Power System Analysis

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POWER SYSTEM ANALYSIS

McGraw-Hill Series in Electrical and Computer Engineering

Senior Consulting Editor Stephen

W.

Director,

Carnegie-Mellon University

Circuits and Systems Communications and Signal Processing

Computer Engineering Control Theory Electromagnetics

Electronics and VLSI Circuits

Introductory Power alld Eflergy

Radar and An tennas

Previous Consulting Editors

Ronald N. Bracewell, Colin Cherry, James F. Gibbons, Willis W Harman, Hubert Heffn er, Edward W Herold, John G. Linvill, Simon Ramo, Ronald A. Rohrer, Anthony E. Siegman,

Charles Susskind, Frederick E. Terman, John G. Truxal, Ernst Weber, and John R.

Whinnery

Power and Energy

Senior Consulting Editor

Stephen W. Director,

Carnegie-Mellon University

Chapman: Ele c tric Machinery Fundamentals

Fitzgerald, Kingsley, and Umans: Electric Machinery Elgerd: Electric Energy Systems Theory Gonen:

Electric Power Distribution System Engineering

and Stevenson: Power System Analysis Krause and Wasynczuk: Electromechanical Motion Devices Stevenson: Elements of Power System Analysis

Gr ainger

Also available from McGraw-Hill Schaum's Outline Series in

Electronics & Electrical Engineering

Most outlines include basic theory, definitions and hundreds of example problems solved in step-by-step detail, and supplementary problems with answers.

Related

titles on the current list include:

Circuit A n alysis

An alog & Digita l Communications

Basic

Basic

Electrical Engineering

Basic Mathematics for Electricity & Electronics

Basic Electricity

Digital Principles Electric Circliits

Electric Machines & Electromt'c/zallics

Electric Power System,\

Eleclrolnagllclics

Llcctl"Ollic Circuits

EleClrollic Commllnic(I{ion

Electronic Dccias & Cirelli/s Eleccronics TecIII IO/OJ:,T)'

Ellgilleering Ecol/omics FCI'dhuck & Conrrol Systems Ill/roell/Clioll 10 Digila! Systems

Aficroprlarquette 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; 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 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'-------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 - .

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