Electrically Parallel Lines

Figure6�10showstwodistributionlinesthatareelectricallyparallel�

Theanalysisoftheelectricallyparallellinesrequiresanextrastepfrom

thatofthephysicallyparallellinessincetheindividuallinecurrentsarenot

known�Inthiscaseonlythetotalcurrentleavingtheparallellinesisknown�

In the typical analysis, the receiving end voltages will have been either

assumedorcomputedandthetotalphasecurrents[IR]willbeknown�With

FIGURE 6.10

Electricallyparallellines�

Thecurrentinline2asafunctionofthetotalcurrentandthecurrentin line 1

Example 6.8

The two lines of Example 6�5 are electrically parallel as shown in

Figure 6�9�Thereceivingendvoltagesaregivenby

kVA at 990PF

Thefirststepinthesolutionistodeterminethetotalcurrentleavingthe

Thecurrentinline2is

When the shunt admittance of the parallel lines is ignored, a parallel

equivalent3×3phaseimpedancematrixcanbedetermined�Sincevery

ExpandEquation6�83tosolveforthevoltagedrops:



vabc Z IR Z IR IR Z IR Z IR

[ ]

⁼

[ ]

^{11 ⋅}

[ ]

^{1 +}

[ ]

^{12 ⋅}

( [ ]

⁻

[ ]

¹

)

⁼

^{[ ]}

^{21 ⋅}

^{[ ]}

¹⁺

^{[ ]}

^{22 ⋅}

( ^{[[ ]}

⁻

^{[ ]} )

[ ]

⁼

( [ ]

⁻

[ ] )

^⋅

[ ]

⁺

[ ]

^⋅

[ ]

⁼

[ ]

⁻

IR

vabc Z Z IR Z IR Z Z

1

11 12 1 12

(

21

[[ ]

22

)

^⋅

[ ]

^IR1+

[ ]

^{Z22 ⋅}

[ ]

^IR _

(6�84) CollecttermsinEquation6�84:



( [ ]

Z11 ⁻

[ ]

Z12 ⁻

[ ]

Z21 ⁺

[ ]

Z22

)

^⋅

^{[ ]}

IR1 ⁼

( ^{[ ]}

Z22 ⁻

^{[ ]}

Z12

)

^⋅

^{[ ]}

IR ^{ (6�85)} Let



[ ]

ZX ⁼

( [ ]

Z^{11 −}

[ ]

Z^{12 −}

[ ]

Z^{21 +}

[ ]

Z²²

)

^{ (6�86)}

SubstituteEquation6�86intoEquation6�85andsolveforthecurrentin

line1:



[ ]

IR1 ⁼

[ ]

ZX^−1⋅

( [ ]

Z22 ⁻

[ ]

Z12

)

^⋅

[ ]

IR ^{ (6�87)} SubstituteEquation6�87intothetoplineofEquation6�83:



[ ]

vabc ⁼

( ( [

Z¹¹⁻Z¹²

] )

^⋅

[ ]

ZX^−1⋅

( [ ]

Z^{22 −}

[ ]

Z¹²

)

⁺

[ ]

Z¹²

)

^⋅

^{[ ]}

IR ^{ (6�88)}

 v Z IR

Z Z Z ZX Z Z

abc eq

eq

[ ]

_{=  ⋅}

[ ]

  =

( [

−

] )

^⋅

^{[ ]}

^{− ⋅}

^{[ ]}

⁻

where 11 12 1

22

[ ]

112 12

( )

⁺

^{[ ]}

(

^Z

)

_ ^(6�89)

TheequivalentimpedanceofEquation6�89isthe3×3equivalentfor

thetwolinesthatareelectricallyparallel�Thisisthephaseimpedance

matrix that can be used in conventional distribution power flow pro-gramsthatcannotmodelelectricallyparallellines�

Example 6.9

Thesametwolinesareelectricallyparallelbuttheshuntadmittanceis

neglected� Compute the equivalent 3 × 3 impedance matrix using the

impedancepartitionedmatricesofExample6�5:



ZX Z Z Z Z

j j

[ ]

⁼

[ ]

⁻

[ ]

⁻

[ ]

⁺

[ ]

=

+ +

11 12 21 22

7 1697. 12 2446. 0 0039. 0 3041. 00 0032 0 7046 0 0039 0 3041 7 1616 13 3077 0 0013 1 9361

. .

. . . . . .

−

+ + −

j

j j j

00 0032. − 0 7046. 0 0013. − 1 9361. 7 1610. + 14 2577.















j j j 



6.6 Summary

Thischapterhasdevelopedthe“exact,”“modified,”and“approximate”line

however, there are situations where the shunt admittances should not be

neglected�Thisisparticularlytrueforlongrurallightlyloadedlinesandfor

manyundergroundlines�

Amethodforcomputingthecurrentflowingintheneutralandground

was developed� The only assumption used that can make a difference in

the computing currents is that the resistivity of earth was assumed to be

100Ω-m�

A simple version of the ladder iterative technique was introduced and

appliedinExample6�5�Theladdermethodisusedinfuturechaptersandis

fullydevelopedinChapter10�

Thegeneralizedmatricesfortwolinesinparallelhavebeenderived�The

analysisofphysicallyparallelandelectricallyparallellineswasdeveloped

withexamplestodemonstratetheanalysisprocess�

Problems

6.1 A2milelongthree-phaselineusestheconfigurationofProblem4�1�

The phase impedance matrix and shunt admittance matrix for the

configurationare

 a� Determinethegeneralizedmatrices�

 b� Forthegivenload,computetheline-to-lineandline-to-neutralvolt-agesatthesourceendoftheline�

 c� Computethevoltageunbalanceatthesourceend�

 d� Computethesourceendcomplexpowerperphase�

 e� Computethepowerlossbyphaseovertheline�(Hint:Powerlossis

definedaspowerinminuspowerout�)

6.2 UsethelineofProblem6�1�Forthisproblem,thesourcevoltagesare

specifiedas 7620 120

7620 120 /

 a� Theloadline-to-neutralvoltages

 b� Theloadcurrents

 c� Thecomplexpoweratthesource

 d� Thevoltageunbalanceattheload

6.3 UseWindmilforProblem6�2�

6.4 ThepositiveandzerosequenceimpedancesforthelineofProblem 6�1are

 z+=0.186+j0.5968 /mileΩ , z0=0.6534+j1 9 7 /mile. 0 Ω Repeatproblem6�1usingthe“approximate”linemodel�

6.5 ThelineofProblem6�1servesanunbalancedgroundedwyeconnected

constantimpedanceloadof

 Zag=15/3 0Ω, Zbg=17/36 87 . Ω, Zcg=2 /25 84 0 . Ω

The line is connected to a balanced three-phase 13�2kV source�

Determine

 a� Theloadcurrents

 b� Theloadline-to-groundvoltages

 c� Thecomplexpoweroftheloadbyphase

 d� Thesourcecomplexpowerbyphase

 e� Thepowerlossbyphaseandthetotalthree-phasepowerloss

 f� Thecurrentflowingintheneutralandground

6.6 RepeatProblem6�3withonlytheloadonphasebchangedto50/36�87Ω�

6.7 Thetwo-phaselineofProblem4�2hasthefollowingphaseimpedance

matrix:



z

j j

j

[ ]

abc ^{= + +}

+

0 4576 1 0780 0 0 1535 0 3849

0 0 0

0 1535 0 3849 0 0 46

. . . .

. . . 115+ 1 0651













 j . 

Ω/mile

Thelineis2mileslongandservesatwo-phaseloadsuchthat Sag=2000kVAat0�9laggingpowerfactorandvoltageof7620/0V Scg=1500kVAat0�95laggingpowerfactorandvoltageof7620/120V Neglecttheshuntadmittanceanddeterminethefollowing:

 a� Thesourceline-to-groundvoltagesusingthegeneralizedmatrices

(Hint:Eventhoughphasebisphysicallynotpresent,assumethatit

iswithavalueof7620/−120Vandisservinga0kVAload�)

 b� Thecomplexpowerbyphaseatthesource

 c� Thepowerlossbyphaseontheline

 d� Thecurrentflowingintheneutralandground

6.8 6.9 Thethree-phaseconcentricneutralcableconfigurationofProblem4�10

is2mileslongandservesabalancedthree-phaseloadof10,000kVA,

13�2kV, 0�85 lagging power factor� The phase impedance and shunt

admittancematricesforthecablelineare

 a� Determinethegeneralizedmatrices�

 b� Forthegivenload,computetheline-to-lineandline-to-neutralvolt-agesatthesourceendoftheline�

 c� Computethevoltageunbalanceatthesourceend�

 d� Computethesourceendcomplexpowerperphase�

 e� Computethepowerlossbyphaseovertheline�(Hint:Powerlossis

definedaspowerinminuspowerout�)

6.10 ThelineofProblem6�9servesanunbalancedgroundedwyeconnected

constantimpedanceloadof

 Zag=15/3 0Ω, Zbg=5 /36 87 0 . Ω, Zcg=2 /25 84 0 . Ω

Thelineisconnectedtoabalancedthree-phase13�2kVsource�Determine

 a� Theloadcurrents

 b� Theloadline-to-groundvoltages

 c� Thecomplexpoweroftheloadbyphase

 d� Thesourcecomplexpowerbyphase

 e� Thepowerlossbyphaseandthetotalthree-phasepowerloss

 f� Thecurrentflowingineachneutralandground

6.11 Thetape-shieldedcablesingle-phaselineofProblem4�12is2mileslong

andservesasingle-phaseloadof3000kVA,at8�0kVand0�9laggingpower

factor�Thephaseimpedanceandshuntadmittancesforthelineare



z

j

[ ]

abc ⁼

+

















0 0 0

0 0 0

0 0 0 5291. 0 5685.

Ω/mile



y

j

  =abc

















0 0 0

0 0 0

0 0 140 39.

µS/mile

Determine the source voltage and the power loss for the loading

condition�

6.12 TwodistributionlinesconstructedononepoleareshowninFigure6�11�

Line#1data:

Conductors:336,40026/7ACSR

GMR=0�0244ft,Resistance=0�306Ω/mile,Diameter=0�721in�

Line#2data:

Conductors:250,000AA

GMR=0�0171ft,Resistance=0�41Ω/mile,Diameter=0�574in�

Neutralconductordata:

Conductor:4/06/1ACSR

GMR=0�00814ft,Resistance=0�592Ω/mile,Diameter=0�563in�

Lengthoflinesis10miles�

FIGURE 6.11

Twoparallellinesononepole�

6.0 ft

2.5 ft 4.5 ft

3.0 ft

b c

n

30.0 ft a

2.0 ft

b

a c

Line 1 Line 2

Balancedloadvoltagesof24�9kVlinetoline Unbalancedloading:

Load#1:Phasea:1440kVAat0�95laggingpowerfactor Phaseb:1150kVAat0�9laggingpowerfactor Phasec:1720kVAat0�85laggingpowerfactor Load#2:Phasea:1300kVAat0�9laggingpowerfactor

Phaseb:1720kVAat0�85laggingpowerfactor Phasec:1370kVAat0�95laggingpowerfactor

Thetwolineshaveacommonsendingendnode(Figure6�6)�Determine

 a� Thetotalphaseimpedancematrix(6×6)

 b� Thea,b,c,dandA,Bmatrices

 c� The sending end node voltages and currents for each line for the

specifiedloads

 d� Thesendingendcomplexpowerforeachline

 e� Therealpowerlossofeachline

 f� Thecurrentflowingintheneutralconductorandground

6.13 ThelinesofProblem6�12donotshareacommonsendingorreceiving

endnode(Figure6�7)�Determine

 a� The sending end node voltages and currents for each line for the

specifiedloads

 b� Thesendingendcomplexpowerforeachline

 c� Therealpowerlossofeachline

6.14 ThelinesofProblem6�12areelectricallyparallel(Figure6�8)�

Computetheequivalent3×3impedancematrixanddetermine

 a� The sending end node voltages and currents for each line for the

specifiedloads

 b� Thesendingendcomplexpowerforeachline

 c� Therealpowerlossofeachline

Windmil Assignment

UseSystem2andaddatwo-phaseconcentricneutralcablelineconnected

tonode2�Callthis“System3�”Thelineusesphasesaandcandis300ft

longandconsistsoftwo1/0AA1/3neutralconcentricneutralcables�The

cablesare40in�belowgroundand6in�apart�Thereisnoadditionalneutral

conductor�CallthislineUG-1�AttheendofUG-1,connectanodeandcallit

node4�Theloadatnode4isdelta-connectedloadmodeledasconstantcurrent�

Theloadis250kWat95%laggingpowerfactor�

Determinethevoltagesatallnodesona120Vtbaseandalllinecurrents�

References

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

PublishingCo�,Boston,MA,1995�

 2� ANSI/NEMA Standard Publication No� MG1-1978, National Electrical

ManufacturesAssociation,Washington,DC�

179

7

In document Distribution System Modeling and Analysis (Page 184-196)