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Analysis of Faulted Power Systems (IEEE Press Series on Power Engineering)
Analysis of Faulted Power Systems Editorial Board J. B. Anderson, Editor in Chief R. S. Blicq R. F. Hoyt S. Blancha
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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-
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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
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2. LOWERCASE
a
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- 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
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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) •
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. . . . . . . . . . . . . .. instantaneous current
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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
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j . k, k' k k .
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3 . UPPERCASE GREEK . .. delta connection; determinant (of a matrix) . . . . . . . . . . .. . . summation symbol n . . .. . . . . . . . . . . . ohm, unit symbol abbreviation for impedance
A �
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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 •
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5. SUBSCRIPTS a A b B B c
C d D
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max . . . min . .
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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) .
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List of Sy m bols
8
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sym . . . . . . T . . . . . . . u
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. . . . .. . Y .. . . . . . . 1 , 3
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 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) = 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>
=
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
=
(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 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
- !�----..
c
Ge,
-- - -
,.,
t --�--.p. -+'-'
v
-
r-.
'-'
+c T
'-'
IIPc
(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 0. 1
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