• No se han encontrado resultados

Fuente; Eurononey, Agosto, 82 de cifras publicadas por el BIS,

Theprocedureforcomputingthephaseimpedancematrixfortwooverhead

parallellinesispresentedinSection4�1�8�Figure4�18showstwoconcentric

neutralparallellineseachwithaseparategroundedneutralconductor�

The process for computing the 6 × 6 phase impedance matrix follows

exactlythesameprocedureasfortheoverheadlines�Inthiscase,thereare

a total of 14 conductors (6 phase conductors, 6 equivalent concentric neu-tral conductors, and 2 grounded neuneu-tral conductors)� Applying Carson’s

equationswillresultina14×14primitiveimpedancematrix�Thismatrix

ispartitionedbetweenthesixthandseventhrowsandcolumns�TheKron

reductionisappliedtoformthefinal6×6phaseimpedancematrix�

Example 4.5

Two concentric neutral three-phase underground parallel lines are

showninFigure4�19�

Cables(bothlines):250kcmil,1/3neutral Extraneutral:4/0copper

Determinethe6×6phaseimpedancematrix�

Solution

FromAppendixBforthecables, Outsidediameter:dod=1�29in�

Neutralstrands:k=13#14copperstrands FromAppendixAfortheconductors,

250kcmilAl: GMRc=0�0171ft, rc=0�41Ω/mile, dc=0�567in�

#14Copper: GMRs=0�00208ft, rs=14�8722Ω/mile, ds=0�0641in�

4/0Copper: GMRn=0�1579ft, rn=0�303Ω/mile, dn=0�522in�

D1–2 D1–3 D2–3 D1–13

D2–13

D3–13 9

3 8

1 2 7

D4–5 D5–6

D4–6 D4–14

D5–14

D6–14

4 5 6

Dnm

10 11 12

14 13

FIGURE 4.18

Parallelconcentricneutralundergroundlines�

10 in.

4 in. 4 in. 2 in.

a b

4 in. 4 in.

c

5 in.

b a c

FIGURE 4.19

Parallelconcentricneutralthree-phaselines�

Theradiusofthecircletothecenterofthestrandsis

 R d d

b= od s = =

24

1 29 0 0641

24 0 0511

. . . ft

TheequivalentGMRoftheconcentricneutralstrandsiscomputedas

 GMReq=kGMR k Rs⋅ ⋅ bk−1 =130 00208 13 0 05111. ⋅ ⋅ . 12 =0 0486. ft ThepositionsofthesixcablesandextraneutralusingCartesiancoordi-nateswiththephaseacableinline1(topline)astheordinate�Notethe

phasinginbothlines�

Phasea,

line1: d1=0+j0 Phaseb,

line1: d2 4 j 12 0

= + Phasec,

line1: d3 8 j 12 0

= +

Phasea,

line2: d4 4 j 12

10

= 12 Phaseb,

line1: d5 0 j10

= − 12 Phasec,

line1: d6 8 j 12

10

= 12 Equivalent neutrals

Phasea,

line1: d7=d1+jRb Phaseb,

line1: d8=d2+jRb Phasec,

line1: d9=d3+jRb Phasea,

line2: d10=d4+jRb Phaseb,

line2: d11=d5+jRb Phasec,

line2: d12=d6+jRb Extra neutral

 d13 10 j

12 5

= 12

Thespacingmatrixdefiningthedistancesbetweenconductorscanbe

computedby



i j

Di j di dj

= =

=

1 13 1–13

,

Thediagonaltermsofthespacingmatrixaredefinedastheappropriate

GMR:

 D1 1, =D2 2, =D3 3, =D4 4, =D5 5, =D6 6, =GMRc=0 0171. ft

 D7 7, =D8 8, =D9 9, =D10 10, =D11 11, =D12 12, =GMReq=0 0486. ft

 D13 13, =GMRn=0 01579. ft

Theresistancematrixisdefinedas

7 93402

Once the primitive impedance matrix is developed, it is partitioned

4.3 Summary

This chapter is devoted to presenting methods for computing the phase

impedancesandsequenceimpedancesofoverheadlinesandunderground

cables�Carson’sequationshavebeenmodifiedinordertosimplifythecom- putationofthephaseimpedances�WhenusingthemodifiedCarson’sequa-tions,thereisnoneedtomakeanyassumptions,suchastranspositionof

thelines�Byassuminganuntransposedlineandincludingtheactualphas-ingoftheline,themostaccuratevaluesofthephaseimpedances,selfand

mutual,aredetermined�Itishighlyrecommendedthatnoassumptionsbe

made in the computation of the impedances� Since voltage drop is a pri-maryconcernonadistributionline,theimpedancesusedforthelinemust

beasaccurateaspossible�Thischapteralsoincludedtheprocessofapply-ingCarson’sequationstotwodistributionlinesthatarephysicallyparallel�

Thissameapproachwouldbetakenwhentherearemorethantwolines

physicallyparallel�

Problems

4.1 The configuration and conductors of a three-phase overhead line are

showninFigure4�20�

Phaseconductors:556,50026/7ACSR Neutralconductor:4/0ACSR

FIGURE 4.20

Three-phaseconfigurationfor

Problem4�1�

4.0 ft

2.5 ft 4.5 ft

3.0 ft

a c b

n

25.0 ft

Determine

 a� Thephaseimpedancematrix[zabc]inΩ/mile

 b� Thesequenceimpedancematrix[z012]inΩ/mile

 c� Theneutraltransformationmatrix[tn]

4.2 Determinethephaseimpedance[zabc]matrixinΩ/mileforthetwo-phase

configurationinFigure4�21�

Phaseconductors:336,40026/7ACSR Neutralconductor:4/06/1ACSR

4.3 Determinethephaseimpedance[zabc]matrixin Ω/mileforthesingle-phaseconfigurationshowninFigure4�22�

Phaseandneutralconductors:1/06/1ACSR

4.0 ft 3.0 ft

a c

n

25.0 ft 7.0 ft

FIGURE 4.21

Two-phaseconfigurationforProblem4�2�

5.0 ft

n

25.0 ft b

0.5 ft

FIGURE 4.22

Single-phase pole configuration for

Problem4�3�

4.4 CreatethespacingsandconfigurationsofProblem4�1through4�3in

theWindmilprogram�Comparethephaseimpedancematricestothose

computedinthepreviousproblems�

4.5 Determinethephaseimpedancematrix[zabc]andsequenceimpedance

matrix[z012]inΩ/mileforthethree-phasepoleconfigurationinFigure

4�23�Thephaseandneutralconductorsare250,000AA�

4.6 Compute the positive, negative, and zero sequence impedances in

Ω/1000ftusingtheGMDmethodforthepoleconfigurationshownin

Figure4�23�

4.7 Determine the [zabc] and [z012] matrices inΩ/mileforthethree-phase

configurationshowninFigure4�24�Thephaseconductorsare350,000

AAandtheneutralconductoris250,000AA�

4.8 Compute the positive, negative, and zero sequence impedances in

Ω/1000ftforthelineofFigure4�24usingtheaverageself-andmutual

impedancesdefinedinEquations4�70and4�71�

4.9 A 4/0 aluminum concentric neutral cable is to be used for a single-phaselateral�Thecablehasafullneutral(seeAppendixB)�Determine

theimpedanceofthecableandtheresultingphaseimpedancematrix

inΩ/mileassumingthecableisconnectedtophaseb�

4.10 Three250,000CMaluminumconcentriccableswithone-thirdneutrals

are buried in a trench in a horizontal configuration (see Figure 4�14)�

Determinethe[zabc]and[z012]matricesinΩ/1000ftassumingphasing

ofc-a-b�

FIGURE 4.23

Three-phasepoleconfigurationforProblem4�5�

2 ft 2 ft 2 ft

25 ft 4 ft

a b

c n

4.11 Create the spacings and configurations of Problems 4�9 and 4�10 in

Windmil�Comparethevaluesofthephaseimpedancematricestothose

computedinthepreviousproblems�Inordertocheckthephaseimped-ancematrix,itwillbenecessaryforyoutoconnectthelinetobalanced

three-phasesource�Asourceof12�47kVworksfine�

4.12 Asingle-phaseundergroundlineiscomposedofa350,000CMalumi- numtape-shieldedcable�A4/0copperconductorisusedastheneu-tral�Thecableandneutralareseparatedby4in�Determinethephase

impedancematrixinΩ/mileforthissingle-phasecablelineassuming

phasec�

4.13 Three one-third neutral 2/0 aluminum jacketed concentric neutral

cablesareinstalledina6in�conduit�Assumethecablejackethasathickness

of0�2in�andthecableslieinatriangularconfigurationinsidetheconduit�

ComputethephaseimpedancematrixinΩ/mileforthiscabledline�

4.14 Create the spacing and configuration of Problem 4�13 in Windmil�

Connecta12�47kVsourcetothelineandcompareresultstothoseof

4�13�

4.15 Twothree-phasedistributionlinesarephysicallyparallelasshownin

Figure4�25�

 Line#1(leftside) Phaseconductors=266,80026/7ACSR

  Neutralconductor=3/06/1ACSR

 Line#2(rightside) Phaseconductors=300,000CONLAYaluminum

  Neutralconductor=4/0CLASSAaluminum

a b

c 2 ft

2 ft n

2 ft

2 ft

25 ft

FIGURE 4.24

Three-phasepoleconfigurationforProblem4�5�

 a�Determinethe6×6phaseimpedancematrix�

 b�Determinetheneutraltransformmatrix�

4.16 Twoconcentricneutralundergroundthree-phaselinesarephysically

parallelasshowninFigure4�26�

Line#1(top) Cable=250kcmil,1/3neutral

  Additionalneutral:4/06/1ACSR

 Line#2(bottom) Cable=2/0kcmil,1/3neutral

  Additionalneutral:2/0ACSR

 a�Determinethe6×6phaseimpedancematrix�

 b�Determinetheneutraltransformmatrix�

n

18 ft

a b c

n

c a

2.5 ft 4.5 ft b 3.0 ft

4.0 ft

25.0 ft

2.5 ft 4.5 ft 3.0 ft

4.0 ft

25.0 ft

FIGURE 4.25 ParallelOHlines�

24 in.

6 in. 6 in. 4 in.

a b

6 in. 6 in. 4 in.

c a

c

b

FIGURE 4.26

Parallelconcentricneutralthree-phaselines�

Windmil Assignment

FollowthemethodoutlinedintheUser’sManualtobuildasystemcalled

“System1”inWindmilthatwillhavethefollowingcomponents:

• 12�47kVline-to-linesource�The“BusVoltage”shouldbesetto120V�

• Connecttothenodeandcallitnode1�

• A5,000ftlongoverheadthreedistributionlineasdefinedinProblem

4�1�CallthislineOH-1�

• Connectanodetotheendofthelineandcallitnode2�

• Awyeconnectedunbalancedthree-phaseloadisconnectedtonode

2andismodeledasconstantPQloadwithvaluesof

• Phasea–g:1000kW,powerfactor=90%lagging

• Phaseb–g:800kW,powerfactor=85%lagging

• Phasec–g:1200kW,powerfactor=95%lagging

Determinethevoltagesona120Vbaseatnode2andthecurrentflowingon

theOH-1line�

References

 1� Glover,J�D�andSarma,M�,Power System Analysis and Design,2ndedn�,PWS

PublishingCo�,Boston,MA,1994�

 2� Carson, J�R�, Wave propagation in overhead wires with ground return, Bell System Technical Journal,5,539,1926�

 3� Kersting, W�H� and Green, R�K�, Application of Carson’s equations to the

steady-stateanalysisofdistributionfeeders,IEEE Power System Conference and Exposition,Phoenix,AZ,March2011�

 4� Kron,G�,Tensorialanalysisofintegratedtransmissionsystems,partI,thesix

basicreferenceframes,AIEE Transactions,71,1952�

119

5

Shunt Admittance of Overhead