voltage VL a n d l in e current 1L > respectively,
IV 1 =
P
I VL I
{3
and
( 1 . 36 )
and substi t uting i n Eq. 0 .35) yields
( 1 . 37)
The total vars are
( 1 . 38)
( 1 . 39)
and the vqltamperes o f the load a r e
( l AO)
Equations ( 1 .37), 0 . 39), a nd ( l AO) a r e used for calculating P, Q, a n d \ 5 \
in b al anced t hree-phase networks since t h e q u antities u sually known are l ine­
to-line voltage, line current, and the power factor, cos 8p " When we speak of a
t h ree-phase system, b a l anced con d itions are assumed u nless described other­
w ise; and the terms voltage, current , and p ower, u n l ess identified otherw is �, are
1 . 10
PER-UNIT QUANTITIES
25
understood to mean line-to-line voltage, line current , and total three-phase
power , respectively.
If the load is connected L1 , the vol tage across each i mpedance is the
l i n e -to-line vol tage and the m agnitude of the current through each impedance is
t h e m agnitude of the l ine current d ivided by 13, or
and
T h e t o t a l t h r e e - p h a s e pow e r i s
P
=
3 1 V/'
( 1 .4 1 )
I II
/'
I cos. ()/'
a n d s u b s t i t u t i n g i n t h i s e q u < t t i o n t h c va i u c s o f
p
=
c
I v) anu i l)
fj I V!. I I I /. l eos f)I'
from EC] .
1.10
( 1 .4 1 ) gives
( 1 .43 )
to E q . ( l . 3 7 ) . I t fo l lows t h a t E q s . ( 1 . 39) and ( l .40) a r e
of w he t h e r a p a rt i c u l ar l o a d is co n n e c t e d .1 or Y .
w h i c h is i clc il t i c a l
v a l i cJ r e g a r d l e :, s
( 1 .42)
PER- � N JT
a l so
QUANTITI ES
o p e r a t e d a t vol t a g e l eve l:- w h e r e the k i lovolt (kY)
most conve n i e n t u n i t to e x p r e ss vol tage. Because of the l arge amount of
p ow e r t r a n s m i t t e d , k i lowa t ts or m e g a w a t t s a n d ki l ovo l t a m pe res or megavoltam­
p e r e s a r e t h e com m o n t er m s . However, these q u an t i t ies as well as amperes and
o h m s a re o ft e n e xp r e s s e d a s a p e rce n t o r p e r u n i t o f a base or reference value
s p e c i fi e d fo r e a c h . Fo r i n s t a n ce , i f a base vo l t age of 1 20 kV i s chosen, volt ages
of 1 0(:; , 1 20 , and 1 26 k Y b e co me 0.90, 1 .00 , a n d 1 .05 p e r u n i t , or 90, 1 00, a n d
1 05 0c. , r e s p e c t i v e l y . T h e per-unit ['allie of a ny q u a n t i ty i s d e fi n e d a s t h e r a t i o o f
t h e q ll �� n t i t y t o i t s h Cr s c n p r c s s e d ( I S a d e c i m a l . T h e ra t i o i n p e rce n t is 1 00 times
t h e v a l u e in p e r u n i t . B o t h t h e p e r c e n t (l n d p e r- u n i t m e t hods o f c a l c u l a t io n are
s i m r ! c r :tnc.l o ft e n m o re i n fo r m a t ive t h a n t h e use of act u a l a m p e res , ohms, and
vo l t s . T h e p e r - u n i t m e t h o c\ h a s a n a d v ( t n t a g e ove r t he perc e n t m e t h od b ecause
t h e p r o d u c t or two q u a n t i t i e s e x r nc s s c d i n fJe r u n i t is e x p r e s s e d in p er u n i t
i t s e l f, h u t t h e p {() d u c t ( ) j [ \\ 0 q u a l l t i t i e s e x p r e ss e d in fJc rc c n t must be d iv i d e d by
P o w ;..; r t r a n s m i s s i o n l i n e s a r e
is t h e
Vo l t age .
W U [ 0 u b [ a i n [ h c r c s u l t i n p e rce n t .
c u ne n t ,
k i \ ovo \ l a rn pc re s .
a n ti
i m r e tl a n c c
a rc s o
v a l u e s of c u rr en t
se k c t i o n o f b � I S C v , d u e s f ur ,I n y t w u 0 1 t h e m d e t e r m i n e s t h e
re l a ted t h a t
b a s e va l u cs o f t h e
and vo l tage, b ase
i m p e d a n c e a n d base k i lovo l t a m p eres can b e determined . The b ase impedance is
t h a t i m p e d a n c e wh i c h w i l l h ave a v ol t a g e d rop across it equal to the base
vo l t a g e w h e n the c u rre n t fl owi n g i n t h e i m p e d a n ce is equal to the base value of
t h e c u r re n t . Th e b a se k i lovo l t a m pe res in s i n g l e - p h ase systems is t h� product of
r c m Cl i n i n g t w o .
If w e
s p e c i fy
the
base
26
CHAPTER 1
BASIC CONCEPTS
b ase voltage i n kilovolts and base current i n a m peres. Usually, base mega­
voltamperes and base vol tage in kilovolts a re the quantities selected to specify
the base. For single-phase systems, o r t h ree-ph ase systems where the term
current refe rs to line current, where t h e term voltage refers to voltage to neutral
and where t h e term kilovoltamperes refers to kilovoltamperes per p hase, the
following formulas re l a te the various quantities:
Base curren t , A
Base impedance , n
Base impedance , n
Base impeda nce , n
Base power , k'W } (P
Base power , MW I 4>
Per-unit impedance of an element
b ase kVA 1 1>
=
base vol tage , k VL N
=
b ase voltage , VL N
base curren t , A
=
( 1 .44 )
( 1 .4 5 )
( b ase vol tage , kVL t\, )
2
base kYA 1 c!>
( base voltage , k YL N )
1 000
( 1 .46)
2
= --------­
MVA I 4>
x
( 1 .4 7 )
=
base kYA I </>
( 1 .4 8 )
=
base MY A 1 4>
( 1 .49)
actual imped a nce , n
= --------­
base i mpedance ,
fl
( 1 .50 )
In t hese equations the subscripts 1 4> a nd L N denote " per phase" and " line to
neutral," respectively, where the equa tions apply to three-phase c i rcu its . I f the
equations are used for a single-phase circu i t , k VL N means the vol tage across the
single-ph ase line, o r l ine-to-gro u n d voltage i f - on e side is grounded.
Since balanced three-phase c i rcuits are solved as a s ingle l i ne with a
neutral return, the bases for quantit ies i n the imped a nce di agram are kilo­
vol tamperes per phase and ki lovolts from line to n e u tra l . Da ta a re usually given
as total t hree-phase kilovoltamperes or m eg avoltamperes and l ine-to-line kilo­
vol ts. Because of this custom of specifying l in e-to-line voltage and total kilo­
vol tamperes or megavoltamperes, confusion m ay arise regarding the relation
between the per-unit value of line volta ge and the per-unit value of phase
vol tage. A l t hough a line voltage m ay be specified as base, the vol t age in the
single-phase circuit requ ired for the solution is still the voltage to neutral. The
base voltage to neutral is the base voltage from line to line divided by f3 . Since
this is also the ratio between l ine-to-line and line-to-neutral voltages of a
b a lanced t h ree-phase system, the per-unit value of a line-to-neutral voltage 9n the
line-la-neutral voltage base is equal to the per-unit value of the line-to-line voltage
l.ID
P E R - U N I T Q U A NTITIES
27
a t (he same poillt on the line-to-line coltage base if the system is balanced.
S i m ilarly, t he three-phase kilovo l ta m pe re s is t h ree t i mes the k il ovolt amperes
p e r p hase, and the three-phase kilovo l t am p e res base is three times the base
kilovoltamperes per phase . The re fore, the per-unit value of the three-phase
k i/ol'oltamperes 011 the three-phase k ilouoltampere base is iden tical to the per-unit
value of the kiloL'oltamperes per phase on the kilovoltampere-per-phase base.
A numerical e x a m p l e c l a r i fi e s t he re l a t ionsh ips. For instance, i f
B a s e k V A 3 e/>
a n cl
w h e re s u b s c r i p t s
B a s e k V/. L
:; '/'
and
IL
l3ase
k V A ' lj'
a c t u a l l i n c - t o- l i n c vo l t a ge
vol t Clg c i s 1 0 0 / /3
linc-to·· neu trai
an
Pe r- u n i t
Fo r t o t a l
] () , ( ) ( )()
=
1 20
M
v :'
and " l i n e
=
69 .2
=
to l in e,"
kV
k V i n a b a l a n ce d
1 08
62 .3
1 20
kW
=
respective ly,
kVA
62.3 kV, a n d
vol t age
power
1 0 ,000
=
o f l OS
t h ree-ph ase powe r o f 1 8,DOO
P e r- u n i t
"'. 1 2 0 k V
" t h r e e - p h a se"
B a s e k V'_ N =
an d
For
mea n
3 0 , 000 k V A
=
-
69.2
=
th ree-ph ase set t h e
0.90
the power per phase is 6000
l R ,OOO
]() ,O(m
6 ,000
1 0 ,000
=
kW,
and
0.6
lllq.':< I vu l Ll ll1 p e re v ;l I l i C S 11l ; I Y he s u bs t i t u t e d ro r k i l o­
W,l t t (l n d k i l uvul t <l rn r cre va l u e s t h ro u g h o u t t h e above U i S c liss i o n . Un less o th e r ­
O f c o u rs e , Il1 e g ; l \V ; l t r a m i
wise
spe c i fi e d ,
l i n c - to - l i n c
a
given
vo l t :l g e ,
; t n cl
value
a
o f base vol t age in a three-rhase sys tem is a
v a l l i e of base k i l ovo\ t a m p e rcs o r b a se mc g a­
given
e n t c a n b e compu ted d i rectly from t h ree­
p h ase val ues of b a s e k i lovo l t s and b a se k i lovoltamperes. I f we i n t e rpret base
ki/o L 'o/({/llIpercs and hem: ( 'o/rage ill kilol'olls to mean base kilol'oltamperes for the
total of rhe th ree phases a n d base I 'o/rage from line to line, we find
v o l t d 1l 1 p c r e s is t h e t o t a l
t h re e - p h ase b a s c o
B Cl s e i m p e d a n c e a n d b a se
curr
B a s e c u r re n t , A
r1�
base
x
kYA ,</>
base vol tage ,
k Vu_
( 1 .5 1 )
28
CHAPTER 1
BASIC CONCEPTS
and from Eq. (1. 4 6)
Base impedance
Base impedance
Base
i m pe d an c e
=
=
=
( base voltage , kVL L / J3 ) 2
( b ase vol tage, kVL L )
2
( base vol tage , kVuJ
2
X
X
1 000
1 000
-------­
base kV A : "/'
base M VA 3</>
( 1 .5 2)
( 1 .5 3 )
( 1 .54)
Except for the subscripts, Eqs. 0 .46) and 0 .47) are identical to Eqs. 0 .53) a n d
( 1 .54), respectively. Subscripts h ave been u s e d in expressing these rela tions i n
order t o emphasize the distinction between working with three-phase q u a nt i ties
and quantities per p hase. We use t hese e quations without the subscrip ts, but we
must
•
•
Use l ine-to-lin e kilovol ts with t h ree-phase kilovol tamperes or megavol tam­
p eres, a n d
U s e l in e-to-neutral kilovol t s w i t h kilovoltamperes or megavoltamperes p e r
phase.
Equation 0 .44) determines the b ase current for single-phase systems or for
three-p hase systems where the bases a re spec ified in total kilovoltamp eres per
p hase a n d kilvolts to neutral. Equa t ion ( 1 .5 1 ) d etermines the base c urre n t for
three··p hase systems where the bases are specified in total kilovoltamperes for
the three phases and in kilovolts from line to line.
Find the solution of Example 1 .3 by working in per unit on a base of
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 e n ter the prob l e m .
Example
4.4 kV, 1 27
Solution.
1 .4.
A
Base impedance is
4400/fS
127
=
20 .0
fl
,
and therefore the magnitude of t he load i mpedance is also 1 .0 per u nit. The line
1.11
impedance i s
Z =
Va n
=
C H A NGI N G T H E B A S E OF PER·UNIT QU ANTITIES
�
1 .4
/750
LQ: + 1 .0/
LQ: + �
1 .0
-
1 .0
=
per u n i t
= 0 .07
20
30°
1 . 0495 + jO . 04 95 = 1 . 05 ]
VI. !.
a rc
1 . 05 1
=
/...
=
,)
4 .4
=
207()
/2.70°
Y, o r
2 .67
per u n i t
kY
4 ' ()2 k Y
more complex, a n d particula rly when
of ca lcu lat i ons in pe r u n i t are more
a p p a r e n t . l m p e c Li n c e v a l u e s a n d o t h e r p �l r (l m e t c r s of a c Ll m po n e n t when g iven
i n per u n i t \\' i t h o u t s pe c i fi ed b a s es , a r e g e n e ra l ly u n d e rs t oo d to be based on t h e
m e gavo l t a m p c r e a n d k i l o vo l t ra t ings o f t h e component.
W h e n t h e p ro b l e m s t o
t r a n s fo r m e r s
1.11
be
X
4400
--
0 / 750
X 0. 7
0 .0 7
VL N = 1 . 0 5 1 x
29
i n vo l v ed thc
CHAN GING
,
s o l v e d a re
adva n t ages
,
THE BAS E
QuANTITIES
O F PER-U NIT
Sometimes the per-u n i t i m pe d a n ce of a component of a system is expressed on
a b d s e other t h an the one s e l ected as b ase for the part o f the sys tem i n w h i c h
t h e c o m po n e n t is located . S ince all im pedances in any one part of a system must
b e expressed on t h e s a m e i mped ance base when making com p u t ations, i t i s
n e c e s s a ry t o h av e a means of convert i n g per-un i t impedances from o n e base to
a n o t h e r . S u b s t i t u t i n g t h e e x p r e s s i o n for h a s e i m ped a n c e g i v e n b y Eq . ( 1 .46) o r
( 1 . 5 ] ) fo r h a s e i m p e d a n c e i l l Eq . ( 1 . 5 ( ) g iv e s fo r a ny c i rc u i t e l e m e n t
P e r - u n i t i m pe d a n c e
( a c t u a l i m ped ance , .0)
( h;lsc
vol t a ge ,
x
kyr
)
( base k YA)
x 1 000
( 1 .55 )
t o h a s c k i lovol t a m ­
p e r e s a n d i n v c l s e l y p r opo r t io n a l t o t h e squ are of t h c base vo l ta g e . The refore, to
c h ,lI1ge from pe r-u n i t impedance on a given base to per-unit i mpedance o n a
n ew b a s e , t h e fo l l ow i n g e q u a t i o n a p p l ies:
w h i c h s h ows t h a t p e r - u n i t i m p e u a n c c i s li i re c t l y p r o p o r t i o n a l
P e r- u n i t 7 n ew
=
per- u n i t
Zgl\'c n
.
2
( baSe kYgjven ) ( baSe kYA new )
base
k Ynew
base k Y A giv en
( 1 .56)
30
CHA PTER I
BASIC CONCEPTS
The reader should note t h a t t h i s equation h a s nothing to do with transferring
the ohmic value of imped ance from one side of a t ransformer to another. The
applic at ion of the equation is in c h a n ging the value of the per-unit impedance
o f a ny component given on a particu l a r base to a n ew b ase.
Rather than using Eq. (1 .56) d irectly, the c hange in base may also be
accomplished by first converting the per-u n i t value o n the given base to ohms
and then d ividing b y the new base i mpedance.
Example 1 .5. T h e reactance of a generator designated X" is gi\ en a s 0.25 p e r unit
based on the generator's nameplate rating of 1 8 k Y , 500 M VA. The base fo r
calculations is 20 kY, 1 00 M Y A. Finu X" o n t he n e w base.
Solution. By Eq.
(1.56)
X"
=
:>
(
1 8 ) ( 100 )
0 .25
20
or b y converting the given va l u e
X II
_
_
500
to ohms
=
0.0405 p e r unit
and d ividing b y t h e n e w base i m p e d a nce,
0.25 ( 182/500)
202/100
=
0 .0405 per unit
Resistance and reactance of a device in percent or per u n i t are usually
available from the manufacturer. The impedance base i s u n derstood to be
derived from the rated kilovol t amperes and k ilovolts of the device. Tables A . 1
and A . 2 i n t he Appendix list s o m e representative val ues o f reactance for
tran s formers a n d generators. Per-unit quantit ies a re fu rther discussed in Chap.
2 a ssociated with the study o f t ransformers.
1 .12
NODE EQUATIONS
The junctions formed when two o r more c i rcuit e l e m e n ts ( R , L, or C, or an
ideal source of voltage or curr e n t) a re connected to each other a t their
termin als are called nodes. Systematic formulation of equations determined at
nodes o f a circui t by applying Kirchhoff's current law is the basis of some
excellent computer solutions of power system p roblems.
In order to examine some fe a t ures of node elJuations, we begin with the
simple circuit d iagram of Fig. 1 .2 3 , which s hows node numbers within circles.
Current sources are connected at nodes @ and @ and all other elemen ts are
represented as admittances. Single-subscript notation is used to designate the
voltage o f each node with respect to the r eference node @ ' A p p lying Kirch­
hoff's current l aw a t node CD with current away from the node equated to
current i nto t he node from t h e source g ives
( 1 .5 7)
1.12
N O D E EQUATI O N S
31
®
Re terence
Fl G L' R E
A c i rc u i t
and
1 .2 3
d i J g r cJn)
fo r n o d e
s h ow i n g ;: u rre n t
Q)
1/,.' Y"
....: .
s o u r c e S :l l n o d e s
( V.'
-
V::' )
Y/)
))
,
+
:I n d
(V
\
.
@) ; a l l
-
j/ 1
oth�r
)Y
=.�
('
c l e ments
a re
J u m i l l a n ces.
( 1 .58)
/�
R e a rr a n g i n g t h c s c e q u a t i o n s y i e l d s
At node
At node
CD :
Q) :
S i l� l i i (1 1 c q U J t i O ll S C d l l b e [o r m e d C m n o d e s
( 1 .59)
( 1 .60)
(j)
and
@) ,
<t n d t h e fou r
C 2 n h e so l v e d s i rn u l l a n e o u :-. l y fo r t h e v o l t a g e s V I ' V2 , V� , a n d V4 .
r
equat ions
Al l b r a nch
c u r r e n t s c a n b e fo u n d w h e n t h e s e v o l t a g e s a rc k n ow n , a n d a n o d e e q u a t i o n
fo r m e d fo r t h e r e fe r e n c e n o d e would y i e l d n o fu rther i n formation. Hence, t h e
re q u i r e d n u m b e r o f indep(,ll dm ( n ( ) d e e q u a t i o n s i s o n e l e s s t h a n the n u mber o f
n o d e :; .
\Ve h a ve n o t w r i t t e n t h e e q u a t i o n s for n o d e s
W
and
@) because w e c a n
a l r e a d y s e c ho\v to fo rm u l a t e n o d e e q u a t i ons i n standard notation. I n both Eqs.
( 1 .59) a n d 0 .60) it is <l p p m c n t t h at t h e current Rowing into the network from
c u r r e n t s o u r c e s con n e c t c d to a n o d e is e q u ated to the sum of several products.
At any node one ' p rod u c t is t he vol tage of t hat node t imes the sum of all the
a d m i t t a n c e s w h ic h t c rm i n a t e o n the n od c . This p;oduct accou nts for the cu rrent
t h a t Rows away from t he node if t h e vol tage is zero at each other node. Each of
t h e o t h e r p r o d u c t s e q u a l s t h e n e g a t i v e of the voltage a t another node times the
a d m i t t a n c e con n e c t e d d i re c t l y b e t w e e n t ha t node a n d t h e n o d e a t which the
32
C H A PTER
I
BASIC CONCEPTS
equ a tion is formulated. For instance , i n Eq. 0 .60) for node Q) a p roduct is
- V2 Yb , which accounts for t he current flow away from node Q) w hen voltages
a re zero at a l l nodes except node (1) .
The u s u a l matrix format of the fou r inde p endent e qu ations for Fig. 1 .23 is
-
CD (1) Q) @
CD
�
Q)
YI I
Yl 2
Y1
Y2 1
Yn
Yn
Y4 1
Y4 2
Y3 1
@
YI 4
3
Y2
4
Y3 4
Y4 4
Y4 J
Y3 2
Y3 J
VI
V2
VJ
12
II
=
V4
IJ
( 1 .61 )
14
The :-;ym m e t ry of t h e e q u a t i o n s i n t h i s fo r m m a k e s t h e m e ll s y t ( ) r e m e m h e r, li n d
t heir e xtension t o a ny number o f nodes i s ap p are n t . T h e o r d e r o f t h e Y
su bscri p ts is effect-cause ; t h a t is, t h e fi rs t subscri p t i s t h a t of the node a t wh ich
the curren t i s being expressed, and t h e second s ubscript is the same as t h a t o f
t h e voltage causing this component of c urrent. T h e Y matrix i s designated Y b u s
a n d called the bus admittance ma trL e The usual rules when forming t h e ty p ical
elem e n ts of Y b u s a r e :
•
•
:
The d i a gonal element
nee ted to node 0 .
1Jj
e q uals the sum of the admittances directly con­
The off- diagona l element 0j e q u a l s the nega tiL'e of the net admittance
connec ted between n o des CD and (j) .
d i agonal admittances are cal l e d t h e self-admittances a t the nodes, and
t h e o ff- d i agonal adm i ttances a re the mutual admittances of t h e n od e s . Some
authors c all the s elf and mutual a dm i ttances of t he nodes t he driving-point and
transfer a dmittances of the nodes. From the above rule s Y bus fo r the ci rcu it of
Fig . 1 .23 becomes
The
-
Y bus
=
CD
CD
Q)
CD
( Yc Yd +
-Y
d
-Y
c
- Yf
@)
+
Yf )
(Yb
- Y"
Yd
- Yb
-Y
e
+
@)
- Yr
Q)
CD
-
+
Ye)
Yc
-Y
b
( Ya Yb Yc )
+
0
+
-Y
c
( Y"
+
0
Yf Yg )
+
( 1 .62)
w here the encircled numbers are node numb ers w h i ch a lmost a lways correSpon d
to the s u bscri p ts of the elements r:j of Y bus ' S e p a ra t i n g o u t t h e e ntries fo r a ny
1 . 12
NODE EQUATIONS
33
one of the admittances, say, Yc ' we obtain
Y bus
=
CD ( Yd + Yf )
Yd
(1)
0
Q)
Yf
@
Q)
(1)
Q)
@)
T h e rn a t r i x fo r
fo ] l O\vs:
CD
(0.
Q)
@)
--J
'ti h i t e
r:.
@
Y,.
CD
+
-Y
d
( Yb + Yd
- Yb
- r:.
0
0
- yr
()
+
Ye)
()
0
()
0
0
r;.
( 1 .63)
0
c a n be \v r i t t en a s s hown i n Eq.
CD
Y
c
- yc
-- yc
0 .63)
®
Q)
(]
:I
y(
0
( Ye + Yf + Y"J
0
@
()
- Ye
- Yb
( Ya + Yo )
CD
0
- Yf
0
- y;
0
®
Q)
@
CD
<==>
.J
CD
CD
more compactly a s
or
-
Q)
Q)
[ -:
�1
Yc
( 1 . 64 )
the l e ft - h a n d s i d e shows t h e a c t u a l m a t rix con trib uted by Yc to Y b u s , we
ca n interpre t the s m a l l e r m a t r ix o n the right a s a compact storage m atrix for the
s a m e c o n t r ib u t i o n . T h e enc ircled n u m bers CD and CD point to the rows a nd
c eJ l u m n :"i o f \' h u � t o w h i c h t h e e nt r i e s r; a n d - r:. b e l o n g . Th e 2 X 2 m at rix
m u l t i plying Yc is an i m p o r t a n t bu ildil/g hlock in rorm ing Yblls [or morc general
we c o n s i d e r in C h a p . 7 .
I n v e r t i n g Ybu\ y i e l d s a n i m po r t a n t l1lJtrix c a l l e d t h e
n e tw o rks, w h i c h
Z I1l: ,
w h i c h h ;l S t h e s t ;l n d ; l rtl fo r m
0)
Z-' b u s
Th e
cons tr uc t i
=
y
l
bus
=
r ZI I
CD
@
�
on a n d p ro r e r t i e s o f
CD
l
CD
Z l2
l
Z 22
Z :I I
Z]2
Z4 1
Z 42
Z
Z hu s
'
bus
impeda n ce ma trix
Q)
®
Z 2}
Z
£ ]]
Z 34
24 3
2 44
Z 13
a re considered
14
Z2
in
4
Cha p . 8.
( 1 . 65 )
34
CHAPTER I
BASIC CONCEPTS
DIAGRAM
1.13
THE SINGLE-LINE OR ONE-LINE
I n Chaps. 2 t h rough 6 we develop the circuit models for transformers, syn­
chronous m achines, and transmission lines. Our p resent interest is in how to
portray the assembla ge of these compon ents to model a complete system. Since
a b a lanced t h ree-phase system is always solved as a single-phase or per-phase
equivalent circuit composed of one of the t h ree lines and a neut ral return, it is
seldom n e cessary to show more than one phase and the neutral return when
d rawing a d i agram of the c i rcuit. O ften the diagram is simplified fu rther by
omitting the completed c ircu i t t h ro u g h t h e n e u t ra l and by i n d i c a l i n g l h e
compone n t parts b y standard symbo l s r a t h e r than b y their e q u iva l e n t circuits.
Circuit para meters a re not shown , and a t ransmission line is represe n ted by a
single line between its two ends. Such a s i m p l i fied diagram of an electric system
is called a single-line or one-line diagram . It i n d icates by a single line a n d
sta n d a rd symbols how the transmission l i n e s and associated apparatus o f a n
e lectric system a r e connected together.
The p u r pose of the one-line d i agram is t o supply in con cise form the
significan t i n formation about the system. The importance of different fea tu res
of a system varies with the problem u n d e r considera tion, and the amou nt of
information i ncluded o n the diagram depends on the pu rpose for which the
diagram i s i ntended. For instance, the location of circu i t breakers and relays is
u n important in m aking a load study. Breakers and rel ays are not shown if the
primary function of the d iagram is to p rovide information for such a study. On
the other h a n d , determination of the stabil ity of a system under transient
conditions resulting from a fault depends on the speed with which relays and
circuit breakers operate to isolate the fau l ted part of the system . Therefore,
info rm ation about the circuit b reakers may be of extreme importa nce. Some­
times o ne l i n e d i agrams include i n fo r m a t i o n abou t the curre n t and p ot e n t i a l
transformers which connect t h e r e l a y s t o t h e sy st e m or w h ic h a rc installed for
metering. The i n formation fou n d on a one-line d i agram must be expected to
vary accord i n g to the problem at h a n d a n d accord ing to the practice of the
particu l a r company preparing the diagram.
The American Nation al Standards Institute (ANSI) and the I nstitute of
Electrical and Electronics Engineers (I EEE) have pub l ished a set of standard
symbols for e l ectrical d iagrams. l Not all au t hors follow these symbols consis­
tently, especially in i n d icating transfo rmers. Figure 1 .24 s hows 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
a nyon e who i s not working constantly with one-line d i agrams, it is clearer to
-
I See G r a p h i c Symbols for Elect rical a n d Electronics Diagrams,
IEEE
S t d 3 1 5 - 1 975.
1.13
M a chine or r o t a t i n g
a r m a t u r e (basic)
-0-
Power ci rcu i t b r e a k e r,
oi l o r o t h e r l i q u i d
--r\--
A i r circu i t b re a k e r
T w o- w i n d i n g power
t r a n sformer
3S
T H E S I N G L E-LIN E OR ONE-LINE D I A G RAM
Three-p h a se, t h ree- w i r e
y
d e l t a co n n ec t i o n
T h ree-wi n d i n g power
t ra n s fo r m e r
T h r e e - p h a se wye,
n e u t ra l u n gr o u n d e d
F u se
Cu r re n t t r an s fo r m e r
ne u t r a l
Th ree - p h a se wye,
m
gro u n d e d
P o t e n t i a l t r a n s fo r m e r
A m mc t c r a n d v o l t me t e r
Fl G l ' l� E 1 . 2 .t
r\ [) p a rd t l l s s y m b o l s
type a n d ra t i n g .
i n d i c a t e a p a rt
i c u l a r mach i n e by t h e b a s i c symbol fol l owed by i n fo r m a t i o n o n i ts
It i s i m po r t a n t t o k n ow t h e lo c a t i o n of po i n ts w h e r e a syst e m is con ne c t e d
p rou il c1 i n o r d c r to c a l c u l a t c t h e a m o u n t o f c u r r e n t flow ing w h e n a n
unsym m e t r i c a l fa u l t i nv o l v i n g g ro u n d occ u rs_ T h e s t a n d a r d symbol t o d es i g n a te
to
wit h t h e n e u t r a l sol i d ly gro u n d e d is shown i n Fig. 1 .24. If a
re s i s t u r o r r e a c t o r i s i nserted between t h e n e u tral of t h e Y a n d grou n d t o limit
the fl o w of c u r re n t to g r o u n d d ur i n g a fa u l t , the approp r i a t e symbol for
r e s i s t a n c e or i n d u ctance may be added to t h e s t a n d a r d symbol for t h e g ro u n ded
y . \1ost t r a n s ro r m c r n e u t r a l s i n t r a n s m i s s i o n s y s t e m s a re so l i d ly grou n d e d .
G c n e r 3 t o r n e u t r a l s a r e u s u al ly grou n d e d t h rough fa i r l y h igh resistances a n d
a
thre e-phase Y
s o m et i m e s t h ro u g h i n ci u c t a n c e
F i g u re
1 .25
is
co i ls .
t h e s i n g l e - l i n c d i ,l g r , l m o r a s i m p l e
g e n e t a to r s , o n e g ro u n d e d t i l ro u g h
iY0-r�
II
-o-� �-----o-i
[1
Lo::d A +--0 -
Fl G li RE 1 .25
pow cr
r e ; l c l o r a n d o n e t h ro u g h
S l n g l e - l i n e J i J g r a ll1 o f a n e l e c t r ica l pow e r sys t e m .
T2
sys t c m .
Two
a rcsistor, a re
36
CHAPTER I
BASIC CONCEPTS
connected to a bus and through a step-up transformer to a transmission line.
Another generator, grounded through a reactor, is c onnected to a bus and
t hrough a transformer to the opposite end of the transmission line. A load i s
connected t o each bus. On the d i agram i n formation about t h e loads, ratings of
the generators and transformers, a n d reactances of the different components of
the circuit is often g iven.
1.14 IMPEDANCE AND REACTANCE
DIAGRAMS
In order to calculate t h e perfo r m a n c e o r a s y s t e m
the
occu r re n ce o f
a
fa u l l , t h e o n e - l i n e d i a g r, l I11
is
L I mi e r load cu n d i t io n s or U [)o I1
used t o d raw t h e s i ng l e-p h asc
or per-ph ase e quival ent circu it of the system. Figure 1 .26 comb i n es the equiva­
lent circuits (yet to be d eveloped) for the various components shown in Fig. 1 .25
to form t he per-phase impedance diagram of the system. If a load study is to
be made, the l agging loads A a n d B are represented by resis t a nc..: and ind uc­
tive reactance in series. The impedance diagram does not include the current­
'
limiting imped ances shown i n the one-line d iagram between the neutrals of the
generators a n d ground because no current flows i n the ground u nder bq.l anced
conditions and the neutrals of the generators a re at the potenti a l of the neutral
o f the system. S ince the shunt current of a transform e r i s usu ally insignifica n t
compared w i t h the ful l-load curre n t , the s h u n t admittance i s usua l ly omitted i n
the equivale n t circuit of the transformer.
Resistance is often omitted w h e n m aking faul t calculations, even i n
computer programs. O f cou rse, omission of resistance i ntroduces some error,
but the results m ay be satisfactory since the inductive reactance of a system is
much l arger than its resistance. Resistance and inductive reactance do not a d d
directly, a n d impedance i s not f a r d ifferent from t h e i nductive reactance if t h e
resistance is s m a l l . Loads w hi c h d o not involve rotating mach inery h ave l ittle
effect on the total line current d u r i n g a fau l t and are usua lly omitte d . Syn­
chronous motor loads, however, are a lways i n cluded in making fau lt calcul ati ons
� � ''---�v�-�_/ '---�v---�- ----��--�.,, '---v-' �
G e n e rators
1 a n'd 2
Load
A
Tra nsformer Tl
Tra nsmission l i n e
Tra n s f o r m e r T2
FIGURE 1 .26
The per-phase impedance diagram corre s p o n d i n g to the singl e-l i n e d i agram o f
Fig,
1 .25 ,
Load Gen. 3
B
P R OBLEMS
37
N e utra l bus
FI G U RE 1 .2 7
P e r- p h as e r e a c t ;] nce d i ;] g r ;] m ;] d a p t e d from F i g . 1 . 2 (1 b y o m i t t i n g
;] d m i t t a n c e s .
;] 1 1 l o ;] d s , r e s i s t a n ce s , ;] n d s h u n t
gen erated emfs con t r i b ute to the short-circuit current. The d i agram
shou ld t a k e i nd uct ion m otor s i n to accou nt b y a gene ra te d emf in s eries with a n
i n d u c t ive reacta nce i f the d iagram is t o b e used t o dete rmine t h e c urrent
i m m e d i a t e l y after t h e occurr ence of a fa u l t . I n d uction m o t o r s are ignore d i n
com p u t i n g t h e c u r r e n t a few cyc l e s a ft e r t h e fa u l t o c c u r s b e c a u se t h e c u r re n t
con t r i b u t e d by a n i n d u c t i o n m o t o r d i e s o u t ve ry q u i c k l y a ft e r t h e i n d u c t i o n
motor is s h o r t - c i rc u i t e d .
If w e d e c i d e t o s i m p l i fy o u r c a l e u l a t i o n of fa u l t c u n e n t b y a m i tti n g a l l
static loads, a l l resistances, t h e s h u n t a d m i t t a n ce o f each transformer, a n d the
capaci t(lnce of the tra n smission l in e , t h e imped a n ce d i agram red u ces to the
per-ph ase r e a c t a n c e d i a g r am of F i g . 1 . 2 7 . These simpl i fic ations a pply to f(lu l t
calcul ations only as discussed i n C h a p . 1 0 a n d n o t t o power-flow stu dies, w h ich
a r e the subject of C h ap. 9 . If a computer is aV(l il able, such s i m p l ifi ca tion is n o t
s i n ce t h e i r
n e c e ss a ry .
The per-phase impe d a nce a n d reactance d i agrams d is c u ss e d here a re
som e t i m es cal led the per-phase posiriL'e-seqllence diagrams s ince t hey show
i m pe dances to b a l a nced c urre n ts in one p h ase of a symme trical t h ree-phase
system . The sign ifica nce of t his designation is apparent in Chap. 1 1 .
1. 1 5
S C \1 MARY
Th i s c h Cl p t c r r e v i e w s fu n cl a lT1 e n Ul i s of s i n g l e - p h (lse (1 1 1(j b ; l l a n c e d t h re e - p h a s e
c i rcu i t s and cxp l a i ns somc o f t h e
t hroughout t h e text.
d i ;l g r ;l fll , (l l o n g with i ts
n o tat i o n t o be llsed
assoc i a t e d i m [J c d ; l rl c e ( 1 i ;I g r ; l l11 , i s d e s c r i h c cl . For fll u l ; t l i o n o r n o l l e e q u ;l t i o n s for
P e l - u n i t c i l c u l ,l t i o n s ,I r e
i n t ro cJ u ced
,I n cl t h e s i n g l e - l i n e
c i r c u i t s w i t h o u t l11 u t l l ; t i c o u p l i n g i s ; l l s o d e m o n s t r a t e d .
PROBLEMS
V a n e! i = 1 1 . 3 1 cos( (tJ {
30" ) A, fin d for e a c h ( a ) t h e
maxi m u m v a l u e ( b ) t h e r m s v a l u e a n d ( c ) t h e p h asor e x p ression i n p o l a r a n d
rect a n gu l a r fo rm i f volt age i s t h e r e fe r e n c e . I s t h e c i rc u i t i n d u c t ive o r c a p a c i t ive?
1 .1 . I f
L"
=
1 4 1 . 4 s i n (, :tJ l
,
-I
]0 )
-
,
1 . 1 co n s i s t s o f a p u rely r e s i s t ive a nd a p ur e l y r e a c t ive
c l e m e n t , A n d n a n d X ( a ) if t h e c l e m e n t s a re in s e r i e s a n d ( h ) if t h e c l e m e n t s a r e
1 . 2 . I f the c i r c u i t of P rob .
i n paral lel.
38
CHAPTER 1
B A S I C CONCEPTS
1.3. In a single-phase circuit
Va
�
Llr
= 1 20
V an d Vb 100
V with ri,;spect to a
Find Vba in polar form.
1.4 . A single- phase ac voltage of 240 V is applied to a series circuit whose impedance is
10
n. Find R, X, P, Q, and the power factor of the circuit.
LiQ:.
reference node
o.
=
1.5. If a capacitor is connected
1.6.
1.7.
1.8.
1.9.
1.10.
1.11.
1.12.
in parallel with the circuit of Prob. 1 .4, and if t h i s
capacitor supplies 1250 var, find the P and Q supplied by the 240- V source, and
find the resultant power factor.
A single-phase inductive load draws to M W at 0.6 power-factor lagging. Draw the
power t r ian g l e and determine the reactive power of a capacitor t o b e connected i n
pa r a ll e l w i t h the load to raise the power factor to 0 . 8 5 .
A single-phase induction motor is o p era t i ng a t a very l i g h t l o a d during a large part
of every day and d raws 1 0 A from the s u p p l y . A device is proposed t o " i n c r e ase t h e
efficiency" of the motor. During a d e m o n s t r a t i o n t h e d e v ice is p l a c e cJ i n para l l e l
with the unloaded motor and the cu r r e n t J rawn from t h e supply drops to 8 A .
W h e n two of t h e devices arc placed in paral le l, t he c u r r e n t drops to 6 A . What
simple device will cause this drop in c u r r e n t ? D iscuss t h e advantages of the device.
Is the efficiency of the motor i n cre a s e d by thc d e v i c e ? ( R c c Zl I I t h il l a n induction
motor draws lagging current.)
If t he impedance between mac h i n e s 1 a n d 2 of Example 1 . 1 is Z = 0 - j 5 D ,
determine ( a ) whether each mach ine is generating or consuming power, ( b )
whether each mach ine is receiving or supplying p o s i t ive reactor power and t h e
amount, and (c) the value of P and Q absorbed b y t h e i mpedance.
�
1.15.
current th rough the source is given
Q and state whether the sou rce is
=
a - I
1 - a2 + a
a2 + a + j
ja + a 2
Three i denti cal i m p ed a nces o f 1 0Llr D are Y-c onn e ct ed to balanced three­
phase line voltages of 208 V. Specify a l l the l ine and phase voltages and the
currents as phasors in polar form with Vca as reference for a phase sequence of
In a balanced three-phase system the Y -connected impedances a re 1 0� D . If
Vbc = 4 16L2.2:. V , specify len in polar form.
The terminals o f a three-phase supply a re l abe l ed a , b, and c. Between any p a i r a
voltmeter measures 1 1 5 V. A resistor of 1 00 D and a capacitor of 1 00 D at the
fre quen cy of the supply are connected in series from a to b with the r es i sto r
connected to a . The point of connection of the elements to each other is labelyd n .
a bc .
1 . 1 4.
=
delivering or receiving each .
Solve Ex a mp l e l . 1 if E, = 1 00& V and £ 2 1 20� V . Compare t h e r e s u l t s
w ith Example 1 . 1 and form some conclusions about t h e e ffe c t of variation of the
mag ni t ude o f £2 in this circuit.
Evaluate the fol lowing expressions in polar form:
(a)
(b)
(c)
( d)
1.13.
�
Repeat P roblem 1.8 if Z 5 + jO fl .
A v o l t a g e source Ea/l = - 1 20
V and t h e
by Ina = 10
A. Fi nd the va lu e s of P and
P R O B LE M S
1.16.
1 . 17.
1 . 1 8.
39
Determine graphically the voltmeter read ing between c and n if phase sequence is
and if phase sequence is ac b .
Determine the current drawn from a three-phase 440-V line by a three-phase
1 5-hp motor operating at full load, 90% efficiency, and 80% power-factor lagging.
Find the values of P and Q drawn from the line.
I f the impedance of each of the three lines connecting the motor of Prob. 1 . 16 to a
bus is 0.3 + J LO .0, find the line-to-line voltage at the bus which supplies 440 V at
the motor.
A balancecl-n load co n s ist i n g of p ur e r e s i s t a l1ces of 1 5 n per p hase is in parallel
wi t h a b a l an ce d- Y l o ad having phase impedances of 8 + j6 n . Identical impedances
of 2 + j5 n are i n eoch o f the three l ines connect ing the combined loads to a
j 1 0- V t h re e - p h a se s u p p l y . F i n d t he c u rr e n t d raw n fro m the supply and line voltage
abc
at t h e com b i n ed l o ad s .
t power
0.707 lagging from a 4 4 0 -V
which draws 6 0 k VA.
1 . 1 9 . A t h re e - p h ase l o a d d r;t\\·s 2 5 0 k W a a
f;\c t o r o f
l i n e . I n p a r;l I l e l w i t h t h i � l o a d i s ; 1 t h rc e - p hase C<l paci t or b a n k
Find t he t o t a l c u r rc n t (l nll resu l t a n t powe r fac t o r .
o
from
t h re e - p h ase m t u r ci r ;l\vs 20 k V A a t 0 . 707 powe r - f,l c t o r l agging
a 220-V
sou rce . D e t e rm i n e t he � i lovol t <1 m pe r c r;l t i n g of c a p a c i t o rs t o m a k c t h e c o m b i ne d
p o w e r [a c t o r 0 . 90 l a g g i n g, a d d e t e r m i n e t he l i n e c l l r r c n t before
1 .20. A
capacitors arc added.
1 .2 .3 .
1 . 24.
and after the
)TI,lchine in a n ope n - p i t mine consu mes 0 .92 M V A at 0 . 8
p ower-hctor l a g g i n g \\ h e n i t digs co a l , a nd i t g e n e r a t e s (delivers to the electric
sys t e m) 0 . 1 0 M VA at O � powe r - fa c t o r l e a d i n g when the l o a d e d shovel swings away
from t h e p i t w a l l . At [ h e e n d of t i l e " d i g " period t h e c h a nge in supply current
m a g n i t u d e c a n cause t r i p p i n g o f a protective rday, which i s constructed of
s o l i d - s t a t e c i rc u i t ry. Th e r e fo r e , i t i s desired to minimize the change in current
m a g n i t u d e . Co n sid e r t h e p l a c e m e n t of capacitors at the machine terminals and find
the amount of capaciti\'c correction (in k"Var) to eliminate the change in steady-state
cu rre n t m a g n i t ud e . The machine is energized from a 36.5 kV, three-phase supply.
S t a r t t he sol u t i o n by l e t t i ng Q b e the t o t a l three-phase megavars of the capacitors
c o n n e ct e d ac ross t he mach i n e terminals, and write an expression for the magnitude
o f t h e l i n e cu rre n t dr{) �VII by t h e m ac h i n e i n t e r m s of Q for both the digging and
y.c ller'l ting ope ra t i o l l s
.J\ g C rl C ril t Oi ( w h i(; h nl d Y be re p res e n t e d by a n e m f i n s e ri e s with an inductive
reactance) is rated 500 M V A , 22 k V . Its Y -c o n n e c t e d windings have a reactance of
1.1 p e r u n i t . F i n d t h e o h m i c va l u e of t h e reactance o f t h e w i n d i n gs.
The g e n e ra t o r of Prol . 1 . 2 2 i s in a c i rc u i t fo r w h i c h the b a s e s arc sp e ci fl e d a s 1 00
M VA, 20 k V . S t a r t i n g w i t h t h e p e r - u n i t v a l u e g i v e n in Prob. 1 .22, find the per-unit
v a l u e of r e a c t Cl n c e of t h e ge n e ra t o r w i n d i n gs on t h e s pe ci fi e d b a s e .
D r aw t he :;ingle-phasc e q u iva l e n t circuit for the motor (an emf in series with
i n d u c t ive r C tl c t a ll C C l a b e l e cl Z,) <l nd i t s con n e c t i o n t u t h e voltage supply described
i n Probs. 1 . 1 6 and
Show on the diagram the per-unit values of the line
i m p e d a n c e a n d t h e volt age at t he motor terminals on a base of 2 0 kVA, 4 4 0 V .
T h e n u s i n g per-unit val ues, finel t h e slJPply voltage i n per unit and convert the
p e r- u n i t va l ue of t h e s u ppb: voHa,ge tD v e i ls
1 . 2 1 . A coal m i n i ng
J .22.
n
"d r 8 g - l i n c "
l
1 . 17.
40
1.25.
1.26.
CHAPTER 1
BASIC CONCEPTS
Write the two nodal admittance equations, similar to Eqs. ( 1 .57) and 0 .58) , for the
voltages at nodes CD and ® of the circuit of Fig. 1 .23. Then, arrange the nodal
admittance equations for all four i ndependent nodes of Fig. l.23 into the Y bus form
of Eq. ( 1 . 6 1 ).
The values for the parameters of Fig. 1 .23 are given i n per unit as follows:
Ya
=
-jO . 8
Y"
YI:
=
-j4.0
= -)·0 .8
1";.
=
-j 4 . 0
Yd
=
-j8 . 0
Yc
= - 5 .0
j
Substituting these v a l u e s in the equations determined in Prob. 1 .25, compute the
voltages at the nodes of Fig. 1 .23. ]\;umcrically determine t h e corresponding Z b us
matrix.
CHAPTER
2
TRANSFO RMERS
Tran sformers a r e t h e l ink between t h e gen erators o f t h e power system a n d the
t r a n s m i s s i o n l i n e s , a n d be tween l ines of d i fferent voltage l evels. Tra nsmission
hnes opera t e a t n o m i n a l voltages u p to 7 6 5 kY l ine t o l i n e . Genera tors are
u s u a l l y b u i l t i n t h e range of 1 8- 24 kV wi t h s o m e a t s l i g h t l y h i g h e r r a t e d
voltages. Tr a n s fo rm e rs (l l s o l ower t h e v o l t a g es t o d i s t r i b u t io n l evels a n d fi n a l l y
fe r re s i d e n t i a l u s e a t 240 / 1 2 () V . They a re highly ( nea rly J OOO/C) efficient a n d
v e ry f c l i a b l c .
t h a t regu l a t e
vLl I ( a t�e fIl :.l g n i t u d c a n d p h d S C s h i ft i n g , a n d i n t h is a n d a l a t e r c h a p t e r w e s h a l l
sec h o w t h e s e reg u l a t i ng t r a n s fo r m e rs a r c Llsed t o co n t r o l t h e fl ow o f real a nd
I n t h i s c h ,t p r c r w e d i s c u s s t h e m o d e l i n g
of
t r a n s fo r m e r s a n d s e e
t h e gre a t
a d v �\ n u\ g (' s o f p e r- u n i t c a l c u l a t i o n s . W e a l so co n s i d er t r <1 n s fo r m e rs
rC d C (
i v c POW c r.
vu l t a ? e o f <1 gC' ne r a tor to t h e t r a n s miss i o n - l i n e vo l t a g e .
75 0 :vI VA, 5 25 / 2 2 . 8 kY.
F i g u re 2 . 1 i s t h e p h o t o g r a p h
of
,\
t h re e - p h a se t r , l n s fo r m e r w h ic h
2.1
THE I D EAL TRAt'; SFORM E R
same
magne t i c
ra ises t h e
T h e transformer i s rated
Transfo r m e rs con sis t o f t w o o r m o r e co i l s p l aced s o t h a t they a re
fl u x . I n
a
p ow e r t r a n s fo r m e r
t h e coils Ll rc
l inked b y t h e
p laced o n a n i ron core
42
CHAPTER 2
TRAN SFOR M ERS
2.1
Photograph of a three-phase t ransformer r a t e d 7 5 0 M V A , 525 /22.8 kV. ( Collrtesy Duke Power
FIGURE
Company .)
in order to confine t h e flux so t h a t a l most a l l of t h e fl u x l i nking any one coi l
l i n ks a l l t h e others. Seve ral coi l s may b e con nected i n series o r parall e l t o furm
one w inding, the coils of which may be s tacked on t h e core al ternately with
th ose of the other winding or windings.
Figure 2.2 shows how two w i n d ings may be p l aced on an i ron core to form
a single-phase transformer of the so-ca l I ed shell type. The number of tu rns in a
winding may range from several h u n d reds u p to several t housands.
W e begin our an alysis by ass u m i n g t h a t t h e fl u x varies s i n usoida l l y i n t h e
core and that the transformer i s ideal, w hi c h m eans t h a t ( 1) the permea bility J..L
of the core is infinite, (2) a l l of the fl u x i s confined to the core and t h erefore
links all of the tu rns of both wind ings, and (3) core l osses and w inding
resistances are zero. Thus, the vol tages e ] a n d e2 i nduced by the changing fl ux
must equal the terminal voltages v 1 a n d v 2 ' respectively.
We can see from the relationsh i p of the w ind i n gs shown in Fig. 2.2 t h a t
i n stan taneous vol tages e l and e2 i n duced b y t h e changing flux are i n p' h ase
2. 1
( . u uCu .�u .. t
/
+
+
+
)0
11
.
.
u
'mum
V
� Nl
c
t
--�----t��-+�
�
t
�--------� -�+--�
T
u
. . .
.
....
43
.
C
el
12
Flux paths
.
.
+ +
VI
+
--
THE I DEAL TRANSFO R MER
�
+ et2
•
J N2
c ��
��
t
t
�-------�--�-----�--��
�
V2
c
FI G VRE 2 .2
Two - w i n d i ng t r a n s fo r m e r .
when defined by the
+
and - polarity m a rks indicated. Then, by Faraday's l aw
d¢
N 1 dt
and
(2.1)
( 2 .2 )
1
w h ere cjJ is t h e i n s ta n t a n eous value o f the flux and N, an d N2 a re t h e n umber
o f t u rns o n w i n d i n gs
a n d 2 , a s s hown in Fig. 2 . 2 . The fl u x ¢ is taken i n the
p o s i t i v e direction o f c o i l 1 a c c o r d i n g to the right-hand rule, wh ich states that i f a
c o i l is grZlsped i n t h e right h a n d w i t h fi ng ers c ur l e d in the direct ion of current
ftmv , t h e t h u m b e xt e n d s in t h e d i rection of t h e fl u x . S i n c e we h ave assumed
s i n u s o i d a l va. r i a t i on of
d ivi d i n g Eq. (2. 1 )
the
by Eq.
we c a n convert
(2.2) t o y i e l d
fl u x ,
V1
U su a l l y ,
w e do not
=
=
E2
t h e vol tages t o
phasor form a fter
( 2 .3 )
kn o w t he d irection i n which the coils o f a transform e r
a re w o u n d . O n e device t o p r ov i d e winding i n formation i s to place a dot at t h e
e n d o f e a c h w i nding s uch t h a t a l l dotted end� o f windi ngs a r e positive a t t h e
same t i m e ; that i s , voltage drops from dotted ( 0 unmarked terminals of all
windings are ill phase. Do t s are s how n o n the two-winding transformer i n Fig. 2.2
44
+
CHAPTER 2
i1
t
TRANSFORMERS
-
vI
+1
-!
e1
FIGU RE 2 .3
t+
!-
e2
N2
Nl
Z
2
-
1+
J-
V2
Schematic represe n t a t i o n of a two-w i n d i n g t ra n sfo rmer.
according to this convention. We also note that the same resu lt is achieved by
placing t h e dots so that current flowing from the dotted term i n a l to the
unmarked terminal of each winding produces a magnetomotive force acting i n
the sam e d i rection i n the m agnetic circuit. Figure 2.3 i s a sch e matic representa­
tion o f a transformer and provides the same i nformation about the transformer
a s that in F ig. 2.2.
To find the relation between the currents i 1 and i2 in the windings, we
apply Ampere's law, which states that the m a gn etomotive force ( m m f) a round a
closed path is given by the line integral
¢ H ' ds
where i
H . cis
H
=
=
=
=
i
(2 .4)
net current that p asses through the area bounded by the closed path
m agnetic field i n te nsity
product of the
t a n ge n t i a l
compon e n t of
H
and
the incre m en t a l
distance ds along the p a t h
I n applying t h e l aw around e a ch o f t h e closed p a t h s of flux shown by dotted
lines in Fig. 2.2, i 1 is endosed N J times and the current i2 is e nc losed N2 times.
However, N, i, and N2 i2 produce m m fs i n opposite directions, and so
(2 .5 )
m inus sign would change t o plus i f w e h a d chosen t h e opposite d irection for
the current i2 • The integral of the field intensity H around the closed path is
zero w h en permeability is infinite. If this were not true, flux density (being equal
to
woul d be infinite. Flux d ensity must h ave a fi nite value so that a finite e
is induced i n each wind ing by the varying flux. So, upon converting the currents
Th e
/-LH)
2.1
THE I D EAL TRANSFO R M E R
45
to p hasor form, we h ave
( 2.6 )
(2.7)
=
I [ and 12 a re therefore i n p hase. Note then that I[ and 12 are in phase if we
choose the current to be positive when entering the dotted terminal of one winding
and Lea ving the dotted terminal of the other . I f the d i rection chosen for either
and
current is reversed, t hey a re 1 800 out of phase.
From Eq . (2.7)
a n d i n t h e i u e a l t r a n s fo r m e r
N2
1 1 = - /�
N[
(2.8)
.'.
I[
be zero i f '2 I S z e ro .
The w i n d i ng across w h i c h a n i mpeda nce or other load m a y b e connected i s
c a l l e d t h e seconda ry wind ing, a n d a n y circuit e l e m e n ts connected t o t h i s
winding are said to be on the secondary s i d e o f t h e transformer. Similarly, the
win d ing which i s toward t he source of energy is called the primary winding on
the primary side. I n t he power system ene rgy often will flow in e i t her d i rection
t hro u gh a transformer a n d the d esign a tion of primary and secondary l oses its
m e aning. These terms arc in genera! use, however. and we shall use them
wherever they do not cause confusion.
If an impedance 2 2 is connected across winding 2 of F igs. 2.2 or 2.3,
must
V2
Z = -
a n d s u b s t i t u ti n g [or V2 a n d 1 2
12
2
the
v a l u es
(2 .9)
fou n d from Eqs.
(2. 3) a n d (2.7) gives
(2. 10)
T h e i ill li c d a n c e a s m e a s u r e ci
< I c ross
2'2
=-
t h e li r i m a ry w i n d i n g i s
.V[
II
=
( N] 1
N2
2
Z')
then
(2.11)
�
the impedance connected t o t h e secondary side is referred to the primary
side by m ultiplying the i mpedance o n the secondary side of the transformer b y
t h e s q u a r e of the r a t i o of p r i m a ry to secondary vol tage.
Th u s ,
46
CHAPTER 2
TRANSFORMERS
We should note also t h a t VJ;+' a n d V2 Ii are e qual, as shown by the
following equation, which a g a in m akes use of Eqs. (2.3) and (2.7):
( 2 . 1 2)
So,
(2.l3)
which m e a ns that the complex power input to t h e primary winding equals the
complex power output from the secon d a ry w i n d i n g s ince we a re consi dering an
ideal transformer.
1200 � V and ' \ = 5/ 30° A w i t h a n i m p e d a n ce 22 c o n e c t e d a c r oss
2, find V2 J '2' Z 2 , a n d t he impeda n c e Z � , w h i c h i s d c l i n cd a s t h e va l u e of
Exa mple 2. 1 . If
VI =
NJ
=
2000
and
N 2 = 5 UO i n
t h e c i rc u i t o f r i g . 2 . 3 , a n d
n
-
winding
22
if
referred to the p r i m a ry side of t h e t r a n sfo r m e r .
Solution
N2
V2 = _ VI =
Nt
500
/ Oc )
2000 ( 1200�
=
300LQ:
V
A l tcrn ativcly,
2.2
MAGNETICALLY C OUPLED COILS
The ideal transformer is a first step in study ing a p ractical transformer, w here
(1) permeability is not infinite a n d inductances are t herefore finite, (2) not a l l
the flux linking any one of t h e windi ngs l in ks t h e other windings, (3) wind ing
resistance is present, and (4) l o ss e s occur in t h e i ron core due to the cyclic
changing o f d irection of the fl ux. As a second step, let us consider the two coils
of Fig. 2.4 which represent the wind ings of a transformer of the core type of
,
+t
L )
VvV
-
+f
Tl
v)
-t
•
c
IJ
"
---�--i
,
.
-
-t
¢2j · · · · · ····· · · · . .
f L 21 \ C1
�
� L' j
�
C ---'-cIi�-h
f :�2 -� S_N2
LII L 21 L22 � :
- .......
e)
.. . . . . . . . . . . . . . . . . . . . . .
"
'.
,.
.
. . . . . . . . .�
rP . . . . . . . . . . . .
L:
-
(a)
.............�
b
..
+
¢ ) 2 . . . . . . . . . .. . . . . . . . . . . .
. . . . - ���
'.
<....
: h
f-- ';--1-'
C
:
I---
1
I
cc)
(b)
Mutu ally co up l ed coils with: ( a ) mutual flux due to currents
( c ) l eakage flux rP2 1 a n d mutual fl ux rP ) 2 d u e to i 2 alon e .
Figure 2.4
I)
and i2;
(b)
lea kage fl ux ¢ ! I and m u t u a l fl u x
4> 2 )
d ue t o
I)
alone;
48
CHAPTER 2
TRANSFO R M E RS
construction. For the moment we con tinue to n eglect l osses i n the iron core, but
t h e o t h e r t hree physical characteristics of the practical transformer a re now
considered.
In Fig. 2.4 the d irection of curren t i 2 is c hosen to produce flux (according
t o t h e right- hand rule) i n the same sense as i I when both c urrents are either
pos itive or negative. This choice gives positive coefficients in the equ a tions
which fol low. Later we return to the d i rection chosen for i 2 in Fig. 2.2. The
current i I acting alone produces flux cP I I ' w h ich has a mutual compon e n t <P 2 1
linking both coils and a small leakage component <P I I linking only coil 1 ) as
shown in Fig. 2.4( b). The flux l i n ka ges of coi l 1 d u e to current i I acting alone
are given by
1.
( 2 . 1 4)
wh ere N I is the number o f turns a n d L I I is the self-inductance o f coil Under
the same condition of i l acting a lone the fl ux l inkages of coi l 2 are given by
( 2 .1 5)
w here N2 i s t h e n umber o f t urns of c oil 2 a n d L 2 1 is the mutual i n ductance
between the coils.
Similar definitions a pply when i 2 acts a lone. It produces fl ux <P 22 ' w hich
also has two components - l eakage flux <P 2 / l in ki n g only coil 2 and mutual flux
cp J 2 l inking both coils, as shown in F i g. 2 . 4( e ). The fl ux l i nka ges of coil 2 due to
i2 acting a lone a re
( 2 . 1 6)
w here L 22 is the self-inductance of coi l 2, a n d the flux l i nka ges of coil
i 2 a lone are
1 due to
( 2 . 17)
. When both currents act together, the flux l i n kages add t o give
( 2 . 1 8)
The order of the subscripts of L j 2 a n d L 2 1 is not i mportant since mutual
i nductance is a single reciprocal property of the coils, a n d so L 1 2 = L 2 l . The
d i rection o f the currents and the orientation of the coils determine the sign of
m u tu al inductance, which i s positive i n Fig. 2.4 because i l and i 2 are ta�en to
m a gnetize in the same sense.
2 .2
MAGNETICALLY COUPLED COILS
49
When the flux linkages change with time, the vol tage d rops across the coils
in the direction of their circulating currents are
( 2 . 19 )
( 2 .20)
The positive signs of Eqs. (2. 1 9) a n d (2.20) are usually associated with a coil that
is absorbing power from a source as i f the coil were a load. For instance, in Fig.
2.4 if both V 2 and i 2 h ave positive values simultaneously, then instantan eous
power is being absorbed b y coi l 2. I f the voltage d rop across coi l 2 is n ow
- v � , we h JVC
r e v e rse d so t h a t v'2
=
( 2 .2 1 )
For positive instantaneous values of v� a n d i 2 power is b e i n g supplied b y coil 2 .
T h u s t h e n e g a t i ve signs of E q . (2.2 1 ) are characteristic of a coil acting as a
generator del ivering power (and energy ove r time) to an external load.
I n the steady state, with ac vol tages and curre n ts i n the coils, Eqs. (2. 1 9)
a n d (2.20) assume the phasor form
,
(2.22)
(2.23)
H e r e we l I s e
l owe rcase
:: ij
to
i m p e d a n c e s /'j ' I n ve c t uI - I 1 1 a t r i x
d is t i ngu ish
t h e co i l
Co rm I � q s . ( 2 .22)
and
i m red a n ce s
( 2 . 2J )
f ro m
become
node
( 2 .24)
that t h e V 's a re t h e vol tage d rops across the term i n a l s of
the coils and the I 's are the circu l a t i ng cu rrents in the coils. The inverse o f t h e
coefficient mat rix is t h e m atrix of a d m ittances d e n o t e d by
We s h o u l d a l so n o t e
[Yll
Y21
( 2 .25 )
50
CHAPTER 2
Multiplying
TRAN S FO R M ERS
Eq. (2.24) by the admittance matrix gives
( 2 .2 6 )
Of course, t h e
and z parameters w i t h the s a m e subscripts are n o t si mple
reciprocals of each other. I f the term i nals of c oil 2 are ope n, then setting 12 = 0
in Eq. (2.24) shows that the open-circuit input impedance to coi l 1 i s
y
( 2 . 27)
If the term inals of coil 2 are closed , then
short-circuit input impedance to coil 1 is
V2
=
0 and Eq. (2.26) shows that the
( 2 .28)
By substituting the expressions defining the z if from Eqs . (2.22) and (2.23) into Eq. (2.28), the reader can show that t h e apparent reactance of coil 1 is r e duced
b y the presence of closed coil 2. In Chap. 3 a similar result is fou n d for the
synchronous m achine under s hort-circuit con ditions.
An important e quivalent c ircuit for the m u t ually coupled coils is shown i n
Fig. 2.5 . The current o n the coil 2 side appea rs a s 1 2 1a and t he term i n a l vol tage
as a V2 , w here a is a positive constant. O n the coil 1 s i de V[ and I[ a re the same
as b e fore. B y writing Kirch hoff's vol t age equa tion around the path o f each of
t h e c urrents 11 and 1 2 1a in Fig. 2.5, t h e reader shou l d find t h a t Eq s . (2.22) and
(2.23) are satisfied exactly. The i nductances i n b rackets in Fig. 2.5 a re t h e
leakage inductances L l l a n d L 2I o f t h e coils i f we l e t a = N , IN2 . This i s shown
FI G U RE 2.S
An ac equivale n t circuit for Fig. 2.4 with seco n d a ry c u rre n t a n d voltage r e d e fi n e d a n d
a =
N1 /N2 .
2.3
THE EQUI VALENT CI RCUIT OF A S I NGLE·PHASE TRANSFORMER
51
FIGURE 2.6
The equiva lent circu i t of Fig. 2.5 with i nductance paramete rs renamed.
from E qs. (2. 1 4) through (2. 17) as foll ows:
L it
�
L I I - aL 2 1
L 2 1 #::. L n
-
L 1 2/a
=
=
N I 4> I l
il
-
-
N2 rP 22
12
N I N2 cP 2 1
N2
II
N2 N l cP 1 2
--
NI
[2
=
'v
I
. ( <P I I - <P 2 I )
{ \
=
"V-,
( 2 .29)
.
q; I {
1 2 "'-
�
( <P 22 - cP I 2 )
( 2 .30)
CP u
wh er e 4> 1 1 and cP 2 1 are the l eakage ftuxes of the coils. Likewise, with a Ni lN2 ,
t h e s h u n t indu ctance aL 2 1 i s a magnetizing inductance associated w i t h t h e
m u t u a l fl u x rP 2 1 l inking t h e coi l s d u e t o i l since
=
( 2 .3 1 )
Defining t h e se ries leakage reactances X l = w L l ! and x 2 w L 2 ! , a n d the shunt
magnetizing susceptance IJ", ( (U a /' 2 1 ) . 1 , leads t o the e quival e n t c ircuit of
F i g . 2 . 6 , \vhich i s the basis of the equivalent circu it of the practical tra nsformer
in S ec. 2 . 3 .
=
2.3 THE EQUIVALENT CIRCUIT
OF A SIN GLE- PHASE TRANSFORMER
=
The equ ivalent circu i t of Fig. 2.6 comes close to matching the p hysical c harac­
teris t i cs of the practical transformer. However, i t has three deficie ncies: (1) I t
does n o t reflect any curre n t o r voltage t ransformation, (2) i t d o e s n o t provi d e
for e lectrical isol ation of t h e primary from t h e secondary, a n d (3) it does n o t
accoun t for the core losses.
52
CH APTER 2
TRANSFORMERS
When a s inusoidal voltage is applied to the primary winding of a practical
transformer on an iron core with the secondary winding open, a small cu rrent
It: called the exciting current flows . The major component of this cur r e n t is
called the magnetizing current , which corresponds to the current t hrough the
magnetizing susceptance B m of Fig. 2.6. The magnetizing current produces the
flux in the core. The much s m a l ler component of ]E ' which accounts for losses
in the i ro n core, leads the magnetizing cu rrent by 90° an d is not represented i n
Fig. 2 . 6 . T h e core losses occur d ue , fi rst, t o t h e fact that t h e cyclic changes o f
the d i r e c t i on o f t h e flux i n the iron require energy which i s d issipated a s h eat
and is called hysteresis loss . The second l oss is due to the fact that circula t i n g
currents a r e indu ced in t h e i ro n due to t h e changing flux, and t hese currents
2
produce an I I I R loss in the i ro n cal led eddy-current loss . Hysteresis loss is
r educed by the use of c e rtain high g r a d e s of a lloy steel for t he core . E d d y- c u r ­
re nt loss is re d uc e d b y b u i l d i ng up the c o r e with l a minated s h e e t s of st e e l . W i t h
the secondary open, t h e t r a n s fo r m e r primary c i rcu i t is s i mply one of very high
inductance due t o the i ron core . In the equivalent c i rcuit IE i s taken fu lly i n to
accou nt by a co n d u c t a n ce G( i n p a ra l l e l with the magnetizing s uscep t a nce Bill >
as shown i n Fig. 2.7.
In a well -designed transforme r the maximum flux density in the core
occurs at the knee of the B-H or satura tion curve of the transformer. So, flux
density is not linear with respect to field intensity . The magnetizin g c u r r e n t
cannot b e s in usoidal if it is to produce sinusoidally varying flux required for
inducing sinuso idal voltages e I an d e2 when the applied voltage is sinusoidal.
The exciting current IE will h ave a t hird harmonic content as h igh as 40% a n d
l esser amoun ts o f higher h armonics. Since JE is s m a l l compared to rated
current, it is treated as sinusoidal for convenience, a nd so use of Gc and Bm is
accept a b l e in the equivalent circuit.
V ol tage and current transformation and electrical isolation of the primary
from t h e secondary can be obtained by adding to Fig. 2.6 an ideal transformer
rl
1
-j
+
VI
FIG UR E 2.7
Xl
a2 X 2
a2 r 2
r
12
a:l
•
T-
�
•
+
1
j-
V2
N1
N2
Ideal
Eq u i va l e n t c i rc u i t for a s i n g l e - ph a s e t r a n sfo r m e r w i t h a n i d e a l t r a n s fo r m e r o f turns r atio a =
+
N,1 /N2.
2.3
53
TH E E Q U I VA L ENT C I R C U I T OF A S I N G LE · P H A S E T R A NSFO R M E R
FI G U RE 2.8
Transfo r m e r e qu iv a l e n t circuit w i t h magnetizing current n e ­
glected.
with t urns ratio a = N I l Nz , as shown in Fig. 2.7. The location of t he ideal
transformer is not fixed. For instance, it may be moved to the left past the series
ele m e n ts a Z r2 and a 2 x 2 ' w hich then become the winding resistance 'z and the
leakage reactance x 2 of the seco n dary w i n d i ng . This is i n keeping with the rule
e s t a b l i s h e d fo r the ideal transformer in Sec. 2 . 1 that whenever a bran ch
i m p e d a n c e is referred from a g i v e n s i d e to the oppos i te side of a n ideal
t r ,1n s fo r m e r , i t s i m p e d a n c e va l u e i s m u l t i p l ie d hy the sqU : HC o f t he rat i o of the
t u r l l S o n t h e o p [! ( ) s i t e s i d e
all
t o t h e t lI r n s
( ) Il
t he g i v e n s i d e .
1l1 ,ly b e o m i t t e d i ll t h e e q u i v ,d e n t c i rcu i t i f w e refer
e i ther t h e h i gh- or the l ow-vol tage s ide of the t ra nsformer. For
i n s t a nce, in F i g . 2.6 we say that a l l vol t ages, curr ents, a n d i mped a nces are
re fe rr e d to the primary cirel.! it o f t h e transformer. W i thout the i d e a l t rans­
fo r m e r , w e h ave t o he ca refu l not t o create u n n ecessary short c i rcuits w h en
developing e q u iva l e n ts for m u l tiwi n d ing transformers.
Ofte :1 we negl ect exc i t i n g current because it is so sma l l compared to the
u c;ual load currents and to s i m p i ify the c i rcuit further, we let
T h e i d e a l t ra n s fo r m e r
q u a n t i t i es t o
( 2 .32)
to obt a in the equ ivalent c i rcu i t of Fig. 2 . S . A l l impedances and voltages in the
part of t h e c i rcuit connected t o the seco n d a ry t erm inals must now be referred to
regulation i s
t h e p r i m a ry s i d e .
f/o/:(/ge
t u d e a t t h e I O (l d k r m i n (l l s
p c rc c n t u f fu l l - l o a d vo l t a g e
c q l l ;l t i O I l
d i fference between t h e voltage magni­
o f t h e t r (l Ils fo rm c r a t fu l l l o a d a n e! a t no load in
w i t h i n pu t v o l t a g e h e l d co n s t a n t . I n t h e fo rm o f a n
dcflned
a s t he
Perce n t regu l a t ion
( 2 .33)
I V::
l i s t h e magni t u d e of l oad voltage V7
. NL
m (l g n i t u cJ e of V2 a t fu l l load with I V I I constant.
w h e re
'
Exa mple 2 . 2 . A s i n g l e - p b ase t r a nsfor m e r h a s
at no load and I Vz
2000
a re X I
= R.O fl
I is
the
t u rn s o n t h e p r i m ary w i n d i n g
a n d 5 0 0 t u r n s on t h e s ec o n d a ry. W i n d i n g r e si s t a n c e s
0 . 1 2 5 n . Leakage r e a c t a n c e s
FL
3re r1
and x2 = 0.50
fl.
= 2.0 fl
an d
r2
=
T h e r e s i s t a nce l oa d
54
CHAPTER 2
TRANSFORMERS
j16 Q
4 Q
FIGURE 2.9
Circuit for Exa m p l e 2.2.
22
12 n . If app l ied voltage a t the terminals of the prim ary w i n d i n g is 1 200
V2 and the voltage regulation. Neglect magnet i z i ng curre n t .
is
find
V,
Solution
2000
N
a = -I = -=4
500
N2
RI
= 2 + 0 .125 ( 4 ) 2 = 4 .0 n
XI
= 8 + 0 .5( 4 ) 2 = 16 n
2�
= 12 X
LQ:
(4) 2
The equivalent circu i t is shown i n F i g .
1 =
I
1 200
1 92 +
a V2 = 6 1 0/
.
V2 =
= 1 92 n
2.9, a n d we c a n calcu late
4 + j1 6 = 6 . 1 O!- 4.6r
4 6r
/
-
.
x
1 1 7 1 .6 - 4 .67 :
Voltage regulation
4
=
1 92
= 1 17 1 .6/
A
-
4 . 6r
= 292.9/ - 4 .6r
1 200/4 - 292 . 9
292 . 9
V
V
= 0 .0242 or 2.42%
The p arameters R and X o f 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. Usually, the low-voltage
side is short-circuited and just enough voltage is applied to the h igh-voltage
terminals to circulate rated current. This is because the current rating of the
source supplying the high-voltage side can be smaller. Voltage, current; and
2 .3
TH E EQUI VALENT CI RCU IT OF A SING LE-PHASE TRANSFO R M E R
55
power input are determined_ Since only a smal l vol tage is required, the exciting
c urrent is i nsignificant, and the calculated impedance is essentially equal to
R + jX.
Exa mpJe 2.3. A single-phase tran sformer is rated 1 5 M VA, 1 1 .5/69 kY. I f the 1 1 .5
kV w i n d i ng (designated winding 2) is short-circuited, the rated current flows w h e n
the vol tage a p p l i e d to wi nding 1 is 5.50 kV. T h e power i n p u t is 1 05 . 8 kW. F i n d R}
a n d X I in ohms referred to t h e h i g h -vol tage wind ing.
Solution.
Rated curre n t for the 69-kY winding has the magnitude
1 5 , 000
69
Th e n ,
( 2 1 7.4r R I
')
RI
IL)
=
1 05 ,800
=
2 . 24 n
5500
=
--
2 1 7 .4
=
=
2 1 7 .4
A
25 . 3 0 D
The example illustrates t h e fact that the w i n ding resistance m ay o ften be
omitted i n the t ransforme r equivalent circuit. Typical ly, R is l ess t h a n 1 %.
Al though exc i t i ng current may be n eglected (as in Exam ple 2.2) for most power
system calculat ions, Gc jBm can be calculated for t h e equivalent circuit by a n
open · circuit test R ated voltage i s applied to the l ow-voltage termin a ls, a n d the
p mver input and cu rrents are measu red . This is because the voltage rating o f
t he source s u p pl yi n g t h e low-vol tage s i d e c a n be sma l l e r. The measured
i m p e d a nc e i n c l u des the resi st a n ce a nd leakage reacta nce of t h e wind ing, but
t h e s e v a l ue s are i nsignificant when compared to l /( Ge jBn).
-
_
-
2.3 the open-circuit test with 1 1 .5
kY a p p l i e d resu l t s in a power i n p u t of 66.7 kW a n d a current of 30.4 A. Find t h e
v a l u e s of Gc a n d E m referred to t h e h igh-vol tage w i n d i n g 1 . W h a t i s t h e efficiency
of t h e tra nsformer for a l o a d of 12 MW at 0.8 power-factor l a g g i n g at rated
vol tage?
Exa m p l e 2.4. For the transfor m e r o f Exa mple
The t u rn s ratio is a = N \ /N2 = 6. Measurements are made o n the
low-vo ltage s i d e . To t r a nsfer shunt admittance Y
G c - jAm from high-voltage
2
side 1 to l ow-vo l tage s i d e 2, m u l t i p ly by a since we wou l d divide by a 2 to tran sfer
Solution.
=
56
CHAPTER 2
TRANSFORMERS
impedance from side 1 to s i d e 2. U n d e r open-circuit t e s t con d i tions
Gc
I YI
=
Bm =
Under rated
I V2 1
1 12 1
V I YI 2
X
-
= 1 4 .0 X 1 0 - 6
30.4
= 1 5
a2
1 , 00
1
G}
10-6
X
S
1
7 3 .4
36 =
-/73 .42
-
X
10 6 S
1 4 . 0 2 = 72 .05
1 0 (, S
'
co n d i t i o n s t h e t o t a l l oss is a p p rox i m a t e l y t h e s u m of s h o r t -c i rc u i t a n d
o p cll-circ u i t test losses, � n d s i nce c t l i c i e ncy is t h c r;l l i o o f
k i l ow a tts, w e h ave
Efficiency
2.4
X
1 2 , 000
=
+ ( 1 05 .8
----
1 2 ,000
+ 66 .7)
X
t i l e u u t r HI l t o t h e i n p u t
1 00 = 9 8 . fi %
This example illustrates the f a c t that G c i s so m u c h s m a l l e r than Bm that i t m ay b e
omitted. B m is a lso very small s o t h a t IE is oft e n n e g l ected e n t irely.
PER-UNIT IMPEDANCES
IN SINGLE-PHASE
TRANSFORMER CIRCUITS
The ohmic values of resistance and l e akage re actance of a transformer depend
on whether t hey are measured on the h i gh - or low-vol tage side of the trans­
former. If they are expressed in per u n i t , t h e base kilovol tamperes is understood
to be the kilovoltampere rating of the transformer. The base voltage is under­
stood to b e the voltage rating of the l ow-voltage wind ing if the ohmic values of
resista nce and leakage reacta nce are refe rred to the low-voltage side of the
transformer. Likewise, the base volt age is taken to be the voltage rating of the
high-vol tage wind ing if the ohmic values are referred to the high-vol tage side of
the transformer. The per-unit impedance of a transformer is the same regard­
less of whether it is determined from ohmic values referred t o the high-voltage
or low-voltage sides of the transfo rmers, as show n by the following example.
2.S. A s i n gle-p h as e t ransfo r m e r is rated 1 1 0/440 V, 2.5 k VA. Leakage
reactance measured from the low-vo ltag e side i s 0.06 fl. Determine leakage
reactance in per u n i t .
Example
Solution. From E q . ( 1 .46) we h ave
Low-voltage base i mp e d a nce
0 . 1 1 0 2 X 1 000
=
2.5
= 4 .84 fl
2.4
P E R - U N IT I M P ED A NCES I N S I N G L E - P HA SE TRANSFO R M E R C I R C U ITS
57
In per u n i t
X
0 .06
= -- =
4 . 84
0 .0 1 24 per u n i t
If l eakage reactance h a d been measured on t h e high-voltage side, t h e value
wo u l d b e
X
I Iigh-vo l t agc b a s e i m pe d a n c e
I n per unit
X
= -- =
0 . 96
7 7 .5
=
0 .0
( )
6 - =
440 2
110
0 .440 2
=
X
1 000
2.5
0 .9
6n
= 77.5 n
0 .0 1 24 pcr u n i t
great advan tage i n m aking per-un i t computat ions i s real ized by the
p roper selection o f different b ases for circuits connected t o each o t h er t hrough
a transfonner. To achieve the adv an tage in a single-phase system, the voltage
bases for the circuits connecred through [he transformer must have the same ratio
as the turns ra tio of the transformer l1:indings. With such a selection of voltage
bases and the same kilovo l t a mpere base, the per-unit value of an i mpedance
will be t h e same when it is expressed on t h e b ase selected for its own side of the
transformer as when i t is referred to the other side of the transform er a nd
expressed on the base of that s i d e.
So, the transformer i s represented completely b y i t s imped a n ce (R + jX )
i n per u ni t when magnetizing curre nt is n eglected . No per-unit volt age transfor­
m a t i o n occurs when t h i s sys t e m is used, a n d the c u rrent wi l l a l so h ave t h e same
p e r - u n i t v a l u c o n b o t h s i d e s o f t h c t r a n s fo r m e r i f m a g n e t i z i n g cu rrent is
A
neglected .
Exa m p l e 2.6. T h r e e p a r t s o f a s i n g l e - D h ase e l e c t r i c system arc desi g n a t e d A , B ,
u n d (' a n d a rc con n ec t e d
2 . 1 0 . T h e t ra nsform e rs a r c
A -B
B-C
1 0,000 kY A,
1 0 , 000 kYA,
t o e a c h o t h e r t h r o u g h t ra n s fo r m e rs, as
rated as
shown i n Fig.
[o l l ows:
13.8/ 138 kY, leakage r e a c t a n c e 1 0 %
138/69 k Y , leakage r e a c t a n ce 8%
138
I f t h e b a se i n c i rc u i t B i s c h os e n a s 1 0 , 000 kYA,
k Y , fi n d t h e per-u n i t
i m p e d a n ce of t h e 300-D r es i s t i v e l o a d i n c i r c u i t C r e fe r r e d to c i r c u i ts C , B , a n d
A . D r aw t he im p e d a n c e diagram n e g l e c t i ng m a g n e t i z i n g current, transformer
r es i s t a n c es, and l i n e i m p e d a n c e s
.
58
CHAPTER 2
TRANSFO RM E RS
1-10
2-1
A
B
A-B
300n
B--{;
FIGURE 2 . 1 0
Circuit for Examp le
2.6.
Solution
Base vol tage for ci rcu i t A :
Base voltage for circu i t C :
Base imped ance of circu i t C :
Per-unit impedance of load i n circ u i t C :
0.1
x
O.S X
138
=
1 3 . 8 kV
138 = 69 k V
692 X 1000
10,000
300
476
- =
=
476 n
0.63 per u n i t
Because the selection of base i n various p arts of the system is d e t e r m i n e d by
the t u r n s ratio of the transformers, and b ecause the base kilovol tamperes i s t h e
s a m e in all parts o f the sys tem, the per-u n i t i mpedance of t h e l o a d re ferred to a n y
p a rt of the sys tem w i l l b e t h e s a m e . This i s verified a s fol l ows:
Base impedance o f c i rcui t B :
Impedance o f load referred to circuit B :
Per-un i t impedance of load refe rred to B :
Base i m pedance of c ircu i t A :
Impedance of load referred t o c i rcu i t A :
Per-un i t imped ance of load referred t o A :
1382 X 1000
10,000
=
1900 n
300 X 22 1200 n
=
1200
1 900
-- =
0. 63 per u n i t
13 .82 X 1 000
10,000
--- -- =
19 n
300 X 22 X 0.12 12 n
12
19
- =
=
0.63 p e r u n i t
=:=10.63+)0
j O. l
2 .5
T H R E E-PHASE TRANSFO R MERS
59
j O.08
Imped ance d i agram for Example 2.6. Impeda n ces are
m a rked in per u n i t .
F I G U R E 2. 1 1
S i nce t h e chosen bases for k i l ovol ts a n d k i l ovol tamperes agree w i t h t h e tra nsformer
rat ings, the t ransform e r reacta nces in pe r u n i t a re 0 . 08 and 0. 1 , r esp e ct iv ely .
Figu re 2. 1 1 IS t h e r e q u i r e d i m p e d ance d i agram w i t h impedances m a rked in per
unit.
Because o f t h e advantage p reviously pointed o u t , t h e p r i n c i p l e demon­
st r a t e d i n t h e preceding exa m p l e fo r se l ecting b ases in various p arts of t h e
s i n g l e - p h ase system is always followed i n m a k i n g co m p u t a t i o n s i n per u n i t . That
is, the k i/olJo/ta mpere base sh o u ld h e the saine in all pa rts of t h e system , a n d the
select ion of ri t e base ki/o (:o/rs il l
ki/o L'o/rs
to
the system d('(ermines the base
ra tios of the rTan'<'jormers , to the
Th i s p r i nc i p l e il i l o w s liS to c o m b i n e on o n e
De assign ed , acc()rdill�
of r h e system .
pa rt of
t o the
OtiC
tUTIlS
d i a g r a rn t h e p c r - u n i t i m p e d a n c e s 0 1 t h e e n t i re sys t e m .
or/lcr parts
2.5
i mpedance
THR.E E - PfLA.S E TRAN SFORM ERS
Th ree i d en t ic a l s i n g l e - p h a s e t ra n sforme rs may be connected so that the t h re e
w i n d i ngs o f o n e voltage rat i ng a re L\-connected and t h e t h r e e w i n d i n g s o f t h e
o t h e r vol t age rcl t i n g (l IT Y - c o nn e c t e d t o form a t h r e e - p h a s e tran sformer. S uc h a
transformer is s a i d to be con nected 'y'-6. o r 6 -Y . T h e o t h e r possi ble connections
a r e Y- Y a n d l'l - L\ . I f e a c h o f the t h ree s i n g l e-ph as e transformers h a s t h re e
w i n d ings ( a primarj, secondary, and t er t i a ry) , two sets m ig h t b e conn ected i n Y
a n el o n e i n 6 , or two cou l d b e 6. - co n n ec t e d with one Y - co n n e ct e d . I nstead
u s i n g t h ree i d e ;l t i c a l s i n gi e - p h a s e t r a n s fo r m e rs, a m o re usual u n i t is a three­
p h a s e t a n s fo r m e r w h e r e a l l t h ree p h a s e s a re o n t h e s a m e iron structure. The
theory is t h e s a m e for a t h ree-phase t r Cl n s fo r m e r a s fo r a t h re e-ph ase b a n k of
s i n io' : e - p h a s e t ra n s fo rrn c r s . T h e t h r e e - p h Cl s e u n i t has t h e a dv a n t a g e of requi r i ng
of
r
co r e ,
more econ omical t h a n t h re e
s i n g le-phase u n i ts h ave t he
a d v a n t a g e of r e pl a c e m e n t o f o n ly one u n i t o f t h e t h ree-phase bank i n case of a
fa i l u re rather t h a n l o s i n g t h e whole th ree-ph ase bank. I f a fai l u re occurs i n a
6 - 6. b a n k c o m p o se d of t h r e e s e p a r a t e u n i t s, o n e of t h e s i n g l e - p h as e t ra nsform­
e rs c a n b e r e m ove u and t h e re m a i n i n g t w o wi l l s t i l l o p e r a t e a s a t h ree-phase
trans former a t a red uced k i l ovol t a m pere . Such an ope r a t i o n is called open delta .
For a s i n g l e -· p h ase t r a n s for m e r we can con t i n u e to p l ace a dot on one e n d
o f e ac h w i n d i n g , o r a l t e r n a t ively, t h e dutted e n d s m a y b e m a r k e d H I for t h e
h i g h -v o l t a g e w i n d i ng a n d X l for t h e low-voltage wind i n g . The oppos i t e e n d s a r e
t h e n l a b eled H :. and X 2 , respect ively.
F i g u r e 2 . 1 2 shows how t hree s i n g l e - phase transformers a re conn ected to
form a Y - Y th r e e - p h as e t r a n sfo r m e r b a n k . In this text \\l e s h a l l u se capital
less
iron
to
fo r m
the
and
is
t h e r e fo r e
s i n g l e - p h a se t1 n i t � (l n u occ u p i e s l e ss s p ' l c e . T h r e e
60
CHAPTER 2
TRANSFORM ERS
IB
A
Hl
C
H2
H3
N
-
•
-
-
-
•
n
x�
b
( a ) Y-Y connection d i a g ram
A �------�
B
-0--..
-
•
N
C
H3
--0-------'
X2
.-------c>-
-
-
b
,----0--- a
n
,-1 -------------� c
(b) Alternate form of c onnection d i ag ram
FI GURE 2 . 1 2
Wiring d iagrams for Y - Y tra nsformer.
l e tters A , B, and C to identify the p hases of the high-voltage wind ings and
lowercase l etters a, b, and c for the low-vol tage w indings. The h igh-voltage
terminals of three-phase transformers are marked H I ' H 2 ' and H 3' and the
low-vol tage terminals are marked X l ' X 2 , and X3. I n Y - Y or � - � transfo rmers
t h e m arkings are such that voltages t o n eutral from terminals H I ' H 2 , and H3
are in phase w ith the voltages to n e u tral from terminals X" X 2 , and X3,
respective ly. Of course, the .6. w i n d ings have n o neutral, but the part of the
system to w h ic h the � winding is c onnected will have a con nection to groun d .
Thus, the g round c a n serve a s t h e e ffective n eutral u n d e r balanced conditions
and voltages to neutral from the term inals of the � do exist.
To conform with the American standard, the terminals of Y-� and � - Y
transformers are l abeled so that the vol t a ges from H I ' 2 ' and H3 to n e utral
lead the voltages to neutral from X l ' X 2 , and X 3' respectively, b y 30° '. W e
consider t hi s p hase shift more fu l ly i n the n ext section .
H
2.5 THREE·PHASE TRANSFO R ME R S
61
'r
I
1. �_�'------'
66 kV
6.6 kV
3.8 1 kV
Fl G U RE 2 . 1 3
y-y t ra n s form e r r a t e d 66/6.6 k V
Figure 2 1 2( h ) p r ov i d e s t h e S il nl C i n fo r m a t i o n as ! -' ig. 2 . 1 2( a ). W i n d i n gs of
t h e p r i m a ry , \ n d s e c on d a ry , w h i c h a r c d ra w n i n p , \ r a l l c l d i r e c t i o n s i n Fig.
2 . 1 2( 6 ) , a r c for t h e same s i n g le-ph a s e t r a n s fu r m e r or o n t h e s a m e leg o f a
t h re e p h a s e t r a n s for m e r . Fo r i n s t a n c e , t h e w i n d i n g from A to N is l i n ke d by
t h e s a m e A u x as t h e w i n d i n g from (l t o 1 1 , , \ Il d V! N is i n p h a se w i t h V;" I . T h e
d iagrams o f F i g . 2 . 1 2 ( b ) a r e wiring diagrams only. They a r e not phasor d i a
-
­
F i g u re 2 . 1 3 i s
-
g r Zl m s .
a schematic met hod of i n dicating w i n d i n g con nections o f a
t h ree phase tra nsformer. Voltages are shown for a 6 6 j 6 . 6-kV, Y-Y t ransfor m e r
supplying 0.6-0 re s is t o rs or impeda nces. Figure 2 . 1 3 sho ws a balanced system i n
which e ach phase can b e tre a t e d s e p a ra t e l y , whether o r not t h e n e u t ra l p o i n t s
a r e c o nn e c t ed . Then , i m p e cJ a n c es wou l d t ransfer f r om t h e low-voltage to t h e
high-vol tage s i d e b y t h e square o f t h e ratio o f l i ne-to-n e u t ral voltages, w h i c h i s
t h e s a m e a s t h e square o f t h e r a t i o of l i n e-to-line vol tages; t h a t is,
0.6
r �)
, �8 1
\
_� . 8 1
2
=
0 .6
( -(i6- )
)
6.6
2
=
60 D
t h e resis to r s
w i t h t i l e S;l llle () () · � V p r i m ; I I'Y, t i l e 6 w i n d i n g s wOl l l d he r;l l e ll (d) k V r(l t h e r t h an
3.8 1 kY. So fa r a s t h e vo l t age m agnitude a t t h e l uw-v o l t a g e t e r m i n a l s is
c onc e rn e d , th e Y - 6 t ra n s fo r m e r cou l d t h e n b e re p l aced b y a Y-Y trans­
for m e r b a n k h avi n g a n e ffe c t ive p h a s e - t n- n e u t ra l t u rns rat i o of 38 . 1 : 6.6/ /3 , o r
I f we h a d l I '; c d ; \ Y - � t r ,\ n S rl ) f fll C r t o o ht a i n (l . () � V , \ c ross
N2 j f3 ,
w o u l d b e s e e n by t h e p r i m a ry . So, we see t h a t t h e c r i t e r i on fo r the selection of
base v ol t a g e i nvolves the square of t h e ratio of r ne-to- line voltages a n d not t he
squ a re of t h e turns ratio of t h e i n d ividu a l windings of the Y -6 transformer.
This d iscussio n l eads to the conclusion that to transfe r the ohmic v a l u e of
i m pedance from the vol t age l evel on o n e side of a three-phase transfor m e r to
the voltage l evel o n the oth er, the m ul t i p l ying factor is t he square of the ratio o f
l i ne-to-line voltages regard l ess o f w h e ther t h e t ransformer con nection
i s Y-Y o r
,
NI
:
as s h o w n i n Ta b l e 2 . 1 ,
so
that the
same
6 0 - n r e s i s t ance p e r p h a s e
TABLE 2.1
Transferring ohmic val ues of per-phase impedances
from one side of a three-phase tra nsformer to another t
I
y-y
I
y-tl
- -+­
: vl n l
_
t_
'-_
_
_
_
_
_
tl -Y
t Secondary
load consists of b a l a nced Y -con n e c l e d i m peda nces Z" .
2.5
THREE-PHASE TRANSFO R M E R S
63
This is shown in Table 2. 1 , w hich summarizes the relations for the effective
turns ratio of the different types of transformer connections. Therefore, in
per-u nit calculations involving transformers in three-phase circuits we 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 kilovoltampere
base is the same on each side .
y-� .
Y-�
Three transformers, e ach rated 25 MVA, 38. 1 /3.81 kV, are
connected
with a bala nced load of three 0.6-.0, Y-connected resistors. Choose
a base of
MVA,
kV for t h e hig h-voltage side of the transformer and specify
the base for the low-voltage side. Determine the per-unit resistance of the load on
the base for the low-vo ltage s i d e . Then, d etermine the load resistance RL i n ohms
refe rred to the high-voltage side and the per-unit va lue of this resistance on t h e
chosen base.
Example
2.7.
75
66
� ,
13 X 38. 1 kV e q uals 66 k V the ra t i n g of the transformer as a
bank is 75 MV A, 66Y /3_8 1 kY. So, base for t he low-voltage side is
75 MY A , 3 8 1 k Y .
B y Eq . (J .54) b a s e i m p e d a n ce on t h e l ow vo l t a g e s i d e i s
.
Solution. S i n c e
t h ree-phase
( base
kVL L ) ='
base M VA J <i>
a n d on
the
Jow-\'oltage side
RL
B ase
i m p e d a nce on t he
=
( 3 .81 ) 2
--75
0 . 1 93 5 .0
.6
-.
0 . 1 935
0
=
h i g h -voltage
= 3 1 0 per u n i t
side
-75 =
( 66)
is
2
58 . 1 .0
-
Th e
=
res i s t a n c e referred to t h e h i gh - v o l t age side is
0.
R j-
6
=
(
66
-
3 .8 1
-1 8 ()
58 . )
)2
=
=
1 80 .0
3 . 1 0 per u n i t
The res istance R and leakage reactance X of a t hree-ph ase transformer
are me asured by the short-circu i t test as discussed for single-phase transform­
ers. In a three-ph ase equivalen t circuit R and X are connected in each line to
an ideal three-phase transformer. Since R and X will have the same per-u nit
value whether on t he low-voltage or the high-voltage side of t he ,transformer,
64
CHAPTER 2
-
TRANSFOR.\il E R S
the per-ph ase equivalent circuit will account for the t ra n s fo r m e r by the p e r uni t
impedance R + jX without the ideal tra nsformer, i f phase-shift is not i m p o r tan t
in t he calculations and all qua nti ties in the circ u i t are i n per u n i t w i t h the
p roper selection of base.
Tab l e A . 1 in the Appendix l ists typical values of transformer impedances,
which are essentially equ al to the lea kage reacta nce si nce the resista nce is
usually less than 0.0 1 per unit.
Exa m p l e 2 . 8 . A t h re e - p h ase t ra n s fo r m CJ' is r a t e d 4 0 0 M YA , 2 2 0 Y
k Y . The
Y - e q u iva l e l l t s h o rt -circ u i t i m pe d a nce measu red o n the l ow-\,ol t<lge side o f t h e
t r;l I l s ru r m e r is 0 . 1 2 1 1 2 , a n d b e c a use o r t i l e l o w res i s t a n c e , t h is v;l i u e m a y h e
con s i d e red e q u a l to t h e leakage rea c t a nce. D e t e r m i n e t h e p e r - u n i t re a c t a n c e o f
t h e t r a n sformer a n u t he va l u e to be used t o represe n t t h is t r a n s ro rmer i n a s y s t e m
whose b a se on t h e h igh-voltage s i d e of the t r a ns fo r m e r is 1 00 M Y
kY.
/22 6.
Solution. On i t s o w n b a s e t h e t r a n s former re a c t a n c e is
0.121
( 22 )2/400
---:;:--
= 0 . 1 0 per
A, 230
unit
On t h e chosen base t he reactance becomes
(
220 ) 2 1 00
0.1
= 0.0228
230
-
2.6
400
-
THREE-PHASE TRANSFORMERS:
SHIFf AND EQUIVALENT CIRCUITS
per u nit
PHASE
As m e n t i o n e d in Sec. 2 . 5 , ; 1 p h ; l s e s h i ft occu rs in Y - 6. t r a n s ro fm e r s . W e n ow
examine p h ase sh i ft i n m o r e d e t a i l , u n d t h e i m p o r t a n c e o f p h a s e s e q u e n c e
becomes a p p a r e n t . L a t e r i n s t u d y i n g fa u l ts w e have to d e a l w i t h b o t h pos i t i v e ­
or ABC-s equence q uantities a n d ne g a l i ve - or A CE- s e CJ u e nc e q u a n t i t i e s . S o , we
need to examine phase s h i ft for both p osi tive and n e ga tive se q ue n c e s . Posi tive­
sequence voltages and currents are identified by the sup e r s c ri p t 1 a n d negative­
sequence voltages and currents by the superscript 2. To avoid too many
subscripts, we so m e t i m es w r i t e Vj l ) i nstead of vj � for t h e vol t a g e d ro p fro m
terminal A to N a n d similarly iden t i fy other voltages a n d c u rr e n ts to neutra l . I n
a positive-sequence set of l ine-to-neutral vol tages VA l ) lags VY ) b y 1 200 ,
whereas V� l) lags V,J I ) by 2400 ; i n a negative-sequence set of line-to-neutral
voltages V(B 2 ) leads VA(2) by 1 200 whereas VC(2) leads VA(2) by 2400 ' La t e r on
when we d iscuss u n bala nc ed currents and voltages ( in Chaps. 1 1 and 1 2), we
must b e c areful to d istinguish between voltages to neutra l and voltages to
ground since they can differ u n d er u nbala nced con d i tions.
Figure 2. 14(a) is the schematic wiring diagram of a Y - 6. t r a ns fo F m e r,
where the Y side is t h e h i gh-voltage side. We reca ll that c a p i t a l l e t t e rs a p p J y to
,
2.6
T H R EE- P H A SE T R A N SFO R M E R S : P H A S E S H I FT A N D EQU I VA L E N T C I RC U ITS
HI
A ---�o----,
�------�-- a
----
----
1...
Hz
•
B -----o---..J
--->-
18
-c
-------.-
Ibc
H3
c -----o----"
lb
C
b
( a) Wiring d i agram
al l
(l)
V/l B
)
Ie
-
-
B( l )
A( l
X3
Xz
----
Ie
65
VBe
)
V(I
A
I)
VlJ( lCl
)
C( I
(b)
VUI
A
2)
VA( ll
bt l
)
l)
Vc(o
)
Positive sequ ence components
C (2
2)
V( (. /\
'
l)
VCi I
VC{ A
Va( �I
)
V( (.'
�l
Vl(J2 )
C
O)
)
VlJ( 2e)
B(2
)
a ( 'L 1
C c) N e g ative s e q u ence components
F1 G u RE 2 . 1 4
W i r i n g d i ag r a m a n d vol t ag e p h a s ors f o r a t h r e e - p h ase
i s t h e h i g h - vo l t a g e s i d e .
t r a n s fo r m e r co n n e c t ed
Y- 6..
w h e re t h e
Y side
66
CHAPTER 2
TRANSFORM ERS
the high-voltage side and that windings drawn in p a raUei are linked by the same
flux. In Fig. 2.14(a) wind ing AN i s the ph ase o n the Y -connected side, which is
linke d magnetically with the phase w i n d i ng a6 on the 6.-connected side. The
location of dots on the windings shows that VA N i s always in phase with Vab
regardless o f phase sequence. If H I i s the terminal to which line A i s
con nected, it is customary to con nect p hases B a n d C to terminals H2 and H3 ,
respectively.
The Ame rican standard for d esigna ting termi nals H I and X l o n y - �
transformers requ i res t h a t thc pos i t i vc-sequence vol tage drop from H I to
neutral lead the posit ive-sequence voltage d rop from Xl to neu tral by 3 0°
regardless of whether the Y or the � w i n d i n g is on the high-vol tage side.
Similarly, the voltage from H 2 to n e u t ra l l e a d s t h e vol t age from X 2 t o n e u t r a l
b y 30° , a n d t h e v o l t a g e from /1 .1 t o n c u t r ; i l I C ; l d s t h c vo l t a g c fro m X � t o n e u t ra l
by 30° . The p h a s o r d i a g ra m s for t h e [)os i t i v c - a n d n e g a t ive-scq u c nc e c o m p o ­
n en ts of voltage are shown i n Figs. 2. 1 4( 6 ) and 2. 1 4( c ), respecti\ c l y .
Figure 2 . 1 4( 6 ) shows the r e lation of the voltage p h asors whe n positive­
sequence volt ages are applied to term i n a l s A , B , a n d C. The vol t ages Vj l ) ( t h a t
is, Vl�) a n d Va(� ) are i n p hase because of the dots, a n d a s s o o n .as we have
d rawn Vl ) in p hase with Va(; ), the other voltages for the phasor diagrams can be
d etermined. For i nstance, on the high-vol tage side VA l ) l ags Vl ) b y 1 20° . These
two vol tages and vt ) meet at the tips of their a r rows. L ine-to-line voltages can
then be d rawn. For the low-voltage d i a gram V�� ) a n d Vc�l ) c a n b e d rawn i n
p h ase w i t h V� I) and vt), respectively, and t h e n t h e line-to-neutral vol tages
fol low. We see that Vl 1 ) leads Va(l ) by 30° and term inal a must be marked Xl
to satisfy the American standard. Term i nals 6 and c a re ma rked X 2 and X 3 '
respectively.
F igure 2 .14(c) shows the relation of the voltage phasors when negative­
sequence voltages are applied to terminals A , B, and C. We note from the dots
on the w i ring d i agram that V,P) ( n ot n ecessarily in p h ase with VJ I » i s in phase
with v}l). After drawing Vj 2 ) i n phase with �IS;), we complete the d i a grams
simil a rl y to the posit ive-se q u ence d i agrams but keep i n g in mind t h at VJ 2) l eads
Vl2) by 120° . The completed d iagrams of F i g . 2. 1 4(c) show that VY) lags VY)
by 30° .
If N I and N2 represent the number of t urns i n the high-voltage and
l ow-vo ltage windings, respectively, of a ny phase, then F i g. 2.l4(a) shows that
�( l ) = ( N l /N2)Va(I� ) and v,f) = ( Nl / N2 ) Va(l) b y transformer action. I t then
follows from the geometry of Figs. 2. 1 4( b ) and 2. 14( c ) that
VAO )
=
N
/ 3 00
_I v3 Vae l )�
N2
VA(2)
=
N
_1· 13 Va(2)/ - 30°
N2
( 2 . 34 )
Likewise, currents i n the Y-6. transformer a re d isplaced by 30° in the d i rection
of the voltages since the p hase angles of the currents with respect to their
associated voltages are determined by the load i mpedance. The ratio o r the
2.6
TH R EE·PHASE TRANSFO R MERS: PHASE S H I FT AND E Q U I VALENT CI RCU ITS
67
rated line-to-line voltage of the Y w in d in g to the rated line-to-line voltage of the
D. wi n d i n g equals fS N I /N2 , so that in choosing t h e line-to-line voltage bases on
the two sides of the transformer in t h e same ratio, we obtain in per u n i t
JA(2}
=
] (2)
a
X
1/_
( 2 .35 )
-
30°
Tran sformer imped a nce and magnetizing cur ren ts are hand led separately from
t h e p hase shi ft , which can be represe n ted by an :deal tra nsformer. This explains
why, a c c ord i n g to Eq. (2.35), the per-uni( magni tudes of voltage and cu rrent a re
exactly the same on both sides of t h e t ransformer (for i n stance I V(l( l ) I = I VY ) I ).
Usual ly, the h igh-voltage w i n d i n g i n a Y -D. t ransformer is Y -connected.
I n s u l a t i o n cos t s fo r , I g i v e n s t e p u p i n v o l ta g e a re thereby red u ced si nce t h is
co n n e c t i o n t a k e s a d v a n t a g e o f t h e fact t h a t t h e vo l t a g e t r a n s form a t ion from the
l o w - v o l t a g e s i d e t o t h e h i g h -v o l t i\ g e s i u e o f t h e t ra n s fo r m e r i s t h e n fS ( N I / N2 ),
,
w h e r e /VI ([ n ei N2 ,I re t h e S , l m e ,IS i n Eq.
w i n u i n gs il re 8. - co n n e c t e d , t h e t ransformat ion ratio of
l i n e v o l t ages i s reduced r a t h e r t h a n i n crease d . F i gu r e 2 . 1 5 is the schematic
diagram for t h e 8. - Y transform e r w here t h e 6 side i s the high-voltage side. The
reader should ve rify that the vol t age p hasors are exactly the same as in Fig s
2 . ] 4( 0 ) and 2. 1 4(c), an d EC]s. (2.34) and (2.35) a re therefore s t i l l v a l i d . These
e q u a t i o n s s t i l l hold if W e reverse t h e d i rect ions of all c u r re n t s on the wiring
diagram.
Under n ormal oper a t i n g con d i t ions only positive-sequence quant I t I es are
i nvolved an d then the general rule Jor a ny Y - 8. or 8.- Y transforme r is that
vol t ag e is a d v a n c ed 3 0 ° when it i s s t e p p ed u p . A s already d iscussed, we can
i n d icate t h is phase shift i n vol tage b y an ideal t ransformer o f c o m p l e x turns
ratio 1 : EJ·; /6 . S i n c e VY ) / J� I ) = V} Ij / /,; 1 ) i n Eq. (2.35), p e r u n i t impedance
If
the
( 2 . 3 4).
h i g h -vol t a g e
.
-
A
Xl
H\
--------�o�--�
•
X3
FI G u RE 2 . 1 5
W i r i n g d i a g r a m for a t h re e - p h a s e t ra n s fo r m e r
side.
Xz
H:;
----j:r=-------------'
--
COIl l l c c t e u 6 · Y w h e re t h e
0.
a
------
C
]a
Ie
B
C
------
Ib
------
b
s i d e is t he h i g h -vo l t a g e
68
CHAPTER 2
TRANSFO R M E RS
values are t h e same when moved from o n e side of t h e ideal transform er to the
other. R e a l a n d reactive power flow i s also not a ffected b y the phase s hift
because the current p hase shift compensa tes exactly for t h e voltage phase s hift
as far as power values are concerned. This is easily seen by writing the per-un i t
complex power for each side o f t h e Y - � ( o r � -Y) transformer from Eqs. (2.35)
a s fol lows:
( 2 .3 6)
Hence, i f only P and Q quantities are required, i t is not necessary t o include
ideal transform ers for the phase s h i rt o [ Y -� and �-Y tra nsformers i n t h e
i m pedance d i a g r a m . T h e o n l y case i n w h i c h t h e i d e a l t r a n s fo r m e r c a n n o t b e
ignored i s i n a ny closed-loop portion o f a system i n which t h e prod uct o f a l l t h e
act u a l transformer voltage ratios is n o t u n i ty around t h e loop. We encounter
one such case i n Sec. 2.9 when parallel con nections of regulating transformers
are considered. In most other situations we c a n eliminate the ideal transformers
from the per-un i t impedance d iagram, a n d then the calculated currents and
vol t ages are proportional to the a c tu a l currents and voltages. P hase a ngles of
t h e a ctual currents and voltages can b e foun d i f n eeded by noting from the
one-l i n e d iagram the positions of the Y -� a n d �-Y transformers a n d b y
applying t h e rules o f E q . (2.35); namely,
When stepping u p from the l ow-vol tage to the h igh-voltage side of a �-Y
o r Y -� transformer, advance positive-sequence voltages and currents by
30° and retard negative-sequence voltages and currents by 30° .
It i s important to note from Eq . (2.36) that
( 2 . 3 7)
w h i c h shows t hat t h e current ratio of a ny transformer with phase shift is the
reciprocal of the cample.:r conjugate of t h e vol tage ratio. General ly, only vol tage
ratios are shown in circuit diagrams, b u t i t is always u n derstood that the curren t
ratio is t h e reciprocal o f t h e complex conj ugate o f t h e voltage ratio.
In Fig. 2 . 1 6( a ) the single-line d i a gram indicates Y -6 transformers to step
up voltage from a generator to a high-voltage transmission l i ne and t o step
down the voltage to a l ower l evel for d i stribution. In the equivalent circu i t of
Fig. 2. 1 6( b ) transformer resistance and leakage reactance are in per unit a n d
exci t i n g current is neglected. B locks w it h i d e a l transformers indicating phase
shift are shown along with the equ iva lent circu it for the transmission line, which
i s d eveloped in Chap. 6. Figure 2 . 1 6 ( c ) is a furt her simplification where the
resistances, shunt capacitors, and ideal transformers are neglected. Here we rely
u p o n the single-line d iagram to rem i n d u s to account for phase shift due to t h e
Y - 6 transformers. W e must remember t h a t positive-sequence voltages a n d
2.6
TH R EE-PH ASE T R A N S FO R M E R S : PHASE S H I Ff A N D EQU IVALENT CIRCU ITS
�
�t
O�E
I
�
3E
3E
Transm ission l i n e
----t----+-
t. - y
Y - t.
(a)
-'
>----I-
69
to load
(b)
XL
Xb
(c )
HGl: RE 2 - 1 6
( a ) S i n gl e - l i n e d i a gr a m ; ( b ) p e r· p h a s e e q u iv a k n t c i rc u i t s w i t h p a r a m e t e rs i n p e r u n i t ; ( c ) p e r -p h ase
e q u i v a l e n t c i rc u i t w i t h r e s i s t a n c e , c a p a c i t a n c e , a n d i d e a l t ra n s formers n e g l e c t e d . T h e p e r - p h ase
e q u i v a l e n t c i rc u i t for t h e t ra n s m i s s i o n l i n e i s d e v e l o p e d in C h a p . 6.
cu r r e n t s in the
higher-yol tClge tran s m ission l i ne lead the corresponding quanti­
t. i c :, i n t h e l ow e r-vo l t a ge ge n e ra t o r a n d d i s t r i h u t i o n c i rc u i t s h y
Exa m p l e 2 . 9 : F i g u r e: 2 . 1 7 s h ows
gC I 1e:r'ltor
3 0° .
]()O M Y 1\, n k Y
s u p p l y i ng a sys t e m l o at.! o f 2 4 () M Y 1\ , D.t) powe r - fa c t or l a g g i n g a t 230 k Y t h ro u g h
a 330 - M V A 2 3 6 j230Y - k Y s t e: p - u p
t r a n s for m e r o f l e a ka ge r e a c ta n c e 1 1 % .
N e g l e c t i n g m a g n e t i z i n g c u r re n t a n d c h oos i n g b a s e v a l u e s a t t h e l oa d o f 1 00 M YA
a n d 230 k Y , fl n d 1/1 , In , a n d Ie s u p p l i e d to t h e load i n p e r u n i t w i t h VA as
I,
a
t h r e: e: - p h ase:
t e r m i n a l vol t a g e .
m t c tl
r e fe r e n c e . S p e c i fy i n g t h e p r o p e r base fo r t h e g e n e rator c i rc u i t , d e t e r m i n e 10 ' Ih '
a n t.!
fro m t h e g e n e r a t o r a n d i t s
Solution . T h e c u r re n t s u p p l i e d to t h e l o a d i s
2 4 0 ,000
.f3
X
230
602 .45 A
70
CHAPTER 2
TRANSFORMERS
To
Or------t---� �>-------f--�LOad
( a)
t::. - y
1 : Bjrr /6
10
--
+,
J
fA
--
r
L
v,.
V"
Load
(b)
FIGURE 2. 1 7
(a) Single-line d i a g ram;
The
base
e q u ivalent c i rc u i t for Example 2.9, a l l param eters in pe r unit.
(b) per-phase
current at the load is
1 00,000
= 251 .02 A
y 3 x 230
h
The power-factor a ngle of t he load current is
0 = cos -- I 0.9 = 25 .84°
Hence, with VA
load are
= 1 .0� as reference
1A =
18
Ie
=
602.45
i n Fig.
lag
2.17(6),
/ - 25 . 84° = 2.40/ - 25 .84°
25 1 .02 -
'----
2 .40/ - 25 .84° - 1 20°
= 2.40/ - 25 .84°
+
1 20°
=
=
the line currents i nto the
per u n i t
2.40/ - 145 .84°
2.40/ 94 . 160
per u n i t
per unit
Low-vol tage side currents further lag by 30° , and so i n p e r u n i t
lb
=
2.40/
1 75 . 84°
2.7
THE AUTOTRANSFORMER
71
The transformer reactance modified for t h e chosen base is
0 .11
and so from Fig.
2. 1 7( b )
x
1 00
1
-
330
=
-
30
per unit
the terminal voltage of the generator is
/ - 30°
=
1 .0
=
0 .9322
-
+
�
x
30
j0.45 5 1
=
2.40/ - 55 .84°
1 .0374L
- 26 .02"
per u n i t
hase g en e r a tor voltage is 2 3 k Y , w h ich m e a n s t h a t t he terminal voltage of t h e
g e n erator is 23 X 1 . 0374 = 23.86 k Y . T h e re a l powe r s u p p l i e d by t h e generator is
The
R c { VJa* }
�c
1 . 0 3 74 X 2 . 4 c o s ( - 26 .02° + 5 5 . 1-\4° )
=
2 . l GO
per
unit
which corresponds t o 2 1 6 M W absorbed by t h e l oad since there are no ] 2 R losses.
The i n t e re s t e d re a d e r wi l l find the same v a l u e for I VI i by o m i t t i n g the phase s h i ft
of the transformer altogether or by recalculating VI w i t h the reactance j /30 per
u ni t o n the high-voitage s i d e of F i g . 2. 1 7(b).
2.7
THE AUTOTRANSFORMER
a u totransformer diffe rs from the ordinary transformer in t ha t t h e windin gs
of t h e a u totransforme r are e l ectrical l y connected as well as coupled by a m u t u al
flux . We exa mine the a utotransformer by electrically connecting t h e windin gs of
an i d eal transformer. Figure 2. 1 8( a ) is a schematic d i agram of a n ideal t rans­
fOlmer, and Fig. 2. 1 8(b) shows how the windings are connected e lectrical ly t o
form a n a u t otrcDsformer. Here t h e w i n d i ngs are shown so that their vol tages
a r e a d d i t ive a l t hough t h ey coul d have been connected to oppose each other.
The great di sadvantage of t h e a u totransformer is t h at electrical isolation is lost,
but the fol l o w i n g cxample dcmonstrates the increase i n pow er r ating obtained.
pJl
\.
f.
T3
/:
-
-l
VI
.
N2
NI
Inl
r
1-
V2
12 !1
+1
VI
-l
l,n
II
j
N2
+
V2
!Ii )
I hl
FIG U RE 2 . 1 8
Sch e m a t ic d i a g r a m of a n i d e a l t ra ns-
( b ) a s a n autptransformer.
fo rmer co n n e c t e d : (a) i n the u s u a l
manner;
72
CHAPTER
2
TRANSFORMERS
Example 2 .10. A 90-MVA s ingle-phase transformer rated SO/ 1 20 kV is connected
as a n a u totransformer, as shown in Fig. 2 . 1 S( b ). Rated vol tage I V} I = SO k V is
applied to the low-voltage winding of the transformer. Co nsider the transformer to
be ideal and the load to be such that currents of rated magnitudes 1 /1 1 and 1 /2 1
flow in the windings. D e term i ne I V2 1 and the kilovoltampere rat i n g of the
a u totransformer.
Solution
111 1
90 ,000
=
1 /2 1 =
I V2 1
=
SO
90 ,000
1 20
80 +
=
1 1 25 A
=
750 A
1 20 = 20 0 k V
The d i rections chosen for I } and 12 in relation to the dotted terminals s how that
these currents are in phase. So, the input current is
I lin I = 1 1 25
+
750 = 1 S75 A
Input kilovoltamperes are
Outp u t kilovoltamperes are
The increase in the kilovoltampere rating from 90,000. to 1 50 000 kVA a n d in the
o u tput voltage from 1 20 t o 200 k V demonstrates t h e adva ntage o f t h e
a u t o t r a n s fo r m e r. T h e a u t o t l " a l 1 S rO fl1l e r prov i d es a higher r a t i ng for t h e s a m e cost,
,
i.Jll d
i t s e t I i c i e n cy i s g r e a t er
since
c on n ec t ioll of the same t ra l l s fo r m e r.
the
l osses a r e t h e same a s i l l
the
o rd i n a ry
Single-phase autotransformers can be co n n e c t e d for Y - Y three-ph a se
operation o r a t hree-phase u n i t can be built. Three-phase autotransformers are
often used to con nect two transmission l ines opera ting a t differen t voltage
levels. If t h e t r a nsformer o f E x a m ple 2 . 1 0 were con n ected as o n e p h ase o f a
three-ph ase y -Y autotransformer, the rating of the three-ph ase u n i t would be
450 MY A, 1 38 /345 k Y (or more exactly 1 3 8 .56/346.4 1 kY).
2.8 PER· UNIT IMPEDANCES
OF THREE-WINDING TRANSFORMERS
Both t he primary and the secondary w i n d i ngs of a two-winding transfo rmer,have
the same k ilovoltampere rating, but a l l three w i n d in gs of a t hree-w i n d i n g
transformer may h ave d iffe r e n t k i lovu] t a m p c r e r a t i n g s . T h e i m p e d a n c e of e a c h
2.8
P
-{
-----h.
PER UN IT
-
n---t---- s
,...L---+---- t
(a)
I M PEDAN CES O F THREE-W I N D I N G TRANSFO R M ER S
73
p --------,
s
t .------�
(6)
FI G U RE 2 . 1 9
T h e e n ) sche m a tic d i a g r a m a n d ( b ) e q u iva l e n t c i rc u i t o f a t h r ee - w i n d i n g t r a n s former. Poi n t s p , S ,
a n d I l i n k t h e c i rc u i t o f t h e t r a n s fo r m e r t o t h e a p p r o p r i a t e e q u iv al e n t c i r c u i t s repre s e n t i ng p a r t s o f
t h e sys t e m con n e c t e d t o t h e p r i m a ry, s e co n d a ry , :l nll t e r t i a ry w i n d i n g s .
w in d i ng of a th ree-winding t ran sformer may b e gIven I n perce n t or per u n i t
b 3 s c d on t h e )",l t i n g o f i t s own w i n d i n g, o r tests m ay b e m a d e t o d eterm i ne t h e
imped ances. I n a n y case, a l l t h e p e r- u n i t impedances m u s t b e expressed o n the
s a m :..: k ilovo l t arn p e re base_
A single-phase thre e-win d i ng transformer is shown schematically in Fig.
2. 1 9( a ), where we designate the t hree w i n d ings as primary , secondary , a n d
tertia ry . Three impedances m ay b e m e asured b y the stan dard short-circui t test,
as follows:
Zps
-
Zp l
Z51
leakage impedance measured 1D primary with secondary short-circuited
and terti ary open
leakage impedance m easured i n pnmary w i t h tertiary short-circ u i t e d
a n d secon d a ry open
lea kage i mpedance measured 1 D secondary with tertiary short-circu ited
a n d primary open
of
t h e w i n d i n g s , t h e i m p e d a n c e s o f e �l c h s e p a r a t e w i n d i n g refe rred t o that s a m e
w i n d i ng arc related to t h e m easured impe da nces so referred a s fol lows:
I f t h e t h rc e i rn r c d a n c e s m c a s l I rc d i n o h m s ,Ire rlf e rr e d to t h e vo l t a g e of one
( 2 .38)
and Z are t h e i m p ed ances of the prim a ry, second a ry, and tertiary
w i n d i ngs, respectively, referred to t h e primary circu it i f Zp s ' Zp I ' a n d Z SI a r e
t h e m e asured impedances referred t o t h e prim ary c i rcuit. Solving ' Eqs. (2.38)
Here
ZP ' Zs '
I
74
CHAPTER 2
TRANSFORMERS
simultane ously yields
( 2 . 3 9)
The impedances of the t h re e w i n d i ngs a re connected to represent t h e
equiva l e n t circuit o f t h e single-phase three-wi n d i n g transformer w i t h magnetiz­
ing c u rr e n t neglected, as s h o w n i n Pig. 2. l 9 ( h ) . T h e common rJoint is fictitioL l s
a n d u n r e l a t e d to t h e ne ut r a l o r t h e s y s t e m . The p o i n t s {J , s , a n d f a re c o n n c c t c tl
to the p a rts of the i mpedance diagrams repres e n t i n g the pa rts of t h e system
connected to the primary, seconda ry , and tertiary windings, res p e c t i v e l y , of the
t ransform er. As in two-winding transformers, conversion to per-unit im pedance
requires t h e same kilovo!tam pere base for a l l t h re e c i rc uits and req u i res vol tage
b ases i n t h e t h ree c irc uits that a re in the same ratio as the rated l i ne-to-line
vol tages of t h e three c ircuits of t h e transformer.
When t hree such t ransformers a re connected for three-phase opera tion ,
t h e primary and secondary windings are usually Y -connected and the tertiary
w in dings are connected in t::. to provid e a p a t h for t h e t h i rd harmonic of the
exciting curren t.
Example
2.11.
The three-phase ratings of a three-winding
15
Primary
Y-connected, 66
Secondary
Y-connected,
13.2 kV, 1 0 MVA
Tertiary
6-connected,
2.3 kV, 5
kV,
transformer are:
MVA
MY A
Neglecting resistance, t he leakage i mp eda nces a re
ZPS
=
Zp I
=
ZS I
=
7% on 15 MVA , 66 kV base
9% on 1 5 MVA, 6 6 kV base
8% on 1 0 MV A, 13.2 kV base
Find t h e per-u nit impedances of t h e per-phase equivalent circuit for a base of
MY A, 6 6 kV i n the primary circ u i t .
15
With a base of 15 MVA, 66 k V i n t h e primary circuit, t h e proper bases
for the per-unit impedances of the equivalent circuit are 15 MVA, 66 kV for
prima ry-circuit quan tities, 1 5 M Y A, 13.2 kV for secondary-circui t quantities, and
15 MYA, 2.3 kY for tertiary-circuit quantities.
SoLution.
2.8
PER-UNIT I M P EDANCES OF TH R E E-WINDING TRANSFO R M E R S
75
Since Z ps and Zpt are m easured in the primary circui t, they are already
expressed on the proper base for the equivalent circuit. No change of voltage base
is required for Zst . The required change in base megavol tamperes for Z S ( is made
as follows :
15
Z SI
=
8% X
10 =
12%
In per unit on the specified base
2. 1 2.
Zp
=
Zs
=
}( j0.07 + jO.09 - j0.12) = jO.02 per u n i t
}- ( j0.07 + j0 . 1 2 - jO.09) jO.OS per u n i t
=
Z, }- ( j0.09 j0.12 - iO.07) jO.07 p e r u n i t
=
=
+
co nst ant-vo ltage source ( i n fi n i t e bus) supplies a purely resistive 5
MW, 2.3 kY three-phase load and a 7.5 MY A , 1 3.2 kY syn chronous motor having
; \ s u b t ra n si c n t r e a c t a n ce o f X " = 20 % . T h e source is con nected to t h e primary of
t h e l h r e e -w i n u i n g t r a n s fo r m e r desc r i b e d i n Exa m pl e 2 . 1 1 . The motor a n d resis t ive
load are connected to t h e secondary a n d tertiary of the t ransformer. D raw t h e
impedance diagram of t h e system a n d mark the per-u n i t impedances for a base o f
6 6 k Y , 1 5 M Y A in t h e p r i m a ry . Neglect exci ting curre n t a n d a l l resistance except
that of the resistive load.
Exa m p l e
A
Solution. The constant-vol tage source can be represented by a generator having no
internal impedance.
The resistance of the load i s 1 .0 per unit on a base of 5 M YA, 2.3
tertiary . Expressed o n a I S M YA , 2.3 kY base, t h e load resistance i s
R
=
1 .0 X
15
5
=
3.0 per unit
The reactance of the motor on a base of 1 5 MYA,
X"
kY in t h e
15
7.5
= 0 . 20 - = 0 . 4 0
per
13.2 kY is
unit
Figure 2.20 is the required d iagram . We must remember, however, the phase s h i ft
which occurs between t h e Y -connected primary and the 6-connccted tertiary.
j0 05
+
E""
F I G UR E 2.20
I m p e d a n c e d i agram for E x a m p l e 2. 1 1 . ,
76
CHAPTER 2
TRANSFO R M ERS
2.9 TAP-CHANGING
TRANSFORM ERS
AND REGULATING
Tra nsformers which provide a sma l l adj u stment of vol tage magnitude, usually i n
the range o f ± 10%, a n d others which s h i ft the p h ase a ngle o f t h e l i n e voltages
are important components of a pow e r system. Some transformers regulate both
the magnitude and phase a ngle.
Alm ost a l l transformers provide t a ps on windings to adjust the ra tio of
transformation b y changing taps when the transformer is deenergized . A change
in tap can be made while the transformer is e n e rgized, and such transformers
a re c a lled load-tap-changing (LTC) transformers or tap-changing-under-load
(TCU L) transformers. The tap changi n g is a utomatic and ope rated by motors
w h ich respond to rel ays set to h o l d t h e vol t age a t the prescribed l evel. Spec i a l
circuits a l low t h e change t o b e made without i n terrupting t h e current.
A type of transformer designed for small adjustments of voltage rather
than l a rge changes in vol tage levels is called a regulating trallsformer. Figure
2.21 s hows a regu lating t ransforme r for control of voltage magnitude, and
F ig. 2.22 s hows a regulating t ransformer for p hase-angle con t rol . The ph asor
d ia gr a m of Fig. 2.23 helps to expla i n the s h i ft in phase a ngle. Each of the three
w i n d in gs to w h ich taps are made is on the same magnetic core as the phase
w inding whose voltage is 90° out of p hase with the voltage from neutral to the
point connected to the center of the t a pped wind ing. For instance, the voltage
to neutral Va n is increased by a component � Va n ' which is in phase or 1 80° out
of phase with Vbc ' Figure 2.23 shows h o w the t h ree l ine voltages are sh ifted i n
p h a s e angle with very l i t t l e change i n m a gnitude.
b .-------�--�--+--�
Series
transformers
FIGURE 2.21
Regula t i n g t ra nsformer for con trol of voltage m a g n i t u d e .
2.9
TAP-CHA N G I N G A N D REG ULATI N G T R A N S FO R M E R S
VCII + � Vcn
77
F I G L R E 2.22
R e g u l a t i n g t r a n s fo r m e r for co n t rol
of phase a n g l e . W i n d i ngs d rawn
D a r a l k l to e a c h o t h e r a re on t h e
s a m e i ron c o r e .
The p roce d u r e to d e t e rm i n e t h e bus a d m i ttance m a t r Lx Y bus i n p e r u n i t for
a network con t ain i ng a regu l a t i n g t r ansformer is the same as t h e p rocedure to
accou n t for any transform e r whose turns ratio is other than t h e ratio used to
select the ratio of base vol tages on the two sides of t h e tran sformer. We d e fe r
co nsider a t ion o f t h e proced ure u nt i l Ch ap. 9 . W e can, howeve r, i nvestigate t h e
u se fu lness of tap-changing and regu lating t ransformers b y a simple exampl e .
I f we have two buses connected by a transforme r . and i f t h e ratio of t h e
line-to-line voltages of t h e transforme r i s the s a m e a s t h e r a t i o o f t h e b ase
vol t ages of the two buses, the per-phase equivalent circuit (wi t h the m ag n e t izing
current n eglected) is simply t h e transformer impedance i n per unit o n the
chosen base connected between the buses. Figure 2 . 24C a ) is a one- p n e d i a gram
of two t r a n s formers in parallel. Let us assume t h a t one of them has the voltage
r a t i o l in , w h i c h i s also t h e r a t i o of base vol t ages o n the two sides o f t he
t r a n s form e r , a n d t h a t the vol t age ra tio of the other i s l in'. The equivalent
c i rcu i t is t h e n t h ,l( of Fig. 2 .24( b ). W e n e e d the ideal ( n o imped ance) t ra ns­
for m e r w i t h t h e r(l t i o 1 / ( i n t h e p e r - u n i t r e ,l c ta n cc d i a g ra m t o take ca re of the
o fT- n o mi n ; 1 1 t u rn s r;d i o o r t h e s e c o n d t r a n s fo r m er b e c a u s e b a s e vol t <lges are
Shd(c·(j
V" ,, ·'
- O r ig i n a l V""
F I G U R E 2.23
P h a s o r d i agram for t h e r e g u l a t i n g t r a n s fo r m e r shown i n
Fig. 2 . 2 2 .
78
CHAPTER 2
TRANSFORMERS
ljn
h
CD
®
l in '
(a)
----I---�--
FIG U RE 2 . 24
Tran sformers w i t h d i ffe r i n g
t u rn s ratio connected i n paral­
lel: (a) the single-line diagram;
(b) t h e per-ph ase reactance d i ­
agram in p e r u n i t . T h e t u rns
ratio I II i s e q u a l to n ln ' .
(b)
determined by the turns ratio o f the fi rs t transformer. Figure 2.24( b ) m ay b e
interpreted a s two transmission l ines i n p a ra l le l with a reg ulating transformer i n
o n e l ine.
Two transformers are con nected i n para l le l to supply a n i m p e d a n c e
to n eutral per phase of 0 . 8 + j O . 6 per unit ·at a vol tage of V2 1 .0ft per u n i t .
Transformer T" h as a vol tage r atio e q u a l to t h e ratio of t h e base vo l tages on t h e
two sides o f t h e t ransformer. T h is transformer h a s an impedance of jO. l p e r u n i t
on the a ppropriate base. T h e second transformer Tb also has a n impedance of j O . 1
per u nit o n the same base b u t h as a step-up toward the load of 1 .05 t imes t ha t of
Ta (seco n d a ry w i n d i n gs on 1 . 0 5 tap).
Figure 2 .25 shows the equ ivale n t circuit w ith transformer Tb r epresented by
its impedance and the insertion of a vol t age tl v. Find the complex power
transmitted to the load t h rough each transformer.
Exa mple 2 . 13.
=
Solution.
Load current
is
1 .0
---- =
+ jO.6
0.8
0.8
-
jO.6
per unit
2.9
TAP-CH A N G ING AND R E G U LATING TRANSFO R M E R S
79
)0. 1
s
lcirc
tN
=
O.OSlQ:.
0.8
)0.6
F1 G C RE 2.25
An equivalent c i rcuit for Exa m p l e 2 . 1 3 .
An ap proxi m a t e sol u t i o n to t h i s prob l e m i s fou n J b y recogn izing t h a I rig. 2.25 with
sw i t ch 5 c l o s c u is a n e CJ l I i va l e n t c i rc u i l ror t h e p ro b l e m i f t h e vol t age � V, which is
i n t he branch of the c i r c u i t equiva l e n t t o transformer Tb , i s equal to I
1 in per
u n it. I n other words, if T" is p roviding a voltage r a t i o S o/c higher than Tb, I equals
1 .05 and �V equals 0 . 05 per u n i t . If we consider that the current set up by �v
circulates arou nd the loop i n dicated b y Ie i r c w ith switch S open, and that with S
closed only a very small fraction of that current goes through the load impedance
(because it is much larger than the transformer impedance), then we can apply the
supcrposition pri n c i p l e to �V and the sou rce voltage. From � V alone we ob tain
-
[e i re
0.4
0.05
jO.2
=
. - -jO .2S
p er unit
and with �V short-circuited, the currcnt in each path is half the load current, o r
- /0.3. Then, superimposing the ci rcul ating current gi\cs
ITo
IT
/.
so that
=
=
0 .4 - jO.3
.-
=
0 4 - j O .OS per unit
.
0 . 4 - j O . 3 + ( -jO .25) = 0 .4 - jO.55 p e r u n i t
ST
o
=
5Th =
Th is
( -) 0 .25)
0 .40
+
jO .OS
per unit
0 .40 + jO.55 per u n i t
example s hows t h a t t h e t r ansformer w i t h the higher t a p s e t t ing i s
supplying most o f t h e react ive power to t h e load. The real power i s d ivid i n g
e q u a l l y b e tween t h e transfo rm ers. Since both t ransformers h ave t h e same
imped ance, they would sh a re both t h e real and re a c t i v e power equ ally if they
h a d the same t urns ratio. In t hat case each would b e represented by the same
per - u n i t react ance of jO. 1 betwe e n the t w o buses a n d wou l d c a r ry e q �l a l c u r r e n t .
80
CHAPTER 2
TRANSFORMERS
When two t r ansformers are in p arallel , we can v ary the d istribution of reactive
power betw een the transformers by a dj usting the voltage-magnitude r atios.
When two p aralleled t ransformers of equal kilovoltamperes do not share t he
kilovoltamperes equally because their impedances differ, t h e kilovoltamperes
may be more nearly equalized by a djustment of the voltage-magnitu d e r atios
through tap c h a n g in g .
ft
Example 2 . 1 3 excep t t h a t Th i n c l u d e s h o t h a t r a n s fo r m e r
h a ving the same turns ratio as Ta a n d a regu lating transformer w i t h a phase shift
of 3° (t £:)'rr /60 = l .O
). The i m pe dance o f the two components of Tb is iO. l
p e r u n i t on the base of Ta '
Exa mple 2 .14. Repe a t
=
Exa m p l e 2 . 1 3, we c a n o bt a i n an a p pr ox i m a t e so l u t i o n o f t h e
problem by inserting a vol tage SOurce � V i n s e r i e s w i t h t h e i m p e d a nc e o f
tra nsformer Tb . The proper per-uni t voltage i s
Solution. As in
[eire =
IT
=
IT"
=
a
So ,
0.0524
/91 .5°
-�-!./::::===- = 0 .262 + iO.0069 p e r unit
0.2 90°
0 .4 - iO.3 - (0.262
+
iO .007)
=
0 . 1 38 - iO .307 per unit
0.4 - iO.3
+
iO .007)
=
0 . 662 - iO.293 per unit
+
( 0.262
51
a
5 .,.
"
=
0 . 1 38
+
iO.307 p e r unit
= 0 . 6()2 + iO.293
per
unit
The example shows that the p hase-sh ifting transformer is usefu l to control
the a m ou n t of real power flow b u t has less effect on the flow of reactive power.
B oth E xa m p les 2.1 3 and 2.14 are i llustrative of two transmission l i nes in parallel
with a regulating transforme r in one of the l i n es .
2.10 THE ADVANTAGES OF PER-UNIT
COMPUTATIONS
W hen b ases are specified p roperly for the various p arts of a circuit connected by
a t r a n sfor m er the per-unit values of impedances d etermined in their own p ar t
of t h e system a r e t h e s ame w hen viewed from another part. Therefore, i t is
n ec essary only to compute e ach impedance on t h e base of its own part of t h e
circuit. The gre a t advantage o f using per-unit values i s t h a t no computati9ns are
req u i re d t o refe r an impedance from one side of a transformer to the other.
,
2. 1 0
1.
T H E A DVANTAGES OF P E R - U N I T COM PUTATIONS
81
The fol lowing points should be kept in mind:
A base k ilovolts and base kilovoltamperes is selected in one part of the
system. The base values for a t h ree-phase system are u n derstood to b e
l ine-to-l i ne kilovolts and t h ree-phase k i lovoltamperes o r megavoltamperes.
2. For other parts of the system, t h a t is, on other sides of transformers, the b ase
kilovolts for each part is d etermined according to the l i ne-to- l i ne volt age
ratios of the transformers. The base kilovoltamperes wi l l be the same in all
p arts of the sys t em. I t will be helpful to mark the base kilovolts of each p art
of t h e system on the one-l i ne d i agram
3 . I mp ed ance i n formation ava ilable for t h ree-ph ase tra nsformers will usually be
i n per u n i t or perce n t on the base determined b y their own ratings.
4. For t hree single-ph ase t ransform ers connected as a t h ree-phase u n i t the
t hree-phase rat i ngs a re determ ined from the si ngle-phase rating of e ach
i n d i v i d u a l t r a n s for m e r . I m p e d a n c e i n p e rce n t fo r t h e t h r e e p hase u n i t is the
s a m e as t h a t fo r each i n divi d u a l t r a nsfo r m e r.
5 . P e r- u n j t imped ance given on a base o ther than t hat determined for the p art
of the system in which the e l e m e n t is l ocated must be c h a nged to the p roper
base b y Eq. ( 1 .56).
.
-
�1aking compu ta tions for e lectric systems in terms of per- u n i t values simplifies
the work great l y. A rea l appreci a t io n of the value of the pe r-un i t method comes
throug h experience. Some of t h e advantages of the method are summarized
b ri efly below:
1 . M a n u fa c ture rs
2.
usually speci fy the i m pedance of a p iece of apparatus in
perce n t or per u n i t on t h e base of t h e nameplate rating.
T h e per-unit impedances of machines of the same type and widely d iffe rent
r a t i n g u s u a l l y l i e w i t h i n a n a rrow range a l t hough the ohmic values d iffer
m a t e r i a l l y for mach ines of d i ffe re n t ratings. Fo r t h i s reason when the
i m p e d a nce i s n o t k n ow n d e fi n i t e l y , i t i s g e ne r a l l y p o ssi b l e t o se lect from
; I ve r a g e
correc t . E x p e r i e n c e i n
tab u l ateu
3.
w i l l be r e a so n a b ly
per- u n i t v a l u e s b r i n gs fa m i l i a r i ty with t h e
p e r- u n i t
i m p e d a n ce w h i c h
working w i t h
proper val ues o f per- u n i t i m p e dance for d i ffe rent t y p e s of appara tus.
When impeda n ce in ohms i s specifi e d in an equivalent circuit, each imped a nc e
must b e r e fe rred t o t h e same c i rc ll i t b y m u l t i p l y i n g it by the square of t he
ra t i o of t h e r a t e d v o l t a g e s of t h e two s i d e s of t h e t ra nsfo r m e r connect i n g t h e
reference c i rc u i t and t h e cir c u i t con t a i n i ng the i mpeda n ce . The per-unit
im p edance once expressed on t h e p roper base, is t h e same referred to e i ther
s i de of any tra nsformer.
The way in which transformers are connected i n t h ree-ph ase circuits does
not affect the per-un i t impedances of the equivale n t circ u i t a lthough the
t r ansformer connection does d e termine the relation between the volt age
bases on t h e two sides of the t ransformer.
,
4.
v;I i u cs a
82
CHAPTER
2.1 1
2
TRANSFORM ERS
S U M MARY
The intro duction in this chapter of t h e s implified equivalent c ircuit for t h e
transfo rm e r i s o f great i mportance. Per-unit calculations a r e useu a lmost
continuously throughout the chapters to fol l ow . We h av e seen h ow t h e trans­
former is e l i mi nated in the equivalen t circui t by the u s e of per-unit calculations.
It i s important to remember t h a t fJ does not enter t h e deta i l e d p er-unit
computations because o f t h e specification of a base l i n e-to-line voltage a nd b a se
line-to-neu tral vol lage related by fJ .
The concept of proper selection of base i n the various parts of a c irc u i t
linked by transformers and t h e calculation o f parameters i n per u n i t o n t h e base
specified for the part or t h e circuit in w hich the parameters exist is fu ndamental
in building a n equivalent c i rc u i t from a single-l ine diagram.
PROBLEMS
single-phase transformer rated 7.2 kY A, 1 .2 kY / 1 20 Y h as a primary winding of
800 t urns. Determine (a) the turns ratio and t h e number of turns in the secondary
winding and (b) the currents carried by the two windings when the transformer
d elivers i ts rated kY A at rated voltages. Hence, verify E q. (2.7).
2.2. The transformer of Prob. 2. 1 is delivering 6 kYA at i ts rated voltages and 0.8
power-factor lagging. (a) Determine the impedance Z 2 connected across i ts sec­
ondary terminals. (b) What is the value of t h is impedance referred to the prim a ry
side (i.e. Z ;)? (c) Using the value of Z; obtained in part (b), determine t he
magnitude of the primary current and the kY A supplied by the source.
2.3. W i t h reference to Fig 2.2, consider that the flux density inside the center-leg of t he
transformer core, as a function of time I , is B ( t ) = Bm sin(27T It ), where Bm is the
peak value of the sinusoidal flux density and I is the operating frequency in2 H z . I f
t h e fl ux density i s uniformly d istribu ted over t h e cross-sectional area A m of t he
center-leg, determine
( a ) The instantaneous flux 4> ( 1 ) in terms of B"" f, /1 , and f .
( b ) Thc instantaneous induced-voltage e / d, according to Eq . (2. 1 ) .
( c ) Hencc, show that the rms magnitude of the induced voltagc of the primary i s
given b y I E ) I = firrfN) BIII A .
100 cm 2 , f 60 Hz, Bm = 1 .5 T, and N) 1 000 turns, compute I E, I .
( d ) If A
2.4. For the pair of mutually coupled coils shown in Fig. 2.4, consider that L ' I
1.9 H,
L ' 2 = L 2 , 0.9 H, L 2 2 = 0.5 H , and r , = r2 = a fl . The system is operated at 60
2.1. A
=
=
=
=
=
H z.
( a ) Write the impedance form [Eq. ( 2.24)] of the system equations.
2.S.
( b ) Write the admittance form [Eq. (2.26)] of the system equ ations.
( c ) D etermine the primary voltage V, and the primary current I I whe n the
.
secondary is
CD open-circuited and has the induced voltage V2 = 1 00
v.
Oi) s hort-circu ited and carries the current 12
A.
For the pair of mutually coupled coils shown in Fig. 2.4, develop an equival en t-T
network i n the form of Fig. 2.5. Use the parameter values given in Prob. 2,4 and
assu me that the tu rns ratio a equals 2. What are the values of the leakage
reactan ces o f t h e windings and the magnetizing susceptance of the cou pled coils?
= 2�
LQ:.
P ROBLEMS
2.6.
2.7.
2.B.
83
A single-phase transformer rated 1 .2 kV / 1 20 V, 7.2 kVA has the following winding
parameters: ' 1 = 0.8 fl, x I = 1 .2 n, r2 = 0.01 .0, and x2 = 0.01 n. Determine
( a ) The combined winding resistance and leakage reactance refe rred to the
primary side, as shown in Fig. 2.8.
( b ) The val ues of the combined paramete rs referred to the secondary side.
(c) The vol tage regulation of the transformer when it is delivering 7.5 kVA t o a
load at 1 20 V and 0.8 power-factor l agging.
A single-phase transformer is rated 440/220 V, 5.0 kVA. When the low-voltage
side is short-circu ited and 35 V is applied to the high-voltage side, rated current
flows in the windings and the power input is 1 00 W. Find the resistance and
reactance of the high- and low-voltage windings if the power loss and ratio of
reactance to resistance is the same in both windings.
A single-phase transformer rated 1 .2 kV / 1 20 V, 7.2 kVA yields the following test
results:
O p e n - c i rcu i t t e s t ( p r i m a ry -o p e n )
V o l t a ge V=,
= 1 20 V ;
c ur re n t I .,
=
1 .2
rower W2 =
A;
S hort-circuit test (secondary-shorted)
Vol tage
2.9.
VI
=
20 V ;
curren t
JI =
power
6.0 A ;
W,
=
4 () W
36 W
Determine
2
2
( 0 ) The para meters R , = ' , + a r 2 , X l = X I + a x 2 , Ge , and Bm referred to the
primary side, Fig. 2.7.
( b ) The values of the above parameters referred to the secondary side.
( c ) The efficiency of the transformer when it delivers 6 kVA at 1 20 V and 0.9
power factor.
A single-phase transformer rated 1 .2 kV / 120 V, 7.2 kV A h as primary-referred
p arame ters R I r l -+- 0 2 ' 2 1 .0 fl and X I X I + a 2 x 2 = 4.0 fl. At rated volt­
age its core loss may be assumed to be 40 W for a l l values of the load current.
( 0 ) De t e r m i ne the eA1cicncy and regu ;ation of the tra nsformer when it de livers 7.2
1 20 V a n d powe r factor of ( i ) 0.8 laggi ng and ( ii) 0.8 leading.
kVA at V2
(b) for a given load vol tage and power factor it can be shown that the efficiency of
; \ t r a n s ro r m e r ;I l t a i n s i t s m ax i m u m v a l u e a t t h e k V A load level which makes
t h e I :' R w i n c.J i ng l o sses e q u a l t o t l t e c o r e l o s s . U s i n g t h i s r e s u l t , d et c r m i n e the
maximum e rti c i e n cy o f the above t ra nsformer a t rated voltage and 0.8 power
factor, and the kV A load level at which it occurs.
A s i n gle -p h a s e sys t e m s i m i l a r to that shown in Fig. 2 . 1 0 has two transformers A -B
and B-C connected by a line B fec d i n g a load a t t he receiving end C. The ratings
a n d p a ra m e t e r v a l u es o f the com ponents a r c :
=
=
=
o c
2 . 1 0.
Transformer A -B : 500 V / 1 .5 kV, 9.6 kVA, leakage reactance
=
5%
Tra n sfo r m e r B- C :
=
4%
Line B:
Lo a d C :
1 .2 kV / 1 20 V, 7.2 k V A , leakage reactance
series impeda nce
=
1 20 V , 6 kVA a t U . S
CO.5
j3.0) n
powe r-factor
-1
lagging'
84
CHAPTER
2 TRANSFO R M E R S
(a)
Determine the value of the load imped ance in ohms and t he a c t u a l ohmic
impedances of the two transformers referred to both their prim ary a n d
secondary sides.
(b) Choosing 1 .2 kV as the voltage base for circuit B and 1 0 kYA as the
systemwide k VA base, express all system impedances in per unit.
( c ) What value of sending-end voltage corresponds to the given loading condi­
tions?
2 . 1 1 . A balanced �-connected resistive load of 8000 kW is connected to the low-vol tage,
6 -connccted s i d e of a Y -6 transformer rated 1 0,000 kYA, 138/ 1 3.8 kY. Find the
load resistance in ohms in each p hase as measu red from line to neutral on the
high-vol tage side o f the transformer. Neglect t ransformer impedance and assume
rated vol tage is appl ied to the t ransformer primary .
2.12. Solve Prob. 2. 1 1 if t h e s a m c r e s i s t a n c e s a r e r ec o n n e c t e d i n Y .
2 . 1 3 . Th ree t r a n s fo rm e rs , e a c h r a t e d 5 kY A , 22() Y o n t h e s e co n d a ry s i d e , , Ir e con n e c t e d
�-� and h ave b e e n su p p l y i n g a b a l a n c e d 1 5-kW p ur e l y resist ive l o a d a t 2 2 0 V . A
change is made which reduces t h e load to 1 0 k W , s til l purely resistive a n d
balanceu. Someone suggests that w i t h two-thirds o f t h e load , one t r a n s fo r m e r can
be removed and the system can bc opera ted open � . B a l an c e d t h r ee - p h a s e
voltages will still be supplied to the load since two of the I ine vol t a g e s (and thus
also the third) will be u nchanged.
To investigate the suggestion fur ther,
( a ) Find each of the line curre nts (magnitude and angle) with the lO-kW load and
the transformer between a and c removed . (Assume Va'" = 220
V, se­
quence abc.)
( b ) Find the kilovoltamperes supplied by each of the remaining transformers.
( c ) What restriction must be placed on the load for open-6 operation with these
transformers?
( d ) Thin k about why the individual t ransformer kilovoltamperc values i nclude a Q
component when the load is purely resistive.
2. 1 4. A transformer rated 200 M YA, 345Y /20.56 kY connects a bala nced load rated
180 MYA, 22.5 kY, 0.8 powe r- fa c t o r lag to a t ra n s m i ssion l i n e . D e t e r m i n e
( a ) The rating of each of three single-phase transformers which when properly
connected will be equivalent to t h e above t h ree-phase t ra nsfo r m e r .
( b) T h e complex impedance of the load in 'per unit i n the impedance d iagram if
the base in the transmission l ine is 100 M YA, 345 kY.
2 . 1 S. A three-phase transforme r rated 5 MY A, 1 15 / 13.2 kY has per-phase series
imp edance of (0.007 + jO.075) per u nit. The transformer is eonnected to a short
d istribution line wh ich can be represented by a series impedance per p hase of
(0.02 + jO. 1 0) per unit o n a b a s e o f 10 MV A , 1 3.2 kY . T h e l i n e supplies a balanced
th ree-phase load rated 4 M VA, 1 3 . 2 kY, with lagging power factor O.�S .
( a ) D raw an equivalent circuit of the system indicating all impedances in per u n it.
Choose 10 M YA, 1 3.2 kY A as the base at the load.
( b ) With the voltage at the p rimary side of the transformer held constant at 1 1 5
kY, the load at the receiving end of the line is d isconnected. Find the voltage
regulation a t the load.
2 . 1 6. Three identical single-phase transformers, each rated 1.2 kY / 1 20 Y, 7.2 kY A a n d
h aving a leakage reactance of 0.05 p e r unit, are connected together t o form a
three-phase bank. A bala nced Y -connected load of 5 n per phase is con nected
across the secondary of the b a n k . D e ter m i n e the Y-equivalent p e r- p h ase i m p e d a n ce
LQ:.
PROBLEMS
2. 17.
85
( in ohms and in per u n it ) seen from the primary side when the transformer b a n k is
connected ( a ) Y-Y, ( b ) Y - � , ( c ) � -Y , and ( d ) �-�. Use Table 2. l .
Figure 2. 1 7( a ) shows a t h ree-phase generator supplying a load t hrough a three­
p hase transformer rated 1 2 kVA/ 600 V Y, 600 k Y A The transformer has
per-phase leakage reactance of 1 0 % . The line- to-line voltage and the line current
at the generator term i nals are 1 1 .9 k Y and 2 0 A , respectively. The power factor
seen b y the generator is 0. 8 l agging and the phase sequence of supply is ABC.
( a ) Determine the l ine curren t and the line-to-line volt age at t he load, and t h e
per-phase C equiva lcnt-Y) i mpedance of t h e l oa d .
( b ) Using t h e li ne-to-neutral v oltag e VA a t t h e t r a n s form e r p r i m a ry a s reference,
draw cO Ill p lcte per-phase phasor diagrams of all vol tages and curren ts. Show
t h e co r r e ct phase re lations be tween primary and secondary q u a n t i t i e s .
( c ) Com p u t e t h e r e a l and r e a c t i v e pow e r supplied by t h e gene rator and consumed
by t h e l oa d .
2 . 1 S . S o l ve P r ob . 2 . 1 7 w i t h p h a s e s e q u e n ce A CE .
2 . 1 9. A s i n g l e -p h ase t ra n s fo r m e r r a t e d 30 k Y A , 1 20 0 / 1 20 V i s
(a)
t r a n s fo r m e r t o s u p p l y 1 3 20 Y f r o m a 1 200 Y b u s .
as a n au to­
marks on
each winding so
D r a w ;\ d i ag r ,l m o f t h e t ra n s fo r m e r con n e c t i o n s sh o\\ i n g t h e p o l a r i t y
the w i n d i ngs
( u ) i'.'1 ark
tha t
(d)
co n n ec t e d
and d i rections c h o se n as
t h e c u r re n t s w i l l be
on t
he
in phase.
pos i
d i ag r a m t h e v a l u e s o f ra t e d
t ive
for
curren t
CU iTent
in the
in
windings and at the
i n p u t and o u tp u t .
Determine t h e rilted ki lovoltampercs of t h e u n i t as an autotransformer.
If t h e efficiency o f t h e trz:nsformer con nected for 1 200/ 1 20 Y operation at
r a t e d load u n i t y p owe r fl1ctor is 97%, d e t e r m i ne its e fficiency as an au totrans­
fo r m e r w i t h r a t e d c u , re n t in t he windings and o p e r a t i ng a t rated voltage to
supply a l o a d at u n i ty power factor.
Sol'/e P r ob . 2 . 1 9 if the t raf'.sformer is t o supply 1 080 Y h o m a 1 200 Y bus.
Two b u s e s n and b a r e con nected to each other through imped ances X l = 0.1 a n d
X 2 = 0 . 2 p e r u n i t i n p a r a l l e l . B u s b is a load bus s u p p l y i n g a cu rrent J = 1 .0 - 300
pcr u n i t . T he per-u n i t b u s vol t ag e Vb is 1 .0 �. Find P and Q into bus b through
C n c h o f t h e pa ra l l e l b r a n c h e s ( (I ) i n the circll it described, ( b ) i f a regulating
t r a n s fo r m er i s c O n lH.: C f l: d < I t b t i S Ii i n t h e l i n l: o f h i g h e r !T � l c t ;l n c c t o g i v e i t boos t of
3 % i n v o l t a g e rn a g n i t l l ci e towa rcl l i l l: l o a d ( 0
1 . 0J), a n d ( c ) i f t h e r e g u l a t i n g
t r ;1 I 1 s fo r l1l e r a d v a n c e s t h e p h a s e 2° ( (I
I' j;l /C)(I). Use t h e c i r c u l a l in g - c u r r e n t m e t h od
fo r p;\ r t s ( h ) a n d ( c), a n d ;\ S S U l1l e t h il t ,<, is a dj u s t e d [o r each p a r t of t h e prob lem
s o t h a t VI) re m a i n s con s t a n t . F i g u re 2 . 2 () i s t h e s i n g l e - l i n e d i a g r a m showing buses a
3 n d b of t h e sys t e m w i t h t h e r e g u l a t i n g t ransfo r m e r i n p l ac e . Neglect t he impeda n ce
(c)
2.20.
2.21.
/
=
= ,
of t h e t r a n s fo r m e r .
,. \
\!)
x
�-,
)0.1
1
---
FI G U RE 2 . 2 6
l i r c u i t fo r P r o b . 2 . 2 1 .
86
CHAPTER 2
TRANSFO R M ERS
�
o reactances X l = 0.08 and X2 = 0 . 1 2 per u9i� _are i n parallel between two
per unit,
buses a and b in a power system. If Va = 1 .05 � and Vb = 1 . 0
w h a t should b e the turns ratio o f t h e re gu l a t ing transformer to be inserted in series
2.22. Tw
with X2 at bus b s o that
2 .23.
no vars flow into bus b from t he branch whose reactance
i s Xl? Use t he circu la t i ng c u rren t me t hod a n d neg l ec t the reactance of the
r egu la t i ng transformer. P a nd Q of t he load and Vb rem a i n constant.
Two t r a n s former s e d e h r at e d 1 1 5Y / 13.2il kY, operate in parallel to supply a load
of 35 MYA, 1 3.2 k Y a t O.H pow c r - fa <.: tor l a g g i n g . T r a ns fo rm<.:r 1 i s r a t e d 20 M Y A
with X = 0.09 p e r uni t , an a t ransformer 2 is r a t e d 15 MY A w i t h X = 0.07 p e r
unit. Find the magni t u d e o f t h e current in per u n i t through each transformer, the
meg av o l ta mpere output of each transfo rmer, a n d the megavol tamperes to wh ich
t h e total load m u s t be limited so that neither transformer is overloaded . If t he taps
on t ra ns fo r m e r 1 a rc s c t at I I I kY to g i v e a 3 . 1l % boost in vo l t a ge t o w a r d t h e
l ow v o l t a g e s i d e o f t h at t rans fo r m e r c o m p a red L O t r Ll I1 s fo rm c r 2, w h i c h r emLl i n s on
t he 1 15-kY tap, find the megavoltampere output of each transformer for the
origin a l 35-MYA total load and the m aximum megavoltamperes of t he t o t a l l o a d
w h i c h w i l l not overload the transformers. Use a base of 35 MYA, 13.2 kY on the
l o w v o l t a ge side. The circulating-curre nt method is sat isfactory for this prob l em
-
,
,
-
-
.
CHAPTER
3
THE
SYNCHRONOUS
MACHINE
syncIHonou s m achine as an ac genera tor, driven by a turbine to convert
m e c h a n i c a l e n e r gy into e l e c trical energy, is the major e l ectric power generat i n g
s o u r c e throughout t h e world. A s a motor t h e machine converts e lectrical e n e rgy
Th e
mech anical energy. \Ve
tOL b u t we shall give som e
are chiefly concern ed with the synchronous genera­
consideration to the synchronous motor. We c a n n ot
treat the synch ronous machine fu lly, but t here are m a ny books o n the subject o f
a c m a c h i n e ry w h i c h provi d e qu ite a d e q u a te a n a l y s i s o f generators a n d m otors . !
O ur i n t e r e s t i s i n t h e J pp! ica t i o n a n d opera t i o n o f the syn chronous m a c h i n e
w i t h i n a l a rg e i n t e rconnected power syste m . Em ph asis i s o n principles a n d
ext e r n a l b e h avior u n d er both steady-state a n d t ransie n t conditions.
The w i n d i ngs of the polyph ase syn c h ronous machine constitute a group o f
ind uctively coupled electric c ircuits, some of which rotate relative to others so
t h a t mu t u a l i n d u ctances are variab le . The general equations developed for the
flux lin k2.ges u f t h e various win d i ngs a r e appl icabl e to both steady-state a n d
tra n s i e n t a n al)'sis. On l y l i n ear m agnetic circuits are con sidered, w i t h saturation
n eglecte d . T his allows us, whenever convenient, to refer separately t o t h e flux
to
I
For
3
m u c h m o re d e t a i l e d discussion of s y n c h r o n o u s m a c h i n e s , co n s u l t any of
m a c h i n e ry s u c h a s A . E . F i t zge r a l d , C. K i n g s l e y ,
M c G r aw-H i l i , I n c , N e w York, 1 98 3 .
the
texts on
electric
J r. , a n d S . D . U m a ns, Electric Mach inery , 4th e d . ,
87
88
CHAPTE R
3
THE SYNCH RONOUS M AC H I N E
and flux l i nkages produced b y a compone n t m a gnetomot ive force (mmn-even
though in a ny e lectric machine there exists only net physical flux due to the
resultant mmf o f a l l t h e magnetizing sources. S implified equ ivalent circuits a r e
developed t hrough which important p hysical relationships within the m a c h i n e
can be visualized. Therefore, our t r e a t m e n t o f the synchronous machine shoul d
p rovide confidence i n the equivalent circuits sufficient f o r un derstanding the
role o f the generator i n our fu rther s t u d i e s of power syste m analysis.
3.1 DESCRIPTION OF THE SYN C H RO N O U S
MACHINE
The two p r i n ci p a l p a rt s o r a syn c h ro n o u s m a c h i n e a rc fe r ro m a g n e t i c s t r u c t ur e s .
The stationary p a r t w h i c h i s ess e n t i a l l y a h o l l ow cy l i n d e r, c a l l e d t h e .\'/(I / oJ' o r
armature , has l ongitud inal slots in w h ich there a re coils -of t h e a rm a t ure
wind ings. These windings carry the current supplied to a n electrical load by a
generator or t he current received from a n a c supply by a motor. The rotor i s the
p a r t of the machine w h ieh is mounted on the shaft and rotates i n s i d e the holl ow
stator. The winding on the rotor, called the field winding , is s u p p l i e d w i t h d e
current. The very high m m f produced b y this curre n t i n the fi e l d w i n ding
combines with the mmf produced b y c urrents i n the a rm a t ure windings. The
resultant flux across the air gap between the stator a n d rotor generates vol tages
in the coils o f the armature w i n di ngs and provides the e lectromagnetic torq ue
between the stator and rotor. Figure 3.1 shows t he t h reading o f a fou r-pol e
cylindrical rotor into t h e stator o f a 1 52 5 - MVA generator.
The dc current is supplied to the field winding by an exciter , which may be
a generator mounted on the same shaft or a separate d c sou rce connected to
the field winding through brushes bearing on slip r i n g s . L a r g e ae generators
usually h ave exc iters consisting of an ac source with solid-state rectifiers.
If the machine is a generator, the sh aft is d riven by a prime mover , w h ich
is usually a steam or hydraulic turbine . The electromagnetic torque developed
in the generator when it delivers power opposes the torque of the p rime m ove r.
The d ifference between these two torques is due to losses in the iron core a n d
friction. I n a motor the electromagnetic t orque developed in t h e machine
(except for core and friction losses) is converted to the shaft torque which drives
the mechanical load.
Figure 3.2 shows a very element a ry three-phase generator. The fi e l d
w inding, indicated b y the j-coil, gives rise to two poles N and S a s m arke d . The
axis of the field poles is called the direct axis or s i mply the d-axis, while the
centerline of the interpolar space is c a l l e d the quadrature a:ris or simply the
q-axis. The positive direction along the d-axis leads the positive d irection along
the q-axis b y 90° as shown. The generator i n Fig. 3 .2 i s called a nonsalient or
round-rotor machine b eca use i t has a cyli n drical rotor l ike that of Fig. 3 . 1 . In the
actual machine the winding has a la rge number of t u rns d istributed i n slots
around the c ircumference of the rotor. The strong magnetic field produce � l in ks
J. I
F1 G lJRE
D ESCR I PT I O N O F THE S Y N C H R O N O U S MACH I N E
89
:U
P h otograph s ho\',ing t h e t h re a d i n g of a fo u r - p o l e cyl i nd r i c a l r o t o r i n t o t h e s t a t o r of a l S2S - M VA
g e n e r a t o r . ( Cou rtesy Utilit}' POIIEr Corpora tioll , Wisconsin .)
t h e s t a t o r coils t o i n u c e vol t a g e i n
d
by
t h e p r i m c m ov e r .
Th e
t h e a rm a t u r e windi ngs as the shaft is turned
s i d e s of a c o i l ,
w h i c h i s a l m o s t r c c t a n g u l ; l r , ;I rc i n s l o t s a a n d a' I SO° a p a r t . S i m i lar c o i l s a re i n
s l ots h a n d h i ,l !ld s l o t s c , I n d ( . Co i l s i d e s i n s l o t s [{ , h , ,l n d c a re 1 200 a part.
T h e c o n u u c t or s s h o w n i n t h e s l o t s i n d i c l t c a co i l o r o n l y Olle t u r n , but slIch a
coil m a y h ave m a ny t u r n s a n d i s u su a l ly i n series with i d e n t ical coils i n a djacent
slots to fo r m a w i n d i n g having e n d s des ignated a and a' . Wind ings w i t h ends
designated b
hi a n d c
c' a r c t h e s a m e a s t h e a
d w i nd i ng except fo r
t h e i r sym m e t r i c a l l o c a t i o n at a n g l e s of 1 2 ()0 a n c i 24() 0 , respectively, a ro u n d t h e
S i ll t o r
i s s h o w n i n cross s e c t i o n i n F i g . 3 . 2 . O p p o s i t e
"
-
-
-
armature.
Figure 3 . 3 shows a sa lien t -pole m a c h in e w h i c h h a s four poles . O ppos i t e
sides of a n arm a t u r e coil a r e 90° apart. So, t here are two coils for each p hase.
Coi l sides a , b , a n d c of adj ace n t coils a r e 60° a p a r t . The two coils of each
p h a se m a y be connected i n series or in parallel.
90
,,-axiS
,
CHAPTER 3
Ai r ga p
THE SYNCHRONOUS M A C H I N E
Stator
FI G U RE 3.2
Elementary three- p hase ac g e n e ra­
Rotor
d-axis
"
tor showing end view of t he two-pole
cyl i n drical rotor and cross sectiOn of
the stator.
mmf of de w i n d i n g
de field
� w i n d i ng coi l s
q-axis
-'.
FIG U R E 3.3
Cross section of an element ary
stator and s a l i e n t-pole rotor.
Although not shown in Fig. 3 . 3 , salient-pole machines usually h ave damper
windings, whic h consist of s hort-circ uited copper b ars t h rough the pole face
similar to part of a " squirrel cage" winding of an induction motor. The purpose
of the d amper winding is to reduce the mechanical oscillations of t h e rotor
about synchronous speed, w h ic h is d etermined by the n umber of poles of t h e
machine a nd the frequency of t h e system to w h ic h the mach ine is connected.
I
3.2
THREE-PHASE G E NERATION
91
I n the two-pole machine one cycle o f voltage is generated for each
revolution of the two-pole rotOL I n the four-pole machine two cycles a r e
generated i n e a c h coil per revolu t io n . Si nce t h e number of cycles p e r revolution
equals the n umber of pairs of p oles, the frequency of the generated voltage is
f=
P N
2 60 = 2 fm
w h e re f
=
e l ectrical frequency in
P
=
number of poles
N
=
fill
=
P
(3.1)
Hz
Hz
rotor speed i n revol utions per minute ( rpm )
Nj6() ,
0. 1 )
the
mcch:tn iCid
I ILlt
freq l l e n cy i n revo l u t i o n s
rer s ec o n d ( r p s ) .
a t J GOO r p m ,
w h e r e a s :l fo ur - p o l e m a c h i n e ope rCl t es : I t I t) ()() r p m . U s u a l l y , foss i l - flred steam
t urbogcn e rators are two- p o l e m a c h i nes, whereas hyd roge ncrat ing u n its a re
slOwer machi nes w i t h many pole p a i rs .
S i nce one cycle o f voltage (3 600 of t h e vol tage wave) is generated every
time a pair of poles passes a co i l , we must d istinguish b e tween electrical degrees
used to express vo l tage and curre n t and mechanical degrees used to express t h e
pc)s i t i o n o f t h e rotor. I n a two-·pole m achine electrical a n d mecha n ical degrees
are e q u a l . I n any other mach i n e the number of e lectrical degrees or radians
e q u a l s P / 2 ti mes the n umber of mechanical degrees or radians, a s can be seen
2
from E q . (3. 1 ) b y multiplying both sides b y 7T . I n a fou r-pole mach ine,
therefore, two cycles or 720 e lectrical degrees are produced per revolution of
360 mechanical d egrees.
I n this chapter all angu l a r measurements are exp ressed in electrical
d egrees u n less otherwise stated, a n d the d i rect axis always leads the quadrature
?,xis b y 90 e l e c t r i c a l d e g rees i n t h e coun terclock wise d i rect ion o f rot a t ion
rcg( "Hdiess of t h e n u m ber of poles o r the type of rotor const ructio n .
Eq u d l i o n
3.2
t e l ls u s
a
t w o - p o l e , ()( ) - I- I z
m ac h i ne
uperates
THREE- PHA S E G EN ERAT IO N
The R e i d and armature w i n d ings o f the synchronous mach ine described in Sec.
3 _ 1 a r e d i s t r i b u t e d i n s l o t s a ro u n d t h e p e r i p h e ry o f t h e a i r gap . Section A. I of
t h e Append ix s hows t h a t t h e se d i s t r i b u ted w i n d i ngs can be replaced a long t h e i r
axe s b y con cen tra ted coils w i th a p p ro p r i a t e s c J f- a n d m u t u a l i nducta nces. Figure
3.4 \hows t h ree s uch coils-a, b , a n d c-whid: represent the three armatu re
w i n d i ngs on t h e stator of t h e rou n d - rotor machine, and a conce ntrated coi l f,
which represents the distri b u t e d fie l d winding on the rotor. The t hree station ary
a r m a t ure coi l s are identical in every respect and each has one of its two
t e r m i n a l s conn ected to , I co m m o n p o i n t o. The other t h ree ter m inals a re
92
CHAPTER 3
THE SYNCH RONOUS M A C H I N E
()d
=0
a- axis
Quad rature axis
-
Direct axis
Rotation
Rotating field
ir/
f
�
� v{f'
"""
b- axis
�
"
.-'
b
+
__
f'
\
\
\
I
I
I
f
R , Le,
.....
c
"-
-....
c - axis
FIG U RE 3.4
Idealized thr ee-p h ase generator s h owi n g i d e n t i cal a r m a t u re c oils a , b, and c . and fieJd coi l
axis leads q u adrature axis by 90° in the a nt i clockwise d i re c t i on of rot a t i o n .
f. D i rect
marked a, b, and c . The axis o f coil a is chosen a t (J d = 0° , and coun terclock­
wise around the air gap are t h e axes of the b-coil at (J d = 1 200 and of the c-coil
at (Jd = 240° . For the round-rotor machine it is shown in Sec. A.1 of t h e
Appendix that:
•
Each of the concentrated coils a , b, and c has self-inductance L which is
equal to the self-inductances Lac/ > Lb1p and Lee of the d istributed a rm a ture
w i n d i ngs w h ich the coils represent s o t h a t
S
'
( 3 2)
.
•
The mutual inductances Lab' Lbe, and Lea between each adjacent p a i r of
concentrated coils are negative constants d enoted by - Ms so that
( 3 .3)
•
The mutual i n ductance between t h e field coi l f and each of the stator coils
v a ries w i th the rotor position (Jd as a cosinusoidal function with m ax'imum
3.2
value
Mf
T H REE·PHASE G ENE RATION
93
so that
( 3 .4)
The
field co i l has a co n s t a n t se l f- i n d u c t a n ce L ff ' T h i s i s because i n t h e
ro u n d - rotor m a ch i n e ( a n d , indeed, i n t h e sal ient-pole machine also), t h e fi e l d
w i n d i n g o n t h e d-axis prod uces fl u x t h ro u g h a s i m i l a r magnetic path i n the
stator for a l l pos it ions o f t h e rotor (neglec ting the small effect of armature
sl o t s ) .
t h e co i l s a , b ,
c u rr e n t a n d t h e cu rr e n t s i n
t h re e
t h e refore w r i t t e n for a l l fo u r c o i l s as fo l l ow s :
of
the
Flux l i n kages with e ac h
Armature :
an d
f a r e due to
own
other coils. Flux- l i nkage equations a re
c,
its
..
Field :
( 3 .6 )
I f i ll ' i i"
a n d I e a rc a
S e t t i n g i ll
balan ced t h r e e - p h a s e set o f c u rr e n t s , the n
( 3 . 7)
( 3 . 8)
For n o w we are i nt erested in steady-state condi tions. We assume, there­
fore, that c u rr e n t if i s dc w i t h a constant value If and that the fiel d rotates a t
,
94
CHAPTER 3
constant
THE SYNCHRONOUS MACHIN E
a ngul ar velocity w so that for the two-pole m achine
d8d
- =w
dt
and
( 3 .9)
i nitial p osition of the fi e l d winding is given by the angle fJdO ' which c a n b e
arb itrarily chosen a t t = O. Equations (3 .4) give the express ions for Luf' Lhf ,
and LeJ i n terms o f 8d• Substituting ( w I + 8d O ) for 8d and using the resu lts
along with i f = If in Eqs. (3 .8), we obta i n
The
Aa
=
( Ls +
Ms ) i a + Mf If cos ( w t +
8d O )
( 3 . 1 0)
The fi rst of t h ese equations shows t h a t A a has two fl u x - li nkage cum p o n e n t s - o n e
due to t h e fi e l d current If a n d t h e o t h e r d u e t o t he armature current i a , which
is flow i ng out o f the machine for generator action. If c o i l a has resistance R ,
then t h e voltage drop v a across t h e coil from terminal a to terminal 0 i n F i g.
3.4 i s g iven b y
The nega t ive signs apply, a s d iscussed i n S e c . 2.2, because t h e machine i s being
treated as a g en erator. The last term o f Eq. (3 . 1 1 ) represents an internal emf,
which we n ow call ea' . This emf can be written
( � . 1 2)
w here the rms m agnitude I Ei l , p ro p o r t i o n a l to the Reid curren t, is defined by
IE I
I
=
f ..
I... f..:....
_
w_
M....:.
fi
( 3 . 13)
action of the field c u r r e n t c a us e s ea' t o a p p e a r across the terminals o f t h e
a-phase when i a is zero, a n d so it i s called by vari ous n ames such as the no-load
voltage, the open-circuit voltage, t h e synchronous internal coltage, or generated
emf of phase a . The angle ()d O ind icates the position of the field w inding (and
the d-axis) rela tive t o the a-phase a t t = O . Hence, 8 � 8dO - 90° ind icates the
position o f the q - axis, which i s 90° behind t h e d-axis i n Fig. 3 . 4 . For l a t e r
convenience we now set 8dO = 8 + 90° , a n d then w e have
The
(,3 . 14)
3.2
TH R E E - P HA S E G E N E RA TI O N
9S
where ed ' w, and 0 h ave consiste n t u n i ts o f angular measurement. Subst i tu ting
from Eq. (3 . 1 4) into Eq _ (3. 1 2) a n d noting that sin(a + 900 ) = cos a , we obtain
for the open-circuit voltage of p hase a
The terminal voltage
V
a
of
( 3 . 15)
Eq . (3 . 1 1 ) i s t h e n g iven by
( 3 . 1 6)
This equ ation corresponds to the a - phase circu i t of Fig. 3.5 i n which the no-load
vol tage c a ' i s t h e s o u r c e c t n d t h e e x t e rn a l l oad i s b a l a n ced across a l l th ree
phases.
(3. 1 0) c a n h e t rea ted i n t h e same
S ince t h e a rm a t u re w i n d i n g s a re identical, r e s u l t s similar to Eqs.
(3. 1 5 ) and ( 3 . 1 6) ca n be fou n d for the no-load vol tages e b , a n d e c' which l ag e a'
T h e fl u x l i n k a g e s A " a n d A c g iv e n h y Eq .
way as A a '
by 1 200 and 2400 , respe ctively, in Fig. 3 . 5 . Hence, e a , e b " and ec' constitute a
b a l a nced th ree-phase set of emfs which give rise to balanced t h ree-phase l i n e
,--- '--�-----:-_.
+
a
- - - - r- - - - - r
L-_____________________l_+ i:_'..... : L
c -�- � -
_
_
Fl G G RE 3.5
Arm a t u r e e q u i va l e n t c i r c u i t
vol t a g e s ea' , E,i . a n d
ee'
of
th e
i d e a l ized
in t h e s t e a d y s t a t e .
n
t h re e - p h ase g e n e r a t o r s h ow i n g b a l a nc e d
no-load
96
CHAPTER 3
THE SYNCHRONOUS MACH I N E
currents, say,
ia
=
i ll
=
ie
=
V2 l /a I cos( w I + 0
(5
n l / l cos ( w t +
a
n I /u 1 cos ( w t
+ (5
-
-
-
ea )
ea
ea
-
120° )
-
240° )
( 3 . 1 7)
where I fa l i s t h e rms v a l u e and Va is t h e phase a n g l e or lag o f t h e c u r re n t i ll
with respect to e,l ' W h e n t h e cm rs a n d t h e c u rre n t s a re expre ssed CiS p h a s o rs ,
Fig . 3 .5 becomes v e ry m uc h l i ke t h e eq u i va l e n t c i r c u i t i n t rod uced i n F i g . 1 . 1 l .
B e fo r e e m p l oy i n g t h e e q u iv;t l c n t c i rcu i t , l e l u s c o n s i d e r t h e n u x l i n k ,l g e s '\ f o r
t h e fi e l d wi n d i n g .
T h e e x p r e ss i o n s for I� I/ ! ' 1> 1>1 ' a ll d L ' I i ll Eq s . 0.4) C l ll h e s u b s t i t u t e d i n t o
E:q . (3 .6) t o y i e l d
The fi r s t t e r m w i t h i n t h e b rackets c a n be exp a n d e d a cco r d i n g to Eqs .
(3. 17) a s
(3. 1 4 )
and
fol l ows :
(3 . 19)
The
E: q.
t ri go n o m e t r i c i d e n t i t y
2 cos
a cos
(3. 1 9 ) yields
(3
=
cos( a
-
(3 ) +
cos( a
+ (3) a p p l i e d t o
( 3 . 20 )
T he i u a n d
Ie
terms in E q . (3 . 1 8) l e a d
I /a l
12
-
s i milar results, a n d
sin(2( wl
+
8)
-
e"
we
-
h av e
1 20° ) } ( 3 .2 1 )
(3.20) t h rough (3 . 22) a r e b a l anced s e c o n d ­
har m o n i c s in u so i d a l q u a n t i ties w h i c h sum t o z e r o a t e a c h p o i n t i n t i m e . H e n c e ,
addi ng t h e b racke ted t e r m s of E: q . (3. 1 8 ) toge ther, w e ob t a i n
The t e r m s i nvolving
2wt
{ - si n e"
to
i n E:q s.
3.2
TH REE-PHASE G EN E RATION
97
and t h e expression for ;\ f t a k e s on t h e s i m p l e r form
( 3 .24)
cu rren t
w h e r e de
by Eq . (3 .23)
/f [ i a cos fid
ld =
+
iI.J COS( (Jd
1 20° )
-
i / cos
+
(Jd
-
240° )] or
( 3 .25)
..
w h i c h i s u s e fu l l a t e r i n
t h i s c h a p t e r. F or n o w l e t u s observe fro m Eq_ (3 . 24) t h a t
t h e fl u x l i n k a ge s w i t h t h e fi e l d w i n d i n g d u e t o t h e comb i n a t ion o f i a ' i b • a n d ie
d o n o t va ry w i t h t i m e _ \Ve c a n , t h e r e fore , r e g a r d those fl u x l i n kages a s com i n g
from t h e s t e a d y d c c u rre n t i " i n � l fict i t i o u s d c c i rc u i t c o i n c i d e n t w i t h t h e d-ax.i s
a n d t h u s s ta ti o n a ry w i t h r e s p e c t to t h e fi e l d c i rcu i t . The t w o c i rc u i ts ro t a t e
t o g e t h e r i n s y n c h ro n i s m em d h ave a m u t u a l i n d u c t a nce
Mf be tw e e n
t h e m , as s h o w n i n F i g . 3 . n . I n g e n e r ;t I , t h e fi e l d w i n d i n g w i t h r e s i s t (l n ce R f a nd
e n t e r i n g c u r r e n t i l b as term i n a l vol t a g e UIt give n by
( /3/2 )
( 3 _26)
�:
a - axis
" /"
Di rect axis
0,
i/&6)�) j 3 :
"
""
"
/
r
I
L
winding rotati n g with ro/
lor)
Arn1;]t u r e e q U i v a l e n t
2-
""""
/: ,
'/;/ (
/
f
I fr
� ".. ".. "..
"
__
_ ".. - l'rr
"...­
Quad rature axis
,
..-' "..
:
M
1
M utual
,, " /
(
J
/
/
/
"
/
"
"
.'
/
"
"
"
/
i n d uctance
"
Field windi n g rota t i n g
with rotor
,,,-
f'
'"
b - axis
c - axis
""-
FI G L RE 3 . 6
/3 / 2 /1f!
R e p r e se n t i n g
in ductance
lhe
a r m a t u re
with the
of
lhe
li e l cJ
s yn c h r o n o u s
m3chine
by
a
d i rect-axis w i n d i n g of m u t u a l
synchro � ism.
w i n d i n g . B o t h w i n d i n gs r o t a t e t o g e t h e r i n
98
CHAPTE R
3
THE S Y N C H R O N O U S M AC H I N E
Because A f is not varying w i t h time i n the steady state, the field vol tage
becomes vff' = R f lf and i f = If can be supplied by a dc sou rce .
Equ ation (3.25) shows t h a t the n u me rical value of id depends on the
magnitud e of the armature curre n t I Ia I a n d i ts phase angle of lag e a relative to
the internal voltage ea" For l agging power factors ea i s positive and so id is
negative, w hich means that the combined effect of the armature currents i l b '
a n d i e is d emagnetizing; t h a t is, i d opposes the magnetizing influence of the
field current If ' To overcome this demagnetizing influence, If has to be
i ncreased by the excitation system of t h e gen erator. At leading power fac tors 8a
takes o n smaller val ues, which me ans t h a t t h e de magnetizing influence of the
a rmature currents (represen ted by i d = - /3 I Ia 1 sin ea) is reduced and If can
then be l owered b y the excitation sys t e m. In an actual machine the effect of the
currents i Q ' ib, and i c is called a rm at u re rea c tio l l a n d the control of the field
current is called excitation system con trol, w h i c h is d iscussed i n Sec. 3 . 4.
a '
Example 3 . 1 . A 60- Hz three-phase synchronous generator with negligible armature
resistance has the following in ductance para meters:
Laa
Lab
=
=
Ls
Ms
2 .7656
=
1 .3828
=
Mf
mH
mH
L
ff
=
=
3 l .6950
mH
433 . 65 69
mH
The machine is rated at 635 MY A, 0 . 9 0 power-factor lagging, 3600 rpm, 2 4 kY.
When operating under rated load co ndi tions, the l ine-to-neutral terminal voltage
and l ine current of phase a m ay be written
Va =
1 9596 cos
wI
Y
in
=
2 1 603 COS ( w l
-
2 5 . 84 1 9° ) A
Determine t h e m a g n i t u u e o r t h e sync i l r o n o u s i n t e rn a l v o l t a ge , t h e fi e l u curren l if '
and t h e fl u x l in k a g e s w i t h t h e fi e l d w inding. C a l c u l a t e t h e v a l u e s o f t h e s e q u a n t i t i e s
wh e n a load of 635 M YA is s e rved a t r a t e d v o l t a ge a n d unity po w e r f a c t o r . W h a t is
the field current for r a t e d a r m a t u r e vo l t a g e on an open ci rcu it?
The given maximum value of Va is )2 (24,000/ [3 ) = 1 9596 Y , the
maximum value of in is )2 (635,000/ fS X 2 4 ) = 2 1 603 A, and the power-factor
angle 0 = cos - 1 0.9 = 25 .84 1 9° l agging.
W ith R = 0, in Eq. ( 3 . 1 6) the synchronous i nternal voltage can be written
Solution.
=
=
Va + ( 2.765 6
+
1 95 96 cos
- ( 4 . 1 484) 1 0 � 3
w(
1 . 3828) 1 0 - 3
di a
--:it
X w X
2 1 603 sin ( w t - 25 . 1)4 1 9° )
3.2
S et t i n g
w = 1 20rr,
w e ob t a i n
e a, = V2 I Ei l cos ( w l
and expanding t h e seco n d
gives
+
8 ) = 1 9596 cos w i - 33785 sin ( w t - 25 _84 1 9° ) V
term
according to s i n( a - (3 )
ea, = V2 I E, l cos( w l
+
=
45855 cos ( (tJ I
a
cos
=
4 5055
-� ---
1 2 ( ) rr x 3 1 . (/Xi x I ( )
·
=
flu x l i n ka g e s w i t h t h e fi e l d w i n d i n g s are g i v e n by Eq.
behind
Va ' w hic h
l a gs
en
I fa l s in e a
4 1 .5384°
= 25 .84 1 9°
=
cos
fi I E, I = 45855
a
sin
(3 .24),
ea, .
b e h i n d e a" it fo l l ows t h a t
+
V and
S i nce In
l a gs
4 1 .5384° = 67.3803°
2 1 603
.fi s m 67 .3803° = 1 4 1 00 .6 A
a n d subst i t u t i n g i n t h e a bove e x p r e ss i o n fo r A f y i e l d s
Af
= ( 433 .656()
x 10
1 ) 3838
- --�.f2=i��3 x
1 664 .38 - 948.06 = 7 1 6 .32
3 1 .695
x j0
.1
x
14 1 00 .6
Wb- t u r n s
R e p e a t i n g t h e a bove s e q u e nce o f c a l c u l a t i o n s a t u n i t y p o w e r fa ctor, we
e , = fi I E I cos( w t
il
I
+
f3
3 � n � ;\
\v here ea is t h e a n gl e of J a g of in m e as u r e d w i t h respect t o
25 .84 1 9°
(3 -
+ 4 1 .5384° ) V
Bence, t h e syn c h ro n o u s i n t e r n a l v o l t a g e has m a g n i t u d e
angle 8
4 1 .5384° . From Eq . 0. 1 3) w e l i n d
{(1 M,
sin
0 ) = 34323 cos w i - 30407 sin w t
=
The
99
T H R E E - P H A S E G E N E RA TION
0)
=
1 9596 cos w I - 33785 s i n w I
= 39057 cos (
(u
( -f
59 .88 )4° )
ob t a i n
100
CHAPTER
3
THE SYNCH RONOUS MACH I N E
Because I Ei l is directly proportional to If' we have from previous calcu la tions
If =
Current t a
is in
I Ja l sin Oa
phase
=
Al =
=
with
VII
3 90 5 7
45855
a n d l a gs
1 5276 sin 5 9 .8854°
(433 .()S()c)
1 4 1 7 .62
-
X
X
=
3838
e ll' by
0/)0 .43
=
3269
A
5 9 . 8854° . Ther e fo re ,
1 32 1 4 A
1 ) 3 2()c)
10
=
3
-
X
3 1 .695
./2
X
10-3
X
1 32 1 4
5 2Sl . 1 Sl Wb- L u rns
Thus, when the power factor of the load goes from 0 . 9 lagging to 1 . 0 under rated
megavoltamperes load ing and voltage cond itions, the field current is reduced from
3838 to 3269 A . Also, t h e net air-gap fl u x l i nking the field winding of t h e generator
is reduced along with the demagnetizing influence of armature reaction.
The field current required to maintain rated terminal voltage in the machine
u nder open-circuit conditions is found from Eq. ( 3 . 1 3), and Eq. ( 3 . l 6) with i a = 0 ,
/f
-
v'2I Ei l
-w Mf
1 9596 X 1 0 3
-
----=
120
X 3 1 .695
..
1 640
A
3.3
SYNCHRONOUS REACTAN C E
A N D EQUIVALENT CIRCUITS
The coupled-circu it model in Fig. 3 . 4 represents the idealized V-connected
round-rotor synchronous machine. Le t us assume that the mach i n e is rotating at
syn c h ronous speed w and that the field current Ii is steady dc. Under t hese
conditions the balanced three-phase circuit of Fig. 3 . 5 gives the steady-s tate
operation of the machine. The no- load voltages are the emfs e a" eh" and e c, ·
Choosing a-phase a s t h e refe r ence p hase for t h e m ac h i n e , we o b t a i n t h e
p er-phase equivalent circuit of Fig. 3 .7 ( a ) with s teady-state sinusoidal currents
a n d voltages which lead the corresponding currents and vol tages of phases b
a n d c by 1 200 a n d 2400 , respectively.
We recall that the phase angle of the current ia in Eq. (3. 1 7) i s chosen
with respect to the no-load v oltage e a , of the a-phase. In practice, e a , cannot be
m e as u red under load, and so i t is prefe rable to choose the terminal voltage Va
as r e ference and to measure the phase angle of the current ia with resp'ect to
3. 3
SYNCH R O N O U S R EA CTANCE A N D E Q U I VA L E NT C I R C U ITS
101
-
1-
o
FI G U R E 3 . 7
1-
E q u i va l e n t
�------ �
o
(b)
Vo.
T he r e fore , w e
circllit
for
re fe r e n c e
p h a s e a of t h e synchronous m a ­
c h i n e s h o w i n g voltages a n d c ur ­
r e n t s a s ( n ) cos i n u so i d a l a n d ( b )
ph asor q u a n t i t ies.
deAne
ta
= Ii I I 1 c o s ( w t - 8 )
a
( 3 .27)
Note that e o, cor resp o n d s to Eq. (3 . 1 5) a n d t h a t ia d iffers from Eq. (3 . 1 7) o n ly
i n t h e r e s p e c t t h a t t h e p h a se a n gle 8 = 80 - 0 i s n ow the a ngle o f lag o f i a
m e a s u r e d w i t h r e s p e c t t o t e r m i n a l v o l t a g e VtI • T h e p h a s o r e q u i v a l e n t s o f E q s .
(3 . 2 7 )
a rc
v11
and t hese
n rc
'-'=
/ 0°
I V{J I �
'
.
m a rk e d on t h e e q u i v a l e n t c i rc u i t o f F i g . 3 .7( 6 ) fo r w h i c h
p h a s o r-vo l t ll g e e q u a t i o n I S
va
=
------fU a
E/
--­
n o load
W h e n the curre n t
a r m a t u re
s i st a nc e
D u e to
GenerJted
at
( 3 .28)
re
10
jw Ls Ia
----­
Due to ar rr:at ure
se I f- re a c t a nce
jw MJa
-------
the
( 3 . 2 9)
'--
Due to arm ature
m u t u a l r e a c t a nce
l e ads Va ' t h e a ngle 8 is num erically negative; and when l a
102
CHAPTER 3
THE SYNCHRONOUS MACHINE
lags Va ' the angle () is numerically positive. S ince symmetrical cond itions apply,
phasor equations corresponding to Eq. 0 . 29) c a n be written for b-phase a n d
c-phase. The combined quantity w(Ls + M) o f Eq. (3.29) has the dim ensions of
reactance and is customarily called t h e synchronous reactance Xd of the ma­
c hine. The synchronous impedance Zd of the machine is defined b y
( 3 .30)
and Eq. (3.29) then can be written in the more compact form
( 3 .3 1 )
from which follows the generator equivalent circuit of Fig. 3 .8( a). The equiva­
lent circuit for the synchronous motor is ide ntical to that of the generator,
except that the di rection of I(2 is reversed, as shown in Fig. 3 .8( b ), which has t he
equation
( 3 . 3 2)
r
Zd
'"
"\
1a
-
1+
Va
0
r
Zd
A
l-
(a)
iQUO'
jXd
\
1a
-
1+
•
Va
0
l..
(b)
FIG U R E 3.8
E q uiv a l e n t c i rcu i ts for ( a ) the syn c h ronous
gen e r ator a n d (b) the synch rono u s mot o r w i t h
const a n t sy nchro no u s i m pedance 2,, '= R +
jX.t.
J.J
S Y N C HR O NO U S R E AC TANCE A N D E O U I VALENT CIRCU ITS
103
(6)
FI G U R E 3 . 9
( 0 ) ovncxc i t e d g e n e ra t or d e l ive r i n g I it g g i ' l g c u r re n !
d r awi n g l a g g i n g c u r re n t la '
P h a� o r d i a g r a m s o f :
I,, ;
( h ) u n d e re x c i t e u m o t o r
P h asor d iagrams for Eqs . 0 . 3 1 ) and (3 .32) are shown i n Fig. 3.9 for t h e c ase of
l agging power-factor angle f) measured wi t h respect to the term i n a l voltage . I n
Fig. 3 . 9( a ) for the generator note t h at Ei always leads Va ' and i n Fig. 3 . 9( b ) for
the motor Ei always lags Va '
Except for the case of an isolated generator supplying its own load, most
synchronous machi nes are con n e cted to la rge interconnected powe r systems
such t h at t h e t e r m i n a l vol tage Va ( soon to b e cal led VI for e mph asis ) is not
a lten.:d by machi ne loa d i ng. I n that case the poi n t of con nection is called a n
infin ice bus, which means that i ts voltage remains constant a n d n o frequency
ch ange occurs regardless of cha nges made in operating the sync h ro nous m a ­
chi ne .
S y n c h ronolls mach i ne rarameters a n d opera ting Cl uan tities such as vol tage
a n d c u r re n t lI r c n o r m a l l y rc r rcs e n t c d i n r c r u n i t or n or m a l i z e d va l u es usi ng
b a s e s c o r r e s po n d i n g to t h e Il < l m c p l (l t c (l a t a o f t h e mach i ne. Such parame ters are
prov i ded by the manufact u rer. Mach i nes of similar design h ave norm a lized
param eters which fal l i n a very n arrow ra nge rega rd less o f size, and this is very
(see Ta b l e
A.2 in
Appe n d ix ) . I n the a rm a t u re or t h e t h r e e - p h a s e m a c h i n e u s u a lly the kilo­
vo l t ampere base correspon d s to the t h ree-phase ra t i ng of the mach ine and base
voltage in kilovol ts correspon ds to the rated l in e- to-li n e voltage i n kilovolts .
Accordi n gly, the per-phase equivalent circuit of Fig. 3 .8 has a k Y A b ase equal
to t h e k ilovol tam pere rat i n g of one p h ase a n d a voltage base eq u a l to the r a te d
l i n e - to-n eutral vol t age o f t h e m achi n e . B ase arm ature i m peda nce i s t herefore
c a l c u l a t e d from Eq. ( 1 .5 4 ) in t h e usual w a y.
u s e fu l wh e n d a t a fo r a p Clr t i c u l a r m a c h i n e are n o t a v a i l ab l e
the
104
CHAPTER 3
THE SYNCH R O N O U S MAC H I N E
Although the generated vol tage Ei is con trol led by the fiel d current,
nonetheless, i t is a per-phase armature voltage which can be normal ized on the
arma ture base. Equa tions (3.3 1 ) and (3 .3 2) are thus d irectly applicab l e I n per
unit on the armature base.
The 60-Hz synchronous gener ator described in Example 3 . 1 is
serving its rated load u nder ste ady-state operating conditions. Choosing the
armature base equal to t he rat i n g of t h e m ac h i n e , determ ine th e valu e of th e
syn c h ronous react a nc e a n d t h e p l1 asor e x p re s s i o n s f or the stator q u a n t i t i e s V:p '
Exa mple 3.2.
a n d Ej i n per u n i t . If t h e b;lse fie l d cu rre n t e q u a l s t h a
t va l u e o r If
",
w h i c h p ro d u c es
rateu t e r m i n a l vol t a ge U IH J cr o p e n - c i rc u i t co n d i t i o n s , d e t e rm i n e the v a l u e o f If
under th e sp ec i fied op e ra t i n g cond i t i o n s .
Solution .
a p
From Ex m l e 3 . 1 w e l i n d for t h e a rma t ur e
B ase kYA
B ase
k VL L
B ase cu r rent
B ase i m pedance
=
=c
635 ,000 kYA
24 k Y
635 ,000
=
=
that
IS X
242
-
635
24
=
= 1 5275 .726 A
0 .9071 n
Using the values g iven for the i nductan ce parameters L J and Ms of the a rmatu re,
we compute
Xd
=
w(L,
+
MJ
= 1 20 .. ( 2 .7656 + 1 . 3 828) 1 0 - 3
=
1 .5 639 n
w h ich i n p e r u n i t i s
X"
1 . 5639
=
0 . 907 1
=
1 .724 1
The load is to be served at rated voltage e q u a l to th e specified base, and so
use the terminal vol t ag e Va as the refere nce p hasor, we obtain
V,l = 1 . 0
L.2: per u ni t
if
we
The load current h as the rms m ag n i t u d e I Ia I
635 ,000j(y3 x 24) A, w h i c h is
also the base arm atu re current. He nce, I Ia I = 1 . 0 per un it, a n d since the power­
factor angle of the load is e = cos 1 0.9 = 25 .84 19° lagging, the p hasor form of
the lagging current fa is
=
-
1 .0
/
-
25 . 84 1 9°
per
unit
Synch ro nous i n t e r n a l
=
=
E, c a n
vo l t a g e
LQ:
1 .0
1 . 75 1 5
+
+
3.4
be
cal cu l a ted
=
from Eq. (3. 3 1 )
1 .0/ 2.340/ 4 1 .53840
j 1 . 724 1 X
j 1 .55 1 7
a rm a t u r e
p roport ion a l t o If ' w e
t h e s p e c i fi c d o p e rat i n
g
volt age)
with R
per u n i t
0,
1.0 pe -u
is 1 640 A . T h e r e fo r e , s i n ce I E, I i s d irec t l y
cu r r e n t of 2 . 34 x 1 640 = 3 8 3 8 A u n d er
to produce
h av e a n exci t a t i o n
co n
=
25 . 84 1 9°
I [1 Exa m p l c 3 . 1 t h e b a se fie l d curre n t (which i s requ i red
o p e n - c i rc u i t
1 05
R E A L AN D R EA Cf I VE P O W E R CONTR O L
r
nit
d i t i ons .
The i n terested reader may wish to d raw a ph asor d i agram for the resu l ts
of t h is exa mple a 'n d comp are t h e phaso r method o f sol u t ion w i th the t i me ­
dom a i n a p p roac h of Example 3 . 1 .
3 .4
REAL A � D REACTIVE P OWER C O),;TROL
When the synch ronous m achine is con n e c t e d t o an i n fi n i te bus, its speed a n d
te rm i n a l voltage a re fued and u n a l terable. Two con t rollable variables, how ever,
a re the field current and the m echanical torque on t he shaft. The variation of
the fi eld curr e n t If ' referred to as excita tion systenl control , i s applied t o ei t her
a g e n e r a t o r or a motor to su pply or ab sorb a variable amount of rea ct ive powe r .
Because t h e synchronous m achine runs a t const ant speed, the only m eans o f
varying t h e real power i s through con trol of t h e torque i mposed o n t h e shaft b y
either t h e prime mover i n t h e case o f a generator o r the m echan ical load in t he
case of a motor.
I t is conven i e n t to neglect resi sta nce a s we consider reactive power con t ro l
o f the rou n d -rotor generator. Assume that t h e genera tor is del ive r ing pow e r s o
t h a t a ce r t a i n a n g l e [) exists b e tween t h e t e rm i n a l v ol t a ge v: a n d t h e gener a t e d
v o l t a g e E, 0 f t h e m achine [sec Fi g . 3 . J O( (l )] . T h e c o m p l ex power d e l ivered t o
t h e system b y t h e generator i s g i ve n i n per u n i t by
s
=
P +
jQ
=
V, f:
=
I V, I
1 1,, 1 ( cos e
+ j sin e)
(3.33)
Equ a t i ng real a n d i mag i n a ry q u a n t i t ies i n t h i s equ a t i on, we obt a i n
P
=
I v: I I J ) cos
e
Q
=
I V, I I IJ s i n e
( 3 . 34 )
We note that Q is posit ive for l agg ing power factors since the a ngle e i s
n u merically positive. I f w e d ecid e t o m ainta in a certa in power del ivery P fro m
the generator to the constant vol t age system , it is clear from Eq. (3 .34) t h a t
I la l eos e m u s t remain cons t a n t . As w e v a ry t h e d e fi e l d c ur r e n t If u n d e r t hese
,
106
C H A PTER 3
THE SY;-"CHRONOUS M ACH I N E
/
I
1- - - - - - - --,I - - - - - - - 1 _
I
1
I
:
:
I
:
I
Ia
�
Constant power
locus of E i
- - - - - - - - - - - _ _ _ _ _ _ _
I
Constant power
l ocus of
"" i
8
I
j1a Xd
VI I
I Ia l Xd
E
'
cos
e
(a)
�----�-+--, ---L----� - - - - - - - - - - - J
I
:
f l1alcos e
4 - - - - - +_ 1
I
1
I
I
I
I
I
1
:
- - - - - - - -
o
I
I
I
I
I
:
�
I
I
Ei
-
l Ie Xd sin e
- - - - - - - - - - -
I Ia l Xd
cos
1I<...__
.__
---L...l.L......__
__
_
__
_
_
_
�'1
0
(b)
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _
FIGURE 3 . 1 0
P h asor d i a g r a m s show i n g c o n s t a n t - powe r l o c i o f a n
« (1 ) ove r e x c i t e d ge n e ra t o r d e l i v e r i ng r e a c t ive
power t o t il e syst e m ; ( I) ) u n d c rexc i t ed g e n c r ; t t o r receivi n g r e a c t i ve rowe r from t h e sys t e m . T h e
powe r delivered by t h e ge n e r<l t o r i s the
same
in
b o t h cases.
cond itions, the generated voltage E, varies proportionally but a lways so as to
keep I fa l cos e consta nt, as shown by the loci of Fig. 3 . 1 0( a ). Normal excitation
is defined as t h e cond ition w h e n
I E · I cos fj
I
=
!VI
I
( 3 . 35)
a n d the machine i s said to be e i t h e r ouerexcited or underexcited acco r d i ng t o
w hether I Ei l cos 8 > I � I or I Ei l cos 8 < I VJ For t h e condition of F i g . 3 . 10(a)
the generator is overexcited a n d supplies reactive power Q to the syste m . Thus,
from the system viewpoint the mach ine is acting l i ke a capacitor. Figure J . I 0( b )
3. 4
gI
of a n
(a)
107
gI
ea)
FIG U RE 3 . 1 1
Phasor d i a g r a m s
R EAL AND REA CfI V E POWER CONTROL
(6)
ovc r n c i t e d a n d
(h)
u n d c rcxci t e d sy n c h ro n o u s m o t or d r aw i n g C lI rre n t
a n d C() n s t �l l l t pow e r a t cOlls t a n t t e r l 1 1 i l l ; d vol t a g e .
fa
is for a n u n d c rexci ted gene rator supplyi ng the same a mo u n t of rea l power a n d a
l e a d i n g current to the system , or i t may be considered to be d rawi ng l aggi n g
c u r r e n t fro m t he syst e m . Th e underexcited gener<1 tor d raws reactive power fro m
the system and i n this respect acts l i ke a n i n ductor. The reader is encouraged to
explain t his a ction in terms of the a rm a tu re react ion discussed i n connect i on
w i t h Eqs. (3.24) and (3.25).
Figure 3 . 1 1 shows ove rexci ted and underexcited synchronous motors d raw­
i n g t h e s a me real power at t h e s a me term ina l vol t a ge. The overexcited motor
d raws leading current and a cts l ike a capacitive circuit when viewed from the
n e twork to which it suppli e s reactive power. The unde rexcited motor draws
l agging current , absorbs rea ctive power, a nd i s acting l i ke an i n d uctive circ u i t
w h e n viewed from the ne twork. B riefly t h e n , Figs. 3 . 1 0 a nd 3. 1 1 show t ha t
overexcited generators and motors supply reactive power t o the system a n d
underexcited generators and motors absorb reactive power from the system .
Now we t u rn our a t t e n tion to re (l l powe r P, which is con t ro l l ed by ope n i n g
o r c losi ng t h e valves t hrou gh which steam (or water) enters a turb i n e . I f t h e
powe r i nput t o t h e gene rator is i ncrease d , t h e rotor speed w i l l s t art t o increase,
a nd i f the field c urrent 1[ and hence l Ei I a rc he ld cons tant, t he a ngle 0
b e tween Ei and V, w i l l i ncrease . I ncreasing 0 resu l t s in a l a rger I f l eos (j , as
may be seen b y ro tati ng the ph asor Ei cou n terclockwise i n Figs . 3 . 1 0( a ) a n d
3 . 1 O( b ) . The genera tor wit h a l a rger 0 , t he refore, del ivers more power t o t h e
n etwork; exerts a higher cou n tertorq u e on the prime mover; a n d h ence, the
i npu t from the prime mover i s reestablished at the speed correspon d i ng to
t he frequency of the i nfinite bus. Sim ilar reason ing applies also to a motor.
The dependence of P on the power angle 0 is also show n as foll ows . I f
a
and
J
108
C HAPTER 3
. where � a n d
Ei are expressed in volts to n e u tral or in per unit, then
I Eil� - I V;I
and
THE SYNCHRO N O US MACH I N E
/ a* -
- )Xd
Therefore, the complex powe r d el ivered to t h e system at t h e
genera tor is given by
S = P + j'Q = VI f*
a
=
of the
I V I I E -I / - o - I V, 1 2
_I _' =
�
==- j'X
I VI I I E; I( cos 8
-j
_
_
_
cl
=
t e rm i n a l s
( 3 . 36)
sin
8)
-
IV/
( 3 . 3 7)
-jXd
The rea l a n d imaginary parts of Eq. ( 3.37) are
( 3 .38)
E;
W he n volts rather than per-un i t values are substituted for V; and Ei i n
Eqs. (3 .38), w e must b e careful t o note t h a t V; a n d
are line-to-neutral
voltages and P and Q will be per-phase quan tities. However, l in e-to- l i n e
voltage val u es substituted for V; and E i will y ield total three-phase values for P
a n d Q . The per-unit P and Q of Eqs. (3. 38) are m u l tiplied by base three-phase
m eg avol tamperes or base megavoltampe rcs per phase depend ing on whether
tot a l t hree-phase power o r power per phase i s wanted .
Equation (3.38) shows very clea rly the dependence of P on the power
angle 8 if I E ; I and I V; I a re consta n t . Howeve r, i f P and V; a re consta nt, Eq .
(3 .38) s hows that 0 must d ecrease i f I E; I is i ncreased by boosting the d c field
excit ation. With P const a n t i n Eq. (3 .38), both a n increase in I E, I and a
d ecrease i n 0 mean that Q will i n c reas e i f i t is a l ready posi t ive, or i t will
d ecrease in magnitude and pe rhaps become pos itive i f Q is al ready negative
b e fore the field excitation i s booste d . These operating characteristics of the
generator are m a d e graph i c a l l y evid ent in S e c . 3 . 5 .
LQ:
The generator o f Example 3 . 1 has synchronous reactance Xd =
per unit and is connected to a very large system. The terminal vol tage i s
p e r u n i t a n d t h e generator i s supply i ng to t h e system a current of 0.8 p e r
unit a t 0.9 power-factor l aggi n g. Al l p e r-unit values a r e on the mach i n e base .
Neglecting resistance, find the m agnitude a n d a ngle of the synchronous i n te rnal
voltage Ei, and
and Q de livered to the i nfi n i te bus. If the real power output of
the generator remains constant but t h e excitat ion of the generator is
i n;:reas e d
Exa mple 3.3.
1 .724 1
1 .0
P
(a)
3.4
by
anu (b)
bus vol r a e ,
20%
REAL AND R EACTIVE POWER CONTROL
E;
decreased by 20%, fi n d the angle 0 between
Q d e l i v ere d to the bus by the generator.
g and
Solution. The power-factor angle is e
0.9 25.8419°
synchro nous internal vol tage given b y Eq . (3.31) is
=
=
Equ a t ions 0 .3 8 )
1 .0LQ:
+
jl .724 1
1 .6012 j l .24 1 4
+
give t h e P a n
d
=
i Vl l
- ( I E I cos 0
Xd
r
I V: 1 )
-
--
I V, I I E, I
Xd
sin 0
i5
t h e:
new value
Q
-
=
of
l .0
=
=
1 .0
-1 .724 1 [ 1 .20
X
p e r u ni t
t h .: g e n e ra t o r.
--
l V; I I E, 1
XJ
--
-
sin -
X
l
.
- sin
�
c o n st a n t gives
1 .2 X 2.0261
1 .7241
( 0.72
sin
l .724 1
1 .20 2 .0261
X
X
era t o r i s
)
0.72
i5
=
=
30.701 6°
2 .0261 cospO.70 1 6° ) - 1 .0J 0.6325 per unit
( b ) Wi th e x c i t a t i o n d ecre ased
-
and so
=
Q s u p p li e d by the g e n
-
lagging,
term inal
1 .0 2.0261
s i n 37.7862°
0.7200 p e r u n i t
1 . 724 1
l.0
l .7241 ( 1 . 6012 1 .0) = 0.3487 p e r u n i t
20% w i t h P a n d
exci t a t i o n b y
-
=
the
0.8! 25 .84 1 9°
2.0261/37.7862°
x
=
=
x
( a ) I n creasi n g
anu
COS - I
Q output of
p =
Q
=
a nd
0
=
o =
=
20%, we o n t a i n
l . ()
X
sin - }
G.oD
)(
2.0261
1 .7241
( 0 .72
1 .7241
0 .80 X 2.0261
X
s i n i5
)
=
=
0 .72
49.98270
109
the
110
CHAPTER 3
THE SYNCH RONOUS M ACH I N E
a n d the value
Q=
of
Q
1 .0
1 .7241
now supp lied
b y the
generator is
[0 .80 X 2 .0261 cos ( 49 .9827° ) - 1 .0]
=
0.0245 per u n i t
Thus, w e sec how excitation con t ro l s the reactive powe r ou tpu t of t h e ge nerat o r.
3.5
LOADIN G CAPABILITY DIAGRAM
Al l the normal o p e r a t i n g c o nd i t i o n s 0 (" t h e ro u n d - ro t o r g e n e r a t o r co n n e c t e d to
a n infi n ite bus c a n be shown on a s i n g l e d iagra m , u su a l ly cal led the loadillg
capability diagram o r opera tioll chart 0 (" t h e m ac h i n e . The ch a r t is i m p o r t clll t to
the pow e r - p l a n t o p e r a t o rs w h o ,I re r e s D o n s i b l e for p ro p e r l o a d i n g a n d o p e r a l i o n
of the gen erator.
The chart i s constructed on the a ss umption that the generator has fix ed
terminal voltage V, a nd negligib l e a r m a t ure resistance . Construct ion begins with
the ph asor d iagram of the mach ine havi ng V; a s the reference phasor, as s h own
in Fig. 3 . 1 0( a ). The m irror image of Fig. 3 . 1 0( a ) can be rotated to give t h e
p hasor d iagram of Fig. 3 . 12, which shows five l oci passing through the opera t i ng
point m . These l oci cor respond to five possible opera t ing modes, in eac h of
w hich one parameter of the generati ng u n i t is kept constant.
The con stant excitation circle has point n a s center
and a radius of length n-m equ a l t o the internal voltage magnitude I E i l , w h ich
can be m ai ntained constant by hol d i ng the dc cu rrent If in the field winding
constant according t o Eq. (3 . 13).
CONSTANT EXCITATION.
I Ia l .
The circl e for constant arma t u re cu rrent has point ° as center
and a rad ius o f length o-m proportional to a fixed value o f l Ia I . Beca u se I Vt 1 i s
fixed , the operating points on t his l ocus correspond t o consta n t megavo ltampere
output ( 1 � I I Ia l ) from the generator.
CONSTANT
CONSTANT POWER. Active power output of the m a ch ine is given by P =
1 l!; I I Ia l cos (J i n per u nit. S ince I � I i s constant, vertical l ine m-p a t the fixed
distance Xd l Ia l eos (J from the vertical axis n-o represents a locus of operating
points fo r cons t an t P . The megawatt o utput of the generator is a lways posit ive
regardless of the power factor of the ou tpu t .
The reactive power output of the m ach i n e is
given b y Q = I VI I l Ia \ sin e in per u n i t w h en t h e a ngle (J i s defined posi tive for
lagging power factors. When I Vt I is consta nt, horizontal l i ne q-m at the fixed
d ista nce Xd l Ia l l sin e I from the hor izo ntal axis represents a locus of oper ating
points for constant Q. For u nity power-factor operation the Q output ? f the
CONSTANT REACTIVE POWER.
Q
I
t r II
---------
I
- .� - - .... ...
� - - - - - - - -
- ....
.... ....
--
I Ic Xd
cos
0:
... ... "
"'
....... ...... .....
,
(a) Constant P
' .....
"
"
...
,
I! /"/
,
,�
- - - - - - -
"
-
l:
>-
- - t- - - ----------------------�-� <
q
�
,,'
\
,
\
/
,,/
-
,
\
/
,
\
\
,
-
\
( e ) Con stant power facto r angle
/ /
/
- - - - -/
,/ /
/
( 6 ) Constant Q
\ C c ) Co n s t a n t � Ei l
\ \�
\
\
\
\
,
\
'
'
\
\
\
C d ) Constant : 10 1
��
\
1
p -------
�
Lead i n g power fact o r
n
Phasor diagram obtained from mirror image of Fi g . 3. 1 0( a ) showing five loci t h rough po i n t m
correspond ing to: ( a ) const ant power P; (b) constant reactive power Q; (c) constant internal voltage
l Ei I ; ( d) constant armature current I /a I ; (e) constant power-factor angle e.
FIGURE 3.12
r
-------__
- .... ...
- ""-
"
"
.... ,
_
"
"
�,
""
"
Constant P
cos 8 1
"
,�,
,
- - - - - - -
,
q
I
/
\ "
\ "
\ ,
,
\
,
\
',
1
:1
l o
_'!...
t
I EY, I
p
�
---L
,, /
�,
I
Constant power factor angle
,/
,
r - - -------------------"'"-a:- -:... /m .
- - - - - - - - - - I � la
-
,
- .... ... ... .. ....
,
"
'
//
,
\\
\
\
\
\
Constant Q
'�
Constant
\
\
�
\
,
,
/'
I EYt l
Xd
nstant ! VI la l
t
�
Lagging power factor
� ----------------- --___
--_____ _ ____----L--+-
Real power P
-
n
Phasor di agram obtained b y m u l t i plyi ng (resca li ng) a l l dista nces i n Fig. 3 . 1 2 by I V,/Xd I .
FIGURE 3 . 1 3
Leading power factor
3 .5
LOA D I NG CA P A B I LITY D I A G R A M
LOAD I N G CAPAB I LITY CURVE
0 .8
113
r
0.7
0.6
E�
(1) U
� Q.J
<fl �
>-. U
<fl
x
(1) Q.J
r.
'_ (1)
>
0 o
�
c�
0.5
0.9 PF
0.4
0.95 PF
Armature
h eating
limit
0.3
0.2
k
0.1 -:
<fl
'-
C1l
>
('Ij
OJ
Cll
Q 0. 0
E
-c
::J
,
'Cll
0.1
0
0.2
0. 1
D--
0.3
'-
'- ...
"-
...
...
0.4
:S
E
0.5
>- . <fl
U
x
�
<lJ
0
'-
LL
<lJ
U
c
:J
�
0 . 58
O.G
"-
0.9
\
\
1
\
l . 0 PF
l.0
Prime
mover
l i mit
00% excitation circle
\
\
0.95 PF
\
\
m'
\
\
0.90 PF
\
n
FI C U R E 3 . 1 4
=
"-
\
Lo a d i l l g c a D a b i l i t y c u rv e ro r
Xci
0.8
'-
0.3
(1) U
(lJ
<fl _
0.7
0.6
I
I
/
P e r - u n it m e gawatts P
0.2
E �
0.5
0.4
P
a
cy l i I1l1 r i c ; d - ro t o r l u rt)()g c n c r a t o r ,
1 72 . 4 % wi t h m a x i m u m t u rb i n e o u t p u t
=
6]:; MV 1\ , 24 kY, 0.9 power fa ctor,
() ] ) M W . P o i n t k
rel a t e s to Ex a mp l e 3 . 4 .
1 14
CHAPTER 3
THE SYNCHRONOUS MACH I N E
generator is zero corresponding to an operating p o i n t o n the horizo n t al axis a-p o
For lagging (leading) power facto rs the Q ou tpu t is positive (negat ive) and the
operating point is in the half-pl ane a bove ( below) the lin e o-p .
CONSTANT POWER-FACTOR. The r a d i a l l i n e o -m corresponds to a fixed
power-fa ctor a ngl e () between the arm'l ture curre n t fa and te rminal voltage � .
In Fig . 3 . 1 2 t h e angle e is for a l agging power-factor load. When 0 = 0° , the
power factor is un ity and t h e opera t ing poi nt is actually on the horizontal axis
o-p . The half plane below the horizo n t a l axis a p p l i c s t o lea d i ng power factors.
Figu re 3 . 1 2 is most usefu l w h e n t he axes arc scaled to in d i c a t e t h e P and
Q load i n g of the generator. Accordi ngly, we rea rrange Eqs. (3.38) to read
p =
I Ei l l V: 1
Since sin 2 0 + cos 2 0
x"
=
.
S l l1 a
( 3 . 3� )
1 , squari ng each s i d e of Eq. (3.39) and add i n g give
( 3 .40)
wh ich h a s the form of ( x - a ) 2 + ( y - b) 2 = r 2 for a circle of center
(x = a , y = b ) and radius r . The locus of P a n d Q is therefore a circle of
2
radius I Ei l I � I /Xd and center (0, - I V, 1 /Xd ). This circle c a n be obtained by
mul tiplying the length of each phasor i n Fig . 3 . 1 2 by I � I /Xd or, equivalen tly, by
rescaling the diagram to conform to Fig. 3. 1 3 , which has axes l abeled P
horizontally a n d Q vertically from the origin a t point o . On the vertical axis of
F ig . 3 . 1 3 the length o-n equals I � 1 2/Xd of reactive power, w here � is the
term i n a l voltage. Usu a l ly, the loading d iagram is constructed for 1 � 1
1 .0 per
u nit, i n wh ich case length O-I ! represents react ive power equal to l /X" per unit.
So, l ength o-n i s the key to setting the sca le for the real and reactive power on
th e P and Q axes .
The loa ding chart of the synchronous generator can be made more
practica l by taking account of the maximum permissible heating ( I 2 R losses) i n
t h e armature and t h e fi eld windings, a s wel l a s the power lim its of the prime
m over a n d heating in the a rma ture core . Using the exa m pl e of a cyl i n d rical­
rotor t urbogen era t ing u n i t rated 635 MYA, 24 kY, 0.9 power factor, Xd =
172.4 1 %, let u s demonstrate the procedure for constructing the loading capabil­
i ty diagram 0 f Fig . 3 . 1 4 as follows:
=
•.
•
Take I � I = 1 . 0 per u n i t on the rated-vo l tage base of the machine.
Using a convenient volt ampere sca le, m ark t he poin t n on the vertical axis so
that length o-n equals 1 /Xd in per u n i t on the rated base of the m achipe. I n
3,5
LOAD I N G CAPAB ILITY D I A GR A M
115
our example Xd 1 .724 1 p e r u n i t , a n d s o t h e length o-n i n Fig. 3 . 1 4
corresponds t o l /Xd = 0.58 per u n i t o n the vertical Q-axis. T h e s a m e scale
obviously appl ies to active power P i n per unit on the horizontal axis.
Along the P-axis, m ark the d istance corresponding to the m aximum power
output of the prime m over. For present p urposes t h e meg awatt l i m it of the
t u rbine is assumed in Fig. 3. 1 4 to be 1 .00 per u nit on the rated m egavoltam­
pere base of the m ach i n e . D raw t h e vertical l ine for P = 1 .00 per unit.
M ark the l en gt h o-m = 1 .0 per u n i t on t h e rad ial l i ne from the origin at t h e
r a t e d powe r-factor a n gl e fJ , which i n t h is case equals cos - 1 0.90. With 0 a s
center a n d l ength o-m a s rad ius, d raw t h e per-unit megavoltampere circular
a rc correspond ing to t h e armatu re-cu rre n t l i m it.
Construct the a rc m -r of m ax i m u m pe rmiss ible excitation using n as center
a n d d istance t/ -m as ra d i us . Th is c i rcu l a r a rc corre sponds to t h e m axi m u m
fiel d-cu rrent l i mi t . T h e co n s t a n t exci t <1 t ion c i rcl e w i t h ra d ius o f l e ngth o-n
u s u a l l y d di n e s 1 0 0 % o r 1 . 0 pe r-u n i t exc i t a t i o n , a n u so rig. 3 . 1 4 s hows the
fi e ld-current l im it occu r r i n g a t 2 .340 per- u n i t exc i t a t io n , t h at is, (length
r-Il ) /(l e n g t h o-n ) o n the Q- axis,
An und erexc i tation l i m i t also app l i es at l ow l evels of excitation when vars are
he i ng imported from t h e system to t h e m ach ine. It is determ ined by the
manufactu rer's design as d iscllssed below.
=
•
•
•
•
I n F i g. 3.1 4 poi n t m corresponds to the megavoltam pere rating o f the
gene ra tor at rated power- factor l agging, The m ach i n e designer h as t o a rrange
sufficient field cu rrent to su pport overexcited ope ration of the generato r at rated
point m . The l evel o f t he fie l d curre n t is l im ited to this m aximum value along
the c i rcu l a r arc m -r , a n d the capab i l ity of t h e generator t o d e liver Q to t h e
system is thereby reduce d . I n actu a l i ty, m ach ine satura tion decreases the v a l u e
o f t h e synch ronous reacta n ce Xd , and for t h is reason most m an u facturers '
cu rves depart fro m t h e t h eoretical fi e l d - heating limits described h ere.
The m i rror image of m is t h e operat i n g poi n t m' in the underexcited
r e g i o n , Power-pl a n t operators try t o avoi d opera ting c o nd i t i o ns i n the u nderex­
c i ted region of the capab i l i ty cu rve fo r two d i fferent reasons. The fi rst rela tes to
steady-state s t ab i lity of the system a n d t h e second rel at es to overheating of the
m achine itself.
Theore t ica l ly, the so-ca l l ed sIet1dy -state sta bility limit occurs when the
angle 8 b e tw e e n E j and VI i n F igs. 3 . 1 2 a n d 3. 1 3 reaches 9 0° . In practice,
however, system dyn a mics e n t e r i n t o the p ictu re to complicate t h e determ ina­
t i o n of the actual 'st abi l i ty l i m i t . For t h is reason power-p l a n t operators prefer to
avoid u n d e r ex c i t e d mach i n e operat ion whenever possible.
As the mach i n e e n ters i n to the u n d e rexcited region of operat ion eddy
cu rre n ts i nduced by the syste m in i ron p a rts of � h e armature beg i n to increase.
Th e accompa nying 1 2 R h e a t i n g also i ncreases in the e nd region of the a r m a­
t u re. To l i m i t such hea t i ng, t he m ach i n e manufact u re rs prepare capab i l ity
116
CHAPTER 3
TH E S Y N CH RO N O U S M A C H I N E
curves specific to t hei r own designs a n d recommend l i mits within wh ich to
operate. In Fig. 3 . 1 4 t h e l i ne m' -n is t h e refore d rawn for il lust rative purposes
only.
To ob tai n m egawa t t and megava r values for any operating point in Fig.
3.14, the per u n i t values of P and Q as read from the chart are m ul t i p l ie d by
the megavolt amp e r e r a t i n g of the m a c h i n e , w h ich i n this case i s 635 M YA . Also,
distance n -m of Fig. 3 . 1 4 i s the per- u n i t mega vol tampere va l u e o f the qua n tity
I Ej V, I IXd at operating poi n t m , as s hown in Fig. 3 . 1 3 . Therefore, we can
calculate t h e value of I E; I in per u n i t o n the rated vol tage base (24 kV in t h is
case) b y mult iply i n g l e ngth n -m ( ex pre s s e d i n per-u n i t vol tamperes) b y th e
per- u n i t ratio Xd l 1 V, I , or s imply by Xd si nce I � I = 1 . 0 per u n i t i n Fig. 3 . 14.
Conversi o n to k ilovolts t h e n req u i res m u l t i p l ica t ion by the vol tage rating of t h e
machine i n kilovolts.
I f t h e act u a l t e rm i n a l vo l t age I V; I i s n o t 1 . 0 D e r u n i t, t h e n t h e pe r - u n i t
value I IXd assigned to d i st a n ce o-n of Fig . 3 . 14 h a s to be changed to I V; 1 2 IXd
i n per u n it, as shown i n Fig. 3 . 1 3 . This c h a n g e a l ters the scale of Fi g . 3 . 1 4 by
1 � 1 2, a n d so the p er- u n i t P a n d Q r e a d i n g s from the cha rt must be firs t
m ul t i p l i ed b y I V; 1 2 i n per u n i t and t h e n by the m egavol tampere base (635 M Y A
in this case) in order to give correct m egawa t t a n d megavar va l u es for the actu a l
operati n g con d itions. For instance, if t h e a c t u a l term i n a l vol tage i s 1 .05 p e r u n i t,
then t h e p o i n t n on t he Q-axis of Fig. 3. 1 4 corresponds to the actual va l u e
0.58 X 0 .05)2 = 0 .63945 p e r u n i t o r 406 Mvar, a n d t h e po i n t shown a s 0 . 9 per
unit on t h e P-axis has an actual va lue of 0 .9 x 0 .05 ) 2
0.99225 p e r unit or 630
MW.
=
To calculate t h e correct excitation v o l ta g e E; c or r e s p o n d i ng to a n operat­
ing poin t m when the termi n a l vol tage i s not exactly equ a l to i ts rated vol tage,
we cou l d first m u l tiply l ength l1 -m o b t a i n e d d irect ly from Fig. 3 . 1 4 by I V; 1 2 i n
per u n it t o correct the scale a n d t h e n b y the ra t io Xul I V; I i n per u n i t to
convert to I E; I , as a lready d iscussed . The n e t res u l t is t h a t l e n g t h n - m obt a i n e d
d irectly from F i g . 3 . 1 4 w h e n m u l t i p l i e d b y t h e a C l u a l p e r- u n i t v a l u e of the
product Xd X I VI I y i e l ds the correct p e r- u n i t va lue of I E; ! . T h e n , i f p h ys i cal
units of k i lovolts are desired, mu lt i plica t io n by the rated k ilovo l t base o f t h e
machi n e follows. I t is i mport a nt to n ote t hat t h e power-fa ctor a ng l e f) a n d
inter nal a ngle 8 a r e t h e same before a n d a fte r the resca l i ng si nce t h e geometry
of Figs. 3 . 1 2 and 3. 13 i s preserved . The reader should note, however, that the
operati n g constrain ts forming the bo u n d a ry of t h e opera ting region of the chart
are p hysical limits. So, the b ou n d a ry of the operating region may be affected
once the scale is a ltered .
The fol lowi n g example ill ustrates t h e p rocedures.
Example 3 .4. A 6 0-Hz t hree-p h as e g e n erator r a t ed a t 635 M VA, 0.90 powe r
factor, 24 kV, 3 600 rpm h a s t h e operat i n g chart shown i n Fig. 3 . 1 4 . The gener a tor
is d el iveri n g 4 5 8 .47 M W a nd 1 14 . 62 Mvar at 22. 8 kV to a n i n fi ni t e bus. Calcu l a t e
t h e exci t a t i o n
vol tage
E;
u sing
(a)
the equ ivale nt
circ u i t of
Fig .
3 . 8( a )
and
( b ) the
I
3 .6
THE TWO-AXIS MACHINE MODEL
loading d i agram of Fig. 3 . 1 4. Th e synchronous reactance is Xd
on the machine b ase and resi s t ance is n egligible.
=
1 17
1 .724 1 per unit
Solution. I n the calc u l ations which follow all per- u n i t val u es a re based o n the
megavoltamp ere a n d ki lovolt rati ngs of the m acp.. i n e .
(a) Choosing the term i n a l
�
P
+
iQ
J(l
Ei
=
vol t age a s the refe re n ce p hasor, w e h ave
22 . 8 /
--L...i?:
24 .0
45 8 .47
=
(\0
+
/
0 . 95LJ?:
p er u n i t
{"\O
=
=
635
0 . 722 - j O. 1 805
=
=
=
=
=
0 .95�
�
0.722 + i O . 1 805 per
j 1 1 4 .62
+ jXd I(l
=
0 .76 - jO . 1 9 p e r u n i t
0 . 95�
1 .2776 + j 1 .3 1 03
=
unit
+ j 1 .724 1 ( 0 . 7h
- jO . 1 9)
1 .830/ 45 .72390 per
u ni t
43 .920/ 45 .7239° kV
( 6 ) The poi n t k corresponding to the a c t u a l operating con d i t i o n s c a n be located
on t h e ch a rt of Fig . 3 . 1 4 as fol lows :
0 . 722 + i O . 1 805
---�-- =
0 . 95 2
+
0.8 + iO.2
per u nit
0 . 78 2 = 1 . 1 1 73 per u n i t w h e n calcu l a t e d o r
The dista nce n -k equ als ../0 . 8 2
measured on t he scale of t he c h a r t o f F i g 3 . 14. T h e actu a l v a l u e o f I Ei l i s t h e n
.
co m p u t e d as
l EI I
3 .6
w h i c h is t h e s a m e
as
=
( 1 . 1 1 73
x
0 . 95 : )
1 . 724 ]
0 . 95
=
o b t a i n e d a b ove . The a n g l e 0
1 .830 per u n i t
=
4 5° c a n be e a s i l y m e asured.
THE TWO-AXIS MACHINE M O D EL
The round-rotor theory already d eveloped in this chapter gives good results for
the steady-state performance of the synchronous machine. However, for tran­
sien t analysis we need to consider a two-axis model . In this section we i ntrod uce
the two-axis mode l by means of the equations of the salient-pole machine i n
which t h e ai r gap i s much narrower along the direct axis t h a n a long the
118
CHAPTER
3
T H E SYNCH RONO US MACH I N E
The largest generating u nits are s team - tu rb ine ­
driven alternators of round-rotor construction; fossil-fired units have two pol es
and nucl ear units have four p ole s for reasons of economical design and
operational efficiency. Hyd roelectric generators usually have more pole-pairs
and are of sal ient-pole construction. These units run a t lower speeds so as to
avoid mechan ical damage due to centri fuga l forces.
The three-phase salient-pole machine, l i ke i ts round-roto:- counterpa rt,
has t h ree symmetrically distributed armature windings a, b, and c, and a field
winding f on the rotor which produces a sinusoidal flux d istribution around the
air gap? In both types of m achines the field sees, so to speak, t h e same air gap
and magnetizing paths in the stator regardless of the rotor position. Conse­
quently, the field winding has constant sel f-inductance Lff o Moreover, both
m ac h i n e t y p e s h ave t h e s a m e cos i n l l s o i d a l m u t u a l i n d u ct a n c e s r il l ' l ' hI ' a n d
L ei with t h e a r m a t u re p h a ses a s g i v e n by Eqs . ( J .4 ) . A d d i t i l l l1 ; t l i y , t h ro u g h o u t
each revolution of the rotor the self- i n ductances L a a , L b b , and L : : o f t h e s t a t o r
windings, a n d t he mutual inductances L a b ' L b O and L e a between them, are not
constant i n the salient-pole machi ne but also vary a s a function of the rotor
angular d isplacement (Jd . The flux linkages of phases a, b, and c are related to
the currents by the inductances so that
quadrature axis b etween poles.
( 3 .41)
These equations look similar to Eqs. 0 . 5 ) for the round-rotor machine but all
the coefficients are variable, as summarized i n Table 3.1.3 A s a result, the
equations for t h e flux linkages A a , A b , and A e of the salient-pole machine are
more difficult to use than their rou nd-rotor counterparts. Fortunately, the
equations of the salient-pole machine can °be expressed in a simple form by
transforming the a , b, and c vari ables of the stator into correspond i ng sets of
new variables, c a l l e d the direct-axis , quadra ture-axis , and zero-sequence quanti­
t ie s which are distinguished by the subscripts d, q , and 0: respectively. For
example, the three stator curren ts i a ' i b ' and ie can be transformed into t hree
equivalent currents, called t h e direct-axis current i d ' t h e qua dra ture-axis current
iq and the zero-sequence current i o . The transformation is made by the matrix P,
2 For further d iscu ssion on the sal i e n t-pole m ac h i n e , s e e P. M . And erson a n d A A. Fo uad, Power
System Control and Stability , Chap. 4, The Iowa S tate U n iversity Press, Ames, Iowa, 1 977.
3
The D-
and Q-damper wind ings refe rred
to in
Table
3 . 1 are d i s cussed in SeCo 3 . 8 .
1J"lE lWO-AXIS MACHINE MODEL
3.6
119
TABLE 3.l
Expressions for the inductances of three-phase salient-pole synchronous
generator with field. D-damper and Q-damper windings on the rotor.
Stator
{
{
Field windin g: Lrf
Self- i n d uctances D da mpc r winding: Lf)
Q-dalllJ)t'r winli i nt!: I. v
Rotor
M u t ual-i nductances
-
P'icld /D-wind ing : M,
F i dd /Q wi nd i ng : 0
D-winding/Q-wind ing: 0
-
Stator-rotor
mutual
i n du c ta n c e s
cal led
Park IS tramformatioJl
P
=
h
®
®
©
I
where
@
cos ()tI
sin ()eI
I
' Ii
®
( ()eI
©
- 1 200 )
cos ( ()eI - 2400 )
s i n « ()d - 1200 )
sin ( ()d - 2400 )
I
1
cos
li
( 3 .42)
{i
which was introduced by R. H. Park in slightly d i fferent form from that shown
here. The matrix P has the convenient property (called
inverse P '
1
orthogonality )
that its
equals its transp([)se P T, which is found simply by interch'lnging rowS
120
CHAPTER 3
THE S Y NCH RONOUS M A C H I N E
in Eq. (3.42). This property is most i mportant as it ensures that
power i n t h e a , b , and c variables is not altered by P , as d iscussed in Sec. 8 . 9 .
The currents, voltages, and flux linkages of p hases a , b , and c are transformed
by P to d, q, and 0 variables as follows:
and columns
ld
lq
-
ta
lb
Ie
P
10
( 3 .43 )
The P-transformation defines a set of currents, voltages, and fl ux l i nkages for
three fictitious coi ls, one of which is the stationary O-coil. The other two coils
are the d-coil and t h e q-coil, which rolate i n syn c h ro n ism with the rotor. The
d- and q-eoils have constant fl u x linkages \v i t h th e fleld a n d a n y other w i n d i ngs
which may exist o n the rotor. Section A . 2 of the Appendix illustrates the
detailed manipulations which transfo rm the currents, vol tages, and fl ux l inkages
of phases a , b , and c into d-q-O quantities according to Eqs. (3.43). The
resulting d, q , and 0 flux-linkage equations are
( 3 .44 )
in
which if is the actua l field curr;e nt, and the inductances are defined by
( 3 . 45 )
Ms
have the same meanings as before and L is a positive
number. The i nductance Ld is c al le d the direc t-axis inductance , Lq is called the
quadrature-axis in duc tance and L o is known as the zero-sequence inductance.
The flux linkages of the field are still g iven by Eq. (3.24), w hich is repeated here
i n the form
Parameters Ls and
m
,
( 3 .46)
Equations (3.44) and (3.46) have constant i nductance coefficients, and thus are
quite simple to use. Physically interpreted, these simpler fl ux-l inkage equations
show that L d is t h e self-indu ctance o f a n equ iv a len t d-axis armature w i'nding
..
...
,
�
.
_
_
......
..L ' _
_
_
&
_ __
_
�_ �
� ...J
�
�
4-
t... .....
.c. ..-. ' ..-J
..-.. -.. ......J
� . ,
1.-. � ....... t...
...... .,.."
..... ...� ; D {.
r- t I t"t"� n t
l'
trl
3. 6
D i rect
axis
a-axis
Quadrature
axis
Rr, L rr
f'
FIG U RE 3 . 1 5
121
All c oils rotate together
R otation
b- axis
THE TWO-AXIS M A C H I N E MODEL
c - axis
R e p r ese n t a t i o n of the s a l i e n t -p o l e synch r o n o u s g e n e ra t o r by a r m a t u r e - e q uival e n t d i rect-axis a n d
q u a d r a t u r e - axis coi l s ro t a t i n g i n synchro n i s m w i t h t h e fi e l d w i n d i ng on t h e ro t o r .
produce the same mmf on the d-axis as do the actual stator currents i a ' ib , and
i Simil arly, Lq and i q apply to the q-axis. Accordingly, i d and i q give rise to
mmfs which are stationary with respect to the rotor. The fictitious d-axis
winding and the f winding representing the physical field can be considered to
act l ike two coupled coils which are stationary with respect t o each other a s they
rotate together sharing the mutual ind uctance kMf ( k = {3/2 ) between them,
as shown by Eqs. 0.44) and 0 . 4 6 ) . Fu rthermore, the ReId and the d-axis coil do
not couple magnet ically with the fkt itioliS q winding on t he q-axis, which lags
t he d-axis in space by 90° . The zero-sequence inductance La is associated with
a stationary fictitious armature coil with no coupling to any other coi ls. Under
balanced condit ions this coil carries n o current, and therefore we omit i t from
further discussion .
The d-axis and q-axis coi l s representing the stator windings are shown i n
Fig. 3 . 15 , w h i ch s h o u l d b e c o m p a r e d w i th t h e s i n g l e - axis d i a g r a m for t h e
r ou n d - r o t o r machine in Fig. 3 .0 .
c '
U n d e r s t eady-state ope r a t i n g condi tions t he a rmature o f t h e sal ient­
p o le sync h ro nous genera to r c arries symm et rical s i n u soidal thre e-phase cu rrents
Ex a m p l e 3 . 5 .
122
CHAPTER 3
TH E SYN CHRO N O U S M A C H I N E
where ()d = w t + 8 + 90° , a s shown in Eq. (3. 1 4). Using the P-transformation
matrix, find expressions for the corresponding d-q-O currents of the a r m a t u r e .
Solution.
From Eqs. (3.42) a n d (3 .43) we h a v e
ld
if
iq
iO
.J
cos () d
sin
1
()d
Ii
cos( () d - 1 200 )
cos ( ed - 240° )
si n e 0d - 1 200 )
s i n e ed - 2400 )
Ii
Ii
1
1
a n d row -by-col u m n m u l t ip l i c a t i o n t h e n gives
I
q
=
Under b a lanced con d itions ia + i b + i c = 0, a n d so
= O. By m e a n s of t h e
trigonom etric i d e n t i ty 2 s i n a cos f3 = si n ( a + f3 ) + s i n ( a - f3), w e o b t a i n
io
Likewise, w e have
,.
'":,. ...
In t he t h re e p rece d i n g trigonometric expa n s i o n s the fi rst terms i n s i d e t h e b rackets
are second harmonic si nusoidal q uant i t i e s which sum to zero a t every i n stant of
3.7
VOLTA G E EQUATIONS: SALI ENT-POLE MACHINE
123
time as in Sec. 3.2, a n d so we obtain
We reca ll from Sec. 3.3 that (Ja = (J + 8, where (J is the phase angle of lag o f
m easured with respect to the term i n a l voltage and (Ja is the p ha se a ngle of lag of i a
with respect to the i n ternal vol tage of the machine.
Accord i ngly,
ia
We c a n
s how
in a si m i l a r m a n n e r t h a t t h e q u a d r a tu re-axis c u r re n t
i '/
=
v'J 1 /0 1 cos 0"
=
If I (, I cos( 0
+
8)
Thus, the expression for i d is exactly the same for the salient-pole a n d the
rou n d-rotor machines. The flux l inkages in the field winding are given by Eq.
(3 .46), which shows that the di rect-axis current i d is directly opposing the
magnetizing influence of the field when ()a = 'TT' /2, and the quadrature-axis
current i q is then zero.
3 .7
VOLTAGE EQUATIONS: SALIE NT-POLE
MACHINE
In Sec. 3.6 the flux-linkage equations for the salient-pole machine are remark­
ably simple when expressed in tenns of the d, q , and 0 variables. We now
consider other important simplifications which occu r when the P-transformation
is also applied to the vol tage equations of the armature.
Using the voltage polarities and cu nent directions of Fig. 3.4, l e t us write
the terminal-voltage equations for the armature wind ings of the salient-pole
machine in the fonn
V a = - Ri a -
J /I. 'I
'
dr '
V"
= - Ri h -
d /l. h
dt '
vc =
- Ri c
d A.c
-
--
dt
( 3 . 4 7)
In these equations the voltages V I P V " , and V c are the line-to-neutral terminal
vol tages for the armatu re phases; the negative signs of the coefficients arise
because currents i i b ' and i are directed out of the generator. While simple
in format, Eqs. (3.47) are in fact very difficult to handle if left i n terms of A. a , A. b ,
a n d /I. e ' Again, a m u c h simpler set o f equations for the voltages V d ' vq • a n d V a i s
fou n d b y e m p l o y i n g t h e P - t r a n s fo rm a t io n . The calculations l eading , to the new
a'
c
124
CHAPTER 3
THE SYNCHRONOUS MAC H I N E
volt a ge equations are straightforward but tedious, as shown in
Appendix, which yields
Sec.
A.2 of the
( 3 .4 8 )
is t h e rotation a l speed de,,1 dl . Equ ation (3.26) for t h e field w i n d i n g i s
not subject t o P- t ra n s ro r rn a t i o n , a n d so a rr a n g i n g t h e d-q-O /l u x - l i n k a g e a n d
vo l tage equations according to their axes gives
d-axis:
w here
w
( 3 .49)
( 3 . 50)
q
-
ax l s :
vq =
- Ri q
dAq
-
dr
--
+
( 3 .5 1 )
WA d
where k = /3/2 . Equations involving i o a n d A D sta nd alone and are not of
interest under balanced conditions. Equa tions (3.49) through (3.5 1 ) are much
sim p ler to solve than their correspondi ng voltage and flux-linkage equations in
a-b-c v a ri ab les Furthermore, a set o f e q u ivalent circuits m ay b e drawn to satisfy
·. these simpler equations, as shown in Fig. 3 . 16 . The f c i rcuit represent s the
actual fi eld since t h e P-transformation affects only t h e armature phases, w hich
are replaced by the d- a n d q-coils. We see that the f-coil is mutually coupled to
'
.
-
the d-coil on the d-axis, and so
ft ux-linkage
and voltage equations can he
3.7
VOLTAGE EQUATIONS: SALI ENT-POLE MACH I N E
125
written to agree with Eqs. (3 .49) and (3.50). The fictitious q-coil is shown
m agnetically uncoupled from the other two windings since the d-axis a n d the
q-axis are spatially in quadrature with one another. However, there is interac­
tion between the two axes by means of the voltage sources - wA q and W A d'
w hich are rotational emfs or speed voltages internal to the machine due to the
rotation of the rotor. We note that the speed voltage i n the d-axis depends on
A q , and similarly, the speed voltage in the q-axis depends on A d ' These sources
represent ongoing electromechanical energy conve rsion. No such energy conver­
sion could occur at standstill ( w = 0) since the field and the other d-axis circuit
would then act l ike a stationary transformer and the q-axis circuit l ike an
ord inary inductance coi l .
To summarize, Park's transformation repl aces the physical stationary
windings of the armature by:
1 . A d i re c t - a x i s c i r c u i t w h i c h r o t a t e s w i t h t h e
AcId
c i rcu i t
and
is
mutually
coupled to it,
2 . A q u a d r a tu r e - a x i s circ u i t w h i c h i s d i s p l a c e d 90° from the d-ax is, and thus
has no mutual inductance with the field or other d-axis circuits al though it
rotates in synchronism with them, and
3. A stationary stand-alone a-coil with no coupling to any other circuit, and thus
is not shown in Fig. 3 . 1 6 .
Figure 3 . 1 6 is most useful in analyzing the performance of the synchronous
machine under short-circu it conditions, which we consider in the next section.
�JV���+r - - - - - -
}-----'--O- - -
----
(b)
FI G U RE 3 . 1 6
E q u iva l e n t c i rc u i t for t h e s a l i e n t - po l e syn c h ro n o u s g e n e rator:
( b ) w i t h a rm a t u re sho r t - c i rc u i t e d .
( a ) w i t h t e rm i n a l vol tages
Vd and
vq ;
126
CHAPTER 3
THE SYN CHRON OUS MACH I N E
Example 3 .6. Direct current
If
is supplied to the field winding of an un loaded
salient-pole synchronous generator rotating with constant angular velocity w .
Determine the form of the open-circu i t armature voltages and their d-q-O
components.
The armature currents
open circuit, and so
Solution.
l a ' i I> ,
and
lc
are zero because o f the armature
Substitut ing these va lues i n Eqs. ( 3 .49) through (3.5 1 ), we obtain fo r k
/3/2
=
0
=
0
(v A "
=
=
a n d from Eqs . (3 .48) we then fi n d
va =
- Ri "
-
dA "
til
+
k w Mf lf
Thus, we see that thc co nstant flux l i n k ages A d on the d-axis give rise to t h e
rota tional emf k w Mf lf o n t h e q-axis. S i n ce p - l = p T, i t foll ows fro m Eqs. (3.43)
that
3. 8
TRANSIENT AND SUBTRANSIENT EFFECfS
127
Therefore, the steady-state open-circuit armature voltages in the idealized salient­
pole machine are balanced sinusoidal q u a ntities of amplitude J2 IEj l = w Mflf, as
obtained previously for the rou nd -rotor machine.
3.8 TRANSIENT AND SUBTRANSIENT
EFFECTS
When a fault occurs in a power network, the current flowing is determi ned by
the internal emfs of the m achines in the network, by their impedances, and by
t he impedances in the network between the machines and the fault. The current
flowing in a synchronous machine immediately after the occurrence of a faul t
differs from t h at flowing a few cycles later and from th e sustained, or steady-state,
value of the fault current. This is because of the effect of the faul t current in the
armature on the flux generating the voltage in the machine. The current
changes re l a t ively slowly fro m it s initial v a l u e t o i ts steady-state value owin g to
the changes in reactance of thc synchronous machinc.
Our immed iate i nterest is in the inductance effective in the armature of
the synchronous machine when a th ree-p hase short circuit suddenly occurs at its
term i n a ls. Before the fault occurs, suppose that the annature voltages are V a '
V b ' and Vc and that these give rise to the voltages Vel ' v q , and V a accordin g to
Eq . (3.43). Figure 3 . 1 6 ( a ) shows the vol tages v d and Vq at the terminals of the
d-axis and q-axis eq uivalent circuits. The short circuit of phases a , b, and c
imposes the condi tions VII = V b = Vc = 0, which lead to the conditions V d =
V
= O. Thus, to simulate short-circuit conditions, the terminals of the d-axis
q
and q-axis circuits in Fig. 3.16(a) must also be shorted. Each of these circuits
has a net terminal vol tage of zero when e q ua l but opposite voltage sources are
connected in series, as shown in Fig. 3 . l 6(b). In that figure the switches S
should be interpreted i n a symbolic sense; namely, when the switches a re both
open, the sources - V d and - V q are in the circuit, and when the switches are
closed, those two sources are removed from the circuit.
The principle of superposi tion can be applied to the series-connected
vol tage sou rces, provided we assu m e that the rotor speed W rema ins at its
p refault steady-state va lue-for Eqs. (3.49) through (3.5 1 ) are then linear. With
both switches closed in Fig. 3 . 1 6( b ), we have the steady-state operation of the
machine since the sources V d and Vq then match perfectly the d-axis and q-axis
voltages at the terminals just before the fault occurs. Sud denly opening the
swi tches S adds the voltage sou rce - V(/ in series with the source V d and - V q i n
series with t h e sou rce Vq to produce t h e required short circuits. Thus, the
sources - V ri and - Vq a r e those d etermining the instantaneous changes from
the steady state due to the sudden short-circuit fault. By superposit ion, we can
calculate the fault-induced changes of all variables by setting the external
sources vjf' , Vd ' and Vq of Fig. 3 . 1 6( b ) equal to zero and suddenly applying the
voltages - Vd and - v q to the u n excited rotating machine, as shown in Fig. 3 . 17.
T h e in ternal speed voltages - W A q and W A d are initially zero because flux
l inkages with all coils are zero i n Fig. 3 . 1 7 before applying the - � d and - vq
128
CHAPTER 3
THE SY N CHRONOUS M ACH I N E
r-
q-axis
WAd
+
=
0
FIGURE 3 . 1 7
Equ ivalent c i rc u i t o f s a l i e n t - p o l e sy n c h ro n o u s g e n e r a t o r r o t a t i n g a t c o n s t a n t , p t: c d w i t h f i e l d
shon-circuited. Clo s i n g sw i l c l l e s : I t ( ( ) c o r r e s p o n d s t o s l I d d c l l a p p l i C : l t i o l l ( , : S I H ) r t c i rc l l i t t o
�
'
machine t e r m i n a l s .
sources. The flux-linkage changes on the d-axis of
Eq. (3. 4 9), which gives
the
machine
af;:
governed by
( 3.52)
where � denotes incremental changes. Since the field winding is a closed
physical winding, its flux linkages cannot change instantaneously according to
the principle of constant flux linkages. Therefore, setting !:l A f equal to zero in
Eq. (3 .52) gives
,and substituting for
!:li
f in the equation for !:l A d yields
( 3 . 5 3)
flux l inkage per unit current I n Eq. (3 . 5 3 ) defines the
inductance I..:d , where
The
d-axis transien t
( 3 . 54)
Since (k Mf) 2 / L If is posItIve, Eq. (3 .54) shows t h a t the direc:-axis transien t
W Ld is always less than the direct-axis synchronou s reac'tance
rea ctance Xd
=
3 .8
T R A N S I ENT A N D S U BTR ANSI ENT EFFECT'S
129
Xd = w L d ' Thus, following abrupt changes at its terminals, the synchronous
machine reflects in -its a rmature the transient reactance X�, which is less than
its steady-state reactance Xd .
In defining X�, we assume that the field is the only physical rotor winding.
In fact, most salient-pole machines of p ractical importance have damper wind­
ings consisting of shorted copper bars through the pole faces of the rotor; and
even in a round-rotor machine, u n d e r short-circuit conditions eddy currents are
induced in the sol id rotor as if in damper windings. The effects of the
eddy-current damping circu i ts are represented by d irect-axis and quadrature-axis
closed coils, which are treated in very much the same way as the field winding
except that they have no applied vol tage. To account for the addition of damper
wind ings, we need only add to Fig. 3. 1 6 the closed D-circuit and Q-circuit of
Fig. 3 . 1 8 , which have self-ind uctances L D and L Q and mutual inductances with
the other wind ings as shown. In the steady state the flux linkages are constant
between all circuits on the same rotor axis. The D- and Q-circuits are then
passive (baving neit her ind uced n o r applied vo l tages) and do not enter into
steady-st ate analysis. Under short-ci rcu it condit ions, however, we ca n d e termine
from F i g . 3 . 1 8 th e initial d - a x is flux- l i n kage changes result ing from sudden
shorting of the synchronous machine with damper-winding effects. The proce­
dure is the same as already discussed. The field and D-dampe r circuits repre­
senting closed phys ical wind ings are mutually coupled to each other and to the
d-coil representing the armatu re al ong the direct axis. There cannot be sudden
change in the flux linkages of the closed windings, and so we can write for the
flux-linkage changes along the d-axis
( 3 .55)
'
a re simil a r to Eqs. (3. 52), but they have ext ra terms because of
the addi tional se l f- and mutual ind ucta nces associa ted with the D-damper
circu i t . Coeffi c i e n t s re flecting stator-to-rotor mutual coupling have the multi­
plier k = V3/2 . Mr re la tes to mutual coupling between rotor-based windings
on t he d - a x i s a n d t h u s h a s no k m U l t i p l ier. S o l v i n g Eqs. ( 3 .55) for t1 i f a n d .j, i D
in terms of D. i d yields
These e q u a t io n s
130
CHAPTER 3
+
TH E SYNCHRONOUS MACHI N E
•
Field wi nding
+
kM
,-!...
d-axis armature equ ivalent w i n d i n g
f'--�--------------�
in
-+
Rn
D - d amper wind ing
=
-I
Vn
Ln
0
+
d-axis
RQ
•
Q-damper w i n d i n g
vQ
0
LQ
-I
Vq
=
Equivalent circuit
FIGURE 3.18
q-axis armat u re equivalent winding
q-axis
o f the s a l i e n t-pole synchronous g e n e r a tor w i t h o o e fi e l d wi n d i n g a n d two d am p e r
windings on the rotor.
and
substituting these results into the
!:l. A d
expression
direct-axis sub transient inductance L,� , defi n ed b y
of Eq. (3 .55) yields the
( 3 .5 6 )
direct-axis subtransient rea c ta n ce X� , defined as X� = w L:� , is considerably
smaller than X�, which means that X � < X � < X d. The reader should check
The
data given by t he machine manufacturers in Table A.2 in the
Appe ndix to confirm these inequal ities. We should note that similar reactances
c an be defined for the q-axis.
We have shown that the synchronous m achine h as d ifferent reactances
when it is subjected to short-circuit faults at its terminals. Immediately upon
occurrence of the short circuit, the armature of the machine behaves with an
effective reactance X�, which combines with an effective r e s i s t a n c e d eterm i n e d
by the dam ping circuits t o define a direct-axis, short-circuit sublransient timethe numerical
.
3.8
T R A N SIENT AND S U BTRANSIENT EFFECfS
131
constant T:; in the range of 0.03 s . The period over which X� is effective is
called the subtransient period, and this is typically 3 to 4 cycles of system
freq uency in duration . When the d a m per-winding currents decay to negligible
levels, the D- and Q-circuits are no longer needed and Fig . 3 . 1 8 reverts t o
Fig. 3 . 16 . The machine currents t h e n decay more slowly with a direct-axis,
short-circuit transient time-constant T� d etermined by X� and a m a chine resis­
tance which depends on R f of the field . The period of effectiveness of Xd is
called the transient period a nd T� is of the order of 1 s. Finally, for sustained
steady-state con di tions the d- and q-axis reacta nces Xd = wLd and Xq = wLq
determine t h e performance of t h e s a l i e nt-pole machine, j u s t as t h e synchronous
reacta nce Xd applies to the rou n d - rotor synchronous machine in the steady
state.
The various reactances supp l i e d by the machine manufactu rers are usually
expressed in per u n i t based on the n a m e p l a t e rating of the m a ch i n e while time
co n s t a n t s a rc g i v e n i n s e co n d s . Tah l c 1\.2 i n t h e A p pe n d i x s e ts forth a summary
oj" typ i c a l p a ra m d � rs fo r t h e s y nc h ro n o u s m a c h i n e s of p r a c t i c a l i m port a nce .
t h e p e r - u n i t v a l u e o f X:, fo r t h e 6 0 - H z synchronous
generator of Exa m p \ c 3 . 1 . Use t h e m ac h i n e rat ing of 635 MVA, 2 4 kV as base.
Exa m p l e 3 . 7 . Ca l c u la t e
Solution . Va l u e s for t h e i n d u c t a n ces o f t h e a r m a t ure and field w i n d i n gs are given
in Example 3 . 1 , a n d Fig . 3 . 1 0 s hows Lt/
Ls + Ms ' Th e r e fore,
L"
=
Ls + Ms
=
=
2 . 765 6
+
1 .3828 = 4 . 1 4 84 mH
The t ransi e n t i n d u c t a nce [;d is n ow c a l c u l a ted from Eq. ( 3 . 5 4 ) :
=
4 . 1 4�4
-
( /3 / 2
X 3 1 . 6950
4 3 3 . 6 5 69
r
')
= 0 . 67 3 6 m H
a n d t he t r a n s i e n t r e a c t a nce is
Xd
=
w L'd = 1 20 7T X 0 . 6736 X 1 0 - 3 = 0 . 2540 n
T h e impe d a n c e base o n t h e m ac h i n e rat i n g e q ua ls (24 2/635) n so that
Xd
Thus,
X:,
0 . 2540 X 6 35
=
242
=
0 . 28 per u n it .
i s m u c h l e s s t h a n t h e syn c h ronous reactance
X"
=
1 .734 1
�er
unit.
132
CHAPTER 3
3.9
SHORT�CIRCUIT CURRENTS
THE SYNCH RONOUS MAC H I N E
When an ac voltage i s a p p l i e d s u d d e n ly across a series R - L c i rc u i t , t h e
current w h i ch flows gen e r ally has two compone nts-a d c compo n e n t , w h i c h
decays according to t h e t i m e con s t a n t
L /R
o f the c i r c u i t , a n d a s teady-s t a t e
sin u s o id a l l y v a ry i n g compo n e n t o f con s t a n t ampl i tu d e .
A
s i m i l a r b u t more
complex p h e n o m e n o n occurs when a short ci rcu i t a p p e ars s u d d e n ly a c ross the
term i n a l s o f a syn c h ronous m a ch i n e . T h e
res u l t i n g p hase cu rrents i n
the
machine will h ave d c compon e n ts, w h i c h c a u s e them t o be offs e t o r asymmetri ­
cal w h e n p l ot t e d a s a fu n c t ion o f t i m e . I n Ch a p . 1 ( ) w e s h a l l d i sc us s how t h e
symmetrical p orti o n of t he s e s h o rt -c i rc u i t c u rre n t s i s u s e d i n t h e r a t i n gs o f
ci rcu i t b r e a k e rs . Fo r n ow l e t liS c o n s i d e r h ow s h o rt c i rcli i ts a ffe c t t h e reac t a n ces
o f th e m a c h i n e .
A good w a y to a n a lyze t h e e ffe c t o f' a t h re e - p h a s e s h o rt c i rc u i t a t t h e
terminals o f a p re v i o u s l y u n l oaded g e n e r a tor i s t o t a ke a n
osci l l ogram of
the vo l ta g e s g e n e r a t e d i n the p hases o f a t h r e e - p h a s e m a c h i n e a r e d i s p l a ced
the
current i n o n e o f t h e p h ases u p o n t h e occu rrence o f such a fau l t . S i n ce
1 20
the voltage wave o f each phase. Fo r t h i s re ason t h e u n i d i r e ct i o n a l o r d c
e l e c trical d eg r ees from each other, t h e s h o r t c i rc u i t occurs a t d i ffe r� n t p o i n ts on
transient compo n e n t of current i s d i ffe r e n t i n each p h as e . 4 I f the dc compo n e n t
o f current i s elim i n a t e d from t h e c u rre n t o f e ach p hase, t h e amp l i t u d e o f t h e a c
component o f e ach p hase cu rrent p l o t t e d v e rsus t i m e , s hown i n F i g . 3 . 1 9 , v a r i e s
approximately a ccordi n g to
le t )
=
1
I E · I - + I E· I
w h e re ei =
I Xd
I
Ii I Ei l cos w t
E q u a t i o n (3 .57)
( 1
X'd
1
Xd
- - -
)
,
c - r / Td
(1 1) "
+ l E I - - - [ - [ I Tt! ( 3 . 5 7 )
I · X"d
X'd
is the syn c h ro n o u s i n t e r n a l or n o - l oad vol tage of t h e
the dc remov e d , h a s t h ree compon e n t s , t w o o f w h i c h d e c a y a t d i ffe re n t r a t e s
over t h e s u b t r a n s i e n t a n d tra n s i e nt p e riod s . N e g l e c t i n g t h e co m p a r a t ively s m a l l
machi n e .
c le a r ly s h ows t h a t t h e a r m a t ure p h a s e c u rr e n t , w i t h
res is tance o f t h e a r m a t u re , t h e d is t ance o -a i n Fig. 3 . 1 9 i s t h e m ax i m u m v a l u e of
the s u st a i n e d s h ort-c i r c u i t current, w i t h t h e rms va l u e I I I given by
III -
( 3 .58)
o -a
If t h e e nvelope o f t h e curre n t wave is exte n d ed back t o z e ro t i me a n d t h e fi rs t
few cycles w h e re the d ecre m e n t a p p e a rs t o b e very r a p i d a re neglecte d , t h e
4
1 983 a n d C h a p . 1 0 o f t h i s book.
For further d iscussion of t h e dc compon ents s e e A . E. F i t zger a l d e t a l . ,
McGraw-Hili, I nc., N ew York,
Electric Machinery ,
4th e d . ,
3. 9
S H O RT-CIRCUIT C U R RENTS
133
c
b
a
o
Time
)0
FlGURE 3.19
Curre n t a s a fu n c t i o n o f t i m e for a sy n c h ro n o u s g e n e r a t o r s h o rt -c i rc u i t e d w h i l e ru n n i n g a t n o l o a d .
T h e u n i d i rect i o na l , t ra n s i e n t com pon e n t o f cu r re n t h a s b e e n e l i m i n a t e d i n red ra w i n g t h e osci l lo­
gram.
i n t e rcept is the d ist ance a -b . The rms v a l u e of the c u r rent represented by this
i n t e rce p t is known as the tran sient current 1 1' 1 , defined by
11' 1 -
o -b
( 3 .59)
Ii -
The r m s va l li e of t h e cu rre n t d e t e rm i n e d b y t h e d is t a nce o-c I n Fig.
ca l l e d
t h e suiJtrtlllsiel l t currel l t I !" I , g i v e n b y
I I" I -
I E, I
o -c
3.19
is
( 3 . 60 )
X d"
S u b transient cu r r e n t is oft e n ca l l e d t h e initia l symmetrical rms current , which is
more d escriptive because
it
conveys t h e i d ea of n eglect i n g the d c compone n t
a n d t a k i n g t h e r m s v a l u e o f t h e a c component o f c u r r e n t i m m e d i a te l y a fter t h e ·
occu rrence of the fau l t . Equ a t i o n s
p a r a meters
Fig.
3 . l9
is
(3.59)
and
(3 . 60)
c a n be used to calcu late t h e
X; o f t he m a c h i n e w h e n an oscillographic r ecor d s u ch as
ava i l ab l e . On t h e other h a n d , Eqs. 0.59) and (3 . 60) also i n d i cate the
X�
and
,
134
CHAPTER 3
TIlE SYNCHRONOUS MACH I N E
(c)
(b)
FI G U RE 3.20
X:i ;
XJ ; C c )
(a) subtransient
Xci. Vol tage E, c h a n ges w i t h
Equ iva l e n t ci rc u i t s for a syn c h r o n o u s g e n e r a t o r w i t h i n t e r n a l vo l t a g e E; a n d
reactance
(b)
10.2.
t ra n si e n t r e a c t ance
load a s discussed i n S e c .
s ync h ro n o u s rea c t a nce
m e t h o d o f d e t e rm i n i n g t h e fa u l t c u r r e n t in a g e n c r a tor w h e n i ts re a c t a n cc s a re
kno w n .
If t h e generator i s unloaded w h e n t h e fa u l t occurs, the m a ch i n e i s
r e pr e s e n t ed by t h e no-load vo l t age t o n e u t r a l i n series with the p r o p e r rc:ac­
tance . T o c a l c ulate currents for s u b t r a n s i e n t co n d i t io n s , we use r e a c t a n c e
, series with the no-load vol t age
Ei,
Xd
X� ,
Xd.
as s hown i n F i g . 3 . 20( b ) . I n t h e
i s u s e d , a s shown i n Fig. 3 . 20( c} The s ub t r a ns i e nt c u r r e n t
is much l arger t h an t h e s t e a dy-state c u r r e n t
than
The i n t e r n a l voltage
Ei
in
as shown i n Fig. 3 .20( a ) and fo r t r a n s i e n t
con d itions w e u s e t h e series reac t a n c e
steady s t a t e
X�
III
because
X:;
1 1" 1
i s m uch s m a l l e r
i s t h e s a m e i n e a c h ci r cu i t of F i g . 3 . 20 b e c a u s e
the generator i s assumed to b e i ni t i a lly u n l oa d e d . I n C h a p . 10 we s h a l l co n s i d e r
how t h e e q u iva l e n t circu i ts a r e a l tered t o acco u nt for l o a d i n g o n t h e m a c h i n e
when t h e s hort c i rcu it occu rs.
Two generato rs are con n e c t e d in p a r a l l e l to the l ow-vo l t age s i d e of a
t h re e-phase Q - Y t ra n sformer, as shown i n F i g . 3 .2 1 ( a ). G e n e r a t o r 1 is r a t e d
50,000 kYA, 1 3 .8 kY. G e n e rator 2 is ra t e d 25,000
k V A , 1 3 . 8 k Y . Each g e ne r a t o r
\
h as a subtransi e n t reactance of 25 % o n i t s own base. The t r a n s former i s rated
75 ,000 kYA, 1 3 . 8 Q / 69Y kY, with a r e a ct a nce of 1 0% . B e fore the fau l t occu rs, t h e
vol tage o n the high -vol tage side o f t h e t r a n s fo r m e r i s 6 6 k V . T h e t ra n s form e r is
u nl o a d e d a n d t her e is no c i rcu l a t i n g c u r re n t b e t\V een t h e ge n e r a t o rs . Fi n d t h e
subtransi e n t eurre n t i n each generator w h e n a t h re e - p h ase short c i r c u i t occurs o n
t h e high-vol tage side o f t h e tran s fo r m e r .
E x a m p l e 3.8.
Solution. S e l e c t 69 kY, 75 ,000 k Y A as b ase i n t h e h i g h -vol t a ge c i rc u i t . Th e n , t h e
b as e voltage o n t h e low-vol tage side i s 1 3 . 8 kY.
Generator 1
X�l
=
0 .25
75 ,000
5 0 , 000
66
E; J
=
-
69
=
=
0 .375 per u n i t
0 . 957 p e r u n i t
SHO RT-CIRCUIT C U R RENTS
3.9
jO.375
jO.10
j O . 75
( a)
p
(6)
( a ) S i n g l e - l i n e d i agram; ( 6 ) react a n c e d i ag r a m for Exam p l e
FIGURE 3 . 2 1
Gen erator
135
3.8.
2
X;;2
=
Ei 2
=
0 .25
75 ,000
=
25 , 000
0 . 750 per
0 .957 per
unit
XI = 0 . 1 0 p er
unit
66
69
=
unit
Transformer
Figure 3 . 2 1 ( b ) shows t h e reacta n c e d i agram before t h e fau l t. A t h re e-phase
faul t at P is sim u l at e d by closing switch S. Th e i n ternal vol tages o f the two
m a c h i nes may be con s i d e red to be in parallel s i nce they are i d e n tical i n
m a g n it ude a n d p h ase a n d no c i r c u l a t i n g current flows betwe en t h e m . The
e q uiva l e n t p a ra l l e l subtransient r e a c t an ce is
X"d
=
X"(/ 1 X"d 2
as
a
J"
=
v;
=
0 . 375
X ti" l + X "" 2
current in t h e s h o r t c i rc u i t is
Th e re fo r e ,
------ =
ph asor w i t h
£I.
J'Xd" + J'X I
X
0 .75
0 . 375 + 0 . 75
Ei
� Ei l = Ei 2
0 .957
j0 . 1 0
------
j O .25
+
as
0 . 25 per unit
r e fe r e n c e ,
the
- -j2 .735 per u n i t
The vo l t age VI on the 6. s i d e o f t h e tran sformer i s
1" X
jXI
c."
( - j 2 .735 ) ( jO . l O)
=
0 . 2735 p e r unit
s u b t r a nsient
136
CHAPTER 3
THE SYNCHRONOUS MAC H I N E
In genera tors 1 and 2
1"
I
I�
3.10
-
-
Ej}
-
�
0 .957 - 0 . 2735
}'Xd" l
-­
Ei 2
VI
-'XIIJ .1 2
jO .375
0 . 957 - 0 .2735
-
=
-j 1 . 823 per u n i t
--
=
jO .75
-
-j O . 9 1 2 p e r
unit
SUMMARY
Simplified equivalent circuits for the synchronous genera tor a re developed i n
this c hapte r for use throughout the rema i n d e r of the text.
We h ave seen that the steady-state performance of the synch ro nous
m ac hine rel ies on the concept of synchronous reactance Xd , which is the basis
of the steady-state equivalent circuit of the machine. In steady-state operation
w e h ave observed that the synchronous generator delivers an i ncreasing amoun t
o f reactive power t o the system t o which i t i s con n e cted as its exci tat ion is
increased. Conversely, as its excitation i s reduced, the synchronous generator
furnishes less reactive power, and when u n derexci ted, it d raws reactive pow�r
from t h e system. All of these normal ste ady-state operating co ndi tions of the
roun d-rotor generator, connected to a l arge system as i f to a n infinite b us , are
-shown by the loading capability d iagram of the machi n e .
Tra nsien t analysis o f the synchronous generator requires a two-axis m a ­
chine model . W e have seen t h a t t h e correspond ing equat ions involving p hysical
a-b-c phase variables can be simplified by Park's transformation, which i ntro­
duces d-q-O currents, voltages, and flux l i n kages. S i m p l ified equiva l e nt circuits
which fol low from the d-q-O equa tions of the mach i n e a l low definition s of the
s ubtransient reactance X:; a n d tra ns i e n t reactance X:, . S u b transi e n t reactance
X� is important in calcul ating currents resulting from short-circuit fa u l ts a t or
near syn ch ronous generators, a s discussed i n Ch ap. 1 0. The tra nsient reactance
X� is used i n stab i l i ty s tud ies, as d emonstrated i n Cha p . 1 6 .
P ROBLEMS
3 . 1 . Determ i n e the h i g h est speed at w h ie h two ge n e r a t o r s m ou n te d on the same s h a ft
can be d riven so t h a t t h e frequency o f o n e generator is 60 Hz a n d t h e freq u e n cy of
the o t h e r is 25 Hz. How many poles does each m a c h i n e h ave?
3.2. The t h re e-ph ase syn c h ronous generator d escribed i n Example 3 . 1 is operated at
3600 rpm and suppl ies a un i ty power-factor load. If t h e terminal voltage of t h e
m achine is 2 2 kV a n d t h e fi e l d current i s 2500 A, d e te r m i n e t h e l i n e curre n t a n d
t h e tot a l powe r consumption o f t h e load .
1.0&
3.3. A t h ree-phase round-rotor sync h ro n o u s g e nerator h a s n e gl i gible arm at u re r esis­
t a n ce a nd a synchronous reacta nce Xd of 1 .65 per u ni t . The mach i n e is con n ected
d i r ectly to a n i nfin ite bus of vol t a ge
per u n i t . Fi n d the i n tern a l voltage Ei
/
of the m achine when it del ivers a current o f (a) 1 .0
137
� per u n i t , (b) L!!:. per
P R OBLEMS
1 .0
u n i t , and (c) 1 . 0 - 300 per u n i t to the i n fi n ite bus. Draw p h a so r diagrams
depicting the operation of t h e m a c h i n e in each case.
3.4. A t h ree-ph ase rou n d-rotor synchronous generator, rated 1 0 k V , 5 0 MV A h as
armat ure resistance R of 0 . 1 per u n i t and synchronous reacta n ce Xd of 1 .65 per
u ni t . The machine operates o n a l O-kV i n fi n i te b u s deliverin g 2000 A a t 0 . 9
power-factor lead ing.
( a) Deter m i ne the i n t e r n a l vol tage E, and the power angle 8 of t h e machine.
D raw a phasor diagram d epicting i t s opera t i o n .
W h a t i s the open-ci rc u i t vol tage o f t h e m ac h i n e a t t h e s a m e level o f exc i t a t io n ?
(c) What i s t he steady-s tate short-circu i t c u r r e n t a t t h e same l evel o f excitati o n ?
Neglect a l l satu r a t i o n e ffects.
(6)
3.5. A t h ree-phase rou nd-rotor syn c h ro n ous gene rator, rated 16 kV and 200 MVA, h as
n eg l i gi b l e losses a n d syn chronous reactance of 1 . 65 per u n i t . It is operated on a n
i n fl n i te b u s having a voltage o f 1 5 kV. Th e i n t e r n a l e m f E j a n d t h e powe r a n g l e 0
o f t he mac h i ne are fou nd to be 24 kV ( l i n e to l i n e ) and 27.40 , respectively.
( a ) Deter m i n e the l i ne c u r r e n t and the t h ree-phase rea l and reactive power b e i n g
d e l ivered t o t h e sys t e m .
( 6 ) I f t he m echanical power i npu t a n d t h e fi e l d c u r r e n t of t h e g e n erator a r e n ow
changed so that the l i n e c u rre n t of t he m achine i s reduced by 25 % at t h e
power factor of part ( a ), fi n d t h e new internal e m f El a n d t h e pow e r a ng l e 8 .
( c ) W h i l e del iveri n g t h e red uced l i n e curre n t o f part ( 6 ), t h e m e c h a n i c a l pow e r
i n p u t and the exci t a t i o n a rc further adjusted so t h a t the m ac h i n e operates a t
u n i ty power fa ctor a t i t s terminals. Cal cu l ate t h e n e w v a l u e s o f E; a n d 8.
3.6. The t hree-phase synchronous gen erator of Prob. 3.5 i s operated o n an i n fin i te bus
of voltage 1 5 k V a n d delivers 1 0 0 MVA at 0 . 8 power-factor l aggi n g .
( a ) Determine t h e i n ternal vol tage El , t he power a n gle 8 , a n d t h e l i n e curre n t o f
t h e machine.
( 6 ) If the field c u r r e n t of t h e m a ch ine is reduced b y 1 0 % , w h i l e t h e mechanical
power i n put to the m a c h i n e is m a i n tain ed const a n t , determine the new value o f
8 an d t h e react ive power d e l i vered to t h e syste m .
( c ) The p r i me mover power i s n e x t adjusted wit hou t changing t h e excitation so
t h a t the machine d e l ivers zero re act ive power to t h e sys t e m . Determine t h e
n ew power a n g l e 0 a n d the real power b e i n g d e l ivered to t h e syst e m .
( d ) Wh a t i s t h e maxi m u m reac tive powe r t h a t t h e m a c h i n e can d e l iver i f the l evel
of exci tat ion is m a i n t a i ned as i n parts (b) and ( c )?
D raw a ph asor d i agram for t h e operation of the mach i n e i n parts ( a ) , ( b), a n d (c).
3.7. S t a rl i n g w i t h Eq . <:1 . ::1 1 ), m od i fy Eq .
')
p =
R-
Q
=
R
2
I�I
+
to show t h a t
') { I E; I ( R cos o + Xrl sin o ) - I � I R )
Xd-
1 y;, 1
+
O . ::IR)
2
Xd
.
{ Xi I E, l cos 8
-
I � I ) - R I E; l sin
{j
w h e n t h e sy nchronous generator has nonzero a r m ature resistan ce
}
R.
138
CHAPTER 3
THE SYNCHRONOUS MACH I N E
3.8. The three-phase synchronous g e n e r a to r described in Exam p l e 3 . 4 is now o p er a t e d
on a 25.2-kV infinite b u s . I t i s fou n d t ha t t h e i n t ernal vol t age m ag n i t u d e
=
49.5 kV a n d that the power a n g l e 8
3 8 . 5° . U sing t h e loading capab i l i ty d i agram
of Fig. 3 . 1 4 , determine graphically t h e real and reactive power d e l ivered to the
system by t h e machine. Verify your a n swers using Eqs. ( 3 .38).
IEi l
=
3.9.
A t hree-phase salient-pole synchronous generator with n egligible armature resis­
t ance has the fol lowing values fo r t h e i n d u c t a nce p a r a m e t e rs spec i fi e d in Tab l e 3 . 1 ;
Ls = 2 .7656 mH
Ms
=
1
. 3 8 28
Mf
mH
=
L II
=
3 1 .6950 m H
433 . 6569 roH
Lm
During balanced stea dy-state o p e ra t i o n t h e /I c 1 d c u r re n t
curr e nt o f t h e m a ch i n e h ave t h e res p e c t i v e v a l u e s
if
=
4000
A
i ll
=
20 ,000 s i n e 0d
-
0 .377 1 m H
=
a n cl (I -phase
30°
)
a r m a t ur e
A
( a)
Using Eq. (3 .41), determine t h e i ns t a n t a n eous values of t h e flux l i n kages A (I '
A b , A c , and A ! when ()d
60° .
(b) Using P ark's t ransformation g iven by Eq s . ( 3 .42) a n d (3.43 ), d e t e r m i n e t h e
instantaneous values of the fl u x l i n kages A d ' A (I ' a n d A o , a n d the c u r re n t s id,
i q , a n d i o when a J 60° .
(c) Verify resul ts using Eqs. (3.45) a n d (3. 46)
=
=
3.10. The a rmature of a t hree-phase s a l i e n t -p o l e g e n e r a to r carries t h e c u r r e n t s
(a) Using the P-transform ation m a trix o f Eq. ( 3 . 42), find t h e d i rec t - a x is c u r r e n t
and t h e quadrature-axis c u r r e n t ill ' W h a t i s t h e z e ro-se q u e nc e curr e n t io?
(b) Suppose t hat the armature curre n ts a rc
fa
id) i q , and
=
fi
X 1 000 sin( O"
- 6;, )
id
A
io.
3.11. Calculate t h e d irect-axis synchronous reactance Xd , t h e d i rect -axis transient reac­
tance Xd, and the d irect-axis subtransient reactance X; of the 60-Hz s a l i e n t-pole
synchronous m achine with the fol l o wi n g p ar a m e t ers:
D etermine
L If = 433.6569 mH
LD
mH
M!
=
Mo = 3 . 1 523
0 . 377 1 mH
M,
=
L s = 2.7656 mH
Ms
L ",
=
=
1 . 3 8 28
3 1 . 6 9 50 mH
37.028 1 mH
= 4 .2898 mH
mI l
P R O BLEMS
139
3.12. T h e single-l ine diagram of an unloaded power system is shown in Fig. 3.22.
Reactances of the two sections of the transmission line are shown on the diagram.
, The generators and transformers are rated as follows:
1:
Generator
G e n erator 2:
Ge nerator 3:
Tran sfo r m e r
20 MYA, 1 3 . 8 kY, X:; = 0.20 per u n i t
30 MY A, 1 8 k Y , X;
0 . 20 p e r unit
30 MYA, 20 kY, X:; = 0 .20 per unit
25 MYA, 220Y / 1 3.8Ll kY, X = 1 0 %
s i n g l e - p h a se u n its, each r a t e d 1 0 MVA, 1 27/1 8 kY,
3 5 M VA , 220Y 122Y k V ,
= 10%
=
T] :
T r a n s fo r m e r T2 :
T r a n s fo r m e r T3 :
X
X
=
10%
D raw t h e i m ped a n c e d i a g r a m w i t h a l l r ea c ta n c e s marked
in per unit and with
e
t
t
e
rs
to
i
n
d
i
c
a
t
e
p
o
i
n
t
s
correspo
n
di
n
g
to
the
single-line
d i agram. Choose a
l
b a s e o f 50 MV A, 1 3 . 8 kV i n t h e c i rc u i t o f g e n e r a t o r 1 .
( b ) S u ppose t h a t t h e sys t e m is u n l oaded a nd t h a t the vol tage throughout the
sy s t e m is 1 . 0 p e r u n i t o n bases c h ose n in part { / . If a three-phase short circuit
o cc u rs rrom b u s C t o g ro u n d , rind t h e p h asor v a l u e of t h e s h o rt-c i rcuit current
( i n a m p e res) i f each g e n e ra to r is r e p r es e n t e d by i t s sub t r a n s i e n t r e a c t a n c e.
( c ) Fi n d t h e m e g a vo l t a m p e res su p p l i e d by e ac h g e n e r a t o r u n d e r the cond itions o f
p a r t ( b ).
(a)
F I G U RE 3.22
O n e - l i n e d i a g r a m for P rob. 3 . 1 2 .
3 . 1 3 . The r a t i n gs
of
t h e g e n e ra tors, m o t o rs, a n d t r a ns fo r m e rs of Fig. 3 .23
G e n erator 1 :
Gen erator 2:
S y n c h ro n ous m o t o r
3:
Th re e-phase Y-Y t r a n s fo r m e rs :
Thre e - phase Y-Ll t r a n s fo r m e r s :
(0) D raw
fo r
20
20
30
20
15
X; 20%
X:; = 20%
1 3 .8 kY, X:; = 20%
1 38Y 120Y kY, X = 1 0%
1 3 8Y 1 1 3. 8 Ll kY, X = 1 0 %
M V A , 18 k V ,
M V A , 18 k V ,
MVA,
MV A,
M VA,
are
=
t h e p o w e r sys t e m . M a r k i m p e q a n ces i n p er
u n i t . N e g l ect r es i s t a nce and use a base o f SO MY A, 138 kY i n the 40-D l i n e .
t h e i m p e d a n c e d i a gr a m
140
CHAPTER 3
TIlE SYNCHRONOUS MACH I N E
(b) Suppose that t h e system i s un loaded and t h a t t h e voltage throughout t h e
system i s 1 .0 per u n i t on b ases chosen i n part ( a ). If a three-phase s h o r t circuit
occurs from bus C to grou nd, find the ph asor value of the short-circu i t current
(in amperes) if each generator is represented by i ts subtransient reactance.
(c) Find the megavoltamperes supplied by each synchron ous machin e under the
cond itions of part (b).
� Y±.
- t �
�
A
__
j40 n
�-
�
..-:J:-·2_0_n£'(-_-y
-'
C
t>
j20 0
y
t:>
--.L--.---L..- c
FIGURE 3.23
One-line d iagram for Prob.
3.13.
CHAPTER
4
S ERIES
IMPEDANCE
OF
TRANS MIS S I ON
LINES
An
electric transmission l i n e has fou r parameters which affect its ability to fulfill
its function as part of a power system: resistance, inductance , capacitance , a n d
con ductance. In t h i s chapter we discuss t h e first two of these parame ters, and w e
shall consider capacitance i n t h e n ext chapter. The fourth parameter, con d uc­
t a n c e , e x i s t s b e tw e e n con ductors or between conductors and t h e groun d . Con­
ductance accounts fo r t h e l eakage current at the insulators of ove rhead l ine�
a n d t h ro u g h t h e i nsu lat ion of cables. S i ncc lea kage at insulators of overh e a d
l i n es i s n e g l i g i bl e , t h e co n d u ct a n c e b e t w e e n conductors of an ove rh e a d line is
u s u a l l y n e g l e c te d .
Another reason for n eglecti ng conducta nce is t h a t since i t i s quite vari able,
t h ere i s n o good way of t a k in g i t i n t o a ccou n t . Le a k a g e a t i n su l a tors, the
princi pal sou rce of conductance, cha nges a ppreciably with atmospheric condi­
t ions and with the conducti n g pro perties of d i rt that coll ects on the insulators.
Corona, which resu lts in l c akage betwee n l ines, is also quite variable with
atmospheric conditions. It is fortunate that the effect of conductance is such a
n egligible component of shuht admittance.
Some of the properties of a n e l ectric circuit can be explained b y the
elec t ric and magnetic fields which accomp a ny i ts current flow. Figure 4.1 shows
a s i n g l e - ph a se line and i ts associ a ted magnetic and electric fields. The lines of
mag n e t ic flux fo rm closed l oops linking t h e circu i t , and the l i nes of' electric flux
ori g i n a t e on the positive charges on one con d u c tor and term i n a t e on the
142
CHAPTER 4
SERIES I M PEDANC E OF TRANSMISSION L I N ES
Magnet
FI G U R E
..U
M a g n e t i c a mI e l e c t r i c t i t: k � s a ssoc i a t e d w i t h
a two-w i r e l i n c .
negative c har ge s 0 11 t h e othcr c o n d u ct o r. V a r i ,l l i o l1 o f t h e c u r re n t i l l t h e
con ductors causes a change i n t h e number of l ines o f m a g n e l l c n u x l i n k i n g
t h e circu i t . Any change i n t h e ft u x l in k i ng a circui t induces a vol tage i n the
circui t which i s proport iona l to the r a t e o f c h a ng e of tl u x . The i n d ucta nce of the
circuit relates t h e vol tage induced b y c h a n g i n g flux t o t h e ra t e of change of
curre nt.
The capaci tance which exists between t h e cond uctors is defined as the
c harge o n the condu ctors per u n i t of poten t i a l d iffe rence between them.
The resistance and i nductance u niformly d istributed along the l i ne form
the series impedance. The conductance and capaci tance exi sting b e tween con­
ductors of a single-phase l i ne or from a con d uctor to neutral of a three-phase
line form the shunt admittance. Alt hough the resistance , inductance, and
capacitance are distribu ted, the equivale n t circui t of a line is made up of
l umped parameters, a s we shall see when w e d iscuss them.
4.1
TYPES O F C ON D U CTORS
In the early days of the transmission of e lectric power conductors were usually
copper, b u t aluminum conductors have completely replaced copper for ove r­
head l i nes because of the much l ower cost and l ighter weight of an a l um i num
conductor compared with a copper con d u ctor of t h e same resistance. T h e fact
that an aluminum con d u c t o r h as a l a rger d i a m e t er t h a n a co p p e r c o n d u c t o r of
t h e same resistance is also an a dvantage. W i t h a l a rger d iame ter, the l i nes of
e l ectric fl ux originating on the cond uc tor will be farther a p a r t at the cond uctor
surface for the same vol tage. This means t h e re is a l ower vol tage gradient at the
conductor surface and l ess tend ency to io nize the air around the con ductor.
Ionization produces the u ndesirable effect called coron a .
Symbols identify ing diffe rent types of aluminum cond uctors are as fol lows:
AAC
AAAC
ACSR
a ll -alum i num cond uctors
all-aluminum- alloy cond uctors
a l u m i n u m co n d uctor, stee l-rei nfo rced
., 1 " rn ;
n"
rn
.... A n
r1 " ,... r n ,-
-:l l l ",,_rp ; n fo n'prJ
4.2
RESISTANCE
143
Al u m i n u m
FIGURE 4.2
Cross section of a s t ee l - r e i n forced conductor, 7 steel
s t r a n d s , a n d 24 a l u m i n u m st rands.
Alu m i num-alloy conductors have higher tens ile stre ngth than the ordinary
electrical -cond uctor grade of a l u m i n u m . ACS R consists of a central core of
steel stran ds su rrou nded b y l a ye r s of a l u m i n u m st ra n ds. ACA R has a central
core of hi g h e r- s t r e n g t h a l u m i n u m s u r ro u nded b y layers of elcct rica l-conductor­
gra d e a l u m i n u m .
Al ternate l ayers o f w i re o f a s t randed cond uctor are spiraled i n opposite
directions to preven t u nw i n d i n g and to make the outer rad i us of one l ayer
coincide with the i n·n er rad ius of t h e n ex t layer. Strand ing provides flexibility for
a l a rge cross-sectional area. The nu mber of strands depends on the number o f
I"ayers a n d o n whether all the strands a re o f t h e s a m e d iameter. T h e total
number of strands in concen t rical ly stranded cabl es, where the total a nnular
space is filled with s trands of uniform d i a meter. is 7, 1 9, 37, 6 1 , 91, or more.
Figure 4.2 s hows the cross section of a typical steel-reinforced alu m inum
cable (ACS R). The conductor shown has 7 steel strands forming a central core,
around which t here are two l ayers of aluminum strands. There a re 24 aluminum
strands in the two outer layers . The c onductor stranding is specified as 2 4 Al/7
St, or simply 24/7. Various tensile strengths, cu rre nt capacities, and cond uctor
sizes are obtained by using different combinations of steel and aluminum.
Appendix Table A�3 gives some el ectrical characte ristics of ACS R. Cod e
names, uniform t h roughout t h e a l u m i n u m industry, have been assigned t o e a c h
cond uctor for easy reference.
A type of conductor known as expanded ACS R has a filler such as paper
separating the inner steel strands from the outer aluminum strands. The paper
gives a larger d iameter (and he nce, l ower corona) for a given cond uctivity a n d
tensile strengt h . Expanded ACS R i s llsed for some extra-high-voltage (EHV)
lines.
4.2
RESISTA N C E
The resistance of transmission-line conductors is the most i mportant cause of
power loss i n a t ransm ission line. The term " resistance," u n less specifically
q u a l ified, means effective resistance. The e ffective resistance of a conductor is
R =
powe r
�
l o ss i n
)J 1 2
------
conductor
----
n
(4 . 1 )
144
CHAPTER 4
S E R I ES I M PEDANCE OF TRANSMISSION LI N ES
where t h e power is i n watts a n d I is the rms cu rrent in t h e conductor i n
amp eres. The effective resistance i s e q u a l to t h e dc resistance of t h e con d uctor
only if the distribution of curre n t t h roughout the con ductor i s uniform. We shall
discuss nonuniformity of curre n t d istribution b ri e fly after reviewing some fun d a­
mental concepts of d c resistance.
Di rect-current resistance is given by the form ula
where
p =
I =
A =
Ro
=
pI
- fl
A
( 4 . 2)
resistivi ty of con ductor
lenoth
I::!
cross-sectional area
Any con s istent set of units m ay be used . I n power work in the enited S tates / is
usually given in feet, A i n circular mils (cmiI), a n d p i n ohm-ci rcu l a r mils per
foot, sometimes called ohms per circular m il-foot. I n S 1 u nits I i s in meters, A
in square m eters and p i n ohm-m eters. l
A circu l a r mil is the a rea of a circle having a d iameter of 1 m i l . A m i l is
equal to 1 0 - 3 in. The cross-sectional area of a solid cyl ind rical con d uctor i n
circular m ils i s equal t o the square o f t he d iameter o f the conductor expressed
in mils. The n umber of circular mils m u ltiplied by 17/4 equals the n umber of
square mils. Since m a n u facturers i n the U nited States identify conductors by
their cross-sectional area in circular m i ls, we must use this unit occasiona lly.
The a re a in square mill imeters e q uals the area i n circular m ils multiplied by
5.067 X 1 0 - 4.
The i n t e rn a t i o n a l s t a n d a rd o r co n d u c t i v i t y i s l i l ; l l o f a n n e a l e d c op p e r.
Com me rc ial h a r d d r a w n cop p e r w i re h a s <J 7 . YYo a n d a l u m i n u m h J S 6 1 % o r t h e
con d uctivity of s t a n d a rd annea led co pper. A t 20°C fo r hard-d rawn copper p is
1 .77 X 1 0 - H fl . m 0 0.66 n c m i l / h ). Fo r a l u m i n u m a t 2 0 ° C p i s 2. 83 X 1 0 - 1)
.n m ( 1 7.00 n c miljft).
The dc resistan ce of s t ra nd e d cond u ctors is greater than the value com­
p uted by Eq. ( 4 .2) because s p i ra l i n g o f t h e s t r a n d s m a kes t h e m longer than the
conductor i tself. For each m i le of cond uctor the current in all strands except the
o n e i n t he center flows i n more than a m i l e of wire. The i ncreased resistance
due to s piraling is estimated as 1 % for th ree-strand cond uctors and 2% for
concentrically stranded conductors.
The variation of resistance of metallic cond uctors with temperature is
practically l inear �ver the norm a l range of ope ration. If temperature is p lotted
o n the vertical axis and resistance on the horizontal axis, as i n Fig. 4.3, extension
-
.
.
lSI
.
is
t h e o ffi c i a l des i g n a t ion for the I n tern a t i o n a l Sys t em o f
U n i ts.
4.2
R ESISTANCE
145
t
R
I
T
1
/
/
/
/
/
FIG U RE 4.3
R e s i s t a nce of a m e t a l l i c con ductor as a fu nction of tem­
per ature.
of the straigh t-line portion of the graph provides a conve nient method of
correcting res ista nce for changes in tem perature . The point of in tersection of
the extended line with the tempera ture axis at zero resistance is a constant
of the materi a l . From the geometry of Fig. 4.3
( 4 .3 )
where R l and R 2 are the resistan ces of the conductor a t t e mperatures t 1 a n d
f 2 ' respectively, i n degrees Ce lsius a n d T i s the constant de termined from t h e
graph . Values o f the constant T i n d egrees Celsius a r e a s fol lows :
T
=
{
234 .5
241
228
fo r anne a l e d copper of 1 00% conductivity
for hard-d rawn copper of 97 . 3 % conductivity
fo r h a rd-drawn alum i n um of 6 1 % conductivity
Uniform distribu tion of c urre n t t h roughout the cross section of a conduc­
tor exists only for d i rect curre n t . As the frequency of a lterna ting current
i ncreases, the no n u n i form i ty of d ist ribu t ion beco m e s more p ronounced. An
increase in frequency causes non u n i fo rm cu rrent dens ity. This p he nomenon is
cal led skin effect . I n a circular co nd uctor the curren t d ensity usually increases
from the interior towa rd the surface. For cond uctors of sufficiently large radius,
however, a cu rrent density oscillat ory with respect to radial d istan ce from t h e
center may result.
A s we shal l see when d iscussing i n d u ctance, some l i nes o f magnetic flux
exist in side 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
fl ux l inking a filame n t in the i n te rior. The alternating flux induces higher
vol t ages acting on the inte r ior fi l aments than are ind uced on fi laments near the
surface of the conductor. B y Le nz's l a w the ind uced vol tage opposes the
146
CHAPTER 4
S ERIES I MPEDANCE OF TRANS M I S SION LINES
changes of current p roducing i t , and the h igher i nduced voltages acting o n
the inner filame n ts cause t h e higher current d e ns i ty i n filaments nearer t h e
surface, a n d therefore higher effective resistance results. Even a t power system
frequencies, skin effect i s a significant factor i n large conductors.
4.3
TABU LATE D R E S I STAN C E VALUES
The dc resistance o f various types of conductors is easily found by Eq. ( 4 . 2 ) , a n d
the increased resistance d u e t o spira l i n g can be estimated. Te m pe ra t u re correc­
tions a re d et ermined by Eq. ( 4 .3). The increase i n resistance caused by skin
effect can be calcul ated for rou nd wires a n d tu bes or solid m a t erial, and c u rves
of R /R o a re ava i lable for these simple cond uctors. 2 This information i s not
necessary, however, s ince m a n ufact ure rs s upply tables of e l ect rical ch aracteris­
tics of the i r con duct ors. Ta b l e A . 3 is a n e x a m p l e o r s o m e o r t h e ( Li l a a v a i l a b l e .
Example 4 . 1 . Tab l es o f e l e c lrical c h aracteri stics o f a l l - a l u m i n u m .\fa rigo!d s t r a n d e d
conductor l i s t a d c resistance of 0 . 0 1 5 5 8 D p e r 1 00 0 ft a t 200e a n d a n a c r e s i s t a n c e
of 0 . 0 9 5 6 Djmi a t 5 0°C. The conductor has 6 1 s t rands and i t s size is 1 , 1 1 3 ,0 0 0
c m i l . Verify t h e dc resistance a n d fi n d t h e ratio of ae to dc res i s t a n c e .
Solution. At 2 0 0 e
Ro
At a t e m p e r a t ur e
from Eq. (4.2) w i t h an increase of 2 % for s p i r a l i n g
1 7 .0
1 1 13
=
x
1 00 0
X 1 03
X 1 .02
=
0 . 0 1 558 D
per
1 0 00 ft
of 5 0 0 e from Eq. ( 4 . 3 )
Ro
R
Ro
=
0 .0 1 5 5 8
228 + 5 0
228 + 2 0
=
0 . 0 1746 n p e r 1 000 ft
0 .0956
------ = 1 . 037
0 .0 1 746 x 5 . 2 8 0
Skin effect causes a 3.7% i n crease in resistance.
4.4 INDUCTANCE OF A C O N D U CTOR DUE
TO I NTERNAL FLUX
The inductance of a tran smission line is calculated as flux linkages per ampere.
If p ermeabi lity IL is constant, sinusoid a l current produces sinusoidally varying
flux in phase with the current. The res u l ting flux lin kages can then be expressed
2 S e e T h e A l u m i n u m Asso c i a t i o n , A luminum Electrical Condu rlnr Handbouk , 2d ed., W a s h i,n g t o n ,
DC, 1 982.
4.4
I N DUcTANCE O F A CONDUcrOR D U E TO I NTERNAL FLUX
147
as a p hasor A , and
L
A
( 4 .4 )
I
I f i , the instantaneous value of current, is subst i t uted for the p hasor 1 in
Eq. (4.4), then A should be the v a l u e of the i nstantaneous flux l inkages
prod uced by i. Fl ux li nkages are meas u red in weber-turns, Wbt.
Only flux lines external to the cond uctors are sh own in Fig. 4. 1. Some of
t h e magnetic field, however, exists i n side the cond uctors, as we mentioned w he n
considering skin effect. T h e changing l i ne s o f fl u x inside the conductors a lso
con trib u te to the induced vol t age of t h e circuit and therefore to the i n ductance.
The correct value of ind uctance d u e to i n ternal A ux can be computed as t h e
ra tio o f fl ux l i n kages t o cu rre n t b y t a k i n g i n t o account the fact t h a t each l i ne o f
i n t e r n a l fl u x l i n ks o n l y ,I fr<'lction o f t he tot a l c u r r e n t .
To obtain a n accurate va l u e for the in ductance o f a transm ission line, i t is
necessary to consider t h e fl ux inside e a c h conductor as wel l as the external fl ux.
Let us consider the l ong cylind rical conductor whose cross section is shown in
Fig. 4.4. We assume t h a t the return path for the current in this c onductor is s o
far away that i t does not appreciably affect the magnetic field of t h e conductor
shown. Then, the l ines of flUX are concentric with the conductor.
By Ampere's law the magnetomotive force (mmi) i n ampere-turn s aro u n d
a n y closed path i s equal t o t h e n e t c u r r e n t i n amperes enclosed b y the p a t h , a s
discussed i n Sec. 2 . 1 . The mmf e q u a l s t h e l i n e integral around the closed p a t h
o f the component of the magnetic field i ntensity tangent to the path a n d i s given
by Eq . (2.4), now written as Eq . (4.5) :
mmf
=
¢ H . ds = J At
( 4 .5 )
//;:-- - -::�
�
�\
- - -
I I
I I
I f
I I
I
\ \
\
,
,
"
"
......
_
�x
\ \
\ \
\
x
f
"'
,/
/
' -�
('
- -Flux
I
...... -- - .....-- _ / ..... /
I I
/ '
/
/
/
FIGURE 4 . 4
Cross sect ion of a cyl i n d ri c a l con d u ctor.
'
148
CHAPTER 4
H
where
SERIES I M PEDANCE OF TRAN S M I S S I O N LINES
= magnetic field intensity , At/m
s = distance along path , m
I
=
curren t enclosed , A
H and I are shown as phasors to represent s i nu soidally a lternating
quantities si nce our work here appl ies e q u a l ly to alternating and d i rect curre n t .
For si m pl i c i ty t h e current I could be interpreted a s a d irect cu rre nt a n d H a s a
real number. We recal l that the dot b e tween H and ds i n d icates t h a t t h e value
of H i s the component of the field i n t e nsity tangent to ds.
Let the field intensity at a d istance x meters from the center of t h e
conductor be designated Hx ' Si nce t h e fi e l d is symmetrical, Hx is constant a t a l l
points equid istant from t h e ce nter o f t h e cond u ctor. I f t h e i n tegrat ion i n d icated
in Eq. (4.5) is pe rformed a round a circ u l a r path co nce n t ric w i t h t h e con d u ctor
at x meters from the center, fIx i s const a n t over the p a th and tange n t to i t .
Equ ation (4.5) becomes
' Note t ha t
( 4 . 6)
( 4 .7)
and
where
Ix is the current enclose d . The n , a ss u m i ng u niform current density,
(4 .8)
I i s t h e total curre nt in t h e con d uctor. T h e n , substit ut i n g Eq . ( 4 . 8) I n
Eq . (4 .7) a n d solvi ng fo r Hx ' we obtain
where
( 4 . 9)
The
flux d e ns i ty x meters from the cent e r o f the con d u ctor is
where J.L
Bx
= J.L Hx
/-L xI
=
2 7T" r 2
Wb/m 2
( 4 . 10 )
i s the permeability of the con ductor. 3
I n the tubular element of thickness dx t h e fl u x d¢ is B times t h e
cross-sectional area of the element normal t o the flux lines, the area being dx
J
ILr = J-L /
1 n SI u n i ts
J-Lo·
x
the permeabi lity of free space
is
/.Lo
�
4 7T
X
10-7
Him, and t h e r e l a t ive p e r m e a b i l i t y
is
4.5
FLUX LI N KAGES BETWEEN TWO POI NTS EXTE RNAL TO AN ISOLATED CONDUCTOR
149
t imes the axial length. The flux p e r m e t e r of length is
J.LxI
dx
2
d¢ = 2 TT'f
(4 . 1 1 )
Wb/m
The flux l inkages d A per meter o f length, which are caused by t h e flux in the
tubular element, a re the p ro duct of the flux per meter o f length and the fraction
of the current l i nke d . Thus,
dA
=
1T X 2
d e/>
rr r 2
=
j.L lx J
2 TT r
4
dx Wbt/m
( 4. 12)
I ntegrating from the cen te r of the conductor to i ts outside edge to fi n d A inp the
total flux li nkages i nside the conductor, we ob tain
A jrll
=
1
r
o
For a relat ive permeab i l i ty of
1,
A int
J.L lx 1
4
2 1T r
L int =
f..L I
=
8 rr
-
= 4 7T X 1 0 - 7
J.L
=
d'(
I
'2
1
-
2
X
X
( 4. 13)
Wbt/m
H i m, a n d
10 - 7
Wb t / m
( 4. 14)
10-7
Him
( 4. 15)
W e have computed the inductance per u n i t le ngth ( h e n rys p e r m e te r) of a
rou n d conductor attri buted only to the flux inside the conductor. Here after, for
convenience, we refe r to inductance per unit length s imply as inductance, but we
m u s t b e ca refu l to use the correct d i m ensional u n i ts.
The validity of com p u t ing the i nternal i n d uctance of a sol i d rou n d wire by
the method of partial flux l i nkages can be demonstrated by deriving the internal
i n d uctance i n a n entirely d ifferent m anner. Equating energy stored in t h e
m a g netic fi e l d wit h i n the co n d uc t o r p e r u n i t length at a ny instant t o L int i 2 /2
and solving for L illt wi ll yield Eq. (4. 1 5 ).
4.5
FLU X LINKAGES B ETWE E N TWO
PO INTS EXTERNAL TO AN I S O IATED
C O N D UCT OR
As a step i n comp uting i n ductance d u e to flux external to a conductor, let us
de r ive an expression for the flux l i n kages of an isol ated conductor due only to
that portion of the extern al fl ux which lies b e tween two points at D I and D 2
m e ters from the center of the cond uctor. I n Fig. 4.5 P I and P2 are two such
points. The cond uctor carries a current of J A . S ince the flux paths are
J
150
CHAPTER 4
S E R IES I M P E D A N C E OF T R A N S M I S S I O N LI N ES
FI C U R E 4.5
A co nductor alld ext e rn a l poi n ts P I and P2 '
concentric circles around the con d uctor, a l l the flux between P I and P 2 l ies
within the concentric cyli n drical surfaces (ind i cated by solid circular li n es) w hich
pass t h rough Pl and P2 ' At the tubular element which is x meters from the
cen ter of the cond u ctor the field i ntensity is Hx ' The mmf around the ele m e n t is
( 4 . 16)
Solving
that
for Hx a n d multiplying b y
yield t h e fl ux density
J.L
Ex
i n t h e e l e m e n t so
( 4 . 1 7)
The
flux dcjJ in the tubular clement of thickness
d(p =
j.L /
--
2 rr x
dx
dx
IS
Wb/m
( 4 . 18)
flux l i nkages di\. p e r meter are n u merically e qual t o t h e flux d¢ s i nce fl ux
extern a l t o t h e con d u ctor l i n ks a l l t h e c u rr e n t i n t h e cond uctor only once. So,
between PI and P 2 the flux l in kages are
The
( 4 . 19 )
or for a rel ative p e rm e a b i l i ty
A 12
=
of
1
2
X
10
7
I
D2
I n - Wbt / m
D,
( 4 .20)
,
4.6
I N D UCTANCE
OF A SINGLE-PHASE TWO-WIRE
LINE
151
The inductance d u e only to the flux included between P I and P2 i s
( 4 .21)
INDUCTANCE
TWO-WIRE LINE
4.6
OF
A SINGLE-PHASE
We can now d e termine the i nductance of a simple two-wire l i n e composed o f
sol i d rou nd conductors. Figu re 4 .6 shows such a line having two conductors o f
rad i i ' 1 and ' 2 ' One cond uctor is t h e return circuit for the other. First, consider
only the fl u x l i nkages of the circu i t caused by the current in conductor 1 . A l ine
of flux s e t up by cu rrent in conductor I at a distance e Cl u a l to or greater than
D + ' f r o m t h e ce n t e r of cond u ctor 1 docs n o t l i n k thc c i rcu i t . A t a d i s t a nce
less t h a n D r2 t h e fr" c l i o n o f t h e t o t a l c ur r e n t l i n ked by a l i n e o f fl ux is 1 .0.
Therefore , it is l ogical w h e n D is much greater than ' I a n d ' 2 to assume t h a t D
can b e used instead of D - ' 2 o r D + ' 2 ' I n fact, i t c a n be show n that
calculations made w i t h t h i s assumption are correct even when D is smal l .
We add i n ductance d u e t o i n ternal fl u x l inkages determ i n e d by Eq. ( 4. 15)
to i n d uctance due to exte rnal fl ux l i nkages determ i ned by Eq. (4 .2 1 ) with ' 1
replacing D 1 and D replacing D2 to obtain
"
-
( 4.22)
which is t h e i n ductance of the circu i t d u e to the curr e n t i n cond u ctor
1
o n ly.
FI GURE 4.6
Con d u ctors of d i ffe r e n t r a d i i a n d t h e mag­
netic fi e l d d u e to cu r re n t In co n d u ct o r 1
only.
152
CHAPTER 4
S E R I ES I M PEDANCE Or: TRA N S M I S S I O N L I N ES
The expression for inductance may b e p u t in a more conCIse form by
factoring Eq. (4.22) and by noting that In £ 1 /4 = 1 14, whence
( 4 .23)
Upon
combining terms, we ob tain
( 4 .24 )
If
we substitute r ; for r 1 f - 1 / 4 ,
L)
=
2
X
10 - 7
In
D
,
r
1
Him
( 4 . 25 )
The radius r 1 is that of a fictitious con d uctor assumed to have n o internal fl ux
but w i th the same inductance as the actual con d uctor of radius r I ' The quantity
£ - 1 / 4 is equal to 0.7788. Equation (4 .25) omi ts t h e t e rm accounting fo r i nternal
flux but compensates for it by using an adjusted value for the rad ius of the
conductor. T h e multiplying factor of 0.7788 , whi c h adj usts the radius in order to
accou n t for i nternal fl ux, applies only to solid rou n d cond uctors. We consider
other con d uctors later.
Since the current i n cond u ctor 2 fl ows in the d i rection opposite to that i n
conductor 1 (or i s 1800 out o f phase with it), the fl u x li nkages p roduced by
current in conductor 2 considered a lo n e are i n the same d i rect i o n t h rough t h e
circui t a s those p roduced b y CU IT e n t i n con d uctor 1 . T h e resul ting fl u x for the
two cond uctors is determined by the sum of the mmfs of both cond uctors. For
constant permeability, however, the fl ux linkages (and l ikewise the i n d ucta n ces)
of the two con d uctors considered separately m ay be a d d e d .
By comparison with Eq . (4.25), the i n d uctance d u e t o current in con d uctor
2 is
L2
and
for
the
=
2
X
D
10 - 7 In -, H i m
r2
( 4 .2 6 )
complete circuit
( 4.27)
.
If r�
= r; =
4.7
FLU X L I N KA G ES OF O N E CONDUcrOR IN A G R O U P
153
r' , the total i nductance reduces to
L =
4
X
10-7
In
D
-
r'
( 4.28)
Him
This value of inductance is someti mes called the inductance per loop meter or
per loop mile to distinguish it from that component of the inductance of the
circui t attributed to the cu rrent in one conductor only. The latter, as given by
Eq . (4. 25) , is one-half the total i n d u ct ance of a single-phase l ine and is called
the inductance per conductor .
4. 7
FLUX L I N KAGES O F O N E COND UCTOR
IN A GROUP
A more general problem than that of the two-wire l i ne is presented by one
cond uctor i n a group o f con d u ctors where the sum o f the currents i n all the
con d u ctors is zero. Such a group of cond uctors is shown in Fig. 4.7. Conduc­
tors 1 , 2, 3, . , n carry the phasor currents 1 1 , 12 , 13 " " , In " The d istances of
t h ese conductors from a remote point P are i ndicated on the figure as
D I P ' D z p , D 3 P " ' " Dn P ' Let us dete rmine A I P I ' t h e flux l inkages of con ductor 1
due to I I i n cl u d ing internal flux l i nkages but excluding all the flux beyond the
point P. By Eqs. (4. 14) and
.
.
(4.20)
(� + 2I l n Dlpl10-7
A I Pl =
2
I
( 4.29)
rI
( 4 .30)
The fl ux linkages A I P 2 with con d u ctor 1 due 10 12 but excluding flux beyon d
point P is equal to t h e fl u x produced by / 2 between t h e point P and conductor
FIGURE 4.7
Cross-sectional view of
of
n
a
group
cond uctors carrying cur­
rents whose sum i s zero. Point
n
F is remote from the con d uc­
tors.
154
CHAPTER 4
SERIES I M PEDANCE OF TRANSMISSION LINES
1 (that is, within the limiting d i st a n ces
D 2 f'
an d
from cond uctor 2), a n d so
D 12
( 4 .3 1 )
The flux l i n k a g e s A I I > w i l h conductor
but excluding flux beyond point P is
due
to aI/ ,lie conductors
I II
t h e g ro u p
( 4 . 32)
w hich becomes, by expanding the loga rithmic terms and regrouping,
A
IP
=
2
X
( -
1 0 - 7 I I In
+ I, In
1
r�
1
I2 In - +
+
D,p
D I2
+ I, I n
D2P
1
I
3
In D 13
+ 13 I n
+
D3P +
... +
In In
. . . + In
1
DIn
In Dn p
)
( 4 . 33)
Since the sum of all the currents in the group is zero,
_
and solving for
I" ,
II +
we obtain
12 + 13 +
...
+
III
=
0
Subst i tu t i ng Eq. (4.34) i n the second t e r m contammg
recombining some loga r i t h m i c terms, we have
+ /1 ln -- + /2 1 n -- + /3 I n -- + . . . +
DIP
D2 P
D 3P
Dn p
Dn P
Dn P
III
In
In - I I n
Eq. (4.33) and
D (n
-
I)?
Dn P
1
( 4 .35)
Now letting the point P move i n fi n itely far away so that the set of t.erms
containing l og a r i th m s of ratios of d istances from P becomes infinitesimal, since
4 .8
I N D U CTA N C E OF COM POSITE-C ON DUcrOR L I N ES
155
the ratios of the distances approach 1 , we obtain
_1_ )
+ I In
n
D In
Wbt/m
( 4 .3 6)
B y letting point P move infin ite ly far away, we have included all the fl u x
l i n kages of con ductor 1 i n o u r derivation. Therefore, Eq. (4.36) expresses all t h e
fl u x l inkages of conductor 1 in a group o f cond uctors, p rovided the s u m o f a l l
t h e currents is zero. ] f t h e currents a re al ternating, they must b e expressed a s
instantaneous currents t o obtain i nstantaneolls flux l i nkages o r as com p l e x r m s
values t o u h t a i n the r m s va l u e of n u x l i n k i t g c s as a co m p l e x n u m her.
4.8
I N D U CTAN C E O F C O M P OS IT E­
CO \l D U CTO R LI N ES
S tra nded co n d u c tors cum e u n d e r the general c lassification of compo site conduc­
tors, wh ich means con d u ctors composed of two or more elements or strands
electrically in parall e l . W e l i m i t o u rselves to the case where all t h e s t r a n d s a re
identical and share the current equal ly. The va lues of internal i n d ucta nce of
specific cond uctors a re generally avail abi e from the various m a n ufactu rers a n d
can be fou nd in ha ndbooks. The method to b e developed indicates t h e a p proach
to the more com plicated problems of nonhomogeneous con d uctors and u n eq u a l
division of curre nt between strands. The tnethod i s applicable to the determina­
t ion of inductance of l i nes consisting ofcircuits electri.cally i n parallel since two
conductors in par all el can be treated as strands of a s i ngle composite conductor.
'
Figure 4.8 shows a single-phase line composed of two conductors. In o rder
to be more general, each conduc tor forming one side of the line i s s hown a s a n
arbitrary arrangement o f a n indefinite n u mber o f con ductors. The o n ly restric­
tions are that the p a r a l l el filamentS are cyl indrical and share the current
equally. Cond uctor X i s composed of n identical, parallel fi la m e nts, e a ch o f
which carries t h e current lin . Conductor Y , which i s the return c ircu it for t h e
current i n conductor X, is composed of m identical, parallel fi.l a me nts, e a c h o f
which carries the current - / Im . Distances between the elements w i l l b e
designated by the letter D w i t h a ppropriate subscripts. A pplying Eq. (4.36) t o
'-�---...V'_-_J
Condo X
FIG U R E 4.8
'----v---"
Condo Y
S i n g l e -ph ase l i n e consis t i n g of t wo compos i t e c o n ductors.
156
CHAPTER
filament a
Aa
=
S E R I ES I M PEDANCE OF TRANSM ISSION LI N ES
4
o f conductor X, we obtain for flux l in kages of filament
2 X 10
-
7
I
-
n
( 1
2 X 10-7
_
r�
In
-
�
m
(
+
In
I n _1_
Doa'
1
-
Da b
+
+
In
1
-
D ac
I n _1_
Doh'
+
+ . . , +
I n _1_
Doc'
1
In
-
Da n
..
+
)
)
+ In
.
a
_1_ )
Da m
( 4 . 37)
from wh ich
Vlb t/ m
( 4 . 38)
Dividing Eq . (4.38 ) by the current lin , we find t h a t t h e inductance of fi lament
a is
L
Aa
=
a
l in
-
=
2n X 1 0 - 7 I n
VDaa' Dab, Dac'
Vr�Dab Dac
l i t .---_
_
_
_
n ,---_
Dam
----
. . .
-- -.
.
Dan
.
Him
( 4 .39)
Similarly, the inductance of filament b is
( 4 . 40)
The
average inductance of the fi la ments of conductor X is
La + L b + L c + .
+ Ln
L av = -------
n
. .
( 4.41 )
Conductor X is composed of n fi laments electrically in p arallel. If a l l the
filaments had t h e same indu ctance, the i n ductance of the con ductor would b e
lin t imes t h e inductance of one filament. H e r e a l l the filaments have d ifferent
inductances, but the inductance of all of them i n p aral lel is l i n times the
average inductance. Thus, the inductance of conductor X is
La + Lb + Lc +
' "
+ Ln
n2
Sub stituting the logarithmic expression fo r inductance
of
( 4 .42)
each filament i � Eq.
4 .R
I N D UCTA N C E OF COM POS I TE·CO N D UcrOR LI N ES
157
(4.42) and combining terms, we obtain
Lx
= 2 X
10-7
( 4 .43 )
where r� , r� , and < h ave been replaced b y Dall , Dhb , and Dnn , respectively, to
make the exp ression appear more symmetric a l .
N ote t h a t the numera tor of t h e a rgument of the logari t h m i n Eq. (4.43) i s
the mn th root o f m n terms, which are the p rod ucts of t h e distances from a l l t h e
n fi l a ments of con ductor X to a l l the In filaments o f cond uctor Y . For each
fi l a m e n t in cond uctor X t h e re a re 11/ d istances to filaments in cond uctor Y, a nd
there a re !l fi l aments i n con d uctor X. The prod uct of m distances for each of n
fi l aments results i n /7111 t e rm s. The m n t h root of the prod uct o f the mn
d i s t a nces is cal led the geometric mean distance between conductor X a n d
conductor Y . I t i s abb rev i a ted D", or G M D and is also called t h e mutual GMD
between the two cond uctors.
2
The denominator of the argu ment of the logarithm in Eq. (4.43) i s the n
root of n 2 terms. There are n filame nts, and for each filament t here a re n terms
consisting of r' for that fi l am ent times the d i sta nces from that filament to every
other fi l ament in conductor X . Thus, we account for n 2 terms. Sometimes r� i s
called t h e d istance from fi l a ment a t o itself, especially when i t i s designated as
Daa . With this in m ind, the terms u nder the radical in the denominator m ay be
described a s the prod uct of the d istances from every fi l a ment in the conductor
to i ts e l f and to eve ry other fi l amen t . The n2 root of these ter m s i s called the self
GMD of conductor X, a nd the r' of a separate filament is ca l l e d the self G M D
of t h e filament. Sel f G M D is also called geometric mean radius , or G M R . T h e
correct m a t he m Cl t i c a l expression is s e l f G M D, b u t com mon practice h as m a d e
G M R more prevalent. W e u s e G M R i n orJ e r to conform t o t h i s practice a n d
ide n t i fy it b y D
I n terms of Dil l a nd D, Eq. (4.43) becomes
\ .
. Lx = 2
x
to
7
f)1Il
In Ds
Hjm
( 4 . 44 )
ThL: reader should compare Eqs. (4.44) and (4.25 ).
The inductance of cond uctor Y i s determined i n a similar m a nner, a n d t h e
indu ctance of the l i n e i s
L
=
Lx
+
Ly
158
CHAPTER 4
SERIES I M PEDANCE OF TRANS M I SS I O N LI N ES
One ci rcuit of a single-phase tra nsmission line is composed of t h ree
solid O.2S-cm-radius wires. The r eturn ci rcuit is composed o f two O.S-em-ra d i u s
wires. T h e arrangement of conductors i s shown i n F i g . 4 . 9 . Fi n d t h e i n d u ct a n c e
due to the cu rrent in each side of the line and t h e i ndu ctance of the com p l e te l i n e
. i n h e n rys per meter (and i n millihe n rys per m i le).
Example 4.2.
So/utioll.
Fi nd t he GMD b e t w ee n sides X and Y:
DIII
Then, find
the G M R
=
=
V92
6
____-----X 15 X 1 1 7 3/2
=
1 0 .743 m
for side X
V(O.25
9 ,--------- ----,.----3
X 0 . 7788 X 1 0 - 2 )
X 64
X
1 2 2 = 0 . 48 1 m
6m
6m
10
c
'--y---'
Side X
'--r--'
Side Y
FIGURE 4.9
Arrangement
of con d u ctor s for Exa m p l e 4.2.
4.9
a n d for side Y
Ds
Lx
= V (O . S
4
=
2
X 10
X 0 . 7788 X 1 0 - 2 )
-7
In
L y = 2 X 1 O - 7 1n
L = Lx
(L
6
=
2
62 =
,-____________________
+
1 0 . 743
6.212
X
0.153
=
X 10-7
H im
0.153
= 8 . 503 X 1 0 - 7
H im
1 0 . 743
1 0 -7
X
1 609
X
Him
103
=
2 .37
OF
TABLES
159
m
0 .4 8 1
L y = 1 4 .7 1 5 X 1 0 - 7
1 4 .7 1 5
X
THE U S E
mH/mi)
I n Exa m p l e 4 .2 t he con d u ct ors i n para ll e l on o n e side o f the l in e are
separated by m, a n d t h c d i st ance between the two sides of the l i ne is 9 m .
Here t he ca lcul ation of m u t u a l G M D is importa n t . For stranded con ductors the
d istance between sides of a l i n e composed of one cond u ctor per side is usually
so great that the mutual G M D can be taken as equal to the center-to-center
d i st ance with n egl igible error.
If the effect of the steel core of ACS R is neglected i n calculating
i n du ct a nce, a h igh degree of accu racy results, provided the aluminum strands
are in a n even n u mber of l ayers. The effect of the core i s more apparent for a n
odd n u mber o f layers o f a l u m in u m strands, b u t t h e accuracy i s good w h e n t h e
calcula tion s are based o n the a l u m inum strands alone.
4.9
THE U S E OF TABLES
Tables l isting values of G M R are generally ava i lable for standard conduc tors
and p rovide other i n formation for calcu lating i nductive reactance as well as
shunt capacitive reactance and resistance. S i nce i ndustry i n the U n ited States
conti nues to use units of i nches, fe e t , and m i les, so do th ese tables. The refore,
some of our exa mples will use fect and m iles, but others will use meters a n d
k ilometers.
I nductive reactance rather than i nductance is usually desired. The i n duc­
t ive reactance of one cond uctor of a single-phase two-conductor l i n e is
( 4 .45 )
or
( 4 . 46)
160
CHAPTER
4
S ERI ES I MPEDANCE OF TRAN SM ISS ION LINES
where Dm is the distance between conductors. Both Dm and Ds must be i n the
same units, usu ally e ither meters or feet. The GMR found i n tables is a n
equ ivalent Ds ' which accounts for s kin e ffect where i t is appreciable enough t o
affect inductance. O f course, skin effect is greater a t h igher frequencies for a
conductor of a given d iameter. Values of Ds listed i n Table A.3 of the Appendix
are for a fre quency of 60 Hz.
Some tables g ive values of i nductive reactance in a d dition to GMR. O n e
method is to expa nd t h e loga ri thmic t e r m o f Eq . (4.46), a s fo l l ows:
XI,
=
2 .022
x
1 0 - :If In
1
D."
-
2.022
+
10
x
:If I n
Dill D j m i
( 4 .4 7 )
I f both Ds a n d Dill a r c in feet, the fi rs t term in Eq . (4.47) is t h l: inductive
reactance of one conductor of a two-conductor l ine h aving a distance of 1 ft
between con d uctors, as m ay b e s e e n by co m p a r i n g Eq . (4.47 / w i t h E q . ( 4 . 4 () ) '
The refore, t h e first term of Eq. ( 4 . 47) i s c a l led the inductive reactance a t l -ft
spacing Xa' I t d epends on the GMR o f t h e con d uctor and the fre quency. The
secon d term of Eq. (4.47) i s called t h e inductive reactance spacing factor Xd .
This second term is independent of the type of cond uctor and depends o n
frequency a n d spacing only. Table A.3 i ncl udes v a lues o f induc[ive reactance a t
I-ft spacing, a n d Tabl e A . 4 l ists va lues o f t h e i n d u ctive reactance spacing factor.
Find the inductive reactance per m i le of a single-p:"ase l ine operating
at 6 0 Hz. The conductor is Partridge, a n d spacing is 2 0 ft between centers.
Exa mple 4.3.
For this conductor Table A.3 lists D s
onc conductor
Solution.
XL
=
=
2 .022
X
1 0-3
X
=
0.02 1 7 ft. From Eq. (4 .46) for
20
60 In 0 . 02 1 7
--
0 .828 fl/ m i
The above calculation i s u s e d only i f Ds is known. Table ..\.3, however, l ists
i nductive reactance at 1-ft spacing Xa = 0.465 fl/m i . From Table A.4 the i nductive
reactance spacing factor is X" = 0.3635 O/ mi, and so the i n d '-l c t ivc reactance o f
one conductor is
0 . 4 65
+ 0 .3635
=
0 .8285 fl/m i
S ince the con ductors composing t h e two s i des o f t he l i ne arc iden tical, the
i nductive reactance of the line is
2 XL =
2
X
0 . 8285
=
1 . 657 fl/mi
4. 1 1
I N D U CTA N C E O F TH R E E - P HASE LI N ES WITH U N S Y M METRICAL S P A C I N G
FI GURE 4. 1 0
Cross-sectional view of t h e equ i l aterally
t h ree-p h ase l i n e .
s pa ce d
161
conductors o f a
4.10
I N D UCTAN C E O F THREE- PHASE L I N E S
WI T H E Q U I LATERAL S PA C I N G
So fa r in our d iscussion w e h ave considered only single-phase l i n es. The
e q u a t ions we have deve loped a re qu ite easily adapted, however, to the cal cula­
tion of the ind uctance of th ree-phase l i nes. Figu re 4 . 1 0 shows the conductors o f
a t h ree-phase l i ne spaced a t t h e corners o f a n equilateral triangle. If we assume
that there is no neutral w i re, o r if we assume balanced t hree- p hase p hasor
currents, I + Ib + Ie = O. Equ ation (4.36) dete rmines the fl ux l inkages of
cond u c tor a :
a
Aa
S ince Ia
Aa
and L a
=
�
=
=
2
- ( Jb +
X 10-7
X
1a I n
1
+
Iii I n
Ie ), Eq. (4.48) becomes
2 X 1O - 7
2
( Ds
(
fa I n
�, -
1 0 - 7 In - Him
fa
In
�1
�
-DI
+
Ie I n
-DI I Wbt/m
2 X 1 0 - 7 fa In
D
Os
� Wbt/m
( 4 .48 )
( 4 .4 9 )
( 4 .50 )
Eq u a tion (4.50) is the same in form as E q . (4.25) for a si ngle-phase line except
that Ds repl aces r ' . Because of symmetry, the ind uctances of cond uctors b a n d
c a re the ,same a s the i n d uctance o f conductor a . S i nce each phase consists o f
only o n e co n d u ct o r , Eq . (4.50) gives t h e i n d u c t a n c e per phase of t h e three-p h ase
line.
4. 1 1
l N D U CTANC E O F THREE-PHASE L I N E S
WI T H U N SYMM ETRI CAL SPAC I N G
When t h e cond uctors o f a t hree-phase l i n e a r e n o t spaced equilateral ly, the
p robl em of finding the inductance becomes more d ifficult. The flux lin kages a n d
i n d ucta nce o f each p hase are not t he same. A different i n ductance i n e ach
. \,
162
CHAPTER 4 SE RIES IMPEDANCE OF TRANS M ISSION LINES
Condo a
Condo
Pos. 2
Condo b
Cond. a
Condo
Pos. 3
Condo c
Co n d o b
Condo a
Pos. 1
Condo b
c
c
FIGURE 4 . 1 1
Transposi t ion cycle.
p hase resul ts i n a n u nbal a nced c i rcu i t . Balance of t h e t h ree ph ases can be
restored by exc h a nging the pos i tions o f t h e conductors at regular i n t ervals a long
t h e line so that each conductor occ u p i es the o r i g i n a l pos ition of every o t h e r
con d uctor ove r a n e q u a l d istance. S uch a n exc ha nge o f conductor posit ions i s
called transposition . A complete t ra nsposition cyc l e is s hown i n F i g .
The
phase conductors a rc designated a , b, a nd c , a n d thc positions occ upied a r e
n umbered 1, 2, and 3, respectively. Transposition results in each con ductor
havin g the same average inductance ove r the whole cycle.
Modern power l i nes are usu a l ly n o t transposed at regular i ntervals al­
though a n i nterchange in the positions of the con d uctors may be made at
switching stations i n order to balance the i n d uctance of t h e phases more closely.
Fortun ately, the dissymmetry between the phases of an un transposed line is
small and n eglected in most calcu l ations of i nductance. I f the dissymmetry is
neglected, the inductance of the u ntransposed l i n e is taken as equal to t h e
average value o f the inductive reacta nce of one p h ase of the same l i ne correctly
transposed. T he derivations to follow a re for tra nsposed l ines.
To find the average i nductance of one con ductor o f a transposed line, we
first determine the fl ux l i nkages of a con ductor for each p osition i t occup i es in
the transposition cycle and then dete r m i ne t h e average fl ux l i nkages. Applying
Eq. (4.36) t o co nd uctor ( / 0 [' r i g . 4 . 1 I l o f I n d t h e r l 1 ; l s or e x p r e s s i o n fo r t h e n u x
linkages o f a i n posit ion 1 w h e n /; is i n pos i t i o n 2 a n d c i s i n posi t ion 3 , w e
o btain
4.1 l.
Aa l
=
2
x
(
1O - 7 Ia
(
1
In _
Ds
+
I" In
With a in position 2, b i n position 3 , a n d
Aa 2 = 2
X
1
10 - 7 Ia I n -
Ds
+
Ib
D 12
+
1
__
c
Ie
In
1
D 23
and, with a i n position 3 , b in position 1 , and
c
+
)
Wbt / m
( 4 .5 1 )
1
Ie In - Wbt/m
( 4 . 52)
i n position
In -
D31
1
__
1,
D !2
)
i n position 2,
( 4,.53)
4. 1 1
I N D U Cf A N C E
OF TH R E E- P H A SE
The average value of the flux l in kages of
With
the rest riction that 1a
A"
=
. =
a n d the
a verage
2 X IO-7 (
3
2 X I ()
'
7
a
163
is
+ J),
= - ( Jb
----
L I N E S WITH U N SYM METRICAL SPAC I N G
1
1 1
] /Ii I n - - I Ii I n
D,
D 1 2 D �.' D J I
-----
.1
�2 /)2J UJI
I Ii I n
Wbt
( 4 .55)
m
ind uctance per phase is
L
a
=
2
X
10-7 In
D
�
Ds
( 4.56 )
Him
where
( 4.57)
and Ds is the GMR of the condu ctor. Deq , the geometric mean of the three
d istances of the u n symmetrical l ine, is the equivale nt equ i lateral spacing, as may
be seen by a com parison of Eq. (4.56) with Eq. (4.50). \Ve should note the
s i m i l arity between all the equations fo r the in ductance of a conductor. I f
the inductance is in henrys per meter, the factor 2 X 1 0 - 7 appears in all the
e q u a t i o n s , and t h e denom i nator of t h e logari t h m ic term i s always the G M R of
t h e co n d u c t o r . T h e n u m e r a t or i s t h e d i s t a n c e b e tween w i res of a two-wire l i ne,
the mutual GMD b e tw e e n si des of a composi te-conductor s i ngle-phase l ine, t h e
d istance b etween conductors of a n equi lateral ly spaced l ine, o r the equivalent
eq u i l ateral spacing of an u nsymm e trical l i n e .
Exa m p l e 4 . 4 . 1\ s i n g l e -c i rc u i t t h re e - p h a se l i n e ope r a t e cJ a t
shown in Fig. 4 . 1 2 . The cond uctors
p e r m i l e p e r ph ase.
60 Hz is
a rc ACS R Drake. Fi n d the
FIGURE 4 . 1 2
a r r a nged, as
inductive reactance
Arra ngem e n t of c o n d u ctors for l::: x ample 4 . 4 .
164
CHAPTER
4
Solution.
S ERIES I M PEDA N CE OF TRAN S M I S S I O N LI N ES
From Table
Ds =
L
XL
Equation
D CQ
0 .0373 ft
= 2
X
1 0 - 7 In
(4.46) m ay
Xd
=
XL =
= �/20
24.8
0 . 0373
= 2 17 60 X 1 609
and by i nter(1ola tio I 1
"
A.3
X
= 1 3 .00 X 1 0 - 7
=
24.8
ft
Him
1 3 .00 X 1 0 - 7 = 0 . 788 .fl/mi per p h ase
b e useu a lso, or
fro m Ta b l e s A.J
XII
for
X 20 X 3 8
=
and
A.4
O . : W'J
24.l'i n
0 .3896
0 . 399
+
0 .3896 = 0 . 7886 .fl/mi
4. 1 2 INDUCTANCE CALCULATI O N S
FOR BUNDLED C ONDUCTORS
per p hase
At extra-high voltages ( EHV) , that is, voltages above 230 kV, coron a w i t h its
. resultant power loss and particularly its i n terference with commu nications is
excessive i f the circui t has only one c on ductor per p h ase. The high-voltage
g�adient at the conductor in the EHV range is reduced considerably by hav i n g
two o r more conductors p e r phase i n close p roximity compa red with the spacing
. between phases. Such a l ine is said to be composed of bundled conducto rs. The
bundle consists of two, three, or fou r cond uctors. Figure 4. 1 3 shows the
a r rangements. The current will not d ivide exactly b e tween the conductors of the
bundle unless there i s a transposi tion o f t h e conductors w i t h i n t h e bundle, b u t
the diffe rence i s o f n o practical import<l 11 ce, a nd t h e G M D m e thod is accu rate
.for
calculations.
;
Reduced reactance is the other equal ly important a dvantage of b u n d l i n g .
I ncreasing t h e number of conductors i n a b u n d l e reduces the effects of coro n a
a n d reduces t h e reactance. The reduction of reactance results from t h e i n ­
creased GMR of the b u n d l e . The calcu l ation of GMR is, o f cou rse, exactly the
same as that of a stranded conductor. Each conductor of a two-co n ductor
bundle, for i nstance, is treated as one strand of a two-strand conductor. If w e l e t
D; i n d icate t h e G M R o f a b u n d l e d conductor a n d Ds t h e G M R of t h e
F I G U R E 4 . 13
B u n d l e a r r 3 n ge m c n ls.
4. 1 2
I N DUcrA N C E CALCULATIONS FOR O U N DLED CONDUcrORS
individual conductors composing the b u n d l e,
we
165
find, referring to Fig. 4 . 1 3 :
For a two-strand bundle
( 4 .5 8 )
For a t hree-strand bundle
( 4 .59)
For a four-strand b u n d l e
D;>
=
V ( D,
]6
x
d X d X
12 d ) 4
L09
=
JD,
X
dJ
( 4 . 60 )
I n comput i ng i nductance using Eq . (4.56), D; of t h e b u n d l e replaces D� o f
a single con ductor. To com p u t e De ll ' t h e d ista nce fro m t h e c e n te r o f o n e b u n d l e
to t h e center o f another bundle i s sufficie ntly accurate for D oln D bc' a n d Dca.
Obtain ing the actu al G M D between conductors of one b u n d l e and those o f
a n o t h e r wou ld be a lmost i n d istinguishable from the ce nter-to-center d is t a nces
for the usual spacing.
Pheasan t .
Exa m p l e 4 . 5 . E a c h con d l l c t or of t h e b u n d l e d -co n d u c t o r l i n e s h own i n F i g . 4 . 1 4 i s
ACSR,
1 , 272,OOO-c m i l
Find
the
k i l o m e t e r ( a n d p e r m i l e ) p e r p h a s e for d
=
i n d uc t ive
re actance
in
ohms
per
45 c m . A l so, fi n d t h e p e r- u n i t s e r i e s
r e a c t a n c e o f t h e l i n e i f i t s l e n g t h i s 1 60 k m a n d t h e base i s 1 00 M Y A , 3 4 5 k Y .
Solution. Fro m Ta b l e A . 3
to m e t e rs .
D"
"
X I.
�
,=
�
=
Base
D,
= 0 . 04 6 6 f t , a n d w e m u l t i p l y [e e t by 0 . 3 04 8 t o convert
/0 .0466
x 0
2 rr o()
2 x 10
U . 365
x
.
.1048 x (l A S
x
iO
f l j k m per p h ase
0 . 3 65 X 1 . 6 0 9
.,
=
"\
In
O . OSO
X=
0 . 36 5 x 1 60
1 1 90
=
m
l O .m�
-­
(1 . OS
0 . 5 8 7 i1 / m i per p h as e
= 1 1 90 n
Z=
(345)�
1 00
7
=
0 . 0 49 p e r u n i t
166
CHAPTER
4
SERIES I M PEDANCE OF TRANSM ISSION LINES
d
4.13
=
45
FI GURE 4.14
Spacing of cond uctors of a bun­
d l ed -conductor line.
elm
SUMMARY
programs are usua lly ava i l ab l e or written rather easily for
calcu l ating i n ductance of a l l k i n ds of l i n e s, some u n d erstanding of the develop­
m e n t of the equations used is rewarding from the standpoint of apprec iat ing t h e
effect of variables i n designing a l in e . However, tabul a te d val ues such a s those
in Tables A.3 and A. 4 make the calc u l a tions q u i te s i m p l e except for parallel-cir­
cuit lines. Tabl e A.3 a lso l ists resistance .
The important equation for i nd u c t a nce p e r p hase of single-circuit three­
phase lines is given here for conve nience:
Although co m p u t e r
L
=
2
10 - 7 I n � H j m per p h ase
D
X
Ds
(4.61)
I nductive reactance i n ohms p e r k i l ometer a t 60 H z i s found b y mul tiplying
i nductance i n henrys per meter by 2 7T 6 0 x 1 000:
XL
or
XL
=
=
0 .0754
X
In
0 .1213
x
In
Dcq
Ds
-
Dc<.J
Ds
-
fljkm per p hase
( 4 . 62)
fl j m i per ph ase
( 4 .6 3 )
Both D C q a n d Ds must be i n t h e same u nits, usually feet. I f the l i ne has one
con ductor per phase, Ds is foun d d irectly from tables. For bundled condu ctors
D:, as defined in Sec. 4. 1 2, is subst i tu t e d for Ds . For both single-conductor and
b u n dled-co n ductor l ines
( 4 . 64)
For bundled-conductor
lines Da b ' Dbc ' and Dca are distances between t h e
centers of the bundles o f p hases a , b , a n d c .
For lines with one conductor per p hase it i s convenient to determine XL
from tables by a dding Xa for the con ductor as foun d in Table A.3 to ){d as
fou n d in T abl e AA correspond ing to DC q .
PROBLEMS
167
PROBLEMS
The all-al u m i n um co nductor (AAC) i d e n tified by the code word Bluebell is
composed o f 37 strands, e a ch h aving a d iameter of 0 . 1 672 in. Tab les o f characteris­
tics of AACs list an area o f 1 ,033,500 emil for t h is conductor ( 1 emil = (7T /4) X
1 0 - 6 in2). Are t h ese v a l u e s cons i st e n t with each other? Find t h e overall area of the
strands i n square m i l l i meters.
4.1.
Determine the dc resistance in o h m s per km of Bluebell a t 20°C by Eq. (4.2) a n d
t h e i nformation i n Prob . 4. 1 , a n d check t h e result against the value l is ted i n tables
of 0.0 1 678 D. per ] 000 [t . Com p u t e t h e d c resistance i n ohms per kilometer at 50 ° C
and compare the resu l t with t h e ac 60- H z resistance of 0. 1 024 D./mi l i sted i n tab les
for this cond uctor at 50°C. Exp l a i n any d i fference in values. Assume t h a t the
increase i n resistance d u e to s p i r a l i n g i s 2%.
A n AAC i s composed o f 37 s t r a n ds , e a c h h a v i n g a d i a m e t e r o f 0.333 c m . Com p u t e
t h e d c r e s i s t a n c e i n o h m s p e r k i l o m e t e r a t 75°C. Assume t h at t he i ncrease i n
r e s i s t a n c e d u e t o s p i r;t l i n g i s 2%.
4.2.
4.3.
4 . 4 . T h e e ne rgy d e n s i ty ( t h a t i s , t h e c l l e rgy pc r L1 ll i t vo l u m e ) a t a p o i n t i n a m a g n e t ie
fi e l d c a n be s h ow n
4.5.
to bl! B 2/2 /-L , w h e r e [] is t h e fl ux d e n s ity a n d /-L is t h e
p e rm e a b i l i t y . U s i n g t h i s r e s u l t a n d Eq (4 . 1 0), show that t h e total magne tic fi e l d
e n e rgy s t o r e d w i t h i n a u n i t l en g t h of s o l i d c i rc u l a r co n d u c t o r carrying curren t I i s
given by JL / 'J. / 1 6 7T . Neglect s k i n e ffect, a n d thus ver i fy Eq . (4. 1 5).
The conductor of a single -phase 60-Hz l i n e i s a sol i d rou n d a l u m i n u m w i re h avi n g a
d i ameter of 0.4 1 2 cm. The conductor spacing is 3 m . Determine t h e inductance o f
the l i ne i n m i l l i henrys per m i l e . How m u c h o f the indu ctance i s d u e t o i n ternal flux
linkages? Assume skin effect i s negl i g i b l e .
4.6.
A singl e-phase 60-Hz overhead power l i n e is symme trically supported o n a horizon­
tal crossarm. Spacing between the cent ers o f the condu ctors (say, a and b ) i s 2.5 m .
A telephone l i ne i s also sym m etrically supported o n a horizontal crossarm 1 .8 m
d irectly below the power l i n e . Spacing between the centers of t hese con d u ctors
(say, c and d) is 1 .0 m.
(a) Using Eq. (4.36), show t h a t the m u t u a l indu ctance per unit length between
circuit a -b and c ircu i t c-d i s given by
Dud [) J,e
Dl lc lJ ,,,,
(b)
(c)
4.7.
where, for exam ple,
a n c! d .
H e nce, co m p u t e
the
Dud d e notes t h e di stance i n meters betwe en conducto rs
m u t u a l i n d uc t a n c e per
a n d t h e t e l e p h o ne l i n e .
Find the
H/m
G O - H z vo l t age
kil ome ter hetween
p e r k i l o m e t e r i n d uc e d i n
pow e r l i n e c a r r i e s 1 5 0 I\.
t h e pow e r
a
line
th e telephone l i n e when t h e
I f t h e pow e r l i n e and t h e t e l e p h o n e l i n e d escribed in P r o b . 4.6 arc in the same
horizontal plane and the distance b etw een the nearest conductors o f the t w o l i n e s
is 18 m , usc t h e resu l t o f Prob . 4 . 6 ( a ) t o find the m u tual inductance between t h e
power and tel ephone circuits. Also, fi n d t h e 60-Hz volt age p e r kilometer i n duced i n
t h e tel ephone l i n e w h e n 1 5 0 A fl o w s i n t h e power l i n e .
168
CHAPTER 4
4.8.
4.9.
S E R I ES I M PEDA NCE OF TRA N S M I S S I O N L I N E S
Find t h e G M R of a three-strand conductor in terms of r of a n i ndividual strand.
Find the GMR of each of the u nconventional conductors shown in Fig. 4 . l 5 in
terms o f the radius r of a n i n d ividual s trand.
83
(b)
(a)
(e)
FIGURE 4. 1 5
C los�-sec li(JIl;t1
(d)
l i () l l �d C O l l u l i C l ur .'
4.12.
4. 13.
4. 14.
4.15.
4.17.
4 .18.
f
U Il C O I 1 V t: I l ­
distance between con d u ctors o r a s i n g l e - p h a se l i n e i s t o I t . E a c h o f i t s
conductors is composed of s i x strands symmetrically placed around o n e center
strand so that there are seven equal strands. The diameter of each strand is
in.
Show t h a t Ds o f each co nductor i s 2 . 1 7 7 times the r a d i us o f e ac h s t ra n d . F i nd t h e
i nductance o f t h e l i n e i n m H/mi.
Solve Example 4 . 2 for t he case where side Y of the single-phase line is identical to
side X and the two sides a r c 9 m apart, as s hown i n Fig. 4 . 9 .
Fin d t h e inductive react an ce o f ACS R Rail i n ohms per ki lometer a t I - m spacing.
Which conductor listed in Table A.3 has an induct ive rea c t a n c e a t 7 -ft spacing of
0 . 65 1 Djm i?
A three-phase line has t hree equilaterally spaced conductors of ACSR Dove. I f the
conductors are 1 0 ft apart, determine the GO-Hz per-phase reactance of the l ine in
D/km.
A t h ree-phase l ine is designed with equ i l ateral spacing of
fl. I t is decided to
build the line w ith horizontal spacing ( D I 3 2 D l 2 2 D n). The conductors are
transposed. What should be the spacing between adjacent conductors in order to
obta i n the same i n d u c t a n c e a s i n t h e o r i g i n a l de s i g n ?
A t h r e e - p h a se 60-Hz t ra ns m ission l i n e h a s i t s con J u c tors a r ra n ge d i n a triangu lar
for mation so that two of the d istances between conductors a r e 25 ft an d t h e t hird
distance is 42 ft. The conductors a rc ACS R Osprey . Determine the inductance and
i nductive reactance per p hase per mile.
A th ree-phase
l i n e h a s fl a t h o r i zo n t a l s pa c i n g . T h e conductors have a G M R
of
m with 10 m between adjacent conductors. Determine the induct ive
react ance per phase in ohms per kilometer. W hat is the name of this cond u ctor?
For short t ransmission l ines if resistance is neglected, the maximum power which
can be transmitted per phase is equal to
=
4.16.
(J
l ur ( ' ro b , 4 . l) ,
0.1
4.10. T h e
4.1 1.
\' i t: w
0.0133
=
16
60-Hz
where Vs and VR are the l ine-to-neutral voltages at the sending and receiving ends
of the line and X is the i n ductive reactance of the l ine. This rel ationship will
become apparent in the study of Chap. 6 . If the m a gnitUdes o f Vs and Vn are held
PROBLEMS
4.19.
4.20.
4.21.
4.22.
169
constant, and i f the cost of a conductor is p roportional to its cross-sectional area,
find the conductor in Table A.3 which has the maximum power-handling capacity
per cost of conductor at a given geometric mean spacing.
A t hree-phase underground d istribution line is operated at 23 kY. The three
conductors are i n sulated w i th 0.5-cm solid black polyethylene insulation and lie
flat, side by side, d irectly next to each other in a dirt trench. The conductor is
circular in cross section and has 33 strands of aluminum. The d iameter of the
cond uctor is 1 .46 cm. The m a n u factu rer gives the GMR as 0.561 cm and the cross
section of the conductor as 1 .267 cm 2 . The thermal rating of the line buried in
normal soil whose maxi mum 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
w hether to consider skin e ffect in calcu lati ng resistance, de termine the percent skin
effect at 50°C in the ACSR conductor of the size nearest that of the u nderground
conductor. Note that the series i mped ance o f the distrib ution line is dominated by
R rather than XI_ because of the very low inductance d u e to the close spacing of
the conductors.
The single-phase power line of Prob. 4.6 is replaced by :1 three-phase line on a
horizontal crossarm i n t h e same posit ion as that of t he original single-ph ase line.
Spacing of the conductors o f the powe r l i ne is D ' 3 = 2 D ' 2 = 2 D 2 3, and equivalent
equi lateral spacing is 3 m. The telephone line re mains i n t he position described in
Prob. 4.6. If the curre n t i n t h e power line is 1 50 A, find the voltage per kilometer
induced in the telephone l i n e . Discuss the phase relation of the induced vol tage
with respect to the power-line current.
A 60-Hz three-phase line composed of one ACSR Bluejay conductor per p hase has
flat horizontal spacing of 1 1 m between adjacent conductors. Compare the induc­
tive reactance i n ohms per ki lometer per p hase of this line with that of a line using
a two-conductor bundle o f ACSR 26/7 conductors having the same total cross-sec­
tional area of aluminum as t h e s i ng le-conductor line and 1 1 -m spaci ng measured
from the center of the bund les. The spacing between conductors in the bundle is
40 cm.
Calculate the inductive reactance in ohms per kilometer of a bundled 60-Hz
three-phase l ine h aving t h ree ACSR Rail conductors per bundle with 45 cm
between conductors of the bundle. The spacing between bundle cen te rs is 9, 9, a n d
1 8 m.
CHAPTER
5
CAP ACITANCE
OF
TRANS MISS ION
LINES
M
w e d iscussed briefly a t t h e beginn i n g o f C h a p . 4, t h e s h u n t admittance of a
transmissio n line consists of con ductance a n d capacitive reactance. We h ave
also men tion e d that conductance is usually neglected because its contribution to
shunt admitt a nce is very smal l . For this reason this chapter has been given the
title of capacitance rather than shunt a d m ittance.
Capacitance of a transmission l i ne is the resu l t of the potential d i fference
betwee n the conductors; it causes t h e m to be c h a rged i n the same m a n n e r as
the p lates of a capacitor when there is a poten t i a l d ifference between t h e m . The
capacitance between conductors is the charge per unit of potential d i ffe rence.
Cap acitance between parallel conductors i s a constant depending on the size
and spacing of the conductors. For power lines l ess than about 80 k m ( 5 0 m i )
long, the effect of capacitance can be slight a n d is often neglected . For longer
lines o f h igher voltage capaci t a nce becomes i ncreasin gly impor t a n t .
An alternating voltage impressed o n a transmission line causes the c h a rge
o n the cond uctors a t a ny point t o i n crease and d e c rease with the i ncrease and
decrease of the instantan eous val u e of the vo ltage between conductors at t h e
point. The flow o f charge i s current, a n d t h e current caused b y t h e a l t e r n a t e
chargin g a n d d ischarging o f a l i n e d u e t o a n a l t er nating voltage is c a l l e d the
charging current of the l i n e. Since capacitance is a s h u n t between con d uctors,
charging c urrent flows in a transmission line even when it is open-circuited. It
affects the voltage d rop a long the lines as well as efficiency and power fact or of
the line a n d the stab i l i ty of the system of wh ich t h e 1 i n c is a D 'l f t .
170
5.1
ELECT R I C F I ELD
O F A LON G ,
STR A I G HT
CONDUCTOR
171
The basis of our analysis of capacitance is Gauss's l aw for electric fields .
The l aw states that the total e l ectric charge with i n a closed surface equals the
tot al electric flux emerging from t h e surface. In other words, t h e total charge
within the closed surface e quals t h e i ntegral ove r t h e su rf a ce of the normal
component of the electric fl ux d ensity.
The l ines of electric flux originate on posit ive charges and terminate on
negative charges. Charge density n ormal to a surface is d esignated Df and
equals kE, where k is the perm ittivity of t h e m aterial su rrounding t h e surface
and E is the electric field i n tensity. I
5.1
ELECTRIC FIELD OF A L O N G , S T RAI G HT
CON D U CTOR
If a long, s t r a i g h t cy l i n d r i c a l conductor l i es in a u n i form medium such as air and
is iso l a ted from other charges so t ha t t he c h a rge is u n i formly distributed around
its periphery, the flux is ra d i a l . All poi nts e q u i d i s t a n t from such a conductor are
poi n ts of equipotential and have t h e same electric fl u x de nsity. Figure 5 . 1 shows
sllch an i sol a ted conductor. T h e electric fl u x d e ns ity at x meters from the
con ductor can be computed by imagining a cyl i n drical surface concentric with
the con d uctor and x meters in radius. S ince a l l p arts of the surface are
equ i d istant from the conductor, the cyl i n drical s u r face is a surface of equipoten­
tial and the electric flux density on t he surface is equal to the flux leav i ng the
conductor per meter of length d ivided by t h e area of the surface in an axial
length of 1 m. The electric fl ux densi ty is
(5. 1)
where q i s the charge o n t h e con d u ctor i n coulombs per meter o f l e ngth and x
is t he d istance in meters from t h e conductor t o t h e p oint where the electric flux
density is comp uted. The el ectric fi e l d intens i ty, or the negat ive of t he potential
g r a d i e n t , is e q u a l t o t h e e l e c t r i c fl u x d e nsity d ivided by the pe rm i t tivity of the
medium. Therefore, t h e e l ectric field i n tens i ty is
E
E
l
q
=
2 TT Xk
Vim
( 5 .2)
and q bot h may be instantaneous, p hasor, or d e expressions.
of f r e e s p a c e ko is 8.85 X 1 0 - 1 2 F 1 m (fa ra d s p er m eter). Relat ive
p e rm i t t iv i t y k r is t h e ratio of t h e act u a l p e rm i t t iv i t y k of a material of t h e p e r m i t tivity of free space.
Th us, k ,
k l k Q • For d ry air k r is 1 . 00054 a n d is ass u m e d equal to 1 .0 in c a l c u l a t i o n s for overhead
I n 5 1 u n i t s t h e p e r m i t t i vity
=
l i nes.
172
CHAPTER
/ .,... -
5
CAPAClTANCE OF TRAN S M ISSION LINES
- ...... ,
FIG U RE 5. 1
Lines o f electric flux origi n a t i n g on the pos i t ive charges u n i ­
formly d istribu ted over t h e su rface o f an isolated cyl i n drical
conductor.
S.2
THE POTENTIAL DIFFERENCE
BETWEEN lWO POINTS DUE TO A CHARGE
The potential diffe rence between two points in volts is numerically equal to the
work i n joules per coulomb necessary to move a coulomb of charge b etween t h e
two points. The electric field intensity is a measure of t h e force o n a charge i n
the field. T h e electric field intensity i n vol ts p er m eter i s equal t o t h e force i n
newtons p e r coulomb on a coulomb of charge a t t h e p o i n t considered. B e tween
two points the line integral of the force i n n ewtons a cting on a coulomb of
positive c harge is the work done i n moving the charge from the point of lower
potential to the point of higher potential and is numerically equal to t h e
poten t i a l d ifference between the two poin ts.
Consider a long, straight wire c arrying a positive charge of q elm, as
shown in Fig. 5.2. Points P I and P 2 are located a t distances D ) and D 2 meters,
respectively, from the center of t h e w i re. The wire is an equipotential surface
and t h e u ni fonnly distributed c harge on the wire is equivalent to a c harge
concen tr ated at t he ce n t e r of the wire fo r ca l c u l a t i n g flux exte r n al to t h e wire.
The positive c h a rge on t h e w i re w i l l exe r t t I r e p e l l i n g fo rce 0 1 1 , \ pos i t iv e c h a rge
placed in the field. For this reason a n d because D 2 in t h i s case i s g re a t e r t h a n
Path of
i ntegration
I
I
,
�-��I P2
/
/
/
I
I
I
/
I
I
I
I
I
I
/
F IG U RE 5.2
Path of i n tegra t i o n be tween two points ext ernal to a
cyl i n d rical conductor h aving a u n i formly d ist.-ibuted
pos i t ive charge.
5.3
CAPACITANCE OF A TWO·WI R E L I N E
1 73
P2 to P I ' and PI is
at a h igher potential than P2 . The d ifference i n potential is the amount of work
done per coulomb of charge move d. On t h e other hand, i f the o n e coulomb o f
charge moves from P I t o P2 , it expends energy, a n d the amoun t of work, o r
energy, in newton-meters i s t h e vol tage drop f r<? m P I t o P2• T h e potent i a l
diffe rence is i ndependent of the path fol lowed. The simplest way to compute
the vol tage drop between two points is to com pute the voltage betwee n the
equipotential surfaces passing t h rough P I and P2 by i n tegrating the field
i n tensity over a radial path between t h e equipotential su rfaces. Thus, the
instantaneous volt age d rop between P I a nd P2 is
D I , work must be done on a posit ive charge to move i t from
(5 .3 )
w h e r e q i s the instanta neous charge on t he wire in coulombs per meter of
length. Note that t he voltage d rop between two poi nts, as given by Eq. (5.3),
may be posit ive or negative depending on whether the charge causing the
pote n t i a l d i ffe r e n c e is positive or negative and on whether the vol t age d rop i s
computed from a point n e a r t h e conductor t o a poin t farther away, o r vice versa.
The sign of q may be either positive or negative, and the logarithmic term is
e i ther positive or negative depending on whether D2 is greater or less than D I .
5.3
CAPACITANCE OF A TWO -WI RE LI NE
Capacitance between t he con ductors of a two-wire l ine is defined as the c harge
on the cond uctors per u n i t of potential d ifference between them. In the form of
an equation capacitance per u n i t l ength of the line is
C
=
q
-
v
F jm
(5 .4)
the potential
d i l le r e n ce b e t w e e n t h e co n d u c t ors i n vo l t s . H e re a ft e r, for c o n ve n i e n ce, we re fe r
to cul)(lcit u l / cC !)(.'r i ll / it h'l I/.',r/t , I S c([pa citallce el ml i n d i c ,l t e the correct d i mens ions
[or t h e e q u a t i ons d e r ived . T h e capaci t a n ce be tw e e n two condu ctors can b e
fou n d by subst ituting in Eq . (SA) th e expression fo r v i n terms of q from E q .
(5 . 3 ) . T h e vol t age U " h b e twe e n t h e two con d uctors of t h e two-wire line shown in
F i g . 5 . 3 c a n be fo u n d by d e t e rm i n i n g t h e p o t e n t i a l d i ffe r e n ce between t h e two
co n d uctors of t h e line, first by comp uting t he vol tage d rop due to the charge q a
on c o n ducto r a a nd then by com puting the voltage d rop d ue to the charge q b on
conductor b . By the principle of s up e rposition the vol tage d rop from cond uctor
a t o conductor b due to the cha rges on both conductors is the sum of the
vol t age drops caused by each c ha rge alone.
The c h a rge q a on conductor a of Fig. 5 . 3 causes surfaces of equipotential
i n the vicinity of conductor b , which are shown in Fig. SA. We avoid the
w h e r e q i s t h e c h ,lr g e
on
t h e l i n e i n co u l o m b s p e r
me t e r
,I n d
v
is
I
174
CHAPTER 5 CAPACITANCE OF TRAN SMISSION L I N ES
FIGURE 5.3
Cross section of a paralle l-w i re l i n e .
distorted equ ipotential surfaces by i n tegrating Eq . (5.3) a long the a l te rnate
rather t ha n t h e direct path o f Fig. 5 .4 . In determ i n i n g V a b due to q u ' w e fo l l ow
the p a t h t hrough the u n distorted region a n d see t h a t d istance D \ o f Eq. (5.3) is
the r a d i u s ra of conductor a and d istance D2 is the cen ter- to-center d i sta nce
between conductors a and b . S i m i l a r ly, in d e t e rm i n ing v a /> due to q" , we fi n d
that the d istances D 2 and D \ a r e t il a n d D , respectively. Conve rt i n g t o p h asor
notation ( q n a nd q ,} become phasors), we obt a i n
and s ince
qa =
-
due to
qb
( 5 .5 )
due to
qa
for a two-wire l in e ,
Va b
=
qb
--qa ( -D - )
2 rr k
In
ra
- In
rb
D
V
( 5 . 6)
Equipotential
suriaces
Direct path of integration
from a to b
FIG U RE 5.4
Eq u i po t e n t i a l s u r faces of a por­
t i o n of the e l e c t r i c fi e l d c a u s e d by
c h a rged c o n d u c t o r a ( n o t
shown). Cond uctor b causes t h e
equipot e n t i a l s u rfaces to become
distorted. A r rows i n d ic a t e op­
tional paths of i n te g r a t i o n be­
tween a p o i n t o n the e q u i p o t e n ­
tial s u rface o f conductor b a n d
t h e con d uctor a , whose c h a rge qa
creates t he equ i pote n t i a l' s u rfaces
show n .
a
5.3
1 75
CAPACITANCE OF A nvO-W I R E L I N E
or by combining the logarithmic terms, we obtain
( 5 . 7)
The capacitance between conductors is
2 7T k
----If r
=
u
rb
In( D2jra rb )
Fjm
( 5 .8 )
= r'
Cu b
=
'TT k
I n ( D jr )
Fjm
( 5 .9)
Equation (5.9) gives the capacitancc betwe en the conductors of a two-wire
l in e . If the l i n e i s supp l i e d by a transformer h av ing a grou nded center tap, t h e
potential difference between each conductor and ground i s h a l f t h e pote nt i a l
d ifference between the two conductors a n d t h e capacitance t o ground, o r
capacitance r o neutral, is
Ca n
=
Cbn
=
=
----
In( Djr )
(5 . 1 0 )
F j m to neutral
The concept of capacitance to neutral is i llustrated i n Fig. 5 . 5 .
Equation (5 . 1 0) corresponds t o E q . (4.25) for inductance. O n e difference
between the equations for capacitance and inductance shou l d b e noted c a re­
fully. The radius in the equation for capacita nce is the actual outside radius of
the conductor and not the geometric mean ra tio (GMR) o f the con d u ctor, as in
t h e i n d u c t a nce fo rm u l a .
Equa t ion (5.3), from \vh ich Eqs. ( 5 . 5 ) t h rough (5. 1 0) were d erived , i s based
on the assumpti on o f u n i form charge d is tribution ove r the su rface of the
conductor. Whcn other charges are present, the d istribu tion of c h arge o n the
surface of the conductor is not u ni fo rm a n d the equations derived from Eq. (5 . 3 )
are not strictly correct. The nonuniformity of charge distrib ution, h owever, c a n
aO
--I1 ,--( ----0
--1
_______
(a)
eM
R e p re s e n lation o f l i n e -to - l i n e c a pa c i t a n c e
b
(b)
R e p rese ntation of l i n e - to - n e utral c a p a c i t a nce
FI G U RE 5.5
R e l a t io n s h i p b e tw e e n t h e c o n ce p t s of l in e - t o - l i n e c a p a ci t a n c e a n d l i n e - t o -n e u t ra l c a p a c i t a n c e .
I
176
CHAPTER 5
CAPACITANCE OF TRAN S M ISSJ O N LINES
be n eglected e n t irely in overhead l in es since t h e error in Eq . ( 5 . 10) is o n l y
0.01 %, even for such a close spacing as that where t h e ratio D Ir = SO.
A question arises about the val u e to be used in the denominator o f t h e
argument o f the l ogar i t h m in Eq. (5 . 10) w h e n the con ductor is a stra nded cable
because the equation was derived for a sol id round conductor. Since electric
flux is p e rp e n d icular to the surface of a pe rfect conductor, the electric fi e l d at
t h e surface of a stranded conductor is not the s a m e a s the field at t h e su rface o f
a cylin drical conductor. Therefore, t h e capaci tance calcula ted for a s t r a n d e d
c o n ducto r by subst i t u t ing t h e ou t side r a d i u s of t h e conductor fo r r in Eq. ( 5 . 10)
will be slightly in e rror because o f t h e d i fference b e tween the field in t h e
n eighborhood o f such a cond uctor anci t he fi e l d ncar a so l id con d u c t o r fo r w h ich
Eq. (5. 1 0) was d erive d . The error is ve ry small, however, si ncc only the fi e l d very
close to t h e surface of the conductor is a ffected . The outside radius of the
stranded conductor is used in calculating the capacitance.
Afte r the capacitance to n e u t r a l h a s been determined, the capacitive
reactance existing between one con d uctor a n d neutral for relative per m i t tivity
k = 1 is fou n d by using t h e expression for C given in Eq. (5 . 10) to y i e l d
r
Xc
-
1
2Tf f C
=
2 .862
f
X
D
r
109 In
-
.n
. m to neutral
(5 . l 1 )
Since C i n Eq. (5. 1 1) is in fa rads p e r m eter, the p roper un its for Xc m u s t b e
ohm-meters. W e should also note t h a t Eq. (5. 1 1) expresses t he reactance from
line to n eutral for 1 m of line. S i nc e capacitance reactance is in para llel a l o ng
t h e line, Xc i n ohm-meters must b e divided b y the length of the l ine i n m e ters
to obtain t h e capaci tive reactance i n ohms to neutral for thc ent ire length of t h e
line.
When E q . ( 5 . 1 1 ) is d ividcd by 1 6 09 t o convcrt t o ohm-miles, we obtain
Xc
1 . 779
=
--
f
X
1 0 () I n
D
-
r
.n
.
O1 i
to neutral
( 5 . 1 2)
Table A.3 l ists the outside d i a meters of the most widely used s izes of
ACSR. If D and r in Eq. (5 . 1 2) a re in feet, capacitive reactance at l -ft spacing
X� is t h e first term a n d capacitive reactance spacing factor X� is the seco n d term
w h e n t h e e q uation is expanded as fol lows :
Xc
=
1 . 779
f
--
X
106 In
1
-
r
+
1 . 779
--
f
X
106 I n D n . m i to neutral ( 5 . 13 )
Table A . 3 i ncludes values o f X� for c o m m on sizes o f ACSR, a n d similar tables
. a re readily available for other types and sizes o f cond uctors. Tab l e A.S i n t h e
Append ix lists values o f X� which, o f course, i s different from t h e synchronous
mach i n e t ransient reactance bearing the same symbol.
5.4
OF
CAPA C I T A N C E
A TH R EE - P HASE L I N E W ITH EQU I LATERAL SPACING
177
Find the capacitive susceptance per mile of a single-ph ase line
operating at 60 Hz. The conductor is Partridge, and spacing is 20 ft between
centers.
Exa mple 5 . 1 .
Solution. For t h i s conductor
Table A.3 lists an o u t s i d e d iameter of 0.642 in, and so
r =
and from
Eq.
Xc
0 .642
2
12
x
=
0 .0268 ft
(5. 12)
1 .779
"60
=
Be =
1
Xc
=
1Q6 ln
x
5.10
X
20
0 .0268
=
0 . 1961
106
x
n .
mi to neutral
1 0 - 6 Simi to neutral
o r i ll t e rm s o f c J p a c i t ivc re a c t a n ce at l - n spacing and c a p ac i t ive reactance spacing
factor from Tables A.3 and A.5
X;
=
X�.
=
Xc
=
0 . 1 074
Mn
0 . 0889
MD
0 . 1 074
+
. mi
. mi
0 .0889
=
0 . 1 963
Mn
. m i per conductor
Line-to-line capacit ive reactance and susceptance are
Xc
Be
=
=
2
x
1
Xc
0 . 1 963
=
2 .5 5
X
X
106
=
0 .3926
X
106
n
. mi
10- 6 Simi
5.4
CAPACITAN CE OF A THREE-PHASE LINE
WITH EQUILATERAL S PACING
5.6 .
The t hree identical conductors of radius r of a three-phase l ine wit h equilateral
s p a c i n g a re shown i n Fig.
E q u a t i o n (5 . 5 ) exp r e s s e s t h e vol t a g e between two
con d u c tors due to the c harges on e a ch one if the charge d istrib ution o n the
cond uctors can be assu med to b e u n iform. Thus, the vol tage Vab of the
t hree-phase line due only to the cha rges on conductors a and b I S
Vab
=
. -�- -
2 Tfk
(q
a
In
D
r
q �)
+ b in
D
y
( 5 . 14)
178
CHAPTER
5
CAPACITANCE OF TRANSM ISS I O N LI N ES
D
a
c
FI G U R E 5 . 6
Cross sec t i o n of a t h r e e - p h ase l i ne w i t h e q u i l a t e r a l s p a c i n g .
Equation (5 .3) enab les us to i nc l u d e the e f fe c t o r q c s i nce u n i fo r m c h 'lrge
d istribution ove r the surface of a con d u ctor is eq uivalent to a conce n t ra ted
charge at the center of the conductor. Therefore, due only to the c h a rge CJ( ,
Vab
=
D
2Tfk D V
qc
--
In
-
w h ich is zero since qc is equi distant from a and b. However, to s how t h a t we
are considering all three charges, we c a n write
Vab
=
2Tfk
1
--
( qa D
In
r
+
qb I n
D
r
-
+
qc
In
D]V
D
(5 . 15 )
( 5 . 1 6)
Adding Eqs. (5 . 1 5) and (5 . 1 6) gives
(5 . 17)
I n deriving these equations, we have assumed t h a t gro und i s far e nough away to
h ave negligible effect. Since the voltages a re assumed to be s i nusoidal and
expressed a s phasors, the charges are sinusoidal and expressed a s p hasors. I f
there a r e n o o ther charges i n t h e vici n i ty , t h e sum of t h e charges on t h e thre e
conductors is zero and we can substitute - qa in Eq. ( 5 . 1 7) for q b + qc and
obtain
( 5 . 18)
Figure 5.7 is the phasor d iagram of voltages. From this figure we obtain the
following relations between the line v ol t ag e s Vab and Vac and t he vol tage Va n
5.4
179
CAPA CITANCE OF A T H R E E , P H AS E L I N E WITH E Q U I LATERAL SPACI N G
F IG U R E 5 . 7
P h asor d i agram of the balanced volt ages o f a t h re e - phase l i n e .
from l ine
a
13- V:1I1�
to the neutral of the th ree-phase circuit:
V:I1>
Val =
Add i ng Eqs. ( 5 . 1 9)
-
=
=
Veil = /3 V:'" I - 30 °
and
13 V:'1I( 0 .866 + )0 . 5 )
=
(5.20) gives
( 5 . 19)
13 Va,, ( 0 .866 - )0.5 )
( 5 .20 )
( 5 .2 1 )
S u hs t i t u t in g 3 Va n for Va b
+
Va c in Eq. (5 . 1 8 ), we obta i n
Va n
qa
=
--
2 TTk
In
D
-
r
V
( 5 .22)
S ince capaci tance to neutral is the ratio of t he charge on a conductor to the
vol tage b e tween that condu ctor and neutra l ,
2 7T k
----
I n ( Dj r )
F j m to neutral
( 5 . 23 )
Comparison of Eq s. ( 5 .23) a n d (5. 10 ) shows t h a t t h e two a re i d e n tica l .
These e q u a tions express t h e capacitance to neu t ral for single-phase a n d equi­
l a t e r a l l y spaced t h ree-phase l ines, respective ly. S i m i l a r ly, we reca l l t h a t the
equ ati ons for inductance per conductor are the same for single-phase a nd
e q u i l at eral ly sp ac�d t h ree-phase l ines.
The t e rm cha rging current is appl ied to the current associated w i t h the
c a p �l c i t a n c e o f a l i n e . F o r
p ro d u c t of
p h asor,
the
a
t h e l i ne- to- l i ne
single-ph ase c i rc u i t t h e ch a rg i n g c u rre n t i s the
l i n c - t o- l i n e v o l t a g e
and
suscepta nce, or
as a
(5 .24 )
For a three-phase line t h e c h a rging current is found by multiplying the voltage
to n e u t r a l by t h e capaci tive susceptance to neutral. This gives t h e charging
I
180
CHAPTER
5
CAPACITANCE OF TRA NSM I SS ION L I NES
curren t per p hase and is in accord with t h e calculation of balanced three-phase
circuits on t h e basis of a single p hase with n eutral return. The phasor c harging
current in p h ase a is
( 5 .25)
Since t h e rms voltage varies along the line, t h e charging current is not t h e same
everywhere. Often the voltage used to obtain a value for charging current is the
normal vol tage for which the line is designed, such as 220 or 500 kY, which is
probably not t he actual vol tage at either a generating station or a load.
5.5 CAPACITANCE OF A THREE-PHASE LINE
WITH UNSYMMETRICAL SPACING
When the conductors o f a t hree-phase l i n e a re not equilaterally spaced, t h e
problem o f calculating capacitance becomes more d i fficult. In the u s u a l u n t rans­
pose d line the capacitances of e ac h p h ase to n e u tral are unequal. In a t rans­
posed line the average capacitance to neutral of any phase for the com plete
t ransposition cycle is the same as the average capacitance to neutral of any
other p h ase since each phase con ductor occupies the same position as every
other p h ase conductor over an equal distan ce along the transposition cycle. The
dissymmetry of the u ntransposed line is slight for the usual configuration, and
capacitance calculations are carried out as though all lines were transposed.
For t h e line shown i n Fig. 5 . 8 three e q u ations are found for Vab for the
three d ifferent parts of t he transposition cycle. With phase a in position 1 , b i n
position 2 , and c i n position 3 ,
V;l /J
With p h ase
a
=
( D1 2
1
q n
2 7T k " I
r-
-
+ q ,)
D21 ) V
r
In D
1
+ (j, I n
D 3"\
2
i n position 2, b i n position 3 , a n d
c
in
( 5 . 26 )
position 1 ,
( 5 . 27)
2
FIGURE 5.8
1
3
Cross s e c t i o n o f a t h ree -phase line with u nsy m m etrical
spacing.
5.)
a n d with
a
CAPACITANCE OF
181
A TH R E E- P H A S E f . I N E WITH UNSYMMET R I CA L SPACING
in posi tion 3, b i n posi tion 1 , and
i n position 2,
c
( 5 .28)
Equations (5 .26) through (5.28 ) a re sim i l a r to Eqs. (4.5 1 ) t hrough (4.53) for
the magnetic flux l i nkages of o n e cond uctor of a transposed line. However, i n
t h e equations for magn e tic fl ux l i n kages w e note t h a t t h e current i n a n y p h ase
is the same in every part of the transposition cycle. In Eqs. (5 .26) through (5.28),
if we d isregard the voltage d ro p along the l ine, the voltage to neutral of a p hase
in one p a r t of a transpos i t io n cycle is e q u a l to the voltage to neutral of t h a t
p h ase i n a ny part of t h e cyc l e . H e nce, t h e vol t age between any two conductors i s
t h e same in a l l p a r t s of t h e t r a n sposition cyc l e . It fo l lows t h at t h e charge o n a
co n d uctor m ust be d i ffe re n t w h e n the pos i t ion of the conductor changes with
respect to other conductors. A tre a t m e n t of Eqs. (5 .26) th rough (5 .28) analogous
to that of Eqs. (4.5 1 ) through (4.53) is not rigorous.
The rigorous solu tion for capacita nces is too i nvolved to be p ractical
except perhaps for fiat spacing with equal d istances bet\veen adjacent conduc­
tors. With t h e usu a l spacings a n d cond uctors, sufficient accuracy is obtained by
assuming t h a t the charge p e r u n i t length on a cond uctor is the same i n every
part of the transposition cycle. When the above assumption is made with rega rd
to charge, the voltage between a pair of cond uctors is different for each p a rt of
the transposition cycle. Then an average val u e of voltage betwee n the cond u c­
tors can b e found and the capacitance calcul ated from the average voltage. We
obtain the average voltage by a d d ing Eqs. (5 .26) through (5 .28) and by dividing
the resu lt by 3. The average voltage between cond uctors a and b, assum i ng
the same charge on a conductor regardl ess of i ts position i n the transposit i o n
1
cyc.e,
IS
•
( 5 29 )
.
( 5 .30)
where
S imilarly, t h e average vol tage d rop from conductor a to conductor
Va c
1
=
2 7T k
--
( Dcq
qa
In
.
-
t
+ q
C
In
1
D V
r
-
eq
C
IS
1
(5 . 3 )
182
CHAPTER 5
CAPACITANCE OF TRANSM ISSION LINES
Applying Eq. (5.21) to fi n d the vol tage to neutral, we h ave
3 Va n - Va b + Vac
_
=
1
2k
Tr
(
D eq
2 qa I n r
+
r
qb I n Deq
+
qc
)
In
r
- V ( 5 .32 )
De q
( 5 .3] )
e"
and
q"
=
=
---(
I n D cq / r )
( 5 .34)
F / m to neutral
Equ ation (5 .34) for capacitance to neutral of a transposed t h ree-phase l i n e
correspo n d s to Eq. (4.56) for the i n ductance p e r p hase of a similar l i n e . I n
fi nding capacitive reactance t o neutral corresponding t o en , w e can spl i t t h e
reactanc e into compone n ts o f capacitive reactance t o neutral a t I - ft spacing X�
and capaci t ive reactance spacing factor X� , as defined by Eq. (5 . 13).
Find the capaci tance and the capacitive reactance for 1 m i o f the
line described i n Example 4.4. I f the length o f the line is 1 75 mi and the norma)
operating voltage is 220 kV, find capacitive reactance to neutral for the entire
leng t h of the line, the charging current per mile, and the total charging
megavoltamperes.
Example S.2.
Solution
r =
1 . 1 08
2 X 12
=
0. 0462
Dcq = 24 .8 ft
ft
-
ell =
2 IT x 8 .85 x 1 0 1 2
= 8. 8466
1 n (24.8/0 .0462)
Xc =
1012
2-rr x 60 x 8 . 8466 x 1609
=
x
1 0- 1 2
0 . 1 864
F/m
X
106
.0.
. mi
or from tables
X�
Xc
=
=
0.0912
X
(0.0912
10 6
+
x�
=
0 .0953
0.0953) x 1 0 6
=
X
106
0 . 1 865
X
1 0 6 .0.
.
m i t o neutral
J
5.6
EFFECf OF EARTH ON TH E CA PACITANCE OF THREE-PHASE TRANSMISSION LINES
For a length of
175
183
mi
Capacitive reactance
I fchg l
=
0 . 1 865 X 1 0
1 75
220 , 00 0
=
13
6
=
1 066 n
13
220 ,000
Xc
to neutral
X
10 - 6
x 0 . 1 865
=
0 .6 8 1
Ajmi
A for the l ine. Reactive power is Q = 13 X 220 X 1 1 9 X
10-3
43.5 Mva r. This amou n t of reactive power a b s o rb e d by t he d istributed
capacitance is negat ive i n keeping with the convention d iscussed in Chap. 1. In
o t h e r w o r d s , p o s i t i v e re(lctive power is b e i n g [?('l I eratecl b y t h e distributed
or
0 . 68 1 x 1 75
=
=
119
C <l fl a c i t ( l I l C C o r t h e l i l l c .
5.6
EFFECT O F EA RTH O N T H E
CAPACITAN C E OF TH REE-PHASE
TRA N S M I S S I O N L I N E S
Earth affects the capacitance of a transmission line because its presence a lters
t he electric field of the l i ne. I f w e assume that the earth is a perfect cond u ctor
in the form of a horizontal p l a n e of i n fi nite extent, we realize that the electric
field of charged conductors above t h e earth is n o t the same as it wou l d b e i f t h e
equ ipotential su rface o f t h e e a r t h were not p resent. The electric fi e l d o f t h e
charged conductors i s forced t o conform t o t h e presence of t h e earth's s u rface.
The assumption of a flat, equ i pot e n t i a l su rface is, of course, l imited by t h e
i rregularity o f terrain a n d the type o f surface o f t h e e a r th. The assumption
enables us, however, to understa n d the effect o f a conducting earth on capaci­
t ance calculations.
Consider a circuit consisting o f a s i ngle overhead conductor with a re turn
p a t h through the earth. In c ha rging t h e cond uctor, charges come from the earth
to reside on the con ductor, and a potential d i fference exists between t h e
conductor and the eart h . T h e e a r t h has a charge e q u a l i n magni tude t o t h a t o n
t h e conductor b u t of oppos ite sign . T h e e l ectric fl ux from the cha rges o n t h e
conductor to the charges o n t h e e a r t h is perpend icu l a r to t h e earth's equipoten­
tiel l su rface since the su rface is assumed to be a perfect conductor. Let us
i m agine a fictit ious con d uctor of t h e same size and shape as the ove r h e a d
cond uctor lying di rectly below t h e o r i g i n a l conductor at a d istance e q u a l to
t w i ce t h e d i s t a n ce o f t h e c o n d u c t o r a bove
the p l ane
o f t h e gro u n d .
The
fi c t i t i o u s
conductor is be l ow the su rface of tbe e a rth by a distance equal to the distance
o f the overhead cond uctor ahove the e a rt h . If the earth is removed and a c h a rg e
e q u a l a n d opposite t o that o n t h e overhead cond uctor i s assume d o n t h e
fictitious conductor, t h e p l a ne m idway between the original conductor a n d t h e
fictitious conductor is an equ i pot e n t i a l surface a n d occupies t h e same position
a s the equipotential surface o f t h e e a r t h . The el ectric flux between the overhead
con d uctor and this eq u ip ot e n t i a l surface is the same as that w � ich existe d
184
CHAPTER
5
CAPACITANCE O F TRANS M I S S I O N LINES
between the conductor and the earth. Thus, for purposes of calcu l a tion of
capacitance t h e earth m ay be replaced by a fictitious charged conductor b e l ow
the surface of the earth by a distance equal to that of the ove rh e ad cond u c t or
above the e arth. Such a conductor h a s a charge equal i n mag n i t u d e a n d
opposite i n sign t o that of the origi n a l con ductor a n d i s ca l l e d t h e image
The method of calculat ing capacit a nce by replacing th e e a r t h by t he i m age
of an overhead conductor can be extended to more than one con d uctor. I f w e
locate a n i m a g e cond uctor for e a c h ove rhead conductor, t h e A u x h c twc c n t h e
original conductors a n d their images i s perpen d icula r t o the p l a n e w h ich
repl aces the earth, and that plane is a n equ i potenti a l su rface . T h e A u x above
the plane is the same as it i s when t h e earth is prese n t i n stead u r t he i m age
conductors .
To apply t h e met hod of i mages to t h e ca l c u la t i o n o f c a r a c i l a n c c ror a
three-phase line, refe r to F i g . 5 .9. We a s s u m e that t h e l i n e is t r a n sposed a n d
conductor .
77/ 7
FIGURE 5.9
T h r e e-phase l i n e and its i m a g e .
5.6
EFFECT OF EA RTH ON T H E CA PA CITA NC E OF T H R E E - P H A SE T R A N S M IS S ION L I N ES
1 85
that conductors, a , b, a n d c c a r ry t h e c h a rges q qb ' and q c a n d occupy
positions 1, 2, and 3, respec tively, in the first p a r t of the transposition cycle. The
plane of the earth is show n , a n d below it a re the conductors with the i mage
charges - q a ' - q b ' and - qc Equ ations for the t hree p arts of the transposition
cycle can be written for the voltage drop from conductor a to conductor b as
determined by the three c harged conductors a n d their images. With conductor
a in position 1 , b in position 2, a n d c in pos ition 3, by Eq. (5.3)
a'
(5 .35)
Similar equ ations for Vall a re written for the other parts of the t ra nsposition
cyc l e . Accepting t h e approxim ately correct assu mption of constant charge per
unit length of each conductor t h roughout the transposit ion cycle allows u s to
obta i n a n average value of the p hasor Va ll ' The equat ior. for the average value
of the p h asor Vac is fou nd in a s i m i l a r m an ner, and 3 Var. is obtained by a d di ng
t he average values of Vlln a n d Vac - Knowing t h a t the sum of the charges is zero,
we then fi n d
cn
=
In
( - ) - ..,; -DCq
r-
In
(5. 3 4)
""\
H l � HZ3 H3 1
J r-
-
yH) H2 H3
1
F1m to neutral
( 5 .36)
Comparison of Eq s .
and (5.36) shows that t h e effect of the earth is
to in c r e a s e th e cI [J <1 c i t il n c c o f ,I I i n c . T o ,ICCO li n t fo r t h e e a r t h , t he denominator
of Eq . (5.34) m u s t !l,IVC s u b t r a c t e d fro m i t t h e term
I f the con ductors are h i g h a bove ground compared with the d istances between
them, the d i agonal distances i n t he numera tor of the correction term a re nearly
equal to the vertical distances in the denominator, and the term is very small.
This is the usual case, a n d the effect of groun d is ge nerally neglected for
186
CHAPTER 5
CAPACITANCE OF TRANSM ISSI ON LI NES
three-phase l ines except for calculations by symmetrical components when t h e
sum o f t h e three l in e currents i s n o t zero.
CAPACITANCE CALCULATIONS
FOR BUNDLED CONDUCTORS
Figure 5 . 1 0 shows a bu ndled-conductor l i ne for which we ca n write an equat ion
for the voltage from conductor a t o con d uctor b a s w e did i n deriving Eq . ( 5 .2fl),
5.7
except t h a t now we must consider t h e c h a r g e s on a l l s i x i n d iv i d u a l conductors.
The conductors of any one bundle a re in para l l el , and we can assume the charge
per bundle d ivides equa l ly between the con d u ctors of the bund I e since t h e
separation between bundles i s usu a l ly m o r e t h a n 15 t i mes the spaci ng between
the conductors of the bundle. A l so, si nce D 1 2 is m uc h greater than d, we can
use D J 2 i n p lace of the d is ta nces D I 2 d and D I 2 + d and m a ke o t h e r s i m i l a r
substitutions o f b u n d l e separation d istances i nstead of u s i n g t h e m o re exact
expressions that occur in fi n ding Val) ' The d i ffere nce due to this a p p roxima tion
cannot be d etected i n the final res u l t for usual spacings even when the
calc ulat ion is carried to five or s ix significant figures.
If charge on p hase a is q a ' each of conductors a and a f has the ch arge
qa/2; similar division of charge is assumed for p hases b a n d c . Th e n ,
-
1
--
2rrk
l(qa
- In
2
+
D 12
r
+ In
D
J2
--
d
�
�
a
a
'
]
+
b
- ( D21 D21 1 j
qc
In
2
-"
f) .1 \
�--
c
+
b'
( 5 .37)
I n -'
D -:, \
-------
c'
The l etters u nder each loga r i thmic term indicate the con ductor whose charge is
accounted for by that term. Combi n i n g te rms gives
( 5 .38 )
a
1�----- D3 1 --------�� 1
0
..--- D 1 2
1....
0
I� d -l
a'
1
----· -----
b o
O o
I- d-\
D 23 -----...
�
c o
o c'
I- d - I
FIGURE S.lO
Cross section of a b u n d l ed-con ­
ductor t h ree-phase l i n e .
5.7
CA P A C ITA N C E CA LCU LAT I O NS F O R B U N D L E D C O N D UCTORS
1 87
Equation (5 .38) is the same as Eq. (5.26), except that VrJ has replaced r. I t
therefore follows that if w e consider t h e l i n e t o b e t ransposed, w e fi n d
e'l
=
--(--D)
-
In
eq
( 5 .3 9)
F 1 m to n eutral
Vrd
,
The Vrd is the same as D:' for a two-conductor bundle except t h a t r has
r e p l a c ed DJ • T h is leads us to the very i mport ant conclusion that a modified
geom e t ric m e a n d istance ( G M D ) m e t h od a p p l ies to t he calculation of capaci­
t a n c e of a h u n d l e d -cond u c t o r I h r e e - r h a se l i ne h a ving two conductors per
h u n d l e . Thc mOll i fi c a t i o n is I h ,1 I w c a rc l I s i n g o u t s i d e r a d i u s in p l a ce o f the
G M R o f a s i n g k conduct o r.
I t i s l og i e <i l to c on c l u d e t h a t t h e m o d i fl e d G M D method a pplies to ot her
b u n d l i n g co n fi g ur a t i o n s . If we l e t f):� s t a n d for t h e mod i fi e d G M R to be used
in capa c i t a n ce cal cu l a t i o n s to d is t i ngu i s h i t from D;" u sed in inductance calcula­
t i o n s . w e h a ve
2 ... k
( )
-�n--D-�qThe n , for a two-strand bund l e
Dsc
F1 m
( S AO)
to neutral
( S Al )
-
fo r a t h r e e s t r a n d b u n d l e
( 5 .42)
a nd fo r
a
four-strand hu n d l e
1 . 09
Exa m p l e 5 . 3 . Fi n d t h e
Example 4 . 5
capac I t I ve reac t a nce
..;rd J
4
. ( 5 .43)
t o n e u t r a l of the l i n e d e s c ribed i n
i n o h m - k i l o m e t e r s ( a n d i n o h m - m i les)
per phase.
188
CHAPTER 5 CAPACITANCE OF TRANSMISSION L I N ES
Solution. Computed from the diameter given i n Table A.3
r =
D:c
Deq
=
=
ell =
(
Xc
=
Xc
=
1 .382
2
X
X
0 .3048
12
JO .01755
""{8 X 8
X
0.45
3 �----=--...,.
.
271
X
16
(
=
)
=
0 .01755 m
0.0889 m
=
1 0 .08 m
8 .g5 X 1 0 - J 2
1 0 . O /)
In
O.08�9
X
10J2 X 10-3
27160 X 1 1 .754
0.2257 X 1 0 6
1 .609
=
=
1 1
.754
0 .2257
0 . 1403
X
X
X 10
1 0 ('
f2
.
J
2 F/m
km
per
p h a s e to n e u t ra l
1 0 6 11 . m i per phase to neutral
)
5.8 PARALLEL-CIRCUIT THREE-PHASE
LINES
If two t hree-phase circuits that are identical i n construction and operating i n
parallel a r e so close together t h a t cou pling exists between them , t h e GMD
method can be used to calcul ate the inductive and capacitive reactances of t heir
equ ivalent circuit.
Figure 5 . 1 1 shows a typical a rrangement of parallel-circui t th ree-phase
l i n es on the same tower. Although the l i ne will probably not be transposed, we
obtain practical values for inductive and capaci tive reactances j f transposition is
assumed. Conductors a and a ' are in p a rallel to compose phase a. Ph ases b a n d
a
O----
18· ---0 �
1 0'
8-----2 1' -----0
b
'.
J
----0
b'
1 81
FIGURE 5. 1 1
Typical arrangement o f co n d u ctors o f a p a r a l lel­
circu i t t h ree-p h a s e l i n e .
5 . l'i
PA RA LLEL-C I R C UIT TH R EE - P H A S E L I N E S
189
c are similar. We assume t h a t a and a ' take the positions of b a nd b' and then
o f c and c' as those conductors are rotated similarly in the transposition cycle .
To calculate Deq the GMD method requires that we use D:b , D bc ' and
D(� , where the superscript indicates that these quantities are for p a rallel l ines
and where D:b means the GMD between the conductors of p hase a and those
of p hase b.
For inductance calculations Ds of Eq . (4.56) is replaced by D/, which is ·
the geometric mean of t h e GMR values of the two cond uctors occu pying fi rst
the p o siti o n s of a and a' ) then the positions of b and b', and fi n a l ly the
positions of c and c'.
Because of the sim i l a ri ty b e t w e e n i n d uctance and capacitance calcula tions,
we c a n ass u m e t h a t the D/c for ca p a c i t a nce i s t h e same as D! for i nductance,
excep t that r i s used ins tead of D} of the i n d ivid ual condu ctor.
Fol lowi n g each step of Ex a m p l e S A is possibly the best means of u nd er­
s t a n d i ng the procedure .
Exa m p l e 5 .4 A t h re e - p h ase d o u b l e - c i rc u i t I i n e i s c o m pos e d o f 300,OOO-cm i l 2 6 / 7
Ostrich cO l l d u c t o r s a r r a ll g e d a s s h o w n i n F i g . 5 . 1 1 . F i n d t h e 6 0 - H z i n d u c t ive
r e a c t a n c e and c a p a c i t i ve su sce p t a n ce in o h m s per m i l e p e r p h ase a n d s i e m e n s p e r
.
.
m i l e pe r p h a s e , respe c t i v e l y .
Solution . From
Table
D i stance
A.3
for Ostrich
a to b :
D is t a n ce a to
b' :
Ds
=
0 .0229 ft
o r igi n a l pos i t ion =
/1 0 2
+ 1 .52
o r i g i n a l p os i t i o n =
/1 02
+ 1 9. 5 2 = 21 . 9 ft
1 0 . 1 ft
T h e G M D s b e t w e e l l p h a s e s a rc
1 8 . 9 7 ft
DC Q =
�1 4 .88
X
1 4 .88
X
1 8 .97
= 1 6 . l ft
For i n d u c t a n c e c a l c u l a t i o n s the G M R fo r t h e p a ral l e l -ci rcu i t l i n e is fou n d after
fi rs t obta i n i n g the G M R v a l u e s for t h e t h r e e posi t i o n s . The a c t u a l d is t a nce from a
190
CHAPTER 5
to
a' is
CAPACITANCE OF TRANSM ISSION L I NES
';20 2
+
18 2 = 2 6 . 9 ft. Then, G M R o f each p hase is
../26 .9
'
a - a :
I n position
../2 1
In position b - b' :
X 0 . 0229 = 0 . 785
X 0 . 0229
-./26 .9 X
In position c - c' :
= 0
ft
. 69 3 ft
0 . 0229 = 0 . 785
ft
Therefore,
D;'
=
L =
XL
�O.n5 X 0 . 693
2
x
X
(�7H5
1 O - 7 1 n -16.1
0 . 753
=
0 .7 53
It
= 6.13 X 10-7
= 2 11- 60 x 1 0 9 x 6 . 1 3 x
6
10 - 7
Him per phase
= 0 . 3 -:- 2
D/ mi
per phase
For capacitive calculations Die is t h e same as t h a t of Dj' , except that the ou tside
radius of the Ostrich conductor is used i nstead of i t s G M R . Th..-: o u ts i de d i ameter
of Ostrich i s 0 . 680 i n :
r =
Dj'e
=
0 . 680
2 x
= 0 . 0283
12
( ";26 .9
x 0 .0283 /21 x 0 . 0283 ";26 . 9 x
= /0 .0283
ell =
Be
2 11"
= 2 11"
X
= 1 1 .4 1
( 26 . 9
� . �5
111
X
ft
x
x 21 x
10
-­
O .�37
1 (> . 1
6 0 x 1 8 . � 07
X
10-6
12
x
26.9) 1 / 6
l H . H07
=
x
1 3
0 .02S3 ) /
0 . 837
ft
12
F
10 -
1m
1 609
Simi per phase to neu tral
5.9 SUMMARY
The similarity between inductance a n d capacitance calculations has been em­
phasized t h roughout our discussions. As in inductance calculations, computer
programs are recommended i f a large n umber of calcul ations of capacitance is
required. Tables l ike A.3 a n d A.S make t h e calculations quite simple, however,
except for p arallel-circuit l ines.
191
P R O B LEMS
The important equation for capacitance to n e utral for a s in g l e-circuit,
three-phase l ine is
.
cn
DSL' is the outside r ad ius
=
Dcq
--=--
In
F / m to n eutral
(5 .44 )
Ds c
-
of the conductor for a l i n e consisting of one
2
cond uctor per phase. For overhead l ines k is 8.854 X 1 0 - 1 since k r for air is
1 .0 . Capacitive reactance i s ohm-meters is 1 /2 1T!C, v,,' here C is in farads per
meter. So, at 60 H z
XC' =
r
4 . 77
X
1 0 4 In
D
eq
n . km to n e u t ral
( 5 ,45 )
Deq
n . mi to neutral
Dsc
( 5 ,4 6 )
--
D
se
o r upon d ivid ing by 1 .609 km / m i , wc have
Xc
=
2 . 965
X
1 0 4 In
-
Values for capacit Ive susce p t ance in siemens per kilometer a n d SIemens per
m i l e are the reciprocals of Eqs. (5 ,45) a n d (5 ,46), respectively.
Both Deq and Dsc must be in the same u nits, usual ly feet. For bund led
conductors DsbC is substituted for Dsc, For both single- and b u n d led-conductor
l ines
(5 4 7 )
,
For bundled-conductor l i nes Da b ' Dhc ' and Dca a re d istances b e tween the
centers of the bundles of phases a , h, and c .
For l i nes with one conductor per phase i t i s convenient to determine X c
by adding X:/ for the conductor as fou n d in Tabl e A . 3 to X:, as found in Table
A . S correspond i ng to D"q '
I nd u c t ance, capaci t a n c e , <l ncl t h e assoc i a t e d re a c t a n ces of p a ra l l e l -c i rc u i t
l ines are found by fol lowing the p roce d u re of Example 5 ,4.
PROB LEMS
5. 1 .
A t h re e - p h a s e t ra n s m i s s i o n l i n e h a s fl at horizontal spacing with 2 m between
adjacent conductors. At a c e r t a i n instant the charge on one of t he outside
condu ctors is 60 ,u C/km, and the charge on the center conductor and on the other
outside conductor is 30 J.L C/km. The radius of each conductor is 0.8 cm. Neglect
the effect of the ground and find the vol tage d rop between the two identically
charged conductors at the i n s t a n t s p e c i fi e d .
-
192
5.2.
5.3 .
5.4.
5.5.
CHAPTER
5
CAPACITANCE OF TRANS M ISS I O N L I NES
The 60-Hz capacitive reactance to neutral of a solid conductor, which is one
conductor of a single-phase line with 5-ft spacing, is 1 96.1 k O-mi. W h at value of
reactance would b e specified in a table l isting the capac itive reactance i n oh m - m i l es
to neutral of the conductor at I -ft spacing for 25 Hz? W hat is the cross-sectional
area of the conductor in circular mils?
Solve Example 5.1 for 50-Hz operation and 1 0 ft spacing .
Using Eq. (5 .23), determine the capacita nce to neutral (in p, F / k m ) of a t h re e - p h ase
l i ne with t h re e Cardinal ACS R c o n d u c t o r s e q u i l a t e ra l l y s pa c e d 2(j f t a p a r t . W h a t is
[he c h a rg i ng current of t h e l i n e (in A / k m ) at 60 H z and 1 00 k Y l i n e t o l i ne ?
A t h ree-phase 60- H z t ra ns m i ss i o n l i n e h a s i t s c o n d u c t o rs a rr a n g e d i n d t r i ,l l1 g u L l r
formation so t h a t two of t h e d is t a n c e s b e t w e e n co n d u c t o rs a rc 25 :'[ a n d t h e t h i r d is
42 [t . The c o n q u c t o r s arc ACS R Os"I'ey . D e t e r m i n e t h e e a p a e i L t nee t o n e u t r a l i n
m icrofa rads p e r m i l e and t h e c a p a c i t i v e r e a c t a nce t o n e u t ra l i n ll h m - l11 i l es . If t h e
l i n e i s 1 50 m i l o n g , fi n d t he c a pac i t a n c e t o n e u t r a l a n d c a pa e i t i e r e a c t a n c e o t' t h e
-
\'
line.
thr ee -ph a s e 60-Hz l i n e h a s fl a t h o r i zo n t a l spac i n g . T h e c o r. J u e t o r s h a v e a n
out s i d e d i a m e t e r o f 3 . 2H c m w i t h J 2 III b e t w e e n c () n d u c t o r � . D e t e r m i n e t h e
c a pac i t i ve r e a c t a nce t o n e u t r a l i n o h m - m e t e rs a n d t h e capac i t ivc r C tl c t d n ce o f t h e
line in o h ms i f its length is 1 25 m i .
5.7. ( a ) Derive an equation for the capacitance to n e u t r a l i n far a d s p e r m et e r of a
single-phase l ine, t a k i ng into account t h e e ffe ct of ground. Use t h e same n o m e n cla ­
tu re as i n the equation derived for t h e c a pac i tance of a t hree-phi:: s e l i ne where t h e
effect of ground is represented b y i m a ge c harges.
(b) Using the derived equation, calculate the capacitance to n e u , r a l i n farads per
meter of a single-phase line c o m p o s e d o f two s o l i d c i rcu l a r co n d u c o rs , each having
a d i ameter of 0.229 in . The conductors are 10 ft apart and 2 5 f t abov e grou nd.
Co m pare the result with the value obtained by applying Eq. (5. 10).
5.8. Solve Proh . 5.6 while t a k i n g i n t o a c c o u n t the e ffe c t o f gro u n d . .-\ s s u m e t h a t the
conductors a r c horizont a l ly p l ace d 20 m above g ro u n d .
5.9. A 60-Hz t hree-phase line composed of one ACSR Blllejay conductor per phase has
flat horizontal spacing of 1 1 m b e tw e e n a dj ac e n t co n d u c t o rs . C o m p a re the c a p a c i ­
tiv e reactance in o h m - k i l o m e t e rs p e r p h ase of th i s l i n e w i t h t h a : o f ,I l i n e u s i n g a
t w o-co n d uctor b undle of ACS R 26/7 conductors having t he s a m e t o t a l cross­
sec t i o n a l area of a l u m i n u m as t h e s i n g l e conduc t or l i ne a n d t he I l -m spacing
measured between b u n dl es. The spacing between conducto r s in the b u n d l e is
40 c m
5.10. C a l cu l a t e the capacitive reactance i n ohm-kilometers of a bund led 60-Hz three­
phase line having three A C S R R ail cond uctors per b undle with 45 em between
conductors of the bundle. The spacing between bundle cen ters is 9, 9, and 18 m.
S . U . S ix conductors of ACSR D ra ke cons titute a 60-Hz double-circu i t three-phase l i n e
arranged a s shown i n Fig. 5 . 1 1 . T h e vertical spacing, howeve r , i s 1 4 ft; t h e longer
horizontal d istance is 32 ft ; and the shorter horizontal d i st ances are 25 ft. Find
(a) The inductance pe r ph ase (in H / mi) and the inductive reactance Un D / mi) .
( b ) T h e capacitive reactance to ne u t ra l (in n · m i ) and the charging current in
A / m i per phase and per con ductor a t 138 kY.
5.6. A
-
.
CHAPTER
6
CURRENT
AND VOLTAGE
RELATIONS
· ON A
TRANSMIS S I ON
LINE
We h ave examined the parameters of a transmission line and a re ready to
con sider the l ine as an element of a power system. Figure 6.1 s hows a 5 00-kV
l in e h aving bundled conductors. I n overhead lines the conductors are suspended
from the tower a n d i nsulated from it and from each other b y insulators, t h e
n u mber of which i s determined b y the vol tage o f the line. Each insulator s t r i n g
i n F i g. 6.1 h a s 22 insu lators. T h e two shorter arms above t h e p h ase conductors
support wires usua l ly made of steel. T h ese wires, much smaller i n diameter than
the phase conductors, are not visible in the p ic ture, but they are e lectrically
connected to the tower and a re therefore at ground potential. These wires are
referred to as shield or ground wires and shield the phase conductors from
l ightning strokes.
A very important problem in the d esign and operation of a power system
i s t h e m a intenance of the vol tage within specified l i mits at various points i n t he
system. I n this chapter we d evelop formulas by which we can calcu l a t e the
voltage, current, and powe r at any point on a transmission l i ne, p rovided w e
know these values at one poin t , usually at o n e e n d o f t h e l ine.
The purpose of this chapter, however, is not merely to develop the
pertinent equ ations, but also to p rovide an opportunity to u n d,erstand the
effects of the parameters of the line on bus voltages and t he flow of power. In
193
194
CHAPTER
6
CUR R ENT AND VOLTAGE RELATIONS ON A TRANSMISSION U i\ E
500-kV transmission line. Conductors are 76/ 1 9 ACS R w i t h a l u m i n u m cross section of 2,5 1 5 ,000
FIGURE 6.1
A
em i l . Spacing between phases is 30 ft
3
i n a n d t h e two con d u ct ors per b u n d l e are 1 8 in a p a r t .
(Courtesy Carolina Power and Light Company.)
6.1
R E P R E S E NTATION OF L I N ES
195
this way we can see the importance of the design of the line and better
understand the developments to come in l ater chapters. This chapter a l so
p rovides an introduction to the study of transients on lossless l i nes in order to
in dicate how problems a rise due to su rges caused by l ightning and switching.
I n the modern power system d ata from all over the system are b e i ng fed
contin uously i n to on-line compu ters for control and information purposes.
Power-flow stu d i es performed by a computer readily supply a nswers to q ues­
tions concern ing the effect of switching l i nes into and out of the system or of
cha nges in l ine parameters . Equa t ions derived in this chapter remain i m portant,
however, i n developing an overall u nderstand i ng of what is occurring on a
system and i n calcu l a t i ng efficiency of transm ission, losses, a n d I i m i ts of power
flow over a line for both s te a d y-s t a t e and transient conditions.
6. 1
REPRESENTAT I O N O F L I N ES
The g e n e r a l e q u a t ions re l a t i n g vo l U t g c (l n u c u r re n t o n a t ransmission l ine
recogn ize t h e f.t c t t h a t a l l fo u r of t h e parameters of a transmission l i n e
d iscussed i n t h e two preced ing c h a p te rs a re u n i formly distributed along t h e l i n e .
W e derive these general equations l a ter, b u t first we use lu mped paramete rs
w hich give good accu racy for short l i nes and for l i nes of medium lengt h . I f a n
overhead line i s classified a s short, shunt capacitance i s so small that i t c a n b e
om itted entirely with l ittle loss o f accuracy, and we need to consider only t h e
series resistance R and t h e series i n d uctance L for t h e total length o f t h e l i n e.
A medium-length line can be represented sufficiently well by R a n d L a s
I u m p e d parameters, as shown i n Fig. 6.2, w i t h h a l f the capacitance to n e u t r a l o f
t h e l ine lumped a t each e n d of t h e equivalent circuit. S h unt conductance G , a s
me ntioned previously, i s usually neglected i n overhead power transmission l ines
when calculating voltage and current. The same circu i t represents the short l in e
i f capacitors are omitted.
I nsofar as the h a n dling of capacitance is concerned, open-wire
l ines
l ess than about 80 km (50 m i ) long are short l ines. Medium -length l i nes a re
roughly betwe en 80 km (50 mi) a n d 240 km 0 50 mi) long. Lines longer than
( I SO m i) req u i re c a lcul a t i o n s i n terms of d istributed constants if a h igh
degree of accu racy is req u i re d , a l t ho u g h fo r som e pu rposes a lumped-parameter
representation can be used for l i nes L I P to 320 km (200 mi) long.
60-Hz
km
+
J I�
R
L
240
+
FI G U RE 6.2
S i ngle-phase equ ivalent of a m e d i u m ­
l e n g t h l i n e . T h e capacitors a r e omitted
fo r a s h or t l i n e .
196
CHAPTER
6
CURRENT AND VOLTAGE R ELATIONS ON A TRANSM ISSIOi' LINE
Normally, transmission l i nes are operated w i t h b a l anced three-phase l oads.
Although the lines are n o t spaced equ ilaterally and not transposed, the resu l t ing
dissymmetry is slight and the p hases are considered to be balanced.
I n o rder to distinguish between the total series impedance of a l i n e and
the s eries i m p edance per u n i t length, the following nomenclature is adopted :
z = series im pedance per u n i t l ength per p hase
y
I
=
=
Z =
6.2
shunt admittance per u n i t l ength per p h ase to neutral
l ength of l i ne
zl
=
Y = yl
=
tot a l series i mpeda nce per pbase
total shunt a d m ittance per phase to neu t r a l
THE SHORT TRANSMISSION LINE
The equivale n t circuit of a short transmission l i ne is s hown in Fig. 6.3, w here Is
a n d IR a re t h e sending- and receivi ng-end currents, respectively, and Vs and VR
are the sending- and receiving-end l i n e-to-neutral vol tages.
The circui t is solved as a simple series ac circuit. So,
(6.1)
( 6 .2 )
where Z is zl, the total series impedance of the line.
The effect of t h e variation of t h e power factor of the load on the voltage
regulation of a line is most easily u nderstood for the short li ne and therefore
will be considered at this time. Voltage regul ation of a t ransmission l i ne is the
rise in voltage at the rece iving end, expressed i n percent of ful l -load voltage,
when full load at a specified power factor is removed while the send ing-end
Z -R +j wL
Gen.
Equ iva l e n t circu i t of a short transmission l i n e w h ere t h e resis t a n c e R and i nd u c t a nce L are values
FIGURE
6.3
for the e n t i re length of the line.
6.2
( a ) Load
p.
f. = 70 %
lag
( b ) L oad
p.
f.
=
1 00 %
TH E S H O RT TRANS M I SS I O N L I N E
(c)
Load p.
1 97
f. = 70 % lead
FI G U R E 6.4
Phasor d i agrams of a s h ort tra nsmission l i n e . A l l d i agrams are d rawn for t he same m ag n i t u des of VR
a n d JR '
vol tage is held constant. Correspond i ng to Eq. (2.33) we can write
Percent regu l a tion
=
I Vn . N f. 1 - I VR . FI. :
I VI( . 1-'1. 1
X
1 00
( 6 .3)
where I VI? N L I is the magn i t u d e of receivi ng-end volt age at no load and I VR. F L I i s
the magnitude o f receiving-end vol t age at ful l load with i Vs l constant. After the
load on a short t ransmission l i n e , rep resented by the circuit of F ig. 6.3, is
removed, the vol tage a t the recei v i ng end is equal to the voltage at the sen d ing
end. I n Fig. 6 .3, wi th the load connected , the receiving-end voltage is designated
by VR , and I VR I = I VR. FL I. The send ing-end voltage is Vs ; and I Vs l = I VR . NL I. The
p h asor d i agrams of Fig. 6.4 are d r awn for the same m agnitudes of the receivi ng­
end voltage and current and s how t h at a l arger value of the se nding-en d voltage
i s req u i red to maintain a given receivi ng-end vol tage \vhen the receiving- e n d
curre n t i s l agging t h e vol t age t h a n when t h e s a m e curre nt and vol tage a re i n
p h ase. A s t i l l s m a l l e r sen d ing-end voltage i s required t o m a i n t a i n t h e g iven
receiving-end voltage when the receiving-end current leads the vol t age. The
voltage d rop i s the same i n the series impedance of the l ine i n all cases; becau se
of the d i ffe rent power factors, however, the vol tage drop is added t o t h e
r e ce i v i n g - e n d v o l t il g e il t a d i ffe r e n t a n g l e i n each c a s e . T h e r eg u l a t i o n i s grea test
fo r l a g g i n g pow e r factors a n d l e a s t , o r e v e n negat ive, ro r l e a d i n g pow e r factors.
The inductive re actance of ( t tran s m is s io n line i s l a rger t h a n the res ista nce, a nd
t h e p rinciple of reg ula t ion i l l u s trated in F i g . 6.4 is true for any load s u p p lied by
a p redominantly indu ctive c i rc u i t . The magn i t udes of the volt age drops fa R a n d
lu X L for a short line have been exagge rated with respect to Va i n drawing t h e
p h a s o r d i a g r a m s i n o r d e r t o i l l u s t ra t e t h e p o i n t m o r e c l e a r l y . The relat ion
between power factor and regu l a tion for longer l i n e s is similar to t h a t for s hort
l i n es but i s not visua lized so easily.
300-MVA 20·kY three-phase generator has a subtransient react­
ance of 20%. The generator supplies a number of synchronous motors over a
64-km transmission line having transformers a t both ends, as shown on the
one-line d iagram of Fig. 6.5 . The motors, a l l rated 1 3.2 k Y , are represen ted by just
Exam p l e 6 . 1 . A
, '198
T�
--o -------'?l�]--.:. , �
o--y �>-l !CHAPTER
( 2 0 kV)
�
6
C U RRENT A N D VOLTA G E RELA T I O N S ON A T R A N S M ISSION L I N E
T
( 1 3 , 8 kV)
( 2 3 0 kV)
FIGURE 6.5
One-line d iagram
for
M
.,r-"r t>
<J q
M
1
Y
Example 6. 1 .
two equivalent motors. The neutral of
M , i s g o u n d d through
reactance. The neutral of the c ond
M2
not
to ground (an
u nusual condition). Rated inputs to the motors are 200 M Y A and 1 00 kYA for M J
and M 2' respectively. For both motors X� = 20% . The
transformer
Tl is
350 MY A, 230/20 kY with l k g e
of 10%.
T2
is composed of
rated 1 27 / 1 3.2 kY, 1 00
MYA with l ea k g
of 1 0 % .
of
t r a n s m i ss i o n
is
0.5 il / km. Draw the reactance d iagram with all reactances marked i n per unit.
Select t h e generator rating as base in the generator
se motooner motis or ctohnnecree-ptrehad se e
rated a therreeeacstiannglcee-phase t raSernseafioeasrmrereraces,actaeanctacncehe the Transforml ierne
c i rc u i t .
is
Solution.
The three-phase rating of transformer T2
3
X
1 00
= 300 kYA
and its l ine-to-line voltage ratio is
I3 x
220
127
= - kY
1 3 .2
1 3 .2
A base of 300 MY A, 20 kY in the generator circu i t requires a 300-MYA b ase i n all
parts of the system and
fol lowing vol
the
In the transmission line:
In the motor circui t:
tage bases:
230 kY ( since
1 3 .2
230 220
=
Tl
is rated 230/20 kY)
1 3 .8 kY
These bases are shown in parent h eses on the one-line diagram of Fig. 6.5. The
reactances of the transformers converted to t h e proper base are
Transformer T1 :
Transformer T2 :
X
X
= 0.1
=
X
300
350
= 0.0857 per unit
( 1 3 .2 ) 2
0 . 1 -1 3 .8
=
0.09 15 per unit
6.2
jO.0857
jO. 1 8 1 5
THE SHORT TRANSMISSION L I N E
199
j O . 09 1 5
m
j O . 5490
FIGURE 6.6
R e a c t a n c e d i agram for Example 6 . l . Reactances a re in per u n i t on the spec i fie d base.
The
base impedance
of the transm ission
(230) 2
300
l i n e is
176.3f1
and the reactance of the line is
0 .5 x 64
1 76 .3
Reactance X� of motor MJ
=
Reactance X� of motor M2
=
JS t h
e
MW, respectiveltyh, e
Figure 6.6
om itted.
0 . 1 8 1 5 per unit
=
0 .2
( 300 ) ( 1 3 .2 )
-
--
-
--
200
1 3 .8
2
(
300 ) ( 1 3 .2 ) 2
0.2
1 00
1 3 .8
p
=
0 .2745 e r unit
=
0 .5490 per unit
required react ance diagram when transformer
operate e
e MW,
p
phase shifts a re
motors M J and M 2 o f Example 6. 1 h a v inputs of 1 20 and 60
at 1 3.2 kV, a n d bot h
at un ity power factor, find the
vo lt age at the termin als of the generator and the vol tage regu l ation of the line.
Exa m p le 6.2. I f
Solution.
Together the motors t a k 1 80
l �O
300
Therefore, with V and I
at
t
he
=
or
0.6 er unit
motors i n per unit,
IVI
x
III
=
0.6 per u n it
200
CHAPTER 6
CURR ENT AND VOLTAGE RELAT I ONS ON A TRANSMISSION L I N E
V
W ith phase-a voltage at the motor terminals as reference, we have
J
=
=
1 3 .2
1 3 .8
/
O.9565L.9.:
no
=
0.6
=
0.9565
unit
per unit
per
0 . 6273LQ:
0"
P h a:; e -a p e r - L1 n i t vo l t a g e s a t nt l l er po i n t s o r F i g . (d) ; \ r e
At m :
At
At
V
I: V
V
k:
=
0 .9565
0 . 9565
=
0.9565
0.9565
=
0 .9565
0.9565
+
+
+
+
+
+
0.6273 ( j0.09 1 5 )
jO .0574
=
0.9582/ 3 .4340
per unit
/ 10. 1 540 uni t
0.9826/ 13.23T uni t
0. 6273 ( J0 .0915
j0 . 1 7 1 3
=
=
j0 . 1 815)
0 .971 7
0 . 6273 ( J0 .091 5
jO.2250
+
+
per
j0 . 1 8 1 5
+
iO.08S7)
per
The voltage regulation o f the l i n e i s
0 .9826 - 0 .9582
0.9582
Percent regulation
and magnitude l generator
s
h
ow
t
h
e
pha
s
e
s
h
i
f
t
s
due
t
o
t
h
e
t
r
a
ns
f
o
r
m
e
r
s
,
t
h
e
an
g
l
e
s
thphae se-a cur elnta ine thae lineanshoulshdoulalsdo bebe iinnccrreeaasseedd bbyy fThenrom the angle of
=
the
or t h e vo t a g e a t t h e
0.9826 X 20
I f it is desired to
of
phase-a vo t g s t
the
6.3
m
d
100
=
2.55%
t erm i n a l s i s
=
1 9 .652 kV
Y
I
x
-
�
30° .
30°
0° .
THE MEDIUM-LENGTH LINE
The shunt admittan ce, usually pure c ap aci t ance, is i ncluded in the calculations
for a line of medium length. I f the total s hu n t admittance of the line is d iv ided
into two equal parts placed at the sendi n g and receiving e nds o f the line, the
circuit is called a nominal 7r . We refe r to Fig. 6.7 to derive equations. To Qbt a i n
a n ' expression for Vs) w e note t h at the c u rr e n t i n t h e capacitance a t t h e
6.3
Tl-I E M E D I U M-LENGTH L I N E
201
FIGCRE 6 . 7
Nom i n a l -;;- circu i t o f a med i u m-length
t ransmission line.
receiving end is VRY/2 and the curre n t in th e series arm is IR + VRY/2. Then,
( 6 .4)
( 6 .5 )
To derive Is , we note th a t t h e cu rre n t i n t h e sh u n t capacitance a t t h e sen d i n g
end i s Vs Y/ 2 , which added t o t h e cu rren t i n the se ries a rm gives
( 6 .6)
Substituting Vs , as given by Eq.
(6.5), in Eq . (6.6) yields
( 6 .7)
Equ ations
(6.5) and (6.7) may be expressed in the ge neral form
( 6 .8)
( 6 .9)
wh ere
ZY
A = D = - + l
2
( 6 . 10)
B = Z
These ABeD constants are sometimes called the generalized circuit constants of
the transmission line. In general, they are complex n umbers. A a n d D are
d im ensionless and equ a l each other i f the l ine is the same when viewed from
either end. The d i mensions of B and C a re ohms a n d m hos or Siemens,
202
CHAPTER 6
C U RRENT AND VOLTAGE R ELATI O N S ON A TR A N S M I S S I O N L I N E
respectively. The cons tants apply to any l inear, p assive, and bilateral four- termi­
nal n e twork h aving two pairs of ter m inals. S uch a netwo rk is c alled a two-port
network .
A p hysical m eaning is easily a ssigned to the constants. By letting 1,< be
zero i n Eq. (6.8), we see t h at A is t h e r atio VS / VR at no load . Similarly, B is t h e
ratio Vs /IR when the receivi ng end is sho rt-ci rcuited. The constan t A is u s eful
in c om p u ting regul ation. I f VI?, '.L i s t h e receiving-en d vol tage at fu [ I load for a
sending-end voltage of ��. , Eq. (6.3) becom es
Percent regulation
=
- I VJ?.F/J
I VU, FL I
I Vs l /l A I
x
1 00
(6.1 1 )
Table A.6 in the Appendix l ists ABeD constants for v a riou s networks and
combinations of n c tworks.
THE LONG TRANSMISSION LINE:
SOLUTION OF THE DIFFERENTIAL
EQUATIONS
6.4
150
The exact sol ution of any transmISSIon l in e and the one required for a h igh
degree of accuracy in calculating 60-Hz l ines more than approximately
mi
long m u s t consider t h e fact that t he parameters of t he : i nes are not lumped but,
rather, a re distributed uniformly throu gh o u t the length of the l ine.
Figure 6.8 s hows one phase and the neut ral connection of a three-phase
line. Lumped p aram e ters are n ot shown because we are ready to consi der the
solution of the line with the impedance and admittance uniform ly d istributed .
I n Fig. 6.8 we consider a d ifferential element of length dx i n the l in e a t a
d ista nce x from the receiving end o f the line. Then z dx and y dx a re,
respectively, the series i mpedance and shu n t ad m it tance of the elemental
section . V and I a re ph asors which v a ry w i th x .
+
I
I
I
Gen.
Vs
V + dV
:
�
I
I
dx --t+-I
----
x --------I
FIGURE 6.8
Sche matic diagram of a transmission l i n e show i n g one phase a n d the neu t ra l ret u r n . Nomenc; l a t u re
for the l i ne and the elemental length a re i n dica t e d .
6. 4
TH E LONG TRAN S M ISSION LI N E : SOLUTI ON OF T H E D I FFER ENTIAL E Q U ATIONS
203
Average line current in the el ement is (J + I + dJ) /2, and the increase of
V i n the distance dx is quite accurately expressed as
dV =
I + 1 + dI
�---
2
z dx
=
Iz dx
( 6 . 1 2)
when products of the differential qu antitie s are neglected. Similarly,
dJ
=
v
+
V + dV
y d.x:
2
-----
=
Vy dx
( 6 . 13)
The n , from Eqs. (6. 1 2) and (6. 1 3) we have
dV
( 6 . 14)
- = fz
dx
and
Let u s differenti a te Eqs.
dI
dx
-
(6. 14) and (6. 1 5) w ith respect t o
d2 V
dx 2
--
and
Vy
=
d2]
dx 2
-
= z
=y
( 6 . 1 5)
x,
dI
dx
d2V
dx
and
dx 2
=
( 6 . 1 6)
-
dV
dx
( 6 . 17)
­
If we subst itute the va lues of dI/dx and dV/dx: from Eqs.
Eqs . (6. 1 6) and (6. 17), resp ectively, we obt a i n
-2 =
and w e obtain
(6. 15) and (6. 14) in
yz V
( 6.18)
yzl
( 6 . 1 9)
Now we h ave E q . (6.18) i n which t h e o n l y vari ables are V and x and Eq. (6. 1 9)
in which the only vari ab l es are I a n d x . The sol u tions of those eq uations for V
and I , respectively, must be expressions which when d ifferentiated twice with
respect to x yield the original expression t i m es the constant yz . For instance,
the solution for V when d i fferent iated twice with respect to x must yie ld yzv.
,
204
CHAPTER
6
C U RRENT AND VOLTAG E RELATIONS ON A TRANS M I SS IO� L 1 � E
This suggests a n exponential form of solution. Assume t hat the solu tion of Eq.
(6. 1 8) is
( 6 .20)
Takin g the second derivative of V with respect to
d2 V
dx 2
�- = yz
[AI
c
jYix
+ A
2
c
x
in Eq. (6 .20) yields
- fYl
x
]
( 6 .2 1 )
which is yz t imes the assumed solut ion for V. Therefore, E q . (6.20) i s t he
solution of Eq. (6. 18). \V hen we subst itute the value given by Eq . (6.20) fo r V i n
Eq. (6 . 1 4), w e obt a i n
J 1
=
vz/y
A c fYlx I
..;
1
z /y
A -, E
· Fzx
( 6 .22)
�
The constants A 1 and A 2 can b e evalu a ted by u sing the con d i tions at the
receiving end o f the line; namely, when x = 0 , V = VR and J = JR' Substitu tion
of these values in Eqs. ( 6.20) and (6 .22) yiel ds
and
Substituting Zc = vz/y and solving for A I gIVe
and
Then, substituting t h e values found for A I and A 2 in Eqs. (6.20) and (6.22) and
letting y
fYZ, w e obtain
=
( 6 . 23 )
( 6 . 24 )
where Zc
fiY
=
vz/y and is called t h e characteristic impedance of the l in e , and
and is called the propagation constant .
Equations (6.23) and (6.24) g ive the rms values of V and J and their p hase
angles at any specified point along the l i n e i n t e rm s of the distance x from the
receiving end to t he specified poi n t , p rovided v�? , JR , and t h e parameters of the
line are known.
y =
6.S
T H E LO N G T R A N S M 1 SS 1 0N Ll N E: l N TE R P R ETATION OF THE EQUATIONS
205
6.5
THE LONG TRANSMI S SI ON LINE:
I NTERPRETATION OF THE EQUATIONS
Both y and Ze are complex q u a ntities. The r e a l part o f the propagation
constant y is called the attenuation constant a and is measu red i n nepers per
unit l e ngth . The quad rat u re p a rt o f y is called the phase constant f3 and is
measu red in rad ians per u nit l e ngt h . Th us,
y
= a
+ jf3
( 6 . 25 )
a n d Eqs. (6.23) and (6.24) become
( 6 . 26 )
and
( 6 .27)
The properties of E ClX a n d E J{3x help to expla i n the vanatlOn of the p h asor
values of vol tage and cu rrent as a fu nction of d istance along the l i ne . The term
E Cl .r changes in magnitude as x c h a nges, but E J{3x ( i dent ical to cos {3x + j sin (3x)
always has a magnitude of 1 a n d cau ses a shift in phase o f {3 rad ians p e r u nit
length of l ine.
The fi rst term in Eq. (6.26), [( Vu + fR Z)/2]E Clx£ i/3x, i ncreases in m agni­
tude and advances in phase as d i s tance x from the receiving e nd incre ases.
Conversely, as progress along t h e l i ne from the sending end toward the
receiving end is considered, the term d i mi nishes i n magnitude and i s retarded in
phase. This is the characteristic of a t raveling wave and is simil a r to the
beh avior of a wave i n water, wh ich va ries in magnitude with time at any poin t,
whe reas its phase is retarded a n d its maximu m val ue dimi nishes with d istance
from the origi n . The var iat ion in i ns tanta neous value is not expressed in the
t e r m b u t i s u n d e rs t o o d s i n ce VI? a nd I N (I r e p h Cl sors . T h e fi rs t term i n Eq. (6.26)
is c a l l e d t h e in ciden t I 'o/tage .
The second term i n E q . ( 6 .2(»), [( Vu - lu Z)/2]E - Ir.I C - ii3 r , d i m i nishes in
magni tude a n d i s retarded i n p hase from the rccc iving end toward the send i ng
end . I t is called the reflecred voltage . At any point along the l ine the voltage is
the s u m of t h e c o m p on e n t i nc i d e n t a n d reflected vol tages at that point.
S i n c e t h e e q u a t i o n fo r c u r r e n t i s s i m i l a r to t h e e q u a t io n for vol t a g e , t h e
cu r r e n t may bc considered t o be com poscci of i ncident and reRected cu rrents.
If a l ine is termi nated in its cha racteristic i m pedance Ze ' rece ivi ng-en d
vol tage VR i s e qual to fR Ze and t h e re i s n o reflected wave of e ither vol tage o r
cu rrent, as may be seen by subst i t u t i ng fR Ze for VR i n Eqs. (6.26) and (6.27). A
line term inated i n its characte rist i c i mpedance is cal led a fiat line or a n infinite
line. The latter term arises from the fact that a l i ne of infinite l ength c a nn ot
h ave a reflected wave . Usu a l ly, power l i nes are not terminated i n t � eir charac­
te ristic impedance, but com m u n i cation l i nes are frequently so terminated i n
206
CHAPT E R 6
CURRENT AND VOLTAGE RELATIONS ON A TRAN S M I S S I O N LI N E
order to eliminate the reflected wave. A typical valu e of Zc is 400 fl for a
single-circui t overhead l in e a n d 200 n for two circuits in p a ra l l e l . Th e ph ase
angle of Zc is usually between 0 and
1 5° . Bundled-conductor lines have lower
values of Zc since such lines have lower L and h igher 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 loss\ess l i n e . I f a l i n e is lossless, its series resistance and shunt
conductance are zero and the characteristic impedance reduces to the real
number ..;L IC , which has the dimensions of ohms when L is the series
inductance of the line in henrys and C is the shunt capacitance in farads. Also,
the propagation constant y = {ZY for the line o f length I red uces to the
im aginary number j{3 = jW'/LC' ll since the attenuation constant a res u l ting
from line losses is zero. When dealing with h igh frequencies or with surges d ue
to lightning, losses are often negl ected and the surge impedance becomes
important. Surge-impedance loading (S I L) of a line is the power delivered by a
line to a purely resistive load equal to its su rge impedance. When so load ed , the
line supplies a current of
-
where I VL I is the line-to-line vol tage at the load. Since the load
resistance,
or with
I V" I
in
k i lovo l ts,
SIL
=
I V 12
'..;i7c
MW
L ie
IS
pure
( 6 .28 )
Power system engineers sometimes find it convenient to express the power
transmitted by a line in terms of per unit of S I L, that is, as the ratio 'of the
power tra nsmitted to the surge-impedance loading. For i nstance, t h e p ermissi­
ble loading of a transmission line may be expressed as a fraction of its S I L, and
S I L provides a comparison of load-carrying capabilities of lines.I
A wavelength � is the d istance along a line between two poin ts of a wave
which d iffer in phase by 360° , or
rad. If {3 is the phase shift in radians per
27T"
I See R. D . D u n lop R. Gu t m a n , a n d P. P. M a rc h e n k o " A n a l ytica l Developme n t of Loa'd ab i l i ty
Cha racteris t ics fo r EH Y a n d U H Y Trunsm ission L i n e s , " IEEE Tra nsactions on Power Apparatus
,
and System s , vol. PAS·98,
,
nO.
2, 1 979, p p . 606- 6 1 7 .
6.6
207
THE LONG TR A N S M ISSI ON L I N E : H Y P E R BOLIC F O R \1 OF TH E E Q U AT I O NS
mile, the wavelength in miles is
2 rr
( 6 .29)
.1 = -
{3
The veloci ty of propagation of a wave i n miles per second is the product of the
wavelength in miles and the frequency in hertz, or
Velocity
=
A. I
For the lossless line of length I meters f3
(6.30) become
I
fiL e
----=
--,
---
m
2 rrI
( 6 30 )
f3
= -­
=
v e l oc i t y
.
2rr tlLC I I and Eqs . (6.29) and
=
I
!LC
--
mls
values of L and C for low-loss overhead l ines are subst ituted i n these
equations, it is found that the wavelengt h is approximately 3000 mi at a
frequency of 60 Hz and the velocity of propagation is \'e ry nearly the speed of
l ight in air (approximat ely 1 86,000 mils or 3 X 1 0 8 m/s).
If there is no load on a line, IR is equal to zero, and as d etermined by Eqs.
(6.26) and (6.27), the incid ent and reA ected voltages are equal i n magn itude and
in phase at the receiving end. In this case the i ncident and reflected currents are
equal i n magnitude bu t are 1 800 out of phase at the receiving end . Thus, the
i ncident and reflected cu rrents cancel each other at the receiving end of an
open line but not at any other poi nt unless the l ine is entirely lossless so t hat the
atten uation a is zero.
Wh e n
6.6
LONG TRANSMISSION LINE:
HYPERBOLI C FORM OF T HE E Q U ATIO N S
THE
Th e n i
i c de n t a n d re fl e c t e d waves of vol t a ge a re s e l d om fo u n d w h e n calculating
the voltage of a power l ine. The reason for discussing the vol tage and the
cu rrent of a l i n e i n terms of the i ncident an d reflected components is that such
an a n alys i s i s helpful i n obt a i n i ng a be t t e r u nd e rs t a n d i ng of some of the
p h e no m e n a o f t ra n sm i s s i o n l i n e s . A m o re c o n v e n i e n t fo r m o f t h e e q u a t i o n s for
c o m p u t i n g c u rr e n t a n d vo ! l <l g c o f ,l powe r l i n e i s fo u nd by i n t roduc i n g h y p e r ­
bo l i c functions. Hyperbolic fu nctions are d e fi n e d i n e xpon e nt i al form
sinh (J
cosh (J
2
= ----
=
2
( 6 .3 1 )
( 6 .32)
208
CHAPTER 6
C U R RE NT AN D VOLTAGE R ELATIONS ON A T RANSMISS ION LIN E
By rearranging Eqs. (6.23) and (6.24) and substitut ing hyperbolic functions for
the expone ntial terms, we find a new set of equat ions. The new equat ions, giving
voltage and current anywhere along the l in e, are
( 6 .33 )
I = Iu cosh "IX
Letting x = /
to
obta in
the
vo l t a g e
and
+
VR
- si n h "IX
( 6 . 34 )
Z(
t h e c u r re n t
(It
t h e s e n u i n g e n u , we
h a ve
( 6 .35 )
Is
=
If(
cosh "1 /
+
VI?
Z
sinh
c
( 6 . 36)
yl
From examinat ion of t hese equ ations we see that the general ized circui t
constants for a long l ine are
A
= cosh yl
B = Zc sinh yl
Solving Eqs. (6.35) and (6 .36) for VR and
sinh 1'1
c = ---
D
IR
( 6 . 37)
= cosh 1'1
in t erms of Vs and
Is ,
we obtain
( 6 .3 8 )
II?
=
Is cosh "1 / -
v:�. .
Zc
S J Il h
( () . 3 9)
"1 /
For balanced three-phase lines the currents in the above equat ions are
l i ne currents and the voltages are li ne-to-neutral volt ages, that is, l ine volt ages
d ivided by 13 . In order to solve t he equ a t ions, t he hyperbolic functions must be
evaluated. S ince "II is usually complex, the hyperbolic functions are a lso
complex and can be evaluated with t he assistance of a calculator or compu ter.
For solving an occasional problem without resort ing to a compu t er t here
are several choices. The following equat ions give the expansions of hyperbolic
sines and cosines of complex arguments in terms of circular and hyperbolic
functions of real arguments:
cosh ( exl + j{3/ )
sinh ( exl + j{3l )
=
=
cosh exl cos {31 + j sinh al sin {31
sinh exl cos {31 + j cosh exl sin {3i
Equa tions (6. 4 0) and (6. 4 1 ) make poss i ble th e
co m p u ta t i o n
( 6 . 40 )
( 6 .41 )
of hype rbolk
fu nc-
6,6
209
TH E L O N G T RA N S M I SS I ON L I N E : H Y P E R B O L I C FO RM OF TH E EQUATI O N S
tions of complex arguments. The correct mat hematical unit for {3/ is the radian,
and the radian is the unit found for {3 1 by computing the quadrature component
of "II. Equations (6 . 40) and (6 . 4 1 ) can be verified by substituting in t hem t h e
exponential forms o f t h e hyperbolic functions a n d t h e similar exponential forms
of the circular functions.
Another method of evaluating complex hyperbolic functions is suggested
by Eqs . (6.3 1 ) and (6. 32). S ubstituting a + j{3 for 0, we obtain
cosh ( a + j{3 )
c "E Jf3 +
=
-
sinh( a + j{3 )
-
£ (r E jf3
-
£ - "£ -jfJ
2
E
2
- a C - jf3
=
1
2
(E"iJ!.
+ £-"
U)
Z (eaL!i - c -aU )
1
( 6 .42)
( 6 .43 )
370 km ( 230 m i ) l ong. The
w i th fl a t ho r i zon t a l spacing and 7.25 m (23.8 ft ) b etween
on the l i n e i s 1 25 MW at 215 kY w i t h 1 00% powe r factor.
cond u c to rs a re Rook
cond uctors, The load
F i n d t h e v o l t a g e , c u r re n t , a n d powe r at t h e se n d i ng e n d a n d the voltage reg u l a t ion
of the l ine . Also, d e te rm i n e t h e wavelength a nd veloci ty of propagation of the l i n e .
60-Hz t ra n s m ission
Exa mp le 6 .3. A s i n g l e -c i rc u i t
line is
Solution.
Feet a n d m i l e s r a t h e r t h a n m e ters a n d ki lome t ers a r e chosen fo r t h e
calcu l a tions i n ord e r to u s e Tab les A.3 t h rough A.5 in the Appendix :
Deq
/23 .8
J �-:------:-=--::--- ---:3
X
==
=
a n d from t h e tabl e s for Rook
z =
y =
"II
VR
iR
=
=
0 . 1 603
+
;Y; I
=
O . 4 772
+
0 .4 1 27)
+
L84 , 5 2°
=
0 .0456
0 . 843 1
=
215 ,000
13
=
i
S mi
/,
/ 79 .04° + 90°
230 0 .843 1 X 5 . 105 X 1 0 - 1> �--2
5 . 1 05 X 1 0 - 6
=
30.0 ft
/
0.843 1 79 .0r n/mi
6
L22:
]
0 . 1 009) X 1 0 - = 5 . 1 05 X 1 0 - 6
j(0,4 1 5
[
j 1 / ( 0 ,0950
2 . 8 X 47.6
=
1 24 , 1 30
j0 .4750
/
+
7 9 . 04°
2
L.2: V to neut ral
1 25 ,000,000
Li2.:
= 335 .7 /
13
3 X 2 1 5 ,000
{\O
A
-
90°
=
406 .4
/ - 5 . 48°
n
210
CHAPTER 6
Eqs.
From
CURRENT AND VOLTA G E R ELATIONS O N A TRANS M ISSION L I N E
( 6.42) and ( 6.43) and noting tha t 0.4750 rad
= 27.22°
= 0 .4654 + j O . 23 94 + 0 .4248 - jO.2185
=
sinh yl
0.8902 + jO .0209
= 0 .4654
+
0.0406
+
=
jO .2394 - 0.4248
j0 .4579
Then, from Eq. (6.35)
Vs =
124 , 1 30 x 0.8904 / 1 .34°
/27.77°
=
1 10,495
=
1 37 ,860
and from
Is
Eq .
+ j2,585
0.H904 / 1 .34°
=
+ 1 1 ,483
+
jO.21�5
= 0 .4597/ 84 .93°
335 .7 x
+
+
406.4/ - 5 .48°
x
j61 ,656
V
( 6.36)
= 335 .7 x 0.8904/ 1 .34°
= 298 .83 + j6.99 - 1 .00
= / A
332.3 1
0 .4597[84 .93°
26 .330
+
+
124,130
406 .4/ - 5 .48°
/t...
....- -
x 0 .4597 84.93°
j 1 40 .4 1
At the sending end
= fS x 137 .86 = 238.8 k V
Line current = 332.3 A
Power factor = co ( 27 . 77 ° - 26 .330 ) = 0.9997 1 .0
Power = fS x 238.8 x 332.3 x l .0 137 ,443
( 6.35 ) we see that a t no load
= 0)
Line voltage
s
=
=
From
Eq.
( JR
V
I?
Vs
- cosh y l
-
--­
kW
6 . 6 THE LONG TRANS MISSI ON L I N E: H YPER BOLIC FORM OF THE E QUATIONS
211
So, the vol tage regulation is
1 37 .86/0 .8904 - 124 . 13
------- X 100 = 24 .7%
124 . 1 3
The wavelength and velocity of propagation are computed as follows:
0.4750
230
{3 =
Vel ocity
{3
=
fA.
=
0.002065 rad / m i
0 .002065
----
=
60
x 3043
=
=
3043 m i
1 82,580 mijs
We note particularly in this example t h at in the equations for Vs a nd Is
the val ue of vol tage m ust be expressed i n vol ts and must be the l ine-to-neu tral
voltage.
Exa mple 6 . 4 .
Solve for t he send ing-en d voltage and t he current found in Example
6 .3 using per-un i t calculations.
Solution . We choose a base of 1 25 M VA, 2 1 5 kV t o achieve the simplest per-unit
values and to compute base impedance and base current as follows:
Base impedance
B ase c u rre n t
So,
Zc
V
=
=
/I
For l i s e i n E q . ( 6 . 3 5 )
VI?
=
/
=
x
-
we
1 .0
c h o s e VI<
/..Q:
as
n
= f3125 ,0002 1 5 =
- 5 .48°
--�===370
v0"
215/
215
215
2 1 5 / 13
406 .4
= 370
2152
-1 25
335 . 7
/
1 .098 - 5 .48° per u n i t
1 .0
per u n i t
t h e r e fe re nce vo l t <1 g e . So,
per unit ( as
a
l ine-Io-neutra l v o ltage )
and s ince the load is at unity power factor,
IR
=
A
/..Q:
337.S
337 .5
/..Q:
l .o
212
CHAPTER 6
CURRENT A N D VOLTAGE R E LAT[ONS O N A TR A N S MISSION LI N E
If the power factor had been less than 1 00%, IR would have been greater than l.0
and would have been at an angle d etermined by the power factor. By Eq. (6.35)
Vs
= 0 .8902 jO.0208 0.0923
= / per u nit
= 1 .0 X 0 .8904
+
+
1 . 1 102
1 .0
1 .098
x
+
/
+
-
1 .0
X
=
+
/
0.4597 84 .93°
27 .7SO
0 .8904 / l . 34°
= 0 .8902
X
j0 .4961
and by Eq. (6.36)
Is =
5 .48°
+
1 .0
LQ:
1 .098L- 5 . 4 8 °
x
0.4597/ 84 .93°
fO .0208 - 0 . 003 1 + j 0 . 4 1 86
0 .990/ 26 .35° per u n i t
At t h e sending end
Line voltage
=
1 . 1 1 02
Line current
=
0 . 990
x
X
2 15
335 .7
=
=
238.7 kV
332 . 3 A
Note that we multiply li ne-to-li ne voltage base by t h e per-unit magnitude of the
voltage t o find the li ne-to-line voltage magnitude. W e could have multipl ied the
line-to-neutral voltage base by the per-uni t voltage to nnd the li ne- to-neu tral
voltage magni tude. The factor 13 does not enter t h e calculations a fter we h ave
expressed all quanti ties in per unit.
6.7
OF
THE EQUIVALENT CIRCUIT
A LONG LINE
The nominal-7T circuit does not represent a transm ission l ine exactly because i t
does n o t account for the parameters o f t h e line be i ng uniformly distributed. The
discrepancy between the nom inal 7T and the actual line becomes larger as the
length o f l ine increases. I t is possible, however, t o find the equivalent circui t o f a
l ong t r a n s m iss io n line and to rep resent t h e l i n e accu rately, in sofar as measure­
men ts at the ends of the line are concerned, by a network of lumped parame­
ters. Let us assume that a 7T circu i t similar to that of Fig. 6.7 is the equivalent
circui t of a long l ine, but let us call the series a rm of our equivale nt-7T circuit Z'
and the shunt arms Y' /2 to distinguish them from the arms of t he nominal-7T
circuit. Equation (6.5) gives the sending-e n d voltage of a symmetrical -7T 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 a n d Y/2 in Eq. (6.5), we obtq.in t he
sending-end vol tage of our equ ivalent circu it in terms of its series and shunt
6.7
THE EQUIVALENT CIRCUIT OF A LONG LI NE
( Z'Y' 1
213
arms and the voltage and current at t h e receiving end:
Vs =
2- + 1
-
VR + Z ' IR
( 6 .44)
For our circuit to be equivalent to t h e long transmission l ine the coefficients of
VR and IR in Eq . (6.44) must be identical, respectively, to t h e coefficients of VR
and IR in Eq. (6.35). Equating the coefficients of JR in the two equations yie l ds
Z'
=
Z'
=
Zc
r;
-
y
z' = z
( 6 .45)
s i nh yl
sinh y l
= z/ -=--
sinh y !
\ zy
I
sinh y l
( 6 .46 )
-­
yl
where Z is equ al to zl, the total series impedance of the l ine. The term
(sinh y /)/yl is t he factor by which the series impedance of the nominal 7T m ust
be m u l tiplied to convert the nominal 7T t o the equ ivalent 7T . For small values of
y/, both sinh y I and y I are almost i d entical, and t his fact shows t hat the
nominal 7T represents the med ium-l ength transmission I ine quite accu rately,
insofar as the series arm is concerned .
To investigate the s hu nt arms of the equivalent-" circu it, we equ ate t h e
coefficients o f VR i n Eqs. (6.35) and (6.44) and obt ain
+ 1 = cosh y !
Z'Y'
--
2
( 6 .4 7 )
S ubstituting Zc sinh yl for Z' gives
Y 'Zc
sinh y !
-----
2
+ 1
=
cosh y l
1 cosh yl
Y'
2
Zc
-
1
( 6 .48 )
( 6.49)
sinh yL
An other form of the expression for the s hunt admittance of the equival ent
circu i t can be found by substituting in Eq. (6.49) the identity
tanh
yl
-
2
cosh y l
=
-
sinh y l
1
( 6 .5 0 )
The i dent ity can be verified by substituting the exponential forms of Eqs. (6.3 1)
,
214
CHAPTER 6
Z'
=
C U R R ENT AND VOLTAG E R E LATI O N S ON A TRANSMISSION L I N E
Z C si n h "'l
I
=
Z
sinh YI
yl
1::
= J....
2
y'
2
Zc
:=: 2
ta n h
!..!.
.2
FIGURE 6.9
Eq u iva J e n t -7T ci rcu i t of a t r ans­
m ission l i n e.
y tanh rl/2
1//2
and (6.32) for the hyperbol ic fu nctions and by recal ling that tan h
sinh () Icosh e . Now
Y'
2
-
2
-
"II
1
- tanh -
Y' Y
Y
2
=
( 6 .5 1 )
2
Zc
()
tanh ( "1112 )
( 6 .52)
"1 112
where
is equ al to yl, the total shunt admittance of the l ine. Equation (6.52)
shows the correction factor u sed to convert the admittance of the shunt arms of
the n ominal 7r to that of the equivalent 7r . S ince tanh( "1112) and "1112 are very
n e arly equ al for small values of "I I, the nominal TT represents the medium-length
transmission l ine quite accurately, for we h ave seen previously that the correc­
tion factor for the series arm is n egligible for medium-length lines. The
equivalent-TT ci rcuit is shown in Fig. 6.9. An equ ivalent-T circuit can also be
found for a transmission line.
6 . 5 . Find the equivalent-7T circuit for the l ine described in Example 6.3
and compare it with the nominal-7T circu it.
Example
Solution. S i nce s i n h y / a n d cos h y / a rc a l re a d y k nown fro m Exa m p l e 6 . 3 ,
( 6 . 45)
Z'
and
=
/ - 5.480
(6.49)
4aC1 . 4
0 .8902
Y'
2
are now used .
+
la.a2a8
/
1 86 .82/ 79 .450
=
0 .4597 / 84 . 930
X
-
1
U36 .82/ 79.45°
/
1 86 .82/ 79 .45°
=
n
Eqs.
in series arm
0 . 1 1 1 8 169 .27°
0.000599 89 .8ZO S in each shunt arm
Using the val ues of z
series impedance of
and
y
from Example 6.3,
Z = 230 X
0.84311
79 . 04°
we
=
find for
the
193.9/79.04°
nominal-rr
circuit a
6.8
and equal shunt arms
y
2
For t h i s
is
5 . 105 10 - 6 L22:
of
230
---===- x
-
2
l ine the i m p e d a n c e
of t h e equiva l e n t 7T
POWER F LOW THRO UGH A TRANSMISSION LINE
by
=
0.000587 L22:
series arm
of t h e
S
i l
of t h e n o m n a 7T
3 . 8 % . The co n d u c t a nce o f t h e s h u n t a rm s of
2 . 0 % l ess t h a n t ha t o f t h e equ i v a le n t
215
TT .
exceeds t h at
the n o m i n a l 7T
We conc lude from the preced ing example that the nomi nal 7T' may
represent long l i nes sufficiently w e l l i f a h i gh degree of accuracy i s not requ i red.
6.8
POWER FLOW THROU G H A
TRAN SMISSI O N LI N E
A l t h o u g h powe r now a t a ny po i n t
along ( \ l ra nsm ission l i n e can always he fou n d
i f t h e voltage , c u r r e n t , a n d power fa ctor a r e k n own o r can b e calcu l ated, very
interesting equations for powe r can be d erived i n terms of ABeD constants.
Th e equations ap ply to any network of two ports or two terminal p aIrs.
Repeating Eq. ( 6 .8) and solving for the receiving-end current JR yields
(6 .53 )
( 6 .54)
Le tt i ng
A
=
IAI�
B =
I B ill
we obtain
( 6 .55)
Then, the complex power VR l� at t h e receiving end is
( 6 .56)
216
CHAPTER 6
CURR ENT A N D VOLTAG E R E LATIONS ON A TRA:"<SM ISSJON L I N E
var
-f------L---l-W
FI G U R E 6. JO
P h asors of Eq. ( 6 . 5 6 ) p lo l l e d i n t h e c o m p k x p l a n e , w i t h
magn i t u des a n d a n g l es a s i n d i c a t e d .
and real and reactive power at the receiving end a re
(6.57)
( 6 .58 )
Noting that the expression for complex power PR + iQ R is shown by Eq.
(6.56) to be t h e resultant of combining two phasors expressed in polar form, we
can plot th ese phasors in the complex plane whose horizontal and vertical
. coordi nates a re in power units (watts and vars). Figure 6. 1 0 s h ows the two
complex quantities and their diffe rence as expressed by Eq. (6.5 6). Figure 6 . 1 1
shows the same p h as o r s w i t h . t h e o r i g i n o f t h e coord i n at e ax es s h i ft e d This
figure is a power d iagram wi t h t h e res u l t a n t w hose m a g n i t u d e is I Pn + )'Q u l, or
I VR I I JRI, at an angle 8 R with the horizontal axis. As expected , the real and
imagin a ry components of I PR + iQul a re
.
( 6 .5 9 )
( 6 .60)
where 8R is t h e p hase angle by whic h VR l eads JR , as discussed in C h ap . 1 . T h e
sign of Q is consistent with t he convention w h ich assigns positive values to Q
when current is l agging the voltage.
Now let us determine some points on the power d iagra'm
various loads with fixed values of I Vs l and I VR I . First, we notice that the position
of point n is not dependen t on the current JR and wil l not ch ange so long as
6.B
p OWE R FLOW TH R O U G H A TRANSM ISS ION L I N E
217
var
+
FI G U R E 6 . 1 1
Power d i agram o b t a i n e Ll by s h i fting t h e origin o f the
coo rd i na t e axes of Fig. 6. 1 0 .
n
kk
i s constant. We note further t hat the d istance from point to point i s
constant for fixed values o f I Vs I a n d I Vu l. Therefore, as the distance 0 to
changes with changing load , the point k , since it must remain at a constant
distance from the fixed point I I , i s const rai ned to move i n a c ircle whose center
is at n . Any change i n P R wi l l req uire a change i n Q R t o keep k on the c i rcle. I f
a d ifferent value of I Vs l is held constant for the same value of ! VR I, the location
of point n is unchanged bu t a new c ircl e of radius
is found.
Examination of Fig. 6 . 1 1 shows that the re is a l imit to the power that can
b e transmitted to the receiving end o f the l ine for specified m agnitudes of
sending- and receiving-end volt ages. A n increase i n power delivered m eans that
the point k will move along the ci rcle unti l the a ngle {3 - [) is zero; that i s,
more power will be del ivered u nt i l (5 = {3. Further i ncreases in (5 result i n l ess
power received. The maximum power is
I VR I
nk
I A I I VR I 2
-- - cos ( f3 - a )
IBI
( 6 .6 1 )
The load mu st draw a la rge l e a d i ng current to achieve t he condition of
maximum power received . Usua lly, operation i s l imited by keeping (5 less t h a n
about 350 and I Vs l/I VRI equal t o o r greater than 0.95. For short li nes thermal
ratings l imit the loading.
In Eqs. (6.53) t hrough (6.61) \ Vs \ and \ VR I are l ine-to-neutral vol tages a nd
coordinates in Fig. 6. 1 1 are watts a n d vars per ph ase. However, if I Vs l and I VR I
are li ne-to-line voltages, each d istance in Fig. 6. 1 1 is i ncreased by a factor o f 3
and the coord inates on t he d iagram a re total three-phase watts and vars. If t h e
voltages are kilovolts, t he coordi nates are megawatts and megavars.
218
OF
6.9
CHAPTER
6
C U RRENT A ND VOLTAGE R ELAT I O NS O N A TRA N SM I SSION L I N E
REACTIVE COMPENSATION
TRANSMISSION LINES
The performance of transmission l in es, especially those of med ium length and
longer, can be improved by reactive compensation of a series or parallel type.
Series compensation consists of a capacitor bank pl aced in series with each phase
conductor of the line. Shunt compensation refers to the placement of inductors
from e ach line to neutral to reduce partially or completely t h e shunt suscep­
tance of a h igh-voltage line, which is particularly important at light loads when
the voltage at the receiving end may othe rwise become ve ry h igh.
Series compensation re duces the se ries i mped ance of the l i ne, w h i ch i s t h e
principal cause of vol tage d rop and the most important fac tor i n d etermi n i ng
the maximum power which thc l ine can transmit. I n ord er to understand the
effect of series impedance Z on maxim u m power transm ission , we exa m i n e Eq.
(6.6 1 ) and see that maximum power transmitte d is d ependent on the reci procal
of the generalized circu i t constant B , which for the nominal-1T equals Z and for
the equivalent-1T equals Z (si nh ),1)/)'1. Because the A , C , and D constants a re
functions of Z , t hey will also c hange in value, but these changes will be small in
comparison to the change in B .
T h e d esired reactance of the capacitor bank can b e determ ined by
compensating for a specific amount of the total ind uctive reactance of the line.
This l eads to the term " compensation factor," which is d efined by Xc iXL >
w h ere Xc is the capacitive reactance of t h e series capacitor bank per phase and
XL is the total inductive reactance of the l i ne per phase.
When the nominal-7T circuit is u se d to represent the l ine and capacitor
bank, the p hysical location of the capacitor bank along the l ine is not taken into
accoun t. If only the sending- and receiving-end conditions of the line a re of
interest, this will not create any significant e rror. However, when the operating
conditions along the l ine are of interest, t h e physical location of t h e capacitor
bank must be taken into account. This can be accomplished m ost e asily by
determining ABeD constants of the porti ons of line on each side · of t h e
capacitor b a n k a n d by representing the capacitor bank b y i ts ABeD constants.
The e quivalent constants of the comb ination (actually referred to as a ca scaded
connection) of l ine-capaci tor�l ine can then be d etermined by applying the
equ ations found in Table A.6 in the Appendix.
In t h e southwestern part of the United S tates series compensa tion is
especially i mp ort a n t beca use l a rge generating p l a n ts a re located hun dreds of
miles from load centers and l arge amounts of power must be transmitted over
l ong d i stances. The lower vol tage drop in t he l ine with series compensation is a n
additi onal a dvantage. S eries capacitors are also useful in balancing the voltage
drop of two p arallel l ines.
Example 6.6. 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 as series compensation is
applied , find the constants for the line of Example 6.3 when uncompensated ,and
for a series compensation of 70%.
6.9
R EA CT IVE COM PENSATION OF TRA N S M ISSION L I NES
The equivalent-7T circuit and quantities found in Exa m p l e s
can be used with Eqs. (6.37) to find, for the uncompensated l ine
Solution.
A
D
=
B
(.'
=
=
cosh y l
2'
=
sinh y'
=
= 0.00 1 1 3 1
/
/
/ - 5 .48°
0.4590/ 84 .940
406 .4
90 .420 S
of
t h e e q u i v ,l I e n t -7T
new se r i es ;I r m i m pe d ; l l1 c e i s a l so t h e ge n e ra l i z e d co n s t a n t D. So,
=
=
and by Eqs. ( 6 . 1 0)
A
c
The
=
=
=
60 .88
2
X
1 86.78/ 79 .46"
34 . 1 7
+
/ 55 .850
j5 0 . 3
x
R
=
0 .000599/89.81°
0.000599/ 89 .8 10 +
0.001 180/ 90.4 10
j O.7 X 230( 0.41 5
60.88/ 55.85°
-
6.5
79 .460 n
T h e se r i e s co m p e n s a t i o n a l t e rs o n l y t h e s e ri e s a rm
J3
and
0 .8 90 4
= 1 86 .78
= _____
/ 1 . 340
6.3
219
+
+
circu it. The
0 .4 1 27)
n
1
=
0.970/ 1 .240
60.88/55.85° (0.000599/ 89 .81° ) 2
S
ex ample shows t hat compe nsation has reduced the const ant B to about
one - t h i rd of its value for the uncompensated line without affecting t h e A a n d C
co nstants appreciably. Thus, maxi mum power which can be transmitted is
i ncreased by ahout ]O() % .
When a transmission l i ne , with or withou t se ries compensation, has the
d es i re d l o a d t ra n s m i s s i o n c a p a b i l i ty, a t t e n t i o n i s t u r n e d t o o p e r a t i on u n d e r l i ght
loads or at no load. Charg i ng cu rrent is an important factor to be considered
and should not be al lowed to exceed the rated fu ll-load current of the l i ne .
Equ ation (5.25) shows us that the charging current is usually d efined as
Bc l V I i f Be is the tot al capacit ive susceptance of the l i ne and I V I is the rated
voltage to neu traL As noted foll owing Eq. (5 .25), this calculation is not an exact
determination of chargi ng cu rrent because of the variation of I V I along t he l i ne.
H we co nnect inductors from l i ne to neutral at various poi nts along the l i ne so
220
CHAPTER 6
C U RRENT A N D VOLTAG E R E LATIONS ON A TRANS M I SSION LIN E
that the total inductive susceptance is B v the charging current becomes
( 6 .62)
We recognize that the charging current is reduced by the term in pare ntheses.
The shunt compensation factor is B,j Be The other benefit of shunt compensation is the reduction of the receivi ng­
e n d voltage of the line which on long high-voltage l i nes tends to become too
high at no load. In the discussion preceding Eq. (6 . 1 1) we noted that I Vsl /1 A I
e quals I VR, N LI. We also have seen that A equals 1 .0 when shunt capacitance is
neglected. In the medium-length and longer lines, howeve r, the presence of
capacitance reduces A . Thus, the reduction of the shunt susceptance to the
value of ( Be - B L ) can limit the rise of the no-load voltage at the receiving end
of the l i n e if shunt inductors are i ntroduced as load is removed.
By applying both series and shunt compensation to long transm ission li nes,
we can t ransmit l a rge amou nts of power efficiently and within the desired
voltage constraints. Ideally, the series and shunt elements should be placed at
int ervals along the l ine. Series capacitors can be bypassed and shunt inductors
can be switched off when desirabl e . As with series compensation, ABeD
constants p rovide a straigh tforward method of analysis of shunt compensation .
Find the voltage regu la tion of the line of Example 6.3 when a shunt
inductor is connected a t t h e receiving end of the line during no-load conditions if
the reactor compensates for 70% of the total shunt admittance of the l ine.
Exa mple 6.7.
Solution.
From Example 6.3 the shunt admi ttance of the line is
y = jS . l OS X 1 0 - 6
Simi
and for the e ntire line
Be = S . 1 0S X 1 0 - 6 X 230 = 0 .001174 S
For 70% compensation
B L = 0 . 7 X 0 .001 1 74
= 0.000822
We know the ABCD constants of t h e l ine from ExampJe 6.6. Table A.6 of the
Appendix tells us that the ind uctor a lone is represented by the generalized
constants
A =D= 1
B=O
C
=
-jBL = -jO.000822 S
The equation in Table A.6 for combi ning two networks in series teJJs us that for
6.10
/ 1 .340
0
1 . 41 1 / - 0 .4
the l in e and i n d uctor
A eq
=
=
0.8904
+
T R A NS M IS S I ON - L I N E T R A N S I ENTS
L
/
186 .78 79 .46° ( 0 .000822
- 90
0
221
)
The vol t age reg u l a t ion w i t h t he s h u n t reac tor con ne c t e d at n o lo ad becomes
1 3 7 . 8 6/ 1 . 0 4 1 1 - 1 24 . 1 3
------ =
] 24 . 1 3
6 . 67%
w h i ch i s a consid e r a ble red u c t io n from t h e v a l u e o f 24 .7% for the regu l a t i o n o f t h e
u n compe n s a ted l i ne .
6. 1 0
TRANSMISSION-LINE TRAN S I ENTS
The transient overvo l tages which occur on a powe r system a re e i ther of external
ori g i n (for example, a l ightning d ischarge) or generated i nte rnally by swi tching
operations. I n general , the t ransients on transmission systems a re caused by any
sudden change in the operating con d i t ion or configu ration o f the syste m s . .
Lightning is always a potent ial hazard to power system equipment, but switching
operations can also cause equ i pment damage. At voltages up to about 230 kV,
the insulation l evel of the l i nes and equi pment is d ictated by the need to protect
against I ightning. On syste ms where voltages are above 230 kV but less t h an
700 kV swi tching operations as wel l as l ightning are poten tially d a m a g i ng to
insulation. At vol tages above 700 kV switching surges are the main d etermi n a n t
o f the level of insu lation .
Of cou rse, underground cables are immune to d irect lightn i ng strokes a n d
can b e protected agai nst t ransie nts o riginating on overhead lines. Howeve r, for
economi c and technical reason s overhead lines at t ransmission voltage l evel s
prevail except u nder unusual c i rcumstances and for short d istances such a s
under a river.
Overhead lines can be p rotected from d i rect st rokes of ligh t n i ng i n most
cases by one or more w i re s at g round potent ial strung above the pow er-l i ne
conductors as mentioned in the d escript ion of Fig. 6. 1 . These protecting wi res,
cal led ground wires , or sh ield wires , are connected to ground t h rough the
transm ission towers supporting the l ine. The zone of p rotection is u s ually
co n s i d ered t o b e 3 o n e a c h s i d e o f v e r t i c a l b e n e a t h a g r o u n d w i r e ; t h a t i s , the
power l i nes must come within this 6 sector. The ground wires, rather than the
power line, rece ive the ligh tning strokes in most cases.
Lightning strokes h i t ting either ground wi res or power conductors c ause
an i njection of current, whi ch divides with h alf the current flow i ng in one
direction and half i n the other. The crest value of current along the stru ck
con ductor varies widely because of t he wide vari a tion in the intensity o f t h e
strokes. Values o f 1 0,000 A and u pward a re typical. I n t h e case w h ,e r e a p ower
222
CHAPTER 6
CURRENT A N D VOLTAGE R E L AT I O N S ON A TRA N S M I S S I O N LI N E
line receives a d irect stroke the d amage to equipment at line terminals i s caused
by the vol tages between the line and t h e ground resulting from the injected
charges which travel along the line as curre n t . These voltages are typically above
a million volts. Strokes to the ground w ires can also cause h igh-vol tage surges
on the power lines by electromagnetic i n d uction .
TRANSIENT ANALYSIS:
TRAVELING WAVES
The study of t ra n s m i s s i o n -l i n e s u r g e s ,
6.11
r e g a r d l e ss o r t h e i r o r i g i n , is very comrlex
and w e can consider here only the case of a l o s s l c s s l ine.2
A lossless line is a good representation for l ines of high frequency w here
wL and we become very la rge compared to R and G . For ligh tning su rges on a
power t ransmission l ine the study of a lossless l i ne is a simplification that
enables us to understand some of the phenomena without becoming too
involved in complicated theory.
Our approach to the problem is similar to that used earlier for deriving
the steady-state voltage and current re lations for the long l ine with d istributed
constants. We now measure the distance x along the line from the sending end
(rather than from the receiving end) to t h e d ifferential element of length � x
shown i n Fig. 6.12. The vol tage v and the current i are functions of both x and
t so t h a t we need to use partial derivatives. The series voltage drop along the
elemental l ength of line is
ai
i( R tl x ) + ( L tl x ) ­
at
and we can write
av
tl x
ax
= -(
Ri + L at
ai
)
tl x
( 6 .63 )
The negative sign i s necessary because v + (a v lax) 6.x must be less t h a n v for
positive v a l ues of
i
and
ai/a t . S i m i la rly,
:� A x
�
-
(
Gv +
C
:� ) A x
( 6 .64 )
For further s tudy, see A. G reenwood, Electrical Tra nsients ill Power Systems, 2d ed., Wiley-Lnters ci­
ence, New York, 1 99 1 .
2
6. 1 1
.
�
+
TRANSIENT ANALYSIS: TRAVELING WAVES
ai
61x
ax
FIGURE 6.12
u
x
1-
223
I
f-- t1 x
Schematic diagram of an elemental section of a
t ransmission line showing on e phase and neu­
tral return. Voltage v and current i are func­
t ions of both x and t. The distance x is mea­
su red from the sending end of the l in e .
I
---.,
We can divide through both Eqs. (6.63) and (6.64) by � x , and s ince we are
considering only a lossless line, R and G wil l equal zero to give
au
(Ix
=
ai
and
ai
at
-L -
( 6 .65 )
au
-c at
ax
( 6 .66)
Now we can e li m ina t e i by taking the partial derivative of both terms in Eq.
(6.65) with respect to x and the partial derivative of both terms in Eq. (6 .66)
with respect to t . This proce dure yields a 2 ijax a t in both resulting equations,
and eliminating this second partial derivative of i between the two equations
yi el d s
( 6 .67)
Equation (6.67) is the so-ca l l ed tra veling-wave equation of a lossless
v I ), a n d the
r an s m i ss ion line. A solution of the equation is a function of (x
vol t age is expressed by
t
-
V = J( X - Jl I )
Th e fu n c t i o n is u n d e fi n e d b u t m u s t b e s i n g l e v a l u e d . The const an t
( 6 .68)
must have
the dimensions of meters per secon d if x is in meters and t is in s econ ds . W e
c a n verify this sol u t i o n by s u bs t i t u t i n g this expression for v into Eq. (6.67) to
d e t erm in e v . Fi rs t we make the change in variable
v
,
u
=
x
-
vt
( 6 .69)
and write
u ( x , t ) = f( u )
( 6 .70)
224
CHAPTE� 6
C U R RENT AND VOLTAGE R E LAT I O N S ON A TRA N S M I S S I O N
LI NE
v
FIGURE 6.13
A voltage wave which is
··1
a
fu nction o f (x
-
v i ) i s s h own for va l u es o f
I
x -
equal to
II
a nd
12,
Then,
au
at
and
=
af ( u ) a u
au at
-
-v
af( u )
au
a 2f( u )
a2u
- = v2
at 2
au2
( 6 .7 1 )
( 6 .72 )
Similarly, we obtain
( 6 .73 )
Substituting these second partial derivatives of
u
i n E q . (6.67) yie1ds
( 6 .7 4)
and we see that Eq. (6.68) is a solution of Eq. (6.67) i f
v=
1
( 6 .75 )
VLC
--=-
The voltage as expressed by Eq. (6.68) i s a wave trave ling in the positive x
direction. Figure 6 . 1 3 shows a function of ( x v t ), wh ich is similar to the shape
of a wave of voltage traveling along a line w hich has been struck by lightning.
The function is shown for two values of time t I and t , where t 2 > t 1 . An
2 nt on the wave sees
observer t raveling with the wave and staying at the same poi
no change in voltage at that point. To the observer
-
x
.
JJ { = a
consta n t
6. 1 1
from which it follows that
dx
-
dt
= v =
TRANSI ENT ANALYSIS: TRAVEL I NG WAVES
1
/LC
--
m/s
225
( 6 .7 6 )
for L and C in henrys per meter and farads per meter, respectively. Thus, the
volt age wave travels in the positive x d irection with the velocity v .
A function of ex + v t) can also be shown to be a solution of Eq. (6.67)
and, by similar reasoning, can be p roperly interpreted as a wave traveling in the
negative x direction. The gen e ral solution of Eq. (6.67) is
( 6 .77)
wh ich is a solu tion for simultaneous occurrence of forward and backward
components on the l ine. I n i tial conditions and boundary (terminal) cond i tions
determine the particular values fo r each component.
If we ex p r e ss a forward t ra v e l i ng wave, also called an inciden t wave, as
( 6 .78)
a wave of current will resul t from the moving charges and will be expresse d by
( 6 .79)
which can be verified by substitut ion of these values of voltage and current i n
Eq. ( 6.65) a n d b y the fact that v i s equal t o 1 / /L C .
S imil arly, for a backward moving wave of voltage where
( 6 .80)
the correspond ing current is
l
"'"'
( 6 .81 )
From Eqs. (6.78) and (6.79) we note that
( 6 .82 )
226
CHAPTER 6
C U R R ENT AND V O LTAGE R E LATI O NS ON A T R ANSM ISSION L I N E
and fro m Eqs. (6.80) and (6. 8 1 ) that
( 6 .83 )
If w e had decided to assume the positive direction of current for i - to be in the
direction of travel of the backward traveling wave the m inus signs would change
to plus signs in Eqs. (6. 8 1 ) and (6.83). We choose, h owe v er to keep the p os i t i ve
x dire ction as the direction for posit ive current fo r both forw a r d a n d b a c kw a r d
traveling waves.
The ratio of v + to i + is called the characteristic i m p e d an ce Zc of the l ine.
We have encountered characteristic impedance p re v i o u s l y i n th e steady-state
solution for the long l i n e w h e re Zc was d e fi n e d as Fly ) w h i c h e q u a l s ';f../C
when R and G are zero.
,
6. 12
TRANSIENT ANALYSIS: REFLECTIONS
We now consider what happens when a vol tage is first applied to the sending
end of a transmission l ine which is termin a ted in an impeda nce Z R ' For our
very si mple treatment we consider Z R to be a pure resistance. If the termination
is other than a pure resistance, we woul d resort to Laplace transforms. The
transforms of voltage, current, and impedance wou ld be functions of the
Laplace transform variable s .
When a switch is closed applying a voltage to a l ine, a wave of vol tage v +
accompanied by a wave of current i + starts to travel along the line. The ratio of
the vol tage VR at the end of the l ine at any time to the current i R at the end of
the line must equal the terminating resistance Z R ' Therefore, the arrival of v +
and i + at the receiving end where their values are v;' and i � must result in
backward traveling or reflected waves v - and i- havi ng values v ii and if;. at
the receiving end such t h a t
( 6 .84 )
where v ii and i'R are the reflected waves v - and i - measured at the receiving
end.
If we let Zc = ,;L IC we find from Eqs. (6.82) and (6.83) that
,
( 6 .85)
(.6 .86)
6. 1 2
T R A N S IENT AN A LY S I S : REFLECTIONS
227
Then, substituting th ese values of i; and i"R i n Eq. (6.84) yields
( 6 .87 )
The voltage vii at the receiving end is evi dently the same function of t as v;
(b ut with d iminishe d magnitud e u nl ess Z R is zero or infinity). The reflection
coefficient P R for voltage at the receiving end of the line is d efined as vii Iv; ,
so, for voltage
( 6 . 88)
We note from
Eq s . ( 6. 85) a n d (6.86) t h a t
=
( 6 .89)
and therefore the reflect ion coefficient for curre nt is always the negative of the
reflection coefficient for voltage.
If the line is terminated in its characteristic impedance Zo we see that the
reflection coefficient for both voltage and current will be zero. There w i l l be no
reflected waves, and the l ine will be have as though it i s i nfinitely long. O n ly
when a reflected wave returns to the se nd ing end does the source sense that t he
l ine is neither infini tely long nor terminate d i n Zc '
Termination in a short circui t results i n a P R for voltage of - 1 . I f t h e
tennination i s a n open circuit, Z R i s i nfi nite a n d P R i s found by d ividing t h e
n umerator and d e nomi nator i n Eq. (6.88) by ZR a n d by allowing Z R to
approach infinity to yield P R = 1 in the l imit for voltage.
We s h o u l d n o t e a t t h i s ro i n t t h eI r w av e s t rave l i n g back t o w a rd the s e n d i ng
end w i l l c a u s e n e w r c l l e c t i o n s as d e t e rm i n ed by th e r e l1c c t i on c oe ffi c i e n t at t he
senJing e n d P ,I ' For impedance a t t h e se n d i n g e n d eq u a l to Z \. Eq. (6.88)
becomes
Ps
(6.90)
W i th sendi ng-end impedance of ZS ' the value of the initial voltage
impressed across the l i ne will be the sou rce volt age m ultiplied by Zc/(Zs + Z).
Equation (6.82) shows that t he inci d e n t wave of vol tage experiences a l in e
imped ance of Zc ' and a t t h e instant when t he source is connected to t h e l in e Zc
and Zs in serie s act as a voltage d ivider.
228
CHAPTER 6
C U R RENT A N D VOLTAGE R E LA TIONS ON A TRAl\SMISSION L I N E
Zc
30 Q
=
(a)
1 20
T
1 20 V
,
,
,
60
2T
: 1 80 V
- 60
3T
1 20 V
- 30
4T
-
5T
- -
-
-
30
,
,
,
,
,
- - + -
90-V
- -
4.25T
1 20 V
135 V
(b)
1 80
1 20 V
- - ----
90,
T
-------
3T
-
135 V
F I G U RE 6 . 1 4
Circuit d i a g r a m , l a ttice d ia g ram, a n d p l o t of v olta g e
versus time for Exa m p l e 6.8, where the receiving-end
5T
resistance is 90 D.
( c ),
A dc source of 120 V with n egligible resistance is connected through
a switch S to a lossless transmission l ine having Zc = 30 11 . The line is termi n a ted
i n a resistance of 90 11. If the switch closes at I 0, plot V R versus time until
t = ST, where T i s the time for a voltage wave to tra\·el the length of the l ine. The
circuit is shown in Fig. 6.14(a).
Exa mp le 6.8.
=
Solution. When switch S is closed, the i ncident wave of vol tage starts to travel
along the line and is expressed as
v
= 1 20U( v t
-
x)
where U(v t x ) is the u n i t step function, which equals zero when (v t x ) is
negative and equals unity when ( v t x) is positive. There can be no reflected
wave u ntil the incident wave reaches the end of the l ine. With impedance to the
i nc ident wave of Zc = 30 11, resistance of the source zero and + = 120 V , the
reflection coefficient becomes
-
-
-
v
PI<
=
90 - 30
90 + 30
1
2
i
6. 1 2
T R A NSI ENT A N A LYS I S : R E FLECfIONS
229
W h e n v + reaches t h e e n d of t h e l i ne , a reflected wave origina tes of value
v
and so
-=
VR
=
(�) 1 20 =
+
1 20
60
60 V
=
1 80 V
W h e n I = 2 T, t h e reflec t e d wave a rrives a t t he s e n d i n g e n d w h e re t he sen d i n g- e n d
r e fl e c t i o n coe ffi c i e n t P s is c a l cu l a t e d b y Eq. (6.90). The l i n e term i n a t i o n for t h e
r e fl e c t e d wave i s
Zs '
t h e i m p e d a n ce i n series w i t h t h e sou rce , o r z e ro i n t h is case.
So,
0 - 30
PR
=
-1
0 + 30
a n cl a rc l1 e e t e u wave o f - 60 V s t a rts t o w a r d t h e r e c e i v i n g e n d to k e e p t h e
se n d i ng e n cl vo l t age e q u a l t o 1 20 V . T h i s n ew w ave r e a c h e s t h e r e c e i v i n g e n d a t
t
=
-
3T a n d re.A ccts tow a r d t h e se n d i ng e n d a wave o f
1
2
( - 60 )
- 30 V
a n d t h e rece i v i n g - e n d vo l t a g e b e comes
la ttice diagra m
VR =
1 80 - 60 - 30
=
90 V
An exce l l e n t m e t h o d o f k e e p i n g t rack of t h e various r e fl e c t i o n s as t hey o c c u r
is the
shown in Fig.
6.14(b).
H e r e t i me is meas ured a l o n g t h e
v e r t i c a l axis i n i n t e rva l s o f T. O n t h e s l a n t l i n e s t h e re a r e reco r d e d t h e v a l u e s of
t h e i n c i d e n t a n d r e fle c t e d waves. In the space be twee n t h e slant l i n es t he re a re
s h o w n t h e s u m of a l l t h e waves a b ove a n d t h e c u r r e n t or \'oltage fo r a p oi n t in t h a t
a r e a o f t h e c h a r t . For i n s t a n ce , a t x e q u a l t o t h r e e- fo u r t h s o f t h e l i n e l e n g t h a n d
t
=
4 . 2 S T t h e i n terse c t i o n o f t h e d as h e d l i n e s t h ro u g h t hese po i n ts i s w i t h i n t h e
a r e a w h i c h i n d i c a t e s t h e vo l t ag e i s 9 0 V .
Figure
6 .14(c)
s h o ws t h e r e c e i v i n g - e n d vo l t a g e p l o t t e d a g a i n s t t i m e . T h e
vol t age is a ppr o a c h i n g i t s s t e a d y - s ta t e va l u e o f 1 20 V .
Lattice diagrams for current m ay also be d rawn. We must remember,
however, that the reflection coefficient for current is always the negative of the
I f the resisumce at t h e end of the l i n e of Example 6.8 is reduced to 1 0 n
as shown in the circuit of Fig. 6 . 1 5 ( a ), the lattice diagram and plot of voltage are
as s hown in Figs. 6. 15(b) a n d 6 . 1 5( c). The resistance of 10 n giv es a negat ive
value for the reflection coefficien t for voltage, which always oc curs for resistance
Z R l ess than Zc- As we see by comparing Figs. 6 . 1 4 and 6 . 1 5 , the negative P R
causes the receiving-end vol tage to build up graduaIIy to 120 V, while a positive
P R causes an initial jump in vol tage to a value greater than that of ! he voltage
orig i n a l ly applied at the sending e n d .
re fl e c t i o n co e ffi c i e n t for v o l t a ge .
811 20
230
CHAPTER 6
Ps
=
Jon
C U R R ENT AND VOLTAGE R E LAT I O N S ON A TRAN S M I S S I ON L I N E
Zc
30 Q
=
(a)
-1
x�
1 20
P R ==
0
- 1-
T - 1 20 V
2T
<:-::
60 V
3 T - 1 20 V
-30
4T
90 V
30
5T
105 V
( b)
1 20 V
60
90- - - - - - - - - 1 0 5
'
- - - --- - - -
T
I
(el
3T
5T
F I G U R E 6. 1 5
Circu i t d iagra m , l a t t ice di agram, a n d p l o t of voltage
versus time when the receivi n g-end res istance for Ex­
a m p l e 6.8 is changed to 10 n .
R eflections d o not necessarily a rise only at the end of a l ine. If one line is
joined to a second l ine of different c h a r a c t e r i s t i c impedance, as i n the case of an
overhead l ine connected to an underground cable, a wave incident to the
junction will behave as though the fl.rst l ine is terminated i n the Z c of the
second line. However, the part of the incident wave which is not reflected will
t ravel (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 t hat a t horough study of t ransm ission -line
transients in general i s a complicated problem. We realize, however, t hat a
voltage surge such as that shown in Fig. 6. 1 3 e ncountering an impedance at the
end of a lossless line (for ins tance, at a transformer bus) will cause a voltag e
w ave of t h e same shape to travel back toward the source of t h e surge. The
reflected wave will be reduced in m agnitude i f the terminal impedance is other
than a short or open circuit, but if Z R is greater than Z c ' our study has shown
that t h e peak terminal voltage will be higher than, often close to d ouble t h e
peak voltage o f t h e surge.
6. 1 3
D I RECf-CU R RENT TRANSMISSION
231
Terminal equipment is p rotected by surge arresters, which are also called
lightning arresters and surge diverters. An ideal arrester connected from the l ine
to a grounded neutral wou l d (1) become conducti ng at a design voltag e above
the arrester rating, (2) limit the voltage across its terminals to this d esign value,
and (3) become nonconducting again when the l i ne-to-neutral voltage d rops
below the design value.
Originally, an arrester was simply an air gap. In this application w h e n the
surge vol tage reaches a value for w h ich the gap is de signed, an arc occurs that
causes an ion ized path to ground, essentially a short circuit. However, when the
surge ends, the 60-Hz current from the generators still flows through the a rc to
ground. The a rc has to be exting u ished by the opening of circui t breake rs.
Arresters capable of ext inguishing a 60-Hz current after conducting s urge
current to ground were deve loped l ater. These arresters are m a d e u s i ng
non l i near res istors in series with a i r gaps to which an arc-q uench ing cap ab i l ity
h a s bee n a d d e d . The n o n l i n e a r r e s i s t a n c e d e c r e a s e s r a p i d l y as the voltage
across it r ises. Typical resistors m ade of sil icon carbide cond uct curre n t p ropor­
tional to approxim ately the fou r t h powe r of the vol tage across the resistor.
\Vhen the gaps arc over as a result of a voltage surge, a low-resistance current
path to ground is provided t h rough the nonlinear resistors. After the s urge ends
and the voltage across the arrester returns to the normal l ine-to-neutral level
the resistance is sufficient to limit the a rc 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 insulat ing p l a tes.
The most recent development in surge arresters is the use of zinc oxid e in
place of silicon carbide. The voltage across the zinc oxide resistor is extremely
constant over a very high range of cu rrent, which means that its resistance at
normal l i ne voltage is so high t h at a series a ir gap i s not necessary to l im i t t h e
drain o f a 60-Hz current a t norm al voltage. 3
6 . 13
DIRECT-CURRENT TRANSMISSION
transmission of energy by d i rect cu rrent becomes economical when com­
pa red to ac t r a n s m i s s i o n o n l y w h e n t h e cxtra cost of the terminal equipment
r e q u i re d for d c l i n es is o frs e t b y the l owe r cost o f h u i l d i n g t h e l i nes. Con v e r t e rs
a t t h e two ends of t h e d c l i n c s ope ra t e as b o t h recti fiers to c h a nge t h e gener a ted
alternating to d i rect curre n t and inverters for converting d i rect to a l ternat ing
current so t h a t pow e r c a n /low i n e i t h e r d i re ct i o n .
T he
Th e yea r
1 95 4 is g e n e ral l y reco g n i zed a s t h e s t a r t i n g d a t e for m o d e rn
h igh-voltage dc t ransmission when a dc line began service at 1 00 kY from
V a s t e rv i k o n t h e m a i n l a n d of S w e d e n t o V i sby on t h e i sl an d o f Gotland, a
J
See E. C. S a k s h a u g , 1 . S. Kre s g e , a n d S . A M is k e , J r. , " A N e w Concept in S t a t ion A r re s t e r
D e s i g n , " IEEE
Transactions o n Power Appara tus and Systems, v o l . PAS-96, no. 2, Marchi A p r i l
1 97 7 , pp. 647 - 656.
232
CHAPTER 6
CU RRENT AND VOLTAGE R E LATIONS ON A T R A N S M ISSION L I N E
distance of 1 00 km (62.5 mi) across the Baltic Sea. Static conversion equipme nt
was in ope ra tion much earlier to transfe r e nergy between systems of 25 and
60 Hz, essentially a d c transmission line of zero length. In the United S tates a
d c line operating at 800 k V transfers power generated in the Pacific northwest
to t h e southern part of California. As the cost of conversion equ ipme nt
d ecreases with respect to t he cost of line const ruction, the economical minimum
length of d c lines also decreases a n d at this time is about 600 km ( 3 75 mi).
Operation of a dc l ine began in 1 977 to transmit power from a mine-mouth
generating plant burning lignite at Center, North D a k o t a , to n e a r Duluth,
M inn eso t a , a d i s t a n c e o r 740 k ill ( 4 ()() ll1 i ). P r c l i m i n ; l r y s t u u i e s s h ow e d t h ; l ! t h e
dc l ine including terminal facilities wo u ld cost about 30% less t han the compa­
rable ac l i ne and auxiliary equipment. This l i n e o p e r a t e s a t ± 250 kV (SOO kV
l in e to line ) and transmits 500 MW.
Direct-current lines usually have one con d u ctor wh ich is at a positive
potential with respect to ground and a second conductor operating at an equal
negative potential. S uch a l i ne is said to be bipola r . The line could be ope rated
with one energized conductor with the return p a t h through the earth, which has
a much lower resistance to direct current than to alternating curre nt. I n this
case, or with a grounded retu rn conductor, the line is said to be monopolar .
I n addition to the lower cost of d c transmission over long distances, there
a re other advantages. Voltage regulation is less of a problem since at zero
frequency the series reactance wL is no longer a factor, whereas it is t h e chief
contributor to voltage drop in an ac l ine. Another advantage of direct current is
the possibility of monopolar operation in an emergency when one side of a
b ipolar l i ne becomes grounded.
Due to the fact that underground ac transmission is limited to about
5 km because of excessive charging current at longer d istances, direct current
was chosen to transfer power under the English Channel between Gre at Britain
and France. The use of direct cu rren t for this installation also avoided the
d ifficulty o f synch ron izing t h e a e systems o f t h e two countries.
No n e twork or d e l i n es i s poss i b l e a t t h i s t i m e b e c a ll s e no c i rcu i t b r e a k e r i s
available for direct current that is com parabl e to the highly developed ac
b reakers. The ac breaker can extinguish the arc which is formed whe n the
breaker opens because zero cu rrent occu rs twice in each cycle. The d i rection
and a mount of power in the d c line is controlled by the converters in w hich
grid-controlled mercury-arc devices are being d ispl aced by the semiconductor
r ectifier (SCR). A rectifier u n i t w i l l conta in pe rhaps 200 SCRs.
Still another advantage of dire ct curre nt is the smal ler amount of right of
w ay required. The distance between the two conductors of the North Dakota­
D ul ut h 500-kV line is 25 ft. The 500-kV ac l i ne shown i n Fig. 6.1 has 60.5 ft
b etween the outside conductors. Another consideration is the peak vol t age of
t h e ac l ine, which is fi x 500 707 kV. So, the line requires more insulation
b etween the tower and conductors as well as greater clearance above the e arth.
We conclude that dc transmission has many advantages over altqnating
current, b u t dc transmission rem ains very li mited i n usage except for long l i n es
=
PROBLEMS
233
since t here is no dc device which can provide the excellent switching operations
and p rotection of the ac c i r c u it breaker. There i s also no simple device to
change the vol tage l evel, which the transformer acco mplishes for ac systems.
6.14 SUMMARY
The long-line equations given by Eqs. (6.35) and (6.36) 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 i n t roduced because of their instructional v a l ue in
showing the maximum power which can be transmitted by a l ine and also in
showing the effect of the power factor o f the load or the addition o f capacitors.
ABeD constants provide a straightforward means of writing equations i n a
more concise form and are v e ry convenient i n probl ems involving n etwork
reduction. Their usefulness is a pparent in the discussion of series and s h u n t
reactive compensation.
The simple discussion of transients, although confined to lossless lines a n d
d c sources, should give some i d e a o f t h e complexity o f t h e study of transi e n ts
which arise from lightning and swi tching in power systems.
PROBLEMS
IS-km, 60-Hz, single-circuit, th ree-phase line is composed of Partridge con d u c­
tors equilaterally spaced with 1 .6 m between cen ters. The line delivers 2500 kW a t
1 1 kV t o a balanced load. Assume a w ire temperature of 5C.
(a) Determine the per-phase series impedance of the line.
C b ) What must be the sending-end voltage when the power factor i s
(j) SO% lagging,
( ii) unity,
(iii) 90% leading?
(c) Determine the percent regu lation of the line at the above power factors.
Cd) Draw phasor d iagrams depicting the operation of the line in each case.
A I OO-mi, single-circuit, t hree-phase t ransmission l ine del ivers 55 MVA at O.S
power-factor lagging to the load at 1 32 kV (line to li ne). The l i ne is compose d of
Drake conductors with flat horizontal spacing of 1 1 .9 ft between adjacent con d u c­
tors. Assume a wire temperature o f 5C. Determine
(a) The s e ri es i m p ed a nc e a n d the s h u n t adm i t t ance o f t h e l i n e .
( b ) The ABeD constants of t h e l i ne.
( c ) The sending-end Voltage, current, real and reactive powers, and the power
factor.
Cd) The percent regul ation of t h e line.
circuit having a 600-.0 resistor for t h e shunt
Find t h e ABeD constants of a
b ranch at the send ing end, a l - k fl resistor for the shunt branch at the receiving
e n d , and an SO-o. resistor for the series branch.
6.1. An
6 .2.
6.3.
7T
234
6.4.
CHAPTER 6
CURR ENT AND VOLTAGE RELATIONS ON A TRA N SMISSION L I N E
A
=
D
B
C
6.5.
6.6.
6.7.
6.8.
/ 0 .98°
142/76.40 D
The ABCD constants of a three-phase tra nsmission line are
=
=
=
0 .936
3 3.5
+
+
iO .016
i1 3 S
( - 5 .] 8
+
=
=
i914)
0 .936
x
The load a t the receiving en d is 50 M W at 220 kV with a power factor of 0.9
lagging. F i nd t h e magnitude o f the sending-end vo l lage and the vol tage reg u l �l t ion.
Assume that the magn itude of the send ing-e n d voltage rem a ins consta n t.
A 70-m i, single-circu it, three-phase line composed of Ostrich conductors is a r­
ranged i n flat horizontal spacing w i th 1 5 ft betwee n adjacent conductors. The line
deli vers a load of 60 MW at 230 kV w i th 0 .8 power-factor l aggi ng.
( a ) Using a base of 230 kV, 1 00 M VA, de termin e the series i mpedance a n d t he
shunt admittance of the line in per u n i t. Assume a wire temperat ure of 500 C.
Note that the base admittance m ust be the reciprocal of base i m p e d a n c e .
(b) Find the voltage, current, real and reactive power, and the power factor a t the
send ing end in both per unit and a bsol u te u n its.
Cc) What is the percent regulation of the line?
A single-circuit, three-phase transmiss ion line is composed of Parakeet con ductors
with flat horizonta l spacing of 19.85 ft between adjacent conductors. Determine the
characteristic impedance and the propagation constant of th e l ine a t 60 Hz a n d
5 0° C temperatu re.
Using Eqs. (6.23) and (6.24), show that if the receiving end of a line is terminate d
by its characteristic i mpedance Zc ' t h e n the impedance seen a t the sen d i ng e n d of
the l i n e is also Zc regardless of l i ne length.
A 200-m i transmiss ion line has the fol lowing pa rameters at 60 Hz:
Resistance r = 0 . 2 1 D/mi per phase
Series reacta nce
x =
Shunt susceptance b
=
0 .78 D / m i
5 .42
X
6.10.
per
phase
1 0 - 6 Simi per phase
Determine the a ttenuation consta nt a, wavelength A, and the veloc ity of
propagation of the line at 60 Hz.
(b) I f the line is ope n-ci rcu ited at the rece iving end and the receiving-end voltage
is maintained at 100 kV line to line, use Eqs. (6.26) and (6.27) to determine the
incident and reflected components o f t he sending-end voltage and current.
( c ) Hence, determ ine the sendi ng-end voltage and current of t he l ine.
.
Evaluate cosh e a n d s inh e for e = 0.5
Usi ng Eqs. (6. 1), (6.2), (6.10), and (6.37), show that the gene ralized circuit consta nts
of all three transmission-line models satisfy the condition that
(a)
6.9.
10-6 S
�
AD - BC = 1
P R O B LE M S
6. 1 1 .
6.12.
6.13.
6.14.
The sending-end voltage, current, and power factor of the line described in
Example 6.3 are found to be 260 kY (line to l i n e), 300 A, and 0.9 lagging,
respectively. Find the corresponding receiving-en d voltage, current, and power
factor.
A 60-Hz three-phase transmission line is 175 mi long. I t has a total series
impedance of 35 +
D. and a shunt admittance of 930 X
S. It
delivers 40 MW at 220 kY, with 90% power-factor lagging. Find the voltage at the
sending end b y ( a ) the short-line approximation, (b) the nominal-'1T approximation,
and (c) the long-line equation.
Determine the voltage regulation for the line described in Prob. 6.12. Assume t h at
the sendi ng-end voltage remains constant.
A three-phase, 60-Hz transmission line is 25 0 mi long. The voltage at the sending
c n d is 2 2 0 kY. The parameters of the line arc R
0. 2 D. / m i , X = 0 8 D./mi, a nd
Y = 5.3 ,u S / m i F i n d th e se n d i n g e n d c u r r e n t w h e n t h e r e is no l o a d on the line.
I f t h e l oad on t h e l i ne d es c r i b ecl i n Prob.
is SO M\V a t 220 k Y , with u nity
pow e r fa ct o r calculate t h e c u rre n t , vol t ag e , and powe r at t h e sending end. Assume
that the sen ding-end voltage is held constant and calculate the voltage regulation
of t h e line for the load specified above.
A three-phase' transmission line is 300 mi long and serves a load of 400 MY A, w i t h
0.8 lagging power factor at 345 kY. The ABCD constants of the line are
j140
1O-6�
.
6.15.
6.14
-
,
6.16.
235
A
B
C
=
=
=
D
=
=
.
�
0 .8180
1 72 .2/ 84 .20 D.
0 .001 933
/90.40
S
( a ) Determine the sendi ng-end lin e-to-neutral voltage, the sending-end current,
and the percent voltage drop a t full load.
( b ) Determine the receiving-end l i n e-to-neutral vol tage at no load, the sending-end
c u rre n t
y
6 . 1 7. J usti f
a t n o l oa d ,
;l n cl t h e
E q . ( C>.50) by s u b s t i t u t i n g
e x po n e n t i a l e x p r e ss i o ns.
6.20.
for
t he
y
h pe r bo l ic
fu nctions the equivalent
et e r m i n e t h e e q u iv a l e n t -7T c i rc u i t fo r t h e l i n c of Prob. 6. 1 2.
Usc Eqs. (C>. J ) and (6.2) to s i mp l i fy Eqs. (6.57) and (6.58) for the short transmission
line with (a) series r e a c ta n ce X and resistance R and (b) series reactance X a n d
neg l i g i b l e r e s i s t an c e .
R ig h t s of w a y fo r t r a n s m ission circuits a rc d i flicu lt to obtain in urban areas, a n d
cxi�ting l ines are often u pgraded by rccond uctoring the line with larger conductors
or by reinsulating the line for operation a t higher voltage. Thermal considerations
and m aximum power which the line can transmit are the important considerations.
A 138-kY line is 50 km long and is composed of Partridge conductors with fiat
horizontal spacing of 5 m between adjacent conductors. Neglect resistance and find
6. 1 8. D
6.19.
vo l t a ge reg u l a t io n .
236
CHAPTER
6
C U R RENT A N D VOLTAGE R ELAT I O N S O N A T R A N SM I S S I O N L I N E
the perce nt i ncrease in power which can be transmitted for constant I Vsl and I VR I
while 0 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,
(b) If a second Partridge conductor is placed in a two-conductor bundle 40 cm
from the original conductor and a center-to-center d ista nce between bundles
of 5 m, and
( c ) I f t h e v oltag e o r t h e o r i g i n a l l i n e is raised to 230 kV with i nc reased conductor
spacing of I) m.
6.21. Construct a receivi ng-end power-circle d iagram similar t o Fig. 6. 1 1 [ r the line o f
Prob. 6. 12. Lo c a t e t h e poi n t corresponding t o t h e load o f P rob. 6 . 1 2 , a d locate lhe
center o f circles for v a r iou s values of I Vs l i f I Vu l 220 kY. Draw the circle passing
through the load point. From the measured r a d i u s of the latter circle determine
Wil l and compare this value with the val ues calculated for P r o 6 . 1 2 .
6.22. A synchronous condenser is connected i n parallel \vi t h the load described in Prob.
6. 1 2 to improve the overall power factor at the receiving end. The sending-end
voltage is always adjusted so as to maintain the receiving-end vol tage fixed a t 220
kY. Using the power-circle diagram constructed for Prob. 6 . � 1 , determine the
sending-end voltage and the reactive power supplied by the synchronous condenser
when the overall power factor at the receiving end i s (a) unity (b) 0.9 leading.
6.23. A series capacitor bank having a reactance of 146.6 n is to be i nstalled a t the
m i d point of the 300-mi line of Prob. 6. 1 6. The ABCD constants for each 1 S 0-mi
portion of line are
on
=
b
.
A = D = 0 .9S34fu
B = 90 .33 �t.
C
=
0 .00 1 0 1 4/ 90 . 1° S
cascade combination
(a)
6.24.
Determine the e q u i v a l ent A B CD c o n s ta n ts for the
l ine-capacitor-line. (See Table A.6 i n the Appendix.)
(b) Solve Prob. 6.16 using these equivalent ABCD constants.
The shunt admittance of a 300-m i t ransmission l i ne is
yc
6.25.
n
=
0
+
of the
}6 .87 X 1 0 - 6 Simi
o.Doni
D etermine the ABeD constants of a shunt reactor that will compensate for 60% of
the total shunt admittance.
90° S is con­
A 2S0-Mvar, 34S-kV shunt reactor whose admittance is
n ected to the receiving end of the 300-mi line of Prob. 6 . 1 6 at no load.
(a) Determine the e quivalent ABCD constants of the line in series with the shunt
reactor. (See Table A.6 in the Appendix.)
(b) Rework part (b) of P rob. 6 . 1 6 using these equivalent ABCD constants and the
sending-end voltage found in Prob. 6. 1 6.
-
PROBLEMS
6.26.
6 . 27.
6.28.
6.29.
6.30.
237
Draw the lattice d iagram for curre n t and plot current versus time at the se n d i n g
end of the line of Example 6.8 for the line terminated in (a) an open circuit (b) a
short circuit.
Plot voltage versus time for the line of Example 6.8 at a point d istant from the
sending end equal to one-fourth of the length of the line if the line is terminated in
a resistance of 10 D.
Solve Example 6.8 if a resistance of 54 n is in series w ith the source.
Voltage from a dc source is applied to a n overhead transmission line by closing a
swi tch. The end of the overh ead line is connected to a n underground cable.
Assume that both the l ine and the cable are lossl ess and that the initial voltage
along the l ine is v + . I f the characteristic impedances of the line and cable are 400
and 50 fl, respectively, and the end of the cable is open-circuited, find in terms
of v +
( a ) T he vol tage at the ju nction of the line and cable i m med iately after the arrival
of the i n c i de n t wave a n d
( b ) The vo l t age at t h e open end of the cable immediately after arrival of the first
voltage wave.
A dc sou rce of voltage VI and internal resistance Rs i s connected through a switch
to a l os s l c s s l i n e having ch a r a c t e r i s t i c i mp e d a nc e R c . The l i n e is terminated in a
r e s i stance R R . The traveling t ime of a vo l t a ge across the line is T. The switch
o.
cl oses at
( a ) D raw a l a t t ic e diagram showing the voltage of the line during the period t
0
to t = 7T. Indicate the voltage components i n terms of � and the reflection
coetflcien ts P R and P
( b ) D e ter m i n e the receiving-end voltage a t t
0, 2T, 4T, and 6T, and hence at
(
2 n T where ! l is any non-negative i nteger.
(c) Hence, determine the steady-state voltage at the receiving end of the l i ne i n
terms o f v" R s ' R R ' a n d R e C d ) Verify the result i n part (c) b y analyzing the system as a simple d c circuit i n the
steady state. (Note that the line is loss less and remember how i nductances and
capacitances behave as short circuits and open circuits to dc.)
(
=
=
S"
=
=
CHAPTER
7
THE
AD MITTANCE
MODEL AND
NETWORK
CALCULATIONS
The typical power transmission network spans a large geographic a rea and
involves a l arge number and variety of network components. The e lectrical
characteristics of the individual components a re d eveloped in previous chapters
and now we are concerned with the composite representation of those compo­
nents when they are interconnected to form the network. For large-scale system
analysis the network model takes on the form of a network matrix. with elements
determined by the choice of parameter.
There are two choices. The current flow t h rough a network component
"
can be related to the voltage d rop across it by either an a dmittance or an
impedance parameter. This c hapter treats t h e a d m ittance representation in the
form of a primitive model which describes the electrical characteristics of the
network components. The primitive model neither requires nor provides any
information about how the components are interconnected to form the network.
The steady-state behavior of all the components acting together as a system is
given by the nodal admittance matrix based on nodal analysis of the network
equations.
The nodal admittance matrix of the typical power system is large and
sparse, and can be constructed in a syste matic building-block manner� The
building-block approach provides insight for developing algori thms to account
for network changes. Because t h e n e t w o r k m a t rices a r e very l a rge, !)pa rsily
7. 1
BR ANCH A N D N O D E
A DMITTANCES
239
techniques are needed to enhance the computational efficiency of computer
programs employed in solving many of the power system problems described in
later chapters.
The particular importance of the present chapter, and also Chap. 8, which
develops the nodal impedance matrix , becomes evident in the course of pow er­
flow and fault analysis of the system.
7.1
BRANCH AND NODE ADMITTANCES
In per-phase analysis the components of the power transmiSSion system are
modeled and represented by passive impedances or equivalent admittan ces
accompanied, where appropriate, by active voltage or current sources. In the
steady state, for example, a generator can be re presented by the circuit of e ither
Fig. 7 . l ( a ) or Fi g . 7 . 1 ( b ) . The circu it having the constant emf Es ' series
impedance Za ' and terminal vol tage V has the vol tage equation
( 7 . 1)
D ividing across by Z a gives the current equation for Fig . 7. l(b)
E
!s = _s = I + VYa
2a
( 7 .2)
where Ya = 1 /2a ' Thus, the emf Es and its series impedance Za can b e
interchanged with the current source Is and its shunt admittance Ya , p rovided
Es
1 = ­
Za
s
I
r
V
Es
(a)
I
1
Ya = ­
Z
and
( 7 .3)
a
I
N
N
e
w
0
Yo
V
e
t
w
0
r
k
r
k
(b)
Fl G U RE 7 . 1
Circ u i t s i l l u strat ing t h e eq uivalence of sources when Is
=
Es/Zu a n d
Y" =
1 /Z/I'
240
CHAPTER 7
THE ADMITTANCE
M O D E L A N D N ETW O R K CALCU LAT I O N S
Sources such as Es and Is may be considered externally applied at the nodes of
the transmission network, which then consists of only passive branches. In this
chapter subscripts a and b distinguish branch quantities from node quantities
which have subscripts m , n, p, and q or else numbers. For network modeling
we may then represent the typical branch by either the branch impedance Za or
the branch admittance Ya , wh ichever is more convenient. The branch i mpedance
Za is often called the primitive impedance, a n d l ikewise, Y{l is called the
primitive admittance. The equations characterizing the branch are
vII
=
Z U Ia
( 7 .4 )
or
where Y:l is the reciprocal of Z{l and V:I is the vol tage d rop across the branch in
the d i rection of the branch current I {l ' Regardl ess of how it is co nnected into
the network, the typical branch has the two associated va riables V{l and fa
related by Eqs. (7.4). In this chapter we concentrate on the branch a dmittance
form in order to establ ish the nod al a d m ittance representation o f the power '
network, and Chap. 8 treats t he impedance form.
I n Sec. 1 . 1 2 rules are given for forming the bus a dmittance m atrix of the
network. Review of t hose rules is recommended since we are about to consider
an a lternative method for Y hus formation. The new method is more general
because i t is easily extended to networks with mutually coupled elements. Our
approach first considers each branch separately and then in combination with
other b ranches of the network.
S uppose that only branch admittance Ya is connected between nodes ®
and ® as part of a larger network of which only the reference node appears in
Fig. 7.2. Current i njected into the network at any node is considered positive
and c ur re n t leaving the network at a ny node is considered nega tive . In Fig. 7.2
c u rrc n l / 1/1 i s that p or t io n o r t h e t o t a l cu rr e n t i njected i nto node ® w h i c h
passes through r:1 ' L i k ewise, I" is that portion of the current i njected into node
® wh ich passes t h ro u g h Y:, . The volt ages V;,/ and V;1 a re the vui t Cl g e s of nodes
1m
@
)0
I
+
I
r+
�-
Vm
fa
�
c==
Va
Ya
)o ®
=
Za
1
-
1+
!-
Vn
I
I"
�
Reference node
Primi tive b ra n c h voltage drop V;" branch c u r re n t III ' i njected cu rrents JI/I and I" , node volt flges Vm
a n d V,I w i t h respect to ne twork refe rence.
FIGURE 7.2
7.1
B RANCH AND NODE A D M ITTANCES
241
@J
and ® , respectively, m e asured with respect to network reference. By
Kj rchhoff's l aw 1m = 1a a t node @) and 1/1 = - la at node ® . These two
current equations arranged in vector form a re
( 7 .5 )
In Eq. (7 . 5) the labels or pointers @J and ® associate the d irection o f I a from
node § to node ® wi t h t h e e nt ries 1 and - 1 , w h ich arc then said to be in
row ® anu row "i!) , r e s p e ct i v e l y I n a s i m i l a r m a n n e r, the vo ltage d rop in the
d i re c t i o n of 1 has t h e equa tion Va = V,II
v, , , O r expressed in vector for m
.
a
-
@)
®
- 1
[ 1
Subst i t u ting t h i s expression for �I In
the
-I]
®
( 7 .6)
]
admittance equation Ya Va
�
[ 1
� I,
=
1a gives
( 7 .7 )
and premul tiplying both si des of Eq. (7.7) by the column vector of Eq. (7.5), we
obtain
®® [ l lYa
-1
which,
s i m p l i fies
[
@)
1
to
@)
®
®
-1]
[ �: 1 [ �: 1
®
( 7 .8)
( 7 . 9)
Ya
and the coefficient m a t rix is
the noda l admittance matrix . We note that t h e off- diagonal e le ments e q u al t he
n egat ive of the branch admittance. The m atrix of Eq. (7.9) is singul a r b ecause
neither node § nor node ® connects to the refe re nce. In the par ticular case
w h e r e one of the two nodes, s ay, ® , is the reference node, then n � d e vol tage
This is the nodal admittance equa tion for b ranch
242
CH APTER 7
TH E ADMITTANCE MO DEL A N D N ETWO R K
Vn is zero and Eq. (7.9) reduces to the 1
x
@)
® [ Ya ] Vm
correspo n d i n g t o re m ova l o r
row ®
CALCULATIONS
1 m atrix equation
=
and
1m
( 7 . 1 0)
co l u m ll ®
fro lll
t h e c o e /'n cie n t
matrix.
Desp ite its s t r a i g h t forw a r d d e r i v a t i o n , Eq . (7.<) ) a n d t h e p mc e d ur e l e a d i n g
t o it arc important in more general s i t u a t i o n s . W e n o t e t h a t t h e b ranch vo l t age
Va is transT'ormed to the node vol tages v,1/ and V, I ' and the b r a n c h cu rrent l a is
l ikewise re presented by the cu rrent injections III/ and 1/1 " The coefficient ma trix
relating the node voltages a n d cu r re nts o f Eq . (7 .9) fol lows from the fact that in
Eq. (7.8)
®
-1
@)
®
(7.11 )
]
This 2 X 2 matrix, also seen i n Eq. 0 .64), is an im portant bllilding block for
represe nting more general n etworks, as we shall soon see. The row and column
pointers identify each entry in the coe fficient m a trix by node numbers. For
instance, in the first row and second column of Eq. (7 . 1 1 ) the entry - 1 is
identified with nodes @) a nd ® of Fig. 7.2 and the other entries are similarly
identified.
Thus, the coefficient m atrices of Eqs. (7.9) and (7. 1 0) a re s i m p l y storage
matrices with row and col umn l abels determ ined by t h e e n d nodes of the
branch. Each branch of t he ne twork has a sim i l ar mat rix labe led according to
t h e nodes of the network to w h ich t hat bra nch i s connec ted. To obtain the
overall nodal admittance matrix of the e n t i re network, w e s i m p l y co m b i n e the
ind ividual branch matrices by a dding together clements with identical row a n d
column labels. Suc h add ition causes t h e sum of the b r a n c h currents flowing
from each node of the network to equal t he total curre nt inj ected into that
nod e, as required by Kirchhotl's current law. I n the overall m a t r i x , the off-d iag­
onal element 0j is the negative sum of the adm ittances con nect ed between
nodes CD and CD and the d iagon al e n t ry Y,i is the algebraic sum of the
admittances connected to node CD . Provided at least one of the networ k
branches is connected to the refe rence node, t h e n et result is Y bu S of the system,
as show n in the following example .
The single-line diagram of a small power system is shown i n Fig. 7. 3.
The corresponding reactance diagram, with re actances specified in per u n i t , is
per u n i t is conllected
shown in Fig. 7.4. A generator with e m f equal to
t hrough a t ransformer t o h igh-voltage node G) , w h i l e a motor w i t h internal
voltage equal to 0 . 85 - 45° is similarly c o n n e c t e d to n o d e GG · Deve l o p t h e n o d a l
Example 7.1.
!
1 .25&
7.1
B R A N C H A N D N O D E A D M ITTANCES
243
®
CD
F I G U RE 7.3
S i n g l e - l i n e d i agram of t he fou r-bus sys­
t e m of Exa m p l e 7.1. R e fe rence n o d e is
n o t shown .
)0.25
)0.2
jO.25
)0.4
@ --,--+--
--+--'-
@
0
R e a c t a n c e di ag r a m fo r F i g. 7.3.
FIG U R E 7.4
Node
@ is reference, react a n ces
and vo l t a ges a te in rer u n i t .
a d m ittance m atrix for e a c h o f t he network b ra n ch es and then write the n odal
a d m i t tance equations of the system.
The re a c t a n ce s o f t h e g e n e r a t o r and t h e m o t o r may b e combined w i t h
t he i r respective step-up t ransformer reactances. Then, by t ransfo rm a tion of sou rces
t h e com bincd rcactanccs a n d the gencrated e m rs a rc repl aced by t h e e quival e n t
curre nt sources a n d s h u nt a d m i t ta nces shown in Fig. 7 . 5 . We wi l l t re a t t h e current
sources as external injections at nodes G) and @ and name the seven pass ive
b ran ches accord i n g to the subscri pts of their currents and vol tages. For example,
the branch between nodes CD and G) will be called branch c . The a d mittance o f
each b ranch is s i m ply the r eciprocal of the branch impedance a n d Fig. 7.5 shows
the resulta nt a d m itt ance d i agram with a l l values i n per u n i t. The twp branches a
a n d g connected to t h e r e ference node arc characterized by Eq. (7. 1 0), w h ile Eq.
Solution .
244
CHAPTER
7
THE A D M ITTANCE M O D E L A N D N ETW O R K CA LCU LA T I O N S
-- v �
+
c
--- v:(�+
_
+
1.00/- 90°
t
V}.'
t
@
FIGURE 7.5
Per- u n i t admit tance d i agram for Fig. 7 . 4 w i t h c u r r e n t sou rces r e p l a c i n g vo l t a g e sou rces. B r a n c h
names a t o g correspond t o t h e subscripts o f b ra n ch voltages and c u rrents.
(7.9) applies to each of the other five branches. By setting
and n in those
equations equal to the node n u mbers at the ends of the individual b ranches of Fig.
7.5, we obtain
0
0
(1)
G) [ l ] Ya
G)
CD
-
1
1
G) CD
[ -: ]
-
1
1
0 G)
[ -: ]
Yd
@
G)
Yb
@
0
CD
-1
1
Yc
@
CD
[ -: ]
G) CD
[ -: ]
-(1)
1 Yc
1
[ -: ]
m
@
@
[1]
Yg
-1
@ CD
1 Yf
The order in which the labels are assigned is not important here, provided t h e
columns a n d rows follow the same order. Howe\ er, for consistency with later
sections l et us assign the node numbers i n the directions of the branch currents of
Fig. 7.5, w h i ch a l s o shows the n u merical v a lues o f the a d m ittances. Comb i n i n g
together those elements of the above matrices ha\'ing identical row and column
labels gives
CD
G)
G)
@
CD
( � + Yd
Yd
Yc
- Y,
+
Yf )
Yd
( Yb + Yd
- Yb
- Y:,
@
G)
G)
-
+
Yc )
Yc
-Y
b
( Ya + Yb + � )
0
(Y
c
- Yf
Yc
0
-;- YJ + YJ
7.2
M UTUALLY CO UPLE D B RANCHES I N Y bus
245
which is t he same as Y hlls of Eq . ( 1 .02) si nce F i g s . 1 .23 and 7.5 me for the same
ne twork. S u b s t i t u t i ng t he n u m e rical val ues of the branch admitta n ces i n t o this
matrix, we obtain for the overall n e twork the nodal admi ttance equations
-j 1 4 . S
j8.0
j8 .0
-j 1 7 .0
j 4 .0
j4 .0
j2.S
j5 .0
j 4 .0
j4 .0
-jS . S
0 .0
j2.5
j) .O
0 .0
-jS .3
VI
V2
V]
/
/ - 900
0 .68 / - 1 350
0
0
=
�
1 .00
/
w h e re V I ' V� , VJ , and Vol <I r e t h e !lode vo l t ages measured with respect to t h e
re fe rence node a n d I I
0 , 1 2 = a I J = 1 .00 - 900 and 14 = 0 . 68 - 1 35° are
t h e ext ernal c urre n t s i nj ected at t h e system n o d es.
=
The coe l fl c i e n t m , l t r i x o o t <l i n e cl i n t h e a h ove e X ; l m p i e
is
the s a m e
(I S t h e b u s (t u m i t l a n c c m�lt l i x ro u n d i n S e c . 1 . 1 2 lIs i n g t h e u s u a l ru l e s fo r Y b u s
fo r m a t i o n . However, t h e approach based on the build ing-block m a trix h as
adv a n t ages w h e n extended to n e tworks w i t h m u t u a lly cou p l e d bra n ches, as w e
n ow d e m on s t r a t e .
exac t l y
M UTUALLY C O U PLED B RA N C H E S
IN Y b u s
7.2
The p roce d u re based on t h e b u i l d i ng-block m a trix is n ow extended to two
m u t u a l ly cou p l e d branches w h i c h a re p a r t of a l a rger n et\vork but w h ich a r e n o t
i n d u c t ively coupled t o a n y othe r b r a n c h e s . I n Sec. 2 . 2 t h e p r i m i t iv e equations of
s u c h m u t u a l ly coupled branches a r e d eveloped in t h e for m of Eq. (2.24) for
i m p e d a nces and Eq . (2.26) for a d m i t t a n ces. The nota t i on is d ifferent h e re
b e c a u s e we a re n ow u s i n g n u m bers to i d e n t i fy n odes r a t h e r t h a n b r a n ches.
Assume t h ; l t h r <l n c h i m p e d a n c e Z" c o n n e ct e d b e t we e n nodes ® a n d ®
i s cou p l e d t h ro u g h m u t u a l i m p e d a n c e LM to b r a n ch i m p e d a n c e Zb con n e c t e d
b e t w e e n node's ® a n d (!t o f F ig. 7 . 6 . The vol tage d rops
a n d VI> due t o t h e
b l a n c h c u rr e n t s I " (l n cl I I ! ;l l e t h e n g i v c n b y t h e p r i m i t ive i m p e d a n ce e q u a t i o n
co r r e s p o n d i n g t o E q . ( 2 .24) i n the fo r m
T<I
( 7 . 1 2)
i ll w h i c h t h e coe fll c i c n t m a t r i x i s sy m m e t r i c a l . As n o t e d i n S e c . 2.2, t h e m u t u a l
a n d II! e n ter the
ter m i nals m a r ked w i t h clots i n Fig. 7 . 6( a ) ; t h e voltage drops Va a n d Vb t h e n
h ave t h e p o l a r i t i e s show n . \1 u l t i p l y i n g E q . ( 7 . 1 2) b y t h e i nverse of t h e prim i t ive
i m p e d ance m a t r i x
i n1 p cc.L. l l1 cc
Z,\/
is
co n s i d e r e d
2M]Zh
1
=
pos i t iv e
when
c u r re n t s
Iii
YM]
Yb
( 7 . 13 )
246
1
CHAPTER
-'"
@)
7
THE ADMITTANCE MODEL A N D NETWORK CALCULATIONS
III
Ib
---'--+-
�
-I
p
+
+
®
®
Zb
Vb
®
In-----
� Iq
(a)
@)
1m---
In--
®
®
III
-1p
---'--+-
VIl
Ya
------ 1q
(b)
FIG URE 7.6
Two
m u tu a lly c o u p l e d
branches with ( a ) i m ped a n ce
p a ra m e t ers a n d ( b ) correspo n d i n g a d m i t t ances.
we obtain the admittance form of Eq. (2.26) for the two branches
( 7 . 14)
which is also symme trical . The admi ttance matrix of Eq. (7.14), ca lled the
primitive admittance matrix of the two coupled branches, corresponds to Fig.
7.6(b). The primitive self-admitta nce "y;, equals Z,,/( Z,, Z,, Z;1 ) and similar
expressions from Eq. (7. 13) apply t o Y,) and the primitive mutual aumitta nce
YM • We may write the vol tage-drop equations V" = ��I/
v" and VI> = Vp - Vq
of Fig. 7.6 in the matrix form
-
-
@ ®
-1
o
o
1
(7.15)
which the first row of the coefficient matrix A is associated with branch
admi tt a nc e Ya and the second row relates to branch admittance Yh. The node
ill
vo l ta g es Vm , Vn , Vp ' and � are measured with respect to the network reference.
In Fig. 7.6 the branch current fa is related to the injected currents by the two
M UTUALLY COU P LE D B RA NCHES IN Ybu .
7.2
247
node equations 1m = la a n d In = - 10 ; similarly, branch current lb is related to
the currents 1p and 1q by t h e two node equations 1p = 1b and 1q = lb' These
fou r current equations arranged in m atrix form become
-
o
1
@)
@
®
®
-
1
o
(7. 16)
o -1
with the coefficient matrix equal to t h e transpose of that in Eq.
(7. 1 5 ) we substitute for the vol tage d rops of Eq . (7. 14) to find
[
Yh
Y
"
�\1
Y"
1
VI I I
V:,
�,
A
[ �;: 1
v:/
(7. 1 5).
( 7 . 17)
and premultiplying both sides of t h is equation by the matrix AT of Eq.
obtain
[
AT
'-.......----' .....
4 X
2
Yo
Y
"
YM
Yb
1
J
2
X
2
A
--
-
2 X 4
From Eq.
1m
In
Ip
Vm
Vn
Vp
(7. 1 6),
we
( 7 . 1 8)
lq
Vq
\Vhen the multipl ications ind icated in Eq. (7. 1 8) are performed, the result
gives the nodal admittance equations of the two mutually coupled branches in
the m atrix form
®
®
®
®
®
Yo
-Y
0
YM
- YM
®
-
Y
Ya
a
- YM
YM
®
YM
- YM
Y"
- Y"
®
- YM
YM
- Y"
Y"
V:'I
v"
�,
VI]
fl/ l
In
=
If!
II]
J
( 7 . 1 9)
The two m utually couple d branch es a re actuall y part of a larger n etwork , a n d so
the 4 X 4 m at rix of Eq. (7 . 1 9) forms part of the larger nodal admitt ance matrix
of the overal l system . The pointe rs @ , @ , 0 , and ® indica te the rows and
colum ns of the system matrix to which the elemen ts of Eq. (7 . 1 9) belong . Thus,
for examp le, the' quanti ty e ntered in row ® and colum n ® of the system
248
CHAPTER 7 THE A D M IIT ANCE
M O D E L A N D N ETW O R K CALCU LATI O N S
nodal admi ttance matrix is - YM and similar e ntries are made from the other
elements o f Eq. (7.19).
The nodal admittance matrix of the two coupled branches may be for med
directly by i nspection. This becomes clear w h e n w e write t h e coefficient m atrix
of Eq. ( 7 . 1 9) in the alternative form
@
®
®
®
®
- 11 ]
®
r:l
®
1
- 1
®
®
1
YM
®
( 7 .20)
To obtain Eq. (7.20), we m ul tiply each element of the pnm l tlve admittance
matrix by the 2 X 2 buildi ng�block matrix. The l abels assigned to the rows and
columns of t h e multipliers in Eq. ( 7.20) are easily determined. First, we not e
that t h e self-admittance Ya is measured between nodes @) and ® with t h e dot
at node @j . Hence, the 2 X 2 matrix multiplying Ya in Eq. (7.20) has rows and
col umns l abeled @) and ® in that order. Then, the self-admit tance Yb
between nodes ® and ® is m ultiplied by the 2 X 2 matrix with labels ® and
® i n the order shown since node ® is marked with a dot. Finally, the l abels
of the matri ces multiplying the m utual admittance �\r are assigne d row by row
and then column by column so as to align and agree with those already give n to
the self-inductances. In the nodal admi ttance m atrix of Eqs. (7. 1 9) and (7.20)
the sum of the columns (and of the rows) adds up to zero. This is because none
of the nodes @j, ®, ®, and ® has been considered as the refe rence nod e of
the n e twork. I n the special case where one of the nodes, say, n ode ® , is in fact
the reference, Vn is zero and col u m n n in Eq. (7. 1 9) does not need to appear;
furthermore, III does not h ave to be expl icitly represented since the refe rence
node current is not an independent quantity. Consequently, when node ® is
the reference, we may eliminate the row and col umn of that node from Eqs.
(7.19) a nd ( 7 .20).
It is important to note that nodes ® , ® , ® , and ® are often not
distinct. For i nstance, suppose that nodes ® and ® are one and t h e same
nod e. In that case columns ® and ® of Eq. (7. 1 9) can be combined together
since � = Vq , and the correspond ing rows can be added because In and Ip are
parts of the common inj ected current. The fol lowing exam ple illustrates this
situation.
Examp]e
7.2. Two branches h aving i m pedances equal t o j O .25 per u n i t' are
co upled through m u t u a l i m p e d a n ce 7M
=
iO. 1 5 per u n i t , as s h own i n Fig.
7.7.
Fi n d
MUTUALLY COUPLED B R A N CHES
7.2
I N Ybu<
249
F I G U R E 7.7
two
The
(a)
mutually
cou p l e d
b r a n c h e s of Example 7 . 2 ,
their
p r i m i t ive i mp e d a n ces a n d
( b ) p r i m i t ive a d m i t t an c e s
(a)
in
pe r u n i t .
(6)
t h e n o d a l a d m i t t a n c e m a t r i x for t h e m u t u a lly c o u p l e d branches a n d w r i t e t h e
correspo n d i ng n od a l a d m i t t a n c e e q u a t i o n s .
Sollllion.
The pri m i t ive
i m pe d a n ce m � l t r i x ror t h e m u t ll ,t l ly c o u p l e d b r a nc hes o f
e ll t i t y t o y i e l d t h e p r i m i t ive t l cl m i t t a n c c s o f Fig.
f i g . 7 . 7( (I ) i s i n v e r t e d d S , I s i n g l e
7 . 7( h ) , I h ; 1 1 i s ,
jO . 1 5
j O . 25
[ iO . 25
jO.IS
] [
-I
=
-j6 .25
i 3 .7 5 ]
· -)6 .25
j3 .75
F i r s t , t h e rows and co l u m ns of t he b u i l d i n g- b l o c k m a t r i x w h i ch m u l t i pl i e s t h e
are l a b e l e d G) a n d
in
p r i m i t ive s e l f-ad m i t t a n ce b e twe e n nodes
and
t h a t o r d e r to corre spo n d to t h e dot m a r k i n g n ode
Next, t h e rows a n d col u m n s
of t h e 2 x 2 m a t rix m u l t i p l y i n g t h e s e l f-a d m i t t a n c e b et w e e n nodes
and
are
l ab e l e d
and
i n t h e o r d e r s hown b e ca u se node G) is marked. Fi n a l ly, the
CD
G)
(1)
®
G) -
CD
G)
(1)
p o i n t ers of the m a t r i ces mu l t ip l y i n g the m u t u a l a d m it t a n c e are a l igned w i t h t h ose
of t h e s e l f- a d m i t t a nces to fo r m t h e 4 x 4 array s i m i l a r to Eq. (7.20) as fol lows:
G)
G)
CD
G)
(1)
-:
CD
[ -: 1
CD
CD
[ : -: 1
( -)6 .25 )
G)
I CD
@
(1)
( i 3 .7 5 )
G)
W
G)
(1)
[ -: -: ]
[ -: : 1
( j 3 . 75 )
( -)6 .25 )
I
i n f i g . 7 . 7 , t i l e r eq u i re d 3 X ] m a t rix is fo u n d by
a d d i n g t h e co l u m ns a n d rows o f c o m m o n node
to obtain
S i n c e t h e re ; I re o n l y t h re e I l o d e s
CD
(1)
G)
�
CD
G)
(1)
G)
-)6 .25
) 3 .75
j3 .75
-j 6 . 25
) 6 . 2 5 -- ) 3 .75
- ) 3 .75
+
j6 .25
)6.25 - )3 .75
-
)3 7 5
.
+
-j6.25 - )6 .25
j6 .25
+
2( j3 .75)
J
250
®
•
Za
CHAPTER 7
� Ia
®
•
THE ADMITTANCE MODEL A N D N ETWORK C ALCU LAT I O NS
� Ib
([)
•
Zc
Zb
'----./
ZM l
®
PC
ZM 2
®
FI G U R E 7.8
Three branc hes w i t h
b ra n c l l L: s 1I a n d h a n d
7.M 2
m u t u a l co u p l i n g 2M I b c twL:L:o
b c tWL:L: 11 brallchcs (I , l i l t ! ( ,
The new diagonal e l e m e n l r e prese n t i ng node CD, for instance, i s t h e s u m o f
the four clements ( -j6.25 - j6.25 + j3.75 + j3.75) in rows G) and co l u mns CD of
the previous matrix. The three nodal admittance equations in vector-matrix form
are then written
[
-j6 .25
j3 .75
j2.50
j3 .75
-j6.25
j2.50
where VI ' V2 • and V3 are the voltages at nodes CD, (1), and G) measured with
respect to reference, while 11 ' 12 , and 13 are the external currents injected at the
respective nodes.
As in Sec. 7. 1 , the coctl1cient matrix or the last equation can be combined
with the nodal adm ittance matrices of the other branches of the network In
order to obtain the nodal admittance matrix of the entire system.
For three or more coupled branches we follow the same proced ure as
above. For example, the three coupled branches of Fig. 7.8 have primitive
impedance and admittance matrices given by
( 7 .2 1 )
in the Z-matrix arise because branches b and c are not directly
coupled. For all nonzero values of the current l a of Fig. 7 .8, branches b and c
are indirectly coupled through branch a , as shown by nonzero YM 3 df the
primitive admittance matrix.
The zeros
7.3
AN E Q U I V A LENT A D M ITT A N C E N ETW O R K
25 1
Therefore, to form Y b u s for a ne twork which has mutually coupled
branches, we do the following in sequence:
1. I nvert the primitive impedance m atrices of the network branches to obtain
the
2.
3.
corresponding p r i m i t ive admittance m a t rices. A s i ng le branch h a s a
1 x 1 matrix, two mut u a l ly coupled branches have a 2 X 2 matrix, t hree
mutually coupled branches h ave a 3 x 3 matrix, and so on.
M u l tiply the eleme nts o f e a c h primitive admittance matrix by the 2 X 2
b u il d i n g b l oc k m a t r i x .
] ,abe l the two rows a n d t h e two col u mns of e a c h diagonal building-block
m a t r i x w i t h t h e e n ei - n o d e n u mb e r s of t h e co r re s p o n d i n g self-admittance. For
I1l L I l u , t l l y c o u p l e el h r a n c i l e s i t i s i m p or t ll n t t o 1 ,I h e l in t h e o r ei e r o f t he marked
-
( do / l cd ) - IIICI l - lll l ! / l(/ril. ('(/
m , l t rix w i t h
w i t h t h e row 1 , l b e l s ,Issignecl
t h e c o l u m l l s co n s i s l e n l w i t h t h e co l u i l l n 1 ; l h e l s o r D ) .
4 . La b e l
I h e t w o rows o r
n u m hers a l igned lind
5.
( line/oiled ) n o d e n u m b e rs .
e ; l c h ojJ'-dia.£.!,ollo/ b u i l d i n g - b l oc k
c() n s i s t e n t
in (3);
node
b
then l a el
C o m b i n e . b y a d d i n g t og e t h e r , t h o s e c l e m e n t s w i t h ident ical row a n d col u m n
l a b e l s t o o b t a i n t h e nod a l a d m i t t a nc e m a t r ix o f t h e overall network. If o n e of
t h e n o d e s encounte reci is the refe rence node , omit i ts row and col u m n to
o b t a i n t h e sys t e m Y I "" .
7.3
A N E Q U IVALE I\'T A D M I TTA N C E
N ETWO RK
\Ve h ave d e monst r a t e d
how to write the nodal admittance equations for one
b r a n c h () r a n u mber of mu t u a l ly coupled branches which are part of a l arger
n e t w o r k . W e n o w s h o w t h a t such equ a t ions can be i nterpreted as representing
a n e q u i va l e n t a dmi t t a n c e network with no m u t u a l l y coupled elements. This m ay
b e useful w h e n fo r m i n g Y b u s for an origi nal network having m utually coupled
d e s c r i b e d i n t e r m s of
n o d e vo l t a g e s a n e! ,I d m i t t a n c e s by Eq . 0 . 1 <)). Fo r e x a m p l e , t h e equation for the
c u r r e n t Jill ,I t n o d e CD i s g i v e n b y t h e f i rs t row o f Eq . (7 . 1 <)) a s fol l ows:
clements.
T h e c u rr e n t s i nj e c t e d i n t o t h c n od e s o f F i g .
7.0
,I rc
( 7 .22 )
A d d i n g tl n d s u b t r a c t i n g t h e
c ( ) m b i ll i n � t e r m s w i t h c o m m o n
e q u Ll t i o n �d l
noLle CD
t h e r i g h t - h a n d s i d e of
coc fllc i c n ts , w e o b t a i n t h e
t e r m YM V:lI
on
Eq. (7.22) and
K i r c h h o ff s c u r re n t
'
( 7 .23)
I�n p
IlIlil
"
,
:.. ;:;:
252
Aq ,
CHAPTER 7 THE ADM ITfANCE MODEL A N D N E TWORK CALCULATIONS
@)
Im p-
Yo Im n
i
®
®
®
Ya
@
(a)
- YM
®
- Ill
-- 'IHII
Ip Q
t
/,
FIGURE 7.9
(c)
@)
®
-Y
M
1m-
In-
YM
®
@)
It
...-
In
Y
In p
- YM
q
(6)
®
-
1,,
®
-
®
-+-- 1 1 1
Ya
Yb
®
t
Yr,
-Y
M
®
-+-- I
q
Cd)
;'
Developing the nodal admittance network of two m u tually coupled branches.
The double subscripts indicate the dire ctions of the currents 1m , Imp ' and Imq
from node ® to each of the other nodes @ , ® , and ® , respectively, of Fig.
7.9(a). A similar analysis of the second and third rows of Eq. (7. 1 9) leads to
equations for the currents In and Ip in the form
I
I
( 7 . 24)
( 7 . 25 )
and these two equations represent the p a rtial networks of Figs. 7.9( 6 ) and ( c ) .
The fourth row of Eq. (7. 1 9) does not yield a separate partial network because it
is not independent of the other rows. Combining the three partial networks
without duplicating branches, we obtain an equivalent circu it in the form of the
lattice network connected among nodes @), @ , ® , and ® in Fig. 7. 9 ( d).
This l attice network has no mutually coupled branches, but it is equivalent in
every respect to the two original cou p led branches of Fig. 7.6 since it sa tisfies
7.�
AN
EOLJ I VALENT ADM rITANCE N ETWO RK
253
®
F I G U RE 7 . 1 0
0) .
T h e n o d a l a o m i l l , m ce n e t work o f two m u t u a l ly cou pled branches c o n n ected to nodes
and
([0 . ® ,
Eq . ( 7 . 1 9). Accord ingly, t h e stand ard rules of circuit analysis may b e applied to
the c q u iv a l c n t . For e x a m p l e , if the two co u p l e d branches are p hy s i c a l l y con­
nect ed among three independent nodes as in Example 7.2, we m ay regard nodes
® and ® of Fig . 7.9(d) as one and the same node, which we simply join
together, as shown in Fig. 7. 10. The th ree-bus equivalent circu i t of Fig. 7 . 1 0
t h e n yields the nodal equations for the original branches.
Thus, each physical branch or mutually coupled pair gives rise to an
equivalent admittance network to w h i ch the usual rules of circuit analysis a pply.
The fol lowing example i l l ustrates the role of the equivalent circuit in forming
You s '
CD G)
Exa m p l e 7 . 3 . R e p l a c e b r a n c h es b <I n d c b e t w e e n node-p a i rs
and
o f F i g . 7 . 5 by t h e m u t u a l l y co u p l e d b r a n c h e s o f F i g . 7 . 7 . The n , fi n d Ybus
W- G)
-
a n d the nodal e q u a t i o n s of t he n e w ne twork.
So/ulion . T h e
a d m i t t a n ce
d i agra m
or
t ile
n ew
n e twork
c ( ) l I r l i n g i s s l w w n i n f i g . 7 . 1 1 . F r o m E X l I m p l c 7.2
c o u D l e d b r a n c h e s have t h e
n o d a l a d m i t t a nce m a t r i x
m[
CV
Q)
CD
-j 6 . 2 5
G)
j3 .75
we
mutual
k n ow t h a t t h e m u t u a l ly
G)
j2 5 0
j3 . 7 5
-j6 .25
j2.50
j2 .50
j 2 50
-js .OO
.
i n cl u d i ng t h e
j
w h i c h corresponds t o t h e e q u iva l e n t c i rcu i t shown e n c ir c l e d i n Fig. 7 . 1 2. The
rem a i n i n g portion of Fig. 7 . 1 2 is d r a w n from Fig. 7.5. S i n c e m u t u a l coupl i n g is n ot
evi d e n t i n Fig. 7 . 1 2, w e m ay a p p l y t h e s t a n d a r d rules o f Y hu s for m a t i o n to t h e
,
254
CHAPTER 7
THE ADMITTANCE MODEL A N D NEnvORK CALCULATIONS
-j8.0
+
1 . 00/- 900
-)0.8
.
jS
-
.
O
�
+
0.68/- 135'
®
FIGURE 7. 1 1
Per-unit a d mi t t a nce d i agram [or Examp l e 7 . 3 .
@
1 . 00 /- 90°
-jO.8
®
0.68/- 135'"
FIGURE 7.12
Nodal admittance ne twork for Example
branches connected betwee n buses CD,
@,
7.3.
The shaded portion represents two m ut u a ll y cou pled
and Q) .
7.4
.•
MODIFICATION O F Ybus
255
overall network, which l eads to the n'odal admittance equations
CD
-j16.75
jl 1 .75
j2.S0
j2.S0
G)
(1)
G)
0
@
jl 1 .75
-j19 .25
j2.S0
jS .oo
®
j2 50
j2.50
-jS .80
0
yield
7.4
=
j2.50
.
N o t e t h a t the two a d m i t t a nces between
Yl 2
0
jS .OO
0
-j8.30
n o d es
CD
and
VI
V2
V3
V4
(1)
0
0
/ - 900
0.68 / - 1350
1 .00
combine in p a rallel to
- ( -j:' .75 - j8.00) = j l 1 .75
MODIFI CATION OF Y bus
The b uilding-block approach and the equivalent circuits of Sec. 7.3 provid e
import a nt insights into the manner in which each branch self- a n d mutual
admi ttance contributes to the entries of Ybus and the corresponding equivalent
nehvork of the overall system. As a result, it is clear that Ybus is merely a
systematic means of combi n in g the nodal admittance matrices of the various
nehvork branches. We simply form a large array with rows and columns ordered
according to the sequence in which the nonrefere nce nodes of the n e twork are
n u mbered, and within it we combine e ntries with matching l abels drawn from
the nodal admittance matrices of the i ndividual branches. Consequently, we can
easily see how to modify the system Y bus to account for branch a dditions or
other changes to the system n etwork. For instance, to modi fy Ybus of the existing
n etwork to reflect the addition of the branch admittance betwee n the n odes @)
and ® , we simply add Ya to the elements Ymm and Ynn of Ybus and subtract Ya
from the symmetrical elements Ymn and Ynm• I n other words, to i ncorporate the
new branch admittance Ya into the network, we add to the existing Yh u s the
change matrix L1Yhus give n by
L1Y h u s =
( 7 .26)
Aga in, we recognize 6.Vhlls as a storage matrix with rows and columns marked @)
a n d ® . Using Eq . (7.26), we may change the admittance value of a single
branch of the network by adding a new branch between the same e nd nodes @)
a n d ® such that the para l l el combination of the old and new branches yields
the desired value. Moreover, to remoIJe a branch admittance Ya already con­
n ected between nodes ® a n d ® of the network, we simply add the branch
admittance Ya between t h e same nodes, w hich amounts to s4btracting the
-
256
CHAPTER
7
THE A D M ITTANCE MODEL AND N ETWO RK CALCULATIONS
elements of t1Ybus from the existing YbuS' Equation (7.20) shows that a pair o f
mutually coupled branches ,can b e removed from the network by subtracting the
entries in the change matrix
@J
Ya
- �l
YM
- YM
@
®
t1Y bus =
®
®
®
®
-Y
®
- YM
YM
- Yh
l';\1
- YM
Yh
a
�l
- Y""I
Y" ,
-
Y"
( 7 .27)
Y"
from the rows a n d co l u m ns o f Yhus co r respo n d i n g to t he e n d nodes @ , ® ,
® , a n d @ . O f c o u rs e , i f o n l y o n e of t he two m u t u a l l y co u p l e d b r a n ch e s i s t o
be removed from the network, we cou ld fi r st remove all t he e n t r i e s for t he
mutually coupled pair from Y bus using Eq. (7 .27) and then add t he ent ries for
the branch to be retained using Eq. (7.26). Other strategies for mod i fy i n g Y b u s to
reflect network changes become clear from the i n sigh ts developed i n Secs. 7 . 1
through 7.3.
Exa mple 7.4. Determine t h e b u s a d m i ttance matrix of the ne twork of Fig.
removi n g the effects of m u t u a l cou p l i n g from Y bus of Fig.
Solution.
found
in
The Y bus for the e n tire sys t e m of Fig.
Exa m ple 7 . 3 to be
CD
CD
-j1 6 .75
j 1 1 .75
CV
Y bus =
j2 .50
G)
j2.50
@
7. 1 1
0
j 1 1 .75
- j 1 9 .25
j2.50
j5 .00
7. 1 1 .
7.5
by
i n cl u d i n g mu tual coupl i n g is
0
j2.50
j2.50
-j5 .80
0
@
j2 .50
j5 .00
0
-jo . J O
T o r e m ov e completely t h e e ffe ct o f m u t u a l cou p l i n g from the n e twork YlJus , we
proceed in two steps by ( a) fi rst remov i n g the two m u t u a l l y coupled br<mches
a l together a n d (b) then restori n g each of the two bra nches w i thout m u tu a l
cou p l i n g betwee n t h e m . ( a ) To r e m ove the two mutually coupled branches from
the network, we s u b tract fr o m the syst e m Y bus t h e e n tries i n
....
" .
,
i:
. l-ll r
. �.
;" �
AYb u s• I =
CD
CV
0
@
CD
-j6 .25
j3 .75
j 2 .50
correspo n d i n g to the encircled portion of
cv
j3 .75
-j6 .25
j2.50
F i g . 7. 12.
G) , @
j2 .50
j2 .50
-j5 .00
7.5
(b) Now we must
257
recon nect to t h e network the u n c oupled branches,
of
1
- j4.0 p e r u n i t . Accordingly, to reconnect
each
w h i c h h a s an admitta nce ( jO.25 ) the b ra n c h between nodes
6Y bus
THE N ETWO R K I NCI D ENCE MATRIX A N D "bus
')
CD
=
and
=
G), we a d d to Y
CD CD
W- I
C1)
G) - 1
@
6Y hu s . J
=
CD
W
G)
@
the change m a trix
G) 8)
-1
( -j4 .0)
1
a n d s i m i l a r l y fo r t h c b r a n c h b c t w e e n nodes
CD
bus
G)
and
G)
we a d d
G) @ @
-1
- 1
( -j4 .0)
1
A p prop riate l y' s u b tract i n g a n d a d d i n g t h e t h ree c hange m a trices a n d t h e o r i g i n a l
Y b u s g ive t h e new b u s a d m i t t ance m a tr i x for t he u n coupled branches
Y hus ( new)
=
CD
G)
G)
@
CD
- j 1 4 .S
G)
@
j2.S
j8 0
j4 .0
j4 .0
jS .O
j4 .0
-j8 .8
0
j8.0
-j 1 7 . 0
j 2 .S
jS .O
j4 .0
@
.
0
-
j8 3
.
w h i c h a g r e e s w i t h E xa m r l c 7 . 1 .
T H E N ETWO R K I N CI D EN C E M ATRIX
A N D Yh u �
7.5
In S e e s . 7. 1 and 7 . 2 nodal ad mittance equations for each branch and m utually
coupled pa i r of b ra n c h e s a re d e r i ved i n d e p e n d e n t ly from those of other branches
in the network. The nodal admittance matrices of the individual branches are
then combined together in order to bu i l d Y hu s of the overall system. Since we
now understand the process, we may p roceed to the more formal approach
which t reats all the equat ions of the system simultaneously rather than sepa­
rate ly. We will use the example system of Fig. 7 . 1 1 to e stablish the general
procedure.
Two of the seven branches in Fig. 7 . 1 1 are mutually coupled as shown.
The mutually coupled p a i r is characterized by Eq. (7 . 14) and the other five
,
258 .
THE ADMI1TANCE MODEL AND N EnvORK CALCULATIONS
CHAPTER 7
by Eq. (7.4 ). Arranging the seven branch equations into an array
fonnat, we ob t a i n
branches
-jO .80
-j6 .2S
Va
fa
Ve
Ie
Vb
j3 .7S
j3 .7S -j6 .25
V:,
-j8 .00
-j5 .00
-j2 . S 0
-jO .80
fb
Ia
Vc
Ie
Vf
If
Ig
VR
( 7 .28)
The coefficient matrix is the p r i m i tive admittance m a trix fo rmed by in sp e c t i o n
o f Fig. 7.11. Each branch of the network contributes a diagonal entry equal to
the simple reciprocal of its branch impedance except for branches b and c ,
which are mutually coupled and h ave entries determined by Eq. (7. 1 3). For the
general case Eq. '(7 .28) may be more compactly written in the form
( 7 . 29)
where Vpr and I pf are the respective column vectors of branch voltages and
cu rre nts , while Ypr represents the primitive admittance matrix of the network.
The primitive equations do not tell how the branches are configured within the
network. The geometrical configuration of the branches, called the topology , is
provided by a directed graph, as s hown in Fig. 7.l3( a ) in which each branch of
the network of Fig. 7. 1 1 is represented between its end nodes by a directed l ine
®
c
d
f
e
g
'to
•
"
ea)
®
@)
®
a
( b)
tree branch
link
The linear graph for Fig. 7. 1 1 showing: (a) directed-line segments for branches; (b) branches ,a, b,
'and f define a tree while bra nches d, e, a n d R are lin ks.
, FIGURE 7.13
c,
7.5
TH E NETW O R K I NC I D ENCE MATRI X A N D Ybus
259
segment with an arrow in the direction of the b ranch current. When a branch
con nects to a node, the branch and node are said to be incident . A tree of a
graph is formed by those branches of the graph which i nterconnect or span all
the nodes of the graph w ithou t forming any closed path. In general, t here a re
m any possible t rees of a network since different combinations of branches can
be chosen to span the nodes. Thus, for example, b ra nches a , b , c, and f in Fig.
7. 1 3 ( 6 ) define a tree. The remain ing branches d, e, and g are called links, and
w hen a l ink is added to a tree, a closed path or loop is formed.
A graph may be described in terms of a connection or incidence ma trix . Of
particular interest is the branch -to-node incidence malrix A, w hich h as one row
for each branch and one col umn for e a c h n o d e with an entry a ij in row i a nd
co l u m n j �Icco rd i ll g to t h e fo l l o wi n g r u l e :
"
(J
o
I)
-1
i f b r:l I1 c h i i s
not con nected to n od e CD
if curr e n t in branch i is directed away from node CD ( 7 .30)
i f c u rr e n t i n bra n c h i i s directed toward node 0
Th is r u l e for m a l izes fo r t h e network as a whole the procedure used to set up the
coefficient matrices of Eqs. (7.6) and (7. 1 5 ) for the individual branches. In
netwo rk calculations we usually choose a reference !lode. The column corre­
spondi ng to the reference nod e is then omitted from A and the resultant matrix
is denoted by A. For example, choosing node ® as the reference in Fig. 7 . 1 3
and invoking the rule of Eq. (7.30), w e obtain the rectangular branch-to-node
m atrix
CD
a
A =
d
c
e
1
-1
1
0
0
- 1
0
1
U
0
-1
-
I I
g
0
0
0
b
Q) G) @
I
- 1
0
J
()
0
()
0
0
0
0
1
(7.31)
1
1
ca l l ed independcn t nodes
or
buses , and when we s a y t h a t the network has N b u ses, we generally mean t hat
there are N independent nodes not including the reference node. The A m a trix
has the row-column dimension B X N for any n e twork with B branches and N
nodes excluding the reference. We note that each row of Eq. (7. 3 1 ) has two
nonzero entries which add to zero except for rows a and g , each of w h ich has
T h e I W ll r e fc r e n c e n o d e s o r
,\
n e t wo r k ; I r e o rt e l l
260
CHAPTER
7 THE ADM I1TANCE M O DEL AND N ETWORK CALCULATIONS
This is because branches a a nd g of Fig. 7 . 1 1 have one
con nected to the reference node for which no column is shown.
The voltage across each branch may be expressed as the difference in i ts
end-bus voltages measured w i t h respect to the reference node. For example, i n
Fig. 7. 1 1 the voltages a t buses CD , (1) , (1) , and @ with respect to reference
node ® ·a re denoted by VI ' V2 , V3, and � , respectively, and so the voltage
drops across the branches are given by
o"n ly one nonzero entry.
end
V = V1
Vb = V3 - V2
v:: = V3 - VI
Vd = V2 - V I
� = V4 - V2
Vf = V - V
Vg = V4
/J
V:,
�)
V:.
V:,
•
4
or
Ve
()
=
Vg
()
0
-1
-1
0
1
-1
0
0
-1
0
0
1
0
0
0
1
-1
V
I
1
()
1
0
1
0
VI
0
V3
0
1
V2
V4
in which the coefficient matrix i s the A matrix of Eq. (7.31). This is one
illustration of the general res u lt for any N-bus network given by
Vp r = AV
( 7 .32 )
Vpr is the B X 1 column vector of branch voltage drops and V is the
1 column vector of bus vol tages measured w ith respect to the chosen
reference node. Equations (7.6) a nd (7. 15) are p a r t icu l a r a p p l i ca t i o n s of Eq .
(7.32) to i ndividual branches. We fu rther n o t e t h a t K i r c h h o ff' s cu rrent law at
nodes CD to @ of Fig. 7.1 1 yields
where
N
X
fa
0
-1
1
-1
1
0
0
0
0
/
0
0
-1
1
-1
-1
0
0
1
0
0
0
0
0
0
1
1
1
/
0
III
Ie
Id
Ie
I
f
Ig
0
-
0
13
14
1 .00 - 900 and 14 = 0 . 6 8 - 1350 are the external currents in­
j ected at nodes (1) and @, respectively. The coeffi cient matrix in this equation
'is AT. Again, this is illustrative of a general result applicable to every elec;rical
:network since it simply states, in accordance with Kirchhoff's current law, that
the sum of all the b r a n c h c u r r e n ts i nc i d e n t to a node of the network equals the
where
13 =
7.5
Ti l L::
N L::T WO R K I N C l D t..:: N C t..::
M ATR I X A N D y�u,
26 1
i njected curre nt at the node. Accord i ngly, we m ay write
A I =I
T
( 7 .33)
pr
where I p r i s the ' B X 1 col u m n vector of branch currents and I is the N X 1
col u m n vector w ith a nonzero entry for each bus with a n external current
source . Equations (7.5) and (7. 1 6) are pa rticular examples of Eq. (7.33).
The A matrix ful ly d escribes t he topology of t h e n e twork and is indepe n­
dent of t h e particular values of t h e branch parameters. The l atter a re suppli ed
by t h e primitive adm ittance rn a trix. There fo re, two dlff eren t network configura­
t i o n s e m p l o y i n g the s a m e b r anc h e s w i l l have d i ffere n t A matrices but the same
Y rJr" On t h e o t h e r h a n d , if c h a n g e s occ u r in b r a nc h p a r a m e t e rs w h i l e mai ntain­
i n g the s a m e n e t w o r k co n fi g u r a t i o n , o n l y Y p r i s a l t e re d b u t not A.
M u l t i p l y i n g E q . ( 7 . i. lJ ) b y A , w e o b t a i n
( 7 .34 )
The rig h t - h a n d side
( 7.32), we
n nd
of Eq . (7.34)
equ als
I
and
s u b s t i t ut i n g fo r Vpr from
Eq.
�.
( 7 . 35)
W e m y \v r i t e
a
where
the N
Eq. (7 . 3 5 )
X
N
in
t h e more concise form
bus admitt ance matrix
Ybus
=
N x N
Y"us h as
so
of
A
Ybus
fo r
( 7 . 3 6)
the
A
T
o n e ro w l l m l o n e c o l u m n fo r c , \ c h o f
t hc N h uscs
t h c s t a n u a rd fo rm o f l h c fo u r i ll ll c [) c n d e n l e q u a t i o n s 0 ["
Fig.
7 . 1 1 is
YI I
Y2 1
Y:; J
Y<1 1
YI 2
Yn
Y:;2
Y42
Yu
Y2J
Y:n
Y4 1
YI 4
Y2<1
Y-, 4
Y4 4
V2
12
11
VI
V:;
V4
( 7 . 37 )
B x N
B x R
N x R
system is given by
=
I :;
and
t h c exa mple system
i n t h e n e t work,
( 7 . 38)
14
four u n knowns are the bus voltages VI ' V2 , V3 , and V4 when the bus-in­
j ected c u rre n t s I I ' 12 , '-:" a nd 14 are specified. Generally, Y pr is symmetrical, in
w h i c h c a se t a k i n g the transpose of each s i de of Eq. (7.37) shows that Yhlls is also
Th e
sy m m e t r i c a l .
262
CHAPTER 7
THE A DM ITTANC E
MODEL AND NETWORK CALCU LATIONS
Example 7.5. Determine the per-unit bus admittance matrix of the example system
of Fig. 7 . 1 1 using the tree shown in Fig. 7.13 with reference node ® .
The (primi tive a dmittance matrix Ypr describing the branch admittances is
given by Eq. 7.28) and ) the branch-to-node incidence matrix A for the specified
tree is given by Eq (7.31 . Therefore, performing the row-by-column mUltiplications
for AT Yp r indicated by
Solution.
.
0
0
-1
0
-1
0
0
-1
0
-1
0
0
-1
1
1
T
0
0
0
0
1
1
1
0
0
0
0
- jO.S
- j6 .25 j3 .75
j3 .75 -j6 .2S
-jS.O
-j5 .0
-j2 .S
- jO .S
we obtain the intermediate result
AT ypr
=
0
0
- jO .8
0
-j3 .75
j6 .25
- j2.5
0
j6 .25
-j3.7S
- j2.5
0
which we may now p ostm u lt i pl y by
Ybus
=
T
A Ypr A
=
Q)
@
0
@
A
jS.O
- jS.O
0
0
jl 1 .75
-j 19 .25
j2 .50
j5 .00
0
0
- j O .S
to calculate
CD
-j 1 6 .75
j 1 1 .75
j2.50
j2.S0
Since currents are injected at o n ly buses
m atrix form are wri tten
- j 1 6 . 75
j 1 1 .75
j2.50
j2.50
0
j2 .5
0
j 0
- 2 .5
0
jS .O
0
-j5 .0
j 2 .50
)2.50
- j5 .S0
0
0
j 1 1 .75
-j19 .25
j2 .50
jS .oo
G)
j2.50
)5 .00
0
- jS .30
and
G)
@
j2.50
jS .oo
0
-j8 .30
j2.S0
j2.S0
-j5 .S0
0
@,
VI
V2
V3
V4
the
I
nodal e qu a tions
In
0
0
1 .00/ - 900
0.68 / - 1 350
Example 7.6. Solve the node equations of Example 7.5 to find the bus vol tages by
inverting the bu s admittance mat rix.
Solution. Premultiplying
both sides of
the m atrix nodal equation
t h e bus admittance matrix (determin ed by using a standard
by the inverse of
progr a m
on a com puter
7.6
THE M ETHOD OF SUCCESS I V E ELIMI NATI O N
263
or calculator) y i e l d s
VI
V2
V3
=
V4
jO.73 1 28
jO . 69 1 4 0
j O . 6 9 1 40
j O . 7 1 966
j O . 6 1 323
jO . 60822
jO.63677
j O . 64 1 7 8
j O . 6 1 323
j O . 6 0 822
jO . 69890
j O . 55 1 1 O
jO .64 1 78 jO.55 1 1 O
jO .69890
jO .63677
! - 1 8.41 1)9°
0 . 96734 / 0 _ 99964/ - 1 5 .37 1 8°
. 9 4 6�
1 .00/ - 90°
0
0
0 .68/ - 135°
Pe rform i n g the i n d i c a t e d m u l t i p l i cations, we o b t a i n the per-u n i t resu l t s
0 . 96903
I H .6()28°
0
7.6
86
2 0 . 74 ()6°
=
I
lI
- j O .30859
0 . 96388 - jO . 2 6 49 9
0.91 939 - j O .3 0 6 1 8
0 . 9 1 680
0 . 887 1 5 - jO .3360S
T H E M ETHOD OF S U C C E SS IVE
ELI M I N ATI O N .
In industry-based studies of power systems the networks being solved a re
geographically extensive and often e ncompass m any hundreds of substations,
generating plants and load centers. The Y b u s matrices for these l arge netw orks
of thousands of nodes have associated systems of nodal equations to be solved
for a correspondingly large numbe r of u nknown bus voltages. For such solutions
computer-based numerical techniques are required to avoid direct matrix inver­
sion, thereby minimizing com putational effort and com puter storage. The
m ethod of successive elimination, called gaussian elimina tion , under! ies many o f
the n umerical methods solving the equations of such large-scale power systems.
We now describe this method using the nodal equations of the four-bus system
( 7 .39)
( 7 . 40 )
(7.41 )
of
( 7 .42)
reduci ng t h is system of fou r equations i n
t h e fo u r u n k n ow n s V I ' V2 , V I , <l n d VI t o , \ sys t e m o r t h re e eq u a t io n s i n t h ree
unknowns and then to a reduced system of two equations in two u nknowns, a n d
so on u n t i l there remains o n l y o n e equation in o n e u nknown. The fi n a l equation
yield s a value for the corresponding u nk nown, which is then substituted back in
the reduced sets of equations to calcu late the remaining u nknowns. The
successive eliminat ion of unknowns in the forward direction is ca\l�d forwa rd
elimination and the subst i t u t ion process using the latest calcu l ated values is
Gaussian elimin ation consists
264
CHAPTER 7
THE ADMITTANCE MODEL AND NETWORK CALCULATIONS
called back substitution . Forward e limination b egins by selecting one equation
and eliminating from this equation one variable whose coefficient is cal l ed the
pivot. We exemplify the overall procedure by first e liminating V I from Eqs.
(7.39) t hrough (7.42) as follows:
Step 1
1.
Divide Eq. (7 .39) by
t h e p ivo t Y I I l o
obt a i n
1
Y1 I
-
2.
[
( 7 .43 )
1
Multiply Eq. (7.43) by Y2 1 , Y3 1 , and Y4 1 and s u bt ract the resu l ts from Eqs.
( 7.40) through (7.42), respect ively, to get
( 7 . 44 )
(7 .45)
( 7 .46)
Equations (7.43) thro ugh (7 .46) may
be
w r i t t e n more compactly in t h e form
( 7 .4 7)
Y22(l)v2 +
y(32l ) v2 +
y(23I) V3
+
y(1)V
33 3
+
( 7 .48 )
y(I)V
24 4 I(1)
2
y(1)V
34 4 -- I(3 1 )
=
( 7 .49)
' ,'
,
( 7 .50 )
',wh ere the superscript denotes the Step 1 set of derived coefficients
Yjk(l) - Yjk
_
_
1] 1 Yl k
Yl l
for j and k
=
2
,
3 4
,
(7 .5 1)
265
T H E METHOD OF S U CCESS I VE E L IM I N AT I O N
7.6
and t h e modified right-hand side expressions
1� 1
)
)
=
I.
-
J
Y,
�J� J
YI I
for j =
I
2, 3, 4
( 7 .52)
Note that Eqs. (7.48) through (7.50) may now be solved for V2 , V3 , and V4 since
VI has been eliminated; and the coefficie n ts constitute a reduced 3 X 3 m atrix,
which can be interpreted as representing a reduced equivalent network with bus
CD absent. In this th ree-bus equivalent the voltages V2 , V3 , and V4 h ave exactly
the same values as in the original fou r-bus system . Moreover, t he effect of the
current injection l i on the network i s taken into account at buses W , ®, and
@ , a s shown by Eq. (7.52). The current I I at bus CD is multiplied by the factor
�- I /YI I before being distributed , so to speak, to each bus CD still in the
network.
\Vc nex t consider t h e e l i m ina t io n or t h e v a r ia b l e V .
-
Step 2
1 . D iv i d e
Eq.
(7.48)
by
the
new p ivot Y2ci l to
obtain
-ym 1
22
1
2.
2
C)
( 7 .5 3 )
Multiply Eq. (7.53) by Y§i l and Y4�1 ) and subtract the results from Eqs. (7.49)
and
(7.50)
In a manner
t o g et
similar t
o that of
S t e p 1,
we rewrite Eqs. (i.53) through (7.55) I n
t h e fo rm
V
2
y(l)
-,
y (2I )
2
")
+
y(l)
--'.\
VJ + -----'0l V4
Y2 2
y33( 2 ) V3
+
y34(2 ) V4
y4(2)
:1 V3
+
y4(42 ) V4
I
=
=
=
J(
2
I
)
(
Y
22
1 ( 2)
3
](2)
4
I)
( 7 .5 6 )
( 7 57 )
.
(7 58 )
.
266
CHAPTER 7 THE ADMITTANCE MODEL A N D N ET WO R K CALClJLATIONS
where the second set of calculated coefficients is given by
Yjk(2 )
_
-
y(1)
_
jk
YJ·2(1) y2(kl )
yii)
and the net currents injected at buses
for j and k = 3 , 4
( 7 .59 )
G) and @ are
for j
=
3, 4
( 7 .60)
Equations (7.57) and (7.58) describe a fu rther r e d u c e d equ ivale n t n e twork
ha vi ng only buses Q) a nd @ . V o l t ages V] a n l! V .I a re exactly t h e s a m e a s t h o s e
of the original four-bus network because the c u rr e n t i nject ions If) a n d I.f 2)
re p re s e n t the effects of al l t h e o r i g i n a l c u rr e n t sources.
We now consider el imination of the variable V3 .
Step 3
1.
Divide Eq. (7.57) by the pivot Y1P to obtain
1
y ( 2 ) 13(
33
2.
( 7 .6 1 )
2)
Multiply Eq. (7. 6 1 ) by Y4�2) and subtract t h e result from Eq. (7.58) t o obtain
y44( 3 ) V4
in which we have defined
y(443 )
=
y(2
44 )
_
y4( 2 ) y(342 )
3
y(2)
=
14( 3 )
( 7 .62)
and
( 7 .6 3 )
33
Equation (7.62) describes the single equivalent branch admittance YN) with
voltage V4 from bus @ to reference caused by the equivalent injected current
/(3)
4
.
The final step in the forward elimination process yields
V" .
Step 4
1.
Divide Eq. (7.62) by ��) to obtain
( 7 .64)
At this
p o i nt
we have
fou n d a va l u e ror bus
vo l t age VI which
C ; 1 1l
be s u b s t i t u t ed
267
7.6 THE METHOD OF SUCCESSIVE ELIM INA TION
back in Eq. (7.6 1 ) to obtain a value for V3• Continuing this process of back
substitution using the values of V3 and � in Eq. (7.56), we obtain V2 and then
solve for V I from Eq. (7.47).
Thus, the gaussian-elimination procedure demonstrated h ere for a four-bus
system provides a systematic means of solving large systems of equations
without having to invert the coefficient matrix. This is most desirable when a
largc:-scale power system is being analyzed. The following example numerically
illustrates the procedure.
Exa m ple 7.7. U s i n g g a uss i a n e l i m i n a t io n , solve t h e n o d a l equat ions of Exa m p l e
7 . 5 to find t h e bus vo l t ages. At e a c h step of t h e so l u t i o n h n d t h e equivalent c i r c u it
m a t rix.
of t h e r e d u c e d c o e fl i c i e n t
Solution. 1 n Exa m p l e 7 . 5 t he nodal
t o be
CD
Q)
CD
-j 1 6 . 7S
] j l 1 .75 ]
G)
@
W
j l 1 . 7S
- · j 1 9 .2S
a cl m i t t a n c e
G)
@
j 2 .S0
jS .OO
0 .0 0
I j2 . S 0 ]
j2 .S0
j2.S0
j2.S0
-jS .SO
j2.S0
jS .OO
0 . 00
e q u a t i o n s in m a t r i x form are fou n d
VI
/ - 900
0.68/ - 1350
0
1 . 00
V2
V3
-j8 .30
V4
0
S tep 1
To eliminate the variable VI from rows 2, 3, and 4, we first divide the first row
by the pivot - j 1 6.75 to obtain
VI - 0 . 70 1 49 V2 - 0 . 14925 V3 - 0 . 1 4925 V4
=
0
We now use this equation to eliminate the j 1 1 .75 entry in (row 2, column 1 ) of
Y b u s ' and in the process all the other elements of row 2 a r e modified. Equation
(7.5 1 ) shows the p roced ure. For example, to modify t he element j2.50 under­
l i n e d in (row 2, co l u m n 3 ), s u b t r ac t from i t t h e p roduct of the elements enclosed
by r e c t a n g l e s d i v i d e d by t h e pivot - j 1 6.75; t h a t i s ,
Y(I) =
2.1
Y2 .1 -
Y2 I YI ''\
YI I
= J'/
� . 5() -
S i m i l a r l y, t h e o t h e r c l e m e n t s o f n e w
Yi� )
=
- j 1 9 . 25 -
y24
( l ),
= )'5 . 00 -
j l 1 . 75
row
2
X
j2,SO
= j4 .25373 per unit
:I r e
j 1 1 .75 x j 1 1 .7 5
- - j 1 1 .00746
- j 1 6 . 75
j 1 1 .75 X j2 .50
- j 1 6 .75
=
j6.75373
per unit
per unit
268
CHAPTER
7
THE ADM ITTANCE MODEL A N D NETWORK CALCULATIONS
--,.---..... ®
�-------�--�
®
-jO.373 13
-jO.S
1.00/- 909
0.68/- 13s<'
®
FIGURE 7.14
The equivalent three-bus network following Step 1 of Example 7 . 7
Modified elements of rows 3 and 4 are likewise found to yield
1
- 0 .70149
- 0 . 1 4925
- 0 . 14925
VI
0
-
0
0
0
-j l l .00746
j4.25373
j6 .75373
j4.25 373
-j5 .42686
jO .373 1 3
current
j6 .75373
jO .373 1 3
-j7.92686
is d i s t r ib u t e d
V2
V3
-
V4
0
1 .00/ - 900
0 .68 / - 1 350
bu s
Ii , a n d
t o the remaini ng buses
@ , G)' and @ , and so t h e c u r re n ts Ii ,
Id in the right-hand-side
vector have the same values before and after S tep 1 . The p a r t i t i oned system of
equations involving the unknown voltages V2 , V3, and V4 corresponds to the
three-bus equivalent network constructed in Fig. 7.14 from the reduced coeffi­
cient matrix.
Because
II =
0,
no
CD
from
Step 2
Forward elimination applied to the partitioned 3
tion proceeds as in Step 1 to yield
1
0
- 0 . 70149
1
- 0 . 14925
- 0 .38644
- 0 . 14925
- 0 .6 1356
0
0
0
0
-j3 .78305
j2.98305
j2 .98305
-j3 .78305
X
3 system of the last equa­
VI
0
V2
V3
V4
-
a
�
1 .OO
.
0 .68 - 1350
7.6 THE M ETHOD O F S UCCESS I VE ELI M I N AT I O N
®
-)2.98305
-�L......,--
1 . 00/- 900
t
269
-)0.8
0.68/- 135"
@
FI G U RE 7 . 1 5
The e q u iva l e n t two - b u s n e twork fo l lowi n g S t e p
2
of Exa m p l e 7.7.
in which the i njected currents n 2 ) and / P) of Step 2 a re also u nchange d
because / � I ) = /2 = O. A t t h is stage w e have e l i minated V I and V2 from the
o r i g i n a l 4 X 4 system of equations and t h ere remains the 2 X 2 syste m i n the
variables V3 and V4 correspond i n g t o Y b u s of Fig. 7 . 1 5 . Note t ha t nodes CD a n d
W are elim in a te d .
Step 3
Con t i n u ing the forward e l i m inatio n , we fi n d
1
0
0
- 0 .70 1 49
1
0
- 0 . 1 4925
- 0 . 38644
1
- 0 . 1 4 925
- 0 .6 1 35 6
- 0 .78853
0
0
0
-j 1 .43082
G)
/ - 90°
3 . 7 R3 ()5 / - (jO°
i n w hich the e n t ry for b u s
a n d the modified current at bus
Figure
=
o 68
.
@
/ - 1 35°
1 . 35738
V3
V4
LQ:
0
0
0 .26434
-
1 .35 7 3 8 / - 1 1 0 .7466°
in the right-hand-side vector is calc u lated to be
1 .00
=
VI
V2
=
O . 2()434
�
per u n i t
is given by
-
j2 .98305
/ - 1 1 0 .7466°
-j3 .78305
1 .00
- 90°
L--
per u n it
7 . 1 6 ( a ) shows the s ingl e a d m i ttance resu lting from Step 3.
270
CHAPTER 7
THE ADM rITANCE M O D EL A N D N ETWO R K CALCULAT I O N S
1@ +
.--..--
1 .35 738/- 1 1 0 .7466"
j 1 . 43 0 82
-
1
@)
+
)4
'-------1...---11 @ -
0.94867/- 20.7466
"--�
(a)
(b)
I
FIGURE 7.16
®
1
r
+
The equivalent circuits following ( 11 ) S t e p 3 and ( h ) Step 4 o f Exa m ple 7.7.
Step 4
The forw a r d e l i m i n a t i on process term i nates with the calculation
-
corresponding to the source transform ation of Fig. 7 . 1 6( 6 ). Therefore, forward
elimination l eads to the triangular coefficient matrix given by
1
0
0
0
- 0 . 70149
1
0
0
- 0. 14925
- 0.38644
1
0
VI
V2
V3
- 0 . 14925
- 0 .6 1356
- 0 .78853
1
=
�
0
0
0. 26434
0 .94867 / - 20 . 7466°
�
Since V4 = 0.94867/ 20 .7466° , we now begin the back-substitution p rocess to
determine V3 u sing the third row entries as follows:
-
. V3
- 0 .78853 V4
=
V3 - 0 .74805 /
-
20 .7466� = 0 .26434�
which yie lds
V3
Back substitution for
V3
=
0 . 99965 ! - 15 .37 16°
per
unit
and V4 in the second-row equation
V2 - 0 .38644 V3 - 0 .61356V4 = 0
yields
V2
= 0 .9 6 734L - 1 8 . 6030° per unit
7.7
N O D E E L I M I NATI O N ( K R O N R E D UCfI O N )
271
When the calculated values o f V2 , V3, and V4 are substit ute d in the first-row
equ a tion
V1
we obtain
- 0 .70149 V2 - 0 . 14925 V3 - 0 . 14925 V4 = 0
VI
and s o t he p e r- u n i t b u s
VI
V2
V,
V-1
=
0 .96903 ! - 1 8 .41 89° per u n i t
vo l t ages a rc
= 0 .06903 / - 1 8 .4 1 890
=
=
=--=
O .96734L - 1 8 . 6030°
O .t)l)t)()4/ - J 5 . 3 7 1 ()0
0 . 94867/ - 20 . 7466°
=
=
=
=
0 . 9 1 9 39
- j O . 3 06 1 8
0 . 9 J 6 80 - j O . 3 0 8 5 9
( ) . 963��
- j O .26499
0 . 887 1 5 - j O . 3 3605
w h ich agree a l most exactly w i th t he resu l ts fou n d in E x am p le
7. 7
7.6.
N O D E ELI M I N ATI O N ( KRO N
RE D U CT I O N )
S e c t i o n 7 .6 s h ow s t h a t gaussian e l i m i n a ti on removes the need for m a trix
-
i nve rsion when solving t h e n o d a l e q u a t ions of a l a rg e - s c a l e powe r system. At t h e
s a m e t i me i t i s a l so s h own t h a t e l i m i n a t i o n o f v a riables i s identical t o network
r e d uction s ince it l e a d s to a s e q u e n ce o f re d u ce d o r d e r network equivalents by
node e l im i n at ion at e a c h s t e p . T h i s may b e i m p o r t a n t in an alyzing a l arge
i n t e rc o n n e c t e d powe r system if t h e re is s p ec ia l i n t e rest in the vol tages a t o n ly
sume of t h e b u se s o f t h e o v e r a l l sys t e m . For i ns t a nce, one e l e c tr i c u t i l i ty
com pany \v i t h i n t e rc o n n e c t i o n s to o t h e r c o m r ; l I 1 i cs may w i s h to con fin e i t s s t u d y
o f v u l t a g e l e v e l s t o t h o s e S L l b s t ; l t i o fl S w i t h i n i t s o w n se rv i ce t e rr itory. By se lect ive
n u m b e r i n g of t h e sys t e m b u s e s , w e m a y a p p l y g a u ss i a n elimi nation so as to
rc cJ u ce t h e Y illl, e q u a t i o n s o f t h e ove r a l l sys t e m t o a se t w h ich con t a i ns o n l y
t h o s e b u s vu l t ;l g e s o r s p e c i ; d i n t e r e s t . T h e coe l I i c i e I l t m a t r i x i n t h e red uced-order
se t o f e q u ; l t i u n s t h e n r e p r e s e n t s t h e Y " , , , fo r ;\ ll e q u iv a l e n t n e twork con t a i n i ng
o l l l y t h ose b u s e s w h i c h ; l l C 1 0 he rc t ' l i n c li . A l l o t h er lnl ses Me c l i m i n a t c u i n t h e
mat hematical s e n se t h a t t h e i r b u s v o l t a g e s a n d c u r r e n t injections d o n o t app e ar
expli c i t l y. S u c h r e u uc t i o n in s i z e of the equ n t io n set l e a d s to efficiency of
com p u t a t ion a n d h elps to foc u s more d i rectly on t h at portion of t h e overall
n e twork w h i c h is of p r i m a ry i n tere s t .
I n gaussian eliminn tion o n e bus-voltage variable at a t i m e is seque n t i a l l y
removed from t h e original system o f N e q u a t i o n s i n N un knowns. Fo ll owi n g
S t e p 1 of t h e p roce d ur e , the v a r i a b l e V I d o es not exp l icitly appear in the
272
CHAPTER 7
THE A DMITIANCE M ODEL AND NETWORK CALCULATIONS
resultant ( N 1) X (N 1) system, which fully represents the original network
if the actual value of the voltage Vl at bus G) is not of direct interest. If
knowledge of V2 is also not of prime in terest, we can interpret the (N 2) X
(N 2) system of equations resulting from S tep 2 of the procedure as replacing
the actual network by an (N 2) bus equivalent with buses CD and CV
removed, and so on. Consequ ently, i n our network calculations if i t is advanta­
geous to do so, we may eliminate k nodes from the network representation by
employing the first k steps of the gaussian-elimination procedure. Of course,
the current i njections (if any) at the eliminated nodes are taken into account at
the remaining ( N k) nodes by successive application of expressions such as
that in Eq. (7.54).
Current injection is a lways zero at those buses of the network to which
there is no external load or generating source connected. At such buses i t is
usually not necessary to calculate the voltages explicitly, and so we m ay
eliminate them from our representation. For example, when 1 \ = 0 in the
four-bus system, w e may write nodal admittance' equations i n the form
-
-
-
-
-
-
CD
@
W
@
CD
W
Y2 1
Y3 1
.Y4 1
Y2 2
Y3 2
Yl l
[
CD Yij ,
G) Y3(i)
@ Y.t�)
@
Y1 3
Y23
Y1 2
Y1 4
Y2 4
Y3 4
Y33
Y4 3
Y4 2
and following elimination of node
@
W
Y44
CD , we
W
y(l)
23
@
Y24( l '
y(l)
yell
ye ll
43
0
-
V3
V4
34
44
12
( 7 .65)
13
14
obtain the 3
y<l)
33
]
VI
V2
X
3 system
[�: J ��: ]
=
( 7 . 66)
in which the superscripted elements of the reduced coefficient matrix are.
calculated as before. Systems in which those nodes w ith zero current injections
a re eliminated are said to be Kron \ reduced. Hence, the system having t he
particular form of Eq. (7.65) is Kron reduced to Eq. (7.66), and for such systems
node elimination and Kron reduction are synonymous terms.
Of course, regardless of w hich node has the zero current injection , a
system can be Kron reduced without having to rearrange the equations as in Eq.
nodal equations of the N-bus system, we
example, if Ip 0 in the
, , (7.65). For
.
=
IAfter Dr. Gabriel Kron (190 1 - 1 968) of General Electric Company, Schenectady, NY, who con­
tributed greatly to power system analysis.
7 .7
N O D E ELI M I N ATION (KRON REDUCfION)
273
may d irectly calculate the elements of the n ew, reduced bus admittance matrix
by choosing Ypp as the p ivot and by e liminating bus p using the formul a
( 7 . 67)
where j and k take on a l l the i nteger values from 1 to N except since row p
and column p are to be eliminated . The subscript (new) d istinguishes the
elements in the new Y of d imension ( N - 1 ) X (N - 1) from the elements in
the origin al Y
p
bus
bu s '
Exa m p l e 7.8. Using Y2 2 as t h e in i t i l pivot, e l i m i n a t e n o d e @
corresponding voltage V, from t h e 4 x 4 syst e m o f Example 7.7
a
The i ot Y22 q u l s
may e l i m i n a t e row 2
col u m
Solution.
1
pv
cl ements
Y 1 1 ( n ew)
YI 3(new)
=
=
and
Y1 1 -
YI J -
YI 4( new) = YI 4 -
e a
-j 1 9.25 . With p se t e q u a l to 2 in Eq. (7.67), we
n 2 from Ybus of Exa m p l e 7.7 to o b t a i n t h e new row
Yl 2 Y2 1
-j 1 6 .75
Y22
Y l 2 Y23
Y22
Y 1 2 Y2 4
Y22
= j2 .50 -
=
j2.50 -
-
( j 1 1 .75 ) ( j 1 l . 7 5 )
-
j 1 9 25
-j9 .57792
.
( j l 1 .75 ) ( j2 'sO)
=
-j 1 9 .25
( j l 1 . 7 5 ) ( j5 . 00 )
j4 .02 597
= j5 .55 1 95
-j1 9 .25
-)0.64935
-)5.55195
-j4.02597
1 .00/- 900
CD
-j O.S
jO 8
-
@
Kron -reduced network of Example 7.8.
FI G l IR E 7 . 1 7
The
t
and the
.
,I,
0.68/- 1 350
274
CHAPTER 7
THE ADM I1TANCE MODEL AND NETWORK CALCU LATIONS
Similar calculations yield the other elements of the Kron-reduced matrix
Q) r
G)
@
7.8
CD
-j9 .5 779 1
j4.02597
j4 .02597
j5 .55 1 95
-j5 .47532
jO .64935
G)
r
/ - 90°
0
1 .00
0 .68
/
- 1 35°
1
Because the coefficient matrix is symmetrical, the e q u iv a l e n t c i rcu i t of Fig. 7 . 1 7
applies. Further use o f Eq. (7.67) to e l i m i n a t e node CD from F i g . 7. 1 7 l e a d s t o t h e
Kron-reduced equivalent circuit shown i n Fig. 7 . 1 5 .
TRIANGULAR FACTORIZATI O N
In p,r actical studies the nodal admittance equations o f a given large-scale power
system are solved under different operating conditions. Often in such studies
the network configuration and parameters are fixed and the operating condi­
tions differ only because of changes made to the external sources connected at
the system buses. In all such cases the same Y bus applies and the problem then
is to solve repeatedly for the voltages corresponding to different sets of cu rrent
injections. In finding repeat solutions considerable computational effort is
avoided if a ll the calculations in the forward phase of the gaussian-elimination
procedure do not have to be repeated. This may be accomplished by expressing
Y bus as the product of two matrices L and U defined for the four-bus system by
1
Yl l
L=
Y2 1
y(l)
22
YJ I
yell
32
y (2)
Y.t l
y(1)
42
y ( 2)
U=
:\3
43
Y1 2
Y! 3
Y14
Y1 1
Y1 1
yell
Y1 1
y(l)
24
y(l)
22
y (2)
34
y ( 2)
1
23
y(l)
22
1
( 7 . 68 )
33
y (3)
1
44
The matrices L and U are called the lower- and upper-triangular factors of
Yb u s because t h ey have zero elements above a nd below t h e respective p r i ncipal
d iagonals. These two matrices have the remarkably convenient property that
their product equals Ybus (Problem 7 . 1 3). Thus, we can write
LU = Y bus
( 7 .69)
The process of developing the triangular matrices L and U from Y bu s is called
triangular factorization since Ybus is factored into the product LV. Once Y b us is
so . factored, the calculations in the forward-elimination phase of ga � ssian
7.8
TR I A N G ULAR FACfORI Z ATION
275
e l i m ination do not h ave to be repeated s ince both L an d U are u n ique a n d do
not change for a given Y b u s ' The entries i n L and U are formed by systematically
recording the outcome of t h e calculations at e ach step of a single pass through
the forward -elim ination p rocess. Thus, in forming L and U, no new calculations
are involved .
This is now demonstr ated for the four-b us system w ith coefficient matrix
Y hlls
=
CD
W
G)
@
CD
YI I
Y2 1
Y,\ 1
Y4 1
C1)
Yl 2
Yn
Y,,\ �
Y4 2
G)
Y I :1
Y2:1
YJ :1
Y4:1
@
YI -1
Y�:4
Y ;.1
Y�-1
( 7 .70)
g;ll1 ss i ;l I1 e l i lll i n il t i o n i s ;t p p l i e d to t he fO l l r 110ti ;t i cql 1 <1 t ions co rrespo n d i ng
to t h i s Ybli S ' w e n o t e t h e fo l l owing.
Step 1 y i e l d s res u l ts g iven by Eqs. ( 7.47) t hrough ( 7 .50) in which:
When
The coefficients Y1 1 , Y2 1 ' Y3 1 , and Y4 1 a re e l i m i n a ted from t h e first column of
t h e original coefficient m atrix of Eq. ( 7.70).
2. New coe ffici e n ts 1, Y I 2/Y1 1 , Y U / Y1 1 , and YI 4/YI I are generated t o replace
t hose in the first row of Eq . ( 7.70 ) .
1.
The coefficients in the o t h e r rows and col u mns a re also altered, b u t we keep a
separate record of only th ose specified i n ( 1 ) and (2) si nce t hese are the only
resu lts from S tep 1 w hic h a re neither used nor a l tered i n Step 2 or subsequent
steps o f the forwa rd -el i m i n a t ion p rocess. Col u m n 1 of L and row 1 of U i n Eqs.
(7.68) show the recorded coefficients.
Step 2 y iel d s resu l ts giv e n b y Eqs. (7.56) t h rough ( 7.58 ) i n w h ic h :
Coc fll c i c n t s Y ii >' Y.� i >' <l n d Y4i1 ) (l rc e l i m i n a t e d from t h e second column o f t h e
r e d u ced coe nic i e n t m a trix corresp o n d i n g t o E q s . (7.47) t hrou g h ( 7.50).
2. New coe ffi c i e n ts I , Y�� )/ Yii l, and Yi,; )/ Yii ) a r c gene ra ted in Eq . ( 7.56) for
t h e sec()nd ro w .
1.
These coefflcients a r c n o t n e e d ed i n t he r e m a i n i n g steps o f forward e l i m i na t ion,
and so w e record them as column 2 of L and row 2 of U to s h ow the Step 2
record . Con t i n u i n g t h is record-keep ing proce d ure, we fo rm col u m ns 3 and 4 of
L and rows 3 and 4 of U u s i n g the res u l ts from Steps 3 and 4 of Sec. 7.6.
Therefore, m a t r ix L is s im pl y a record of those col umns w hich a r e
succe ssive ly e l i minated a n d m a t r ix U records t h ose row e n t ries w h i c h are
s u ccessive ly generated at each step of the forw a rd stage of gaussian el imination.
276
CHAPTER 7
THE ADM ITI'ANCE MODEL AND N ETW O R K CALCU LATI ONS
We may use the triangular factors to solve the original system of equations
by substituting the product LU for Y bus in Eq. ( 7.38) to obtain
L
NxN
As
u
NxN
v
I
Nx l
(7.71)
Nx l
an intermediate step in the solution of Eq. (7.71), we
p ro duct UV by a new voltage vector V' such that
L
V'
I
NxN Nx 1
u
and
N x 1
N x N
V
N x !
m ay replace
V'
the
( 7 .72)
.V x 1
Expressing Eq. (7.72) in full format shows that the original system of Eq. (7.38)
is now replaced by two triangular systems given by
Y1 1
Y2 1
y(l)
22
Y3 1
y(l)
32
y (2 )
33
Y4 1
y(l)
42
y ( 2)
43
1
and
Y1 2
Y1 1
1
y (3 )
44
Y1 3
YI 4
YI I
YI I
y (3l )
2
y(l)
22
1
y24( l )
V'1
II
V'2
12
V'3
13
V'4
14
VI
V'I
yO)
22
y3( 2 )
4
y33
(2 )
V2
V3
V'3
1
V4
V'4
-
V'2
( 7 .73 )
( 7 . 7 4)
The lower triangular system of Eq. (7.73) is readily solved by forward substitu­
-tion beginning with V;. We then use the calculated values of V; , V� , V;, and V;
to solve Eq. (7.74) by back substitution for the actual unknowns VI ' V2 , V3 •
� and �.
.
Therefore, when changes are made in the current vector I, the solution
vector V is found in two sequential steps; the first involves fonvard substitution
using L and the second employs back substitution using U.
7.8
Example
277
T R I A N G ULA R FACTOR I ZATION
G)
0.60/ - 1200
7.9. U s i n g t h e t r i a n g u l a r factors of Y nUS ' d e t e r m i ne t h e voltage at bus
of F i g . 7. 1 1 when the c u rre n t sou rce a t b u s
7. 1 1
per u n i t . A l l o t h e r con d i ti o n s of F i g .
Ynu s
Solution . T h e
@
is c h anged to
are u n c h a n g e d .
for t h e n e two rk of Fi g . 7 . 1 1
14
=
is given in Example
7.3.
The
correspond ing m a t r i x L may b e a ss e m b l e d col u m n by col u m n from Example
7.7
s i m ply by reco r d i n g t h e col u m n w h i c h is e l i m i n a t e d from the coe ffi c i e n t m a trix a t
e a c h step o f t h e forw a rd - e l i m i n a t i o n p rocedure. Th e n , subs t i t u t i n g for L a n d t h e
n e w c u r r e n t vector
I
i n t h e e q u a t i o n LV '
=
I , w e ob t a i n
-j I 6 .7 5
j 1 1 .75
-j 1 1 .00746
j2.50
j 6 . 75 3 73
j4 .25373
j2 .50
-j3 .78305
j2
.
V;
=
V2
=
.
- 900
0.60/ - 1200
V(
1 .00/
�
yields
900
3 . 78305 / - 900
-
-
- ( j2 .9S305 ) V;
-j 1 .43082
=
The m a t ri x U i s shown d i re c t l y fol l o w i n g S t e p
4
.
r/I
-j 1 .43082
0.60,[ - 1 20°
=
1 0 0/
V:.l
9 83 0 5
V'' -
0;
0
V2
S o l u t i o n by forw a rd s u b st i t u t i o n b e g i n n i n g w i t h
V(
0
V I'
=
0.26434
0.93800 /
- 1 2.9163°
"----o f t h e forward e l i m i n a t i o n i n
Ex a m p l e 7 . 7 . Subs t i t u t i n g U a n d t h e c a l c u l a t ed e n t r i e s o f V ' i n t h e e q u a t i o n
U V = V ' g i ve s
1
- 0 .70 149
-
0 . 14 25
9
- 0 .38644
-
0
.
1 4 92
5
- 0 . 6 D5 6
- 0 .7885 3
VI
0
0
0 . 264 34
0 . 9 3 80 0 / - 1 2 9 1 630
V2
V:l
V4
.
0 . 9 3 8 00/ - 12.9163°
w h ich \Ve m ay s o l ve by b a c k s u b s t i t u t i o n to o b t a i n
V4
Vl
=
V;
= 0 . 6434 - ( - 0 .7885 3 ) V4
=
2
=
0 . 99904 /
per u n i t
- 9 .5257°
per u n i t
I f d e s i r e d , w e m a y co n t i n u e t h e b a ck s u b s t i t u t ion u s i n g t h e v a l u e s for
eva l u a t e
unit.
V2
=
0.96 1 18/- 1 1 .555 10
p e r u n i t and
VI
=
V)
and
V4
0.96324/ - 1 1 .43880
to
per
278
CHAPTER 7
THE ADMITTANCE MODEL A N D NETW O R K CALCULATIONS
When the coefficient matrix Ybu s is symmetrical, which is almost always the
case, an important simplification results. As can be seen by inspection of Eq.
(7.68), when t h e first column of L is divided by Yl 1 , we obtain the first row of U;
when the second column of L is divided by Yi�\ we obtain the second row of U ;
and so o n for the other columns and rows of Eq. (7.68), provided �j = �' i '
Therefore, dividing the entries in each column of L by the principal diagonal
element in that column yields U T whenever Y "us is symmetrical. We can then
write
YI I
Y2 I
L = uTn
YI I
=
Y3 1
YI I
�l
Yl
l
y(l)
32
y(l)
22
yO)
42
yell
22
y(l)
22
( 7 .75)
y ( 2)
1
33
y (2 )
43
y (2 )
1
33
y (3 )
44
where diagonal matrix D contains the d iagonal e lements of L. Substituting i n
Eq. (7.71) for L from Eq. (7.75), we obtain t h e nodal admittance equations in
the form
Y busV =
u Tnuv =
( 7 .76)
I
Equation (7.76) may be solved for the u nknown voltages V in three consecutive
steps as follows:
( 7 .77)
nv' =
V"
( 7 . 7 8)
=
V'
( 7 .79)
UV
These equations will be recognized as an extension of Eqs. (7.72). The interme­
diate resu l t V" i s first found from Eq. (7.77) by forward substitution. Next, each
entry in V' is calculated from Eq. (7.78) by dividing the corresponding element
of V" by the appropriate diagonal element of n. F i na l l y, the solution V is
; ob taine d from Eq. (7.79) by b ack substitution as demonstrated in Example 7.9.
Example 7.10.
Using Eqs.
(7.77)
through
(7.79),
d etermine the solution vector V of
u nknown voltages for the system a n d operating cond itions of Example
7.9.
·
7.1)
Solution. S u b st i tu t i ng i n Eq.
(7.77)
S P A R S ITY A N D N EA R - O PTI M A L O R D E R I N G
fro m the c u rre n t vector
Ex a m p l e 7.9, w e obt a i n
1 .00/
/
V�'
- 0.70 149
- 0 . 1 4925
- 0 .38644
- 0 . 1 4925
-
0
.
V":1
1
6 1 356
V"
4
- O . 7RR53
i
and
matrix U of
0
V"1
1
I
279
0
0 . 60
- 9 00
- 1 200
S t ra i g h t fo rward sol u t i o n o f t h i s sys t e m of e q u a t i o n s y i e l d s
v;'
v;'
=
V�'
O.(loL- .� :20:
' c
v;'
=
0
-I-
O . 7XX:'i l l "-;'
S u b s t i t u t i n g for Y " i n E q . ( 7 . 7 8 ) l e n c i s
-
j 1 6 75
I .()O
=
=
I
. 342
9()0 p e r u n i t
rO�02 .9 1 64°
-
t o t h e d i ag o n a l sys t e m
V'1
.
-j 1 1 . 00 7 4 6
V'3
-j 3 .78305
- j 1 .43082
V'4
pe r u n i t
o
o
/
1 .342 1 0/
1 . 00
-
- 90°
1 02 . 9 1 64°
t h e s o l u t i on of w h i c h is e x a c t l y e q u a l to V' of Example 7 . 9 , a n d so the re m a i n i n g
s t e p s o f t h i s exa m p l e m a t c h t h o s e o f Ex a m p l e 7 . 9 .
7.9
SPARSITY
A ND NEA R - O PT I M AL
O RD E RI N G
La r g e - sc a l e p o w e r
con n e c t e d
to
systems
a small n u mher of transmiSSion l ines
s u b s t a t i o n I n t h e n e tw o r k g r a p h fo r stich systems
h ave onl y
each h u l k - rower
.
a bou t 1 .S
and
the
c o r r e s p o n d i n g Y \1us h a s 1ll ;l i n l y ze ro c l e m e n t s . 1 1 1 fa c t , i f t h e r c a re 750 h r a n c h e s
i n (l S O( ) - no d e n e t w o r k ( e x c l u d i n g t h e r de rc n c e n o d e ), t h e n s i n c e each n o d e has
,I n ( l s s o c i ,l [ e d d i a g o n a l c l e m e n t a n d e a c h b r(l n c h gives r i s e t o t wo sym metr ically
p l a c e d o tf- d i a gona l elements, t h e t o t a l n u m b e r of nonzero e l e ments is (500 + 2
X 750) = 2000. Th i s comp a res to a total of 250, 000 e l e m e n ts in Y h ilS ; that is,
o n l y O.R% o f t h e c l e m e n t s i n Y blls ar c n o n z e r o . Bec a u s e o f t he s m a l l number o f
n o n z e ro e l e m e n t s, such m a t rices a r e said to b e spa rse . F r o m the viewpoint o f
computational speed, accuracy, and storage i t i s desi rab l e to process only the
n o n z e ro entries in Y h u s a n d to avoid filling-in n e w nonzero elements in the
co u r s e o f g a u s s i an e l i m i n a t i o n and triangular factorization . Orde r ing refe rs to
t h e s e q u en c e i n w h i c h t h e e q u a t i o n s o f a s y s t e m a re p r o c e s s ed When a sparse
t h e r ; l l i o 0 1' t h e n u m b e r o r b r ;l I1 c h e s to t h e n u m b e r o f n o c i e s i s
.
280
CHAPTER 7
THE ADMITTANCE MODEL AND N ETWORK CALCULATIONS
matrix is triangularized, the order in which the unknown variables are elimi­
nated affects the accumulation of new nonzero entries, called fill-ins, in the
triangular matrices L and U. To help minimize such accumulations, we may use
ordering schemes as described in Sec. B.1 of the Appendix.
7.10
SUMMARY
Nodal representation of the power transmission network is developed i n this
chapter. The essential background for understanding the bus admittance matrix
and its formation is provided. I ncorporation of mutu ally coupled branches into
Y bus can be handled by the bu ilding-block approach described here. Mod i fica­
tions to Y bus to reflect network changes arc thereby facil itated.
Gaussian elimination offers an alternative to matrix i nversion for solving
large-scale power systems, and triangular factorization of Y bus enhances compu­
tational efficiency and reduces computer memory requirements, especially when
the network matrices are symmetric.
These modeling and numerical p rocedures underlie .the solution ap­
proaches now being used i n daily practice by the electric power industry for
power-flow and system analysis.
PROBLEM S
7.1.
Usin g t h e building-block procedure described i n Sec. 7 . 1 , d e t e r m i n e Y bus
circuit of Fig. 7.18, Assume there is no m u tual coupling between any
branches.
fo r t h e
of t h e
CD
)0. 125
+
®
1.10LQ:.
•
®
+
0.90/- 300
F1GURE 7.18
Value's shown are voltages and impedances in per u nit. Dots represent mutual coupling between
,
branches u nless otherwise stated in problems.
P R O BLEMS
281
Y bus modification proce d u re d escribed in Sec. 7.4 and ass u m i n g no
m u t u a l coupling between branches, m o d i fy t h e Y b u s obtained in Prob. 7 . 1 to reflect
removal of the two b ra nc hes CD - G) a n d 0 - G) from the circuit of Fig. 7 . 1 8.
7.2. Using the
7.3. The circuit of Fig. 7 . 1 8 h as t h e l i n e a r graph shown i n Fig. 7 . 1 9, with ar rows
i n dicating d irections ass u m ed for t h e branches a to h . Disregarding all m u tu al
coupling between branches:
(a) Determi � e the branch-to-node incidence matrix
as reference.
( 0 ) Find the circu i t Y bus u s i n g Eq. (7.37).
a
b
g
A
for the circuit with node
®
®
d
c
r
e
®
®
h
FI GURE 7 . 1 9
@
7.4.
L i n e a r g r a p h for P rob. 7 . 3 .
Co n s i d e r t h at only t h e two bra nches C!) - G) and 0 - G) in the circ u i t o f Fig.
7. 1 8 are m u t ua l l y co u p l e d as ind icated by t h e d o t s besi d e them and t h a t their
m u t u a l impeda nce is j 0 . 1 5 per u n i t (that is, ignore the dot on branch 0- G) ).
D e t e r m i n e t h e c i rcu i t Y b u s by the p roce d u re d escribed i n Sec. 7.2.
7.5. Solve Prob.
fr o m t h e
7.6. U s i ng
u s i n g E q . 0 . 3 7). D e t e rm i n e the b ra nch-to-node incidence m a trix
l i n e ,l r g r a p h o f F i g . 7 . 1 9 w i t h n o d e @ as r e fe r e n c e .
7.4
A
t h e m od i fi.c a t i o n proce d u re of S c c. 7 . 4 , mod i fy the Y bus sol u t i o n of Prob. 7.4
(o r P r o b . 7 . S ) t o re flect re moval of the branch W - G) [rom t h e circ u i t .
7.7 . Mo d i fy the Yb ll S d e t e r m i n e d i n Exa m p le 7.3 to reflect r emoval of the m u tu ally
co upled b r a n c h CD - G) from the c i rc u i t of Fig. 7. 1 1 . Use t he modification
p rocedu re of Sec. 7.4.
7.8.
7.9.
u n i t i s a d d e d between nodes 0
a n d G) in t h e c i rc u i t o f Fig. 7 . 1 1 . M u t u a l i m p e da n ce of jD. l per u n i t couples t h is
n ew b ranch to the branch a l ready exist i n g between nodes 0 and G) . M o d i fy the
Y b u s o b t a i n e d in Example 7.3 to acco unt for the a d dition of the n ew branch.
A n e w b r a n c h h a v i n g a se l f- i m pe d a n ce o f
j O.2
per
S u ppo se t hat m u tual cou p l i n g exists p a i rw i se b etween bra nches CD - G) a n d
W- G), and a lso between branches W - G) a n d 0 - G) of Fig. 7.18, a s shown
b y t h e dots in t h a t fi g u re . T h e m u t u a l i m p e d a n ce be tween the former p a i r o f
282
CHA PTER 7
THE ADMITTANCE M O DEL A N D N ETWORK CALCU LATI ONS
branches is jO. 1 5 per unit (the same as i n Prob. 7.4) and between the latter pair is
jO. l per u nit. Use the procedure of Sec. 7.2 to find Y bus for the overall circui t
including the three mutually coupled br anches.
7.10. Solve for the Yb us of Prob. 7.9 u sing Eq. (7.37). Use the l in e ar graph of Fig. 7 . 1 9
with reference node ® to d etermine the branch-to-node i ncidence matrix A.
7.1 1 . Suppose that the direction of branch d in Fig. 7. 1 9 is reversed so that it is now
dire ct e d from node 0 to node G) . Find the bra n c h -to-node inciden ce matrix A
of th is modified graph and t h e n so lve for t h e Ybus of P rob 7.9 using Eq. (7.37).
7.12. U s i n g the Yhus nl odi tica t i o ll p roce d u r e d escri bed in Sec. 7.4, rcmove b r a n c h
(1)- G) from the Y t,lIS so l u t i o n o b t a i n e d in P ro b . 7.9 (or P ro b . 7. 1 0 or P rob. 7. 1 1 ).
7.13. Write n o d a l a d m i t t a nce e q u a t i o n s for t h e c i rc u i t o r rig. 7. 1 i) d i s re g a rd i ng a l l
m utual c oupling. Solve t h e r esu l t a n t e q u a t i o n s for t h e bus volt ages b y t h e me thod
of gauss ian e limination.
7.14. Prove Eq. (7.69) based on Eq. (7.68).
7.1S. Using the g aussian-el i minat ion calcu l ations o f Prob . 7 . 1 3 , fi n d the t ri a n g u l a r factors
of Y bus fo r the circui t of Fig. 7.18.
7.16. Use the triangular factors obtained i n Prob . 7 . 1 5 to calculate new bus vol tages for
Fig. 7.18 when the voltage source at bus G) is changed to
per u n i t .
Follow the p rocedure of Example 7.9.
7. 17. Using the triangular factors obta ined in Example 7.9, fi n d t h e voltage a t b u s G) o f
t h e circuit o f Fig. 7.1 1 w h e n an additional curre n t of 0.2/ - 1200 pe r u n i t is
injected a t bus 0 . All other conditions of Fig. 7. 1 1 are unchanged.
.
1 .0�
G) .
reduce Y bus of the c ircui t of Fig.
to reflect elimination of node
( b ) Use t h e Y
transformation
o
f
Table
to eli minate node
from the
6.
circuit o f Fig.
and find Y b u s for the resul t i n g reduced network. Compare
results of parts ( a ) and ( b ) .
7.18. (a) Kro n
7. 18
1 .2
-
7.18
7.19.
Find the L a n d
U
triangular factors o f t h e symmetric m atrix
Verify the result using Eq. (7.75).
0
CHAPTER
8
THE
I MPEDANCE
MODEL AND
NE'rWO R K
CALCULA'rI O N S
The bus admittance matrix o f a large-scale interconnected power syste m i s
typically very sparse with mainly zero elements. I n Chap. 7 w e saw how Y b u s i s
constructed branch by branch from primitive admittances. I t i s conceptually
simple to i n v e r t Ybu s to find the bus impedance matrix Z b u � ' but such d i rect
inversion is rarely employed when the systems to be analyzed are large scale. In
p r a c t i ce Z bus is r a r e l y e xp l i c i t l y required, and so the tria ngular factors of Y b u s
a re u s e d to ge n e r(l t e c l e m e n t s of Z h l l S o n l y as t h ey are n e e d e d s i n ce t h is i s often
,
as b e i n g a l r e a d y constructed and
a n a l y s t can de r ive a great deal of i n s i g h t .
r e g ar d i n g Z h u s
t h e m o s t C O I l 1 I Hl t i l t i o l l ; t l l y d J i c i e n t m d i l oli . U y se t t i n g c O IH r l l t a l i o n a l co n s i d e r a ­
t ions aside,
however,
and
is t h e approach t a k e n in t h i s c h a p t e r .
The bus impedance matrix c a n be d i re c t l y constructed element by ele m e n t
using simple algorithms to incorporate one element at a time i n to the syste m
representation. T h e w o r k e n t a i l e d i n const ructing Z bus is much greater than
that required to construct Y b u s ' but the informat ion content of the bus impeda nce
rna trix is far greater than that of Y b us ' We shall see, for example, that each
diagonal element of Z b u s reflects i mportant characteristics of the entire system
in the form of the Thevenin impedance at the corresponding bus. Unlike Y bus '
the b u s impedance matrix of an interconnected system is never. sparse a n d
contains z e ros only when t h e s y s t e m i s regarded a s being subdivided i nto
ex p l i c i t l y a v a i l a b l e , t h e rowe r sys t e m
Th i s
283
284
CHAPTER 8
THE I M PEDANCE MODEL AN D N ETWO RK CALCULATIONS ·
independent parts by open circuits. In Chap. 1 2, for instance, such open circuits
arise in the zero-sequence network of the system .
The bus admittance matrix is widely used for power-flow analysis, as we
shall see in Chap. 9. On the other hand, the bus impedance matrix is equally
well favored for power system fault analysis. Accordingly, both Ybus and Z bus
have important roles in the analysis of the power system network. I n this
chapter we study how to construct Z b us directly and how to explore some of the
conceptual insights which it offers into the characteristics of the power transmis­
sion network.
8.1 THE BUS ADMITTANCE AND
IMPEDANCE MATRICES
In Example 7.6 we inverted the bus admittance matrix
resultant the bus impedance matrix Z b u s ' By definition
Y hus
and called the
(8.1)
and for a network of three independen t nodes the standard form is
CD
Z b us = Q)
G)
CD Q) G)
2 1 1 2 12 213
2 2 1 222 223
( 8 .2)
2 3 1 2 3 2 2 33
Since Y bus is symmetrical around the principal diagonal, Z bus must also be
symmetrical . The bus admittance matrix need not be determined in order to
obtain Z bus > and in another section of this chapter we see how Z b us may be
formulated directly.
The impedance elements of Z hus on the principal diagonal are called
driving-point impedances of the buses, and the off-diagonal elements are called
the transfer impedances of the buses.
The bus impedance matrix is important and very useful in making fault
ca1cula tions, as we shall see l ater. In order to understand the physical signifi­
cance of the various i mpedances i n the m a trix, we compare them with the bus
admittances. We can easily do so by looking at the equations at a particular bus.
For instance, starting with the node equations expressed as
I = Y busV
( 8 .3 )
we have at bus (I) of the three independent nodes
'( 8 . 4)
H. I
THE
BUS ADM ITTANCE A N D I M PEDANCE M AT R I C ES
285
J
VI and V3 are reduced to zero by shorti n g buses CD a n d Q) to t h e
reference n o d e , and voltage V2 is a p p l i e d a t b u s (I) s o t h a t c urren t 2 e n t e r s a t
bus W , t h e self-adm i ttance at bus (I) i s
If
( 8 .5 )
T h u s , the s e l f-ad m i tt a n ce of a particu l a r bus cou l d be m easured by
short ing all other b u ses to the reference node and then finding the ratio of the
cu rren t i njected a t t h e bus to the vol tage applied at t h a t bus. Figure 8 . 1
illustra tes t h e m e thod for a t h ree-bus reactive network. T h e result i s obviously
eq u iv a l e n t t o add i n g a l l the a d m i t ta nces d i rectly connected to the bus, w h i c h is
t h e p rocedure up to now w h e n m u t ua l l y coupled branches are abs e n t . Figure
8.1 a l s o s e rv e s to i l l u s t r3 t e the off-di agonitl a d m i t t a nce t e rm s of Ybll s ' At bus CD
t h e equ a t ion obta i ned by expan d i n g Eq . (8.3) is
( 8 .6)
from \vh i ch w e s e c t h a t
( 8 .7)
Thus, the m u t ual a d m i t t a n ce term Yl 2 i s m easu red b y shorting all b u ses exc e p t
bus G) to the refe rence n o d e a n d b y app lyin g a vol tage V2 at bus Cl), a s s h ow n
i n Fig. 8 . 1 . The n , YI 2 is the ratio of the n egat ive of the curre n t l eavi n g t h e
n e t work i n t h e short c i rcu i t a t n o d e CD to t h e vol tage V:, . T h e n eg a t ive o f t h e
curre n t l e aving t h e network a t node CD is used s i n ce I I i s d e fi n e d a s
t h e c u rrent entering t h e n e twork. The resultant admitta nce i s t h e n egative
®
CD
®
..
12
+
V2
~
FI G C RE 8 . 1
Circ u i t fo r m e a s u r i n g Yn . Y1 2 • a n d
Y32 .
286
CHAPTER 8
THE I MPEDANCE MODEL A N D N ETWORK CALCULATIONS
of the adm;ttance directly connected between buses CD and @ , as we would
expect since m u tu a lly coupled branches are absent.
We have made this detailed examination of the bus admittances in order
to differentiate them clearly from the impedances of the bus impedance matrix.
Conceptually, we solve Eq. (8.3) by premultiplying both sides of the
equa tion by Y;;:I� = Z hus to yield
( B .B)
and we must remember w h e n dea l i ng w i t h Z hus t h a t V a n d I a re col u m n vec t o rs
of the bus voltages and the c u r r e n ts e n t e r i n g the buses from c u r r e n t s o u rc e s ,
respectively. Expanding Eq. (8.8) for a network of three in d ependent nodes
yields
( 8 .9)
( 8 . 1 0)
(8 . 1 1 )
From Eq. (8. 1 0) we see that the driving-point impedance 222 is deter­
mined by open-circuiting the current sources at buses CD and G) and by
injecting the source current 12 at bus W . Then,
V2
222 = -12
Figure 8.2 shows the circuit described . Si nce 2 2 2
®
+
FIGURE 8.2
Circuit for measuring
2 22 , 21 2 ,
and
232.
( 8 . 12)
IS
defined by openmg the
THEVENI N'S THEOREM A N D Z bus
8.2
287
cur re n t sources coimected to the other buses whereas Y22 is found with the
oth er buses shorted, we m ust not expect any reciprocal relation between t h ese
two quantities.
The c i rcuit of Fig. 8.2 a lso e nables us to m easure some transfer impedances,
for we see from Eq. (8.9) t h a t with cu rrent sou rces 1\ an d 1 3 open-circuited
( 8 . 1 3)
a n d fro m Eq. (8. 1 1 )
Zl�
.
-
=
v�I I
�
�
I,
-
I,
-
II
( 8 . 1 4)
Th us, w e can meas u re t h e tran sfer i m peda nces Z l 2 and Z:l 2 by i nj ecting c u r rent
at b u s W a n d by fi n d i n g t h e ratios of VI a n d V� to f; w i t h the sources open at
a l l buses except bus W . We note that a m u t u a l adm i tt a n ce is measured wit h a l l
b u t on e b u s s h o r t circ u i t e d and that a tra nsfe r i mpedance i s measure d wit h a l l
sources open-circu ited exce p t o ne.
Equ a t io n (8.9) tells us t h a t i f we i nject cu rrent into bus CD with current
sources at buses @ and G) ope n , the only impedance t h rough which 1\ fl ows
is Z 1 1 ' U n d e r the same con d i t ions, Eqs. (8. 1 0) a n d (8 . 1 1 ) show that 1 1 is causing
vo l t ages at buses @ and G) expressed by
'
-
and
( 8 .15)
I t is i m portant t o re a l ize the i mp l i ca t ions o f t h e p reced ing d iscussion, for Z bus is
som e t i m e s u sed i n power-flow studies a'ld is extremely valuab l e in fau lt calcula­
tiuns.
8.2
TH EV E N I N 'S T H EO R E M A N D Z hus
/
Th e b u s i m p e d a n c e m a t r i x p r ov i d e s i m p o rt a n t i n form a t i o n rega rd i n g the p ower
sys t e m n e t wo r k w h i c h w e C <l n u s e t o Cl d v a n t a gc in n e t work c a l cu l a t ions. In t h i s
s e c t i o n w e e x a m i n e t h e re l a t i o n s h i p b e tw e e n t h e c l e m e n ts o f Z blls a n d the
Theve n i n i mp e d an ce p r e s e n t ed b y t h e n e two rk a t e a c h o f its b uses. To establish
n o t a t ion, lct us d e n ote t h e b u s v o l t a g e s co r re s p o n d i n g t o the i n itial val ues 1° of
t h e b u s c u r r e n t s I by y O
Z bus 1 u . The voltages V IO to V,$ are t he effect ive
open -circuit vol tages, w h ich can be measured by vol tmeter between the buses of
the n e twork and the ref e r e n c e n o d e . When t he bus currents are changed from
t h e i r initial values to n ew values I [J = 6 I, t h e new bus voltages are given by the
.
=
288
CHAPTER 8
THE IM PEDANCE MODEL AND N ETWO R K CALCULATIONS
superposition equation
v = Z bus( I o + � I) = Z bus I o + Z bus � I
�
VO
(8 .16)
t:N
where �v represents t h e changes in t h e bus voltages from their original values.
Figure 8.3( a) shows a large-scale system in schematic form with a repre­
.
sentative bus ® extracted along with the reference node of the system.
InitiallY, we consider the circuit not to be energized so t h a t t h e bus c u r r e n t s 1 °
and voltages V O a re zero. Then, into b u s ® a curre n t of 11 ft-; a m p ( o r 11 ft-; per
unit for Z hus in per unit) is injected in to the sys te m fro m a c u r r e n t s o u rce
connected to the reference node. The resulting v olt a g e c h a nges at t h e buses of
- �v
3
Il @.
2 II ®
�V I CD
1I
�V
--4
Original network
Zbus
� Vn
®
I
I
®
Reference
� Vk
cD
(a)
Original network
Zbus
V;.G
+
®
Z th
:0=
Zk k
®
t
+
!-
6 Ik
VII
Reference
: FIGURE 8.3
(a) Original network with bus
®
(b)
and reference node extracted . Voltage �Vn at bus
by curre n t tlI k entering the network. (b) Theven i n equivalent circu it at node CD ·
®
is s:aused
8.2 THEVEN IN'S THEOREM
the
n e twork, indicated by t h e incr emental q u an tities 6 V1 to 6 VN ,
il V I
il V2
CD
W
CD
W
®
Z21
Z 22
Zl 1
are give n by
ZlN
0
Z2k
Z2 N
0
Zlk
Z12
®
il Vk
®
Zk L
Zk 2
Zk k
Zk N
il VN
®
ZN l
ZN 2
Z Nk
Z NN
289
AND z.,...
tJ.
Ik
0
( 8 . 1 7)
w i t h the only nonz ero e n t ry in t h e cu rre n t vcctor equal t o Ll lk i n row k .
Row-by-col u m n m U l t i p l i c a t i o n i n Eq. (8. 1 7) y i e l d s the i ncreme n ta l bus vol tages
il V,
Ll V
2
Ll V"
il VN
CD
CV
®
Z lk
Z 2/.:
®
Zu
®
Z Nk
Ll Ik
( 8 . 1 8)
w h ich are n u merically equal to t h e ent ries in col u m n k of Z bus m u l tiplied by t h e
c u r r e n t Ll Ik . Ad d i ng t hese voltage c h anges to the original vol tages at t h e b uses
according t o Eq. (8. 1 6) yields a t bus ®
(8.19)
Th e c i rcu i t correspo n d i n g t o t h is cq u a t ion i s shown i n Fig . 8.3( b ) from w h i c h i t
i s evident that t h e Theve n i n imped a n ce Z th a t a represe n t a tive b u s ® o f t h e
syst e m i s given by
( 8 . 20)
where Z k k is the d i a go n a l e n t ry in row k and col u m n k of Z bu s ' W i t h k s e t
equal to 2 , t h is i s esse n t i al ly t h e s a m e resu l t obtained i n Eq. (8. 1 2) for the
d riving-p oint impe dance at bus W of Fig. 8.2.
In a s i m i l a r m a n n e r, we can d e te rm i n e the Theve n i n impedance b e tween
any two b uses (j) and ® o f t h e n e twork. As s hown i n Fig. 8.4(a), t h e
otherwise dead n etwork i s e n e rgized by t h e current i nj ections 6 Ij .a t b u s 0)
290
CHAPTER 8
THE I MPEDANCE MODEL A N D NETWORK CALCULATIONS
and I:1 Ik at bus ® . Denoting the changes in the bus voltages resulting from the
combination of these two current inj ections by � Vl to � VN ' we obtain
CD
� Vl
� v1.
� Vk
-
� VN
(j)
OJ
CD
ZI I
Zj t
®
Z lj
Zt k
ZJJ. .
Zj k
®
ZI N
0
Z1 v
6 /1.
I
®
Zk l
Zk j
Zk k
Zk N
� lk
®
ZN I
Z Nj
ZN k
Z NN
0
� l1" + Z1"k � lk
Z k j � Ij + Z kk � Ik
ZlJ"
=
( 8 . 21 )
in which the right-hand vector is nume rically equal to the product of � Ij and
column j added to the product of � lk and column k of the system Z bus ' Adding
these voltage changes to the orig inal bus voltages according to Eq. (8.1 6), we
obtain at buses OJ and ®
( 8 .22 )
: Adding and subtracting Zjk \ � Jj in Eq.
(8.23), give
v.
1
=
V1 O + ( Z11" "
(8.22), and l ikewise,
- Z1"k. ) � I1"
Zk j � lk in Eq.
+ Z1"k ( � I1 + � lk )
(8 .24)
( 8 .25 )
Since Z bus
is symmetrical, Zjk equals Zk j and the circuit corresponding to these
two equations is shown in Fig. 8.4Cb), which represents the Thevenin equivalent
circuit of the system between buses (j) an d ®. Inspection of Fig. 8.4Cb ) sp ows
that the open-circuit voltage from bus ® to bus OJ is VkO Vi o , and the
-
8.2
THEVEN lr-;'S THEOREM A N D Zbus
291
O r i g i n a l n etwork
l ®
2I
6 V I CD
1 I
Zbus .
LiV
®
o
Ll JJ.
R efere n c e
(a )
Jj)
O r i g i n a l n e tw o r k
+
jk
=
VkO
Zk k - Z k ;
I
"
V) O
Z )
k
®
®
�
...--
Zbus
Z
®
:'J. lk
+
J)
Zjj - Z) k
0(
0
� I)
(
Ib
�
Reference
FI G C R E 8 . 4
e q u i v a l e n t c i rc u i t :
Z
b
(d)
(b)
O r i g i n a l n e t wo r k w i t h ( a ) cu r re n t so u rc e s
(
6 I} a t b u s
CD
and
([); (b) Thevenin
b u s e s CD and ([) .
6 1� a t b u s
(c) s h o r t - c i rc u i t c o n n e c t i o n ; ( tI ) i m p e d :1n c e Z " b e t w e e n
i m pe d a n c e e n co u n t e re d by t h e sil ort -cif'C[ {il Cl I rre n t I I (
i n F i g . H . rJ ( ( ' ) i s e v i d e n t l y t h e T h c VL' l l i l 1 i l l l p e t \ ; I I 1l' c .
Crom
bus
®
to b u s
(j)
( 8 . 26)
This
res u l t is rea d i l y con fi rmed by subs t i t u t i n g 1,.(, = 6. lj = 6. /" i n Eqs. ( 8 .24)
Vk between the resu ltant e q u a ­
a n d (R.2S) a n d by s e t t i n g the d i ffe rence �.
t i o n s e q u a l to z e r o. A s fa r a s exte r n a l con n ections t o buses (]) a n d ® a re
concerned, Fig. 8A( b ) represe nts t h e e ffect of the o riginal syste m . From bus 0
to the refere nce node w e can t race t h e Theven i n i mpedance 2jj ( 2jj - 2j k )
+ Zjk a n d t h e open-circ u i t vol tage Vj Q ; from b u s ® t o t h e refe rence node we
h ave the Theve n i n i mp ed a n ce Zkk
( Zkk Z k, ) + Z kj and the open-circuit
-
-
=
=
-
292
CHAPTER 8
THE IMPEDANCE MODEL AND N ETWOR K CALCULATIONS
voltage VkO; and between buses ® and Q) the Thevenin i m p ed a nce of Eq.
(8.26) and the open-circuit voltage VkO - VjO is evident. Finally, when the
branch impedance Zb is connected between buses (]) and ® of Fig. 8.4( d ),
the resulting current Ib is given by
( 8 .27 )
We u s e this e q uat ion in Sec. 8.3 to show how to modify
imp e d an ce is added between two buses of the network.
Z bus
when a branch
capacitor having a reactance o f 5.0 p e r u n i t is connected between
the reference node and bus @ of the circu i t of Examples 7.5 and 7.6. The original
emfs and the corresponding external current i njections a t buses CD and @ are
the same as in those examples. Find the current d raw n by the capacitor.
Example 8 . 1 . A
L.
Solution. The Thcvenin equ ivalent c i rc u i t at bus
@
has an emf with respect to
reference given by V40 = 0 . 9 4 866 - 20 .7466° p er u n it, wh ich is the vol tage at bus
@ fou n d in Example 7.6 before the capacitor is connected. The Thevenin
impedance Z44 at bus @ is calculated i n Example 7.6 to be Z44 = jO.69890 per
unit, and so Fig. 8.5(0) follows. Therefore, the current leap drawn b y the capacitor
IS
Ieap
=
0.94866/ - 20 .7466°
-j5 .0 + jO.69890
=
0.22056/
- 69 .2534°
If an additional current equal to
injected into the network at bus @ of Example
buses CD, (1), Q), a n d @ .
Example 8.2.
per u n it
- 0.22056/69.2534: per umt IS
7.6, find the resulting voltages a t
.
Solution The voltage changes at the buses d u e to t h e additio nal i nj e cted current
. .
can be calculated b y making use of the bus impedance matrix found i n Example
The required impedances are in colu m n 4 of Z b uS ' The voltage changes due to
the added current i njection at bus @ i n per u n i t are
7.6.
�Vl
� V2
� V3
� V4
=
=
=
=
- IeapZl 4
=
-leapZ24
=
- lcapZ34
=
- lcapZ4 4
=
- 0.22056�.2534°
X
jO.63677
- O.22056�?534° x jO .64178
- 0.22056/ 69.2534° x jO .55110
-
O
.
22 0
56�?534� X jO.69890
=
=
=
=
0 .14045/ - 20.7466°
0 .14155/ - 20 .7466°
0.12155/ - 20 .7466°
0.15415/ - 20.7466°
8.2 THEVENI N 'S THEOREM A N D
o
I
I
V4O
293
�
/'
�
jO.69890
0.94866/- 20.7466°
Z hus
'i' -j5 .0
Reference
(a )
- 2 0 . 7 4 6 G"
(b)
Exam ples 8 . 1 and 8.2 s h o w i n g : ( 0 ) Theve n i n e q uivalent c i r c u i t ; ( b )
FI G L' RE 8.5
0.
Ci rc. u i t fo r
bus
p h asor d i a g r a m a t
(R. 1 6)
B y s u p e rp o si t i o n t h e r e s u l t i n g vo l t ag e s a re d e t e r m i n e d from E q .
a d d i n g t h e s e c h a n g e s to t h e o r i g i n a l bus vol t ag e s fo u n e! in Ex a m p l e
bus volt ages
VI
V2
VJ
V4
=
=
=
=
per u n i t a re
0.96901/
in
- I R . 4 1 R9°
+ 0 . 1 4045/ - 20 . 746(')°
O .96734� P 3 .6028°
+
0 . 99964
+
0 .94866
/ - 15 .371 8°
/ - 20 .7466°
+
0.14155
O . l 2 1 55
0 . 1 54 1 5
/ - 20 .7466°
=
1 . 1 0938
7.6.
L- I R .7 1 3 5°
= 1 . ] () 8 80
� R7 64°
.
�20.7466° = 1 .12071/ -- 1 5 .9539°
/ - 20.7466°
=
by
The new
1 . 10281/ -;- 20 .7466°
·294
CHAPTER 8
THE I MP EDANCE M O D EL A N D NETWORK CALCU l;ATIONS
S ince the changes in voltages due to the injected current are all at the
same angle shown in Fig. 8.SCb) and this angle d iffers little from the angles of
t h e original voltages, an approximation will often give satisfactory answers. The
change in voltage magnitude at a .bus may be approximated by the product of
the magnitude of the per-unit current and t he magn itude of the appropriate
driving-point or transfer i mpedance. These values added to the original voltage
magnitudes approximate the magni tudes of the new voltages very closely. This
approximation is valid here because the network is p urely reactive, but it also
provides a good estimate where reactance is consid erably larger than resistance,
as is u sual in transmission systems.
The last two examples illustrate the importance of the bus impedance
matrix and incidentally show how adding a capac itor at a bus c a u s e s a rise i n
bus voltages. The assumption t h a t t h e an gles o f 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 conside r such system
operation in Chap. 9 using a compu ter power-flow program.
8.3
MODIFICATION OF AN EXISTING Z bU S
Z bus
In Sec. 8.2 we see how to use the Theven in equivalent circu it and the exist ing
to solve for new bus voltages in the network following a branch addition
without h av ing to develop the new Z b u s ' S ince Z b u s is such an i mportant tool in
power system analysis we now examine how an existing Z bus m a y be modified to
add n ew buses or to con nect new lines to established buses. Of cou rse, we cou ld
create a new Y bu s and invert it, but direct methods of modi fying Z b u s are
available and very much simpler than a matrix i nversion even for a small
number of buses. Also, when we k n ow how to m o cJ i fy Z htl � ' we c a n see how to
build it directly.
We recognize several types of modifications in which a branch having
impedance Z/J is added to a network with k nown Z b\l s ' The origi n a l bus
impedance matrix is identified as Z ori!o: ' an N X N m a trix.
In the notation to be used in our analysis existing buses will be identified
by n umbers or the letters h, i, j, and k. The l etter p or q will designate a new
bus · to be added to the network to convert Z orig to an e N + 1 ) X e N + 1 )
m atrix. A t b u s ® t h e original voltage w i l l be denoted by VI?' t h e new voltage
after modifying Z bus will be Vk , and .1 Vk Vk - VkO will denote the voltage
change at that bus. Four cases are considered in this section.
=
CASE
1.
bus ® to the reference node.
addition of the new bus ® connected to the reference node through
Adding Zb from
a
new
The
Zb wit hout a connection to a ny o f the buses of the original network cannot alter
the o riginal bus voltages when a current Ip is i nj ected at the new bus. The
Ik
Ip
�
8.3
®
I
I
®
", I
I
M O D I FICATION OF AN EXISTING Z bus
295
�
Original
network with
bus
Zb
®
and the
reference node
extracted
f--
A d d i t ion of new bus
@
to exis t i ng bus
0
0
Z Orig
=
®
0
con­
®.
I(lZ" , Then,
VIlJ
VIS
�)
®
nected t h rough i m p e d ance Zb
Reference
voltage �) a t the new b us i s e q u a l to
V.,O
FI G U RE 8.6
0
0
Z bus(ncw)
;" J
II
/2
IN
( 8 .28)
lp
We note that t h e column vector of c u rren ts m u l t i p l i e d by the new Z bu s w i l l not
alter the vol t ages of the ori g i n a l n e twork and w ill result i n the correct vol tage a t
t h e n ew b u s ® .
CAS E 2 .
A dding
Zb
from
a
new bus
®
t o an existing bus
®
The addition of a n ew bus ® connected through Zb to a n existing b u s ®
i njected at b u s ® will cause t h e c u rren t e n t e ring t he original n etwork
with
at bus ® to beco me t he s u m o f lJ..:. i nj e c t e d at bus ® plus t h e curren t 1p
com i n g t h ro u g h Z" , a s s h ow n i n r i g . H . 6 .
T h e c u r re l l t 1 " I l ow i ng i n t o t i l L: n e t wo r k
; 1 1 l H I S ® w i l l i n c r L: a s L: t h e
IJ
o r i g i n a l vo l t a g e V" by t h e vol t a ge lp ZkJ..:. j ust l i k e i n Eq. ( O . P »); t h a t i s ,
Ip
a n d �) w i l l be l a rg e r
( 8 . 2 9)
t h a n t h e n ew
V"
by t h e v o l t a ge
I{) 2". S o ,
( 8 .30)
and
su
bst i t u t i ng for V"o, w e obt a i n
( 8 .3 1 )
Vu
I;
296
CHAPTER 8
THE I MPEDANCE MODEL A N D N ETWO RK CA LCULATIONS
We now see that the new row which must be a dded to
� is
Z ori g I n
order to find
Since Z bus must be a square matrix around the principal diagonal, we must add
a new column which is the transpose of the new row. The n ew column accounts
for the increase of all bus voltages due to I", as shown in Eq. (8 . 1 7). The matrix
equation is
VI
V2
®
I
Vp
/2
LU
-
Z o rig
VN
I -
/1
Z I t.:
®
Zk l
Z Nk
Zk 2
Zk N
Zk k
+
Zb
Iv
.( 8. 3 2 )
lp
z hUs( nCw)
Note that the first N elements of the new row are the e lemen ts of row k o f Z ori g
and the first N elements of the n ew column are the elements of column k of
Z orig'
CASE 3. A dding Zb from existing bus
®
to the reference node
To see how to alter Z Orig by con necting an impedance Z/J from an existing
bus ® to the reference node, we add a new bus ® connected through Z b to
bus ® . Then, we short-circuit bus ® to the refe rence nod� by letting Vp equal
zero to yield the same matrix equat ion as Eq. (8.32) except that Vp is zero. So,
for the modification we proceed to create a n ew row and new column exactly
the same as in Case 2, but we then elimi nate the ( N + 1 ) row and ( N + 1 )
column b y Kron reduction, wh ich i s possible because o f the ze ro in the column
matrix of voltages. We use the method developed in Eq. (7 .50) to find each
element Z"i(IICW) in the new matrix, where
Zh i
Zhi(new) -
CASE
4.
_
Zh ( N + l ) Z ( N + l ) i
Zk k
Adding Zb between two existing buses
+
CD
Zb
®
buses 0
and
( 8 .33 )
and ® already
To add a branch impedance Zb between
established in Z on g , we examine Fig. 8.7, w hich shows these buses extracted
,
from the original network. The current Ib flowing from bus ® to bus CD is
similar to that of Fig. 8.4. Hence, fro m Eq. (8.2 1 ) the change i n voltage a t each
.
MODIFICATIO N OF AN EXISTING Z bu'
8.3
(])
I
J
)0
I
I
®
�
bus
16
I
I
®
1
Ij
1
+
Ib
297
)-
Orig i n al
network with
Zb
1
1k
buses
([) , ® and
the reference
node extracted
Addition of i m pedance Z6 between existing buses OJ and
@.
FI G U R E 8.7
)0
-
1b
®
caused by the i njection
Reference
(j)
a t bus
I,)
and
-
Ih
at bus
®
is given by
( 8 .34)
which means t h a t the ve c t o r 6V of hus vol t a ge c h a n g es is found by subtracting
col u m n k from col umn j of Z orig a n d by m u l t i p l yi n g the result by I,) . Based on
the d efin i t ion of vol tage c h a n g e we now write some equa tions for the bus
vol t ages as fo l lows:
,
( 8 .35)
and using Eq . (8.34) g i v e s
(j)
Similarly, at buses
v )
Zj j 1j +
Vk = Z k j l j
.
. .
+
and
®
2jj Ij
+
Zjk
I
k
+
V} O
+
.
.
.
+ Z k Jj + Z k k Ik
VIiO
+
.
.
.
+
2jN J N +
( 2jj
+ Z k N IN + ( ZkJ
-
Zjk ) Ib
( 8 . 3 7)
6. V}
-
2kk ) l b ( 8 . 38 )
6. Vk
\Ve need one more equation s i n ce Ib is u nknown. Th is is supplied b y Eq. (8 . 27),
which can be rearranged in to the form
( 8 .39)
From Eq. (8. 3 7) we note tha t V, 0 equals the prod uct of row j of Z orig and the
col u m n of bus curre nts I; l i kewise, VkO of Eq. (8.38) equals row , k of Z o rig
mul tiplied by 1 . Upon subs t i t u t i n g t h e expressions for � o and VkO i n E q . (8.3 9),
298
CHAPTER 8
THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS
we obtain
o = [ ( row j - row k ) of Z Orig ]
By examining thc cocHlcicnls
can write the matrix equation
or Eqs .
W.3() )
( 8 .40)
t h ro u g h
(C;.3C;)
a n d £4.
VI
11
V-)
IJ,
Vk
VN
0
( col . j
Z ori g
=
( row j
-
-
col . k )
(c;.40), wc
Ik
( 8 .41 )
of Z orig
row k ) of Z Orl' g
IN
Z bb
-
lb '
in which the coefficient of lb in the last row is denoted by
(8 .42 )
The ncw column is column j m i n lls col u m n k 0 [" Zorig w i t h ZJ,h in t h e ( N + 1 )
row. The ncw row i s the t ra n spos e o f th e ne w column. El i m i n a t i ng t he ( N + 1 )
row and ( N + 1 ) col u m n o f t h e square matrix of Eq. (8. 4 1) i n t h e same manner
as previously, we see that ea e h element Z" i(llcw) i n the new matrix is
2h i(ne w)
=
Z"i
-
2Jz( N + 1 ) 2( N + I )i
211 + 2 kk - 2 2jk + Zb
( 8 .43 )
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.
Removing a branch. A single bra nch of impedance Z b between two nodes
can be removed from the network by adding the negative of Zb between the
, same terminating nodes. The reason is, of course, that the parallel combination
o f the existing branch (Z b) and the added branch ( - Z b) amounts to an effective
,
, open circuit.
Tabl e 8 . 1 summarizes the procedu res of Cases 1 to 4.
8.3
Modification of existing Z bU8
MODIFICATION OF A N EXISTING Zbwo
TABLE S. 1
Case Add branch Z b fro m
®
Refe re nce node to new bus
1
®
Existing bus
®
to new bus
®
Existing bus
Zorig
3
®+
®
®
to reference node
f-+1-C==::J--ih
®
( N ode
®
®�
( Node
®
2 and
- Remove row p and column p
by Kron reduction
is temporary . )
Existing b u s
Zorig
- Repeat Case
Zb
L-____________�
4
1
col . k
2
(j)
to existing bus
®
r+
H��
lZ--b --r- lh---1
®
(j)
l
) H,f---
@
•
®
--------'
----
is temporary . )
[
Fo r m the
matrix
Z o ,',
row j
where Z th , jk
=
-
row k
Z j + Z kk
j
and
®
col . j
Z th , jk
-
col . k
+ Zb
2 Zjk
-Remove row q and column q
by Kron reduction
-
-
,
-
299
300
,
CHAPTER 8
THE I M PEDANCE MODEL A N D N ETWORK CALCULATIONS
Modify the bus impedance matrix of Example 7.6 to account for the
connection of a capacitor having a reactance of 5.0 per unit between bus @ and
the reference node o f the circuit o f Fig. 7.9. Then, fi n d V4 using the impedances o f
the new matrix a n d t h e current sources of Example 7.6. Compare this value o f V4
with t h a t foun d in Example 8.2.
Example 8.3.
and recognize that Z O rig is the 4 X 4 matrix of
that subscript k = 4, and that Zb = - j5.0 per unit to find
Solution. We use
Example
7.6,
�
Eq. (8.32)
=
jO . 6 3 6 77
j O . M l 7!)
jO .55 I 1 0
j ( J . ()l)!)lJO
jO.63677
jO.64178
jO.55 1 10
)0 .69890
12
-j4 .30 l l 0
'"
11
13
14
The terms i n the fi fth row and column were obtained by repeating the fourth row
and column of Z orig and noting that
Z44
+
Zb
=
jO .69890 - j5 .0
=
-j4 .301 10
Then, eliminating t h e fi fth row and column, we obtain for
Zl 1(new) jO.73 1 28 =
ZZ
4{ ne w)
=
jO.64178 -
jO.63677 X jO.63677
-}4. .30110
jO.69890 X jO.64 1 78
-}4 .3011 0
-
.
Z b u S( new)
=
=
from
Eq. (8.33)
jO.82555
jO.74606
and other elements i n a similar m a n n e r to give
Zb us(nc w)
=
jO.82555
jO.78641
jO.69482
jO.74024
jO.78641
jO.81542
jO .69045
jO.74606
jO.69482
jO.69045
jO.7695 1
jO.64065
jO.74024
jO .74606
jO.64065
jO.8 1 247
The column matrix of currents by w hich the new Zb us is mUltiplied to obtai n the
new bus voltages is the same as i n Example 7.6. S ince both II and 12 are zero
while 13 a n d 1 4 are nonzero, we 0 bta in
�
=
=
=
jO.64065 (1 .00/ - 900 ) + jO.81247 ( 0.68/ - 1350 )
1 .10281 L - 2 0.7466°
1 .03131 - j0.39066
as fou n d in Example
8.2.
t
"
per unit
IJ
S. 4
D I R ECT D ETE R M I NATION OF Z bw
301
=
It is of i nterest to note t h at V4 may be calculated d irectly from Eq. (8 .27) by
setting node (J) equ al to t h e reference node. We then obtain for k
4 and
2 th = 2 44
since
8.4
=
1 . 1 028 1
/
-
20 .7466° p e r
Ic3P i s already calculated i n Exam p l e
unit
8.1.
D I RECT D ETERMI NATI O N O F Z bus
We could determine Z o us by fi rs t fi nd in g Y b u s a n d then i nverting i t, but this is
not convenient for large-scale systems as we h ave seen. Fo rtunate ly, formulation
o f Z hus u s i n g a d i rect b u i l d i n g a l go r i t h m is a straightforward p rocess on the
com pu ter.
At the outset we h 8ve a l i st of the branch impeda nces showing the buses
to which t hey are connected . We start by writing the e q uation for one bus
connected through a branch i m ped ance Zu to the refe rence as
CD
[ VI ] = G) [ Z a ] [ I I ]
(8. 44 )
a n d t his can be considered as an equation i nvolving three m a trices, e a ch of
w hich h as one row and one colum n . Now we m ight a d d a new bus connected to
the fi rst bus or to the reference node. For instance, if t h e second bus I S
connected to the refere nce node through Z b ' we h ave the matrix equation
( 8 .4 5 )
a n d we p roceed to modify the evolving Z bus matr ix by adding other buses a n d
branches following the p roced ures described i n Sec. 8.3. The combin ation of
t hese procedures cons t i tutes the Z bus building a lgorithm . U s u a l ly, the buses of a
network m ust be renumbered intern a ll y by t h e computer algorithm to a gree
with the order in which they a re to be added to Z b u s as it is b u i l t u p .
Example 8.4. D e t e rmi ne Z bu s for the n e twork shown in Fig. 8.8, where t h e
impedances l abeled 1 through 6 a r e s hown i n p e r u nit. Preserve a l l buses.
The branches are added i n t h e order of their l abe l s and numbered
s u b s cr i p t s o n Z h us wi l l i n d i c a t e i n te rmediate steps of the so lution. 'We start by
Solution .
302
CHAPTER 8
THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS
@
�-r-
j 1 .25
(4)
@
Reference
establishing bus
CD
FIGURE 8.8
Ne lwork for Examples 8 . 4 a n d 8 . 5 .
B r a n c h i m ped a nc e s a rc i n pe r u n i l
; 1 I l l! b r a n c h Il u m b e rs a r c i n p ; l r C I l ­
I h eses.
with i ts impedance to the refe re nce nod e and write
CD
[ VI ] = CD U l .2S ] [ 11 ]
We then have the
1
x 1 bus imped a nce matrix
Zbu s , 1
To establish bus
CV
=
CD
CD [ J 1 .25 ]
with i ts i mpedance to bus
[
CD i l .2S
Zb us, 2 =
Cl) il .25
CD ) we
i 1 .25
j l .SO
]
follow Eq. (8.32) to write
The term j 1 .50 above is the sum of j l .2S and iD.2S. The el ements j l.2S in t h e n ew
row and column are the repetition o f t h e clemen ts of row 1 and co lumn 1 o f the
matrix be ing modified.
Bus G) with the impedance con necting it to bus CV is established by
writing
[
CD
CD j l .2S
Zbus , 3 = CV il .2S
Q) il .2S
CV
i 1 .2S
iI .SO
i I .SO
Q)
j. 1 2S
iI .SO
i I .90
J
Si nce the new bus Q) is being con nected to bus CV , the te rm i 1.90 above is t h e
s u m o f 222 of the m atrix being modified a n d t h e imped ance 2 b of t h e branch
being connected to bus CV from bus Q) , The o t he r e l e m e n ts of the new r9w a n d
column are t h e repetition of row 2 a n d column 2 of the matrix be i n g modified
since t he new bus is being con ne c t e d to bus CV .
8 .4
Dl R ECf D ETE R M I NATION
OF Z bua
303
If we now decide to add the impedance Zb = j l .25 from b us G) to the
re ference n ode, we fol l o w Eq . ( 8 . 3 2 ) to conn ect a new bus ® t h ro u gh Z b a n d
o b t a i n the i mpedance matrix
j.
Z bu s . 4 =
CD
CD
@
G)
3
®
W
j l .25
j l . 25
j 1 .25
j I .50
j l .25
j l .25
+
G)
®
j. 1 25
j 1 .5 0
j 1 .50
j 1 .25
j l .90
j 1 . 90
j 1 .50
j 1 . 90
j3.15
j . I 50
w here 3 1 5 a bove i s t h e s u m o f Z :1 3
T h e o t h e r c l e m e n t s i n t h e n ew row a n d
col u m n a re t h e re p e t i t io n o f row 3 a n d col u m n 3 of t h e m a t r ix b e i n g m odi fi ed
s i n c e bus
is be i n g con n e ct e d to t h e reference nod e t h roug h
We now e l i m i n a te row p a n d co l u m n p by Kron red u c t i o n . Som e o f the
c l e m e n ts o f t he n e w m a t rix fro m Eq. ( � . 3 3 ) a r e
G)
ZI
Z22(new)
Z2.\(new)
j l . 25
J ( new) =
= Z 3 2(new)
=
j 1 . 50
.
=)1
Z" .
Zb'
( j l . 25 ) ( j l . 2S )
-
-
.5 0 -
=
j3 . 1 5
( j l .50) ( jl .50)
=
j3 . 1 5
( j 1 . 50) ( j 1 .90)
j3 . 1 5
j O . 75 3 97
j O .7857 1
= j O . 59524
�jOO.75Y7n jO.65477() jO .496U3
J
W h e n all the e l em e n t s are d e t e rm i n e d , we have
Z b " , . :;
=
CD
Q) j
G)
u s i n g E q . (8.32), a nd we obt a i n
We
lJOw
decide
10
Z bus. 6
add
=
I he
CD
@
G)
@
CD
. 65 4 6
j O .7H5
CD
1
j O .59 5 2 4
j O .49 60 3
i m ped a n ce Z"
G)
W
�
jO . 5 9524
)0 .75397
jO .20 fro m b u s
@
CD
t o e st a b l is h bus
G)
@
jO.59524
jO .49603 I jO . 49603
jO. 753 97
jO.65476
j O . 65476
jO.785 7 1
jO.59524
j O.49603
jO. 59524
jO. 753 97
j O .49603
j O . 5 9524
jO.75397
I
jO . 75397
jO.95397
@
304
CHAPTER 8
THE IMPEDANCE MODEL AND N ETWORK CALCULATIONS .
The off-diagonal el ements of the new row and column are the re p e ti t i o n of row 3
and column 3 of the matrix being modified because the new bus @ is being
connected to b u s ® . The new diagonal element is the sum of Z33 of the previous
matrix and Zb = jO.20.
Finally, we add the impedance Zb = jO.l25 between buses G) and @ . If
we l et j a n d k in Eq. (8.41) equal 2 and 4 , respectively, we obtain the e l ements for
row 5 and column 5
Z I5
=
ZI 2 - ZI 4
=
j O . 65476 - j0.49603
Z2 5
=
Z22 - Z24
=
jO. 7 85 7 1 - jO.59524
Z 35
=
Z45
=
Z32
- Z34 = j O . 5 95 24 - jO .75397
Z42 - Z4 4 = j O . 5 95 24 - jO .95397
=
jO. 1 5873
=
jO . 1 9047
=
=
- jO. ) 5873
-jO .35873
and f r o m Eq. (8.42)
=
So, em ploy ing
j { 0 .785 7 1
Z b uS. 6
® jO. 1 5873
+
0.95397 - 2 ( 0 .5 9524 ) }
+
previously fou nd , we write the 5
jO . 1 25 = j O . 6742 1
x
5
m a t r ix
®
jO. I 5873
j O. I 9047
-jO. I 5873
-j O.35873
jO. 1 9047
-jO. 1 5 873
- jO.35873
jO.67421
a n d from Eq. (8.43) we fi nd by Kron red u ction
CD
Z bu s
=
G)
®
@
@
j O .60992
®
j O .53340
jO.58049
jO .60992
j O .73 1 90
jO . 64008
jO .6965 9
jO .53340
j O . 64008
j O . 7 1 660
jO.6695 I
jO.58049
jO .69659
jO .6695 1
jO .763 1O
CD
G)
j O .7 I 660
which is the bus impedance matrix to be determined. All calcu lations ha,-:e been
rounded off to five decimal places.
8.4
D I R ECT DETE R M INATION OF Z bu.
305
Since we shall again refer to these results, we note here that the reactance
diagram of Fig. 8.8 is derived from Fig. 7. 1 0 by omitting the sources and one of
the mutually coupled branches. Also, the buses of Fig. 7. 1 0 have been renum­
bered in Fig. 8.8 because the Z b us building algorithm must begin with a bus
connected t o the reference node, as previously remarked.
The Z bus bu ilding procedures are simple for a computer which first must
determine the types of modification involved as each branch impedance is
added. However, the opera tions must follow a sequence such that we avoid
connecting an impedance between two n ew buses.
As a matter of interest, we can check the impedance values of Z b us by the
network calculations of Sec. 8 . 1 .
Exa m p l e 8 . 5 . Find
ZII
m e a s u red be t w e e n b u s
W , CD,
Solution .
a nu
Q)
ar�
o f t he
CD
c ircu it 0 r Exam p l e 8.4 by d e t e rm i n i ng the i mpeda nce
t h e re ference nod e wlH ; n cu r rents i nj ected at buses
and
z e ro .
T h e e q u ati o n co r respond i n g t o E q . ( 8 . 1 2) i s
W e r e cog n i ze tw o p a ra l l e l pa t hs
8.8 w i t h
b e tw e e n
b u ses
the resu l t ing i m p ed a n c e of
+ jO . 20 ) ( j 0 .40)
j ( 0 . 1 25 0 .20 + 0 .40)
( jO . 1 25
-�-----
+
This i mp e d ance in ser i es
w i t h ( j0.25
y i e ld
+
j 1 . 25 )
(1)
and
G)
of
the c ircuit
of Fig.
= j O . 1 793 1
comb i n e s in
parallel
w i th
j 1 .25 to
j 1 .25 ( j0 .25 + j l .25 + j O . 1 793 1 )
= jO . 7 1 660
=
ZI I
j( 1 .25 + 0 .25 + 1 .25 + 0 . 1 793 1 )
which
is i d e n t i c a l
w i t h t h e val u e
fo u n d i n
E x a m p l e 8.4 .
Although the network reduction m e t hod of Example 8.5 may appear to b e
s i m p l e r b y com parison wi t h othe r methous of form i n g Z bus ' such is not the case
because a different network reduction is requ ired to evaluate each element of
the matrix. In Example 8.5 the n etwork reduction to find 244 , for instance, is
more difficult than that for find ing 2 1 1 , The computer could make a network
reduction hy node e l i m i n ation bu t wou ld have to repeat the process for each
n o d e.: .
306
CHAPTER 8
THE IMPEDANCE MODEL AND N ETWO RK CALCULATIONS
CALCULATIO N OF
FROM Ybus
8.5
ELEME NTS
Zbus
When the full numerical form of Z b us is not explicitly required for an applica­
tion, we can readily calculate elements of Z b us as needed if the upper- and
lower-triangular factors of Y bus are available. To see how this can be done,
consider postmultipiying Z h u s by a ve c to r with only one nonzero element 1 = 1
in row m and all other elements cqual to zero. When Z o u s IS an N X N matrix,
we have
m
CD
@
@
®
-
CD
Zl 1
Z2 1
Zm l
ZN l
@
Z 12
Z22
Zm 2
ZN 2
@)
®
Zmm
Zm N
ZIN
Zim
Z2N
Z2m
C§
-
1m
Z NN
ZNm
CD
0
0
0
®
0
Z b us
C§
Zim
Z2 m
Zmm
( 8 .46)
ZNm
-----
m)
z (bus
Thus, postmultiplying Z bus by the vector shown extracts the m th column, which
we have called the vector Z h�;; that is
z (m)
bus £
column
of
m
Z bu s
CD
0
®
®
-
@)
Zim
Z2 m
Zm m
Z Nm
Since the product of Y b us and Z bus e qual s the unit matrix, we have
o
o
Y bus Z bus 1 m
o
-
) Y bus z(m
bus -
o
o
1
(8. 47)
m
o
I
If t h e lower-triangular matrix L and the upper-triangular matrix U of Ybus are
CA LCULATION OF Zbu. ELEMENTS FROM Ybu£
8.5
307
available, we can write Eq. ( 8 .47) in the fonn
0
0
1m
=
Luz(m)
bus
( 8 .48 )
0
I t i s now a pparent t ha t t h e e l e m ents i n t he col u mn ve ctor Zb':� can be fou n d
from Eq . (8,48) by forw a rd e l im i n a t ion and back subs t i tu tion, as ex p l a i n e d i n
S e c . 7.8. I f o n ly som e o f t h e e le m e n t s o f z b:l a re req u i re d , t h e calcu lations c a n
b e r e d uc e d accord ingly. For exam p l e , su ppos e t ha t w e wish t o generate Z 3 3 a nd
243 o f Zbus for a fou r- b u s sys t e m . Us i n g conve nient notatio n for t h e e le m e n ts o f
L an d U , w e have
1
III
12 t
/ 22
/3 2
/ 3
3
'4 1
[ 42
1 4 :'
'3 1
U 13
Ul2
il 2
1
3
Ul4
.2 1 3
U 24
Z2 3
U 34
Z 33
1
Z4 3
1
14 4
0
0
1
0
( 8 .49)
�
We c a n solve this equat ion for
i n two st eps as follows:
XI
'I I
12 1
1 22
1 '\ 1
/32
{4 1
1 2
4
U l2
w here
Z�Js
(3)
Z bus
I 33
1 4 .1
un
u 23
1
1
1 44
U I4
U 24
U 34
1
0
0
1
X2
X3
zu ]
0
X4
223
2 33
XI
x2
=
x]
24 3 J
L
------
By forward
( 8 . 50)
(3)
Z bllS
( 8.51)
X4
substitution Eq . (8.50) i m m ediately yie lds
XI
=
0
X2 -
a n d by b a c k s u b s t i t u t i o n o f
0
1
X4
-
t hese i n t e rm e d i a t e re s u l t s i n Eq. ( 8. 5 1 ) we fi n d the
308
CHAPTER 8
THE IM PEDANCE MODEL A N D N ETWORK CALCU LATIONS
required elements of column 3 of Z bus ,
If
all elements of
z i;�s
are required, we can contin ue t he calcu l ations,
The computational effort in gene rating t he required clements can be red uced by
judiciously choosing the bus numbers.
In l ater chapters we shall find it necessary to evaluate terms like ( Z illl Zjn) involving differences b etween columns (if!) and ® of Z b I f the elements
o f Z bus are not available exp licitly, we can calculate the required differences by
stS1ving a system of equations such as
us '
o
n
Luz(m
bus- ) =
( 8 .52)
- 1n
o
where Zb�s- n) = Z�:l - Z���� is the vector formed by subtracting column n from
column m of Z bus , and 1 /11 = 1 in row m and - I " = - 1 in row n of the vector
s hown .
In large-scale system calculations considerable computational efficiency
can be realized by solving equations in the' triangu larized form of Eq. (8.52)
while the ful l Z bus need not be developed. Such computational consider(\tions
underlie many of the formal developments based on Z b in this text.
,
'
The five-bus system shown in Fig . 8.9 has per-unit impedances as
marked. The symmetrical bus a dmittance m atrix for the system is g iven by
us
Example 8.6.
Ybus
=
CD
CV
®
0
G)
CD
CV
®
0
0
G)
0
0
-)30.0
) 1 0 .0
)1 0 .0
-)26 .2
) 1 6 .0
0
) 1 6 .0
-)3 6 .0
0
0
-j20 .0
0
0
-f20 .0
0
)20.0
0
0
0
f20.0
)20 .0
)20 .0
8.5
CALCULATION OF Z tKu ELE M ENTS FROM Ybus
309
a n d i t is fou n d t ha t t h e triangu l a r factors of Y bus are
-
j 30
.
0
}1O.0
L=
0
}20 .0
-} 2 2 . 866667
} 1 6 .000000
-}24 .804666
}6 . 66 66 67
}4 .664723
0
1
0
- 0 .333333
-j3 .84579 3
j3 .761 164 -jO.195604
} 20 .000000
o
- 0 .69970 8
U =
0 666667
- 0.291545
- 0. 1 88051)
-
.
o
o
- 0 . 806300
- 0 . 977995
1
Use the t ria ngu l a r fact ors to c a l c u l a t e Z lh. 45 = ( Z 4� - 245) ( Z54 - Z55), the
Thcvenin i m pedance l ooki n g i n t o t h e system between b uses @ a n d G) of F i g .
-
8 .9.
Solution. S i n c e Y bus is symm e t r i ca l , t h e reader should c heck t h a t t h e row
of U equ al the col u m n e l eme n ts of L d ivi ded
t h e i r corresponding
elements
d i a gonal
elements. With l's represen ting the n u merical values of L, forward solu tion of the
system of equa tions
by
'1 1
'21
'22
1 32
1 33
141
1 42
14 3
1 44
15 1
1 52
15 3
1 54
131
15 5
Xl
0
X2
0
X3
0
x4
1
-1
Xs
®
®
) 0.0625
)5.0
)0.1
FIG U RE 8.9
R e a ct a nce d ia g r a m for Exa m p l e 8.6, a l l
imped a nces.
,
va l u e s
are p e r-unit
310
CHAPTER 8 THE I MPEDANCE MODEL AND NETW O R K CALCULATIONS
yie lds the intermediate
X4 =
1«\
=
values
( -j3.845793) - \
=
Backsubstituting in
1
=
jO.260024
- 1 - j3 .76 1 1 64 x j0.260024
-j0 . 1 95604
-j0. 1 12500
t h e system o f eq u at i o n s
U
12
1
UD UI4 1 5
u2) U24 u25
[ZC4 - 5 l]
U
1I 45
0
0
0
jO.260024
-jO.1 12500
It
:14
I t :IS
hu�
1
where u 's represent the numerical values of U, we
fi n d
from the l ast two rows that
-jO. I l 25 per u nit
Z54 - Z55
=
Z44 - Z45
=
jO.260024 -
=
jO. 1 500 per u nit
U45 ( Z54 - Z55 )
=
jO.260024 - ( - 0 .977995 ) ( -jO. 1 125)
Th e desi red Thevenin impedance is therefore calculated as follows :
Z th . 45
=
( Z44 - Z 45 ) - ( Z54 - Z55 )
I nspection of Fig.
8.9 verifies
=
jO. 1500 - ( -jO.l 125)
=
jO.2625
per u n i t
this resu lt.
8.6 POWER INVARIANT
TRANSFORMATIONS
The complex power in a network is a physical quantity with a value which
should not change simply because we change the way we represent the network.
For example, in Chap. 7 we see t h a t t h e n e twork cu rrents a n d v ol t a g es may be
chosen as branch quantities or as bus quantities. In either case w e should expect
the power in the branches of the network to be the same regardless of which
quantities are used in the calculation. A transformation of network variables
which preserves power is said to be
For such transformations
involving the bus impedance matrix certain general relationships must be
satisfied, which we now establish for use in l ater chapters.
Let V and I describe the set of bus voltages and currents, respectively, i n
the network. The complex power associated with these variables is a
power invariant.
I
scalar
8.6
quantity ,
311
P O W E R I N V A R I ANT T R A N S FO RM A:r I O N S
w h i ch w e may rep r e s e nt b y
( 8 .53 )
o r in matrix form by
v2
'
"
VN ]
1 1*
/ 2*
( 8 .54)
1,".,.
*
S u p p ose t h a t
we t r a n sfo r m t h e bus c u r r e n t s I to a new set of
u s i ng the t ransformat i on m a trix C s u ch t h a t
I
=
bus c u r r e n t s
I new
( 8 . 55 )
C 1 n ew
we shall s e e b e l ow s u c h a tran s fo r m a t i u n occu rs, for i n s tance, w hen the
re fe r e n ce node of t he n e t work i s cha nged a n d it i s req u ired to compute t h e new
bus i m ped a n ce matrix, wh ich we c a l l Z b ll S(lleW) ' The bus v ol t age s in terms of the
existing and the new variab les are repre se n ted by
As
,
( 8 .56)
an d
and we now seek t o establ ish the co n d i t ions to be satisfied b y Vn ew and Z bUS(neW)
so t h a t the power rema ins inva r i a n t w h e n the curren ts are c h a nged ac c o r d i n g to
Eq . (8 .55).
S u b s t i t u t in g V from Eq . (8.56) in E q . (8.54) gives
where
obt a i n
Z h m i s sy m m e t r i c a l . F r o m Eq .
(K5S)
we su bs t i t u t e fo r I
( 8 .57)
i n Eq . (8 .57)
to
( 8 .5 8 )
fro m
which
it
fol l ows t h a t
( 8 .5 9)
Z huS(nCw )
Compa ring Eqs. (8.57) a n d (8.59), we see t h at the complex p ow er will be
i nvariant in t erms of the n ew variables, p rovid e d the new bus impe 9 a nce m atrix
312
CHA PTER 8
TH E IMPEDANCE M O D EL AND NETWORK CALCU LATIONS
is calculated from the relationship
( 8 .60 )
This is a fundamental result for constructing the new bus impedance matrix.
From Eqs. (8.56) and (8.59) we find that
( 8 .6 1 )
It also follows from
Eq.
(8.54) that
( 8 .62 )
and we m ay then conclude from Eqs. (8.6 1 ) and (8.62) that the new voltage
variables Vnew must be related to the existing voltage variables V by the
fundamental relationship
( 8 .63 )
In many transformations, especially those involving the connection matrices of
the n etwork, all the entries in C are real, and in such cases we drop the complex
conjugate superscript of C* .
Equation (8.53) is the net sum of all the real and reactive power entering
and leaving the buses of the network. Hence, SL represents the complex power
loss o f the system and is a phasor quantity with real and reactive parts given by
Eq. (8.59) as
The complex conj ugate of the
S*L
-
PL
transpose
-
J'Q L
-
of Eq. (8.64)
is
I Tnew CTZT*
b us C* 1 *n ew
·
( 8 .64 )
( 8 .65 )
Adding Eqs. (8.64) and (8.65) together and solving for PL yield
PL
=
[
1
Z b us + Z Tb us*
IT CT _
C* 1*new
new
2
( 8 .66 )
When Zbus is symmetrical, which is almost always the case, we may write
Z b us = R b us +
jXbus
( 8 .67)
where both R b us and X b us are symmetrical. We note that R b us and X b u s are
available by inspection after Z bu s has been constructed for the ne�ork.
Substituting from Eq. (8.67) in Eq. (8.66) cancels out the reactance part of Z bu S '
POWER I N VARIANT TRANSFORMATIONS
8.6
II
CD
------+-
12
®
®
Vl
Zbus
Vz
t
Initial
Reference
®
FI G U RE 8.10
----:----:
a n c.!
we t he n
3 13
II n cJ
Changing
Z bus ·
I"
I)L
-
IF
n l' W
e r R e* l *
bus
the
re fe re nc e
of
( 8 .68 )
new
wh ich simplifies the n u m e ri cal calcu l a t i o n of PL smce o nly the resistance
p o r t io n o f Z bu s is i nvolved.
An i mpor t ant a p p l i ca t i o n o f Eqs . (8.60) and (8.6 3 ) a ri ses when the re fe r­
ence n o d e for the Z bu s represe n t a t i o n of the system is change d . Of course, we
cou l d again use the b u i l d i n g a lgori t h m of Sec. 8.4 to rebu i ld completely t he n ew
Z bus starting with the new reference node. However, this wou l d be com p u ta­
ti o n a lly i n efficient, a n d we n ow s how h ow to mod ify the existing Z b us to accou nt
for the ch ange of reference n od e . To i ll us trate , consider t h a t Z b us has already
b een constructed for the five-nod e system of Fig. 8 . 1 0 based on node ® as
reference . The standard b us e q u a t i o n s are t hen written as
VI
V2
V:l
V4
CD
@
G)
@
CD
211
Z21
212
Z 22
Z4 1
Z 32
Z42
Z :lI
a n e!
®
213
Z 23
Z 33
Z4 3
@
214
224
Z34
Z 44
II
12
( 8 .69)
I."
14
V4
a re measu red w ith respect to node
as reference, a n d the c u rre n t i nj ections 1 1 , /2 , /3 , and /4 a re i ndepend ent.
i n w h i c h t he b u s vo l t a ge s VI ' V2 , V:l '
®
@
Ki rchhoff's curre n t l aw fo r Fig. 8 . 1 0 shows t hat
( 8 . 70 )
If we n ow change the refe rence from n o d e ® to nod e @ , for inst ance, t h e n 14
is no longer i n d epen d e n t s i nce i t c a n be expressed in terms of the other four
314
CHAPTER 8
TH E IMPEDANCE MODEL A N D NETWORK CALCULATIONS
node currents. That is,
(8 71)
.
From Eq. (8.71) it fol lows that the new independent current vecto r I new is
related to t h e old vector I by
I}
12
I."
14
0
0
0
1
0
0
1
0
-1
1
-1
1
-
--
I
0
0
0
-1
I}
12
( 8 . 72)
13
1/1
-
I ncw
C
Equation (8.72) is merely a statement that II > 12 , and 13 remain as before but
that 14 is rep l aced by the inde p endent current In ' which app e a rs in the new
current vector I n ew as shown. In the tra nsformation matrix C of Eq. (8.72) all
the entries are real, and so substit u tin g for C and Z b u s in Eq. (8.60); we fin d that
Zb us(n ew ) =
1
0
0
0
0 0 -1
1 0 -1
0 1 -1
0 0 -1
c
211
22 1
23 1
24 1
212
2 22
2 32
2 42
T
2 13
223
2 33
243
1
0
0
214
2 24
2 34
244
-1
0
0
1
-1
0
1
0
-1
c
Z bus
0
0
0
-1
( 8 .73 )
The matrix m u ltiplications in Eq. (8.73) arc accompl ished in two easy stages as
fol lows. First, we compute
2 1 1 - 241
2 2 1 - 24 1
C T Z buS =
23 1 - Z4 1
- Z4 1
2 12 - 242
2 22 - 242
2 32 - 242
- 2 42
2 1 3 - 2 43
22 3 - 2 43
233 - 2 43
- 243
2 } 4 - 244
224 - 244
234 - 2 44
- 244
( 8 .7 4)
which for convenience we write in the form
C TZ bu S =
2� 1
2;1
2; 1
2� I
212
2 ;2
2 ;2
2 �2
2 i3
2 ;3
2 ;3
2 �3
2� 4
2 ;4
2 ;4
2'44
( 8.75)
lH,
P O W E R I N V A R I A NT T R A N S F O R M AT I O N S
3 15
I t is evi d e nt from Eqs. (8.74) a nd (8.75 ) t h at t h e e l em ents w i t h p r i m e d super­
scri p ts a re fou nd by sub tract i ng t h e exist i ng row 4 from e ach of t h e other rows
of Z b u s a n d by changing t h e sign of existi ng row 4. S econd, we postm u l t i ply Eq.
(8.75 ) by C to obta i n
=
CD
@
G)
®
CD
Z; I Z 2' 1 Z; I L 4' -- 1
@
Z'
Z ; 2 - Z;4
Z �2 - Z 24
Z;2 - Z;4
Z�2 - Z�4
Z �4
Z ;4
Z �4
W
Z;3 - Z;4
Z;3 - Z;4
Z ;l
34
Z�3 - Z�-l
-
@
Z� 4
- Z;4
( 8 .76 )
- Z; 4
- Z� 4
w h i c h i s e q u a l l y s i m p l e t o c o m p u t e from C " Z i JUS b y s u b t r a c t i n g t h e forth colu m n
o f Eq. (8.75) fro m each o f i t s o t h e r col u m ns a n d b y changing t h e s i g n o f the
fou rt h colu m n . I t is worthw h i l e n o ting that the fi rst d iagonal e l e m e n t of
Z b u s(ne w) ' expressed in t e rms of t h e original Z b us e nt ries, has t h e va lue ( Z ; 1
Z ; ) = ( Z I I + Z-l 4 - 2 Z I ), w h ich is the Th e ven i n i mpedance between nod es
CD a n d @ J a s we wou l d expect from Eq . (8. 26). S i m i l a r rema rks a p p ly to e a c h
of t h e o t h e r d iagona l e le m e n ts o f Z b u s ( ncw ) '
The bus vol tages with respec t to t he n ew reference node @ are given by
Eq . ( 8 . 6 3 ) as fol lows :
-
Vnew
=
VI
new
V2.
new
v.I . n e w
v'1 .
new
1
0
0
0
1
0
0
0
0
0
1
()
"
-}
-1
- }
-- 1
VI
V2
V3
V4
VI - V4
V2 - V4
V3 - V4
- V4
( 8 .77 )
C'
Therefore, i n the ge n e ra l c ase whe n b u s ® of a n ex is t i ng Z h u s i s c hosen
as the n ew re fe re nce node , we m ay d e t e rm i ne t he n ew b u s i m pe d a n ce m a t r ix
Z b uS(n cw) i n two consecll t ilJe s t e p s :
1. S u b t ra ct the ex ist ing row k from each of the oth e r rows i n Z bu s a n d c h ange
t he sign of row k. The resu l t i s C T Z bus '
T
2. S u b tract col u m n k of t h e r e s u l t a n t m a t r ix C Z bu s from e ach of i ts other
c o l u m n s and c hange the s ign of col umn k. The result is C T Z b u C = Z h u s( new)
with row k and col u m n k now r e p r e se n t i n g t h e node which was t h e p revi o us
S
refe re nce node.
316
CHAPTER 8 TH E IMPEDANCE MODEL AN D N ETWORK CALCU LATIONS
We use these procedures, for example, when studying economic operation In
Chap. 13.
I N Z bus
8.7
MUTUALLY COUPLED BRANCHES
So far we have not conside red how to inco rporatc mutually couplcd c lcments of
the network into Z hu s ' The proccdures for doing so are not difficult. B u t they
a re somewhat unwieldy as we now d emonstrate by extend i ng the Z h us buildi ng
algorithm to provide for addit ion to thc netwo rk of one pa ir of mu tua lly coupled
branches. l One of the bra nchcs may be a d u cd to Z u rig using thc appropr i a te
procedurc of Scc. K3, a nd thc qucstion rcma in ing i s how to add thc sccond
branch so that it mutually cou p lcs with thc branch already i nc luded in Z u ri g ' W c
consider bus ® already established within Z u rig in the fol lowi ng four cases,
which continue the numbering system introd uced in Sec. 8.3.
CASE
5.
Adding mutually coupled Zb from existing bus
®
to new bus ®
. Let u s assume that the branch impedance Z a is a lready added to the energized
·n etwork between nodes @) and ® of Fig. 8.1 1 . The bus impedance rn a trix
Z orig then includes Z a and the existing buses @) , ® , a !1 d ® as shown.
Be.tween bus ® and the n ew bus ® of Fig. 8. 1 1 it is required to add the
branch impedance Zb ' which is mutually coupled to Za through mutu a l
impedance ZM ' The voltage-drop equations for the two coupled branches are
given i n Eq. (7.9) and repeated here as
( 8 .78 )
( 8 .7 9 )
where fa is the branch current flow in Z a from bus @) to bus ® and t he
c urrent fb from bus ® to bus ® equals the negative of current injection fq .
Solving Eq. (8.78) for fa and substituting t he result a long w i th Ih = - Iq in Eq.
(8.79) give
( 8 .80)
In terms of bus voltages the voltage d rops across the branches are given by
Va = ( Vm - Vn ) and Vb = ( Vp - Vq ), and substituting these relationships in Eq.
I For a more general treatment, see
System Analysis, McGraw-H i l l , I n c . ,
G. W.
Stagg and A. H. EI-Abiad, Computer Methods in Power
1 968 .
New York,
,
8.7
M UT U A LLY COUPLED B R A N CH ES IN Z bu ,
317
Zor ig ® -.---1----,
@)
•
I
,
I
®
FIG L' RE 8. 1 1
A d d i n g m \l t u a l l y co u p l ed bra n c h Z" to
Reference
(8.80)
yie l ds
ZM
Z
q
V = V · - - ( Vm - Vn ) p
-
a
This eq u a t i o n gives t he vo l t a g e a t n e w b us
[ Z�
Z 1
n
®
- zb 1
( 8 .8 1 )
q
with
both mu tually cou p l e d
b ra nches now i ncluded i n the n etwork. T h e b us i mpedance m atrix for t h e
sys tem augme nt ed b y n e w b u s ® i s g iven by
®
Z lq
Z 2 Q 12
ZNq
Zq f',' Zqq Iq
VI
V2
VN
Vq
®
Zq l Z q 2
Z o r ig
11
IN
( 8 . 8 2)
a n d it is now req u i r e d to n n d e x p re ss i o n s ro r t he new c l e m e nts w i t h subscrip ts q
i n row q and co lumn q . A typical row i of Eq. (8 .82) may be wr i t te n i n t he for m
( 8 . 83)
V/
where
for conven ience we h ave
O
d e n oted
VO r
N
"
L Z .I.
j� )
I} }
( 8 . 8 4)
318
CHAPTER
8
THE I MPEDANCE MODEL A N D NETWORK CALCULATIONS
Setting i equal to m, n, p , and q in Eq. (8.83) gives expressions for Vm , Vn , V ,
p
and Vq , which can be substit uted in Eq. (8.8 1 ) to obtain
_
(
2 1�
2 II
_
)
2 b Iq
( 8 .85 )
Equation (8.85) is a general equ ation fo r the augmented network regardless of
the particular values of the cu rrent inj ections. Therefore, when 1 q = 0 , it must
follow from Eq. (8.85) that
Vqa
=
Va
p
-
2
�
2
a
( Vam
-
Va)
( 8 .86)
n
Substituting i n Eq . (8.86) for v,�, Vno , Vpo , and VqO from Eq. (8.84), collecting
terms, and eq uating coefficients o f lj on both sides of the resultant equation, we
find tha t
( 8 .87)
1
for all values of j from to N but not including q. Thus, excep t for 2qq, Eq .
(8.87) tells us how to calcu la te the entries in the new row q of the bus
i mpedance matrix using the known val u es of 2 M , ZU , and certain el ements of
Z o rig' Indeed, to obtain the entry in row q , each c lement of row n is subtrac ted
from the corresponding eleme n t of row m and the difference mul tiplied by
2m /Za is t hen subtracted from the corresponding el ement of row p . Thus, only '
rows m , n , and p of Z orig enter into the calculations of new row q . Because of
symmetry, the new col umn q of Eq. (8.82) is the transpose of the new row q,
and so Zq J Zjq . The expression for the d iagona l element Zqq is determined by
considering a l l currents except lq set equal to zero and then eq uating t h e
coefficients of 1q o n both sides o f Eq. (8.85), which gives
=
>
( 8 .88)
This eq uation shows that there is a seq ue nce to be fol lowed in determining the
new el ements of the bus impedance matrix. First, we calculate the elements 2qj
of the new row q (and thereby 2jq of col umn q ) employing Eq. (8.87), and then
we use the newly calculated quantities 2mq , Znq ' and 2pq to find 2qq froI? Eq.
(8.88).
H.7
M UT U A LLY COU P L E D I3RAN C H ES I N Z bul
319
T here are three o the r cases of i nterest i nvolving mutually coupled
b ra n ch es.
Adding mutually coupled
CAS E 6.
Zb
from existing bus ® to reference
For t hi s case t he procedure is b as i cally a special application of Case 5 . First,
betwee n bus ® and new bus ® w e add i mped ance Zb cou p l e d by m u t u a l
i m p e d ance Z M t o t he i m p e dance Za a lready i n cluded i n Z orig ' Then, we
shor t-circu it bus ® t o the refe rence node by setting Vq equal to zero, w hich
yie l ds the same m at r ix e q u ation as Eq. (8.82) excep t that Vq is zero. Thus, for
t h e modified bus i m p e d ance matrix we proceed to cre ate a n ew row q a n d
colu m n q exactly t h e s a m e as i n Case 5 . Then, w e el im inate t h e newly forme d
row a n d col u m n b y t h e stand ard tech n i q u e of Kron reduction si nce VI] i s zero i n
t h e column o f vol tages 0 f Eq. (8.8 2 ).
A dding mutually coupled Zh bctween cxisting buses ® and CD
CA S E 7 .
The p r oc e d u r e i n t h i s case essen t i a lly combines Cases 5 a n d 4. To begin,
we follow the proce dure of Case 5 to add the mutually cou p led bra n ch
i mp edance Zh from existing bus ® to a new temporary bus ® , recogn iz ing
that Z a is a lready part of Z orig ' Th e resu l t i s the augmented matrix of Eq. (8 .82)
w h ose q-elements are given i n Eqs. (8.87) and (8.88). N ext, we short-circu i t bus
® to bus ® by a d d i n g a b ranch of zero i mp edance between t hose buses. To
do so, we apply Case 4 to Eq . (8 .82) as follows. S i nce ( Vq - Vk ) is require d to b e
zero, w e fi n d a n expression for t h a t quantity b y subtracting row k from row q in
Eq . (8.82) and t hen u se the resu lt to replace the existing row q of Eq. (8.82). By
symmetry as in Case 4, a n ew colu m n follows d i re ctly from the t ranspose of t h e
n ew row, and w e obt a i n
( co l . q - col . k )
Eq .
( row
q
-
row
of
( 8 .82)
k ) o f Eq . ( 8 . 82)
( 8 .8 9 )
wh ere Zc equals t h e su m C Z q q + Zk/.: - 2 Zqk ) of elements d rawn from Eq.
(8 .82). Becau se ( Vq - V,) e quals 0, we m ay e l i m i nate the n ew row a n d n ew
co l u m n of Eq. (8 .89) by Kron re d u c tion to fi n d the final form of t he N X N b u s
i mp e dance mat rix.
To remove a m u t u a l ly coupl e d b ranch , we must mod ify t h e above proce­
du res as now expl ai n e d .
., j
320
CHAPTER 8 THE I MPEDANCE MOD\ L A N D N ETWORK CALCU LATI ONS
CASE 8.
Removing mutually coupled Z" from
existing buses ® a nd ® .
"
A single u ncoupled branch of impedance Zb can be removed from t he
network model by adding the negative of imped ance Zb between the same
terminating buses. When the impedance Z b to be removed is a lso mutually
coupled to a second branch of impedance Za ' the rule for mod i fy i n g Z b us is to
add between the end buses of Z" a bra nch, which has n egat i ve i m p eda nc e - Z h
and the sam e m u t u a l co u p l i ng to Z" as t he o r ig i nal Z" . The p r oc e d u r e is
i l l u str a t ed in Fig. 8. 1 2. As shown, the m u t u a l ly cou p l e d branches ZI/ a n d Z"
with mutual impedance Z M a re a l re a d y i n cl u d e d i n Z orig ' I n acco r d a n c e w i t h
the a bove rule, we add the impedance - Z/; having m u t u a l coupl i n g Z M w i th
the branch Za ' and the voltage-drop eq u a t i on s fo r t h e three m u t u a l l y cou pled
branches then are
( 8 .90 )
( 8 .9 1 )
( 8 .92)
where the branch Currents la' Ib , and I� are as shown in Fig. 8 . 1 2 . S u bt r a c t i n g
Eq. (8.92) from Eq. (8.9 1 ) yields
( 8 .93 )
which shows that ( Ib + I� ) is zero, and substituting this result in Eq. ( 8 .90) gives
Because
Zorlg
( J"
( 8 . 94 )
+ 1;,) equals zero, the net cou plin g effect of the two pa ral lel
®
@
I.
J
Za
ZM
Zb
®
•
FIG U RE 8. 1 2
Removing mutua lly cou pled
branch Zb from Z orig ' Initially,
the switch is considered open
and the branch Zb is added
from existing bus ® to tem­
-
ZM
®
1 I.
•
•
porary bus ® . The swi{ch is
t hen closed to con nect bus ®
to bus
®.
117
M UTUALLY CO U P LED BRANCHES IN Z bus
321
branches between buses ® and ® is zero. Consequen tly, the im pedance Z a
between buses @) and ® can be made to stand alone as evide nced by Eq.
(8 .94). To do so, we fo llow exactly the same proced ure as i n Case 7, except that
the eleme n ts o f t h e n ew row and col umn of the temporary bus ® are
calc u l a ted sequentially lI s i n g the m od i fi e d fo r m u l � s
( 8 .95 )
with index j
ranging
�
L'I'I
=
fro m 1 to N and
- - Z,H [ Y., ! YM ] [ - Z 1 - ( y" Z;,
-
LillI
Z ""I
z
1"1
_
z
" '1
-' k 'i
+
2/,)
( 8 .96)
Usi n g t h ese clements i n Eqs . (RR2) and (R.R9), (l n d Kron red ucing the l atter,
gi v e s the des i red new b u s i m p e cJ: i n c e m a t r i x fro m w h i c h t h e m u t u a l ly coupled
branch b e tw e e n b uses ® a n cl ® i s o m i t t e d . Equ a tions (8 . 95 ) and (8.96)'
( d eveloped in Prob. 8 . 1 9) h ave admittances Ya and YM , which a re cal cu l ated
from the impedance pa rameters Z(I ' 2 ,> , and 2 M accord i n g to Eq. (7. 1 0).
Table 8.2 summarizes the p rocedures of Cases 5 t hrough 8.
E x a m p l e 8 . 7 . Betwt.:en buses CD a n d @ o f F i g . 8.8 i m pe d a nce Zb e q ua l t o jO.25
per u n i t is connected so that it cou p l es t h rough m u t u a l i m pedance j0. 1 5 p e r u n it
to the branch impedance a l ready con nected between buses CD « n d 0 . Modify
t h e b u s i m pe d a nce m a t rix o f Example 8.4 to i ncl u d e the a d d i ti o n of Zb to Fig . 8.8.
Zb between buses CD and @ correspo n d s to the condit ions
of Case 7 above. Our calcu l a t i on s beg i n with Z orig ' t h e 4 x 4 m a trix solution of
Examp l e 8.4 which i n c l u de s the br anch between buses CD and 0. T o estab l is h
row q a n d col u mn q fo r temporary b u s ®, w e fol low Eq. (8.87), recog n izing t h a t
subscript m = 1 , su bscri pt n = 2, s ubsc r i pt (J = 1 , a nd we find t h a t
Solution. Con n ec t i n g
/. /1{
--Z 1.1
�
The r a t i o ( Z M /Z) is g i v t.: n by ( jO . 1 5 /j O . 25 ) � ·
a long w i t h t h e t.: l e l11 e l 1 t s o f ro w I i l l 1 d row 2 0 1"
obt a i n
Zq l
=
0 .4 Z 1 1
Z,/ 2
=
0 . 4Z 1 2
Zq)
=
0.4Z13
Z I/ 4
=
+
+
+
0 .6Z:; 1 = 0 .4 ( j O . 7 1 660)
+
)
Z 1 /·
-
I
- Z -.
ZM
/. (I
J
-J
s u bst i t u t i ng t h is v a l u e
in t h e «bove e q u a t i o n, we
0.6, a n d
Z orig
O .6 ( j0 .60992) = jO .65259
0 . 6 Z 22 = 0 . 4 ( JO.60992) + 0 .6( j 0 .73 1 90) = jO .683 1 1
0 . 6 Z 2:1 = 0 .4 ( jO.5J340) + 0.6( j 0.64008) = jO .59741
0 . 4 Z 1 4 + 0 . 6 Z 24
=-
0 .4 ( jO .58049 )
+
O .6 ( j 0 .69659) = jO . 650 1 5
ZbU8 modifications; mutual coupling
TABLE 8.2
Case
Add mutually coupled Z b from
Existing bus
Z bU8(new)
® to new bus ®
•
®
®
J
I
5
Existing bus ®
F or m the matrix
•
[
®
Zmi,
row q
col . q
1
refe rence node
10
®
T
I
•
Repeat Case 5 and
•
Eliminate row
•
6
® a nd column ®
by Kron reduction
Existing bus
® to
exis ting bus
®
®
•
•
Repeat Case 5
Then, form the matrix
l
col . q
7
row q
-
row k
and
•
E l i m in ate last row a n d column
by Kron reduction
-
col . k
1
Removal of mutually coupled line
8
zorig @·12 .
Za
®
-----
•
Zb ZM
_�e-----=--t
�-)c:>
I..:
@ -,-
I..::
Zb
•
�-- -
Continue as in Case
7.
-- --- -----�--- - --- ---- , . -.- -------'
1l.7
323
MUTUA LLY COU PLED BRANCHES I N Zb...
These ele ments const i t u t e n ew row q and new colu mn q except for the d i agonal
element
which is fou nd from Eq .
by s e t ting the subscripts m , n , a n d p
as a bove to yield
Zqq'
(8.88)
S u bstituting in this e q u a t i o n for
Z M jO.IS, Za = jO.2S and
=
Zb
=
jO.2S gives
Zq q = 0 .4Z 1 <7 + 0 .6Z21} + j0.16
= 0.4( j0 .65259) + 0 .6( j0.683 1 l ) jO. 1 6 = jO.83090
+
Assoc i a t i n g
E xa m p l e
(0;.4
the
n e w l y ca l c u l a t e d c l e m e n t s of
res u l t s in t h e
5
x
5
row
m a t rix
q a n d col u m n q
Z orig
jO.7 1 660
jO.60992
jO.53340
jO.58049
® jO.65259
jO.60992
jO.73 1 90
jO.64008
jO.69659
jO.683 1 1
jO.53340
jO.64008
jO.71 660
jO.6695 1
jO.59741
jO.5�049
jO.69659
jO.66951
jO.76310
jO.65015
with
Z ori g
from
-®
jO.65259
jO.683 1 1
jO.59741
jO.6501 5
jO.83090
At t h i s point in our sol u t i o n t h e m u t ually cou p l ed b ranch i mpedance
has been
i ncorpora ted i n t o t h e n e twork b e t w e e n bus CD a n d b u s
T o complete the
c o n n e c t i o n o f Zb to b u s 0 , w e m u s t fi n d ( v:,
V4 ) and then set i t e q u a l t o zero.
The fi rst of th ese steps is accom p lished by s e t t i n g subscript k equal to 4 i n Eq.
subt ract ing row
from row q of the a b ove
x
m a t rix, and u s i n g the
re s u l t t o r e p l ac e t h e e x i s t i n g r o w q a n cl col u m n q to obt a i n
Zb
®.
-
(8.89),
4
5 5
Z orig
jO.072 1 0 -jO .0 1 348
Th e
new
The only
-jO.072 1 0 -jO. 1 1 295
jO.07210 1
-jO.0 1348
-jO.0721O
-jO.1 1295
jO.29370
d i ag onal e l e m e n t is c a l cu l ated from
re m a I n i ng c a l c u l a t i o n s i n vo l v e
se t t i
ng
e v:,
--
V4 ) equal to' zero
In
Eq.
324
CHAPTER 8 THE I M PEDANCE MODEL AND N ETWORK CALCULATIONS
(8.89) and elim inating
Z bus
=
CD
CV
G)
@
the new row and col u m n by Kron reduction to g i ve
CD
j O .69890
CV
jO . 6 1 32 3
G)
j O .55 1 1O
j O. 60822
j O . 6 1 323
j O .73 1 28
j O .63677
jO . 69 1 40
j O .55 1 1 0
j O . 63677
j O . 69890
j O .64 1 78
j O .60822
j O . 69 1 40
jO.64 1 78
j O .7 1 966
0
is t h e d esired bus imped ancc m a t rix. T h is res u l t is the same as t h a t shown i n
Exam ple 7.6 except for t h e ch ange i n l hc b u s nu mbers of F i g . 7.9 t o t h o s e 0 f F i g .
8.8.
which
8.8
_
SUMMARY
This chapter introduces the importan t Z bus building algorithm, which starts by
choosing a branch tied to the reference from a node and adding to this node a
second branch connected from a new node. The starting node and the new node
then have one row and one column each i n the 2 X 2 bus imped ance matrix
which represents them. Next, a third branch connected to one or both of the
first two chosen nodes is added to expand the evolving network and i ts Z bus
representation. In this manner the bus impedance matrix is built up one row
and one column at a time u ntil all b ranche s of the physical network have been
i ncorporated i nto Z bu s . Whenever possible at any stage, it is compu tationally
m or e efficient to select the next branch to be added between two nodes with
rows a n d columns already included in the evolving Z hu S . The elements of Z hus
can also be generated as n ee de d using the triangu l a r factors of Yhll s , which i s
often the most attractive method computationally.
The variables used to analyze a powe r network can take on many different
forms. However, regardless of the particu l ar choice of representation, powe r i n
the physical network must not be arbitrarily altered when representing currents
and voltages in the chosen format. Power i nvariancy imposes requirements
whe n transforming from one set of curren ts and vol tages to another.
Mutually coupled branches, which do not generally arise except under
u nb a l a n ced short-circuit (fault ) con di tions, c a n be a dd e d to and removed f r o m
Z b u s b y algorithmic methods.
PROBLEMS
8.1.
Form Z bu s for the circu i t of Fig. 8. 1 3 a fter removing node G) by converting t h e
voltage source t o a current source. Dete rmine t h e voltages with respect t o refe r­
ence node at each of the four other nodes when V 1 .2
and t h e load currents
are IL l
-j O. l , IL2
jO. I, Iu = jO.2, and IL 4 = -jO.2, all i n per u n i t .
=
=
=
-
-
�
325
PROBLEMS
®
)0.6
tID
)0.5
I1A
IL3
Load 3
Load 4
F I G U R E 8.13
Load 2
Load 1
C I rc u i t d i a g ram show i n g con­
stant current l o a d s s u pp l i e d by
an i d e a l vol t a g e sou rce. C i rc u i t
ILZ
IL l
8.2.
®
)0.4
CD
Reference
p a ra m e t e rs
a re i n
per u n i t .
t h e sol u t i o n o f P r o b . 8 . 1 , d raw t h e T h cve n i n e q u i v a l e n t c i r c u i t a t b u s
of
f i g . 8 . 1 3 lI n u u s e i t t o d c t e r m i n e t h e c ur re n t d r a w n hy i \ c a p a c i to r o f r e a c t a n c e 5 . 4
@
r
F om
@
p e r u n i t c o n n e c t e d b e tw e e n b u s
lI n u r e fe r e n ce . F o l l o w i n g t h e p roced u re o f
EX�l m p l e R . 2 , c a l c u l a t e t h e v o l tage ch anges a t e a c h o f t h e b u s es d u e t o t h e
capacitor.
8 . 3 . M o d i fy Z b us o f Pro b . 8 . 1 to i n c l u d e a c a pa c i t o r of r e a c t a n ce 5 . 4 per u n i t c o n n e c t e d
@
from b u s
t o refe r e nce a n d t h e n calcu l a te t h e new b u s v o l t ages u s i n g t h e
m o d i fi e d Z b u s ' Check your a nswers u s i n g t h e resu l t s o f Probs. 8.1 and 8.2.
8.4. M o d i fy t he Z bu s d e t e r m i n e d i n Ex a m p l e 8.4
G)
n ew node c o nnected t o b u s
fo r t h e c i r c u i t o f Fig.
8 . 8 by ad d i ng a
t h ro u g h an i m pe d a nce of jO.S per u n i t .
8 . 5 . Modify t h e Z bus d e t e r m i ned in Exa m p l e
.' per u n i t between b u se s
CD
and
@
8.4 b y a d d i n g a b ranch o f i m p e d a n c e jO.2
of t h e circ .J i t o f Fig. 8.8.
8.6. M o d i fy the Z b u s d e t e r m i n e d i n Exa m p l e 8 . 4 by removing t h e i m pe d a n c e con n e c t e d
b e t we e n buses
and
o f t h e c i rc u i t o f F i g . 8.8.
W
G)
8 . 7 . F i n d Z b us for the c i rcu i t of Fig. 7 . 1 8 by t h e Z O ll S bui l d i n g a l gori t h m d iscussed 1 0
S e c . 8 . 4 . Assu me t h ere i s no m u t u a l co u p l i n g between branches.
'.
8.8. For t he reactance n e twork of Fig,
( a ) Z b u s by d i rect for m u l a t io n ,
(0 ) The v o l t a g e
at
8 . 1 4 , fi n d
each bus,
( c ) T h e c u r r e n t d ra w n b y a c a p ll c i t o r I l ; \ v i n g , \ r C ll c t il l1 C C o f 5 . 0 p e r u n i t con n e c t e d
f ro m b u s
G)
t o n e u t ra l ,
o
jO.5
j0 2
)2.0
jO.4
FIG U R E 8.14
Circ u i t fo r Prob. 8.8. Vol t a g e s a n d
impeda nces a re in per u n i t .
326
CHAPTER 8 TH E I MPEDANCE MODEL A N D NETWO RK CALCULATIONS
(d)
The change in voltage at each bus when the capacitor is connected at bus G) ,
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.
Find Zb us for the three-bus circuit of Fig. 8.15 by the Z b us building algorithm of
Sec. 8.4.
8.9.
�-@+I-j-H�
R eference
8.10.
8.1 1 .
FIG U R E 8. 1 5
Ci rc u i l for Probs. K . () , X . I
reac t a nces i l l pe r u n i l .
Find Z bus for the four-bus circuit of Fig. 7. 12, which has per-unit admittances as
marked.
The three-bus circuit of Fig. 8.15 h as per-unit reactances as marked. The symmetri­
cal Yb us for the circuit has triangular factors
L
=
r
l
-
j6 .0
j5�0
-j21 .633333
j20.0
-j 1 .5 1 0038
8. 13.
8.14.
8.1 5 .
8.16.
8 . 1 7.
1
- 0 .833333
1
� 1
- 0.92 499
Use L and U to calculate
(a) The elements 2 1 2 , 223 , and 233 of the system Z bus and
(b) The Thcvcn i n i m p e d a nc e 2 th ' 1 1 looking into thc circuit of Fig. 8. 1 5 between
buses CD and G) .
Use the Yb us triangular factors of Prob. 8 . 1 1 to calculate the Thcvenin impedance
222 looking into the circuit o f Fig. R. 1 5 b e tween bus W a n d the reference . C h e c k
your a nswer by i n spec t i o n o r rig. �. I ) .
The Ybus for the circuit of Fig. 7. 1 2 has triangular factors L and U give'n in
Example 7 .9. Use t he triangular factors to calculate the Thcvehin impedance Zlh 24
looking into the ci rcu i t of Fig. 7. 1 2 betwe e n buses @ and @ ' Check your ans�er
using the solution of Prob. 8 . 1 0.
Using the notation of Sec. 8.6, prove that the total reactive power loss is given by
the formula QL = 1 T X bus I * .
Calculate the total reactive power loss in the system of Fig. 8 . 1 3 using Eq. ( 8.57).
Using the procedure discussed in Sec. 8.6, modify the Z bus determined in Example
8.4 to reflect the choice of bus CV of Fig. 8.8 as the reference.
(a) Find Z b us for the network of Fig. 8 . 1 3 lIsing node G) as the reference. Change
the reference from node G) to node @ and determine the new Zb u s of the
.
8.12.
I , a n d H . 1 2 . V a l u c s s h ow i l a r c
327
PROBLEMS
n e twork using Eq. (8.60 ) . Use the n u m er i c a l values of the load currents ILi of
Prob. 8 . 1 to d e t e rm i n e I n e w by E q. (8.55) a nd Vnew by E q. (8.56).
( b ) Change the Z bus refe re n c e from node
b ack to node
and u sing Eq.
(8.63), d e t ermine t h e voltages a t buses
and
relative to node
What
are the values of these b u s vol t ages with respect to the gro u n d reference of
Fig. 8 . 1 3?
@
CD
W,
@
W.
n e w b ranch havi n g a n imp e d a n c e o f iO.25 p e r u n i t is connected b etween nodes
and
of t h e circ u i t of Fi g . 8 . 8 i n p a ra l l e l w i t h t h e exist i n g i m p e d a nce o f jO.2
per u n it between t h e s a m e two n o d e s . These two bra n c h e s have m u tu al imp e d a nce
o f jO .1 per u n i t . Modify t h e Z bus d e te r m i n e d i n Examp l e 8 . 4 to account f o r t h e
a d d ition of the n e w b r a n c h .
8 .18. A
G)
8)
8.19.
D e rive E q s . ( 8 . 9 5 ) a n d (8.96).
8.20.
M od i fy t he Z bus deter m i n e d in Ex a mp l e 8.7 to remove t h e b r a n c h b e tween b u ses
and
c o u p l e d by m u t u a l i mp e d a n c e i O . l s per u n i t to t h e b r anch
and
@ already
between buses CD
8) .
CD
8 . 22.
8.23.
CD - CD
0 - CD
t h e o n l y m u t u a l ly co u p l e d b ra n c h e s ( as i n d i c a t e d by t h e dots) w i t h a mu tual
i m p e d a n c e o f jD. I S p e r u n i t b e tw e e n them. Find Z bus fo r t h i s c i r c u i t by t h e Z b us
bu i l d ing algori t h m .
8.2 1 . !\ s s u m e t l) ; l t t h e t wo b r a n c h e s
and
i n t h e c i rcu i t o f Fig. 7 . 1 8 are
Prob.
0 - G) ,
M od i fy t h e Z hu, ob t a i n e d i n
8 . 2 1 t o re move b ra n c h
w h i c h is
c ou p l e d to b r a n c h
t h ro u g h a m u t u a l i mp e d a nce of i O . I S per u n i t .
CD - CD
I n Fig. 8. 1 6 a new b u s ® i s t o be c o n nected t o a n exi s t i n g b u s ® t h ro u g h a n ew
bra nch c. New b r a n c h c is m u t u a l l y cou p l e d to b ra n ches a a n d b , w h i c h are
a l ready m u t u a lly coupled to one a no t h e r as show n . The pri m i t ive imp e d a nce
m a t rix d e fi n i n g se l f- a n d m u t ual i m pe d a n c e s o f t h ese t h ree m u t u a l l y coupled
branches and its rec iproca l , the p ri m i t ive a d m i t t a nce ma trix, h ave the forms
To acco unt for the a d d i t i o n of new bus ® , prove t h at the exist i n g bus imp e d ance
ma t r ix of t h e n e twork m u s t b e a u g m e n t e d by a n e w row q a n d col u m n q w i t h
a
@ i(, ®
(JJ
0 --
•
-.
@
@
New bus
'--------f----..
FI G U RE 8. 1 6
New b ranch c from new
and b.
bus ®
i s m u t u a l l y coupled to b ranches
a
328
CHAPTER 8 THE IMPEDANCE MODEL AND NETWO R K CALCULATIONS
elements g iven
8.24.
by
YCb ] Z .. � Z k "
[ �m i ]
Zn i
JI
I
for i
=
1,
. .,N
.
Note that these equations are a g e n e ra l i z a t i o n of Eqs. (8.87) and (8.88).
Branch @ - G) of the circuit of Fig. 7. 1 8 is m utually coupled to two bra n ch e s
CD - G) and C.V - G) through m ulual impedances o f j O . I S pe r u n i t and jO . l p e r
unit, respectively, as indicated by dots. Using the formulas gi e n in Prob. 8.23, fi n d
Z bus for the circuit by the Z bus bui l d i n g algorithm.
v
CHAPTER
9
P OWER-FLOW
S OLUTIONS
Power-flow studies a r e o f great i mportance i n p l an n i n g a n d desig n i n g t h e futu re
exp a nsion of power systems as w ell as in d etermin ing the best operation o f
existin g systems. The p r i n c i p a l i n formation obtained from a powe r-flow study i s
t h e m agnitude a n d ph ase a n g le of the vol t age a t e ach bus a n d t h e real a n d
reactive power fl owing i n e ach l i n e . Howeve r, m u c h additional i n forma t io n o f
val u e i s provided by t h e p r i n tout of the so lution from computer p rograms u s e d
by t h e electric u tility compani es. M o s t o f t h e s e fe atu res are made evid e n t i n o u r
d i scussion o f power-flow studies i n t h i s c hapter.
W e shall exami n e some of t h e meth ods upon w hich s o l utions to the
pow e r - fl ow problem ar c b a s e d . Th e g rc::lt va l u e of the pow e r-flow com p u t e r
prog ram i n pow e r sys t e m d e s i gn a n d o p e r a t ion w i l l become appare n t .
9. 1
THE POWER-FLOW P R O BLEM
E i t h e r the bu s s e l f- and m u t u a l a d m i t tances w h i ch com pose t he b u s a d m i t t a n c e
m a t rix Y b u s or the driving-po i n t and t ra n sfer i m p e d a nces w h i ch co mpose Z hus
may be used in solving t h e power-flow p rob l e m . We confine o u r study to
m e thods using admittances. The s t a r t i n g poi n t in obtaining t h e d a t a w h ich m u s t
b e furnished to t h e compu t e r i s the single-line d iagram o f t h e system. Tra n s m is­
sion l i n es are represented by t he i r p e r-phase nominal-Ii equiv a l e n t c ircuits l ike
t h a t s hown in Fig . 6.7. For e a ch line n um e r i ca l val u es for the serie s i mped ance
( usually in terms of l i n e-charging
Z and the total line-charg i n g a d m i tt a n ce
megavars at nom i n a l vol tage of t h e system ) a re ne cessary so t h a t t h,e computer
can d et e rm ine a l l t h e e l e m ents of t h e N X N b u s a d m i ttance m a tr ix of w h ich
Y
329
330
CHAPTER 9 POWER·FLOW SOLUTI ONS
the typical elemen t �j is
Other essential information includes tra nsformer ratings and impedances, shunt
capacitor ratin gs, and transforme r tap s e tt i n gs I n a d v a n c e o f e ach power - fl o w
study certain bus vol ta ges and power injections m u s t b e g iven k n ow n v a l u e s as
d iscussed b elow.
The vol ta ge a t a typical bus CD of the system is g i v e n in po l a r coord i n a t e s
.
,
by
( 9 .2)
and the vo l t age a t a n o t h e r bus (J) i s s i m i l a rl y w r i t t e n
from i to j . The net current inj ected i n to t h e netw o r k
elements � n of Y bus is given by the summa tion
Ii = Y; I VI + � 2 V2 + . . . + Y;N VN =
at bus
t h e s u b sc r i p t
i n te rms o f t h e
b y cha n g i n g
CL
N
L t:n Vn
( 9 .3)
n= i
Let Pi and Q ; denote the net real and reactive powe r entering the n etwork at
the bus CD . Then, the complex conjugate of the power i njected at b u s CD IS
N
Pi - jQi = V:* L
Y; n Vn
=
11
( 9 .4 )
1
in which we substitute from Eqs. (9. 1 ) and (9.2) to obtain
Pi
-
jQ i
N
=
L
II
=
1
1 Y;1IV;V,, 1 / 8ill + 011 - 0 i
Expanding this equation and equating
Pi
=
N
.
Qi =
.
a n d r e a c t ive p a rt s , we obt a i n
-
+
On
-
0i)
( 9 . 6)
J
N
L I Y;n V: Vn I s i n e 8i n + On
II = 1
.
al
L 1 �,Y:· � l cos ( 8ill
n =
.
re
( 9 .5 )
-
0i)
( 9 .7)
Equations (9.6) and (9.7) constitute the polar form of the power-flow equations;
they provide calculated values for the net real power Pi and reactive power Qi
entering the n etwork at typical bus CD . Let Pg i denote the scheduled power
being generated at bus CD and Pdi den ote the sch eduled power demand of the
9. 1
TH E POWE R-FLOW P R O B L E M
(a)
331
(b)
Notation fo r ( a ) a c t i v e a n d ( b ) react ive p ow e r a t a typ ical bus
FIGURE 9 . 1
CD
i n pow e r-ftow s t u d i e s .
0,
load at that bus. The n , Pi. sch = P i - Pdi is the net scheduled power b e i n g
i nj e cted i n to the n e twork a t b u s
as i l l ustrated in F i g . � . l ( a ) . D e noting t h e
cal c u l ated va l u e of Pi by Pi . ca l c leads to the definit ion of mismatch 6 Pi as t h e
schedu led value Pi . seh m i n u s the ca l c u l a te d val u e Pi. c�ie '
6 P,
Li kewi s e , fo r
r eac t ive
=
Pi . seh
-
Pi. ca l c
=
( P);i
-
power a t b u s CD w e h ave
Pi/i )
-
Pi. c alc
( 9 .8)
( 9 .9 )
a s s hown i n Fig. 9. l ( b ) . M ismatches occu r i n t h e cou rse of solvin g a power-flow
probl e m w he n calcu l a t e d v al u es o f Pi and Q i do not coi ncide w i t h t h e s ch e d­
u le d valu e s . If the calc u l ated values Pi. calc and Qi . OIC mat ch the s ch ed ul e d
val u es Pi, sch a n d Q i. sch p erfectly, then we say that the mism atches 6 Pi a n d D. Q i
are zero a t b u s CD , and we write t h e power-balance equations
( 9 . 1 0)
(9.11)
s h a l l s e e i n S e c . 9 . ::\ , t h e fu nct i o n s g i' a n cl gi" ,I r e conve n i e n t fo r w r i t in g
c e r t a i n equat ions i nvolving t h e mismatches 6 Pi a n'd 6 Qi' I f b u s CD has n o
g e n e r a t io n or load, t h e a p p ro p r iate t e rms a re se t e q u a l t o z e ro i n Eqs. (9. 1 0)
a n d ( 9 . 1 1 ) . Each bus of the n e twork has two s u ch e q u a t ions, and the power-flow
p rob lem is to solve Eqs. ( 9 . 6 ) and (9.7) for va l u es of t h e u n known b u s vol t ag e s
w h i c h c a u s e Eqs. (9. 1 0) and (9. 1 1 ) t o be n u m erically sat isfied a t each b us. I f
there i s no sc heduled val u e Pi. sell for bus CD , then the m ismatch D. Pi = Pi , sc h
Pi, ca l c c a n not be defi ned a nd t he re is no requ i rement to sat isfy t h e correspond­
i n g Eq. (9. 1 0) i n t h e co urse of solv i n g the power-flow prob l e m . S i m i l ar ly, if
Q i. sch is not specified at b u s CD , t h e n Eq . (9. 1 1 ) does not have to be satisfied .
Four potentially u n know n q u anti ties associated with each b u s CD a r e
Pi , Q i ' voltage a n g l e 8 i , a n d vol tage magnitude I v: I . At most, t h e re a r e two
e q u a t ions like Eqs. (9. 1 0) and (9. 1 1 ) ava i l ab le for e ach node, and so w e m u st
c o n s i de r how t h e n u m b e r of u n k n own q u a n t i t ies can be reduced to .agree w i t h
As
we
-
332
CHAPTER 9
POWER· FLOW SOLUTIONS
the number o f available equations before beginning to solve the power-flow
problem. The general p ractice in power-flow studies is to identify three types of
buses in the network. At each bus CD two of the four quantities 8 i , I V; I , Pi ' and
Qi are specified and the remaining two a re calculated. Specified quantities a re
chosen accor ding to the fol lowing d iscussion:
Load buses. At each nongenerator bus, called a load bus, both PK i and Qg;
are zero and the real power Pd; a nd reactive power Qdi drawn from t he
system by the load (negative inputs into the system) are known from histori­
cal record, load forecast, or measu reme n t . Quite often in practice on ly reaJ
power is known and the reactive powe r is then based on an assumed power
factor such as 0.8 5 or higher. A load bus CD is often ca lled a P-Q bus
because the scheduled val ues Pi , sell
Pcli and Qi , sch = Q dj are kn own
and mismatches 6. Pj and 6. Qi can be defined. The corresponding Eqs. (9. 1 0)
and (9. 1 1 ) are then explicitly incl uded in the st atement of the power-flow
prob lem, and the two unknown quantities to be determined for the bus are
0 i and I V; I .
2. Voltage-controlled buses. Any bus of the system at which the voltage magni­
tude is kept constant is said to be voltage con trolled . At each bus to which
there is a generator connected the megawatt generation can be controlled by
adj usting the prime mover, and the voltage magnitude can be controlled by
adj usting the generator excitation. Therefore, at each generator bus CD we
may p roperly specify Pg i and I v: I . With Pdi also known , we can define
mismatch f1 Pj according to Eq. (9 .8). Generator reactive power Q g i required
to support the sched u l ed voltage I �I can not be known in advance, and so
mismatch 6. Qi is not d efi n e d . T h e re fore, at a ge nerator bus CD voltage
angle 0i is the unknown quantity to be determ ined and Eq. (9. 1 0) for P i is
the available equ ation. After the powe r-flow problem is solved, Qi can be
calculated from Eq. (9.7).
For obvious reasons a generator bus is usu ally called a voltage-con­
trolled or PV bus. Certain buses without genera tors may have vol tage
control capability; such buses are a lso design ated vol tage-con trol led buses at
which the real power genera tion is simply zero.
3. Slack bus. For convenience throughout this chapter bus CD is almost always
designated as the slack bus. The voltage angle of the slack bus serves as
reference for the angles of all other bus voltages. The particular angle
assigned to the slack bus voltage is not important because voltage-angle
differences d etermine the calculated values of Pi and Q i in Eqs. (9.6) and
(9. 7). The usual practice is to set 8 )
0° . Mismatches are not defined for
the s lack bus, as explained below, and so voltage magnitude I VI I is specified
as the other known quantity along with ° 1 0° . Then, there is no require­
ment to includ e either Eq. (9. 1 0) or Eq. (9. 1 1 ) for the slack bus in the
power-flow probl em.
1.
=
-
-
=
=
.
.
9.1
THE POWER-FLOW P R O B L E M
333
To understand why P I and Q 1 are not sched u l ed at the slack bus, consider that
at each of the N buses of the syst e m an equation similar to Eq. (9. 10) can be
written by l e t ti ng i range from 1 to N. When the resul ting N e q u a t ions are
a d d e d toget h e r, we obt a i n
PL
- [ Pi =
N
i
�
=
I
R e a l powe r loss
N
L Pg i
i
I
=
To t a l ge n e ra t i on
N
[, Pdi
i= 1
-- -
( 9 . 1 2)
To t a l l o a d
The t e r m PI- in this equation is evi d e n t l y the tota l / 2 R loss i n the t ransmission
I i n e s and t ransformers of the n e twork. The i nd iv i d u a l currents in the various
t r a n s m ission l i n e s of the network cannot be calcu l a ted u n t i l a fter the vol ta ge
m a g n i t u d e a n d a n gl e a r e k nown a t e ve r y b u s o r t h e sys t e m . The r e fore, PL is
i n i t i a l ly u n k nown and i t i s not poss i b le to prespecify all the qua n t i t ies i n t h e
s u mm ations of Eq . (9. 1 2) . I n t h e fo rmulation of t he power-flow p roblem we
choose one b u s , the slack bus, at which PI; i s not sch e d u l e d or othemise
p respecified . A ft e r t h e power-flow p roblem has been solve d , the d i ffe re n ce
( slack ) between the total specified P going into the system a t all the other b uses
a n d t h e total output P plus / 2 R losses are assigned to the slack bus. For t his
reason a generator bus must be s e lected as t h e slack bus. The difference
between the total me gavars s u p p l i e d by the generators at the buses a n d the
megavars received by the loads is given by
N
[ Qi =
N
L Qg i
i= I
N
- L Q di
i= I
( 9 . 1 3)
This equation is satisfied on an ind ividual bus basis by satisfyin g Eq .
(9. 1 1 ) at
in the course of solving the power-flow p roblem. Ind iv i d u a l Q i c a n
each bus (J)
be eval u ated from Eq. (9.7) aft e r t h e power-flow solution becomes ava ilable.
Th us, t he q u a nt i ty on the l e ft- h a n d s i d e of Eq . (9. J 3) accounts for the c ombi n e d
Jl1 c g <l v a r s assoc i ,l t c d w i t h l i Il e c h a r g i n g , s h u n t c a p a c i t o r s and r eactors i nsta l le d
,I t t h e h u s e s , ; l Il d t h e so-ca l l e cJ 1 2 X l o s s i n t h e s e r i e s reacta nces o f t h e
n s c h e u u l e d h u s- vo l t agc m a g n i t u d e s and a n g l e s i n t h e i n p ut d a t a o f
t h e p o w e r - n ow s t u d y , t f e c a l l ed s l U l c l!{I ria h!cs or dependent variables s i nce t heir
v a l u e s , which describe t h e sta l e of t h e system, depend on the q u a n t ities
s pec i fi e d at all the buses. H e nce, the power-flow p ro b le m i s to de t e r m i n e val ues
for all s t a te va riables b y so l v i n g an e q u a l n u mb e r of power-flow equ a t ions base d
on the i nput data specificat i o n s . If t here a re NI: voltage-controlled buses (not
cou n ti ng the sl ack bus ) i n the syste m of N buses, t here will be (2 N - N - 2)
g
e q u a t ions to be solved fo r (2 N - NI: - 2) state variables, as shown in T able 9 . 1 .
O nce t h e state variables h ave b e e n calculated, t h e complete state of t h e system
is known a n d all other qu a n t i t ies w hich depend on the state variables can b e
t r : m s rn i s s i o n l i n e s .
The
u
334
CHAPTER 9 POWER-FLOW SOLUTIONS
TABLE 9.1
Summary of power-flow problem
Bus
type
=
S l ack: i
=
No. of buses
1
2, . . . , Ng +
Volt age con trol led
(i
(i
=
Nx
Load
+
2,
N
Total:;
° 1 , I VI I
I)"
1)
. .. , N)
Q uantities
specified
-
N,, -
N
I V, I
P, . Q ,
2N
I Vj I
state variables
available
eq u a t i ons
No. of
No. of OJ,
0
0
NI:
2( N - N�
-
I)
2 N - NI: - 2
2( N
2N
N!]
-
-
N" -
I)
tv',
2
-
d etermined. Quantiti es such as P I a n d Q 1 at the sl ack bus, Qi at each
voltage-controlled bus, a n d the power loss PL of the system are examples of
depe n d e nt functions.
The functions Pi and Qi of Eqs. (9.6) and (9.7) are nonl i near functions of
the s t a t e variables 8i and \ V, \ . H e nce, power-flow calculations us u ally employ
i terative t echniques s uch as the Gauss-Sei d e l and Newton-Raphson proce d ures,
which are d escribed in this chapter. The Newton-Raphson method solves t h e
polar form of t h e power-flow equations u n t i l t h e 11 P a n d I1Q mismatches a t all
buses fal l within specified t olerances. The Gauss-Seidel method solves the
power-flow equations in rectangul a r (complex variable) coordi nates u n t i l d if­
ferences i n bus voltages from one i t e ration t o another are sufficiently smal l .
Both met hods are b ased o n bus admitta nce equations.
Suppose that the P-Q load is known at each of the nine buses of a
small power system and that synchronous generators are connected to b u ses CD,
0 , G) , a n d (j) . For a power-flow study, identify the t1P and t1Q m ismatches
and the state v ariables a ssoc i a t e d with each b u s . Choose bus CD as t h e slack b us.
Example
9.1.
Solution. The
nine
buses
of t h e sys t e m
P-Q buses :
a re
categorized as follows:
G) , @ , (0 , @ , a m i ®
P-V buses: 0 , G) , and (j)
Slack bu s :
CD
The mismatches corresponding to the specified P and Q arc
At P-Q buses :
t::. P3 , t1 Q3; t1 P... , 6. Q4 ; 6. P6 , t::. Q 6 ; t1 P13 , t1 QI3; t1 P9 , t1 Q9
At P-V buses :
9.2
THE GAUSS·S E I D E L METHOD
335
and the s t ate variables a re
S i n c e N = 9 a n d N = 3, t h e re a re 2 N
g
t h e 1 3 s t a t e variables s h own .
9.2
-
Ng
-
2
= 1 3 equations to be solved for
THE GAUSS-SEIDEL METHOD
The complexi ty of ob t a i n i n g a fo rmal sol u t ion for power flow in a pow e r syst e m
arises because of t he d iffer ences i n t h e type o f d a t a specined for t h e d ifferent
k inds of buses. A lthough the for m u l a t ion of s u ffi c i e n t e q u a t i o n s to match the
n u m b e r of u n k n o w n s t a t e v a r i a b l e s i s n o t d i Hi c u l t , a s we h a ve see n, the closed
form of sol u t ion i s not practica l . D i g i tal sol u tions of the powe r-flow pro b l e m s
fol l ow a n i t e r ative p rocess b y assigning estimat e d va l u es to t h e u nk nown bus
vol t ages a n d by calc u l a t i n g a n ew val u e f or e a c h b u s v o l t a g e from the e st im a t e d
v a l ues a t t he o t h e r b u ses a n d t h e r e a l a n d reactive pow e r s p e c i n e d . A new s e t o f
v a l u e s for t h e volt age a t e a c h b u s is t h u s obt a i ne d a n d used to calcu l a t e s t i l l
a n o t h e r s e t o f bus vol t ages. E a c h ca lcu l a t io n o f a n e w s e t o f voltages i f c a l l e d a n
iteration . The itera t ive p rocess is r e p e a ted u n t i l t h e cha nges a t e a c h bus a r e less
t h a n a specined minimum v a l u e .
We d e rive equat ions for a fou r-bus system and w r i t e t h e gene ral equa tions
l a t e r . With the slack bus designated as number 1 , comput a t ions start with bus
W . If P2. sch a n d Q 2. SCh are the sch e d u le d real a n d reactive power, respectively,
e n te r i n g t h e n e twork a t b u s @ , i t fol lows from Eq. (9.4) with i se t e qu a l t o 2
a n d N equal to 4 t h a t
( 9 . 1 4)
Sol v i n g for Vz gives
( 9 . 15)
us ass ume t h a t buses CD a n d @ a rc also l o a d buses w i t h re a l a n d
re a c tive power specified. Express i o ns simi lar t o t h a t i n Eq. (9. 1 5 ) m a y b e wri t t e n
for e a c h b u s . At bus G) w e have
For nOw
let
( 9 . 1 6)
I f we were to e q u a t e r e a l a n d i m a g i n a ry p ar t s o f Eqs. (9. 1 5), (9. 1 6), a n d t h e
336
CHAPTER
9
POWER-FLOW SOLUTIONS
02 04
V21
V41 .
similar equation of bus @ , we would obtain six equations in the six state
variables
to
and I
to I
However we solve for the complex voltages
directly from the equations as they appear. The solution proceeds by iteration
based on the scheduled real and reactive power at buses 0, ® , and @ , the
scheduled slack bus voltage VI = I
and initial voltage estimates
0
an d V4( ) at the other buses.
Solution of Eq. ( 9 . 1 5) gives the corrected vol tage VP) calcu lated from
VJO),
el)
V2
=
y22
_1_
[P2. Kh
VI I&,
j Q2 , seh.
V2(0 )*
-
_
viO),
( Y2 1 V
1
+
Y
2J
V (O) + Y
J
2.1
V4(O» )
]
( 9 . 17)
]
( 9 . 1 8)
]
( 9 . 1 9)
i n w hich a l l quantities in the right-hand-side express ion are either fixed s pe c ifi ­
cations or initial estimates . The calculated value V? ) and the estimated value
'
will not agree. Agreement would be reached to a good degree of accu racy
after several iterations and would be the correct value for V2 with the esti mated
voltages but without regard to power at the other buses. This value wou ld not
be the sol ution for V2 for the specific powe r-flow conditions, however, because
the voltages on which this calculation for V2 depends are the estimated values
and V4(0 ) at the other buses, and the actual vol tages a r e not yet know n .
As the corrected voltage is found at each bus, it is used to calculate the
corrected voltage at the next bus. Therefore, substituting V�- I) i n Eq. ( 9 . 1 6), we
obtain for t h e first calculated value at bus G)
V�O)
vjO)
vel)
3
= _1_
y33
[ P3 , sel V
-
jQ J , seh
0*
3( )
_
31
(Y
V1
+ Y32 V2( l ) + Y34 V4(O» )
The p rocess is repeated at bus @ a n d at e ach b u s consecu tively throughout t he
network (except at the slack bus) to complete the first iteration in w hich
calcu lated values are found for each state variable. Then, the entire p rocess is
carried out again and again until the amou nt of correction in voltage at every
bus is less than some predetermined precision index. This process of solving the
power-flow equations is known as the Gauss-Seidel iterative method.
Convergence upon an e rroneous solution is usually avoided if the in itial
values are o f reasonable magnitude and do not differ too widely in p hase. It is
common p ractice to set the initial estimates of the u nknown voltages at all load
buses equal to 1 . 0
per unit. Such initialization is called a fiat start because
of the uniform voltage profile assumed.
For a system of N buses the general equation for the calculated voltage a t
a ny b u s CD where and Q a re scheduled i s
L!:.
1
P
VI ( k ) = y .
_
II
[PI, sel
-
I, seh,
'Q .
}
V (k - I )*
I
_
i- J
'\"
l...J
J= I
YIJ. VJ ( k )
_
�
L.J
j=i+ 1
YIJV) (k - I )
The superscript ( k ) denotes the number of the iteration in which the voltage is
9.2
TH E GA U S S -S E I D E L
M ET H O D
337
c urre n tly being calcu l ated a n d ( k - 1) i n d icates the n umber of the p receding
i te r a t i o n . Th us, we see t h a t the val ues for the vol t ages on the right-ha n d side of
t h i s e q u at ion a re the most recently calcul ated v alues for the correspo n d i n g
b uses ( or t h e estimated v o l t a g e if k is 1 a n d n o i teration h a s y et b e e n m a d e at
that p a rticular bus ) .
S i nce Eq. (9. 1 9) a p p l i es only a t load buses w here re al a n d reactive powe r
a r e specified, a n addi tional s t e p i s n ecessary a t vol t age-contro l l e d buses w h e re
vol tage magnitude is to r e m ain cons t a n t . Before investigating t h is a d d i t io n a l
ste p , l e t u S look a t a n exa m p l e o f t h e cal c u l a t ions a t a load b u s .
E xa m p l e 9 _ 2 . F i g u r e 9 . 2 s h ows t h e o n e - l i n e d i a g r a m o f
G e n e ra t o r � ,I r e co n n e c t e d a t b u s e s
(I)
and
@
(1 s i m p l e p ow e r sys t e m
w hi le l oads arc i nd ica t e d a t a l l
fo u r h u se s . Rise v,! l lI e s fo r t h e t r , l I1 s m i s s i o n syst e m ,I re I ( ) ( ) M VA , 2 3 0 k Y . T h e
data
of
Tab l e <;.2
.
fine
g i ve p e r - u n i t s e r i e s i m p e d a n c e s ,Ill el l i n e -c h a r g i n g s l i s ce p t a nces
blls da ta
Q,
w hich
each bus.
for t h e n o m i n ,t \ - 7T e q ll i " , t \ e n t s o f t h e fo u r l i n e s i cJ e n t i fl c c! by t h e b u ses at
Q values
t hey t e r m i n a t e . The
The
i n Ta b l e 9 . 3 l is t v a l u e s for P ,
the
load
b uses
(1)
and
G) .
G e n e ra t e d
Q
CD
!
Qi. SCh '
o r load (l rc c a l cu l a t e d from t h e c o r rcspo l l ci i n g
Q i; i
powe r fa c t o r o r 0 . 8 S . T h e n e t s c h e d u l e d v a l u e s , P' . ,ch a n d
Birch
is
and
P
V
at
va l u e s a s s u m i n g a
a r e n e g a t ive a t
n o t spec i fi e d w h e re vo l t age
®
Elm
Maple
FI G U RE 9_2
O n e - l i n e d i ag r a m for Exam p l e
s h ow i n g t he b u s n a mes a n d
n u moers.
9.2
TA B L E 9 . 2
Li n e d a t a fo r Exa m p l e
Series Z
Line,
Series Y
=
Z -
I
T ot a l
Shunt Y
Y / Z
per unit
II
c h a rg i n g
per unit
G
per u n i t
0 . 0 1 000
3 . 8 1 5629
- 1 9. 07 8 1 44
] 0.25
0.05 1 25
0.0074 4
0.0 504ll
0.03720
5 . 1 69561
- 2 5 . 8 4 7 809
7 . 75
0 . 03875
2-4
0 . 00744
0.03720
5 . 1 69 5 6 1
- 2 5 . 8 4 7 809
7 . 75
0. 03875
3 -4
0 . 0 1 272
0.06360
3 .023705
- 1 5 . 1 :. 8528
1 2. 7 5
0.06375
R
per unil
1 -2
1 -3
bus to
bus
t B a s c 1 00 M V A , 2 3 0 k V .
tAt
9.2t
230 k V .
X
;\ l v a rt
per u n i t
338
CHAPTER 9 POWER-FLOW SOLUTIONS
data for Example 9.2
Bus
TABLE 9.3
Generation
Bus
P, MW
Q, Mvar
1
2
0
0
P, MW
Load
Q, Mvart
50
3 0 .99
1 70
1 05 . 3 5
V, per
I .()()
u nit
�
I . ()O�
R emarks
Slack
bus
Load b u s
( i n d u c t ive)
3
0
0
200
1 23.94
1 .00�
Loao b u s
( i n d u c t i ve)
4
3 18
49.58
80
1 .02�
V o l t age con t ro l l e d
tThe Q values o f load a re ca lcu l a ted from t h e correspond i n g P v a l u e s assu m i n g a pow e r fa c t o r o f 0 . 8 5 .
m agnitude is constant. In the voltage colum n the values for the load buses are
flat-start estimates. The slack bus voltage magnitude I VI I and angle 8 ) , and also
magnitude I V4 1 at bus @ , are to be kept constant at the values listed. A
power-flow study is to be made by the Gauss-Seidel method. Assuming that the
iterative calculations start at bus CD , find the value of V2 for the first iteration.
In order to approach the accuracy of a digital computer, we perform the
computations indicated below to six decimal places. From the line data given in
Table 9.2 we construct the system Y bus of Table 9.4. For example, associated with
bus @ of Fig. 9 . 2 are the nonzero off-diagonal elements Y2\ and Y24 , which are
equal to the negative of the respective line admittances
Solution.
Y2 1
=
- ( 3 .8 1 5 629 - j 1 9 .078 1 44) ;
Y24 =
-
( 5 . 1 695 6 1 - j25 . 847809)
Since Y22 is the sum of all the admittances connected to bus (1) , including the
TABLE 9.4
Bus a d mitt ance m a t ri x fo r Exa mple
Bus
no.
CD
@
®
8)
CD
9.2"\"
(i)
8.985 1 90
- 3. 8 1 5 629
-j44.835953
+ j 1 9.078 1 44
- 3 . 8 1 5629
8.98 5 1 90
+ j I 9.078 1 4 4
-j44.835953
- 5. 1 69561
+j25. 847809
0
0
(J)
- 5 . 1 69 5 6 1
+ j 2 5 . 847809
0
@
0
- 5 . 1 6956 1
+ j25.847809
8 . 1 93267
- 3.023705
-j40.863838
+ j 1 5. 1 1 8528
- 5 . 1 69 5 6 1
- 3. 023705
8 . l 93267
+j25.847809
+ j 1 5 . l 1 8528
-j40.863838
t Per-unit val ues rou nded t o s ix deci m a l p l aces
T H E GAUSS-S E I D E L M ETHOD
9.2
s h u n t susceptances for l i n e charging of l i n e s
Y22 = e Y2 1 ) + j O .05 1 25
-
Substit u t ion in
( 1)
V2
_
-
Eq. (9. 17)
1
-
Y22
[-
+
e
-
Y2 4 )
+
@ CD
and
-
jO.03875
1 .7 + j l .0535
- 1 .00 e - 3 .8 1 5 629 + j 1 9 .078144)
1 .0 + j O . O
Y"2 [ 1 .7
-
7 . 3�05S I
have
yi e l d s the per-un i t voltage
1
-
we
8 .985 1 90 - j44 .835953
=
- 1 0 2 ( - 5 . 1 69 5 6 1
=
(1)- @ ,
. 33 9
+
-
j l . 0535 +
9 .088581 - j 4 5
j44 . 3Sl)40l)
------ = (l . lJS 3 5 64
0 . 985 1 90
j44 . 8 3 5 9 5 3
- -
�-
.
+
j 2 5 .847809)
44 2 9 0 9]
1
jO . OJ23 1 6
Exper ien ce with the Gauss- S e i d e l met hod of solution o f power-flow p rob­
lems has shown t ha t the nu mber of i te r a t ions r e q u i re d m ay be reduced consid­
e rably i f the co rrection in voltage a t e a ch b u s i s m u ltiplied by some con s t a n t
t h a t i ncreases t h e amou n t o f correct i o n t o b r i n g the voltage closer to t h e v a l u e i t
i s approachi ng. Th e m u l t i p l i e r t h a t a ccompl ishes t h i s improved convergence is
c a l l e d an acceleration jactor_ Th e d i ffe rence b e tween t h e n ewly calc u l a t e d
voltage a nd t h e best previous voltage a t t h e b u s is multiplied b y the a ppropri a t e
a cceleration factor to obt a i n a b etter correc tion t o b e a dded to the previous
val u e . For example, a t bus W in the fi rs t i teration we h ave the acce l e r a t e d
value VF1cc defined b y the stra igh t - l i n e formu l a
Vel)
2 . ace
=
(1
-
a ) V2(O)
+
a V2( I )
=
V2(O)
+
a ( V)( I )
_
-
V (O»
2
In which a is the acce l e ra t ion factor. More generally, for bus
i teration k the llcce lerated val u e is given by
V(k)
/ . acc
=
(1
-
a ) V( k -
I . acc
1) +
a V( k )
/
=
V(k � l )
/ • acc
+
( 9 .20)
)
CD
d uring
a ( V(k ) - VY - 1 » ) ( 9 .21)
/
/ . acc
If a = 1 , t h e n t h e Gauss-Se i d e l compu t e d val u e of Vi is stored as t he c urrent
val u e . I f 0 < a < 1, then the v a lue to be stored is a weighted average of t h e
Gauss-Seidel val u e and t h e v a l ue s t o r e d from t h e p revious i teration. If 1 < a <
2, t h e n the value to be stored is esse n t i a l ly a n extrapolated val u e . I n power-flow
s t u d ies a is ge nerally set a t abo u t 1 . 6 a n d c a nnot exceed 2 i f convergence i s t o
occ u r .
340
CHAPTER 9
POWER· FLOW SOLUTIONS
Substituting the results of Example 9.1 and an acceleration factor of 1 .6 i n
Eq. (9.20), w e find that
v?�cc
U SIng
·
V (l)
2, acc
=
=
1 + 1 .6 [ ( 0 .983564 - jO .0323 1 6) - 1 ]
0. 973703 - jO .05 1706 per u n i t
In
. SImI
. ' 1 ar calculations for bus r:l\
(V g ives fi rst-i terat ion value
V{I:'CC
=
0 .953949 - jO .066708 per u n i t
Since bus @ i s v o l t ag e con t ro l l e d , w e m l l s t t r e ,\ t i t d i ffe r e n t l y a s e x p l a i n e d
below. The acce l e r a t ion fac t o r fo r t il e r e a l co m p o n e n t o r t h c c or r ect i o n lll a y
d iffer from that [or t h e i m a g i n a ry co m po n e n t . For a ny sys t c m o p t i m u m v a l u e s
for acceleration factors exist, and poor choice of factors may result in less rapid
convergence or make convergence i mpossible. An acceleration factor of 1 . 6 for
both the real and imaginary components is usually a good choice but studies
may be made to determine the best choice for a part icul ar system.
Voltage-controlled buses. When vol tage magn itude rather than reactive
power is specified at bus CD , the real and imaginary components of the vol tage
for each i teration are found by fi rst computing a value for the reactive power.
From Eq. (9.4) we have
Q I.
=
{
- Im vI *
�
L
j= l
y. v }
IJ
J
( 9 .22 )
which h as the equivalent algorithmic expression
( 9 . 2 3)
where 1m means " imaginary part of" and the superscripts indicate the relevant
iteration. Reactive power Q (I k ) is eva l u ated by Eq. (9.23) for the best p. revious
voltage values at the buses, and this value of Q � k ) is substi tuted in Eq. (9. 1 9) to
find a new value of v:( k ) . The components of the n ew v:( k ) are then multiplied
by the ratio of the specified constant magnitude , V: , t o the magnitude of v:( k )
found by Eq. (9. 19). The result is the corrected complex voltage of the specified
magnit ude. In the fou r-bus example if bus @ is voltage controlled, Eq. (9.23)
yields the calculated value
( 9 .24)
wherein the calculated voltages of
buses 0 a n d Q) are accelerated v alues of
the first iteration. Substituting Q �l) for Q 4 sch in the equ ivalent of Eq. (9.19) for
9.2
bus
@
THE GAUSS-SEIDEL M ETHOD
341
yields
( 9 .25 )
a n d a l l quantities o n t h e right-hand s i d e a re now known . S ince I V4 1 is s pecified,
we correct t h e magnitude of VP ) as fol l ows:
( 9 .26)
a n cl p roceed
to t h e n e x t s t e p w i t h stored v a l u e V4:�Hr u f bus @ vo l ta g e h av ing
t h e speci ned magn itude i n t he rem a i n ing calculat ions of the i t e ration.
As discussed in Sec. 9 . 1 , either voltage magnitude or reactive power must
be specified at every bus except the slack bus, where voltage is specifi e d by b o t h
vol tage magn itude and angle. At b u s e s with generation t h e vo l t a g e m a g n i t u d e is
specified as well as the real power Pg supplied by the generator. The reactive
power Q g entering the n e twork from the generation is then determi ned b y the
comput e r in solving the power-flow p roble m. From a practical v iewpoin t the Qg
output of the generator must be w i t h i n d efinite l imits given by the i ne quality
w here Q min is the m J n l m u m and Q max is t h e maximum l im i t imposed o n the
react ive power output of the generator a t t h e bus. I n the cou rse of power-flow
sol u t ion if the calcu l ated va l u e of Qg is outside e i t h e r l im i t , t h e n Qg is se t equal
to the l imit violated, t h e origina l ly specified vol t a g e magni t u d e at the b u s is
rela." ed, and the bus is then treated as a P-Q bus for which a n ew vol t a g e is
calc u l a te d by t h c com p u t c r p r o g r a m . I n s u b s e q u c n t i terat i o n s t h e p ro g ram
c n d eavors to sust a i n the origi n a l l y spcci flcd vo lt age at t h e bus w h i l e ensu ring
t h a t Q R is w ithin the permitted ra nge of values. This cou ld w e l l be possible
s i n ce other changes may occu r elsewhere in the system to support the l ocal
action of the generator excita tion a s it adj usts to sat isfy the spe c i fi e d terminal
volt age.
9.3. To complete the first iteration of the G a uss- S e i d e l proce d u r e , fi n d
the v o l t g e a t b u s @ o f E a m p l e 9 . 2 co m put e d with the origi nally estimated
voltages at buses W and G) rep l aced by t h e acce l e r a t e d v a l u e s i n d i c a t e d above.
Example
a
Solution.
x
T a b l e 9.4 shows
Q(l)
4
-
-:-
_
Y4 1
Im
e q u a l to zero, and so Eq.
{ V(O)* [ Y
4
42
V2,e l a) cc
+
Y Vel)
3 , acc
43
(9.24)
+
gives
Y V(O)] }
44
4
342
CHAPTER 9
POWER·FLOw SOLUTIONS
{
Substituting values
Q�l) =
=
This
=
_
for
}
the indicated quantities in this equation, we obtain
1 .02[( - 5 . 169561 + j25 .847809)(0.973703 - jo.05 1 706)
1m
+ ( - 3 .023705 + j 15 . 1 18528)(0.953949 - jO.06670t;)
+ (8.193267 - j40 .863838) ( 1 .02) ]
- Im{ 1 .02[ - 5 .573064 + j40.059396 + (8. 193267 - j40.863838) 1 .02 ])
1 .654 151 per u n i t
val u e
for Q �l ) i s n ow s u bs t i t u te d i n t o
=
=
1
Y44
E(j .
(9.25) t o y i e l d
[ 2.381 .02- j1-.6541
51
jO.O - ( - 5 .573066 + j40.059398) 1
j
7.906399 - j41 .681 1 1 5.
8 . 1 93267 - }. 40. 8 63 8 38
H ence, I VP ) I equals
=
l .017874 jO.Ol0604 p e r u n it
-
1 .017929, and so we must correct the magnitude
V}, 12orr =
=
to
1 .02,
1 .02
( 1 .017874 - jO.010604)
1 . 0 17929
1 .019945 - jO.010625 per u n i t
In t h is example Q�I ) is found t o be 1 .654 1 5 1 per u nit i n the first i teration. I f the
reactive power generation at bus @ were l imited below 1 .654 1 5 1 per unit, then
the specified l i m i t value would be used for Q�l ) and bus @ in that case would
be treated as a load bus within the i te ration. The same strategy is used w i t h i n
any o t h e r i teration i n which t h e generator Q -limits are violated .
The G au ss-Seidel p rocedure is one method of solving the power-flow
p roblem. H owever, today's industry-b ased studies generally employ the alterna­
tive Newton-Rap hso n iterative method, which is reliable in convergen ce, com­
puta tionally faster, and more economical in storage requirements. The con. verged solution of Examples 9.2 and 9.3 agrees with t he results given in Fig. 9.4
fou n d later by the Newton-Raphson method.
9.3
THE NEWfON-RAPHSON METHOD
Taylor's series expansion for a function of two or more variables is the b � sis for
t he N ewto n-Raphson method of solvin g the power-flow p roblem. Our study o f
9.3
THE N EWfON-R A P H S O N METHOD
343
the method begins by a d iscussion of the �olu t:on of a p roblem i nvolvin g o nly
two equations and two var iables. The n , we see h ow to extend t h e a nalysis to the
sol u t io n of power-flow e qu a tions.
Le t us consider the equation of a fu n ction h I of two variables x 1 and x2
equal to a constant b 1 expressed as
( 9 .27)
and a second equation involving a n o t h e r funct ion h 2 such that
( 9 .2 8 )
w h ere 6 2 is also a con stan t . The symbol II re pr e se n ts an indep e n d e n t con trol,
w h i c h is con s i d e re d consta n t in t h is c h a p t e r . As in E q s . (9.9) and (9. 1 0), t h e
fu n c t i on s 6" 1 a n d g 2 a re i n t ro d u ce d for c o nv e n i e n c e t o a l l ow u s t o d iscuss t h e
d i ffe r e n c e s b e t w e e n c alcu l a t e d v a l u e s o r h i a n d 11 2 a n c! t h e i r r e s p e c t ive speci­
fied values 6 1 a n d 6 2 •
Fo r a s p e c i fl e d va l u c o f I I l e t l i S e s t i m a t e t h e sol u t ions o f t he s e equ a t ions
to b e x \O) a n d x �()). T h e zero superscripts i n d icate t h at these va lues are i n itial
est imates and not the act u a l sol u t ions x :i and x; . We designate the correc tions
� x iO) and � x �O) as the values to be added to x \O) and x �O) to yield the correct
sol ut i o n s x7 and x� . So, we can write
( 9 .29)
( 9 .30)
Our problem now is to solve for � x iO) and .6. x iO), which we do by expa n d i n g
E q s . ( 9 .29) a n d ( 9 .30) i n Tayl or's series about the assumed solu tion to give
( 9 .3 1 )
= 0
( 9 .32)
where the pa rti a l derivatives of ord e r greater t h a n 1 i n the series of t e rms of t h e
exp a n sion have n o t been listed. T h e t e rm 3 g 1 /3x 1 1 ( 0 ) ind i ca tes t h a t t h e p a rtial
d e rivative is evaluated for the estim a t e d values of x \O ) and x �O). Other such
terms are eva l u ated si m i l a rl y .
344
CHAPTER 9 POWER-FLOW SOLUTIONS
If we neglect the partial derivatives of order greater than 1 , we can rewrite
Eqs. (9.3 1 ) and (9.32) in matrix form_ We then h ave
0 -
o
g I ( x(O)
J '
x(O)
2 ' u)
- .:, 2 ( x(O)
I ' x(O)
2 ' u)
(1
=
bI
- h I ( x eoI ) ' x(O)
2 , u)
b2
- h 2 ( x CI() ) ' x(O)
2 ' u)
where the square matrix of partial derivatives is c a l l e d the jacobian J or in this
case J ( O) to indicate t hat the initial estim ates x �O) and x�O) have been used to
compute the n u merical va lu es o f the partial derivatives. W e note that
g l(X�O ), x�O), u) is the calculated value of g I based on the estimated values of x\O)
and x�O), b u t t his calculated value is not the zero value specified by Eq . (9.27)
unless our estimated values x\O) and x�O) are correct. As before, we designate
the specified value o f g I mi n u s the calculated value of g I as the mismatch i1 g �O)
and define the mismatch i1 g�O) simi l arly . We then have the linear system of
mismatch equations
J (O)
[ i1 X C O) ]
J
i1 x(O)
2
=
[ i1 g eo
]
1 )
( 9 . 34)
i1 g(O)
2
By solving the mismatch e quat i o n s , either by triangular factorization of the
jacobian or (for very smal l problems) by finding its inverse, we can determine
Llx\O ) and i1 x �O). However, since we tru ncated t h e series expansion, these v al ues
added t o our initial guess do not determine for us the correct solution and we
m ust try again by assuming new estima tes x i I ) and x �J ), where
X CI I )
=
+ i1 x CO1 ) .,
x (O)
1
( 9 . 35 )
We repeat the p rocess u ntil the corrections become so sma ll in magnitude that
they satisfy a chosen pr eci s i o n index c > 0; that is, until l i1 x J I and l i1 x 2 1 are
both less than c . The concepts und erlying the N ewt on - R a phs o n method are
now exemplified n umerically_
Example 9.4. Using the Newton-Raphson method, solve for
nonli n ear equations
Xl
and
x2
of the
9.3
Treat the parameter
con d i t ions x iO)
=
a
u
=
.
og l
3g 1
<i g 2
dg2
aX I
t r e a t e d as a
=
XI
cos
u
Setting
has a fixed
0 - R2. calc
bl -
=
b2 -
=
1 .0
u =
4 u sin x 1
4 w: 2 sin X I
value equ a l
specifiab le o r cont ro l vari abl e .
ll g iO) = 0 - g 1 . catc
ll g �O )
4 uX 2
oX2
w e calculate t h e mismatches
First i tera t i o n .
with respect to the x 's yie lds
aX 2
aX I
l Iere t he p a ra m e t e r
be
345
as a fixed number e q u a l to 1 .0, and choose t he initial
x�O) = 1 0 . The precision i n d e x � is 1 0 - 5 .
rad and
Solution. Partial differen t iation
J
T H E N EWTON·RAPHSON M ET H OD
8x2
l .0,
to
but in
some s t u d ies i t co u l d
a n d u si ng t h e i n i t i a l estim ates
h \O) = - 0 . 6 - 4 s in ( O )
h �O)
4 11 cos X I
-
=
- 0 .3 - 4
X
=
of
and x2'
X1
-0.6
2
( 1 .0 ) + 4 cos( 0 )
- 0 .3
which we use i n Eq . ( 9.34) to y i e l d the m ismatch e q ua t ions
[
4 si n ( 0)
4 COS( 0 )
4 si n ( O)
8 x 2 - 4 cos ( O )
1[llX}O) 1 [ 1
L1 X �O)
-
- 0 .6
- 0 .3
I nverting this s i m ple 2 X 2 m a trix, we determine t h e initial corrections
[ 1 [ ][ 1 [ 1
� x �O)
. 6. x �O)
=
4
0
0
4
..
w h i c h p rov i d e fi r s t i t e r a t i o n v a l u e s of
X \ I ) = x \O)
x ii ) :.= x �()
+
+
L1 X\()
6 X �»
=
=
- 0 . 1 50
- 0.6
0.0
1 .0
I
XI
+
+
-
0.3
(tnd
=
-
X2
a s fo l l ows:
( - O . I SO)
( -- 0 .075)
=
=
T h e c o r r e c t i o n s exce ed the s p e c i fi e d t o l e r a n c e , a n d so
[LlLl ggi�ll )) 1 [
S ec o n d i t e r a t i o n .
=
The n ew m ismatches are
0 .07 5
- O . l S0 rad
0 . 925
we
- 0 .6 - 4 ( 0 .925 ) s i n ( - 0 . 1 5 )
2
- O J - 4 ( 0 .925 ) + 4 ( 0 .925 ) cos ( - 0 . 1 5 )
con t i n u e .
1 [
- 0 .047079
- 0 .064047
,
1
346
CHAPTER 9
POWER·FLOW SOLUTIONS
[aX�a X�ll)) 1 [
][
] [-
and updating the jacobian, we compute the new corrections
=
3 .658453
-
- 0 .597753
0 .55292 1
3 .44491 6
-1
-
0.047079
- 0 .064047
These corrections also exceed t h e p recision index, a n d
next iteration with t h e new corrected values
X �2 )
x�2 )
ax�3)
-
=
=
0 . 1 50 +
0 .925 +
( - 0 .0 1 (335 )
( - 0 .021 2 1 4 )
=
=
-
so
=
0 .0 1 6335
- 0 . 02 1 2 1 4
]
we move on t o t h e
0 . 1 66335 rad
0 . 903786
Continuing on to the th ird iteration, we find that the corrections 6. x\ J ) a nd
are each smaller in magnitude than the stipulated tolerance o f 1 0 - 5 .
Accordingly, we calculate the solution
X \4)
=
- 0 . 1 66876 rad ;
X �4 )
=
0 . 903057
The resultant mismatches a re insignificant as may be easily checked .
I n this example we have actually solved our first power-flow p roblem by
the Newton-Raphson method. This is because the two nonlinear equations of
t h e example are the power-flow equations for the simple system shown i n
Fig. 9.3,
( 9 . 3 6)
( 9 . 37)
Pg 1
-
Qg l
CD
I Vl l !!!
--+--+-
)6.0
®
)0.25
To load
FIGURE 9.3
The
system
eq u a t i o n s
with
power-flow
c o r r e s p o n d in g
t hose o f Exa m p l e 9 . 4 .
to
T H E NEWTON- RAPHSON POWER- FLOW SOLUTI O N
9.4
347
w he re x I represents t h e a n g le O 2 a n d x 2 represents the vol tage magn itude I V2 1
at bus G) . The co n t ro l u d e n o tes t h e vol tage magn i t u de I VI I of the slack bus,
and by changing i ts val u e fro m the specified val u e of 1 .0 per u n i t , we m ay
control the sol u t i on to t h e prob l e m . I n this textbook we do not i nvestigate this
con t rol c haracteristic b u t , r a t her, concen trate on the application of the
Newton-Raphson proced u re in p ower-flow stud ies.
9.4
T H E N EWTO N - RAPH S O N POWER-FLOW
SOLUTI O N
T o apply t h e N ewton - R ap h son m et hod t o t h e so l u t ion o f t h e power-flow
e q u a t io n s, we express bus vol tages a n d l i ne a d m i t t a n c e s in pol a r form . W hen n
is s e t e q u a l t o i i n ECl S . (9.6) a n d (9.7) a n cl t he corresponding terms a re
s c r ; l r a t c c..! rro m t h c S U ll1 m ; l { i o n s , wc o bt<l i n
/J;
Q;
= I �f ( ; u
=
-
+
I
/\'
II
L
- iI
1/ I
I V, i :: B i i -
Vi v't Y, ,, l cos( (Ji"
N
L. I V, V,) �
11 = 1
t1
1/
+
l si n ( 8iI /
() 1/
- ('5J
+ 0 1/ - 0 ; )
( 9 .38 )
( 9 .39)
*" i
These e q u a t ions can be rea d i ly d iffe re n t iated with respect t o voltage a ngles a n d
m ag n i t u d es. T h e t e rms invo l v i n g Gi; a n d Eii come from the defin ition o f }�j i n
E q . ( 9 . 1 ) and t h e fact t ha t t h e a n g l e ( 0 1 / - 0 ) is zero whe n n = i .
Let u s p ostpone cons i d e ra t i on o f vol tage-con trolleu buses for now a nd
reg a rd a l l buses (except t h e sl a ck bus) as load b uses w it h known dema n d s Pdi
a n d Qdi' The slack bus has sp ecified values for o ( and I VI I , and each of the
o t h er buses in t he n etwork h as t h e two state v a r i a bles 0i and
I V,I t o b e
c a l c u lated i n the pow er -flow sol u t ion . T h e known values o f Pdi a n d Q d i
correspond to t he n ega t ive of t h e b const a n t s shown in Eqs. (q . 27) a n d (9.28), as
d e monstrated i n Ex a m p l e 9 . 4 . A t c a c h or t h e nonsl ack buses e s t i m a t e d val ues of
0/ a n d I Vi i c orrespond to t h e c s t i m a t e s X �") and x �)) i n the p rece d i n g section.
Co r responde nce to t h e 6 g m i s m a t c h e s or E q . (9.34) fol lows from Eqs. (9. 8) a n d
(9.9) b y writing t he powe r m ism a t ch es for t h e typical load bus CD ,
6. Q i
( 9 .40 )
=
Q' . sch - Qi, calc
( 9 .4 1 )
Fo r sim p l ic ity sake, we n ow w r i te m i s m a tch equations for a fou r-bus system, a n d
i t w i l l be come obvious h ow t o exte n d t hose equ ations to systems w ith more than
fou r buses.
348
CHA PTER
9
POWE R-FLOW SOLUTIONS
For real power
Pi we have
( 9 A2)
The last t h ree te rllls c a n be
magnitudes without a l t e r i n g
so
m u l t i p l i e d a n d d i v i d e d hy t h e i r r e s p e c t ive v o l l cl g c
t h e i r va l u c s , a n u
wc
o b ta i n
( 9 .43 )
There are advan tages to this form o f the equation as we shall soon see. A
similar m ismatch equation can be written for reactive power Qi'
( 9 .44 )
Each nonslack bus of the system has two equations like those for 6. Pi and
Collecting all the mismatch equations into vector-matrix form yields
a P2
a04
a P2
a0 2
aP4
a0 2
a Q2
a0 2
I
a Q4
a0 2
JII
aP4
a 04
a Q2
a0 4
aP
I V4 1 a l v2
41
aP2
V
1 2 1 a l V2 1
ap4
1 V21 3 1 V2 1
aQ
I V21 al V2
21
JI2
J22
J 21
aQ 4
a0 4
I V2 1
aQ 4
a l v2 1
Jacobian
ap
I V4 1 a l v4
41
aQ
I � I al v2
41
a Q4
I V4 1 a l v4 1
6.0 2
6. P2
6.0 4
6. P4
6. 1 V2 1
6. Q 2
--
I V2 1
il l
V4 1
I V4 1
--
----- � ----
Co rrect ions
6. Qj.
6. Q4
---I
M is m a t c h e s
( 9 .45)
9.4
T H E N EWTO N - R A PHSON POWER- FLOW SOLUTIO N
349
We c a nn o t i ncl ude mismatches for t he slack bus si nce 6. P I a n d 6. Q I are
u nd e fi n ed when P I and Q I a re not scheduled. We a lso omit all terms involving
6. 01 and 6. 1 VI I from t h e equ ations because t hose corrections a re b o t h zero at
the s lack bus.
The parti tioned for m of E q (9.45) emphasizes t h e fou r d ifferent types of
partial d erivatives which e n te r i n t o the j a cobian J . The elements of J 1 2 a n d J 2 2
h ave volt age-magnitude m u l t i p l i e rs because a simpler and more symmetr i c a l
j acob i a n resu l ts. I n choos i n g t h is format, w e h ave used the i d e n tity
.
------- -----lVI
c7 PI,
j aI
X
VJ I
E l e me n t of J 1 2
L1 1 � 1
I VJ I
=
(7 p.
-I
() Vj' I
X
( 9 .46)
�I �I
Co rrec t io n
and t h e correct ions become 6 1 � 1 / I � I a s s h ow n r a t h e r t h a n 6. I Vj I .
T h e so l u t i o n of Eq. ( 9 . 4 5 ) is fo u n d by i t e ra t i o n a s fo l l ows:
•
•
•
•
E s t i m a t e values ofO) and I � I (O) fo r the state variables,
Use the estima tes to c a l c u l a t e :
p/:Ojotc a n d Q ���alc from E q s . ( 9.38) a n d (9.39),
mismatches 6. p/O) and 6. Q�O) from Eq s . (9.40) and (9.41), and
the p a rt i a l derivat ive e l e m e n ts o f the jacobian J .
Solve E q . (9.45) for the i n i t i a l corrections 6. 0�0) a n d � I v: I ( 0 ) / I v: I (0).
A d d t h e solved corrections to the i n i t i a l estim a tes to obtain
I VI I ( I ) = I VI,' I ( 0 ) + L1 1 VI, I erl)
•
U s e the new v a l u es
()� I ) and I � 1 ( l l
co n t i n u e .
more general
v a r i a b l e s are
In
t e rm s , t h e
+
8(k
I
I)
=
+
6. 8 I( k )
(
I VI I (0) 1 +
..'l l v: I (0)
I VI 'i ( O )
)
as s t <l rt i n g v a l u e s fo r i te r a t i o n
u p da t e fo r m u l a s
8 ( 1.: )
I
=
( 9 .4 7 )
( 9 .48)
2 and
fo r t h e s t a r t i n g v a l u e s o f t h e s t a t e
( 9 .49 )
350
CHAPTER 9
POWER-FLOW SOLUTIONS
For the fou r-bus system subm a trix J I I has the form
JI I
a P2
a02
a p:,
a0 2
a P4
a0 2
=
a P2
a 04
a p-"
a 04
aP4
a 04
a P2
a03
a p:,
a03
aP4
dO"
(9.51)
Expressions for the c l e m e nts of t h i s e q u a t i o n ( I re e a s i l y found by d i ffe re n t i a t i n g
the appropriate n umber of t erms in Eq . ( � . 3 8 ) . W h e n the v a r i ab l e ! l e q u a l s t h e
p articu l ar value j , only o n e o f t h e cos i n e terms i n t h e s u m m a t i o n o f E q . (9 .3R)
contains OJ , and by partial d iffere n t i a t i n g that single t e rm with respect t o Dj , we
obtain the typical off-d iagon al elem e n t of J I I '
( 9 . 52 )
On t h e other h a n d , ecery term i n the summation of
so the typical d i agonal element of J I I I S
a p1aD·
L. I V; V,I Y,) s i n ( 8ill + D I I - Di )
N
I
I
*i
n ="
n
By comparing this expression and t h a t for Q i i n
ap
a D 1.
1
- Q I.
Eq.
- I vf B -I
II
Eq. ( 9 .38) contains 0;,
N
a p1
- L. -
nn
I
oF ;
=
d Oli
a nd
( 9 .53)
(9.3 9 ) , we ob t a in
( 9 .54)
J 2 1 as follows:
In a q uite similar manner, we can d e rive form ulas for t h e elements of submatrix
=
ao I·
- I VI VJ YIJ· i cos ( 8 IJ. .
+
0J - 0 I )
L. 1 V; � �n I cos ( 8i n + 0"
nn
N
�
1
oF ;
-
( 9 .55 )
oJ
( 9.56)
9.4
THE N EWfON-RAPHSON POWER-FLOW SOLUTION
351
Comparing this equation for aQjaoi with Eq. (9.38) for Pi ' we can show that
aQ i
2
- = p. - I V I G /..I
a DI,
I
(9.57)
I
The e lemen ts o f submatrix J 1 2 a r e easily found by first fi n d i ng the expression for
t h e d er i vat i ve CJPja l Vi i and then multiplying by I Vi i to o b ta in
o P'
I Vj l -oI VI
)
Cu mparisol1 wi t h Eq _ ( 9 . ) ) )
=
I Vj l I V; Y) cos( 81) + 0) - 01 )
( 9 .58 )
yields
( 9 .59)
T h i s is a most u s e fu l resul t , for it re d u c es the compu tation i nvolve d i n forming
the j acobian since the off- d i agona l c l e m e n ts o f J 1 2 are now simply the negatives
of the cor r es p o n d i n g e l e m e n t s in J 2 1 - Th is fact wou ld not have become ap parent
if we had not mu l t i p l i e d JP';J I Vi I by t h e ma g n itu d e I L� I i n Eq. (9.43). In an
a n a logous manner, the d iago n a l e l e m e n ts of J 1 2 are found to be
( 9 .60)
a n d comparing this resu l t w i t h Eqs. (9_56) and (9.57), we arrive at the formula
I V I -71 V1
a Pr
I
(
I I
=
a Qi
a oI
-
2
+ 21 VI G ,
I
1/
=
p. + I VI I GI'I'
I
2
(9.61)
Fin al l y , t h e off-diagon a l a n d d i agonal eleme nts o f submatrix J22 o f the jacobian
a re d etermined t o be
( 9 . 62)
( 9 .63)
352
CHAPTER 9
J
POWER-FLOW SOLUT IONS
Let US now bring togeth er the results develo ped above In the follow ing defini­
tions:
Off-diagonal elements, i
=1=
j
ap.I
M I. J �
i =
j
d OJ
aQI
NI J. �
Diagolla l c!CJ1lCIl ls ,
lVI
a Q I.
( 9 . 64 )
<I I V; I
-
( 9.65)
j
M/ I �
D PI
ao·
I
I
- Mii
apI
l VI a I
l V:
-
=
N/ I
- .'
? I Vi I 2 S il
2 '' VI I
+
( 9 . 66)
-
2
G / I·
( 9 . 67)
interr elation ships among the eleme n ts in the four subma trices of the jacobi an
are more clearly seen if we use the definit ions to rewrite Eq. (<).45) in the
follow ing form:
M 22
M 23
M 24
M32
M 33
M 34
N22 + 2 1 V2 1 2 C22
- N3 2
M4 2
M4 3
M4 4
- N4
N22
N 3
2
N3 3
N24
N3 2
N4 3
N4 2
x
N34
N44
Ll o 2
22
-M
i103
i1 0 4
i1 I V2 1 / I V2 1
i1 1 V3 1 / 1 V3 1
i1 1 � 1 / I V4 1
-
M3 2
M4 2
N33
1 2 B 22
-S
2
N 3
+ 2 1 v3 1 2 C 33
- N4
2
2 1 V2
-
3
24
- N34
N.!, :,
+
2
2 1 VI 1 C4 4
M 23
M �.!,
- M33 - 2 1 V3 1 2 B':l
M� 4
M4 3
2
- M')4 - 2 1 V4 1 B 44
Ll P2
i1 P3
i1 P4
i1 Q 2
i1 Q3
i1 Q 4
( 9 .68 )
T H E N EWTO N - R A P HSON POW E R - FLOW SOLUTION
9.4
353
So fa r we h ave reg arded a l l nons lack buses as l o a d buses. We now consider
vol ta ge-cont rol led buses a lso.
Voltage-con trolled b u s e s . I n the polar form o f the powe r-flow equations
volt age-con trolled buses a re easily t aken into account. For example, i f bus @
of t h e fou r-bus system is vol tage con t rol l e d , then I V4 1 has a specified constant
v a l u e and the voltage corr e ct ion � I V4 1 / I V4 1 must a lways be zero. Conse­
q uently, the s i xth column of the jacobian of Eq . (9. 68) i" a lways mul tiplied by
zero, and so it may be removed. Fu rthermore, since Q-\ is not specified, the
m i sm a tch � Q 4 cannot be d e fi n e d , a n d so we must omit the s ixth row of
Eq. ( 9 . 6 8 ) corresponding to Q4 . O f cou rse , Q4 c a n be calcul ated after the
power-flow sol u tion becomes ava il a b l e .
I n t h e g e n e r a l case if th ere a re N,,, vol tage-contro l l e d buses besides t h e
s l a c k b u s , a row a n d co l u m n fo r e a c h s u c h b u s i s omitted from t h e p o l a r form o f
t h e sys t e m jacob i a n , which t h e n has (2 N - Ng - 2 ) rows a n d (2 N - NK
2)
co l u m n s c o n s i s t e n t w i t h Ta h l e 9. 1 .
-
Ex a m p l e 9 . 5 . T h e s m <l l l powe r sys t e m o f Ex a m p l e 9 . 2 h a s t h e l i ne d a t a a n d b u s
cl a t a g i v e n i n T8 b l e s 9 . 2 a n d 9 . 3 . A power- Aow s t u d y o f t h e sys t e m i s t o b e m a d e b y
t h e N e w t o n - R a p h s o n m e t hod u s i n g t h e p o l a r fo rm o f t h e e q u a t ions for
P
a n d Q.
De t e rm i n e t h e n u m b e r of rows a n d co l u m n s i n t h e j a cob i a n . C a l c u l a t e t h e i n i t i a l
t1 PjOJ
mismatch
a n d t h e i n i t i a l v a l u e s o f t h e j a cob ian e l e m e n t s o f t h e (second row,
t h i rd c ol u m n ) ; of t h e ( s e c o n d row, s e c o n d co l u m n ) ; a n d of t h e ( fi f t h row, fi ft h
co l u m n ) . U s e t h e sp e c i fi e d v a l u e s a n d i n i t i a l vo l t age e s t i m at e s shown i n Ta b l e 9 . 3 .
Solution . S i nce t h e s l a c k b u s h a s n o rows or col u m ns i n t h e j a c o b i a n , a 6 X 6
m a t rix wou l d be n e c e s s a ry i f
P
a n d Q w e re s p e c i fi e d for t h e rem a i n i ng t h ree
b u s e s . In fa c t , h oweve r , the v o l w g e m Cl g n i t u d e is s p e c i fi e d ( h e l d c o n s t a n t ) at bus
@,
and t h u s the j a cob i a n w i l l b e a 5
based on
x
t h e e s t i m a t ed ;1 11t1 t h e spcc i fi c Ll vol t a ges
26. 3 5969:')/ 1 0]
fo r m o f t h e o fI- d i u gon a J e n t r i e s o f Ta b l e 9 .4,
Y:l I
8 �1I)
-
(j �())
a n d t h e (kl go n<l l
val ues
,I n c.!
c l e l l l e n t Y.1.1
,l IT
all
. 3 0993° ;
,,�
ze r o ,
0 . \ lp, 2() 7
f ro m Eq .
( 1 . 0 ) 2 8 . 1 93 2 6 7 + ( 1 . 0 X
+ ( 1 .0
=
-
X
1 .02
P3. c alc
o r Ta h l e 9.3, we n e e d t he p o l a r
5 m a t r i x . In ord e r t o c a l c u l a t e
X
(9.30) we obt a i n
-
1 .0
j 4 0 . (; () J H 3 t: . S i n c e
X
and the i n i t ial
26 . 359695 ) cos ( 1 0 1 .30993°
1 5 A l 7934 )cos( 1 0 1 .3 0993°
0 . 060 4 7 p e r u n i t
Y.12
)
)
- 354
CHAPTER 9
POWER-FLOW SOLUTIONS
Scheduled real power into the network at bus ® is 2.00 per unit, and
initial mismatch which we are asked to c al c u l a t e has the v a l u e
-
tl
P���alc =
- 2 . 00 -
From Eq_ (9.52) the (second
=
-
a n d from Eq. (9.53) t h e
For
1 5 .420B9B
element
1 .0
X
per
=
4 1 . 26B707 per u n it
1 5 .4 1 7934 ) s i n ( l O I .30993° )
unit
2 6 . 35 9695 ) si n ( 1 0 1 .30993° )
r o w fi ft h
,
unit
e
c leme n t is
o f t h e (second row, second
( 1 .0
the clement o f t h e (fi fth
x
per
=
X
- 1 .93953
c
1 . 02
X
=
t h i rd co l u m n ) j a o b i a n
row,
- ( 1 .0
( - 0 .06047)
so t h
co l u m n ) i s
+ 1 5 .420898
co l u m n ) E q . (9.63) yields
- 4 l .268707 - 2 ( l .0) \ - 40J) 63838)
= 40 .458969 p e r u n i t
Using t he initial i n p ut data, we c a n simil a r ly ca l c u l at e initial v a l u e s o f t he other
elements of the jacobian and o f the power m i s m a tche s a t a l l the b u s es of the
system.
For the system of the preceding example the numerical values for i nitial­
ization o f the mismatch equations are now shown, for convenience, to only t hree
9.4
T H E N EWTO N - RA P HS O N POWER- FLOW SOLUTION
355
decimal places as fol lows:
@
G)
@
G)
W
G)
G)
45 A43
0
@
4 1 .269
0
- 26 . 3 65
- 1 5 .42 1
8 . 882
- 1 5. 4 2 1
4 1 . 786
- 5.273
- 9.089
0
5 . 273
44. 229
- 8. 254
- -- I
3 . 084
i1 0 2
i1 o)
0
i1 04
8 . 1 33
0
- 26.365
0
G)
G)
i1 1 V2 1
- 3.084
1 V2 1
0
i1 1 V3 1
1 V) I
40.459
0
.597
- 1 . 94U
2 .2 1 3
- 0 . 447
- 0 . 835
Th i s system of equations y i e l d s v a l u e s for t h e vol tage corrections of t h e first
i t e r a t ion wh ich are needed to u p d a te the s t a t e variables a ccord ing t o E qs. (9.49)
and (9.50). A t the end of the first i te ra t ion the set of upda t e d vol tages at t h e
buses is:
B u s no. i
=
I V, I (per u n i t )
0,
( d e g.)
CD
0
l . 00
W
- 0 .93094
0 . 98 335
G)
- l . 78790
0 .97095
CD
- l . 54383
1 .02
Th ese updated voltages a re t h e n used to reca l cu l a t e t h e jacobi a n a n d m is­
ma t ches of t h e second i t e r a t i o n , a n d so on . The i t e ra t ive proc e d u re con t i n u e s
u n t i l e i t h e r t h e m is m a t ches 6. Pi a n d 6. Q, b e c o m e l ess t h a n t he i r stip u l a t e d
al lowable val ues or a l l 6.8, and 6. 1 V: I a re l ess t h a n t h e chosen p recision index.
When the solu t ion is comp l e t e d , w e can use Eqs. (9. 38) and (9.39) to calc u l a t e
re a l and reactive powe r, P I a n d Q I ' at t h e s l ack b u s , and t h e reactive power Q 4
a t vo l t age-con t rol led bus @ . L i n e flows can also be computed from the
d ifferences i n bus vol t ages a nd t he known p a ramet ers of t h e l i n es. Solved values
of the b u s voltages and I ine fl ow s for the system of Example 9.5 a r e shown in
F i g s . 9.4 and 9.5.
The number of i t e r a t i o n s req u i re d by t h e N ew ton- Raphson m e t hod using
bus a d m i t t a nces i s practic a l ly i n d e p e n d e n t of t h e n u mber of buses. The t i m e for
t h e Gauss-Seidel method ( e m ploy i n g b u s a d m i ttances) in creases a l most d i rectly
with t h e n u mber of b u ses. O n t h e other h a n d , com p u ting the e lements of t h e
356
CHAPTER 9
POWER-FLOW SOLUTIONS
j acobian is time-consuming, and the time per iteration is considerably longer for
the Newton-Raphson method. When sparse matrix tech niques are employed,
the advantage of shorter computer time for a solution of the same accu racy is in
favor of the Newton-Raphson method for all but very small systems. Spars i ty
features of the j acobian are d iscussed in Sec. B . 2 of the Append ix.
9.5
POWER-FLOW ST U D I ES
D E S I G N AND OPERATION
IN
SYS TEM
Electric u t l l ity companies use very elaborate programs for pow e r - n o\\' studies
a imed at evaluating t h e adequacy of a c om p l ex, i n t e rcon n e c t e d n e t w o r k I m por­
tant i n fo rm a t i o n i s ob t a i n e d c o n c e rn i l l g t h e d e s i g n a n d o p e ra t i o n o f sy s t e m s
t h a t h a ve n o t y e t b e e n h u i l t a n d t h e e l rc e t s 0 [" c l 1 ; l l l ges o n e x i s t i n g s ys t e m s . A
power-now study ["o r a sys t e m o p e r �l t i l l g LI n d e r a c t u a l m p roj c c t c u n o r m a l
o perating cond itions i s ca l led a base case. The results from t he base case
constitute a benchmark for comparison of changes in network R o w s and vol tages
u nder abnormal or con tingency cond itions. The transmission pl anning engineer
can d iscover system weaknesses such as low vol tages, line overloads, or loading
con d i tions deemed excessive. These weaknesses can be removed by m a k i n g
design studies i nvolving changes and/or add itions to the base case system. The
system model is then subjected to computer-based contingency testing to
discover whether weaknesses arise under contingency condit ions involving ab­
normal generation schedules or load levels. I nteraction between t he system
designer and the compu ter-based power-flow program continues until system
p erformance satisfies local and regional planning or operating criteria.
A typical power-flow program is capable o f 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 faci lities are
sufficiently large.
Data supplied to the computer must include the n u m e r i c a l val u es of the
l i n e and bus data (such as Tables 9 . 2 and 9 . 3 ) and an indication of \v h e t h e r a
bus is the slack bus, a regulated bus where voltage magn itude is held constant
by generation of reactive power Q , o r a l o a d bus with fixed P a n d Q. Where
values a re not t o b e h e l d c o n st a n t t h e q u a n t i t i e s given in t h e t ab l e s a re
interpreted as initial estimates. Lim i ts of P and Q generation usually must be
specified as well as the lim its of line kilovoltamperes. Unless otherwise speci­
fied, programs usually assume a base of 1 00 MY A.
Total line-charging megavars specified for each line accoun t for shunt
capacitance and equal 13 times the rated line voltage in kilovolts times Ic hg '
d ivided by 1 0 3. That is,
.
( 9 .6 9 )
w here
I VI
IS
the rated line-to-line voltage in k ilovolts, ell
IS
line-to-neutr al
9.5
P O W E R · FLOW STU D I ES IN S YSTEM DES I G N A N D OPERATION
357
c apacitance in fa r a d s for the e n t i re l e ngth of t h e l i ne , a n d Ichg is d e fi n e d by
Eqs . (5 .24) and (5.25). The p rogram creates a n o m i n al -7T representation of t h e
l i n e similar to Fig. 6 . 7 b y d iv i d i n g t h e capacitance computed from t h e g iven
v a l u e of charging megavars equ a l ly between the two ends of the line. It is
evi d e n t from Eq. (9.69) that l i ne-c h a rg i ng m egavars in p e r u n i t equals the
per-un i t shunt susceptance of the l i n e a t 1 .0 per- u n i t vol tage. For a l o n g l i n e the
comput e r could be programmed to compute the equivalent 7T for capacitance
d i s t r i buted even ly along t h e l i n e .
The p r i ntout o f resu lts p rov i d e d b y t h e computer consists o f a n u m b e r o f
t a b u l a t i o n s . Usual ly, t h e most i mp o r t a n t i n form ation t o b e con s i d e red fi rs t is
the table which l ists e ach bus n u m b e r and name, b us-voltage magn i tude in p e r
u ni t and p hase a ngle, g e n e ra t i o n a n d load a t each b u s i n m egawa tts a n d
megava rs, and megavars of s t a t i c c a p a c i tors or r eactors on t h e o u s . Acco m p a ny­
i n g the bus i !l format ion i s the fl ow of meg a w a t t s a n d megava rs from t h a t b u s
ove r e a c h t r , t n s m i ss i o n l i n e c o n n e c t e d t o t h e b u s . T h e t o t a l s o f system ge n e ra­
tion and l oads a re l isted in megaw a t t s a n d megavars. T h e t a b u l a t ion described
is shown in Fig. 9.4 for t h e fou r-bus system of Example 9. 5 .
A sys t e m m a y be d i vi d e d i n t o a re a s , o r o n e s t u d y m a y i n c l u d e t h e sys t e m s
u f severa l com p a n ies with each designated as a d i ffere n t a r e a . The computer
p rogram wi ll exam i n e the flow between areas, and d ev i a t ions from t h e pre­
scri b e d flow will be ove rcome by c a u s i n g the approp riate change i n ge nerat ion
o f C:! sel ected gene rator i n e ach a re a . In a c t u a l system operat ion i ntercha nge of
power between a reas is monitored to determine whe t h e r a g iven a r e a is
p rodu cing that amount of pow e r w h i ch will result i n the desired i nterc h a n g e .
Among other i n for mation t h a t may be obt a i n e d is a l isting of a ll b u s e s
where the pe r-unit volt age m a gn i t ud e is above 1 .05 or below 0 . 9 5 , o r o t h e r
l i m i t s t h a t ma y b e spec i fi e d . A l ist of l in e l o a d i n gs i n megavoltampe res c a n b e
2
o b t (1 i n e d . Th e p r i n t o u t w i I I a l s o l i s t t h e t o t a l me g a w a t t ( \ / \ R ) l osses a n d
2
m e g av a r ( \ / \ X ) r eq u i r e men ts of t h e syst em, and b oth P a n d Q m isma tch a t
e ach bus. M ismatch is a n i n d i c a t ion of t h e preciseness of the solut ion and i s the
d i fference between P (and also u s uatly Q) entering a n d leaving each bus.
T h e n u m c r i ct l resu l t s i n t h e p r i n t o u t o f Fig. 9.4 a rc from ,I N c w t o n ­
R a p hson powe r-fl ow s t u d y of t h e s y s t e m described i n Ex a m p l e 9 . 5 . T h e sy s t e m
l i n e d a t a a n u b u s d a t a LI re p ro v i d e d i n Ta b l e s 9 . 2 .I n d 9 . 3 . Th ree N ew t o n ­
R a p hson i t e r; \ t i o n s w ere req u i re d . S i m i l a r st u d ies e m p l o y i n g t h e G a u ss - S e i d e l
p ro c e d u re re q u i r e d m a ny m o r e i t e r a t i o n s , ;I n d t h i s i s a c o m m o n obs e rva t i o n i n
com p a r i n g t h e two i t e ra t iv e m e thods . I n spe c t i o n o f t h e p r i n t o u t reve a l s t h a t t h e
4.81 MW.
1 1 1 2 R l osses of t h e sys t e m a r e (504 . 8 1 - 5 0 0 .0)
Figure 9 . 4 m ay b e exa m i n e d for more info rm a t i o n t h a n just t h e tab u l at e d
results, and t h e i n form a t ion p rovi d e d c a n b e d isplayed on a one-line d iagram
showi ng the ent ire system o r a portion of t h e system such a s a t load bus G) of
F i g . 9 .5. Transm ission design e n g i n e e rs and system operators usually can call for
video d isplay of such selected power-flow resu l ts from computer-interactive
te rm inals or workstations. The m e g aw a t t loss i n a ny of the l ines can be fou n d by
com p aring the values of P at t h e two e n ds of the I i n e . A s an ex?mple, from
=
x - - - - - - - - - - - - - - - - - - - - - - - - - -Bu s in fo rm ati on - - - - -
Bus
Volts
Angle
(degJ
no.
Name
(p. u . )
1
Birch
1 .000
2
Elm
0.982
3
4
Pine
Maple
0.969
1 . 020
Area totals
FI GURE 9.4
O.
- 0.976
- 1 .872
1.523
-
X -
-
-
-
-
--
-
-
-
(MW)
(Mvar)
- -
- - - - - - - - - - -X - - - - - - - - - - - L in e fl o w
186.81
1 14.50
50.00
O.
O.
1 70.00
O.
318.00
504.81
O.
(MW)
181.43
295.93
230 kV
30.99
SL
2
3
1 05.35
PQ
49.58
500.00
To
B u s Name
123.94
80.00
Bu s
type
(Mvar)
200.00
Newton-Raphso n powe r-flow sol u t i o n for the sys tem of Example 9 . 5 . B a s e is
respe c t ive ly.
-
- - G eneration - - - - X - - - - - Load - - - - X
PQ
PV
Elm
Pi n e
1
Birch
Maple
1
- - - - - - - - - -X
Line fl o w
(MW)
4
4
2
3
-
Birch
Maple
Elm
Pine
(Mvar)
22.30
38.69
61.21
98. 12
- 38.46 - 31.24
- 131.54 - 74. 1 1
- 97.09 - 63.57
- 102. 9 1 - 60.37
74.92
133.25
104.75
56.93
309.86
and
1 00
MVA. Ta b l e s 9.2 a n d 9.3 show t h e l i n e d a t a a n d bus d a t a ,
9.5
P O W E R -FLOW STU D I ES IN SYSTEM D ES I G N A N D O P E R ATION
! t
MW flow
-
98. 1 2
61.21
97.09
63. 5 7
®
359
1 t
! t
200
Mvar flow
102.91
1 04 . 75
-
-
----+--
----+--
60.37
56.93
123.94
V:I
=
0.969/- 1 . 87°
V4
=
W
1 .02/1 .52°
To IC3d
of P
FI G U RE 9.5
fl o w
Flow
anu Q at bus
G)
fo r the sys t e m of Exa m p le 9 . 5 . N u m b e rs b e s i u e t h e a rrows show t h e
o f P 3 n u Q i n me gawa t t s a ll d m e g a v a rs. T h e b u s vol t age i s s h o w n i n p e r u n i t .
Fig. 9 . 5 w e see that 98. 1 2 MW flow from b u s CD i n to l i n e (1)- G) a nd 97.09
M \V flow into bus G) from t h e same l i n e . Evidently, the 1 1 1 2 R loss i n a l l t h ree
phases of the l i n e is 1 .03 M W .
Consideration of t h e megavar fl o w on the l i n e b e tween b us CD and b u s ®
is s l ig h tly more com p l ic a t e d because of t h e charging m e g avars. The computer
con s i d e rs the distributed capacitance of the line t o be concen trated, h a l f at one
e n d of the l i ne a n d half at t h e o t h e r e n d . In t h e l i n e d at a g iven i n Tab l e 9.3 the
line charging for line CD - G) i s 7 .75 Mvars, b u t t h e computer recog n izes t hi s
value to be t h e va l ue w h e n t h e voltage i s 1 .0 p e r u n i t . S ince charg i n g m e g avars
vary as the square of the vol tage accor d i n g to Eq. (9.69), vol t ages a t b us e s CD
a n d G) o f 1 . 0 p e r u n i t a n d 0.969 p e r u n i t , respec t ively, m ake t h e c h a rg i n g at
those buses eq u a l t o
7 .75
-2
7 .75
-- X
2
X
( 1 .0 )
2
( 0 .969) 2
=
=
3 . 875 Mvars at b us W
----
---3 . 638 Mva rs a t b u s U)
Figure 9.5 shows 6 1 .2 1 Mva rs going from bus CD i n t o the l i n e to b u s G) a n d
63.57 Mva rs received a t b u s ® . T h e i ncrease i n m e gavars is d u e t o the l in e
chargi ng. T h e t hree-ph ase fl o w o f megawatts a n d m e g avars i n t h e l i n e i s s hown
in the single-line d iagram of Fig . 9.6.
360
CHAPTER 9
�
1.0
POWER-FLOW SOLUTIONS
�
98 2
98 . 2
9 7 .0 9
9
1
1
t-__
- __.--___________
_____----r-_
61.2
1
t
FIGURE 9.6
65. 0 85
59 . 9 3 2
3.875
3 . 638
I
®
09
;
63 . 5 7
l
Single-line diagram showing now o[ mcgawa t t s <l n d Illcga v a rs in t h c l i n e con n e c t i n g
of the system of Exa m p les 9.5 a m I 9.6.
0 969
buses
/- 1 . 8 72'
CD a n d G)
Exa m ple 9.6. From t h e l i l l e ! l ows s i 1 n w l l i l l F i g . I) . h , c a l c u la t e thL C l I rr e n t ilow i ng
in t he equivalent
of Fig.
9.2.
c i rc u i t of
from b u s
the l ine
CD
to b u s
G)
i n t h e 2 3 0 - k Y sys t e m
Usi ng t h e ca l c u l a t e d c ur re n t a n d t h e l i n e p a r a m e te rs g i ve n i n T a b l e
c a l c u l a t e the l i n e
9.2,
loss and c o m p a r e t h i s v a l u e w i t h t h e d i ffc r e n c e b e t w e e n t h e
pow e r i n t h e l i n e from b u s
a n d the power out a t b u s
S i m i l a r l y , fi n d J 2 X
{ 2R
G) .
CD
i n t h e l i ne a n d compare w i t h t h e v a l u e w h i c h cou l d b e fou n d from t h e d a t a i n
Fig.
9.6.
Solution. Figure 9 . 6 s h ows t h e megawa t t a n d m e g ava r flow
e q u i v a l e n t c i rc u i t of l i ne
X of
all
t h ree p h ases is
or
CD - G) .
S
=
S
= 97 . 0 9
98 . 1 2
and
or
calculated
using I I I
+
1 17.744/33.56°
j59 .932 = 1 1 4 .098 / 3 1 . 69°
_
1 1 7 ,744
h
---,-----
I II
-=
I Vt
X
13
230
x
230
x
/ 2R
R
1 .0
I
-
x
/ 2 X of l i n e
the
per-ph ase
MVA
MVA
A
= 295 . 5 7 A
0 . 969
i n the s e r i e s R +
jX of l i ne
CD - G)
V3 1 / I R + JX I . T h e base i mp e dance
3
X
( 295 . 5 6 )
=3
X
( 295 .56)
=
295 . 5 6
1 1 4 ,098
(23 0 ) 2
1 00
a n d X p a r a m e te rs of T ab l e
loss
=
-----
Z base =
and using t h e
=
j65 .085
III
The m ag n i t u d e of c u rren t
=
+
in
The t o t a l m e g av o l t am p e r e flow t h ro u g h R a n d
=
529
9.2,
can a lso be
is
n
w e h ave
2
X
0 . 00744
X
529
X
10-6
2
X
0 .03720
X
529
X
10-6
= 1 . 03
=
MW
5 .157
Mvar
T h ese v a l u e s com p a re w i t h (98. 1 2
5 . 1 5 3 Mva r from Fig. 9 . 6 .
9.6
- 97.09)
9.6
R EG U LAT I N G TRANSFO R M ER S
1 .03
=
361
M W a n d (65.085 - 5 9 . 932)
=
REGULATING TRANSFORM ERS
As we h ave seen in Sec. 2.9, regu l at i n g transformers can be used to control the
real and react ive power flows i n a circu i t . W e now d evelop the bus a dm it t ance
e quations to i nc l u d e such transformers i n power- fl ow s t u d ies.
Figure 9.7 i s a more d e t a i l e d repres e n t at ion of the regu l a t i n g t r a n s form e r
o f Fig. 2.24( b). The a d m i t t a nce Y i n per u n i t is t h e reci procal of t h e p er-unit
imped ance of the transformer which has the transform a t i on ratio 1: t as s hown.
T h e a d m it tance Y is shown on t h e s i d e of the ideal t ra nsform e r nearest node CD ,
w h i ch is t h e tap-changing s i d e . T h i s design a t ion is i m port a n t i n u s i n g the
e q u a tions which are to be d e r i ve d . I f w e a re co n s i d e r i ng a t r a n s fo r m e r with
off- n o m i n a l turns rat io, t m ay be rc a l or i m a g i n a ry. s u c h a s 1 . 02 for an
approxi mate 2% boost in vol t age magnitude or f irr / 6 0 for a n a pproxim a t e 3°
s hift p e r phase.
Figure 9.7 shows curr e n t s I i and Ij e n t e r i n g t h e two bus es, ancl the
voltages are V; n nd Vj refe n' c d t o t h e refe rence node. T h e compl e x expressions
for power i n to the ideal t ra n s former from bus CD and bus CD are, respective ly,
5I
=
Sj.
VJ*
I
I
=
t VI /J*
( 9 . 70 )
S ince we are assu m i n g t h a t we have an i d e a l transformer w i t h n o losses, t h e
power 5 i i n to t h e i d e a l transformer from bus CD m ust e q u a l t h e power - Sj out
of the i de a l transformer on t h e bus CD side, and so from Eqs . (9. 7 0) w e obtain
JI
The curre n t Ij can b e expressed by
=
- ( *1
( 9 .7 1)
)
( 9 . 72)
+
�
CD
1
�
I,
Ideal transformer
--1
:
t
+
VI
1
-.L
•
�
/.
J
t
tV,
t
t
(j)
�
�
�
.
+
t- -
v-j
F I G U R E 9.7
More
detailed
l a nce
d i a g r a m fo r the
pe r-u n i t
t u rn s ra t i u is 1 It.
fo r m e r o f Fig.
2.24( b )
react r answ hose
362
CHAPTER 9
Vi
-t
POWER-FLOW SOLUTIONS
tY
t ( t - l)Y
CD
(1 - t ) Y
!-
Vj
FIGURE 9.8
C i rcuit h av i n g the node a d m i t t a nces o f Eq.
( 9.74) when , is real.
.---�---- --------�-
Multiplying by
Setting [ [ *
we have
-
t * and substituting
/I.
=
CD
(JJ
=
f;
for
1 1 * Y VI
-
-
t * Ij yield
, * Y V)
2
1 / 1 and rearranging Eqs. (9.72) and (9.73 )
( 9 . 73 )
i n t o Yhu,
matrix form,
[Y;i Y:i] [�l [l t l 2 Y
CD (JJ
yjl.
yJJ.
VJ
(])
CD
(jJ
- (Y
( 9 .74)
The equivalent-7T circuit correspon d i ng to these values of node admittances can
be found only if t is real because then Yi j
�' i ' Otherwise, the coefficient
matrix of Eq. (9.74) and the overall system Y bus ar e not symmetrical because of
the phase shifter. If the transformer is changing magnitude, not phase s h i fting,
the circuit is that of Fig. 9.8. This c ircu it cannot be realized if Y has a real
component, which would require a negative resistance in the cir cuit.
Some texts show admittance Y on the side of t h e t r a n s fo r m e r opposite to
the tap-changing side, and often the t r a n s [o r m a t i o n ratio i s c x p r e s s e d as 1 : a , as
shown in Fig. 9.9( a). Analysis sim i l a r to that developed abovc shows that the
bus admittance equat ions for Fig. 9 . 9( 0 ) take the form
=
0J ] [�] [
(]J
�)
which c a n be verified
j
CD
CD
-
CD
Y
Y/a*
OJ
- Y/ a
2
Y/ i a 1
] [V:]
�'
:...
[I, ]
( 9 .75 )
lj
from Eq. (9.74) by interchanging bus numbers i and j and
setting t = l /a . When a is real, the equivalent circu it is that shown in
Fig. 9.9( 6). Either Eq. (9.74) o r Eq. (9.75) can be used to incorporate the m odel
of the tap-changing trans former into the rows and columns marked i and j in
Yb u s of the overall system. Simpler equations follow if the represen ta tion o f
Eq. (9.74) i s used, but the importan t factor, however, i s that w e can n ow account
for magn itude, p hase shifting, and off-nominal-turns-ratio transformers in calcu­
lations to obtain Y bus and Z b u s '
I f a pa rticular transmission line i n a system is carrying too small o r too
large a reactive power, a regulating trans fo rmer can b e p rovided at one end of
9.6
--
I d eal transformer
l:a
-�
R E G U LATI NG T R A N S FO R M E R S
363
JJ.
- (})
y
VI
(a)
(1)
+
1
l
V,
a
- 1
)'/0
-- y
a
(j)
--
t
1 - a
-- y
[< I C U R E 9 . 9
P n - r h ase repres e n t a t ion
t he
t-
VJ
lal2
(6)
+
regu l a t i n g
(a)
of
t ra n s fo rm e r
show i n g :
per- u n i t a d m itt a n ce Y oppos i t e to t h e tapc il ,u gi n g s i d e ; ( b ) per-u n i t
equ Iva l e n t circu i t w h e n a i s
r ea l .
t h e l i n e to make t h e l ine tran smit a larger o r smaller reactive power, as
d e monstrated in Sec. 2.9. And any appreciable drop in voltage on the primary of
a tra nsformer due to a change of load may make it desirable to change the t a p
setting on transformers provided w i th adjustable taps in o r d e r to maintain
p roper volt a ge a t the loa d . W e can i nvestigate voltage-magnitude adj u stment at
a bus by means of the a u tomatic ta p-ch anging feature in the power-flow
prog r a m . For i n s t ance, in t h e fou r-bus system of Example 9.5 suppose we wish
to raise the load volt age at bus G) by inserting a magnitude-regu l a ti n g trans­
for mer between t h e load and t h e b us. With t real, in Eq. (9.74) we set i = 3 a n d
a s s i g n t h e n u mb e r 5 to b u s 0 from w h i c h t h e load i s n ow to be served . To
a c c o m m o d a t e t h e reg u l a tor in t h e p ow e r- fl ow e q u a t i o n s , Y b u s of the n e twork is
exp a n d ed by one r o w and o n e col u m n for bus G) , and the e l ements o f buses G)
a n d 0) i n t h e matrix of Eq. (9.74) a re then added to t h e p revious bus
ad m i t t a n c e m a t r ix t o g i v e
Y b u s( n e w )
=
CD
W
W
@
G)
W
Q)
@
W
Y2 1
Yn
U
0
Y3 1
0
Y3 3 + t 2 y
Y21
Y3 4
- tY
0
Y4 2
Y4 3
Y4 4
0
0
0
- tY
0
Y
CD
Y1
1
Yl 2
Y1
:l
()
0
( 9 .76)
364
CHAPTER 9
POWER-FLOW SOLUTIONS
The �j elements correspond to the parameters already in the network before
the regulato r is added. The vector of state variables depends on how bus G) is
treated within the power-flow model. There a re two alternatives: e ither
•
•
Tap t can be regarded as an independent parameter w ith a prespecified value
before the power-flow solution begins_ Bus G) is then tre ated as a load bus
with angle Os and vo l t age magn i t u d e I Vs l to be determined along with the
other five state variables represented in Eq. (9.68). I n this case the state­
variable vector is
or
Vol tage IllLlgnitude at b u s G) ca n b e prespeeilicd . T a p I tlH.:n rep l aces I V) ! as
the state variable to be determined along with 85 at the voltage-controlled bus
G) . In this case x = [ 82 , 03, 84, 85, I V2 1 , I V3 1 , t V and the jacobian changes
accordingly.
VI
=
! t
1.0LQ:.
--,.--
-
CD
MW flow
98.29 69.02
Mvar flow
97.17 70.94
102.83
104.74
---
---
64.62
6 l .59
-+-+--
! t
200
200
V3
135.56
0 .966/ - 1 .85°
=
V4
=
@)
1 .02/1 .52"
123.94
®
I 1 123.94
t � V5 0.976/ - 4 . 19°
=
To load
FIGURE 9.10
Flow of P a n d Q at bus G) of the system of Fig. 9.5 when a regu lating t ra nsformer is in t\irposed
betwee n the b u s and the load.
9.6
R EG U LATING TRANSFO R M E R S
365
In some studies the variable tap t is considered as an i n dependent control
variable. The reader is e n couraged to write t h e j acob i a n matrix and the
m ismatch equations for e ach of t h e above a l te r n at ives (see P robs. 9.9 a n d 9 . 1 0).
Because there i s a definite step between tap settings, discrete control
action occurs when regulators are used to boost voltage at a bus. The results of
r egu l a t in g the voltage of t h e load p reviously a t bus G) of t h e system of Fig. 9.5
a re s hown in the one-l i n e d i ag ram of Fig. 9 . 1 0 . A per- un i t reactance of 0 .02 was
assumed for the load-tap- c h a n g i n g (LTC) transformer. W h e n t h e voltage a t the
load is raised by se tting the LTC ratio t equal to 1 .0375 , t h e voltage at bus ®
is s l i g htly lowered compared t o Fig. 9.5, resu l t i n g in slightly higher vol tage d rops
across the l ines CD G) and @ - G) . The Q supplied t o t h ese l i n e s from
huscs CD a n d @ is i ncreased ow i n g to t h e reactive power required by t h e
r e g ul a tor, but t h e re a l power fl o w is r e l a t i v e l y u n a ffect ed. The i ncreased
me gavars on t h e l i n e s cause the losses to i n c r e a s e and the Q contributed by the
c h a rg i n g capac i t a n ces a t bus G) is decre ased .
To d e t e r m ine t h e effect of p h ase-shi ft i ng t ransfo r m e rs, we l e t t b e
complex wi t h Ll m agni t ude of u n i ty i n Eq. (9.74).
-
Exa m p l e 9.7. S o l v e Ex a m
the two
re s u l t s .
parallel
Hence, the currents in
unit.
1(0)
jX
1
------+-
11
----J-
+ �
Y
=
using t he Y b u s model of Eq. (9.74) for each of
and co m p a re the solu t io n w i t h the approximate
2. 1 3
t r a n s fo r m e r s
Solution . T h e a d m i t t n nc e
�
ple
each
is given by 1 /jO. 1 -jlO per
Fig. 9. 1 1 can be determ ined from
of
t r a ns for m e r
of
t r a n s fo r m e r
Ta
jO.1
Ta
lIb)
1
=
1'2
1 / 1 .05
jX
,. ,
)0. 1
I(b)
�
+
T"
VI
F I G U RE 9. 1 1
C i rc u i t for Exa m p l e 9.7. V a l u e s ,lT e i n per u n i t .
j O .6
366
CHAPTER 9
POWER-FLOW SOLUTIONS
the bus admittance equation
CD
[ ;j:: 1 � [
�
-
(1)
� -� ] [ �: 1
CD
[
W
�
and the currents in transformer T" with ( = LOS,
by Eq. (9.74), w hich takes the numerical form
CD
W
[
w
CD
-;:� 1 [�: 1
-jlO
jlO
as
shown i n Fig. 9. 1 1 ,
w
CD
j 1 0 J OO
-j 1 1 .025
-j 1 0 .000
j 1 0 .500
are given
1 [VV21 1
In F i g . 9. 1 1 current 11 = ( f� a ) + I}b» , and likewise, 12 ( Jia) + lib » , w h i c h
m eans t h a t t h e prece d i n g two m a t r i x e q u a t ions can be d i rect l y added ( l i k e
admittances i n pani.llel) to obtain
=
[ II ]
12
G)
@
=
[
CD
- j21 .025
j20 .500
W
j20 .500
-j20 .000
][ ]
LQ:.
VI
V2
I n Example 2. 1 3 V2 is the reference vo l t a g e 1 .0
and the current 12 is
ca l c u l a te d to be - 0.8 + jO.6. Therefore, [rom the second row o f t h e p r e ce d i ng
e q u a t i o n we have
12 = - 0 .8 + jO.6 = j20 .5 V1 - j20( 1 .0)
w h i c h g ives t h e
per-u n i t
VI =
vo l t age a t b u s
- 0 .8 + j20 .6
---j20.5
1 .0049 + jO .0390 per unit
--
Since
VI and V2 are now both
fo r t r a n s fo r m e r Ta to obt a i n
Ia
CD
known, we can return to the admittance equation
i ) = j l O V1 - j l 0 V2 = j1 0 ( 1 .0049
=
jO .0390 - 1 .0)
= - 0 .390 + jO.049 per unit
and from the admittance matrix for transformer
l� b )
+
j 1 0 . 5 VI - j10V2
= - 0 .41
+
=
T"
j 1 0 . 5 ( 1 .0049 + jO .0390) - jlO
jO.55 L pe r
unit
9.6
367
R EG U LATIN G T R A N SFO R MERS
H ence, the complex power outputs of the transformers are
= - V2 Iib)*
5Tb
=
0.41
+
j0.55 1 per u n it
Th e results fou nd by the approximate circulating c u rre n t method of Example
compare favorably with the above exact solution.
2. 1 3
Solve the phase-sh ifter problem of Example 2 . 1 4 by the exact Y bus
model of Eq. (9.74 ) and compare resu lts.
E x a m p l e 9.8.
Solution .
The
I = E j rr / 60
=
l .O
w hi c h c a n be
i n Exa m p l e
ft
(9.74)
bus a d m i ttance equation
i s g i v e n by E q .
for
t h e p h ase - s h i ft i n g t r a n s for m e r
obtain
[II ] [ 12
=
w ith
as
a d d e d d i re c t l y t o t h e a d m i t t a nce e q u a t i o n fo r
9.7 to
Tn
tra nsformer
0 .5234 -: j 1 9 .9863
-}20.0
-j20 .0
0 .5234 + j 1 9 .9863
To
given
][ ]
VI
V2
Fol lowing the proced u re of Example 9.7, we have
- 0 . 8 + jO.6
w h ic h
( - 0 .5234
=
yields the voltage a t bus
VI
=
- 0.8
- 0 . 5 234
CD ,
=
/ �h) =
=
j l O( VI
I :.
j 1 9 .9863
+
- O J)7
V2 )
-
. - I i")
=
+
a !1d t h e complex powel' o u t p u t s
S Th
=
=
- O tI
j O .2 9
=
1 . 03 1
+
- j20 ( 1 .0)
jO.013 per unit
- 0 . 1 3 + j O . 3 1 per
.
+ jO.6
-
( - 0.13
unit
+
jO .3 1 )
rer u n i t
,Irc
- V2 /i h )*
A g a i n , t h e a p p r o xi m a t i o n v a l u e s o f
t h i s examp l e .
j 1 9 .9863) VI
j20 . 6
+
W e t h e n d e t e r m i n e the c u r re n t s
l�a)
+
=
0 . 67
+
j O 2 9 per u nit
.
E x a m p l e 2. 1 4 arc
close t o t h e
ac tu a l values o f
368
CHAPTER 9
POWER-FLOW SOLUTIONS
THE DECOUPLED POWER-FLOW
METHOD
9.7
the s trictest u s e of the Newton-Raphson procedu re, the jacobian is calcu­
lated a n d triangularized in each itera tion in order to update the LU factors. In
practice, however, the jacobian is often recalculated only every few iterations
and this speeds up t h e ove ra l l sol u l i o n r:>rocess. T h e l i n a l so l u l i o n i s J C l e r m i n e cJ ,
of course, b y the allowable power mismatches a n d voltage tolerances a t the
In
buses.
When solving
egy for i m p rov i n g
require m e n t s
ing t h e
is
l a rge-s c ; d e D o
•
•
co m p u t <t t i o n;t \
the
a p p rox i m a t e v e r s i o n
we r
t ra n s m i s s i o n sys t e m s , an a l t e rn a t ive s t r a t ­
l
e l ic ie nc
y ancJ red ucing
dccoup/ct/ 1)()w{'/"�f!()1I'
or the
d e c o ll p l e d <I p p m <l c l l i s
w h i c h m a k e s u s e o f , I ll
rHocc <.i ur e . The r r i n c i p \ c u n cl e rly­
o\)sc rv ; t l i O l l s :
m et/l O d ,
N c w t o l1 - R ,l phsO I l
h;lsed U il t wu
c o mp u t e r s t o r a g e
Change in the voltage angle 0 at a b u s primarily affects the flow of real power
P in the transmission l ines and leaves the flow of reactive power Q rel a tively
unchanged.
Change in the voltage magnitude I V I at a bus primarily affects the flow of
reactive power Q in the transmission lines and leaves the flow of real power P
relatively u n c hanged.
We h ave noted both of these effects in Sec. 9.6 when studying the phase shifter
and voltage-magnitude regulator. The fi rst observation states essentially that
aPjaD j is m uch larger than a Qja Dj , which we now consider to be approxi­
mately zero. The second observation states that a Q ja l V; I is much larger than
8Pja l V; I , which is also considered to be approximately zero.
Incorporation of these approximations into the jacobian of Eq. (9.45)
makes the elements of the s ub matrices J 1 2 and J 2 1 zero. W e are then l eft w i th
two separated systems of equations,
a P2
a02
a P2
Ao4
6. 0 2
a P4
a84
6. 8 4
( 9 . 77 )
J11
aP4
a 02
aQ2
I V4 1
a l V4 1
a Q2
1 V2 1
a l V2 1
and
aQ4
V
I 21 al
v2 1
J 22
a Q4
I V4 ! \
a V4 1
6. V2
--
I V2 1
6. V4
I V4 1
--
( 9 .7 8 )
These e q u ations a re
decoupled
9.7
369
THE DECOU PLED POWER-FLOW M ETHO D
i n the sense that the vol tage-a ngle corrections
/1 P ,
/1 8 a r e c a l c u l a ted u s i n g o n l y r e a l powe r mismatches
w h i l e the voltage-m ag­
n i t u d e c o r r e c t i o n s are c a l c u l a t e d u s i n g o n ly 6. Q m i sm a tc h e s . However, th e
whereas the
co efficien t m a t ri c es J I l a n d J22 a re sti ll i nte r d e p e n d en t because the e le m e n t s of
J 1 1 d e p e n d o n t h e vo l tage ma g n i t u des b e i n g solved i n E q .
(9.78),
equ ati ons c o u l d be solved a l t e r n a t e ly, u s i n g in one set the most re c en t
e l e m e n t s of J 22 depend on the a ngles of Eq.
(9 .77). Of
course, the two sets of
solutions
of t h e two c oe ffi c ie n t m a t r i c es at e a c h i te r a t ion . To avoi d such com pu t a t i o n s , we
fro m the o t h e r set. B u t this s c h e m e w ou l d s t i l l requ i re eva l u a t i o n a n d f actoring
s i o n - l i n e pow e r fl ow , as we now exp l a i n .
i n t r o d u c e fu r t h e r s i m p l i fic a t i o n s , w h i c h a re j u s t i fi e d by t h e p hysics of t r a n s m is­
I n a w e l l - d e s i g n e d a n d p ro p e rly o p e r a t e d power t ra n sm i s s i o n syst e m :
•
1 y so s m a I I t h a t
T h e a n gu l a r d i ffe re n c es ( 8 , - 8) b e tw e e n typ i c a l bu ses o f t h e system
1I S Ll a 1
are
( 9 . 79 )
•
Th e l i n e susc e p t a n c e s
C;i j so t h a t
Bli
a r e m a ny t i m e s l a rger t h a n t h e l i n e co n d uc t ances
G I ) s i n ( 8 I.
•
Qi
8j )
-
BI· ).
«
( 9 .80 )
cos ( 8 I - 8). )
(J)
o p e ra t i o n i s m uc h l es s t h a n the r e a c tive powe r w h i c h would fl ow if a l l l i nes
The r e a c ti ve pow e r
i nj e c t e d i n t o a n y bus
of t h e system d u r i n g n o rmal
fro m t h a t bus we re s h o r t - c i rc u i t e d to r e fe r e nce . Th a t i s,
( 9 .8 1 )
T h e s e a p p roxi m a t i o n s c a n b e u s e d t o s i m p l i fy t h e c l e m e n ts o f t h e j a cob i a n . I n
Eq.
(9. 62)
t h e off- d i a go n a l c l e m e n t s o f J I
ii P,
lVI
G O)
Using t he i d e n t i ty
apI
3 81·
wherc
B, ;
=
sinea
3Q .
j
+
rl Q
I
J i V) I
I
a n d J 22 a re g ive n by
( 9 .82)
-
(3)
=
sin
I
- l Vj I
d l Vj I
a
cos
{3
+ cos
a
sin
(3
in Eq.
(9.82)
g ives u s
-
I Y;) s i n
e'i
a n d Gij
=
I �) cos eii'
T h e a p p roxi m a t i o n s l i s t e d ab ov e
370
CHAPTER 9
POWER-FLOW SOLUTIONS
then yield the off-diagonal elements
apI.
JQi
- J VJ ao}.
J a I Vj I
�
- I VI V) I EIJ. .
( 9 . 84)
The diagonal elements of J I I and J 22 have the2 expressions shown in Eqs. (9 . 54)
and ( 9 .63). Applying the inequality Qi « I V;· 1 Bii to t hose expressions yields
iJ Q ;
I V.I I --,
a D I.
By substituting the approximations of
m atrices J I I and J 22 , we obtain
==
( 9 .8 5 )
- I VI I BI. I.
Eqs. (9.84)
2
and (9.85 ) in
the
coe fficient
( 9 .86)
and
- I V2 V2 1 Bn
- I V2 V:\ I B21
- I V2 V4 1 B 24
- I V2 V3 1 B 3 2
- 1 V3 V3 1 B 33
- I V3 V4 1 B 34
- 1 V2 V4 1 B 4 2
- I V3 V4 1 B4 3
- 1 V4 V4 1 B44
� I V2 1
1 V2 1
� I V3 1
�Q2
1 V3 1
� Q3
I V4 1
L1 Q4
� I V4 1
( 9 . 87)
To show how the voltages arc removed from the entries in the coefficient matrix
of Eq. ( 9 .87), let us multiply the first row by the correction vector and then
d ivide the resultant equation by I V2 1 to obtain
( 9 . 88 )
The coefficients in this equation are constants e qual to the negative of the
susceptances in the row of Ybu S corresponding to bus @ ' Each row of Eq. (9. 87)
can be similarly treated by representing the reactive mismatch at bus CD by the
quantity � Qj I V: I . All the entries in the coefficient matrix of Eq. (9.87) become
constants given by the known susceptances of Y lJUS' We can also modify
Eq. (9.86) by m ultiplying the first row by the vector of angle corrections and
9.7
THE D ECOUPLED POWER-FLOW M ET H O D
371
rearranging the res u lt to obtain
( 9 . 89 )
The coefficients i n t h i s equation can be m a d e the same a s those i n Eq. (9. 88) by
setting I V2 1 , I V3 1 , and I � I e q u a l to 1 .0 p e r u nit in the left-hand-s i de expression.
Note that the q u a ntity I1 P 2/ I V2 1 represents the real-pow e r mismatch in
Eq. (9.89). Treating all t h e rows of Eq . (9.86) in a simi l ar manner leads to two
d ecoupled systems of e q u ations for the fou r-bus network
- B .."
- 8 24
t1 8 ..,
- /J ., -)
- B"
- 13
14
t1 [) :.
- ·- B
t1 8 -1
t1 P4
Q2
--
'
- B.
I
J
�
p
_.'
. "> .'
-
B � ."
Ii
.
and
t1 P2
-I V2 1
- B"..,
.
44
- Bn
- B 23
- B2
4
6 1 V2 !
- B 32
- B 33
- B :' 4
11 1 V3 !
- B4 2
- B4
-B
11 1 V4 1
Ii
3
44
=
t1
PJ
-I VJ I
( 9 . 90 )
I V4 1
I1
I V2 1
Q3
-6
=
I V3 1
( 9 .9 1 )
Q4
-I1
I V4 1
Matrix B is general ly symmetri c a l a n d s parse w i t h nonzerO elements, which a re
co nstan t , real numbers exact ly e q u a l to t h e n ega tiue of the suscept a nces o f Y bus '
Consequen t ly, m a t r ix B i s easily formed a n d i ts t riangular factors, once com- '
p u ted at the beginning of t he sol u tion , d o not h ave to be recomputed, which
l eads t o very fast iterations. A t voltage-co n t rolled buses Q is not specified a n d
t1 1 V I is zero; t h e rows a n d c o l u m n s correspond i n g t o slI ch buses a rc omitted
from Eq . ( 9 . 9 1 ) .
One typi c a l sol u t io n strategy is to:
2.
1.
3.
4.
5.
Calcu l ate the i n itial m i s matches t1 P / I VI ,
Solve Eq. (9.90) for 11 0 ,
Update the a ng les 0 a n d use t hem to calcul ate m ismatches I1 Q / I V I ,
Solve Eq. (9. 91 ) for 6. 1 VI a n d u pdate the m a gn itu des I VI , a n d
Return to Eq_ (9 . 90) t o rep e at t h e it erat ion u ntil a l l mismatches are w ithin
specified tolerances.
372
CHAPTER 9
POWER-FLOW SOLUTIONS
Using this decoupled version of the Newton- Raphson proced ure, faster power­
flow solutions may be found within a specified d egree of precision.
Example
Using the deeoupled for m of the N ewton-Raphson method,
d e t e r m in e the first-iteration s olu t i o n to the power-flow problem of Example 9.5.
9.9.
Solution. The B
l
Tab l e
ma t r i x can b e read directly from
9.4 and t h e m is m a tches
co r re s po n d ing to t h e initial voltage estimates are a l ready calcula ted in Exa m ple 9.5
so that Eq. (9.90) becomes
44 .835953
()
()
4() . H(,3H38
- 25 . 8471)09
- 1 5 . 1 1 H5 2H
Solv ing this equation gives t h e angle corrections in radians
6 82
683
- 0 .0205 7 ;
=
Add i ng these results to the
=
-
6 84
0 .0378 1 ;
fl at st ar t estimates of Table
=
0 .02609
updated
values of 82, 83, and 84, which we then use along w ith the e lements of Y hus to
calculate t he reac tiv e m is ma t c h e s
1
=
=
Ll Q 3
1 V3 1
9.3 givc:: s
.0 1
-11-
-
-
- 1 .0535
+
1
2
.0 ( - 44 .835953)
=
=
OJ)
1 9 .455965
1T/ 1 80 + 0 + 0 .02057 ) + 2 6 . 35 9695
1 .02 s i n ( 1 0 1 . 3 0993 x 1T/ 1 80 + 0 . 02609 + 0 .0205 7)
si n ( 1 0 1 .30993
x
+
X
- 0 . 80370 p e r u n it
1
=
the
2
Q 2 , SCh - [ - I V2 1 B22 - I YI2VIV2 1 sin(812 + o J
.
+ 84
- I Y24V2V4 I sm(824
02)]
I V2 1
l�
_
=
{
{
-
(3)}
3
V3 1 { Q , sCh - Q 3. calc }
_1_ {Q3. 5Ch
1
1 V3 1
1
--
1 1 .0 1
{
-
2
[ - I V3 1 S33
-
- 1 .2394
+
I YI 3 V\ V3 1 si n ( 0 I 3 + 8 \ . 8 + 4_
34 8 03)]
1 Y34V3V4 1 SIn(
1 .0 2 ( - 4 0 . 863838)
sin ( 1 0 1 .30993
X
-
x
7T / 1 80
1 .02 sin ( 10 1 .30993
- 1 .27684 per u ni t
X
+
0
+
+
7T / 1 80
26 .359695
0 . 0378 1 )
+
+
0 .0260 9
1 5 . 417934
+
}
}
0.0378 1 ) ,
1).7
T H E D ECOU PLED POWER-FLOW M ETHOD
A r eac t i v e m is m a t c h c a l cu l a t i o n is not required for bus
c o n t ro l l e d _ Accord i n g ly, in t his example Eq. ( 9 . 9 1 ) b e comes
[44.8305953
which y i elds the so l u t i o n s
t
0
40 .863838
t. I V2 1
vo l t age mag n i u d e s a t b u s e s
W
=
] [ V21 ] [
11 1
11 1 V) l .
- 0 . 0 1 793
and
G)
a re
=
@,
- 0 .80370
- 1 .27684
and 6. I V3 1
1 V2 1 = 0 . 98207
=
373
w h i c h is vol t age
]
and I
- 0 . 0 3 1 25 . T h e n ew
V3 1
=
0 . 9 6875 ,
m i s m a tches for t h e s e c o n d i te r a t i o n o f
E q . (9.90) a re c a l c u l a t e d u s i n g t h e n ew vol t a g e v a l u e s . Repe a t i ng t h e p r oce d u r e
over a n u mb e r o f i t e r a t i o n s y i e l d s
s a m e so l u t i o n as t a b u l a t e d i n Fig. 9 . 4 .
w h i c h c o m p l etes the fi rs t i t e r a t i o n . U p d a t e d
the
O ft e n i n industry-based programs c e r t a i n m o d i n c a t i o n s a re m a d e i n Eqs.
( 9 . 9 0 ) a n d ( tJ . 9 J ) . The m o d i fi c a t i o n s to TI i n Eq. ( 9 . 9 1 ) a re g e n e r c.l l 1 y as fol lows:
•
LQ:..
O m i t the angl e-sh i ft i n g e ffects of phase s h i ft e rs from 13 by setting t = 1 . 0
W h e n rows and col u m n s for voltage-contro l l e d b uses are a lso o m it t e d as
p reviously i n d icated , the res u l t i n g ma trix is cal led B" .
The coefficient matrix in Eq . (9 .90) is gen e r a l ly modified as fol l ows:
•
O mit from B t hose elements that m a i n l y affect megavar flows such as shunt
capaci tors and reactors, and set t a ps t of off-nominal transformers e qu a l to 1 .
Also, ignore series resistances i n t h e eq uivalent-Ii circuits o f the t ra ns mission
l in es in forming Y b u s from which B in Eq. (9.90) is obt a i n e d . The resul t i n g
m at r ix i s ca l led B' .
n e two r k .
When B i n Eq. (9.90) is repl aced by B', t h e model becomes a l ossl ess
I f, in a d d i t i o n , a l l bus vo l t ages a rc a s s u m ed to re m a i n const a n t at
n o m i n a l values of 1 . 0 per u n i t , t h e so-cc.l l 1 e d de power-flow model res u lts. U nder
t h e s e a d d i t i o n a l assum pt ions, Eq . (9.9 1 ) is n o t necessary (since � I � I = 0 at
e a c h bus CD ) and Eq . (9 .9CJ) for the d c p ow e r flow hecomes
( 9 . 92)
B'
where it is und erstood t h a t the elements of B' are calculated ass u m i n g all l i n es
a re lossless. Dc power-ftow a n a l ysis can b e used where approximate solutions
are acceptable, as i n the con t i n g e n cy studies d iscussed in Chap. 1 4.
374
CHAPTER 9
9.8
SUMMARY
POWER-FLOW SOLUTIONS
This chapter explains the power-flow problem, which is the determination of the
voltage magnitude and angle at each bus of the power network under specified
conditions of operation. The Gauss-Seidel and Newton-Raphson iterative pro­
cedures for solving the power-flow problem are described and numerically
exemplified.
In addition to d iscussing how power-flow studies are undertaken, some
m etho d s of real and react ive power-flow con t rol are prese n t e d . The resu l t s of
paralleling two transformers when the voltage-m agn i tude ratios arc d i fferent or
when one transformer provides a phase s h i ft are examined. Equations are
d eveloped for the nodal adm ittances of these transformers and equivalent
circuits which allow a n a lysis of reactive-power control are i n t roduced.
The computer-based power-flow p rogram can b e used t o study the appli­
cation of capacitors a t a load bus simply by in corpora t i ng the sh u n t admittance
of the capacitor i nto the system Y bus ' Vol tage control at a generator bus can also
be i nvestigated by specifyi ng the values of the P V bus vol tages.
Fast, approximate methods for solvi n g the power-flow problem are intro­
duced by means of the dc pow�r-flow model, which depends on the linkages
b etwee n real power P and voltage angle 8 , and reactive power Q and vol tage
m agnitude.
Table 9.5 summarizes the equations for each of the methods of power-floW
a n alysis.
Summary
TABLE 9.5
of power-flow equations a n d solution m ethods
Power-flow equ ations
Pi
L,
N
I Y" , vi V" l co s(Oill
Pi
Qi
0" - 0, )
+
=
L
N
I Y,1l V; V" I s i n (8in
+
On
0)
!1 Q i = Qi. sch - Q i . calc
!1 p.I =
• calc
�------r- ------ ------ --- ' ------ ---'
Gauss-Seidel m ethod
=
11 = 1
pI. , .seh
To obt a i n V a t
VI < k )
a
-
bus i w i t h known P a n d Q :
=
I
_
Y/ I
[ P.
V< k
I . se
h
-
jQ
l , seh
-
I
To obt a i n Q at a regul ated bus
I)
_
i- I
\'
L.. Y')V<
) k)
j� 1
=
_
1
N
\'
L..
j=i+ !
YI)V(
k- l)
)
1
i:
U s e 01 acceler atio n factor a a t b u s i in i t er a tio n k .
VI,( ace
k)
11 0 ,
( 1 - a ) V(k
- l)
I. ace
+
a V(k)
I
=
V<k - I ) +
I, ace
a (VY
)
I
-
VCk - I »)
I . ace
-
TABLE 9.5
9.8
( Continued)
il P2
Newton-Raphson met hod
a P2
a0 2
a PN
(J O N
.
.
.
.
a Q2
.
.
.
.
.
.
.
.
.
.
(J Q "2
aoz
(IO N
,) ° 2
.
..
(J Q N
aDN
- I I ; 1 -; Y" I \ i r l ((J'J
,1 1 '
i
'I ,')j
.. i v
I
(lP'
IVI -
vJ y. /
I V2 1
a 1 V1 1
J12
a PN
aON
J 21
r
I Vl l
a W2 1
a PN
a0 2
.
a P2
aON
J1j
! cos(
. () lJ, .
.
.
.
.
I V" I
il l V:1
.
(J Q "
.
.
.
.
.
.
.
.
.
=
I V I -iJ
a Q,
)
l V; 1
=
I
O e co u p l c d cJ pow er-flow
sol u t i o n t e c h n i q u e
-8
+
,�
(/ 1 ',
. 6, )
.I
f)J. . /) )
•
- B :.1
I
- Ii "
-, /
1 2"
- B .'
l
- BN
- IJ N
J
Bii
2
.
.
.
.
'
. .
.
.
_
[,
II
Il
-�
I
:F I
rJo
r) r'i ,
a p'
' --
;, , V. I
,
�Q
i
I V, : ("
. V, I
l
IJ :' N
- B NN
- iJ 2 N
- /J.' N
- B NN
6Q
.
.
1/
iJ Q i
')
+ - + 2 I V I - CI I·
QO •.
tJ. P2/
6.1 ve l
6. ° N
tJ. IV, 1
6. I VN I
iJ Pi
aa),
1 V2 1
6. PN/ I VN I
tJ. Q2 / 1 V1 1
6. 0.1 / I V, I
L tJ. QN/ I VN I
a r e t h e i ma g i n a ry p a r t s o f t h e correspo n d i n g Yblls e l e m e n t s .
V o l t a ge-co n t ro l l ed b u ses arc n o t re prese n tt:d i n
6. 1 V I
'
- - - 2 1 VI I B·-I I
tJ. PJ/ I V;\ I
- n2
N
-
.
il Pr
I
- BN
- 1J e."2
.
8 (k) + 6 8 ( k )
Z
- BN
.
r7 Q ,
a Pi
- Bl l
2
.
il il ,
O () }
+
.
(i( J N
: v\ '
a WN i
- !l I V" 1
I
.
W,
-B
D
n
.
i1 Q ",
S t a t e v a r i a h l e u pd a t e f orm u la s :
a(k + I )
.
I v, I --- - -
a VI I
j
375
SUMMARY
e q u ations.
2
376
CHAPTER 9
POWER-FLOW SOLUTIONS
PROBLEMS
9.1 .
9.2.
9.3.
In Example 9-3 suppose that the generator's maximum reactive power generat ion
a t bus @ is limited to 1 25 Mvar. Recompute the first-iteration value of the voltage
at bus @ using the Gauss-Seid el method.
Fo r the system of Fig. 9.2, complete the second iteration of the Gauss-Se idel
p rocedure using the first-iteration bus voltages obtained i n Examples 9.2 and 9 . 3 .
Assume an acceleration factor of 1 .6.
A synchro nous condenser, whose n ..: a c t ivc po e
assumed
be
u n l i m i t d i s i n s t a l l e d at l o a d b u s 0) or t h c s s t c rn or Exa m p l e tJ . 2 t o hold
b u s vol t a g e m g n u d e a t 0 . 99 p n u n i t . U s i n g
(J ; ! u ss-Se i d c l m e t h o d , f i n d t h e
v l t e s buses CD a n d CD for t h e !i rst i t e ra t i o n .
Take Fig. 9.12 t e equ ivalent-7T representation of the t a s m is s i
bus 0 and bus @ of the system of F i . 9 . 2 .
v n
in Fig. 9 . 4 , d e t e rm i n e a n d i u i c
on f7ig. 9 . 1 2 t e v a l u e s o f ( a ) J> and Q l e a v i n g
buses 0 a nd @ on line 0- @, ( b ) charging megavars o f t h e e qu i va l e t
l i n e 0 - @, a n d ( c ) P a d Q at both ends o f t h e series part o f t h e e q u v a e t
of line G) - @ .
w
r
capabi
l
i
t
y
i
s
t
o
c
y
t
i
l
e
a
i
t
t
h
c
o ag at as h
r
n
on
l
i
n
e
bet
w
een
g
Us
i
n
g
t
h
e
powe
r
f
l
o
w
s
o
l
u
t
i
o
n
gi
e
nn atc h
ni l n of
,
-
9.4.
"
IT
®
+-----�--�---+
0
Diagram for Prob. 9.4.
FIG U RE 9 . 12
9.5.
From the l i n e
F g . 9 .4,
determ i n e / 2 R l ss each of the f u r t ra n s m issi n l i n e s and ver i fy t h a t t e
the
4.8 1 M W .
o f th ese l i n e losses i s e q u l
Suppose t hat a s h u t c p c i tor b a n k
1 8 Mv r is
bus G)
in
and the reference node
t he system of Example 9.5.
Ybus
Table 9.4 to account for this
and estimate th actual megavar reactive
power i nj ected into the system from the capacitor.
For the system of Example 9.5 a ugmented
a synchronous condenser as
d escribed in Prob. 9.3, find the j acobia n calculated at the i n i al e s t i m a t e s . Hin t :
i n Sec. 9.4 fol lowing
Example 9.� t h a to start calculations from the
Suppose t hat i n Fig. 9.7 the t a p i s on t h e side of node CD so that t h e transforma­
tion ratio is t : 1. Fin d elements of Y bus similar to those in Eq . (9.74), and d raw t h e
equivalent-1T representation similar t o Fig. 9.8.
In the four-bus system of Example 9.5 suppose that a magnitude-regulating
transformer with 0 . 2 per-unit reactance is i nserted b e tween the load and the bus a t
bus G), a s shown in Fig. 9 . 1 0 . The variable tap i s o n the load side of t h e
transformer. If t h e voltage magnitude at t h e new load bus G) is prespecified, and
oflowininfoarmtato ion oftootatlhesyspowetem lro-sflooofw solution given in hi sum
n a ian capacitorra,ted a coeModinnectfeydthbeetwe n given
wi
t
h
It
would be simplenr to modify the jacobian matberigxinsnihownng. it
,
9.6.
9.7.
9.8.
9.9.
PROBLEMS
9.10.
therefore is not a state variable, t he variable tap t of the transformer 'should be
regarded as a state variable. The Newton-Raphson method is to be applied to the
so l u t i o n of the power-flow e q u atio n s .
( a ) Write mismatch equations for t h is problem in a symbolic form similar to
Eq. (9.45).
(b) Write e q u a ti o n s o f the j acob i a n elements of the co lumn corresponding to
variable t (that is, part i a l derivatives w ith respect to t ), and evaluate them
using the i n it i a l voltage estimates shown in Table 9.3 and assuming that the
voltage magnitude at bus G) is specified to be 0.97. The i nitial estimate of 0 5
is O.
C c ) Write equat ions of P and Q m ismatches at bus G) and evaluate the m for the
first i teration. Assu me the i n i t ial estimate of variable t is 1 .0.
I f the tap setting of the transformer of Prob. 9.9 is prespecified instead of the
voltage magnitude at bus G) , then V5 should be regarded as a state variable.
Suppose that t he tap se t t i n g ( i s spec i fied to be 1 .05.
( II ) In t h i s ease w r i t e m i s m a tch e q u a t ions i n a s y m ho l i \.' for m s i m i lar to E q . (9.45).
( b ) Write eC]uations of t h e j acob i a n c lements which Zlre partial derivatives with
respect to I Vs I , and eva luate t h e m using t he i n i tial estimates. The i n i tial
est imate of V5 i s 1 . 0
( c ) Write equat ions fo r t h e P a n d Q m i s m a t c h e s at bus G) , and evaluate them for
the fi rst i teration.
Redo Example 9.8 for ( = 1 .0
, and compare the two results as to cha nges in
real and reactive power flows.
The generator at bus @ of the system of Example 9.5 is to be represented by a
generator connected to bus @ t h rough a generator step-up transformer, as shown
in Fig. 9. 1 3 . The reacta nce of this transformer is 0 . 02 per unit; the tap is on the
h igh-voltage side of the tra nsformer with the off-nominal turns ratio of 1 .05.
Evaluate the jacobian elements of the rows corresponding to buses @ and G) .
iQ:.
9. 1 1 .
9.12.
377
�
---rr --- I
..----- @
®
F I GURE 9 . 1 3
G e n er a t o r s t e p - tl iJ t r a n s fo r m e r fo r Prot.
9.13.
9. 1 4.
9.12.
For the system of Prob. 9. 12, find matrices B ' and B " for use in the decoupled
power-flow method.
A five-bus power system is shown i n Fig. 9.14. The line, bus, transformer, and
capacitor data are g iven in Tables 9.6, 9.7, 9.8, and 9.9, respectively. Use the
Gauss-Seidel method to find the bus voltages for the first iteration.
378
CHAPTER 9
POWER-FLOW SOLUTIONS
for Probs. 9 . 4
FI G U R E
System
9. 1 8 .
9_ 1 4
Th e lin e
1
t h ro u g h
giv e n i n Tables 9 . 6 t h ro u g h
Line
data for the system of Figure 9 . 1 4
TABLE 9.6
Per-unit
Series Y
Per· u n i t
Line
bus to bus
CD
CD
@
G)
@
R
@
@
G)
G)
G)
Series Z
Charging
Mvar
X
G
0.0108
0.0649
2.5
- 15
6.6
0. 0235
0.09 4 1
2.5
- 10
4.0
0.0 1 1 8
0.0471
5.0
- 20
7.0
B
0.0 1 47
0.0588
4.0
- 16
8.0
0.01 1 8
0.0529
4.0
- 18
6.0
Bus data for the system of Figure 9 . 1 4
TABLE 9.7
Q (Mvar)
Generation
Bus
CD
@
®
G)
CD
P (MW)
P
(MW)
Q (Mvar)
Lo a d
V
p.u.
L.Q:
1 .0LQ:
1.0La"
l .olit
1 . 0 LQ:
1 .0 1
1 90
60
35
70
42
80
50
65
36
Transformer data for the system of Figure 9.14
TABLE 9.8
Transformer b u s t o bus
Per-unit reactance
Tap setting
0.04
0.975
9.9.
a n d b u s d a t a a re
Remark
Slack bus
PV bus
PROBLEMS
Ca pacitor data for the
system of Figure 9.14
379
TABLE 9.9
Rating i n
Mvar
Bus
G)
o
18
15
p o w e r fl o w s o l u t i o n o f t h e sys t e m of
F i g . 9 . 1 4 , d e t e r m i n e ( (/ ) Yb"s of t h e sys t e m , ( lJ ) t h e mismatch equa tion at bus @
e v a l u a t e d a t t h e i n i t i a l v o l t a g e e s t i m ,l t e s of T a b l e 9 . 7 fo r t h e fi r s t i t e ra t i o n , a n d ( c )
w r i t e m i s m a t c h eq u a t i o n s in a for m s i m i l a r t o Eq . ( 9 . 4 5 ) .
r m t h e sys t e m o r r i g . 9 . 1 4 . f i n d m ,l l r i c e s ll' a n d n " for l i s e i n t h e decou pled
ro w e r I l o w m e t h od . A l s o , d e t e r m i n e t h e f i rs t - i t e r ,l t i o n I) a n d Q m i s m a t ch equa­
t i o n s ,I t b u s @ , a n d f i n d t h e vo l t age 1 l 1 ;l g n i t u li e a t b u s @ a t t h e e n d o f t h e fi rst
l
9 . 1 5 . To a p p y t h e [\; c w t o n - R a p h s o n m e t ho d t o t h e
9. 1 6.
-
iteration .
9.17.
-
Suppose t h a t
s h i ft e r w h e re
in
Fig.
9 . 1 4 t he t ra n s fo r m e r
G) is a phase
Find Y ou s o f th is
and
t i s now a complex variable and is 1 .0
. (a)
s y s t e m . ( b ) W h e n c o m p a r e d w i t h the power- flow sol u t ion of Prob. 9 . 1 5 , w i l l t h e
real power flow i n t h e l i n e fro m b u s
to
i n c r e a se or d e cr e a s e ? What
a b o u t t h e r e a c t ive power flow? E x p a i n why q u alitatively.
l
W
bus G)
9. 1 8. To a pp l y t h e d e c o u p l e d pow e r - flow m e t h o d
m at r i c e s B' and B" .
to t he
sys t e m
of Prob. 9 . l 7 ,
find
an 1 8-Mvar shunt capacitor b a n k IS added to b u s G) .
I n a pp l y i ng the N ew t o n - R a p h s o n m e t h o d i f th e amount of reactive power r e q u ir e d
t o m a i n t a i n t h e sp e c i fi e d voltage at a PV bus ex c e e d s the m ax i m u m l i m i t o f its
react ive power generat ion capability, the r e ac t i v e pow e r a t that b u s i s s e t t o t h a t
l i m i t and t h e t yp e of bus b e c o m e s a load bus. Su ppo s e t h e maxi m u m r e a c t ive
power g e n e r a t io n at b u s 0 is l i m i t e d to 150 Mvar in t h e system of Ex a m p l e 9.5.
Usinf! the first-iteration r e s u l t g i v e n i n S e c . 9.4 fol low i n g Ex a mp l e 9.5, d e t er m in e
w h e t h e r or n o t t h e t y p e o f b u s (j) s h o u l d be conve r t e d to Cl l o ad bus at t he s t a rt of
t h e s e co nd i t e r a t i o n . If so, c a l c u l a t e t he react ive power mismatch at b us 0 t h a t
9 . 1 9 . Redo Exam p l e
9.20.
H
between buses (2)
9. ] 0
when
,
s h o u l d be u sed i n t h e s e c o n d - i t e r , l t i o n m i s m a t c h e q u a t i o n .
iCHAPTER
10
SYMMETRI CAL
FAULTS
A fault in a circuit i s any failure which i n terferes with the normal flow of
current. Most faults on transmission lines of 1 15 k V and higher are caused by
lightning, which results in the fl ashover of insulators. The high voltage betw e e n
a conductor and the grounded supporting tower causes ionization, which pro­
vides a path to ground for the charge induced by the l ightning stroke. Once the
ionized path to ground i s 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 o f a transformer or generator, thus completing
the circuit. Line-to-line faults not involvi n g ground are l ess common. Open i n g
Circuit b reakers to isolate the faulted portion of the l i ne from t h e rest of the
system i nterrupts the flow of current i n the ionized path and allows deionization
to take place. After a n i nterval of about 20 cycles to allow deionization,
breakers can usually be reclosed without reestablishing the arc. Experience i n
the operation o f transmission lines has shown that ul tra-high-speed reclosing
breakers successfully rec10se after most faults. Of those cases where reclosure is
not successful, many are caused by permanent fa ults where rec\ o s u re wou l d b e
impossible regardless of the interval between opening and reclosi ng. Permanent
faults are caused by lines being on the ground, by insulator strings breaking
because of ice loads, b y permanent d a m age to towers, and by surge-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. Roughly 5 % of all faults i nvolve all three p hases.
These are the so-called symmetrical three-phase faults which are considered i n
this chapter. Other types of transmission-line faults are line-to-line faults, which
do not i nvolve ground, and double line-la-ground faults. All the above fau l ts
381
T R AN S I ENTS IN RL S E R I ES C I RCU ITS
10.1
exc e p t t h e t h re e - ph ase type c a u s e a n i m b a l a n ce b e tween t h e p h a s e s , a n d
they a r e c a l l e d
unsymmetrical faults.
T h e s e a r e c o n s i dered i n C h a p .
so
12.
T h e c u r r e n ts w h i c h fl o w i n d i ffe r e n t p a r t s o f a p ower s y s t e m i m med i a tely
a ft e r t he occ u rrence of a f a u l t d i ffe r from t h ose fl owi n g a few cyc l e s l a t e r just
b e fore c i rc u i t b reakers a re c a l l e d upon to o p en the line on both s i d es of the
f a u l t . A n d a l l o f t h ese c u r re n ts d i ffe r w i d e l y f ro m the c u r r e n ts w h i ch wou l d
flow
u n d e r s teady-state con di t i o n s i f t h e f a u l t w e r e n ot i s o l a t e d from t h e rest of the
s y s t e m by the opera t i o n of c i rcu i t b r e a k e r s . Two o f the f a c t o r s on w h i c h the
p ro p e r selection of c i rc u i t b re ak e rs d e p e n d s are the c u r r e n t fl ow i ng i m m e d i ately
a ft e r t h e fa ul t occ u rs and the c u rr e n t w h i c h the b r e a k e r m u s t i n terru p t . I n
fault
a n a lysis va l u es of t h e s e cu r r e n t s a re c a l c u l a t e d for t h e d iffe r e n t types of fau lts
at va r i o u s locations in the s ys t e m . T h e da ta o b t a i n ed from fa u l t c a l c u l a t i o n s a l so
s e rv e to d e t e r m i n e t h e s e t t i n gs o f r e l a y s w h i c h c o n t r o l t h e c i rc u i t b re a ke r s .
TRA \l S I ENTS I N R L S E R I E S C I R C U I TS
10.1
The s e l e c t i o n o f a c i rc u i t b r e a k e r for a p ow e r sys te m d e pe n d s not o n l y on the
c ur r e n t t h e b r e a k e r i s to c a rr y u n cl e r n o r m a l o p e r a t i n g co n d i t i o n s , but a l so on
t h e m a x i m u m c u rre n t i t m ay h a v e t o c a r ry mom e n ta rily a n d t h e c u r r e n t i t may
h a ve to in terrupt at the vol t ag e of t h e l i n e i n w h i ch it is p l a c e d .
I n o r d e r t o a p p ro a c h t h e p r o b l e m o f c a l c u l a t i ng t h e i n i t i a l c u r ren t w h e n a
to a c i rc u it con tai n i n g c o n s t a n t v a l u e s of r e s is t a n ce a n d i nd u c t a n ce. Let t h e
s y s t e m i s short-ci rcu i t e d , con s i d e r w h a t h a p p e n s w h e n an a c v o l t a g e is a p p l i ed
a p p l i e d vol tage b e Vmax s i n ( w t
vol t a g e . The n ,
a
+ a),
where
is zero a t the t i m e of a p p l y i n g t he
t
d e termi n e s t h e m a g n i t u d e of the vol t a g e w h en t h e c i rc u i t i s
c l o s e d . I f t he i n s t a n ta n eo u s v o l t a g e i s z e r o a n d i nc r e a s i n g i n a p o s i ti ve d i rect ion
m ax i m u m i n s t a n t a n eo u s v a l u e , a is
whe n it i s a p p l i ed by c l os i n g a s w i t c h ,
Vm�lx
sin (
i s z e ro . I f the vol tage is at i t s p o s i t ive
a
11/2.
wt +
a
)
The d ifferen t i a l e q u a t i o n is
Ri + L ­
di
=
( 10.1)
dt
T h e so l u t i o n o f t h i s e q u a t i o n I S
i
w here
12 1
=
=
�;
31x
[sin(
VR2 + ( w L ) 2
T h e first t e rm of Eq.
is no nperiodic
and
(ll
t
+
and
0 0.2)
d ec a y s
a
e
=
e)
- E -R
I
/ L s i n ( (l'
-
e)]
( 1 0 .2)
t a n - 1 ( w L /R ).
v a r i e s s i n u soi d a l ly w i th t i m e . T h e second t e r m
exp o n e n t i a l l y w i t h a
t im e con s t a n t
L/ R.
This
s i n u so i d a l t e rm a s t h e s t e a dy - s t a t e v a l u e o f t h e c u r r e n t i n a n RL c i rc u i t for t h e
n o n p e r i o d i c t e rm i s c a l l e d t h e
=-: 0,
dc component
o f t h e curren t . W e recognize t h e
g iv e n a p p l i e d vol tage. I f t h e v a l u e o f t h e s t e a d y - s t a t e term i s n o t z e r o w h e n
t
t h e d c com p o n e n t a p p e a r s i n t h e s ol u t i o n i n o r d e r t o s a t i s fy , t h e p h y s i c a l
382
CHAPTER 10
SYMMETRICAL FAU LTS
(al
FI G U R E 1 0 . 1
t i m e i n a n R L circu i t
0 = - rr/2, w h e re
0; (h) a
l ( w L / !n. T h e vo l t a g e i s V" "" s i n ( ll) / +
Cu rre n t as a fu n c t i o n of
Time
for: (0)
() = t a l l
IY )
( b)
a
-
(]
=
a p I1 1 i c d < I t / C"
-
() .
cond ition of zero current at the instant of closing the swi tch . Note t hat the dc
term does not exist if the ci rcui t is closed at a point on the vol tage wave such
t h a t a e = 0 or a e = rr . Figure 1 0 . I ( a ) shows the variation of curre n t w i th
time according to Eq. 00.2) when a e = O. If t h e switch is closed at a point
on the voltage wave such that a e = ± rr / 2, the de component has i ts
maximum initial value, which is equal to t he maxim u m value of the sinusoidal
component. Figure 1 0.l(b) shows curre n t versus time when a e = - rr /2.
The d c component may have any value from 0 to Vmax / I Z I , depending on the
instantaneous value of the vol tage when the circuit is closed and on the power
factor of t he circu i t . At the instant of applying the vol tage the dc and steady-state
components always have the same magnitude but are opposite in sign i n order
to express the zero value of current then existing.
In Chap . 3 we discussed the principles of operation of a synchronous
generator consisting of a rotating magnetic field which generates a voltage in an
armature winding having resistance and reactance . The current flowi ng when a
generator is short-ci rcu ited is similar to that flowing when an alternating voltage
is s uddenly applied to a resistance a n d an inducta nce in series. There are
important differences, however, because the curren ts in the damper w indings
and the armature affect the rotating field, as d iscussed i n Secs. 3 . 8 and 3.9. I f
the d c component of current i s eliminated from t h e short-circuit curre n t of each
a rm ature phase, the resulting plot of each p hase current versus time is t h at
shown in Fig. 3 . 19. Comparison of Figs. 3 . 1 9 a nd 1 0 . l ( a ) shows the d ifference
between applying a voltage to the ord i n a ry RL circuit and applying a short
circuit to a synchr<?nous machine. There is no dc component in either of t hese
figures, yet the current envelopes are quite d ifferent. In a synchronous machine
the flux across the a ir gap is not the same a t the instant the short circuit occurs
as it is a few cycles later. The change of flux is determ ined by the combin e d
action of t h e field, t h e armature, a n d t h e d a mper windings or iron parts of t h e
round rotor. After a fault occurs, the subtransient , transient , and steady-state
periods are characterized by the subtransient reactance X;, the transient
-
-
-
-
-
,
1 0. 2
383
Xd , respectively. These re a c t a nces
h ave i ncreas i n g values ( t h a t is, X; < X:t < Xd ) a n d t h e corresponding compo­
n e n t s of the shor t - c i rcu i t c u rrent have d e c r e a s i n g m a gn i t u d es ( 1 1" 1 > 1 1' 1 > I I I ).
W i t h t h e d c com ponent re moved , t h e initial symmetrical rms current i s t h e rms
reactance
va l u e
of
X� ,
I NTERNAL vOLTAGES OF LOADED MAC H INES UNDER FAULT CON DITIONS
a nd t h e s t e a d y-state r e a c t a nce
the ac
component of
t h e faul t
current immediately a fter the fault
occurs.
In a n a l y t i cal work t h e i n t e r n a l vol t a ge to t h e m a c h i n e a n d t h e s u b t r a n ­
s i e n t , t ra n s i e n t , and steady-state c u rre n ts m a y be e x p r e s s e d a s p h a s o r s . T h e
vu l t age i n d u ce d i n t h e a r m a t u re w i n d i n g s j u s t a ft e r t h e fa u l t o c c u rs d i ffe rs from
t h a t w h i c h exists a ft e r s t e ady s t a t e is rea c h e d . We acc o u n t for t h e d i ffe r e n c e s in
i n d uced \'o l t a ge by u s i n g the d i ffe r e n t reactances
s t e a d y - s t a t e cond i t i o n s . 1 f
( X;, X� ,
and
Xd )
in s eries
w i t h the i n t e r n a l vo l t a g e t o c a l c u l a te c ur rents fo r s u b t ra n s i e n t , t ra n s i e n t ,
a
and
m a c h i n e i s r e p r e se n t e d hy the n o - l oa d vol t a g e t o n e u t r a l in s e r i e s w i t h the
gen e ra t o r i s u n l oa d e d w h e n t h e fau l t o cc u rs , the
r ru p e r re :l c t a n ce , ; I S s h o w \l in f7 i g , l , 20 , The r e s i st a n c e i s t a k e n i n t o a c c o u n t i f
t e r m i n a l s and t h e s h o r t c i rc u i t, t h e e x t e r n a l i m pe d a n ce m u s t b e
g rc a t e r a c c u r;ICY i s d e s i re d . H t h e r e is i m p e d a n ce e x t e r n a l t o t h e g e n e ra tor
between i t s
i n c l u d ed in t h e c i rc u i t . \V e s h a l l ex a m i n e t h e t ra n s i e n t s fo r m a c h i n e s c a rr y i n g a
l o a d in t h e n e xt sect i o n ,
Al t h o u g h mach i n e r e a c t a n c es a r e n o t t r u e co n s t a n ts o f t h e m a c h i n e a n d
d e p e n d o n t h e degree o f s a t u r a t i o n o f t h e m a g n e t i c c i rc u i t , t h e i r v a l u e s u s ua l l y
l i e w i t h i n ce r t a i,n l i m i t s a n d c a n b e pred i c ted fo r va r i o u s t y p e s o f m a c h i n es.
Ta b l e A.2 in the Appe n d i x g ives t y p i c a l va l u es o f m a c h i n e r e a c t a n c e s t h a t a re
n e e d e d i n m a k i n g fa u l t c a l c u l a t i o n s a n d i n s t a b i l i ty s t u d i e s . I n g e n e r a l , s u b t r a n ­
s i e n t reactances o f gene r a tors a n d motors a re u s e d t o d e t e r m i n e t h e i n i t i a l
c u rr e n t flowing on the occu rrence o f a s h o r t c i r c u i t . For d e t e rm i n i n g t h e
i n t e r ru p t i ng c a p a c i ty o f c i rcu i t b re a k e rs, exc e p t t h ose w h i c h o p e n i n s t a n t a ­
n e o u s l y , s u b t r a n s i e n t r e a c t ance i s u s e d fo r g e n e rators a n d t r a n s i e n t rea c t a n c e i s
u s e d f o r synch ronous m o t o rs . In s t a b i l i ty s t u d i e s , wh ere t h e problem i s t o
d e t e r m i n e whether a fa u l t w i l l c a u se a m a c h i n e to l os e syn c h ro n ism
reactances apply,
with
t he
rest of t h e system if the fa u l t is r e m oved a ft e r a cert a i n t i m e i n t e rv a l , t r a n s i e n t
10.2
I NTERNAL VO LTAGES O F LOAD E D
MAC HINES U N D E R FAU LT C O N D ITIONS
Lc t
li S
co n s i d e r a g e n e r a t o r t h a l i s l o a d e d w h e n a fa u l t occ u rs . F i g ur e I O . 2 ( a ) is
I n t e r n a l \'ol t a ges a n d r e £.l c t a n c es o r t h e g e n e ra to r a re n ow i d e n t i fi e d by t h e
t h e c q u i v ,l i c l1 l c i rc u i t or <I g C I l C r <l [ O r l h a l
has
,\
b a l a nced t h r e c - p h a s e l o a d .
s u b s c r i p t g s i n c e some o f t h e c i rcu i t s to be c on s i d e red a l so h ave m o tors.
Exte r n a l i m p e d a n c e is shown b e tw e e n the g e n erator t e rm i n a l s a n d the p o i n t
w h e r e t h e fa u l t occurs. The c u r re n t f1 0w i n g b e fore t h e fau l t o c c u r s a t po i n t
IL , the vo l tage at t h e fa u l t i s
Vf '
P
P is
a n d the term i n a l vo l t age of the g e n e ra t o r i s V; .
T h e s t e a dy- s t a t e e q uiva i c n t c i rc u i t o f t h e synchronous g e n e r a t o r i s i t s n o - l o a d
vo l t a g e E � i n s e r i e s with i ts sy n c h ro n o u s r e a c t a nce
Xdg .
If a thre�-phase fault
384
C HAPTER 10
SYMMETRICAL FAU LTS
Zext
p
"
jXdg
r
V(
(a)
L
FIG U RE 10.2
Z ext
p
+
'X"
J de
ZL
S
E g"
(b)
V(
L
Equiva lent c i rc u i t for a generator s u pplying a b a l anced t h ree-ph ase l o a d . A p p l i c a t ion of
a
Zt.,
t h rcc­
p h ase fa u l t a t J> is s i m u lated by clos ing :iw i t cl! S : (a) usual s t eady-s t a l e g e n e r a t o r e q u i va l e n t c i rcu i t
wi t h load;
( b ) c i rc u i t for calcu lation o f 1" .
occurs at point P, we see t hat a short circuit from P to neutral i n the equivalent
circuit does n o t satisfy the conditions for calculating subtransient current, for
the reactance of the generator m ust be X�g i f we are calculating subtransient
current 1" or X�g if we are calculating transient current J' .
The circuit shown in Fig. lO.2( b ) g ives us the desired result. Here a
voltage E� i n series with X�g supplies the steady-state current fL when swi tch
S is open and supplies the current to the short circuit through X�g and Zext
when switch S is closed. If we can determine E� , the current through X�g will
be 1" . With switch S open, we see that
( 10.3)
and this equation defines E� , which is cal1ed the subtransient internal voltage.
Simi larly, when calculating transient current I ' , which must be su pplied t h rough
the transient reactance X:fll , the d rivi ng vo l tage is the transient internal voltage
E� , where
( l O A)
Thus, the value of the load current iL d etermines the values of the vol tages E�
and E� , w hi ch are both equal to the no-load voltage Eg only when IL is zero so
that Eg is t hen equal to v, .
At t h is point it is important to note that the particular value of E� in
series with X�g represents the generator imme d iately before and immedi ately
after the fau lt occurs only if the prefa u l t current in the generator has the
correspondi n g value of fL ' On the other hand, Eg i n series with the syn­
chronous reactance Xd8 is the equivalent circuit of the machine under steady­
state condit ions for a ny value of the load current. The magnitude of 1;g is
determined by the field current of the machine, and so for a different value of
·
IL
E�
1 0. 2
I NTERNAL V O LTA G ES OF LOA D E D M A C H I N ES U N D E R F A U LT CO N D ITI O NS
i n the c i rc u i t of Fig.
1 0.2( a )
l E/: I
385
wou l d rem a i n t h e same b u t a new v a l u e
wo u l d be req u i r e d .
of
S y n c h r o n o u s motors h ave r e a c t a n c e s o f t h e same type a s g e n e r ators.
W h e n a motor is short-c i r c u i t e d , it n o l o n g e r r e c e ives e l e ctric e n e rgy fro m t h e
p o w e r l in e , b u t i t s field r e m a i n s e n ergized a n d the i n e r t i a o f i t s r o t or a nd
c o n n ec t e d l o a d k eeps i t r o t a t i n g for a s h o r t p e r i o d of t i m e . The i n t e r n a l v o l t age
o f a sy n c h ro nous motor c a u ses it to co n t ri b u te c u rr e n t to the sys t e m , for it i s
t h e n a c t i ng l i ke a g e n e r a t o r . By c o m p a r i s o n w i t h t h e c o rrespon d i n g form u l a s for
E;�
a n d t r a n s i e n t i nt e r n a l vol t age
" IL
f/I - J'X dm
( 10 .5)
for a synchronous motor are g i v e n by
a g e n e r a t o r t h e s u b t ra n s i e n t i nt e r n a l vo l t a g e
E;"
£"! I I
l·� '
� I II
- f/I - }'X dill
'
--
/I.
( 1 0 .6)
w h e re VI i s now the t e rm i n a l v o l t a g e o f the m o t o r . Fau l t c u rr e n t s in systems
by c a l c u l c H i ng t h e s u b t r a n s i e n t ( o r tra n s i e n t ) i n t e r n a l vol t a g e s o f t h e
c o n t a i n i n g g e n e r a t o rs and m o t o rs u n d e r l o a d m a y be s o h' ed in e i t h e r o n e of two
ways :
(1)
machi nes or
(2)
b y u s i n g T h eve n i n ' s t h e o r e m .
A
s i m p l e e xa m p l e w i l l i l l u s t r a t e
t h e two a p p ro a c h e s .
S u p p o s e t h a t a s y n c h r o n o u s g e n e r a t o r i s c o n n e c t e d t o a s y n c h ro nous
m o t o r by a l i ne o f e x t e r n a l i m p e d a nce Zexl ' Th e m o t o r is d ra w i n g l o a d c u r r e n t
f r o m t h e g e n e rator w h e n a s ym m e t r i c a l t h re e - p h ase fa u l t o c c u r s a t the m otor
Ii"
t e r m i n a ls. Figure
1 0.3
s h ows the e q u i v a l e n t c i r c u i t s and c u rr e n t fl ows of the
syst e m immed i a t e l y b e fo r e a n d i m m e d i a t e l y a ft e r the fau l t occurs. By r e p l a c ing
the
synchronous
s h o w n in Fig.
r e a c t a n c e s of the m a c h i n e s by t h e i r
1O.3( a),
r e a c t a n c e s as
sub transient
we c a n c a l c u l a t e the s u b t r a n s i e n t i n t e r n a l v o l t a g e s o f the
m a. c h i n e i m m ed i a t e l y be fo r e t h e f a u l t o c c u r s by s u bs t i t u t in g the v a l u es of
( j O. 20)
+
p
(jO. l0) 1"
(jO.20)
-----+-
( j O.20)
E"8
Neutral
( a ) Before the fau lt
g
jXd m
E"In
En
e
VI
p
--
1"m
p;
(jO.20)
Neut ral
jXdm
E"
m
(b) After the fault
FIG URE 10.3
E q u i v a l e n t c i rc u i t s a n d c u r r e n t flows b e fore a n d a ft e r a fau l t at t h e t e r m i n a l s of a synchronous
motor con n e c t e d to a synchro n o u s g e n e r a t o r by l i n e i m p e d a nce 20,1 ' N u m erical va l u e s are for
Exa m p l e 1 0 . 1 .
386
and
CHAPTER 10
IL
SYMMETRICAL FAULTS
i n the equations
( 10 .7)
E"
'm
=
V1 - )'X"dm 1L
( 10 .8)
When t h e faul t is on the system, as s hown In Fig. 1 0 3C b ) the subtransie nt
currents I; out of the generator and I;:, out of the m otor arc fou n d from the
rel ations
.
J�"
/"
+ !I .
E"
=
==
m
-
--�
' --
'Xtil'
"
L
--' e x l -I- J
"
Enr
_
_
_
_
,
( 1 0 .9)
( 10 . 10)
"
J'Xdnz
These two currents add together to give the total symmetrical fault current 1;
shown in Fig. 1 0.3(b). That is,
I"
1
=
I" + I"
8
m
=
Z ext + J'Xdg
"
/"gf
+
jXdm
( 10 . 1 1 ) '
["m!
where 1;1 and I; I are the respective contributions of the generator and motor
to the fau l t current If. Note that the fa ult current does not i nclude the prefault
(load) current.
The alternative approach using Thcvc n i n 's theorem i s based on the
observation that Eq. (10. 1 1 ) requires a knowledge of only VI ' the prefault
voltage of the fault point, and the parameters of the network with the subtran­
sient reactances rep resenting the machines. Therefore, 1; and the additional
currents produced throughout the network by the fault can be found simply by
applying voltage VI to the fault point P i n the dead subtransient network of the
system, as shown i n Fig. 1 0.4( a). If we redraw t h a t network as shown in Fig.
lO.4(b), it becomes clear that the symmetrical values of the subtransient fault
curren ts c a n be found from t h e Thevenin equiva lent circuit of the subtransient
network a t the fau l t point. The Thevenin equivalent circu i t is . a single generator
and a single impedance terminating at the point of application of the fault. The
equivalent g enerator has an internal volt age equal to VI ' the voltage at the fault
point before the fault occurs. The impedance is that measured at the point of
application of the fault looking back into the circui t with all the generated
voltages short-circuited. Subtransient r eactances are used since the initial sym-
1 0.2
I NTERNAL VOLTAG E S OF LOADED MACHINES U N D E R FAULT CON DITIONS
Z ext
I"g (
-
P
"g
J'X d
I"m (
-
jX d m
I If
Vr
(a)
C i rc u i t s i l l u s t ra t i ng t h e a d d i t i o n a l
+
Vr
+
FI G L R E 1 0.4
dead
'X"
J dg
Z ext
g(
jXd m
-
[If
[If
(b)
c u r re
nt
( /J )
! l ows d u e: to t h e t h re e - p h a s e fa u l t at
T h c \'t� n i n e: q u iv ;l I e n t loo k i n g i n t o t h e
m e t r i c a l f{l u l t c u r r e n t i s d e s i re d . ) n f i g .
I O. 4C h )
m(
/If
P: ( a ) a pplying VI to
circu i t at po i n t P .
the T h c v e n i n i mp e d a n c e
. X' ''
X"
)
J. dm ( 2 " X I -+- ] dli
Zth
P
-
r
---
n e t w o r k t o s i m lJ l a t e t h e fa lI l t :
387
Z th
is
( 1 0 . 1 2)
U p o n t h e occu rrence of a t h re e - p h a se s h o r t c i rcu i t a t P , s i m u l a t e d by c l o s ing
sw i t c h S , t h e s u b t ra n s i e n t c u rre n t i n t h e fau l t is
Vf [ ZCXI + j( X;;R + X;;m ) ]
jX:;m ( Z exI + jX�g )
I"f
( 1 0 . 13)
T h u s , t h re e - p h a s e s ym m e t r i c a l fa u l ts o n systems containing g e n e r a t o r s a n d
vol t ag e s o r Theven i n ' s t h e o r e m , a s i l l u s t r a t e d i n the fo l l ow i n g exa m p l e s .
m o to rs u n d e r l o a d m a y b e a n a l y z e d b y e i t h e r the use of s u b t ra n s i e n t i n t e rn a l
Exa m p l e 1 0 . 1 . A sy n ch ro n o u s ge n e ra t o r a n d m o t o r a re ra t e d 3 0,000 k Y A , 1 3 . 2 kY,
a n d bo t h h ave
kW
s u b t r;\ I1 s i e n t re a c t a nc e s
o f 20%. T h e l i n e co n n e c t i ng t h e m has a
r e a c t a n ce o f 1 0 So o n t h e b a s e o f t h e m a c h i n e ra t i ngs. T h e m o t o r is d raw i n g 2 0,000
at
0 . 0 powe r - f a c t or
s Y ll l l l 1 c l r i c t i l l m: c - p i 1 : t S C f; 1 \ I I I
n1 ; 1 c h i n e s .
c u rre n l s
the
kY w h e n a
O C C l \ I'S , I I l i l e J l l o t o r l eI' ll 1 i l1 : t i s . r i n d t h e s u b t r ll n s i e n t
Ill t l i O f , , l l l l l t h e ra l l l t h y l I s i t 1 g l he i n l e r n a l v o l t a ges o f
l e a cl i n g a n d
i n t i l l' g e n e f' : l i ( lI . t he
,\
l e rm i n ;\ 1 vo l t a ge of 1 2 . 8
SO/Illion. T h e p r c fa u l t e q u i va l e n l c i rcu i t o f t h e sys l e m curresponds t o Fig.
C hoo s i n g a base o f 3 0 , 0 0 0 k Y A , 1 3 . 2 k Y a n d u s i n g t h e v o l t age
a s t h e r e ference p h asor, w e o b t a i n
1 2 .8
V = -
f
1 3 .2
0 . 970
L..2.:
per unit
Vi
10.3(a).
a t t h e fa u l t poi n t
388
CHAPTER 10
S Y MMETRICAL FAU LTS
B ase current
IL
=
=
=
30,000
13 X 1 3.2
=
1312 A
20,000/ 36.90
0.8 X fS3 X 12.8
O .�() ( O .K
+
jO J) )
=
1 1 28
O N)
=
/ 36 .9 A
�---0
+
jO .52 p e r u n i t
For the g enerator
�
E;
Ig"
0 .970
=
0.918
=
0 .8 14
= -
=
+
+
jO . l (0 .69
jO.069
jO.S2)
jO .2(0 .69
jO.207
= 0.69
jO.3
+
1312(0.69
Fv.
+
+
-
j2.71)
-
+
0.918
-
+
j O .52)
}· 2 . 7 1 p e r
90 5
=
=
jO.069 p e r u n i t
= 0.814 +
jO .207
per u n i t
unit
j3SS0 A
"r
�
£;;,
=
�
.
0 .970
=
+
/ 0° per u nit
') . 6 9
J
= 1 .074 - jO. l 38
1"
'"
=
=
1 .074 - jO . 1 38
jO .2
+
jO .S2)
= 0 . 970
- -
jO. I 3R
+ 0 . 1 04 per u n i t
1--
- 0 .69
1 3I2( - 0.69 - is .37)
=
-
uni t
-90 5 - j7'v_
In the fault
If
= I; + I�
=
-
j8.0S
=
X
0.69 - j2.7 1 - 0 .69 - jS .37
1312
=
-
jl0,600 A
Figure IO . 3(b) s hows the paths o f 1:, I�" an d 1[.
7=
-j8 .u,
10.2
I NTER N A L
E xample 1 0 .2.
Solution.
VOLTAGES OF LOA D ED M ACH I N ES U N D ER F A ULT CO N DI TION S
389
Solve Example 10. 1 by the use of Thcvenin's t heorem .
The Thevenin equivalent circuit corresponds to Fig. 10.4.
Zth =
VJ
In the fau l t
1 '/
f -
_
=
VJ
_
= . 1 2 per u nit
iO
i O . 3 X iO . 2
'
'
J 0 .3 + J 0 .2
0 .970
L.Q:
per unit
0 . 97 + i O
Zlh
-i8 . OS
iO . 1 2
per u n i t
t h e p a r a l l e l c i rc u i ts o f t h e m ac h i n e s i n ve r s e l y
s i m p l c current divis ion we o b t a i n t h c fa u l t c l l r re n t s
T h i s b u l t c u rre n t i s c1i \ i d e d b e t w ee n
�\S
t h c : � i m p c ( Li n c c s By
From gener a tor: 1;r
From motor: 1�'r
=
=
-
is . 08
iO.2
X
- i3 .23 p e r
-
iO.S
iO.3
- i 8 .08
X
-i 4 .SS
-­
j O .S
N e g l e c t i n g load c u r r e n t g ives
Fau lt curren t from generator
Fau l t current fro m motor
Curre nt i n fault
=
3 .23
X
131 2
=
4 .8S
x
1 3 12
=
8 .08 x 1 3 1 2
=
unit
p e r unit
4240 A
6 360 A
=
1 0 , 600
=
A
The current i n t he fa ult is the same whether or not load current is considered, but
the currents in the l i nes differ. When load curre nt I L is i ncluded, we fi nd from
Exa m p l e 1 0. 1 that
I:
1;:,
=-
=
':1 +
1::,!
-
fL
11.
=
=
-
i3 . 23 + 0 .69 + iO .5 ?
- i4 .85 - 0 . 69 - i O . 5 2
=
=
0 . 69 - i 2 .7 1 p e r u n i t
-
0 . 69
-
is .37 per unit
Note that I/. is i n the sa m e d ir ect i o n a s I ; b u t opposite to ,;: . T h e p e r un i t v a l u e s
Ii , 1; , and 1;;, a r e t h e s a m e as i n Exa m p l e 1 0 . 1 , a n d s o t h e am pere
values w i l l also be the s ame,
,
fo u n d fo r
Fault current from generator
Fault curre nt from motor
=
=
1 905 - j3SS0 1
1
-
=
905 - j70S0 1
-
3 600 A
=
72 0 0
A
The sum of the magnitudes of the generator and motor c urrents does not e q ual t h e
fa u l t current because the currents from t h e g ene r a t or and motor ar � not i n phase
when load current i s included.
390
CHAPTER 10
SYMMETRICAL FAULTS
Usually, load current is omitted in determining t h e current in each l i ne
upon occurrence o f a fault. I n the Thevenin method n eglect of load current
means that the prefaul t current in each l ine is not added to the component of
current flowing toward the fault in the l ine. The method of Example 1 0 . 1
neglects load current i f the subtransient internal voltages o f all machines are
assumed equal to the voltage VI at the fau l t before the fault occurs, for such is
the case if no current flows anywhere in the network prior to the fau l t .
Resistances, charging capacitances, a n d off-nominal tap-changing o f transform­
ers are also usual ly omitted in fault studies since they are not l ikely to influence
the level of fault current significantly. Ca l cu l a t i on o f th e fa u l t c u r re n t s i s
thereby simplified s i nce t h e n etwork m o d e l b e c o m e s an in t e r co n n e c t i o n of
inductive reactances and all currents throughout t he fa u l t e d sys t e m are t h e n I n
phase, as demonstrated in E x a m p l e 1 0.2.
10.3
FAULT CALCULATIONS USING Z bus
Our discussion of fault calculations has been confined to s i m p l e c i rc u i t s , b u t
now w e extend o u r study to general networks. W e proceed t o t h e general
equations by starting with a specific network with which we are already familiar.
In the circu it of Fig. 7.4 if the reactances i n series with the genera ted vol tages
are ch anged from synchronous to subtransient values, and if the generated '
volt ages become subtransient interna l vol tages, we have the network show n i n
Fig. 1 0.5. This n etwork can be regarde d a s the · per-phase equivalent of a
balanced three-phase system. I f we choose to study a fau l t a t bus (1) , for
example, we can follow the notation of Sec. 1 0.2 and designate Vf as the actual
voltage at bus 0 before the fault bccurs.
+
® -'--11---
--I-.L...
G)
FI GURE 10.5
E"a
+
+
E'b
React a n ce d i agram obta ined from
Fig. 7.4 by subst i t u t i ng subtransient
values for sync h ronous re acta n!;:es
and synchronous i n ternal voltages
o f t he m a ch ines. Reactance values
are m arked in per u n i t .
1 0. :;
®
FAU LT CALCULATIONS U S I N G
@
CD
+
+
'l
E
�u
Vi
-
Vi
F I G U R E 10.6
C i r c u i t o f Figure 1 0.5 w i t h a
E,:
+
A t h r e e - p h ase fa u l t a t b u s
w h e r e t h e sou rce vo l t a g e s
Vf
@
Cl n d
fa li l t
t i 1 r � e - p h a s t:
, i ll 1 l 1 i ;t t l: d by
on
V and
I
bus
-
V
f
S t.:r I l: S .
i s s i m u l a t e d b y the network of Fig.
-
391
Z b us
Vf
(1)
In
1 0 . 6,
i n s e r i e s con s t i t u t e a s h o r t - c i rc u i t
b ra n c h . S o u r ce v o l t a g e Vr a c t i n g a l o n e i n t h i s branch wo u l d m a t c h t h e p r e fa u l t
(1) ,
v o l t a g e a l re a d y a t b u s
b ranch . W i t h
Vf
a n d t h e re fore w o u l d n o t c a u s e c u r r e n t t o fl o w i n t h e
a n d - Vf i n s e r i e s , t h e b r a n c h becomes a s h o r t c i rc u i t a n d t h e
Ii
branch current is
t h e 8. d d i t i o n of t h e -
a s s h own .
VI
It
source. The current
Ii
sys t e m from t h e refe re n ce n o d e b e fo r e fl o w i n g
E� , E;; ,
a c t a l o n e and - Ii in fO b u s
W
fro m e x t e r n a l sou rc e s . W i t h -
V;
Vf
and
is c a u s e d
by
d istributes i tsel f t h roughout the
out
of bus
W
t hrough t h e -
changes
so u rc e . I n d o i n g so, i t p r o d u ce s w h a t ev e r b u s vol t age
sys t e m d u e t o t h e fa u l t . I f
Ii
i s evi d e n t , t h e r e fore, t h a t
t h a t occur i n t h e
a r e s h o r t -ci rcu i t e d , t h e n -
is t h e n t h e only c u r r e n t
Vf
entering
Vf
i s l e ft to
t h e n e tw o r k
a s t h e o n l y source, t h e n e t w o r k h a s t h e n o d al
i m p e d a n ce e q u a t i o n s i n t h e Z bl l S m a t r i x fo rm
6. V
I
6. V,
6 V,
6. V
-1
6. V
t
- V
J
6 V,
� V.J
CD
(1)
CD
CD
21t
.% t 2
Ln
2 :' 1
L 12
G) L , t
@
2 ,\
24 1
The p re fi x 6. i s chosen to i n d i c a t e t h e
t o t h e c u rrent -
Ii
i nj ec t e d i n t o b u s
Zt
2
changes
W
G)
Z2
:-
�
L �l
Z . I :1
@
2 1 .1
2 2 '1
/ :\.1
7,44
()
-
II
"
()
( l O . 1 4)
0
i n t h e vo l t a g e s a t t h e b u s e s d u e
b y t h e fa u l t .
Th e Z b l l s b u i l d i n g a l g o r i t h m , o r some o t h e r m e a n s such a s
Y bus
t r i a n g u l a r­
i z a t i o n a n d i n v e r s i o n , c a n be u s e d to eva l u a t e t h e b u s i m p e d a n c e m a t r ix for t h e
,
392
CHAPTER
1 0 SYMMETRICAL FAU LTS
network of Fig. 10.6. The numerical values of the elements of the matrix will be
different from those in Example 7.6 because subtransient reactances are now
being used for the synchronous machines. The changes in the bus vol tages due
to - If a r e given by
� V1
� V2
� V3
�
�
The second row of this
� V1
- VI
- Ij'
=
� V:I
� V4
e q u a t i o n s h ow s
l
Cul u m n
ot
Z hll.\
21
- Z 1 2 If
=
-L
22 I"
/
- 2 .1 2 If
- 2'1 2 1j'
( 10. 1 5)
t hat
( 1 0 . 1 6)
We recognize 2 22 as the diagonal element of Z bl ls representing the Theveni n
impedance o f t h e network a t bus @ . Substituting t h e expression for Ii into
�q. 00.15) gives
� Vl
� V2
� VJ
� V4
-
,
-
-
2 12
2 22
J
V
f
-
v
Z 32
V
Z 22 f
Z4 2
v
Z 22 I
( 1 0 . 17)
-
-
When the genera to r vol tage V is s h o r t -ci rell i t e d in t h e n e tw o r k of Fig. 1 0.6
and the sources E�, E;; , and Vf Iare rei nserted i nto the network, the currents
and vol tages everywhere in the network will be the same as those existing before
the fault. By the principle of superposi t ion these prefault vo l t a ge s a d d t o the
changes given by Eq. (10. 1 7) to yield the total vol tages existing after the faul t
occurs.
The faulted network is usually, but not a lways, assumed to be without load
before the fault occurs. In t h e absen ce of loads, as remarked previously, no
prefault currents flow and there are no voltage differences across t he branch
impedances; all bus voltages throughout the network are then the same as Vf'
the prefault voltage at the fault point. The assumption of no prefault current
simplifies our work considerably, and by applying the principle of s uperpos it ion,
-
lU.3
FAULT CALCULATIONS U S I N G
Z bu s
393
we obtain the bus voltages
VI
VI
6. V
V2
VI
6. V
2
+
1
V�
VI
6. V
3
V4
VI
6. V4
VI - 2 1 2 Ii
VI - VI
VI - 2 :12 I]'
VI - 2 42 IJ
2 12
Z 22
1 0
=
VI
2 :n
1 2 22
2 42
1 2 22
( 1 0 . 1 8)
-
Thus, the vol tages at all buses of t h e ne twork can be c al c u lat e d using t h e
prefa u l t vo ltage VI o f the fa u l t bus a n d the c l ements i n t h e co lumn of Z bus
correspo n d i n g to t h e fa u l t bus. The calcu lated values of the bus vo l t a g e s w i l l
y i e l d t h e subtransicnt cu rre n ts i n t h e b r a n c h e s o f t h e ne t w o rk i f t h e sys t e m Z bus
h a s been fo rmed w i t h s u b t ra n s i e n t va l u es fo r the mac h i n e r e acta n c e s.
In more ge neral terms, when the three-phase fa u l t occ urs on bus ® of a
l arge-scale n e twork, we h ave
V
- _I_
II" 2" "
( 10 . 1 9 )
a n d n e g lectin g prefau l t l o a d currents, we can t h e n wri te fo r t h e voltage a t a ny
b u s 0 d ur i n g the fau l t
( 1 0 .20 )
w h ere 2j " a n d 2 u a re e l ements i n col umn k of the s y s t e m Z OllS . I f the p re fa ul t
vo l t age of b u s (j) is not t h e same a s t h e prefa u l t vol tage of fau l t bus CD, t h e n
we s i m p l y re p l a ce V o n t h e l e ft in Eq . ( I O . 2()) hy t h e a c t u a l prcfa u l t vo ltage of
I
bus (j) . Knowing the bus vol t ages d u ring the fau l t , we can calculate t he
subtran s i e n t current I:�. from bus CD to bus (j) i n the l i n e of i mp edance 2b
con n ecting t hose two buses,
( 1 0 .21 )
This e q u ation shows I:; as the fraction of t h e fau l t c u rr e nt Ii appearing as a
line flow from b us CD to b u s (]) i n the fau l te d n e twork. I f b u s 0 is d i rectly
con n e cted to the fa ulted bus CD by a l i n e of series i mpedance 2b, then the
current contributed from bus (]) to the current i n the fau lt at b u s CD is s i m pl y
Vj/ 2 b ' w here Vi i s g iv e n by Eq . 0 0 .20).
394
CHAPTER 10
SYMMETRICAL F AULTS
The d iscussion of this section s hows that only column k of Z bu s , which we
now denote by Z b:�, is required to evaluate the impact on t h e sys t e m of a
symmetrical three-phase fault at bus ® . I f n ecessary, the e l e m ents of Zb:� c a n
b e generated from the triangular factors of Y bus ' as d emonstrated in Sec. 8.5 .
Example
three-phase fau l t occurs a t bus W of the network of Fig. 10.5.
Determine the initial symm etrical rms currc n t (that is, thc subt ra n s i e n t c u r r e n t ) i n
the fault; the voltages a t buses CD , G) , a n d @ d u r i n g the fault; the current flow
in the line from bus ® to bus CD ; and the cu rr e n t contributions to t he fault from
10.3. A
1 .0ZQ:. p e r u n i t
lines G) - W, CD - W,
e q u a l to
(l n o
@ - (1) .
Ta k e t h e p r da u l t vll l t ;\ge
a n o n e g l e c t ;1 1 1 p r da u l t c u rr e n t s .
Solution. Applying t h e Z hllS b uil d i n g a l gor i t h m t o f i g .
Z bus
(1)
CD
=
jO. 1 938
jO.22Q5
jO. 1 494
jO.1506
CD jO.2436
CD jO.1938
G) jO. 1 544
@ jO. 1 456
G)
1 0.5, we
jO. 1544
jO . 1494
jO. 1 954
jO . l 046
.�
�
1 .0
-
Z22
=
1 .0
jO .2295
---
bus is
-j4 .3573 per unit
and [rom Eq. (10.18) t h e vo l ta g e s d ur i n g t h e fa u l t a r c
VI
1 -
V2
=
V3
�
The current flow in
l i ne
11-
jO . 1 938
jO.2295
0
jO.1494
® - CD
jO .2295
jO.1506
jO.2295
is
0.3490 - 0 . 1 556
jO.25
0. 1556
o
per
unit
-jO.7736
pe r
0.3490
at bus
G)
that
jO . 1 456
jO .1 506
jO.1O-l6
jO. 1 954
Since l oa d currents are n eglecte d , the pr e fault voltage at each
unit, the same as Vf at bus @ . W h e n the fault occu rs,
[" =
f
fi n el
T'J
0.3438
unit
l .Oft p e r
FAU LT CALCULATIONS USING Z bw EQUIVALENT CIRCUITS
I DA
Faul t currents contributed to bus
From bus CD :
Exce p t ror
10.4
From bus
G) :
From bus
@:
395
W by the a dj a c e n t unfaulted buses are
Vj
0 . 1 55 6
Zbl
VJ
0 .3490
Zb 3
jO .25
V4
Zb 4
round-off e rr o r s , the sum
-j l .2448 per unit
jO.125
=
0 .3438
=
of
jO .20
=
-j l .3960 per unit
-jl .7 1 90 per unit
these current contributions equals If.
FAU LT CALC U LAT I O N S U S I N G Z blls
EQ U IVALE NT . C I R C U ITS
We
cannot devise a physica l l y re a l izable n e twork which d irec tl y i n c orpora te s all
the i n d iv i d u a l e l e m e n t s of the bus i mpedance m a t r i x . However, Fig. 8.4 shows
t h a t we can use the m a tr ix elements to construct the Thevenin equival ent circuit
between any pair of b uses i n the n etwork that may be of interest. The Thevenin
e q u ivale nt circuit is very helpfu l for illustrating the symmetrical fault equations,
which are developed in Sec. 10.3.
In the Theven i n e q u i val e nt c i rc u i t of Fig. lO.7( a ) b u s ® i s assumed to be
the fau l t bus and bus (j) is u n f au lt ed . The i mpedances shown correspond
directly to the eleme n ts of t h e n e twork Z bus and all the pre fa u l t bus voltages are
the same as Vf of t he fau l t b u s if load currents a re neglected . The two points
U nfaulted bus
U nfaulted bus
}
Bus to be faulted
Z· · - Z k
v
r = _
f
'-!...
""'"
II;
(a)
(])
))�
�
(])
Z · }· - Z · k
..r---- ��
x
Faulted bus
®
�
s
Zk k
�jli
(b)
a n d ® of sys t e m w i t h no p r e fa u l t 1 0 <l u cur r e n t s :
t h c Ll u l t ( S o rc n ); ( 1) ) u u r i n g t h e fa u l t (S c l osed).
FI G U RE 1 0 . 7
Thev e n i n e q u ival e n t b e t w e e n b u s e s
CD
( a ) befor e
396
CHAPTER 10
SYMMETRICAL FAU LTS
marked x h ave the same potential, a n d so they can b e joined together to y i e ld
the equivalent circuit of Fig. 1 0. 7 ( b ) w i th a single voltage source Vr as shown. If
the switch S is open between bus ® a n d t h e reference node, there is n o short
circuit and no current flows in any branch of the network. When S is closed to
represen t the fault on bus ® , c urre n t flows in the circuit toward bus ® . This
current is If = VrIZk k ' which agrees w i t h Eq . C 1 0. 1 9), and it i n d uces a voltage
d rop ( Zjk l Zk k )Vr in the direclion ji'o/11 t he r e fe r e n c e node towa rd b u s CD . The
voltage from bus (]) to the reference changes therefore by the amou n t
- (Zjk IZ kk ) Vf s o t h a t t h e voltage a t b u s (]) d u ring the fau l t is VI (Zj k IZkk ) Vr , which is co ns i s t ent with Eq. ( 1 0 . 20).
Thus, by substituting appropriate n u m e r i c a l v a l u e s for the i m p e d a n ces i n
the simple e quivalent circ uit of Fig. 10.7( b ), we can calculate the bus vol tages of
the system before and after the fau l t occurs. With swi tch S open in the circu i t ,
the voltages at b u s ® a n d the represe n tative bus (J) are equal to Vf . The s a m e
u niform voltage profile occurs i n Fig. 1 0.6 i f there a r e n o prefa u l t curren ts so
that E� and E� equal Vf ' If S is closed in Fig. 1 0.7(b), the circuit reflects t h e
voltage of representative b u s CD w i t h respect to reference while the fau l t i s o n
bus ® . Therefore, i f a three-ph as e short-circu i t fau l t occurs a t bus ® of a
large-scale network, we can cal culate the c urrent i n the fau l t and the vol tage at
any of the u n faulted buses simply by i nserting the proper impedance values i n to
elementary circuits like t hose i n Fig. 10.7. The fol lowi ng example i l l ustrates t h e
procedure.
10.4. A five-bus network has generators at buses CD and G) rated 270
and 225 MVA, respectively. The generator subtransient reactances plus the
reactances of the transformers con necting them to the buses are each 0 . 3 0 per u n i t
on t h e generator rating a s base. T h e t u rn s ratios o f the t ransformers are such that
the voltage base in each generator circuit is e qual to t h e v o l t age rating of the
generator. Line impedances in per u n it o n a 1 00-MVA system base are shown in
Fig. 1 0.8. Al l resistances are neglected. U sing the bus impedance m a t r i x for the
network which includes t h e g e n e r a l o r a n d t r a n s fo rm e r re a c t , I I 1 C C S , fi n d t h e
Example
1.0�
®
)0. 168
jO. 126
®
)0. 126
) 0.210
®
1 . 0�
FIGURE 10.8
Impedance diagram for Example 1 0.4. G en erator react ances i nc l u d e subtransient value6 p l u s
reactances of s et-up tra nsformers. A l l values i n p e r u n i t on 3 1 00-MYA b a s e.
\ 0 .4
FAULT CALCULATIONS USING
Z bus
EQUI VALENT CIRCUITS
397
subtransient current in a t hree-phase fault at bus @ 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 p er unit before the fault occurs.
Converted to the lOO-MVA base, the combined generator and
transformer reactances are
Solution.
Generator at bus
CD :
X
Generator a t bus
G) :
X
100
=
0 .30
X
-
=
0 .30
X
-
270
=
0.1 111
per unit
=
0 . 1 333
per unit
100
225
These valu es, a long with the line impedances, are marked in per unit in Fig. 10.8
from which the bus i mped an ce matrix can be determined by the Z bu s building
algorithm to yic ld
Z bus
=
CD
W
G)
@
G)
CD
W
j O .0793
W
jO .0558
0)
jO.0382
C1)
jO . 05 1 1
j O.0608
j O .0558
j O . 1 338
j O . 0664
j O .0630
j O .0605
jO .0382
j O .0664
jO .0875
jO .0720
jO .0603
j O .05 1 1
j 0 .0630
jO .0720
jO .232l
jO . 1 002
j O . 0 608
j O .0605
jO .0603
jO . 1 002
jO. 130l
Since we are to calcu late the currents from buses
G) and G) i nto the fault at bus
Unfaulted bus
U nfaulted b us
Z 33 - Z 34 ®
V(
_.
l . O�O°
+
.
j O.0720
If
Z44� S
=
j O . 1601
-j4 .308
@
Vr
=
l.oLQ:,
Z4 5
=
_
Z54 jO.l 002
If
--
G).
U s e of Theve n i n e q u i v a l e n t c i rc u i t s to c a l c u l a t e voltages at
at b u s
(a)
-
Z5'
+
ea)
Fl G U RE 1 0 . 9
®
-;.02�9 I
Z 1)1)
bus
G)
(b)
and
=
jO.1319
-j4.30B
s
--
(b) b u s
@
d u e to fault
398
CHAPTER 10 SYMMETRICAL FAULTS
need to know V3 and Vs during the fault. Visualizing equivalent circui ts
like those of Fig. 10.9 helps in finding the desired currents and voltages.
The subtransient current in the three-phase faul t at bus 8) can be calculated
from Fig. 1 O.9(a). Simply closing switch S gives
0 , we
1 .0
V _
IfII - I Z44
_
From
j O.2321
Fig. 1O.9( a ) the voltage at bus
VJ
From Fig.
=
Vf - IlZJ4
1 O.9( b )
=
1 .0
-
G)
=
-j4 .308
per u n i t
during the fault is
( - j4 . 3 08 ) ( J0 .0720)
t he voltage a t bus G)
Vs = Vf - fl Zs4
=
@
0 . 6891-: p e r u n i t
d u r i ng t h e fa u l t is
1 . 0 - ( -j4 .3(8) ( j0 . 1 002)
Currents i nto the fault a t bus
=
=
0 .568 3 p e r u n i t
over t h e line impedances Zb are
From bus
@:
-
From bus
G) :
-
VJ
Z b3
Vs
Zb S
0 . 6898
=
jO.336
0 .5683
=
jO .252
Hence , total fault current at bus
@
=
=
-j2 . 05 3
per u n i t
- j 2 .255
per unit
- j4 .308 per u n i t
Other equivalent circuits based on t h e given bus impedance matrix c a n b e
developed for three-phase faults o n a n y o f the other buses o r transmission lines
of the system. A specific application w i l l demonstrate how this is accomplished.
Three-p hase fau lts occur more often on transmission l ines than on substa­
tion buses because of the greater exposure of the l ines to storms and acciden tal
disturbances. To analyze a line fault, the point of fault on the l i ne can be
assigned a new bus n umber and Z b u s for the normal configuration of the
network can then be modified to accommodate the new bus. Sometimes t he
circuit breakers at the two ends of the l i n e do not open simultaneously when a
line fault is being cleared. If only one circ u i t breaker has opened and the fau l t is
not fully c leared, shor t-circuit curre n t persi sts . T he so-ca l l e d line-end fa ult
represents the p articu l a r situation where the three-phase fault occurs very close
to one of the terminating buses of the l i n e , on the line side of the first breaker
(nea r the fault) to open. The line breaker near the fault is called the near-end
breaker and that at the end away from the faul t is called the remme-end breaker .
The single-line d iagram of Fig. 1 0 . 1 0 shows a four-bus network with a
line-end fau lt a t point P on the line con necting b uses CD and 0 The line has
series impedance Zb o The near-en d breaker at bus @ is open a n sI the
remote-end breaker is closed, leaving the fault still on at point P , which we now
°
1 0 .4
CD
®
@
o
F I GLRE 1 0 . 1 0
L i n � · e nd fa u l t
Fig
399
FAU LT C A LCULATIONS U S I N G Z buS EQU I VA L E N T C I RCUITS
at
I O.H.
point
P
on l i n e o f s e r i e s i m pe d ;tnce
-
Closed breaker
-
Open breaker
Z" b e t w e e n
b J se s
CD
nnd
G)
of sys t e m of
call bus ® . I n order to study t hi s fa u l t con d i t ion, we need to modify the
existing b us impedance matrix Z orig for the normal con figuration of the system
t o reflect the near-end breaker o p e r ation. This is accomplished in two steps:
1.
2.
Establish the new b us ® b y a d d in g a line of series i mped ance Z b b etwe e n
bus CD a n d bus ®.
Remove t h e line between b u s CD a n d b u s W b y add ing l i n e impe da nce
Zb b etween t hose two buses i n t h e manner explained in Sec. 8.4.
-
The first step fol lows the proce d u r e given for Case 2 in Table 8.1 a n d y i e l ds, i n
terms o f the e lements Z ij o f Z orig , t h e first five rows and columns o f t h e
®
ZII
221
.% � I
Z4 1
211
+ Zb
( Z I I - 22 1 )
Z21
®
-
Z I I - 2 12
.1..1 1
-
-
222
2:12
24 1 - Z42
ZII
-
Z I I1 . 1 2
2 12
-
2b
-
( 1 0 .22)
whe r e Z th. 1 2
=
Z I I + Z22 - 2 Z 1 2 w h e n
be accom p l i s h e d
by
fo r m i n g
Z ori g is sym met ric. The second step can
row ® a mI co l u m n ® as shown a n,d then Kron
400
CHAYTER 10
SYMMETRICAL FAULTS
reducing the matrix Z to obtain the n ew 5 X 5 matrix Z bu s . new including bus ® ,
as explained for Case 4 of Table 8.1. However, since 2 k , n ew is t he only element
required to calculate the current in the fault at bus
( point P of Fig. 1 0 . 1 0),
we can save work by observing from Eq. 00.22) that the Kron reduction form
glves
®
( 1 0 .23)
LQ:
Again, w e n o t e t h a t 212 2 2 1 a n d 2(h, 1 2 = 2 I I + 2 22 - 2 Z 1 2 ' By neglecting
prefault currents and assigning pre fault voltage VI = 1 .0
per unit to the
fault point P, we find the line-end fau l t cu rrent 1; out of b u s ® as fol l ows:
=
I"
f
- 1 .0
- 2
k k , ne w
1 .0
=
Z b - ( 2 1 1 - Z 21 ) 2 / ( Zth
- Zb )
------
2\ !
+
12
( 1 0 .24)
Thus, the only el em e n ts of Z orig entering into the calculation of I; are 2 1 1 ,
Z 12 = Z 21' and Z22 '
It is worthwhile observing that the same equation for the line-end faul t
current c a n be found directly by inspection of Fig. 1 0 . l l (a ), which shows t h e
Theveni n equivalent circuit between buses CD a n d W o f the prefa u l t network.
The impedances Zb and - 2b are con nected as shown in accordance with steps
1 and 2 above. Circuit analysis then shows in a straightforward manner that the
imp ed a n ce looking back into the circu it from the terminals of the open switch
S is
( 2 1 1 - 2 1 2 ) ( 222 - 22 1 - 2b )
+ 2 12
2 k k , new = 2b + 2 1 1 2 1 2 + 2 22 2 21 2 b
_
_
( 1 0 .25 )
_
S ince 2 1 2 = 2 2 1 and 2(h 1 2 = 2 1 1 + 2 22 - 2 2 12 , Eq. 00.25) can be reduced to
gIve
( 1 0 .26 )
Thus, simply by closing switch S as shown in Fig. 10. 1 1 ( b ) and using elementary
circuit analysis, we can ca l cu l a te the line-end faul t current I; in agreement with
Eq. (10.24). Of course, the circuit approach using the Thevenin equivalent , m ust
yield th e same results as the matrix manipulations of Eq. 00.22)-for the same
1 0.4
FA U LT CA LCU LATIONS U S I N G :Z bU,
EQUIVALENT CIRCUITS
®
401
s
p
-z
(a)
�
--' --- -----
ZH , ,, � w
=
Zi t
+
Zb
L--____
V{
=
1.0iQ:.
�
-__
.:..
_
Z 21-.)_
_
l
l_
_
Z
�
-
ZLh , 12 - Zb
___
®
L I J\'
! 1 t
S
P
(b)
Fl G U RE 1 0. 1 1
S i m u l ating t h e l i n e - e n d fau l t o f Fig. 1 0. 1 0 by Theve n i n equivalent c i rcu i t : ( a ) w i t h l i n e
open before t h e fau l t ; ( I» d u ri n g t h e fau l t ( S close d),
CD - (1)
external connections a rc being m a d e to the l a rger system model a s to its
Theve n i n equiva l e nt.
Other uses of the equ ivalent circuits based on the bus impedance matrix
are poss i b l e .
Exa m p l e 1 0 . 5 . In the five-bus system o f F i g . 1 0.8 a l i ne-end, short-circu i t fau l t
occurs on I inc CD - W , on the line side of the breaker at bus W . Neglecting
p rc fa u l t c u r re n t s a n d assu m i ng r a t e d sys t e m vo l t age a t the fault poi n t , calculate
the subtransicnt current into the faul t when only the ncar-end breaker a t bus @
opens.
Figu re 1 0.8 shows that the imped ance of line J) - (l) is Z b jO. 1 68 per
unit a nd the required elem ents of Z hus a re given in Example 10.4. The Thevenin
equivalent circuit looking into t h e intact system between buses ,CD a n d @
Solution.
=
402
CHAPTER 10 SYMMETRI CAL FAU LTS
corresponds to Fig. 1O.1 1(a). The numerical values of the im p e dan c e s shown i n
parallel are calculated a s follows:
Z11
222 - 2 2 1 - 2b
=
=
jO .0793 - jO .0558
=
jO .0235
jO. 1338 - jO.0558 - j0. 1 68
=
-
jO .09
The new Thcvenin impedance seen looking back i n t o t h e fa u l t e d system between
the fault point P and the reference is therefore g iven by Eq . ( 1 0.25) as
.
2H , n cw = 1 0 . 1 68 +
=
Thus,
( jO.0235) ( -jO . ( 9 )
( j0 .0235 jO .09)
_
+
j O.255() per u n i t
jO .0558.
l he s u b t ra l l s i c l l t c u r re l l t i l l t o t l l c l i l l c -c n J h u l t is
=
1; jO . 255 6
1
=
- j3.9 1 2 per uni t
10.5 THE SELECTION OF CIRCUIT
BREAKERS
The electric u ti lity company furnishes data to a customer who must determine
the fault current in order to specify circu i t breakers p roperly for an i nd ustrial
plant or industrial power distrib ution system connected to the u ti l i ty system at a
certain point. I nstead of providing the Theve n i n i mped ance of the system at the
point of connection, usually the power company i nforms the customer of the
short-circui t megavoltamperes which can be expected at nominal voltage; that
IS,
S hort-ci rcuit
MYA
=
13 X ( no m i n a l kY) X I isc i
X
1 0 -- 3
( 1 0 .27)
where I Isc i in amperes is the r ms magni t u de of the short-ci rcu i t current i n a
three-phase fault at the connection point. Base megavoltamperes are rel ated to
base kil ovolts and base amperes I/basc i b y
Base
MYA
=
13 X ( base kY)
1 1 baSe I X 1 0 - 3
( 10 .2 8 )
l I se I in per unit
( 1 0 .29)
x
If base kil ovol ts equ al nominal kilovol ts, t h e n divi d i ng Eq. 00.27) by Eq. 0 0.28)
converts the former to per u n i t , and we obtain
Short-circuit
MY A in per u ni t
=
�
At nomi nal vo l ta g e the Thevenin equivalent circuit l ooking back into the system
from the point of connection is an emf of 1 .0
per unit in series with the
per-u n i t impedance
Zth. Therefo re ,
1 .0
-- per u n i t
l Is e !
1 0.5
THE SELECTION OF CIRCUIT BREAKERS
403
u n d e r short-circuit cond itions,
1 .0
=
short-circuit
MVA
per u n i t
( 1 0 .30)
O fte n resistance and shunt capacitance are n eglected, i n which case Zth = Xth .
Thus, by specifying short-circu it megavoltamperes a t the customer's bus, the
e l ectric u t i l i ty is effectively describing the short-ci rcu i t curre n t at nominal
vol tage and the reciprocal of the Th e venin impedance of the system at the point
of connection.
M uch study has been g iven to ci rcu i t-b reake r ratings a n d applications, a n d
our d iscuss ion h e re gives so me i n troduct ion to t h e subject. T h e p resentat ion is
n o t i n t en d e d Zl S a s t u cl y of b r e a k e r a p p l i ca t i o n s b u t , ra t h e r, to i n d i c a t e t h e
i m p o r t an c e o f u n d e rs t a n d i n g fa u l t c Zl l c u l a t i on s . For additional gui dance in
s p e c i fy i n g b re a k e rs t h e (T i l ( i e r s h o u l d c ( ) l l s u l t t h e A N S I p u b l i c a t i o ns listed in
t h e footnot es w h i c h a cc o m p a n y t h i s s e c t i on .
From t h e current viewpoi n t two factors to be considered i n selecting
c i rc u i t b re a k e rs a r e :
•
•
The maximum instantaneous cu rr e n t which the breaker m ust carry ( withstand )
and
Th e total cu rrent when the breake r contacts part to in terrupt the circui t.
Up t o this point we h ave devote d most o f our attention to t h e subtransient
current call ed t h e initial symmetrical current , which does not i nclude t h e d e
component. Inclusion of the dc com ponent res u l ts i n a rms val u e of c u rrent
immediately aft e r the fau l t, which is h igher t h a n the subtransient current . For
oi l circu i t b reakers above 5 kV t he subtran s i e nt c u rrent m u l tipl ied by 1 .6 is
considered to be the rms v a l ue of the current whose d isruptive forces the
bre a ke r must withstand during the fi rst half cycle a fter the fa u l t occurs. This
c u r r e n t i s ca l l e d t h e momentary current , a n d for many years c i rcu i t breakers
we re ra t e d i n t e rms o f t h e i r mo m e n t a ry c u r re n t as w e l l Zl S o t h e r criteri a . I
T h e ill t amptill/{ ra t il/g o f a c i rc u i t b re ,l ke r w a s s I1 ec i fi e d in k i lovo l tam­
peres or m e gavol t a m pe re s . T h e i n t e r rupting k ilovoltamperes equal 13 X ( the
k i l ovo l t s of the b u s to w h i ch the b rc , l k e r i s c o n n e c t e d ) X ( t h c c u rr e n t wh ich the
b r e a k e r m u s t be c l ]J a b l c o r i n t e r r u p t i n g w h e n i t s c o n t a cts p a r t ) . This interrupt­
ing c u rren t i s , o f cou rse , l ow e r t h a n t he m o m e n t a ry C U ITe n t a n d d e p e n d s on t h e
spe ed of the breaker, such as 8, 5, 3 , o r 2 cycl es, w hi ch is a m easu re of t h e t i me
from the occu rrence o f t h e fa u l t to t h e ex tinct ion of the a rc . Breakers of
different speeds are class ified by t h e i r rated interrupting times. The rated
l See
Lester, " H igh Yoltilge Ci rcu i t Bre a k e r S t a n d a rds i n t h e USA: P a s t , Prese n t, a n d
IEEE Transactions on Power Apparatus a n d Systems, v o l . 93, 1 974, p p . 5 90,600.
G . N.
Fu t u re , "
404
CHAPTER 10
SYMMETRICAL FAULTS
Extinction of
arc on
Initiation of
primary contacts
short ci rcu it
Energization
of t ri p circuit
Parti ng of
primary arcing
contacts
Time �
;
Inte rrupti ng time
{
•
Tripping Opening
time
d e l ay
(
I
•
)
Arcing
time
�
)
Contact
parting ti m e
(
l
FIG U RE 1 0. 1 2
D e fi n i t io n
of
i n t e r ru p t i n g
time
give n
tion Guide for A C High VO/lage Circuit
Rated Of! a Symmetrical Currell t Basis .
In
ANS I / I EEE S t a n d a r d C.37.0 1 0 - 1 979, Applica ­
Bre a k ers
interrupting time of a circuit breaker is the period between the instan t of
energiz ing the trip circuit and the arc extinction on an opening operation, Fig.
10. 12. Preceding this period is the tripping delay time, which is usu ally assumed
to b e � cycle for relays to pick up.
The current which a breaker must inte rrupt 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 cu rrent
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 m ay also be made on the basis of total current (dc component
inclu ded). 2 W e shall l im i t o u r discus s ion t o a brief treatment of the symmetrical
basis of breaker selection.
a Symmetrical Current Basis, ANSI C37.06 - 1 987, a n d Guide for Calculation of Fault Currents for
Application of A C High-Vo/tage Circuit Breakers Rated on a Total Current Basis, ANSI C37. � - 1 979,
2See Preferred Ratings and Related Required Capabilities for A C High-Voltage Circuit Breakers Rated
on
American National Standards Inst i t u t e, New York.
10.5 THE SELEcnON OF CIRCUIT BREAKERS
405
Breakers are identified by nominal-voltage class, such as 69 kY. Among
oth er factors specified a re rated continuous current, rated maximum voltage,
vol tage range factor K, and rated short-circuit current at rated maximum
kilovol ts. Th e rated maximum voltage of a circuit breaker is the highest rms
vol tage for wh ich th e circuit breaker is designed. The rated voltage range factor
K is the rat io (rated maximum voltage -7- t he lower l imit of the range of
operating voltage). K determines the range of voltage over which the product
(rated short-circuit current X operating voltage) is con s t a nt . In the application
of circuit breakers it is important not to exceed the short-circu it capabilities of
the breakers. A br e ake r is req ui red to have a maximum symmetrical internlpting
capability equa l to K x rated short -circuit current. Between the rated maximum
voltage and 1/ K times the rated maximum voltage the symmetrical interrupting
capability is defined as the prod uct [rated short-circu it current X (rated m axi­
mum vol tage/operating voltage)].
Exa m ple 10.6. A 69-kV circuit b reaker having a voltage range factor K of 1 .2 1 and
a con ti nuous curre nt rati ng o f 1 200 A has a rated short-circuit current o f 1 9,000 A
at the m aximum rat ed voltage of 72 .5 kV . Determine the maximum symmetrical
i nt errupt ing capab il ity of the brea k er and expl ain its significance at lower operat ing
voltages.
Solution .
The max imum symmetrical i nterrupting capab i l i ty is given by
K x
rated short-circuit current
=
1 .21
x
1 9 ,000
=
22 ,990 A
This value of symmetrical i nterrupting current must not be exceeded. From the
definition of K we have
Lower l i mit of operating voltage
rated maximum voltage
=
K
=
72.5
1 .21
-
=
60 kV
Hence, in the operating voltage range 72.5-60 kV, the symmetrical interrupting
current may exceed the rated short-circu it current of 1 9,000 A, but it is lim ited to
22,990 A. For example, at 66 kV the interrupting curre n t can be
72 .5
66
-
x
1 9,000
=
20,87 1 A
B re a kers of the 1 1 S -kV class a n d h i gh e r have a K of 1 .0.
A simplified procedure for calcu J at ing the symmetrical
short-circuit cur­
rent , called the E/X method, 3 d isrega rds all resistance, all stat ic load, and all :
3 S e e Application Guide for A C High- Voltage CirCLIit B reakers Rated on a Symmetrical Current Basis,
is also
A NS I C37.0 1 O - 1 979, A m e ri c a n N at iona l S tand ards I nstitute, New York. This publication
I EEE Std. 320 - 1 979.
.
'
... - � . �
406
CHAPTER 1 0 SYMMETRICAL FAULTS
current. Subtransient reactance is used for generators in the E jX
method, and for synchronous motors the recommended reactance is the X� of
the motor t imes 1 .5 , which is the approximate value of the transient reactance
Xd of the motor. Induction motors below 50 hp are neglected, and various
multiplying factors are applied to the X� of larger i nduction motors depending
o n their size. If no motors are present, symmetrical short-circuit current equals
subtrans ient current.
The impedan ce by which the vol tage Vf at the fau l t is divided to find
short-circ u it current must be examined when the E jX method is used. In
specifying a bre aker for bus ® , this impedance is Zk k of the bus impedance
matrix with the proper machine reactances si nce the short-circuit current is
expressed by Eq . ( 1 0. 1 9). If the radio of XjR of t h i s i mpedan ce i s 1 5 or le ss a
breaker of the correct voltage and kilovoltamperes may be used if its i n terrupt­
ing current rating is eq ual to or exceeds the cal culated current. If t h e Xj R r a t i o
is unknown, the calcul ated curren t should be no more than 80O/C of the allowed
value for t h e bre aker at the existing bus voltage. The ANSI application g uide
s pecifies 'a corrected method t o accou nt for ac and dc ti me constants for the
decay o f the current amplitude i f the XjR ratio exceeds 15. The corrected
method also considers breaker speed.
prefault
,
10.7. A 25,000-kVA 1 3.8-kV generator with X;; = 1 5 % is con nected
through a transformer to a bus which s upplies fou r identical motors, as shown in
Exam ple
Fig. 1 0 . 13 . The s ubtransient reactance X; of each motor i s 20% o n a base of 5000
kVA, 6.9 kV. The three-phase rating of the transformer is 2 5 ,000 k YA, 1 3 .8/6.9
kV, with a leakage re actance of 10%. The bus voltage at the motors is 6.9 kV when
a three-phase faul t occurs at point P . For the fau l t specified , determine (a) t h e
subtra nsien t current i n t he fault, ( b ) t he subtra nsicnt curre nt i n breaker A , and CC)
the symmetrical short-c ircuit interru pting current ( a s defined fo r circuit-breaker
appl ica tions) i n the fault and in breaker A .
(a) For a base o f 25,000 kV A, 1 3 .8 kV i n the generator circui t the base
for the motors is 25,000 kVA, 6.9 kV. The subtransicnt reactance o f each motor is
Solution.
X"d
=
0 .20
25 ,000
5000
=
1 .0 per u n it
Figure 1 0. 14 is the diagram with sub transient values of reactance marked. For a
FIGURE 10. 13
One-l i n e
diagram for Exa mple 1 0,7.
THE SELECfION OF CIRCUIT BREAKERS
1 0.5
407
FIGU R E 10. 14
Reactance d i agram for Example 10.7.
fault at P
VI =
J .O�
I .()�
/" J -
The base current i n the 6 .9
jO . 1 25
circuit is
kY
i Il i
(b)
=
-jR .n per
jO . 1 25
u
per
nit
unit
25 ,000
= 2090 A
3 X 6. 9
f3
=
and so
Zlh
per u n i t
2090
8 X
=
=
16,720 A
Through breaker A comes the contribution from the generator and three
of the four motors. The generator con tributes a current of
- j 8 .0 X
0.25
0 .50
--
=
-j
4.0 per unit
Each motor contributes 25% of t he remaining fault current, or
amperes each. breaker A
J"
(c)
=
-j4 .0
To
+-
� ( - j l .O)
=
-j 7 0
.
per u
ni t
or
7
x
2090
=
-
j l .O
per-unit
1 4 ,630 A
the
j 1 .5 in the motor
c o m p u t e t h e c u r re n t to be i n t e rr u p t e d by b re a k e r A . r e p l ace
c n
j l .O
circuits of Fig. 1 0. 1 4. Then,
s u b l r a n s i e n t re a t a c e of
Zlh
=
j
b y t h e t ra ns i e n t r e ac t a nce of
0 .375
0 . 375
X
+
0.25
0.25
The generator contributes a cu rrent of
1 .0
jO. 1 5
--
x
0.375
0.625
=
jO . l S
per unit
-j4.0 per unit
·
408
CHAYfER 10
SYMMETRICAL FAU LTS
Each motor contributes a current of
1
- X
4
1 .0
-jO.15
X
0 .25
-0 . 625
=
-jO.67
per unit
The symmetrical short-circuit curren t to he interrupted is
( 4 0 + 3 X 0 .67) X 2090
.
1 2,560 A
=
Suppose that all the breakers connected to the bus are rated on t h e basis of t h e
current into a fault on the bus. In t hat case the short-circuit curren t interrupting
rat ing of the breakers connected to the 6.9 kY bus must be at l east
4 + 4 X
or
0 . 67 = 6 .67 per unit
6 . 67 X 2090
1 3 ,940
=
A
.
A 14.4-kV circuit breaker has a rated maximum voltage of 1 5.5 kY and a K
of 2 .67. At 15.5 kY its rat ed short-circuit interrupting current is 8900 A. This
breaker is rated for a symmetrical sh ort-circuit interrupting current of 2 6 7 X 8900
23, 760 A, at a voltage of 15.5/2.67 5 8 kY. This current is the maximum that
can b e interrupted even though the breaker may be in a circui t of l ower voltage.
The short-circuit interrupting current rating at 6.9 kY 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 t h e
breaker i s suitabl e with respect to short-circuit current.
The short-circui t current cou l d have been found by usin g the bus
impedance matrix. For this purpose two buses CD a n d @ a re identified in Fig.
10. 1 4 . Bus CD i s o n the low-vo l tage s i d e o f t h e t ra ns former and b u s 0 i s o n
the high-voltage side. For motor reactance of
per unit
Y =
\I
Y1
2
The
= jl0
-
' 10 +
)
Y
22
=
j1.5 /4
1
-j l 0 -
1.5
-j 1 2
j 6 . 67
=
.
67
- j 16 .67
node admittance matrix and its inverse are
Y bus
=
CD:(l}[
CD
- j12.67
jI0 .00
@
j l0 .00
-jI 6.67
1
Z bus
CD
=
CD[j O. 150
@ j O.090
@
1
j O .090
jO . 1 1 4 ,
THE SELECTION OF CIRCUIT BREAKERS
10.5
[
1
)0. 15 - )° .09 CD
11--__-+__
Vf
=
l.O�
0
.
� J.o�
l l 4 - ) o. 9
__ L___�__
jO.09
Fig u re
Z !HIS o f
s,
tl 1 s,
®
__
FIG U RE 1 0. 1 5
Bus i m p e d a nce e q u i v a l e n t c i rc u i t fo r t h e
409
__
_
f'ig. 1 0. 1 4 .
1 0. 1 5 is t he n e twu r k corresponding to Z ou s a n d VI = 1 .0 per unit. Closing
5[ with 5 2 open represents a fau lt on bus CD .
The symme trical sho rt-circuit i n terrupting current in a three-phase faul t at
bus CD is
I')c
l .0
=
--
jO . 1 5
=
-j6 .67 per unit
w h ich agrees with our previous calculati ons. The bus impedance matrix also
gives us the voltage at bus Cl) w i th the fau l t on bus CD ·
and
V2
=
1 .0 -
1St Z 2 J
=
1 . 0 - ( -j6 .67) ( j0 .09 ) = 0 .4
since the admittance between buses
the fau l t from the transformer is
( 0 .4 - 0 .0) ( -jI 0)
CD
and
W
is
- j l 0, the cur rent i nto
- j4 .0 per unit
which also agrees with ollr p revious resu l t .
We a lso know immediately the sho rt-ci rcu it current in a th ree-phase fault
a t b u s W , which, by referring to Fig. 1 0. 1 5 with S I open and S2 closed, is
1sc
l .0
=
jO.1 14
=
-j8.77 per unit
This simple example ill ustrates the value of the bus impedance matrix
, where the
410
CHAPTER 1 0 SYMMETRICAL FAULTS
effects of a fault at a number of buses are to be studied. Matrix i nversion is not
necessary, for Zbus can be generated d irectly by compute� using the Z bu s
building algorithm of Sec. 8.4 or the triangular factors of Y bu s ' as explained I n
Sec. 8.5.
of the network of Fig . 1 O. 1 6( a )
have s y n ch ro no u s reacta nces X" I = X" 2 = j l 7 0 r>er u n i t (as marked) a n d
subtransient r e ac t a n ce s X� = X:; = j O .2S p e r u n i t . If a t h re e - phase s ho r t - c i rcu i t
faul t occurs a t bus
wh � n the :e is no load (al l b u s vol t ages e q u a l 1 .0 0° pe r
un it), find the i n i ti a l symmetrical (sub t r a n s i e n t ) c u rr e n t ( 0 ) i n t h e fau l t
l i ne
and ( c) the voltage a t b u s
Use t r i a n g u l a r fac tors of Y bus i n t h e
10.8. The generators at b u s es CD
Example
G)
CD G) ,
and
(1)
.
/
U.0m"
@.
-
calculations.
fig. I O. I ()( h),
Solution. Fo r the give n ra u l t co n d i t i o n s t I l e n e t wo r k l I a s t he s u h t r ; l Il s i e n t re a c t ;I I1CC
diagram
shown
factors
i ll
and
-0.1
tIl(': co r re spo n d i n g
[
h a s t h e l r i �I I l g ll b r
- O.�4304
- 0 .5
1
-j7.9
j3 .5
L
Y I )II'
u
. ][X J [OJ
1
Since the faul t is at bus 0, Eqs. (10. 1 9) through 0 0.21 ) show that the calculation s
involve column 3, Z�Js' o f the subtransient Z bu s ' which we n ow generate as follows:
-jl O
jl
jS
jO.2 ® )O.3333
1.0LQ:.
...L
(a)
-
FIGURE 10.16
l.O�
-j7 .9
j3.S
1
.
-j3 .94937
X2
X3
=
0
1
)0.2 ® )0.3333
j O.25
j O.25
(b)
t
-
The reactance d iagram for Example 1 0.8 with g en erators represented
behind Xd ; (b) e q u ivale n t cu rre n t sou rce i n parall e l w i t h X:;.
by:
(a) series voltage �ou rce
10.6
Solving, we obtain
1
= ---=j .
-j3 .94937
O 25 3 20 per
X3
and so
the e l e m e n t s
of Z b�s
given by
are
- 0.5
- 0.1
(u)
2:122..11
2
t hat
Acco r d i n g t o EC] .
2� ,
u
(
2
I" _ _ V
I
b
_
(c)
1_
_
jO . 2
During
I
0
=
"
jO.25320
2 33
= j O . 25 3 20 per u n i t
= jO . 1 1 2 1 t{ pcr u n i t
= j O . 1 3 782 p e r u n i t
jO .25 320
w ri t e
2 1 3 - 233
(
1
---
( b ) From Eq . ( 1 0.21) we c a n
D
u nit
( J 0. 1 9), t h e s u b t ra n s i c n t cu r r e n t i n t h e fa u l t i s
VI
" - _
J1 -
-
411
0
- 0 .44304
We And
][222I)3 ] [ ]
SUMMARY
Z �3
)
for
-
=
j3 . c)493 7 per u n i t
t h e current i n l i n e
jO . 1 3 782 - jO . 25320
j O .25320
V=, = Vr
W
( - 7."�)
%::.
t h e f(l u l t t h e vo l t a ge at b u s
I
)
0 - G)
= - j 2 . 27844 per u nit
(
)
is g iven by Eq . OU.20)
= 1 .0
jO . 1 1 2 I H
j O . 25320
as fol lows:
= 0 .55695 p e r u n i t
10.6
SUMMARY
The cu rrent flowing immed iate ly afte r a fau l t occu rs in a power network is
dete rmined by the impedances of the netwo rk components and the synchronous
machines. The initial symmet rical rms fault cu rrent can be determined by
representing each machi ne by its subtrans ient reactance in series with its
subtransient internal vo ltage . S u btransient cu rrents are larger than" the transient
,
.
412
CHAPTER 1 0
SYMMETRICAL FAU LTS
and steady-state currents. Circuit breakers h ave ratings determined by the
maximu m i nstan taneous current whi ch the breaker m ust withstand and later
interrupt. Interrupting currents depend on the speed of breaker operation. ·
Proper selection and application of circu it breakers should follow the recom­
mend ations of ANSI standards, some of wh ich are re ferenced in this chapter.
S implifying assumptions usually made in i n dust ry-based fault stud ies are:
•
•
•
•
•
All shunt connections from system buses to the reference node (neutral) can
be n eglected in the equivalent circu its representi ng trans miss ion lines and
transformers.
Load ·impedances are much larger t han those of network co mponents, and so
they can be neglected in system mode I i ng.
All b u ses of t he sys t e m have ra ted /nom i n a l vo l tage or I .()� pe r u n i t s o t h a t
no prcfaul t currents tlow in the network.
Syn chronous machines can be represe nted by vol tage of l . 0
per unit
behind subtransient o r transient reactance, depe nding on t h e speed of t he
circu it breakers and whether the mom en tary or interrupting fau l t curren t is
being calculated (ANSI standards should b e consulted).
The voltage-source-plus-series-impeda nce equivalen t circu it of each syn­
c hronous machine can be transformed to an equivalent current-source-plus­
shunt-im pedance model. Then, the shunt i mpedances of the machine models
represent the only shunt connections to the reference node.
�
The bus impedance matrix is most often used for fault current calcu la­
tions . The elements of Z bu s can be made available explicitly using the Z bus
building algorithm or they can 'be generated from the triangular factors of Y bus '
Equivalent circuits based �:m the elements of Z b u s can simpl i fy fault-current
cal culations as demonstrated in this chapter for the line-end fault.
PROBLEM S
10. 1.
A 60-Hz alternating voltage h aving a rms value of 1 00 V is applied to a series RL
circui t by closing a switch. The resistance is 1 5 n 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)
W h a t i s the
i n s t a n t a n e o u s va l u e
of t h e v o l t age w h i ch w i l l p r o d uce t h e
maximum dc component of current u pon closin g the swi tch?
What is the instantaneous val ue of the voltage which will result in the absence
of any dc component of c urrent u p on closing the switch?
(d) If the switch is closed when the i nstantaneous voltage is zero, find the
instantaneous current 0.5 , 1 .5, and 5.5 cycles l ater.
A generator connected through a 5-cycle circuit b reaker to a transformer is rated
1 00 MVA, 18 kV, with reactances of X� 1 9 %, Xd 26%, and Xd = 1 30 � . I t
(c)
10.2.
=
=
P R OBLEMS
1 0.3.
413
"
is operating a t no load and rated vol tage when a three-phase short circui t o ccurs
between the breaker and the transformer. Find (a) the sustained short-circu it
current in the breaker, ( b ) the initial symmetrical rms current in the breaker, and
(c) the maximum possible d c component o f the short-circuit current in the
breaker.
The three-phase transformer connected to the generator described in Prob. 1 0.2 is
rated 1 00 M YA, 240Y / 1 8 6 k Y X = 1 0%. If a three-phase short circ u i t occurs
on the high-voltage side of the transformer at rated vol tage and no load, find (a)
the initial symmetrical rms current in the transformer windings o n the h igh-volt­
age side and ( b) the initial symme trical rms current in the line on the low-vol tage
side.
A 60-Hz generator is rate d 500 M Y A , 20 k Y , w i t h X':, = 0.20 per u n i t. It supplies
a purel y resist ive load of 400 M W at 20 kY. The load is connected d irectly across
the terminals of t he generator. If all three phases of the load are short-circuited
s i m u l t a neou sly. fi nd t h e i ni t i a l sy m m e t r i c a l rms current in t h e generator i n per
u nit on a base of SOO M Y A, 20 k Y .
A g e ne r a to r is c onn e c t e cJ t h r o u g h a t ra n s fo r m e r to a s y n c h ro n o u s m ot o r. Reduced
to the same base, the per-unit subtransient reactances of t h e generator and motor
a r e 0 . 1 5 and 0.35, respect ive l y, and the leakage reactance of the transformer is
0 . 1 0 p e r u n i t . A t h r ee p h a s e fa u l t occu rs a t t he ter m i n (l l s of the motor when the
ter m i nal voltage of t h e ge n e ra t o r is 0 . 9 p e r u n i t a n d the o u t p u t current o f the
ge n e r a t o r is 1 .0 per u n i t at 0 . 8 powe r factor l e a d ing. Find th e subtransient current
in per u n it in t h e fau l t in t he generator, and in the motor. Use the terminal
voltage of the generator as the reference phasor and obta in the sol u tion (a) by
computi ng the voltages beh ind subtransient reactance in the generator and motor
and (b) by using Thevenin's theore m.
Two synchronous motors having subtransient reactances of 0.80 and 0 . 2 5 per u nit,
respectively, on a base o f 480 Y , 2000 k Y A are connected to a b u s. Th is motor is
connected by a lin e having a reactance of 0.023 fl t o a bus of a power system. At
t he power system bus t h e short-circui t megavoltamperes of the power system are
9.6 M Y A for a nomi nal vol tage of 480 Y . When the vol tage at the motor bus is
440 Y , neglect load current and find the initial sym metrical rms current i n a
three-phase fault at t he motor bus.
T h e bus impedance m a t r i x o f a four-bus network with va lues i n p e r unit IS
,
1 0 .4.
1 0.5.
-
,
10.6.
10.7.
Z bus =
jO . 1 5
jO .08
jO .04
j O .07
jO .08
jO . 1 5
jO .06
jO .09
jO.04
jO .06
jO . 1 3
jO .OS
jO .07
jO .09
jO .05
jO . 1 2
Generators connected to buses CD and 0 have their sub transient reactances
incl uded i n Z bu s ' If prefa u l t current is neglected , find the subtransient curren t in
per u n i t in 19c fault for a three-phase fault on bus @ . Assume the vol t � ge at the
fault is 1 .0L.2.:. per unit before the fault occurs. Find a lso the per-unit current
from generator 2, whose sub transient reactance is 0.2 per u n it.
414
10.8.
CHAPTER 10 SYMMETRICAL FAULTS
-
For the network shown in Fig. 1 0 . 1 7, find the subtransient c urrent in per u nit
from generator 1 and in line CD @ and the voltages at buses CD and G) for a
t hree-phase fault on bus CD . Assume that n o current is flowing prior to the faul t
a n d that t h e prefault voltage a t b u s @ is 1.0& per unit. Use the bus
impedance matrix i n the calculations.
FIGURE 1 0. 1 7
N etwo rk for P robs. 1 0 .8 a n d 1 0 . 9 .
: . 10.9.
10.10.
10. 1 1 .
10. 12.
10.13.
For the n etwork shown in Fig. 1 0. 17, determine Y bus and its triangular factors.
Use the triangular factors to generate the elements of Z b u s needed to solve Prob.
10.8.
If a three-phase fault occurs at bus CD of the network of Fig. 1 0.5 when there is
no load (all bus voltages equal 1 .0
per unit), find t h e subtransient current in
the fault; the voltages a t buses @ , ® . and @ ; and the current from the
generator connected to bus @ . Use e quivalent c ircu its based o n Z bu s of Example
1 0.3 and simil ar to those of Fig. 1 0 . 7 to illustrate you r calcu l a t ions.
The network of Fig. 1 0.8 has the bus i m pe d a nce m a t rix given in Exa mple l O A . I f a ·
short-circuit fault occurs a t bus @ o f t h e netwo rk w h e n t h ere i s n o load (al l bus
voltages equal 1 .0
per u n it), find the subtransicnt c u rrent i n t he fau l l, the
voltages at buses CD and G) . and the current from the generator con nected to
bus CD . Use equivalent c ircuits based on Z bu s and similar to those of Fig. 1 0.7 to
illustrate your calculations.
Z bus for the network of Fig. 1 0.8 i s g iven in Example 10.4 . If a l i ne-end
short-circuit fault occurs on line ® G) of the network on the l ine side of the
breaker at bus G) , calculate the subtransient current i n the fault when only t h e
near-end breaker at bus G) has opened. Use the equivalent c ircuit approach of
Fig. 1 0. 1 1 .
Figure 9.2 shows the one-line d iagram o f a single power network which has the
line data given in Table 9.2. Each generator connected to buses CD and @ has a
s ubtransient reactance of 0.25 per unit. M aking the usual fault study assumptions,
s ummarized in Sec. 1 0.6, determine for the network ( a ) Y bU5 ' ( b ) Z bus ' (c) the
s ubtransient current in per unit i n a t hree-phase fault on bus ® and Cd ) the
contributions to the fault current from line CD G) and from line @ G) .
&
&
-
-
-
PROBLEMS
10.14. A 625-kV generator with X'd
415
0.20 per unit is connected to a bus through a
circuit breaker, as shown in Fig. 1 0. 18. Connected through circuit breakers to the
same bus are three 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 ful l bad,
u nity power factor, and rated voltage, with the load equally divided among the
machines.
(a) Draw the impedance diagram with the impedances marked i n p e r unit on a
base of 625 kVA, 2.4 kY.
( b) Find the symmetrical short-circuit current in amperes, which must be i nter­
rupted by breakers A and B for a three-phase fault at point P. Simplify the
calcu lations by neglect ing the pre fault cu rrent.
(c) Repeat part ( b ) for a three -phase fau l t at point Q.
Cd) Repeat part ( b ) for a three-phase fau l t at point R .
FI G U RE 1 0. 1 8
O n e · l i n e d i a g ra m for P rob. 1 0 . 1 4 .
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 vol tage is
38 kV and the rated short-circu it current at that vol tage is 22 kA Find (a) t he
voltage b elow which rated short-circuit current does not increase as operat ing
vol tage decreases and the value of that current and (b) rated short-circuit current
at 34.5 kY.
10.15. A
CHAPTER
11
SYMMETRI CAL
COMPO NENTS
AND SEQUENCE
NETWORKS
One of the m os t p owerful tools for dealing wi th unbalanced polyphase circui ts
is the method of symmetrical components in troduced by C. L. Fortescue. !
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 componen ts are equal in
length , and the angles between adjacent phasors of the set are equal. Al though
the m ethod is applicable to any unbalanced polyp hase sys tem , w e con fine our
d iscussion to three-phase systems.
I n a three- phase system which is normally balanced, unbalanced fault
conditions generally cause u nbalanced c urren ts and vol tages to exist in e ach of
the p hases. I f the curren ts and vol tages are related by constant impedances, the
system is said to b e linear and the principle of superposi tion applies. The
voltage response of the linear system to the u nbalanced cu rrents can b e
determined by considering the separate responses of the individual elements to
the symmetrical components of the currents . The system elements of interest
are the machines, t ransformers, transmission lines, and loads connec ted to 6.. or
Y configurations.
I e. L . For t escue, " M ethod of Symmetrical Coord i n at es Applied to the Sol u t ion of Polyph ase
Networks, " Transactions of AlEE, vo l. 37, 1 9 1 8, p p. 1 027- 1 1 40.
4H l
1 1.1
SYNTH ESIS OF UNSYMMETRICAL PHASORS FROM TH EIR SYMMETRICAL COMPONENTS
417
In this chapter we study symmetrical components and show that the
response of each system element depends, in general, on its connections and the
component of the cu rren t being considered. Equivalent circui ts, called sequence
circuits , will be developed to refl ect the separate responses of the elements to
each current component. There are three equivalent circuits for each element of
the three-phase system. By organizing the individual equivalent circuits into
n e tworks accord ing to the interconnections of the elements, we arrive a t the
concept of three sequence net wo rks Solving the sequence networks for the fault
con d i tions gives symmetrical cu rrent and voltage components which can be
combined together to reflect t h e effects of the original unbalanced fault currents
on t he ove ra ll system .
Ana lys is by symme trical compon ents is a powerful tool which makes the
calculation of unsymmetrical faults almost as easy as the calculation of t hree­
phase fau lts. Unsymm etrica l faults are stud ied in Chap. 1 2.
.
S Y NTHES I S OF UNSYM M ETRI CAL
PHASORS FROM THEI R S Y M M ETRI CAL
COMPO NENTS
11.1
Accord ing to Fortescu e's t h eorem, th ree unba l anced phasors of a three-ph ase
system can be resolved into th ree balanced systems of phasors . The balanced sets
of components are:
Positive-sequence components consisting of three phasors equal in magnitude,
d isplaced from each other b y 1 200 in phase, and havi ng the same phase
sequence as the original phasors,
2 . Negative-sequence components consisting of three phasors equal in magni­
tude, displaced from each o ther by 1200 in phase, and having the phase
sequ ence opposite to that of the original phasors, and
3 . Zero-sequence components consisting o f t hree phasors equal in magn itude
and with zero ph ase d ispl acemen t from each other.
1.
It is customary w h e n solvi ng a p rob lem by sym metrica l components to
designate the th ree phases of t he syslem as a, b, and c i n such a manner that
the phase sequ ence of the voltages and cu rrents i n the system is abc. Thus, the
ph ase sequence of the pos itive-sequence components of the unbalanced p h asors
is abc, and the p h ase seq u e n ce o f the n e g a t i ve-se q u e n c e comp o n e n t s is acb . I f
the ori ginal phasors are voltages, they may b e designated Va ' Vh ' and VC . The
three sets of symmetrical components are designated by the additional s uper­
script 1 for the positive-sequence components, 2 for the negative-sequence
components, and 0 for t he zero-sequence components. Superscripts are chosen
so as not to confuse bus numbers with sequence indicators later on i n this
chapter. The positive-sequence components of Va ' Vh ' and Vc are Va( l ) , Vb( I ), and
Vc( l ) , respectively. Simi larly, the negative-sequence components are VY), VP),
,
418
CHAPTER 1 1
SYMMETRICAL COMPONENTS A N D SEQUENCE NETWORKS
y (l)
y (2 )
a
a
VP) -------{
Zero-sequence
components
N egative-seq u ence
com ponents
P os itive-s equence
c omponents
FIGURE 1 1 . 1
Three sets of b a lanced p h asors which a re the sym m e t r i c a l compon ents of th ree u n b a l a nced ph asors.
and Vc(2 ), and the zero-sequence components are Va(O ), V� O ), and Ve(O), respec­
tively_ Figure 1 1. 1 shows three such sets of symmetrical components. Phasors
represen ting currents will be design a ted by J with superscripts as for voltages.
S in ce each of the original unbalanced phasors is the sum of its compo­
nents, the original phasors expressed in terms of their components are
va
vc
=
Va(O) + Va( l ) + Va(2 )
(11 .1)
( 1 1 .2)
=
V (O) + V ( I ) + V (2)
c
c
( 1 l .3)
c
The synthesis of a set of three unbalanced phasors fro m the three sets of
symm etrical componen ts of F i g . 1 1 . 1 is shown in F i g . 1 1 .2_
The many advantages of analysis of power systems by t he method of
symmetrical components will become apparent gradually as we apply the
method to the stu dy of unsymmetrical fa ults on otherwise sym metrical systems.
It is sufficient to say here that the method consists i n finding the symmetrical
compon e n ts of c u rre n t a t the fa u l t . Th e n , t h e v a l u e s o f c u r r e n t a n d vol t a g e at
various poi n ts in the system can be fou n d by means of the bus impedance
matrix. The method is simple and leads to accu rate predictions of system
b ehavior.
1 1 .2 THE SYMMETRICAL COMPONENTS
OF
UNSYMMETRICAL PHASORS
In Fig. 1 1 .2 we observe the synthesis of three u nsym metrical phasors from th ree
sets of symmetrical p h a so rs . The syn t h e s i s is made in accor d a n ce w i t h EQs.
1 1 .2 THE SYMMETRICAL COMPONENTS OF U NSYMMETRICAL PHAS ORS
VCl( O l
Vc( 2)
vc( l
VCl(2
419
)
)
FI G U R E 1 1 .2
showll ill Fig. 1 1 . 1
G raphical a d d i t i o n of t he compo nents
to o b t a i n three u n ­
balanced phasors.
( 1 1 . 1 ) t h rough ( 1 1 .3). Now l e t us examine t hese same equations to determ i ne
h ow to resolve t h ree unsymmetrica l phasors i nto the i r sym metrical compone nts.
F i rst, we note that the n u m be r of unknown quantities can b e reduce d by
exp ressing each componen t of Vb and Vc as t h e p roduct of a compon e nt o f Va
and some function of the ope rator a = 1
, which was i ntroduced i n Chap.
1 . Refe re nce to Fig. 1 1 . 1 verifies t he fol lowing relations:
�
V (O)
=
b
VhOl
Va(O)
V ( O)
c
V(l)
=
=
C
=
Va(O)
a V(I)
( 1 1 04)
11
a V (2)
a
R C f) C (l t i n g Fq . ( 1 1 . 1 ) a n d s u bs t i t u t i n g Eqs. ( I J . 4 ) i n Eqs. ( 1 1 . 2 ) a n d ( 1 1 .3 )
V( I )
/I
+
yield
( 1 1 .5 )
( 1 1 .6)
( 1 1 .7)
or in m a t r ix form
V
a
Vb
Vc
=
1
1
1
1
a2
a
Va(O )
Vel)
a2
)
V (2
1
a
a
a
=
A
V eO )
a
)
V(l
a
2)
V(
a
( 1 1 J8)
420
CHAPTER 1 1
SYMMETRICAL COMPON ENTS A N D SEQUENCE NETWO R KS
where, for convenience, we let
( 1 1 .9 )
Then, as m ay be verified easi ly,
( 1 1 . 10)
a nd premul tiplying both sides
Va(O)
V(l)
Va(2)
a
1
-
3
0
f
Eq. ( 1 1 .8) by
1
1
1
1
a
a2
1
a2
a
A-
Va
Vb
Vc
I yic I d s
A- I
=
Va
Vb
( 1 1 .11)
Vc
which shows u s how to resolve three unsymmetrical phasors into their sym metri­
cal components. These rel ations are so i mportant that w e write the separate
equ ations in the expanded form
V(a O)
Va( l )
Va(2)
=
=
-
�3 ( V +
( 1 1 . 12 )
II
�3 ( V + a Vh + a 2 V )
( 1 1 . 13)
+ a 2 v.,) +
( 1 1 . 14)
a
�
3
(V
a
C
,/
2
I f required , the components v,(0)
b , Vb( l ) ' Vb(2)' Vc( O )' Vc( l ) ' and V ( ) can b e fou nd by
Eqs. ( 1 1 .4). Similar resul ts apply to l ine-to-line voltages simply by replacing Va '
Vb ' a nd � in above e q u a t ions by Vab , V"c , a n d VC t" respectively.
Equ ation ( 1 1 . 12) shows that no zero-sequence components exist if the sum
o f the unbal a nced ph asors i s zero. Since t h e sum of t h e line-t6-line voltage
p hasors i n a three-phase system is alw ays zero, zero-sequ ence compon ents are
never present i n the l ine voltages regardless of the degree of unbal ance. The
s u m of the three line-to-line neutral volt age p hasors is not necessarily zero, and
voltages to neutr al may cont ain zero-sequence components.
The preceding equ ations could h ave been written for any set of rel ated
phasors, and we m ight have written them for cu rrents instead of for vol tages.
They m ay be solved either analytica l ly or grap hic(l lly. Because some of the
C
1 1 .2 THE SYMMETRICAL COMPONENTS OF UNSYMMETRICAL PHASORS
421
preceding equations are so fund amental, they are summarized for currents:
Ia
Ib
Ic
J (O)
a
Ja( 1 )
J a( 2 )
=
1a(0 ) +
I� O)
+
a 2 1a( 1 )
+
aI(2)
a
1a(0) +
a 1(a 1 )
+
a2I(2)
a
=
=
1a(2)
1 a( 1 ) +
=
1..3 ( Ia +
=
t ( la
=
1..
:I
(1
+
IJ
Ib +
alb + a2/c )
+
u
( 1 1 . 15)
( 1 1 . 1 6)
a 2 Ib + alc)
Finally, these results can be extended to phase currents of a � c ircuit [such as
t h a t of Fig. 1 1 .4( a )] by rep lacing 111 ' 1,." and 1, by I,,/) , I;" . and ICIl ) respect ively.
Exa m p l e 1 1 . 1 . One conductor o f
t hrec -phnse l i n e is o p e n . The current fl ow i n g to
the 6.-connccled load t h ro u g h l i n e ( [ is l O A . W i t h the curre n t in l i ne (/ as
rcfcn:nce and assu m i n g that l i n e c i s o pe n f i n d the sym m e t r i c a l c o m po n e n t s o f
the line cu rrents.
a
,
Solution. Figu re 1 1 . 3 is a di agram of the circu i t . T h e line currents are
1c
Fro m Eqs.
0 1 . 1 6)
l�O)
1� I)
=
=
=
1� 2 )
=
=
H 10L2:
H 1 0L2:
5
-
j2 .89
HlOLo
5
+
+
+
=
+
j2 .89
=
1 0�
10L 1800
+
+
5 .78L - 300
10/ 1800
+
5 . 7 8�
0)
=
1 200
0
+
0)
A
2400
+
A
a --------�
c ----
=
FIGURE 1 1 .3
Circuit for Exampl e 1 1 . 1 .
0)
OA
422
CHAPTER 1 1
SYMMETRICAL COMPONENTS A N D SEQUENCE NETWORKS
From Eqs. ( 1 1 .4)
[g I )
The result I�U)
If!.)
= 5.78/ - 1500
=
S.7H� A
= J�U) = I�O) =
leO}
c
=0
A
IP)
=
5.7B/ - <JO° A
0 holds for any t h ree-wi r e sys t e m .
In Example 1 1 . 1 we note t h a t components 1� ' ) and 1�2) have nonze ro
values although line c is open a n d can carry no net current. As is expected ,
therefore, t he sum o f t h e components in l ine c i s ze ro. O f cou rse, t h e s u m o f
t h e components i n line a i s 1 0� A, a n d t h e s u m o f the components i n line b
is 10� A.
1 1 .3
SYMMETRICAL Y AND d CIRCUITS
In three-phase systems circuit elements are connected between l ines a , b , a n d c
in either Y or fl. configuration. Relationships between the symmetrical compo­
n e nts of Y and fl. currents and volt ages can be established by referring to
Fig. 1 1 .4, which shows symmetrical i mped ances connected in Y and fl. . Let us
agree that the reference phase for fl. qu antities is branch a -b . The p articular
choice of reference phase is arbitrary and does not affect the results. For
currents we h ave
( 1 1 . 17)
a �--�----�
+
Zy
+
C ---1_---'-______
FIGURE 11.4
(a)
-----'
Symmetrica l impedances: (a) �-co n nected; ( b) Y -co n nected.
(6)
1 1 .3 SYMMETRICAL Y AND A CIRCUITS
423
Adding all three equations together and invoking the definition o f zero­
sequence current, we obtain I�O) ( Ia + Ib + 1)/3 0 , which means that line
currents into a D.. -connec(ed circuit have no zero-sequence currents. Substituting
components of current in the equation for Ia yields
=
=
=
( J�g)
-
o
15�»)
+
( /�b)
15�»)
-
+
( /!�)
-
11�»)
( 1 1 . 1 8)
Evidently, if a nonzero value of circulating current 1��) exists i n the D.. circu it, it
cannot be determined from the l ine currents alone. Not i ng that 15� ) a"�h) and
1 ��) a 211�t), we now write Eq. ( 1 1 . 1 8) as follows:
=
=
1(1)
a
+
f (2)
a
=
(1
-
(1)
a ) 1 ab
+
(1
- a2
) laC�)
�
( 1 1 . 1 9)
A s i m i l a r equat ion for phase b is 1 h( 1 ) + 1b( 2) = ( 1 - a ) Ib( eI ) + ( 1 - a 2 ) Jb(2)
e , and
2
1
2
2
(
1
1
express ing 1b( 1 ) ' 1b( ) ' /b( e) ' and 1 be
( ) in terms' of 1a ) ' 1(2) ' 1(1)
a D ' and a( b) ' w e obtain a
result ant equat ion which can be solved along with E q . ( 1 1 . 1 9) to yield the
important results
a
ja( 1 )
=
13 /
-
30°
X
1a( b1 )
( 1 1 .20 )
These results amount to equ ating cu rrents of the same sequence in Eq. ( 1 1 . 1 9).
Complete sets of posi t ive- and negative-sequence components of currents are
shown in the phasoT diagram of Fig. 1 1 . 5 ( a).
I n a sim ilar. manner, the li ne-to-line vol tages can be written in terms of
l ine-to-neutral voltages of a Y -connected system,
vu f)
V" ,.
v;,· u
=
:co
=
Vo n
-
V,,,, -
v:."
-
Vb n
v: "
( 1 1 .21)
�,"
Ad d i n g together a l l three equat ions shows that V:,(j/) e V:,h + V"c + V,, )/3 = O.
J n words, lille-Io-line uolluges l/{fpe 1 1 0 zero-sequence comp on e nt s . Substituting
components of the vol tages in the e quation for Vu h yields
=
V0e0l )
+ V ( 2)
at>
=
( VeOl
an
+
()
V(anI ) + V ez»)
Ill!
-
( V(O)
bll
+ V ( l ) + Vb('�»
on
,
424
CHAPTER I I
SYM METRICAL COMI'ON ENTS AND SEQUENCE N ETWO R KS
/ (1 )
e
/ (1 )
ea
(2
1be
)
[(2 )
a
Y-----
/ (1 )
b
[( l )
[��)
[ (' 2 )
be
ca
Positive-sequence
components
Neg ative-sequence
components
(a )
v(l)
ab
va( n\ )
Ol
Vea
VbOl
( l)
ven
)
vbe( l )
2)
Va(n
Positive-sequence
components
va(b2 )
Negative-sequence
com pon e nts
(b )
FIGURE 1 1 .5
Posit ive- a n d nega t ive-sequence components of (a) l i n e a n d d e l t a currents a n d ( b ) l i ne-to- I i n e a n d
line-to- n e u tral voltages o f a t h ree-phase syst e m .
Therefore, a nonzero value df the zero-sequence vol tage v}�) can not be deter­
mined from the l ine-to-line voltages alone. Separating positive- and negative­
sequence quan tities in the manner exp l ai n ed for Eq. (1 1 . 1 9), we obta in the
importa n t voltage relations
VaCI)
b
=
( 1 - a 2 ) V C1 )
an
( 1J. .23 )
I I .J
SYMMETRICAL Y A N D 6 CIRCU ITS
425
Complete sets of positive- and n egative-sequence components of voltages are
shown in the phasor d iagrams of Fig. 1 1 .5(b). If the voltages to neutral are in
per u ni t referred to the base voltage to neu tral and the line voltages are in per
unit referred to the base vol tage from l ine to l ine, the 13 multipliers must be
omi tted from Eqs. ( 1 1 .2 3 ). If both vol tages are referred to the same b ase,
however, the equations are correct as given. Similarly, when l ine and 6. currents
are expressed in per unit, each on i ts own base, the 13 in Eqs. ( 1 1 .20)
disappears since the two bases are related to one another in the ratio of f3 : l .
When the currents are exp ressed on the same base, the equation i s correct a s
written.
From Fig. 1 1 .4 we note that �"jlab = Z 6. when there are no sources or
mutual cou p l i ng inside the c. c i r cuit. When positive- and negative-sequence
q u a n t i t i es are both prese n t , we have
1 (1)
VII(hI )
=
IIi>
S u bst i t u t i n g from
Eqs. ( J J .20)
a n c.J ( J J . 2 3 ),
13- Vu(nI )�
13 �
1(\)
II
_
(I)
- Z6. -
Va(nl )
so t hat
Z:-,
=
---
fa
Z().
-
3
Va(i,2 )
( 1 1 .24)
1 (2)
a i,
we
obta i n
13 V}�)/
1a( 2 )
13
_
-
30°
/ - 300
5
v ( 2)
( 1 1 .2 )
an
1a( 2 )
which shows that the c. c on nec t e d i mped ances Z6. are equivalent to the per­
phase or Y -co n nected impedances Z y = Z6./3 of Fig. 1 1 .6(a) insofar as posi-
(a)
(b)
FI G U RE 1 1 .6
( (I ) S y m m e t r i c ; 1 i CI - cO I l Il e c t e d i m r e d ; l!l ccs and t h e i r V-co n n e c t e d e q u iva l e n ts re l a t e d
( b ) Y - c o n n e c t e d i m p e d a nces w i t h n e u t ra l c o n n e c t i o n to g ro u n d .
by
'
2y
=
26/3;
426
CHAPTER 1 1
SYMMETRICA L COMPONENTS A N D S EO U EN C I : N ETWO R KS
tive- or negative-sequence currents are
have been anticipated from the usu a l
Of
concerned.
course, this resul t could
/1 -Y transformations o f Tabl e 1 .2. The
Z/l/3 is correct whe n the impedances Z/l and Zy are both
Zy
expressed in ohms or in per unit on the same kilovo)tampere and voltage bases.
relation
=
1 1 .2. Three identical Y -con ne cted resistors form a load bank wi th a
three-phase rating of 2300 V and 5 00 kVA. If the load bank has applied vol tages
Exa mple
I V',c l
=
I V:'11 I
2760 V
= 230() V
find the line voltages and currents i n per unit i n t o the load. Assume that the
neutral of the load is not connecte d to the neutral o f the system a nd select a base
of 2300 V, 500 kV A.
The r a t i n g of the l o a d bank c o i n c i d e s w i t h the s p e c i f i e d base, and s o t he
resistance values are 1 .0 per u nit. O n the same base the g ive n line vo l tages i n per
unit are
Solution.
I
Vca I
=
1 .0
Assuming an angle of 1 80° for Vc a a n d using the law of cosines to fi nd
of the other l ine vol tages, we find the per-u n i t values
Vab
=
/ 82 .8°
=
=
=
1 (0 . 1003 + jO .7937
0 . 2792
angles
0 .8
The symmetrical components of the l ine vol tages
=
the
+
� (0 . 1 003
jO .9453 =
+
jO .7937
- 0 . 1 790 - jO .l51 7
+
0 . 2 3 72 + j l . 1 7 63
O.9t{S7/ 73 '()o
0 .2346
+
0 .5
+
jO . 8(60)
p C I' u n i t ( I i n c - to - l i n e
- 1 . 1 373 - jO . 3 8 28
=
a rc
/ 220 .30
+
O .S
--c
v o l t age base )
jO .8660)
per u n i t ( line-to-l ine voltage base )
The absence of a neutral connection means that zero-sequence currents a rc not
present. Therefore, the p hase voltages at the load contain positive- and negative­
sequence components only. The p hase voltages a re fou nd from Eqs : ( 1 1 .23 ) with
the {3 factor omitted since the line voltages are expressed in terms of the base
voltage from line to line and the p hase voltages are desired in per u nit of the' base
1 1 .4 POWER IN TERMS OF SYMMETRICAL COMPONENTS
427
voltage to neutral . Thus,
=
==
0.9857/ 43 .6° per unit (line-to-neutral voltage base)
/ 250 .30
0 .2346
per unit (Iine-to-neutral voltage base)
S i ncc e a c h resistor has an i m p e d a nce
J�I )
/ (2 )
a
=
v(l)
1 . 0/�
V (2 )
=
1 .0L.Q:
�
=
=
of 1 .0LQ:.
0.9857L
pc r u n i t ,
43 . 6° p e r u n i t
O.2346i.3S0.3°
per
unit
The pos itive direction of current is chosen to be from the supply toward the load.
1 1 .4
POWER IN TERMS OF SYMMETRICAL
COMPONENTS
If the sym metrical components of current and voltage are known, the power
exp ended in a three-phase circuit can be computed directly from the compo­
nents. Demonstration of this statement is a good example of the mat rix
manipulation of symmetrical components.
The total complex power flowing into a three-phase circui t through three
lines a , b, and c i s
S·3</.
"- ,
jJ
+ JI)
�: =
V" ! "*
+
V" I I>*
+
Vt · jC*
( 1 1 .26)
where Va ' Vb ' and Ve are the vol tages to reference at the terminals and la ' lb '
and Ie are the currents flowing into the circuit in the three lines. A neutral
conn ection may or may not be present. If there is impedance in the neu tral
con nect ion to grou n d , then the vol tages V;/ ' VI" a n d V;. must be in terpreted as
vol tages fro m the l i ne to grou nd ra ther t h a n to n e u t r a l . I n matrix notation
( 1 1 .27)
428
CHAPTER 1 1
SYMMETRICAL COMPONENTS AND SEQU ENCE N ETWORKS
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 Eq. ( 1 1 .8) to obtain
( 1 1 .28)
] (0)
va( O)
a
Va( 2 )
where
and
( 1 1 .29)
(Z )
Ia
The reversal rule of matrix al gebra states that the transpose of the prod uct
of two matrices is equal to the p roduct of the transposes of the matrices in
reverse order. According t o this rule,
( 1 1 . 30)
( 1 1 .3 1 )
and so
Noting t h at AT
s 3 </>
=
=
A and that a
[ V (O)
and a Z are conjugates, we obtain
Va( l )
a
1
1
1
1
1
1
1
1
( 1 1 . 32)
1
a
or smce
1
1
1
a
534>
=
3
v(O)
a
o
1
o
1
1
V(l)
a
V ( Z)
a
1a( 0
)
U
0
1
*
( 1 1 . 33 )
So, complex power is
which shows how complex power (in voltamperes) can be computed from the
symmetrical components of the voltages to reference (in volts) and line currents
1 1 .5
SEQUENCE CIRCUITS OF Y AND A IMPEDANCES
429
(in amperes) of an unbalanced three-phase circuit. It is important to note that
the transformation of a-b-c voltages and currents to symmetrical components is
power-invariant in the sense d iscussed in Sec. 8.6, only if each product of
sequence voltage (in volts) times the complex conjugate of the corresponding
sequence current (in amperes) is multiplied by 3, as shown in Eq. ( 1 1 .34). When
the complex power S 3 1> is expressed in per unit of a three-phase voltampere
base, however, the multiplier 3 d isappears.
Example 1 1 .3 . Using symmetrical components, calcu late the power absorbed i n
the load o f Example 1 1 .2 a n d check the answer.
Solution.
In per unit of the three-phase 500-kYA b ase, Eq . ( 1 1 .34) becomes
Substitut ing the components
obtain
S ) <b
=
=
=
0 +
O.9S57[ 4 3 .�".
X
( 0 .9857) 2 + ( 0 .2346) 2
of
vol tages
0.9857[ . . 4 3 .6°
=
cu rre nts from Example 1 1 .2 , we
,md
+
0.2346/>50 .3"
X
0.2346/ - 250 .3"
1 .02664 per unit
5 1 3 .32 kW
The per-unit ya)ue o f the resistors in each phase of the Y-connected load b a n k is
1 .0 per unit. In ohms, there fore,
2
( 2300)
=
= 1 0 .58 D
R
Y
500 ,000
and the equivalent 6.-connected resistors are
R Cl.
=
3Ry
=
3 1 .74
D
From the given line-to-line vol tages we calculate directly
=
11. 5
.
( 1 840) 2
+
(2760) 2
3 1 .74
+
(2300) 2
- -
SEQUENCE CIRCUITS
OF Y AND �
'
IMPEDANCES
= 5 1 3 .33 kW
impedance Zn is inserted between the neutral and ground of the Y �co nnected
impedances shown in Fig. 1 1 .6( b ), then the sum of the line currents is equal to
If
430
CHAPTER 11
SYMMETRICAL COMPONENTS AND S EQUENCE NETWORKS
the current /n in the return path through the neutral. That is,
( 1 1 .35 )
Expressin g the u nbalanced l ine currents in terms of their symmetrical compo­
n ents gives
( /a( 0)
=
+
/b(0 )
+
1(0» )
c
+ ( / ( 1 ) + /b( 1 ) +
fI
1(1» )
c
�---- --�
o
3 1 (0)
+
( 1(2)
a
+
1 (2)
h
()
a
+
1 (2» )
c
(II
.36)
Since the positive-seq uence and nega tive-sequence cu rrents add sep a rately to
zero a t neu tral point n, there cannot be any positive-sequence or negative­
sequence currents in the connections fro m neutral to ground regardless of the
value of Zn ' Moreover, the zero-sequence currents combining together at n
become 3 /�0) , which produces the vol tage drop 3 1� 0) Z /l between neutral and
ground. It is important, therefore , to d ist inguish between voltages to neutral
and voltages t o ground under unbal anced cond itions. Let us designate volt ages
of phase a with respect to neutral and ground as Va ll and � , respectively. Thus,
the voltage of phase a with respect to ground is given by Va = Van + v,, , where
� = 31�0)ZTI ' Referring to Fig, 1 1 .6( b ), we can write the vol tage d rops to
ground from each of the l ines a, b, and c as
( 1 1 .37)
symmetrical compon ents a s fo l l ows:
The a-b-c voltages
a n d cu rr e n t s i ll
A
Va(O)
V}t)
t h is equation
be
1 (0)
rep laced by t he ir
a
=
ZyA
J! I )
( 1 1 .38)
1a(2)
Va(2)
M U ltiplying across by the inverse matrix
· v<a O )
c a ll
1(2)
a
A- I,
we obtain
1
+
3 1a(0)Zn A - I 1
1
1 1 .5
Postmultiplying
of A- I ,
1
A - 1 by [1
SEQU ENCE CIRCUITS OF Y AND A IMPEDANCES
'
I V amounts to adding the elements in each row
v (O )
1
Va( I )
a
and so
431
Zy
=
V( 2)
+
3 1a(0)Zn 0
( 1 1 .39)
o
a
In expanded form, Eq. ( 1 1 . 39) becomes three separate or decoupled equations,
v (O )
a
vel)
va( 2 )
a
( Z Y + 3Z
=
n
) 1(0)
a
ZY/ ( 1)
=
a
Z Y 1 (2 )
=
a
=
=
=
Z O 1(0)
a
( 1 1 AO)
Z I 1(J( 1 )
( l 1 A1 )
Z 2 /a(2)
( 1 1 .42 )
It is customary to use the symbols Z o , Z l ' and Z 2 as shown.
Equations ( l 1 AO) throug h ( 1 1 .42) could have been easily developed in a
less formal manner, but the matrix approach adopted here will be usefu l in
developing other important rel ations in the sections wh ich follow. Equati ons
( 1 1 .24) and ( 1 1 .25) combine with Eqs. ( 1 1 .40) through ( 1 1 .42) to show that
currents of one sequence cause voltage drops of only the same sequence in /:::,. ­
or Y -connected circuits with symmetrical impedances i n each phase. This most
important result allows us to draw the three single-phase sequence circuits
shown i n Fig. 1 1 .7. These three circuits, considered simultaneously, provide the
sam e information as the actual circuit of Fig. 1 1 .6 ( 6), and they are independent
of one another because Eqs . ( l 1 AO) through ( 1 1 .42) are decoupled . The c ircuit
of Fig. 1 1 .7( a ) is called the zero-sequence circuit because it relates t he zero­
sequence voltage Va( O ) to the zero-sequence current 1� O), and thereby s e rv e s to
define the impedance to zero-sequence curren t given by
v(J«( I )
1 (0)
=
(1
a �
+Vt iOl
-�
,0
J (O)
Zy
a
n
3 Z"
Zo
(a)
Reference
FI G U R E 1 1 .7
+t
VO)
"
J(l)
a
-----
Z'
I
Zo
Zy
=
2y +
3 2 1/
( 1 1 A3)
n
I
+
Reference
Reference
(6)
(c)
Z e r o - , posi t iv e - , a n d n e g a t i v e - s e q u e n c e c i r c u i t s for Fig.
l 1 .6(b).
432
CHAPTER 1 1
SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS
impedance to positive-sequence current , whereas Fig. 1 1 .7( c ) is t h e negath'e­
sequence circuit and Z 2 is t h e impedance to negative-sequence current . The
Likewise, Fig. 1 1 .7(b) i s called the positive-sequence circuit and Z I is called the
names of the impedances to currents of the d ifferent sequences are usually
shortened t o the less descriptive terms zero-sequence impedance 2 0 , positive­
sequence impedance 2 I ' and negative-sequence impedance 22 , Here the positive­
and n e g at iv e -s e qu e nc e i m p e d a n ces Z I < t n d Z2 ' r e s r c c t i v c l y, a rc /1 o t h fou n d t o
b e equal t o the usual per-phase i m p e d a n ce Z y , w h i c h i s ge n e r a l l y t h e c a s e for
stationary sy mme t ric a l circuits. Each or the three seque nce c i rc u i ts re p r e sen t s
one p hase of th e actual three-phase circuit when t h e l a t t e r c a r r i e s c u r r e n t o f
only t h a t seq u e nce . W h e n t h e t h r e e s e q u e nce c u r r e n t s a r c s i m u l t a n e o u s l y
p re s e n t , all three sequence c i rcuits a re n e e d e d t o fu l ly r e p r e se n t t h e o r i g i n a l
circuit.
Voltages in the positive-sequ ence and n e g at ive -s e q u e n c e c i rc u i t s can be
regarded as vol tages measured with res p e c t to e ither neutral or grou nd whet h e r
o r not there i s a connection of some finite value of i mpedance Zn b e tw e e n
neut r a l and grou nd. Accordingly, in t h e positive-sequence circuit there is no
d ifference between V ( I ) and V ( I ) and a s i m i l a r s t a t e m e n t a p p l i e s t o V ( 2 ) and
Va�) in t h e negat ive-sequ ence circu it. However, a voltage d i fference can exis t
between the n eutral and the reference of the zero-sequence circuit. I n the
circuit o f Fig. 1 l .7( a) th e current 1� 0) flowing through impedance 3Z n produces
t h e same voltage drop from neutral to ground as the current 3 1�0) flowi ng
through impedance Z n in the actual circuit of Fig. 1 1 .6(b).
If the neutral of t h e Y -conn ected circuit is grounded through zero
impedance, we set Zn = 0 and a zero- imped ance connection then joins the
neutral point t o the reference node of t he zero-sequence circuit. If th e re i s no
connection between neutral and ground, there cannot be any zero-sequence
current flow, for then Zn =
which is indicated by the open circuit between
neutral and the reference node i n the zero-sequence ci rcui t of Fig. 1 1 .8( a).
Obviously, a L1-connected circu i t cannot provide a path through n e utral,
and s o line currents flowi ng into a L1-con nected load or its equivalent Y circuit
cannot cont ain any zero-seq uence components. Consider t h e symmetrical L1 connected circui t of Fig. 1 1 .4 with
U II
U
a
,
00 ,
( 1 1 .44 )
Adding the three prece d i n g
equations together, we obtain
( 1 1 .45)
and
since the
sum of the
l i ne - t o- l i n e voltages is always zero, we therefore have
v <O)
ab
=
[a(0b)
=
0
( 1 1 .46)
Thus, in �-connected circuits with impedances only and no sources or m'u tual
coupling t h e re cannot be any circulating c u r r e n t s . S o m e t imes s ing l e -p h a s e
1 1 .5
1( 0)
�
SEQUENCE CIRCUITS OF Y AND A IMPEDANCES
433
Z
a +jl
:
y
I_ �
Reference
z,..
FI G U R E 1 1 .8
( a ) Un grounded Y-co nnected and
(b) 6-connected circuits and their
zero-sequence circ uits.
Reference
( b)
circula ting currents can be produced in the 11 circuits of transformers and
generators by either induction or zero-seque nce generated voltages. A 11 circuit
and its zero-sequence circuit are shown in Fig. 1 1 .8(b). Note, however, that even
if zero-sequence voltages were generated in the phases of the 11, no zero­
sequence vol tage could exist between the 11 term inals, for the rise in vol tage in
each phase would then be matched by the vol tage drop in the zero-sequence
impedance of each phase.
Three equal impedances of j21 .n are connected in .1 . Determine
the sequence impedances and circuits of the combination. R epeat the solution for
the case where a mutual i mpedance of j6 n exists b e tw e e n each pair of adj acent
Example 1 1 .4.
branche� in the
6. .
Vl an 1 lj2 1
1
2
j
l
j
l V}2)
Solution. The line-to-line volt ages a rc related t o the 6. currents by
VI,..-
r< "
0
0
j2 !
0
0
Transform ing to symmetrical c ompon e n ts of voltages and curren ts gives
A
VoOj,. )
2)
va(t>
=
0
0
o
j2 1
a
434
CHAPTER
11
j7
: --++0
SYMMETRICAL COMPONENTS AND SEQUENCE N ETWORKS
�
�t �
. Reference
+
I
t -](l)
a
V ( 2)1
V (l)
Positive-sequence
Zero-sequence
�t :
,
](2 ) .
a
'7
J
'
Reference
Negative-seq uence
�
�
+t l[j]
'�
+t 'D
+! D
(a)
V(O):
�+ :
j21
j3 3
[(0)
.
V ( l ):
I (1
�t :
Refere nce
,
)
j5
j5
V ( 2 }1
Reference
Positive-sequence
Zero-sequence
II
�t :
I ( 2)
,
R e fere n c e
N eg ative-seq uence
(b)
FIGURE 1 1.9
Zero-, positive-, a n d negat i ve-se quence ci rcu i ts for Cl-con nected i m p e d a n ces o f Ex a m p k 1 1 . 4 .
a n d premuitiplying ea c h side by A - I ) we obtain
o
j2 l
o
Thc pos i t ivc a n d negat ive-se q u e nc e c i rcu i t s have p e r- p h ase i m p e d a nces Z =
22 = i7 fl, as shown i n Fig, 1 1 .9( a ), a n d s i nce V"cJ})
0 , t h e zero-se q u e nc e c u r r e n t
I��) = 0 so that the zero-sequence circuit is a n o p e n c i rc u i t . T h e j 2 1 -fl res i s t a n c e
in the zero-sequence network has s i g n i A c an ce o n l y w h e n t h ere is an i n ternal
sou rce in the or i g i n al t1 c i rc u i t .
When there is mutual i nductance j6 n b e t w e e n phases,
I
-
=
A
[V}2) l lj2 1
(1l
.Vnb
V (2l
ab
=
i6
j21
)' 6
j6
)' 6
[1
6
j ] [/ (0) ]
j6 A
2)
J'21
1a( b
I�i)
The coefficient matrix can be separated into two parts as follows:
i 21
, [ j6
i6
i6
j21
i6
i6
j6 ] = j1 5 0
a
i 21
o
1
a
1
1
1
1 1 :6
and substituting into
SEQUENCE CIRCUITS OF A SYMMETRICAL TRANSMISSION LINE 435
the previous equation,
( O)
Vab
Va(bl )
2)
v(
ab
::::
j 15 A - JA
we
1
1
+ j6 A - [
obtain
1
1
1
1
1
1
1
)
1 (0
ab
A
1
j
6
r�
° lW;� l
{rT
'D)
n3
WJ
L
�
0
=
115
0
0
jlS
�
)
Ia( bI
( 2)
1ab
0
j15
o
o
j 15
0
0
0
+
0
o
o
/(1)
ab
/ (2 )
ab
II h
/(1)
!�;)
li b
The positive- and negative-seque nce i m p e d a nce s Z 1 and Z 2 now take on t h e value
jS n , as shown in Fig. 1 1 .9(b), and since �il?) 1��)
0, t h e zero-sequence ci rc u i t
is open. Ag(}in, we note t h a t the j33-[2 resistance in the zero-sequence network
has no significance because there is no i nterna! source i n t h e o r igin a l 6 circuit,
=
=
The matrix manipu lations of this exa mple are useful in the sections which
fol low .
1 1.6
SEQUENCE CIRCU ITS O F A
SYMMETRICAL TRAN SMISSI O N LIN E
W e are con cerned primarily with systems that are essentially symmetrically
hal anced and which become unbalanced only u pon the occurre nce of an
u n symme trica l fau l t. I n p ractical t ransmission systems such complete symmetry
is more i d e a l t h a n realized, but s i nce the effect o f the departure from symmetry
is usually smal l , perfect bala nce between ph ases is often assumed especially if
the l i nes a re t ransposed a long their l e ng ths. Let us consider F i g 1 1 . 1 0, for
i ns l Ll n c c , w h i c h s h ow s o n e s e c t i o n o f a t h r e e p h a se transm ission l ine with a
n e u t ral conductor. The se lf-imp ed ance Z a a is the same for each phase conduc­
tor, and t h e neu t ral conductor has sel f- i mpedance Z"" . When currents la ' 1b ,
a n d Ie i n t h e p h a s e co n d u c t o rs a rc u n b a l a n ce d , t h e n e u t r a l c o n d u c t o r serves as
a ret urn path. A l l the currents are assumed positive in the d i rections shown
even though some of their numerical values may be negative under unbalanced
conditions caused by faul ts. Because of mutual coup l ing, curren t flow in any one
of the phases induces volt ages in each of the other adjacent p hases and i n the
neu tral conductor. Similarly, In in the neutral conductor induces voltages in
each of the phases. The coupl i n g between all three phase conductors is
regarded as being symmetrical and m u tu a l impedance Z a b is assumed between
.
-
436
CHAPTER 1 1 SYMMETRICAL COMPON ENTS AND SEQUENCE N ETWO R KS
Zo b Zo b Z o
Z
o
b
Zo
I
e
l Vb!n +t
�+�-----
10
a'
a __
Ib
b'
b�
Ie
V• •
_
Vc n
_
_
!
�
I
n>..==n=
,�
Z
--C==
n n�1
_
_
--------r+�
'
c
L
I
Zo n
n
FIGURE 1 1. 10
Flow of u n b a lanced cu rre nts in a symmetrical t h rce- phasc - l i n e
'
�I +
±-
Ve . n .
s ec t i o n w i t h
n e utral
conductor.
each pair. Lik ewise, the m u t u a l i m pe d a nce b e t w e e n t h e n e u t r a l co n d u c t o r a n d
each of t h e p hases i s taken to b e Za n "
The voltages induced i n p hase a , for example, b y currents i n the other two
phases and the neutral conductor are shown as sources in the loop circu i t of
Fig. 1 1 . 1 1 , along with the similar voltages induced in the neutral con ductor.
Applying Kirchhoff's voltage law around the loop circuit gives
( 1 1 .47)
from which voltage drop across the line section is found to be
v:,n - Va'n' = ( Zaa - Zan ) la
+
( Za b - Z an ) ( lb + IJ + ( Zan - Znn ) In
( 1 1 .48)
Similar e q uations can be written for p hases b and
V::n - �' n ' = ( Zaa - Zan ) Ie
T
�
a
10
�
Va n
_
n
In
�
Zab 1b
+
Z an Ia
+
Za n In
+
c
as follows:
( Zab - Z an ) ( la + Ib ) + ( Zan - Znn ) ln
Zo
Zn ,.
( 1 1 .49)
t
T
n'�
a
Va' '
n
F1 G U RE 1 Ll 1
W r i t i n g Ki rchhoff's voltage
e q u a t ion a r ou n d t h e loop
formed by line a and the. n eutral conductor.
1 1 .6 SEQUENCE CIRCU ITS OF A SYMMETRICAL TRANSMISSION LINE
437
Wh en the line currents la ' Ib, and Ie return together as In in the neutral
conductor of Fig. 1 1 . 10, we have
( 1 1 .50)
Let us now substitute for In in Eqs. ( 1 1 .48) and ( 1 1 .49) to o b ta i n
+
( 7" /1
+
21111 -
( 1 1 .51 )
2 211/1 ) I,.
T h e coefficients in these equations show t hat the presence
o f the n eu tral
conductor changes the self- and mutual impedances of the p h ase conductors to
the fol lowing effective values:
( 1 1 .52)
� 'J l
�z
Using these definitions, we can rewrite Eqs. ( 1 1 . 5 1 ) in the convenient m atrix
form
Va a
V ,
. 1> / ,
V;· r'
=
V
(i l i
V/>11
V;
"
-_
vo "
n
V/! "
V;·'n'
/I
J
=
2/11
.r
( 1 1 .5 3 )
Z/II
w h e re the vol tage drops across the p hase conductors a r e n o w denoted by
V , ,@ V
UU
(l lP
- vII " n
Vee'
£
V:·n
-
v:.,
n'
( 1 1 .54)
S ince Eq. ( 1 1 .53) does not explicitly include the neutral conductor, Z s and Zm
can b e regarded as parameters of the phase conductors alone, w ithout any self-'
or m u t u a l indu ctance being associated with the return path.
The a -b-c vol tage d rops and currents of the line s ec t i on can be written in
terms of their symmetrical components according to Eq. ( 1 1 .8) so
, that with
438
.
CHAPTER
II
SYMMETRICAL COMPONENTS A N D SEQUENCE NETWO RKS
phase a as the reference phase, we h ave
A
Va(0)
a
Va(l)
a
Va(a2)
( 1 1 . 55 )
This pa rticular form of the equation m akes calcula tions easier, as demonstrated
i n Example 1 1 .4. Multiplying across by A I , we obtai n
-
v(O)
00
Ve l)
1
1
v (2)
aa
1
1
aa
( 1 1 .5 6 )
The matrix multiplication here i s t h e s a m e as in Exa mple 1 1 .4 and yields
va(0)
a
V(I)
( 1 1 .57)
aa
va(a2)
Let us now define zero-, positive-, and negative-sequence impedances in
terms of 2s and 2m introduced in Eqs . ( 1 1 .52),
( 1 1 . 58)
From Eqs. ( 1 1 .57) and ( 1 1 .58) t h e sequence components of the voltage drops
between the two ends of the line section can be written as three simple
equations of the form
v(�)
aa
v(�)
aa
va(a2')
=
=
=
Van(O)
-
V el) an
Va�o�
11
Va� nl �
Van( 2) - Va�2n�
=
=
=
1(0)
2 1(1 )
2 1(2)
2
0
a
I
a
2
a
( 1 1 . 5 9)
Because of the assumed symmetry of the circuit of Fig. 1 1 . 1 0, once again we see
1 1 .6
+r
V O)
-t
[(0)
a
a
--
SEQUE:--I C E CI RCUITS OF A SYMMETRICAL TRANSMISSION L INE
Zo
I
Va��'
�
a
(
a ll
+r
V(l'
-t
n
a
-+
I '
n
/(1)
�
ZI
-r
V
I
a
�
a('�)'
a ll
+tV (2)
439
n
a
n
1(2)
a
-
Z2
j+
V
n�
I
a
(?),
on
I
a II
n
FIGURE 1 1 . 1 2
Seque ncl: c i rc u i t s for t h e symm e trical l i n e section of Fig.
1 1.10.
that the zero-, posltlve-, and negat ive-sequence equa tions decouple from one
another, and correspond ing zero-, posit ive-, and negative-sequence circuits can
be drawn without any mutual cou pling between them, as shown in Fig. 1 1 . 12.
Despite the simplicity of the line model in Fig. 1 l . 1 0, the above development
has demonstrated importa n t characterist ics of the sequence impedances which
apply to more elaborate and pract ical line models. We note, for instance, that
the positive- and negat ive-sequence impedances a re equ al and that they do not
i nclude the neu tral -con du ctor i m pedances Zn Jl and Zan ' which enter into the
calcul ation of only the zero-sequ ence imped a nce Zo, as shown by Eqs. ( 1 1 . 58).
In o t h e r wo r d s , i mped a n ce p a r a m e t e rs of the retu rn-pa th conductors enter i nto
the values of t he zero-sequence impe dan ces of transmission lines, but they do
not affect e i t h e r the p os i t i v e or negative-sequence impedance.
Most aerial tra nsmission l i nes have at least two overhead condu ctors
cal led ground wires, which are grou n d ed at u n i form intervals along the length of
the l i ne. The ground wires combi n e with the earth re t ur n path to constitute an
e ffe c tive n e u t r a l co n d u c t o r w i t h i m ped a n c e p a ra m e t e r s , l i k e ZlI n a n d Z an '
which depend on the re sistivity of the earth. The more spec i alize d literature
shows, as we have demonstrated here, that the parameters of the return path
are i nclu ded in the zero-sequence impedance of the line. B y regarding the
neu t ral conductor of Fig. 1 1 . 1 0 as the effect ive return path for the zero­
sequence compone nts of the unbal anced currents and including its parameters
in the zero-sequence impedance, we can treat the ground as an i deal conductor.
The voltages of Fig. 1 1 . 1 2 are then i nterpreted as being measured with respect
-
I
440
CHAPTER 1 1
S Y MMETRICAL COMPONENTS AND S EQ U ENCE N ETWORKS
to perfectly conducting ground, and we can write
VCO)
V CO) - V CO)
'
=
aa
a
V ( l)
V ( l ) - V ef l )
=
aa
V ( 2)
oa
a
a
a
V (2 ) - V ('2 )
a
Il
=
=
=
=
Z /a( 0)
0
Z I /a( 1 )
( 1 1 .60 )
Z 2 [( 2)
a
where the sequence com ponents or the vol tages V:, a n d Vu' arc n ow w i t h respect
to ideal ground.
In derivi ng the equations fo r ind uctance and capacitance of tra nsposed
transmission lines, we assumed balanced th ree-phase currents and did not
specify p hase order. The resu l t i ng p a r a m e t e r s are therefore val i d fo r both
positive- and negative-seque nce impeda nces. When only ze ro-sequence current
flows in a transmission line, the cu rrent in each phase is identical. The current
returns through the ground, through over head ground wires, or t hrough both.
Because zero-sequence current is identical in e a ch phase condu ctor ( r a th e r than
equal only in magnitude and d isplaced in p h a s e 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 negat ive-sequence current. The
d i fference in magn etic field results in the zero-sequence i n d u c t ive react ance of
ove rhead t ransmission lines b e in g 2 to 3 .5 times as large as the posi tive­
sequence reactance. The ratio i s toward the h igher portion of the specified
range for double-circuit l ines and l i nes without ground wires.
Example 1 1.5. In Fig.
the terminal voltages at the left-hand and right-hand
e nds of the line are given by
11.10
Val +.170.0
Vbl
170.24
Zaa j60 2ab j20
fa' Ib'
Zo Za 2Zab 32 -6Za
ZI Z2 Zao -Zab j60 -j20
= 1 82 . 0
kY
= 72 . 24 - j32.62 k Y
V::n
-
=
Va'n' 154.0 +
Vb'I '
Ve'd +
Znn
= 44 .24 - j 74 . 62 k Y
=
+ j88 .6 2 kY
The line impedances in ohms are
=
j28 .0 kY
=
=
- 1 98 . 24
= j80
Z
j46 .62 kY
a ll
=0
D etermine the line currents
a nd Ie using symmetrical components. Repeat
the solution without using symmetri cal components.
Solution.
The sequence impedances have calculated values
=
=
+
=
+
nn
=
n
=
=
+
0 j40 + j240 -j180 j1 60
j40
j6
'
n
=
n
1 1.6 SEQUENCE C I RCUITS OF A SYMMETRICAL TRANSMISS I ON LINE
[ VaVa��/1 l [VaVbnn -Vb-Va''nn'' l
V};I v:.n -v:.'n'
[228.8.00 ++ jj442.2.00 1
441
[ (1(72.82.204 --44.154.204)) +- j(j(730.2.062-28.-74.0)62) 1
-(170.24 - 198.24) + j(88.62 -46.62)
[28.0 +0j42.0 1
The sequence components of the voltage drops in the line a re
= A-
=
A-
= A-I
)
I
28.0 j42.0
+
=
kV
0
S u bs t i t u t i n g i n Eq. ( 1 1 .59), we obt a i n
V«a (l�) = 20 , 000 + ;'42 , 000 = j' 1 6 0 !(/(O)
j40/( l )
()
=
()
= j40 ! (2)
a
II
from which we d e te r m i n e t h e symmet rical components of t h e
I� O)
=
262.5 -j1 75
currents in phase
a,
A
a b 262.(1 1.552)-j175
+ Znn -2Z(1l j60 + j80 -j60 j80
Z"" - 2ZIlf + j60 j40
(1 1.53)
j 10) [j80j40 j40j80
j40 j40
j40j40 j - l [2828 +j+ j4422 j 103 [262.262.55 -j175
j
-j175
The l i n e c u rrents are therefore
I
=I
=
Ie
=
A
The s elf- a n d m u t u al impeda n ces of E q .
Zs
Zm
= Za a
=
[V(1V"(1'' j [ 2828
. 28
h ave val u e s
=
= j20
Zil " +
JRO -
�e'
n
=
n
without symmetrical
a n d so l i n e curren ts can be c al c u l a ted from Eq,
components as fol l ows:
=
=
+ j42
+
j42
+ j42
X
=
j80 28 + j42
X
=
262.5 -j175
A
442
CHAPTER I I
S Y MM ETRI CAL COMPONENTS AND SEQUENCE NE1WORKS
--
a .,....-
Z"
______
I"
FIG U RE 1 1 . 1 3
b
c
C i rc l l i t d i a g r < l m o f
Eall , E"n ,
t h ro u g h
a
�!
g e n e r a t or g ro u n d e d
Ecn arc
reac t a n c e .
and
The
r h asc
e m fs
po� i t ivc seq u e n c e .
11.7
SEQUENCE CIRCUITS OF THE
SYNCHRONOUS MACHINE
A synchronous generator, grounded through a reactor, is shown in Fig. 1 1 . 1 3 .
Whe n a fau l t (not indicated in t h e figure) occurs a t t h e term inals o f t h e
generator, currents la ' lb ' and Ie flow in t h e lines. If the fa u l t invulves ground,
the current flowing i nto the neutral of the generator is designated In and the
line currents c a n b e resolved into their symm etrical com ponents regard less of
how u nbal anced they m ay be.
The equations developed in Sec. 3.2 for the idealized synchronous ma­
chine are all based on t h e assumption of b alanced i nstantaneous armature
currents. In Eq. (3 .7) we assumed that i a + i b + ie 0, and then set i a
- (i b + i) i n Eq. (3.5) i n order to arrive a t E q . (3. 1 ] ) for the terminal voltage of
p hase a i n t h e form
=
=
( 1 1 .61)
The steady-state counterpart of t h is equation i s given i n Eq. (3 .24) as
( 1 1 .62)
where Eo n is the synchronous internal voltage of the machine. The subscripts of
the voltages in Eqs. (1 1 .61) and (1 1 .62) d i ffer slightly from those of Chap. 3 so as
to emph asize that t h e vol tages are with respect to neutral. If the subs t it u tion
La = - (ib + iJ had not b een made as i nd icated, then we would h ave found
Va n
=
�
Ri a
-
di a
LS - +
dt
d
M - ( ib
5 dt
+ ie) +
ean
( 1 l .63 )
Assuming for now t ha t steady-state s i n usoidal curren ts and voltages of nominal
l i .7
443
SEOUENCE C[ RCUITS O F THE SYNCH RONOUS MACH INE
system frequency w continue to exist i n the armature, we can write Eq. (1 1 .63)
i n the phasor form
( 1 1 .64 )
where E a n again designates the phasor equivalent of e a n . The armature p h ases
b a n d c of the idealized machine h ave similar equations
V[ I/ I
1
( 1 1 .65)
We can a rrange
�>n
=
-
�
Eq s .
[lu 1 [1
( 1 1 .64) and 0 1 .65) in vector-matrix form as follows:
[ R + j w ( L , + MJ ]
Ih
�.
+ j w Ms 1
1
1
1
1
( 1 1 .66)
Followi ng the procedure demonstrated in the two preceding sections, we now
express the a -b-c q u a n t i t i e s of the m achine in t er m s of symmetrical components
of p h a s e a of the armature
VV( I)
(2)
vall( O)
all
all
=
- [ R + jw ( L s + MJ ]
+
j(u M 1 A - 1
1
1
1
1
1
E'III
1 (0)
A I II
1/
(I)
1 II( 2 )
2£
+ A- I a
.
( 1 1 .67)
1111
a E O II
S ince the synch ronous g e n e r a tor is designed to su pply balanced three-phase
vol tages, we have shown t h e generated vol tages Ea ll • E'm ' and Ec n as a
posit ive-sequence set of phasors i n Eq . ( 1 1 .67), w h e r e the operator a = 1 /1 200
anu
a2
V ( 2)
=,
/2400 . T h e
m ;l t r i x
of Eqs . ( 1 1 .56), anu so we
va(nO)
V(I
an
an
)
=
In
li l t i p l i c a t i o n s o f Eq .
o b ta i n
- [ R + jw ( Ls + Ms ) ]
1 (2)
/aCO)
I a(
a
I)
3
+ jw Ms
( 1 1 . 67)
ar e
s i m i l a r t o t h ose
1a(0 )
1a( 1 )
1a(2 )
0
+
Ean
0
( 1 1 . 68)
,.
444
'
CHAPTER 1 1
SYMMETRICAL COMPONENTS AND SEQUENCE NETWORKS
The zero-, positive-, a n d negative-sequence equations decouple to give
v(O
an )
Va(nl )
v ( 2)
all
=
=
=
-
RIa( O) - J'w( L s - 2 Ms ) 1a(0)
- RIaO ) - J'w ( L s + M ) 1a( 1 ) + Ea n
- Ri 1Ia)
- J· w ( / .
.\"
+
M.,. ) /,(, :!)
Drawing the corresponding s e q u e n c e c ircu i ts is
( 1 1 .69) in the form
a(n0 )
vel) - E
v
=
Gil
vall(2 )
=
(l ll
( 1 1 .69)
5
m ade simple by w r i t i n g
--- E
a ll
-
j (O ) Z
-
1a( 1 ) ,c'
7l
(l
Eqs.
!; o
- 1a(2)2 2:
( 1 1 .70)
where Zg O ' Z I ' and Z2 are the zero-, posltlve-, and negative-sequence
impedances, respectively, of the genera tor. The sequence circuits shown in Fig.
1 1 .14 are the single-phase equivalent circuits of the balanced three-phase
m a ch i n e t h rough which the symmetrical components of the unbalanced currents
are considere d to flow. The sequence components of current a re flowing
through i mped ances of their own sequence o nly, as ind icated b y the appropriate
subscripts on the impedances shown in the figure. This is because the machine
is symmetrical with respect to phases a , b, and c . The positive-sequence circuit
is comp osed o f an emf in series w ith t he positive-sequence impedance of t he
generator. The negative- and zero-sequ ence circuits con tain no emfs b u t i nclude
the i m p e d a nces of t h e g ene r ato r to negative- a n d zero-sequence c u rren ts,
respectively.
The reference node for the posi tive- and negative-sequence circ u i ts is the
neutral o f the gene rator. So fa r as p os itive- and negat ive-seque nce com ponents
a r c co n c e rn e d , t h e n e u t r a l o f t h e g e n e r a t o r is a t g ro u n d p o t e n t i a l i f t h er e is a
connection between neutral a n d ground h aving a finite or zero i mpedance s ince
the connection will carry no positive- or negative-sequence curre nt. Once again,
w e see that there is no essential difference betwee n V} l ) and Va� ) i n the
positive-sequence circui t or between VP) and Va<'� ) in the nega tive-sequence
circuit. This explains w hy the posit ive- and negative-sequence voltages v}l ) and
of Fig. 1 1 . 14 are written without subscript n .
The current flowing i n the impedance Zn between neutral and ground is
3I�O), By referring to Fig. 1 1 . 1 4( e ), we see that the voltage drop of zero
sequence from point a to ground is - 3 /�O) Zn - I�O )Zg O ' where Zg O is the
zero-sequence impedance per phase of the generator. The �ero-sequence cir­
cuit, which is a single-phase circui t assumed to carry only the zero-sequence
VY)
1 I .7
( a) Positive-sequence current paths
a
SEQUENCE CI RCUITS OF THE SYNCHRONOUS MACHINE
44S
Reference
( b ) Positive-sequence network
.------'/ (2)
a
n
c
(c) N e g ative-seque nce current paths
Reference
( d ) Negative-sequence network
'-_--..
---,
n
Reference
( e ) Zero-sequ ence current paths
(n Zero-s equence network
FI G U R E 1 1 . 1 4
P a t h s f o r c u r r e n t o f each sequence i n a g e n e r a t o r and t h e correspo n d i n g s eq u e nce n e tworks.
cu rrent of one phase, m u s t t here fore have an impedance of 3 Zn + Zg a , as
shown in Fig. 1 1 . 1 4( / ). The total ze ro-sequence impedance through which J�a)
fl ows is
( 1 1 . 71)
U s ually, the components o f current and vol tage for phase a are fou n d
from equations determined b y t h e sequence circu its. The equa ti ons for the
components of vol tage drop from point a of phase a to t he reference node (or
446
CHAPTER I I
S Y M M ET R I CA L CO M PO N ENTS A N D S EO U FN ( ' I ;. N I ;:rWO I { K S
ground) a re written from Fig. 1 1 . 1 4 as
v(O)
- Q
V (I )
a
Va( 2 )
=
=
=
£ a ll
-
/(l)Z
a
1
( 1 1 . 72)
where £ /1/1 is the positive - seq u e nce voltage to n e u t ra l , Z 1 a n d Z 2 arc the
positive- and negative-sequence i m p e d a nces of t h e generator, respectively, and
2 0 is d e fi n e d by Eq. ( I 1 .7 1 ).
The equations d eveloped to this p o i n t are b a s e d o n a s i m p l e m J c h i n e
model which assumes the cxistence o r o l 1 l y rU I l lL l I ll c n l a l c o m p o n e n t s o r c u r ­
rents; o n this basis the positive- and negative-sequence impedances are fou n d to
be equal to one another b u t qu ite different from the zero-sequence impedance.
I n fact, h owever, the impedances of rotating machi nes to currents of the th ree
sequences will generally be different for each sequence. The m m f produced by
negative-sequence armatu re current rotates in the direction opposite to that of
the rotor wh ich has the dc field winding. U n l ike the flux produced by pos i t ive­
sequence current, which is stationary with respect to the rotor, the flux pro­
duced by the negative-sequence current is sweeping rapidly over the face of the
rotor. The currents induced in the field and damper windings cou nteract the
rotating mmf of the armature and thereby reduce the flux penetrating the rotor.
This condition is similar to the rapidly changing fl ux immediately u pon the
occurrence of a short circu i t a t the terminals of a machine. The flux paths are
the same as t hose encou ntered in eva luating subtransient reactance. So, in a
cylindrical-rotor m achine subtra nsient and negative-sequence re act ances are
equal. V a l ues given in Table A.2 in the Appendix confirm this statement. The
reactances in both the positive- and negative-sequence circuits are often taken
to be equal to the subtransient or transient reactance, depending on whether
subtransient o r transient conditions a re being studied.
When only zero-sequence c u rrent flows in the armature wind ing of a
three-phase m ach ine, the current and m m f of one phase are a maxi mum at the
same t i m e as t h e current and mmf of each of the other phases. The windings are
so distributed around the ci rcu m ference of the armature that the point of
maxi m u m m m f produced by o n e phase is d i s p l ac e d 1 20 e l e ctrical d eg r ee s i n
space from the point of maximu m mmf of each of the other phases. If the m m f
p roduced b y t h e current of each p h ase had a perfectly s i nusoidal distribution i n
space, a plot o f m m f around t h e armature wou ld result in three s in usoidal
curves whose sum would be zero a t every point. No flux would be produced
across the air gap, and the only reactance of any phase winding woul d be that
due to l eakage and end turns. In an actual machine the winding is not
distributed to p roduce perfectly s i nusoida l mmf. The flux resulting from the,sum
o f the m m fs is very small, wh ich makes the zero-sequence reactance the smallest
SEOUENCE CIRcu i TS OF THE SYNCHRONOUS MACHINE
1 1 .7
447
of the m achine's reactances-just somewhat higher than zero of the ideal case
where there is no air-gap flux d u e to zero-sequence current.
Equations (1 1 .72), wh ich apply to any generator carrying unbalanced
currents, are the starting points f or the derivation of equations for the compo­
nents of current for different types of faults. As we shall see, they apply to the
Th6venin equivalent circuits a t any bus of the system as well as 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 £' or E" is
substituted for
Eo n '
salient-pole generator without dampers is rated 20 MVA,
1 3.8 kY and has a direct-axis subtransient reactance of 0.25 per unit. The negative­
and zero-sequence reactances are, respectively, 0.35 and 0. 1 0 per unit. The neutral
of the generator is sol idly grounded. With the generator operating unloaded at
rated vol tage with E a n l .Oft per u nit, a single l i ne-ta-ground fault occurs at
the m achine terminals, which then have per-unit vol tages to ground,
Example
1 1 .6. A
=
v"
=
0
v"
=
1 .01 3/
-
1 02 .25°
Determine the subtransient current in the generator and the line-to-line vol tages
for sub transient conditions due to the fault.
Figure 1 1 . 1 5 shows the line-to-ground fault on phase
In rectangular coordinates Vb and Vc are
Solution.
Vb
JJ;..
=
-
=
-
0 . 2 1 5 - j O .990
0 .2 1 5
a
of the machine.
per unit
+ jO.990 per unit
a .------,
I
z"
FIGURE 1 1 . 1 5
Circu i t
gro u nd
nals of
neutral
tance.
d iagram for a single l i n e-to­
fau l t on p hase a at the termi­
an u n loaded generator whose
is grounded through a reac­
448
CHAPTER 1 1
S Y M M ETRICAL CO M PO N ENTS A N D S EQ U ENCE N ETWO R KS
] [-0..1433 ]
The sym m etrical components of the voltages at the fault poi n t are
[V}OV(l)) ]
V
a
( 2)
a
From Eqs.
1a( 0)
/(/( 1 )
1a( 2)
13 [11 1 1 ][
1
( 1 . 7 2)
=
0
- 0 . 2 1 5 - }0 .990
a2
-
- 0 .215 +
a
and Fig. 1 l . 1 4 w i t h 2 "
=
=
=
V eil)
a
- --
ZgO
VZ(,2)
0
=
=
we ca l c u l a t e
( - 0 . 1 43 + }O)
)0 . 1 0
=
E a " - V11( I )
( 1 .0
ZI
a
- --
}0 .990
+ }O
0 64 + }O per u n i t
- 0 .500 + }O
+
- j l '+ 3
.
+
(0 .643
jO .25
}O)
-
-j l . · n pcr
jO)
( - 0 .500 + }O )
jO. 5
=
per u ni t
=
3
unit
-j 1 . 4 3 per u nit
Therefore, the fau l t current into the ground is
I
a
=
1a( 0
)
+
13
1 (1)
a
The base current is 20,000/(
i n l ine a is
fa
+ [a(2)
X
3 / ( 0) =
a
13.8) 837
837
=
-}4 .29
=
=
=
X
-)'4 .29 per u n i t
A, and so the subtransient curre n t
-j3 ,590 A
Li ne-to-line vol tages during the fau l t arc
vti ll
V"C
�'1I
=
OC�,
=
Va - V"
VI,
V
0.2 1 5
=
V
c -
=
=
- 0.21 5
+
1 .0 1 / 77.T per u n i t
1 . 980�
- j l .980
0
v:, =
c
+ jO.990
1 .0 1 i..l2 . r
jO.990
per
unit
per u n i t
Since t h e generated voltage-to-neutral E a ll was taken as l .0 per u nit, the above
l ine-to-l i ne voltages are expressed i n per unit of the base vol tage to neutral.
Expressed i n vo l ts , t h e pos t fa u l t l i n e vo l t ages a r c
V . 1 1 3 .8 / 77.70 8 / 77.70 kY
1 3 .8
1 . 1 1 3 .8 8 k Y
a
.
b
Vbc
=
l 0
=
1 .980
Vc a =
0
X
X
X
13
= . 05
_
/
..
13 LE!!.
., .., ,,0
13
/
1 02 .30
=
=
/
15 .78LE!!.
..
" .., ,...., 0
.05 L/_ 102 .30
kY
SEQUENCE CI RCUITS OF Y-A TRANSFORMERS
1 1 .8
449
b
b
a �---+
c
( a ) Prefa u l t
Postf a u l t
FIG U RE 1 1 . 1 6
P h asor d iagrams of the l i n e voltages of Example 1 1 .6
be fo re a n d after the fau l t .
t h e ra u l t t h e l i n e vo l t a g e s w e re b " la n c e d l i n d e q u a l t o
B e fore
l�l fl
(b)
c
c o m p a r i s o n wi l h l h e l i n e vo l l a ges a fte r t h e f,l ll l t occu rs, t h e prcfa u l t
=
F.i/ I/
�,"
=
as
rdc re nce ,
! J .8! 3 (jO
; I IT
= 13.8�
g ivCI! as
kV
v" , .
kY
v:"
=
1 3.8 kY. For
vol tages, with
13.R�
kY
F i g u re 1 1 . 1 6 s h ows p h ll s o r d i ag r a m s o f p rc fa u l t a n d pos t fa u l t voltages.
The p reced ing example s hows that I(�(J) I�l) I�2) in the case of a single
l ine-to-ground fault. This is a general result, which is established in Sec. 1 2.2.
=
=
1 1 .8
S EQUENCE CIRCUITS OF V-A
TRANS FORM ERS
The sequ ence equivalent circu its of three-phase transformers depend on the
con nections of the primary and seco n d a ry wind ings. The d ifferent combi nations
of L1 and Y windings determine the configurat ions of the zero-sequence circuits
and the phase shift in the positive- and n egative-sequence circu its. On this
acco unt the re ader may wish to review some port ions of Chap. 2, notably Secs.
2.5 a n d 2 . 6 .
We remember t h a t n o cu rre n t flows in the primary of a transformer u n ­
less cu rrent fl ows in the seco nd a ry , i f we n eglect the relatively s m a l l m agnetizing
curre n t . We know also that the primary current is determined by t h e secondary
current and the turns ratio of the windings, again with magnetizing current
n e g l e c t e d . These p r i n c i p l e s guide us in the a n alysis of i n d ividual cases. Five
possible connections of two-w i n d ing tra nsfo rmers will be d iscussed. These
connections are summarize d , along w i t h t heir zero-sequence circuits, in Fig.
1 1 . 17. The arrows on the connection d i agrams of the figures to fol l ow show the
poss ible paths for the flow of zero-seq u ence current. Absence of an a rrow
ind icates that the transformer connection is such that zero-sequence current
cannot flow. The zero-sequence equival ent circuits are a pproximate a s shown
since resist ance and the magnetizing-cu rren t p a t h a re omitted from each circuit.
The l e t t ers P and Q ide n t i fy correspo n d i ng points on the connection d iagram
f
450
CASE
1
2
3
CHAPTER 1 1
SYMMETRICAL COM PONENTS AND SEQUENCE N ETWORKS
SYMBOLS
� �.
CONNECTION DIAG RAMS
.FY �
ZERO-SEQUENCE EQUIVALENT C I R C U ITS
pi
I
I
I
•
Reference bus
Zo
p
�
�y
•
Reference b u s
-- --
��
I
I
:Q
I
6
p
--
•
Zo
n
Reference bus
Zo
�
p
4
•
p
•
Reference b u s
Zo
�
5
Q
.�----�r-t�----�.
��
6
Zo
p
I
•
Reference b u s
•
Q
•
•
Q
-
•
Q
•
•
Q
•
•
FIGURE 1 1 . 1 7
Zero-seq u e n ce e q u iv a l e n t c i rcuits o f t h ree-phase t r a n s form e r b a n k s , toge t h e r w i t h d i ag r a m s o f
con n ections a n d t h e symbols for o n e - l i n e d i a g r a m s . I m pe d ance 2 0 accou n t s for t h e l ea k a ge
impe d a n ce 2 a n d t h e n eu t ral i mpedances 3 2 N and 3 2" where a pp l i cab l e .
and equivalent circuit. The reaso n i ng to justify the equivalen t c i rc u i t for each
connection follows.
CASE
1. Y-Y
Ban k , Both Neutrals Grounded
Figure 1 1 . 1 8(a ) shows the ne u t ra l s of a Y - Y b a n k gro u n d e d t hrou g h
imped an ce ZN on t h e high-vol t age s i d e a n d 211 on t h e low-vol tage s i d e . The
arrows o n the diagram show the d i rections chosen for the cu rrents. W e first
treat the trans formers as ideal and add series leakage impeda nce later when the
shunt m agnetizing current can also be i ncluded i f necessary . We continue to
designate voltages with respect to gro u n d by a single subscript such a s VA ' VN ,
and Va' Voltages with respect to neutral h ave two subscripts such as VA N and
�n' Capital letters are assigned to t h e hi gh-voltage and lowercase l etters are
assigned on the other side. As before, windings that are d rawn in paral lel
1 1 .8 SEQU ENCE CiRCUITS OF Y-L\ TRANSFORM ERS
]
A
]
B
]C
=
A
=
B
=
c
] ( 0 ) + ] ( 1 ) + ] ( 2)
A
A
A
]
b
-,
_
__
__
__
__
_
_
_
_
] (0) + ] ( 1 ) + ] ( 2 )
B
B
�
[ (0)
c
+
] (0) + ] ( 1 ) + ] (2)
b
b
b
,
�-----
]
B
a
•
] ( 1 ) + ](2)
c
=
]
C
c
-------'
=
=
451
b
::;:::=)oo� a
] ( 0 ) + ] ( 1 ) + ] ( 2)
a
a
a
](0) + ]( 1 ) + 1 (2 )
c
,
c
c
'----- c
(a)
----�
i\ -,-------.--,
+
+
.-----.----,-- a
+
+
NI : N 2
3
[ (0)
"
N
I
3 ] a( 0
ZN Zn
)
•
Va
n
(b)
FIGURE 1 1 . 1 8
(a)
Y-Y
connected
tra nsfo r m er
bank
wi t h
bot h
n e u trals
grounded
t h rough
impeda nces;
(b) a pair o f t h e m a g n e t ical ly l i n k e d wi n d i n g s .
d i rections are t h ose l i n ked m ag n e t i c a l l y o n the same core. Two such windings
t a ke n fro m Fig. 1 1 . 1 8( a ) a rc s h ow n i n Fig. 1 1 . 1 8(b ). The vol tage m easured w i t h
respect t o g r o u n d o n t h e h i g h -v o l t a g e s i d e i s g i v t.: n b y
( 1 1 .73)
S ubstit uting the symmetrical compon e n ts of each vol tage g ives
( 1 1 .74)
and equating quantities o f
again confirms the fact t h a t
a r e equal t o posit ive- a n d
sequence voltage difference
t h e s a m e sequence, as explained for Eq. ( 1 1 . 1 9) ,
positive- and negative-sequence voltages to gro u n d
negative-sequence vol t ages to n eutral. The zero­
between neu tral and ground is equal t o (3ZN ) I�O).
J
452
CHAPTER I I
SYMMETRICAL COMPONENTS AND SEQUENCE N ETWO RKS
., Similarly, o n the low-voltage side we h ave
V CO)
a
+
V C l ) + V( 2 ) = ( vCO)
a
a
all
V C I ) + V ( 2 ) ) - 32
+
an
a ll
fl
] (0)
( 1 1 .75 )
a
There is a minus sign in this equation because the d i rection of 1 (0) is out of the
transformer a n d into the l ines on the low-voltage side. Voltages ; n d cu rre n ts on
both sides of t h e transformer are related by turns ratio N I /N2 so that
Multiplying across by N, /N2 gives
N
-l ( V (O)
N2
a
+
Vo)
a
+
v (2 » ) = ( VACO)N + VAe l ) + V ( 2 » )
N
AN
-
a
3 21 1
(N
1
_
N2
J
, 2
JA( O ) ( 1 1 . 77)
and substituting for ( Vl� + Vl� + Vl� ) fro m Eq. ( 1 1 .74), we u b t a i n
N
l ( VCO) + V(l) + V ( 2» ) = ( V CO) + V e l ) + V (2))
A
A
A
N2
a
a
a
_
3 2 N JA(O)
_
( )
N
j
3 2n
N2
2
1A(0)
( 1 1 .78)
By equating voltages of the same sequence, we can write
NI ( )
_
V I =
N2
<l
( )
VA I
N
V CO ) = VA(O)
N2 a
_
1
NI
N2
_ _
_
[
V
( 2)
=
V
3 2N + 3 2
(2)
(NI )2] ] «()
A
11
II
N2
A
( 1 1 .7 9 )
( 1 1 , 80)
The positive- and negative-sequence relations of Eqs. ( 1 1 .79) are exactly the
same a s i n Chap. 2 , a n d the usual p e r-phase e q u iva l e n t circuit o f the trans­
former therefore applies when posi tive- or negative-sequence voltages a n d
currents a r e present. The zero-seq uence equiva l e n t ci rcuit representing Eq.
( 1 1 .80) is d rawn in Fig. 1 1 . 19. We h ave a d ded the l eakage impedance 2 of the
transformer i n series on t h e high-voltage s i d e as s hown so t h a t the total
impedance to zero-sequence current is now 2 + 3 2 N + 3(Nl/N2 )22n referred
to the h igh-voltage side. It is appa rent t h a t the s h u n t magnetizing impedance
could also be added to the circuit of Fig. 1 1 . 1 9 if so d esired. When voltages
, on
both sides of the transformer are expressed i n per u ni t on kilovolt line-to- l i n e
1 1 .8
O)
VA( '
SEQUENCE CI RCUITS OF
y-� TRANSFORMERS
VA( Ol
FIG l'RE 1 1 . 1 9
Zero - s e q u e nce c i rc u i t of Y- Y co n n e c t e d t r a n sform e rs of Fig . 1 1 . I R . I mpedance
re a c t a nc e
bases
453
,IS
measl i red
Oil
the
h i g h-vo l t ;l g e side or
t i l e t r; l I 1 s fo r m e r .
Z is t h e
le akage
in accorda nce with the rated voltages, the t u rns ratio in Fig. 1 1 . 1 9
unity a n d N t /N2 d i s a p p ears, a n d we arrive a t the zero-seq uence
circu it shown under Case 1 in Fig. 1 1 . 1 7, where
chosen
becomes
Zo
=
Z
+
3 ZN + 3Z n per u n i t
( 1 1 .8 1 )
Aga in, we note that impedances connected from neutral t o ground i n t he actual
circu it are multiplied b y 3 i n the zero-sequence circui t . Where both neutrals of a
y -Y bank are grounded d irectly or through impedance a path t hrough t h e
t ransformer exists for zero-sequence currents i n b oth wi ndings. Provided t h e
zero-sequence cu rrent can fol l ow a complete circuit o utside t h e transformer o n
both sides, it can flow in both windings of the transformer. I n t h e zero-sequ ence
circu it points on the two sides of the transformer are connected b y t h e
zero-sequence impedance of the transform er i n the same manner as in t h e
positive- a n d negative-sequence n etworks.
CASE
2. Y-Y
Bank , One
Neutral Grounded
If either one of the neutrals of a Y Y bank is u ngrounded, zero-sequence
current cannot flow in either winding. Th is can b e seen b y setting either Z N or
Z n equal to in Fig. 1 1 . 1 9 . The absence of a path t h rough one winding prevents
cu rren t i n the other and a n open ci rcu it exists for zero-sequence current
between the two p a rts of the system connected b y the transformer, a s shown in
F i g . 1 1 . 1 7.
-
cc
CASE
3.
6-6 Bank
The phasor sum o f the l i n e-to-line voltages e q u a l s zero on each s i d e of t h e
6-6 t ransformer of Fig. 1 1 . 20, and so V,J<]i = vaCE) O. Appl ying t h e rules o f t h e
=
,
454
II.
A
CHAPTER I I
=
11°)
�
+
111)
+
S Y MMETRICAL COM PONENTS A N D SEQUENCE N ETWORKS
I
112)
=
a
1(0) + 1(l)
+ 1(2)
a
a
a
�
r------ a
+
B -.1.....----411'--. -..
In
Ie
1�0) + Ij/) + I�2)
O
= I� ) + Ig) + I�2)
�
=
.
�
FIGURE 1 1 . 2 0
Wiring d i a �raJ1l of l i ll:
C -------�
-
�------ b
I l l rl: l: · p l l ; I � C � . � C U I 1 1 I C l' l c d I l' a 1 1 S I( 1 I 1 l 1 C I.
conventional dot notation for coupled coils to that figure, we have
N]
N2
VA B = - V" ....,.
V (l) +
AB
VA(2)B
=
N1
_
N2
( V(l) +
ab
( 1 1 .82)
V ( 2»
ab
The l ine-to-line voltages can be written as l i n e-to-neutral voltages according to
Eqs. ( 1 1 .23) giving
and so
VAON)
=
I v( l )
N ""
N
_
2
v ( 2)
AN
=
N
I v (2)
N :>. {/II
_
( 1 1 .84)
Thus, the positive- and negative-sequ ence equivalent circu its fo r the 6. - 6.
tra nsformer, like those for the Y Y con nection, correspond exactly to the usual
per-phase equivalent circuit o f Chap. 2. Since a 6. circuit provides no return
p at h for zero-sequ ence current, no zero-sequ ence current can flow i nto either
side of a Ll-Ll bank a lthough i t can somet imes circulate within the Ll windings.
Hence, I�O) = J�O) = 0 in Fig. 1 1 .20, a n d we obtain the zero-sequence equivalent
circuit shown i n Fig. 1 1 . 1 7.
-
CASE 4. Y-� Bank, Grounded Y
I f the n eutral of a Y-Ll bank is grounded, zero-sequence currents h,ave a
p at h to ground through the Y b ecause corresponding i nduced currents can
1 1 .8
I
I
1 ( 0) + 1 (1 ) + 1 ( 2 )
=
A
A
A
SEQUENCE CI RCU ITS OF Y·I!> TRANS FORMERS
A
�
c
Ie
=
Ib
-
•
=
] (0)
c
-----
+
](1)
c
+
+
1a( 1 )
,----- a
A
•
Ic
1a( 0)
=
a
](2)
c
455
+ 1d:(2)
le(1) + 1�2 )
I�O)
+
�
C
1b(0)
+
1b(1)
�----- b
+
1(2)
b
FI G U RE 1 1 .2 1
W i r i n g d i agram
i m pe cJ a nc e ZN '
Y-O
for a t h ree-p hase
transfo r m e r bank wit h
neutral
gr ounded t h rough
c i rcu l a t e in t h e 6. . The ze ro-seque nce cu rrent c i rcu l a t i ng in t h e 6. magnetically
b Cl l d n c e s t h e z e r o - se q u e n ce curre n t i n the Y b u t c a n n o t flow i n the l ines
connected to the 6. . Hence, l�O) = 0 in Fig. 1 1 . 2 1 . P h a se-A voltage on the Y side
can be written t h e same a s in Eq. ( 1 1 .74 ) from w h ich we obtain
y eo) + V ( I ) + V ( 2 )
A
A
A
gIves
=
N
I
2
_
N
y ebo )
n
+
N
1
_
N2
Vn(1)
b +
N
1 ( 2)
2 Vab
_
N
+
3 2 1 (0)
N A
( 1 1 .85)
Equating correspond ing sequence components, as expla ined for Eq. ( 1 1 . 1 9),
VA((l)
-
3 2 1 (0)
N A
=
N
I
_
N2
VaeO)
h
-
N
V ( I ) = _I V ( I ) =
,
,.
VA(2)
N2
NI
( :2 )
:2 Vtil!
= _
N
ill>
=
0
( 1 1 . 86)
N
_1 /3 / 3(f
N:;>,
N
1
_
N2
13/
y .)
_
-
X
30°
V(I)
1I
X
.
vtI( 2 )
( 1 l .87)
Eq uation ( 1 l .8 6 ) enables us to draw t h e zero-sequence circuit s hown i n
F i g . 1 1 .22( a ) i n w h i c h Zu = 2 + 3 2 N when l eakage impedance 2 is referred t o
the high side of the transformer. T h e equ ivalent circuit p rovides for a zero­
sequence cu rrent path from t h e line on t h e Y side through t h e equivalent
resistance and leakage reacta nce of the transformer to the reference node . An
open circuit must exist between t h e line and the reference node o n the 6. s i de.
W hen the connection from neutral to grou n d conta i n s impedance 2 N as s h own,
456
CHAPTER
SYMMETRICAL COMPONENTS A N D SEQUENCE N ETWORKS
II
] (0)
A
z
-
1 (0)
N
= 0
a
-
( a)
1---------'
I
z
+
I
I
13 _1 /30" 1
N
N
2
I
II
:
Ideal
:
+
I
Z
--�c===J
t\
I
Vf" :
I
-
- - - - ,
- /- 30
N2
Nj
I
0
:
I
I
11
] (2)
t+
1
a
_
:
V(2)
:
� �
_____-�r�i _�1 1 -�i--�!-1
I
J
'-
I V3
1
I
I
_ _ _ _ _ _ _ _ _ _
(b)
- - - - -
I
I
,_
FI G U RE 1 1.22
�-4
1--�
�
(c)
Ideal
_ _ _ _ __ _ _ _ _
I
I
J
( a ) T h e zero-sequence circ u i t of Y -� t r a nsfo r m e r b a n k w i t h g r o u n d i n g i m p e d a n c e Z N a n d
correspo ndi ng (b) pos i t ive-sequence a n d ( c ) n e g a t ive-sequ e n ce c i r c u i t s .
t h e zero-sequence equivalent circ u i t m u s t have impeda nce 3 ZI\. in series with
t h e equivalent resistance and leakage reactance of the t ransformer to con­
n ect the l i n e on the Y side to grou n u .
CASE 5. Y-il Ban k , Ungrounded
Y
u ngrounded Y is a special case where the impedance Z N between
ne utral a n d ground is infinite. The impedance 3Zn in the zero-sequence
equivalent circui t of Case 4 becomes i n fi nite and zero-sequ ence current cannot
flow in the transformer windings.
An
T h e positive- and negative-sequence equival en t circuits of the Y- Ll trans­
former shown i n Figs. 1 1 .22(b) a n d 1 l .22( c ) are b ased on Eqs . ( 1 1 .87). We recall
from Sec. 2.6 that the multiplier 13 Ni lN2 in Eqs. ( 1 1 .87) is the ratio of the two
rated line-to-line (and a lso line-to-neutraU voltages of the Y-Ll transformer. In
p er-unit calcu l ations, therefore, Eqs. ( 1 1 .87) become exactly the same as Eqs.
(2.35), a n d again we have the rules
V(I)
A
=
l)
V(
a
X 1�
/ 300
I�I )
I�2)
=
=
I� ! )
I� 2 )
X
X
1�
1/ - 300
( 1 1 .88)
1 1 .8 SEQUE NCE CIRCUITS O F Y - A TRANSFO RMERS
457
That is,
When stepping up from the low-voltage side to the high-voltage side of a
� -y or Y - � transformer, advance the positive-sequence voltages (and
currents) by 3 0 ° and retard the negative-sequence voltages (and currents)
by 3 0 ° .
The next example shows the numerical application of Eqs. (1 1 .88).
1 1 .2
The resistive Y -connected load bank of Example
i s supplied
from the low-voltage Y-side of a Y-tl transformer. The voltages a t the load are the
same as i n that example. Find the line vol tages and currents i n per u n i t on the
high-voltage s i d e o f t h e t ra n s form e r .
E x a m p l e 1 1 .7.
Sollliion . In Exa m p l e 1 1 .2 w e
c urrents flowing
fo u n d
t h e pos i t i v e - a n d n e g a t i v e - se q u e n c e
l hat
0.9857/ 43. °
t o w a r d t h e re s i s t ive load a re
f,� ' )
I(�!.)
=
=
6
0 . 2346
/ 250 .3°
per u nit
per
unit
while the correspond ing vol tages on the low-voltage Y side of the transformer are
/ 43 .6°
per unit ( I ine-to-neutral voltage base )
0.9857
0.2346/250 ,3°
30°
0.9857�6 + 30° 0.9857//73.6° 0.2783 + }0.9456
0.2346/ 250 .3 -30° 0.2346 2 0 .3° -0.1789 -}0 . 15 1 7
0.0994 jO.7939
0.9857! -46.4° 0.6798 -}0.7138
O.2346i..=- 19.7° 0.2 09 -}0.0791
-0.9581 -}0.2318
-0.0419 + }0.2318
-1.0 +}O 1.0�
Va� )
=
Va\�)
=
p e r unit ( line-to-neutral voltage base )
Advancing the phase angle of t h e low-voltage positive-sequence vol tage by 30° and
retarding the negative-sequence voltage b y
give on the high-voltage side
Vj I )
vY)
VA
=
=
=
=
V� I )
=
VII =
+
Vc
=
lTvy)
V,� I )
V�I)
=
=
vY)
+
= ,
VJ I ) = a !' VJ I ) =
Vf)
=
+
+
=
=
v,F)
VP)
=
=
=
=
0 . 9 007
-
jO .7929
=
=
(U�/ 82 .80
per
1 .20i..=- 4 1 .4°
per u n i t
unit
per u n i t
458
CHAPTER
II
SYM M ETRICAL COM PON ENTS A N D S EO U E N C' I : N ETWO R K S
Note that the line-ta-neutral vol tages on the h igh -voltage
a rc equal in p er unit to h e lillc-to-lillc voltages
low-vol tage Y side. The line-to-l i n e vol ta ges are
t
found insi of for
0.09 4 jO.7939 -0.9007 jO .7929 -0.8013 jl .5868
1 .78/ 1 16.8" p
( i n a voltage base)
1 .7S
0.90 7 -jO.7939
2.2.0066//-22 .7° per ni / in n ra vo b s
-22 .7° l.1 9 -22.7° per unit (line-line voltage base)
-1.0 -0.0994 -jO.7939 l.09 4 - jO .7939
pe (line-neutral voltage base)
/
1.356 / 0.783/ 215 ,8" per unit ( line-line voltage base)
pha
s
e
i
s
a
r
e
s
i
s
t
a
nc
e
of
1
.
0
/
Q:
per
uni
t
,
and
p
i
va
l
u
e
s
i
n
t
h
i
s
pr
o
bl
e
m.
Li
k
e
w
i
s
e
,
and
in p Therefore, must be identical to VA expres ed in
VA B VA - VB
=
=
=
Vn c
=
=
=
Vc 'A
=
=
=
r:;
vJ
/
=
I I I () X)
I .02XL
-----
=
f3
+
+
the transformer
1 1 .2
the
Exam p l e
=
+
pc r u n i t ( I i n c - l i ne vo l L l g c h a s e )
I .n =
u t ( l e - eut l
1 .9007 -
l t age a
jO .7l)]l)
e)
=
Ve - VA
f3
er u n it l e - neu t r l
L L 6 .H." =
Vn - Vc
1 .35 6
+
de
6.
=
2 1 5 .80
2 1 5 .8"
=
r unit
=
I� l )
If-)
Since the load impedance in each
VY> are fou n d to h ave ident ical er - u n t
VY> are i dentical
er unit.
IA
per un it. Thus,
in
Ie
=
=
1.20/
-
4 1 4°
.
p
er
unit
1.0� per ni
u t
example emphasizes t h e fact t h a t i n going from one side of the 6. - Y o r Y - 6.
transformer to the other, the posi t ive-sequence components of curr e n ts a nd
voltages on one side must be phase shifted separately from the negative­
sequence components on the same side before combining them toget her to fo rm
the actual voltage on the other side.
Remarks on Phase Shzft. The American National Standards I nstitute
(ANSI) requires connection of y-� and � -y transformers so that the positive­
sequence voltage to neutral, VH N ' on the h igh-voltage side leads the positivesequence vol tage to neutral, VX )n' on the low-voltage side by 30° . The wiring
This
I
-�'
1 1 .9
A
8 HI
X1
B
8 H2
X2
c
-O H3
X30
(a)
V)l)
(}
()
a
A
C HI
XI
UNSYMM ETRICAL SERIES I MPEDANCES
�
b
B
-o
H2
X2 8
c
C
8 H3
X3 0
le ads Va( l ) by 300
459
b
c
a
FIGURE 1 1.23
Labeling of l ines connected to
a t h re e - p h a s e Y -� t r a n s ­
former.
di agram of Fig. 1 1 .2 1 and the con nection diagram of Fig. 1 1 .23( a ) both satisfy
the ANSI requi rement; and b ecause the connections of the p hases to the
transformer term inals H I ' H 2 , H3 - X I ' X2 , X3 are respectively m a rked
A , B, C a, b, c as shown in those fi gures, we find that the positive-sequence
vol tage to neutral Vl� leads the positive-sequence vol tage to neutral Va�)
b y 3 0° .
I t is not abso l u tely necessa ry, however, to l abe l l i nes attached to t h e
transformer terminals Xl ' X2 , a n d X 3 b y t h e letters a , b, a n d c , respectively, a s
w e have done, since n o standards have been adopted for such labeling. I n fact,
in ca lcu l a t ions the designa tion of l i nes could be chosen as show n in Fig.
1 l .23( b ), \\'h ich shows the letters b, c, and a associated, respectively, with X l '
X 2 ; and X:; . If the scheme of Fig. 1 1 .23( b ) is preferred, i t is n ecessary o n l y t o
exchange b for a , C for b , and a for c in the wiring and phasor di agrams of Fig .
1 l .2 1 , which wou ld then show that VaS: ) l e ads V)� b y 90° a n d t h a t V};) l ags
V);� by 90° . It is easy to show that similar statements also apply to t h e
correspond ing currents.
We shall continue to fol low the l abel ing scheme of Fig. 1 1 .23( a ), a n d Eqs.
( 1 l .88) then become identical to the ANSI requirement. When problems i nvolv­
ing u n symmetrical fau lts a re solved, positive- and nega tive-sequence compo­
nents are found sep a rately and phase shift is taken into account, if n ecessary, by
applying Eqs. ( 1 1 .88). Computer programs can be written to i n corporate the
effect s of phase shift.
A transformer in a three-phase circu it m ay co nsist of three indivi d u a l
s i n g l e - p h a s e u n i ts , o r i t m a y b e a t h re e - p h ase t ra n sfo r m e r . A l t ho u gh t h e
zero-sequence series imped ances of three-phase units may d i ffer slightly from
the positive- and negative-sequence v a lues, it is customary to assume that series
i m p e dances of a l l se q u e n c e s are e q u a l r e g a rd l ess o f t h e type o f t ransformer.
Tab l e A . 1 in the Appendix l ists tranformer reactances. Reactance and
i mpedance are almost equal fo r transformers o[ 1 000 kYA or l arger. For
s i m p l i c i ty in our c a l c u l a t i o n s we o m i t the s h u n t a d m i t t a nce which a cc o u n t s for
excit i ng current.
-
1 1 .9
U N SYM M ETRI CAL SERIES
IMPEDANCES
In the p revious sections we have been concerned particularly with systems t h a t
are normally balanced. Let us l ook, however, a t the equa t ions o f a three-phase
,
460
a
CHAPTER
1 1 SYMMETRICAL COMPONENTS AND S EOUENCE NEnVO RKS
-
10
Za
Ib
Zb
b
-
c
-
Ie
a'
b'
Zc
c'
flGURE 1 1 .24
Portion of a three-phase system showing th ree u n equal
series i mped a n ces.
circuit when the series impedances are unequal. We shall reach a conclusion
that is important in analysis by symmetrical components. Figure 1 1 .24 shows the
unsymmetrical part of a system with three u nequal series impedances Z a ' Z b '
and Zc ' If we assume no mutual inductance (no coupling) among the three
impedances, the voltage drops across the part of the system shown are given by
the matrix equation
( 1 1 .89 )
and i n
t enns of the symmetrical components of voltage and current
vaa'
( O)
A Vaa'(I )
where A
equation
Za
0
0
0
Zb
0
0
0
Zc
(2 )
vaa'
( 1 1 .90 )
a
I
1a(2)
is the matrix defined by Eq. ( 1 1 .9). Premuitiplying both sides of the
by A - I yields the matrix equation from which w e obtain
vaa( O)
=
13 1a(0) ( Za
+
Z b + Z c ) + 1.3 1a( 1 ) ( Z a
( + aZ b
+ 11(2
3 a ) Za
vaael)
V{2}
ad
I f the
A
1a(0)
1(1)
=
=
11(
0) ( Z a
3 a
) ( Za
l[(O
3 a
+
+
a2Zb + aZ )
c
a 2 Zc )
+
aZ b + a2 Z c ) + 1.[
3 a( 1 ) ( Za + Zb + Z )
+
a2Zb
c
+
aZ c )
+
i mpedances are made equal (that
1 )( Za + aZb
11(
3 a
IS,
if Za
=
Zb
+
=
( 1 1 .91)
a 2Zc )
ZJ, Eqs. ( 1 1 . 9 1 )
1 1.10 SEQUENCE NETWORKS
461
reduce to
v(O)
aa
=
1(0)Z
a
a
( 1 1 .92)
If t h e imped ances are
un e qu a l , however, Eqs. ( 1 1 .91) s h ow that the voltage drop
of a ny one sequence is dependent on the currents of all three sequences. Thus,
we conclude that the symmetrical components of unbalanced currents flowing in
a balanced load or in balanced series impedances produce voltage d rops of like
sequence only. If asymmetric coupling ( such as unequal mutual inductances )
existed among the three impedances of Fig. 1 1 .24, the square matrix of Eqs.
( 1 1 . 89) and ( 1 1 .90) would contain off-diagonal elements and Eqs. (1 1 .91) would
h ave additional terms.
A l though current i n any conductor of a th ree-phase transmission line
ind uces a voltage in the other phases, the way in which reactance is calculated
eliminates consideration of coupling. The self-ind uctance calculated on the
basis of complete transposition i ncludes the effect of mutual reactance. Th e
assumption of transposi tion yields equal scries impedances. Thus, the compo­
nent currents of any one sequence produce only voltage drops of l i ke sequence
in a transmission line; that is, positive-sequence currents produce positive­
sequence voltage drops only. Likewise, negative-sequence currents produce
negative-sequence voltage drops only and zero-sequence currents produce zero­
sequence voltage drops only. Equations 0 1 .9 1 ) apply to u nbalanced Y l oads
because points a' , b', and c' may be connected to fonn a neutral. We could
study variations of these equations for special cases such as single-phase loads
where Zb = Zc = 0, but we continue to confine our discussion to systems that
are balanced before a fault occurs.
11.10
SEQUENCE N ElWORKS
In the preceding sections of this chapter we have developed single-phase
equivalent circuits in the form of zero-, positive-, and negative-sequence circuits
for load impedances, transformers, transmission l ines, and synchronous ma­
chines, all of which constitute the main parts of the three-phase power transmis­
sion network. Except for rotating machines, all parts of the network are static
and w ithout sources. Each individual part is assumed to be l inear and t hree­
phase symmetrical when connected in Y or fl configuration. On the basis of
these assumptions, we have found that:
•
•
n any part of the network voltage drop caused by current of a certain
sequence depends on only the impedance of that part of the network to
current flow of that sequence.
The impedances to positive- a nd negative-sequence currents, Z I and Z2' are
equal in any static circuit and m ay be considered approximately equal i n
synchronous machines u nder subtransient conditions.
I
462
•
•
•
•
•
CHAPTER 1 1
SYMMETRICAL CO M PO N ENTS AND SEQU ENCE N ETWO RKS
I n any part of the network impedance to zero-sequence current, 2 o , is
generally different from 21 and 22.
Only positive-sequence circuits of rotating machines contain sources which are
of positive-sequence voltages.
Neutral is the reference for voltages in posltlve- and negative-sequence
circuits, and such voltages to neutra l a re the same as vol tages to grou nd if a
p hysical connection of z e ro or o t h e r li n i t e i m ped a nce e xists between n eutral
and ground in the actual circuit.
No positive- or negative-sequence c u r r e n ts How b e tw e e n n e u t r a l p o i n t s a n d
ground.
Impedances 2 n in the physical connections between neutral and ground are
not included in positive- an d negative-sequence circu its but arc represen ted
b y impedances 3 2" b e tw ee n t h e p o i n t s fo r n e u t r a l a n d g ro u n d in t h e z c r o ­
sequence circuits only.
These characteristics of individual sequence circuits guide the construction of
corresponding sequence networks. The object of obtaining t he values of the
sequence i mpedances of the various p a rts of a power system is to enable us to
construct the sequence networks for the complete system. The network o f a
particular sequence-constructed by joining together all the correspond ing
sequence circuits of the separate parts-shows all the paths for the flow of
current of that sequence in one phase of the actual syste m .
In a balanced t hree-phase system t h e currents flowing in the three p hases
under normal operating conditions constitute a symmetrical positive-sequence
set. These positive-sequence currents cause voltage drops of the same sequence
only. Because currents of only one sequence occurred in the preced ing chapters,
we considered them to flow in an i ndependent per-phase network which
combined the positive-sequence emfs of rotating m a c h i n e s a n d t h e impedances
of other static circuits to positive-sequence currents only. That same per-phase
equivalent network is now called the positive-sequence network i n orde ' to
d istinguish it from the networks of the other two sequences.
We have discussed the construction of impedance and admittance rep re­
sentations of some rather complex positive-sequence networks in e a rl i e r ch ap­
ters. Generally, we have not included the phase s h i ft ass o c i a t e d w i t h 6. - Y and
y-� transformers in positive-sequence networks since p ractical systems are
d esigned with such p h ase s h i ft s su m m i ng to z e ro a r o u n d all loops. I n detailed
calculations, however, we must remember to advance all positive-sequ ence
voltages and currents by 30° when stepping up from the low-voltage side to the
high-voltage side of a !:J.-Y or Y -� transformer.
The transition from a positive-sequence network to a nega tive-sequence
network is simple. Three-phase synchronous generators and motors have inter­
nal voltages of positive sequence only because they are designed to generate
balanced voltages. Since the positive- and negative-sequence impedances are
the same in a static symmetrical system, conversion of a positive-sequence
I LID SEQUENCE NETWORKS
463
network to a negative-sequence network is accomplished by changing, if n eces­
sary, only the impedances that represent rotating m a c h i ne ry and by omitting the
emfs. Electromotive forces are omitted on the assumption of balanced gener­
ated voltages and the absence of negative-sequence voltages induced from
o u tsi d e sources. Of course, in using the negative-sequence network for detailed
calculations, we must also remember to retard the negative-sequence voltages
and currents by 30° when stepping up from the low-voltage side to the
high-voltage side of a D.-Y or Y -Do transformer.
Since all the neutral points of a symmetrical three-phase system are at the
same potential when b a l a nced t h ree-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 and is the reference node of the positive- and negative-sequence n et­
works. Impedance connected between the neutral of a machine and ground is
not a p a rt 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 previ­
ous chapters, may con tain the exact equivalent circui ts of parts of the system or
be s imp l ified by omitting series resistance and shunt admittance.
Draw the negative-sequence network for the system described i n
Example 6 . l . Assume t hat t he negat ive-seque nce reactance o f each m achine is
equal to its sub transient reactance. Omit resistance and phase shifts associated
with the transformer connections.
Exa m p l e 1 1 .8.
Since all the negative-sequence reactances of the system are equal to the
positive-sequence reactances, t he negative-sequence network is identical to the
positive-sequence network of Fig. 6.6 except for the omission of emfs from
t he negative-sequence network. The required n etwork is drawn without tra nsformer
ph ase sh i fts i n F i g . 1 1 .25 .
Solution.
d e t e r m i n e d fo r t h e
various separate
parts of the system are readily com bined to form the complete zero-sequence
network . A three-phase system operates single phase insofar as the zeroZe ro-se q u e n ce e q u iva l e n t c i rc u i t s
) 0 . 08 5 7
l
jO. 1 8 1 5
j O.20
m
j O . 09 1 5
jO. 5490
Reference
FIGURE 1 1 .25
Negative-sequence network for
Exa mple 1 1 . 8.
464
CHAPTER 1 1
R
SYM METRICAL COMPON ENTS A N D SEQUENCE N ETWORKS
T
T
s
M
N
-
-
Reference
FIGURE 1 1.26
One-l i n e d iagram of a sma l l power sys t e m a n d t h e correspon d i ng zero- s e q u e n c e n e twork.
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 cu rrents may be flow i n g i n the ground,
the ground is not necessarily at the same potential at all points and the
reference node of the zero-sequence network does not represent a ground of
unifonn potential. We have already d iscussed the fact that 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 node of the system. It is
because the impedance of the ground is included in the zero-sequence
impedance that vol tages measured t o the r eference node of the zero-sequence
network give the correct voltage to equivalent ideal ground. Figures 1 1 .26 and
1 1 .27 show one-line diagrams of two small power systems and their correspond­
ing zero-sequence networks simplified by omitting resistances and shunt admit­
tances.
The analysis of an unsymmetrical fault on a symmetrical system consists i n
finding the symmetrical components o f the u nbalanced currents that a r e flow­
ing. Therefore, to calculate the effect of a fault by the method of symmetrical
1 1 .10
g�
N
:�
�
8.
8.
8.
N
Q
p
46S
z
� y
g�
� fi J: ut � �
8. r Tu
6�
SEQUENCE NElWORKS
Y
X
� �
w
z
[] _�
R
T
S
Cfj
Reference
FI GURE 1 1 .27
O n e - l i n e d iagram of a s m a l l power s ys t e m and t he corres p o n d i n g zero-sequence network.
components, it is essential to determine the sequence impedances and to
combine them to form the sequence networks. The sequence networks carrying
the symmetrical-component currents l�O), I� l ), and 1�2) are then interconnected
to represent various unbalanced fault conditions, as described in Chap. 1 2.
Draw the zero-sequence network for the system d escribed in
Assu me zero-sequence reactances for the generator a n d motors of
Exa mple 1 1 .9.
Example
6. 1 .
The
ze ro-se q u e n c c r e a c t a n c e
0 . 0 5 p e r u n i t . A c u r re n t - l i m i t i n g r e a c t o r o f 0 . 4 n i s in each o f t h c ncutrals of the
ge ne r a t or a n o the l a rg e r motor.
l i n e is 1 .5 n/km.
of the
transmission
The zero-sequence l eakage reactance of transformers is equal to the
positive-sequence reactance. So, for the transformers X 0 = 0.0857 per unit and
0.09 1 5 per unit, as i n Example 6 . 1 . Zero-sequence reactances of the generator and
Solution.
466
CHAPTER I I
S YMMETRICAL COM PONENTS A N D SEQ UENCE NE1W O R KS
motors are
Generator :
Motor 1 :
Motor 2 :
I n t h e generator circui t
Xo
0.05
0.05 ( -200300 ) ( --1 3.3 .82 ) 0.0686
D.05 ( 300 ) ( 11 33 ..28 ) 0. 1372
(32000)2 1.3 3
300
3Zn 3 ( 1.0.333 ) 0.900
3Zn 3 ( 0.0.6435 ) 1.890
176.3 1 .2�.
=
per u n i t
Xo =
Xu
1
=
2
1 00
=
'2
-
--
'-=
Base Z =
per u n i t
per
n
a n d i n t h e motor circuit
Base Z
=
( 1 3 K) 2
=
0 . 635 H
I n t h e i mpedance network for the ge nerator
=
and for the motor
=
4
--
=
per u nit
per u ni t
--
For the transmission line
1 . 5 x ()4
�
0 .5445 per
u ni t
T h e zero-sequence network i s shown i n F i g .
k
jO.0857
l
jO.5445
m
jO.0915
n
r
,-_L-_ r
jO.06B6
)1.890
Reference
FlGURE 1 1.28
Zero-s eque n ce n etwork for Exam p l e
1 1 .9.
jO. 1372
unit
PROBLEMS
11.11
467
SUMMARY
Unbalanced voltages and currents can be resolved into their symmetrical
components. Problems are solved b y treating each set of components separately
and superimposing the results.
I n balanced n etworks h aving strictly symmetrical 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 power-flow
stu d i e s, faul t calculations, and stabil i ty studies. I f the fau lt calculations or
s t a b i l ity studies involve unsymmetrical fau l ts on otherwise symmetrical systems,
the negat ive- and zero-sequence networks are also needed. Synthesis of the
zero-sequence network requires particular care because the zero-sequence
network may diffe r from t h e others considerably.
PRO BLEMS
1 1 .2.
1 1.3.
1 1 .4 .
1 1 .5.
&
�
10�
, V2) = 20
, and V};;) =
V , determine analytica l ly
Va� ) = 50
the vol tages to neutral Van ' Vb" , and Veil ' and also show graphically the sum of the
given symmetrical components which determine the l ine-to-neutral voltages.
When a generator has term inal a open and the other two terminals are connected
to each othcr with a short c i rcuit from this connection to ground, typical values
for the symmetrical com ponents of current i n phase a are 1� 1)
I� 2) =
and I� O) =
A. Find the cu rrent into the ground a n d the
current in each phase of the generator.
Determine the symmetrical components of the three currents I = 10&, Ib
and Ie =
A.
The currents flowing in the l i nes toward a balanced l oad connected i n � are
10 =
Ib =
a nd Ie
Find the symmetrical compo­
n e n ts of the g i v en l i n c c urre n t s and d raw ph asor diagrams of the positive- and
1 1 .1. If
250L22:., 350�
10/230 , 10�
100&, 141 .4/2250 , 100�.
Thc voltagcs there is 100&, .sO.SL-1 2 1 .4 ° ,
1O-D.
=
=
a
=
n e g a t i v e - s e q u e n ce l i m: (l n d p h a se c u r r e n t s . W h a t is luI, i n a m pe res?
at the t e r m i n a l s of a bal anced load consisting of t h r e e 1 O-D. resistors
in Y are Va l> =
and v;.a =
V/Jc =
V.
Assuming t h a t
no connection to the neutral of the l oad, find the l ine
cu rrents from the symmetrical components of the given l i ne voltages.
Find the power expended in the t h ree
resistors of Prob. 1 1 .5 from the
symmetrical com p o n e n t s o f curre n t s and voltages. Cheek t h e answer.
If therc is impedance in t h e neutral connection to ground of a Y -connected load,
then show that the vol t ages Va ' Vb ' and v;. of Eq. ( 1 1 . 26) must be interpreted as
voltages with respect to grou n d .
A bal anced three-phase l o a d consists of �-connected impedances Z 6 i n p a rallel
with solidly grounded Y -con nected i mpedances Z y .
( a ) Express the currents la ' 1b, a n d Ie flowing i n t h e lines from the supply source
toward the l oad in terms of the source vol tages Va ' Vb) and �.
90�
con ne c t e d
1 1 . 6.
1 1 .7 .
1 1 .8.
60 /-90°,
,
468
CHAPTER
11
SYMMETRICAL COM PO NENTS AND S EQ UENCE N ETWORKS
,
,
(b) Transform the expressions of part ( a ) i nto their symmetrical component
equivalents , and thus express 1(0)
and [(a2 ) in terms of Va(O) Vel)
, and
a , 1(1)
a
a
vy).
(C) Hence, draw the sequence circuit for the combined load.
11.9. The Y-connected impedances in parallel with the t.-connected impedances
Ze:,. of
Prob. 1 1 .8 are now grounded thro ugh a n i m pedance Zg .
( a ) Express the currents la '
a n d Ie flowing i n the lines from the supply source
t ow ard the load in te rms of the source voltages Va ' Vb ' a nd � and the voltage
� of the neutra l point.
( b) Expressing Vn in terms of I�() , 1� 1), 1�2), and Z}; , fi nc.l the e q u a t i o n s fo r these
currents in terms of V(O)
a , and Va( 2 ) ·
a , V(I)
(c) Hence, draw the seq u en c e circuit for the co mb in e d l oa d
Ib,
1 1. 10.
.
impedances Zo , Z ] ,
1 1. 1 1 .
11.12.
( b ) Hence, determine t h e line curren t IL i n amperes.
(c) D etermine the open-circuit vol tages to n eu tra l of phases b and c a t the
receiving end.
( d ) Verify your answer to part (c) w i thout using symmetrical compon ents.
and Z2 o f t h e l i n e .
Solve Prob . 1 1 .10 if the same 420-D i nductive load is connected b etwe e n phases a
a nd b at the receiving end. In part ( c ) find the open-circu i t voltage of phase c
only.
A Y-connected synchronous generator has sequence rea c tan ces Xo = 0.09, X I =
0.22, and X2 = 0.36, all in per u n i t. The neutral p oi n t of the machine is grounded
through a reactance of 0.09 per unit. The machine is running on no load with
rated terminal voltage when i t suffers a n unbal anced fault. The faul t currents out
of the machine are la
0,
= 3.75
, and Ie = 3.75� , all in per u n i t
with respect to p h a s e a l i n e to -n e u t r a l voltage. Determine
( a ) The terminal voltages i n each p hase of the machine with respect to ground,
( b ) The voltage o f the neutral point of the machine with respect to ground, and
( c ) The nature (type) o f the fault from the r esu l ts of part ( a).
Solve Prob. 1 1 . 1 2 i f the fault currents i n per u n i t are la
0, Ib = - 2.986
,
=
1 1.13.
and
1 1.14.
1 1.15.
-
Suppose t ha t the l ine-to-neutral voltages at the send ing end of the l i n e described
in Example 1 1 .5 can be maintained constant at 20()-kY and t h a t a s ingl e p h a s e
inductive load of 420 n is con nected b e tw e e n p h as e a and n eu t r a l at t he
receiving end.
(a) Use Eqs. ( 1 1 .5 1 ) to express n u merically the receiving-end sequence voltages
VY;/., Va�� , and V}�l in terms of the load current IL and the sequence
Ie
=
2.986
&.
Ib-
�
=
&
Assume that the currents specified i n P r o b . 1 1 .4 a r e flowi ng toward a load from
lines connected to the Y s i d e of a Ll-Y t ransformer rated 10 MY A, 1 3 . 2 6.j66Y
kY. Determine the currents flowi ng in the l i n e s on the t. s i de by converting the
symmetrical components of the currents to per unit on the base of the trans­
former rating and by shiftin g the components according to Eq. ( 1 1 .88). Check the
results by computing the current s in each p h ase of the t. windings i n amperes
d irectly from the currents on the Y side by mUltiplying by the turns ratio of the
windings. Complete the check by co mp u ting the line currents from the phase
currents on the � side.
Three single-phase transformers a re connected as shown in Fig. 1 1.29 to f� r m a
Y-Ll transfo rmer. Th e hig h -v o l t a g e w i n d ings are Y-connected with polarity marks
PROBLEMS
as ind icated. Magnetically coupled wind ings are d r awn
469
in parallel directions.
Determine the correct placement of polarity marks on the low-voltage windings.
Identify the numbered terminals on the low-voltage side (a) with the letters a, b,
and c , w he re l� l ) l e a d s 1;1) by 30, and ( b) wi th the l etters a', b', and c ' so t hat
I�}) is 90 out of p hase with I�l) .
d
- -
FIGURE 1 1 .29
Ie
Circu it for Prob. 1 1 . 1 5 .
- -----+-
1 1 . 1 6.
1 1 . 17 .
1 1 . 1 8.
1 1 .19.
Bal anced three-phase voltages of 1 00 V l i n e t o l i n e are applied t o a Y-conn ected
load consisting of three resistors. The neutral of t h e load is not grounded. The
resistance in p hase a is 10 D , in p h ase b is 20 D , and in p hase c is 3 0 D . Select
voltage to neutral of the three-phase l in e as reference and determine the current
in phase a and the volt age Va n .
Draw the negative- and zero-sequence impedance networks for the power system
of Prob. 3 . 1 2 . Mark t h e values of all reactances i n pe r unit on a base of 50 MVA,
13.8 kV in the circui t of generator 1 . Letter the networks to correspond to the
single-line diagram. The neutrals of generators 1 and 3 are connected to ground
through current-limiting reactors h aving a reactance of 5%, e ach on the b ase of
the machine to which it is connected. Each generator has negative- and zero­
sequence reactances of 20 and 5 % , respectively, on i ts own rating as base. The
z e ro - s e q u ence r e a ct a n c e of the transm ission line is 2 1 0 D fro m B to C and 250 D
from C to E.
Draw the negative- and zero-sequence impedance networks for t he power system
of Prob. 3 . 1 3 . Choose a base of 50 MVA, 1 38 kV in the 40-fl transmission line
and mark all reactances in per unit. The negative-sequence reactance of each
synchronous machine is equal to i ts subtransient reactance. The zero-sequence
reactance of each m a c h i ne is 8% based on its own rati ng. The neutrals of the
m a c h i n e s a re con n e c t e d to ground t hrough c u r re n t -l i m i t i n g reactors h aving a
r e a c t a n c e of 5 % , each on the base of the machine to w hich i t is connected.
Assume that the zero-sequence reactances of the transmission l i n e s are 300% of
their positive-sequence reactances.
Determine the zero-sequence Theven i n impedance seen at bus © of the system
'
described in Prob. 1 1 . 1 7 if t ransformer T) has (a) one u ngrounded a n d one
solidly grounded neutral, as s hown 10 Fig. 3.23, and (b) both neutrals are solidly
grounded.
CHAPTER
12
UNSYMMETRICAL
FAULTS
M ost of the faults that occur on power systems are u nsymmetrical faults, which
may consist of unsymmetrical s hort circuits, unsymmetricar faults through
impedances, 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 unsymmet rical faults, through
either the breaking of one or two conductors or the action of fuses and other
d evices that may not open the three phases simultaneously. S ince any u nsym­
m etrical 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 ocCurrence of the fault. We
consider faul ts on a power system by applying Thevenin's theorem, which allows
us to find the current i n the fault by replacing the entire system by a single
generator and series impedance, and we show how the bus impedance matrix is
applied to the analysis of unsymmetrical faults.
12.1
U NSYMMETRICAL FAULTS
O N POWER SYSTEMS
J
In the derivation of equations for the symmetrical components of currents and
voltages if! a general network the curren t s flowing o u t of the original balanced
1 2. 1
UNSYMMETRICAL FAU LTS ON POWER SYSTEMS
-----.-1�Irll
b------�i-F-rb------r
Ij
a
c
471
­
------------------
-
If'
F I G U RE 12.1
Three conductors of a t h ree- p h a se system.
The stubs ca rrying currents '/0 ' I[h ' and l[e
may be i n t erconnected to represe n t d i ffer­
ent types of fa u l t s .
system from p h ases a, b, and c at the fau l t point wil l be designated a s If lib '
a n d l[c ' respect ive ly. We can visualize these cu rrents by refe rring to Fig. 1 2 . 1 ,
wh ich sh ows the th ree l i n es a , b , and c o f t h e t h ree-phase system a t t h e p a r t o f
t h e network where the fa u l t occurs. The flow of current from e a c h l i n e i nto the
fa u l t is indicated by arrows sh own on the d i agram beside hypothetical stubs
con nected to each l i ne at t h e fau l t l ocation. Appropriate connections of the
s t u bs represent the various types of fa u l t . For instance, d i rect con n e ction of
stubs b and c produces a l i ne-to-l ine fau l t t h rough z ero impedance. The current
in stub a is then zero, and lib equ a l s l[c '
The l i ne-to-grou nd vol tages at any bus (J) of the sys tem Juring t h e fau l t
will be designated V}a ' �h' a n d � c ; and w e s h a l l con tinue to u s e s u perscripts 1 ,
2 , a n d 0 , respect ively, to d e note positive-, negative-, and zero-seque nce quanti­
t ies. Thus, for example, Vj.�l ), V}�12 ), and Vj�O) will denote , respective ly, t h e
posi tive-, negat ive-, a n d zero-seque nce components of the l i ne-to-gro u n d volt­
age Vj·a at bus CD during t h e fau l t . The l ine -to-neutral vol tage o f p hase a at t h e
fau l t point before t h e fau l t occurs w i l l be designated s imply by VI ' w hich i s a
positive-sequence voltage si nce the system is ba lanced. We m e t t h e p re faul t
voltage VI previously i n Sec. 1 0.3 when calcu l at ing the curre n t s i n a p ower
system with a symmetrical th ree-phase fau l t app lied.
A singl e-l ine d iagram o f a power system conta ining two synchronous
Ill � l c h i il c s i s s h ow n i ll Fi g . 1 2 . 2 . Such ,l sys t e m i s s u ffi c i e n t l y ge ne ral for
e q u a t i o ll s de rived t h e re fro m t o be , l p p l i c ( l b l e t o , I ll Y b;I I Cl llccd sys t e m regardless
o f the com plexity. F i g ur e 1 2 . 2 a l so s h o w s t he seq u e nce n e t w o r k s o f the system.
Th e poi n t wh ere a fa u l t i s a s s u m e d to occ u r is m a r k e d P , a n d in this particular
exam ple it is cal led bus CD on the single-l ine di agram and in the s equence
ne tworks. Machines are represented b y their subtransient i n te r n a l voltages in
series with t h e i r subtrans ient reactances when subt ransient fau l t condit ions a re
b e ing studied.
I n Sec. 1 0.3 we u s e d the bus impedance matrix composed o f positive­
seque nce impedances to d e t e r m ine currents and vol tages upon t h e occurre nce
of a symmet rical th ree-phase fau l t . The method can be eas i ly extended to apply
to unsymmetrical fau l ts by realizing t h at the negative- and zero-sequ ence
ne tworks also can be represented by bus impedance matrices. The bus imped ance
m a t rix w i l l now be wri tten symbolica lly for t h e posit ive-sequence n e twork i n t h e
a,
-
,
472
CHAPTER
12
UNSYMM ETRICAL FAULTS
following form:
CD
Z (buI )s -
W
®
(l)
( l)
®
CD
W
)
Z(l
11
(I)
Z 21
Z 12
®
(I)
Zk l
(I)
Zk 2
Zk( Ik)
Z k( IN)
®
)
ZN( I I
Z N( I 2)
( \k)
ZN
(I)
Z NN
(\)
Z 22
Z lk
Z 2( Ik)
)
Z I( IN
)
Z 2( IN
( 12. 1 )
Similarly, the bus impedance matrices for the negative- and zero-sequence
n etworks will be written
)
Z (b2us
O
Z (bu
s
)
Thus,
CD
W
CD
@
®
)
Z (I 2I
)
Z (2
)
Z 1( 22
( 2)
Z 22
)
Z (I 2k
Z 2( 2k)
21
®
)
Z (I2N
)
Z 2( 2N
( 1 2 .2)
=
®
)
Zk( 21
Zk( 22)
Z k( 2k)
Z k( 2N)
®
Z N( 2)I
Z N(22)
Z N( 2k)
( 2)
Z NN
CD
W
®
®
CD
W
Z (I OI )
Z 2( O1 )
Z (1O2)
( O)
Z 22
Z (I Ok)
Z ( O)
®
)
Z k( Ol
Z k(O2)
Z k( Ok)
Z k( ON)
@
( Ol
ZN
)
Z N( O2
(O
Z Nk
)
(O )
Z NN
2k
)
Z (I ON
)
Z 2(ON
=
)
ZW, zg>, and ZfJ) denote representatiye elements of the bus impedance
matrices for the positive-, negative-, and zero-sequence networks, respectively.
I f s o desired, each of the networks can be replaced by i ts Thevenin equivalen t
b etween a ny one of the buses and the reference node.
12.1
UNSYMM ETRICAL FAULTS ON POWER SYSTEMS
473
y
(a) Single-line diagram of balanced three-phase system
p
( b ) Positive-seq uence n etwo rk
+
®
( e ) The venin equivalent of the
positive-seq uence network
fa
-
1(2)
Reference
Cc) Negative-sequence n etwork
Reference
( d ) Zero-sequence n etwork
p
+0
( f) The venin equivalent of the
negative-sequence n etwork
( g ) Th eveni n eq uivalent of the
zero-seq uence network
FlGURE 1 2.2
S i n g l e - l i n e d i a g r a m of a t h re e-phase sys t e m , the t h ree sequence networks of the system, a n d the
Thev e n i n e q uivalent of each n e twork fo r a fa u lt at P , whieh is cal led bus CD.
The Thevenin equivalent circuit between t h e fault point P and the
reference node in each sequence network is shown adjacent to the diagram of
the corresponding n etwork in Fig. 1 2.2. As in Chap. 10, the v o l t ag e source in the
positive-sequence network and i ts Thevenin equivalent circuit is VI ' the prefault
voltage to neutral at the fault point P, which happens to be bus ® in this
illustration. The Thevenin impedance measured between point P and the
reference node of the positive-sequence network i s zW, and its value depends
474
CHAPTER
12
UNSYMMETRICAL FAU LTS
on the values of the reactances used i n the network. We recall from Chap. 1 0
that subtrans ient reactances of generato rs and 1 .5 t imes the subtrans ient
reactances (or else the t ransient reactances) o f synchronous motors a re the
values used i n c a l cu l a t i n g the sym metrical current t o b e interrupted.
There a re no negative- or zero-sequence curre nts flowing before the fau l t
occurs, and the prefa u l t voltages are zero a t a l l buses o f t h e n e g a t i ve a n d
zero -sequence n e tworks . Therefore, t h e prefa u l t voltage between p o i n t P a n d
the reference node is zero i n t h e ne ga t i ve - a n d zero-se q u e n c e �e tw o rks a n d n o
electromotive for c e s (emfs) a ppear i n t h e i r Thcv e n i n e q u iv a l e n t s . T h e n e g a t ive­
a n d zero-s equence i mped ances be t w e e n p o i n t P at bus CD J n cJ the re fe re n c e
n o de i n the respective networks are r e p r e s e n t e d by t h e Thc\ e n i n i m p e d a n c e s
Zk� and Z fOJ d i a go n a l e lements o f Z \��s and Z \��S ) r e s p c c t i v c : y .
Since Ifl1 i s t h e c u r re n t n ow i n g froll/ t h e sys t e m il / to t h e fa u l t , i t s
symmetrical components 1fa
( 1 ) ) 1fa( 2 ) ' ' ll1 d 1I(0) How O l l l o f t h e r c '-' p c c t i v e s e Cl u c n e e
networks and t h e i r equivalent circu i ts a t point P , as s h own i n F i g . 1 2 . 2 . T h u s ,
t h e currents I}�), - Ij;), an d - I}�) represent injected currems ill lO the fa u l ted
bus ® of the positive-, negative-, and zero-sequence networks due to the fa u l t .
These curren t i nj ections cause voltage changes a t t h e b u ::, c ::, u f t h e p o s i t i v c - ,
negative-, a n d zero-sequence networks, which can b e calcul ated from t h e b u s
impedance m a t rices in t h e manner d emonstrated in Sec. 10.3. For instance, d u e
t o t h e inj ection I}�) i n t o b u s ®, t h e voltage changes in the pos itive-seq uence
network o f the N-bus system are given i n general terms by
-
-
(
iJ
�
-
-
®
Ll V1Oa )
6. Vk( al )
zf\) zW
Z � II )
zW
=
®
Z k( Il)
)
Z k( !2
)
Z ( I,\'
I
I
Z( )
2 1\'
o
o
..- I ( aI )
f
(I)
ZH
(I)
Z Nk
®
®
(I)
Z NN
o
- 2 ( 1 )1 ( 1 )
l k fa
2 k fa
- Z ( I )1 ( 1 )
k k fa
- Z ( I )1 ( 1 )
fa
- 2(1) 1 (1)
Nk
( 12.3)
1 2.1
UNSYMMETRICAL FAULTS ON POWER SYSTEMS
475
This equation is quite similar to Eq. ( 1 0. 1 5) for symmetrical faults. Note t h at
only column k of ZblJs enters into the calcu lations. I n industry practice, it is
customary to regard all p refault currents as being zero and to designate the
voltage Vf as the positive-sequence voltage at all buses of the system before the
fault occurs. Superimposing the changes of Eq. ( 1 2.3) on the prefault volt ages
then yields the total positive-sequence voltage of phase a at each bus during the
fau l t,
Vl(a\ )
Vf
Vf
V2( a\ )
=
Vk( lIl )
VI'
.1 V2( aI )
+
�(
VNil
(I)
- Z ( k1 ) 1f( a1 )
I
(1)
Vf L 2( \k) 1fa
Vf
.1 Vl(al )
-
.1 VA( llI )
V/
-
'kli. fa
7(1)[(I)
( 12 .4)
6. VNil
(I)
Th is equation is similar to Eq . 0 0. 1 8) for symmetrical faults, the only difference
being the added superscripts and subscripts denoting the positive-sequence
components of the phase a quanti ties.
Equations for the negative- a nd zero-sequence voltage changes due to the
fau l t at bus ® of the N-bus system are similarly written with the superscripts
in Eq. 0 2. 3 ) changed from 1 to 2 and from 1 to 0, respectively. Because the
prefault voltages are zero in the negative- and zero-sequence networks, the
voltage changes express the total negat ive- and zero-sequence voltages during
the fau lt, and so we have
V ICIIO )
V I(2)
{I
V"( a2 )
( 2)
VNa
=
- Z(0)
(0 )
I k 1fa
V2(O)11
- _ 7 (O) j ( O}
' 2k fa
- Z k( 2k) Jfa( 2 )
Vk(O)
a
- Z(O)
kk 1(0)
fa
( 2") [fa
(2)
ZN
vNil
(O )
- Z Nk
(0)
( O ) 1fa
_
( 12 .5)
When the fault is at bus ® , note that only the entries in col umns k of Zb2Js and
ZhoJs are involved in the calcul ations of negative- and zero-sequence voltages.
Thus, knowing the symmetrical components 1}� ) , 1}�), and I};) of the fault
curre n ts at bus ®, we can determine the sequence voltages at any bus CD of
the system from the jth rows of Eqs. 0 2.4) and ( 1 2.5). That is, during the fault
476
CHAPTER
12
UNSYMMETRICAL FAU LTS
at bus ® the voltages at any bus
vJa(O )
(]) are
=
(O )
- Zj(Ok )/fa
( 12 .6)
V(2)
Ja
=
I f the prefault voltage at bus CD is not Vf ' then we simply replace Vf in Eq.
02. 6 ) b y the actual value of the prefault (posi tive-sequence) vol tage at th at bus.
Since Vf is by definition the actual prcfault voltage at the faulted b u s CD , we
always have at that bus
V(O)
ku
V(I)
ka
=
=
V(2)
ka -
_
Z(O)
kk 1fa(0)
Vf - Z(l)
kk /(1)
fa
( 1 2 .7)
- Z (2) / (2)
kk fa
and these are the terminal voltage equations for the Thevenin equivalents of the
sequence networks shown in Fig. 12.2.
It is important to remember that the currents Ij�), Jj�), and I};) are
. symmetrical-component currents i n the stubs hypothetically attached to the
system at the fault point. These currents take on values determined by the
particular type of fault being studied, and once they have been calculated, they
can b e regarded as negative injections into the corresponding sequence net­
works. I f the system has .1 -Y transfo rmers, some of the sequence voltages
calculated from Eqs. (1 2.6) may have to be shifted in phase angle before being
combined with other components to form the new bus voltages of the fau l ted
system. There are no phase shifts involved in E q . 02.7) when the voltage Vr at
the fault point is chosen as reference, which is customary.
In a system with .1-Y transfo rmers the open circuits encountered in the
zero-sequence network require carefu l consideration in computer appl ications
of the Z b us building algorithm. Consider, for instance, the solidly grounded Y-tJ.
transformer con nected between buses § and ® of Fig. 12.3(a). The positive­
and zero-sequence circuits are shown in Figs. 12.3( b ) and 1 2.3( c), respectively.
The negative-sequence circuit is the same as the positive-sequence circuit. It is
straightforward to include these sequence circuits in the bus impedance matri­
ces Z�Js ' Z��s ' and Zb2�s using the pictorial representations shown in the figures.
This will be done in the sections which follow when Y-/l transformers are
present. Suppose, however, that we wish to represen t removal of the trans­
former connections from bus ® i n a computer algorithm which cannot avai l of
pictorial representations. We can easily undo the con nections to bus ® in the
positive- and n egative-sequence networks by applying the building algorit nm to
1 2. 1
z
'OOlJ'
?
�
Reference
Z
(b)
Reference
Z
UNSYMMETRICAL FAULTS O N POWER SYSTEMS
'-
Reference
477
Z
(c)
Z
�
Reference
Cd)
(e)
FIGURE 12.3
( a ) 6 -Y grou nded transformer w i t h I e a kage impedance Z ; ( b ) posItIve-sequence circu it;
(c) ze ro-seq uence circuit; (d) posit ive-sequ ence c i rcuit w i t h i nternal node; ( e ) zero-seq u e n ce circ u i t
w i t h i n ternal node.
the matrices ZbIJs and zb22s in the usual manner-that is, by adding the negative
of the leakage impedance Z between buses @) and ® in the positive- and
negative-sequence networks. However, a similar strategy does not apply to t h e
zero-sequence matrix ZboJs if i t has been formed di rectly from t h e pictorial
represcntation shown i n Fig. 12.3(c). Adding - Z between buses @) and ®
d o c s n ot re move t h e zero-sequence connect ion from bus ® . To permit uniform
procedures fo r all sequence networks, one strategy is to include an internal
node ® , as shown in Figs. 1 2.3(d) and 1 2.3(e ). 1 Note that the leakage
i m p e d a n c e is now s u b d i v i d e d into two parts between node ® and t h e o t h e r
n o d e s as s h o w n . Con n e c t i n g - Z /2 between b u s e s ® and ® in each of the
s e q u e n c e c i rcu i t s o f Figs. 1 2. 3 ( d ) a n d 1 2.3( c ) will o p e n the transformer con nec­
t ion s t o bus ® . Also, t h e o p e n circuits c a n b e re p resen ted w i t h i n t h e com p u t e r
algori thm by br a nches of arbitrarily large impedances (say, 106 per u n i t).
Intcrn al nodes of transformers can be useful in p ractical computer applications
ISee
New
H. E.
York,
Brown, Solution of Large Networks by Matrix Methods , 2 d e d., J o h n Wi ley & Sons, I nc.,
1 9R5.
478
CHAPTER
12
U N SYMM ETRICAL FAU LTS
of the Z bus building algorithm. The reader is re fe r red to the reference cited in
footnote 1 for further guidance i n handlin g open-circui t and short-circuit (bus
tie) branches.
The faults to be discussed in succeedi ng sections may involve impedance
Zt between lines and from one or two l i n es to ground. When Zj = 0, we have a
d irect short circuit, which is called a bolted fault . Although such direct short
c ircuits result in the highest value of fault current and are therefore the most
conservative values to use when determ i n i n g the effects of anticipated faults, the
faul t impeda n ce is seldom zero . Most fa u l ts a re the result o f i n su l a t o r flashovers,
where the impedance between the l i n e and g ro u n d d e pe n d s o n t h e re s i s ta nce o f
Ir a
a
b
Ifb
c
Ire
t
j
�
a
Zr
O z/
b
Irb
c
Zr
Ira
b
Ir b
c
Ire
t
t
�
a
r
I
T
t
j
l£l
,
\
-
S i n g l e l i n e-to- g r o u n d fau l l
------------��-------------Ira
t
�
r
b ------------�
• --------------
Zr
(c) Llne-to- line fau l t
FIGURE 1 2.4
Ire
(6)
(a) Three-phase fau lt
a
Ira
irb
-�
Zr
c --I----�.-- ----Cd) Double
l i n e-lo- g ro u n d
Connection d iagrams of the hypot hetical stubs for various fau lts t hrough i m p e d an ce.
fault
1 2. 1
UNSYM M ETR ICAL FAU LTS ON POWER SYSTEMS
479
the arc, of the tower itself, and of t h e tower footing if ground wires are not
used. Tower-footing resistances form the m ajor part of the resistance between
line and ground and depend on t h e soil conditions. The resistance of dry earth
is 10 to 100 times the resistance of swampy ground. Connections of the
hypothetical stubs for faults through impedance Z f are shown in Fig. 1 2.4.
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 positive-sequence currents flow. With t h e fault impedance Zf equal in all
phases, as shown in Fig. 12.4(a), we simply add impedance Zf to the usual
(positive-sequence) Theve n i n equivalent circuit of the system at the faul t bus ®
and calculate the fault curre n t from the equation
( 1 2 .8)
Fur each of the other fau l ts shown in Fig. 1 2.4, formal derivations of the
equations for the symmetrical-compone nt currents I}�), I};>' and If;) are pro­
v i d e d i n the sections wh ich follow. I n e ach case the fault point P is designated
as bus CD .
Two synchronous mach ines are connected t hrough three-phase
transformers to the transmission l i ne shown in Fig. 1 2.5. The ratings and reactances
of the machi nes and transformers are
Example 12. 1 .
Machines
Transformers
On
T)
1
and
and
2:
T2 :
1 00 M V A , 20 kV ;
XI
=
X2
networks a nd find t h e
b u i lding algorithm.
=
15%
Xo
1 00 M V A, 20�/345Y kY ;
a n d X()
z e r o - s eq u e n c e
FI G U RE 12.5
S i n g l e - l i n e d i agram of the system of Exa m p l e 1 2. 1 .
=
in
=
=
XI
X2
=
4%, Xn
X
=
=
=
8%
20% ,
5%
circuit the line
D raw c <l c h o f t h e t h r e e sequence
bus im pedance matrix by means of the Z hUS
a c h o s e n b a se o f 1 00 M Y A , l 4 5 k Y
r e a c t a n ce s a r c
x;;
t h e t r a ns m i s s i o n - l i n e
50%.
480
CHAPTER
12 UNSYMMETRICAL FAU LTS
Reference
(a)
jO.04
(2)
® �1
(6)
Reference
7 jO. 15
(b)
FIGURE 12.6
(a ) Positive-sequence and ( b ) zero-sequence n e t works o f the system of Fig. 1 2 . 5 . Buses
are internal nodes of t h e transformers.
0
and
®
The given per-unit impedance values correspond to the chosen base, and
so they can be used directly to form the sequence networks. Figure 1 2.6( a ) shows
the positive-sequence network, which is identical to the negative-sequence network
when the emfs are short-circuited; Fig. 1 2.6( b) shows the zero-sequence network
with reactance 3 Xn =
per u n i t in the neutral connection of each machine.
Note that each transformer is assigned a n i n t ernal node-bus W for transformer
T J and bus ® for transformer T2 • These i nt ernal nodes do not have an active role
in the analysis of the system. In order to apply the Z bus building algorithm, w h ich
is particularly simple in this example, let us l abel the zero-sequence branches from
1 to 7 as shown.
Solution.
0.15
Step 1
Add b r a n c h
1 t o t h c r c fc rencc node
CD
CD
[jO. 1 9]
Step 2
Add branch 2 to the reference node
CD
G)
CD
[jO.019
W
0jO.04 1
UNSYM METRICAL FAULTS ON POWER SYSTEMS
12.1
Step 3
G)
A d d branch 3 between buses
[ G)
CD
(])
Step 4
A d d branch
4
between buses
[0
CD
CD
CD
G)
W
[0
®
and
0
jO.04
jO . 04
Q)
O
jO O4
jO.os
[
W
jO '()4
i
G)
0
iO .04
iO.OS
iO 04
iO .OS
iO .OS
iO .5S
G) and ®
G)
j
�
0
0
jO . 04
iO .04
0
!
0
0
J-
I
G)
@
®
0
jO.04
jO .04
jO .04
0
j O .04
jO .08
j O .08
0
jO .04
jO .08
jO .58
0
j O.04
jO .08
jO.58
0
j O . 04
jO .08
jO .58
jO.66
Add branch 6 from bus
CD
@ to the reference
CD - jO . 1 9
G) 0
G) 0
®
@
@
I
@
jO . 1 9
0
0
0
Step 6
0
@
jO . 1 9
0
0
jO . 1 9
0
()
w
Add branch 5 between buses
CD
CD
CD
Step 5
f
and
G)
G)
G)
©
@
jO . 1 9
0
0
0
j O . 04
j O . 04
jO .04
jO .04
0
jO .04
jO .08
jO .08
jO .08
0
jO.04
jO .08
jO .58
jO.58
0
jO .04
jO .08
jO .58
jO.66
0
0
0
0
0
0
0
0
0
0
481
482
CHAPTER
12
UNSYMMETRICAL FAULTS
Buses G) and ® are the fictitious internal nodes of the transformers which
facilitate computer application of the Z bu S building algorithm. We have not
shown calculations for the very high impedance branches representing the open
circuits. Let us remove the rows and columns for buses G) and ® from the
matrix to obtain the effective working matrix
CD
Z(O) - W
G)
@
tHIS -
CD
jO . 1 <)
W
G)
@
0
j O . O�
jO .Oc;
0
0
j O . OS
j O .58
0
()
()
n
()
0
0
jO . 1 9
The zeros i n Z ��s show that z e r o - se q u c n ce c ur re n t i nj ec t c c.J i n t l ! b u s CD o r b u s
@ of Fig. 1 2.6( b ) cannot cause vol tages at the other b u s e s b C G.l u S C o f the o pen
circuits i n t roduced by the 6-Y transfo rmers. Note also that the jO.08 per-unit
reactance in series with the open circu it between buses ® a n d @ d o e s n o t
affect Z��s s in ce it cannot carry current.
By applying the Z bus building algorithm to the positive- and negative­
sequence n e tworks in a similar manner, we obtain
Z (bIu)s
=
Z(b2u)s
=
CD
W
®
@
CD
j O . 1 437
j O. 1 2 1 1
j O .0789
j O .0563
W
jO.121 1
j O . 1 696
j O . 1 1 04
jO .0789
G)
j O .0789
jO . 1 1 04
jO . 1 696
j O . 1 21 1
@
jO .0563
jO .0789
jO . 1 21 1
jO . l 437
We use the above matrices i n the examples which ful low.
12.2
S IN GLE LINE-TO - G RO U N D FAU LTS
The single l ine-to-ground fau lt, the most common type, is caused by lightning or
by conductors making contact with grounded structures. For a single l ine-to­
ground fau l t through impedance Zf the hypothetical stubs on the three l i lies are
connected, as shown in Fig. 1 2.7, w here phase a is the one on which the faul t
occurs. The relations to be developed for t h i s type of fau l t will apply o nly when
t h e fau l t is on phase a, but this should cause no difficu lty since the phases are
lab eled arbitrarily and any phase can be designated as phase a . The con d itions
at t h e fau l t bus ® are expressed by the fol lowing equations:
( 12.9)
12.2 SINGLE L1NE-TO-GROUND FAULTS
a
b
483
®
��
�-----------------
I,o t
®
r
Irb{ -'-
--__
__
__
__
__
__
�
�
-
®
� I
c --------------�.
__
Ir,
With
by
If b
=
ffe
-=-
----
--
Con nection diagram of the hypothetical
st u bs for a s i ngle l i n e-to-gro u n d fau l t . The
fa u l t poi n t is ca lled bus
CD .
0, t h e sym metrical c ompon e nts o f the stub c u r rents a r e given
=
1 (0
fa
)
(1)
1fil
1 (2
fa
and
FI G U R E 1 2.7
--__
__
__
___
1
3
=
)
1
1
1
1
a
a
1
a2
a
Ifa
2
0
0
p e rforming the m u l t i p l ication yields
1 (0 )
fa
Substituting I}:�)
for
ollLl i n
=
1( 1 )
Ja
=
1 (2)
fa
I}I; ) a n d 1/(;) shows t h a t
Vk(2)
a
=
H
V
f
-
=
V1..(O)
0
+
Vkeal )
+ V ( 2)
/; a
=
V
f
( 12 .10 )
3 1}�:), and from Eqs. 0 2.7) we
=
fa
( 0)
Z k( lk) 1fa
fa
( 12.11)
( 2) / ( 0 )
- Z kk
=
S u m ming these equa tions a n d n o t i n g t h a t
vk a
I a
f
Ifa
3
- Z (0) I (0)
v (O) ka -
V1..e 11l )
=
-
Vk o
=
3Zf IX?
g ive
1(0)
( Z{O) + Z { l ) + 2(2))
kk
fa
kk
kk
=
3 2f ffa
rO}
484
C HAPTER
12
UNSYMMETRICAL FAULTS
Solving for /J�) and combining the result with Eq. 0 2.10), we obtain
/f(0)a
=
/f(a1 )
=
/f(a2 )
=
v
f
'
( 12 . 12)
Zk( Ik) + Z(2)
kk + Z(O)
k k + 3 Zf
Equations 02. 12) are t h e fau l t current equations particular to t h e single
l i n e to groun d fault through impe d a nce Zf ' a n d they are used with t h e symm et­
rica l compon ent relations to d e te rm i n e all the voltages and currents at t h e fau l t
point P . I f the Thevenin equivalent c ircu its o f t h e t hree sequ ence ne tworks o f
the system a r e connected in series , as shown i n F i g . 12.8, we s e e t h a t the
-
-
-
current s a n d voltages resulting therefrom satisfy t he above e q u a tions-for the
Theve n i n impedances looking into the three sequence networks a t fau l t bus CD
a re t he n in series w i t h t h e fau l t i m p e d a nce 3Zf a n d t h e p rc fa u l l vul t age sou rce
Vf ' W i t h the equivalent circuits so connected, the volt age across each sequence
network is the corresponding sym metrical component of the vol tage Vk a at the
fault bus ® , and the current i nj ected into e ach sequence n etwork at bus ® is
the negative of the cOfn;spondin g sequence current i n the fZlUl t . T h e s eries
connection of the Thevenin equivalents of the sequence networks, as shown i n
Fig. 12.8, i s a convenient m eans of remembering the equations for the solution
of the single line-to-ground fault, for a l l the necessa ry equa tions for t h e fault
] ( 1)
fa
zi�
Zi�
•
]( 2 )
fa
,
t+
®
1 (0)
fa
t
Vk(2)
a
1 CD )
fa
,
®
+
Vk(O)
a
=
](\)
fa
=
]('2)
fa
3Zr
FlGURE 12.8
Con n ection of the Thevenin equivalents of the sequence networks to simulate a single l i n e-to1ground
fault on phase a at bus ® of the syste m .
12.2
SIN G LE L1NE·TO·G ROU N D FAULTS
485
point can be determined from the sequence-network connection. Once the
currents I}�), I}�), and I};) are known, the components of voltages at all other
buses of the system can be determined from the bus impedance matrices of the
sequence networks according to Eqs. (12.6).
Two synchronous machines are connected t h rough three-phase
transformers to t he transmission line shown in Fig. 1 2.9(a). The rati ngs and
reactances of the m achines and t ra nsformers are
Example 1 2 . 2.
Machines
Tra nsfo r m ers T,
1
and 2 :
a n d T.... :
1 00
100
MVA ,
20
kV;
MVA, 20Y /34SY
Bo t h t r a n s fo r m e rs a r c s o l i d l y g ro u n d e d
on
x;;
=
XI
=
X2
=
X"
20% ,
=
5%
X = R%
kV;
e b ase o f 1 00
two s i d e s . O n (\ c h os n
:145 kY i n t h e t r a n s m i s s i o n - l i n e c i r c u i t t h e l i ne r e a c t a n ces are X I = X2 =
1 5 % a n d XI) = 50%. T h e sys t e m i s o p e ra t i n g a t n o m i n a l v o l t a g e w i t h o u t p r c fa u l t
curre n ts w h e n a bo l t e d ( Zr = 0 ) s i n g l e l ine -to -grou n d fau l t occ u rs o n phase A a t
MV 1\ ,
bus G) . Using t h e b u s i mpcdance m a t r i x for e a c h o f t h e three sequence networks,
determine the subtransient curre n t to ground at the fault, the li ne-to-g round
voltages at the tcrminals of machine 2 , and the subtransient current out of p h ase c
of machine 2.
The system is the same as in Example 1 2. 1 , except that the transformers
are now Y Y connected. Therefore, we can conti nue to use Zb1Js and Z\,2Js
corresponding to Fig. 1 2.6( a), as given in Example 1 2 . 1 . However, because the
Solution.
-
Mac;:�
oH�>---+---<
-+-I
1 �Ho �
CD
TI
®
@
n Y ·.I.
( a)
T2
@
n �
ne 2
)0. 15
Reference
(b)
FI G U RE 1 2.9
( a ) T h e s i n gl e · l i n e
d iagram
and ( b ) z e r o - se q u e nc e n e tw o r k of t h e sys t e m o f Exam p l e
1 2. 2 .
486
CHAPTER 12
UNSYMMETRICAL FAU LTS
I (l)
fA
V{
= 1.0�
+
®
I
j O . 1696
It
j O. 1696
+
V3( 2,1)
®
[ ( 0)
fA
jO. 1 999
,
--
FIGURE 12. 1 0
Series connecti on o f the Thevenin equiva len ts o f the sequence ne tworks for the single l i n e-to-gro u n d
fault o f Exa mple 1 2.2.
transformers are solidly grounded on both sides, the zero-sequ ence network is fu lly
connected, as shown in Fig. 1 2.9( b ), and has the bus impedance matrix
CD
Z(O) - CV
hll:-i
-
G)
G)
CD
(1)
iO . l S S 3
i O . 1 407
j O . 1 999
j O . 07 0 I
jO .049]
jO . 1 407
jO .0493
iO .O]47
G)
j O 04 9 3
jO .070 1
j O . 1 099
j O . 1 407
.
@
j O 0 3 47
jO .0493
j O . 1 407
i O . 1 SS ]
.
Since the l ine-to-ground fau l t is at bus G), we must connect t h e Thevenin
equivalent circuits of the sequence n etworks i n series, as shown in Fig. 1 2. 1 0. From
this figure we can calculate the symmetrical components of the current IfA out of
the system and into the fault,
] (0) - 1 ( 1 )
fA - fA
-
1 (2 ) fA
Z( I )
-
1 .0
j(0 . 1 696
L2Q:
33
+
+
0 . 1 696
Zm
33
+
+
Z «l)
33
0 . 1 999)
- - j 1 .8549 per u n i t
487
12.2 SINGLE LlNE-TO-GROUND FAULTS
The total current in the faul t is
=
=
-j5 .5643
1
0
,
0
00/
45 6 5
-j5 .5643 6 5 93 1/ 270
2
,
( 1 2.6)
j 4,
-( J O . 1 407 ) ( -J· I .8549) -0.26 0
- 1 - ( J·0. 1 2 1 ) ( -J· 1 .8549) 0.7754
(j0 . 1 2 11 ) ( -j1 .8549) -0.2246
IjA
3 I}?J
per unit
/3
and since the base cu rre nt in the high-voltage transmission line is
3 = 1 7 3 A, we have
.
IrA
X
=
1 73
.
- Z� 'I((J)� /J(IJ)/ 1
V-I(III )
=0
V
J
i n vo l v e d .
=
- zf:, ) /J:i
=
(/
unit
per unit
d e n o t e v o l t ages a n d c u r re n l s i n t h e h i g h -volta g e
r e s p e ct i v e l y ,
From t h e a b o ve
l i n e - t o -grou n d v o l t a g e s a t bus
per
=
-
are
1 pe r u ni t
=
low-vo l t a ge c i rc u i ts ,
is
=
the terminals of machi ne
=
J /1
� o t e t h a t s u bsc r i p t s /1 a n d
s h i ft
@,
=
Z ( 1 )/ ( I )
-1�
A
=
The phasc-a sequence voltages at bus
calcu lated from Eqs.
with k = 3 a nd
X
@
and
Y-Y con nected transformer. No phase
sym m e t r i c a l compo n e n t s w e can calculate a -b -c
of the
as fo l l ows:
-0.0.72675104 -0.0.52386984 -jjOO..O36 0
-0.2246 -0.5364 jO.36 0
0.1.20819887 -121.3°
L .
1
+
a
+
=
"
a-
a
[Q:
L
I () un
20/
Ii ,
V-l iI
=
1 2 I �o
--
To e x p r e ss t h e l i n e - t o - g r o u n d vo l t a g e s
3.346L.Q:
which gives
kV ·
V4 h = 1 1 . 763
of
/
m ach i n e i n t h e
Ollf
-
1 2 1 .8°
m a ch i n e
V4IOl
u
--
=
=
=
1 1 .763
�.8Q
kY
must first calculate t h e
b r anch e s repre senting the
t h e zero-sequence current
we
t h e p h ase-a c u r re n t i n t h e
s e q u e n c e networks. From Fig.
o f the machine is
-
V4 c
kV
21 2.9,(b)
0.jO.261040 -j6 .525
To d e t e r m i n e p h a se -c c u r re n t o u t of
sy m m e t r i c a l c o m p o n e n t s
of m achine 2 in k i l o vo l t s, we mult iply by
per
unit
488
CHAPTER 1 2 UNSYMMETRICAL FAULTS
and from Fig. 12.6(a) the other s equence curr ents
/a( 1 ) =
/a( 2 ) =
f
v
-
-
V4(la )
"
jX
1 .0 - 0 . 7754
=
v4(a2 )
jX2
--
jO . 2
0 2246
0
are calculated as
-j 1 . 1 23 per u n i t
- -jl . 1 23 per u n i t
.
j O . 20
Note that the mach i ne currents a rc s h o w n w i t h o u t subscript J , w h i ch i s rese rved
exclusively for the (stub) currents a n d voltages of th e fa u l t poi n t . The phJse-c
currents in machine 2 are now easily calcul ated,
=
-j 6 . 5 2 5
+
a ( -j 1 . 1 23 ) + a 2 e -j1 . 1 23 )
= -j5 .402
per
unit
The base current i n the machine circu i t s i s 100,OOO/({3 X 20) = 288 6 . 7 5 1 A , a n d
so I Ie l = 1 5,594 A . Other voltages a n d c u rrents i n t h e system can b e calculated
s imi la rly
.
12.3
LINE-TO-LI NE FAULTS
To represent a l i n e-to-lin e fau l t through i m pedance Zf ' the hypothetical stubs
on t he t h re e l i n es at the fault a re con n ected, as s hown in Fig. 12. 1 1 . Bus ® is
aga i n the fault point P, and without a n y loss of generality, t h e li ne-to-l i ne fa u l t
i s regarded a s being o n phases b and
satisfied a t t h e fau l t point
c.
The
following rel a tions must b e
( 1 2 . 1 3)
12.3 LlNE-TO-LINE FAULTS
Vr
®
z( l)
kk
+
®
Zr
+
Vk( al )
2)
Zk( k
Vk( 2a)
Reference
489
FIGURE 1 2 . 1 2
Con n ection of t h e Theven i n equ ivalents of t he positive- and negative-sequence networks for
line-to-line fa u l t between phases b a n d c at bus ® o f the system.
Since
If b
=
- Ife
and
Ifa �
0, the sym metrical com ponen ts of curre n t are
0)
If( li
If( a
I)
J (2)
fa
a
1
a
1
3
1
a
2
a
2
a
0
Ifb
- lfb
and m u l tiplying t h rough i n t h i s equation shows t h a t
( 1 2 .14)
1(1)
fa
=
-
( 12.1 5 )
/ (2)
fa
The voltages t h roughout t h e zero-sequence n e twork must b e zero since there
are no zero-sequence sourc es, a n d because 1J�) 0, cu rrent is not being
i nj ected into t h a t network d u e to t h e fault- Hence, line-to-line faul t c alcula tions
do not i nvolve the zero-seq uence n e twork, which remains the same as before
the fa u l t - a dead network.
T o satisfy the req u i re m e n t t h a t I}(�) = - l}�>' let us connect the Thevenin
e q u ivalents of the pos i t ivc- and negat ive-seque nce networks in parallel, as
shown i n Fi g. ] 2 . ] 2 . To show t h a t t h is con nection of the networks a lso satisfies
t h e voltage e q u a t ion V" h - Vkc 111> /, f , we now exp a nd e ach side of that
e q u ation separately as fol lows:
=
� ,
1 Z
fb f
= ( 1f(b1 ) + 1fb(2» ) Zf -- ( a 21fa( 1 ) + a lfa(2» ) Zf
490
CHAPTER 12
Equating
UNSYMMETRICAL FAU LTS
both terms and setting I};)
=
- IJ�) as in Fig.
( a 2 - a ) ( Vk(al ) - V<k a2 »
or
=
Vk( al ) - Vk( a2 )
=
1 2 . 1 2, we ob tain
( I )Z f
( a 2 - a ) Ifa
Ifa( I ) Zf
( 1 2 . 1 6)
which is p recisely the vol tage-drop e q u a t ion for i m p e d a n c e Zf i n F i g . 1 2 . 1 2.
Thus, all the fa u l t cond i t i o n s o f Eq s . ( 1 2. 1 3 ) a r c s a t i s fi e d by co n Il e c t i ng the
positive- a n d negat ive-seq u e n c e n e t w o r k s il l pam llel l h ro u g h i m p c J ; l l1 c e Zr ' as
s how n in Fig. 1 2. 1 2. T h e z e ro - se q ll c l l c � I l c l w o r k i s i n a c l i v e il n u dm:s n o l e n L e r
into t h e l i n e - to- l i ne fau l t c a l c u l a t i o n s . T h e e q u a l i o n fo r t h e p os i t i ve - s e q u e n c e
cu r r e n t in t h e fa u l t ca n b e d e t e r m i n e d d i r L' L" t l y f ro m F i g . 1 2 . 1 2
1fa( 1 )
For
=
- 1f(2)
a
so
t h:lt
( 1 2 . 1 7)
_
a bolted line-to- l i n e fau l t we set Lf O .
Equations ( 1 2 .1 7) are the fau l t c u r re n t e q u a t i o n s fo r a l i n e - t o - l i n e fau l t
t h rough impedance Zf ' Once I}�) and Ii;) a re know n , they can h e t re a t e d as
i nj ections I}2 and -: IJ�) into the p o s i t ive- and negative-sequence networks,
r e sp e ctively, and the changes i n t he s eq uence vol tages at the buses o f the sys tem
due to t h e fau l t can be ob tained from the bus impedance m a t rices, as p r e v i o u s l y
demonstrated. When 6. - Y transformers a re present, p h as e s h i ft o f the pos i tive­
and negative-sequence currents and voltages must be t a ken i n t o acco u n t in t h e
calculations. The following example shows how t h i s i s accomplish ed .
=---'
-
Example 12.3. The same system as i n Example 1 2 . 1 is opera ting at nominal sys tem
voltage without prefault cu rrents when a bolted l i ne-to-li ne fault occ urs at bus G) .
Using the bus impedancc matrices 0 f the seque nce networks for subtransient
conditions, determine the currents ill the rau l t, t h e l i n c - t o - l i n c Yo l t (l g e s a t t h e Ll u l t
bus, a n d t h e line-to-line vol tages a t t h e t er minals o r machine 2 .
z�)12s a n d Z\;J, a rc a l re a d y s e t fo r t h i n Example 1 2 . 1 . A l t h o u g h Z\:�s is
also given, we arc not concerned w i th the zero-sequence ne twork in this solu t ion
since the fault is line to line.
To simulate the fault, the Thcve n i n equivalent circuits at bus G) of the
positive- and negative-sequence networks of Example 1 2. 1 arc connected in
parallel, as shown in Fig. 1 2. l 3 . From this ligure the seque nce curre n ts arc
calculated as follows:
Solution.
1 + jO
j O . 1 696
+ jO . 1 696
- j2.948 1 p e r
unit
Uppercase A is'used here because the fau l t is in the high-voltage t ransmission-line
1 2.3
j O. 1696
v(
=
1 . 0�
®
(1)
1(A
�
+
+
Fl C U RE 1 2 . 1 3
1f( 2A)
t+
491
-
jO. 1696
)
V3(2A
)
Vel
-l
®
L1NE-TO-L1 NE FAU LTS
3A
I
Reference
Conn ection of t h e Theve n i n equiva l e n t c i rcuits for t h e l i ne - t o - l i ne fa u l t of Example 12.3.
circuit. Since
If
A
Ifll
=
0, the com po ne n ts of currents in the fault are calculated from
Ij�) + IJ�) =
=
=
Ij�)
a 2 IJ�) + alJ;; =
-
j 2. 948 1
+
j2.9481 = 0
-j2 . 948 1 ( - 0 .5 - jO .866) + j2 .9481 ( - 0 .5 + jO . 866)
- 5 . 1 06 1 +
jO per u n i t
5 . 1 06 1 + j O per unit
As in Example 1 2 .2, base current in the transmission line
ffll
Irc
=
=
Vj�)
=
=
1 67.35 A , and so
�A
- 5 . 1 06 1 X 1 67 .35 = 8ssLQ: A
- 5 . 1 061
X
1 67 .35
= 855
Symmet rical compon ents o f phase-A vol t age to ground
vJ(O)A
is
at
bus
®
are
0
Vj�) =
1
Zi� /}�)
-
=
1 - ( j0 . 1 696)( -j 2 . 948 1 ) = 0 . 5 + jO per u n i t
Line -to-ground vol tages a t fa ult b u s G) are
V3 A - V3(O)
1'1 +
-
V3 C = V3 8
,;"
vj�)
0 .5
+
�
vj�)
= 0 +
per unit
0 .5 + D .S
=
LQ: per unit
1 .0
492
CHAPTER 12
UNSYMMETRICAL FAU LTS
G)
Line-to-line voltages at fault bus
V3, A B V3A - V3B
=
V3 , BC = V3B - V3C
V3, CA
( - 0 .5 + jO) - ( - 0.50 + jO )
=
V3C - V3 A =
=
1 .0 + jO) - ( - 0.50 + jO)
(
=
( - 0 .5 + j O) - (
l .0 + jO )
Expressed i n volts, these line-to-line vol tages a re
V3 . A ll
=
V3 , CA
=
1 .5
LQ:
1 .5/
1 .5L.! p er unit
are
X f3
= 299
345
1 80'" x
345
",'
v3
=
=
0
=
1 .5
=
L..Q:
�
pe r u n i t
kY
299L 1 800
kY
._
For the moment, let us avoid phase shifts due to the Ll-Y transformer connected to
m achine 2 and proceed to calcu late the sequence voltages of phase A at bus @
using the bus impedance matrices of Example 1 2 . 1 and Eqs. ( 1 2.6) with k = 3 a nd
j = 4
V4(AO)
V4�)
V4�)
=
=
=
-
2 (0)1
3 (0)
4
fA
VI - Zg>lj�
-
=
=
0
1 - ( j0 . 1 2 1 1 ) ( -j 2 . 94 8 1 )
ZWI};J =
- ( j0 . 1 2 1 1 ) ( j2 .948 1 )
=
0 . 643 p e r u n i t
= 0 .357 per u n i t
T o account for phase shifts i n stepp i n g down from t h e high-vol tage t ra nsmission
line to the low-voltage termi nals of machine 2, we must retard the posi tive-sequence
voltage and advance the negative-sequence voltage by 30° . At machine 2 terminals,
ind icated by lowercase a, t h e voltages are
V4';) ��)�
=
=
V4 a
=
��) + �(� )
+
�
0 .3 5 7
V4(;)
=
=
0
+
=
0 . 3092
+ jO . 1 785 per unit
(0 . 5569 - j O . 3 2 1 5 )
0 .866 1 - jO . 1 430
+ (0 .3092 + j0 . 1 785)
= 0.87781 - 9 .40
per u n i t
1 2.3 LINE-TO-UNE FAULTS
493
Phase-b voltages at terminals o f machine 2 are now calculated,
V4b
and for
v4((l)
c
V4\1 )
V4�) + V4(� )
=
ph ase
=
=
V4(O)
"
=
=
=
+
+
+
-
-
=
=
=
=
=
=
(1/ 2400
�(j )
+
V2)
0 .357L
=
+
=
+
V4 a - V4b
70 60 per u n i t
V4b - V4c
V4(
_...
V4 0
per unit
=
-
+
per unit
are
=
=
-
0 . 8661
= 0 . 9665
V4 , ca
=
at
+
per unit
pcr unit
tcrm i n als arc
x
X
153 .65°
jO . 1 43)
=
machinc
�
0 , 9665
=
-
- ( 0 .866 1
+
v o l t s , l i n c - t o - l i n c v o l t ages
V4 , bc
+
( - 0 .866 1 -
=
� . a h = 1 . 7322
per unit
+
=
In
jO. 1 785)
per u n i t
=
=
V4. CiJ
-
(1� )(0.643/ -300 ) 0.643�
)(0 .357� ) -90
0 (j0.643) ( 2jO.357) 0 jO.286
(1.07.8322661 -jjOO. 143) -(0.8661 -jO. 143)
j0.143) - (0 jO.286)
j0.429 0.96 5/ -153 .650
(0 jO.286) -CLr;661 j0.429 0.96 5/ 153.65"
2 20
/ -1 5 3. 65° 20 11. 2 / -1 5 3. 65°
1 1 .2/153.650
/
=
V4 be
0.8778/
+
2
Line-to-line voltages at the terminals of m achine
V4. ab
per u n i t
+
0
a V}� ) =
��)
V4<;)
c of machine
V2) = a 2 V4(;)
V4c
+
-0.3092 j0.1785
0 ( 0.5569 -jO_3215) ( - 0.3092
0.8661 -jO. 143 1 .
X
fS
-
=
13
=
20
13
=
20
�
kY
kV
kV
494
CHAPTER 12
U NS YMMETRICAL FAULTS
Thus, from the currents Ijr;t, Ij!), and Ij;; of the faul t and the bus impedance
matrices of the sequence networks we can determine the unbalanced bus
voltages and branch currents throughout the system due to th e l i ne-to- line fau l t .
12.4
DOUBLE LINE-TO-GROUND FAULTS
For a double line-to-ground fau l t the hypothetical stubs are connecte d , as
shown in Fig. 12.1 4 . Again, the fa u l t is taken to be on phases b and c, and t h e
relations now existing a l t he fa u l t b u s ® a rc
Since Ifa i s zero, the ze ro - s eque nce cu rre n t is given by
t h e vol tages of Eq. 0 2 . 18 ) t h e n become
Ij�)
=
( Jjb +
( 12. 10)
If) 1 3, a n d
( 1 2 . 19)
Substituting Vk b for Vk c in the symmetrical-component transformation, we fi n d
that
Vk( aO )
1
1
Vka( l ) - - 1
(2)
vka
a
b
c
Ir<l�
Irbt
Ire
t
3
1
1
1
a
a
a
2
2
a
Vk a
Vk b
Vk b
( 12 .20)
®
i
®
•
®
FIGU�E 12. 14
Connection diagram for the hypot h e t i c a l
stubs for a double line-to-grou nd fau l t .
The fau l t po int i s called b u s
®.
1 2.4
bOUBLE LlNE-TO-G ROUND FAULTS
495
The second and third rows of this equation show that
v: ( l )
=
ka
Vk(2)
a
( 12.21)
while t h e first row a n d Eq. (12.19) show that
Collecting zero-sequence terms on o n e side, setting
Vk(;), we obtain
Vel)
ka
VJ;)
=
Vf�),
and s o l v i n g for
Vk(O)u - 3 Zf 1f(0a )
=
Bringing together Eqs. 0 2 . 2 1 ) a n d 0 2 . 22), and again noting t hat
arrive at the resul t s
V(I)
ka
=
Vk( a2 )
+
1f(a0 )
V (O)
=
ka
1f(a1 )
+
-
1 (2)
fu
Ifa
( 1 2 .22 )
=
0, w e
3 Zf 1fll(0 )
=
0
( 1 2 .23 )
These characterizing equations o f the double l in e- to-ground faul t are satisfied
when all t hree of the sequence n etworks are con nected in parallel, as shown in
Fig. 12.15. The d iagram of n e twork c onnections shows t ha t the positive­
sequence current I}�) is determ i n e d by applying p re fault voltage Vf ac ross the
total impedance consisting of Zk� in series with the parallel comb i nation of zi�
and (Zko; + 3Zf ). That is,
1f(a1 )
] ( 1)
9
�
Z(I)
kk
fa
®
-! +
vkOa}
�-
=
Z k( Ik)
+
[
Z k( 2k) ( Z k(O)k + 3 Zf
Z k( 2k) + Z k( Ok) + 3 Z J.
)
fa
®
-] ( 2)
�
Z( 2 )
kk
-t +
vk(2)
a
�-
1
�
] (0)
-c::::J-
Zk(O)
k
( 1 2. 24 )
®
-\ +
Vk(O)
a
I•
FIG U RE 1 2 . 1 5
Co n nection of t h e Theve n i n equivalents of t h e sequence networks for a double l i n e-to-ground fault
on phases b and c at bus ® of t h e system.
496
CHAPTER 12
UNSYMMETRICAL FAULTS
The negative- and zero-sequence currents out of the system and into the fault
can be determined from Fig. 12. 15 by simple current division so that
[(III2) _
_ /(1)
III
fa
1 (0)
_ 1(1)
=
la
[ Z(O)
[
3Z
Z (2) + Z (O) + 3 2
k /.;
kk
f
kk
+
Z [�
2 k( 2k) + L k(Ok)
I
1
( 1 2 .2 5 )
( 1 2 .26)
+
For a bolted fault Zf is set e q ua l to 0 in the above e q u a t i o n s . \\' h e n LI =
the
zero-sequence circuit b e c o m e s a n opc n c i rc u i t ; 1 1 0 z e ro - se q u e nc e c u rr e n t c a n
then flow and the equations rcvc r L back t o t h o s e fo r t h e l i n c - t o - l i n e fa u l t
discussed in t h e preceding section.
Again, we observe that the sequence currents I}�), I}�l, a n d I}�l, once
calcu lated , can be treated as negative inj ections into the sequ e nce n e tw o r k s at
the fault bus ® and the sequence voltage changes at all buses of the system
can then be calcul ated from the bus i mpedance matrices, as we have done i n
p receding sections.
0::- ,
12.4. Find the subtransie n t currents and the l ine-to-l ine voltages at the
fa u l t under subtransient conditions when a doublL: line-to-ground fault with Z 1 = 0
Example
occurs at the term inals of machine
i n the syste m of Fig. 1 2 . 5 . Assume that the
system is u nloaded and operating a t rated voltage when the fault occurs. Use the
bus impedance matrices and neglect resistance.
2
.
.
.
SoIutwn.
·
Th e b us Impe. d ance matnces Z<hIu)s ' Z(hu2 )s ' an d Z(O)
h us a rc t h · same as In
Example 12.1, and so t he Theven in i mp ed anc es at fau l t bus 0 arc equal in per
unit to the diagonal clements Z��) = j O . 1 9 and Z��) = Z��) = j O . 1 437. To simulate
the double l ine-to-ground fault at bus @, we connect the Thevc n i n e q u i va l e n t s of
a l l three sequence networks in parallel, as shown in Fig. 1 2 . 1 6 , from wh ich we
t;
VI
=
/,.,
l . OL..9�
+ ,--j O . 1437 V e l ) -{=::Jr-----t-t:--:-'+
\ -
40
j O . 19
FIGURE 12.16
Connection of the Theve n i n e quivalents of t h e s e q u e nce n e t works for t h e double
Example 1 2.4.
l i n e -lo- I i ne fault o f
497
1 2.4 DOUBLE LlNE-TO-GROUND FAULTS
obtain
I (a! )
f
--
r-V
_f___..,,-
_
_
_
z(l)
44
+
[ Z(2+)Z(0) 1 .
44
ZUl
44
Z��)
j O . l 437
[
+
1 +
jO
( j0 . 1 437) ( j0 . 1 9 )
( j 0 . 1 43 7
+
jO . 19)
-j4 .4342 per unit
Therefore, the sequence voltages a t the fault are
V(I)
4
a
=
V Ol
� II
=
V (O)
4 (/
=
V
j
- If(aI )Z ( I )
4-1
-
1
=
( -JA 4342 ) ( J' 0 . 1 437 )
•
Cu r re n t i nj e c t i o n s i n t o t h e n e ga t iv e - a n d ze ro-se q u e n c e
arc
networks
c a l c u l a t e d hy c u rre n t d i v i s i o n a s fo l l ow s :
/ (0)
fa
=
_ f( l )
fa
[
Z�)
1
Z�) + Z��)
=
-
'4 434 ?
J .
[
jO. l 437
j(0. 1437
0 . 1 9)
+
1
=
The currents out of the system at the fa u l t point are
IIG Ij�) Ij�) + Ij;)
=
=
=
+
j 1 . 9095
j 1 .9095 - j4 .434 2
(1/240'" )(4.4342/
- 90"
+
- 6 .0266 j 2 . 8642 0,6726/ 1 54 .6"
= j 1 .9095
=
+
=
+
+
=
and the current
=
6 .6726
/ 25 .40
If i n t o t he ground is
If Ifb + fie
=
=
3 Ij�)
=
=
0 3 6 2 8 per
•
unit
a t t h e fa u l t bus
j 1 . 9095 per u n it
0
) + ( 1 LE2: ) ( 2 .5247� )
(l� ) (4 .4342/ - 90CJ )
6.0266 + j 2 . 8 6 42
j2 . 5247
=
per u n i t
+
( !� ) ( 2 . 5247L22: )
p e r unit
j5.7285 per unit
498
CHAPTER 12
UNS Y M M ETRICAL FAU LTS
Calculating a-b-c voltages at the fault bus, we find that
V4 • co
=
V4 C - V4 0
Base current equals
and s o w e find that
lib
Ifc
If
=
1 00
X
2887
lO:l/(!X
=
X
-
1 .088 4 per u n i t
20)
X
=
6 .6726/ 154 .6°
2887
=
= 2887 X 6.6726/ 25 .40 =
=
2887
X
5 .7285LiQ:
=
The base line-to-neutral voltage in machine 2 is
V4 • ub
V4• cu
=
=
1 .0884
X
- 1 .0884
f3
20
X
f3
20
=
=
A i n t h e circuit o f m a c h m e
2,
19,262/ 154 .6° A
/
1 9 ,262
25 .4° A
16 ,538LiQ:
20/ 13
-A
kV, and so
1 2 .568LQ: kV
1 2 .568� kV
Examples 12.3 and 12. 4 show that phase shifts due to /l- Y transformers do
not enter into the calculations of seq uence cu rrents and vol tages in that part of
the system where the fault occurs, provided Vf at the fault point is chosen as the
reference voltage for the calculations. However, for those parts of the system
which are separated by /l - Y transformers from the faul t point, the sequence
currents,
and voltages calculated by bus impedance matrix must be shifted in
phase before being combined to form the actual voltages. This is because the
bus impedance matrices of the sequence networks are formed without consider­
a t i o n of phase shifts, and so they consist of per-unit impedances referred .to the
part of the network which includes the fault point.
1 2.4
DOUBLE L1NE-TO-GROUND FAULTS
499
@, the end of
the transmission l ine remote from the double l ine-to-ground fault, in the system of
Example 1 2.4.
Example 12.5. Solve for the sub transient voltages to ground at bus
Numerical values of the faul t-current components are given in the
solution of Example 1 2.4 and the elements of zb11s, Z �Js ' and zbols are provided in
the solution of Exampl e 1 2. 1 . Neglecting phase shift of the �-y transformers for
the moment and substi tuting the appropriate values in Eq. ( 1 2 .6) , we obtain for the
voltages at bus @ due to the fault at bus 0 ,
Solution.
Vi�)
Vi�)
- I}�)Z��)
=
=
Vi�) =
v, - IJ�)Z��)
- Ig)z W
- ( j l .9095 ) (0)
=
=
1
=
�
V2(� = Vi�)/ - 300
=
Vi� )
=
=
0
- ( -j4 .4342) ( j0. 0789) = 0 . 65 0 1 per unit
- ( -j2.S 247 ) ( j O .0789)
Account i n g for ph ase s h i ft in stepp i ng
fau l t at b u s G) , we have
Vi �
=
up
=
0 . 1 992 per unit
to the transmission-line
circu it from the
.
�
0 . 1 992/ - 300
0 .6501
=
0 .5630
+
jO.325 1 per unit
= 0 . 1 725 - jO . 0996 per unit
The required vol tages can now be calculated:
v2 A
V2 B
=
=
V2(O)
A + V2( AJ )
V2A
( O)
+
V2( 2A)
+
a 2 V2( AJ )
+
a V2(2A)
=
(0 .5630 + j O .325 1 )
= 0.7355
=
(O
V2C - V2 A)
+
aV(I)
2A
+
a 2 V2(2A) -
=
0 .7693
- jO .0996 )
/ 17 .0° per unit
( 0 . 1 725
/ 30° )
( 1 / 2400 )( 0 . 653 1 �
�)
- 0 . 1 725 - j O .5535 = 0 .5798/ 1 07 .30
+
=
+ jO.2255
+
( 1 LE2: ) (
0 . 1 992
per unit
(1/ ] 20(> )(0 . 6531/30C' )
( 1 / 2400 )( 0 . 1 992� )
- 0 .5656 j O . l 274 0 .5798/ 1 67 .30
_
+
+
=
per unit
These per-unit values can be converted to volts by multiplying by the line-to-neutr al
base vol tage 345 / f3 k V of the transmission 1 ine.
SOO
12.5
CHA PTER 12
UNSYMMETRICAL FAU LTS
DEMONSTRATION PROBLEMS
Large-scale computer programs b ased on t h e bus i mpedance matrices of the
sequence networks are generally used to analyze faults o n electric u til ity
transmission systems. Three-phase and single l ine-to-ground fau l ts are usually
the only types of faul t studied. S ince circuit-breaker applications are made
according to the symmetrical short-circuit current that m ust be i n terru pted, this
current i s calculated for the two types o f fault. The p r i n tout includes the total
fault current and the contributions from each l ine. The results a lso l ist those
quantities when each l ine co n n ecte d to the fau l ted b u s is opened i n t u rn w h i l e
a l l others are i n opera tion.
The program uses the impedances for the l i nes as provided i n t h e l i n e d a t a
for t h e power-flow program and inc l u d e s the appropriate r e a c t a n c e for each
m achine in forming the positive- and zero-sequence bus impedance matrices. As
far as impedances are concerned , the n egative-sequence network is t a ke n to be
the same as the positive-sequence network. So, for a single line-to-ground fault
a t b us ®, I}�) i s calculated in p er unit as 1 .0 d ivided b y the sum (2 Zk� + Zk�
+ 3 Zr). T he bus voltages are included in the compu ter printout , i f called fOf, as
well as the current in l ines other than t ho s e connected to the faul ted bus since
this information can easily be found from the bus imped ance matrices.
The fol lowing numerical examples show the a nalysis of a single l ine-to­
ground fault on (1) a n industrial power system and (2) a small electri� u ti l i t y
system. Both of these systems are quite small in extent compared to the
large-scale systems normally enco untere d . The calculations are presented with­
out m atrices in order to emphasize t h e circuit concepts wh ich underlie fau l t
analysis. The presentation should allow t h e r e ader to become more fam i l i a r
wit h the sequence networks and how t h ey a r e used to analyze f aul ts. The
principles demonstrated here are essentially t h e same as those employed within
the l arge-scale compu ter programs used by industry. The same examples are to
be solved by the bus impedance matrices in the pro blems at the end of t his
chapter.
A group of identical syn chronous motors is con nected t h rough a
transformer to a 4 . l 6-kV bus a t a loca tion remote from the generating pl ants of a
power system. The motors are rated 600 V a n d operate at 89.5% efficiency when
carrying a ful l load at u n ity p ower factor and rated voltage. The sum of their
output ratings i s 4476 kW (6000 h p ) . T h e r e actances i n per u n i t of e a c h motor
based on its own input kiJovoltampere rat i n g are X� = X I 0.20, X 2 = 0.20,
Xo 0.04, and each is grounded through a reactance of 0.02 per unit. The motors
are con nected to the 4.l6-kV bus .t hrough a transformer bank composed of t hree
single-phase units, each of which is rated 2400/600 V, 2500 kVA. T h e 600-V
windings are connected in 6. to the m otors and the 2400-V windings are connected
in Y. The leakage reactance of e ach transformer is 10% .
The power system which supplies t h e 4.16-kV b u s is represented by a
Thevenin equivalent generator rated 7500 kVA, 4 . 1 6 kV with reactan<;es of
Example 12.6.
=
=
1 2.5
DEMONSTRATION PROBLEMS
501
CD
F IGURE 1 2 . 1 7
Single-line d iagram of the system of Example 1 2.6.
Motors
x;;
X2 = 0. 1 0 per u n i t , Xo = 0.05 per u n i t a n d Xn from neutral to grou nd
eq u a l to D.05 p e r u n i t .
Each o f t h e i d e n t i c a l m o t o rs i s s u pp l y i n g <I n e q u a l s h a re o f a total load o f
3730 k W (5000 h p ) (\ n d i s o p e r a t i n g a t rated vol t age , 85% powe r-fa c t o r lag, and
88% e f li c i e n cy w h e n , I s i n g l e l i ne-to-ground fault occurs on the l ow-voltage side of
t h e t ransfo r m e r bank. Tre a t the g ro u p o f motors as a s i n g l e equivalent motor.
D r aw t h e se q u e n c e n etworks show i n g val ues of the impedances. Determine the
=
,
subtransien t l i ne currents in all parts of the system with prefault current n e gl ected.
The single-l i ne diagram of t h e system i s shown in Fig. 1 2 . 1 7 . The 600-V
bus and the 4 . l 6- k Y bus are nu mbered CD an d (1), respectively. Choose the
rating of the equivale n t generator a s base: 7500 k Y A , 4 . 1 6 kY at the system bus.
Since
Solution.
f3
X 2 4 00
=
3
4 1 60 Y
X
2500
7500
=
kYA
the three-phase rating of the transformer is 7500 kYA, 41 60Y /600� V.
base for the motor circuit is 7500 kVA, 600 Y.
The input rating of the single equivalent motor is
6000 X 0 .746
0 .895
=
5000
So,
the
kYA
re;lctances or t i l C e q u i v a l e n t motor in perce n t arc t h e s a m e on t h e base of
t h e com b i n e d ra t i ng a s t h e reac t a nc e s of t h e i n d i v i d u a l motors on the base of the
ra t i n g o r an i n d i v i d u ;t 1 motor. The re actanccs of t h e e q u i v a l e n t motor in p e r unit
on t h e s e l e c t e d base a rc
<t mJ t h e
X:;
=
Xl
=
In t h e z e ro-se q u e nc e
X2
=
0 .2
7500
5000
n e tw o r k
the
3 XII
X
equ ival ent motor is
=
3
Xo
0.3
=
=
0.04
7500
--
5000
=
0 .06
reactance between neutral and ground of the
0 .02
7500
5000
=
0 .09 p e r u nit
502
CHAPTER 12
UNSYMMETRICAL FAULTS
and for the equivalent generator the reactance from neutral to ground is
3Xn 3 0. 05 0. 15
X
=
=
per unit
Fi gure 1 2.18 shows the series connection o f the sequence networks.
S ince the motors are operating at rated vol tage equal to the base voltage of
t he motor circuit, t h e prcfault volt a ge of ph ase a at the fault bus CD is
,
1.0
7,500,060000
Vf
=
per u n i t
B ase current for the motor circu i t is
= 72 1 7 A
13 X
---=---
and the actual motor current is
746
5000
0.88 600 0.85
0.85 0.665 '0 .565
X
13 X
-----;0=------ =
X
X
Current drawn by the motor through line
--I
7217 L------
4798
-
cos - 1
!
1-
=
a
before the faul t occurs is
3 1 .80
-
4798 A
=
-
jO.350 per unit
1.0&
If prefaul t current is neglected, E; and E;� are made equal to
in Fig.
12.1 8. Thevenin imped ances a re compu ted at bus CD in each sequence network as
follows:
(i O . l +
j(O. l +
- jO0..1l )( j0 .3) jO.12
1.
0
jO. 12 jO. 1 2
1(2) 1(1 ) -)'2 564
3 1}�) 3( -j2.564) -j7.692
-j2.5jO.6450 jO.30 -j1.538
Z( I )
11
=
Z (2)
11
_
+
0.3)
Fau lt curre n t i n t he series con nection
V,_
..;; _
____..
If(a1 ) _
- zg) + zg) + z \7)
---=-
fa
=
1 (0a)
f
=
fa
Curr ent i n the faul t
=
the sequcnCL: networks is
+
+
per u n i t
sequence network the portion of
=
of
-------
.
=
found by current division to be
X
I}�)
z\�) = jO. 15 per unit
per unit
=
jO . 1 5
=
1 .0
jO.39
----
=
-j2.564
per unit. In the posltlve­
flowing toward P from the t r ansformer is
=
per unit
12.5
®
CD
0.565 - j 1 .SSS
-
j O. 3 0
Positive-seq uence
network
CD
-j 1 .538
rv-"'><><-..
-
j O. 30
N eg ative-sequence
netwo rk
®
jO.IO
CD
p
1
-j l . 026
p
jO. IO
503
0.565 - jO. 676
p
jO.lO
DEMONSTRATI ON PRO B LEMS
-j2.S64
-j 2.564
) 0 .05
jO.06
Zero-sequence
network
jO.09
jO. I S
FI G U RE 1 2 . 1 8
Con n e c t i o n of t h e s e q u e n c e n e twor k s of Exam p l e 1 2 .6. S ub t ra n s i e n t currents a re m a rked i n per u n i t
f o r a si n g l e l i ne - t o- g r o u n d faul t a t
P.
P refa u l t c u r r e n t i s i nc l ud e d .
P
and the portion of flowing from the motor ward is
- j 2.5j6O4.50 jO.20 -j 1.026 per unit
Sicfroommpone
imlarly,nmottthofeoporr. tifornomofthe motfroomr ist h-je t1r.a0n26sroperrmerunit. of perflouniwst,toandward
Cur
r
e
nt
s
a
t
t
h
e
f
a
u
l
t
,
s
h
own
wi
t
h
out
s
u
b
s
c
r
i
p
t
ar
e
:
To from the t ransformer in per unit:
7
0
.
6
0
1l 1 [- 1
I}�)
to
X
Ii;')
t he
P
is
I};)
in th e fines
1
a
a1
a2
j 1 .538
- j 1 .538
-
=
- 1 .538
j
All
3
J}�)
jj1 .538
j 1 .538
j,
the
P
S04
CHAPTER
12
UNSYMMETRICAL FAULTS
To P from the motors in per unit:
a
1 ][ -j2.-jl.506426 ] [-j4 .616
-jl.026
lL23(a )
a
a2
-j 1 .538
=
-jL538
Our method of l abeli ng the l ines is the same as in Fig.
s u c h t h a t c u rre n ts
2
I
and � ) in the lines on the high-voltage side of t h e transformer are related to
the currents 1� 1 ) and I�2) in the l ines o n t h e l ow-vo l t a g e side by
I�l)
Hence,
I�l) =
I�2) =
( -J l .538) �
- j 1 . 3 32
3 / -60° 0.769
( -jL538)/ -30° 1 .538/ -1 2 0°
12.18
0
(0 .769 -Jl.332) ( -0.769 -jl.332) j2.6b4
(1/2400 )(1 .538/ -60° ) -1. 5 31)
)(1.538/ -1200) 1.538
0
= 0. 769 j1. 3 32
( 1�
j2.664
.
= 1 5
8
=
=
=
- 0 .769 -
j 1 .332
and from Fig.
we note that I�O) = in the zero-sequence network. S i nce
there are no zero-sequence currents on the high-voltage side of t h e t ra n sform e r ,
we have
fA
I�l )
If)
I8
IP)
It2)
Ie
=
=
=
I�I) + I�2)
a 2 I�1 )
8
= a
=
=
+
l�l )
a 2 I�2)
/ 8(2 )
Itl) + 12) =
per u n i t
+
) ( 1 .538/ - 600 )
( 1 / 2400 ) ( 1 .538/ - 1 200 )
-
+ jO
=
-
=
=
+ jO
=
=
a I�2 ) = ( 1�
= /(\)
=
+
=
=
- 0 . 769
+
j1 .332
per unit
If voltages throughout t h e system a re to be fou n d by circuit analysis, their
components at any point can be calculated from the currents and reacta n ces of the
sequence networks. Components of voltages o n the high-voltage side of the
transfo rmer are found first without regard for phase shift. Then, the effect of
p hase shift must be determined .
B y evaluating the base currents o n the two sides of the transformer, we can
convert the above per-unit currents to amperes . Base current for the motor circu it
12.5
was found previously a n d equals 721 7
IS
A.
13 X 4 1 60
Curre nts in
50S
Base current for the h igh-voltage circuit
7 ,500,000
Current in the fa ul t
DEMONSTRATION PROBLEMS
---=---
=
7 . 6 9 2 x 72 1 7
=
1 04 1 A
is
55 ,500 A
the l i n es bet wee n the transformer a n d the
3 .070
X
72 1 7
In line h :
1 .538
721 7
In l ine
X
1 .5 38 x
In line
a:
c:
721 7
= 22,200 A
=
=
1 1 , l OO A
1 1 , 1 00
Currents in the lines between the motor and the fault
In l i n e
a :
In line b :
In line
c:
4 . () l(j
l .538
l .538
X
x
x
727 1
72 1 7
fa u l t a re
=
=
A
are
:r� , 3rn \
1 1 , 1 00 A
72 1 7 = 1 1 , 1 00 A
Cu rrents in the lines between the 4 . 1 6 kY bus and the transformer are
In line A : 2 . 664
x
1 04 1 = 2773 A
In l ine B :
I n line
C:
0
2 . 664
x
1 04 1
=
2773
A
The cu rre n t s we h ave calculated i n the above example a re those which
w o u l d fl ow u po n the occu r r e n c e of a s i n g l e l i n c - t o- g ro u n cl fau l t whe n t here is no
load on the motors. These c u rrents a re correct o n ly if the motors are drawing
no current whatsoever. The statement of the problem specifies the l oad condi­
tions at the time of the fa u l t , however, a n d the load can b e considered. To
acco u n t fo r t h e load, we a d d t h e per-unit cu rre n t d rawn by t h e motor through
l ine a before the fa u l t occu rs to the port ion of IJ�) flowi ng toward P from the
transformer and subtract t he same cu rrent from t h e portion of I};) flowing from
the motor to P. The n ew va l u e of posit ive-sequence current from the trans­
forme r to the fau lt in phase a i s
0 .5 65 - jO . 350 - j1 .538
=
0 .565 - j1 .888
a n d the n ew val u e of posit ive-sequence curren t from the motor to t h e fault in
506
CHAPTER 1 2
UNSYMMETRICAL FAULTS
- j 2 .664
j 1 . 538
,
)
.
)
} 1 . 538
)
- j 3 . 07 6
,
+j 2 . 6 6 4
FIGURE 1 2. 1 9
Per-u nit va lues
o f subt ransient l i ne currents i n a l l pa rts of
current neglected.
phase
t h e sys t e m of Ex a m p l e 1 2 . 6 , p refa u l t
a IS
- 0 .565
+
jO.350 - j l .026
=
- 0 .5 65 - j O . 676
These values are shown in Fig. 1 2 . 1 8 . The remainder of the ca lc u la
t hese n ew values, proceeds as in the example.
Figure 12. 1 9 gives the per-unit values of subtransient line currents i n all
parts of the system when the fault occurs at no load. Figu re 12.20 shows the
values for the fault occurring on the system when the load specified i n the
exam p l e is considered. I n a larger system where the fault current is much higher
in comparison with the load current the e ffect of neglecting t h e load current is
less tha n is indicated by comparing Figs. 12. 1 9 and 1 2.20. In the la
however, the prefault currents determined b y a power- fl ow study cou ld simply
be added to the fault current found with the load neglected.
0 . 6 6 5 -j 2 . 68 5
-0. 586 +j l . 2 2 4
-
>
---:==�=-_-----' - 0.35 1 - j 0. 565
-
....__
•
- 0.314 + j 3.250
O.0 2 1 +j 2 . 20 2
•
-- �
0. 565-j3.42 6
+
-n .692
FlGURE 1 2.20
Per-unit values of subtransient line currents i n a l l p a rts of the sys tem of Example 12.6, prefa u l t
current considered.
1 2.5
507
DEMONSTRATION PROBLEMS
The single-l ine di agram of a small power system is shown in Fig.
1 2.21 . A bolted single line-to-ground fault a t point P is to be analyzed. The ratings
and reactances the generator and the transformers are
Example
12.7.
of
G ene r a
tor
:
Transformers T I and T2 :
On a choseanr of
reactances
base
e
X"
1 00 MVA , 20 kV ;
X2
=
1 00 M VA , 206. /345Y k V ;
From T2
t
to
o P:
=
4% ,
1 0%
kY n
i the transmission-l ine circuit the l ine
1 00 MV A , 345
From T I
x
Xo =
= 20% ,
XI
=
X�
= 20 % ,
X() = 5 0 %
P:
Xo =
r i c s , as s h ow n i n F i g .
30%
the with the valures of
e
c r n flows maapplrkyed. in
the transformers
vaoreltage ofeWip thedae ntcce poiseopeennt lno,oking eirnettaoaktehnecaussertqheuencentrseafeerreneezntecwreorovokasnltdaatgtehte
per
t
The sequence cur e
t of phase
t
reactances marked
the
l i g u l e a n u d r aw a co m p l e t e t h r e e - p h ase c i rcu i t diag a m with
per u n i t . Ass u m e that
are lettered so
To s i m u l a t e t h e fa u l t , t h e s e q u e n c e n e t wo r ks o f
in
rer
u n i t : l rc con n e c t e d i n s
all u e t
t h at Eqs. 0 1 .88)
c u r re n t s
Solu tion.
u n i t . Th
S l lOW Il in the
sw i h S
h as A at
i mp
s
P
Z pp
(I) =
1(1)
-
A - fA -
�
T
CD \ ®
-
open-ci rcuit
1.0 + jO.O per
( j0 .6) ( j0 .4)
jO.6
+
jO.4
2
Z""
( )
/f{2}
A
-
1 .0
=
jO .24
= jO.S
s ub
per u ni
A
+ jO .O
j O . S + jO .5 + j O . 24
unit
=
at P are
.
-jO.8065 p e r u n i
® Tz @)
l3f'---lI ---------� ---t- +33E
-tE r/_
t- S -
VDE
.6 h
G enerator n
t he p f ult
can b
n t s i n t h e hypot h e t i c a l
-
1 2 . 2 2 . Ve r i fy
h e faul t point
Z�� =
l}()
system
0 .6
Switch open
FI G U RE 1 2.2 1
Single-l i n e diagr a m of t he sys t e m of Ex a mple 1 2 .7. S i ngle l i n e - to-ground fault
is
at poi nt P.
508
CHAPTER 12
U NSYMMETRICAL FAU LTS
t Ii� =
jO.2 + tp jO.l
-�
t Ii�
@)
@ Tz
@)
-
�......--.
t
VP(2)
A
1
jO.B065
=
jO.2 + P jO. l
] (0 )
fA
=
IrA
3
= -jO .B065
-j O . 3226 j 0.48�
@
Reference
® T2
V(l)
PA
Reference
Reference
j O.B065
-
t li�
.
jO.5 + P jO.3
_
)·0 . 1
-
t
VP(O)
A
J
@
G)
-J
Tz
J'0 . 1
FIGURE 12.22
Con nection of t h e sequ ence networks for t h e system of Fig . 1 2. 2 1 to s i m u l a t e si ngle l i n e-to-ground
fault a t poi n t P .
The total c urrent in the fau lt is
If
A
= 3 1}�)
In the stub of phase B at point P
-) 2 . 4 1 95 per unit
=
have
we
I}V = a 21}:.J = 0.8065 / - 900
I}�) = aI};/ = 0 . 8065 /
-
+ 2400 = 0 . 8065�
90° + 1 20°
- 1f(0)
1f(0)
B
A
=
=
-
Likewise, i n the stub of phase C a t point
IIe
-
�
1(0)
Ie
+
P
0.8065
0. 8065
we have
1 ( 1 ) + 1 (2)
Ie
Ie
=
0
�
/
-
90°
1 2.5
509
D E MONSTRATION PROBLEMS
In the zero-sequence n etwork the cu rrents are:
Toward P from T]
1A(0) =
=
In the
jO.4
( 0 .8065 /
jO .6 + JO A
O . 3226
L - 900
t r a n s m i s sion
Tow a r d P from T I
O .J22Cl
90°
L...-_
_
)
P from T2
jO.6
(
0 .8065
.
j O .6 + j O .4
I�Ol = .
per unit
=
0 .4839
/
/ - 90° )
L...-__
900 per unit
-
l ine the currents arc:
/-
I n l ine
A:
In l ine
B:
0.3226/ - 9 00
+
I n line
C:
0 .3226
+
Toward P from
�
Toward
90
0
/ - 9 00
/ 90
O .8005 �
0 .8065 L2Q:
O . K065
+
-
0
L 90
0.806S L2Q:
+ 0.8065 �
+
O .S06S
-
+
0
= -j I .93S6 per uni�
= jO .4839 per unit
= j0 .4839 per unit
T�
IA
In l ine A :
In l i n e B :
=
-
j0 .4839 per u n i t
I B = -j0 .4839 per u n i t
Tn line C:
Ie
=
-j0.4839 pe r u n i t
Note that POS ] t lve -, negat ive-, a n d zero-sequence components o f cu rrent flow i n
l ines A , B , and C from T ] but o n l y zero-seq uence components flow i n these l ines
from T2 • Kirchhoff's curre n t l aw i s fu l fil led, howeve r.
In the generator t he cu rrents arc
= - j l . l 969
1,
=
1a(0 )
+ a
J( I )
(1
= j 1 .3909
+
0 2 Ja( 2 )
=
0
+
0 .8065/ -. 1 200
_
+
1 200
+-
0.8065/ - 600
_
+ 2400
=0
Th e three-phase circui t diagram of Fig. 1 2.23 shows all the current flows
in
per
Ia
=
-j 1 .3969
a
c
P
B
Ib
FIG URE 1 2.23
Current flows i n
=
j 1 .3969
b
j0 .4839
j0.4839
-
Ira
Ire
P
=
I
1
0
=
a
IrA
= -j 2 .4 1 95
.
-j0.4839
-j0.4839
b
0
the system of Fi:
1 2.5
DEMONSTRATION PROBLEMS
511
u nit. From this diagram Dote that:
•
•
•
•
•
•
•
10
Lines are lettered and pol a rity marks are pl aced so that Eqs. ( 1 1 .88) are valid.
Stubs are connected to each l in e at the fault.
For a single line-to-ground fault stub curre nts IB = Ie = 0, b u t IhO),
Ii:?) , It) , and It2) in the stubs all have n onzero values .
Fa u l t current flows out of s tub A , then partly to T) and partly to T2 .
I n the generator only positive- and n egative-sequence currents are flowing .
In the Ll windi ngs of T2 only zero-sequence currents are flow ing
I n the Ll windings of T\ each phase wind ing contains positive-, n egative-, and
zero-sequence com ponents of cu rrent. These components are shown i n Fig.
.
�
=
1 . 3969/- 90c
Ie
n�)
l��)
l��)
lea
0
=
=
=
=
0. 1 863/- 90°
0. 4656�
0 . 4 656�
•
0.2 794�
c
I6�) 0 . 1 863/ - 90°
I6�) 0.4656/ 1500
I6�) 0.4656/ 30c
Ibe O.2794�
1.3969/900 --=
-..
-----
a ------....
=
=
=
16
c
•
=
=
•
Fl G U RE 12.24
I��)
I!V
I!i)
la b
=
=
=
=
0 . 1863/- 90"
0 .4656/ gOo
0.4656/ 90°
-
-
1 . 1 1 75/ - 900
b
------::..
3 I�O)
•
=
B
0 .9678/- 900
IE
=
0.4839/ 900
Sym m e t rica l components of c u rre n t s i n t ra n s fo r m e r 7, of Fig . 1 2.23.
512
CHAPTER
12
UNSY M M ET R I CA L r:'A U LTS
12.24 and yield
12.6
OPEN-CONDUCTOR FAULTS
When one ph ase of a bal
created and asymmetrical currents flow. A s im i !
w hen any two of the three phases a re opened while t h e t h i rd p h a s e re m a i n s
closed. These u nbalanced conditions are caused, fo r example, \v h e n o n e - or
two-phase conductors of a transmiss ion l i n e are p hy s i cal
storm. I n other circu its, due to current overl oad, fuses o r o t h e r switching
d evices may operate in one or two conductors and fail to
conductors. Such open-conductor faults can be analyzed by means of tbe bus
impedance matrices of the sequence networks, as we now demonstrate.
Figure 1 2.25 depicts a section of a t hree-phase c i rcu i t i n w h i c h t h e l i n e
currents in the respective p h ases are la ' Ib , and Ie ' with pos i t i
bus @ to bus ® as s hown. Phase a is o p e n b etw e e n p o i n ts p and p i i n Fig .
12.25(a), whereas phases b and c are open b etw e e n t h e sam e t\VO po i nts i n Fig.
12.25(b ). The same open-cond uctor faul t
phases are fi rst opened between points p and p i , and short circuits are then
applied in those phases which are s hown to be closed i n F i g . 1 2.25. The e ns u i ng
developmcnt fo ll ows t h i s reason i n g .
8 - ® a l toge t h e r
@ a n d ® t o t h e po i n ts
O pe n i n g t h e t h re e p h ases i s t h e s a m e a� r e m o v i n g l i n c:
and then adding approp riate i mpedances from b u s es
p and p ' . I f Ii
simulate the opening of the three phases b y a d d i ng t h e n ega t ive i mp e d a nces
- Zo , - Z I ' and - Z 2 between buses @ a n d ® i n t h e co rrespond i ng Thevenin
equivalents of t h e t h ree s e q u e nce n e tworks of t h e intact sys t e m . T o exem p l i fy,
, consider Fig. 12.26(a), which s hows the connection of Z 1 to
sequence Thevenin equivalent between b uses @J a n d ® . The im p e d a nc es
shown are the elements Z m( I m) ' Z n( ln) ' and Z m( I n) Zn( lm) of t h e p os i t ive- s e q u e nce b us
ZJ� � + Z,� � - 2Z���
impedance m atrix Zb1Js of the intact system, and Z(��mn
is the corresponding Thevenin impedance between buses @) a n d ® . Vol tages
Vm and Vn are the normal (positive-seque nce) voltages of phase a at b uses ®
and @ before the open-conduc tor faults occur. The p os i t i v e -s eq ue n ce
imp edances kZ I a n d ( 1 k )Z l ' where 0 � k � 1 , a re a d d e d as s h own t o
-
=
=
-
12.6 OPEN-CONDUCTOR FAULTS
@
a
10
1 -
C
@) Ie
1 -----
p
P
I
@ Ib
b 1 -----
@)
a
I
fa
--
C
®
I
®
I
®
1
(a)
pi
p
®
1
®
@) f6
6 1 @)
1
513
I
Ie
(6 )
®
1-------11
FIGURE 1 2.25
Open-cond uctor fau l t s on a sect ion
of a t h ree-phase syst e m between
buses § a n d ® : (a) conductor a
open ; ( b ) can ductors b and c ope n
between points p a n d p'.
represent the fractional lengths of the b roken line @ - ® from bus @) to
point p and bus ® to point p ' , respectively. To u s e a convenien t notation , let
the vol tage VY ) denote the phase-a positive-sequence component o f th e voltage
drops Vp P '. a ' Vpp' b ' and Vp P'. from p to p' in the phase conducto rs. We shall ,
soon see that Va(l ); and the corresponding negative- and zero-seque nce comp o­
nents Va(2) and Va(O ), take on d ifferent values depending on which one of the
open -conductor fau It
By sou rce transformation we can replace the voltage drop Va( l) in series
wi t h t h e i m p e d a n c e [ kZ 1 + ( 1 - k ) Z I ] in Fi
in p a r a l l e l w i t h t h e i m pe d a n ce Z I ' as s h o w n in Fig. 1 2.26( 0 ). I n t h i s latter figure
t h e p a r a l l e l comb i n a t i o n o f - Z 1 a n d Z 1 can be canceled, as shown in Fig.
1 2. 2GC c ).
The above considerations for the positive-sequence network also apply
d i rect l y to the n e g a t ive- and zero-sequence n etworks, but we must remember
t h a t the l a t t er ne tworks do not contain any i nternal sou rces of their own. I n
drawi
understood that the currents VY)jZ and Va( O )jZo , l ike the current Va( l)j Z l of
2
Fig. 1 2. 26C c ), owe their origin to the open-cond uctor fault between points p a n d
p ' i n the system. I f there is no open conductor, the voltages V} l), VP), a n d Va(O)
are al
each of the sequence currents V} O)jZo , V} I )jZ \ , and �(2)jZ2 can be regarded
in turn as a pair of injections into b uses ® and ® of the corresponding
c
z(l)
mm
-
Z( 1 )
mn
®
p
kZ 1
+
(
Vm
)
Z( l
th ,
z( l )
nm
=
ZmO)n
Z(l)
nn
_
z( 1)
nm
-Z
mn
z ( 1 ),
®
(1
pp
-
k )Z I
r
VaO
0(
l
p
I
)
1-
....
N
(a)
( l) - z( 1 )
Zm
m
mn
+
Z mC I m) - Z m( I rt) @
I
+
Z ( l)
th o
Vn
+ Zn( l)
n
Z n( I)
m
=
-
Z ( l)
nm
( I)
Zm
n
t
mn
V(!( l
)
®
I
2(1)
rl r11
(b)
FlGURE 1 2.26
Simulating the opening of line
svstem: ( b) transforma t ion to current source; ( c ) res u l t a n t e q u iv a l e n t circu i t .
( l )o
Z th
Vl\
_
__
-
,%" ( 1 )
� (l f n
(c)
t
m il
1 2.6
O PEN-CONDUCTO R FAULTS
SIS
sequence network of the intact system. Hence, we can use t he bus impedance
m atrices Z�Js ' ZblJs ' and Z�Js of the normal configuration of the system to
determine the vol tage changes due to the open-conductor faults. But fi rst w e
must find expressions for t h e symmetrical components V}O), V} l ), and VP) of the
vol tage drops acrOss the fault points p and p i for each type of fault shown in
Fig. 1 2.25 . These voltage d rops can be regarded as g iving rise to the following
sets of inject ion currents into the sequence networks of the n ormal system
configurati on :
Positive
Sequence
At bus
At
bus
Va( l )
@) :
-
Va(O)
Va(2)
2
@:
Zero
Sequence
Negative
Sequence
20
2
1
V1I( l )
--
-
21
2
V1I(2)
-
--
2
V (0)
a
--
20
2
as shown i n Figs. 1 2.26 and 1 2.27. By multiplying t he bus impedance m atrices
ZboJs ) Z blJs , and Zb2Js by current vectors containing only these current i njections,
we obtain the fol lowing changes in the symmetrical components of the phase-a
voltage of each bus (J) :
6. VI (O) =
Zero sequence :
Positive sequence :
6. VI ( l )
N egative sequence :
6. V(
2)
,
2 1(0)
- 2 I�0)
n
m
2°
2(1)
=
=
Va(O)
2 I�n1 )
Va( 1 )
- 2m
2 1(2)
III
Ir/
Va(2)
-
1m
71
2
2
( 12 .27)
Before developing the equ ations for Va(O), v:,( 1 >, and VP ) for each type of
open-con ductor fault, let us de rive express ions for the Thevenin equivalent
'
i m p e d a nces o f the s eq u e n ce n etwo rks, as seen from fa u l t points p a n d p .
Looking into the positive- sequence networ k of Fig. 1 2.26( a) between p
and p' , we see the imped ance 2��, given by
Z ( l ),
pp
=
kZ +
1
Zth , m ,, ( - Z I )
z(l)
(Il
th . mn
;
- Z
1
+
(1
-
k)Z
1
2
=
_ 21
Z
___
_
Z t( hI ). m "
-
1
( 1 2 .28)
�2)
mm
_
Z(2) @)
mn
p
kZ2
...
)
Z(2
th ,
Z(n2n)
Z(2 )
=
m n
_
mn
-
Z2
®
Z ( 2)
nm
•
(1
( 2)'
Zpp
k )Z2
-
Zn(2)m
p
kZo
�
Z (O)
mn
=
Z(O)
Zth
( O)
o
Z( O ) - Z (O) ®
ll n
I
mm
r
- Zo
...
t
Z (2 )
mn
Z (2)
@)- ®
Zn( 2)n - Z(n2)m
(1
-
k )Zo
Z2
®
�
()
�
tTl
:;c
....
N
c
Z
til
-<
::
::
�:;c
)r
."
Zm( Om)
-
(0 )
2 fli
T!
Z(O)
Z,<,�!
Zm( On)
E
@)
I(O)
-
)C
Z th .
Z;I�:I. ®
t
m n
nm
(6)
'
)
(II
.e'\
()
r
p'
Zth o m n
t
Va( 2
--
nm
)
Zp( Op)' y e o
nm
F1GURE 12.27
Simulating the opening of line
I( 2)
a
mn
nm
mn
Va( 2)
( a)
§
(O)
Zm(O)m - Zm
n
p
Z(2) - Z(2) @)
between points p and p : ( a ) negative-sequence a n d (h) zero - se q u e nce e q u iva l e n t c i rcu i t s .
yae o )
Zo
1 2.6
OPEN-COND UCTOR FAULTS
517
a nd from p to p ' the open-circuit v o lt ag e obtained by voltage d ivision is
Open -circuit voltage from p to p' =
2(1)
- zI
th , mn
( Vm
Z1
_
( 1 2 .29)
Before any con d uctor opens, t h e c u rren t Imn in p hase
positive sequence a nd is given by
I
II l it
'""c
v"'
-
1""1 i n Eq .
Subst i t ut i ng this expressio n for
a
of the l ine
V
/I
@) - ®
is
( 1 2 .30)
( 1 2.29), we obt a i n
Open-cir cu i t vol tage from p to
p
'
=
Jmn Z��,
( 1 2 .3 1 )
Figure 12.28(a) s hows t he res u l t i ng positive-sequ ence equivalent ci rcu it between
points p and p'. Analogous to Eq. ( 1 2.28) we h ave
Z(2� =
PP
Z t(2)
h mn
Z
(O), =
zPP
a nd
2
- Z o2
Z (O)
th mn
-
20
( 1 2 .32)
which a re the negative- and zero-sequence impedances, respective l y, between p
and p i i n Figs. 1 2 .28( b ) and 1 2 .28( c). We can now p roceed to d evelop expres­
si ons for t h e sequence voltage d rops Va( O ) , Va(I ) , and vy ).
One open con d uctor
Let
us
co n s i d e r
c i rc u i t i n p h a se
(1 ,
o n c o pe n
t h e cu r re n t
c o n d u c t o r as . i n f i g .
1 11 = 0 , a n d
J {.I(O)
w h e re
IIi ' '1 nd
v o l t a ge
(l
Ic
.,
J(I)
a
+
1 (2 )
iI
=
0
( 12 .33 )
=0
( 1 2 .34)
a r e t h e s y m m e t r i c a l c o m p o n e n t s of the line c u r ren t s la '
'
from p t o p . S i nce p hases h a nd c arc closed, we a l so have t he
J ( O ) J( I )
a
+
J 2.25 ( a ). Ow i n g to t h e o p e n
so
a
,
and
1 ( 2)
a
drops
VP P ' , c
Reso lving the series vol t age d rops across t he fau l t point into their sym metrical
518
CHAPTER 12
U NS YMMETRICAL FAULTS
1a( 1
\.
)
----+-
]
mn
Z( l )'
+
p
O l'
Zpp
]a(2)
----+-
r
I-
r
Z (2),
pp
Va( l)
pp
p'
(a)
-----i
p
r
l-
Va( 2
p'
I
(6)
)
] (0 )
a
-
p
)
Zpe Op '
r
yaeO
)
p'
(c)
FIGURE 12.28
Look i ng into t h e system between points
(c) zero-sequ ence equiva l e n t ci rcu its.
[
I [
p
and
com ponents, we obtain
V} D )
V(l)
a
Va(2)
1
=
-
3
1
1
1
That is,
=
Va( I )
( a ) pos i t ive-sequence, ( b ) negative·se que nce, a n d
[\ I I [ I
a
a
1
Va(D)
'
p :
=
�)0P ' .
� VpP' . a
a
3
0
Va( 2 )
=
Vpp' . a
V,) ,
p .
V ,
pp , a
3
( 1 2 . 35 )
a
( 1 2 36)
.
I n words, this equation states tha t the o pen conductor i n phase a causes equal
voltage d rops to appear from p to p' in each of the sequence n etworks. We can
satisfy this requirement and that of Eq. ( 1 2.33) by connecting the Theve,nin
equ ivalents of the sequence n etworks in parallel at the points p a n d p ' , as
shown in Fig . ] 2 .29. From t h is figure the expression for the positive-sequence
1 2.6 OPEN-CONDUCTOR FAULTS
]( 1 )
a
Z(I ),
]m n Z(pI)p '
pp
p
i
p
Vae l )
t+
Z( 2 ),
](2 )
� p
p
pp
i
t
Z(O),
ya( 2)
](0)
� p
I
t+
yae O)
P .
pp
519
-
Fl G U RE 1 2 .29
Co n ne c t i o n of t h e sequence networks of t h e
a nd p' .
c u rr e n t
1� 1 )
i s found to be
1( 1) - 1
1I
7( I �
!! 1 11
PI'
+
m, Z I'(O){' ,
z IIII
Z ( 2 ),
�
" f'
+
Z ( ( I ),
1'1'
( 1 2 .3 7 )
The seq u e nce vol t age d rops V} O ), Va( I ), a n d Va( 2 ) are t he n given by Fig. 1 2.29 as
- V (O)
a
=
V (2 )
=
a
=
1
III n
Va( I )
=
1 ( 1 ) Z (2 ), +
zpp
(2� Zpp
(O),
a
pp
Z (O),
pp
Z ( O ), Z ( I � Z ( 2)'
pp
pp
pp
O
,
l
+
)
(
I
),
)
Z(
Z
Z( , Z m, + Z m, Z( O ),
pp
pp
pp
pp
pp
( 1 2 .38 )
pp
The q u antIt Ies on t he right-hand side o f t his equation a re known from t h e
i m p e d a nce p aramet e rs of t he sequence n e tworks and t he p refa u l t current i n
p h ase a of t h e l i n e ®- @ . Thus, t he curre n ts v'1(O ) / Z U ' VY ) / Z I ' an d VY)/ Z 2
for i nject ion i nto the correspon d i ng seq u e n ce n etworks can be determined from
Eq. 0 2 .38).
Two open conductors
When two con ductors a re ope n , as shown in Fig. 12.25( b), we h ave fau l t
cond i t i ons which are t he dua/5 2 o f t hose i n Eqs. ( 1 2 .33) a nd ( 1 2. 34); namely,
( 1 2 .39 )
1
b -
2Dualiry
is
t reated i n m a ny textbooks
0
1
=
c
0
( 1 2 . 40)
520.
CHAPTER 12
UNSYM METRI CAL FAU LTS
Resolving the line currents into their symmetrical components gives
( 12 .4 1 )
Equatio ns ( 1 2.39) and ( I 2.4 1 ) are both sat isfied by co nnect ing the Thcven in
equ ival ent of the negative- and zero-seq uence networks in series be tween points
p and pi , as shown in Fig. 1 2.30. The seque nce currents a re now expressed by
-I-
Z ( I ),
'1'1'
+
( l 2 .42)
Z ( :'1 ")
� /'
where I is again the prefault current i n p h ase a of the line ® - ® before
the open circuits OCCur in ph ases b and c. The sequence voltage drops are now
given by
m il
Vel)
a
=
(
1 ( 1 ) Z (2), + Z ( O ),
a
pp
pp
)
=
I
=
I
=
m il
mn
I11 1 /1
(
Z ( I ),
pp
pp
pp
+
pp
Z ( 2), +
pp
)
Z ( O ),
Z ( l ), z (2 ), + Z (O ),
pp
- Z ( 1 ), Z (2),
pp
pp
Z ( I ), + Z (2), + Z ( O ),
pp
pp
pp
Z ( I >,
pp
( 1 2.43)
- Z ( I ), Z (O),
-; .
'
pp
pp
Z (2), + Z ( O ),
pp
pp
In e ach of these equat ions t he right- hand side q u a n t i t i es a re al l known befo re
the fault occurs. Therefore, Eq. ( 1 2 .38) can be used to eva l uate t he symmetrical
components of the vol tage drops between the fault poi n ts p and pi when an
open-conductor fau l t occurs; and Eq . ( 1 2.43) Can be similarly useJ when a fa u l t
due to two open co nductors occurs.
The net effect of the open conductors on the positive-sequence network is
to i ncrease the transfer impedance across the line in which the open -cond uctor
fault occurs . For one open co nductor this increase in impedance equ als the
parallel combination of the negative- and zero-sequence ne tworks between
points p and pi; for two open conductors t he increase in impedance equals the
series combination of the negative- a n d ze ro-sequence netwo rks between p.oints
p a n d p'.
1 2.6
I
m it
Z( l )
p p'
+
Ia( 1
�
)
p
Z( l ),
Z (2
1a( 2
Vae l
)
p
�
pp'
)
1a( 0
�
p
I
j
t+
'
p
�
)
t-
p'
)
521
t+
pp
-
OPEN·CONDU CTOR FAULTS
)
Va( 2
t-
1 (1)
a
=
1 (2)
=
<l
[(0)
a
t+
1
Z (O),
pp
yae O
t
pi
)
-
FlG URE 12.30
Co n n ec t i o n of t h e
po i n t s
p
a nd
'
p .
Example
12.8. In the system of Fig. 1 2.5 consider that machine
Solution .
All t
2 is a motor
drawing a load equivalent to 50 MVA at 0.8 power· factor lagging and nominal
system voltage of 345 kV at bus G) . Determi ne the change i n voltage at bus G)
when the transmission line undergoes ( a ) a one-open-conductor fault and (b) a
two-open-conductor fault along its span between buses W and G) . Choose a base
of 1 00 M V A , 345 kV i n the transmission l i n e .
current in line W- G)
choos i n g t h e
12
J
=
as
fo l l ows:
P - jQ
V:l*
The sequence networks of F i g .
Z\ =
2
2
=
------ =
1 .0 + jO .O
0 .5(0.8 - jO.6)
1 2.6
0 .4
- jO .3 per unit
show that line (1) - G)
jO . l 5 per unit
Zu
has
parameters
= jO .50 per unit
The bus impedance matrices Z�oJs and Z\,IJs = Z�Js are also given in Example 12. l .
Designating the open-circuit poi nts of the line as p and pi, we can calculate from
522
CHAPTER 1 2
U NSYMMETRICAL FAU LTS
Eqs.
and 02.32)
0 2. 28)
- ( j0 . 1 5)
2
------- -------
j 0 . 1 696 + jO. 1 696 - 2( j0 . 1 l 04) - j O . 1 5
Z
pp'
(U)
-
_
=
_
------- Z2
--:-:-:-__
u
---:-:::-_
=
jO . 7 1 20
pc r
--::-:-__
(O, + Z(O) - 2 Z (O) - Z
Z 22
()
33
23
uni t
"1
- ( f 0 .5 0 ) '
j O . 08 + jO .58 - 2 ( f 0 . 08 ) - jO.50
=
00
Thus, if the line from bus @ to bus G) is opened, then a n infinite impeda nce is
seen looking into the zero-sequence network between points p a n d p' of the
opening. Figure 1 2.6( b) confirms this fact si nce bus G) would b e i sol a t e d fro m t h e
reference by openi ng the connection betwee n bus CV and bus 0.;. .
One open conductor
In this example Eq. 0 2. 38) becomes
=
=
( j0 .7 1 20) ( j0 . 7 1 20)
( 0 . 4 - jO . 3) -. ----- -j 0 . 7 1 2 0 + j O-.7-1 20
0 . 1 068 + j O . 1 424
per unit
and from Eqs. 0 2.27) we now calculate thc sym m e t r i c a l com ponents o f the voltage
at bus (1) :
Ll V ( 1 )
3
=
Ll V (2 )
3
=
Z ( I ) - Z (l)
33 V e l )
32
a
ZI
=(
=(
=
Ll V (0)
3
=
Z (O) - Z ( O )
32
33 V (O)
a
Zo
jO . 1 1 04 - jO . 1 696
.
j0.15
)
- 0 . 0422 - j O .0562
jO . 08 - j O . 5 8
.
j O .50
)
( 0 . 1 068 + jO. 1 424)
per unit
(\ 0 . 1 068 + j O. 1 424)
- 0 . 1 068 - jO . 1 424
per unit
- 0 . 1 068 - j O . 1424 - 2 ( 0 .0422 + jO .0562)
.
- - 0 . 1 91 2 - j O . 2548 p e r
unit
1 2.7
Since the prefault voltage at bus G) equals
1 . 0 + jO.O,
IS
Vj
1 2.7
=
V3
=
0 . 8088 - jO .2548 = 0 .848
+
tl V3
=
SUMMARY
523 .
the new voltage at bus G)
( 1 .0 + jO.O) + ( - 0 . 1 9 1 2 - jO.2548)
/
- 1 7 .5°
per unit
Two open conductors
Inserting t he infi n ite impedance of the zero-sequence network in series
between points p and p' of the positive-sequence network causes an open circuit
in the l a t t e r . No power transfer can occur in t he system-confirmation of the fact
that power cannot be transferred by on ly one phase conductor of the transmission
l ine in t h i s case since t h e zero-sequence network offers no return path for current .
S U MMARY
in
shown i n Fig. 1 2.2 a re
repl aced by s h o r t c i r c u i t s , t h e i m pe d a nce b e t we e n the fau lt bus ® and the
r e fe r e n ce node is t he posi t ive-se que nce i mpeda nce ZJ� in t h e equation devel­
oped for fa u l t s on a power system and is the series imped ance of the Thevenin
equivalent of the circu it between bus ® and the reference node. Thus, we can
regard Zk� as a single im pedance or the entire positive-sequence network
between bus ® and the refe rence with no emfs present. If the voltage Vf is
connected in series with this mod ified positive-sequence network, the resu lting
circu it, shown in Fig. 1 2. 2( e ), is the Thevenin equivalent of the original
posit ive-sequence network. The circu its shown in Fig. 1 2.2 are equ ivalent only in
their effect on any external connections made between bus ® and the refer­
e nce node of the original networks. We can easily see that no curren t flows i n
t h e branches o f the equivalent circuit i n the absence o f an external connection,
b u t current will flow in the branches of the originaL positive-sequence network if
any difference exists in the p hase or magni tude of the two emfs in the network.
I n Fig. 1 2 .2( b ) t h e cu rrent flowing in the branches in the abse nce of an exte rnal
c o n n e c t i o n i s t h e prefa u l t load cu rrent.
W h e n t h e o t h e r seq uence n e t w o r k s a r e intercon n ected with the positive­
sequence network of Fig. 1 2 .2( b ) or its equivalent shown in Fig. 1 2.2C e ), the
current fl owing out of the network or its equivalent is I};l and the voltage
betwee n bus ® and the reference is Vk((� l. With such an external connection,
the cu rrent in any branch of the original positive-sequence network of Fig.
1 2 . 2 ( b ) is the positive-sequence cu rrent in phase a of that branch during the
fau l t . The prefault component of this cu rrent is included. The current i n any
branch of the Thevenin equ ivalent of Fig. 1 2.2( e), however, is only that portion
of the actual positive-sequence current fou nd by apportioning J}�) of the fault
among the branches represented by zi� according to their impedances and
does not i nclude the prefault component.
In the p r e cedi n g sect ions w e h ave s e en that the Thcvenin equivalents of
the sequ ence networks of a power system can be interconnected so' that solving
I I t h e e m rs
a
p o s i t i v e - s e q u e n ce n e two rk l i k e t h a t
+ t1l
I
__
•
Z(
Pos.-seq. n etwork
Vk( a
�
®/ + ] (1)
•
fa
Po s . - s e q . n etwork
+t )
V(l
-t
ka
®I Fi�)
.
P o s . - s e q . network
I
�
vk( a2
-t
)
S H OR T-C I R C U IT F AU LTS
I
G
·
fa
N e g . - s e q . n e twork
I
Zf
+t
VkC oO )
-+
Dou b l e line-to-ground f ault
®l f 1i�)
�
I
N e g . - s e q . n e twork
Line-to-line fault
Vh( 2a )
-+
I
Three-phase fau lt
Z{
+t
® I t ] ( 2)
3
+t
)
v(l
-t
+t)
ka
vk( a2
-t
+ t O)
V(
_ tk "
•
P o s . -s e c . n e twork
N e g . - s e q . n e tw o r k
®
.
t
1 (0)
fa
Ze r o - s o q . n e t w o r k
S i n g l e lin e-to-g round fau l t
® I + ]j�)
�
Zero-s e q . n e twork
I
N
+ �
+ -1
� vy )
1�1) + p
p '.!
.
Po s . -seq . n e twork
2)
�Va(
1� 2 ) t p
p' l
•
N e g . - s e q . network
I
One open conductor
O P EN-CON DUCTO R FAULTS
+ -1
O)
Va(
l�O ) t
�
p
P' l
•
Zero - s e q . n e tw o r k
a
�
IeI
)
-=j
I'-!- V} I )
IP
I
t ip
I Neg�-Seq.
I
· i
I � Va(2)�
Pos.-seq.
(2)
1a
p' ,--
p' ,.....-
�
1 + Va(O) �
-+-
t IP
zer�-seq
1(j( 0 )
n e two r k
net ork
-I
I
net:ork I
I
P r--
Two open conductors
vI�I, VI!), a nd
Va(OI, Vyl, a n d Va(2) are t h e
FlGURE 12.31
Summary of the connections of t h e seque nce networks to simula te various t ypes of short-circ u i t fa u l ts t h rough impeda nce Zf.
Vk(;) a re the symmetrical compone nts of p h a se a voltage a t the fault bus ® with respect to referen ce.
symmet rical components o f t h e phase a voltage drops a cross t h e open-circ u i t points p a nd p ' .
TABLE 12.1
Summary of equations for s equence voltages and currents at the fault
point for various types of faults
1
Line-to-ground fault
�
I::
III
I.
I.
::l
to>
IW
( I ) + Z(2)
k k + Z k(Ok) + 3 Zf
= Z kk
I};)
I::
e:r
III
VJ
I---
V(ll
Vl
III
OJ)
<0
-
k il
I}�)
=
"0
I}�
- 'h
'k
"
IOl
Vk "
Vl�)
VPl
=
fa
Ij�)
)
Jj' )(zF) + 710) +
ka -
III
u
I::
III
::l
0OJ
VJ
V��), Vl�),
V;O), Va(J ),
=
kk
- 1?" )2 k( Ok)
17 r )
-
-
V
f
Zk( 2k)
+
+
Z
f
1 (2) - - I}!)
fa
I}�)
V (2) - - / ( 1 )2 (2)
...
NOle : " II"
=
Iflal ) - ( I )
Zk k
VII)
k"
=
Vka(2)
=
V1�)
I}� )
1 (0)
+
7�r )
f"
z i�
-
+
Vf
Z i�ll ( Z k�
/ (a1 )
f Z (2)
kk
(l
V ( I ) - V}�)
-
0
and
a rc the sym m e t r i c a l co m po n e n t s o f p h os e
a r e the sym m e t r i c a l compon e n t s o f
the
a
vol t age at
phase
a
+
-
Zf )
3 Zf )
Z k( Ok) + 3 Z
+
(2 )
ZU
Z f.:I.:
(O)
+
1a(2)
f
1 (0 )
3 Zf
(l
:' 1}�)Zr
=
Z
=
Vk(O)
a
) Z ( II )
- I?(l - " "
V"I
the fa u l t b u s
{/
®
w i t h respect to
)
"
IO\
-,- ZpI p). 'I Zpp
)
a
I
=
Z( l \
pp
)
I))
II I I
"
=
=
/ ( 2)
pp
pp
'
f'f'
.
-;-
a
pp
Z l e ),
1 (0)
z ((l),
(l
pf'
Z ( 2 ), Z (0),
!'�. �
__
7. 1 2 )
,
- /1/'- T
7(11)
�J)/"
I 2 ),
- 1Ll( 2 )Zpp
- I " ) ) L I I l ),
iJ
t h e: refe r e n c e .
vo l t a ge d rops a c rOSs t h e oDe n - c i r c l i i t ro i n t , '
Ill
pp
pp
"
I
I mn Z l l )'
- /� ' , Z I � I,
V (2)
=
pp
Two open conductors
Z ( O ),
" Z (�), -'- Z tD),
=
v( I )
(l
II)
ll
_I
=
- IP )Z(2)
kk
i m p l ies p a r a l l e l comb i n a t i o n of i m p e d a n ces,
and
+
+
1a( 1 )
V (2) -
k"
a
(ZiT
iF Z (2)
) kk
_
-
ka
JP )Z kk
( 2)
=
=
1(2)
_
fa -
= 0
'jll(
,, 7(2)
-kk
O n e open conductor
Double l i n e-to-gro u n d fa u l t
Line-to- l i n e fault
Vf
III
to>
III
::l
Open-circuit fa ults
Short-circuit faults
:Inu p'.
PI)
I
I
V( I ) =
n
( I )'
Imn Zpp
+
=
=
Z (l2fJ),
Va(2)
=
PP
a
1(1)
a
PP
+
Z ( O), )
_ / (2)2 (2),
(1
f'f'
1 , ( 1 ) ) ,., _ / ( I I )L (O),
"
Z(O),
/(1)
I l l )( Z (2 ),
(l
+
f
(/
PI'
PP
527
PROBLEMS
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 short-circuit fau lts, including a symmetrical three-phase fault, are shown in
Fi g. 12.3 1 . The sequence networks are indicated schematically by r ec t a n gl es
enclosing a h e avy line to represent the reference of the network and a point
marked bus ® to represent the point in the network where the fault occurs.
The positive-sequ ence network con tains emfs that represent the internal volt­
ages of the machines.
Regard less of the prefault voltage p rofile or the particular type of short­
circu i t fau lt occu rring, the only current which causes positive-sequence voltage
changes at the buses of the system is the symmetrical component I}�) o f the
current 'Ia coming out of phase a of the system at the fault bus ® . These
posit ive-sequence vol tage changes can be calcu lated simply by multi plying
col umn k of the pos itive-sequence bus impedance matrix Z �\�, by the injected
current II,;)' Simila rly, t h e neg at ive- and ze ro-sequence components of the
volt age cha nges due to the short-circ u i t fa u l t on the sys t e m a rc caused, respec­
t ively, by the symmetrical components I}�) and IJ�) of the fau lt c ur rent If out
of bus ® . These sequ ence voltage changes a re also calculated by mu l tiplying
columns ® of Zi,2Js and Z��s by the respect ive current i njections - Ij�) and
-
a
.
- 1 (0)
I"
In
a very real sense , t herefore, th ere is only one proced u re for calculating
the symmetrical components of the voltage chan ges at the buses of the system
when a short-circuit fault occu rs at bus ® -that is, to find I}�), I}�>' and I}�)
and to mult iply columns ® of the corresponding bus impedance matrices by
the negative values of these currents. For the common types of short-circuit
faults the only differences i n the calcu l ations concern t he method of simulating
the fau lt a t bus ® and of formu lating the equations for I}�), I}�), and I};). The
connections of the Theven in equivalents of t he sequence networks, which
provide a ready means of deriving the equ ations for I}�), I}!>, and I};), are
summ arized in Fig. 1 2 . 3 1 and the equations t hemselves are set forth in Table
12.1.
Fa u lts due to open con ductors i nvolve two inj ections into each s e q u e n ce
ne twork a t t h e buses nea rest the ope n i ng in t h e con ductor. Otherw ise, the
procedure for caku lating sequence voltage changes in the system is the same as
that for short-circuit faults. Equati ons for the sequence volt ages and currents at
t h e fa ult are also summarized i n Tab le 1 2. 1 .
The reader is rem i nded of the n e e d to a dj u s t the phase a ngles of the
sym m e t ri c a l compo n e n t s o f t h e c ur r e n t s and t h e v o l t ages in t h ose p a rts o f the
sys tem which are separated from the fau I t bus by �-Y transformers.
PRO B LEMS
60-Hz
groundedluandrbogeopencraralotirng at rated voltage load.
12. 1 . A
is
is
r a t e d 5 0 0 MY A , 2 2 kY .
at no
It
i s Y-connected and solidly
I t is discon nected from the
,
528
12.2.
CHAPTER 12
U NSYM METRICAL FAULTS
rest of the system. Its reactances are X� = X I = X2 = 0 . 1 5 and Xo = 0 . 05 per
unit. Find the ratio of the sub transient line current for a single l ine-to-ground
fault to the sub transient line curre n t for a symmetrical t h ree-phase fault.
Find the ratio of the sub transient l i ne current for a line-to-l ine faul t to the
subtransient current for a symmetrical th ree-phase fa u l t on the genera tor of Prob.
12.1.
i n o h m s t o b e inse rteu i n th e neu tral connec­
tion of the generator of Prob. 1 2 . 1 t o limit the sub transient line cu rrent for a
single l ine-to-ground fau l t to t hat for a th ree-phase fau l t .
12.4. With t h e inductive reac tance fou n d i n Prob. 1 2 . 3 i n s e r te d i n t h e n e u t r a l o f t h e
gener a tor o f Prob. 12. 1 , find t h e ra t i os o f t h e s u b t r a n s i e n t l i n e c u r r e n t s for t h e
fo l lowing fa u lts to t h e s u b t ra n s i e n t l i ne c u r re n t fo r a t h rec-rhasc fa u l t : ( a ) s i n g l e
line-to-ground fau l t , ( b ) l i n e - t o - l i n e fa u l t , a n d C C ) d o u b l e l i n e - t o - g ro u n d fa u l t .
12.5. H