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Unformatted text preview: 52 Basic Principtas Chap. 2 ﬁgure P24 5 The load is symmetrical and Z is not given. The sources are not baiancett, but we note
that Ea + Er, + Ec "e 0‘ Find Sills "www.m 0 CHAPTER 3 TransmissionsLine Parameters 3.0 INTRODUGWN In Chapters 3 and 4 we consider the modeling of transmission lines and develop use
ful lumpedcircuit models In these models a line is characterized by four distrib
uted parameters: series resistance, series inductance, shunt conductance, and shunt
capacitance. The series inductance and shunt capacitance represent the effects of
the magnetic and electric fields around the conductors. The shunt conductance ac
counts for leakage currents atong insulator strings and ionized pathways in the air; this
icakage is usuaily a smaii effect and is frequently ignored. The remaining parame
ters, however. are needed to develop a useful transmissionline model for use in gen—
eral power system studies We are interested in two kinds of questions. The ﬁrst is descriptive: How are
the parameters particularly inductance and capacitance, reiated to tine geometry
(in, wire size and configuration}? Systems designers have some choice in the selec
tion of this geometry. One way the selection affects system performance is through
the parameter vaiues. We areY therefore, interested in how inductance and capacitance
depend on the line geometry. A second question concerns modeling. In Chapter 2 the advantages of per
phase analysis of batancert threephase circuits were emphasized, the expectation
being that this useful too! would be applicable to practical systems, including three
phase transmission lines This implies that in the circuit model there is no mutual
inductance between phases If one considers the physical threephase transmission
tine. there clearly is magnetic coupling between the phases How,then, does one get
rid of the mutuai inductance in the circuit model? Similar questions can be asked
about the shunt capacitances For the reader who has studied transmission tines solety from a ﬁeid point of
view, we note that the circuits approach we Will adopt is equivalent for a lossless line
For the practical case of lines with email losses and operating at low frequencies, the
use of the circuits approach is completely justifiable. In the foilowing sections we derive expressions for inductance and capacitance
of three«phase lines in terms of line geometry. 5:! 54 Yrsnsmissiontine Parameters Chap. 3 3.1 REVIEW OF MAGNE'I’ICS We start by recalling some basic facts from physics.
Ampére’s circuftal law thHdlrie (3.2}
l.
where
F m magnetomotive force (mmf), ampereturns
H m magnetic field intensity, ampereturnsfmeter
d: m differential path length, meters
‘9 m total instantaneous current linked by the closed path F
H and d! are vectors in space. In {3.2} we can replace the dot product:
H'didelcose (3.2) where 8 is the angie between H and di, and we drop the vector notation to indicate
scalar quantities The reference direction for EC is related to the reference direction
for T by the familiar righthand rule. In words, (3.1} states that the line integral of H around a closed path I‘ is equal
to the total current linked. The totai current iinited may be found unambiguousiy by
considering the algebraic sum (or integral) of the currents crossing any surface, or cap,
bounded by the closed path I‘; such a surface is called a Gaussian surface. Even though {3.1) is an implicit relation in H, it is useful for actuaiiy calculat
ing H in special cases where we have symmetry. Such a case will be considered in
Section 3.2. We note that there is an expiicit expression for 1!, one form, known as
the Biothavart law, which shows that at any given point H is a linear function of all
the differential current elements of the circuit. Thus, we can caiculate H using su
perposition. Magnetic ﬂux 45. if a linear relationship between ﬂux density and magnetic
ﬁeld intensity is assumed, seen 83
where
B m ﬂux density, wehers/‘tueter2 p, w permeabiiity of medium Sec. 3.1 Review of Magnetics  55 The total flux through a surface area A is the surface integral of the normai comet)»
nent of B; that is, safea go
A where that is a vector with direction normal to the surface element do and with may
nitnde equal to do. If B is perpendicular to, and uniform over an area A, (3.4) simpiifies to 95 3 BA (3.5)
where the quantities B and A are scalars Flux linkages A. The use of ﬂux linkages is very familiar to electrical en
gineers Faraday's law gives the relationship between generated electroruotwe‘force
(emf) in a loop and the time rate of change of ﬂux linkages. The intmuvc notton of
ﬂux linkages is very simple and can be illustrated by consrdcrmg ﬁgure 3.]. Some
of the flux links all the turns; Iii; is a ﬂux line illustrating this case. Some of the ﬂux,
cailed leakage ﬂux. links only some of the turns; all} and 1113 illustrate these components If all the ﬂux to links all the N turns of the coil, then imNa am where A m flux tinkages, in webcr turns If. as in the figure, there is leakage ﬂux, we
add the ﬂux iinkage contributions turn by turn. in this case N
A w 2 «a (3.?) i=1 where of is the ﬂux linking the ith turn of the coii. Figure 3.1 Flux linkages. 55 _ TransmissionLine Parameters Chap. 3 Inductance. Assuming (3.3) to be true, the reiationship between flux linit
ages and current is linear and the constant of proportionaiity is cailed the inductance. .t a Li {3.3) Exmpie 3.1 Caientate the inductance of a coit wound ciosely on a toroidal iron core (Figure
Ell). Assume  1. Small cross—sectional area.
2. At] the ﬂux iinks all the turns. Crosssectioeat area
A Centerlirs:
orcoﬂ N turns Figure E33! Solution This is a classic example the reader may recall from physics From physics
we can assume that the ﬂux is inside the core and the flux ﬁnes form concentric circles.
For the choice of reference direction {or current shown and using the right~hand rttle.
we find that the associated reference direction for ﬂu): is counterclockwise. Taking a
counterclockwise path F. aligned with one of the ﬂux lines at a radius r, and applying
Ampere's circuits] law {3.1), we get FmﬂfﬁNi where t’ a 2m is the length of the path 1". The magnetomotive force {mmf ) is N times
1‘ because the Gaussian surface bounded by I‘ is crossed N times by the multitum Winding. Using {3.3) in the preceding expression,
3 * uh‘ * #7
From this expression and assumptioa Lwe can see that B does not vary much across the crossmscctional area, and using a value of 8 corresponding to 8 == 27:}? gives a very
good approximation. Then Sec. 3.2 Fiux Linkages of Inﬁnite Stasight Wire . 51 We note that it. is proportional to F = Ni and inversely proportional to a quantity
C/oA involving only the geometry and material of the core. This quantity is called re
luctance. In Appendix 1 the reader will find a short introdccticn to the more general
use of this quantity. Returning to our calculation of inductance, we have, finaliy, using the assump
tion that all the ﬂux it: links all N turns, 3.2 FLUX LINKAGES 0F It‘ll:le mutter MRI! We next consider a case of great interest to us in modeling transmission lines This
is the case of the ﬂux linkages of an inﬁnite straight wire. The inﬁnite wire is an ap
proximation of a reasonably long wire and is simpier to consider for the articulation
of parameters. We should note at the outset that a single inﬁnite wire is certainly
not an approximation of a practical transmission line but its consideration is impor
tant to us as a step in using superposition to find inductance parameters for realistic
multicondtictor tines. The notion of flux linkages for a single straight wire needs some justification.
More commonly, flex linkages are defined for a coii where the interlocking of the
turns of the coii and the lines of flux are easy to visualize. The present case is Zess fa»
miliar but the interlocking is also clear; there is no way to visuaiize how the lines of
flux may be detached from the infinite wire. Altemativety, and consistent with this,
we may imagine the wire to close at infinity { oo and + co the same point}, thus es»
tablisin'ng a kind of “oneturn coil“ with the return path at infinity. Assume then a 1. Straight inﬁnitely iong wire of radius r.
2. Uniform current density in the wire. Total current is i. In the case of dc or lowfrequency ac, assttmption 2 is justiﬁed in practice. From elementary physics we know that the ﬂux lines form concentric circles
{i.e., ii is tangential). Assume the current in the wire is coming out of the page. Then
the associated reference direction for the flux is shown in Figure 3.2. Because of am
guiar symmetry it suffices to consider H(x). Case 1: Assume that x > r (Le, points outside the conductor}. Ampere’s cir—
cttital law appiied to the path I] gives H'JIRH‘Z’rrxﬂi r‘ i. (3.9a)
H z: .—._._
277x 58 , TransmissionLine Parameters Chap. 3 Figure 3.2 Wire carrying our“
sent 1‘. We note that for current out of the page, H is directed upward for x positive. Case 2: Assume that x 5 r {i.e., points inside the conductor). Ampére’s law
applied to F3 gives where we have used the assumption of uniform current density, Then x .
H m 5132»; (3.92») To find the flux density we use {3.3} but write p. : mug, where in the SI system
of units, in, = 47: X 10'? is the permeabiiity of free space and n, is the reiative per
meahiiity. Thus 3 a omit {310) Outside the conductor, in air, it, as 1, and inside it, a= 1, if the conductor is nonmag—
netic (eg, copper or aluminum). In most cases, however, steet strands are inciuded
for strength and 31., b 1. We next consider the flux linkages associated with the inﬁnite wire. It is evident
that the ﬂux linkages of the inﬁnite wire are inﬁnite, and we are led to consider, in
stead, the flux linkages of the wire per meter of Iength and consider oniy the flux
within a finite radius R. To ﬁnd these ﬂux iinkages, consider Figure 3.3. The ﬂux
per meter surrounding the wire out to a radius R is the same as the ﬂux crossing the
1 meter X R meter rectangle. For convenience the rectangle is taken to be in the
horizontal plane. There are two components to the ﬂux linkages They are the linkages due to
the ﬂux outside the wire and some additional linkages due to ﬂax inside the wire.
We consider the two cases separately and then add the results Sec. 3.2 Flux Linkages of infinite Straight Wire . 53 2/ 3
Infinite wire /
radius r 4“
Slot or ”
are: do /
1 at): “R I Figure 3.3 Flux crossing rantsngte. 1. Flux linkages outside the wire: Because each line of flux interlocks with the wire
only once, the total ﬂux linkages are numerically the same as the toad flax cross ing the rectangie in Figure 3.3 from the radius r to R. Thas, using the subscript
1 for the external ﬂux contributions, R g .
A1m¢ImIBnaeJ3(x}dxmﬂaj_£mdx A r I 2111 to R {3”} mv—m—‘mh i
2w 3’ Here, we have integrated over the differential: areas do 2 1 dx shown in Fig
ure 3.3, with B found from (3.93) by using 8 a right. 2. Flux linkages inside the wire: This portion of the ﬂux links only part of the wire
and it would not he reasonahie to assign it full weight in caicuiating ﬂax link:
ages. A carefui analysis based on energy justifies the idea that each flux con
tribution to the ﬂux iinkages shouid be weighted by the fraction of the total
current linked. The result, although intuitiver reasonable, is by no means ob— vious: Then using (3.91:), (3.10), and the assumption of uniform current densi~
ty, an integration over the rectangle inside the wire yields r eminent?”
2 ' ,2?"sz ‘
new {3‘12}
811' The result is independent of wire radius Adding {3.11} and {3.12} the total
flux linkages per meter are i (3.13} 80 Transmissiooiine Parameters Chap. 3 in the second line we have substituted the numerical value of in} to get a form
more useful for calculations We will freely exdtange between the two forms
but prefer the ﬁrst form for development of the theory. Regarding (3.13),
note that )1 increases without bound as R goes to infinity. This is a difﬁculty
we will dispose of when the practical multicondnctor case is considered. 3.3 FLIJX LINKAGES; MANYcﬂﬂlmk GASE We turn now to the practical case. Suppose that instead of one conductor we have it
round conductors as suggested by Figure 3.4. Some of these conductors are the “re
turn” conductors for the rest, but it is equivalent and more convenient to think of
each wire as having its return path at infinity (Le, the case we considered in the pre»
ceding section). Strictly speaking, the assumption of uniform current densit}r in each
inductor needs to be justiﬁed once the ﬁelds and currents interact. However, if the
distance between conductors is fairly large compared with their radii, the assump»
tion is justified and is convenient because we can then use superposition. We are still interested in caicniating the ﬂux linkages of conductor 1 up to a
radius R, from the origin, but now we add the contributions due to conductors
2, 3, . . . ,n. We use superposition. Consider the contribution of the kth conductor
current as shown in Figure 3.5. Assume that ali other currents are zero. The ﬂux
lines due to i}, are concentric circles, as shown in Figure 3.5. The ﬂux line tit; does
not link conductor 1. The flux line (113 does. The ﬂux line #195 links conductor 2 but is
beyond the radius R, :it does not qualify. The flux lines tire and 9'14 mark the approxe
imate extremes of flux Zines that qualify. Thus, we are interested in all the ﬂux gen
erated by a}, that passes between the points 1; and c on the x axis. The calculation is
much simpler, however, if we find the flux crossing the surface deﬁned by the points
a and c because then the ﬂux and surface are perpendicular. Defining the distance
from the center of the ktlt conductor to the point c as Rhthe ﬂux to be calculated is
simply that between a radius (in, and Rh. But this calculation was already done can
lier; the result given in (3.11) is directly applicable to the radii presently under core
sideration. We get as the contribution of ik to the ﬂux linkages of coil 2, ﬁgure 3.4 Many conductors. Sec. 3.3 Flux Linkages; Manyﬂonductor Case _ 61 Conductor
I Figure 3.5 Conwihution of ii. ’lLoI‘k Rk
A m __ in...
it 2” 15 d” (314) in the calculation we mate a reasonable approximation about the partial flux link—
ages of conductor 1. We sum the remaining contributions similarly to get the total flux linkage of
coil 2 up to a radius R, from conductor 1. Fe . a. R . R R
A} == + 1n + izln 93“ +   i + iﬂln (3.15)
12 in We note that in general A, w) on as R, «a on, but in the practical case we make a ltcy
assumption regarding the instantaneous currents in the wires. Assume that f1+i}+"‘+i,,wﬁ (3.16) in which case we can show that (3.15) does not diverge as R1 «we on. The assumption
{3.16} is reasonable for a transmission line under normal operating conditions It is
certainly the intention of the system designers that all the currents should flow in the
wires of the transmission Iine. This condition, which may be violated under abhor»
trial (fault) conditions, is reconsidered in Section 3.6. Rewriting (3.15) in a way that permits the assumption (3.16} to he introduced,
#11 . it, 1 . 1 t
A m m we. m m . . . ' .w.
1 217i11(4 mrt)+ hind}: + +rnlﬁdia
“a (3.17)
+£(i1mR1+1QIBR2+' ' '+EHIDRR) 82 _ Transmissientine Parameters Chap. 3 To the second part of (3.27) add
“gmhtR; + sztnR; + ‘  +:,.121R1) This is legitimate since. by (3.16), we have added zero. Then the second part of
(3.1?) becomes “.9 v R} _ R2 ‘ R“)
m + «h... + t v v + 2170} R; £211} R1 in R; The first term is zero and in the limit as R, + 00, each other term Rk ,2“ R1 tends to 1, and each logarithmic term then tends to zero. Thus in the limit as R1 —»eo,
(3.17} reduces to , 3
A] w ﬂag. t he.) + r . v + inmfm] (3.28)
in Equation (3.18} can be made more symmetrieal by noting the identity #9 :1" m InaW" (3.29)
Thus
, 1
5+1n+=meef‘+1nim2n3~emmln% (329)
4 r! r1 r1 3*,
where
r; g new"
Then (3.18) becomes
.uo . 1 . 1 . z )
a “ —“ + m + ' ' ' m
A} 21r<h 11! :1 £211} (in + In In d1” For a nonmagnetic wire, pt, w I, and r; w rle'W m {3.7738r, a 0.78m. Interpreted
physicaiiy r; is the radius of at} equivalent hoHOW conductor with the same ﬂax link
ages as the solid conductor of radius ri. Amphre’s eircttital few makes it clear that
the helmet conductor has no internal ﬂax. However. sinoe r; is less than 1'1 , the ex—
ternal ﬂux is increased to compensate for this lack. Sec. 3.3 Fiux Linkages: Manyﬂanduetor Case . 33 Note that the flux linkages of conductor 1 depend on all the currents i] ,
i2, . . . ,fn. The equation thus is in the form A1 2: Inf, + Imiz +   ‘ + (mimwhere
the indoctanee parameters I” depend on geometry. Similarly, with obvious modif
cetions, the flux ﬁnkages per meter of the genera? kth conductor are 1 i
Akw%(ilmgg+ . .  ‘ ’ ‘ +isln‘aiw) in the form At : {Hi} + fut; +    4» thin. Thusﬂt is deer that there is mutuaz in
ductance in the circuit description.  Example 3.2
Caicuiete the inductance per meter of each phase of a threephase transmission ﬁne (Figure E32}. Assume that in; Figure 33.2 1. Conductors are equally spaced,D,and have equal radii r.
lin+ib+h¢l Solution Using (3.21) {or phase a, we have He 1 . 1 . 1
Aﬂ—2W(rah1r,+lbhlp+rrlnb)
oﬂv 1w 1
m2#(t,1l'lr’ titan)
mating
217“ 1"
Thus,
3.
1a=4=ﬂm2=2xw41n3 (3.23}
1‘, 21: r r In the derivation use has been made of the equiiaterai spacing and the fact that
in e —(:‘h + it) Because of symmetry we gel the same results for A5 and Ac. Thus, we
get the self—inductanees L, w I» == 1‘, w (proﬁle) ht (Djr') henrys per meter. We also
note that A” depends on i, atone {ie, there are no mutual inductance terms in this. pat“
ttcalar desa‘iptioa). The same is true for b and e phasea 34 Tcansmiaslorrtina Parameters Chap. 3
The preceding exampie iliustrates some important points: 1. Although physically there is magnetic coupling between phases, if
in 4— ii, + it 2 0 and with equilateral spacing, we can model the magnetic effect
using only selfinductances. The self‘inductances are equal. This then permits
the use of per phase anaiysis 2. To reduce the inductance per meter we can try to reduce the spacing between
conductors and increase their radii. Reducing spacing can only go so far because
of considerations of voltage flasbover. On the other hand, there are cost and
weight problems associated with increasing the radii of solid condtictors and
problems of ﬂexibility and ease of handling as weli as cost in the case of hollow
conductors. The problem is neatly solved by the practice of conductor
“bundling,” which is considered next. CONDUCTOR 3UNDLING Suppose that instead of one conductor per phase there are in conductors in close pros»
imity as compared with the spacing between phases Such a composite conductor is
said to be made up of bundled conductors The conductors within a bundle are sup»
ported at frequent intervals by a conducting frame, as suggested in Figure 3.6, and thus
these conductors are effectively in parailel. In the ﬁgnre is shown an equilaterally
spaced set of phase conductors each composed of four bundled conductors Typicaliy,
two, three, or four conductors per bundle are used. We wili now consider the configuration of Figure 3.6 in more detail. Suppose
that the spacing between bundle centers is D, where D is large relative to the space
ing between the conductors of the same phase. Assume that alt the conductors have
the same radius r. Consider the ﬂux linkages of conductor 1 in the phase a bundle. Assume for
simplicity that the current in each phase spiits equally among the four parallel branch»
es We can use {3.21) and get the flux Zinkages of conductor 1, 8 7 Conducting frame
l’hasc b supporting
5 ‘5 centimeters
D D
4 3 12 I}
D i 2 9 :0
Phase 4 Phase c ﬁgure 3.3 Bundied conductors. Sec. 3.4 Conductor Bundling  35 a, i, 1 1 1 1 )
a __._.._ _._ n... is. w. + w.._. + ._
A; 2%”[4011 r, in (In lad13 1nd14 + £30an + lav—i" + ln—l— + luv—L)
4 dis dis d1? dis {324} 1 1 i
a, In W» + i In wwmwm.» + r, in ~~~~~~~~~~~~~ my]
217' l {’1 diZdlﬁdliyjm h (dl5dlﬁd11‘dlﬁy'ﬁ {dl9d1.lﬂdl.ildi.12)li4 My . Z , 1 , 1 )
m _._._ __. + l W. 4}. l W
2 (1,, In b rt, n m 1,, n k where we have introduced the definitions manages“
m geometric mean radius (GMR) of bundle mﬁwemmm‘
2 geometric mean distance {GMD} from conductor 1 to phase b Dir: é {disdmodmdmzlm
m GMD from conductor 1 to phase c Noting that D“, e Dk *3 D, and assuming that i, + it, + i, z 0, we get, approxi tl,
maey #3 D A, m 2’” a, 1n Rb (3.25)
Comparing (3.25] with {3.23}, we see that the only difference is that r' is replaced by
the (much larger) geometric rrtean radius of the bundle, Rb.
In calculating the inductance per meter of phase a, we can proceed intuitively.
First, calculate the inductance per meter of conductor 1. Noting that the current in
conductor 1 is i,i4,we are led to the no.2
1, “fat” 4(217 1n Rb {3.26} Next, calculating A2, we find the same value of Rt, as previously; the reason is the
symmetrical spacing of conductors In calcuiating GMDs from conductor 2 to phas»
es 2) and c, we ﬁnd that £32,, a: D2, e= D because of the large spacing between phases.
Thus, 12 es ii. In this manner we find that l} a £2 a: £3 a [4. Then, since we have four
approximately equal inductors (I, , l2, I3, 11,) in parallel, enema D mZXIO‘ll m 3.27
4 21: R, I’R, ( ) Jae BB ‘tcanemissioniine Parameters Chap. 8 Checking the steps in the catcuiation, we see also that In 2 (b w 10 We have been considering the casa of four bundled conductors per phase. The
reader should check that (3.27) holds also for any other bundled configuration in
which the bundled conductors are smmetricaliy spaced around a circle and the dis«
tances between phases are equai and iarge compared with the spacing of conductors
within each phase. Of course, the GMR must be defined appropriately; With E) con
ductors we get Rb % (was. . . . . do)“ b a 2 (3.28) Note: We can define Rb z r’ if b w 1 {i.e., the nonbundled case}, and then
(3.2?) includes {3.23) as a special case. Empte 3.3 Find the GMR of three symmetrically spaced conductors {Figure 1333}. Assume that
r = 2 cm and r‘ m 22“" = 1.56 cm. so cut Figaro £33 Solution
Rb a (1.56 x 50 x 50}15 = 15.7 cm Example 3.4 Find the GMR of four symetricaily spaced suboondnctors (Figure 83.4). Assume that
7 = 2 cm and r’ z 1.56 cm. 50 cm Figure 33.4 80an
R» z {2.55 x so x 50 x sev’iyc z 22.9crn Sec. 3.5 Transposition 87  In Example 3.4 the inductance of the line has been rednced by a factor 2.69
compared with a line with singie conductors of the same radios or by a factor 1.44 com
pared with singie conductors with the same amount of material as in the bundles. Three additional comments about bundling are as foliows: 1. If we View the brindle as an approximation of a hollow conductor, the reason
for the increased “radius” is intnitively ciear. 2. The larger “radius” heips in another respect. At high voltages, above approx
irnately 230 kV, the electric field strength near conductors is sufficiently high to
toniae the air nearby. This phenomenoncalied corona, has an undesirable cf
{ect since it is associated With line losses. radio interference, and audible noise.
Ail other things being equal, the larger the conductor radius, the less the elec
tric field strength at the surface of the conductor. Bundhng is beneﬁcial since
it effectivety increases the conductor radius 3. Compared with a single condnctor of the same crosssectional area, bundled
conductors, having a larger surface area exposed to the air. are better cooled. rThus, higher currents may be carried without exceeding theme? limits. More
on this point later. Exercise 1. In checking the validity of comment 1, consider the GMR of a
sequence of b symmetrically spaced snbcondnctors alt at a ﬁxed radius 1 (Figure
EX 1). As b—>oo the spacing between conductors necessarily tends to zero (as
does r’) and the configuration looks like a holiow conductor of radius 1. In this case
we get! I (none) in (19,31), and we expect that R}, r— 1. Do you believe it? As a
check, pick r z 0.1, pick b 2 31, obtain the values of (iij graphically, and caictriate R1,.
Note that {or b t 32 we approximate a holiow conductor. ® O O The simple result of Example 3.2 has been extended in one direction; with mul
tipie conductors per phase instead of a singie conductor, we use the GMR, Rb in
stead of r’. In Section 3.5 we extend the result in another direction. 3.5 WSWSH’IDN In practice the equilateral arrangement of phases discussed in the preceding two sec—
tions is usually not convenient. It is nsuaiiy more convenient to arrange the: phases
in a horizontal or vertical conﬁguration. In this case symmetry is lost. One way to regain the symmetry and restore
balanced conditions is to use the method of transposition of lines. This is iliustrated 58 Transmissiontine Parameters , Chap. 3
Position
.7 b 1 13 b
I " c
3 Section 2 Section 2 Section 3 Figure 3.7 Tronspased lines. in Figure 3.7. We can think of this as a top view of three condactors in the same hor
izontal piane. Equaliy weil, it could be a side view of three conductors in the same
vertical piane or ever: a view of three condnctors not in the same plane. In this case,
to visuaiize the transposition, think of a ﬁne crew that sets out the insulators on the
poies or towers in an arbitrary but fixed configuration labeied 1, 2, 3 on every pole.
Another crew then strings the wires, totaling each phase, at intervals, throogh each
insoiator position. For exaropie, phase it starts in position 1. gets rotated to position
2, then position 3, and then the cycle repeats. From the way the tine is constructed it seems physicaiiy reasonabie that the
average indectance of each phase wiil he the same. We now wish to calculate that
inductance. Assume that 1. Each phase occupies each position for the same fraction of the total length of Zinc.
2. Each phase is a single {nonbundled} conductor of radius r.
3. iﬂ+ib+ic=£1 Next. we calculate the average flux linkages per meter of phase a. This is
X. a in“? + 3‘3“ + hi?) {3.29) where RE? is the ﬂex linkage per meter in the ith section,£ m 1. 2, 3. Substitnting in
{3.29} from {3.21} anti using the transposition cycie in Figure 3.7, we get _ 1 1 . 1 . 1
A z..me lam+ 1 _+ ——.—
‘7 3 2a“ 1" r’ lb 31d}; 1‘ In dig 1 Z 1 1 Z 1
+.m+. “ﬁrm” . m . M . m
tail] r, wind23 rein dig + tain r, + ibln dB + i. In “'23)
Each of the rows in {3.30) is the appiication of {3.21) to conductor a as it ﬁnds itself
saocessively in positions 1, 2, and 3. We can group terms, as foiiows: {333} — ,ug _ I . 1 . i)
2...... _. _+ —
In, 2 (15,.an + thin Dm QIan g M m“...
211'!“ r’ (3.31) Sec. 3. 5 Transposition 55 Q . .
where D", m (dudzgdnyﬂ : geometne mean of the distances between positions 1,2
and 3. Thus we get the average inductance per meter: 3 fnxﬂ‘imyﬂ 2” r, {3.32) The formuia is the same as {3.23) except that D is repiaeeci by Dm . It is clear, because
of the transposition, that L, m 1,, n it. We note that this result is consistent with the
use of per phase anaiysis Suppose now that instead of being single round conductozs, the conductors are bundled. ﬂow does this affect (3.32)? Comparing (3.2?) and (3.23), we can guess that
the result wiii be     o
ismismlrwfmﬂinmﬂr—ZX10‘7ln—w “m which, in fact, is correct. This can be strewn as follows. If we consider:
the transposed line [using appropriate notational and other modifications U3. wan, to accommodate the unequal spacing of phases], we get the approximate females for
the flux linkages of conductor 1: no . 1 . 1 i
it}; a E0“ in E; + {bin Egg + i. In dm) {3.34) where (11,? and dﬁ} are the section 1 spacings between phases a and b. and a and c, re« spectivety. Equation {3.34} corresponds to {3.24} with the approximations (1213 Sr— D”.
d3? e: D1... We next calcolate and get — i no . i. 1 1 1
m x . . .
A“. 3 25.1511: Rb + 1b in d“ + {(1%} def] 1?» rain Rh
1 . . 1 . 1
+ lb in dd 4» z. in da) + 1.1!! Rb + thin + icin (:3) Equation (3.35} corresponds to (3.30} with appropriate changes in notation. Re
groaping terms gives ns {3.35}  Fe . 1 . 1 . 1 pi} D
Ana—tainwt» Emmir «w n—' “~31
1 2w( Rb lg n Dm t in D”) 2” “in Rh (3.36}
where BM % {diﬂdﬂ§d§§}m m [dﬁﬂdﬁhﬁﬁﬁ] 1" 3. The notation obscures the eqeaiity of
the two bracketed terms; the reader should check that they are equal because of the
transposition of phases By exactiy the same techniqne used in going from {3.25) to
{327), we go From {3.36) to {3.33). It is again evident that I. m L, a it in thiscase also. 7D Transmission—Line Parameters Chap. 3 . We note in practice, even when the lines are not consistently transposed,
it is convenient and causes littie error to caicuiate the inductance as if they were
transposed. Ne‘: also that with the deﬁnition Rt, 2 r’, for b w 1. (3.33} covers all
combinations of we have considered. Example 3.5 Find the in:....ance per meter of the 3d: line shown in Figure E35. The conductors
are aiumint: . a, z 1), with radius r w 0.5 in. «I M 40 fl
damn“"20 ft “fww«QO iii—M—v—b
C O 0 O O O — }*“—Eﬁin. Figure 53.5 Suluﬁon among}: for simplicity the problem is stated without precisionjt is generaiiy
understood :1: what is required is the average inductance of the transposed line. Thus 3,” = {20 ft X 20ft X 4028’?“ a 25.211
R5 2 {0.73 X {3.5 x 18)“2 a 2.65 in. m 022 ft ' #0 D»: _ f a. — _.... 2: ? .W m “' y
a 2 in Rh 2 X It] It: [l 9.47 X 10 ﬁlm Note: .LZhough line dimensions are in feet. the result of the calculation is in
henrys pet cater. We have used no r— 41: >< 10' '. In introduczg the calculation of inductance, we have simpliﬁed matters by mak— ing some reasonable assumptions and approximations. In most cases we can extend
the anaiysis and tittain more accurate results without undue cxtra labor. A case in
point is the caZCLLation of the geometric mean radius (GMR) for stranded conduc«
tors This is an izmsttant practicai case since almost invariany conductors are strand
ed rather than soli. The method used to calcuiate the GMR for bundled conductors
extends easily to 12:5 application. For exampie, for the stranded conductor, iabeled
“Blucjay,” with ‘ :cntral) strands of steel and 45 strands of ainrninnm, the outside
diameter is 1.25.L :2. and the GMR is 0.4992 in.‘ For stranded conductors we get
more accurate toils by substituting the conductor GMR in piace of r' in at} our for
malas. However. even if we ignore the stranding, our rcsuits are usually quite accu
rate. In the Bluejzj examptejust quotedji we negiect the stranding (and the presence
of iron} and assucsc a solid round aluminum conductor of the same outside diame
ter, we get r' == t. _83 X 1.259,!2 m 0.4903 in, compared With the GMR of 0.4992.
This small error is spice} for ACSR conductors ' See’Fabte A8; 11 Appendix 8. Sec. 3.8 impedance of hired Phase Lines including Ground Return 71 A similar extension permits the use of an exact geometric mean distance {GMD}
between phases instead of the approximation used here. For the detaiis of these
caiculations, the reader is referred to Anderson {Chapter 4), and Grainger and
Stevenson (Chapter 4). 3.6 IMPﬁDANGE OF matte PHASE “NEE INCLUMNG
GROUND RETURN In some cases we cannot assume haianced operation. There may be Zack of
transmissiontine transposition or of load balance. More extreme cases of unbalance
are treated in Chapter 22, where we analyze the steadyetate performance of power
systems during faulted conditions We consider, for example, single phase to ground
faults. The current now ﬂows in the neutrai conductor (if present} and the groans!
(or earth) return as tired as in the phase a. b, and c transmissiondine conductors As
a result, we will need to model the effect of earth and neutral retnrn on the imped
ance of a transmission line. In this section, we present the essential features of the approach. In a land»
mark paper in 1923, J. R. Carson developed a procedure to describe the imped
ances of overhead conductors with an earth return. This paper, with certain
modifications, has served as the basis for transmissiomline impedance calcu»
Zations in such cases. The matrix equation (3.37} is used to relate the voit
age drop from one end of each line to the other end of the line in terms of
iine and neutrai currents. The model also assumes a singie isoiated neutral
conductor. Vt Zn Zn Za Zn In
V Z Z
assist {m
a Zn an Zoe Zn In
where
V! : phasor voltage drop of phase 1’, t' m a, b, c, n
1, K phase: current ﬂowing in phase i Zh m selfimpedance of conductort' including the effect of ground return
3 Zn X E f m Iength of transmission line in meters ,. D
m . f I _? E
2;. (t': +313} 4 ;m2 X It] in R‘ ﬂftn i is.
it!!!“va {L 9‘} F.__. ‘12 TransmissionLine Parameters Chap. 3 i; a: resistance of phase i in {this
rd m 9.869 X 10”?)6 is the earth resistance in ﬁlm f = operating frequency in hertz 1)., w 658.368 J? m p as resistivity of the earth in the
GMR m geometric mean radius of conductor 1‘ in meters Zif m zir X t? is the mutual impedance between conductor iand conductor 1'
including the efiect of the gronnd return De
Zgj 3: rd + x 10m? u‘
d5, z distance between conductors i and j in meters Some additioeai points foiiow: L Carson determined that the earth resistance rd is a function of the frequency and
derived the given empiricai formula. 2. in the formula for DN if the actuai earth resistance is unknown, it is common
to assume ,0 to be 108 011}. 3‘ De, 8MP.“ and d4) are only invotved as ratios in the formulas for 23.; and 2‘}.
As iong as the units are consistent, either feet or meters can be used. 4. For further details on the evaluation of seiﬁirnpedances and mutual imped»
ances, see Anderson, Chapter 4, pages 78433. In (3.37), the effect of the neutral conductor is included. However, if 1,, m {}
(i.e., neutral conductor is open circuited), or V" a 9 {e.g., neetrai conductor impeti
ance is zero). (3.37} may be simplified. Consider the case 1,, a 0. In this case we
may solve for V“, Vb, Vc in terms of IQ, {milr alone. in matrix terms we may eliminate
the right hand column and bottom row of the 2 matrix in {3.37}. Next, consider the case V" x 0. Then Va: x 0 "v: Zulu + 2121):}: + zrscfr + ZR”!!! which impiies + Z + I
I” 3 “(Zena an’h ch c) Sec. 3.8 Impedance of Three Phase Lines lncéuding Ground Return 73 Substituting {3.39) in the ﬁrst three equations of (3.37) yields
Vs Zia Zia Zhr L
V}: 2 25»: Zia! Zinc It: (349)
Vc Zia :23 Zia“ Ir where
I , Zi'n‘Zm' .
Z“. x Z“. » ; r w a,b,c {341)
Zf “ zinznj ‘ .
i}. .s 2“. w Z ; ;, Jr 2: a, b, c {342} “11 Under additional assumptions equation (3.40} inciudes as a special case that of
a balanced and compietely transposed line connected to a symmetricai ioad. In this
case {3.40) reduces to that given by (3.33). See Anderson, Chapter 4, for details. Example 3.6 Compute the phase so]? and mutual impedances of the 161 kV line shown in Figure
23.6. The condactors are “Hawk” ACSR 261’”, with a resistance of 0.3933 {Umi
and a GMR of 0:0289 ft. Assume that the frequency is 60 Hz {as 2 377} that the
neutral {current is zero {Lew the ground wires are open circuited), and’that the
phase wires have the conﬁguration shown in Figure 353.6 for the entire iength of
the line. Assume that the earth resistivity p is 100 (1‘11: and that the Linc is 68
miles long. mi In? Figure 53.8 T See Terrie A83 in Appendix 3. 74 TransmissionUna Parameters Chap. 3 1
Solution {a = rh 3 rr 3 0.193}. ﬂfmi
(Ht/IR,I m GMRb m GMRr = (113289151 9 Moo
3: w 2 we a 90ft
D, 2160\/;ft 2168 \y 60 27 A:6t}tiz,r, m 9.869 x 10”; ﬂfm z 592.; x 10" turn.
Using the conversion factor 1 mite 2 2.609 kilometers, we get rd 3 0.09528 ﬂ/rni Then . 31 Dr {I
zmmmzzﬂﬁrwrn m2><19 {tam Km 2790 _
a (9.2931 + 009523} + gm x 2 x mUmaﬁg x 1.609 x 103mm] = 0.2884 + J£1,392? mm; The mutual impedanccs are computed as foilows: . _.7 D:
zabzrdt ;wZX 10 in}: 21'90 ,.3
e ‘ '? ~W»~ X 1.609 X 10
w 0.09528 + 1(377 X 2 x 10 ) In 1861,12 = 0.99528 + j 0.6301 ﬂfmi It is important to note that, as suggested eartjer, in calculating the natural logarithm,
we have converted the spacing between conductors a and E: into feet. Thus, 2,, m z,, = 0.09523 + [0.5301 ﬂfmi Finaliy, we calcuiate a“, noting that at? quantities except for allﬂr are the same;
d," z 372 inches a: 31 feet. ZN = 0.09528 + $546 ﬁlm For 60 miles of line, we multiply the foregoing vaines by 60 to write, in matrix notation, (27.394 + 183.562} {5.717 + 137.31) (5.717 + j32.76)
2m x {5.711 + 187.81) (17.304 + j83.562) (5.717 + 13731) 0
{5.727 + j32.76) {5.717 4» 137.31} (17.304 + j33.562) , Sec. 3.7 Review of Seenit: Fields 75 Unzii new we have been considering magnetic effects, ﬁne indeetanoe in Sections
3.1 through 3.5, and sexies impedance (series resistance and series inductive reaca tanoe) in Section 3.6. We turn next to the calculation of (shunt) capacitance, anoth
er important transmissiomliee parameter. 3.? “NEW OF ELEM“: FIELDS We start with some basic facts from physics. Gem's law. Gauss’s law states that J D  d8 a" at (3.43) where i} m electric flux density vector. ermlombstrn2
do w differentiai area do with direction normai to the surface, to2
A r: total closed surface area, In2 q, a algebraic sum of alt charges enciosed by A, couiombs As in the case of Ampére’s circnital 23w, Ganss's iaw is implicit in D, the variable of interest, but can he usefnliy applied when symmetries axe present, as in the £0}!on
ing example. Example 3.7 and the ﬁeld of an inﬁnite uniformly charged straight round wire. Solution Draw a cylindrical Gaussian surface concentric with the wire and 12 meters
long {Figure E3.7,where the charge on the wire is q din of iength). Considerations of
symmetry indicate that D is radial and constant in magnitude ova the curved portion
of the cylinder (it is zero on the end caps). Thus, using Gauss's law, 1B _ TransmissionLina Parameters Chap. 3
Then
I) = NEW R a: r {3.44}
27:}?
where D is the seats: version of D. Knowing that I) is directed radially, we also have
n z a,—‘~"w R .2: r (3.45)
217R where a, is a radially directed unit vector. Electric ﬁeld E. In a homogeneous medium the electric ﬁeld intensity E is
related to D by a w as (3.46} where the units of E are voits per meter, and in free space, in the 51 system of units,
E a E9 3 8.854 x 18"” farads per meter. In other media we may write 5 3 6,69.
where gr is called the retative permittivity. The relative permittivity for dry air may
be taken to be 1.0 with negligible error. Voltage difference. We can next ﬁnd the voltage difference between any
two points PR and P5 by integrating E (in voits per meter) along any path joining the
two points. From physics, P!‘
359“ e as m up“ n *J E  :11 (3.47)
Po
We will now apply the resuit in Exampte 3.? to ﬁnd an expression for Zine capaci»
tance. 3.3 LINE WAGITANGE Roughly speaking. capacitance rotates charge to voitage, and hence we consider next
the voltage differences associated with the inﬁnite charged tine of Example 3.7. Since
the eiectric ﬁeld is radial, we can think of a path in which we get from P,I to P3 by ﬁrst
moving parailei to the wire and then at a ﬁxed radius until we can reach P3 by a pure»
1y India} path. The integrand in (3.47) will be zero for the first two segments, and for the third segment we get Rs a
"' titanium‘1 (3.48} “a” t um N v?" m “J 211's}? Zara R3 R... Sec. 3.8 Line Capacitance , 17 where Rt, is the (radial) distance of PI, from the wire and R3 is the (radial) distance
of P5 from the wire. Note that if q is positive and P3 is cioser to the wire than F
zts potential will be higher than that of Pa. Note that we have not yet considered a}?
"voltage" of the tine. To do this we could let PB be a point on the line and let P be
the reference point (i.e., the point whose voltage is the datum or reference voitaage}
An attractive reference point would be the point at an infinite radios. Unfortunate:
iv. in this case Ra —~> co and from {3.48) we see that the voltage diverges This diffim
catty, sunllar to the one discussed in Section 3.2 in connection with ﬂux ﬁnkages, has
the same resolution; in the manywcooductor case, under reasonable assumptions, the
lmetodatum voltages are welt deﬁned, even if the datum point is at inﬁnity. Constder next the molticondactor case ilinstzated in Figure 3.8. Shown here
are cross sections of an inﬁnite line with n conductors, the charges per meter of these
condoctors, and a possible path of integration to evaluate the potential of P relative
to P1 With conductor radii smaﬂ compared with the spacing betwoen condtfctors we
can assame that the effect of the interaction between the charges on different oonv
doctors is negligible. In this case we can evaluate (3.47) using superposition We are
thus led to the muiticondtzctor version of (3.48): I _ 2 n Rm.
“so ‘“ “Pa " “Pa = 5;; qr ln “1;; {3.49} ‘ We next make a key assumption regarding the instantaneous charge densities
tn any cross section of the line. Assume that 91+92+"'+q,,==0 (350) which is analogous to (3.16}. Proceeding as in the development foilowjng {3. 26), we
then rewrite {3.49) in a way that faciiitates the introduction of (3.50}: v we ,.;~— ~— . .
B“ 2‘? “Rﬁth EatinRa. (3.51} Conductor q 1
1 ®
Pastime path of
4‘ integration ﬁsm 3.3 Conductor cross
of E sections. 78 _ Transmissionwune Parameters Chap. 3 To the second part of (3.51) add Z 1’!
— W in R 3.52
21,”, qr o1 { }
Because of (3.50) we have added zero. 0n the other hand, the second part of (3.51)
now becomes Rai 1 Pl
M .; 3.53
21“ q, rt R“: i ) Now we are in a position to pick point B, as reference and move it out toward ity. As we do this, alt the ratios RamXRG; mi» 1 and each iogarithmic term individualiy
tends toward zero. In the limit {3.53} becomes zero. We can use the notation of; to
indicate voltage at any point I; with respect to the reference. 113 w mm 2 q. in —I {3.54) We note that the formula holds for an arbitrary point it; as tong as it is not inside a
wire. R5, is the distance between the (otherwise) arbitrary point ES and the center
of the ith conductor. If the point ,‘3 is on the ith conductor surface, then Rm, the dis
tance between the surface of the ith conductor and its center, reduces to the radius. The main appiication of {3.54} is to relate transmissiondine voltages and
charges. For exampie, we find the voitage at a poirtt ori the surface of conductor 1
itt Figure 3.8. i 1 1 1 vrﬂ%(qtin}:+q2mE+‘+q,,li't R1”) {3.55)
where Ru is the radius of conductor 1, R}: the distance from the point to the center
of conductor 2, and so ort. Now while physicai considerations indicate titat the sur
face of conductor 1 is an equipotential surface, application of (3.55) gives slightiy dif
ferent resuits for different choices of points on the surface. This discrepancy is
expected in the light of the approximations made in the derivation. A more detaijed
investigation of the potential ﬁelds indicates that Rf} should be replaced by d0, the
spacing between centers of conductors. With small radii compared with intercon—
dactor spacings, the differences are insignificant. Thus we have
1 1 1 1
x mm — + —— + v  i + m .5
273,6(‘Irinra grin du 9a 111 d”) (3 6)
With obvious modiﬁcations the equation for the kth condttctor voltage is
1 1 1 1
z W. ._ + . . . + _ . . . m _
12k Ewe (qt In a,“ qt in rk + + q" 111 dk”) (3 57} Sec. 3.8 Line Bapacitenee _ 13 1a matrix notation v m Fl; {3.58}
where v is an nveetor with components 121 , v3, . . . v,i . q is an nvector with com
ponents q1, q3, . . . , q,” and F is an n X a matrix with typicai element ff}. m (HZtre) In {Help}. To ftnd the capacitance parameters, we need the inverse re
lattonshtp, q n {Iv {3.59}
where C = F". Note that, in general. there is mutual capacitance between conductors; that is, the charges depend on all the voitages. and vice versa. However, in an important
special case we can model with onty seifcapacitances Empie 3.8 {Salcuiate an expression for the capacitance per meter of a three—phase transmission
line [Figure ES.8{a}}. Assume that ‘i_\..
a» 4—3 to ﬁgure lasers; 1. Conductors are equaﬂy spaced,D, and have equal radii r.
Zqi+qt+qrs0 Solution Using a, b, c notation instead of 2.2.3, we find ea by using (3.56): n_1_ 1 i. 1 t5, 2m(qainr +qbin—5 + 93115)
2 9.: m2 (360)
2on5 r where we have used the assumption on + qt, + qr m D. In the same way we ﬁnd iden»
treat reiationships between 1);, and qb, and between at and qt. We are now ready to cal— culate the three capacitances To interpret the resuit physicaliv, however. we digress brieﬂy. Consider the potential of a point p equidistant from the conductors a,b, and
e. {332113954} gives as pa Rte Rpr Transmisstor'iLina Parameters Chap. ‘3 Thus the point p is at the same potential as the point at inﬁnity and may be taken as
the datum point. In fact. construct an imaginat'y geometric lane paraltel to the cons
doctors passing through the point P. Every point on thts lute ts at the datum poten‘
tial. Picking this also to be the neutra? potential. we get the relattonsths between
phase—ttnneutrai voltages and phase charges via three equal phase~nentraz {dtstnb uted} capacitances: 21H? — «—  —~ 3.61
came—ewcmlnwm ( ) Fftn to neutral arranged as shown in Figure 83.303). This symmetric circuit they also be taken to be
a representation of the physical cross section of the trattsnussmn lme. /\ c b Fauna 33.301) In the more general case of unequai interphase spacings and bundled subcon—
ductors as described in Section 3.4, we assume that the Lines are transposed and get _ 2m: “ a. z ' z m ~—~«~—— 362}
‘“ “" °’* “t " momma ‘ Fiat to neutrai where D," is the OMB between phases and
Rt: % {rdiz ' ' ‘ dulyb We note that the only difference between (3.63} and {3.28} is that ris used instead of
r’. If E: m 1, it is convenient to deﬁne R3 = r so that we can use {3.62} to cover the non bundled case. b a 2 (3.63) Empie 3.9 Find phaseneutral capacitance and capacitive reactantx per mile for a three~phase line with D,” m 35.3 ft. Conductor diameter 2 1.25 in. Solution In seine a are] m 1.0920 = 8.854 X 10'" ‘2. Using this vaiuein {3.52),we have 12
2cm 271* X 3354 X 10 u 853 X 10.41 Rim E ""M moms) ” in [(3513 x 12),r{1.25;2}} Sec. 3.9 Determination of Line Parameters Using Tables I 81 We next calculate the susceptance. Br 2 we : 27: x an x 8.53 x 19"” a 3.216 x 10"9121110/m
m 3.215 x 109 x 1609.34 3 5.175 >< 10"3 mhofml Finaﬂy, we have the phaseneutral reactance in 0mm; 1 2
m z—w = WWW—m z .9 me '
B. ammo'6 m 3 m‘ Exercisa 2. Comparing ! in (3.33} and E in (3.62}, we see that there is an ap
proximately inverse relationship Assume that b m 2, u, m 3., D,"be z 15, and
show that IE "~= 31.06“. We note that (pose)— W m 2.993 X 108 mlsec is the velocity of
light in a vacuum (to. a universal constant.) Before Zeaving the subject of line capacitance. it:
glecting the effect of the (conducting) earth under t
are induced in the earth, and these have some effect on the cammmw ......_
pacitance. To study this effect further, see Andersen (Chapter 5) and Grainger and
Stevenson (Chapter 5}. The effect is usually quite small for Lines of reasonable height
operating under normal noofault conditions 3.9 DETERMINATION OF LINE PMETEﬂS USING TABLES Values of GMR for the important case of stranded conductors without bundling
are available in tables provided by manufacturers The names given in Appendix
8 also provide values of inductive reactance, shunt capacitive reactance. and resis
tance. These tables continue to use units of inches. feet, and miles based on the
practice in the ciectﬁc utility industry in North America. In much of the analysis using iiztc parameters, we utilize the values of inductive
{cadence and shunt capacitive rcactance rather than the values of inductance and
shunt capacitance. Using (3.32) the per phase inductive reactance for a single solid conductor of
radius r is given by XL = ani m 41rf X 19'71n%ilfm D (3.64)
m 2.022 x 10*3 x fin:13: Qfmi where r’ m ire—“’4. In the case of stranded conductors we would use the GMR rather
than r' although, as pointed out on page TO, the use of 3" provides a good approxi
mation for the GMR. Using the GMR rather than r' we get 82 TransmissionLine Parameters Chap. :3
X m 2022 ><1{)'"3 >< fln D“ mm {3 65)
L " ‘ GMR ' Finding the GMR by using Table A3.1 in Appendix 8 we can solve for X L using
(3.65). There is another way to find X L directly from the tables in Appendix 8. Be
cause of a logarithmic property we can replace (3.65) as foiiows: 1 .
.. 4; ﬂ, ,3
XL~2.022><10 xfin GMR + 2.022x10 x111:me (Vim .m—v—V—Nw—“‘N'—‘J wpv—‘N
. Xd Xe If the units of GMR and D". are in feet, the ﬁrst term in (3.66} represents the in
ductive reactance per phase of a line having a distance of 1 f1 between the conduc
tors This term. X“, is referred to as the inductive reactance or 1 ft spacing. It is a
function of the frequency and the GMR. The second terrain (3.66) is catled the in
ductive reactance spacing factor X g. This term is independent of the type of con
doetor and is a function of the frequency and spacing between conductors The
values of X... are given in Tables ASJ and the values of Xd are given in Tabie A82
(both in Appendix 8). In a simiiar fashion, we can aiso obtain the shunt capacitive reactance to neu»
tral using tables Using (3.62) as it applies to the eonbundled case. we can caicuiate the shunt caﬁ
paeitive reactance to {tantra} as {366) \H... 1
X6. :— 5??? {km to neutrai {3.67)
where
w 211's
Dm
in ~—
r
As a result,
I Dm .
X6 = ? x 1379 >< 1961:; w?» ﬂml to neutral (3.68} Foiiowing the procedure applied in the case of inductive reactance, we can rewrite
(3.68) in the form r x J X: m x1379 x 106159} + x 2.799 x 196 111(1)...) Q—mi to neutral (3.69)
we “Y
X5 X3
For a line with one stranded conductor per phase, X; and X; can be found directly
from Table ABA and A83. respectively. See. 3.9 Determination of Line Parameters Using Tables 83 Exampk 3.10 21d 11:6 inductive reactance per phase in ohms per mile, and the capacitive reactaece
e1: ta to ohm—miles for a threephase line that has three equilaterain spaced con» doctors of ACSR D . I
is 60 HZ ove The conductors are 10 ft apart, and the operating frequency Soletioe We wit] solve this exampte using both the approximate expressions derived in the chapter for a singie soiid conductor ‘
\ 0 .
described in this section, f radlus r and the method 95ng the tables Dove GMR = 00313 ft, D e 10 ft, r = 0.4635 inches 0.4635 r 2 M“? m Inductive Reactertce Calculation From {3.65). D X; “ x x In ““‘““ l
W X X 2] w" ' From Table A8.1, we have for the Dove conductor X“ m 0.420 [mm
From Table A82 for a spacing of 10 ft. we have
X5 = 0.2?94 11/1112
From{3.65) XLEX + Xd=0420+02?94 '
. ‘ a . . # 0.6994$ {Una which is
the value obta: d ' ‘ ‘ ' ‘ ‘ very time to
dius r1 ne prevrousiy usmg the expresswri for a sartgle sohd conductor of ra— Shtmr Capacitive Reacmnce Calculation From (3.68},
1
X6 m — x1279 x 10“2n(&)
f r
_ L 10
f X 1.779 X log in # 0.1648 Mﬂ—mi to [[31183] From Table A8.1, we have for the Dove conductor
X; x 0.0965 Mﬂmi
From Tame A33, we have for a 10~ft spacing
X g e 0.0683 M04211 From(3.68) XcmX’+X;=00965t00683 ‘
A. _ a . . ===0.1648M0m, ' ‘
vaiue obtmned asmg the expression for a single solid conductor. t mush is the same 34 TransmissionLine Parameters Chap. 3 3.19 more. FWMETER VALUES in this section we provide some parameter valoes for highvoltage lines Consider the
three iines in Figure 3.9. Some typical data concerning the lines are given in Table
3.}. A few comments on the tabulated values foliow. 1. The spacing between the phases increases with the voltage rating. However, by
increasing the number of conductors per bundle, the inductance actually de
creases with voltage rating. 2. Line resistance increases with temperature and with frequency. In Tabie 3.]
are shown the tine resistance for do and 60 Hz. At dc the distribution of cur
rent across each conductor cross section is eniforrtt. Because of inductive effects,
the current distrihetion changes as the frequency is increased with higher cur
rent densities at the surface. Aithough the current averaged over the cross sec
tion remains the same, the PR tosses increase, and this may be viewed as the
reason for the increased R. Checking the tabie, we see that the resistance at 60 Hz is oniy 5% higher
than at dc. This is evidence that there is oniy a slight departtzre front the uni
form current distribution assumed in out caIcetation of inductance. 3. The resistance is relatively small compared with the inductive reactance. In
caiculating power ﬂows, vohages, and currents, the resistances have a minor ef~
fect compared with the inductive reactanees. In some caiculations they are ne~
glected. However, in other calculations they are of vitai importance. Bxarnpies
are the calcaiation of line thertnai limits considered in Chapter 4§and problems
of economic operation considered in Chapter 1}. ﬁgure 3.9 Typical higMoltaga transmission lines. Copyright ©
1977. Electric Power Research institute. EPFIiEMEBS. Synthetic
Electric Utility Systems for Evatuan‘ng Advances Technologies.
Reprinted with permission. 5 . 3r 1
at: ‘1 Summary 35 TﬂaLE 3.1 LENS: EJATA Line voltage (RV) 138 345 765
Conductors per phase (18411. S cin
Number of strands aluminamfsxiiet 8} sin 452i? :1”?
Diameter {in} 097? 1.165 1 4224'
Conductor GMR (ft) 0.8329 {10386 00479
Currentcarrying capacity per conductor (A) m 1010 1,258
Bundie GMRAfih (f1) 0.0329 0.2406 0 6916
Flat phase sparing (ft) 1?.5 26.0 450 phasekspacittg (£th 22.05 32.76 563K}
lnoactance {rum x to '} i3.02 9.83 8281 XL {Qfmj} {17?20 n‘sor. n ma Resistance {fl/mi dc 50°C 0 18 G 0
Resntanee {fl/rail 61} Hz, 50°C 0.1688 0.0563 3 Sarge meedance loading (MVA) Si} 415 4. for future reference we list surge impedance loading for the three iines This
Is a measure of hne power~handhng capabiﬁty considered in Chapter 4. 5. On each transmission tower in Figure 3.9 the reader may have noticed two tri»
angular structures above the horizontal cross member. These structures sup»
port "ground" wires, which are electricain connected to ground or to each tower
and thereby to ground. The purpose of these wires is to shieid the phase cons doctors in the event of lightning strikes and to provide a towimpedance path
In the event of a phasetogrottnd fault. In Chapter 4 we consider the modeling of transmission lines using the distrib~
eted parameters described in this chapter. 3.11 SUMMARY For a threephase ﬁne with transposition and handling, the average per phase in
ductanee is given by _ £0. ?a _ M Dr»
1' 2w2n Rh a 2 x 10 tax a Him (3.33) where for convenience we have dropped the i' notation D is the OMB
‘ I I . m between
phases lie. Dm u (arcade,ch VJ}. Rf, Is the GMR of each phase bundie given by Chap. 3 85 Transmissionlioe Parameters ‘ . . . 3.3.
_ _ d“, )ifb, whefe r" w , gxp ("pr/4} aﬁd dlJ IS the 61331108 be‘wegn “)3 doctors i and 1'. We are assuming that each phase has the same synnnetrical arrange»
ment of bundled conductors all of the same radius r. if there is no bundling, Rb w r .
If the GMR of the {stranded} conductors is given, it should be used in the formulas in place of r’. ‘ I
The formula for average capacitance to neutral 15 R3. a {Ed}: ' 3.5. _ 211's
'“ inlel’Rt) ‘ ' ‘ d been defined, and h re a w c m 1.96 a 8.854 X 10 “in air, [3,,1 has alrea y f I
:3: (H112  dip)“ is? the same as Rb except : is used instead of r . We are again as
suming a symmetrical arrangement of bundled subconductors all of the same radius . Without bundling R; w r.
r Equations {3.33} and (3.62) make the dependence of l and c on geometry quite clear. In this connection the approximately inverse relationship of I and c should be noted. ' A I I I
In this chapter we have also provided a description of the series impedance, in cluding the effect of ground return using the classical approach of Carson. An al
ternate approach to calculating both inductive reactance and capacitive shunt reactance using tables is also given. . c Ffrn to neutral (3.62) 3.6. 3.7. Hebe Regarding the Problem Stet With the proliferation of digital computers in power system analyses, the expressmri
for the calculation of line parameters can be directly and convenieth programmed.
To this effect, in the following problem section, there is a special desrgn exerclsc using
Marten to develop a program to calculate the line parameters This design exercise will continue through the various chapters as we consider dif
ferent topics Problems . at a oilHz sin le base wer line and an openwire telephone line are pan
3.1 iﬁgﬂeath other and of; samgltorizontal plane. The power line spacing is 5 ft;
the telephone wire spacing is 12 in. The nearest conductors of the two lines at? 20 f:
apart. The power line current is 190 A. Find the'magnitude of the induced loop
or “roundmtrtp” voltage per mile in the telephone line. flint: Followmg the approach
in (3.11),ﬁnd the ﬂux linkages of the telephone Wire pair due to the power line con
tents. Use Faraday‘s law to find the loop voltage. Assume that the telephone more
radius is negligible. I I
Repeat Problem 3.1 but asstune that the telephone line has been displazoed vertically by
it} it lie, the nearest wnductors of the two lines are now (292 + 10’) it apart]. 3.10. EEE§ Problems 87 Repeat Problem 3.2 but now assume a threephase power line with a ﬂat horizontal spacing of 5 ft between phases The power line currents are balanced threephase each
of magnitude 190 A. Hint: Use phaaors to add the ﬂux contributions Suppose that in Problem 3.1 the telephone wires are transposed every 1000 ft. Calcu
late the induced loop voltage in a lmile length of telephone wire  luatify the formula I =3 4 x 10‘" In (Dir? henrys per meter for the round~trip induc
tance of a singlephase line made up of round wires of radius r separated by a distance
D between centers For simplicity, assume that D >> r and make reasonable simpliﬁ~
cations Note that the inductance “per conductor" is the same as the per phase induc tance of a threepliase line. but the inductance is doubled because there are two
conductors. Reconsider Example 3.2 but assume that each of the conductors is hollow. Find an ex»
pression for the inductance fin this case. Reconsider Example 3.2but assume that each of the conductors is stranded with a cen«
tral wire of radius r surrounded by six wires each of the same radius Assume that
D >> r and justify the approximate formula D
{=2 X lﬂqlnm
R 3 where
R! “ [{dudrz' ' 'ditlt’qidzadzz' ‘ ‘dz‘rlm' ‘ '{dndiz' ' 'drrlmlm and where d,, $ 3" and d5 is the distance between centers of the an and jth strands. RE
is called the geometric mean radius {GMR} of the stranded conductor. Hint: Notice
the similarity with Figure 3.6 and proceed by analogy with the derivation of (3.27). The
only difference is that since the seven inductances 1,,12, . . . , I, are not all equaL we
need to use the more general rule for combining parallel inductors: namely,
I"M{*+I§1+ v+t;'1. ' Given an aluminum 32.6%de mil conductor composed of seven strands, each strand
with a diameter of 0.0867 in. and an outside diameter of 0.2601 in, ﬁnd 12,, the GMR, using the fonnula in Problem 3.? and compare with the manufacturer's figure of
0.00787 ft. Calculate the per phase inductance (per meter) of the 7646! line described in Section
3.10. Note the flat horizontal spacing of 45 it between phases Assume that four con~
doctors per bundle are placed at the corners of a square (with 28 in. on a side). Use the
values of GMR specified in place of 1". Repeat Problem 3.9, neglecting the stranding and using the speciﬁed (outside) diame
ter of 1.424 511.33 if each individual conductor were solid aluminum. What is the percent
error compared with the result in Problem 3.9? Calwlate the per phase inder reactance of the line in Example 3.5 in ohms per mile.
Repeat l’roblem 3.9 for the 345kV line in Section 3.10. Repeat Problem 3&2, neglecting the attending (Le, using r’ instead of the GMR}.
Calculate the per phase inductive reactance ofthe line in Problem 3.12 in ohms per mile. Calculate the phaseneutral capacitance per meter of the 7651“! line described in
Section 3.16. 3.16. 3.17. 3.1&
3.19. 3.2]. TransmisséonLine Parameters Chap. 3 Calculate the capacitive reactance of the fine it: l’rebtem 3.15 in megohm‘miles Him.
First calculate the susoeptance per mile and then take the reciprocal to ﬁnd 5X3. Calculate the phaseneutral: capacitance per meter of the 345~kV line described in
Sectictz 3.16. Calcuiate the capacitive reactance of the line in Problem 3.17 in megohmqniles Calculate the product of inductance and capacitance values found for the 765~kv Zine
in Problems 3.9 and 3.3.5 and compare with the value of Mg. Repeat Problem 3.29 {or the 345kV line using the results of Problems 3.12 and 3.17. Compute the phase impedance matrix 2.,“ for the tine described in Figure P321. Asm
sume that the line is 30 miles long. The conductors used are Grosbeak, and the pa~
rameters for this conductor are given in Table A83. Figure “.21 3.22. Compute the phase impedance matrix Zm for the line described in Figure P322. As« sume that the line is 40 miles long. The conductors used are Ostrich, and the parame
ters for this conductor are given in Table A8.1. Figure 93.22 Sec. [33.2 Design Exercise as 93.1 DESIGN EXERCISE Develop a MATLAB program to determine the per phase series impedance and the
shunt capacitance of a given transmission line configuration in an appropriate set of
units Assume that the program input will consist of the following:  Radius or diameter of the conductor i1} appropriate units 0 Bundling information if the conductor is bundled  Conductor configuration in terms of distances dab, d,“ and db“
' Conductor resistance per unit iength ...
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This note was uploaded on 10/21/2010 for the course ECE 501 taught by Professor H.b during the Spring '10 term at Yildiz Teknik Üniversitesi.
 Spring '10
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