The Velocity Factor of an Insulated Two-Wire Transmission Line

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1 The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ Mrch 7, 008 Estimte the velocity fctor F = v/c nd the impednce Z of two-wire trnsmission line mde of cylindricl conductors of rdius whose centers re seprted by distnce d, when ech wire is insulted by lyer of reltive dielectric constnt ɛ of thickness t, sshownin the figure below. The thickness t is of the sme order s rdius, but + t d. The spce outside the insulted wires hs unit reltive dielectric constnt. All medi in this problem hve unit reltive mgnetic permebility. Solution. Velocity Fctor The propgtion speed v of wves on trnsmission line is v = LC, where L nd C re the inductnce nd cpcitnce per unit length of the two-line system. If there is no insultion on the wires, the propgtion speed is the speed of light c. Thus, c = L0 C 0, where L 0 nd C 0 re the inductnce nd cpcitnce per unit length of the two-line system without insultion on the wires. Since the inductnce is unffected by the presence of See, for exmple, sec. of See, for exmple, pp. 6-7 of

2 insultion ssumed to hve unit mgnetic permebility, the velocity fctor of the insultedwire trnsmission line cn be written F = v c = C0 C. 3 The pproximtion employed here is tht for d + t the potentil difference between the two wires cn be clculted s twice the potentil between rdii nd d when only single wire is present. In this pproximtion the electric field E outside n uninsulted wire is in Gussin units E = Q r, 4 where Q is the electric chrge per unit length on the wire. Then, the potentil difference between rdii nd d is Q lnd/, so the cpcitnce C 0 of n uninsulted two-wire line is C 0 = Q V. 5 4ln d To convert to SI units, replce the fctor /4 by πɛ 0 =7.8 pf/m. An exct expression for C 0 is 3 C 0 = 4ln d+ d 4 4ln d d + 4ln d which shows tht the pproximtion 5 is rther good when d. Similrly, the electric displcement D outside n insulted wire is, 6 d ln d/ D = Q r, 7 nd hence the electric field is E <r<+ t = Q ɛr, 8 Er >+ t = Q r, 9 where the lyer of dielectric constnt ɛ extends from rdius to +t. The potentil difference between rdii nd d outside single such wire is Q ɛ ln + t +Qln d + t. 0 From this we estimte the cpcitnce C to be C 4 +t ln +4ln, d ɛ +t 3 See, for exmple, prob. of

3 nd therefore the velocity fctor 3 is F ɛ ln +t ln d +ln d +t An Excel spredsheet tht implements eq. is vilble t Impednce, The chrcteristic impednce Z of trnsmission line is relted to its inductnce nd cpcitnce per unit length ccording to L Z = C. 3 The inductnce L per unit length of trnsmission line tht is surrounded by medi of unit mgnetic permebility cn be relted to the cpcitnce C 0 for this cse by eq.. Thus, Z = c C 0 C. 4 In prticulr, the impednce Z 0 of n uninsulted two-wire trnsmission line is Z 0 = = 4 cc 0 c ln d + d 4 4 c ln d = 0 ln d Ω, 5 reclling tht /c = 30 Ω. Then, using eq., the impednce of n insulted two-wire trnsmission line is Z 0 ln d ɛ ln + t +ln d Ω. 6 + t Furthermore, eliminting C in eq. 4 by use of expression 3 for the velocity fctor, we find Z = F = FZ 0. 7 cc 0.3 Estimte of the Cpcitnce by n Energy Method An lterntive computtion of the cpcitnce per unit length C cn be bsed on the reltion for stored electrosttic energy U per unit length when chrge ±Q per unit length is plced on the two wires, U = Q ED D D D C = 8π dare = 8πɛ dare = 8π dare + ɛ 8π dare = Q +t 8π ɛ 4Q r [ 4 ɛ ln + t +4ln d ] + t Q + C 0 πr dr Q [4ln d +4 ɛ ln + t. 8 3 ]

4 This yields the sme estimte for C s eq.. We cn lso use the energy method to find the next correction to eq. for the cpcitnce. For this we note tht the electric displcement D when dielectric is present is the sme s the electric field E if the dielectric constnt were unity. For the ltter cse, the corresponding electric potentil V cn be determined from n pproprite complex logrithmic function, 4 V = Q ln x c + y x + c + y, 9 in coordinte system x,y whose origin is midwy between the two conductors, s sketched below. The conductors re centered t x = ±b = ±d/, nd the prmeter c is given by c = b b b = b d. 0 For use in energy expression 8, we wish to estimte the integrl +t π ɛ dr rdφ D +t π 0 8π = ɛ dr rdφ E 0 8π, in coordinte system centered on the righthnd wire. So, we mke the chnge of vribles x = x + b, leding to V = Q ln x + b c + y x + b + c + y. The electric field E then hs components E,x = V x + b c x + b + c = Q Q x x + b c + y x + b + c + y x + /d Q r + x/d x + d x + /d Q d +dx r + x/d d x, 3 d E,y = V y = Q y x + b c + y Q y x + b + c + y y Q r + x/d y, 4 d 4 See, for exmple, pp. 4-6 of 4

5 where r = x + y. Then, x E 4Q + x/d + y x + /dd x+y r + x/d d r + x/d 4Q x r + x/d d y d +x d 4Q r x d x r d d + 4 x y r d +x d. 5 The integrl over φ in eq. leds to the cnceltion of ll terms in eq. 5 except the first, using x = r cos φ nd y = r sin φ. Hence, we find no corrections to the cpcitnce t either order /d or /d. Tht we find no corrections t these orders is surprising in view of numericl computtions for the cse = t = d/ in which the insulted wires touch, 5 indicting cpcitnce bout twice tht predicted by eq...4 Estimte of the Cpcitnce Supposing the Insultion Follows Equipotentils The equipotentils of the potentil 9 re circles of the form x c coth V + y = c csch V Q Q, 6 nd the corresponding fields lines re lso circles, chrcterized by prmeter W ccording to x + y + c cot W = c csc W Q Q, 7 s sketched in the figure below. 5 See, for exmple, J.C. Clements, C.R. Pul nd A.T. Adms, Computtion of the Cpcitnce Mtrix for Systems of Dielectric-Coted Cylinders, IEEE Trns. Elec. Compt. 7, , 5

6 If the region inside n equipotentil were filled with dielectric of permittivity ɛ, then the bove forms describe the displcementfield D rther thn the electric field E. In generl, one cnnot write D = V, becuse D is nonzero t the interfce between regions of differing permittivity. However, if D nd E re everywhere perpendiculr to such interfces, then D = 0 everywhere, nd the displcement field cn be deduced from sclr potentil. This provides nother method of estimting the cpcitnce of n insultted pir of wires. We suppose tht the surfce of the insultion is circulr, but the center of this circle is displced from the center of the wire, such tht the surfce of the insultion is on n equipotentil. The equipotentil 6 intercepts the positive x-xis t x = c coth V x Q ± c csch V x Q, i.e., x = c tnh V x 4Q nd c coth V x 4Q, 8 where V x is the potentil t x when ɛ = everywhere. In prticulr, if x = b is on the surfce of wire, the potentil of tht wire is V wire =4Q tnh b c =Q ln b + c = V 0 = Q C 0, 9 where V 0 =V wire is the voltge difference between the wires, so the cpcitnce of the bre wires is C 0 = =, 30 4ln b+c 4ln d+ d 4 s previously stted in eq. 6. If insted the wires re surrounded by circulr cylinders of insultion of permittivity ɛ tht extend from x = c tnh V x /4Q to c coth V x /4Q, then the electric field for x <= c tnh V x /4Q is the sme s for bre wires, but the field for c tnh V x /4Q < x <b is smller thn for bre wires by fctor /ɛ. Hence, the voltge difference between the wires is now V = V x + V 0/ V x ɛ = V 0 ɛ +V x = Q +4Q ɛ ln c + x ɛ ɛ c x, 3 nd the cpcitnce is now C = +4ɛ C 0 ln c+x. 3 c x The ctul wires hve concentric lyer of insultion of thickness t. Setting x = b t in eq. 3 corresponds to incresing the mount of insultion until if fills the equipotentil tht psses through this point, which overestimtes the cpcitnce: C< +4ɛ C 0 ln c+b t. 33 c b++t On the other hnd, we could remove insultion from the wires until it fills the equipotentil tht psses through point x = b + + t, in which cse eq. 3 would underestimte the 6

7 cpcitnce. Reclling eq. 8, the surfce of this insultion lso psses through the point x = c /b + + t, so we hve tht C> ɛ C 0 ln b++t+c b++t c We might then tke re our estimte to be kind of verge of eqs. 33 nd 34, such s C = +ɛ C 0 ln c+b t c b++t +lnb++t+c b++t c +ɛ C 0 ln c+b tb++t+c c b++tb++t c. 35 The pproximtion 35 ppers to be significntly better thn tht of eq. for the cse tht b = + t. 7

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IMPORTANT. Read these directions carefully: Physics 208: Electricity nd Mgnetism Finl Exm, Secs. 506 510. 7 My. 2004 Instructor: Dr. George R. Welch, 415 Engineering-Physics, 845-7737 Print your nme netly: Lst nme: First nme: Sign your nme: Plese