Analysis of Faulted Power Systems
Editorial Board J. B. Anderson, Editor in Chief R. S. Blicq
R.
F. Hoyt
S. Blanchard
S. V. Kartalopoulos
M.Eden
P. Laplante
R. Herrick
J. M. F. Moura
G. F. Hoffnagle
R. S. Muller I. Peden
W. D. Reeve
E. Sanchez-Sinencio D. J. Wells
Dudley R. Kay, Director of Book Publishing Carrie Briggs, Administrative Assistant Lisa S. Mizrahi, Review and Publicity Coordinator Valerie Zaborski, Production Editor
IEEE Power Systems Engineering Series Dr. Paul M. Anderson, Series Editor Power Math Associates, Inc.
Series Editorial Advisory Committee Dr. Roy Billinton
Dr. Charles A. Gross
Dr. A. G. Phadke
University of Saskatchewan
Auburn University
Dr. Atif S. Debs
Dr. G. T. Heydt
Virginia Polytechnic and State University
Georgia Institute of Technology
Purdue University
Dr. M. El-Hawary Technical University of
Dr. George Karady
Texas A & M University
Nova Scotia
Mr. Richard G. Farmer Arizona Public Service Company
Arizona State University Dr. Donald W. Novotny
University of Wisconsin
Dr. Chanan Singh
Dr. E. Keith Stanek University of Missouri-Rolla
Dr. J. E. Van Ness
Northwestern University
Analysis of Faulted
Power Systems PAUL M. ANDERSON Power Math Associates, Inc.
IEEE PRESS Power Systems Engineering Series Paul M. Anderson, Series Editor
+IEEE
T he Institute of Electrical and Electronics Engineers. Inc .• New York
mWILEY �INTERSCIENCE
A JOHN WILEY & SONS. INC.. PUBLICATION
New York' Chichester' Weinheim
•
Brisbane' Singapore' Toronto
!D 1995 THE INST I TUT E OF ELECTRICAL AND ELECTRONICS ENGINEERS, INC. 3 Park Avenue, 17th Fl oor , New York, NY 10016-5997 !D 1973 Iowa State University Press All rights reserved.
No part of this publication may be reproduced , stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections \07 or 108 of the 1976 United States
Copy righ t Act, without either the prio r written permission of the Publisher, or
authorization through payment of the appropriate per-copy fee to the Copyright
Clearance Center, 222 Rosewoo d Drive, Danvers, MA 01923, (978) 750-8400, fax
(978) 750-4470. Requests to the Pu bl is her for permi s sion should be addressed to the
Permissions Dep artment , Joh n Wiley & Sons. Inc., III River Street, Hoboken, NJ 07030,
(201) 748-6011, fax (201) 748-6008.
For ordering and customer service, call1-800-CALL-WILEY.
Wiley-Interscience-IEEE
ISBN 0-7803-1145-0
Printed ill the United States of America.
10 9
8
7
Library of Congress Cataloging-in-Publication Data Anderson, P. M. (Paul M.)
Analysis of faulted power systems / Paul M. Anderson. p. cm. - (IEEE Press power system engineering series) "An IEEE Press classic reissue."
Reprint. Originally published: Iowa State University Press, 1973. Includes bibliographical references and index. ISBN 0-7803-1145-0 1. Short circuits.
2. Electric circuit analysis.
power systems-Mathematical models. TK3226.A55 1995 621.319-<1c20
I. Title.
3. Electric II. Series. 95-15246 CIP
Table of Contents Preface to the IEEE Reissued Edition
Preface List of Symbols Chapter 1 . General Considerations 1.1 1 .2 1.3 1 .4 1.5 1.6 1 .7 1 .8
Per Unit Calculations . . . . . . . . . . . . . . . . . . . . . . Change of Base . . . . . . . . . . . . .. Base Value Charts. . . . . . . . . . . . . . . . . . . . . . . . . . Three-Phase Systems . . . . . . . . . . . . . . . . . . . . . . . Converting from Per Unit Values to System Values Examples o f Per Unit Calculations . . . . . . . . . . . . . Phasor Notation . . . . . . . . . . . . . . . . . . . . . . . . . . The Phasor a or a-Operator . . . . . . . . .. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . . . . . . . . .
. . . . . . .. . . ............. .
. . .. . . . . . . . .
. . . . . .
.
. . . .
. . . .
. . . .
. . . .
.
. . . . ... . . . . . . .
.
. . . .
. . . .
. . . . . . .
.
.
. . . . . .
. . . .
. .
5 7
7 7 10 10 15 16 17
Chapter 2 . Symmetrical Components 2.1 2.2 2.3 2 .4 2.5 2. 6 2.7
Symmetrical Components of an n-Phase system . . . . . . . . . . . . . . . . Symmetrical Components of a Three-Phase System . . . . . . . . . . . . Symmetrical Components of Current Phasors . . . . . . . . . . . . . . . . . Computing Power of Symmetrical Components . . . . . . . . . . . . . . . . Sequence Components of Unbalanced Network Impedances . . . . . . Sequence Components of Machine Impedances . . . . . . . . . . . . . . . . Definition of Sequence Networks ...................... . . .. Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19 23 25 25 27 30 31
33
Chapter 3. Analysis of Unsymmetrical Faults: Three-Component Method I.
Shunt Faults The Single-Line-to-Ground (SLG) Fault . . The Line-to-Line (LL) Fault. . . . . . . . . . The Double Line-to-Ground ( 2LG) Fault . The Three-Phase (3t/» Fault . . . . . . . . . . Other Types of Shunt Faults . . . . . . . . . Comments on Shunt Fault Calculation .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . .. .
. . . . . .
. . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . .
. . . . .
.
.
. . . . . .
37 42 44 49 52 53
II.
Series Faults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
3.7 3.8 3 .9
Sequence Network Equivalents for Series Faults . . . . . . . . . . . . . . . Unequal Series Impedances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . One Line Open (lLO) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55 61 63
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . . . . .
. . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . .
. . . . . .
. . . . .
. . . . .
. . . . .
37
.
3 .1 3.2 3 .3 3.4 3.5 3. 6
.
.
.
.
.
.
.
v
vi
Ta ble of Contents
3 .1 0 Two Lines Open (2LO) . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . 3.11 Other Series Faults . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
64 66 66
Chapter 4. Sequence Impedance of Transmission Lines
4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.1 0 4.1 1 4.1 2 4. 1 3 4.1 4 4.1 5 4.1 6
Positive and Ne gative Sequence Impedances of Lines . . . . . . . . . . . . Mutual Coupling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . S elf and Mutual Inductances of Parallel Cylindrical W ri es . . . . . . . . Carson 's Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Three-Phase Line Impedances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Transpositions and Twists o f Line Conductors . . . . . . . . . . . . . . . . Completely Transposed Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Circuit Unbalance Due to Incomplete Transposition . . . . . . . . . . . . Sequence Impedance of Lines with Bundled Conductors . . . . . . . . . Sequence Impedance of Lines with One Ground W ri e . . . . . . . . . . Sequence Impedance of Lines with Two Ground Wires . . . . . . . . . . Sequence Impedance of Lines with n Ground Wires . . . . . . . . . . . . . Zero Sequence Impedance of Transposed Lines with Ground Wires . Computations Involving Steel Conductors . . . . . . . . . . . . . . . . . . . . Parallel Transposed and Untransposed Multicircuit Lines . . . . . . . . . Optimizing a Parallel Circuit for Minimu m Unbalance . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 73 75 78 81 84 98 1 02 106 1 12 123 1 28 1 29 1 33 1 37 1 43 145
Chapter 5. Sequence Capacitance of Transmission Lines
5.1 5.2 5. 3 5.4 5.5 5.6 5 .7 5.8
Positive and Negative Sequence Capacitance o f Transposed Lines . . Zero Sequence Capacitance of Transposed Lines . . . . . . . . . . . . . . . Mutual Capacitance of Transmission Lines . . . . .. . . . . . . . . . . . . . . Mutual Capacitance of Three-Phase Lines wit hout ground Wires . . . Se quence Capacitance of a Transposed Line witho ut Ground W ire s . Mutual Capacitance of Three-Phase Lines with Ground Wires . . . . , Capacitance of Double Circuit Lines . . . . . . . . . . . . . . . . . . . . . . . . Electrostatic Unbalance o f Untransposed Lines . . . . . . . . . . . . . . . , Problems . . . . . . . . . . . . . .. . . . . . . . . . . . .. . . . . . . . . . . . . . . . . .
1 52 1 56 1 58 1 63 1 66 1 68 1 72 1 77 181
Chapter 6 . Sequence Impedance o f Machines 1.
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10
Synchronous Machine Impedances .. . . . . . . . . . . . . . . . . . . . . . . . . 1 83 General Considerations . . . . . . . . . . . . . . . . . . . . . . Positive Sequence Impedance . . . . . . . . . . . . . . . . . . Negative Sequence Impedance . . . . . . . . . . . . . . . . . Zero S equence Impedance . . . . . . . . . . . . . . . . . . . . Time Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . Synchronous Generator Equivalent Circuits . . . . . . Phasor Diagram of a Synchronous Generator . . . . . . Subtra nsient Phasor Diagram and Equivalent Circuit Armature Current Envelope . . . . . . . . . . . . . . . . . . . Momentary Currents . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . . . . . . . .
. . . . . .
. . . . . . II . Induction Motor Impedances . . . . . . . . . . . . . . . . . . . . 6.11 General Considerations .. . . . . . . . . . . . . . . . . . . . . . . 6.1 2 In duction Motor Equivalent C ircuit . . . . . . . . . . . . . . . .
. . . . .
. . . . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . ..
. . . . . . . . .
. . . .
.. .. .. .. .. .. .. .. .. .. .. .. ..
.. .. .. .. .. .. .. .. ., .. .. .. .,
183 1 89 1 99 201 202 205 207 214 21 9 221 222 222 222
vii
Table of Contents
6.1 3 Induction Motor Subtransient Fault Contribution . . . . . . . . . . . . . . 226 6 .1 4 Operation with One Phase Open . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 Chapter 7 . Sequence Impedance o f Trans formers I.
7 .1 7.2 7 .3 7.4 7 .5 7 .6 II .
Sin gle-Phase Trans formers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 31 Single-Phase Trans former Equivalents . . . . . . Trans former Impedances . . . . . . . . . . . . . . . . Trans former Polarity and Terminal Markin gs . Three-Windin gTrans formers . . . . . . . . . . . . . Autotrans former Equivalents . . . . . . . . . . . . . Three-Phase Banks o f Single-Phase Units . . . .
. . . . . . Three-Phase Trans formers . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
7 .7 7 .8 7 .9 7 .10 7 . 11
Three-Phase Trans former Terminal Markin gs . . . . . . . . . Phase Shi ft in Y - t::. Trans formers . . . . . . . . . . . . . . . . . . Zero Sequence Impedance o f Three-Phase Trans formers Grounding Trans formers . . . . . . . . . . . . . . . . . . . . . . . . The Zi gzag- t::. Power Trans former . . . . . . . . . . . . . . . . .
III .
Trans formers in System Studies . . . . . . . . . . . . . . . . . . .
7 .1 2 O f f-Nominal Turns Ratios . . . . . . . . . . . . . . . . . . . . . . . 7 . 1 3 Three-Winding O f f-Nominal Trans formers . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . . . .. ..
. . . . . . . . .
. . . . . . . . .
2 31 233 2 34 236 2 39 243 247 247 248 251 255 257 260 260 265 2 65
... ... . . . ... ... ... ... ... ...
. . . . . . . . .
. . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
273 273 278 280 284 286
Chapter 8. Changes in Symmetry Creating Symmetry by Labelin g. . . . . . . . . . . . . . . . . . . . . . Generalized Fault Diagrams for Shunt (Transverse) Faults . . Generaliz ed Fault Diagrams for Series (Longitudinal) Faults Computation o f Fault Currents and Voltages . . . . . . . . . . . . A Fundamental Result : The Invariance o f Power . . . . . . . . Constraint Matrix K . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 . 7 Kron's Primitive Network . . . . . . . . . . . . . . . . . . . . . . . . . . 8.8 Other Use ful Trans formations . . . . . . . . . . . . . . . . . . . . . . . 8.9 Shunt Fault Trans formations . . . . . . . . . . . . . . . . . . . . . . . . 8.1 0 Trans formations for Shunt Faults with Impedance . . . . . . . 8.11 Series Fault Trans formations . . . . . . . . . . . . . . . . . . . . . . . . 8.1 2 Summ ary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8.1 8.2 8.3 8.4 8.5 8.6
... ... . . . ... ... ...
. . . . . .
. . . . . .
2 88
. . . . . .
. . . . . .
291 294 298 304 304 305
Simultaneous Faults by Two-Port Network Theory . . . . . . . . . . . . .
308 308 31 9 323 325 330 334 336
.
. . . . . .
. . . . . .
. . . . . .
. . . . . .
Chapter 9. Simultaneous Faults I.
9 .1 9.2 9.3 9 .4 9.5 9.6 II .
Two-Port Networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . Interconnection o f Two-Port Networks . . . . . . . . . . . . . Simultaneous Fault Connection o f Sequence Networks . Series-Series Connection ( Z-Type Faults ) . . . . . . . . . . . Parallel-Parallel Connection (Y-Type Faults) . . . . . . . . . Series-Parallel Connection ( H-Type Faults) . . . . . . . . . .
. . . . . . . . .
... ... . . . ... ... Simultaneous Faults by Matrix Trans forma tio ns . . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
.. ..
. .
.. .. ..
Ta ble of Contents
viii
9.7
9.8
Constraint Matrix for Z-Type Faults . . . . . . . . . . . . . 337 Constraint Matrix for Y-Type and H-Type Faults . . . . . . . . . . . . . . 339 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 .
.
.
.
.
.
.
.
.
.
.
Chapter1 0. Analytical Simplifications 1 0.1 I. 1 1 1 1
0.2 0. 3 0.4 0.5 II . 1 0.6 1 0.7 III. 1 0.8 1 0.9 1 0.1 0 1 0.11 IV. 1 0.1 2 1 0.1 3
Two-Component Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345 .
Shunt Faults . .. . . . . . . . . . . . Single-Line-to-Ground Fault . . . . . . . .. . Line-to-Line Fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Double Line-to-Ground Fault . . . . . . . . . . . . . . . . . . . . . . Three-Phase Fault . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Series Faults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two Lines Open (2LO) . . . . . . . . . . . . . . . . . . . . . . . . . . One Line Open (l LO) .......................... Changes in Symmetry with Two-Component Calculations . Phase Shifting Transformer Relations. . . . . . . . . . . . . . . . . SLG Faults with Arbitrary Symmetry . . . . . . . . . . . . . . . 2LG Faults with Arbitrary Symmetry . . . . . . . . . . . . . . . Series Faults with Arbitrary Symmetry . . . . . . . . . . . . . . . Solution of the Generalized Fault Diagrams . . .... Series Network Connection-SLG and 2LO Faults . . . . . . . Parallel Network Connection-2LG and1 LO Faults . . . . Problems .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.. . .. .. .. .. .. .. .. .. .. .. ..
.
. . .. ... ... ... ... .. . ... ... ... ... ... ... .
.
. . . . . . . . . . . . . ..... .....
. . . . . . . . . . . . . .. ..
.
.
.
.
.
.
.
.
.
.
.
.
.
,
347 347 349 350 352 353 353 354 355 356 357 358 360 362 362 363 363
Chapter11 . Computer Solution Methods Using the Admittance Matrix 11 .1 11.2
11 .3 11 .4 11 .5
Primitive Matrix . . . . . . . . . . . . . . . . . . . . . Node Incidence Matrix . . . . . . . . . . . . . . . . Node Admittance and Impedance Matrices Indefinite Admittance Matrix . . . . . Definite Admittance Matrix . Problems . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.... .... .... . .. .... .
.
.
.
.
.... .... .... .. . .. .... .
.
.
.... .... .... . . .... .
.
.
Chapter1 2. Computer Solution Methods Using the Impedance Matrix 1 2.1 Impedance Matrix in Shunt Fault Computations . . . . . . . . . . . 1 2.2 An Impedance Matrix Algorithm . . . . . . . . . . . . . . . . . . . . . . . 1 2.3 Adding a Radial Impedance to the Reference Node. . . . . . . . . 12.4 Adding a Radial Branch to a New Node . . . . . . . . . . . . . . . . . . 1 2.5 Closing a Loop to the Reference . . . . . . . . . . . . . . . . . . . . . . . 1 2.6 Closing a Loop Not Involving the Reference . . . . . . . . . . . . . . 1 2.7 Adding a Mutually Coupled Radial Element . . . . . . . . . . . . . 1 2.8 Adding a Group of Mutually Coupled Lines . . . . . . . . . . . . . . 1 2.9 Comparison of Admittance and Impedance Matrix Techniques Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix A . Matrix Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix B. Line Impedance Tables . . . . . . . . . . . . . .. . . . . . . . . . . . Appendix C. Trigonometric Identities for Three-Phase Systems . . . . .
. . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.... .... . .. . . .. ....
. . . 366 . . . 369 . . . 373 . 375 . 386 . . . 389 .
.
. . . . . . . . .
. . . .
.
.
. . . . . . . . . . . . .
. . . . . . . . . . . . .
393 401 401 402 403 404 408 41 5 41 8 41 9 421 436 467
ix
Ta ble of Contents
Appendix D. Self Inductance of a Straight Finite Cylindrical Wire Appendix E. Solved Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . Appendix F. A-Y Transformations . . . . . . . . . . . . . . . . . . . . . . . . Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . .
. . . .
. . . .
. .. .. ..
.
. . . .
. . . .
, . . .
470 474 502 503
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . , . . . . . . . . . . . . . . . . . . . . . . . . . . 507
Preface to the IEEE Reissued Edition Many textbooks have been written to describe a power system operating under balanced or normal conditions.
A few texts also deal, at least in an elementary
manner, with the unbalanced system conditions. The primary object of this book is to provide a text to be used by graduate students and to challenge the student to draw upon a background of knowledge from earlier studies. Since this is intended for advanced studies, the student reader should be able to use circuit concepts at the advanced undergraduate or beginning graduate level and should have a work ing knowledge of matrix notation. Particular stress here is placed upon a clear, con cise notation, since it is the author's belief that this facilitates learning. Although the thrust of the book is toward the solution of advanced problems, a thorough background is laid in the early chapters by solving elementary configura tions of unbalanced systems. This serves to establish the algebraic style and nota tion of the book as well as to provide the reader with a growing knowledge and facility with symmetrical components. It also introduces the elementary concepts for those not familiar with this discipline. In either case, this background information of the early chapters should be studied since it establishes certain conventions used throughout the book. The text contains sufficient material for an entire year of study on the subject. The first three chapters are introductory and may be covered rapidly in a graduate level class where the students have already been introduced to symmetrical compo nents. This material should be reviewed, at least in a quick reading, as the matrix The middle portion of the book, Chapters 4 through 7, treats the subject ofpower Here, the sequence impedances of transmission lines, machines,
notation used throughout the book is introduced in these chapters.
system parameters.
and transformers are developed in detail. This is important material and is often omitted in the education of power engineers. Methods are used here that permit ex act solutions of very general physical problems, such as finding the impedance of un
transposed or partially transposed lines. Matrix methods are used to clarify computations and adapt them to computer usage. The final portion of the book presents the application of symmetrical components to a variety of problems and provides an introduction to computer solution methods for large networks. Here, one learns to appreciate the use of matrix algebra in the solution of complex problems. This section also reinforces the engineer's appreciation of symmetrical components as a problem-solving technique. In preparing the manuscript of a book of this type, one stands on the shoulders of giants. Several excellent books introduced these concepts in the first half of the xi
xii
Preface
twentieth century, in particular those by C.
F . Wagner and R. D. Evans, G. O. Cal
abrese, and the books by Edith Clarke, are classics on symmetrical components and are still used by many of us. Their basic ideas are enlarged upon here, and the pre sentation simplified by the use of matrix methods, thereby aiding understanding and making computer solution much easier. The author's colleagues at Iowa State Uni versity played an important role in the development of the book. They used the book in its early stages, found many problems, and offered helpful suggestions for areas needing improvement. The electrical engineering department at Iowa State Univer sity was also helpful and understanding of the need for this effort and provided sup port in many ways. This new printing of the book by IEEE Press provides software that was espe cially developed at Iowa State University for the solution of exactly the kind of prob lems introduced in this book. The program PWRMAT was developed by the author, fellow faculty members, and graduate students for the purpose of providing the en gineer with a convenient method of solving the many problems associated with ma trices of complex numbers. The program has been improved over the years by Iowa State University and has been used and enjoyed by many students for at least twenty years. Iowa State University has graciously agreed that the program could be dis tributed with this printing of the book to make problem solving easier. This is im portant, since the drudgery of solving these complex problems by calculator distracts the engineering student from the objective ofleaming the concepts. The software and text file versions of the user's manual are attached to the book on diskettes. The soft ware is easy to master and to use. It permits the user to write small programs in a simple language that can read a user�created data file to solve the problem at hand and print the computed results in an orderly matrix notation. Operations such as matrix inversion or reduction are accomplished with ease. It is hoped that the addi tion of this software will help many new readers in their desire to become more pro ficient with the important subjects covered in this book. P. M. Anderson San Diego, California
Preface
Many textbooks have been written dealing with the power system that operates under balanced or normal conditions. A few, particularly the excellent recent text by W. D. Stevenson [ 9 ] , deal in an elementary way with both normal and faulted systems. Most of the books treating unbalanced and faulted systems, however, have been in print for years and are inadequate for several reasons. Nevertheless, in spite of their date of copyright, the serious student should be come familiar with the famous works of C. F. Wagner and R. D. Evans [10] , the outstanding volumes by Edith Clarke [ 1 1 ] and the more recent work of Cala brese [24] . The goal here is to produce a text to be used by graduate students, one that can draw upon a background of knowledge from previous courses. Since this is an advanced text, the student should be able to employ circuit concepts not usually taught to undergraduates and should recognize the beauty and simplicity of matrix notation. Particular stress is placed upon a clear, concise notation since it is the author's belief that this facilitates learning. Although the thrust of the book is toward the solution of advanced problems, a thorough background is laid by solving elementary configurations of unbalanced systems. This serves to establish the algebraic style and notation of the book. It also serves to introduce the elementary concepts to the uninitiated. Thus for some it will be an organized review and for others an introduction to the solution of faulted networks. In either case this background should be studied since it establishes certain conventions used later. The text contains sufficient material for a two-semester or three-quarter treat ment of the subject. The first three chapters are introductory and may be covered rapidly in a graduate class where the students have already been introduced to symmetrical components. This material should be reviewed, at least in a quick reading, as the matrix notation used throughout the book is introduced in these chapters. The middle portion of the book, Chapters 4 through 7 , treat the subject of power system parameters. Here the sequence impedances of lines, machines, and transformers are developed in detail. This is important material and is often omitted in the education of power engineers. Methods are used here which per mit exact solutions of very general physical problems such as finding the impe dance of untransposed or partially transposed lines. Matrix methods are used to clarify these computations and adapt them to computer usage. The final portion of the book presents the application of symmetrical comxiii
xiv
Preface
ponents to a variety of problems and provides an introduction to computer solu tion of large networks. Here one learns to appreciate the use of matrix algebra in the solution of complex problems. This section also reinforces the engineer's appreciation of symmetrical components as a problem-solving technique. At Iowa State University we have found it convenient to cover most of the first 10 chapters in a two-quarter sequence, leaving computer applications as a separate course. This means that some of the sections in Chapters 1- 1 0 must be omitted, but the student is encouraged to pursue these on his own. In this two quarter presentation Chapters 1-3 are skimmed quickly since the course carries an undergraduate prerequisite which introduces symmetrical components. Then the balance of the first quarter is spent on power system parameters, leaving the applications for the second quarter. This book would not have been possible without the unique contribution of many individuals to whom the author is greatly indebted. Several Iowa State Uni versity colleagues, particularly W. B. Boast, J . W. Nilsson, and J . E . Lagerstrom (now of the University of Nebraska) , are largely responsible for the author's interest in the subject. These three were also responsible for the organization and teaching of a short course in symmetrical components, taught in connection with the Iowa State University A-C Network Analyzer for 10 years or so. The author's interest in this course, first as a student and later as a teacher, helped him gain competence in the subject. Indeed, many ideas expressed here are taken directly or indirectly from the short course notes. The influence of the late W. L . Cassell must also be mentioned, for his insistence on a clear notation has contributed to the education and understanding of many students, the author included. The author is particularly indebted to David D. Robb who used much of the book in a graduate class and made countless valuable suggestions for improvements. Por tions of the computer solutions presented are the work of J. R . Pavlat and G. N . Johnson, and these contributions ar e gratefully acknowledged. Finally, I wish to express my thanks to the Electrical Engineering Department of Iowa State University and to W. B. Boast, head of the department, for giving me the opportunity to prepare this material. Special thanks are due to my wife, Ginny, who provided both moral support and expert proofreading, and to my editor, Nancy Bohlen, who is a marvel with both mathematical notation and eccentric authors.
List of Symbols
1. CAPITALS
A . . . . . . . . . . . . . . . ampere, unit symbol abbreviation for current A . . . . . . . . . . . . . . . complex transformation matrix; transmission parameter matrix; node incidence matrix A . . . . . . . . . . . . . . . magnetic vector potential a . . . . . . . . . . . . . . . inverse transmission parameter matrix B . � Y, susceptance B . . . . . . . . . . . . . . . complex transformation matrix; shunt susceptance matrix C . . . . . . . . . . . . . . . capacitance C . . . . . . . . . . . . . . coulomb, unit symbol abbreviation for charge C . . . . . . . . . . . . . . complex transformation matrix; Maxwell's coefficients; capacitance coefficients D . . . . . . . . . . . . . . . distance or separation E . . . . . . . . . . . . . . source emf; voltage E . . . . . . . . . . . . . . . primitive source voltage vector F . . . . . . . . . . . . . . . farad, unit symbol abbreviation for capacitance F, F'. . . . . . . . . . . . . fault point designation F-D-Q .. . . . . . . . . . rotor circuits of a synchronous machine G . . . . . . . . . . . . . . . = me Y, conductance G . . . . . . . . . . . . . . . inverse hybrid parameter matrix GMD, GMR . . . . . . mutual geometric mean distance, geometric mean radius H . . . . . . . . . . . . . . . henry, unit symbol abbreviation for inductance Hz . . . . . . . . . . . . . . hertz, unit symbol abbreviation for frequency H . . . . . . . . . . . . . . . hybrid parameter matrix I . . . . . . . . . . . . . . . rms phasor current Iabe ' . . . . . . . . . . . . . = [ la Ib Ie ] t, line current vector = [ lao Ia I Ia2 ] t, sequence current vector 101 2 , 18, base line current, A J . . . . . . . . . . . . . . joule, unit symbol abbreviation for energy J . . . . . . . . . . . . . . primitive current source vector K . . . . . . . . . . . . . . . dielectric constant K . . . . . . . . . . . . . . Kron's transformation or connection matrix L . . . . . . . . . . . . . inductance LL . . . . . . . . . . . . . . line-to-line LN . . . . . . . . . . . . . line-to-neutral L . . . . . . . . . . . . . . inductance matrix M . . . . . . . . . . . . . . . = 106 mega, a prefix .
.
.
.
.
.
.
.
.
.
.
.
.
.
=
.
.
.
.
.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
.
.
.
,
xv
xvi
Ust
Moo MiJ Moo
N N N
0
p P
(J>
Q
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
R oo Roo
S
SB
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
SLG T oo TB
o
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
T!/>o
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
U
V V VA Vab c VO l 2 VB W 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
.
•
•
•
•
.
•
0
•
•
•
•
.
Wb X 'Y
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
•
•
•
0
•
0
•
0
•
0
0
0
0
0
0
•
0
0
0
0
0
0
0
0
0
0
0
0
•
0
0
•
0
0
•
0
0
0
•
0
0
0
0
•
0
0
•
•
•
0
0
0
0
0
0
Y YB oo 0
y
0
0
•
•
•
•
•
•
•
•
•
0
0
Zoo
0
0
0
0
0
0
0
0
0
0
0
•
0
0
0
0
0
•
•
0
0
0
0
.
0
•
0
0
0
0
0
•
0
0
0
0
•
0
0
0
•
•
•
0
0
0
0
Z
.
0
0
ZB
Z
.
of Symbols
mutual inductance minor of a matrix two-port network vector newton, unit symbol abbreviation for force zero potential bus designation two-port network vector phasor operator average power; transformer circuit designation Vandermonde matrix; potential coefficient matrix; Park's transformation matrix average reactive power; transformer circuit designation; total charge; phasor charge density = ate Z, resistance; transformer circuit designation resistance matrix = P + jQ, complex apparent power base apparent power, VA single-line-to-ground time; time constant; torque; equivalent circuit configuration base time, s twist matrix unit matrix rms phasor voltage volt, unit symbol abbreviation for voltage voltampere, unit symbol abbreviation for apparent power = [ Va Vb Vc) t , phase voltage vector = [ Va o Va l V( 2 ) t , sequence voltage vector o. base voltage, V watt, unit symbol abbreviation for power weber, unit symbol abbreviation for magnetic flux = g". Z, reactance primitive admittance matrix = G + jB, complex admittance base admittance, mho admittance matrix primitive impedance matrix = R + jX, complex impedance base impedance, n impedance matrix 0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
•
•
•
•
0
0
•
0
0
0
0
0
0
0
0
0
0
0
0
.
0
0
0
0
•
•
2. LOWERCASE
a
0
0
ac a-b-c adj
0
0
0
0
0
0
0
0
0
•
0
•
0
- e 12,, /3 , 1 200 operator -
alternating current stator circuits of a synchronous machine; phase designation adjoint (of a matrix) = we, line susceptance per unit length b ber, bei . real, imaginary Bessel functions c capacitance per unit length dc . . . . . . . . . . . . . . . direct current 0
.
.
0
0
0
0
•
0
0
0
0
0
0
0
0
0
0
0
0
•
•
•
•
•
•
•
•
•
•
0
•
•
0
•
•
•
•
0
0
0
•
0
0
0
0
0
0
0
0
0
0
0
•
•
•
•
0
•
•
•
•
0
•
0
•
•
0
•
0
0
•
•
•
0
0
•
0
0
•
•
•
List of Symbols
xvii
do, d2 zero, negative sequence electrostatic unbalance factors d, q stator circuits, referred to the rotor det . . . . . . .. . . . . . . determinant ( of a matrix) e ... . . ... . .. . ... base for natural logarithms f ...... . . . . . . . .. frequenc y fit kth fraction of total line length g . . ground terminal h . .. . two-port hybrid parameter designation h .. . . . . . . . . . . . . . a constant (1 or V3) •
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
.
.
.
.
.
.
.
.
.
•
•
•
•
.
•
•
.
.
.
.
.
.
.
.
.
•
•
•
•
•
.
.
.
.
.
.
.
.
•
•
•
.
.
.
.
.
.
.
. . . . . . . . . . . . . .. instantaneous current
.
instantaneous current vector = A, 900 operator constants used in computing L, C .. = 103 , kilo, a prefix . = oJ 3/2 , a constant used in synchronous machine theory . e inductance per unit length; leakage inductance .. In, log . . . . . . ... . . natural (base e) , base 10 logarithms = 10-3 , milli, a prefix; a constant used in computing skin m . effect rn mutual inductance per unit length rno, rn2 zero, negative sequence electromagnetic unbalance factors ... rn . . complex transformation ratio n . . . ..... . . . .. .. number of phases; number of nodes; = 10-9 , nano, a prefix n . . neutral terminal; neutral voltage; turns ratio; number of . .. turns pu . .. .... .... . . . per unit p .. . ... . . . instantaneous power q . . . . .. . ... . linear charge density of a wire q .. . . . .. . . vector of linear charges on a group of wires r . radius; internal (source) resistance; resistance per unit length . . .. s . . . .. . . . . . . . .. . line length, length of section k, slip of an induction motor s .. . . . . . . . . . . . . . speed voltage vector t ... .. . time u . . ... unit step function v . . . .... instantaneous voltage v . . . . . . . ... . . . .. instantaneous voltage vector x . . line reactance per unit length; internal (source) reactance . .... y two-port admittance parameter designation z . .. two-port impedance parameter designation; internal (source) impedance; impedance per unit length Z . . transmission line primitive impedance i
.
.
j . k, k' k k .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
•
•
•
•
•
.
.
.
.
.
.
.
.
.
.
•
.
.
•
.
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
•
.
.
.
.
•
3 . UPPERCASE GREEK . .. delta connection; determinant (of a matrix) . . . . . . . . . . .. . . summation symbol n . . .. . . . . . . . . . . . ohm, unit symbol abbreviation for impedance
A �
.
.
.
.
.
.
.
.
.
.
.
.
.
xviii
list
of Sym bols
4. LOWERCASE GREEK Ci
phase angle ac/dc skin effect ratios /j torque angle of a synchronous machine /j jj Kronecker delta = €oK , permittivity € () . . . . . . . . . . . . . . . phase angle; rotor angle of a synchronous machine K dielectric constant . . . . element of Vandermonde matrix; flux linkage A = Ilollr , permeability (Ilo , free space; Ilro relative) Il = 10-6 , micro, a prefix Il 1T pi, 3.141 59265 . .. p resistivity T time constant cp magnetic flux; phase angle . . . . w radian frequency; synchronous machine speed •
•
•
CiR, CiL
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
.
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
.
.
.
.
.
.
.
.
.
5. SUBSCRIPTS a A b B B c
C d D
e eq
.
.
. •
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
. . . . . .
.
.
.
.
•
. . .
.
.
.
.
.
.
.
.
.
.
. . . .
.
.
. env . . . . F . f . g . .
.
.
.
.
LN LL
m
.
.
.
.
H
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
.
.
.
max . . . min . .
M n q
Q r R s
.
.
.
.
.
.
.
. . .
phase a; armature phase a . . phase b phase b . • . . . . . . . . base quantity phase c; core loss quantity phase c; transformer circuit designation . direct axis; direct axis circuit quantity . direct axis damper winding quantity . excitation quantity, of a transformer . equivalent circuit quantity; equivalent spacing . . . . . . . . . . . . envelope of an ac wave referring to the fault point; field winding . . referring to the fault point . referring to the fault point . . .. . . transformer winding designation line-to-neutral . . . . line-to-line magnetizing quantity ( in a transformer); motor quantity; . . mutual (coupling or GMD) . . . . . . . . . . maximum . minimum . . . mutual (frequently MO , M1, M2) . . neutral . quadrature axis; quadrature axis circuit quantity . . quadrature axis damper winding quantity . rotor quantity . . . rotor quantity . . . . source quantity; stator quantity; self (GMD) .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. .
.
. .
.
.
.
.
.
•
.
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
List of Sy m bols
8
.
.
.
.
.
.
.
.
sym . . . . . . T . . . . . . . u
X
.
.
.
.
.
.
.
.
.
. . . . .. . Y .. . . . . . . 1 , 3
•
•
•
•
•
.
.
.
.
.
.
.
.
.
( )t
.
.
. stator quantity; transformer circuit designation; self (frequently 80, 81 , 82) . . . . . . . symmetrical . . . . . . . transformer circuit designation . per unit . . . . . . . transformer winding designation . . . . . . . transformer winding designation single-phase; three-phase zero, positive, negative sequence quantity . . . . . . . zero, sum, difference sequence quantity . initial condition; change condition
.
.
.
.
.
.
.
.
.
.
.
.
•
•
•
•
•
•
•
.
.
.
.
.
.
.
.
.
.
.
.
.
6 . SUPERSCRIPTS
( t1 . . . C) . . .. C) . r) ( )*. . . . . . . . . . . . .
transpose ( of a matrix) inverse (of a matrix) (tilde), distinguishing mark for various quantities (circumflex) , distinguishing mark for various quantities = d/dt, derivative with respect to time conjugate, of a phasor or a matrix ( ) , . . . . . . . . . . . . . . ( prime), transient ( ) " . . . . . . . . . . . . . . ( double prime), subtransient •
.
.
.
•
.
.
.
.
.
.
.
.
.
•
•
•
•
•
•
•
•
•
•
•
•
•
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
7. NUMERAL SYMBOLS 1. . . . . . . . . . . . . . single-phase 2LG . . double-line-to-ground . . 3
.
.
.
.
.
.
.
.
.
•
•
.
•
xix
.
.
.
.
•
.
.
.
.
•
•
•
Analysis of Faulted Power Systems
chapter
1
General Considerations
The analysis of power systems usually implies the computation of network voltages and currents under a given set of conditions. In many cases the computa tion is organized to give a particular kind of information for a special purpose. For example, it may be desirable to determine the current flowing through the ground in d particular situation to facilitate the setting of a ground relay. Fig ure 1.1 presents an organization for power system computation. The class of problems to the left in the figure are called "steady state" because they are to be solved by algebraic equations. This is not to imply that the system is static or un changing at the moment in time for which a solution is sought. On the contrary, the system may be undergoing very rapid changes, for example, in a faulted condi tion. The point is that algebraic equations are much easier to solve than differen tial equations. We therefore have learned to make good use of the steady state solutions in system planning and for determining system protection. This is like taking a set of photographs of the system behavior under certain specified condi tions. From these photographs we can design system additions and protective schemes and can learn a great deal about the system strengths and weaknesses. The dynamic problems shown to the right in Figure 1.1 are what we usually call "stability problems" in power system work. Here we solve a set of differen tial equations to determine the behavior of voltages, currents, and other variables as functions of time. Both the steady state and dynamic problems are usually of large dimension for power system work. Networks of a few hundred nodes are quite common and many machines may be represented in dynamic problems. Thus in both kinds of problems we must be oriented toward computer solutions of some kind. The emphasis in this text is on the faulted network, both balanced and un balanced. As will be shown later, the techniques used for unbalanced faulted net works will work equally well for unbalanced normal networks. In most cases we will consider the system to be linear and will use algebraic equations exclusively to take "photographs" of voltage-current relationships under certain given condi tions. There is much we can do with these photographs, and these ideas will be explored in some depth. Unbalanced system operation is one we would usually prefer to ignore. Often when dealing with normal or near-normal operation, this is precisely what we do ignore it. That is, we assume balanced operation, solve the network on a per phase basis, and extrapolate to obtain information for the remaining two 3
C hapter 1
4
POWER SYSTEM ANALYSIS
STEADY STATE
NETWORK
DYNAMIC
NETWORK FAULTED
NORMAL
BALANCED UNBALANCED BALANCED
I. system assumed linear
2. ins tantaneous photograph In time
- a l gebraic equations
Fig.
1.1.
NETWORK NORMAL
NETWORK FAU LTED
I
UNBALAN C ED
I I I
I. network balanced or unbalanced
2. system linea r or nonlinear
I 3. solution in time domain I I I _,_ differentia l equations . ..0.------...... __
An organization of power system analysis problems.
phases. I This results in a great saving in time and effort and usually gives results of reasonable accuracy if the system is nearly normal, i.e . , nearly balanced. When the system is obviously unbalanced, other methods must be used. The method generally favored is that of "symmetrical components" as proposed by Fortescue III 191 8 [1] . This method permits us to extend the per phase analysis to systems with unbalanced loads or with unbalanced termination of some kind, such as a short circuit or fault. The method is of limited value when the system itself is un balanced. Our analysis will make extensive use of symmetrical components and will do so with the aid of matrix notation whenever possible. In doing so, we will assume a balanced system (Za = Zb = Zc ) with unbalanced loading unless special note is made of a system or network unbalance. The only reason for doing this is to extend the simplicity of the per phase representation to unbalanced system terminations. It may be possible in the future by the use of a large computer to completely represent all three phases of a system and to handle unbalances in either the system or the load in a more direct manner. The method of symmetrical components has led to the development of useful apparatus to measure or use symmetrical component variables. One example is the negative sequence relay used to protect generators from overheating in the event of unbalanced loading. The positive sequence segregating network is some times used to supply the sensing voltage to generator voltage regulators. Certain connections of instrument current and potential transformers develop zero se quence quantities that are used in protective ground relaying schemes. Finally, a comment concerning the use of digital computers is necessary. IA three-phase system is assu med here since they are by far the most common. When un specified, a three-phase system will always be implied.
5
General Considerations
Since power systems are really very large electric networks, it is usually impossible to perform a complete solution by hand computation because of the size of the problem. Thus any solution or technique which we devise must be applicable to the digital computer. The power system engineer should be capable of bridging this gap between theory and computer utilization, and this subject will be empha sized throughout the text. 1.1
Per Unit Calculations
The calculation of system performance conveniently uses a per unit represen tation of voltage, current, impedance, power, reactive power, and apparent power (voltampere). 2 The numerical per unit value of any quantity is its ratio to the chosen base quantity of the same dimensions. Thus a per unit quantity is a "normalized" quantity with respect to a chosen base value. Any quantity can be converted to a per unit quantity by dividing the numerical value by a selected base quantity of the same dimensions. In an electrical network five quantities are usually involved in the calculation of networks. These are shown in Table 1 . 1 together with their dimensions. Our use of the term "dimenTable 1.1. Electrical Quantities and Their Dimensions Quantity
Current, A Voltage, V Voltamperes, S Impedance, 12 Phase angle Time, sec
[I] [V] [VI] [VII]
I V
Dimension
1/>, (J, etc.
dimensionless
Sy mbol S -P + jQ Z = R + jX
t
[T]
sion" here is admittedly loose since I and V can be specified dimensionally in terms of force, length, time, and charge, for example. The meaning is clear, how ever. In calculations of steady state phenomena the time dimension is suppressed in phasor notation. Of the five remaining quantities one is dimensionless and the other four are completely described by only two dimensions. Thus an arbitrary choice of two quantities as base quantities will automatically fix the other two. In power system calculations the nominal voltage of lines and equipment is almost always known, so the voltage is a convenient base value to choose. A second base is usually chosen to be the apparent power (voltampere). In equipment this quan tity is usually known and makes a convenient base. In a system study the voltam pere base can be selected to be any convenient value such as 1 00 MVA, 200 MV A, etc. The same voltampere base is used in all parts of the system. One base voltage is selected arbitrarily. All other base voltages must be related to the arbitrarily selected one by the turns ratios of the connecting transformers. Special treatment must be given to load tap changing transformers and to network loops wherein the net product of turns ratios around the loop is not equal to unity. This matter is treated in Chapter 7. In three-phase networks the turns ratios used to relate the several base voltages are those of Y- Y or equivalent Y- Y transformers. 2Portions of this section are derived from [ 2 ] , prepared by several authors, particularly J. E. Lagerstrom, and J. W. Nilsson.
W. B. Boast,
6
Chapter 1
If we designate a base quantity by the subscript B, we have on a per phase basis and
base voltamperes= SB base voltage
=:
VB
VA
( 1 . 1)
V
(1.2)
Then the base current and base impedance are computed as S Base 1= IB =.-J! A VB V V2 Base Z= ZB = �= � n SB IB
( 1 . 3) (1.4)
Similarly , we define a base admittance as Base Y=: YB =:
SB mho V�
(1.5)
Having defined the base quantities, we can normalize any system quantity by dividing by the base quantity of the same dimension. Thus the per unit impedance Z is defined as Z ohm - pu ZB
Z=:
--
(1.6)
Note that the dimensions (ohm) cancel, and the result is a dimensionless quantity whose units are specified as per unit or pu. At this point one might question the advisability of using the same symbol, Z,. for both the per unit impedance and the ohmic impedance. This is no problem, however, since the problem solver always knows the system of units he is using and it is convenient to use a familiar nota tion for system quantities. Usually we will remind ourselves that a solution is in per unit by affixing the "unit" pu as in equation ( 1.6). Furthermore, where there may be a question of units, we will always identify a system quantity by adding the apprppriate units, such as the notation (Z ohm) of equation ( 1.6). Since we may write Z= R + jX in ohms, we may divide both sides of this equa tion by ZB with the result Z=R + j X pu
(1.7)
R ohm pu ZB
(1.8)
X ohm pu . ZB
(1.9)
where R= and X=
In a similar manner we write S=P + jQ voltampere, and dividing through by SB , we have S=P+ jQ pu
(1. 10)
7
G enera l Considerations
where P watt
pu
(1.11)
= QSBvar
pu
( 1 . 1 2)
P=
and Q 1 .2
Change of Base
SB
The question sometimes arises that given a pu impedance referred to a given base, what would be its pu value if referred to a new base? To answer this ques tion, we first substitute equation ( 1 . 4) into (1.6) with the result
Z = (Z
ohm )
8� V
(1. 13)
pu
Two such pu impedances, referred to their respective base quantities, are now written, using the subscript 0 for old and n for new.
Z (Z Zn = (Z °
=
ohm)
8Bo Vjo 8 Vjn
ohm) �
( 1 . 14)
But the system or ohmic value must be the same no matter what the base. Equat ing the (Z ohm) quantities in ( 1 . 14 ), we have
Zn = Zo ( VVBBnO) 2 ( 88Bn) Bo
pu
( 1 . 1 5)
Equation ( 1 . 1 5) is very important since i t permits us to change base without knowledge of the ohmic value (Z ohm ). Note that the pu impedance varies directly as the chosen (new) voltampere base and inversely with the square of the voltage base. 1.3
Base Value Charts
In most power system problems the nominal transmission line voltages are known. If these nominal voltages are chosen as the base voltages, an arbitrary will fix the base current, base impedance, and base admittance. choice for the These values are tabulated in Table 1.2 for several values of system voltage and for several convenient MV A levels. A more extensive list of base values is given in Appendix B.
8B
1.4
Three-Phase Systems
The equation derived for pu impedance ( 1 . 1 3 ), or its reciprocal for pu admit tance, is correct only for a single-phase system. In three-phase systems, however, we often prefer to work with three-phase voltamperes and line-to-line voltages. We investigate this problem by rewriting ( 1 . 1 3 ) using the subscript LN for "line to-neutral" and 14> for "per phase, " with the result
Table
1.2.
Base current in amperes
Base impedance in ohms
Base admittance in micromhos
B as e Current , B as e Impedance , and B as e Admittance for Common Transmission Voltage Levels a n d for Selected M V A Levels
Base Kilovolts
5. 0
1 0. 0
20. 0
34.5 69.0 115.0 138.0 161.0 230.0 345.0 500.0 34.5 69.0 115.0 138.0 161.0 230.0 345.0 500.0 34. 5 69.0 115 0 138.0 161.0 230.0 345.0 500.0
83.67 41.84 25.10 20.92 17.93 12.55 8.37 5.77 238.05 952.20 2645.00 3808.80 5184.20 10580.00 23805.00 50000.00 4200.80 1050.20 378.07 262.55 192.89 94.52 42.01 20.00
167 35 83.67 50.20 41.84 35.86 25.10 16.74 11.55 119.03 476.10 1322.50 1904.40 2592.10 5290.00 11902.50 25000.00 8401.60 2100.40 756.14 525.10 385.79 189.04 84. 02 40.00
334.70 167.35 100.41 83.67 71.72 50.20 3 3.47 23.09 59.51 238.05 661.25 952.20 1296.05 2645.00 595 1.25 125 00.00 16803.19 4200.80 1512.29 1050.20 771.58 378.07 168.03 80.00
.
.
Base Megavol t·A mperes 25. 0
5 0. 0
1 0 0. 0
200. 0
250. 0
418.37 209.19 125.5 1 104.59 89. fi5 62.76 41.84 28.87 47.61 190.44 529.00 761 76 1036.84 2116.00 4761.00 10000.00 21003.99 5251.00 1890.36 1312.75 964.47 472.59 210.04 100.00
836.74 4 18.37 251.02 209.18 179.30 125.51 83.67 57.74 23.81 95.22 264.50 380.88 5 18.42 1058.00 2380.50 5000.00 42007.98 10502.00 3780.72 2625.50 1928.94 945.18 420.08 200.00
1673.48 836 74 502.04 418.37 358.60 251.02 167.35 1 15.47 1 1.90 47.61 132.25 190.44 259.21 529.00 1190.25 2500.00 84015.96 2 1003.99 7561.44 5251.00 3857.88 1890.36 840.16 400.00
3346.96 1673.48 1004.09 836.74 717.21 502.04 334.70 230.94 5.95 23 81 66.13 95.22 129.61 264.50 595.13 1250.00 168031.93. 42007.98 15122.87 10502.00 7715.75 3780.72 1680.32 800.00
4183.70 2091.85 1255. 11 1045.92 896.51 627.55 418.37 288.68 4.76 19.04 52.90 76. 18 103.68 211.60 476. 10 1000.00 210039.91 5 2509.98 18903.59 13127.49 9644 69 4725.90 2100.40 1000.00
.
.
.
.
G enera l Considerations
and
9
Z = 8� (Z ohm) pu V80LN
( 1 . 16)
V82 -LN ( Y mho) pu 880 1 1/>
( 1 . 1 7)
= V8-LL .j3
( 1 . 18)
Y=
But using LL to indicate "line-to-line" and 31/> for "three-phase, " we write for a balanced system
V80LN and 88_1 1/>
_ -
V
8 8-3 1/> VA 3
( 1. 19)
Making the appropriate substitutions, we compute Z=
V80LL
(Z ohm) pu
( 1 . 20 )
V802 LL
( Y mho) pu
(1.21)
8 3 �_ ¢
and Y=
8 8-3 C/J
These equations are seen to be identical with the corresponding equations derived from per phase voltampere and line-to-neutral voltages. This is fortunate since it makes the formulas easy to recall from memory. A still more convenient form for ( 1 . 20) and ( 1 . 2 1 ) would be to write voltages in kV and voltamperes in MV A. Thus we compute Base MVA3p (Z 0h m ) pu Z = (Base kV d 2
( 1 . 22)
L The admittance formula may be expressed two ways, the choice depending upon whether admittances are given in micromhos or as reciprocal admittances in megohms. From (1.21) we compute Y
=
(Base kVLd 2 ( Y M mho) pu (Base MVA31/» ( 1 0 6 )
( 1 . 23 )
( Base kV Ld2 ( 1 0- 6 ) pu (Base MV A 3
( 1. 24 )
or Y=
Equations ( 1 . 23) and ( 1 . 24) are both useful in transmission line calculations where the shunt susceptance is sometimes given in micromhos per mile and sometimes in megohm-miles. Usually the subscripts and 31/> can be omitted without ambiguity, but it is wise to use caution in these simple calculations. Many a system study has been
LL
10
Chapter 1
made useless by a simple error in preparing data, and the cost of such errors can easily run into thousands of dollars. For transmission lines it is possible to further simplify equations ( 1.22) to (1.24). In this case the quantities usually known from a knowledge of wire size and spacing are
1. the resistance R in ohm/mile at a given temperature 2. the inductive reactance XL in ohm/mile at 6 0 Hz 3. the shunt capacitive reactance Xc in megohm-miles at 60 Hz We make the following arbitrary assumptions: Base MV A 3 tP
line length
= =
100 MVA 1.0 mi
( 1.25)
Values thus computed are on a per mile basis but can easily be multiplied by the line length. Likewise, for any base MVA other than 100 the change of base for mula ( 1 1 5 ) may be used to correct the value computed by the method given here. Thus for one mile of line .
Z
Kz
=
=
( Z n /mi) (Base MVA3 tP) (Base kVLL ) 2
Base MV A 3 tP (Base kVLd 2
==
=
(Z n /mi) Kz
pu
100
( 1.26) (1.27)
(Base kVLd 2
Similarly, we compute (1.28)
where KB
=
(Base kVLL ) 2
100
(10 -6 ) = 10 -8 (Base kV
LL )
2
( 1.29)
The constants Kz and KB may now be tabulated for commonly used voltages. Several values are given in Table 1 . 3. 1 .5 Converting from Per Unit Values to System Values
Once a particular computation in pu is completed, it is often desirable to con vert some quantities back to system values. This conversion is quite simple and is performed as follows : (pu I ) (Base I) = I A (pu V ) (Base V) = V V (pu P) (Base VA) = P W (pu Q ) (Base VA ) = Q var
(1.30)
Usually there is no need to convert an impedance back to ohms, but the procedure is exactly the same. 1 .6
E xamples of Per Unit Calcu l ations
To clarify the foregoing procedures some simple examples are presented.
Table Base k V
2.30 2.40 4.00 4.16 4.40 4.80 6.60 6.90 7.20 11.00 11.45 12.00 12.47 13.20 13.80 14.40 22.00 24.94 33.00 34.50 44.00 55.00 60.00 66.00 69.00 88.00 100.00 110.00 115.00 132.00 138.00 154.00 161.00 220.00 230.00 275.00 330.00 345.00 360.00 362.00 420.00 500.00 525.00 550.00 700.00 735.00 750.00 765.00 1000.00 1100.00 1200.00 1 300.00 1400.00 1500.00
1.3. Values of Kz
and KB for Selected Voltages
KZ
18.903592 17.361111 6.250000 5.778476 5.165289 4.340278 2.295684 2.100399 1.929012 0.826446 0.762762 0.694444 0.643083 0.573921 0.525100 0.482253 0.206612 0.160771 0.091827 0.084016 0.051653 0.033058 0.027778 0.022957 0.021004 0.012913 0.010000 0.008264 0.007561 0.005739 0.005251 0.004217 0.003858 0.002066 0.001890 0.001322 0.000918 0.000840 0.000772 0.000763 0.000567 0.000400 0.000363 0.000331 0.000204 0.000185 0.000178 0.000171 0.000100 0.000083 0.000069 0.000059 0.000051 0.000044
KB
0.0529 X 10- 6 0.0576 0.1600 0.1731 0. 1936 0. 2304 0.4356 0.4761 0.5184 1.2100 1.3110 1.4400 1.5550 1.7424 1.9044 2.0736 4.8400 6. 2200 10.8900 11.9025 19.3600 30. 2500 36.0000 43.5600 47.6100 77.4400 100.0000 121.0000 132. 2500 174. 2400 190.4400 237. 1600 259.2100 484.0000 529.0000 756. 2500 1089.0000 1190. 2500 1296.0000 1310.4400 1764.0000 2500.0000 2756. 2500 3025.0000 4900.0000 5402.2500 5625.0000 5852. 2500 10000.0000 12100.0000 14400.0000 16900.0000 19600.0000 22500.0000 X 10- 6
Chapter 1
12
Example 1 . 1
Power system loads are usually specified in terms of the absorbed power and reactive power. In circuit analysis it is sometimes convenient to represent such a load as a constant impedance. Two such representations, parallel and series, are possible as shown in Figure 1. 2. Determine the per unit R and X values for both the parallel and series connections. Loa d
Rp
Fig. 1 . 2.
BUI
Loo d
BUI
Xp
Constant impedance load representation : left , parallel representation; right, series represen tation .
Solu tion Let P = load power in W
Q = load reactive power in var
Rp or R, = load resistance in n
Xp or X, = load reactance in n V = load voltage in V
Parallel Connection. From the parallel connection we observe that the power absorbed depends only upon the applied voltage, i.e. ,
( 1. 3 1 )
From equation ( 1 . 1 3) we have Ru =
Rp (SB ) ( VB ) '
where the value subscripted u is a pu value. compute
( 1 . 32)
pu
Substituting Rp from ( 1 . 3 1 ), we
( 1 . 33 )
and we note that ( 1.33) is the same as ( 1 . 3 1 ) except that all values are pu. Similarly, we find the expression for pu X to be Xu = ( VI VB ) ' ( S B /Q ) = VJ / Qu
pu
( 1 .34)
Series Connection. If R and X are connected in series as in Figure 1.2 b, the problem is more difficult since the current in X now affects the absorbed power P. In terms of system quantities, I = VI (R s + jX.) . Thus P + jQ = VI * =
VV * .
RB - J X,
=
Iv I2
R B - jX,
(1.35)
13
G enera l Considerations
Multiplying ( 1 . 3 5) by its conjugate, we have
p2 + Q2
Also, from ( 1 . 3 5)
. P + JQ
=
=
Iv l4
( 1 . 36 )
R� + X;
Ivl2 (R, + jX,)
( 1 . 37)
R ,2 + X.2
Substituting (1.36 ) into ( 1.37), we compute
Rearranging,
. _ (R, + jX,) (P2 + Q 2 ) P + JQ Iv l2 (1. 38)
Equation (1. 38) is the desired result, but it is not in pu. Substituting into ( 1.13), we have
(R, + j X,) SB R u + J· Xu -_ V� Then we compute from ( 1 .38)
Example
- V� SB (P watt)
Ru
-
Xu
=
p2
+
Q2
( 1 . 39)
pu
V� SB ( Q var) pu p2 + Q2
( 1.40)
1.2
Given the two-machine system of Figure 1 .3, we select, quite arbitrarily, a base voltage of 161 kV for the transmission line and a base voltampere of
��
l l �L
�
' "'�-""""" + ""'j-:-IO""'O-O""'h--� 5 Om y Fig.
1.3.
ood
A two-machine system.
20 MV A. Find the pu impedances of all components referred to these bases. The apparatus has ratings as follows : Generator : 15 MVA, 13.8 kV, x = 0. 1 5 pu Motor : 10 MVA, 13.2 kV , x = 0.15 pu T1 : 25 MV A, 13.2-161 kV, x = 0.10 pu T2 : 1 5 MVA, 1 3.8-161 kV , x = 0 . 1 0 pu Load : 4 MVA at 0 .8 pf lag
Solution Using equation ( 1 . 1 5 ) , we proceed directly with a change in base for the apparatus.
Chapter 1
14
( 20 ) ( 1 3.8) 2 = 0.2185 pu Generator: x = (0.15) 15 13.2 Motor: x = (0.1 5 )
(20) (1 3 .2) 2 = 0 . 2745 pu 10 13.8
( 20) (161) 2 = 0.08 pu 25 161 (20 ) ( 16 1) 2 = 0.1 333 pu T2 : x = (0.10) 15 161 T1 : x = (0.10)
(1 .41 )
For the transmission line we must convert from ohmic values to pu values. We do this either by dividing by the base impedance or by application of equation ( 1 .22 ). Using the latter method, Z=
(50 + j���1�m) � (20 )
= 0.0386 + jO.07 71 pu
( 1 .42 )
For the load a parallel R-X representation may be computed from equations ( 1 .32 ) and (1 .34) S = P + jQ = lsi (cos 0 + j sin 0 ) = 4 (0 .8 + jO.6)
= 3 .2 + j 2 .4 MV A
Then
Ru =
VJ SB P
=
VJ (20)
Similarly, Xu = 8 .33 VJ pu.
Example
3.2
=
6 . 25 Vu2
pu
( 1 .43)
1.3
Suppose in Example 1 .2 that the motor is a synchronous machine drawing at 0.9 pf lead and the terminal voltage is 1 . 1 pu. What is the voltage at MVA 10 the generator terminals?
Solution as
First we compute the total load current. For the motor, with its voltage taken the reference, i.e. , V = 1 .1 + jO, we have 0 P jQ 9 - j( - 10 sin 25.9 ) = 0 . 409 + J· 0 • 198 5 pu IM = = V* 20( 1 . 1 ) -
For the static load
3.2 - j2 .4 I = L 20( 1 . 1 )
Then the total current is 1M + h or
= 0 .1455
-
J·0 . 1 09 pu
1 = 0 .5545 + jO.0895 pu
( 1 .44)
From Example 1 .2 we easily find the total pu impedance between the buses to be
15
General Considerations
the total of T1 , T2, and Z (line ) ; Z = 0 + jO.213 pu. Note that the transmission line impedance is negligible because the base is small and the line voltage high for the small power in this problem. Thus the generator bus voltage is Vg = 1 . 1 + jO
+
(0 + jO.213) (0.5545 + jO.0895)
= 1 .1 - 0.0191 + jO.118 = 1 .08 + jO . 1 1 8 pu
= 1 .087 /6 .24° pu on 13.2 kV base = 14 .32 kV
1 .7
Phasor Notation
In this book we will deal with voltages and currents which are rms phasor quantities. This implies that all signals are pure sine waves of voltage or current but with the time variable suppressed. Thus only the magnitude (amplitude) and relative phase angle of the sine waves are preserved. To do this we use a special definition. Definition : A phasor A is a complex number which is related to the time domain sinusoidal quantity by the expression 3 a(t) =
( 1 .45)
If we express A in terms of its magnitude IA I and angle a, we have A = IA lejO! and a(t) =
( 1 .46)
Thus equations ( 1 .45 ) or ( 1 .46 ) convert the rms phasor ( complex) quantity to the actual time domain variable. We seldom have occasion to do this since it is generally preferable to work exclusively with complex rms phasor quantities. Note the simplicity of the definition. The phaso\" is not a "rotating vector. " It is simply a complex number having the same dimensions (ampere, volt) as the time domain quantity. Note that a single constant frequency (constant w ) is im plied by the definition . Since we are dealing with steady state solutions of net works, w is constant and impedance is a constant ratio of V/1 and is itself a phasor (complex) quantity. There is, however, no occasion to convert impedance to the time domain since this is meaningless. From our phasor definition we may write v'2 IAle i (wt+a)
=
=
v'2 IA I cos (w t + a ) + j.J2 I A l sin ( w t + a ) a ( t) jb(t)
+
( 1 .47 )
If we differentiate ( 1 .46 ) with respect to time, we have ci ( t ) = - w J2 IAlsin (w t
or from ( 1 .47 ), ci ( t) = - w b( t )
+
a)
( 1 .48)
Substituting b ( t ) from ( l .48 ) into ( 1 .47 ), w e have the differential equation ci ( t) + jwa( t) = j W v'2A e i w t 3The definition could just a s well u s e the imaginary part o f (V2Aeiw t ), b u t either definition should be used consistently .
C ha pter 1
16
which has the solution
[
aCt) = a (o) This can be rearranged to the form
A=
�] e-i w t + � e i w t
aCt) - a(O) e -i w t j y'2 sin w t
(1 .49)
which is a formula for computing a phasor A, based on definition ( 1 .45) and given the time domain sinusoid aCt). It is convenient to think of (1 .49) as a "phasor transformation" which we designate by the script letter
(1.51)
Either the symbolic notation o f ( 1 .50) o r the exact formula (1 .49) may b e used. Note that the expression ( 1 .50) is a linear transformation [ 4 ] . 1 .8
The Phasor a or a-Operator
In working with three-phase quantities, it is convenient to have a simple phasor operator which will add 1200 to the phase angle of a phasor and leave its magnitude unchanged. A common notation for this is to define a = 1L1.2!L = e i 2 1T/3 (1.52) a
Fig.
02 1.4.
Phasor diagram of the a-operator.
It is convenient to think of the a-operator as one which rotates a phasor by + 1200 and the a2 -operator as one which rotates a phasor by - 1200 • It is also clear that (see Fig. 1 .4) a3 = 1L1L. , a4 = a a6 = 1 ,
etc.
(1.53)
At times it is convenient to know various linear combinations of a-operators. Some familiar combinations are given in Table 1.4. 4This notation i s due to Nilsson [ 3 ] .
17
G enera l Considerations
Table
1.4. a-Operator Functions
Function
Polar
a a2
1/120° 1/240° ll5L 1/120° 1/60° 1/-60° va/- 30° va� va/150° va/- 150° va!!ML va/- 90° 1/180°
a3
a4 1 + a = - a2 1 + a2 = - a
11aa2 aa2 a+ 1+
a
a2
1 1
a2 a
a2 a + a2
Rectangular
- 0.5 + jO.866 -0. 5 - jO.866 1.0 + jO -0.5 + jO.S66 0.5 + jO.S66 0.5 - jO.S66 1.5 - jO.S66 1.5 + jO.S66 - 1.5 + jO.S66 - 1.5 - jO.S66 0.0 + j1.732 0.0 - j1.732 - 1.0 + jO o
o
Problems
1.1. Convert all values to pu on a 10 MV A base with 100 kV base voltage on the line.
Gfl. ��-------50� � G
TI
T2
L I N E'
AY
YA
M
Fig. Pl . l .
1 5 MVA, 13.S kV, X - 0.15 pu 10 MV A, 12 kV, X - 0.07 pu T1 : 20 MVA, 14-132 kV, X - 0.10 pu T2: 15 MVA, 13-115 kV, X " 0.10 pu Line : 200 + j500 il
Generator: Motor:
1.2.
Prepare a per phase schematic of the system shown and give all impedances in pu on a
100 MVA, 154 kV transmission base.
20+ J 80 o h m
10 + j 4 0 ohm
10 + j 40 0 h m L oa d
Fig. Pl . 2.
G1: G2: T1: T2:
50 MVA, 13.8 kV, X " 15% 20 MV A, 14.4 kV, X .. 15% 60 MVA, 13.2-161 kV, X = 10% 25 MVA, 13.2-161 kV, X = 10% Load : 15 MV A, SO% pf lag
1.3. Draw a per phase impedance diagram
for the system shown. Assume that the load impedance is entirely reactive and equal to j1.0 pu. Find the Thevenin equivalent, looking
18
Cha pter 1
CD
01
X- 0. 1
�' O.l
@
X - 0. 1
®
I 0
X = 0.0 5
Fig. P1 . 3 .
into this system from an external connection at bus 3 if (a) Generated voltages VI and V2 are equal. (b) Generated voltages V I and V2 are not equal. 1.4. The following table of values has been prepared for the various line sections in a small electric system. Find the total pu impedance and shunt susceptance of each line on a 10 M VA base, using the line nominal voltage as a voltage base.
Nominal Voltage (k V)
Line Length (mi)
Wire Size
13.8 13.8 13.8 13.8 13.8 69.0 69.0
5.0 2.0 3.9 6.2 7.3 10.0 25.0
4/0 cu 4 cu 4/0 A 336.4 A 556.5 A 4/0 A 336.4 A
R
X
Xc
(n/mi)
(n/mi)
(Mn-mi)
0.278 1.374 0.445 0.278 0.088 0.445 0.278
0.690 0.816 0.7 11 0.730 0.330 0.711 0.730
0.160 0. 193 0.157 0.172 0.142 0.157 0.172
1.5. Verify equation (1.49) and compute A for the following time domain sinusoids: (a) Am cos w t (b ) A m cos (wt + a) (c) Bm s in w t (d) Bm sin (wt + (j) 1.6. Prove that the transformation (J> [aCt) ] is a linear transformation. 1.7. Given that aCt) is a sinusoid given by equation (1.46), find the phasor transform of b et) where b et) is the derivative of a (t) with respect to time.
1.8. Compare the phasor transform (J> to the Laplace transform .c.
chapter
2
Sym metri ca l Co m po n e nts
It was pointed out in Chapter 1 that a per phase representation of power sys tems is preferred because of its simplicity. In solving problems of balanced three phase networks this is easily accomplished by changing all delta connections to equivalent wye connections and solving one leg (one phase) of the resulting net work. Since the system is balanced, the results for the remaining two phases differ from the first by ± 1 20° in phase and the problem is solved. Such a solution avoids the complexity of detailing each terminal and makes use of inherent system symmetry. The method of symmetrical components provides a means of extending per phase analysis to systems with unbalanced loads. This is possible because of a property of unbalanced phasors discovered by Fortescue [ 1 ) . He observed that a system of three unbalanced phasors can be broken down into two sets of balanced phasors plus an additional set of single-phase phasors. If the voltages and currents of the problem are represented in this way, a per phase representation is adequate for each component and the desired simplification has been achieved. Such a sys tem will be analyzed in detail later. First, as an introduction to the subject con sider the more general problem of an n-phase system. 2. 1
Symmetrical Components of an n·Phase System
In a brilliant paper published in 191 8 [ 1 ] , C. L. Fortescue proposed a system whereby an unbalanced set of n phasors may be resolved into n - 1 balanced n-phase systems of different phase sequence and one zero-phase sequence system. By his definition a zero-phase sequence system is one in which all phasors are of equal magnitude and angle or they are all identical. We denote Fortescue's system mathematically as follows. Consider the n-dimensional system of phasors defined by the following equa tions:
Va = Val + Va2 + Va 3 + . . . + Van Vb = Vbl + Vb2 + Vb3 + . . . + Vb n ( 2. 1 ) where Va , Vb , . . . , Vn = an un balanced set of phasors Vah Vbi t , Vnl = the first set of n balanced phasors with an angle 2-rr In between components a, b, . . . , n •
•
•
19
C hapter 2
20
, Vn = the second set of n balanced phasors with an angle 4rr ln be Va 2 , Vb 2 , 2 tween components a, b , . . . , n •
..
..
..
..
..
..
..
..
..
•
..
•
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
.
..
..
..
..
..
..
..
..
..
..
..
..
,.
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
..
Va (n - J ) , Vb (n- l ) , , Vn ( n- J ) = the (n - 1 )st set of n balanced phasors with an angle 2rr (n - 1 )/n between components a, b, . , n Va n , Vb n , , Vnn = the final or nth set of n balanced phasors with an angle n ( 2rr In ) = 2rr between components a, b, . . . , n ; or the components of this last set are all identical •
•
•
.
•
•
.
•
Equation ( 2. 1 ) may be simplified by adopting a notation similar to the a-operator of section 1.8. Suppose we generalize the definition of a to be
(2. 2 ) where n
=
the system dimension or number of phases. Then n a2 = � 4 1r/ rotates a phasor by 4rr In radians n a 3 = ei 61r/ rotates a phasor by 6rr In radians n a = ei 2 1r
rotates a phasor by 2rr radians.
Negative powers o f a are also easily defined and amount to negative (clockwise) rotation of phase angles. Using the definition ( 2. 2), we operate on ( 2. 1 ) by multiplying the equations n , a with the result by 1, a, a2 , •
•
•
Va a Vb a2 Vc
= = =
+ Va n Val + Va 2 + + a Vbn aVb l + aVb 2 + + a2 Vcn a2 VCl + a2 Vc 2 + .
.
.
.
.
.
.
.
.
( 2.3) B y our definition of the phase sequence systems 1, 2 , . , n w e immediately rec ognize that .
.
(2.4) This is because we defined this sequence to have exactly an angle 2rr In between components and, since the components are balanced (equal magnitudes and phase displacements), rotating Vbl by 2rr ln radians makes it coincide with Val t etc. Note also that the sum of all the components of equation ( 2.4) is n Val . Now add the equations (2.3) together and consider this sum. For the right hand side we have n Val for the first (subscript 1 ) group as previously noted. All other groups add to zero. This can be shown by adding vertically the subscript 2 group as an example. Va 2 aVb 2 a2 VC 2
= = =
aa - 2 Va
2 a2 a-4 Va 2
= = =
Va 2 a - I Va 2 a- 2 Va2
21
Sym m etrica l Components
an- I Vn 2
=
an- I a - 2 ( n- I ) Va 2 = a -( n- I ) Va 2
( 2. 5 )
The sum i s zero because the right-hand side o f ( 2. 5 ) i s a balanced set o f n voltages whose sum is zero. In a similar way we show that all groups on the right side of ( 2.3) except the first add to zero since they are all balanced sets of voltages. Add ing equations ( 2. 3 ) then gives the result (2.6 )
or (2.7) I n a similar way we show that
( 2.8) Matrix notation provides a convenient way of writing equations (2.7) and ( 2.8). We list the equation for Va n first since it is the simplest, and we call it Va O ' Then VaO Val
Va l
Va (n-I )
1
n
1 1 1 1
1 a a2
1 a2 a4
1 an-I a2 ( n-t )
an - I
a2 ( n- l )
a(n-
I ) ( n-t )
Va Vb Vc Vn
( 2.9)
which we may write in matrix notation as y = C V
( 2. 10)
Here we use the notation V-slash (1) to indicate the voltage V split into sym metrical components of phase a and where V is an array of the original n un balanced phasors. The matrix C is a matrix of operators, or it is a transformation whereby a set of phasors V may be resolved into a new set of phasors which define the symmetrical components V. Actually, C determines only the symmetrical components of phase a, namely Vao , Va l > ' . . , Va ( n-I h but from these the com ponents of the other phases are easily found by symmetry. Since ( 2. 1 0) defines a transformation, the uniqueness of this operation is of immediate interest. The transformation is unique if and only if C is nonsingular [ 5, 6] . If C is nonsingular, its inverse C-I = A
(2. 1 1 )
22
C hapter 2
exists and we may write from ( 2. 10) (2.12)
V=AY
Equation ( 2. 1 2) indicates that the phase voltages may be synthesized from the symmetrical components of phase a, a most important result. Our problem, then, is to show that C is nonsingular. It is helpful to recognize that C is a special case of the Vandermonde matrix [ 4, 7, 8) P where 1 1 1 1
( 2. 13) In the case of matrix C we have exactly the form P with X I of P is given by det P = IT (X i - Xj ) = (X2 - X I ) . . . (X n - X d (X 3 - X2 ) i>j
.
•
=
•
1. The determinant
(X n - Xn-I )
( 2. 14)
and we see that P is nonsingular as long as the X i are distinct. In the case of C (2. 1 5) are all distinct, being complex quantities of magnitude 1 but differing in argument by 21r In radians, as shown in Figure 2. 1. Thus the transformation ( 2. 10) is in deed unique, and its inverse transformation ( 2. 1 2 ) exists. 03
a
I I I
Rad i a n s
o n-I
Fig. 2. 1 . Complex .operators from row 2 o f C.
To find the inverse transformation, we examine equation ( 2. 1 ) where Vao = Since all symmetrical components may be written in terms of the phase a components, we rewrite ( 2. 1 ) entirely as a function of phase a. Thus
Va n .
Va = VaO + Val Vb = VbO + Vbl Ve = VeO + Ve l
Va 2 + Vb 2 + Ve2 +
Va (n-I ) + . + Vb (n-I ) + . . . + Ve (n- J )
+
.
.
.
.
+
.
( 2. 16)
23
Symmetrical Components
First, we note that Va O = Vb O = VeO = . . . = Vn O ' Also, from ( 2.4) Val = a Vbl = a 2 Ve l = . . . = a n-I Vnl Similarly, Va 2 = a 2 Vb 2 = a 4 Ve 2 = . . . = a 2 ( n- J ) Vn 2 with similar expressions apply ing for the remaining sequences. These relationships are established by definition of sequence quantities. Substituting into ( 2. 16), we have Va = VaO + Vb = VaO + Ve = Va o +
Va (n- I ) Va 2 + . . . + ) (n 2 a- Va 2 + . . . + a - -J Va ( n- I ) a -4 Va2 + . . . + a- 2 ( n-J ) Va (n l ) -
Val + a- I Val + a- 2 Val +
(2.17) This equation may b e simplified b y changing the negative exponents o f "a" to positive exponents by the relation a - k = ai n - k , 0 < k < n, i = 1, 2, . . . which amounts to adding 3600 to a negative argument to make it positive. If this is done, ( 2. 1 7 ) becomes Val + Va = VaO + Vb = Va O an-I Val + Ve = Va O an-2 Val + Vn = VaO or in matrix form
Va Vb Ve
++ +
Va (n- I ) Va 2 + . . . + n 2 + . . . a a Va2 + Va (n-l ) n 2 . . . a -4 Va 2 + + a Va ( n- I )
a Val + a2 Va 2 + . . . + an- I Va ( n- J )
1 1 1
1
1
an - I an - 2
an - 2 an - 4
1 a a2
. . . . . . .. . . . . . . . . . . . . . .
VaO Va l Va2
an - I a a Vn 1 Va (n - I ) This is identical with ( 2 . 1 2 ) V = A V from which we conclude that 2
1 1
A= 1 1
1
1
an - I an - 2
an - 2 an - 4
a
( 2. 1 8)
(2.19)
1
(2 .20)
2.2 Symmetrical Components of a Three-Phase System
The n-phase system described in the preceding paragraph is of academic inter est only . We therefore move directly to a consideration of the more practical three-phase system . Other useful systems, such as the two-phase system, may be described as well but are omitted here in order to concentrate on the more com mon three-phase problem . If n = 3, the phasor a-operator rotates any phasor quantity by 1 200 and the identities of section 1 .8 apply. We are directly concerned with the "analysis and
24
Chapter 2
synthesis equations, " (2.10) and (2.12) respectively in the three-phase system. From section 2 . 1 , with n = 3 , we have for the analysis equation
(2.21) or VOl2 = C Vabc
(2.22)
where we have written Y as VOl2 and V as Vabc since the latter notation is more descriptive and is not unduly awkward when n = 3. We will use either notation, however, depending upon which seems best in any given situation. Also note that the subscript 012 establishes a notation used by many authors [9, 10, 1 1 ] , for example, where
o refers to the "zero sequence " 1 refers to the #1 or "positive sequence " 2 refers to the #2 or "negative sequence" The names zero , positive, and negative refer to the sequence (of rotation ) of the phase quantities so identified . This is made clear if typical sequence quantities are sketched as shown in Figure 2.2. There it is quite plain that the positive sequence
800 -.::: :-�-\--_ o
0
VoO = VbO= VcO
Fig. 2. 2.
A typical set of positive, negative, and zero sequence voltages.
set ( Va l ! Vb " Ve d is the same as the voltages produced by a generator which has phase sequence a-b-c, which we denote here as positive . The negative sequence set ( Va2 ' Vb 2 , Ve 2 ) is seen to have phase sequence a-c-b which we denote as nega tive. The zero sequence phasors ( VaO , VbO, VeO) have zero-phase displacement and thus are identical. The synthesis equation for a three-phase system, corresponding to equation ( 2 1 9 ) is
.
(2.23)
Sym metri ca l Compone nts
25
or
Vabc = A V012 By direct application of (2.23) we "synthesize " the phase quantities as graphically in Figure 2.3. Note that we could also apply equation (2.21 ) unbalanced phasors of Figure 2.3 by either analytical or graphical means to the symmetrical component quantities.
(2 .24) shown to the obtain
Vc
Fig.
2.3.
Synthesis of sequence quantities to give phase quantities.
2.3 Symmetrical Components of Current Phasors
Up to this point we have considered the symmetrical components of line-to neutral voltage phasors only. Symmetrical components of line-to-line voltages may also be computed and used. These same techniques, however, apply to any unbalanced set of three-phase (or n-phase) quantities. Thus for currents we have relations identical to (2.22) and (2 .24) , viz. ,
( 2. 25) and
(2. 26)
2.4 Computing Power from Symmetrical Components
For any three-phase system the total power at any point is the sum of the powers computed in the individual phases, i.e., with P3 t1> indicating the average value of the three-phase power where
( 2 . 27)
V�be = transpose of Va b e = [ Va Vb Ve ] , a row vector I:• •
m
[�J
·
the conjugate of I• • •
But from ( 2. 24) and ( 2. 26) we may write the power in terms of symmetrical com ponents as P 3 t1>
=
At A * I�1 2
(2. 28)
Since A is a symmetric matrix, At = A. We examine the matrix product AA*
Chapter 2
26
where we note that a * = a2 ,
( 2. 29)
Performing the operation indicated by the matrix product, we conclude that AA* = 3 U where u
=
[� � �] -
(2.30)
�e unit maw
Substituting this result into (2.28), we have (Re
V�1 2 U I:1 2 = 3 (Re ( Vao I:O + Va l I: + Va2 /:2 ) = 3
Plq) = 3
(2.3 1 )
This is an important result and shows that the total three-phase power is a func tion of the symmetrical components of voltage and current of the same phase sequence. Thus there is no "coupling " of power from the negative sequence cur rent reacting with the positive or zero sequence voltage, for example. Equation (2 . 3 1 ) shows clearly that a change in the power relationship has oc curred by virtue of the transformation from the a-b-c frame of reference of the 0-1-2 frame of reference. At times it would be convenient to work with a power invarian t transformation which we may denote as A. For power invariance we re quire that A A* = U
such that PS,p
(2.32)
= 9le(Vao1:o + VaII:I + Va2I:2 ) . We assume that A differs from A we let A = A/h then we easily show that A-I = hA- 1 • We
only by a constant. If also compute
A A 1 1 AA * = 2 AA * = 2 3 U h h -
-
( 2.33)
But from ( 2. 32) this quantity should be equal to the identity matrix. From this we conclude that h = va for power invariance. This means that we have the transformation A=
1 -
va
and its inverse 1 A- = va A
1
� :J [� :J 1
a2 a
1 a a2
(2.34)
(2.35 )
Transformation (2 .34) i s often found i n recent power literature. This result suggests that a new definition of the A matrix may be desirable. Suppose we define
Sym metri ca l Compone nts
A=
[� � � [� :J
�
such that
27
1
a2 a
(2 . 36)
1
A- ' =
a
a2
Then we may set
h = 1 for the Fortescue transformation = va for the power invariant transformation
(2.37 )
(2.38)
In this book definition (2. 36) will be used. The reader may substitute his own value of "h" from ( 2 .3 8 ) . In this way we let the same transformation serve both
those who prefer the older definition and those who insist upon power invariance. Using (2 .36) then, we compute P31/1
_
-
3 ( Vao la*O + Va l 10*1 + Va 2 Ia*z ) h 2 (Re
( 2 . 39)
2.5 Sequence Components for U nbalanced Network I mpedances
Consider the three-phase system shown in Figure 2.4, where each current en counters an impedance in its phase conductor and where, in general, the self and mutu al impedances are unequal, i.e., (2.40)
Thus both the self and mutual impedances constitute sets of unbalanced or un equal complex impedances. Even balanced currents will therefore produce un e qual voltage drops between m and n . 10
Fig. 2 . 4 .
m
.
I
Z aa
� ; zl- ) il rc
zcc
-
Vm: "I/l1'o
+
-
Z'bc Zc a
I "
.1
n
A three-phase system with series Z.
Using equation (2.22 ) , we were able to find symmetrical components for a set of three unbalanced (complex) phasor voltages. We apply a similar technique to investigate the possibility of finding symmetrical components of the unbalanced impedances. I I We emphasize the word of because later we will define and use sequence impedances Zo o Z l o and Zz which will express impedance to the flow of laO . lat , and lal respectively. These are not the impedances under investigation here.
Chapter 2
28
We begin by writing the voltage drop equation from m to n of Figure 2.4, which we will call V mn .
( 2. 4 1 )
where V mn is actuall y V mn-a bc- Applying the synthesis transform ( 2. 24) to both sides of ( 2.4 1 ), we have A V m n -0 1 2 = Z A 10 1 2
(2. 42)
or the symmetrical components of the voltage drop are given by Vmn - 0 1 2
Suppose we define
A - I Z A 10 1 2
( 2. 4 3 )
Vmn -0 1 2 = Zm n -0 1 2 10 1 2
( 2. 44 )
Zm n-0 1 2 = A - I Z A
( 2. 4 5 )
=
Then Zm n - 0 1 2 is defined by the similarity transformation [ 7 ] The concept of the similarity transformation gives us mathematical insight into the operations being performed. We interpret A as a linear operator which trans forms currents or voltages from one coordinate system (0- 1-2) to another (a-b-c). Z is another transform which takes a current vector la be into a voltage drop vector Vm n ' both being in the a-b-c coordinate system. The operator A, since it is non singular, may be used to find the transformation of similarity to take currents into voltage drops in the 0- 1 -2 coordinate system. Since At * A -I , the transformation is not orthogonal [ 7] . The new matrix Zmn - 0 1 2 may be found directly as indicated by equation ( 2. 45) with the result Z mn -0 1 2 =
[
ZM2 )
(ZS O + 2ZMO )
(ZS 2 -
(ZS I ( ZS 2 -
( Zso - ZMO ) ( ZS I + 2 ZM d
where we have defined
ZM t l ZM2 )
(2.46)
ZS O = ( 1 / 3 ) ( Zaa + Zb b + Zee) 2 ZS I = ( 1 /3 ) (Zaa + a Zb b + a Zee ) ZS 2 = ( 1 / 3 ) (Zaa
and
+
a2 Zb b + a Zee )
ZM O = ( 1/ 3) (Z b e + Zea + Zab ) 2 ZM 1 ( 1 / 3 ) ( Zb e + aZe a + a Zab ) ZM2 = ( 1 /3 ) ( Zb e + a2 Zea + a Zab- )
( 2. 47)
==
( 2.48)
In these definitions we have taken advantage of the fact that mutual impedances of passive networks are reciprocal [ 12] , Le., Za b = Zb a , etc. Note that we have
followed the form of equation ( 2. 2 1 ) in grouping terms for definitions ( 2.47) and ( 2. 48). We find a curious symmetry in these definitions. Since equation ( 2.44)
29
Symmetr i ca l Compone nts
relates symmetrical components of phase a, similarly ( 2.47) and ( 2.48) pivot on phase a quantities as a point of symmetry. Note that these results are identical for either h from ( 2. 38). (Why?) Substituting the matrix ( 2.46) into equation ( 2.44), we see immediately a problem which we should like to avoid. Examine, for example, the equation for the positive sequence component of the voltage drop V mn -O l 2 : Vm n- I
=
(ZS I - ZM I ) lao + (Zso - ZMO ) la l + ( ZS 2 +
2ZM2) la2
(2.49)
Obviously, the positive sequence voltage drop depends upon not only la l but la o and la 2 as well. This means there is mutual coupling between sequences-a rather disturbing result. Furthermore, Zm n - o l 2 is not symmetric, therefore this mutual effect is not reciprocal. There are several special cases in which the matrix Zmn - o l 2 is simplified. Specifically, there are three special cases of both self and mutual impedances which cause significant changes in this matrix. Special case 1-zero impedance. The absence of any impedance in either the self or mutual case is obviously a simplifying assumption. In many cases the mutual impedances are neglected since they are often small compared to self impedances. Note however that the matrix Zmn-o l 2 is nonsymmetric with respect to bo th Zs and ZM terms and is not made symmetric by eliminating either the self or mutual terms. Since elimination of self impedance terms must generally be rejected , this is not a satisfactory simplification to make. Special case 2--equal impedance. In many problems the self or mutual im pedances may be equal in all three phases. In such cases equations (2.47 ) and ( 2. 48) become ZSO =
Zaa
ZS l = ZS2 = 0
( 2 . 50)
and ZMO
= Zb c
ZM I =
ZM2
=0
( 2. 5 1 )
Referring to (2 .46), w e note that the simultaneous application o f ( 2 . 50 ) and (2. 51 ) eliminates the off-diagonal terms of Zm n-o l 2 ' This means that Zmn - o l 2 is not only reciprocal but that there is zero coupling between sequences. Note that the application of either (2.50) or ( 2.5 1) alone will still result in a nonreciproc al Zm n- O I 2 ' Special case 3-symmetric impedance. A less restrictive case than those above is one in which the self or mutual impedances are symmetric with respect to phase a , Le. , Z b b = Zcc Za b =
Zc a
(2.52)
With these restrictions the self impedances become
(1/3) (Zaa + 2 Zb b ) = ZS 2 = (1/3 ) (Zaa - Zb b )
Zs o = ZSI
(2.53)
Chapter 2
30
and the mutual impedances are ZM O = (1 / 3 ) (Z b c + 2 Zab ) ZM l = ZM 2 = ( 1 / 3 ) (Z b c - Zab )
(2 .54)
This restriction, applied with respect to self or mutual impedances, makes the Zmn - 0 1 2 matrix symmetric with respect to that kind of impedance. However, the application of one without the other stilI leaves the matrix nonsymmetric. Summary. Since we have four possible patterns for two kinds of impedances there are 16 different combinations of interest. These are shown in Table 2.1 where we use the notation :
zero = all matrix elements are zero diag = the matrix is diagonal sym = the matrix is symmetric non = the matrix is nonsymmetric
Table 2. 1. Summary of Matrix Conditions Based on Impedances Self Impedances
Mu tual Impedances
Zero
Equal
Symmetric
A rb itrary
Zero Equal Symmetric Arbitrary
zero diag sym non
diag diag sym non
sym sym sym non
non non non non
If the impedance matrix is zero, the voltage drops vanish and the problem is a degenerate one. A diagonal matrix means that the sequences are uncoupled, Le., currents from one sequence produce voltage drops only in that sequence-a very desirable characteristic. A symmetric matrix means that there is mutual coupling between sequences but that it is reciprocal, e .g., the coupling from positive to negative sequences is exactly the same as the coupling from negative to positive. This situation can be simulated by a passive network. A nonsymmetric matrix means that the mutual coupling is not the same between two sequences. This situ ation generally requires controlled sources for laboratory simulation, but its mathematical representation is no more difficult than the symmetric case. It does require the computation of all matrix elements, whereas the symmetric case re quires the computation of only the upper (or lower) triangular matrix because of symmetry. In most of our computations we will consider the self impedances to be equal. Thus, except for the case of arbitrary (nonsymmetric) mutuals, the problem is one of a diagonal or symmetric matrix representation. If the self impedances are only symmetric or are arbitrary, we often conclude that the method of symmetrical components is too complicated to be of value. This was particularly true in the days of network analyzer simulation. With digital com puter solutions the representation of nonreciprocal networks should be a much less formidable problem, and we may soon be dealing with arbitrary impedances as a commonplace occurrence. 2.6
Sequence Components of Mach ine Impedances
In the case of synchronous or induction machines we have a very special case of section 2. 5 . Lewis and Pryce [ 1 3] have asserted that in this case the impedance
Sym m etrica l Compone nts
31
matrix corresponding to the Z matrix of (2.42) is the "circulant matrix"
(2.55) This matrix form is the result of the self impedances being equal and the mutual impedances being circular, Le. , Zm is clockwise (a-b-c) and Zn is counterclockwise (a-c-b ) . It is easy to show that a similarity transform Z012 = A - I Z A diagonalizes Z, i.e., Z,,,
�
K' ZA
•
[�' ; �J ,
(2.56)
This means that transforming from a-b-c coordinates to 0.. 1 -2 coordinates com pletely decouples the sequences, a fact that makes symmetrical components ex tremely useful in dealing with machines. Performing the operation indicated by (2.56), we easily show that Z o = Zk + Zm + Zn
Z 1 = Z k + a2 Zm + aZn
Z2
= Z k + a Z m + a2 Z n
(2.57 )
It is also shown in [ 13] that Zo , Z h and Z2 are the eigenvalues of Z. Thus we note that the eigenvalues are distinct and that the columns of A are the eigenvectors of Z. 2.7 Definition of Sequence N etworks
In many problems the unbalanced portion of the physical system can be iso lated for study , with the rest of the system considered as balanced. This is the case for an unbalanced load or fault supplied by a system of balanced or equal phase impedances. In such problems we attempt to find the symmetrical compo nents of voltage and current at the point of unbalance and synthesize these se quence quantities to determine system (a-b-c) quantities. To find the sequence quantities is the major objective in many problems, and it is helpful to in troduce the concept of sequence networks to do this. Before defining the sequence network, we define the fault point of a network. The fault point of a system is that point to which the unbalanced connection is attached in an otherwise balanced system . Thus, a one-line-to-ground fault at bus K defines bus K as the fault point. Similarly , an unbalanced three-phase load at bus M defines bus M as the fault point. We take a liberal definition of what con stitutes a "fault," it being any connection or situation which causes an unbalance among the three phases. A sequence network is a copy of the original balanced system to which the fault point is connected and contains the same per phase impedances as the physi cal balanced system, arranged in the same way, the only difference being that the value of each impedance is a value unique to each sequence . This really is not as complicated as it sounds. Since the positive and negative sequence currents are both balanced three-phase current sets, they see the same impedance in a passive
32
Chapter 2
three-phase network. The zero sequence currents generally see a different imped ance from positive and negative sequence currents and may sometimes see an infi nite impedance. Machine impedances are usually different for all three sequences. The topic of sequence impedances will be greatly expanded upon later. Each seq�ence network, once the fault point is determined, can be analyzed by Thevenin 's theorem by considering it to be a two-terminal (one-port) network where one terminal is the fault point, designated F, and the other terminal is the "zero�potential bus," designated N. This is done in the usual way, looking back from terminals F and N to find the open circuit voltage and the short circuit cur rent or equivalent impedance (for that sequence ). The impedance seen is usually designated Zo, Z i t and Z2 for the zero, positive, and negative sequence networks and represents the Thevenin equivalent impedance to the flow of lao , lal ' and la2 respectively. '1. The Thevenin equivalent voltage in the positive sequence network is the open circuit voltage at the fault point and is a phasor quantity. Thevenin equivalent or open circuit voltages in the negative and zero sequence networks are zero because by definition the only voltages generated in the three-phase system are positive sequence (sequence a-b-c ) voltages. Sequence networks are usually designated schematically by boxes in which the fault point F, the zero-potential bus N, and the Thevenin voltage and impedance are shown. Such a set of boxes is shown in Figure 2. 5 for the zero, positive, and negative sequences.
roo t
IF
Vo O
lk-
102
Fo
t
IF -lk
Zo
F2 Z2
Vo 2
NO
i---=-
N EG ATIVE
PO S I T I V E
Z E RO
N2
Fig. 2. 5 . Sequence networks with defined sequence quantities.
Several important definitions are shown in Figure 2.5. First, the direction of sequence currents is assumed to be away from ( or out of) the F terminal. This is because the unbalanced connection is to be attached at F, external to the boxes, and we assume currents as flowing toward this unbalanced connection. Second, the voltage across the sequence network is defined to be a rise from N to F as this makes Val positive for a normal system. These assumptions are arbitrary but are necessary in order to proceed with problem solving. Note that the Thevenin equivalent voltage is shown as VF which is the prefault (F open) voltage of phase a at F. (This voltage is sometimes designated E al or simply as Eg. ) From Figure 2. 5 we write the important equations for the voltage drop from F to N as
I��:�J ��J [ J =
a
0
r; � �J ��::] 0
0
Z2
a2
(2. 58)
2 Note that these are not the same impedances as Zso. ZS h and ZS 2 or ZMO. ZM I o and ZM 2 of section 2. 5 and are not the machine impedances alone of section 2.6.
33
Sym metrica l Com ponents
or ( 2 . 5 9) Example 2. 1 Suppose that a certain unsymmetrical condition gives the following data at the fault point.
llaaol ZIZo
la2 Z2 a O v [ VOl2 -J [ J [ L!L] [- J Vaa b VOl2, [�Vej � I�l� �, �J [�::�J 1l� J Va Vb Ve Vabb VaVb - VbVe Va VaVbbe Vea VF = 1.0 L!L , =
(0.2& )h,
=
=
(0.2L!L )h
Find the phase and line-to-line voltages at F. Solu tion From equation ( 2. 58) Va l
=
=
Va2
0
h 1 . 0L!L
0. 6 h 0. 2 �
-
0
e =A
Then, from ( 2. 24)
=
=
=
- 0. 2/0
a
h
0.6
L!L
0.8 m:
- 0.2[!L
0.2[!L
h
a
(0.6m: )h
�
.251- 1 36.� . 25/136.2
( 2.60)
We see that = 0, which indicates that phase a is shorted to ground at the fault point F. Note that both and are 125% of normal. This condition results in line-to-ground faults whenever Zo is greater than Z I ' To compute the line-to-line voltages, use the results of ( 2.60) and write =
V e Vea
= Vc
�
=
-
=
0 - 1 . 2 5/- 136.2° = 1 . 25/43.8° pu = (- 0. 9 - jO. 8 6 6) - (- 0.9 + jO. 866) = - j1. 732 = 1 . 73 2/- 90° pu = Vc = 1 . 25/13 6 . 2° pu
where these values are all in pu on a phase voltage base. On a line-to-line voltage base, which may be more meaningful, we divide by V3 to get =
=
=
0. 7 22/43.8° pu 1. 0/- 90° pu 0. 7 22/136. 2° pu
(2.61)
Problems
2. 1. Sketch a set of symmetrical component phasors for a four-phase system, i.e., n = 4. Indi cate clearly the angle between components Vax, Vbx, Vex' and Vdx in each set. Compute the a-operator angle and show that equations (2.4) and (2.5) are satisfied. 2.2. Repeat problem 1 but with n = 5. 2.3. Show that C of equation (2.10) is nonsingular by performing elementary operations [ 7 ] on its rows and columns. 2.4. Evaluate a, C, det C, and C- I = A for n = 2, 3. 2.5. Write out matrix C for n = 2, 3, 4, 5, 6 , . and reduce each element to its simplest form; .
.
Chapter 2
34
for example, when n 5, set a6 a, at! = a3 , a 9 a4 , etc. Can you see a patten. emerging as to the construction of C insofar as a-exponents are concerned? 2.6_ Show that the A matrix defined by equation (2.20) has rank n . 2.7. Apply the analysis equation (2. 21) to the unbalanced phasors of Figure 2. 3 to obtain the symmetrical component phasor quantities. 2.8. Given that Va - 100 + jO, Vb " j80, and Vc = - 60, determine the symmetrical compo nents Val t Va2 , and VaO ' 2.9. Given that V00 .. 100 + jO, Vat 200 - j100, and Va2 - 100 + jO, find the phase voltages Va, Vb , and Vc' Let h - 1 . 2.10. If the load is unbalanced, neutral current will exist. Find the relation between neutral cur rent In and phase a zero sequence current laO�
=
=
�
IO
=
__ ......
J
I b - --I Ic In -�--/ Fig. P2. 1 0. __
2. 11. If the load is unbalanced, its neutral n ' will not be at the same potential as the source neutral n. Find the relation between the neutral shift Vn ' n and the zero, i.e., Va n ' O ' b
c
c
Fig. P2. 1 1 .
2. 1 2. Each current transformer h as 400/5 A ratio. The currents i n the ammeters are as shown on the phasor diagram. Find all five remaining currents. Relate 14 , 1m and 100, In
n ··�-----..!..g. . • a �--------
Fig. P2. 1 2.
Symmetrical Components
35
2.13. This isolated neutral system has a ground fault o n phase Q . When ungrounded, ali line-to neutral (also line-to-ground) voltages are balanced. If helpful, the balanced line capaci tances to ground may be used to aid in visualizing the line-to-neutral voltages of this delta circuit. Find the symmetrical components of the phase Q line-to-neutral voltages when the ground fault is applied, i.e. , VanO , Van l t Van2 •
GENERATOR OR TRANSFORMER WINDINGS
Fig. P2. 1 3 .
2.14. Prove that power i s invariant under the transformation to the 0-1-2 frame o f reference i f the value o f "h" i s taken to be v'3. 2. 15. Consider an impedance matrix which is symmetric with all off-diagonal terms equal. Find a transformation which will diagonalize this impedance matrix and show that such a transformation is real. Hint: Find the eigenvalues and eigenvectors of the Z matrix and examine the unitary matrix of eigenvectors. (See [4, 5, 6 ) .)
chapter
3
An a lys i s of U nsym metri ca l Fa u l ts : Th ree - Co m po n e nt M ethod
Having introduced the symmetrical component notation and defined the se quence networks, we are ready to evaluate the way in which unsymmetrical condi tions may be represented. In doing so, we will proceed in a very orderly manner, evaluating the conditions at the fault point and then deriving the exact circuit representation in the 0-1-2 coordinate system. We call this type of analysis the three-componen t method to distinguish it from other methods to be introduced later. In what follows we will refer to the unbalanced condition at the fault point as a "fault. " It should be understood that this term is intended to mean any un balanced situation and may be an unbalanced load or other unsymmetrical condition. It is also convenient to distinguish between shunt and series unbalances or faults. A shunt fault is an unbalance between phases or between phase and neu tral. A series fault is an unbalance in the line impedances and does not involve the neutral or ground, nor does it involve any interconnection between phases. We will consider these items separately. Our objective here is to determine exactly how the sequence networks are re lated or how they are interconnected for various kinds of fault situations. Since we must do this for several different situations we establish the following pro cedure.
36
1 . Sketch a circuit diagram of the fault point showing all phase connections to the fault. Label all currents, voltages, and impedances, carefully noting assumed positive directions and polarities. Such a sketch is shown in Figure 3 . 1 . It is assumed that a "normal" system consisting of only balanced im pedances is connected to the left and right of the fault point and that the Thevenin equivalent looking in at this point is known. Note that phase voltages are defined as drops from line to ground at this point and that cur rents are defined as flowing from the system toward the fault. 2 . Write the boundary conditions relating known currents and voltages for the type of fault under consideration. 3. Transform the currents and/or voltages of 2 from the a-b-c to the 0-1-2 co ordinate system by use of the transformation A or A - I . 4. Examine the sequence currents to determine the proper connection of the F or N terminals of the sequence networks, satisfying 3.
Ana lysis of U n sy m m etrica l Fa u l ts : Three-Component M ethod
a
I--- THE "F�LT POINTII -1
a
b -t--.----t--
c
37
b
-t--+-.-----t--l--
Fig. 3. 1 .
FAULT
Circuit diagram of the fault point.
5. Examine the sequence voltages to detennine the connection of the remain ing terminals of the sequence networks, adding impedances as required to satisfy 3 and 4. These five steps will be followed rigorously for each type of fault except where noted to the contrary. I. SHUNT FAU LTS
Shunt faults are an important class of faults and include various kinds of "short circuits" as well as unbalanced loads. 3. 1
The Single Line-to-Ground (SLG) Fault
==
1. Circuit diagram : See Figure 3.2. 2. Boundary conditions : By inspection of Figure 3.2,
I/}
and Va
0
( 3. 1 )
Z, Ia
( 3. 2)
e I
=
3. Transformation : From equation (2.25) we write 1012 ( 3. 1 ), with A defined from (2.36)
i" , a
b
=�
= l� : :'] [!] � m
c
,
.f
+v +y +y _ 0 _ b_ c
41 Fig. 3 . 2.
I .
F
a
b
c
Zf
Io! -
I I
.... lifo Ie= o
Diagram of a SLG fault at F.
=
A - I la b e, or from
(3.3)
Chapter 3
38
or all sequence currents are equal. Also, we have from (3.2) and (3.3), Va = Z, 10 =
which w e may write as
h Z, /o l 3
(3.4) 4. Sequence currents: From (3.3) we note that the sequence currents are equal. This implies that the sequence networks must be connected in series, as shown in Figure 3. 3.
Fig. 3 . 3 .
Sequence network partial connection specified by the current equation.
5. Sequence voltages : From (3.4) we see that the sequence voltages add to 3Z, /o l • This requires the addition of an external impedance as noted from Figure
Fig. 3 . 4 .
Sequence network connection for a SLG fault.
3.3. The final connection is shown in Figure 3.4. With this connection we com pute h VF �O = � 1 = � 2 = Zo + Z 1 + Z2 + 3 Zf
(3. 5)
and knowing the sequence currents we easily find the sequence voltages from equation ( 2. 58). Example 3. 1 The simple power system shown in Figure 3 . 5, consists of a generator, trans former, transmission line, load transformer, and load. Consider a SLG fault at bus
Ot€�>-�81 1 0 kV
A
GEN
�
K
6�
20 kV
L_INE
___
2 Ct�
___
2+ j4
ohm
�
SLG
0
6Y
FAULT
Fig. 3.5. Power system for Example 3 . 1 .
LOAD
P - IO MW Q a 5 M VAR
Ana lysis of U nsym metrica l Fa u lts : Three -Component M ethod
39
C with a fault resistance of 4 ohms. The following data concerning the system is known.
Generator: T1 : Line : T2: Load :
25 MVA, 10 kV, x = 0. 125 pu , connected Y-grounded 30 MVA, 10-20 kV, x = 0. 105, connected � -Y-grounded Z = 2 + j4 U 20 MVA, 5-20 kV, x = 0.05 pu, connected Y-� static (constant z ) load of 10 + j 5 MVA at 5 kV
Solution
Select So = 20 MVA, a load voltage of 5 kV, and compute all system impedances. Let h = 1 . x =
Generator: T1 : Line : T2 :
x =
z =
x =
(0.1 25) ( 20/25) = 0. 10 pu (0.105) ( 20/30) = 0.07 pu [ ( 2 + j4 ) (20) ] /(20) 2 = 0. 1 + jO.2 pu 0. 05 pu
Load (as series impedance) :
R
( Vu ) 2 ( SB ) P p2 + Q 2
=
( 1 . 0) 2 ( 20 X 106 ) ( 1 0 X 106 ) 200 = = 1 . 6 pu 125 ( 1 0 X 106 )2 + ( 5 X 106 )2
=
Similarly, X 1 00/1 25 0.8 pu. Then the positive sequence network for a fault at bus C is represented as shown in Figure 3.6. =
=
Fig. 3 .6.
SLG
One-line diagram of the positive sequence network.
The load current IL is (with V as the reference phasor)
IL
=
p
jQ
v* -
=
10 - j 5 20
=
0 . 5 - J'0 . 25 pu
The Thevenin voltage at bus C is
VF
=
=
1 . 0 + jO + (0.5 - j O.25) (j0. 05) 1.0125 + jO.025 = 1.0125/1.27°
pu
We set VF = 1.01 25l!L and it becomes the reference phasor in the fault calcula tions. The impedance seen loo king in at F1 with E shorted is 0. 1 + jO. 37 on the left in parallel with 1.6 + jO. 8 5 on the right, or
40
Chapter 3
Zl
=
(0. 1 + jO.37)(1.6 + jO.85) 1.7 + j 1 .22
0.33 19/67. 1905
=
. 0. 1287 + JO.3059
=
pu
This is the impedance to the flow of both positive and negative sequence currents. The zero sequence current, for reasons to be investigated later, sees an open circuit to the left of bus A , with A grounded (because of the Y-grounded connec tion) and an open circuit to the right of bus C. Thus Zo is the sum of the line and T1 impedances, or Zo = 0. 1 + jO. 27 pu.
HO
"
,(1.012S&- ) h
FO � I , 0.1y + J 0. 27
_
+
FI
, O.l2 8 +jO.307
HI
�y
•
- VaO -
-
•
- Va l --
-J
tal
,A
I QI28+i
H2
-
a, f2
ta2
•
- V o2 -
--
r.OO- 1 0 , - 1 02
3Zf - 0. 6+ J O
Fig. 3 . 7 . Sequence networks for SLG fault a t bus C.
The complete connection of sequence networks is shown in Figure 3.7, where Z, is shown to be Z,
+ jO = --z;= 20 = 0 . 2
4
4
The total circuit impedance is Zt
=
Zo + Z l + Z2
+
3Z,
From (3. 5) we compute Ia l Then Ia
=
=
3Ial
0.3 =
!:;/����
05
=
= 0. 9 56 + jO.884 = 1.3/42.8°
0.7779/-42.6496
2.3338/- 42.6496
pu
=
0.5722 - jO.5271
pu
pu
pu
Since the fault occurs on a 20 kV bus, IB and Ia
=
=
SB/ VB
J3
20
=
X
3(20
106 x
103)
=
577.3503
A
1347.4 A. We may also synthesize the phase voltages by first computing
the sequence voltages.
VaO
Va l Va2
Thus
= = = = = =
-Zola o
=
-(0.2879/69.677)(0.7779/- 42.6496)
0.2240/- 152.9727
=
-0.1995 - jO. 1018
VF - ZlIa l = (1.0125 + j O) - (0.3319/67. 1905)(0. 7779/- 42.6496) (1.0125 + jO) - (0.236 + jO. 1075) = 0.7765 - jO. 1075
-Z2Ia2
=
-(0.33 19/67. 1905)(0.7779/- 42.6496)
-0.2348 - jO. 1072 Va
=
VaO
+
Va l
+
=
Va2
0.2582/- 155.4591 =
0 . 3433 - jO.3 162
=
0.4668/- 42.6496
Analysis of Uns y m metrical Fa u lts : Three - Component M ethod
41
(0 ) b
o
(b) Fig.
3.8.
Currents and voltages for the voltages at the fault.
c
SLG fault : ( a ) sequence quantities (b) postfault LL
which should check with
Va
=
Zf1a l
=
(0. 6)(0. 7779/- 42.6496)
=
0.4667 /- 42 .6469
This is a good check considering the algebra involved. From ( 2. 23 ) Vb
=
=
=
Vc
=
=
=
VaO + a2 Vai + aVa2
-0. 1995 - jO. 1018 - 0.4817 - jO.6199 + 0.2102 - jO. 1497
-0.4709 - jO.8714
VaO + a Val + a 2 Va2
=
0. 9905/- 1 18.3879
-0. 1995 - jO. 1018 - 0.2960 + jO.7271 + 0.0246 + jO.2569
-0.4709 + jO.8823
=
1.0001/+ 1 18.0913
These voltages are shown in Figure 3.8 together with the line-to-line voltages at bus C, where
Chapter 3
42
The Line-to-Line ( L L) Fault 1. Circuit diagram: See Figure 3 . 9 and note that the LL fault is placed be tween lines b and c to retain symmetry with respect to phase a. 3. 2
b c
o
F
o
b
t
�Vo :Vb:Vc
Fig.
10=01
Ib
l
IJ
c
b Zf
3 . 9 . Diagram o f a LL fault a t F.
2. Boundary conditions: By inspection of Figure 3 .9 ,
10 = 0
Ib = - Ie
Vb - Ve = Ib Z,
(3.6)
(3.7 )
( 3.8) I 3 . Transforma tion: From the analysis equation lOll = A - lobe and incorporat ing ( 3 . 6 ) and ( 3 . 7 ) , =
4n � Also from (3 . 8 ) we write
[�
1
a
�]J a
[�b ] = � [ �J -1 - Ib
(3.9)
1 1 Ve _- h ( Vao + a1 Va l + aVa 2 ) - h ( Vao + aVa l + a2 V01 ) (a - al ) ( al - a) 1 1 Va l Va l + h Z, (lao + a la i + ala l ) h h
Z, lb _ - Vb
-
_
From ( 3 . 9 ) lao = 0 and 10 1 = 101 • Substituting this information into the above, we have for any h , -
or ( 3. 1 0)
4 . Seq uence curren ts: From (3.9), lao = 0 , so the zero sequence network is open. Also from (3.9) la J = 10 2 , which requires the connection shown in Figure -
3 . 1 0.
5. Seq uence voltages: From equation ( 3 . 1 0 ) and Figure 3 . 1 0 we see that the remaining connection must be as shown in Figure 3. 1 1 . Then Ia I
h VF
= ---"---
Z J + Z l + Z,
(3.11 )
�
43
Ana lys i s of U n sy m metrica l Fa u lts : Th re e - Co m ponent M ethod
IO O= O , FO NO Fig. 3 . 1 0.
10 l f FI Vol N I
+
'-----'
�
Sequence network partial connection specified by the current equation.
IOO & O
Fig. 3 . 1 1 .
Zf +
V02
FO NO
Sequence network connection for a LL fault.
Example 3. 2
Compute the phase voltages and currents for a LL fault at bus C of Figure 3.5, where a fault impedance of 4 ohms is assumed between phases b and c. Let h = 1. Solution
The sequence networks are exactly as shown in Figure 3.7, but their inter connection is that of Figure 3 . 1 1 . With the new connection the total impedance is Z t = ZI + Z2 + Z, = 0.456 + jO.614 = 0.765/53 .4° pu Then 1 .0125i.!L Ial = - Ia2 = hVF /Z t = 0.765/53.4 0
= 1 . 3251- 53.4° = 0.788 - j 1 .065
From equation (3.9) with h Ib
=
pu
1
= - Ie = - j y'3 Ial = - 1 .86 - j 1 .38 = - 2. 32/36.6°
pu
= 1 320
A
This system voltages may also be synthesized from a knowledge of the sequence currents and sequence network connections. Thus with h = 1 ,
Val = h VF - ZI Ial = 1 .0 1 2 5 - (0.332/67 .3° ) (1 .325/- 53.4° ) = 1 .0125 - 0.427 - jO.l 055 = 0.5855 - jO.1055 pu Va2 = - Z2 Ia 2 = (0.332/67 .3° ) (- 1 .325/- 53 .4° ) = 0 .44[1 3 .9° = 0.427 + jO.1055 pu -
and we compute
Va = Val + Va2 = 1 .01 25 + jO Vb = a2 Va l + a Va2 = - 0.389 - j0.448 - 0.310 + jO.31 7 = - 0.699 - jO.131 = - 0 .705/10 .6° pu Ve = a Va l + a2 Va2 = 0 . 1 96 + jO .554 - 0 . 1 1 8 - j0 .42 3 = 0.314 + jO.131 = 0.342/- 22.6° pu -
-
-
Chapter 3
44
The line-to-line voltages on a phase voltage base are Db
V Va - Vb Vbc Vb V Ve - Vo Vbc Vb - Ve IbZ, =
=
Ql
We make a check on
Vbc =
-
=
= 1 .7 1 2
+ j O . 1 31 pu
Vc = - 0 . 385 - jO .262 pu = - 1 . 326 + j O . 1 3 1 pu
from equation ( 3 .8) =
= (- 1 .86 - j 1 .38 ) ( 0 . 2 ) =
-
0.372 - jO.276 pu
which is a good check, considering the algebra involved . Ie.!..
_
_
/
I
Ie
I
Ie2
Va
REF
(0)
(b)
Fig. 3. 1 2 . Phasor diagrams for a LL fault : (a) currents ( b ) voltages.
Phasor diagrams of currents and voltages are shown in Figure 3 . 1 2. Note that if Z, were zero the voltage "triangle " would collapse to zero on the b-c side and the Vo l - V0 2 phasor would go to zero . 3.3
T he Double L ine-to-G round ( 2LG ) Fault
1 . Circuit diagram : The fault connection is shown in Figure 3 . 1 3 . Note that symmetry with respect to phase a is obtained by faulting phases b and c. 2 . Boundary conditions: By inspection of Figure 3 . 1 3 we have
10 = 0
Vb (Z, Zg ) Ib Zg le Ve (Z, Zg) Ie Zglb =
+
+
=
+
+
3 . Transforma tion : From ( 3 . 1 2 ) we write
(3.12)
(3.13)
(3.14) ( 3. 1 5 )
Ana lysis of U n sy m m etrica l F a u lts : Th ree-Com ponent M ethod
45
a -r------�- a
F
b -f--.----+--...--
b c
Fig. 3 . 1 3 .
Diagram of a 2LG fault at
F.
From (2 . 24 ) we write Vb
Ve
=h 1
=
and find the difference Vb
-
2 ( VaO + a Va l + a Va2 )
� ( VaO
Ve
2
+ aVa l + a Va2 )
va -- h
j
_
( Va l - Va2 )
But from ( 3 . 1 3 ) and ( 3 . 1 4 ) we also find that Vb - Ve
= Z, (lb
- Ie )
(3 .16 ) (3.17)
( 3 . 1 8)
(3 .1 9)
Substituting ( 3 . 1 8 ) and a similar relation for Ib - Ie into ( 3 . 19) , we have for any h, Va l - Val = Z,(la l - lal ) , or Va l - Z,la l
= Va2 -
Z,lal
(3.2 0 )
Also, adding equations ( 3 . 1 6 ) and (3 . 1 7 ) , we find the sum Vb + Ve =
h 1
[ 2 VaO - ( Va l + Va 2 ) ]
(3 . 2 1 )
which we equate to the sum of ( 3 . 1 3) and ( 3 . 1 4) Vb + Ve =
Z � [ 2 1aO -
( la l + lal ) ] +
Z � [ 4 lao - 2 (la l + la2 ) ]
(3 .22)
Since ( 3 . 2 1 ) and ( 3 . 22) are equal , we collect terms to write for any h,
2 VaO - 2Z,lao - 4Z, lao = Va l + Va2 - Z, (la l + la2 ) - 2Z, (Ia l + la2 ) This may be simplified by noting that la l + la2 = - lao and using the result ( 3 .20 ). Rearranging then, we have
( 3 .2 3 ) VaO - Z,lao - 3Z, lao = Va l - Z,la l 4 . Seq uence curren ts: From ( 3 . 1 5 ) we see immediately that the N terminals of the sequence networks must be connected to a common node as shown in Fig ure 3 . 1 4 .
46
C h apter 3
Fig. 3 . 1 4.
Sequence network partial connection specified by the current equation.
5. Sequence voltages: From ( 3.20) we observe that the voltages across the positive and negative sequence networks are equal if an external impedance of Z, is added in series with each network. Similarly, (3.23) requires that an external impedance of Z, + 3 ZIl be added to the zero sequence network. The final con nection is shown in Figure 3 . 1 5 . Then ( 3.24)
Fig. 3. 1 5. Sequence network connection for a 2LG fault.
Example 3. 3
Refer again to the system of Example 3 . 1 , this time with a 2LG fault and with fault impedances of Z, = 4 ohms and ZII = 8 ohms. The fault is at bus C, so the se quence networks are internally the same as those of Figure 3.7 . Let h = 1 .
Solution
The pu impedances are ZI
Zo = 0 .1 + j O .27 ,
=
Z2 = 0 . 1 28 + jO.307
Zg = 0 .4 + j O
Z, = 0 .2 + jO,
As before, h F = 1 .0125 + j O pu. The network to be solved appears schematically as shown in Figure 3 . 1 6 . The total impedance seen by current fa l is
V
Zt
Then fa l is
.
(0.328
= 0.328 + J O . 307 + -
= 0.617
Ia l =
+
jO.515
1 .012 5 ffi:. 0 . 805139 .80
=
=
j O . 307)( 1 . 5 + jO.27) 1 .828 + jO.577 +
0 .805/39.8 °
1 . 2 59 /- 39 8 ° _
.
pu
=
0 .965 - J· 0 805 pu •
47
Ana lysis of U nsymm etri ca l Fa u lts : Three-Component M ethod
loof
1 .4
0. 2
0.2
FO
FI
F2
101
0.1 + j
0.27
t
h VF� ()
0. 1 2 8 + J 0 . 307
102t
0. 1 2 8 + j 0. 307
No- NI= N 2 Fig. 3 . 1 6 .
A 2LG fault network.
By inspection of Figure 3 . 1 6 we compute la o = =
-
-
Z 2 + Z, I0 1 Z 2 + Zo + 2Z, + 3ZIl 0 .449 � ( 1 . 2 59 /- 3 9 . 8 ° ) = - 0 .2 9 4/- 14. 2 ° = 1 . 9 2 / 1 7 . 5°
-
(0.285 - J. O . 072) pu
Similarly, 1 .525 � Zo + Z , + 3ZIl ( 1 .259/- 3 9 . 80 ) 10 = 1 .92/1 7 . 5 ° Z 2 + Zo + 2Z , + 3ZIl 1 = - 1 .00/- 47 .1 ° = - (0.68 - j O . 7 3 2 ) pu
10 2 =
-
Checking these results, we compute 10 = laO + 10 1 + 10 2 = 0 - j O.001 sizing the other phase currents, we have
�
O. Synthe
Ib = la O + a2 1a l + ala 2 = - 0 . 285 + jO.07 2 - 1 . 1 8 1 - j0 .432 - 0 . 293 - j O .955 - 1 . 759 - j 1 . 3 1 5 = 2 . 1 9 5/36.8 ° pu =
-
and Ie = laO + ala i + a2 Ia 2 = - 0 . 285 + jO.072 + 0.21 5 + j 1 .24 + 0.977 + j O . 2 1 0 = 0 . 9 0 7 + j 1 .522 = 1 . 7 74/59.2° pu
Thus that
Ib
+ Ie = - 0 .852 + j O .207 pu. But we may also show from (2.26) and ( 3 . 1 5 )
Ib + Ie = 3 /00 = 0.855 + j O . 2 1 6 -
which is a good check.
( 3 .25 )
Chapter 3
48
The sequence voltages are
VaO = - Zo lao = (0.1 + jO.27 ) ( - 0.285 + jO.072 ) -
= 0 .048 + jO.07
pu
Va l = h VF - Z I Ia l = 1.01 25 - ( 0. 1 28 + jO.307) (0.96 5 - jO.805) = 1 .0 1 2 5 - (0 .370 + jO.193) = 0 .6425 - jO.193 pu
Va2 = Z2 102 = (0.128 + j O.307 ) ( - 0.68 - jO.732) -
= 0.311
-
+ jO.1 1 5 pu
and the drop across parallel networks is
Va l - Z, la l = 0 .6425 - jO.193 - (0.2) (0 .965 - j O .805) = 0 .4495 - jO .032 pu
The phase voltages are synthesized from equation (2 .24).
Va = Vao + Va l + Va2 = 1 .001 - j O.008 pu
which is close to the desired value of 1 . 01 25 + jO. 2 Vb = VaO + a Va l + a Va 2
= 0 .0479 + j O .07 - 0 .488 - j0 .4595 - 0 . 2 552
= - 0 .69 53 - jO.1775 = - 0 .7 1 7 �
+ jO.212
pu
Ic2 I��: :::-'--'---
Va
(0)
Fig.
3.17.
( b)
Current and voltage phasors for a
2LG faul t : ( a ) currents (b) voltages.
Ana lysis of U nsym metrical Fau lts : Three -Component M etho d
49
2 Ve = Va o + aVa l + a Va2 = 0.04 9 7 + j O .0 7 - 0 . 1 54 + jO.6525 - 0.0558 - jO.3270
=
-
0.1619 + jO.3955 =
-
0.427 /- 67.7°
pu
As a check we compute from ( 3 . 1 3 )
Z,Ib + Z, (Ib + Ie ) ( 0 . 2 ) (- 1 .7 59 - j 1 .31 5) + (0.4) (- 0 .852 + jO.207)
Vb = (Z, + Z, ) Ib + Z,Ie =
= (- 0 .3 5 2
-
=
j O . 263) + (- 0 .341 + jO.083 ) = - 0 .693 - jO . 1 80 pu
a close check. The current and voltage phasors are plotted in Figure 3 . 1 7 . 3.4 The Three-Phase (31)) Fault
The three-phase fault, although not an unbalanced one, is analyzed here to complete the "family " of shunt unbalances of greatest common interest. This fault is important for several reasons. First, it is often the most severe type l and hence must be checked to verify that circuit breakers have adequate interrupting rating. Second, it is the simplest fault to determine analytically and is therefore the only one calculated in some cases when complete system information is lack ing. Finally, it is often assumed that other types of faults, if not cleared promptly, will develop into 31> faults. It is therefore essential that this type of fault be com puted in addition to the other types. a
F _.----....... -
a
b -+-r------+-_.- b
c -+-.--.----+---t- c
Fig. 3. 1 8.
Diagram of a 3 1> fault at F.
1 . Circuit diagra m : The 31> fault connection is shown in Figure 3.18. The fault is shown as a 31> -to-ground fault with individual phase impedances Z, and neutral-to-ground impedance Z, . 2 . Boundary conditions:
Va
=
Zf Ia + Z, (Ia
Ib + Ie )
(3.26)
Vb
=
Zf Ib + Z, (Ia + I" + Ie )
(3 .27 )
+
1 Exceptions : The SLG fault is more severe than the 3ct> fault in cases where ( 1 ) the genera tors have solidly grounded neutrals or low-impedance neutral impedances and ( 2 ) on the y-grounded side of A -Y-grounded transformer banks.
50
Cha pter 3
(3.28)
Ve Z,lc Zg(/a Ib Ic) (3.26)-(3. 2 8) Ib 10 Ie (3/h)/ao , =
3.
+
+
Transformation: First, w e write components of phase a, recognizing that Thus
+
+
in terms o f the symmetrical = by definition .
+
(3.29) (3. 3 0) Ve _ h1 ( VaO Val Va2 ) _ h1 Z,(Iao la 102 ) h3 Zg I00 (3. 31) ( 3. 3 1) (3.30), Vb c 2 - a2 )/a2l (3.32) Va2 Z, ( Vbc Vb - Ve Val - jy'3, (3. 3 2) Vbc W3 Val W3 Va2 Z,(- 101 102 ) Val - Va2 Z'(/al - 102 )' (3.33) Val - Z,lal = Va2 - Z,la2 1 (3.29) (3. 3 0) Va Vb � (2VaO - aVal - a2 Va2 ) � Z,(21ao - alai - a2/a2 ) � Zglao (3.34) (3.3(3.4) 3 3) (3.34) ( Va l - Z,lad 2 ( VaO - Z,lao - 3Zglao ) -1, (3. 3 5) 2 (VaO - Z,Iao - 3Zglao) ( Val - Z,Ia l ) (3.30) and (3. 3 1) Vb Vc Vb Vc 2VaO - Va l - Va2 Z, (21ao - 102 ) 6 Zglao 2 (VaO - Z,Iao - 3Zglao) (Val - Z,la l ) ( Va2 - Z,la2 ) (3 . 33) (3. 36) VaO - Z,lao - 3Zglao Vat - Z,lat 4. (3 . 33) and (3. 36) +a
-
Subtracting
from
= (a2 - a)
=
Since (a2 - a) =
+a
-
we find + (a - a2 )
i
+
+ a2
for any h to be
[ a - a )/a l + (a
=
may be simplified as follows :
=
-
=
or
+ a2
=
+
jy'3
+
jy'3
This may be written more conveniently as
Now add equations 1 + a = a2 • Then
and
and recognize that
+ a2 = a and -
-
=
+
=
+
Rearranging and canceling the h factor, we have From of
we see that the two quantities in parentheses on the right-hand side are equal. Thus simplifies to = ( a + a2 )
Since a + a2 =
this may be written as
=
Now add equations a + a2 = - 1 . Then
+
-
+
to find
=
=
lo t
-
and recall that
+
which we rearrange to write
=
and utilizing
+
again,
=
Sequence curren ts : There are no sequence current equations for this fault. we deduce that each
5. Seq uence voltages : From equations
51
Ana lysis of U nsym m etrical Fau lts : Three -Component M et hod
Z, 3Z" Z" Z, 3.15. (3.35) (3.36), (3.37) (3.38) Val - Z,lal Vao - Z,lao 3Z, laO (3.33) VaO (3.39) Va l - Z,lal Va2 Z, l Va3.19. 2
sequence network with series impedances of + positive, and negative sequences are in parallel, exactly ever, upon further examination of equations and =
as
and for the zero, in Figure How
-
we see a contradiction. Obviously, these equations can be satisfied simultaneously only if = 0, in which case = 0 also. But this requires that also be zero, i.e. ,
laO
=
-
a2 = 0
=0
If we short out the negative sequence network, la2 = 0 and our sequence network connections are those shown in Figure
as
well. Thus
Zft3Z, 100 +
VoO
Fig. 3 . 1 9.
Sequence network connections for a 3� fault.
Note that the result in Figure 3 . 1 9 could have been obtained by inspection of Figure 3 . 1 8. Since the "load" is in each phase and the applied voltages are balanced, the currents are obviously balanced. Then
Z,
( 3/h )lao = la + lb + Ie = 0
[�
Also, since the currents are balanced, I, u =
�
3 )
( . 40
1
a ( 3. 4 1 )
= h la , lao = la2 = O. is really the load impedance of a balanced three-phase load_ The impedance If this impedance is small, however, we would consider this situation to be a short circuit. In either event the computation is the same.
or
lal
Z,
Example 3. 4 Consider a 3� fault with fault impedance of 4 ohms Figure 3 . 5 . Let h = 1 . Solu tion The pu fault impedance is
= 4/
Z, Z Z l Z,
la l = la =
B
h VF +
(Z,
= 4 ohm) at bus C of
= 4/20 = 0. 2 pu. Thus =
1 . 0 1 2 5 + jO 0. 328 + jO. 307
Cha pter 3
52
or
lal = 01.0125lOC . 449/43.10 2. 255/- 43. 10 1.647 - J.1 . 54 pu Va Val = Z,lal (0.2) (1.647 - j 1.54) 0.329 - jO. 308 = 0.451/-43. 10 1200 and +1200 . 3.1-3.4 0.4 0.22LG Z, Z, =
=
The voltage at the fault is =
=
=
pu
Currents and voltages in phases b and c are found by applying phase rotations of respectively to the above results
-
It is interesting to compare the results of Examples to get some idea as to the relative severity of the various faults. This comparison is somewhat arbitrary because of the choice of the fault impedances. For example, in the fault the impedance was chosen to be pu, whereas was chosen as pu in all cases. With this understanding that the results are subject to choice of im pedance, we have the comparisons shown in Table 3. 1. It is clear from this table
Table 3. 1. Comparison of Fault Currents and VoJtages Current in Faulted Phase
Type of Fault
SLG * LL* 2LGt 34> * *z,
(pu )
(pu )
2 � 34 2 .3 2 2 1 95 2 2 55
0.468 0. 342 0.303 0.451
.
.
=
Lo west Voltage at Fault
0. 2 pu, tZ,
=
0. 2 pu and Z,
=
0.4 pu.
that the SLG fault is the most severe from the standpoint of current magnitudes. This is not to imply that this would always be the case by any means. The circuit parameters of the examples used here are not chosen to be typical of any physical system but merely to illustrate the procedure. In a physical system each fault at each fault location must be computed on the basis of actual circuit conditions. When this is done, it is often the case that the SLG fault is the most severe, with the 3
Zg
3.5 Other Types of Shunt Faults
The four types of faults discussed above are the types of most gen eral interest. The SLG fault is usually assumed to be by far the most prevalent, making up per haps 70% of all transmission line faults. Occasionally, a fault configuration other than these four is of interest, e.g. , in special cases where it is considered necessary to analyze a given unusual occurrence. In such cases a straightforward application of steps 1-4 will permit the evaluation of any situation. Reference [ 14 ] gives the sequence network connections for a number of these special cases, and some of them are recommended for study in the problems at the end of the chapter.
2
2 The " Westinghouse Transmission and Distribution Reference Book" [ 1 4, p. 3 58 ) gives a typical frequency of occurrence for 3tj), 2LG , LL, and SLG faults as 5%, 1 0%, 1 5%, and 7 0% respectively.
Ana lysis of U nsym metrica l F a u lts : Three -Component M ethod 3.6
53
Comments on Shunt F ault Calculation
In all the above the computation of fault currents and voltages are based upon finding the open circuit (Thevenin) voltage h VF at the fault point F. This voltage and the Thevenin impedance are then connected to external faults, i.e., to fault connections external to the positive sequence network. This equivalent circuit gives correct results for the current flowing external to F, namely 10 1 , If the distribution of 100 , 10 1 , and la'! inside the sequence networks is desired, this must be found by Kirchhoff's laws applied to the actual branches within the three sequence networks. Then the currents in any branch may be synthesized by lob e = A 101 2 to find the branch phase currents at that location which contribute to the fault. Note that this branch current does not include the balanced load current which was flowing prior to the fault. However, since the network is linear, the fault and load currents may be added by linear superposition to obtain the total current flowing after the fault is applied. Example 3. 5
Compute the total current flowing from bus B to bus C in the system of Figure 3 . 5 for the SLG fault condition of Example 3. 1 .
Solu tion
In Example 3 . 1 we computed the load current to be IL = 0.5 - jO. 2 5 pu. This is based on a reference voltage at bus D of 1 . 0 + jO pu. The SLG fault current is 10 = 1 . 71 3 - j 1 . 96 5 pu, based on a reference voltage h VF = 1 .0125 + jO at bus C. Actually there is only 1 .27° difference in the reference voltages and this will be ignored. If it were significant, the fault current could easily be rotated to agree with the load reference. The load and fault currents are superimposed as shown in Table 3. 2 . Table 3.2. Phase a b c
Total Current Leaving Bus B for a SLG Fault at Bus C
Load Current
0.5
- jO.25
- 0.467 - jO ..308 - 0.033 + jO.558
Fault Current
Total Current
1 . 4 34 - j 1 .554 0 . 1.42 - jO.008 0.142 - jO.008
1.934 - j 1 . 804 - 0.324 - jO.316 0.109 + jO.550
I I . SE R I ES FAU LTS
Next we consider a group of unbalanced conditions which do not involve any connection between lines or between line and neutral at F. These are referred to as series faults because there is generally an unbalanced series impedance condition. Since the unbalance for this type of problem is in series with the line, there is no "fault point" in the sense described in Part I. Rather, there are two "fault points, " one on either side of the unbalance. Thus the sequence network is still that of a completely symmetric system, and the unbalanced portion is isolated ou tside the sequence network. Our notation for this situation will be to consider the system connections to the point of unbalance as shown in Figure 3. 20, where the two sides of the fault point are F and F' . Current direction is assumed to be from F to F' , and a voltage drop is shown in the assumed direction of current.
Cha pter 3
54
F
a
+ Vaa' _
10
----
-
c
-
Vbb' _ Zb + Vee ' _ l Ze
Ie
.f
Fig. 3. 20.
za
+
Ib
b
F'
a' b'
.f
e'
:::;:
'Voltages and currents at the fault point F-F' .
The sequence networks contain the symmetrical portions of the system, look ing back to the left of F and to the right of F' . These symmetrical portions may or may not be interconnected. We shall represent the sequence networks sche matically as shown in Figure 3.21. Note that the voltage polarities and current directions are consistent with those defined in Figure 3. 20. As shown in Chapter 2 the general case of series unbalance (i.e. , with Za ::1= Z b ::1= Zc ) generates a condition where coupling exists between sequence networks. Such a condition would require a special connection from, say, la t to the network of Ja l , with possibly an isolation and phase shifting transformer. We will generally avoid such situations since they introduce as many problems as they solve. The problems under study here are the simple cases of series unbalance, but they in clude some very important special cases. One class of special problems which are of particular interest is that of open lines. This is easily solved by our usual technique, as shown below. + vaa'O -
FO
+
�O
l al
t�4
FI
+
Val
poO
NO
pa l
+ Vaa' I
FI'
NI
+
Vo ' l
P0 2 Fig. 3 . 2 1 .
Sequence networks for series faults between
F
and
F '.
Ana lysis of U nsym m etrica l F a u lts : Three -Compo n e nt M ethod
55
3.7 Sequence Network Equivalents for Series F aults
Before proceeding with the analysis of series faults, we examine briefly the nature of the sequence networks of Figure 3 . 2 1 . For a shunt fault the situation is simpler since only one point in the balanced network, namely F, is of interest, and a straightforward application of Thevenin 's theorem derives a simple equivalent circuit. With series faults the situation is changed. Now there are two points in the balanced (unfaulted) system which are of direct interest, F and F'. Hence we need a two-port Thevenin's equivalent. Since Thevenin 's theorem is not com monly used for more than one port, we digress briefly to establish a method of obtaining such an equivalent. 3 The zero and negative sequence networks-being entirely linear, bilateral, and passive-are no problem insofar as their two-port description is concerned. Many authors, [ 5] for example, examine two-port parameters of such networks in great detail, and six possible sets of parameters are derived-Y, Z, H, G, ABeD, and (l
V=ZI
and
( 3.42)
I=YV
( 3 . 43)
We establish the nature of these parameters by referring to the general two port network shown in Figure 3. 22, where we define the currents to be entering
Fig. 3. 22.
General two-port network.
the network as shown and the voltages to be drops in the direction of the current through the network. Then we write,
[�J [�:: �::J [�:] ] rv ] [II] = [ =
or
12
Yl l Y2 1
Y1 2 Y22
(3.44)
1
�V2
(3.45)
The parameters of (3. 44) are called the "open circuit impedance p aram eters" since they are found by leaving the ports "open. " Thus
( 3 . 46) 3 This subject i s explored in greater depth in [ 15 ] where Ward applies Thevenin 's theorem to n ports and derives quite general results. Our analysis is restricted to two ports.
56
Chapter 3
and these equations show exactly how the parameters may be found. For exam ple, to find . Z apply a current, say ampere, at port and leave port 2 open. Then measure or compute to find the ratio Note that a current source has infinite impedance, so that both ports are open ; hence the name "open circuit Z parameters. " Similarly, the parameters of (3.45) are found by shorting terminals. Thus we have
11 ,
1
VI
Y
VI /II' 1
Y11 VIII ] V2=O Y21 12 ] V2=O Y22 V12]2 VI=O = -
= -
= -
VI
(3.47)
and we note that in this case we apply a voltage source of zero impedance to one port while shorting the other port. Note that the networks with which we deal are linear, bilateral, passive net works. In such cases we are assured that Z = We also note and = that the lower (negative) terminals are common to ports and 2 in power net works since this connection represents the system neutral. A port description often ignores impedances in this "neutral" in any event [ 5] . Also, it is easy to show that Y = Hence the name "short circuit parameters. " The parameters are really more useful for finding an equivalent circuit since the elements of the Y matrix give the admittance connected between nodes of the network. This is because each is found with the opposite port grounded so that any impedance at that port is shorted out. Thus we may redraw Figure 3.22 in a general way, using admittances computed in (3.47) to find the equivalent circuit is the sum of admittances connected to the of Figure 3. 23. We see that
I2 Z21 Y112 Y2 1 • Y
I. ZY
Y YI 1
+1
r - - - - - - - -, I
1 1 2 1 -
� : '-2 2 : ;
II
Ya
- I
I
0
.
Y,
2
I
I L _ _
I 1
_
_
_
Ya Yb
•
=
- Y1 2 YI I +YI 2
Yc · Y 2 2 + Y12
I
Fig. 3. 23. Two-port equivalent circuit.
Y22
Y1 2
terminal, is is the sum of admittances connected to the +2 terminal, and the negative of the admittance between and +2. Returning to the sequence networks of Figure 3 . 2 1 , we see that for the zero sequence network,
+1
( 3. 48) and similarly for the negative sequence network. Thus, for example, we write
rL
lao] = [1(1 1-0 Y1 2-01 [Vao ] laO 1(21-0 1(22-0J Va,o
( 3 . 49)
where we continue to associate the left (number 1 ) terminal with F and the right (number 2) terminal with F' .
Analysis of U n sym metri c a l Fa u lts : Th ree - Component M ethod
57
The positive sequence network description is the same as that given above ex cept that the positive sequence network has internal sources. If we consider these sources as independent,4 the network equations become [ 5]
V = Z I + hV.
(3.50)
1=YV
(3.51)
and
+ I. h
where V8 and Is are due exclusively t o the internal sources, and Z an d Y are the same as before. More specifically V. in equation (3.50) consists of the open cir cuit voltages measured at ports 1 and 2 (F and F' ), such voltages being due to in ternal (generator) sources. Likewise Is in (3.51) consists of the currents which would flow in to ports 1 and 2 (F and F' ) if these ports were shorted. Since
Y = Z-l
(3.52)
Is = Y VB
(3.53)
we easily show that -
It is reasonable that these two source descriptions should be related. Note that the scalar multiplier h is required on the source terms due to the way the sym metrical component transformation A was defined in ( 2 . 36) , T o construct an equivalent circuit for the positive sequence network, it is easiest to work with the admittance description. Thus from equation (3.51) we write
(3. 54)
Fig.
3. 24.
Equivalent circuit for positive sequence network.
The equivalent circuit corresponding to equation ( 3. 54) is shown in Figure 3. 24, where we use the simplification
Ya = Y12- 1 , -
Yp
= Yl l- 1
+ Yll- 1 •
Y-y = Y22 - 1
+ Yl l- 1
( 3. 55 )
where Ya • Yp • and Y-y are elements of the equivalent circuit but the quantities on the right of ( 3. 5 5) are matrix elements.
Y
Example
3. 6
The system shown in Figure 3 . 2 5 is to be studied for a series fault between buses F and F' . Find the equivalent circuits for the positive and negative sequence networks.
4 It could be argued that the terminal voltage of a synchronous generator depends both upon its current and perhaps upon some remote voltage. For simplicity we assume constant voltage sources. Otherwise the Y and Z matrices become complicated and are no longer recipro cal [ 5 ] .
C ha pter 3
58
®
+
1 .0
F
G
I
F'
®
t¥.: G :l F
3
F'
2
!5
Fig. 3 . 2 5.
ohrns8voIts
+
1.5
System diagram for Example 3.6.
Solu tion Following the notation of Figure Figure
3.26.
Fig. 3 . 26.
3.22,
we sketch the system
as
shown in
Currents and voltages of the system under study.
Y
To find the elements of the matrix, we first remove the internal sources. Since these are voltage sources, they are shorted. Then, with apply and compute
VI = 1.0
V2 = 0,
Yll = /1 = 3 1. 05/6 = 236 Y21 = 12 = - (1 /6) 11 = - 1 /23 VI = 0 V2 = 1.0, 23/4 = 463 1 Y22 = 12 = 22 (23/4) +
and
Now, with
and
mho
compute
+
and
mho
mho
YI2 = II = - "41 (2 223/4) (3461) = - 231 [ Y - 6 /23 - 1 / 23] - 1 /23 31 /46 +
Thus
mho (check)
mho
Now, with both F and F' shorted, we compute the source currents
as
follows
Ana lysis of U n symmetrical Fa u lts : Three -Component M ethod
for h =
1 (using superposition) :
)-
(
+ IS I ( 1. 5 ) 1 0 = 5 5 1.0 =8 1 + 15/8 4 ( ) 23 A = Is 2 ( 1 . 0 ) + IS 2 (1 . 5 ) 1.0 3 1.5 3 3 81 = = = - 8 1 + 15/8 2 - 23 - 4 - 92
lS I = Is l ( l . o)
IS 2
(
59
-
)-
A
Thus the positive sequence equivalent circuit is shown in Figure 3.27. The nega tive sequence network equivalent is the same as Figure 3.27, assuming that the negative sequence impedances are the same as positive sequence impedances, ex cept the current sources are missing.
+
v
/' 01 l h230 jP
1/23 tr
5
�tr Htr NI
----....
- 0'1 �P h �o +
l
V.
Fig. 3 . 2 7 . Positive sequence network equivalent.
One additional important result may be established for the two-port Thevenin equivalent. Our two-port network is a special one since if any connection is to be made at all between F and F ' , it is made in such a way that la = la' . Then 10 1 2 = 10 1 2 , or 1 01 2 at F = 10 1 2 at F' , and the currents entering the ports of the sequence networks have a special constraint. Thus, for example, la l leaves the network at F and enters at F' . Note that this constraint is not due to any internal network con dition but is due entirely to the anticipated external connection. In other words, we assume that the phases do not become crossed in establishing the series un balance . (See [ 1 6 ] for a case where the phases do become crossed. ) This equality of sequence currents is recognized in equation ( 3 . 54 ) and may also be noted in connection with the Z description ( 3 . 56 )
[ ] [ ] [
but this equation simplifies immediately to
Va l VS I =h Va ' i VS 2
_
(Zl l_1 - Z I2-d /a l - ( Z22-1 - ZI2-. ) la l
]
(3 . 57 )
Thus the voltage Va l may be written in terms of the equivalent voltage VSI and the current la I , exactly as in the case of shunt faults where we write
Va l = h VF
-
Z I la l
This means that the ports of the network are completely uncoupled and may be represented by the circuit of Figure 3 . 2 8 .
60
Chapter 3
Using the Y parameters, which are preferred when finding the equivalent cir cuit, we solve for and by premultiplying by y- I . The result is given by
(3.54) Val Va' i
=
1
det Y
[h( YI1-I IS2 - Y21-1 lS I ) - ( Y I1-1
( l
h YI1- lS I - Y II_I Is2 )
+
( YI1-1
+
Y21-1 ) 101]
+
YIl-l ) Ia l
(3.58)
where det Y = YU• l Y21- 1 - Yl1- l . Thus for the uncoupled representation of Figure the Z parameters are preferred.
3.28
Fig. 3 . 28.
Uncoupled positive sequence network.
Example 3. 7 Find the uncoupled representation similar to Figure Example
3. 6 .
3.28
for the system of
Solution The determinant of Y is
det Y = Thus from
(3.58)
(6/23)(31/46) - (1/23)2 = 4/23
[Val] = 23 [h[ (- 1 /23)(- 81/92) - (31/46)(- 5/23) ] - [(-1/23) + (31/46)] lOll Va' 4 h [(-1/23)(-5/23) (6/23)(- 8 1 /92)] [(-1/23) (6/23)] 101 J [1.063h - 3.625 10 1] = 1.375h + 1.250 10 1 I
-
+
+
Ana lysis of U n sy m metrica l F a u lts : Three -Component M ethod 3.8
61
U nequal Series I mpedances
One case of unequal series impedance where phase a is symmetric is that shown in Figure
3. 29.
F'
F
Q
.!L
b
�
c
Fig. 3 . 29.
1.2.
Za
a'
+ Vbb' Zb + Vcc' -
b'
-
Zb
f
c'
t
=!"""
Circuit diagram for unequal series impedances at
3. 29
Circuit diagram : See Figure and note that Boundary conditions: By inspection of Figure
Za Zb ' 3.29 *'
or
the
F-F' .
(3.60) = VOl2 Z A-I ZabcA Za 2 Zb Za - Zb Za - Z ZOl2 � Za - Zb Za 2 Zb Za - Zb Za - Zb Za - Zb Za 2 Zb
(3.59) (3.60)
from the a-b-c coordinate system to 3 . Transforma tion: We transform coordinate system by a similarity transformation. Thus
0-1-2
where
V�12
Voo'''()12
�
=
Z Ol2 1012
Ol2 =
Performing the indicated matrix multiplication , we find
=
45..
+
+
+
�
(3.6 1) (3.6 2) (3.63)
Seq uence curren ts: There are no sequence current equations for this case. Seq uence voltages: Since there are no sequence current equations, we must completely determine the sequence network connections by considering only the voltage equation From equation we compute row row to find
or
(3.(3.6 1).6 1)
1- 2 V - V Zb (lao - lal ) 00 '0
00 ' 1
=
(3.64)
62
Chapter 3
2 3 , l V Z"la2 Z"lal Vaa' 2 (3.64) (3.6 5) v00'0 - Z"lao = V00' 1 - Z"la l = V00 '2 - Z"la2 1 2 Vaa,o Vaa (1/3)(2Za Z,,)(Iao lal) (2/3)(Za - Z/» la2 (3.64) V00'0 (3.6 6) (3.65) (3.63.6)30. 3.238)0 3. la l'
Similarly , from row - row we see that combines with for the important equation
Also, taking row
+
Substituting for
+
row 'i
=
which
aa '
we find that
=
+
+
the value taken from
+
and simplifying, we find that
From the two equations we see that the sequence networks and must be connected as shown in Figure From Figure (knowing the equivalent impedance in each sequence network from Figure we may com pute Before doing so, however, we simplify the notation.
l oo t
+
FO + VaO
l ad
FI
voa'O -
Va' O
NO
+
+
Val
p ao FO' +
Vaa' l -
NI
Zb
+
� l al FI '
Zb
Va ' i
Zb
F2
+
�2 N2 Fig. 3. 30.
F-F' .
In working with shunt faults, we defined Z I to be the impedance to the flow in the positive sequence network. Similar definitions applied for Z2 and If we follow this same rule for series faults,
Zo .la l of
Sequence network connection for unequal series impedances at
Zo = (Zu.o
-
Z I2-0 ) + (Z22-0 - Z I2-0 )
(3. 6 7)
Ana lysis of U nsym metrica l F a u lts : Th re e - Co m ponent M ethod
63
Similarly ,
(3.68) and
(3.69) We also defined h VF as a voltage rise in the direction of la ! flow for shunt faults, or the open circuit voltage drop from F to neutral. In the case of series faults we have
h VF = VS !
-
VS 2
or h VF is the open circuit voltage drop from F to F ', or h VF =
Vaa ,! ] la ! = 0
(3. 7 0) (3. 7 1)
With this notation established, we compute
Ia !
=
h VF
Zt
=
h (open circuit VFF')
Zt
(3. 7 2)
where
( 3. 7 3) and where
- Zb )(Zb + ZO)(Zb + Z2 ) 3.7 Z = (Z b + Z2)(Za - Z b(Za ) + (Zb + ZO)(Za - Zb ) + 3 (Zb + ZO)(Zb + Z2 ) ( 4 ) By inspection of Figure 3.30 we see that Ia 2 = - /a I Z/ (Zb + Z2 ) ( 3.75 ) and
laO - la I Z/(Z b + Zo) where Z is defined as in ( 3 . 7 4 ) . 3.9
=
( 3 .76)
One Line Open ( 1 LO)
One open line conductor i n phase a i s a special case o f the previous case in which
Za
= 00 ,
Zb
is finite
( 3. 77)
No additional computation is necessary to show that the sequence network con nections are as shown in Figure 3.31. In this case the branch ( 1/3 )(Za - Z b ) which shunts all three sequence networks is missing. The sequence network connection of Figure 3.31 is similar to the 2 LG shunt fault connection except the parallel connections are made between F and F ' rather than between F and N. By inspection of Figure 3.31 we compute
( 3 .78) where
(3.7 9)
Chapter 3
64
loo
+ Voo'o -
t
FO
+ VoO
FI
+
NO
+ Vo l
102 ' F2
Zb
Vo'O
+ VoO' I -
lo l t
Wv-
hoo FO'
t 101
+
-Wv-
FI '
Zb
Vo'l
NI
+ Voo' 2 -
. 1 02
'VVvZb
F2'
+ VO
+ Vo 2 N2
'2
' Sequence network connection for line a open at F-F .
Fig. 3 . 3 1 .
and
Z
Then,
=
(Z b + ZO )(Z b + Z2 ) 2 b Z2 + Zo
( 3.80)
Z+
(3.81 )
( 3 .8 )
and
2
3. 1 0 Two l ines O pen (2LO )
of Figure 3.30 will If two lines are open, Z b = 00 . Note that the conne ction que. techni ical analyt usual not suffice for this situation . We adopt the = . 00 = c b Z that Z 1 . Circuit diagram : See Figure 3 .32 and note F
a
�O
b c
1 0 + Voo 'Z
+
+
Ie -
t
o
a'
+ Vbb' -
b'
+ Vec' -
+
V-a Vb Ve -
• f t
Fig. 3 . 3 2.
=0
F'
:or:
+
t
t
• t
e'
+ + V a' - -Vb' -Ve'
Circuit diagram for two lines open.
Ana lysis of U nsym m etrica l Fau lts : Th ree -Component M ethod
2. 3.= rIa 0 0]
Boundary conditions: By inspection of Figure
I.
V00 '
Ib = Ie = 0,
=
ZaIa
Transforma tion: The sequence currents are reduces to t S O that
10 1 2
3. 3 2
65
we write
1012 = A- I labc '
But from
(3.83) (3.8 3) (3.84)
and the sequence currents are all equal. This is somewhat analogous to the SLG fault, where we found this same condition to be true . Also from V(J(z ' = or
(3.8 3)
ZaIa
h ( V00 '0 + V00 ' 1 1
Rearranging, we have for any
4. 5. 3 Za 3. 3 3.
( V00 '0 -
+
h + (V
ZaIaO )
V00 '2
) hZa (lao I Ia 2 ) =
- ZaIad (3.84) (3. 8 5)
' 00 1
+ al +
+ ( V00'2
- ZaIa2 ) = 0
(3.85)
requires that the three sequence net Seq uence curren ts: Equation works be in series. Seq uence voltages: Equation requires that an impedance with total value be inserted in series with the sequence networks. The equation also could be associated with each network as shown in suggests that an impedance Figure
Za
1 00
t
FO Za
Zo
�
f f
Fig. 3.33.
Fa'
+
Vo O
1 0l f
NO
+
Vol -
2
+
Vo'O
+ Voo'l -
+
f
NI
Voo'2 -
+
-
+
-
Va '
+ 1 02
F2'
F2 �
V0 2
l l o.
FI '
F. ,
I0
h oo
+ vaa'O -
N2
-
Vo'2
Sequence network connections for two lines open.
Chapter 3
66
To compute the sequence currents, we write from Figure
lao -la la2 - Zo Z hVFZl 3 Za _
3.1 1
Other Series Faults
_
1
3.3 3
(3.86)
_
+
-
1 +
+
Nearly all cases of series faults of general interest are special cases of the three types considered in the preceding paragraphs. The one notable exception is the case where a neutral impedance exists in the zero sequence network. If it is de sirable to account for the effect of such neutral impedance separately from a separate impedance may be placed in series with the zero sequence network. The multiplier is required because a neutral current, if it exists, is always and since we define the zero sequence network to have only a current we may associate the factor with the impedance . Then the voltage drop across the zero sequence network and 11 is
"3" 3Z,l
31ao
3Z VaO 3 Z,l lao (Zo 3 ZIl) lao 3 Z,l
Ordinarily, the impedance
Vaa'o Zolao· =
-
Zo , lao , (3.87)
+
=
+
-
will be a part of Zo , in which case we write simply
Problems
3 . 1 . Repeat Example 3 . 1 for a SLG fault at bus C but wi th zero fault impedance. 3.2. Repeat Example 3 . 1 for a SLG fault at bus C but with a fault impedance of 8 ohms.
a
Plot f versus fault impedance.
3 . 3 . Analyze the system of Figure 3.5 for a SLG fault at bus B with zero fault impedance. 3.4. Repeat Example 3. 2 for a LL fault at bus C but with zero fault impedance. 3.5. Repeat Example 3.2 for a LL fault at bus C but with a fault impedance of 8 ohms. Plot
fb versus fault impedance.
3.6. Analyze the system of Figure 3.5 for a LL fault at bus B with zero fault impedance. 3.7. Repeat Example 3 . 3 for a 2LG fault at bus C but with Zg O. 3.8. Repeat Example 3.3 for a 2LG fault at bus C but with Z, O. 3.9. Repeat Example 3. 3 for a 2LG fault at bus C but with Z, Zg O. 3.10. Analyze the system of Figure 3.5 for a 2LG fault at bus B with Z, Zg O. 3 . 1 1 . Graph the results of Example 3.3 and problems 7-9 plotting fb versus Z" then fb versus Z,l ' Also plot a series of phasor diagrams similar to Figure 3 . 1 7 , showing how the various =
=
=
-
=
=
phasors are changed by fault impedance.
3 . 1 2 . Repeat Example 3 .4 for a &/> fault at bus C but with a fault impedance of 0 and then 8
a
ohms. Plot f versus Zf.
3.13. Analyze the system of Figure 3.5 for a 31> fault at bus A, bus B , and bus D. 3.14. Compute the total current leaving bus B for a LL fault at bus C in Example 3.2. 3 . 1 5 . Determine the sequence network connections for the network condition shown in Figure P3. 15 .
0
F Zf
b
c -
Fig. P3. 1 5.
Analysis of U n sym m etrical Faults : Th re e - Co m ponent M ethod
67
3.16. Determine the sequence network connections for the network condition shown in Figure P3 . 16 . F 0 _-----
... -..--
b -t-
-
-
c -+----f---_
Zb
Zo
Fig. P3. 16. 3 . 1 7 . Considllr an unbalanced � load at point F. Under what conditions can such a load be
represented by symmetrical components? Explain fully.
3 . 1 8. Determine the sequence network connections for the network condition shown in Figure P3.1S.
o
F
�-----
b -+---t---c +---+---1--
Fig. P3 . I S. 3. 19. Consider the problem of Figure P3 . 18 but where the phase impedances are Za . Zb . and Ze . with Z b "* Ze as in problem 3.18. How can such a problem be solved by symmetrical components?
3.20. Find the two·port Z parameters for the circuits shown in Figure P3.20 . All impedances
are in ohms.
+ VI
+ Z V2
3
+
2
VI 3
(0 ) 4
+ V2
( b) 2 +
+
�2
VI
(d)
(c )
Fig. P3 . 20.
68
Cha pter 3
3.21. Find the Y parameters for the circuits shown in Figure P3. 20. Show that
Y Z- I in -
every case. 3.22. Determine the two-port equivalent for the circuit of Figure P3.22 with terminals F and ' F brought out as the only external connections. Find both the positive and negative se quence equivalents, with your result similar to Figure 3.24.
Ol---+-I ----f-I ---f-1----I- tOG
'
F
F
F
F'
Fig.
H
P3 . 22.
3.23. Repeat problem 3 . 22, with the impedance between F and F ' removed. 3.24. In the case of series faults, we propose a connection be made from F to F ' external to the normal system in such a way that lobe � . That is, if lobe leaving F has any value, it will have that same value upon entering at F' . Use this to establish that the sequence cur -
rents are also equal , i.e., 1012 lim . 3.25. Suppose that (contrary to the usual assumption with a series fault that lobe = I�be ) we let 10 I� and permit phases b and c to become crossed, i.e., Ib I; and Ie Ii, . Derive the sequence network connection to represent such a condition. See [ 16 ] . 3 . 26. Find the uncoupled two-port equivalent, similar to Figure 3.28, for the system of Figure ..
-
=
=
P3. 22. 3.27. Find the uncoupled two-port equivalent, similar to Figure 3.28, for the system described in problem 3.23. 3.28. Consider the system of Example 3.6 in Figure 3.25 and let it be terminated with unbal anced impedances Zo .. 1 ohm and Z b 3 ohm " Ze ' Suppose further that the positive, -
negative, and zero sequence system impedances are identical. Compute the system cur rents from F to F' and the voltages at these nodes for all three phases. 3 .29. In the system of Figure P3.22 suppose that unbalanced impedances Zo .. j 1 and Z b j5 are connected between F and F ' . Compute the currents lo be flowing toward the fault at ..
F.
3.30. The system of Figure P3.22 has one line open between F and F ' . Compute the system
voltages and currents in the vicinity of the fault.
3 . 3 1 . The system of Figure P3.22 has two lines open between F and F ' . Compute the system
voltages and currents in the vicinity of the fault.
3.32. Consider the 3q, fault shown in Figure P3.32, where each line at F is faulted through an
impedance Z, to a point W and then to ground G through impedance Z, . (a) Determine the sequence networks for this fault, clearly identifying points F, W, and G in each network. (b) Compute the voltage VFW - VAF - VWG ' (c) Distinguish between the ground G and the neutral W in each sequence network.
Analysis of U n sy m metrica l Fau lts : Three -Component a
b
c
- - -
+
-
of:
+.
-
-
Vat Vbf �
taf Ibf
I
It
,
Z f It
zt
I cf
M ethod
69
F
I
---w
- - -8
Fig. PS . S 2.
T. In order to limit the SLG fault current for faults on the generator leads, the generator neutral is "high-resis tance" grounded through the grounding transfonner TG and resistor R. The balanced bank of capacitors C represents surge protection capacitors plus the lumped capacitances to ground of the generator leads, generator windings, and transfonner T windings. See Figure P3.33.
3.33. Generator G supplies its load through a power transfonner bank
IFf
a
b
IN
@
R G)
t
c
TiT c
NI
-
c
@CC
Fig. PS.S3.
In order to minimize the danger of damaging transient overvoltages, the following criterion for sizing R is sometimes advocated : Select R so that the power dissipated in it during a SLG fault on the generator leads is equal to �Qc Where �Qc is equal to total reactive power (magnetizing vars) generated in all capacitances during the SLG fault. Capacitance : C
=
0.40 J.LF/phase
Generator:
13 ,800 V (LL) 25 ,000 kW (31/» 3 1 ,250 kVA (31/»
60 Hz
X l = 1 . 20 pu based on generator ratings X2 - 0 . 1 0 pu based on generator ratings Xo 0.05 pu based on generator ratings =
70
Chapter 3
Grounding transformer TG : Rated HV = 13,800 V Rated LV 480 V Turns ratio Q Nl/N2 = 1 3 ,800/480 Neglect leakage impedance of transformer -
=
Generator excitation is such that generated positive sequence voltage is Ea! = 1.00 pu. The SLG is assumed to occur at no load. (a) Draw the sequence networks properly interconnected to represent the SLG fault on phase Q. Evaluate all impedances either in ohms or in pu. (b) Calculate the value of the grounding resistor R that will satisfy the given criterion. (c) Calculate the fault current IF and the generator neutral current IN , that correspond to the value of R found in (b). (d) Find the minimum kVA rating of the distribution transformer TG .
chapter
4
Sequence I m peda nce of Tra nsmission U nes
I n the previous chapter the various common shunt and series fault connec tions are analyzed and sequence network connections are derived for each situa tion. The application of these ideas to physical systems requires detailed knowl edge of the actual impedances which make up each of the sequence networks. This means that each component of the power system-the lines, machines, trans formers, and loads must be analyzed to determine the impedance of each to the flow of positive , negative, and zero sequence currents. The approach taken here to the understanding and computation of sequence impedances is a practical and at times heuristic one . A more rigorous and exhaus tive treatment of the subject is to be found in many references, particularly those by Fortescue [ 1 ] , Wagner and Evans [ 10 ] , Clarke [ 1 1 ] , Kimbark [ 1 7, 18, and 19] , the Westinghouse Corporation [ 14 ] and the General Electric Company [ 2 0 ] . Most of these resources provide additional references. Stevenson [ 2 1 ] also gives an excellent brief review of the subject. Since the subject is so well documented, the emphasis here will be on providing insight and understanding and on establish ing correct, expedient methods for determining sequence impedances. The subject is conveniently divided as follows. Chapters 4 and 5 consider the sequence impedance and admittance of transmission lines. Machines are discussed in Chapter 6 and transformers in Chapter 7 . This analysis o f transmission line impedance is restricted to the case o f three phase lines unless otherwise noted . Even with this restriction there are several important areas to investigate. The most obvious is the self impedance of the line to the flow of balanced three-phase currents and to uniphase or zero sequence currents. The effect of line transpositions is important. Since it is becoming more common to not transpose lines, what effect does this have on the phase imped ances? The computation of mutual impedances is discussed, and the effect of currents in nearby circuits or in the ground below an overhead line is also examined. 4. 1
Positive and Negative Sequence I m pedances of L ines
A transmission line is a passive device, and if transposed it presents identical impedances to the flow of currents in each of the phase conductors. Further more, the phase sequence of the applied voltage makes no difference since the voltage drops are the same for an a-b-c sequence as for a-c-b . Therefore the 71
Cha pter 4
72
positive and negative sequence impedances are identical, or Z I = Z2
=
R I + jX I
n /phase
(4.1 )
Usually we compute the impedance on a per unit length basis and multiply by the line length I (4.2 ) where r l = line resistance to positive sequence currents of each phase, n /unit length x I = line reactance to positive sequence currents of each phase, n /unit length s =
line length
The resistance r l is simply the resistance of one phase conductor or bundle per unit length . It is usually tabulated by wire size as a function of temperature and for various frequencies of interest. We assume that this resistance is the same in all three phases. The reactance x I is often thought of as having two components : one due to all flux linked internally and externally out to a radius of one unit of length (one foot, for example ) and the other due to flux external to one unit of length and out to the (geometric mean) center of the conductor carrying the return current. Thus, from [ 9] , using English units, 2 eI
D = 0 .3219 In m mH/mi Ds
or XI
=
4.657 X 10 - 3 f lOg lO
=
2.020 X 10 - 3 f In
�m
�ms n /mi
(4.3)
8
where
D m = mutual geometric mean distance (GMD ) Ds = self geometric mean distance, or geometric mean radius (GM R ) These terms need further amplification (see Stevenson [9] and Woodruff [ 2 2 ] ). In general terms when a phase conductor consists of several cylindrical, nonmag netic strands, we compute D m and Ds between two conductor groups x ( consist ing of m strands) and y (consisting of n strands) as follows :
Dm = D8 =
( (
)
prOduct of distances from all m strandS l Imn of X to all n strands of y
) 11m
prOduct of distances from all m strands of x to itself and to all other strands of x
2
where the distance from a strand "to itself" is sometimes called the self GMD and equals 0 . 7 788r (r = radius) for a cylindrical wire. To be more specific, for three I
We use the common notation here of capital letters lower case r, x, and z for ohms/unit length. 2 We use the notation "In" for "log to the base e. "
R,
X,
and Z for total line ohms and
Sequ e n ce I m pedance of T ra n s m i ssion U ne s
73
phase conductors separated by distances Dab , Dbc , and Dca center to center, we compute (4.4) D m = (Da b Db c D ca ) I /3 � D e q which is sometimes called the "equivalent spacing. " the self GMD is
Also, for a transposed line (4.5)
where D.i = self GMD of phase a in position i of a transposition and where the units of D. and D m must be the same. For example, if the frequency is 60 Hz and length is in miles, equation ( 4.3 ) may be written in the more useful forms XI
7
1
= 0.2 94 og lo
Dm
D
il /mi
(4.6)
8
or X l = 0 .1 2 1 3 In
Then we define at 60 Hz xa
=
XI
Dm il /mi DB
( 1 ft spacing) = 0 .1 2 1 3 In
(4 .7 )
�
il /mi
(4.8 )
•
Xd = X I (spacing factor) = 0 . 1 2 1 3 In D m
il Imi
(4.9)
Obviously equation (4.8) is a function only of the conductor and conductor ar rangement if bundled. For single-wire conductors xa is tabulated in wire tables as shown in Tables B.4 to B .1 5 of Appendix B. Equation (4.9), on the other hand, does not depend on the conductor type at all and is a function only of D m . Values for X d are tabulated in Table B.24 and are referred to as "inductive re actance spacing factors:' These values need only be added to find X I , i.e., X I = Xa 4.2
+ Xd
(4.10)
il Imi
Mutual Coup l i ng
One of the problems inherent in representing transmission lines is the situation wherein any wire of a group of parallel wires carries a nonzero current. In such cases any conductor which parallels the given current-carrying wire will experience an induced voltage for each unit of length of the parallel because the flux linkages of that nearby circuit are not zero. In terms of field concepts we write the flux linkages A 2 1 , or the flux linking circuit 2 due to a current in circuit 1 , as [ 1 2 ] A2 1 =
where
f A ' ds, C2
A = magnetic vector potential
>
=
(4.1 1)
0 Wb tum
1J. o I 1 41T
ds, = element of length along circuit 2 dSI = element of length along circuit 1
J
CI
.! dS I r
Wb/m
( 4. 1 2)
Chapter 4
74
Obviously, a mutual coupling exists any time the magnetic vector potential A is greater than zero. When circuit 1 (the inducing circuit) is a balanced three-phase circuit and II is considered the superposition of the three phase currents, A is zero and no mutual induction takes place. In the case of zero sequence currents, however, I 1 = 3 /ao and the mutually induced voltage may be large. We represent this mutual coupling as a mutual inductance M where M = ).. 2 1 //1
(4.13)
H
and treat this problem circuitwise in much the same way that we analyze a trans former. Suppose that two parallel circuits are designated a and b , as shown in Figure 4.1 , and have self impedances Zaa and Z" " respectively and mutual impedance Za" . Let currents la and I" enter the unprimed ends of each circuit. We assume that the circuit representation of "ground" implies a perfectly conducting plane. Also, as noted in Figure 4 . 1 , we designate voltage drops to ground at each end of
+
v.00,
/ IY +
-t
Vo
"
Z bb
Vb
-t
� /
+ vbb'
__
+
\ \\ \ \ \
b'
'
_�z.. 0
,
Fig. 4 . 1 . Two circuits a and b with mutual coupling.
the two circuits and also identify a voltage drop in the direction of the assumed current. The equation for these voltage drops along the wires is
(4 .14 )
where we recognize that Za " = Zba in a linear, passive, bilateral network. Equa ti on (4.1 4) can be justified and further insight gained by viewing this situation as shown in the one-turn transformer equivalent of Figure 4.2. Here the wires a-a ' and bob ' are viewed as turns of a one-turn (air core) transformer the "core" of which is shown. If we apply a voltage Vaa ' with polarity shown and this causes a current la to increase in the direction shown , a flux CP"a (flux linking coil b due to la ) will be increasing in the direction shown . Then , by Lenz's law a flux CP a " will be established to oppose CP " a ' This require!! that the b terminal be positive with respect to b ' , or an induced current I� will flow if the circuit bob ' is closed through a load. This is indeed a voltage drop in the direction bob ' and adds to the self im pedance drop I" Z"" of that circuit as in the V" equation (4.14). A similar argu ment establishes the validity of the Va equation (4.14). Following the usual dot convention, we can dot the a and b ends of the two lines to indicate the polarity of the induced voltage. The dot convention is particularly convenient in situations where the polarity is not obvious. Figure 4. 1 or 4.2 and equation (4.14) are all
Sequence I m pedance of T ra nsm ission Li nes a
I
75
b' 7 ' t, : I N DUC E D CUR R E N T
a
1··'
-
T O PRODUCE
IF
CIRCUIT
"' ob b - b ' IS
CLOS E D T H ROUGH A L O A D I M P E DANCE
10
� ASSUMED POS I T IVE C U R RE N T DIREC T IO N S
Fig. 4 . 2 . A one·turn transformer equivalent.
that we need to establish the correct voltage and current relationships i n mutually coupled circuits. 4.3 Self and Mutual I nductances of Parallel Cyl ind rical Wires
Inductance is usually defined by dividing the flux linkages by the current. For self inductance the flux linkages of a given circuit are divided by the current in that circuit, i.e. , (4. 1 5 )
where A 1 1 is the flux linkage in weber turns of circuit 1 due to current II in am peres. Mutual inductance is similarly defined, as given by equation (4 . 1 3 ) . The computation of the self inductance of a straight finite cylinder is shown in several references ( [ 2 3 ] , for example) and is usually divided into two compo nents (4 . 1 6 )
where Lj i s the partial self inductance o f the wire due to internal flux linkages and Le is the partial self inductance due to flux linkages outside the wire. Then it can be shown that for uniform current density (4.1 7 )
where =
P- w
=
s=
permeability of the wire Him 41T X 10- 7 Him for nonferrous materials length of wire in meters
-- (
The external partial flux linkages are [ 2 3 ] Ae =
where
P- m I l 2 1T
P- m
=
s In
8 +� r
-
)
.J8 2 + r2 + r
Wb tum
permeability of the medium surrounding the wire 41T X 1 0 - 7 Him for air r = radius of the wire in meters =
(4 . 1 8)
Chapter 4
76
If r « 8, as is always the case for power transmission lines, (4.18) may be simplified to compute
L
e
=
Il m 8
211'
(
In
28
r
)
- 1 H
(4.19)
Combining ( 4. 1 7 ) and (4. 19), we have for the inductance of a cylindrical wire meters long L=
Il w 8
811'
+
J.lm 8
211'
(
In
28
r
)
- 1 H
8
(4.20)
(A more detailed derivation is given in Appendix D . ) In most cases we are con cerned with nonferrous wires in an air medium such that Il w and Il m are both equal to 411' X 10- 7 henry/meter. This assumes, for composite conductors like ACSR, that the ferrous material carries negligible current. With this approxima tion we write the inductance as
(4.21 ) But 1 1 1 4" = In - 1/4 = In e 0.779
(4.22)
and we recognize for cylindrical wires that 0.779r = DB . Using this relationship, we further simplify the inductance formula to write f
�: - 1) Him
(4.23)
( �: - )
Hlunit length
(4 2 4 )
(In �: - )
H/unit length
(
= 2 X 10 - 7 In
where both D B and 8 are in meters or any other consistent units of length . Some engineers prefer the inductance to be specified in henry/mile, while some prefer henry/kilometer. Also, many prefer using base 10 logarithms instead of the natural (base e) logarithms. Both of these choices affect the constant in equation (4.23 ) . Suppose we replace the quantity 2 X 1 0 - 7 by a constant "k" to write f
= k In
1
.
then the constant k can b e chosen according t o the user's preference for English or metric units and for base e or base 10 logarithms. Several choices of the con stan t k are given in Table 4 .1. Obviously, s and D, must always be in the same units It seems odd that this inductance should be a function of s, the line length. We will show later that the numerator 2 s cancels out in every practical line con figuration where a return current path is present. Following a similar logic, we define mutual inductance to be [ 23, 24]
.
m=k
1
(4.25 )
Sequence I m peda nce of T ransmission U nes
Table Constan t
4. 1. Inductance Multiplying Constants
Uni t of Length
Natural Logarithm
(In)
Base 1 0 Logarithm (l o gl O )
km mi km mi
0.2000 X 10-3 0.3219 X 10-3 1.257 X 10-3 2.022 X 10-3
0.4605 X 10-3 0.7411 X 10-3 2.893 X 10-3 4.656 X 10-3
km mi km mi
0.01000 0. 01609 0.06283 0. 10111
0.02302 0.03705 0.1446 0.2328
km mi km mi
0.01200 0. 01931 0.07539 0. 12134
0.02763 0.04446 0. 1736 0.2794
k 211'k f = 50 Hz fk wk
f=
77
60 Hz fk wk
Note : 1. 6093 km = 1. 0 mi; f = 50 Hz, w = 3 1 4. 1 59 rad/sec ; f = 60 Hz, w = 376.991 rad/sec.
where D m is the geometric mean distance between the conductors. This defini tion comes directly from equation (4.13) where the flux linkage term is similar to (4.18) except r (radius) is replaced by D (the distance between wires). Boast [ 1 21 shows that this distance is actually Dm , the geometric mean distance or the dis tance between the centers of cylindrical wires. All of the foregoing assumes that the current has uniform density, an assump tion that is often accepted. This assumption introduces a slight error for large conductors, even at power frequencies. This complication, due to skin effect and proximity effect, is discussed in many references such as Rosa and Grover [ 2 5 1 , Stevenson [9 1 , Calabrese [ 241 , Attwood [ 2 3 1 and Lewis [261 . Briefly stated, skin effect causes the current distribution to become nonuniform, with a larger current density appearing on the conductor surface than at the center. This re duces the internal flux linkages and lowers the internal inductance as compared to the uniform current density (dc) case given by ( 4 . 17 ) . It also increases the re sistance. These effects are summarized in Figure 4.3 for solid round wires. Stevenson [ 9 1 gives convenient formulas for both the resistance and internal in ductance ac to dc ratios as (subscript 0 indicates dc value ) ex
a
�
RoLi L Li o R
=
=
=
=
l �[ mr 2
mr
ber mr bei' mr (ber' mr)2
-
bei mr ber' mr
+ (bei' mr)2 ber mr ber' mr + bei m r ber mr (ber' mr) 2 + (bei' mr)2
J ]
(4.26) (4.27)
where r is the conductor radius and m , in units of (lengthr 1 , is defined as where 11 =
I1rl1o .
m
Stevenson [91 writes
= -J Wl10
mr = 0 .0636 -J
I1rf/Ro
(4.28) (4.29)
where is the frequency in Hz, I1 r is the relative permeability, and R o is the dc re sistance in ohm/mile . The Bessel functions used in (4.26) and (4.27 ) were origi-
f
78
Cha pter 4 1 .7
RO R
1.6
0.95
1 .5
0.90
1.4
Li
L iO
1. 3
0.8 5 0.80
1.2
0.75
1.1
0.70
1.0
2 .0
mr
3.0
4.0
(0 ) Fig. 4 . 3 .
( b) mr
Ratios of (a) ac to dc resistance and ( b ) internal inductance as a function of the parameter mr. ( From Elements of Power System A nalysis by W. D. Stevenson. Copyright McGraw-HilI, 1962. Used with permission of McGraw-HilI Book Co. )
nally derived by Lord Kelvin [ 2 5 ] and are defined as follows [ 9 ] :
8 1 - 2(mr 2 . 6) 2 82 2 . 4t2 + 22 . 4(mr (m r ) 2 - (mr )6 + . . . . bel m r = -22 2 2 . 42 . 6 2
ber m r =
and
ber
/
. bel
/
.
--
-
(4.30)
d ber mr = 1 d mr = d(mr) ber mr m dr d 1 d mr = -d(mr) bel mr = m dr bel mr '
(4.31)
'
Using the above refinements for internal inductance we may rewrite (4. 1 7 )
=
Ci L J.l.w/8rr kCi / 4 H/unit length This changes the inductance formula (4.24) to = k ( In 2s;�L - 1 ) H/unit length �i
=
as
L
(
�
4 .32 )
For many computations sufficient accuracy may be obtained by ignoring this ef fect. 4.4 Carson ' s line
A monumental paper describing the impedance of an overhead conductor with earth return was written in by Carson [ 27 ] . This paper, with certain modifi cations, has since served as the basis for transmission line impedance calculations in cases where current flows through the earth . Carson considered a single conductor a one unit long and parallel to the ground as shown in Figure The conductor carries a current with a return
1923 4.4.
fa
Sequence I m peda nce of T ra n s m i ssion Li nes
10
a
Va
LOCAL
d
z oo
t
+
SUR FACE OF REMOTE EARTH
Do d
•
I-
-
[d " -10
-I
I UNIT
Fig. 4 . 4 .
79
FICTI TIOUS GROUND RE TURN ·CONDUCTOR "
Carson's line with ground return.
through circuit d-d ' beneath the surface of the earth. The earth is considered to have a uniform resistivity and to be of infinite extent. The current a in the ground spreads out over a large area, seeking the lowest resistance return lpath and satisfying Kirchhoff's law to guarantee an equal voltage drop in all paths. Wagner and Evans [10] show that Carson's line can be thought of a single return con ductor with a self GMD of 1 foot (or 1 meter), located at a distance Dad feet (or meters) below the overhead line, where Dad is a function of the earth resistivity The parameter Dad is adjusted so that the inductance calculated with this config uration is equal to that measured by test. From equation (4.1 4) we write for Carson's line, Z ad ] [_ la ] V /unit length _ Z dd (4.33) la where Va , Va' , Vd and Vd , are all measured with respect to the same reference. Since V 0 and Va ' - Vd , 0 , we solve for Va by subtracting the two equations to find d as
p.
=
=
Va
=
( zaa
+ Zdd
- 2 za d ) la
=
z aa 1a
(4.34)
by definition, where we clearly distinguish between Z and z . Thus Zaa zaa Zdd - 2Zad Q /unit length (4.35) Using equations (4 .24) and (4.25 ) and ignoring skin effect, we may write the self and mutual impedances of equation (4.33) follows. The self impedance of line a is Zaa = ra + j wQ a = ra jwk ( In �:a - 1) Q junit length (4.36) Similarly, Zdd rd + j w k (In �:d - 1) Qjunit length ( 4.3 7 ) where we arbitrarily set D.d 1 unit length. Carson found that the earth resis tance rd is a function of frequency and he derived the empirical formula rd = 1. 588 1 0- 3 f Q /mi (4.38) 9. 869 1 0-4 f Q /km +
�
as
+
=
=
X
=
X
80
Cha pter 4
which has a value of 0.09528 ohms/mile at 60 Hz. Finally, from (4.25) the mu tual impedance is (4.39) Za d = j W m a d = j wk (In �:d - 1 ) n/unit length Combining (4.36), (4.37), and (4.39) as specified by formula (4.35), we com pute the impedance of wire a with earth return as � Zaa = zaa + Zdd - 2 Za d = ( ra + rd ) + jwk In � d n/unit length (4.40) sa As pointed out in (4.37), it is common to let D d = 1 unit of length (in the same units Dad and Dsa ) such that the logarithm terms is written as In DD�saDsd d = In ��L (D sa ) ( l ) The argument of this logarithm has the dimension (length2 jlength2 ) or it is dimen sionless. However it appears to have the dimension of length. For this reason it has become common practice to define a quantity De as (4.41) De D �d /D sd (unit length) 2 /unit length Then we write (4.40) as e (4.42) Zaa = (ra + rd ) + j wk In � njunit length sa This expression, or one similar to it, is commonly found in the literature. It could be argued that De is not a distance since it is numerically equal to D�d ' which has dimensions (unit length)2 . We will take equation (4.41) as the definition of De . The self impedance o f a circuit with earth return depends upon the impedance of the earth which in turn fixes the value of De . Wagner and Evans [10] discuss this problem in some detail and offer a physical explanation of Carson's original work. Table 4.2 gives a summary of their description for various earth conditions. as
=
Table 4.2. De for Various Resistivities at 60 Hz
Retu rn Earth Conditio n Sea water Swampy ground
A verage damp earth
Resistivity
(nm)
De (ft )
Dad
0. 01-1. 0 1 0- 1 00
27. 9-279 882-2790
5. 28-16.7 29. 7-52. 8
1 000 10 7 1 09
8820 882,000 8,820, 000
9 3. 9 939 2970
1 00
Dry earth Pure slate Sandstone
( ft )
2790
52. 8
p
The quantity De is a function of both the earth resistivity and the frequency t and is defined by the relation (4.43) De = 2160 .J pit ft If no actual earth resistivity data is available, it is not uncommon to assume p to be 100 ohm-meter, in which case the italic quantities in Table 4.2 apply. Wagner and Evans [10] provide data for at various locations in the United States. p
81
Seque nce I m pedance of T ra n s m i ssion li nes
4.5 Three-Phase Line I mpedances
To find the impedance of a three-phase line we proceed in exactly the same way as for the single line in the previous section. The configuration of the circuits is shown in Figure 4.5 where the impedances, voltages, and currents are identified. G'
I/ "?"'"7"7'::Hf:S"7
,
ALL WIRES GROUNDED HERE TO LOCAL EARTH POTENT IAL
d'
REF .. ....- I UN IT
--------11
Fig. 4 . 5 . Three-phase line with earth return.
Since all wires are grounded at the remote point a' -b' -c', we recognize that (4.44) Then, proceeding as before, we write the voltage drop equations in the direction of current flow as follows:
[ [ ][ J :
vaa
V bb Vee
:
va - va'
=
Vb Ve
�
Vb Ve
=
Id = - (/a + Ib + Ie )
zaa
:a b Za e
r]
Za b
zae
"d
Zbb
z-b e
Zb d
Ib
Zee
_
Ie
Zb e
Ze d
'
V
/unit length
(4.45) We call these equations the "primitive voltage equations." The impedance of the line is usually thought of the ratio of the voltage to the current seen "looking in" the line at one end. We select a voltage reference at the left end of the line and is We can do this since curre nt and solve (4 .45 ) for the voltages known and because we may write (4.46) for the condition of the connection at the receiving end of the line. Since 0, we subtract the fourth equation of (4.45) from the first with the result: Vdd '
Vd - Vd '
Zad
Zb d
Ze d
Zd d
as
Va , Vb ,
Ve '
Id
Id
Vd =
Va - ( Va' - Vd ' ) = (zaa - 2Zad + zdd ) Ia + ( Za b - zad - Zb d + zd d ) Ib + (za e - za d - Zed + Zd d ) Ie
For convenience we write this result as where we have 0, is defined new impedances and Note that when exactly the impedance for the single line with earth return (4.40). If we repeat the above operation for the b and phases, we have the result Za a , Za b ,
Va = Z a a Ia + Za b Ib + z a e Ie , Zae ' I b = Ie =
z aa
c
V
/unit length (4.47)
C h a pter 4
82
where we recognize the reciprocity of mutual inductances in a linear, passive, bilateral network (Zob = Z bo , etc.). The impedance elements of (4.47) are readily found to be Self impedances Zoo Zb b zee
Mutual impedances Zob Zb e Zo e
= Zoo = Zb b , = zee = zob
2 Zod + Zdd 2 Zb d + Zdd 2Zed + Zdd
-
-
-
O/unit length O/unit length O/unit length
(4.48)
Zod - Zb d + Zdd Zbd - zed + Zdd zod Ze d + Zdd
O/unit length = Zb e O/unit length (4.49) = Zo e O/unit length To examine these impedances further we use (4.36), (4.37), and (4.39) to identify elements similar to zoo , Zdd and Zod respectively. We shall call these impedances the "primitive impedances." All are listed below in terms of the physical distances involved. -
-
-
-
+ jWk (ln �: j wk (In ��b jWk (ln �: (In �:d
Primitive self impedances Zo o
= ro
Zb b = rb + ze c = re +
Zdd = rd + j wk
1) O/unit length 1) 0/unit length - 1) 0/unit length 1)0/unit length
-
-
-
jWk (ln �:b 1) O/unit length zb e = jWk (ln �: 1) O/unit length e zeo = jWk (ln �:d 1 ) O/unit length
(4.50)
Primitive line-to-line mutual impedances zo b =
-
-
-
(In J;: ) 0 /unit length (In ;;'d 1) /unit length (In � ) 0 /unit length
( .5 )
4 1
Primitive line-to-earth mutual impedances zod
= j wk
Zb d
= jwk
Zed
= jwk
d
-
1
0
-
d
-
1
(4. 52)
From these primitive self and mutual impedances we compute the circuit self and mutual impedances using equations (4.48) and (4.49). For simplicity we use
Sequence I m pedance of T ra n s m ission Li nes
the approximations
..;JJ: = D ad
= D b d = D cd ,
D s = D Ba
= Dsb = Dsc
1. Then we compute e n /unit length zaa = (ra + rd ) + j wk ln� s Z b b = (r b + rd ) + j w k In �e n /unit length s e n /unit length Zcc = (rc + rd ) + j wk In � s I
and use the definition D Bd =
De Z ab = rd + J· w k ln D ab
De Z bc = rd + J· W k ln Dbc De Zc a = rd + J· W k ln Dc a
83
(4. 5 )
3
(4. 54)
n /unit length n /unit length n /unit length
(4. 55)
This result is interesting since the mutual impedance terms have a resistance component. This is due to the common earth return. Example 4. 1 Compute the phase self and mutual impedances of the 69 kV line shown in Figure 4.6. Assume that the frequency is 60 H z (w = and the phase wires are 19-strand 4/0, hard-drawn copper conductors which operate at 25 C. Ignore the ground wire entirely and assume that the phase wires have the configuration shown for the entire length of the line. Assume that the earth resistivity p is 100 ohm-meter and that the line is 40 miles long.
377)
f
0.375 E BB ___ , ___ GROUND WIRE STE E L f ' 15 1 0 ' ,-- PHASE 4/0 Cu --- 1 0 WIR ES ..0 19 STRAND 0a b c
45'
~
Fig. 4.6. Line configuration of a 69 kV circuit.
Solu tion From Table BA we find the conductor values
ra D Ba
= rb = rc = 0.278 n /mi = D sb = DBc = 0.01668 ft
84
Chapter
4
From Table 4. 2 we find De = 2790 ft. At 60 Hz, rd = 0.09528 n /mile and the constant wk is, from Table 4. 1, w k = 0. 12134. Then, from (4. 54 ) the self im pedance terms are
. = ( 0. 278 + 0.09528) + J (0. 12134) In
2790 0. 01668
= 0. 3733 + j 1. 4 594 n /mi The mutual impedances are computed from (4. 55) .
as
follows.
De
Za b = rd + J W k ln Dab = 0. 09528 + j (0 . 12134) In
2
��0 = 0. 0953 + jO.6833
n /mi
2790 . Zae = 0. 09528 + J (0. 1 21 34) In 2() = 0. 0953 + jO.5992 n /mi Z be = Z ab = 0. 0 953 + jO.6833 n /mi For 40 miles of line we multiply the above values by 40 to write, in matrix notation
[
]
( 14. 932 + j 58. 376) (3.812 + j 27. 332) (3.812 + j 23.968) Zabe = (3. 8 1 2 + j 27.332) ( 14.932 + j 58. 376 ) (3.812 + j 27.332) n (3. 81 2 + j 23.968) (3.81 2 + j 27.332) ( 14.932 + j 58.376) It is now appropriate to ask the question, What is the impedance of phase a? From the matrix equation ( 4.47) it is apparent that the voltage drop along wire a depends upon all three currents. Thus the ratio of Va to fa does not give the en tire picture of the behavior of phase a. The question then is best answered by equation ( 4.47) since this gives the complete picture of the line behavior for all conditions. 4.6 Transpositions and Twists of line Conductors
From equation (4.4 7) it is apparent that the phase conductors of a three-phase circuit are mutually coupled and that currents in any one conductor will produce voltage drops in the adjacent conductors. Furthermore, these induced voltage drops may be unequal even for balanced currents since the mutual impedances depend entirely on the physical arrangement of the wires. From equation (4. 55) we note that the mutual impedances are equal only when D ab = Db e = Dae , an equilateral triangular spacing. In practice such a conductor arrangement is seldom used. One means of equalizing the mutual inductances is to construct transpositions or rotations of the overhead line wires. A transposition is a physical rotation of the conductors, arranged so that each conductor is m oved to occupy the next physical position in a regular sequence such as a-b-c, b-c-a, c-a-b, etc. Such a trans position arrangement is shown in Figure 4. 7, where the conductors begin at lower
Seque nce I m pedance of T ransmission U nes
Fig. 4 . 7 .
85
An overhead line transposition or rotation.
right in an a- b-c horizontal arrangement and emerge at upper left in a b-c-a hori zontal arrangement. If a section of line is divided into three segments of equal length separated by rotations, we say that the line is "completely transposed." Under this gement the current in conductor a would see the impedances in the first column of the impedance matrix of (4.47) for one-third the total length but would then see the impedances of the second column and finally of the third column, all in equal amounts. The usual terminology for this rotation of conductors is to refer to each rota tion such as that in Figure 4 . 7 as a "transposition" and to a series connection of three sections, separated by two successive rotations, as a "transposition cycle" or a "completely transposed line." We will find it convenient to refer to the rolling of three conductors as in Figure 4. 7 by the name "rotation" to distinguish it from a situation where only two wires are transposed. This latter arrangement will be called a "twist." Both operations are identified in the literature as "transposi tions" but we must distinguish more clearly between the two if we are to accu rately compute the exact phase impedances. arran
4.6. 1
R otation of l i n e conductors u sing Rep
Mathematically we may introduce a rotation by means of a simple matrix operation. We define for this purpose a "forward (or clockwise) rotation matrix" Rep where 2 3 1
(4.56) R<=m � �] Premultiplying an impedance matrix by this rotation matrix has the effect of
"rotating" the elements as shown in Figure 4.8; i.e., the self and mutual im pedances for the conductor in position 1 are moved in the matrix to the positions initially occupied by the self and mutual impedances of conductor 2, etc. In applying the rotation matrix, we shall consider only the case where all phase wires are identical. Then ( 4 .57 ) Under these conditions the impedances depend, not on the phase designation, a-b-c, but only on the p osition the wire occupies on the tower. We designate these positions 1-2-3 as in Figure 4.8.
Chapter 4
86
a •
2
3
b
c
•
•
<:9 •
c
•
a
•
b
POSITION BEFORE ROTAT ION
t
,
ROTAT ION AFTER ROTAT ION
Fig. 4.8. Rotation due to ReP premultiplication.
to
It is helpful to recognize that the inverse of the matrix ReP exists and is equal
=:G � �]=� 3 1 2
RO '
(4. 58)
From Figure 4. 8 we recognize this to be a reverse or counterclockwise rotation since it rearranges the matrix elements in exactly the reverse of the reordering produced by ReP Also, we may compute the quantities '
ReP2 -
R-1 eP '
(4. 59)
Thus two rotations produce the same effect as one rotation in the opposite direc tion. This fact may be verified intuitively by an examination of Figure 4. 8. Mathematically, a rotation is the result of a linear transformation. If equation is rewritten with numerical subscripts to reference the quantities to a phys ( 4.4 ical position rather than a terminal ( a-b-c) connection, we have
7)
( 4. 60) or in matrix form, (4 . 6 1 ) This equation may be transformed by the linear transformation (4.5 6 ) to compute (4.6 2 ) or
The result of this operation is to place the bottom row of Z 1 2 3 on top and move the other two rows down . Note however from (4 .62 ) that we may compute (4. 6 3 )
Sequence I m peda n ce of T ransmission Unes
87
Carrying out the indicated operation, we have (4.64) or V
312 = Z 312 1312
(4 .6 5 ) The result of postmultiplying a matrix by R� 1 is to move the third column to the first position. These operations are noted compactly by (4.65). In a similar way we may show that the transformation R�1 produces the fol lowing result.
(4.66) or (4.67) The effect of these various matrix manipulations are shown in Table 4.3. Table 4. 3. The Effect o f Transposition Matrix Operations Premultiply Move 3rd row by RcJ> to position 1
(
by R; l
Postmultiply
Move 3rd column
Premultiply by R; l R�
Move 1st row to position 3
Postmultiply by RcJ>
Move 1st column to position 3
=
B
)
to position 1
We now apply the rotation matrix to the problem of computing the imped ance of a line which includes transpositions or rotations. Consider the line shown in Figure 4.9 where each current sees the impedance of each wire position for a portion of the line length. , In sec tion 1 we observe that positions 1-2-3 correspond to phases a-b-c such that the voltage equations (4.68) apply for the section.
88
:_10
POSITION I
::�: : : :::
Chapter 4
;(
c
;(
:
b
:
: :
SECTION I I SECTION 2 I SECTION 3 I $ I MILES + S 2 MILES -+ $3 MILES }
1-----
Fig. 4.9.
s MILES ------1 -1
A complete transposition or rotation cycle.
(4.68) or
(4.69)
where the impedances here are the total impedances for line section notation
1. We use the (4.70)
where fie
= sle ls
i, j =
1,
2 , 3 = position indicator = 1 , 2 , or 3 = line section identifier Sle = length of line section k = total length of line
k
s
H ere Zij is the total line impedance corresponding to positions i, j and fie is the fraction of the total line length in section k . For section 2 phases c-a-b correspond to positions 1-2-3 such that the voltage equation may be written as
Veab = V123 = Z I l 3 I 123 = Z 123 Ieab V Applying the transformation R� 1 , this equation becomes Vabe = Z 231 labe
[� : [ J Va
Z22_2 - Z3 2-2 Z 12-2
_
Z23-2 Z33-2 Z1 3-2
�
Z21_ Z3 1-2 Zl 1-2
(4.71)
(4 7 2) .
(4.73)
This equation may be verified by an inspection of Figure 4.9. Section 3 has the correspondence b-c-a to 1-2-3 or
Vbea = V1 23 = Z 123 1 123 = Z 123 I bea V Transforming by R.p = (R� 1 ) 2 , we compute Vabe = Z 312 Iabe V
(4.74 ) (4.75 )
Sequence I m pedance of T ransmission Lines
89
or (4.76)
[]�
[/oJ
To compute the total voltage drop along the line, we add the drop in each and (4.76), with the result
section given by equations (4 .68), (4.73), t Va
t V" t Ve
=
Z l 1 _ 1 + Zll-l + Z33-3 ) (Z n_ l + Z13-1 + Z3 1_3 ) (Z1 3- 1
(Zl l - l + Z31-1
+
Z 1 3_ 3 ) (Z22- 1 + Z33-1
+
Zl 1_3 ) (Z23- 1
+
+
Zl l-l Z31-l
+ +
J
Z31_3 )
Z n_3 )
(Z3 1 - 1 + ZIl-2 + Z13_3 ) (Z32_ 1 + Z 1 3-2 + Z2 1- 3 ) (Z33- 1 + Z l l-l + Z21-3 )
I"
Ie
V
(4.7 7 )
of the total line impedance in by appropriate fraction fit defined in equation (4.70).
It may b e convenient to write (4.77 ) i n terms each position multiplied the The result is
From (4 . 5 4 ) and (4.55)
(4.78)
Zij = (ro + rd ) S + jwkS ln �e
n, i =j
s
(4.79)
It is apparent that for identical conductors n
Zll = Z22 = Z33 � Zs a linear, passive, bilateral network Zij = Zji. Using this fact we may define Also, inkinds three of mutual impedance terms as follows: + + ZZhhl2 == f1f1 ZZZ1132 ++ ff22 ZZZ2123 ++ f3Z13 (4.8 1 ) Zh3 = f1 2 3 f2 13 f3f3ZZ2123 (4.80)
n n n
Then (4.78) becomes (4.82)
In terms of the geometry of the line, impedances (4.8 1 ) may be written as De + f3 In De ) n De + f2 InZhl = rd S + J. wks ( fl ln DD D 13 12 23
Chapter 4
90
(4.83) Example
4. 2
Compute the total impedance of the line in Example 4.1 if the line is trans posed by two rotations such that SI = 8 mi 'I = SI /s = 0.2 S2 = 1 2 mi '2 = S2 /S = 0 .3 83 = 20 mi '3 = S3 /8 = 0.5 Solution The impedance computed in Example 4 . 1 for the configuration given in Figure 4 .6 will be taken the impedance per mile of the first section 'I s = 8 miles in length. Then from (4.68) we premultiply by the length, 8 miles, to compute (2 .986 + j l l .675) (0.762 + j 5 .466) (0.762 + j4.794) for 8 1 : Z 1 23 = (0.762 + j 5 .466 ) (2 .986 + j 1 1 .675 ) (0 .762 + j 5 .466) n L( 0.762 + j4.794) ( 0 . 76 2 + j 5 .466) ( 2 .986 + j l l .675) The second section is 12 miles long and is transposed to a 2-3-1 configuration given by (4.73 ) . Thus, for this section premultiplication by 1 2 R� 1 gives (4.479 + j 1 7 . 5 1 3 ) (1 .143 j8.199) (1 .143 + j8.199) for S2 : Z23 1 = ( 1 .143 + j8 .199 ) (4.479 + j17.513 ) ( 1 .143 + j7 .190 ) n (1.143 j8.199) (1.143 j7.190) (4.479 j17.513) The third section is 20 miles long and is transposed to a 3-1-2 configuration in 2 I gives ( 4.76 ). Then premultiplication by 20 (R� ) ( 7.466 + j29.188) (1.906 + jl1. 9 83) (1.906 + j13.665) for 83 : Z3 1 2 = (1.906 + jll.983) (7.466 + j29.188) (1.906 + j13.665) n (1.906 + j1 3.665 ) (1.906 + j13.665) ( 7.466 + j29.188 ) Then the total impedance is the sum of the impedances of the three sections, viz., as
r
J
[
Za b c
=
[
+
+
+
+
[
+
+++
J ]
as
as
]
(14.932 j58.376) (3.812 + j25.650) ( 3 .812 j 26.659) ( 3 .812 + j25.650) (14.932 j58.376) ( 3 .8 1 2 j26.323) n (3.812 + j 26.323) ( 1 4.932 j 58. 376) (3.812 + j26.659)
+
Note that the diagonal terms are the same as for the untransposed line. Similarly, the off-diagonal resistances are the same as in Example 4.1 . The off-diagonal reactances, however, are more nearly equal or balanced than before. In Example 4.1 the difference between the largest and smallest mutual reactance is diff = 27.332 - 23.968 = 3. 364 n In the transposed line this largest difference is diff = 26.659 - 25.650 = 1.009 Thus, even very unequal transpositions help a great deal in equalizing the mutual impedances. n.
91
Chapter 4
Example
4. 3
Consider another case of transposing the line of Example 4.1 but using only two transposition sections rather than three. Let the sections be defined by = 0.4 = 16 mi, f2 = 0.6 24 mi, =0 Solution This example is solved in the same way as the previous one except that only two impedance matrices need to be computed. For the first section we have ( 5.973 + j23.350) (1.525 + j10.933) (1.525 + j9.587) for S I : Zabc (1.525 + j10.933) (5.973 + j23.350) (1.525 + j10.933) n (1.525 + j9.587) (1.525 + j10.933) (5.973 + j23.350) And for the second section, again assuming a rotation R� I , we have 8.959 + j35.025) (2.287 + j16.398) (2.287 + j16.398) for S2 : Zabc (2.287 + j16.398) .<8.959 + j35.025) (2.287 + j14.380) n (2.287 + j16.398) (2.287 + j14.380) (8.959 + j35.025) Adding the two sections, we have ( 14.932 + j58.376) (3.812 + j27.332) (3.812 + j25.986) Za b c = (3.812 + j27.332) (14.932 + j58.376) (3.812 + j25.314) n (3.812 + j25.986) (3.812 + j25.314) (14.932 + j58.376) The largest reactance difference now is diff = 27.332 - 25.314 = 2.018 n , which is still a substantial improvement over the case with no transpositions. Suppose that instead of a rotation R;I we rotated in the opposite. direction, i.e ., RI{> . Then the impedance of the second section would be, from (4.64) (8.959 + j35.025) (2.287 + j14.380) (2.287 + j16.398) for S2 : Zabc (2.287 + j14.380) (8.959 + j35.025) (2.287 + j16.398) n (2.287 + j16.398) (2.287 + j16.398) (8.959 + j35.025) Then 1 4 .932 + j 58 . 376) ( 3 .8 1 2 j 2 5 . 3 1 4) ( 3 . 8 1 2 j25 .986) Za b c = (3.812 + j25.314) (14.932 + j58.376) (3.812 + j27.332) n (3.812 + j25.986) (3.812 + j27.332) (14.932 + j58.376) This result is the same as the previous one except that the matrix elements have been moved. Equation (4.82) is the desired equation for a line which has undergone a com plete transposition or rotation cycle. Note that the line impedance includes the effect of any return current through the earth. In the case of balanced currents, as in the positive and negative sequence cases, this earth impedance will disappear since no current flows through the earth. When zero sequence currents flow, how ever, all three currents return through the earth and the earth impedance is very important. To compute the sequence impedances, we tum to the definitions of equations (2.46)-(2.48) to compute Zs o = Z. = (ra + rd ) s + jwks In �e n
fl
=
=
�
�
�
]
�
J
=
�
f3
=
�
+
+
•
�
]
92
4
(phase wires identical) Zs � ZMO ZM I . k [ 3 ) n +3 a2 + afl3 a2 f3 ] [(f3 + a23f,. af. ) n Z l + a2 3 af2 ) n - + a2 l + af3 n - ] 3 3 3 where in Z O we use the quantity the "equivalent spacing," of three conduc tors definedM as the GMD of the three distances or Chapter
I = ZS 2 = 0
e n eq 2 (f3 + af2 + a f l
= rd s + j wks ln
= JW S +
( fl + af3
-
'
+
(f
f2 ) In De D2 3 +
k
M2 - JW s
f +
I
I
I
De D2 3
De D I2
(f2 +
+
)
De DI3
In
n
De
D 12
(f2 +
f
)I
De DI
n .a
(4.84)
Deq ,
13 D e q = (D I 2 D2 3 D I 3 ) /
(4.8 5 )
and where all distances are in the same units throughout. Applying equation ( 2 .46) we write the sequence impedances as
- ZM 2 (Zs - ZM O) 2 ZM I
Since ZM and ZM '1. are nonzero, there is coupling between the sequence networks in the general case of a transposed line. The self impedance to the flow of laO and Ia l are equal respectively to I
(4.86)
Zo = Zs + 2 ZMO [(ra + 3rd) + jWk ln D�2 ] Zs - ZMO �a jwk In �sq) where we note that the earth resistance has vanished in Z as predicted. Later we will show that is simplified in the special case where 3 =s
eq
B
Z I = Z2 =
=S
+
n
n
I
(4.86)
(4.87)
fl = f2 = f
•
ROl2 The method used above to compute the sequence impedance of a transposed line was to first identify the phase impedances of each line section, add these sec tion impedances, and finally transform the sum to the 0-1-2 frame of reference. Stated mathematically, this operation consisted of the following: 4.6. 2
Com putation of sequence i mpedances by
I � V0 1 2 = A - I � Vabc = SA- ( fI Z I 2 3 + f2 Z2 3 1 + f3 Z 3 12 ) Iabc
V
(4.88)
Expanding this equation, we write 1 I I l � V0 1 '1. = s A- ( fl z I 2 3 + f2 R; Z 1 2 3 R41 f3 R41 Z 1 2 3 R; ) AA- labc = s [ fl A- I Z 12 3 A + f2 (A- I R; I A ) (A- l Z 1 2 3 A) (A- l R41 A ) l l l + f3 (A- l R41 A ) (A- Z 1 2 3 A ) (A- R� A) ] 10 1 2
+
(4.89)
Sequence I m peda nce of T ransmission ti nes
Suppose we define the quantity
� a �J A ' [� �
93
0
R, , ,
�
A- ' Ro A =
0
Then we readily verify that
(4.90)
0
R,I, =
A-'
RO'
a2
(4.9 1 )
0
We also recognize that for the first line section, using (2.45) I Z0 1 2 = A- Z I 23 A n /mi Using ( 4 .90 )-(4.92), equation ( 4.89) becomes 1: VOl2 = 8«(1 ZO l2
+ (2 R -0\2 ZOl 2 Ron
+
(3 RO l2 Z012 R -612 ) 10 12
(4. 92 )
(4.9 a )
This result is better understood when it is recognized that 2 (Rii�2) = ROl2
R � 12 = R -612 ,
(4.94)
so that the third term may be thought of as two rotations of the initial configura tion. One may easily verify that a2 z O I a: " a Z I2 RO I2 Zo 1 2 R 112 = a : lo Z11
["[
j
a Z20
a Z2 1
Z22
00 2 R-612 Z0 12 R0 1 2 = a z I0
a ZO I
a" '
Zl1 2 a z2 1
a Z I2
aZ20
where from ( 2 .46) we compute
Z22
J
In �e n/unit length ZSI = ZS2 = 0 n /unit length e n/unit length ZM O = rd j wk In � eq Wk ( In �:3 a In �:3 a2 In �:J n /unit length ZM 1 = j a wk (In �:3 + a2 In �:3 + a In �:J n /unit length ZM 2 = j a such that the elements of Z0 12 are defined to be Zoo = (ra + a rd ) + j wk :;2 n/unit length eq Zso = (ra
+
rd ) + j wk
(4.95)
(4.96)
II
+
+
+
In
II
(4.9 7 )
94
C h a pter
4
n /unit length Z l 1 = Z22 = ro + j k In ;9 n/unit length (4.98) Z2 1 = + 2 ZM I n/unit length Z I 2 = + 2 ZM Combining as specified in (4.93) gives the same results for the entire line as (4.84 )-( 4.86) of the previous section. = Z20 ,
ZOI = - ZM2
Z02 = - ZM I = Z IO
w
8
2'
Example 4 . 4
Compute the sequence impedance matrix ZO l2 for the untransposed trans mission line of Example 4.1 and also for the transposed lines of Examples 4.2 and 4.3. Solution
Using the computer programs of Appendix A , we compute the sequence im pedances by the similarity transformation (4.92). 1. For the line of Example 4 . 1 with no transpositions, we compute
[
= 1 .0 ,
fl
(22.556 + j l l 0.80)
Z0 1 2 =
13 = 0
(0.971 - jO.561 )
( 1 1 . 12 0 + j32.165)
(- 0.971 - jO.561 )
( 1 .942 + j 1 . 1 2 1 )
(0.971 - j O . 561 )
2.
12 = 0 ,
(- 0 .971 - jO.561 ) (- 1 .942 + j 1 . 1 21 )
(11 .120 + j 3 1 . 165)
For the line of Example 4 .2 with two rotations
�
fl
f2 = 0.3 ,
= 0 . 2,
f3 = 0 . 5
we compute Z0 1 2 =
3.
22.556 + j l l 0.80)
(0.291 - jO.056 )
( 0.291 - 0.056)
( 1 1 . 1 20 + j32 . 1 6 5 )
(0.583 + jO.1 1 2 )
(- 0 .291 - jO.056)
(- 0 .583 + j O . 1 1 2 )
( 1 1 . 1 20 + j32.165 )
For the line of Example 4.3 with transposition
[
= 0.4,
f2
= 0.6,
f3 = 0
we compute ZO l 2 =
J
(- 0.291 - jO.056)
II
J
n
(22.556 + j 1 10.80)
(0.388 + j0 .449 )
(- 0 .388 + j0.449 )
( 1 1 .120 + j32.165) (- 0.777 - jO.89 7 )
(0.388 + j0.449)
(0.777 - jO .897 )
(- 0.388 + j0.449)
J
n
n
( 1 1 . 1 20 + j32.165)
The transpositions are effective in reducing the coupling between sequence networks. Even one rotation produces a substantial reduction in the magnitude of the off-diagonal terms of Zo n and two rotations practically eliminate this coupling. a later example we show that a completely transposed line has zero coupling between sequences. In
Sequence I m pedance of T ransm issi i n Li nes 4.6.3
95
Twisting of line conductor pairs
Occasionally only two of the three line conductors may be transposed as shown in Figure 4.10. This may be thought of as a single-phase transposition and results in a reversal of phase sequence from a-b-c to a-c-b. Mathematically this re0
•
2
3
b
c
POSITION BEFO R E T W I ST ING
•
•
•
•
•
•
•
•
•
,....--..
'-./
0
AFTE R TWISTING
b
c
T W I ST 2 a 3
Fig. 4 . 10. Twisting of conductors in positions 2 and 3.
gement may be accomplished by means of a twist matrix T� , where we define 1 3 2 0 T�2 3 2 0 0 (4. 99) 3 0 1 such that arran
= T �] = Actually there are three twist matrices, the other two being defined as T� 2 3 la e b
2
la b e
1
( 4. 1 00)
3
T 3 �] and 3 2 1� 0 = 3 �] Obviously, twisting any pair twice restores the configuration to its original state, i.e., � �] = u, (4.103) [0 1
T� 1 2
=2
1
0
LO
0
(4.101)
1
T� 1 3
2 0
1
1
0
T�2 1 2 - T�2 2 3 - T�2 1 3 _
_
( 4. 1 0 2 )
0
_
1
We also readily verify that each twist matrix is equal to its own inverse, i.e.,
(4.104 )
96
Chapter
4
This is logical since it makes no difference which direction the two wires are twisted, the results being the same in either case. If a matrix equation is transformed by the twist matrix, a change in the equa tion occurs which depends upon which of the three twist matrices is used. For example, suppose the transformation Ttl> 2 3 is used to premultiply both sides of the equation, V I 2 3 = Z 1 23 1 1 2 3 . Then TtI>23 V 1 2 3 = (TtI>23 Z 1 2 3 T;� 3) (Ttl> 2 3 1 1 23) or (4.105)
which may be written out as
(4. 1 06 ) I n general, we may easily verify the following twist rules : ( 1 ) premultiplication by Tt/>iJ interchanges rows i and j ( 2 ) postmultiplication by Tt/>i/ interchanges col umns i and j. Note that ( 2 ) is the same as postmultiplication by T � h . POSITION POSITION 2 POSITION 3
:
•
__ IC
X � I �ECTION -+
SECTION $1 = fl$
1-----
2
$ 2 = f2$
:
S
Fig. 4 . 1 1 . A line with wires 2 and 3 twisted.
As an example of twisting consider the line shown in Figure 2 and 3 are twisted. In the section 1 we write
[�oJ [�:] [�:: �:: �:� [�: : �:J Ve
=
V3
= (1
Z3 1
Z32
Z3J
13
= Ie
4.11
where wires
(4. 107)
Where Zij is the total impedance of s miles, based on the configuration of sec tion 1 . For section 2 we begin b y writing
Voeb
= V123
which we transform by TtI>23 to compute
= (2 Z 1 2 3 loeb
I�:J [�:: �:: �e
= f2
Z2 1
Z23
(4. 1 08)
(4. 109)
( 4. 1 10)
97
Sequence I m pedance of T ransm ission Unes
where the reciprocity of mutuals has been recognized to write Zii = ZiJ"in all cases. Obviously, any twisted pair could be analyzed in this same way. Combining the rotation and the twist, we may compute the line impedance for any physical situation. Note that this method allows an exact accounting of every wire con figuration and the length of line with a particular arrangement. It therefore com prises an exact method of finding not only phase impedances but sequence im pedances as well, taking into account the exact amount of mutual coupling between phases or between sequence networks.
Example
4. 5
Repeat Example 4.3 with line sectioning 1 1 = 0.4,
'2 = 0.6,
'3 = 0
except twist the wires by applying Tt/l 23 instead of transposing by R� I . Solu tion. The impedance for the first section is exactly the same as that computed in Example 4.3, viz. , for
�
[
( 5. 973 + j23. 3 50 ) ( 1 . 52 5 + j 1 0. 933) ( 1. 525 + j9. 587) 8 1 : Z l 2 3 = ( 1 . 525 + j 1 0. 933) ( 5.973 + j23.350) ( 1 . 525 + j 1 0. 933) n ( 1 . 52 5 + j9. 587) ( 1 . 52 5 + j10.933) ( 5.973 + j23.350)
�
According to (4.1 09) the second section has the impedance matrix for
82 : Z 1 3 2 Then
�
=
J
8. 959 + j 3 5. 0 26 ) ( 2. 287 + j14. 381 ) ( 2. 287 + j16.399) ( 2. 287 + j14.381 ) (8.954 + j35.026 ) ( 2. 287 + j 16.399) n ( 2. 2 87 + j16.349) ( 2. 287 + j16. 399) ( 8.959 + j 3 5. 026)
J
( 1 4.932 j58.376) ( 3.8 1 2 + j25. 3 1 4 ) ( 3 . 8 1 2 + j25.986) ( 14.932 + j58. 376) (3.812 + j27.332) n Zabc = (3.81 2 + j25.314) (3.812 + j27.332) ( 14.932 + j58. 3 76 ) ( 3 . 8 1 2 + j25. 986)
+
This result i s the same numerically as to a-cob on the pole.
a
rotation Rt/l , but the phase position changes
In the foregoing analysis we have considered three different twist matrices Tt/l 2 3 , Tt/l 1 2 ' and Tt/l 1 3 . It can be easily shown that only one is necessary since the other two may be defined in terms of a rotation-and-twist or twist-an d-rotation combination. Suppose we define the only twist matrix to be Tt/l where
Tt/I = Tt/I:l3
( 4. 1 1 1 )
Then it is easy to show that
Tt/l 12 = Rt/I Tt/l = Tt/l R� ' ,
Tt/l 1 3 = R� ' Tt/l
=
Tt/l Rt/l
(4. 1 1 2 )
Furthermore, i f it i s desirable t o compute the 0-1-2 impedances for section 2 directly, we may write (4. 1 13 ) where we define (4.114)
Chapter 4
98
It is easy to show that (4. 1 1 5) Had we defined Til> to be a choice other than (4. 1 1 1 ), then (4. 1 14 ) would be complex. Using identities (4. 1 1 2) we may compute the effect of twisting any pair of wires. Thus we shall hereafter refer to only one twist matrix (4. 1 1 1 ), although in practice the use of all three may be preferred by some. 4. 7 Completely Transposed Lines
In some problems the lines may be completely transposed or may be assumed so to simplify computations. In such problems the impedance calculation is a special case of equation (4.78) where
Zu Z"2 Z"3 (Z12 Z23 Z13)
= Then, from ( 4. 8 1 ) = where we compute, from (4.83),
= ( 1 /3 )
+
+
(4. 1 1 6 )
= Zit b y definition, (4 1 1 7) .
and write
�VVaaolJ ro Zl Va2 Zo a
This simplifies the sequence impedances since from ( 2.48) write
0 , and we
0
0
( 4. 1 19)
0
where from ( 4. 87 ) for one unit length line
= (r
=
0
_
-
ZM I ZM2 =
(4. 1 1 8 )
+ 3 rd ) + j w k In
D�2 D s
eq
n lunit length
(4. 1 20 )
Observe that when the line i s completely transposed, the impedance matrix (4. 1 1 9) is diagonal and there is no coupling between sequence networks. Note also that the positive and negative sequence impedances are equal, a fact that is intuitively obvious. The zero sequence impedance is sometimes seen in the literature in a form somewhat different from that of (4. 1 20). Suppose we write the zero sequence impedance as
Zo (ra =
+ 3rd )
+
j 3 w k In
D I��2/3 s
eq
n /unit length
( 4. 1 2 1 )
�D'
Sequ en ce I mpeda n ce of T ra n s m i ssion Li nes
/
I
\
-
./'
0 "'-
s
-
013
---
-
2
-
---
"'"
0 23
3
Os
./
/'
99
V' COMPOSITE
CONDUCTOR
/
"a "
\
WITH SELF GMD 000
Fig . 4 . 1 2 . A fictitious composite wire a.
We can write the denominator of the logarithm in the form
2 1 3 = D I 1 3 (D 12 D 2 D 1 ) 2 19 = [D 3 (D D 3 D 1 3 ) 2 ] 1 19 D sI 1 3 D eq 8 12 2 s 3 3
( 4 . 122)
D aa = D 8I 1 3 D e2q13
( 4. 123)
This quantity is the self GMD of a composite conductor, which we shall call "a, " consisting of three single wires of GMR equal to D 8 and separated by distances D u , D2 3 , and D 1 3 as shown in Figure 4.12. Then the quantity (4.1 22) is clearly the self GMD of this composite conductor or
The zero sequence impedance may then be written as
Zo
= ( ra
+ 3rd ) + j3 w k In
gaae
Q /unit length
(4. 1 24)
This view of a transposed line behaving as a composite conductor is a useful con cept. In fact, the zero sequence may be computed directly from this idea. Sup pose that zero sequence currents flow in the three phase conductors and return in the earth as shown in Figure To compute the inductance of the composite line conductor a under these conditions, we recognize two components of induc tance-that due to Ia and that due to Ie as in Figure The inductance due to Ia is
4.13.
= Ia
Since Ie
o
Fig. 4 . 1 3 .
La = k In
and D se =
-::::-
I
H/unit length
(4 . 1 2 5 )
8
1.0, the inductance due to Ie is, L e = k In ��� H/unit length
100/ �/ �
�m = k n �aeaa
4.13.
( 4. 1 26 )
SURFACE
3100 / /'
(0)
(b)
C ircuit for which zero sequence impedance i s desired : ( a ) p ictorial view ( b ) equiva lent arrangement for computing inductance.
1 00
Cha pte r 4
Adding, we have the total inductance for phase a which we call L aa . Laa = La + Le
aae
D2 D k In � = k In D D aa
=
H/unit length
( 4. 1 2 7 )
Then the total self impedance of line a and earth return is D Zaa = raa + j x aa = (r + rd ) + j wk ln e Daa
If f = 60 Hz and the length is in miles, Z aa = (r + 0. 09528) + jO. 1 2 1 3 In
Q /mi
(4.128)
geaa n/mi
This is the zero sequence impedance of one line with ea rth return. In this problem the composite line a consists of three conductors, each carry ing a current la o . If we consider the three conductors to be arranged as shown in Figure 4 . 1 2, then Daa for equation ( 4. 127 ) is given by equation ( 4. 1 23). If we let r be the resistance in ohms per mile of any one conductor a, b, or c of composite conductor a, then r = ral3 and the impedance seen by the current 31a o is
a
(; ) + rd
Zaa =
+ j w k In
�aea n/unit length
(4. 1 29 )
O n a per phase basis, since each phase i s i n parallel with the other two, the im pedance seen by la o is Zo = 3za
a
=
(ra + 3rd ) + j3 wk In
�eaa
n /unit length
(4. 1 3 0 )
For example with f = 60 H z and the length in miles, Zo
= ( ra + 0 . 2 6 ) + jO. 3 6 4 In �e
8
aa
n /mi/phase
Equation ( 4. 1 30) is exactly the same as (4.124) which was derived by a different method. The GMD method is much simpler and is usually recommended for com pletely transposed lines. The positive ( or negative) sequence impedance is also computed using the GMD method for transposed lines, and the subject is discussed thoroughly in many references, for example, Stevenson [ 9] , Westinghouse [ 14] or Woodruff [ 22] . W. A. Lewis introduced a useful notation [ 10, p. 4 1 9 ] as follows. Define the following quantities : r re
Xe
::: resistance of one phase,
:::
3rd
= 0. 004764f
= 3 w k In De'
Xa
n /mi
n /unit length
= wk In
D' 1
Xd = w k In Deq
n /unit length
( 4. 1 3 1 )
s
Then we may write, from (4. 1 3 0 )
Z o = (r + r ) + j (xe +
e
Xa - 2 Xd)
n /unit length
(4.132)
Sequ e n ce I m pedance
Example
of T ransm ission
1 01
Li nes
4. 6
Compute the sequence impedance of the 40-mile transmission line of Example 4. 1 assuming that the line is completely transposed, i.e., fl = f2 = f3 = 1 /3.
Solution The self impedance term Z. in ( 4. 1 1 8 ) is exactly the same as that computed in Example 4 . 1 .
Zs =
14.932
+
j58.376 n
Zk
The off-diagonal terms are all equal to of (4. 1 1 7). To find this value, we first compute D eq = [ ( 1 0) ( 1 0) ( 20 ) ] 1 / 3 = 1 2.6 ft. Then = rd s + jw ks In
Zk
geeq
= ( 0. 09 528) (40) + j(0. 1 2 1 34) (40) In
= 3 .812 + j 26 .21 1
�
In matrix notation
Zabc
=
n
( 14.932 + j 58 .376 ) (3.812 + j 26.21 1 ) ( 3 . 8 1 2 j26.21 1 ) ( 1 4.932 + j58. 376) ( 3 .812 j 26.21 1 ) ( 3.81 2 + j 26.21 1 )
++
;���
J
(3.81 2 + j 26.211 ) ( 3 .8 1 2 + j 26.2 1 1 ) n ( 1 4.932 + j58.376)
It is interesting to compare this result with those computed previously for the partially transposed line. Only the off-diagonal reactance changes, and these changes are shown in Table 4.4. The sequence impedances are computed from ( 4. 1 19) or by a similarity transformation of Z abc with the result ZO l 2
=
[
( 22 . 5 54 + j 1 1 0. 79) (0. 000 + jO.OOO) ( 0. 000 + jO.OOO)
+
4.4. Comparison of ResultlrExamples 4. 1 to 4.6 Section Factors Transposition Xca X bc Xab Type (ohms) (ohms) (ohms) f3 f2 fl Table
Ex ample Nu m ber
n o ne �I
4. 1 4. 2 4. 3 4.3 a 4. 5 4. 6
1. 0 0. 2 0. 4 0. 4 0. 4 0. 33
RiP l RIP TIP'l 3 �I
J
(0. 000 + jO.OOO) (0.000 + jO.OOO) ( 1 1 . 120 + j32. 166) (0. 000 + jO.OOO) n (0. 000 + jO.OOO) ( 1 1 . 1 20 j32. 166 )
0 0. 3 0. 6 0. 6 0. 6 0. 3 3
0 0. 5 0 0 0 0. 33
27.332 25.650 27.332 25. 3 14 25 . 3 14 26. 2 1 1
27.332 26.323 25.314 27. 3 3 2 27.332 26. 2 1 1
Max AX
(ohms) 3.364 1.009 2.018 2.018 2.018 0
23.968 26.659 25.986 25.986 25.986 26. 2 1 1
Table 4. 5. Comparison of Sequence Coupling Impedances
Section Factors
Transposition Typ e
fl
f2
f3
none �I RiPl RiPl
1. 0 0. 2 0. 4 0. 33
0 0. 3 0. 6
0 0. 5 0 0. 33
0. 3 3
ZOb Z 10. Z02 . or Z20 ±0. 97 1 ±0. 29 1 ±0. 388 + 0. 000 +
jO. 56 1 jO. 056 jO.449 jO. OOO
Z I 2 or Z 2 1 ±1.942 + ±0. 583 + ±0. 7 7 7 0. 000 + -
j1. 1 2 1 jO. 1 1 2 jO. 897 jO. OOO
1 02
Cha pter
4
Note that for this completely transposed line the off-diagonal terms are all zero. Comparison of the off-diagonal sequence impedances computed in Examples 4.4 and 4.6 is shown in Table 4.5. 4.8 C i rcuit U n balance Due t o I ncomplete T ransposition
Section 4.7 notes that a complete transposition of the phase wires, with = eliminates coupling between sequence networks because the sums of = all phase-to-phase mutuals are equal. Suppose that the transposition is only :1= :1= or omitted entirely. What effect does this have partially complete on the sequence coupling and on the system operation? To answer this question it is useful to examine what we mean by the equation for the voltage drop along a line. Equation (4.82) is the voltage at the "sending = = end" of a line which is transposed but not such that In fact these factors may take on any value between 0 and 1 as long as their sum is unity. Equation ( 4. 82) is therefore a perfectly general equation for a transmission line, providing the line has equal self impedance in each phase. If we omit the factor s, the line length, we may write the voltage drop per unit length as
fl fa f3 , (f fa f3) l
fl f2 f3 '
(4. 133) where
Zs
and
Zkl Zk2 Zk 3
= ( ra
rd rd rd
+ jwk
+ + rd )
j w k In
fl f2 f3 f2 f3 fl f3 fl f2
�
e
il /unit length
(4. 1 34)
s
DeDn D2DeDe3 DI3
In In In
il /unit length (4. 135)
Obviously the voltage drop in each phase involves the impedance of the earth, even though the voltage of the earth " conductor" at the point of voltage meas urement is zero. It is convenient to think of impedances ( 4. 1 34) and (4. 1 3 5 ) as the effective phase self and mutual impedances of a physical system in which the ground is a perfectly conducting plane, as shown in Figure 4. 14. Here the earth impedances are included as a part of the phase impedances. Then the impedance of the line may be computed by placing a three-phase short circuit on the remote end of the line and observing the voltages and currents at the sending end. Equation ( 4 . 1 3 3 ) expresses this relationship, and from this equation we may extract the impedance data we seek. To determine the amount of unbalance in a given line, it is convenient to solve ( 4.1 33) for the currents since we often measure unbalance by specifying the per
Seq u e n ce I m pedance of T ra n s m ission Lines
1 03
0'
PERFECT CONDUCTOR
1----- 1 M I L E -----I
Fig. 4 . 1 4 .
Equ ivalent circu it for equation ( 4 . 1 3 3 ).
unit negative or zero sequence currents when the applied voltage is strictly positive sequence. Transforming ( 4 . 1 33 ) , we write
where
V O l 2 = ZO l 2 1 0 1 2
V lunit length
Z0Z122 Z2 J
(4. 136)
n lunit length (4. 1 37)
The elements of ZO l 2 are completely defined by ( 2. 46 )-( 2.48). Applying these formulas we compute
ZZSsOI == zZ.S2 = 0 ZMZMOI == (1/3)(zklZkl+ Zk+ 2aZk+ 2Zk+3)Zk3) ZM2 = + a2 Zk2 + Zk3) ZZlo1 =Z2= Z. 2+=z. (Zkl +(ZZkkl2 +Zk+ Zk23)+Zk3) ZZ0O2I == ZZ2IO0 == ((aa2zkzkl l ++aaZk2zk22 ++ ZkZk33)) ZZ2121 == -2-2zoz021 1012 :: YOl2 V012 n lunit length
n lunit length
2 ( 1 / 3) (a
n lunit length n lunit length n lunit length
( 1 / 3 ) (a zkl
(4.138)
From these quantities we apply ( 2 .46 ) to compute ( 2/3)
n
- ( 1/3)
lunit length n /unit length
- ( 1 /3 )
n / unit length
( 1 /3)
n /unit length
-
n lunit length
n /unit length
(4. 1 39)
Since the amount of unbalance is determined from the unbalanced current flowing when balanced voltages are applied, it is convenient to invert the voltage equation ( 4. 1 36 ) to find A
where
( 4 . 1 40)
Chapte r 4
1 04
Y0 2 ] YOl2 [�:Y2:0 �::Y2l YnY 12
If we define
=
mho ' unit length
(4. 142)
then every admittance element is easily identified by the corresponding impedance quantities of
( 4. 141).
Since the unbalance i s t o b e measured with only positive sequence voltage ap plied , we write ( 4 . 1 40) as
Y0Y121 [Ol lYOI] Yn Y2l Va l = y "
Va l (4. 143)
0
W e then define the per unit unbalances for zero sequence and negative sequence currents as [ 28 ]
(4. 1 44) In terms of the
ZOl 2
matrix elements these factors are written as
( 4. 1 45)
Gross and Hesse [ 28 ] note that in physical systems
If w e divide the
mo
proximate formulas
and the m 2 equation by
equation by Z22
mo ZZoOoI mo (az(2saz+zkkl l (ZaZkkal22z+k2ZkZk2 3Zk+) 3Zk) 3) m2 3zs - (Zkl Zk2 Zk3) m2 m2 , � -
-
,
and su bstituting from ( 4 . 1 39 ) we write �
�
+
+
-
2
3
2
+
+
+
we have the ap
(4. 146)
pu
(4. 147 )
pu
+
Gross and Hesse [ 28 ] give values of m o and
Zoo ,
for various configurations and
mo
also examine the effect of ground wires on these unbalance factors. These com putations show that for most commonly used configurations, stant at about
is nearly con
1 % but is increased a great deal by the addition of ground wires, the
effect of which should not be neglected. The factor
on the other hand, varies
over a wider range, say from about 3 to 20%, but is only slightly affected by ground wires.
Gross and Nelson [ 2 9 ] give quick estimating curves for these fac
tors at a variety of spacings and configurations. vestigation to include double circuit lines.
Gross et al. [ 30 ] extend this in
Seq u e n ce I m pedance of T ransm ission Li nes
1 05
(4.144)-(4.147)
The unbalance factors computed in equations are due to the line only. This is much greater than the total unbalance of the system, assuming that only this one line is unbalanced. To compute the system unbalance, we should include the effect of the Thevenin equivalent impedance, looking into the network at each end of the line. If we designate these Thevenin impedances and voltages by the primed and unprimed notation of Figure we have for the sequence voltage equation
4.14,
where the Thevenin voltages E and E; are strictly positive sequence and the I Thevenin equivalent impedances and are assumed to be balanced and therefore uncoupled. Since the unbalance factors have the Z matrix diagonal terms in the denominator, the system unbalance is smaller than the un balance due to this one line alone. However, since the Thevenin impedances are functions of the system condition, it may be preferable to compute the unbalance for the line only, as in or with the understanding that the values computed will be pessimistic. If done in this way, the unbalance factor may be considered as a parameter of the line itself. It may be desirable, however, to per form computations of system unbalance in a particular situation.
t lZ to , Z tl, Z tl
Z ;o , Z;\o Z ;2
(4.144) (4.145),
Example 4. 7 Compute the unbalance factors of previous examples using both the exact and the approximate formulas. Solu tion From Examples
4.4 and 4.6 we find Z0 1 2 matrices which we invert as follows : 1. From Example 4.1 with no transpositions and fl = 1.0, f2 = 0, f3 = 0 we compute 768 - j8.671) (0.311 jO.010) (- 0.146 - jO.274) Y Ol 2 (1.(- 0.146 - jO.274) (9.717 - j27.846) (- 1.933 - jO.284) mmho (0.311 jO.010) (0.7 20 j1.816) (9.717 - j27.846) 2. From Example 4. 2 with two rotations and fl 0.2, f2 = 0.3, f3 = 0.5 we compute - j8.667) (- 0.057 - jO.05 2 ) (0.074 jO.026) Y01 2 (1.764 (0.074 jO.026) (9.609 - j27.776) (0.336 + jO.388) mmho (- 0.057 - jO.05 2) (- 0.456 - jO.234) (9.608 - j27.776) 3. From Example 4.3 with one rotation and fl = 0.4, f2 = 0.6, f3 = 0 we compute 765 - j8.668) (0.024 jO.148) (- 0.144 + jO.054) Y O l2 (1.(- 0.143 jO.053) (9.633 - j27.791) ( 0.046 - j1.023) mmho (0.023 jO.148) (1.006 - jO.197) (9.633 - j27.791) =
=
==
[
[ [
]
+
+
+
=
+
+
+
+
+
-
]
�
Cha pte r 4
1 06
Table 4. 6. Comparison
Example Section Factors Nu m ber f, f3 f2 4. 1 4. 2 4. 3 4. 3a 4. 5
1. 0 0. 2 0. 4 0. 4 0. 4
0. 0 0. 3 0. 6 0. 6 0. 6
0. 0 0. 5 0. 0 0. 0 0. 0
�(1.
% mo
of
Sequence Unbalance Computations
Approximate Method % m2 % mo
Exact Method % m2
1. 054/72.680 0. 265[- 67 . 000 0. 5 10l1 5 1.89° 0. 521l- 9. 55° 0. 521[- 9.550
0. 992/7 1. 5 10 6. 624/ 1 39. 140 0. 262l- 67.6 0° 1. 746/- 81. 860 0. 525l150. 6 1° 3. 484l59. 83° 3. 484l- 1 4 1. 7 1° 0. 525l- 7.60° 3. 434l- 141. 7 1° 0. 525l- 7.60°
4. From Example 4.6 with complete transposition and compute Y0 1 1 =
+
6. 590l1 39. 07° 1.7 4 3l- 8 1. 860 3.487l59. 96° 3. 487l- 141. 82° 3.487l- 141. 82°
]
f, f2 f3 =
=
= 1 /3 we
( 0. 000 + jO.OOO) 764 - j8.66 7 ) (0.000 jO. OOO) mmho ( 0. 000 jO. OOO) ( 9.608 - j27. 775) ( 0. 000 jO. OOO) (9.608 - j27. 715) ( 0. 000 + jO.OOO) (0. 000 + jO.OOO)
+
+
The unbalance factors are now easily computed from equations ( 4. 1 44 ) and (4. 146). Results are shown in Table 4.6 and show that the approximate formulas (4.146) for computing the unbalance factors are quite accurate and will suffice for most computations. 4.9 Sequence I mpedance of Lines with Bundled Conductors
Consider the transmission line shown in Figure 4. 1 5 consisting of four over head wires with common earth return. This system is similar to that of the three phase line of Figure 4. 5, and the various primitive impedances are computed from relations similar to (4. 50)-(4. 52) which may be written by inspection. Also, as in the previous case, we let (4.148) 10 + Ib + Ic + Ix = - Id A Then we may write the primitive equation Voo '
Vo - Vo '
Zoo
Zob
Zoc
Zox
Zad
10
Vb b ,
Zbo
Zbx
Zb d
Ib
Zea
Zb b Ze b
Zb c
Vee '
Vb - Vb ' Ve - Vc '
Zce
Zcx
Ze d
Ie
Vxx '
Vx - Vx '
zx o
Zx b
zxc
zxx
Zx d
Vdd ,
0
Vd�
Zd o
Zd b
Zdc
Zd x
Zdd
Ix
-
=
x
1fT . IT.:' -
Vx
� !slIt, Ic
-
(4.149)
Id
'f..
x'
too
'f
V /unit length
bb
icc
_Vb _Vc
i dd
t--- ONE UNIT
------I-I
Fig. 4 . 1 5 . Four conductors with earth return.
d'
1 07
Seq u e n ce I m pedance of T ransm ission Li nes
matrix are all defined by where the elements of the primitive impedance
(
zpp = rp + j wk In p = a, b, c, X and
Zp q
= j wk
�:p
)
- 1 n /unit length (4. 1 5 0)
( �: ) In
q
- 1 n /unit length
p, q = a, b, c, x, d
(4. 1 5 1 )
p =i= q
Id , the last equation may be sub Since the line currents sum to the negative of ions tracted from the other four to write the line equat Va Vb Ve Vx
=
Z aa
Zab
Zae
Zax
Z ba
Zb b
Zb e
Z bx
Z ea
Zeb
Zee
Z ex
Zxa
Zx b
Zx e
Z xx
Ia Ib
Ie Ix
V /unit length (4.15 2)
primitive impedances , viz., where the matrix elements are defined in terms o f the Zp q = Zp q - Zpd - Zdq + Zdd n /unit length
(4. 1 5 3)
p , q = a , b , c, x
wire a such that their N ow suppose that wire x is connected in parallel with find that voltage drops are identical or Vxx ' = Vaa ' from which we
(4. 1 54 )
equation in ( 4. 152) b y a new T o make use o f this property , we replace the Vx es equation computed from (4 . 1 54). Then ( 4.152) becom
[fl
Z xa - zaa )
(Zx b - Zab )
(zx e - Zae )
(Zxx - Zax )
Ix
(4.15 5)
phase a compo site o r Since wires x and a are in paralle l, they form a new the new phase a define e w where "bundled" conductor a s indicated in Figure 4 .16
TE (COMP CONDOSIUCTOR
"a"
C} .)1 7 7 7 7 7 7 7 7
/ 7 7 7 7 7 7 7 7
a composite conduc tor. Fig. 4 . 16. Bundlin g of a and x to form the new phase
Cha pter
1 08
current to be
4
fa = Ia + Ix
( 4 . 1 56 )
A
Now a zIx product i n each o f the a-b-c equations may b e added and subtracted, leaving the equations unchanged . This amounts to replacing fa in ( 4 . 1 55 ) by ( 4 . 1 56) and replacing the fourth column of the impedance matrix by the dif ference between column 4 and column 1. The result is
Z aa
Va Vb Vc
Z ba =
Z ea
I
Zac
Zab
Zb b Z cb
I
Z bc
I
Zee
t
I I
I
- - - - - - - -- - - - - - - - - - - - - - - - -
0
(Z xa - Zaa )
(Z xe - Z ae )
(Zx b - Zab )
(Z ax - Z aa ) (Z bx - Z ba ) (Z ex - Z ea ) - - - - - - - ·
I
Zxx
where zxx = Zxx - Zax - Zxa + Zaa ' Writing ( 4 . 1 57 ) in partitioned matrix form we have
[Vabcl =
J
o
Iz ,
z 21
lz3 Z4J
I
Ia + Ix fb Ie - - -Ix -
(4.157)
[Iabc] (4. 158)
Ix
We then compute by matrix reduction ,
Vabc = ( z ,
- Z 2 z:; ' Z 3 ) la b e
(4.1 59)
The effect of adding the one wire x to phase a is to greatly increase the GMR of phase a . This reduces the impedance of phase a but also reduces all other self and mutual impedances. The amount of the reduction is given by the matrix z:;' Z 3 each term of which, for this simple case, may be computed from the formula
Zl
( Z 2 Z4
-1
Z3 )
(Z Xq - Zaq ) _ - (zpx - zpa )
Zx x - Za x - Zxa + ' zaa p, q = a, b, c pq
( 4. 160 )
This same idea can be extended to any number of added wires which may then be paralleled with any phase. These additions may be made either one at a time or simUltaneously. Of particular interest is the case where three wires are added to the a-b-c configuration with one wire to be added to each phase. The circuit is shown in Figure 4.17 where wires r, s, and t are bundled with wires a, b , and c respectively. Before bundling, we have the general voltage equation similar to ( 4 . 1 5 2 ) but expanded to include all six wires. o·
"r b'
Fig\ 4 . 1 7 .
Bundled conductors in all three phases.
1 09
Sequence I m peda n ce of T ra n s m ission li nes
Va Vb
Ve Vr Vs Vt
Zaa Z ba
Zb b
Z ra Zs a
Z rb
Z ea
Z ta
Z eb Z sb Z tb
Zar
Zas
Zee
Z er
Zes
Z re
Z rr
Z te
Zsr Z tr
Z ae
Zab
Zb e Zse
Zb r
Zat
Zbs
Zb t
z rs
Z rt
Zss Z ts
Ze t
Ia Ib Ie Ir
V
/unit length
Is It
Zst Ztt
(4.16 1 )
After bundling, the constraining equations may be written, (4.162)
and we define (4.163 )
Then by the technique used before we alter ( 4 . 1 6 1 ) to write Z..
Zob
Zoe
Zbo
Zbb
Zbe
Ze.
Zcb
Zcc
I (zo , - Z.o ) ; (Zb ' - Zbo )
- - - - - - - - - - - - - - -- - - - -- - - - -
(z,. - Zoo )
(z,. - Z bG )
(ZtG - ZCG )
(Z,b - Zob )
(z,c - Zoc )
(Z,b - Z b b )
(z" - Zbc )
(Ztb - Zcb )
(Ztc - Zee )
: (Z., -
:
�
Zco )
(z., - Z.b )
(Z. t - Z.c )
(Zb' - Z b b )
(Zb t - Zbe )
(Zu - Zcb )
(Zc t - Zcc )
- - - - - - - - - - - - - - - - - - - - - - -
i"
I
: :
i ,t
i"
i"
i ..
i,t
i t,
i ll
i lt
' I".
ib
ic I,
I, It
v /unit
length
(4.164 )
where all elements in the lower right position, labeled may be written from the formula z,
Zpq = Zp q - Z iq - Zp h i, h = a,
b,
c
+ Zih
(4.165) Then, following the partitioning of (4.1 64) we apply (4.158) to find the new impedance matrix from (4.159) or p , q = r, s, t
(4.166 )
Here we must invert a 3 X 3 matrix Z4 , whereas this partition was a complex scalar in the previous case. The above technique will permit the computation of the a-b-c impedance matrix of a bundled conductor line where each phase consists of two-wire bundles. This same result could be obtained by applying (4.158) with appropriate subscripts and adding the second wire to each phase one at a time. The amount of work involved is the same in either case. Once the impedance matrix for the bundled conductor line is known, the sequence impedances are computed by similarity transformation. (4.167) A-I Znew A n /unit length ZO l 2 =
Compute the impedance matrices Zab and Z O l2 for the bundled conductor 345 kV line shown in Figure 4.18. There aree two ACSR conductors in each phase bundle spaced 18 inches apart as shown in the insert of Figure 4.18. Ignore the
Examp le 4 . 8
1 10
Chapter 4
--rt-----
u
Steel Ground ,- Wires
R z 4 n / mi GM R ,. 0. 001 '
I S'
�
a
�
a'
o
12'
12'
b
bls
Phase Wires : 795000 CM 2617 ACS R
w
b'
-+- 1 2 '
Fig, 4 . 18 . Conductor arrangement for a 345 kV bundled·conductor line.
ground wires for this example. Let the line length be 40 miles so that results can be compared with previous computations. Solution From Appendix B we find the following wire data for 795000 CM ACS R :
r = 0 .1 1 7 n /mi at 6 0 H z (small currents) Ds = 0 .0375 ft
From (4.161 ) , using a ' , b' , and c' for r ,
Va Vb
Ve Va ,
Zaa Z ba
Zab Zb b
Zae Zb e
Za 'a
Za 'b
Za'e
Z e 'b
Ze 'e
Z ea
Vb ' Ve ,
Z b 'a Ze'a
Z eb
Z b 'b
s,
Zaa' Z ba'
Zea' Za ,a ,
Zee
Z b 'a' Ze "a
Z b 'e
and t, we write
Z ab '
Zae'
Z eb'
Zee' Z a "e
Zb b ' Za'b '
Z b 'b '
Ze'b '
Zb e'
Z b 'e' Ze 'e ,
Ia Ib
Ie Ia ,
n /mi
Ib '
Ie ,
But, in our problem, the self impedances are, from (4.54) and ( 4.165)
Z jj
= Zpp = ( ra
p = a, b ,
and since
ra Dsa
= rb
= Dsb
=
+ rd ) + j O . 1 2 1 34 In
c, a' , b' , c' re = ra '
= Dse
=
= rb '
= rc ,
Dsa' = Dab '
�esp
n /mi
= 0 . 1 1 7 n /mi = Dsc' = 0 .0375 ft
we conclude that the diagonal terms of the z matrix are all equal. Furthermore, we may write, for any i and h or p and q , from (4.55) and (4.165)
Zi h = Zp q = rd
+
j O . 1 21 34 In
gep q
n /mi
111
Sequence I m pedance of T ransm ission Li nes
,' p q and the off-diagonal terms are symmetric. It is helpful to write the voltage matrix equation in partitioned form, taking advantage of symmetry in the way we write the matrix. Thus we have
p, q = a, a' , b, c,
b' c
#:
o.!'' �' _], fY'V!!�..c! j [ 'o b 'c,� = [�'ztm- -�-� Z I , . obc where each partition of z is 3 3. If the bundled conductor were made of two unlike conductors, theobctwo z. matrices would be different. From (4.164) we I
X
-
write
_ _
(0.0953 Zm = [(0.0953 (0.0953
+ + +
+
+
+
+
+
+
]
jO.9135) (0.0953 + jO.5697) (0.0953 + j0.4892) jO.5849) (0.0953 + jO.9135) (0.0953 + jO.5697) n/mi j0.4968) (0.0953 + jO.5849) (0.0953 + jO.9135) From these matrices the new matrices (zm - z,) and (z� - z.) can be easily com puted. Then we compute ZIt = (z. - Zm ) - (z� - zs) - 0.2340 - jO.8 952) (0.0000 jO.0005) (0.0000 jO.0001) = - (0.0000 (0.0000 + jO.0005) (-0.2340 - jO.8952) (0.0000 jO.0005) n/mi + jO.0001) (0.0000 + jO.0005) (- 0.2340 - jO.8952) Inverting, we compute + + +
�
+
+
+
]
�
- 0.2733 + j1 .0456) (- 0.0003 + jO .0005) (- 0 .000 1 + jO .000 1 ) (- 0.0003 + jO .0005 ) (- 0.2733 + j 1 .0456) (- 0 .0003 + jO.0005) (- 0.0001 + jO .0001 ) (- 0.0003 + jO.0005) (- 0 .2733 + j 1 .0456)
Z�I = -
To complete the matrix reduction we compute
ZN = (zm 0.0585 jO.2239) = �(0.0001 0.0000 - jO.0001) - jO.0001) - zs ) Zk l ( z� - zs )
+
(
-
Finally then, from (4.166)
mho · mi
�
(- 0.0000 - jO.0001) (0.0001 - jO.0001) (0.0585 + jO.2239) (- 0.0000 - jO.0001) n /mi (-0.0000 - jO.0001) (0.0585 + jO.2239)
�(0.0953 0.1538 + j1.1372) (0.0953 jO.5772) (0.0952 j0.4930)� jO.5772) (0.1538 j1.1372) (0.0953 jO.5772) n /mi +
=
J
+ +
+ +
+ + +
(0.0953 j0.4930) (0.0953 jO.5772) (0.1538 j1.1372)
1 12
Chapter 4
�
From this result we readily compute
Zo n
=
and
You
�
( 0.0675 - jO.4375 )
=
�
0.3444 + j 2.2354) (0 .0243 - jO.0140 ) (- 0 .0243 - jO.0140) (- 0 .0243 - jO.0140) (0 .0585 + jO .5881 ) (- 0 .0486 + jO.028 1 ) ( 0 .0243 - jO.0140) (0.0486 + jO.0280 ) (0.0585 + jO.588 1 )
�
(0.0224 - jO.0060 ) (- 0.0164 - jO.0164)
( 0.0164 - jO.0164) (0.1725 - j1 .6996) (- 0.1 548 + jO.0517 ) -
(0 .0224 - jO .0060)
�
n /mi
(0.1222 + jO.l082) (0.1725 - j1 .6996)
mho ' mi
Then the total impedance for the 40 miles of line is
ZOl l =
J
1 3 .775 + j89 .4 1 7 ) (0 .970 - jO .561 ) (- 0.971 - jO.560) (- 0.970 - jO.560) ( 2 . 3409 + j23.523) (- 1 .943 + j 1 .122) ( 0.970 - jO.561 ) ( 1 .943 + j 1 .1 21 ) ( 2 . 3409 + j23.523)
We may also compute the unbalance factors two ways.
rno
= �
m
Yl1
=
_
Zoo
'l =
Yl1
�
0 .0224 - 0 .0060 0.1725 - J 1 .6996
YOI ZO I = _ Y'l l =
=
0.01 36 / 6 9 .2°
0.0243 - jO.0140 0.3444 + j 2 .2354
�
n
= 0 . 012 4 �
0.122 2 + 0 . 1 082 0 .096 / 1 2 5 .7° 0 .1 7 25 - J 1 .6996 0 .0486 + jO.0280 0 .095 / 125 .6° 0.0585 + j O . 5881 =
=
It is interesting to compare these results with those of the previous examples. First, comparing against the of the untransposed line of Example 4.4, we see that the off-diagonal terms are almost exactly the same for the two lines even though the conductor size and spacing are greatly different for the two cases. From this we observe that the coupling between sequences is not critically de pendent on phase spacing, bundling, or wire size. The diagonal terms Zoo , Z l 1 and Z'll are reduced, both in the real and imaginary parts, by bundling. The unbalance factors may be compared against those of a similar untrans posed line in Example 4.7. Since the denominator terms in the unbalance equa tions, o 0 , , and Y I I are all reduced by bundling, we would expect the unbalance factors to increase.
ZOIl
Z Z'l 2
4. 1 0 Sequence I m pedance of lines with One G round Wire
In many physical transmission lines, wires are added above the phase wires to "shield" the line against direct lightning strokes. Determining the location of such wires is beyond the scope of our concern . We shall concentrate on the effect such ground wires have on the line impedances (also see [ 10 , 1 1 , and 24] ). Consider the line arrangement shown in Figure 4.19 where one ground wire, labeled w, is shown with each end connected solidly to the local ground point. Clearly, the voltage equation of this arrangement is exactly the same as that of Figure 4 . 1 5 given by equation ( 4 .149) except that in this case VoX = Vw = O. Thus
1 13
Sequence I m peda n ce of T ra ns m ission Li nes
�
� Zoo
a
I
b
. .1 :r: Va
Vb
d
+
l ob
i bb
.1£.
i cc
Iw
z ww
1;
Idd
Vc
a
�
,
I bc
l ad
d'
----·-11
ONE UNIT
Fig. 4 . 19 . A three-phase line with one ground wire.
we write the primitive equation
Voo ' Vb b , Vee ' Vw w ' Vdd ,
Vo - Yo' Vb - Vb ' Ve - Ve ' 0 - Vw , 0 - Vd ,
zoo Zbo Zeo zw o Zdo
zoe Zbe Zee zw e Zd e
Zob Zb b Ze b Zw b Zd b
Zow Zb w Zew zw w Zd w
Zod Zb d Zed zw d Zdd
10 Ib Ie Iw Id
V /unit length (4.168)
Now with ground wire w in parallel with the earth conductor d, the return cur rent will divide between the two paths or 10 + Ib + Ie = - (/d + Iw ) ' Rearranging, we write
Id = - (/0 + Ib + Ie + Iw )
(4 .169)
Using this result for Id in (4.168), we subtract the Vdd , equation from each of the others to compute
Vo Vb Ve
=0
- - - - -
Vw
Zoo Z ba Zea
Zab Zb b Ze b
Zae II Zaw Z b e II Z b w Zee II Zew
- - - - - - -- - - -
Zw b
zw a
Where as before ,
Zp q p,
zw e
-+ - - - I
I
zw w
la Ib Ie
V /unit length
Iw
= Zp q ZP d - Zdq + Zdd
( 4 . 1 70)
-
q = a, b, c, w
(4. 1 7 1 )
Note that Zp q is defined to include ra or r w when p = q but is purely imaginary when p "* q . From ( 4 . 45)-( 4.52) we compute
Zp q = ( ra + rd ) + j wk ln
= rd + j. W k In
De
pq D
'
�peq ,
p=q
p "* q il /unit length
(4.172)
Chapter 4
1 14
V
w
Since
r
=
0, (4.170) may be reduced immediately to the form
= (Z l
Vabe
l - Z2 z;j Z 3 ) labe
][] [
= Zabe la b e
(4.1 7 3)
where the Z partitions are defined in (4.1 58 ) . Performing the operation indicated in (4.173) , we have Zabe =
oo
Zab
Zoo
Zba
Zbb
Zbe
Zea
Zeb
Zee
a
=b C
( ( �
zo w
_
) ( ) ( ) �
a
Zew
b
Zaw Zw a Zw w
Zab -
zb a -
Zb w Zw a zw w
Zb b -
Zae
Zew Zw a zw w
Zeb
Zaa
_
_
�:J
Zbw
_
[zwa Zw b zw e ]
) ( ) � ) �
Zaw Zw b
Zae
_
Zb W ZW b Zw w
Zbe
_
Zew Zw b Zw w
Zee
_
Zw w
) ) )
c Zaw Zw e Zw w
Zb w Zw e Zw w
n /unit length
Zew Zw e Zw w
or each element of the reduced matrix is of the form
p , q (row, col) = a , b , c
(4.174)
(4.175)
Each element of the matrix is smaller by a correction factor involving the mutual impedances to the ground wire w . In most lines of interest we may assume that the three phase wires have equal self impedances or Zaa = Z b b = Zee . This makes the leading terms on the diagonals equal, but the subtracted corrections are still unequal since the mutuals from w to a , b , an d c are usually unequal. 4. 1 0. 1
Sequence impedance of an untransposed line with one ground wire
To compute the sequence impedances, we again use equations ( 2.46)-(2.48). Since the matrix (4.174) is symmetric but with unequal mutual terms, the sequences will, from (2.46), be coupled by nonreciprocal terms. The compu tation is straightforward, however, and involves only the application of (2.46 ) (2.48 ) or ZOl 2 = A- I Za be A. This result is too laborious to write out in detail. An example will illustrate the procedure.
Example
4. 9
Compute the sequence impedance of the 40-mile line of Example 4 . 1 , taking into account the presence of the ground wire. Assume that the ground wire has a resistance of 4 ohm/mile and a self GMD of feet. There are no transpositions.
10""3
Solution We are given that rw
= 4.0
U /mi,
Dsw
= Dw w
==
10-3 ft
115
Sequence I m pedance of T ra n s m ission Li nes
Then the self impedance term is
Zww rw rd =
+
= 4 .095
+
+
DDwwe =
j wk In
4.0
+
27 90 . 13 ) In 0 .09528 + J(0.12 0 .001
j 1 .800 n /mi
= 16 3 . 80 j 7 2 .00 n . and for 40 miles of line To compute the mutuals , we need the distances
+ Zww DDbaw De w = w Zew = rd �aew
2 2 V 1 0 + 1 5 = 1 8.03 ft
=
= 1 5 .0 ft
Then Za w =
= 0.09528 + j(0 . 1 2 1 3 ) In
+ j wk In
= 0 .09528
+
jO.612
= 0 .09528
+
j ( 0 . 1 2 1 3 ) In
n /mi
and
Zb w
:���
���� = 0 .09528 + jO.634
n / mi
The total mutual impedances for the 40 mile line are
Z w Zew = Zbw =
a
�
3.81 2 + j 24.470 n
= 3.812 + j 25 . 3 6 3
n
In matrix form
Zobcw =
1 4'932 + j58.37 6) ( 3.81 2 + i27. 332) �3.:f!.1� � !�.:�6�l ( 3.81 2 + j 24.470)
J
( 3.81 2 + j27.332) (3.812 + j23.968) : ( 3.812 + j24 . 470) ( 1 4.932 + j58. 376) (3.812 + j27.332) 1 ( 3.812 + j25.363) (��.!:� ���.��21 _ J!.����:A�!.�!�U (�.!��j!��7�L_ ( 3.81 2 + j25.363) (3.812 + j24.470) 1< 163.81 2 + j72.036U
_ _
�
Reducing along indicated partitioning, we compute A
Za be
=
++ +
�
++
Then by digital computer we find the sequence impedances
Z0 12 =
J
1 7 .500 j 56 . 1 07 ) (6.484 + j 24 .996) (6 .380 + j21 .698 ) (6 .484 j 24.996 ) ( 1 7 . 7 1 2 j55 .972 ) (6 .484 + j24.996 ) (6 .380 j 2 1 .698) (6.484 j 24 .996) ( 1 7 . 500 + j56.107 )
( 30.470 + j l0 3 . 8 5 ) (0 .860 - jO.618 ) (- 0.966 - jO.436 ) 0 .966 j0 .436 ) ( 1 1 .1 2 1 + j 32 .1 6 5 ) (- 1 .944 + j 1 .121 ) ((0.860 - jO .619 ) ( 1 .943 j 1 . 1 23 ) ( 1 1 .121 + j32.165 )
+
The unbalance factor may be easily computed as Z l _ mo = - O = Z oo
m2
= - Z21 = Z 22
_
0.860 - � 0.6 18 30.470 + J 103.85
�
1 .943 + 1 . 123 1 1 . 12 1 + J 3 2 . 165
= 0 . 009 78 17 0 . 650
= 0.06 5 9 1 1 3 9 . 1 10
J
n
n
n
1 16
Chapter 4
Comparing this result with that of Example 4.4, we observe the following:
1 . The positive and negative self impedances Z 1 1 and Z22 are unchanged by the addition of the ground wire. 2. The zero sequence impedance is changed in a remarkable way. viz. , (a) the real part R oo is increased by about 40% and (b) the imaginary part Xoo has de creased by about 7%. 3 . The off-diagonal terms are changed but not by a significant amount. 4. The zero sequence unbalance has been increased by the addition of the ground wire but the negative sequence unbalance is unchanged.
Example
4. 1 0
Repeat the computations of Z 0 1 2 made in Example 4.9 but with a ground wire of higher conductivity. Compare the results with those of Example 4.9.
Solution For this example let us choose two different ground wires as follows : 1 . 3 /8-inch copperweld, 40% conductivity
2.
D. rw
=
Zw w
=
=
0.00497 ft 1 .264 51 /mi 1 .359
+
jO.12134 In
0
1F copperweld-copper
DB rw Zw w
=
����7 = 1 .359 + j 1 .606
51 /mi
0.00980 ft
= 0.705
51 /mi
= 0.8003
+
jO. 1 2 1 34 In
O
.����O
=
0.8003 + j 1 . 524 51 /mi
Then, by digital computer we compute the following sequence impedances for the 40 miles of line. For the line with a 3/8-inch copperweld ground wire the impedance for 40 miles of line is given by Zo n
=
�
J
(- 0.844 - jO.327 ) 31 .207 + j90 . 1 48 ) (0.706 - jO.568 ) ( 0 .844 - jO.327) ( 1 1 . 122 j32.163 ) (- 1 .946 + j 1 .121 ) (0.706 - jO .568 ) ( 1 .944 + j 1 . 1 2 5 ) ( 11 . 122 j32.163 )
+
-
+
51
For the line with a 1F copperweld-copper ground wire the impedance for 40 miles is given by Z0 1 2
=
�
+
J
27 .403 j83.866 ) (0.652 - j0.499 ) (- 0.758 - jO.316) ( 1 1 . 1 2 2 + j32.162) (- 1 .947 j 1 .122) ( 0.758 - jO.316) ( 1 1 . 122 j32.162) ( 1 . 945 + j 1 . 1 2 5 ) (0.652 - j0.499) -
++
51
It is apparent that the zero sequence reactance reduces appreciably as the ground wire impedance goes down. The zero sequence resistance remains high compared to the case with no ground wire. It is also interesting to compare the unbalance factor for the three cases.
Sequence I m peda nce
of T ransm ission
Lines
1 17
Table 4.7. Comparison of Zero Sequence Unbalance, m o , in Percent, as a Function of Ground Wire Selection IF Cw-Cu No Ground Wire 3/B-inch Cw 3/B- inch Steel 1. 054 0.978 0.950 0.931 I mo l 72.680 angle 70.650 70.67 0 7 0.280 Obviously the negative sequence unbalance is unchanged (Why? ) . sequence unbalance is tabulated for comparison in Table 4.7.
The zero
Example 4. 1 1 Repeat the computation made · in Example 4.9 except consider the configura tion of Figure 4 .20 where the ground wire is located at the same height as before but 5 feet to the right of center.
0.375"
GROUND WI R E STEEL
o
4/0
0
PHASE WI RES COPPER
°1-10· -+,5' Fig.
4.20.
Line configuration for Example 4. 1 1 .
Solution
W e proceed exactly as i n Example 4 . 1 0 , the only difference being in the dis w to a , b , and c. This makes the reactance terms different than before by a very small amount in the fourth row and column. Thus we compute
�
tance from
Zabcw
=
14'931 + j58.37 5)
(3.812 (3.81 2
+
+
j27 .331 ) j23 .966)
( 3.812 + j27 .331 )
( 1 4 .931 + j58 .375) ( 3. 8 12 + j27.33 1 )
(3 . 812 + j23.966) 1 (3.812 + j 2 3 . 680 ) ( 3.812 + j27 .331 ) : ( 3.812 + j25.107) ( 14.931 + j58.375) 1 ( 3 8 1 2 + j25.107)
:r
.
(i.8;i;j2i.680) - - (3�12�j25�07) - - (3.812;j25�107) �63�i2�j72.036)
�
Reducing to find
Zabc =
Zabc ,
we compute
J
1 7 _319 + j56.223) (6.360 + j 2 5 .075) (6 .247 + j 2 1 .783) ( 6 . 360 + j25.07 5 ) ( 1 7 _651 + j56.0l l ) ( 6 .41 1 + j 2 5 .043) ( 6.247 + j 2 1 .783) (6.41 1 + j 2 5 .043) ( 1 7 .651 + j 56.01 1 )
and converting to the 0-1-2 frame of reference we have
Z012
]
K 30 .374 + j 10 3 .92) = \ (- 1 .1 39 - j0.454) � 0.804 - jO .454)
( 0.804 - j O .454) (- 1 .1 39 - jO.454) ( 1 1 . 1 23 + j32 . 1 6 5 ) (- 1 .939 + j1 .120 ) ( 1 .946 + j 1 .120) ( 1 1 .123 + j32.164 )
J
The unbalance factors for this case are
mo = -
Z OI
Zoo
=_
0 .804 - jO.454 30.374 + j103.92
0 .00853 176.840
n
n
n
1 18
Chapter
m2 = -
ZZ l ZZ 2
==
-
4
1 .946 + j 1 . 1 20 1 1 . 123 + j32 . 164
= 0 . 0 6 5 9 113 8.9 9
0
Comparing results, we note that all terms of the Zob c matrix have been changed by moving the ground wire, with the row and column for wire a seeing the greatest change. In the Z0 12 matrix the zero sequence row and column is af fected. The zero sequence unbalance is reduced slightly, which may come as a surprise. 4. 1 0.2 Current division between the ground wire and the earth
It is apparent that the addition of a ground wire to an untransposed line section reduces the zero sequence reactance Xoo and increases the zero sequence resistance roo . This is due in part to the relatively high impedance of the ground wire compared to the earth and in part to the fact that the ground wire is phys ically nearer the phase wires than the ground return path. Because the ground wire is near the phase wires, the current induced in this wire tends to be large and if the wire has a large resistance, this makes roo appear to be great. The closeness of this coupling, however, reduces the inductance about in proportion to the current lw • The proportion of the total zero sequence current which flows in the ground wire may be determined from equation (4.168). If3/0we0 set 10 = lb = lc = /oo
(4.176)
0 = [ (zw o + Zw b + zw c ) - (Zda + Zd b + Zdc ) ] la o + (zw w - Zd w ) lw + (Zw d - Zdd ) /d
(4. 1 7 7 )
in (4.168) and subtract the d equation from the w equation Also,
from (4.169) and (4.175)
Id = -
(3/0 0
+ Iw )
If we substitute (4.1 78) into ( 4 1 77), we compute
(4.178)
.
_
lw
310 0
__
) - _-=... + Zd 3Zw d:o. - (Zdo +--, (zw o + zw b + zw, c ) _.!-"'-,,,- Zdb ,,-,<--_ -=..o:.c.:-..: +_ .- 3Zdd -= = �"-"-_=-",----,� 3 (Zww - Zwd - Zd w + Zdd )
(4.179)
The primitive impedances in this expression may be combined according to (4. 1 7 1 ) to write (4.180)
Also, from (4.178), we compute _
�=13/0 0
Z w o + z w b + zw c 3 zw w
(4.181 )
The two foregoing expressions give the ratio of currents in the ground wire and earth. respectively to the total zero sequence current. From (4.180) and (4.181 ) we compute the ratio of ground wire to earth current.
1 19
Sequ e n ce I m peda n ce of T ransm ission Li nes
Suppose we define
Zag � ( 1/ 3 ) (Z w a + Z w b + Z w c ) = rd + j wk In where
�eag
Dag = (Dw a Dw b Dw c )1/3
(4.182)
(4.183)
is the GMD between the ground wire and the phase wires. Then
Iw = Zag Id Zw w - Zag
(4.184 )
Obviously as Z w w becomes smaller, the fraction of the total current flowing in the ground wire becomes large. Note that it is not possible for the denominator to go to zero. We have defined Z w w as
Therefore
Da Z w w - Zag = rw + j wk ln g > 0 Dw w for any conceivable situation . Obviously Iw can be greater than Id • This occurs when the following con ditions are met. If we compute the ratios of the phasor currents, we have
( 4.185 ) then it is apparent that Iw is greater than Id when
I z w w - zag l < I zag i
(4.186)
Since the currents in (4 . 1 85 ) are both phasors, it may be more meaningful to take the absolute value of these expressions. 4. 1 0.3 Sequence impedance of a transposed line with one ground wire
The mutual coupling between sequences may be reduced and even eliminated by transposing the line . Suppose that the line is transposed according to the rotation sequence in Figure 4.9. Let the three wire positions be identified as a , (3, and 'Y such that the arrangement in Table 4.8 i s observed. Then the impedance of each transposition section may be determined from ( 4.174 ) with the result shown in Table 4 .9 . The total impedance in ohms per unit length is always the sum of the three matrices of Table 4.9 or
(4.187 )
Chapter 4
1 20
Table 4.8.
Conductor Arrangement in Transposition Sections
Section
Length
1
fl s f2s f3 s
2 3 Table 4.9.
Section
Z, l -
&
(
( ( ( ( ( ( ( (
Z'Y 'Y
S3
Z,3 -
zw w
Z OI 'Y
ZQ j1
ZI!WZ W Ol zw w
Zj1 'Y -
Z'Y w Z w l! Zw w
Z'Y'Y
_ ZQ W Z W I! •
Zw w
ZOI'Y
_ Z'Y w Z W 'Y Zw w
_ Z OI WZW Zw w
Z(I'Y -
ZI!W Z w Zw w
ZQ W Z W I!
Z'Y OI
ZOIOI
zQ 'Y
zw w
Z'Y j1 -
_ Zl!w Z W I! zw w
_
zj1j1 -
ZW Q _ Z'Y w . zw w
Z'Y j1 . ZOI j1
) ( ) � ( ) ) (( �)I!) ( ) ( ) ( ( ) ( ) ( ) ( ) ( ) ( ) ( �) ( 'Y) (�flOl ) (
_ ZQ W Z W Q
Zj101 -
Zj1j1
Zf2 -
a-b-c c-a-b b-c-a
Im edance Matrix ( /unit length )
z'YOI
$2
Phasing
Impedances of Transposition Sections
ZOIOI
$1
0:-(3-1 0:-(3-1 0:-(3-1
Position
Zj1 'Y
zw w
Z'Y w Z W
z'Y 'Y
zw w
zl! w z w l zw w
Z'Y OI
Zw w
_ ZOI W Z W 'Y
ZOIOI
Zw w
Z'Y w Z W Q
Z'Y j1
Zw w
_ ZOI WZ W OI
ZOI(I
? I# W
-
zw w zw w
�w w
'Y)
) ) ) ) ) )
_ Z'Yw Z W 'Y zw w
ZI!W ZW OI Zw w
_ Z'Y W Z W OI Zw w
_ Z OIW Z W OI Zw w
- Z'Y w Z W I! Zw w
-
Z OI WZWI! Zw w •
Zl!w Z w Z(l fl Zw w
ZPW Z W I
)
ZQ W Z w 'Y
- ZI! W Z w
zj1 Q -
_ Z'Yw Z W 'Y
_
_
p)
When the line is completely transposed, fl f2 = f3 liS, and some simplified equations for the sequence impedances may be derived. Firlit we observe that the three diagonal terms are equal, thus we define =
21, � Zoo = 7.b b = Zcc
:;:: 1.
( 210101
:;::
z. M
- 1.
(
�w'Y)
/unit length Similarly, all off-diagonal terms are equal such that we may define 21m as 3
=
1 3
-
) + z flj1 + z 'Y'Y
(ZOIj1 + zj1 'Y
+ 21 '1 01 ) -
3
1 3
-
ZOI W ZW OI ZW W
(21
These relations are true since
+
OIW 21 W (3 ZW W
z PQ
z w zw z p + 'Y w p zw w ZW W
+
21(3 W 21 W 'Y ZW W
+
n
) /UnI.t length
21'YW 21W OI Zw w
n
(4 . 1 88 )
( 4. 1 8 9 )
is reciprocal. Then the impedance to currents
Sequence I m pedance of T ransm ission Lines
lo b e may be written in matrix form as
j
zm Zm z,
121
n /unit length (4. 190)
Because of the symmetry of (4. 1 90) the sequence impedances are much simpli fied. From equations ( 2.47) and ( 2.48) we compute
Zs o = Z. , ZM O = Zm ,
ZS I = ZS 2 = 0 ZM I = ZM 2 = 0
Then
Z I I = Z22 = z. - Zm ,
Zoo = z. + 2z m
(4. 191 )
and all the off-diagonal terms in the Z0 1 2 matrix are zero. From ( 4. 1 91 ) the positive sequence impedance is found to be
Z 1 1 = "3 (z�� + z(3(3 + z'Y'Y ) 1
1 "3 (z�(3 + z(3-y + z-y� )
n /unit length ( 4. 1 92) We examine this equation term by term. For the first term we compute
� (z�� + z(3(3 + z'Y'Y ) = (ra + rd ) + j w k In �e
n /unit length
(4.193)
B
where Ds � D �� = D(3(3 = D'Y'Y ' For the second term we compute
� (z�(3 + z(3-y + z-y� ) = rd + j w k In �eq n /unit length (4. 1 94) e The third and fourth terms may be combined to where D e q � (D�(3D(3-yD-y� / )1 3 .
write
1
3--
Zw w
[ (z�w zw � + z(3w zw(3 + z-yw zw-y) - (z�w zw (3 + z(3w zw -y + z-yw zw � )] = -
w 2 k2 M 3
Zww
(4. 195)
where we let
(4. 196 )
(4. 197)
1 22
Z
11
Cha pter 4
In most physical configurations is simply
De » DOI.w , Dpw , D'Yw
so that M
to write
Zl 1
Ii
1
B
Xd = wk In
'
O. Then
(4.198) (4.199 ) (4.200)
For convenience we define the quantities Xa = w k In
�
Deq
= ra + j (xa + xd ) fl /unit length
All quantities needed (ra , Xa , and Xd ) may be determined by the tables of Appen dix B. Note also that equations and are identical to previous re sults for a completely transposed line , the only approximation necessary being that of M S:! O . For the zero sequence impedance we write , from
(4. 1 98) (4 . 1 99) (2.46), Zoo z, + 2 zm (1/3)(zOI.OI. + zpp z'Y'Y) (2/3)(zOI.p zP'Y z')'OI. ) - (1 /3zww ) [(ZOl.W ZWOI. + zpw zwp Z')'W zw')') 2(zOl.w zw(J + zpw zW'Y + z'Yw ZWOl. ) ] (4.201) (4. 1 93) (4.Z1x94). w Zwx 2 C2 2AB 2BC 2CA = (A B + C)2 (4 .202) 0d jwk g;Jr 3Z� w [� + j wk �eJ � j wk �:) 13 w (3 j3wk De j3wk D1 ) 2 (4.203) � � (DOI.w Dpw D')'w 4. 1 0. (4. 201). +
+
+
=
=
+
+
+
The first two terms are known from and last term requires more detailed study . We know that Thus the numerator is of the polynomial form
A + B2 +
or we may write the third term
In
d
=
-
rd +
+
+
The numerator of for x = a , {3, "( .
+
as
+
In
+
=
+
In
d +
+
+
In
In
�
) 1 /3 which is the GMD between the phase
where we define Da, = wires and the ground wire . Now for convenience we define the quantities shown in Table fined quantities may then be used to write all of We define
Table 4. 10. Definitions of Useful Quantities Value in n/mi
Quan tity
3wk - wk wk wk - wk
3rd
IIInnn D,DeDeq In Da, InDww
Designation
e
=
0. 2858 2. 888
These de
Sequence I m pedance of T ra n s m ission Li nes
1 23
ZO(o) � (1 / 3 )(z aa + zfJfJ + zn) + ( 2/ 3 )(zafJ + zfJ-y + z-ya) D3 = ( ro + 3 rd) + j wk In � Ds D e q = (ro + re ) + j (xe + xo - 2 xd ) !1 /uniqength
(4 . 204 )
ZO(m) � [ zaw zwa + zfJw zwfJ + z-yw zw-y + 2 (zaw zwfJ + zfJw z w -y + z-yw Zwa)] 1/2 = 3 rd + j 3 wk In D e = re + j (X e
-
3 xo, )
-
j 3 wk In Do,
( 4.205 )
ZO(6) � 3 zww = ( 3 rw + re ) + j (Xe + 3 x,) !1 /unit length
( 4 . 206)
Using these definitions, we write
Zoo - ZO(o) _
-
2 ZO(m) z 0 (6)
( 4.207 )
!1 / umt length .
This formula agrees with that of [ 14 ] for the case of one ground wire and is a commonly used expression for computing zero sequence impedances. It may be used when the following conditions are true . 1 . The line is completely transposed. 2. The three phase wires are identicaL 3 . Either D e » Da w , DfJ w , D -yw , or Da w � DfJ w � D-yw so that M ( 4.1 9 6) . 4. The phase wires ar e all the same height above the ground, i.e.,
D e
o = D b e = Dee
4. 1 1
�
0 in
� De
Sequence I m pedance of Lines with Two G round Wires
A system of three phase wires and two ground wires is analyzed in exactly the same way as the case for one ground wire. Consider the system shown in Figure 4.21 where ground wires u and w in parallel with the phase wires are solidly 10
a
-
b
-
11:
lao
a' low
Ib
Ie ...!ll..
�
z uu i
: Va : Vb:Ve : Vu - Vw -O
w
lad
l bd _
: Vd -O
Zed
zud z wd
d'
--
I�
Id
dd ONE
UNIT
.1
Fig. 4 . 2 l . A three-phase line with two ground wires.
Chapter 4
1 24
grounded at each end of the line to the local ground point. We write the primitive voltage equations as follows :
Vaa '
Va - Va '
Zaa
Zab
Zac
Zau
Zaw
Zad
Vbb , Vcc
Vb - V,,, Vc - Vc '
Z"a
Zb b
Zbu
Z" W
Zbd
Zca zu a
Zcb
Zbc Zcc
Zcu zu u
Zc w zu w
Zcd Zud
'
Vu u Vww ' '
Vdd '
=
0
Vu ' Vw '
0
Vd '
0
=
zu b Zwb
zwa zda
Zd b
zu c zw c
zwu
Zdc
Zdu
Since the return current divides between paths
d,
zww
Zwd
Zdw
Zdd
U,
and
Ia
Ib Ic
Iu Iw
V /unit length
Id
(4.208)
w, we have
10 + Ib + Ic = - (ld + Iu + lw ) or
Id = - (la + Ib + Ic + lu + lw )
(4.209)
Making this substitution into (4.208) and subtracting Vdd ' from each of the other voltages, we compute
Va
Z oo
Vb Vc
Z bo
-----
Zo b
Zb b
= Zc a
Zc b
za c
Z bc
Vu = 0 Vw = O
Zua
Zu b
Z wa
Z wb
zou
Zaw
la
I
Zb u
Z bw
Zcu
Ib Ie
I I
Z cc
-- - - -- - - - -
I I I
-
Zu e
Z wc
I + I I I I ,
-
Zc w
------
Zu u Zw u
V /unit length
Iu Iw
:::j
(4.210)
where we define Zp q = Zp q - Zp d - Zd q + Zd d ; p , q = a , b , c, U , W exactly as in (4. 1 7 1 ) . W e immediately recognize that this matrix equation may b e reduced t o a third-order system in variables sUbscripted a, b , and c. Calling the resulting im pedance matrix Za b c , we have
[
zoZbecj - [zZbauu Z c Zc u
where we have defined
Yuu Y wu
Yuw
]
Y ww
=
rzu u lzwu
Zu w Z ww
]
Yuw] Yww
-\
=
[ZZua wa
[
Z ww 1 det Z u w - Z wu
n
/unit length
( 4.21 1 )
(4.212)
and where det Z u w = Z u u Z ww - Z u w Z w u ' W e may easily show that any element o f (4. 2 1 1 ) may b e written as
p , q ( row, col) =
a, b,
c
This result is similar to (4.174), the result for one ground wire.
(4.21 3 )
Sequence I m peda n ce of T ra n s m i ssion Unes
4. 1 1 . 1
1 25
Sequence impedance of an untransposed line with two ground wires
If the line is untransposed, we may prepare a table similar to Table 4 . 9 and use appropriate values of fl ' f2 ' and f3 to describe the line sections. In particular, for f2 = f3 = 0 we have a matrix of terms which may be computed from ( 4 .2 1 3 ) with p, q = a , (3 , 'Y where we again use the Greek letters to indicate the wire posi tions. The resulting matrix would be exactly the same as ( 4.211 ) , and a similarity transformation will transform this result to the symmetrical component imped ances. Since this result may not be written in a more compact notation than ( 4.21 1 ) , it will be left this way .
Exa mple
4. 1 2
Consider the bundled conductor line shown in Figure 4.18 of Example 4.8. Compute the impedance of this line , including the effect of the two ground wires, and compare the results with those of Example 4.8. The ground wires are 3/8-inch EBB steel.
Solution The impedance matrix of the five conductors is computed first from (4.210 ), using ( Z new )a b c as defined in Example 4.8 as the three equivalent single-conductor phase conductors. b
a
a
b
Z. b c u w
=
c
u w
u
c
w
�
(0.1 54 + j1.l37) (0.095 + jO.57 7 ) (0.095 + j0.493) : (0.095 + jO.604) (0.095 + jO'518 (0.095 + jO.577) ( 0 . 1 54 + j 1 .l37) (0.095 + jO.577 ) : (0.095 + jO.604) (0.095 + jO.604) (0.095 + j0.493) (0.095 + jO.57 7 ) (0.154 + j 1 . l 3 7 ) : (0.095 + jO.518) (0.095 + jO.604) - - _ J...
_ _ _ _ _ _ _ _ _ ___ _ _ _____ __ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ __ ____ __ __ _ _ _ _ _ _
.
n /ml
(0.095 + jO.604) (0.095 + jO .604) (0.095 + jO.518) : (4.095 + j1 .801) (0.095 + jO.57 7 ) (0.095 + jO.518) (0.095 + jO .604 ) (0.095 + jO.604 ) I (0.095 + jO.577) (4.095 + j1 .801 I
� �
�
J
Reducing along the indicated partitions to reduce to the a-b-c matrix, we have
a
Zab c = b
a 0 . 24 3
+
b
j l .036) (0.192
(0.192 + j0.47 0 ) (0.259
c (0.183
+
+
+
c
j0 .470) (0.183 + jO .392 j 1 .023)
(0.192
jO.392) (0.192 + j0.470)
+
j0.470) Q /mi
(0.243 + j 1 .036 )
Finally , transforming and multiplying by 40 miles we have 25.035
+
j 7 6 . 7 3 7 ) (0.582 - jO.7 1 1 )
Z0 1 2 = (- 0 .905 - jO.149 ) ( 0 .581 - jO .709)
(- 0 .906 - j O . 1 49
�
( 2 . 392 + j23.505)
(- 1 .923 + j 1 .139) Q
( 1 .948
( 2 .392 + j 23 .505)
+
j 1 .096)
The unbalance factors are mo
=-Z
0.582 - jO.7 1 1
ZO l oo
=
Z2 m2 = - 1 = Z22
- 25 .03 5 + j76.737 1 . 9 48 + .j 1 .0 9 6
_
2 .392 + J23.505
= 0 .01 14 / 57 .37
0
= 0 09 46 / 125 17 0 .
.
As in the case of one ground wire, the presence of two ground wires increases Roo
1 26
Chapter 4
and reduces Xoo . This results in a slight increase in the zero sequence unbalance factor. 4. 1 1 .2 Sequence impedance of a transposed line with two ground wires
If the line is transposed, the three matrices similar to those of Table 4.9 must be added to find the total impedance Za bc . This result may then be transformed to Z0 1 2 by a similarity transformation. When the line is completely transposed such that '1 = '2 = '3 = 1/3, the result may be reduced to a more compact formula. Since this formula often appears in the literature , it will be derived. For any transposed line we may write
'� [�
'2 '1
'3
and
J
zaa
'2
iJiJ
'1
Zn
n lunit length
(4.214)
(4.215)
where the Greek-subscripted quantities are defined by (4 .21 3 ) with p, q = a , (3, '1 . In the special case where '1 = '2 = '3 = 1 / 3 , the three equations (4.214 ) are all equal so we define
Z. � Zaa = Z b b = Zcc = (1 /3) ( zaa + ZiJiJ + zn)
Then from (4.213) with p,
q = a , (3 , '1
Z. = ( 1 /3 )( zaa + ziJ(1 + zn) - ( 1 /3 K) ZL
(4.216)
K � Zuu Zww - z 3 w = denominator of (4.21 3 )
(4.217 )
where we define
and
ZL � + zau zww zua - Zau Zu w Zwa - zO/w zwu zua + zO/w zuu zwa + ziJu zww zuiJ - ziJu zu w zwiJ - ziJw zwu zuiJ + zPw zu u zwp + z"(u zww zu"( - z"(u zuw zw"( - z"(w zwu zu"( + z"(w zuu zw"(
(4.218)
Moreover the three equations (4.21 5 ) are all equal such that we may define D.
_
Zm = Zab - Zbc - Zca A
A
_
A
= ( 1 / 3)(zaiJ + ZP'Y
_ 1 (
- 3"
ZaiJ + ZP'Y + Z"(a A
A
A
)
+ Z"(a) - ( 1/ 3 K) ZM
(4.219)
where K is defined in (4.217 ) and
ZM � + zau zww zu(1 - zau zu w zwp - zaw zwuzuiJ + zaw zu u zwiJ + ziJu zww zu"( - zpu zuw zw'Y - zpw zwu zu"( + ziJw zu u zw "( + z'Yu zww zua - z'Yu zuw zwa - z-yw zwu zua + z"(w zu u zw a
(4.220)
Sequence I m pedance of T ra nsmission Lines
1 27
Then, by symmetry
Zso = Z . , (4.221 ) and the sequence impedances are
Z I I = Z 22 = Z. - Zm ,
Z oo = Z , + 2 zm
(4.222)
as often noted before . The positive and negative sequence impedance may be written in terms of ( 4.216 ) and ( 4.21 9 ) with the result
Z I l = Z2 2 = (1 / 3 ) (zoa + zfjfj + z'Y'Y) - ( 1 / 3 ) (zafj + zfj-y + z-ya ) (4.223 )
- ( 1 / 3 K)(ZL - ZM ) n lunit length The first two terms are known from (4 . 193 ) and (4.194 ) as follows :
3" 1
(zaa + zfjfj
+
1 z'Y'Y) - 3" (zafj
+
Deq . zfj-y + z-ya) = ra + J wk DB = ra + J (xa + Xd) •
The third term of (4.22 3 ) may be expanded to compute
( 1 / 3 K)(ZL - ZM ) = ( 1 / 3 K )(- zww + zuw + zwu - zuu )(- w 2 k2M ) where M is defined by ( 4 . 1 96). In most overhead lines this term may be neglected such that we write
(4. 224) This is the same result as previously derived for the case of one ground wire in ( 4. 200) and for no ground wires in (4.120). The zero sequence impedance, also defined by ( 4.222), may also be rearranged into a familiar form. From (4.216 ) and (4. 219 ) we compute
Zo o = z. + 2zm = ( 1 / 3) ( zaa + zfjfj + z'Y'Y ) + ( 2 / 3) ( zafj + zfj-y + z-y a ) - ( 1 / 3K) ( Z L + 2zM ) n lunit length
(4. 225)
The first two terms are known from (4. 204) and are identified as the quantity ZO(a )' The third term requires more detailed study. We may use (4 . 202) to compute 2 2 ZL + 2 ZM = Zuu (zaw + zfjw + Z-yw ) + Zw w ( zau + zfju + Z-yu ) - 2 zuw (zau + zfju + z-y u ) (zaw + zfjw + z -y w) (4. 226 ) If the two ground wires are symmetrically located with respect to the three phase wires, as is often the case, then (zau + zfju + z-y u ) = (zaw + zfjw + z-y w ) and (4. 226) becomes ZL + 2 ZM = ( zuu + Zw w - 2 zuw) (zaw + zfjw + Z-y w } 2 . Moreover, if the ground wires are identical,
(4. 2 27) (4. 228) We also evaluate 3K as 3K � 3 ( zuu z w w - z! w ). But with identical ground wires ( 4. 227) applies, or
3K = 3 (z� w - z!w } = 3 (zw w + zuw } (z w w - zu w )
Chapter
1 28
4
Then the third term of ( 4. 225) may be written as
--
1 (zaw + z@w + Z:YW ) 2 6 Z �(m ) (Z L + 2 ZM ) ( 3/2) (zww + zuw ) 3K ZO (6 ) _
Then
(4. 229)
(4.230)
(
)
which is exactly the same as before. The remaining impedance Z O(6) is defined as
Z O(,) = = where we define
3 "2 (Zw w + Zu w ) = "2 Tw
(�
Tw
) (
3
+ Te + j X e +
Xd , = w k In Thus if we write
Z oo = ZO(o) -
�
1
D uw
+ 3 Td
' 3 + J "2 w k In
x,
Xd '
-� )
D� D w w Du w
n /unit length
n /unit length
2 ZO m ) n /unit length ( ZO(,)
(4.231)
(4.232)
(4.233)
the only difference between one ground wire and two ground wires is in the term as follows :
Z O(,)
ZO(, ) = 3 rw + Te + j (x e + 3x, ) one ground wire = ( 3/2)rw + re + j [xe + ( 3/2)x, - (3/2)Xd, ] two ground wires
(4. 234)
Reviewing the above derivation, we note the following restrictions of the use of (4. 233) in the computation of the zero sequence impedance where two ground wires are present.
1. The line is completely transposed, i.e., fl = f2 = f3 = 1 / 3. 2. The two ground wires are symmetrically located, or Dau D@u D'Yu = Daw D@w D'Yw 3. The two ground wires are identical. 4. The three phase wires are identical and at the same height above the ground. 4. 1 2
Sequence I mpedance of Lines with
n
G round W i res
Lines are seldom constructed with more than two ground wires, but the computation of line impedances may be extended to any number of ground wires by the methods used in the previous sections. The most general case is that in which n ground wires are used, where n is any integer. This problem is of aca demic interest only but will be examined briefly. If there are n ground wires, the primitive impedance matrix will be of order 3 + n + 1 and the earth conductor may be eliminated to reduce the system to
3
Sequence I m pedance of T ra n s m i ssion Li nes
1 29
order + n. Then we may reduce the impedance matrix to the usual third-order form by the equation
Zde = r::: ::: ::� r::: L�eo Zeb Ze:J �el
(4. 235)
where
(4. 236)
is the matrix of ground wire self and mutual impedances and is surely nonsingular. The result of this computation is certainly not obvious, even for a completely transposed line. When the line is completely transposed the result is given in [ 14]
as
3 n 1) ZO(R) = (3rw -;- + reJ\ + J rXe + -;;3 Xo - ( n Xct'J1 n /UnIt. Iength (4. 237) L with Z O(o ) and Z O(m) being defined as before. The proof of (4 . 2 3 7 ) is left as an -
•
exercise.
4. 1 3 Zero Sequence I mpedance of Transposed Lines with Ground Wires
The zero sequence impedance of a completely transposed line with ground wires may be computed by the method of sections 4.10-4.1 2. However, if the line is completely transposed the zero sequence impedance may be computed directly using the GMD method (see Wagner and Evans [ 10] and Stevenson [ 21 ] for a discussion of the GMD method). Since this method is often used, it will be derived below. 4. 1 3. 1
Single-phase circu it with earth return
First. consider the mutual impedance between circuits with common earth re
turn as shown in Figure 4.22. Recall that the mutual inductance between two cira
T� Jo;
I --L- ' ,: Ia + 1 b _ .
"
Fig. 4.22.
CARSON 'S CONDUCTOR
•
... - \. os - 1.0
UNI T
Single-phase circuit with earth return.
cuits is defined in terms of the flux linkage X . Thus if X 1 2 = flux linking circuit 1 due to 12 • then M1 2 = X 1 2 /12 H where (see [ 1 2] )
Chapter 4
1 30
A 12 = 2 X
1 0 - 7 12 In
�:
Wb tum/m
(4.238)
Also, since the system is passive and linear, M = M 12 == M 2 1 In our case the flux linkage of circuit ae due to Ib in conductor b is •
A Oe- b = 2 X 1 0 - 7 Ib
In
Db e
D bo
(4. 239)
Wb tum/m
where we have defined conductor 2 of (4.238) to be b and conductor 1 to be a. Similarly, the flux linkages of circuit ae due to current Ib in conductor e is A oe-e
Adding
as
( 4. 239)
and
-
( 4. 240),
A oe
and if Do e
==
:!! D be :!! VDe
2 X
1 1 0 7 Ib In
D oe
-
(4. 240)
Wb tum/m
we find the total flux linkages of circuit ae due to Ib
= 2 X 1 0 - 7 /b In
Do e D b e D ob
Wb tum/m
(4. 24 1 )
(the conductors are at equal heights), A oe � 2 X 1 0 - 7 /b
In
De Dob
Then the mutual inductance between circuits ae and m o e - be = mob
=
A oe lb
==
2 X 10-7
In
be
is
De Do b
(4. 242)
HIm
or mo b =
In Figure
4. 23
De
k In D
ob
H/unit length
(4. 243)
the circuit is shown with polarizing dots similar to Figure
4.2.
Fig. 4.23. Single-phase circuit with earth return, one unit long.
Current Ib , since it flows in the same direction as circuit ae of Voe- b = ( 1 / 3 ) re 1b + j wm o b 1b = zob 1b
where we have defined the mutual impedance
10 ,
causes a
voltage drop
V lunit length
in
(4. 244)
Sequence I m pedance of T ra n s m i ssion li nes
131
( 1/3 ) re + j w mab n /unit length
(4. 245)
Za b
=
For example, if the unit of length is the mile
Za b = ( 1 . 588 X 10 - 3 f) + H 2 1T k) f In
De n /mi Da b
( 4. 246)
Now consider a one unit length section of the two mutually coupled circuits a and b with earth return, as shown in Figure 4.23. The voltage drop in the direc tion la and Ib in one unit length of line and earth return is
(4.247) If we define
Zaa Zb b Za b
= = =
self impedance of one unit length of circuit ae self impedance of one unit length of circuit be zba == mutual impedance between ae and be for one unit length
then we may write
(4. 248) 4. 1 3. 2
Single circuit with ground wire and earth return
Now modify the circuit of Figure 4. 23 to that of a m ore useful configuration, as shown in Figure 4. 24 where the following changes are noted : the circuit b has
Io
.1
a
\I
f .: �" t ,
+vae +_ vge
_
\\ \\
...!i..
_
1
" I" \\ \ , \\ , , 1 I -I
e - - - t-
I
-
�'
-
1-- 1.0 UN IT
-
:
f
aI
• \I' +_ va , e, +_ v9 , e,
,, '"
,t \�
-I 1
\
\ , \\
- - e
I
�
Fig. 4 . 24 . Single circuit with ground wire and earth return.
been designated as the ground wire g and the current Ib = - 16 , or the return cur rent is divided between the earth and the ground wire. Since circuit g is a ground circuit, this implies that its voltage drop to ground is zero. Mathematically, the changes made are
( 4 . 249) We may now rewrite (4.248)
as
(4. 250) Solving ( 4 . 250) for la , we have A
(4. 2 5 1 )
I f we define the equivalent impedance o f one unit length o f circuit with ground wire and earth return as z� , then by inspection of (4.251)
1 32
Z� Zoo =
where
Chapter 4 -
� n /unit length 2
z"
Zoo = 0 + �) + �:a zo. = � + �e %:. Z" = �. +�) j w k In
0
j w k In
n lunit length
n lunit length
a.
+ j w k In
n /unit length
(4.252)
(4.253) (4.2 54 ) ( 4.255)
It is often convenient to interpret conductor a of Figure 4.24 as a composite consisting of all three phase wires and conductor g as a composite of n ground wires. In this case we modify these last three equations so that each of the three phase wires carries a current la o and the ground wires carry a total current I.. The earth returns 3100 - I. amperes. A typical arrangement is shown in Figure 4.25
o.�
� ° 99 ', 0 1 9 0)
/���;�: ��-�;��:io.. �
_ _ _ _ _ _ _
... .... ,'''' ',... '''
� 3 1 0�:1 9 �.\
De
... ... ... ... ... .... "
, _ ,'
Fig.
Ds = I.O f t
4 . 2 5 . A typical arrangement o f phase and ground wires.
where there are two ground wires (n = 2). For the three-phase circuit with n ground wires we have (with 10 designating the current in composite conductor a ) Zo
s o that Z oo = 3
=-
V
�
la o
=
V ,
�= 3 za'
10/3
(Zoo �) -
z"
n lunit length
( 4.256)
Note that this defines zo as the impedance seen by la o. which is consistent with our previous definition. We refer to this as "z o on an la o basis. " as noted later. In equation (4. 256) we require that the resistance terms be that of the composite conductors a or g. Thus we have
Zoo (� +�) + zo. � + ge
j w k In
=
=
jw k In
a.
�:o
n /unit length
n lunit length
Sequence I m peda n ce of T ransm ission Unes
Z., =
(: �) +
+ j w k In
�:,
n /unit length
1 33
(4. 257)
4. 1 3.3 The basis for zero sequence impedance : 1. 0 or 31.o ? From the foregoing discussion of zero sequence impedance it is seen that there are two ways in which Zo could be defined The z� in ( 4.252) gives the total im pedance of a circuit with earth and ground wire return and is the impedance seen by ( 31C10 ) . Equation (4. 256 ), on the other hand, gives the impedance seen by lCio . We prefer the latter definition as one which is consistent with our definition of zero sequence impedance as "the impedance seen by the current lC1o " (see Chapter 2). Actually, either definition will suffice to compute the impedance. The prob lem arises at the end in knowing whether or not the result should be multiplied by three. Adding to the confusion is the fact that some references use a notation dif ferent from that used here such that the results appear to have the factor 3 missing. For example, in sections 4.10 and 4. 1 1 the following expression was developed for computing the zero sequence impedance of a transposed line (see equations 4.207 and 4. 233) Z oo = Z O(CI) - ( z � ( m ) /Z O(f » where we have determined that
(4.258)
��e2/q3
Z O(CI) = (ra + re ) + j3 w k In l D ZO( m )
=
re + j 3 w k In
�ea,
Z O(, ) = ( 3 rw + re ) + j3 w k In =
(�
)
rw + re
s
ge
ww
for one ground wire
D
+ j 3 w k In D ww
�
uw
for two ground wires
(4. 259)
Comparing ( 4. 2 59) with ( 4. 257), we verify that
Z O(a ) = 3 zClCI•
Z O(m ) = 3 zo, .
Z O ( , ) = 3 z. ,
and the two methods of computing Zoo are identical . Impedances ZO (a ) . ZO( m ) ' and Z O(f) are preferred for use over Z aa , za" and z., since these latter quantities are described on a 31C10 basis which is contrary to our definition of a zero sequence impedance .
4. 1 4 Computations I nvolving Steel Conductors It is common practice in the construction of high-voltage lines to use steel cored aluminum phase conductors called ACSR 3 and either high-strength steel or a steel composite such as copperweld or alumoweld4 for the ground wires. Both types of conductor contain magnetic material and therefore have a nonlinear flux linkage-current relationship and a nonlinear self inductance which is a function of the current in the conductor. In the case of ACSR phase conductors the change in permeability is not great. and the reactance is usually taken as a constant. Thus in Appendix B the conductor tables for ACSR show the inductive reactance at 1 foot
3 ACSR is an abbreviation for " aluminum conductor, steel reinforced. "
4 See Appendix
B
for properties of these conductors.
1 34
Chapter 4
Grode - Ordlnory Seven Strond Coble , 60 H . r t z V4
In
Resistonce
1 - 13 0
9/32 1 n
' � /4 In
...... _ - ';1'
/
I&J
/
3
o o I
I
15
t::-G.M . R .-
114
In
10 - 6
- - -
dlometer
In
9/ 32
Resistonce
20
10
30
40
50
AMPERE S P E R C AB L E
60
1 0- 4
J
I&J
1 0- 6
3
II: I&J
is
8
Z ex I&J l:
l: 0 I&J C)
"'" /�
2
2
�
ct t-
� �-
Grode - Ordlnory SevenStrond Coble , 25 Hertz
4
VI ::l
I&J
.....
<" \
�
U
..... '"
12
/-
1 0- 1
'-
12
L-�--�--�����-�� · 0 6
.... I&J
....... .....
I
13
dlometer
5/ 1 6 1n "
14
14
a..
10- 10
l: :r 0
VI
1 0- 12
I&J u Z
5 Resistonce 1 0- 1
0
10
20
AMPE RES
30
- 40
'"
O I
50
60
�
iii
VI
II:
I&J
1 0-
�
14
10 - 16 Resistonce
10- 18
0 0
60
PER C A B L E
'\
, ...
�V \\
\
\\
\
,
-
-
VI ::l is ex
II:
Z ex I&J 2 u
ii:
.... I&J l: 0 I&J C)
1 0 -6
1/2 In diometer
\ \ , .... _ M.R \ \ \- :1 G.
z
10 - 4
Grode E . B. B. Seven - S trond ....... Coble , 25 H . r t z .....
.... I&J � I&J
1 0- 8
-
1 0- 10 10- 1
,---�
2
0-1 4 ><--::-:-:::-:--"!..!7-:�-=:l 1
_-==';=--£..!:;�-::::l
��:---...c.-._L-1
10 - 16 10 - 18
1 0-
20
L-�__���__4':'--L.__--' 1 0 - 2 2
AMPERES P E R CABLE
AMPERES
PER
C AB L E
60
F ig. 4 . 26 . Electrical characteristics of steel ground wires. ( From Sy mmetrical Components by C. F. Wagner and R. D. Evans. Copyright McGraw-Hill, 1933. Used with permission of McGraw-Hill Book Co. )
Sequence I m pedance of T ransm ission Unes
1 35
spacing as a constant for all currents. This is because the current is largely con fined to the aluminum strands which, of course, are nonmagnetic. This is not the case, however, with steel conductors where both the resistance and the self inductance are strongly dependent upon the current magnitude. H. B. Dwight compiled data on the resistance and reactance of steel cables [ 3 1 ] in 1919. This data was converted by Wagner and Evans [ 10] to the more con venient form shown in Figure 4.26, where the resistance and D. (or GMR) are plotted as a function of current. Since in any given fault computation the current is usually not known until the computation is completed, one possible procedure is to carry along two or three values of impedance for circuits with ground wires. Then, once the current is found, the problem as to the ' correct impedance may also be resolved. An example will illustrate the procedure.
Example
4. 1 3
Compute the zero sequence impedance of the circuit of Example 4. 1 , assum ing that the line is completely transposed and that the ground wire is 0.37 5-inch EBB steel.
Solution From equation ( 4.207 ) we have for the zero sequence impedance of a line with one ground wire , z2 Q(m ) n /ml Z oo - ZO(a) z O(g ) •
_
The line is 40 miles long. To illustrate the procedure, we compute each quantity and Z OIg) ' both by formula, and also by table where appropriate.
z O(a ) ' zO(m ) '
1 . Zero sequence self impedance of the phase wire, z O(a) ' From (4.204)
;
D n /mi 2 D s eq = (ra + re ) + j (x e + xa - 2Xd ) n /mi
ZO(a) = (ra + re ) + j w k In
where, from Table 4. 1 , w k = 0 . 1 2 1 3 when f = 6 0 Hz. The first expression is some times written as ZO(a) = (ra + re ) + jO. 3639 In where
geaa
n /mi
Daa = (DsD;q ) 1 / 3 = (Dsa Dsb Dsc D a b D a cDb a Dbc Dc a Dcb) 1 /9 2 = [ (0 .01 688 )3 ( 1 0 ) ( 20 ) 1 1 /9 = 1 .393 ft
4
Then with De = 2790 , we compute
2790 = 2 .770 n /mi 1 . 393 From Tables B . 1 and B.9 we also compute, with an equivalent spacing ( 10 X 1 0 X 20) 1 /3 = 1 2 . 6 ft,
XO(a) = 0 . 3 6 39 In
XO(a) = Xe
+
X a - 2 Xd
= 2 .888 + 0 .497 - 2 ( 0 . 3074 )
= 2 . 7 70
n /mi
De q of
Ch a pter 4
1 36
The resistance is
Then
rO (a) = ra + re = 0 .278 + 0 .286 = 0.564 ll /mi ZO (a) = 0.564 + j2 .770 ll /mi
or
ZO(a) = 22.56 + j 1l0.8 II for 40 mi
zO(g).
2. Zero sequence self impedance of the ground wire, (4.197) and (4.206)
ZO (,) = ( 3 rw + re ) + j 3 wk In
ge
From equations
= ( 3 rw + re) + j(xe + 3 x,) ll /mi
[ � �� WUl
From Figure 4.26 for 3/8-inch = 0.375-inch EBB steel at 60 Hz we have 1 X 1 0 -3
Dww =
1
5 X 10 - 1 2 1 .5 X 1 0 - 10
ft at 30 A 60
Then , as a function of ground wire current, we have
[ �
2790
ww
5.40
=
�] [3.�
----n- = 2 . 888 + 0.3639 In
XO (f ) = 0.3639 In
1
1 2.29
II /mi at 30
1 1 . 10
60
1
D ww
A
The resistance is also a function of current. From Figure 4.26 we find rw
=
7 .8 ll /mi
6. 0
which is added to re = 0.286 to write, after rounding,
rou, = Then
��::�
� 9.1 �
Il lmi at
Ia�
��
A
� � � � �� � 3 � � � �� 1.6
5.4
Z O(I) = 24. 5 9 + j 1 2 . 29 II /mi at 30 A 19.19
1 1 .10
60
For the 40 miles of line we have
67 .
16.
R O(I) = 98 . 6 + j 49 1 . 6 II at 30 A 767 .6
444.0
60
1 37
Sequence I m pedance of T ransmission U nes
�
�
or
515.1{24.8 ZO(, ) = 1099.6{26.6° in polar form 886.8{30.0° Here we note that the resistance is about one-half the reactance whereas for the copper phase wires the resistance was only about one-seventh the reactance.
3. Zero sequence mutual impedance, zO( m ) ' From (4.205) Z O (m) = re + j3 wk In
�ea, = re + j(xe - 3xa,)
First, by slide rule we compute (see equation
Da, = (Daw Dbw Dcw ) 1/3
==
4.203)
[ (18.03)(15.0)(18.03)] 1/3
Then
Zo(m) = 0 .286 + jO.3639 In
:��� = 0.286
As a check we repeat this computation by table. Da, = 16.95 we see that x a, = 0.3435 n /mi. Since X e = 2.888, we compute
Z O(m ) = re + j (x e
-
+
==
j1 .855
16.95 ft
n /mi
From Table
B.24(a) with
3xa,) = 0.286 + j 1 .857 n /mi
which is an excellent check. The total impedance for the and Z �m ) =
n /mi
40 mile line then is
Zo( m ) = 1 1 .44 + j74.24 = 75.2/81.25° n
5660/162.5°.
4. The total zero sequence impedance, Zoo . The total zero sequence Zoo may be computed from ( 4.2 07 ) as
�
Zoo - ZO(a) _
Z5 ( m ) Z0(, )
3 0.66 + j l 03.4
= 26.25
+
�
1
j l 0 7 . 2 3 n at 30 A
26.85 + j 1 06. 1 0
60
4. 1 5 Parallel Transposed and Untransposed Multicircuit Lines
A study of parallel untransposed multicircuit lines by Hesse [32, 33] has shown that the unbalance between parallel lines must be examined for two sepa rate types : the net effect or "through unbalance " and the circulating currents due to the unbalanced voltages being out of phase in the parallel circuits. This re search has revealed that the in-phase unbalanced currents which contribute to the overall through unbalance are usually quite low. The circulating currents, on the other hand, may be much greater. These latter currents are usually important where the parallel lines are bused at both ends such that the circulating current
Chapter 4
1 38 m
l o be
Z abe
-
lo "' e ' �
n
Z a' b' e '
Fig. 4.27 . A double circuit transmission line from m to n .
d by the impedance sees only the impedance of the parallel lines and is not affecte capacitors will seen looking into the system . Any line compensation by series the relocation have an adverse effect on these circulating currents and may dictate examine the exact of such capacitors to another circuit. It is also important to configurations are some since s circuit l paralle phase arrange ment of untransposed than others. nce unbala t curren ting significantly worse with respect to circula ient to ne conven is it s, circuit nced In performing the calculations of unbala either end at system the into �lect the Thevenin eauivalent impedan ces, looking n buses m betwee section of the parallel line section , and consider only the line m and n is the same and n . Also, we assume that the voltage drop between buses these bus voltages that for each circuit since they have commo n termination and true for physical are positive sequence (balanced) voltage s. This is approximately system s. ting the un With these assumptions, Hesse [33] gives a method of estima o can be d ne for all balance due to line currents and circulating currents. This elimination of possible phasing arrangements . Making such estimates permits .the few configu ly relative a only leaving the most undesirable phasing arrangements, rations for detailed study. 4.27, where Consider the double circuit line shown schematically in Figure be untrans may ' circuit Each one circuit is designated a-b-c and the other a -b' -c' . m to n. from length its in posed or may have any number of rotations and twists nted by a voltage An y line section , and therefore the total length, may be represe the form of line equation expressing the voltage drop in the given length of Ia Zac II Zaa' Za b ' Zac' -Zaa Zab 1: Va I Ib Zba Zb b Zbc I Z ba' Z",, ' Z bc' 1: Vb I Ic Zcc I Zca' Zcb ' Zcc' Zcb Zca 1: Vc I V -+= I 1a , Za'a Za 'b Za'c II Za 'a , Za 'b ' Za "c 1: Va ' Zb 'a Zb 'b Z b 'c II Zb 'a' Zb 'b ' Z b 'c' Ib ' 1: Vb ' I ( 4.260 ) Ic , Zc 'b ' Zc "c Zc 'a Zc'b Zc'c I Zc "a l: Vc ' --- - - - - - - - -
- -- - - - - -
- - -
columns of the matrix If there are ground wires, it is assumed that the rows and is a true represen 4.260) that so ( ons have been reduced to eliminate these equati wires. If there ground of effect the tation of the actual voltage drop, including may be repre drops e voltag such are several transposition sections, the sum of all sented by equation (4.260 ). Using matrix notation , we may write ( 4.260 ) as
J [
r1: Vabc L1: Va'b 'c'
=
Zaa Za'a
( 4.261 )
Sequence I m pedance of T ransm ission li nes
1 39
Since the circuit unbalance is expressed in terms of sequence curre nt unbalance, we solve (4 .261 ) for the currents.
[ 'e '] [ Iab c la' h
Yaa Yo'a
=
(4.262)
Since the Z matrix of (4 .261 ) is symmetric , we can show that [ 7 ]
Yaa = Z -a1a + KL- 1 K t , Ya 'a = Y �a ' ,
where
Yaa ' = - KL - 1 Ya'a' = L- 1
-I , K t - Za 'a Zao
(4.263) (4.264 )
From ( 4 .263 ) and (4.264) we observe that the 6 X 6 Z matrix may be inverted by performing two 3 X 3 inversions , one on Zaa and one on L. The inversion is then completed by performing the indicated matrix multiplications. Having computed the phase currents by (4 .262 ) , we need a method whereby these currents may be transformed to the 0-1 -2 frame of reference. To do this, we introduce the transformation A 2 defined by
A, = such that
IL�o � ��JJ _
:
( 4.265)
(4.266)
and where, obviously ,
A -2 I -
[
[
[o
A- I (4.267 )
Premultiplication of ( 4 .262) by A 2 1 gives the result
J
10 1 2 A- I Yaa A = A- I Ya'a A 10' 1 ' 2 '
]
A- 1 Yaa' A A- I Ya'a ' A
( 4 .268)
We now observe that the sequence voltage at each end of the line is a positive se quence voltage by definition . Therefore the sequence voltage drops expressed in ( 4.268) are entirely positive sequence or
����j
0
0
k Va l
1
0
0
=
=
k Va l
0
0
k Va ' 1
1
0
0
where we note also that Va l = Va' i since the lines are bused at each end.
(4.269)
1 40
Chapter 4
Expanding ( 4.268 ) to the full 6 X 6 array and recognizing ( 4 .269 ), we may write
Yoo Y10 la2 Y2 la'o YO'0 la ' l Y1 ,0 Y2,0 10 0 10 1
Y02 Yoo' Y01' Y02' Y1 2 Y10' Yll , Y12 ' Yu Y20, Y2 1 , Yu' YO'2 Yo'o' YO' I ' YO'2 ' YI '2 YI,o' YI ' I ' YI '2 ' Y2 '2 Y2 ,0' Y2 ' 1 ' Y2 '2 ' 1 1 YYOII1 YYOIl1,' Y Y 2 21 1 , 10'0 YO' 1 YO' I ' � Va l Y1' 1 Y1' 1 ' Y2'1 Y2' 1 '
Y01 Yll Y21 YO' 1 Y1'1
1 1 1 1 1 0 1 = ------------1 -----------1 1
10'2
1
1
1
Y2' 1
1
0 1 0
� Va l 0 1 0
(4.270)
Having done this, we easily compute
10 0
+
0 102
+
+
=
+
10 '1 10 '2
+
+
(4.271 )
and we note that only the second an d fifth columns o f the Y matrix are used. Then we define the net through unbalances as 6.
10 0 + 10 '0 pu, + I0 '1
mOt a l mOe lao - la,o mOt YOI m2t Y2 1 mOe YOI m2e Y21
(4.27 2 )
=I
and the net circulating current unbalance as
[ ][ =
I0 1 + I0 '1
(4 . 2 7 3 )
pu ,
o r substituting from these equations, we may write
=
+ + + +
YOI' YO' 1 Y21, Y2' 1 Y01' - YO' I Y21' - Y2' t +
+
where ( 4.274 )
mOe mOt
Hesse [ 33 ] shows that, for parallel lines in flat configuration, » and mlt > > often by about a factor of ten or so in each case . This is not obvious from an inspection of ( 4.274 ) because it is hard to appreciate the vast difference in phase angle of the admittances.
m2c
Example
4. 1 4
Compute the through and circulating current unbalances for the double cir cuit 345 kV untransposed line shown in Figure 4. 2 8 .
1 41
Sequence I m peda n ce of T ransm ission lines 10" 10'
UrjiW at + Ig' ;tC
;.. on
19 '
N
,
Ground Wi,": 79' kCM, 2617 ACSR G M R · 0. 037' ft . ro 0. 1 1 7 A / m i
,
•
b'
Pho. . Wi", :
I . 7' in Expanded ACSR G M R . 0 . 06 4 0 " ro .
0.0663 A Imi
Fig. 4 . 2 8 . A 34 5 kV double circuit configuration.
Solution First we must determine the impedance matrix of self and mutual impedances defined by equation ( 4. 79). These 64 impedances are given in Table 4.11, and the 36 impedances which result after reducing this matrix to eliminate rows u and w are given in Table 4.12. The inverse of the matrix given in Table 4.12 is computed by digital computer with the result Yoo = Yoo ' =
Yo' 0 ' =
[ r-
+ + ( - 0. 001 1 + jO.0894) (- 0. 0000 + jO.0827) (- 0.0039 + jO. 1099)
++ � (- 0.0087 + jO.1416 � (- 0.0011 + jO.0894) (- 0.0011 + jO.0877) (- 0.0056 + jO.1 195)� (- 0.0214 + jO.2131)
( 0. 1057 - j 1 . 1 573) (- 0.0214 jO.2131) (- 0.0056 jO.1 195 (- 0.0214 jO. 2131) (0.0983 - j1.1178) (- 0.0245 jO.2348) (- 0.0056 + jO. 1 195) (- 0.0245 jO.2348) (0.0954 - j1. 1006)
+ 0.001 1 + jO.0877 ) ( 0.0039 + jO. 1 099) -
(- 0.01 79 + jO. 1950)
[
+
(0.0954 - j1.1006 ) (- 0.0245 jO. 2348) (- 0.0245 + jO. 2348) (0.0983 - j1.1178) (- 0.0056 jO. 1195) (- 0.0214 + jO. 2131) (0. 1057 - j1.1 573)
+
Each partition above is now transformed to the ified by (4.268), with the following results A- I Yoo A =
[ [([([
0-1-2 frame of reference as spec
+
+ � (0.0329 + jO.0010)� (0.0417 - jO.0161)
+
+
( 0.0655 - jO. 7469) (0.0287 - jO.0351) (- 0.01 54 - jO.0426 (- 0. 01 55 - jO.0421) (0. 1 1 70 - j1.3156) (- 0.0634 jO.0200) ( 0.0286 - jO.0346) (0.0546 + jO.0400) (0.1171 - j1.3156)
0.01 29 + jO.33 1 1 ) (- 0.0309 jO.0113) (0.0018 - jO.0376) (- 0. 01 56 jO.0288) (- 0.0348 - jO.0281 ) (0.0018 - jO.0355)
A- I Yo' o A
= (0.0057 jO.0323)
A- I Yoo, A
= (0.0330 jO.0013)
A- I Yo' o, A =
++
� j
0.01 79 + jO.3311) (- 0.0156 jO.0290) (0.0056 jO.0325) (0.0089 - jO.0355) (0.0417 - jO.0162) (- 0. 0310 - jO.01 16) (- 0.0348 - jO.0281 ) (0.0018 - jO.0376)
+
(0.0655 - jO. 7469) (0.0446 + jO.0079) (- 0.0448 - jO.0072 ( - 0.0449 - jO.0067 ) (0. 1 1 70 - j1.3156 ) (- 0.0617 0.0278) (0.0446 jO.0085) (0.0487 jO.0455) (0.1171 - j1.3156)
+
+
+
Table 4. 11. Impedance Tabulation for the Circuit of Figure 4. 27 (8 X 8 matrix of phase wire and ground wire impedances) tJ
b
C
tJ
,
b'
C
,
U
W
0.16158 +j1 .296235
0. 09528 +jO. 5 5 3453
0.09528 +jO.488002
0. 09528 +jO. 460342
0. 09528 +jO.468743
0. 09528 +jO. 521302
0. 095 2 8 +jO. 5617 39
0. 09528 +jO. 5 1 89 34
0. 09528 +jO. 553453
0. 16158 +j 1. 296235
0. 09528 +jO. 553453
0. 09528 +jO. 4687 4 3
0. 09528 +jO. 45 0692
0 . 09528 +jO. 468743
0. 09528 +jO.474048
0. 09528 +jO. 452254
0. 09528 +j0. 488002
0. 095 28 +jO. 55 3453
0. 16158 +j 1. 296235
0. 09528 +jO. 521302
0. 09528 +j O. 4687 4 3
0. 095 28 +j O. 460342
0. 095 2 8 +j O.436824
0. 09528 +jO. 429368
,
0. 09528 +j0. 460346
0. 09528 +jO. 468743
0. 09528 +jO. 521302
0. 16158 +j 1. 296235
0. 09528 +jO.553453
0. 09528 +j O.488002
0. 09528 +jO.429368
0. 09528 +jO. 436824
b'
0. 09528 +jO.468743
0. 09528 +jO.4 50692
0. 09528 +jO.468743
0. 09528 +jO. 5 53453
0. 16158 +j 1.2962 35
0. 09528 +j O. 5 53453
0.09528 +jO. 452254
0. 09528 +j O. 474048
,
0. 09 5 28 +j0. 5 2 1 302
0. 09528 +jO. 468743
0. 09528 +jO.460342
0. 09528 +jO. 488002
0. 09528 +jO. 553453
0. 16158 +j 1. 296235
0. 09528 +jO. 5 1 8934
0. 09528 +jO. 56 17 39
0. 09528 +jO. 5617 39
0. 09528 +jO. 474048
0. 09'5 28 +jO. 436824
0. 09528 +jO. 429368
0. 09528 +j O. 452254
0. 09528 +j O. 5 1 89 34
0. 21228 +j 1. 361096
0.09528 +jO. 599185
0. 09528 +jO. 518934
0. 09528 +jO. 452254
0. 09528 +jO. 429368
0. 09528 +jO. 436824
0. 09528 +jO. 474048
0. 09528 +jO. 5 6 17 39
0. 09528 +jO. 599185
0. 21 228 +j 1. 361096
a
b c
tJ
c
u w
1 43
Sequence I m peda n ce of T ra n s m ission Li nes
Table 4.12. Impedance Tabulation for the Circuit of Figure 4.27 (6 X 6 matrix of phase impedances after matrix reduction) a
b
c a
,
b' c
,
b
c
a
0.1034 +jO.9973 0.0379 +jO.2978 0.0381 +jO.2493 0.0380 +jO.2220 0.0377 +jO.2142 0.0368 +jO.2247
0.0379 +jO.2978 0.1058 +j1.0775 0.0402 +jO.3494 0.0402 +jO.2648 0.0395 +jO.2327 0.0377 +jO.2142
0.0381 +jO.2493 0.0402 +jO.3494 0.1073 +j1.1055 0.0410 +jO.3307 0.0402 +jO.2648 0.0380 +jO.2220
0.0380 +jO.2220 0.0402 +jO.2648 0.0410 +jO.3307 0.1073 +j1.1055 0.0402 +jO.3494 0.0381 +jO.2493
Then fro m laO la l
la2
la 'o
la ' i
la '2
,
a
b
0.0377 +jO.2142 0.0395 +jO.2327 0.0402 +jO.2648 0.0402 +jO.3494 0.1058 +j1.0775 0.0378 +jO.2978
c
,
0.0368 +jO.2247 0.0377 +jO.2142 0.0380 +jO.2220 0.0381 +jO.2493 0.0378 +jO.2978 0. 1034 +jO.9973
(4.27 1 ) we combine these results to write
(0.0286 - jO.0351 ) + (- 0 .0349 + jO.0133) (0.1169 - j1 .3144) + (0.0059 - jO.0386) (0.0545 + jO.0396) + (- 0 .0541 - j O.0437) V (- 0 .0 1 1 6 + jO.0044) + (0.0446 + jO.0079) l: a l (0.0047 - jO.0345) + (0.1 169 - j 1 .3144 ) (- 0.0154 - jO.0 1 2 2 ) + (0.0487 + jO.0451 )
Then, from
'
=
- 0.0131 - jO.0061 0 . 1 2 59 - j1.3511 0.01 97 + jO.01 1 9 0 .0 1 37 - jO.00 34 0 . 1 1 88 - j 1 . 3532 0 .0 1 39 + jO.01 7 5
l: Va l
, (4.272) we compute the "through unbalances. ;
mO t m2 t
Similarly , from
=
=
la o + 10'0 10 1 + 10'1
102 + 10'2
=
=
la l + la ' i
0.0268 - � 0.0095 0 .2447 - J2.7043 0.0336 + � 0.0294 0.2447 - ]2.7043
=
0 .0104 { 65.2 0
=
0.0163 / 1 26.0 0
(4 .273) we find the "circulatory unbalances. "
mOe m2e
=
la o - 10'0 10 1 + 10' 1
=
102 - 10'2 10 1 + 10'1
=
=
0 .0006 - �0 .0027 = 0 .00102 /- 17.7 0 0.2447 - J2.7043 0.0058 - � 0.0209 = 0.00296 / 40.8° + 0.2447 - J2.7043
_
For the vertical configuration of this problem the negative sequence unbalances about equal but not much greater than
are
m Oe '
4. 1 6 Optimizing a Parallel Circuit for Minimum Unbalance
In the previous section each parallel circuit line was treated in terms of its total impedance characteristic from one terminal to the other. Obviously, it would be possible to write separate equations for each line for each section of like configuration and then apply the transposition matrices Rtil and Ttil to each circuit to reorder the equations in an a-b-c-a ' -b' -c' sequence. Once each section has been
Cha pter 4
1 44
so arranged, the matrices may be added to find the total impedance of each in dividual line. In many practical situations the lines are not transposed at all. This is because the transpositions are costly and, perhaps even more important, they are often the cause of circuit failure due to either mechanical or electrical weakness at the trans position structures [ 34, 351 . The general problem of the effect of circuit un balance due to lack of transpositions has been studied in detail by E. T. B . Gross [ 28, 29, 30, 34, 35, 36] and is an excellent resource for the interested reader. In double circuit lines which are not transposed, Hesse [331 has shown that there is an optimum conductor arrangement which will minimize the unbalance factors. There is value, therefore, in developing a method whereby the unbalance factors can be determined in a straightforward way, while making sure that all possible configurations are examined. We shall do this for lines which are them selves untransposed but which can have any desired phase identification. From equation (4.268) we have
=
][ ]
[Yoo
V012 VO' I ' 2 '
Yoo' Yo ' o'
Yo' o
phased a-b-c (4 . 275)
phased a'-b'-c'
where we have defined new admittance submatrices Y00 , Y 00' , Yo ' 0 and Yo ' 0' as a matter of convenience. We have also indicated the phase designation quite ar bitrarily as a-b-c and a' -b' -c' for the configuration used in writing these equations. Suppose, however, that the second (primed) circuit had been phased b' -c'-a' ( on wires 1-2·3). This means that the voltage (or current) equation must be pre multiplied by R41 to rearrange the elements to the desired a-b-c order. However, since equation ( 4. 27 5) is already in the 0-1-2 frame of reference, we may use the rotation matrix R 0 12 defined by equation (4.90). Similarly, multiplication by Rol2 defines a rotation in the opposite direction. These rotations of only the primed circuit may be performed mathematically by transformations
l�
and
[�
o
R0 12 R0- I
o
]
]
a-b-c (4.276)
b'·c'·a '
a-b-c
12 c -a -b I
"
] r V0 12 ]
(4.277)
For example, premultiplying both sides of (4.275) by (4.276) gives
[10 12
] = [Yoo R
R��i;:-;';
Yoo, Rol2 C o;;Yo� �Ro�Yo';Ro I2
1 I
L-Ro�V�:�;
a-b-c b' -c' -a '
(4.278)
This operation is equivalent to multiplying every element in the fifth row and the sixth column of (4.270) by a and multiplying elements of the sixth row and fifth column by a2 • This result, with only positive sequence voltages applied, is
1 45
Sequence I m pedance of T ransmission Unes
100 10 1 10 2
=
10' 0 10' 1 10 ' 2
YO I + al YO I' Y l l + al Y l l, Yl l + a2 Yl I,
phased a-b-e � Va l
- - - -- - - - - - -
YO' I + a2 YO' I' a ( Y I' 1 + a2 Y I ' I ' ) phased b ' -e' -a' a2 ( Y2 ' 1 + a2 Y1' 1' )
(4.279)
Another rotation of the second circuit would give a similar result, but with oper ators a and a1 interchanged. If instead the second circuit is transposed according to transformation To 1 2 , the result is equivalent to interchanging the fifth and sixth rows of (4.270) and then interchanging the fifth and sixth columns. The result is
100 10 1 10 2 10 ' 0 10 ' 1 10' 2
YO I + Y0 2' Yl l + Y 1 2, =
phased a-b-e
Y2 1 + Y2 1,
� Va l
-------
YO' I + YO' l' Y2 ' 1 + Y1' 2 ' ' Y I' I + Y I' 2 '
phased a' -e ' -b'
(4.280)
If this arrangement is then changed by a rotation of the second circuit to ( 4.280) are multiplied by a and a2 exactly as in (4. 279). In all, six arrangements are possible, with the first circuit remaining un changed. These are
a-b-e : b'-a'-e' , the Y elements of
II �
3l
u
2 ROl2
a-b-e
a-b-e
a- b-e
a-b-e
a'-e'-b'
a' - b ' -e' b '-e'-a'
a-b-e
e -a - b "
,
6
[U
b ' -a' -e'
R0 l 2 To 1 2 '2
a-b-e '
U ,
e -b -a
,
(4. 281 )
for the first circuit re and another six relationships may be obtained with the placed by T O I 2 ' These are the only unique arrangements-a fact that may require a little careful thought. This is due to the fact that which circuit is unbalanced with respect to its neighbor is immaterial. Thus 12 arrangements in all must be examined. In general, it is possible to have 6" /3 significantly different phase ar rangements for an n-circuit system. Problems
Some of the problems suggested here should be solved by digital computer to provide high accuracy, to minimize labor, and to correctly display the desired impedance characteristics. Some helpful programs are available in Appendix A. 4.1. Compute the per mile positive and negative sequence impedance for the line configura tion shown in Figure P4. 1 where the conductor is 410, 7-strand copper. Assume that the
1 46
Chapter 4 a
Fig. P4. 1 .
line i s transposed such that the methods o f section 4 . 1 ar& applicable. Compute Xl by equation (4. 3) and use equation (4. 10) as a check. 4.2. Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.2 where the conductor is 3/0 ACSR. Use the method of section 4.1. b
Fig. P4.2.
4.3. Compute the per mile positive and negative sequence impedance for the line confIgura tion of Figure P4.3 where the conductor is 500,000 CM, 61% conductivity, 37-strand, hard-drawn aluminum.
45'
Fig. P4. 3 .
4.4. Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.4 where the conductor is 336,400 CM, 26/7-strand ACSR. --,r--__ 9
Fig. P4.4.
Sequence I m pedance of T ra nsm ission Unes
4.5.
1 47
Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.5 where the conductor is 397,500 OM , 26/7-strand ACSR. Q
Q
Fig. P4.5.
4.6.
Write the phasor equations for a two-winding transformer with self inductances Lp and L. in primary and secondary windings respectively and mutual inductance M. Compare
these equations with equation (4.14) for two mutually coupled, parallel wires. Review the concept of placing polarity markings on the transformer coils. 4.7. Verify that the four-terminal equivalent circuits shown in Figure P4. 7 a and Figure 4.7b accurately represent the equations for two mutually coupled wires given by equation (4. 14). Circuit b is due to Starr [ 37 ] . a
b'
,
r -
Vaa '
1
•
Zoo -l o b a
t 10
•
l ob
(a)
t
1
'
loa
b' - Z ob
V b b'
l bb -Za b lb
a
b
1
a
(b)
b
Fig. P4.7 .
Find the internal inductance in millihenrys/mile of 106 circular mil, all aluminum (61% conductivity) conductor when the operating frequency is 60 Hz. 4.9. Find the internal inductance in millihenrys/mile of a certain iron wire having a dc resis tance of 2.0 ohms/mile and a relative permeability of 100. The frequency is 60 Hz and the wire is cylindrical with a diameter of 0.394 inches (4 / 0 SWG pure iron wire). 4.10. Verify the impedance for Carson's line, given by equations (4.35)-(4.40). 4. 11. Compute the voltage drop Vm n across the circuit of Figure P4. 11 and compare with equation (4.35). Neglect the resistance of the windings.
4.8.
n
m
Fig. N . l l .
Chapter 4
1 48
4. 12. Consider a single conductor plus earth return circuit. The conductor is size 4/0, 7-strand hard-drawn copper at 50 C suspended 30 ft above the earth; f = 60 Hz. (a) Find the self impedance of the circuit, roo + jxoo ' in ohms/mile, for ( 1 ) P .. 10 n . m and (2) P = 1000 n . m (b) What is the percent change in Xoo and in Zoo for this 100-fold increase in earth resistivity? 4 . 1 3. Compute the impedance matrix for two wires a and b with earth return d and sketch an equivalent circuit for this circuit similar to Figure P4. 7(a) where the effect of the earth is included in the circuit parameters. 4. 14. Verify equation (4.47) for a three-phase line with earth return and compute all self and mutual impedances where the assumptions made in equation (4.53) are not true. 4. 1 5. Illustrated in Figure P4. 1 5 is a 4-wire circuit which is used to serve a three-phase, 4-wire, ' ' ' Y-connected load. The line consists of three phase conductors a-a , b-b , and c-c sepa rated by distances Dab , Dbe , and D ca as shown. Also shown is the neutral conductor non ' separated from the phase conductors by distances Do" , Db" , and Dc" . The neutral con ductor is not connected to the earth, and its Impedance does not involve the earth in any way. One end of the line is short circuited in order to facilitate finding the line impedances. Find zo, the impedance of this line to the flow of zero sequence currents. Assume that the four conductors are identical with resistance r ohms per unit length and a GMR of D• . a
b c
n
zbb
Dab
Icc
Dbc
J
iob z bc
Den Icn �
In ..
REFERENCE FOR ALL VOLTAGES
�\ \
D
J
0' b'
ico
�
Don
J
zan
,
Dbn
,
I
ibn
c' n'
Fig. P4 . 1 5 .
4. 16. The line configuration shown i n Figure P4. 16 i s similar to that used in certain overhead circuits in the 5-1 5 kV class for primary distribution. Phases a, b, and c are insulated but not shielded cables. The neutral conductor n is a grounded bare messenger which sup ports all wires mechanically and also carries any unbalanced neutral current (3100 ), The frequency is 60 Hz.
a
o------o b
Fig. P4 . 16.
Sequence I m peda nce of T ransmission Li nes
1 49
Phase conductors: 3/0 hard-drawn aluminum, 19 strand D B 0.0 1483 ft 0.178 inches r " 0.558 fl/mi @ 25 C -
..
Neutral conductor: I/O ACSR, 6/ 1 strand 0.00446 ft 0.0535 inches 0.888 fl/mi @ 25 C
D. r
-
-
..
(a) Ignore the ground wire and determine the phase impedance matrix Zabc for this circuit. (b) Assume balanced applied voltages and unbalanced loading as follows: Va ... 8000l!L line-ta-neutral Vb - al Va
la
'"' 200/- 30 A
Ib - 200/- 150 A
Ve '"' a Va
Let Ie be variable with the following values: lIe I
Case 1 2 3 4
100 100 100 100 100
5
A ngle of Ie 60° 9 00 1200 1500 1800
Case 6 7
8 9
10
lIe I
200 200 200 200 200
A nllle of Ie 600 900 1200 1500 1800
Find the line-ta-neutral voltages two miles from the sending end of the line for cases
1- 10.
(c) Tabulate the results of part (b) and also the power factor of phase c and the percent voltage drop in each phase. (d) Plot the percent voltage drop in each phase versus the phase c power factor: (1) for Ile l - 100 A and (2) for lIe I 200 A. Identify the one point which represents a balanced load. Compute the phase impedance matrix Zabe for the line described in problem 4. 1 . As sume that the line is 25 miles long and is not transposed. Compute the phase impedance matrix Zabc for the line described in problem 4.2. As sume that the line is 35 miles long and is not transposed. Compute the phase impedance matrix Zabc for the line described in problem 4.3. As sume that the line is 50 miles long and is not transposed. Ignore the ground wire. Compute the phase impedance matrix Zabe for the line described in problem 4.4. As sume that the line is 50 miles long and is not transposed. Ignore the ground wire. Compute the phase impedance matrix Zabe for the line described in problem 4.5. As sume that the line is 60 miles long and is not transposed. Ignore the ground wires. Compute the total impedance matrix Zabe for the lines of problems 4. 17-4.2 1 with the following transposition arrangements. '"'
4.17.
4.18. 4. 19. 4. 20. 4.21. 4.22.
Fraction Configu ration fl 0.20 a-b-c fl 0.80 b-c-a f3 0.00 c-a-b (b) fl - 0.25 a-b-c fl 0 . 35 b-c-a b-a-c f3 - 0.40 (c) fl 0. 30 a-b-c c-a-b fl 0.60 f3 0. 10 c-b-a (a)
..
..
'"
..
'"'
.. '"
1 50
Cha pter 4
(d) fl - 0.33
f2 = 0. 33 f3 - 0.33
a-b-c c-a-b b-c-a
4.23. Verify the results given in Table 4.3 by perfonning the indicated operations. 4. 24. (a) Repeat problem 4. 22(a) for any given configuration and let fl be variable from 0 to 1 in steps of 0.2. Plot Xab as a function of fl ' (b) Repeat problem 4.22(a) with f3 - 0.5 and let fl be a variable from 0 to 0.5 in steps of 0. 1. Plot Xab as a function of f1 and compare results with part (a). 4. 25. Verify equations (4.78) and (4.82) for a transposed line. 4.26. Compute the sequence impedance matrix Z0 12 for each line of problem 4.22. 4. 27. Compute the sequence impedances for the study specified in problem 4.24 and detennine the way in which the sequences are coupled together as a function of fl ' 4.28. Verify the twist equation (4. 110). 4.29. Compute the unbalance factors of each line of problem 4.22, (a) Using the exact fonnula (4. 144). (b) Using the approximate fonnula (4. 146). 4. 30. Write the matrix equations for the flux linkages in each section of the transposed line of Figure 4.9. 4.31. Consider the four circuits a, b, c, and d shown in Figure 4.5. (a) Write the instantaneous voltage equations of the fonn 11 - ri + cD.. / dt for each circuit. (b) Apply the phasor transfonnation (1.50) to the equations of (a). (c) Compute the phase impedance matrix Zabc from the result of (b). 4. 32. Verify the fonnula for unbalance factors given by (4. 144) by perfonning the computa tion implied. 4.33. Consider the line configuration of problem 4.3 shown in Figure P4. 3. A ground wire of I/O K copperweld-copper is present but is used as a second conductor for one of the phase wires (assume here that the ground wire is fully insulated). Compute the phase im pedance matrix Zabc if this extra wire is connected as follows: (a) In parallel with phase a. (b) In parallel with phase b. ( c) In parallel with phase c . 4.34. Consider the line configuration shown in Figure P4.4. Instead of using a single conductor of 336,400 CM ACSR in each phase, with current-carrying capacity of 530 amperes, sup pose that each phase consists of a two-conductor bundle of two 3{0 ACSR conductors with capacity of 300 amperes/conductor. Let the two conductors of each bundle be separated by 1.0 ft vertically. (a) Compute the phase impedance matrix Zabc for the bundled conductor configuration and compare with the previous solution (problem 4.20). (b) Compute the sequence impedance matrix for both the new and old conductor ar rangements and compare. 4.35. Let the circuit described in problem 4.5 and Figure P4.5 be altered to consist of a two conductor horizontal bundle of 397,500 CM ACSR phase wires. Compute phase im pedance matrix Za bc and sequence impedance matrix ZO l lo Bundles with 1.2 ft spacing. 4.36. Verify equation (4. 164) for a bundled conductor line. Then show that the necessary matrix operations are all 3 X 3 matrix operations as suggested in Example 4.8. 4. 37. Verify equation (4. 174) for the phase impedance matrix Zabc of a line with one ground wire. Then extend this analysis to the case of a partially transposed line where fl ::/:. f2 ::/:. f3. 4.38. Verify the computations which result in equations (4. 198) and (4.201) for a completely transposed line with one ground wire. 4.39. Consider an untransposed line described in problem 4.3 and Figure P4.3. Let the ground wire be I/O K copperweld-copper and recalculate the phase impedance matrix Zabc , the sequence impedance matrix Z012 , and the unbalance factors rno and rn 2 ' Compare with previous results from problem 4. 19 for the same line without the ground wire.
Seq uence I m pedance of T ra n s m i ssion Li nes
1 51
4.40. Consider an untransposed line described in problem 4.4 and Figure P4.4. Let the ground the sequence im wire be I/O ACSR and recalculate the phase impedance matrix pedance matrix and the unbalance factors. Compare with previous results from problem 4. 20 for the same line without the ground wire. 4.41. Consider an untransposed line described in problem 4.5 and Figure P4.5 with two ground wires of 2/0 ACSR. Compute the phase impedance matrix the sequence impedance matrix and the unbalance factors m o and m2 ' Compare with the results of problem 4. 2 1 for the same line without ground wires. 4.42. Verify carefully the result given by equation (4.230)-(4.2 34) for impedance of a trans posed line with two ground wires. 4.43. Derive the general expression for a completely transposed line with n ground wires in such a form that (4.233) may be used. 4.44. Repeat the computation of the sequence impedances for problem 4.40 if the ground wire is taken to be 1 / 2-inch EBB steel messenger. Perform this computation for 1, 30, and 60 amperes in the ground wire. 4.45. Repeat problem 4.44 for the case where the line is completely transposed. 4.46. Compute the sequence impedance matrix for the line described in problem 4.4 and Figure P4.4. Let the ground wire have a GMR of 0.001 feet and let the resistance of the ground wire vary from 0 to 5 ohms/mile. Plot the ratio of Iw to 3IaO and to Id as a func tion of rw for nxed GMR. 4.47. Repeat problem 4.46 for a range of GMR's from 0.001 to 0. 10 and plot the current ratios Iw /Id and Iw / 3IaO for the entire family. 4.48. Equation (4. 184), giving the ratio Iw /Id ' indicates that if zw w " Zag then Iw/Id � oo. Show that this is impossible for the range of rw and GMR encountered in physical situations. How high does this ratio become as a practical matter? 4.49. Suppose that the lines shown in Figure P4.4 and P4. 5 (with ground wires specifled in problems 4.40 and 4.41 respectively) are located on the same right-of-way and separated by a horizontal center-to-center distance of 50 feet. (a) Compute the matrix of phase impedances including all ground wires and reduce this matrix to a 6 X 6 phase impedance matrix similar to equation (4.260). (b) Invert and transform the matrix of (a) to compute the four unbalance factors mO h mu, mOe , and m2e ' (c) Recalculate (b) for all 1 2 possible conductor arrangements for the 2 circuits. 4.50. Prove that there are 6"/3 unique arrangements of n parallel circuits which must be examined to determine the optimum phase connguration. 4.51. Repeat the computations of Example 4. 14 for all conngurations specified in equation (4. 281).
Zabe,
Z012 ,
Z012 ,
Zabe,
chapter
5
Seq u e n ce Ca pa c ita n ce of Tra n s m i ss i o n Li nes
This chapter focuses upon the shunt admittance of the transmission line. In overhead lines this shunt admittance is a pure susceptance since the conduction current between wires or between wires and ground is negligible. Furthermore, this susceptance is purely capacitive. We begin with the computation of the capacitance to neutral of an isolated transposed line and review the method, adequate for most problems, which is used for this situation. In later sections, we examine capacitance in greater detail and find the capacitance between wires (mutual capacitance ) and the capacitance un balance in un transposed lines. 5.1
Positive and Negative Sequence Capacitance of Transposed Lines
The capacitance to ground of each phase of a transposed line is computed by finding the ratio of the linear charge density to the voltage , averaged for each sec tion of the transposition. From field theory we recall that the voltage drop from is given by point 1 to point 2, both external to a linear charge density
Dx2 V1 2 = 2qx1T € Dx l In
qx ,
V
(5.1)
Note that we can keep the subscripts straight if we follow this notation carefully, always interpreting the distance from a charged line to itself, i.e. , as the con ductor radius. Figure 5.1 illustrates the distances expressed in equation ( 5 . 1 ) . Note that if > 0, and > then i s a positive drop in potential of polarity indicated. Since a transmission line is passive, the capacitance is the same for positive and negative sequence systems because the physical parameters do not change with a change in sequence of the applied voltage. Stevenson [9] uses a convenient "mod ified GMD method " for finding the capacitance to neutral of transposed three phase lines with the result,
qx
Dx2 Dx 1 ,
en
=
In (Dm Ir h'
V1 2
) nF/unit length, to neutral
Dx x
(5.2)
where i s the same GMD computed in equation ( 4 . 4 ) , r i s th e radius o f the phase conductor, and h ' is a constant depending on units of length given by Table 5 . 1 . Note the similarity in the logarithm term to the inductance equations where
Dm
1 52
1 53
Seque nce Capacita nce of Tra n s m ission lines
Fig. 5 . 1 . Configuration for equation ( 5. 1 ).
Table 5. 1. Capacitance Multiplying Constants for C in nF/Unit Length * Constan t
k' (1/3)k' f-
50 Hz
f- 60
Hz
Natural Logarithm
Unit of Length
( In )
BaBe 1 0 Logarithm
(Iog lo)
55.630 89.525 18. 543 '29.842
·km mi km mi
24.159 38.880 8.053 12.960
fk ' wk '
km mi km mi
2781.49 4476.24 17476. 57 28125.04
1207.97 1943.99 7589.90 12214.42
f k'
km mi km mi
3337.78 5317�49 20971.89 33750.07
1449.57 2309. 33 9107.88 14657. 32
wk'
3
if C is desired in microfarads; multiply *n nano 1 0- 9 ; multiply table values by 1 0 table values by 1 0- 9 if C is desired in farads; 1 . 0 mi .. 1 . 6093 km ; f .. 50 Hz, w 3 1 4. 1 9 5 rad/sec; f 60 Hz, w 376. 991 rad/sec ; eo ( 1 /lloc 2 ) .. 8 . 8 54 X 1 0- 1 2 F / m. ..
-
=
=
...
=
=
D.
we wrote In(Dm /D,) . Here the D m is the same, but has been replaced by the conductor radius since all charge on a conductor resides on its surface. If we in terpret the of each phase to be based on the radius r instead of for each wire, we may generalize the equation to write
D,
D,
h'
( 5.3) D /D� ) nF/unit length and b n = W C n nmho/unit length where D� = (D�I D� 2 D� 3 ) 1 / 3 and D�I = the geometric
Cn = In(
m
mean of distances from conductor center to outside radius in part i of a transposi tion . Usually b n is in the range 4.8 to 5.5 micromho/mile for single circuit, 60 Hz overhead lines. This can best be illustrated by means of an example .
Example
5. 1
Find the capacitance to neutral of the double circuit line shown in Figure 5.2, where the notation in parenthesis, (a, c, b ) for example , indicates that this position is occupied in turn by phases a , c, and then b in the three parts of the transposi tion cycle . All conductors are 477 ,000 circular mil ACSR , 26/7 strand . It is as = sumed that fl = or each transposition section is one-third the total line length .
f2 f3 ,
Solution
Since the line of Figure 5.2 is a double circuit line, we assume that conductors
a and a' are connected together at each end and form a parallel phase a conductor,
1 54
Chapter
5
,
5'
C-t ('.b�
"",*__
1 0'
40 '
.---- 1 4 '---4001
I
Fig. 5 . 2 . A double circuit line configuration .
and similarly for phases
D� = (D: I D�2 D� 3 ) 1/3 , where D8' 1 - (raDaa 'ra'Da 'a ) 1/4 D�2 = (r"D""'r,, , Db 'b ) I/4 D'83 -- (rc D cc 'rc ,Dc 'e )1/4
b and c. Then
Daa ' = Dec ' and, assuming that all wires have the same radius, we D� = (r 3/1 Daa, DVl,) 1/3 = rIll D!�� D Vg, The GMD may be computed as (for position 1, shown in Figure 5.2) Dm - (D- ab Db e Dca ) We note that have
_
-
-
1 /3
where
But
Then
D -- (Dab Dab , ) 1/4 (D bc D bc ' Dea De'a , ) 1/1 2 DcoI/6' m
5.2 we compute Daa ' = ( 101 + 261) 1/2 = 27.8 ft, D bb ' = ( 1 02 + 22 ) 112 = 1 0 .2 ft = Dca ' Dab = 14 ft, Dab ' = ( 1 02 + 122 )1/2 15.65 ft Dbc = 10 ft, Db e ' = ( 1 02 + 162 ) 1/2 18 .9 ft
From Figure
=
=
Sequence Ca pacita nce of Tra nsmission Lines
Dea = 2 4 ft, r=
1 55
De'a' = 28 ft
0.858
2 X 12
(from table B.8)
Substituting, we compute
12
= 0.0 35 7 ft
13
D: = ( 0 .0 35 7 ) / ( 27 .8) / ( 1 0 . 2 ) 1 /6 = ( 0 . 1 89)( 3.029)( 1 . 47 3 ) = 0.843 ft Dm = ( 1 4 X 15.65)1/4 ( 10 X 1 8 .9 X 24 X 28 ) 1/ ( 1 0. 2 ) /6 = ( 3.847)(2.663)(1 .473) = 1 5 .086 ft Then from tral is
12
1
(5.3) the capacitance per phase (or per equivalent conductor) to neu Cn =
89.5 89 .5 = In ( 1 5 .086/0.843) 2.8 8
and we may compute the
= 31 . 0 3 n F/ mi
60 Hz susceptance as
be = 21T (60) Cn = 11.70 � mho/mi!phase The preceding computations ignore the effect of the conductor height above the ground. Stevenson [ 9 ] shows that (5.3) is modified slightly when conductor height is taken into account. If conductor heights are measured to the image con ductors as shown in Figure 5 .3, we have
Cn =
In(Dm /D: )
- In(HI 2 H23H3dH1 H2H3) 1 /3 k'
nF/unit length
(5 .4)
From this equation we observe that the effect of taking conductor height into con sideration amounts to subtracting from the denominator the term In
(H1 2 H'1 3 H3 1 ) 1/3 H1 H2 HS
(5.5)
If the conductor height is large compared to the spacing between wires, this term is nearly zero . It is therefore often omitted . Usually, the actual height of attachment to poles or towers is modified for these calculations. A figure often used is the attachment height minus one-third the sag [ 38 ] . 2
2
b'
Fig. 5 .3 . Three phase conductors and their images.
Chapter 5
1 56
5.2 Zero Sequence Capacitance of Transposed Lines
The shunt admittance between line connections and the zero-potential bus in the zero sequence network depends upon the capacitance to neutral as seen by zero sequence voltages. To compute the zero sequence susceptance , we use the method of images as applied to zero sequence charges and compute the capaci tance of a transposed line. Using this method, we consider first a line without ground wires and later add the ground wire. (Also see Stevenson
[21].)
Capacitance with no ground wires. Consider the system shown i n F igure 5 .3 where charges and reside on conductors a, b , and c respectively , while the negative of these charges resides on the image conductors. Then the volt age drop from wire a to neutral is half the voltage drop or, if e is the permit tivity (see
qaO , qb O , qeO Vaa, [12] ), D f VaO = (1/2)Vaa' (1/2) L 2 2 qi L 4q, �2I �qao HraI + qb o HD I 2 qeO H3D-1 1 2 D 31 D - qaO In HraI - qbO I122 qeO H-31)3 1 VaO 2 �qaO HraI + qbO DH1I 22 + qeO H331)1 H3 ) H H 12 2 ( 2 eo o qa qb VbO O q D D 2 32 12 rb 12 �qao HD-1 33 + qbo H2D-3 qeo H3e ) Veo = r 1 23 VaO = Vb O = Veo (5.8 )-(5.1 0) qaO q b O q eO =
i
In -
= 4
1
1r e
1r ex
D I
dx =
In - +
In - H
i
1r e
In
(5.6)
In
In
(5. 7 )
which simplifies to
Similarly ,
In -
1
=
1r e
=
1
In -
In
1r e
In -
+
In
In
1r e
In D
+
+
V
In
V
In -
By definition we know that
so we conclude from
that
*
*
V
(5.8 ) (5 .9 ) (5. 1 0) (5. 1 1) (5.12)
i n any transposition section , but for the usual spacings these charges are nearly equal . If we assume equal charges over the full transposition cycle, or
qaO = qbO = qeo V "'" VaO + V3bO VeO
the voltage becomes an average of the three voltages given by av
00 -
+
2 H ) 3 3 2H2 1 1 V 3qao [HrarI Hb2re(H3(H D 12 D23 D31 )2 ] 1/9
(5. 8 )-(5. 1 0),
or
Performing this operation, we compute
00
=
21r e
In
V
( . 3
51 )
Sequence Capacitance of Tra nsm ission Lines
1 57
or we observe that
(5.14) where
Haa = GMD between the three conductors and their images Daa = self GMD of the overhead conductors as a composite group, but with D, of each wire taken as its radius
Then
qaO Co = - = Va o =
33
211' €
In(Haa IDaa )
( 1 / )k '
In (Haa ID aa )
F1m/phase
nF /unit length
(5.15)
As a rule of thumb b c o has values i n the range o f 2. 5-3.5 ILmho/mile for single cir cuit 60 Hz overhead lines. Capacitance with ground wires. To solve the capacitance problem with ground wires, we first analyze the case of a single conductor a with one ground wire g as shown in Figure 5.4. This may later be extended to a more general situation. g
qg Hg
0' Fig. 5 . 4 .
One conductor with one ground wire.
Referring to Figure 5.4, we write
Likewise ,
Va = .! 2 Vaa , = � 2 1T€
(q
H Ha In ag + a In g Da ra g
q
)
( 5 . 1 6)
(5.17) Solving ( 5 . 16 ) and ( 5 .1 7 ) for qa ' we have (5.18)
Chapter 5
1 58
Now suppose that conductor a is the composite of three phase wires and con ductor g is the composite of n ground wires. Then qG is the total charge of three conductors or, if all wires have the same charge, all
qGO = (1 / 3) qG
Also, since VGo = VbO = Veo , we have VG O = VG From (5.18)-(5 .20 ) we compute (1 / 3)k' ln(H,, /D,, ) = =--G_G-In-H�'-='-_--="='(-ln-H=-G'-)""2 " nF/unit length/phase Co ltl--H D" DG, DGG where same as in equation (5.14) self GMD of ground wires with D. = r, DDDGG"G, === the GMD between phase wires and ground wires HGG = GMD between phase wires and their images H" = GMD between ground wires and their images HGI = GMD between phase wires and images of ground wires 5.3
(5.19) (5.20)
(5.21)
Mutual Capacitance of Transmission Lines
In the preceding paragraphs we considered the calculation of capacitance be tween a charged conductor and ground for positive, negative, and zero sequence charges. We now expand our consideration to the capacitance between nearby conductors and shall refer to the result as "mutual capacitance." The subject of self and mutual capacitance is treated in many references. Rather than treat every possible physical configuration, we will investigate the transposed single circuit three-phase line in some detail and then consider briefly the effect of ground wires. Dissymmetry due to untransposed lines will be cov ered in section 5 . 8 . Consider a group of n conductors carrying linear charge densities qG ' qb ' . . . , qn ' located above the ground plane as shown in Figure 5 . 5 . Using the method of images (see [ 1 2 ] ), we compute the voltage drop between two points 1 and 2 by superposition (since the system is linear). From ( 5 . 1 ) we write 1
(5.22)
where the D's are distances between conductors or points above the ground and the H's are distances between conductors or points in space and image charges. In this way we compute the voltages of all conductors to ground by noting that is referred to Clarke [11 ] , Wagner and Evans [ 10 ] , and Lyon [ 3 9 ] for additional reading o n the subject. Our approach follows Calabrese [ 24 ] which i s recom mended for its clarity and complete coverage of the subject.
1 The interested reader
Seque nce Ca pacita n ce of Tra nsm ission Lines
1 59
G ROU N D
Fig . 5 . 5 .
Group of n charged lines and their images.
(5.23)
(qa
Then, for example ,
V =
a
Ha + q b In Hab Han ra D-ab + . . . + q n In D-an Dab - . . . - q n In D-an) - qa In Hraa - qb In Ha b Han 1 1 V ,=41T € 2
aa
-
In
-
and combining terms,
V =
a
1 -
21T€
Similarly ,
Vn
=
�qa
-
In
-� 1 21T €
-
} qa n
Ha + qb In Hab + . . . + q n In Han ) r Da b D-an V -
-
a
) Han + b In H-b n + . . . + q n n Hn - V r Db n Dan n -
}
q
(5.24)
(5.25)
( 5.26)
Equations (5 . 2 5 ) and (5 .26 ) are often written in matrix form as V
=
Pq
(5.27)
where V is the voltage vector, q is the charge vector, and P is a matrix of coeffi cients called "potentia} coefficients" [ 2 4 ] where p
.. -
Hj ' rj 1 H. . = - In � , i * j 21T € D ij
=
I}
1 } l =] n 21T € •
•
F- 1 m
( 5 .28)
Chapter 5
1 60
and we note that P is symmetric. Note also that V and q are not restricted in any way but may be sinusoidally varying time domain quantities or transformed quantities such as phasors. One can think of the matrix P as being a chargeless or "zero charge matrix." For example, suppose we set
1 o
0
q =
elm
(5.29)
o
i.e . , q a is the only charge present and its value is 1 coulomb/meter. Then
Va
Paa :::
1
-
211'€ V
Pna
In H-raa In HD-aabb In
Ha n Dan
V (5.30)
or P le a is the potential in volts assumed by wire k due to linear charge on wire a alone ( of 1 coulomb/meter) with all o ther charges zero. All these coefficients are positive quantities since Ha n > Da n , these being equal only if charge a is on the ground plane . With charge qa acting alone as in and (5.30), the potential Va is the greatest of all potentials in ( 5. 30 ) , but all are positive. In a similar way we can define each column of P as a set of potentials derived by setting the charges equal to 1 coulomb/meter and with all other charges equal to zero. Thus each column of P is independent of all others depending only upon the geometry of the system, and the columns of P comprise a basis for an n dimensional vector space [ 40] . This being the case, P is nonsingular and has an inverse P - I . Thus from we may compute
(5.29)
(5.27)
q = cV
(5.31 )
c = p-I
(5.32 )
where 2
The elements of the matrix c are called "Maxwell's coefficients" or more specifi cally are called "capacitance coefficients" when referring to diagonal terms and "coefficients of electrostatic induction" when referring to off-diagonal terms. These coefficients may be thought of as short circuit parameters which relate the charges to the system voltages. For example, with 2 Obviously, this is not the same matrix C as defined in Chapter 2. There will be no ambiguity, however, as we usually use for the symmetriCal component matrix. Here the c is written lowercase for capacitance per unit length. The uppercase C is for total capacitllnce of the line.
A-I
Sequence Capacitance of Tra nsmission lines
1 61
1 o V=
0
1
o
(5.33)
i.e. , with Va = and all other voltages zero (short circuited), w e find the first column of c to be
elm
=
-C
qn
(5.34)
na
The negative signs from the off-diagonal terms arise due to the fact that the coef ficients of P in ( 5.27 ) are all positive. Then the inverse of P is adj P [PiJ ) t = = p- l = det P det P
c
(5 .35)
where Pij is the cofactor of element Pij . But
Pij
=
(- 1 )i +JM1j
where Mij is the minor of element Pij . We set
(5.36)
[24]
Cij = :!� P for i = j or (i + j ) even =
Then c=
-
t
O
-C
-C
M ij
det P
-C
.
. #= ' ] or ( l + ]. ) 0dd
for l ab
-C
j
"
ba
+C bb
-C
na
- Cn
+C nn
b
bn
F/m
(5.37 )
(5.38)
and only the diagonal terms have positive signs. The coefficients (elJ ) themselves are all positive quantities and are all capacitances. Note that all coefficients may be found by superposition, i.e . , by applying 1 volt to wire a, b, etc., always with every other wire shorted to ground. The negative sign really means, then� that applying a positive potential to one wire induces a negative charge on the other wires. This is intuitively correct since the application of a dc voltage to a capa�· itor makes one terminal (or plate) of the capacitor positive but also makes the other terminal ( plate) negative. Equation (5.31 ) is the desired relationship for self and mutual capacitances of an n wire system. It is not in the best form for physical interpretation, however. To make this clearer, suppose that both q and V are sinusoidal time varying quantities. Specifically , let one of the charge densities be
1 62
Chapter 5
q = Qm cos w t
(5.39 )
Then, since current is the time rate of change of charge, we compute i = dq/dt = - w Qm sin w t
(5.40)
Now transform both (5.39 ) and (5.40) into phasors. Then the phasor charge density Q and phasor charging current I are defined as ( 5.41 ) and 1- w Qm e h r/2
�
(5.42 )
--
or
1 = w Qe h r/2 = jw Q
(5.43)
If we write ( 5 .31) in terms of phasor charge density and voltage Q and V , we have Q = CV
(5.44)
which we change to a current equation by multiplying by jw as in (5.43), or (5.45)
1 = jw Q = jwCV
But, from circuit theory we write the charging current as I = YV
(5.46)
Y = jwC
(5.47 )
or the phasor admittances are For example, from (5.46) we interpret that Ykit = sum of all admittances connected to k = jW Ck k , a capacitive susceptance
(5.48 )
Yk m = the negative of all admittance connected between
k and m (5 .49 )
= - jw ck m
or the actual admittance between k and m is (using lowercase for the actual quan tity and uppercase for the matrix element) (5.50 ) which is again a capacitive susceptance. From (5.47 )-(5.50) we visualize an equivalent circuit as shown in Figure 5.6. Since YkIt is defined in (5.48), the capacitances to ground are Cag
=
Caa - Ca b - Cae - • • •
Cbg
=
- Cba + Cb b - Cbe
Cng
=
- Cna - Cn b - Cne
-
-
-
Can
F/m
Cbn
F/m
• • . + Cn n
F/m
•
•
•
-
(5.51 )
Sequence Capacita nce of Tra n s m ission Lines
\
1 63
, , \
Fig. 5 . 6. Self and mutual capacitances of an n-phase system .
With capacitances so defined, it is easy to see that the sum of admittances con nected to node k is indeed Yk k • The equivalent circuit of Figure 5.6 can also be obtained by algebraic manipulation of ( 5 .47). (See problem 5 .2. ) 5.4 M utual Capacitance of Three-Phase L i nes without G round Wires
We now consider a special case of the general mutual capacitance problem where there are only the thref> charged conductors of a transposed three-phase line with no ground wires. From ( 5 .25 ) and (5.26) the potential equations are, in terms of potential coefficients, V = Pq, or
pa
j
Pbe
5)
(5. 2
Pee
�aJ [
where the elements of P are given by ( 5 .28 ) . The charge equation is found by solving ( 5 . 5 2 ) for q. Thus, in terms of Maxwell's coefficients, q = cV , or
qb
_
-
qe
where the elements of
caa - C ba
- Cab Cb b
- Cea
- Ceb
c are given by ( 5. 3 7 ) .
Let
[�j
elm (5.53)
det P = Paa ( PbbPee - P�e ) - P�bPee + 2PabPbePae - P�ePbb Then compute the minors (noting that P il. = P k i )
Maa = Pbb Pe c - P�e Mb b
= PaaPee
- P�e Mee = PaaPbb - P�b
c
Mab Mac
= PabPee
- PaePbe
= Pab Pbe - PbbPae M be = PaaPbe - PabPae
( 5.54)
Then the elements of are computed from ( 5 . 3 7 ) . For the capacitance to ground we compute from ( 5 .5 1 )
Call
= Caa
- Cab - Cae
F /m
C bll = Cbb - Cab - Cbe F/m Cell = Cee - Cae - Cbe
F/m
( 5.55)
These various capacitances may be viewed symbolically as shown in Figure 5.7 where the various capacitances are shown as circuit elements.
1 64
C h a pter 5
Fig. 5 . 7 . Capacitances of a three-phase line with no ground wires.
If the line is transposed in sections of per unit length f, , f2 ' and f3 the voltage equation (5.52) may be written using the notation of Chapter Thus we write
4.
( 5. 56) where
C2 3 1
=
R� l C m RI/>'
C3 1 2 = RI/> C I 23 R� '
and C m is defined to be the Maxwell's coefficient matrix for section fl ' The potential matrix is then found by inverting the entire matrix expression in ( 5. 56) . If the line is completely transposed, each phase occupies each position for one-third the total line length. Since each section is one-third the total length, the capacitance per phase per meter consists of one-third meter of each of and or
Cab ' Cb c .
Cac
CMO
=
( 1/3)( Cab +
Similarly, for the capacitance to ground c,.o = ( 1 /3 )( ca,. + ClI,. + where
CsO
=
Cbc + Cac) F/m
cc, )
( 1 /3)( caa +
s
=c
o
- 2 CMO
( 5. 57 )
F/m
Cbb + Ccc) F/m
( 5. 58)
( 5 . 59 )
Actually, these "transposed capacitances" are just averages o f th e capacitance seen by each phase in each transposition section. The capacitance to ground can be thought of as a combination of a self capacitance and the mutual capaci tance
Cs
CM'
Example
5.2
Find the mutual capacitance and capacitance to ground of the lower circuit only (ignore upper circuit and ground wires) of Figure 5.2. Compare the capaci tance to ground computed in this way to that obtained by application of ( 5. 3).
Solution First, we form the P matrix, the coefficients of which upon geometry. In our case (see Figure 5 . 2 )
are
entirely dependent
D a 28. 0 ft Dab D bc 1 4 .0 ft, Ho b Hbc = ( 802 + 142 ) 1 /2 = 8 1 . 2 ft Ha Hb Hc 80.0 ft, 0.0357 ft Hac ( 80 2 + 28 2 ) 1 2 84. 8 ft, =
= =
c
=
=
=
/
=
=
=
r =
1 65
Sequ ence Capacit a n ce of Tra nsm ission Lines
Then from ( 5 . 28)
H 1 In -.!! P aa = P"" = P cc = 21T f ra and if we let
where "
f = fo"
= " /( 36
Ha ra
F- 1 m
= 1 for air dielectric, w e have P aa = 18 X 10 + 9 ln
= 11.185 In Ha ra
1T
X
MF" l mi
= 11.185 In 0. :�5 7 = 86. 28
1 Ha b Pab = Pbc = 1 1 . 1 8 5 n Da b
Pac = 1 1 . 1 85 In Then p=
[
109)
M F- 1 mi
1.3 = 19.66 = 11.185 In 814.0
�:: = 1 1 . 1 85 In �::� = 12.39
[
86. 3 19. 7 1 2. 4
19. 7 86.3 1 9. 7
1
MF" mi
MF- 1 mi
�
1 2.4 19. 7 MF- 1 mi 86. 3
Direct inversion o f P by digital computer (see Appendix A) gives the result
c=
J
1 2. 34 - 2. 54 - 1.19 - 2. 54 1 2. 7 5 - 2. 54 - 1.19 - 2. 54 12.34
nF/mi
Since a digital computer is not always available, we confirm this result by hand computation. From
(5.54) we compute,
A = d et P = Paa Pb bPcc - PaaP�c - P�bPcc + 2P a bPcaPb c - P�c Pb b
=
86.33 - 86.3(19. 7)2 - ( 19.7)2 (86.3) + 2(19. 7 )2 ( 1 2.4) - ( 1 2.4)2 (86.3) e! 647,500 - 33,600 - 33,600 + 8,330 - 13,300 = 575,330
Also from
( 5. 54), Maa
= 7091 Mab = 1459
= 7091
Mc c
M b b = 7326 from which we compute trom
= 1 2.31 C b b = 1 2. 72 Caa
Ma c = - 685 M bc = 14 59
( 5.37),
= 12.31
nF/mi
Ccc
nF/mi
Cac = 1.190 nF/mi
nF/mi
Ca b = 2.535 nF/mi
C b c = 2.535 nF/mi
These values are seen to be very close indeed to the digital computer results. Since
1 66
C h a pter
5
the more accurate digital computer solution is available, we use it as a basis for computations which follow. The capacitances to ground are, from ( 5. 55), Cag = Caa - Cab - Cae Cbg = Cb b - Cab - Cb e ceg = Cee - Cae - Cb e
= 8. 6 06 = 7. 66 7 = 8 . 6 06
nF /mi nF/mi nF /mi
For a transposed line the capacitance to ground is the average of these three, Le., from ( 5 . 58 ), CgO
= ( 1 /3)( cag + Cbg + ceg ) = 8. 2 93
nF/mi
The average ( transposed) mutual capacitance is, from ( 5. 57), CMO
= ( 1 /3 )( cab
+ Cb e + cae )
= 2.091
nF/mi
The average (transposed) self capacitance is, from ( 5. 59),
Cso = ( 1 /3 )(caa + Cb b + Cee ) = 1 2. 47 5
nF/mi
Now compare Cgo computed above to the capacitance to neutral computed by equation ( 5 . 3 ) which neglects the effect of the earth and is computed under the assumption of equal charge density in each part of a transposition cycle [ 9] . From ( 5.3)
Cn = In(D89.m /5D., )
where
nF/mi
= Deq = (Da b DbeDea ) I/3 D; = r = 0.0357 ft
Dm
=
1 7. 5 ft
Then en = 1 4 .4 nF/mi. How do we reconcile this difference? If we imagine an equivalent circuit similar to Figure 5. 7, the .6 -connected mutual capacitors all have a value of CMO 2.091 nF /mi. Convert this .6 to a Y. Then each capacitor the in the Y has a value cy 6.273 nF/mi. Since this is in parallel with total capacitance to neutral per phase is the sum Cn Cg O + c y = 14. 566 nF/mi. If the computed by ( 5. 3 ) is corrected by its height above the ground, these values will be in quite close agreement.
=
=
=
Cn
Cgo ,
5.5 Sequence Capacitance of a Transposed line without Ground Wires
Consider a three-phase line, the capacitance of which is described in terms of its Maxwell coefficients as q = CV, where q and V are phasors. Then by ( 5.45), 1 = j w CV or, to emphasize the a-b-c coordinate system, lab e
= j wCVabe
(5. 6 0)
But this is easily changed to a 0- 1 -2 coordinate system by a similarity transforma tion. Thus ( 5.61 ) or (5.62)
1 67
S e q u e nce Capacitance of Tra nsmission Lines
where we have defined
( 5.63) Performing the indicated transformation, we compute
CO l l
=
o
�
o
CSO - 2 CMO ) 1 (CS I + cM d 2 (CS2 + CM 2 )
j
1
2
( CS 2 + C M 2 )
(CS I + CM d (CS 2 - 2 CM2 )
(cs o + C MO ) (CS I - 2CM d
( cs o + CMO )
( 5.64)
where we define CSO and CMO as in ( 5. 59) and ( 5.57) respectivelyl and where
2 CS I = (1 /3 )( caa + aCb b + a ccc ) 2 CS 2 ( 1 /3 )( caa + a cb b + a ccc ) 2 CM I (1 /3 )( cbc + a ca c + a ca b ) 2 CM2 ( 1 /3 )( cbc + a cac + aCa b )
= = =
( 5.6 5)
If the line is transposed, the values in ( 5. 6 5) are all zero because each phase oc cupies each position for an equal distance, thereby acquiring a multiplier of 2 ( 1 + a + a ) for each capacitance term . With these sequence mutuals all zero, ( 5.64 ) becomes
CO ll
=
t
0
s o - 2 CMO 0
CSO + CMO
o
0
( 5.66)
and the mutual coupling between sequence networks is eliminated. Note from ( 5 . 66 ) that the zero sequence capacitance is much less than the positive and negative sequence capacitances. Also note that the positive and nega tive sequence capacitance to neutral is given by
( 5.67) which should agree closely with ( 5. 3 ). Similarly, for the zero sequence
Coo
= Cs o -
2 CMO
( 5.68)
which should check with ( 5. 1 5).
Example
5. 3
Compute the matrix CO l 2 for the transposed line of Figure 5. 2, data for which is computed in Example 5.2. Compare with results computed from ( 5.3) and
( 5. 15).
3 Note that the signs i n C O l 2 are not in the same pattern as in the ZOl 2 matrix of the last chapter. This is because of the negative signs in the Ca bc matrix of equation ( 5 . 53 ).
Chapte r 5
1 68
Solution From Example 5.2 we have
Cso
=
CM O
12.475 nF/mi,
=
2.091 nF/mi
Then C l l = C22 = Cso + CM O = 14 . 5 nF/mi which checks exactly with the Cn computed in Example 5. 2. Also Coo 8. 293 nF/mi. For the zero sequence capacitance we have from 5 . 1 5 ,
66
(
)
Co = where
=
29.8
1n(Haa /Daa )
n F /mi
Haa [HaHbHc (HabHbcHac ) 2 r/9 ( 80 ) 1/3( 81 . 3 X 81 .3 X 84.6)2/9 = (4.3 )( 1 8.9) Daa [ r3 (Dab Dbc Dca) 2 ] 1/9 ( 0.03 57) 1 /3 ( 1 4 1 4 X 28) 2 /9 = (0.33)(6.75) =
=
=
81 .3 ft
=
X
=
Thus
Co =
29.8 1n ( 81 .3/2. 22)
=
=
2. 22 ft
8.28 nF/mi
which checks very well with Coo .
5.6 Mutual Capacitance of Three-Phase Lines with G round Wires
Having computed the self and mutual capacitance of a circuit without ground wires, we now consider the additional complication added by the ground wire and study its effect upon capacitance of a line. As before, we consider the line as being transposed, leaving the problem of unequal phase capacitances for a later section. Using the subscript n to denote the ground wire, we may write the equation of potential coefficients for the four-wire system shown in Figure 5.8. We write V = P q or, in more detail and with Vn = 0 , a
b C
a
Ca b
Cog
0
b
c oo
Fig. 5.8. A three-phase line with one ground wire n .
C 0
Sequence Capacitance of Transm ission Lines
Va Paa Pab Pac Vb Pba Pbb Pbe Ve Pea Pcb Pee Pna Pn Pne =
----------
b
0
I I I I I ..1-
I I
1 69
qa qb qe qn
Pan Pbn Pen Pnn
---
( 5.69 )
But this matrix can be reduced to three equations by eliminating the fourth row and column. Solving the last equation and substituting back, we have
Va �Paa PanPna) Pnn Vb (Pba - P�nn:na) Vc (pea PenPna) Pnn
�ab - P;�nb) (Pbb - P�nn:nb) oeb - PenPnb) Pnn
_
=
_
�ae - PanPnc) Pnn ) 0be - PbnPne Pnn (Pee - PenPne) Pnn
qa qb qe
( 5.70)
This could be simplified slightly by taking advantage of the fact that P is sym metric. Since the elements of P are all positive, the new P of ( 5.70) contains elements all of which are smaller than corresponding elements for the same line with no ground wires. Note that the "correction factor" for each element is a positive quantity which depends only upon the geometry of the system (see (5.28». If the ground wire is moved toward infinity, this correction factor approaches zero since the entire row and column associated with the ground wire vanish. Using the corrected P matrix of ( 5 . 70), we again find the capacitance matrix V, where Ca be is the of Maxwell's coefficients by matrix inversion, i.e . , q = inverse of the 3 X 3 matrix of ( 5.70). Since the presence of the ground wire makes the elements of P smaller, we would expect the elements of C to be larger than the case of the same system with no ground wires. The sequence capacitance matrix CO l 2 is again found by a similarity transfor mation, exactly as in the case of no ground wire, by ( 5.63). If it is desirable to consider the capacitance to ground separately from the capacitance to neutral, as shown in Figure 5.8, this can be done by inverting the 4 X 4 P matrix of ( 5 .6 9 ) . This would permit separate identification of elements such as Ca n , C b n , and C e n as sh own in Figure 5.8. Since these capacitances are in parallel with call ' C bll ' and Cell ' it is apparent that the capacitances are increased by the ground wire . If there are two ground wires m an d n , the procedure is exactly the same. In this case the equations involving potential coefficients are
Cabe
Paa
Pa b
Pac I
Pam
Pba
Pb b
Pbe I
Pb m
Ve = Pea o
o
which we rewrite as
Pe b
Pma
Pm b
Pna
Pn b
I
Pee
II
� Pm m
Pme I I Pne I
-- ---------
Pem
Pan Pbn Pe n
Pm n
- - - -- - -
Pn m
P
nn
(5.71)
Chapter 5
1 70
( 5. 7 2) where the matrices P I , P2 , P 3 , and P4 are defined according to the partitioning of ( 5 . 7 1 ). Solving for Va ll c we have (5.73) where
Pa llc
=
P I - P2 P:;j I P3
( 5. 74 )
Thus, P2 P:;j l P 3 may be thought of as a correction due to ground wires. The Max well coeffiCients are found by inverting ( 5. 74), i.e. ,
Cabc = P;�c
( 5. 75)
and the sequence capacitances are found by applying the similarity transformation ( 5.63) to Ca llc .
Example
5. 4
Repeat the capacitance calculation of Example 5.2, this time including the ef fect of the ground wires shown in Figure 5.2. Assume the ground wire is 3/8-inch EBB steel.
Solution First we compute the potential coefficients for the ground wires. Thus
Pmm
= Pnn
H = 1 1 . 1 85 In --.!!!. rm
where
rm
==
Hm
( 1/2) ( 3/8) ( 1 /1 2 ) = 0.01562 ft,
or
Pmm
= Pnn =
1 1 . 185 In
Also,
Pmn
==
0.
�;�6 2
==
=
2 ( 55)
=
1 1 0 ft
99. 096 MF-' mi
Hm n Pn m = 1 1 . 185 In D mn
where
or
Pmn
= Pnm
==
1 1 . 1 85 In ( 1 1 1 /1 4 )
=
23. 1 4 7 MF- ' mi
We also compute
Pan
= Pbn = Pb m = Pc m
H = 1 1 . 185 In an Da n
Sequence Capacita n ce of Tra nsm ission Lines
and
Pam = Pen = where
Then
Ha n = (
11.185 In DHaamm Ham = (
952 + 21 2 ) 1 /2 = 97. 29 ft Dam = ( 1 5 2 + 21 2 ) 1 /2 = 25. 81 ft
952 + 72 ) 1 /2 95.26 ft, Da n = ( 15 2 + 7 2 ) 1 / 2 = 16.55 ft, 5!!
Pan =
11 . 185 In 95.26 16.55 = 19 . 574
Pam =
11 . 185 In 97.29 25.81 = 14. 844
and
Thus
a
a
b P= c
m
n
b
c
m
MF-1
MF-1
mI·
mI·
n
86.3 19.7 12.4 I 14.8 19.6 19.7 86.3 19.7 I 19.6 19.6
���_��� _ ��: �_��� _ ��!
14.8 19.6 19.6 99.1 23.1 19.6 19.6 14.8 23.1 99.1 I I
I
Now P4 1 = adj P4/det P4, so we compute
99.1 2 - 23.P = 9821 - 534 = 9287 1 r 99.1 - 23.1] r 0.0107 - 0.00251 928 7 t 23.1 99.1 = l:- 0.0025 0.010 7J
det P4 = p;; 1
4
Then P2 P4 1 P 3
_
�19.614.8 19.619.6� [- 0.0025 0.0 107 - 0.002 5] �14.8 19.6 19.6 0.0107 19.6 19.6 14.8J 19.6 14.8 �- 5.51.99 6.275.51 4.705.51 4.70 5.51 4.99J �819.76.3 19.786.3 �::� I:::� :::� ::�� -
=
_
and
Pabc = PI - P2 P4 ' P3 =
12.4 19.7 86. �J L�. 70 5.51 4.9�J -
171
1 72
=
Finally,
Co b c =
[
�
Chapter 5
�
1.30
14. 1 5
7.6
14. 1 5
80.02
14. 15
7.69
14. 1 5
81.30
�
P;; lc , which we compute by digital computer to be cobc =
12.747
- 2.106
- 0.839
2. 106
13 . 242 - 2.106
- 2.106
-
- 0.839
nF/mi
12.747
The capacitances to ground are
co.
=
coo - Cob - coc
C b.
=
C b b - cob - Cbc = 9.030 nF/mi
=
9.802
n F Imi
cc• = Ccc - Coc - Cbc = 9.802 nF lmi For the transposed line the capacitance to ground in each phase is the average of these three values, i.e.,
C.o
=
( 1 /3) ( co. + Cb. + cc. )
=
9 . 545
nF/mi
which is considerably larger than the 8. 293 nF/mi for this same line without ground wires. The average (transposed) mutual capacitance is
CM O
=
which is smaller than the
( 1 /3) ( Co b + Cbc + coc )
=
1 . 683
nF/mi
2.091 nF /mi for the same line without ground wires.
The average (transposed) self capacitance is
cs o
CCC ) 12.912 12.475 Coo = CSO - 2CMO C ll C22 Cs o + C O 1 4.595 8.293 14. 566 =
( 1 /3 ) (coo + Cb b +
which is greater than the
=
nF/mi
nF/mi for the same line without ground wires.
The sequence capacitances are =
=
= 9. 546 nF/mi and
M
These values are both greater than the
=
nF/mi
and
computed for the same
line without ground wires.
5.7
Capacitance of Double Circuit L ines For the case of a double circuit line with or without ground wires, the problem
becomes more complicated because of the presence of so many charges . Consider
5.9
the configuration of Figure where the distances from one conductor c' to all other conductors and to all images are shown . Using the method of section 5 . 4 , w e may write the voltage equations i n terms of potential coefficients as V Pq or
=
1 73
Seque nce Capacitance of Transm ission Lines
-qb
Fig. 5 .9. Configuration of a double circuit line.
Va
Paa
Pab
Pac
Vb
Pba
Pb b
Pb c
Pca
Pcb
Pcc
Vc Va ,
=
Vb ' Vc ,
-----------
Pa'a
Pa' b
Pb ' a
Pb ' b
Pc' a
Pc' b
I I I
Paa'
Pa b '
Pac'
qa
P ba '
Pb b '
Pb c'
qb
Pcb '
Pcc'
qc
I I Pca' I
T
------------
Pa' c I Pa' a' I Pb ' c I P b ' a ' I Pc' c I P c ' a ,
Pa ' b ' Pb ' b '
Pa' c' Pb ' c'
qa'
Pc' b '
Pc ' c ,
qc'
qb ' ( 5. 7 6)
which we may write as (5.77) The capacitance matrix C is the inverse o f P , which may b e computed according to the partition ing of ( 5. 7 7 ) as ( see [ 7 ] ) ( 5.78) where we define so that Ft
= P2 1 Pl l
F = P,l P 12
( 5.79 )
since P is symmetric, and E = P22 - P2 1 P, l P 12 = P2 2 - P2 1 F
Once C is determin ed, we may write I =
jw
( 5. 80 )
C V, where both I and V are 6 X
1 . It
1 74
Chapter
will be helpful to write
5
[ ] IO b C
lo' b ' c'
where C " C2 , Ca , and C4
are
( 5.81 )
dermed in ( 5.78). Then we compute
lob c = j w (C 1 Vob c + C2 Vo' b ' c ' ) lo' b ' c' = j w ( C a Vob c + C4 Vo' b ' c' )
( 5.82)
Since the lines are in parallel ( 5. 83 ) or lob c = j w (C1 + C2 ) Vobc Io' b ' c ' = jw (Ca + C4 ) Vob c
( 5.84)
and the total charging current is Ich• = Iobc + Io' b ' c ' = j w (C I + C2 + Ca + C4 ) Vob c
( 5. 85 )
Then the parallel circuit line behaves like a single circuit line with capacitance Ceq , where ( 5.86) The sequence capacitance may be found for each circuit or for the parallel equiva lent by similarity transformation, i.e. , CO l 2eq =
A - I Ceq A
( 5.87 )
Also, from ( 5 . 86 ) and ( 5 . 78) we compute Ce q =
Pi t
+
(F
-
U) E- I (F
-
U)t
( 5 .88)
where U is the unit matrix. Thus we see that the effect of the second circuit on the first depends strongly upon the m atrix (F U). Since P I I =1= P 1 2, we write -
F
-
U
= Pi l P 1 2
-
U
( 5 .89)
If F U is nearly zero, Ceq becomes nearly Pi l , which is the value for Cd c alone and would indicate no effect at all due to the second circuit. This will be the case if P 1 2 9!! P l l such that F U = Pi t Pl l U = U - U = O. This will be approxi mately true if the height is large compared to the distance between conductors and if the conductors are arranged in vertical phase configuration such that Ho b ::!! Ho b ' , etc. The matrices P i l and P 1 2 will never be exactly equal since this would require the wire radius to equal the distance between phases, or r = D 00 ' such that Poo = P oo' . For the case of double circuits with ground wires, the equation involving P will be similar to ( 5.76) but will have added rows and columns for the ground wires. These can be eliminated by matrix reduction since the voltage of the ground wires is zero. The result then can always be reduced to a 6 X 6 P matrix equivalent. -
-
-
Sequence Capacitance of Tra nsm ission Lines
Example
1 75
5. 5
Compute the capacitance matrix for the double circuit line of Example 5.1 and Figure 5.2, including the effect of the ground wires which are assumed to be 3/8-inch EBB steel. Examine only one transposition section.
Solution
First we compute the P matrix. If we arrange as shown in ( 5.76), we know the lower right partition from Example 5.4, which involved only the lower a' -b'-c' circuit and the ground wires. For the upper circuit
D b e :;: 10 ft,
Da b :;: 14 ft,
Dea :;: 24 ft
and
Ha Hab Hbe He a
:;: Hb = He = 100 ft :;: (100 2 + 14 2 ) 1 / 2 = 100.98 ft :;: ( 100 2 + 10 2 ) 1 /2 = 1 00. 50 ft = (100 2 + 24 2 )1 /2 = 102.84 ft r = 0 . 0 35 7 ft
Then
Paa :;: Pb b :;: P ee :;: 1 1 . 1 85 In P a b :;: 1 1.185 In
1
�:
:;: 11. 185 In
��: �8 :;: 22.100
��:�0 :;: 25.810 1 4 Pea :;: 11. 1 85 In �!: � :;: 16.276
Pbc :;: 1 1 . 1 85 In
1
0.
���7 :;: 88. 784
MF- 1 mi
MF- 1 mi MF- 1 mi MF- 1 mi
Between the a-b-c circuit and the ground wires we compute
Dam Dan Ham Ha n
:;: :;: :;: :;:
7.07 ft 19.65 ft 105. 1 2 ft 106. 71 ft
Dbm Dbn Hbm Hb n
:;: :;: :;: :;:
10. 30 ft 7.07 ft 105.39 ft 105. 1 2 ft
Dem Den Hem Hen
:;: :;: :;: :;:
19.65 ft 7.07 ft 106.71 ft 105. 12 ft
such that
105. 12 :;: 3 O. 1 89 M F- 1 mI' P am :;: P en :;: P bn :;: 1 1 ' 185 In 7.07 106. 7 1 _ - 18.9 2 7 MF - I ml· Pa n -- P em - 1 1. 1 85 In 19.65 105.12 . Pb m - 11 1 85 In 10.30 - 26.0 1 5 M� I ml
.
Between the a-b-c circuit and the a' -b' -c' circuit we compute
C h a pter
1 76
5
Daa ' = 27.86
ft
D b a ' = 15.62
ft
Dca ' = 10. 20
ft
Dab ' = 1 5.62
ft
D b b ' = 10. 20
ft
De b ' = 15.62
ft
D o c ' = 10. 20
ft
D be ' = 18.87
ft
De c ' = 27.86
ft
Haa ' = 93.68 Ho b ' = 90. 80
ft
Hb a ' = 90. 80
ft
He a ' = 90.02
ft
ft
ft
He b ' = 90.80
ft
Hac ' = 90.02
ft
Hb b ' = 90. 02 Hb e ' = 91.41
ft
Hee ' = 93.68 ft
such that
93.68 = 1 3 56 5 MF- ' mI· 27.86 . 90.80 = 19 . 686 MF- ' mI· Pa b ' = Pba ' = Pc b ' = 1 1 . 185 In 1 5.62 90 .02 = 24 . 359 M F- I mI· Pac ' = P b b ' = Pe a ' = 1 1 . 185 I n 10. 20 91.41 = 1 7 . 649 MF- 1 mI· Pbe ' = 1 1 . 185 1n 18.87 Paa ' = Pee ' = 1 1 . 185 I n
Rounding to the nearest
a
b c
P=
a'
b' c
,
m
n
a
b
88.8 22.1 16.3
22. 1 88.8 25. 8
0.1, we write a
c
,
16.3 II 13.6 25.8 : 19. 7 88. 8 24.4
----- -----
-
1 I
,
b'
c
19. 7 24.4 19. 7
24.4 1 7.6 13.6
---- - -
-
13.6 19. 7 24.4
19.7 24.4 1 7.6
24. 4
86.3 I 19. 7 I 19. 7 I 13.6 I 12.4
19.7 86. 3 19. 7
30. 2 1 8.9
26.0 30. 2
18.9 30. 2
1 9.6 19.6
14.8
19.6
--
m
n
30.2 26.0 18.9
1 8.9 30.2 30.2
,
--
-
-
-
-
--
--
1 2.4
14.8
19.7 86.3
19.6 19.6
19.6 19.6 14.8
19.6 14.8
99. 1 23. 1
23. 1 99. 1
MF- 1 mi
We eliminate the two " outside" rows and columns delineated by the solid parti tion lines to compute a "correction matrix "
6
X
6
matrix is
Pab e = P - Pc .
Pe =
Finally,
10.70 10.98 9.03 -
10.98 13.04 1 1.60
9.03 11.60 10. 70
--------- -
6. 56 7. 86 7.27
8.04 9.00 7.78
--
Pc , I I I I
:
-,
where we define
6.56 8.04 7. 27 -
-
--
7. 27 I 4.99 I 7.86 I 5. 51 I 6.56 I 4.70
7.86 9.00 7.86
7.27 7.78 6.5 6
- - - -
-
5.51 6.27 5.51
Pc such that the
-
-
4. 70 5. 5 1 4.99
1 77
Sequence Capacitance of Tra nsm ission Lines
a
Po bc
=
b c a
,
b' c'
a
b
c
78.08 11.12 7 . 24
11.12 7 5.75 14.21
7.24 14. 21 78.08
7.00 11.82 17.09
1 1 .64 1 5. 36 9. 87
17.09 11.82 7.00
a'
b'
c
I I I I I
7.00 11.64 17.09
1 1.82 1 5.36 11.82
1 7.09 9.87 7.00
I I I I I
81.30 14. 15 7.69
14. 1 5 80.02 14.1 5
7.69 14.1 5 81.30
'
--------------+--------------
which is inverted by digital computer to find
a
b
c=
c a'
b' c'
a
b
c
13.80 - 1 . 29 - 0.55
- 1. 29 14.43 - 1.88
- 0. 55 - 1.88 13.93
--------------
- 0.45 - 1. 20 - 2.45
- 1. 14 - 1.95 - 0.87
We also compute
Pi } P1 2 and
F
-
U
=
=
� [
- 2.38 - 1. 1 2 - 0.44
. 0526 0. 1091 0.1 939 0.9438 0. 1091 0. 1939
I I I
f I I I I I
a'
b'
c
- 0.45 - 1.14 - 2.38
- 1. 20 - 1.95 - 1. 1 2
- 2.45 - 0.87 - 0.44
'
- - - -- - - - - -- -- -
13.34 - 1.63 - 0. 54
0.1177 0. 1648 0. 1 105 0. 1177 - 0.8352 0. 1105
- 1.63 13. 79 - 1.66
- 0. 54 - 1.66 13.30
0. 201 0.0905 0.0546
=F
�
nF /mi
�
0.2010 0.0905 - 0.9454
Since F - U is quite different from 0, we conclude that the second circuit has a substantial effect upon the capacitance of the first. This is also evident if one compares the a-b-c partition of C with the C matrix of Example 5.4.
5.8
E lectrostatic Unbalance of U ntransposed Lines
If transmission lines are left untransposed, a practice which is becoming rela tively common, an electrostatic unbalance exists in addition to the electromag netic unbalance studied in Chapter 4. Any unbalance in transmission line charging currents results in the flow of neutral ''residual'' current in solidly grounded sys tems, and this current flows at all times, independent of load current. If the un balance is great and these residual currents are large, they could possibly affect system relaying or cause the voltages to become unbalanced. This problem has been studied extensively and methods have been developed for computing the amount of unbalance in a given situation (see [ 28-30, 34-36,
1 78
C h a pter 5
41-43 ] ) . Having established a definition of the "unbalance factor, " different line configurations may be examined in detail to optimize the line design.
[34-36]
Ground displacement of lines. In the early 19 50s, Gross and others developed a definition for the electrostatic unbalance of a line. This definition is established with reference to a line supplied from a Y -connected transformer bank as shown in Figure 5 . 1 0 where we recognize the presence of capaoitance between
- Van - V en
a
+
10
-
a Cea
+
+
+
b
e
+
Va Vb V c
III
Fig. 5 . 1 0 . Transmission line supplied from a Y·connected transformer.
wires and capacitance to ground. The neutral connection may be closed (grounded) or open, but in many modern systems it is grounded. The unbalance factor is defined differently for each connection, i.e., for the system neutral either grounded or ungrounded. The system shown in Figure 5. 1 0 is conveniently defined electrostatically by ( 5. 2 7 ) , i.e . , Vabc = P«Jab c, where we assume that the effect of ground wires is in cluded according to the method discussed in section 5.6. Then the charging cur rents flowing at no load are, from (5.60),
labc = j w C Vabc = j B Vabc where B is the shunt susceptance matrix. Also from ( 5 . 6 2)
1012
= j BOl 2
V012
( 5.90)
(5.91)
where
BO l 2
=
A- I B A = A- I (w C) A
(5.92)
Since the applied transformer voltages are balanced, positive sequence voltages, we write
then
Vabc = V + Vn and VOl 2 = A- I Vabc = A- I ( V + Vn)
where
V is strictly positive sequence. Expanding
(5.93), we compute
(5.93)
1 79
Seq u ence Capacitance of Tra nsmission Li nes
Vn (5.91)
or
Van laO = j (Boo VaO + BOI Val ) lal = j (BIO VaO + Bl 1 Va l ) la2 = j (B20 VaO + B2 1 Val )
is a zero sequence voltage, may be written as
is positive sequence, and
Va2
(5.94) =
O.
Then
(5.95) lao O.
Neutral ungrounded. If the system neutral is not grounded, the neutral = will usually be nonzero and the neutral current will be zero , or voltage and w e define the neutral "displace + Then from ment" or unbalance as
Vn
(5.95), Boo VaO BOI Val = 0, d o VaO/ Val = BOI /Boo = COl /COO (5.64) CM2 ) d0 - -CSO( CS2- +2CMO (5.58) d 0 -- - ( cnCg+o CM2 ) 2 do = cacgag+ +a CbgCbg ++ceacgeg cag + a32cgCbog + aceg Vn VaO (5.95), laO jBOl Vah lal jBl 1 Vah la2 jB2 1 Val d o = lao /lai = Bot lBl 1 COdCl l (5.64) 2 CS 2 + CM2 d o CCSSO2 ++ CM CMO CgO + 3CMO cag + a2 Cbg + aceg cag + a2 Cbg + aceg ( cag + Cbg + ceg ) + 3 (cab + Cbe + Cea) 3 (cgo + 3cMO ) =
(5.96)
-
we also write
From equation
Also from
-
(5.97) (5.98)
or, writing the numerator in terms of capacitances to ground,
(5.99)
_
Neutral grounded.
If the system neutral is grounded,
write from
=
=
=
In this case we define the displacement or unbalance as =
=
=
0
and we
(5.100) (5.101)
Then from
=
=
=
=
(5.102)
This expression may be simplified to neglect the capacitance between conductors since they are considerably smaller than the capacitance to ground. If this is done, we write
(5.103)
C h a pte r 5
1 80
(5.99).
which is exactly the same as Thus we have a convenient expression for the electrostatic unbalance which is independent of the system neutral grounding.
Example
5. 6
5.2,
Compute the displacement or unbalance of the lower circuit of Figure using computed values of Example 5.4 where possible, where we now assume the line to be untransposed.
Solution From Example 5.4 we have
C' O
= 9.545
nF/mi
1.683 nF /mi CS O = 12.912 nF/mi
CM O
=
= cbe = 2.106 nF/mi Coe = 0.83 9 nF /mi co, = ce, = 9.802 nF /mi
Cob
Cb , =
9.030
nF /mi
Then
c , + aCe, ( 1 + a) co, + a2 cb, a2 (co, - Cb, ) 3c,o ac,o ac,o = �(��747�) = 0 . 0269 {- 60° or 2.69% the system had a grounded neutral and the capacitance between phases is not neglected, we compute from (5.102) (co, - Cb, ) = a2 (0.772 ) = 0.0177 crur. or 1. 7 7% od = - 3(a29.545 3 (14.594) + 5.049) do
e!!
Co, + a2 b
=
=
-
---=---='-
-
If
0
-
Obviously, in this case the use of the approximate equation gives a very pessimistic result.
[34 ]
(5.99)
Reference gives examples of similar computations using with various wire sizes, spacing, and conductor heights and also shows the effect of ground wires. The results may be summarized as follows :
1 . Electrostatic unbalance may be reduced by the addition of ground wires and by increasing the spacing between wires. Electrostatic unbalance may be reduced by changing the arrangement of phase and ground wires, e.g. , by lowering the middle conductor of a flat configuration or by arranging the wires a-c-b , rather than in a vertical configuration.
2.
a-b-c ,
(5.95)
Note that from we may also define a negative sequence unbalance. In the case of the grounded neutral system we write
d2 or
= 10 2 1/0 1
= B2 l IB l l
=
-
C 2 1 /C l l
(5.104)
Sequence Capacitance
of Trans m issi o n
In the case of flat horizontal spacing with wire d
- a( c b b 2 -
b
1 81
Li nes
in the center, this reduces to
- Coo + cbe - cae ) 3( cso + CMO )
(5.105)
This unbalance factor i s small an d i s usually ignored. Problems
5.1.
5. 2. 5. 3. 5.4. 5.5.
5.6. 5.7. 5.S. 5.9.
Compute the positive and negative sequence capacitance to neutral for the line con figuration indicated below. Then compute the 60 Hz susceptance and the charging kVA per mile, assuming the line to operate at the nominal voltage Indicated (neglect the effect of conductor height). (a) Configuration of Figure N.1, 34.5 kV. (b) Configuration of Figure P4.2, 34.5 kV. (c) Configuration of Figure P4. 3, 69 kV. (d) Configuration of Figure P4.4 , 69 kV. (e) Configuration of Figure P4.5, 161 kV. Derive the equivalent circuit for self and mutual capacitances shown in Figure 5.6 by algebraic manipulation of (5.4 7). Compute the change in capacitance in Example 5 . 1 if the height above the ground is considered. Verify .(5.57)-(5.59) by using (5.56) as a starting point. Show that the total capacitance to neutral in the positive sequence network may be com puted by converting the mutual capacitance Ll to a Y and adding the per phase capaci tance to cgo (see section 5.4). Compute the positive and negative sequence capacitance for the circuit of Figure 4.6, using the method of section 5 . 1. Compute the zero sequence capacitance for the circuits of Figure 4.6, using the methods of section 5. 2. Repeat the computation of positive and negative sequence capacitance of problem 5.6, this time taking into account the height of the conductor above the ground. Examine the circuit of Figure 4.6 and compute the following (neglecting ground wires). (a) The matrix P of potential coefficients. (b) The matrix C of Maxwell coefficients. (Use of a digital computer is recommended for this step, but manual methods may be used. ) (c) The matrix COl2 of sequence capacitances for a transposed line.
5.10. Repeat problem 5.9 for the circuit of Figure 4.1S. Assume wire height is 70 ft. 5. 1 1. Repeat problem 5.9 for the upper circuit of Figure 5.2, ignoring ground wires.
5.12. Examine the circuit of Figure 4.6 and compute the following, including the effect of ground wires. (a) The matrix P of potential coefficients. (b) The matrix C of Maxwell coefficients. (c) The matrix COl2 of sequence capacitances for a transposed line. 5. 1 3. Repeat problem 5 . 1 2 for the circuit in Figure 4. 1S. Assume the phase wires to be 70 ft above the ground. 5. 14. Repeat problem 5 . 1 2 for the upper circuit in Figure 5.2, including the effect of ground wires. 5 . 1 5. Repeat the computations of Example 5.4 by inverting the 5 X 5 P matrix to obtain a new 5 X 5 C matrix. Explain the meaning of each term of C and label these capacitances on a sketch. 5. 16. Repeat the computations of Example 5.5, omitting the effect of the ground wires. Com pare results of the two computations and justify the change in capacitance by physical reasoning. 5.17. Compute the capacitance (a-b-c coordinate system) for a double circuit line consisting of two identical lines like that of Figure 4.6 and separated by a distance of 25 ft, assuming
1 82
5.18.
Chapter 5
the two circuits operate in parallel at 66 kV. Suggestion: Set up the P matrix by hand computation but use a digital computer to invert the matrices (see Appendix A). Compute the electrostatic unbalance factor do for the upper circuit of Figure 5.2. (a) Neglecting the ground wires. (b) Including the effect of the ground wires.
chapter
6
Seq u e n ce I m peda n ce of M a ch i n es An important problem in the determination of sequence impedances of a power system is concerned with machines. This problem is especially difficult since machines are complex devices to describe mathematically, requiring that many assumptions must be made in deriving expressions for impedances. For example , the speed, degree of saturation , linearity of the magnetic circuit, and other phenomena must be considered . Our discussion here is divided into two parts; synchronous machines and induction machines. In these devices the several circuits are coupled inductively and are therefore related by differential equa tions. Having established the appropriate equations, however, we will immediately assume that the load is constant but unbalanced . Thus we will again be con cerned with algebraic equations, and phasor notation will be used. This treatment should not be considered exhaustive by any means, and the interested reader should consult the many excellent books on the subject. 1 I . SYNCH R ONOUS MACH I N E IMPEDANCES
6.1 G eneral Considerations
A synchronous machine is sometimes called a "dynamic circuit" because it consists of circuits which are moving with respect to each other, and therefore the impedance seen by currents entering or leaving the terminals is continually changing. There are several complications here . First, there is the problem of changing flux linkages in circuits where the mutual inductances change with time (Le . , with changing rotor position ). There is also the problem of dc offset when a fault occurs. This is due to the shift in the ac envelope required, since there can be no discontinuity in the current wave of an inductive circuit. Since the normal (pre fault ) currents differ in phase by each phase current ex periences a different dc offset. The concept of constant flux linkages over the period from just prior until just after the fault also requires certain fast reactions
1200,
in the coupled circuits which generate large but rapidly decaying alternating cur rents whose magnitudes must be estimated. There is also the consideration of the machine speed. Since the machine sees a faulted condition, the load (active power) that it can deliver is changed suddenly. The prime mover requires a finite time to
sense this change in load , so the rotating mass responds by changing its speed and allowing energy to be taken from or supplied to its inertia. Thus a suddenly apI For example, see [ 1 0 , 1 1 , 1 4 , 1 9, 39, 44 and 4 5 J and examine the excellent list of references given by Kimbark [ 1 9 J . 1 83
C hapter 6
1 84
plied fault sets up a dynamic response in a machine. This response must be esti mated to permit computation of fault currents. All these questions require elaboration .
6. 1 . 1
Mach ine dynamics
First, consider the problem of the change in speed of the generator due to a sudden change in load . Consider a single machine which is supplying a passive load when a three-phase fault is applied at its terminals. Since the voltage of all three phases becomes zero, the three-phase electrical power leaving the machine suddenly becomes zero . But the input power supplied by the prime mover is the same as before the fault. Thus all the input (mechanical) power is available to accelerate the machine. Note there can be no discontinuity in the angle of the machine rotor. If this angle is (J , write (J
=
W I t + li +
1r
/2
(6.1)
where W I i s the synchronous angular frequency, li i s the torque apgle, and the constant 1r /2 is added to conform with the usual convention, as noted later. Thus the angle advances linearly with time up to the time to when the fault is applied. During this pre fault period the speed of the machine is a constant, and the torque angle li is a constant or
e
=
WI
rad/sec
t < to
After the fault occurs, the speed changes t o a new value constant,
t > to
(6.2) W
which i s usually not
(6.3)
and the angle advances at a new rate given by (6 . 1 ) as the shaft accelerates at a rate
if
=
w = S·
(6.4)
This problem of solving the differential equations of a machine following a dis turbance, even a balanced three-phase disturbance, is a formidable problem in itself and involves the solution of differential equations of the machine, the prime mover, and its control system. As explained in Chapter 1 , our goal is to simplify the solution of a faulted system to an assumed steady state condition such that algebraic equations involving phasor quantities may be used . The application of a fault near a machine is obviously not a steady state condition and requires rationalization . Actually the severity of the fault is the key to this problem. If the fault is not severe but is an unbalanced load or other permanent condition, we cotnpute the fault voltages and currents, using phasor notation , after all transients have died away and the system is in a steady ( w = a constant) condition. However, if the fault is severe (such as a short circuit), we assume the fault will be removed before the frequency has changed appreciably . We will, however, include all known ma chine circuit responses required to maintain constant flux linkages over the dis continuity . Thus we will solve a fictitious circuit problem in which we replace the generator by a Thevenin equivalent wherein both the voltage and impedance are
Seq uence I mpedance of M a chines
1 85
arbitrary quantities intended to represent the (worst) condition immediately after the fault occurs. We then proceed with an algebraic solution. The assumption that this procedure will give usable results to compute the settings of relays has been established through years of experience. 6. 1 .2 Direct cu rrent
At the instant a fault occurs, the generator currents change to new values which depend on the new value of impedance seen at the generator terminals. However, since the circuit is inductive, there cannot be a discontinuity in the current. That is, the current just prior to to equals the current just after to in each phase or, mathematically,
i"bc ( to )
= iabc ( t�)
(6.5)
The current just after to i s composed o f two components, a d c component which dies out exponentially and an ac component. The rms value of the ac component after the fault is different from that before the fault. The exact amount of dc offset depends on the exact time in the current (or voltage) wave at which the disturbance appears and on the angle of the impedance seen by the generator. A typical offset for the three phase currents is shown in Figure where the response to a 31/J fault is illustrated. Note that the sum of the dc components in the three phases is zero . Kimbark [ 19 ] shows that the amount of the dc offset can be found by taking -12 times the projection of the negative of the phasor fault currents on the real axis in the complex plane (see problem 6 . 2 ) .
6.1
6. 1 . 3
I n itial value of fault currents-the flux linkage equations
6.1,
As seen in Figure the initial value of the fault current has both a dc and an ac component, with the dc component decaying to zero in a short time and the
thfin",
UUUI'"
fYmmh
.lfUUUU
UUlUUI tfllffftft
Fig. 6. 1 . Short circuit currents of a synchronous generator. Armature currents i" . ib • and ic ; field current if. ( From Kimbark [ 1 9 ] . Used with permission. )
Chapter 6
1 86
ac component decaying slowly to a much lower steady state value. This phenom enon can be explained by the principle of constant flux linkages (see [19] ). Assume that prior to the fault the generator field is energized but the machine is It helps to visualize this situation as that of six separate but unloaded (i = mutually coupled circuits consisting of three phase windings (a, b, c), a field wind ing (F), and two equivalent damper windings (D and Q) for which we may write the flux linkage equation1
0).
LaLbce Le a Le b Le e AFAD LDaLFa LDbLFb LDeLFe AQ LQa LQb LQe
i
I I 1I 1
LaFLbF LbLaDD LbLaQQ ['.ab LcFLFF LFLCDD LFLeQQ �:iFe LDFLQF LDDLQD LQQLDQ i'DQ .
----------- ------- - ----
I
I
Wb turns
(6.6 )
The notation adopted here is t o use lowercase subscripts for stator quantities and uppercase subscripts for rotor quantities. Most of the elements of the inductance matrix are functions of the rotor angle 8 . These inductances may be computed by carefully examining Figure 6.2 which has been suggested by the IEEE as the d AXIS
DIRECTION OF ROTATION
(
q AXIS
sb
c AXIS
Fig. 6 . 2 . Reference for measurement of machine parameters.
standard definition for the several physical parameters involved. These induc tances may be computed as follows. The stator self inductances (diagonal elements) are functions of twice the angle 8 , 3
LaLba LLss LLmm - Ls s ee m , m L L L L L L, Lm . A = = =
where
2
>
+
+
+
cos 28 H
2 (8 120) = + Lm cos ( 28 + 120) H cos 2 ( 8 + 120) = + cos ( 28 - 120) H
cos
(6.7)
We use the symbol for flux linkage in accordance with the American National Standard, ANSI Y l O . 5 , 1 968. 3 Here we adopt the useful convention of designating a constant self or mutual inductance by a single subscript.
1 87
S eq uen ce I mped a n ce of M a c h i nes
The stator-to-stator mutual inductances are also functions of 28 . Lab = Lba = - Ms - Lm cos 2 (8 + 30) = - Ms + Lm cos (28 - 120) H Lbc = Lcb = - Ms - Lm cos 2 (8 - 90) = - M. + Lm cos 28 H Lca = Lac = - M. - Lm cos 2 (8 + 1 50)
=
-
M.
+ Lm cos ( 28 + 120) H
(6.8)
where 1M. I > L m . The ro tor self inductances are all constants, so we redefine these quantities to have a single sUbscript. LFF = LF H LDD = LD H
(6.9)
L QQ = L Q H The rotor mutual inductances are also constants. L FD = L DF = M R H LFQ = L Q F = 0 H LD Q = L Q D = 0 H
(6.10)
Finally, the stator-to-rotor mutual inductances are functions of the rotor position 8 . La F = L Fa = MF cos 8 H LbF = L Fb = MF cos (8 - 120) H
(6. 1 1 )
LcF = LFc = MF cos (8 + 120) H La D = L Da = MD cos 8 H LbD = Lnb = Mn cos (8 - 1 20 ) H
(6.12)
Lc D = Lnc = Mn cos (8 + 120) H La Q = L Q a = M Q sin 8 H L b Q = L Q b = M Q sin (8 - 1 20 ) H
(6.13)
L C Q = L Qc = M Q sin (8 + 1 20 ) H
Now consider the problem o f a generator with negligible load currents (compared to the fault currents ) which is to be faulted symmetrically on all three phases at t = O . At t = 0 - the currents are ia = ib = ic
:!!
0,
in =
iQ
=0
and the flux linkages are computed to be Aa
L aF
M F cos 8
Ab
LbF
Ac
LcF
MF cos (8 - 120) MF cos (8 + 120)
AF
LFF
An
LnF
MR
AQ
LQ F
0
iF =
LF
iF
(6.14)
1 88
C ha pter
00
6
But at t = 0, 0 = + 11" /2 , and the flux linkages are functions of the torque angle. Consider now the sequence of events associated with a three-phase fault on the unloaded generator. Before the fault occurs, the field with flux linkages A F produces an air gap flux CPag , leaving the N pole of the field and entering the armature, thereby establishing an armature S pole which moves with respect to the armature and produces time varying flux linkages expressed by (6.6). At the instant t = 0, the fault is applied, at which time the flux linkages are given by (6.14) and (by the principle of constant flux linkages) must remain at this value (at least for an instant) . Thus exactly the same flux CPag which entered the arma ture S pole at t = 0 - continues to do so, and this S pole remains fixed at the exact (stationary) location it occupied at t = 0 - , even though the field winding con tinues to rotate . The field N pole similarly continues to produce an air gap flux CPag which emerges from the field winding as before and is fixed with respect to the field winding. Obviously, similar statements could be made concerning the field S pole and armature N pole but they will be omitted here. We summarize the flux condition at the time of the fault as follows: 1 . At the field N pole-(a) CP O g = a constant, leaving N, and (b) N fixed with respect to the field winding. 2. At the armature S pole-(a) CPag = a constant, entering S, and (b) S fixed with respect to the armature winding. As Kimbark puts it, it is "as if the poles had stamped their images upon the armature" at t = O. Now both the armature and the field windings will react by inducing currents to maintain the flux linkages of (6.14) as described above. They do so as follows : 1 . To maintain the stationary armature poles and force the flux CPag to enter S as described requires a dc component of current flow in the armature circuits as shown in Figure 6.1 , with a different dc magnitude in each phase winding, depending on the rotor position 0 corresponding to t = o. 2. To counteract the production of armature flux linkages due to the spinning rotor field requires that alternating currents flow in the armature windings. These currents are positive sequence armature currents which are of a magnitude sufficient to hold the armature flux linkages at the prefault value specified by (6.14). The MMF produced by these currents rotates with synchronous speed, is stationary with respect to the field winding, and opposes the field MMF. The field winding reacts in turn with an increased current, the two forces balancing each other such that the flux linkages remain constant. 3. The stationary armature field, viewed from the field winding, appears as an alternating field and induces an alternating current in the field winding. Such an alternating current produces a pulsating MMF wave which is sta tionary with respect to the rotor. If one thinks of this pulsating wave as being composed of two moving MMF waves, one going forward and one backward, the backward wave will be such as to oppose the stationary armature field . The forward wave moves at twice synchronous speed with respect to the armature and induces a second harmonic current in the armature circuit. In machines with damper windings this second harmonic induction is small.
1 89
S eq uence I mpeda n ce of M a chi nes
If the flux linkages of the machine were to be constant for all time, the situa tion would remain exactly as just described . However, as noted in Figure 6.1 , the induced currents decay to new, lower values at different time constants. These time constants will be described in more detail in section 6.5. 6. 2 Positive Sequence I mpedance
The occurrence of a fault on a synchronous machine causes many different responses in the machine windings, each circuit responding to physical laws in known ways. The result is a response which is difficult to express mathematically so that sequence impedances can be defined in the usual way. For passive net works we write (6.1 5 ) where we assume that the elements o f Za b c are complex numbers and the problem is stated algebraically in phasor notation. This is not possible for a synchronous machine, as will be shown . Instead, we usually work problems either in the time domain, as in the case of stability problems, or make certain bold assumptions and use the phasor domain with the impedances assumed constant. This practice re sults in a whole family of positive sequence impedances, depending upon the exact condition under study, but only one negative and one zero sequence im pedance . Different time constants are associated with the different impedances. Since all these quantities are in common use, their definitions are reviewed here. For a more thorough treatment see Kimbark [ 1 9 ] and Prentice [46] . 6.2. 1 Park's transformation
It has been shown that the equations for a synchronous machine can be greatly simplified if all variables are transformed to a new coordinate system defined by the transformation where
P=
143 I!�fi �in 0
1 /..;2
cos (0 - 120) sin (0 - 120)
1 /..;2
�
cos (0 + 120 ) sin (0 + 120)
(6.16)
(6.1 7 )
We call this transformation the d-q transformation or Park's transformation, named for its early proponent R . H. Park [ 47 , 48] . The transformation (6. 1 7 ) is different from the one used by Park in the constant y' 2/3, used to make P orthogonal. When this transformation is used, the current id may be regarded as the cur rent in a fictitious armature winding which rotates with the field winding and with its MMF axis aligned with the field d axis. The MMF thus produced is the same as that produced by the actual phase currents flowing in their actual armature wind ings. The q axis current iq is similarly interpreted except that its fictitious wind ing is aligned with the q axis (see Fig. 6.2). The third component (usually called io) is really iao , the zero sequence current, expressed in instantaneous rather than phasor form. It is not difficult to show that the transformation P is unique and that its inverse is given by
Chapter 6
1 90
ia b c
= p - I iodq
�
where sin 8 sin (8 sin (8 +
cos 8 cos (8 cos (8
p- I
120) + 120)
(6.18)
120) 120)
(6.19)
Park's transformation is used to simplify the usual expressions for the phase volt we write an equation ages of a synchrono us machine as follows. From Figure ous generator in the synchron ed -connect Y a of for each phase-to-neutral voltage form v = - ri - X +
6.3
Vn
•
i
o -
rF
'F:[�}F {� rO
}D
'D '
'��QJLQ
n'
L bb fb
rn Ln
rb
i
b -
sb
+ Vn -
+
_Yo
I -n
e + + vb v _ _ c
y
The voltage equation for the six coupled circuits of Figure written in matrix form as follows, with = rb = =
a Vb Vc - VF - VD = 0 - VQ = 0
--- -
re r.
ra
V
rotor
n
IQ
Fig. 6 . 3 . Schematic diagram of a synchronous generator.
stator
b
ie
rQ
-
a
= -
l J
0 0 0 r 0 I 0 0 r I Il rOF 0rD 00 II 0 0 rQ
r
--
I I I
-
---
o
where
I '
I
0
-- - -- - -
ia ib ic iF iD iQ
6.3 may be
Xa Xb Xc
XF
XD XQ
+
[�Ol (6.20 ) (6.21)
6. =
- n lo be - Ln lo' be R '
191
Seq ue nce I mpedance of M a chi nes
[ �Jl , Rs
[�
This equation could be in volts or in pu. If we define
VF
VFDQ then
=
=
r U,
RR
F
=
=
rD o
ILoRs RR I�UaFDQbc - [��abcFDQ [vnJl J J (6.23)J P (6.16).
(6.20) may be written in matrix form as
abc J [vVFDQ
0
-
(6.22)
+
0
0
O-d-q
(6.23)
It is convenient to transform the a-b-c partition of to the frame of reference by the transformation as in This greatly simplifies the flux linkage equation since it removes all time varying inductances. To do this we premultiply both sides of by the transformation
(6.6)
(6.23 )
which by definition transforms the left side to
O-d-q voltages.
Thus
(6.24)
rRs RR [iiFDQ abc ] [Rs RR fLiFDiodq ] J Q J IP ol r�.�FDabc J IP � abcD l LO � L Q L �F QJ 1t (6.16 ) P abc 1t a . abc abc + bc P1t abc
The resistance term becomes
rp 01 Lo uJ Lo
0
=
0
0
(6.25)
The flux linkage term requires careful study. Applying the transformation, we write =
We can evaluate as follows. From the definition Then, taking the derivative, � Odq = it A P find �Odq = P �
= � Odq - P ��bC = � Odq - p p - l � Odq
We easily show that
p P-'
so that the last term of
=
[� � -�]
(6.26)
we write from which we
(6 . 27)
(6.27), which we shall call s, becomes s � P p-1 >' Odq =
'- : J L�w � x x
and this is recognized to be a speed voltage term. The last term of is transformed as follows.
(6.23)
(6.28)
1 92
Chapter 6
where we have defined the voltage
(6.29)
Substituting all transformed quantities into ( 6 . 2 3) we have the new voltage equations,
0 ] [iOdq ] [AOdq ] [S ] [nOdq] [VOdq ] [RS 0 0 0 VFDQ iFDQ - XFDQ =
+
RR
-
+
( 6 .30 )
We shall now show that this is much simpler than the original voltage equation etc. ) . which included many time varying coefficients I t i s convenient t o think o f the O-d-q partition o f (6.30) a s the stator voltages referred to or seen from the rotor. Since fundamental frequency ac quantities in the stator appear to the rotor to be dc quantities, we would expect these voltages to be constants under steady state operation. Because of the transformation of stator quantities to the rotor or O-d-q quantities, two remarkable things happen. First, the inductances which were so complicated and time varying in (6.6) have been transformed into constants. Second, a speed voltage term s has appeared, which adds voltage components to Vd and Vq proportional to w , the rotor angular velocity. Looking at this result another way, we have replaced a linear system with time varying coefficients by a nonlinear system (because of s) with constant coefficients. Since under most con ditions the speed w is nearly constant, the nonlinearity is of little concern . The important change is in the inductances. If we examine the transformation of the flux linkage equation, we have the following. By definition of the P transformation,
(Laa , Lab ,
=
c_ [!_: _�] -I rlo!_J�Jr�� [!-��_I_��J UJ L IFDQ] LRR J 0
r..!-i-�I [0 I UJ [LRa
I u
I
I
where the partitions of the inductance matrix are defined in (6.6). Then
[ LO La b c � �
1
( 6 .3 )
By straightforward computation of the partitions of (6 .31 ) we may show that 0
P
p- I =
d
1 93
Seq uence I mpeda n ce of M a chi nes
where
Lo = L, - 2M" Ld =
L, +
M, (3/2) Lm , Lq +
=
La
+ M. - (3/2) Lm
and these inductances are all constants. We also compute
P
LaR =
0 0 0 IIMF /fMD 0 � Lm 0 0 IIMQ
and
0 � MF 0 0 � MD 0 = L � 0 0 A MQ
LRa P - 1
Thus we may write the transformed flux linkage equation as
0 kMF 0 LF MR 0
0 0 1 Lq : 0 1 0 11 kMQ where for convenience we set k � .J 3/2. Lo 0 Ld 0 kMF kMD 0
I I
o
o
1
o
I
o
1
0 kMD 0 MR LD 0
io 0 id kMQ iq 0 iF iD 0 LQ iQ
0
------------ -------------
o
(6.32)
Observe now that every element in this inductance matrix is constant. Furthermore, the Ao equation is completely un coupled from the other equations and may be discarded when balanced conditions is constant, the time derivative are under study . Since every inductance in is easily found. of this equation , required in If the voltage is now written in expanded notation , it is instructive to see the way in which the transformed equations are coupled .
(6.30)
(6.32)
(6 .30),
r O O 11 o o r 0 1 1 o 0 r 1 ------1- -------1
o
l
I
iq rF 0 0 iF O rD 0 0
0
Chapter
1 94
-
I I
Lo
0
0
0
Ld
0
0
0
Lq
I
I I I
6 ·
io id
0
0
0
k MF
k MD
0
0
0
k MQ
·
iq ·
r-----------I 0 LF MR I I 0 LD I MR I
I
-----------
0
kMF
0
0
k MD
0
0
0
k MQ
I
-
o
- w(Lq iq + kMQ iQ) W(Ld id + k MF iF + k MD iD)
+
---
------------- -
·
iQ
LQ
0
0
·
iF iD
3rn
io
-
0
3Ln fo
o
+
o
o
o
o
o
o
(6.3 3 ) Or, written in a more compact form
r
+ 3rn
0
0
o
r
wLq r
-
I I I I
!
0
0
0
0
0
k wMQ
-- o ----- o --- o -- -- � ----- o ----- o T r o 0 rD : 0 0 0 o rQ 0 : 0 0 0 o
WLd
- kwMF
- k wMD
0
.
io .
iD
(6 .34 ) By proper choice of rotor and stator base quantities, all the foregoing equa tions may be written in exactly the same way in pu or in system quantities (volt, ohm, ampere, etc.). Equation (6.34) is very unusual as a network equation because the "resis tance " matrix is not symmetric. This is because the network is an active one which contains the controlled source terms due to speed voltages. We now investigate the meaning of the newly defined inductances of (6.32) and some related quantities.
S eq uence I mpedance of M a ch i nes
1 95
6. 2. 2 Direct axis synchronous inductance, Ld
If we apply positive sequence currents to the armature o f a synchronous ma chine with the field circuit open and the field winding rotated at synchronous speed with the axis aligned with the rotating MM F wave as shown in Figure 6 .4a,
d
Ld =
where we require that
Aa/ia Ab/ib Ac/ic =
=
=
L, + M, +
�:GcJ [�: :::;0(0 1200 120J =
Under these conditions only
d
(3/2)Lm
(6 .35)
-
../2 1 cos
+
axis current exists or
.j3
where the facto r is due to the arbitrary constant multiplier chosen to make the P transformation orthogonal. It is known that positive sequence currents flowing in the annature produce a space MMF wave which travels at synchronous speed. However, the flux pro duced by this rotating MMF wave depends upon the reluctance of the magnetic circuit. The reluctance is greatly influenced by the air gap. In cylindrical machines, therefore, the reactance seen by positive sequence currents is almost a constant irrespective of rotor position. In salient pole machines this is not true since the reluctance, and therefore the flux, depends strongly upon the relative position of the MMF wave and the protruding pole faces of the rotor. This is shown in Figures 6.4a and 6.4b. Thus we would expect the inductance of a salient pole machine to be a function of rotor position. This is true not only of the self inductance but the mutual inductance as well. The inductance will be a maximum when the rotor position is as shown in Figure 6.4a, and this maximum inductance is called Ld , the d axis synchronous inductance. The d axis synchronous reactance is defined as
wh ere w is restricted to be the sy n chro nous spee d actance is me aning le ss
WI
(6.36)
since the concept of re
o therwise. Kimbark [ 1 9 ] p oints out that (6 . 3 5 ) is valid whether instantaneous, maxi mum, or e ffe ctive values are used for the flux linkage and current This flux link age appears to the armature as a sinuso idal ly varyin g linkage and induces an EMF in quadrature with the flux and current, the ratio of this induced voltage to also being equal to Xd .
ia
•
6. 2. 3 Quadrature axis synchronous i nductance,
Lq
Proceeding as before , we may de termine th e q a is synchronous inductance by apply in g positive sequence currents to the armature with the field circuit open and the field winding turning at synchronous speed w ith the rotor q axis align ed
x
Chapter 6
1 96
$1.101 m.m.l. w.v. (sl•• dy sl,I.) �
(d )
:tq'
Fig. 6 . 4 . Flux patterns under synchronous, transient, and subtransient conditions. � From Prentice [ 46 ] . )
with the rotating MMF wave . Under these conditions the currents are shifted by 90° in phase or
[:::lJ [�� ::: �: :�)- J =
i
...[2 1 cos (8
�
-
120�
90 + 120)
This makes io and id go to zero , so that only iQ LQ
= A a / ia = A b /ib
=
A c /ic
=
L.
= ..;31 exists. +
M.
-
Then
( 3/2) Lm
(6 . 37 )
With the rotor in the position specified, the synchronous inductance will assume its lowest value, since the reluctance of the flux path is a maximum, as seen in Figure 6 .4b. Corresponding to this rotor position we also define the inductive reactance
1 97
Seq uence I mpeda n ce of M a chines
(6.38)
In cylindrical rotor machines
(6.39)
but for all machines
Xd > xQ
(6.40)
with the inequality being much more pronounced in salient pole machines. 6. 2. 4 Direct axis subtransient inductance,
d
Ld
The axis transient and subtransient inductances are defined with the field circuits shorted , Le. ,
[Va] [Y2V Y2 V Vc Y2 V
� - 120)
an d with positive sequence voltages applied suddenly at t if u ( t ) is the unit step function,
=
Vb
or
id
Vd
cos 6
cos (6 cos (6 +
120 )
= 0 to the stator.
u( t)
0
(6.41) Thus
(6.42)
vaV.
(6.43)
and only exists after t = and it jumps suddenly from zero to Similarly , is the only current component, but it must build up more slowly according to the time constant of the d axis circuit. Since all currents are zero at t = 0 - , the flux linkages are also zero and by the law of constant flux linkages must remain zero at t = At this instant we may write
0+.
(6.44)
iF and iD as a function of id with the result - MDMR) . k (LFMD - MFMR ) . 'F - - k(LDMF LF LD - MA ld ' 'D - - LFLD - M:h ld (6.45) Substituting these currents into the flux linkage equation for X d, we have X d [Ld - LFL:z_ MA (LDMJ LFMfi - 2MFMDMR)] id � L�id (6.46)
from which we may find .
=
Thus
_
.
+
_
1 98
- - LF L D
L"d L d -
Chapter
k2 _
M�
6
(LD MF2 + LF MD2
which we call the sUbtransient inductance . Often a subtransient
reactance
- 2MF MD MR)
(6.47)
is used and this is defined as (6 .48)
As illustrated in Figure 6 . 4e, very little flux is established initially in the field winding and the subtransient inductance is due largely to the damper windings. These windings have a very small time constant, and the subtransient currents die away fast leaving only the so-called transient currents. Since the air gap
flux
prior to applying the stator voltage was zero , the damper
winding currents try to maintain this no-flux condition .
The result is that the
flux path established is a high-reluctance air gap path, as in Figure 6 . 4e, and very small.
6. 2. 5 Direct axis transient inductance,
L�
is
Ld
If we examine the transient situation just described only a few cycles after the stator voltages are applied , the damper winding currents have decayed to zero and we are in the so-called transient period where induced currents in the field wind ing are important.
This situation also exists if there are no damper winding so
that the air gap flux links the field winding as shown in Figure 6 .4c. To compute the
d
axis inductance seen under this condition , we can let
in ( 6 . 4 4 ) to compute .
IF
or
-
kMF .
L
k2 M;'
L d' = L d and
(6.49)
ld
LF
( d - 4 ) 'd. =
Then from ( 6 .3 2 ) we may compute
Ad =
=
-
iD = 0
L'::.
k2 MF2
L' .
d ld
(6.50)
__
LF
(6.51 ) This transient inductance is determined under the same rotor condition as the
d
axis synchronous inductance , the only difference being in the fact that it is meas ured immediately after the sudden application of the three-phase (positive se quence ) voltages.
The sudden establishment of flux across the air gap is opposed
by establishing a current in the field winding, tending to hold
AF
at zero . Thus,
as shown in Figure 6 .4c, the only flux established is that which does field winding, and this is a small flux . Hence
The
d
q
not link the
is small but is greater than
Ouadrature axis subtransient and transient inductances,
6. 2. 6
to the
L�
L� and L�
L� .
axis subtransient and transient inductances are defined in a similar way
axis inductances except that the rotor in this case is positioned with its
q
Sequence I mpeda nce of M a ch i nes
1 99
axis opposite the spatial MMF wave of the stator as shown in Figures 6.4d and 6.4f. The flux in this case is not much different than the synchronous case for the salient pole machine pictured in Figure 6 .4 . In round rotor machines the synchronous case results in a greater flux similar to that pictured in Figure 6.4 for the axis. Thus we see the need to carefully distinguish between salient pole and round rotor machines in determining q axis inductances. If we consider the rotor as spinning with the correct alignment and positive sequence voltages applied suddenly , we have a situation similar to equations (6.42) and (6.43) except for a 90° phase lag in the applied voltages to get the proper alignment with the q axis. Thus
d
[�:J [��: �9 [uo] [ =
Uc
or
1 20 .../ 2 V sin (8 + 120
Ud
=
-
0 0
� u(t) ;J
J
y'3 Vu( t)
uq
(6 .52)
(6.53)
Since the flux linkages in the q axis are zero both before and after the voltage is applied, we compute A Q = 0 = kM Q iq + L Q iQ or . kM Q . IQ - lq LQ
(6 54) •
Then
where we define (6.55) and (6.56 ) In round rotor machines th e same effect as just described is also noted in the transient period because of the rotor iron acting much like a field winding. Thus we often assume for round rotor machines that
L q » L�
�
L�
>
L;
�
L�
(6.57)
In salient pole machines the presence of a damper winding makes a great deal of difference . Thus for salient pole machines we estimate that
with dampers : L q = L'q > L"q > L"d without dampers : L q = L'q = L q" 6.3
(6.58)
Negative Sequence Impedance
If negative sequence voltages (sequence a-c-b ) are applied to the stator wind ings of a synchronous machine with the field winding shorted and the rotor spin-
6
Chapte r
2 00
ning forward at synchronous speed , the currents in the stator see the negative se quence impedance of the machine . Mathematically , the boundary conditions are iab c
=
[�; ::: �o J + 1 20 )
V21 cos (0 - 120 )
(6 .59)
and vF = O . Then by Park's transformation we compute [ 19]
iOdQ d
id
=
iQ
[..j3..j311
1 sin 20J o
cos 20
id
(6.60)
and observe that both and are second harmonic currents. Since acts only on the axis of the rotor, it induces a second harmonic voltage in the field wind ing. If we assume that at double frequency the field reactance is much greater than its resistance, we have VF 0 = � which requires that AF be a constant, or in the steady state
rF iF + �F �F
=
( 6.61 ) But this is exactly the transient condition described in section 6.2.5. Thus (6.49) applies, and the flux linkages are found from (6 .32) to be
'71. 0
Ad AQ AF AD AQ
where as
0
=
L�id LQ iQ 0 0
(6.62)
0
L� is defined in (6.50). Then the flux linkage of phase a, Ao' is computed (6 .63) Ao = V2 I [L'd L cos + L'd - L Q cos ( 30 + 2a ) +
2
Q
0
2
Ao/io
]
which is observed to have both a fundamental and a third harmonic component, and is not a constant. Kimbark [ 19] defines the fundamental frequency component ratio as the negative sequence inductance, i.e . , by definition
L� + L Q = Ao (fundamental) io (fundamental) 2 and this definition applies for the case of no damper windings . L2
=
If damper windings exist on the rotor, we compute
(6 .64)
S eq ue n ce I mped a n ce of M a ch i nes
20 1
and
= (L� + L� ) / 2
L2
(6.65)
as
We also define the negative sequence reactance
(6.66) The relationship between X2 and the subtransient reactances x � and x� depends strongly upon the presence of damper windings, as shown in Figure 6.5 where measurements are recorded with the rotor blocked or stationary .
1 .0
0.1
�
'\
f- - .... -
QUAORATURE AX IS K q
1\
0.6
V l
i-- 0 IRECT A X IS II;
1\
�
V
N� O�IAPE�s
J d /V
-
NEGAT IVE SEQ EN
'\
'\.
0.4
-,..-
1/ /
V
8LOCK E O - ROTOR IAET HOO O. 2 0 o
COPPER OAIAPERS I
30
ANGUL A R
60
P O S I T ION
90
OF ROTOR
1 20 IN
I
150
OEGREES
I
180
Fig. 6 . 5. Relationship between subtransient and negative sequence reactances. (From Westing house Electric Corp . [ 1 4 ] . Used with permission . )
6.4 Zero Sequence I mpedance
If zero sequence currents are applied to the three stator windings, there is no rotating MMF but only a stationary pulsating field. The self inductance or re actance in this case is small and is not affected by the motion of the rotor. The pulsating field is opposed by currents induced in the rotor circuits, and very little air gap flux is established . Thus L o is very small, generally smaller than L� . The boundary conditions for this situation are iab c
=
I�� ::: :l L�/ �J
0,
["...16/0 OJ
where iF = iD = iQ = and 0 , the rotor angle , may be taken by Park's transformation
.
Io dq and
(6 . 67)
cos
as
any value . Then
cos
=
o
(6.68)
Chapte r 6
202 Table
6. 1.
Typical Synchronous Machine Constants
Water-Wheel Generators ( with dampers) t
Turbcr generators (solid rotor) Low Reactances in pu
x fl
Xd Xd xq'
xa xq" xp X2 xo *
0.95 0.92 0. 12 0. 12 0. 07 0. 10 0. 07 0. 07 0. 01
A vg.
High
1. 10 1. 08 0. 23 0.23 0. 12 0. 15 0. 14 0. 12
1.45 1.42 0.28 0.28 0. 17 0.20 0.21 0. 17 0. 10
Resistances in pu 0.0015 ra (dc) r(ac) 0. 003 0. 025 r2 Time constants in seconds
T'do T'd T;; - T� Ta
2. 8 0.4 0. 02 g. 04
5.6 1. 1 0.035 0. 16
Low
0.60 0.40 0.20 0.40 0. 13 0. 23 0. 17 0. 13 0.02
High
1. 15 0.75 0. 37 0.75 0.24 0. 34 0. 32 0. 24
1.45 1. 00 0.50 :1: 1. 00 0. 35 0.45 0.40 0.35 0. 21
Avg.
High
Low A vg. High
1.80 1. 15 0.40 1. 15 0.25 0.30 0.34 0. 24
2.20 1.40 0.60 1.40 0.38 0.43 0.45 0.37 0. 15
0.80 0.60 0.25 0.60 0.20 0. 30
1.50 0.95 0.30 0.95 0. 18 0.23 0.23 0. 17 0.03
9.5 3. 3 0.05 0. 25
6.0 1. 2 0.02 0. 1
1.20 0.90 0. 35 0.90 0. 30 0.40
1.50 1. 10 0.45 1. 10 0.40 0.50
0.25 0. 35 0.45 0. 27 0.04
0.015 0.010 0.070
0.020 0.002 0.015 0.004 0.200 0.025 5.6 1. 8 0.035 0. 15
1. 5 0.5 0.01 0.03
Synchronous Motors (general purpose)
Low
A vg.
0.005 0.003 0.008 0.003 0.045 0. 012 9. 2 1.8 0.05 0. 35
Synchronous Condensers
9. 0 11.5 2.0 2.8 0.035 0. 05 0. 17 0. 3
Source: Kimbark [ 1 9 J . Used with permission of the publisher. * x o varies from about 0 . 1 5 to 0.60 of X d , depending upon winding pitch. tFor water-wheel generators without damper windings, Xo is as listed and Xd
0.8 5 x� ,
x�
x�
X2
xq •
(Xd
xq )/2
:j: For curves showing the normal value of Xd of water-wheel-driven generators as a function of kilovolt-ampere rating and speed, see [ 50 J . =
=
=
0 0
Ad Aq =
AD
0 0
( 6.69)
0
AQ
From (6.69 ) we compute
+
Lo io
Ao
AF
=
(6.70 )
where, from (6 .32 ) (6.71 ) La Lo and is small compared to and We also define (6.72 ) Lo The actual value of Xo depends upon the pitch of the windings but usually in = A a /ia =
Lo
Ld
Lq •
Xo
=
WI
-
2Ms
is
203
S eq uen ce I mpeda nce of M a chi nes
x .
x�
the range [ 19 ) of 0.1 5 < Xo < 0.60 � Typical values of synchronous machine reactances are given in Table 6 .1 which also gives values of the various time con stants of the machine circuits defined in section 6 .5 .
6.5 Time Constants
If a 3cp fault is applied to an unloaded synchronous machine, an oscillogram of the phase current appears as in Figure 6. 1 . If we replot the current in one phase with the dc component removed, the result is shown in Figure 6 6 A careful examination of this damped exponential reveals that the current envelope has an unusually high initial value O-c and that it decays in a few cycles to a lower rate of decrement. The current then continues to decay at this lower rate until it finally reaches its steady state value, represented by the peak value O-a. The ac com ponent of current in the field circuit decays over a very long period of time, as noted in Figure 6.1 . We investigate these various decrements in greater detail since the time constants are of direct interest in faulted systems. In particular, we need to establish the approximate point in time on Figure 6.6 ( or 6.1 ) at which the circuit relays will be likely to open. This is the value of fault current we would like to compute if possible or, alternatively, we seek a value which will give results to provide relaying margins on the safe side. There are several time constants associated with the behavior noted in Figures 6.1 and 6.6. Since these are often quoted in the literature, they are reviewed here briefly. .
.
c b a
o
TIME
-
Fig. 6 .6 . The ac component of a short circuit current applied suddenly to a synchronous gen erator. ( From Elements of Power System A nalysis, by William D. Stevenson, Jr. Copyright McGraw-Hill , 1 9 6 2 . Used with permission of McGraw-Hill Book Co . )
6.5. 1
Direct axis transient open circuit time constant,
TdO
Consider a synchronous machine with no damper windings which is operating with the armature circuits open. Then a step change in the voltage applied to the field is unaffected by any other circuit, and the response is a function of the field resistance and inductance only. From ( 6. 3 4 ) with id = iD = we write
0
If iF is initially zero and we let unit step function, the result is
iF
VF
(6.73)
= ku (t), where
= � (1 rF
-
k is a constant and u ( t ) is the
e-rFtI LF) u(t )
(6.74)
Chapter
204
6
Thus the time constant of this circuit, defined
T�o = LF /rF
as
T�o, is
sec
(6.75)
Since the open circuit armature voltage varies directly with iF , this voltage changes at the same rate as the field current. Typical values of are quoted [ 1 9] as from 2 to 1 1 seconds, with 6 seconds an average value. The large size of is due to the large inductance of the field.
T� o
6.5.2
Direct axis transient short circuit time constant,
T�o
Td
A synchronous machine operating with both the field and armature circuits closed is different from the preceding case. If currents flow in both windings, each induces voltages in the other, which in turn cause current responses. If the rotor were stationary, the coupled circuits would behave as a transformer with currents of frequency f in one circuit inducing currents of this same frequency in the other. In the case of a machine, however, the frequency of induced currents is different because of the rotation of the rotor. Thus a direct current in the rotor causes a positive sequence current in the armature, whereas a direct current in the armature is associated with an alternating current in the field winding. This alternating component is clearly shown in Figure 6.1. Each of these induced currents changes at a different time constant depending upon the resistance and inductance each current sees. We call the time constant which governs the rate of change of direct current in the field the But this must correspond to the rate of change of the amplitude (or current envelope) in the armature, which we like to call the We call the time constant which governs the rate of change of the direct current in the armature the and this is the same as the time constant of the envelope of alternating currents in the field. The field or axis transient time constant depends upon the inductance seen by which in turn depends on the impedance of the armature circuit. With the armature open we have found that this time constant is T�o . If the armature is shorted, the inductance seen by iF is greatly reduced. We have already computed the required values. From (6.3 5 ) we compute the armature inductance to be Ld with the field open. With the field shorted as noted by (6.50), the armature in ductance is computed as Ld . Clearly then, shorting the armature as viewed {rom the field will change the inductance seen by the ratio Ld /Ld , and the d axis transient (short circuit) time constant T� is defined by
constant,
field time constant. direct axis transient time constant.
armature time
d
iF ,
T� = (Ld /Ld ) T�o sec
(6.76)
Kimbark [19] notes that the resistance seen in the two cases is practically the same since with the armature shorted its resistance is negligible compared to its inductance. Typically, is about 1 /4 that of or approximately 1 .5 seconds. Thus and are the two extremes for the field time constant. With the machine loaded normally or faulted through a finite impedance, the time constant will be somewhere between these extremes. If a known external inductance Le exists between the machine and the fault point, this inductance may be added to both Ld and Ld in (6.76).
T�o
6.5 . 3
T�
T�
T�o ,
Armature time constant, Ta
As explained above, the armature time constant applies to the rate of change of direct currents in the armature or to the envelope of alternating currents in the
S eq uence I mped a n ce of M a chines
205
field winding. It is equal to the ratio of armature inductance to armature re sistance under the given condition. To determine the armature inductance we note that this situation, with rotor currents of frequency t, is analogous to the case of negative sequence armature currents (and inductance L1) which produced field currents of frequency 2 t. In both cases the rotor flux linkages are constant, and the flux is largely leakage flux. Thus with alternating currents in the rotor, the stator inductance is essentially L2 , and the armature time constant Ta is given by A typical value of 6.5.4
Ta
= L2 /r sec
Ta
(6.77)
is 0. 1 5 seconds for a fault on the machine terminals.
D i rect axis subtransient time con stants,
TdO Td and
In a machine with damper windings there is an additional coupled circuit in the d axis of the rotor, namely , the damper winding. This is a low impedance winding, and currents induced in its conductors may be large but decay rapidly to zero. Viewed from the armature, the direct currents in both the field and damper windings appear to the armature as positive sequence currents whose magnitudes reflect the coupling to both rotor circuits. Thus, as shown in Figure 6.6, there are two distinct time constants apparent in the alternating current wave, one with a time constant much shorter than the other . The shorter of these time constants is due to the damper winding and is identified by double-primed notation such as Here, as in the case of no damper appli es with the armature circuits open, whereas applies = 0.125 sec ( 7. 5 cycles) and with the arm atur e shorted. Typical = 0.035 sec (2 cycles) .
T�
windings, T�o
6.5.5
Quadrature axis time consta nts,
T�. values are T�o T'qo, T'q, T�o, and T�
T�
with the q axis in a way similar to no field winding on the q axis. It is also important to recognize the significantly different structure of the q axis between salient pole and cylindrical rotor machines . Thus in salient pole ma chines , has no meaning since there is no quadrature rotor winding; but with We
can identify time constants associated
the treatment of the
d
axis except that there is
T�
damper windings present , a time constant of
�
T� T�
is used .
In cylindrical rotor machines the q axis flux has a lower reluctance path than in salient pole machines, and currents may be established in the steel which decay at various time constants depending upon the impedance of the current path. It i s
by representing the with time constants and with time constants and = 0.8 sec. = 0.035 sec and
usually observed that this situation may be approximated
armature
T�T�
6.6
current as the sum of two exponentials , one
corresponding to a reactance x
�
�
and another
corresponding to x . Typical values are Synchronous Gene rator E q u ivalent C i rcu its
T� T� =
T�T�oo T�
of a synchronous generator, we now circuit. Referring to (6.33 ) , we note first of all that the zero sequence equation is uncoupled from the others. A simple passive R -L network will satisfy this equation . The remaining five equa tions are more difficult and require further stUdy. From (6.34) we write by Having carefully defined the parameters
consider the construction of an equivalent
rearranging,
206
Chapter
6
- W �q o o
--
= -
kMF
+
kMD I
MR LD
I I I I
o
I
- -------------,-- - ----kMQ I Lq o
I I
kMQ
d
. iq iQ
LQ
•
(6.78)
These equations represent a reciprocal set of coupled circuits and coupled q circuits with controlled sources w �q and w �d as shown in Figure 6.7, where the controlled source terms are speed voltages and depend on currents in the other circuit. Note that the zero sequence circuit is completely uncoupled and is passive.
r + 3 rn
:¢ - i if F �� kM MRkM0 y i�4lr,-VD>D: I : 1 ';0) lO)--- kMo �
LO + 3 L n
rF
UF
ro
w
"
F____
+
io
Uo
_
r
i
d
f.
Ld
ud •
wXq r
ra
ua 8 o
_
J-
_ iq t.
Lq
+
-
uq
\-
w Xd
Fig. 6 . 7 . Synchronous generator equivalent circuit.
Lewis [48] shows that the d and q equivalents may be greatly simplified if a T circuit is used to represent the mutual coupling. This requires that ( 1 ) all cir cuits in the equivalent be represented in pu on the same time basis, (2) all circuits in the equivalent be represented ill pu on tne same voltage basis (base voltage), and ( 3 ) all circuits in the equivalent have the same voltampere base. If we assume that these requirements are all met, the equiValent circuit may be redrawn as in Figure 6.8 where, because of the restrictions in the choice of base quantities, the off-diagonal mutual inductances are equal. Then we define
Seq uence I m pedance of M a chines
(
r + 3r"
207 1
0
_:�
-.y.
L O + 3 L,
IF
rF
-
id
Id
f+
uF +
rQ U q sO +
Fig. 6 . 8 .
lq
IQ
iQ
wXq
+
_ Iq
�
t+
Mq
r
iq + i Q
+ w),
r:
d
Equivalent T circuits for a synchronous generator.
(6.79) M q � kM Q pu We also define the leakage inductances, designated by script f , according to the equation pu,
Md � kMF = kMD = MR
-
Md = Ld - fd = LF fF = LD - fD Mq = L q - f q = LQ - f Q pu
pu (6.80 )
The equivalent circuit of Figure 6.8 is often used computation because of its simplicity. It is also convenient for visualizing transient and subtransient con ditions. For example, the subtransient impedance and time constant in the d and q axes is that seen looking into the d and q circuits from the terminals on the right and with all voltage sources shorted . The transient condition is determined in the same way but with the damper circuits open. The concept of leakage inductance is also useful since these inductances are linear. Thus in Figure 6.8 only the mutual inductance Md would normally be come saturated and it should be considered nonlinear. 6.7
in
Phasor Diagram of a Synchronous Generator
In fault studies we prefer to work with phasor quantities since phasor solu tions require less work than solving the nonlinear differential equations for a synchronous machine. But phasors usually are assumed to represent a steady state condition. We therefore derive the phasor diagram for a generator operating in the steady state . We may then examine this diagram to determine the operat ing condition immediately following a fault. A three-phase fault or load may be studied by making the appropriate im pedance connection to the positive sequence network only . But we may also
C hapter 6
208
view anyIn other fault condition this way, toaddibengadded a faulmayt impedance at the fauloft point. the general case the impedance be a combination In alenti l casesrelytheby posi tposi ive tsequence cur the negative andin thezero generator sequence arenetworks. rents flowing determined the i v e sequence network and this fault impedance if weposineglect the loadnetwork. currents.ThisThuspermits we limitus our consideration at this point to the t i v e sequence toancederitheve diagram only onederived phasorherediagram. Wey recognize thatposifortivesome kindsquantities. of unbal may appl only to the sequence If we assume positive sequence, balanced currents, we may write = �:: : : : :J cos and = Then by direct application of the transformation we compute
[;:l [:: �l
i
e
W i t + 0 + 90.
) - 120) (w 1 t + 1/J + 1 20 )
(6 .82 )
P
- I/J ) (0 - I/J )
lq
wiEquation th the result
(6.8 1 )
[::J [- sincos�o J may be manipulated to define the angle in terms of and cos _ - sin cos = ..j31
iq
J
(6.82 )
id
I/J
I/J -
.
id
0
0 + iq ..j3f
_ iq -
0 + id va l
0
sm i/J sin cos These values may be substituted into to write = cos cos and, sincetransformation the phase currents arewebalacoul nced,d writeand are known also. Applying the phasor immediately as a phasor quantitmight y. Before doing this,tohowever, which be convenient use. we digress to examine the reference systems In synchronous machines thereshallarecalatl theleast two convenient andframe. widely4 used frames of reference. One we reference The secondUsingis thean arbi d-q reference frame discussed in the preceding sections. trary reference frame we can write a phasor current as e = cos sin ia
( 6.81 )
v'273 [ id
(w I t + 0 + 90) + iq
(6.84)
(w I t + 0 )]
ic (6.84)
ib
( 1 .50),
6.7 . 1
(6.83)
Phasor frames of reference for synchronous machines
arbitrary
�
-
la
�
�
lar + lax = la I/J + j Ia = fa
j
la
tP
rfJ
= lar + j lax
(6.85)
4 The "arbitrary reference" used here i s a phasor reference and should n o t be confused with the arbitrary reference frame defined by Krause and Thomas [49 ] which is a rotating reference that revolves at an arbitrary angular velocity and is used in the study of induction motors.
2 09
Seq ue nce I mpedance of M a chines
(a )
,..,
1 ax REF
d 10
( b)
Id
Fig. 6.9.
q - REF
Two frames of reference for phasor quantities : (a) arbitrary reference , (b) d-q refer ence.
where thisquantity phasorsymbol is pictured in Figurethat6.9thisa. Note thatty iwes based use theon titheldearbitrary over the phasor to indicate quanti frame. Usualbetween ly such attenti on to detaisystems. l is unnecessary, but8 5)herethe wecompo need reference tonentsclearly di s tinguish two reference From (6. of _ are (6.86) NoteThecareful ly that the quantity (without the tilde) is a scalar quantity. d-q frame of reference is pictured in Figure 6. 9b. Here we use a bar over Thus (6.87) Then (6.88) is nowtypossible phasorbasisexpressed one reference frame to the sameItquanti expressedto onrelathete asecond by the informulas (6 .89) From this formula we compute cos sin ) cos sin Ia
Ia
the phasor symbol to indicate that this phasor is based on the d-q reference frame.
Iar + j Iax = (lq
[5] .
{)
-
Id
{)
+ j(Id
{)
+ Iq
{)
5 For a more detailed discussion of the subject of transformation of bases, see Hueslman
210
Chapter
6
Fig. 6 . 1 0 . 10 components from two reference frames.
or = cos sin = cos sin These quantities are shown in Figure Fromty in termsweof have the generator current expressed as aequivalents time domainof quanti the magni t udes and i q , whi c h are rotor the instantaneousmagnitudes stator currents. rotor-referenced as We now define rms stator equivalents of these = 1v'3 , = 1v'3 rms A or rms pu wherearitheses asv'3a scale comesfactor fromforthe wayand the Using transformatiweon wriwastedefinedasand there fore = cos cos I t Applying the phasor transformation we compute e i6 = )ei6 = la e l6 L = e J(6 + 90) as giThe ven bymagnitude of should be the same in any frame of reference. Thus lo r
6.7.2
fj - Id
lq
fj ,
lax
(6.84 )
id
ia
lq
id
id
y'2 Id
ia
(6.9 1 )
iq
iq •
P
(6.91 ),
+ fj + 90) + y'2 lq
(w
(6.90)
S
6.10.
Phasor transformation of generator quantities
Id
fj + lq
Id
(6.84)
(w I t + fj )
( 1 .50),
+ lq
Id
(lq + j ld
(6.92)
(6.89) .
10
1a
= II I = II I = v'1q2 a
a
+ IJ
= ..j3 � V'I2+I2 'Ii 'd
(6.93)
T
la ,
couldspeci alsofibeed.computed from the components of but these valTheuThe esmagnitude areforegoing not usually d-q frame permi t s us to represent a machine i n the convenient andin thethensystem. easily transfer to thea more general reference which is used for d-q voltages and currents, we may by allof reference machines Knowing write these quantities as phasors. (6.92)
S eq uence I mpeda n ce of M a chi nes 6.7.3
21 1
The steady state phasor d iagram
To derive the steady state phasor diagram, we write the d-q voltage equations for the balanced case and then transform them to the complex domain. Since we are considering the balanced , positive sequence case, the zero sequence volt age and current are both zero , i.e . ,
V o = io
=0
(6.94)
Also, since the system is in steady state, the speed is constant and the damper currents and all current derivatives are zero, i.e.,
w = WI iD = iQ = 0 id = iq = iF = iD = iQ = 0 axis voltage i s Vd = - rid - wLq iq - k wM Q iQ - �d
From (6 .34) the d porating (6.95), this voltage may be written as
(6.9 5 ) and, incor
Vd = - rid - W I L q iq
(6.96)
Also from (6.34 ) the q axis voltage is
V q = - riq + WLd id + kwMF iF + kwMD iD - �q
which may be written as
Vq = - riq + W I Ld id + kw l MF iF
(6.97 )
IF = iF/V3
(6.98 )
We now define an rms stator equivalent of the rotor field current to be Then, using this definition and definition (6.9 1 ) of d and q axis currents, we de fine rms equivalent d and q axis voltages by dividing (6.96) and (6.97 ) by V3, with the result
Here we recognize mutual reactance
Vd = vd /V3 = - rId - xq Iq Vq = vq /V3 = - rIq + xd Id + k XmFIF the previously defined reactances X d and Xq
(6.99 ) and define the (6.100)
Using (6 .92 ), we may compute the rms phase a terminal voltage on an arbitrary reference basis to be Va = ( Vq + j Vd ) e il) = [ (- rIq + xd Id + kx mF IF ) + j(- rId - Xq Iq )] e il)
We recognize the quantity in brackets to be Va on a d-q basis, according to (6.89). Thus using the bar notation for this basis and rearranging, we have
Va
= =
Vq + j Vd = - r(Iq + jId ) - jxq Iq + xd Id + kXmFIF - rIa - jXq Iq + xd Id + kXmF IF
(6.101 )
It is convenient to define a field excitation voltage
EF = Eq + Ed = Eq + jEd =
kXmFIF
(6.102)
212
C hapter 6
such that wriThetereason asis zero is that there is no field winding on the q axis. Then we may whicyhingis stil where equationwe ofseemithatxed phasor and scalar notation. We clarify this by appl (6.103)
Ed
(6.101 )
(6.104)
an
(6.88),
Iq
(6.104 )
= lq ,
Id
=
-
j 1d
Then in d-q phasor notation becomes = 6 or,reference. multiplying by e 1 , we can transform this equation to an arbitrary frame of Equationin the steadydefines theTwophasortypical diagramphasorof adiagrams, synchronousonemachine when operating state. for a leading power factor and one for a lagging power factor, are shown in Figure In Eq
Va + rIa + j xq lq + j Xd 1d
(6.105)
(6.105)
6.1 1 .
d
d
(b) Fig . 6 . 1 1 . Phasor diagram of a synchronous generator with arbitrary reference : (a) leading power factor, (b) lagging power factor.
S eq uence I mped a n ce of M a ch i nes
213
either case we can define power factor = cos where Figure is definedy which to be Va leads 10 - Th voltage V",q shown in Fp
e
=
= r/> v
angle b
6.11
V",q
Eq
=
r/>l '
e
jXd Id �
�
�
(6.106)
e
=
-
(6.107 )
ThisInvolmany tage willreferences be neededtheinphasor a laterdidevelopment. a gram of a synchronous generator is drawn inaddeda sland ightlsubtracted y differentfrom way than shown Figure If a quantity jXq Id is we canin develop Va rIa jXq 1a j(Xd - x q) Id This form is interesting because the last term is approximately zero for phasor round rotor generator since Xd Xq for these machines (see Table diagram which combines and is shown in Figure Eq -
6 .11 .
(6.105),
=
-
+
......
......
+
+
......
(6.108)
a
�
6 .1 ) . A 6.12.
(6.108 )
(6.105 )
d
Fig. 6 . 1 2 . Another form of phasor diagram of a sy nchronous generator.
(6.105)
(6.108)
The condition described in and or by the phasor diagrams of erator Figurecircuitandreduces toUnder describedequivalent above, thecircuiequits ofvalent thattheof steady Figure state condition where all Figures 6 . 1 1 and 6 . 1 2 can also be visualized in terms of the synchronous gen
6.8.
6.7
6.13
Id
-
r
Iq
-
+
Eq-t'3 0>1 ). d Fig. 6 . 1 3 . The d·q equivalent circuits i n the steady state case.
214
Cha pter
6
v'3
voltages andvaricurrents have been divided by to use the equivalent rms stator (uppercase) a bl e s as in (6. 9 9). d and equations to obtain (6. 1 08), we can draw the combientningcircuithet shown simpleAfterequival in Figure 6. 1 4, where an EMF due to saliency of q
Io
r
-
+
t+
Vo
Eq
tr+:----o
j ( X d - Xq ) I d
Fig. 6 . 1 4 . Equivalent circuit of a synchronous generator.
jti(Xveld y small, thiis shown asis asmalseparate volalways tage. inSinphase ce thewiquantity ibys relthea s EMF l and is t h as indicated phasor diagram of Figure 6. 1 2. -
6.8
x q ) Id
(Xd - Xq)
Eq
Subtransient Phasor Diagram and Equivalent Circuit
In fault studiestheweincidence are usualoflytheinterested inwetheexamine subtransi eassumptions nt period imme diately following fault. If the used ibetween n developing the steady state phasor diagram, we observe certain di f ferences the statement steady stateof theand steady subtransient conditions.In theEquation (6. 9 5)period is a mathematical state condition. subtransient wecannot can assume usually that assumethe asdamper beforewinding that thecurrents speed isarenearl yasconstant. However, we zero in (6. 9 5). The third condition, setting the current derivatives to zero, iswords, equivaltheentfluxto linkages are the This sameisimmedi aassumption tely followingto make the faultfor asthetheysubtransient were just period before andthe faul t occurred. a good follows from Lenz'ofs thelaw volin tthat the currents(6.do3 4)notforchange instantaneousl ydepends . The solution age equations a transient condition upon thechinitial condifaulttionin(atour probl 0-) and the nature of the change or driving func tion whi is the ete solution also requiofrestime.the solution of a torque or inertia equatioenm.to fincompl d the speed as a function (0, written complete final solution state steady solution transient solution Ine .,ourthecasesolution we areat usually satisfi ednotif aswedifficul can findt asonly the forsubtransient solution, 0+. This is sol v ing the compl ete solu i.tion as a function of ti m e, but i t still involves a considerati o n of the same equations. The desired subtransient solution may be written in words as five subtransient at t 0+solution initial condition delta condition or change(6.109) stating that the rate of change of flux linkages is zero. In other
t
=
A
w
F,
Even if we assume balanced conditions and constant speed, we are faced with the simultaneous solution of six equations
solution may be
as
d, q ,
D, and Q). Then the complete
+ complimentary or
=
t=
=
at t = O -
+
from t = O - to t = O+
S eq ue nce I mpedance of M a chines
21 5
The initial or pre fault condition is the steady state solution discussed in the pre vious section , where the voltages and currents depend on the machine load. These prefault (t = 0 - ) conditions are : are usually nonzero and are a function of load. All current derivatives are zero . All flux linkage derivatives are zero due to item 2 . All damper currents are zero, iD = iQ = O . The field current i s constant, iF = a constant . All zero sequence quantities are zero . 7 . The speed is constant, W = W I ' 1. 2. 3. 4. 5. 6.
Vd , vq , id , iq
Immediately after the fault (t = 0+ ) we assume : 1 . The fault is represented by a step change in id and iq . 2 . The current derivatives are nonzero (an impulse) at t = 0 but are zero at t 0+ . 3. The flux linkage derivatives are zero at t 0+ . 4 . The damper currents are nonzero at t = 0+ . 5 . The field current changes to a new value at t = 0+ . 6. The zero sequence quantities are zero . The speed is constant, W = W I ' =
=
7.
We shall represent all variables by denoting the initial condition plus the change as indicated in (6.109), using the notation (6.110) where
io = initial condition , t 0it. = change from t = 0 - to t = 0+ =
(6.111)
At t = 0+ we may write the flux linkages from (6.32) in matrix form as " = Li, or
A o + �t.
= L(io + it. )
(6 . 1 1 2 )
Then the initial condition establishes the prefault flux linkages as
Ld id O + kMF iF O Ad O Lq iq o Aq O kMF id O + LF iF O AFO kMD id O + MR iFO A DO (6.1 1 3 ) kMQ iq o AQ O Now, following a change in currents idt. and iqt. the currents in the coupled cir cuits will adjust themselves so that the flux linkages remain constant, in the F, D , an d Q circuits, i.e . , from ( 6.32) and ( 6 . 1 1 2 ) Adt. A q t. A Ft. A D t. A Q t.
Ld idt. + kMF iFt. + kMD iDt. L q iqt. + kMQ iQt. = kMF idt. + LF iFt. + MR iD t. kMD idt. + MR iF t. + LD iDt. kMQ iqt. + L Q iQt.
? ?
0 0 0
( 6 .1 1 4)
216
Chapter
6
(6.Thi1s14)equation sets up a constraint among the current variables, we have LFiFtl. + MR iDI1 = - kMF idl1 , MR iFI1 + LD iDI1 = - kMD idl1 which we can iFI1 iDI1 it the result . - - kLDMF kMDMR idl1 lFI1 LFLD MA kMFMR id l1 (6.1 15) 'D I1 - - kLFMD LFLD MA Also, from the (6. 1 14) we compute . = - kML Q • (6. 1 16) 'QI1 Q Currents from (6.1 15) (6.116) now n (6.1 14) 2 J 2 ) MFMDMR ] '. L'" d (6. 1 17) dl1 -- [Ld - k (LD Mi LLFFM 2 LD - MR where (6.47) computeL; - (L - kL2M3 ) . (:;= L'" (6. 1 18) Q and circuit flux linkages terms we write from (6.34) Vd = VdO Vdl1 - r( ido i l1 ) - k WI MQ ( Q o - ),d = 0+, ),d = O. (Sinhce ) iQ O O. (6.116). Making these VdO = -ridO ' - W I L'" Vdl1 - - 'd - WI L ' W lkL2QM§ . -- -ndl1 ; - rid W- L;lL i I o (6. 1 19) (6. 1 20) (6.96) edo = 0 (6.34) for
solve for
and
from
w h
+
_
_
.
+
_
_
last equation in
Iq l1
and
may
b e used i
+
�
is defined in
t o compute
dl1 �
dl
11
to be the quantity in the brackets. Similarly, we
'\
"' ql1 -
q
Iql1
q lq l1
Thus we have identified the changes in d q in of the change in currents. The voltage equations may also be written in terms of initial-plus-change no tation. For the d axis voltage
+ idl1) - W I Lq (iq o + q
=
+
i
+ iQ I1)
this equation must represent the solution at t we note that W y ? Also, b y definition the initial condition i s steady state s o that Finally, we note that iQ I1 may be written in terms of iql1 from equation changes, we write w I L q iq o and - r1 l1
q lql1 +
or, adding the two equations, Vd = Adding and subtracting a quantity
Iql1
w
=
q lq l1
q q o - w 1 L iq l1 . iq gives the result
where we define
From it is apparent that since this component does not exist in the steady state. For the q axis we similarly compute from
Vq
=
VqO + Vq&
+
=
-r(iqo + iq&) + ([)lLd(idO + id&) + kw 1 MF(iFO + in)
kw1MD(iDO + iD&) - lq
217
S eq uence I mpedance of M a ch i nes
In this case Then
where from
iFO = a constant, iDO = 0, �q = 0, and iFtJ. and iD tJ. are given by (6. 1 15).
(6. 1 21) (6. 1 22) (6. 1 23) (6.1 24) (6. 1 25) (6. 1 26) (6. 1 03) (6. 1 27) (6. 1 28)
(6. 1 03)
eqo = k w1MFiFO Making substitutions of iFtJ. and iD tJ., we compute Vq tJ. = - riqtJ. + wl L; idtJ. Equations
and
(6. 1 21) (6. 1 23)
are combined to write
where we define and
e�
It is interesting to note that the definitions for and e; parallel similar def initions for the steady state condition. In particular, we compute from and - eq o = W 1 L or
(6. 1 26) e;
( d - L;) idO '
We may also compute
ed O
Vxq = eq o + Xd idO = e; + x; idO
-
e; =
Wl
(Lq - L; ) iq o or
ed O
Vd
since = O. To construct a phasor diagram for the subtransient case, we divide the and equations by v'3 to rewrite the equations as rms stator equivalents. Thus
Vq
(6.1 29) (6.99), (6. 1 30) (6. 1 31) (6. 1 32)
where we note the striking similarity to the steady state equations the only difference being the addition of the double primes (" and the appearance of the voltage E� . As before , we compute on a d-q reference,
)
where we define E"
Equation
(6. 1 30)
= E; + jE;
may be changed t o the form E"
= Va + rIa + jx;fa + j (x; - x; )
Iq = E; + jE;
This form is convenient since the quantity (x; - x � ) is positive o n both round rotor and salient pole machines but is usually quite small. The phasor diagram for the subtransient condition is constructed according to in Figure where the condition of Figure is used as an initial condition. Several observations are in order concerning this important phasor diagram . We note that the new (fault) current is larger than the initial current, and it
(6. 1 32)
6. 1 5
6. 1 2
Chapte r 6
218
d
Iq
Vd
Eq
Vq Vxq
Vx d
Eq O
q
J X dIdo
Fig. 6 . 1 5 . Subtransient phasor diagram of a synchronous generator.
lags the terminal voltage by a larger angle () , which is nearly 90° for a fault on the machine terminals. This means that Iq is small and Id is large. Since the speed is assumed constant, 8 has not changed. An importanfpoint on the diagram is the encircled point located at
(6.1 33) But from (6.127) and (6.128) VX q
+ Vxd = (Eq o - jXd 1d o ) + (- xq lq o) = (E� - jx;ld o ) + (E� - x� lqo ) (6. 1 34)
or this voltage is a constant and depends only upon the pre fault or initial condi tion. Therefore, the location of the circled point is unchanged from t = 0- to t = 0+ and forms a kind of pivot about which the other phasors change. Thus VX q + Vx d = Va + rIa = Va(O ) + rla( o )
= a constant
(6. 1 35)
Referring again to the phasor diagram, we note that the phasor j(x; - x� ) lq is very small since from Table 6.1 (x; - x � ) is small, and Iq is usually small under fault conditions. This quantity might be neglected as an approximation, or set We observe that
J.(X q" - X d" ) Iq
-
=
E; » E �
0
(6.136) (6.137)
or approximately,
(6.138 ) This is quite apparent from the phasor diagram but could also b e concluded from the definitions of E; and E� , where Eq o will be the dominant component. If approximations (6.136) and (6.1 38) are allowed , the voltage equation (6.1 32) becomes
(6.1 39) Both the exact equation ( 6.1 32) and the approximate equation (6.139 ) can be
Seq uence I mpedance
of M a chines
219
represented equivalfroment(6.circuits, n Figure E" areditions. constantsbysince 1 25) andas (6.shown 1 26) ithey depend6. 1 6.onlyNoteon that the initial conE; circuitneglected of Figuresin6.ce1 6bit isis small. widely Weusedmayfor thifaulnkt calcul the The resistance of thisations, circuitoften as a wisub-th � Il; and
r
-
(0 ) +
(b )
q
E"
Il"d
+
j (x'q -Ild U q r
I"
�
+ v. B
,�
Fig. 6 . 16 . Equivalent circuits of the subtransient condition : (a) exact equivalent, ( b ) approxi mate equivalent.
transi e nt Thevenin equivalent since it is clearl y a constant vol t age behind a "con stant"ofimpedance. Obviously, is valid only at the instant but this is the time greatest interest in faultitcalculations. t=
6.9
0+ ,
Armature Current E nvelope
It is convenient toFigures have a6.mathematical expression foris thetheenvelope of the armature currents of 1 and 6. 6 . The envelope sum of several with ofitstheowncurrent time asconstant, whichof time. combine to determine the peak valterms, uThe e oreach envelope a function final valuecomponent of the current inenvelope Figure andhasmaya bepeakplotted value alone as isthecalled the synchronous of the line a-d. Since the current of interest is a fault current, Iq wil be small and la (6. 1 40) Making this substitution into (6. 1 08), we compute for r negligible, (6. 1 41) Va orfora threep hase faul t on the machine terminals (Va 0) the current magnitude Id shown in Figure 6.1 7 is given by (6. 1 42) Superimposed on the synchronQus componentinofFigure current6.17.is Equations the transientfor component which is given by the quantity theof section transient6. 8condition have not beennotation derivedisherereplaced but arewithidentical toprimes. the results i f the double-prime single The result for a 31P fault wi t h negligible resistance can be computed from (6. 1 39), with the result O-a . It
6.6
�
Eq
=
Id
+ jXd 1d
=
(i� - id )
220
Chapter
J2 I d
---
-
Id
6
- - ��----
o �-----.- t
Fig. 6 . 1 7 . The armature current ac envelope, excluding id e '
- The transientcomponent. current envelope is thely locus intoFigure and includes the synchronous It is usual convenient wri t e the transient current envelope as and The this envelope decayscomponent with time ofconstant T tois seconds). subtransient faul t current superimposed ongenerator the pre viterminals ously defined components. Its i n i t i a l rms value for a fault on the is found from with neglected. Thus - Id" Xd The subtransient component envelope is written as e i� - i;' (I� and thicomponent s component decays theveryinfast wirmsth current a time ifconstant ofnoT�dc component. or cycles). ThisThe determines i t i a l there is the current envelope is the sum of the synchronous, transient,acandcomponent subtransientof components I,a
0.
Id' - Eq' /Xd'
( 6.1 43)
6.6
b-d
i� - id = v'2(I�
Ia"
2
= Eq" / "
0.
+ (Id
(6.1 44)
3f/>
= V2
i. e ., la c = ienv /...[2.
� (1
r
( 6.139 )
ienv = ...[2 [Id
Id ) e-tl Td
-
- I� )
( 6.145)
-
tlTd
(6.146)
(2
Id ) e- t lTd + ( 1;
I� ) e- tlT3 ]
-
3
(6.147)
or the rms value ofInth�termscurrent as machine a functionreactances of time theis rms value timesofthisalternati amount,ng of the and a function that of time may be estimorated from wicurrent th theasassumption [1:. (� 1:.) ( � �) ] Usually is assumed toalsobehasin thea dcrangecomponent of towhichpudepends for faulont computations. The total current the switching angle (radians) given by (see cos Eq
la c !?:! Eq
a
Xd
�
E�
+
Xd
Eq
_
�
-
(6. 142), (6.143),
E;
Xd
e-tlTd +
Xd
1.0
1.10
[ 19] )
Id e
-
1 /...[2
...[2 Eq
"
Xd
a -t lT a e
-
Xd
e- tIT a
(6.145 )
(6.1 48)
(6.1 4 9)
Seq uence
I mpedance of M a chines
22 1
Then the effective value of armature current is
+ lJc ) I /2
(6.1 50 )
lm ax = v'3 Eq Ix�
(6.1 51 )
I = (l;c
which has a maximum value at
t=0
corresponding to a = 0 of
Values of Xd , x� , x � , T� , � , and To may all b e found from an oscillograph of armature currents similar to Figure 6.1 . The technique for doing this is explained fully in [ 19] and will not be repeated here. These values, expressed in pu, tend to be nearly constant for a particular type of machine and are often tabulated for use in cases where the actual machine parameters are unavailable. Reference values are given in Table 6 . 1 . Since our motivation i s t o perform fault calculations, w e are often concerned with fault analysis under extremum conditions ; i.e. , we are trying to determine either the maximum or minimum value of fault current seen by a particular relay or protective device. For maximum current we usually take the curre nt l� as the rms value of fault curre nt, or we consider that the generator has reactance x� in the positive sequence network, with X 2 and X o in the other networks. For other than maximum fault conditions the value of reactance used may be changed to a value which will give the correct current at the time the protective device operates. Thus the time may be substituted into (6. 148), and the current and hence the equivalent reactance found . 6. 1 0 Momentary Currents
The equivalent circuit developed in section 6.8 will permit the solution of the subtransient current l� corresponding to the highest point in the current envelope of Figure 6 . 1 7 . This current is the symmetrical current, i.e. , it includes no dc off set at all and corresponds to the highest rms value of Figure 6.6. If our purpose in computing the fault current is the selection of a circuit breaker to interrupt l� either at the generator or some other location, we should include the dc offset. Thus the total effective value of the current must be computed as in (6. 1 50 ) . If this is done, for the worst case the result is that of (6.1 5 1 ) for 100% dc offset in a given phase , or
( 6. 1 52)
where l� sy m = the symmetrical part of the total current. Actually , the current to be interrupted by the circuit breaker will never be this great since this would re quire zero fault detection time (relay time) and zero breaker operating time. The IEEE, recognizing that some consideration should be given to the inclu sion of dc offset in the selection of breakers, has formulated a recommended method for computing the value of current to be used in these computations. The computed current, called the momentary current, depends upon the time after initiation of the fault at which the circuit will be likely to be interrupted. Thus the natural decrement of the dc offset, which occurs at time constant To, is ac counted for approximately and provides a reasonable compromise between the limits of l� S ym and 1 .732 l� Sym ' In the recommended method of computation, the momentary duty of the breaker is based upon the symmetrical current l;sym, with multipliers to be applied according to the breaker speed . These multipliers are given in Table 6.2 which is taken from [ 5 1 ] . These multipliers determine the so-called "interrupting duty " of the circuit breaker. Another rating called the
222
Chapter
Table 6. 2.
6
Current Multiplying Factors for Computing Cireuit Breaker Interrupting Rating
Breaker Speed or Location
IdsYm Multiplier M
8 cycle or slower
1. 0 1. 1 1. 2 1.4
5 cycle
3 cycle 2 cycle Located on generator bus If sym fault MV A > 500
add 0. 1 to multiplier add 0. 1 to multiplier
"momentary duty" is found by multiplying the symmetrical current by 1 .6, and this may be checked against a published breaker momentary rating. (The present practice is to use only the symmetrical interrupting duty, and breakers are so rated. ) In summary the two breaker ratings which must be checked are
1.6 I� sym interrupting rating > I� momentary rating >
M
where
M
(6.153)
is found from Table 6.2. II.
6. 1 1
sym
I N DUCT I ON MOTOR IMPEDANCES
General Considerations
The striking difference between induction and synchronous machines, insofar as their response to faults is concerned, is in the method of machine excitation. A synchronous machine obtains its excitation from a separate dc source which is virtually unaffected by the fault. Thus as the prime mover continues to drive the synchronous generator, excited at its prefault level, it responds by forcing large transient currents toward the fault. The induction machine, on the other hand, receives its excitation from the line. If the line voltage drops, the machine .excita tion is reduced and its ability to drive the mechanical load is greatly impaired. If a 3t/> fault occurs at the induction motor terminals, the excitation is completely lost; but because of the need for constant flux linkages, the machine's residual excitation will force currents into the fault for one or two cycles. During these first few cycles following a fault the contribution of induction motors to the total fault current should not be neglected. However, it is somewhat unusual to find an induction motor large enough to make a significant difference in the total current. For example, if the base MV A is chosen to be close to the size of the average synchronous generator, an induction motor would have to be larger than about of base MVA to be of any consequence. When induction motors are included in the computation, a subtransient reactance of about 25% (on the motor base) is often used. The transient reactance is infinite.
1%
6. 1 2
I nduction Motor Equivalent Circuit
The induction motor is usually represented by a "transformer equivalent," i.e. , a T equivalent where separate series branches represent the stator (primary ) circuit and the rotor (secondary) circuit, with a shunt branch to represent iron losses and excitation. At standstill or locked rotor the induction machine is indeed a trans-
223
S eq uence I mpeda n ce of M a chi nes
the secon ofFigure impedance theofequivalent however, turning, With cithercuirotor former.or rotor in noted as s slip the function a be to seen is t dary If the rotor induced is Er at standstill, it is sEr when the rotor is turning6.18.at EMF
X.
R. •
Val
r - RsRr
Xr
lal
-
I rl Rc
}(I- S) r -•
Fig. 6 . 1 8 . Positive sequence induction motor equivalent circuit.
inducedof of slipthe. Since a function isstator inductance rotor apparent , the frequency Similarlyhave slip s. currents reactance frequency, the is f where sf rotor the rotor or reactance at standstill. Then, we com at anywillslipbes, sXIr'r where pute sEr, /(XRrr is thejsXr)rotor (6.154) Ir' = --�-(R r/s)Er' jXr Thus the rotor impedance is (6.155) Zrl (R r/s) jXr as noted in Figure 6.18. We usually write (6.155) as Zrl = (Rr j Xr) 1 s- s Rr a function and is not standstill impedance term is thethe shaft where is of s. The delivered the power load toatwhich secondthetermfirstrepresents =
+
+
+
�
+
+
--
(6.156)
The shaft or mechanical torque is (6.157) ml PWml W I P(1ml- s) Nm/phase I;, Rr Nm/phase (6.158) where rad/secin rad/sec speed inspeed synchronous w,Ws = shaft slip in pu of terms in torque base the pu, wew ,select torque is expressed mechanical Ifbasethepower 6.157) and baseinspeed (base voltamperes) . Then from ( /w,(l - s ) P1 m1u- s 1;1sR r pu (6.159) pu Pm ,SB/W, which has units of pu power/pu slip or pu torque. Some authors state pu in T
=
=
= --
=
=
Tm ' =
= -- = --
Tm l
Chapte r
224
6
PER U N I T TOR O U E
_I 2
Fig. 6 . 1 9 .
BACKWARD
0
FORWARD
0
-I
•
Torque of an induction motor as a function of speed or slip. ( From Kimbark [ 19 ) . Used with permission. )
"synchronous watts" (see [ 1 9 ] ) . The average torque-speed characteristic o f an induction motor is shown in Figure 6.19. If negative sequence voltages are applied to the induction motor, a revolving MMF wave is established in the machine air gap which is rotating backwards, or with a slip of 2.0 pu. Then the slip of the roto r with respect to the negative sequence field is 2 8 when the rotor is moving with forward rotation. Since 8 is small, the approximation is sometimes made that this negative sequence slip is 2.0 rather than 2 8. If 82 is the negative sequence slip, 82 = 8 == 2. The equivalent circuit for negative sequence currents is the same as that of Figure 6.18 with 8 replaced by 8 as shown in Figure 6.20. The mechanical power
-
-
2-
2
Rc
Rr
-( l-s ) (2- s)
Fig . 6 . 20 . Negative sequence induction motor equivalent circuit.
Pm2 - -Ir2 R r -_ Tm 2 -- 1';2 R r
associated with the negative sequence rotor current (1 (2
_
8) 8)
Ir2
is (6.160)
W/phase
the negative sign indicating that this would cause a retarding torque w d 2 - 8)
Tm2
Nm/phase
where (6.161 )
Then the net torque per phase is
Tm Tm l Tm2 = Rr (1;1 _ 1;2 ) =
+
WI
s
2- 8
N m/phase
(6.162)
which is a smaller torque than that present when only positive sequence voltages are applied . The net effect of unbalanced voltages applied to an induction motor is to reduce the mechanical torque. Since the speed of the motor depends upon the simultaneous matching of motor and load torque-speed characteristics, the effect of the unbalance depends on the type of load served by the motor. Should the motor applied voltage change but remain balanced, the torque 1 will vary Depending varies directly with as the square of the applied voltage since
Irl
Va lT'm
225
S eq ue nce I mped a n ce of M a chi nes
15
�
'----������ 1 .0 0. 50 0.75 0.25 o PER U N I T S P E E D
Fig. 6 . 21 . Torque-speed relationships for motor and load : (A) motor torque-speed curve at rated voltage , ( B ) constant torque load , ( C ) constant power load, ( D ) typical load torque-speed curve , ( E ) motor torque-speed curve with unbalanced applied voltages. ( From Clarke [ 1 1 , vol . 2 ] . Used with permission . )
oncausethetheloadmotorcharacteri stic,Aasystem 3ct> fauldisturbance t near the which motor unbalances or a reducedthevoltage may to stall. voltage will also cause the torque of the motor to be reduced, as shown by curve of Figure In thispeed s casecharacteristic the motor wiwill tslow down andIf theseekloada new operatintorque g pointsuchon the6. 2a1.torqueh the load. is constant conveyor or load, is constant powermaysuchstallas aifregulated dc generator driving a con stant impedance the motor i t is unable to deli v er the necessary For a load such as a fanof orFigpump the2 1. torque variloades torque atthethe square lower speed. about as of the speed, such as curve ure 6. Such a woultod continue tounless be served at a reduced speedtorque if the drops motorpractically torque is reduced due any cause, the reduction in motor to zero. Bythesethetorques use of can equivalent ci r cui t s for the posi t i v e and negati v e sequence networks, be determined and the motor performance may be evaluated if theSince torque-speed characteri sticusual of thely shaft loadeitheris known oror can be estimated.con induction motors are wound for ungrounded nection,for thea zerozerosequence sequenceequivalent currentscircuit. in the motor are always zero and there is no need s areis gigivvenen byin Approximate valuandes forcredited induction motor references. equivalent ciThisrcuitdata Clarke [11, vol. 2] to several Tablso ebe determi Clarkenedsuggests that theratedposivoltivetagesequence impedance forpu current any loadormayby alcomputing by di v i d ing by the assumed the bemachine driving-point impedance. However, as forthe theactualequicurrent woul d seldom known, Fi g ures and 6. 2 0 may be used valent circuits of positive and negative sequence networks respectively. E
as
D
�
Y
6 .3.
6 . 18
Table 6. 3.
Rating
Full Load Efficiency
(HP)
(%) 75-80 80-88 86-92 91-93 93-94
Up t0 5 5-25 25-200 200-1000 Over 1000
Approximate Constants for Three-Phase Induction Motors R and X in per Unit* Full Load Full Load Power Factor Slip X. + xt Xm R, Rr (%) (%) pu) ( (pu) (pu) (pu) 7 5-85 3.0-5.0 0. 10-0. 14 1.6-2.2 0.040-0.06 0.040-0.06 82-90 2.5-4.0 0. 12-0.16 2.0-2. 8 0.035-0.05 0.035-0.05 84-9 1 2.0-3.0 0. 15-0. 17 2.2-3.2 0.030-0.04 0.030-0.04 85-92 1.5-2.5 0. 15-0. 17 2.4-3.6 0.025-0.03 0.020-0.03 88-93 "'1.0 0. 15-0. 17 2.6-4.0 0.015-0.02 0. 015-0.025
Source : Clarke [ 1 1, vol. 2 ] . Used with permission of the publisher. * Based on full load kVA rating and rated voltage. t Assume that X. Xr for constructing the equivalent circuit. =
Cha pter
226
6
6. 1 3 I nduction Motor Subtransient Fault Contribution
The application ofexcitation a short ciforrcuithet near theandtermiits nfield als ofcollapses an induction motor removes the source of motor very rapi d ly. Clarke [11, vol. 2] gives the approximate time constant of the decay of rotor flux
as
(6.163)
where theto bequantities usedHowever, in the siequation areodefiof Xnedto inR isFigure 6.18 and are as sumed ohms. n ce the rati used in (6.163), these quanti t i e s may be in pu. From Tabl e 6.3, if we take 0.16 and 0.035 as average values which of X. isX,.lessandthanR,.onerespecti vel(1ycycle , we compute = 0 . 0121 sec for a 60 Hz motor cycle 0.01667 sec if f = 60 Hz ) . The cur rentststo2-4be cycl interrupted by thefaulcircuit breaker ins case transmission systems is that which exiinduction e s after the t occurs. In thi the current contribution from motors may be negl e cted. In industrial pl a nts, however, l o w-vol tage, icycle. nstantaneous, circuit breakers are often used which clear faul t s in about onebe In such cases the faul t contribution from induction motors shoul d not neglHowever, ected. if the maximum value of current is required for computing fuse melting or cibercuifound t breaker mechanical stresses,thisthesubtransient contributionperiod fromasanainduction motor may by treating i t during synchro nous machine. That is, we will compute a generated EMF behind a reactance, and this will constitute the subtransient equivalent circuit as shown in Figure 6.22. in
t,.
+
=
air
j ( x . + xr ) •
Vo l
1-
101
FOSITI VE SEQUENCE
NEGATIVE SEQ l£ N CE
Fig. 6. 2 2 . Sub transient induction motor equivalent circuits.
Em
Here, we assume that From duringTable the subtransi entXrtime 0.16 intervalpu., Weactscompute throughfromthe reactance X. X,.. 6.3, Figure 6 .22, (6.164) where 101 is the prefault motor current and Vol the prefault motor voltage. Approximate thefaulsubtransient valtageue ofand drawi for nag100ratedhpcurrent inductionat 88% motor,power op erati n g before the t at rated vol factor. Then find the subtransient contribution of the motor to a 3f/J fault at the motor terminals. From (6.164), taking Vol as the reference phasor, 1 .0i!t' - j(1.0/- 28.5° )(.16) = 0.9235 - j O.1406 = 0.935/- 9.35° pu +
Example
X. +
6. 1
Em
Solution
Em =
�
227
S eq ue nce I mpedance of M a chines
Therefore, for a 3ct> fault on the motor terminals, ° = 0.935/ 9 . 35 = 5 . 85 L- 99 • 35° 0.1 6/90 °
Im
pu where the current is in pu on the rated base kVA of the motor. 6. 1 4
Operation with One Phase Open
Ifresulting one of seritheethree leads supplying an uinduction motor shouldcomponents. become open,In the s unbalance may be eval ated by symmetrical such a case the boundary conditions are = ( 6.165) = 0, = From the first two boundary conditions we compute = A- I or (let h 1) l l lLll aa, a�JJ � a I: Ib
Io
-
Ie,
Vb - Ve
Vb e
[ J ["� "J 101 2
l�: J
=
b
Io l
=
O
-
m ila
Ib
=
a.
=
lobe
-
(6.166)
= from whichnetworks we seemust thatbe connected This means that the positive and negative sequence as shown in Figure 6.23. - Io2 .
I l
. Ib
3
Yal I
• •
(Val - Va2) " j Vbc
Xr
Xs
Rs
Rr
PO S ITI VE
101
Rc
Xm
Rr
SEQ UENC E
NI
(I -s ) s
Ial " - Ia2 = j � ..
3
Rs •
Va2
10
Xs
Xr
Rr
NEGATIVE SEQUENCE
..
2
Rc
Xm
- Cl-s ) R r
2-s
N2
Fig. 6 . 23 . Sequence network connections for an induction motor with line a open.
The only known voltage is, from (6.165), with h = 1 = = a a2 Summarizing from (6.1 66) and (6.167 ) Vbe
Vb - Ve
=
-
( Voo +
j � ( Vo l - Va l )
Va l + aVo2 ) - ( VoO + Vo l +
al
Va l )
( 6.167)
228
C h a pter
Ial =
. Ib I0 2 = J � '
6
. Vb e V02 = J �
(6.168) and if is known, may be determined from Fi g ure 6. 2 3. Thi s compl e tel y es the sequence networks (assuming that is known), and the net torque may bethisolvscomputed from (6. 162). Wagner and Evans [10] show that the net torque in siltouati omust n remai nbes positootilarge, ve as orlongtheasmotor the slwilip lisstall. smallClarke . This [11, meansvol.that2] dis the shaft ad not cusses ttheor bank case ofis inan paral openleconductor inmotor. the supply tos ancaseinduction motor wheremaya capaci l wi t h the In thi a negati v e torque resul the induction condittiwhion cwillh willnotcause be discussed here. motor to reverse direction of rotation. This Vbe
-
V0 1
Ib
-
s
vet)
Problems
t
Consider a series R-L circuit supplied from an ac voltage source Vm sin ( w + a:) and assume that initially the current is zero because of an open switch in the circuit. If the switch is closed at 0, find the current as a function of the series impedance magni· ° ° tude and angle, Z - I Z I &.. Sketch the resulting current for a: (J 0, 45 , 8,Ild 90 . 6.2. Express the voltage and final value of current for problem 6 . 1 as phasors. Show how the dc offset can be found by projecting the current phasor on the real axis. 6.3. Verify the inductances given by (6. 7)-(6. 13). 6.4. Show that p- 1 of (6.19) is indeed the inverse of P given in (6.17). 6.5. In Figure 6.3 consider that all self and mutual inductances are constants instead of the time varying inductances given by (6.6)-(6. 14). Write the equations for the synchronous machine voltages, then transform by the phasor transformation, and transform again to the 0·1-2 coordinate system. 6.6. Repeat 6.4 with an impedance ZPI rPI + jwLPI in the neutral and a current In entering the neutral. Write the equations of phase voltage to ground rather than to neutral. How does the neutral impedance appear in the 0·1·2 coordinate system? 6.7. (a) Verify (6.28) and explain th e meaning of the results. What k i n d of induced voltage is s? (b) Verify (6.32), making use of the trigonometric identities of Appendix C. 6. 8. Explain why the voltage nOdq affects only the zero sequence. 6.9. Show that by proper choice of base values (6.34) may be written exactly the same whether in volts or in pu. What restriction does this place on the selected base values? 6. 10. Prove that (6.35) is true under the conditions specified. Do this for only one phase, e.g. ,
t
6.1.
co
-
-
..
co
vf
Xb lib ·
6. 1 1. Given (6. 20), reduce by Kron reduction to write vabe in terms of only by eliminating the rotor current variables. Is the result Laplace transformable? 6.12. Examine the following special cases for the Park's transformation iodq ... Plo be with (J + l) + 90. (a) iobe is strictly positive sequence, i.e., ..
WI t
lobe ViI -
[:: (WI�Ittt Jl [COS Wi(WItt � -
cos
120)
+ 120
find iOdq • (b) lo be is strictly negative sequence, i.e.,
iobe find �q.
=
ViI cos cos
(Wit
+ 120)
- 120)
S eq uence I mpeda nce
of M a ch i nes
(c) iubc is strictly zero sequence, i.e.,
� - �l .� w, t
[
find iOdQ . 6. 1 3. Given the unbalanced set of currents
(W I t Ibm cos (W I t lem cos (W I t
lam cos
iobe "
+ -
+
229
m
a)
J
120 + (3)
120 + 1)
compute iOdQ . 6. 14. Verify equation (6.46) and explain the meaning of this result in two or three sentences. 6. 15. Compute vOdQ from equation (6. 34) under the condition that (a) VF 0 and (b) VF - rF iF = o . 6. 16. Compute vOdQ by making a Park's transfonnation of the result of problem 6. 1 1. 6. 17. Verify that the equivalent T circuit of Figure 6.8 can be derived from the coupled cir cuits of Figure 6.7. 6. 18. Show how (6.79) can be satisfied by proper choice of base. 6. 19. Construct equivalent T circuits for a typical synchronous machine, using values from Table 6. 1 where the machine is (a) round rotor and (b) salient pole. 6. 20. Find the impedance seen looking into the generator d and q tenninals of the equivalent T circuit under (a) steady state conditions, (b) transient conditions, and (c) subtransient conditions 6.21. Compute numerical values for the impedances found in problem 6. 20 using the data from problem 6. 19. 6. 22. Find the time constant of the impedance seen looking into the generator tenninals under (a) steady state conditions, (b) transient conditions, and (c) subtransient conditions. Use data from problem 6. 19. 6.23. A synchronous generator is operating in the steady state and supplying a lagging power factor load. Assume that the machine is connected to a "solid" bus with several other generators. Explain, using before and after phasor diagrams, what happens if the operator increases the field voltage (at constant power) (a) gradually and (b) suddenly. 6. 24. Repeat problem 6.23, but this time assu me that the load is increased while the excitation remains constant. 6.25. Verify (6.82) by perfonning the P transfonnation. 6.26. Find 10 if id 0. 5, iQ = 0.8, and io o. Sketch a phasor diagram. Find lor and lax if 6 = 60° . 6.27. Explain in words how we can interpret kx mF IF as an induced EMF in equation (6. 104). 6.28. Use data from Table 6. 1 for a synchronous machine loaded such that on an arbitrary basis we have ° ° 10 0.8/- 15 Va 1. 0/ 30 , =
=
=
6. 29. 6. 30.
6.32.
=
Then compute Eq and construct a phasor diagram similar to Figure 6 . 12 if the machine is (a) round rotor and (b) salient pole. Construct the equivalent circuit similar to Figure 6. 13 for the machines of problem 6.28. Using the condition of problem 6.28 as the initial condition for a 3t/> fault, compute the initial flux linkages at time t 0+ . Compute all necessary quantities to sketch the phasor diagram immediately following a 3t/> fault on the generator tenninals, using the condition of problem 6.28 as the prefault condition. Compute the maximum cu rrent two cycles after the fault of problem 6.31 is applied. Repeat for 70 cycles and 150 cycles. =
6.31.
-
230
C ha pte r
6
6. 33. Estimate the interrupting and momentary rating of the circuit breaker required to inter rupt a 3ljJ fault on the terminals of the following round rotor generators: (a) 150 M V A, (b) 600 MVA, and (c) 900 MVA. 6. 34. Justify the approximations specified in equations (6.57) and (6.58). 6.35. A cylindrical rotor, synchronous machine having (average) machine constants given in Table 6. 1 is operating at rated current (1.0 pu) and 90% lagging power factor when a 3ljJ fault occurs on the machine terminals. Compute: (a) The voltage E behind the synchronous impedance. (b) The voltage E" behind the subtransient impedance. (c) The initial symmetrical sub transient fault current. (d) The peak symmetrical current after 5 cycles and 10 cycles. (e) The maximum asymmetrical current after 5 cycles and 10 cyeles. 6. 36. Repeat problem 6.35 for a salient pole machine. 6. 37. A 1000 hp induction motor operates at a slip of 2.0% and with 93% efficiency while driving a rated shaft load. Use average values from Table 6.3 to construct a positive sequence equivalent circuit for this machine when the base MVA is the machine base. The motor is rated at 4 kV. 6. 38. If the motor of problem 6.37 operates in a large system which is under study, recompute the equivalent circuit if the base for the system study is 100 MV A, 200 MVA, and 500 MVA. 6.39. Suppose that the voltage supplying the motor of problem 6.37 has a 5% negative sequence component. Compute the positive, negative, and net torque in pu. 6.40. Compute the time constant for the decay of rotor flux for the motor of problem 6.37 by applying (6. 163). What is the time constant in cycles at a frequency of 6 0 Hz? 6.41. Suppose that line a of the supply to the motor of problem 6. 37 is opened. Find the se quence voltages and currents if Vbe is 1.0 pu on a line-to-line basis.
chapter
7
Seq u e nce I m peda n ce
of Tra nsforme rs
An examination of the transformer completes our study of the major compo nents of the power system. Insofar as the study of faulted networks is concerned, we need to solve two kinds of problems. First, given a particular transformation, we must determine the sequence network representation of the physical device. This requires a knowledge of transformer equivalents, standard terminal markings, and the various kinds of transformer connections. The second problem is the estimation of reasonable transformer parameters for installations still in the plan ning stage. In this case a knowledge· of average impedance values is required, and the way in which the various transformer connections affect the sequence net works is helpful. Since the type of transformer connection often influences the type of protective scheme , these concepts must be studied in system planning to assure workability and reliability in the overall design. I . S I N G LE·PHASE TRANS F O R M E R S
Single-phase transformers, connected to form three-phase banks, were in wide general use for transforming both high and low voltages in the first half of the twentieth century . One reason for this was that a fourth transformer could be purchased and installed with a three-phase bank for later connection in the event one single-phase transformer failed. Recently the trend has been toward the use of three-phase transformers because they are cheaper and more efficient and are generally regarded as very reliable . Still, however, many single-phase units are in service , and large transformers are often single phase because of limitations in shipping sizes and weights. 7.1
Single-Phase Transformer Equivalents
The equivalent circuit often used for a single-phase transformer is shown in Figure 7 .1 a where the high and low voltage leakage impedances ZH and Zx are given in ohms. The transformer core losses are assumed to vary as the square of the H·winding voltage and are represented by Rc . The rms value of magnetizing current is represented by the reactance Xm • The turns ratio is defined as
n
=
nH /nX
(7.1)
where 231
C hapte r 7
232
nH nx
= =
number number ofof turns turns iinn winding winding HX We usual ly elimZxinateto thethe Hmagnetization branch sinshown ce Ie «in FiIHgurande 7.1b. also transferat theopen impedance si d e of the ci r cui t as circuit Thus
(7.2)
and, the ampereturns of the transformer windings are equal except for the excitatisinceon MMF nHI. . we have nHIH = nxIx or (7.3) Modem10substation Ie of less than 1 % in sizes up to about MVA and hitransformers gh-voltage ratiusualngslyuphave to about kV. 69
HI
+t -, Ii2
VH
-I H
lHX
� ZH
n
2
:;:
)(
(b) pu Z H X
HI
� � "
-
+t -,
pu I
p u VH H2-
"{::J� Vl(
Ie)
�-
X2
, :1
pu V)(
1 )(-
•
2
Fig. 7 . 1 . Single-phase, two-winding transformer equivalents : (a) equivalent in system quanti ties, (b ) simplified equivalent with all series impedance referred to the H winding, (c) pu equivalent.
Any ZxX-winding current Ix is seen by the primary side as Ix In and the im pedance appears to IH to be Zx . Thus the total transformer impedance as seen by the H-winding current is (7 . 4 ) which shownshuntin branch Figure i7.s eliminated. 1b as the total transformer impedance where, since Ie is smalIt iliss, theconvenient to change ily select the base quanti ties as Figure 7.1b to a pu equivalent circuit. We arbitrar n2
233
S eq uence I mpedance of T ra n sformers
SB transformer A rating VHB rated VH VX B rated Vx =
V
=
(7. 5 ) Then (7.6 ) and (7. 7 ) Fromratio (7.2 )isand we compute in pu at no load , pu Vx pu VH, and the turns pu n 1.0 (7.8 ) Alwinding so, fromto(7.be7 ) we compute the total transformer impedance as viewed from the pu ZHX = (ZH + n2 Zx )/ZHB (7.9 ) Finally , from (7.3) we compute puIH pulx (7 .1 0) From (7.8 )-(7. 1 0) wealmost establish the useequivalent circuit offault, Figureand7.1cstabiandlitythisstudies.the equivalent we will always for load flow, study of reactance ferroresonance, surges, traveling waves, harmonics, etc., the(In the magnetizing shouldswitching not be ignored.) =
(7.5)
pu
=
=
H
=
is
7.2 Transformer Impedances
are nearly the Transformer transformer ratiimpedances ng. Thus from (7.9 ) always given in pu (or percent) based on PU ZH X = (ZH + n2 ZX ) /ZHB = PU ZH + n2 Zx /ZHB = PU ZH + pu Zx (7.1 1) This result may be verified by dividing ZHB by ZX B to compute (7. 1 2) ZHB/ZXB = (VHB /VX B )2 = n2 nearlarey constant forfromtransformers of a givenand size may and debe Values of ofpuaverage ZHX arevalues sign. Tables avail a bl e the manufacturers used when actualdistribution nameplate data are not known. Table 7and.1 gives typicalNotevalthatues forin thetwo-winding transformers rated 500 kV A below. lapplications arger sizes thewhere impedance is almost entirely inductiveused,reactance. In subis station these larger sizes are sometimes the resistance oftenciency, neglected entirely.studiUsually thengresistance is considered only when losses, effi or economic e s are bei considered. Tableis convenient 7.2 gives typical values forimpedances large two-wofinding power transformers. This table for estimating transformers where different methods of cool i ng may be under consideration. Note that a range of impedances is specified shoul for each voltaken tage from class. the lower the transformer isrange self specified. cooled (OA),Forced the impedance d be end of the cooling allows a transformer of a given size to dissipate heat faster and operate at If
Chapter 7
234 Table 7.1.
Distribution Transformer Impedances, Standard
Reactances and Impedances for Ratings 500 kVA and below (for 60 Hz transformers) Rated-Voltage Class in kV
2.5
Phase
Single-
kVA
Ratmg*
15
Average Reactance
Average Impedance
%
Average Reactance
%
%
3 10 25 50 100 500
1.1 1.5 2.0 2.1 3.1 4.7
2.2 2.2 2.5 24 3.3
0.8 1.3 1.7 2.1 29
.
.
4.8
4.9
25
Average Imped-
ance % 2.8 2.4 2.3 2.5 3.2 5.0
69
Average Impedance
Average
Average
Reactance
Impedance
Average Reactance
%
%
%
%
4.4 4.8 4.9 5.0 5.1
5.2 5.2 5.2 5.2 5.2
6.3 6.3
6.5 6.5 6.5
6.4
Source: Westinghouse Electric Corp. [14]. Used with permission. *For three-phase transformers use 1/3 of the three-phase kVA rating, and enter table with rated line·to-line voltages.
a greater voltampere loading and therefore should be represented by a greater impedance than a similar rating in a self cooled unit. Resistance is usually neglected in power transformers. 7.3
Transformer Polarity and Terminal Markings The terminals of a single-phase transformer manufactured in the United
States are marked according to specifications published by the American National Standards Institute [52], often called ASA Standards. These ASA Standards specify that the highest voltage winding be designated HV or H and that numbered subscripts be used to identify the terminals, e.g. , HI and H2 • The low-voltage winding is designated
LV
or
X
and is subscripted in a similar way. If there are
Z
more than two windings, the others are designated Y and with appropriate subscripts. This same marking scheme for windings permits the identification of taps in a winding. Thus in a given winding the subscripts 1, 2, . . , n may be used to identify all terminals, with one and n marking the full winding and the intermediate
3,
3,
numbers 2,
.
.
.
, n
-
1 marking the fractional windings or taps. These numbers
are arranged so that when terminal terminal
i,
the
i
.
i+1
is positive (or negative) with respect to
is also positive (negative) with respect to i
-
1. The specifications
further require that if HI and Xl are tied together and the H winding is energized, the voltage between the highest numbered H winding and the highest numbered
X
winding shall be less than the voltage across the H winding.
The standards also specify the relative location of the numbered terminals on
the transformer tank or enclosure. The two possibilities are shown in Figure 7.2. Examples of transformers with tapped windings are shown in [ 14] and [52]. We have avoided use of the terms "primary" and "secondary" here since these terms refer to the direction of energy flow and this is not specified when deriving gen eral results. Figure 7.2 also illustrates what is meant by the terms "additive" and "sub-
235
Seq uence I mped a n ce of Transfo rme rs
Table 7.2. Standard Range in Impedances for Two-Winding Power Transformers Rated at 65 C Rise (Both 25- and 60-Hz transformers) Impedance L imit in Percen t HighVoltage Winding Insulation Class kV
15 25 34.5 46 69 92 115 138 161 196 230
L o wVoltage Winding Insulation Class kV
16 15 15 25 25 34.5 34. 5 46 34.5 69 34.5 69 92 34.5 69 115 46 92 1 38 46 92 161 46 92 161
Class OA OW OA/FA * OA/FA/FOA *
Class FOA FO W
Min
Max
Min
Max
4.5 5.5 6.0 6.5 6.5 7.0 7.0 8.0 7.5 8.5 8.0 9.0 10.0 8.5 9.5 10.5 9.5 10.5 11.5 10.0 11.5 12.5 11.0 12.5 14.0
7.0 8.0 8.0 9.0 9.0 10.0 10.0 11.0 10.5 12.5 12.0 14.0 15.0 13.0 15.0 17.0 15.0 16.0 18.0 15.0 17.0 19.0 16.0 18.0 20.0
6.75 8.25 9.0 9.75 9.75 10.5 10.5 12.0 1 1.25 12.75 12.0 13.5 15.0 12.75 14.25 15.75 13.5 15.75 17.25 15 .0 17 .25 18.75 16.5 18.75 21.0
10.5 12.0 12.0 13.5 13.5 15.0 15.0 16.5 15.75 18.75 18.0 21.0 23.25 19.5 22.5 25.5 21.0 24.0 27.0 22.5 25. 5 28.5 24.0 27. 0 30.0
Source : Westinghouse Electric Corp. [ 1 4 ] . Used with permission. *The impedances are expressed in percent on the self-cooled rating of OA/FA and OA/FA/FOA. Definition of transformer classes : OA-Oil-immersed, self-cooled OW-Oil-immersed , water-cooled. OA/F A-Oil-immersed, self-cooled/forced-air-cooled. OA/FA/FOA-Oil-immersed, self-cooled/forced-air-cooled/forced-oil-cooled. FOA-Oil-immersed, forced-oil-cooled with forced-air cooler. FOW-Oil-immersed, forced-oil-cooled with water cooler. Note : The through impedance of a two-winding autotransformer can be estimated knowing rated circuit voltages, by multiplying impedance obtained from this table by the factor (HV - L V/HV).
tractive" polarity. The lower of the three sets of drawings shows a core segment with voltage polarities and terminals labeled. Obviously there are two ways to orient two windings on a core and these possibilities are shown in Figures 7.2a and 7.2b. If Ix is increasing in the circuit on the left, this would cause a flux to tend to increase in the upward direction. But by Lenz's law a current IH will flow to oppose this change in core flux and hold the flux linkages constant. This establishes H I as a positive terminal. Since the HI and X bushings have the 1
236
C hapte r SUBTRAC TIVE
ADDITI VE
� XI
7
X2
Fig. 7 . 2. Standard polarity markings for two-winding transformers : (a) subtractive, (b ) addi tive. (From Westinghouse Electric Corp . [ 1 4 ] . Used with permission. )
same relative position on the tank, as noted on the left-hand set of drawings, this is by definition a "subtractive polarity." Thus for the transformer on the left the polarity may be determined by connecting adjacent terminals together (viz., HI and X I ), applying a voltage between H I and Hz , and then checking that the volt age between adjacent connections Hz and Xz is less than the applied voltage. If so, the polarity is subtractive. Obviously, the reverse is true on additive polarity transformers. Additive polarity is standard for single-phase transformers of 500 kVA and smaller when the H winding is rated 8660 volts and below [14] . All other transformers are normallyV subtractive polarity, although the nameplate should be checked before connecting any transformers in a three-phase bank. From the viewpoint of circuit analysis, we often identify the coils of a coupled circuit by polarity markings or dots. By this convention we easily estab lish that if HI is a dotted terminal, X is likewise dotted shown in Figure 7 .3. This fact follows from the definition ofI additive polarity where we establish that the instant H I is positive with respect to Hz , then XI is likewise positive with respect to Xz • If we choose IH as entering the dotted terminal (HI ) and Ix as leaving the dotted terminal (XI ), these currents are in phase if exciting current is negligible. Equations (7.2) to (7.10) are written on this basis. as
IH
-
+t
•
I X
HI
XI
H2
X2
VH
-I Fig. 7 . 3. 7.4
Dot convention for a two-winding transformer.
Three-Winding Transformers
is quite common in power systems to utilize transformers with more than two windings.z This is especially true in large transmission substations where voltIt
z This analysis follows closely that of [ 1 4 ] to which the interested reader is referred for additional information.
237
S eq u e n ce I mpeda n ce of T ransfo rmers
ages are transformed from high-voltage transmission levels to intermediate sub transmission levels. In such cases a third voltage level is often established for local distribution, for application of power factor correcting capacitors or reactors, or perhaps simply to establish a connection to provide a path for zero sequence currents. Although such transmission substations are usually equipped with three phase transformers, the theory of the three-winding transformer is more easily understood by examining a single-phase unit. These results may later be used on a per phase basis in the study of three-phase applications. The windings of a three-winding transformer are designated as H for the high est voltage, X for the intermediate voltage, and for the lowest voltage. We also assume that exciting currents are negligible and will use equivalent circuits similar to Figure 7.1b and 7.1c, where the excitation branch is omitted. Consider the winding diagram of a three-winding transformer shown in Figure 7.4a where the windings H, X, and Y are shown to have turns nH , nx , and l!.
Y
Z HI H ) H 1 l·.... -
+
' n ri
Z Y I HI
VH
- !�----<.
.''[]:
(b)
p ul H
"2=
nH
nx "H ny
pu1 x
pu Zx (0)
nl=
p u Vx +
PU Z H
+
P U VH
puzy
puly +
p u Vy
(e) Fig. 7 . 4 .
Three-winding transformer : ( c ) equivalent circu it in pu. permission. )
( a ) winding d iagram , ( b ) equivalent circuit in ohms, ( From Westi n ghouse Electric Corp. [ 1 4 ] . Used with
ny respectively. In testing such a transformer to determine its impedance we
make short circuit tests between pairs of windings. This is done with the third winding open. Applying such a test between the H and
X (7.4)
gives the ohmic impedance ZHX ' Following the procedure of tion on the H winding and the
or,
in
X
H
with instrumenta
winding shorted, we have
ZH X = ZH + n� Zx pu based o n the
windings with Y open
H
(7.13)
winding voltage and voltampere capacity w e write
(7. 14)
238
Chapte r 7
where the basis is indicated by the notation (H). We shall refer all impedances arbitrarily to the H winding where SHB = VA rating of winding H (7.15) VHB = rated voltage of winding Then (7.16) ZHB = VkB /SHB and if we multiply (7.14) by ZHB , we have on H base (7.17) ZHX(H) ZH(H) + ZX(H) where ZH(H) is the same ZH as in (7.13) but (7.18) ZX( H) = n r Zx where n l = nH/nx (7.19) To complete the equivalent circuit of Figure 7 Ab, we also define n 2 = nH/n y (7.20) and (7.21) such that (7.22) on H base ZH Y(H) = ZH(H) + ZY(H) A third measurement may be made from winding X to winding Y, with H open, to determine Zxy . We compute (7.23) where (7.24) Rewriting (7.23), we have H
11
=
11
11
11
ZXY( X)
= n� Zx n+21 n� Zy
ZX(H) + Z Y(H) 11 nr
on X base (7.25) We have now established three equations in three unknowns-viz., (7.17), (7.22), and (7.25)-which we rewrite as =
r�:::::J rL�� 11n�� OJ [ ]
L�xY(X)
=
1
ZH( H) ZX(H)
n
on mixed X and H base
(7.26) Note that the column vector on the left is known or can be determined by tests on the transformer. The column vector on the right contains the desired quanti1 1ni
ZY(H)
239
S eq ue n ce I mped a n ce of T ra n sfo r m e rs
ties for the equivalent circuit. Since the determinant of the coefficient matrix is nonzero (det = 2 /nn, a unique solution exists,
[�:::�J [� -
Z Y(H
= 1 /2
-1
-
� :� r:;::� � � ;J -
1
n
XY(X
n
on mixed X and H base
(7.27)
These values are inserted into the equivalent circuit of Figure 7 Ab . We convert the equivalent circuit to a pu equivalent by dividing (7 .27 ) by the base ohms of the H winding. We also note that n�
= (nH/nx )2 = ( VHB/ VXB)2 = SHB ZHB /SXB ZXB
[�:� [ � -� -�l � ::::j)
(7 .28 )
Incorporating ( 7 .28 ) , we divide (7 .27 ) by ZHB and find the pu expression
Zy
= 1 /2
-1
1
1
.....
where
ZX Y (H)
pu (7 .29 )
ZXY(H ) = (SHB /SXB ) ZXY(X)
These values are used in the pu (or percent) equivalent circuit of Figure 7 .4c. Note carefully that these values are computed from pu winding-to-winding im pedances where ZHX and ZHY are on an H base and ZXY is on an X base (both voltampere and voltage ) . The above procedure i s necessary in transformers with more than one winding since the pu impedances must be carefully identified as to the base used in specifying each value. The values of equation (7 .29) could be based upon the rating of any winding, but the H winding is often selected. It is important to realize that when fully loaded the H winding divides its capacity between X and Y. It is reasonable then that the ratio SH B /SX B affects the impedance ZX Y which in turn affects the equivalent circuit. Note also that the effect of SH U /SX B on ZX Y(X) is to convert this quantity to an H-based pu impedance. Observe that the "T equivalent" of Figure 7 .4 has a node at the junction of ZH , Zx , and Z y which is fictitious and has no physical significance whatever. The voltage at this point is not usually computed. This node can be eliminated by changing the T to a ..1 . Equation ( 7 . 29 ) contains negative signs and the com puted results are not necessarily positive. Indeed , it is rather common for one leg of a T equivalent to be negative or zero. Equivalent circuits are developed in the literature (e.g. , see [ 1 1 , vol. 2 ] and [14] ) for transformers with four or more circuits. Since these devices are not in common use , they will not be studied here. 7.5 Autotransformer E qu ivalents
In the modern power system a large number of major transformations are ac complished by autotransformers rather than separate-winding transformers.3 One reason for this is the lower cost of autotransformers. There are, however, other 3 This development follows closely that of [ 1 4 ] but [ 1 0 ] and [ 1 1 . vol. 2 ] are also recom mended for further reading.
2 40
C hapte r
7
significant differences [14] . The impedance of an autotransformer is smaller than that of a conventional transformer of the same rating. This is both an advantage and a disadvantage. The lower impedance means lower regulation, but it also means lower fault limiting capability and sometimes necessitates the inclusion of external impedance. Autotransformers also have lower losses and smaller exciting currents than two-winding transformers of the same rating. They are physically smaller and generally have higher efficiencies. Consider the autotransformer circuit of Figure 7.5 where we denote the ratio
Fig. 7 . 5 . Circuit for a two-winding autotransformer.
of H to X voltages by the turns ratio n , i.e .,
(7.30) Then, according to this definition n2/n l = n - 1, and the ratio of turns in the two windings is not equal to the no-load voltage ratio n . When the transformer is loaded, the MMF of the two windings must be the same (if ideal), i.e., (7.31 ) and since VH /Vx
=
( n l + n 2)/n ,
=
n
(7 .32)
we compute
(:: 1) =
(7.33) Comparing (7.30) and (7.33) with similar equations (7.2 ) and (7.3) for a two winding transformer, we see that the autotransformer looks exactly like a two winding transformer with turns ratio n where n is defined by (7.30). From transformer theory we know that the total circuit voltamperes in a transformer must be the same in both the input and output (for an ideal trans former). For the autotransformer we compute, using subscript c for "circuit" voltamperes, (7.34) Se = VH IH = Vx 1x but the winding voltamperes are S, Vx Il Vx IH (n2/n ,) for winding 1 S2 ( VH - Vx ) IH = Vx IH (n2/n ,) S, for winding 2 (7.35) The ratio of winding to circuit voltamperes is (7.36) SdSe n2 /( n , + n2 ) = ( n - l )/n Ix
=
=
+
IH
nIH
=
=
=
=
24 1
S eq ue n ce I mped a n ce of Tra n sforme rs
n
This amounts to a considerable "savings" in winding capacity . For example, if the transformation is two to one, i.e . , = 2 , the ratio (7 .36 ) is one-half, or the autotransformer windings have twice the voltampere capacity of the equivalent
two-winding transformer. The autotransformer impedance may be measured, as in the case of the two winding transformer, by making a short circuit test. The circuit for performing this test is shown in Figure 7 .6 where we compute the leakage impedance H
IH
--
�p: VH
•
n2
IX
-
•
nl
XI
HO-+--..J----..... X o
Fig. 7 .6.
Short circuit test circuit.
(7 .37 )
two-winding
This is the impedance shown in the equivalent of the autotransformer of Figure 7 .7 where we depict the autotransformer as a transformer with turns ratio and impedance ZHX .
n
n = t + ..!!!. "I
Fig. 7 . 7 . Conventional equivalent circuit for an autotransformer. 7.5. 1
Three·winding autotransformers
Three-winding autotransformers are also widely used . Because of the physical arrangement of autotransformer circuits, three transformer windings are usually connected Y -Y with grounded neutral. Since the Y Y c onnection contains no path for third-harmonic exciting currents, a third winding is provided in each phase , which can be Ll connected . This winding, called the Ll -tertiary , provides a path for exciting currents and can be used to transform power as well. In three-winding autotransformers it is necessary to recognize clearly whether one is computing impedances on a basis or on a basis. Care must also be used in applying formulas given in the literature si nce there cines not ap pear to be a commonly accepted practice (e.g . , [ 14 ] uses a circuit basis whereas [ 1 1 ] uses a winding basis ) . Wagner and Evans [ 1 0 ] give an excellent treatment of the comparison be tween the impedances of a three-winding autotransformer computed on a circuit basis and a winding basis . Three-winding autotransformers usually consist of two physical windings , one of which is tapped to form two subwindings which are separately identified . In the following analysis we shall call the and t (for series, common , and tertiary ) and the H, X, and Y (sometimes called
-
circuit
winding
circuits
windings s, c,
C hapter 7
242
M,
H, L or H, L , T by other authors) . These circuit and winding configurations are identified in Figure where the three-terminal equivalent circuits are also
7.8
WI N D I NG BAS I S
C I RC U I T BAS I S
( b)
(a )
\I �
'" Ia(equiv I " nla(aclual l
Ia ( equlv ) "In-tl I a ,o c t ua l l I • Zc
5
c ';(
I c C equlv) 0 I c ( aclua I L/
t
I b ( a c lua l )
Z s o l a lacl u o l )
H
•
Z + l c C a C l u o l ) Zx I I x�,
Ic ( a c l u a l )
o I bCoc lual ) .j. l a C a e l u a l ) y Z y
ZH
I b l e q",'v l ' l b ( a� luall
I e)
(d)
Fig. 7 .8. Autotransformer connections and equivalents for analysis o n a winding or circuit basis. ( From Sy mme trical Components by C.. F. Wagner and R. D. Evans. Copyright McGraw-Hill, 1 9 3 3 . Used with permission of McGraw-Hill Book Co. )
shown for each case . These are the equivalent circuits for the positive and nega tive sequence networks . The zero sequence equivalent depends upon the nature of the connection, as discussed in section Note, however, that since three circuits to which external connections are to be made a.re present in either case (a )
7.6.
(a)
Table 7.3. Conversion Formulas for Autotransformer with Tertiary Winding* Circuit Basis Winding Basis
Zt Zc
=
=
1/2 (Zt c + Zt s - Zcs) 1/2 (Zc t + Zcs - Zs t )
Zs = 1/2 (Zs t + Zsc - Zc t ) (c)
-( f
ZXY = Zc t n- I ZHX n-
Zc =
-n
n -
1
Zsc
=
-
n n- I
--
(-y --- - ---
= ZX Y
Zsc = __ n 1
Zts =
n
I
n
n
ZHX
ZHY
1
n
1
ZXY +
n Z (n - 1) 2 HX
1
(f) Zx = -- Zc
Zx
n
Z
Z
( r
=
n
c Z y = Zt + n
x Zt = Z y - -n- I
Zs
Z y = 1/2 (Zyx + Z YH - ZXH ) Zx 1/2 (ZXY + ZXH - Z YH ) ZH = 1/2 (ZHY + ZHX - Z yx )
(d ) Zc t
n- l n- I Z ZHY = .-l£ + -- Zts -2- Zcs n n n (e)
(b )
ZH +
n Zx (n - 1 ) 2
( r --
n- I ZH = -n
n- 1 ZS - n 2 Zc
Source : Wagner and Evans [ 1 0 ) . Used with permission. *c-winding or X circuit used as voltage base. For pu use arbitrary
VA
base. n
=
VH IVx
243
S eq ue n ce I mped a n ce of T ra nsformers
or (b) of Figure 7 .8, the equivalent circuit must have three terminals as in (c) and (d) . Note also that in Figure 7.8 we use the ratio
n = VH,/ Vx
(7.38)
which agrees with (7.30) for the two-winding autotransformers. Table 7.3 gives the equations which convert from one basis to another with all values based on either the c winding or the circuit voltages, and where all im pedances are in pu on these bases, with an arbitrary voltampere base. Then equations (a) and (b) are the same as (7.29) and the various conversion equations of (c), (d), (e) , and (f) are all in pu on these same bases. The only exception to this is that the current relationships given in the Figure 7.8 equiva lent circuits (c) and (d) are ampere relations . The equations shown are developed in [10] . Table 7.4 gives typical values of autotransformer impedances.
X
Table 7.4. Typical Characteristics of Autotransformers
Losses! Percent on Rated k VA Baset
Reactances * t Low to High to Recom- tertiary tertiary Range mended average High to lo w
Ratings
No load losses Range
Total losses
Recommended average
Range
Recommended average
0. 09 0. 08 0. 07 0. 09 0. 08 0.08 0. 07
0. 25-0. 40 O.21}-0. 35 0. 1 5-0. 30 0. 30-0. 50 0.25-0.45 0. 20-0.40 0. 15-0. 35
0. 33 0. 30 0. 26 0. 38 0. 35 0. 30 0. 26
Three-phase:
300-600 MVA
Three single-phase:
601}- 1 200 MV A 345-138 kV 345-161 kV 345-230 kV 501}- 138 kV 500- 161 kV 500-230 kV 500-345 kV
1 1}- 1 3 8- 1 1 5-9 15-20 1 2- 1 8 1 1}- 1 3 6- 1 0
12 10 7 17 15 12 8
20-30 18-28 15-25 13- 2 1 10-20
25 23 20 17 15
1 1-19 20-25 5-15 7-18 15-25 23-25
24-34 0.07-0. 10 1 0-20 0.06-0. 10 0.06-0.10 21}-30 0.07-0.1 1 20-30 0.06-0.10 20-30 0.06-0 . 10 30-40 0.06-0. 10
Three single-phase: 300- 1 200 MV A 7 00-138 kV 7 00-161 kV 7 00-230 kV 7 01}-345 kV 7 00-500 kV
0. 1 0 0. 09 0. 09 0.08 0. 07
0. 55 0.48 0. 37 0. 28 0. 22
Source : Federal Power Commission [ 53 ] . '" All reactances in percent on rated voltage and kV A base. tThe above transformer reactances and loss data are based on 1 050 kV BIL for 345 kV transformer windings, 1 5 5 0 kV BIL for 500 kV transformer windings, and 2 1 7 5 kV BIL for 7 00 kV transformer windings. Increasing or decreasing the BIL values results in a corresponding increase or decrease in transformer reactances and losses of up to approximately 10% of the suggested values. 7.6
Three-Phase Banks of Single-Phase Units
Since our concern is primarily with three-phase power systems, we are natu rally interested in the application of single-phase transformers connected to form three-phase banks.4 In this study we will limit our consideration to three iden-
4
This discussion follows closely that of Clarke [ 1 1 , vol. 2 ] to which the interested reader referred fOf further information.
IS
244
Chapte r
7
tical single-phase units and will develop the sequence network representations for commonly used connections. The use of dissimilar units introduces a series un balance which may be treated separately as such once the basic sequence network configurations are known. One problem in the computation of transformer impedances is the selection of the base for converting ohmic values to pu. For transformers it is common to choose the voltampere rating of a single-phase unit as the base or, in the case of three-winding transformers, we refer all impedances to the voltampere rating of the one winding. If the windings are Y connected, the LN voltage is taken as the base voltage, and this is the winding rated voltage. If windings are A connected the LN voltage is 1 /...(3 the winding rated voltage, and this should be computed as the base value. But as noted in ( 1 .16) and (1.20), one may choose the voltampere base as the total bank rating, or three times the rating of each unit, and use the LL voltage as base voltage. Note that once a voltage base is selected for one winding, the basevis automatically fixed, through the turns ratio, for the other windings. If these alues do not coincide with the arbitrarily chosen base voltages, the off nominal ratio must be accounted for. This may be done by inserting an ideal transformer with a turns ratio equal to the off-nominal ratio of these voltages. Another method for doing this will be given in Part III of this chapter. It should be emphasized here that all A -connected transformer banks will be represented as an equivalent (ungrounded) Y insofar as the positive and negative sequence networks are concerned. For the zero sequence network, however, we must also recognize that the A connection permits zero sequence currents to flow, whereas theY ungrounded Y does not. This concept of thinking in terms of an equivalent is quite acceptable insofar as the relationship between phase quanti ties is concerned. Thus the total current entering the bank is correct and phase relationships are correct among quantities on either side of the transformer. If we wish to compare the phase relationship of quantities across a Y-A bank, this must be done as a separate computation since the equivalent Y representati on destroys this relationship. Using this equivalent Y idea, we see that the transformer equivalent of Figure
7.1c is the (per phase) equivalent for the positive and negative sequence networks for any transformer connection of two-winding transformers. For three-winding transformers the T of Figure Ac is adequate. Thus with the base quanti discussed the positive and negative r en a i ties chosen are identical and are the same as the s n e phase transformer The zero equivalent circuit depends upon t n ec n and the method of grounding. In a -connected winding, can flow, but n each leg of the the trans former terminals. Thus the in ng as viewed from the as an mpedan e with both sh rte to ground leO ternal circuit. In grounded Y circuit the three currents laO ' lb O , produce 31ao flowing to (or from ) ground. Since the zero sequence network represents only one phase with current la O ' we include any impedance Zn as 3Zn in the sequence network so that the neutral voltage be computed cor rectly. The ungrounded appears open to zero sequence currents the three phase currents add to zero, from which we conclude that laO zero. These concepts are illustrated in the zero sequence equivalent circuits of two and three-winding transformers shown in Figure where the letters P, Q, and R
circuit 7 above, i gl sequence A since they are equal i Aendsw dio d c a
as
i
Y
-
sequence rep es t t ons equivalent. he type of co n tio sequence currents A, they zero do not leave external circuit appears and an open circuit to the ex and add to neutral will since is 7.9
245
S eq ue nce I mped a n ce of Transforme rs TRAN SFORMER C O N NECTION
(a)
P
( b)
P
(e )
P
(d ) (e )
(f)
(0 ) (h)
P
Z ERO SEQUENCE CIRCUIT EQUIVA L E NT
G:}' �
� BJ
p
_ � Z PQ
Q
E? �, 8J �
P _ ..,.......,.. .... Q Z PO
P ... � Q
o
ZP O
....
O I:}? � O � O Ptif n?l � ::P tr IT?t � � P�IT?i� p
P � ZP Zo Zp P "" Q
Q
R ...
p",
p
o
Zp
R ...
P
o
Zp
R
R
Fig. 7 .9.
_ O
...
Z
R
Z
R
ZR
ZQ
ZO
Q
Q
Zero sequence equivalent circuits for transformer banks of three identical single phase units_xciting currents neglected. (From Clarke [ 1 1 , vol. 1 ] . Used with permission. )
designate the transformer circuits. Once the voltage levels are specified, these designations may be replaced by H, and for the appropriate windings. In this figure the ground symbols represent the zero potential bus No . Two additional special cases of zero sequence equivalent circuits are shown in Figure 7.10 where a neutral impedance is present in (a) and a corner of the A im-
X,
Y
Df. : � � 3 ' P
:
Q
N
P
Zt
Zn
I
,
(a)
,0
Z ERO POTE N T I A L BUS
P
'Zt -
( b)
Fig. 7 . 1 0 .
V-A
Q
N
Zn
Z ERO
Z
;
[
(e)
0
N
'
POTEN T IAL BUS
(d)
A,
bank with (a) neutral grounded through Z n ' (b ) Z n in a corner of the (c) and ( d ) zero sequence equivalent circuits for (a) and (b ) respectively , where Z t leakage impedance between windings. (From Clarke [ 1 1 , vol. 2 ] . Used with permission . ) =
246
Chapte r 7
pedance exists in (b). The equivalent circuits are shown in Figure 7.10c and 7.10d respectively for these special cases. It is assumed that the neutral impedance Zn in (a) and (c) is expressed in pu on the system base voltamperes per phase and voltage of the Y-connected windings (the same as the transformer impedance Z t ). The corner of the A impedance is assumed to be expressed in pu on the system base voltamperes per phase and the base LL voltage of the circuit. It is also possible to form three-phase banks from single-phase, two- or three winding autotransformers. Such banks are commonly Y connected, often with the neutral grounded, and with a third (tertiary) winding connected in A to supply third-harmonic MMF for excitation. Such a transformer connection is shown in Figure 7.11a where the circuits are designated H, X, and Y. Clarke [11, vol. 2] LN
A
H
IX
Iy
Z lCO
ZH
ZlC lC
l
Zy y
1H
H
IT '
-IH
zzoI-=
H
ZERO POTENTIAL BUS
(c )
(b)
(0 )
EX
Z HO
Fig. 7 .1 1 . Equivalent circuit for a Y- A-Y connection of three identical three-winding auto transformers : (a) autotransformer connection, ( b ) positive and negative sequence equivalent, (c) zero sequence equivalent. (From Clarke [ 1 1 , vol. 2 ] . Used with permission. )
shows that the equivalent circuit for the positive and negative sequence networks is that shown in Figure 7.11b where the T circuit parameters are taken from Table 7 .3(b) to be 1
-1 1
( 7 .39 )
which is seen to be exactly the same as ( 7.29). Here all impedances are in pu on the same base. The zero sequence impedances for Figure 7 . 1 1 c are also given by Clarke [11, vol. 2 ] to be 1 - 1 1 (n - l)/n ZHX ZXO ZH O = 1/2 1 1 - 1 (n - 1)/n2 ZHY pu - 1 1 1 lin ZXY ZnO
[
J
[
-
]
6Zn (7.40) where n is defined by equation (7.30) to be n = (nl + n 2 )/n l ' The proof of ( 7 40) is left as an exercise (see problem 7 . 1 3). If the tertiary winding is missing, ZnO = 00 and the impedance from H to X is the sum of Zxo and ZHO ' In the case of an ungrounded bank, Zn = 00 and all the impedances in ( 7.40) go to infinity. One way to solve thisA problem is to convert the Y equivalent (con sisting of Zxo , ZHO, and ZnO) to a equivalent and then take the limit as Zn O ap proaches infinity. If we call these" A impedances ZHX A , ZHYA, and ZX Y A , we .
compute the so-called "resonant equivalent to be A
[ZHXZHYtJ.tJ.] r ] ZXYtJ.
S eq ue n ce I mped a n ce of Tra n sforme rs
where
Zst
=.
247
n - 1 )2 In - (n - 1 ) Zst pU (n - l )/n
( 7.41 )
is the series-tertiary impedance defined (winding basis) in Figure 7.8. I I . TH R E E-PHASE TRANSFO R M E R S
Much of the information needed to perform studies of a faulted system and to properly represent the transformers in such studies is given in the preceding sec tions. It is often of little concern whether the transformer bank consists of three single-phase units or one three-phase unit. There are, however, certain differences which may be important under some conditions. Among these are the standard practice of marking the terminals of three-phase units, the impedance differences due to different core designs, and the phase shift across transformer banks. 7.7
Three-Phase Transformer Terminal Markings
The terminals of three-phase transformers are labeled H, X, Y, etc. , exactly as for the single-phase transformers discussed in section 7.3. In this case, however, Group 3 Autotransformers
22 X4 x 2
MI OTAP BROUGHT OUT OF ONE PHASE ON THE LOW·VOLTAGE SIDE
MIDTAP BROUGHT OUl OF ONE PHASE ON THE LOW·VOlTAGE SIDE
Fig. 7 . 1 2 .
Angular displacement and terminal markings for three-phase transformers. Dashed lines show angular displacement between high- and low-voltage windings. Angular displacement is the angle between a line drawn from neutral to HI and a line drawn from neutral to XI ' ( Courtesy USAS Institute [ 5 2 ] . )
248
Chapte r 7
each voltage level will have three terminals, one for each phase, and may have four if the neutral is brought out as an external connection . These are labeled by sub scripts 1, 2, and 3, with 0 designating the neutral. The subscripts are intended to designate the time order of the instantaneous voltages if connected in a logical manner. Thus if the ph ase sequence is a-b-c and these phases are connected to H1 , Hl , and H3 • then Xl > Xl , and X3 will be of phase sequence a-b-c. The USAS r 52] also specify a definite phase relationship between windings of a three-phase transformer. For Y-Y and A -A connections the phase angle between like-numbered terminals is 0° , whereas for V-A or A -Y connections the H-terminal voltage leads the corresponding X-terminal voltage by 300 . This is shown in Figure 7 . 1 2 which also shows the USAS arrangement for terminal positions on the transformer tank . From Figure 7 . 1 2 we observe that the voltage associated with phase a on one side of the transformer may have many different phase relations with the phase a voltage on the other side, depending on the way the phases are labeled. Thus, for Y - Y or A -A transformers this phase relationship may be 00 , +1 200 , or 1 200 • In Y -A or 06 -Y connections, the relation between phase a windings may be ± 30° , ± 1 50° , or ± 90° , again depending on labeling. The transformer itself, if of standard design and marking, is consistent in the relationships shown in Figure 7 . 1 2. -
7.8
Phase Shift in
V-A Transformers
In computing unbalanced currents, the A -A or Y - Y transformer presents no special problem since the currents are transformed as mirror images . Thus, a fault between phases b and c on one side of the transformer appears as a similar pair of currents on the other side and will also flow in lines b and c if so labeled (and we always assume the lines are similarly labeled ). In a V-A or A -Y transformation , however, this is not the case . Here, for ex ample , a SLG fault on phase a on the (lUounded ) Y side appears as a current flow in both lines B and C on the A side. Obviously , this requires special treatment . The equivalent circuits for positive , negative, and zero sequence currents of a Y -A transformer bank are shown in Figure 7 . 1 3 , where we note there is a trans formation in both voltage magnitude and phase . ( Figure 7 . 1 3 shows the A side leading the Y side by 300 , which would be the case if the A winding were the H winding . ) Usually , however. we do not take this phase shift into account in our per phase equivalent circuit because we are not usually interested in the phase re lationships across the transformer but need to examine only the phase relation ships between voltages and currents on one side or the other. Indeed, we would not learn anything of value by shifting these quantities in phase by ± 30° . We therefore reduce the A winding to an equivalent Y winding with neutral un grounded. This permits us to change the positive and negative sequence equivalent circuits to a simple series impedance . However, we account for the fact that zero sequence currents can circulate in the A winding by retaining the zero sequence representation of Figure 7 . 1 3d. Although w e find i t convenient t o ignore the phase shift i n a V-A transforma tion for most problems, we need to establish a procedure for calculating the phase relationship of all voltages and currents when required to do so . As mentioned previously , the actual phase relationship depends upon the labeling of the three phases and may be ± 30° , ± 1 50° , or ±90° as shown in Figure 7 .14. However, there is no need to exhaustively analyze all six arrangements shown in Figure 7 .14.
LL
249
S eq ue n ce I mped a n ce of T ra nsformers
00------. E 09
� f EAG
.--___--=�
t
?
?
e E el d
'iI"
"
.:,' !� ., !1�
E02
:.:t
"2 . -"3 n l
(0)
Z PO
'III/'v
I bl
Ie )
I
(d)
IC
j 30
rtJ�,
C
t E CG
..
- j 30
.
Z PO 'III/'v
Z PO 'III/'v
A
fb: o
0
:Fo
Fig. 7 . 1 3 . Equivalent circuits o f a Y-fl transformer bank : (a) schematic diagram, (b ) positive sequence, (c ) negative sequence, ( d ) zero sequence. Corp. [ 1 4 ] . Used with p erm issi o n . )
(From Westinghouse Electric
Indeed, if we analyze only one, we may make any desired computation and can then relabel the results to fit any of the six possible configurations. The problem is choosing arbitrarily the most convenient of the six phase-shift arrangements. Stevenson [ 9, 21 ] has analyzed the 90° phase shift. This is very convenient to work with and will be chosen as the basis for all computations here. The basic arr angement is shown in Figure 7 .1 5 where several important features should be noticed. First, note that the phase labeled a is connected to HI and that labeled A is connected to X3 • Since by definition the positive sequence voltage is a-b-c, once we choose to connect b to H2 , we must connect B and C to Xl and X2 re spectively in order that the phase sequence be a-b-c (A -B-C) on each side of the transformation. Comparing with Figure 7 .14, we see that this is a 90° connection (or labeling) of terminals. This is verified in the positive sequence phasor diagram in Figure 7 .1 5b where we see that Va l lags VA l by 90° . Figure 7 .1 5 also follows the convention that windings drawn in parallel (e.g., bn and CA ) are linked mag netically and that labels are arranged such that a-n on the Y side is linked to B-C on the fl side . Lowercase letters are used on one side, uppercase on the other.
C hapte r 7
2 50
c
B
b
A
o
a
B A
c
c
a
B
b
C
A
E a g LEADS EA G BY 0
b
B
C
c
A
B C
c
a
Eao LEADS EAG BY � a
X3
C
B
c
b
A
Fig. 7 . 14. Angular phase displacements obtainable with three·phase y-� transformer units. ( From Westinghouse Electric Corp. [ 1 4 ] . Used with permission. )
Our notation is such that Va l is in phase with VBC I This being the case, these coils are linked magnetically so that a current en tering a would produce a current IBC which would be leaving B, as in Figure 7.3 (where IH = Ia , lx = - IBC )' Thus Ian and IBC are 1800 out of phase, as shown in Figure 7.16, and this is true for both the positive and negative sequences. From the voltage and current relation ships of Figure 7.15 we write '
VA l
and
= +jVa l '
b
c
-
HI
H
IA
Ie
IC
(a )
01 el
B
)( 3
H3
bl
(7. 4 3 )
XI
10
Ib
Vbl Vel
c2
CI
BI
( 7 . 42 )
= -j Va2 pu
IA 2 = - j Ia 2 pu
IA t = +jIa t • a
VA2
02 AI
�2
C
� A2
VA
Vbc2 "'b2
b2
A
B2
VB2
VBC2
2
Ve 2
C2
NEGATIVE SEQUENCE
POSITIVE SEQUENCE
( b )
Fig. 7 . 1 5. Wiring diagram and voltage phasors for a three-phase transformer connected y-�: (a) wiring diagram, ( b ) voltage components. ( From Elements of Power System Analysis by William D. Stevenson, Jr. Copyright McGraw-Hill, 1962. Used with permission of McGraw-Hill Book Co. )
� ' :,
251
S e q u e n ce I mped a n ce of T ra n sfo rmers
I.O'
I O'
[ C A2
[A2
Posit ive sequence com ponent s
Fig. 7 .1 6.
Nevot ive sequence components
Current phasors o f a three-phase transformer connected Y- !::. . ( From Elements of Power System A nalysis by William D. Stevenson, Jr. Copyright McGraw-Hill, 1 9 6 2 . Used with permission of McGraw-Hill Book Co . )
Obviously , this tells us nothing about the phase relationship between sequence quantities since this depends entirely on the boundary conditions at the fault.
Example
7. 1
As an example of the use of ( 7 .42 ) and ( 7 .43) we compute the currents flow ing on both sides of transformer T1 in Example 3 . 1 , where we have already estab lished that the total fault current is la = 3la l = 2 . 34/- 42 .8° pu and lb = Ie = O.
Solution
We easily compute from la o = la l = la 2 = 0 .78 /- 42 .8° = 0 . 5 7 2 - j O . 5 30 pu
that IA I = jO.78/- 42 .8° = 0 .78/+47 .2° = 0 .530 + j O . 5 7 2 pu
and TA 2 = - jO.78/- 42 .8° = 0 .7 8 /- 1 3 2 .8° = - 0 . 5 30 - jO .572 pu
and by inspection, lA O = O. Thus TA = IA O + IA I + IA2 = 0 2 IB = lA O + a lA I + alA 2 = 0 + (0.230 - jO.745 ) + (0.760 - j O . 1 7 3 ) = 0 .990 - j O .9 1 8 = 1 . 35/- 42 .8° pu
and Ie = lA O + alA I + a 2 TA 2 = 0 + (- 0.760 + j O . 1 7 3 ) + (- 0 .230 + jO .745 ) = - 0 .990 + j O .9 1 8 = 1 .35/180° - 42 .8° pu
Thus the generator views the SLG fault as a fault in lines B and C and sees no zero sequence current at all. Voltages at the generator bus can be calculated by a simi lar procedure and this is left as an exercise for the interested reader (see problem
7.16).
7 . 9 Zero Sequence I mpedance o f Th ree-Phase Transformers
In analyzing the sequence impedances of three-phase transformer banks con sisting of three single-phase units, we used the same impedance for the zero se quence network as for the positive and negative sequence networks. This is be cause the individual transformers present a given impedance to any applied
C hapte r 7
2 52
voltage. Indeed, the individual transformer is not aware of the sequence of ap plied voltages. This is not the case with three-phase transformers where the fluxes of each phase winding share paths of a common magnetic circuit. Here the device re sponds differently to positive and zero sequence voltages, and these differences may be important in obtaining a correct sequence network representation. If precise information is lacking, these refinements may be neglected (and often are) in system studies. Even if this is the case, however, the intelligent engineer should be able to estimate the possible error in making a simplifying assumption. There are two basic designs commonly used for three-phase transformers-the core-type design and the shell-type design. These core configurations are shown in Figure where fluxes are shown due to positive sequence applied voltages.
7. 1 7
8,
A
IPA
�
ale,
B
.------h
C
.p.
9
Ge,
IPc IPc ' -IPA -
9,
r------t.
.p.
(a) A <>
-
8, ----
/"'\
+0 2
,-..
--t--
'-' '-'
I<>IPA
B 2 a 8, - - -<> .C'L
+. 2"
L"'>..
c
Ge,
<>
-- - -
,.,
<>t --�--.p. -+'-'
v
-
r-.
'-'
+c T
'-'
I<>IPc
(b)
Fig. 7 . 1 7 . Positive sequence fluxes in core-form and shell-form transformers : (a) core-form, ( b ) shell-form. ( From Westinghouse Electric Corp. [ 54 J . Used with permission. )
Note that the core-form windings (only the primary windings are shown) are all wound in the same sense. In the shell-form transformer the center leg is wound in the opposite sense of the other two legs to reduce the flux in the core sections be tween windings. Since nearly all flux is confined to iron paths, the excitation cur rent is low and the shunt excitation branch is usually omitted from the positive and negative sequence transformer equivalent circuits for either core-type or shell type designs.
Seq uence I mped a n ce of T ra n sfo rmers
253
The zero sequence impedance of a three-phase transformer may be found by performing open circuit and short circuit tests with zero sequence voltages ap plied. If this is done, the short circuit test determines leakage impedances which are nearly the same as positive sequence impedances, providing the test does not saturate the core. The open circuit test, however, reveals a substantial difference in the excitation (shunt) branch of the zero sequence equivalent for core-type and shell-type units. This is due to the different flux patterns inherent in the two designs, as shown in Figure 7 . 1 8 where zero sequence voltages are applied to c
IPo �
,----n
I
(al A ·0
IPo
8 I ..
- - - <>
t
r-.
- -
'-"
�
- -
'-"
fo IPo
� - -J '"'
'-"
'"'
\,J
Of
c 0
-
..
- - - - - Of ,..
-1-- f<> IPo '"'
v
\,J
"1 <> - - - •
(bl
Fig. 7 . 1 8 .
Zero sequence fluxes in core-form and shell-form transformers : (a ) core-form, (b) shell-form . (From Westinghouse Electric Corp_ [ 54 ] . Used with permission. )
establish zero sequence fluxes. In the core-type design the flux in the three legs does not add to zero as in the positive sequence case. Instead, the sum 3"'0 must seek a path through the air (or oil) or through the transformer tank, either of which presents a high reluctance . The result is a low zero sequence excitation im pedance, so low that it should not be neglected in the equivalent circuit if high precision is required in computations. The shell-type design may also present a problem if the legs between windings become saturated. Usually, however, the elCcitation impedance of the shell-type design is neglected. The excitation of either the core-type or shell-type design is dependent upon
C hapte r 7
254
the magnitude of the applied zero sequence voltage, as shown in Figure 7 .19, but the shell-type design is much more variable than the core-type due to saturation of the shell-type core by zero sequence fluxes. Figure 7 .1 9a also shows how the transformer tank acts as a flux path for zero sequence fluxes in core-type trans formers and is sometimes treated as a fictitious tl tertiary winding of high impedance [ 1 0 ] .
25
50
75 100
PERCENT Z ERO SE�UENCE VOLTAGE
100
PERCENT ZERO SE�UENCE VOLTlIGE
I
(0 ) Fig. 7 . 1 9 . Typical zero sequence open circuit impedances : (From Clarke [ 1 1 , vol. 2 ) . Used with permission . )
50 501 r-l1ij/trllOOrlt50 100 100 50tl-YfJJJgg�_1!50 50 50 50 50 50 tLJlOOLJfI,7.L-J 100
(b)
(a) core-type, (b ) shell-type.
In any situation where high precision is required in fault computation, the transformer manufacturer should be consulted for exact information on zero sequence impedances. Where precise data is unavailable or high precision is not required (as is often the case), the zero sequence impedance is taken to be equal to the positive sequence impedance and the zero sequence equivalent is taken to be the same as that developed for three single-phase units. This problem of finite excitation impedance is more pronounced in small units than in large three-phase designs. Three-phase distribution transformers, rated below 500 k V A and below 79 kV, are always of core-type design. Many but not all of the large power transformers are of shell-type design. There is also a five-legged core-type design which has an excitation impedance value between the three-legged core-type and the shell-type. This treatment of the subject of excitation of three-phase transformers is cer tainly not exhaustive, and the interested reader is referred to the many references available, particularly [ 1 1 , vol. 2 ] , [ 14 ] , and [ 5 5 ] . Reference [ 2 0 ] gives data for representation of three-phase, three-legged, core-type transformers as shown in Figures 7 .20 and 7 .2 1 , where the notation Q IIN indicates that terminals Q and N would be connected together, or ZPQ l IN is the impedance between P and the par-
S eq ue n ce I mped a n ce of T ra n sformers Z E R O SEQ UE N C E EQUIVALENT CIRCUIT
T R A N S FORMER CONNECTION
I
2
P
3
0
Q
NO
NO
p
0
No
0
NO
P-
p
Z PQ-O " ZPQ-I Z P N-O " 5 Z pQ- 1 � Q N - O = 6 Z rQ - 1
Z P N - O " 5 Z PO - I
N
-
.;:
Z PQ-O " co
0
P -
6
No
QII N
P
5
N
p -
P
4
APPROXIMATE ZERO SEQUENCE REACTANCE
:57 � 'IO :57� '=:L: :Y C:C '�O 5? � �� ]5J � P
255
Q
NO
Z f'Q II N-O " . 8 5 ZPQ_I
NO
- 0
NO
- 0
NO
Fig. 7 . 20. Zero sequence equivalent circuits for three-phase, two-winding, core-type trans formers. ( From General Electric Co. [ 20 ) . Used with permiBBion. )
allel combination of Q and N. Note that these data apply for core-type units only. Transformer circuits are labeled P, Q , and R and these labels may be replaced by and when voltage levels are known.
H, X, 7. 1 0
Y
G round i ng T ransformers
Grounding transformers are sometimes used in systems which are ungrounded or which have high-impedance ground connections. Such units serve as a source of zero sequence currents for polarizing ground relays and for limiting overvolt ages. These transformers must have some connection to ground, and this is usually through some sort of Y connection. As a system component the ground ing transformer carries no load and does not affect the normal system behavior. When unbalances occur, the grounding transformer provides a low impedance in the zero sequence network. Two kinds of grounding transformers are used, the Y-Il and the zigzag designs. These will be discussed separately . The Y -Il grounding transformer is an ordinary Y-Il transformer connection but with the Il winding isolated as shown in Figure 7 .22a. Viewed from the Y side, the impedance is the excitation impedance which is usually taken to be in-
C hapte r 7
2 56 TRANSFORM E R C O N N ECTION
7
� �� Q
P
8
0
9
0
Z pQ _O · ·8 5 ZpQ _ 1
PI' �_ -,
Z p R UN -O= .75 Z PQ_1
N0
RlIN
NO
- R
R II N
Z PR liN _0 ' ·75 ZpR-1
NO
NO
P
Q
- R
NO
R
ZQRIIN -C · . ilZ OR - 1
,
P
� �� P
Fig. 7. 21 .
R
REAC TAN C E
-R
R
� CC� P
APPRO X I M ATE
Z E R O S E Q U E NCE
Z E RO SEOUENCE E OUIVALENT C I RCUIT
•
Z PQ/IRI1N-O··85ZPOIR-1
QURIIN
NO
Zero sequence equivalent circuits for three-phase, three-winding, core-type trans formers. ( From General Electric Co. [ 20 ) . Used with permission. )
finite. Since the side is not serving any load, the presence of this transformer does not affect the positive or negative sequence networks in any way. The zero sequence network, however, sees the transformer impedance Zt from point Po to No since currents laO may flow in all Y windings and be balanced by currents circulating in the winding. The zero sequence equivalent is shown in Fig ure 7.22b. �
�
P
a
b
c
Fig. 7 . 22 .
(0)
f>
Po
~
NO
(b)
y-� grounding transformer : ( a ) connection a t P, ( b ) zero sequence network representation.
The zigzag grounding transformer is a connection of 1 autotransformer windings, where primary and secondary windings are interconnected as shown in Figure 7.23. When positive or negative sequence voltages are applied to this con nection, the impedance seen is the excitation impedance which is usually consid ered to be infinite. Thus the positive and negative sequence networks are unaf fected by the grounding transformer. When zero sequence currents are applied, it is noted that the currents are all in phase and are connected so that the MMF produced in each coil is opposed by an equal MMF from another phase winding. These ideas are expressed graphically in Figure 7.24 where (b) shows the normal positive sequence conoltion and (a) shows the opposing sense of the coil con nections. (Note that Figure 7.24a is valid for three single-phase units or one three phase unit such as Figure 7.23.) Thus it appears to winding a 1 that it is "loaded" in winding a2 by "load current" Ie which is equal to la . The impedance seen by 1:
S eq uence I mped a n ce of T ra nsformers o
c c
� =�
--)
b
o
c
� -p
r--
-� - >>
257
=�
b
��
c �-
� =�
� '-p
r -P
-
(0)
(b)
Fig. 7 . 23. Zigzag grounding transformer connections : (a) winding arrangement on a core form magnetic circuit, ( b ) schematic arrangement where parallel windings P and Q share core 1 . ( From Westinghouse Electric Corp. [ 1 4 ] . Used with permission. )
Vc
c
b o
Fig. 7 . 24 .
(0)
(b)
Zigzag grounding transformer : ( a ) wiring diagram, ( b ) normal voltage phasor dia
gram. ( From Clarke [ 1 1 , vol. 2 ] . Used with permission. )
la , then, is the leakage impedance between a 1 and a2 , or Zt per phase. Thus the zero sequence representation is exactly the same as Figure 7 .22b . 7.1 1
The Z i gzag-Do Power Transformer
If another winding connected in Do is wound on the same core as the zigzag connected windings, a transformer connection exists which is capable of power transmission with grounding and with no phase shift. 5 Such a transformation might be useful to parallel an existing Do -Do bank and supply grounding at the same time. If the added winding is Y connected the phase shift of the bank is 30° , exactly the same as for the Y-b. connection . Both connections are shown in Figure 7 .25 where windings subscripted 2 and 3 are connected in zigzag. The windings subscripted 1 are not connected in the figure, but parts (b ) and (c) show the voltage phasor diagrams which result from b. and Y connections respectively 5 Some authors refer to this connection as the " interconnected star-delta" connection. We reject this label in preference for the shorter "zigzag-b. " or "zigzag-Y" name.
C hapter 7
2 58
��] 0 2 01
Va £
03
Va
( 0)
Vc (C )
(b)
Fig. 7 . 2 5 .
Zigzag-Ll and zigzag-Y transformer banks : (a) wiring diagram o f connections with windings of Ll or Y indicated but not connected, (b ) and (c) normal voltage phasor diagrams of zigzag-Ll and zigzag-Y banks respectively. (From Clarke [ 1 1 , vol. 2 ] . Used with permission . )
of a . , b. , and C . . Clarke [ 1 1 , vol. 2 ] analyzes the sequence impedances accord ing to three-winding transformer theory to develop the three-legged equivalent similar to that of Figure 7 .4c. Following Clarke's notation, we analyze the zigzag-Ll connection by defining the three-winding leakage impedances for core a as Z 1 2 = leakage impedance between a . and a2 Z . 3 = leakage impedance between a 1 and a3 Z23 = leakage impedance between a 2 and a3
(7 .44)
where all impedances are in pu based on rated voltamperes per phase and rated voltage of the windings. We take rated voltage of a . to be the Ll LL voltages, and for a l and a3 (which have the same number of turns) we use 1 /.,[3 times the base voltage of the zigzag side. Since all impedances are in pu on the same volt ampere base, we compute from (7 .29), Z ZIl 1 -1 1
LN
�
J
Zy
Zz
= 1 /2
[ _
1 1
-1 1
l�
J
1
Z1 3
1
Z1 3
pu (7 .45)
where Zx , Zy , and Zz are the impedances of the three-legged equivalent as shown in Figure 7 .26a, and with similar results for the b and c windings of Figure 7.25 given by Figure 7.26b and 7 .26c respectively. The notation and current directions
2 59
Seq ue n ce I mped a n ce of T ra n sformers VA
Zx Zy
(0
Vo 3
I
V
�I B= lO-IC
Z.
ZZ
�C
Vb2
VB
( bI
I�
k'IOI, c
i
Vb3 Vc 2 Ie
Ve 3
I
Fig. 7 . 26. Identical equivalent circuits to replace each of the three-winding transformers, with currents and voltages indicated. All values in pu. (From Clarke [ 1 1 , vol. 2 J . Used with permission. )
for Figure 7 .26 correspond to those of Figure 7 .2 5 . From Figure 7 .25a But from Figure 7 .26b and 7 .26c
=
Va =
Vb 3 - Ve 2
(7.46)
VB - IB Zx - 10 Zz Ve - Ie Zx + la Zy
(7 .47 )
VB - Ve - 10 ( 2 Zx + Zy + Zz ) + (Ib + Ie ) Zx
(7.48)
Vb 3 Ve2
=
combining (7 .46 ) and (7 .47 ), we compute Va
=
Now let VA , VB , and Ve be a positive sequence set of voltages, VA Vel ' Then VBl - Ve l - j V3 VA l , and we also note that =
h
VB h
and
and ( 7 .48) becomes (7 .49)
But this is computed for Va l based on 1 /V3 times the rated LN voltage (or based on winding or voltage). Based on the system LN voltage, Val would be 1 /V3 times this amount. Similarly, 1 is based on rated voltamperes and rated voltage; and if we choose a new01 base voltage va times as large as the old base, the new pu current is V3 times as large as the old pu current. Using a bar to signify the new pu value we have a2
a3
a2
(7 .50)
which changes (7 .49) to V0 1
or, rearranging,
=
_
.
J VA 1
_
I 3 Zx + Zy + Zz 01 3
(7.51 )
(7 .52)
C ha p te r 7
2 60
where 3Zx + Zy + Zz pu 3
(7.53)
Equation (7.52) is satisfied by the equivalent circuit of Figure 7 .27a. We may also let VA , VB , and Ve be a zero sequence set VAO , VBo , and Yeo . In this case, VBO - Veo = 0 and (7 .48 ) becomes (7.54) VaO = (Zy + Zz ) /aO -
After the change of base in (7 .50) we compute +Z z la o Va O - - Zy 3
-
_
=
(7.55)
Zo la o -
-
where Z0
-
-
Zy + Zz 3
Z2 3 pu 3
= -
(7 .56)
Equation (7.55) is satisfied by the circuit of Figure 7 .27b.
IAI
+ VAl
-
ZI
TTi 1 : .-1 2
0 NI -
t
4 S IDE
(0)
-
+.
-1
VAO
Z IG Z AG SID E
4 SIDE
i
Zo
�
NO
(b I
•
+-
O Va+
ZIGZ AG SIDE
Fig. 7 . 27 . Equivalent circuits of a zigzag-.1. transformer : (a) positive sequence, (b ) zero sequence. III. 7. 1 2
T R ANS F O R M E R S IN SYSTE M STU D I ES
Off-N ominal Turns R atios
In the study of very small radial systems there is no problem in representing the transformer, following the guidelines previously presented, and the chosen base voltages are conveniently taken to be the rated transformer voltages. This was the case in Example 1 .2. In multiply interconnected systems involving two or more voltage levels, however, it is not always possible to choose the base voltage as the transformer rated voltage because the transformer voltages are not always rated the same. Consider, for example, the simple system shown in Figure 7 .28 where three systems 81 , 82 , and 83 are interconnected as shown and where the three trans formers may have different voltage ratings. Yet the three transformers operate essentially in parallel, interconnecting systems nominally rated 69 kV and 161 kV. Such a connection of transformers presents two problems, an "operating" prob lem and a "mathematical" problem. If the transformers are of different turns ratio, even though fairly close to the 69-161 kV nominal transformation ratio, an interconnection like that of Figure 7 .28 will cause currents to circulate and reac tive power to circulate in the interconnected loops. This is the operating problem, which is often ignored for fault computation, although we realize that an off-
S eq ue n ce I mped a n ce of T ra nsfor m e rs
261
69 k V S 3 �r-----� L4
Fig. 7 . 28 . System with unmatched transformations.
nominal transformer tap ratio may be employed, either in fixed taps or in load tap changing equipment, to eliminate any circulating currents. The mathematical problem is that of deriving a correct representation for a system such as that in Figure 7 .28. This problem is of interest since it involves the correct system representation for an assumed operating condition. In other words, if our chosen base voltages do not coincide with the transformation ratio of the transformers, how do we compensate for this difference? We illustrate this problem by an example. Example 7. 2 Given the following data for the system of Figure 7 .28, examine the trans former representations for arbitrarily chosen base voltages of 69 kV and 161 kV and for a base MVA of 100 if the transformers are rated as follows:
T1 : X = 10%, 50 MVA, 1 61 (grd Y)-69 ( A ) kV T2 : X = 10%, 40 MVA, 1 61 (grd Y)-66 ( A ) kV T3 : X = 10%, 40 MVA, 1 54 (grd Y)-69 (A ) kV
Discussion We compute the voltage transformation ratios as follows : T1 : Ratio T2 : Ratio T3 : Ratio
=
=
=
1 6 1 /6 9 = 2 . 3 3 (nominal) 161/66 = 2. 4 4 ( 1 1 % high) 1 54/69 = 2.23 (10% low)
If we view each transformer from the 1 61 kV system only, we compute the following impedances. T1 : X = (0.1 )(100/50)(161/161) 2 = 0 2 0 pu T2 : X = (0.1 )(1 00/40)(161/161 ) 2 = 0 . 2 5 pu T3 : X = (0.1 )(100/40)(1 54/161 ) 2 = 0 22 9 pu .
.
Suppose now that we have exactly 1 .0 pu voltage on the 161 kV system with all transformer low-voltage connections open. Then the voltages on the transformer low-voltage buses would be as follows: T1 : V = 69/69 = 1 .0 pu T2 : V = 66/69 = 0 .957 pu T3 : V = (161/1 54)(69/69 )
=
1 .045 pu
Thus we could correct for the off-nominal turns ratio by inserting an ideal trans-
Chapte r 7
2 62
former
on the low-voltage side of
T2 and T3 with turns ratios 0.957 :
1
and
1 . 045 : 1 respectively and would close the connection to the bus in this way. This
is exactly the way this problem is solved in the laboratory. The resulting positive and zero sequence networks are shown in Figure
7.29 where the intercon-
L4 (0 I
LI TI
L3
T2
NO
NO
T3
NO
L4 ( bl
Fig. 7 . 29. Positive and zero sequence networks for the system of Example 7 . 2 : (a) positive sequence, (b) zero sequence.
necting systems 81, 82 , and 83 are not shown in detail. The negative sequence network is exactly like the positive sequence network in this problem since there are no generators in the system under consideration. If the tap changing mecha nisms were located on the high side (the 161 kV side) on the actual transformers, the transformer impedances should be computed on the low-voltage base and the ideal transformer shown on the side corresponding to the physical tap changer. Actually, in a Y-d transformer i is quite likely that the tap changer will be located near the neutral end of the Y side. t
In analytical work it is somewhat awkward to deal with an ideal transformer. Instead, a passive network equivalent is sometimes used where the transformer tap changer (ideal transformer) and series impedance are replaced by a pi equivalent circuit. Such an equivalent circuit may be derived with reference to Figure 7.30 where an LTC (load tap changing) transformer is represented between nodesj and k and with tap changing equipment on the j side. Note also that the tap ratio is indicated as from k toward j, e.g ., a tap ratio n > 1 .0 would indicate that j is in a boost (step-up) position with respect to k . The admittance Y is the inverse of the usual transformer impedance Z and is used as a matter of convenience. With currents defined in Figure 7 .30 we compute (7.57) llr Y ( Vm - Vir) as
=
_
2 63
S eq ue n ce I mped a n ce of T ra n sfo rme rs
n:1
Ij
\r---1
k
Fig. 7.30. Equivalent circuit of an LTC transformer with tap changer on node j side.
But for the ideal transformer we have and we eliminate V from (7.57) by substitution to compute m
Multiplying by l in , we have
Ik = Y I} =
(�
� (�
)
(7.58) (7.59)
Vi - Vk
)
(7.60) We now establish similar equations for the pi equivalent circuit of Figure 7.31 Vi - Vk
Fig. 7 . 3 1 . Pi equivalent circuit for Figure 7.30.
where we write by inspection
Ii = Y 1 ( Vi - Vh ) ,
From (7.61) we compute and
Ii = Y1 Vi + Ii ,
Ii = Y3 Vk + Ik
(7.61) (7.62)
(7.63) We now compare (7.59) and (7.60) with (7.62) and (7.63) respectively to write 1 1 - 2 n Y' y3 = n - 1 y (7.64) Y1 = -Y1 = - Y' n n n Note that all three of the admittances (7.64) are functions of the turns ratio n . Furthermore, the sign associated with the shunt components Y1 and Y3 are al ways opposite so that Y2 and Y3 are either inductive or capacitive for Y representing a pure inductance, depending entirely on n . This is shown in Table 7.5 and is further illustrated by Example 7.3.
C ha p te r 7
2 64
Table 7.5. Nature of Circuit Elements of the Pi Equivalent for Y -jB, B > O. -
n>I
n
inductance inductance capacitance
inductance capacitance inductance
Element
Example
7. 3
Compute the pi equivalent representation of transformer T3 of Example 7 .2 . Solution If the node which corresponds to node j of Figure 7 .30 is arbitrarily set at 1 .0 pu voltage (this is the S3 node of Figure 7 .29 ), we set n = 0 .957 pu. Then com pute .1 . 1 = '4 Y= J = J J . 36 pu X 0 .229 -
l In
Y1
=
-
-
--
-
1 .046
= ( 1 .045)(- j4.36) = - j4.56 pu (ind)
pu (ind) pu (cap) Y3 AB impedances the inverses of these quantities are computed as Zl = +jO .219 pu Z2 = +j48.8 pu Z3 - j50.9 pu Since we usually work with impedances rather than admittances and where Zt is the transformer impedance, we compute from (7 .64) Y2
=
=
( 1 .045) 2 (0 .043)(- j4.36 ) = jO .0205 ( - 0 .043)( - j4.36) = +jO.01965 -
=
Zl
= n Zt
Z3
= n -n 1 Zt
( 7 .65)
--
The above conclusions are summarized in Figure 7 .32. Note that as n approaches unity, the shunt branches of the pi equivalent both approach an infinite impedance and the series branch approaches Zt. Also note that if transformer resistance is included in Z one of the shunt impedances has a t,
lEO t
n: 1
k
ttk_+
+YJ
-,
(0)
I
t
.t
k
Yk+
vJ
-�
(b)
l.
-
Fig. 7 . 3 2 . The transformer pi equivalent : (a) transformer equivalent, (b ) pi equivalent.
Seq u e n ce I m pedance of Tra nsformers
265
negative real part. This is of no concern in analytical work but would be difficult to simulate on a network analyzer. 7.13
Three-Winding Off-N omi nal T ransformers
A three -w in ding transformer is usually represented by a Y equivalent similar to
Figure 7 . 4c. If, however, one or more of the three windings differs from the chosen base voltage or if a winding operates at a tap position which causes that winding to differ from base voltage , that leg of the Y equivalent must be changed to account for this voltage error. Suppose that the equivalent circuit of a transformer is computed in pu on the transformer base or, given ZHX , ZHY , and ZXY , we compute (on the H winding base voltamperes and voltage ) from ( 7 . 29 )
r�:l
-lJ �
1 /2 r � - 1 1 �J �1 1 1 1
=
J
HX
ZH Y
pu
ZX Y
Now suppose that one of the windings, say winding Y, has a voltage rating differ ent from the chosen base voltage for the Y voltage system . Then to match the transformer to the system , it is necessary to add an ideal transformer of voltage ratio n where n is defined as n
Base Vy --=transformer rated
= -----
(7. 29).
V
(7.66)
Y
---
This connection is shown in Figure 7 . 33 where the Y impedances are taken from x X
n: 1
H
TeASE
ZH
n 2 Zy I-n
n Zy
Y eASE
nZ y HI
n-I
-=
(0)
(b)
Fig. 7 . 3 3 . Three-winding transformer equivalent with off-nominal Y winding: (a) ideal trans former, ( b ) pi equivalent.
Problems
7 . 1 . A single-phase distribution transformer is rated 50 kVA, 60 Hz, 2400-240 V and
has
impedances as follows: ZH = 0.7 + j 1 . 0 n Zx = 7 + j 1 0 m n Ym l/Zm 3 - j20 mmho =
=
referred to
LV winding
Compute all parameters needed to represent this transformer according to equivalent cir· cuits in Figures 7 . 1a, 7 . 1b, and 7 . 1c . Compare your solution to the values given in Table 7.1.
C hapter 7
266
7.2. For the transformer of Figure P7.2 determine the terminal markings X I and X2 for the right-hand winding.
Fig. P7.2.
7.3. The terminals of the transformer of Figure P7.3 are assumed to be unlabeled and the polarity is not known. As a test, terminal HI is connected to and 200 V are applied between HI and H2 . Under this test condition a voltmeter connected from H2 to reads 220 V. What is the transformer polarity and what is its turns ratio?
Xn
Xm
Fig. P7.3.
7.4. Verify (7.23)-(7.25). 7.5. The following is taken from the nameplate of a three-winding transformer. Rated VH 1 6 1 kV, Rated Vx - 69 kV, Rated Vy - 13 . 8 kV ZXY = 1.92% ZHX - 8.69%, ZH Y 5 . 33%, Sy 10.5 MVA Sx = 10.5 MV A, SH 30 MVA, Compute the values ZH, Zx, and Zy in percent to be used in the equivalent circuit of Figure 7 Ac. 7.6. Given a three-winding autotransformer whose H, X, and Y windings are rated 200 kV, 100 kV, and 10 kV respectively and with circuit impedances of =
=
=
=
10% on a 30 MV A base - 9% on a 10 MVA base - 15% on a 30 MVA base
ZHX = ZXY
ZH Y
7.7. 7.8. 7.9. 7.10. 7.11.
7.12.
Compute the following in pu on a lOu MV A base. (a) Equivalent circuit impedances ZH ' Zx , and Zy . (b) Equivalent circuit impedances Ze , Zt . and Z, . (c) Circuit impedance Zei t Zse . and Zts . Verify equations (a) and (b) of Table 7.3. Verify equation (c) of Table 7.3. Verify equation (d) of Table 7.3. Verify equations (e) and (f) of Table 7.3. Three 10 MVA, 100-15 kV transformers have nameplate impedances of 10% and are con nected fl-Y with the high voltage side fl. Find the zero sequence equivalent circuit. (a) If the neutral is ungrounded. (b) If the neutral is grounded solidly. (c) If the neutral is grounded through a 5 n resistance. (d) If the neutral is grounded through 5000 J.LF capacitance. Find the resistance in ohms which must be placed in the corner of the fl of the trans former bank of problem 7.11 to result in exactly the same zero sequence circuit as in problem 7.11a-7.l1d.
S eq ue n ce I mped a n ce of T ransforme rs
267
7.13. Verify (7 .40) . 7.14. Suppose that three identical autotransformers are connected Y-a-Y as shown in Figure 7.11a. If the transformers have the impedances specified in problem 7.6, find the equiva lent circuits: (a) For positive and negative sequence networks, corresponding to Figure - 0.1 pu on an base.) (b) For the zero sequence network, correspond 7.11b. (Let ing to Figu re 7 .11 c . 7.15. Reconnect the transformer windings of Figure 7.15 so that VAl - -iVai . Is this an ac ceptable phase relationship with which to perform all other computations? Draw phasor diagrams to show all sequence voltages and currents. 7.16. Extend the computation begun in Example 7.1 to include the voltages on both sides of transformer Tl. Note that this requires the computation of the voltage drop from bus B to the fault point. 7.17. Show that third-harmonic exciting currents can be supplied in the zigzag grounding trans former connection of Figure 7 . 2 2 . 7.18. Consider a bank of 3-winding autotransformers connected zigzag-a and connected in a system with voltages of 34.5 kV ungrounded and 69 kV grounded. The transformer is to be connected to provide a ground to the 34.5 kV system. The transformer impedances and ratings are (for each of the three units), Zll " jO.15 pu on a 5 MVA base Z 1 3 - jO.15 pu on a 5 MVA base Z13 - jO .1 0 pu on a 2.5 MV A base Find the positive and zero sequence impedances for a problem in which the Base MV A of the study is to be 25 MV A. 7.19. Compute the pi equivalent representation of transformer T2 of Example 7.2. 7.20. Verify the circuit of Figure 7 . 33b. 7.21. Given a 3-phase, wye-connected autotransformer with delta tertiary winding and with
Zn
X
ratings and known data as follows, find the positive and zero sequence equivalents for this transformer: Winding
H:
Winding X: Winding
Y:
ZHX(){) ZHY(H) Zxy(){)
8HB VHB given 8XB VXB given 8YB VYB given
8HB and VHB in per unit based on 8HB and VHB in per unit based on 8XB and VXB in per unit based on
7.22. Figure P7.22 shows three single-phase transformers connected Y-A with the transformer terminals marked with both the IEEE H-X markings and the a-b-c phase designations.
HI�H2 HlbJ X I pX2 xlr b
a
1
HIL1iLJ
Xip
---
A B
---
C - IEEE STANDARD
B
A_AR B ITRARY LABalNG
C
Fig.
P7.22.
Note i n particular the fl. winding where a n arbitrary labeling i s specified which does not agree with the IEEE terminal marking.
(a) Sketch two phasor diagrams of the H·winding voltages, assuming the applied voltages to be first positiye then negative sequences. Let Va n be the reference voltage.
(b) Sketch phasor diagr�s of the X-winding positive and negative sequence voltages, . usmg the IEEE labeling. Show that HVLN leads LVLN by 30° in the positive sequence case but lags by 300 in the negative sequence.
(c) Sketch phasor diagrams of the X-winding positive and negative sequence voltages,
C hapte r 7
2 68
7.23.
using the arbitrary labeling. Show that HVLN lags LVLN by 90° i n the positive se quence but leads by 90° in the negative sequence. In the three·phase transformer connection shown in Figure P7.23 all three windings of each transformer are assumed to be in use but at essentially no load. Given that VAB '"' 69,OOOL.!L V and the sequence is A-B-C on the H winding side. Given that ' n - nH /nX - 144, n - nH /n y - 16.5.
H,
I�
' ''' U u
JH 2
a'
b' b
a
c'
n
c
Fig . P 7 . 2 3 .
(a) Sketch three vol tage p hasor diagrams, one each showing all six (3 LL and 3 LN) voltages for the three voltages designated H, X, and Y. L abel each diagram to show all voltage magnitu des, phase angles, and the A-B-C, a-b-c, or a ' -b' -c' phase designa tions. Orient the three diagrams to show the phase shifts between VA N , Va n , and Va ' n ' . (Where is n ' ?) (b) Assume that the H-windings are energized with sinusoidal LL balanced three-phase voltages. List the odd harmonic orders through n - 11 which exist in (4) The Y-winding voltages. ( 1 ) The R-winding LN voltages. (2) The X-winding LN voltages. (5) The H-winding curre n ts . (6) The Y-winding currents. (3) The X-winding LL voltages. (c) Repeat (b) but with the Y-winding open. (d) Repeat (b) with the Y-windings open and a high-voltage neutral wire connecting N to the high-voltage source. (e) Are any of the connections (b), (c), er (d) unsuited for practical application? Why? 7.24. In the transformer connection of Figure P7. 24 each single-phase transformer has a turns
:: I.L .6N .6 ,, ' T T ·r A
--
--
rOI !
I02 a
--
b
� :
a
,..
� " -Vae
n2
c
b
( b)
nl
,
POS I T I V E S E Q U E N C E
VOLTA G E S - a bc S I D E
- - - - - 0° R E F e
Fig. P7. 24 .
Seq uence I mped a n ce of Transformers
2 69
ratio of n = n t /n 2 = 10. The low-volt,e side carries an unbalanced load with sequence currents given as la t = - la2 = 100/- 30 A. Find the sequence currents IA t and IA 2 as phasors . 7 .25. Consider the single-phase three-winding transformer shown in Figure P7 . 2 5 , with the windings labeled arbitrarily as s-c-t and with ratings given for each winding. The follow ing partial short circuit test data is known. s
c
Fig. P 7 . 2 5 . Excited Winding
Shorted Winding
Applied Volts
A mperes in Excited Winding
c c
s t t
252
62.7 62.7 208.0
s
770 217
Note the absence of wattmeter data for copper-loss measurements. Assume that winding resistances are negligible. Consider a three-phase connection of identical units with windings c-s-t connected Y-Y-Ll with both Y's ungrounded. Let the base kVA be 3000 kVA. (a) Determine the equivalent circuit parameters ZH(H) , ZX(H) , and Z Y(H) in ohms re ferred to the H-winding. (b) Determine ZH , Zx , and Zy in pu . 7 . 26. Consider a three-phase bank of transformers identical to those of problem 7.25 but con nected as autotransformers (Figure P7 . 26) and again connected Y-Y-Ll with the t-windings connected in Ll and the c-windings connected to a common neutral. s
I.... .�
e
.. .. .. .. .. ..
Fig. P 7 . 26.
t
I n
(a) Determine the s-c-t parameters of the winding base equivalent circuit for positive and negative sequences, viz., values Zs( c ) , Zc (c ) , and Zt(c ) in ohms referred to the c winding. (b) Convert the s-c-t parameters to pu on a 3 . 0 MV A base . 7.27 . Consider the autotransformer connection of problem 7 . 26. (a) Find the parameters of the H-X- Y equivalent circuit on a circuit basis for positive and negative sequences; viz . , find the values ZH(X) , ZX(x» and Zy(X) in ohms re ferred to the X circuit . (b) Convert the H-X- Y parameters to pu on a 3.0 MVA base. 7 .28. Consider the three-phase autotransformer connection of Problem 7 . 26 with the neutral grounded through an impedance ZN = 0 + j 1 .0 n. Determine the T circuit parameters in pu for the zero sequence network . 7 .29. Let the three-phase transformer bank T1 in Figure P7 . 29 be the bank described in Prob lem 7 . 28 . Transformer bank T2 has reactances X t = X2 = 0.10 pu based on the trans former rating of 79.6 / 1 38-17 . 95 kV and 30 MV A. Thevenin impedances Z t = Z2 for the 1 38 kV supply system are such that the three-phase fault MV A at bus 1 is 600 MVA. (a) Draw and label the three sequence networks and mark the pu values of all imped ances. Assume the Thevenin equivalent voltage is 1.0 pu. (b) Compute the symmetrical SLG and 3¢ fault currents in amperes for a fault on bus 3.
2 70
C hapter 7
Q) 79. 6 / 1 3 8 k V ® 1 0 .3 6 / 1 7. 9 5 k V @ 7. 96 / 1 3. 8 k V
Fig. P 7 . 29 . 7 .30. A three-phase 34.6 kV subtransmission system (Figure P7.30) supplies several distribu
tion sUbstations (which are not shown) all of which are II connected on the high-voltage side. (a) Sketch the zero sequence network for this system and note the position of the Peterson coil, labeled PC. (b) The purpose of the Peterson coil is to reduce the SLG fault currents to small, harm less values and to avoid service outages when such faults occur. Describe how this could be possible and suggest any problems you can think of which might limit the effectiveness of this device.
v.L
A PC 'f-""'.l. X = 8% .,.. �-----J O �----� 1 5 MVA
7
8
2
3
9 6
4
5
Fig. P7.30.
7 . 3 1 . Let the line configuration for the subtransmission network of problem 7.30 be that given in Figure P7 . 3 1 , where the line data is given as follows: Line Section 0-1 1-2 1-3 1-4 4-5 4-6 0-7 7-8 7-9 7-4
Conductor Size
Type
STR
Radius
266.8 2/0 2/0 266.8 2/0 2/0 266.8 2/0 2/0 266.8
ACSR ACSR ACSR ACSR ACSR ACSR ACSR ACSR ACSR ACSR
26/7 6/1 6/1 26 /7 6/1 6/1 26/7 6/1 6/1 26 /7
0.0268 ft 0.0186 0.0186 0.0268 0.0186 0.0186 0.0268 0.0186 0.0 186 0.0268
Section Length
10 mi 15 5 10 15 5 10 15 5 10
Ground Wire Size " 1/2
none none none none none 1 /2 " none none none
Type
SM steel
SM steel
S eq ue n ce I mped a n ce of T ra n sfo rmers g O
1
I
s'
I
,
o a
HI
50'
271
50.3'
50. 1'
1 � H� .� 0'
bl
I
I I
000 = DeO = 9 . 2 7 '
D bg = S . IO'
,
O
c'
g'
Fig. P7.3 1 .
(a) Compute the zero sequence capacitance and 6 0 H z susceptance per phase per mile for all line sections. (b) Specify the following parameters for the Peterson coil: ( 1 ) inductance (4) continuous kVA rating (2) reactance (5) continuous voltage rating (3) continuous current rating 7.32. Consider the subtransmission bus shown in Figure P7.32 where the secondary winding of the supply transformer is connected in d. A grounding transformer bank made up of three identical single-phase transformers is connected to the bus to provide a ground con nection for the Peterson coil. Assume that these single-phase transformers have a leakage reactance of 6% based on their ratings. 1 5 MVA
DELTA - CO N N ECTED
X=8%
SECONDARY O F S U PP LY B A N K
--�-r_+-'---- O ----�_+_+--�---- b ------�_;--_+---+--- c
PC
Fig. P 7 . 3 2 .
t>
GROU N D I N G BANK
C hapter 7
2 72
We wish to determine the required kV A rating of the single-phase grounding trans formers from the list of standard available sizes provided below. Assume that the bus a-b-c is to serve the subtransmission network of Figure P7 .30 in place of the fl.-Y bank originally specified for that system. Assume that a SLG fault may persist indefinitely to fix the grounding transformer ratings. Available single-phase transformer sizes (kVA) are 50, 7 5 , 100, 167, 250, 333, and 500.
chapter
8
Cha n ges i n Sym m etry
In previous chapters we introduced the concept of solving an unsymmetrical system by a symmetrical per phase technique known as symmetrical components. We followed this by examining the system components, transmission lines, ma chines, and transformers to determine the impedance of these components ·to the flow of sequence currents. We should therefore be capable of representing any power system and analyzing all elementary series and shunt unbalances. We now examine other interesting and challenging problems. 8. 1
C reating Sym metry by labeling
It has been carefully noted that our calculations of symmetrical component currents and voltages are always to be in terms of phase a as the symmetrical phase . Should an unbalance occur which is not symmetrical with respect to phase a, such as a SLG fault on phase b , we have suggested that the phases of the system simply be relabeled. For example , if phase b is the phase of symmetry , let (a, b, c) b e labeled (C, A, B), and proceed i n the usual manner. I n simple cases of one unbalance this is often the best and easiest way to proceed. In some cases, however, it may be desirable to compute an unbalanced fault where phase a is not the symmetrical phase. There are at least two ways to do this and both will be explored . The first method is due to Atabekov and is based on the construction of a generalized fault diagram . The second method is an analytical one based on the work of Kron [ 5 7 ] and others.
[56]
8.2 G eneral ized F ault D iagrams for Shunt (T ransverse ) Faults
A systematic way of expressing fault information which is perfectly general insofar as phase symmetry is concerned is through the "generalized fault diagram" of Atabekov [ 56 ] . For shunt or transverse faults we begin with the general con dition at the fault point shown in Figure 8.1 where, depending upon the values assumed by Za, Zb , Ze , and Zg (including 0 and 00 ) , the fault could be any of the types usually considered. We write by inspection
Zg Zb + Zg Z,
(8.1 ) 273
C h a pte r 8
2 74
FAULT POINT -r-------.- O
-+��-----+--_.- b -+--��----�--�� C
Fig. 8 . 1 .
A general shunt ( transverse) fault condition at F.
which is easily transformed to the symmetrical component domain by premulti plying both sides of by K t . In matrix notation
(8. 1 )
(8.2 )
Then
VO l2 = A- II Vabc
= A- Za b c A A- I lab c = Z
Ol
Then we may define
= = =
1,
as
= = =
+ + + + a a c + +a + =
a times row
a
=
+a
=
+
By adding the rows of the three rows directly, we have
Adding row
(8.4)
(8.4) Zb2 Zc) 3Zg ZIO (1/3)(Za aZb 2 Zc) Zb 2 Z ) Zll (1/3)(Za Zb Zc) aZb Zc) Z2 1 2 (1/3)(Za Zb Zc) Z20 (1/3)(Za + Zb 2 Zc) Z21 (1/3)(Za Zb a Zc) (8. 5) Z22 (1/3)(Za Zb Zc) (8.3) (8.1 ), (8.6) 2 3 (8.7 )
where we compute the elements of
Zoo (1/3)(Za ZOl (1/3)(Za Z02 (1/3)(Za
(8.3)
2 10 12
+
+ a2
+
+a
+a
+
+
we may write
and a2 times row
+a
+
in a different form. Adding
gives
Cha nges in Symm etry
275
Finally , adding row 1 , a2 times row 2 and a times row 3 gives
Va2
=
ha2
"3 ZaIa + 3 Zb Ib h
+
3 Zc Ic
ha
(8.8)
These three equations ( 8 .6)-(8.8) may be used to construct a diagram which completely defines the condition described in Figure 8 . 1 . To do so, however, re quires the use of a phase shifting device which may be thought of as a transformer with complex turns ratio of (1 , a, or a2 ) . Such a device rotates both the current a n d voltage by a given phase angle (00 or ± 1 200 ) without changing the magnitude of either quantity . We need not think of this as a physical device but merely as a symbolic phase shifter (we will have no occasion to be concerned with the phys ical realizability of the network we are constructing). The phase shifter is char acterized by the equations
V2 = n V. , (8.9) 12 = n 11 j6 where n = e and where a shift of () radians is apparent. The circuit is shown in Figure 8 .2 and is called the generalized fault diagram for shunt faults. It is easily verified by noting that the points Q , Q o , Q l , and Q 2 are all at the same potential (and they could be connected together) . The voltages at Po , Pi t and P2 are
and these voltages appear across the top, center, and lower transformers re spectively, viewed from the right side . Taking the transformer phase shifts into account gives the voltages on the right side of (8.6 ) , (8.7 ) , and (8.8) for the zero, positive, and negative sequence networks respectively. Usually we are interested in two types of unbalanced faults, a SLG fault and a 2LG fault (where we may consider the LL fault as a special case of the 2LG ) . Atabekov [ 56 ] examines these two cases i n detail and develops special, simplified, generalized fault diagrams for these two cases. SLG
fault. For a SLG fault on phase a we set the network conditions as
Za
=
0,
Zb
=
Zc = 00 ,
Z, =1= 0
(8.10)
and we easily show that since (8.1 1 ) then
(8. 1 2 ) Through the boundary conditions (8.1 1 ) w e establish a simpler form o f general ized fault diagram for the SLG fault as shown in Figure 8 . 3 . The phase shifts of the transformers (specified as n o , n 1 , and n2 ) are given by (8.1 3 )
C h a pter 8
2 76
l aO ----
h
� ! Zala : F
1 aO
-
O
+
1
� l Zbl b�
VaO NO
� iI Zc 1c _ +
�,iI Zala
I
+
+
� iZ b lb:
Va l
NI
°1
a �i Zc'c :
l a2
� iI Z a1a +
h'1
-
3I
---...
h a2 I ai
-
ha l
al
-
�
h�a Za/3h 2
Po
h la l
--
-
_
a2
FI
�
00 h I aO
1
lal -
h1a O
PI
h�lb Z b /3 h2
-
tl-1c P2
-
t Q
ZC '3 h2
hla2
-
_
F2
�,, 'jI Zb1b...,
+
a
Va2 -N2
+
°2
a2
�.L Zc1c +_ 5
ha 1 0 2
•
ha 2 1a 2
-
Fig. 8.2. Generalized fault diagram for shunt (transverse ) faults.
for a SLG fault on phase a. Phase shifts for faults on phases b or c may be easily verified. Note that the factor h is not required in Figure 8.3 since equation (8.12) is valid for any h, but a factor h/3 is required if phase current is used.
Example
8. 1
Verify the phase shifts given in Figure 8 . 3 for a SLG fault on phase b.
Solution For a SLG fault on phase b we have the network conditions
Zb
=
giving boundary conditions
0,
Za
=
Zc = 00 ,
Z, :1= 0
277
C h a nges i n S y m m etry
Then
1 1 2 h (la o + 10 1 + 102) = h (laO + a la i + a 1a2)
0
This verifies that if n o = 1 , then for
from which we compute la O = a2 1a , = a la 2 ' any h
FAULTED P HASE SH IFT " P HASE 2 "0 " , I I A
10 0 -
FO
=
02
B C
0
O
a2
I : nO +
NO
-
VoO
3Z g 1 01 -
FI
\
NI
1
;-,
102
-
1 : n2
F2 +
-
V0 2
N2
Fig.
2LG fault. to be
8.3.
Generalized fault diagram for a SLG fault.
For a 2LG fault on phases b and c we set the network conditions
Za
=
00
,
ZII =1= 0
(8. 1 4)
The LL fault will be a special case with ZII = 00 and will give an impedance 2Z be tween lines. The boundary conditions corresponding to (8.14) are
10 = 0,
(8. 1 5 )
In this case the generalized fault diagram reduces to the configuration shown in
278
C h a pter 8
Figure 8.4 where phase shifts are all 1 : 1. If phases b or c are not the faulted phases, other turns ratios are required as noted in Figure 8.4. The two diagrams given in Figure 8 . 3 and 8.4 enable the engineer to solve all common shunt fault configurations for any faulted phase arrangement.
FAULTED
PHASE C -A A-B
•
: no
I
+
nO
B -C
laO
FO
PHASE SHIFT n,
n2
02 02 I
o
0
VoO NO
:-
3Z,+Z
l al FI
I
+
Va l NI
-
Z
la2
-
F2
,
+
n2
Va 2 N2
-
Z
Fig. 8.4. Generalized fault diagram for a 2LG fault. 8.3
G eneral ized F ault Diagrams for Series ( L ongitud inal) Faults
We may denote a general series or longitudinal fault by the set of unbalanced series impedances of Figure 8 . 5 , where we conceive of three impedances connect ing network points f and m with currents defined as flowing in the f-m direction. The impedances Za , Zb , and Zc are free to take on any value including 0 and 00 , thereby representing all common series fault conditions such as open lines. This network condition is described by (8.16) which is easily transformed to the symmetrical component equation (8.17)
Cha nges in Symmetry f -� .
Ia
+ v.fm-a
Ib
+ Vfm-b -
Ie
+ Vfm-c Ze
-
-
J
-
+
+
279 m
Za
a b
Zb
+
J
+
e
=:=
Fig. 8 . 5 . General series ( longitudinal ) fault condition between ( and m.
[
This was done in section 2 . 5 for the general case where Zab e includes mutual impedances. In our case Zmn -o l l is a circulant matrix,
Z , m .OI 2
=
ZsO Zs l
Za 2 Z.o
Zs 2
ZS I
where Z.o , Zs l , and Z.2 are given by (2 .47 ) . If we write voltage equation ( 8 . 1 7 ) i n terms o f Za , Z b , and Ze , the result i s similar i n form t o (8.6)-(8 .8) , the only difference being the absence of the Zg term. Thus these equations may be repre sented by a generalized fault diagram very similar to Figure 8 .2. This diagram is given in Figure 8.6 and may be verified by writing ( 8 .1 7 ) in the form Va O Va l Va 2
= "3h Za 1a + "3h Zb 1b + "3h Ze1e
ha = "3h Za1a + "3 Z b 1b + ha 3 Ze1e 2
h ha ha = "3 Za1a + 3 Zb 1b + "3 Ze1e 2
(8 . 1 8 )
The striking similarity between the generalized fault diagrams o f Figures 8.2 and 8 .6 is apparent. Note that in each diagram, the symmetrical components of fault current the networks at F and return at N or M. We have defined the se quence voltages in the two cases as a voltage through the symmetrical (unfaulted ) network from F to N or M. Thus F is labeled positive in both cases. Two cases of series or longitudinal faults are generally of greatest interest, one line open and two lines open. It is easy to show that these two cases reduce the generalized fault diagram of Figure 8.6 to the diagrams of Figure 8.7 for two lines open and to Figure 8.8 for one line open. These two diagrams should be com pared immediately with Figures 8. 3 and 8.4 for shunt (transverse) faults. The similarity is obvious. In fact, we can generalize the results for the two most common shunt and series fault connections by only two diagrams, as long as we make a proper interpretation as to the physical meaning of points F and M in each case. The phase shifters required to represent these faults on the various phases are the same in the sense that the phase shift depends only on the choice of By this we mean the phase that is symmetrical with respect
leave
symmetrical phase.
drop
2 80
Chaoter 8 1 00
-
h: 1
FO
h I 00 .. h l OO
--
+
NO Vfm-O ---
1 01
---
h: I
h lo l --
�I
FI
NI
+
MI
--
102
h:I
--
PI P2
ZO/3 h2
h2I b
Z b/ 2 3h
h2IC
Z C/ 2 3h
t
---
-
h I 02
---
ho I 0 2
F2
-
+
N2 Vfm-2
Fig. 8.6.
OI
ha l ol
--
h2 IO
Po
--
Vfm - I
M2
3 1 00h
h l oO
MO
h 0 2 IO ,"
tleneralized fault diagram for series (longitudinal) faults.
to the other two, as shown in Table 8 . 1 . Thus if only one phase is involved in the fault, that phase is symmetrical. If two phases are faulted, the third (unfaulted ) phase would be considered symmetrical. In summary, then, two sequence network connections can be used to repre sent all common faults. The series connections of Figure 8.3 and 8.7 represent the SLG fault and two lines open respectively . The parallel connections of Figure 8.4 and 8 .8 represent the 2LG fault and one line open respectively. These ideas are summarized in Figure 8.9, which may be used in the computation of all common faults irrespective of symmetry. Note that in using Figure 8.9 in computations involving longitudinal faults, Zg is not involved and should be set to zero in the diagrams. The impedance Z is used both for series line impedance and for the impedance from line to Zg in the case of the 2LG fault. 8.4 Computation of Fault Cu rrents and Voltages
It is useful to establish a definite procedure for the computation of fault currents for series or shunt faults which are determined by series or parallel con nections of the sequence networks. (Note the dual use of the word "series. " The meaning i s usually clear from the context o f use, however. ) For both the series and parallel network connections of Figure 8 .9 we will view the system as that of three one-port networks, one active and two passive. Then taking the
C h a nges i n S ym m etry OPEN
LI NES BBC CBA AB B
100
-
FO NO
281 PHAS E SHI FT
nO n, " 2 I
I
02 0 0 02
1 : no
+
Vfm-O MO -
Z
FI NI
1
+
Vfm-I -
MI
Z 10 2
-
1 N2 Fig. 8. 7 .
F2
I : n2
+
Vfm-2 M2 Z
Generalized fault diagram for two lines open and an i mpedance Z in the sound line.
external transformations and terminal connections into account, we may derive equations for the fault currents or voltages. Referring again to Figure 8.9, note that a consistent identity has been estab lished for all port voltages and currents external to the transformers. We arbi trarily define voltages Vo , VI , and V2 and prescribe that currents 1o , II , and 12 agree direction with lao , la l ' and la 2 and enter the + terminal. We also identify voltage V between terminals cp -cp and specify current I to enter the + V node. Since the transformers terminating the networks are considered ideal phase shifters, we write
in
i = 0, 1 , 2 This defines a new voltage V
(8. 1 9 )
for each sequence network, which is a voltage drop
Ki the drop across the fault impedance and across in the direction of lai and includes 2
the sequence network. The values of ni are always 1 , a, or a as specified in Figure 8 .9 and depend only on the fault symmetry. We easily compute VK i by
Chapter 8
2 82
LINE
PH A SE SHIFT nO n l
OPE N
NO
I
A B C
1 00
-
FO
nZ
oZ 0 oZ o
I : no
+
V'moO -
MO
Z
lal
-
FI NI
+
V'm- I
MI Z I aZ FZ NZ
V'm- Z +
MZ
Z
Fig. 8 . 8 . Generalized fault diagram for one line open and an impedance Z in the two sound lines.
inspection to be
o
Z
o
(8.20)
The voltage vector [ VaO Va l Va 2 ] t depends on the kind of fault, i.e . , whether a shunt or series unbalance . In every case, however, the zero and negative se quence networks consist of passive impedances Zo and Z2 , and the positive se quence network consist'S of an EMF E in series with an impedance Zl ' Thus it is Table 8. 1. Definition of the Symmetrical Phase
Phase Shift
Fault Location
Sy mme trical Phase
no
nl
n2
a or b-c
a
1 1 1
12 a a
1 a a2
b or c-a c or a-b
b c
C h a nges in S y m m etry SERIE S
NETWORK
2 83
PARALLEL NETWORK
CONNECTION
CON NECllON
K2
- Z---'IN'v--J\III'Y--
-- -- Z
---J\Nv-
=r.= K- N
�
Z
K -M Z gO - Z
K-M
Z gO- Z
Zg
SYMMETRICAL PHASE SH I FT PHASE "0 " I
Z QO - 3 Z g
0
I
C
I
b
I
I 02
0
"2
K - N
I 0
ZgO - 3 Z g + Z
02
Fig. B.9. Summary of network connections for common fault conditions.
always possible to write (8.21) and for any fault we define E to be the value of when O. Obviously, the values of Zo , ZI , and depend on the fault type and location but are always the driving point impedance between terminals Fi and Ki for i 0, 1, 2. Incorporating (8.21) into (8.20), we write 3Z6 + Z + Zo ) o 0 (Z + ZI ) 0 0 (8.22) Premultiplying (8.22) by a diagonal matrix, diag ( o, and using (8.19), we have 0 o o hE (Z + Zd o 11 o (Z + Z2 ) 0 (8.23)
[VVK1Kj [ VK2
Z2
Val Ial
=
=
_
n nJ o n2) ,
J
[IoJ tnl � 12
C h a pter 8
2 84
For the series connection we have
V = Vo + VI + V2 = 0,
1 = 10 = II = 12
(8.24)
or and
(8.25) Knowing this current, we compute the individual sequence currents from (8 . 1 9 ) and may then completely solve the networks. For the parallel connection it is convenient to write the inverse of (8.23) or
[�J � �. �] mJ f+� '
=
+
where
yo =
1 3Zg + Z + Zo '
____
For this connection we note from Figure 8 .9 that Then I
1 = 10 + II + 12 = 0 ,
=
V=
0 = ( Vo + VI + V2 ) V + V I nl hE and
Vo VI = V2 =
V = Y.-o Y+I YIn l hE + Y2 A
A
(8.26)
(8.27) ( 8 .28 ) (8.29)
Knowing this voltage, we may compute the individual sequence voltages from ( 8. 19) and completely solve the networks. The methods described above, culminating in the diagrams of Figure 8.9, are extremely useful in computing common faults with any possible symmetry. We now explore an entirely different method based almost exclusively on matrix algebra. This method is due originally to Kron [ 57 ] who used tensor analysis to solve a wide variety of network problems including those where the circuits are in motion, such as machines. He treated stationary networks as a special case in his general scheme of analysis but still used tensor notation. It is our view that the methods of tensor analysis, although entirely adequate for network problems, are not required and the more familiar matrix notation will be of greater value to en gineers. Several authors have adapted portions of Kron 's work to network prob lems using matrix algebra. One such work is LeCorbeiller [ 58] , which is recom mended reading. Our approach follows more closely that of Lewis and Pryce [ 1 3 ] , which is more directly aimed at our problem. Other works from this area which are worth investigating include Hancock [ 59] and Stigant [60] . 8.5
A F undamental R esult: The I nvariance of Power
Let X I be an n vector and let and B be n ties are related by the homogeneous equation
A
(A - B) X I
=
0
X
n matrices, where these quanti ( 8. 3 0 )
285
C h a nges i n Sym m etry
(8.30) A- B det (A - B) = 0 Note that (8.30) does not imply that A = B. Suppose that we now find additional x vectors
But from matrix theory we learn that the system has a nontrivial solution if and only if the coefficient matrix is singular, i.e. ,
(8.31)
X2, X3, , Xn such that the n x vectors are linearly independent (in which case we say that they "span" an n-dimensional vector space) so that we have the set of equations
(A - B) X I = (A B ) = (A - B) X2
-
•
•
•
0
0
(8.32)
xn = 0
We may write so that
(8.32) in matrix form by defining the matrix X where X=[ xn ]
(8.32) becomes
XI X2
•
•
(8.33)
•
(8.34) (A - B) = Since X is composed of linearly independent columns, det X 0 and X- I exists. If we postmultiply (8.34) by X- I , we have A - B = 0 or A=B (8.35) X
0
=1=
Note that the foregoing argument is not changed if A and B are row vectors of n elements. Suppose now that a network is under consideration for which we have a volt age equation (in matrix form)
V=ZI
(S.36)
but for some reason we require that the currents be transformed to a new set of currents which are linearly related to I by the transformation
I'
( S.37)
1 = K I'
We now seek a voltage transformation such that we may write
V' Z' I' =
(8.38)
where this transformation is subject to the restriction that the power is invarian t, i.e. ,
(S.39)
V t K*I'* V' t I' * (8.39), (V t K* - v' t ) 1' * (S.40) and since this is valid for all current vectors 1 ' , we conclude from (8.35) that (S.41)
This restriction will assure us that the law of conservation of energy is preserved. = Substituting into we have or
(8.37)
= 0
Transposing and rearranging, we have
V' = K* t V
(S.42)
C h a pter 8
286
as the voltage constraint which must accompany the current transformation (8.37) if power is to be invariant. Example 8. 2 When the phase currents are transformed into symmetrical component cur rents according to the equation Iabe = A 10 1 2 determine the corresponding voltage transformation if power is to be invariant. Solu tion From (8.37) we write I = K I' where 1 = lab e,
K = A,
I ' = 1 0 12
Then from (8.42) V01 2 = K *t Va be = A*t Va be• Since A is symmetric, A = A t and V0 1 2 = A *Vab e • Since A=
[� :J �� :J
�
1
a2
a 1
A' =
a
a2
=� A-I h2
Thus, we have V01 2 = (3 /h 2 ) A- I Vab e which agrees with the conclusions of Chap ter 2 derived by a different method. 8.6 Constraint Matrix K
It is appropriate to recognize that the operation implied by equation I = K I' is not a new concept to the student of circuit theory . This is just a way of re arranging the available information into a different and perhaps more useful form. For example, if I represents the branch currents in a network, then I' might repre sent the loop or mesh currents. In such a case K would be rectangular, and the transformation ( 8.37) is not unique since K is singular. This simply means that the meshes are not always defined in the same way. The matrix K which expresses the relationship between the old and new cur rent vector is called the constrain t matrix or the connection matrix. It is also re ferred to as Kron 's transformation matrix. It determines exactly how I and I' are related and is usually composed only of the real elements +1, 1 , and O. (How would this simplify (8.42)?) In the literature one often finds the Kron transfor mation designated by the letter C. We prefer K here to avoid confusion with the C used for the capacitance matrix . of Chapter 5. If ( 8.37) is written in elementary form, we have in the case where I represents branch currents and r the mesh currents, �
(p = 1 , 2, . . . , B)
(8.43)
Cha nges i n Symmetry
2 87
where
Kp Q = +1 if branch p is contained in mesh q and has the same sense = - 1 if branch p is contained in mesh q and has the opposite sense = 0 if branch p is not contained in mesh q B = the total number of elements in I, i.e. , the number of branches M = the total number of elements in I' , i.e., the number of meshes If there are N nodes, we know that [ 6 1 ]
( S.44)
M=B- N+ l
so that M < B, sometimes by quite a margin, which means that K is rectangular of dimension B X M. The transformation may also be used to change from one set of mesh currents to another since mesh currents are arbitrarily defined. In this case K is nonsingular and the transformation is unique. To gain experience with the constraint matrix we solve a simple example.
Example 8. 3
Given a network with N = 4 nodes, B = 6 branches, and connected as shown in Figure S. 1 0, find the constraint matrix K and determine the voltage relation which guarantees the invariance of power. Note the sense markings on all branches.
I' 3
Fig. 8 . 1 0. Circuit for Example 8 . 3 .
By inspection of Figure S. 1 0 we write I = K I'
SoUt tion
II
12 13 14
Is
16
2'
3'
1
0
1
2
-1
1
0
3
0
-1
-1
4
5
1
0
0
0
1
0
6
0
0
1
1
=
l'
�J
288
Chapter 8
Then the voltage equation is
V' = K* t V =
I'
2' 3'
1
2
�
-1
3 0
4
5
6
1
0
�}
0 1 0 -1 0 0 -1
1
V
in a network are re If we can say that the original currents I and voltages lated by an impedance matrix Z, we write = Z I, and for the transformed case we have V' = Z'I'. Since = K*tV, we have V' = K* tZ I and substituting
V'
I = K I' ,
Then, apparently,
V
V' = K * t Z K I'
(B. 45)
Z' = K*tZ K
(B. 46)
expresses the relationship between Z and Z'. Synge [6 1 ] proves that Z' is non singular by considering the network to be dissipative. If (ZI )-l exists, (and it always which we premultiply by K to obtain will) we may write I' = (Z' t i
V'
I = K (K*tZ Kfl V'
Finally, we eliminate V' to compute I as 1 = K (K* t Z Kfl K*tV
(B.47)
In many cases K is simply constructed, and the computation of (B.47) is straight forward. Using ( 8.47), we may compute the branch currents directly. 8.7 Kron's Primitive Network
In our discussion up to this point we have carefully avoided defining exactly what we mean by the vectors V and I in a network such as that of Figure B. I0. The quantities V' and I' are easier to visualize for the engineer accustomed to solving networks by mesh currents, and most engineers could write Z' by in spection. The real problem lies in identifying what we mean by the equation V = Z I in a physical network. Kron visualized a collection of individual branches for which he could write branch voltage equations. The collection of these equations he called the primitive equations and the disconnected "network" he termed the "primitive network" shown in Figure 8. 1 1 . The primitive equations are easily written as V =ZI
-
E
Fig. 8 . 1 1 . Kron's primitive network for a B-branched circuit.
( B. 4 8 )
289
Cha nges i n Sym m etry
K
Since the matrix specifies the assembly of the primitive branches into a net work , it is apparent from Kirchhoff's voltage law that = 0 , which simply says that the total voltage around any mesh specified by is zero. Then may be written as if the primitive network were shorted , i.e. ,
K* t V K
(8.48) (8.49)
E =ZI
with I the shorted current. Since the branch impedances here are considered to be independent , Z is diagonal , i.e. , 1
2
o
l[Zl
Z=2
B
B
.� .
Zl
o
0
Furthermore , E and I are B-dimensioned vectors of branch EMF 's and branch cur rents , with all branch currents defined as leaving the positive terminal of each voltage source. In the remainder of this chapter we use E for the EMF vector. This is a remarkable way to solve a network. The equation (8.49 ) which re lates branch EM F's to branch currents in the individual shorted branches does not describe any network at all ! That is to say , the primitive network equation gives absolutely no information as to how the individual branches are interconnected to form a useful network. This "connection " information is conveyed exclusively through the constraint matrix An example will illustrate the method .
K.
Example
8. 4
The circuit in Figure 8. 1 2 has three branches and two nodes and is labeled as specified in Figure 8 .1 1 . Find the matrix which will specify the connection indicated and transform branch currents to mesh currents.
K
2
Fig. 8 . 1 2 .
Network with three branches and two nodes.
Solu tion We write the primitive network equation from (8 .48 ) as E =Z I where
C h a pter 8
290
From (B .37) 1 = K I' where, by inspection
b'
a'
or
Z'
=
-1 1
o
] �a
o
K* t ZK or 0 �b
Then from (B.46) Z' = [1
[ �]
K = -�
-1
o
-1
ZJ J O
O
°
-1
-1
°
0
[ 1
c
We also write V' = Z'I' or
1 -
_
[Za
+ Zb - Zb
which the engineer familiar with writing mesh current equations will verify by inspection of Figure B.12 to be correct. We also note that from (B .44)
M
=B
-
N
+
1
=
3
-
2 + 1 = 2 meshes
which agrees with our solution. We extend this example by one additional step to compute the branch cur rents . From (B.47) we compute 1 = K (Z' ) - l K * t E
We define !l
=
(Za + Zb) (Zb + Zc ) - Z6
( Z' ) - l
=
det
= _!l1 [Zb Zb+ Z
Z' to write
c
and
Finally, we note that we can verify by inspection that
V'
=
K * t E = K t E or
Cha nges i n Sym metry
8.8
291
Other Useful Transformations
In section 8.7 , through the mechanism of the primitive network, we see the way in which branch currents of the primitive network may be transformed into mesh currents of the connected network. This transformation is accomplished through the constraint matrix We now investigate other applications of the idea of a transformation of voltage and current under the constraint of power in variance . The first transformation we consider is one which takes one set of mesh cur rents into a new set of mesh currents. This is possible since mesh currents are arbitrarily defined and the transformation technique permits us to discard one set of network equations in favor of another, presumedly more useful, set of equa tions. We will illustrate by example.
K.
Example 8. 5 The network shown in Figure 8 . 1 3 is correctly described by the mesh current equations
- Zs 4 (Z + Zs + Z6) - Z6
in terms of la , Ib , and Ic . Change the current variables to the new mesh currents 1. , 12 , and 13 and find the new mesh equations.
1 3 ------/ Fig. 8 . 1 3 .
Network for Example 8 . 5 .
Solution By inspection of Figure 8 . 1 3 find the relationship between the old currents I in this case la , I b , and Ic ) and the new currents I ' which are I . . 12 , and 13 • By ( selecting an appropriate tree , we relate the mesh currents by the equations
or, in matrix form I = K I I or
C h a pter 8
2 92
V V l ] I [] ' [ J O = [; ; V.V.l2 - -Vd [ + + Z2 ++ Z2)+
The new voltage vector is computed by V'
Kot E =
_
o 1
which is easily verified by inspection . The new impedance matrix is Z' =
K*t Z K =
(Z l
Z2 + Zs )
0
-
(Z l
(Z l + Z2)
(Z l
Z4
Z6)
(Z l + Z4)
Zl
which is also verified by inspection . Finally, we write V' = Z' I ' and the transfor mation is complete . Note that in this case K is nonsingular and the transformation is unique. We could therefore reverse the process, transforming V' , I ' and Z' back to E, I, and Z through the matrix K - l
,
•
Observe that just as in the transformation from branch currents to mesh currents, the process described above is based upon the idea
(8.50)
or I K I' . Furthermore, if the power is to be invariant, the old and new voltages are related by
=
(8.51 )
The symmetrical component transformation is an arbitrary transformation where the new and old currents are related by the matrix K A or from (8 .50 ) with lold lobe and Inew
= 10 1 2
=
==
lo be
A lo u
=
(8.52)
If the power invariant form of A (h v'3 ) is used , the voltage transformation is given by (8.51 ) . The Fortescue transformation ( h = 1 ) requires special treatment as noted in Example 8.2. Other transformations can be devised which relate currents of one network to those of a second network, thus specifying mathematically the connection or constraint K between the networks. An example will illustrate this procedure.
Example 8. 6 Consider the two networks of Figure 8.14 where the network mesh equations are given by
[VlV2 ] _[ZI Z2 OJ [/112J Z3 13 V3 1113 ==1;/�,, 12 =/=1;� +/� +/� 0
-
and Vn
= Znln.
0
0 0
0
The two networks are to be combined with the constraints In
293
Cha nges in Sym metry II
ZI 12
V':7 � I Zn
Z2 13 13 1 2 II
Z3 ( 0)
Fig. 8 . 1 4 .
(b)
Networks for Example 8 . 6 : (a) three-mesh circuit, ( b ) one-mesh circuit.
Find the network connection and examine the special case where Vn
=
O.
Solution
The relationship between the "old" voltages and currents is E Z I or VI ZI 0 0 0 II V2 = 0 Z2 0 0 12 V3 0 0 Z3 0 13 Vn 0 0 0 Zn In The constraints between old and new currents completely specify K, i.e ., =
or
I Old = K Inew ,
11 12 13
1
0
1 = KI '
0 0
�;J
1 0 1 1 1 1 In or 1 0 0 0 1 0 K= 0 0 1 1 1 1 The voltages are related by V' = K*t E or V I V. + V 0 0 V2 = [V2 + Vn:J 1 0 � � 0 1 VnV3 V3 + Vn The new network is shown in Figure 8.15. In the special case where Vn 0, we note that V = V ' and the new connection is that of a three-phase generator with phase impedances Z , Z2 , and Z3 and with neutral impedance Zn. Since the orig inal network of FigureI 8.14a was that of a three-phase generator with zero neutral impedance, we have added an arbitrary neutral impedance through the choice of a partiCUlar constraint K .
[VJ = II
0 0
�
=
Chapter 8
294
Z2 I'
-L. Zn
+Yn -
Fig. 8. 1 5 . Solution for Example 8.6. 8.9 Shunt Fault Transformations
We now consider the application of Kron's technique to the solution of faulted networks. Suppose that the network is to be faulted at a fault point F and that the positive, negative, and zero sequence Thevenin equivalents are known. These equivalents, shown in Figure 8.16, are described by the sequence network FO
FI
F2
NO
NI
N2
!i® Ii: � i:� Fig. 8 . 1 6 . Sequence networks.
equations
(8.53) But by the method of Kron we need to consider the sequence networks as primi tive ones. Then we may specify the connection of these primitive networks by developing a constraint matrix K which describes the fault condition under study. From Figure 8.11 and (8.49) we see that the primitive sequence networks are those in which (8.54) as shown in Figure 8.17. From (8.53) and (8.54) we have the primitive network equation E =Z I or
t
YaO
r
FO
l ao
101
FI
E] [J
NO
-
I. t
Yal
-
! YF
+
NI
I. E] F2
Va 2
r
N2
Fig. 8. 1 7 . Primitive sequence networks.
10 2
-
295
C h a nges i n Sym m etry
o
(8.55)
The application of (8.55) to specific fault conditions K.
SLG
faults
.
requires the specification of
For a SLG fault on phase a we have the condition (8.56)
where we use the prime notation to indicate sequence currents sequence networks or where I = K I ' with
I
in the connected
= sequence currents of primitive networks l ' = sequence currents of interconnected sequence networks
For the SLG fault on phase a we require the networks to be connected such that
[alaolJ [�I�ooJ [lJ I�o la2 lao = 1
=
( 8 .57)
1
Then
and I ' =
ldo ,
a scalar. We also compute
V'
=
K*' E = [ 1
1 I]
�vj
=
h V,
(8.58) Thus both I ' and V' are scalars, i.e. , 1 X 1 matrices. For the connected network we write so we must also compute Z where
V' = Z 'I'
I
z ' = K *'ZK =
also a scalar. which
=
[1 1 I]
r�� o
(8.59) o o
(Zo + ZI + Z2) = (ZO+ZI +Z2) Ifault�o impedance is either zero or Zo , Zo ZO(sY8tem) + We examine fault
Then (8 .59) may be written as
(8 .60)
h VF
is obviously correct for the case where the is included as a part of i.e . , = 3Z, . impedance problem in greater detail later.
will
the
C h a pter 8
296
For a SLG fault on phase b we have a different constraint among the sequence currents. In this case 10 = = 0, and we compute
Ie
�,, = A- ' �., = �
l:J I.
(8 .61 )
In other words the relationship among sequence currents corresponding to (8.56) the previous case is from (8 .61 ) ,
in
( 8.62)
Note that this agrees with the solution obtained from the network of Figure 8.3. From (8 .62) and Figure 8.3 we equate primitive to connected currents to write
[I1a0O1J = [I171:J = ��a:O ] = �Ja I�o a2 0 1 0 Ia 1 2 2 or = l: J Note that since a = ei2n/3 then = e-i2n/3 = a2 and, similarly, (a2 * = a. Then 2 = [1 a a] and �vj = = [1 Finally, o fr 2 Z' = a a] � Z Zl Z o 2 To compute the fault current, we write = or a2 h = (Zo Z Z2 ) Ida and aO = Zo Zl Z2 which is seen to the same as the current computed from except the re �
a
�
(8.63)
( 8.64)
K
)
a*
K*t
.'
V'
= K*t Z
=
+
[1
K
+
V'
/'
be +240°
,' bV,
'J
Z'I'
a2 h VF + +
VF
+
1
+
(8.65) ( 8.66)
(8.60)
sult is rotated by as it should be for a fault on phase b. faults. another example of the Kron method applied to faulted net works we examine a 2LG fault on phases b and c with zero fault impedance. For this condition we recall that the sequence networks are connected in parallel as shown in Figure 3.15. With zero fault impedance we have, for the connected net work, Ia = 0, or 2 LG
As
(8.67 )
� j ��j ll�l:lal I�J I -� [I� ] I 1 [ � -OIJ The voltage equation is [-1 1 �] �vj [�VF] V'
297
C h a nges in Sym metry
Then we determine the constraint matrix K from a Ia la Ia2' Ia2 Ia2, or =
1
1
=
=
1 0
0
,
I
Ia2
(8.68)
1
K=
= K*tE =
=
0
-1
The transformed impedance is Z' = K*t Z K
=
tl
-1
1 0
[, �J [� -� 0
�]
ZI
0
=
[
(8.69)
Zo + Z I Zo
(
Then the connected network may be solved for 1�1 and 1�2 from V' = Z' I' or
8 70 ) .
(8.71)
Solving for the currents, we have
8.7 2 ) which from (3.22 ) are obviously correct. Then I�o may be computed from (8.67) and la be computed in the usual way. If the 2LG fault is on phases other than b and c, the K matrix becomes complex. For example, if the fault is on phases a and b , then Ie = 0, or (8.73) Then al� l - a2 1� a - a [Ia']l I'a l = 1 0 la, 2 = 1�2 0 1 (8.74) (
[
and
J
l
J
[ �
C h a pte r 8
298
K= which is complex.
a
-a
1 o
0 1
( 8.75)
From the foregoing brief experience in solving faulted networks by Kron's method , we note the following features concerning the formation of K.
1 . In the case of SLG faults it is always possible to write two current con
straints, e.g., for a SLG fault on phase b , II - 2 1 ' 00
-
a
ah
In such cases when we write I = K I' , there are two constraints among the three variables of I ' and we can always eliminate two currents. This means that K will always be 3 X 1. 2 . In the case of 2LG faults it is possible to write only one current constraint. For example, if the fault involves lines a and b , we have I�o + aI� l + a2 I� 2 = 0 Here we have one constraint for three variables and can eliminate only one variable. Thus the K matrix is always 3 X 2. 3. In all cases if the fault is symmetric with respect to phase a, then K is real and K* = K. If any other symmetry is presented, K is complex but it in volves only the cOll)plex p aram eters 1, a, and a2 • 4. In cases where two current constraints exist, the K matrix may be written such that any one of the currents I�o. I� h or I� 2 is explicitly in the solution. In the SLG cases above we solved for I�o, but we could have chosen I� l or I� 2 ' In cases where only one current constraint exists, the K matrix may be written such that any two currents are explicitly in the solution. In the 2LG faults above we solved for I� l and I� 2 ' but we could have chosen any two currents.
5.
8. 1 0
Transformations for Shunt F aults with I mpedance
In Figure 8. 1 a generalized shunt fault is shown with arbitrary values of fault impedance. The equations describing this situation are given in ( 8. 1 ) in terms of phase quantities and in ( 8. 3 )-(8.5) in terms of sequence quantities. The various shunt fault conditions of interest � then specified by prescribing values to the impedances Zo , Zb , Zc, and Zg . This in turn gives certain constraints among the sequence voltages and currents which form the boundary conditions for the dif ferent fault situations. We may take advantage of this analysis and adapt it to the Kron analytical technique by enlarging our view of the primitive network to include certain fault impedances as well as the sequence impedances for the normal (symmetrical) portion of the network. Since we are usually concerned with only the SLG, LL, and faults, with appropriate impedance, we focus our attention on the following special cases :
2LG
SLG fault: Zg * 0
Zo = Zb = Zc = 0
Z. Z. "*
299
Cha nges i n Symmetry
LL fault:
=
00
any two of
2LG fault:
0
any two of
Zo, Zb, Zc Z Zo, Zb, Zc Z =
and
and
=
These are exactly the special cases studied in section 8. 2 where we developed the sequence network connections given in Figure 8.3 for the SLG fault and in Figure 8 .4 for the LL and 2LG faults. The LL and 2LG faults differ only in the specifica tion of and we note that 100 = 0 for the LL fault such that the zero sequence network is isolated from the other networks (see Figure 8.4). For any symmetry, Figures 8. 3 and 8.4 provide us with the desired constraints for determining the K matrix if we interpret the Kron primitive networks as those of Figure 8. 1 8.
Z.'
F2 FI ZIO O la2
lal
3Zg+Z
Z2
hV�
Z
Z
Fig. S . l S . Primitive sequence networks for shunt faults with impedance.
We must also decide, as discussed in section 8.9, which sequence currents in the connected network are to be specified. The choice here is arbitrary. It seems reasonable, however, ·that in any fault involving the ground the ground current 3/� o is of direct interest. Hence I� o should be the current to find in both SLG and 2LG faults. In LL faults, which are a special case of 2LG, either I� l or 1�2 is re quired. Let us arbitrarily choose I� l . Then we may specify I� o and I� l for the 2LG fault and can easily identify the LL case as that for which I�o o. For any fault we have the primitive equation E Z I or
[�vJJ o
=
=
rzo + �. + Z) L 0
=
( Z l O+ Z )
0
which is transformed by the current constraint I = K
V'
=
Z 'I'
� l �:� (Z2 Z� �oJ +
( 8. 76)
I' to a new equation ( 8. 7 7 )
The constraint matrix may b e determined from the connections o f Figures 8 . 3 and 8.4 which are tabulated in Table 8.2. Note that it is unnecessary to provide a separate tabulation for LL faults since the constraint matrix in this case is im mediately available from the 2LG results by eliminating the column of K corre sponding to I�o. To solve for the sequence currents, we must solve (8.77) for the current I' given in the second column of Table 8. 2. This requires two additional items V' and Z ', which are readily computed once K is known. These results are given in Table 8.3. We illustrate the use of Tables 8.2 and 8.3 by the following examples.
300
Chapter 8
Table
Type Fault
8.2. The Constraint Matrix Solution Equation
SLG or two lines open I -
KI�o
K
b I�o -I�1
a
I
b
Ia I�o
-0 Ie - 0
-
Ia
Ib
c
I-K
[I�o] ,
la l
a
(b-c fault )
b
(a-c fault ) C
(a-b fault ) 1-
LL
Example 8. 7
lao
and Ial
c ______
Current Constraints Symmetrical PhtJse System Seq uence
Ie
2LG or one line open
in Terms of
-
0 0
- 0 -
0
Ia - 0
Ib Ie
K I�1
-0
-0
-
-
a 2 I� 1
I�
-
Constraint Matrix K
0'
2
aI�2
0'
I� o - a I� 1 - a2 I�2
I�o I�1 I�2 +
+
-
I�o
+ + a I� 2 - 0
I�o
a I� 1 + a 2 I�2
+
0 0 1
0 1
[
0' 1 0 -1 0' 1 0
[
[
2 - a2
-0
same as 2LG but with I -0
�o
0'
2
a2 1� 1
WJ WJ �[:j
0 1 2
0' 1 0
-a
I'
-n I'
J -! ] l'
,
use 2nd column of 2LG results
Consider a SLG fault on a system for which the sequence networks are given as those of Figure Since the fault is SLG, we let Z = 0 and Z, =!: O. Find fault currents for a SLG fault on phase a, then on phase b. Compare With results of similar computations from section
8.18.
8.9.
Solution For a SLG fault on phase a we have from Table 8.2
K=m
C h a nges i n S y m m etry
30 1
Table 8.3. Transformed Voltages and Impedances Corresponding to Constraint Matrices of Table 8.2 * Type Fault
Sy m me trical Phase
SLG or two lines open
a
V'
[V��] � V�I J 0 hVF
K* t Z K t
Z ' = Zo + ZI + Z2 + 3 Z" =
same
z'
[
same
]
does not apply
1
same
change k only
a2
c (a-b fault)
same
change k only
a
any
V� I = hVF
=
b (a-c fault)
_
Z'
I
-
ZI I
' * Values for V ' and Z in this table correspond to solution for currents laO and 10 1 • ' t Z components are defined as Zoo Z o + Z2 + 3 Z, + 2 Z, Z 1 1 Z + Z2 + 2 Z Z 22 and Zgg Zg (SLG) Z ( 2LO ). =
=
k
k Z22 Zoo 2 k Z22 Zl 1
a (b-c fault)
Z,
z'
V�o = ahVF
c
LL
K*thE
V�O = hVF V�o - a2 hVF
b
2LG or one line open
=
=
=
and I' = I�o, a scalar. Also from Table B.3, V' since V' = Z' I' , we have h VF
=
=
I�o
,
=
Z2
+
h VF , a scalar, and Z ' = Zo + Z I + Z2 , + 3Z, . Then, (Zo + Z I + Z2 + 3Zg ) I� o and =
h VF Zo + ZI + Z2 + 3Z,
This is exactly the same result given in equation (3.4). By symmetry we know that a SLG fault on phase b will be the same as on phase a except rotated by 1 200 ( or +2400 ). We check this by taking values from the tables. From Table B. 2 -
and from Table B. 3 so that
I� o = Zo
I
a2 h VF
+ Z + Z2 + 3Z,
which is obviously correct. In both cases for faults on phase a and phase b, these results agree with equations ( B.60) and ( B.66) except that the impedance has been increased by the amount 3Zg • Example B. B Consider a 2LG fault on a system for which the sequence networks are given as those of Figure B. 1 B where Z =1= 0 and Z, =1= O. Find the sequence current and compare with equation ( B. 72 ) for a 2LG fault on phases b and c.
I� I
302
C h a pter 8
Solution
From Table 8. 2 we have K=
and
[l�o] lal V' = lb10 VFJ I' =
From Table 8. 3 we find that and
Zll] Z' = [Zoo Zll Z u
Since V' = Z'I' we compute Then l� = I
[� J
det
Zoo 0 Z�l=l'_:_h-V..:;F = Zoo h VF det Z' Zoo Z Z� l
____
But, from Table 8.3 Zoo = Zo + Zl + 3Z, + 2Z, and
.....
II
Zi l
= ZI
+
= (Z I + Zl + 2Z) - Z�2hVF /(Zo
If Z = Z, = 0, this reduces to ( 8. 72) exactly. Example
-
Zl + 2 Z, Zll = Zl + Z
+ Zl
+ 3 Z, + 2 Z)
8. 9
Repeat Example 8. 8 except that this time let the fault be applied to phases a and c. Then simplify to the case of a LL fault, also on phases a and c.
Solu tion
For a 2LG fault on phases a and c we have from Table 8. 2
C h a nges i n Symmetry
[I�O]
and I
Then from Table 8.3 V
= Ial
'
]
J
, = rV�O = fO LV�l Lh VF
and z'
where
303
[
= Zoo aZ22
Zl l = Zl + Z2 + 2 Z,
Zoo = Zo + Z2 + 3Zg + 2Z,
We compute the sequence currents from Then
Zoo ] [ = Zoo hVF det de t a Z2 2 h VF det Z' From Table 8.2 we have the sequence current constraint I�o + a2 I� l + aI� Ia2' = a2 IaO' a a 1 = a(Z22 - Zoo) hVF - a (Zo + 3 ZgZ'+ Z) hVF All currents may be found from I�o, I� and I�2 ' In the case of a LL fault, I�o = 0 or
and
,
Ia l
_
=
o
1
Z'
_
I
'
..
I
and
or
we have
-
_
det Z '
, [0 ] =
I�l
det
2
=
0 or
304
C h apter 8
I� , = hVF/ZII = hVF/(Z, + Z2 + 2 Z) From Table 8. 2 with I�o = 0, we also compute a2 I�, + aI�2 = 0 or I�2 = aI�l. Again, by knowing the sequence currents, we can solve for all currents. -
8.1 1
Series Fault Transformations
It is apparent that a one-to-one relationship exists between the generalized fault diagram of Figure 8 .9 and Kron's analytical technique of section 8.10. We recognize this relationship in developing an analytical technique for series faults which will be based upon the network unbalance described in Figure 8.5 and (8.17 ). The similarity between series faults involving one or two open lines and shunt faults of the 2LG and SLG conditions is noted in Figure 8.9. In the case of series faults the impedance Zg is always taken as zero but the impedance Z is not zero in general. Thus we may take the primitive sequence networks to be those of Figure 8 . 1 9 except that Zg = O. Two In the case of two lines open we may write two current constraints, or if the open lines are b and c, Ie = 0 Ib = 0 , Since we have two constraints, we need to solve for only one of the three se quence currents. Let us arbitrarily choose I�o . Then this case is mathematically identical to that of a SLG fault or phase and the entries in Tables 8.2 and 8 . 3 apply. Similar statements could also be made for two lines open involving other phases. One line open. If only one line is open, there is only one current constraint which we can write, and we must therefore solve for two sequence currents. If we arbitrarily choose I�o and I� l ' the entries under the 2LG fault of Tables 8 .2 and 8.3, with Zg always taken as zero, apply. Because of the unique similarity in the symmetry of shunt and series faults there is no need to develop separate tables for the series fault situation. In using Tables 8 .2 and 8.3, the only caution is that Zg must always be set to zero for a series fault. Obviously, h V has a different meaning in the two situations also. lines open.
a,
F
8. 1 2
Su mmary
We have presented two methods for dealing with symmetries different from phase symmetry. The first method is valuable since it places emphasis on circuit concepts with which most electrical engineers are familiar. It results in a quite general pair of circuit diagrams given in Figure 8 .9 , from which the engineer can solve nearly any fault which may be of interest. The second method, due to Kron, is more analytical than the first in the sense that it is based upon topology and matrix techniques. It is readily adapted to computer analysis because of its orderly method of computation and because it does not rely on the drawing of a circuit diagram. Both methods have their unique advantages, and both can be extended to help us solve even more difficult problems. a
305
Cha nges in Sym metry
Problems
8.1. 8.2. 8.3. 8.4. 8.5.
Verify the network connections of Figure 8.3 by network reduction of Figure 8.2 for the condition of a SLG fault on phase a. Verify the transfonner phase shift of Figure 8.3 for the case of a SLG fault on phase c. Verify the network connections of Figure 8.4 by network reduction of Figure 8.2 for the condition of a 2LG fault on phases b and c. Verify the transfonner phase shift of Figure 8.4 for the case of a 2LG fault on phases c and a. Repeat for a 2LG fault on phases a and b. Given a system with Thevenin equivalent sequence network representations shown in Figure P8.5, sketch the generalized fault diagram for a SLG fault on phase b at F and compute the fault currents in all phases. Let the fault impedance be jO.2 pu.
Fig. PS. 5 .
8.6.
8.7 .
Using the sequence networks of problem 8.5 , sketch the generalized fault diagram for a 2LG fault between phases a and b with fault impedance of jO. 1 in each line (Le., Z = j O.1) and grounding impedance Zg of j O. 1. Compute the total fault current in each phase. Given a system with two-port Thevenin equivalents at points F and M as shown in Figure P8.7, sketch the generalized fault diagram for the case of two lines a and b open between F and M, and line c having an impedance of jO.1 pu. Compute the three line currents. FI
MI
F2
Fig.
M2
FO
MO
PS.7 .
Using the sequence networks of problem 8.7, sketch the generalized fault diagram for the case of line b open and with lines a and c each having an impedance of jO.2 pu. 8.9. Find the constraint matrix K if the mesh of I� in Example 8.2 is taken to be the mesh Z6 , -Zs , -Z4 ' (This is the mesh in the "window" defined in the clockwise sense.) 8.10. Given a connected network with B branches and N nodes (see [ 57 ] ), (a) Show that there are N(N 1) possible voltages between nodes, taking polarity into account. (b) Show that of these N(N 1) voltages only N 1 are independent. (c) Show that if there are S subnetworks, the number of independent voltages is N S, where N is the total number of nodes in all networks. 8.11. Given a connected network with B branches and N nodes and P = N 1 independent node voltages (see problem 8. 10), (a) How many independent node equations can we write? (b) flow many independent branch voltage equations can we write? (c) How many independent mesh current equations can we write? Call this number M. 8.8.
-
-
-
-
-
C h a pter 8
306
(d) How many independent mesh current equations can we write if there are S subnetworks? 8.12. Given the circuit of Figure P8. 12 with branch currents, in the directions indicated, (a) Write the primitive network equation E Z I. (b) Transform the primitive network equation into a mesh equation where the meshes are defined to be 1-2-4, 2-3-6, and 4-5-6 in the order indicated. (c) Transform the mesh equations of (b) to a new set of equations, viz., 1-2-4, 2-3-6, and 2-3-5-4. (d) Compute I from (8.47). In both parts (b) and (c) carefully define the constraint matrix K. Note that in part (c), K should be nonsingular. What does this mean? -
Fig. PB . 1 2 .
8.13. Repeat problem 8 . 1 2 where a mutual impedance Z 3 5 exists between branches 3 and 5. (a) Define a new Z matrix which includes Z35 ' Consider the mutual impedance voltage drop as positive and explain what this implies. (b) , (c), (d). Transform from branch to mesh currents and compute branch currents as in 8. 12. 8.14. Given the circuits shown in Figure PS. 14 with mesh currents defined as flowing through impedances (a) a-n , bon , con and (b) d-e and f-e, (a) Write the mesh equations for circuit (a) and circuit (b). (b) Define a new network by combining (a) and (b) according to the constraint
fa = 10 1 = 102 ,
tb = Ib l ,
I� - lei = le2
Write the constraint matrix. (c) Assuming power is to be invariant, write the voltage relationship between new and old networks. (d) Sketch the new network. 10 2 +
Va l
+
Va2
Zd Z. Zf
Zn
(a )
(b ) Fig. PB . 1 4 .
8.15. (a) Write the mesh current equations for the circuit of Figure PS. 15. (b) Consider the circuits of Figures PS. 14a and P8. 15 with the current constraint given in problem 8. 14b. Find the voltage relationship between the given networks and the combined networks. (c) Find the combined impedance network, i.e., K *t Z K. (d) Sketch the new network.
Z'
-
Cha nges in Sym m etry
307
Zg
9 Zh Fig. PB .I 5.
8.16. What fault condition does Figure 8.17 and (8.55) describe? Give proof for your answer. 8.17. Find the constraint matrix K which will transform the primitive sequence currents into SLG fault on phase c currents, where I�1 is to be solved explicitly. 8.18. Find the constraint matrix K which will transform the primitive sequence currents into 2LG fault currents on phases a and c where I�o and I�l are to be found explicitly. 8.19. Find the constraint matrix K which will transform the primitive sequence currents into LL fault currents where I�2 is to be found explicitly and where the faulted phases are (a) b and c, (b) a and c, and (c) a and b . 8.20. Repeat problem 8.5, using Kron's analytical technique and Tables 8 . 2 and 8.3. 8.21. Repeat problem 8.6, using Kron's analytical technique and Tables 8.2 and 8.3. 8.22. Repeat problem 8.7, using Kron's analytical technique and Tables 8.2 and 8.3. 8.23. Repeat problem 8.8, using Kron's analytical technique and Tables 8.2 and 8.3.
chapter
9
S i m u lta n e o u s Fa u l ts
One oftwotheormostmoredifficul ts problems in thesimultaneously. solution of faultSuch ed networks is that involving faul t whi c h occur an occurrence may be which the resulforces t of more some than eventonesuchfaulastaatstroke of location lightning. orOna theman-caused ac cident, a single other hand, aconditions simultaneous fault(orsituation may arise fromUsual the lchance occurrence of twofaulfaultst at two more) remote points. y onl y two simultaneous aretwoconsidered. Thisfaulistsa ipracti ceallloiwmitation because the jointas probabil ity ofofeventhe simul t aneous s qui t si n ce it i s computed the product twotsingl e-event probabilities. In a welweather l-designed circuitoccurtheonly likelihood of a single faul i s fai r l y remote even in stormy and may once every yearsInperthe100casemilesof two of line.simultaneous faults there are four cases of interest. 5-10 For faults occurring at points A and B these four cases are: AA shunt shunt faul faultt atatAA and and aa shunt faultt atB. at B. seri e s faul A seri serieess faul faultt atat AA and and aa seri shuntes fault fault atat Actual canrequire viewexactly this as theonlysamethreecomputationfaulscheme. t configurations since situ ationsWelywil,andwel develop achmethod for computing twoth thesimultaneous faultsnoted by twofor methods, each of whi wi l l permit us to deal wi three si t uations common fault gconfigurations. Thesecond first method is oriented toward the general ized network confi urati o ns, and the i s an extensi o n of Kron' s analyti c al tech nique. Both methods are valuable but either will suffice. 1. 2.
B. B.
3. A
4.
differen t
2
4
I. 9. 1
SIMU LTAN EOUS FAU LTS BY TWO·POR T N ETWOR K THEORY
Two-Port Networks
Beforein aproceeding toofa two-port consideration of thetheory. simultaneous fault problem we engage brief review network This wi l l enable us to lay thetroduce groundwork for subsequent applications ofnstheto two-port methodOurandtreatment will in the two-port conventi o ns and notati o be used later. subjectreferences will be brief,on theandsubject the interested reader isanadvi[5]sedortoGuiconsult some of theof theAexcellent such as Huesl m l e mi n [62]. network, as theontermthe isnetwork to be used here, is oneWewitmake h twoapaispecial rs of termintwo-port als emerging as shown of Figure 308
9. 1 .
309
S i m u ltaneous F a u lts
Fig. 9 . 1
A two-port network.
distinction between a "port" and a "pair of terminals" by requiring that in a port the same current must leave one of tl�e terminals (the bottom terminal) as enters the other (the top terminal). This is not a restriction on the network itself but on the external connections made the network. Thus we must always be certain that no matter how complicated the connections among a group of two-port net works, this current requirement is always preserved. Often we guarantee this by placing 1 : 1 transformers on the ports before making any external connections (see [5] for a discussion of this technique). We note that two-port networks are sometimes called "four-terminal networks" or "two terminal-pair networks," but we reject these names being inadequate for the reasons discussed above. Passive two-port networks are commonly specified in terms of the network p eters defined in Table 9. 1. Huelsman [5] reviews the method for finding all p eters by appropriate tests on the networks. In every case the tests consist of placing appropriate terminations on the ports, preserving the port current cri teria and leaving the network unchanged test conditions changed. That is, if a port is shorted for the first test, it must be shorted for subsequent tests as well. The port terminations used are reviewed in Table 9. 2. Note especially the to
as
aram
aram
are
as
Table 9. 1. Two-Port Network Parameters Equation
Designa tion
Z (impedance) parameters
Y (admittance) parameters
H (hybrid)
parameters*
G (inverse hybrid) parameters A (transmission) parameterst
=
=
=
C h a pter 9
310
Table 9. 2. Port Terminations for Parameter Determination
Type
Description
1
voltage source
2
current source
3
short circuit
4
open circuit
Diagram
I
Termination Impedance
zero
� , flj � jj
infinite
I
zero
1: = 0
infinite
specified tennination impedance. Thus if a port has a current source attached for one test, its impedance is infinite, and either a current source or an open circuit must be used at that port in subsequent tests. An example will illustrate the pro cedure.
Example
9. 1
Compute the Z parameters for the network shown in Figure 9.2. R
R
Xz
Fig. 9.2. Network for which Z parameters are required.
Solu tion
It is apparent from the way the Z parameters are defined in Table 9.1,
that (9. 1 ) and that Z1 2 = ( VdI2 )I, =O
and Z22 = ( V2/12)I, =O
(9.2)
Note that conditions (9.1 ) require that port 2 be open (12 = 0) and conditions ( 9. 2) require that port 1 be open (1. = 0). This explains why the Z parameters are sometimes called the "open circuit Z parameters. " That is, these p aram eters are detennined with both ports tenninated in an infinite impedance.
31 1
S i m u ltaneous F a u lts
:It - I
_
IA
1
+VI
R
R R
R
�
I.m
R
I�O -
I.2"' 1 -
trr f: t
V2+
��.
,-
IA
( b)
(0 )
Fig. 9 . 3 . Network test conditions for Z parameters : (a) network condition for ( 9 . 1 ), ( b ) net work condition for ( 9 . 2 ).
To satisfy ( 9.1), we visualize the network as shown in Figure 9.3a, where we arbitrarily set II = 1 ampere ( or 1 pu). Then from ( 9. 1 ) with = 1 the voltage VI is numerically equal to % 1 1 or 2 % 1 1 = VI = R + 2 R /3R = 5R/3
II
and , similarly , %2 1 = V2 = RIm = R (1 /3) In a similar way we use the network of Figure 9 .3b to compute the second column of the Z matrix. Here we set 12 = 1 arbitrarily and compute % 1 2 = VI = RIn = R ( 1 /3) and 2 %22 = V2 = R + 2R /3R = 5R /3 Thus we have the Z matrix
Z=
r
R 3
R 3
which we observe is symmetric. This is always the case in a reciprocal network and will always be the case in a power system. In complicated networks it is often helpful to sketch figures similar to Figure 9. 3 as an aid in computation and as a visual check that the port impedance is the same for both tests. All two-port network descriptions given in Table 9. 1 show the relationship be tween pairs of variables. In each case there are two equations in two unknowns. Thus it is apparent that the equations can be rearranged to determine unique rela tionships among the various port parameter sets. These relationships are given in Table 9. 3. Note that some sets of parameters may not exist for a given network. For example , if %2 2 = 0 (see Table 9 .3, row H, column Z), the H matrix for that network is infinite in all its terms and does not exist. The use of Table 9. 3 will be illustrated by means of an example.
Example ple
9. 2
Compute the Y and H parameters from the Z parameters found in Exam
9.1.
Table 9. 3.
z
Y
Zll
Z 12
z2l
Z 22
Y 22 -det
Z
Z
n --
Y
H
det Z
Z
Z 22
Yll
1
Z 22
Zll Z 21 Zll Z ll -
_
_
det
Y21
det
hll
h l2
h 21
h 22
Y
Y22
det
21 - Y -
-
1
- --
Y 22
Y 22
det
H
22 - Y -
-
det
H
det Z
ZII det Z
- --
1
Y2I
Y21
Y --
Yll
Z 21
Y21
Yu
det Z
YII
- --
Z I2
ZI2
1
ZII Z 12
-
det
Y I2 det
h
Y
- --
Yu
H
h 21
- --
hZI _
hn
h 21
1
1
-
-
Y I2
h l2
h 22
Y 22 Y I2 -
H
hu
h 12
gi l
A 21
A21
1
H
!!.!!
A 22
g22
g22
A I2
- gu -
-
-
ll - -
g22
Au
A I2
A 12
--
A
al2
A 22
A 22
all
al l
1
gl2
gi l
g2 1
det G
gl l de t G
gl2
g22
_
g22 g2 1
1
gi l
h 21
g21
det
H
- --
h 12
Ci 2 1
gl 2
det G
h l2
Ci l l
Ci 2 1
det G
--
gu
hll
det Ci
A 21
1
-
Ci 21
A 2l
-
h21
Ci 2 1
gl l
det G
hll
-
gl l
g2 1 - --
--
A
1
Ci
22 - -
22 - -
det G
det H
det
- --
-
g22
h
det
A ll -
--
g22
12 - --
A
l2 - g det G
g21
--
Y22
--
-
1
h hll
h 21 -
Yll
Y
-
l2 - -
Zl l
det
gl l
hll
--
Yll
h 22
1
hll
Yll
-
h 22
-
Y I2
1
h
12 -
Ci
A
G
hn
22 --
_ Z 22
-
Y
Y I2
1 -
Z I2
h
21 - -
--
Z21
Z 22
h 22
Y
Z12
Z2 1
Z 21
a
1
Z 22
-
det
H
--
Yll --
Y 22
-
Z 22
det
Yu
Z I2
det
Y I2 - --
Y I2
det Z
Z21
H
YII
--
1
A
Z
det Z
Y
det
det Z
det Z
-
G
Y2l
Z
lI --
Y
- --
I2 - --
u - --
_
Relationships among Two- Port Parameters
det G
- --
gl2 gl l g1 2
- --
gu g22 gl2
1
-
g12
A
det
- --
A I2
1
A
det
--
-
1
all
-
Ci 1 2
al2
--
det Ci
a 22
a12
Ci l 2
1
-
1
A 21
det <1
<1 u
A 22
A 22
<1 1 1
Ci l l
-
A 21
Ci 2 1
All
a 22
a 22
det <1
<1 1 2 a 22
All 1
_
-
A
--
det
A I2
All
A I2
Au
A 22
A 22 det
_
A
Au de t
--
All
A ll
A
(1 22 <1 22 det Ci <1 2 1 det a
A 12
A
det
AI1 det
- ---� - --
1
-
A -
_
al2 det <1
all det a
<1 1 1
a l2
<1 2 1
a 22 ----
313
S i m u ltaneous F a u lts
Solution
z = [; � � _� ��1 _ �� �:l : z :i�J J�� 5�J:1
From Example 9.1 we have
Then we compute det Z
. 1 [
y - -det Z
=
( 2 5R 2 /9) - (R 2 /9)
Z 22
=
- Z2 1
We also compute H=
9 24R 2
(24R 2 /9) and from Table 9.3
=
R 3
3
8R
--
-
--
R 3
5R 3
Z 2�
>22
8R
8R
-
5
8R
5
which can be checked by computing again from the Y parameters.
Two-port networks with internal sources. As a final consideration in the de scription of two-port networks we consider the effect of internal sources. Internal sources can be of two kinds, dependent or independent. Dependent or controlled sources develop an output voltage or current which depends on some other net work voltage or current . For example , the d axis voltage in a synchronous ma chine depends upon the q axis current. Controlled sources affect the matrix description of the two-port network by altering one or more of the elements. Usually if the matrix is initially symmetric, as with the Z or Y matrix of a passive network, the controlled source causes the matrix to be unsymmetric. This will be illustrated by means of an example.
Example
1=
9. 3
The two-port network shown in Figure 9.4 is identical with the network of Example 9 .1 except that a controlled current source k V 1 has been added. Find the Z parameters of this two-port network.
Solu tion
We solve the network by node voltages, applying test conditions as before. In doing so, note that any impedance in series with a current source is negligible com-
�R VI -
Fig. 9 . 4 .
i
R R
�
(Dtz kvl
'14 "
R R
I2 -
t
+ 2
V
--+--�--...---' --,-_ -
_
Two·port network with controlled source.
314
C h a pter 9
pared to the infinite internal impedance of the source. Thus with II 12 we have the nodal equations
=
0
2 V3 -
1 2 = 1 + k Vh V3 + R V4 = 0 R R R and we note that VI = R + V3, V2 = V4 • Solving, we find that 2R ( 1 + kR) V3 = 3 - 2kR 1
V4
and
=
1 ampere,
-
V3 R (1 + kR) 4 = 2 = 3 - 2kR
v
Then
Zl l
--= VI. = R + V3 = -3--5R 2kR
Z2 1
kR) = V2 = V4 = R3(1- +2kR
and
Similarly, with 12
= 1, II
2 V3
R
=
1
0
we write
from which we find that
R - kR) 4 = 3(2- 2kR
V3 = 3 - R2kR '
Then
Z 12
and
Z22
Finally then, we write
- R-1 V3 + R-2 V4 = 1
= k V, = k V3, R V4
-
=
v
= V I = V3 = 3 R2kR _
- 3_k_R�) V2 = V4 + R = R�3( 5__2kR _
[ 3 - 2kR R
R 1 ( 1 + kR) R (5 - 3kR )J which reduces to the previous result when k = O. We also note that if the network is not connected to any external active network, all currents and voltages are zero. This is because the dependent source 1 = , VI depends upon the existence of a Z=
1
5R
voltage VI .
The presence of a dependent source changes the two-port network descrip-
S i m u ltaneous F a u lts
31 5
tion . As seen in Example 9.3, the source parameter k is present in all matrix elements. We now consider an independen t source located within the two-port network. Such a source will hold its output to a specified value irrespective of external connections to the two-port network. Indeed, if the two-port network is completely isolated, the internal independent source will cause measurable open circuit voltages or short circuit currents to appear at the two ports. Thus for in dependent sources within the network we write in the case of Z parameters,
and we interpret the equation
as
follows.
(9.3)
1 . All port voltages and currents are defined as before. 2. When 1 1 = 12 = 0, the port voltages are
(9.4)
and these open circuit voltages are due to the internal source or sources. 3. We may determine the Z matrix of (9. 3) by removing the internal indepen dent sources where we are careful to (a) short out all independent voltage sources and (b) open all independent current sources. Having done this, equation (9.3) reduces to (9. 5)
In a similar way we can establish equations similar to (9.3) for networks de scribed by any other set of two-port parameters. In general we must add to each equation a source terin of the same dimension as that equation, and we determine the size of the source term by testing at the appropriate terminal with all ports properly terminated. The "proper" termination depends upon the equations but will be either short or open circuits. Mathematically, we write
U = P W + Us
where
(9.6)
U, W are vectors containing port voltages and/or currents P is the 2 X 2 matrix of two-port parameters Us is the independent source term, each element having the same dimension as U
Then we find Us by the test
U = Us when W = 0
(9. 7 )
U = P W when Us = 0
(9.8)
and we compute P by the test
In (9.8) we set Us = 0 by properly removing the internal sources. We note also that test (9. 7) will ordinarily be impossible for transmission p arameters (A or (1) since it is not usually possible to set 12 = V2 = 0 simultaneously. Thus a two-port description including the effect of an independent source is limited to the Z , Y ,
C h a pter 9
31 6
Desigrwtion
Table 9.4-
Equations of Two-Port Networks with Independent Sources
[ ] � [z u [ ]_ [YU [ ]_ [hu [ II ]_ ['U
] [II ] ]� ] ] el ] ] [VI ]
Equation
[Vd ] [IYI ] [ Vhl] Ih2 [ ]
Test for Source Term
:0: 11 Y 12 VI Y C[JJ 12 IY2 Y21 Y22 V2 h 12 VI :[JJ h21 h22 V2 12 112 11 G co: V2 121 122 12 Vg2 H, and G descriptions. These are tabulated in Table 9.4. The computation of the independent source term is illustrated by an example. le Examp Consider the current source in Figure 9.4 to be an independent source Is and compute the independent source term associated with the Z parameter description of the two-port network. Solution As indicated in Table 9.4, to determine Vz I and Vz2 we test the two-port net work with II = 12 = 0, or with both ports open. Then we compute VI and V , which result from the presence of current source I,. By inspection of Figure 9.42 it is apparent that the source current sees two paths, R (to the left) and (to the right). This divides the current inversely the impedance II 12 VI V2
Z
H
Z21
Z 12 Z22
12
+
+
+
+
Vz 1
6
9. 4
h = ( 2 / 3) I"
as
18 = ( 1 / 3 ) Is
or, with
=
2R
= 0,
and we compute, with II = 0 and 12 = simultaneously, 0
[�::] [�:] [��:] ���:� ��:] =
=
=
Thus the complete description for the two-port network with independent current source I, is
The two-port parameters are related to one another in a definite way as noted in Table 9.3. It is not surprising, therefore, that the independent source terms of Table 9.4 are also related. If we write the Z and Y parameter equations matrix form, we have = Z I + Vz (9.9) in
v
317
S i m u ltaneous F a u lts
1 =
For convenience we also write and where we arbitrarily define
(9 . 1 0 )
Y V + Iy
M = H N + Mh
( 9 .1 1 )
N = G M + Ng
(9 . 1 2 )
(9 .1 3 )
and (9 . 1 4 )
manipulation reveals the relationships among the Then straightforws.ardForalgebraic we write example, source parameter vZ = U V Z =
�
y - l IY =
-
o
(9 . 1 5 )
Table 9. 5. Relationship among Source Parameters'"
[ ]
1
0
0
1
Z 22
Z l2
det Z
det Z
VZ I Vz = Vz 2
[ ]
IY I Iy = IY2
[ ]
Z 21
1
[�:]
0
-
-
1
-
Zl l Z21 Zl 1
Y l2
det Y
det Y
Y21
Yll
det Y
det Y
1
0
0
1
Zll
-
-
Z 12 Z 22
--
1
-
1
-
-
-
Yll Y21
-
Z 22
Yll
0
1
-
I
0
-
-
* ( Vector in first column )
=
1
-
0
-
1
-
-
_
--
det Z
det Z
Vh l Mh = Ih 2
Ng =
-
Y22
hll h 21 hl l
h 12 h 22 1
gi l
0
1
1
0
1
0
I
0
1
Y12 Y22
h 22
h 12
det H
det H
1
h 21
hll
Y22
det H
det H
-
0
gl l S21
h 22
-
0
-
1
--
_ g22 det G
1
-
-
gl2 g22 1
-
g22
� det G
g21
gl l
det G
det G
1
0
0
1
(matrix from table) X (vector above table entry ).
C h a pter 9
318
where = the unit matrix. Exploring all possible combinations, we may develop the matrices of Table 9.5. We illustrate the use of Table 9.5 by means of an ex ample. U
Example Use
9. 5
the value of V. computed in Example 9.4 and the Z parameters found in Example 9.1 to determine the H parameter equation for the active network of Example 9.4. Solution From Example 9.4 we have v •
and from Example 9.1
3
RI.
[2] 1
[3 51 1]5 parameter description so, using Table 9 3 we have Z1111 l.. fde t Z11 L- Z1 1 1 J 2 (25R /9) - (R1 /9) 2 4 R /9, we compute Z=
We desire the H
=
R
.
,
Z
H=
and since det Z =
=
H-
[ : :J
Also from Table 9.5 we have Mh =
1
o
c
�ils 5I:ld
-
1
--
%12
r� -�3 Jl 5R -
V
[! ] ��1�1] Ie
RI. 3 RI.
=
5
These results may be easily verified by computing the H equation directly. The result is
319
S i m u ltaneous F a u lts
I nterconnection of Two-Port Networks
9.2
One of the principal reasons for describing two-port networks so carefully is to simplify the problem of interconnecting two-port networks to form larger, more complicated systems. Furthermore, in doing this we preserve a well-defined notation in the interconnected system. There are several ways two-port networks may be interconnected. These will be examined briefly here, but the interested reader should consult other references such Huelsman [5] on the subject. as
Ixl;- - - - -I2c0I
:1 1
l + v l°
1 r
I 1
I lb
1 +_ -I
1
� I
b
Vl b
-
V2 °+ 1 _1 I
a
1_
v1 +
�I
V2 b
v2+
I
L
r
n
L
_ _ _
_
_
_
.J
_
Series connection of n two-port networks.
Fig. 9 . 5 .
b,
I2
-
Consider a group of two-port networks, labeled a , . . which are connected in series as shown in Figure 9 . 5 where we define n
Series connection. , n, .
Vk = Ik =
and
l
tv::] [ ]
k = a, b ,
Il k 12 k
k = a, b ,
J Then we write for each two-port network Zk =
Zl Ik Z2 1 k
Z I2k Z22 k
Vk = Z k 1k
.
.
,n
.
. . . ,n
k = a, b,
k = a, b , . .
.
.
. ,n
(9 .16) (9 .1 7 )
. ,n
(9.18)
(9.19)
We now seek a description of the interconnected group of two-port networks designated in Figure 9.5 by the terminal symbol and defined by I/>
V=ZI
where v=
[�:1 ,
1=
(9 .20 )
G:J
(9.2 1 )
C h a pter 9
320
and where (9.22) describes the relationship between these "external" quantities. works are in series, we observe that
Since the net (9.23)
and
I = Ia = Ib = . . .
=
From (9.19), (9.20), (9.23 ) , and (9 .24) we write
V
In
(9.24)
Va + V b + . . . + Vn = Za 1a + Z b 1b + . . . + Zn 1n = (Za + Zb + . . . + Zn) I =
(9.25 )
Then from (9 .20) and (9.25) we see that
Z = (Za + Z b + . . . + Zn)
(9.26)
This means that we can easily find the Z parameter description of any number of two-port networks which are connected in series by first finding the Z parameters of the individual two-port networks and adding these together as in (9.26 ) . It must be understood, however, that port descriptions must be preserved in the interconnection . Parallel connection.
. . . , n,
Consider a group of
n
two-port networks, labeled
b, connected in parallel as shown in Figure 9 .6. We again define the port voltage and current vectors as in (9.16 ) and (9. 1 7 ) respectively. For the parallel connection , however, we do not use the Z parameter description but instead
a,
define
l k = a, b, . . J r--- - - - - - - - - -1 Y 1 2k YZ2k
.,n
12 0
I Ia
-
Vlo
I
VI
I I I
M
+
I
I I I I
J L
I lb
--
A..
+
..
...!!n� In
_
_
12b
-
Y2b +
z
,
+
_
�2 0 +
b
V lb
I I-- 2 n +
n
_
J
---
a
+
_
I I
I
_
_
I2
I --A- -
�
V2 +
I I I
V2 n
_
(9.27 )
_
I
..J
Fig. 9 . 6 . Parallel connection of two-port networks.
S i m u ltaneous F a u lts
32 1
such that we may write for any two-port network, k = a, b, . . . , n
I k = Yk Vk
(9.28)
If we again define voltage and current vectors as ( 9 .2 1 ) for the cf> -¢ network, we may write
I=YV where
Y=
fY I l 11 2 1
]
Y12
Y 22
(9.29) (9 .30)
describes the relationship between V and I . For the parallel connection we ob serve that
(9.31 )
and
V = Va = V b = " ' = Vn Combining these last two equations with (9.28 ), we write
I = Ia + Ib + . . . + In = Ya Va + Y b V b + . . . + Y n Vn = (Ya + Y b + · · · + Yn) V
Comparing this result with (9.29), we see that
Y = Ya + Yb + . . . + Y n
(9.32)
(9.33) (9 .34)
and the two-port description for this connection is easily found by adding the Y parameters of the individual networks. We again require that two-port descrip tions are preserved in the interconnection. Note that it is not convenient to find a Z parameter description for the parallel connection. It is preferable to first con vert the Z parameters of the individual two-port networks to Y parameters and add these as in (9 .34 ) . H ybrid connection. Consider a group o f n two-port networks, labeled a, b, . . . , n , connected in series at one port and in parallel at the other as shown in Figure 9.7 . We refer to this as the hy brid connection. Here, instead of the voltage and current parameters of ( 9 . 1 6 ) and (9. 1 7 ) , we define the port hybrid voltage-current vectors
and the hybrid matrix
Hk =
[
h l lk h 2 1k
k = a, b, . . . , n
(9.35)
k = a, b, . . . , n
(9.36)
k = a, b, . . . , n
(9.37)
Then from Table 9.1 we write
Mk = Hk Nk
k = a, b, . . . , n
(9.38)
C h a pter 9
322
--1I
.1
r- - - - -I 10 -
� I
�
Vlo
I
I
j
F--
I J J
l ib
-
tYlb
J
I I
� In +
oJ '
,
L
b
r �
I
""'
a
+
I
_
_
_
_
I20 +
_
I
12b
I
I
_
,
z
-
I2 n
'{ 2n +
_
_
_
_
I2
�
W
V 2 b j1J
n _
V2 0
I I
2
I
_
_
I J
Fig. 9 . 7 . Hybrid (series-parallel ) connection of two-port networks.
We now describe the interconnected system according to the equation M=HN
where M and N are defined according to "external" quantities,
(9.39) (9.40)
and (9.41) and where
�
J is the hybrid parameter matrix for the interconnection. H=
From Figure 9.7 we observe that
and
hll
h 11.
h2 1
h 22
M = Ma + M b + .
. . + Mn
N = Na = Nb = . . . = N n
Combining these observed results with (9.38 ) we write M = Ma + Mb + . . . + Mn = Ha Na + H b Nb + . . . + H n N n = (Ha + H b + . . . + H n) N Then comparing (9.45) and (9.39), we have H Ha + H b + . . . + H n Again we require that port descriptions be preserved. =
(9.42 ) (9.43) (9.44) (9.45) (9.46)
S i m u ltaneous Fau lts
323
Note that a similar result could be obtained by reversing the series-parallel connection of Figure 9.7 form a parallel-series connection. In such a case= the G parameters will be found to add and a description corresponding to N G M derived. This description is redundant in a sense since we are at liberty to number the ports any way we choose, and if we would exchange the 1 and 2 subscripts, we would have a parallel-series (G matrix) hybrid. Thus, the hybrid connection of Figure 9.7 will suffice for either situation requiring a series and a parallel connection. Cascade connection. Finally, we review the cascade connection of two-port networks as shown in Figure 9.8 where we define [VlIl k] [AA2111kk AAI222�] [V122k] � k = a, b, . . . , n (9.47) k k k to
=
�o �
II- 1 1 0
-
VI - Vi a
+
+
b
+
V2a Vl b
a
Fig . 9 . 8 .
We seek a description
r
b +
+
V2 b �In -
n
-
+
V2n- V2
Cascade connection of two-port networks.
] [V2] [VIII] [A2A 11l A12 Au 12
(9.48) which describes the connection. If we observe the constraints of the network interconnection, we write [V� [Vial , [_Vla] [VlblJ , . . . (9.49) Id IlaJ Ila Il b Combining with (9.47), we have =
=
=
1 = [�1] ��:: ���::J ��::J [��:: ��:::] [� �:: ���::] [-�::] fA 11a -AI2aJ [A11b �A 1 2b� ' " rA 11n �Alln] [_ VlJ LA 21a -Ana A21b A 2lbJ LA lln A21n 12 (9.50) =
=
Comparing (9. 50) with (9.48), we see that the composite A matrix is the product of the individual A matrices with the second column of each changed in sign. These negative signs arise because the currents between two-port networks flow in opposite directions. Using the ABeD parameter description, with I directed in of the two-port network, the cascade connection is simply a matrix I with no sign changes. We define the currents as entering the network conform with established usage [5, 63] .
and
l
OUt
product
9.3 Simultaneous Fault Connection of Sequence Networks
I
to
In Chapter 8 we developed the interconnection of sequence networks for common shunt and series faults. These results, summarized in Figure 8.9, show all
C h a pter 9
324
the sequence networks as one-port networks, with the current always entering the network at K and leaving at F. We further defined K to be the zero potential bus N for shunt faults and the current-receiving terminal M for series faults. (See Figures 8.3, 8.4, 8.6, and 8.7.) The point F is always defined to be the "fault point." We note also that the positive sequence network is an active one-port network since it contains independent sources, but the negative and zero sequence networks are passive one-port networks. Suppose that two faults occur simultaneously on a system. Using the two port network theory developed in sections 9.1 and 9 .2, we can represent the two faults by finding the two-port parameters of the positive, negative, and zero sequence networks and then combining them in a way which will properly repre sent the common fault configurations of Figure 8.9 [63] . Considering the four common types of faults which can occur at two different locations, there are a total of 16 possible problems which require our attention. These are summarized in Table 9 .6. Since it is immaterial whether a given fault point is represented as the left or the right port, one could eliminate the G matrix and always use the H matrix for hybrid combinations. We will do this in our analysis, but this restric tion is not necessary in practice. Table 9.6. Two-Port Fault Combinations R igh t Port (primed) Shun t fault (K
Fault Type Kind
(K = N)
Shunt fault Left Port (unprimed)
(K = M)
Series fault
SLG
=
2LG
N)
Series fault (K
l LO
=
M)
2LO
(series)
(parallel)
(parallel)
(series)
ZNN
HNN
HNM
ZNM
(parallel)
GNN
YNN
YNM
GNM
lLO (parallel)
GMN
YMN
YMM
GMM
2LO (series)
ZMN
HMN
HMM
ZMM
SLG (series) 2LG
Reference to Figure 8.9, which shows the various port connections, reveals the several pertinent facts concerning fault analysis. First, note that each se quence network is always terminated in an ideal transformer so that port currents and voltages are preserved for any network interconnection. This means that the two-port network theory of sections 9.1 and 9.2 will always apply. Note also that the added impedances Z and ZII ' used to represent impedance at the fault, are considered here to be a part of the network and must be added to the usual Thevenin equivalent sequence impedance. We also recognize that for simulta neous faults the sequence networks will have two fault points and two-port networks will be defined for each sequence network, with points F and K being defined according to the fault location and type. The simultaneous fault problems to be solved can be generalized by three forms: (1) a series-series fault, (2) a shunt-shunt fault, and (3) a series-shunt fault, where the terms "series" and "shunt" refer to the type of network inter connection shown in Figure 8.9. If we view the sequence network interconnec-
325
S i m u lt a neous F a u lts
-
��i.t+
-
K'i Ki S EQU E N C E 2- P O R T N E T W ORK
_ I: n
VK ' i
��
- �. -
Fig. 9.9 . Two-port network representation for simultaneous faults at F and
F '.
tion from the external 1/>-1/> connections, the network is shown in Figure 9.9 where the network voltages and currents are defined for simultaneous faults at F and F '. (Note that there should be no confusion between II and 12 as defined in Figure 9.9 and the sequence currents Ial and lal)' This figure will be used to define the current directions and voltage polarities to be considered in subsequent develop ment. The transformations shown are composites of the actual transformers or phase shifters required for various faults. Since these phase shifters are ideal, we .define the transformation ratios (9.51) These quantities will always b e subscripted with i equal to 0, 1, or 2 for the vari ous sequences and will take a value given in Figure 8.9, for a particular fault type and symmetry. 9.4
Series-Series Connection ( Z-Type Fau lts)
A series-series connection of two-port sequence networks is required to represent (see Table 9.6): 1. Simultaneous SLG faults at F and F ' (ZNN ) . 2 . A SLG fault at F and two lines open at F ' (ZNM )' 3. Two lines open at F and a SLG fault at F ' ( ZMN )' 4. Two lines open at F and two lines open at F ' ( ZMM ) ' The sequence network connection is shown in Figure 9.10. For the positive
sequence network we write
(
where
is
the independent source term viewed from the Kl - K' I transformers) . But from with i =
nKI =
(9.51)
V l l l VK l
Premultiplying (9.52) by
= I I l IIK 1 '
1
9.5 2 )
terminals (inside the (9.53)
C h a pter 9
326
�
I KO
"KO"
KC> K'
-
+
V KO
VK'O
0
I
K'I
KI
+
V, = O
+
K' ,
VK' ,
I
..1 2
' : " K',
+
V2 a O
--t-4K2 K' 2 e---t-��
__
2
-----.
-
0, 1 . 2
Fig. 9 . 1 0 .
Sequence network connection for simultaneous Z-type faults.
we get For the negative sequence network we write
(9.55)
which we premultiply by to obtain And
rV1 21 LV22J
=
(9.54)
r
(nKl /nK'l) Z12(2 )! ZI 1(l) knK '2 /nK2 ) Z2 1 (2) Z2 2 (2) J
for the zero sequence network we write rvKO
]
=
] [IKO Z 22(0) IK,oJ
rz 1 1(0) Z 1 2(0)
LVK 'O LZ2 1(0) which transforms (since nKO = nK'O = 1 ) to
(9.56)
(9.57 )
(9.58)
327
S i m u ltaneous Fau lts
But from Figure 9 .10 we observe that for a series-series connection, (9.59 ) and (9.60 ) Performing the addition indicated in (9 .59 ) and making the substitution (9.60), we write (9.6 1 ) where Z I 1 = Z I 1 (O) + Z l 1 ( l ) + Z l 1 (Z ) Z I2 = Z IZ(O) + ( n K ! l n K'l) Z IZ( l ) + ( n KZ/ n K'Z ) Z 1 2 (Z) ZZ l = ZZ l (O) + ( n K'J !n K l) ZZl(l) + ( n K'Z/ n KZ) ZZl(Z) ZZZ = ZZZ(O) + ZZZ( l ) + ZZZ(Z)
(9.62)
Equation (9.6 1 ) may be written as V = O = Z I + Vz
(9.6 3)
I = - Z - I Vz
(9.64)
Then
is the solution of the series-series connection . Knowing this current, we know all currents in the outside transformer windings, and we may compute all voltages from (9 .54 ), (9 .56 ), and (9 .58) . Then the individual sequence network currents and voltages may be found by applying Kirchhoff's laws to the individual net works. We may write out the solution (9.63 ) as
[�:J
=
:� Z
d
[-::: -:::J [::� �J
(9 .65)
I
In this form, we see that the current at each fault depends upon the independent source voltages at both fault locations . We illustrate this method by means of an example .
Example 9. 6 A simple power system is shown in Figure 9 . 1 1 with simultaneous faults in dicated by X's at fault points F and F ' . The following system data is known : Generator L : Z� = Zz = jO.12 , Zo = j O.10, EL = 1 .1{30° Generator R : Z� = Zz = jO.15, Zo = jO.13, ER Transformer AB : Zl = Zz = Zo = jO.10
Transmission line Be : Zl = Zz = j O .5 0 , Zo = j 1 .00 Transformer CD : Zl
=
Z2
=
Zo
=
jO.12
=
1 .0�
Chapter 9
328
Fig. 9 . 1 1 . Power system for Example 9.6.
We assume that faults are applied as follows : Fault F: lines a and b open Fault F ' : SLG fault on phase b Solve the system for the currents and voltages at the circuit breaker positions at B and C . Assume that all fault impedances are zero .
Solution First, we note that the two faults specified require a series connection of se quence networks. Thus a series-series two-port description is applicable. Refer ring to Table 9 .6 , we must determine the Z parameters ZM N for each sequence network . The sequence networks are sketched in Figure 9 .1 2, where it is noted that the series fault requires the insertion of an open section adjacent to breaker B . We arbitrarily define the current as positive in the B C directiol\ (IK) since this would constitute positive tripping current for that breaker, as noted at F. We also iden tify the positive tripping current direction for breaker C as away from bus C and toward the faulted line. To solve the system, we must first find the Z parameters. We easily compute the following.
[VKOJ VK'O
=
r j 1 . 35 L- jO.25 F
I I«)
] [IKO ] IK,o
- jO.25 jO.25
F'
I K,O ,...-----f'-K ' 0 a NO
'--":"'+-_ F' O
'--"-_ F ' I
2
I K' ,..-----.,...-- K ' 2 < N 2
VK'2 .
C
B
_
��- F' 2
Fig. 9 . 1 2. Sequence two-port networks for Example 9 . 6 .
�VVK'Ktl] �VKVK2]'2
] rKt[K't] [ lK2]2
]
S i m u ltaneous Fau lts
[ jO.99 -jO.27 jO.99 [-jO.27 =
329
0.047 - jO.550 -jO.27 jO.27 - 1.0L!L -jO.27] jO.27 IK Fault symmetry is the next consideration. From Figure 8.9 we confirm the following: Fault two lines open on a and b nKO = 1, nKl a, nK2 = a2 Fault F ' : SLG fault on b =
+
'
F:
=
The impedance matrix is then computed from (9.62) with the result [ j3.33 jO.02 ] Z= jO.02 jO.79 Finally, from (9.65) we compute [It] det- 1 Z [ (jO.79) aVK1 - (jO.02 ) VK' l ] (j0.02) a VK t + (j3.33) VK' t 12 [- 0.088 + jO.1221] [0.1589/123.76 ] - 1.094 + jO.630 1.2622/150.08 and we note that
aa22
=
-
=
=
From (9.51) we write =
Currents and voltages at breaker laO = IKo = 0.1615{123.75° B:
lat IKl = a2 laO la2 IK2 alaO la laO lal2 la2 0 Ib laO a /at a/a2 3 ao 0.477{123.76° =
=
=
+
=
=
Ie
=
+
1
=
+
=
+
=
°
[0.1589/3.76° ] 1.262/270.08°
330
Chapter 9
VaO = IKO = 0.0134 + jO.00897 Va l = 1. / 3 ° - (j0. 22) IK I = 0.954 + jO. 514 Va2 = (j0. 22) IK2 = 0.0381 +
-
(jO.10) 1 0
jO.0157
-
Va = VaO + Va l + Va2 = 1.08/29.9° Vb = VaO + a 2 Va l + a Va2 = 1.11/269. 1° Vc = VaO + a Va l + a 2 Va 2 = 1.06/145. 8°
Currents and voltages at breaker C:
laO = IK,o - IKO
Then
101 = IK't 10 2 = IK '2
- IK I
- IK 2
= = =
jO.630 0.002 - j1.262 1 .092 jO.633
- 1.094 + -
+
r��:] L�r� �2 �J lL �:1.092��: � jO.63�J ��:::�l = [�. ;�;150.1� jO J =
c
We also compute
a2
a
VaO = - laO Va l =
0+
+
(j0.25) = 0.124 + jO.251 1.0 � lal (j0.27) = 0.656 jO.042 Va2 = la (j0. 27) = 0.209 - jO.314 2
[ ]=
+
-
-
so that at breaker C
Va Vb
Yc
9.5
[0.990/- 1.160] AVO l 2 = 0 jO 0.990/128.5° +
Parallel-Parallel Connection (V-Type Faults)
A parallel-parallel connection of two-port networks is required to represent the following simultaneous fault conditions (see Table ' 1. Simultaneous faults at F and F (YNN )' A fault at F and one line open at F ' ( YNM )' One line open at F and a fault at F ' ( YM N ). 4. One line open at F and one line open at F ' ( YMM )'
2. 2LG 3.
9.6):
2LG
2LG
The sequence network connection with parallel-parallel termination is shown in Figure where the ideal transformations are phase shifters with voltage and current relations as given by We write the two-port equations for the three sequence networks as follows. For the positive sequence network
9.13
(9.51).
[ �::. J = �::::: �:::::] [�::.] [�::] +
(9.66)
where I y is the independent source term viewed from the K1 and K ' l ports. Simi-
33 1
S i m u lt a n e o u s F a u lts
-
r VKO
X.I
K' O
KO
-
-
- I 'n
K'2
VK' 2��
2
:� �: 1:
-
-
K' I
K2
��V K2
I'
VK'O:�
0
-,-�" KI
'1
-
12
0
-
0
...__ ' __ ' _ 2 ....
O I
Fig. 9 . 1 3 . Sequence network connection for simultaneous V-type faults.
larly , for the passive sequence networks we have
[
IK21 IK, J
Y 1 1 (O ) Y2 1 (O )
Y12(O� Y22 ( O ) J
Y 1 1 (2 ) Y2 l (2 )
Y 12 (2 ) Y22 (2 )
[ [
=
][
(9.67 )
J
VK2 VK'2
(9 .68)
Using (9.51 ) for each sequence, we reflect the two-port expressions across the transformations to obtain new two-port equations which include the transformers. This is done by premultiplying (9 .66)-(9 .68) by
for i
[�
= 1 , 0 , and 2 respectively.
[
J
/l l 12 1
=
[
�
K
J
[ J
The result for the positive sequence network is
Y l l (l ) ' (nK l /nKdY2 1 (1 )
Similarly , since n K O
Ki
(nK dnK ' d Y 1 2 ( l � Y22 ( 1 ) J
Vl l V2 l
+
J
rnK1 /Y1 LnK'l /Y2
= nK'O = 1 in all cases, for the zero sequence we have
B::J [;::::: ;:::::J [�:J
(9 _ 69 )
=
[ J=r L
And for the negative sequence network we compute
/12 122
(9.7 0 )
][ ]
( nK2 /nK '2 ) Y 1 2 (2 ) Y l l (2 ) (nK'2 /n K 2 ) Y21 (2 ) Y22 (2 )
VI 2 V:z:z
(9.7 1 )
C h a pter 9
332
From Figure 9.13 we observe that (9.72)
and
(9.73) Substituting (9.73) into (9.69)-(9.71) and then substituting these results into (9.72) gives (9.74)
where YI I YI2 Y2 1 Y22
= = = =
Y I I(O) + Y I I( I ) + Y I I(2) Y I2(O) + (n Kdn K' d Y I 2( 1 ) + (n K2/n K ' 2 ) Y I 2(2) Y21(O) + (n K d n K' d Y 2 1 ( 1 ) + (n K ' 2 /n K2 ) Y21(2) Y22(O) + Y22( I ) + Y22(2)
In matrix form we write (9.74) as which we solve for V to obtain or
(9.75)
yV+
I.
(9.76)
V = y- I
Is
(9.77)
0=
-
(9.78) In this case we solve for the external port voltages which are easily converted to port voltages inside the transformations. Then, if the sequence network vol!;ages are known, the networks can be completely solved. We illustrate the solution of a parallel-parallel fault condition by means of an example. Example 9. 7
Consider the system of Figure 9.11 for which the sequence impedances and source voltages are given in Example 9.6. Find the phase currents and voltages at the breaker locations for the following fault conditions: Fault F: line b open Fault F' : 2LG fault on b and c Solu tion As in Example 9.6 we have a series fault at F and a shunt fault at F' . There fore, the sequence networks are the same as before and are given by Figure 9.12. Here the similarity in the two problems ends. The symmetry in the fault condi-
333
S i m u lta neous Fau lts
tion of this example is given by the following: Fault F: line b open
nKO =
Fault F' :
1,
n K I = a2 ,
n K2 = a
2LG fault on b and c n K 'O = n K ' 1 = n K'2 = 1
For the sequence networks themselves we need to write current equations similar to (9 .66)-(9 .68). We again use the results of Example 9 .6 to good advantage. From Table we note that the Y matrix is the inverse of the Z matrix, or
9.3
-jO.25] -1 = [ - jO.91 -jO.911 jO.25 -jO.91 -j4.90J jO.99 -jO.27]-1 = [- j1.39 -j l .391 Y I = Z -I I = -j1.39 -j5.09J -jO.27 jO.27 and Y2 = Y 1 • Also , from Table 9.5 we note that -j1.39] [0.047 - jO.55] I = - Z I V = Y I V - [-j1.39 - 1.0 -j1.39 -j5.09 [ +0.764 -j1.32] [1 . 5 28/-60.00° ] = 5.085/- 81 . 36° +0.764 -j5.03 Yo
= Z-0 I =
j1.35 [-jO.25
[
_
I
y
z
z
=
Then we write
� .91 -jO.911 rv l [-j4.90J J] [--JO.91 �1.39 -j1.391 I -31. 3 9 J L J G:�J G��:!: ���:�:J [�:�J (9.74) 11 [ LoJ
flKo LIK' o IIKI l!K' 1
Ko VK' o
0
=
=
- j 5.09
VK I
VK' I
+
11.5 28/- 60.000 L5.085/- 81.36°
=
J
These current equations are reflected across the phase shifters by ( . 6 )and combined according t o with the result 0
=
Yll Y2 1
where from (9.75)
Y l I = Y I l (O) + Y I l (I) + Y l l (2) = =
-jO.91 -j1.39 -j1.39 -j3.69
- .9 1 1 39 a(-j1.39) - jO.91 + j1.39
Y I2 = Y I2(O) + ( n Ktl n K' I ) Y I2(1 ) + ( n K2/nK'2 ) Y I2(2) + a2 (- j . ) + = jO = = +j0.48
9 9 (9.71)
C h a pter 9
334
Y:u == Y21(0) (nK' t !nKI) Y21(1)a2 (nK'2 /nK2) Y21( 2 ) (- = Y22 == Y22(O) Y22(1) Y22(2) = = det = Yll Y22 - Y12YZl Then from ] ][ [VI] = [ V2 ] = [VI 0] [ Vll] = [V12] =[ V20 VZI V2Z From we compute [ VKO ] = [Yto] = r VK, o o L J 1 [VKI ] = ra 1 ] = [ J VK' I L ] [VK2 ] = ra2 ] = rL VK'2 L +
+
- jO.91 + a(- j 1 .39) +
+
+jO.48
j 1 . 39)
+
- j4 .90 - j5 .09 - j 5 .09 =
Y
-
- j 1 5 .08
55.65 + 0.23
-
55.41
(9.78)
- j 1 5.08 1 __ 55.41 - jO.48
- j O .48 - j 3 .69
1 . 528/- 60.000 + 240.00 5.085/- 81.36°
0.41 1 8/96.06°
=
0.3365/- 173.59°
(9.51 )
0.4 1 1 8/96.06°
Vi
O. 3365/- 1 7 3. 59°
Yt Vi I
1
0 . 41 1 8/216.060
0.3365/- 1 7 3 . 59 °
Yt z Viz
0.41 1 8/- 23.94°
0. 3365/- 1 7 3. 59°
Finally then we compute the voltages at F to be (using h = 1 )
[��Jl r�:L� :��:�:�:.�:JJ �L � J [Va'] � [ �J J =
- A
and at F'
.41 1 8/- 25.26°
v" , Vc'
9.6
.235/ 83.930
0. 3365/186.270
- A 0. 3365/1 86.27° 0. 3365/186.27°
Series-Parallel Connection (H-type Faults)
0
=
1 .009
°
0
The third and last connection required to solve for common simultaneous fault conditions is the series-parallel connection shown in Figure 9.14. This con nection may be used for all series-parallel faults and, by reversing the defined left port and right-port definitions, for all parallel-series faults as well. From Table 9.6 the series-parallel H-type fault configurations include: 1 . A SLG fault at F and a 2LG fault at F ' (HNN ) . 2. A SLG fault at F and one line open at F ' (HNM ) . 3. Two lines open at F and a 2LG fault at F ' (HMN) . 4 . Two lines open at F and one line open at F' (HMM ) .
335
S i m u ltaneous F a u lts 110 -
_
ft
KO
'1 1 KO
..-___...
,...... -+.. KO _
I
KI
-
tVKI
K'O
IK' O _
l 'ft K ' O
1 20 _
o
KI
I '
K' f
Kf
-
I'
VK'I � 1
I
�t---it,;�-... '-(I�-+
+
V�
..
__
I22
-
K' 2 �--r\
,-r---+" K 2 2
0, f ,
V2
2
Fig. 9 . 1 4 . Sequence network connection for simultaneous H-type faults.
For these conditions we write the two-port hybrid equations. For the active posi tive sequence network we have
]
h12(1) h22(I)
(9.79)
where the last term is the independent source term viewed from the Kl and K ' 1 ports. For the passive sequence networks we have
and
[VIKoJ�� [h1�1(O)1(O) [VIK��2J [h21(2) hl l(2)
]
=
hI2(O) �2(O)
=
h12(2� rIK � h22(2J LVK 2
( 9 .80 )
]
(9.81 )
Proceeding as before, we now write these equations in terms of the quantities out side the transformers by premultiplying by
for i = 1 , 0 , and 2 respectively . The result for the positive sequence equation is
For the zero sequence network with
nKO = nK'O
rIl ll LV2� =
1,
+
J
rnKl Vh l LnK'l Ih2
(9.82)
] [;::] [�:::: And for the negative sequence network, rVI� [ hl l(2) (nK2 /nK'2) h12{2� LI22J (nK'2/nK2) �1(2) h22(2) J C h a pter 9
336
h12(O) �2(O)
-
(9.B3)
J For the series-parallel connection of Figure 9.14 we observe that =
fI12 LV22
(9.B4) (9.B5)
and
(9.B6) Therefore we may substitute (9.B2)-(9.B4) into (9.B5), making use of (9.B6) write
to
(9.B7)
where
hll h 12 h21 hZ2
In matrix form
= hl l(o) + hl 1(l) + hl1(2) = h12(o) + (nKt lnK'I) hI2(1) + (nK2 /nK'2)h12(2) = h21{o) + (nK't lnKl ) h21(1) + (nK'2 /nK2) h21(2) = h22(O) + h2Z(I) + h22(2)
(9.B7) may be written as
(9.BB)
(9.B9) where N is defined by (9.14), M. by (9.B7), and the matrix H by (9.BB). Solving (9.89), we have
0=
or
HN + M,
[iJ = ;.: [-::; -::: ] [:�. ;:,j
(9.91) Knowing II and V2 , we may solve completely for all port quantities and then for internal network quantities. H
I I . S I M U LTAN EOUS FAU LTS BY MAT R I X TRANSFORMATI ONS
We now develop a second method for computing simultaneous faults which is based on the matrix transformation technique developed for a simple fault in sec tions B.9-B.11. This method is also described briefly by Lewis and Pryce [13] , which is recommended for further reading on the subject.
S i m u ltaneous F a u lts
337
9.7 Constraint Matrix for Z-type Faults
Z-type faults include any combination of SLG shunt faults and series faults with two lines open and are characterized by a series-series connection of the sequence networks. This connection is conveniently described by the two-port Z parameters according to the equations
[VKi ] [Z l I(i) ZI2(i)] VK'i Z2I(i) Z22(i)
G::] [�:] ,
_
where
o
But, by definition
H"
Ii
i = 0, 1 , 2
(9.92)
i
=
0 , =f: 1 1, i = 1
=
i
=
0, 1, 2
(9.93)
and the port equations can be written in terms of the sequence currents . From section S.S we write I Old
= K 1new
and , if we can find a matrix K which constrains I old , the sequence currents, to a series connection then we can use this same constraint matrix to find the new vol tage vector from V new = K *tVold ' For the problem at hand we define
I Old = I
= (9.94)
and write (9.92) in 0-1-2 order as
VKO VK I VK2 VK,o VK' I VK'2
Z 11(0) o
o
Z21(0)
o
Zll(l) 0
0
o o
o
Z21(1) 0
1
1 1
ZI2(0)
ZII(2) � 0
--- - -- - -- - - --
O
!
0
Z21(2)
I
11 I I
o
0 0
Z 12(0 I)
Z22(0)
0
o
Z12(2)
- - - - - - - - --
0
0
If we define , according to Figure 9 . 5,
Z22(I) o
o
0
o
laO la I la2 la'o la' i la'2
0
VzI 0 +
0
Vz2 0
(9.95 )
(9.96) where we use the circumflex to identify Inew since the prime has a different mean-
Chapter 9
338
ing here. Then I = Ki , or for a series-series connection
1=
la o la ' la 2 la' o la " la'2
0 n2 n-,- - -0-
1
2 l'
0
2' 1
0 0
0
1
n2 , n,
0
,
and n t = n 2 so that
K*t =
1
o
(9.97)
n2 n, 0 ----0
[
n ,2 n�
0
K=
nt = n,
1
0
0'
Therefore, obviously,
We easily show that
0
1
0 =
0'
0
n, 0
Then
n2 0
I
I
0 1
0
n�
0 n�
]
(9.98)
(9.99) (9.100 )
or
n2 I 0 I o I 1
VKO VK , VK2 VK,o VK " VK'2
0 , n,
(9.101)
Now (9.95) may b e written in matrix form, using I and V as defined in (9.94) and (9.100) respectively, as Premultiplying by K*t gives
V = Z I + Vz
V = K*tZ 1 + K*tVz = (K* tZ K) i + K*tVz = z i + Vz But, for a Z-type fault, V = 0 so that we compute , assuming Z-, exists, i - Z-, V'z =
(9.1 02) (9.1 03) (9.1 04 )
339
S i m u ltaneous F a u lts
We easily show that
where
Z- I
does exist since
Zll,Z1 2 ,Z2 h Z 22 and
Z
=
rLZ21Z l l Z12Z22]
K*t Z K =
(9.105)
are defined by equation ( 9 .62) in section 9.4. Also 0
"9'z
=
Vz1
0- -
K* t
0
=
Vz2
0
[n: VZI] nl VZ2
(9.106 )
The sequence currents at the fault are then easily found by premultiplying ( 9.104) by K, i.e . , (9.107 )
9.8
Constraint Matrix for V -type and H -type F aults
Proceeding in exactly the same manner as above , we may easily establish the constraint matrix K for Y -type faults. For this type of fault it is convenient to write
[Iai ] = [IKi ] = [Y II(i) Y12(i)l la'i IK't Y21(i) Y22(i)J
i = 0, 1 , 2 (9.108)
We then develop the constraint
V=KV or
VKO VK1 VK2 VK � VK'1 VK'2
0 1
2 =
(j
0
(9.109)
0'
1
n2 nl
0 0 0
-----
0
1
n2 2' nl [VKO] [VI] VK,
0
,
0
,
( 9. 1 10)
where we define according to Figure 9 . 1 3 ,
=
Using V as defined in (9.1 1 0 ) , we write ( 9 . 1 0 8 ) as
( 9 .1 1 1 )
C h a pter 9
340
0 I Y12(O) 0 YI l(l) 0 I 0 Y 12(l) 0 I 0 0 Y1 2(2) 0 Y I l(2) �
Yl l(O) la2
0
o 0 =
0
------------
--------------
Y21(O)
0
Y21(1)
o o
o
I I I
0
0
Y2 1(2) I
Y22(O) 0
Y22(1)
0
o
VK � VK 'l VK'2
0
0
o
Y22(2)
+
--
0
IY2 0 ( 9.1 1 2)
From ( 9 . 1 1 0 ) we note that the constraint matrix K is exactly the same as that of (9.98) for the Z-type fault. For power invariance we compute
(9.1 1 3) and write
i = y v + iy = 0
(9.114)
v = - y- I i y
(9 . 1 1 5 )
V = - K y- q y
(9. 1 1 6 )
Then
and from (9.1 09 )
Thus, as long as we use the Y parameter description, the V-type fault is described by equations quite similar to those for the Z-type fault condition . For the H-type fault it is easy to show that the constraint m atrix K is again the same as in (9.98 ). In this case we write
[ V�i ] [ IK i
h 1 1 (i) h2 1 ( i)
h l 2 ( i)
h 1 1(0)
0
0
0
h 1 1(1)
0
0
0
h 1 1(2)
=
or
VKO VKI VK2
h22(i)
] [IVK i ] + i [ V ]
-------------
la 'o la 'i
h 2 1 (0)
0
0
0
h21(1)
0
la '2
0
0
h 2 1 (2)
K'i
I
I I I I I T I I I I I
hi
c5 l
Ih 2
i = 0, 1 , 2
'
h 12(0) 0
0
0
h 1 2( 1 )
0
0
0
h 1 2(2)
-------------
h22(0)
0
0
0
h22( 1 )
0
0
0
h22(2)
(9. 1 1 7 )
laO la l la2
VK,o VK 'I VK 'J
0
Vh 1 0 +
0
Ih2 0 (9. 1 18 )
Then we compute
S i m u ltaneous F a u lts
341
(9.1 19) and K is exactly the same as before. Using the notation of (9.1 1 ), we write M = H N + Mh
(9.120)
N = KN
(9.121 )
But
and for power invariance (9.122) Premultiplying (9.120 ) by K*t , we compute
M
where
=
UN
+ Mh
(9.123 )
(9 . 1 24)
But M = 0 for this type fault, so that
an d N =
-
K
A- I �h
N=
or
-U-I Mh
(9.125)
(9 . 1 26)
Problems 9.1. 9.2. 9.3.
Compute the Y parameters for the two-port network of Figure 9 . 2 . Check against the Z parameters of Example 9 . 1 . Compute the H parameters for the two-port network of Figure 9. 2. Check against the Z parameters of Example 9. 1. Compute the A parameters for the two-port network of Figure 9 . 2. Check against the Z parameters of Example 9 . 1 .
342
9.4.
C h a pter 9
Compute the Z, Y, H, and A parameters for the two-port networks of Figure P9.4. z •
(a )
( b)
� r' : z
z
, ...
z
(c )
(d )
•
•
(. )
•
(f )
Fig. P9.4.
9.5.
Compute the A parameters of the two-port networks of Figure P9.5.
(a)
IDEAL TRANS FORMER (n rea l )
(b) PH ASE S HIFTING TRANSFaWER (n a ' e J8)
(c )
(d )
~ (e )
Fig. P9.5.
9.6.
Compute the Z and Y parameters of the following two-port networks. Z3
ZI
� � ( b)
+
VI •
(el
-
•
Fig. P9.6.
Z,
Z
2
(d)
343
S i m u lta neous F a u lts
9.7. 9.8. 9.9. 9 . 1 0. 9.11. 9. 1 2. 9.13. 9.14.
9. 15. 9. 16.
Compute the G and H parameters of the two-port networks of Figure P9.6. Compute the Z and Y two-port equations for the networks of Figure P9.6 if the sources shown in each network are regarded as independent sources of magnitude Vs or I•. Repeat problem 9.8, solving for the G and H equations. Repeat Example 9 . 3 if the controlled source has a value 1 = kI2• Verify the first column of Table 9.5. Verify the second column of Table 9 . 5 . Verify the results of Example 9.5 by direct computation of the H equation. Sketch the series connection of two-port networks of Figure P9.4 and determine the Z matrix of the interconnected network for the following combinations: (a) in series with (b). 4. (c) in series with (d). 1. (a) in series with (c). 5. (a), (c), and (e) in series. 2. (b) in series with (c). 6 . (d) in series with (f). 3. Investigate the parallel connection of the combinations named in problem 9. 14. Investigate the hybrid (series-parallel) connection of the combinations named in problem 9. 14.
9 . 1 7 . Compute the transmission parameters for a cascade connection (left to right) of the fol lowing two-port networks from Figure P9.4. 1. (a) and (b) with impedances labeled Z 1 and Z2 respectively. 2. (b) and (a) with similar labeling. 3. (a), (b), and (a). 4. (b), (a) , and (b). 5. (a) and a two-port network with matrix A.
6.
A two-port network with matrix A and (a).
9 . 1 8. Consider the positive sequence network only and identify the currents and voltages de fined in Figure 9.9 in terms of sequence currents and voltages of phase a.
' (a) For a shunt unbalance at F, shunt unbalance at F (b) For a series unbalance at F, shunt unbalance at F' .
.
9. 19. Verify equation (9.54). 9. 20. Beginning with the currents entering the series-series connection given by (9.65), develop
an expression for the currents entering each of the ports of each sequence network (in side the transformers), and simplify as much as possible. 9.21. Repeat Example 9.6 if the fault condition is given as:
Fault F : lines b and c open ' Fault F : SLG fault on a
9. 22. Repeat Example 9.6 if the fault condition is given as:
Fault F: SLG fault on phase b Fault F' : SLG fault on phase c 9. 23. Repeat Example 9.6 if the fault condition is given as:
Fault F: lines a and b open ' Fault F : lines b and c open
9. 24. Repeat Example 9.6 if the fault conditions F and F' are reversed. 9.25. Verify (9.69). 9. 26. Beginning with (9.78) compute the voltages at each port of the sequence networks (in
side the transformers) and show how these voltages can be used to completely solve the sequence networks. 9.27. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: 2LG fault on phases a and b ' Fault F : line b open
C h a pte r
344
9
9. 28. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: 2LG fault on phases a and b Fault F' : 2LG fault on phases b and c
9. 29. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: line a open Fault F' : line c open
9. 30. Beginning with (9.91) compute the sequence voltages and/or currents inside the trans
formers and show how these quantities can be used to completely solve the sequence networks. 9. 31. Repeat Example 9.6 for the series-parallel fault condition: Fault F: SLG fault on phase c Fault F' : 2LG fault on phases b and c
9.32. Repeat Example 9.6 for the series-parallel fault condition:
Fault F: SLG fault on phase b Fault F' : line b open 9.33. Repeat Example 9.6 for the series-parallel fault condition:
Fault F: lines a and b open Fault F' : line b open
9 . 3 4 . Repeat Example 9.6 for the series-parallel fault condition:
Fault F: lines a and b open Fault F' : line c open
9.35. Compute the constraint matrix K for the fault condition of Example 9.6 using definition (9.98) for K. Then solve for the symmetrical components of the fault currents by using (9. 1 07). 9. 36. Compute the constraint matrix K for the fault condition of Example 9.7. Then solve for the symmetrical components of the voltages aUhe faults by using (9. 1 16). 9.37. Compute the constraint matrix K for the fault condition of problem 9.31. Then solve for the symmetrical component quantities at the fault by using (9. 126). 9. 38. Compute the inverse of the Z matrix of (9.95). Is this the same as the Y matrix of (9. 1 1 2)? 9. 39. Suppose that for the Y-type fault we construct a constraint matrix based on the current equations of Table 8.2. Investigate using this constraint matrix to solve for the sequence
quantities in a Y-type fault.
chapter
10
A n a l yt i c a l S i m p l ifi c a t i o n s
The preceding material on symmetrical components is based upon what is often called the three-component method. i.e there are three sequence networks and three sets of sequence voltages and currents. This is emphasized in the com putation of simple faults in Chapter 3 and is continued for more complex situa tions in Chapter 9. There is one concern, however, in using the three-component method. and that is the large number of network compon.ents required to describe the system. Recall that we begin with a three-phase system and simplify it by means of a single-phase (or per phase) representation. But in order to completely describe unbalanced situations, we need three such single-phase representations, namely the positive, negative. and zero sequence systems. What then has been gained? Perhaps we would have been just as well off to have retained the three phases a . b . and c . This problem of the number of components needed to represent a system is crucial, whether the problem is to be solved by digital computer. by network analyzer, or by hand. In computer solutions, whether analog or digital. the net work representation occupies the machine hardware or memory. and this is al ways limited by the size of the machine itself. Thus an arbitrary upper limit is always placed on the size of system which can be solved, and this limit is related to the computing device to be used. .•
1 0. 1
Two-Component Method
The two-component method for solving unbalanced system conditions is based on the idea that the positive and negative sequence networks have nearly the same impedance. Recall that the positive sequence voltage is defined to have the phase sequence of the system generators. We arbitrarily label this sequence a-b-c . The negative sequence then has phase sequence a-c-b. and this is the only difference between the two. Both are balanced sets of voltages (or currents). It is intuitively obvious that any passive network components, such as trans mission lines or transformers. present the same impedance to either positive or negative sequence currents. Thus the impedances Z and Z2 for these compo nents are equal. Only the rotating machines have different positive and negative sequence impedance. In synchronous machines. which contribute most of the fault current in a power system, these sequence impedances are nearly the same if we consider only the subtransient condition. Indeed. from Table 6.1 we see that for both round rotor and salient pole machines the subtransient and negative sequence reactances are exactly equal, i.e ., 1
345
346
Chapter
10
(10. 1 )
This is not exactly true for synchronous condensers or synchronous motors. However, these devices are usually small compared to the generator ratings, and some error in their fault computation can be tolerated. Because of the equality or near-equality of all positive and negative sequence impedances we make the assumption that Z2 Z l ( 10.2 ) for all circuit components; i.e ., we will replace all negative sequence impedances by the corresponding positive sequence values. I Referring to Figure 2 . 5 with Z2 = Z l , we may rewrite the primitive sequence network equation (2.58) as =
(10.3)
If we add and subtract the second and third rows of ( 10.3 ) and replace these rows by the resulting sum and difference quantities, we have (10.4)
These equations are interesting because the last two rows are both positive se quence equations. This is confirmed by the fact that each contains VF , a posi tive sequence source, and Z., a positive sequence impedance. These equations, therefore, describe the positive sequence Thevenin equivalent of Figure 2 . 5 . If + and the this is true, the sum and difference quantities corresponding currents are also positive sequence quantities. Thus under the assumption ( 1 0 . 2 ) we have created a new mathematical description consisting of one zero sequence equation and two positive sequence equations as in ( 10.4). Because of the interest in positive-plus-negative and positive-minus-negative quantities it is also convenient to similarly rearrange the analysis and synthesis equations (2.21 ) and (2.23). For convenience we define the sum and difference quantities Va l
Va2 , Va l - Va2 ,
with a similar definition for currents. Then from we add and subtract the second and third rows to derive the new equation I I Some engineers use the average of Z and Z2 for machines.
(1 0. 5 )
(10.6)
Ana l ytical S i m p l ifi cations
347
(10.7)
or in matrix form V 0 1: 4 =
B- 1 Vabc
(10.8)
where V 0 1: 4 is defined as noted and 1
-1
-1
j y'3
- j y'3
From ( 1 0 . 8 ) we compute
[�
where
1
B 6= !h
J
1
1
0
(10.9)
(10.10)
J
- 1 /2 - j y'3 /2
+j y'3 /2
- 1 /2
(10.1 1 )
Using these relations, we may solve for any of the shunt or series unbalances as before since we may still write the same number of equations. The number of networks, however, is reduced from three to two since both the sigma and delta quantities are defined as voltages or currents associated with the positive sequence network. The two primitive sequence networks are shown in Figure 1 0 . 1 . FO
·
_
Fig. 1 0 . 1 .
FI
loot lli=F=-' va O
Zo
1aI
I04 or
HI
1
Z,
O�hVF •
,
L
Va I
Va6 or
r
Two sequence networks for the two-component method.
I.
SHUNT FAU LTS
We begin with an analysis of shunt faults which will parallel the three-compo nent development of Chapter 3 . Frequent comparison with the results of Chapter 3 is recommended . 10.2 Single-Line-to-Ground F ault
From Figure 3.2 we note that phase a is the symmetrical phase and that the boundary conditions are (10.12)
348
Chapter
From (10.10) and (10.12) we compute 0 = h1 100 1 Ie = 0 = h 100 -
Ib =
10
1 . vr:3 2 h 10 '1:. - J 2h lo A 1 1 + . va l 2h 0'1:. J 2h oA
(10.13)
Adding these equations, we compute for any h and subtracting, we have
2 100 = 10'1:.
(10.14)
lo A = 0
(10.15)
From the second equation of (10.12) we write for any h o 2 ) = Z, (loo
Voo + ( VO l + V
and substituting lao from (10.14 ) ,
a
+l l +
la 2 )
aO + Va '1:. = (3/2) Z,la '1:.
( 10.16)
VaO = - Zo lao = - Z"2o ( 2 1aO)
(10.17)
V
Equation (10.15) requires that the positive sequence network be open at the fault point. This network is shown in Figure 10.2a and is recognized to be the normal (unfaulted ) positive sequence network. All voltages measured in this normal network are to be labeled delta quantities as noted. Equations (10.14) and (10.16) dictate that the positive and zero sequence networks be connected in series to compute the zero sequence and sigma quantities. This connection is shown in Figure2 10.2b. We also note that since the current through the zero sequence network is 1ao , we satisfy the first equation of (lO A) as follows: Thus we use only one-half the actual zero sequence impedance values. FI
FI
:tot. = 0
Z,
hvF.
t
(0 )
t.
VoA HI
,-
t
1: oI
Z, •
hV(
2 Iool ZO/2
-
HI FO
f--
NO
1''' -''''''
t.
Val
r
1.
VoO
3Zf
"2
r
( b )
Fig. 1 0 . 2.
Sequence network connections for a SLG fault on phase a with fault impedance Z, : ( a ) delta, (b) sigma and zero.
3 49
Ana lytica l S i m p lifications
From the networks of Figure 10.2 we easily compute the sequence voltages and currents at any point in the network. Then we apply equation ( 1 0 . 1 0 ) to synthesize phase quantities. At the fault the total current is 10 = 3 /00 • By inspec tion of Figure 10.2 we compute h VF (Zo /2 ) + Z l + ( 3 Z, /2 )
2/0 0 =
a
or
I
-
-
o
h VF Zo + 2 Z 1 + 3 Z,
(10.18)
This i s exactly the same a s ( 3 . 5 ), the current computed b y the three-component method , with Z , set equal to Z 1 1 0.3
•
L ine-to-Line Fault
From Figure 3 .9 we note that phase a is the symmetrical phase. Also , from ( 3.6)-( 3 .8 ) and Figure 3 .9 we write the boundary conditions 10 = 0,
Vb - Ve = Vbc
Ib = - Ie ,
�aoJ � [1 -1
= Z , /b
(10.19)
Substituting (10.19) into the current version of ( 1 0 . 7 ) , we compute 1
lo T.
=
2
0
lo t..
j V3
[��VJ � [�
The known voltage is
Vb e
which we compute from ( 1 0 . 1 0 ) as follows :
t
1
Then
(10.20)
1
� J r�:� � �a�
_ 1 /2
-j
- 1 /2
+j V3 /
/
laF.
( 1 0 .2 1 )
are zero . Thus the sigma cur From ( 1 0 .20) we learn that both 100 and rents and voltages are found from the normal network as shown in Figure 1 0 . 3a. F
I o t -O
t .
Ia
t
t..
FI
ZI •
h"F. (0)
Fig. 1 0. 3 .
f+ ,.
V a t.. HI
Z, / 2
(b)
Sequence network connections for a LL fault on phases b-c with fault impedance Z , : ( a ) sigma, ( b ) delta.
3 50
Cha pte r 1 0
From the third boundary condition we compute for any h ,
(10.19)
and equations
(10.20) and (10.21) (10.22)
This equation is satisfied by the network of Figure 10.3b. From the two net works of Figure 10.3 the sequence voltages and currents may be completely determined . The total sequence current at the fault may also be computed from Figure 10.3. From Figure 10.3a
(10.23)
and from Figure 1 0.3b
( 10 .24 )
la /1 = Z, + Ztl2
Adding
(10.23) and (10.24), we easily compute Ial = 2Z1h V+F Zf
(10.25)
This is exactly the same result found by the three-component method if we re place Zl = in ( 3 . 11 ).
ZI
1 0.4
Double L ine-to-Ground Fault
From Figure 3 . 1 3 we again note that phase a is the symmetrical phase. From ( 3 . 1 2 )-(3 . 1 4 ) and Figure 3 . 1 3 the boundary conditions are
la = 0, Vb = ZfIb + Z,(lb + Ie), Since Ia = 0, we compute
Ve = ZfIe + Z,(lb + Ie) (10.26)
(10.27) from which we observe that all sequence quantities will usually be nonzero since and are gen the right-hand side of will be nonzero . This is because erally neither equal nor opposite for this type of fault. We also observe from ( 1 0 .2 7 ) that for any h,
(10.27)
Ib
Ie
(10.28)
and this establishes one of the required network connections. From the second and third boundary conditions combined with ( 1 0 . 2 7 ) we write
But from
(10.10)
351
Ana lytical S i m pl ifications
Combining, w e have for an y h ,
(10.29)
Va6 = Z,Ia6
This completely determines the delta network connection shown in Figure 1 0.4a.
FI
-
FI
ZI
hVf':O NI �
t.
VoA
l-
Zf
hV
NI
(a)
Fig. 10.4.
fl ol
ZI t. � O Vol l-
l: o A
FO
Ioo
t
f.
2Z 0 2yae 1-
NO
(b)
Sequence network connections for a 2LG fault on phases b-c with phase impedance
g
Z, and ground impedance Z (see Figure 3 . 1 3 ) : (a) delta, ( b ) sigma and zero.
(10.26) and (10.27) we compute Vb + Ve = Z, (lb + Ie) + 2 Zg(lb + Ie) = (Z, + 2Zg)(/b + Ie) = (Z, + 2 Zg) � lao But from (10.10) Vb + Ve = h2 VaO h1 Val: Again from
( )
-
Combining, we have for any h
2 VaO - Val; = (3Z, + 6Zg) lao
(10.30)
This equation is satisfied by the network connection of Figure 10.4b where twice the normal zero sequence impedance is set in order to force the voltage relation ship of From Figure and Figure 1 0 Ab we compute
(10.30).
10.4a h VF Ia 6 Z I + Z, '
Adding, we compute
2 /a l
(
V Z I +1 Z, + Z 1 + 2Zo +1 3Z, + 6Zg
=h F
which can be reduced to
la l: Z 1 + 2Z h V+ F3Z, + 6 Zg 0
__
)
(10.31) 3_Z� (Z Z ) Z + , + , ) + I _ �0� � -,-� g � _ _ Z I + Z, + -,--(Z - Z o + Z + 2Z, + 3Zg This equation is exactly the same as (3. 2 4) which was derived from the three component method if Z2 in ( 3. 2 4) is replaced by Z I . __
1
3 52 1 0.5
Chapter 1 0
Three-Phase Fault
3lP
The general fault condition is shown in Figure boundary conditions
3.18 where we note the
Va = Zfla + Zg (Ia + Ib + Ie) Vb = Zflb + Zg(Ia + Ib + Ie ) Ve = Zfle + Zg(Ia + Ib + Ie) Replacing Ia + Ib + Ie by (3 /h ) Iao and adding, we compute Va + Vb + Ve = 0 = (3/ h ) Zflao + (9/h )ZgIa o or
� ] �1
(10.32)
la O = 0
�J �
�
(10.33)
(10.7) we compute sigma and delta quantities 1 1 0 ao h h = = 2 1 - (Ib + Ie ) I T, 3 2 - 1 -1 I 3 _ 0 j y'3 j y'3 I: I: A j � (Ib le ) But, for this connection since Ia + Ib + Ie = 0, then Ia = - (Ib + Ie) and 0 Ia O h Ia T, = "3 3 Ia (10.34) Ia A j y'3 (Ib - Ie) From the first of equations (10.32) we have, with Ia + Ib + Ie = 0, Va = Zfla = (Z,!h)IaT, but Va = ( l /h ) Va O + ( l /h ) Va T, = ( l /h ) VaT, since la O 0, or (10. 3 5) VaT, = Z,IaT, Subtracting the second two of equations (10.32), w e compute, with Ia + Ib + Ie = 0 and incorporating (10.34), Also, from
J
[j [
_
J
=
But from
(10.7)
or for any h,
(10.36) Thus the equations for the sigma and delta networks are exactly the same and are as shown in Figure
10. 5 .
3 53
Ana lytical Si mp lifications
-
FI
-
FI
l oA
I a l:
t
Va l
t.
VaA
Zf
t
,-
(a)
Fig. 1 0 . 6 .
If
(b)
Sequence network connections for a 34> fault with impedance Z, in each phase : (a) sigma, ( b ) delta.
By inspection of Figure 1 0 .5 we compute lor. = hVF /(Z 1 + Z , ) ,
loA = hVF /(Z 1 + Z, )
(10.37)
and adding we have 10 1 = h VF /(Z 1 + Z, ) . I I . S E R I ES F AU LTS
We next consider the application of the two-component method to series or longitudinal faults. Again, the network currents and voltages at the fault point are defined as indicated in Figure 3.20 with sequence networks as defined in Figure 3 . 21 . 10.6 Two L ines O pen ( 2LO)
a
The series unbalance for 2LO is shown schematically in Figure 3 . 32 for which the boundary conditions are seen to be
Ib = Ie = 0,
�aj [
Vaa, = Zl
J [/aJ l/a�
( 10.38)
where Z is the impedance in line a . Then from the current version of (10.7) lor.
loA
=
:
1
1
1
2
-1
-1
0
j v'3
- J. v'3
0 0
=
� 2/0 0
( 10 .39 )
Thus the delta network must b e open for this type o f fault. Also, w e observe that ( 1 0 .40)
the same as for the SLG fault. From the second boundary condition (10 .38) incorporating ( 1 0 .40 ) we write for any h , ( 1 0 .4 1 ) Equations (10 .40) an d ( 1 0 .4 1 ) fix the connection o f sigma and zero sequence net works to be that of Figure 1 0 .6. This connection has a striking resemblance to Figure 10.2 for a SLG shunt fault. If we interpret h VF to be the Thevenin equivalent open circuit voltage gen erated in the sigma network, we compute 210 0
- Io r. - Z l
h VF + Zo /2 + ( 3/2 ) Z
3 54
Ch a pter 1 0 FI
loI NI
,
F' I
Fig. 1 0. 6.
Z,
21 0 0
I
I
1.
Voa ' I F" FO
t
Zo /2
r t.
Voa 'O
l
1 F' O
3 Z
'2
r
Sequence network connections for phases b and c open and with impedance Z in phase a.
But we also know that la A
= O.
Adding these two equations, we have
= 2 /a l = Z
2/a O
h VF
I + Zo /2 + ( 3 /2 ) Z
or la o = la J
=
h VF + 2 Z J Zo + 3 Z
(10.42)
which agrees exactly with ( 3.86), derived by the three-component method, if we set Z 1 = Zl in that equation. 1 0.7 O ne L ine Open ( 1 LO)
If only one line is open and we assume that is line a , we have the boundary conditions ( 10.43) where we assume an impedarice Z in the sound lines. Since la = 0 , ( 1 0 . 2 7 ) applies and we immediately establish the similarity be tween this fault and the 2 LG shunt fault where for any h, (10.44) From the second two equations of (1 0.43) we compute the sigma and delta quantities
Vbb' + Vee '
= Z (lb
+ Ie ),
( 1 0 .45)
But from ( 1 0 . 1 0 ) we compute
( 10.46) and from ( 1 0 .2 7 )
Ib + 1e
=
-
3 /a � h
-
...
'
Ib
- -I
e
=
_ l' va Ia ... ..
h
( 10 .47 )
355
Ana lytica l S i m pl ifications
Substituting ( 1 0 .46) and (10 .47 ) into ( 10 . 4 5 ) , we compute for any h, 2 Vaa, o - Vaa'l:
=
Vaa '4
=
-
3 Zla l:
(10.48)
Z la 4
( 1 0.49 )
These two equations completely specify the network connections of Figure 10.7. fl
N'I Fig. 1 0 . 7 .
1
l
-
L o1:
lot.
I
f.
ZI
r
I
F, J
H ' z,
Z
Voo't.
Sequence network connections for line
F" a
3Z
F'O
open and an impedance Z in lines b and c.
This network connection is a parallel connection similar in construction to that derived for the 2LG shunt fault. By inspection we compute
and adding we find , after algebraic reduction, la 1 = Z1
+
Z
+
(Z l + Z)(Zo + Z) Zo + Z l + 2 Z
( 1 0.50)
"-'--"---"-'---=-
This is exactly the same result computed by the three-component method and given by ( 3 .7 8 )-(3.80) if we set Z2 = Zl . I I I . CHANGES I N SYMMETRY WITH TWO-COMPO N ENT CA LCU LATIONS
An obvious similarity exists between certain shunt and series fault connections computed by the two-component method. If we take advantage of these similar ities, we should be able to extend the two-component computations to situations in which any phase is the symmetrical one. It is again convenient to limit our analysis to common fault connections, namely , the shunt SLG and 2LG and the series 1 LO and 2LO. These situations are shown in general form by Figures 8 . 1 and 8 .5 for which we write the general equations, similar to (8 .6)-(8 .8) and ( 8 . 18) , as
[
J
Va o - 3 Zg la
where Zg
=
va .. '<'
Va 4
=
=
h 3
-
r
B- 1
za
2 Za
� L� �b 0
Za
0
0 for the series fault case. Then from ( 1 0 . 1 0 ) we may write
(10.5 1 )
3 56
Cha pter 1 0
(10.52 ) where w e define Vo z = Voo - 3 Z. Ioo , The values of Zo , Zb , Ze , and Z, depend upon the type of fault, but in all cases (10.52) can be further simplified. In particular, if we restrict our consideration to the case where finite nonzero values of Zo , Zb , and Ze are equal, there is no mutual coupling between the zero, sigma, and delta networks. We will make this simplification since it greatly reduces the network complexity while still retaining a realistic fault situation. The faults to be considered may be recognized as of two types, those which have a single fault current of zero and those which have two zero currents. This information alone, substituted into ( 1 0 . 7 ) , provides the interesting relations shown in Table 10.1 . Table 10. 1. Sequence Current Relationships Due to Zero Phase Currents Current
Constrain t
Simple zero
Current Relationships
10 0 + 10E
-iva loA
=
...
0
h (lb - Ie)
10 0 + 10E = h/o j va loA - hIe 10 0 + 10E - h /o -j va /oA - h lb Double zero
10 0 + /oE 0 loA - j y'3 la O -
la O + 10E - 0 loA '" -i va 10 0 1 0.8
Phase Shifting Transformer R elations
It is necessary to examine once again the use of a phase shifting transformer. In the cases in Chapters 8 and 9 the phase shifters were special in the sense that they always had a turns ratio of unity. Thus any voltage V l on the primary was changed in phase only but not in magnitude to a secondary value of V2 = e J6 Vl where (J is the phase shift in radians or degrees. Currents were shifted in phase by exactly the same amount. In the case of zero, sigma, and delta networks it is necessary to use phase shifting transformers which change both the magnitude and angle of the primary voltage and current. Referring to Figure 1 0 .8 , let the turns ratio m be a complex
Ana lytica l S i m p l ifications
357
12
II
ero I D EAL
Fig . 1 0 . 8. A n ideal transformer with complex turns ratio m.
number. Then by inspection we write V2 = m V I = lt ei8 v l
(10.53)
and if VI = I V l l e jll l then ( 1 0 . 5 3 ) may be written as V2 = 1t I VI I e j(6 1+ 8 )
(10 .54)
The output V2 is changed in magnitude by a factor It and shifted in phase an amount +8 radians . Since the transformer is ideal, we may also equate the input apparent power to the output apparent power, or S I = S2 , which we write as Vl n = V2 If
(10.55)
Then , combining ( 10 .53 ) and ( 1 0 .55 ) V2 1 VI = nllf = m or 12 = ( l /m * ) I I
(10 .56)
If we let I I = 1ll l e j O! I , we compute 12 = ( l llte -i 6 ) l ll l e jO! I or
11 = !.. 11 , l e i(0! , +6)
(10.57 )
It
Thus the current phasor 12 is changed in magnitude an amount lIlt as in an ordi nary transformer. It is also shifted in phase by +8 radians, which is exactly the same phase shift as the voltage phasor of ( 1 0 .54). 1 0.9 S L G Fau lts with Arbitrary Symmetry
The SLG fault on phase a was investigated in section 1 0 .2 with the solution shown in Figure 1 0 .2 . If the fault is on phase a , we may write the constraints from Figure 8 .1
Za = 0,
Z b = Ze = 00 ,
ZII =1= 0
(10.58)
and write the boundary conditions
Va = ZgIa ,
Ib = Ie = 0
( 1 0 .59)
Then from Table 1 0. 1
( 1 0 .60) Also from ( 1 0 .59 )
Va = Zg Ia ( l /h) ( VaO + Va l: ) = (Zg /h) (lao + Ial: ) or incorporating ( 1 0 .60 ) , we have for any h ,
VaZ + Val: = 0
( 1 0 .6 1 )
3 58
10
Chapter
If the fault is on phase b , the constraints are (10.62) and the boundary conditions become
Vb
10 = Ie = 0
= Zglb ,
Then from Table 10.1
( 1 0 .63)
( 1 0.64 ) Also from ( 1 0.63)
Voo
-
Vb
= Zglb
� Vol: j V; Vo� = ZIl �OO � 10l: j V; -
-
-
Combining with ( 10.64), we compute
Voz
-
�
Vol:
-
j
V;
I�)
Vo� = 0
(10.65)
Results similar to ( 1 0 .64) and ( 1 0.65) may be obtained for a SLG fault on phase c. Upon closer examination of ( 1 0 . 60), (1 0 .61 ) , ( 1 0 .64), and ( 1 0 .65 ) a pattern is observed which leads us to the network connections of Figure 10.9. Note that for phase a symmetry m z = 0, which shorts the right side of that transformer and leaves the delta network open as in Figure 1 0 . 2 .
S Y M M E T R I C AL
PH A S E
mo
COMPLEX
m,
m2
RATI O
TURNS
- 112 -tv'!/2 - 112 j ,.,fi/2 0
a
b
c
•
+ m l Vol + m 2 Va ll - 0
CON STR AINTS ' - V · mO
'4:Jz
.� mt m1
t . 2 100
_
106
-;;r
Fig. 1 0.9. Generalized fault diagram for a SLG fault on a zero-sigMa-delta basis.
1 0_ 1 0 2lG Faults with Arbitrary Symmetry
Double line-to-ground faults for any symmetry may also be analyzed on a zero-sigma-delta basis. For the case of a 2LG fault on phase b and c we have the
Ana lytica l S i m pl ificat ions
3 59
following constraints from Figure 8 . 1
Zb = Ze = Z,
Za = 00 ,
Zg =1= 0
(10 .66)
and may write the boundary conditions
la = 0,
(1 0.6 7 )
Then, from Table 10.1 for la = 0 we have
la o + la 'E = 0
- j y'3 /at:. = h (/b - Ie)
(10.68) (10.69 )
Subtracting the voltage relations of ( 1 0 .67 ), we compute
Vb - Ve = Z(/b - Ie) Then from ( 1 0 . 1 0 ) and ( 1 0 .69 ) - j y'3 Vat:. = - jy'3 ZIat:. or
Vat:. - ZIat:. = Vzt:. = 0
(10.70 )
where we define the quantity ( 1 0 . 70 ) to be Vzt:. . Adding the voltage equations of (10.67 ) , we compute
Vb + Ve = Z(lb + Ie) + 6Zg laO which is rearranged to form
or
2 [ Vao - (Z + 3 Zg) /ao] = Va'E - Zla'E
(10.7 1 )
2 Vzo = VZ 'E
(10 .72 )
where the quantities Vzo and VZ'E are defined by ( 1 0.71 ). If the 2 LG fault is on phases a and c, we write the constraints for Figure 8 . 1 as
Zb = 00 ,
Za = Ze = Z,
Zg =1= 0
(10.73 )
from which we have the boundary conditions
(10 . 74) From Ib = 0 we have from Table 1 0 . 1
h /a = lao + la'E ,
hIe = j y'3 Iat:.
(10.7 5 )
But
or for any h,
Ia O -
! L� 2
'"'
. J y'3 Ia.. = 0 2 A
From the first voltage equation of (10. 74) we write for any h,
VaO + Va'E = Z(lao + la'E ) + 3Zg Ia o which is easily rearranged as
( 1 0.76)
Cha pte r 1 0
3 60
(10.7 7 ) (10.78)
or
VZO
=
-
VZ l:
From the remaining boundary condition on Vc we write
VaO
2"
1
-
. v'3 Val: + J 2 VoA = Z laO
( 2"1 -
from which we compute
[ VaO - (Z + 3Z,) lao]
lal: + J
) . v'3 2
loA + 3Z,la o
� ( VoA - Zlaa. )
= - j ..;
or
(10.79) (10.80) 0
If the fault is on phases a and b , a similar computation can be made. These for Note that when m 2 results may be combined as shown in Figure phase a symmetry, this shorts the left side of that transformer leaving the delta network shorted on itself as in Figure 10 .3 . Also note the slight difference in defining Vz o , VZ l: , and Vz a. compared to Va Z , Val: ' and Va a. which is necessary due to the presence of the impedance Z in Figure
10. 1 0.
=
10. 1 0.
S Y M M ET R I CAL PH A S E
-112
b
l=O
- 1 12
c
•
m2
m,
mO
a
V
R AT I O
TUR N S
COMPL E X
-J +
0
-/3/2
i .)3/2
C O N S T R AI N T S -V=
2 VZ O
VZ I
mO
__
m,
.
I = m*O Io O + m,* IoI + m2
VZ6
m2 *
I o a.
Fig. 1 0. 1 0. Generalized fault diagram for a 2LG fault on a zero-sigma-delta basis. 1 0. 1 1
Series Faults with Arbitrary Symmetry
1LO
and 2LO, can be analyzed for any symmetry in exactly Series faults, both the same way as shunt faults. Details of this development are left to the interested reader. The results of this effort reveal the expected similarity to the shunt fault case .
Ana lytical S i m pl ifications
361
For 2LO the network connections are similar to the SLG shunt fault situation except that the external impedances are different. In the 2LO case there is no impedance Zg, so we set this to zero in Figure 1 0 .9 . There is an impedance Z, however, which is included as Z /2 in the zero network but as simply Z in the sigma and delta networks. Points labeled NO and N1 are relabeled MO and M1 where m is defined as the current input terminal (see Figure 8. 5). For the 1 LO case the network connections are similar to Figure 10.10 except that the impedance Zg is zero . Again , the points NO and N1 must be relabeled MO and M1 to indicate the point at which sequence currents enter the network. The generalized fault diagrams for both shunt and series fault conditions are shown in Figure 1 0 .1 1 . Here , as in the three-component method, all common shunt and series faults may be described by two diagrams. For the series network connection we may write
( 1 0 .81 )
PARALLEL NETWORK CONNECTION
S E R I E S NETWORK CONN ECT ION
v +
1=0
I
Z
Z AVI
-.1M/'--
K=M Zgo = Z K=N Z go = 3 Z g Z oo
S Y M M ETRI CAL PH ASE a
b
c
Fig. 1 0. 1 1 .
C O M PL E X mO
TUR N S RAT I O
ml
"' 2
0 - 1 /2 - j ./3i2 - 1/2 + j ./3/2
K
N
Z gO = 3 Z g+ Z =
Summary of zero-sigma-delta network connections for common fault conditions.
Cha pte r 1 0
3 62
mo , m l , and m2
where tions we have
are defined in Figure 1 0 . 1 1 . For parallel network connec
- V = 2Vzo/mo = Vz'£. /ml = VZA /m l 1 = m: lao + mf la'£. + m2 laA = 0
(10.82 )
Equations ( 1 0 .81 ) and ( 1 0 .82 ) are easily verified in the cases examined in sections 1 0 .9 and 1 0.10 . Other cases are easily verified too , but this is left as an exercise for the reader. Figure 1 0 . 1 1 is the result which permits us to analyze all common fault condi tions by first making the assumption that Then if we define the primitive zero , sigma, and delta networks as in Figure 1 0 .1 1 , we may solve these networks as an interconnection of one-port networks.
Z 1 = Z2 .
IV. SO LUTION OF TH E G E N E RA L I Z E D FAULT D I AG R AMS
Having developed the generalized fault diagrams of Figure 1 0 . 1 1 , we may solve these networks for the sequence voltages and currents for any symmetry . 1 0. 1 2
Series N etwork Connection-S LG and 2LO Faults
From Figure 1 0 . 1 1 and equation ( 1 0 .81 ) we have
mo Vzo + m I Vz'£. + m 2 VZA = 0 or (10.83)
VaO , Va '£. ,
Va ll.
But, from the primitive network equation ( 1 0.4) we write in and terms of the prefault voltage h and the Thevenin impedances. Substituting ( 1 0 .4 ) into ( 1 0.83 ) gives
VF
mo (- Zo Iao - ZgoIao) + mdh VF - ZI Ia'£. - ZIa E) + m 2 ( h VF - ZI Ia A - ZIa A ) = O
Rearranging, we have
(ml + m 2 ) h VF = mo (Zo + Zgo) lao + ml (ZI + Z) /aI; + m 2 (ZI + Z) la A 2 1ao /m � = la I; / m t = la A / mf or laI; = (2 mt /m�) /ao , la A = (2mflm�) lao
( 1 0.84)
But from 1 0.81 we compute
( 10.85)
Substituting ( 1 0.85) into ( 1 0.84), we have,
2 mt (ml + m 2 ) h VF = mo(Zo + Zg o) lao + 2mmol m* t (ZI + Z) /ao + 2mmo (Zl + Z) /ao * Solving for laO ' h (m l + m 2 ) VF ( 10.86) IaO = mo(Zo + Zgo) + (2/mt)(ml mt + m m f)( Z + Z) 2
We note that for any symmetry
mo = m t = 1 , m z m t = Imzl2 ,
m 1 m f = I m JI 2 Im l 1 2 + I m 2 1 2 = 1
I
Ana lytica l S i m plifications
so that
I
a
0
=
363
( m 1 + m2 ) VF (Zo + ZII O ) + 2 (Z l + Z ) h
( 10.87)
which agrees with previous results. From ( 10.85) we also compute
* - m* 2 I mo* 00
- m1
I02 -
( 10.88)
and the sequence currents are easily determined. Equation ( 1 0.87) may be com bined with either ( 1 0.85) or ( 10.88) to synthesize the phase currents. Voltages may then be found from (lOA) and ( 10.9). 10. 1 3 Parallel Network Connection-2LG a n d 1 LO Faults
The parallel connection of Figure 1 0 . 1 1 may be analyzed from the basic equation ( 10.82) ( 10.89)
la o
by combining the primitive equation We simplify this equation to solve for ( lOA) with the first of equations ( 10.82), viz. ,
VZ1:: = (2mdmo )Vzo,
to compute the relations
I01:: 1
a
.6
h VF + 2ml Z + Z mo h VF + 2mf = Zl + Z mo
=
1
(ZZo ++ZZ,o\") I (ZoZl ++ZZ,o\] lao 1
00
( 1 0.90)
Then substituting ( 1 0.90) into ( 1 0.89) , we have
(10.91)
Substituting back into ( 1 0.90) , w e have the complete set o f sequen ce quantities from which phase currents can be found. Problems
10. 1. 10. 2. 10. 3. 1 0. 4. 10. 5. 10.6.
Verify ( 10.8) and (10.10), particularly the matrices B and B- 1 , by starting with ( 10.5). Rework Example 3.1 for a SLG fault using the two-component method of section 10.2. Show that the results are equivalent. Rework Example 3.2 for a LL fault using the two-component method of section 10.3. Rework Example 3.3 for a 2LG fault using the two-component method of section 10.4. Rework Example 3.4 for a 31/) fault using the two-component method of section 10.5. Using the network of Figure 3.25 with series unbalance between F and F ', use the twocomponent method to solve this system when (a) Lines b and c are open with 1 ohm in line a (see section 10.6). (b) Line a is open with 1 ohm in lines b and c (see section 10.7).
3 64
10.7. 10.8. 10.9. 10.10. 10.11. 10.12. 10. 13. 10. 14. 10. 15. 10.16. 10. 17.
Chapter 1 0 Repeat problem 10.2 using the two-component method if the SLG fault is on (a) phase b and (b) phase c. Repeat problem 10.4 using the two-component method if the 2LG fault is on (a) phases a and c and (b) phases a and b. Repeat problem 10.6(a) if the open lines are (a) lines a and c and (b) lines a and b. Repeat problem 10.6(b) if the open line is (a) line b and (b) line c. Verify Figure 10.9 for a SLG fault on phase c. Verify Figure 10.10 for a 2LG fault on phases a and b. Verify the series connections of Figure 10. 1 1 for 2LO in any symmetry. Verify the parallel connections of Figure 10.1 1 for 1LO in any symmetry. Show that the results in (10.88) are identical with the currents computed by the three component method for any symmetry. Show that the results of ( 10.90) and (10.91) verify (3.24) for phase a symmetry. Show that the results of (10.90) and (10.91) are identical with the currents computed by the three-component method for any symmetry.
chapter
11
Co m p u te r S o l u t i o n M et h od s U s i n g t h e Ad m i tt a n ce M a t rix
The electrical networks which power engineers must solve are usually large. This means that any solution ''by hand" will often be out of the question and a computer will be used. Usually this will be a digital computer, and our emphasis in
this chapter is on techniques which would be useful in performing a digital solution. However, the same network may also be solved by constructing a
laboratory model of the physical network. A network analyzer is such a model and analyzers were used for many years in solving small to moderately sized
networks.
There are several important differences in solving network by digital com puter as opposed to analyzer techniques. In the digital computer we solve equa tions. We are not concerned about the physical realizability of these equations. In the network analyzer, however, physical realizability is sol te y essential. The phase shifting transformers used in Chapters 9 and 10, for example, may be very difficult and expensive to build. In digital solutions this is of no concern whatever since solve the equations which describe the system model with little regard to the concept of physically modeling the equations. The digital computer has another distinct advantage in that it provides a means of solving large networks. Even the largest network analyzers are limited to net works of hundred nodes or so. There is also a limit to the size of system which can be solved by given digital computer. However, computers are commonly available which will solve networks of several hundred nodes. Furthermore, by clever usage of auxiliary memory in the form of magnetic discs, tapes, and drums the size of problem to be solved can be further expanded. The digital computer is therefore a flexible tool for the solution of large systems. The size of the network should not be dismissed as unimportant, however. Large problems are costly to solve, and the engineer should know how to compute network equivalents or be able to use judgment to reduce any problem to its bare essentials. Finally, the digital computer provides a means of solving large networks ac curately. The hand solution of large networks, or even those of three or four nodes, involves considerable rounding error and often includes computational mis takes. These problems are not eliminated entirely by the digital computer, but they may be attacked in a systematic way to reduce errors to a minimum. Power system problems do not require 1 0 or 1 5 place accuracy; but being able to coma
ab
u
l
we
a
a
365
Cha pter 1 1
366
pute with such accuracy makes it possible to solve large systems with tolerable error, and this is important. Moreover, once a computer program is thoroughly "debugged " there is little likelihood that the solution obtained will include me chanical mistakes. Because of the many advantages of digital computer solution of networks, we will consider in this chapter the way in which the system data and equations may be arranged for computer solution. The treatment here is not exhaustive by any means, and the interested reader is urged to consult the many fine references on the subject (see [64-76 ] , for example). 1 1. 1
Primitive Matrix
A power transmission or distribution system is a linear, passive, bilateral net work of impedances (or admittances) which are interconnected in some specified way at various points called nodes (or buses) . The impedances represent the per phase impedance of transmission lines and transformers. Usually we represent only one phase in three-phase systems since the impedance to the flow of posi tive, negative, or zero sequence currents is the same in each phase except at points of system unbalance, and these unbalances require special treatment. The rest of the network is made up of balanced impedances such that a per phase representa tion is possible. Indeed, if this were not the case, there would be no advantage in using symmetrical components. The generators and loads are also connected to the nodes or buses exactly as these terminations occur on the physical system . In some studies the loads may be represented as constant impedances, in which case a per phase impedance (or admittance) may also be specified for the loads. Often , as in short circuit studies, the load currents are negligible when compared to the fault currents, and the load impedances are taken to be infinite. As pointed out in Chapter 6, the generators have a different impedance to the flow of positive, negative, and zero sequence currents . This is not the case with transformer and load impedances which are usually the same in each sequence network. Line impedances are the same in the positive and negative sequence networks but, as noted in Chapter 4 and 5, are dif ferent in the zero sequence network. Also, the topology of the sequence net works is often different in the positive and zero sequence cases. In order to solve the network, it is necessary to organize the system impedance data and the network topology data in such a way that this information may be conveniently introduced and stored in the computer memory. The primitive net work matrix provides a simple way of accomplishing the desired organization. The primitive matrix also defines an orientation or positive direction for each matrix element, and this too is useful. In section 8.7 we defined Kron 's primitive matrix for a network in which the branches were not mutually coupled . In pO'.ver systems the branches or the net work are often mutually coupled , especially in the zero sequence network. In such cases the primitive matrix contains off-diagonal terms for those mutually coupled elements. We write the primitive equation from (8.48) as v
where
= ZI
-
E
(11.1)
Com puter Solution M ethods U s i n g the Adm itt a n ce M atrix
367
E = the column vector of branch source voltages, defined as a voltage rise in the
direction of I I = the column vector of branch currents with the direction chosen arbitrarily but thereafter used to define the branch orientation Z = the primitive impedance matrix of self (diagonal) and mutual (off-diagonal) branch impedances
Note that we have used a script Z to indicate the primitive impedance. This will help us to distinguish clearly between primitive impedances and other important impedance quantities to be defined later. If we number the primitive impedances in an arbitrary way, Z will appear as follows for a network with b branches.
Z =
1
2
b
1 1-1 1 2 "2 1
"1 2
"Ib
,, 22
"2 b
" b2
"bb
b "b l
( 1 1 .2 )
The elements o f ( 1 1 .2 ) are defined as follows : .."" = self impedance of branch k ; k = 1 , . . . , b ""t. = mutual impedance between branches k and e ; k , e = 1 , . . . , b
In many networks the mutual impedance would be simply a mutual inductance . For mutually coupled transmission lines, however, every ,," l will have both real imaginary parts as noted in Chapter 4 . Notice that the diagonal elements of ( 1 1 .2 ) will always exist and will be finite and nonzero . Furthermore , if ( 1 1 . 1 ) represents the primitive equation for the positive, negative, or zero sequence network elements of a power system , the mu tual impedance terms will always be smaller than the diagonal elements (Why?). Since the diagonal elements of Z are all nonzero , the inverse of Z exists. Mul tiplying ( 1 1 . 1 ) by Z - I , we have where
1 = 'Y V - J
J = - 'Y E
(11 .3)
( 1 1 .4)
Equation ( 1 1 .3 ) is the Norton equivalent of ( 1 1 . 1 ) with the current source term defined with positive direction as shown in Figure 1 1 .1 (see [ 64 ] ) . Once the con straint matrix K for any network connection is specified , it is easy to show that
Fig . 1 1 .1 . Primitive network elements defined.
3 68
C h a pter 1 1
( 1 1 .5 ) which asserts mathematically that the sum o f the currents entering any node is zero . Therefore, when a transformation of the form V = KV ' ,
( 1 1 .6)
is used , the matrix expression I' = y'V'
( 1 1 .7 )
will be found from the primitive equation ( 1 1 .4 ) and the constraint ( 1 1 .5). Be cause of ( 1 1 .5 ) we may write the primitive equation as ( 1 1 .8)
J = 'Y V
This is the admittance equivalent of the problem explored in section 8.6 where a voltage equation was derived which was the inverse of ( 1 1 .7). Usually for power systems the primitive matrix is sparse, i.e., it contains many zero elements (but never a zero-diagonal element). This is because the lines in a power system are mutually coupled only to those lines which are physically parallel and located in close proximity . This condition usually exists where sev eral circuits share a common right-of-way.
Exa mple
1 1. 1
Compute the primitive 'Y and Z matrices for the small power system shown in Figure 1 1 .2 . The nodes are numbered in an arbitrary way and these numbers are
Fig. 1 1 .2 . A small power system with 5 nodes.
circled on the diagram. The circles numbered 4 and 5 are generators. The im pedances, including generator impedances, are also numbered in an arbitrary way, beginning in this case with the generators and proceeding through the lines. The impedances are taken to be pure inductive reactances for this example. These are all specified in Table 1 1 .1 , and all connecting nodes are also specified. This tabulated information completely describes the network for all sequences. Table 1 1. 1. Reactances of the System of Figure 1 1 .2 Self Impedance
Impedance Num ber
Connecting Nodes
Positive
Negative
Zero
1 2 3 4 5 6
4, 5, 1, 2, 2, 1,
0.25 0.20 0.08 0.06 0.06 0.13
0.15 0.1 2 0.08 0.06 0.06 0.13
0.03 0.02 0.14 0. 10 0. 1 2 0 . 17
1 3 2 3 3 3
Zero Sequence Mutual Impedance
Mu tual Element
0.05 0.05
5 4
Com puter Solution M ethods Using the Ad m itta n ce M atrix
369
We must somehow convey all this information to the computer in order to solve a problem . Solution
We may write the primitive impedance matrices for all sequences by inspection . For the zero sequence we have
1 1
Zo =
jO.03
2
0
3
0
4
0
5 6
0 0
2
3
0
0 0
jO .02
4
j O .14 0 0 0
0 0 0 0
5 0 0 0
0 0 0 jO.10
jO .05
jO .05
jO.12
0
0
6 0 0 0 0 0 jO.17
and we note that the matrix is very sparse, having only two off-diagonal entries. Except for the 2 X 2 matrix in position 4-5 , the inverse may be found by taking the reciprocal of each diagonal entry , i .e . ,
'1kk = 1 /J-kk
] [
]
where element k k is not mutually coupled. For position 4-5 we readily compute
[0.1 0
0.05 - 1
0.05
0 .1 2
=
1 2 .632 - 5 .263
- 5 .263 10.526
to write the primitive admittance matrix for the zero sequence as 1
3
0 0
0 - j50.000 0
4
0
0
5 6
0 0
0 0
1 2 'Yo =
2
- j 33.333
3
0 0 - j 7 .143 0 0 0
4 0 0 0 - jI2.632 j 5.263 0
5 0 0 0 j5.263 - j10.526 0
6 0 0 0 0 0 - j5.882
The primitive matrices for the positive and negative sequences are diagonal. These are left as an exercise (see problem 1 1 .10). Since the primitive matrices are so sparse, we usually store only the nonzero terms in the computer memory . The tabulated form of the data in Table 1 1 .1 is more efficient in terms of storage requirements and is the preferred storage for mat. It is implied that any element not specified is taken to be zero . 1 1 .2 Node I ncidence Matrix
We now develop a procedure whereby the primitive impedance or admittance matrix may be transformed into a form more useful for solution of the network. The primitive matrix itself provides no information concerning the way the var ious branches are connected to form a network. The connection information is
370
Chapter 1 1
known , however, and is easily tabulated as in column 2 of Table 1 1 . 1 . We now arrange this tabulated information in matrix form to indicate the nodes to which the various branches are connected . We shall call this connection matrix the "augmented node incidence matrix" and denote this array by the symbol where we define the b X m matrix
l A= 2
l
1
2
m
1
au a2 1
al 2 au
a, m aZ m
b
abl
ab2
ab m
A
( 1 1 .9)
The rules for locating the nonzero elements of A are simple. We begin with a b X m array of zeros. Then we make nonzero entries for each row in columns corresponding to the two nodes to which that branch number (row number) is connected . For example, in Example 1 1 .1 branch number 4 is connected to � nodes two and three . Then in row 4 (branch 4) of A the only nonzero elements will be those in columns two and three, i.e., a4 2 and a43 It is convenient at this point to record an orienta tion for each branch as well as its connection . This means that we record the arbitrarily chosen positive cur rent direction from Figure 1 1 .1 for each branch . These current directions may be noted on the diagram or may be chosen according to some arbitrary convention such as from the smaller to the larger node number. In any event these directions or branch orientations may be recorded in the matrix in the following way . Let •
apq
== =
A
+ 1 if curren't in branch p is leaving node q - 1 if current in branch p is entering node q 0 if branch p is not connected to node q
=
=
( 1 1 . 1 0) 1, b, q 1, m The matrix A is recognized to be similar to the connection matrix defined by (8.43) except that it conveys nodal rather than mesh information.
Exa mple
p
1 1.2
Form the augmented incidence matrix for the system of Example 1 1 . 1 . Take the positive direction of branch current always to be from the smaller to the larger node number.
Solution By inspection of Figure
1 1 .2 or Table 1 1 .1 we write
bm
2 1
A=
1
2
3
4
5
1
0
0
-1
0
0
0
1
0
-1
3 4
1
-1
0
0
0
0
1
0
0
5
0
1
-1 -1
0
0
6
1
0
-1
0
0
371
Computer Solution M ethods Using the Adm ittance M atrix
We now rewrite the primitive equation ( 1 1 .3 ) with the subscript b to indicate that these are branch currents and voltages. I b = " Vb - Jb
(11.11)
If we premultiply by A..t , we have the interesting result
At
I b = O = At " Vb
This result is the zero (null) vector since
At
-
At
Jb
( 1 1 .1 2 )
Ib = sum of currents leaving each node =
0
Again we illustrate by means of an example .
Example
Compute the product A t Ib for the network of Example 1 1 .l . 1 1.3
Solution 1
At I b
2 =
3 4 5
1
2
3
4
5
6
1 0 0 -1 0
0 0 1 0 -1
1 -1 0 0 0
0 1 -1 0 0
0 1 -1 0 0
1 0 -1 0 0
Ib l
1
Ib 2
2
Ib3
=3 4 5
Ib 4 fb S
fb l + Ib 3 + Ib6 - Ib3 + Ib4 + hs
Ib 2
-
Ib4 - Ibs - Ib6
- Ib l - Ib 2
fb 6
The first three rows of the result are obviously zero . Note that if nodes 4 and 5 are not connected to anything, Ib l and fb 2 are also zero . The product At Jb is the vector of all source currents entering each node. If we think of the network as being passive , these currents are usually thought of as external currents inj ected at each node, as indicated in Figure 1 1 .3. These curPASSIVE
N ETWORK
Fig . 1 1 .3 . Injected currents At Jb
=
1m .
rents are Ausually called simply the "node" currents and are conveniently desig nated by 1m . Thus ( 1 1 .1 3 ) where w e use the subscript m to remind u s that there ar e m o f these currents. Ob viously the sum of these m nodal currents is equal to zero , so they are not inde pendent. We will explore this property more fully later. We may compute the apparent power delivered to the network in two ways.
3 72
C h a pter 1 1
From Tellegren's theorem [ 7 7 ] we may compute the power entering the terminals of Figure 1 1 .3 or we may find the source power of each branch . These relations are given by
t j*m sm = Vm
S b = VP :
Since the network power is invariant then 8m
t jm* vm
(11 .14) =
Sb or
Vtb J b* (11.15) But from Jll�.13), �in<1e A i s real, j ! � A �J t and , substituting into ( 1 1 .15), we compute V!n I! = V!n AtJ: = V P : or V!n At = vt . Then the branch and node =
voltages are related by .
( 1 1 . 1 6) and from
( 1 1 .13) ( 1 1 .17 )
Equation (11 .17) is called the indefinite admittance matrix description o f the system [ 5 ] . It is indefinite because the voltage reference is not physically con nected to the network. This description has some other interesting properties which will be investigated later [ 5 ] . In equation ( 1 1 . 1 7 ) we define the indefinite admittance matrix as the m X m matrix
(11 .18) This computation i s illustrated by an example.
Example
1 1. 4
Compute the indefinite admittance matrix for the zero sequence admittances of the network of Example 1 1 . l o
Solution
The primitive admittance matrix was computed in Example 1 1 .1 and the node incidence matrix was found in Example 1 1 .2. Using these results, we easily compute the indefinite admittance matrix Y' o
1
Yo
= -j
1 2 3 4 5
2
- 7 .143 46 .359 - 7 .143 19.774 - 5 .882 - 12.632 0 - 33.333 0 0
3 - 5.882 - 1 2.632 68. 5 1 4 0 - 50.000
4
- 33.333 0 0 33.333 0
5
0 0 - 5 0.000 0 5 0.000
In forming the indefinite admittance matrix, we have found a network de scription which involves only the external network connections. In doing this, we have completely suppressed the details of what exists within these external con nections. This is a desirable feature since it permits us to describe the constraints
Computer Solution M ethods Using the Ad m itta nce M atrix
373
among the external ( or nodal ) voltages and currents in a most efficient way . Since power networks are large and computer memory is limited, this efficiency is an important consideration . We also recognize that the indefinite admittance matrix description is not exactly that required for a power system . Referring to Figure 1 1 .3 , we observe that the voltage reference is external to the network and the currents are not in dependent. Usually we would prefer to name some node in the network as the reference and measure all voltages with respect to that point. Having named a reference node, it is convenient to leave the current entering this node unspecified . Then the remaining m- 1 nodal currents are independent. 1 1 .3
Node Admittance and I mpedance Matrices
Consider a network consisting of n+ 1 nodes, including the reference node as shown in Figure 1 1 .4 where node 0 is chosen as the reference node . This net-
T;. +
Fig .
n+ 1
NO DE
-1.2
1!
NETWORK
'---r-r-----'
1 1 .4 . An n + 1 node network with node 0 as reference .
work has the same primitive (impedance or admittance) matrix whether a refer ence is chosen or not. It has a different node incidence matrix, however, which we shall designate as A (without the hat) . This network differs from the unrefer enced case in that the n voltages and currents subscripted 1 , 2 , . . . , n are inde pendent sets. The reference node current is ( 1 1 .1 9 ) and this current i s usually not required explicitly . The node incidence matrix for the network o f Figure 1 1 .4 i s formed using exactly the same rules as before except that no entries are made for the reference node at all , i.e.,
1 1 A=2 b where
]
2
n
al2
a,
a2 1
a22
a2n
ab l
ab2
ab n
["
( 1 1 .20)
apq = +1 i f current in branch p i s leaving node q = - 1 if current in branch p is entering node q = 0 if branch p is not connected to node q p = 1, b;
q = 1,
n
( 1 1 .2 1 )
C h a pter 1 1
374
This is the same as the augmented node incidence matrix A with all rows and columns deleted corresponding to nodes connected to the reference. The forma tion of A will be illustrated by an example.
Example
1 1. 5
Find the node incidence matrix for the zero sequence network o f the system specified in Example 1 1 . 1 .
Solution First we sketch the zero sequence network for the system of Figure 1 1 .2 where we shall consider the NO bus as the reference . Note that generator 4 ( on the left), represented by impedance element 1, is Y-ungrounded and is therefore not con nected to NO. Generator 5- is Y -grounded so its impedance, element 2, is con nected to NO. See Figure 1 1 .5 .
NO
Fig. 1 1 . 5. Zero sequence diagram for the network o f Example 1 1 . 1 .
Proceeding according t o the rules ( 1 1 .2 1 ) , we write
1
2
3
2
0
0
1
3
1
-1
0
0 5 0 6 1
1 1
-1
A=4
0
-1 -1
where we have ignored branch 1 entirely since no zero sequence current will flow in this branch (actually , branch 1 could be omitted from the diagram as it does not enter into any computation ) . Since branch 1 is eliminated, the branches copld be renumbered , but this is not essential . Note that this result is the same as A with row 1 and columns 4 and 5 deleted .
To compute the node admittance matrix, we proceed exactly as prescribed by ( 1 1 . 1 8 ) where we define the n X n matrix ( 1 1 .22) where now the primitive matrix must be that which actually includes only those elements used in the network. In Example 1 1 .5 we note that branch 1 is not re quired , so this branch (row, column ) must be eliminated from the primitive ma trix and the incidence matrix (or set 411 4 = 411 = 0 in all computations) .
375
Co m puter Solution M ethods Using the Ad m ittance M atrix
Example
1 1.6
Compute the node admittance matrix for the network of Example 1 1 .5 .
Solution
[
Equation ( 1 1 .22) may be applied directly to compute
1 Yo = - j
2 3
1
2
3
1 3.025
- 7 .143
- 5 .882
- 7 . 1 43
19 .774
- 12 .632
- 5 .882
- 1 2 .632
68.514
Yo
J
It is interesting to compare this result with computed inA Example 1 1 .4 . This result i s observed t o be the same a s the 1 -2-3 partition o f Yo if '11 = '11 4 = - 33 .333 is subtracted from position ( 1 , 1 ) . This is necessary since '11 is not re quired in the zero sequence network and c guld have been omitted from the be ginning. Rows ( 4, 5) and columns ( 4, 5 ) of Yo are eliminated since nodes 4 and 5 have become the reference node . It is possible to develop ( 1 1 .22) in a similar way to that by which ( 1 1 . 1 7 ) was derived. This is left as an exercise. (See problem 1 1 .1 9 . ) The result is important however and may be stated as I = YV
( 1 1 . 2 3)
where Y is the node admittance matrix defined by ( 1 1 .2 2 ) and V and I are column vectors of node voltages and currents defined in Figure 1 1 .4 . Since these voltages and currents are independent sets of variables, the inverse of Y exists, or we may write V = ZI
( 1 1 .24)
y- I
( 1 1 .2 5 )
where Z=
Note that this last operation was not possible i n the case o f the indefinite admit tance matrix since Y is singular. Many authors call Z the matrix of "open circuit driving point and transfer im pedances" and the matrix of "short circuit driving point and transfer admit tances . " They are also called Z-BUS and Y-BUS by some authors [ 64] . Before examining these important matrices in detail, however, it is instructive to study the indefinite admittance matrix to learn the important properties of this matrix network description.
Y
1 1 .4
I ndefinite Admittance Matrix
Consider the n + 1 terminal network of Figure 1 1 .6 where the network be havior is to be described in terms of the currents entering at nodes 0, 1 , . . . , n and the voltages measured as voltage drops to an external reference point labeled REF . As noted previously , only n of the voltages and n of the currents are inde pendent. From ( 1 1 .1 7 ) we write the indefinite admittance equation i = yv
( 1 1 .26 )
Chapter 1 1
376
...-! n �
..,
__
Fig. 1 1 .6 . An n
+
1 terminal network with external voltage reference.
which provides information concerning the terminal behavior of the network. Expanding ( 1 1 .26) we have
II
YO n Yi n
Vo
YlO
YO I Yl l
Yn O
Yn l
Ynn
Vn
Yo o
10 =
In
VI ( 1 1 .2 7 )
This equation shows clearly the way i n which the elements o f Y ar e defined . Each element is a "short circuit admittance . " For example, the kth column is defined by
( 1 1 .28) with all terminals except k shorted. For a given network we can let Vk 1 . 0, for example, and short all other ter minals. Then the kth column of admittances is equal to the currents entering the n + 1 nodes as specified by ( 1 1 .2 8 ) . This is a relatively simple task for most net works and can often be done by inspection . In general w e may state the rules for finding the elements o f Y as follows : =
Y'k I
=
I· Vf Vk '
, _
=
0 , f *- k
( 1 1 .29)
This technique is more direct than forming the augmented node incidence matrix. The procedure will be illustrated by an example.
Example
1 1. 7
Find the indefinite admittance matrix for the network of Figure 1 1 .7 by the application of ( 1 1 .29 ) .
Solution Figure 1 l .8a shows the network arranged with the condition
VI
=
1 .0 ,
V2
=
V3
=
V4
=
0
According to ( 1 1 .29), the four terminal currents should be numerically equal to column 1 of Y . By inspection
Computer Solution M ethods Using the Ad m itta n ce M atrix
4
Ion
2
Fig. 1 1 .7 .
Network for Example
II 12 13
Yl l
=
Y2 1 Y3 1
1 .5
=
Y4 1
14
- 1 .0 - 0.5
The second column of Y is found by setting V2 = 1 .0 ,
1 1 .7 .
0
VI = V3 = V4 = 0
as shown in Figure n .8b. Then we easily determine
II 12 13
=
14
Columns 3 and
�
- 1 .0
Y1 2 Y22 Y3 2
=
Y4 2
1 .25 - 0 .25 0
4 are found in exactly the same way with the result 4 3 2 1 Y=
1 2
3 4
1 .5 0
- 1 .00
- 0 .50
- 1 .00
1 .2 5
- 0 .25
- 0 .50
- 0 .25
0.85
0
0
- I
2.11
II
( a ) I.O V-
+
- 0.10
-
LJ 0.10 4
1 0.11
II _ I
114
4
10 .11 (b)
Fig. 1 1 .8 . Two steps in the solution of Figure 1 1 .7 .
371
378
Chapter 1 1
Note that the matrix is symmetric. Several other striking features of this matrix will now be discussed. 1 1 .4. 1
I ndefinite admittance matrix properties
The indefinite admittance matrix has several properties which are of consid erable interest in applications involving power systems. There are five major prop perties, all of which will be stated without proof [ 5 ] . 1 . The summation of elements in any row or column is equal to zero . 2 . If the jth terminal of an m terminal network is grounded (connected to the reference), the resulting admittance network description is that of an m 1 terminal network and is found by deleting the jth row and jth column of the indefinite admittance matrix of the m terminal network. This resulting matrix description is simply the admittance matrix (or definite admittance matrix) of the j-referenced network. 3. Connecting any terminals j and k together in a given network yields a new network whose indefinite admittance matrix may be formed by adding rows and columns j and k of the original indefinite admittance matrix. 4. The effect of connecting an admittance y between any two terminals j and k of a network is to add y to the major diagonal elements jj and kk , and to subtract y from the off-diagonal elements jk and kj. 5 . Any terminal k may be suppressed or eliminated from consideration (open circuited ) by performing a matrix reduction on the indefinite admittance matrix, pivoting on element kk, with the result being an m 1 dimensioned indefinite admittance matrix. This property can be easily visualized by re arranging the original matrix until the kth equation is the last (bottom ) one. Then -
-
( 1 1 .30) where the original matrix is partitioned as y
=
r���l. -��l LY2 1 I y2 2J
( 1 1. 3 1 )
would b e the element Y" " , ( 1 X 1 ), Y 1 2 i s m 1 X 1 , Y� 2 ' and Yl l i s a square (m 1 ) dimensioned array. Actually any number of nodes, say k nodes, k < m , may be eliminated simultaneously
Then Y2 2
Y2 1
=
-
-
by this technique.
The proof of the five properties will be left as an exercise (see problem 1 1 . 23). Instead of a detailed mathematical proof, the properties will be demonstrated by extending Example 1 1 . 7 to show the effect of each property.
Example
1 1. 8
Use the network of Example 1 1. 7 to demonstrate the five matrix properties.
Solution Property be
1.
The indefinite admittance matrix of Example 1 1 . 7 was found to
�
Computer Solution M ethods Using the Ad m itta nce M atrix
1
y=
2 3
4
2
3
1 . 50
- 1. 00
- 0. 50
- 1. 00
1.25
- 0. 2 5
- 0. 50
- 0. 2 5
0. 85
0
0
1
379
4
q
-O
- 0. 1 0
0. 1 0
A careful inspection of this matrix verifies that the sum o f elements in each row and each column is exactly zero. This is due to the fact that the four currents are not independent since they are constrained by Kirchhoff's law such that - 14 11 + 12 + 13 • Thus the determinant of Y is zero. Property 2. Suppose we ground node 2 of Example 1 1 . 7 as shown in Figure 1 1 . 9. Then the (definite) admittance matrix may be found by deleting row 2 and =
Fig. 1 1 .9 . Example network with node 2 grounded.
t
column 2 of the indefinite admittance matrix with the result
Y ( 2 gTd)
=
1
3 4
-
3
4
1. 50
- 0. 50
0. 50
0.85
- o lo
1
0
- 0. 1 0
�
l O. l �J
This result may be readily checked by inspection of Figure 1 1 . 9 and by applying the equation I Y V to that network. Property 3. To illustrate property 3 , we examine the network which results if nodes 1 and 3 are shorted together as shown in Figure 1 1 . 1 0. The third prop=
I
3
V
4
2
Fig. 1 1 . 1 0 . Example network with nodes 1 and 3 connected.
erty indicates that Y for this network may be obtained by replacing rows 1 , 3 and columns 1 , 3 by a new row and column formed by adding elements of these two rows and columns. The result is
3 80
Cha pter 1 1
1, 3
Y new = 2
4
t
1, 3
2
1.35
- 1.25
- 1.25
1.25
- 0. 1 0
�
4
J
- O 10
0
0. 10
Note that the result is still indefinite and all rows and columns sum to zero. Property 4. To demonstrate this property we consider the addition of a new admittance between nodes 2 and 4 as shown in Figure 1 1 . 1 1. According to I
2n
,on
:5
4
� 2
Fig. 1 1 . 1 1 . Example network with 5 ohm resistor added between nodes 2 and 4 .
�
property 4 w e simply add 0.2 mho t o elements 2 , 2 and 4, 4 and then subtract 0 .2 mho from elements 2, 4 and 4, 2 . The result is
Ynew
_
1
-3
2
4
1 1 . 50
2
3
- 1 .00
- 0. 50
- 1. 00
1
. 45
4
�
-O2
- 0. 2 5
�
- 0. 50
- 0. 2 5
0. 8 5
- 0. 10
0
- 0. 20
- 0. 10
0. 30
This result may b e readily verified by inspection o f Figure 1 1 . 1 1. Prop erty 5. This property gives the rule for suppressing a node of the original matrix. Suppose that the external current entering node 3 of the original matrix is zero. Then we may reduce the indefinite admittance matrix description by a matrix reduction, pivoting on element 3 , 3 . The new circuit will be that defined by terminals 1, 2, 4 as shown in Figure 1 1 . 1 2. We may write the original matrix with ordering 1 , 2, 4, 3 or 1
2
1
1 . 50
- 1.00
2 Y= 4
- 1. 00
1.25
o
0
I - 0. 50 I I - 0. 25 0 I 0. 1 0 I - 0. 1 0
0
-------- --- - - -
3
- 0. 50
- 0. 25
3
4
+
- 0. 1 0 I
----
0. 85
=
[��-�-Y,1 Y2 1
I
Fig. 1 1 .1 2. Example network with node 3 suppressed.
Y 22
Co m puter Solution M ethods Using the Ad m itta nce M a trix
381
where submatrices Y l I , Y 1 2 , Y 2 h and Y 22 defined i n ( 1 1 . 3 1 ) have been identified. Then we compute
Ynew = Y l l - Y 1 2 Y2� Y 2 1
[
2
4
1 . 50
- 1. 00
= 2 - 1.00
1.25
°
1
[ t
0
0
1 . 50
- 1.0 0
- 1 . 00
1 . 25
4
= =
1
0
0
J[ J J �
o
-
0. 1 0 °
o
0.10
1 . 2059
- 1. 1 470
- 1 . 1 470
1 . 1 76 5
- 0. 0588
- 0. 0 294
[
0.5:
- 0.25
- 0.10
-
[ �] 0. 5
�
[- 0. 50
2941
. 1470
.058
. 1 470
.0735
. 0294
. 0588
.0294
. 01 1 7
- 0. 2 5
- 0.10 ]
. 0.058
- 0. 0294 0.0883
The rows and columns still add to approximately zero but some round-off error is evident. 1 1 .4. 2
The indefinite admittance description of a power system
In the preceding section we developed a method for forming the indefinite ad mittance matrix of a network by constructing the primitive admittance matrix 'Y and the augmented node incidence matrix A with the result
( 1 1. 3 2 ) This method proved t o b e wasteful o f valuable computer memory since both A and 'Y are extremely sparse matrices. Furthermore, the matrix multiplication in dicated by ( 1 1 . 32) consists of a great many multiplications by zero. A much more economical way of forming the indefinite admittance matrix is possible by the direct application of property 4. This property asserts that the ef fect on the matrix of adding any element to a network is to increase the diagonal elements and reduce the off-diagonal elements by exactly the branch admittance added. This suggests a direct method for building the indefinite admittance matrix, one element at a time, by starting with a matrix of zeros and simply add ing each branch admittance as required by property 4 (see [ 6 5 ] ) . Consider the network of Figure 1 1 . 1 3 which may be thought of as one of the sequence networks of the system at any stage of its development in computer memory, and where the element Y is to be added between nodes r and s. For the original n + 1 terminal network prior to adding Y we write 10 = Yoo Vo + . . . + YOr Vr + YOs Vs + . . . + YOn Vn
Ir = YrO Vo + . . . + Y" Vr + Yrs Vs + . . . + Yrn Vn Is = Yso Vo + . . . + Ysr Vr + Yss Vs + . . . + Ysn Vn ( 1 1 . 33 )
Cha pte r 1 1
382
+
0
0-1 "
� -
It ----
, ,
1 0
In 10-1
-
n+1
I,
TER M I N A L
---
I,
N ETWORK
-
II -
10
Fig . 1 1 .1 3 . Adding a new admittance Y from r to s.
But we may also write the indefinite equations for the element Y which is to be added as
Ira = YVra - YVsa Isa = - Y Vra + YVsa
( 1 1. 34)
Now if the two networks are connected along the dashed lines of Figure 1 1. 13, the node voltages and currents are constrained as
Vra = Vn
V,a = Vs
I; = Ir + Ira,
( 1 1. 35)
I; = Is + Isa
( 1 1 . 36)
where we now define new terminals r' and s' with node currents I; and I� . Adding ( 1 1. 3 3 ) and ( 1 1.34) according to ( 1 1.36) and making the voltage substitution ( 1 1 . 3 5 ) into ( 1 1 . 34), we have
10 = Yoo Vo
+. . .+
I; = YrO Vo + I; = Ys o Vo +
.
.
. .
.
+
. +
YOr Vr
+.
.
. +
Yon Vn
( Yrr + Y) Vr + ( Yrs - Y) Vs + . . . + Yrn Vn ( YBr - Y) Vr + ( YSB + Y) VB + . . . + Ysn Vn (11.37)
Equation ( 1 1. 3 7 ) verifies property 4. Note that Y was added t o the network from r to s. This changed Y by adding Y to locations r, r and s, s and subtracting Y from locations r, s and s, r. This procedure is easily programmed for computer formation of Y, given only the tabulated line ( or branch) data similar to that of Table 1 1 . 1 . Furthermore, the line data can be in any arbitrary order which avoids costly sorting. Once Y is formed, the row and column corresponding to the reference node (NO, Nl , or N2) may be deleted to find the Y matrix, or these rows and columns could be ignored in the matrix formation . 1 1 .4. 3
Correcting for mutually coupled lines
Occasionally transmission lines are mutUally coupled, especially in the zero sequence network. This situation presents no particular problem in the formation of the indefinite admittance matrix. In the development which follows only two
Co mputer Solution M ethods Using the Adm itta nce M atrix
383
mutually coupled lines are considered. The method is easily extended to any number of coupled lines, however, with no increase in the complexity of the matrix formation algorithm. Consider the two mutually coupled transmission lines shown in Figure 1 1 . 14. Z rs
Irs
J::; r
Vr
-
Zpq
Vp
-.
s
'1
+ q +V s _
R EF
t
_
t
Fig . 1 1 .1 4 . Two mutually coupled lines.
[� ] []
From Chapter 4 we write the voltage drop equation
Vr Vp
Vs Vq
=
Vrs rZ rs = Vpq LZm
Inverting and solving for the currents, we have
] fIrs]
Zm Zpq
Llpq
( 1 1 . 38)
( 1 1 . 39) where the admittances are uniquely defined by the matrix inversion. Expanding, we have
Irs Ipq
= ( Y;s Vr - Y;s Vs) + ( Ym Vp - Ym Vq ) = ( Y;q Vp - Y� Vq) + ( Ym Vr - Ym Vs)
( 1 1 . 40)
or in general for a coupled line, Irs = (self admittance term) + (mutual admittance term), where the self admittance term is exactly the same as (11.34) for the un coupled line. The mutual term specifies that Ym be added in the rth row to the column having the same polarity as r, namely p, and subtracted from the column having opposite polarity, namely q. Equation ( 1 1 . 40) may be expanded to the full indefinite admittance form by adding two more equations which are the negative of ( 1 1 . 40). The result is
Irs = ( Y;s Vr - Y;s VS) + ( Ym Vp - Ym Vq) Isr = (- Y;s Vr + Y;s Vs) + ( - Ym Vp + Ym Vq) Ipq = ( Y;q Vp - Y� Vq) + ( Ym Vr - Ym Vs) Iqp = (- Y� Vp + Y� Vq) + (- Ym Vr + Ym Vs)
(11.41)
The network with elements to be added is shown in Figure 1 1 . 1 5. When the coupled elements are added t o the network, the following con straints are observed. ( 1 1 . 42 ) and
Chapter 1 1
384
n
..!!!-
q � p� • r
.!.!.. .!!-
n
+ I
TERMI NAL NETWORK
I �
o
..!o.
Fig. 1 1 . 1 5. Adding two mutually coupled elements.
I; I;
= Ir + Irs> = Ip + Ipq,
= Is + Isr I� = Iq + Iqp I;
( 1 1. 43)
Combining these constraints with the original indefinite admittance equations, we have 10 = YOO Vo +
.
.
.
+
Yo r Vr
+ Yo. V.
+
.
.
.
+ YO P Vp
I; = YrO Vo + · · · + ( Yrr + Y;') Vr + ( Y,.. - Y;') V. + · · · + ( Yrp + Ym ) Vp + ( Yrq - Ym) Vq + · · · + Yrn Vn
I; = Y.o Vo + · · · + ( Y.r - Y;.) Vr + ( Y.. + Y;.) V. + , · · + ( Y,p - Ym ) Yp + ( Y.q + Ym ) Vq + · · · + Y.n Vn ' + ( Ypp + Y;q ) Vp + ( Ypq - Y;q) Vq + · · · + Ypn Vn
I; = YpO Vo + "
' + ( Ypr + Ym ) Vr + ( YJII - Ym ) V. + "
I� = Yqo Vo + "
' + ( Yqr - Ym ) Vr + ( Yqo + Ym ) V. + · · · + ( Yqp - Y;q) Vp + ( Yqq + Y;q) Vq + , · · + Yqn Vn
(11.44)
From Figure 1 1 .1 5 we observe that terminals r and p are the dotted or polar ized terminals. Therefore Ym is added to all like (Le . , dotted or not dotted) loca tions namely rp and pr and also to locations sq and q s. By the same reasoning we sub tract Ym from all unlike locations, namely rq , qr, sp , and ps (see [65] ) . Extension of the above to any number o f mutually coupled elements follows exactly the same rules. Note that the mutual admittance always occurs between pairs of lines, and the rule for adding Ym to the matrix is the same for every pair of mutually coupled branches.
Example
1 1. 9
Compute the zero sequence indefinite admittance matrix of the system of Example 1 1 .1 by the direct method discussed above .
Solutio n From the data of Table 1 1 .1 and the network topology of Figure 1 1 . 5 (with
'114
= 'I t = 0) we record the self admittances by inspection as follows :
[V'
Computer Solution M ethods Using the Adm ittance M atrix
NO
NO 1
Y self -
0
0
2
2
3
0
0
- 41 2
41 3
V. - 41 4 - 41 5
413 + 41 6 - 1# 3
- 1# 2
3
1
-
- 1# 6
]
o
41 3 + 41 4 + 41 5 - 1# 4 - 41 5
41 2 + 1# 4 + 415 + 41 6
where from Example 1 1 .1 41 3 = j 7 .143,
41 2 = - j 50 .000, 41 5
41 4
-
= - j 1 0 .526,
385
=
-
j 1 2 .632
416 = - j 5 .882
Then NO NO - j 50.000 Yself =
1
2
3
0
0
j 50. 000
1
0
- j13.025
j 7 . 143
j 5. 882
2
0
j7. 143
- j30.301
j23. 1 58
3
j 50. 000
j5. 882
j23. 1 58
- j79.040
The mutual admittance between branches 4 and 5 is V m +j 5 .263. This admit tance should be added and subtracted according to ( 1 1 .44) , i.e . , =
Add V m to (r , p ), (p , r) , (s ,
q ), ( q , s ) Subtract v m from (r , q ), ( q , r ) , (s, p ) , ( p, s ) In this example r = p = 2 and s = q = 3. Therefore we must
NO 1
Y=
2 3
NO = j -
1 2
3
[ [
Add 2 41 m to ( 2 , 2 ) , ( 3 , 3 )
Subtract 2 V m from ( 2 , 3 ) , ( 3 , 2 ) NO 50 .000
3
2 0
0
j50 .000
0
- j 1 3 .025
0
j7 . 1 43
- j30.301 + 2 41 m
j 5 . 88 2
j 2 3 . 1 58 - 2 41 m
j 5 0 .000
NO
50 .000
-
1
1 0
j 7 .143
2 0
j 5 .882
3 0 50 .00
]
0
1 3 .025
- 7 .143
- 5 .882
0
- 7.143
19.775
- 1 2 .632
- 5 . 882
- 1 2 .632
68 . 5 1 4
50 .000
J
j 2 3 . 1 58 - 2 41m
- j79.040 + 2 y m
This is the same as the result in Example 1 1 .4 if branch 1 is neglected in that ex ample . Property 4 provides a straightforward algorithm for computing the indefinite (or definite ) admittance matrix for any network, including the effect of mutually
Cha pte r 1 1
386
coupled lines. There are other methods of accomplishing this same result such as the incidence matrix transformations ( 1 1 . 18 ) and ( 1 1 . 22). 1 1 .5
Definite Admittance Matrix
The indefinite admittance matrix defines the relationship between node volt ages and currents of a network in such a way that the details of the primitive admittances are suppressed. This is desirable since we seek a network solution where the total memory requirement for matrix storage is minimized. The indefinite admittance matrix is an efficient way of relating node voltages and currents and it may be formed with a minimum of effort. These are important considerations. Another important feature of the indefinite admittance matrix is its sparsity. This depends entirely on the network represented, but power systems are usually quite sparse, having between 1 .0 and 1 . 5 branches per node. This means that the indefinite admittance matrix itself has many zero elements and that considerable savings in memory requirements may be gained by storing only the nonzero ele ments in some simple tabular form. In power system analysis we are interested in the definite admittance matrix or simply the node admittance matrix Y defined by the equation I=YV
(11.45)
The admittance matrix is formed by crossing out the rows and columns of all nodes connected to the reference node. In a power system this will usually be the zero potential nodes NO, N1 , or N2. Moreover the n currents and voltages of ( 1 1 .45) are independent sets of complex quantities. Given the n independent voltages V, one can solve for the currents I by direct application of ( 1 1 .45). Also, since Y is nonsingular, one can find V when I is given by computing V=
y- I I Z I
( 1 1 .46)
=
In applying ( 1 1 .45) and ( 1 1 .46) to the solution of faulted networks, we usually make the simplifying assumptions given in Table 1 1 .2. Table 11.2. Simplifying Assumptions Used in Fault Calculations Assumption
Comment
1. All load currents negligible.
2. All generated voltages are equal in phase and magnitude to the posi tive sequence prefault voltage, h V,.
3. The positive and negative sequence networks are identical. 4. The networks are balanced except at fault points. 5. All shunt admittance (line charging susceptance, etc.) negligible.
Zabc are symmetric in all references for all lines. The only network reference connection is through generator impedances.
Computer Solution M ethods Using the Ad m itta nce M atrix
387
Since all generated voltages are equal in phase and magnitude, all generator nodes are connected together and a voltage h V, is applied at that point as indi cated by Figure 1 1 . 1 6 where only the positive sequence network is shown. Also, I u a Ou
� QQ - 199
POSITI V E SEQUEN
•
•
Q2
IQ2
_
QI IQI
N ETWORK
:Ii ,aO
- ll
l Q2
!!t-
ff
-.!L fI I ·
l QI
NO CONNECTION TO N I SINCE ALL S H UNT PATH S OPEN -----""
}iL� }
I FAIA.TS
Vo l- f l
+
NI B U S Fig. 1 1 . 1 6 . Positive sequence network which results from the assumptions of Table 1 1 .2 .
V, [ IIf,] lVaVall-_',J Ig Val-g Vr• Vk Val-k V"
since loads and shunt admittances are all neglected, the generator connections are the only connections to N1 , the reference node . The negative and zero sequence networks are similar except that the voltage h is zero and the generator nodes are connected directly to the reference . Since there are f fault nodes, e load nodes and g generator nodes, we may write the indefinite admittance equati o n for Figure 11.16 in terms of the positive sequence voltages as
=Y
( 11.47)
We now consider a change in reference . Since the only reference connections are made through the generators, suppose we select node 0 as the reference (this is already the case in the negative or zero sequence networks) . In terms of this reference all generator node voltages are zero and all load and fault voltages are the positive sequence voltage less h If we identify the node voltages with re spect to node 0 by a single subscript, the node voltage at any node k tnay be written as
- h
=
or in vector form V
Then adding Y VF
=
==
Va l - [ 1
0
k = 1 , 2, . . . , n
1
( 1 1 .4 8 )
from each side of ( 1 1 .4 7 ) , we compute
( 1 1 . 49 )
Chapter 1 1
388
Since nodes gl , g2, . . . , gg are connected to the reference, the last g rows and col umns of Y may be eliminated (property 2) to write the definite admittance equation
]
r- I, L If
=
Y
[V,V ] I
=
[
y" YI,
V� [ YuJ V,j Y,�
( 1 1 . 50 )
where the order of ( 11 .50) is (n g). Furthermore, since It = 0, we may reduce this expression to find I, with the result -
V,
I, = - ( Y,, - Y" Yi l yr ,) V,
( 1 1 . 51 )
where is the vector o f voltages applied t o the faulted nodes, measured with re spect to node O. For a 3rt> fault with zero fault impedance the sequence voltage Va l., is zero at the faulted buses. Then from ( 1 1 .48) the vector of voltages at all faulted buses is 1 where V, is the scalar prefault voltage, usually taken to be unity. Su bstituting into ( 1 1 .5 1 ) , we compute = 1 l ] t hV, ( 1 1 . 52) I, + (Y,, - Y,f Yr, I Yr ,) [ l The magnitude of I, is simply scaled up or down in pro portion to the magnitude of h V, Equation ( 1 1 . 5 2 ) provides a compact form of the solution for all fault cur rents I, under 3d> faul t conditions. If there is only one fault, this equation is scalar. Note that the procedure of finding I, for each of the n nodes requires n different complex matrix inversions each of order ( n - 1 ) . This is a serious dis advantage for this method and makes its use very expensive, even for small net works. For this reason an alternate method is sought. One alternative is to de velop an iterative technique which will converge on the solution If in a stepwise procedure. A more direct alternative is to invert the admittance matrix and find the fault currents by working with the impedance matrix. Often the matrix inversion is avoided by forming the impedance matrix without first developing the admittance matrix [64] . This procedure has its own special problems as noted in Chapter 1 2 . .
Example
1 1. 10
Compute the positive sequence admittance matrix for the network described in Example 1 1 . 1 and Table 1 1 . 1 . Then compute the total fault current 1. 2. 3. 4.
For a 3rt> fault at node 1 . For a 3rt> fault at node 2 . For a 3rt> fault at node 3 . F o r simultaneous 3 rt> faults at nodes 1 and 2 .
Assume zero fault impedance i n each case . Let h
= 1 and V, = 1 . 0 .
Solution From Example 1 1 .6 we have for the positive sequence network
389
Computer Solution M ethods Using the Adm itta n ce M atrix
1 Y= 2 3
[
]
3
1
2
j 2 4 . 19
j 1 2 .50
j 7 .69
j 1 2 .50
- j45.84
j 33.34
j 7 .69
+j33.34
- j 46.03
Since the faults are at different locations, ( 1 1 .52) must be solved separately for each case. F or example, for case 1 we compute If! = ( Y l l - Y t Yt, l Y r dh Vf l
or In = - j 2 4 . 1 9 - [j 1 2 . 5 0
j 7 .69]
[
In
= - j45 .8 4 - [ 12 . 50
j 33 .34]
and for case 3 If3
= - j46.03 - [j 7 .69
[ { [i;] [- ]
j 33 . 3 4 ]
]
j 3 3. 3�
j 33 . 34
- j 46.03
j24.19
j 7 . 6�
[-
Similarly for case 2 we have
r
[
r ] r r [ ] r [ J
- j45 .84
j 7 .69
- j 46.03
j24.19
j 1 2 . 50
j 1 2 .50
- j 45.84
]
l
1 2.50
j 7 .69
l
j 12.5� j 33. 34
1
j 7 .69
j 33.34
. = - J 7 .852
. = - J 7 . 436
= - j 8. 230
Case 4 is a simultaneous fault on nodes 1 and 2 , so the solution is =
=
j 1 2 . 50
- j 24. 1 9 j 1 2. 50
j 4. 8 3 5
_
j45.84
_
j 7 . 69
j 3 3 . 34
[ _ J' 46 . 0 3 ] - 1 [ J' 7 . 69 J' 33 . 34 ]
}[ ] 1 .0
1.0
- j 3. 6 2 2
The simultaneous faults are less severe than the isolated faults o n buses 1 or 2 because one contribution from the adjacent bus is zero . Example 1 1 . 1 0 illustrates the maj or difficulty of using the admittance matrix to solve for fault currents, namely , that every different fault location requires the inversion of a different matrix . To avoid this numerical problem, in Chapter 1 2 we shall examine the impedance matrix as a means of finding fault currents and develop a method of finding the impedance matrix which avoids a matrix inver sion . Problems
1 1. 1. 1 1.2. 1 1. 3.
Construct the positive sequence primitive impedance matrix for the network of Figure P 1 1 . 1 , using symbols z 1 3 , etc. , for the branch impedances. Construct the positive sequence primitive impedance matrix for the network of Figure P11.2, using symbols Z 1 3 , etc., for the branch impedances. Construct the positive sequence primitive impedance matrix for the network of Figure PH.3, using symbols Z 1 3 , etc., for the branch impedances. Repeat for the zero sequence and assume that Z24 and Z 2 S are mutually coupled by an amount Zm '
Chapter 1 1
390
4 2
Fig. P 1 1 . 1 .
'J------ 2
5
3
Fig . P l l . 3. 1 1. 4.
Construct the positive and zero sequence primitive impedance matrices for the network of Figure P 1 1 .4 , using the line data tabulated below and ignoring all resistances.
Branch Number
Branch Terminals
1 2 3 4 5 6 7
0-3 1-3 3-4 2-3 1·4 2-4 1·5 2·5 5-6
S
9
Rl
1 1. 5.
Ro Xo
Zero Sequence
0 0.10 0. 1 1 O.OS 0.10 0.20 0.15 0.25 0.35
0 0.30 0.40 0.30 0.35 1.00
jO.05 jO.90 jO.SO jO.75 jO.35 j 1.20
00
00
® \ 1 1.6.
Xl
Positive Seq uence jO.20 jO.30 jO.25 jO. 20 jO. 15 jO.50 jO.30 jO.40 jO.90
' �.
0.70 1.00
j 1 . 10 j2.50
Mu tual
JO.20
S
jO.20
6
• •
Fig. Pl l . 4 .
Repeat problem 1 1 .4, including the effect of resistances. Invert the positive and zero sequence primitive reactance matrices of problem 1 1.4 to find the primitive admittance matrices 'YI , and 'Y o respectively.
Co mpute r Solution M ethods Using the Ad m itta nce M atrix
11.7. 11.8.
391
Invert the positive and zero sequence primitive impedance matrices of problem 1 1.5 to find the primitive admittance matrices '11 and 'lo respectively. Find the zero sequence primitive impedance matrix for the network of Figure P11.4 if the lines 1-3, 3-4 , and 3-2 are all mutually coupled as follows: Zm(2,3)
= jO. 1 5,
zm(2,4)
=
j O .10,
Z m(3.4)
= j O.20
1 1 .9. Invert the zero sequence impedance matrix found in problem 11.8 to find the new 'Y o . 1 1 . 10. Compute the positive and negative sequence primitive Z and 'Y matrices for the system of Example 1 1 . 1 . 1 1 . 1 1 . Develop the augmented node incidence matrices A for the following networks with positive branch orientation from smaller to larger node numbers. (a) The circuit of Figure P I L L ( b ) The circuit o f Figure Pl1.2. (c) The circuit of Figure P l 1 .3. (d) The positive sequence circuit of Figure P 1 1 .4. (e) The zero sequence circuit of Figure P 11.4. Use data from problem 1 1 .4 for all networks. 1 1 . 1 2 . Develop a zero sequence equivalent network for the system of Figure 1 1.14 where the mutually coupled line pair is replaced by an equivalent circuit of self impedances. 11.13. Compute the indefinite admittance matrix Y for the following networks using the augmented node incidence matrices found in problem 1 1 . 1 1 and applying ( 1 1 . 18). (a) The circuit of Figure P I L L ( b ) The circuit o f Figure P11.2. (c) The circuit of Figure P11.3. (d) The positive sequence circuit of Figure P 1 1 .4. (e) The zero sequence circuit of Figure P11.4 . 1 1 . 14. Check the result o f Example 1 1 .4 by applying ( 1 1 . 18). 1 1 . 15. Verify ( 1 1. 17) beginning with ( 1 1. 13) and ( 1 1 . 14 ) . 1 1 . 16. Compute the node incidence matrices A for the zero sequence networks given by (a) Figure P1 1 . 1 , (b) Figure P11.2, (c) Figure P11.3, and (d) Figure P1 1.4. 1 1.17. Using the results of problem 1 1 . 16 , compute the corresponding zero sequence definite admittance matrices Y, using ( 1 1 .22). 11.18. Verify the computed result of Example 1 1 .6. 1 1 . 19. Verify ( 1 1 . 22). 1 1 . 20. Using a digital computer, invert the admittance matrices of each circuit of Figures P11. 1-P l l .4 to find the zero sequence impedance matrices for these circuits. 11.21. Compute the positive sequence indefinite admittance matrices of the following net works by applying ( 1 1.29). Neglect all resistances. (a) The circuit of Figure P I L L ( b ) The circuit of Figure P l 1 . 2 . ( c ) The circuit of Figure P 1 1 . 3 . ( d ) The circuit o f Figure P11.4. 11.22. Repeat problem 11.21 including the effect of resistances. 1 1 .23. Prove the five properties of the indefinite admittance matrix stated in section 1 1 .4 . 1 (see [ 5 ] ). 1 1.24. 1}eginning with the indefinite admittance matrices computed in problem 1 1 . 13, find the Y matrix description after the following circuit changes are made: (a) Grounding node 2 of Figure P11.2. (b) Grounding node 5 of Figure P11.4; (c) Connecting nodes 1 and 5 of Figure P11.3. (d) Connecting nodes 2 and 6 of Figure P11.4. (e) Connecting an impedance jO.1 between nodes 1 and 5 of Figure P11.3. (f) Suppressing terminal 4 of Figure 1 1.3. (g) Suppressing terminals 4 and 5 of Figure 1 1 .4 . 1 1 . 25. Compute the positive sequence indefinite admittance matrices for the following net-
3 92
Cha pte r 1 1
works, using "property 4" and ( 1 1.37) to build the matrix one element at a time. Compare the result with that of problem 1 1 . 13. (a) The circuit of Figure PILL (b) The circuit of Figure P11.2. (c) The circuit of Figure P11.3. (d) The circuit of Figure P11.4. 1 1 . 26. Compute the zero sequence indefinite admittance matrices for the network of Figure P11.4, using the result (1 1.44). Build the matrix one element at a time in the order given in problem 1 1.4. 1 1 . 27. Compute the 3ct> fault current using the admittance matrix technique ( 1 1.52) for the following systems (neglect resistance). (a) Figure PI LI with a fault at node 4. (b) Figure PI L I with faults at nodes 1 and 4. (c) Figure P11.2 with a fault at node 2. (d) Figure P11.3 with a fault at node 4. (e) Figure P11.3 with a fault at node 5. (f) Figure P11.4 with a fault at node 4. (g) Figure Pll.4 with a fault at node 5. (h) Figure P1 1.4 with faults at node 4 and 6. (i) Figure P1 1.4 with another generator at node 6 with impedance 0 + jO.1 and a fault at node 4. (j ) Same as (i) with faults at nodes 4 and 5. (k) Same as (i) with a fault at node 4 with fault resistance of 1.0 pu.
chapter
12
Co m p u te r S o l u t i o n M et h o d s U s i n g
t h e I m ped a n ce M a t rix
I n Chapter 1 1 an admittance matrix method was devised which could be used to compute fault currents in a power system. This method, although simple to implement, was observed to have certain disadvantages in its application to large networks. It was also observed that the impedance matrix , although more diffi cult to derive , has certain advantages for fault computations. This is primarily due to the impedance matrix being an "open circuit" network description , and this coincides with the open circuit approximation usually used in fault studies. This chapter will be devoted to a study of impedance matrix methods for use in fault studies. We shall also develop an algorithm for finding the impedance matrix which is more direct and cheaper to implement than performing an inversion of the admittance matrix. 1 2. 1
I mpedance Matrix in Shunt Fault Computations
Consider the network shown in Figure 1 2 . 1 where the network could be the positive, negative, or zero sequence network but where the reference is always node 0 of Figure 1 1 .16 or the common generator node. We arbitrarily define all . , n and all voltages to be the curren ts to be entering the network at nodes I , 2 , voltage drops from each node to the reference. Then we define the impedance matrix by the equation .
.
V=ZI
(12.1)
In expanded form ( 1 2 . 1 ) i s written as Vt V2
=
Vn
Zl 1
Z 12
Z tn
It
Z2 1
Z 22
Z 2n
12
Z nt
Z n2
Z nn
In
( 1 2.2)
If every current except that entering the kth node is zero , we have from ( 1 2.2) VI V2 Vn
Z u! =
Z2k Z nk
Ik ,
Ij =
0 , j =fo k ( 1 2 .3) 393
3 94
Cha pter 1 2
n
In
_
n-PORT
12
N E TWORK
O
GENERATOR NOOE
NODE VOL TA8 E REFERENCE
Fig. 12. 1 .
An n·port sequence network .
.. .
This defines the kth column of the impedance matrix since we may solve ( 1 2.3) to compute
Zih = (VdIh )rro , i,ph ,
i = 1, 2,
,n
( 1 2.4)
Since all impedance elements are defined with all nodes open except one, the im pedance elements are called the "open circuit driving point and transfer imped ances" or
Zih = open circuit driving point impedance, i = k = open circuit transfer impedance,
i =1= k
For small networks these impedances may conveniently be found by injecting = 1 .0 ampere (or 1 .0 pU), and solving ( 1 2 .4) for the 1 .0 ampere at node k , i.e., voltages. The impedance matrix can be used directly for fault computation if we apply the generator voltage to the faulted node in the positive sequence network as noted in Figure 1 1 .16.
Ih
1 2. 1 . 1
Three-phase fau lt computations
N1
Three-phase faults are computed by connecting the faulted node to the bus through a suitable fault impedance as noted in Chapter 3 and in Figure 3.19. When the generator node 0 is used as a reference, the positive sequence net work is as shown in Figure 12.2. Note that the fault current Ifh is the negative of
Zt
- () Zf
n
In - O _
- -..... r--
Ik Ifk n - PORT k _ _---f PO S I T I V E _-"""',.,.. NI SEQUENC BUS '-..._VO l + N ETWOR K I -0 2 .!.... +
/1 + 1 +;
Vk _V, _VI _
+
II- O
'----r-....
GENERATOR
NOD E
NO DE VOLTAGE REFE R E N C E
Fig. 1 2. 2.
Network connections for a 3IP fault on node k .
Ih , the current entering node k.
Since all currents except Ih are zero, ( 1 2 .4) de scribes the network exactly. Solving this equation for we have
Ih ,
Computer Solution M ethods U si ng the I m peda nce M a trix
= Vk /Zkk =
Ik
-
-
I'k
( 1 2.5)
From the way in which node k is terminated we also may write
Vk
= Z,IfIe
h V,
which we may substitute into (12 .5) to eliminate Vk and solve for Ifle or
I'k
395
= h Vr /ZT
(12.6) (12.7)
where ZT = Zkk + Z, . Thus the 3 � fault current at node k depends upon the open circuit driving point impedance at node k . This is the diagonal impedance element Zkk of the impedance matrix and may be thought of as the impedance seen look ing into the network at node k with all nodes except the kth node open. Once the total fault current I,k is known, we may readily find all the node voltages from the impedance matrix equation or, more simply , from ( 12 .3). Thus for any node i we may write
Vi
= Z ik lk
-
= Zik h Vr/ZT ,
i = 1 , 2, .
..
,n
(12.8)
These are the node voltages measured with respect to the reference node O. The positive sequence voltages are measured with respect to node Nl and are found by adding h V, to (1 2.8), or at node i we have
Vo l - i
( i;)
= h V, 1 -
(12.9)
Knowing the voltage at each node, we may compute the current flowing in all lines in the network. This requires knowledge of the primitive impedances. For any pair of nodes i and i we may compute the current flowing in line 'Jij as
Vi - Vj Ii ' =
= h V, (Zjk -
Zik )
(12.1 0)
'Jij ZT 'J 1i Note carefully that the capital letter Z with two subscripts refers to matrix ele ments whereas the script "li is the primitive impedance element between nodes i J
and i.
In summary we note that the matrix diagonal element is used in finding the total fault current, and the other elements of the kth column are used to find node voltages and branch currents.
Example
1 2. 1
Find the positive sequence impedance matrix for the system of Example 1 1 . 1 . Then compute the total fault currents for faults o n each bus. The fault impedance is zero and V, is unity . Let h = 1 .
Solution The positive sequence network is shown in Figure 12.3 where the two genera tor impedances have been connected to node O. According to ( 1 2.3) we can find the first column of the Z matrix by injecting 1 .0 pu current into node 1 . Then the three resulting voltages VI ' V2 , and V:3 will correspond to the first column im pedances Zl l , Z21 , and Z:31 respectively. This situation is shown in Figure 12.4 with the result
Chapter 1 2
396
J O. 0 6 3
2
)0.08
JO.20
0
j O. 2 5 _
REF
Fig. 1 2 . 3. Positive sequence network.
��
+ \3 - J O. 098
REF
�J �IJ �
Fig. 1 2. 4. Network solution when II
V2
=
V3
�
Z2 1
=
=
: 0.127 4
J
JO . 1 061
jO.0981
Z3 1
1 . 0.
�
The second and third columns may be found in the same way with the result 1
2
0 . 1 274
jO.1061
jO.098
jO.1061
jO.1 345
jO . 1 1 51
3 jO .0981
jO.11 51
jO. 1 2 1 5
1 Z=2
.
3
The 3 1/> fault currents are computed from ( 1 2 . 7 ) . With Vf = 1 .0, h = 1 , and Z, = 0 we compute 1(1 1(2 1(3
= 1 .0/Z1 1 = 1 .0/Z22 = 1 . 0/Z3 3
= 1 .0/jO.1274 = - j 7 .852 pu = 1 . 0/jO . 1 345 = - j 7 .436 pu = 1 .0/jO.1215 = j8.230 pu -
These results check exactly with the results of Example 1 1 .10. Equations ( 1 2.9) and ( 12.10) could now be used to compute the system positive sequence voltages and line contributions for each fault location . This is left as an exercise. 1 2. 1 .2
Single·line·to·ground fault computation
We may extend the impedance matrix solution to situations which involve the negative and zero sequence networks as well as the positive sequence network. In all sequence networks we assume that the common generator connection or node o of F igure 1 1 . 1 6 is taken as the refArence. Then we write V = Z I for all networks where I is the vector of currents entering each node in the sequence networks and
Com puter Solution M ethod s U si ng the I m peda nce M a trix
397
the voltages are the voltage drops from each node to the nodal reference bus 0 as shown in Figure 1 2. 5. The notation used to distinguish between sequences is also given in Figure 1 2 . 5 . In general the nodal subscripts read (sequence , node) in all :;:ases. On
I().n
10-2
10 - 1 01 o
In
_ r----,
-
-
r----,
1 2 I�
0
1;.
VI_ I
NODAL REF- N O
Fig. 1 2. 5.
I I_n
2n
I2-n
-
r---...,
2 2 I�
0
2 1 I�
2
V2-I - Val- I 0 1 - .1 NODAL REF - N 2
+t
ll:!NI BUS
Sequence quantities defi ned for nodal equations.
We further identify the se q uence voltages by the subscript a where necessary, especially in the positive sequence network . Using this notation, we may write the nodal equations in matrix form as (12.11) where the meaning o f these equations is clear from F igure 1 2 . 5 . S pecifically , the positive sequence equation is
I I-I
V i ol
ZH I
Z 101 2
Z Hn
V I _2
Z
1-2 1
Z lo22
Z I-2n
11-2
V I_ n
Z I -n l
Z I-n 2
Z I -nn
II- n
( 1 2. 1 2 )
We note one additional problem concerning notation . Here we define the se quence currents to be those en tering the network. At the faulted node , however, we identify the sequence currents lao , la l ' and la 2 to be those leaving the network. Where necessary to identify these fault currents we shall add the subscript a in the usual way . For a SLG fault on node k the sequence networks are arranged as in Figure 1 2 . 6 . Since the driving point impedance at k is known , we write
la o -k = la l -k = la 2 - k =
h V, h V, ;::, Z O - k k + Z I - k k + Z 2 - k k + 3Z , = ZT
( 1 2 .1 3)
where we define (12.14) Then the voltages at all nodes may b e computed from the currents o f ( 1 2 . 1 3 ) and the matrix. From ( 1 2 . 8 ) for any sequence s we may write
Z
V.oj = - Z.oj k h V, /ZT
(12.1 5)
This equation gives the sequence voltage i n the zero and negative sequence net-
398
C h a pter 1 2
On
l aO- k
-
+
Io-k -
Ok
-o _ 0 1 lCH
T+
r
� k vlQ.- 1
NO
3
Iol-k
-
l
Z
I I- k
-
t
1 -II- l'u2-k
I
1 ,_,. 0
II
f
0
-
Ik
1 +
-
VJ:: I
+ + VO I _ V'to!!. !!.
0
1,-n= O
In
k-�k
_
0
N I
12·n - O n _
«-,
I02- k 2 k 12-11 -
f.
'
-
V2-k V2 -1 1 - -r -
N2
2
12 _1-0
'+
0
Fig. 1 2.6. Sequence network connections for a SLG fault on bus k .
as in works. In the positive sequence network the voltage h V, must be added are voltages ( 1 2 .9). The resulting sequence
Va H Va 1-2 Va l-n
VaO-1 VaO-2
VO_1
Va o - n
Vo-n
VO-2
VI -2
+h
=
VI-n
Va2- n
=
( 12.16)
Zo - n k V,
V,
VI-I
VaH Va 2-2
= -
ZO _l k V, ZO -2 k h ZT
V,
=
h
V2-n
_
h V, ZI-2k
ZT V,
V, V2_1 V2_2
V,
Z l-I k
Z I- nk
(12.17)
Z 2_l k =
_
h V, Z2-2 k
ZT Z2- n k
( 12 .18)
3 99
Computer Solution M ethods U si n g the I mpeda nce M a trix
The currents flowing in the branches of the positive and negative sequence net works may be computed from equation ( 1 2 .10) . To compute the current in branch i-j in the zero sequence network which is not mutually coupled, again use (12.10). If this element is mutually coupled to another element or elements, ( 1 2 . 1 0) must be solved simultaneously for all coupled currents. In full double subscript notation the solution becomes very messy to write. For example if ele ments i-j, m-n , and p-q are mutually coupled in the primitive matrix, the solution for all coupled currents involves the inverse of that partition of the primitive im pedance matrix with the result
[ ] �.iJ-iJ
. i}- mn
ii
_
Im n
Ip q
- .mn -ii 'fp q - il
i j [lie i j
. J -pq
-
I
� Z le
. mn-mn
.mn-p q
Znle
f'p q -m n
.pq -pq
Zqle - Zple
Z
Zmk
h
V,
ZT
( 1 2 . 1 9)
where all impedances denoted l' are primitive zero sequence impedances. As in the 3 rt> fault we note that the diagonal element of the matrix helps us find the total fault current, and the remaining matrix elements from the k th column help us find the node voltages and hence the branch currents. Usually we save matrix storage by assuming that the positive and negative se quence networks are identical. This changes Z T to and changes the
Example
Z
Z T = Z O - Iele + 2Z l -k le + 3Z,
( 1 2 .20)
subscripts in ( 1 2 . 1 8) to indicate positive sequence impedances.
Z
1 2. 2
Compute th e zero sequence impedance matrix for the system of Example 1 1 .1 . ·rhen, using the positive sequence matrix from Example 1 2. 1 , compute the total fault current for a SLG fault on bus 2 and find all voltages and branch line cur rents. Let h = 1 .
Solution
The zero sequence network is shown in Figure 1 2 . 7 . We find the impedance matrix tliis time by inverting the admittance matrix of Example 1 1 .6 with the result
Zo
1
�
3
1
1
2
001157
j O .0546
J Oo02
j O.0831
jO .0200
j O.0200
j O .0200
= 2 jO .0546
3 jO .0200
jO. IO 2
jOl 4 REF
Fig. 1 2 . 7 .
Zero sequence network.
Chapter 1 2
400
Notice that the technique of injecting 1 .0 pu current into node 1 to find column 1 poses a more difficult problem than in Example 1 2 . 1 because of the mutually coupled elements. The solution to this problem will be considered in sections 12.7 and 1 2.8. In the following computations we assume that the fault impedance is zero and the negative sequence impedances are equal to the corresponding positive sequence . impedances. Then from ( 12.14) with k = 2
ZT =
ZO -22 +
2Z1- 22 = = 02 =
j O.0831 + j O.2690
= jO.3 5 2 1
Then from ( 1 2. 1 3 ) the sequence fault currents are 1
�VoOVoo•.2 � � [- � VoO-3J OI . � � � [ � � �VV0ol1••32J �V02-02.2j � � [ � 100- 2
= 101•2
-
2
1 .0/jO.3 521
=
-
j 2 .841 pu
The sequence voltages are computed from ( 1 2.16)-( 12 .18) as ' 0 .054
= j 2.841
0.1551
� 0.0831 =
- 0.2360 - 0.0568
J O .0200
j o . 106
.O
=
pu
.6986
1 . 0 + j2.841 jO.1345 = 0.6 1 80 1.0
jO. 1 1 51
V02•3
0.6730
0'301
'0' 106
= j 2. 841
pu
� 0. 1345 = JO. 1 1 51
pu
0.3820
-
- 0.3270
There are three lines contributing to the total fault current at bus 2, one line from bus 1, and two parallel lines from bus 3 ( see Figures 1 2. 3 and 1 2. 7 ) . The contribu· tion from bus 1 is computed from ( 1 2. 10).
1-12 2-12 Vf (ZZI-22TI-I.12- ZI _12) Zo-d 0-12 - V, (ZO-22 ZTi-0-12
1
=I
=
h
-h
1
-
-_
j2.841 (j0. 0284) J'0 08
_ =
= - J' 1 . 008
pu
•
j2.84 1 (j0.0285) - - J' 0 . 578 pu ' J 0 14 •
For the parallel lines in the positive sequence network we again apply ( 1 2. 1 0). Since the positive sequence line impedances for are equal, the current divides equally. 1
ZI -32) 1-32 2-32 V,(ZI_22 Z i-32 =1
=
h
T.
=
i-23
_ j2.841 00. 0194) J'0 06
= _ J' 0 . 916
pu
•
In the zero sequence network the self impedances are unequal and the lines are
mutually coupled. We use the 4-5 position of the primitive admittance matrix to compute from ( 1 2. 1 9)
[10-32(4)J 10-32(5)
-
'
- - J2.841
l
j12.632 . J 5. 263
j5. 263 .
J
- J1 0 . 5 26
[
]-[
]
jO.06 3 1 j 1 . 320 . . - JO.943 JO.063 1 -
pu
Com puter S o l u tion M ethods U si ng the I m pedance M a trix 1 2. 2
40 1
An I mpedance Matrix A lgorithm
The impedance matrix is an excellent network description to use in the solu tion of faulted networks. If load currents are neglected, the fault information needed, both for total fault current and branch currents, is easily computed from the impedance matrix elements. Although it is easy to use for fault computations, the impedance matrix is not as easy to form as the admittance matrix. Furthermore, whereas the admittance matrix is extremely sparse, the impedance matrix is completely filled. It is sym metric, however, so that only n(n + 1 )/2 impedances need to be saved in computer memory. This requires n(n + 1) memory locations if the impedances saved are complex. However, many systems have high x/r ratios so that sufficient accuracy is obtained by using only the reactances. One difficulty associated with forming the impedance matrix is the ordering of the primitive elements. Ordering is no problem in forming the admittance matrix. This is because the indefinite admittance matrix exists and all matrix elements are defined when the voltage reference is arbitrarily selected. This is not true for an impedance matrix description. Even the simplest possible network consisting of one element has no finite impedance matrix description unless the element is connected to the voltage reference. This is because of the way the elements of the matrix are defined as given by the equation
i = I, 2, . . . , n
( 1 2. 21 )
I f an impedance element i s not connected in any way t o the reference, Vi is clearly infinite when III is given any value ( such as 1 . 0). This characteristic of the impedance matrix suggests that the primitive matrix impedances should be sorted beginning with an element which is connected to the voltage reference (bus 0) and with each succeeding element connected directly to one of the previously entered elements. Thus an algorithm for building the impedance matrix in a step by-step manner is required. Much has been written on this subject, as noted in [ 67-72] which describe early attempts at implementing the impedance matrix in fault studies. These works are recommended reading on the subject, particularly the work of Brown et al. [ 70, 7 1 ] and E I-Abiad [ 72 ] . The analysis that follows should be regarded as tutorial and not necessarily the best or the only way to con struct an impedance matrix. In the remainder of this chapter we will consider the various ways a network may be changed by adding a new radial branch or closing a loop thereby changing the "previously defined" Z matrix made up of elements added prior to the one under consideration.
1 2. 3
Adding a Radial I mpedance to the R eference N ode
Consider a network in which p nodes have already been defined and the Z matrix has been computed. We now introduce a new branch of im pedance f- from the reference to a previously undefined node q as shown in Figure 1 2. 8. Applying ( 1 2. 2 1 ) with III = Iq we can determine the qth column of the Z matrix. Since the impedance .. is connected only to the reference and all currents except Iq are zero, all voltages are zero except Vq which is given by ( 1 2. 2 1 ) ; or the new impedance matrix is
(p X p )
C h a pter 1 2
402
-
2 P
q
!.tIp -
P -PORT
NETWORK
�1
0 REF
Fig. 1 2.8. Adding a new branch radial from the reference.
2
p
q
ZI l 2 Z2 I
Z 12 Z22
Z IP Z2p
0
Zp l
Zp2
Zpp
0
0
0
0
..
1 1 Z=
P
q
0
( 1 2. 22)
We summarize as follows : Conditions: 1 . Branch added from reference to q. 2. q previously undefined. 3. New branch not mutually coupled to other elements. Rule : 1 . Set Zqq = ,.. 1 2. 4
Adding a R adial Branch to a New N ode
We now consider the addition of a radial branch with self impedance ,. from a previously defined node k to form a new node q as shown in Figure 1 2. 9. Again we use the defining equation ( 1 2. 2 1 ) to find the qth column. If we let Iq = 1 . 0 , the voltages VI ' V2 , , Vp , Vq will be numerically equal to the qth column im pedances. But letting Iq = 1 . 0 gives the same voltage distribution for the new net work as letting Ik = 1 . 0 in the old network (without the added branch). These voltages are precisely the kth column impedances. Element Zqq is the driving •
•
•
IL
q
� , l. . IP k _\. k
2 ,
Fig. 1 2.9.
.!L ..!!..-
P-PORT
NETWORK
10
Adding a new branch radial from node k .
Computer Solution M ethods U si ng the I mpedance M atrix
403
point impedance at q which by inspection is ( Zkk + ,.). Therefore the new matrix is given by 1
2
k
p
q
1
Zl I
ZI2
Zlk
Z Ip
Zlk
2
Z2 1
Z22
Z2k
Z 2p
Z 2k
Z= k
Zk l
Z k2
Zkk
Z kp
Zkk
P
Zp l Zk l
Zpk Zkk
Zpp Zkp
Zpk
q
Zp2 Zk 2
(Z kk + ,. )
( 1 2. 2 3 )
Conditions : 1. Branch added from k to q. 2. k previously defined. 3. q previously undefined. 4. k is not the reference node. 5. p nodes previously defined. 6. New branch not mutually coupled to other elements. Rules : 1. Set Zi q = Z i k , i = 1, 2, . . . , p . 2. Set Zqi = Zk h i = 1 , 2, . . . , p. 3. Set Zqq = Z kk + ,.. 1 2. 5
Closing a Loop to the R eference
Now consider a case in which an element is added from the reference to a node k which has been previously defined. In other words, the element being added is a link branch since its addition will close a loop in the network. The added element is not mutually coupled to other network elements. We shall do this in two steps. First we add the branch radially from node k to a new node q. The result of this step is given by ( 1 2.23). Then as a second step we connect node q to the reference as indicated by the dashed line in Figure 1 2. 1 0. If the impedance matrix for the original p-port network was p X p, the P
Iq -
lq I
I I
I
I
I
1
k
2
Ip
-
L\
�
P - PORT
N ETWORK
II
-
I I Fig. 1 2 . 1 0.
to
Adding a link branch from k to the reference.
branch added radially gives the impedance matrix ( 1 2. 23 ) which is (p + 1 ) X (p + 1 )
Cha pte r 1 2
404
If we write out the voltage equations in matrix fonn, we have following the first step
( 1 2. 24 )
Vq
Then, upon connecting node q t o the reference, = 0 , a Kron reduction may be perfonned to eliminate Iq as a variable , reducing the matrix once more to p X p . In matrix notation ( 1 2.24) may be written as
[�::Zpl . . . ZZppIP] - [.ZZpk.1.12] Zd1 [Zkl . . . Zkp Zd Zkk
Then the new impedance
from which we compute matrix may be written as Z=
where
=
•
.
•
•
.
.
.
.
.
] ( 1 2 .2 5 )
+ ,. .
Conditions: 1 . New link branch added from k to reference. 2 . k previously defined as one of p nodes . 3 . New branch not mutually coupled to other network elements. Rules: 1. Set q = p + 1 . = i = 1, . . , po 2 . Set = i = 1, . . . ,po 3 . Set + •. 4. Set Zqq = 5 . Eliminate row q , column q by Kron reduction . 1 2.6
Z'Zq'q ZkZ'kl t, Zkk
.
Closing a Loop Not I nvolving the R eference
Consider the addition o f a new element between nodes i and k where nodes i and k are both previously defined as part of a p-port network shown in Figure 1 2 . 1 1 . Assume that the added node is not mutually coupled to any other net work element. Assume further the total number of nodes defined to be p ;;., i, k and that neither i nor k is the reference node . In this case as in the previous one it is convenient to think in terms of adding the new branch from k to a new tem porary node q. Then as a final step we may connect q to i. The first step, being the same as that of the previous section, results in precisely the same equation ( 1 2 .24 ) . In this case , however, the second step of connecting q to i gives
Vq
-
V,
=0
( 1 2.26)
Therefore, we replace the qth equation of ( 1 2 .2 4 ) by the difference between the
40 5
Computer Solution M ethods U si ng the I m peda nce M a trix
P
Iq :.!L
-
q
Ii
11
--
"' - - -1� -
Iq
Ik
I
k
-
1.
I
Fig. 1 2. 1 1 .
P - PORT NETWORK
--
to
Adding a link branch from i to k .
qth and ith equations. At the same time we recognize that the current I, shown in Figure 1 2 . 1 1 is not the new nodal current at this node . Once q is connected to i the total current entering this node is (Iq + Ii ). Changing all curre nt variables in the ith row from Ii to (Iq + Ii ) requires subtracting column i from column q . The result of both operations on ( 1 2 .24) is 1
V,
V. Vp V. - Vi
ZI I
k
p
q
Z"
Z ,.
Z 'P
(Z , . - Z I i )
I,
ZI/
Z,.
Zip
(Z,. - ZI/ )
1. +1,
Z.,
Z••
Z."
(Z.. - Z., )
I.
( Zp. - Z,., )
Ip
k
Z. ,
P q
Zp ,
Z,.,
Zp.
Zpp
(Z. , - ZI ' )
( Z.i - ZI/ )
(Z•• - ZI. )
(Z"' - Zip )
Zd
( 1 2.27)
I.
where Zd = Zit + Zkk - Zik - Zki + .. = Z jj + Z q q - Ziq - Zql (of 1 2 . 2 3 ) . = 0, the last row and column are eliminated by Kron reduction Since q with the result
V - Vi
][
Z IP . . . Zpp
_
Zlk - ZIi . . . Zpk - Zpi
]
Zd- l [ ( Zkl - Zu ) . . . (Zkp - Zip ) ]
( 1 2 .28)
Conditions: 1 . Branch added from i to k . 2 . i and k both previously defined . 3 . p nodes defined in all. 4. Added branch not mutually coupled to other network elements. Rules: 1. Set q = p + 1 . 2 . Set ZJq = Zjk - Zjj , j = 1 , . . . , p . 3 . Set Zqj = Zkj - Zij , j = 1 , . . . , p o 4 . Set Zqq = Zjj + Zkk - Zik - Zk/ + .. = Z / j + Zqq - Zjq - Zq j . 5 . Eliminate row q and column q by Kron reduction.
Cha pter 1 2
406
This completes the four basic configurations needed to add a new element to the positive sequence network. In the zero sequence network, however, the added branch may be mutually coupled to some other network element. Since con nections to the reference node 0 are all generator impedances, we need not con sider mutual coupling for these elements. The cases of interest then are the addi tion of a radial line (tree branch) and the addition of a line which closes a loop (link branch) where the reference is not involved. These cases are investigated following Example 1 2 .3 which involves no mutual coupling.
Example
1 2. 3
Consider the positive sequence network given in Table 1 1 .1 and shown in Figure 1 2 .3. Take the branches in the order given in Table 1 1 . 1 and construct the impedance matrix using the rules of sections 1 2 .3-1 2 .6 .
Solution
From Table 1 1 .1 we have the positive sequence data as follows: Branch Num ber
Connecting Nodes
Positive Sequence Impedance
1 2 3
0-1 0·3 1-2 2-3 2-3 1-3
jO.25 jO.20 jO.08 jO.06 jO.06 jO. 1 3
4 5 6
If we introduce the elements in the order 1 , 2 , . . . , 6 , we will build up the network piecemeal as shown by the six steps of Figure 1 2 . 1 2 . Sketching the net work is not necessary , but it is helpful in describing the process. The actual STEP 1
. 1_
=1-0
2 ,.3
-4---1
STEP 4
-1
2
0
---
3
ST E P
2
2
------+ 0 STEP
3
2
Fig. 1 2 . 1 2.
3
2
3
-------t o
3
� 2 �_--=-__�
STEP
5
3
2
3 0
-::-
2 tf--�--� 3 STEP 6
----'----+ 0
Step-by-step construction of the positive sequence network.
Computer Solution M ethods Using the I m pedance M a trix
407
decision-making process needed at each step can be done by computer program ming techniques . We begin with a computer impedance array which is initialized to zero . Since the resistance of all elements is negligible, only one such array is required and it should be at least 4 X 4. Step 1 . Introduce element 1.
"0 1 = j O. 25 Since this is the first element, it is a radial element from the reference to node 1 as shown in step 1 of Figure 1 2 . 1 2 . If this first element were not con nected to the reference, it would be necessary to sort the elements to find one which was connected to O. The rule for introducing an element radial from the reference is given in ( 1 2. 2 2). Thus we complete step 1 by adding 0.25 in location q-q or 1-1 with the result. 1
3
3
� � L �
� r:
z=j
2
5
0
0
Step 2. The addition of "0 3 = jO.20 is performed exactly the same way as the previous step except that the diagonal term affected is element 3-3.
1
Z =j
2 3
r
1
2
3
.25 0 0
Step 3. We now add a radial element "12 = jO.08 as shown in Figure 1 2 . 1 2. The rule for doing this is given by ( 1 2 . 2 3 ) where k = 1 , q = 2 . Thus the first row is added to the second row and the first column to the second column. Then Z 22 is set equal to Z 1 1 + 0.08. 1
Z=j
� �: � : 3
L
0
2 0.25 0.33
o
Step 4. This addition of the fourth element of h3 = j O.06 closes a loop not involving the reference. The rule for this situation is given by ( 1 2 .28 ) which is a Kron reduction of ( 1 2 .27 ) . We begin with the 4 X 4 array where k = 2 , i = 3 . Thus we have an augmented matrix 1
2
3
0 .25
0.25
0
2 0.25
0.33
0
1
Z =j 3
o
0
0.20
-------------
q 0.25
0.33
- 0.20
q I I I I
0.25 0.33
1 - 0.20
J I I
----
0.59
408
Cha pter 1 2
where (col 4) = (col 2 ) - ( col 3) and similarly for row 4. Element 4-4 is Zd = Z 22 + Z 33 - Z23 - Z 32 + jO.06 = j O . 59
�
�
Performing the Kron reduction to eliminate row q and column q , we have 1
2
3
0 . 1441
0. 1 1 02
0.0847
2 0 . 1 1 02
0.1454
0.1119
3 0.0847
0. 1 1 19
0.1322
1 Z =j
Step 5 . The fifth element "23 = jO. 06 calls for exactly the same operation as the previous step. We begin with the 4 X 4 array. Again we have k = 2, i = 3 .
Z=j
�
1
2
3
1 0 '1441
0 . 1 1 02
0.0847
2 0.1 102
0.1454
0. 1 1 1 9
I I
q 0.025
i
0.0335
3 -------------0.0847 0 . 1 1 1 9 0 . 1---+----322 I 0.0203 q 0.0255 0.0335 - 0.0203 I 0.1 1 38
�
-
This array is reduced to find the matrix at this step to be
Z =j
2
1
3
J
1 0 . 1 384
0. 1027
0.089
2 0. 1027
0.1 356
0 . 1 1 78
3 0.0893
0. 1 1 78
0.1286
The last element to be introduced is "1 3 = j O. 1 3. This element also closes a loop, so we begin with the 4 X 4 array, with k = 1, i = 3 .
Step
6.
1
Z
=
3
2
1 0.1 384
0 . 1 027
2 0 1027
0.1 356
:
j 3 0 0893 q 0.0491
0 . 1 1 78
0.0893 I 0. 1 1 78
0.1286
�
I
0.0491 - 0.01 51
1 - 0.0393
T - 0.0393 I 0.2184
--------- ---------
- 0.0151
I
q
-----
�
After Kron reduction, pivoting on element q-q , we have the positive sequence impedance matrix : Zl
=
j
0 .1273
0 . 1 06 1
0.098
0 . 1 061
0.1 345
0.1151
0.0981
0.1 1 5 1
0.1215
1 2.7 Adding a Mutually Coupled R adial E lement
We now consider the addition of a radial element from node p to a new node q where the added element is mutually coupled to the network element from m to n. We assume that nodes m, n, and p have been previously defined in a network with p total nodes. (N ote there is no loss in generality by letting p be the highest
Com puter Solution M ethods Using the I m pedance M atrix
r� ,pq
409
•
q _"'Nv---.....
P - PORT
NETWORK
o
Fig. 1 2 . 1 3.
Adding a mutually coupled radial element.
numbered node .) Assume t.hat there is an impedance lin n between nodes m and n and the new impedance .pq is mutually coupled to 'l- m n by a mutual impedance I'M . The network connection is shown in Figure 1 2. 1 3. The voltage drop for two mutually coupled lines is given by ( 4 .1 4 ) . For the two elements under consideration here we write
[
l Vp q J
Vm n
= rL p
] [f-M
Vm - Vn V - Vq
Solving for the currents, we write
=
.m n
J
f- M
.p q
r Im n �p q = - Iq
J
( 1 2.29)
( 12 . 30) where the admittance matrix of ( 1 2. 30 ) is the inverse of the impedance matrix of ( 1 2 .29) . Suppose we write the equation for the new network of Figure 1 2 . 1 3 in symbolic form, i.e . , 1
k
m
n
p
q
-
VI
1
Zl l
Zl k
Zlm
Z ln
Z iP
Z lq
Vk
k
Zk l
Zk k
Zk m
Zk n
ZkP
Zlrq
Zm lr Z nk
Zmm
Zm n
Znm
Znn
Zm p
n
Zm l Zn l
P q
Z-p l Zq l
ZP Ir Zq k
Z-pm Zq m
Z-pn Zq n
Vm
Vn Vp
Vq
=m
Znp Zpp
Zq p
-
-
Z-m q Znq -
Zpq Zqq
II IIr 1m In Ip Iq (12.31)
where the elements of the qth row and column are completely unknown as yet. This fact is acknowledged by the tilde placed over these elements. The matrix elements of Z's without tilde ' s are the original Z ' s. Finally we note that caution must be observed to clearly distinguish between branch impedimces f- m n and f- p q and matrix elements Zm n and Zp q .
C h a pter 1 2
410
To determine the qth row and column we make the following test. Let
1/2 =
1,
k
j ", k ,
Ij = 0 ,
=
1 , 2, . . . , p
( 12.32)
1/2
= 1 at This requires that we methodically move from port to port injecting each port with all other ports open. Then for injection at the kth node we com pute the voltages at all nodes as
VI V/2 VVnm �p Vq
�.'�
Zu = Zm /2 Zn/2 Zp/2Zq/2 J
Iq mn
. . .
..
( 12.33)
Note that = 0 for all tests. Therefore, the voltage drop from m to n is a func or tion only of I
1/2 lImnJ1 = L"Mfllmn
(12.34)
Imn 'fmn . Vpq ]lIpq [VVpqmn] = ["mn lIlIMpqJ [VmVp � VqVnJl
= 1 may cause a current The injection of to flow through If so, this current will induce a voltage because of the mutual coupling. From ( 1 2. 30) with q = 0 we compute
I
11 M
o
11M
( 12.35) The second equation of ( 1 2 .35) may be written as
'Ipq Zq/2) Zq/2 Zpk IVpq)(Zm/2 - Zn/2) Vmn _'fpq 1Ipq
0 = 11M ( Zm k - Znk) +
( ZPk
-
from which we compute
=
+ ( 11M
(12.36)
However, from ( 12.29) and ( 12 . 30) we have the relations -
�,
_� -
Ll
( 12.37)
from which other forms of the solution are possible. One convenient form is the following:
( 1 2.38) This equation may be used to find the first q 1 elements of the qtQ ro w and, by symmetry, the qth column. The only unknown then is the -
element Zqq .
Com puter Solution M ethods Using the I m pedance M atrix
41 1
To find ZQQ ' we devise the following test. Let Iq = 1 ,
/ l = I" = "
' = lp = 0
(1 2 .3 9)
Then from ( 1 2.31) we have Vl
ZlQ
Vm
Zm q
Vn
= Znq
Vp
Zpq ZQQ
VQ
( 1 2 . 4 0)
where the tilde's have been removed from the first p impedance elements as they are now known. Also, since Ipq = - IQ = - 1 , we may write from ( 12.30)
[ � lJ / n
1
=
fVmn lv M
l fVm � V� V� V � LVp VM
or - 1 = Y M ( Vm - Vn) + Y PQ ( Vp - VQ) ' Then
VQ = Vp + 1 /vpq + (YM ly pq )( Vm - Vn) This may be changed to the form VQ = Vp combining with ( 1 2.40) ,
+
( 1 2 . 41)
fpQ - (j·M !,. m n ) (f-M + Vm - Vn ) or, ( 12. 42)
This completes the formulation of the new impedance matrix. We should be able to test this result to see if it agrees with the case of adding IW uncoupled radial line wh�e f-M = O. In this case, from ( 12 . 38) we have ZQk = Zpk and from ( 1 2.42) ZQQ = ZPQ + f-pq . This is the same result as found earlier for the uncoupled radial line. Conditions : 1 . Branch added radially from p to q . 2 . Branch m n previously defined. 3 . Branch pq previously undefined. 4. p is not the reference node . 5 . Branch pq mutually coupled to branch m n . Rules: 1 . Set Zq/t = Zp/t - (f-M If-m n)(Zm /t - ZRIt) , k *' q . 2. Set Z/tQ = ZQ/t ' k *' q . 3 . Set ZQq = Zpq + J.pq - (f-M If-m n ) (f-M + Zm Q - Znq ) .
Having considered the addition of a mutually coupled radial element, there is no need to study separately the addition of a mutually coupled link branch. This case can always be handled by first adding the mutually coupled branch radially to a fictitious node q and then closing the loop from q to the appropriate node by means of a Kron reduction to eliminate row q and column q . The case just considered will solve a great many o f the mutually coupled zero sequence network problems. Occasionally, however, a situation is encountered where several lines occupy the same right-of-way and a number of circuits are mutually coupled. This case is considered in section 1 2.8,
Chapter 1 2
412
Example
1 2. 4
Consider the zero sequence network given in Table 1 1 . 1 and shown in Figure 12.7. Consider the branches in the order listed in the table and construct the zero sequence impedance matrix.
Solution From Table 1 1 .1 an d Example 1 1 . 1 we find the following zero sequence data. Branch Num ber
Connecting Nodes
1 2 3 4 5 6
open
Zero Sequence Impedance
0·3 1·2 2·3 2�3 1 ·3
jO.02 jO.14 jO.10 jO. 12 jO.17
Mutual Impedance
jO.05 jO.05
We proceed in much the same fashion as for the positive sequence network until we reach the mutually coupled element. Step 1 . Introduce element 2.
�
�3 = jO.02
1
1
Z =j
2.
3
O
2 0 3
Step
2
Introduce element 3.
0
in = jO. 14
Neither of the nodes 1 or 2 have been entered yet. Since a radial element must be connected to the network in some way, we must skip this element and pick it up later. Step 3. Introduce element 4 . .2 3 = jO. 10
This is a radial element which connects node 2 to a new node 3. Thus with k = 3, q = 2 we have
� o.�� 1
1
Step
4.
2
0
Z=j 2 0
0. 1 2
3 0
0.02
Introduce element 5.
3
. 2 3 = jO.12 with J- M
0. 02
=
jO.05
This is a mutually coupled element which closes a loop. However we add it as a radial element from node 2 to node q and will later eliminate node q .
Com puter Solution M ethods U s i ng the I m peda nce M atrix
41 3
The qth row and column elements are computed from ( 1 2 .38). Thus -
Zo ic = ZPIc - ( J-M IJ-mn )(Zmlc - Zn/c )
with
p = 2,
n = 3,
m = 2,
k = 1, 2, 3
For k = 1 ,
k
= 2, Z 2 = Z 2 2 - ( J-M lJ-mn)(Z22 - Z3 2) = j O . 1 2 - j q
�:�� ( 0 . 1 2
Z 3 = Z 23 - (J-M /t:. m n )(Z23 - Z33) = jO.02 - j q
�:�� (0.02 - 0.02 ) = jO.02
-
0.02 ) = jO.07
k = 3,
For the diagonal term we use ( 12.42) to compute Zqq = Z 2 q
+
J-2q - (J-M /J-m n )("M + Z 2 q - Z3 q )
= j O.07 + j O . 1 2 - j
�:�� (0.05 + 0.07 - 0.02) = j O . 1 4
W e now have the matrix 1
1
2
3
0
0
0
0 .1 2
0 .02
0 .02
0.02
2 0 Z=j 3 0 A
q
I I
----------
q 0
0 .07
0
I 0 .07 Ir 0 .02
---
I 0.14
0.02
where a new node q extends radially from node 3 . We close q to 2 by the follow ing steps: (1 ) Replace row q by (row 2 ) - (row q ) , ( 2 ) replace column q by (col 2 ) - (col q ) , and ( 3 ) replace element q q by Zq q + Z22 - Z2 - Z 2 as noted in q q (12.27) for closing a loop. This gives us the new matrix 1
2
3
1 0
0
0
2 0 Z= 3 0
0.12 0 .02
I
0
Now reduce Z , pivoting on Zq q to compute
Z=j
2
� r� � l�
0 .0 92
3
0
I
0 .02 I,
0 .0 5 1
I
0 .02 I 0.05
----------
q 0
q
0 .02
0
, I 0.12 ----
3
�l o.o �J
0 2
414
Chapter 1 2
Step 5 . Introduce element 6.
1-1 3
= jO.17
This element is radial from node 3 to the newly defined node 1 . Therefore by inspection of (12.23) with k = 3 , q = 1 we have
Step
6.
=j
�
2
3
0 .1900
0 .0200
0 .020
2 0 .0200
0.0992
0 .0200
3 0 .0200
0 .0200
0 .0200
1 Z
�
1
Introduce element 3 which was passed earlier. 1- 1 2
= jO.14
This element closes a loop from node 1 to node 2 . From (12.23) with k i = 2 we compute 3
1
2
0.1900
0 .0200
0 .0200 1 0 .1900
2 0.0200
0 .0992
Z = 3 0.0200
0 .0200 1 0.0200
0 .0200
0.0200 1 0 .0200
1
q 1
1
-----------------
q
0.1900
= 1,
0 .0200
1
-----
0 .0200 1 0 .3300
This adds the new element radially from node 1 to a new node q . Then from ( 1 2 . 2 7 ) we form the intermediate result 1 1
Znew = j
2
3
q
I 0.1 700
0.1900
0 .0200
0 .0200
2 0.0200
0.0992
0.0200
3 0.0200
0.0200
0.0200
I - 0 .0792
1
I
------------______
q 0.1 700
[
- 0 .0792
0
1 1 1
0
_____-
0.3892
Finally we make a Kron reduction, pivoting on element qq to compute 2
3
1 0.1 1 57
0.0546
0 .0200
Zo = j 2 0.0546
0.0831
0 .0200
3 0 .0200
0.0200
0 .0200
1
Example 1 2. 5
J
Using the impedance matrices of Examples 1 2 . 3 and 12.4, compute the total fault current fa for a SLG fault on phase a of each node with zero fault im pedance . Let h = 1 .
Solution The total fault current is computed from ( 1 2 . 1 3 ) as la k = 3lao-k = 3 h Vr /ZT
Computer Solution M ethods Using the I m pedance M a trix
41 5
Setting h V, = 1 .0, we easily select the appropriate diagonal impedances from the positive and zero sequence impedance matrices to compute the following values. Faulted Node
VaO
fa
1 2 3
Va l
Vca
(pu)
(pu)
(pu)
(pu)
- j8.098 -j8.522 - j l l .406
- 0. 3 1 24 - 0.2360 - 0.0760
0.6562 0.6 180 0.5 380
- 0.3438 - 0.3820 - 0.4620
is interesting to note that these fault currents are greater in every case than the corresponding 31/> bus fault currents computed in Example 1 2 . 1 . This is not un usual for systems with no neutral impedance in the generators. It
1 2.8
Adding a Group of Mutually Coupled L ines
We expand the foregoing analysis to the case where a group of mutually coupled lines are to be added to a network. We will arbitrarily assume that the first self impedance "Ii of the group has been introduced. The remainder of the mutually coupled group is to be added as radial connections to nodes k . . n. These radials may later be connected to other nodes by Kron reduction, thereby forming closed loops if this is necessary . Consider the network shown in Figure 12.14 where the branch 'II} has been .
n'
I n'
Ip
�
p ----i
P - PO R T
NETWORK
II - I, 1 ---1 -
o
Fig. 1 2 . 1 4.
Adding a group of mutually coupled elements.
incorporated into the p-port network and we now wish to add the mutually coupled branches 'lil li ' • • • and where this entire group of elements is a mu tually coupled group consisting of, say, m elements. Taking the group of m mu tually coupled elements from the primitive matrix, we write this submatrix as
j- nn' 1
f"] 1 �J Vkk, . . Vnn' .
2
= 2 hi m
"' 11 '
'1kll' '1ni '1nll' .
.
.
So� l [ � l m
·
.
.
·
.
.
·
.
.
n � :, 'Inn'
IJ
IlIk' �II1' . Inn' = - In'
( 1 2.43 )
41 6
C h a pter 1 2
'Yb
where we recognize that branch currents in the radial elements can be related to the nodal currents entering. Since the inverse of this primitive matrix exists, we compute it, writing the result in partitioned form as 1 2
Ii}
����
['Y�- i �: �����, t
rV ii
-
j
m - In '
Vi - Vj
Vn
-
Vn '
(12 .44)
The impedance matrix which we wish to find may be written in symbolic form with a tilde over all unknown elements as follows. V, Vi
Vi
V� Vn Vp V�' Vn'
Z" Zj J
Z/ I
Z'i Zii
Zjl
Z 'j Zii
Z,�
Z' n
Zi�
Ziti
Zi�
Zu
Z,p Zip
Zin
Zip
Z� n
Z�p
I I I I I I I
I
Z , �,
Z, .
I,
Z Ut'
Z;n'
II
Zi� '
Z In'
Z�n' I Z��, I I Znp I Zn� Znn Z ni Zroj Zn ' ZM Znn' ' I I Zpi Zp ' Zpp I Z�, Zpn Zp/ Zp� ZPn' I :; - -- - - ---_--- :;----- - :; - - - - - - --:;.- -- -- --;;; - -t-- -_-- - --- --:;; Z�' , Z�' n Z�' I Z�'j Z� ' p I Z�' �' Z�' � Z� ' . I , . . . . . I . . . . . . . . . in. n. " . Zn'n " . in' p II Z" �, " . i' l in' 1 in' � Zn' ,
Z� ,
Z�i
Z��
ZIrj
I
Ii
I�
In
Ip
I�' In'
( 1 2.45)
We can find the unknown elements by perfonning two tests. First, we inject 1.0 unit of current at each of the nodes 1 . . . p while holding all other currents at zero, including the currents entering the primed nodes, i.e., Is = 1 . 0, It = 0, lk' = . . . = In' 0
t "* 5,
5 = 1 , 2, . . . , p ( 1 2. 46)
=
By inspection of ( 1 2.45) this test equates the port voltages to the impedance ele ments of the 5th column, i.e., VI
Zis
Vi Vj
Zis
Vk
Z/u
Vn
Zjs
=
Zns
Vp
Zps
Vk ,
Zk' s
Vn,
Zn' s
, 5 "*
k ', .
.
. , n'
-
( 1 2.47)
Cha pter 1 2
41 7
Now, returning to ( 1 2.44), we note that the currents entering the primed nodes are all zero and we may write the (m - 1 ) equations
�J �:J
Solving for the primed voltages, we have =
+
�.' � ( V, - V,)
Now, from ( 1 2. 47) we substitute the appropriate impedance values in place of the voltages to write ( 1 2. 4 8 )
This determines all unknown elements except the lower right partition of ( 1 2. 45). The second test consists of causing a current of 1.0 to enter each of the primed nodes in tum while holding all other port currents to zero. Let It' = 0, Is' = 1, ' ' s = k' , . . . , n
t' =1= s ' ( 12.4 9 )
Then ( 1 2 .45) becomes VI
Z ls'
Vi Vj
Zjs'
Vit
Zit.'
Vn Vp VIt,
Vn ,
Zj,'
=
Z"',
, s' = k ' ,
. . .
,n
'
Zps,
Z�"
Zn' .,
( 1 2. 5 0 )
where the upper p elements are shown without the tilde since these are now known. From the primitive equation ( 1 2. 44) we have under test conditions ( 12. 4 9 )
C h a pter 1 2
41 8
Iij
I V ii i
Iij
I
0 0 - Is' = - 1
'Yb
- - --1 - - I I
6
- Is'
=
0 0
r:J- �J
Solving for the primed voltages, we have
'Ye
I I I
I I
I I I I I
'Yd
+ � ' I.. + �d ' �' ( V,
( 1 2. 5 1 )
-
V, )
Finally, substituting the impedances ( 1 2 .50 ) for these voltages, we determine the s'th column of unknowns
(12.52) ' Now repeat with S taking all values from k I to n ' to complete the impedance matrix . The process described for mutually coupled elements is certainly not the only one which may be devised . A number of excellent papers have been written on this subject and the interested reader is encouraged to study some of these. Recent "build and discard " techniques have also been devised which attempt to retain only certain required nodes of very large systems [ 7 3-7 5 ] . An interesting review of the literature has been made by Kruempel [76] who carefully traces the development of these techniques. 1 2.9
Comparison of Admittance and I mpedance Matrix Techniques
From the discussion presented in Chapters 11 and 12 it should be apparent that power system faults may be solved by computer, using either an admittance or an impedance matrix technique. It is also apparent that each technique has its problems : the admittance method in requiring a new inversion for each new fault simulation and the impedance method in its difficult and time-consuming matrix formation method . But there are other important considerations which should be mentioned such as computer memory utilization and the methods of changing the network representation corresponding to network changes. In our formulation of the admittance matrix and in its use for fault calculation we made no use of the matrix sparsity . Tinney and others [ 81 , 82 ] have made a thorough analysis of sparsity in their optimally ordered triangular factorization technique . This technique permits the use of the admittance matrix, stored in sparse triangular factored form, to solve network problems. This formulation al ways requires much less computer storage than the impedance matrix, an advan tage that increases as the square of the network size. Furthermore, their studies show that the sparSe matrix technique exhibits a definite advantage in computer
Com puter S o l ution M ethods Using the I m pe d a n ce M at rix
41 9
time required on all problems except faults on small networks where there is only one nonzero current in the I vector. In such cases the impedance matrix is faster. Since the trend is toward the solution of very large networks, the optimally ordered triangular factorization technique has definite advantages because it can effectively and economically solve networks at least 10 times as large as the im pedance matrix technique [ 83 ] . Another important practical consideration in solving faulted networks by computer is the ability to change the network to represent added or removed lines without completely rebuilding the stored admittance or impedance matrix. As noted in Chapter 1 1 , the addition or removal of a branch or group of branches in the Y matrix formulation, whether mutually coupled or not, is an almost trivial problem of adding or subtracting a few primitive admittances. A number of algo rithms have been devised for making changes to the impedance matrix [ 84-86] which make it possible to consider limited changes in the network without alter ing the Z matrix. A method has also been devised [ 8 3 ] to permit similar changes in an ordered triangular factorization without changing that factorization. A detailed consideration of the method of computer solution of faulted net works is beyond the scope of this book on symmetrical components, but the dis cussion of Chapters 11 and 12 should serve as an introduction. Problems
12. 1. Extend Example 12. 1 to find the voltages at each node and all Une contributions (a) for a 3tf> fault on node 1, (b) for a 3t/> fault on node 2, and (c) for a 3t/> fault on node 3. 12.2. Repeat Example 12.1 if the fault impedance Z, is 0.1 pu (pure resistance) and with a prefault voltage of 1.05 pu. 12.3. Use the method of "unit current injection" described in Example 12.1 to find the im· pedance matrices for the following positive sequence networks : (a) Figure Pl l . l , (b) Fig· ure P11.2, (c) Figure Pl l.3, and (d) Figure P11.4. Use data given in problem 1 1.4. Neglect the resistance. 12.4. Repeat problem 12.3 to find the zero sequence impedance matrices. 12.5. Consider the network of Figure P 1 1 .4. Add a a generator at node 6 with Z 1 = 0 + jO. l and find the impedance matrix by the unit current injection method. Neglect resistance. 12.6. Compute the total 3tP fault currents for the case of zero fault impedance, considering each node in tum as a faulted node, for the networks of problem 12.3. 12.7. Compute the total 3t/> fault current at each node of the system of problem 12.5. Com· pare results with those of problem 12.6(d). 12.8. Compute the line current contributions and bus sequence voltages for the following faults. The fault is a 3t/> bus fault in each case. (a) Fault on node 2, Figure P I L L (b) Fault on node 1 , Figure Pll.2. (c) Fault on node 4 , Figure Pll.3. (d) Fault on node 5 , Figure P11.3. (e) Fault on node 5 , Figure Pll.4. (f) Fault on node 2, Figure P l 1 .4 . (g) Fault on node 2, system of problem 12.5. 12.9. Repeat Example 12.2 to find the complete network solution for a SLG fault (a) On bus 1 with Z, - 0 . 1 + jO, V, 1.05 + jO. (b) On bus 2 with Z, - 0 . 1 + jO, V, - 1.05 + jO. (c) On bus 3 with Z, - 0.05 + jO.05, V, - 1.10 + jO. (d) On bus 3 with Z, 0.1 + jO , V, - 1. 05 + jO. 12. 10. Compute the total SLG fault current for the fault conditions described in problem 12.8. -
-
420
Cha pter 1 2
12. 11. Compute the sequence voltages of all buses adjacent to the fault and all contributing line currents for the faults of problem 12.10. 12.12. Verify equation (12.27). In particular. show that changing the ith row current from Ii to Iq + Ii requires taking impedance differences as shown in column q . 12.13. Repeat Example 12.3 except introduce the elements in the order (a) 1, 3 , 4, 5, 6, 2. (b) 1 , 6 , 2, 4, 5, 3. (c) 2, 4, 5, 6, 1, 3. (d) 1 , 4, 6, 5, 3, 2. 12. 14. Use the Z matrix algorithm of sections 12.3-12.6 to find the positive sequence imped· ance matrices of the following networks: (a) Figure P11.1, (b) Figure P11.2, (c) Figure P 1 1.3, (d) Figure P1 1.4, and (e) systems of problem 12.5. 12.15. Use the algorithms of sections 12.3-12.8 to find the zero sequence impedance matrices of the networks specified in problem 12.14. 12.16. Show that (12.38) may also be written as (a) ZqIe
(b) Zq�
12.17. 12. 18. 12. 19. 12.20.
-
-
Z"Ie
(1 - tmn¥mn)(Zmle - Znle)
fmnVM tMVmn (Zmle - Znle) Z"Ie 1 tMYM _
_
-
Verify the steps leading to (12.42). Verify (12.48). Verify (12.52). Show that (12.48) and (12.52) reduce to (12.38) and (12.42) when there are only two mutually coupled lines. 12.21. Develop formulas similar to (12.48) and (12.52) for the special case of three mutually coupled lines.
appendix
A
M a t rix A l gebra
The analysis of power systems under faulted conditions by analytical tech niques is complex because of the large number of simultaneous equations that must be formed and solved. Matrix algebra, with its capability for the writing and manipulating of large numbers of equations in terms of a small number of sym bols, is a desirable, almost indispensable mathematical tool. The purpose of this appendix is to review the appropriate concepts of matrix algebra. I First, we consider simultaneous linear equations and their solution by means of determinants. A. 1
Simultaneous L inear Equations
We often work with equations of the form
X I = a l l Y I + a l2 Y2 + . . . + alnYn X2 = a 2 1 Y I + a22 Y 2 + . . . + a2 n Y n (A. l )
Here the x 's and y 's are variables and the a 's are constant (either real or complex) coefficients.
Note partiCUlarly the m ethod of subscripting. In the problem s of
interest, the x's will often be voltages (or currents), the y 's will be currents (or voltages), and the a's will be impedances (or admittances). Equation (A.l ) can be solved for the y 's by means of determinants to get Y l = f l 1 X I + f1 2 X2 + " ' + f l n Xn Y2 = f2 1 X I + f22 X2 + . . . + f2n Xn
(A.2)
where (A.3)
I Much of the material for this appendix is taken from notes, "Review of Matrix Algebra, " prepared b y H . W . Hale and used for teaching short courses in Power Systems Engineering at Iowa State University, Ames. It is used with permission of the author. 42 1
422
where
[
Appendix A
al l
�
= det
n
=
.. .
L ai/ Gij /= 1
(A.4)
a nI
and
= the cofactor ij = ( - 1 ) i+jMij Mij = the minor determinant resulting from the deletion of row i and column j Gij
(A.5 ) The general form o f (A. l ) and o f its solution (A.2) will b e pertinent t o some of the following material. A.2
Definitions Pertaining to Matrices
Following are some of the basic definitions and terminology pertaining to matrices. 1 . Definition of a ma trix . A matrix is a rectangular (including square as a spe cial case) array of elements (symbols or numbers) arranged in rows and columns. For example
is a matrix of two rows and three columns. Note that in this general form the subscripts attached to the elements (a 's) indicate the row (first subscript) and col umn (second subscript) in which it is found . Thus , the term " a ij " or "element ij " specifies a definite location at which the element is found . For a specific matrix of numerical values such as
subscripts are not attached to the elements; however, it is apparent that all
= 1,
a12
=
-
2,
a21 =
3,
an =
2
2 . No ta tion . We will indicate arrays as matrices by enclosure in brackets as shown in definition 1 . Other conventions are in use and may be encountered in the literature. We can also indicate a matrix symbolically by a single symbol such as
A
=
[
]
al l
a12
a2 I
a2 2
Although other conventions, such as underscoring or script type, are used, we will use boldface type for matrices. In most cases it is clear from the context which quantities are matrices and which are not. We will see later that we can write equations in terms of symbolic representation of matrices and manipUlate the symbols as long as the appropriate rules are followed.
M atrix Algebra
42 3
3. Order of a matrix. A matrix with m rows and n columns is referred to as an m X n or ( m , n ) matrix. In the special case that the matrix is square, m = n , it is
referred to as an nth order matrix. 4. Column and row matrices. A matrix with only one column is referred to as a column matrix, while one with only one row is a row matrix. They are also of ten referred to as column and row vectors respectively. 5 . Transpose of a matrix . The new matrix formed by interchanging the rows and columns of a given matrix is called the transpose of the original matrix. Thus
� � _�] -
�
[� �l [ -�J
the transpose of
2
Obviously, an m X n matrix has a transpose that is an n X m matrix. We indicate the transpose of a matrix A by A' or AT or A t . 6. Conjugate of a matrix . If all elements of a matrix are replaced by their complex conjugates, the new matrix is the conjugate of the original one. Thus
�
(1 - j1)
1
J
(3 - j 3 ) (2 + j 2 )
'
IS
the conJ ugate of
.
�
1 (1 + j1) ( 3 + j 3 ) ( 2 - j2)
J
The conjugate of a matrix A is indicated symbolically as A * . 7 . Symmetric matrix . A square tnatrix that has aj i au for all i and is sym metric. Thus
[� �]
=
is symmetric, while
[� :]
j
is not symmetric
Another way of stating the definition is that a symmetric matrix is equal to its own transpose. Only square matrices can be symmetric. S . Principal diagonal. The elements a 1 1 , a 22 , a 33 , . . . , ann of a square matrix form its principal diagonal. 9 . Unit ma trix . A square m atrix that has + 1 's on its principal diagonal and ze ros elsewhere is a unit matrix. It is denoted symbolically as I, U , E, or 1. Some times a subscript is attached to indicate the order. Thus,
U - l = E, -
[� �] o
1
o
(A. 6 )
In electrical engineering we usually avoid using I for the unit matrix. 10. Null matrix . A matrix with all elements equal to zero is a null matrix. 1 1 . Scalars. The elements of matrices are called scalars. A scalar is a 1 X 1 matrix. 12. Su b matrix . If some rows and columns of a matrix are deleted, the result is a submatrix. For example, if A
=
[� � � �]
has columns 2 and 4 deleted, the result is a new matrix
Appendix A
424
that is a submatrix of A. 13. Determ inant of a matrix . It is important to note that a matrix is not a determinant. However, the determinant of a square matrix has meaning and is taken to be the (scalar) quantity det A
[: :] =
= det
12 - 20
=
-
B
The determinant of a matrix that is not square is not defined. 14. Nonsingular and singular matrices. If a square matrix has a nonzero determinant it is nonsingular. All other matrices, including those that are not square, are singular. 1 5 . Rank of a matrix. The rank of a matrix is the order (number of rows or columns ) of the largest nonsingular square submatrix. Thus
A= has rank 2 because
while
has rank 1.
[�
B= G
2 1
1 1
]
-1 -1
16. Adjoint of the matrix A. The adjoint of the matrix
defined as
A, written adj A, is
(A.7 )
where Cu
1 7 . Inverse of the matrix defined by the equation
A.
= cofactor of a u
The inverse of a square matrix A of order n is (A.B)
In terms of elementary operations the inverse of A may be computed from the formula K'
=
adj Afdet A
Only nonsingular matrices (det A =1= 0) have inverses.
(A.9)
42 5
M a trix Algebra
18. Skew-symmetric matrix . A skew-symmetric matrix A is defined by the relation (A.10) The maj or diagonal elements of a skew-symmetric matrix must be zero since for i = i the only way (A.10) may be satisfied is for aiJ = O. An example of a skew symmetric matrix is
�o
+ jO
�
3-j
2+j
A = -3 +j
0 + jO
4 + jO
-2 - j
- 4 + jO
0 + jO
1 9 . Hermitian matrix. A hermitian matrix A is defined by the relation (A. l l ) The maj or diagonal elements o f a hermitian matrix must b e real, for with i = i the only way (A. l l ) can be satisfied is for aiJ to be real. Only square matrices may be classified as hermitian. An example of a hermitian matrix is
� o
2 + jO
A=
2- j - 5 + jO 1 - j3
2 +j
+ jO
�
0+j 1 + j3 0 + jO
20. Skew-hermitian matrix . A skew-hermitian matrix A is defined by the
rel ation
(A.12)
and this property applie s only when A is square. The terms of the major diagonal i = i may be zero or p ure imaginary , but not real. An example of a skew-hermitian matrix is given by A
=
[
5 + jO
0 + j2 - 5 + jO - 3 + j2
0 + jO 1
-
j7
2 1 . Dominant matrix . A dominant matrix which has the additional property that ajj
�
1
3 + j2 - 1 - j7 0 - j6
A is defined as a symmetric matrix
L lajj I , j :#= i n
j= 1
[ : � -��
(A. 1 3)
for all values of i. An example of a dominant matrix is given by A=
-1
3
where (A . 1 3) is satisfied or the elements of the major diagonal are greater than the sum of the absolute value of all elements in that row (or column). For exam ple, the Y matrix of a power system is always dominant.
Appendix A
42 6
A.3 R u les of Matrix Algebra
Matrix algebra has rules that must be followed explicitly. The basic rules are: 1 . Equality of matrices. Two matrices are equal if and only if each of their corresponding elements are equal. Thus G = H only if gjj = hjj for all i and j. Only matrices having the same number of rows and columns can be equal. 2 . A ddition of matrices. Two matrices are added by adding corresponding elements to obtain the corresponding element in the sum. Thus
[
1
0
2
-1
-1 3
-1 2
�]
Addition is defined only when both matrices have the same number of rows and columns. Addition is commutative, i.e., A+B=B+A and is also associative, Le., A + (B + C ) = (A + B ) + C
3. Subtraction. A matrix is subtracted from a second matrix by subtracting the elements of the first from the elements of the second. Subtraction is defined only when both matrices have the same number of rows and columns. 4. Multiplication of a matrix by a scalar. A matrix is multiplied by a scalar by multiplying each element of the matrix by the scalar, e.g., k
[_ � �] = [�: :] 3
5 . Multiplication of matrices. Multiplication is defined by the following: G = A X B (or A B) implies that the elements of G are related to the elements of A and B by n
gjj = L a jk b k j k 1 =
(A.14)
where n is the number of columns of A and the number of rows of B. Multiplica tion is defined if and only if A has the same number of columns as B has rows. Two such matrices are said to be conformable. Thus
2X 2
2X 3
where, for example,
g l 2 = L a lk b k 2 = a l 1 b l2 + a l 2 b n + a l 3 b 3 2 k 1 3
=
427
M atrix Algebra
It follows that in the product (A. 1 5 )
C=AX B
where A is an m X n matrix and B i s a n n X p matrix, C i s an m X p matrix. That is, the number of rows and columns of C will be equal to the number of rows of A and the number of columns of B respectively. It is sometimes convenient to check conformability by writing the order (row X column) above each matrix. Thus (A. 1 5 ) would be written m X p
m X n
C
A
nX p X
B
(A .1 6 )
The adjacent terms of the superscripted numbers must agree (n ::: n ) to have con formability, and the outer terms determine the order of C, viz. , m X p. This will work for products involving any number of matrices. It is also apparent that multiplication is not commutative, Le ., AX B* BX A
(A. 1 7 )
It is, however, associative and is distributive over addition, i.e., A(BC) = (AB)C
(A. 1S)
A X (B + C) = AB + AC
(A. 19)
and Because multiplication is not commutative, it is vital that the order of multiplica tion be clear. Thus in the product A X B we say that B is premultiplied by A or that A is postmultiplied by B . 6. The product VA. From the multiplication rule it i s apparent that the prod uct of an identity matrix of order n and an n X p matrix A is the matrix A, Le. , VA = A
(A.20 )
If A is square, i.e . , of order n , then V X A = A = A X V. Thus in this very special case multiplication is commutative . 7 . The inverse matrix and the solu tion of equa tions. Suppose that given the matrix A there exists a matrix A - I such that A - I X A = V. Then A - I is called the inverse of A. The use of and the properties of the inverse are pointed out in terms of (A.1 )-(A. 5 ) . Consider (A.1 ) which can be written in matrix from as x = Ay where x, A , and y are n X l , n X n , and n X 1 matrices respectively. Assuming that the inverse of A exists, we may pre multiply both sides of the equation by A - I , with the result that
A- I X = A- I Ay = Vy = Y
Now (A.2 ) can be written as y = Fx. It follows that F = A - I and the rules for determining the inverse of a square matrix result from a comparison of F (as in equations A.3 , A.4 , and A.5) and A- I . Thus given
428
Appendix A
the inverse of A is
]
C3 1
C 32
C3 3
=
adj A det A
(A.2 1 )
where tJ. i s the determinant of A, an d Cij i s the appropriate cofactor. The exten sion to higher order matrices is apparent. It is also apparent that only square matrices can have an inverse and that a square matrix has an inverse if and only if it has a nonzero determinant, Le., if it is nonsingular. A.4 Some Useful Theorems on Matrices
We state without proof several theorems which are useful, even indispensable, in dealing with matrices and matrix equations.
1 . The inverse of the product of two square matrices is equal to the product of the inverse taken in reverse order, i.e . , (A.22) 2 . The transpose of the product of two matrices is equal to the product of the transposes taken in reverse order, i.e . , (A.23) 3. The transpose of the sum of two matrices is equal to the sum of the trans poses, Le., (A.24) 4. The inverse of the transpose of a square matrix is equal to the transpose of the inverse, i.e . , ( At )- I = ( A - I ) t (A.2 5 ) 5 . If a square matrix is equal to its transpose, it is symmetric,
= At -- A is symmetric A
(A.26)
6 . If a square matrix is equal to the negative of its transpose, it is skew symmetric, i.e., A = - At -- A is skew-symmetric
(A.27 ) 7 . Any square matrix A with all real elements may be expressed as the sum of a symmetric matrix and a skew-symmetric matrix (this is also true if the elements are functions of the Laplace transform variable s ). Stated mathematically , A is square -- A = B + C
where B is symmetric C is skew-symmetric
(A.28 )
42 9
M atrix Algebra
From theorems 5 and 6 we compute
At = B t + Ct = B - C
(A.29 )
Then we may add and subtract (A.28) and (A.29) to obtain, respectively , the symmetric matrix
B = ( 1 /2) (A + N )
(A.30)
C ;::: ( 1/2) ( A - At )
(A.3 1 )
and the skew-symmetric matrix
8. A hermitian matrix is a square matrix which is equal to the conjugate of its transpose and vice versa, i.e . ,
A = (At )* -- A is hermitian
(A.32 )
9. A skew-hermitian matrix is equal to the negative of the conjugate of its transpose and vice versa, i .e . ,
A = - (A t ) * -- A is skew-hermitian
(A.33 )
1 0 . Any square matrix A with real and complex elements may be written as the sum of a hermitian matrix and a skew-hermitian matrix, i .e . ,
A i s square and real o r complex -- A = B + C where B is hermitian C is skew-hermitian
(A.34)
From theorems 3, 8, and 9 we easily show that we may write the hermitian matrix as
B
=
( 1 /2)
[A + (At )* ]
(A.35)
[A - (At )* ]
(A.36)
and the skew-hermitian matrix as
C = ( 1 /2)
A.S
Matrix Partitioning
When equations involving large matrices are to be manipulated, it is frequently useful to partition these matrices into smaller submatrices. Partitioning may simplify the procedures and quite frequently makes the proof of general results much simpler. Consider the following example :
n [u
C = C2
C3
=
.'J
a 12
a13
a2 1
an
a2 3
a2 4
a3 1
a32
a33
a 34
dl d2 = A D d3 d4
(A.37)
Appendix A
430
If we partition A and D by dashed lines as shown we can write C = A D = [A I A 2 1
[�J
(A.38)
= [ A I D I + A2 D2 1
( A. 3 9 ) and now write ( AAO) or tent; i.e. , all products and Again the partitioning of the matrices must be consis of A by colum ns determines sums must be defined. As before, the partitioning of A by rows deter ioning the partitioning of D by rows. In additi on, the partit . mines the partit ionin g of C by rows or vice versa partitionings as sh own below . ple multi into alized gener be can s These result 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5
1 2
3
4 5 6
1 1 1 1 Cll Cu 11 1 1 -----,--
7
8
A.6
C2 1
1 1 I I
en
1
2
3
4 =5
Au -_-----
6
7
8
A2 l
Matrix Reduc tion ( Kron Reduc tion)
1 1 1 1 1 1 1 1 1 I
1
1 1
r
1 1 1 AI2 1 1 1 - 1 1 1 I A22 1 1
----
1 1 1 Du 1 DI2 3 1 ----- 11 --1 1 6 D2 1 I D22 I 1 1 8 1 D3 1 1 D32 1
1 2
AI3 ___ _ _--
A23
4 5
7
9 10
· - - ---
---
( AA 1 )
ion, after rearranging and In physi cal problems the following type of equat partitioning, is often obtained.
M a trix Algebra
43 1
(A. 42)
Here, the nature of the problems are such that A n is usually nonsingular and the variables in O 2 are unwanted and can be eliminated. If we expand the above equation, the result is
Cl = A l l 0 1 + A l 2 O2 0 = A2 1 0 1 + A22 O 2
( A.43)
Remembering that the quantities in these equations are matrices and must be treated as such, we can solve the second equation for O 2 with the result that O 2 = A2� A u 0 1 and substitute into the first of equations ( A.43) with the result -
C l = ( A 1 1 - A l 2 A2� A 2 t l 0 1
(A.44)
This last result is particularly useful and is a graphic illustration of the utility of matrix methods. A.7
T ransformation of Coordi nates
It is often desirable in electrical engineering problems to transform a matrix equation from one coordinate system to another in which a solution may be more readily obtained. For example, we often find it convenient to transform from the a-b-c phase domain into the 0- 1-2 symmetrical component domain or the O-d-q Park domain. The result is a sim ilarity transformation of the impedances as will be shown. Let T be any unique (Le., nonsingular) transformation which relates currents ( and voltages) between the a-b-c domain and the O:-�-'Y domain according to the equation
(A.45)
and similarly for voltages. We write (A. 4 5 ) in matrix notation as lab e = T la/3'Y
(A.46 )
Suppose we are given a matrix voltage equation, expressing the voltage drop in lines a, b, and c as (A. 47 )
Premultiplying both sides of (A. 4 7 ) by T- l we have (since T- l exists)
r- l Va b e = T- l Z ab e 1ab e
If we now insert the product T T- l = U between Z ab e and lab e, we have T- I Vab e = T- l Zabe T T- l lab e But by
(A.46) this equation is (A.48)
Appendix A
432
Thus under the (linear) transformation T, the impedance has been changed to (A.49) or Zabc has undergone a similarity transformation. This is an important trans formation of linear algebra, and we say that the two matrices Zabc and ZQ/J'Y are similar. Computer Programs
A.8
In working with matrices of order greater than two it is extremely laborious to perform any computations by hand. Matrix inversions, for example, require something greater than n3 multiplications for an n X n matrix. Very clever manipulations can save perhaps half the work, but nothing like an n-fold saving is possible. Similarly, a great deal of computation is required for matrix multiplica tions and even additions. Our attitude here will be that a digital computer is usually necessary and is available for matrix problems. Since FORTRAN is nearly a universal computer language, solutions will be given in this language. Two difficult matrix computations often required are the matrix inversion and the matrix reduction. Both operations can be performed by the same FORTRAN subroutine, and an example of a program to perform these operations follows. 2 ::
� l I N P P OG R A � F OR S � I P l E Y - C O l E " A N - J O H N S ON C OM P L E X I � V E R T E � - R E O � C E � . � I S O N � D t � E N S I O N O F T H E S O U A � E I NP U T " A T R I X - " X I N . K � O I S T H E A R R AY OF R O W S A N D/ � � C J l U M � I � O t C E S J F � O W S l � O "' O T U S E D F !J R I N \f E R S I O � . : J l 'JM N S T O 8 E E L I M I N A T E D I N T H E R E D U C T I O N . I N V . 0 F O R R E D UC T I O "' . ' �V = 1 FOR I NVER S I ON , C OM P L E X . 8 MX I N C I O O ' , M X O U T C Y " ' , I N V E R T , R E D U C E , O U T I N T E r. E R • 4 N , K � O ( l � ' , I N V n AT 4 I N V E � T I ' I NV E R T E O ' I , P. E D U C E I ' R E D UC E � ' I O UT = I N V E R T
,.
c C r
l !'
20 �o c
P EA D ( 5 , ! O ' N ,
I NV
F I)R � A T ( 2 1 5 ' N � = N .. N I F f I NV . E � . l ' � O T � 3n OUT = R E D U C E P E A D ('; , 2 0 ' K G O F OP � A T ( 1 0 I 5 ' ( "X I N I I ' ,
P EA O ( S , 4 0 ' �X I N
I N R OW OR D E R
� OR � A T C @ F I 0 . 0 '
50
W R I T E C 6 , -: O ' O J T F OP M A T ( ' l T H E M A T P I X
or 1 01
=
1 , NN '
C � EAl , IMAG ' .
40
�O 70 �O
1
8E I �G
J = 1 ,N 0" 80 J M! N • ( J - l ' • N I . 1 , N DO 70 I J = I + J "1 1 N W R I T F. ( 6 , 6 0 1 J , I , M X t N ( I J 1
"
� OR M A T C ' M X I N C ' , 1 2 , ' , I , C ON T t N U E C ON T I N U E C A l l C S H P ( "1 X I N , M X O U T , N , WRITE I 6 , l 00 1 F OP M A T C " O IJ P S ' ,
12 ,
2A�/ '
KGO,
"
=
I NV ,
"
I PE I 4 . 6 ,
&1 1 0 ,
'
+J
•
&90'
2 This program was written by G. N. Johnson and is used with his permission.
"
1 PE ! 4 . 6 '
43 3
M a trix Algebra IJ 1 121
, ':l ) ! 40 ! "i '
GO
TO
'
WQITE ( 6 , t��I � OR � A T ( "
T �E
OUT ' , 2�4, 1, N
'
MATR I X '
DO 1 �0 J = J �l � ., ( J - 1 1 t � DO 1 40 I " 1, N I J ., I .. J � l � W R ! T E ( b , ! 3 � 1 J , I . M XO UT ( I J I � OR � � T ( ' '" l( 0 UT ( ' . I 2 , ' , ' . ! 2 ,
C ON T I N U E C ON T I N U E GO TO 1
II
' I
.,
1"E1 4• 6,
"
' "J
•
"
1 p :: 1 4 • S I
E"O
C S H P ' ''' X I 'II ,
S UR R O U T I 'II E
.,
Y NV "' X l " "' X I N
"
C ('
IS
\
L
IS
16
22�
230 240 250 1
4
B I G STA ,
R EDUC E
e
A
G I VE N
ce,
BB.
DO ,
CO"JG
DONE O J R I " ;
NTH ORD E R
IN
THE
D�URL E I I 0 0 1
01
GO
TO
E L I MI NATE
E L I M I N A T I NG
BY
KGOTOT ROWS
ARR AY KGO f I I . TO 2 5 0
C ON T I N U E W R I T E 1 6 , 2 40 '
F OR M A T ( ' O • • • R ET U RN2 K GO T O T .,
DO
I
5
=
MATR I X
D OU B L E I I . = DO 1 0 0 I I I I F I I NV . EQ. L AR G E S T
B IGSTA DO 20 DO 1 0 IFIJ I FI I
WANT
= =
.EO. .EO.
1, 1 ,
I NTO
D O UB L E
=
THE
M AT R I X ?
••• ' I
W O R K I NG
MATR I X.
MXI N I I I
= 1 , KGOTOT O ' GO T O 35 U NU S E D
D I AG O N A L
EL E MENT
N I I I
I I I J GO TO 1 0 l I J I I GO TO 2 0
I TRY
OF
P R E C I S I ON
C ON T I N U E D I AG E L = DOUB L E I I I - i l • N + I f T RY = R E A L I O I A G E L • C O N J G I O I A G E L I I I F I B I G S TA . G T . TRY I GO TO 20 K = I Lill I ) B IGSTA
ALL
1
0 .0
., I J
YOU
KGOTOT 1 , NN
I NPUT
I "VER S I J" .
GO TO 2��
OF
C
10 30
1.
AA,
MATRI X
P ICK
C
OI AGEL ,
TRY
A N D C OL U M N S S P E C I F I E D D O 2 3 0 K G O T O T ., 1 , N I F I KGO I KGOTOT I . E Q .
M OV E
5 9
.,
THE
P Y V I NV ,
. N
• 'II . EO. K G O T O T ., N GO TO 1
C C
.,
A WOR K AR R A Y F O R K E E P I �G T � AC K OF R O W S
I F ( I NV
C
ARE
9 �X I N ( l � O I , M X O U T I I 0 n l , L ( 1 0 1 . K , KGO ( l O I
C OM P L E X . R EAL NN =
I NV ,
I "V= R S I ON,
M X JUT M X OUT
I N T E r. F R t 4
c
K�O,
I N V ., 0 � OR K R O N R E O U C T T n " . I � PUT A � ) OUT PUT M A T � I C E S R E S "E : T I V E L V . A NO A R E S I NGL E S U R S C R I PT E D A N D I " R O W OR � E R . ONE D I ME N S I ON O F THE S�UAR E MATR I X MX I N . ANQ
C OM P L E X .
c
C e C
F��
N,
"' X O U T ,
FOR
P I VOT
FOR
I NVE R S I ON .
Appendix A
43 4 20
C ONT I NU E I F C B I GSTA
. LT.
I . E-26 t
GO T O 1 2 0
C
C C
L AR G E S T
D I AG O N A L
E L E M ENT
C HO S EN .
GO TO 37
C C
G OI N G
TO
DO A
K R O N R E D UC T I ON W H E N
I NV .. 0 •
• •
C 35
37
C
K .. K GO C I I I t DO = DOUB L E C C K-I )
� O ST IF A
C C C
40 50
* N
P I V I N V = 1 . 0 00 0 I DO K MI N = C K- l t • N DO 50 I " I, N I F ( I . EQ . K t GO T O 50 B 8 • DOU B L E C K M I N + I ) D O 40 J c I , N I F C J . E O . K t GO TO 40 ACC U R A T E
WAY
+
K)
TO CALC
W A S T H E M A TR I X B E I NG
I J = C J- l t • N + I A A = D O UB L E C I J t C C = OOU B L E C I J - r • K t A A = C A A • DO - B 8 • CC t D O U B L E C I J t .. A A C ONT I NU E C ON T I N U l: I F C I NV
.EO.
Ot
A C I , J t .. A f I , J t - C A C I , K J . A C K , J t t / A t K , K ) I N V E R T E D OR R E DU C E D .
• P I V I NV
TO 1 0 0
GO
C
C C
C
C
D I VI D E
55
ROW
BY - A C K , K t
DO 55 I = 1, N I F C I • E Q . K ) GO TO K MI N I = K M I N + I A A = DO U B L E C K MI N I J A A = -AA * P I VI NV D OU B L E ( K M I N I J '"' AA C ON T I N U E I) I V I D E
C
P I V OT
P I VOT
C O L U MN
ON
I N V E R S I ON ONLY .
55
B Y - A I !'. , , , )
O�
I NV E R S I ON
ONL Y .
00 60
60 C C C
J .. I , N I F ( J . E O . K t GO TO KJ = C J-l t * N + K A A = DO U B L E I K J t A A = -AA • P ! V I NV D OU B L f C K J ) .. A A C ON T I N U E R E PL A C E A C K , K I
1 00
D OU B L E C K M I N C ON T I NU E I FC INV .EO.
BY
+ KJ 0'
GO
60
-1 . O / A C K , K t =
ON
I NV E R S I ON O N L Y .
-P I V I NV TO
150
C
C
N E GA T E
C 110
12'
1 30
ENTI RE
R E S UL T A N T
MATR I X
DO 1 10 I = 1 , NN M XO U T C I t .. - DO U B L E C I ' R ET U R N I WRI T E C 6 , 1 30 t F OR M A T ( ' O ** . T H I S M A T R I X R ET U R N2
IS
T O R EAL I Z E
S I NGULA R• •• · '
I N V ER S E .
M atrix Algebra c C C C C C
43 5
C O MP L E X S H I P L E Y M A TR I X I N V E R TOR M OD I F I E D
153
NM =
N
-
I< G O T O T
I< RON R E DUC T I ON
160
170
180
& P G MO BY G . N . J OHNS ON P . E .
-
R E GR OUP I NG OUT P UT M A TR I X .
I< = 0 00 170 I = 1, N 00 160 J = 1 , I< GOT OT I F ( I . E O . I
+
JL )
appendix
B
Li n e I m p ed a n ce Ta b l es
The following tables provide all the data normally required for determining the normalized impedance of overhead transmission or distribution lines. 1 . Tables B.1-B. 3 give base currents, base impedances, and base admittances for a wide range of voltages and for four convenient MVA bases. 2 . Tables B.4-B. l l give the characteristics of overhead wires, presented in a convenient format. They are used with permission of the Westinghouse Corporation and are taken from [ 14] and [ 5 4 ] . 3 . Tables B .12-B.23 give characteristics often needed for bundled conductor EHV lines. These tables also use the convenient form of the previous tables. They are used with permission of the Edison Electric Institute and come from [ 7 8 ] . 4. Tables B.24 and B.25 give the 60 Hz spacing factor for inductive reactance Xd and capacitive reactance x� respectively for a wide range of wire spacings.
436
Line I mped a n ce Tables
SA S E K I L OVOL T S
2. 3 0 2 . 40 4. 0 0 4. 1 6 4. 40 4. 8 0 6. 60 6. 90 7. 2 0 1 1 . 00 11 . 45 1 2. 00 1 2 . 47 13. 20
1�. § 0
14. 40 22 . 0 0 24. 94 33 . 0 0 34 . 5 0 44 . 0 0 55 . 00 60. 0 0 66 . 0 0 69 . 0 0 88 . 0 0 I QQ. 00 1 1 0. 0 0 1 1·5 . 0 0 U2 . 0 0 1 3 8 . 00 1 54. 0 0
Ul. OQ
2 20 . 0 0 2 30 . 0 0
21:2. Q2 3 30 . 0 0
3 45 . 0 0 HQ. og 3 62 . 0 0 420. 00 5 00. 0 0 5 25 . 0 0
5 50 . 00 11212. 0 12 7 3 5 . 00 7 50 . 0 0
16:2. QQ 1 0 00 . 0 0
1 1 00. 0 0 1 2 00 . 00 1 3 00 . 0 0 1 4 00 . 0 0
nQQ. Qg
43 7
Table B. 1. Base Current in Amperes
BAS E M E GAV OLT-AM P E R E S
50. 00 1 2 5 5 1 . 09 2 8 1 2 0 2 8 . 1 3 06 7 2 1 6. 8 7 84 693 9. 30 61 6 5 60 . 7 9 8 5 6 0 1 4. 06 5 3 43 73 . 8 6 5 7 4 1 8 3 . 6976 4 0 0 9. 3 769 2 6 2 4 . 3 1 94 2 5 2 1 . 1 8 02 2 4 0 5. 6 2 6 1 2 3 1 4. 9 5 70 2 1 86. 932 8 �09 1. 8 4 8 8 2 004. 6 8 84 1 3 1 2 . 1 5 97 11 5 7 . 4 7 8 5 8 74 . 77 3 1 8 36 . 7 3 9 5 6 5 6 . 0 7 99 524. 8639 4 8 1 . 1 2 52 4 3 7 . 3 8 66 4 1 8 . 3 6 98 3 2 8 . 03 99 288.6751 2 6 2. 4 3 1 9 2 51 . 02 1 9 U 8. 6 9 3 3 2 0 9 . 1 8 49 1 87. 451 4 1 79. 3 0 1 3 1 3 1 . 2 1 60 1 2 5. 5 1 09 lQ� . 9 I2 8 8 7. 4773
1 00 . 0 C 2 51 02 . 1 8 56 2 4 0 56 . 2 6 1 2 1 443 3 . 7 5 6 7
2 8 . 86 7 5 26. 2432 2 4. 0 5 6 3 2 2 . 20 5 8 2 0 . 61 9 7
1 3 87 8 . 6 1 2 2 1 31 2 1 . 5 9 70 1 2 02 8 . 1 3 06 8747. 7 3 14 8 3 6 7. 3 9 5 2 8 0 1 8. 7 5 3 7 5248. 6 3 8 8 5 0 4 2 . 3 6 04 481 1. 2522 46 2 9 . 9 1 3 9 4 3 7 3 . 86 5 7 4 1 8 3 . 69 7 6 40 09 . 3 76 9 2 6 2 4 . 3 1 94 2 3 1 4. 9 5 7 0 1 7 49 . 5 4 6 3 1 6 7 3 . 4 79 0 1 3 1 2 . 1 5 97 1 0 49 . 7 2 7 8 96 2 . 2 50 4 8 74 . 77 3 1 8 36 . 7 3 9 5 6 5 6 . 0 799 5 7 7. 3 5 03 5 2 4.• 8 6 3 9 5 0 2 . 043 7 43 7. 3 866 4 1 8. 3698 3 74 . 9 0 2 8 3 5 8. 6 0 2 7 26 2 . 431 9 2 5 1 . 02 1 9 209. 94�6 1 74. 9 546 1 6 7 . 3 4 79 160 . 3 7 5 1 1 59 . 4 8 9 0 1 3 7 . 46 4 3 1 1 5 . 47 0 1 1 09 . 9 7 1 5 1 0 4 . 97 2 8 82 . 4 7 8 6 78. 5 5 1 1 76. 9 80 0 1�l i 7 � 57. 73 5 0 5 2 . 4864 48. 1 1 2 5 44. 4 1 1 6 41. 2393
1!. Z!�g
�11 !9QQ
8 3. 6740
1112. 1 8 7 5 7 9 . 7 44 5 68. 7322 57. 7 3 50 5 4. 9 8 5 7 5 2 . 4 8 64 4 1 . 2� 9 3 3 9. 2 7 5 5 3 8 . 49 00
�1. n��
2 00 . 0 0 5 0 20 4 . 3 7 1 2 481 1 2. 5224 U86 7 . 5 l 3 5 2 7 7 5 7. 2 2 4 5 2 6 243. 1 94 1 24056. 2 6 1 2 1 74 9 5 . 46 2 7 1 6 7 3 4. 7904 l 6 03 7 . 5 0 7 5
1 0497. 2 7 76 1 0084. 7 2 0 9 9 6 2 2 . 5 045 9 2 5 9. 8 27 9 8747. 7 3 14 8 367. 3 9 5 2 8 0 1 8. 7 5 3 7 5 2 ,. 8 . 6 3 8 8 462 9. 9 1 3 9 34 99 . 0 9 2 5 3 346 . 9 5 81 2 6 2 4 . 3 1 94 2 099 . 4 5 5 5 1 9 24. 5 0 0 9 1 749. 5463 1 6 7 3 . 47 9 0 1 3 1 2 . 1 5 97 1 1 54. 7 00 5 1 04 9 . 7 2 7 8 1 004. 0 8 74 8 74 . 7 7 3 1 83 6. 7 3 95 74 9 . 8 0 5 5 7 1 7. 2 0 5 3 524. 86 39 50 2 . 0437 4 1 9. 11 9 11 3 49 . 9 0 9 3 33 4 . 6 9 5 8 3 2 0. 7 5 01 3 1 8 . 9780 2 74 . 9 2 8 7 2 3 0. 9 40 1 2 1 9 . 9 43 0 2 09 . 9 4 5 6 1 6 4. 9 5 7 2 1 5 7 . 1 02 1 1 5 3 . 9 60 1 1 5 0. 9412 1 1 5. 4701 1 04 . 9 7 2 8 96 . 2 2 5 0 88. 8231 8 2 . 4 7 86
7t. nQO
2 50 . 00 62 7 5 5 . 4640 6 0 1 40 . 6 53 0 3 6 084. 3 91 8
3 46 9 6 . 5 3 06 3 2 8 0 3 . 99 2 6 30070. 3265 2 1 8 6 9 . 1 2 84 2 091 8. 4 8 8 0 2 0 04 6 . 8 8 4 3 1 3 1 2 1 . 5 97 0 1 2 605 . 901 1 1 2 0 2 8 . 1 3 06 1 1 5 7 4 . 7 84 9 1 09 34. 6 642 1 0459. 2 440 1 002 3 . 4 42 2 6 5 60 . 798 5 5 7 8 7 . 392 4 4 37 3 . 8 6 57 4 1 8 3 . 6 97 6 3 2 8 0 . 3 993 2 6 24. 31 94 2405. 6261 2 1 8 6 . 93 2 8 2 0 9 1 . 848 8 1 640. 1 996 1 443 . 3 757 1 3 1 2 . 1 59 7 1 2 55 . 1 09 3 1 09 3 . 4664 1 04 5 . 9244
9 3 7 . 2 56 9 89 6 . 5 066 6 5 6 . 07 9 9 6 2 7 . 5.546 5i 4. 86 3 9 4 3 7 . 3 866
4 1 8 . 3698
400. 9 3 7 7 39 8 . 7 22 6 3 4 3 . 6 6 09 2 8 8. 6 1 5 1 2 7 4 . 92 8 7 2 6 2 . 43 1 9 2 06· 1 96 5
1 96 . 3 176
1 92 . 4 5 0 1 1 8 8. 6 1 66 1 44. 3 3 7 6 1 3 1 . 2 1 60 1 20. 2 8 1 3 1 1 1 . 02 89 1 0 3 . 0983
!tI ZZ�O
Appendix B
438
Table
B.2. Base Impedance in Ohms
8A s e K I l OV O L T S 2.30 2 . 40 4. 0 0 4. 1 6 4. 40
4. 80 6. 60
6. 90 7. 2 0 1 1 . 00 11 . 4 5 12. 00 12 . 4 7 13. 20 13. 80 14. 40 22. 0 0 24. 9 4 33 . 0 0 34. 5 0 44 . 0 0 55. 0 0 60. 00 66. 0 0 69. 0 0 88 . 00 1 00. 0 0 1 10 . 00 1 15.00 1 32 . 0 0 1 38 . 00 1 54 . 0 0 1 61 . 0 0
2 20. 00
2 30 . 0 0 2 75. 0 0 3 30 . 0 0 3 45 . 0 0 3 60. 0 0 3 62 . 0 0 4 20. 0 0 5 00 . 0 0 5 2 5. 0 0 5 50 . 0 0
700. 00
7 3 5 . 00 750. 00
765. 0 0 1 0 00 . 0 0 1 1 00 . 0 0 1 2 00. 0 0 1 3 00 . 0 0 1 4 00 . 0 0 1 5 00. 00
BAs e 50. 0 0
M E GAVOL T - AM P E R E S
1 00 . 0 0
8 7 . 1 2 00 9 5. 2 2 00 1 5 4 . 8 800 2 00 . 0 0 0 0 242. 0000 264 . 5000 34 8 . 4 8 00 3 8 0 . 8 8 00 474. 3 2 00 5 1 8 . 42 00 9 6 8 . 00 00 1 05 8 . 00 00 1 5 \2 . 5 0 0 0 2 1 7 8 . 0 0 00 2380. 5000 2 592. 0000 2 6 2 0 . 8 8 00 3 5 2 8 . 00 0 0 5000. 0000 5 5 1 2. 5000 6 0 5 0 . 00 00
0. 0529 0 . 0 5 16 0. 1 600 0. 1731 0 . 1 93 6 0 . 2 3 04 0 . 43 5 6 0. 4761 0 . 5 1 84 1 . 2 1 00 1 . 3 1 10 1 . 44 0 0 1 . 5550 1 . 74 2 4 1 . 9 01t4 2 . 0 73 6 4 . 8 1t 0 0 6 . 2200 1 0. 8 9 0 0 1 1 . 902 5 1 9 . 3600 3 0 . 2 500 36. 0000 43. 5600 4 7 . 6 1 00 7 7 . 44 0 0 1 00 . 0 0 0 0 1 2 1 . 0000 1 32. 2 500 1 7 4 . 2 40 0 1 90. 4400 2 3 7 . 1 6 00 2 5 9 . 2 1 00 4 84 . 0 0 0 0 5 2 9 . 0000 7 5 6 . 2 5 00 1 08 9 . 0 0 0 0 1 l 9 0 . 2 50 0 1 2 96. 0000 1 3 10. 4400 1 7 64. 0 0 0 0 2 5 00 . 0 0 0 0 2 756. 2 500 302 5 . 0000
9 8 00 . 00 0 0 1 08 0 1t. 5 0 00 1 1 2 5 0 . 00 0 0 1 1 7 04. 5 0 0 0 2 0 0 0 0. 0 0 0 1) 2 4 2 0 0 . 0 0 00 2 8800. 0000 3 3 8 0 0 . 0 0 00 3 9 2 0 0 . 0 0 00 45000. 00 00
4900. 0 0 0 0 5402. 2 500 5 6 2 5 . 0 0 00 5 8 52 . 2 500 1 0 00 0 . 0 0 0 0 1 2 1 00 . 0 0 0 0 1 44 0 0 . 0 0 0 0 1 6 9 0 0. 0 0 0 0 1 96 0 0 . 0 0 00 2 2 500. 0000
0 . 1 0 58 0. 1 1 52 0.3200 0. 3461 0 . 3 8 72 0. 4608 o. 871 2 0.9522 1 . 0 3 68 2 . 1t2 00
2.6221
2 . 8 8 00
3 . 1 1 00
3. 4848 3. 8088
4. 1472
9. 6800
1 2 . '" 0 1 2 1 . 7800 2 3. 8 0 5 0 3 8 . 7 2 00 6 0. 5 0 00 7 2 . 0000
2 00. 0 0 0 . 0 2 64 0. 02 88 0. 0800 0 . 0865 0. 0 9 6 8 0. 1 152 0. 2 1 78 0. 2 3 8 1 0 . 2 59 2 0. 6 0 5 0 0. 6 5 5 5 0 . 7 2 00 0. 7 7 7 5 0. 8712 0. 9522 1 . 0 3 68 2 . 4 2 00 3 . 1 1 00 5. 4450 5. 95 1 3 9 . 6 8 00 1 5 . 1 2 50 1 8 . 0 0 00
2 5 0. 00 0 . 02 1 2
6 6 . 1 2 50 8 7 . 1 2 00 9 5 . 2 2 00 1 1 8 . 5 8 00 1 2 9. 6050 2 4 2. 0 0 0 0 2 6 4 . 5 0 00 378. 1 2 50 544. 5 0 00 5 9 5 . 1 2 50 64 8 . 0 00 0 6 5 5 . 2 2 00 8 8 2 . 0 00 0 1 2 5 0. 0 000 1 ;' 7 8 . 1 2 5 0 l !i l 2 . 5 0 00
0. 0230 0. 0640 0 . 0692 0. 0774 0. 0922 0 . 1 742 0 . 1 9 04 0. 2074 0 . 4840 0 . 5 244 0. 5760 0. 6 2 2 0 0. 6970 0 . 761 8 0. 8294 1 . 9360 2 . 48 8 0 4 . 3 5 60 4. 761 0 7 . 7440 1 2 . 1 00 0 1 4 . 4 00 0 1 7. 4240 1 9 . 044 0 3 0. 9760 4 0 . 0000 4 8 . 4 000 5 2 . 9000 69. 6960 76. 1 760 94. 8 6 4 0 1 0 3 . 6840 1 9 3 . 6 00 0 2 1 1 . 6 0 00 3 02 . 5 0 0 0 4 3 5 . 6 00 0 4 76 . 1 00 0 5 1 8 . 4 0 00 5 24. 1 7 60 7 0 5 . 6000 1 000. 0000 1 1 0 2 . 5000 1 2 1 0. 0000
2 4 5 0 . 0 000 2 7 01 . 1 2 5 0 281 2 . 5000 2 9 2 6 . 1 2 50 5 0 00. 0 0 0 0 6 0 5 0. 0000 1 2 0 0 . 0 0 00 8 4 5 0. 0 000 9 8 0 0 . 0000
1 96 0 . 0 0 0 0 2 1 60. 9000 2 2 5 0 . 000 0 2 34 0 . 9 0 0 0 4 0 0 0 . 0 000 4 8 4 0 . 0000 5 1 6 0 . 0000 6 76 0 . 0000 7 84 0 . 0000
2 1 . 7 8 00 23. 8050 3 8. 7200 5 0. 0000 6 0. 5 0 00
1 1 2 50. 0000
9000. 0 000
439
L i n e I m peda nce Tables
BA S E K I L OVOL T S
2. 30 2 . 40 4. 00 4. 1 6
4. 40 4. 80 6. 60
6 . 90
7. 2 0 1 1 . 00
11 . 45
Il. !.}!.} 12 . 47
1 3 . 20 1?! 80
14. 40 2 2 . 00 24 . 2 4 33 . 0 0 34 . 5 0 44 . 0 0 55 .
00
6 0 . 00
6� . 00 69. 0 0 88 . 0 0 1 00 . 0 0
1 10. 00
1 1 5. 00 1 32 . 00 1 38. 00 1 54 . 0 0
1 61 . 0 0 2 20 . 00 2 30. 00 2 75 . 0 0
3 3 0. 0 0 3 45 . 0 0 3 60 . 0 0 3 62 . 0 0 4 20. 00
� OQ. O Q 5 2 5. 0 0
5 50. 0 0
1 00. 0 Q
135. 00 7 5 0 . 00 765. 00 1 0 00 . 0 0
1 1 00 . 0 0 12 00. 0 0 1 3 00 . 0 0 1 400 . 0 0
l � QQ. Qg
Table B.3. Base Admittance in Micromhos B A S E M E GA V O L T- AM P E R E S
50 . 0 0
1 00 . 0 0
8 6 8 0 5 55. 5 5 56
1 890 3 59 1 . 6 8 2 4 1 73 6 1 1 1 1 . 1 1 1 1
2 5 8 2 644. 62 8 1
5 7 7 847 6 . 3 3 1 4 5 1 6 52 8 9 . 2 562
94 5 1 7 95 .
84 1 2
3 1 2 5 0 0 0. 0 0 0 0 2 8 89 2 38 . 1 6 5 7
2 1 10 1 3 8 . 8 8 8 9 l 1 4 7 8 42 . 0 5 6 9 1 0 5 0 1 9 9 . 5 3 19 9 64 5 06 . 1 7 2 8 4 1 3 2 2 3 . 1 40 5 3 8 1 3 B O. 9 8 0 5 �4 7 l 2 ' . ' 2 " 3 2 1 54 1 . 54 7 3
2 8 69 60. 5 1 42 2 6 2 549. 88 4 5 2 4 1 1 2 6. 5 4 3 2 1 03305. 1851
8 0 � � 5 . 3 8 6 ft 4 5 9 1 3 . 68 2 3 42001 . 9 8 1 5 2 5 8 2 6 . 44 6 3
16528. 1 3 8 8 8.
92 56 8889 l 1 418 . 42 06 1 0 5 0 1 . 99 5 4 64 56 . 6 1 1 6 5 000 . 00 0 0 4 1 32 . 2 3 1 4 3 1 80 . 7 1 8 3 2 8 69. 60 5 1 2 6 2 5. 4 9 8 8
2 1 08 . 2 8 1 3 1 9 2 8. 9 3 7 9 1 03 3 . 0 57 9 9 4 5 . 1 19 6 661 . 1 5 70 4 59. 1 3 6 8 4 20. 019 8 3 8 �. 8 0 2 5 3 8 1 . 5 51 2 2 8 3 . 446 1 200. 0 0 0 0 1 8 1 . 40 5 9 1 6 5 . 2 89 3 1 0'. Q4Q8 9 2 . 5 5 40
88. 8889 8 5 . 437 5 0 . 0000 41 . 32 2 3 3 4. 72,2 29. 5 8 5 8 2 5 . 5 1 02
ZZ. ZZZZ
6 l 5 0 0 0 0. 0 0 0 0
4 3 40 2 1 1. 1 7 1 8 2 2 9 5 6 84 . 1 1 3 9
2 1 0 03 9 9. 0 7 5 8 19�90 1 2. 31t 5 7 82 6 44 6 . 2 8 1 0 16 2 7 6 1 . 96 l 1
�2H4!t.!t!t!t!t 64 3 0 8 3 . 0947 5 1 3 9 2 1 . 02 8 5 5 2 5 0 9 9. 7 6 9 0 4 8 2 2 53 . 0 8 64 2 0661 1 . 5702
1� 0 77 Q. II 3 7 9 1 8 2 7 . 3646 840 1 5. 9 6 3 0 51 6 5 2 . 89 2 6 3 3 0 5 7. 8 5 1 2 217 77. 7778 2 2 9 5 6 . 84 1 1 2 1 003. 9908 1 29 1 3. 2 2 3 1 1 0 00 0 . 0 0 0 0 8 2 6 4 . 4 6 28 7 5 6 1 . 4367 5 1 3 9. 2 1 0 3 5 2 5 0 . 99 7 7
42 1 6 . 5 6 2 7 3 8 5 7. 8 7 5 9 2 066. 1 1 5 7 1 890. 3 5 92 1 32 2 . 3 1 40 91 8. 2 73 6 8 40 . 1 5 9 6 7 7 ).1 6 04 9
7 63 . 1 0 2 5 566. 8 9 3 4
!tOO. OQOO 362. 81 1 8
3 3 0. 5 7 8 5 Z Q4. QI U 1 8 5. 1 0 8 1 1 17. 1 7 7 8 7 0 87" 1 00. 000 8 2 . 446 6 9. 4444 5 9. 1 7 1 6
5 1 . 0 2 04
��. !��!
2 00 . 0 0
2 50. 0 0
4591368 . 2 2 17 42 0 019 8 . 1 5 1 6 3 8 5 8 0 2 4. 6 9 1 4 1 65 2 8 9 2 . 5 62 0 1 5 25 5 2 3 . 9 2 2 1
8 6 8 0 5 5 5. 5 5 5 6
4 72 5 8 979. 2 06 0 43 1t02 7 77 . 7 7 7 8 1 5 6� 5 000. 0000 1 4446 1 90. 82 81t 1 29 1 3 2 2 3 . 1 4 0 5 1 08 5 0694. 1t1t1t4 5 7 39 2 1 0 . 2 84 7 5 2 5 09 9 7 . 6896 4 8 2 2 5 3 0 . 861t22 066 1 1 5 . 7 0 2 5 1 906 9 04 . 9 0 2 7
UUIUUI. 1I1U12 1 2 8 6 1 6 6 . 1 894
U�U U. UU
3 7 8 07 1 8 3. 3 64 B 3 1t 7 22 2 2 2 . 2 22 2 1 2 5 00000. 0 0 0 0 l 1 5 5 6 9 5 2 . 66 2 7 1 03 3 057 8 . 5 1 24
1 1 4 7 8 42 . 0 5 6 9 1 0 5 0 1 99 . 5 379 9 64 5 0 6 . 1 1 2 8 4 1 3 2 2 3. 1 40 5 �ZU!tl. �!tU 1 8 3 6 54 . 7 2 9 1 1 6 8 0 3 1 . 9 2 61 103305.7851 66l 1 5 . 7025 5 5 5 55 . 5 5 56 1t 5 9 1 3 . 6 8 2 3 42 0 07 . 98 1 5 2 5 826. 4463 2 0000. QOOO 1 6 52 8 . 9 2 5 6 1 5 1 2 2 . 87 3 3 &1478. 4 2 06 1 0 50 1 . 9 9 5 4 843 3 . 1 2 53 1 71 :! . I 5 ). 7 1t 1 32 . 2 3 1 4 3 78 0 . 7 1 8 3 2 64 4 . 6 2 8 1 1 836. 5473 1680. 3 193 1 54 3. , 099 1 5 2 6 . 2 049 1 1 3 3. 7 8 6 8 8 0 0. QOOO 725.6236 66 1 . 1 5 7 0
4121. 1633
37 0 . 2 1 6 1 3 5 5. 5 5 5 6 4 1489 2 00 . 0 0 0 0 16 5 . 2 89 3 131. 8 8 89 1 1 8. 3 4 3 2 1 0 2 . 01t08
111.11112
1 60 7 7 0 7. 7367 1 43 4 8 02 . 57 1 2 1 31 2 749. 4 2 24 1 20 5 6 3 2 . 1 1 6 0 516528. 9256
!Q1 2Z6. 2UZ
229568. 4114 2 1 003 9. 9076 1 2 9 1 32. i3J.4 8 2 6 44 . 6 2 8 1 6 9444. 4441t 5 7 3 92 . 1 0 2 8 5 2 5 0 9 . 9769 3 2 2 83 . 0579 � 5 0 00. 0000 2 06 6 1 . 1 5 7 0 1 8903. 5917 1 43 4 8 . 02 5 1 1 3 1 2 7 . 4942 1 0 5 4 1 . 4066 9 644. 6 826 5 1 6 5 . 289 3 4 7 2 5 . 8979 3305.1851 2 2 9 5 . 6 84 1 2 1 00 . 399 1 ). 9 , 9 . 0 11 3 1 90 7 . 7 5 6 2 1 4 1 7 . 2336 lQOO. QQOQ 9 0 7 . 029 5 8 2 6 . 446 3
HQ. ZQ!J,
462 . 77 0 1 4 4 4. 1t1t1t 4 427 8 2 5 0 . 0000 2 06. 6 1 1 6 173. 61U 1 4 7 . 9 2 90 1 2 7. 55 1 0
lU.Un
j j ��r
Table m. d
linI!
Characteristics of Copper Conductors , Hard Drawn ,
lam pro". . eler Outeigh Cu rciS of aide Breakin rent P :; 'S I !'di- Di.m Strenlth �r C� rry. iii VIdual eler Pounds Inl i l. M ---j Straade I ncb� .., Incb
Conductor
'
C'1fC)Jlar MIle l!t1Zi e
I
B.4.
..:
i
aeo:-
97.3
:c. I nductive Reactance
r. Reeiatance r Obm. pe Cond uctor per Mile
me tric M ea n Radiua at 60 ,..•. _._-
5O"C.
25"C. (77"1'.)
Ohm.
re� ��r: uc tor
At 1 Ft. Spacinl
( l 22°F.)
300 300
250 250 211 211 211 167 1 67 1 33 105 83
62
. 21 6 . 2 20 . 224 . 226
31 1 70 27 020 22 5 1 0 2 1 590
. 422 . 4 32 . 44 3 . 445
77 0 . 726 . 679 . 7 10
19 17 15 15
75C 560 590 140
. 45 1 4 58
. 229 . 23 5 . 24 1 . 24 1
. 629 . 657 . 574 . 600
13 13 11 11
510 1 70 360 1 30
. 4 76 . 470 . 4 87 . 481
. 259 . 256 . 266 . 263
. 528 . 552 . 522 . 492
9 9 9 7
617 483 1 54 556
. 497 . 49 1
. 272 . 269
. 464 . 4 14
7 5 4 3
366 926 752 804
3 3 2 3
620 045 9 13 003
1 351 1 1 07 1 1 061
. 443 0 . 532 . 4 55 . 546 . 467 . 560
. 289 . 298 . 306
2 2 2 1
433 359 439 879
858
52
52 41
. 368
. 328
� 1T 1 1 7
3 3 3 4
.0 3 0 . 1 32 1 . . . . . . 3 0. 1 1
8 1'_ OOIMluew
at
At 1 Ft S pacina
. 400 . 406 . 4 13 . 4 17
.
. 260 . 285 . 229 . 254
1 970 1 506
850
667 534 529
170 1 50 140
84 1 674
1 205
424
1
420
333 264
692 . 8111 . 873
270 230 240 220
200 200
1 59 1
280 1 030 826
. 349 . 440 . 555 . 699
420 360 3 10 270
.
. 864
. 699
. 555
. 6De
. 350 . 440 . 555 . 6De
. 692 . 882
. 692 . 882
. 692 . 882
. 349 . 440
Same u d-e
1 90 180
O . M' 0 . 481 . 606 0 . 765
0 . 757 0 . 964 . 9S5 . 946 1 . 216
2 . 39
1 20 1 10
3 . 80
90
3 . 01
per
bour (2 IV_) , frequency-60 cyel..
. 090 1 09 1 6 . 09 34 . 0943
.
. 0954 . 09 77 . 1004 . 1005
. 1020 . 1038
. 245 0 . 24 9 . 254 . 25 1
. 1058
. 1044
. 1 080 . 1 068 . 1 108 . 1 094
� M ' � w � . � r ' " r 1 '1 "" I
L� I 1 . 518
1 . 503 1 . 9 14 1 . 895 2.41
130
n"C., air at 25ec., wind 1.4 mil.
. 349 . 440 . 555
. 48 1 . 607
. 48 1 . 607
. 48 1 . 607
0. '" 0 . 222 0 . 227 0 . 233
0 . 232 0 . 239 0 . 238 0 . 242
. 464 . 557 . 47 8 0 . 574 . 476 . 57 1 . 484 . 58 1
. 299 . 314 . 307 . 323
. 1 1 32 . 1 1 19
. 1 44 . 14 . 1 52
. . . .
149 1 57 1 53 1614
. 1 205 . 1 240 . 1 274
. . . .
1 246 1308 1 281 1346
s.me u d-e
rrrrr; . . 0 . 256 0 . 260 0 . 262
. 51 1 . 519 . 523
. 6 13 . 623 . 628
. 332 . 348 . 34 1
. 1 66 1 . 17 . 17
0 . 265 0 . 27 1
. 53 1 . 542 . 554
. 637 . 65 1 . 665
. 356 . 364 . 37 2
. 17 . 1 82 1 .1
. 277
t
a
:c.' Sbunt C.p.citive R ea c ta n ce Melohma per C o nduc tor Per M ile
43 830 39 .� 1 0 3 5 1 20
33 fOO
450 fOO 350 350
@
Percent Conductivity
. 1 384 . 1 449 . 1419
. 1483
. 1 552
. 1517
»
'C 'C (1) ::l a.
)C .
to
Table B.S.
Characteristics of Anaconda Hollow Copper Conductors
Wires
Design Number
Size of Conductor Circular Mils or
A.W.G.
Outside Diameter Number
Diameter
I
Inches
Breaking Strain Pounds
I ��m'
(Anaconda Wire & Cable Company)
Weight Pounds per Mile
Inches
Mean Radius at 60 Cycles Feet
ra
Resistance Ohms per Conductor per Mile
Approx. Current Carrying Capacity Amps*
25°C. ( 77°F. )
d-c 25 cycles
--
966 96R1 939 360R 1 938 4R5 892R3 933 924 925R1
565R1 936 378R1 954 935
903R1 926 915R1 24R1 923 922 50R2 158R1 178R2
495R1 570R2 909R2 412R2 937 930 934 901
890 750 6 50
500 000 000
28 42 50
500
000 000 000
50 50 18 21
600 550 510
450
400 380 350 350 350
321 3 00
300
300
000 000
000 500 000 000 000 000 000 000
000 000 000 250 000 410 410 410 3/0
250 250
3/0 3/0 2/0 2/0 2/0 125 600 1 2 1 300 1 19 400
50
21 22
21
15 30 22 18 15 12 18 15 12 18
15
14 16
15 12 15 14 13 14 15 12
--
o�i61O 0. 1296 0. 1097 0 . 1 053 0 . 1 009 0.0970 0 . 1 558 0. 1353
0 . 1227 0.1211 0 . 1 196 0. 1444 0 . 105 9 0 . 1 1 13 0 . 1 205 0 . 1 338
--
1 . 650 1. 155 1. 126 1 . 007 1.036 1. 000 1 . 080 1 . 074 1 . 0 14 1. 003
0.950
0.860 0.736
0.920
0.839 0.797
0 . 150 7
0 . 750
0. 1368 0. 1005 0. 1 109 0.1 152 0.0961
0.683 0.700 0.663 0.650 0.606
0.1123 0.0880 0.0913 0.0950 0.0885 0.0836
0. 595
0.1100
0. 12 14
0.0996
0.0936
0. 766 0.725
0.560 0.530 0.515 0.505 0.500 0.500 0.470
I
36 34 29 27 25 22 21 19
000 200 500 500 200 700 400 300
15 12 10 9 9 8 8 7
085 345 761 905 103 485 263 476
0.0612 0.0408 0.0406 0,0387 0 , 0373 0.0360 0 . 0394 0. 0398
1395 1 160 1060 1020 960 910 900 850
17 16 15 15 16 13 13 13
200 300
642 331 813 776 739 343 984 953
0.0376 0 . 0373 0 .0353 0.03 1 1 0.0253 0 .0340 0.0307 0.0289
810 780 750 740 700 700 670
200
6 6 5 5 5 5 4 4
660
950 000 000 9 300 9 300 9 300 7 500
4 4 4 4 3 3 3 2
937 155 148
0.0279 0.0266 0.0245 0.0255 0.0238 0.0234 0.0221
600 590 580 530 520 520 460
7 7 5 6 6 5 5 5
2 2 2 2 2 2 2
13 10 11 11
100 100
400
100
850
050
600 600 000 950
650 400 300 000
1
133 521
510 785
510
772 785
207 203 083
2 13
979 015
'For conductor at 75°C., air at 25°C., wind 1.4 miles per hour (2 ftlsec), frequency
�
0.0266
0.0214 0.0201 0.0191 0.0184 0.0181 0.0180 0.0179 0.0165
650
460
370 370 370 360 350 340
450
1
50 cycles 60 cycles
25 cycles
Shunt Capacitive
Xa
50°C. ( 122°F. )
d-c
xa '
50
cycles 60 cycles
Inductive Reactance Ohms per Conductor per Mile at 1 Ft. Spacing
25 cycles
50 cycles
I
0 . 0676 0.0791 0.0915 0 . 0991 0. 1081 0 . 1 178 0 . 1 184 0. 1324
0 . 0734 0 . 0860 0.0994 0 . 1077 0 . 1 17 7 0 . 1283 0 . 1289 0. 1443
0 .0739 0. 0865 0 . 1001 0. 1084 0 . 1 183 0 . 1289 0 . 1296 0 . 1448
0 . 1412 0.1617 0 . 1621 0. 1644 0 . 1663 0. 1681 0 . 1630 0. 1630
0.282 I 0.323 0.324 0.329 0.333 0.336 0.326 0.326
0.339 0.388 0.389 0.395 0.399 0.404 0.391 0.391
0. 1485 0 . 1565 0 . 1695 0. 1690 0 . 1685 0, 1851
0 . 1491 0 . 1572
0 . 1624 0 . 1 7 12 0 . 1854 0 . 1849 0 . 1843
0 . 163 1 0.1719
0. 1658 0 . 1663
0.332 0.333
0 . 1854 0 . 1849 0 . 2 03
0, 1754 0 . 1860 0.1710
0.351 0.372
0.398 0.399 0.406 0.421 0.446
0.352 0.359
0.423 0 . 430
0. 1980
0 . 1695 0. 1690 0 . 1856 0. 1985
0 . 1969
0. 1975
0.238 0.237 0.237 0.281 0.281 0.280 0.354
0,239 0.238 0.238 0.282 0.282 0.281 0.355
0. 1964
0.353 0.352 0.446 0.446 0.446 0.473 0.491 0.507
60 cycles, average tarnished surface.
0, 1969
0.354
0.446 0.446 0.446 0.473 0.491 0.507 0.353
-
0.202
0.215
0.216
0.215 0.260 0.259 0.259 0.307 0.307 0.306 0.387 0.386 0.385 0.487 0.487 0.487 0.517 0.537 0.555
0. 1860
0.217 0.216
0.261 0.260 0.260 0.308 0.308 0.307 0.388 0 . 2 16
0,387 0.386
0.487 0.487 0.517 0.487
0.555
0.537
0. 1691 0.1761
0 . 1 793
0. 1810 0. 1834 0. 1876 0. 1855 0. 1889 0. 1898 0. 1833
0 . 1928
0. 1976 0.200 0.202 0.203 0.203 0.203 0.207 0 . 1943
0.338
0.342 . 0.410 0.367 0.362 0.367 0.375
0.371 0.378 0.380 0.386
0.389 0.395 0.400 0.404 0.406 0.406 0.407 0.415
Reactance Megohms per Conductor per Mile at 1 Ft. Spacing 50 cycles
60 cycles
0 . 1907 0.216 0.218 0.221 0. 224 0.226 0.221 0.221
0 . 0953 0 . 1 080 0. 1089 0 . 1 105 0.1119 0 . 1 13 1 0 . 1 104 0. 1 106
0.0794 0.0900 0. 0908 0.0921 0.0932 0.0943 0 . 0920 0.0922
0.225 0.226 0.230 0.237 0.248 0 . 232 0 . 239
0 . 1 126 0 . 1 130 0 . 1 15 0 0 . 1 185 0. 1241 0 . 1 16 1 0 . 1 194
0 . 0939 0.0942 0 , 0958 0,0988 0. 1034
0.247
0.245 0.249 0.253 0.252 0.256 0.257 0.262
0 . 1234
0. 1028
0.263 0.268 0.271
0. 1226 0. 1246 0. 1267 0. 1258 0. 1278 0. 1285 0.1310 0.1316 0. 1338 0. 1357 0. 1368 0. 1375 0. 1378 0. 1378
0. 1097 0.1115 0.1131
25 60 cycles i cycles
0.0671 0 .0786 0.0909 0.0984 0. 1076 0 . 1 173 0 . 1 1 78 0. 13 19
0.1700
o
0.434 0 .440 0.450 0.445 0.453 0.455 0.463 0.440
0.466 0.474 0.481 0.485 0.487 0.487 0.488 0.498
I
0.242
0.275 0.276
0.274
0.280 0.276
0. 1212
0 . 1400
0.0968 0.0995 0,1010 0. 1022 0. 1038 0. 1066 0.1049 0. 1065 0. 1071 0. 1091
0.1146 0.1149 0 , 1 140
0.1167
0. 1 149
Table B.6. ConducOuttor Wall side(1) Size ThickbiamCircular ness eter Mils or Inches Inches A. W. G.
Weight Pounds per Mile
Characteristics of General Cable Type HH Hollow Copper Conductors (General Cable Corporation)
Approx. GeoCurrent Breakmetric Carrying Mean ing trength Radius CapaPounds Feet city(2) Amps
I
ra
Resistance Ohms per Conductor per Mile
�
25°C. (77°F.)
50°C. ( 122°F.)
:Re
Inductiv actance Ohms per Conductor per . . Mile at 1 Foot SpaCIng
d-c
25 cycles
50 cycles
60 cycles
d-c
25 cycles
50 cycles
60 cycles
25 cycles
50 cycles
60 cycles
I
I
��� Xa '
o
Shunt Capacitive React c gohms per on uc or per Mile at 1 Foot Spacing 25 cycles
50 cycles
60 cycles
0. 1 734 0. 1757 0 . 1 782 0. 1805
0.0867 0.0879 0.0891 0.0903
0.0722 0.0732 0.0742 0.0752
0.0917 0 .0953 0.0932 0.0945 0.0962 0.0974 0. 0992 0. 1012
0.0764 0.0794 0.0777 0.0788 0 .0802 0.08 1 1 0.0827 0.0843
1 000 950 900 850
000 000 000 000
2 . 103 2.035 1 .966 1.901
0. 150* 0. 147* 0. 144* 0. 140*
16 15 14 13
160 350 540 730
43 41 38 36
190 030 870 7 10
0 .0833 0 .0805 0. 0778 0.0751
1620 1565 1505 1450
0.0576 0.0606 0.0640 0.0677
0.0576 0.0606 0.0640 0.0678
0.0577 0.0607 0.0641 0.0678
0.0577 0.0607 0.0641 0.0678
0.0630 0.0663 0.0700 0.0741
0.0630 0.0664 0.0701 0.0742
0 . 0631 0.0664 0.0701 0.0742
0.063 1 0.0664 0.0701 0.0742
0. 1257 0. 1274 0 . 1 29 1 0. 1309
0.251 0.255 0.258 0.262
0.302 0.306 0.310 0.314
800 790 750 700 650 600 550 512
000 000 000 000 000 000 000 000
1 .820 1.650 1 . 750 1.686 1.610 1.558 1.478 1.400
0. 137* 0.131t 0. 133* 0. 130* 0. 126* 0. 123* 0. 1 19* 0. 115*
12 12 12 11 10 9 8 8
920 760 120 310 500 692 884 270
34 34 32 30 28 25 23 22
550 120 390 230 070 910 750 110
0.0722 0.0646 0.0691 0.0665 0.0635 0.0615 0.0583 0.0551
1390 1335 1325 1265 1200 1 140 1075 1020
0.0720 0. 0729 0.0768 0.0822 0.0886 0.0959 0. 1047 0 . 1 124
0.0720 0.0729 0.0768 0.0823 0.0886 0.0960 0. 1048 0 . 1 125
0.0720 0.0730 0.0768 0. 0823 0.0886 0. 0960 0. 1 048 0 . 1 125
0.0721 0.0730 0.0769 0.0823 0. 0887 0.0960 0. 1048 0 . 1 125
0.0788 0.0797 0.0840 0. 0900 0.0969 0. 1050 0. 1 146 0. 1230
0.0788 0.0798 0.0840 0.0900 0.0970 0. 1051 0. 1 146 0. 1230
0.0788 0.0799 0.0841 0.0901 0.0970 0. 1051 0. 1 147 0. 1231
0.0788 0.0799 0.0841 0.090 1 0.0970 0. 1051 0. 1 147 0.1231
0. 1329 0 . 1385 0 . 135 1 0. 1370 0 . 1394 0. 1410 0. 1437 0. 1466
0.266 0.277 0.270 0.274 0.279 0.282 0. 287 0. 293
0.319 0.332 0.324 0.329 0.335 0.338 0.345 0.352
0. 1833 0. 1906 0 . 1864 0. 1891 0.1924 0. 1947 0. 1985 0.202
500 000 500 000 500 000 500 000 450 000 450 000 400 000 400 000
1.390 1.268 1. 100 1.020 1.317 1 . 188 1.218 1 . 103
0 . 1 15* 0. 109t 0. 130t O. l44t 0. 1 1 1* 0. 105t 0. 106* 0. 100t
8 8 8 8 7 7 6 6
076 074 068 063 268 266 460 458
21 21 21 21 19 19 17 17
590 590 590 590 430 430 270 270
0 .0547 0.0494 0.0420 0.0384 0.0518 0.0462 0.0478 0.0428
1005 978 937 915 939 910 864 838
I
0.1151 0.1151 0.1 150 0 . 1 150 0. 1 279 0. 1278 0. 1439 0. 1438
0. 1151 0 . 1 152 0 . 1 151 0 . 1 150 .01280 0. 1279 0. 1440 0. 1439
0 . 1 152 0. 1 152 0 . 1 152 0 . 1 152 0. 1280 0. 1279 0. 1440 0. 1439
0 . 1 152 0 . 1 152 0 . 1 153 0 . 1 152 0. 1280 0. 1280 0. 1440 0. 1440
0. 1259 0. 1259 0. 1258 0. 1258 0. 1400 0. 1399 0. 1575 0 . 1574
0 . 1 260 0. 1260 0. 1259 0.1259 0. 1401 0. 1400 0. 1576 0. 1575
0. 1260 0. 1260 0. 1260 0. 1260 0. 1401 0. 1400 0 . 1 576 0. 1575
I 0.1261
0. 1469 0. 1521 0. 1603 0. 1648 0 . 1496 0. 1554 0. 1537 0. 1593
0 .294 0.304 0.321 0.330 0.299 0. 3 1 1 0.307 0.319
0.353 0.365 0.385 0.396 0.359 0.373 0.369 0.382
0.203 0.209 0.219 0.225 0.207 0.214 0.212 0.219
0. 1014 0. 1047 0. 1098 0 . 1 124 0. 1033 0 . 1070 0. 1061 ' 0. 1097
0.0845 0.0872 0.0915 0.0937 0.086 1 0.0892 0.0884 0.0914
1. 128 1.014 1.020 0.919 0.914 000 , 0.818 000 0. 766 500 0.650
0. 102* O.096t 0 . 096* 0.09lt 0.091* O. 086t 0.094t 0.098t
5 5 4 4 4 4 4 3
653 650 845 843 03 7 036 034 459
15 15 12 12 10 10 10 9
110 110 950 950 790 790 790 265
0.0443 0.0393 0 . 0399 0.0355 0 . 0357 0.0315 0.0292 0.0243
790 764 709 687 626 606 594 524
0. 1645 0. 1645 0. 1919 0. 1918 0.230 0.230 0.230 0.268
0. 1645 0. 1645 0. 1645 0. 1646 0. 1919 0 . 1919 0 . 1918 0. 1919 I 0 . 230 I 0.230 0 .230 0.230 0.230 0.230 0.268 0.268
0. 1799 0. 1799 0 . 2 10 0.210 0 . 252 0 . 252 0.252 0.293
0. 1800 0. 1800 0.210 0.210 0.252 0.252 0.252 0.293
0. 1800 0 . 1800 0.210 0.210 0.252 0 . 252 0.252 0.293
0. 1800 0. 1576 0 . 180 1 0 . 1637 0.210 0 . 1628 0 . 2 10 0 . 1688 I 0.252 I 0 . 1685 0 . 1 7<8 0.252 0 . 1787 0. 1879 0.294
0.315 0.328 0.326 0.338 0.337 0.350 0.357 0 .376
0.378 0.393 0.391 0 .405 0.404 0.420 0.429 0.45 1
0.218 0.225
i
0. 1644 0. 1644 0 . 1918 0. 1917 0.230 0.230 0.230 0.268
0 . 232 0.233 0.24 1 0.245 0.257
0 . 1089 0 . 1 127 0 . 1 124 0 . 1 162 0 . 1 163 0 . 1203 0 . 1226 0. 1285
0.0907 0.0939 0.0937 0.0968 0.0970 0. 1002 0. 1022 0.1071
0.082t 0.080t 0.080t
3 41 5 2 707 2 146
9 140 7 240 5 750
0.0281 0. 0230 0.0186
539 454 382
0.272 0.343 0.432
0.272 0.343 0.432
0.272 0.343 0.432
0.297 0.375 0.472
0.297 0.375 0.473
0.298 0.375 0.473
0.298 0.375 0.473
0.361 0.381 0.403
0.433 0.458 0.483
0.248 0. 262 0.276
0. 1242 0. 1309 0. 1378
0. 1035 0. 1091 0 . 1 149
350 350 300 300 250 250 250 214
000 000 000 000 000
410 3/0 210 Notes:
1
0.733 0.608 0.500
0.272 0.343 0.432
----.l...- ----
0. 1260
0. 1260 0. 1261 0. 1401 0. 1401 0. 1576 0. 1576
o."u
----
0. 1806 0. 1907 0.201
1
- --- --- -
*Thickness at edges of interlocked segments. tThickness uniform throughout. ( 1 ) Conductors of smaller diameter for given cross-sectional area also available; in the naught sizes, some additional diameter expansion is possible. (2) For conductor at 75°C., air at 25°C., wind 1.4 miles per hour (2 ft/sec), frequency = 60 cycles.
I 0 . 225
'
Table B.7. Size of Conduc-
tor
-
Circular Mils or
Diameter No.
ofIndi-
of
vidual
Strands
Strands Inches
A. W. G.
Characteristics of Aluminum Conductors, Hard Drawn, Goo-
Outside Diameter Inches
Intimate Strength Pounds
Weight Pounds Per Mile
@
61 Percent Conductivity
(Aluminum Company of America) metric Mean Radius at 60
Cycles Feet
To Resistance
%0 Inductive Reactance
%0 ' Shunt Capacitive Reactance
Ohms per Conductor per Mile
Ohms per Conductor per
Megohms per Conductor per
Approx. Current Carrying city* d..
Amps
Mile a t 1 F t . Spacing
50°C. ( 1 22 °F. )
25°C. (77°F.)
Caps-
25
50
60
cycles
cycles
cycles
d..
Mile at 1 Ft. Spacing
25
50
60
25
50
50
60
cycles
cycles
cycles
cycles
60
25
cycles
cycles
cycles
cycles
cycles
0.0612 0.0772 0.0867 0.0974 0.1094
0. 184 0.232 0.260 0.292 0.328
528 826 1022 1266 1537
130 207 261 329 414
0.00556 0.00700 0.00787 0.00883 0.00992
100 134 155 180 209
3.56 2.24 1.77 1.41 1.12
3.56 2.24 1.77 1.41 1.12
3.56 2.24 1.77 1.41 1 . 12
3.56 2.24 1.77 1.41 1 . 12
3.91 2.46 1 .95 1 .55 1 . 23
3.91 2.46 1 .95 1.55 1 . 23
3.91 2.46 1.95 1 . 55 1.23
3.91 2.46 1 .95 1.55 1 .23
0.2626 0.2509 0.2450 0.2391 0.2333
0.5251 0.5017 0.4699 0.4782 0.4665
0.6301 0.6201 0.5679 0.5739 0.5598
0.3468 0.3302 0.3221 0.3139 0.3055
0.1734 0. 165 1 0.1610 0. 1570 0. 1528
0. 1445 0. 1376 0. 1342 0. 1306 0. 1273
7 19 7 19 7
0. 1228 0.0745 0. 1379 0.0837 0 . 1 548
0.366 0.373 0.414 0.419 0.464
1865 2090 2350 2586 2845
523 523 659 659 832
0.011 13 0.01177 0.01251 0.01321 0.01404
242 244 282 283 327
0.885 0.885 0.702 0.702 0.557
0.8851 0.8651 0.7021 0.7021 0.5571
0.8853 0.8853 0.7024 0.7024 0.5574
0.885 0.885 0.702 0.702 0.558
0.973 0.973 0.771 0.771 0.612
0.9731 0.9731 0.77 1 1 0.77 1 1 0.6121
0.9732 0.9732 0.7713 0.7713 0.6124
0.973 0.973 0.771 0.771 0.613
0.2264 0.2246 0.2216 0.2 188 0.2157
0.4528 0.4492 0.4431 0.4376 0.4314
0.5434 0.5391 0.5317 0.5251 0.5177
0.2976 0.2964 0.2890 0.2882 0.2810
0. 1488 0. 1462 0.1445 0. 144 1 0. 1405
0. 1240 0. 1235 0 . 1 204 0. 1201 0.1171
310 410 410 250 000 266 800
19 7 19 37 7
0.0940 0. 1739 0. 1055 0.0822 0. 1953
0.470 0.522 0. 528 0.575 0.586
3200 3590 3890 4860 4525
832 1049 1049 1239 1322
0.01483 0.01577 0.01666 0.01841 0.0177 1
328 380 381 425 441
0.557 0.441 0.441 0.374 0.350
0.5571 0.44 1 1 0.44 1 1 0.3741 0.3502
0.5574 0.4415 0.4415 0.3748 0.3506
0.558 0.442 0.442 0.375 0.351
0.612 0.485 0.485 0.411 0.385
0.6121 0.485 1 0.4851 0.41 1 1 0.3852
0.6124 0.4855 0.4855 0.4115 0.3855
0.613 0.486 0.486 0.412 0.386
0 . 2 1 29 0.2099 0.207 1 0.2020 0.2040
0.4258 0.4196 0.4141 0.4040 0.4079
0.5110 0.5036 0.4969 0.4848 0.4895
0.2601 0. 2726 0.2717 0.2657 0.2642
0. 1400 0. 1363 0. 1358 0.1328 0. 1321
0. 1167 0 . 1 136 0 . 1 132 0 . 1 107 0.1101
266 300 300 336 336
800 000 000 400 400
37 19 37 19 37
0.0849 0.1257 0.0900 0.1331 0.0954
0.594 0.629 0.630 0.866 0.668
1322 1487 1487 1667 1667
0.01902 0.01983 0.02017 0.02100 0.02135
443 478 478 514 514
0.350 0.3 1 1 0.3 1 1 0.278 0.278
0.3502 0.3 1 12 0.3112 0.2782 0.2782
0.3506 0.3117 0.3117 0.2788 0.2788
0.351 0.312 0.312 0.279 0.279
0.385 0.342 0.342 0.306 0.306
0.3852 0.3422 0.3422 0.3062 0.3062
0.3855 0.3426 0.3426 0.3067 0.3067
0.386 0.343 0.343 0.307 0.307
0.2004 0. 1983 0. 1974 0. 1953 0. 1945
0.4007 0.3965 0.3947 0.3907 0.3890
0.4809 0.4756 0.4737 0.4688 0.4668
0.2633 0.2592 0.2592 0.2551 0.2549
0.1316 0. 1296 0. 1296 0. 1276 0.1274
0.1097 0. 1080 0.1080 0.1063 0. 1062
350 397 477 500 500
000 500 000 000 000
37 19 19 19 37
0.0973 0. 1447 0 . 1 585 0 . 1 623 0 . 1 162
0.681 0.724 0.793 0.812 0.813
5180 5300 5830 5940 6400
6680 6880 8090 8475 9010
1735 1967 2364 2478 2478
0.02178 0.02283 0.02501 0.02560 0.02603
528 575 648 664 664
0.267 0.235 0.196 0. 187 0.187
0.2672 0.2352 0. 1963 0. 1873 0. 1873
0.2678 0.2359 0. 1971 0. 1682 0.1882
0.268 0.236 0.198 0. 189 0. 189
0.294 0.258 0.215 0.206 0.206
0.2942 0.2582 0.2 153 0.2062 0.2062
0.2947 0.2589 0.2160 0.2070 0.2070
0.295 0.259 0.216 0.208 0.208
0. 1935 0.1911 0. 1865 0. 1853 0.1845
0.3870 0.3822 0.3730 0.3707 0.3689
0.4644 0.4587 0.4476 0.4448 0.4427
0.2537 0.2491 0.2429 0.2412 0.2410
0.1268 0. 1246 0.1214 0. 1206 0. 1205
0. 1057 0.1038 0.1012 0.1005 0. 1004
556 636 715 750 750
500 000 500 000 000
19 37 37 37 61
0.1711 0. 1 3 1 1 0.1391 0.1424 0.1 109
0.856 0.918 0.974 0.997 0.998
9440 11240 12640 12980 13510
2756 3152 3546 3717 3717
0.02701 0.02936 0.03114 0.03188 0.032 1 1
0.168 0. 147 0. 137 0.125 0. 125
0.1683 0.1474 0.1314 0.1254 0.1254
0. 1893 0. 1484 0.1326 0. 1267 0.1267
0. 170 0. 149 0. 133 0. 127 0.127
0. 185 0. 162 0. 144 0. 137 0.137
0. 1853 0. 1623 0. 1444 0. 1374 0. 1374
0. 1862 0.1633 0. 1455 0. 1385 0. 1385
0. 187 0 . 164 0. 148 0. 139 0. 139
0. 1826 0. 1785 0. 1 754 0 .1 743 0.1739
0.3652 0.3589 0.3508 0.3485 0.3477
0.4383 0.4283 0.4210 0.4182 0.4173
0.2374 0.2323 0.2282 0.2266 0.2263
0 . 1 187 0 . 1 162 0. 1 141 0 . 1 133 0. 1 132
0.0989 0.0968 0.0951 0.0944 0.0943
795 000 874 500 954 000 1 000 000 1 000 000
37 37 37 61 91
0. 1468 0. 1538 0.1606 0. 1280 0.1048
1 .026 1.077 1.024 1. 152 1 . 153
13770 14830 16180 17670 18380
3940 4334 4728 4956 4956
0.03283 0.03443 0.03596 0.03707 0.03720
710 776 817 664 664
897 949 1000 1030 1030
0.117 0.107 0.0979 0.0934 0.0934
0 . 1 175 0.1075 0.0985 0.0940 0.0940
0 . 1 188 0.1089 0. 1002 0.0956 0.0956
0. 120 0 . 1 10 0.100 0.0966 0.0966
0. 129 0.118 0. 108 0. 103 0. 103
0. 1294 0 . 1 185 0. 1085 0. 1035 0. 1035
0. 1306 0 . 1 198 0 . 1 100 0. 1050 0. 1050
0. 1 31 0.121 0. 1 1 1 0.106 0 . 106
0. 1728 0. 1703 0. 1682 0. 1668 0.1664
0.3455 0.3407 0.3363 0.3332 0.3328
0.4146 0.4088 0.4036 0.3998 0.3994
0.2244 0.2210 0.2179 0.2182 0.2160
0.1 122 0. 1 105 0.1090 0. 1081 0.1080
0.0935 0.0921 0.0908 0.0901 0.0900
1 033 500 1 113 000 1 192 500
37 61 61 91 61
0. 1672 0.1351 0.1398 0. 1 145 0. 1444
1 . 1 70 1.216 1.258 1 . 259 1.300
18260 19660 21000 21400 22000
5122 5517 5908 5908 6299
0.03743 0.03910 0.04048 0.04062 0.04180
1050 1 1 10 1 160 1160 1210
0.0904 0.0839 0.0783 0.0783 0.0734
0.0910 0.0845 0.0790 0.0790 0.0741
0.0927 0.0864 0.0810 0.0810 0.0762
0.0936 0.0874 0.0821 0.0821 0.0774
0.0994 0.0922 0.0860 0.0860 0.0606
0.0999 0.0928 0.0886 0.0886 0.0813
0.1015 0.0945 0.0884 0.0884 0.0832
0. 102 0.0954 0.0895 0.0895 0.0843
0. 1661 0. 1639 0 . 1 622 0. 1620 0 . 1606
0.3322 0.3278 0.3243 0.3240 0.32 1 1
0.3987 0.3934 0.3892 0.3888 0.3853
0.2150 0.2 124 0.2100 0.2098 0.2076
0. 1075 0. 1062 0.1050 0.1049 0. 1038
0.0896 0.0885 0.0875 0.0874 0.0885
1 351 1 431 1 510 1 590 1 590
61 61 61 61 91
0.1489 0. 1532 0. 1574 0.1615 0. 1322
1.340 1.379 1.417 1.454 1.454
23400 24300 25600 27000 28100
6700 7091 7487 7883 7883
0.04309 0.04434 0.04556 0.04674 0.04691
1250 1300 1320 1380 1380
0.0691 0.0653 0.0618 0.0587 0.0587
0.0699 0.066 1 0.0627 0.0596 0.0596
0.0721 0.0685 0.0651 0.0622 0.0622
0.0733 0.0697 0.0665 0.0636 0.0636
0.0760 0.0718 0.0679 0.0645 0.0645
0.0767 0.0725 0.0687 0.0653 0.0653
0.0787 0.0747 0.0710 0.0677 0.0677
0.0798 0.0759 0.0722 0.0690 0.0690
0. 1590 0 . 1 576 0. 1562 0. 1549 0. 1547
0.3180 0.3152 0.3123 0.3098 0.3094
0.3816 0.3782 0.3748 0.37 18 0.3713
0.2064 0.2033 0.2014 0. 1997 0. 1997
0. 1027 0.1016 0. 1007 0.0998 0.0998
0.0856 0.0847 0.0839 0.0832 0.0832
6 4 3 2 1
110 110 210 210 310
g� ggg 500 000 500 000 000
7 7 7 7 7
- -
'For conductor at 75°C, wind 1.4 miles per hour (2 ftJsec), frequency = 60 cycle•.
--
-
---
Circular Mils or A.W.G. Aluminum 1 1 1 1 1 1
590 000 510 500 43 1 000 35 1 000 272 000 192 500
,
*
Aluminum
"
�
i
�
j
54
54 54
54
54
54
Steel
Table B.S.
ti
r.
a 1 ". -. �". """ �-" �. � � i5 '" tO o � ]
"" �
.0
Copper Equivalent* illtimate Strength Circular Mils or A. Pounds W. G.
e
i:! =i
19 19 19 19 19 19
0. 1030 0 . 1 004 0.0977 0.0949 0.0921 0.0892
1.545 1.506 1.465 1.424 1.382 1.338
0.1436 0.1384 0.1329 0. 1291 0.1273 0. 1214
19 7 7 7 7 7
0.0862 0. 1384 0. 1329 0. 1291 0. 1273 0.1214
1.293 1.246 1.196 1 . 162 1 . 146 1.093
700 000 650 000 600 000 566 000 550 000 500 000
3 3 3 3 3 3
0.1716 0. 1673 0. 1628 0.1582 0. 1535 0. 1486
1 00 0 950 900 850 800 750
000 000
000 000 000
Weight
Pounds per Mile
Approx. 0.0Current metric Carrying Mean Radius at Capa· cityt 60 Cycles Amps Feet
Resistance Ohms per Conductor per Mile
d-o
25 cycles
50 cycles
60 cycles
per Conductor per Mile at 1 Ft. Spacing
Mile at 1 Ft. Spacing All Currents
50°C. (122°F.) Current Approx. 75% Capacity;
25°C. (77°F.) Sman Currents
Shunt Capacitive Reactance Megohms
Inductive Reactance
Ohms per Conductor per
d-o
25 cycles
50 cycles
60 cycles
25 cycles
50 cycles
60 cycles
25 cycles
50 cycles
1
cycl
60 es
000 200 400 600 800 100
10 777 10 237 9 699 9 160 8 62 1 8 082
0.0520 0.0507 0.0493 0.0479 0.0465 0.0450
1 1 1 1 1 1
3SO 340 300 250 200 160
0.0587 0.0618 0.0852 0.0691 0.0734 0.0783
0.0588 0.0619 0.0653 0.0692 0.0735 0.0784
0.0590 0.0621 0.0655 0.0694 0.0737 0.0786
0.0591 0.0622 0.0656 0.0695 0.0738 0.0788
0.0646 0.0680 0.07 18 0.0761 0.0808 0.0862
0.0656 0.0690 0.0729 0.0771 0.0819 0.0872
0.0675 0.0710 0.0749 0.0792 0.0640 0.0894
0.0684 0.0720 0.0760 0.0803 0.0851 0.0906
0.1495 0. 1508 0 .1522 0. 1536 0.1551 0 . 1568
0.299 0.302 0.304 0.307 0.310 0.314
0.359 0.362 0.365 0.369 0.372 0.376
0.1953 0. 1971 0. 1991 0.201 0.203 0.206
0.0977 0.0986 0.0996 0. 1006 0. 1016 0.1028
40 200 37 100 34 200 32 300 3 1 400 28 500
7 544 7 019 6 479 6 1 12 5 940 5 399
0.0435 0.0420 0.0403 0.0391 0.0386 0.0368
1 1 10 1 060 1 010 970 950 900
0.0839 0.0903 0.0979 0.104 0.107 0.117
0.0640 0.0905 O.09SO 0. 104 0 . 107 0.118
0.0842 0.0907 0.0981 0. 104 0. 107 0 . 1 18
0.0844 0.0908 0.0982 0. 104 0.108 0 . 1 19
0.0824 0.0994 0.1078 0.1 145 0 . 1 178 0.1288
0.0935 0. 1005 0.1088 0 . 1 155 0 . 1 188 0.1308
0.0957 0 . 1025 0 . 1 1 18 0 . 1 175 0.1218 0 . 1358
0.0969 0.1035 0 . 1 128 0 . 1 185 0.1228 0. 1378
0.1585 0.1603 0.1624 0.1639 0.1646 0.1670
0.317 0.321 0.325 0.328 0.329 0.334
0.380 0.385 0.390 0.393 0.395 0.401
0.208 0.211 0.214 0.216 0.217 0.220
0.1040 0.1053 0. 1068 0.1078 0.1083 0.1100
0.0867 0.0878 0.0890 0.0898 0.0903 0.0917
200 400 300 100 600 500
5 770 6 5 17 4 859 5 193 5 865 4 527
0.0375 0.0393 0.0349 0.0355 0.0372 0.0337
900 910 830 840 800
0.117 0 . 1 17 0.131 0 . 131 0.131 0.140
0.117 0.117 0.131 0.131 0.131 0. 140
0.117 0.117 0.131 0.131 0.131 0.141
0.117 0.117 0.132 0.131 0.131 0.141
0.1288 0.1288 0.1442 0.1442 0.1442 0.1541
0.1288 0.1288 0 . 1452 0.1442 0 . 1442 0.1571
0.1288 0.1288 0. 1472 0.1442 0 . 1442 0. 1591
0.1288 0 . 1288 0.1482 0.1442 0.1442 0. 1601
01680 0.1637 0.1697 0 . 1887 0.1664 0.1715
0.332 0.327 0.339 0.337 0.333 0.343
0.399 0.393 0.407 0.405 0.399 0.412
0.219 0.217 0.224 0.223 0.221 0.226
0.1095 0.1085 0. 1 1 19 0 . 1 1 14 0.1 104 0 . 1 132
0.0912 0.0904 0.0932 0.0928 0.0920 0.0945
56 53 50 47 44 43
000
'.'
'.
Characteristics of Aluminum Cable, Steel Reinforced (Aluminum Company of America)
0.0814 0.0821 0.0830 0.0836 0.0647 0.0857
1 1 13 000 1 033 500 954 000 900 000 874 500 795 000
54 54 54 54 54
3 3 3 3 3 3
795 000 795 000 715 500 715 500 715 500 666 600
26 30 54 26 30 54
2 2 3 2 2 3
0.1749 0 . 1628 0 . 1 151 0.1659 0.1544 0.1111
7 19 7 7 19 7
0.1360 0.0977 0 . 1 151 0. 1290 0.0926 0.1111
1. 108 1 . 140 1.036 1.051 1.081 1.000
500 500 450 450 450 419
636 000 636 000 636 000 605 000 605 000 556 500
54
26 30 54 26 26
3 2 2 3 2 2
0.1085 0.1564 0.1456 0.1059 0 . 1525
7 7 19 7
7
0.1085 0.1216 0.0874 0.1059 0 . 1 186 0 . 1 136
0.977 0.990 1.019 0.953
0.966 0.927
400 000 400 000 400 000 380 500 380 500 350 000
23 600 25 000 3 1 500 22 500 24 100 22 400
4 3 19 4 616 5 213 4 109 4 391 4 039
0.0329 0.0335 0.0351 0.0321 0.0327 0.0313
770 780 780 750 760 730
0.147 0.147 0.147 0.154 0.154 0.168
0. 147 0. 147 0. 147 0 . 155 0.154 0. 168
0 . 148 0. 147 0. 147 0 . 155 0.154 0.188
0.148 0. 147 0.147 0.155 0. 154 0.168
0 . 1618 0.1618 0. 1618 0.1695 0. 1700 0.1849
0.1638 0 . 1618 0 . 1618 0.1715 0. 1720 0.1859
0 . 1678 0.1618 0. 1618 0.1755 0. 1720 0. 1859
0. 1666 0. 1618 0 . 1618 0.1775 0. 1720 0.1859
0 . 1726 0.1718 0 . 1693 0.1739 0. 1730 0.1751
0.345 0.344 0.339 0.348 0.346 0.350
0.414 0.412 0.406 0.417 0.415 0.420
0.223 0.227 0.225 0.230 0.229 0.232
0. 1140 0 . 1 135 0 . 1 125 0 . 1 149 0 . 1 144 0 . 1 159
0.0950 0.0946 0.0937 0.0957 0.0953 0.0965
500 30 000 30 000 , 26 000 30 500 26 500 30
2 2 2 2 2 2
350 000 314 500 300 000 300 000 250 000 250 000
27 200 24 400 19 430 23 300 16 190 19 980
4 588 4 122 3 462 3 933 2 885 3 277
0.0328 0.03 1 1 0.0290 0.0304 0.0265 0.0278
730 690 670 670 590 600
0.168 0 . 187 0.196 0.196 0.235 0.235
0.188 0.187 0.196 0.198
0. 168 0.187 0 . 196 0.196
0.168 0.187 0.196 0.196
0.1849 0.206 0.216 0.216 0.259 0.259
0.1859
0.1859
7 7 7 7 7
0.953 0.904 0.858 0.883 0.783 0.806
0.1859
0. 1291 0.1355 0. 1261 0.1236 0 . 1 151
0 . 1362 0 . 1291 0. 1054 0.1261 0.0961 0.1151
0. 1728 0 . 1754 0 . 1790 0 . 1 768 0. 1836 0. 1812
0.346 0.351 0.358 0.353 0.367 0.362
0.415 0.42 1 0.430 0.424 0.441 0.455
0.230 0.234 0.237 0.235 0.244 0.242
0.1 149 0 . 1 167 0 . 1 186 0 . 1 176 0 . 1219 0.1208
0.0957 0.0973 0.0988 0.0980 0 . 1015 0.100s
26 30 26 30 26
2 2 2 2 2
0 . 1 138 0.1059 0. 1074 0 . 1000 0. 1013
7 7 7 7 7
0.0885 0. 1059 0.0835 0. 1000 0.0788
0.721 0.741 0.680
410
050 040 650 430 250
2 442 2 774 2 178 2 473 1 936
0.0244 0.0255 0.0230 0.0241 0.0217
530 530 490 500 460
0.278 0.278 0.311 0.3 1 1 0.350
0. 1872 0.1855 0.1908 0. 1883 0. 1936
0.376 0.371 0.382 0.377 0.387
0.451 0.445 0.458 0.452 0.465
0.250 0.248 0.254 0.252 0.258
0.1248 0 . 1238 0.1269 0.1258 0.1289
0.605 0.581 0.62 1 0.641 0.656 0.665
0.259 0.267 0.275 0.284 0.292 0.300
0.1294 0.1336 0. 1377 0.1418 0. 1480 0.1500
0.1079 0 . 1 1 13 0.1147 0 . 1 182 0.1216 0.1250
0.665 0.642 0.861 0.659 0.655 0.665 0.673
0.308 0.306 0.317 0.325 0.323 0.333 0.342
0.1542 0. 1532 0. 1583 0.1627 0.1615 0.1666 0.1708
0.1285 0.1276 0.1320 0.1355 0.1346 0.1398 0.1423
556 500 477 477 397 397
336 400 336 400 300 000 300 000 266 800
54
� :!: I �
g �� I
31 38 26 28 34 24
000 000 000 000 000 000
14 17 12 15 11
4/0 188 700 188 700 3/0
840
arne as d·c
arne as d-c
0.306 0.306 0.342 0.342 0.385
��
1 2 3
9 645 8 420 6 675 5 345 4 2SO 3 480
1 802 1 542 1
4 4 5 6 6 7 8
2 3 2 1 2 1 1
790 525 250 830 288 460 170
484 568 384 304 358 241 191
268 900 4/0 3/0 'lJ0 JlO 1
6 6 6 6 6 6
1 1 1 1 1 1
0.2109 0.1878 0. 1672 0. 1490 0. 1327 0.1182
7 1 1 1 1 1
0.0703 0. 1878 0 . 1672 0 . 1490 0. 1327 0 . 1 182
0.633 0.563 0.502 0.447 0.398 0.355
3/0 2/0
2 2 3 4 4 5 6
6 7 6 6 7 6 6
1 1 1 1 1 1 1
0.1052 0.0974 0.0937 0.0834 0.0772 0.0743 0.0661
1 1 1 1 1 1 1
0. 1052 0. 1299 0.0937 0.0834 0.1029 0.0745 0.0681
0.316 0.325 0.281 0.250 0.257 0.223 0. 198
1/0
769 610
I
"' �
o .!i
LO
0.00446 0.00418
0.351 0.441 0.556 0.702 0.885 1.12
0.351 0.442 0.557 0.702 0.885 1 . 12
0.00418 0.00504 0.00430 0.00437 0.00452 0.00416 0.00394
180 ISO 160 140 140 120 100
1.41 1.41 1.78 2.24 2.24 2.82 3.56
1.41 1.41 1.78 2.24 2.24 2.82 3.56
g �n
. �i
.
I
0.351 0.444 0.559 0.704 0.887 1.12
0.352 0.445 0.560 0.706 0.888 1 . 12
0.386 0.485 0.612 0.773 0.974 1.23
1.41 1.41 1.78 2.24 2.24 2.82 3.56
1.41 1.41 1.78 2.24 2.24 2.82 3.56
1.55 1.55 1.95 2.47 2.47 3.10 3.92
will
I
"' e.
;,
0.450 0.514 0.642 0.806 1.01 1.27
0.510 0.567 0.697 0.866 1.08 1.34
0.552 0.592 0.723 0.895 1 . 12 1.38
0. 194 0.218 0.225 0.23 1 0.237 0.242
1.59 1.59 1.95 2.50 2.50 3.12 3.94
1.66 1.62 2.04 2.54 2.53 3.16 3.97
1.69 1.65 2.07 2.57 2.55 3.18 3.98
0.247 0.247 0.252 0.257 0.257 0.262 0.268
,
�
... .!i
o .!i
C'II t-
LO
0.388 0.437 0.450 0.462 0.473 0.483
0.466 0.524 0.540 0.554 0.588 0.5SO
0.252 0.242 0.259 0.287 0.273 0.277
0.504 0.484 0.517 0.534 0.547 0.554
0.493 0.493 0.503 0.514 0.515 0.525 0.536
0.592 0.592 0.604 0.611 0.618 0.630
0.277 0.267 0.275 0.274 0.273 0.279 0.281
0.554 0.535 0.551 0.549 0.545 0.557 0.561
0.643
.
Current Appros.. 75% Capacityl:
Small Currents
460 340 300 270 230 200
0.00684 0.00814
0. 1039 0. 1032 0. 1057 0.1049 0.1074
Single Layer Conductors
For Current Approx. 75% Capacity
*
1
!
I
o .!i
"' e.
t For conductor at 7S°C., air at 25°C., wind 1.4 miles per hour (2 ft/sec), frequency = 60 cycles Based on copper 97 percent, aluminum 61 percent conductivity. produce 50°C. conductor temp. (25°C. rise) with 25°C. air temp .. wind 1.4 miles per hour. 1: "Current Approx. 75% Capacity'" is 75% of the "Approx. Current Carrying Capacity in Amps." and is approximately the current which
•
Table B.9.
Circular Mils or A.W.G. Aluminum
�
850 g 1 150 0 1 338 0
Filler Section
Steel
Aluminum
" " �- 1 1l � 1.0 � � � � � ,li i � ] '" Aluminum
1'"
54 54 66
..,� 2 2 2
lj
0.1255 0.1409 0. 1350
�
19 19 19
Paper
"" c
0.0834 0.0921 0. 100
4 4 4
0 . 1 182 0. 1353 0. 184
23 24 18
�
� �.
Goo-
Characteristics of "Expanded" Aluminum Cable, Steel Reinforced (Aluminum Company of America)
j 2 2 2
a.
]0 �..9
L38 1.55 1.75
Copper Equivalent Circular Mils or A. W. G. 534 000 724 000
840 00()
ih
.. .
We g t
35 371 41 900 49 278
7 200 9 070 1\d40
5 .£
] "§
( 1 ) Electrical Characteristics not available until laboratory measurements are completed.
Pounds per Mile
metric Mean Radius atSO Cycles Feet (1)
(1)
Resistance Ohms per Conductor per Mile
1
l
l
25°C. (77°F.) Small Currents d-c
25 cycles so cycles 60 cycles ( 1)
x. '
x.
r.
Approx. Current Carrying Capacity Amps
50°C. (122"F.) Current Appro•. 75% Capacit� d-c
1
!
1
Inductive Reactance Ohms per Conductor per Mile at 1 Ft. Spacing All Currents
Shunt Capacitive Reactance Megohms per Conductor per Mile at 1 Ft. Spacing
25 CYcles 50 cycles 60 cycles 25 cycies 50 cycles 60 cycles 25 cycles 50 cycles 60 (1)
(1)
(1)
(1)
( 1)
(1)
cycles (1)
Table B.IO.
Characteristics of Copperweld-Copper Conductors (Copperweld Steel Company) Approx.
Size of Conductor Copper Number and Nomi-
Diameter of
nal
Wires
Designation
w
Copper.
eld
Copper
Rated
EquivaOut-
lent
side
Circular
Diam-
Mils or
oler
A.w.n
Inches
Breakiog Load Lbs.
Weight Lbs. per Mile
Geo·
Current
metric
Carry-
Mean
iog
Radius
Capa-
at 60
city
Cycles
at 60
Feet
Cycles
°
Ohms per Conductor
d·c
Inductive Reactance
Capacitive
Ohms per Conductor
Reactance Megohms
Current Approx. 75% of
Sman Currents
Amps'
Resistance Ohms per Conductor
25
50
60
cycles
cycles
d·c
25
50
cycles
cycles
per Conductor
One ft. Spacing
Average Currents
Capacity**
cycles
per Mile
per Mile at 50 C . (122°F.)
per Mile at 25°C. (77°F,)
xa '
xa
ra ra
Resistance
per Mile One ft. Spacing
60
25
50
60
25
50
60
cycles
cycles
cycles
cycl ..
cycles
cycles
cycles
0.204
350 E
7 •. 1576"
12 •. 1576"
0.788
350 000 32 420
7 409
0.0220
660
0.1658 0 . 1 728 0 . 1 789 0.1812 0.1812 0.1915 0.201
0.1929
0.366
0.463
0.243
0.1216 0.1014
350 EK
4•. 1470"
15x. 1470'"
0.735
350 000 23 850
6 536
0.0245
680
0.1658 0. 1682 0. 1700 0. 1705 0.1812 0.1845 0. 1873 0.1882 0.1875
0.375
0.450
0.248
0. 1241 0.1034
350 V
3'.1751"
9x.1893"
0.754
350 000 23 480
6 578
0.0226
650
0.1655 0 . 1 725 0.1800 0.1828 0.1809 0.1910 0.202
0.206
0. 1915
0.383
0.460
0.248
0.1232 0. 1027
300 E
7 •. 1459"
12x.1459'
0.729
300 000 27 770
6 351
0.0204
600
0.1934 0.200
0.209
0.2 1 1
0.222
0.232
0.235
0. 1969
0.394
0.473
0.249
0.1244 0.1037
300 EK
4 •. 1361"
1 5x.136J"
0.680
300 000 20 960
5 602
0.0227
610
0. 1934 0.1958 0.1976 0. 198
0.211
0.215
0.218
0.219
0. 1914 0.383
0.460
0.254
0.1269 0 . 1 057
WO V
3x. 1621"
9x.1752"
0.698
300 000 20 730
5 639
0.0209
590
0 . 1930 0.200
0.208
0.210
0.211
0.222
0.233
0.237
0.1954
0.391
0.469
0.252
0.1259 0.1050
0.207
250 E
7•. 1332"
12x. 1332"
0.686
250 000 23 920
5 292
0.01859
540
0.232
0.239
0.245
0.248
0.254
0.265
0.275
0.279
0.202
0.403
0.494
0.255
0.1276 0.1604
250 EK
4x.1242"
15x. 1242"
0.621
250 000 17 840
4 669
0.0207
540
0.232
0.235
0.236
0.237
0.254
0.258
0.261
0.261
0.1960
0.392
0.471
0.260
0.1301 0.1084
250 V
3x.I480'
9x. 16()(Y
0.637
250 000 1 7 420
4 699
0 . 0 19 1 1
530
0.232
0.239
0.246
0.249
0.253
0.264
0.276
0.281
0.200
0.400
0.480
0.258
0.1292 0.1077
4/0 E
7•. 1225"
12x.1225'
0.613
4/0
20 730
4 479
0.017 1 1
480
0.274
0.281
0.287
0.290
0.300
0.312
0.323
0.326
0.206
0.4 1 1
0.493
0.261
0. 1306 0.1088
4/0 G
2•. 1944"
5•. 1944"
0.583
4/0
15 640
4 168
0.01409
460
0.273
0.284
0.294
0.298
0.299
0.318
0.336
0.342
0.517
0.265
0. 1324 0 . 1103
4/0 EK
4x. 1 143"
15•. 1 143"
0.571
4/0
15 370
3 951
0.01903
490
0.274
0.277
0.278
0.279
0.300
0.304
0.307
0.308
0.200
0.401
0.481
0.266
0.1331 0 . 1 109
4/0 V
3•. 136 1"
9x.1472"
0.566
4/0
15 000
3 977
0.0 1758
470
0.274
0.281
0.288
0.291
0.299
0.3 1 1
0.323
0.328
0.204
0.409
0.490
0.264
0 . 1 322 0 . 1 1 0 1
4/0 F
1 •. 1833"
6x.1833"
0.550
4/0
12 290
3 750
0.0 1558
470
0.273
0.280
0.285
0.287
0.299
0.309
0.318
0.322
0.210
0.421
0.505
0.269
0.1344 0 . 1 220
3/0 E
7x.1091"
12x.1091"
0.545
3/0
16 800
3 552
0.01521
420
0.346
0.353
0.359
0.361
0.378
0.391
0.402
0.407
0.212
0.423
0.508
0.270
0.1348 0 . 1 123
3/0 J
3•. 185 1"
4x.185 \ "
0.555
3/0
16 170
3 732
0 . 0 1 156
410
0.344
0.356
0.367
0.372
0.377
0.398
0.419
0.428
0.225
0.451
0.541
0.268
0.1341 0 . 1 1 18
3/0 G
2'. 1731"
2x. 173 1 "
0.519
3 305
0.01254
0.344
0.355
0.365
0.369
0.377
0.397
0.423
0.221
0.443
0.531
0.273
0.1365 0.1137
3/0 EK
4x.1018"
4x. I 0 1 8"
0.509
3/0
12 370
3 134
0.01697
420
0.346
0.348
0.350
0.351
0.378
0.382
0.386
0.386
0.206
0.412
0.495
0.274
0. 1372 0 . 1 143
3/0 V
3• . 1 3 1 1"
9x. 1 3 1 1 "
0.522
3/0
12 220
3 154
0.01566
410
0.345
0.352
0.360
0.362
0.377
0.390
0.403
0.406
0.210
0.420
0.504
0.273
0.1363 0 . 1 136
3/0 F
lx.1632"
6x.1632"
0.490
3/0
9 980
2 974
0.01388
410
0.344
0.351
0.356
0.358
0.377
0.368
0.397
0,401
0.216
0.432
0.519
0.277
0 . 1 385 0 . 1 155
2IO K
4x.1780'"
3x. 1 780"
0.534
210
17 600
3 41 1
0.00912
360
0.434
0.447
0.459
0.466
0.475
0.499
0.524
0.535
0.237
0.475
0.570
0.271
0. 1355 0 . 1 129
210 J
3•. 1648'
4x.I648"
0.494
210
13 430
2 960
0.01029
350
0.434
0.446
0.457
0.462
0.475
0.498
0.520
0.530
0.231
0.463
0.555
0.277
0 . 1 383 0 . 1 152
210 G
2x.I542'"
5x.1542"
0.463
210
1 0 510
2 622
0.0 1 1 19
350
0.434
0.445
0.456
0.459
0.475
0.497
0.518
0.525
0.227
0.454
0.545
0.281
0.1406 0 . 1 1 7 1
210 V
3•. 1080"
9x. 1 1 67"
0.465
210
9 846
2 502
0.01395
360
0.435
0.442
0.450
0.452
0.476
0.489
0.504
0.509
0.216
0.432
0.518
0.281
0.1404 0.1 170
210 F
Ix.1454"
6x.1454"
0.436
210
8 094
2 359
0 01235
350
0.434
0.441
0.446
0.448
0.475
0.487
0.497
0.501
0.222
0.444
0.533
0.285
0.1427 0 . 1 189
I/O K
4•. 1585"
3x.1585"
0.475
I/O
14 490
2 703
0.00812
310
0.548
0.560
0.573
0.579
0.599
0.625
0.652
0.664
0.243
0.487
0.584
0.279
0.1397 0 . 1 164
l/O J
3x.1467"
4x.1467"
0.440
I/O
10 970
2 346
0.00917
310
0.548
0.559
0.570
0.576
0.599
0.624
0.648
0.659
0.237
0.474
0.569
0.285
0.1423 0 . 1 186
I/O G
2x.1373"
5x. 1373"
0.412
I/O
8 563
2 078
0.00996
310
0.548
0.559
0.568
0.573
0.599
0.623
0.645
0.654
0.233
0.466
0.559
0.289
0.1447 0.1206
I/O F
Ix. 1294"
6x.1294"
0.388
I/O
6 536
1 870
0 01099
310
0.548
0.554
0.559
0.562
0.599
0.612
0.622
0.627
0.228
0.466
0.547
0.294
0.1469 0. 1224
IN
5x.1546"
2x.1546"
0.464
1
15 410
2 541
0.00638
280
0.69 1
0.705
0.719
0.726
0.755
0.787
0.818
0.832
0.256
0.512
0.614
0.281
0.1405 0 . 1 1 7 1
1K
4•. 1412"
3x. 1 4 1 2"
0.423
1
11 900
2 144
0.00723
270
0.691
0.704
0.716
0.722
0.755
0.784
0.813
0.825
0.249
0.498
0.598
0.288
0.1438 0 . 1 198
1 861
0.008 1 7
270
0.691
0.703
0.714
0.719
0.755
0.820
0.243
0.486
0.583
0.293
0. 1465 0.1221
12 860
3/0
400
0.416
0.215
0.431
1J
3x.130'-
4 x . 1 307"
0.392
1
9 000
0.783
0.806
1G
2•. 1222"
5x.1222"
0.367
1
6 956
1 649
0.00887
260
0 691
0.702
0.712
0.716
0.755
0.781
0.805
0.815
0.239
0.478
0.573
0.298
0.1488 0.1240
I F
1 • . 1 153"
6x. 1 I 53"
0.346
1
5 266
1 483
0.00980
270
0.691
0.698
0.704
0.705
0.755
0.769
0.781
0.786
0.234
0.468
0.561
0.302
0. 1509 0.1258
2P
6• . 1540"
Ix. 1540"
0.462
2
16 870
2 487
0.00501
250
0.87 1
0.886
0.901
0.909
0.952
0.988
1.024
1.040
0.268
0.536
0.643
0.281
0. 1406 0 . 1 172
2N
5x.137'-
2x. 1 377"
0.4 13
2
1 2 680
2 015
0.00568
240
0.871
0.885
0.899
0.906
0.952
0.986
1.020
1.035
0.261
0.523
0.627
0.289
0.1446 0.1205
2 K
4•. 1257"
3x.1257"
0.377
2
9 730
0.871
0.884
0.896
0.902
0.952
0.983
1.014
1.028
0.255
0.510
0.612
0.296
0.1479 0 . 1 232
3'. 1 164"
4x. 1 1 64"
0.349
2
7 322
1 701 0.00644
240
2 J
1 476
0.00727
230
0.871
0.883
0.894
0.899
0.952
0.982
1.010
1.022
0.249
0.498
0.598
0.301
0.1506 0. 1255
2A
1 •. 1699"
2x. 1 699"
0.366
2
5 876
1 356
0.00763
240
0.869
0.875
0.880
0.882
0.950
0.962
0.973
0.979
0.247
0.493
0.592
0.298
0.1489 0.1241
2G
2x.l089"
5x.l089"
0.327
2
5 626
1 307
0.00790
230
0.871
0.882
0.892
0.896
0.952
0.980
1.006
1.016
0.245
0.489
0.587
0.306
0.1529 0. 1275
2F
b. 1026"
6x.l026"
0.308
2
4 233
1 176
0.00873
230
0.8? 1
0.878
0.884
0.885
0.952
0.967
0.979
0.985
0.230
0.479
0.575
0.310
0.1551 0.1292
3P
6x.137 1 "
lx.137 l "
0.4 1 1
3
13 910
1 973
0.00445
220
1.098
1.113
l . l27
l . l36
1 .200
1.239
1 .273
1.296
0.274
0.547
0.657
0.290
0. 1448 0.1207
3N
5x.1226"
2x.1226"
0.368
3
10 390
1 598
0.00506
210
1.098
1 . 1 12
l . l26
1 . 133
1.200
1.237
1.273
1.289
0.267
0.534
0.641
0.298
0.1487 0.1239
3K
4x. l 1 20"
3x. 1 I 20"
0.336
3
7 910
I 349
0.00574
210
1.098
1.111
1 . 123
1 . 129
1.200
1.233
1.267
1.281
0.261
0.522
0.626
0.304
0. 1520 0 . 1 266
3
1
5 955
1 171
0.00648
200
1.098
1 . 1 10
1.121
1 . 126
1.200
1.232
1.262
1.275
0.255
0.509
0.6 1 1
0.309
0.1547 0.1289
4 810
1 075
0.00679
210
1096
1 . 102
1 . 107
1 . 109
1 . 198
1211
1.225
1 . 229
0.252
0.505
0.606
0.306
0.1531 0. 1275
! 11 420
1 564
0.00397
190
1385
1 . 400
1.414
1.423
1.514
1 .555
1.598
1.616
0.280
0.559
0.671
0.298
0.1489 0 . 1 24 1
8 460
1 267
0 00451
180
1 .385
1.399
1.413
1.420
1.514
1 .554
1593
1.610
0.273
0.548
0.655
0.306
0.1528 0. 1274
4
7 340
1 191
0.00566
190
1.382
1.389
1.396
1.399
1.51I
1 . 529
1.544
1 .542
0.262
0.523
0.628
0.301
0.1507 0.1256
0.290
4
3 938
853
0.00604
180
1 . 382
1.388
1 . 393
1 .395
1.511
1.525
1.540
1.545
0.258
0.517
0.620
0.314
0.1572 0.1310
lx.I087"
0.326
5
9 311
1 240
0.00353
160
1 .747
1 . 762
1.776
1 .785
1.909
1.954
2.00
2.02
0.285
0.571
0.685
0.306
0.1531 0. 1275
Ix. 1438"
0.3 1 0
5
6 035
944
0.00504
160
1.742
1. 749
1.756
1.759
1.905
1.924
1.941
1.939
0.268
0.535
0.642
0.310
0.1548 0. 1290
2x.1200"
0.258
5
3 193
676
0.00538
160
1 . 742
1 . 748
1 . 753
1 .755
1.905
1 .920
1.936
1 .94 1
0.264
0.528
0.634
0.323
0.1614 0. 1345
0.273
3J
3x.l036H
4x.I036"
0.311
3A
Ix. 1 5 I 3"
2x. 15 i 3"
0.326
3
4P
6x. 122 1 "
lx. l n l "
0.366
4
4N
5x.l092"
2x.l092"
0.328
4
4D
2x.16IS"
lx. 1 6 1 5"
0.348
4A
Ix. 1347"
2x.1347"
5P
6x.1087"
5D
2x.1438'
5A
1 •. 1200"
6D
2 •. 128 1 "
lx.128 ! "
0.276
6
4 942
749
0.00449
140
2.20
2.21
2.21
2.22
2.40
2.42
2.44
2.44
0.547
0.656
0.318
0.1590 0 . 1 325
6A
lx.l068"
2x.1068"
0.230
6
2 585
536
0.00479
140
2.20
2.20
2.21
2.21
2.40
2.42
2.44
2.44
0.270
0.540
0.648
0.331
0.1655 0.1379
6C
lx. l046"
2x.l046"
0.225
6
2 143
514
0.00469
130
2.20
2.20
2.21
2.21
2.40
2.42
2.44
2.44
0.27 1
0.542
0.651
0.333
0.1663 0.1386
7D
2x. 1 I 4 t "
lx. 1 I 4 1 "
0.246
7
4 022
594
0.00400
120
2.77
2.78
2.79
2.79
3.03
3.05
3.07
3.07
0.279
0.558
0.670
0.326
0.1631 0. 1359
7A
lx.1266"
2x.OS9S"
0.223
7
2 754
495
0.00441
120
2.77
2.78
2.78
2.78
3.03
3.05
3.07
3.07
0.274
0.548
0.658
0.333
0.1666 0. 1388
8D
2x. 10 1 6"
lx.1016"
0.219
8
3 256
471
0.00356
1 10
3.49
3.50
3.51
3.51
3.82
3.84
3.86
3.86
0.285
0.570
0.684
0.334
0. 1672 0 . 1 393
8A
Ix. l l lT
2x.0797"
0. 199
8
2 233
392
0.00394
100
3.49
3.50
3.51
3.51
3.82
3.84
3.66
3.87
0.280
0.560
0.672
0.341
0 . 1706 0 . 1 422
8C
lx.OSOS"
2x.OS34"
0. 179
8
1 362
320
0.00373
100
3.49
3.50
3.51
3.51
3.82
3.84
3.86
3.86
0.283
0.565
0.679
0.349
0 . 1 744 0. 1453
2 •. 0808"
Ix.0808"
0.174
9'h
1 743
298
0.00283
85
4.91
4.92
4.92
4.93
5.37
5.39
5.42
5.42
0.297
0.598
0.712
0.351
0 . 1754 0.1462
91h D
*Based on a conductor temperature of 75°C. and an ambient of 25°C., wmd 1 . 4 miles per hour (2 ft/sec.), frequency
=
60 cycles, average tarmshed surface.
**Resistances at 50°C. total temperature, based on an ambient of 25°C. plus 25°C, rise due to heating effect of current. The approximate magnitude of current necessary to produce the 25<>C. rise is 75% of the "Approximate Current Carrying Capacity at 60 cycles."
Table B. l l .
Nominal Number Conand Size ductor of Wires Size
Area of Outside ConDiamductor eter CirculRT Inches Mils
Rated Breaking Pounds Load Strength
High
Extra High
Characteristics of Copperweld Conductors (Copperweld Steel Company) Geo-
Approx. Current Weight Radius Carrying Pounds Capaat 60 per Cycles city"' Mile Amps and Average at Currents 60 Cycles Feet
ra
Resistance Ohms per Conductor per Mile at 75°C. (167°F.) Current Approx. 75% of Capacity"''"
ra
metric Mean
Resistance Ohms per Conductor at 25°C. (77°F.) Small Currents 25 cycles
d-c
50 cycles
60
cycles
d-c
"a
Inductive Reactance Ohms per Conductor per Mile One Ft. Spacing Average Currents
..
"a '
Capacitive Reactance Megohms per Conductor per Mile One Ft. Spacing
cycles
25
50 cycles
cycles
60
25 cycles
50 cycles
60 cycles
cycles
25
50 cycles
60 cycles
30% Conductivity 7/8" 13116" 23132"
19 No. 5 19 No. 6 19 No. 7
0.910 0.810 0.721
628 900 55 570 498 800 45 830 395 500 37 740
66 910 9 344 55 530 7 410 45 850 5 877
0.00758 0.00675 0.00601
21132" 9/16" 518"
19 No. 8 19 No. 9 7 No. 4
0.642 0.572 0.613
248
313 700 3 1 040 800 25 500 292 200 24 780
37 690 4 660 30 6 10 3 696 29 430 4 324
9/16"
7 No. 5 7 No. 6 7 No. 7
0.546 0.486 0.433
231 700 20 470 183 800 16 890 145 700 13 9 1 0
318" 11132" 5116"
7 No. 8 7 No. 9 7 No. 10
0.3SS 0.343 0.306
3 No. 5 3 No. 6 3 No. 7
3 No. 5 3 No. 6 3 No. 7
0.392 0.349 0.3 1 1
99 310 78 750 62 450
3 No. 8 3 No. 9 3 No. 10
3 No. 8 3 No. 9 3 No. 10
0.277 0.247 0.220
49 530 39 280 3 1 150
112"
7116"
470
540
0.306 0.386 0.486
0.316 0.396 0.496
0.326 0.406 0.606
0.331 0 . 4 11 0.511
0.363 0.458 0.577
0.419 0.518 0.643
0 .476 0.580 0.710
0.499 0.605 0.737
0.261 0.267 0.273
0.493 0.505 0.517
0.592 0.606 0.621
0.233 0.241 0.250
0 . 1 165 0. 1206 0. 1248
0.097 1 0. 1005 0. 1040
0.00535 0.00477 0.0051 1
410 360 410
0.613 0.773 0.656
0.623 0 .783 0.664
0.633 0 . 793 0.672
0.638 0.798 0.676
0. 728 0.917 0 .778
0.799 0.995 0.824
0.872 1 .075 0.870
0.902 1 . 106 0.887
0.279 0.28�:, 0.281i
0.529 0.541 0 .533
0.635 0.649 0.640
0.258 0.266 0.261
0. 1289 0. 1330 0 . 1306
0. 1074 0 . 1 109
24 650 3 429 20 460 2 7 19 16 890 2 157
0.00455 0.00405 0.00361
360 310 270
0.827 1.042 1.315
0.835 1 .050 1 .323
0.843 1.058 1.331
0.847 1 .062 1 .335
0.981 1 . 237 1 . 560
1.030 1 .290 1.617
1 .080
1.343 1 . 675
1.099 1 .364 1.697
0.28.1 0.293 0.299
0.545 0.557 0.569
0.654 0.668 0.683
0.269 0.278 0.286
0. 1347 0. 1338 0. 1429
0 . 1 122 0 . 1 157 0 . 1 191
13 890 11 280 9 196
1 710 1 356 1 076
0.00321 0.00286 0.00255
230 200 170
1.658 2 .09 2.64
1.666 2 . 10 2 .64
1 .674 2.11 2.65
1 .678 2.11 2 .66
1 .967 2.48 3.13
2.03 2.55 3.20
2.09 2.61 3.27
2.12 2 .64 3.30
0.305 0.311 0.316
0.58 1 0.592 0.604
0.697 0.711 0.725
0 .294 0.303 0.311
0. 1471 0.1512 0. 1553
0 . 1 226 0 . 1 260 0 . 1 294
9 262 7 639 6 291
11 860 9 754 7 922
1 467 1 163 922.4
0.00457 0.00407 0.00363
220 190 160
1 .926 2.43 3.06
1 .931 2.43 3.07
1 .936 2.44 3.Q7
1 .938 2.44 3 . 07
2.29 2.38 3.63
2.31 2.91 3.66
2.34 2 .94 3.70
2.35 2.95 3.71
0.289 0.295 0.301
0.545 0.556 0.568
0.654 0.668 0.682
0.293 0.301 0.310
0. 1465 0.1506 0. 1547
0.1221 0. 1255 0. 1289
5 174 4 250 3 509
6 282 5 129 4 160
731.5 580 . 1 460.0
0.00323 0.00288 0.00257
140 120 110
3.86 4.87 6.14
3.87 4.87 6 . 14
3.87 4.38 6.15
3.87 4.38 6.15
4.58 5.78 7.28
4.61 5.81 7.32
4.65 5.85 7.36
4.66 5.86 7.38
0.307 0.313 0.319
0.580 0 . 59 1 0.603
0.696 0.710 0. 724
0.318 0 .326 0.334
0 . 1 589 0. 1629 0.1671
0. 1324 0.1358 0.1392
1 15 600 11 440 91 650 9 393 72 680 7 758
620
O.lOSS
40% Conductivity
7/8" 13116" 23132"
19 No. 5 19 No. 6 19 No. 7
0.910 0.810 0.721
628 900 50 240 498 800 41 600 395 500 34 390
. . . . . . . . . . . . . . . . . .
9 344 7 410 5 877
0.01175 0.01046 0.00931
690 610 530
0.229 0.289 0.365
0 .239 0.299 0.375
0.249 0.309 0.3SS
0.254 0.314 0 .390
0.272 0.343 0.433
0.321 0.396 0.490
0.371 0. 450 0.549
0.391 0.472 0.573
0.236 0.241 0.247
0.449 0.461 0.473
0.539 0.553 0.567
0.233 0.241 0.250
0 . 1 165 0. 1206 0. 1248
0.0971 0. 1005 0. 1040
21132" 9/16" 518"
19 No. 8 19 No. 9 7 No. 4
0.642 0.572 0.613
313 700 28 380 248 800 23 390 292 200 22 310
4 660 3 696 4 324
0.00829 0.00739 0.00792
470 410 470
0.460 0.580 0.492
0.470 0.590 0 .500
0.480 0.600 0.508
0.485 0.605 0.512
0.546 0.638 0.584
0.608 0. 756 0.624
0.672 0.826 0.664
0.698 0.753 0.680
0.253 0.259 0.255
0.485 0.496 0.489
0.582 0.595 0.587
0. 258 0 .266 0.261
0. 1289 0 . 1330 0. 1306
0. 1074 0 . 1 109
9116"
7 No. 5 7 No. 6 7 No. 7
0.546 0.486 0.433
231 700 18 510 183 800 15 330 145 700 12 670
. . . . . . . . . . . . . . . . . .
3 429 2 719 2 157
0.00705 0.00628 0.00559
410 350 310
0.628 0. 790 0.994
0.636 0.798 1.002
0.640 0.802 1 .006
0 .736 0 .928 1 . 170
0.780 0.975 1 .220
0.843 1.021 1.271
0.840 1.040 1 .291
0.261 0.267 0.273
0.501 0.513 0.524
0.601 0.615 0.629
0 .269 0.278 0.286
0. 1347 0. 1338 0. 1429
0 . 1 122 0 . 1 157 0 . 1 191
11132" 5/16"
0 .3SS 0.343 0.306
1 1 5 600 10 460 9 1 850 8 616 72 680 7 1 2 1
. . . . . . . . . . . . . . . . . .
1 7 10 1 356 1 076
0.00497 0.00443 0.00395
27 0 230 200
0.620 0.782 0.986
7 No. 8 7 No. 9 7 No. 10
. . . . . . . . . . . . . . . . . .
1 .244 1.568 1.978
1 .252 1 .576 1.986
1 .260 1 . 584 1.994
1 .264 1 .538 1.998
1 .476 1.861 2.35
1 .530 1.919 2.41
1 . 584 1.978 2.47
1.606 2.00 2.50
0.279 0.2SS 0.291
0.536 0.548 0.559
0.644 0.658 0.671
0 . 294 0 .303 0.311
0. 1471 0 . 15 12 0. 1553
0. 1226 0 . 1260 0. 1294
3 No. 5 3 No. 6 3 No. 7
3 No. 5 3 No. 6 3 No. 7
0.392 0.349 0.3 1 1
99 310 78 750 62 450
8 373 6 934 5 732
1 467 1 163 922.4
0.00621 0.00553 0.00492
250 220 190
1.445 1 .821 2.30
1 .450 1.826 2 .30
1 . 455 1 .831 2.31
1 .457 1 .833 2.31
1.714 2.16 2.73
1 .738 2.19 2.75
1.762 2.21 2 . 78
1 . 772 2.22 2 . 79
0.269 0.275 0.281
0.514 0.526 0.537
0.617 0.63 1 0.645
0 .293 0 .301 0 .310
0. 1465 0 . 1 506 0. 1547
0.1221 0. 1255 0 . 1289
3 No. 8 3 No. 9 3 No. 10
3 No. 8 3 No. 9 3 No. 10
0.277 0.247 1 .220
49 530 39 280 31 150
4 730 3 898 3 221
. . . . . . . . . . . . . . . . . .
731.5 580 . 1 460.0
0.00439 0.00391 0.00348
160 140 120
2.90 3.65 4.61
2 .90 3.66 4.61
2.91 3.66 4.62
2.91 3.66 4.62
3.44 4.33 5 . 46
3.47 4.37 5.50
3.50 4.40 5.53
3.51 4.41 5.55
0.286 0.292 0.297
0.549 0.561 0.572
0.659 0.673 0.687
0.318 0.326 0.334
0 . 1 589 0.1629 0 . 1671
0. 1324 0. 1358 0. 1392
0.310
0.596
0.715
0.351
0 . 1754
0 . 1462
112" 7/16" 318"
. . . . . . . . . . . . . . . . . .
8.78 8.77 7 .34 7 .32 ,-7 .� 19 590 2 23� . . . . . . 289.3 0.00276 7 .33 8.73 90 8.69 3 No ...!!. 0 . 1 74 *Based on conductor temperature of 125°C. and an ambient of 25°C. '"'"Resistance at 75°C. total temperature, based on an ambient of 25°C. plus 50°C. rise due to beating etrect of current. The approximate magnitude of current necessary to produce the 5(»C. rise is 75% of the "Approximate Current C arrying Capacity at 60 Cycles."
3 No. 12
O. lOSS
Line I mped a n ce Tables
447
Table B.12. Inductive Reactance of ACSR Bundled Conductors at 6 0 Conductors per Bundle
COOl
UO
U I I \.
$ f l lI O $ ST ' I.
OJ I. ,_.
GM'
" .
1 1 ' ''' O l D
, . 0.000
6211
19
2 . �OO
0 . 0900
£ I'JI,IO
2 2 .. . 0 0 0
61/6
It
O . OIS!
' .'uon
1 11 " 1 0 0 0
2 . UO 1 . 1S0
It
1 2 '\000
\ l I lt S O / II
I I W,
Z 1 6 7000
72
IlU! l 1 1 D
' 1 \1000
, ..
( 1"'0(1
"
o .• h.
�
0 . 2922
r--:-. '. I 0. 1881
0 . 1910
0. 1910
0 .
0 . ))]6
.. ,
---- 6 0
$ , I'l( COlO.
l
I I O V ( 1 1 '1' ( .
9
( 'I. )
S'lt l l '
12
I S
0 . 1 6 ) !,
0 . 1 16 6 /
0. 1 )26
• 6 6 1.
D . ' '' ' O
D . ' ) '> S
I
I
O .. ..
S / .. I l (
OI
I
H
0 . 1 1""
O . I H"
D . I 2 2 t.
0 . • 67]
O. '
0 . 2011
0 . • "'2
0 . 1 661
0. 1�]2
0 . 1 11 2 2
1 . 6 00
O. os ' e
O . ) II H
0 . 2 1 !» 0
O . l tO "
0
0 . 1 .. ...
0. 1 "
0 . ) ,, 711
0 . 2 1 \1
0. 1'12
. 1 7 10
o . • !» " ,
o . as 7 I
0. 1 1]7
0 . 1 60 2
0.
1 . 111
O . OSII
0 . ) 11 , . O . ]H I
0 . 2 ' "0 0.21"
0 . ' 1'" 0 . 1 9S0
3. 1 7 1 ' 0. 1 1H
0 . 1 \'"
0.
•
11 ' ,
S
==<:lI
I I D l li S _____-,
J OO T
20 '
1 . 131
J
0 . 09 ' "
0 . 0 791
O . 0 99
O . ot l l
I
0. t i . 2
O.OtJI
1
'1
O . OhS
O
O U 'J
0 . 0 1 , ..
'"
O . • JII
O. I
�)
0 . Ot 7 )
O . O lS l ')
• 719
O. I U I
O. f l U
0 . 09 7 7
O . Ouo
0 . 1 ] 1' 0 . 1 11 ' 6
0. 1 t" 0 . " .. .
0 . 0966 0 . 1 00 1
0 . 01 ' 1 0 . 0 1 ., 1
e MU ' "
1 710000
h
.,
" L UI
I SIOOOO
I . S II \
0 . 0\ 1 1
O . UIO
O. U I
O . l l tO
O
US
O . I S""
C . l l S 11
0 . " '26
0. 1 I1l
O. I O I J
1
I . \02
0 . 0 11 "
0 . ] 61 1 0
O . U II I
0 . 1 '9\
O . ' UO
0 .. 1 6 1 \
0 . I S 1 ..
0 . 1 1 1 1$
O. IOll
o . OI n
.9
I . S06
0 , OS06
O . J6 2 1
0. 22 ) 1
o . I 'n
0. 1110
O . I 6 1S
O . I \h
0. "
"
O . I UI 0 . 1 11'1 0
0. 1 2 1 J
!l it
I
O . I II S
I noooo t \ l OUD
S it It \
It
L I '. ' "
0 . 1 20 1
O . I O li
0 . 0119
P " IO l . U l .... T ' "
It
1 . 602
0 . 0 11 1 1
0 . 1 6 10
0. 22\\
0 . 2001
0. 1 11\
0 . 1 1"
O . I SI9
0 . 1 ' '' '
0 . ' "\6
0 . ' 22 1
0. 10.)
0 . 0 " .,
0 . 0"9"
0 . )6\0
0 . 2 2 11 \
0 . 1 991
O. IIU
O . I 6It
0 . I S19
0 . 1 771
O . I U'
0. 1 2 1 '
0 . 1 0 )6
0 . 01 1 9
.�
7
1 , "27
0 . 0" '0
0 . J1 I 0
0. 22"
0 . 2UO
O . I IS\
0 . 1 720
0 . 1 601
0 . 1 7t 7
O . ' U'
0 . 1 2 11
0 . 1 0 ."
0 . "09
�.
It
' . " 2 '1
0 . 0 '1 ' 2
O . lUO
0. 2260
0 . 10 1 "
O . l hO
0 . 1 7 0 11
O . I Sh
0 . 1 11 1
O . I U'
0. 1 227
0 . 1 0 116
0 . 01"
7
I . JU
0 . 0"S9
0. )711
0 . 2190
0 . 2 0 ....
0. 11"
0. 1 ')"
0 . 1 62 1
0. "
0 . 1 11 7 '
0 . 1 2" 6
0.
066
0 . 0" 1
�u
0 . 0 ' 16 6
0. )721
O . UI I
0 . 1 1 10
0 . 1 7 2\
0. 16'"
0. 1'01
0 . 1117 )
O . l hO
0 . 1 060
0.0912
0 . 0 ""' .
0. )779
a.ulo
0 . 2 0 6 11
0 . ' "0
0 . 1 7\ "
O . I UO
0 . 1 0 19
a . Ot l l
0 . ) 7 60
O . UO I
0 . 20\\
0 . 1 1 10
o. l n\
• 1 ....
0 . 1 "1 2
O . O" S '
0.
O . I UO
I . JU
O . I U"
0. '1'"
0 . 1 " 16
O. 'U)
0. 101J
O . otH
1 . 10 2
O , O" U
O.lUI
0.2])1
0 . 101S
0.
0 . 1 77\
0 . 1 6 6 1&
0 . 1 1 '"
0. 1\"
0 . 1 2'"
0 . 1 0')
0 . 0 111 6 O . Gt U
1 0 1 0 \. ' 1 1 'l
I U I \ OO
0 1 "U
I n uo o
' '' U U I T
1 / 72000
.
I I T 1 ( 1I
1 2 7 2000
.�
.,
I . J.S
"'CI l ( T I I' I
I l t UOO
19
I l u nD
.. .�
It
f I le w
O . Of "
I . �U
�.
U
.
0 . 1 101 0 . 1 7 11 11
1 . �66
it S
, u . ooo . ., , , 00 0
•alT
0 . 1 6 11 0
0 . ' 1& 1 ] 0 . 1 � 29
It
I S I OS O O
, ,, 0 '1 ( 1
O. as 1 6
I
J
with 1, 2, and 3
� r--;- ) - C O ' O U C f O I V I C . I ' ( 1 1 . I .1 • '2 IS I I 0. 1 2 ' " O . • � )'> 0. 1
('� I I
' ( IC "I ( (
COIDUC TO'
Hz
�.
19
I .
0 . lOU
•910
07
•
I l l l OOO
�.
1 . 293
0 . 0" 1 5
0. )101
0.2UI
0 . 1 1 6�
O.
US
O. IIU
O . I S OO
0 . 1 1t '
0. 1017
1 1 1 1000
.�
I . Ut
O . O 'l I S
0 . 111 '
0 . 21\ 1
0 ' 207S 0 . 2 10\
0 . 1 '0 1
IlUfJU (UIUW
O. I U I
0 . 1 1'\
O. IUS
0 . ' 1"
O . I UO
0 . I U7
0 . 1 1 07
..
1 . 2 '"
O . O'UO
0 . JU7
O . un
O . JOti
O. lt23
0 . 1 '"
0 . 1 677
0 . ' 116 1
0. 1\1\
0 . 1 212
0. 1 102
O . O U II
OIlOlAl
l o nsoo
t O l UO O
1.211
0 . U70
0 . 2 1 2.
O . I IS O
0. 11'\
0 . 1 70 "
0 . 1 11 1
0 . 1 \)]
O . l lOO
0. 1 I 1t
0. 0'72
I O USOO
"
1
16
0 . 0• • 2
0 . 1900
UI',U
.-
0 , " 1'1
O . 19S\
O . Utl
0 . 2 1 \2
0. 1,71
0 . ' 111 2
0. 1732
0 . 1 17 '
O. ISSI
O. I J II
0 . 1 1 11
O . OttO
�.
1 . 1 96
0 . 0.02 o . onl
O . lIOO O . lt'"
0 . U70 O . Uts
0 . 2 1 1 '6 O . J ' '1 '
O . I 'S O 0. 1" \
0. 1 1 1 \ 0. IU9
0 . 1 70" 0 . 1 1 21
0 . 1 '" 0. 1171
O. IU) O. IU'
0 . 1 1.0 O. llIt
0. 1 1 1' 0 . 1 1 11
0 . 0" 2 O . Otll
I . ' 'I ' 1 . 1 12 1 . 1'1
0 . 0)10
0 . 1 t0 0 0
0 . 2 16 2 1
0 . 2 1 1\
0 . 2000
O . I IU
O . l n.
0 . 1 1t '
O . IUI
O. l lU
0. 1 1\1
O . I OO�
M A U .. I ,
1ISOOO
.� ,. .. .
! . I IS
IUDDY
tuooo .uooo nuoo tooooo
)0
I'
7
( UOt llL
un
C A l I l lO U'UY
c u el
100000
. •
1 . 1 .0 1 . 1 0'
t , OU2
O . lI]O
O . U"
0 . 2 . .. 0
0 . 1 16 \
0 . 1 1 10
0. 1"11
0. 1111
O . I \'J
O . l l fO
0 . 1 1 10
O . UI 1
0 . 2 . 1 11
0.2'"
0 . 1 '1 "
O . I 'U
0 . 1 7. ,
0 . 1 ,'0
0. IU2
0 . I U9
0 . , , 1 It
0 . 1 00 1
O . JUO
O . U"
O . J I ltO
o . I 'n
0 . 1 110
0. 111'
0. 1'"
0 . 1 \ '1 1
0 . 1 1 10
' . 1 1 10
0 . 0112
O.
•)1.
0 . 01" 0 . 0)1'
0 . 2 11 1 1
0 . 2 1 10
0 . 2U I
0 . 2 1 1S
0 . 2000
O. IU\
0 . 1 1\"
• h.
0 . , 1 t,
a . I Ut
O . I IU
0. 1 1 \ 1
O . I OO S
0.211'
0 . l007
0. 1"
1
0 . ' "' '
0 . 1 11'
0. 1571
O . I U'
0. ' 1\1
0. 1010
0 . 2 2 0\
0 . J030
O . ' l ts
0 . 1 ""
0. 1'"
O . fUI
O . I U II
'. "
0 . 1026
O.lIU
O . r tl t
O . I IU
O . , "U
0 . 1 '"
'. IU'
O . I UI
O . I . .. S
O.U'I
0 . 2 1 10
0 . 1 ''' '
0 . 1 110
0 . 1 7 119
O. 'ltl
O . I S61
O . I UO
O . I I SO
0 . 1 00 1
0 . 2"'1
0 . 1 2 00
0 . 202\
0 . 1 1t0
0 . 1 17'
0. 191 1
O. IUl
O . l lSO
0 . 1 1 70
0 . 1 022
0 . 1 10 1 1
0 . 2.10
0.22 1.
0 . 2011
0 . 1 90 .
0 . 1 71 l
0 . 1 1 10
0 . 1 \12
O . I U'
0 . 1 1 1'
0. 101/
0.
0. 2""
0.2UI
0 . 2 0 111
O. ltl l
0 . 1 1 00
O . I IlS
0 . 1 \"
0 . 1 1 111
0 . 1 1 '"
0 . 1 01 6
0 . 2"1 1 0 . 2\12
o.un 0 . 1216
0 . l06 1 0.21 1 1
0 . 1 92\ 0 . 1 17 1
O. II'S O . I II S
0 . 1 91' 0 . 1 "1
0. 11" 0 . 1 11 1 0
0 . 1 1 7 11 0 . 1 ."
0. 1 19) 0. 1 22 7
0 . 1 0' " 0 . 1 0'0
1ISOOO
U
1UOOO
�.
L OU
0 . 0)10
0
Oo
7noao
1 . 09 1
.�
I . OU
O . .,U
0 . .. 0 " .
" 50 0 0
z
0 . .,S2
(OOT
" \0 0 0
16
1 . 0 11 0
0 . 0' 1 7
I ( DW I "
7 1 SSOO
).
1 . 0' 1
, . .IU "
71
U
1 . 0\ .
,T I l '
ssoo
7 1 \\00
z
1 . 0)6
U
1 .01"
h
1 . 000 O . 'S ]
' .... f T f UM I " O
" . . AD ....00
u n oo
01
••
It
0 . 0912
0 . 2 It H
o . n"
DlAi f
19
0 . 0'\ ,
0•
( OI D O I
1 (11
•
.• 000
0 . ' '' 1
0 . 2'S I
0 . l t 1l
0 . 2"0'
0 . 0]1)
O . U' I
o . ons
O . I 6O S I
O . O]H O . O J IIl O . OHS
0 . 0)0'
0 . "2 2 )
0.
•01l •
'21
O . l tt \
O . I UO
O. I U I
O. lnl
0 . 1 1 30
0 . 1 1 \0
ll
0 . 1 00 2
O . O t. .
( GU T
I no o o
0 . OlS2
0 . 11 0 1 .
O . UO\
' . 2 0 10
0 . 1 19\
0 . 1 7'"
0. 191.
O . I UI
o . l n.
0. 1 1 7)
0 . 1 02'
S lO S l ( U 100'
UIOOO
10
1
O , ttO
O . OUS
0 . .. . 2 1
0.2"1
o.un
IUOOO
2.
1
O . t71
0.0]11
O . II ' S O
O . Utl
0 . 2UO
O . JOI I 0 . 201S
0 . 1 92\ 0 . 1 "' 0
O. IIIS 0 . 1 1 l'
0 . 1 91" 0 . 1 ,. "
0 . 1 606 0. 1116
0 . 1 1 7 '1 0 . 1 11 1
0 . 1 1 91 0 . 1 20 ]
O . I O IIE 0 . 1 0\\
I I IGI I ID
UIOOO
"
I
0 . t' l 0
0 . 0 ) 0 11
0.
)'
0 . 2 \ 1 60
0. U9"
0. 2 ' "
0 . 1 9 111
0. 117]
0 . l t 7 1&
0 . ' 61 1 6
0 . 1 11 ' )
0. IH2
O . ' O'�
'W I F T
6 UO O O
16
I
O . UO
0 . 0101
0 . II 1 S I
0 . 2 \ 11 6
0 . 2 1 00
0 . 2 1 2\
0. ' 990
O. II"t
0. 1" 1
O . I ISO
0.1.1'
0 . 1 2 16
0 . 1 01 t
H Al. I ). " 'UCO(_
lOSOOO
••
19
O . th
0 . 0). 1
0 . "0,'
0 . 2 . ;-0 0
0 . 222'
0 . 20S0
0. 192'
O. ,�U
O . I J6 6
0. 1 1"
O . I O lt
2'
1
O . tll
0 . OU 7
O
' SO
0 . 2"'1
0 . 2UO
0 . 201S
0 . 1 9 1 11 0 . 1 ., 0
0 . 1 ' 0.
IOs000
o. I'n
0 . 1 9' "
0.1616
0 . 1 111
0. 1201
O . IO'�
10SOOO
2<
1
O . ts l
O . O) l t
0 . " 1 10
o. un
0 . 20tO
0 . 1 '\ \
0 . 1 1""
0 . 1 9\"
0.
o.
n)
0.1211
••
•. UI I
0 . 1 '. 0
a . I , tt 0 . 1 1\" 0 . 1 '"
O . I 'n 0 . ' 16 1 0. 1" 1
0. ' 1 ' 6 0. 11]) O . I 'U
0 . 1 11 ) 0 . 1 ,, 00 0 . ' '' ' 0
0 . 1 203
O . , OS �
O . l t ll l O . l tl O
O. IJlt 0 . 1 . -, 0
'. 1072 ' . 1 01 1
O. "
' S
0 . 2 00 1
0. 161)
0 . 1 .... 0
O . l t tO
0. 1 1 12
0. '111
0 . 1 , 7.
0 . ' U6
0. 1."
0. 1212
UCl(
!t sn o o
1 .01'
•2
.•
0 . 2.S I
•
111
•
• . 1 0' \
.,
O . IS )
0 . 012 7
5SISOD nnoo SUiGO
21 2' "
7 .1
0 . 12 7 0 . 1 1 '1
• . II I S 0 0 . •.2: 0 0 O.lUll
•. Utl 0 . U20 O. UlI
0 . 2 UO 0. 22'" 0. 1210
O. J075 0 . 2 1 00 0. 2 1 1 S
I
0.'"
0 . 0l 1 . 0 . 0)" 0 . 021.
0 . 11 1 2 1
O . U' I
O. UJS
0. 2 1 ' 1
. n ooo
lO
1
0 . 11 1
O . O J O ..
o.un
O . U"O
0 . 2 2 1 11
0. 2 1 1'
••••
" '0 0 0 111 7 70 0 0
U
7
O.
0 . 2\"
O.21n
0 . 2 1 S0
O . 20 I S
0 . 1 .0 ..
O . l tt .
0. 16"
O . I 'I U
O . I JU
o . l on
2<
7
O."UI
O . UI I
o.un
0. 2 1 6 1
O . lOU
0. 1'1\
0 . 200 1
0. 1171
0 . ' .' 0
0 . 1 21 0
0. 1 1 12
'( L I CU
U 7000
"
0 . 11$ 6 0.11"
o . on•
0 . 11 1 0 0
' L I CU '
0 . .. . , 0
O . UU
O . UIO
O . UOS
0. 2070
O . I IS '
0 . 20 1 1
0 . 1 70 1
' . ' lt 7 0
O . I Ut
0 . ""2
0 . 20 1 1
0. '"1
,uuu'
•
••
U
I
0 . 021" 0 . 02 1 "
••
l
OU
0 . 1 ".
O . I IOS
O . rol
0 . 02 7 7
0 . US2
0 . 2 0 11 0
O . I ' JO
O. I IH
0 . 0 2 6 14
0.
10
0 . 2626
0. 2UO
0 . 22OS
0 . 2010
0 . 1 '\ 1
0 . 10l 1
0 . 1 70 1
A " ''' '
0 . 1 2 10
0 . 711
0 . 1 11 "
0. 172
o . onl
0 . 11'1 1 1
o. un
0 . 2 1'.
O.22lt
0 . 20"
0. 1911
0 . 1 0lt 0
0. 1712
O . I It "
' . I nt
' . ' U'
0. 1 1\1
lU.
317500
IIIS
3tH.O
21
lUI '
3 1 7 50 0
0 . 7 11 1
0 . 0 2 11 1
0
1
O . UI I
0 . 2U\
O . UIO
0 . 2 1 lS
0 . 20 H I
O . JOU
0 . 1 7"
0 . I S0 7
0 . 1 126
0 . 1 1 7,
0 . 7 11 '
o . OlSS
O . nSl
0 . 2"'7
0 . 2 11 0 1
o. un
0 . 20"
O . I U.
0 . 20.S
0. 1 7 1 7
0 . 1 ., .
0 . 1 10 .
O . I I St
(I I I ( U . f f
U 7 S00
z"
7 I
,I I OL f
UI.O.
lO
7
l 1 .l f T
J J IIt . o
U
7
MUl l l
nuoa
a S T l I ( 1I
100000
"
"
• 11
. •\2
o . n ..
O . USO
0 . 2 1 11
O. I IU
0.121
0 . 02 ' U
0 . 1611
0 . 2UO
0. 2 120
0 . 200'
0 . 206.
0 . 1 7 16
O . I SO Il
O . 1 32 J
0 . 1 1 16
0 . ". 0 . '"
0 . 0222
0 . "'20
0.2131
0 . 2."
O.UIO
0 . 2 1 1\
0 . 2 0 6 ..
0.2101
0. 1 713
O . I UO
O . I UO
0. 1 2 1 2
O . 02 H
O.
CI . 2 1 1 2
0 . 1""
0.2UI
0 . 2 ' \1
0 . 2 01 1 \
0 . ;051
0 . 1 7tO
O. ISH
0 . 1 ; 11 7
O . I , {(i J
O . "S I I •SI
1
o . un
Appendix B
448
Table B . 1 3. Inductive Reactance of ACSR Bundled Conductors at 60 Hz with 4 and 6 Con· ductors per Bundle
COOl
Uf a CN l l
Ua
S Il
iD .OS
Sf
0'1. ,..
, .. . f1.
0 . 0'00
-
II
",
I I OUC T , y (
C O I Du t T O.
6 9 F 0 . 1 2 �' 0 . 0, ' 7
. ( A C T 'I C (
SP A t " G
t ll. )
'2
r !»
0 . 0 6 2 !t
0 . 0 11 2 2
1 ( ) 1 1
� II
I
0 . 0 2 1) 6
I
' I
a
.. .. s / .. l l l 6
-
J O.
I
fOOt
C O I OU C T O .
I, 0 . 0' 2 6
0 . 0" 1 6
' A O ' U S
$P&( I I'
IS
O.OIU
SD
( I. )
-0. 0 1 0 1
II
- O . O l l !»
"'''0(0
) . 0 8 00 0
6111
19
2 . S00
"'''0(0
2 2 9 0 00 0
61/6
19
2 . )20
o. O I S I
0. 1 271
0 . 0902
0 . 0 6 11 0
0 . 0 11 3 7
0 . 017 1
- 0 . 00' 1
-0. 017'
' '' 1 1' 0 0 0
19
1 . 7�O
0 . 06"0
0 . 1 ) 60
0 . 09 9 1
0 . 071'
0 . 0 !t 2 6
0 . 0 ) 60
0 . 0 1 '"
- 0 . 00 1 2
. 0 . 0' "
1 1 7 5 000
It
1 . .00
0 . 0" "
0 . 1 190
0. 101 1
0 . 0760
0 . 0 !} !} 7
0.0],1
0 . 0 11 ' "
0 . 0 .. . ..
0'''0£ 0
'!t l / ll '!t o , I ,
o . o .. n
0 . 0 1 1"
( I ' U O( O
O . D� I "
0 . 0 !t 0 1)
0 .02 " 1
- 0 . 00 1 1
-0.0"
1 . 7]7
0 . O !» 7 1
0 . 1 190
O . I OB
0.076)
O . OHO
o . o no
0.0""
O . O�O.
0 . 02 1 1
- 0 . 000'
.0 . 0 I tl . 0 . 0 1 91
0 . 0 1 1 1)
6
I 'W,
I l u t l ' II O
2 ' HOOO
10
.,
1 . 162
O . O�..
O . I ) I�
0. 10'6
0 . 07S'
O . O� � 1
O . O l t�
0.09'2
O. 0�02
0 . 02 1 1
- O . OO I S
CMUU.
1 7 10000
10
19
1 . 602
0 . 0� 1 6
O. '"'1
0 . 1 0 .. ..
0. 0113
O . OH,
0 . 0 14
0 . 09)0
0 . 0000
·0 . 0 "
,
�o
19
I.
0 . 1 '1 2 1
O . l OSt
0 . 0190
0 . OH 1
0 . 0 11 2 '
0 . 09 l S
O . O� H
0. 01l0
0 . 0009
-0.0"
6
I S90000
!» II IS
O . Ci !t 2 )
l lP.' "G
• S90000
o. n�20
0. 0229
f l l C Ol
lit
j . �02
0 . 00 "
'URO l
l u T tt A l C M
1 5 1 0.1) 0 0
'l0" .
I")
' 1 6 1 00 0
l '!t I 0 5 0 0
12
o� SO
0\
D . t .. U
D . 1067
O . OIOS
0 . 0602
0 . 0 0 )6
0 . 0 ' '' ')
O . OSH
0 . 0200
0 . 00 1 9
- 0 . 0 1 66
I . S 06
0 . OS06
D . HU I
0 . 1 06 2
0 . 0100
0 . OS91
O . O lf l
o . 0� ) 2
0 . 02" ,
0 . 00 1 �
- 0 . 0 1 69
0 . 0 .. 8 '
0 . 1 "")
0 . 01 1 /
0 . 060'
0 . 0 .. ..
0 . 0 ' '' '
0 . 00 2 "
.0 . 0 1 6 1
I o tt 6 S
O . O tt ' 't
O . I II U
0 . 01 0 7
0 . 06o"
0 . 0 14 1 1
o . O,�O
O . Osoo
I'
0 . 1 07.
I )
0 . 0901
, , " ,6
o. 0�)1
0 . 0206
0 . 00 2 0
- 0 . 0 1 60
1 . 't 2 7
0 . O tt 7 0
0 . 1 " 1) )
0 . 1 01"
0 . 01 2 2
0 . 06 1 9
0 . 0 " ,) ]
0 . 09 S '
0 . 0 ') . 7
0 . 00)0
- 0 . 0 1 \_
.9
1 . 11 2 "
0 . 0 111 ' 2
0 . t "J ,, 6
0. 1077
0 . 0 ' 1 1}
0 . 06 1 2
0. 0""6
0 . 09U
O. 0""2
O . OH I
O . OO H
- 0 . 0 1 �9
I . J l I}
o . o .. !» .
0 . 06 2 1
0 . 0 11 6 1
0 . 0962
O . O�U
0 . 026 1
O . O O J I)
- 0 . 0 1 '"
0 . 0 11 6 6
0 . 1 0' 1
0 . 01 10
l . lI 2
O.
0 . 1 0' 7
O . OI I S
0 . 06 / 1
O. O,�,
0 . 0 1) 11 9
0 . 02�1
0 . 00 1 l
·O . O I S i
1 . ) .. 1}
0 . 0 " 1' "
0 . 0 14 1 1
0 . 0961
O. O��I
0 . 0161
0 . 0002
-0 . 0 , "2
0 . 0 .. ,, 1
0 . 0 11 .. 0
0 . 06 ] 7
1 . ))1
0 . 1 1"6
. 1 101
0 . 0 .. ') 6
0. 1097
0 . O "J 2 9
0. 1
0 . 0 '1 6 6
0 . " '1 1
0 . 01 1 \ 0 . 08S0
0 . 06 J 2
1 . 102 1 . 29)
0 . 0 11 3 6
O. "'"
O. ' 101
0 . 0 1 .. -;
o . a t. . �
0 . 1 .. ' 1
0 . 1 1 22
0 . 0" 2 0
0. 1""
0. 1 1 1'
19
1 1 5 1 !:I O O
�o o�
,P(.
1 ] 5 1 500
0\
'"( A S U T
1 2 1 1 000
I'
ti l l H U G'A(l l £ aul T il'
1 1 72000
\0
\0 os
I'
IOIOl '
NU" . '"
••
, I lil C H IlU(JU
I 000
I Il l l O O O
1 1 9 2 5 00 , 1 9 2 1j O O
1 1 1 ]000
�o
os
\0
"
• 11 , 0
O . " ' 1) 6
O
.
I
I HO
0 . 1 06'
0
1 11
0 . 0'" 7
0 . O l !} 6
O. 096�
O. O��S
0 . 0260
0 . 00 1 9
- 0 . 0 . .6
1
0 . 01 1 �
O . OU S
0 . 0 2 7 ..
0 . 00 "
-O . O t h
0 . 06 . ,
0 . 0 11 1 6
0 . 09 1 2
O. OS6 2
0 . 02 7 1
0 . 00 " 6
� 0 . 0 1 J9
0 . 0160
0 . 06S 1
O.
o . O� 1 2
0 . 028 1
O . OOH
-0 . 0 1 29
0 . 06�1
0 14 ' 1
0 . 0912
O . OIH
0 . 0081
0 . 0910
O . OHO
0 . 02 7 9
O . OO S )
-0.01 ) '
0 . 0 6 ,, 7
0 . 0 .. 8
1 1 1 3000
o�
1 0 ) HOO
�o
I . 2 !t 9
1 0 ) HO O
o�
1 . 2 1 )
0 . 0 11 0 2
0 . 1 1) 0 1
0. 1 132
0 . 08 70
0 . 06 6 7
0 . 0 ') 0
0 . 0S ' 9
0 . 02"
0 . 0062
-0. 0 1 22
)6
0 . 0 3 8 11
O. IS I�
O . I , .. 6
0 . 0 . . ..
0 . 06' 1
O . OS
0 . 0991
O . OS"
0 . 02,7
0 . 00 7 1
-O.Ol l l
C ! II D I I l l
9 S 0 000
So
I. I"
I . 1 96
0 . 0002
0 . 1 '!» 0 1
0 . l I l2
0 . 0110
0 . 0661
I I"
0 . 09 1 9
l o nsoo
O . O !t O I
0 . 0919
0 . OS19
0 . 02n
0 . 00 6 2
•' l l
HoOOO
o�
1 . 1 6S
0 . 0)16
O. I�
' !» . O D O
16
I . 1 1. 0
O . 0370
0 . 1 1) 2 '
CURL ( .
O . T O l .... T UAGU
C'"'IO
1 . 2 11 6
I)
· 0 . 0 1 22
I ' .. ..
0 . 0112
0 . 06 19
O . O� I I
0 . 09 9 1
O . OSI1
0 . 0296
0 . 00 1 0
0 . 019�
0 . 0692
0.
0� 16
-0.01 , "
O. 1 1�7
O . 1 00 �
o . OUS
0 . 0 300
0 . 00 1 9
· 0 . 0 1 06
O.
C .. . . . '
900000
�o
1 . 1 62
0 . 0392
G . I !t O I
0 . 1 1 19
0 . 0' 7 7
0 . 0 6 7 ..
RUDO"
900000
-\
1 . 1 11
0 . 037"
0 . 1 !t 2 l
O . I I �"
0 . 0892
0 . 061'
o . 0�01
0 . 0990
0 . 0�10
0 . 0� 2 )
O . 1 00 )
0 . 02U
0 . 00 6 1
-0.01 1 1
O. O�tl
0 . 0 302
0 . 00 1 1
- 0 . 0 1 01
"
0 . 00 6 1
-0.01 " · 0 . 0 1 01
MAl l U D
7 9 !t O O O
)0
l " , ' tt O
0 . 0192
O . 1 ') 0 1
. O . I 1 ]9
0 . 08 7 7
0 . 0.01
O . O� . .
0 . 021 )
D"H (
19S000
26
1 . 1 08
0 . 03 7 1
O . I !t 2 l
O . I I ') "
0 . 019)
0 . 0 6 7 ..
0 . 0"
0 . 0619
0 . 0 ') 2 14
0 . 1 0 0 1&
O . O�'o
0 . 0)03
0 . 00 1 1
COl 00'
1HOOO
�.
1 . 09 1
0 . 0170
0 . 1 ') 2 6
0. I IS7
O . ol'�
0 . 06 9 2
0 . OU 6
O . 1 00 �
0 . 0�9�
0 . 0 10 "
0 . 00 7 9 0 . 00 1 '
· 0 . 0 1 01 · 0 . 00 9 6
.,
· (' . 0 1 06
( 0 ( 0 00
"
� OOO
20
1 . 0 '1
0 . 0366
0 . 1 ') 2 9
0 . 1 1 60
0 . 0191
0 . 069�
O. 0�29
0 . 1 00 1
O . O�"
0 . 0301
f U JI (00 '
1 9 5 00 0
o�
• .
06)
O . O)H
0 . 1 ,) " 1
0. 1 1 72
0.0910
0. 0707
0 . 0 ') 14 1
0 . 1 0 1 ')
0 . 01 . .
0 . 00 8 9
7 9 5 000
)6
1 . 0 .. 0
0.OJ77
0 . 1 ,20
0 . 1 1 !t 1
0 . 081'
0 . 0616
O . OU O
O . t OO I
o . 060� O. O�"
0 . 0 )0 1
0 . 00 1 S
- 0 . 0 1 09
7 1 1j lj O O
)0
1 . 01 '
0 . OJ7)
0. 1 \2]
0 . 1 1 \ ..
0 . 08 9 )
0 . 061'
O . OHO
O . I Da "
o. OS,"
0 . 0303
0 . 00 1 1
- 0 . 0 1 01
0 . 0) 1 3
0 . 00 1 7
I ( OW ' I ' S U " l I lG S" l T ,U"( l
••
� � OO
16
I . O !» I
O . D ) !t !t
0 . 0901
20
1 . 0)6
0 . 01.7
O . I !» "
0 . 1 1 69
7 1 !t S O O
O . I S""
0 . 1 1 76
0 . 0"
"
0 . 0 7 0 .. 0.07 1 1
0 . 0 \ 11 \
0. 101'
lit
0 . 0601
0 . 0) 1 1
0 . 00 1 2
- 0 . 00 9 3
6 6 6 60 0
26
1 . 0 1 11
O . Ol'U
0 . 1 ') '"
0 . 1 1 10
0 . 09 "
0 . 0 7 1 !»
0 . 0 ') '"
0. 1021
0 . 06 1 1
O.OUO
0 . 0010
- 0 . 00 ' 0
0 . 00 9 9
- 0 . 00 1 6
7 1
2_
f l 'N t "GO
666 600 6 !t ) 9 00
..
( GU T
6 ) 6000
)0
.9
6 ) 6 0 00
26
1
(i Il O U ( U
SWI f ' T ( Al
S�O . . ' ( A co e .
( I li l l
OOH
O sP' l ( ,
N(I .. . ..
' l I CI [I
'f l l ca. lUI
6 ) 6000
20
6 )6000
oe
7
- 0 . 00 9 7
0 . 09 \ I
0 . 0"'1
0 . 0 ') 1 2
o . ' 0 '"
o. lon
O . 0 6 1 !.
0 . 032<
O . OI U
0 . 0 3' "
0 .0 1 1 6
· 0 . 00 6 '
O. I O H .
0 . 060S
O . OJ ' "
0 . 00 1 1
· 0 . 00 9 6
0 . I SS6
0. I 1,7
O . o'H
0 . 0122
O . OS�6
, o. ;OH
O. 06 1 �
O . OU<
0 . 0011
· 0 . 00 1 6 · O . OD ' I · 0 . 00 "
0 . I ,) S 6
0 . OH2
O . I ') ' "
0 . 990
O . OHS
0. 1 1'7
0. 1 1 72
O . O'H 0 . 09 1 0
0. 0122 0. 0707
O . OS " I
h
0.'71
0 . OJ 2 1
O . I SU
0 . 1 1 9 11
0 . 0')2
0 . 0129
O . OS 6 )
0 . 1 0 30
0 . 0620
O . OU'
0.010"
0 . 9.0
0 . 010.
O . I SI S
0. 1216
O . O'H
0 . 01U
0 . 0 ') "
O . I O. S
O . 06 3 S
O.O
"
0.0 1 11
0 . 9 10
0 . 0)0 1
O . I !» I I
0. 1 2 '9
o. O,�.
O . 01�S
o . on,
0 . 1 0 11 7
0 . OU 1
0 . 0 ) '6 '
0 . 0 1 10
- 0 . 00 "
30
,9
D . 9'"
0 . 0"' 1
G . I ,) ,) I
0 . 1 1 '2
0 . 0'20
0.07 , 7
O . o !» !t
0 . 1 02 2
0 . 06 I I
0 . 012 1
O . OO I S
· 0 . 00 1 1
6 0 � O OO
26
1
0 . 966
0. 0321
O . 1 S6 ]
0. ' ,9.
0 . 0132
0 . 0 1 29
I
0 . 1 030
0 . 0620
o . on.
0 . 0 1 0<
60S000
20
1
O . 'H
0.031'
0 . 1 ,) 1
t
0 . 1 202
0 . 0 9 tt O
0 . 07 ) 1
G . OS 7 1
O. 1 0 3�
O . 062S
o � 1) 3 ) 111
0 . 0 1 0'
_ 0 . 00 1 5
0 . OU 2
0 . 0721
O . OSU
0 . 1 0 30
0 . 06 2 0
0 _ 0 1 0<
- 0 . 00 "
O . O'o�
O . Oh,
o . on.
0.0 1 1 2
· 0 . 00 1 2
• ) 6 000
60S000
16
O . OH )
· 0 . 00 1 1
0 . '� 3
0 . 0321
O . I �U
O. ' I '"
2&
0 . '2 1
0.03"
0 . 1 \7'
0 . 1 201
O . OS ' I
o. l O l l
20
0.9..
0 . 0 )06
O . I �I )
O. I'".
O. O.� 3
0. 01U
0 . 0 ') "
0 . 1 0 ....
O . OU O
0 . 030 )
0.01 11
- 0 . 00 1 7
0 . 01'"
0 . ' 606
0. 1 1l1
O . 091�
0.0112
0 . 0601
O . l OS.
0 . 0 6 '"
o . on.
o.oln
- 0 . 00 5 2
O . OI ) S
0 . 0 3. .
0.0 1 1'
- 0 . 0066
O . 060S
O . O)S<
- O . OO S I
0 . 03S1
o.oln
O.OIU
· 0 . 00 5 2
o. 0�1 3
0.0'"
- 0 . 00 3 1
1
UISOO
30
s�noo ��6�00
n 6 �00
0.1"
., 1000
.1
)0
0. 113
0 . 0 ) 0._
o. lsn
0. 1211
o . o'n
0 . 01U
O . O�U
O . I OO S
t1 7 7 0 0 0
26
O . IS '
0 .021'
0 . 1 60 1
0 . 1 2 ll
0 . 09 1 0
0. 0111
0 . 01 0 1
O . l OS S
tt 1 7 0 00
2'
0.1"
0 . 0210
O . 1 60 6
0. 1 1l1
O . Ot1S
0 . 0112
0 . 0101
O. Ion
0 . 0'"
1 7 00 0
II
0 . 8 ' 1&
0 . 02"
O
0 . 1 2S'
0 . Ott1
0 . 0 7 ' ..
0 . 01 1 '
0 . 1 0 7 11
0 . 06"
191500
)0
0 . 106
0 . 02 7 7
. • U.
0. '61t
O . 1 2" �
0 . 01 1 '
0 . 0110
0 . 06
" o . on.
0 . 1 010
0 . 0150
0 . 1 01<
0 . 0"0
0 . 0311
0.0'"
. 0 . 0031
O . OI3S
0 . 101'
0 . 016'
0.0371
0 . 0 1 52
. 0 . 00 3 1
tt
1
1 1
O . OUI
o . ou.
0 . 0363
0 . 0 1 31
- 0 . 00 < 1
0 . 113
0 . 015<
O . I ll'
o . l n'
0 . 0911
0. 01"
1
0 . 112
O. 02�'
O . I IH
0 . 1 111
0 . 1 00 ..
0 . 01 0 1
0 . 71U
0 . 0 2 11 1
O . l nl
0 . 1 211
o. lon
0 . 0' 2 2
0 . On6
O . l Ol l
0 . 06 1 l
O.OUI
1
0 . 1< 1
o . ons
O . l Ilt
0 . 1 2 10
0 . 1 00 1
O . OIOS
O . OU'
O . '0"
0 . 06 1 1
O . OllO
O . O U<
· 0 . 00 30
0 . 0' "
0 . 015<
0 . 1 0' 0
0.01"
. 0 . 00 2 1
0 . 0' "
o . on l
0 . 1 10'
o . ono
0 . 0 31 1
0 . 06 99
0 . 000'
0.011l
· 0 . 00 0 2
O. I IOl
0 . OU 2
0 . 0<0 1
0 . 0 1 11
. 0 . 00 0 '
' I ' S
1 9 1 �0 0
26
",: 6 1 1
) 9 7 S00
20
("' C·.O((
1 9 1 \0 0
.1
0' 1 Ol(
H6000
)0
l I llI r NUl t i
l U 'OO
26
H6000
"
0 . "0
0 . 02 2 2
G. 161'
O. ' ' ' 2
O . I OS O
)00000
26
0 . 610
0 . 0229
0. 161 1
0 . 1 )02
0 . 10"
0 . 01 ) 1
O S TII I C "
0 . OH6
O. 060.
O. 1212
O . O)�S
1 .019
0 . 0)01
0. 1 0
0. 1\11
1 . 000
O . ' !» l
O. OS 19
1
0 . 12 0
0 . 0203
0. 1 15'
O . l lIO
O. lOll
0 . 06 1 2
o.oln
· 0 . 00 1 '
449
Line I mped a n ce Tables
Table B.14.
61(1
6H
S TUIOS
( C l UO .
z o o 3000
72
Z " �OOO
n
e M il
U UOOO
�.
U'7COO 2 1 1 2 00 0
h ••
U UOOO
'2
U OOO
.,
11
so •• oZ
. .. 7000
' U 7000
oz
' Z 0 30 0 0
30
'Z"
,. ••
OOO
• • 7toOO
• • nooo I I UOOO • • 0'000
10
1 010000
0. 1 1 1 1
O . I � 76
a . I " s
0 . ' "6
0. 1 7 1 1
O. 'HI
o
O. "
0 . 1 70'
O.
0
U
0.2'U
0. "
'2
0 . 2 ' �0 0 . 2 ' �0
0.
0.
I . 7)\
0 . 0�6 1
0 . 3.n
0 . 2 ' 61
O . • •l l
1 . 7 )\
O . O HO
O . UfO
O . H03
• 900
0. "
• '5 7
I
.. tt ' ./ M I L. l
r-;-I I
' .
fO'
O I U S
"
. oo y
I
COIOUC T " ,
' 2
" I t "' ( I I . ) .�
0 . 1 10 1
O. I J7J
. • •n . • • 62
0 . 1 70 '
0. ' '' 3
0. 1 1"
0 . 1 6"
0. "
0 . 1 1 11 ' 0. 1 1 "
0 . 0'60 0 . 0160
3
� � 0 . 0"
1
0 . 0' "
0 . ""
0. 1 7 ••
O. ' U'
O . OUO
0 . . ...
0. 1 7 "
0 . ' U6
I IU
0 . 09 7 7
O . ' �"
O. I IU
0 . 09 7 '
O . 1 7 JO
O . 'U'
0 . . .. .
0. 1 7 "
O . ' U'
O. I I U
0 . 09 7 '
0 . 01 2 � O . OI U
O.
O. I 6 I Z
O . ' SO l
0 . ' 72 6
O . ' lI'
0 . OU7
0 . ' U6
0 . ' 7 09
0 . 1 11 2 1
o. • •n
O . O' U
0. 16",
0 . 1 00 1
0 . 0110
0.
• 737
• 701
0 . 1 60 2
I . 7H
O . O�U
0 . 1010
0.216.
O. "1�
O
I . 60 Z
O . OS "
0 . 3\10
O.U.S
o . ""
0
II
' . 60 2
0 . 0010
O . 'UO
0 . 2UO
o. ZOo.
0 . ' 110
0 . 1 6"
..
' . 60 2
O . OHI
O . lS,"
0 . 210 ' O . UU
O.
O. I U '
7
O.
•161
O
with 1, 2, and
O . I 7 JO
•H,
O.
0 . 1 712
. • 700
. • 7U
• 717
"
O.
0. "
"
o . on.
0 . 01 1 1
0 . 1 60 �
O. '0"
0. 1711
0 . ' 11 l
O . ' 1 60
0 . 0110
0 . 01 3 2
O . ' 6 60
O.
o
1 751
O . "21
O . I I 97
0. 1016
o . on,
O . , 7' 1
O. ' O� I
O . ' llO
o.
0 . 01 1
O . • S"
O . I 7U
O . / OZo
O. ""
O. '01 '
o . OI n O . OIlS
• s., O . • �..
•on
1
' . S O'
0 . 0" 6
0 . '670
0. l ll3
O • • 00 1
0 . 00 1 1
0. 373'
0. 1217
O . ZO"
0 . 1 167
0. 1 71 '
• u,
O. /OU
I . �O'
O•
O. ' 7 "
' 1
O . I IOS
0 . 1 1' 7 7
O•
0 . 01 . 7
I . �O '
o . oo n
0 . 360 7
0. 22"
0. "
O
O . ' UI
O . ' H'
0 . 1 776
O . . .. .. .
O . 'Z 1 6
o.
• ou
0 . 011'
1 . 3'�
0 . 00 )6
O . UO '
0.2U'
O . ZO>5
• • u.
• z.�
0 . '0"
.,
O . ' U'
O . ' �OO
0 . ' 217
O . '017
0 . 01 l '
0 . 37t0
O.U"
O . Z070
• 90 '
O . I U�
0 . 0" "0
O.
o. "n
t . ) 11 '
0 . 1 60 '
0 . 1 12 '
0 . . ...
O.
O . OU �
0 . 1 . .. .
0 . U'2
O. zou
O . ' IZ Z
0 . ' 71 1
0 . 1 676
O . ".Z
O . I S I ft
O•
UI]
O . ' OU
O . Oo Z '
0.
•u.
0. 1 10I
O . OI� I
"
1 . 30 2
U"
O . ZOOI "
O.
• •U
O . I IU
O . ""
• 710
O, 'U'
0 . ' 1 21
0 . ' 7' 2
O . "U
O . I I. ,
O.
0
' 2U
O . I I O�
o . on '
O.
0. 17"
0.
0. '117
O
0 . ' 2 1'
0 . 1 0"
0 . 0'"
I I�' o. lIn O. I U I
O•
o . un
O. I t ' "
0 . 011 7
O. "02 0 . 1 61 '
1 . 102
0 . 00 1 7
O . 'U�
O.
1 . 30 2
0 . 0026
O . U2I
o . un
7
I . Ut
0 . 0007
O . UU
0 . 2 31 ]
0. Z I I 7
O. "
'Z
O . ' 10 7
I . n.
0 . 0001
0 . 3112
O . UI I
o .Z. IS
o. "
o . 1 6n
O . O" Z
0 . 31 70
0 . 2U'
O. Z I IO
••
0 . 1 10 1
I . U'
o. lIn
O . ' 100
O . ""
• . 2.6
0 . 00 0 '
0 . n0 3
0 . 2 372
O . 2 I lI
O. ""
0. 1111
O.
I. Z06
O . OO O J
0 . llt7
O . llli
0 . 2 'l l
O . . .. .
O. ' I l l
1 . 2 0'
O . O OO S
o . UI I
0. 2361
0 . 2 ' 20
O. I I O�
O. '1 .0
I . 2..
0 . 0007
I l .,
O . UU
O . UI)
.
O . Z I OZ o . zon
'3 ..
30 II
0. 1" 6
0. 2 ' U 0 . 2 ' 11
0 . 3'�'
13
20
0 . 2 1 32
O . JoU
O . Jo�'
"
• • OtOOO
•
o . Jon
O . OHI
••
U
•
O . OS"
O . OH'
20
' ! l 1 00 0
1 ' � 3 000
.�
'1
0 . 0�96
• • 762
S.
. U 6000
'Z
---:-:-l
(il i a
S'lt l l , ( I I . )
1 . '2 '
I . 712
"
. ( I ( l ._C£
I .U'
"
OJ
1 6 7 30 0 0
,
•J
"tlOOO
<2
'
O . l lt l '
z . UOOO
7 0000
r-:;--
I
COIOUCIOI
•
O . ,. '"
I l
' / 00 0
C Ol D .
l 'OUC f I W ( 2
O . OH '
••
"
S I IGL(
O . O�"
�.
"
,•• ft.
I .U'
Zl
So ••
0.1. '1.
1 . 762
37
2 2 2 7000
/ 12 1 0 0 0
It ,
,..---- 6 0
(q.
1( . ..
Hz
Inductive Reactance of ACAR Bundled Conductors at 60 Conductors per Bundle
•IIS
0. Z 1 I 7
O. "'2
0 . '107
• 66'
O . ""
• 70S
0 . 1 6"
O.
IS•• . • SO l • �U
o. U "
O. I I I I
0 . 0'"
0 . ' 51 1
0 . 1 2 10
O . l lot
0 . 0"2
0 . ' 112
O • • �,.
0 . 1 30 1
0 . 1 1 20
0 . 01 7 3
O. ' r lO
0 . ' U2
O. " "
0 . ot7 '
O . ' U'
U'
O. '2"
O . I S lO
0 . 1 29 7
0. 1 1 16
O . otll
o . un
O. I I I "
0 . 016 7
' �' 2
O . ' U'
0. ' 517
O. I
' 0 7 7000
JO
7
' .Z'Z
0 . 011l
0 . 3921
0 . 23"
O . Z ' lI
0 . ' '' '
O•
O. ' ' ''
O . ' 170
O.
O . l lO'
0. 1 12'
0 . 0 11 .
l O o tooO
20
!l
1 . 212
O . OJII
0 . 3 9. 0
0 . 2 3 90
D . l l '"
O. ' 970
0. "
O.
O. ' 17,
O • • �U
0. 1 l 1 l
O. ' I U
O . Ol l �
I O n ooo
II
"
1 .212
0 . 0111
O . lt • •
O . l llO
0 . 2 ' 30
O . , ,� .
O . I U'
O
0 . 1 117
O . I � lI
O.
o. l ln
0 . 01 7 1
' . "6
0 . 0311
O. "
.3
0 . 2UZ
0 . Z "6
O. ,,7'
0 . ' 1 1'
O . " IS
O. I " �
O . I S",
0. 13"
O . f l J'
0 . 011'
• . ". • . ".
O . OllO
o . n��
O . UII
O. Z. Sl
O
O . . ..Z
0
O . ' �S '
O. ' l l'
0 . 1 1 11
0 . 0110
0 . 3931
O. U17
O. 2
O. '17Z
O . ' S. .
O. I J' I
0. 1 13.
o . OIn
1 . . 6,
0 . 0311
0 . 311 '
0 . 2. 1 1
O.Z'U
O
O . ' ".
O . ' S 60
O. ' 327
0. ' '''
O . Ott'
O.
. • •t o
• 7U O . I 7Z •
0. 11"
o . on .
• • ,"
' O� O O O O
30
I OU O O O
Z.
'1
,16000
II
"
"••00
10
lit I
O . "" o
• •u "
0. 113'
. •IU
• 7Z. . •71l •
0 . 1 7 .. ..
1 101
nuoo
30
I . .,.
0 . 0)6'
O . Z I 7,
o. ZOOZ
0 . 1 1"
O. IIU
o . l ln
O . I I , ,,
2.
I)
' . 1 6S
0 . 031"
• 0 0 ..
O .Z022
' ' ' lOO
0 . 3917
0 . 2 11 1 "
O . Z . ..
o . , t . ..
O. 'UI
O . , 7.,
0 . 1 1 90
O. ' S 6 Z
0.
• 12.
O . I I ',
O. ' 0 0 1
nlOoo
13
I . UI
0 . 031 '
O . 399'
0 . 1' . ,
0 . 2 ' 73
O. " "
O. 'UI
0 . 1 75 3
0 . 1 19 3
o . •u�
0 . ' U2
1 . ' 6'
0 . 03"
0 . 396S
0 . 2003
O.Z ' 51
o . ". Z
a . ' I' H
0 . 1 111
O. ' S� '
O . l ll l
' . I IU
0 . 1 00 ..
' 0 1900
2. II
O. I U I
O . Otl.
toO l O O
30
0 . 11 0 '1 0
0 . 2 .. .. .
O.l.n
0 . 20Z0
O. '"S
0 . 1 7 7.
0 . 1 901
O . ' UO
O . I l.,
0. "
66
0. 1011
30
,
' . • 01
O . OH'
7nooo
O . OUo
O."U
O . ZOl3
O. ll37
O . Z06Z
0 . 1 91 '
0. 1 ' 1 ,
0 . ' 936
I . . ot
O . OHS
0 . 2 " '"
O . UOO
o . 10ZS
0 . 1 190
0. 1 1 7'
O.
O
0 . OU9
0.2 ""
O . llll
o . 10� 3
0. 19••
O
O.
• 60Z
O . ' 3"
. • 1 70 • • "' O• • •U
O
. 3
O.
•"e
0 . 1 0" ,
'3
O.
0 . ' I '"
h
O
0 . 1 0 .. '
0. 1 177
0 . 1 02 9 0. 1016
' 7 7 30 0
••
' . OO Z
•OS .
O. '1"
O.
•736
II
"
0 . ' "0
o . • 769
,nooo
••
••
• . OS� ' . • 01
' . Olt
0 . 0'"
O . '0"
O . ,.S6
0.22'0
0 . 1036
O . 1 9 00
U 90 0 0
,0
1
1 , 0'l
0 . 0 1 11 '
0 . 0092
0 . 2"'7
O.ZU'
0 . 2 0 11 '
O . I Si l l
O.
10 7 700
10 ••
'1
1 . 06 )
O . OJ'Z
0
0.1 222
O . ZO "
O . ' 9 1 .)
O
"
I . on
0 . 0 l Oe
.• 096
0 . 1061
7 I n oo
O . ,, 0 7 S
O . Z.S'
0.22' Z
O . Z03 7
O
O
0.03"
0 . " 1.0
0 . 2\
O. llU
0 . 1 0 10
" "0 0
'0
0 . 990
O . OHO
O . 1t , ' 7
I I
• • 90Z
O . Hot
0. 2163
0 . 1011
o.
O . "H
700000
,.
.1
0 . "0
0 . 01 0 '
0 . " . ,.
O.H'S
0 . 2269
0 . 10,"
6 " 600
II
.,
O.
6 1 2000
.�
7nooo 1�' 200
' 1 7 � OO
2 .
U
0 . "0
0. 036 1
0 . 11 1 0 7 0 . 16 0 ) 0
0 . 1036
O . Z . 90
0. 20'S
o . tt O
G. OUo
0 . " 1 62
O . zso .
O . l US
O. ZOI '
0 . '2 7
0 . 0116
O. "27,
0 . lSS6
O . lJ l O
O . Z ' JI
• ,u
• ,U o . " .. ,
0 . 2000
. • eo,
• 710
o. , eo o
• • I OZ . • 7'1
' 1". O• • hZ O.
O . , I .. .
O • • I IS O.
• ltO
0. " " O
. • UO
0. "
0"
0. 'U7
O . ' SI !
0 . ' \ 16
o.
O.
• 1 7S • uo I ). ,
O. ' 9 "
O . ' UO
0. ' n7
0 . 1 '2'
O . I !t ' 7
O.
O. ' 12 6
O.
O . ' S"
G . ' US
O . ! lU
I I I .. O . I I I !»
O . I I� '
0 . ' 12 6
0.
• JII
O. ' Z' 3
O. "
0. "
"
• s,.
I ', ,,
O.
0 . 1 1 78
0 . 1 00 7
. • Oll
0. 10',
0. 1037
� . 1030
H
O . ' US
O . 1 l .2
O. '2'Z
o.
0.
• tS7 O • • ,..
0 . ' 6 7'
O. "
O
O . I OU
o . ' UO
O . ' 11 7
O . ' 20 7
O . I OU
0 . 1 9"
O . I ' �'
0 . "2'
O . l l ft '
0 . 1 0"
16
.•2'S
• on
O. ' 0 10
6 " 200
'2
0. 029'
0. '292
O . Z�66
0 . "00
o
O . I IU
0. '03'
O . ' UO
0, , '0]
.. . h O O
I !I
0....
0 . 0 160
O . ""29
o . lIn
0 . 2 3"
0 . 12 ' 0
0 . 10"
0 . 1 1"
0. 2037
0 . 1 7"
0. "76
0 . 1 211
0 . 1 1 0'
" HOO
'2
0 . 1 ' 11
0 . (1 2 6 .
O . '... 2 .
0. 2632
0 . 1 ]86
O.21'Z
O. Z077
0. "
0 . 20 H
0 . 1 70 '
, . ,.00
.�
0 . 61.
0 . 0 1l '
0 . 'U6
O. l'U
0 . 1'"
0 . 21 ' 1
O . Z I 7'
O . Z067
0. 210)
0 . 1 175
0 . 1 ' 11 1
0 . 1 '"
0 . 1 2 1 11
U�OOO
'2
0 . 680
0 . 02 "
0 , '637
0 . Z 7 11
0 . 1093
O.ll'l
0 .1 0 ' 3
0 . 10 ' 1
0 . 2 1 06
0 . 1 7 7'
O. I�"
0 . ' 3U
0. 1 2 "
0 . 127
O. lUO
0 . 2 ' 06
0 . 10 1 1
66
. • tt .
O' ''H
O . ' Z"
O. ""
4 50
Appendi x B
60 H z with
Table B.15. Inductive Reactance of ACAR Bundled Conductors at Conductors per Bundle at(A
12'
(C
•
[Q. At
C ll l t
F,
IO
Oil.
SUUOS ( C / 620 I
.
..
G" . FT.
6
II
I I OU C T I V (
I £ A ( l U ( ( (I) ' A
• • C Ol OU C T O l S ' I C I I G ( I I ' ) 12
I�
0 . 07�0
0 . 0� . 7
,
I
--:-:-l "
I
DIMS/ M i l l 6 6
•
f O.
I
'DOT
COI DUC T O . ,
1 2
I�
0 . 0'0'
D . O."
0 . 0201
· 0 . 00 1 1
· 0 . 0202
D . O.'.
0 . 0 20 7
· 0 00 1 '
· 0 . 02 0 ]
O . l lI l
0. 1 0 1 2
n 7 �000
U
0 . 1 31 1
0. 1012
0 . 07�0
O . O� . ,
0 . 0 31 1
O . I UO
0. 1 0 1 1
0.0'"
o . o�"
0 . 0)10
�.
I .UI I .UI
0 . 0�99
nt 1000
U H
O. 0�96
n nooo
1 . 71 l
O . �H I
o . l n'
o . l on
0. 0763
O . OH O
O . 03n
0 . 0" 1
O . OSOI
0 . 02 1 7
U 6 2 000
1 . 712
O . OV'
O . I UO
0 . 1 02 1
0 . 0760
0 . OH 7
0 . 039 1
O . OSGS
0 . 02 "
HUOOO
�.
2 1 9 1 000
.,
1 1
2 ' StOOD
'2
19
1 1 9 90 0 0
�.
.2
1 9
"
0 . 0'01
O . 0�96
· 0 . 00 1 '
· 0 . 0 202
0 . 0201
1 .121
2 2 2 1000
�
0 . 0'9'
19
1 1
.ID I U S
0 . 0'0'
12
"'
6
S'IC I IG ( I I . )
1 0 1 ]000
�.
0. 031 1
I I
4 and
.
- 0 . 0 00 '
. 0 . 00 I I
·0 . 0 11l
- 0 . 0 1 16
1 . 712
O . OHI
O . I 4.U
0 . 1 02 1
0 . 0760
0 . OH7
O.OHI
O . Ot l � o . 0" �
O . OSGS
0 . 02 1 0
· 0 . 00 1 1
7 '�
O . O�"
O . , .0 0
O . lvl l
0. 0769
O . OHI
0 . 0.00
0 . 09 2 1
O . OS I I
0. 0220
· O . OOOS
· 0 . 0 1 '0
I . US
O . OU O
O. .. 1 7
0 . 1 0.'
0 . 0116
O. OSU
0 . 0. . 7
O . otJJ
O . OS l l
0 . 02 J 2
0 . 0001
· 0 . 0 "1
1 . 7H
O . OUI
0 . 1 396
O . 1 02 7
O . 076S
0 . 0�62
0 . 0 3 96
0 . Ot l 9
0 . OS09
0 . 02 1 1
· 0 . 000'
·0 . 0 112
1 . 602
O . O� "
O . . . 23
O . I OH
0 . 0 792
O . OS"
0 . 0. 2 )
0 . 0937
0 . OU 7
0 . 0236
0 . 00 1 0
· 0 . 0 1 7.
I
•
·0.01"
1 1 10000
'1
1 1
1 . 60 2
0 . 0.10
O. . .. ,
0 . 1 07 2
O . Ol i O
0 . 0107
0 . 0' . '
0 . 0,.1
0 . 0�lI
0 . 02"
0 . 00 2 2
· 0 . 0 1 62
" 0 1 000
.,
I t
1 . 60 2
o. ,.,,
O . I OU
0 . 0788
O . oses
0 . 0" 9
O. on.
0 . OS 2 '
0 . 023l
0 . 0001
·0.0 1 77
I . U'
0 . OU6
O . O'U
O. Oi l 2
0 . 060'
O . OS'O
0 . 02"
0 . 00 2 '
-0. 0 1 6 1
0 . 0. "
o . I Hf
0 . 1 090
0 . 01 2 1
0 . 062S
o . o.�,
0 . 0'6 1
O . OSS I
0 . 0210
0 . 00 3 '
· O . O I SO
1 6 1 2 000
I . �O.
0 . 0• •]
0 . 01S0
11
O . "'J
0 . 1 070
••
I . So.
O . O..�
0 . 1 061
0 . 0107
0 . 060.
0 . 0.31
0 . 0'"
O . OS3I
0 . 02.6
0 . 00 2 0
. 0 . 0 1 60
1 3 ] 1 000
�.
O . . . 76
O. I 1 0 7
O. Ol'�
0. 06'2
0. 0076
0 . 09 7 2
0 . Ost2
0.0271
0 . 0006
· 0 . 0 1 39
'2
O. , . 7 3
0 . 1 1 0'
0 . 0" 2
0 . 06 3 9
0 . 0. 7 3
0 . 09 7 0
O . OSIO
0 . 0 26 9
0 . 00 . '
·0 . 0 ' "
1 2' JOOO
10
I . 3'� I . J'�
0 . 0. 3 6
1 296000
1
O . . .1'
O . "17
0. 1 1 "
O . 01 � 6
O . 06S )
0 . 0.17
0 . 0' 7 9
O . OS 6 9
0 . 027'
O . OO S )
· 0 . 0 1 32
" 1 3000 " 0 7 000
�. '2
It
1 9
I . J02
0 . 0••0 0 . 0' 2 1
1 . 30 2
0 . 0" 7
O . 08�'
0 . OlS6
0 . 0"
0
0 . 098 1
O . OS T I
0 . 0 2'0
O . OO S S
· 0 . 0 1 )0
O . O. U
O . . . '0
0. 1 12 1
1 . 30 1
O . . .1 1
O. I I . .
O . DIS<
0 . 060 '
0 . 0'1 3
0 . 09 7 7
0 . OS67
0 . 0216
O . OOSO
· 0 . 0 1 )'
I . 2�t
0 . 0.07
O . . .97
0 . 1 1 21
0 . 0166
0 . 0663
0 . 0.97
0 . 0911
0 . OS 7 6
0 . 02U
O . OO S '
0 . OH6
0 . 02'S
0 . 00S9
· O . O l lS
O. os 10
0 . 0213
O. DOS 7
·0.0127 ·0. 0 1 22
1 2 1 1 000
2'
1 1 7 90 0 0
"
1 1 6 ] 000
10
I I ] 3000
2.
1 3
I . 2�9
0 . 0.01
O.
0 . 016�
0. 0662
D . O."
0 . 0986
"
19
I
2�9
0 . 0" 2
. .96
0. 1 1 27
I 1 0 .0 0 0
0 . 1 09)
0 . 1 1 2.
0 . 0162
0 . 06st
0. 0.93
0 . 091.
1 . 206
I J
19
•
·G . 0 1 2S
0 . 0'0 1
0 . 1 �0 1
O. I I U
0 . 01 7 1
0 . 06 6 7
0 . OS02
0 . 0'19
O . OH'
0 . 0 21 1
0 . 00 6 2
1 1 38000
10
1 . 2.6
0 . 0.03
O . I �OO
0. 1 1 11
0 . 0'69
0 . 0661
O . OSOO
0 . 0'"
O . OH I
0 . 0217
0 . 00 6 1
·0.0123
1 1 09 000
2'
I ]
1 . 2'6
o . o.o�
O . . . 91
0 . 1 1 29
0 . 01 6 8
0 . 06U
0 . 0.19
0 . 0917
O . OS 7 7
0 . 0216
0 . 0060
· 0 . 0 1 2'
1 0 1 00 00
1 8
19
1 . 2'6
0 . 0. 0 7
O . . .,7
0 . 1 1 11
0 . 01 6 6
0 . 0916
0 . OH 6
0 . 0 21 S
O . OO S '
· 0 . 0 1 2S
10
1 . 2 1 2
0 . 0393
O . 1 �0 7
0 . 1 1 38
0 . 01 7 7
0 . 066 3
0 . 0. 9 7
1 0 1 7 00 0
0 . 0610
O . OSOI
0 . 0993
O . OS I l
0 . 0292
0 . 00 6 7
·0.0I II
1 0. 9000
2'
0. 03"
O. I�I I
0 . 06 7 7
O. OS I I
o . o,n
O . OSIS
0 . 02"
0 . 0069
.0.0 1 16
18
I . 21 2
0 . 0396
O . I �O�
0 . 1 1 02
0 . 0110
1 01 1000
1 . 212
O . I 1 16
0 . 017'
0 . 06 7 1
O . OSOS
0 . 0992
O . OS I 2
0 . 02 9 1
O. 006�
·0 . 0 1 1 9
I O�OOOO
30
1 . 1 96
0. 0311
O. I � I I
O. "
0 . 011 1
0 . 06 7 7
O . OS I 2
0 . 0996
O . OSI6
0 . 0 2' S
0 . 0069
·O.O I l S
1 0HOOO
2'
I J
1 . 1 96
0 . 06 1 1
O . OS I S
0 . 0'91
O . OSII
0 . 0297
0 . 00 7 1
·0.0 1 1 )
1 9
1 . 1 96
0 . 039 1
O. I�I�
0 . 011'
I I
0 . 0'"
0. 1 '"
9 9 6 0 00
O . I �O'
O. "
.0
0 . 01 7 1
0 . 067�
O . OSO'
0 . 0'"
O . OS "
0 . 0 29 3
0 . 0061
·0. 0 1 1 7
99.100
JO
I . 1 6�
0 . 0 3 16
O. IHI
0 . 0190
0 . 06 8 7
0 . OS 2 1
0 . 1 00 2
O . OS 9 2
0 . 030 1
O . OO I S
· 0 . 0 1 09
9SHOO
)0
0. 0369
O . I �27
0 . 069]
0 . OU 7
0 . 1 00 6
O . O�"
O . O )O S
0 . 00 7 9
· O . O IOS
1 6�
0 . 0]7"
0 . 08 9 2
0 . 068'
O . OSB
0 . 1 00 3
O . OS93
0 . 0 30 2
0 . 0071
1 3
I . I �I
0.037 1
O . I SZ3
0 . 089.
O.OUI
O. OS2�
O . I OO S
O . O US
0 . 0 )0 0
0 . 00 7 1
· 0 . 0 1 01
9. ]900
2 '4 2'
1 9
0 . 0)8 1
O. I S1 7
O . 1 , .1
0. 0186
0. 0'1l
O . OS 1 7
0 . 099'
O . OSI9
0 . 0 291
0 . 00 7 3
·0.01 12
90 0 3 0 0
)0
I . , ,�
I SI O . I I !t 'l O. I lSI
0 . 0196
969 300
I . "0
O. I I S2
1 . 1 08
0. OH8
0 . I H6
O. I 167
O . otos
0 . 0702
0 . OS ) 6
0. 1012
0 . 0602
0 . 03 1 1
o . ooes
· 0 . 0099
I I � JOOO
9HOOO
JJ
1 8
I] I'
13
I
•
O . I �2S
O.
.2
I
·0.0 106
79�000
30
1 . 0' 2
O.OJ"
O . I S�7
0 . 1 1 11
0 . 0926
0. 072]
O . OS S !
0 . 1 026
0 . 06 1 6
O . OllS
0 . 0099
· O . OO I S
8 7 1 JOO
2 '1
O . O)H
O . I S lI
0 . 1 1 69
0 . 0'01
0. 070.
0 . OH 9
0. 10..
0 . 060'
0.0)1 J
0 . 00 8 7
· 0 . Oot7
..
, :1
1 . 1 01
19�000
0. 0339
O. I SH
O. I 1 8 )
0 0' 2 2
0 . 07 1 8
0 . 06 1 3
0. 0322
0 . 0096
· 0 . 0�81
1 8
19
1 . 1 01
0.0361
O. I S33
0 . 1 1 60
0 . 0'0 2
0 . 01"
0. 1010
0 . 0600
0 . 0 ]0 '
0 . 00 "
·0.0101
7 9�000
1 8
1 9
1 . 069
0 . OJ'9
O. I I 7S
0 . 09 1 )
0.0710
O. 1017
0 . 06 0 7
0.0316
0 . 00 9 1
· 0 . 00 9 .
I . 063
O . I �••
O . I I 80
0 . 09 1 8
0.071S
O . OS S ) o . os n o . o s• •
O . lOll
1�'200
I . O��
0. 102 1
0 . 06 I I
0 . 0 3 20
0 . 00 ' "
· O . Ooto
1 . 06J
0 .0302
O . I SSO
0. 1 111
0 . 09 1 '
0.0716
O . OSSO
O. 1021
0 . 06 1 1
0 . 0 3 20
0 . 009S
· 0 . 00'0
1 . 06 3
0 . 03'8
O . I � " '"
O . I I 7S
0 . 09 . .
0 07 1 0
o. O�'S
O. 1018
0 . 0608
0.0317
0 . 00 9 1
· 0 . 00 9 3
o. OS7 1
O . 1 0)S
o.e62�
0 . 0 3 3'
0.0 109
· 0 . 0076
I J
.
0 . OS .9
1 2 9 00 0
JO
8 0 7 7 00 186�00
l'
7 7 7�00
3 3
0 . 990
0.0319
0 . 1 ·S 7 1
0 . 1 202
0 . 0'"0
0 . 0 7) 7
7 1 8 JOO
)0
0 . 990
0 . 0320
O . I S 70
0 . 1 20 I
0 . 0919
0. 0736
O . OHO
O. I OlS
0 . 06 2 S
0 . 0 3 3'
0 . 0 1 01
· 0 . 00 7 6
0 . 990
0 . 0) 1 7
O. ISTJ
0 . 1 20'
O . 0 7 3t
O . OS 7 ]
O. 10)1
0. 0627
0 . Oll6
0.01 10
· 0 . 00'
0 . 032'
O . I S66
O. I 1 9 7
0 . 0"
2
0 . 990
0. 0732
0 . OS66
0. 1032
0 . 06 2 2
O . Ol l l
0.0106
· 0 . 00 7 9
0 . I S93
0 . 1 22'
o . o )S 0 . 0' 6 3
0 . 0760
O. os,"
O . I OSO
0 . 0'"0
O . OlSO
0. 0 1 2.
· 0 . 0060
0 . 06 • •
18
,)
1 9
O . OJ')
O . I �.9
.
700000
2'
68 1 600
1 8
6 3 2000
I�
0. 977
0 . 0296
. , 6200
1 2
0 . 92 7
0 . 029 1
O . I S'9
0 . 1 2 )0
0 . 0968
0 . 076S
O . OS99
O. I DS .
087000
O. 1 2 6 '
0 . 1 002
0 . 0799
0 . 06 3 )
O. 1077
0.8"
O. O
HI
0. 1633
1 1
o.e ..
0 . 0160
. 7 � 1 00
I�
0 . 1 6)2
0 . 1 263
0 . 1 00 I
0 . 0798
0 . 06 3 2
). 3600
I �
0 . 680
0 . 02 2 1
O . 1 68 2
O. I JI J
O . I OS I
0 . 0a.1
] ] �OOO
12
0 . 68.
0 . 02 1 9
0 . 1 6SS
O. 1 3 0 6
O . I DS'
O . OIS I
I J
1 9
.
O . OlH
0 . 0 1 27
· 0 . 00 S 7
0 . 0667
0 . 0 376
O . O I SO
· 0 . 00 3 '
O. 1076
0 . 0666
0 . 017S
O.OIU
· O . OOlS
0 . 0682
0 . 1 1 09
0 . 0699
0 . 0'09
0.0113
· 0 . 0 00 1
O . 068�
0. 1 1 1 1
0 . 070 I
0 . 0"
O . O l lS
0 . 0000
0
451
Line I mped a nce Tables
Table B . 1 6 Capacitive Reactance o f ACSR Bundled Conductors at 60 H z with 1 , 2 , and 3 Conductors per Bundle
S I .Glf
01&, II .
C OO l
(0- 0 .
(II , , ,,
( (i O I. .. . lII l l ( S
' . O I U S
�
r-:-- 2 C O I O li C l O I S P t ( H I (i ( I I ) � � 3 ( 0 I O " C I 0 �' S P A ( I ' (i ( . 1 ' 2 9 I� ' 1 0 . 0 6"4 ]_ 1 0 . 0 ) 1 ' G . O' Il l 6 O . OI ttH l �)_ ' 0��1 7� S_ l I 0A . 0"�� 0 01 2 " 0.0 " 0 0 . 028 . 361 60
S' IAlOS
HI
C Io , a C I T J V (
I ( Io C T " , C !
•
_
_
o
o ...
1II
I I
f OI
I
J OO '
•
J-
-
-
,.
] 1 01000
6 2 ( 11
I'
1 . ... 0 0
0 . 06 7 ,
2 2 hOOO
HI6
1 11
' . 120
0 . 06')
0 . 0 ) ... 1
P,Oll)
Q . 01h
0 , 0)61
0 , 0281
0 . 02 ) 1
0,0111
0.01\1
'.t ' / II
19
1 . 7 ... 0
O. 0 7 1 7
'n
0 . 0389
I " ' '' 0 0 0
0 , 0 ... "
0 . 0 ", ) 1
O . Oll.
0. 0311
0
0 )16
0.0)'6
O . O]S,
0 . 01 1 \
0,0"
( _'UOlO
1 1 ' '' 0 0 0
\ 0 / ..
1 . 6 00
0 . 08 0 ]
o . O\OS
0 . 0 .. .. .
0 . Oh1
0 . 0 11 0 \
O. O)l�
0 , 02 6 '
0. 011"
0 . 0 ' 1'
2 1 6 1000
72
I . 7)7
0 . 0.02
o . O U IT
• IWI
I'
O . O �H
2 1 \ 6000
(MU'"
, 1 1 0'0 0 0
' It
I .
, AleOI
1 \'0000
\ ..
I' ••
0 . 0 )9 1 0 . 0l9� 0 . 0 " 0 0;
l "W ' " ' ' .. . . 0 1 IU,",tt"
1 \90000
1 \ I D 'J O O
'4 "
1 \ 1 0 ., 0 0
It \
( I'nor , ( " "' 3 1 0 ( l'IIOfO IlU f l "
O
'lhU ,O IO l t " MUT•• D I "('
. ..
\ 11
, ,, ] 1 00 0
\ "1
II' )
1 000
.. \
I H I 'J O O
!;, 16
C UO , Ull .ll l
(&11"
CUUT 'UOD'
0
0.01"
I "
O . O"S"
0 . 0" "
0. 0378
O.OHI
It
I . ... 0 6
O . 01"
O . O S , ...
O . O ll S )
0 . 0 11 1 1
0 . 0311
0 . 01 / 9
O . O�
I'
o . o ,, ' p
O . O" , S
0. 0)82
I' 7
0. OH 1
0.0191
0 . 0 1 '" 0.01'"
I
t.
O . O H I)
0 . 0" I .
O.Olll
0 . 0" ' 9
2
0 . 0/ 7 7
0 . 0 38 6
0. 0lS9
0 . 0" 1 6
0 . 0 J l6
0. 0/19
O . O/l�
0 . 0 11 1 9
0 . 0386
0 . OJS9
0 . 0" 1 6
0 . Oll6
0 . 0/19
O .O/l�
0 . 0 1 9'
0 11 6 £
0 . 0" 2 3
0 . 0)90
O . 036J
0 . O il 1 9
0.0 lI9
0 . 011/
0 . 0/11
0 . 010/
0 . 0 190
1 . " '\
0 . 01 10
O.O� ••
0 . 0 ,, \ 1
O . O" I �
0 . 0 )'
1 . "27
0 . 0' 3 7
0 . 0 0; 2 2
a . 0 1t 6 .
" , 1&
0 . 01 3 1
0 . 0"61
0 . 0 ' '' '
0 . 0\ 2 2
0
I'
1
1 . 382
0 . 0 1 '"
O . O�IO
0 . 0 11 6 6
0 . 0"2 )
0. 0163
0 . 0 .. 1 9
0 . 0 1 19
0 . 0/1 /
0 . 0 / 11
0 . 0/0/
I . ) .. ...
O . OH S
O. 0�10
0 . 0 11 1 0
0 . 0"2 7
0 . 0 .. 2 2
0. 0)"2
O . 018S
0 . 0 2 11 1
O . O/O�
O . OIH
O . O�l/
0 . 0" ' 1
0. 0"29
O . 0 )9.
0. 0361
I . )))
0 . 0 3 'Ho
0. 0169
0 . 0 ,, 2 )
0.Oh3
0 . 0/16
0 . 01' /
0 . 0106
0 . 0 8 6 '"
O . O S ] I)
0 . 0 11 1 0;
0 . 0" 1 2
0 . 0 199
O . 0) 11
0 . 0 ,, 2 1)
o . O JII S
O . OU'
0 .02""
0 . 0/01
26
1.
.
11\
I . ]0 2
16
0 . 0 1) 2 6
1 . 293
0 . 01 '
7
0 . 0\ )6
0 . 0" "
O . O" l l
0 . 0"00
0.0313
O . 0"
0 . O l lt 6
0 . 0 /1 9
o . o/.�
0 . 0109
I . H' I • , .. 6 I .2I3 t • • 86
o 01 '\
0 . 0 1) ,, 0
o
0 ... 1 0
O . 0 11 ) 1
0 . 0 14 0 "
0 . 0371
0 . 0 "' 2 9
0 . 0 ) 1t 1
0 . 0292
0.on1
0 . 02 I I
0 . 08 7 8
0 . 0 \ 14 ]
0 . 0.1 '
0 . 0 11 3 9
0 . 0 .. 0 6
0 . 0)79
0 . 0 " 30
0 . 0 3 '"
0 . 0111
0 . 0/'1
0 . 01 1 2
O . OH I
0 . 0/ 1 ,
0 . 0116
o O S II 6
O . OUS
O . OU )
0 . 0" 1 0
0 . 0113
0 . 0' 1 /
0 . 01�/
0 . 019�
0 . 01 9 /
0 . 0 1) 11 '
0 . 0 11 1 9
0 . Ou 6
0 . 0"
I)
0 . 0 316
O . O ll H
o . on"
0 . 0/91
O . O/�l
O . 02 1 �
0 . 0190
0 . 0 \ 14 1
0 . 0 1ll "
0 . 0 1 114 50
0 . 0 1l 1 2
O . Oll�
0 . 0 " 3 11
0 . 0 1� 1
0 . 0/91
O . OH /
9�.000
, . 1 '\
0 . 0191
O . O�H
0 . 0 11 9 1
0 . 0 14 " 9
0 . 0 .. "
0 . 0119
0 . 0" 3 6
0 . OH6
0 . 0 /91
O . O/H
0.0119
,� , o o o
I .
, "0
0 . 09 0 "
o. O���
0 . 0 .. 9 S
0 . O ll S 2
0 . 0" 1 9
0 . 039/
0 . 0 .. ] 1
0 . 01�1
0 . 0 10 I
O. 02� 1
0.0//1
900000
I . 162 I . I) I
0 . 0 .. 1 6
0 . 0 11t
0 . 0" 3 1
O . OlH
0 . 0/99
O . O/S�
0 . 01 1 9
O . ° J9 )
0 . 0 ,, ) 9
0 . 0)S9
0 . 010/
O . O/H
0 . 0//2
JI O . A '''' . O.OU)
O . O lH
0 . 0 10 I
0 . 0/�1
0 . 0 2' 2 '
0 . 0161
0 . 0 10 "
0 . 0260
0 . 0 1 2 11
1 . 1 96
9 \ .. 0 0 0
'00000
... . l l . . O
1t�OOO
10
DU.(
1 9�000
Z6
C O l001
0.0
0 . 0/1/
0 . 0 1 '6
o 0 1)
1 9
1 0 1 HOO
0 . 0/1/
0 . 0122
,, "
1 0 ) )\00
0 . 0 1 l0
0 . 0/16
I . .,. 0 2
1 1 91\00
OlU
O . Op"
O.Olll
7
19
1 &1 ' C. ( '
O.OllI
0 . 0 " 1 II
O . 0 3 1 ,.
"
g"
0 . 0" "
O . OHS
0 . 0 11 0 1
""
It " 'J .. It "
0 . 0230
O . O ll S O
'l it !:I II
0 . 0 2 1 \1
O . O� 1 0
'l it
I I )000
0.0))1
0 . 01
1 2 7 2 000
I
'. 0111
I . \ 16 \
1 1 'HaD
1 0 ) 1 '.1 0 0
0 . 012 7
I'
O . o lit I
1 2 7 1000
1 1 1 )000
O . OH I
0. 0)27
0 . 0168
f l(ll
tiC.
0 . 0111
O.OH I
0 . 0 .. 0 . O . a ll I I
. OU1
O . O H II
0 . 0 11 0 2
'''ti l (
Ilur J" (UI L I .
0 . 0 1 71
0 . 0//'
o . O J 29
0
O . O Il U
' N ( i s .'' f
f
0 . 02 "
0 . 0 ' 7,
0 . 01 1 "
0 . 0161
0 . 03St
0 . 0 11 ) 0
'4 \
lul l l l G
O . OHO a . 02S'
0 . 0 190
0 11 9 0
O . O � O ..
I H I I!I O O
I'
0. 0 ) 1 1
0 . 0 .. ) 2
o
0 . 01 0 1
1 . 1& "
,
0 . 0 ) I I)
0 . 0 11 9 1
O.011\
' 16
0.0".)
0 . 01/�
0.0779
76'
1 . 60 1
7
.
1 9 �000
I'
�.
(UUOO
7 9�OOO
/.
"n
7 ' !a Q O O
.�
0 , 01 9 '
O . O��/
0 . 0"'2
0 . On9
0 . 0906
O . D \ !t 6
0 . 0"16
0 . 0 " 0; )
O . 0 1t 2 0
0'0"
OSS\
0 . 0 14 ' 0;
0 . 0 " 0; 2
0 . 0 11 . ,
09 1 2
0��9
0 . 0 14 "
O . 0 14 50 6
0 . 0" 2 3
O . Olt6
O . O"lS
0 . 0III
I . ' .. a I . l oa
0. Olt2
0 . 0 ..
0 . 0 16/
0 . 01 1 6
o. 0/2�
O . O !lo O I
O . OIt s .
a . 0 50 0 I
0 . 0 " 50 '
O . o" H
0 . 0 1 11
0 . 0 . .. )
0 . 0 16 /
0 . 010'
0 . 026 1
O . O/I�
O . osos
O . 0" 6 2
0 . 0 11 . 9
0 . 0 .. 0 2
O . OlH
0 . 0 10 1
0 . 026'
0 . 0121
O. O S6,
o . oso.
0 . 0" "
0 . 0 .. 3 3
0 . 0 .. 0 50
o . o .. .. s
0 . 0""8
0. 0)61
O. O l i O
0 . 0/66
0 . 0 2 10
0 . OS 6 1
0 . 0�01
0 . 0 .. 6 0
0 . 0" 2 7
0 . 0"00
0 . 0 " " ..
0 .0361
0 . 0101
0 . 016/
0 . 0126
0 . 0 11 0 "
0 . 0 "' "
0 . 0166
0 . 0 10 9
O . O/B
0 . 0129
O . OS' I
1 . 09)
09 1 6
1 . 0'2
O.091 7
0 . OS 6 1
1 . 06 3
0 . 0 9 ' '''
O . O !o 6 S
, . '"0
O . 09) I
ai '
0 . 09/0
0 . 010�
0 . 026 1
c oo ,
7 9 �OOO
16
I l O. ' U "
7 . \ ';, 0 0
10
7 1 \\00
Z6 /.
1 . 0\ ,
0 . 09 2 1
0 . 0\67
0 . OS07
O . O U 14
0 . 0" ) I
I .OU
0 . 09 1 1
O . OS"
O . os o ,
0 . 0.66
0 . 0 14 3 3
0 . 0"06
0 . 0 .. .. .
0 . 0 161
0.0)1
26
1 .0
p,
0 . 09 1 9
0 . DH 2
o . OS ' 2
0. 0"6'
0 . 0 ..
0 . 0"0'
O . O"SO
0 . 0 ) 70
0.03
0 . 0269
O . O/ l l
666600
/.
1 . 000
0 . 0 ' 14 3
O . OS h
O . O S I 14
0 . 0 ,, 7
)6
I
0 . 026'
6 6 6600
O . O ll ] .
0 . 0" I I
o . O"S I
0.0)1 1
0.011"
0 . 0 / 10
0 . 0 2 1' 0 . 02 39
S U ' l l l e. St I I I GU'l l , l l. I I G O f
GI £ T
7 . ... ... 0 0
6� lIOO
'I
6 3 6000 6 1 6000 60�OOO
O . O ' �1
0.031'
0 . 0/ 1 �
0 . 09 J 1
0 . OS 7 1
0 . 0 50
O . 0" )6
0 . 0 .. 0 1
O . a li s o
O . O J6I
0.03 ' 2
0 . 0/61
0.0211
0 . 0\ 76
II
O . 0 111 6 '
0 . 0 9 11 6
O .O �1 6
0 . 0 1& 7 )
o . Ouo
0 . 0 .. ' )
O . O.� /
0.011/
0 .01 1 �
0 .027 1
1. ,I
1
0.971
O . OHO
O . O� 1 1
O.O�18
O , O .. ' S
0 . 0 '111 2
0 . 0 .. .
s
O . O il S "
0.OH3
O.OJ 1 7
0 . 021/
o . o/ n
10
19
16
1 .0
O . OH I
0 . 0""\
0 . 0236
O. 0�1 l
0 . 0\ 1 l
0 . 0" 1 '
0 . 0 .. ,, 1
O . 0"20
O . O'�I
0. 0177
0. 01/0
C . 0211
0 . 0200
0 . 0 9 6 11
O . OH�
O . O� H
0 . 0 11 ' 2
0 . 0 .... 9
0 . 0" 1 1
O . O ll S 9
0 . 0 111
0.01/1
0 . 0/ 1 1
0 . 0/"
O. o,. � O . 09� 3
0 . 0� 7 S
0 . 0 50 I �
O . 0"'2
o. o"lt
0 . 0. . 1 2
0 . 0 .. � 2
0 . 0)72
0.031 S
0 . 02 7 1
O . 02 n
0 , 0 1) 7 '
O . O� I t
0 . 0' P 1
0 . 0" " 3
0 . 0" 1 6
O . O" S S
0 . 0 ) 7 0;
O.Olll
0 . 02
O. OSI 1
0 . OS 2 1
O . O. "S
0 . 0"S6
0 . 0176
0.01"
0 . 02 1 �
0 . 0211
O . O, H
O . O il • •
lit
0 . 021.
O . o,� 1
O.027�
0 . 0 1 1'
60\000
/.
60S000
/.
O. 'S)
uca r
... ... ' ... 0 0
10
O . ,\)
0 . 0 50 1 '
O . OH I
O.
DOW(
... ... ' ... 00
Z6
O . Ot'�
0 . OS 1 6
O . O�H
0 . 0"' 3
''''' H I OS'" '
... ... 6 \ 0 0
/.
O . '21 O.tI"
0 . 0" 0
O . OS J 7
O . O" I S
0 . hS2
... ... ' ... 0 0
O. "
9
0 . 0' 1 1
o . O� I I
0 . 0 .. ,
.. 1 1 00 0
'I
O . OHI
10
7
0 . 11 3
O . OU O
OBl
O . O\ l l
,
7 7000
Z6
1
O
0 . 0111
os,,
O . OS H
" 7 7 000
h II
1
. •� .
0 . ' ' ''
0 . 0" /
o . nn
O . O� l I
O.I'"
0 . 1 0 0 11
0 . 060S
o . OS"S
0 . 1007
H I CI ( ' '(L ICU
0.0316
0 . 09 6 1
S OU "
lit
0 . O ll S 6
O . , .. 0
'( ICOC I
7
0 . 0"
0 . 9 10
o . " .. 0 . 966
-I' ft l .,
I
I'
6 16000 D
,I
0 . 02 1
O . 'H
'001
T
0 . 0 ,, 7 ,
3
0 . 990
16
T ( Il
O . OS I '
t
I
7
10
6 )6000
Sw . '
I .
I'
6 ) 6000
" o $ lr a. I . IGI"
I'
O . O\U
� . 0 .. " a""
O.O.'I
0 . 0 11 50 6
0 . 0)16
0.031'
0 . 0 11 2 )
0 . 0 ,, \ .
0. 0)7,
0 . 01l/
0 . 01 1 1
0 . 0202
0 . 0 14 6 0
0 . 0 11 0
0.01l1
0 . 0219
0 . 020 1
O . 0 14 S 1
O . o .. u
O . O U II
0 . 0 1 1'1
0. 01l'
0 . 011l
0 . 02 0 1
0 . 0"'0
O . O ll V
0 . 0 ", ) 0
0 . 0 1t 6 "
0 . 0 11 l
0.OU1
O .GZlI
0 . 020'
0 . 0 .. . .
O . O" i '
0 . 0 11 ) .
0 . 0 ,, ' 7
0 . 0 11 6
0 . 0 121
0 . 01 1 S
0 . 010'
0 . 0 11 "
0 . 0 ,, 6 3
0 . 0 .. ) 6
0 . 0 ""
1
0 . 0 31 1
o .on
I
0 . 0217
0 . 0\0/
0 . 0 1l 6 t
0 . 0 .... 2
0 . 0" 7 2
0 . 0 111
o . oln
0 . 0 29 1
0 . 0 1611 50
O . O .. S O
0 . 0 .. ) 0
l AI.
1 1 1 \00
1
0 . 106
0 . 0\01
0 . 021 1
I I' S
111\00
26
1
0 . 7'3
0. 10 I �
0 , 06 1 1
a s . ..
O . OU I O . OU'
o. O��O
o. 0�01
0 . 0",\
O . 0""1
0 . 0 111 "
o . o ln
O . OllS
) 9 7 \00
h ,.
7
0 . 7 72
0 . 1 0 20
0 , 06 ' )
O . O�H
O . os 1 0
0.OU1
O . O .. � o
0 . 0 .. 7 1
0 . 0 11 1
0 . 0 )11 0
0 . 0 211
o . o no
0 . 0110
0 . 1 111 )
0. 103 1
0 , 06 ' 1
O . OH I
0 . 0\ "
0 . 0.1/
0 . 0 .. \ \
0 . 0" 1 1
0 . 0 .. 0 .
0 . 03""
0 . 0 100
0. 0263
10
7
O.
0. 1 012
0 . 01 1 ,
O . OH '
0 . 0\ 1 '
0 . 0"1)
0 . 0.0
0 . 0 100
0. 12 1
0 . 1 0 11 0
O . UIl
0 . 0\61
o . n /o
0 . 0" , 7
0 . 0 166 0
0 . 0" "
I
0 . 0'",
1
O . O .. U
0 . 0" 1 '
16
H,
O . 0 ,, 0 11
o 03.'
0 . 0 10 1
0 . 0111
I OH
0 . 06 3 1
0 . 0\ 1 0
O . O\ l a
0 . 0 .. "
0 . 0' "
0 . 0.0'
o.on/
0 . 0 10 '
0. 0111
0 , 06 3 ,
0 . 0 50 7 1
O . O\ / t
O . a ll , s
0 . 0 .. . .
0 . 0 111 ' 0
0 . 0"0'
o . on /
0 . 0 10 1
0 . 011/
IIUT e l ' ( U OH O' I OU
ll" IT
"UL ' I O S l t l ( 1I
."
000
, , ' \ 00
UhOO
n ' .. o o u , .. o o 100000
10
II
16
O. "
,
0 . 610
O.
O. '0\7
0 . 0101
0.
0 . 0"70
0 . 0 14 "
0 . 0 " .. )
0 . 0 .. 7 )
0 . 0111
o . OIn
o . ons o . onl
0 . 02"
4 52
Appendix B
Table B.17. Capacitive Reactance of ACSR Bundled Conductors at 60 Hz with 4 and 6 Conductors per Bundle
E o
Al I I
COOl
tN ' l
2 . �00
U 9. 0 0 0
16/6
.,
2. UO
19
' 2 7�000
ut. �O ' 0
"
• • 7�0 I . iDa
"".OU
. . "000
. . ."
.. 0 1 0
2 1 6 7 000
IlUI I . RO
2 1 � 1 00 0
..
tOUI "
l 7 eGOOO
' H CO.
lAPW I . G
••.
.9
1 1 OeGOO
( 1,
01 &.
HI'
! I'UOIO ( I'"OIO
5 "liDS H 5'
0,
6
CAPac l t ' r!
. " --;:---, r:-' 5 �� I I . ' O . OOH o.O" S 0 . 0092
I(OC''"((
• - t O l Ou t t Ot S " t ' O G ( ' 9
'2
• •
(" ' &
_
_
M' GOOM- M . U S 6
-
•
_
0 . 0296
0 . 020'
0.0"2
0 . 0302
0 . 02 ' 2
0.0'"
0 . 0098
O . O OS '
O. Olll
0 . 02H
0 . 0 1 69
0 . 0 1 1 9
0 . 007'
' 0'
I
f OO t
to.Ou t l Ot
1 l0 ' US
'''C I .G
' 2
(
.�
�
••. ,
- O . OO U
II
0 . 00"
O . OO U
0 . 0 1 98
0 . 0098
0 . 00 2 7
- 0 . 0028
- 0 . 00 7 ]
0 . 02 ' 2
0.01 1 2
0 . 00"
- 0 . 00 .
- o . oon
.
- 0 . 00 7 7
O . O U'
0 . OU9
0 . 0 1 75
0 . 0 1 26
O . oo . �
0 . 02 1 7
0.01 16
0 . 00'5
- 0 . 00 1 0
- O . OOH
0 . 0323
0 . Ol l 3
0 . 0 ' 69
0.01 1'
0 . 00 7 9
0.02 1 ]
0.01 1 2
0 . 00"
- 0 . 00 "
- o . oo n
'9
• . 717
1 . 762
0 . 0 32 2
0 . Ol l 2
0 . 0 1 6'
0.01 1'
0 . 0011
0.02 1 2
0.01 1 2
0 . 00"
- 0 . 00 1 5
- 0 . 0060
,.
'9
'
0. 0329
0 . 02 l t
0 . 0 . 75
0 . 0 1 2\
0 . 0015
0 . 02 . 7
0.0 1 1 6
0 . 00'5
- 0 . 00 1 0
- O . OO H
1 � 90000
�'
"
••0 2
" So�
0. 0332
0. 02'2
0 . 0 1 7'
O . O I lI
0 . 0011
'0 . 0 2 1 '
0. 0 ' "
0 . 00 ' 7
- 0 . 0001
- O . OOU
' � 90000
.�
7
1 . �02
o . on.
0.0/"
0 . 0 1 10
0 . 0 ' 30
O . OOqO
0. 0220
0 . 0 1 20
0 . 00"
- 0 . 00 0 7
- O . OO U
72
'&UO'
. � . O SOO
S.
.,
. . S06
0. 033' .
0 . 02"
0 . 0 1 10
0 . 0 1 ]0
0 . 0019
O . OUO
0.0 ' "
0 . 00 0 '
- 0 . 000 7
- O . OO U
'U'",'C"
. s . OS O O
.s
7
. . 016
0 . On6
0 . 02'6
0.0112
0.0'32
0 . 00 9 1
0 . 022 1
0.0121
O . OOSO
- 0 . 0006
- 0 . 00 5 '
. . .n
0 . OH6
0. 02'6
0 . 0 112
0 . 0 ' 32
0 . 00 9 1
O.OU I
0. 0 1 2 1
0 . 0050
- 0 . 0006
- 0 . 00 \ 1
. . 027
o . on,
0 . 02"
0.01"
0 . 0 1 )'
0 . 0093
0 . 0222
O.OIU
O . OO S '
- 0 . 00 0 0
- 0 . 00"
1 . '2'
O . OHI
0 . 02"
0 . 0 . ..
O . O ' lO
0 . 0 0"
0 . 0222
O . O I U
0 . 00 5 1
- 0 . 000 '
- 0 . 000'
.S
19
I . US
0 . 0300
O. OlSO
0.0 '86
0 . 0 1 ]6
0 . 0096
O. OU'
0 . 0 1 2'
O . OO H
- 0 . 000 l
- 0 . 00"
1 2 1 2 000
S.
19
1 . 31 2
0 . 03.0
0 . 0 ' ]6
O . 0096
0 . 00 S 3
- 0 0003
- 0 . 00 0 '
1 . 3'S
0 . 0 ) 11 1
O . 02S 2
0.0'"
0.0 I
0 . 00 9 8
o . 02�s
O . O I l !..
0 . 00 � .
- 0 . 00 0 '
- 0 . 00"
• . 1l3 • . 307
0. 0)'3
O. OlS)
0 . 0 . 11
II
O . a l i II
0 . 0 1 2'
'S
O. 0250
O.O'U
1 2 7 2000
0 . 0 ' 19
O. 0098
0 . 0226
0 . 0 1 2\
0 . 00 5 '
- 0 . 000 1
- 0 . 00 "
'lO".
" ' 1 000
S.
IUO U "
. . J I OOO
oS
.. . . " .
I ] $ I S OO
S.
O "' ! I
I ] $ I S OO
'" ! A SlI ' In
GR A C l l I
1 1 92�00
S.
" "
I U . ' ' '' G
"
9 2 S 00
'S
J ' ICH
"
"000
So
IlU! JAY CUIUW 'OlU HUGI R
l onsoo
• • • JOOO
S.
' 0 ] ]$00
'S
01
CI.D 'I Al
• OBSOO
19
19
19
'S
. . 293
0.0"
0.0'"
O . o u ,
0 . 0 1 93
0 . 0 ' 0 1
0.0 '03
o . a 1 9.
O . O H I I&
0.0'0]
O . 02 2 9 0. 0229
0 . 0 1 29
0 . 00 5 '
0 . 0003
- 0 . 00 ' 3
O . a 1 96
O.
aI••
O . O ' OS
O . O l lO
0 . 0 1 30
0 . 00 \ 9
0 . 000.
- 0 . 00"
• . '86
0.0'97
0.0'"
0.0 107
0 . 0231
O. a 1 3 '
0 . 0060
0 . 0005
-0. 0 0. 0
o.a .97
0 . 0 1 11 1
O. a '07
O.OUI
0 . 0 1] 1
0 . 0060
O . OOOS
- 0 . 00"
O . 026 3
0 . 0 ' 99
0.0'"
0.0 101
0 . Oll2
0 . 0 1 32
0 . 006 1
0 . 00 0 6
- 0 . 0019
0 . 02"
O . 0200
G.OI S I
0.01 10
0.02H
0 . 0 ' 33
0 . 0062
0 . 000 7
- 0 . 00 3 1
0 . 0263
0 . 0 ' 99
0 . 0 1 11 9
0 . 0 ' 0'
0 . 02 3 2
0 . 0 1 32
0 . 00 6 1
O . 0006
- O . OO l t
O . 020 .
0.0 I� I
0.0
II t
0 . 0l ] '
O.
a 1 3'
0 . 00 6 2
O . 000 7
- O . OO l l
0 . 03\)
900000
S.
• . 162 t. I II
O. 03S�
26
tOOOOR c u e ' 00 ' f 'l
�.
- 0 . 00. )
0 . 026 '
I . ' "0
7 9 S 000
0 . 000 2
O. 026 '
1 . 1 6S
7 9�000
0 . 00 S 7
0. 03S2
]6
0' ' ' 1
0 . 0 ' 28
o.onl
'S
"
- O . OO.s - 0 . 00. .
0 . 0260
' S < OO O
]0
0 . 0000 0 . 00 0 1
0. 0257
9$0000
oS
O . OOSS 0 . 00 S 6
0 . 02S'
II I l
7 ' � 000
0 . 0 1 27 0.0. 27
O. 03.7
U t l l ID
900000
O. 0227 0.OU7
0 . 03"
1 . 1 96
JUOOY
0 . 0 1 00 0.0'01
o . o no
S.
Mal l "'0
0
l . lS'
9$'000
( , . .. . ,
1t61
O. OlS' 0 . 02H
• . 2.6
. . 2 . 3
]6
0. 0
0 . 03'S O . 03'S
O. 0353
o . ons
O. 026S
0 . 035\
0 . 02"
0 . 0200
O.OIS'
0.0 "
0 . 02 J ]
0 . 0 1 J]
0 . 00 6 7
0 . 000 7
I . ' 0'
0.on7
O . 0266
0 . 0202
O.a'\3
0.0 1 12
0.02U
0 . 0 1 35
0 . 00 6 3
0 . 000'
- 0 . 00 3 7
1 . 09 3
o . 035'
0 . 02 6 7
O . 020]
0 . 0 1 5.
O.0 "
3
0 . 0216
0 . 0 1 1\
0 . 006'
0 . 0009
- 0 . 0036
1 . ' "0
0
- O . OO l l
7 9 S 00 0
2.
' . 092
o . on'
0. 026 7
O . 020)
G . O I !. "
O.0 "
)
0 . 02 l 1
0 . 0 1 ]$
0 . 00"
O . ooot
- 0 . 0036
7 ' SOOO
.S
. . 06]
o . 0 360
0 . 0169
o. a20S
o. O.�6
0.01 15
O. Oll7
O. a 1 ]7
O. 0066
0 . 00 ' 0
- 0 . 00 3 5
COO,
7 9 S 000
]6
1 1 0_ ' "
7 . SSOO
30
S" "
l ' IG
7 1
$1 1 l
T
"
1 . 0 .. 0
0 . O l6 '
0 . 027 1
0. 0207
0.0'$7
0.01 1 7
0. 0231
0 . 0 ' 38
0 . 00 6 7
0 . 00 1 '
-0. 003'
•
O . 0 3se
0 . 0268
O . 020.
O.O .S\
O . OllS
0 . 0 ' 36
0 . 00 6 5
0 . 00 1 0
- 0 . 00 3 6
0 . 010'
O . O l l li
0. 0 1 37
•
D• •
� !t O O I SSOO
26
0. 0237
0 . 00 6 6
O . 00 I I
- 0 . 00 3 '
O . 0362
0. 027 1
0 . 0107
G.O'S7
0.01 16
. . 03'
o. 016 1
0 . 0770
2'
o.a.SI
0. 0 1 1 1
O . 0 2 11
0 . 0 1 JI
0 . 00 6 7
0 . 00 1 2
- O . OO H
T f L IN t " O
.... 00
26
I . 0 ' 16
O. 0361
0. 0273
0 . 0 20'
o. o.s'
0.0 1 "
0. 0219
0 . 0 1 Jt
0 . 00 6 8
0 . 00 1 3
- 0 . 00 3 2
6 6 1 600
2'
' . 000
0 . 0360
0. 0270
O. 0 2 ' a
o . a 1 10
0 . 0 1 20
0 . 01.0
0 . 0 .. 0
O . 0069
0 . 00 1 3
- 0 . 00 3 2
0 . 9S 3
0 . 0 ] 6'
0. 0271
0 . 02 "
0 . 0 1 6 11
O.a '2]
0. 02'2
0.0"2
0 . 00 7 1
0 . 00 1 6
- 0 . 00 2 '
!GU '
6 1 6000
30
I .0 I'
0. 0363
O. 0273
0 . 020'
O . O I S'
0 . 0 1 1'
0. 0239
0 . 0 1 Jt
0 . 00 6 8
0 . 00 1 2
- 0 . 00 3 3
'IOSI( ..
0 . 990
0 . 0365
G'.'(
7
6 S 1 900
••
6 1 6000
26
ROO_
6 1 6 000
1 ' 1'1 ' 1 0
6 ] 6000
2' •• ]6
SWI J '
U 6000
tr'l
6 O S000
30
SOu • • PI .tOO
6 0 S000
26
6 0 S 000
2'
I . O !lo 1
19
19
0 . 00 6 9
0 . 00 "
- 0 . 00 3 1
0 . 02'0
0 . 0 "' 0
0. 977
0 . 0 366
O. 0276
0.02 ' 2
0.0 162
0 . 0 ' 22
0. 02"
0 . 0 I I. .
0 . 00 70
0 . 00 1 \
- 0 . 00 3 1
0 . 9.0
O . 0369
0 . 02 7'
O . 02 ' S O . Ol ' �
o.o.n
O . a 1 2'
0.020]
0.0'03
0 . 00 7 2
0 . 00 1 6
- 0 . 0029
O. 027S
0.02"
0.0161
O. a '2'
O . Ol"
0.0..3
0 . 0072
0 . 00 1 7
- 0 . 00 2 8
0 . 0 . .0
0 . 00 6 9
0 . 00 "
- 0 . 00 3 1
0. 0202
0.0'"
0 . 00 70
0 . 00 1 \
- 0 . 00 3 0
0. 0202
0 . 0 . .2
0 . 00 7 '
0 . 00 1 6
-0. 002'
- o . oon
0 . 930
0 . 0370
0 . 0 2 7'
0 . 0 1 66
o. a ' l�
0 . 99.
O . O)6S
0.0270
O . Ol i O
0.0'61
0 . 0 1 20
0 . 0200
0 . 966
O . 0367
0 . 02 7 7
0 . 02 ' )
0.0163
O . O ' ll
0 . 9S 3
0 . 0361
O . 027'
0 . 02 "
O . a 1 6'
0.0 '23
S
"Gl!
��6S00
]0
0 . IS3
0 . 0 ]6 1
O . 02 7 '
0 . 02 "
0 . 0 1 6'
0 . 0 "2
0 . 00 7 1
0 . 00 1 6
� S 6 S00
26
0 . 927
0 . 0 3 70
0 . 02'0
0 . 02 ' 6
O . a 1 16
o.a 123
0 . 02'2
OO V !
0 . 0 1 2\
0 . 01"
0.0 . .3
0 . 00 7 2
0 . 00 1 7
0 . 0 ' 26
0 . 02..
0.0'"
O . 00 7 3
0 . 00 1 8
- 0 . 00 2 7
- 0 . 00 2 8
''' ' ' H '
S 5 6S00
2'
O. ".
0.0371
0 . 0 28 '
O . 01 "
O.a 167
OsPRP
S S 6 S00
"
o
1 71
0. 037'
0 . 02"
0 . 0220
O . 0 . 70
O . O l lt
0. 02'6
0 . 0 1 '6
O . 0 07 S
0 . 00 2 0
-O
010
. 7 7000
30
0. 183
0.037'
0 . 02U
O . 02 "
O . a 1 70
0 . 0 1 29
0 . 02"
0 . 0 . .6
0 . 00 7 \
0 . 00 2 0
- 0 . 00 2 6
" I••
. 7 7000
26
0. 021S
0 . 02 2 '
0 . 0 1 72
0.01 3 '
0. 02.7
0.0"7
0 . 00 7 6
0 . 00 2 1
- 0 . 00 2 '
2'
o .•u
0. 0376
. 7 7000
•••6
0. 0377
0. 0286
0 . 0222
0.0.7]
0 . 0 1 32
0 . 02"
0.0'"
O . 00 7 7
0 . 00 2 2
- 0 . 00 2 3 - 0 . 0022
[l
J I I C1('
'
I
lAlU
C '"
O
a . 76
. OO l
. 7 7000
"
0. . . .
O . OliO
0 . 0289
0.02lS
O.
o . a ' 3\
0 . 0250
0 . 0 1 50
0 . 00 7 9
O . OOZ.
1 9 7 500
)0
0 . 80 6
O. OliO
0 . 0290
0 . 0226
0 . 0 1 76
0 . 0 1 36
0 . 025 1
0 . 0 1 \0
0 . 00 7 9
O . 002.
- 0 . 00 2 1
0 . 0292
0 . 022'
O . a "9
o . a 1 1'
0. 02U
O.
aIS2
0 . 00 1 1
O . OO l S
- 0 . 00 2 0 - 0 . 00 1 t
I I' S
19 7500
26
0 . 713
0 . OU2
. " ... T
J 9 7 S00
20
0. 772
O . O�I J
O. 021l
0. 022'
0 . 0 1 80
0 . 0 I Jt
0. 02\3
O.OIU
0 . 00 1 1
0 . 00 2 6
tO I tlOOI£
1 9 7 5 00
0. 70]
O . 0 1 16
O . 0296
O.Oll2
0.0'12
0 . 0 .. 2
O. OHS
O.
0 . 00 2 1
- 0 . 00 1 7
J J6000
30
0 . 7"
0 . 0386
0 . 0296
0.0232
0.0183
0.0"2
0 . 02H
a I S'
0 . 00 8 3
OR ' O U
••
0 . 0 1 �'
0 . 00 8 3
0 . 00 2 1
- 0 . 00 1 7
n 6000
26
0.721
0 . 0 31 9
0 . 0291
0. 021'
0 . 0 1 1S
0 . 0 . ..
O. 0256
0 . 0 1 56
O . OOIS
0 . 00 3 0
- 0 . 00 1 6
) ] 6 .00
••
0 . 61 '
a
0392
0. 0302
0 . 0 2 l1
0.0 111
0 . 0 . ..
0 . 02\'
O . O l st
0 . 00 1 7
' . oo n
- 0 . 00 1 3
0 . 610
0. 0]93
O. 030)
0 . 0219
O. O . U
0.0
• ••
0 . OlS9
0.0 159
0 . 0011
0 . 00 3 2
- 0 . 00 1 3
1. 1 • • ( T N(Il ' 1 05 f t 1 C "
] 00000
2'
4 53
Line I mped a n ce Tables
Table B.1S. Capacitive Reactance of ACAR Bundled Conductors at 60 Hz with 1, 2, and 3 Conductors per Bundle AlIA (Q. ! C - Ol eM i l
.----- 6 0
621
S I IGl ! COlO.
01 &.
S T RAIOS
II.
H/ 620 1
3000
72
1 9
1 . '2 I
2 1 7�000
63
21
1 . 82
)1
1 . 82 I
1 3
1 . 161
I'
1 . 162
1 3
I . 13�
19
I . 1)�
2"
2 1 11000 2 2 9 7000
�.
2 2 6 2000
.1
2 2 2 6000 2 2 2 7000
�.
2 1 93000
.1
2 1 �9000
I
1 . 762
1 . 1 3 !t
1 1 99000
�.
7
1 . 602
, ' 70000
.1
1 3
1 . 601
Hh l DOQ
11 2
1 9
1 . 601
1 6 7 3000
�o
1 60 7000
.8
1 3
1 6 2 2 000
01
I9
I I I 7000
�o
I . S O li I . 'J O ..
1 1 96000
.1
I . S O li
1 . 3 1i S
1 9
I . J 4 1)
1 3
1 . 30 2
1 9
1 . 302
1 . 301
1 2HOOO
30
1 2 1 1 0 00
I 1 7 9000
l' I.
1 1 6 3 000
30
1 1 3 3 000
10
IJ
1 . 2 1) 9
IB
19
1 . 2 1) 9
1 1 � 3000
33
1 . 2 li 6
I 1 38000 I 1 09000
30
1 . 2 11 6
I 1 0 1l 0 a o
1 080000
10
I.
I J 19
1 0 7 1000
30
7
1 0"9000 1 0 2 2000
l' 1 8
I )
1 9
I O�OOOO
30
1
1 0 2 3000
10
I J
99.0CO
I.
9 9 0 8 00
30
9�' 600
30
1 9
7
I J
969300
20
9� 8000
10
IJ
90 3 900
II
1 9
900300
30
7 9�000
30
8 7 7 3 00
10
1 3
I . 1�9
2
-
I�
0 . 0 392
0.01 1 2
O . OH�
0 . 02 1 1
O . O , ,�
0. 0327
O . 039�
0 . 03 I �
O . 02�1
0 . 02 1 0
0 . 0 "1
O . OBI
0 . 0 3 30
0 . 0397
0.03 I 7
0 . 0260
0 . 02 1 6
0 . 0 1 10
0 . 0)68
0 . 0]'
0. 032�
0 . 0268
0 . 022.
0.0181
0 . 0 3 18
O.OH I
O . 0 11 1
I
0.0331
0 . 021.
0 . 0230
O . O l go
0 . 0 \1 2 7
0 . 0390
0 . 0361
0. 0022
0. 0302
O . 028�
0 . 02 0 1
0 . 020�
0 . 00 3 1
0 . 0399
0.0312
O . OOH
O. 03'�
0 . 0288
0 . 02"
0 . 0201
0 . 0311
0. 0'29
0 . 03'.
0. 0292
0 . 0207
0.02 "
I II ( ' . .. ( I I . ' --:-:---l I. I
ROC TlOn
C O I OU C I O R S . A t I I G I� 1 2 9
o . 016�
O . 008�
O . O'B
0. 0383
0. 0309
O.03p
0. 011�
0 . 0'90
O. O. 30
0. 0387
O . O H'
0.0119
0. 0093
O . O. 32
0 . 0 3 90
0 . 0803
O . O�O.
0 . 0 11 " '"
O . 0'01
o . O ,, � '"
0 . 0821
0 . 08��
O. O� 30
0 . 086�
O. O� 3�
O. 08 1 �
O. 0�00
0 . 0818
0 . 00 70
' OR
M ( GOIM-M l l I S
I
3
-
9
6
"
I
. OOT
UO I US
COI OUC I OR S'&( l lG ( I I . '
II
0 . 0'80
0 . 00 3 1
O . 0 '1 8 1
0 . 009
0 . 0006
0 . 0319
0 . 0030
0 . 0309
0 . 0293
0 . 0101
0 . 02 1 2
0 . 0032
0 . 03�2
0 . 029S
O . OlS l
O . 02 1 �
O. 03� 3
0 . 029 1
O . O'�1
0.0216
1 . 2 11 6
I2
0 . 0886
0 . 0�06
0 . 0086
0 . 0'03
0.0 .. 0
0 . 0383
1 . 1 96
0 . 0890
0 . 0� 0 8
0 . 0088
O . Ooos
0.0"2
0 . 038S
0 . 0898
O. 0�S2
0 . 00 9 1
0 . 00.9
0.0"6
0 . 0389
0 . 0'16
0 . OH6
0 . 0299
0 . OlS5
0 . 02 1 9
0 . 090"
o . OS��
0 . 009�
0 . 00�1
0 . 00 1 9
0 . 0192
0 . 00 3 8
0 . OH8
0 . 010 1
0. 01S1
0. 021 1
O . OS�1
0 . 00 9 1
0 . 0009
0 . 00 1 6
0 . 0389
0 . 00 3 6
0 . 0 3�6
0 . 0 299
0 . 0255
0 . 02 1 9 0 . 0210
1 . 2
1 . 2 I 2 1 . 2
I2
I . 1 96 I . / 96 I . 1 6 1) I I I I
. . . .
' 'l ,
1 6 1j
0 . 089.
1 1) 8
0 . 0899
O . OO�O
0 . 0" 7
0 . 0 3 90
0 . 00 1 7
O . OH I
0 . 0300
0 . OlS6
0 . 0898
o . O�SI
0 . 0091
1 6 1)
o. 0��1
0 . 0. 9 1
0 . 00'9
0 . 00 1 6
0 . t lB 9
0 . 0' 36
0 . 03�6
0 . 02 99
0 . 02 � �
0 . 02 1 9
I . 1 08
0 . 09 1 1
0 . OS�9
0 . 0099
0 . 00S6
0 . 00 2 3
0 . 0396
0 . 0" " '
0 . 0 36 1
0 . 0300
0 . 0260
0 . 021'
1 . 0 11 2
0 . 09 3 I
o. 0�68
O. O�O.
0 . 006S
0 . 0032
O.
0. 0361
0.0]
0 . 0��9
0 . 0099
0 . 00�6
0 . 00 1 3
0 . 0396
O . O li li
I
0 . 0 36 1
I0
0 . 0266
0 . 09 1 1
Ooo�
0 . 0 '111 7
1 . 1 08
0 . 0300
0 . 0260
0 . 0110
0 . 0]66
0 . 0309
O . 026S
0 . 0229
I
0 . 0 36 ,
0 . 010.
0 . 0260
0. 0220
0 . 02 30
I . O !, !, I . 108
0 . 09 2 1
0. O�66
O . OS06
0 . 0 \1 6 3
0 . 0 11 3 0
0 . 0 11 0 1
0 . 0 " ", 6
0 . 09 1 1
0 . 0��9
0 . 00 9 9
0 . 0056
0 . 00 2 3
0 . 0 3 96
0 . 0 " 11
' - 069
0 . 0 92 3
O . O S tI "
0 . 0 '; 0 \1
0 . 0 \1 6 2
0 . 0 11 2 8
0 . 0 11 0 1
0 . 0""5
O . 036S
0 . 0 30 8
0 . 0160
0 . 0211
I . 063
O . 091�
0 . OS6S
O . OSO�
0 . 00 6 1
0 . 0019
0 . 0'01
O . O Il " S
0 . 0165
0 . 0 30 8
o.ono
0 . 0218
0 . 0 9 11 6
O. OS 16
o.O S1 6
O . 0 11 7 ]
0 . 0 "' ''' 0
n . 0 11 1 ]
O . 00� 2
0.0311
O.031�
0.021 1
O . OI B
O. 096�
O . 0�86
0 . 0� 1 �
0 . 00 8 1
O . OoSO
0 . 00 2 3
0 . 0.�9
0. 0319
0. 0311
0 . 02 1 8
0 . 02 0 1
O . 1 0 0 ..
0 . 060S
O . O�oS
O . OSOI
. 0069
0 . 0,"'
0 . 00 1 2
0 . 0392
0 . 0 3 35
0 . 029 1
0 . 02�0
I 0�6
0.06 3 1
0 . 0�10
0 . 0�28
. 009S
0 . 0068
0 . 0089
0 . 0009
0.OH2
0 . 0 30 1
0.0212
1 g�OOO
10 18
13
19�OOO
I.
,9
8 1 90 0 0
30
1
8 0 1 100
10
1 3
7 8 6 1;, 0 0 1 ] 7 � 00
I.
J3
.. '"
0 . 990
J 1 8 )00
10
1
0 . 990
100000
10
I ]
0 . 990
6 8 1 600
1 8
,9
6 3 2000 6 ' 6 2 00
I�
11 8 7 11 0 0
I �
0.81
II
" 7� lOO
1 1
0. 8 1
It
h )60Q
I S
0 . 6 8 '"
3 3 � 000
11
0 . 6 8 '1
1 1
1 2
�
CAPAC I T I ¥(
1 . 2 1& 6
8 '.> 11 1 0 0
19
Il
. 06 ]
. 06 3
0 . 990 0 . 92 1 .927
O .
4 54
Appendix B
Table B. 19. Capacitive Reactance of ACAR Bundled Conductors at 60 Hz with 4 and 6 Conductors per Bundle U[A 621 (Q. Al It
.
(M i l
S T RIUS £t/620 1
01&. II.
72
1 . 12 1
h i )000
2 ) 1 �000
6J
21
1 . 82 1
2 H'OOO
S.
J 7
2 2 9 7000
S'
• .62.
225 2000
••
2 2 26000
'2
2 7 2 7000
S.
"
O. O. IS
O . 001�
0 . 02 1 7
0.01 16
O . OOOS
- 0 . 00 1 0
- O . OO� �
0 . 02"
0 . 0 1 10
0 . 0 1 )0
0 . 0090
0 . 0220
0.01 1 9
0 . 0001
- 0 . 0007
- O . OOS!
0 . 03'2
O .02�2
0 . 0 1 81
0 . 0 ' )1
0 . 0091
0. 022S
0 . 0 1 2S
O . OOSO
- 0 . 00 0 1
- 0 . 0006
O.OhS
0 . 02 S '
0 . 0 1 90
0.0'"
0 . 0 ' 00
0 . 0 2: 7
0.0127
O . OOS�
0 . 0000
- O . Ooos
0. 0307
0.OH7
0 . 0 1 91
O.O,"J
0 . 0 ' OJ
0 . 0219
0 . 0 1 28
O. OO� 7
0 . 0002
- O . OO O J
0. 01"8
0 . OH8
0 . 0 . to
0 . 0 111114
0.0101
0. 0279
0 . 0 1 2'
0 . 00S8
0 . 0003
- 0 . 00 0 3
O . OlSO
0 . 0260
O . D . 96
0 . 0 "6
0 . 0 ' 06
0 . 0 2 30
0 . 0 1 30
O . OOH
0 . 0000
- 0 . 00"
O . Ols I
0 . 026 1
0.0197
0.0'07
0 . 02) I
0.0I I'
0 . 0060
O . OOOS
- 0 . 00 " ,
I . • 6S
0 . 03S3
0 . 0263
0 . 0 ' 99
0. 02J2
0 . 0 06 1
0 . 0006
- 0 . 00 3 9
0. 016'
0 . 0200
0 . 0 • •9
0 . 0 ' 08
O . Ols'
O.OI� 1
0.01 10
0. 02)3
0.0131
0 . 006 2
0 . 000 7
- 0 . 00 11
I . 1 6S
0 . OlS3
0. 0213
0 . 0 . 99
0 . 0 . 08
0 . 0232
0 . 0 1 12
0 . 00 6 .
0 . 0006
- 0 . 00 3 9
I . 1 !"J 8 I . 1 6S
0 . 03� 3
0 . 0263
0 . 0 1 99
0 . 0 . .9
0 . 0 1 ,, 9
0 . 0 1 09
0. 0211
0 . 0 1 12
0 . 00 6 1
0 . 0006
- 0 . 00 3 9
0. 03S3
0 . 026 3
0 . 0 1 99
0.01.9
0 . 0 '0'
0 . 02 ) 2
0 . 0, )2
0 . 00 6 1
0 . 0006
- 0 . 00 3 9 - O . OO P
'2
H
1 3
I . So .
I . SO.
1 6 2 2 00 0
'2
1 3 ) 7000
�.
1 2 9 6 00 0
•2
•9 7
1 . 302
2'
1 3
1 . 30 2
1 1 79000
"
1 9
1 . 302
I H lOOO
30 20
1 3
1 . 2H
IB
19
20
•
2 �9
I . 2�9
Jl
1 . 206
JO
1 . 206
I J
1 022000
I I
30
1 0 2 3000
2'
1 3
•I
19
9,.,00
30
9�0600
30
1 9
1 . 206
1 . 2 ' 2 1 . 2 1 2 I . 196
96UOO
20
1 1
9HOOO
2.
1 3
9. )900
"
, 9
900100
10
79�000
30
1 7 1 30 0
2' 20
' . 20 6
1 . 2 ' 2
I O�OOOO
79�OOO
I . 30S
I
1 0. 9 0 0 0
996000
I . SO.
1 . 3' S
30
1 3 1 3
0 .0330
O . O . 7S
1 . 60 2
1 0 7 7 000
0 . 0239
- O . OO S '
1 . 60 2
19
0. 0329
I'
- 0 . 00 . .
1 . 602
II
0 . 0 1 69
l S
0 . 00"
"
I I
. OU 3
�
( II. )
0.01 1 2
' 3
2'
0
1 2
.AOI US
0 .02 1 J
'1
1 0'0000
0.0323
9
1
SH( I I '
0 . 00 7 9
1 JS
S.
1 1 09000
0.0 1 18
' 00
0. 0 1 1'
I . 73S
"
1 1 00000
0 . 0 1 61
I
- 0 . 0060
t 9
I I � IOOO I 1 18000
0 . 02 3 2
6
'0'
e o . OU ( T O I
- O . OO I S
1 119000
I l l lOOO
0. 0322
I
-
0 . 00. .
1 .
30
0.01 16
6
0.0 1 1 2
I . 73S
1 2 1 1 000
0 . 0 1 66
M ( GO'M. M l l ( S
0 . 02 I 2
I . 762 I . 762
. 3
1 2' )000
0 . 0230
"
0 . 00 7 1
1 . 762 1 3
t 9
IS
0. 0320
'I
-::r . " I 0 . 007S
- 0 . 00 6 1
.,
'1
(I)
SPAe I IG ( I I . )
- 0 . 00 1 6
'8
1 60 7000
1 2
HAC TAle(
0 . 0039
'2
1 6 7 1000
,
.
0.0 1 10
2 1 9 )000
70000
eO'oue T O
•
0.0210
2 1 � 90 0 0
I I" 000
F-
(HAC I T I ¥(
•
1 . 1 96 1 . 1 96
I . lit '
0.0 1 32
I • • 08
0 . OlS7
0 . 0266
0 . 0202
0 . 0 ' SJ
0.0' 1 2
0 . 0 2 1S
0 . 0 . ls
0 . 006 1
0 . 00 0 '
I . 0' 2
0 . 0 )6 1
0 . 02 7 1
0 . 0207
0.0'S7
0.0' 1 7
0 . 0238
0 . 0 ' 38
0 . 0067
0 . 00 1 1
- 0 . 00 1 '
I . ' 08
0. 03ST
0 . 0266
0. 0202
O.OISJ
0.0' 1 2
0 . 021�
O . O . lS
0 . 0063
0 . 0008
- O . OO P
0 . 00 1 1
- 0 . 00 1 0
I . OSS
0 . 0160
0 . 0 2 70
0 . 0206
0.0 1S'
0.0' 16
0. 0237
0.0 ' 37
0 . 0066
'�0 200
I I
1 9
1 . 1 01
0. 01S7
0 . 0266
0 . 0202
O.O ' �J
0.0"
2
O. 023S
O . O . lS
0 . 006 ]
0 . 0008
- 0 . 00 3 7
T 9 � OO O
I I
t9
1 . 069
0 . 0)S9
0 . 0269
0 . 020S
O . O ' SS
O.O I IS
0.0237
0 . 0 1 J6
0 . 006S
0 . 00 ' 0
- O . OO lS
• • 063
0 . 0]60
0 . 0169
0 . 020S
0 . 0 ' S6
O.O"
S
0.02J7
0.0117
0 . 0066
0 . 00 1 0
- O . OO l S
O . O ]6 S
O . 0 27S
0 . 02 I I
0.016 •
0.0' 21
0 . 02 00
0 . 0 "0
0 . 0069
0 . 00 "
- 0 . 00 3 1
0 . OJ70
0 . 0280
0.0216
0 . 0 1 66
0 . 0 1 IS
0 . 0200
0.0"1
O . 0072
0 . 00 1 7
- 0 . 0021
8 19000
10
7
107 700
20
II
716�00
1 8
7 2 7�00 7 1 ' 100
JJ
19
1 . 06 j
1 . 063 0 . 990 0 . 990
10
700000
20
I I
6 1 1 60 0
II
.9
6 3 2000 6 1 6 200
I�
. 1 7 000
I S
0 . 0219
o . ons
0 . 0 1 76
0 . 0 ' ls
0 . 02�0
O . O ' SO
0 . 00 7 9
0 . 002'
- 0 . 00 2 2
1 2
0 . 8 II.
0 . 0 110
. 7S200
0. ' "
h 3 600
•S
0 . 6"
0 . 0392
0 . 0102
O . OU'
0 . 0 1 89
0.0'"
0 . OlS9
O . O I �I
0 . 001 7
0 . 00 3 2
- 0 . 00 1 )
l l � 00 0
. 2
1 2
0 . 990 0 . 990 0 . 927 0. 927
0 . 6"
Table 8.20. Resistance of ACSR Conductors (ohms per mile) (Courtesy Aluminum Company of America)
CODE
60
D I A .
DC
I N .
2 5°C
2 5°C
5 0 0C
7 5 °C
1 00°C
1 9
2 . 500
0 . 0 2 9 ij
0 . 0333
0. 0362
0 . 0389
0 . 04 1 8
ARfA
STRANDS
CM I L
AL
S;
�c
-
H l -------,
E XPANDED
3 1 08 0 0 0
62/8
E XPANDED
2 2 9 ij O O O
66/5
1 9
2 . 320
0 . 0399
0 . 04 1 2
0 . 0453
0 . O ij 9 3
0 . 0533
E XPANDED
l ij l � O O O
58/ "
1 9
0 . 0 6 ij ij
0. 0663
0 . 0728
0 . 0793
0 . 0859
E XPANDED
1 2 7 5000
50/u
1 9
I . 750
I . 600
0 . 07 1 6
0. 0736
0 . 0808
0 . 088 1
0 . 09 5 3
K I WI
2 1 6 7000
72
737
0 . 0�2 1
O . O�73
0 . 05 1 5
0. 0552
0 . 0593
B LU E B I RD
2 1 56000
8 ij
1 9
0 . 0�20
0 . 0464
0 . 0507
0 . 0 5 ij 5
0 . 0586
CHU KAR
1 780000
8�
1 9
I . 762
O . OS 1 0
0. 0548
0 . 0599
0 . 0 6 ij 7
0 . 0696
0 . 06 5 3
0 . 0707
0 . 0763 O . 0 7 7 ij
•
1 . 602
I . 5 ij 5
0 . 0567
0 . 0594
0 . 05 7 1
0 . 060B
0 . 0664
0 . 0 7 1 9
1 9
I . 502
1 . 506
0 . 0597
0. 0625
0 . 0686
0 . 0 7 ij ij
0 . 0802
7
I . ij 6 6
0 . 0602
0. 0636
0 . 06 9 7
0 . 0 755
0 . 08 1 3
F A L CON
1 590000
5 ij
1 9
L A PW I NG
1 5 90000
ij 5
7
PARROT
1 5 1 0500
5 ij
NUTHATCH
1 S 1 0500
ij 5
PLDHR
l ij 3 1 0 0 0
5�
BOBOL I NK
1 � 3 1 000
�5
MART I N
1 35 1 500
5 ij
D I PPER
1 35 1 500
ij 5
PHEASAN T
1 2 72000
o�
1 9
B I T T E RN
1 2 72 0 00
�5
7
GRACKL E
1 1 9 2 5 00
1 9
BUN T I NG
1 1 92500
5 ij
�5
F I NCH
I I 1 3000
5 ij
1 9
SLUEJAY
I I 1 3000
ij 5
1 03 3500
5�
ORTOLAN
1 03 3500
TANAGE R
1 0 :; 3 5 0 0
CAR D I NAL
9 5 ij O O O
5 ij
RA I L
9 5 ij O O O
45
CA T B I RD
9 5 ij O O O
36
CANARY
900000
5 ij
RU DDY
900000
ij 5
�ALLARD
795000
30
DRAKE
795000
26
CONDOR
795000
5�
CUCKOO
795000
TE RN
CURLEW
I
�65
0 . 0630
0. 0657
0 . 072 1
0 . 0782
0 . 0 8 ij 3
I . ij 2 7
0 . 0636
0 . 0668
0 . 0733
0 . 0 7 9 ij
0 . 0856
I . �2�
0 . 0667
0. 0692
0 . 0760
0 . O B 2 !>
0 . 0890
I . j85
0 . 067 2
0 . 0 705
0 . 07 7 1
0. 0836
0 . 090 1
I . 382
0 . 0 70 9
0 . 0732
0 . 0805
0 . 0 8 7 ij
0 . 0 9 ij ij
3 ij 5
0 . 07 1 5
0 . 0746
0 . 08 1 7
0 . OB86
0 . 0956
I . 333
0 . 0756
0. 0778
0 . 0855
0 .0929
O . 1 00 0
I . 302
0 . 0762
0 . 0792
0 . 0867
0 . 0 9 ij 2
O . 1 00 2
I . 293
0 . 08 1 0
0 . 0832
0 . 0 9 1 ij
0 . 0993
O . 1 080
I . 259
0 . 08 1 8
0 . 0844
0 . 0926
O. 10 I 0
O. 1 090
0 . 08 7 1
0. 0893
0 . 0979
0 . 088 1
0 . 0905
0 . 0994
O . 1 080
0. 1 1 70
36
I . 2 I 3
o . 1 0 70
0 . 1 1 00
�5
I . 2�6
1 . 1 86
0 . 0885
0 . 09 1 5
O. 1 090
O. 1 1 60
I . 1 9 <;
0 . 09��
0 . 0963
O. 1 0 I 0
O . 1 060
O. " 50
O. 1 250
I . 1 65
0 . 0 9 5 ij
0 . 0978
O . 1 080
O . I 1 70
O . 1 260
I . I ij O
0 . 0959
0. 0987
O . 1 090
O . " BO
0 . 1 270
I 62
O. 1 000
O. 1 020
O . I 1 20
O . 1 220
O. 1 320
I . I 3 I
O. 1 0 I 0
O . 1 030
o . I I 30
O . 1 230
O. I 3�0
O. 1 37
O . I ij 7 O . I ij 9
1 9
1 9
I
I
I 1 9
•
•
•
I . I �o
O . I I I
0 . 1 25
O . I ij 7
O . 1 37
O . " 5
O . 1 27
O . 1 38
O . I I 3
O. 1 1 4
O. 1 37
O . I ij 8
O . " 4
O. I 1 6
o . 1 27
O . 1 39
O . 1 50
O . " 5
O . I I 7
o . 1 29
O . I ij I
O. 1 52
O. I 2u
O. 1 5 1
O . 1 6 ij
O . 1 25
O . 1 26
O . I 39
O. 1 5 1
O . 1 6 ij
O . 1 26
0 . 1 27
O . I � I
O . 1 53
O. 1 65
I .0 I �
O . I 35
O . 1 76
O . I 35
O . 1 37
o . I ij 9
O . 1 62
1 . 000
o . 1 34
O . 1 6 ij
O. 1 77
0 . 95 3
O . 1 40
O . 1 42
o. 1 5 I
O . 1 56
O. 1 7 1
O . 1 84
1 . 0 1 9
O . 1 39
O . I �3
O . 1 57
0. 1 72
O. 1 86
O . 1 56
O . 1 70
O . 1 84
O . I ij 3
O. 1 72
O . 1 86 O . 1 8 li
O . I I 2
O . " 3
2 ij
1 . 092
795000
ij 5
C OO T
795000
36
I . 063
H OW l N G
7 1 55 00
30
1 . 08 I
S TARL I NG
7 1 5 500
26
1 . 093
1 . 0 ij O 1 9
o. I 1 4
o . 1 25
I . 1 08
I . OS I
I . 036
O . " 4
O . 1 26
o . I 28
O . 1 39
ST I L T
7 1 5500
2 ij
GANNE T
666600
26
F L A M I N GO
666600
2�
65 3900
1 8
E GR E T
6 3 6000
30
GROSBEAK
6 3 6 0 00
26
0 . 990
O . 1 40
ROOK
636000
2�
0 . 977
O . 1 42
K I N GB I RO
636000
1 8
0 . 9 ij O
O . 1 45
O . 1 60
O . 1 7 ij
SW 1 F T
6 36000
36
0 . 930
O . I ij 3
o . I S7
O . 1 44
O . I �6
O . 1 6 I
O . 1 75
0 . 1 89
T E A L
605000
30
0 . 9 9 ij
O . 1 46
O. 1 50
O . 1 65
0 . 1 80
0 . 1 95
o . I �9
1 9
1 9
o . 1 42
S QU A B
605000
26
0 . 966
O . 1 6 ij
O . 1 79
O . 1 93
PEACOCK
605000
2 ij
O . 1 65
O . 1 80
O . 1 95
EAGLE
30
O . I �9
O . I SO
556500
0 . 953
0 . 2 1 2
26
0. 927
O . 1 60
o . 1 63
O . 1 96
5 5 6 5 00
O. 1 58
0 . 1 79
DOH
O . 1 62
O . 1 78
O . 1 9 ij
0.9 I �
O . 1 62
O . 1 63
O . 1 79
O . 1 96
0 . 2 I 2
O . 1 83
O . 1 99
0 . 2 1 5
0 . 88 3
O . 1 85
O . 1 90
0 . 209
0 . 228
0. 247
O . 1 88
0 . 207
0 . 226
0 . 2 ij 5
0 . 8 ij 6
0 . 1 89
O . 1 90
G . 209
0 . 2 28
0 . 247
O. 1 93
0 . 2 I 2
PARAKEf T
OSPR f Y
556500
5 56500
2� 1 8
� 17000
30
H Awr
ij 7 7 0 0 0
26
f L I CK E R
� 7 7000
2�
PfL i CAN
H f N
0 . 953
O. 879 0. 858
O. I �7
O . 1 63
O . 1 87
o. 1 66
0. 232
0 . 250
0 . 250
0 . 273
0 . 295
0 . 226
0 . 2 ij 9
0. 27 I
0 . 2 9 ij
O . 2 2 'i
0 . 227
0 . 250
0. 273
0 . 295
0 . 2 2 ;)
o. 1 9 1
� 7 7000
1 8
0 . 8 1 "
LARK
39 1500
30
0 . 806
0 . 222
0 . 227
I B I S
397 500
26
O. 783
O . 2 2 '1
BRANT
397500
2�
0 . 772
OR I OLE
3 36 � 00
30
L I NN f T
3 3 6 � 00
.HRL I N
336�00 300000
26
CH I (kADU
O S T R I CH
397500
0 . 2 1 I
0. 23 I
0 . 25�
0. 277
0. 300
O . 262
0 . 268
0 . 295
0 . 322
0 . H9
26
O . 7" I
O. 72 I
0 . 265
0. 267
0 . 2 9 ij
1 8
0 . 68�
0 . 270
0 . 273
0 . 300
0 . 32 I
O . 3 ij 7
0 . 680
o. 297
0. 299
0 . 329
0. 359
0 . 389
1 8
O. 7�3
O . 328
0 . 355
Table B.21. Resistance of ACAR Conductors (ohms per mile) (Courtesy Kaiser Aluminum and Chemical Sales Inc.) AREA 6H Ee
EO. -
Al
eM I L
DI A.
STRANDS
De
I ".
200e
1 . 82 1
0. 0373
[e/620 1
I
.-------- A t sooe 3 0 0e 0 . 0 '1 5 6
0 . 0� 8 3
-
60
" I
1 00 °C
0 . 05 1 6
0 . 0565
72
1 9
2 3 7 5 00 0
63
28
1 . 82 1
0 . 0379
0. 0�62
0 . 0�88
0 . 05 2 1
0 . 0555
2 3 38000
5�
37
1 . 82 1
0 . 0385
0 . 0 '1 6 7
0 . 0� 9 3
0 . 0527
0 . 05 6 1
2 � 1 3000
2 2 9 7000
S�
7
1 . 762
0 . 0392
0 . 0 11 7 '1
0 . 05 0 2
0 . 05 3 8
0 . 0573
2 2 6 2000
11 8
1 3
1 . 76 2
0 . 0399
0 . 0�79
0 . 050 7
0 . 0 5 11 3
0 . OS80
2 2 2 6000
11 2
1 9
1 . 76 2
0 . 0 11 0 5
0 . 0�85
0 . 05 1 3
0. 05�9
0 . OS85
2 2 2 7000
5�
7
1 . 7 35
0 . 0�05
0 . 0�85
0 . 0 5 1 '1
0 . 05 5 1
0 . 05 89
2 1 9 30 0 0
�8
1 3
1 . 735
O . O� I I
0 . 0 '1 9 1
0 . 05 2 0
0 . 05 5 7
0 . 0595
2 1 5 9000
�2
1 9
1 8 9 9000
5�
1 8 70000
11 8
1 . 735
0 . 0� 1 7
0 . 0�97
0 . 05 2 6
0 . 05 6 2
0 . 060 1
1 . 60 2
0 . 0 1l 7 �
0 . 0550
0 . 0585
0 . 06 2 9
0 . 067�
1 3
1 . 60 2
0. 0�82
0 . 0557
0 . 0592
0 . 06 3 6
0 . 0682
1 8 � 1 000
1 9
1 . 602
0 . 0�89
0 . OS6�
0 . 0599
0 . 06 � �
0 . 0689
1 6 7 3 000
7
I . 50�
0 . 05 39
0 . 06 1 1
0 . 06 5 1
0 . 0702
0 . 075�
1 6 � 7 000
1 3
I . 50�
0 . 05�6
0 . 06 1 9
0 . 0659
0 . 07 1
0 . 0762
1 6 2 2000
1 9
I . S O li
0 . 05 5 5
0 . 0627
0 . 0667
0 . 07 1 9
1 3 3 7000
7
1 . 3 '1 5
0 . 06 H
0 . 07� 2
0 . 0 7 9 '1
0 . 08 6 0
0 . 0925
1 296000
1 9
I . H5
0 . 0695
0 . 0763
0 . 08 1 5
0 . 08 8 1
0 . 09�7
1 . 302
0 . 07 25
0 . 0793
0 . 0 8 '1 9
0 . 09 1 9
0 . 0989
1 2 1 1 000
1 3
1 . 302
0 . 07��
0 . 08 1 2
0 . 0868
0 . 0937
0 . 1 00 8
I 1 7 9000
1 9
1 . 30 2
0 . 07 6 �
0. 083 1
0 . 0887
0 . 0957
0 . 1 028
1 1 6 3 000
7
1 . 259
0 . 0775
0 . 08�2
0 . 0902
0 . 0977
0 . 1 05 2
1 1 3 3 000
1 3
1 . 259
0. 0795
0. 0862
0 . 09 2 2
0 . 0997
0 . 1 07 3
I I O�OOO
1 9
1 . 25 9
0 . 08 1 6
0 . 0882
0 . 09�2
0. 1 0 1 8
0 . 1 09 5
1 1 5 3 000
�
1 . 2�6
0 . 078 1
0 . 0850
0 . 09 1 0
0 . 09 8 7
O. 1 06 �
1 1 3 8000
7
1 . 2�6
0 . 079 1
0. 0859
0 . 0920
0 . 0997
0 . 1 0 7 11
1 2 Q 3000
I
0 . 07 7 1
I 1 09000
1 3
1 . 2�6
0 . 08 1 2
0 . 0880
0 . 09� 1
0. 1 0 1 7
0 . 1 09 5
1 0 80000
1 9
I
2�6
0 . 08 3 �
0 . 0900
0 . 0962
0 . 1 0 39
0. 1 1 1 7 0. 1 1 32
•
7
1 . 2 1 2
0 . 08 3 6
0 . 090�
0 . 0969
0 . 1 05 0
1 0� 9000
1 3
1 . 2 1 2
0 . 0859
0 . 092 6
0 . 09 9 1
0 . 1 07 2
1 02 200e
1 9
1 . 2 1 2
0 . 08 8 2
0 . 0 9 '1 8
0. 1 0 1 3
0 . 1 096
0 . 1 1 77
1 0 7 7 00 0
O . I I S 'I
7
1 . 1 96
0 . 0859
0 . 092 6
0 . 0993
0 . 1 07 6
0 . 1 1 60
1 0 2 3 000
1 3
1 . 1 96
0 . 08 8 1
0. 09�8
0. 1 0 1 5
0 . 1 09 9
0. 1 1 83
9 96000
1 9
1 . 1 96
0 . 090�
0 . 09 7 1
0 . 1 0 38
0. 1 1 23
0 . 1 20 7
1 . 1 65
0 . 0906
0 . 09 7 �
0 . 1 0 11 '1
0 . 1 1 33
0 . 1 22 1
1 050000
99�800 9S� 600
7
I . I �I
0 . 0 9 � 'I
0. 1 0 1 2
0 . 1 08 6
0 . 1 1 78
0 . 1 27 1
9 6 9 300
1 3
1 . 1 65
0 . 0 9 29
0 . 0997
0 . 1 068
0 . 1 1 56
0 . 1 2 11 6
9 5 8 (1 0 0
1 3
I . 1 58
0 . 0 9 11 1
0 . 1 00 8 '
0 . 1 08 0
0 . 1 1 70
0 . 1 260
9� 3900
1 9
1 . 1 65
0 . 09 5 5
0. 1 02 1
0. 1092
0. 1 1 82
0. 1 27 1
900300
7
I . 1 08
0 . 1 00 I
0 . 1 070
0 . 1 1 �8
0 . 1 2 11 6
0 . I JII 5
7 9 5000
1 . 0� 2
O. I 1 33
0 . 1 20�
0 . 1 293
0 . 1 '1 0 '1
0. 1 5 1 6
877 300
7
1 3
I . 1 08
0. 1 0 27
0 . 1 096
0 . 1 1 7 '1
0 . 1 272
0 . 1 372
7 9 5 0 00
1 3
1 . 055
O. I 1 33
0 . 1 2 0 '1
0 . 1 290
O . I �OO
O . I S O !f
8 5 � 2 00
1 9
I . 1 08
O . I OS�
0. 1 1 23
0 . 1 20 1
0 . 1 3 00
O . I �OO
7 95000
1 9
1 . 069
O. I I H
0 . 1 203
0. 1 288
0 . 1 3 9 '1
O . I SO I
8 2 9000
7
1 . 063
O . 1 08 7
0. 1 1 57
0. 1 2�3
0 . 1 350
0 . 1 '1 5 6
807700
1 3
1 . 06 3
O. 1 1 15
0 . 1 27 1
0 . 1 378
0 . 1 '1 8 6
78 6 500
1 9
1 . 063
O.
I I
0 . 1 1 85
�S
0. 1 2 1 6
0 . 1 30 2
O. I � I O
0. 1 5 1 8
7 2 7500
'I
0 . 990
O . 1 238
0. 1 3 1 2
0 . 1 '1 1 1
0 . 1 5 3 '1
0 . 1 658
7 1 8 3 00
7
0 . 990
0 . 1 2 5.."
0. 1 327
0 . 1 '1 2 7
0 . 1 550
0 . 1 675
700000
1 3
0 . 990
0 . 1 28 7
0 . 1 360
0 . I �S9
0 . 1 5 8 '1
0 . 1 709 0 . 1 7�3
6 8 1 600
1 9
0 . 990
O. 1 32�
0 . 1 3 9 '1
0 . 1 11 9 11
0. 1 6 1 9
6 3 2 000
'I
0 . 927
O . I �2S
0 . 1 50 3
0. 1 6 1 5
0 . 1 756
0 . 1 897
6 1 6 200
7
0 . 927
O . I �62
0 . 1 539
0 . 1 65 2
0 . 1 7 9 '1
0 . 1 935
�8 7�00
�
0 . 8 1 �
O . 1 8�9
0 . 1 9 38
0 . 2085
0 . 2268
� 7 5 200
0 . 2 '1 5 3
7
0.8 1 �
0 . 1 896
0 . 1 986
0. 2 1 33
0 . 23 1 7
0 . 250 1
H 3600
�
0 . 68�
0 . 2623
0. 2739
0 . 2 9 '1 8
0 . 3 20 9
0 . 3�70
3 3 5000
7
0 . 6 8 '1
0 .2 690
0 . 2806
0 . 30 1 5
0 . 3 27 8
0 . 3 5 '1 0
456
I
--------.
7 '; ° C
Table B.22. Electrical Characteristics o f Overhead Ground Wires
P A R T A : A LUMOWE L D � ) S T R A N D RESI STANCE
60
( OHMS/M I L E )
FOR SMA L L
2 50C
OF
CAP .
I
60
REACTANCE FOOT
Hz
GEOME T R I C
RAD I US
I N DU C T I V E
CAPAC I T I VE
MEAN
RAD I US
750C 60
OHMS/ M I L E
M E GOHM- M I L E S
FEET
1 . �32
1 . 669
0 . 707
0. 1 1 22
0 . 002958
1 . 536
1 . 773
2. 0 1 0
0.72 1
O . I 1 57
0 . 00 2 6 3 3
1 . 937
2 . 2�0
2 . �70
0. 735
O.
1 9 1
0 . 0 0 2 3 11 5
2 . �00
2 . ��0
2 . 820
3 . 060
0 . 7�9
0 . 1 226
0 . 002085
9
3 . 0 20
3 . 080
3 . 560
3 . 800
0. 763
0 . 1 2 60
0 . 00 1 8 S 8
NO. 1 0
3. 8 1 0
3 . 880
� . �80
� . 7 30
0. 777
O. 1 2 9�
0 . 00 1 6 5 8
0 . 707
0. 1 22 1
0 . 0029�0 0 . 00 2 6 1 8
60
DC
(
AWG
)
7
NO .
5
I. 2 1 7
1 . 2�0
7
NO.
6
1 . 507
7
NO .
7
1 . 900
7
NO.
8
7
NO.
7
75'
750C
CU R R E N T S
2 50C
STRANO
Hz
DC
Hz
HZ
I
3
NO.
5
2 . 780
2 . 780
3 . 2 70
3 . 560
3
NO.
6
3. 5 1 0
3 . 5 1 0
�. 1 30
� . � I O
0.721
0 . 1 255
3
NO .
7
� . �20
�. �20
5 . 2 1 0
5 . �70
0. 735
0 . 1 289
0 . 00 2 3 3 3
3
NO.
8
5 . 580
5 . 580
6 : 570
6 . 820
0 . 7�9
0 . 1 3 2 11
0 . 002078
3
NO.
9
7 . 0�0
7 . 0�0
8 . 280
8. 520
0. 763
0 . 1 3 58
0 . 00 1 85 3
3
NO. 1 0
8 . 870
8 . 870
1 0 . � ij O
1 0 . 670
0. 777
0 . 1 392
0 . 00 1 650
PART
rI
RES I STANCE
S I N G L E L A Y E R A C !,; R tz)
�
noc , I
60
( OHMS/M I L E ) 60
1 =0 (3 )
CODE
B:
Hz
&
1 = 1 00
HZ
OHM S / M I L E
1 =2 0 0
q
REACTANCE I N DU C T I V E AT
1 =0
1 = 1 00
FOR
I
F OO T
RAD I US
CAPAC I T I VE
75°C
M E GO HM- M I L E S
1 =200
0 . 500
0. 520
0 . 5 11 5
0 . 1 0�3
C OC H I N
0 . 11 0 0
0 . �80
0 . 520
0 . 590
0 . 505
0. 5 1 5
0 . 550
O . 1 065
DORK I NG
0 . 1I � 3
0. 535
0 . 575
0 . 650
0. 5 1 5
0 . 5 30
0 . 565
O. 1 079
DOTlE R E l
0 . 11 7 9
0 . 565
0 . 620
0 . 705
0.5 1 5
0 . 5 30
0 . 575
O. 1 09 1
GU I NEA
0 . 53 1
0 . 630
0 . 685
0 . 780
0 . 520
0. 5�5
0 . 590
O . 1 1 06
8 R A HM A
0. 5 1 0
0.565
L EGHORN
0 . 630
0 . 760
0 . 8 1 0
0 . 930
0 . 530
0. 550
0 . 605
O. I 1 3 1
M I NORCA
0 . 765
0 . 9 1 5
0 . 980
1 . 1 30
0. 5�0
0. 570
0 . 6�0
O . I 1 60
PETREL
0 . 830
1 . 000
1 . 065
1 . 220
0 . 550
0 . 580
0 . 655
0 . 1 1 72
GROU S E
1 . 080
1 . 295
I
�20
1 . 5 20
0 . 570
0 . 6�0
0 . 675
0 . 1 2 11 0
.
PART
� I (3)
RES I STANCE
( 7- STRAN D )
o I A .
ORDI NARY
1 / 11
ORDI NARY
GRADE
60
C:
STEEL
�
( OHMS/ M I L E ) H Z
1 =0
I =30
9 . 5
I I . �
9/32
7. 1
9 . 2
9 . 0
ORDI NARY
5/ 1 6
5 . 11
7 . 5
7 . 8
ORD I NARY
3/8
11 . 3
6 . 5
6 . 6
ORD I NARY
1 /2
2 . 3
� . 3
1 /�
8 . 0
9/32
E. E. E. E.
B. B. B. B.
[. 8.
B. B.
I N .
CONDUCTORS
I =6� 1 1 . 3
I
�
6 0 Hz
I =-0
�)
�
R E ACTANCE I N DUC T I H
O HMS / M I L E 1 =30
FOR
I
F OO T
RADI US
CAPAC I T I VE ME GOHM- M I L E S
1 =6 0
1 . 3970
3 . 7 11 3 1
3 . �379
O . I 35�
1 . 2027
3 "0 7 J �
2 . 5 1 11 6
0. 1 3 1 9
0 . 8 38 2
2 . 5 1 �6
2 . 0 11 0 9
0 . 1 288
0. 8382
2 . 2352
1 . 9687
0 . 1 2 3 11
5 . 0
0 . 70�9
1 . 6893
1 . 11 2 3 6
O . I I �8
1 2. 0
10. I
1 . 2027
3. 1 565
O. 1 35�
6 . 0
1 0 . 0
8. 7
I
II. ij 7 0 ij
3 . 7783
2 . 6255
0. 1 3 1 9
5/ 1 6
� . 9
8 . 0
7.0
0 . 98�3
2 . 9 �0 1
2 . 5 1 11 6
0 . 1 288
3/8
3 . 7
7.0
6 . 3
0 . 8382
2 . 5997
2 . 11 3 0 3
O. 1 2 3 �
1 / 2
2 . I
11 . 9
5.0
0 . 7 0 ij 9
1 . 87 1 5
1 . 76 1 5
O . I I ij 8
•
1 305
1 / 11
7 . 0
1 2 . 8
10.9
1 . 6 7 6 ij
5 . l ij O I
3 . 9 11 8 2
O. 1 35�
E. 8. 8.
9/32
5 . �
1 0 . 9
8.7
1 . 1 305
� . �833
3 . 7783
0. 1 3 1 9
E.
5/ 1 6
11 . 0
9.0
6 . 8
0 . 9 8 &1 3
3 . 6322
3 . 073�
0 . 1 288
3/8
3 . 5
7.9
6 . 0
0 . 8382
3 . 1 1 68
2 . 79�0
O. 1 23�
1 / 2
2 . 0
5. 7
11 . 7
0 . 7 0 &1 9
2 . 3 &1 6 1
2 . 2 352
O. I I �8
E.
E. E .
B . 8. B. B. B. 8.
30 1 5
I .
DATA
COMP I L E D
F R OM
E . D.
2 .
DATA
COMP I L E D
f R OM
" R E S I STANCE
3 .
CONDUCTOR
II .
DATA
CURRENT
COMP I L E O
f ROM
IN
-
COPPE RWE LD
AND
STEEL
R E ACTANCE
OF
COMPANY . ALUM I NUM
CON DUC T O R S "
-
A L CO A .
AMPE R E S .
" S Y MM E T R I C A L
COMPON E N T S "
WAGN E R
'
EVANS
( 800K )
M c G R AW- H I L L .
4 57
Table B.23. Typical EHV Line Characteristics Line to Line Voltage (kV)
Cond o per Phase at 1 8 (I nches)
345
1
345
2
500
500 500
1 2
Cond uctor Code and Diameter ( I nc hes)
Phase SpaCing (Feet)
6 0 Hz I n d uctive Reactance i n Ohms per M il e GMD (Feet)
X ..
Xn
X .. + Xo
60 Hz Capacitive R eactance in M egohm·Miles
X.
XI>'
X .. ' + X o'
Z.
(Ohms)
SIL (MVA)
Expanded - 1 . 7 5 0
28
35.3
0.3336
0.4325
0 . 766 1
0.07 7 7
0. 1 057
0. 1 834
374.8
3 18
- 1 . 246
28
0. 1 6 7 7
0.4325
0.6002
0.0379
0. 1 05 7
0. 1 436
293.6
405
2 . 500
38
35.3
0.4694
0.76 1 6
0.067 1
0. l l 47
0. 1 8 1 8
372 . 1
672
38
47 .9 47.9
0.2922
- 1 .602
0. 1 529
0.4694
0.6223
0.034 1
0. l l47
0. 1 488
304.3
822
2 78.6
897
·Curlew Expanded ·Ch u kar
-
500
- 1 . 1 65
38
47.9
0.0988
0.4694
0. 5682
0.02 1 9
0. l l 47
0. 1 366
4
· Para keet - 0;9 1 4
38
47.9
0.0584
0.4694
0. 5 2 78
0.0 1 26
0. 1 147
0. 1 273
259.2
965
735
3
Expanded - l . 7 50
56
70.6
0.0784
0. 5 1 66
0. 5950
0.0 1 79
0. 1 263
0. 1 442
292.9
1 844
735
4
Pheasant - 1 .382
56
70.6
0.0456
0. 5 1 66
0. 5622
0.0096
0. 1 263
0. 1 3 59
276.4
1 955
3
Rail
• N •• r••t to ba •• ·c••• diamet.r .
0.0 0
1 2 3
4 5 6
7
8 9
-- C. O 0 . C 84 1 0 . 1 333
0. 1 6 82 0 . 19 53
0 . 2 1 74 0 . 2 3 t: l
0 . 2523 0 . 2 6 66
10
0 . 21 94
1 1
C . 29 1 0
12 13 14
15 16 17 18 19
20
0 . 30 1 5
0.3112
2 79 4 011 6 0900 1373
0. 1 71 2 0. 197 7 0 . 2 1 C; 4 0. 237 8 0. 2538
0. 2
- 0 . 1 C; 5 3 0 . 0 22 1 0.OS51 u. 0. 0. 0.
l�ll 1 14 1 2 00 1 2214
0 . 2 39 5
0 . 2 55 3
0 . 27 1 9
0 . 3045
0 . 3 149
0. 0. 0. 0.
3 30 2 3 379
0 . 36 35
0 . 3 58 t O . 3 t: 4 7
0 . 3 7G C 0. 3 156
0. 3 762
0 . 3999
0 . 2429 0 . 2 58 2
0.3035
0. 3519 0. 3t4 1
0 . 39 53
0 . 24 12 0 . 2568
0 . 2046 0 . 22 5 2
0 . 3 13 1 0. 3219
0. 3 12 2
C. 3514
27
0 . 2024 0 . 22 33
0 . 1 48 5
0 . 1 79 8
0 . 2 95 3
0. 3501
26
0. 1 770
0 . 2 94 2
C . 344 5
0 . 3 8 56 0 . 3906
0. 1011 0 . 1 4 49
0 . 2 C; 3 2
0 . 34 3 8
C.38C5
0 . 04 0 8 0 . 1 06 2
C . 2 <; 2 1 0 . 30 2 5
0. 3 3 72
23
-0. 1112
0 . 031 8
0
0. 3364
C . 3 6 94
- 0 . 1 46 1
0 . 2 106 0 . 2830
0. 321 1
0 . 37 5 1
0.4
0 . 2 69 3 0. 281 8
0 . 3294
0 . 3 5 13
0. 3
0. 268 C 0 . 2 80 6
0 . 32 0 2
21
25
-0. 0. 0. 0.
0 . 32 8 6
22 24
Table B.24. Inductive Reactance Spacing Factor Xd (ohms/conductor/mile) at 60 Hz
0. 1
0. 381C
0. 3 86 1
0. 391 1 0 . 3 <; 5 8
3452 3 52 1
0 . 3 70 6
0 . 3 81 5 0 . 3 E6 6 0 . 3 C: H 0 . 3 S6 3
28
0 . 40 4 3
C . 4004 0 . 404 8
0 . 4 00 8 0 . 4 C5 2 0 . 4 C9 4
29
30
C . 4086
0. 4090
0.4121
0. 413 1
0 . 4 13 5
3 1
C.�H7
0. 4 1 7 1
0 . 4 17 5
0 . 3 1 40
.
2 84 2
0 . 305 5
0.5
-0. 0. O. 0. O. 0. 0.
0 84 1
-0. 062 0
049 2 1112
0 . ll 5 1 0 0 . 1 159
1 82 5
0. 1 8 52 0. 2090
1520
0 . 29 9 5
0. 3 06 5 0. 3158
0 . 3 0 14 0. 3167 0 . 3253
0 . 31 76 0 . 326 1
0 . 3 3 34
0 . 334 1
0 . 3 40 2
0. 347 3
0 . 340 9 0. 3480
0 . 3 54 0
0. 3541
0. 3604
0. 361 1
0. 36 53
0 . 3 598 0 . 365 9
0. 366 5
0 . 36 7 1
0. 371 1 0 . 3 76 7
0 . 371 7
0 . 3 92 0
0. 0. 0. 0.
3967
O. 372 3
0 . 4 1 39
0. 4 14 3
0 . 406 5 0. 4 10 7 0. 4 14 7
0. 4 1 79
0 . 4 1 13 2
0. 4 1 86
34 24
349 4
0. 3560
0 . 36 1 7
0 . 362 3
0 . 367 7
0 . 36 8 3
0. 3734
0. 0. 0. 0.
335 1
343 1 3500 3 5 66
0 . 36 2 9 0 . 36 8 8
0 . 42 0 2
0. 3783
C . 4 06 9 0.41 1 1
0 . 3789
0 . 384 1 0 . 389 1 0 . 393 9 0 . 398b 0 . 40 3 0
r :r (I)
3"
'C (I) Co QI ::J n (I) -I QI
iii e!II
0. 3745
0 . 4 19 8
0 . 39 8 1 0 .4026
0 . 406 1 0.4103
321 0 3 34 9
0 . 30 0 5 0.3103 0. 3 1 94 0 . 32 7 8
0 . 4 194
0 . 402 1
40 56 4098
3094 318 5
0. 2782 0 . 2899
0 . 4 1 90
0. 3836 0 . 3886 0 . 39 3 5
0. 3 97 7
0 . 34 1 6 0 . 34 8 7 0 . 35 54
0. 0. 0 . 0. 0. 0.
0 . 26 5 3
0.4163
0 . 3 88 1 0. 3 93 0
0 . 40 1 7
0 . 30 84
0 . 26 3 9 0 . 2169
0 . 2 3 44
0. 415 1
0 . 3870 0 . 3972
40 1 3
262 5
275 7
0 . 192 8 0. 21 54
0 . 40 7 3 ;) . 4 1 1 5 0 . 41 5 5
0. 3778
0. 3 8 3 1
2 30 8
24 7 7
0 . 190 3 0 . 2133
0 . 3 740 J . 379 4 0 . 3 84 6 0 . 3 896 J . 394 4 0 . 39 9 0 0 . 40 3 5 0 . 40 7 8 0. 41 19 0 . 41 59
0 . 3128
0 . 3 77 3
0 . 392 5
0 .2508
0 . 29 8 5
0. 2461
0 . 3534
0 . 3 82 &
0 . 232 6 0 . 2 49 3
0 . 2974
0 . 3 32 6
0 . 3 820 0. j871
-0 . 0 1 2 8 0. 0719 0. 12 9 2 0. 165 1
0 . 2 96 4
0 . 2 ic; 0
1878
2112
0.9
02 7 1 071 3 1249 1 62 0
-0 . 0 . 0. 0.
0 . 288 7
0. 3245
0 . 3 5 92
043 3
0 . 2876
0 . 32 1 6
0. 3527
.
0. 2 86 5
0. 2 85 3
0. 3318
0 . 3459
0
0 . 06 4 4 0 . 1205 0. 1588
0 . 26 1 1 0 . 2 744
0 . 2 13 2
0. 3�28
0 . 3 394 0 . 346 6
0. 1 554
-
0. 8
0. 0. 0. 0. 0. 0 .
206 9 227 1 0 . 2 44 5 0. 259 1
J. 3310
0 . 3 38 1
0.7
0.6
0 . 3 7 99
0 . 38 5 1
0 . 390 1
0 . 39 4 9 0 . 39 9 5
0 . 40 3 9 0 .40 82 0.41 2 3
� UI co
32
33 34
0.4205
0. 4243
Table
De l
0.0
0. 420 9 0 . 4 2 4 f:
B.24
(con tinued)
0. 2
0.3
0.4
0 . 5
0 .6
0 .7
0.8
0 . 4 21 3
0. 42 17 ) . 4 2 54
0. 4220
0. 422 4 0 . 426 1
0.422 8 0 . 42 6 5
0 . 42 3 2
0 . 42 5 7
0 . 42 3 5
0 . 42 6 8
0 . 42 7 2
0. 429 7
0.4300
0 . 4304
0. 433 1
0 . 4 36 5
0.4335
0 . 43 3 8 0 . 437 2
0. 4250
35 36
0 . 4 2 79
0. 4283
0. 4 28 t
0 . 4 2 90
0 . 43 14 C . 4 348
0 . 42 9 3
0 . 43 1 8
0 . 4 32 1
J . 4 324
0 . 432 8
C. 4 3 5 2
0. 4 35 5
0. 4358
0 . 4 36 2
0. 9 0 . 42 3 9 0 . 42 7 5
37
0 . 4382
0.4385
0 . 4 38 8
0 . 4 39 1
0 . 4395
C. 4401
0 . 44 04
38
C . 44 1 4
0. 44 1 7
0 . 4 39 8
0.4368
0. 442 0
0 . 443 0
0 . 44 3 3
0 . 44 3 6
0 . 44 4 5
C . 4 4 4 <;
0 . 446 4
0 . 446 7
0 . 44 7 0
0 . 44 7 6
0 . 4 4 7 <;
0 . 448 5
0 . 44 5 3 0 . 44 8 8
0 . 446 1
40
J. 4452 0 . 4 432
0 . 442 7
39
0 . 4 42 3 0 . 44 5 5
0 . 430 7 0 . 4 34 2 ,) . 4 3 7 5 J . 440 8 0 . 44 3 9
0 . 449 1
0 . 44 94
0 . 44 9 7
0 . 4442 0 . 44 7 3
0. 450 C
0.45 03
4 1
C . 4 5 C6
0 . 4 5 (' 9
0.4512
0 . 45 1 5
0 . 451 8
0. 452 1
0.4535
0. 4538
0 . .. 5 2 4
0 . 45 2 7
0. 4 54 1
0. 453 0
0. 45 3 2
42 43
0 . 4 5 44
0 . 4 54 7
0. 4 5 5 0
0.45 5 3
0 .455 5
0.4559
0 . 45 6 1
0 . 4564
0 . 4 56 7
0 . 4 57C
0 . 4 5 72
0. 4575
0. 457 8
O. 4� 8 1
0 . 4 5 84
a . 45 8 6
44
C . 4 592
C. 4 59 5
0 . 4 59 7
0 . 4600
0 . 460 3
0 . 4606
0 . 46 1 1
0 . 46 1 4
45
C . 46 19
46
0 . 46 4 6
0 . 4 62 2 0 . 464 8
0 . 45 8 9 0 . 46 1 6
47
0 . 46 7 2
0 . 43 1 1 0. 434 5 0 . 43 7 8 0 . 44 1 1
0 . 4 t: 2 4
0 . 46 2 7
0 . 46 3 0
0. 463 2
0 . 46 0 8
0 . 40 3 5
0 . 46 3 8
0 . 46 4 0
0 . 46 4 3
0 . 46 5 1
0 . 46 5 4
0 . 46 5 6
0 . 46 5 9
0 . 466 1
0 . 4664
0 . 46 6 7
0 . 46 6 9
48 49 50
0. 4674
0 . 467 7
0. 4680
0 . 46 8 2
0. 468 5
0 . 4b 8 7
0 . 46 9 0
0 . 46 Q 2
0. 470C
0 . 4 70 2
0 . 4 7 05
0. 471 0
0. 4 72 � 0. 4749
0. 4 72 7
0 . 473 2
0 . 413 5
0. 47 1 2 0 . 473 7
0 . 471 5
0. 4722 0 .4747
0 . 4740
0 . 4 74 2
0 . 4 744
0. 475 2
0 . 4730 J. 4754
0 . 470 7
J. 471 7
0 . 46 9 5
0 . 4 6 <; 7
0. 4 75 9
0. 4761
0 . 4764
O . 4 7 !:J 6
0 . 4769
51
0 . 47 7 1
0. 4773
0. 479 7 0 . 4 82 0
0 . 4 77 6
0 . 4 7 78
0 . 4780
4783
0 . 4787
0. 479 0
0 . 4 801
0. 4804
4806
0.4808
0 . 48 1 1
0 . 48 1 3
0 .4822
0 . 4 8 24
0 . 482 7
0 . 4836
0 . 48 3 8
0 . 4 84 7
0 . 4849
0.4831 0. 4854
0 . 4834
0 . 4 1: 4 5
482 9 485 1
0 . 4792 0 . 48 1 5
3 5
0. 0. 0. 0.
0.478 5
0 . 4 79 9
0 . 4856
0 . 48 5 8
0. 486Q 0 . 4891
0. 4871 0 . 4893
0. 4874 0 . 4 89 5
0.4876 0 . 4 8 <; 7
0 . 487 8
0 . 488 0
7
0.4E67 0 . 488 9
0 . 49 0 0
0 . 49 0 2
0 . 4 9 04
1:
0. 491 0
0 . 49 1 2
0 . 491 7
0 . 49 1 9
0 . 4 92 1
0 . 492 3
0 . 49 2 5
0 . 493 1 0 .495 2
0 . 49 3 3
0 . 49 3 5
0. 493 7
0. 4940
0 . 4 9 4; 2
0 . 49 44
0 . 49 54
0 . 495 6
0 . 495 8
0 . 49 4 6
0 . 4962
0. 4964
0 . 49 6 6
0. 4974
0 . 4 9 76
0 . 49 6 0 0 . 49 80
0 . 49 8 2
0 . 49 8 4
0 . 49 8 6
0 . 5 0 04 0. 502 3
0 . 5006 0. 502 5
0 . 504 3
0 . 50 4 5 0 . 5063
0 . 475 7
56 57
C . 4 7 <; 5 0 .48 1 8 0 . 4 84 0 0 . 486 3 0 . 4 8 84 0 . 49 06
58 59
0 . 49 2 7 0 . 4948
0 . 4 � 2 C;
0 . 4 <; 6 8
0 . 49 7 C
0 . 4 <; 7 2
0 . 4 9 1: 8 0. 5008
0. 499C 0 . 501 0
0 . 4 99 2
0 . 4 9 94
0 . 4996
0 . 4 99 8
0. 50 0 0
0 . 5002
0 . 5 C1 2
0 . 50 1 4
0 . 50 1 6
0 . 50 1 8
0 . 50a
0 . 50 2 7 0 . 5 0 46
0. 5C2 9
0 . 5 C3 1
0. 5035
0 . 5 03 9
0 . 504 8
0.5C50
0. 5 03 7
0 . 5020
0 . 5054
0. 5056
0. 5058
0 . 5041 0 . 5 06 0
52 53 54 55
60 6 1 62 63 64
0. 0. 0. 0.
484 4 86 488 4�0
0 . 49 5 0
0. 5033
0 . 5052
0 . 491 4
0. 4978
0 . 5 06 2
� Ol 0
0 .472 0
0 . 4860
0 . 48 8 2
l>
"0 "0 co ::::l
x'
0.
CD
65
C. 5065
66
0 . 5 0 84
67 6 8
0 . 5 1 02 0 . 5 1 20
69
C. 5122
0 . 5 1 38
0. 5 139
70
0. 5155
C. 5 1 5 7
11 72 73 74
0 . 5 1 72
C. 5174
0.5 176
0 . 5 1 78
0 . 5 17 9
0. 51 89
O. 5 1 8 1
0 . 5 2 06
0. 51 9 1 O. 5 2 C 8
0. 5 193
0 . 5 1 94
0 . 5 1 96
0. 5198
0 . 5 20 9
0 . 52 1 1
0. 5223
0 . 52 24
0 . 5 22 t
0 . 5228
0. 52 1 3 0 . 5229
C. 506 7
O . 5C 8 6 C . 5 1 04
0 . 5 06 9
0 . 5 0 71
0. 5C87
0 . 5 0 89
0 . 509 1
0. 5107
0. 5109
0.5 0. 5 0.5 0.5
106 124 14 1 15 9
0 . 50 7 3
0. 5125
0 . 512 7
0 . 5 1 43
0 . 5 145
0 . 5 1 60
0. 5162
0. 50 7 5
0. 50 9 3 0. 5 1 1 1 0. 5 12 9 O. 5 1 4 7 0 . 5 164
0 . 5 0 76
0 . 5078
0. 5080
0. 5095
0 . 5097
0 . 5098
0.5100
0. 5 1 1 3 0.5131
0. 511 5 0 . 5132
0. 5116 0 . 5 1 34
0. 51 1 8 0. 5 1 36
0.5152
0. 5169
0.5153 0. 5 1 7 1
0. 5 1 48
0.5150
0 . 5 1 66
0 . 5 16 7 0 . 5 1 84
0. 521 4
0. 5183 0 . 5 1 99 0 . 52 16
0 . 5218
0. 523 1
0 . 5 2 32
0 . 5 24 7
0 . 5 24 9
0 . 5 26 3
0 . 50 8 2
0 . 5 1 86
0.5188
0. 5203
0 . 52 04
0 . 5234
0 . 52 1 q
0. 52 36
0 . 52 3 7
0 . 52 5 0 0 . 5266
0. 5252
0 . 52 5 3
0. 526 5
0 . 526 8
0. 5269 0 . 52 8 5
0 . 5 20 1
0 . 52 2 1
75
C . 5 2 39
0. 524 1
0 . 5 24 2
0 . 5 2 44
76
0 . 52 55
0 . 5 25 7
0 . 5 25 8
0 . 5 26 0
77
0 . 5271
0. 5272
0 . 5 27 4
0 . 5 2 76
0 . 527 7
C. 52a E
0. 5280
0. 5283
C. 5287
0 . 5279
0 . 5282
78
0 . 5 291
0 . 5293
0. 5294
0 . 5 2 96
0 . 52 9 7
0 . 5 29 9
0. 5300
79
0 . 5 302
0 . 5 30 4
0 . 5 29 C 0 . 5 30 5
0. 5307
0. 5 308
0. 5 31 0
0. 5 3 1 1
0. 531 3
0. 5314
80
0.5317
0. 531 S
0. 5320
0 . 5322
0 . 532 3
0. 5325
8 1 82
0 . 5 3 32
0 . 5 334
0.5335
0. 5 337
0. 5347
0. 5 34 9
0 . 5 350
0 . 5 352
83
0 . 5 3 l: 2
0. 5 36 3
0 . 5 36 5
0 . 5 366
0 . 5 3 76
0. 5378
0 . 5 379
0 . 5381
C . 5 3 <;I
0. 5 392
0 . 5 394
0 . 5 3 95
0 . 5 40 5
0. 5 406
0 . 5 4C 8
87
0 . 54 19
0 . 5 42 C
88
C. 5 4 33
0. 5434
84 85 86
0 . 5 24 5
0 . 5 26 1
C. 5338
0. 5 34 0
0 . 5 36 8
0 . 535 5
0. 5353
0. 5326
0 . 5328
0. 532 9
0 . 53 1 6
0 . 5 34 1
0 . 5343
0 . 5 34 4
0. 5346
0. 5 3 3 1
0.5356
0. 5358
0. 5359
0. 5360
0 . 5 38 2 0 . 5 396
0 . 5 36 9
0. 5 371
0 . 5372
0 . 53 7 5
0. 5 3 8 4 O . 5 39 8
0. 5385
0. 538 7
0 . 5374 0. 538 8
0 . 5 399
0 . 54 0 1
0 . 54 0 2
0 . 54 0 4
C. 5409
0. 541 1
0 . 54 1 2
0 . 54 1 3
0 . 54 1 5
0 . 54 1 8
0 . 5 42 2
0 . 5423
0 . 542 5
0. 542 6
0 . 542 7
0. 543 0
0 . 54 3 2
0. 5 436
0 . 5437
0 . 5 42 9
0 . 54 1 6
0 . 543 8
0 . 544 0
0 . 5 44 2
0 . 5444
0 . 544 5
0. 5389
89
0 . 5447
C . 544 E
0 . 5 44 9
0 . 54 5 1
0 . 5 4 :> 2
0. 5455
0 . 5456
0 . 54 5 7
0 . 54 5 9
90
O . 5 45 3
0 . 544 1
C . 5460
0. 5 4 6 1
0 . 5 46 3
0 . 5464
0 . 5 46 6
0 . 5 46 7
0 . 5 4 l: 8
0 . 54 7 0
0 . 547 1
0 . 54 7 2
91 92
0 . 547 4
0 . 547 5
0. 5 47t
0 . 54 78
0 . 5479
0. 548 0
0 . 5484
0 . 54 8 6
C . 5 4 8 1:
0 . 54 82
0 . 548 3
C. 548 7
0. 548 9
0 . 5 4 91
0 . 5492
0 . 549 3
0 . 5495
0 . 549 1
0 . 549 9
93
0 . 5 5 00
C . 5 50 1
0 . 5 50 3
0 . 5 5 04
0 . 5 5J 5
0.5510
0 . 55 1 2
0.5513
0. 5514
0. 551 5
0. 551 7
0. 5521
0. 5522
0 . 5 52 3
0. 5524
95
C . 5526
0. 552 1
0. 5 528
0 . 5 5 30
0. 5518
0. 5 506 0. 551 9
0 . 5 5 09
94
0. 55C8
0 . 5496
C . 5 531
0. 5532
0 . 5535
0 . 5536
0. 5537
0 . 5 54 4
98
0 . 5563
0 . 5 56 5
0 . 5 56 6
0. 5567
0. 5556 0. 5568
0. 5 57 0
0. C. 0. 0.
0 . 5572
0. 551 3
99
0. 5 5 76
C. 5 57 7
0 . 5 5 79
0 . 5 58 1
0 . 5582
0. 5583
0 . 5584
0 . 5 586
0. 5 5 87
100
0. 5588
0 . 5 5 8 <;
0 . 5 578 0 . 5 59 C
0 . 5 5 92
0 . 5 593
0. 5 594
0 . 5 5 95
0 . 5 596
0 . 559 8
0. 5599
96
97
0 . 5538
0. 5540
0. 554 1
J . 5 5 42
0.5551
0. 5552
0. 5 �54
0. 5555
See note p . 4 6 2 .
0. 554 5 0. 555 7
55 3 3 5546 55 59 5571
0. 5547
0. 5549
0. 55 5 0
0 . 556 0
0 . 556 1
0. 5562 0 . 55 7 5
3'
r-
3"
(I)
'tI (I) Co I» :J n CD
-I I» 0-
ar II>
.j:I. en ...
Appendix B
462
Table B.24
(continued)
Zero-Sequence Resistance " Inductive Factors (re . xe )* (ohms/conductor /mile)
p
(ohm-meter) All
re
r.,. x 'il (f = 60 z) 0. 2860
1 6 10 60
xe
2.0�0 2.343 2.469 2.762 2.888t 3.181 3.307 3.600 3. 7 26
lOot
�OO 1 000 5000 10.000
*From formulas:
re = 0.00476 4f
xe = 0.006986f 10110 where
4.66 5.600
f - frequency = resistivity ( ohm-meter)
p
p ,
tThis is an averaae value which may be used in the absence of definite information. Fundamental Equations:
% 1 = Z3 = ra + j (xa + xd) Zo = ra + re + j (xa + xe - 2xd) xd = wk In d d = separation. ft
Table B.25.
Shunt Capacitive Reactance Spacing Factor x� (megohms/conductor/mile) at 60 0.2
0. 3
0.4
0.5
0 .6
0. 7
0. 8
0. 9
0 . 0000 0 . 0 2 06 0 . 0 326 0 . 041 1 0 . 0417 0 . 0532
- 0 . 0 6 83 0 .0028 0 . 022 0 0 .0336 0 . 0419 0 . 0483 0 . 0 536
- 0 . 047 7 0 .0054 0 . 02 3 4 0 . 0 34 5 0 . 0426 0 . 0489 0 . 0541
-0. 0357 0 . 0 0 78 0 . 0 2 47 0 . 0 3 54 0 . 0 433 0 . 0495 0 . 0 546
-0. 0272 0 .0100 0 . 02 6 0 0 . 0363 0 . 0440 0. 0500 0 . 05 5 1
- 0 . 02 06 0 . 01 20 0 . 0272 0. 0372 0 . 0446 0 . 0 5 06 0 . 0555
-0 . 01 5 2 0 . 01 3 9 :> . 0 2 8 3 0. 0380 0 . 045 3 0 . 05 1 1 0 . 0560
- 0 . 0 1 06 0 . 01 5 7 0 .0295 0 . 0388 0 . 04 5 9 0 . 0516 0 . 0 5 64
-0 . 0066 0. 0 174 0 . 0 30 5 0 . 0 396 0 . 046 5 0 . 0 5 21 0 . 0 56 9
- 0 . 0 0 31 0 . 01 90 0 . 0 3 16 0 . 0 404 0 . 0 4 71 0 . 0 521 0 . 0 5 73
0 . 0 5 17 0 . 061 7 0 . 06 52 0 . 0683
0 . 0581 0 . 06 2 1 0 . 065 5 0 . 0686
0. 0586 0 . 0624 0 . 06 5 8 0 . 06 8 9
0 . 0 5 90 0. 0628 0 . 0 662
0 . 0598 0 . 063 5 0 . 06 6 8 0 . 06 9 8
0 . 0602 0. 0638 0 . 0&71
0 . 0645 0 . 0677
0 . 06 13 0 . :> 6 4 9 0 . 06 80
0 . 0 10 0
0 .0606 0 . 0642 0 . 06 74 0 .0103
0 . 0609
0 . 0 6 92
0. 0. 0. 0.
0 . 0 706
0 . 0 109
0 . 071 1 0 . 07 3 7 0 . 0 16 1 0 . 07 83 0 . 0803 0 . 0 823 0 . 0 84 1 0 . 0857
0 .0114 0 . 0 74 0 0 . 0163 0 . 0185
0 . 071 7 0 . 014 2 0 . 016 5 0 . 07 8 7 0. 0801 0 .0826 0 . 0844 0 . 08 6 1 0 . 0877 0 . 089 2
0. 0. 0. 0.
0719 0 74 5 0 76 8 0189
0 . 0 122 0 . 0741 0 . 0770 0 . 0191
0. 0. 0. 0.
072 5 0 1 49 071 2 0193
0. 0. 0. 0.
0121 '152 0 714 0195
0. 0. 0. 0.
0 . 0 13 5
0 . 0 80 9 0 .0 828 0 . 0846
0 . 081 1 0 . 0830 0 . 0847 0 . 0864 0 . C880 0 . 0 895
0. 0. 0. 0. 0. 0.
081 3 0 8 32 0849 0866 0881 08 9 6
0. 0. 0. 0. 0. 0.
0815 0833 0851 0 861 0883 0 898
0 . 0817 0 . 0835 0 . 08 5 2 0 . 0869 0 . 0 8 84 0 . 08 99
0. 0132 0 . 0 756 0 . 0 119 0. 0799 0 .0819 0 . 0 83 1 0 . 0 8 54 0 . 0810 0 . 0886 0 . 0 90 0
0 . 09 0 9 0 . 09 2 2 0 . 09 3 5 0 . 0 948 0 . 0 96 0
0. a91 0 0 . 0924 0 . 093 1 0 . 094 9 0 . 096 1 0 . 0972
0 . 0 91 2 0 . 0925 0 . 0938 0 . 09 5 0 0 . 0962 0. 0913 0 . 09 8 4 0 . 0995 0 .1005 0. 1015
0 . 091 3 0 .0926 0 . 0939 0. 0951 0 . 0963 0 . 0914 0 . 0985 0 . 0996 0 . 1006 0. 1016
0 . 0 91 4 0 . 0 92 8 0 . 0 940 0 . 095 3 0 . 0 9 64 0 . 0 976 0 . 0 9 86 0 . 0997 0 . 1 007 0 . 1 01 7
0 . 1025
1 0 26
0.0 0
1
2 3 4 5 6
7 8 9 10 11 12
13
14 15
16
17 18 19 20
21 22 23 24 25 26 27 28 29 30 31
-----
0 . 0 874 0. 0889 0 . 0903 0 . 09 1 7 0 . 0930 0 . 0 943 0 . 095 5 0 . 0967 0 . 0978 0 . 0989 0 . 0999 0 . 1009 0 . 1 01 9
0.1
0 . 0805 0 . 0 8 24 0 . 0842 0 . 0 8 59 0 . 0 87 5 0 . 0890 0 .0905 0 . 091 8 0 . 0931 0 . 0 9 44 0 . 0956 0 . 0 96 8 0 . 0 919 0 . 0990 0 . 1 000 0 . 1 01 0 0 . 1 02 0
0 . 0 9 116 0 . 09 2 0 0 . 09 3 3 0 . 0945 0. 095 7 0 . 09 6 9 0 . 0980 0 . 0991 0. 100 1 0 . 101 1 0. 1 02 1
0 . 0 86 2 0 . 0 87 8 0 . 0893 0 . 0 90 7 0 . 0 92 1 0. 0934 0 . 0 94 1 0 . 0 95 8
0. 0. 0. 0. 0.
0 97 0 0981 0 992 1 002 1 01 2
0 . 1 02 2
0 5 94 06 31 06 6 5 06 9 5
0 . 0971 0 . 09 8 2 0 . 0993 0 . 1 00 3 0 . 101 3 0 . 1 02 3
0 . 0983
0 . 0994 0 . 1004 0 . 1 01 4 0 . 1023
) . 1024
0130 0754 0 176 0797
Hz
0.
0 . 0 1 59 0 . l 181 0 . 0 8 01 0 . 0 821 0 . 0 8 39 0 . 0 856 0 . 0 8 72 0 . 0 887 0 . 0 90 2
c: ::l CD
3"
CD c. I» ::l 0 CD -I
"0
I» C"
CD f/J
0 . 0916 0 . 0 9 29 0 . 0 9 42 0 . 0 954 0 . 0965 0 . 0 977 0 . :> 9 8 7 0 . 0 99 8 0 . 1 008 0 .1 018 0 . 1 0 27
� en
c,,)
Table B.25 0.0
0.1
32 33 34 35 36 37 38 39 40
0 .1028 0 . 1 037 0 . 1 046 0 . 1055 0 . 1 06 3 0 . 1071 0 . 1 0 19 0 . 1 08 7 0 . 1 094
0 . 1 029 0 . 1 03 8 0 . 1 04 7 0 . 1 056 0 . 1 0b 4 0 . 1 072 0 . 1 080 0 . 1 08 8 0 . 1 095
41 42 43 44 45 46 47 48 49 50
0 . 1 1 02 0 . 1 109 0.1116 0. 1 123 0 . 1 1 29 0.1136 0 . 1 1 i.. 2 0 . 1 1 48 0 .1155 0 . 1 1 61
0 . 1 1 02 0 . 1 1 10 0 .1117 0.1123 0 . 1 1 30 0 . 1 1 36 0 . 1 1 43 0 . 1 1 49 0.1155 0. 1 161
51 52 53 54 55 56 57 58 59 60
0 . 1166 D . l 172 0. 1 1 18 0 .1183 0 . 1 1 89 0 . 1 1 94 0 . 1199 0 . 1205 0 . 1210 0 . 1 21 5
0 . 1 1 67 0 . 1 1 73 0. 1 1 78 0 . 1 1 84 0 .1 189 0 . 1 195 0 . 1 2 00 0 .1205 0 .1210 0 .1215
61 62 63
0 . 1 220 0 . 1 224 0 . 1 22 9
0 . 1220 0 .1225 0 . 1 23 0
(continued) 0. 5
n.3
0.4
0 . 1030 0 . 1 03 9 0 . 1 048 0 . 10 5 6 0 . 1 06 5 0 . 1073 0 .108 1 0.1088 0 . 1 09 6
0 . 1 03 1 0 . 1 0 40 0 . 1 049 0 . 1 057 0 . 1 0b 6 0 . 1 0 74 0 . 1081 0 . 1 089 0 . 1 0 97
0. 1032 0 . 1041 0 . 1 05 0 0 . 1058 0 . 1066 0 . 1074 0 . 1082 0 . 1 09 0 0 . 1097
0. 0. 0. 0. 0. 0. 0. 0. 0.
0 . 1 1 03 0 . 1110 0 .1117 0. 1124 0 . 1131 0 .1137 0 . 1 1 43 0 . 1150 0 . 1 1 56 0 . 1 1 62
0 . 1 1 04 0. 1 111 0.1118 0. 1 125 0 . 11 3 1 0 . 1 1 38 0 . 1 1 44 0 . 1 1 50 0 . 1 1 56 0 . 1 1b2
0 . 1105 0. 1112 0 . 1119 0.1125 0 . 1 1 32 0 . 1138 0 . 1145 0 . 1 1 51 0.1157 0 .1163
0 . 1 168
0. 1 1 68 0 . 1 1 74 0 . 1 1 80 0. 1 1 85 0 . 1 1 90 0 . 1 1 96 0 . 1 201 0 . 1 206
0 .2
O .HB
0.1179 0 . 1 184 0 . 1 1 90 0. 1195 0 . 1 200 0 . 1206 0 . 1 21 1 0 . 121 6 0. 1221 0 . 122 5 0. 1230
0.6
0. 7
0. 8
0. 9
1 03 4 1 043 1051 1 060 1 0b 8 1 07b 1 0 84 1 091 1 09 9
0 . 1 035 0 . 1 0 44 0 .1052 0 . 1 06 1 0 . 1 06 9 0 . 1 017 0. 1085 0 . 1 092 0 . 1 1 00
0 . 1 035 0 . 1 044 0 . 1 053 0 . 1 06 1 0. 1 070 0 . 1 01 8 0 . 1 085 0 . 1 0 93 0 . 1 1 00
0 . 1 036 0 . 1 04 5 0 . 1 054 0 . 1 062 0 . 1 07 0 0 . 1 018 0 . 1 086 0 . 1 094 0 . 1 1 01
0 . 1 1 05 0.1112 0. 111 9 o . 1 12b 0. 1133 0. 1139 0 . 1 l45 0 . 1 1 52 0 . 1 1 58 0 . l 1 b4
0 . 1 1 0b 0.1113 0 . 1 1 20 �.1127 0 . 1133 0 . 1 1 40 ) . 1 1 46 0. 1152 0. 11 58 0 . 1 1 64
0 . 1 1 07 0.1114 0 . 1121 0 . 1 1 27 0 . 1 1 34 0 . 1 1 40 0 . 1 1 41 0 . 1 1 53 0 . 1 1 59 0.1165
0 . 1 107 0 . 1 1 14 0. 1 121 0 . 1 128 0 . 1 1 35 0. 1 141 0 . 1 1 47 0 . 1 1 53 0 . 1 1 59 0. 1165
0 . 1 1 08 0.1115 0 . 1 1 22 0 . 1 1 29 0.1 135 0 . 1 1 42 0 . 1 1 48 0 .1 1 54 0 . 1 16) 0. 1 166
0. 1 21 1 0 . 1 216
0. 1169 0 . 1 1 74 0. 1180 0. 1186 o . 11 91 0 . 1 1 96 0 . 1202 0 . 1207 0.1212 0 . 1 21 7
0. 1169 0. 1175 0 . 1 1 81 0 . 1 1 86 0 . 1 1 92 0 . 1 1 97 0 . 1 2 02 0 . 1 207 0.1212 0. 1217
0. 1 1 70 0 . 1 1 76 0. 1181 0 .1187 0 . 1 1 92 0 . 1 1 97 0 . 1 203 0 . 1 208 0.1213 0; 1218
0 . 1 1 70 0 . 1 1 16 0 . 1 1 82 0 . 1 1 87 0 . 1 1 93 0 .1 198 0 . 1 203 0 . 1 2 08 0. 1213 0 . 1 21 8
0 .1171 0 . 1 1 71 0 . 1 1 82 0 . 1 1 88 0 . 1 1 93 0 . 1 198 0 . 1 2 04 0 . 1 2 09 0 . 1 214 0 . 1 21 9
0 . 1 1 72 0 . 1 1 77 0 . 1 1 83 0 . 1 1 88 0 . 1 1 94 0 . 1 1 99 0 . 1 204 0 . 1 2 09 0 . 1 214 0 . 1 219
0 . 1 221 O . 1 22b 0 . 1 231
0 .1221 0.1226 0 . 1 231
0.1222 0. 1 22 7 0. 1 231
:> . 1 2 2 2 0. 1 22 7 0 . 1 2 32
0 . 1 223 0 . 1 22 8 0 . 1 2 32
0 . 1 223 0 . 1 228 0 . 1 233
0 . 1 224 0 . 1 229 0 . 1 233
1 03 3 1 042 1050 1059 1 0b 7 1075 1083 1 09 1 1 098
0. 0. 0 . v. 0. 0. 0. 0. 0.
""
""
en
» '0 '0 CD ::J
Co x'
m
64 65 66
67 68 69 70
71
7'
7� 74 75 7e 77 78 79
�o
91 �2 8� e4
95 86 87
88 ��
;�
;!
;2 9! ':) 4
? ':
96 ;7
'; 1 '; ;
� ') ':
See note
0 . 1 234 0 . 1 23 8 0 . 1 243 0 . 1 24 7 0 . 1 252 0 . 1 2 56 C . 1 260 0. C.
1 265 1 26 9
0 . 1 21 3 0 . 1 2 11 0 . 1 2 81 0. 0. 0. 0.
1 28 5 1 289 1 292 1 29 6
0 . 1 300 C. 0. 0. 0. 0. 0. 0. 0.
1 304 1 301 1311 1 31 4 1 31 8 1321 1 32 5 1 32 8
0 . 1 33 2 0 . 1 33 5
. 1 234 .1239 . 1 243 . 1 248 . 1 2 52 . 1 2 57 . 1 261
0 . 12 3 5 0 . 1239 0 . 1 2 44 0 . 1 248 0 .1253 0. 1257 0 . 1261
0. 1 235 0 . 1 2 40 0 . 1 2 44 0 . 1 249 0 . 1 2 53 0. 1 2 51 0 . 1 262
0. 1236 0 . 1 240 0 . 1 245 0 . 1 2 49 0 . 1 2 54 0.1258 0. 1262
0. 0. 0. 0. 0. 0. 0.
1236 1 2 41 1 245 1250 1 2 54 1 2 58 1 262
0 .1237 0 . 1 241 0 . 1 246 0 . 1 250 0 . 1 2 54 0 . 1 2 59
0 . 1 265 0 . 1 269
0. 126 5 0 .1270
0. 1 266 0 . 1 270
0 . 1 266 0 . 1 270
0. 1267 0 . 1 271
0. 1274 0 . 1 278 0 . 1282 0 . 1 286 0 . 1289
0 . 1 2 74 0 . 1 278
0 . 1 2 74 0 . 1278 0 .1282 0. 1286 0 . 1 290 0 . 1 2 94 0 . 1 29 8 0 . 1301
0. 0. 0. 0. 0. 0. 0. 0.
1 27 5 1 279 1 2 83 1 2 87 1 291 1 294 1 298 1302
0 . 1 261 0 . 1 27 1 0 . 1 275 0. 1 279 0. 1 283 0 . 1 287 0 . 1 2 91 0 . 1 2 95 0. 1 299 0 . 1 302
0 . 1 268 0 . 1 2 72 0 . 1 2 76 0.1280 0 . 1 2 84 0 . 1 288 0 . 1 2 91 0 . 1 295 0 . 1 2 99
0 . 1 329 0 . 1 333 0 . 1 336
0 . 1 30 5 0 . 1 309 0.1312 0 . 1 31 6 0 . 1319 0 . 1 323 0 . 1326 0 . 1 330 0. 1333 0 . 1 336
0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
1 3 06 1 309 1313 1316 1320 1 323 1 327 1330 1 333 1 337
0 . 1 3 06 0 . 1 309 :>' 1 3 1 3 0. 1311 0. 1 320 0 . 1 324 0 . 1 32 7 0. 1330 0 . 1 334
0 . 1 339 0 . 1 34 2 0 . 1 3 46 0 . 1 349 0 . 1 352 0 . 1 355 0. 1 3 58 0 . 1 361 0 . 1 364 0 . 1 367
0 . 1340 0 . 1 3 ". 3 0 . 1 346 0 . 1 3 ". 9 0 . 1 35 2 0.1355 0 . 1358 0 . 1 36 1 0 . 1 364 0 . 1 36 7
0. 0. 0. 0.
1 3 40 1 3 ". 3 1 3 ". 6 1 349
0 . 1 3 40 0 . 1 343
0. 0. 0. 0. 0. 0.
1 353 1356 1359 1 36 2 1 36 5 1 36 8
0 0 0 0 0 0 0
0 . 1 2 73 0 . 1 271 0.1 281 0 . 1285 0 . 1 289 0 . 1 2 93 0 . 1 297 o
. DOC
0 . 1 293 0 . 1297 0.1301
0 . 1 3 04 0 . 1 308
0 . 1 30 4 0. 1308
0 . 1 31 1
0. 1 31 2 0 . 1 31 5 0 . 1 31 9 0 . 1322 0 . 1 32 6 0. 1329
0 0 0 0
.1315 . 1 31 8 . 1 32 2 . 1 325
0. 1 329 0 . 1 332 0 . 1 335
0 . 1 33 2 0 . 1336
0 . 1 33 8 � . 1 341 0 . 1 345 0 . 1 348 0 . 1 351 0 . 1 3 54 0 . 1357 0 . 1 360
0. 0. 0. 0.
0 . 1 361
0. 1339 0 . 1 3 42 0 . 1 345 0 . 1 348 0 . 135 2 0. 1355 0. 1358 0 . 1 36 1
':' . 1 3 6 3
0 . 1 364
0 . 1 3�4 0 . 1367
:>
•
1 366
p . 466.
1 3 3 <1 1 3 42 1 345 1 348
0 . 1 3 51 0 . 1 3 54 0 .1357
0 . 1 366
0 . 1 282 0 . 1. 2 8 6
0. 0. 0. 0.
1 290 1 29 4 1 2 97 1 30 1
0. 0. 0. 0. 0. 0. 0.
1 1 1 1
305 308 31 2 316 1 31 9 1 322 1 326
0. 1263
0. 1 33 7
0 . 1 3 47 0.1350 0. 1353
) . 1 356 0. 1 3 5 9
0 . 1 36 2
0 . 1 365 0. 1 368
0 .1237 0 . 1 2 42 0 . 1 2 46 0 . 1 2 50 0. 1255 0 . 1 2 59 0 . 1 263
0 . 1 237 0 . 1 2 42 0 . 1 247 0 . 1251 0. 1255 0 . 1 2 60 0 . 1 264
0 . 1 238 0 . 1 2 42 0 . 1 2 41 0 . 1 2 51 0 . 1 2 56 0 . 1 26 0 0 . 1 264 0 . 1 26 8 0 . 1 2 72
0 . 1 3 03
0 . 1 268 0. 1 21 2 0 . 1 276 0 . 1 2 80 0 . 1 2 84 0 . 1 288 0 . 1 29 2 0 . 1 296 0 . 1 299 0 . 1 3 03
0 . 1 306 0 . 1 31 0 0 .1313 0 . 1 31 7 0. 1 320 0 . 1 3 2 '" 0 . 1 32 7
0. 0. 0. 0. 0. 0. 0.
0 . 1 307 0 . 1 31 1 0 . 1 3 14 0.1 318
0 . 1 3 31 0 . 1 334 0 . 1 331
0. 1 331 0 . 1 334 0. 1 33 8
0 . 1 321 0 . 1 325 0 . 1 328 0 . 1 33 1 0 . 1 335 0 . 1 3 38
0 . 1 3 40 0 . 1 3 "' ''' 0 . 1 3 ". 7 0 . 1 350 0 . 1 3 53 0 .1356 0 . 1 359 0 . 1 362 0 . 1 365 0 . 1 368
0 . 1 3 ". 1 0 . 1 3 44 0 . 1 341 0. 1350 0. 1 353 0 . 1 357 0 . 1 360 0 . 1 36 3 0 . 1 366 0 . 1 36 9
0 . 1 3 41 0 . 1 3 44 0 . 1 348 0 . 1 3 51 0. 1 3 54 0 . 1 3 57 0 . 1 360 0 . 1 363 0 . 1 366 0 . 1 3b9
1 1 1 1 1 1 1
301 310 31 4 31 7 321 3 2 '" 32 8
0 . 1 2 76 0 . 1 280 0 . 1 2 84 0 . 1 288 0 . 1 2 92 0 . 1 296 C . 1 300 0 . 1 3 03
C :J CD
3 'C CD
Q. I»
:J n CD
..... I» c-
CD en
� 0) C1I
466
Appendix B
Table B.25 (continued)
Zero-Sequence Shunt Capacitive Reactance Factor % (meeohms/conductor/mi)
�
Co n duc tor Hei,h t a boH Ground ((t) 10 U 20 25 30 40 60 60 70 80 90 1 00 ,
Xo
where
h
(t
%' .. 6 & Hz) 0. 267 0.303 0.328 0. 3 1 8 0.364 0.390 0.410 0. 426 0. 440 0.462 0.462 0.47 2
1 2. 30
= -(- 10110 2 h
= height above around ( .. frequency
Fundamental Equati ons: , XI
x�
,
,
= x , = xa = %d x� + x � - 2 xd ,
==
%d .. ( 1 /wk' ) In d d .. separation. It
appendix
c
Tri g o n o m etri c I d e n t i t i es fo r Th re e - P h a se Syste m s
In solving problems involving three-phase systems, the engineer encounters a large number of trigonometric functions involving the angles ± 1200 • Some of these are listed here to save others the time and effort of computing these same quantities over and over. Although the degree symbol e ) has been omitted from angles ± 1200 , it is always implied.
sin' (8
sin (8
±
120) = - ! sin 8 2
±
cos 8
(C . 1 )
cos (8
±
120) = - ! cos 8 2
v'3
=1=
v'3
sin 8
(C . 2 )
±
120) = =
cos' (8
±
120)
±
±
!+! 2
4
cos 28
±
v'3 4
sin 28
2
120)
4
=
1 20) =
120)
4
-� sin' 8 � sin 8 cos 8
(C.4)
±
-
!+! 4
4
cos 28
±
va sin 28 4
-.� cos' 8 � sin 8 cos 8
(C.5)
=1=
-! - !
= -
=
(C.3)
±
! - ! cos 28 ± v'3 sin 28
4
8 cos (8
=1=
� cos' 8 � sin' 8 V; sin 8 cos 8
=
sin
=1=
+
=
cos 8 cos (8
2
� sin' 8 � cos' 8 V; sin 8 cos 8 +
=
=
sin 8 sin (8
2
4
cos 28
=1= ..[3 4
sin 28
� /lin 8 cos 8 V; sin' 8
(C.6)
=1=
- ! sin 28 ± v'3 4
4
cos 28
+
v'3 4
(C.7) 467
Appendix C
468 ±
cos 8 sin (8
-� sin 8 cos 8 V; cos2 8 ±
=
120)
= - .! sin 28 -+ va cos 28 ± va 4 4 4 sin (8 + 120) cos (8 + 120)
= - .! sin 8 2
- va
cos 8
cos2 8
4
va
+
4
= - .! sin 28 - va cos 28 4
sin (8
+
120) cos (8 - 120)
sin (8 - 120) cos (8
+
1 20)
(C.S) sin2 8 (C.9)
4
va = .! sin 28 va
(C.10)
= .! sin 28 + va 2 4
(C.1 1 )
= sin 8
cos 8
= sin 8
cos 8 + va 4
_
-
4
2
4
(C.12) sin (8
+
120) sin (8 - 1 20)
cos (8 + 120) cos (8 - 1 20)
= .! sin2
8
= 41 cos2
8 -
4
sin ( 28
±
120)
=
cos ( 28
±
120)
=
sin 8 cos 8
+ +
-
-
-
� cos2 4
8 =
4 sm2
8
3 .
- .! - .! cos 28 4
2
= - 41 + 2"1
� sin 28 V; cos 28 � cos 28 V; sin 28
cos 28 (C.14)
±
(C.15)
+
(C.16)
=0 + 120) = 0
(C.17)
sin (8 - 120) + sin (8 + 120)
cos (8 - 120) + cos (8
(C.13)
sin2 8 + sin2 (8 - 120) + sin 2 (8 + 120) cos2 8 + cos2 (8 - 120) + cos2 (8 + 120)
=
(C. 1S)
�
(C.19)
= 2"3
sin 8 cos 8 + sin (8 - 120) cos (8 - 120) + sin (8 + 120) cos (8
(C.20) +
1 20)
=0 (C.21)
In addition to the above the following commonly used identities are often
required. sin2 8
+
cos2 8
sin 8 cos 8
=1 =
� sin 28
Trigonometri c Identities for Three- Phase Systems
cos 2 8
-
sin2 8
= cos 28
8
= 1 + cos 28
8
=1
-
2 cos 28 2
469
appendix
D
Se lf I nd u cta n ce of a Stra i g ht Fi n i te Cyl i nd ri ca l Wi re
In Chapter 4 much use is made of the primitive inductance of a straight finite line of cylindrical cross section which appears in equation ( 4. 20) as' L =
I1 w S 81T
+
I1 m S 21T
( 2 s - 1) In
H
r
(D. 1 )
I1 w = permeability of the wire, H/m 11 m = permeability of the surrounding medium, H/m
This expression is due to Rosa and Grover [ 25] who made an intensive study of the inductance of common physical circuit arrangements. Although their work is recommended reading, the modern engineer will be troubled by its use of the CGS system of units. A more modern treatment of the subject leading to the same re sult is that of Attwood [ 23 ] . Here the MKS rationalized system of units is used, and the subject is thoroughly explored. In studying material like that of Chapter 4, the engineer often rebels at having unfamiliar facts given without proof. Since (D. 1 ) is not familiar to all, it will be developed here, following a derivation similar to that of Attwood [ 23] . The self inductance of a wire is divided into two parts-that due to flux inside the wire, called L" and that due to flux external to the wire, called L e . Consider a long, straight, cylindrical wire of radius r and length both in meters, made of a material having permeability I1 w henry/meter. To compute the internal induc tance, we assume that the current is uniformly distributed such that by the Biot Savart law we may write the field intensity at a point u meters from the center as
s,
Hu
=
i enclosed � = A/m 21Tr2 21TU
Then Bu = I1 w Hu = where I1 w = 41T X 1 0- 7 11 w r
-
iU l1w 2 1Tr2
(D.2)
Wb/m 2
(D.3)
H/m
J.l w r = relative permeability of the wire material , Both w and 11m may be written as the product o f 0 4 1T 11 I1 of free space and a relative permeability J.lwr or I1mr. a numeric. =
470
X
1 0-7 HIm, the permeability
Self I nd u cta nce of a S traight Finite Cylindri ca l Wire
47 1
Since the wire is s meters long, it contains a finite amount of energy which can be related to the inductance by the familiar expression W = ( 1/ 2 ) Lj i2
J
(D.4)
If we can compute W, we can find Lj • Attwood [ 23 ] points out that the energy density if. proportional to B2, which is known, and is often given by energy density = B2/2tJ. w
J/m3
(D.5)
Suppose we consider a thin cylindrical shell of the conductor with radius u, thick ness du, and length s. Then ( D. 6 )
dW = (B�/2tJ. w ) ( 21T US d u )
or, substituting for B u from (D.3) ''2 tJ. w S dW - -u 3 du 41T r4
and W
- i2 tJ. w S 41T r4
- --
I 0
Tu
_ ;2 tJ. w s 3 du - -161T
J
(D. 7)
But from (D.4) the inductance due to internal flux is Lj = 2 W/i2 = - tJ. w S/81T
H
(D. 8)
The external partial inductance is easier to compute by the familiar expression
flux
Le = >" e /i = NtPe /i
(D.9)
where >" e is the external linkage, N is the number of turns, and tP e is the ex ternal flux. Since we are considering the inductance of only one long straight wire, the concept of a flux linkage is troublesome to many engineers. This may be ex plained by the circuit configuration shown in Figure D.l where the length s of the wire under consideration is part of a loop of wire. However, it is apparent that if we can compute the self inductance of only that length s we have found the in ductance of an "isolated" wire. The inductance of such an isolated wire is of no value whatever since the current i in (D.9) must begin and end at the same point (the generator). However, as shown in Chapter 4, the concept of an isolated wire is extremely valuable in determining the impedance matrix of a group of parallel wires. To find the external inductance, we must first find the density Bp at the external point P, u meters perpendicular from the wire and a meters from one end,
t[p- T e'
----�-' j - --
-----0__----,.' _ -
- -
t_
_
A� u
flux
e
_ _ _
L
Fig. D. l . Inductance of wire section AA ' of length s. (From Electric and Magnetic Fields, 3rd ed. , Stephen S. Attwood, John Wiley and Sons, Inc., New York, 1 949. Used with permission. )
Appendix 0
472
o
Fig.
E
D.2. Intensity of a straight current section. (From Electric and Magnetic Fields, 3rd ed. , Stephen S. Attwood, John W iley and Sons, Inc. , New York , 1 949 . Used with permission. )
shown in Figure D.l. This may be determined in turn by finding the intensity due to i which is equal numerically to the force on the wire due to a unit magnetic pole at P [ 23] . From Figure D.2 we write the force dF due to i in element dx as as
Hp at
P
dF
But if there is a unit pole at
= ( i sin 0 ) dx B
P, the density at line segment dx is 1 cos 2 a ____ B = 1.- X = 4tr x2 + u2 4tru 2 flux
Then, since sin 0 = cos a, we write (D.10) term by term as
dF = (cos 2 a/41fu 2 ) X (i) X (cos a ) X (uda /cos2 a )
i [I cos a da = -' (sin a , F=4tru 4tru
and
(D.10)
•
O!2
±
= i cos a da /4tr u
sin a 2 ) = Hp
(D. ll )
iPm ( s - a2 + 2a ) Wb/m2 (D.12) u 2 ya 4tru y (s - a ) u2 where Pm 41f X 10-7 Pm r is the permeability of the medium surrounding the wire. Here is a vector pointing downward in Figures D. l or D.2 and normal to the plane containing the wire and P. The total in the area of Figure
Choosing the plus sign, (D.l l ) may be written as Bp
=
+
+
=
Bp
AB B' A '
flux
D.l may be found by integrating Bp over this area, noting carefully that Bp is everywhere normal to this plane. Also note that we are finding only the flux in a strip s meters long and extending radially from the wire. This is precisely the flux tP e needed in (D.9) to compute the external inductance. Integrating (D.12) we have tP e
=
= = JUu=r oo 1°0=0 8
Bp
d u da Wb
s + Y S2 + U2 ys2 + u 2 - U - s in --'---2tr u
ip. ! = ---1!
(
) u=oo u=r
(D.1 3 )
S e lf I nd uctan ce of a S traight Finite Cylindrica l Wire
(
Attwood shows that this expression may be written as 4>
i e = /J....2!L. 21T
-1 82 - 8 In �-:--;:;:� :::; =� Y 82 + U 2 + U (8/ U ) - Y (82 /u 2 ) + 1
(-
and the limits then inserted to find 4> e
i = /J....2!L. 21T
8 In
Then
Le = /J.-1!!. 21T
(
S
In
S
-
+ Y 82 + r2 r
-
8 + "';82 + r2 r
)
y 82 + r2 + r
)
y 82 + r2 + r
)
473
u =oo u =r
Wb
(D. 14)
H
and since 8 > > r in every case of interest, we write this equation as
Le = (/J.m 8/ 21T ) or per unit of length,
(� ) 8
In
-1
H
(D. 1 5)
( D. 16) It seems strange that the inductance per unit of length is a function of the length 8. However, as shown in Chapter 4, the argument of the logarithm may have any numerator since it always disappears when circuits containing a return path for the current are used. From (D.8) and (D.16) we have the desired result, viz.,
f fl + fe =
= (/J. w/81T ) + (/J.m/21T )
(
In
)
28 -1 r
Him
(D. 1 7)
The mutual inductance is computed in 'a similar way except in this case the limits of integration are from D to 00 , where D is the distance to the linked current i2 • Then (D.18) or (D.19) These results for the self and mutual inductance for a cylindrical wire of length 8 are both strange and interesting. They are strange because portions of the solution such as the 28 in the logarithm and the (- 1 ) in the parentheses never ap pear in any practical problem involving a "return" wire. The results are of in terest, however, because these primitive building blocks provide powerful tools for analyzing complex problems involving many wires.
appendix
E
S olved Exa m ples
It is often useful for the student of symmetrical components to apply the techniques being studied to a typical system. The purpose of this appendix is to specify small sample systems along with their complete solutions for both normal and faulted conditions. Following are solutions for 3-node, 6-node, and 14-node networks. The 3node network used here is the same system shown in Figure 11.2 and used in the examples of Chapters 11 and 12. The 6-node network is a well-known circuit introduced by Ward and Hale [79] to which a table of zero sequence data has been added. The 1 4-node network was adapted from the IBM assembler load flow described in [80] . These three networks will provide useful data for checking hand computations and computer programs. E.1
A 3-Nocle Network
Consider the 3-node network shown in Figure E . ! . This is the same network as that of Figure 1 1 .2 and is redrawn here for convenience. A normal load flow
for the 3-node network is given in Table E . l where an arbitrary load and genera tion pattern is assumed.
0 , ®
<;Yi
Fig.
I
..
5
6
@
1"0
E . l . A a-node network.
'\
The primitive impedance data is given in Table 1 1 . 1 , where it is noted that the positive and negative sequence data are not equal for the two generators. The usual practice, especially in digital computer solutions, is to assume that the positive and negative sequence impedances are equal and to provide only one set of positive sequence data. When these impedances are not equal, as in the system of Figure E .l , one can either ignore the negative sequence data and set Z2 = Zl or average the two impedances. The first alternative will give correct results for 3<; faults but will result in slight errors on all computations which involve the nega tive sequence network. Averaging Zl and Z will make SLG faults correct and 3<; 2 faults incorrect. Here we shall average the positive and negative sequence data and will call this result the positive sequence data. 474
Table E. 1. I CW A S T AT E UN I VE R S I T Y V E R S I O N OF
T t Tl E-?
8US
L INE
AND
X--------
0
P
1 2
N O.
2 3
2 1
3 �
BASE
C A SE
3 6 0 1. 0AO F l ON PROG RAM
E X AMPL E
T R Ar� S M 1 S S I ON
L INE
A 3 -Node Network Load Flow Study
0 0
1
0 8US
R I PCT I 0. 0 0.0
0. 0 0.0
DA T A
T' U N S F OR M ER
A C T UA L
D AH
-------
X ( P CT '
8 . 00 6 .·00
u . OO
6 . 0 1)
A S S E M8 L Y
- - X X - - - C ONV E R T E D N O T A P
K V AC
8C 1 2 1 PU t
G C PU I
0. 0 0. 0 0. 0 0. 0
0. 0 0. 0
0. 0
0. 0
0. 0 0. 0 0. 0 0. 0
E F F E C T ----X
8 I PU .
1
3
I\U� 1
610
3
64
8 U';
B US 2
64
O. O. O. O.
- 1 6. 6 6 6 7 - 1 6. 6 6 6 7 - 1 2 . 5000
- 7. 6 9? 3
TAP
L I MITS TM I N TMA X
PHAS E SH I F T
0.0 0.0 0. 0 0.0
X----- M A P D A T A - - - - - - X TaP F L OW R E V �LOW PG LOC 0 lOC PG L O C Q
0
0
0
0
0 0 0 0
0
0
0
0
G 0 0 0
E NT E � f D
X ------ B U S ------ X X - - -- VOL T A G E ----- x- - - L OAD - - - - X X - G E N E R A T I ON - x re V A R MW MV AR NAM E MW A R E A R E G M A G I PU . ANG I OE G . NC.
2
TAP MV A R A T I NG U T I O
2 0 1
1 . 03 0 1 . 00 1) 1 . 020
0. 0 0. 0
0. 0
0. 0 50 . 0 80. 0
0. 0 20 . 0 40. 0
0. 0 0. 0 1 00. 0
0. 0 0. 0 0. 0
QM I N
"VA R
QMA X
MVAR
0. 0 0. 0
0. 0
0. 0
0.0
70. 0
R EA C TOR
X - - - - - - - --- � AP D A T A - - - - -- - - - -- X LOAD GEN VO L T R E ACTe" KVAR P A G E L OC A l OC Q LOC OS LOC 0
o. O. O.
0 0 0
0
0 0
0 0
0
0 0
0
0 0 0
CJ) 0
<" (I) 0-
m
)(
3 'C CD III
IJJ
SU"" A�Y
L I �� A N D B U S T O T A L S ACTUAL ---- --- - - - - - ----- -TO A N �� f S S f O N L I NE S .. T P AN SF OR � E P S - F I X E D 0 - LTC 0 P H A S !: $H f F T f R S 0 T � T A L L I N E S ------ - - 4 B US E S - NON R E G ( I NC L UD I NG S W I NG . 2 - G E N E R A T OR 1 T eT A L � US !: S --------3 C A P A C I T O R S OR P E A C T O P S 0
T HE ' E
A� � NO VOL TA GE S
UNDE R 0. 950
THE R E
ARE
OVER
NO VOL T A G E S
1 . 0 50
MA X
750 2 50 2 50 0 750
TOTAL L O A D TOTAL l O S S E S l I NE C HA R G I N G F I X E D C A P I R E AC
1 QO 50 0
TOTA L
S Y S T E M M I SMATCH
MW - --- - -- - -1 30 . 0 0 0 0 . 000
MYAP ------ - - - -
6 0. 000 0. 929 0. 0 0. 0
- 0. 000
0. 006
1 30. 000
60. 9 3 4
500
70
G E N E RA T I ON
I T ER A T I ONS B E T W E E N BU S-OR D E R
SORT S
50
M f SC E L l ' NE OU S C O N S T A N T S
4 A C T U A L I TE R AT I ONS MA X I M UM I T f R AT I r-N S 1 (1 T O L E R ANC E - R E AL 1 . 000 E - 0 3 - I M IG I . OCO E-03 AC C F A C T . - � E A l 1. T 1. 1 - l M AG 1 . 8 1.4 LTC S T A R T 10 4 SKI P 400 E NO ACe F A C T 1. 2
.j:I. -..I C1I
�
..... C)
r OW A
S T AT E
T I T l E -]
UN r V E � S r T Y VE � S r ON OF
8US
3 6 0 l OA D F l D W PR OG RAM
E X A M PL E
X --------------------------------- 8 U S R E PO R T
OF l O A D F l O W C A L C UL A T I ON S
AREA NO.
8US
61t
t4
N A "' E
.I l S I4 A T CH IIW MV A R
0. 0
8US
64
Table E.1. (continued)
0. 0
TOTAL A T
D
VOL TS
ANGl E
1 . 03 0
0. 0
A
lOA D
MW
MVAR
0. 0
0. 0
G E N E U T I ON MW "VAil. 3 0. 0
4 S W I NG 8US · 1 8 U S 1 , AR E A 64. l t N E - D A T A -- ------ - - - - - - X x- - - - - - - - -- - - - - - - - T 0 - - - - - -- - - - -- - - - - - - x
I T E R AT I ON S .
----- - - - - - - - - - - - - - - - - -- - - - - -- - - - -
C A P / !I. E A C MVAR
X- - - - - - - - - - - - -
8U S
NAME
23. 1 2 - 8US 2 3 - 8 US 3
2
8US 2
3
8US
3
- 0. 0 0 0
0. 004
1. 018
-1. 0
50. 0
2 0. 0
0. 0
0. 0 0 2
1 . 020
-0. 5
80. 0
END OF
1t0. 0
R EPOR T
FOR
1 00 . 0
64
0
64
64 64
o o 1
64
0
64
0
0.0 1 - 8 US 1 3- 8 US 3 3 - 8US 3
0. 0 0 0
PAR NO. NO .
AR E A
3 7 . 8 11. 1 - 8U5 1 2 - 8U5 2 2 - 8 US 2
THI S C A S E
64
64
0 1
"'W
l i NE
22. 5 3
FL OW MVAR
7. 4 7
15.16 7 . 96
- 1 � . 74 - 1 3 . 74
- 1 4 . �0 -2. 70 -2 . 7 0
-2 2 . �3
- 7. 47 1 3. 74 1 3. 71t
- 7. 8 1 2.81 2. 81
CAP
PCT
T AP
:g
'0 ('I) :l Co x·
m
Table ,
E XA " PL=
" :.I S
T� ! � � � I S� ! ��
L I �E
A� �
T � A N S F OP M E P D A T A
DATA
FOR
x - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - x -- - - - - -- -- p
0
2
0.0 0. 0 O. a 0. 0
1
2
0
2
3
0
3 3
1 a
X I PCT I
Q �P C T I
a
0
�
0
E
L
F
S E QU E � C E
•
20. 00
0. 0
0 . 08 0 0 0 . 06 0 0 O. O e O O 0. 1 300
0 O. a o. a
0. 1 6 0 0
O. J
0.
8. 00 6. 0 0 6. 0 0 1 3. 00
C O � V E RT E o
( PU I
0. 2000
0 . 0-
le. OO
O. Q
X
I PU I
0. 0
G
X
-- - - - - - - - - - - - - -
( PU I
8
0.0
0. 0 0.0
o. a
P
Q
0
3
1
3 3
e x ! S '!"l
AC T U A L
0 0 a 1 0
"UTU AL
'l G
X - -- - - -
2. 00
0. 0 0.0 0. 0 0. 0
----- - X
BU S
2 Z
3
3 3
fil a .
a 1
0
X
( PU I
0. 0 O. a 0. 0 O.a
0. 1 00 0 0. 1 2 0 0 0. 1 7 0 0
M AT R I X
x--
( PU I
0. 0200 0 . 1 '>0 0
0. 0
10.00 1 2 . 00 17.00
A D � I T T A NC E
S EL F
R
1 4. 0 0
0. 0
B US 2
X ( PC T I
C OU P L E D L I �E
BUS
BUS
2
3
--x
�O.
R
1 1 2
1 2
0. 0
0 . 1 0 4e 7
O. C
3
0. 07e2 7 0 . 1 1 2 ;:: 1 0 . 0 9 2 7 <; 0 . 0 8 40 1
0. 0 0.0 0.0 0. 0
2 3 3
"'� : T ", C' :; <: X
0 . 0 9 ii 9
- 1 0. 6 6 6 7
0. 0
G
S E Q UE NC E
POSI T I VE
BUS
3
- 7. 6 9 2 3
F ::' "-
( PU I
B
0. 0 0
x - - - - - - - - - - - - - - - - - -- -
2
2 -5 . 3 8 2 4
0. 0
X-------G
0. 0 O. a
S
E
L
F
-- - - - - - - X 8
AC TU A L
�D.
Q
P
- 5 0. 0 0 0 0 - 7 . 1 429
0.
� 3
X ------- ----- -
( �U I
C/l o <" m x Q)
S ;o Q U e "l c !:
IEQO
S E L F --- ---- - -------- - - -- - - - - - - - - - - - - X - - - - - - - - - -- - X -- - - - - -- - - - - - - C D� V E R T E D - - - - - - - - - - - - - - - x
� ( PCT I
"I D .
T HE
B US
2
x - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - x -- - - - - -- - - -
FOQ
1
-5. 0000 -0 . 2 5 0 0 - 1 2. 5000 - 1 6. 0667
0. 0
DA T A
x
( PU I
MA TR I X
I
EX I ST ING
- - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -
- -- - -- - - - - - - X - - - - - - - -- - - - - -
AC T U A L
�O.
A S S = �8LY
PO S I T I V E S
E.2. A 3 -Node Network Fault Study
3
3
GM
XNI PCT )
R M ( PC T I
1 a
X------
U T U A L -- - - - - - - - - - - - - - - - - - - X - - - - -- - - -- - - - X - - - C O "l V E P T E D - - - - X
M
5 . 00
0.0 0. 0
M
U
T
U
5. 0 0
A
L
Q�
( PU I
0. 0
0.0
XM
iD
"0 Ul
( PU I
0. 0 5 0 0 0 . 0 � 00
------ X 8 ..
- 1 2. 6 3 1 5 8 0.0 - 1 0. 5 2 6 3 3
5. U 3l6
• "'" "'"
E �I STI�G
! WS
Z
MA T � I X
1
� U '4 P L � � : o � � � O F F AU L T �
Q .: �
2 2 !
F O R THE Z E P O S E OU E N C E N E T WORK
BUS
�
1 2 3 2 3
0. 0 0.0 0. 0 0. 0 0. 0 0. 0
3
fable E . 2 .
1
�A '4 e
SUS
1
�
.....
X
IX)
0. 1 1 574
0. 0545 8 0. 02000 0. 0 8 3 06 0 . 0 20 00 0. 02000
C A L C U L A T I ON S
� -- - - - - - - - - -- - - - - - - - 8 U S - D A T A - ---- - - - ----- - - - - - - - X X - - 3 - P H A S E - - - x - - - - - - -- - - S - L - G ------- - - - x
GUS
(continued)
AMPS
9. 5 5 4
DEGR E E S
- 9 0. 0
V OL T S
D E GR E E S
0.6780 0. n 2 0 0 . � 56 0
0.0 1 8 0. 0 1 80 . 0
AM P S
9. 2 2 9 3.076 3. 076 3. 07 6
O e GR � E S
x - - -- - - - - - - - - - - - - - - - - - - - l x - - - -- - - - - 3 - P H A S E 8US
-90. 0
0.6�51
C . : 610 9 0. 2701
0.0 1 8 0. 0 1 80. 0
9. 756 3.252 3.252 3. 2 5 2
DEGRE E S
AMPS
-90. 0 N -90.0Z
2-
8. 91 2
VOL T S
D EGRE E S
VOLTS
D E GR E e S
AMPS
D E GR E E S
- 9 0 . 0 T ----------- - - - - - -- - -- - - -- - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - ----- - - -9 0 . 0 P
0-
BU S 2
NO.
I N E - D A T A - - - - - - - - - -- - - - - - - - - - - - - x - - - - -- - - -- X - - - - - - - - - - S - l - G - - - - -- - - - - - X
0
0
1 . 0000
0. 0
0. 1 9 7 3
0. 0
5. 000
2 . 467
-9 0 . 0 T 3.220 -90. 0 P 0. 0 1. 610 1 . 000 0 -90. 0� 0. 0 0.0 1. 610 0. 0 0. 0 0. 0 O. OZ -- . - - - - - - - - - - - - - - --- - - -- - - -- 90. 0 -9 0 . 0 T 2 . 932 -90. 0P 0. 0 0. 74 1 6 0. 7 9 4 0 . 7 94 1 8 0. 0 0 . 2 5 84 90. 0N 0. 1 6 7 9 1 8 0. 0 1 . 3 4 4 - 90 . 0 1 - -- - - - - - - - - - - - - - - - - - - - - -- - - - 90. 0
3-
0
0. 2 7 1 3
0. 0
2 . 087
-90 . 0
3-
0
0. 1 7 3 1
0. 0
2 . 885
-90 . 0
3- 1
-90 . 0T 3.617 -9 0 . 0 P 0. 6 9 8 2 0. 0 1. 053 9 0 . 0" 0 . 30 1 8 1 8 0. 0 le053 -9 0 . 0 l 0. 0650 1.511 1 8 0. 0 - - - - - - - - - ----- - --- -- - ----- - -
0. 1 7 3 1
0. 0
2. 885
-90. 0
-90. 0T 3.185 -90 . 0 P 0. 6 9 8 2 0. 0 1.053 - 9 0 . 0N 0. � 0 1 8 1 8 0. 0 le 053 -9 0 . 0 Z 0 . 0650 180.0 1 . 079 -- - ----- - - - - -- - ---- - - - - - - - - -
3 . 077 -90 . 0T -9 0 . 0 P 0 . 7 6 54 0. 0 0.672 - 9 0 . 0 '"' 0 . 2346 1 8 0. 0 0. 6 7 2 0 . 061 5 180.0 1.732 90. 0Z - - - ----- - - - - -- -- - - - - - - . - - - -- 9 0 . 0 T - -- - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -- - - - - - - - -- - - - - - - - - - - - - - - - - - ------ - - - 9 0 . 0P -9 0 . 0 N - 9 0 . !) Z -90 . 0 T - 90. 0 0. 0 3 . 1 42 0. 2 5 1 3 1 - !) 2.955 - 90. 0P 0. 0 0. 7 2 6 8 1 . 1 46 - 9 0 . 0 '"' 180.& 0 . 2132 1 . 1 46 -9 0 . 0 Z 0. 1 775 1 80. 0 0. 66 2 - - - - - - - - - -- -------- - --------
):0
'C 'C CD � Q.
XO m
? � U S E XA M P L E R o P O P T OF F AU L T
C A L C U L A T r ON S
x - - - - - - -- - - -- - - - - - - - 8 U S - D A T A - - - -- - - - - - - - - - ---- - - X X - - 3 - P HA S E - - - X - - - - - - - - - - S - L -G - - - - - - - - - - x
8US
3
NA � E
'I U S
!
A�PS
1 :1 .
D E GR E E S
102
-9 0. 0
V OL T S
DE GREES
0 . 5 10 5 9 0. 454 1 C. 091 8
1 "0. 0 1 8 0. 0 0. 0
AMPS
1 3. 763 10 . 5 8 8 10 . 5 8 8 10 . 5 8 8
DE GRE E S
L 3 - I �A S E
x - - - - - - - - - - - - -- - - - - - - - - -
8US
X---------
NO .
VOL T S
D E GR E E S
r
N
E
-
D A T A
-- - - - - - - - x
S - L - G -- - - - - - - - - - x A MPS D E GRE E S D E GR E E S
- - - - - - - - - -- - - -
---------- X - -- - - - - - - -
AMPS
D E GR E E S
VOL T S
- O O . O T--- - - - - - - - - - - - ----- - - - - - - - - - - ---------------------- _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
-90. 0P -9 0 . 0 N -90. 0l
0-
0
1. 0 000
0. 0
6 . 2 50
- 90 . 0
1-
0
0. 2 2 9 5
0. 0
1 . 76 6
-9 0 . 0
- 90 . 0 T 1 0 . 2 610 - 90. 0P 1 . 0000 2. 8 ! 8 0. 0 - 9 0 . 0N 0. 0 2. 8 3 8 0.0 -9 0 . 0 l 0. 0 10 . 5 88 0. 0 - ----- -- - - -- ---- - - - - - - - - - - - 0. 6 5 0 1
2-
2-
0
1
0. 0 6 2 6
0. 0626
0. 0
0. 0
1 . 0 10 3
1 . 0 43
0. 0 1 8 0. 0
1 . 6 0 10 0.802
- 90 . 0T -9 0 . 0 P
-9 0. ON 0 . 8 02 0. 3499 O. Ol 0 . 000 180.0 0 . 091 8 - -- --- - - - - ---- - - - - --- - - -- - - - 90 . 0 0 . 9 10 8 - � O . O T 0 . 5 143 - 90 . 0 P 0 . 4 14 0. 0 0 . 4 74 180. 0 0 . 42 5 7 O.Ol 1 8 0 . 0 0 . 000 0 . 09 1 8 - - -- - - - - - - - - -- -- - -- ---------
en
-90. 0N �
- 90 . 0
0 . 5 7 10 3 0 . 10 2 5 7 0 . 09 1 8
0.0 1 80. 0
180.0
0. 948 0 . 4 14 0 . 4 74
0. 000
-90. 0T -90 . 0 P -9 0 . 0 N
O. Ol
_ _ _ _ _ _ _ _ _ __ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
0
�
rn
� 3
ar
� 0
� ...,
CD
Appendix E
480
A complete 3lP and SLG fault study for the 3-node network is given in Table E .2. This table includes all primitive impedance data, the impedance matri ces for both sequences, and the detailed fault data.
E.2 A 6·Node Network
Consider the 6-node network shown in Figure E.2. This network is more completely specified than the 3-node network. The system data includes load data and line susceptances in addition to the line, transformer, and generator impedances. The line impedance data is given on a 50 M VA base in Table E .3. The computer program used to solve this network will change the base to 100 MV A before solving.
®t 4 @
��------------------�� 2 7
® 5 Fig. E . 2 . A 6-node network [ 7 9 ) .
Table E.3. 6-Node Network with Impedances in pu on a 50 MV A Base Impedance Num ber
Connecting Nodes
1 2 3 4 5 6 7 8 9 10 11 12
1-6 1-4 4-6 5-6 2-5 2-3 3-4 0-4 0-1 0-6 Gen 1 Gen 2
Mutual Impedance
Self Impedance
Z 1 = Z2
0.123 + jO.518 0.080 + jO.370 0.097 + jO.407 0.000 + jO.300 0.282 + jO.640 0.723 + j 1.050 0.000 + jO.133 0.000 - j34.100 0.000 - j29.500 0.000 - j28.500 0.010 + jO .120 0.015 + jO. 240
Zo 0.492 + j 1.042 0.400 + jO.925 0.450 + j 1.030 0.000 + jO.300 1.410 + j 1.920 1.890 + j 2.630 0.000 + jO.133
0 .000
+
ZM
Branch
0.250 + jO.475 0.250 + jO.4 75
3 2
jO .016
A complete load flow result is given in Table E .4 where all line and generator flow data is given in megawatts and megavars. All voltages are given in pu and angles are in degrees. A complete bus fault study is given in Table E.5.
I OW A
S T AT E
b
2 4 4
1 1 2
Q
5
OF
0. 0 0. 0 1 52 5 . 80 0. 0
0. 0
1 406. 80 1 983. 0 0
C ON V E R T E O
0. 0 0. 0 0 . 00 7 6 0. 0
0 . 007 0 0. 0 0 9 9 0. 0
BC 1 2 ( PU '
-------- - X X---
KVAe
PROGR A M
ASS EMBLY
L OA D F L JW
D AT A
=60
EA.
1 . 050
( I NC L UD I NG S W I NG . - G E N E P. A T O R T O T A L B U S E S -- - - - - -- C A PA C I T ORS O R R E A C T O RS
0
6 0
5 1
S OO 1 90 500 70
2 50 2 50 0 750
0 7
7�0
2 0
MAX
0. 0
0.
0. 0
0. 0 0. 0
0. 0
5
AC T U A L
1 . 00 0
1 . 000 1 . 0 00
1 . 100 1 . 000
T P A N SM I S S I O N L I N E S T R A N S F OR M E R S - F I X E D - LTC PHA S E SHI F T E R S T OT A L L I N e s - - - - - - - -BUSE S - �ON R EG
L I N E A N D BU S T O T A L S -- -- -- - - - --- - - - - - - -
0
0 0
64 64 64
F OU Q F I VE
0
2
1
64 b4
2 5. 0
0.0 15. 0
0. 0 27. 5
0 • • J
T OT A L
L O AD
G E N E R AT I ON
M I S M AT C Ii
C A P / R c.: 4 C
LOSSE S O I AR G I NG
S Y S T E II4
F ixeD
L I filE
TOTAL TOTAL
0. 0
9.0
2. 5
25.0 0. 0 0. 0
0. 0
0. 0
14 1/
' 72. 6 1 0
-0 . 00 2
1> 7 . 5 0 0 5.112
M il - - - - - - - - --
0. 0
0. 0
0. 0 0. 0 0. 0
0. 0
MVAR
-x
0. 0
0. 0
0.0
12. 5 0. 0 0. 0
0.0
QMA X MV AR
0 . 97 6
0 . 909
3 1 . 047
- 0. 00 1
1 8 . 000 1 7 . 666 - 4. 6 1 8 0.0
MV A P
O.
o. O. O. O. O. O.
TAP RAT I O
- - -- -- - - - -
0.0 0.0
-12. 5 0. 0 0. 0
-0 . 6542 - 3 . 7 5 94 - 1. 1 6 2 5 -1 . 6667
- 1 . 29 1 0 -0. 9 1 3 7 - 0. 3 2 3 0
R A T I NG
MVA
QMI N MVAR
----X
8 ( PU ,
E FF E C T
G E � E R A T I ON
0. 2 7 7 1 0. 0
0. 2 3 9 3 0. 0
0 . 2 1' <; '. 0. 2 1 " � 0 . 2 2 24
0.0 6. 5 0.0
0. 0
TAP
G ( PU '
NO
A 6 -Node Network Load Flow Study
-- - - - - X X - - - - VOL T A G E - - - - - x - - - L Q A O - - -- X X �w A R F. A R E G M A G ( PU ' A N G I DE G . MVAR
64
S IX
8 1 . 40 6 0 . 00
74 . 0 0 1 0 3 . 60 2 1 0 . 00 1 2 8 . 00 2 6 . 60
X ( Pc n
ACTUAL
E NT E R IO O
T W(' T HI' E e
O 'l �
NAM E
BUS
� U 'I4 M AQ Y
6
It
1 2
Ioj o .
x -- - - - -
DAT A
0 0
�
BUS
1 9 . 40 0. 0
0 0
:
6
5
R I P CT I
1 6 . 00 24.60 1 44 . 6 0 5 6 . lt O 0.0
3
NO.
V E R S I ON
A N D T R A N S F OR M E R
X - -- - - - - -
L I NE
0 0 0
4
. 'S� C A se
p
l I NE
T � A N S M I S S I ON
SYSTEM
U�I V ER S I T Y
T I TL E - S I X B U S
Table
0
0 0
0 0
0
PAGE
0
0 0
I T E R A T I ON S
DATA
0 0 0 0
0 0
0
PG
1 . 0 00 e - 0 3
0 0
0
0 0
0
0 0
0
0 0
0
R E AC TO R LOC QS
0 0 0 0
0 0
C
;,
F L OII L OC
--- -- - X
REV
- - - - ---- - - - X
0
0
LOC
TAP
GEN L OC Q
Q
MAP
= 10 1 . c o o e -0 3
C ON S T A N T S
M A X I M U M I T E R A T I ON S TO L E R A N C E - R E AL - I M AG
AC TU A L
0
0 0
0
0 0
0
0 0
0 0
0 0
0 0 0
DAT A LOAD L OC Q
MAP
0 0
0 0
0 0 0
X- - - - F LOW PG LOC
VOLT L OC A
M I s c eL L A N E O U S
O.
o.
O.
O. O. O.
R E ACTOR K VAR
0.0 0. 0
0. 0 0.0
0. 0 0. 0 0.0
SH I FT
PH A S E
X - - - - - - - - --
T AP L I M I T S TMAX TMI N
� 00
VI
Q)
3 "C CD
<" (II Co m )(
0
en
, T a· C
0"
4
' .
l ! �� !
1. 100
VO L T S
A II E A NO.
1> 4
0. 9 3 0
VO L T S
64
NO .
UEA
t: v E D L O A 'J E O
�US �A�e
2 T W t:
�C .
C: ':: IJO
"! IJ S "I') . I\; A" e
_!.4
A "I G L E
-9 . 8
A�GLE
- 0 . 000
-0. 0 0 1
SI X
6
� ..
- 0 . 000
(, ..
O.
- 1) . 0 ' 1
;
(, 4
0 . 00 0
< l VE
F D U II
4
6"
-0. 002
0. 000
0. 0
-o . C O l
T Iot " !: E
�
- 0. 0 0 0
o. a
" I S ,", A T C H "'V Ail "W
OCI
TwO
2
64
III A " e
O '-i E
'I U S
64
'. :1 .
4�EA
-
0
OF
25.0
1�. 0
H I GH
FIVE
NO.
A II E A YOL T S
ANGLE
--
- 1 2 . :!
S U 14J111 A R Y
0 . 91 9
--
BUS
aus
SIX
BUS III A "' E NO.
VOLTAGES
I!>
BUS "IA "' E NO.
VJL T A G E S
TH I S C A S E
0. 0
0. 0
0. 0
0.0
0.0
0.0
9 . 3 11
21.8
0.0
0. 0
25.0
4 7. 6
A "IG l E
S U � "' A R Y
VO L T S
V O L T AG E
64
AR E A "10 .
F OR
2. 5
9. 0
0. 0
II !: O a R T
L O W V OL T A G E
NA Jill E
BUS
-12. 2
-12. !
-9. 8
0. 0
6. 5
27. 5
-12. 8
0.0
0.0
0. 0
MV A R
G E !\I E � A T I J "I " V Ail lO W
"' 0 .
ARE A
A B OV E
64
A R. E A "1 0 .
8 E l OW
"I A 14 E
NO .
T
"10 .
P A il
0
L I lli E 14 101
FL OW "' V A il
l>e T CAP
TA P
64
0
i.58
- 0 . 00
VO L T S
1 . 05 0
0. 9 1 9
VOL T S
A N GL E
- 1 2. 2
A � GL E
I - O lli E 4 - F OU R 5-F I V E
64
64
E: 4
"1 0 .
NO.
a 0
C
[I U S lII A "' E
8US NA Jill E
0. Z4
- ZO. 8 ! -4. 41
"'0.
A II E A
NO.
AR E A
YO L T S
VO L T S
3 . 56
- O . E: &
-5. 3'
AN G L E
ANGL E
0. 976
- 5 . 52 64 0 - 1 4. 76 2 - T WO 6-S l K 6 .. 0 - ! . 4& - 0 . 24 - - - - - - - - - - - - -- - - - - - - - - - -- --- - - --- - - -- - - - --- - - -
� ..
� tit
-9 . 5� 0 - 2 .. . 2 5 1 9 . 79 8 . 98 0 0. 9 0 9 6 .. .. . .. 6 0 0. "3 - - - - - - - -- - - - - -- - -- - - -- - ---- - - - - - - -- - - - - - - - - - - -
3 - T I'1 Q = � 6-S I X
l - : ... e
2 - T WC 64 a -7. 70 1 . 28 -7. 7 9 64 0 - 1 9 . 79 4 - F D U II - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - --- - - - - - - - - - - - - -
b4 a 5-F I V E : 6 . 42 9. 28 - - - - - - -- - - - - - - - - -- - - -- - - - - - - - -- - - - - - - - - - - - - - - -
3 - T HII E E
64 C 2 � . 40 1 2 . 74 64 6-S I X a 22. 1 5 9 . 02 -- - - -- --- - - - - -- - -- -- - - --- -- - - ------ - - - - - - - - - - -
- - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -- - - -
4 - F OU R
BUS
AREA
1 T
. AR E A ON E E: 4 . A ---- ---- - - - - - - X -- - - - - - -- - -- - - - -- - X
"IG S U S . I N E - 0 A
� - -- ---- - - - - - - - - - -
0 . 9 50
CAo'�EAC "V AP
TO T A L ! T = � H r ·: � � = 3 SWI - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - � - - - - - - - - - - - - - L
L O AD
-
-3. 4
14 11
A
0. 0
E�D
A T
P ' C o; llA �
0. 0
A Ili G L E
S
BUS "10. ili A 14 E
5
NO.
0. 9 1 �
U
L C A O F l O ..
0.0 10
0. 9 :! 0
1 . 001
1 . 1 00
1 . 0� 0
VOLTS
B
36C
D ,! O :'P 'T �� L C A O F L O W C A L C U L A T I ON S x -- - - - - -- - - -- - - - - - - - - - - - -- - - - - -- - -
SYSTEM
'J '. ! v � . S I"' Y V = ; S I O'"
· !·l : - � t x eus
! :A�
Table EA. tcontznuea J
)( m
:I
g.,
l> '0 '0 CD
•
�
SI x aus
SYS" M
A ND
MVA
D A TA
F OR
SE Q UE NCE
A S SE MSLY
PO SI T I VE
IC V
o.
I NPUT
S AS E
D ATA
0. 0
T R A NSF OR ME R
1 00.
0 .0
SASE
IC V
M VA
CONV ER TED
A 6 -Node Network Fault Study
5
2
0
0
0
0
0
0
0
0
0
74. 0 0
0 . 02 1 1 1
0 . 0 1 13 1
5
6
I I
0 . 16 2 4 .
0 . 1 4333
0 . 06506
0 . 0688 1
4
5
6
3
4
5
6
4
5
6
2
2
3
3
3
3
4
4
4
5
6
6
5
5
6
0 . 46538 0 .6 1 1 1 9
0 . 1 32 5 6
0 . 8 06 4 9
0 . 34726
0 . 2 78 1 8
0 . 57694
0. 2 7 0 0 7
0 . 32786
0 . 53368
0 . 739 1 2
0 . 1 52 2 "
0 . 2 2306
0 . 1 34 3 4
0 . 1 57 1 3
0 . 3809 4
0 . 1 3 03 4
0 . 1 6 569
0 . 1 3 26 9
0 . 0 7 29 5
0 . 06 1 9 2
0 . 00 0 2 3
- 0 . 00698
- 0 . 00 786
3
2
2
0 . 04 4 2 2
- 0 . 0 1 59 4
2
2
0 . 1 63 79
0 . 1 21 2 5
0 . 1 7266
0 . 04974
X
N ET WORK
0 . 60 0 0
0 .8 1 4 0
0 . 161 1 7
0 . 02 25 4
SE QU E NCE
0. 0
0 . 1 9 40
0 . 2660
1 . 28 0 0
2 . 1000
0 .2 1503
4
0 . 56 4 0
1 . 4460
0 .0
1 . 0 36 0
0 .74 0 0
0 . 02635
1
0. 1 6 0 0
0 . 24 6 0
0 . 24 0 0
0 . 4 80 0
0 . 02 25 3
3
0. 0 2 0 0
0 .0300
- 0. 00609
2
I
R
P OS I T I V E
60 . 0 0
81 .4 0
2 6 .6 0
1 2 8. 0 0
2 1 0 .0 0
1 0" .6 0
I
T HE
24.0 0
4 8 .0 0
SUS
FOR
0 .0
1 9. 4 0
0 .0
56 . 4 0
1 44 . 60
2 4 . 60
1 6. 0 0
2.00
3 . 00
aus
S I X aus S YSTE.. E X I ST I NG Z MATR I X
5
6
6
3
4
2
"
4
4
I
6
0
1
1
2
0
- 1 . 1 6 25
0. 0
- 1 . 6667
- 0 . 65 4 2 - 3 . 7 5 94
- 0 . 32 3 0
0 .2771
0.0
0 . 2 8 83
0 . 2 2 24
- 1 . 29 1 0 - 0 . 91 37
0. 2 1 70
- • • 1 3 79
- 2. 0 7 5 2
0 . 2791
0 . 1 297
0. 3 4 4 S
X-------------------- - S E L F ---- - - -- - - - - - - -- - - - - - - X X-------- - A C TUA L ------- x------ --- C I»I V E RT E O ----- ------- X x C PU ) P Q NO. R I P CT ) X C peT ) R IPU ) G C PU) S C P U)
LI NE
GROUP
TR ANSM I S S ION L I NE
Table E.5.
� co Co)
(II
3 'tI CD
�
m
� Co
f/) o
8US
O.
M VA
R
II
C PU ,
A S SE Iil8 LY
C PU,
OATA
I NPUT 8 AS E 0. 0 IC V
TRA N SI" OR M E R
R ' PCT I
" I\C
O AT A
FOR
8 A SE
M VA
KV
4
6
0
3
5
0
6
1
2
98 . 4 0
0
2
0. 0
8 0 . 00
0
6
4
I
90 .00
0
au s
4
6
2
3
4
5
6
3
4
5
4
6
5
1
2
2
2
2 2
3
3
4
3
5
6
5
6
6
6
5
4
5
4
3
3
1
1
I
0 . 03. 4 9
0 .0
2 . 8200 0
- 0 . 0 1 706
0.0
0 . 0 0 75 6
0. 0
-0 . 0 0 0 0 0
0.0
.1. 780 0 0
0 .0
- 0 . 00 0 0 0
0 .0
- 0 . 0 0 00 0
- 0. 00 0 0 0
0 . 0 2 52 6
0.0
-0. 01 1 2 0
0.0
0. 36392
0.0
1
2
S E Q U E NCE
I
R
ZERO
I
MA T R I II
6
0
0. 47472
0.0
.1 . 8 7 2 0 0
0 . 05554
0 .0
0 . 2 4 1 31
0. 0
0 . 0 32 0 0
0.0
5 . 29200
0 .0
0. 03200
0.0
0 . 0 .1 2 0 0
0 . 0 32 0 0
0 . 3 4 1 20
0.0
0 . 1 1 4 74
0 .0
1 . 1 1 336
0. 0
II
NET W O A IC
4 0 . 20 99 5
0 . 23 1 8 4
10
6
-0 . 5 55 2 1
- 0 . 62 0 1 2
-- ---- -- II 8
2 . 060 0
0 . 90 0 0
1. F
3. 8400
2. 8 2 0 0
E
- 0 . 1 6 92
0 . 1 2 42
5 . 26 0 0
3. 7800
S
-0 . 1 2 5 4
0 . 09 0 1
2 . 0140
0 . 98 4 0
G
- 0 . 39 2 4
0. 1 8 53
1 . 85 0 0
0. 8 0 0 0
11--------
- 3 . 7 5 94
0.0
0 .6 0 0 0
0 .0
NO .
- l e 66 6 7
0. 0
-- II
-3 1 . 25 0 0
0. 0
C PU ,
0 . 2660
8
0 . 0 32 0
C PU,
0.0
G
0. 0
11-- COUPL ED L I NE IU S IUS
au s
Z
STSTEN
THE
4
FDA
0
4
6
0
0
4
1
1
------II 8US NO.
MA TA I II
2 06 . 0 0
38 4 . 0 0
526. 0 0
208 .40
1 85 . 0 0
60 .00
26 . 6 0
3. 2 0
" ' PC T '
au s
E II I ST I NG
S I ll
A D M I TTA NC E
28 2 . 0 0
3 78 . 0 0
0
11 ----- - S E L F
8US
0. 0
0 .0
0
0
0
2
4
0
NO.
0
Q
51 11 8US S Y S T E M E II I ST I NG MU TUAL
10
Q
NO.
0
0
N
U
- 0. 0 8 8 9 5
11----- GM
4
4
M U
U
A
50 . 0 0
50 . 0 0
T
T
U
A L
L
8M
0 . 3 0 3 .19
------ II
95 . 0 0
95 . 0 0
ICJI C p C T ' C PU ,
0 . 50 0 0
0 . 500 0
RM
11 C PU ,
0 . 9 50 0
0. 9500
11M
11 -- - C ONV E R TE O ----II
- - -- ---- -- - - - - - - - - - -
- - - -- ------
AM C p cr ,
11---- - - -------------11- - - - - - - - - - - - - A C T UA L
ZEAO S EQU ENCE
1 0 0.
0 .0
(continued)
CONVERTEO
Table E . S .
11--- - -- ------ -- ---- ------ ------ - S E L F ---- -- - - -- - - ---- ---- ---- ---- -- - - 11 11----------- AC T U AL -----..,.--- -- - 11 -- --------- --- C o. V E RT E D - -- - - -- - - - - - - 11
L I NE
GA OU P
1. 1 NE
SY S TE II
TR A N S M I S S I ON
Si ll
m
)C"
5-
:c "C CD
� CD �
au s
F AULT
S Y S T EM
OF
TWO
THR E E
3
ON E
NA NE
2
BUS
X--
1.321
2 . 608
4 .625
A MPS
U
V OL TS
D E GR E ES
A MPS
D E G RE ES
- 7 7. 6
-83. 4
- 8• • 0 0. 626
0. 1 26
1 75 . 7
0. 1 2 6
0 . 82 1 4
0.379 0 . 1 26
- 1 .9 - 1 6 1 .2
0.9101
1 . 252
0 .0956
- 1 73 . 6
1 .252
1 79 . 7
0.4801
0 . 0 40 1
1 . 252
3 . 75 6
0. 626
0 .2
1 76 . 7
0 . 7336
0 .626
0 .5 199
- 1 •• - 1 71 . 1
0 . 1 35 .
0 . 8665
1 . 879
8US
NO.
VOL. T S
D E GR ! E S
AMPS
D E G RE E S
VOL T S
DEGREE S
A MPS
DE GREES
x -- - -- - - -- --------- - - - I.. J N E - D A T A -- -------------- --X x -------- 3 - PHA S E ------- --x ------- -- S-L - G ----- - - - - X
0. 2 5 9
0 . 22 3
6. 0
1 .3
0 . 1 96 0
0.2378
0
6-
4. 1 5 2
0
0.0
.-
1 .0 0 0 0
0
00 . 562
0 .0 0. 0
0.0 0. 0
1 . 1 25 0 . 562
0. 0
1 . 0 0 00
-76 • • N
-76. 4 T - 7 6. 4P
- 75 . 4
-7 2 . IT
-66 .5P - 66. 5N -7 3 . 6 Z
0 . 2 89
0 . 030
0 . 0 30
0 . 1 0 32 - 1 7 1 . 5
0. 8980
-10 0
0 . 1091
- 6 2 .9N - 1 72 . 6 0 . 035 -r6 . 1 Z 0. 0 7 2 2 - 1 59. 6 0 . 397 - -- - --- - - -- -- - --- -- ---- - - - --
O. O Z 0. 0 - --- - ---- ------- - -- -- --- - - - - 7 1 . 87• • 1 T 0 •• 66 - 62 . 9P 0 . 8920 - 0. 9 0. 035
- 85. 2
0
0
0
0-
3-
5-
0 •• 3 2 8
0 . 6 04 3
1 . 0000
0 . 30 9
0 . 23 7
-8 • •
-1 1 .3
2 . 079
0. 0
-77 .6
-6 3 . 9
- 86 . 4
0 .99a
0.0
0.0
- 86. 7P - 86 . 7H
-a5 .5T
-3. 0 - 1 67 . 9 - 1 73 . 6
-6 • • I P - 6 •• ' N O .O Z
0• • •• 0 .0 00
0. 1 1.
-6 • • 1 '
-3 . 1 - 1 7 10 9 - 1 73. 6
0 .0.0 1 - -- -- -- -
0 . 2 7 9.
0 . 72 . 5
0 . 0 00
0 . 01
-77 . ap
-77. 8N
0 . 1.9
- 77. 8T 0 . 1 .9
0 . 297
- - - - - -- - -- -- - ------ -- ------
0 . 040 1
0 . 1 9 77
0. a 0 78
0 . 2 2a
1 . 252 -a 3 .6 Z 0. 0 0. 0 - ------ - - -- - - - --- --- --- -----
0 . 998
0. 0 1 . 00 0 0
3 .2.7
- 58 . 8 P - 58 . 8 N - 58 . 8 Z 0
0
2-
4-
0 . 2 72 6
0 . 8 04 .
7 . 1
-4 . 6
1 . 0 25
0 .31 5
- 82 . 9
- 60 . 1 0 . 0 . 99
0 . 1 85
0 . 1 96
0 . 030
- 64 • • '
-. 1 .3 P
-- 5 3 . I T
0 .0
0.0
0 . 0 98 0.0
0 .0 98
- 1 .2
0 . 9 3 32
O. 0698 - 1 6 30 8
0. 01
... 4. I N
-64 • • P
0 . 030 - 4 1 . 3M 0 . 0040 - 1 4 8 . 8 0 . 1 26 -58 .8Z -- - -- -- - --- -- ------- ----
-0. 7 - 1 .3. 0
0 . 98 . 1
- 58 . 8 T - - - ------ - - - - ------- -- - - - - - - - - - - - - - - - - - - - - - - - ---- ---------------
- 83 . 6 Z
-83 .6N
- 83 . 6P
0 .2 1 . 3 - 169 • • 0 . 229 - - - - --- - ---- - -- - - ------ --- 8 3 . 6 T--- ---- -- - ----- ----- - - ----- -- - --- - - - - -- - - - - - - -- - - -- - - --- - - - ---- ------
-7 5 . 2 N - 7 5 . 2Z
- 75 . 2 T- ---- - -- - - - - -- ----- - - - ----- -- -- ----- ------- ----- ---------- --- --- 75 . 2P
S - D A T A ------ ----------- X --- X-------- 5- L. - G - -- - -- - X
DEGR E E S
� P HAS E
8
C ALCU LAT J ONS
x- - - ----- - - - -- - - - - -
REPORT
SIX
3
� 00 C7I
(I)
"t:I
CD
I»
m X
CD Co
<"
0
rn
F I VE
51 X
6
!"OUR
5
4
-7 8 • •
-7 a . 0
1 . 6 0 0.
-77.0
l .al5
1 . 6 ••
Table (con tinued)
0 . 7 0.7 .
o . 0 7' 5 a
0 . 40.27
0
0
1-
3-
6 - 0.
-a.6
26 ••
-2 . 3
0.0
0. . 4 8 1
0 . a8 3
0 • •3 5
0. 0
- 7• • a
- 6 3 .a
- 80 . 1
0. 0 0 .0.
0 . 0.
0 . 0.
0. .0. 1 .
G . G8 U
G . a4.G
0 . 5 8 0.
0. 580 0.580
1 .2 1 7a . 0
-1 7 4 . 6
0 . 63 7 7
0. . 2 764
-aG . o z
- e O . ON
-aG . o p
- 77 . 7
- 77 . 1
0 .518
0. 3a.
-1 . 1
l a• •
0 . 4 337
0 . 2 33 4
0.
0
4-
5-
- 7a . 7
0. . 6 . 3
-2 . 0 0 . 7 38 0.
0 . 0.
- 83 • •
1- 0
0 .0
0 . 6.4
-71 . 1
0 .0.
0 .0
6. 1
0 . 52 .
0- 0
0 .4 163
-4 • •
0. . 2 0. 0.
G . aOG
0 . 44.
-a l . 1 N
- 8 1 . 1P
-7 .. . 1'
0. . 01
-65. I N
-6 5 . 1 P
-65. 1 1'
0.. 0.
- 1 66 . 0.
-10 1
-1 . 1
- 14" 3 - 1 50. . 1
0. . 0.
0 . 0. 9 0.
0. . 1 8 1
0. . 0.90.
0. . 1 58
0. . 06.
0 . 0.6.
0 . 295
0. . 0.%
-a 5 . 7' N
-65. 7T - 65. ,..
-6G. IZ
-5a . 8 N
-s a . a p
- 56. 71'
G .GP
-a G . 7 N - 6G . 4Z
0. .2 5 1 0. . 0. 34
- 1 76 . 4
0. . 0. . 5 6
0.. 0.
0. 0.
0. . 0.
G.OZ
- 79. I P - 7• • IN 0. . 1 4 1 174. 1 G . 2 aGa
-7 • • 11' 0 .2.a
0. . 1 4 1
-7 • • 7N
20 3
-76. 6 1'
0. 0 1 ..
0 . 7a 1 3
1 7 .. .
- 79. ,..
0. . 466
0. . 1 ..
-u. a 0. . 0. 93 ------
0. 2 0. 154
0. . 3 0. . 0.337 - 1 5 2 • •
0. . 7946
--- ----- - ---- ------ -- ----- --
-0. . 4
0 •• 0.46
0 . 1 . 85 - 1 74. 2
G.GN - a4. M
-7 •• 41'
0. . 46 1
- a4. 6T
- 8 0.. ,..
0. . 0.
0. . 0.
0. . 46 1
0.0
0. .0.
0. . 25 1
0. 0.
0.. 0
0. . 535
0. . 0. 0. . 0
--- ---------- ------------ - - -
0 . 0.
0. . 0. 765
G •• a60
0 . 0 0. 5 1
0. . 0.353
0 . '7 0 a
- 8O .0 T - -- ---- - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - -- ------
0.
6-
0 .7 3. 0
0.0
G. l la
0. . 1 1 8
-5 • • 3%
- aa. ON
-a a . G P
- . 1 . 5T
0. . 040.. - l S I • • 0. . 0. 50. - 6a . a --------------------- -------
0. . 0. 1 79 • •
0. . 7 5 1 0.
1 . 74 1
0
2-
0. . 0.
1 76. 0.
a.5
------- -------- -
0 . 0.
G . 3 aa7
0. . 6 1 28
- - - -- -- --
G . a36
G . 3a.
- 1 63 . 4
G .3 a.
0. . 7'.5
- 0.. 5
G . 1 2a5 - 176. 4
G. a7'7a
-60 . 1 P - 60. 1 N -60 . 1 Z
0 . 15a
G .G N
G . OP
-a G .'n
0.. 0. 0 . 0. 0. . 63 . -a G . 7' Z - --- - -- -- -- ----- - ---- -- -- - --
0 . 0.
0 .63. 0 . 0.
0.0
- 60 . 1 T - -- -- -- -- -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -------
0.. 0
0
0-
0. 158
0 . 1 5a
0 .3626
1 7 3 .8
0 . 7 57 a
0 . 1 30 2 - 1 6 1 . 7
-2 . 7
0 .4 7 4
- 7a . ' N
- 7 •• • Z
0. 703
- 78 • • P
0 . 70 3
1 .3
1 7a . 1
0 . 4 1 64
0. . 1 6• • - 1 7 0 . 7
0 . 8 774
E.5.
- 78 • • T ------- ---- ---- ---------- - --- -------------- ---------------- ------- - ____
0 . 5 84 0
0 .703
2. 1 1 0
:J
e: )C m
CD
> " "
�
co 0)
48 7
Solved Exa mples
E.3
A 14-Node N etwork
The final example network is the 14-node network given in Figure E.3. A load flow study for this network, including the given line, load, and generation data, is given in Table E .6. A fault study is given in Table E.7. ISB 138
AAA 1 38
AAA 69
cce 1 38
DDD �_-+69
III 69
E E E --i--t19
JJJ 69
GGG 69
HHH 1 38
G GG 1 38
Fig. E . 3 . A 14-node network.
FFF 1 18
! -.s
. , �.=
2 2
7
8 9 10 11 12 13 14
7
5 6
4
1 2 3
AN�
1 9�
6 . 70
0. 0 0. 7 0
1 2 . 00 4 2 . 00 20. 00 2 0 . 00 1 2 . 00 4 2 . 00 2 0 . 00 1 2 . 00 1t 2 . 0 0 1 0. 00 1 0. 00 0 . 00 3 5 0. 0 1 0 5 0. 0 1 0 5
0.0210 0. 0
0.0 0. 00 3 5
0. 02 1 0 0.0 0. 003 5 0.0210 0.0210 0. 0 2 1 0 0. 0 0 . 00 3 5 0. 0 2 1 0 0.0210
0. 0 2 1 0
G GG
E Ee
H HH
HZ
F FF 1 3 8 F F F "9 J JJ 1 3 8 J JJ 6 9 B 8B C CC
D DD b 9
12
213 213 212 21 2 212 213 212 213 212 213
212 213 213
2
2 0 0 0 0 0 0 0
0
1 0 0 1 1
1 . 020 1 . 000 1 . 00 0 1 0 000 1. 020 1 . 00 0 1 . 0 1t 0 1 . 0 00 1 . 000 1 . 00 0 1 . 000 1 . 000 1 . 00 0 1 . 0 00
0. 0
0. 0 0. 0 0. 0 0.0 0. 0 0.0 0. 0 0.0 0. 0 0. 0 0. 0 0. 0
0. 0
1 00 . 0 0. 0 1 00 . 0 0. 0 1 0 0. 0 0.0 100. 0 0. 0 50. 0 50. 0 25. 0 25. 0 50. 0 50. 0 5 0. 0 0. 0 5 0. 0 0. 0 5 0. 0 0. 0 2 5. 0 2 5. 0 20. 0 2 0. 0 2 5. 0 2 5. 0
5 0. 0 0. 0
0. 0 200. 0 0.0 2 00 . 0 0. 0 0. 0 0. 0 0.0 0. 0 0.0 0. 0 0. 0 0. 0
200. 0
- -- - x
- 4. 49 5 5
-8. 9&38 -8 . 9 b 3 8
0.0 56. 2 0. 0 0. 0 0.0 0. 0 0. 0 0. 0 0. 0 0.0 0. 0 0. 0 0.0
0.0
0.0 0. 0 0.0 0. 0 0. 0 0.0 0. 0 0.0 0. 0 0. 0 0. 0 0. 0 0. 0
0.0
o.
O. O.
O.
o. o.
O. O. o.
O. O. O. O.
O.
O. O.
O.
O.
OMAX MVAR
1 . 00 0
1 . 000
0. 990
0 . 990
R AT I O
T AP
1 00 . 0 0.0 0. 0 1 00 . 0 1 00 . 0 0. 0 0.0 0. 0 0. 0 0. 0 0.0 0.0 0. 0 0.0
MVA R A T I NG
OMIN MV AR
- 4. 4 9 5 5 -8 . 3 3 3 3 - 1 . 40 5 2
-8. 3333 - 1 . 4 05 2
- 4. 4 9 5 5
- 1t. 1t9 5 5 - It. 10 9 5 5 -8. 3 3 3 3 - 1 . 10 0 5 2 - 4. 1t 9 5 5 - 1t . 4 9 5 5 - 4. 4 9 5 5 -�. 3333 - 1 . 40 5 2
" I PU I
EFFECT
G E N ER AT I ON - x M il III V A R
3 . O lt 7 7
1. 1 710 3 . 0 1t 7 7
1 . 5 0 60 1 . 5 0 60 0. 0 1. 1 7 1 0 1 . 5 0 60 O. a
1 . 5 060 1 . 5 0 bO 0.0 1. 1 7 10 1 . 5 0 60 1 . 5 0 60 1 . 5 0 60 0. 0 1. 1 710
G I PU I
T AP
1 1 ZOo o.
10. 10 .
1 1 1
1 1
1 1 1 1 1 1
1
O.
o. o. o. o. o. o. o. o. 20.
M VAR
0. 0 0. 0 0. 0 0. 0 0. 0 0. 0 0.0 0. 0 0. 0 0. 0 0. 0 0.0 0. 0 0. 0 0.0 0. 0 0.0
0.0
PHASE SHIFT
331 8 30 878 764 651 86 8 806
220
94
322 84 2 l1t 296 272 46
1 1
88 872
1 ��3
1 1 1 1
1
1 1
1 1 1
1 1 1 1
1
X---- F L O ll P G l OC 0
MA P
DA T A
7 5 "1
717
173
2�'
L OC
TAP
- -- - - -
1t42
938 808
501
26
22
7 1 It
930 712
9 lt2
32 282
18
216
0 8 111 0 1 50 1 54 1t 5 1t 397 93 1t 940
941t
0 176 0
146
a 0 0 0 0 0 0
8 82
82 0 1 12 2 34 8 80 0
X
1 1
i
1
535 '1 0 <1 7 ..
e 04
7b6
82B 8U
U2 ] q�
752 44 '7 2
86 212 1t !l 6
706
1 491 1 8 70
1 1 1
1
1 1 1 1 1 1 1 1
R E V F l O il PG L O C Q
936 0
477
0 0 0 0 0 0 0 0 152 a 537
X -- - - - - - - -- M A P D A T A - - -- -- - - - - - X V OL T R E A C T O l' LOAD GE N P A G E L OC A LOC O S L OC 0 L DC 0
1 . 1 00
1 . 100
R E A C TO R
0. 900
0. 900
TAP L I � I T S TMA X TMI N
A 1 4-Node Network Load Flow S tudy
C ON V E R T E O N O
BC / 2 1 P U I
x x -- -
B U S - - - - - - X X - - - - V O L T A G E - - --- X --- L O A D - --- X X N A� e Mil III V A R A R E A R E G M A G I PU I A N G I O E G I
0. 7 0 4.20 0. 0 0. 7 0 2. 10 2. 1 0
c. o
4. 2 0 4. 2 0
4. 20
4. 20 4. 2 0
4. 2 0 0. 0 0. 7 0
2 0. 00 1 2 . 00 1t 2 . 0 0
2 0 . 00 2 0 . 00 20. 00
4. 20
BC I PCT I
--- - - - - - -
D ATA
PA�GRA�
A S S E MB L Y
3 e o L CA D F l OW
lOF
2 0 . 00
X I PC T I
ACTUAL
T R A � S F CR � E P
�F
A S S E�BLER
V E R S I ON
E NT E R E D
6.70 0.0 35.00 3. 4 0 3.40
&. 70 0.0 3 5 . 00
0.0 35. 00 b. 70
6.70 b.70 0. 0 3 5 . 00 6.70 6. 7 0
R I PCT I
x - - -- - - - -
L ! �E
BUS DATA
a
0 0 0 0
a
a
a
1 0 0
0
0 0
0
0
0
N O.
" "A 1 3 8 AUb9 0 00 1 3 8
X----- "10 .
14
C A SE
13
B AS E
13 7
12
12
!4 14
10 10
�
�
1 11
Q
11 10
8
b 7
5 � b
4 4
3
� 3
1
1
8 9
P
L I�E
T ' A � 5 � I SS I :�
FRO�
u\ I V E � S I T Y
T ! T L o - gQ : 3L t �
Table E.6.
m
:l
0.o ;C
l> "C "C CD
� = =
� T AT E
I B�
OF
A S S E �e L E R
V F R S I CN
,2
CFF1?8
9
212
21
3
212
213
212
213
JJJ6Q
8
212
III
1 4 GGG
1 3 HHH
1 2 E EE
11
1 0 eee
eBB
7 JJJ1 3 �
000
0. 0 0 0
0. 000
-0 . 0 0 2
0.
-0.001
-0 . 00 1
0. 0 0 0
0. 0
-0. 001
0 . 000
-0. 002
0 . 000
-0. C02
0 . 000
0 . 00 1
c. o
0. 001
- 0. 0 0 0
- 0. 0 0 2
- 0. 0 0 0
o. 0 0 0
- 0. 0 1 3
0. 001
-0. 000
�V AR
o. 001
0. 0 0 3
0001 3 8
J0 06 9
0. 0 0 0
-o. o�o
�W
M I S � ATCH
A A A 69
AAA1 38
�A�E
e < C F 69
4
2
�US
212
213
21:
21�
2 13
212
2
�O.
AO�A
0
1. 020
0.982
0. 988
0.928
0. 931
0. 974
0 . 975
1 . 0CO
1 . 04 0
3.1
0. 1
6. 2
0. 3
5.7
10 . 1
- 0 . 10
0. 0
7. 7
8. 5
1 . 000
1 . 000
8. 8
9. 8
1.013
6. 6
-
A NG L E
S
5. 2
J
L OA DFLOw
1. C09
1. 02�
VOLTS
B
LOF
36C
< � P CR T CC L C A O F L 0W C A L C U L AT I O N S x - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - -
F R C�
U� l V E � S l T Y
T I T L � - P Rc 3 L E �
l �W �
�w
A
50. 0
50.0
25. 0
25. 0
50. 0
5 0. a
0. 0
100. 0
0. 0
1 00 . 0
0.0
1 00 . 0
0.0
TOTAL
25. 0
25.0
20. 0
20. a
25. 0
2 5. 0
0. 0
50. 0
0. 1
50. 0
0.0
0. 0
0.0
0. 0
0. 0
0.0
0.0
63.0
0.0
2 00 . 0
0.0
2 00 . 0
5 0. 0
0.0
0.0
200. 0
0. 0
5 0. 0
�V A R
0.0
0. 0
0. 0
0. 0
0. 0
0. 0
0. 0
112.2
0.0
6 7 . 8 <1.
2 . 8R
56.2
0.0
3 8 . 0 11.
GENER A T I ON �w � V AR
3
SW L I
I NG
-8. 6 1
-8. 66
19. 0 1
NA�E
0 0 1
212 212 212
213
0
0 0
0 0
0
3 3 3
0 0
0 1
a
0
0 0
o
o a
3
213
21 21 21 21
213 213
T
PAP NO.
N
2 6 . 73
2 0 . 66 52. 62
�W
6. 41 7 . 02
- 5 0 . 10 5
- 1 1 . 97 1 1 . 97
-9. 2 7 1 1 . 97 4S . 65 48 . 65
- 1 4. 5 6 1 10 . 5 6
6. 28 3 8 . 04 3 8 . 04
28. 52 1 1 . 01 22. 65
- 6 . '1 6 6 . 96
4 . 70 7. 19 2. 9 5 2.95
4. 96
- 2 . 1 '1
6. 2 8
2 . 4 10 -8.78
1 . 68 1 . 68
1 4. 56 9. 3 5
-
- 2 6. 5 2 12. 38
2. 1 8
�VAR
peT
CAP
O. 'l 8 8 R
O . <:' 9 C
0. 990
TAP
J J J 1 3 8 , AR E A 2 1 2 . A - - - - - - - - - - - - -- X
L I NE F L OW
7 A T
- - - - - - - - - - -- - - - - - - X
- 2 0 . 66 2 0. 66
0
BUS = E - 0
l-AAA 1 38
212 213
0 0
-26. 13 -23. 87
- 1 4 . 78 8. 78
5 - F F F 1 38
213
0
-4 7 . 1 1
- 2 . 56
213 0 - 1 3. 7 2 -4. 61 4- 00069 6- FFF 6 9 213 0 -11. 28 -6. 79 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - 6. 60 - 2 5. 73 7 - J J J 1 38 212 0 - 4 3 . 40 0 1 4- G GG 213 2 0 . 24
- 7 . 46 o -3 7. 0 5 213 3 - 0 00 1 3 8 - 37 . 05 3 -000 1 3 8 1 7 . 46 213 o 24. 1 1 212 9 888 1 0 . 08 - - - - - - - - - - - - - - - --- - - - - - - - - - - - - - - - - - - - - - - - - - - - -0. 5 5 2 - AAA 6 'l 212 0 - 1 9. 1 7 - 5 . 83 - 10. 78 212 0 8-J JJ 69
1 0-e e e
-
7 - J JJ 1 3 8 -6. 4 1 0. 9 7 3R - 1 0. 8 3 212 0 1 1-111 6. 4 1 21 2 0 1 0. 8 3 - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - ---- - - - - - - - - -
8 - J J J 6 'l 1 3 - HHH
l-AAAue
1 2 - E EE
5-FFF 138
1 4 -GGG l Io - G G G
3 - 0 C0 1 3 8 6- F F F 6 9
3 - 0 00 1 3 8 12-EEE
213 213
1 0- e e e
213 21 3
212 212
212
212 212
1 0- C e e
10 - 0 00 6 9 5-F F F 1 3 8
l - A AA I 38
1 1-111
2-AU69 7-JJJ 1 38 9-B BB
BUS
AREA NO.
X - - - - - -- - - - - - - - - - -
- · �. 5 1
-
e A P /R E A � �VAR
=
X- - -- - - - - - - - - -
I T E R A T I ON S
----------------- - - - - - - - - -- - - - - - -
LOAD
1 00. 0
A T
PROGR A �
� CD co
V>
m )(
QI 3 'C CD
(1) Cl.
en o <"
l �wA
S T AT E
I B '"
OF
A S SE �BL E R
V E R S I O�
360
NO
12
NC.
VOLTS
1.050
0. 3
ANGLE
OV E R
0. 93 1
VOL T S
"I I S M A T C H "IVAg "I W
VO L T A G E S
212
AP E A NJ.
� A "' E
N � L I � E S O V ER L O A DE D
ARE
ZlZ
N A "I E
9US
�US
T HE R E
11
NO .
AO E A ��.
B D
EEE
T
�W
A
E ND O F
A
AR E A
213
NO.
TOTAL
E.S. (continued)
0. 92 8
VOL T S
--
6. 2
BUS
THI S
�O.
BUS NA "' E
VOL TA G E S
CA S E
GE�E�AT IO� MVAR "I W
A NG L E
S U M "I A R Y
FOR
"IV A R
R E P O RT
LOAD
�O .
AR E A
B ELOW
N A "I E
VOL T S
ANGL E
5-F FF 1 �8 1 3 - HHH
0. 950
C A P / R E AC ... V A R BUS 213 212
AR EA NO.
PAR
NO.
1 0
- 47. 1 1 44 . 2 1
MW
l l �E
BUS �AME
NO.
AR E A NO.
peT
tAP
VOL T S
- 19.88
- 2 . 56
F L OW "' V A R
ANGL E
TAP
7 J J J 1 3 S , AR E A 2 1 2 . 3 S w i NG B U S . I T E R A T I ON S = X - - - - - - - - - - - - - L I N E - D A T A - - - - -- - - - - - - - - X x - - - - - - - - - -- - - - - - - T 0 - - - - - - - - - - - - - -- - - - x
- - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - -
P R O GRA �
L O W VOL T A G E
A N GL E
-
D F L OW
BUS N A "I E
U S
L OA
LDF .
R E P O R T O� L O A D F l � W C A L C U L A T I ON S x -- - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - -
F � O "l
U� I V E � S I T Y
T J T L � - P RC � L E "I
Table
m
x'
(1) :l �
l> 'tl 'tl
� co o
A
LOAD
000 1 3 8 eee e .0 0 e • • • • • • - 3 7 . 0 5 • • • • • • • • • • • � 8 . 04 • • • o
II
1.013
8
J
JJ J6q
*
0 . 9 B R T 4P >
- 1 9. t 7 -0. 5 5 •
* • •
* •
20.66 1 . 68 • • *
•
00069
-2 . 1 9
LO�D 2 5. 00P 2 0. 0 0 0
< <
0. 9 8 8 RT4P
-8. 610 SHUNT
2 . 44
JJ J1 3 8
•
•
•
* •
•
•
•
•
• •
•
•
•
•
• • • • •
•
• • • •
GGG
-2 . 56
2.95
FFF1 38
F
G E OI
1 0 0. 0 0 50.00
LOAD
200. 0 67.8R
•
*
- e . 78
• •
LOAD 1 0 0 . 00 50. 00
56. 2
0* • • • • • • • • • • 0 C. . 3 5
0
0 0 o* * * p
F -9. 2 7 < 4. 10 ·· * · · - 1 1 . 9 7 · · · ·> · · · .· . 1 1 . 9 7 • • • F • < F - 6 . 96 7. 19 < F······*··· F .0 G G • • • • • • - 47 . 1 1 • • • • • • •• • • • 4 8 . 6 5 * * * F 5
GGGGGGGGG • 6 •
1 1 . 97 6.96 • • GG G 6 9
-11. 28 - 6 . 79 • • • • • • • •
•
-4. 6 1 •
• =EE EEEEEEEE EEEEEEEEEEEEEE • • 12
-13. 72
HHH
14
1 . 0 0 0 11 7. 70
* • •
1 4. 5 6 4.96 • •
G
0. 10
0 . 9 8 811
S HVNT
- 8. 660
3
0. 990 T A P < 0 < 0 · · · · * - 1 4. 5 6· * * · > · * * * · · 1 4 . 5 6 . · * 0
10
• 4 0000000 • •
H H H
H
1 . 0 0 0 11 -0 . 4 0
11
0 . 9 2 8 11 6. 2 0
8. 8 0
1 . 0 0 0 11
-0. 0 p•• 2 . 8 1\ 0 •
GEN
LOAD C 5 0 . 00P * *C 2 5. 000 C
G U F J * . * * * . * * 7 . 0 2 * * * . * * * * * * . - 6 . 6 0 * • • H * • • • * * - 4 3 . 4 0 • • • • • • * * • • • 44 . 2 1 * * . G . * * . * . - 4 7 . 1 1 * * • • • * • • • • • 4 8 . 6 5 • • • F • • • P GEN H 6 3 . 0 P• • J 20. 24 - 1 9. 8 8 G -2 . 56 2.95 F 0 -25.73 22.65 H F 1 1 2. 2 0 J G L OA D H SHUNT G L OA D F J 7 1 . 0 4 0 11 J 5 0 . 0 0 P • • H• • • • * * - 1 9 . 5 1 1 . 0 2 0 11 F ••• P 0 . 9 8 2 11 G • • •• • • * 5 0 . 0 0P H 13 -0 . 0 0 J 25. 000 3. 10 G 2 5 . 000 8. 50 F 0
J
• •
••*** •
8
6. 4 1 10.83 •
• • • •
JJJJJJJJ
J • • • * * * • • 6 . 4 1 . * • • • • <* • • •- 6 . 4 1 • • 11. 0 1 > - 1 0. 8 3 >
1 0 0 . 0 0 P* . J 5 0 . 000 J
LOAD
. ** *• • J
-50. 45 2 8. 5 2 * •
*
•
•
• • •
•
5. 20
A* . * • • •• • * * •
A
•
J
2
4 4
• *
0. 30
2 S. 00P 20. 000 •
LOAD
-1. 68
1 . 0 0 8 11
ZZZ ZlZZ lZZZZZZZZZZ ZZZZZZZZZZZZZZZZZ
T4P
0 . 9 3 ! 1I
0. 9 9 0
> >
•
2.18
-5.83 -1 0 . 7 6 �
•
*
•
•
• *
• •
• •
5 2 . 62 - 26. 5 2 * *
*
>
2 5 . 00 0
A > 44A69 4 · · · · * ··2 0 . 6 6 **· · · *< · · · - 2 0 . 6 6 ** · 4
A 4
4 4 4 . *. * . * , • A
5 0. 000
1 0 0 . 0 0 P* . A
S S HU N T 50. 0 0 P• • S• • • • * . - 1 9 . 0 1 0
LOAD
8 8 G E l>; 200. 0
9
0 . 9 7 4 11 5. 70
9. 8
888 8
0
0. 9 7 5 11 4. 1 0
e -7. 46 6. 28 0 GEN 4 8 C U D 2 0 0 . 0 p • • 4 • • • • • • • 2 6 . 7 � • • • • • • * • • • - 2 6 . 1 3 • • • 8 • • • • • • - 2 3 . 8 7 • • * • • • • • • • • 2 4 . 1 1 * • • C • • • • • • - 3 7 . 0 5 • • • • • • • • • • • 3 8 . 04 • • • 0 D* * * P ! 8 . 0 1l 0 4 12. 38 - 1 4. 7 8 8 8. 7 8 - 1 0. 0 e c -7.46 6. 2e 4 0 Q 8 C
� 4
4
44A1 3 8
1 . 0 2 0 11 �. � o
� to
Ul
(J) 0 <" (1) Q. m x III 3 "'C CD
FOR
5 10 10
'3 3
!> 6 7
5
5
to
3
4
1
11
1
5
12 13 7 11 10 1 10
14 14
3 12
9
7 7
3 5
2
0
0 0 0 0 1 1 2
8 B 9 13
P
X -- ---------
0 0 0 1 0 0 0 0 0 0 0
1
0 0 0 0 0 0 0 0 0 0
0. 0 0. 0 0. 0 0 . 20 6. 70 6. 70 0. 0 '3 5 . 0 0 6.70 6.70 6. 70 0. 0 3 5 . 00 6.70 6. 7 0 0.0 3 5 . 00 6. 70 0.0 3 5 . 00 3 . 40 3 . 10 0
A C TUA L
NO.
PO S I T I V E
S E Ou E NC E
ASSEMBLY
1 . 00 1 . 00 1 . 00 1 0 . 00 20. 00 20. 0 0 1 2 . 00 10 2 . 0 0 20.00 20. 0 0 20 . 0 0 1 2 . 00 10 2 . 0 0 20. 0 0 2 0. 0 0 1 2 . 00 42 . 00 2 0. 0 0 12 . 00 10 2 . 0 0 1 0. 0 0 10.00
0. 0 0.0 0. 0 0. 0 0 2 0 0. 0 6 7 0 0. 0 6 7 0 0. 0 0 . 3 5 00 0. 0 6 7 0 0. 0 6 7 0 0. 0 6 7 0 0. 0 0. 3 5 00 0. 06 7 0 0. 0 6 7 0 0. 0 0 . 3 5 00 0. 0 6 7 0 0. 0 0. 3 5 0 0 0. 0 310 0 0 . 0 34 0
0. 0100 0 . 0 1 00 0. 0 1 00 0 . 1 0 00 0. 2000 0. 2000 0. 1 2 00 0 . 10 2 0 0 0. 2000 0 . 2 00 0 0. 2 00 0 0. 1 2 00 0 . 10 2 0 0 0 . 2000 0 . 2 00 0 0. 1 2 0 0 0 . 10 2 0 0 0. 2000 0. 1 2 0 0 0 . 10 2 0 0 0. 1 00 0 0. 1 0 0 0
0. 0 0. 0 0. 0 0. 1 999 1 . 5 0 60 1 . 5 0 60 0. 0 1. 1 710 1 . 5 06 0 1 . 5 0 60 1 . 5 0 60 0. 0 1. 1 7 1 0 1 . 5 060 1 . 5 0 60 0. 0 1 . 1 7 10 1 . 5 0 60 0. 0 1. 1710 3 . 0 10 7 7 3 . 010 7 7 - 10. 4 <; 5 5 -8. 3333 - 1 . 40 5 2 - 4. 10 9 5 5 - 10. 4 9 5 5 -8. 3333 - 1 . 100 5 2 - 10. 10 9 5 5 - 8. 3 3 3 3 - 1 . 100 5 2 - 8 . 96 3 8 -8. 9638
-10. 109 5 5
- 9. 9 9 6 0 - 10 . 10 9 5 5 - 10. 10 9 5 5 -8 . 3! 3 3 - 1 . 10 0 5 2 - 10 . 109 5 5
- 1 0 0. 0 0 0 0 - 1 0 0. 0 0 0 0 - 1 00 . 0 0 0 0
S E L F - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - X ------------X-------------- C ON V E R T E D ------- ---- - - - - X G C PU I X I PU ' e I PU I II. I PU ) X C PCT ) R I PC T I
DAT A
DATA
A 14-Node Network Fault Study
I B M A S S E MB L E R LDF L I N E A N D T R A N S F OR M E R
E.7.
x ------------------- ------------
D�ceL EM F�QM T R A NS M I S S I ON
Table
x· m
::::J 0.
(1)
'tl 'tl
»
� co N
13
3
3
3
5
4
0 . 0 09 3 5 0. 0 0 864 0.00126
0 . 0 0 02 1 0.00063
-0. 0001 9
5
6 7
'3 '3
0. 0006 8 0 . 00093
0. 0 0 0 1 5 - 0 . 0 0 00 6 0. 00003 - 0. 0 0 0 1 0 -0. 0001 3 0. 0070 1 - 0. 0 0 0 5 7 -0 . 0 0 694 - 0. 0 0 0 1 3 -0. 0 00 1 2 0. 00029 0. 00047 - 0 . 0 0 009 0 . 00003 - 0. 0003 5 - 0. 0 0 0 4 6
10 1 1 12 13 14 4 5 6 7 8 9 10 0. 1 1858 0. 0 01 1 7 0 . 0 1 1 20 0. 00020 0. 0 0 0 2 0 0 . 0044 0 0 . 0 06 5 1 0. 0002 0 0 . 06489
0. 00029 0. 00037
0. 00705 0. 00016 0 . 0 04 8 9
0. 00007
9
4
4
0. 0 0 0 2 2 0 . 0 0 0 64 -0. 0 0 0 1 5 - 0. 0 0 0 5 7 0. 0001 2 0. 0001 2 0 . 00476
8
0. 00933 0. 00862 0 •. 0 0 0" 6 0.001 1 7
-0 . 00 1 3 2 -0. 00072
- O. 0 0 0 0 8
0. 0 0 0 1 9 0 . 0 0 00 6 -0 . 0 0 1 0 5
0. 00149 0. 00078 0. 1 1926 0. 00020 0 . 000 1 9 0 . 0 00 1 5 0. 0 0 0 1 6 0. 00762 0 . 0 1 72 6
0 . 00944 0. 00890 0 . 00020 0 . 00 0 1 9 0. 00007 0. 00008 0 . 0 0291 0. 00345 0 . 0048 2 0 . 002 51 0 . 0 06 1 7 0 . 00 0 1 3
X
N E T W O RK
0 . 0 0" 5 5 0. 00237 0 . 06 e 2 6 0. 0 0 0 1 7 0. 0 0 3 8 9 0 . 00202
0. 00635 - 0 . 00 006 - 0 . 0 0 00 6 -0. 0 0 0 1 1 - 0. 0 0 0 1 1 -0.00253 - 0 . 0 0 84 5
- 0. 0 0 0 2 8 -0.0001 6
- 0 . 0 0 0 1 e; -0. 00005
0 . 0 0 00 5 - 0. 0 0 0 0 1
- 0 . 0 0 004 - 0 . 00 0 0 5 -0. 00053 - 0. 0008 0
0. 0 0 0 1 6 0 . 0 0 04 3 -0 . 00006 - 0. 0 0 0 0 5
R
S EQ U E NC E
-0 . 00006 -0 . 00006
1 1 12 13 14
4
4
4
4 4 4 4
:: 3 4
3 3
3 ::
::
5 6 7
4
3
3
14
12 13
10 1 1
9
8
7
5 6
2
2 2
2
2 2
2
2
2 2
2 2 2
lit 2 3 4
1 1
1
10 1 1 12
7
8 9
5 6
3 4
1 2
1 1
1 1
1 1 1
1
1
1
1
BUS
F R e � I B M A S S � M B l ER l D F I M A T R I X F O R THE POS I T I V E
B US
PRe8 l E �
� x I S T I NG
1 1 12 13
5 5
9 10 10 10 10 10 11 11 11 11 12 12 12 13 1 :: 14
9
9 c; 9
9
8 8
e
8
e
8
7
B
7 7
7
7
7
E: 6 7 7
R
14
14 13 14
1:
14 12
1 1 12 13
1 1 12 13 14 10 11 12 13 14
0. 02 5 7 5
0. 0 0 0 9 5 0 . 1 7 50 : -0. 00005 - 0. 0 0 0 0 1 0 . 0 :: 5 8 3 0 . 0 1 77 8
-0. 0 0 0 1 9 -0 . 0 0 0 1 4 0. 02 52 7 - 0. 00009 0. 0000 1 - 0. 0 0 0 1 5 -0.00013 0 . 1 7 70 0 - 0·. 0 0 00 9 0 . 0 0 2 05
-0. 0002 5 0. 0 0 5 0 5 - 0 . 0 0 00 9 0. 00 542 0 . 00 2 6 2 0.03369 0. 0 1 6 7 5 -0.0001 2 - 0. 0000 1 10 11 12 13 14 9 10
0. 0 0 2 0 7 0. 0 1 85" -0. 00043
0 . 0 0 02 5 0 . 0 0 0 .... 0. 00889 0. 01 090 -0.00029 - 0. 0 0 0 1 8 0. 0 0 " 1 8 -0 . 0 00 1 2 0. 00"39
0.00003 0. 0 0 0 0 0 0. 0 0 0 1 1 0. 00700 -0. 0001 2 -0 . 00006 -0 . 0 0 0 3 1 - 0 . 0 0 0 .... - 0. 00008 0. 00003
- 0. 0 0 0 1 2 - 0 . 0 0 009 -0 . 0 0 0 1 2 - 0. 0 0 0 1 1
14
8 9
12 13 14 7 8 9 1 0 1 1 12 13
8 9 10 11
14 6 7
8 9 1 0
5 5 5
5 5 6 6 6 6 6 c 6
BUS
8 US O. O O C U
X
0. 0. O. 0.
00: 85 1 1722 O e l Z I: 0843 0
0.00230
0 . 0 07 0 8
0. 2 74 9 0
0. 0 1 5 3 1
0. 00059 0. 00047 0 . 28806 0. 0004 1
0. 00166 0 . 00370
0. 00057 0. 0809 1
0 . 0 0 :3 1 7 0.00251 0. 0 0 0 8 9
0 . 1 0 .. 79 0 . 0 5" 7 8
0. O U 9 5
0. 0 8 7 8 7 0 . 00065 0. 02673
0. 001 79 0 . 0 0 0 '1 5
0.00151 0. 0008 1 o. 0 2 9 9 � 0 . 0 0 06 9 0. 0 29 1 3 0. 0 1 5 1 9 0 . 1 5 84 8
0. 0 5 2 2 8
0 . 0 04 9 1 0 . 0 06 7 7 0 . 0 56 9 9
0 . 06490
0. 001 1 8 0. 0 0 1 1 0 0. 00062 0. 00090 0. 00063
0.00531 0. 00733 0 . 1 1 860
0 . 000 3 6 0. 0 0 0 6 7 0 . 0 0" 9 0
0 . 0 0 1 l l!
� <0 w
3 "C CD
m >< Cl
0 <" CD Co
CJ)
D AT A
A S S E M BL Y
D AT A FO R
Table E . 7 .
lE RO
4 5 5
3 3 4
3
2 2
1
0 0 0
14
14 0 12 13 0 11 10
12 14
1 3 5 7 9 0 11 5 10 10 0
Q
0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0
NO.
0.10 0. 1 0 0. 1 0 1 3 . 50 1 3. 40 0.0 1 3 . 40 1 7 . 00 1 5. 0 0 15.00 1 . 00 70 . 0 0 1 5.00 1 7. 50 1. 50 75 . 0 0 1 5. 0 0 0. 0 70 . 0 0 8. 50 7 . 50
R I PC T I
6. 00 7. 0 0 5. 00 1> 5 . 0 0 60. 00 12.00 6 0. 0 0 60.00 70. 00 70. 0 0 30. 00 1 2 0. 0 0 6 5. 00 70. 0 0 35. 0 0 1 20. 00 6 5. 5 0 12.00 1 2 5. 0 0 3 5. 0 0 30. 0 0
X ( .. C T I
5
3 3 5 5
3
10 10 10 14 14 14
0 0 1 0 0 1
X- ----- S E L F ---- - - X N O. BUS 8US
5
3
BUS
X--
PROBL E M F R O M I B M A S S E M B L E R L D F E X I S T I N G MUTUAL A DM I T T ANC E M AT R I X
6 6 7 8 8 9 13
P
1 . 2000 0. 6 5 0 0 0 . 7 00 0 0. � 5 0 0 1 . 2000 0. 65 5 0 0. 1 2 0 0 1 . 2 50 0 0.3500 0. 3000
O. �OOO
0 . 2 7 77 0 . 2 0 40 0. 3 9 9 8 0. 3063 0 . 3 545 0. 0
0. 06 00 0. 0700 0. 050 0 0. 6500 0. 6000 0. 1 20 0 0. 6000 0 . 6000 0 . 7 00 0 0. 7 0 0 0
l it
10 1
1 222 3 71t 5 3322 0 3 1t l 0 6 5 52 7 843
S
0 . 4 42 2 2
0 . 36 5 3 2 0 . 44964
0 . 36 53 2
G
X--- -----
0. 0. 0. 0. 0. 0. 0.
0. 1 1 1 0 0. 3 6 2 7
0. 4371
0. 3 54 5
G I PU I
I PU I
x
C OU P L E D L I N E - - x NO. BUS
0 . 00 10 0. 0 0 1 0 0. 0 0 1 0 0. 1 350 0. 1 3 4 0 0. 0 0. 1 34 0 0. 1 700 0 . 1 5 00 0. 1 5 0 0 0. 0 1 0 0 0. 7 0 0 0 0. 1 500 0. 1 750 0. 0 1 5 0 0. 7 5 0 0 0. 1 5 0 0 0. 0 0 . 7 00 0 0. 0 8 5 0 0. 0 7 5 0
R I PU I
B
E L
P
5 5
3 3
- 1 . 45462
- 1 . 4 5 7 97 -1. 58181
-1. 45797
F -------- X B
-2. 8519 - 0. 5 9 9 3 - 1 . 4506 -8 . 3 3 3 3 - 0. 6 0 9 0 -2. 6980 -3. 1373
- 3 . 3 2 96 -0. 6 2 1 8
- 1 6. 1> 6 2 0 - 1 4. 2 B 2 8 -19.9920 - 1. 4748 -1. 5875 -8. 3 33 3 -1. 5875 - 1 . 542 8
I PU I
M
M
U
U
- 0 . 249 9 8
T
L
--------------------
T U A
0.0 0. 0
0.0 0.0
X
8"
0 . 43 4 6 5
0 . 3 7 6 89
L -- - --- X
2 2 . 00 2 2 . 00
20. 00 20 . 0 0
0.0 0. 0
0. 0 0.0
0 . 2 2 00 0. 2200
0. 2 0 0 0 2 0 00 0.
X --- C ON V E R T E D - - - - x R M ( PU I X M ( PU I X M I PC T I
U A
---- ---------
R M ( PCT I
-0. 1 8 5 1 4
GM
1 0
1 0
NO.
X --- - --
l it 14
10 1"0
Q
X - - - - - - - - - -- - - --- - - - X - - - - - - - - - - -- - A C T U A L
S E QU E N C E
(con tin ued)
x - - - - - - -- - - -- - - - - - - - -- - -- - - - - -- - S . E L F ----------- ---- - - - - - - - - - - - - - -- - - X x - - - - - - ----- A C T U A L - - - - ----- - - - X----- - -------- C ON V E RT E D - - - - - - - - - - - - - - - x
T R A N5 � I S S I CN
A S S E � B L E R l DF l 1 N � A N D T R A N S F OR M E R
pq�8L E� FqO� IBM
l> "C "C (I) ::J Co x' m
.j:>. (0 .j:>.
O RO B L E �
11 12 13 14
1 1 1
II
5 5
4 5
4 4
4
4
4 4
"
4
4
= = 4
=
3
3
?
2 2 3 ? 3 3 =
2
II
8
14 5 � 7
13
9 10 11 12
5 6 7
12 13 14 4
9 1 0
8
7
6
12 13 14 3 4 5
1 0
8 9
2 2
2
2 3 4 5 6 7
2 2 2 2 2 2 2
1
1
1 1
6 7 8 9 10
1
1
1 1 1
5
1 2 3 "
1
1
BUS
S E QU E NC E
0.0 0 . 00 1 7 9 0. 0 0 . 0003 1
0. 0 0. 0 0. 0 0.0 0 . 0 0 63 3 0. 0
0. 0 0 . 00 2 66 0. 0 - 0. 00085 0.0 -0.00052 0. 0 0 .00 1 26 0. 0 0 2 4 8 0. 0 0. 0 -0. 00067 - 0. 0 0074 0 . 0 1 99 6 0. 0 - 0. 0 1 1 3 5 0.0
0 . 00 1 6 0 0. 0 - 0. 00036 0. 0 -0. 00024 0.0 0. 001 1 4 0. 0 0 . 0 0 04 6 - 0 . 0 0 04 6 0. 0 0. 0 0 . 0 0 03 1 - 0. 0 0 0 2 3 0 . 00 2 3 7 0. 0 0. 0 0.0 0. 0 0. 0 -0.00237 0.0 0. 0 0. 00 7 5 9 0.0 0.0
R
F R O � I B � A S S E M B L E R L DF Z M A T R I X F O R T H E Z E RO
B US
E X I S T I NG
0. 0 8 1 6 9
0. 0 0 . 0 1 54 0
0.0 0.0 0. 0 0. 06093 0.0 0 . 0 044 1 0. 0 0 . 00309 0. 0 0. 02758 0. 0422 7 0. 0 0. 0 0 . 00 3 7 1 0 . 0 04 0 0 0 . 2 7 534 0.0 0 . 02 8 89 0.0 0.0 0. 0 O. C 0.0 0. 1 5301 0. 0 0. 0 0 . 04 5 7 0
0.05622 0. 0 0 . 00248 0.0 0. 001 3 7 0.0 0. 03 887 0. 0 0 . 03314 0 . 01 963 0. 0 0.0 0. 02 1 3 2 0. 0 1 3 2 5 0 . 1 1 40 6 0. 0 0. 0 0.0 0. 0 0. 0 0 . 00594 0. 0 0. 0
X
N E TWORK
.. .. .. 9 10 10 10 10 10 11 11 11 11 12 12 12 13 13 14
8 9 9
8 8
8
8
8
8
7
7
7
7 7 7
7
6 6 7
B 9 10 11 12 13 14 6 7
5 5 5 5 5 5 5 6 6 6 6 6 6 6
14
10 11 12 13 14 10 11 12 13 14 11 12 13 14 12 13 14 13 14
9
10 11 12 13 14
9
7 8 '? 10 11 12 13 14 8
lit
9 10 11 12 13
8
BUS
BUS
0. 0 0.0 -0 . 00 7 5 9 0. 0 0.0 0. 0 0. 0 7 3 9 1 0 . 03 5 6 2 0. 0 0.0 -0. 000 2 1 - 0. 00 0 5 0 0. 05 895 0.0 0.0 - 0. 0 0 0 5 9 - 0. 0007 0 0 . 1 3 67 8 0. 0 0. 0 0.0 0 . 36 8 3 7 0. 0 0.0 0 . 1 0 1 64 0 . 0 5 34 2 0 . 06 7 9 1
0. 0 - 0. 0 0 04 9 -0 . 0 0 0 6 2 0. 0 0. 0 0 . 00 1 1 5 0 . 0 0 1 66 0 . 0 2 79 2 0.0 0. 0 0.0 0.0 0. 0 0 . 00 5 c H 0. 0 0.0 0. 09463 0.0 0 . 0002 3 - 0. 0 0 04 7 0.0 0.0 0. 04 8 2 3 0 . 0 2 49 1 0. 0023 7
R 0.0 0. 002 6 8 0 . 00345 0.0 0. 0 0. 0295 8 0. 03610 0. 3 1 6 1 6 0.0 0. 0 0. 0 0.0 0. 0 0. 1 7 1 50 0.0 0.0 0. 47554 0. 0 0 . 0 2 3 50 0. 01451 0. 0 0.0 0. 2 6 0 5 1 0. 1 6200 0 . l l 40 6 0. 0 0. 0 0. 03831 0. 0 0.0 0. 0 0 . 3 73 6 5 0. 2 2 231 0. 0 0. 0 0 . 0 1 376 0. 0 09 2 8 0 . 34063 0.0 0.0 0 . 0 09 3 3 0 . 0 06 9 5 0 . 4 8 1 66 0. 0 0. 0 0. 0 0 . 762 36 0. 0 0. 0 0 . 50 1 5 8 0. 3 1 2 01 O . '3 8 0 7 7
X
CD
.j:>.
CD UI
CII
1:1
0 <" CD c.. m )( III 3 CJ)
OF
F AU L T
F R �M l e M
AA A1 3 8
NA M E
3
0 0 01 3 8
Z A A A69
�US
1 07 . 1 6 2
8 . 37 3
1 0 5 . 90 3
A MP S
1
-88.7
-87. 0
-89.
DEGR E E S
VOLTS
D E GR E E S
AMPS
D E GR E E S
0.3 179. 4 - 1 7 8. 7
0. 8829 -0. 1 0 . 1 1 7 2 - 1 79 . 1 0 .765 7 179. 7
0. 3233
0.6616 0 . 3 3 84
0 . 6 74 3 -0. 1 0 . 1 2 5 7 - 1 7 9. 5 0 . 74 8 6 1 7 9. 8
501 834 83 4 834
3� 6M 1 2. 5 5 5 12.555 1 2. 5 5 5
8. 2. 2. 2.
3 9. 9 31 13.310 13.310 1 3. 3 1 0
,�o.
VOL T S
D E GR E E S
AMPS
t N E
D
O E GR E � S
--- - - - - - --
A
A
D E GR E t S
AMP S
DE GR E E S
- - - - - - - - - - - - - - -- - - - - - - - X S - L - G - - - - - - -- - - - X
VOLT S
T
X----------
0 . 0649
0. 6 9 5 8
2- 0
0
0
7-
90. 4 8 9 2
1 . 0000
0
0-
3 . 2 99
2 . 3 19
-0 . 3
0 . 541
-5. 0
27. 9
0 . 0 1 00 . 0 0 0
90 . 0
-71 . 8
- 76 . 5
- 62 . 1
-
0. 2 1 78. 7 0. 0
0. 0 0. 0 0.0
-0. 1 0. 9 3 5 8 0 . 0 6 04 2 - 1 7 9 . 2 0 . 404 1 2 - 1 7 9 . 3
0. 9 6 1 5 - 0. 5 0. 0393 -168. 3 0. 5 1 7 6 179. 8
0.8815 0. 1 1 8 5 0. 0
1 . 000 0 0. 0 0. 0
1 . 08 1 0.291 0. 2 9 1 0. 500
1 . 1 77 0. 4 1 5 0.415 0. 348
0. 1 3 6 0. 0 6 8 0. 068 0. 0
37. 6 1 2 12. 568 12. 568 1 2 . 4 76
6 1 . 6P -6 1 . 6 N O.Ol
-78 . 8 l
8T - 7 1 . 3 1> - 7 1 . 3 >;
-7 a . 3 l
- 74 .
- 76 . 0N
- 76 . 7 T - 76 . 0 P
-
-61 . 6T
89. 4T - 89 . 5P - 89 . 5 N -89. 2 l
0. 9 2 54
0. 04 3 1 5
0
0
1-
1 1-
0. 0
0
0-
- 5. 2
-0 . 0
0.0
- 90 . 0
-55. 4
0 . 7 89
0. 0
7e712
0. 0
0.0 179.7 0. 0
0.0 0. 0 0.0
0. 8070 -0. 8 0 . 1 9 3 04 - 1 7 6 . 7 0. 2 3 2 5 1 77. 1
0. 9748 0. 0 2 5 2 0. 0
0.0 0. 0 0.0
-56. 0P - 5 6 . 0N 0. 1 5 1
-65 . 8 l
-58.2T 0. 2 6 7 0.267 0.684
O.Ol
-9 0 . 6 T - 90 . 6 P -90 . 6N
-88. 7Z
O . Op O.ON - 8 8 . 7T
5. 220 2. 6 1 0 2.610 0. 0
2 . 6 94 0.0 0.0 2. 694
- 8 7 . 8 T-------- -- - - - ---- - ------------ - -- - - -- - - - - - - ------------- --------------- 87 . 8 P - 87 . 8 N - 87 . 8l 1 . 0000 - 90 . 0 0 . 0 1 0 0 . 0 00 304. 3 69 - 8 9 . 2 r 0- 0 1 . 0000 - 89 . 1 P 0. 0 1 1 . 7 1 6 - 89 . 1 N 0.0 0 . 0 1 1 . 7 16 0. 0 -89. 5 Z 0. 0 1 0 . 9 3 8
-87. 6N 87. 6 l
-
8 7 . 6 T - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - 87. 6P
-88.5P -8 8 . 5 N -88. 5 1
- 8 8. 5 T -------- --------------------------------------------- -------- -- --------
BUS
(continued)
X - - - - -- - - - - - - - - - -- - - - - - - L X - - - - - - - - - 3 - I> H A S E
Table E . 7 .
S - D A T A - ------------ ----- - - X - - - x -- -- - - - - - - S - L - G - - - - - - - - - - x
A S S E � BL E R L DF C A L C UL A T I ON S
x -- -- - - -- - - -- - - - - - - - B U x-- 3 - P H A S E
� cPGRT
OQ06 L = �
l> '0 '0 (II ::::I Q. x· m
� co Ol
5
4
BUS
F F F1 3 8
D D D6 9
�AM E
1 06 . 9 1 8
8.41 8
AMP S
V OL T S
DEGREES
AMPS
D E GR E E S
-88. 7
- 86. 6
46. -0 . 1 1 5 . 0. 8549 0 . 1 10 5 2 - 1 7 9 . 3 1 5 . 1 79. 7 1 5. 0 .7097
0. 7687 -0. 1 0. 2 31 3 - 1 79. 6 0. 5375 1 79. 6
557 519 51 9 519
5. 841 1 . 947 1 . 947 1 . 94 7
0
0
0
1
5-
1 0-
1 0-
NO.
4-
BUS
0 . 2 44 5
0 . 2 44 5
0. 9 5 1 6
0. 0 8 8 7
VOL T S
AMPS
1 . 1 59
1 . 1 59
-0 . 3
-0 . 3
4. 5 1 2
-1 . 0
31.9
VOL T S
D E GR E E S
AMPS
DEGRE E S
-5 7 . 2 T 0. 1 73 -5 7 . 2 P 0. 8 9 1 6 0. 2 0. 08 7 - 57 . 2N 0. 1085 178. 0 0. 0 8 7 0. 0 O.Ol 0.0 0. 0 - - ------ - - - - - - - -- - - - ---- - - --72 . 5 7 3 . 6T 2 . 1 96 -7 1 . 6 P - 0. 1 0. 9943 0.529 -7 1 . 6N 0 . 0060 - 1 5 9. 8 0. 5 2 9 -7 5 . 5 Z 0 . 0 5 6 4 - 1 66 . 8 1 . 1 40 - --- ---- - - ------ - ----------- 71 . 8 - 74. 8 T 0. 5 27 -70 . 9 P 0. 9 1 1 5 0 . 1 36 -0. 1 0. 0 8 8 5 179. 0 0. 1 36 70. 9N - 78. 9l 0. 53 1 6 0. 2 5 7 17 8. 9 - ------- - - -- -- - - -- - - - - - ----- 71 . 8 -74. 8 T 0. 527 -0. 1 -70. 9 P 0. 9 1 1 5 0. 1 3 6 - 7 C. . � N 0. 0 8 8 5 1 79. " C. 1 36
-58. 1
DEGRE E S
I N E - D A T A - - - - - - - - - - - - - - - - - - --- -- X - - - - - - - - -- x---------- S - L - G - - - - -- - -- - - X
0. 7 3 9
L 3 - P HA S E
D E GR E E S
X -- - - - - - - -
X-------------------- ---
0. 4 5 5 8
a
1 2-
-4 . 0
0. 1
0. 0
0 . 8 3 10
7. 7 2 7
0. 0
1 . 791 0.0 O. Op 0.0 0. 0 0.0 O . ON 0. 0 0. 0 -8 8 . 4 l 0. 0 0. 0 1 . 791 - - -- -- --- -- - ---- -- - - -- - - - - - -8 9 . 5 T - 89. 9 3 . 574 0. 9 8 3 2 O. a 1 . 7 8 7 · 8 9 . 5 P 89. 5 N 0.0168 179. 6 1 . 7 87 O.Ol 0. 0 0.0 0. 0 - - - - - ----- - - -- - - --------- - -- 5 6 . Io T - 54. 2 0 . 5 57 -5 3 . 8 P 0 . 8 7 40 0. 1 9 3 · 0. 5 · 5 3 . 8N 0. 1263 17&. 2 0. 193
0. 0
0-
0
1 . 0000
0. 0 1 00. 000
- 90. 0
-89 . 2 T 10 3 . 2 2 2 1 . 0000 0 . 0 1 10. 5 1 5 · 89 . 3 P -89. 3 N 0. 0 0 . 0 1 4. 5 1 5 0. 0 0 . 0 1 10 . 1 9 2 ·89. 1 Z - - - - - - -- - - - - - ------- - - --
·U . 3 l 0 . 2 9 82 · 1 7 8 . 6 0. 1 7 2 - --- - - - - -- -- - --- -- - - - ----- -- 8 8 . 0 T - - - - - - - - - - - - - - - - - - - - - - - - - -- -- -- - -- - -- - - - · - - --------------- - --------- ---88.0P - 88. 0N - 88. 0l
0. 9 2 7 3
0
3·
0. 0
0
0-
-88.4T
-7a. 9l 0. 53 1 6 1 7 8. 9 0. 2 5 7 - -- -- ---- -- - - - - - - - - - - - - - - - - - 8 6 . 2 T - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - -- -- - - - -------- - - ---------- - - . - 86 . 2 P - 86. 2 N - 86. 2 Z
U S - D A T A - -- - - - - - -- - - - - - - - - - - X S E - - - X -- - - - - - - - - S - L - G -- - --- - - - - X
D E GR E E S
x - - - - - - -- - - - -- - - - - - - B x -- 3 - P H A
P R r B L E � F R O M 1 9 M A S S E M B L E R L DF R E P a � T OF F AU L T C A LC U L A T I ON S
.ra.....
CD
III
3 'tI CD
III
m )(
III Q.
en 0 <"
NA 'I E
7
J J J1 3 8
6 �F �69
BUS
1 7. 3 3 8
8 . 41 7
I� PS
V OL T S
DE GRE E S
AMPS
D E GR E E S
-8 1 . 1
- 8 6 . I>
1 79. 3
-0. 3 - 1 79. 0
-0.2
1 79. 5
0 . 9 04 0
0 . 8079
0 . 096 1 - 1 7 8. 1
0 . 5 71 9
0 . 2 1 41
0. 7859 5. 4 0 6
4. 999 1 . 66 6 1. 666 1 . 666
1 . 8 02 1 . 80 2 1 . 802
0. 2 1 6 3
0
1
1 4-
1 4-
L
1. 5
1 . 026
1 . 026
-1 . 5
-
0. 739
4. 5 1 2
A M PS
D E GR E E S
-
72 . 9
-72 . 9
-
- 7 1t . 7 T 0 . 1t 7 0 0. 8 8 6 3 - 0. 1 0 . 1 1t 9 - 72 . 2 P 0. 1 138 -1 7 8 . 9 0 . 1 1t9 72 . 2N -78 . 9 1 0. 5608 179. 3 0. 1 7 3 - - . - - - - - - - - --. - - - - - - - - - -- - - -
-5 7 . 4T 0.215 0 . 8657 - 5 7. 4P 0. 3 0. 1 07 0 . 1 3 4 10 - 5 7 . Io N 1 77 . 8 0. 10" 0. 0 0. 0 0. 0 0. 0 1 - - - - -- - - - -- - -- - - . - -- - -- - - - - -
AMP S
- 58. 1
D E GR E E S
2 . 3 38 - 73 . 6 T -H . 8P -0. 1 0 . 99 3 0 0.655 -7l . 8N 0. 00 7 5 - 1 6 0. 1 0.655 0. 0698 - 1 6 7. 1 -7 5. 91 1 . 029 - - ----- - - - - - -- - - - -- -- - - - - - - -
V OL T S
72 . 5
DEGR E E S
I N E - D A T A - - - - - - - - - - - - -- - - - - - - - - - X - - - - - - - - - - X- -- - - - - - - - S - L - G - - - - - - -- - - - X
31. 9
-1 . 0
DEGR E E S
3 . P HA S E
- 10 . 0
0. 4 5 5 8
1 2- 0
0.0
0. 1
0. 0
0.9271
a
5- a
0-
- 89 . 9
- 510 . 2
0 . 8 3 1t
0.0
7 . 726
0.0
5 6 . It T 0. 5 4 2 -0. 6 0 . 8 8 34 - 5 3 . 3P 0. 1 7 8 0. 1 1 7 0 - 1 7 5 . 7 0. 1 7 8 -53 . 3 N -6 2 . 3 1 0. 3 0 9 2 - 1 7 7 . 6 0. 1 86 -------- ------------ - ---- - - -
3. 3 0 8 89. 0T -0 . 0 0 . 9 8 4 1t -89. 0 P 1 . 6 5 10 0. 0 1 5 6 - 1 79 . 8 1 . 6 5 4 - 89 . 0 N 0. 0 0. 0 0. 0 0. 01 - --- - -- - - ----.------ - ---- - --
1 . 6 33 - 8 8 . 3 T 0. 0 0. 0 0.0 O.oP 0. 0 0. 0 0. 0 O. ON 0. 0 0.0 10633 88. 3I - -- - --- - - - - - -- - - - - - - -- - --- - -
7 9 . 2 T - - - - - - - - - - - - - - - - - - - - - - - - -- - - - -- - - - - - - - - - - - - - - -- - - - - -- - -- - ----- - --------H . 2 P - 79. 2 N -79. 2 Z - 88. 9 - 8 6 . 9T 0- a 9 . 99 8 1 . 0000 0.0 1. 922 -8 6 . 9 P 1 . 0000 0. 0 0.961 -8 6 . 9 N 0. 0 0. 0 0. 9 6 1 0.0 0. 0 0. 0 0. 01 - --- - --------- - ---------- ---
-8 5. 7 I
- 85 . 7 P - 8 5 . 7N
- 7 1o . 3 T 0 . 1t 5 2 -0. 1 0. 8 863 0 . 1 109 -72 . 2 P 0. 1 1 38 - 1 78. 9 0. 1 49 -72. 2N 0 . 5 6 08 -78. 1 I 1 79 . 3 0.155 - - - - - - - - - - - - - -- - - - - . - - - -- - - - 8 5 . 7 T - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - - --- - -- - - - -- - - -- - - - -- -
0. 2 1 6 3
0. 0 8 8 7
6- a
VOL T S
0.9517
NO.
3- a
BUS
x - - - - - - -- -
(con tinued)
x-------- ---------------
Table E . 7 .
S - 0 A T A - -- -- - -- --- - - - - - - - - - X - - - x - - - - - - - - - - S - L - G ---------- X
D E GR E E S
x -- -- - - -- - - -- - - - - - - - B U x-- 3-PHA S E
D � " 5 L : " F R O ,," I S 'I A S S E � o L E R L O F � = P :: P T 'CF F AU L T C A L C U L A T I ON S
:> "C "C (I) ::l Q. x' m
.j:>.
�
J J J69
B8B
9
NA � E
8
BU S
9. 065
6. Z67
AMPS
S
- D
A T
-7 2 . 2
-63. 3
0. 9 0. 8 1 7 2 0 . 1 83 4 1 75. 8 0 . 6 34 7 - 1 7 7 . 6
0. 8 0. 631 5 178. 6 0. 3687 0 . 2 63 6 - 1 76. a
V OL T S
1 . 66 6 1 . 666 1 . 666
4 . 99 9
2.310 2. 3 1 0 2. 3 1 0
b.931
BUS
NO. VOL T S
L I N E - D A T A --------- - -- -- ------- -- X 3 - P HA S E - - - - - - - - -- X - -- - - -- - - - S -L - G ------- - - - - X A M P S DE GR E E S D E GR E E S VOL T S D E GR E E S A M P S D E G R E E S
X---------
x -----------------------
a
0. 9 5 1 B
-1 . 0
4. 5 1 2
- 12 . 5
1 1- a
0. 4 5 06
-4 . 1
0 . 8 24
-54. 3
0. 01 0. 0 0. 0 0 . 0 - - - - - - - - - - - - - - - - - - - - --- - - - - -
0 . 796 8
-57. 0T 0. 7 3 0 - 5 5 . 8P 0. 304 -0. 5 -55 . 8N 0 . 3 04 0. 2 0 3 3 - 1 7 8 . 1 -63. O Z 0. 1 2 3 0. 0902 - 1 6 3 . 6 -------- - -- -- - - - - - - - - -- - - - - - 7 6 . 4 T - - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - - - - - - - - -- - - - - --------- - - - - - - ------ - - -76. 4 P - 7 6 . 4N -76. 4Z -0 . 8 -76. 3 T 4 . 5 43 2. 6 1 1 0. 9 5 8 2 -72. 3 1- a -76 . 5 P 0.833 -0. 1 0. 9 9 2 2 76. 5N 0. 0080 - 1 6 7 . 0 0. 833 0 . 0552 - 167. 2 0. 944 - 76. 0Z - ---- - - - - - - --- - -- - - - - -- - -- - - 72 . 1 -7 6 . 5 T -0 . 9 2 . 3 88 4. 542 0. 4797 1 0- 0 -76 . 3P 0. 90lt7 0.4 0 . 8 33 -76 . 3 N 1 76. 6 0. 0955 0 . 8 33 0. 7 2 2 0. 3 7 52 - 1 75 . 5 - 76 . 9 1 - -- - - - - - --- - - - - - - - - - - - - - - - - -
1-
- 75 . 6T 1.982 - 70 . 6 P 0. 9954 0 . 4 34 - 0. 1 - 7 0. 6 N 0. 434 0 . 0 049 - 1 5 8 . 9 -79. 6 1 0 . 0648 - 1 7 0 . 9 1.121 - ---- - - - . - - - -- - - - - - - - ---- - - -56. 1 T -58. 1 0 . 740 0 . 1 42 8- a 31.9 0. 0887 -56. 1 P 0. 9 1 1 0 0.071 0.1 56. 1 N 1 79. 0 0. 0 8 9 0 0. 0 7 1 0. 01 0 . 0 0. 0 0.0 - - - -- - - - - -- - -- - - - - - -- --- - - - -74. 6T -0 . 3 -71 . 8 0. 9 8 9 1 3- a 2 . 3 20 0.4893 - 69 . 9P 0. 95 10 -0. 1 0. 2 2 3 -69 . 9 N 0 . 04 9 1 - 1 7 7 . 8 0.223 -78. 5 Z 0. 5 lt 5 0 . 4lt 1 5 - 1 7 9. 7 - - - - - - -- - - - - - - - - - --- -- - --- - - 84 . 6 T --- - - - --- - - - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - ----- - --- - -- - - - - - - ------ - - -64. 8 P - 84. 8 N - 84. 8 Z -86 . 0T 2. 197 0.0 0.0 0 - a 0. 0 0.0 O.OP 0. 0 0.0 0.0 O. ON O. a 0. 0 0. 0 86. 0Z 0. 0 0 . 0 2. 1 97 - --- - - - - - - - - -- - - - -- - - -- - - - --8 8 . 9T -87. 4 7- 0 4 . 1 00 5. 561 2.6 0. 6673 -8 8 . 9 P 0. 8775 2.050 0. 9 -8 8 . 9 N 2 . 050 0. 1 2 34 1 73 . 5
S - L -G - - -- - - - -- - X D E GR E E S A M P S D E GR E E S
A - -- - - - - - --- - - - - - - - - - X
- - - x------ - - - -
DEGR E E S
x -- - - - - -- - - - -- - - - - -- B U x-- 3-PHA S E
O � D � L � M F Q C M I � � A S S E M B L E R L DF R E P O R T OF F A U L T C A L C U L AT I ON S
� co co
I»
3 '0 m'"
m
)(
<" CD Co
(J) 0
��
I��
F AU L T
�� �M
EEE
HHH
l�
ZZ Z
CCC
N'�E
12
11
10
gU S
'�PS
8. 1 5 8
3. 069
2 . 95 e
-
U
-
7 3. 0
- 57. 5
- 58. 4
-7 2 . 7
1. 0
2. 7 173. 3
1 76 . 2
1. 1
0 . P' 3 � 3 0. 7 0 . 1 62 1 176. 3 0 . 6 76 8 - 1 7 8 . 2
0. 5660 - 1 77. 1
C.7a28 0.2178
C. 4294 - 1 71 . 0
0.7128 0. 2 S99
0 . 1 64 8 1 7 5. 0 0 . 6 72 2 - 1 77 . 5
O . Il ? 5 9
e34
1. 322 1. �22 1. 322
3 . 9b 7
O. b� S
2 . 00 5 0.6b8 0.6b6
0. 8 5 8 0. 8 5 8 0. 8 5 8
2. 5 7 3
1 . 94 5 1 . 94 5 1 . 945
�.
NO .
VOL T S
D E GR e e S
A M PS
D E GR E e S
VO L T S
D E GR E E S
A MP S
DEGR E E S
D A T A ------ ------ - - - - - - - - - -- x - - - - - - - - -- X - -- - - - -- - - S - L - G - - - - - - - - - -- X
L I N E
3 - P HA S E
0
0
0. 3 2 4 3
0. 9 2 0 4
0. 9 2 04
- 0. 7
-1.4
-1 . 4
- 72 . 9
-71 . 9
3 . 07 0
- 72 . 9
4 . 3 64
4 . 3 64
-
-
-
-78 . 3 T 2. 086 - 7 8 . 0P -0. 2 0. 7 1 9 -78 . 0N 0. 0 1 37 - 1 6 8 . 9 0. 7 1 9 0 . 08 2 3 - 1 7 1 . 1 -79 . 0Z 0 . 64 7 -- -- ----- - -- - - --- - - -- - -- - -7b . 3 T 1 . 66 3
0. 9866
-
2.086 -78 . 3 T -78. 0P 0. 9 8 66 0.719 . -0 . 2 -78 . J N 0. 0 1 3 7 - 1 6 8 . 9 0. 7 1 9 0. 0823 - 1 7 1 . 1 0 . 647 -79. 0l -- - - - - - -- - - - - - - - -- - ---- - -
0. 8 3 6 7
0 . 7 8 06
0
s- 0
2-
-9. l
-7.4
1 . 42 8
l . 530
- 59 . 3
-57.6
-
-
1 . 102 - 62 . 91 0. 0 3 02 -1.7 -6b . OP 0. 414 0. 0 7 5 5 - 1 5 8 . 4 -66 . CN 0. 414 - 5 3. 9Z 0 . 0 3 3 5 - 143. 9 0. 219 - - --- - - - - - - - --- - - --- - -- -- - - -
-b6 . 7 T 1 . 472 - 64 . 3 P 0 . 444 -1.5 - 1 5 4. 2 0 . 444 64. 3 N 0 . 0 7 04 - 1 6 0 . 4 0. 5 86 - 7 0 . '- Z - - - ---- - ---- -- - --- - - - - - - - -
0 . 9476 0.0585
- - -- -
4-
0
0. 8 3 8 8
-7. 3
-57. 5
- 57 . 5
1 . 5 34
1 . 5 34
-
-6 1 . 5T 1 . 010 -6 1 . 3 P 0. 9 6 2 2 -1.2 0 . 3 34 0 . 0 43 .. - 1 5 1 . 3 -61. 3N 0. 3 34 0. 1 024 - 1 5 3 . 7 -6l . 8 Z 0 . 34 1 - ---- - - - -- - -- - - - - - - - -- - - -- -
- 76 . 7 P - 76 . 7 N -76 . 7 Z
6-
a
0. 8 3 8 8
-1 . 3
--
- -
-
-61 . lT 0 . 9 96 0. 9 6 2 2 -61 . 3 P 0 . 3 34 -1.2 -6 1 . 3 N 0. 04 3 4 - 1 5 1 . 3 0 . 3 34 -60. 8 Z 0. 1 1 .. 7 - 1 5 3 . 3 0. 3 27 - ----- - - - - - -- - -- --- - - - 7 6 . 7 T -- - -- - - - --- -- - -- - - - - - - - - - - - - - - --- - ---- - - --------- - --- - -- -- ---- - ----- - - -
- b 1 . 3N -6 1 . 3 Z
::i: ;!----------------------------------------------------- ------------------
- 6 5. 1 P -6 5 . 1 N -65. 1 l
-
- 7 7. 0P 0. 8 8 9 0 0. 6 0. 5 0 6 - 7 7 . 0N 1 75. 3 0. 1 1 1 4 0 . 5 06 -75 . 2 Z 0 . 43 7 8 - 1 7 6 . 8 0.651 ---- - 6 5 . l T - - - - - - - - ----- --- - - - - - - - - - - - - - - --------- - - - - - - -- - -- - - - - - - - - - - - - - - - - -- - - -
9-
3-
3-
- 7 7 . 7 T - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -- -- - - - - - -- - - - - - - - - -- - -- - - - - - - - - - - - - - - - 77 . 7 P - 7 7. 7 N - 77. 7 Z
SUS
x -- - - - - - - -
(continued)
x - - - - - - - - - -- - - - - - - - - - - - -
Table E . 7 .
S D A T A - - - - - - - - - - - - -- - - - - - - X - - - x - - - - - - - - - - S - L - G - - - ----- - - X DEGR�E S V OL T S D E GR E E S A M P S DE GR E E S
�
� -p�4 S E
1 1 . 79 7
x--
l DF
C � LC U L A T I Q� S
A S S £ �Al Eq
t -- - - - - - - - - _ _ _ _ _ _ _ _ _
; �PG� T
O��� l � �
m
»
'tI 'tI CD :::l 0x·
o
�
p qa � L ! M
14
BUS
GG G
NA M E
AMPS
-73 . 0
DEGR E E S
B U � - PH AS E
1 1 . 34 5
x --
V OL T S
D E GR E E S
AMPS
DE GRE E S
0 . 84 3 8 0.9 0 . 1 56 8 1 1 5. ! 0 . 6 8 8 0 - 1 1 1. 8
5.336 1 . 119 1. 119 1 . 179
1-
BU S 0
NO. 0. 1 6 3 1
VOL T S -2. 6
- 14 . 1
-72. 1
3 . 61 8
1t . 5 1t2
-11 . 4 T 1 . 6 59 - 0. 2 -11.9P 0 . 5 86 0. 9 6 1 2 - 1 1 . 9N 0. 586 0 . 0390 115. 3 -16. 3 Z 0 . 1t 8 6 o . ! 5 0 1t - 1 7 7 .. 2 - - - - - - --- - - - -- - - - -- - - -- - - -- -
L I N E - 0 A T A --- - -- - - - - - - - - - - - - - - - - - X 3 - P HA S E - - - - - - - - - - X - -- - - - - - - - S -L - G --- --- -- - - - X A MP S DEGR E E S D E GR E E S A�PS DEGRE E S V OL T S D E GR E E S
X---------
X - - - - - - - - - - - - - - - - - - - - - --
- 17. n
0
1
0
5-
1 3-
0
5-
1 4-
0. 2765
0. 9 2 0 1t
0. 9 2 0 4
0. 4 191
2. 618
4 . 364
- 1 . 1t
-2. 1
1t . 3 61t
-1. 4
-0. 9
- 73 . 3
- 72 . 9
- 72 . 9
- 77. 2T 1. 225 -1 8 . 1 P 0 . 4 10 0 . 8868 0.5 -7 8 . 1 N 0. 4 1 0 176. 1 0. 1 1 3 5 - 75 . 6 Z 0 . 1t01t 0 . 5631 - 177. 5 - --- - - - - - - -- - -
-78 . 0T 2 . 09 5 - 77. 6P -0 . 1 0. 6 8 1t 0. 9 8 1 2 -77. 6N 0 . 6 81t 0 . 0 1 30 168.6 18. 1 Z 0.727 0 . 061t3 170. 4 - - - - - - - - - - - - -- - - - - - - ----- - - - 77 . 7 T 2 . 0 16 -71.6P -0 . 1 0 . 6 84 0. 9 8 7 2 - 17. 6 N 0 . 6 8 1t 0 . 0 1 30 - 1 6 8 . 6 77. 9Z 0 . 6 1t 8 0 . 06 4 3 - 1 7 0. 1t - - - - - -- - - - - - - - - - - - - - - - - -
-16. 3 T 2 . 3 09 - 15. 9P 0 . 1 36 0.9158 0. 3 -15 . 9N 0. 136 0 . 0 8 1t 1t 111. 1 -11. 0Z 0 . 4 1 86 - 1 7 6. 5 0. 8 3 1 - - - -- - -- - - - - - - - - - - - -- ---- - -- 1 7 . 7 T - - - - - - - - - - - - - - - - - - - - - - -- - - --- - - - - - - -- - - - - - - -- - - - -- - - - - -- - -- - - - --- - - - - - -77. 7P - 77 . 7 N
S - 0 A T A - - - - - - - - -- - - - - - - - - - - X - - - x -- -- - - - - - - S -L - G - - - - - - - - - - X
1 9 M A S S E �P L E R L O F F A U L T C A L C U L A T I ON S
F�OM
OF
x -- - - - - -- - - -- - - - - - - -
R E PO� T
�
UI
3 " CD
)(
II)
m
[
rn o
appe n di x
F
d -Y Tra n sfo rm a t i o n s Conversion of impedances from t:. to Y and Y to t:. are computed so often that the formulas for these operations are included here for convenience. F.1
V to t:. Transformation
Given three impedances Za . Zb , and Zc connected in Y, the equivalent
t:. impedances Zoo , Zbc ' and Zca are computed as follows:
(F .l )
where Yl: = l/Za + l/Zb + l/Zc In the special case where Za = Zb = Zc = Z, then Zoo = Zbc = Zca = 3 Z. F .2 t:. to Y Transformation
Given three impedances Zoo , Zbc, and Zca connected in t:. , the equivalent Y impedances Za , Zb , and Zc are computed as follows:
[�j i. If � =
where
oo
Zl: = Zoo + Zb C + Zca
In the special case where Zoo = Zbc = Zca = Z, then Za = Zb = Zc = Z/3.
502
(F.2)
PER UNIT CALCU LATIONS
z=
S� l rp
VB-L N
( Z oh m )
=
S�.3tp
VB-LL
(Z oh m ) p u
=
Base MVA 3� (Z ohm ) pu (Base k V L d
P HASOR TRANSFORMATION =
+
a)]
=
IA le i a
SYNTHES I S EQUATION
[1� 1 1 1
Vabc = A V011 =
a2 a
ANALYSI S EQUATION Vou = A - I Vabc
=
A
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and test ing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or aris ing out of, the furnishing, performance, or use of these programs.
WARNING:
Seal may not be broken prior to purchase. If the seal on this pouch is broken, product cannot be returned.
B i b l i og ra p hy
1. Fortescue, C. L. Method of Symmetrical Coordinates Applied to the Solution of Polyphase Networks. Trans. AlEE 3 7 : 1 027-1 1 40 , 1918.
2. Electrical Engineering Staff, Iowa State University. Symmetrical Compo nents. Unpubl. power system short course lecture notes, Ames, 1952-1968. 3. Nilsson, James W. Introduction to Circuits, Instruments, and Electronics. Harcourt, Brace, & World, New York, 1968 . 4. Kom, Granino A., and Kom, Theresa M. Mathematical Handbook for Scientists and Engineers, 2nd ed. McGraw-Hill, New York, 1 968. 5. Huelsman, L. P. Circuits, Matrices and Linear Vector Spaces. McGraw-Hill, New York, 1 963. 6. Marcus, M., and Minc, H. Introduction to Linear A lgebra. Macmillan, New York, 1965. 7. Hohn, Franz E. Elemen tary Matrix A lgebra. Macmillan, New York, 1958. 8. Ogata, Katsuhiko. State Space A nalysis of Control Systems. Prentice-Hall, Englewood Cliffs, N.J., 1965. 9. Stevenson, W. D., Jr. Elements of Power System A nalysis, 2nd ed. McGraw Hill, New York, 1 962. 1 0. Wagner, C. F., and Evans, R. D. Symmetrical Components. McGraw-Hill, New York, 1 933. 11. Clarke, Edith. Circuit Analysis of A-C Power Systems, 2 vols. General Elec tric Co., Schenectady, N.Y., 1950. 1 2. Boast, W. B. Vector Fields. Harper and Row, New York, 1964. 13. Lewis, W. E ., and Pryce, D. G. The Application of Matrix Theory to Electrical Engineering. E. and F. N. Spon, London, 196 5. 14. Westinghouse Electric Corporation. Electrical Transmission and Distribu tion Reference Book, 4th ed. East Pittsburgh, Pa. , 1 950. 1 5. Ward, J. B. Equivalent Circuits in Power System Studies. Res. Ser. 1 09, Eng. Exp. Sta., Purdue University, Lafayette, Ind. , 1949. 16. Ferguson, W. H. Symmetrical Component Network Connections for the Solution of Phase-Interchange Faults. Trans. AlEE 7 8 (pt. 3 ) : 948-50, 1 959. 1 7. Kimbark, E. W. Power System Stability, vol. 1. Elements of Stability Calcula tions. Wiley, New York, 1 948. 1 8. Kimbark, E. W. Power System Stability , vol. 2. Power Circuit Breakers and Protective Relays. Wiley, New York, 1950. 1 9. Kimbark, E. W. Power System Stability, vol. 3. Synchronous Machines. Wiley, New York, 1 9 56. 503
504
B i b l iography
20. General Electric Company. GE Network A nalyzers (manual). Schenectady, N. Y. , 1950. 21. Stevenson, W. D., Jr. Elements of Power System Analysis, 1st ed. McGraw Hill, New York, 1955. 22. Woodruff, L. F. Principles of Electric Power Transmission, Wiley, New York, 1 948. 23. Attwood, Stephen S. Electric and Magnetic Fields, 3rd ed. Wiley, New York, 1 928. 24. Calabrese, G. O. Symmetrical Componen ts Applied to Electric Power Net works. Ronald Press, New York, 1959. 25. Rosa, E. B., and Grover, F. W. Formulas and Tables for the Calculation of Mutual and Self-Inductance. Scientific paper 169, vol. 8, U.S. Bureau of Standards bull., Washington, D.C., 1912. 26. Lewis, W. W. Transmission Line Engineering. McGraw-Hill, New York, 1928. 27. Carson, John R. Wave Propagation in Overhead Wires with Ground Return. Bell System Tech. J. 5 : 5 3 9-54, 1 926. 28. Gross, E. T. B., and Hesse, M. H. Electromagnetic Unbalance of Untransposed Lines. Trans. AlEE 72: 1323-36, 1953. 29. Gross, E. T. B.t and Nelson. S. W. Electromagnetic Unbalance of Unt-TaIlS posed Transmission Lines. II. Single Lines with Horizontal Conductor Ar rangement. Trans. AlEE 74: 887-93, 1954. 30. Gross, E. T. B . , Drinnan, J. H., and Jochum, E. Electromagnetic Unbalance of Untransposed Transmission Lines. III. Double Circuit Lines. Trans. AlEE 79: 1 36 2-71, 1 959. 31. Dwight, H. B. Resistance and Reactance of Commercial Steel Conductors. Elec. J. , p. 25, January 1919. 32. Hesse, M. H. Circulating Currents in Parallel Untransposed Multicircuit Lines. I. Numerical Evaluations. Trans. IEEE PAS-85 : 802-1 1, 1966. 33. Hesse, M. H. Circulating Currents in Parallel Un transposed Multicircuit Lines. II. Methods for Estimating Current Unbalance. Trans. IEEE PAS-85: 812-20, 1 966.
34. Gross, E. T. B., and Weston, A. H. Transposition of High-Voltage Overhead Lines and Elimination of Electrostatic Unbalance to Ground. Trans. AlEE 70: 1 837-44, 1 951. 3 5. Gross, E. T. B. Unbalances of Untransposed Overhead Lines. J. Franklin lnst. , pp. 487-97, December 1952. 36. Gross, E. T. B., and McNutt, W. J. Electrostatic Unbalance to Ground of Twin Conductor Lines. Trans. AlEE 72: 1288-97, 1953. 37. Starr, F. M. Equivalent Circuits. I. Trans. AlEE, vol. 51, June 1932. 38. Hesse, M. H. Electromagnetic and Electrostatic Transmission Line Parameters by Digital Computer. Trans. IEEE PAS-82: 282-91, 1963. 39. Lyon, W. V. Applications of the Method of Symmetrical Components. McGraw-Hill, New York, 1 937. 40. Finkbeiner, Daniel T., II. Introduction to Matrices and Linear Transforma tions, 2nd ed. W. H. Freeman, San Francisco, 1 966. 41. Crary, S. B.t and Saline, L. E. Location of Series Capacitors in High Voltage Transmission Systems. Trans. AlEE 72: 1140-51, 1953. 42. Holley, H., Colemen, D. , and Shipley, R. B. Untransposed EHV Line Compu tations. Trans. IEEE PAS-83 : 291-96, 1 964.
50 5
B i b l i ograph y
43. Hesse, M. H. Electromagnetic and Electrostatic Transmission Line Parameters by Digital Computer. Trans. IEEE PAS-82: 282-91, 1963. 44. Neumann, R. Symmetrical Componen t A nalysis of Unsymmetrical Polyphase Systems. Sir Isacc Pitman & Sons, London. 1939. 45. Concordia, Charles. Synchronous Machines-Theory and Performance. Wiley, New York, 1951. 46. Prentice, B. R. Fundamental Concepts of Synchronous Machine Reactances. Trans. AlEE 56(suppl. ) : 1-2 1 , 1937. 47. Park, R. H. Two-Reaction Theory of Synchronous Machines. I. Generalized Method of Analysis. Trans. AlEE 48: 716-30, 1 929. 48. Lewis, W. A. A Basic Analysis of Synchronous Machines. I. T'rans. AlEE PAS-77: 436-55, 1 958. 49. Krause, P. C., and Thomas, C. H. Simulation of Symmetrical Induction Machinery. Trans. IEEE PAS-84: 1038-53, 196 5. 50. Floyd, G. D., and Sills, H. R. Generator Design to Meet Long Distance Trans mission Requirements in Canada. CIGRE rept. 131, 1 948. 51. AlEE Switchgear Committee. Simplified Calculation of Fault Currents. Trans. AlEE 67 (pt. 2) : 1433-35, 1948. 52. C57 . 12. 70-1978 Connections
for
American
National
Distribution
Standard
Terminal
and Power Transformers,
Markings
IEEE
and
Standards
Association, P. O . Box 133 1 , Piscataway, NJ 08855- 133 1 , USA.
53. Federal Power Commission. National Power Survey. I and II. USGPO, Washington, D.C. , 1964. 54. Westinghouse Electric Corporation. Electric Utility Engineering Reference Book, vol. 3. Distribution Systems. East Pittsburgh, Pa., 1959. 55. Garin, A. N. Zero-Phase-Sequence Characteristics of Transformers. I and II . General Electric Reu. 43 : (March, April) : 131-36, 1 74-79, 1940. 56. Atabekov, G. I. The Relay Protection of High Voltage Networks. Pergamon Press, New York, 1 960. 57. Kron, G. Tensor A nalysis of Networks. Wiley, New York, 1939. 58. LeCorbeiller, P. Matrix A nalysis of Electric Networks. Harvard University Press, Cam b dge, Mass . , 1950. 59. Hancock, N. N. Matrix A nalysis of Electrical Machinery . Pergamo Press, London , 1964. 60. Stigant, S. Austin. Modern Electrical Engineering Mathematics. Hutchinson, London, 1946. 61. Syn ge , J. L. The Fundamental Theorem of Electrical N etworks . Quart. Appl. Math. , vol. 9, no. 2, July, 1951 .
ri
n
62. Guillemin, Ernst A. Communication Networks, vol. 2. Wiley, New York, 1935 . 63.
64.
Anderson, P. M. Ana y sis of Simultaneous Faults by Two-Port Network The ory. Trans. IEEE, PAS-90 (Sep t. fO c t. ) : 2 1 9 9-22 0 5 , 1 9 71 . Stagg, Glenn W. and EI-Abiad, Ahmed H. Computer Methods in Power Sys tem A nalysis. M c G w- Hi l, New York, 1968. Anderson, P. M., Bowen, D. W., and Shah, A. P. An Indefinite Admittance Network Description for Fault Computation. Tra ns. IEEE, PAS-89 (Julyl August) : 1 2 1 5-19, 1970. Shildneck, L. P. , Synchronous Machine Reactances, a Fundamental and Phys ical Viewpoint. General Electric Reu., 3 5 ( N ovember) : 560-6 5, 1932. G mn , A. F. , Habermann, R., Jr. , Henderson. J. M., and Kirchmayer, L. K.
l
ra l
65. 66. 67.
li
506
68. 69. 70. 71 . 72. 73. 74. 75. 76 . 77. 78. 79. 80. 81. 82. 83. 84. 85. 86.
B i bliography
Digital Calculation of Network Impedances. Trans. AlEE 74 (pt. 3) : 1 285-97, 1 955. Hale, H. W., and Ward, J. B. Digital Computation of Driving Point and Trans fer Impedances. Trans. AlEE 76 (pt. 3 ) : 4 76-81, 1957. Toalston, A. L. Digital Solution of Short-Circuit Currents for Networks In cluding Mutual Impedances. Trans. AlEE 78 (pt. 3 B) : 1 720-23, 1 959. Brown, H. E., and Person, C. E. Digital Calculations of Single-Phase to Ground Faults. Trans. AlEE 79 (pt. 3 ) : 657-60, 1 960. Brown, H. E., Person, C. E., Kirchmayer, L. K., and Stagg, G. W. Digital Cal culation of Three-Phase Short Circuits by Matrix Method. Trans. AlEE 79 (pt. 3 ) : 1277-82, 1960. EI-Abiad, A. H. Digital Calculation of Line-to-Ground Short Circuits by Matrix Method. Trans. AlEE 79 (pt. 3) : 3 23-32, 1960. Baumann, R. Some Remarks on Building Algorithms. Proc. Power Systems Computation Conference, London, 1 963. Baumann, R. Some New Aspects of Load Flow Calculations. I. Impedance Matrix Generation Controlled by Network Typology. Trans. IEEE PAS-85 : 1 164-76, 1 966. Brown, H. E., and Person, C. E. Short Circuit Studies of Large Systems. Proc. Second Power Systems Computation Conference, pt. 2, report 4. 11, Stockholm, 1 966. Kruempel, K. C. A Study of the Building Algorithm for the Bus Impedance Matrix. Unpubl. Ph.D. thesis, University of Wisconsin. University Microfilms, Ann Arbor, Mich., January 1970. Penfield, Paul, Spence, Robert, and Duinker, S. Tellegren 's Theorem and Electrical Networks. MIT Press, Cambridge, 1970. General Electric Company. Project EHV. EHV Transmission Line Reference Book. Edison Electric Institute, New York, 1968. Ward, J. B., and Hale, H. W. Digital Solution of Power-Flow Problems. Trans. AlEE 75 (pt. 3) : 398-404, 1956. Allen, R. F., Zakos, R. J., and Lowd, R. IBM System/360 Electric Power Sys tem Load Flow Program. Operating System Version, Order File 360D16.4.005 and 360D-16.4.006. IBM Corp. , Chicago, Ill. Tinney, W. F., and Walker, J. W. Direct Solution of Sparse Network Equa tions by Optimally Ordered Triangular Factorization. Proc. IEEE 55 (Novem ber) : 1801-09, 1967. Sato, N., and Tinney, W. F. Techniques for Exploiting the Sparsity of the Network Admittance Matrix. Trans. IEEE PAS-82 (December) : 944-50, 1 963. Tinney, William F. Compensation Methods for Network Solutions by Opti mally Ordered Triangular Factorization. Trans. IEEE PAS-91 (January/ February) : 123-27 , 1972. Dy Liacco, T. E., and Ramarao, K. A. Short-Circuit Calculations for Multi-Link Switching and End Faults. Trans. IEEE PAS-89 (July/August) : 1226-37, 1970. Reitan, D. K., and Kreumpel, K. C. Modification of the Bus Impedance Matrix for System Changes Involving Mutual Couplings. Proc. IEEE, 57 (August) : 1432-33, 1 969. Storry, J. 0., and Brown, H. E. An improved method of incorporating mutual couplings in single-phase short-circuit calculations. Trans. IEEE PAS-89 (January ) : 71-77, 1970.
I nd ex
ACSR effect of magnetic material, 1 3 3 inductance, 7 6 Admittance matrix definite assumptions in fault calculations, 386-87 fault calculations, 388 power system description, 3 86-89 indefinite, 3 7 2, 3 7 5-76 building algorithm, 3 81-82 correction for mutual coupling, 3 8 2-85 example illustrating properties, 3 7 8-81 finding elements, 3 7 6 power system description, 3 8 1 - 8 2 properties, 3 7 8 node, 374-75 making changes, 4 1 9 sparsity, 4 1 8 primitive, 367 Algebraic equations machine calculations, 1 83 , 1 84 power system computation, 3 , 4 Alumoweld conductor effect of magnetic material, 1 3 3 A matrix n-phase system, 21-23 power invariance, 26, 27 three-phase system, 25 American National Standards Institute ( AN SI). See Standards, U. S.A. Analysis equation for three-phase systems, 24 a-operator, 1 6- 1 7 , 20 n-phase system, 20 table, 1 7 Atebekov, G. I . , 2 7 3 , 2 7 5 Attwood , Stephen S . , 7 7 Base admittance, 6, 7 , 8 Base current, 6, 7, 8 Base impedance, 6, 7, 8 Base quantity pu calculations, 5, 6 selection, 5 Base value, 5-7 change, 7 table, 8 three-phase systems, 7-10 Base voltage, 6, 7
Base voltampere, 6 Bessel functions in inductance calculations, 78 B matrix, two-component method, 347 Boast, W. B. , 5 Boundary conditions for shunt faults, 36 Brown, H. E., 401 Bundled conductors, impedance of lines, 1 06-1 2 Calabrese, G. 0 . , 7 7 , 1 5 8 Capacitance coefficients, 160, 1 6 1 effect o f conductor height, 1 5 5 to ground, transposed lines positive and negative sequence, 1 5 2-55 using conductor height, 1 5 5 zero sequence with ground wires, 1 57-58 zero sequence without ground wires, 1 56 method of images, 1 58-59 ' multiplying constant, k , 1 52-5 3 mutual, 1 5 8-63 self and mutual, 1 58-7 7 double circuit lines, 1 7 2-77 three-phase line with ground wires, 1 6 870 three-phase line without ground wires, 163 sequence similarity transformation, 166-67 , 1 7 0 three-phase line with ground wires, 1 7 0, 172 three-phase line without ground wires, 1 66-68 three-phase lines double circuit lines, 1 7 2-7 7 phase, with ground wires, 1 6 8-7 2 phase, without ground wires, 1 6 3-66 sequence, with ground wires, 1 7 0, 1 7 2 sequence, without ground wires, 166-68 Carson, John R. , 7 8 Carson's line, 7 8-80 Charging current of transmission lines, 1 6 2, 1 74 , 1 7 7 , 1 7 8 Circuit diagram for shunt faults, 36-37 Circulant matrix in machine calculations, 31 Clarke, Edith, 71, 1 58, 225, 226, 228, 243, 245, 246, 254, 2 5 7 , 258, 2 5 9 607
508
C matrix, symmetrical component trans formation. See also Capacitance n-phase systems, 21 three-phase systems, 24 Coefficients of electrostatic induction, 1 60 Cofactor of a matrix element, 161 Completely transposed line. See Impedance, transposed lines Computer solutions, 3, 365-66 using the admittance matrix, 365-89 using the impedance matrix, 393-4 1 9 Constraint matrix, K, 286-88 Copperweld conductor, effect of magnetic material, 133 Current, momentary, synchronous machine, 221-22 Current envelope, synchronous machine, 2 1 9-21 Current unbalance, incomplete transposi tions, 102-6. See also Unbalance Dm . See GMD Ds ' See GMD
dc offset, synchronous machines, 183, 1 8 5 , 1 88 , 221-22 Delta network, 347 quantities, 346-47 Diagonal matrix, impedances, 30 Differential equations machine calculations, 1 83 power system calculations, 3, 4 Digital computer applications, 4, 5, 365-4 1 9 Dot convention, mutually coupled circuits, 74 d-q transformation. See Park's transforma tion Dwight, H . 8. , 1 3 5 Dynamic problems, 3 , 4 Earth resistivi ty, effect on inductance, 7 9-80 Eigenvalues, circulant matrix, 3 1 Eigenvectors, circulant matrix, 3 1 EI-Abiad, A . H . , 401 Electromagnetic unbalance. See Unbalance Electrostatic unbalance. See Unbalance Equivalent circuit induction motor, negative sequence, 2 24 induction motor, positive sequence, 223 synchronous machine, d-q , 205-7 synchronous machine, d-q , steady state, 213, 214 synchronous machine, d-q , subtransient, 219 synchronous machine, T , 207 Equivalent spacing, 73 computing sequence line impedance, 92 Evans, R. D., 71, 7 9, 80, 1 29, 134, 1 3 5, 1 58, 228, 241 , 242 Fault, 31, 36 analysis procedure, 36-37 assumptions, 3 86 diagram, series and parallel connections, 283
I ndex
longitudinal. See Fault, series point series faults, 53-54 shunt faults, 31, 37 series, 36, 53-66 arbitrary symmetry, two-component method, 360-62 circuit diagram , 6 1 general ized fault diagram, 27 8-82 1 LO, generalized fault diagram, 282 1 LO, three-component method, 63-64 1 LO, two-component method, 3 54-55 sequence network connections, 53-66 sequence two-port networks, 55-60 three-component method, 53-66 two-component method, 3 53-55 2LO, generalized fault diagram, 281 2LO, three-component method, 64-66 2LO, two-component method, 3 53-54 two-port Thevenin equivalent, 55-60 unequal Z, three-component method, 61-63 severity, 49, 5 2, 1 84 shunt, 36-53 circuit diagram, 36-37 frequency of occurrence, 5 2 generalized fault diagram, 273-7 8 LL, three-component method, 4 2-44 LL, two-component method, 349-50 relative severity, 49, 5 2 sequence network connections, 37-53 SLG, arbitrary symmetry, two-component method, 357-58 SLG, generalized fault diagram, 27 5-77 SLG, three-component method, 37-41 SLG, two-component method, 347-49 3 , generalized fault diagram, 27 3-7 5 3 , three-component method, 49-52 3 , two-component method , 3 5 2-53 2LG, arbitrary symmetry, two-component method, 358-60 2LG, generalized fault diagram, 277-7 8 2LG , three-component method, 44-49 2LG, two-component m ethod, 3 50-5 1 simultaneous, 308-44 connection of sequence networks, 3 2336 constraint matrix, H-type faults, 340-41 constraint matrix, V-type faults, 339-40 constraint matrix, Z-type faults, 337-39 four cases of interest, 308, 3 24 matrix transformation, 336-4 1 parallel-parallel connection, 3 3 0-34 series-parallel connection, 334-36 series-series connection, 3 25-30 two-port combinations, 3 24 two-port network theory applied, 30836 transformations series, 304 shunt, summary, 300-301 shunt, with fault impedance, 298-304 shunt, without fault impedance, 294-98 transverse. See Fault, shunt Faulted network, 3, 3 1 , 36 Federal Power Commission (U.S.A. ), 243
I ndex
Flux linkages constant, concept, 1 8 3 , 1 8 4 , 1 86, 1 8 7 , 1 88 , 2 1 4 , 222 cylindrical wires external, 7 5-76 internal, 7 5 synchronous machines, 1 8 5-89 Fortescue, C. L., 4 , 19, 27, 7 1 Frame o f reference, for synchronous ma chine phasors arbitrary, 208-9 d-q , 209 General Electric Co., 7 1 , 255, 256 Generalized fault diagram solution, two-component method, 362-63 three-component method, 283 two-component method, 361 Geometric mean distance. See G MD GMD composite conductors, 99 self and mutual, 7 2 GMD method for transposed lines capacitance positive and negative sequence, 1 5 2-55 zero sequence, with ground wires, 1 5758 zero sequence, without ground wires, 1 56-57 impedance positive and negative sequence, 7 1-73, 98 zero sequence, with ground wires, 1 2933 zero sequence, without ground wires, 98-1 00 GMR. See GMD Gross, E. T. B., 1 04, 144, 1 7 8 Ground wires. See also Impedance, sequence current division with earth, 1 1 8- 1 9 effect o n sequence impedance, 1 1 4- 1 8 Grover, F . W . , 7 7 Guillemin, E . A., 308 Hancock, N. N., 284 Hesse, M . H., 104, 1 3 8 Hueslman, L. P . , 209, 3 0 8 , 3 0 9 , 3 1 9 Impedance effect of steel ground wires, 1 3 3-37 induction motor, table, 2 2 5 primitive, untransposed lines, 8 2-83 self, untransposed lines, 8 2-83 sequence completely transposed lines, 9 8 lines with bundled conductors, 1 06-1 2 lines with n ground wires, 1 28-29 lines with one ground wire, 1 1 2- 1 8 lines with two ground wires, 1 23-28 transposed line with ground wires, 1 1 923 transposed line with two ground wires, 1 26-28 untransposed lines with one ground wire, 1 1 4-1 8
509
untransposed lines with two ground wires, 1 25 transposed lines general equation, 98 positive and negative sequence, 7 1-73, 98-100 u ntransposed lines primitive equations, 81-83 sequence currents, 104 unbalance due to incomplete transposi tions, 1 0 2-6 zero sequence current basis, 1 3 3 effect o f steel ground wires, example, 1 35-37 GMD method for transposed lines, 991 00, 1 29-33 Impedance matrix building, 401 adding group of mutually coupled lines, 4 1 5-1 8 adding mutually coupled radial branch, 408-1 1 adding radial branch, 402-3 adding radial branch to reference, 401-2 closing loop not involving reference, 404-8 closing loop to reference, 4 03-4 example, positive sequence, 4 06-8 example, zero sequence, 4 1 2- 1 4 sorting o f primitive impedances, 401 node, 3 7 5 , 3 93-94 advantages for fault computation, 3 93 building algorithm, 401-18 branch currents for three-phase faults, 395 comparison with admittance method, 4 1 8-1 9 currents in mutually coupled branches, 399-400 node voltages for three-phase faults, 395 making changes in, 4 1 9 open circuit, 3 93-94 sequence quantities for SLG faults, 39798 shunt fault calculations, 393-400 SLG fault calculations, 396-4 00 three-phase fault calculations, 394-96 primitive Z, 367 sequence, unequal series Z , 27-30 table of special cases, 30 Incidence matrix. See Node incidence matrix Indefinite admittance matrix. See Admit tance matrix, indefinite Inductance leakage, synchronous m achine, 207 mutual, parallel cylindrical wires, 76 self, parallel cylindrical wires, 7 5-76 synchronous machine direct axis subtransient, 1 97-98 direct axis synchronous, 1 9 5 direct axis transient, 1 98 negative sequence, 200-201 phase domain, 1 86-87 quadrature axis subtransient, 1 99
510
I ndex
Inductance ( continued) quadrature axis synchronous, 1 9 5-97 quadrature axis transient, 1 99 table, 203 zero sequence, 201, 202 Inductance multiplying constant, k, 7 6-77 Induction motor, 222-28 equivalent circuit, 222-24 fault contribution, 2 26-27 general considerations, 222 operation with one phase open, 227-28 torque-speed characteristic, 2 24-25 Initial-plus-change notation, machine subtransient condition, 2 1 6 Internal sources in two-port networks, 5 7 Interrupting duty for circuit breakers, 2 2 1
Maxwell 's coefficients, 160, 1 8 1 capacitance calculations, 160, 163, 166, 169 Minor of a matrix element, 1 6 1 , 163 Momentary duty for circuit breakers, 222 Motors, induction, 222-28 Mutual coupling admittance matrix building, 3 8 2-85 equations, 74 nonreciprocal, between sequences, 30, 1 1 4 parallel wires, 73-75 between primitive matrix elements, 366 between 'lequences, 29, 98, 1 1 9 Mutual impedances passive series networks, 27-30 symmetrical component calculations, 28
Kimbark, E . W . , 7 1 , 1 83, 185, 1 88, 1 89, 1 96, 200, 203, 204, 224 Kirchhoff's law, 63 , 79 Krause, P. C., 208 Kron, G . , 2 28, 273, 284, 286, 288, 294, 3 04 , 308, 366 Kron reduction. See Matrix reduction Kron's constraint matrix, connection matrix, 2 86-88 Kron's primitive network, 288-89 Kron 's transformation matrix, 286-88 Kruempel, K. C., 4 1 8
Negative sequence, 24 Nelson, S. W . , 1 04 Network primitive Kron's, 288-89 sequence, 294, 299 two-port. See Two-port network Nilsson, J. W . , 5, 1 6 Node admittance matrix. See Admittance matrix, node Node impedance matrix. See Impedance matrix, node Node incidence matrix, 373-74 augmented, 369-7 3 Nonreciprocal coupling between sequences, 29-30 Nonsymmetric matrix of impedances, 30 Normalization of system quantities, 5-1 0. See also pu calculations n-phase system, 1 9-23
Lagerstrom, J . E., 5 LeCorbeiller, P., 284 Lenz's law mutually coupled wires, 74 synchronous machines, 214 transformer windings, 236 Lewis, W . A., 1 00, 206 Lewis, W. E . , 30, 284 Lewis, W. W . , 7 7 Line impedance, sequence, 7 1-1 5 1 Load representation, 1 2, 1 3 Lyon, W . V . , 1 5 8 Machine dynamics, 1 84 excitation, 222 impedances, symmetrical components, 3 0, 31 circulant matrix, 3 1 induction, 2 22-28 symmetrical components, 30, 3 1 synchronous, 1 83-222 Machines induction, 222-28 synchronous, 1 83-222 Matrix augmented node incidence, A, 3 7 0 constraint, Kron's, K, 286 constraint, table for shunt faults, 300-3 0 1 Kron 's connection. See Matrix, constraint Kron 's transformation. See Matrix, constraint primitive admittance, 'Y , 367 primitive impedance, Z , 288-90, 367 Matrix reduction, 108, 109, 1 14, 1 24, 1 7 0, 3 7 8 , 3 8 1 , 388, 4 1 1 , 4 14 , 4 1 6, 4 1 8
Open circuit Z parameters, two-port net works, 55-56 Optimally ordered triangular factorization, 4 1 8-19 Orthogonal transformation, 28, 1 89 Parallel lines effect of transposition, 1 37-43 optimizing wire arrangement, 1 4 3-45 Park, R. H., 1 89, 190, 200, 20 1 , 228, 229 Park's transformation, 1 89 inverse defined, 1 90 synchronous machines, 1 89-94 Per phase representation, 4, 1 9 Per unit. See pu Peterson coil, 270-72 Phasor, 1 6-16 diagram frames of reference in machines, 208- 1 0 synchronous machine, steady state, 21 113 synchronous machine, subtransient, 214-18 notation, 15 quantity, 15 transformation,
I ndex
Positive-plus-negative quantity. See Sigma Positive sequence, 24, 3 2 Potential coefficients, 1 5 9-60, 1 6 3 , 1 7 2 matrix for untransposed line, 1 6 3 Power coupling between sequences, 26 invariance matrix calculations, 284-86 symmetrical components, 26 transformation for, 26, 27, 285 from symmetrical components, 25-27 Prentice, B. R., 1 89, 1 96 Prime mover, 1 84 Primitive matrix, Kron 's, 366-68 Primitive network. See Network Proximity effect, in inductance calculations, 77-78 Pryce, D. G., 30, 284, 336 pu admittance on 1 0 0 MVA base for 1 . 0 mi, 1 0- 1 1 three-phase system, 9 calculations, 5-1 0 examples, 1 2- 1 5 conversion t o system values, 1 0 impedance, 6, 7 . on 1 00 MBA base for 1 . 0 mi, 1 0- 1 1 three-phase system, 9 quantity, 5 value, 5 Reactance, synchronous machine direct axis, 1 95 negative sequence, 1 99-201 quadrature axis, 1 95-97 subtransient, 1 97-99 transient, 1 97-99 zero sequence, 201-42 Reciprocity in sequence impedances, 29-30 Residual current, 1 7 7 Rosa, E. B . , 7 7 Rotation cycle, of transmission line, 8 7 - 9 0 linear transformation, 86 line conductors, 85-94 matrix R , clockwise rotation, 85
"J"
R counterclockwise rotation, 86 RO l 2 , sequence domain, 93 Self impedance of passive networks, 28 Sequence current distribution in sequence networks, 53 impedance. See Impedance, sequence machines, 3 0-3 1 passive series networks, 28 to flow of sequence currents, 3 2 network, 3 1 , 3 2 primitive, with fault impedance, 299 primitive, without fault impedance, 294 . SerIes fault. See Fault Series impedance unbalanced, 27 sequence components, 28 Short circuit. See Fault
51 1
Short circuit Y parameters, two-port net works, 56 Shunt fault. See Fault Sigma network, 347 quantities, 3 46-47 Similarity transformation capacitance calculations, 166, 167, 1 7 0 circulant matrix, 3 1 impedances, 2 8 , 3 1 Simultaneous faults. See Fault Skin effect, in inductance calculations, 7 7 , 78 Sources, internal , in two-port networks, 5 7 Spacing factor, 7 3 Sparsity methods of exploiting, 4 1 8 node admittance matrix, 3 86 primitive matrix, 368-69 Speed voltage, in synchronous machines, 1 9 1 , 192 Standards I.E.E.E., circuit breakers, 221-22 I.E.E.E., synchronous machine parameters, 1 86, 188 U.S.A., machine letter symbols, 1 86 U.S.A. , transformer terminal markings, 234-36 U.S.A., three-phase transformer terminal markings, 247-48 Steady state equations, 3 , 4 Steel conductors, 1 3 3-37 Stevenson, W . D . , Jr. , 72, 77, 7 8 , 1 00, 1 29, 1 52, 1 5 5 , 1 56 , 203 , 249, 250, 251 Stigant, S . , 284 Subtransient component of machine currents, 220 solution of synchronous machine equa tions, 214-19 Susceptance, capacitive, transmission lines, 1 5 2, 1 5 3 , 1 5 5 , 1 5 7 , 1 7 8 Symmetrical components analysis equation, 24 currents, 25 LL voltages, 25 n-phase system, 1 9-23 three-phase LN voltages, 23-25 Symmetry changes in, 273-307 two-component calculations, 3 5 5-62 creating by labeling, 273 generalized 1 LO, 282 series faults, 278-84 shunt faults, 273-78 SLG faults, 277 summary of network connections, 283 2LG faults, 278 2LO faults, 281 transformations. See Fault Synchronous component of machine currents, 2 1 9 machine analysis, 1 83-222 Synthesis equations, 24, 25 sequence quantities, 25
51 2
Tellegren's theorem, 3 7 2 Terminal pair. See Port Thevenin equivalent delta network, 347 sequence networks, 3 2, 3 24 sigma and delta networks, 346-47 sigma network, 347 synchronous machine, 1 84-8 5, 2 1 4 , 2 1 9 two-port, 5 5-60 impedance to compute system unbalance, 1 05 to flow of sequence currents, 3 2 theorem, 3 2 , 5 5 , 2 1 9 voltage at fault point, 53 positive sequence network, 3 2 prefault a t F, 3 2 system adjacent t o unbalance, 1 0 5 Thomas, C . H., 208 Three·component method, 36-66 Time constant, synchronous machine armature, 204 field, 204 T , 204
�
T�o , 202-4 T�o , 205 Td , 205
� � T�o , 205 T o , 205
T , 205
" Tq , 205 table, 203 Tinney, W. F., 4 1 8 Torque equation, induction motor, 224 Torque·speed characteristics, 225 Transformation sequence Kron 's technique, 292 series faults, 304 shunt faults, 294-304 shunt faults with impedance, 298-99 SLG, by Kron 's technique, 295-96 2LG, by Kron 's technique, 296-98 similarity, 28 Transformer, 23 1-65 auto, 23 9-43 equivalent circuit, 240-4 1 impedances, table, 243 three-winding, 24 1-4 2 banks, o f single-phase units, 243-47 three·phase � .y connection, 244-45 three-phase, single-phase units, 243-47 three-phase, three-winding autotransformers, 246 three-phase, zero sequence equivalents, 244-45 three-winding autotransformers, 246 core losses, 2 3 1 - 3 2 excitation, 2 3 1 -3 2 grounding, 255-57 y-� , 255-56 zigzag, 257
I ndex
interconnected star-delta, 257 load tap changing, 26 2-63 off-nominal turns ratio, 260-6 5 phase shifting complex turns ratio, 275 parameters derived, 3 56-57 pi equivalent, 26 2-65 single-phase. See Transformer, two winding three-phase, 247-60 core-type, 2 5 2-53 grounding, 255-57 shell-type, 253-54 terminal markings, 247 y-� phase shift, 247-51 y-y and � -� phase shift, 247 zero sequence equivalents, 25 5-56 zero sequence impedance, 251-55 zigzag·� , 257-60 three-winding, 236-39 equivalent circuit, 236-37 impedances, 238 normalization, 23 8-3 9 turns ratio, 231-32 off-nominal, 260-65 two-winding, 23 1-36 equivalent circuit, 23 2 impedance, table, 234, 2 3 5 normalization, 2 3 3 polarity, 234-36 single-phase, 23 1-3 2 terminal markings, 234-36 zigzag-� , 257-60 Transient component, machine currents, 21 9-20 Transmission line impedance. See Impedance Transposition, 84-98 rotation, 84-94 twisting, 95-98 Triangular factorization, 4 1 8- 1 9 Twisting o f line conductors, 85-9 5 Twist matrix phase domain, 95 sequence domain, 97-98 T!p , 97 T0 l 2 , 97-98 three definitions, 95 Two-component method fault calculations, 345-47 sequence current constraints, 3 5 6 sequence networks, 3 4 7 series faults, arbitrary symmetry, 360-61 SLG fault on any phase, 3 57-5 8 solution o f generalized diagram , 36 2-63 2LG fault on any phase, 3 58-60 Two-port network, 3 08-23 cascade connection, 323 hybrid connection, 3 2 1 - 2 2 independent sources, 5 7 , 3 1 6 internal sources, 5 7 , 3 1 3- 1 8 open circuit Z parameters, 55-56 parallel connection, 3 20-2 1 parameters, table, 309 relationship among parameters, 3 1 2 relationship among source parameters, 3 1 7 series connection, 3 1 9-20
I ndex
513
short circuit Y parameters, 56 simultaneous faults, 3 23-25 termination impedance, 3 1 0 test for finding parameters, 3 1 5- 1 6 uncoupled, i n positive sequence, 59-60 Two-port parameters, 55-56 Two-port Thevenin equivalent. See Thevenin, equivalent
Vandermonde matrix, 22 Voltage drop line with rotation cycle, 89 mutually coupled wires, 74 unequal phase impedances, 28 equation, primitive, 81 positive sequence, 32
Unbalance electromagnetic double circuit, circulating, 1 3 7-43 double circuit, through, 1 3 7-43 examples, 1 06 factor, 1 04 optimizing, in parallel circuits, 1 43-45 electrostatic factor, 1 7 9 ground displacement, 1 7 8-79 summary of factors affecting, 1 80 transmission lines, 1 7 7-81 ungrounded neutral, 1 7 9 factors, unequal series impedances, 1 04-5 series impedance, 27-30 Untransposed lines. See Impedance, un transposed line U.S.A. Standards. See Standards, U.S.A.
Wagner, C. F., 7 1 , 79, 80, 1 29, 134, 1 3 5 , 1 58 , 2 2 8 , 241 , 242 Ward, J. B . , 55 Westinghouse Electric Corp . , 52, 7 1 , 1 00, 20 1 , 234, 23 5 , 236, 23 7 , 249, 250, 2 5 2 , 253, 257 Woodruff, L. F . , 72, 1 00 Y-bus matrix. See Admittance matrix, node matrix . See Admittance matrix, node
Y
Z-bus matrix. See Impedance matrix, node O-d-q frame of reference, 1 9 1 Zero potential bus, 3 2 Zero sequence, o r zero-phase sequence, 1 9, 24 Z Matrix. See Impedance matrix, node
