bergen 3 - 52 Basic Principtas Chap. 2 figure P24 5 The...

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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 lumped-circuit 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 transmission-line 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 three-phase 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 thH-dlrie (3.2} l. where F m magnetomotive force (mmf), ampere-turns H m magnetic field intensity, ampere-turnsfmeter 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 right-hand 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, safe-a 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 _ Transmission-Line 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 Sta-sight 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 “one-turn 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 low-frequency 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 , Transmission-Line 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 5-1-32»; (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 {3-10) 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; MANY-cflfllmk 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+1-QIBR2+' ' '+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 Ina-W" (3.29) Thus , 1 5+1n+=meef‘+1nim2n3~emmln% (329) 4 r! r1 r1 3*, where r; g new" Then (3.18) becomes .u-o . 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 self-inductances. 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 diZdlfidl-iyjm 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 cross-sectional 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 (ii-j 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 Transmission-tine 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: {3-33} — ,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 sol-i. 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 transmission-tine 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 self-impedance 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 Transmission-Line 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 0-11}. 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: (3-49) Vc Zia :23 Zia“ Ir where I , Zi'n‘Zm' . Z“. x Z“. » ; r w a,b,c {3-41) 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 Transmission-Una 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 m-Umafig x 1.609 x 103mm] = 0.2884 + J£1,392? mm; The mutual impedanccs are computed as foilows: . _.7 D: zabzrd-t- ;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 Seen-it: 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 _ Transmission-Lina 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 lme-to-datum 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 Ram-XRG; 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 n-veetor with components 121 , v3, . . . v,i . q is an n-vector with com- ponents q1, q3, . . . , q,” and F is an n X a matrix with typicai element ff}. m (HZt-re) 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 seif-capacitances 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. Z-qi+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'i-Lina 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 Faun-a 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 ' ' ‘ duly-b 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 phase-neutral 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 10-9 x 1609.34 3 5.175 >< 10"3 mhofml Finafly, we have the phase-neutral 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 Transmission-Line 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 {3-66) \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?» fl-ml 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 three-phase 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 Mfl-mi From Tame A33, we have for a 10~ft spacing X g e 0.0683 M04211 From(3.68) XcmX’+X;=00965-t00683 ‘ A. _ a . . ===0.1648M0-m, ' ‘ vaiue obtmned asmg the expression for a single solid conductor. t mush is the same 34 Transmission-Line Parameters Chap. 3 3.19 more. FWMETER VALUES in this section we provide some parameter valoes for high-voltage 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. EPFIi-EM-EBS. 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 0-0479 Current-carrying 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 45-0 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 tow-impedance path In the event of a phase-to-grottnd 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 three-phase 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 dl-J- 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 oil-Hz sin le- base wer line and an open-wire 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 three-phase power line with a flat horizontal spacing of 5 ft between phases The power line currents are balanced three-phase 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 l-mile 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 single-phase 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 345-kV 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 phase-neutral capacitance per meter of the 765-1“! line described in Section 3.16. 3.16. 3.17. 3.1& 3.19. 3.2]. Transmisséon-Line 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 phase-neutral: 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 345-kV 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.

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bergen 3 - 52 Basic Principtas Chap. 2 figure P24 5 The...

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