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.
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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
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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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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
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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 write
_ _
(0.0953 Zm = [(0.0953 (0.0953
+ + +
+
+
+
+
+
+
]
jO.9135) (0.0953 + jO.5697) (0.0953 + j0.4892) jO.5849) (0.0953 + jO.9135) (0.0953 + jO.5697) n/mi j0.4968) (0.0953 + jO.5849) (0.0953 + jO.9135) From these matrices the new matrices (zm - z,) and (z� - z.) can be easily com puted. Then we compute ZIt = (z. - Zm ) - (z� - zs) - 0.2340 - jO.8 952) (0.0000 jO.0005) (0.0000 jO.0001) = - (0.0000 (0.0000 + jO.0005) (-0.2340 - jO.8952) (0.0000 jO.0005) n/mi + jO.0001) (0.0000 + jO.0005) (- 0.2340 - jO.8952) Inverting, we compute + + +
�
+
+
+
]
�
- 0.2733 + j1 .0456) (- 0.0003 + jO .0005) (- 0 .000 1 + jO .000 1 ) (- 0.0003 + jO .0005 ) (- 0.2733 + j 1 .0456) (- 0 .0003 + jO.0005) (- 0.0001 + jO .0001 ) (- 0.0003 + jO.0005) (- 0 .2733 + j 1 .0456)
Z�I = -
To complete the matrix reduction we compute
ZN = (zm 0.0585 jO.2239) = �(0.0001 0.0000 - jO.0001) - jO.0001) - zs ) Zk l ( z� - zs )
+
(
-
Finally then, from (4.166)
mho · mi
�
(- 0.0000 - jO.0001) (0.0001 - jO.0001) (0.0585 + jO.2239) (- 0.0000 - jO.0001) n /mi (-0.0000 - jO.0001) (0.0585 + jO.2239)
�(0.0953 0.1538 + j1.1372) (0.0953 jO.5772) (0.0952 j0.4930)� jO.5772) (0.1538 j1.1372) (0.0953 jO.5772) n /mi +
=
J
+ +
+ +
+ + +
(0.0953 j0.4930) (0.0953 jO.5772) (0.1538 j1.1372)
1 12
Chapter 4
�
From this result we readily compute
Zo n
=
and
You
�
( 0.0675 - jO.4375 )
=
�
0.3444 + j 2.2354) (0 .0243 - jO.0140 ) (- 0 .0243 - jO.0140) (- 0 .0243 - jO.0140) (0 .0585 + jO .5881 ) (- 0 .0486 + jO.028 1 ) ( 0 .0243 - jO.0140) (0.0486 + jO.0280 ) (0.0585 + jO.588 1 )
�
(0.0224 - jO.0060 ) (- 0.0164 - jO.0164)
( 0.0164 - jO.0164) (0.1725 - j1 .6996) (- 0.1 548 + jO.0517 ) -
(0 .0224 - jO .0060)
�
n /mi
(0.1222 + jO.l082) (0.1725 - j1 .6996)
mho ' mi
Then the total impedance for the 40 miles of line is
ZOl l =
J
1 3 .775 + j89 .4 1 7 ) (0 .970 - jO .561 ) (- 0.971 - jO.560) (- 0.970 - jO.560) ( 2 . 3409 + j23.523) (- 1 .943 + j 1 .122) ( 0.970 - jO.561 ) ( 1 .943 + j 1 .1 21 ) ( 2 . 3409 + j23.523)
We may also compute the unbalance factors two ways.
rno
= �
m
Yl1
=
_
Zoo
'l =
Yl1
�
0 .0224 - 0 .0060 0.1725 - J 1 .6996
YOI ZO I = _ Y'l l =
=
0.01 36 / 6 9 .2°
0.0243 - jO.0140 0.3444 + j 2 .2354
�
n
= 0 . 012 4 �
0.122 2 + 0 . 1 082 0 .096 / 1 2 5 .7° 0 .1 7 25 - J 1 .6996 0 .0486 + jO.0280 0 .095 / 125 .6° 0.0585 + j O . 5881 =
=
It is interesting to compare these results with those of the previous examples. First, comparing against the of the untransposed line of Example 4.4, we see that the off-diagonal terms are almost exactly the same for the two lines even though the conductor size and spacing are greatly different for the two cases. From this we observe that the coupling between sequences is not critically de pendent on phase spacing, bundling, or wire size. The diagonal terms Zoo , Z l 1 and Z'll are reduced, both in the real and imaginary parts, by bundling. The unbalance factors may be compared against those of a similar untrans posed line in Example 4.7. Since the denominator terms in the unbalance equa tions, o 0 , , and Y I I are all reduced by bundling, we would expect the unbalance factors to increase.
ZOIl
Z Z'l 2
4. 1 0 Sequence I m pedance of lines with One G round Wire
In many physical transmission lines, wires are added above the phase wires to "shield" the line against direct lightning strokes. Determining the location of such wires is beyond the scope of our concern . We shall concentrate on the effect such ground wires have on the line impedances (also see [ 10 , 1 1 , and 24] ). Consider the line arrangement shown in Figure 4.19 where one ground wire, labeled w, is shown with each end connected solidly to the local ground point. Clearly, the voltage equation of this arrangement is exactly the same as that of Figure 4 . 1 5 given by equation ( 4 .149) except that in this case VoX = Vw = O. Thus
1 13
Sequence I m peda n ce of T ra ns m ission Li nes
�
� Zoo
a
I
b
. .1 :r: Va
Vb
d
+
l ob
i bb
.1£.
i cc
Iw
z ww
1;
Idd
Vc
a
�
,
I bc
l ad
d'
----·-11
ONE UNIT
Fig. 4 . 19 . A three-phase line with one ground wire.
we write the primitive equation
Voo ' Vb b , Vee ' Vw w ' Vdd ,
Vo - Yo' Vb - Vb ' Ve - Ve ' 0 - Vw , 0 - Vd ,
zoo Zbo Zeo zw o Zdo
zoe Zbe Zee zw e Zd e
Zob Zb b Ze b Zw b Zd b
Zow Zb w Zew zw w Zd w
Zod Zb d Zed zw d Zdd
10 Ib Ie Iw Id
V /unit length (4.168)
Now with ground wire w in parallel with the earth conductor d, the return cur rent will divide between the two paths or 10 + Ib + Ie = - (/d + Iw ) ' Rearranging, we write
Id = - (/0 + Ib + Ie + Iw )
(4 .169)
Using this result for Id in (4.168), we subtract the Vdd , equation from each of the others to compute
Vo Vb Ve
=0
- - - - -
Vw
Zoo Z ba Zea
Zab Zb b Ze b
Zae II Zaw Z b e II Z b w Zee II Zew
- - - - - - -- - - -
Zw b
zw a
Where as before ,
Zp q p,
zw e
-+ - - - I
I
zw w
la Ib Ie
V /unit length
Iw
= Zp q ZP d - Zdq + Zdd
( 4 . 1 70)
-
q = a, b, c, w
(4. 1 7 1 )
Note that Zp q is defined to include ra or r w when p = q but is purely imaginary when p "* q . From ( 4 . 45)-( 4.52) we compute
Zp q = ( ra + rd ) + j wk ln
= rd + j. W k In
De
pq D
'
�peq ,
p=q
p "* q il /unit length
(4.172)
Chapter 4
1 14
V
w
Since
r
=
0, (4.170) may be reduced immediately to the form
= (Z l
Vabe
l - Z2 z;j Z 3 ) labe
][] [
= Zabe la b e
(4.1 7 3)
where the Z partitions are defined in (4.1 58 ) . Performing the operation indicated in (4.173) , we have Zabe =
oo
Zab
Zoo
Zba
Zbb
Zbe
Zea
Zeb
Zee
a
=b C
( ( �
zo w
_
) ( ) ( ) �
a
Zew
b
Zaw Zw a Zw w
Zab -
zb a -
Zb w Zw a zw w
Zb b -
Zae
Zew Zw a zw w
Zeb
Zaa
_
_
�:J
Zbw
_
[zwa Zw b zw e ]
) ( ) � ) �
Zaw Zw b
Zae
_
Zb W ZW b Zw w
Zbe
_
Zew Zw b Zw w
Zee
_
Zw w
) ) )
c Zaw Zw e Zw w
Zb w Zw e Zw w
n /unit length
Zew Zw e Zw w
or each element of the reduced matrix is of the form
p , q (row, col) = a , b , c
(4.174)
(4.175)
Each element of the matrix is smaller by a correction factor involving the mutual impedances to the ground wire w . In most lines of interest we may assume that the three phase wires have equal self impedances or Zaa = Z b b = Zee . This makes the leading terms on the diagonals equal, but the subtracted corrections are still unequal since the mutuals from w to a , b , an d c are usually unequal. 4. 1 0. 1
Sequence impedance of an untransposed line with one ground wire
To compute the sequence impedances, we again use equations ( 2.46)-(2.48). Since the matrix (4.174) is symmetric but with unequal mutual terms, the sequences will, from (2.46), be coupled by nonreciprocal terms. The compu tation is straightforward, however, and involves only the application of (2.46 ) (2.48 ) or ZOl 2 = A- I Za be A. This result is too laborious to write out in detail. An example will illustrate the procedure.
Example
4. 9
Compute the sequence impedance of the 40-mile line of Example 4 . 1 , taking into account the presence of the ground wire. Assume that the ground wire has a resistance of 4 ohm/mile and a self GMD of feet. There are no transpositions.
10""3
Solution We are given that rw
= 4.0
U /mi,
Dsw
= Dw w
==
10-3 ft
115
Sequence I m pedance of T ra n s m ission Li nes
Then the self impedance term is
Zww rw rd =
+
= 4 .095
+
+
DDwwe =
j wk In
4.0
+
27 90 . 13 ) In 0 .09528 + J(0.12 0 .001
j 1 .800 n /mi
= 16 3 . 80 j 7 2 .00 n . and for 40 miles of line To compute the mutuals , we need the distances
+ Zww DDbaw De w = w Zew = rd �aew
2 2 V 1 0 + 1 5 = 1 8.03 ft
=
= 1 5 .0 ft
Then Za w =
= 0.09528 + j(0 . 1 2 1 3 ) In
+ j wk In
= 0 .09528
+
jO.612
= 0 .09528
+
j ( 0 . 1 2 1 3 ) In
n /mi
and
Zb w
:���
���� = 0 .09528 + jO.634
n / mi
The total mutual impedances for the 40 mile line are
Z w Zew = Zbw =
a
�
3.81 2 + j 24.470 n
= 3.812 + j 25 . 3 6 3
n
In matrix form
Zobcw =
1 4'932 + j58.37 6) ( 3.81 2 + i27. 332) �3.:f!.1� � !�.:�6�l ( 3.81 2 + j 24.470)
J
( 3.81 2 + j27.332) (3.812 + j23.968) : ( 3.812 + j24 . 470) ( 1 4.932 + j58. 376) (3.812 + j27.332) 1 ( 3.812 + j25.363) (��.!:� ���.��21 _ J!.����:A�!.�!�U (�.!��j!��7�L_ ( 3.81 2 + j25.363) (3.812 + j24.470) 1< 163.81 2 + j72.036U
_ _
�
Reducing along indicated partitioning, we compute A
Za be
=
++ +
�
++
Then by digital computer we find the sequence impedances
Z0 12 =
J
1 7 .500 j 56 . 1 07 ) (6.484 + j 24 .996) (6 .380 + j21 .698 ) (6 .484 j 24.996 ) ( 1 7 . 7 1 2 j55 .972 ) (6 .484 + j24.996 ) (6 .380 j 2 1 .698) (6.484 j 24 .996) ( 1 7 . 500 + j56.107 )
( 30.470 + j l0 3 . 8 5 ) (0 .860 - jO.618 ) (- 0.966 - jO.436 ) 0 .966 j0 .436 ) ( 1 1 .1 2 1 + j 32 .1 6 5 ) (- 1 .944 + j 1 .121 ) ((0.860 - jO .619 ) ( 1 .943 j 1 . 1 23 ) ( 1 1 .121 + j32.165 )
+
The unbalance factor may be easily computed as Z l _ mo = - O = Z oo
m2
= - Z21 = Z 22
_
0.860 - � 0.6 18 30.470 + J 103.85
�
1 .943 + 1 . 123 1 1 . 12 1 + J 3 2 . 165
= 0 . 009 78 17 0 . 650
= 0.06 5 9 1 1 3 9 . 1 10
J
n
n
n
1 16
Chapter 4
Comparing this result with that of Example 4.4, we observe the following:
1 . The positive and negative self impedances Z 1 1 and Z22 are unchanged by the addition of the ground wire. 2. The zero sequence impedance is changed in a remarkable way. viz. , (a) the real part R oo is increased by about 40% and (b) the imaginary part Xoo has de creased by about 7%. 3 . The off-diagonal terms are changed but not by a significant amount. 4. The zero sequence unbalance has been increased by the addition of the ground wire but the negative sequence unbalance is unchanged.
Example
4. 1 0
Repeat the computations of Z 0 1 2 made in Example 4.9 but with a ground wire of higher conductivity. Compare the results with those of Example 4.9.
Solution For this example let us choose two different ground wires as follows : 1 . 3 /8-inch copperweld, 40% conductivity
2.
D. rw
=
Zw w
=
=
0.00497 ft 1 .264 51 /mi 1 .359
+
jO.12134 In
0
1F copperweld-copper
DB rw Zw w
=
����7 = 1 .359 + j 1 .606
51 /mi
0.00980 ft
= 0.705
51 /mi
= 0.8003
+
jO. 1 2 1 34 In
O
.����O
=
0.8003 + j 1 . 524 51 /mi
Then, by digital computer we compute the following sequence impedances for the 40 miles of line. For the line with a 3/8-inch copperweld ground wire the impedance for 40 miles of line is given by Zo n
=
�
J
(- 0.844 - jO.327 ) 31 .207 + j90 . 1 48 ) (0.706 - jO.568 ) ( 0 .844 - jO.327) ( 1 1 . 122 j32.163 ) (- 1 .946 + j 1 .121 ) (0.706 - jO .568 ) ( 1 .944 + j 1 . 1 2 5 ) ( 11 . 122 j32.163 )
+
-
+
51
For the line with a 1F copperweld-copper ground wire the impedance for 40 miles is given by Z0 1 2
=
�
+
J
27 .403 j83.866 ) (0.652 - j0.499 ) (- 0.758 - jO.316) ( 1 1 . 1 2 2 + j32.162) (- 1 .947 j 1 .122) ( 0.758 - jO.316) ( 1 1 . 122 j32.162) ( 1 . 945 + j 1 . 1 2 5 ) (0.652 - j0.499) -
++
51
It is apparent that the zero sequence reactance reduces appreciably as the ground wire impedance goes down. The zero sequence resistance remains high compared to the case with no ground wire. It is also interesting to compare the unbalance factor for the three cases.
Sequence I m peda nce
of T ransm ission
Lines
1 17
Table 4.7. Comparison of Zero Sequence Unbalance, m o , in Percent, as a Function of Ground Wire Selection IF Cw-Cu No Ground Wire 3/B-inch Cw 3/B- inch Steel 1. 054 0.978 0.950 0.931 I mo l 72.680 angle 70.650 70.67 0 7 0.280 Obviously the negative sequence unbalance is unchanged (Why? ) . sequence unbalance is tabulated for comparison in Table 4.7.
The zero
Example 4. 1 1 Repeat the computation made · in Example 4.9 except consider the configura tion of Figure 4 .20 where the ground wire is located at the same height as before but 5 feet to the right of center.
0.375"
GROUND WI R E STEEL
o
4/0
0
PHASE WI RES COPPER
°1-10· -+,5' Fig.
4.20.
Line configuration for Example 4. 1 1 .
Solution
W e proceed exactly as i n Example 4 . 1 0 , the only difference being in the dis w to a , b , and c. This makes the reactance terms different than before by a very small amount in the fourth row and column. Thus we compute
�
tance from
Zabcw
=
14'931 + j58.37 5)
(3.812 (3.81 2
+
+
j27 .331 ) j23 .966)
( 3.812 + j27 .331 )
( 1 4 .931 + j58 .375) ( 3. 8 12 + j27.33 1 )
(3 . 812 + j23.966) 1 (3.812 + j 2 3 . 680 ) ( 3.812 + j27 .331 ) : ( 3.812 + j25.107) ( 14.931 + j58.375) 1 ( 3 8 1 2 + j25.107)
:r
.
(i.8;i;j2i.680) - - (3�12�j25�07) - - (3.812;j25�107) �63�i2�j72.036)
�
Reducing to find
Zabc =
Zabc ,
we compute
J
1 7 _319 + j56.223) (6.360 + j 2 5 .075) (6 .247 + j 2 1 .783) ( 6 . 360 + j25.07 5 ) ( 1 7 _651 + j56.0l l ) ( 6 .41 1 + j 2 5 .043) ( 6.247 + j 2 1 .783) (6.41 1 + j 2 5 .043) ( 1 7 .651 + j 56.01 1 )
and converting to the 0-1-2 frame of reference we have
Z012
]
K 30 .374 + j 10 3 .92) = \ (- 1 .1 39 - j0.454) � 0.804 - jO .454)
( 0.804 - j O .454) (- 1 .1 39 - jO.454) ( 1 1 . 1 23 + j32 . 1 6 5 ) (- 1 .939 + j1 .120 ) ( 1 .946 + j 1 .120) ( 1 1 .123 + j32.164 )
J
The unbalance factors for this case are
mo = -
Z OI
Zoo
=_
0 .804 - jO.454 30.374 + j103.92
0 .00853 176.840
n
n
n
1 18
Chapter
m2 = -
ZZ l ZZ 2
==
-
4
1 .946 + j 1 . 1 20 1 1 . 123 + j32 . 164
= 0 . 0 6 5 9 113 8.9 9
0
Comparing results, we note that all terms of the Zob c matrix have been changed by moving the ground wire, with the row and column for wire a seeing the greatest change. In the Z0 12 matrix the zero sequence row and column is af fected. The zero sequence unbalance is reduced slightly, which may come as a surprise. 4. 1 0.2 Current division between the ground wire and the earth
It is apparent that the addition of a ground wire to an untransposed line section reduces the zero sequence reactance Xoo and increases the zero sequence resistance roo . This is due in part to the relatively high impedance of the ground wire compared to the earth and in part to the fact that the ground wire is phys ically nearer the phase wires than the ground return path. Because the ground wire is near the phase wires, the current induced in this wire tends to be large and if the wire has a large resistance, this makes roo appear to be great. The closeness of this coupling, however, reduces the inductance about in proportion to the current lw • The proportion of the total zero sequence current which flows in the ground wire may be determined from equation (4.168). If3/0we0 set 10 = lb = lc = /oo
(4.176)
0 = [ (zw o + Zw b + zw c ) - (Zda + Zd b + Zdc ) ] la o + (zw w - Zd w ) lw + (Zw d - Zdd ) /d
(4. 1 7 7 )
in (4.168) and subtract the d equation from the w equation Also,
from (4.169) and (4.175)
Id = -
(3/0 0
+ Iw )
If we substitute (4.1 78) into ( 4 1 77), we compute
(4.178)
.
_
lw
310 0
__
) - _-=... + Zd 3Zw d:o. - (Zdo +--, (zw o + zw b + zw, c ) _.!-"'-,,,- Zdb ,,-, > often by about a factor of ten or so in each case . This is not obvious from an inspection of ( 4.274 ) because it is hard to appreciate the vast difference in phase angle of the admittances.
m2c
Example
4. 1 4
Compute the through and circulating current unbalances for the double cir cuit 345 kV untransposed line shown in Figure 4. 2 8 .
1 41
Sequence I m peda n ce of T ransm ission lines 10" 10'
UrjiW at + Ig' ;tC
;.. on
19 '
N
,
Ground Wi,": 79' kCM, 2617 ACSR G M R · 0. 037' ft . ro 0. 1 1 7 A / m i
,
•
b'
Pho. . Wi", :
I . 7' in Expanded ACSR G M R . 0 . 06 4 0 " ro .
0.0663 A Imi
Fig. 4 . 2 8 . A 34 5 kV double circuit configuration.
Solution First we must determine the impedance matrix of self and mutual impedances defined by equation ( 4. 79). These 64 impedances are given in Table 4.11, and the 36 impedances which result after reducing this matrix to eliminate rows u and w are given in Table 4.12. The inverse of the matrix given in Table 4.12 is computed by digital computer with the result Yoo = Yoo ' =
Yo' 0 ' =
[ r-
+ + ( - 0. 001 1 + jO.0894) (- 0. 0000 + jO.0827) (- 0.0039 + jO. 1099)
++ � (- 0.0087 + jO.1416 � (- 0.0011 + jO.0894) (- 0.0011 + jO.0877) (- 0.0056 + jO.1 195)� (- 0.0214 + jO.2131)
( 0. 1057 - j 1 . 1 573) (- 0.0214 jO.2131) (- 0.0056 jO.1 195 (- 0.0214 jO. 2131) (0.0983 - j1.1178) (- 0.0245 jO.2348) (- 0.0056 + jO. 1 195) (- 0.0245 jO.2348) (0.0954 - j1. 1006)
+ 0.001 1 + jO.0877 ) ( 0.0039 + jO. 1 099) -
(- 0.01 79 + jO. 1950)
[
+
(0.0954 - j1.1006 ) (- 0.0245 jO. 2348) (- 0.0245 + jO. 2348) (0.0983 - j1.1178) (- 0.0056 jO. 1195) (- 0.0214 + jO. 2131) (0. 1057 - j1.1 573)
+
Each partition above is now transformed to the ified by (4.268), with the following results A- I Yoo A =
[ [([([
0-1-2 frame of reference as spec
+
+ � (0.0329 + jO.0010)� (0.0417 - jO.0161)
+
+
( 0.0655 - jO. 7469) (0.0287 - jO.0351) (- 0.01 54 - jO.0426 (- 0. 01 55 - jO.0421) (0. 1 1 70 - j1.3156) (- 0.0634 jO.0200) ( 0.0286 - jO.0346) (0.0546 + jO.0400) (0.1171 - j1.3156)
0.01 29 + jO.33 1 1 ) (- 0.0309 jO.0113) (0.0018 - jO.0376) (- 0. 01 56 jO.0288) (- 0.0348 - jO.0281 ) (0.0018 - jO.0355)
A- I Yo' o A
= (0.0057 jO.0323)
A- I Yoo, A
= (0.0330 jO.0013)
A- I Yo' o, A =
++
� j
0.01 79 + jO.3311) (- 0.0156 jO.0290) (0.0056 jO.0325) (0.0089 - jO.0355) (0.0417 - jO.0162) (- 0. 0310 - jO.01 16) (- 0.0348 - jO.0281 ) (0.0018 - jO.0376)
+
(0.0655 - jO. 7469) (0.0446 + jO.0079) (- 0.0448 - jO.0072 ( - 0.0449 - jO.0067 ) (0. 1 1 70 - j1.3156 ) (- 0.0617 0.0278) (0.0446 jO.0085) (0.0487 jO.0455) (0.1171 - j1.3156)
+
+
+
Table 4. 11. Impedance Tabulation for the Circuit of Figure 4. 27 (8 X 8 matrix of phase wire and ground wire impedances) tJ
b
C
tJ
,
b'
C
,
U
W
0.16158 +j1 .296235
0. 09528 +jO. 5 5 3453
0.09528 +jO.488002
0. 09528 +jO. 460342
0. 09528 +jO.468743
0. 09528 +jO. 521302
0. 095 2 8 +jO. 5617 39
0. 09528 +jO. 5 1 89 34
0. 09528 +jO. 553453
0. 16158 +j 1. 296235
0. 09528 +jO. 553453
0. 09528 +jO. 4687 4 3
0. 09528 +jO. 45 0692
0 . 09528 +jO. 468743
0. 09528 +jO.474048
0. 09528 +jO. 452254
0. 09528 +j0. 488002
0. 095 28 +jO. 55 3453
0. 16158 +j 1. 296235
0. 09528 +jO. 521302
0. 09528 +j O. 4687 4 3
0. 095 28 +j O. 460342
0. 095 2 8 +j O.436824
0. 09528 +jO. 429368
,
0. 09528 +j0. 460346
0. 09528 +jO. 468743
0. 09528 +jO. 521302
0. 16158 +j 1. 296235
0. 09528 +jO.553453
0. 09528 +j O.488002
0. 09528 +jO.429368
0. 09528 +jO. 436824
b'
0. 09528 +jO.468743
0. 09528 +jO.4 50692
0. 09528 +jO.468743
0. 09528 +jO. 5 53453
0. 16158 +j 1.2962 35
0. 09528 +j O. 5 53453
0.09528 +jO. 452254
0. 09528 +j O. 474048
,
0. 09 5 28 +j0. 5 2 1 302
0. 09528 +jO. 468743
0. 09528 +jO.460342
0. 09528 +jO. 488002
0. 09528 +jO. 553453
0. 16158 +j 1. 296235
0. 09528 +jO. 5 1 8934
0. 09528 +jO. 56 17 39
0. 09528 +jO. 5617 39
0. 09528 +jO. 474048
0. 09'5 28 +jO. 436824
0. 09528 +jO. 429368
0. 09528 +j O. 452254
0. 09528 +j O. 5 1 89 34
0. 21228 +j 1. 361096
0.09528 +jO. 599185
0. 09528 +jO. 518934
0. 09528 +jO. 452254
0. 09528 +jO. 429368
0. 09528 +jO. 436824
0. 09528 +jO. 474048
0. 09528 +jO. 5 6 17 39
0. 09528 +jO. 599185
0. 21 228 +j 1. 361096
a
b c
tJ
c
u w
1 43
Sequence I m peda n ce of T ra n s m ission Li nes
Table 4.12. Impedance Tabulation for the Circuit of Figure 4.27 (6 X 6 matrix of phase impedances after matrix reduction) a
b
c a
,
b' c
,
b
c
a
0.1034 +jO.9973 0.0379 +jO.2978 0.0381 +jO.2493 0.0380 +jO.2220 0.0377 +jO.2142 0.0368 +jO.2247
0.0379 +jO.2978 0.1058 +j1.0775 0.0402 +jO.3494 0.0402 +jO.2648 0.0395 +jO.2327 0.0377 +jO.2142
0.0381 +jO.2493 0.0402 +jO.3494 0.1073 +j1.1055 0.0410 +jO.3307 0.0402 +jO.2648 0.0380 +jO.2220
0.0380 +jO.2220 0.0402 +jO.2648 0.0410 +jO.3307 0.1073 +j1.1055 0.0402 +jO.3494 0.0381 +jO.2493
Then fro m laO la l
la2
la 'o
la ' i
la '2
,
a
b
0.0377 +jO.2142 0.0395 +jO.2327 0.0402 +jO.2648 0.0402 +jO.3494 0.1058 +j1.0775 0.0378 +jO.2978
c
,
0.0368 +jO.2247 0.0377 +jO.2142 0.0380 +jO.2220 0.0381 +jO.2493 0.0378 +jO.2978 0. 1034 +jO.9973
(4.27 1 ) we combine these results to write
(0.0286 - jO.0351 ) + (- 0 .0349 + jO.0133) (0.1169 - j1 .3144) + (0.0059 - jO.0386) (0.0545 + jO.0396) + (- 0 .0541 - j O.0437) V (- 0 .0 1 1 6 + jO.0044) + (0.0446 + jO.0079) l: a l (0.0047 - jO.0345) + (0.1 169 - j 1 .3144 ) (- 0.0154 - jO.0 1 2 2 ) + (0.0487 + jO.0451 )
Then, from
'
=
- 0.0131 - jO.0061 0 . 1 2 59 - j1.3511 0.01 97 + jO.01 1 9 0 .0 1 37 - jO.00 34 0 . 1 1 88 - j 1 . 3532 0 .0 1 39 + jO.01 7 5
l: Va l
, (4.272) we compute the "through unbalances. ;
mO t m2 t
Similarly , from
=
=
la o + 10'0 10 1 + 10'1
102 + 10'2
=
=
la l + la ' i
0.0268 - � 0.0095 0 .2447 - J2.7043 0.0336 + � 0.0294 0.2447 - ]2.7043
=
0 .0104 { 65.2 0
=
0.0163 / 1 26.0 0
(4 .273) we find the "circulatory unbalances. "
mOe m2e
=
la o - 10'0 10 1 + 10' 1
=
102 - 10'2 10 1 + 10'1
=
=
0 .0006 - �0 .0027 = 0 .00102 /- 17.7 0 0.2447 - J2.7043 0.0058 - � 0.0209 = 0.00296 / 40.8° + 0.2447 - J2.7043
_
For the vertical configuration of this problem the negative sequence unbalances about equal but not much greater than
are
m Oe '
4. 1 6 Optimizing a Parallel Circuit for Minimum Unbalance
In the previous section each parallel circuit line was treated in terms of its total impedance characteristic from one terminal to the other. Obviously, it would be possible to write separate equations for each line for each section of like configuration and then apply the transposition matrices Rtil and Ttil to each circuit to reorder the equations in an a-b-c-a ' -b' -c' sequence. Once each section has been
Cha pter 4
1 44
so arranged, the matrices may be added to find the total impedance of each in dividual line. In many practical situations the lines are not transposed at all. This is because the transpositions are costly and, perhaps even more important, they are often the cause of circuit failure due to either mechanical or electrical weakness at the trans position structures [ 34, 351 . The general problem of the effect of circuit un balance due to lack of transpositions has been studied in detail by E. T. B . Gross [ 28, 29, 30, 34, 35, 36] and is an excellent resource for the interested reader. In double circuit lines which are not transposed, Hesse [331 has shown that there is an optimum conductor arrangement which will minimize the unbalance factors. There is value, therefore, in developing a method whereby the unbalance factors can be determined in a straightforward way, while making sure that all possible configurations are examined. We shall do this for lines which are them selves untransposed but which can have any desired phase identification. From equation (4.268) we have
=
][ ]
[Yoo
V012 VO' I ' 2 '
Yoo' Yo ' o'
Yo' o
phased a-b-c (4 . 275)
phased a'-b'-c'
where we have defined new admittance submatrices Y00 , Y 00' , Yo ' 0 and Yo ' 0' as a matter of convenience. We have also indicated the phase designation quite ar bitrarily as a-b-c and a' -b' -c' for the configuration used in writing these equations. Suppose, however, that the second (primed) circuit had been phased b' -c'-a' ( on wires 1-2·3). This means that the voltage (or current) equation must be pre multiplied by R41 to rearrange the elements to the desired a-b-c order. However, since equation ( 4. 27 5) is already in the 0-1-2 frame of reference, we may use the rotation matrix R 0 12 defined by equation (4.90). Similarly, multiplication by Rol2 defines a rotation in the opposite direction. These rotations of only the primed circuit may be performed mathematically by transformations
l�
and
[�
o
R0 12 R0- I
o
]
]
a-b-c (4.276)
b'·c'·a '
a-b-c
12 c -a -b I
"
] r V0 12 ]
(4.277)
For example, premultiplying both sides of (4.275) by (4.276) gives
[10 12
] = [Yoo R
R��i;:-;';
Yoo, Rol2 C o;;Yo� �Ro�Yo';Ro I2
1 I
L-Ro�V�:�;
a-b-c b' -c' -a '
(4.278)
This operation is equivalent to multiplying every element in the fifth row and the sixth column of (4.270) by a and multiplying elements of the sixth row and fifth column by a2 • This result, with only positive sequence voltages applied, is
1 45
Sequence I m pedance of T ransmission Unes
100 10 1 10 2
=
10' 0 10' 1 10 ' 2
YO I + al YO I' Y l l + al Y l l, Yl l + a2 Yl I,
phased a-b-e � Va l
- - - -- - - - - - -
YO' I + a2 YO' I' a ( Y I' 1 + a2 Y I ' I ' ) phased b ' -e' -a' a2 ( Y2 ' 1 + a2 Y1' 1' )
(4.279)
Another rotation of the second circuit would give a similar result, but with oper ators a and a1 interchanged. If instead the second circuit is transposed according to transformation To 1 2 , the result is equivalent to interchanging the fifth and sixth rows of (4.270) and then interchanging the fifth and sixth columns. The result is
100 10 1 10 2 10 ' 0 10 ' 1 10' 2
YO I + Y0 2' Yl l + Y 1 2, =
phased a-b-e
Y2 1 + Y2 1,
� Va l
-------
YO' I + YO' l' Y2 ' 1 + Y1' 2 ' ' Y I' I + Y I' 2 '
phased a' -e ' -b'
(4.280)
If this arrangement is then changed by a rotation of the second circuit to ( 4.280) are multiplied by a and a2 exactly as in (4. 279). In all, six arrangements are possible, with the first circuit remaining un changed. These are
a-b-e : b'-a'-e' , the Y elements of
II �
3l
u
2 ROl2
a-b-e
a-b-e
a- b-e
a-b-e
a'-e'-b'
a' - b ' -e' b '-e'-a'
a-b-e
e -a - b "
,
6
[U
b ' -a' -e'
R0 l 2 To 1 2 '2
a-b-e '
U ,
e -b -a
,
(4. 281 )
for the first circuit re and another six relationships may be obtained with the placed by T O I 2 ' These are the only unique arrangements-a fact that may require a little careful thought. This is due to the fact that which circuit is unbalanced with respect to its neighbor is immaterial. Thus 12 arrangements in all must be examined. In general, it is possible to have 6" /3 significantly different phase ar rangements for an n-circuit system. Problems
Some of the problems suggested here should be solved by digital computer to provide high accuracy, to minimize labor, and to correctly display the desired impedance characteristics. Some helpful programs are available in Appendix A. 4.1. Compute the per mile positive and negative sequence impedance for the line configura tion shown in Figure P4. 1 where the conductor is 410, 7-strand copper. Assume that the
1 46
Chapter 4 a
Fig. P4. 1 .
line i s transposed such that the methods o f section 4 . 1 ar& applicable. Compute Xl by equation (4. 3) and use equation (4. 10) as a check. 4.2. Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.2 where the conductor is 3/0 ACSR. Use the method of section 4.1. b
Fig. P4.2.
4.3. Compute the per mile positive and negative sequence impedance for the line confIgura tion of Figure P4.3 where the conductor is 500,000 CM, 61% conductivity, 37-strand, hard-drawn aluminum.
45'
Fig. P4. 3 .
4.4. Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.4 where the conductor is 336,400 CM, 26/7-strand ACSR. --,r--__ 9
Fig. P4.4.
Sequence I m pedance of T ra nsm ission Unes
4.5.
1 47
Compute the per mile positive and negative sequence impedance for the line configura tion of Figure P4.5 where the conductor is 397,500 OM , 26/7-strand ACSR. Q
Q
Fig. P4.5.
4.6.
Write the phasor equations for a two-winding transformer with self inductances Lp and L. in primary and secondary windings respectively and mutual inductance M. Compare
these equations with equation (4.14) for two mutually coupled, parallel wires. Review the concept of placing polarity markings on the transformer coils. 4.7. Verify that the four-terminal equivalent circuits shown in Figure P4. 7 a and Figure 4.7b accurately represent the equations for two mutually coupled wires given by equation (4. 14). Circuit b is due to Starr [ 37 ] . a
b'
,
r -
Vaa '
1
•
Zoo -l o b a
t 10
•
l ob
(a)
t
1
'
loa
b' - Z ob
V b b'
l bb -Za b lb
a
b
1
a
(b)
b
Fig. P4.7 .
Find the internal inductance in millihenrys/mile of 106 circular mil, all aluminum (61% conductivity) conductor when the operating frequency is 60 Hz. 4.9. Find the internal inductance in millihenrys/mile of a certain iron wire having a dc resis tance of 2.0 ohms/mile and a relative permeability of 100. The frequency is 60 Hz and the wire is cylindrical with a diameter of 0.394 inches (4 / 0 SWG pure iron wire). 4.10. Verify the impedance for Carson's line, given by equations (4.35)-(4.40). 4. 11. Compute the voltage drop Vm n across the circuit of Figure P4. 11 and compare with equation (4.35). Neglect the resistance of the windings.
4.8.
n
m
Fig. N . l l .
Chapter 4
1 48
4. 12. Consider a single conductor plus earth return circuit. The conductor is size 4/0, 7-strand hard-drawn copper at 50 C suspended 30 ft above the earth; f = 60 Hz. (a) Find the self impedance of the circuit, roo + jxoo ' in ohms/mile, for ( 1 ) P .. 10 n . m and (2) P = 1000 n . m (b) What is the percent change in Xoo and in Zoo for this 100-fold increase in earth resistivity? 4 . 1 3. Compute the impedance matrix for two wires a and b with earth return d and sketch an equivalent circuit for this circuit similar to Figure P4. 7(a) where the effect of the earth is included in the circuit parameters. 4. 14. Verify equation (4.47) for a three-phase line with earth return and compute all self and mutual impedances where the assumptions made in equation (4.53) are not true. 4. 1 5. Illustrated in Figure P4. 1 5 is a 4-wire circuit which is used to serve a three-phase, 4-wire, ' ' ' Y-connected load. The line consists of three phase conductors a-a , b-b , and c-c sepa rated by distances Dab , Dbe , and D ca as shown. Also shown is the neutral conductor non ' separated from the phase conductors by distances Do" , Db" , and Dc" . The neutral con ductor is not connected to the earth, and its Impedance does not involve the earth in any way. One end of the line is short circuited in order to facilitate finding the line impedances. Find zo, the impedance of this line to the flow of zero sequence currents. Assume that the four conductors are identical with resistance r ohms per unit length and a GMR of D• . a
b c
n
zbb
Dab
Icc
Dbc
J
iob z bc
Den Icn �
In ..
REFERENCE FOR ALL VOLTAGES
�\ \
D
J
0' b'
ico
�
Don
J
zan
,
Dbn
,
I
ibn
c' n'
Fig. P4 . 1 5 .
4. 16. The line configuration shown i n Figure P4. 16 i s similar to that used in certain overhead circuits in the 5-1 5 kV class for primary distribution. Phases a, b, and c are insulated but not shielded cables. The neutral conductor n is a grounded bare messenger which sup ports all wires mechanically and also carries any unbalanced neutral current (3100 ), The frequency is 60 Hz.
a
o------o b
Fig. P4 . 16.
Sequence I m peda nce of T ransmission Li nes
1 49
Phase conductors: 3/0 hard-drawn aluminum, 19 strand D B 0.0 1483 ft 0.178 inches r " 0.558 fl/mi @ 25 C -
..
Neutral conductor: I/O ACSR, 6/ 1 strand 0.00446 ft 0.0535 inches 0.888 fl/mi @ 25 C
D. r
-
-
..
(a) Ignore the ground wire and determine the phase impedance matrix Zabc for this circuit. (b) Assume balanced applied voltages and unbalanced loading as follows: Va ... 8000l!L line-ta-neutral Vb - al Va
la
'"' 200/- 30 A
Ib - 200/- 150 A
Ve '"' a Va
Let Ie be variable with the following values: lIe I
Case 1 2 3 4
100 100 100 100 100
5
A ngle of Ie 60° 9 00 1200 1500 1800
Case 6 7
8 9
10
lIe I
200 200 200 200 200
A nllle of Ie 600 900 1200 1500 1800
Find the line-ta-neutral voltages two miles from the sending end of the line for cases
1- 10.
(c) Tabulate the results of part (b) and also the power factor of phase c and the percent voltage drop in each phase. (d) Plot the percent voltage drop in each phase versus the phase c power factor: (1) for Ile l - 100 A and (2) for lIe I 200 A. Identify the one point which represents a balanced load. Compute the phase impedance matrix Zabe for the line described in problem 4. 1 . As sume that the line is 25 miles long and is not transposed. Compute the phase impedance matrix Zabc for the line described in problem 4.2. As sume that the line is 35 miles long and is not transposed. Compute the phase impedance matrix Zabc for the line described in problem 4.3. As sume that the line is 50 miles long and is not transposed. Ignore the ground wire. Compute the phase impedance matrix Zabe for the line described in problem 4.4. As sume that the line is 50 miles long and is not transposed. Ignore the ground wire. Compute the phase impedance matrix Zabe for the line described in problem 4.5. As sume that the line is 60 miles long and is not transposed. Ignore the ground wires. Compute the total impedance matrix Zabe for the lines of problems 4. 17-4.2 1 with the following transposition arrangements. '"'
4.17.
4.18. 4. 19. 4. 20. 4.21. 4.22.
Fraction Configu ration fl 0.20 a-b-c fl 0.80 b-c-a f3 0.00 c-a-b (b) fl - 0.25 a-b-c fl 0 . 35 b-c-a b-a-c f3 - 0.40 (c) fl 0. 30 a-b-c c-a-b fl 0.60 f3 0. 10 c-b-a (a)
..
..
'"
..
'"'
.. '"
1 50
Cha pter 4
(d) fl - 0.33
f2 = 0. 33 f3 - 0.33
a-b-c c-a-b b-c-a
4.23. Verify the results given in Table 4.3 by perfonning the indicated operations. 4. 24. (a) Repeat problem 4. 22(a) for any given configuration and let fl be variable from 0 to 1 in steps of 0.2. Plot Xab as a function of fl ' (b) Repeat problem 4.22(a) with f3 - 0.5 and let fl be a variable from 0 to 0.5 in steps of 0. 1. Plot Xab as a function of f1 and compare results with part (a). 4. 25. Verify equations (4.78) and (4.82) for a transposed line. 4.26. Compute the sequence impedance matrix Z0 12 for each line of problem 4.22. 4. 27. Compute the sequence impedances for the study specified in problem 4.24 and detennine the way in which the sequences are coupled together as a function of fl ' 4.28. Verify the twist equation (4. 110). 4.29. Compute the unbalance factors of each line of problem 4.22, (a) Using the exact fonnula (4. 144). (b) Using the approximate fonnula (4. 146). 4. 30. Write the matrix equations for the flux linkages in each section of the transposed line of Figure 4.9. 4.31. Consider the four circuits a, b, c, and d shown in Figure 4.5. (a) Write the instantaneous voltage equations of the fonn 11 - ri + cD.. / dt for each circuit. (b) Apply the phasor transfonnation (1.50) to the equations of (a). (c) Compute the phase impedance matrix Zabc from the result of (b). 4. 32. Verify the fonnula for unbalance factors given by (4. 144) by perfonning the computa tion implied. 4.33. Consider the line configuration of problem 4.3 shown in Figure P4. 3. A ground wire of I/O K copperweld-copper is present but is used as a second conductor for one of the phase wires (assume here that the ground wire is fully insulated). Compute the phase im pedance matrix Zabc if this extra wire is connected as follows: (a) In parallel with phase a. (b) In parallel with phase b. ( c) In parallel with phase c . 4.34. Consider the line configuration shown in Figure P4.4. Instead of using a single conductor of 336,400 CM ACSR in each phase, with current-carrying capacity of 530 amperes, sup pose that each phase consists of a two-conductor bundle of two 3{0 ACSR conductors with capacity of 300 amperes/conductor. Let the two conductors of each bundle be separated by 1.0 ft vertically. (a) Compute the phase impedance matrix Zabc for the bundled conductor configuration and compare with the previous solution (problem 4.20). (b) Compute the sequence impedance matrix for both the new and old conductor ar rangements and compare. 4.35. Let the circuit described in problem 4.5 and Figure P4.5 be altered to consist of a two conductor horizontal bundle of 397,500 CM ACSR phase wires. Compute phase im pedance matrix Za bc and sequence impedance matrix ZO l lo Bundles with 1.2 ft spacing. 4.36. Verify equation (4. 164) for a bundled conductor line. Then show that the necessary matrix operations are all 3 X 3 matrix operations as suggested in Example 4.8. 4. 37. Verify equation (4. 174) for the phase impedance matrix Zabc of a line with one ground wire. Then extend this analysis to the case of a partially transposed line where fl ::/:. f2 ::/:. f3. 4.38. Verify the computations which result in equations (4. 198) and (4.201) for a completely transposed line with one ground wire. 4.39. Consider an untransposed line described in problem 4.3 and Figure P4.3. Let the ground wire be I/O K copperweld-copper and recalculate the phase impedance matrix Zabc , the sequence impedance matrix Z012 , and the unbalance factors rno and rn 2 ' Compare with previous results from problem 4. 19 for the same line without the ground wire.
Seq uence I m pedance of T ra n s m i ssion Li nes
1 51
4.40. Consider an untransposed line described in problem 4.4 and Figure P4.4. Let the ground the sequence im wire be I/O ACSR and recalculate the phase impedance matrix pedance matrix and the unbalance factors. Compare with previous results from problem 4. 20 for the same line without the ground wire. 4.41. Consider an untransposed line described in problem 4.5 and Figure P4.5 with two ground wires of 2/0 ACSR. Compute the phase impedance matrix the sequence impedance matrix and the unbalance factors m o and m2 ' Compare with the results of problem 4. 2 1 for the same line without ground wires. 4.42. Verify carefully the result given by equation (4.230)-(4.2 34) for impedance of a trans posed line with two ground wires. 4.43. Derive the general expression for a completely transposed line with n ground wires in such a form that (4.233) may be used. 4.44. Repeat the computation of the sequence impedances for problem 4.40 if the ground wire is taken to be 1 / 2-inch EBB steel messenger. Perform this computation for 1, 30, and 60 amperes in the ground wire. 4.45. Repeat problem 4.44 for the case where the line is completely transposed. 4.46. Compute the sequence impedance matrix for the line described in problem 4.4 and Figure P4.4. Let the ground wire have a GMR of 0.001 feet and let the resistance of the ground wire vary from 0 to 5 ohms/mile. Plot the ratio of Iw to 3IaO and to Id as a func tion of rw for nxed GMR. 4.47. Repeat problem 4.46 for a range of GMR's from 0.001 to 0. 10 and plot the current ratios Iw /Id and Iw / 3IaO for the entire family. 4.48. Equation (4. 184), giving the ratio Iw /Id ' indicates that if zw w " Zag then Iw/Id � oo. Show that this is impossible for the range of rw and GMR encountered in physical situations. How high does this ratio become as a practical matter? 4.49. Suppose that the lines shown in Figure P4.4 and P4. 5 (with ground wires specifled in problems 4.40 and 4.41 respectively) are located on the same right-of-way and separated by a horizontal center-to-center distance of 50 feet. (a) Compute the matrix of phase impedances including all ground wires and reduce this matrix to a 6 X 6 phase impedance matrix similar to equation (4.260). (b) Invert and transform the matrix of (a) to compute the four unbalance factors mO h mu, mOe , and m2e ' (c) Recalculate (b) for all 1 2 possible conductor arrangements for the 2 circuits. 4.50. Prove that there are 6"/3 unique arrangements of n parallel circuits which must be examined to determine the optimum phase connguration. 4.51. Repeat the computations of Example 4. 14 for all conngurations specified in equation (4. 281).
Zabe,
Z012 ,
Z012 ,
Zabe,
chapter
5
Seq u e n ce Ca pa c ita n ce of Tra n s m i ss i o n Li nes
This chapter focuses upon the shunt admittance of the transmission line. In overhead lines this shunt admittance is a pure susceptance since the conduction current between wires or between wires and ground is negligible. Furthermore, this susceptance is purely capacitive. We begin with the computation of the capacitance to neutral of an isolated transposed line and review the method, adequate for most problems, which is used for this situation. In later sections, we examine capacitance in greater detail and find the capacitance between wires (mutual capacitance ) and the capacitance un balance in un transposed lines. 5.1
Positive and Negative Sequence Capacitance of Transposed Lines
The capacitance to ground of each phase of a transposed line is computed by finding the ratio of the linear charge density to the voltage , averaged for each sec tion of the transposition. From field theory we recall that the voltage drop from is given by point 1 to point 2, both external to a linear charge density
Dx2 V1 2 = 2qx1T € Dx l In
qx ,
V
(5.1)
Note that we can keep the subscripts straight if we follow this notation carefully, always interpreting the distance from a charged line to itself, i.e. , as the con ductor radius. Figure 5.1 illustrates the distances expressed in equation ( 5 . 1 ) . Note that if > 0, and > then i s a positive drop in potential of polarity indicated. Since a transmission line is passive, the capacitance is the same for positive and negative sequence systems because the physical parameters do not change with a change in sequence of the applied voltage. Stevenson [9] uses a convenient "mod ified GMD method " for finding the capacitance to neutral of transposed three phase lines with the result,
qx
Dx2 Dx 1 ,
en
=
In (Dm Ir h'
V1 2
) nF/unit length, to neutral
Dx x
(5.2)
where i s the same GMD computed in equation ( 4 . 4 ) , r i s th e radius o f the phase conductor, and h ' is a constant depending on units of length given by Table 5 . 1 . Note the similarity in the logarithm term to the inductance equations where
Dm
1 52
1 53
Seque nce Capacita nce of Tra n s m ission lines
Fig. 5 . 1 . Configuration for equation ( 5. 1 ).
Table 5. 1. Capacitance Multiplying Constants for C in nF/Unit Length * Constan t
k' (1/3)k' f-
50 Hz
f- 60
Hz
Natural Logarithm
Unit of Length
( In )
BaBe 1 0 Logarithm
(Iog lo)
55.630 89.525 18. 543 '29.842
·km mi km mi
24.159 38.880 8.053 12.960
fk ' wk '
km mi km mi
2781.49 4476.24 17476. 57 28125.04
1207.97 1943.99 7589.90 12214.42
f k'
km mi km mi
3337.78 5317�49 20971.89 33750.07
1449.57 2309. 33 9107.88 14657. 32
wk'
3
if C is desired in microfarads; multiply *n nano 1 0- 9 ; multiply table values by 1 0 table values by 1 0- 9 if C is desired in farads; 1 . 0 mi .. 1 . 6093 km ; f .. 50 Hz, w 3 1 4. 1 9 5 rad/sec; f 60 Hz, w 376. 991 rad/sec ; eo ( 1 /lloc 2 ) .. 8 . 8 54 X 1 0- 1 2 F / m. ..
-
=
=
...
=
=
D.
we wrote In(Dm /D,) . Here the D m is the same, but has been replaced by the conductor radius since all charge on a conductor resides on its surface. If we in terpret the of each phase to be based on the radius r instead of for each wire, we may generalize the equation to write
D,
D,
h'
( 5.3) D /D� ) nF/unit length and b n = W C n nmho/unit length where D� = (D�I D� 2 D� 3 ) 1 / 3 and D�I = the geometric
Cn = In(
m
mean of distances from conductor center to outside radius in part i of a transposi tion . Usually b n is in the range 4.8 to 5.5 micromho/mile for single circuit, 60 Hz overhead lines. This can best be illustrated by means of an example .
Example
5. 1
Find the capacitance to neutral of the double circuit line shown in Figure 5.2, where the notation in parenthesis, (a, c, b ) for example , indicates that this position is occupied in turn by phases a , c, and then b in the three parts of the transposi tion cycle . All conductors are 477 ,000 circular mil ACSR , 26/7 strand . It is as = sumed that fl = or each transposition section is one-third the total line length .
f2 f3 ,
Solution
Since the line of Figure 5.2 is a double circuit line, we assume that conductors
a and a' are connected together at each end and form a parallel phase a conductor,
1 54
Chapter
5
,
5'
C-t ('.b�
"",*__
1 0'
40 '
.---- 1 4 '---4001
I
Fig. 5 . 2 . A double circuit line configuration .
and similarly for phases
D� = (D: I D�2 D� 3 ) 1/3 , where D8' 1 - (raDaa 'ra'Da 'a ) 1/4 D�2 = (r"D""'r,, , Db 'b ) I/4 D'83 -- (rc D cc 'rc ,Dc 'e )1/4
b and c. Then
Daa ' = Dec ' and, assuming that all wires have the same radius, we D� = (r 3/1 Daa, DVl,) 1/3 = rIll D!�� D Vg, The GMD may be computed as (for position 1, shown in Figure 5.2) Dm - (D- ab Db e Dca ) We note that have
_
-
-
1 /3
where
But
Then
D -- (Dab Dab , ) 1/4 (D bc D bc ' Dea De'a , ) 1/1 2 DcoI/6' m
5.2 we compute Daa ' = ( 101 + 261) 1/2 = 27.8 ft, D bb ' = ( 1 02 + 22 ) 112 = 1 0 .2 ft = Dca ' Dab = 14 ft, Dab ' = ( 1 02 + 122 )1/2 15.65 ft Dbc = 10 ft, Db e ' = ( 1 02 + 162 ) 1/2 18 .9 ft
From Figure
=
=
Sequence Ca pacita nce of Tra nsmission Lines
Dea = 2 4 ft, r=
1 55
De'a' = 28 ft
0.858
2 X 12
(from table B.8)
Substituting, we compute
12
= 0.0 35 7 ft
13
D: = ( 0 .0 35 7 ) / ( 27 .8) / ( 1 0 . 2 ) 1 /6 = ( 0 . 1 89)( 3.029)( 1 . 47 3 ) = 0.843 ft Dm = ( 1 4 X 15.65)1/4 ( 10 X 1 8 .9 X 24 X 28 ) 1/ ( 1 0. 2 ) /6 = ( 3.847)(2.663)(1 .473) = 1 5 .086 ft Then from tral is
12
1
(5.3) the capacitance per phase (or per equivalent conductor) to neu Cn =
89.5 89 .5 = In ( 1 5 .086/0.843) 2.8 8
and we may compute the
= 31 . 0 3 n F/ mi
60 Hz susceptance as
be = 21T (60) Cn = 11.70 � mho/mi!phase The preceding computations ignore the effect of the conductor height above the ground. Stevenson [ 9 ] shows that (5.3) is modified slightly when conductor height is taken into account. If conductor heights are measured to the image con ductors as shown in Figure 5 .3, we have
Cn =
In(Dm /D: )
- In(HI 2 H23H3dH1 H2H3) 1 /3 k'
nF/unit length
(5 .4)
From this equation we observe that the effect of taking conductor height into con sideration amounts to subtracting from the denominator the term In
(H1 2 H'1 3 H3 1 ) 1/3 H1 H2 HS
(5.5)
If the conductor height is large compared to the spacing between wires, this term is nearly zero . It is therefore often omitted . Usually, the actual height of attachment to poles or towers is modified for these calculations. A figure often used is the attachment height minus one-third the sag [ 38 ] . 2
2
b'
Fig. 5 .3 . Three phase conductors and their images.
Chapter 5
1 56
5.2 Zero Sequence Capacitance of Transposed Lines
The shunt admittance between line connections and the zero-potential bus in the zero sequence network depends upon the capacitance to neutral as seen by zero sequence voltages. To compute the zero sequence susceptance , we use the method of images as applied to zero sequence charges and compute the capaci tance of a transposed line. Using this method, we consider first a line without ground wires and later add the ground wire. (Also see Stevenson
[21].)
Capacitance with no ground wires. Consider the system shown i n F igure 5 .3 where charges and reside on conductors a, b , and c respectively , while the negative of these charges resides on the image conductors. Then the volt age drop from wire a to neutral is half the voltage drop or, if e is the permit tivity (see
qaO , qb O , qeO Vaa, [12] ), D f VaO = (1/2)Vaa' (1/2) L 2 2 qi L 4q, �2I �qao HraI + qb o HD I 2 qeO H3D-1 1 2 D 31 D - qaO In HraI - qbO I122 qeO H-31)3 1 VaO 2 �qaO HraI + qbO DH1I 22 + qeO H331)1 H3 ) H H 12 2 ( 2 eo o qa qb VbO O q D D 2 32 12 rb 12 �qao HD-1 33 + qbo H2D-3 qeo H3e ) Veo = r 1 23 VaO = Vb O = Veo (5.8 )-(5.1 0) qaO q b O q eO =
i
In -
= 4
1
1r e
1r ex
D I
dx =
In - +
In - H
i
1r e
In
(5.6)
In
In
(5. 7 )
which simplifies to
Similarly ,
In -
1
=
1r e
=
1
In -
In
1r e
In -
+
In
In
1r e
In D
+
+
V
In
V
In -
By definition we know that
so we conclude from
that
*
*
V
(5.8 ) (5 .9 ) (5. 1 0) (5. 1 1) (5.12)
i n any transposition section , but for the usual spacings these charges are nearly equal . If we assume equal charges over the full transposition cycle, or
qaO = qbO = qeo V "'" VaO + V3bO VeO
the voltage becomes an average of the three voltages given by av
00 -
+
2 H ) 3 3 2H2 1 1 V 3qao [HrarI Hb2re(H3(H D 12 D23 D31 )2 ] 1/9
(5. 8 )-(5. 1 0),
or
Performing this operation, we compute
00
=
21r e
In
V
( . 3
51 )
Sequence Capacitance of Tra nsm ission Lines
1 57
or we observe that
(5.14) where
Haa = GMD between the three conductors and their images Daa = self GMD of the overhead conductors as a composite group, but with D, of each wire taken as its radius
Then
qaO Co = - = Va o =
33
211' €
In(Haa IDaa )
( 1 / )k '
In (Haa ID aa )
F1m/phase
nF /unit length
(5.15)
As a rule of thumb b c o has values i n the range o f 2. 5-3.5 ILmho/mile for single cir cuit 60 Hz overhead lines. Capacitance with ground wires. To solve the capacitance problem with ground wires, we first analyze the case of a single conductor a with one ground wire g as shown in Figure 5.4. This may later be extended to a more general situation. g
qg Hg
0' Fig. 5 . 4 .
One conductor with one ground wire.
Referring to Figure 5.4, we write
Likewise ,
Va = .! 2 Vaa , = � 2 1T€
(q
H Ha In ag + a In g Da ra g
q
)
( 5 . 1 6)
(5.17) Solving ( 5 . 16 ) and ( 5 .1 7 ) for qa ' we have (5.18)
Chapter 5
1 58
Now suppose that conductor a is the composite of three phase wires and con ductor g is the composite of n ground wires. Then qG is the total charge of three conductors or, if all wires have the same charge, all
qGO = (1 / 3) qG
Also, since VGo = VbO = Veo , we have VG O = VG From (5.18)-(5 .20 ) we compute (1 / 3)k' ln(H,, /D,, ) = =--G_G-In-H�'-='-_--="='(-ln-H=-G'-)""2 " nF/unit length/phase Co ltl--H D" DG, DGG where same as in equation (5.14) self GMD of ground wires with D. = r, DDDGG"G, === the GMD between phase wires and ground wires HGG = GMD between phase wires and their images H" = GMD between ground wires and their images HGI = GMD between phase wires and images of ground wires 5.3
(5.19) (5.20)
(5.21)
Mutual Capacitance of Transmission Lines
In the preceding paragraphs we considered the calculation of capacitance be tween a charged conductor and ground for positive, negative, and zero sequence charges. We now expand our consideration to the capacitance between nearby conductors and shall refer to the result as "mutual capacitance." The subject of self and mutual capacitance is treated in many references. Rather than treat every possible physical configuration, we will investigate the transposed single circuit three-phase line in some detail and then consider briefly the effect of ground wires. Dissymmetry due to untransposed lines will be cov ered in section 5 . 8 . Consider a group of n conductors carrying linear charge densities qG ' qb ' . . . , qn ' located above the ground plane as shown in Figure 5 . 5 . Using the method of images (see [ 1 2 ] ), we compute the voltage drop between two points 1 and 2 by superposition (since the system is linear). From ( 5 . 1 ) we write 1
(5.22)
where the D's are distances between conductors or points above the ground and the H's are distances between conductors or points in space and image charges. In this way we compute the voltages of all conductors to ground by noting that is referred to Clarke [11 ] , Wagner and Evans [ 10 ] , and Lyon [ 3 9 ] for additional reading o n the subject. Our approach follows Calabrese [ 24 ] which i s recom mended for its clarity and complete coverage of the subject.
1 The interested reader
Seque nce Ca pacita n ce of Tra nsm ission Lines
1 59
G ROU N D
Fig . 5 . 5 .
Group of n charged lines and their images.
(5.23)
(qa
Then, for example ,
V =
a
Ha + q b In Hab Han ra D-ab + . . . + q n In D-an Dab - . . . - q n In D-an) - qa In Hraa - qb In Ha b Han 1 1 V ,=41T € 2
aa
-
In
-
and combining terms,
V =
a
1 -
21T€
Similarly ,
Vn
=
�qa
-
In
-� 1 21T €
-
} qa n
Ha + qb In Hab + . . . + q n In Han ) r Da b D-an V -
-
a
) Han + b In H-b n + . . . + q n n Hn - V r Db n Dan n -
}
q
(5.24)
(5.25)
( 5.26)
Equations (5 . 2 5 ) and (5 .26 ) are often written in matrix form as V
=
Pq
(5.27)
where V is the voltage vector, q is the charge vector, and P is a matrix of coeffi cients called "potentia} coefficients" [ 2 4 ] where p
.. -
Hj ' rj 1 H. . = - In � , i * j 21T € D ij
=
I}
1 } l =] n 21T € •
•
F- 1 m
( 5 .28)
Chapter 5
1 60
and we note that P is symmetric. Note also that V and q are not restricted in any way but may be sinusoidally varying time domain quantities or transformed quantities such as phasors. One can think of the matrix P as being a chargeless or "zero charge matrix." For example, suppose we set
1 o
0
q =
elm
(5.29)
o
i.e . , q a is the only charge present and its value is 1 coulomb/meter. Then
Va
Paa :::
1
-
211'€ V
Pna
In H-raa In HD-aabb In
Ha n Dan
V (5.30)
or P le a is the potential in volts assumed by wire k due to linear charge on wire a alone ( of 1 coulomb/meter) with all o ther charges zero. All these coefficients are positive quantities since Ha n > Da n , these being equal only if charge a is on the ground plane . With charge qa acting alone as in and (5.30), the potential Va is the greatest of all potentials in ( 5. 30 ) , but all are positive. In a similar way we can define each column of P as a set of potentials derived by setting the charges equal to 1 coulomb/meter and with all other charges equal to zero. Thus each column of P is independent of all others depending only upon the geometry of the system, and the columns of P comprise a basis for an n dimensional vector space [ 40] . This being the case, P is nonsingular and has an inverse P - I . Thus from we may compute
(5.29)
(5.27)
q = cV
(5.31 )
c = p-I
(5.32 )
where 2
The elements of the matrix c are called "Maxwell's coefficients" or more specifi cally are called "capacitance coefficients" when referring to diagonal terms and "coefficients of electrostatic induction" when referring to off-diagonal terms. These coefficients may be thought of as short circuit parameters which relate the charges to the system voltages. For example, with 2 Obviously, this is not the same matrix C as defined in Chapter 2. There will be no ambiguity, however, as we usually use for the symmetriCal component matrix. Here the c is written lowercase for capacitance per unit length. The uppercase C is for total capacitllnce of the line.
A-I
Sequence Capacitance of Tra nsmission lines
1 61
1 o V=
0
1
o
(5.33)
i.e. , with Va = and all other voltages zero (short circuited), w e find the first column of c to be
elm
=
-C
qn
(5.34)
na
The negative signs from the off-diagonal terms arise due to the fact that the coef ficients of P in ( 5.27 ) are all positive. Then the inverse of P is adj P [PiJ ) t = = p- l = det P det P
c
(5 .35)
where Pij is the cofactor of element Pij . But
Pij
=
(- 1 )i +JM1j
where Mij is the minor of element Pij . We set
(5.36)
[24]
Cij = :!� P for i = j or (i + j ) even =
Then c=
-
t
O
-C
-C
M ij
det P
-C
.
. #= ' ] or ( l + ]. ) 0dd
for l ab
-C
j
"
ba
+C bb
-C
na
- Cn
+C nn
b
bn
F/m
(5.37 )
(5.38)
and only the diagonal terms have positive signs. The coefficients (elJ ) themselves are all positive quantities and are all capacitances. Note that all coefficients may be found by superposition, i.e . , by applying 1 volt to wire a, b, etc., always with every other wire shorted to ground. The negative sign really means, then� that applying a positive potential to one wire induces a negative charge on the other wires. This is intuitively correct since the application of a dc voltage to a capa�· itor makes one terminal (or plate) of the capacitor positive but also makes the other terminal ( plate) negative. Equation (5.31 ) is the desired relationship for self and mutual capacitances of an n wire system. It is not in the best form for physical interpretation, however. To make this clearer, suppose that both q and V are sinusoidal time varying quantities. Specifically , let one of the charge densities be
1 62
Chapter 5
q = Qm cos w t
(5.39 )
Then, since current is the time rate of change of charge, we compute i = dq/dt = - w Qm sin w t
(5.40)
Now transform both (5.39 ) and (5.40) into phasors. Then the phasor charge density Q and phasor charging current I are defined as ( 5.41 ) and 1- w Qm e h r/2
�
(5.42 )
--
or
1 = w Qe h r/2 = jw Q
(5.43)
If we write ( 5 .31) in terms of phasor charge density and voltage Q and V , we have Q = CV
(5.44)
which we change to a current equation by multiplying by jw as in (5.43), or (5.45)
1 = jw Q = jwCV
But, from circuit theory we write the charging current as I = YV
(5.46)
Y = jwC
(5.47 )
or the phasor admittances are For example, from (5.46) we interpret that Ykit = sum of all admittances connected to k = jW Ck k , a capacitive susceptance
(5.48 )
Yk m = the negative of all admittance connected between
k and m (5 .49 )
= - jw ck m
or the actual admittance between k and m is (using lowercase for the actual quan tity and uppercase for the matrix element) (5.50 ) which is again a capacitive susceptance. From (5.47 )-(5.50) we visualize an equivalent circuit as shown in Figure 5.6. Since YkIt is defined in (5.48), the capacitances to ground are Cag
=
Caa - Ca b - Cae - • • •
Cbg
=
- Cba + Cb b - Cbe
Cng
=
- Cna - Cn b - Cne
-
-
-
Can
F/m
Cbn
F/m
• • . + Cn n
F/m
•
•
•
-
(5.51 )
Sequence Capacita nce of Tra n s m ission Lines
\
1 63
, , \
Fig. 5 . 6. Self and mutual capacitances of an n-phase system .
With capacitances so defined, it is easy to see that the sum of admittances con nected to node k is indeed Yk k • The equivalent circuit of Figure 5.6 can also be obtained by algebraic manipulation of ( 5 .47). (See problem 5 .2. ) 5.4 M utual Capacitance of Three-Phase L i nes without G round Wires
We now consider a special case of the general mutual capacitance problem where there are only the thref> charged conductors of a transposed three-phase line with no ground wires. From ( 5 .25 ) and (5.26) the potential equations are, in terms of potential coefficients, V = Pq, or
pa
j
Pbe
5)
(5. 2
Pee
�aJ [
where the elements of P are given by ( 5 .28 ) . The charge equation is found by solving ( 5 . 5 2 ) for q. Thus, in terms of Maxwell's coefficients, q = cV , or
qb
_
-
qe
where the elements of
caa - C ba
- Cab Cb b
- Cea
- Ceb
c are given by ( 5. 3 7 ) .
Let
[�j
elm (5.53)
det P = Paa ( PbbPee - P�e ) - P�bPee + 2PabPbePae - P�ePbb Then compute the minors (noting that P il. = P k i )
Maa = Pbb Pe c - P�e Mb b
= PaaPee
- P�e Mee = PaaPbb - P�b
c
Mab Mac
= PabPee
- PaePbe
= Pab Pbe - PbbPae M be = PaaPbe - PabPae
( 5.54)
Then the elements of are computed from ( 5 . 3 7 ) . For the capacitance to ground we compute from ( 5 .5 1 )
Call
= Caa
- Cab - Cae
F /m
C bll = Cbb - Cab - Cbe F/m Cell = Cee - Cae - Cbe
F/m
( 5.55)
These various capacitances may be viewed symbolically as shown in Figure 5.7 where the various capacitances are shown as circuit elements.
1 64
C h a pter 5
Fig. 5 . 7 . Capacitances of a three-phase line with no ground wires.
If the line is transposed in sections of per unit length f, , f2 ' and f3 the voltage equation (5.52) may be written using the notation of Chapter Thus we write
4.
( 5. 56) where
C2 3 1
=
R� l C m RI/>'
C3 1 2 = RI/> C I 23 R� '
and C m is defined to be the Maxwell's coefficient matrix for section fl ' The potential matrix is then found by inverting the entire matrix expression in ( 5. 56) . If the line is completely transposed, each phase occupies each position for one-third the total line length. Since each section is one-third the total length, the capacitance per phase per meter consists of one-third meter of each of and or
Cab ' Cb c .
Cac
CMO
=
( 1/3)( Cab +
Similarly, for the capacitance to ground c,.o = ( 1 /3 )( ca,. + ClI,. + where
CsO
=
Cbc + Cac) F/m
cc, )
( 1 /3)( caa +
s
=c
o
- 2 CMO
( 5. 57 )
F/m
Cbb + Ccc) F/m
( 5. 58)
( 5 . 59 )
Actually, these "transposed capacitances" are just averages o f th e capacitance seen by each phase in each transposition section. The capacitance to ground can be thought of as a combination of a self capacitance and the mutual capaci tance
Cs
CM'
Example
5.2
Find the mutual capacitance and capacitance to ground of the lower circuit only (ignore upper circuit and ground wires) of Figure 5.2. Compare the capaci tance to ground computed in this way to that obtained by application of ( 5. 3).
Solution First, we form the P matrix, the coefficients of which upon geometry. In our case (see Figure 5 . 2 )
are
entirely dependent
D a 28. 0 ft Dab D bc 1 4 .0 ft, Ho b Hbc = ( 802 + 142 ) 1 /2 = 8 1 . 2 ft Ha Hb Hc 80.0 ft, 0.0357 ft Hac ( 80 2 + 28 2 ) 1 2 84. 8 ft, =
= =
c
=
=
=
/
=
=
=
r =
1 65
Sequ ence Capacit a n ce of Tra nsm ission Lines
Then from ( 5 . 28)
H 1 In -.!! P aa = P"" = P cc = 21T f ra and if we let
where "
f = fo"
= " /( 36
Ha ra
F- 1 m
= 1 for air dielectric, w e have P aa = 18 X 10 + 9 ln
= 11.185 In Ha ra
1T
X
MF" l mi
= 11.185 In 0. :�5 7 = 86. 28
1 Ha b Pab = Pbc = 1 1 . 1 8 5 n Da b
Pac = 1 1 . 1 85 In Then p=
[
109)
M F- 1 mi
1.3 = 19.66 = 11.185 In 814.0
�:: = 1 1 . 1 85 In �::� = 12.39
[
86. 3 19. 7 1 2. 4
19. 7 86.3 1 9. 7
1
MF" mi
MF- 1 mi
�
1 2.4 19. 7 MF- 1 mi 86. 3
Direct inversion o f P by digital computer (see Appendix A) gives the result
c=
J
1 2. 34 - 2. 54 - 1.19 - 2. 54 1 2. 7 5 - 2. 54 - 1.19 - 2. 54 12.34
nF/mi
Since a digital computer is not always available, we confirm this result by hand computation. From
(5.54) we compute,
A = d et P = Paa Pb bPcc - PaaP�c - P�bPcc + 2P a bPcaPb c - P�c Pb b
=
86.33 - 86.3(19. 7)2 - ( 19.7)2 (86.3) + 2(19. 7 )2 ( 1 2.4) - ( 1 2.4)2 (86.3) e! 647,500 - 33,600 - 33,600 + 8,330 - 13,300 = 575,330
Also from
( 5. 54), Maa
= 7091 Mab = 1459
= 7091
Mc c
M b b = 7326 from which we compute trom
= 1 2.31 C b b = 1 2. 72 Caa
Ma c = - 685 M bc = 14 59
( 5.37),
= 12.31
nF/mi
Ccc
nF/mi
Cac = 1.190 nF/mi
nF/mi
Ca b = 2.535 nF/mi
C b c = 2.535 nF/mi
These values are seen to be very close indeed to the digital computer results. Since
1 66
C h a pter
5
the more accurate digital computer solution is available, we use it as a basis for computations which follow. The capacitances to ground are, from ( 5. 55), Cag = Caa - Cab - Cae Cbg = Cb b - Cab - Cb e ceg = Cee - Cae - Cb e
= 8. 6 06 = 7. 66 7 = 8 . 6 06
nF /mi nF/mi nF /mi
For a transposed line the capacitance to ground is the average of these three, Le., from ( 5 . 58 ), CgO
= ( 1 /3)( cag + Cbg + ceg ) = 8. 2 93
nF/mi
The average ( transposed) mutual capacitance is, from ( 5. 57), CMO
= ( 1 /3 )( cab
+ Cb e + cae )
= 2.091
nF/mi
The average (transposed) self capacitance is, from ( 5. 59),
Cso = ( 1 /3 )(caa + Cb b + Cee ) = 1 2. 47 5
nF/mi
Now compare Cgo computed above to the capacitance to neutral computed by equation ( 5 . 3 ) which neglects the effect of the earth and is computed under the assumption of equal charge density in each part of a transposition cycle [ 9] . From ( 5.3)
Cn = In(D89.m /5D., )
where
nF/mi
= Deq = (Da b DbeDea ) I/3 D; = r = 0.0357 ft
Dm
=
1 7. 5 ft
Then en = 1 4 .4 nF/mi. How do we reconcile this difference? If we imagine an equivalent circuit similar to Figure 5. 7, the .6 -connected mutual capacitors all have a value of CMO 2.091 nF /mi. Convert this .6 to a Y. Then each capacitor the in the Y has a value cy 6.273 nF/mi. Since this is in parallel with total capacitance to neutral per phase is the sum Cn Cg O + c y = 14. 566 nF/mi. If the computed by ( 5. 3 ) is corrected by its height above the ground, these values will be in quite close agreement.
=
=
=
Cn
Cgo ,
5.5 Sequence Capacitance of a Transposed line without Ground Wires
Consider a three-phase line, the capacitance of which is described in terms of its Maxwell coefficients as q = CV, where q and V are phasors. Then by ( 5.45), 1 = j w CV or, to emphasize the a-b-c coordinate system, lab e
= j wCVabe
(5. 6 0)
But this is easily changed to a 0- 1 -2 coordinate system by a similarity transforma tion. Thus ( 5.61 ) or (5.62)
1 67
S e q u e nce Capacitance of Tra nsmission Lines
where we have defined
( 5.63) Performing the indicated transformation, we compute
CO l l
=
o
�
o
CSO - 2 CMO ) 1 (CS I + cM d 2 (CS2 + CM 2 )
j
1
2
( CS 2 + C M 2 )
(CS I + CM d (CS 2 - 2 CM2 )
(cs o + C MO ) (CS I - 2CM d
( cs o + CMO )
( 5.64)
where we define CSO and CMO as in ( 5. 59) and ( 5.57) respectivelyl and where
2 CS I = (1 /3 )( caa + aCb b + a ccc ) 2 CS 2 ( 1 /3 )( caa + a cb b + a ccc ) 2 CM I (1 /3 )( cbc + a ca c + a ca b ) 2 CM2 ( 1 /3 )( cbc + a cac + aCa b )
= = =
( 5.6 5)
If the line is transposed, the values in ( 5. 6 5) are all zero because each phase oc cupies each position for an equal distance, thereby acquiring a multiplier of 2 ( 1 + a + a ) for each capacitance term . With these sequence mutuals all zero, ( 5.64 ) becomes
CO ll
=
t
0
s o - 2 CMO 0
CSO + CMO
o
0
( 5.66)
and the mutual coupling between sequence networks is eliminated. Note from ( 5 . 66 ) that the zero sequence capacitance is much less than the positive and negative sequence capacitances. Also note that the positive and nega tive sequence capacitance to neutral is given by
( 5.67) which should agree closely with ( 5. 3 ). Similarly, for the zero sequence
Coo
= Cs o -
2 CMO
( 5.68)
which should check with ( 5. 1 5).
Example
5. 3
Compute the matrix CO l 2 for the transposed line of Figure 5. 2, data for which is computed in Example 5.2. Compare with results computed from ( 5.3) and
( 5. 15).
3 Note that the signs i n C O l 2 are not in the same pattern as in the ZOl 2 matrix of the last chapter. This is because of the negative signs in the Ca bc matrix of equation ( 5 . 53 ).
Chapte r 5
1 68
Solution From Example 5.2 we have
Cso
=
CM O
12.475 nF/mi,
=
2.091 nF/mi
Then C l l = C22 = Cso + CM O = 14 . 5 nF/mi which checks exactly with the Cn computed in Example 5. 2. Also Coo 8. 293 nF/mi. For the zero sequence capacitance we have from 5 . 1 5 ,
66
(
)
Co = where
=
29.8
1n(Haa /Daa )
n F /mi
Haa [HaHbHc (HabHbcHac ) 2 r/9 ( 80 ) 1/3( 81 . 3 X 81 .3 X 84.6)2/9 = (4.3 )( 1 8.9) Daa [ r3 (Dab Dbc Dca) 2 ] 1/9 ( 0.03 57) 1 /3 ( 1 4 1 4 X 28) 2 /9 = (0.33)(6.75) =
=
=
81 .3 ft
=
X
=
Thus
Co =
29.8 1n ( 81 .3/2. 22)
=
=
2. 22 ft
8.28 nF/mi
which checks very well with Coo .
5.6 Mutual Capacitance of Three-Phase Lines with G round Wires
Having computed the self and mutual capacitance of a circuit without ground wires, we now consider the additional complication added by the ground wire and study its effect upon capacitance of a line. As before, we consider the line as being transposed, leaving the problem of unequal phase capacitances for a later section. Using the subscript n to denote the ground wire, we may write the equation of potential coefficients for the four-wire system shown in Figure 5.8. We write V = P q or, in more detail and with Vn = 0 , a
b C
a
Ca b
Cog
0
b
c oo
Fig. 5.8. A three-phase line with one ground wire n .
C 0
Sequence Capacitance of Transm ission Lines
Va Paa Pab Pac Vb Pba Pbb Pbe Ve Pea Pcb Pee Pna Pn Pne =
----------
b
0
I I I I I ..1-
I I
1 69
qa qb qe qn
Pan Pbn Pen Pnn
---
( 5.69 )
But this matrix can be reduced to three equations by eliminating the fourth row and column. Solving the last equation and substituting back, we have
Va �Paa PanPna) Pnn Vb (Pba - P�nn:na) Vc (pea PenPna) Pnn
�ab - P;�nb) (Pbb - P�nn:nb) oeb - PenPnb) Pnn
_
=
_
�ae - PanPnc) Pnn ) 0be - PbnPne Pnn (Pee - PenPne) Pnn
qa qb qe
( 5.70)
This could be simplified slightly by taking advantage of the fact that P is sym metric. Since the elements of P are all positive, the new P of ( 5.70) contains elements all of which are smaller than corresponding elements for the same line with no ground wires. Note that the "correction factor" for each element is a positive quantity which depends only upon the geometry of the system (see (5.28». If the ground wire is moved toward infinity, this correction factor approaches zero since the entire row and column associated with the ground wire vanish. Using the corrected P matrix of ( 5 . 70), we again find the capacitance matrix V, where Ca be is the of Maxwell's coefficients by matrix inversion, i.e . , q = inverse of the 3 X 3 matrix of ( 5.70). Since the presence of the ground wire makes the elements of P smaller, we would expect the elements of C to be larger than the case of the same system with no ground wires. The sequence capacitance matrix CO l 2 is again found by a similarity transfor mation, exactly as in the case of no ground wire, by ( 5.63). If it is desirable to consider the capacitance to ground separately from the capacitance to neutral, as shown in Figure 5.8, this can be done by inverting the 4 X 4 P matrix of ( 5 .6 9 ) . This would permit separate identification of elements such as Ca n , C b n , and C e n as sh own in Figure 5.8. Since these capacitances are in parallel with call ' C bll ' and Cell ' it is apparent that the capacitances are increased by the ground wire . If there are two ground wires m an d n , the procedure is exactly the same. In this case the equations involving potential coefficients are
Cabe
Paa
Pa b
Pac I
Pam
Pba
Pb b
Pbe I
Pb m
Ve = Pea o
o
which we rewrite as
Pe b
Pma
Pm b
Pna
Pn b
I
Pee
II
� Pm m
Pme I I Pne I
-- ---------
Pem
Pan Pbn Pe n
Pm n
- - - -- - -
Pn m
P
nn
(5.71)
Chapter 5
1 70
( 5. 7 2) where the matrices P I , P2 , P 3 , and P4 are defined according to the partitioning of ( 5 . 7 1 ). Solving for Va ll c we have (5.73) where
Pa llc
=
P I - P2 P:;j I P3
( 5. 74 )
Thus, P2 P:;j l P 3 may be thought of as a correction due to ground wires. The Max well coeffiCients are found by inverting ( 5. 74), i.e. ,
Cabc = P;�c
( 5. 75)
and the sequence capacitances are found by applying the similarity transformation ( 5.63) to Ca llc .
Example
5. 4
Repeat the capacitance calculation of Example 5.2, this time including the ef fect of the ground wires shown in Figure 5.2. Assume the ground wire is 3/8-inch EBB steel.
Solution First we compute the potential coefficients for the ground wires. Thus
Pmm
= Pnn
H = 1 1 . 1 85 In --.!!!. rm
where
rm
==
Hm
( 1/2) ( 3/8) ( 1 /1 2 ) = 0.01562 ft,
or
Pmm
= Pnn =
1 1 . 185 In
Also,
Pmn
==
0.
�;�6 2
==
=
2 ( 55)
=
1 1 0 ft
99. 096 MF-' mi
Hm n Pn m = 1 1 . 185 In D mn
where
or
Pmn
= Pnm
==
1 1 . 1 85 In ( 1 1 1 /1 4 )
=
23. 1 4 7 MF- ' mi
We also compute
Pan
= Pbn = Pb m = Pc m
H = 1 1 . 185 In an Da n
Sequence Capacita n ce of Tra nsm ission Lines
and
Pam = Pen = where
Then
Ha n = (
11.185 In DHaamm Ham = (
952 + 21 2 ) 1 /2 = 97. 29 ft Dam = ( 1 5 2 + 21 2 ) 1 /2 = 25. 81 ft
952 + 72 ) 1 /2 95.26 ft, Da n = ( 15 2 + 7 2 ) 1 / 2 = 16.55 ft, 5!!
Pan =
11 . 185 In 95.26 16.55 = 19 . 574
Pam =
11 . 185 In 97.29 25.81 = 14. 844
and
Thus
a
a
b P= c
m
n
b
c
m
MF-1
MF-1
mI·
mI·
n
86.3 19.7 12.4 I 14.8 19.6 19.7 86.3 19.7 I 19.6 19.6
���_��� _ ��: �_��� _ ��!
14.8 19.6 19.6 99.1 23.1 19.6 19.6 14.8 23.1 99.1 I I
I
Now P4 1 = adj P4/det P4, so we compute
99.1 2 - 23.P = 9821 - 534 = 9287 1 r 99.1 - 23.1] r 0.0107 - 0.00251 928 7 t 23.1 99.1 = l:- 0.0025 0.010 7J
det P4 = p;; 1
4
Then P2 P4 1 P 3
_
�19.614.8 19.619.6� [- 0.0025 0.0 107 - 0.002 5] �14.8 19.6 19.6 0.0107 19.6 19.6 14.8J 19.6 14.8 �- 5.51.99 6.275.51 4.705.51 4.70 5.51 4.99J �819.76.3 19.786.3 �::� I:::� :::� ::�� -
=
_
and
Pabc = PI - P2 P4 ' P3 =
12.4 19.7 86. �J L�. 70 5.51 4.9�J -
171
1 72
=
Finally,
Co b c =
[
�
Chapter 5
�
1.30
14. 1 5
7.6
14. 1 5
80.02
14. 15
7.69
14. 1 5
81.30
�
P;; lc , which we compute by digital computer to be cobc =
12.747
- 2.106
- 0.839
2. 106
13 . 242 - 2.106
- 2.106
-
- 0.839
nF/mi
12.747
The capacitances to ground are
co.
=
coo - Cob - coc
C b.
=
C b b - cob - Cbc = 9.030 nF/mi
=
9.802
n F Imi
cc• = Ccc - Coc - Cbc = 9.802 nF lmi For the transposed line the capacitance to ground in each phase is the average of these three values, i.e.,
C.o
=
( 1 /3) ( co. + Cb. + cc. )
=
9 . 545
nF/mi
which is considerably larger than the 8. 293 nF/mi for this same line without ground wires. The average (transposed) mutual capacitance is
CM O
=
which is smaller than the
( 1 /3) ( Co b + Cbc + coc )
=
1 . 683
nF/mi
2.091 nF /mi for the same line without ground wires.
The average (transposed) self capacitance is
cs o
CCC ) 12.912 12.475 Coo = CSO - 2CMO C ll C22 Cs o + C O 1 4.595 8.293 14. 566 =
( 1 /3 ) (coo + Cb b +
which is greater than the
=
nF/mi
nF/mi for the same line without ground wires.
The sequence capacitances are =
=
= 9. 546 nF/mi and
M
These values are both greater than the
=
nF/mi
and
computed for the same
line without ground wires.
5.7
Capacitance of Double Circuit L ines For the case of a double circuit line with or without ground wires, the problem
becomes more complicated because of the presence of so many charges . Consider
5.9
the configuration of Figure where the distances from one conductor c' to all other conductors and to all images are shown . Using the method of section 5 . 4 , w e may write the voltage equations i n terms of potential coefficients as V Pq or
=
1 73
Seque nce Capacitance of Transm ission Lines
-qb
Fig. 5 .9. Configuration of a double circuit line.
Va
Paa
Pab
Pac
Vb
Pba
Pb b
Pb c
Pca
Pcb
Pcc
Vc Va ,
=
Vb ' Vc ,
-----------
Pa'a
Pa' b
Pb ' a
Pb ' b
Pc' a
Pc' b
I I I
Paa'
Pa b '
Pac'
qa
P ba '
Pb b '
Pb c'
qb
Pcb '
Pcc'
qc
I I Pca' I
T
------------
Pa' c I Pa' a' I Pb ' c I P b ' a ' I Pc' c I P c ' a ,
Pa ' b ' Pb ' b '
Pa' c' Pb ' c'
qa'
Pc' b '
Pc ' c ,
qc'
qb ' ( 5. 7 6)
which we may write as (5.77) The capacitance matrix C is the inverse o f P , which may b e computed according to the partition ing of ( 5. 7 7 ) as ( see [ 7 ] ) ( 5.78) where we define so that Ft
= P2 1 Pl l
F = P,l P 12
( 5.79 )
since P is symmetric, and E = P22 - P2 1 P, l P 12 = P2 2 - P2 1 F
Once C is determin ed, we may write I =
jw
( 5. 80 )
C V, where both I and V are 6 X
1 . It
1 74
Chapter
will be helpful to write
5
[ ] IO b C
lo' b ' c'
where C " C2 , Ca , and C4
are
( 5.81 )
dermed in ( 5.78). Then we compute
lob c = j w (C 1 Vob c + C2 Vo' b ' c ' ) lo' b ' c' = j w ( C a Vob c + C4 Vo' b ' c' )
( 5.82)
Since the lines are in parallel ( 5. 83 ) or lob c = j w (C1 + C2 ) Vobc Io' b ' c ' = jw (Ca + C4 ) Vob c
( 5.84)
and the total charging current is Ich• = Iobc + Io' b ' c ' = j w (C I + C2 + Ca + C4 ) Vob c
( 5. 85 )
Then the parallel circuit line behaves like a single circuit line with capacitance Ceq , where ( 5.86) The sequence capacitance may be found for each circuit or for the parallel equiva lent by similarity transformation, i.e. , CO l 2eq =
A - I Ceq A
( 5.87 )
Also, from ( 5 . 86 ) and ( 5 . 78) we compute Ce q =
Pi t
+
(F
-
U) E- I (F
-
U)t
( 5 .88)
where U is the unit matrix. Thus we see that the effect of the second circuit on the first depends strongly upon the m atrix (F U). Since P I I =1= P 1 2, we write -
F
-
U
= Pi l P 1 2
-
U
( 5 .89)
If F U is nearly zero, Ceq becomes nearly Pi l , which is the value for Cd c alone and would indicate no effect at all due to the second circuit. This will be the case if P 1 2 9!! P l l such that F U = Pi t Pl l U = U - U = O. This will be approxi mately true if the height is large compared to the distance between conductors and if the conductors are arranged in vertical phase configuration such that Ho b ::!! Ho b ' , etc. The matrices P i l and P 1 2 will never be exactly equal since this would require the wire radius to equal the distance between phases, or r = D 00 ' such that Poo = P oo' . For the case of double circuits with ground wires, the equation involving P will be similar to ( 5.76) but will have added rows and columns for the ground wires. These can be eliminated by matrix reduction since the voltage of the ground wires is zero. The result then can always be reduced to a 6 X 6 P matrix equivalent. -
-
-
Sequence Capacitance of Tra nsm ission Lines
Example
1 75
5. 5
Compute the capacitance matrix for the double circuit line of Example 5.1 and Figure 5.2, including the effect of the ground wires which are assumed to be 3/8-inch EBB steel. Examine only one transposition section.
Solution
First we compute the P matrix. If we arrange as shown in ( 5.76), we know the lower right partition from Example 5.4, which involved only the lower a' -b'-c' circuit and the ground wires. For the upper circuit
D b e :;: 10 ft,
Da b :;: 14 ft,
Dea :;: 24 ft
and
Ha Hab Hbe He a
:;: Hb = He = 100 ft :;: (100 2 + 14 2 ) 1 / 2 = 100.98 ft :;: ( 100 2 + 10 2 ) 1 /2 = 1 00. 50 ft = (100 2 + 24 2 )1 /2 = 102.84 ft r = 0 . 0 35 7 ft
Then
Paa :;: Pb b :;: P ee :;: 1 1 . 1 85 In P a b :;: 1 1.185 In
1
�:
:;: 11. 185 In
��: �8 :;: 22.100
��:�0 :;: 25.810 1 4 Pea :;: 11. 1 85 In �!: � :;: 16.276
Pbc :;: 1 1 . 1 85 In
1
0.
���7 :;: 88. 784
MF- 1 mi
MF- 1 mi MF- 1 mi MF- 1 mi
Between the a-b-c circuit and the ground wires we compute
Dam Dan Ham Ha n
:;: :;: :;: :;:
7.07 ft 19.65 ft 105. 1 2 ft 106. 71 ft
Dbm Dbn Hbm Hb n
:;: :;: :;: :;:
10. 30 ft 7.07 ft 105.39 ft 105. 1 2 ft
Dem Den Hem Hen
:;: :;: :;: :;:
19.65 ft 7.07 ft 106.71 ft 105. 12 ft
such that
105. 12 :;: 3 O. 1 89 M F- 1 mI' P am :;: P en :;: P bn :;: 1 1 ' 185 In 7.07 106. 7 1 _ - 18.9 2 7 MF - I ml· Pa n -- P em - 1 1. 1 85 In 19.65 105.12 . Pb m - 11 1 85 In 10.30 - 26.0 1 5 M� I ml
.
Between the a-b-c circuit and the a' -b' -c' circuit we compute
C h a pter
1 76
5
Daa ' = 27.86
ft
D b a ' = 15.62
ft
Dca ' = 10. 20
ft
Dab ' = 1 5.62
ft
D b b ' = 10. 20
ft
De b ' = 15.62
ft
D o c ' = 10. 20
ft
D be ' = 18.87
ft
De c ' = 27.86
ft
Haa ' = 93.68 Ho b ' = 90. 80
ft
Hb a ' = 90. 80
ft
He a ' = 90.02
ft
ft
ft
He b ' = 90.80
ft
Hac ' = 90.02
ft
Hb b ' = 90. 02 Hb e ' = 91.41
ft
Hee ' = 93.68 ft
such that
93.68 = 1 3 56 5 MF- ' mI· 27.86 . 90.80 = 19 . 686 MF- ' mI· Pa b ' = Pba ' = Pc b ' = 1 1 . 185 In 1 5.62 90 .02 = 24 . 359 M F- I mI· Pac ' = P b b ' = Pe a ' = 1 1 . 185 I n 10. 20 91.41 = 1 7 . 649 MF- 1 mI· Pbe ' = 1 1 . 185 1n 18.87 Paa ' = Pee ' = 1 1 . 185 I n
Rounding to the nearest
a
b c
P=
a'
b' c
,
m
n
a
b
88.8 22.1 16.3
22. 1 88.8 25. 8
0.1, we write a
c
,
16.3 II 13.6 25.8 : 19. 7 88. 8 24.4
----- -----
-
1 I
,
b'
c
19. 7 24.4 19. 7
24.4 1 7.6 13.6
---- - -
-
13.6 19. 7 24.4
19.7 24.4 1 7.6
24. 4
86.3 I 19. 7 I 19. 7 I 13.6 I 12.4
19.7 86. 3 19. 7
30. 2 1 8.9
26.0 30. 2
18.9 30. 2
1 9.6 19.6
14.8
19.6
--
m
n
30.2 26.0 18.9
1 8.9 30.2 30.2
,
--
-
-
-
-
--
--
1 2.4
14.8
19.7 86.3
19.6 19.6
19.6 19.6 14.8
19.6 14.8
99. 1 23. 1
23. 1 99. 1
MF- 1 mi
We eliminate the two " outside" rows and columns delineated by the solid parti tion lines to compute a "correction matrix "
6
X
6
matrix is
Pab e = P - Pc .
Pe =
Finally,
10.70 10.98 9.03 -
10.98 13.04 1 1.60
9.03 11.60 10. 70
--------- -
6. 56 7. 86 7.27
8.04 9.00 7.78
--
Pc , I I I I
:
-,
where we define
6.56 8.04 7. 27 -
-
--
7. 27 I 4.99 I 7.86 I 5. 51 I 6.56 I 4.70
7.86 9.00 7.86
7.27 7.78 6.5 6
- - - -
-
5.51 6.27 5.51
Pc such that the
-
-
4. 70 5. 5 1 4.99
1 77
Sequence Capacitance of Tra nsm ission Lines
a
Po bc
=
b c a
,
b' c'
a
b
c
78.08 11.12 7 . 24
11.12 7 5.75 14.21
7.24 14. 21 78.08
7.00 11.82 17.09
1 1 .64 1 5. 36 9. 87
17.09 11.82 7.00
a'
b'
c
I I I I I
7.00 11.64 17.09
1 1.82 1 5.36 11.82
1 7.09 9.87 7.00
I I I I I
81.30 14. 15 7.69
14. 1 5 80.02 14.1 5
7.69 14.1 5 81.30
'
--------------+--------------
which is inverted by digital computer to find
a
b
c=
c a'
b' c'
a
b
c
13.80 - 1 . 29 - 0.55
- 1. 29 14.43 - 1.88
- 0. 55 - 1.88 13.93
--------------
- 0.45 - 1. 20 - 2.45
- 1. 14 - 1.95 - 0.87
We also compute
Pi } P1 2 and
F
-
U
=
=
� [
- 2.38 - 1. 1 2 - 0.44
. 0526 0. 1091 0.1 939 0.9438 0. 1091 0. 1939
I I I
f I I I I I
a'
b'
c
- 0.45 - 1.14 - 2.38
- 1. 20 - 1.95 - 1. 1 2
- 2.45 - 0.87 - 0.44
'
- - - -- - - - - -- -- -
13.34 - 1.63 - 0. 54
0.1177 0. 1648 0. 1 105 0. 1177 - 0.8352 0. 1105
- 1.63 13. 79 - 1.66
- 0. 54 - 1.66 13.30
0. 201 0.0905 0.0546
=F
�
nF /mi
�
0.2010 0.0905 - 0.9454
Since F - U is quite different from 0, we conclude that the second circuit has a substantial effect upon the capacitance of the first. This is also evident if one compares the a-b-c partition of C with the C matrix of Example 5.4.
5.8
E lectrostatic Unbalance of U ntransposed Lines
If transmission lines are left untransposed, a practice which is becoming rela tively common, an electrostatic unbalance exists in addition to the electromag netic unbalance studied in Chapter 4. Any unbalance in transmission line charging currents results in the flow of neutral ''residual'' current in solidly grounded sys tems, and this current flows at all times, independent of load current. If the un balance is great and these residual currents are large, they could possibly affect system relaying or cause the voltages to become unbalanced. This problem has been studied extensively and methods have been developed for computing the amount of unbalance in a given situation (see [ 28-30, 34-36,
1 78
C h a pter 5
41-43 ] ) . Having established a definition of the "unbalance factor, " different line configurations may be examined in detail to optimize the line design.
[34-36]
Ground displacement of lines. In the early 19 50s, Gross and others developed a definition for the electrostatic unbalance of a line. This definition is established with reference to a line supplied from a Y -connected transformer bank as shown in Figure 5 . 1 0 where we recognize the presence of capaoitance between
- Van - V en
a
+
10
-
a Cea
+
+
+
b
e
+
Va Vb V c
III
Fig. 5 . 1 0 . Transmission line supplied from a Y·connected transformer.
wires and capacitance to ground. The neutral connection may be closed (grounded) or open, but in many modern systems it is grounded. The unbalance factor is defined differently for each connection, i.e., for the system neutral either grounded or ungrounded. The system shown in Figure 5. 1 0 is conveniently defined electrostatically by ( 5. 2 7 ) , i.e . , Vabc = P«Jab c, where we assume that the effect of ground wires is in cluded according to the method discussed in section 5.6. Then the charging cur rents flowing at no load are, from (5.60),
labc = j w C Vabc = j B Vabc where B is the shunt susceptance matrix. Also from ( 5 . 6 2)
1012
= j BOl 2
V012
( 5.90)
(5.91)
where
BO l 2
=
A- I B A = A- I (w C) A
(5.92)
Since the applied transformer voltages are balanced, positive sequence voltages, we write
then
Vabc = V + Vn and VOl 2 = A- I Vabc = A- I ( V + Vn)
where
V is strictly positive sequence. Expanding
(5.93), we compute
(5.93)
1 79
Seq u ence Capacitance of Tra nsmission Li nes
Vn (5.91)
or
Van laO = j (Boo VaO + BOI Val ) lal = j (BIO VaO + Bl 1 Va l ) la2 = j (B20 VaO + B2 1 Val )
is a zero sequence voltage, may be written as
is positive sequence, and
Va2
(5.94) =
O.
Then
(5.95) lao O.
Neutral ungrounded. If the system neutral is not grounded, the neutral = will usually be nonzero and the neutral current will be zero , or voltage and w e define the neutral "displace + Then from ment" or unbalance as
Vn
(5.95), Boo VaO BOI Val = 0, d o VaO/ Val = BOI /Boo = COl /COO (5.64) CM2 ) d0 - -CSO( CS2- +2CMO (5.58) d 0 -- - ( cnCg+o CM2 ) 2 do = cacgag+ +a CbgCbg ++ceacgeg cag + a32cgCbog + aceg Vn VaO (5.95), laO jBOl Vah lal jBl 1 Vah la2 jB2 1 Val d o = lao /lai = Bot lBl 1 COdCl l (5.64) 2 CS 2 + CM2 d o CCSSO2 ++ CM CMO CgO + 3CMO cag + a2 Cbg + aceg cag + a2 Cbg + aceg ( cag + Cbg + ceg ) + 3 (cab + Cbe + Cea) 3 (cgo + 3cMO ) =
(5.96)
-
we also write
From equation
Also from
-
(5.97) (5.98)
or, writing the numerator in terms of capacitances to ground,
(5.99)
_
Neutral grounded.
If the system neutral is grounded,
write from
=
=
=
In this case we define the displacement or unbalance as =
=
=
0
and we
(5.100) (5.101)
Then from
=
=
=
=
(5.102)
This expression may be simplified to neglect the capacitance between conductors since they are considerably smaller than the capacitance to ground. If this is done, we write
(5.103)
C h a pte r 5
1 80
(5.99).
which is exactly the same as Thus we have a convenient expression for the electrostatic unbalance which is independent of the system neutral grounding.
Example
5. 6
5.2,
Compute the displacement or unbalance of the lower circuit of Figure using computed values of Example 5.4 where possible, where we now assume the line to be untransposed.
Solution From Example 5.4 we have
C' O
= 9.545
nF/mi
1.683 nF /mi CS O = 12.912 nF/mi
CM O
=
= cbe = 2.106 nF/mi Coe = 0.83 9 nF /mi co, = ce, = 9.802 nF /mi
Cob
Cb , =
9.030
nF /mi
Then
c , + aCe, ( 1 + a) co, + a2 cb, a2 (co, - Cb, ) 3c,o ac,o ac,o = �(��747�) = 0 . 0269 {- 60° or 2.69% the system had a grounded neutral and the capacitance between phases is not neglected, we compute from (5.102) (co, - Cb, ) = a2 (0.772 ) = 0.0177 crur. or 1. 7 7% od = - 3(a29.545 3 (14.594) + 5.049) do
e!!
Co, + a2 b
=
=
-
---=---='-
-
If
0
-
Obviously, in this case the use of the approximate equation gives a very pessimistic result.
[34 ]
(5.99)
Reference gives examples of similar computations using with various wire sizes, spacing, and conductor heights and also shows the effect of ground wires. The results may be summarized as follows :
1 . Electrostatic unbalance may be reduced by the addition of ground wires and by increasing the spacing between wires. Electrostatic unbalance may be reduced by changing the arrangement of phase and ground wires, e.g. , by lowering the middle conductor of a flat configuration or by arranging the wires a-c-b , rather than in a vertical configuration.
2.
a-b-c ,
(5.95)
Note that from we may also define a negative sequence unbalance. In the case of the grounded neutral system we write
d2 or
= 10 2 1/0 1
= B2 l IB l l
=
-
C 2 1 /C l l
(5.104)
Sequence Capacitance
of Trans m issi o n
In the case of flat horizontal spacing with wire d
- a( c b b 2 -
b
1 81
Li nes
in the center, this reduces to
- Coo + cbe - cae ) 3( cso + CMO )
(5.105)
This unbalance factor i s small an d i s usually ignored. Problems
5.1.
5. 2. 5. 3. 5.4. 5.5.
5.6. 5.7. 5.S. 5.9.
Compute the positive and negative sequence capacitance to neutral for the line con figuration indicated below. Then compute the 60 Hz susceptance and the charging kVA per mile, assuming the line to operate at the nominal voltage Indicated (neglect the effect of conductor height). (a) Configuration of Figure N.1, 34.5 kV. (b) Configuration of Figure P4.2, 34.5 kV. (c) Configuration of Figure P4. 3, 69 kV. (d) Configuration of Figure P4.4 , 69 kV. (e) Configuration of Figure P4.5, 161 kV. Derive the equivalent circuit for self and mutual capacitances shown in Figure 5.6 by algebraic manipulation of (5.4 7). Compute the change in capacitance in Example 5 . 1 if the height above the ground is considered. Verify .(5.57)-(5.59) by using (5.56) as a starting point. Show that the total capacitance to neutral in the positive sequence network may be com puted by converting the mutual capacitance Ll to a Y and adding the per phase capaci tance to cgo (see section 5.4). Compute the positive and negative sequence capacitance for the circuit of Figure 4.6, using the method of section 5 . 1. Compute the zero sequence capacitance for the circuits of Figure 4.6, using the methods of section 5. 2. Repeat the computation of positive and negative sequence capacitance of problem 5.6, this time taking into account the height of the conductor above the ground. Examine the circuit of Figure 4.6 and compute the following (neglecting ground wires). (a) The matrix P of potential coefficients. (b) The matrix C of Maxwell coefficients. (Use of a digital computer is recommended for this step, but manual methods may be used. ) (c) The matrix COl2 of sequence capacitances for a transposed line.
5.10. Repeat problem 5.9 for the circuit of Figure 4.1S. Assume wire height is 70 ft. 5. 1 1. Repeat problem 5.9 for the upper circuit of Figure 5.2, ignoring ground wires.
5.12. Examine the circuit of Figure 4.6 and compute the following, including the effect of ground wires. (a) The matrix P of potential coefficients. (b) The matrix C of Maxwell coefficients. (c) The matrix COl2 of sequence capacitances for a transposed line. 5. 1 3. Repeat problem 5 . 1 2 for the circuit in Figure 4. 1S. Assume the phase wires to be 70 ft above the ground. 5. 14. Repeat problem 5 . 1 2 for the upper circuit in Figure 5.2, including the effect of ground wires. 5 . 1 5. Repeat the computations of Example 5.4 by inverting the 5 X 5 P matrix to obtain a new 5 X 5 C matrix. Explain the meaning of each term of C and label these capacitances on a sketch. 5. 16. Repeat the computations of Example 5.5, omitting the effect of the ground wires. Com pare results of the two computations and justify the change in capacitance by physical reasoning. 5.17. Compute the capacitance (a-b-c coordinate system) for a double circuit line consisting of two identical lines like that of Figure 4.6 and separated by a distance of 25 ft, assuming
1 82
5.18.
Chapter 5
the two circuits operate in parallel at 66 kV. Suggestion: Set up the P matrix by hand computation but use a digital computer to invert the matrices (see Appendix A). Compute the electrostatic unbalance factor do for the upper circuit of Figure 5.2. (a) Neglecting the ground wires. (b) Including the effect of the ground wires.
chapter
6
Seq u e n ce I m peda n ce of M a ch i n es An important problem in the determination of sequence impedances of a power system is concerned with machines. This problem is especially difficult since machines are complex devices to describe mathematically, requiring that many assumptions must be made in deriving expressions for impedances. For example , the speed, degree of saturation , linearity of the magnetic circuit, and other phenomena must be considered . Our discussion here is divided into two parts; synchronous machines and induction machines. In these devices the several circuits are coupled inductively and are therefore related by differential equa tions. Having established the appropriate equations, however, we will immediately assume that the load is constant but unbalanced . Thus we will again be con cerned with algebraic equations, and phasor notation will be used. This treatment should not be considered exhaustive by any means, and the interested reader should consult the many excellent books on the subject. 1 I . SYNCH R ONOUS MACH I N E IMPEDANCES
6.1 G eneral Considerations
A synchronous machine is sometimes called a "dynamic circuit" because it consists of circuits which are moving with respect to each other, and therefore the impedance seen by currents entering or leaving the terminals is continually changing. There are several complications here . First, there is the problem of changing flux linkages in circuits where the mutual inductances change with time (Le . , with changing rotor position ). There is also the problem of dc offset when a fault occurs. This is due to the shift in the ac envelope required, since there can be no discontinuity in the current wave of an inductive circuit. Since the normal (pre fault ) currents differ in phase by each phase current ex periences a different dc offset. The concept of constant flux linkages over the period from just prior until just after the fault also requires certain fast reactions
1200,
in the coupled circuits which generate large but rapidly decaying alternating cur rents whose magnitudes must be estimated. There is also the consideration of the machine speed. Since the machine sees a faulted condition, the load (active power) that it can deliver is changed suddenly. The prime mover requires a finite time to
sense this change in load , so the rotating mass responds by changing its speed and allowing energy to be taken from or supplied to its inertia. Thus a suddenly apI For example, see [ 1 0 , 1 1 , 1 4 , 1 9, 39, 44 and 4 5 J and examine the excellent list of references given by Kimbark [ 1 9 J . 1 83
C hapter 6
1 84
plied fault sets up a dynamic response in a machine. This response must be esti mated to permit computation of fault currents. All these questions require elaboration .
6. 1 . 1
Mach ine dynamics
First, consider the problem of the change in speed of the generator due to a sudden change in load . Consider a single machine which is supplying a passive load when a three-phase fault is applied at its terminals. Since the voltage of all three phases becomes zero, the three-phase electrical power leaving the machine suddenly becomes zero . But the input power supplied by the prime mover is the same as before the fault. Thus all the input (mechanical) power is available to accelerate the machine. Note there can be no discontinuity in the angle of the machine rotor. If this angle is (J , write (J
=
W I t + li +
1r
/2
(6.1)
where W I i s the synchronous angular frequency, li i s the torque apgle, and the constant 1r /2 is added to conform with the usual convention, as noted later. Thus the angle advances linearly with time up to the time to when the fault is applied. During this pre fault period the speed of the machine is a constant, and the torque angle li is a constant or
e
=
WI
rad/sec
t < to
After the fault occurs, the speed changes t o a new value constant,
t > to
(6.2) W
which i s usually not
(6.3)
and the angle advances at a new rate given by (6 . 1 ) as the shaft accelerates at a rate
if
=
w = S·
(6.4)
This problem of solving the differential equations of a machine following a dis turbance, even a balanced three-phase disturbance, is a formidable problem in itself and involves the solution of differential equations of the machine, the prime mover, and its control system. As explained in Chapter 1 , our goal is to simplify the solution of a faulted system to an assumed steady state condition such that algebraic equations involving phasor quantities may be used . The application of a fault near a machine is obviously not a steady state condition and requires rationalization . Actually the severity of the fault is the key to this problem. If the fault is not severe but is an unbalanced load or other permanent condition, we cotnpute the fault voltages and currents, using phasor notation , after all transients have died away and the system is in a steady ( w = a constant) condition. However, if the fault is severe (such as a short circuit), we assume the fault will be removed before the frequency has changed appreciably . We will, however, include all known ma chine circuit responses required to maintain constant flux linkages over the dis continuity . Thus we will solve a fictitious circuit problem in which we replace the generator by a Thevenin equivalent wherein both the voltage and impedance are
Seq uence I mpedance of M a chines
1 85
arbitrary quantities intended to represent the (worst) condition immediately after the fault occurs. We then proceed with an algebraic solution. The assumption that this procedure will give usable results to compute the settings of relays has been established through years of experience. 6. 1 .2 Direct cu rrent
At the instant a fault occurs, the generator currents change to new values which depend on the new value of impedance seen at the generator terminals. However, since the circuit is inductive, there cannot be a discontinuity in the current. That is, the current just prior to to equals the current just after to in each phase or, mathematically,
i"bc ( to )
= iabc ( t�)
(6.5)
The current just after to i s composed o f two components, a d c component which dies out exponentially and an ac component. The rms value of the ac component after the fault is different from that before the fault. The exact amount of dc offset depends on the exact time in the current (or voltage) wave at which the disturbance appears and on the angle of the impedance seen by the generator. A typical offset for the three phase currents is shown in Figure where the response to a 31/J fault is illustrated. Note that the sum of the dc components in the three phases is zero . Kimbark [ 19 ] shows that the amount of the dc offset can be found by taking -12 times the projection of the negative of the phasor fault currents on the real axis in the complex plane (see problem 6 . 2 ) .
6.1
6. 1 . 3
I n itial value of fault currents-the flux linkage equations
6.1,
As seen in Figure the initial value of the fault current has both a dc and an ac component, with the dc component decaying to zero in a short time and the
thfin",
UUUI'"
fYmmh
.lfUUUU
UUlUUI tfllffftft
Fig. 6. 1 . Short circuit currents of a synchronous generator. Armature currents i" . ib • and ic ; field current if. ( From Kimbark [ 1 9 ] . Used with permission. )
Chapter 6
1 86
ac component decaying slowly to a much lower steady state value. This phenom enon can be explained by the principle of constant flux linkages (see [19] ). Assume that prior to the fault the generator field is energized but the machine is It helps to visualize this situation as that of six separate but unloaded (i = mutually coupled circuits consisting of three phase windings (a, b, c), a field wind ing (F), and two equivalent damper windings (D and Q) for which we may write the flux linkage equation1
0).
LaLbce Le a Le b Le e AFAD LDaLFa LDbLFb LDeLFe AQ LQa LQb LQe
i
I I 1I 1
LaFLbF LbLaDD LbLaQQ ['.ab LcFLFF LFLCDD LFLeQQ �:iFe LDFLQF LDDLQD LQQLDQ i'DQ .
----------- ------- - ----
I
I
Wb turns
(6.6 )
The notation adopted here is t o use lowercase subscripts for stator quantities and uppercase subscripts for rotor quantities. Most of the elements of the inductance matrix are functions of the rotor angle 8 . These inductances may be computed by carefully examining Figure 6.2 which has been suggested by the IEEE as the d AXIS
DIRECTION OF ROTATION
(
q AXIS
sb
c AXIS
Fig. 6 . 2 . Reference for measurement of machine parameters.
standard definition for the several physical parameters involved. These induc tances may be computed as follows. The stator self inductances (diagonal elements) are functions of twice the angle 8 , 3
LaLba LLss LLmm - Ls s ee m , m L L L L L L, Lm . A = = =
where
2
>
+
+
+
cos 28 H
2 (8 120) = + Lm cos ( 28 + 120) H cos 2 ( 8 + 120) = + cos ( 28 - 120) H
cos
(6.7)
We use the symbol for flux linkage in accordance with the American National Standard, ANSI Y l O . 5 , 1 968. 3 Here we adopt the useful convention of designating a constant self or mutual inductance by a single subscript.
1 87
S eq uen ce I mped a n ce of M a c h i nes
The stator-to-stator mutual inductances are also functions of 28 . Lab = Lba = - Ms - Lm cos 2 (8 + 30) = - Ms + Lm cos (28 - 120) H Lbc = Lcb = - Ms - Lm cos 2 (8 - 90) = - M. + Lm cos 28 H Lca = Lac = - M. - Lm cos 2 (8 + 1 50)
=
-
M.
+ Lm cos ( 28 + 120) H
(6.8)
where 1M. I > L m . The ro tor self inductances are all constants, so we redefine these quantities to have a single sUbscript. LFF = LF H LDD = LD H
(6.9)
L QQ = L Q H The rotor mutual inductances are also constants. L FD = L DF = M R H LFQ = L Q F = 0 H LD Q = L Q D = 0 H
(6.10)
Finally, the stator-to-rotor mutual inductances are functions of the rotor position 8 . La F = L Fa = MF cos 8 H LbF = L Fb = MF cos (8 - 120) H
(6. 1 1 )
LcF = LFc = MF cos (8 + 120) H La D = L Da = MD cos 8 H LbD = Lnb = Mn cos (8 - 1 20 ) H
(6.12)
Lc D = Lnc = Mn cos (8 + 120) H La Q = L Q a = M Q sin 8 H L b Q = L Q b = M Q sin (8 - 1 20 ) H
(6.13)
L C Q = L Qc = M Q sin (8 + 1 20 ) H
Now consider the problem o f a generator with negligible load currents (compared to the fault currents ) which is to be faulted symmetrically on all three phases at t = O . At t = 0 - the currents are ia = ib = ic
:!!
0,
in =
iQ
=0
and the flux linkages are computed to be Aa
L aF
M F cos 8
Ab
LbF
Ac
LcF
MF cos (8 - 120) MF cos (8 + 120)
AF
LFF
An
LnF
MR
AQ
LQ F
0
iF =
LF
iF
(6.14)
1 88
C ha pter
00
6
But at t = 0, 0 = + 11" /2 , and the flux linkages are functions of the torque angle. Consider now the sequence of events associated with a three-phase fault on the unloaded generator. Before the fault occurs, the field with flux linkages A F produces an air gap flux CPag , leaving the N pole of the field and entering the armature, thereby establishing an armature S pole which moves with respect to the armature and produces time varying flux linkages expressed by (6.6). At the instant t = 0, the fault is applied, at which time the flux linkages are given by (6.14) and (by the principle of constant flux linkages) must remain at this value (at least for an instant) . Thus exactly the same flux CPag which entered the arma ture S pole at t = 0 - continues to do so, and this S pole remains fixed at the exact (stationary) location it occupied at t = 0 - , even though the field winding con tinues to rotate . The field N pole similarly continues to produce an air gap flux CPag which emerges from the field winding as before and is fixed with respect to the field winding. Obviously, similar statements could be made concerning the field S pole and armature N pole but they will be omitted here. We summarize the flux condition at the time of the fault as follows: 1 . At the field N pole-(a) CP O g = a constant, leaving N, and (b) N fixed with respect to the field winding. 2. At the armature S pole-(a) CPag = a constant, entering S, and (b) S fixed with respect to the armature winding. As Kimbark puts it, it is "as if the poles had stamped their images upon the armature" at t = O. Now both the armature and the field windings will react by inducing currents to maintain the flux linkages of (6.14) as described above. They do so as follows : 1 . To maintain the stationary armature poles and force the flux CPag to enter S as described requires a dc component of current flow in the armature circuits as shown in Figure 6.1 , with a different dc magnitude in each phase winding, depending on the rotor position 0 corresponding to t = o. 2. To counteract the production of armature flux linkages due to the spinning rotor field requires that alternating currents flow in the armature windings. These currents are positive sequence armature currents which are of a magnitude sufficient to hold the armature flux linkages at the prefault value specified by (6.14). The MMF produced by these currents rotates with synchronous speed, is stationary with respect to the field winding, and opposes the field MMF. The field winding reacts in turn with an increased current, the two forces balancing each other such that the flux linkages remain constant. 3. The stationary armature field, viewed from the field winding, appears as an alternating field and induces an alternating current in the field winding. Such an alternating current produces a pulsating MMF wave which is sta tionary with respect to the rotor. If one thinks of this pulsating wave as being composed of two moving MMF waves, one going forward and one backward, the backward wave will be such as to oppose the stationary armature field . The forward wave moves at twice synchronous speed with respect to the armature and induces a second harmonic current in the armature circuit. In machines with damper windings this second harmonic induction is small.
1 89
S eq uence I mpeda n ce of M a chi nes
If the flux linkages of the machine were to be constant for all time, the situa tion would remain exactly as just described . However, as noted in Figure 6.1 , the induced currents decay to new, lower values at different time constants. These time constants will be described in more detail in section 6.5. 6. 2 Positive Sequence I mpedance
The occurrence of a fault on a synchronous machine causes many different responses in the machine windings, each circuit responding to physical laws in known ways. The result is a response which is difficult to express mathematically so that sequence impedances can be defined in the usual way. For passive net works we write (6.1 5 ) where we assume that the elements o f Za b c are complex numbers and the problem is stated algebraically in phasor notation. This is not possible for a synchronous machine, as will be shown . Instead, we usually work problems either in the time domain, as in the case of stability problems, or make certain bold assumptions and use the phasor domain with the impedances assumed constant. This practice re sults in a whole family of positive sequence impedances, depending upon the exact condition under study, but only one negative and one zero sequence im pedance . Different time constants are associated with the different impedances. Since all these quantities are in common use, their definitions are reviewed here. For a more thorough treatment see Kimbark [ 1 9 ] and Prentice [46] . 6.2. 1 Park's transformation
It has been shown that the equations for a synchronous machine can be greatly simplified if all variables are transformed to a new coordinate system defined by the transformation where
P=
143 I!�fi �in 0
1 /..;2
cos (0 - 120) sin (0 - 120)
1 /..;2
�
cos (0 + 120 ) sin (0 + 120)
(6.16)
(6.1 7 )
We call this transformation the d-q transformation or Park's transformation, named for its early proponent R . H. Park [ 47 , 48] . The transformation (6. 1 7 ) is different from the one used by Park in the constant y' 2/3, used to make P orthogonal. When this transformation is used, the current id may be regarded as the cur rent in a fictitious armature winding which rotates with the field winding and with its MMF axis aligned with the field d axis. The MMF thus produced is the same as that produced by the actual phase currents flowing in their actual armature wind ings. The q axis current iq is similarly interpreted except that its fictitious wind ing is aligned with the q axis (see Fig. 6.2). The third component (usually called io) is really iao , the zero sequence current, expressed in instantaneous rather than phasor form. It is not difficult to show that the transformation P is unique and that its inverse is given by
Chapter 6
1 90
ia b c
= p - I iodq
�
where sin 8 sin (8 sin (8 +
cos 8 cos (8 cos (8
p- I
120) + 120)
(6.18)
120) 120)
(6.19)
Park's transformation is used to simplify the usual expressions for the phase volt we write an equation ages of a synchrono us machine as follows. From Figure ous generator in the synchron ed -connect Y a of for each phase-to-neutral voltage form v = - ri - X +
6.3
Vn
•
i
o -
rF
'F:[�}F {� rO
}D
'D '
'��QJLQ
n'
L bb fb
rn Ln
rb
i
b -
sb
+ Vn -
+
_Yo
I -n
e + + vb v _ _ c
y
The voltage equation for the six coupled circuits of Figure written in matrix form as follows, with = rb = =
a Vb Vc - VF - VD = 0 - VQ = 0
--- -
re r.
ra
V
rotor
n
IQ
Fig. 6 . 3 . Schematic diagram of a synchronous generator.
stator
b
ie
rQ
-
a
= -
l J
0 0 0 r 0 I 0 0 r I Il rOF 0rD 00 II 0 0 rQ
r
--
I I I
-
---
o
where
I '
I
0
-- - -- - -
ia ib ic iF iD iQ
6.3 may be
Xa Xb Xc
XF
XD XQ
+
[�Ol (6.20 ) (6.21)
6. =
- n lo be - Ln lo' be R '
191
Seq ue nce I mpedance of M a chi nes
[ �Jl , Rs
[�
This equation could be in volts or in pu. If we define
VF
VFDQ then
=
=
r U,
RR
F
=
=
rD o
ILoRs RR I�UaFDQbc - [��abcFDQ [vnJl J J (6.23)J P (6.16).
(6.20) may be written in matrix form as
abc J [vVFDQ
0
-
(6.22)
+
0
0
O-d-q
(6.23)
It is convenient to transform the a-b-c partition of to the frame of reference by the transformation as in This greatly simplifies the flux linkage equation since it removes all time varying inductances. To do this we premultiply both sides of by the transformation
(6.6)
(6.23 )
which by definition transforms the left side to
O-d-q voltages.
Thus
(6.24)
rRs RR [iiFDQ abc ] [Rs RR fLiFDiodq ] J Q J IP ol r�.�FDabc J IP � abcD l LO � L Q L �F QJ 1t (6.16 ) P abc 1t a . abc abc + bc P1t abc
The resistance term becomes
rp 01 Lo uJ Lo
0
=
0
0
(6.25)
The flux linkage term requires careful study. Applying the transformation, we write =
We can evaluate as follows. From the definition Then, taking the derivative, � Odq = it A P find �Odq = P �
= � Odq - P ��bC = � Odq - p p - l � Odq
We easily show that
p P-'
so that the last term of
=
[� � -�]
(6.26)
we write from which we
(6 . 27)
(6.27), which we shall call s, becomes s � P p-1 >' Odq =
'- : J L�w � x x
and this is recognized to be a speed voltage term. The last term of is transformed as follows.
(6.23)
(6.28)
1 92
Chapter 6
where we have defined the voltage
(6.29)
Substituting all transformed quantities into ( 6 . 2 3) we have the new voltage equations,
0 ] [iOdq ] [AOdq ] [S ] [nOdq] [VOdq ] [RS 0 0 0 VFDQ iFDQ - XFDQ =
+
RR
-
+
( 6 .30 )
We shall now show that this is much simpler than the original voltage equation etc. ) . which included many time varying coefficients I t i s convenient t o think o f the O-d-q partition o f (6.30) a s the stator voltages referred to or seen from the rotor. Since fundamental frequency ac quantities in the stator appear to the rotor to be dc quantities, we would expect these voltages to be constants under steady state operation. Because of the transformation of stator quantities to the rotor or O-d-q quantities, two remarkable things happen. First, the inductances which were so complicated and time varying in (6.6) have been transformed into constants. Second, a speed voltage term s has appeared, which adds voltage components to Vd and Vq proportional to w , the rotor angular velocity. Looking at this result another way, we have replaced a linear system with time varying coefficients by a nonlinear system (because of s) with constant coefficients. Since under most con ditions the speed w is nearly constant, the nonlinearity is of little concern . The important change is in the inductances. If we examine the transformation of the flux linkage equation, we have the following. By definition of the P transformation,
(Laa , Lab ,
=
c_ [!_: _�] -I rlo!_J�Jr�� [!-��_I_��J UJ L IFDQ] LRR J 0
r..!-i-�I [0 I UJ [LRa
I u
I
I
where the partitions of the inductance matrix are defined in (6.6). Then
[ LO La b c � �
1
( 6 .3 )
By straightforward computation of the partitions of (6 .31 ) we may show that 0
P
p- I =
d
1 93
Seq uence I mpeda n ce of M a chi nes
where
Lo = L, - 2M" Ld =
L, +
M, (3/2) Lm , Lq +
=
La
+ M. - (3/2) Lm
and these inductances are all constants. We also compute
P
LaR =
0 0 0 IIMF /fMD 0 � Lm 0 0 IIMQ
and
0 � MF 0 0 � MD 0 = L � 0 0 A MQ
LRa P - 1
Thus we may write the transformed flux linkage equation as
0 kMF 0 LF MR 0
0 0 1 Lq : 0 1 0 11 kMQ where for convenience we set k � .J 3/2. Lo 0 Ld 0 kMF kMD 0
I I
o
o
1
o
I
o
1
0 kMD 0 MR LD 0
io 0 id kMQ iq 0 iF iD 0 LQ iQ
0
------------ -------------
o
(6.32)
Observe now that every element in this inductance matrix is constant. Furthermore, the Ao equation is completely un coupled from the other equations and may be discarded when balanced conditions is constant, the time derivative are under study . Since every inductance in is easily found. of this equation , required in If the voltage is now written in expanded notation , it is instructive to see the way in which the transformed equations are coupled .
(6.30)
(6.32)
(6 .30),
r O O 11 o o r 0 1 1 o 0 r 1 ------1- -------1
o
l
I
iq rF 0 0 iF O rD 0 0
0
Chapter
1 94
-
I I
Lo
0
0
0
Ld
0
0
0
Lq
I
I I I
6 ·
io id
0
0
0
k MF
k MD
0
0
0
k MQ
·
iq ·
r-----------I 0 LF MR I I 0 LD I MR I
I
-----------
0
kMF
0
0
k MD
0
0
0
k MQ
I
-
o
- w(Lq iq + kMQ iQ) W(Ld id + k MF iF + k MD iD)
+
---
------------- -
·
iQ
LQ
0
0
·
iF iD
3rn
io
-
0
3Ln fo
o
+
o
o
o
o
o
o
(6.3 3 ) Or, written in a more compact form
r
+ 3rn
0
0
o
r
wLq r
-
I I I I
!
0
0
0
0
0
k wMQ
-- o ----- o --- o -- -- � ----- o ----- o T r o 0 rD : 0 0 0 o rQ 0 : 0 0 0 o
WLd
- kwMF
- k wMD
0
.
io .
iD
(6 .34 ) By proper choice of rotor and stator base quantities, all the foregoing equa tions may be written in exactly the same way in pu or in system quantities (volt, ohm, ampere, etc.). Equation (6.34) is very unusual as a network equation because the "resis tance " matrix is not symmetric. This is because the network is an active one which contains the controlled source terms due to speed voltages. We now investigate the meaning of the newly defined inductances of (6.32) and some related quantities.
S eq uence I mpedance of M a ch i nes
1 95
6. 2. 2 Direct axis synchronous inductance, Ld
If we apply positive sequence currents to the armature o f a synchronous ma chine with the field circuit open and the field winding rotated at synchronous speed with the axis aligned with the rotating MM F wave as shown in Figure 6 .4a,
d
Ld =
where we require that
Aa/ia Ab/ib Ac/ic =
=
=
L, + M, +
�:GcJ [�: :::;0(0 1200 120J =
Under these conditions only
d
(3/2)Lm
(6 .35)
-
../2 1 cos
+
axis current exists or
.j3
where the facto r is due to the arbitrary constant multiplier chosen to make the P transformation orthogonal. It is known that positive sequence currents flowing in the annature produce a space MMF wave which travels at synchronous speed. However, the flux pro duced by this rotating MMF wave depends upon the reluctance of the magnetic circuit. The reluctance is greatly influenced by the air gap. In cylindrical machines, therefore, the reactance seen by positive sequence currents is almost a constant irrespective of rotor position. In salient pole machines this is not true since the reluctance, and therefore the flux, depends strongly upon the relative position of the MMF wave and the protruding pole faces of the rotor. This is shown in Figures 6.4a and 6.4b. Thus we would expect the inductance of a salient pole machine to be a function of rotor position. This is true not only of the self inductance but the mutual inductance as well. The inductance will be a maximum when the rotor position is as shown in Figure 6.4a, and this maximum inductance is called Ld , the d axis synchronous inductance. The d axis synchronous reactance is defined as
wh ere w is restricted to be the sy n chro nous spee d actance is me aning le ss
WI
(6.36)
since the concept of re
o therwise. Kimbark [ 1 9 ] p oints out that (6 . 3 5 ) is valid whether instantaneous, maxi mum, or e ffe ctive values are used for the flux linkage and current This flux link age appears to the armature as a sinuso idal ly varyin g linkage and induces an EMF in quadrature with the flux and current, the ratio of this induced voltage to also being equal to Xd .
ia
•
6. 2. 3 Quadrature axis synchronous i nductance,
Lq
Proceeding as before , we may de termine th e q a is synchronous inductance by apply in g positive sequence currents to the armature with the field circuit open and the field winding turning at synchronous speed w ith the rotor q axis align ed
x
Chapter 6
1 96
$1.101 m.m.l. w.v. (sl•• dy sl,I.) �
(d )
:tq'
Fig. 6 . 4 . Flux patterns under synchronous, transient, and subtransient conditions. � From Prentice [ 46 ] . )
with the rotating MMF wave . Under these conditions the currents are shifted by 90° in phase or
[:::lJ [�� ::: �: :�)- J =
i
...[2 1 cos (8
�
-
120�
90 + 120)
This makes io and id go to zero , so that only iQ LQ
= A a / ia = A b /ib
=
A c /ic
=
L.
= ..;31 exists. +
M.
-
Then
( 3/2) Lm
(6 . 37 )
With the rotor in the position specified, the synchronous inductance will assume its lowest value, since the reluctance of the flux path is a maximum, as seen in Figure 6 .4b. Corresponding to this rotor position we also define the inductive reactance
1 97
Seq uence I mpeda n ce of M a chines
(6.38)
In cylindrical rotor machines
(6.39)
but for all machines
Xd > xQ
(6.40)
with the inequality being much more pronounced in salient pole machines. 6. 2. 4 Direct axis subtransient inductance,
d
Ld
The axis transient and subtransient inductances are defined with the field circuits shorted , Le. ,
[Va] [Y2V Y2 V Vc Y2 V
� - 120)
an d with positive sequence voltages applied suddenly at t if u ( t ) is the unit step function,
=
Vb
or
id
Vd
cos 6
cos (6 cos (6 +
120 )
= 0 to the stator.
u( t)
0
(6.41) Thus
(6.42)
vaV.
(6.43)
and only exists after t = and it jumps suddenly from zero to Similarly , is the only current component, but it must build up more slowly according to the time constant of the d axis circuit. Since all currents are zero at t = 0 - , the flux linkages are also zero and by the law of constant flux linkages must remain zero at t = At this instant we may write
0+.
(6.44)
iF and iD as a function of id with the result - MDMR) . k (LFMD - MFMR ) . 'F - - k(LDMF LF LD - MA ld ' 'D - - LFLD - M:h ld (6.45) Substituting these currents into the flux linkage equation for X d, we have X d [Ld - LFL:z_ MA (LDMJ LFMfi - 2MFMDMR)] id � L�id (6.46)
from which we may find .
=
Thus
_
.
+
_
1 98
- - LF L D
L"d L d -
Chapter
k2 _
M�
6
(LD MF2 + LF MD2
which we call the sUbtransient inductance . Often a subtransient
reactance
- 2MF MD MR)
(6.47)
is used and this is defined as (6 .48)
As illustrated in Figure 6 . 4e, very little flux is established initially in the field winding and the subtransient inductance is due largely to the damper windings. These windings have a very small time constant, and the subtransient currents die away fast leaving only the so-called transient currents. Since the air gap
flux
prior to applying the stator voltage was zero , the damper
winding currents try to maintain this no-flux condition .
The result is that the
flux path established is a high-reluctance air gap path, as in Figure 6 . 4e, and very small.
6. 2. 5 Direct axis transient inductance,
L�
is
Ld
If we examine the transient situation just described only a few cycles after the stator voltages are applied , the damper winding currents have decayed to zero and we are in the so-called transient period where induced currents in the field wind ing are important.
This situation also exists if there are no damper winding so
that the air gap flux links the field winding as shown in Figure 6 .4c. To compute the
d
axis inductance seen under this condition , we can let
in ( 6 . 4 4 ) to compute .
IF
or
-
kMF .
L
k2 M;'
L d' = L d and
(6.49)
ld
LF
( d - 4 ) 'd. =
Then from ( 6 .3 2 ) we may compute
Ad =
=
-
iD = 0
L'::.
k2 MF2
L' .
d ld
(6.50)
__
LF
(6.51 ) This transient inductance is determined under the same rotor condition as the
d
axis synchronous inductance , the only difference being in the fact that it is meas ured immediately after the sudden application of the three-phase (positive se quence ) voltages.
The sudden establishment of flux across the air gap is opposed
by establishing a current in the field winding, tending to hold
AF
at zero . Thus,
as shown in Figure 6 .4c, the only flux established is that which does field winding, and this is a small flux . Hence
The
d
q
not link the
is small but is greater than
Ouadrature axis subtransient and transient inductances,
6. 2. 6
to the
L�
L� and L�
L� .
axis subtransient and transient inductances are defined in a similar way
axis inductances except that the rotor in this case is positioned with its
q
Sequence I mpeda nce of M a ch i nes
1 99
axis opposite the spatial MMF wave of the stator as shown in Figures 6.4d and 6.4f. The flux in this case is not much different than the synchronous case for the salient pole machine pictured in Figure 6 .4 . In round rotor machines the synchronous case results in a greater flux similar to that pictured in Figure 6.4 for the axis. Thus we see the need to carefully distinguish between salient pole and round rotor machines in determining q axis inductances. If we consider the rotor as spinning with the correct alignment and positive sequence voltages applied suddenly , we have a situation similar to equations (6.42) and (6.43) except for a 90° phase lag in the applied voltages to get the proper alignment with the q axis. Thus
d
[�:J [��: �9 [uo] [ =
Uc
or
1 20 .../ 2 V sin (8 + 120
Ud
=
-
0 0
� u(t) ;J
J
y'3 Vu( t)
uq
(6 .52)
(6.53)
Since the flux linkages in the q axis are zero both before and after the voltage is applied, we compute A Q = 0 = kM Q iq + L Q iQ or . kM Q . IQ - lq LQ
(6 54) •
Then
where we define (6.55) and (6.56 ) In round rotor machines th e same effect as just described is also noted in the transient period because of the rotor iron acting much like a field winding. Thus we often assume for round rotor machines that
L q » L�
�
L�
>
L;
�
L�
(6.57)
In salient pole machines the presence of a damper winding makes a great deal of difference . Thus for salient pole machines we estimate that
with dampers : L q = L'q > L"q > L"d without dampers : L q = L'q = L q" 6.3
(6.58)
Negative Sequence Impedance
If negative sequence voltages (sequence a-c-b ) are applied to the stator wind ings of a synchronous machine with the field winding shorted and the rotor spin-
6
Chapte r
2 00
ning forward at synchronous speed , the currents in the stator see the negative se quence impedance of the machine . Mathematically , the boundary conditions are iab c
=
[�; ::: �o J + 1 20 )
V21 cos (0 - 120 )
(6 .59)
and vF = O . Then by Park's transformation we compute [ 19]
iOdQ d
id
=
iQ
[..j3..j311
1 sin 20J o
cos 20
id
(6.60)
and observe that both and are second harmonic currents. Since acts only on the axis of the rotor, it induces a second harmonic voltage in the field wind ing. If we assume that at double frequency the field reactance is much greater than its resistance, we have VF 0 = � which requires that AF be a constant, or in the steady state
rF iF + �F �F
=
( 6.61 ) But this is exactly the transient condition described in section 6.2.5. Thus (6.49) applies, and the flux linkages are found from (6 .32) to be
'71. 0
Ad AQ AF AD AQ
where as
0
=
L�id LQ iQ 0 0
(6.62)
0
L� is defined in (6.50). Then the flux linkage of phase a, Ao' is computed (6 .63) Ao = V2 I [L'd L cos + L'd - L Q cos ( 30 + 2a ) +
2
Q
0
2
Ao/io
]
which is observed to have both a fundamental and a third harmonic component, and is not a constant. Kimbark [ 19] defines the fundamental frequency component ratio as the negative sequence inductance, i.e . , by definition
L� + L Q = Ao (fundamental) io (fundamental) 2 and this definition applies for the case of no damper windings . L2
=
If damper windings exist on the rotor, we compute
(6 .64)
S eq ue n ce I mped a n ce of M a ch i nes
20 1
and
= (L� + L� ) / 2
L2
(6.65)
as
We also define the negative sequence reactance
(6.66) The relationship between X2 and the subtransient reactances x � and x� depends strongly upon the presence of damper windings, as shown in Figure 6.5 where measurements are recorded with the rotor blocked or stationary .
1 .0
0.1
�
'\
f- - .... -
QUAORATURE AX IS K q
1\
0.6
V l
i-- 0 IRECT A X IS II;
1\
�
V
N� O�IAPE�s
J d /V
-
NEGAT IVE SEQ EN
'\
'\.
0.4
-,..-
1/ /
V
8LOCK E O - ROTOR IAET HOO O. 2 0 o
COPPER OAIAPERS I
30
ANGUL A R
60
P O S I T ION
90
OF ROTOR
1 20 IN
I
150
OEGREES
I
180
Fig. 6 . 5. Relationship between subtransient and negative sequence reactances. (From Westing house Electric Corp . [ 1 4 ] . Used with permission . )
6.4 Zero Sequence I mpedance
If zero sequence currents are applied to the three stator windings, there is no rotating MMF but only a stationary pulsating field. The self inductance or re actance in this case is small and is not affected by the motion of the rotor. The pulsating field is opposed by currents induced in the rotor circuits, and very little air gap flux is established . Thus L o is very small, generally smaller than L� . The boundary conditions for this situation are iab c
=
I�� ::: :l L�/ �J
0,
["...16/0 OJ
where iF = iD = iQ = and 0 , the rotor angle , may be taken by Park's transformation
.
Io dq and
(6 . 67)
cos
as
any value . Then
cos
=
o
(6.68)
Chapte r 6
202 Table
6. 1.
Typical Synchronous Machine Constants
Water-Wheel Generators ( with dampers) t
Turbcr generators (solid rotor) Low Reactances in pu
x fl
Xd Xd xq'
xa xq" xp X2 xo *
0.95 0.92 0. 12 0. 12 0. 07 0. 10 0. 07 0. 07 0. 01
A vg.
High
1. 10 1. 08 0. 23 0.23 0. 12 0. 15 0. 14 0. 12
1.45 1.42 0.28 0.28 0. 17 0.20 0.21 0. 17 0. 10
Resistances in pu 0.0015 ra (dc) r(ac) 0. 003 0. 025 r2 Time constants in seconds
T'do T'd T;; - T� Ta
2. 8 0.4 0. 02 g. 04
5.6 1. 1 0.035 0. 16
Low
0.60 0.40 0.20 0.40 0. 13 0. 23 0. 17 0. 13 0.02
High
1. 15 0.75 0. 37 0.75 0.24 0. 34 0. 32 0. 24
1.45 1. 00 0.50 :1: 1. 00 0. 35 0.45 0.40 0.35 0. 21
Avg.
High
Low A vg. High
1.80 1. 15 0.40 1. 15 0.25 0.30 0.34 0. 24
2.20 1.40 0.60 1.40 0.38 0.43 0.45 0.37 0. 15
0.80 0.60 0.25 0.60 0.20 0. 30
1.50 0.95 0.30 0.95 0. 18 0.23 0.23 0. 17 0.03
9.5 3. 3 0.05 0. 25
6.0 1. 2 0.02 0. 1
1.20 0.90 0. 35 0.90 0. 30 0.40
1.50 1. 10 0.45 1. 10 0.40 0.50
0.25 0. 35 0.45 0. 27 0.04
0.015 0.010 0.070
0.020 0.002 0.015 0.004 0.200 0.025 5.6 1. 8 0.035 0. 15
1. 5 0.5 0.01 0.03
Synchronous Motors (general purpose)
Low
A vg.
0.005 0.003 0.008 0.003 0.045 0. 012 9. 2 1.8 0.05 0. 35
Synchronous Condensers
9. 0 11.5 2.0 2.8 0.035 0. 05 0. 17 0. 3
Source: Kimbark [ 1 9 J . Used with permission of the publisher. * x o varies from about 0 . 1 5 to 0.60 of X d , depending upon winding pitch. tFor water-wheel generators without damper windings, Xo is as listed and Xd
0.8 5 x� ,
x�
x�
X2
xq •
(Xd
xq )/2
:j: For curves showing the normal value of Xd of water-wheel-driven generators as a function of kilovolt-ampere rating and speed, see [ 50 J . =
=
=
0 0
Ad Aq =
AD
0 0
( 6.69)
0
AQ
From (6.69 ) we compute
+
Lo io
Ao
AF
=
(6.70 )
where, from (6 .32 ) (6.71 ) La Lo and is small compared to and We also define (6.72 ) Lo The actual value of Xo depends upon the pitch of the windings but usually in = A a /ia =
Lo
Ld
Lq •
Xo
=
WI
-
2Ms
is
203
S eq uen ce I mpeda nce of M a chi nes
x .
x�
the range [ 19 ) of 0.1 5 < Xo < 0.60 � Typical values of synchronous machine reactances are given in Table 6 .1 which also gives values of the various time con stants of the machine circuits defined in section 6 .5 .
6.5 Time Constants
If a 3cp fault is applied to an unloaded synchronous machine, an oscillogram of the phase current appears as in Figure 6. 1 . If we replot the current in one phase with the dc component removed, the result is shown in Figure 6 6 A careful examination of this damped exponential reveals that the current envelope has an unusually high initial value O-c and that it decays in a few cycles to a lower rate of decrement. The current then continues to decay at this lower rate until it finally reaches its steady state value, represented by the peak value O-a. The ac com ponent of current in the field circuit decays over a very long period of time, as noted in Figure 6.1 . We investigate these various decrements in greater detail since the time constants are of direct interest in faulted systems. In particular, we need to establish the approximate point in time on Figure 6.6 ( or 6.1 ) at which the circuit relays will be likely to open. This is the value of fault current we would like to compute if possible or, alternatively, we seek a value which will give results to provide relaying margins on the safe side. There are several time constants associated with the behavior noted in Figures 6.1 and 6.6. Since these are often quoted in the literature, they are reviewed here briefly. .
.
c b a
o
TIME
-
Fig. 6 .6 . The ac component of a short circuit current applied suddenly to a synchronous gen erator. ( From Elements of Power System A nalysis, by William D. Stevenson, Jr. Copyright McGraw-Hill , 1 9 6 2 . Used with permission of McGraw-Hill Book Co . )
6.5. 1
Direct axis transient open circuit time constant,
TdO
Consider a synchronous machine with no damper windings which is operating with the armature circuits open. Then a step change in the voltage applied to the field is unaffected by any other circuit, and the response is a function of the field resistance and inductance only. From ( 6. 3 4 ) with id = iD = we write
0
If iF is initially zero and we let unit step function, the result is
iF
VF
(6.73)
= ku (t), where
= � (1 rF
-
k is a constant and u ( t ) is the
e-rFtI LF) u(t )
(6.74)
Chapter
204
6
Thus the time constant of this circuit, defined
T�o = LF /rF
as
T�o, is
sec
(6.75)
Since the open circuit armature voltage varies directly with iF , this voltage changes at the same rate as the field current. Typical values of are quoted [ 1 9] as from 2 to 1 1 seconds, with 6 seconds an average value. The large size of is due to the large inductance of the field.
T� o
6.5.2
Direct axis transient short circuit time constant,
T�o
Td
A synchronous machine operating with both the field and armature circuits closed is different from the preceding case. If currents flow in both windings, each induces voltages in the other, which in turn cause current responses. If the rotor were stationary, the coupled circuits would behave as a transformer with currents of frequency f in one circuit inducing currents of this same frequency in the other. In the case of a machine, however, the frequency of induced currents is different because of the rotation of the rotor. Thus a direct current in the rotor causes a positive sequence current in the armature, whereas a direct current in the armature is associated with an alternating current in the field winding. This alternating component is clearly shown in Figure 6.1. Each of these induced currents changes at a different time constant depending upon the resistance and inductance each current sees. We call the time constant which governs the rate of change of direct current in the field the But this must correspond to the rate of change of the amplitude (or current envelope) in the armature, which we like to call the We call the time constant which governs the rate of change of the direct current in the armature the and this is the same as the time constant of the envelope of alternating currents in the field. The field or axis transient time constant depends upon the inductance seen by which in turn depends on the impedance of the armature circuit. With the armature open we have found that this time constant is T�o . If the armature is shorted, the inductance seen by iF is greatly reduced. We have already computed the required values. From (6.3 5 ) we compute the armature inductance to be Ld with the field open. With the field shorted as noted by (6.50), the armature in ductance is computed as Ld . Clearly then, shorting the armature as viewed {rom the field will change the inductance seen by the ratio Ld /Ld , and the d axis transient (short circuit) time constant T� is defined by
constant,
field time constant. direct axis transient time constant.
armature time
d
iF ,
T� = (Ld /Ld ) T�o sec
(6.76)
Kimbark [19] notes that the resistance seen in the two cases is practically the same since with the armature shorted its resistance is negligible compared to its inductance. Typically, is about 1 /4 that of or approximately 1 .5 seconds. Thus and are the two extremes for the field time constant. With the machine loaded normally or faulted through a finite impedance, the time constant will be somewhere between these extremes. If a known external inductance Le exists between the machine and the fault point, this inductance may be added to both Ld and Ld in (6.76).
T�o
6.5 . 3
T�
T�
T�o ,
Armature time constant, Ta
As explained above, the armature time constant applies to the rate of change of direct currents in the armature or to the envelope of alternating currents in the
S eq uence I mped a n ce of M a chines
205
field winding. It is equal to the ratio of armature inductance to armature re sistance under the given condition. To determine the armature inductance we note that this situation, with rotor currents of frequency t, is analogous to the case of negative sequence armature currents (and inductance L1) which produced field currents of frequency 2 t. In both cases the rotor flux linkages are constant, and the flux is largely leakage flux. Thus with alternating currents in the rotor, the stator inductance is essentially L2 , and the armature time constant Ta is given by A typical value of 6.5.4
Ta
= L2 /r sec
Ta
(6.77)
is 0. 1 5 seconds for a fault on the machine terminals.
D i rect axis subtransient time con stants,
TdO Td and
In a machine with damper windings there is an additional coupled circuit in the d axis of the rotor, namely , the damper winding. This is a low impedance winding, and currents induced in its conductors may be large but decay rapidly to zero. Viewed from the armature, the direct currents in both the field and damper windings appear to the armature as positive sequence currents whose magnitudes reflect the coupling to both rotor circuits. Thus, as shown in Figure 6.6, there are two distinct time constants apparent in the alternating current wave, one with a time constant much shorter than the other . The shorter of these time constants is due to the damper winding and is identified by double-primed notation such as Here, as in the case of no damper appli es with the armature circuits open, whereas applies = 0.125 sec ( 7. 5 cycles) and with the arm atur e shorted. Typical = 0.035 sec (2 cycles) .
T�
windings, T�o
6.5.5
Quadrature axis time consta nts,
T�. values are T�o T'qo, T'q, T�o, and T�
T�
with the q axis in a way similar to no field winding on the q axis. It is also important to recognize the significantly different structure of the q axis between salient pole and cylindrical rotor machines . Thus in salient pole ma chines , has no meaning since there is no quadrature rotor winding; but with We
can identify time constants associated
the treatment of the
d
axis except that there is
T�
damper windings present , a time constant of
�
T� T�
is used .
In cylindrical rotor machines the q axis flux has a lower reluctance path than in salient pole machines, and currents may be established in the steel which decay at various time constants depending upon the impedance of the current path. It i s
by representing the with time constants and with time constants and = 0.8 sec. = 0.035 sec and
usually observed that this situation may be approximated
armature
T�T�
6.6
current as the sum of two exponentials , one
corresponding to a reactance x
�
�
and another
corresponding to x . Typical values are Synchronous Gene rator E q u ivalent C i rcu its
T� T� =
T�T�oo T�
of a synchronous generator, we now circuit. Referring to (6.33 ) , we note first of all that the zero sequence equation is uncoupled from the others. A simple passive R -L network will satisfy this equation . The remaining five equa tions are more difficult and require further stUdy. From (6.34) we write by Having carefully defined the parameters
consider the construction of an equivalent
rearranging,
206
Chapter
6
- W �q o o
--
= -
kMF
+
kMD I
MR LD
I I I I
o
I
- -------------,-- - ----kMQ I Lq o
I I
kMQ
d
. iq iQ
LQ
•
(6.78)
These equations represent a reciprocal set of coupled circuits and coupled q circuits with controlled sources w �q and w �d as shown in Figure 6.7, where the controlled source terms are speed voltages and depend on currents in the other circuit. Note that the zero sequence circuit is completely uncoupled and is passive.
r + 3 rn
:¢ - i if F �� kM MRkM0 y i�4lr,-VD>D: I : 1 ';0) lO)--- kMo �
LO + 3 L n
rF
UF
ro
w
"
F____
+
io
Uo
_
r
i
d
f.
Ld
ud •
wXq r
ra
ua 8 o
_
J-
_ iq t.
Lq
+
-
uq
\-
w Xd
Fig. 6 . 7 . Synchronous generator equivalent circuit.
Lewis [48] shows that the d and q equivalents may be greatly simplified if a T circuit is used to represent the mutual coupling. This requires that ( 1 ) all cir cuits in the equivalent be represented in pu on the same time basis, (2) all circuits in the equivalent be represented ill pu on tne same voltage basis (base voltage), and ( 3 ) all circuits in the equivalent have the same voltampere base. If we assume that these requirements are all met, the equiValent circuit may be redrawn as in Figure 6.8 where, because of the restrictions in the choice of base quantities, the off-diagonal mutual inductances are equal. Then we define
Seq uence I m pedance of M a chines
(
r + 3r"
207 1
0
_:�
-.y.
L O + 3 L,
IF
rF
-
id
Id
f+
uF +
rQ U q sO +
Fig. 6 . 8 .
lq
IQ
iQ
wXq
+
_ Iq
�
t+
Mq
r
iq + i Q
+ w),
r:
d
Equivalent T circuits for a synchronous generator.
(6.79) M q � kM Q pu We also define the leakage inductances, designated by script f , according to the equation pu,
Md � kMF = kMD = MR
-
Md = Ld - fd = LF fF = LD - fD Mq = L q - f q = LQ - f Q pu
pu (6.80 )
The equivalent circuit of Figure 6.8 is often used computation because of its simplicity. It is also convenient for visualizing transient and subtransient con ditions. For example, the subtransient impedance and time constant in the d and q axes is that seen looking into the d and q circuits from the terminals on the right and with all voltage sources shorted . The transient condition is determined in the same way but with the damper circuits open. The concept of leakage inductance is also useful since these inductances are linear. Thus in Figure 6.8 only the mutual inductance Md would normally be come saturated and it should be considered nonlinear. 6.7
in
Phasor Diagram of a Synchronous Generator
In fault studies we prefer to work with phasor quantities since phasor solu tions require less work than solving the nonlinear differential equations for a synchronous machine. But phasors usually are assumed to represent a steady state condition. We therefore derive the phasor diagram for a generator operating in the steady state . We may then examine this diagram to determine the operat ing condition immediately following a fault. A three-phase fault or load may be studied by making the appropriate im pedance connection to the positive sequence network only . But we may also
C hapter 6
208
view anyIn other fault condition this way, toaddibengadded a faulmayt impedance at the fauloft point. the general case the impedance be a combination In alenti l casesrelytheby posi tposi ive tsequence cur the negative andin thezero generator sequence arenetworks. rents flowing determined the i v e sequence network and this fault impedance if weposineglect the loadnetwork. currents.ThisThuspermits we limitus our consideration at this point to the t i v e sequence toancederitheve diagram only onederived phasorherediagram. Wey recognize thatposifortivesome kindsquantities. of unbal may appl only to the sequence If we assume positive sequence, balanced currents, we may write = �:: : : : :J cos and = Then by direct application of the transformation we compute
[;:l [:: �l
i
e
W i t + 0 + 90.
) - 120) (w 1 t + 1/J + 1 20 )
(6 .82 )
P
- I/J ) (0 - I/J )
lq
wiEquation th the result
(6.8 1 )
[::J [- sincos�o J may be manipulated to define the angle in terms of and cos _ - sin cos = ..j31
iq
J
(6.82 )
id
I/J
I/J -
.
id
0
0 + iq ..j3f
_ iq -
0 + id va l
0
sm i/J sin cos These values may be substituted into to write = cos cos and, sincetransformation the phase currents arewebalacoul nced,d writeand are known also. Applying the phasor immediately as a phasor quantitmight y. Before doing this,tohowever, which be convenient use. we digress to examine the reference systems In synchronous machines thereshallarecalatl theleast two convenient andframe. widely4 used frames of reference. One we reference The secondUsingis thean arbi d-q reference frame discussed in the preceding sections. trary reference frame we can write a phasor current as e = cos sin ia
( 6.81 )
v'273 [ id
(w I t + 0 + 90) + iq
(6.84)
(w I t + 0 )]
ic (6.84)
ib
( 1 .50),
6.7 . 1
(6.83)
Phasor frames of reference for synchronous machines
arbitrary
�
-
la
�
�
lar + lax = la I/J + j Ia = fa
j
la
tP
rfJ
= lar + j lax
(6.85)
4 The "arbitrary reference" used here i s a phasor reference and should n o t be confused with the arbitrary reference frame defined by Krause and Thomas [49 ] which is a rotating reference that revolves at an arbitrary angular velocity and is used in the study of induction motors.
2 09
Seq ue nce I mpedance of M a chines
(a )
,..,
1 ax REF
d 10
( b)
Id
Fig. 6.9.
q - REF
Two frames of reference for phasor quantities : (a) arbitrary reference , (b) d-q refer ence.
where thisquantity phasorsymbol is pictured in Figurethat6.9thisa. Note thatty iwes based use theon titheldearbitrary over the phasor to indicate quanti frame. Usualbetween ly such attenti on to detaisystems. l is unnecessary, but8 5)herethe wecompo need reference tonentsclearly di s tinguish two reference From (6. of _ are (6.86) NoteThecareful ly that the quantity (without the tilde) is a scalar quantity. d-q frame of reference is pictured in Figure 6. 9b. Here we use a bar over Thus (6.87) Then (6.88) is nowtypossible phasorbasisexpressed one reference frame to the sameItquanti expressedto onrelathete asecond by the informulas (6 .89) From this formula we compute cos sin ) cos sin Ia
Ia
the phasor symbol to indicate that this phasor is based on the d-q reference frame.
Iar + j Iax = (lq
[5] .
{)
-
Id
{)
+ j(Id
{)
+ Iq
{)
5 For a more detailed discussion of the subject of transformation of bases, see Hueslman
210
Chapter
6
Fig. 6 . 1 0 . 10 components from two reference frames.
or = cos sin = cos sin These quantities are shown in Figure Fromty in termsweof have the generator current expressed as aequivalents time domainof quanti the magni t udes and i q , whi c h are rotor the instantaneousmagnitudes stator currents. rotor-referenced as We now define rms stator equivalents of these = 1v'3 , = 1v'3 rms A or rms pu wherearitheses asv'3a scale comesfactor fromforthe wayand the Using transformatiweon wriwastedefinedasand there fore = cos cos I t Applying the phasor transformation we compute e i6 = )ei6 = la e l6 L = e J(6 + 90) as giThe ven bymagnitude of should be the same in any frame of reference. Thus lo r
6.7.2
fj - Id
lq
fj ,
lax
(6.84 )
id
ia
lq
id
id
y'2 Id
ia
(6.9 1 )
iq
iq •
P
(6.91 ),
+ fj + 90) + y'2 lq
(w
(6.90)
S
6.10.
Phasor transformation of generator quantities
Id
fj + lq
Id
(6.84)
(w I t + fj )
( 1 .50),
+ lq
Id
(lq + j ld
(6.92)
(6.89) .
10
1a
= II I = II I = v'1q2 a
a
+ IJ
= ..j3 � V'I2+I2 'Ii 'd
(6.93)
T
la ,
couldspeci alsofibeed.computed from the components of but these valTheuThe esmagnitude areforegoing not usually d-q frame permi t s us to represent a machine i n the convenient andin thethensystem. easily transfer to thea more general reference which is used for d-q voltages and currents, we may by allof reference machines Knowing write these quantities as phasors. (6.92)
S eq uence I mpeda n ce of M a chi nes 6.7.3
21 1
The steady state phasor d iagram
To derive the steady state phasor diagram, we write the d-q voltage equations for the balanced case and then transform them to the complex domain. Since we are considering the balanced , positive sequence case, the zero sequence volt age and current are both zero , i.e . ,
V o = io
=0
(6.94)
Also, since the system is in steady state, the speed is constant and the damper currents and all current derivatives are zero, i.e.,
w = WI iD = iQ = 0 id = iq = iF = iD = iQ = 0 axis voltage i s Vd = - rid - wLq iq - k wM Q iQ - �d
From (6 .34) the d porating (6.95), this voltage may be written as
(6.9 5 ) and, incor
Vd = - rid - W I L q iq
(6.96)
Also from (6.34 ) the q axis voltage is
V q = - riq + WLd id + kwMF iF + kwMD iD - �q
which may be written as
Vq = - riq + W I Ld id + kw l MF iF
(6.97 )
IF = iF/V3
(6.98 )
We now define an rms stator equivalent of the rotor field current to be Then, using this definition and definition (6.9 1 ) of d and q axis currents, we de fine rms equivalent d and q axis voltages by dividing (6.96) and (6.97 ) by V3, with the result
Here we recognize mutual reactance
Vd = vd /V3 = - rId - xq Iq Vq = vq /V3 = - rIq + xd Id + k XmFIF the previously defined reactances X d and Xq
(6.99 ) and define the (6.100)
Using (6 .92 ), we may compute the rms phase a terminal voltage on an arbitrary reference basis to be Va = ( Vq + j Vd ) e il) = [ (- rIq + xd Id + kx mF IF ) + j(- rId - Xq Iq )] e il)
We recognize the quantity in brackets to be Va on a d-q basis, according to (6.89). Thus using the bar notation for this basis and rearranging, we have
Va
= =
Vq + j Vd = - r(Iq + jId ) - jxq Iq + xd Id + kXmFIF - rIa - jXq Iq + xd Id + kXmF IF
(6.101 )
It is convenient to define a field excitation voltage
EF = Eq + Ed = Eq + jEd =
kXmFIF
(6.102)
212
C hapter 6
such that wriThetereason asis zero is that there is no field winding on the q axis. Then we may whicyhingis stil where equationwe ofseemithatxed phasor and scalar notation. We clarify this by appl (6.103)
Ed
(6.101 )
(6.104)
an
(6.88),
Iq
(6.104 )
= lq ,
Id
=
-
j 1d
Then in d-q phasor notation becomes = 6 or,reference. multiplying by e 1 , we can transform this equation to an arbitrary frame of Equationin the steadydefines theTwophasortypical diagramphasorof adiagrams, synchronousonemachine when operating state. for a leading power factor and one for a lagging power factor, are shown in Figure In Eq
Va + rIa + j xq lq + j Xd 1d
(6.105)
(6.105)
6.1 1 .
d
d
(b) Fig . 6 . 1 1 . Phasor diagram of a synchronous generator with arbitrary reference : (a) leading power factor, (b) lagging power factor.
S eq uence I mped a n ce of M a ch i nes
213
either case we can define power factor = cos where Figure is definedy which to be Va leads 10 - Th voltage V",q shown in Fp
e
=
= r/> v
angle b
6.11
V",q
Eq
=
r/>l '
e
jXd Id �
�
�
(6.106)
e
=
-
(6.107 )
ThisInvolmany tage willreferences be neededtheinphasor a laterdidevelopment. a gram of a synchronous generator is drawn inaddeda sland ightlsubtracted y differentfrom way than shown Figure If a quantity jXq Id is we canin develop Va rIa jXq 1a j(Xd - x q) Id This form is interesting because the last term is approximately zero for phasor round rotor generator since Xd Xq for these machines (see Table diagram which combines and is shown in Figure Eq -
6 .11 .
(6.105),
=
-
+
......
......
+
+
......
(6.108)
a
�
6 .1 ) . A 6.12.
(6.108 )
(6.105 )
d
Fig. 6 . 1 2 . Another form of phasor diagram of a sy nchronous generator.
(6.105)
(6.108)
The condition described in and or by the phasor diagrams of erator Figurecircuitandreduces toUnder describedequivalent above, thecircuiequits ofvalent thattheof steady Figure state condition where all Figures 6 . 1 1 and 6 . 1 2 can also be visualized in terms of the synchronous gen
6.8.
6.7
6.13
Id
-
r
Iq
-
+
Eq-t'3 0>1 ). d Fig. 6 . 1 3 . The d·q equivalent circuits i n the steady state case.
214
Cha pter
6
v'3
voltages andvaricurrents have been divided by to use the equivalent rms stator (uppercase) a bl e s as in (6. 9 9). d and equations to obtain (6. 1 08), we can draw the combientningcircuithet shown simpleAfterequival in Figure 6. 1 4, where an EMF due to saliency of q
Io
r
-
+
t+
Vo
Eq
tr+:----o
j ( X d - Xq ) I d
Fig. 6 . 1 4 . Equivalent circuit of a synchronous generator.
jti(Xveld y small, thiis shown asis asmalseparate volalways tage. inSinphase ce thewiquantity ibys relthea s EMF l and is t h as indicated phasor diagram of Figure 6. 1 2. -
6.8
x q ) Id
(Xd - Xq)
Eq
Subtransient Phasor Diagram and Equivalent Circuit
In fault studiestheweincidence are usualoflytheinterested inwetheexamine subtransi eassumptions nt period imme diately following fault. If the used ibetween n developing the steady state phasor diagram, we observe certain di f ferences the statement steady stateof theand steady subtransient conditions.In theEquation (6. 9 5)period is a mathematical state condition. subtransient wecannot can assume usually that assumethe asdamper beforewinding that thecurrents speed isarenearl yasconstant. However, we zero in (6. 9 5). The third condition, setting the current derivatives to zero, iswords, equivaltheentfluxto linkages are the This sameisimmedi aassumption tely followingto make the faultfor asthetheysubtransient were just period before andthe faul t occurred. a good follows from Lenz'ofs thelaw volin tthat the currents(6.do3 4)notforchange instantaneousl ydepends . The solution age equations a transient condition upon thechinitial condifaulttionin(atour probl 0-) and the nature of the change or driving func tion whi is the ete solution also requiofrestime.the solution of a torque or inertia equatioenm.to fincompl d the speed as a function (0, written complete final solution state steady solution transient solution Ine .,ourthecasesolution we areat usually satisfi ednotif aswedifficul can findt asonly the forsubtransient solution, 0+. This is sol v ing the compl ete solu i.tion as a function of ti m e, but i t still involves a considerati o n of the same equations. The desired subtransient solution may be written in words as five subtransient at t 0+solution initial condition delta condition or change(6.109) stating that the rate of change of flux linkages is zero. In other
t
=
A
w
F,
Even if we assume balanced conditions and constant speed, we are faced with the simultaneous solution of six equations
solution may be
as
d, q ,
D, and Q). Then the complete
+ complimentary or
=
t=
=
at t = O -
+
from t = O - to t = O+
S eq ue nce I mpedance of M a chines
21 5
The initial or pre fault condition is the steady state solution discussed in the pre vious section , where the voltages and currents depend on the machine load. These prefault (t = 0 - ) conditions are : are usually nonzero and are a function of load. All current derivatives are zero . All flux linkage derivatives are zero due to item 2 . All damper currents are zero, iD = iQ = O . The field current i s constant, iF = a constant . All zero sequence quantities are zero . 7 . The speed is constant, W = W I ' 1. 2. 3. 4. 5. 6.
Vd , vq , id , iq
Immediately after the fault (t = 0+ ) we assume : 1 . The fault is represented by a step change in id and iq . 2 . The current derivatives are nonzero (an impulse) at t = 0 but are zero at t 0+ . 3. The flux linkage derivatives are zero at t 0+ . 4 . The damper currents are nonzero at t = 0+ . 5 . The field current changes to a new value at t = 0+ . 6. The zero sequence quantities are zero . The speed is constant, W = W I ' =
=
7.
We shall represent all variables by denoting the initial condition plus the change as indicated in (6.109), using the notation (6.110) where
io = initial condition , t 0it. = change from t = 0 - to t = 0+ =
(6.111)
At t = 0+ we may write the flux linkages from (6.32) in matrix form as " = Li, or
A o + �t.
= L(io + it. )
(6 . 1 1 2 )
Then the initial condition establishes the prefault flux linkages as
Ld id O + kMF iF O Ad O Lq iq o Aq O kMF id O + LF iF O AFO kMD id O + MR iFO A DO (6.1 1 3 ) kMQ iq o AQ O Now, following a change in currents idt. and iqt. the currents in the coupled cir cuits will adjust themselves so that the flux linkages remain constant, in the F, D , an d Q circuits, i.e . , from ( 6.32) and ( 6 . 1 1 2 ) Adt. A q t. A Ft. A D t. A Q t.
Ld idt. + kMF iFt. + kMD iDt. L q iqt. + kMQ iQt. = kMF idt. + LF iFt. + MR iD t. kMD idt. + MR iF t. + LD iDt. kMQ iqt. + L Q iQt.
? ?
0 0 0
( 6 .1 1 4)
216
Chapter
6
(6.Thi1s14)equation sets up a constraint among the current variables, we have LFiFtl. + MR iDI1 = - kMF idl1 , MR iFI1 + LD iDI1 = - kMD idl1 which we can iFI1 iDI1 it the result . - - kLDMF kMDMR idl1 lFI1 LFLD MA kMFMR id l1 (6.1 15) 'D I1 - - kLFMD LFLD MA Also, from the (6. 1 14) we compute . = - kML Q • (6. 1 16) 'QI1 Q Currents from (6.1 15) (6.116) now n (6.1 14) 2 J 2 ) MFMDMR ] '. L'" d (6. 1 17) dl1 -- [Ld - k (LD Mi LLFFM 2 LD - MR where (6.47) computeL; - (L - kL2M3 ) . (:;= L'" (6. 1 18) Q and circuit flux linkages terms we write from (6.34) Vd = VdO Vdl1 - r( ido i l1 ) - k WI MQ ( Q o - ),d = 0+, ),d = O. (Sinhce ) iQ O O. (6.116). Making these VdO = -ridO ' - W I L'" Vdl1 - - 'd - WI L ' W lkL2QM§ . -- -ndl1 ; - rid W- L;lL i I o (6. 1 19) (6. 1 20) (6.96) edo = 0 (6.34) for
solve for
and
from
w h
+
_
_
.
+
_
_
last equation in
Iq l1
and
may
b e used i
+
�
is defined in
t o compute
dl1 �
dl
11
to be the quantity in the brackets. Similarly, we
'\
"' ql1 -
q
Iql1
q lq l1
Thus we have identified the changes in d q in of the change in currents. The voltage equations may also be written in terms of initial-plus-change no tation. For the d axis voltage
+ idl1) - W I Lq (iq o + q
=
+
i
+ iQ I1)
this equation must represent the solution at t we note that W y ? Also, b y definition the initial condition i s steady state s o that Finally, we note that iQ I1 may be written in terms of iql1 from equation changes, we write w I L q iq o and - r1 l1
q lql1 +
or, adding the two equations, Vd = Adding and subtracting a quantity
Iql1
w
=
q lq l1
q q o - w 1 L iq l1 . iq gives the result
where we define
From it is apparent that since this component does not exist in the steady state. For the q axis we similarly compute from
Vq
=
VqO + Vq&
+
=
-r(iqo + iq&) + ([)lLd(idO + id&) + kw 1 MF(iFO + in)
kw1MD(iDO + iD&) - lq
217
S eq uence I mpedance of M a ch i nes
In this case Then
where from
iFO = a constant, iDO = 0, �q = 0, and iFtJ. and iD tJ. are given by (6. 1 15).
(6. 1 21) (6. 1 22) (6. 1 23) (6.1 24) (6. 1 25) (6. 1 26) (6. 1 03) (6. 1 27) (6. 1 28)
(6. 1 03)
eqo = k w1MFiFO Making substitutions of iFtJ. and iD tJ., we compute Vq tJ. = - riqtJ. + wl L; idtJ. Equations
and
(6. 1 21) (6. 1 23)
are combined to write
where we define and
e�
It is interesting to note that the definitions for and e; parallel similar def initions for the steady state condition. In particular, we compute from and - eq o = W 1 L or
(6. 1 26) e;
( d - L;) idO '
We may also compute
ed O
Vxq = eq o + Xd idO = e; + x; idO
-
e; =
Wl
(Lq - L; ) iq o or
ed O
Vd
since = O. To construct a phasor diagram for the subtransient case, we divide the and equations by v'3 to rewrite the equations as rms stator equivalents. Thus
Vq
(6.1 29) (6.99), (6. 1 30) (6. 1 31) (6. 1 32)
where we note the striking similarity to the steady state equations the only difference being the addition of the double primes (" and the appearance of the voltage E� . As before , we compute on a d-q reference,
)
where we define E"
Equation
(6. 1 30)
= E; + jE;
may be changed t o the form E"
= Va + rIa + jx;fa + j (x; - x; )
Iq = E; + jE;
This form is convenient since the quantity (x; - x � ) is positive o n both round rotor and salient pole machines but is usually quite small. The phasor diagram for the subtransient condition is constructed according to in Figure where the condition of Figure is used as an initial condition. Several observations are in order concerning this important phasor diagram . We note that the new (fault) current is larger than the initial current, and it
(6. 1 32)
6. 1 5
6. 1 2
Chapte r 6
218
d
Iq
Vd
Eq
Vq Vxq
Vx d
Eq O
q
J X dIdo
Fig. 6 . 1 5 . Subtransient phasor diagram of a synchronous generator.
lags the terminal voltage by a larger angle () , which is nearly 90° for a fault on the machine terminals. This means that Iq is small and Id is large. Since the speed is assumed constant, 8 has not changed. An importanfpoint on the diagram is the encircled point located at
(6.1 33) But from (6.127) and (6.128) VX q
+ Vxd = (Eq o - jXd 1d o ) + (- xq lq o) = (E� - jx;ld o ) + (E� - x� lqo ) (6. 1 34)
or this voltage is a constant and depends only upon the pre fault or initial condi tion. Therefore, the location of the circled point is unchanged from t = 0- to t = 0+ and forms a kind of pivot about which the other phasors change. Thus VX q + Vx d = Va + rIa = Va(O ) + rla( o )
= a constant
(6. 1 35)
Referring again to the phasor diagram, we note that the phasor j(x; - x� ) lq is very small since from Table 6.1 (x; - x � ) is small, and Iq is usually small under fault conditions. This quantity might be neglected as an approximation, or set We observe that
J.(X q" - X d" ) Iq
-
=
E; » E �
0
(6.136) (6.137)
or approximately,
(6.138 ) This is quite apparent from the phasor diagram but could also b e concluded from the definitions of E; and E� , where Eq o will be the dominant component. If approximations (6.136) and (6.1 38) are allowed , the voltage equation (6.1 32) becomes
(6.1 39) Both the exact equation ( 6.1 32) and the approximate equation (6.139 ) can be
Seq uence I mpedance
of M a chines
219
represented equivalfroment(6.circuits, n Figure E" areditions. constantsbysince 1 25) andas (6.shown 1 26) ithey depend6. 1 6.onlyNoteon that the initial conE; circuitneglected of Figuresin6.ce1 6bit isis small. widely Weusedmayfor thifaulnkt calcul the The resistance of thisations, circuitoften as a wisub-th � Il; and
r
-
(0 ) +
(b )
q
E"
Il"d
+
j (x'q -Ild U q r
I"
�
+ v. B
,�
Fig. 6 . 16 . Equivalent circuits of the subtransient condition : (a) exact equivalent, ( b ) approxi mate equivalent.
transi e nt Thevenin equivalent since it is clearl y a constant vol t age behind a "con stant"ofimpedance. Obviously, is valid only at the instant but this is the time greatest interest in faultitcalculations. t=
6.9
0+ ,
Armature Current E nvelope
It is convenient toFigures have a6.mathematical expression foris thetheenvelope of the armature currents of 1 and 6. 6 . The envelope sum of several with ofitstheowncurrent time asconstant, whichof time. combine to determine the peak valterms, uThe e oreach envelope a function final valuecomponent of the current inenvelope Figure andhasmaya bepeakplotted value alone as isthecalled the synchronous of the line a-d. Since the current of interest is a fault current, Iq wil be small and la (6. 1 40) Making this substitution into (6. 1 08), we compute for r negligible, (6. 1 41) Va orfora threep hase faul t on the machine terminals (Va 0) the current magnitude Id shown in Figure 6.1 7 is given by (6. 1 42) Superimposed on the synchronQus componentinofFigure current6.17.is Equations the transientfor component which is given by the quantity theof section transient6. 8condition have not beennotation derivedisherereplaced but arewithidentical toprimes. the results i f the double-prime single The result for a 31P fault wi t h negligible resistance can be computed from (6. 1 39), with the result O-a . It
6.6
�
Eq
=
Id
+ jXd 1d
=
(i� - id )
220
Chapter
J2 I d
---
-
Id
6
- - ��----
o �-----.- t
Fig. 6 . 1 7 . The armature current ac envelope, excluding id e '
- The transientcomponent. current envelope is thely locus intoFigure and includes the synchronous It is usual convenient wri t e the transient current envelope as and The this envelope decayscomponent with time ofconstant T tois seconds). subtransient faul t current superimposed ongenerator the pre viterminals ously defined components. Its i n i t i a l rms value for a fault on the is found from with neglected. Thus - Id" Xd The subtransient component envelope is written as e i� - i;' (I� and thicomponent s component decays theveryinfast wirmsth current a time ifconstant ofnoT�dc component. or cycles). ThisThe determines i t i a l there is the current envelope is the sum of the synchronous, transient,acandcomponent subtransientof components I,a
0.
Id' - Eq' /Xd'
( 6.1 43)
6.6
b-d
i� - id = v'2(I�
Ia"
2
= Eq" / "
0.
+ (Id
(6.1 44)
3f/>
= V2
i. e ., la c = ienv /...[2.
� (1
r
( 6.139 )
ienv = ...[2 [Id
Id ) e-tl Td
-
- I� )
( 6.145)
-
tlTd
(6.146)
(2
Id ) e- t lTd + ( 1;
I� ) e- tlT3 ]
-
3
(6.147)
or the rms value ofInth�termscurrent as machine a functionreactances of time theis rms value timesofthisalternati amount,ng of the and a function that of time may be estimorated from wicurrent th theasassumption [1:. (� 1:.) ( � �) ] Usually is assumed toalsobehasin thea dcrangecomponent of towhichpudepends for faulont computations. The total current the switching angle (radians) given by (see cos Eq
la c !?:! Eq
a
Xd
�
E�
+
Xd
Eq
_
�
-
(6. 142), (6.143),
E;
Xd
e-tlTd +
Xd
1.0
1.10
[ 19] )
Id e
-
1 /...[2
...[2 Eq
"
Xd
a -t lT a e
-
Xd
e- tIT a
(6.145 )
(6.1 48)
(6.1 4 9)
Seq uence
I mpedance of M a chines
22 1
Then the effective value of armature current is
+ lJc ) I /2
(6.1 50 )
lm ax = v'3 Eq Ix�
(6.1 51 )
I = (l;c
which has a maximum value at
t=0
corresponding to a = 0 of
Values of Xd , x� , x � , T� , � , and To may all b e found from an oscillograph of armature currents similar to Figure 6.1 . The technique for doing this is explained fully in [ 19] and will not be repeated here. These values, expressed in pu, tend to be nearly constant for a particular type of machine and are often tabulated for use in cases where the actual machine parameters are unavailable. Reference values are given in Table 6 . 1 . Since our motivation i s t o perform fault calculations, w e are often concerned with fault analysis under extremum conditions ; i.e. , we are trying to determine either the maximum or minimum value of fault current seen by a particular relay or protective device. For maximum current we usually take the curre nt l� as the rms value of fault curre nt, or we consider that the generator has reactance x� in the positive sequence network, with X 2 and X o in the other networks. For other than maximum fault conditions the value of reactance used may be changed to a value which will give the correct current at the time the protective device operates. Thus the time may be substituted into (6. 148), and the current and hence the equivalent reactance found . 6. 1 0 Momentary Currents
The equivalent circuit developed in section 6.8 will permit the solution of the subtransient current l� corresponding to the highest point in the current envelope of Figure 6 . 1 7 . This current is the symmetrical current, i.e. , it includes no dc off set at all and corresponds to the highest rms value of Figure 6.6. If our purpose in computing the fault current is the selection of a circuit breaker to interrupt l� either at the generator or some other location, we should include the dc offset. Thus the total effective value of the current must be computed as in (6. 1 50 ) . If this is done, for the worst case the result is that of (6.1 5 1 ) for 100% dc offset in a given phase , or
( 6. 1 52)
where l� sy m = the symmetrical part of the total current. Actually , the current to be interrupted by the circuit breaker will never be this great since this would re quire zero fault detection time (relay time) and zero breaker operating time. The IEEE, recognizing that some consideration should be given to the inclu sion of dc offset in the selection of breakers, has formulated a recommended method for computing the value of current to be used in these computations. The computed current, called the momentary current, depends upon the time after initiation of the fault at which the circuit will be likely to be interrupted. Thus the natural decrement of the dc offset, which occurs at time constant To, is ac counted for approximately and provides a reasonable compromise between the limits of l� S ym and 1 .732 l� Sym ' In the recommended method of computation, the momentary duty of the breaker is based upon the symmetrical current l;sym, with multipliers to be applied according to the breaker speed . These multipliers are given in Table 6.2 which is taken from [ 5 1 ] . These multipliers determine the so-called "interrupting duty " of the circuit breaker. Another rating called the
222
Chapter
Table 6. 2.
6
Current Multiplying Factors for Computing Cireuit Breaker Interrupting Rating
Breaker Speed or Location
IdsYm Multiplier M
8 cycle or slower
1. 0 1. 1 1. 2 1.4
5 cycle
3 cycle 2 cycle Located on generator bus If sym fault MV A > 500
add 0. 1 to multiplier add 0. 1 to multiplier
"momentary duty" is found by multiplying the symmetrical current by 1 .6, and this may be checked against a published breaker momentary rating. (The present practice is to use only the symmetrical interrupting duty, and breakers are so rated. ) In summary the two breaker ratings which must be checked are
1.6 I� sym interrupting rating > I� momentary rating >
M
where
M
(6.153)
is found from Table 6.2. II.
6. 1 1
sym
I N DUCT I ON MOTOR IMPEDANCES
General Considerations
The striking difference between induction and synchronous machines, insofar as their response to faults is concerned, is in the method of machine excitation. A synchronous machine obtains its excitation from a separate dc source which is virtually unaffected by the fault. Thus as the prime mover continues to drive the synchronous generator, excited at its prefault level, it responds by forcing large transient currents toward the fault. The induction machine, on the other hand, receives its excitation from the line. If the line voltage drops, the machine .excita tion is reduced and its ability to drive the mechanical load is greatly impaired. If a 3t/> fault occurs at the induction motor terminals, the excitation is completely lost; but because of the need for constant flux linkages, the machine's residual excitation will force currents into the fault for one or two cycles. During these first few cycles following a fault the contribution of induction motors to the total fault current should not be neglected. However, it is somewhat unusual to find an induction motor large enough to make a significant difference in the total current. For example, if the base MV A is chosen to be close to the size of the average synchronous generator, an induction motor would have to be larger than about of base MVA to be of any consequence. When induction motors are included in the computation, a subtransient reactance of about 25% (on the motor base) is often used. The transient reactance is infinite.
1%
6. 1 2
I nduction Motor Equivalent Circuit
The induction motor is usually represented by a "transformer equivalent," i.e. , a T equivalent where separate series branches represent the stator (primary ) circuit and the rotor (secondary) circuit, with a shunt branch to represent iron losses and excitation. At standstill or locked rotor the induction machine is indeed a trans-
223
S eq uence I mpeda n ce of M a chi nes
the secon ofFigure impedance theofequivalent however, turning, With cithercuirotor former.or rotor in noted as s slip the function a be to seen is t dary If the rotor induced is Er at standstill, it is sEr when the rotor is turning6.18.at EMF
X.
R. •
Val
r - RsRr
Xr
lal
-
I rl Rc
}(I- S) r -•
Fig. 6 . 1 8 . Positive sequence induction motor equivalent circuit.
inducedof of slipthe. Since a function isstator inductance rotor apparent , the frequency Similarlyhave slip s. currents reactance frequency, the is f where sf rotor the rotor or reactance at standstill. Then, we com at anywillslipbes, sXIr'r where pute sEr, /(XRrr is thejsXr)rotor (6.154) Ir' = --�-(R r/s)Er' jXr Thus the rotor impedance is (6.155) Zrl (R r/s) jXr as noted in Figure 6.18. We usually write (6.155) as Zrl = (Rr j Xr) 1 s- s Rr a function and is not standstill impedance term is thethe shaft where is of s. The delivered the power load toatwhich secondthetermfirstrepresents =
+
+
+
�
+
+
--
(6.156)
The shaft or mechanical torque is (6.157) ml PWml W I P(1ml- s) Nm/phase I;, Rr Nm/phase (6.158) where rad/secin rad/sec speed inspeed synchronous w,Ws = shaft slip in pu of terms in torque base the pu, wew ,select torque is expressed mechanical Ifbasethepower 6.157) and baseinspeed (base voltamperes) . Then from ( /w,(l - s ) P1 m1u- s 1;1sR r pu (6.159) pu Pm ,SB/W, which has units of pu power/pu slip or pu torque. Some authors state pu in T
=
=
= --
=
=
Tm ' =
= -- = --
Tm l
Chapte r
224
6
PER U N I T TOR O U E
_I 2
Fig. 6 . 1 9 .
BACKWARD
0
FORWARD
0
-I
•
Torque of an induction motor as a function of speed or slip. ( From Kimbark [ 19 ) . Used with permission. )
"synchronous watts" (see [ 1 9 ] ) . The average torque-speed characteristic o f an induction motor is shown in Figure 6.19. If negative sequence voltages are applied to the induction motor, a revolving MMF wave is established in the machine air gap which is rotating backwards, or with a slip of 2.0 pu. Then the slip of the roto r with respect to the negative sequence field is 2 8 when the rotor is moving with forward rotation. Since 8 is small, the approximation is sometimes made that this negative sequence slip is 2.0 rather than 2 8. If 82 is the negative sequence slip, 82 = 8 == 2. The equivalent circuit for negative sequence currents is the same as that of Figure 6.18 with 8 replaced by 8 as shown in Figure 6.20. The mechanical power
-
-
2-
2
Rc
Rr
-( l-s ) (2- s)
Fig . 6 . 20 . Negative sequence induction motor equivalent circuit.
Pm2 - -Ir2 R r -_ Tm 2 -- 1';2 R r
associated with the negative sequence rotor current (1 (2
_
8) 8)
Ir2
is (6.160)
W/phase
the negative sign indicating that this would cause a retarding torque w d 2 - 8)
Tm2
Nm/phase
where (6.161 )
Then the net torque per phase is
Tm Tm l Tm2 = Rr (1;1 _ 1;2 ) =
+
WI
s
2- 8
N m/phase
(6.162)
which is a smaller torque than that present when only positive sequence voltages are applied . The net effect of unbalanced voltages applied to an induction motor is to reduce the mechanical torque. Since the speed of the motor depends upon the simultaneous matching of motor and load torque-speed characteristics, the effect of the unbalance depends on the type of load served by the motor. Should the motor applied voltage change but remain balanced, the torque 1 will vary Depending varies directly with as the square of the applied voltage since
Irl
Va lT'm
225
S eq ue nce I mped a n ce of M a chi nes
15
�
'----������ 1 .0 0. 50 0.75 0.25 o PER U N I T S P E E D
Fig. 6 . 21 . Torque-speed relationships for motor and load : (A) motor torque-speed curve at rated voltage , ( B ) constant torque load , ( C ) constant power load, ( D ) typical load torque-speed curve , ( E ) motor torque-speed curve with unbalanced applied voltages. ( From Clarke [ 1 1 , vol . 2 ] . Used with permission . )
oncausethetheloadmotorcharacteri stic,Aasystem 3ct> fauldisturbance t near the which motor unbalances or a reducedthevoltage may to stall. voltage will also cause the torque of the motor to be reduced, as shown by curve of Figure In thispeed s casecharacteristic the motor wiwill tslow down andIf theseekloada new operatintorque g pointsuchon the6. 2a1.torqueh the load. is constant conveyor or load, is constant powermaysuchstallas aifregulated dc generator driving a con stant impedance the motor i t is unable to deli v er the necessary For a load such as a fanof orFigpump the2 1. torque variloades torque atthethe square lower speed. about as of the speed, such as curve ure 6. Such a woultod continue tounless be served at a reduced speedtorque if the drops motorpractically torque is reduced due any cause, the reduction in motor to zero. Bythesethetorques use of can equivalent ci r cui t s for the posi t i v e and negati v e sequence networks, be determined and the motor performance may be evaluated if theSince torque-speed characteri sticusual of thely shaft loadeitheris known oror can be estimated.con induction motors are wound for ungrounded nection,for thea zerozerosequence sequenceequivalent currentscircuit. in the motor are always zero and there is no need s areis gigivvenen byin Approximate valuandes forcredited induction motor references. equivalent ciThisrcuitdata Clarke [11, vol. 2] to several Tablso ebe determi Clarkenedsuggests that theratedposivoltivetagesequence impedance forpu current any loadormayby alcomputing by di v i d ing by the assumed the bemachine driving-point impedance. However, as forthe theactualequicurrent woul d seldom known, Fi g ures and 6. 2 0 may be used valent circuits of positive and negative sequence networks respectively. E
as
D
�
Y
6 .3.
6 . 18
Table 6. 3.
Rating
Full Load Efficiency
(HP)
(%) 75-80 80-88 86-92 91-93 93-94
Up t0 5 5-25 25-200 200-1000 Over 1000
Approximate Constants for Three-Phase Induction Motors R and X in per Unit* Full Load Full Load Power Factor Slip X. + xt Xm R, Rr (%) (%) pu) ( (pu) (pu) (pu) 7 5-85 3.0-5.0 0. 10-0. 14 1.6-2.2 0.040-0.06 0.040-0.06 82-90 2.5-4.0 0. 12-0.16 2.0-2. 8 0.035-0.05 0.035-0.05 84-9 1 2.0-3.0 0. 15-0. 17 2.2-3.2 0.030-0.04 0.030-0.04 85-92 1.5-2.5 0. 15-0. 17 2.4-3.6 0.025-0.03 0.020-0.03 88-93 "'1.0 0. 15-0. 17 2.6-4.0 0.015-0.02 0. 015-0.025
Source : Clarke [ 1 1, vol. 2 ] . Used with permission of the publisher. * Based on full load kVA rating and rated voltage. t Assume that X. Xr for constructing the equivalent circuit. =
Cha pter
226
6
6. 1 3 I nduction Motor Subtransient Fault Contribution
The application ofexcitation a short ciforrcuithet near theandtermiits nfield als ofcollapses an induction motor removes the source of motor very rapi d ly. Clarke [11, vol. 2] gives the approximate time constant of the decay of rotor flux
as
(6.163)
where theto bequantities usedHowever, in the siequation areodefiof Xnedto inR isFigure 6.18 and are as sumed ohms. n ce the rati used in (6.163), these quanti t i e s may be in pu. From Tabl e 6.3, if we take 0.16 and 0.035 as average values which of X. isX,.lessandthanR,.onerespecti vel(1ycycle , we compute = 0 . 0121 sec for a 60 Hz motor cycle 0.01667 sec if f = 60 Hz ) . The cur rentststo2-4be cycl interrupted by thefaulcircuit breaker ins case transmission systems is that which exiinduction e s after the t occurs. In thi the current contribution from motors may be negl e cted. In industrial pl a nts, however, l o w-vol tage, icycle. nstantaneous, circuit breakers are often used which clear faul t s in about onebe In such cases the faul t contribution from induction motors shoul d not neglHowever, ected. if the maximum value of current is required for computing fuse melting or cibercuifound t breaker mechanical stresses,thisthesubtransient contributionperiod fromasanainduction motor may by treating i t during synchro nous machine. That is, we will compute a generated EMF behind a reactance, and this will constitute the subtransient equivalent circuit as shown in Figure 6.22. in
t,.
+
=
air
j ( x . + xr ) •
Vo l
1-
101
FOSITI VE SEQUENCE
NEGATIVE SEQ l£ N CE
Fig. 6. 2 2 . Sub transient induction motor equivalent circuits.
Em
Here, we assume that From duringTable the subtransi entXrtime 0.16 intervalpu., Weactscompute throughfromthe reactance X. X,.. 6.3, Figure 6 .22, (6.164) where 101 is the prefault motor current and Vol the prefault motor voltage. Approximate thefaulsubtransient valtageue ofand drawi for nag100ratedhpcurrent inductionat 88% motor,power op erati n g before the t at rated vol factor. Then find the subtransient contribution of the motor to a 3f/J fault at the motor terminals. From (6.164), taking Vol as the reference phasor, 1 .0i!t' - j(1.0/- 28.5° )(.16) = 0.9235 - j O.1406 = 0.935/- 9.35° pu +
Example
X. +
6. 1
Em
Solution
Em =
�
227
S eq ue nce I mpedance of M a chines
Therefore, for a 3ct> fault on the motor terminals, ° = 0.935/ 9 . 35 = 5 . 85 L- 99 • 35° 0.1 6/90 °
Im
pu where the current is in pu on the rated base kVA of the motor. 6. 1 4
Operation with One Phase Open
Ifresulting one of seritheethree leads supplying an uinduction motor shouldcomponents. become open,In the s unbalance may be eval ated by symmetrical such a case the boundary conditions are = ( 6.165) = 0, = From the first two boundary conditions we compute = A- I or (let h 1) l l lLll aa, a�JJ � a I: Ib
Io
-
Ie,
Vb - Ve
Vb e
[ J ["� "J 101 2
l�: J
=
b
Io l
=
O
-
m ila
Ib
=
a.
=
lobe
-
(6.166)
= from whichnetworks we seemust thatbe connected This means that the positive and negative sequence as shown in Figure 6.23. - Io2 .
I l
. Ib
3
Yal I
• •
(Val - Va2) " j Vbc
Xr
Xs
Rs
Rr
PO S ITI VE
101
Rc
Xm
Rr
SEQ UENC E
NI
(I -s ) s
Ial " - Ia2 = j � ..
3
Rs •
Va2
10
Xs
Xr
Rr
NEGATIVE SEQUENCE
..
2
Rc
Xm
- Cl-s ) R r
2-s
N2
Fig. 6 . 23 . Sequence network connections for an induction motor with line a open.
The only known voltage is, from (6.165), with h = 1 = = a a2 Summarizing from (6.1 66) and (6.167 ) Vbe
Vb - Ve
=
-
( Voo +
j � ( Vo l - Va l )
Va l + aVo2 ) - ( VoO + Vo l +
al
Va l )
( 6.167)
228
C h a pter
Ial =
. Ib I0 2 = J � '
6
. Vb e V02 = J �
(6.168) and if is known, may be determined from Fi g ure 6. 2 3. Thi s compl e tel y es the sequence networks (assuming that is known), and the net torque may bethisolvscomputed from (6. 162). Wagner and Evans [10] show that the net torque in siltouati omust n remai nbes positootilarge, ve as orlongtheasmotor the slwilip lisstall. smallClarke . This [11, meansvol.that2] dis the shaft ad not cusses ttheor bank case ofis inan paral openleconductor inmotor. the supply tos ancaseinduction motor wheremaya capaci l wi t h the In thi a negati v e torque resul the induction condittiwhion cwillh willnotcause be discussed here. motor to reverse direction of rotation. This Vbe
-
V0 1
Ib
-
s
vet)
Problems
t
Consider a series R-L circuit supplied from an ac voltage source Vm sin ( w + a:) and assume that initially the current is zero because of an open switch in the circuit. If the switch is closed at 0, find the current as a function of the series impedance magni· ° ° tude and angle, Z - I Z I &.. Sketch the resulting current for a: (J 0, 45 , 8,Ild 90 . 6.2. Express the voltage and final value of current for problem 6 . 1 as phasors. Show how the dc offset can be found by projecting the current phasor on the real axis. 6.3. Verify the inductances given by (6. 7)-(6. 13). 6.4. Show that p- 1 of (6.19) is indeed the inverse of P given in (6.17). 6.5. In Figure 6.3 consider that all self and mutual inductances are constants instead of the time varying inductances given by (6.6)-(6. 14). Write the equations for the synchronous machine voltages, then transform by the phasor transformation, and transform again to the 0·1-2 coordinate system. 6.6. Repeat 6.4 with an impedance ZPI rPI + jwLPI in the neutral and a current In entering the neutral. Write the equations of phase voltage to ground rather than to neutral. How does the neutral impedance appear in the 0·1·2 coordinate system? 6.7. (a) Verify (6.28) and explain th e meaning of the results. What k i n d of induced voltage is s? (b) Verify (6.32), making use of the trigonometric identities of Appendix C. 6. 8. Explain why the voltage nOdq affects only the zero sequence. 6.9. Show that by proper choice of base values (6.34) may be written exactly the same whether in volts or in pu. What restriction does this place on the selected base values? 6. 10. Prove that (6.35) is true under the conditions specified. Do this for only one phase, e.g. ,
t
6.1.
co
-
-
..
co
vf
Xb lib ·
6. 1 1. Given (6. 20), reduce by Kron reduction to write vabe in terms of only by eliminating the rotor current variables. Is the result Laplace transformable? 6.12. Examine the following special cases for the Park's transformation iodq ... Plo be with (J + l) + 90. (a) iobe is strictly positive sequence, i.e., ..
WI t
lobe ViI -
[:: (WI�Ittt Jl [COS Wi(WItt � -
cos
120)
+ 120
find iOdq • (b) lo be is strictly negative sequence, i.e.,
iobe find �q.
=
ViI cos cos
(Wit
+ 120)
- 120)
S eq uence I mpeda nce
of M a ch i nes
(c) iubc is strictly zero sequence, i.e.,
� - �l .� w, t
[
find iOdQ . 6. 1 3. Given the unbalanced set of currents
(W I t Ibm cos (W I t lem cos (W I t
lam cos
iobe "
+ -
+
229
m
a)
J
120 + (3)
120 + 1)
compute iOdQ . 6. 14. Verify equation (6.46) and explain the meaning of this result in two or three sentences. 6. 15. Compute vOdQ from equation (6. 34) under the condition that (a) VF 0 and (b) VF - rF iF = o . 6. 16. Compute vOdQ by making a Park's transfonnation of the result of problem 6. 1 1. 6. 17. Verify that the equivalent T circuit of Figure 6.8 can be derived from the coupled cir cuits of Figure 6.7. 6. 18. Show how (6.79) can be satisfied by proper choice of base. 6. 19. Construct equivalent T circuits for a typical synchronous machine, using values from Table 6. 1 where the machine is (a) round rotor and (b) salient pole. 6. 20. Find the impedance seen looking into the generator d and q tenninals of the equivalent T circuit under (a) steady state conditions, (b) transient conditions, and (c) subtransient conditions 6.21. Compute numerical values for the impedances found in problem 6. 20 using the data from problem 6. 19. 6. 22. Find the time constant of the impedance seen looking into the generator tenninals under (a) steady state conditions, (b) transient conditions, and (c) subtransient conditions. Use data from problem 6. 19. 6.23. A synchronous generator is operating in the steady state and supplying a lagging power factor load. Assume that the machine is connected to a "solid" bus with several other generators. Explain, using before and after phasor diagrams, what happens if the operator increases the field voltage (at constant power) (a) gradually and (b) suddenly. 6. 24. Repeat problem 6.23, but this time assu me that the load is increased while the excitation remains constant. 6.25. Verify (6.82) by perfonning the P transfonnation. 6.26. Find 10 if id 0. 5, iQ = 0.8, and io o. Sketch a phasor diagram. Find lor and lax if 6 = 60° . 6.27. Explain in words how we can interpret kx mF IF as an induced EMF in equation (6. 104). 6.28. Use data from Table 6. 1 for a synchronous machine loaded such that on an arbitrary basis we have ° ° 10 0.8/- 15 Va 1. 0/ 30 , =
=
=
6. 29. 6. 30.
6.32.
=
Then compute Eq and construct a phasor diagram similar to Figure 6 . 12 if the machine is (a) round rotor and (b) salient pole. Construct the equivalent circuit similar to Figure 6. 13 for the machines of problem 6.28. Using the condition of problem 6.28 as the initial condition for a 3t/> fault, compute the initial flux linkages at time t 0+ . Compute all necessary quantities to sketch the phasor diagram immediately following a 3t/> fault on the generator tenninals, using the condition of problem 6.28 as the prefault condition. Compute the maximum cu rrent two cycles after the fault of problem 6.31 is applied. Repeat for 70 cycles and 150 cycles. =
6.31.
-
230
C ha pte r
6
6. 33. Estimate the interrupting and momentary rating of the circuit breaker required to inter rupt a 3ljJ fault on the terminals of the following round rotor generators: (a) 150 M V A, (b) 600 MVA, and (c) 900 MVA. 6. 34. Justify the approximations specified in equations (6.57) and (6.58). 6.35. A cylindrical rotor, synchronous machine having (average) machine constants given in Table 6. 1 is operating at rated current (1.0 pu) and 90% lagging power factor when a 3ljJ fault occurs on the machine terminals. Compute: (a) The voltage E behind the synchronous impedance. (b) The voltage E" behind the subtransient impedance. (c) The initial symmetrical sub transient fault current. (d) The peak symmetrical current after 5 cycles and 10 cycles. (e) The maximum asymmetrical current after 5 cycles and 10 cyeles. 6. 36. Repeat problem 6.35 for a salient pole machine. 6. 37. A 1000 hp induction motor operates at a slip of 2.0% and with 93% efficiency while driving a rated shaft load. Use average values from Table 6.3 to construct a positive sequence equivalent circuit for this machine when the base MVA is the machine base. The motor is rated at 4 kV. 6. 38. If the motor of problem 6.37 operates in a large system which is under study, recompute the equivalent circuit if the base for the system study is 100 MV A, 200 MVA, and 500 MVA. 6.39. Suppose that the voltage supplying the motor of problem 6.37 has a 5% negative sequence component. Compute the positive, negative, and net torque in pu. 6.40. Compute the time constant for the decay of rotor flux for the motor of problem 6.37 by applying (6. 163). What is the time constant in cycles at a frequency of 6 0 Hz? 6.41. Suppose that line a of the supply to the motor of problem 6. 37 is opened. Find the se quence voltages and currents if Vbe is 1.0 pu on a line-to-line basis.
chapter
7
Seq u e nce I m peda n ce
of Tra nsforme rs
An examination of the transformer completes our study of the major compo nents of the power system. Insofar as the study of faulted networks is concerned, we need to solve two kinds of problems. First, given a particular transformation, we must determine the sequence network representation of the physical device. This requires a knowledge of transformer equivalents, standard terminal markings, and the various kinds of transformer connections. The second problem is the estimation of reasonable transformer parameters for installations still in the plan ning stage. In this case a knowledge· of average impedance values is required, and the way in which the various transformer connections affect the sequence net works is helpful. Since the type of transformer connection often influences the type of protective scheme , these concepts must be studied in system planning to assure workability and reliability in the overall design. I . S I N G LE·PHASE TRANS F O R M E R S
Single-phase transformers, connected to form three-phase banks, were in wide general use for transforming both high and low voltages in the first half of the twentieth century . One reason for this was that a fourth transformer could be purchased and installed with a three-phase bank for later connection in the event one single-phase transformer failed. Recently the trend has been toward the use of three-phase transformers because they are cheaper and more efficient and are generally regarded as very reliable . Still, however, many single-phase units are in service , and large transformers are often single phase because of limitations in shipping sizes and weights. 7.1
Single-Phase Transformer Equivalents
The equivalent circuit often used for a single-phase transformer is shown in Figure 7 .1 a where the high and low voltage leakage impedances ZH and Zx are given in ohms. The transformer core losses are assumed to vary as the square of the H·winding voltage and are represented by Rc . The rms value of magnetizing current is represented by the reactance Xm • The turns ratio is defined as
n
=
nH /nX
(7.1)
where 231
C hapte r 7
232
nH nx
= =
number number ofof turns turns iinn winding winding HX We usual ly elimZxinateto thethe Hmagnetization branch sinshown ce Ie «in FiIHgurande 7.1b. also transferat theopen impedance si d e of the ci r cui t as circuit Thus
(7.2)
and, the ampereturns of the transformer windings are equal except for the excitatisinceon MMF nHI. . we have nHIH = nxIx or (7.3) Modem10substation Ie of less than 1 % in sizes up to about MVA and hitransformers gh-voltage ratiusualngslyuphave to about kV. 69
HI
+t -, Ii2
VH
-I H
lHX
� ZH
n
2
:;:
)(
(b) pu Z H X
HI
� � "
-
+t -,
pu I
p u VH H2-
"{::J� Vl(
Ie)
�-
X2
, :1
pu V)(
1 )(-
•
2
Fig. 7 . 1 . Single-phase, two-winding transformer equivalents : (a) equivalent in system quanti ties, (b ) simplified equivalent with all series impedance referred to the H winding, (c) pu equivalent.
Any ZxX-winding current Ix is seen by the primary side as Ix In and the im pedance appears to IH to be Zx . Thus the total transformer impedance as seen by the H-winding current is (7 . 4 ) which shownshuntin branch Figure i7.s eliminated. 1b as the total transformer impedance where, since Ie is smalIt iliss, theconvenient to change ily select the base quanti ties as Figure 7.1b to a pu equivalent circuit. We arbitrar n2
233
S eq uence I mpedance of T ra n sformers
SB transformer A rating VHB rated VH VX B rated Vx =
V
=
(7. 5 ) Then (7.6 ) and (7. 7 ) Fromratio (7.2 )isand we compute in pu at no load , pu Vx pu VH, and the turns pu n 1.0 (7.8 ) Alwinding so, fromto(7.be7 ) we compute the total transformer impedance as viewed from the pu ZHX = (ZH + n2 Zx )/ZHB (7.9 ) Finally , from (7.3) we compute puIH pulx (7 .1 0) From (7.8 )-(7. 1 0) wealmost establish the useequivalent circuit offault, Figureand7.1cstabiandlitythisstudies.the equivalent we will always for load flow, study of reactance ferroresonance, surges, traveling waves, harmonics, etc., the(In the magnetizing shouldswitching not be ignored.) =
(7.5)
pu
=
=
H
=
is
7.2 Transformer Impedances
are nearly the Transformer transformer ratiimpedances ng. Thus from (7.9 ) always given in pu (or percent) based on PU ZH X = (ZH + n2 ZX ) /ZHB = PU ZH + n2 Zx /ZHB = PU ZH + pu Zx (7.1 1) This result may be verified by dividing ZHB by ZX B to compute (7. 1 2) ZHB/ZXB = (VHB /VX B )2 = n2 nearlarey constant forfromtransformers of a givenand size may and debe Values of ofpuaverage ZHX arevalues sign. Tables avail a bl e the manufacturers used when actualdistribution nameplate data are not known. Table 7and.1 gives typicalNotevalthatues forin thetwo-winding transformers rated 500 kV A below. lapplications arger sizes thewhere impedance is almost entirely inductiveused,reactance. In subis station these larger sizes are sometimes the resistance oftenciency, neglected entirely.studiUsually thengresistance is considered only when losses, effi or economic e s are bei considered. Tableis convenient 7.2 gives typical values forimpedances large two-wofinding power transformers. This table for estimating transformers where different methods of cool i ng may be under consideration. Note that a range of impedances is specified shoul for each voltaken tage from class. the lower the transformer isrange self specified. cooled (OA),Forced the impedance d be end of the cooling allows a transformer of a given size to dissipate heat faster and operate at If
Chapter 7
234 Table 7.1.
Distribution Transformer Impedances, Standard
Reactances and Impedances for Ratings 500 kVA and below (for 60 Hz transformers) Rated-Voltage Class in kV
2.5
Phase
Single-
kVA
Ratmg*
15
Average Reactance
Average Impedance
%
Average Reactance
%
%
3 10 25 50 100 500
1.1 1.5 2.0 2.1 3.1 4.7
2.2 2.2 2.5 24 3.3
0.8 1.3 1.7 2.1 29
.
.
4.8
4.9
25
Average Imped-
ance % 2.8 2.4 2.3 2.5 3.2 5.0
69
Average Impedance
Average
Average
Reactance
Impedance
Average Reactance
%
%
%
%
4.4 4.8 4.9 5.0 5.1
5.2 5.2 5.2 5.2 5.2
6.3 6.3
6.5 6.5 6.5
6.4
Source: Westinghouse Electric Corp. [14]. Used with permission. *For three-phase transformers use 1/3 of the three-phase kVA rating, and enter table with rated line·to-line voltages.
a greater voltampere loading and therefore should be represented by a greater impedance than a similar rating in a self cooled unit. Resistance is usually neglected in power transformers. 7.3
Transformer Polarity and Terminal Markings The terminals of a single-phase transformer manufactured in the United
States are marked according to specifications published by the American National Standards Institute [52], often called ASA Standards. These ASA Standards specify that the highest voltage winding be designated HV or H and that numbered subscripts be used to identify the terminals, e.g. , HI and H2 • The low-voltage winding is designated
LV
or
X
and is subscripted in a similar way. If there are
Z
more than two windings, the others are designated Y and with appropriate subscripts. This same marking scheme for windings permits the identification of taps in a winding. Thus in a given winding the subscripts 1, 2, . . , n may be used to identify all terminals, with one and n marking the full winding and the intermediate
3,
3,
numbers 2,
.
.
.
, n
-
1 marking the fractional windings or taps. These numbers
are arranged so that when terminal terminal
i,
the
i
.
i+1
is positive (or negative) with respect to
is also positive (negative) with respect to i
-
1. The specifications
further require that if HI and Xl are tied together and the H winding is energized, the voltage between the highest numbered H winding and the highest numbered
X
winding shall be less than the voltage across the H winding.
The standards also specify the relative location of the numbered terminals on
the transformer tank or enclosure. The two possibilities are shown in Figure 7.2. Examples of transformers with tapped windings are shown in [ 14] and [52]. We have avoided use of the terms "primary" and "secondary" here since these terms refer to the direction of energy flow and this is not specified when deriving gen eral results. Figure 7.2 also illustrates what is meant by the terms "additive" and "sub-
235
Seq uence I mped a n ce of Transfo rme rs
Table 7.2. Standard Range in Impedances for Two-Winding Power Transformers Rated at 65 C Rise (Both 25- and 60-Hz transformers) Impedance L imit in Percen t HighVoltage Winding Insulation Class kV
15 25 34.5 46 69 92 115 138 161 196 230
L o wVoltage Winding Insulation Class kV
16 15 15 25 25 34.5 34. 5 46 34.5 69 34.5 69 92 34.5 69 115 46 92 1 38 46 92 161 46 92 161
Class OA OW OA/FA * OA/FA/FOA *
Class FOA FO W
Min
Max
Min
Max
4.5 5.5 6.0 6.5 6.5 7.0 7.0 8.0 7.5 8.5 8.0 9.0 10.0 8.5 9.5 10.5 9.5 10.5 11.5 10.0 11.5 12.5 11.0 12.5 14.0
7.0 8.0 8.0 9.0 9.0 10.0 10.0 11.0 10.5 12.5 12.0 14.0 15.0 13.0 15.0 17.0 15.0 16.0 18.0 15.0 17.0 19.0 16.0 18.0 20.0
6.75 8.25 9.0 9.75 9.75 10.5 10.5 12.0 1 1.25 12.75 12.0 13.5 15.0 12.75 14.25 15.75 13.5 15.75 17.25 15 .0 17 .25 18.75 16.5 18.75 21.0
10.5 12.0 12.0 13.5 13.5 15.0 15.0 16.5 15.75 18.75 18.0 21.0 23.25 19.5 22.5 25.5 21.0 24.0 27.0 22.5 25. 5 28.5 24.0 27. 0 30.0
Source : Westinghouse Electric Corp. [ 1 4 ] . Used with permission. *The impedances are expressed in percent on the self-cooled rating of OA/FA and OA/FA/FOA. Definition of transformer classes : OA-Oil-immersed, self-cooled OW-Oil-immersed , water-cooled. OA/F A-Oil-immersed, self-cooled/forced-air-cooled. OA/FA/FOA-Oil-immersed, self-cooled/forced-air-cooled/forced-oil-cooled. FOA-Oil-immersed, forced-oil-cooled with forced-air cooler. FOW-Oil-immersed, forced-oil-cooled with water cooler. Note : The through impedance of a two-winding autotransformer can be estimated knowing rated circuit voltages, by multiplying impedance obtained from this table by the factor (HV - L V/HV).
tractive" polarity. The lower of the three sets of drawings shows a core segment with voltage polarities and terminals labeled. Obviously there are two ways to orient two windings on a core and these possibilities are shown in Figures 7.2a and 7.2b. If Ix is increasing in the circuit on the left, this would cause a flux to tend to increase in the upward direction. But by Lenz's law a current IH will flow to oppose this change in core flux and hold the flux linkages constant. This establishes H I as a positive terminal. Since the HI and X bushings have the 1
236
C hapte r SUBTRAC TIVE
ADDITI VE
� XI
7
X2
Fig. 7 . 2. Standard polarity markings for two-winding transformers : (a) subtractive, (b ) addi tive. (From Westinghouse Electric Corp . [ 1 4 ] . Used with permission. )
same relative position on the tank, as noted on the left-hand set of drawings, this is by definition a "subtractive polarity." Thus for the transformer on the left the polarity may be determined by connecting adjacent terminals together (viz., HI and X I ), applying a voltage between H I and Hz , and then checking that the volt age between adjacent connections Hz and Xz is less than the applied voltage. If so, the polarity is subtractive. Obviously, the reverse is true on additive polarity transformers. Additive polarity is standard for single-phase transformers of 500 kVA and smaller when the H winding is rated 8660 volts and below [14] . All other transformers are normallyV subtractive polarity, although the nameplate should be checked before connecting any transformers in a three-phase bank. From the viewpoint of circuit analysis, we often identify the coils of a coupled circuit by polarity markings or dots. By this convention we easily estab lish that if HI is a dotted terminal, X is likewise dotted shown in Figure 7 .3. This fact follows from the definition ofI additive polarity where we establish that the instant H I is positive with respect to Hz , then XI is likewise positive with respect to Xz • If we choose IH as entering the dotted terminal (HI ) and Ix as leaving the dotted terminal (XI ), these currents are in phase if exciting current is negligible. Equations (7.2) to (7.10) are written on this basis. as
IH
-
+t
•
I X
HI
XI
H2
X2
VH
-I Fig. 7 . 3. 7.4
Dot convention for a two-winding transformer.
Three-Winding Transformers
is quite common in power systems to utilize transformers with more than two windings.z This is especially true in large transmission substations where voltIt
z This analysis follows closely that of [ 1 4 ] to which the interested reader is referred for additional information.
237
S eq u e n ce I mpeda n ce of T ransfo rmers
ages are transformed from high-voltage transmission levels to intermediate sub transmission levels. In such cases a third voltage level is often established for local distribution, for application of power factor correcting capacitors or reactors, or perhaps simply to establish a connection to provide a path for zero sequence currents. Although such transmission substations are usually equipped with three phase transformers, the theory of the three-winding transformer is more easily understood by examining a single-phase unit. These results may later be used on a per phase basis in the study of three-phase applications. The windings of a three-winding transformer are designated as H for the high est voltage, X for the intermediate voltage, and for the lowest voltage. We also assume that exciting currents are negligible and will use equivalent circuits similar to Figure 7.1b and 7.1c, where the excitation branch is omitted. Consider the winding diagram of a three-winding transformer shown in Figure 7.4a where the windings H, X, and Y are shown to have turns nH , nx , and l!.
Y
Z HI H ) H 1 l·.... -
+
' n ri
Z Y I HI
VH
- !�----..
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
V=ZI
where v=
[�:1 ,
1=
(9 .20 )
G:J
(9.2 1 )
C h a pter 9
320
and where (9.22) describes the relationship between these "external" quantities. works are in series, we observe that
Since the net (9.23)
and
I = Ia = Ib = . . .
=
From (9.19), (9.20), (9.23 ) , and (9 .24) we write
V
In
(9.24)
Va + V b + . . . + Vn = Za 1a + Z b 1b + . . . + Zn 1n = (Za + Zb + . . . + Zn) I =
(9.25 )
Then from (9 .20) and (9.25) we see that
Z = (Za + Z b + . . . + Zn)
(9.26)
This means that we can easily find the Z parameter description of any number of two-port networks which are connected in series by first finding the Z parameters of the individual two-port networks and adding these together as in (9.26 ) . It must be understood, however, that port descriptions must be preserved in the interconnection . Parallel connection.
. . . , n,
Consider a group of
n
two-port networks, labeled
b, connected in parallel as shown in Figure 9 .6. We again define the port voltage and current vectors as in (9.16 ) and (9. 1 7 ) respectively. For the parallel connection , however, we do not use the Z parameter description but instead
a,
define
l k = a, b, . . J r--- - - - - - - - - -1 Y 1 2k YZ2k
.,n
12 0
I Ia
-
Vlo
I
VI
I I I
M
+
I
I I I I
J L
I lb
--
A..
+
..
...!!n� In
_
_
12b
-
Y2b +
z
,
+
_
�2 0 +
b
V lb
I I-- 2 n +
n
_
J
---
a
+
_
I I
I
_
_
I2
I --A- -
�
V2 +
I I I
V2 n
_
(9.27 )
_
I
..J
Fig. 9 . 6 . Parallel connection of two-port networks.
S i m u ltaneous F a u lts
32 1
such that we may write for any two-port network, k = a, b, . . . , n
I k = Yk Vk
(9.28)
If we again define voltage and current vectors as ( 9 .2 1 ) for the cf> -¢ network, we may write
I=YV where
Y=
fY I l 11 2 1
]
Y12
Y 22
(9.29) (9 .30)
describes the relationship between V and I . For the parallel connection we ob serve that
(9.31 )
and
V = Va = V b = " ' = Vn Combining these last two equations with (9.28 ), we write
I = Ia + Ib + . . . + In = Ya Va + Y b V b + . . . + Y n Vn = (Ya + Y b + · · · + Yn) V
Comparing this result with (9.29), we see that
Y = Ya + Yb + . . . + Y n
(9.32)
(9.33) (9 .34)
and the two-port description for this connection is easily found by adding the Y parameters of the individual networks. We again require that two-port descrip tions are preserved in the interconnection. Note that it is not convenient to find a Z parameter description for the parallel connection. It is preferable to first con vert the Z parameters of the individual two-port networks to Y parameters and add these as in (9 .34 ) . H ybrid connection. Consider a group o f n two-port networks, labeled a, b, . . . , n , connected in series at one port and in parallel at the other as shown in Figure 9.7 . We refer to this as the hy brid connection. Here, instead of the voltage and current parameters of ( 9 . 1 6 ) and (9. 1 7 ) , we define the port hybrid voltage-current vectors
and the hybrid matrix
Hk =
[
h l lk h 2 1k
k = a, b, . . . , n
(9.35)
k = a, b, . . . , n
(9.36)
k = a, b, . . . , n
(9.37)
Then from Table 9.1 we write
Mk = Hk Nk
k = a, b, . . . , n
(9.38)
C h a pter 9
322
--1I
.1
r- - - - -I 10 -
� I
�
Vlo
I
I
j
F--
I J J
l ib
-
tYlb
J
I I
� In +
oJ '
,
L
b
r �
I
""'
a
+
I
_
_
_
_
I20 +
_
I
12b
I
I
_
,
z
-
I2 n
'{ 2n +
_
_
_
_
I2
�
W
V 2 b j1J
n _
V2 0
I I
2
I
_
_
I J
Fig. 9 . 7 . Hybrid (series-parallel ) connection of two-port networks.
We now describe the interconnected system according to the equation M=HN
where M and N are defined according to "external" quantities,
(9.39) (9.40)
and (9.41) and where
�
J is the hybrid parameter matrix for the interconnection. H=
From Figure 9.7 we observe that
and
hll
h 11.
h2 1
h 22
M = Ma + M b + .
. . + Mn
N = Na = Nb = . . . = N n
Combining these observed results with (9.38 ) we write M = Ma + Mb + . . . + Mn = Ha Na + H b Nb + . . . + H n N n = (Ha + H b + . . . + H n) N Then comparing (9.45) and (9.39), we have H Ha + H b + . . . + H n Again we require that port descriptions be preserved. =
(9.42 ) (9.43) (9.44) (9.45) (9.46)
S i m u ltaneous Fau lts
323
Note that a similar result could be obtained by reversing the series-parallel connection of Figure 9.7 form a parallel-series connection. In such a case= the G parameters will be found to add and a description corresponding to N G M derived. This description is redundant in a sense since we are at liberty to number the ports any way we choose, and if we would exchange the 1 and 2 subscripts, we would have a parallel-series (G matrix) hybrid. Thus, the hybrid connection of Figure 9.7 will suffice for either situation requiring a series and a parallel connection. Cascade connection. Finally, we review the cascade connection of two-port networks as shown in Figure 9.8 where we define [VlIl k] [AA2111kk AAI222�] [V122k] � k = a, b, . . . , n (9.47) k k k to
=
�o �
II- 1 1 0
-
VI - Vi a
+
+
b
+
V2a Vl b
a
Fig . 9 . 8 .
We seek a description
r
b +
+
V2 b �In -
n
-
+
V2n- V2
Cascade connection of two-port networks.
] [V2] [VIII] [A2A 11l A12 Au 12
(9.48) which describes the connection. If we observe the constraints of the network interconnection, we write [V� [Vial , [_Vla] [VlblJ , . . . (9.49) Id IlaJ Ila Il b Combining with (9.47), we have =
=
=
1 = [�1] ��:: ���::J ��::J [��:: ��:::] [� �:: ���::] [-�::] fA 11a -AI2aJ [A11b �A 1 2b� ' " rA 11n �Alln] [_ VlJ LA 21a -Ana A21b A 2lbJ LA lln A21n 12 (9.50) =
=
Comparing (9. 50) with (9.48), we see that the composite A matrix is the product of the individual A matrices with the second column of each changed in sign. These negative signs arise because the currents between two-port networks flow in opposite directions. Using the ABeD parameter description, with I directed in of the two-port network, the cascade connection is simply a matrix I with no sign changes. We define the currents as entering the network conform with established usage [5, 63] .
and
l
OUt
product
9.3 Simultaneous Fault Connection of Sequence Networks
I
to
In Chapter 8 we developed the interconnection of sequence networks for common shunt and series faults. These results, summarized in Figure 8.9, show all
C h a pter 9
324
the sequence networks as one-port networks, with the current always entering the network at K and leaving at F. We further defined K to be the zero potential bus N for shunt faults and the current-receiving terminal M for series faults. (See Figures 8.3, 8.4, 8.6, and 8.7.) The point F is always defined to be the "fault point." We note also that the positive sequence network is an active one-port network since it contains independent sources, but the negative and zero sequence networks are passive one-port networks. Suppose that two faults occur simultaneously on a system. Using the two port network theory developed in sections 9.1 and 9 .2, we can represent the two faults by finding the two-port parameters of the positive, negative, and zero sequence networks and then combining them in a way which will properly repre sent the common fault configurations of Figure 8.9 [63] . Considering the four common types of faults which can occur at two different locations, there are a total of 16 possible problems which require our attention. These are summarized in Table 9 .6. Since it is immaterial whether a given fault point is represented as the left or the right port, one could eliminate the G matrix and always use the H matrix for hybrid combinations. We will do this in our analysis, but this restric tion is not necessary in practice. Table 9.6. Two-Port Fault Combinations R igh t Port (primed) Shun t fault (K
Fault Type Kind
(K = N)
Shunt fault Left Port (unprimed)
(K = M)
Series fault
SLG
=
2LG
N)
Series fault (K
l LO
=
M)
2LO
(series)
(parallel)
(parallel)
(series)
ZNN
HNN
HNM
ZNM
(parallel)
GNN
YNN
YNM
GNM
lLO (parallel)
GMN
YMN
YMM
GMM
2LO (series)
ZMN
HMN
HMM
ZMM
SLG (series) 2LG
Reference to Figure 8.9, which shows the various port connections, reveals the several pertinent facts concerning fault analysis. First, note that each se quence network is always terminated in an ideal transformer so that port currents and voltages are preserved for any network interconnection. This means that the two-port network theory of sections 9.1 and 9.2 will always apply. Note also that the added impedances Z and ZII ' used to represent impedance at the fault, are considered here to be a part of the network and must be added to the usual Thevenin equivalent sequence impedance. We also recognize that for simulta neous faults the sequence networks will have two fault points and two-port networks will be defined for each sequence network, with points F and K being defined according to the fault location and type. The simultaneous fault problems to be solved can be generalized by three forms: (1) a series-series fault, (2) a shunt-shunt fault, and (3) a series-shunt fault, where the terms "series" and "shunt" refer to the type of network inter connection shown in Figure 8.9. If we view the sequence network interconnec-
325
S i m u lt a neous F a u lts
-
��i.t+
-
K'i Ki S EQU E N C E 2- P O R T N E T W ORK
_ I: n
VK ' i
��
- �. -
Fig. 9.9 . Two-port network representation for simultaneous faults at F and
F '.
tion from the external 1/>-1/> connections, the network is shown in Figure 9.9 where the network voltages and currents are defined for simultaneous faults at F and F '. (Note that there should be no confusion between II and 12 as defined in Figure 9.9 and the sequence currents Ial and lal)' This figure will be used to define the current directions and voltage polarities to be considered in subsequent develop ment. The transformations shown are composites of the actual transformers or phase shifters required for various faults. Since these phase shifters are ideal, we .define the transformation ratios (9.51) These quantities will always b e subscripted with i equal to 0, 1, or 2 for the vari ous sequences and will take a value given in Figure 8.9, for a particular fault type and symmetry. 9.4
Series-Series Connection ( Z-Type Fau lts)
A series-series connection of two-port sequence networks is required to represent (see Table 9.6): 1. Simultaneous SLG faults at F and F ' (ZNN ) . 2 . A SLG fault at F and two lines open at F ' (ZNM )' 3. Two lines open at F and a SLG fault at F ' ( ZMN )' 4. Two lines open at F and two lines open at F ' ( ZMM ) ' The sequence network connection is shown in Figure 9.10. For the positive
sequence network we write
(
where
is
the independent source term viewed from the Kl - K' I transformers) . But from with i =
nKI =
(9.51)
V l l l VK l
Premultiplying (9.52) by
= I I l IIK 1 '
1
9.5 2 )
terminals (inside the (9.53)
C h a pter 9
326
�
I KO
"KO"
KC> K'
-
+
V KO
VK'O
0
I
K'I
KI
+
V, = O
+
K' ,
VK' ,
I
..1 2
' : " K',
+
V2 a O
--t-4K2 K' 2 e---t-��
__
2
-----.
-
0, 1 . 2
Fig. 9 . 1 0 .
Sequence network connection for simultaneous Z-type faults.
we get For the negative sequence network we write
(9.55)
which we premultiply by to obtain And
rV1 21 LV22J
=
(9.54)
r
(nKl /nK'l) Z12(2 )! ZI 1(l) knK '2 /nK2 ) Z2 1 (2) Z2 2 (2) J
for the zero sequence network we write rvKO
]
=
] [IKO Z 22(0) IK,oJ
rz 1 1(0) Z 1 2(0)
LVK 'O LZ2 1(0) which transforms (since nKO = nK'O = 1 ) to
(9.56)
(9.57 )
(9.58)
327
S i m u ltaneous Fau lts
But from Figure 9 .10 we observe that for a series-series connection, (9.59 ) and (9.60 ) Performing the addition indicated in (9 .59 ) and making the substitution (9.60), we write (9.6 1 ) where Z I 1 = Z I 1 (O) + Z l 1 ( l ) + Z l 1 (Z ) Z I2 = Z IZ(O) + ( n K ! l n K'l) Z IZ( l ) + ( n KZ/ n K'Z ) Z 1 2 (Z) ZZ l = ZZ l (O) + ( n K'J !n K l) ZZl(l) + ( n K'Z/ n KZ) ZZl(Z) ZZZ = ZZZ(O) + ZZZ( l ) + ZZZ(Z)
(9.62)
Equation (9.6 1 ) may be written as V = O = Z I + Vz
(9.6 3)
I = - Z - I Vz
(9.64)
Then
is the solution of the series-series connection . Knowing this current, we know all currents in the outside transformer windings, and we may compute all voltages from (9 .54 ), (9 .56 ), and (9 .58) . Then the individual sequence network currents and voltages may be found by applying Kirchhoff's laws to the individual net works. We may write out the solution (9.63 ) as
[�:J
=
:� Z
d
[-::: -:::J [::� �J
(9 .65)
I
In this form, we see that the current at each fault depends upon the independent source voltages at both fault locations . We illustrate this method by means of an example .
Example 9. 6 A simple power system is shown in Figure 9 . 1 1 with simultaneous faults in dicated by X's at fault points F and F ' . The following system data is known : Generator L : Z� = Zz = jO.12 , Zo = j O.10, EL = 1 .1{30° Generator R : Z� = Zz = jO.15, Zo = jO.13, ER Transformer AB : Zl = Zz = Zo = jO.10
Transmission line Be : Zl = Zz = j O .5 0 , Zo = j 1 .00 Transformer CD : Zl
=
Z2
=
Zo
=
jO.12
=
1 .0�
Chapter 9
328
Fig. 9 . 1 1 . Power system for Example 9.6.
We assume that faults are applied as follows : Fault F: lines a and b open Fault F ' : SLG fault on phase b Solve the system for the currents and voltages at the circuit breaker positions at B and C . Assume that all fault impedances are zero .
Solution First, we note that the two faults specified require a series connection of se quence networks. Thus a series-series two-port description is applicable. Refer ring to Table 9 .6 , we must determine the Z parameters ZM N for each sequence network . The sequence networks are sketched in Figure 9 .1 2, where it is noted that the series fault requires the insertion of an open section adjacent to breaker B . We arbitrarily define the current as positive in the B C directiol\ (IK) since this would constitute positive tripping current for that breaker, as noted at F. We also iden tify the positive tripping current direction for breaker C as away from bus C and toward the faulted line. To solve the system, we must first find the Z parameters. We easily compute the following.
[VKOJ VK'O
=
r j 1 . 35 L- jO.25 F
I I«)
] [IKO ] IK,o
- jO.25 jO.25
F'
I K,O ,...-----f'-K ' 0 a NO
'--":"'+-_ F' O
'--"-_ F ' I
2
I K' ,..-----.,...-- K ' 2 < N 2
VK'2 .
C
B
_
��- F' 2
Fig. 9 . 1 2. Sequence two-port networks for Example 9 . 6 .
�VVK'Ktl] �VKVK2]'2
] rKt[K't] [ lK2]2
]
S i m u ltaneous Fau lts
[ jO.99 -jO.27 jO.99 [-jO.27 =
329
0.047 - jO.550 -jO.27 jO.27 - 1.0L!L -jO.27] jO.27 IK Fault symmetry is the next consideration. From Figure 8.9 we confirm the following: Fault two lines open on a and b nKO = 1, nKl a, nK2 = a2 Fault F ' : SLG fault on b =
+
'
F:
=
The impedance matrix is then computed from (9.62) with the result [ j3.33 jO.02 ] Z= jO.02 jO.79 Finally, from (9.65) we compute [It] det- 1 Z [ (jO.79) aVK1 - (jO.02 ) VK' l ] (j0.02) a VK t + (j3.33) VK' t 12 [- 0.088 + jO.1221] [0.1589/123.76 ] - 1.094 + jO.630 1.2622/150.08 and we note that
aa22
=
-
=
=
From (9.51) we write =
Currents and voltages at breaker laO = IKo = 0.1615{123.75° B:
lat IKl = a2 laO la2 IK2 alaO la laO lal2 la2 0 Ib laO a /at a/a2 3 ao 0.477{123.76° =
=
=
+
=
=
Ie
=
+
1
=
+
=
+
=
°
[0.1589/3.76° ] 1.262/270.08°
330
Chapter 9
VaO = IKO = 0.0134 + jO.00897 Va l = 1. / 3 ° - (j0. 22) IK I = 0.954 + jO. 514 Va2 = (j0. 22) IK2 = 0.0381 +
-
(jO.10) 1 0
jO.0157
-
Va = VaO + Va l + Va2 = 1.08/29.9° Vb = VaO + a 2 Va l + a Va2 = 1.11/269. 1° Vc = VaO + a Va l + a 2 Va 2 = 1.06/145. 8°
Currents and voltages at breaker C:
laO = IK,o - IKO
Then
101 = IK't 10 2 = IK '2
- IK I
- IK 2
= = =
jO.630 0.002 - j1.262 1 .092 jO.633
- 1.094 + -
+
r��:] L�r� �2 �J lL �:1.092��: � jO.63�J ��:::�l = [�. ;�;150.1� jO J =
c
We also compute
a2
a
VaO = - laO Va l =
0+
+
(j0.25) = 0.124 + jO.251 1.0 � lal (j0.27) = 0.656 jO.042 Va2 = la (j0. 27) = 0.209 - jO.314 2
[ ]=
+
-
-
so that at breaker C
Va Vb
Yc
9.5
[0.990/- 1.160] AVO l 2 = 0 jO 0.990/128.5° +
Parallel-Parallel Connection (V-Type Faults)
A parallel-parallel connection of two-port networks is required to represent the following simultaneous fault conditions (see Table ' 1. Simultaneous faults at F and F (YNN )' A fault at F and one line open at F ' ( YNM )' One line open at F and a fault at F ' ( YM N ). 4. One line open at F and one line open at F ' ( YMM )'
2. 2LG 3.
9.6):
2LG
2LG
The sequence network connection with parallel-parallel termination is shown in Figure where the ideal transformations are phase shifters with voltage and current relations as given by We write the two-port equations for the three sequence networks as follows. For the positive sequence network
9.13
(9.51).
[ �::. J = �::::: �:::::] [�::.] [�::] +
(9.66)
where I y is the independent source term viewed from the K1 and K ' l ports. Simi-
33 1
S i m u lt a n e o u s F a u lts
-
r VKO
X.I
K' O
KO
-
-
- I 'n
K'2
VK' 2��
2
:� �: 1:
-
-
K' I
K2
��V K2
I'
VK'O:�
0
-,-�" KI
'1
-
12
0
-
0
...__ ' __ ' _ 2 ....
O I
Fig. 9 . 1 3 . Sequence network connection for simultaneous V-type faults.
larly , for the passive sequence networks we have
[
IK21 IK, J
Y 1 1 (O ) Y2 1 (O )
Y12(O� Y22 ( O ) J
Y 1 1 (2 ) Y2 l (2 )
Y 12 (2 ) Y22 (2 )
[ [
=
][
(9.67 )
J
VK2 VK'2
(9 .68)
Using (9.51 ) for each sequence, we reflect the two-port expressions across the transformations to obtain new two-port equations which include the transformers. This is done by premultiplying (9 .66)-(9 .68) by
for i
[�
= 1 , 0 , and 2 respectively.
[
J
/l l 12 1
=
[
�
K
J
[ J
The result for the positive sequence network is
Y l l (l ) ' (nK l /nKdY2 1 (1 )
Similarly , since n K O
Ki
(nK dnK ' d Y 1 2 ( l � Y22 ( 1 ) J
Vl l V2 l
+
J
rnK1 /Y1 LnK'l /Y2
= nK'O = 1 in all cases, for the zero sequence we have
B::J [;::::: ;:::::J [�:J
(9 _ 69 )
=
[ J=r L
And for the negative sequence network we compute
/12 122
(9.7 0 )
][ ]
( nK2 /nK '2 ) Y 1 2 (2 ) Y l l (2 ) (nK'2 /n K 2 ) Y21 (2 ) Y22 (2 )
VI 2 V:z:z
(9.7 1 )
C h a pter 9
332
From Figure 9.13 we observe that (9.72)
and
(9.73) Substituting (9.73) into (9.69)-(9.71) and then substituting these results into (9.72) gives (9.74)
where YI I YI2 Y2 1 Y22
= = = =
Y I I(O) + Y I I( I ) + Y I I(2) Y I2(O) + (n Kdn K' d Y I 2( 1 ) + (n K2/n K ' 2 ) Y I 2(2) Y21(O) + (n K d n K' d Y 2 1 ( 1 ) + (n K ' 2 /n K2 ) Y21(2) Y22(O) + Y22( I ) + Y22(2)
In matrix form we write (9.74) as which we solve for V to obtain or
(9.75)
yV+
I.
(9.76)
V = y- I
Is
(9.77)
0=
-
(9.78) In this case we solve for the external port voltages which are easily converted to port voltages inside the transformations. Then, if the sequence network vol!;ages are known, the networks can be completely solved. We illustrate the solution of a parallel-parallel fault condition by means of an example. Example 9. 7
Consider the system of Figure 9.11 for which the sequence impedances and source voltages are given in Example 9.6. Find the phase currents and voltages at the breaker locations for the following fault conditions: Fault F: line b open Fault F' : 2LG fault on b and c Solu tion As in Example 9.6 we have a series fault at F and a shunt fault at F' . There fore, the sequence networks are the same as before and are given by Figure 9.12. Here the similarity in the two problems ends. The symmetry in the fault condi-
333
S i m u lta neous Fau lts
tion of this example is given by the following: Fault F: line b open
nKO =
Fault F' :
1,
n K I = a2 ,
n K2 = a
2LG fault on b and c n K 'O = n K ' 1 = n K'2 = 1
For the sequence networks themselves we need to write current equations similar to (9 .66)-(9 .68). We again use the results of Example 9 .6 to good advantage. From Table we note that the Y matrix is the inverse of the Z matrix, or
9.3
-jO.25] -1 = [ - jO.91 -jO.911 jO.25 -jO.91 -j4.90J jO.99 -jO.27]-1 = [- j1.39 -j l .391 Y I = Z -I I = -j1.39 -j5.09J -jO.27 jO.27 and Y2 = Y 1 • Also , from Table 9.5 we note that -j1.39] [0.047 - jO.55] I = - Z I V = Y I V - [-j1.39 - 1.0 -j1.39 -j5.09 [ +0.764 -j1.32] [1 . 5 28/-60.00° ] = 5.085/- 81 . 36° +0.764 -j5.03 Yo
= Z-0 I =
j1.35 [-jO.25
[
_
I
y
z
z
=
Then we write
� .91 -jO.911 rv l [-j4.90J J] [--JO.91 �1.39 -j1.391 I -31. 3 9 J L J G:�J G��:!: ���:�:J [�:�J (9.74) 11 [ LoJ
flKo LIK' o IIKI l!K' 1
Ko VK' o
0
=
=
- j 5.09
VK I
VK' I
+
11.5 28/- 60.000 L5.085/- 81.36°
=
J
These current equations are reflected across the phase shifters by ( . 6 )and combined according t o with the result 0
=
Yll Y2 1
where from (9.75)
Y l I = Y I l (O) + Y I l (I) + Y l l (2) = =
-jO.91 -j1.39 -j1.39 -j3.69
- .9 1 1 39 a(-j1.39) - jO.91 + j1.39
Y I2 = Y I2(O) + ( n Ktl n K' I ) Y I2(1 ) + ( n K2/nK'2 ) Y I2(2) + a2 (- j . ) + = jO = = +j0.48
9 9 (9.71)
C h a pter 9
334
Y:u == Y21(0) (nK' t !nKI) Y21(1)a2 (nK'2 /nK2) Y21( 2 ) (- = Y22 == Y22(O) Y22(1) Y22(2) = = det = Yll Y22 - Y12YZl Then from ] ][ [VI] = [ V2 ] = [VI 0] [ Vll] = [V12] =[ V20 VZI V2Z From we compute [ VKO ] = [Yto] = r VK, o o L J 1 [VKI ] = ra 1 ] = [ J VK' I L ] [VK2 ] = ra2 ] = rL VK'2 L +
+
- jO.91 + a(- j 1 .39) +
+
+jO.48
j 1 . 39)
+
- j4 .90 - j5 .09 - j 5 .09 =
Y
-
- j 1 5 .08
55.65 + 0.23
-
55.41
(9.78)
- j 1 5.08 1 __ 55.41 - jO.48
- j O .48 - j 3 .69
1 . 528/- 60.000 + 240.00 5.085/- 81.36°
0.41 1 8/96.06°
=
0.3365/- 173.59°
(9.51 )
0.4 1 1 8/96.06°
Vi
O. 3365/- 1 7 3. 59°
Yt Vi I
1
0 . 41 1 8/216.060
0.3365/- 1 7 3 . 59 °
Yt z Viz
0.41 1 8/- 23.94°
0. 3365/- 1 7 3. 59°
Finally then we compute the voltages at F to be (using h = 1 )
[��Jl r�:L� :��:�:�:.�:JJ �L � J [Va'] � [ �J J =
- A
and at F'
.41 1 8/- 25.26°
v" , Vc'
9.6
.235/ 83.930
0. 3365/186.270
- A 0. 3365/1 86.27° 0. 3365/186.27°
Series-Parallel Connection (H-type Faults)
0
=
1 .009
°
0
The third and last connection required to solve for common simultaneous fault conditions is the series-parallel connection shown in Figure 9.14. This con nection may be used for all series-parallel faults and, by reversing the defined left port and right-port definitions, for all parallel-series faults as well. From Table 9.6 the series-parallel H-type fault configurations include: 1 . A SLG fault at F and a 2LG fault at F ' (HNN ) . 2. A SLG fault at F and one line open at F ' (HNM ) . 3. Two lines open at F and a 2LG fault at F ' (HMN) . 4 . Two lines open at F and one line open at F' (HMM ) .
335
S i m u ltaneous F a u lts 110 -
_
ft
KO
'1 1 KO
..-___...
,...... -+.. KO _
I
KI
-
tVKI
K'O
IK' O _
l 'ft K ' O
1 20 _
o
KI
I '
K' f
Kf
-
I'
VK'I � 1
I
�t---it,;�-... '-(I�-+
+
V�
..
__
I22
-
K' 2 �--r\
,-r---+" K 2 2
0, f ,
V2
2
Fig. 9 . 1 4 . Sequence network connection for simultaneous H-type faults.
For these conditions we write the two-port hybrid equations. For the active posi tive sequence network we have
]
h12(1) h22(I)
(9.79)
where the last term is the independent source term viewed from the Kl and K ' 1 ports. For the passive sequence networks we have
and
[VIKoJ�� [h1�1(O)1(O) [VIK��2J [h21(2) hl l(2)
]
=
hI2(O) �2(O)
=
h12(2� rIK � h22(2J LVK 2
( 9 .80 )
]
(9.81 )
Proceeding as before, we now write these equations in terms of the quantities out side the transformers by premultiplying by
for i = 1 , 0 , and 2 respectively . The result for the positive sequence equation is
For the zero sequence network with
nKO = nK'O
rIl ll LV2� =
1,
+
J
rnKl Vh l LnK'l Ih2
(9.82)
] [;::] [�:::: And for the negative sequence network, rVI� [ hl l(2) (nK2 /nK'2) h12{2� LI22J (nK'2/nK2) �1(2) h22(2) J C h a pter 9
336
h12(O) �2(O)
-
(9.B3)
J For the series-parallel connection of Figure 9.14 we observe that =
fI12 LV22
(9.B4) (9.B5)
and
(9.B6) Therefore we may substitute (9.B2)-(9.B4) into (9.B5), making use of (9.B6) write
to
(9.B7)
where
hll h 12 h21 hZ2
In matrix form
= hl l(o) + hl 1(l) + hl1(2) = h12(o) + (nKt lnK'I) hI2(1) + (nK2 /nK'2)h12(2) = h21{o) + (nK't lnKl ) h21(1) + (nK'2 /nK2) h21(2) = h22(O) + h2Z(I) + h22(2)
(9.B7) may be written as
(9.BB)
(9.B9) where N is defined by (9.14), M. by (9.B7), and the matrix H by (9.BB). Solving (9.89), we have
0=
or
HN + M,
[iJ = ;.: [-::; -::: ] [:�. ;:,j
(9.91) Knowing II and V2 , we may solve completely for all port quantities and then for internal network quantities. H
I I . S I M U LTAN EOUS FAU LTS BY MAT R I X TRANSFORMATI ONS
We now develop a second method for computing simultaneous faults which is based on the matrix transformation technique developed for a simple fault in sec tions B.9-B.11. This method is also described briefly by Lewis and Pryce [13] , which is recommended for further reading on the subject.
S i m u ltaneous F a u lts
337
9.7 Constraint Matrix for Z-type Faults
Z-type faults include any combination of SLG shunt faults and series faults with two lines open and are characterized by a series-series connection of the sequence networks. This connection is conveniently described by the two-port Z parameters according to the equations
[VKi ] [Z l I(i) ZI2(i)] VK'i Z2I(i) Z22(i)
G::] [�:] ,
_
where
o
But, by definition
H"
Ii
i = 0, 1 , 2
(9.92)
i
=
0 , =f: 1 1, i = 1
=
i
=
0, 1, 2
(9.93)
and the port equations can be written in terms of the sequence currents . From section S.S we write I Old
= K 1new
and , if we can find a matrix K which constrains I old , the sequence currents, to a series connection then we can use this same constraint matrix to find the new vol tage vector from V new = K *tVold ' For the problem at hand we define
I Old = I
= (9.94)
and write (9.92) in 0-1-2 order as
VKO VK I VK2 VK,o VK' I VK'2
Z 11(0) o
o
Z21(0)
o
Zll(l) 0
0
o o
o
Z21(1) 0
1
1 1
ZI2(0)
ZII(2) � 0
--- - -- - -- - - --
O
!
0
Z21(2)
I
11 I I
o
0 0
Z 12(0 I)
Z22(0)
0
o
Z12(2)
- - - - - - - - --
0
0
If we define , according to Figure 9 . 5,
Z22(I) o
o
0
o
laO la I la2 la'o la' i la'2
0
VzI 0 +
0
Vz2 0
(9.95 )
(9.96) where we use the circumflex to identify Inew since the prime has a different mean-
Chapter 9
338
ing here. Then I = Ki , or for a series-series connection
1=
la o la ' la 2 la' o la " la'2
0 n2 n-,- - -0-
1
2 l'
0
2' 1
0 0
0
1
n2 , n,
0
,
and n t = n 2 so that
K*t =
1
o
(9.97)
n2 n, 0 ----0
[
n ,2 n�
0
K=
nt = n,
1
0
0'
Therefore, obviously,
We easily show that
0
1
0 =
0'
0
n, 0
Then
n2 0
I
I
0 1
0
n�
0 n�
]
(9.98)
(9.99) (9.100 )
or
n2 I 0 I o I 1
VKO VK , VK2 VK,o VK " VK'2
0 , n,
(9.101)
Now (9.95) may b e written in matrix form, using I and V as defined in (9.94) and (9.100) respectively, as Premultiplying by K*t gives
V = Z I + Vz
V = K*tZ 1 + K*tVz = (K* tZ K) i + K*tVz = z i + Vz But, for a Z-type fault, V = 0 so that we compute , assuming Z-, exists, i - Z-, V'z =
(9.1 02) (9.1 03) (9.1 04 )
339
S i m u ltaneous F a u lts
We easily show that
where
Z- I
does exist since
Zll,Z1 2 ,Z2 h Z 22 and
Z
=
rLZ21Z l l Z12Z22]
K*t Z K =
(9.105)
are defined by equation ( 9 .62) in section 9.4. Also 0
"9'z
=
Vz1
0- -
K* t
0
=
Vz2
0
[n: VZI] nl VZ2
(9.106 )
The sequence currents at the fault are then easily found by premultiplying ( 9.104) by K, i.e . , (9.107 )
9.8
Constraint Matrix for V -type and H -type F aults
Proceeding in exactly the same manner as above , we may easily establish the constraint matrix K for Y -type faults. For this type of fault it is convenient to write
[Iai ] = [IKi ] = [Y II(i) Y12(i)l la'i IK't Y21(i) Y22(i)J
i = 0, 1 , 2 (9.108)
We then develop the constraint
V=KV or
VKO VK1 VK2 VK � VK'1 VK'2
0 1
2 =
(j
0
(9.109)
0'
1
n2 nl
0 0 0
-----
0
1
n2 2' nl [VKO] [VI] VK,
0
0
h 2 1 (2)
K'i
I
I I I I I T I I I I I
hi
c5 l
Ih 2
i = 0, 1 , 2
'
h 12(0) 0
0
0
h 1 2( 1 )
0
0
0
h 1 2(2)
-------------
h22(0)
0
0
0
h22( 1 )
0
0
0
h22(2)
(9. 1 1 7 )
laO la l la2
VK,o VK 'I VK 'J
0
Vh 1 0 +
0
Ih2 0 (9. 1 18 )
Then we compute
S i m u ltaneous F a u lts
341
(9.1 19) and K is exactly the same as before. Using the notation of (9.1 1 ), we write M = H N + Mh
(9.120)
N = KN
(9.121 )
But
and for power invariance (9.122) Premultiplying (9.120 ) by K*t , we compute
M
where
=
UN
+ Mh
(9.123 )
(9 . 1 24)
But M = 0 for this type fault, so that
an d N =
-
K
A- I �h
N=
or
-U-I Mh
(9.125)
(9 . 1 26)
Problems 9.1. 9.2. 9.3.
Compute the Y parameters for the two-port network of Figure 9 . 2 . Check against the Z parameters of Example 9 . 1 . Compute the H parameters for the two-port network of Figure 9. 2. Check against the Z parameters of Example 9. 1. Compute the A parameters for the two-port network of Figure 9 . 2. Check against the Z parameters of Example 9 . 1 .
342
9.4.
C h a pter 9
Compute the Z, Y, H, and A parameters for the two-port networks of Figure P9.4. z •
(a )
( b)
� r' : z
z
, ...
z
(c )
(d )
•
•
(. )
•
(f )
Fig. P9.4.
9.5.
Compute the A parameters of the two-port networks of Figure P9.5.
(a)
IDEAL TRANS FORMER (n rea l )
(b) PH ASE S HIFTING TRANSFaWER (n a ' e J8)
(c )
(d )
~ (e )
Fig. P9.5.
9.6.
Compute the Z and Y parameters of the following two-port networks. Z3
ZI
� � ( b)
+
VI •
(el
-
•
Fig. P9.6.
Z,
Z
2
(d)
343
S i m u lta neous F a u lts
9.7. 9.8. 9.9. 9 . 1 0. 9.11. 9. 1 2. 9.13. 9.14.
9. 15. 9. 16.
Compute the G and H parameters of the two-port networks of Figure P9.6. Compute the Z and Y two-port equations for the networks of Figure P9.6 if the sources shown in each network are regarded as independent sources of magnitude Vs or I•. Repeat problem 9.8, solving for the G and H equations. Repeat Example 9 . 3 if the controlled source has a value 1 = kI2• Verify the first column of Table 9.5. Verify the second column of Table 9 . 5 . Verify the results of Example 9.5 by direct computation of the H equation. Sketch the series connection of two-port networks of Figure P9.4 and determine the Z matrix of the interconnected network for the following combinations: (a) in series with (b). 4. (c) in series with (d). 1. (a) in series with (c). 5. (a), (c), and (e) in series. 2. (b) in series with (c). 6 . (d) in series with (f). 3. Investigate the parallel connection of the combinations named in problem 9. 14. Investigate the hybrid (series-parallel) connection of the combinations named in problem 9. 14.
9 . 1 7 . Compute the transmission parameters for a cascade connection (left to right) of the fol lowing two-port networks from Figure P9.4. 1. (a) and (b) with impedances labeled Z 1 and Z2 respectively. 2. (b) and (a) with similar labeling. 3. (a), (b), and (a). 4. (b), (a) , and (b). 5. (a) and a two-port network with matrix A.
6.
A two-port network with matrix A and (a).
9 . 1 8. Consider the positive sequence network only and identify the currents and voltages de fined in Figure 9.9 in terms of sequence currents and voltages of phase a.
' (a) For a shunt unbalance at F, shunt unbalance at F (b) For a series unbalance at F, shunt unbalance at F' .
.
9. 19. Verify equation (9.54). 9. 20. Beginning with the currents entering the series-series connection given by (9.65), develop
an expression for the currents entering each of the ports of each sequence network (in side the transformers), and simplify as much as possible. 9.21. Repeat Example 9.6 if the fault condition is given as:
Fault F : lines b and c open ' Fault F : SLG fault on a
9. 22. Repeat Example 9.6 if the fault condition is given as:
Fault F: SLG fault on phase b Fault F' : SLG fault on phase c 9. 23. Repeat Example 9.6 if the fault condition is given as:
Fault F: lines a and b open ' Fault F : lines b and c open
9. 24. Repeat Example 9.6 if the fault conditions F and F' are reversed. 9.25. Verify (9.69). 9. 26. Beginning with (9.78) compute the voltages at each port of the sequence networks (in
side the transformers) and show how these voltages can be used to completely solve the sequence networks. 9.27. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: 2LG fault on phases a and b ' Fault F : line b open
C h a pte r
344
9
9. 28. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: 2LG fault on phases a and b Fault F' : 2LG fault on phases b and c
9. 29. Repeat Example 9.7 for the parallel-parallel fault condition:
Fault F: line a open Fault F' : line c open
9. 30. Beginning with (9.91) compute the sequence voltages and/or currents inside the trans
formers and show how these quantities can be used to completely solve the sequence networks. 9. 31. Repeat Example 9.6 for the series-parallel fault condition: Fault F: SLG fault on phase c Fault F' : 2LG fault on phases b and c
9.32. Repeat Example 9.6 for the series-parallel fault condition:
Fault F: SLG fault on phase b Fault F' : line b open 9.33. Repeat Example 9.6 for the series-parallel fault condition:
Fault F: lines a and b open Fault F' : line b open
9 . 3 4 . Repeat Example 9.6 for the series-parallel fault condition:
Fault F: lines a and b open Fault F' : line c open
9.35. Compute the constraint matrix K for the fault condition of Example 9.6 using definition (9.98) for K. Then solve for the symmetrical components of the fault currents by using (9. 1 07). 9. 36. Compute the constraint matrix K for the fault condition of Example 9.7. Then solve for the symmetrical components of the voltages aUhe faults by using (9. 1 16). 9.37. Compute the constraint matrix K for the fault condition of problem 9.31. Then solve for the symmetrical component quantities at the fault by using (9. 126). 9. 38. Compute the inverse of the Z matrix of (9.95). Is this the same as the Y matrix of (9. 1 1 2)? 9. 39. Suppose that for the Y-type fault we construct a constraint matrix based on the current equations of Table 8.2. Investigate using this constraint matrix to solve for the sequence
quantities in a Y-type fault.
chapter
10
A n a l yt i c a l S i m p l ifi c a t i o n s
The preceding material on symmetrical components is based upon what is often called the three-component method. i.e there are three sequence networks and three sets of sequence voltages and currents. This is emphasized in the com putation of simple faults in Chapter 3 and is continued for more complex situa tions in Chapter 9. There is one concern, however, in using the three-component method. and that is the large number of network compon.ents required to describe the system. Recall that we begin with a three-phase system and simplify it by means of a single-phase (or per phase) representation. But in order to completely describe unbalanced situations, we need three such single-phase representations, namely the positive, negative. and zero sequence systems. What then has been gained? Perhaps we would have been just as well off to have retained the three phases a . b . and c . This problem of the number of components needed to represent a system is crucial, whether the problem is to be solved by digital computer. by network analyzer, or by hand. In computer solutions, whether analog or digital. the net work representation occupies the machine hardware or memory. and this is al ways limited by the size of the machine itself. Thus an arbitrary upper limit is always placed on the size of system which can be solved, and this limit is related to the computing device to be used. .•
1 0. 1
Two-Component Method
The two-component method for solving unbalanced system conditions is based on the idea that the positive and negative sequence networks have nearly the same impedance. Recall that the positive sequence voltage is defined to have the phase sequence of the system generators. We arbitrarily label this sequence a-b-c . The negative sequence then has phase sequence a-c-b. and this is the only difference between the two. Both are balanced sets of voltages (or currents). It is intuitively obvious that any passive network components, such as trans mission lines or transformers. present the same impedance to either positive or negative sequence currents. Thus the impedances Z and Z2 for these compo nents are equal. Only the rotating machines have different positive and negative sequence impedance. In synchronous machines. which contribute most of the fault current in a power system, these sequence impedances are nearly the same if we consider only the subtransient condition. Indeed. from Table 6.1 we see that for both round rotor and salient pole machines the subtransient and negative sequence reactances are exactly equal, i.e ., 1
345
346
Chapter
10
(10. 1 )
This is not exactly true for synchronous condensers or synchronous motors. However, these devices are usually small compared to the generator ratings, and some error in their fault computation can be tolerated. Because of the equality or near-equality of all positive and negative sequence impedances we make the assumption that Z2 Z l ( 10.2 ) for all circuit components; i.e ., we will replace all negative sequence impedances by the corresponding positive sequence values. I Referring to Figure 2 . 5 with Z2 = Z l , we may rewrite the primitive sequence network equation (2.58) as =
(10.3)
If we add and subtract the second and third rows of ( 10.3 ) and replace these rows by the resulting sum and difference quantities, we have (10.4)
These equations are interesting because the last two rows are both positive se quence equations. This is confirmed by the fact that each contains VF , a posi tive sequence source, and Z., a positive sequence impedance. These equations, therefore, describe the positive sequence Thevenin equivalent of Figure 2 . 5 . If + and the this is true, the sum and difference quantities corresponding currents are also positive sequence quantities. Thus under the assumption ( 1 0 . 2 ) we have created a new mathematical description consisting of one zero sequence equation and two positive sequence equations as in ( 10.4). Because of the interest in positive-plus-negative and positive-minus-negative quantities it is also convenient to similarly rearrange the analysis and synthesis equations (2.21 ) and (2.23). For convenience we define the sum and difference quantities Va l
Va2 , Va l - Va2 ,
with a similar definition for currents. Then from we add and subtract the second and third rows to derive the new equation I I Some engineers use the average of Z and Z2 for machines.
(1 0. 5 )
(10.6)
Ana l ytical S i m p l ifi cations
347
(10.7)
or in matrix form V 0 1: 4 =
B- 1 Vabc
(10.8)
where V 0 1: 4 is defined as noted and 1
-1
-1
j y'3
- j y'3
From ( 1 0 . 8 ) we compute
[�
where
1
B 6= !h
J
1
1
0
(10.9)
(10.10)
J
- 1 /2 - j y'3 /2
+j y'3 /2
- 1 /2
(10.1 1 )
Using these relations, we may solve for any of the shunt or series unbalances as before since we may still write the same number of equations. The number of networks, however, is reduced from three to two since both the sigma and delta quantities are defined as voltages or currents associated with the positive sequence network. The two primitive sequence networks are shown in Figure 1 0 . 1 . FO
·
_
Fig. 1 0 . 1 .
FI
loot lli=F=-' va O
Zo
1aI
I04 or
HI
1
Z,
O�hVF •
,
L
Va I
Va6 or
r
Two sequence networks for the two-component method.
I.
SHUNT FAU LTS
We begin with an analysis of shunt faults which will parallel the three-compo nent development of Chapter 3 . Frequent comparison with the results of Chapter 3 is recommended . 10.2 Single-Line-to-Ground F ault
From Figure 3.2 we note that phase a is the symmetrical phase and that the boundary conditions are (10.12)
348
Chapter
From (10.10) and (10.12) we compute 0 = h1 100 1 Ie = 0 = h 100 -
Ib =
10
1 . vr:3 2 h 10 '1:. - J 2h lo A 1 1 + . va l 2h 0'1:. J 2h oA
(10.13)
Adding these equations, we compute for any h and subtracting, we have
2 100 = 10'1:.
(10.14)
lo A = 0
(10.15)
From the second equation of (10.12) we write for any h o 2 ) = Z, (loo
Voo + ( VO l + V
and substituting lao from (10.14 ) ,
a
+l l +
la 2 )
aO + Va '1:. = (3/2) Z,la '1:.
( 10.16)
VaO = - Zo lao = - Z"2o ( 2 1aO)
(10.17)
V
Equation (10.15) requires that the positive sequence network be open at the fault point. This network is shown in Figure 10.2a and is recognized to be the normal (unfaulted ) positive sequence network. All voltages measured in this normal network are to be labeled delta quantities as noted. Equations (10.14) and (10.16) dictate that the positive and zero sequence networks be connected in series to compute the zero sequence and sigma quantities. This connection is shown in Figure2 10.2b. We also note that since the current through the zero sequence network is 1ao , we satisfy the first equation of (lO A) as follows: Thus we use only one-half the actual zero sequence impedance values. FI
FI
:tot. = 0
Z,
hvF.
t
(0 )
t.
VoA HI
,-
t
1: oI
Z, •
hV(
2 Iool ZO/2
-
HI FO
f--
NO
1''' -''''''
t.
Val
r
1.
VoO
3Zf
"2
r
( b )
Fig. 1 0 . 2.
Sequence network connections for a SLG fault on phase a with fault impedance Z, : ( a ) delta, (b) sigma and zero.
3 49
Ana lytica l S i m p lifications
From the networks of Figure 10.2 we easily compute the sequence voltages and currents at any point in the network. Then we apply equation ( 1 0 . 1 0 ) to synthesize phase quantities. At the fault the total current is 10 = 3 /00 • By inspec tion of Figure 10.2 we compute h VF (Zo /2 ) + Z l + ( 3 Z, /2 )
2/0 0 =
a
or
I
-
-
o
h VF Zo + 2 Z 1 + 3 Z,
(10.18)
This i s exactly the same a s ( 3 . 5 ), the current computed b y the three-component method , with Z , set equal to Z 1 1 0.3
•
L ine-to-Line Fault
From Figure 3 .9 we note that phase a is the symmetrical phase. Also , from ( 3.6)-( 3 .8 ) and Figure 3 .9 we write the boundary conditions 10 = 0,
Vb - Ve = Vbc
Ib = - Ie ,
�aoJ � [1 -1
= Z , /b
(10.19)
Substituting (10.19) into the current version of ( 1 0 . 7 ) , we compute 1
lo T.
=
2
0
lo t..
j V3
[��VJ � [�
The known voltage is
Vb e
which we compute from ( 1 0 . 1 0 ) as follows :
t
1
Then
(10.20)
1
� J r�:� � �a�
_ 1 /2
-j
- 1 /2
+j V3 /
/
laF.
( 1 0 .2 1 )
are zero . Thus the sigma cur From ( 1 0 .20) we learn that both 100 and rents and voltages are found from the normal network as shown in Figure 1 0 . 3a. F
I o t -O
t .
Ia
t
t..
FI
ZI •
h"F. (0)
Fig. 1 0. 3 .
f+ ,.
V a t.. HI
Z, / 2
(b)
Sequence network connections for a LL fault on phases b-c with fault impedance Z , : ( a ) sigma, ( b ) delta.
3 50
Cha pte r 1 0
From the third boundary condition we compute for any h ,
(10.19)
and equations
(10.20) and (10.21) (10.22)
This equation is satisfied by the network of Figure 10.3b. From the two net works of Figure 10.3 the sequence voltages and currents may be completely determined . The total sequence current at the fault may also be computed from Figure 10.3. From Figure 10.3a
(10.23)
and from Figure 1 0.3b
( 10 .24 )
la /1 = Z, + Ztl2
Adding
(10.23) and (10.24), we easily compute Ial = 2Z1h V+F Zf
(10.25)
This is exactly the same result found by the three-component method if we re place Zl = in ( 3 . 11 ).
ZI
1 0.4
Double L ine-to-Ground Fault
From Figure 3 . 1 3 we again note that phase a is the symmetrical phase. From ( 3 . 1 2 )-(3 . 1 4 ) and Figure 3 . 1 3 the boundary conditions are
la = 0, Vb = ZfIb + Z,(lb + Ie), Since Ia = 0, we compute
Ve = ZfIe + Z,(lb + Ie) (10.26)
(10.27) from which we observe that all sequence quantities will usually be nonzero since and are gen the right-hand side of will be nonzero . This is because erally neither equal nor opposite for this type of fault. We also observe from ( 1 0 .2 7 ) that for any h,
(10.27)
Ib
Ie
(10.28)
and this establishes one of the required network connections. From the second and third boundary conditions combined with ( 1 0 . 2 7 ) we write
But from
(10.10)
351
Ana lytical S i m pl ifications
Combining, w e have for an y h ,
(10.29)
Va6 = Z,Ia6
This completely determines the delta network connection shown in Figure 1 0.4a.
FI
-
FI
ZI
hVf':O NI �
t.
VoA
l-
Zf
hV
NI
(a)
Fig. 10.4.
fl ol
ZI t. � O Vol l-
l: o A
FO
Ioo
t
f.
2Z 0 2yae 1-
NO
(b)
Sequence network connections for a 2LG fault on phases b-c with phase impedance
g
Z, and ground impedance Z (see Figure 3 . 1 3 ) : (a) delta, ( b ) sigma and zero.
(10.26) and (10.27) we compute Vb + Ve = Z, (lb + Ie) + 2 Zg(lb + Ie) = (Z, + 2Zg)(/b + Ie) = (Z, + 2 Zg) � lao But from (10.10) Vb + Ve = h2 VaO h1 Val: Again from
( )
-
Combining, we have for any h
2 VaO - Val; = (3Z, + 6Zg) lao
(10.30)
This equation is satisfied by the network connection of Figure 10.4b where twice the normal zero sequence impedance is set in order to force the voltage relation ship of From Figure and Figure 1 0 Ab we compute
(10.30).
10.4a h VF Ia 6 Z I + Z, '
Adding, we compute
2 /a l
(
V Z I +1 Z, + Z 1 + 2Zo +1 3Z, + 6Zg
=h F
which can be reduced to
la l: Z 1 + 2Z h V+ F3Z, + 6 Zg 0
__
)
(10.31) 3_Z� (Z Z ) Z + , + , ) + I _ �0� � -,-� g � _ _ Z I + Z, + -,--(Z - Z o + Z + 2Z, + 3Zg This equation is exactly the same as (3. 2 4) which was derived from the three component method if Z2 in ( 3. 2 4) is replaced by Z I . __
1
3 52 1 0.5
Chapter 1 0
Three-Phase Fault
3lP
The general fault condition is shown in Figure boundary conditions
3.18 where we note the
Va = Zfla + Zg (Ia + Ib + Ie) Vb = Zflb + Zg(Ia + Ib + Ie ) Ve = Zfle + Zg(Ia + Ib + Ie) Replacing Ia + Ib + Ie by (3 /h ) Iao and adding, we compute Va + Vb + Ve = 0 = (3/ h ) Zflao + (9/h )ZgIa o or
� ] �1
(10.32)
la O = 0
�J �
�
(10.33)
(10.7) we compute sigma and delta quantities 1 1 0 ao h h = = 2 1 - (Ib + Ie ) I T, 3 2 - 1 -1 I 3 _ 0 j y'3 j y'3 I: I: A j � (Ib le ) But, for this connection since Ia + Ib + Ie = 0, then Ia = - (Ib + Ie) and 0 Ia O h Ia T, = "3 3 Ia (10.34) Ia A j y'3 (Ib - Ie) From the first of equations (10.32) we have, with Ia + Ib + Ie = 0, Va = Zfla = (Z,!h)IaT, but Va = ( l /h ) Va O + ( l /h ) Va T, = ( l /h ) VaT, since la O 0, or (10. 3 5) VaT, = Z,IaT, Subtracting the second two of equations (10.32), w e compute, with Ia + Ib + Ie = 0 and incorporating (10.34), Also, from
J
[j [
_
J
=
But from
(10.7)
or for any h,
(10.36) Thus the equations for the sigma and delta networks are exactly the same and are as shown in Figure
10. 5 .
3 53
Ana lytical Si mp lifications
-
FI
-
FI
l oA
I a l:
t
Va l
t.
VaA
Zf
t
,-
(a)
Fig. 1 0 . 6 .
If
(b)
Sequence network connections for a 34> fault with impedance Z, in each phase : (a) sigma, ( b ) delta.
By inspection of Figure 1 0 .5 we compute lor. = hVF /(Z 1 + Z , ) ,
loA = hVF /(Z 1 + Z, )
(10.37)
and adding we have 10 1 = h VF /(Z 1 + Z, ) . I I . S E R I ES F AU LTS
We next consider the application of the two-component method to series or longitudinal faults. Again, the network currents and voltages at the fault point are defined as indicated in Figure 3.20 with sequence networks as defined in Figure 3 . 21 . 10.6 Two L ines O pen ( 2LO)
a
The series unbalance for 2LO is shown schematically in Figure 3 . 32 for which the boundary conditions are seen to be
Ib = Ie = 0,
�aj [
Vaa, = Zl
J [/aJ l/a�
( 10.38)
where Z is the impedance in line a . Then from the current version of (10.7) lor.
loA
=
:
1
1
1
2
-1
-1
0
j v'3
- J. v'3
0 0
=
� 2/0 0
( 10 .39 )
Thus the delta network must b e open for this type o f fault. Also, w e observe that ( 1 0 .40)
the same as for the SLG fault. From the second boundary condition (10 .38) incorporating ( 1 0 .40 ) we write for any h , ( 1 0 .4 1 ) Equations (10 .40) an d ( 1 0 .4 1 ) fix the connection o f sigma and zero sequence net works to be that of Figure 1 0 .6. This connection has a striking resemblance to Figure 10.2 for a SLG shunt fault. If we interpret h VF to be the Thevenin equivalent open circuit voltage gen erated in the sigma network, we compute 210 0
- Io r. - Z l
h VF + Zo /2 + ( 3/2 ) Z
3 54
Ch a pter 1 0 FI
loI NI
,
F' I
Fig. 1 0. 6.
Z,
21 0 0
I
I
1.
Voa ' I F" FO
t
Zo /2
r t.
Voa 'O
l
1 F' O
3 Z
'2
r
Sequence network connections for phases b and c open and with impedance Z in phase a.
But we also know that la A
= O.
Adding these two equations, we have
= 2 /a l = Z
2/a O
h VF
I + Zo /2 + ( 3 /2 ) Z
or la o = la J
=
h VF + 2 Z J Zo + 3 Z
(10.42)
which agrees exactly with ( 3.86), derived by the three-component method, if we set Z 1 = Zl in that equation. 1 0.7 O ne L ine Open ( 1 LO)
If only one line is open and we assume that is line a , we have the boundary conditions ( 10.43) where we assume an impedarice Z in the sound lines. Since la = 0 , ( 1 0 . 2 7 ) applies and we immediately establish the similarity be tween this fault and the 2 LG shunt fault where for any h, (10.44) From the second two equations of (1 0.43) we compute the sigma and delta quantities
Vbb' + Vee '
= Z (lb
+ Ie ),
( 1 0 .45)
But from ( 1 0 . 1 0 ) we compute
( 10.46) and from ( 1 0 .2 7 )
Ib + 1e
=
-
3 /a � h
-
...
'
Ib
- -I
e
=
_ l' va Ia ... ..
h
( 10 .47 )
355
Ana lytica l S i m pl ifications
Substituting ( 1 0 .46) and (10 .47 ) into ( 10 . 4 5 ) , we compute for any h, 2 Vaa, o - Vaa'l:
=
Vaa '4
=
-
3 Zla l:
(10.48)
Z la 4
( 1 0.49 )
These two equations completely specify the network connections of Figure 10.7. fl
N'I Fig. 1 0 . 7 .
1
l
-
L o1:
lot.
I
f.
ZI
r
I
F, J
H ' z,
Z
Voo't.
Sequence network connections for line
F" a
3Z
F'O
open and an impedance Z in lines b and c.
This network connection is a parallel connection similar in construction to that derived for the 2LG shunt fault. By inspection we compute
and adding we find , after algebraic reduction, la 1 = Z1
+
Z
+
(Z l + Z)(Zo + Z) Zo + Z l + 2 Z
( 1 0.50)
"-'--"---"-'---=-
This is exactly the same result computed by the three-component method and given by ( 3 .7 8 )-(3.80) if we set Z2 = Zl . I I I . CHANGES I N SYMMETRY WITH TWO-COMPO N ENT CA LCU LATIONS
An obvious similarity exists between certain shunt and series fault connections computed by the two-component method. If we take advantage of these similar ities, we should be able to extend the two-component computations to situations in which any phase is the symmetrical one. It is again convenient to limit our analysis to common fault connections, namely , the shunt SLG and 2LG and the series 1 LO and 2LO. These situations are shown in general form by Figures 8 . 1 and 8 .5 for which we write the general equations, similar to (8 .6)-(8 .8) and ( 8 . 18) , as
[
J
Va o - 3 Zg la
where Zg
=
va .. ' = Ip + Ipq,
= Is + Isr I� = Iq + Iqp I;
( 1 1. 43)
Combining these constraints with the original indefinite admittance equations, we have 10 = YOO Vo +
.
.
.
+
Yo r Vr
+ Yo. V.
+
.
.
.
+ YO P Vp
I; = YrO Vo + · · · + ( Yrr + Y;') Vr + ( Y,.. - Y;') V. + · · · + ( Yrp + Ym ) Vp + ( Yrq - Ym) Vq + · · · + Yrn Vn
I; = Y.o Vo + · · · + ( Y.r - Y;.) Vr + ( Y.. + Y;.) V. + , · · + ( Y,p - Ym ) Yp + ( Y.q + Ym ) Vq + · · · + Y.n Vn ' + ( Ypp + Y;q ) Vp + ( Ypq - Y;q) Vq + · · · + Ypn Vn
I; = YpO Vo + "
' + ( Ypr + Ym ) Vr + ( YJII - Ym ) V. + "
I� = Yqo Vo + "
' + ( Yqr - Ym ) Vr + ( Yqo + Ym ) V. + · · · + ( Yqp - Y;q) Vp + ( Yqq + Y;q) Vq + , · · + Yqn Vn
(11.44)
From Figure 1 1 .1 5 we observe that terminals r and p are the dotted or polar ized terminals. Therefore Ym is added to all like (Le . , dotted or not dotted) loca tions namely rp and pr and also to locations sq and q s. By the same reasoning we sub tract Ym from all unlike locations, namely rq , qr, sp , and ps (see [65] ) . Extension of the above to any number o f mutually coupled elements follows exactly the same rules. Note that the mutual admittance always occurs between pairs of lines, and the rule for adding Ym to the matrix is the same for every pair of mutually coupled branches.
Example
1 1. 9
Compute the zero sequence indefinite admittance matrix of the system of Example 1 1 .1 by the direct method discussed above .
Solutio n From the data of Table 1 1 .1 and the network topology of Figure 1 1 . 5 (with
'114
= 'I t = 0) we record the self admittances by inspection as follows :
[V'
Computer Solution M ethods Using the Adm ittance M atrix
NO
NO 1
Y self -
0
0
2
2
3
0
0
- 41 2
41 3
V. - 41 4 - 41 5
413 + 41 6 - 1# 3
- 1# 2
3
1
-
- 1# 6
]
o
41 3 + 41 4 + 41 5 - 1# 4 - 41 5
41 2 + 1# 4 + 415 + 41 6
where from Example 1 1 .1 41 3 = j 7 .143,
41 2 = - j 50 .000, 41 5
41 4
-
= - j 1 0 .526,
385
=
-
j 1 2 .632
416 = - j 5 .882
Then NO NO - j 50.000 Yself =
1
2
3
0
0
j 50. 000
1
0
- j13.025
j 7 . 143
j 5. 882
2
0
j7. 143
- j30.301
j23. 1 58
3
j 50. 000
j5. 882
j23. 1 58
- j79.040
The mutual admittance between branches 4 and 5 is V m +j 5 .263. This admit tance should be added and subtracted according to ( 1 1 .44) , i.e . , =
Add V m to (r , p ), (p , r) , (s ,
q ), ( q , s ) Subtract v m from (r , q ), ( q , r ) , (s, p ) , ( p, s ) In this example r = p = 2 and s = q = 3. Therefore we must
NO 1
Y=
2 3
NO = j -
1 2
3
[ [
Add 2 41 m to ( 2 , 2 ) , ( 3 , 3 )
Subtract 2 V m from ( 2 , 3 ) , ( 3 , 2 ) NO 50 .000
3
2 0
0
j50 .000
0
- j 1 3 .025
0
j7 . 1 43
- j30.301 + 2 41 m
j 5 . 88 2
j 2 3 . 1 58 - 2 41 m
j 5 0 .000
NO
50 .000
-
1
1 0
j 7 .143
2 0
j 5 .882
3 0 50 .00
]
0
1 3 .025
- 7 .143
- 5 .882
0
- 7.143
19.775
- 1 2 .632
- 5 . 882
- 1 2 .632
68 . 5 1 4
50 .000
J
j 2 3 . 1 58 - 2 41m
- j79.040 + 2 y m
This is the same as the result in Example 1 1 .4 if branch 1 is neglected in that ex ample . Property 4 provides a straightforward algorithm for computing the indefinite (or definite ) admittance matrix for any network, including the effect of mutually
Cha pte r 1 1
386
coupled lines. There are other methods of accomplishing this same result such as the incidence matrix transformations ( 1 1 . 18 ) and ( 1 1 . 22). 1 1 .5
Definite Admittance Matrix
The indefinite admittance matrix defines the relationship between node volt ages and currents of a network in such a way that the details of the primitive admittances are suppressed. This is desirable since we seek a network solution where the total memory requirement for matrix storage is minimized. The indefinite admittance matrix is an efficient way of relating node voltages and currents and it may be formed with a minimum of effort. These are important considerations. Another important feature of the indefinite admittance matrix is its sparsity. This depends entirely on the network represented, but power systems are usually quite sparse, having between 1 .0 and 1 . 5 branches per node. This means that the indefinite admittance matrix itself has many zero elements and that considerable savings in memory requirements may be gained by storing only the nonzero ele ments in some simple tabular form. In power system analysis we are interested in the definite admittance matrix or simply the node admittance matrix Y defined by the equation I=YV
(11.45)
The admittance matrix is formed by crossing out the rows and columns of all nodes connected to the reference node. In a power system this will usually be the zero potential nodes NO, N1 , or N2. Moreover the n currents and voltages of ( 1 1 .45) are independent sets of complex quantities. Given the n independent voltages V, one can solve for the currents I by direct application of ( 1 1 .45). Also, since Y is nonsingular, one can find V when I is given by computing V=
y- I I Z I
( 1 1 .46)
=
In applying ( 1 1 .45) and ( 1 1 .46) to the solution of faulted networks, we usually make the simplifying assumptions given in Table 1 1 .2. Table 11.2. Simplifying Assumptions Used in Fault Calculations Assumption
Comment
1. All load currents negligible.
2. All generated voltages are equal in phase and magnitude to the posi tive sequence prefault voltage, h V,.
3. The positive and negative sequence networks are identical. 4. The networks are balanced except at fault points. 5. All shunt admittance (line charging susceptance, etc.) negligible.
Zabc are symmetric in all references for all lines. The only network reference connection is through generator impedances.
Computer Solution M ethods Using the Ad m itta nce M atrix
387
Since all generated voltages are equal in phase and magnitude, all generator nodes are connected together and a voltage h V, is applied at that point as indi cated by Figure 1 1 . 1 6 where only the positive sequence network is shown. Also, I u a Ou
� QQ - 199
POSITI V E SEQUEN
•
•
Q2
IQ2
_
QI IQI
N ETWORK
:Ii ,aO
- ll
l Q2
!!t-
ff
-.!L fI I ·
l QI
NO CONNECTION TO N I SINCE ALL S H UNT PATH S OPEN -----""
}iL� }
I FAIA.TS
Vo l- f l
+
NI B U S Fig. 1 1 . 1 6 . Positive sequence network which results from the assumptions of Table 1 1 .2 .
V, [ IIf,] lVaVall-_',J Ig Val-g Vr• Vk Val-k V"
since loads and shunt admittances are all neglected, the generator connections are the only connections to N1 , the reference node . The negative and zero sequence networks are similar except that the voltage h is zero and the generator nodes are connected directly to the reference . Since there are f fault nodes, e load nodes and g generator nodes, we may write the indefinite admittance equati o n for Figure 11.16 in terms of the positive sequence voltages as
=Y
( 11.47)
We now consider a change in reference . Since the only reference connections are made through the generators, suppose we select node 0 as the reference (this is already the case in the negative or zero sequence networks) . In terms of this reference all generator node voltages are zero and all load and fault voltages are the positive sequence voltage less h If we identify the node voltages with re spect to node 0 by a single subscript, the node voltage at any node k tnay be written as
- h
=
or in vector form V
Then adding Y VF
=
==
Va l - [ 1
0
k = 1 , 2, . . . , n
1
( 1 1 .4 8 )
from each side of ( 1 1 .4 7 ) , we compute
( 1 1 . 49 )
Chapter 1 1
388
Since nodes gl , g2, . . . , gg are connected to the reference, the last g rows and col umns of Y may be eliminated (property 2) to write the definite admittance equation
]
r- I, L If
=
Y
[V,V ] I
=
[
y" YI,
V� [ YuJ V,j Y,�
( 1 1 . 50 )
where the order of ( 11 .50) is (n g). Furthermore, since It = 0, we may reduce this expression to find I, with the result -
V,
I, = - ( Y,, - Y" Yi l yr ,) V,
( 1 1 . 51 )
where is the vector o f voltages applied t o the faulted nodes, measured with re spect to node O. For a 3rt> fault with zero fault impedance the sequence voltage Va l., is zero at the faulted buses. Then from ( 1 1 .48) the vector of voltages at all faulted buses is 1 where V, is the scalar prefault voltage, usually taken to be unity. Su bstituting into ( 1 1 .5 1 ) , we compute = 1 l ] t hV, ( 1 1 . 52) I, + (Y,, - Y,f Yr, I Yr ,) [ l The magnitude of I, is simply scaled up or down in pro portion to the magnitude of h V, Equation ( 1 1 . 5 2 ) provides a compact form of the solution for all fault cur rents I, under 3d> faul t conditions. If there is only one fault, this equation is scalar. Note that the procedure of finding I, for each of the n nodes requires n different complex matrix inversions each of order ( n - 1 ) . This is a serious dis advantage for this method and makes its use very expensive, even for small net works. For this reason an alternate method is sought. One alternative is to de velop an iterative technique which will converge on the solution If in a stepwise procedure. A more direct alternative is to invert the admittance matrix and find the fault currents by working with the impedance matrix. Often the matrix inversion is avoided by forming the impedance matrix without first developing the admittance matrix [64] . This procedure has its own special problems as noted in Chapter 1 2 . .
Example
1 1. 10
Compute the positive sequence admittance matrix for the network described in Example 1 1 . 1 and Table 1 1 . 1 . Then compute the total fault current 1. 2. 3. 4.
For a 3rt> fault at node 1 . For a 3rt> fault at node 2 . For a 3rt> fault at node 3 . F o r simultaneous 3 rt> faults at nodes 1 and 2 .
Assume zero fault impedance i n each case . Let h
= 1 and V, = 1 . 0 .
Solution From Example 1 1 .6 we have for the positive sequence network
389
Computer Solution M ethods Using the Adm itta n ce M atrix
1 Y= 2 3
[
]
3
1
2
j 2 4 . 19
j 1 2 .50
j 7 .69
j 1 2 .50
- j45.84
j 33.34
j 7 .69
+j33.34
- j 46.03
Since the faults are at different locations, ( 1 1 .52) must be solved separately for each case. F or example, for case 1 we compute If! = ( Y l l - Y t Yt, l Y r dh Vf l
or In = - j 2 4 . 1 9 - [j 1 2 . 5 0
j 7 .69]
[
In
= - j45 .8 4 - [ 12 . 50
j 33 .34]
and for case 3 If3
= - j46.03 - [j 7 .69
[ { [i;] [- ]
j 33 . 3 4 ]
]
j 3 3. 3�
j 33 . 34
- j 46.03
j24.19
j 7 . 6�
[-
Similarly for case 2 we have
r
[
r ] r r [ ] r [ J
- j45 .84
j 7 .69
- j 46.03
j24.19
j 1 2 . 50
j 1 2 .50
- j 45.84
]
l
1 2.50
j 7 .69
l
j 12.5� j 33. 34
1
j 7 .69
j 33.34
. = - J 7 .852
. = - J 7 . 436
= - j 8. 230
Case 4 is a simultaneous fault on nodes 1 and 2 , so the solution is =
=
j 1 2 . 50
- j 24. 1 9 j 1 2. 50
j 4. 8 3 5
_
j45.84
_
j 7 . 69
j 3 3 . 34
[ _ J' 46 . 0 3 ] - 1 [ J' 7 . 69 J' 33 . 34 ]
}[ ] 1 .0
1.0
- j 3. 6 2 2
The simultaneous faults are less severe than the isolated faults o n buses 1 or 2 because one contribution from the adjacent bus is zero . Example 1 1 . 1 0 illustrates the maj or difficulty of using the admittance matrix to solve for fault currents, namely , that every different fault location requires the inversion of a different matrix . To avoid this numerical problem, in Chapter 1 2 we shall examine the impedance matrix as a means of finding fault currents and develop a method of finding the impedance matrix which avoids a matrix inver sion . Problems
1 1. 1. 1 1.2. 1 1. 3.
Construct the positive sequence primitive impedance matrix for the network of Figure P 1 1 . 1 , using symbols z 1 3 , etc. , for the branch impedances. Construct the positive sequence primitive impedance matrix for the network of Figure P11.2, using symbols Z 1 3 , etc., for the branch impedances. Construct the positive sequence primitive impedance matrix for the network of Figure PH.3, using symbols Z 1 3 , etc., for the branch impedances. Repeat for the zero sequence and assume that Z24 and Z 2 S are mutually coupled by an amount Zm '
Chapter 1 1
390
4 2
Fig. P 1 1 . 1 .
'J------ 2
5
3
Fig . P l l . 3. 1 1. 4.
Construct the positive and zero sequence primitive impedance matrices for the network of Figure P 1 1 .4 , using the line data tabulated below and ignoring all resistances.
Branch Number
Branch Terminals
1 2 3 4 5 6 7
0-3 1-3 3-4 2-3 1·4 2-4 1·5 2·5 5-6
S
9
Rl
1 1. 5.
Ro Xo
Zero Sequence
0 0.10 0. 1 1 O.OS 0.10 0.20 0.15 0.25 0.35
0 0.30 0.40 0.30 0.35 1.00
jO.05 jO.90 jO.SO jO.75 jO.35 j 1.20
00
00
® \ 1 1.6.
Xl
Positive Seq uence jO.20 jO.30 jO.25 jO. 20 jO. 15 jO.50 jO.30 jO.40 jO.90
' �.
0.70 1.00
j 1 . 10 j2.50
Mu tual
JO.20
S
jO.20
6
• •
Fig. Pl l . 4 .
Repeat problem 1 1 .4, including the effect of resistances. Invert the positive and zero sequence primitive reactance matrices of problem 1 1.4 to find the primitive admittance matrices 'YI , and 'Y o respectively.
Co mpute r Solution M ethods Using the Ad m itta nce M atrix
11.7. 11.8.
391
Invert the positive and zero sequence primitive impedance matrices of problem 1 1.5 to find the primitive admittance matrices '11 and 'lo respectively. Find the zero sequence primitive impedance matrix for the network of Figure P11.4 if the lines 1-3, 3-4 , and 3-2 are all mutually coupled as follows: Zm(2,3)
= jO. 1 5,
zm(2,4)
=
j O .10,
Z m(3.4)
= j O.20
1 1 .9. Invert the zero sequence impedance matrix found in problem 11.8 to find the new 'Y o . 1 1 . 10. Compute the positive and negative sequence primitive Z and 'Y matrices for the system of Example 1 1 . 1 . 1 1 . 1 1 . Develop the augmented node incidence matrices A for the following networks with positive branch orientation from smaller to larger node numbers. (a) The circuit of Figure P I L L ( b ) The circuit o f Figure Pl1.2. (c) The circuit of Figure P l 1 .3. (d) The positive sequence circuit of Figure P 1 1 .4. (e) The zero sequence circuit of Figure P 11.4. Use data from problem 1 1 .4 for all networks. 1 1 . 1 2 . Develop a zero sequence equivalent network for the system of Figure 1 1.14 where the mutually coupled line pair is replaced by an equivalent circuit of self impedances. 11.13. Compute the indefinite admittance matrix Y for the following networks using the augmented node incidence matrices found in problem 1 1 . 1 1 and applying ( 1 1 . 18). (a) The circuit of Figure P I L L ( b ) The circuit o f Figure P11.2. (c) The circuit of Figure P11.3. (d) The positive sequence circuit of Figure P 1 1 .4. (e) The zero sequence circuit of Figure P11.4 . 1 1 . 14. Check the result o f Example 1 1 .4 by applying ( 1 1 . 18). 1 1 . 15. Verify ( 1 1. 17) beginning with ( 1 1. 13) and ( 1 1 . 14 ) . 1 1 . 16. Compute the node incidence matrices A for the zero sequence networks given by (a) Figure P1 1 . 1 , (b) Figure P11.2, (c) Figure P11.3, and (d) Figure P1 1.4. 1 1.17. Using the results of problem 1 1 . 16 , compute the corresponding zero sequence definite admittance matrices Y, using ( 1 1 .22). 11.18. Verify the computed result of Example 1 1 .6. 1 1 . 19. Verify ( 1 1 . 22). 1 1 . 20. Using a digital computer, invert the admittance matrices of each circuit of Figures P11. 1-P l l .4 to find the zero sequence impedance matrices for these circuits. 11.21. Compute the positive sequence indefinite admittance matrices of the following net works by applying ( 1 1.29). Neglect all resistances. (a) The circuit of Figure P I L L ( b ) The circuit of Figure P l 1 . 2 . ( c ) The circuit of Figure P 1 1 . 3 . ( d ) The circuit o f Figure P11.4. 11.22. Repeat problem 11.21 including the effect of resistances. 1 1 .23. Prove the five properties of the indefinite admittance matrix stated in section 1 1 .4 . 1 (see [ 5 ] ). 1 1.24. 1}eginning with the indefinite admittance matrices computed in problem 1 1 . 13, find the Y matrix description after the following circuit changes are made: (a) Grounding node 2 of Figure P11.2. (b) Grounding node 5 of Figure P11.4; (c) Connecting nodes 1 and 5 of Figure P11.3. (d) Connecting nodes 2 and 6 of Figure P11.4. (e) Connecting an impedance jO.1 between nodes 1 and 5 of Figure P11.3. (f) Suppressing terminal 4 of Figure 1 1.3. (g) Suppressing terminals 4 and 5 of Figure 1 1 .4 . 1 1 . 25. Compute the positive sequence indefinite admittance matrices for the following net-
3 92
Cha pte r 1 1
works, using "property 4" and ( 1 1.37) to build the matrix one element at a time. Compare the result with that of problem 1 1 . 13. (a) The circuit of Figure PILL (b) The circuit of Figure P11.2. (c) The circuit of Figure P11.3. (d) The circuit of Figure P11.4. 1 1 . 26. Compute the zero sequence indefinite admittance matrices for the network of Figure P11.4, using the result (1 1.44). Build the matrix one element at a time in the order given in problem 1 1.4. 1 1 . 27. Compute the 3ct> fault current using the admittance matrix technique ( 1 1.52) for the following systems (neglect resistance). (a) Figure PI LI with a fault at node 4. (b) Figure PI L I with faults at nodes 1 and 4. (c) Figure P11.2 with a fault at node 2. (d) Figure P11.3 with a fault at node 4. (e) Figure P11.3 with a fault at node 5. (f) Figure P11.4 with a fault at node 4. (g) Figure Pll.4 with a fault at node 5. (h) Figure P1 1.4 with faults at node 4 and 6. (i) Figure P1 1.4 with another generator at node 6 with impedance 0 + jO.1 and a fault at node 4. (j ) Same as (i) with faults at nodes 4 and 5. (k) Same as (i) with a fault at node 4 with fault resistance of 1.0 pu.
chapter
12
Co m p u te r S o l u t i o n M et h o d s U s i n g
t h e I m ped a n ce M a t rix
I n Chapter 1 1 an admittance matrix method was devised which could be used to compute fault currents in a power system. This method, although simple to implement, was observed to have certain disadvantages in its application to large networks. It was also observed that the impedance matrix , although more diffi cult to derive , has certain advantages for fault computations. This is primarily due to the impedance matrix being an "open circuit" network description , and this coincides with the open circuit approximation usually used in fault studies. This chapter will be devoted to a study of impedance matrix methods for use in fault studies. We shall also develop an algorithm for finding the impedance matrix which is more direct and cheaper to implement than performing an inversion of the admittance matrix. 1 2. 1
I mpedance Matrix in Shunt Fault Computations
Consider the network shown in Figure 1 2 . 1 where the network could be the positive, negative, or zero sequence network but where the reference is always node 0 of Figure 1 1 .16 or the common generator node. We arbitrarily define all . , n and all voltages to be the curren ts to be entering the network at nodes I , 2 , voltage drops from each node to the reference. Then we define the impedance matrix by the equation .
.
V=ZI
(12.1)
In expanded form ( 1 2 . 1 ) i s written as Vt V2
=
Vn
Zl 1
Z 12
Z tn
It
Z2 1
Z 22
Z 2n
12
Z nt
Z n2
Z nn
In
( 1 2.2)
If every current except that entering the kth node is zero , we have from ( 1 2.2) VI V2 Vn
Z u! =
Z2k Z nk
Ik ,
Ij =
0 , j =fo k ( 1 2 .3) 393
3 94
Cha pter 1 2
n
In
_
n-PORT
12
N E TWORK
O
GENERATOR NOOE
NODE VOL TA8 E REFERENCE
Fig. 12. 1 .
An n·port sequence network .
.. .
This defines the kth column of the impedance matrix since we may solve ( 1 2.3) to compute
Zih = (VdIh )rro , i,ph ,
i = 1, 2,
,n
( 1 2.4)
Since all impedance elements are defined with all nodes open except one, the im pedance elements are called the "open circuit driving point and transfer imped ances" or
Zih = open circuit driving point impedance, i = k = open circuit transfer impedance,
i =1= k
For small networks these impedances may conveniently be found by injecting = 1 .0 ampere (or 1 .0 pU), and solving ( 1 2 .4) for the 1 .0 ampere at node k , i.e., voltages. The impedance matrix can be used directly for fault computation if we apply the generator voltage to the faulted node in the positive sequence network as noted in Figure 1 1 .16.
Ih
1 2. 1 . 1
Three-phase fau lt computations
N1
Three-phase faults are computed by connecting the faulted node to the bus through a suitable fault impedance as noted in Chapter 3 and in Figure 3.19. When the generator node 0 is used as a reference, the positive sequence net work is as shown in Figure 12.2. Note that the fault current Ifh is the negative of
Zt
- () Zf
n
In - O _
- -..... r--
Ik Ifk n - PORT k _ _---f PO S I T I V E _-"""',.,.. NI SEQUENC BUS '-..._VO l + N ETWOR K I -0 2 .!.... +
/1 + 1 +;
Vk _V, _VI _
+
II- O
'----r-....
GENERATOR
NOD E
NO DE VOLTAGE REFE R E N C E
Fig. 1 2. 2.
Network connections for a 3IP fault on node k .
Ih , the current entering node k.
Since all currents except Ih are zero, ( 1 2 .4) de scribes the network exactly. Solving this equation for we have
Ih ,
Computer Solution M ethods U si ng the I m peda nce M a trix
= Vk /Zkk =
Ik
-
-
I'k
( 1 2.5)
From the way in which node k is terminated we also may write
Vk
= Z,IfIe
h V,
which we may substitute into (12 .5) to eliminate Vk and solve for Ifle or
I'k
395
= h Vr /ZT
(12.6) (12.7)
where ZT = Zkk + Z, . Thus the 3 � fault current at node k depends upon the open circuit driving point impedance at node k . This is the diagonal impedance element Zkk of the impedance matrix and may be thought of as the impedance seen look ing into the network at node k with all nodes except the kth node open. Once the total fault current I,k is known, we may readily find all the node voltages from the impedance matrix equation or, more simply , from ( 12 .3). Thus for any node i we may write
Vi
= Z ik lk
-
= Zik h Vr/ZT ,
i = 1 , 2, .
..
,n
(12.8)
These are the node voltages measured with respect to the reference node O. The positive sequence voltages are measured with respect to node Nl and are found by adding h V, to (1 2.8), or at node i we have
Vo l - i
( i;)
= h V, 1 -
(12.9)
Knowing the voltage at each node, we may compute the current flowing in all lines in the network. This requires knowledge of the primitive impedances. For any pair of nodes i and i we may compute the current flowing in line 'Jij as
Vi - Vj Ii ' =
= h V, (Zjk -
Zik )
(12.1 0)
'Jij ZT 'J 1i Note carefully that the capital letter Z with two subscripts refers to matrix ele ments whereas the script "li is the primitive impedance element between nodes i J
and i.
In summary we note that the matrix diagonal element is used in finding the total fault current, and the other elements of the kth column are used to find node voltages and branch currents.
Example
1 2. 1
Find the positive sequence impedance matrix for the system of Example 1 1 . 1 . Then compute the total fault currents for faults o n each bus. The fault impedance is zero and V, is unity . Let h = 1 .
Solution The positive sequence network is shown in Figure 12.3 where the two genera tor impedances have been connected to node O. According to ( 1 2.3) we can find the first column of the Z matrix by injecting 1 .0 pu current into node 1 . Then the three resulting voltages VI ' V2 , and V:3 will correspond to the first column im pedances Zl l , Z21 , and Z:31 respectively. This situation is shown in Figure 12.4 with the result
Chapter 1 2
396
J O. 0 6 3
2
)0.08
JO.20
0
j O. 2 5 _
REF
Fig. 1 2 . 3. Positive sequence network.
��
+ \3 - J O. 098
REF
�J �IJ �
Fig. 1 2. 4. Network solution when II
V2
=
V3
�
Z2 1
=
=
: 0.127 4
J
JO . 1 061
jO.0981
Z3 1
1 . 0.
�
The second and third columns may be found in the same way with the result 1
2
0 . 1 274
jO.1061
jO.098
jO.1061
jO.1 345
jO . 1 1 51
3 jO .0981
jO.11 51
jO. 1 2 1 5
1 Z=2
.
3
The 3 1/> fault currents are computed from ( 1 2 . 7 ) . With Vf = 1 .0, h = 1 , and Z, = 0 we compute 1(1 1(2 1(3
= 1 .0/Z1 1 = 1 .0/Z22 = 1 . 0/Z3 3
= 1 .0/jO.1274 = - j 7 .852 pu = 1 . 0/jO . 1 345 = - j 7 .436 pu = 1 .0/jO.1215 = j8.230 pu -
These results check exactly with the results of Example 1 1 .10. Equations ( 1 2.9) and ( 12.10) could now be used to compute the system positive sequence voltages and line contributions for each fault location . This is left as an exercise. 1 2. 1 .2
Single·line·to·ground fault computation
We may extend the impedance matrix solution to situations which involve the negative and zero sequence networks as well as the positive sequence network. In all sequence networks we assume that the common generator connection or node o of F igure 1 1 . 1 6 is taken as the refArence. Then we write V = Z I for all networks where I is the vector of currents entering each node in the sequence networks and
Com puter Solution M ethod s U si ng the I m peda nce M a trix
397
the voltages are the voltage drops from each node to the nodal reference bus 0 as shown in Figure 1 2. 5. The notation used to distinguish between sequences is also given in Figure 1 2 . 5 . In general the nodal subscripts read (sequence , node) in all :;:ases. On
I().n
10-2
10 - 1 01 o
In
_ r----,
-
-
r----,
1 2 I�
0
1;.
VI_ I
NODAL REF- N O
Fig. 1 2. 5.
I I_n
2n
I2-n
-
r---...,
2 2 I�
0
2 1 I�
2
V2-I - Val- I 0 1 - .1 NODAL REF - N 2
+t
ll:!NI BUS
Sequence quantities defi ned for nodal equations.
We further identify the se q uence voltages by the subscript a where necessary, especially in the positive sequence network . Using this notation, we may write the nodal equations in matrix form as (12.11) where the meaning o f these equations is clear from F igure 1 2 . 5 . S pecifically , the positive sequence equation is
I I-I
V i ol
ZH I
Z 101 2
Z Hn
V I _2
Z
1-2 1
Z lo22
Z I-2n
11-2
V I_ n
Z I -n l
Z I-n 2
Z I -nn
II- n
( 1 2. 1 2 )
We note one additional problem concerning notation . Here we define the se quence currents to be those en tering the network. At the faulted node , however, we identify the sequence currents lao , la l ' and la 2 to be those leaving the network. Where necessary to identify these fault currents we shall add the subscript a in the usual way . For a SLG fault on node k the sequence networks are arranged as in Figure 1 2 . 6 . Since the driving point impedance at k is known , we write
la o -k = la l -k = la 2 - k =
h V, h V, ;::, Z O - k k + Z I - k k + Z 2 - k k + 3Z , = ZT
( 1 2 .1 3)
where we define (12.14) Then the voltages at all nodes may b e computed from the currents o f ( 1 2 . 1 3 ) and the matrix. From ( 1 2 . 8 ) for any sequence s we may write
Z
V.oj = - Z.oj k h V, /ZT
(12.1 5)
This equation gives the sequence voltage i n the zero and negative sequence net-
398
C h a pter 1 2
On
l aO- k
-
+
Io-k -
Ok
-o _ 0 1 lCH
T+
r
� k vlQ.- 1
NO
3
Iol-k
-
l
Z
I I- k
-
t
1 -II- l'u2-k
I
1 ,_,. 0
II
f
0
-
Ik
1 +
-
VJ:: I
+ + VO I _ V'to!!. !!.
0
1,-n= O
In
k-�k
_
0
N I
12·n - O n _
«-,
I02- k 2 k 12-11 -
f.
'
-
V2-k V2 -1 1 - -r -
N2
2
12 _1-0
'+
0
Fig. 1 2.6. Sequence network connections for a SLG fault on bus k .
as in works. In the positive sequence network the voltage h V, must be added are voltages ( 1 2 .9). The resulting sequence
Va H Va 1-2 Va l-n
VaO-1 VaO-2
VO_1
Va o - n
Vo-n
VO-2
VI -2
+h
=
VI-n
Va2- n
=
( 12.16)
Zo - n k V,
V,
VI-I
VaH Va 2-2
= -
ZO _l k V, ZO -2 k h ZT
V,
=
h
V2-n
_
h V, ZI-2k
ZT V,
V, V2_1 V2_2
V,
Z l-I k
Z I- nk
(12.17)
Z 2_l k =
_
h V, Z2-2 k
ZT Z2- n k
( 12 .18)
3 99
Computer Solution M ethods U si n g the I mpeda nce M a trix
The currents flowing in the branches of the positive and negative sequence net works may be computed from equation ( 1 2 .10) . To compute the current in branch i-j in the zero sequence network which is not mutually coupled, again use (12.10). If this element is mutually coupled to another element or elements, ( 1 2 . 1 0) must be solved simultaneously for all coupled currents. In full double subscript notation the solution becomes very messy to write. For example if ele ments i-j, m-n , and p-q are mutually coupled in the primitive matrix, the solution for all coupled currents involves the inverse of that partition of the primitive im pedance matrix with the result
[ ] �.iJ-iJ
. i}- mn
ii
_
Im n
Ip q
- .mn -ii 'fp q - il
i j [lie i j
. J -pq
-
I
� Z le
. mn-mn
.mn-p q
Znle
f'p q -m n
.pq -pq
Zqle - Zple
Z
Zmk
h
V,
ZT
( 1 2 . 1 9)
where all impedances denoted l' are primitive zero sequence impedances. As in the 3 rt> fault we note that the diagonal element of the matrix helps us find the total fault current, and the remaining matrix elements from the k th column help us find the node voltages and hence the branch currents. Usually we save matrix storage by assuming that the positive and negative se quence networks are identical. This changes Z T to and changes the
Example
Z
Z T = Z O - Iele + 2Z l -k le + 3Z,
( 1 2 .20)
subscripts in ( 1 2 . 1 8) to indicate positive sequence impedances.
Z
1 2. 2
Compute th e zero sequence impedance matrix for the system of Example 1 1 .1 . ·rhen, using the positive sequence matrix from Example 1 2. 1 , compute the total fault current for a SLG fault on bus 2 and find all voltages and branch line cur rents. Let h = 1 .
Solution
The zero sequence network is shown in Figure 1 2 . 7 . We find the impedance matrix tliis time by inverting the admittance matrix of Example 1 1 .6 with the result
Zo
1
�
3
1
1
2
001157
j O .0546
J Oo02
j O.0831
jO .0200
j O.0200
j O .0200
= 2 jO .0546
3 jO .0200
jO. IO 2
jOl 4 REF
Fig. 1 2 . 7 .
Zero sequence network.
Chapter 1 2
400
Notice that the technique of injecting 1 .0 pu current into node 1 to find column 1 poses a more difficult problem than in Example 1 2 . 1 because of the mutually coupled elements. The solution to this problem will be considered in sections 12.7 and 1 2.8. In the following computations we assume that the fault impedance is zero and the negative sequence impedances are equal to the corresponding positive sequence . impedances. Then from ( 12.14) with k = 2
ZT =
ZO -22 +
2Z1- 22 = = 02 =
j O.0831 + j O.2690
= jO.3 5 2 1
Then from ( 1 2. 1 3 ) the sequence fault currents are 1
�VoOVoo•.2 � � [- � VoO-3J OI . � � � [ � � �VV0ol1••32J �V02-02.2j � � [ � 100- 2
= 101•2
-
2
1 .0/jO.3 521
=
-
j 2 .841 pu
The sequence voltages are computed from ( 1 2.16)-( 12 .18) as ' 0 .054
= j 2.841
0.1551
� 0.0831 =
- 0.2360 - 0.0568
J O .0200
j o . 106
.O
=
pu
.6986
1 . 0 + j2.841 jO.1345 = 0.6 1 80 1.0
jO. 1 1 51
V02•3
0.6730
0'301
'0' 106
= j 2. 841
pu
� 0. 1345 = JO. 1 1 51
pu
0.3820
-
- 0.3270
There are three lines contributing to the total fault current at bus 2, one line from bus 1, and two parallel lines from bus 3 ( see Figures 1 2. 3 and 1 2. 7 ) . The contribu· tion from bus 1 is computed from ( 1 2. 10).
1-12 2-12 Vf (ZZI-22TI-I.12- ZI _12) Zo-d 0-12 - V, (ZO-22 ZTi-0-12
1
=I
=
h
-h
1
-
-_
j2.841 (j0. 0284) J'0 08
_ =
= - J' 1 . 008
pu
•
j2.84 1 (j0.0285) - - J' 0 . 578 pu ' J 0 14 •
For the parallel lines in the positive sequence network we again apply ( 1 2. 1 0). Since the positive sequence line impedances for are equal, the current divides equally. 1
ZI -32) 1-32 2-32 V,(ZI_22 Z i-32 =1
=
h
T.
=
i-23
_ j2.841 00. 0194) J'0 06
= _ J' 0 . 916
pu
•
In the zero sequence network the self impedances are unequal and the lines are
mutually coupled. We use the 4-5 position of the primitive admittance matrix to compute from ( 1 2. 1 9)
[10-32(4)J 10-32(5)
-
'
- - J2.841
l
j12.632 . J 5. 263
j5. 263 .
J
- J1 0 . 5 26
[
]-[
]
jO.06 3 1 j 1 . 320 . . - JO.943 JO.063 1 -
pu
Com puter S o l u tion M ethods U si ng the I m pedance M a trix 1 2. 2
40 1
An I mpedance Matrix A lgorithm
The impedance matrix is an excellent network description to use in the solu tion of faulted networks. If load currents are neglected, the fault information needed, both for total fault current and branch currents, is easily computed from the impedance matrix elements. Although it is easy to use for fault computations, the impedance matrix is not as easy to form as the admittance matrix. Furthermore, whereas the admittance matrix is extremely sparse, the impedance matrix is completely filled. It is sym metric, however, so that only n(n + 1 )/2 impedances need to be saved in computer memory. This requires n(n + 1) memory locations if the impedances saved are complex. However, many systems have high x/r ratios so that sufficient accuracy is obtained by using only the reactances. One difficulty associated with forming the impedance matrix is the ordering of the primitive elements. Ordering is no problem in forming the admittance matrix. This is because the indefinite admittance matrix exists and all matrix elements are defined when the voltage reference is arbitrarily selected. This is not true for an impedance matrix description. Even the simplest possible network consisting of one element has no finite impedance matrix description unless the element is connected to the voltage reference. This is because of the way the elements of the matrix are defined as given by the equation
i = I, 2, . . . , n
( 1 2. 21 )
I f an impedance element i s not connected in any way t o the reference, Vi is clearly infinite when III is given any value ( such as 1 . 0). This characteristic of the impedance matrix suggests that the primitive matrix impedances should be sorted beginning with an element which is connected to the voltage reference (bus 0) and with each succeeding element connected directly to one of the previously entered elements. Thus an algorithm for building the impedance matrix in a step by-step manner is required. Much has been written on this subject, as noted in [ 67-72] which describe early attempts at implementing the impedance matrix in fault studies. These works are recommended reading on the subject, particularly the work of Brown et al. [ 70, 7 1 ] and E I-Abiad [ 72 ] . The analysis that follows should be regarded as tutorial and not necessarily the best or the only way to con struct an impedance matrix. In the remainder of this chapter we will consider the various ways a network may be changed by adding a new radial branch or closing a loop thereby changing the "previously defined" Z matrix made up of elements added prior to the one under consideration.
1 2. 3
Adding a Radial I mpedance to the R eference N ode
Consider a network in which p nodes have already been defined and the Z matrix has been computed. We now introduce a new branch of im pedance f- from the reference to a previously undefined node q as shown in Figure 1 2. 8. Applying ( 1 2. 2 1 ) with III = Iq we can determine the qth column of the Z matrix. Since the impedance .. is connected only to the reference and all currents except Iq are zero, all voltages are zero except Vq which is given by ( 1 2. 2 1 ) ; or the new impedance matrix is
(p X p )
C h a pter 1 2
402
-
2 P
q
!.tIp -
P -PORT
NETWORK
�1
0 REF
Fig. 1 2.8. Adding a new branch radial from the reference.
2
p
q
ZI l 2 Z2 I
Z 12 Z22
Z IP Z2p
0
Zp l
Zp2
Zpp
0
0
0
0
..
1 1 Z=
P
q
0
( 1 2. 22)
We summarize as follows : Conditions: 1 . Branch added from reference to q. 2. q previously undefined. 3. New branch not mutually coupled to other elements. Rule : 1 . Set Zqq = ,.. 1 2. 4
Adding a R adial Branch to a New N ode
We now consider the addition of a radial branch with self impedance ,. from a previously defined node k to form a new node q as shown in Figure 1 2. 9. Again we use the defining equation ( 1 2. 2 1 ) to find the qth column. If we let Iq = 1 . 0 , the voltages VI ' V2 , , Vp , Vq will be numerically equal to the qth column im pedances. But letting Iq = 1 . 0 gives the same voltage distribution for the new net work as letting Ik = 1 . 0 in the old network (without the added branch). These voltages are precisely the kth column impedances. Element Zqq is the driving •
•
•
IL
q
� , l. . IP k _\. k
2 ,
Fig. 1 2.9.
.!L ..!!..-
P-PORT
NETWORK
10
Adding a new branch radial from node k .
Computer Solution M ethods U si ng the I mpedance M atrix
403
point impedance at q which by inspection is ( Zkk + ,.). Therefore the new matrix is given by 1
2
k
p
q
1
Zl I
ZI2
Zlk
Z Ip
Zlk
2
Z2 1
Z22
Z2k
Z 2p
Z 2k
Z= k
Zk l
Z k2
Zkk
Z kp
Zkk
P
Zp l Zk l
Zpk Zkk
Zpp Zkp
Zpk
q
Zp2 Zk 2
(Z kk + ,. )
( 1 2. 2 3 )
Conditions : 1. Branch added from k to q. 2. k previously defined. 3. q previously undefined. 4. k is not the reference node. 5. p nodes previously defined. 6. New branch not mutually coupled to other elements. Rules : 1. Set Zi q = Z i k , i = 1, 2, . . . , p . 2. Set Zqi = Zk h i = 1 , 2, . . . , p. 3. Set Zqq = Z kk + ,.. 1 2. 5
Closing a Loop to the R eference
Now consider a case in which an element is added from the reference to a node k which has been previously defined. In other words, the element being added is a link branch since its addition will close a loop in the network. The added element is not mutually coupled to other network elements. We shall do this in two steps. First we add the branch radially from node k to a new node q. The result of this step is given by ( 1 2.23). Then as a second step we connect node q to the reference as indicated by the dashed line in Figure 1 2. 1 0. If the impedance matrix for the original p-port network was p X p, the P
Iq -
lq I
I I
I
I
I
1
k
2
Ip
-
L\
�
P - PORT
N ETWORK
II
-
I I Fig. 1 2 . 1 0.
to
Adding a link branch from k to the reference.
branch added radially gives the impedance matrix ( 1 2. 23 ) which is (p + 1 ) X (p + 1 )
Cha pte r 1 2
404
If we write out the voltage equations in matrix fonn, we have following the first step
( 1 2. 24 )
Vq
Then, upon connecting node q t o the reference, = 0 , a Kron reduction may be perfonned to eliminate Iq as a variable , reducing the matrix once more to p X p . In matrix notation ( 1 2.24) may be written as
[�::Zpl . . . ZZppIP] - [.ZZpk.1.12] Zd1 [Zkl . . . Zkp Zd Zkk
Then the new impedance
from which we compute matrix may be written as Z=
where
=
•
.
•
•
.
.
.
.
.
] ( 1 2 .2 5 )
+ ,. .
Conditions: 1 . New link branch added from k to reference. 2 . k previously defined as one of p nodes . 3 . New branch not mutually coupled to other network elements. Rules: 1. Set q = p + 1 . = i = 1, . . , po 2 . Set = i = 1, . . . ,po 3 . Set + •. 4. Set Zqq = 5 . Eliminate row q , column q by Kron reduction . 1 2.6
Z'Zq'q ZkZ'kl t, Zkk
.
Closing a Loop Not I nvolving the R eference
Consider the addition o f a new element between nodes i and k where nodes i and k are both previously defined as part of a p-port network shown in Figure 1 2 . 1 1 . Assume that the added node is not mutually coupled to any other net work element. Assume further the total number of nodes defined to be p ;;., i, k and that neither i nor k is the reference node . In this case as in the previous one it is convenient to think in terms of adding the new branch from k to a new tem porary node q. Then as a final step we may connect q to i. The first step, being the same as that of the previous section, results in precisely the same equation ( 1 2 .24 ) . In this case , however, the second step of connecting q to i gives
Vq
-
V,
=0
( 1 2.26)
Therefore, we replace the qth equation of ( 1 2 .2 4 ) by the difference between the
40 5
Computer Solution M ethods U si ng the I m peda nce M a trix
P
Iq :.!L
-
q
Ii
11
--
"' - - -1� -
Iq
Ik
I
k
-
1.
I
Fig. 1 2. 1 1 .
P - PORT NETWORK
--
to
Adding a link branch from i to k .
qth and ith equations. At the same time we recognize that the current I, shown in Figure 1 2 . 1 1 is not the new nodal current at this node . Once q is connected to i the total current entering this node is (Iq + Ii ). Changing all curre nt variables in the ith row from Ii to (Iq + Ii ) requires subtracting column i from column q . The result of both operations on ( 1 2 .24) is 1
V,
V. Vp V. - Vi
ZI I
k
p
q
Z"
Z ,.
Z 'P
(Z , . - Z I i )
I,
ZI/
Z,.
Zip
(Z,. - ZI/ )
1. +1,
Z.,
Z••
Z."
(Z.. - Z., )
I.
( Zp. - Z,., )
Ip
k
Z. ,
P q
Zp ,
Z,.,
Zp.
Zpp
(Z. , - ZI ' )
( Z.i - ZI/ )
(Z•• - ZI. )
(Z"' - Zip )
Zd
( 1 2.27)
I.
where Zd = Zit + Zkk - Zik - Zki + .. = Z jj + Z q q - Ziq - Zql (of 1 2 . 2 3 ) . = 0, the last row and column are eliminated by Kron reduction Since q with the result
V - Vi
][
Z IP . . . Zpp
_
Zlk - ZIi . . . Zpk - Zpi
]
Zd- l [ ( Zkl - Zu ) . . . (Zkp - Zip ) ]
( 1 2 .28)
Conditions: 1 . Branch added from i to k . 2 . i and k both previously defined . 3 . p nodes defined in all. 4. Added branch not mutually coupled to other network elements. Rules: 1. Set q = p + 1 . 2 . Set ZJq = Zjk - Zjj , j = 1 , . . . , p . 3 . Set Zqj = Zkj - Zij , j = 1 , . . . , p o 4 . Set Zqq = Zjj + Zkk - Zik - Zk/ + .. = Z / j + Zqq - Zjq - Zq j . 5 . Eliminate row q and column q by Kron reduction.
Cha pter 1 2
406
This completes the four basic configurations needed to add a new element to the positive sequence network. In the zero sequence network, however, the added branch may be mutually coupled to some other network element. Since con nections to the reference node 0 are all generator impedances, we need not con sider mutual coupling for these elements. The cases of interest then are the addi tion of a radial line (tree branch) and the addition of a line which closes a loop (link branch) where the reference is not involved. These cases are investigated following Example 1 2 .3 which involves no mutual coupling.
Example
1 2. 3
Consider the positive sequence network given in Table 1 1 .1 and shown in Figure 1 2 .3. Take the branches in the order given in Table 1 1 . 1 and construct the impedance matrix using the rules of sections 1 2 .3-1 2 .6 .
Solution
From Table 1 1 .1 we have the positive sequence data as follows: Branch Num ber
Connecting Nodes
Positive Sequence Impedance
1 2 3
0-1 0·3 1-2 2-3 2-3 1-3
jO.25 jO.20 jO.08 jO.06 jO.06 jO. 1 3
4 5 6
If we introduce the elements in the order 1 , 2 , . . . , 6 , we will build up the network piecemeal as shown by the six steps of Figure 1 2 . 1 2 . Sketching the net work is not necessary , but it is helpful in describing the process. The actual STEP 1
. 1_
=1-0
2 ,.3
-4---1
STEP 4
-1
2
0
---
3
ST E P
2
2
------+ 0 STEP
3
2
Fig. 1 2 . 1 2.
3
2
3
-------t o
3
� 2 �_--=-__�
STEP
5
3
2
3 0
-::-
2 tf--�--� 3 STEP 6
----'----+ 0
Step-by-step construction of the positive sequence network.
Computer Solution M ethods Using the I m pedance M a trix
407
decision-making process needed at each step can be done by computer program ming techniques . We begin with a computer impedance array which is initialized to zero . Since the resistance of all elements is negligible, only one such array is required and it should be at least 4 X 4. Step 1 . Introduce element 1.
"0 1 = j O. 25 Since this is the first element, it is a radial element from the reference to node 1 as shown in step 1 of Figure 1 2 . 1 2 . If this first element were not con nected to the reference, it would be necessary to sort the elements to find one which was connected to O. The rule for introducing an element radial from the reference is given in ( 1 2. 2 2). Thus we complete step 1 by adding 0.25 in location q-q or 1-1 with the result. 1
3
3
� � L �
� r:
z=j
2
5
0
0
Step 2. The addition of "0 3 = jO.20 is performed exactly the same way as the previous step except that the diagonal term affected is element 3-3.
1
Z =j
2 3
r
1
2
3
.25 0 0
Step 3. We now add a radial element "12 = jO.08 as shown in Figure 1 2 . 1 2. The rule for doing this is given by ( 1 2 . 2 3 ) where k = 1 , q = 2 . Thus the first row is added to the second row and the first column to the second column. Then Z 22 is set equal to Z 1 1 + 0.08. 1
Z=j
� �: � : 3
L
0
2 0.25 0.33
o
Step 4. This addition of the fourth element of h3 = j O.06 closes a loop not involving the reference. The rule for this situation is given by ( 1 2 .28 ) which is a Kron reduction of ( 1 2 .27 ) . We begin with the 4 X 4 array where k = 2 , i = 3 . Thus we have an augmented matrix 1
2
3
0 .25
0.25
0
2 0.25
0.33
0
1
Z =j 3
o
0
0.20
-------------
q 0.25
0.33
- 0.20
q I I I I
0.25 0.33
1 - 0.20
J I I
----
0.59
408
Cha pter 1 2
where (col 4) = (col 2 ) - ( col 3) and similarly for row 4. Element 4-4 is Zd = Z 22 + Z 33 - Z23 - Z 32 + jO.06 = j O . 59
�
�
Performing the Kron reduction to eliminate row q and column q , we have 1
2
3
0 . 1441
0. 1 1 02
0.0847
2 0 . 1 1 02
0.1454
0.1119
3 0.0847
0. 1 1 19
0.1322
1 Z =j
Step 5 . The fifth element "23 = jO. 06 calls for exactly the same operation as the previous step. We begin with the 4 X 4 array. Again we have k = 2, i = 3 .
Z=j
�
1
2
3
1 0 '1441
0 . 1 1 02
0.0847
2 0.1 102
0.1454
0. 1 1 1 9
I I
q 0.025
i
0.0335
3 -------------0.0847 0 . 1 1 1 9 0 . 1---+----322 I 0.0203 q 0.0255 0.0335 - 0.0203 I 0.1 1 38
�
-
This array is reduced to find the matrix at this step to be
Z =j
2
1
3
J
1 0 . 1 384
0. 1027
0.089
2 0. 1027
0.1 356
0 . 1 1 78
3 0.0893
0. 1 1 78
0.1286
The last element to be introduced is "1 3 = j O. 1 3. This element also closes a loop, so we begin with the 4 X 4 array, with k = 1, i = 3 .
Step
6.
1
Z
=
3
2
1 0.1 384
0 . 1 027
2 0 1027
0.1 356
:
j 3 0 0893 q 0.0491
0 . 1 1 78
0.0893 I 0. 1 1 78
0.1286
�
I
0.0491 - 0.01 51
1 - 0.0393
T - 0.0393 I 0.2184
--------- ---------
- 0.0151
I
q
-----
�
After Kron reduction, pivoting on element q-q , we have the positive sequence impedance matrix : Zl
=
j
0 .1273
0 . 1 06 1
0.098
0 . 1 061
0.1 345
0.1151
0.0981
0.1 1 5 1
0.1215
1 2.7 Adding a Mutually Coupled R adial E lement
We now consider the addition of a radial element from node p to a new node q where the added element is mutually coupled to the network element from m to n. We assume that nodes m, n, and p have been previously defined in a network with p total nodes. (N ote there is no loss in generality by letting p be the highest
Com puter Solution M ethods Using the I m pedance M atrix
r� ,pq
409
•
q _"'Nv---.....
P - PORT
NETWORK
o
Fig. 1 2 . 1 3.
Adding a mutually coupled radial element.
numbered node .) Assume t.hat there is an impedance lin n between nodes m and n and the new impedance .pq is mutually coupled to 'l- m n by a mutual impedance I'M . The network connection is shown in Figure 1 2. 1 3. The voltage drop for two mutually coupled lines is given by ( 4 .1 4 ) . For the two elements under consideration here we write
[
l Vp q J
Vm n
= rL p
] [f-M
Vm - Vn V - Vq
Solving for the currents, we write
=
.m n
J
f- M
.p q
r Im n �p q = - Iq
J
( 1 2.29)
( 12 . 30) where the admittance matrix of ( 1 2. 30 ) is the inverse of the impedance matrix of ( 1 2 .29) . Suppose we write the equation for the new network of Figure 1 2 . 1 3 in symbolic form, i.e . , 1
k
m
n
p
q
-
VI
1
Zl l
Zl k
Zlm
Z ln
Z iP
Z lq
Vk
k
Zk l
Zk k
Zk m
Zk n
ZkP
Zlrq
Zm lr Z nk
Zmm
Zm n
Znm
Znn
Zm p
n
Zm l Zn l
P q
Z-p l Zq l
ZP Ir Zq k
Z-pm Zq m
Z-pn Zq n
Vm
Vn Vp
Vq
=m
Znp Zpp
Zq p
-
-
Z-m q Znq -
Zpq Zqq
II IIr 1m In Ip Iq (12.31)
where the elements of the qth row and column are completely unknown as yet. This fact is acknowledged by the tilde placed over these elements. The matrix elements of Z's without tilde ' s are the original Z ' s. Finally we note that caution must be observed to clearly distinguish between branch impedimces f- m n and f- p q and matrix elements Zm n and Zp q .
C h a pter 1 2
410
To determine the qth row and column we make the following test. Let
1/2 =
1,
k
j ", k ,
Ij = 0 ,
=
1 , 2, . . . , p
( 12.32)
1/2
= 1 at This requires that we methodically move from port to port injecting each port with all other ports open. Then for injection at the kth node we com pute the voltages at all nodes as
VI V/2 VVnm �p Vq
�.'�
Zu = Zm /2 Zn/2 Zp/2Zq/2 J
Iq mn
. . .
..
( 12.33)
Note that = 0 for all tests. Therefore, the voltage drop from m to n is a func or tion only of I
1/2 lImnJ1 = L"Mfllmn
(12.34)
Imn 'fmn . Vpq ]lIpq [VVpqmn] = ["mn lIlIMpqJ [VmVp � VqVnJl
= 1 may cause a current The injection of to flow through If so, this current will induce a voltage because of the mutual coupling. From ( 1 2. 30) with q = 0 we compute
I
11 M
o
11M
( 12.35) The second equation of ( 1 2 .35) may be written as
'Ipq Zq/2) Zq/2 Zpk IVpq)(Zm/2 - Zn/2) Vmn _'fpq 1Ipq
0 = 11M ( Zm k - Znk) +
( ZPk
-
from which we compute
=
+ ( 11M
(12.36)
However, from ( 12.29) and ( 12 . 30) we have the relations -
�,
_� -
Ll
( 12.37)
from which other forms of the solution are possible. One convenient form is the following:
( 1 2.38) This equation may be used to find the first q 1 elements of the qtQ ro w and, by symmetry, the qth column. The only unknown then is the -
element Zqq .
Com puter Solution M ethods Using the I m pedance M atrix
41 1
To find ZQQ ' we devise the following test. Let Iq = 1 ,
/ l = I" = "
' = lp = 0
(1 2 .3 9)
Then from ( 1 2.31) we have Vl
ZlQ
Vm
Zm q
Vn
= Znq
Vp
Zpq ZQQ
VQ
( 1 2 . 4 0)
where the tilde's have been removed from the first p impedance elements as they are now known. Also, since Ipq = - IQ = - 1 , we may write from ( 12.30)
[ � lJ / n
1
=
fVmn lv M
l fVm � V� V� V � LVp VM
or - 1 = Y M ( Vm - Vn) + Y PQ ( Vp - VQ) ' Then
VQ = Vp + 1 /vpq + (YM ly pq )( Vm - Vn) This may be changed to the form VQ = Vp combining with ( 1 2.40) ,
+
( 1 2 . 41)
fpQ - (j·M !,. m n ) (f-M + Vm - Vn ) or, ( 12. 42)
This completes the formulation of the new impedance matrix. We should be able to test this result to see if it agrees with the case of adding IW uncoupled radial line wh�e f-M = O. In this case, from ( 12 . 38) we have ZQk = Zpk and from ( 1 2.42) ZQQ = ZPQ + f-pq . This is the same result as found earlier for the uncoupled radial line. Conditions : 1 . Branch added radially from p to q . 2 . Branch m n previously defined. 3 . Branch pq previously undefined. 4. p is not the reference node . 5 . Branch pq mutually coupled to branch m n . Rules: 1 . Set Zq/t = Zp/t - (f-M If-m n)(Zm /t - ZRIt) , k *' q . 2. Set Z/tQ = ZQ/t ' k *' q . 3 . Set ZQq = Zpq + J.pq - (f-M If-m n ) (f-M + Zm Q - Znq ) .
Having considered the addition of a mutually coupled radial element, there is no need to study separately the addition of a mutually coupled link branch. This case can always be handled by first adding the mutually coupled branch radially to a fictitious node q and then closing the loop from q to the appropriate node by means of a Kron reduction to eliminate row q and column q . The case just considered will solve a great many o f the mutually coupled zero sequence network problems. Occasionally, however, a situation is encountered where several lines occupy the same right-of-way and a number of circuits are mutually coupled. This case is considered in section 1 2.8,
Chapter 1 2
412
Example
1 2. 4
Consider the zero sequence network given in Table 1 1 . 1 and shown in Figure 12.7. Consider the branches in the order listed in the table and construct the zero sequence impedance matrix.
Solution From Table 1 1 .1 an d Example 1 1 . 1 we find the following zero sequence data. Branch Num ber
Connecting Nodes
1 2 3 4 5 6
open
Zero Sequence Impedance
0·3 1·2 2·3 2�3 1 ·3
jO.02 jO.14 jO.10 jO. 12 jO.17
Mutual Impedance
jO.05 jO.05
We proceed in much the same fashion as for the positive sequence network until we reach the mutually coupled element. Step 1 . Introduce element 2.
�
�3 = jO.02
1
1
Z =j
2.
3
O
2 0 3
Step
2
Introduce element 3.
0
in = jO. 14
Neither of the nodes 1 or 2 have been entered yet. Since a radial element must be connected to the network in some way, we must skip this element and pick it up later. Step 3. Introduce element 4 . .2 3 = jO. 10
This is a radial element which connects node 2 to a new node 3. Thus with k = 3, q = 2 we have
� o.�� 1
1
Step
4.
2
0
Z=j 2 0
0. 1 2
3 0
0.02
Introduce element 5.
3
. 2 3 = jO.12 with J- M
0. 02
=
jO.05
This is a mutually coupled element which closes a loop. However we add it as a radial element from node 2 to node q and will later eliminate node q .
Com puter Solution M ethods U s i ng the I m peda nce M atrix
41 3
The qth row and column elements are computed from ( 1 2 .38). Thus -
Zo ic = ZPIc - ( J-M IJ-mn )(Zmlc - Zn/c )
with
p = 2,
n = 3,
m = 2,
k = 1, 2, 3
For k = 1 ,
k
= 2, Z 2 = Z 2 2 - ( J-M lJ-mn)(Z22 - Z3 2) = j O . 1 2 - j q
�:�� ( 0 . 1 2
Z 3 = Z 23 - (J-M /t:. m n )(Z23 - Z33) = jO.02 - j q
�:�� (0.02 - 0.02 ) = jO.02
-
0.02 ) = jO.07
k = 3,
For the diagonal term we use ( 12.42) to compute Zqq = Z 2 q
+
J-2q - (J-M /J-m n )("M + Z 2 q - Z3 q )
= j O.07 + j O . 1 2 - j
�:�� (0.05 + 0.07 - 0.02) = j O . 1 4
W e now have the matrix 1
1
2
3
0
0
0
0 .1 2
0 .02
0 .02
0.02
2 0 Z=j 3 0 A
q
I I
----------
q 0
0 .07
0
I 0 .07 Ir 0 .02
---
I 0.14
0.02
where a new node q extends radially from node 3 . We close q to 2 by the follow ing steps: (1 ) Replace row q by (row 2 ) - (row q ) , ( 2 ) replace column q by (col 2 ) - (col q ) , and ( 3 ) replace element q q by Zq q + Z22 - Z2 - Z 2 as noted in q q (12.27) for closing a loop. This gives us the new matrix 1
2
3
1 0
0
0
2 0 Z= 3 0
0.12 0 .02
I
0
Now reduce Z , pivoting on Zq q to compute
Z=j
2
� r� � l�
0 .0 92
3
0
I
0 .02 I,
0 .0 5 1
I
0 .02 I 0.05
----------
q 0
q
0 .02
0
, I 0.12 ----
3
�l o.o �J
0 2
414
Chapter 1 2
Step 5 . Introduce element 6.
1-1 3
= jO.17
This element is radial from node 3 to the newly defined node 1 . Therefore by inspection of (12.23) with k = 3 , q = 1 we have
Step
6.
=j
�
2
3
0 .1900
0 .0200
0 .020
2 0 .0200
0.0992
0 .0200
3 0 .0200
0 .0200
0 .0200
1 Z
�
1
Introduce element 3 which was passed earlier. 1- 1 2
= jO.14
This element closes a loop from node 1 to node 2 . From (12.23) with k i = 2 we compute 3
1
2
0.1900
0 .0200
0 .0200 1 0 .1900
2 0.0200
0 .0992
Z = 3 0.0200
0 .0200 1 0.0200
0 .0200
0.0200 1 0 .0200
1
q 1
1
-----------------
q
0.1900
= 1,
0 .0200
1
-----
0 .0200 1 0 .3300
This adds the new element radially from node 1 to a new node q . Then from ( 1 2 . 2 7 ) we form the intermediate result 1 1
Znew = j
2
3
q
I 0.1 700
0.1900
0 .0200
0 .0200
2 0.0200
0.0992
0.0200
3 0.0200
0.0200
0.0200
I - 0 .0792
1
I
------------______
q 0.1 700
[
- 0 .0792
0
1 1 1
0
_____-
0.3892
Finally we make a Kron reduction, pivoting on element qq to compute 2
3
1 0.1 1 57
0.0546
0 .0200
Zo = j 2 0.0546
0.0831
0 .0200
3 0 .0200
0.0200
0 .0200
1
Example 1 2. 5
J
Using the impedance matrices of Examples 1 2 . 3 and 12.4, compute the total fault current fa for a SLG fault on phase a of each node with zero fault im pedance . Let h = 1 .
Solution The total fault current is computed from ( 1 2 . 1 3 ) as la k = 3lao-k = 3 h Vr /ZT
Computer Solution M ethods Using the I m pedance M a trix
41 5
Setting h V, = 1 .0, we easily select the appropriate diagonal impedances from the positive and zero sequence impedance matrices to compute the following values. Faulted Node
VaO
fa
1 2 3
Va l
Vca
(pu)
(pu)
(pu)
(pu)
- j8.098 -j8.522 - j l l .406
- 0. 3 1 24 - 0.2360 - 0.0760
0.6562 0.6 180 0.5 380
- 0.3438 - 0.3820 - 0.4620
is interesting to note that these fault currents are greater in every case than the corresponding 31/> bus fault currents computed in Example 1 2 . 1 . This is not un usual for systems with no neutral impedance in the generators. It
1 2.8
Adding a Group of Mutually Coupled L ines
We expand the foregoing analysis to the case where a group of mutually coupled lines are to be added to a network. We will arbitrarily assume that the first self impedance "Ii of the group has been introduced. The remainder of the mutually coupled group is to be added as radial connections to nodes k . . n. These radials may later be connected to other nodes by Kron reduction, thereby forming closed loops if this is necessary . Consider the network shown in Figure 12.14 where the branch 'II} has been .
n'
I n'
Ip
�
p ----i
P - PO R T
NETWORK
II - I, 1 ---1 -
o
Fig. 1 2 . 1 4.
Adding a group of mutually coupled elements.
incorporated into the p-port network and we now wish to add the mutually coupled branches 'lil li ' • • • and where this entire group of elements is a mu tually coupled group consisting of, say, m elements. Taking the group of m mu tually coupled elements from the primitive matrix, we write this submatrix as
j- nn' 1
f"] 1 �J Vkk, . . Vnn' .
2
= 2 hi m
"' 11 '
'1kll' '1ni '1nll' .
.
.
So� l [ � l m
·
.
.
·
.
.
·
.
.
n � :, 'Inn'
IJ
IlIk' �II1' . Inn' = - In'
( 1 2.43 )
41 6
C h a pter 1 2
'Yb
where we recognize that branch currents in the radial elements can be related to the nodal currents entering. Since the inverse of this primitive matrix exists, we compute it, writing the result in partitioned form as 1 2
Ii}
����
['Y�- i �: �����, t
rV ii
-
j
m - In '
Vi - Vj
Vn
-
Vn '
(12 .44)
The impedance matrix which we wish to find may be written in symbolic form with a tilde over all unknown elements as follows. V, Vi
Vi
V� Vn Vp V�' Vn'
Z" Zj J
Z/ I
Z'i Zii
Zjl
Z 'j Zii
Z,�
Z' n
Zi�
Ziti
Zi�
Zu
Z,p Zip
Zin
Zip
Z� n
Z�p
I I I I I I I
I
Z , �,
Z, .
I,
Z Ut'
Z;n'
II
Zi� '
Z In'
Z�n' I Z��, I I Znp I Zn� Znn Z ni Zroj Zn ' ZM Znn' ' I I Zpi Zp ' Zpp I Z�, Zpn Zp/ Zp� ZPn' I :; - -- - - ---_--- :;----- - :; - - - - - - --:;.- -- -- --;;; - -t-- -_-- - --- --:;; Z�' , Z�' n Z�' I Z�'j Z� ' p I Z�' �' Z�' � Z� ' . I , . . . . . I . . . . . . . . . in. n. " . Zn'n " . in' p II Z" �, " . i' l in' 1 in' � Zn' ,
Z� ,
Z�i
Z��
ZIrj
I
Ii
I�
In
Ip
I�' In'
( 1 2.45)
We can find the unknown elements by perfonning two tests. First, we inject 1.0 unit of current at each of the nodes 1 . . . p while holding all other currents at zero, including the currents entering the primed nodes, i.e., Is = 1 . 0, It = 0, lk' = . . . = In' 0
t "* 5,
5 = 1 , 2, . . . , p ( 1 2. 46)
=
By inspection of ( 1 2.45) this test equates the port voltages to the impedance ele ments of the 5th column, i.e., VI
Zis
Vi Vj
Zis
Vk
Z/u
Vn
Zjs
=
Zns
Vp
Zps
Vk ,
Zk' s
Vn,
Zn' s
, 5 "*
k ', .
.
. , n'
-
( 1 2.47)
Cha pter 1 2
41 7
Now, returning to ( 1 2.44), we note that the currents entering the primed nodes are all zero and we may write the (m - 1 ) equations
�J �:J
Solving for the primed voltages, we have =
+
�.' � ( V, - V,)
Now, from ( 1 2. 47) we substitute the appropriate impedance values in place of the voltages to write ( 1 2. 4 8 )
This determines all unknown elements except the lower right partition of ( 1 2. 45). The second test consists of causing a current of 1.0 to enter each of the primed nodes in tum while holding all other port currents to zero. Let It' = 0, Is' = 1, ' ' s = k' , . . . , n
t' =1= s ' ( 12.4 9 )
Then ( 1 2 .45) becomes VI
Z ls'
Vi Vj
Zjs'
Vit
Zit.'
Vn Vp VIt,
Vn ,
Zj,'
=
Z"',
, s' = k ' ,
. . .
,n
'
Zps,
Z�"
Zn' .,
( 1 2. 5 0 )
where the upper p elements are shown without the tilde since these are now known. From the primitive equation ( 1 2. 44) we have under test conditions ( 12. 4 9 )
C h a pter 1 2
41 8
Iij
I V ii i
Iij
I
0 0 - Is' = - 1
'Yb
- - --1 - - I I
6
- Is'
=
0 0
r:J- �J
Solving for the primed voltages, we have
'Ye
I I I
I I
I I I I I
'Yd
+ � ' I.. + �d ' �' ( V,
( 1 2. 5 1 )
-
V, )
Finally, substituting the impedances ( 1 2 .50 ) for these voltages, we determine the s'th column of unknowns
(12.52) ' Now repeat with S taking all values from k I to n ' to complete the impedance matrix . The process described for mutually coupled elements is certainly not the only one which may be devised . A number of excellent papers have been written on this subject and the interested reader is encouraged to study some of these. Recent "build and discard " techniques have also been devised which attempt to retain only certain required nodes of very large systems [ 7 3-7 5 ] . An interesting review of the literature has been made by Kruempel [76] who carefully traces the development of these techniques. 1 2.9
Comparison of Admittance and I mpedance Matrix Techniques
From the discussion presented in Chapters 11 and 12 it should be apparent that power system faults may be solved by computer, using either an admittance or an impedance matrix technique. It is also apparent that each technique has its problems : the admittance method in requiring a new inversion for each new fault simulation and the impedance method in its difficult and time-consuming matrix formation method . But there are other important considerations which should be mentioned such as computer memory utilization and the methods of changing the network representation corresponding to network changes. In our formulation of the admittance matrix and in its use for fault calculation we made no use of the matrix sparsity . Tinney and others [ 81 , 82 ] have made a thorough analysis of sparsity in their optimally ordered triangular factorization technique . This technique permits the use of the admittance matrix, stored in sparse triangular factored form, to solve network problems. This formulation al ways requires much less computer storage than the impedance matrix, an advan tage that increases as the square of the network size. Furthermore, their studies show that the sparSe matrix technique exhibits a definite advantage in computer
Com puter S o l ution M ethods Using the I m pe d a n ce M at rix
41 9
time required on all problems except faults on small networks where there is only one nonzero current in the I vector. In such cases the impedance matrix is faster. Since the trend is toward the solution of very large networks, the optimally ordered triangular factorization technique has definite advantages because it can effectively and economically solve networks at least 10 times as large as the im pedance matrix technique [ 83 ] . Another important practical consideration in solving faulted networks by computer is the ability to change the network to represent added or removed lines without completely rebuilding the stored admittance or impedance matrix. As noted in Chapter 1 1 , the addition or removal of a branch or group of branches in the Y matrix formulation, whether mutually coupled or not, is an almost trivial problem of adding or subtracting a few primitive admittances. A number of algo rithms have been devised for making changes to the impedance matrix [ 84-86] which make it possible to consider limited changes in the network without alter ing the Z matrix. A method has also been devised [ 8 3 ] to permit similar changes in an ordered triangular factorization without changing that factorization. A detailed consideration of the method of computer solution of faulted net works is beyond the scope of this book on symmetrical components, but the dis cussion of Chapters 11 and 12 should serve as an introduction. Problems
12. 1. Extend Example 12. 1 to find the voltages at each node and all Une contributions (a) for a 3tf> fault on node 1, (b) for a 3t/> fault on node 2, and (c) for a 3t/> fault on node 3. 12.2. Repeat Example 12.1 if the fault impedance Z, is 0.1 pu (pure resistance) and with a prefault voltage of 1.05 pu. 12.3. Use the method of "unit current injection" described in Example 12.1 to find the im· pedance matrices for the following positive sequence networks : (a) Figure Pl l . l , (b) Fig· ure P11.2, (c) Figure Pl l.3, and (d) Figure P11.4. Use data given in problem 1 1.4. Neglect the resistance. 12.4. Repeat problem 12.3 to find the zero sequence impedance matrices. 12.5. Consider the network of Figure P 1 1 .4. Add a a generator at node 6 with Z 1 = 0 + jO. l and find the impedance matrix by the unit current injection method. Neglect resistance. 12.6. Compute the total 3tP fault currents for the case of zero fault impedance, considering each node in tum as a faulted node, for the networks of problem 12.3. 12.7. Compute the total 3t/> fault current at each node of the system of problem 12.5. Com· pare results with those of problem 12.6(d). 12.8. Compute the line current contributions and bus sequence voltages for the following faults. The fault is a 3t/> bus fault in each case. (a) Fault on node 2, Figure P I L L (b) Fault on node 1 , Figure Pll.2. (c) Fault on node 4 , Figure Pll.3. (d) Fault on node 5 , Figure P11.3. (e) Fault on node 5 , Figure Pll.4. (f) Fault on node 2, Figure P l 1 .4 . (g) Fault on node 2, system of problem 12.5. 12.9. Repeat Example 12.2 to find the complete network solution for a SLG fault (a) On bus 1 with Z, - 0 . 1 + jO, V, 1.05 + jO. (b) On bus 2 with Z, - 0 . 1 + jO, V, - 1.05 + jO. (c) On bus 3 with Z, - 0.05 + jO.05, V, - 1.10 + jO. (d) On bus 3 with Z, 0.1 + jO , V, - 1. 05 + jO. 12. 10. Compute the total SLG fault current for the fault conditions described in problem 12.8. -
-
420
Cha pter 1 2
12. 11. Compute the sequence voltages of all buses adjacent to the fault and all contributing line currents for the faults of problem 12.10. 12.12. Verify equation (12.27). In particular. show that changing the ith row current from Ii to Iq + Ii requires taking impedance differences as shown in column q . 12.13. Repeat Example 12.3 except introduce the elements in the order (a) 1, 3 , 4, 5, 6, 2. (b) 1 , 6 , 2, 4, 5, 3. (c) 2, 4, 5, 6, 1, 3. (d) 1 , 4, 6, 5, 3, 2. 12. 14. Use the Z matrix algorithm of sections 12.3-12.6 to find the positive sequence imped· ance matrices of the following networks: (a) Figure P11.1, (b) Figure P11.2, (c) Figure P 1 1.3, (d) Figure P1 1.4, and (e) systems of problem 12.5. 12.15. Use the algorithms of sections 12.3-12.8 to find the zero sequence impedance matrices of the networks specified in problem 12.14. 12.16. Show that (12.38) may also be written as (a) ZqIe
(b) Zq�
12.17. 12. 18. 12. 19. 12.20.
-
-
Z"Ie
(1 - tmn¥mn)(Zmle - Znle)
fmnVM tMVmn (Zmle - Znle) Z"Ie 1 tMYM _
_
-
Verify the steps leading to (12.42). Verify (12.48). Verify (12.52). Show that (12.48) and (12.52) reduce to (12.38) and (12.42) when there are only two mutually coupled lines. 12.21. Develop formulas similar to (12.48) and (12.52) for the special case of three mutually coupled lines.
appendix
A
M a t rix A l gebra
The analysis of power systems under faulted conditions by analytical tech niques is complex because of the large number of simultaneous equations that must be formed and solved. Matrix algebra, with its capability for the writing and manipulating of large numbers of equations in terms of a small number of sym bols, is a desirable, almost indispensable mathematical tool. The purpose of this appendix is to review the appropriate concepts of matrix algebra. I First, we consider simultaneous linear equations and their solution by means of determinants. A. 1
Simultaneous L inear Equations
We often work with equations of the form
X I = a l l Y I + a l2 Y2 + . . . + alnYn X2 = a 2 1 Y I + a22 Y 2 + . . . + a2 n Y n (A. l )
Here the x 's and y 's are variables and the a 's are constant (either real or complex) coefficients.
Note partiCUlarly the m ethod of subscripting. In the problem s of
interest, the x's will often be voltages (or currents), the y 's will be currents (or voltages), and the a's will be impedances (or admittances). Equation (A.l ) can be solved for the y 's by means of determinants to get Y l = f l 1 X I + f1 2 X2 + " ' + f l n Xn Y2 = f2 1 X I + f22 X2 + . . . + f2n Xn
(A.2)
where (A.3)
I Much of the material for this appendix is taken from notes, "Review of Matrix Algebra, " prepared b y H . W . Hale and used for teaching short courses in Power Systems Engineering at Iowa State University, Ames. It is used with permission of the author. 42 1
422
where
[
Appendix A
al l
�
= det
n
=
.. .
L ai/ Gij /= 1
(A.4)
a nI
and
= the cofactor ij = ( - 1 ) i+jMij Mij = the minor determinant resulting from the deletion of row i and column j Gij
(A.5 ) The general form o f (A. l ) and o f its solution (A.2) will b e pertinent t o some of the following material. A.2
Definitions Pertaining to Matrices
Following are some of the basic definitions and terminology pertaining to matrices. 1 . Definition of a ma trix . A matrix is a rectangular (including square as a spe cial case) array of elements (symbols or numbers) arranged in rows and columns. For example
is a matrix of two rows and three columns. Note that in this general form the subscripts attached to the elements (a 's) indicate the row (first subscript) and col umn (second subscript) in which it is found . Thus , the term " a ij " or "element ij " specifies a definite location at which the element is found . For a specific matrix of numerical values such as
subscripts are not attached to the elements; however, it is apparent that all
= 1,
a12
=
-
2,
a21 =
3,
an =
2
2 . No ta tion . We will indicate arrays as matrices by enclosure in brackets as shown in definition 1 . Other conventions are in use and may be encountered in the literature. We can also indicate a matrix symbolically by a single symbol such as
A
=
[
]
al l
a12
a2 I
a2 2
Although other conventions, such as underscoring or script type, are used, we will use boldface type for matrices. In most cases it is clear from the context which quantities are matrices and which are not. We will see later that we can write equations in terms of symbolic representation of matrices and manipUlate the symbols as long as the appropriate rules are followed.
M atrix Algebra
42 3
3. Order of a matrix. A matrix with m rows and n columns is referred to as an m X n or ( m , n ) matrix. In the special case that the matrix is square, m = n , it is
referred to as an nth order matrix. 4. Column and row matrices. A matrix with only one column is referred to as a column matrix, while one with only one row is a row matrix. They are also of ten referred to as column and row vectors respectively. 5 . Transpose of a matrix . The new matrix formed by interchanging the rows and columns of a given matrix is called the transpose of the original matrix. Thus
� � _�] -
�
[� �l [ -�J
the transpose of
2
Obviously, an m X n matrix has a transpose that is an n X m matrix. We indicate the transpose of a matrix A by A' or AT or A t . 6. Conjugate of a matrix . If all elements of a matrix are replaced by their complex conjugates, the new matrix is the conjugate of the original one. Thus
�
(1 - j1)
1
J
(3 - j 3 ) (2 + j 2 )
'
IS
the conJ ugate of
.
�
1 (1 + j1) ( 3 + j 3 ) ( 2 - j2)
J
The conjugate of a matrix A is indicated symbolically as A * . 7 . Symmetric matrix . A square tnatrix that has aj i au for all i and is sym metric. Thus
[� �]
=
is symmetric, while
[� :]
j
is not symmetric
Another way of stating the definition is that a symmetric matrix is equal to its own transpose. Only square matrices can be symmetric. S . Principal diagonal. The elements a 1 1 , a 22 , a 33 , . . . , ann of a square matrix form its principal diagonal. 9 . Unit ma trix . A square m atrix that has + 1 's on its principal diagonal and ze ros elsewhere is a unit matrix. It is denoted symbolically as I, U , E, or 1. Some times a subscript is attached to indicate the order. Thus,
U - l = E, -
[� �] o
1
o
(A. 6 )
In electrical engineering we usually avoid using I for the unit matrix. 10. Null matrix . A matrix with all elements equal to zero is a null matrix. 1 1 . Scalars. The elements of matrices are called scalars. A scalar is a 1 X 1 matrix. 12. Su b matrix . If some rows and columns of a matrix are deleted, the result is a submatrix. For example, if A
=
[� � � �]
has columns 2 and 4 deleted, the result is a new matrix
Appendix A
424
that is a submatrix of A. 13. Determ inant of a matrix . It is important to note that a matrix is not a determinant. However, the determinant of a square matrix has meaning and is taken to be the (scalar) quantity det A
[: :] =
= det
12 - 20
=
-
B
The determinant of a matrix that is not square is not defined. 14. Nonsingular and singular matrices. If a square matrix has a nonzero determinant it is nonsingular. All other matrices, including those that are not square, are singular. 1 5 . Rank of a matrix. The rank of a matrix is the order (number of rows or columns ) of the largest nonsingular square submatrix. Thus
A= has rank 2 because
while
has rank 1.
[�
B= G
2 1
1 1
]
-1 -1
16. Adjoint of the matrix A. The adjoint of the matrix
defined as
A, written adj A, is
(A.7 )
where Cu
1 7 . Inverse of the matrix defined by the equation
A.
= cofactor of a u
The inverse of a square matrix A of order n is (A.B)
In terms of elementary operations the inverse of A may be computed from the formula K'
=
adj Afdet A
Only nonsingular matrices (det A =1= 0) have inverses.
(A.9)
42 5
M a trix Algebra
18. Skew-symmetric matrix . A skew-symmetric matrix A is defined by the relation (A.10) The maj or diagonal elements of a skew-symmetric matrix must be zero since for i = i the only way (A.10) may be satisfied is for aiJ = O. An example of a skew symmetric matrix is
�o
+ jO
�
3-j
2+j
A = -3 +j
0 + jO
4 + jO
-2 - j
- 4 + jO
0 + jO
1 9 . Hermitian matrix. A hermitian matrix A is defined by the relation (A. l l ) The maj or diagonal elements o f a hermitian matrix must b e real, for with i = i the only way (A. l l ) can be satisfied is for aiJ to be real. Only square matrices may be classified as hermitian. An example of a hermitian matrix is
� o
2 + jO
A=
2- j - 5 + jO 1 - j3
2 +j
+ jO
�
0+j 1 + j3 0 + jO
20. Skew-hermitian matrix . A skew-hermitian matrix A is defined by the
rel ation
(A.12)
and this property applie s only when A is square. The terms of the major diagonal i = i may be zero or p ure imaginary , but not real. An example of a skew-hermitian matrix is given by A
=
[
5 + jO
0 + j2 - 5 + jO - 3 + j2
0 + jO 1
-
j7
2 1 . Dominant matrix . A dominant matrix which has the additional property that ajj
�
1
3 + j2 - 1 - j7 0 - j6
A is defined as a symmetric matrix
L lajj I , j :#= i n
j= 1
[ : � -��
(A. 1 3)
for all values of i. An example of a dominant matrix is given by A=
-1
3
where (A . 1 3) is satisfied or the elements of the major diagonal are greater than the sum of the absolute value of all elements in that row (or column). For exam ple, the Y matrix of a power system is always dominant.
Appendix A
42 6
A.3 R u les of Matrix Algebra
Matrix algebra has rules that must be followed explicitly. The basic rules are: 1 . Equality of matrices. Two matrices are equal if and only if each of their corresponding elements are equal. Thus G = H only if gjj = hjj for all i and j. Only matrices having the same number of rows and columns can be equal. 2 . A ddition of matrices. Two matrices are added by adding corresponding elements to obtain the corresponding element in the sum. Thus
[
1
0
2
-1
-1 3
-1 2
�]
Addition is defined only when both matrices have the same number of rows and columns. Addition is commutative, i.e., A+B=B+A and is also associative, Le., A + (B + C ) = (A + B ) + C
3. Subtraction. A matrix is subtracted from a second matrix by subtracting the elements of the first from the elements of the second. Subtraction is defined only when both matrices have the same number of rows and columns. 4. Multiplication of a matrix by a scalar. A matrix is multiplied by a scalar by multiplying each element of the matrix by the scalar, e.g., k
[_ � �] = [�: :] 3
5 . Multiplication of matrices. Multiplication is defined by the following: G = A X B (or A B) implies that the elements of G are related to the elements of A and B by n
gjj = L a jk b k j k 1 =
(A.14)
where n is the number of columns of A and the number of rows of B. Multiplica tion is defined if and only if A has the same number of columns as B has rows. Two such matrices are said to be conformable. Thus
2X 2
2X 3
where, for example,
g l 2 = L a lk b k 2 = a l 1 b l2 + a l 2 b n + a l 3 b 3 2 k 1 3
=
427
M atrix Algebra
It follows that in the product (A. 1 5 )
C=AX B
where A is an m X n matrix and B i s a n n X p matrix, C i s an m X p matrix. That is, the number of rows and columns of C will be equal to the number of rows of A and the number of columns of B respectively. It is sometimes convenient to check conformability by writing the order (row X column) above each matrix. Thus (A. 1 5 ) would be written m X p
m X n
C
A
nX p X
B
(A .1 6 )
The adjacent terms of the superscripted numbers must agree (n ::: n ) to have con formability, and the outer terms determine the order of C, viz. , m X p. This will work for products involving any number of matrices. It is also apparent that multiplication is not commutative, Le ., AX B* BX A
(A. 1 7 )
It is, however, associative and is distributive over addition, i.e., A(BC) = (AB)C
(A. 1S)
A X (B + C) = AB + AC
(A. 19)
and Because multiplication is not commutative, it is vital that the order of multiplica tion be clear. Thus in the product A X B we say that B is premultiplied by A or that A is postmultiplied by B . 6. The product VA. From the multiplication rule it i s apparent that the prod uct of an identity matrix of order n and an n X p matrix A is the matrix A, Le. , VA = A
(A.20 )
If A is square, i.e . , of order n , then V X A = A = A X V. Thus in this very special case multiplication is commutative . 7 . The inverse matrix and the solu tion of equa tions. Suppose that given the matrix A there exists a matrix A - I such that A - I X A = V. Then A - I is called the inverse of A. The use of and the properties of the inverse are pointed out in terms of (A.1 )-(A. 5 ) . Consider (A.1 ) which can be written in matrix from as x = Ay where x, A , and y are n X l , n X n , and n X 1 matrices respectively. Assuming that the inverse of A exists, we may pre multiply both sides of the equation by A - I , with the result that
A- I X = A- I Ay = Vy = Y
Now (A.2 ) can be written as y = Fx. It follows that F = A - I and the rules for determining the inverse of a square matrix result from a comparison of F (as in equations A.3 , A.4 , and A.5) and A- I . Thus given
428
Appendix A
the inverse of A is
]
C3 1
C 32
C3 3
=
adj A det A
(A.2 1 )
where tJ. i s the determinant of A, an d Cij i s the appropriate cofactor. The exten sion to higher order matrices is apparent. It is also apparent that only square matrices can have an inverse and that a square matrix has an inverse if and only if it has a nonzero determinant, Le., if it is nonsingular. A.4 Some Useful Theorems on Matrices
We state without proof several theorems which are useful, even indispensable, in dealing with matrices and matrix equations.
1 . The inverse of the product of two square matrices is equal to the product of the inverse taken in reverse order, i.e . , (A.22) 2 . The transpose of the product of two matrices is equal to the product of the transposes taken in reverse order, i.e . , (A.23) 3. The transpose of the sum of two matrices is equal to the sum of the trans poses, Le., (A.24) 4. The inverse of the transpose of a square matrix is equal to the transpose of the inverse, i.e . , ( At )- I = ( A - I ) t (A.2 5 ) 5 . If a square matrix is equal to its transpose, it is symmetric,
= At -- A is symmetric A
(A.26)
6 . If a square matrix is equal to the negative of its transpose, it is skew symmetric, i.e., A = - At -- A is skew-symmetric
(A.27 ) 7 . Any square matrix A with all real elements may be expressed as the sum of a symmetric matrix and a skew-symmetric matrix (this is also true if the elements are functions of the Laplace transform variable s ). Stated mathematically , A is square -- A = B + C
where B is symmetric C is skew-symmetric
(A.28 )
42 9
M atrix Algebra
From theorems 5 and 6 we compute
At = B t + Ct = B - C
(A.29 )
Then we may add and subtract (A.28) and (A.29) to obtain, respectively , the symmetric matrix
B = ( 1 /2) (A + N )
(A.30)
C ;::: ( 1/2) ( A - At )
(A.3 1 )
and the skew-symmetric matrix
8. A hermitian matrix is a square matrix which is equal to the conjugate of its transpose and vice versa, i.e . ,
A = (At )* -- A is hermitian
(A.32 )
9. A skew-hermitian matrix is equal to the negative of the conjugate of its transpose and vice versa, i .e . ,
A = - (A t ) * -- A is skew-hermitian
(A.33 )
1 0 . Any square matrix A with real and complex elements may be written as the sum of a hermitian matrix and a skew-hermitian matrix, i .e . ,
A i s square and real o r complex -- A = B + C where B is hermitian C is skew-hermitian
(A.34)
From theorems 3, 8, and 9 we easily show that we may write the hermitian matrix as
B
=
( 1 /2)
[A + (At )* ]
(A.35)
[A - (At )* ]
(A.36)
and the skew-hermitian matrix as
C = ( 1 /2)
A.S
Matrix Partitioning
When equations involving large matrices are to be manipulated, it is frequently useful to partition these matrices into smaller submatrices. Partitioning may simplify the procedures and quite frequently makes the proof of general results much simpler. Consider the following example :
n [u
C = C2
C3
=
.'J
a 12
a13
a2 1
an
a2 3
a2 4
a3 1
a32
a33
a 34
dl d2 = A D d3 d4
(A.37)
Appendix A
430
If we partition A and D by dashed lines as shown we can write C = A D = [A I A 2 1
[�J
(A.38)
= [ A I D I + A2 D2 1
( A. 3 9 ) and now write ( AAO) or tent; i.e. , all products and Again the partitioning of the matrices must be consis of A by colum ns determines sums must be defined. As before, the partitioning of A by rows deter ioning the partitioning of D by rows. In additi on, the partit . mines the partit ionin g of C by rows or vice versa partitionings as sh own below . ple multi into alized gener be can s These result 1 2 3 4 5 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5
1 2
3
4 5 6
1 1 1 1 Cll Cu 11 1 1 -----,--
7
8
A.6
C2 1
1 1 I I
en
1
2
3
4 =5
Au -_-----
6
7
8
A2 l
Matrix Reduc tion ( Kron Reduc tion)
1 1 1 1 1 1 1 1 1 I
1
1 1
r
1 1 1 AI2 1 1 1 - 1 1 1 I A22 1 1
----
1 1 1 Du 1 DI2 3 1 ----- 11 --1 1 6 D2 1 I D22 I 1 1 8 1 D3 1 1 D32 1
1 2
AI3 ___ _ _--
A23
4 5
7
9 10
· - - ---
---
( AA 1 )
ion, after rearranging and In physi cal problems the following type of equat partitioning, is often obtained.
M a trix Algebra
43 1
(A. 42)
Here, the nature of the problems are such that A n is usually nonsingular and the variables in O 2 are unwanted and can be eliminated. If we expand the above equation, the result is
Cl = A l l 0 1 + A l 2 O2 0 = A2 1 0 1 + A22 O 2
( A.43)
Remembering that the quantities in these equations are matrices and must be treated as such, we can solve the second equation for O 2 with the result that O 2 = A2� A u 0 1 and substitute into the first of equations ( A.43) with the result -
C l = ( A 1 1 - A l 2 A2� A 2 t l 0 1
(A.44)
This last result is particularly useful and is a graphic illustration of the utility of matrix methods. A.7
T ransformation of Coordi nates
It is often desirable in electrical engineering problems to transform a matrix equation from one coordinate system to another in which a solution may be more readily obtained. For example, we often find it convenient to transform from the a-b-c phase domain into the 0- 1-2 symmetrical component domain or the O-d-q Park domain. The result is a sim ilarity transformation of the impedances as will be shown. Let T be any unique (Le., nonsingular) transformation which relates currents ( and voltages) between the a-b-c domain and the O:-�-'Y domain according to the equation
(A.45)
and similarly for voltages. We write (A. 4 5 ) in matrix notation as lab e = T la/3'Y
(A.46 )
Suppose we are given a matrix voltage equation, expressing the voltage drop in lines a, b, and c as (A. 47 )
Premultiplying both sides of (A. 4 7 ) by T- l we have (since T- l exists)
r- l Va b e = T- l Z ab e 1ab e
If we now insert the product T T- l = U between Z ab e and lab e, we have T- I Vab e = T- l Zabe T T- l lab e But by
(A.46) this equation is (A.48)
Appendix A
432
Thus under the (linear) transformation T, the impedance has been changed to (A.49) or Zabc has undergone a similarity transformation. This is an important trans formation of linear algebra, and we say that the two matrices Zabc and ZQ/J'Y are similar. Computer Programs
A.8
In working with matrices of order greater than two it is extremely laborious to perform any computations by hand. Matrix inversions, for example, require something greater than n3 multiplications for an n X n matrix. Very clever manipulations can save perhaps half the work, but nothing like an n-fold saving is possible. Similarly, a great deal of computation is required for matrix multiplica tions and even additions. Our attitude here will be that a digital computer is usually necessary and is available for matrix problems. Since FORTRAN is nearly a universal computer language, solutions will be given in this language. Two difficult matrix computations often required are the matrix inversion and the matrix reduction. Both operations can be performed by the same FORTRAN subroutine, and an example of a program to perform these operations follows. 2 ::
� l I N P P OG R A � F OR S � I P l E Y - C O l E " A N - J O H N S ON C OM P L E X I � V E R T E � - R E O � C E � . � I S O N � D t � E N S I O N O F T H E S O U A � E I NP U T " A T R I X - " X I N . K � O I S T H E A R R AY OF R O W S A N D/ � � C J l U M � I � O t C E S J F � O W S l � O "' O T U S E D F !J R I N \f E R S I O � . : J l 'JM N S T O 8 E E L I M I N A T E D I N T H E R E D U C T I O N . I N V . 0 F O R R E D UC T I O "' . ' �V = 1 FOR I NVER S I ON , C OM P L E X . 8 MX I N C I O O ' , M X O U T C Y " ' , I N V E R T , R E D U C E , O U T I N T E r. E R • 4 N , K � O ( l � ' , I N V n AT 4 I N V E � T I ' I NV E R T E O ' I , P. E D U C E I ' R E D UC E � ' I O UT = I N V E R T
,.
c C r
l !'
20 �o c
P EA D ( 5 , ! O ' N ,
I NV
F I)R � A T ( 2 1 5 ' N � = N .. N I F f I NV . E � . l ' � O T � 3n OUT = R E D U C E P E A D ('; , 2 0 ' K G O F OP � A T ( 1 0 I 5 ' ( "X I N I I ' ,
P EA O ( S , 4 0 ' �X I N
I N R OW OR D E R
� OR � A T C @ F I 0 . 0 '
50
W R I T E C 6 , -: O ' O J T F OP M A T ( ' l T H E M A T P I X
or 1 01
=
1 , NN '
C � EAl , IMAG ' .
40
�O 70 �O
1
8E I �G
J = 1 ,N 0" 80 J M! N • ( J - l ' • N I . 1 , N DO 70 I J = I + J "1 1 N W R I T F. ( 6 , 6 0 1 J , I , M X t N ( I J 1
"
� OR M A T C ' M X I N C ' , 1 2 , ' , I , C ON T t N U E C ON T I N U E C A l l C S H P ( "1 X I N , M X O U T , N , WRITE I 6 , l 00 1 F OP M A T C " O IJ P S ' ,
12 ,
2A�/ '
KGO,
"
=
I NV ,
"
I PE I 4 . 6 ,
&1 1 0 ,
'
+J
•
&90'
2 This program was written by G. N. Johnson and is used with his permission.
"
1 PE ! 4 . 6 '
43 3
M a trix Algebra IJ 1 121
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GO
TO
'
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OUT ' , 2�4, 1, N
'
MATR I X '
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BY
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ARR AY KGO f I I . TO 2 5 0
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F OR M A T ( ' O • • • R ET U RN2 K GO T O T .,
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MATR I X
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I TRY
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C ON T I N U E D I AG E L = DOUB L E I I I - i l • N + I f T RY = R E A L I O I A G E L • C O N J G I O I A G E L I I I F I B I G S TA . G T . TRY I GO TO 20 K = I Lill I ) B IGSTA
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C
ARE
9 �X I N ( l � O I , M X O U T I I 0 n l , L ( 1 0 1 . K , KGO ( l O I
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I "V= R S I ON,
M X JUT M X OUT
I N T E r. F R t 4
c
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C OM P L E X .
c
C e C
F��
N,
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FOR
P I VOT
FOR
I NVE R S I ON .
Appendix A
43 4 20
C ONT I NU E I F C B I GSTA
. LT.
I . E-26 t
GO T O 1 2 0
C
C C
L AR G E S T
D I AG O N A L
E L E M ENT
C HO S EN .
GO TO 37
C C
G OI N G
TO
DO A
K R O N R E D UC T I ON W H E N
I NV .. 0 •
• •
C 35
37
C
K .. K GO C I I I t DO = DOUB L E C C K-I )
� O ST IF A
C C C
40 50
* N
P I V I N V = 1 . 0 00 0 I DO K MI N = C K- l t • N DO 50 I " I, N I F ( I . EQ . K t GO T O 50 B 8 • DOU B L E C K M I N + I ) D O 40 J c I , N I F C J . E O . K t GO TO 40 ACC U R A T E
WAY
+
K)
TO CALC
W A S T H E M A TR I X B E I NG
I J = C J- l t • N + I A A = D O UB L E C I J t C C = OOU B L E C I J - r • K t A A = C A A • DO - B 8 • CC t D O U B L E C I J t .. A A C ONT I NU E C ON T I N U l: I F C I NV
.EO.
Ot
A C I , J t .. A f I , J t - C A C I , K J . A C K , J t t / A t K , K ) I N V E R T E D OR R E DU C E D .
• P I V I NV
TO 1 0 0
GO
C
C C
C
C
D I VI D E
55
ROW
BY - A C K , K t
DO 55 I = 1, N I F C I • E Q . K ) GO TO K MI N I = K M I N + I A A = DO U B L E C K MI N I J A A = -AA * P I VI NV D OU B L E ( K M I N I J '"' AA C ON T I N U E I) I V I D E
C
P I V OT
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C O L U MN
ON
I N V E R S I ON ONLY .
55
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ONL Y .
00 60
60 C C C
J .. I , N I F ( J . E O . K t GO TO KJ = C J-l t * N + K A A = DO U B L E I K J t A A = -AA • P ! V I NV D OU B L f C K J ) .. A A C ON T I N U E R E PL A C E A C K , K I
1 00
D OU B L E C K M I N C ON T I NU E I FC INV .EO.
BY
+ KJ 0'
GO
60
-1 . O / A C K , K t =
ON
I NV E R S I ON O N L Y .
-P I V I NV TO
150
C
C
N E GA T E
C 110
12'
1 30
ENTI RE
R E S UL T A N T
MATR I X
DO 1 10 I = 1 , NN M XO U T C I t .. - DO U B L E C I ' R ET U R N I WRI T E C 6 , 1 30 t F OR M A T ( ' O ** . T H I S M A T R I X R ET U R N2
IS
T O R EAL I Z E
S I NGULA R• •• · '
I N V ER S E .
M atrix Algebra c C C C C C
43 5
C O MP L E X S H I P L E Y M A TR I X I N V E R TOR M OD I F I E D
153
NM =
N
-
I< G O T O T
I< RON R E DUC T I ON
160
170
180
& P G MO BY G . N . J OHNS ON P . E .
-
R E GR OUP I NG OUT P UT M A TR I X .
I< = 0 00 170 I = 1, N 00 160 J = 1 , I< GOT OT I F ( I . E O . I > r in every case of interest, we write this equation as
Le = (/J.m 8/ 21T ) or per unit of length,
(� ) 8
In
-1
H
(D. 1 5)
( D. 16) It seems strange that the inductance per unit of length is a function of the length 8. However, as shown in Chapter 4, the argument of the logarithm may have any numerator since it always disappears when circuits containing a return path for the current are used. From (D.8) and (D.16) we have the desired result, viz.,
f fl + fe =
= (/J. w/81T ) + (/J.m/21T )
(
In
)
28 -1 r
Him
(D. 1 7)
The mutual inductance is computed in 'a similar way except in this case the limits of integration are from D to 00 , where D is the distance to the linked current i2 • Then (D.18) or (D.19) These results for the self and mutual inductance for a cylindrical wire of length 8 are both strange and interesting. They are strange because portions of the solution such as the 28 in the logarithm and the (- 1 ) in the parentheses never ap pear in any practical problem involving a "return" wire. The results are of in terest, however, because these primitive building blocks provide powerful tools for analyzing complex problems involving many wires.
appendix
E
S olved Exa m ples
It is often useful for the student of symmetrical components to apply the techniques being studied to a typical system. The purpose of this appendix is to specify small sample systems along with their complete solutions for both normal and faulted conditions. Following are solutions for 3-node, 6-node, and 14-node networks. The 3node network used here is the same system shown in Figure 11.2 and used in the examples of Chapters 11 and 12. The 6-node network is a well-known circuit introduced by Ward and Hale [79] to which a table of zero sequence data has been added. The 1 4-node network was adapted from the IBM assembler load flow described in [80] . These three networks will provide useful data for checking hand computations and computer programs. E.1
A 3-Nocle Network
Consider the 3-node network shown in Figure E . ! . This is the same network as that of Figure 1 1 .2 and is redrawn here for convenience. A normal load flow
for the 3-node network is given in Table E . l where an arbitrary load and genera tion pattern is assumed.
0 , ®
0 0
0. 0
10.00 1 2 . 00 17.00
A D � I T T A NC E
S EL F
R
1 4. 0 0
0. 0
B US 2
X ( PC T I
C OU P L E D L I �E
BUS
BUS
2
3
--x
�O.
R
1 1 2
1 2
0. 0
0 . 1 0 4e 7
O. C
3
0. 07e2 7 0 . 1 1 2 ;:: 1 0 . 0 9 2 7
