notes10 example

8/13/2019 Notes10 Example

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Notes 10: Conductor sizing & an example

10.0 Conductor sizing

Since it is impractical to manufacture an

infinite number of wire gauges, standards

have been adopted for an orderly and simple

arrangement of such sizes.

The American Wire Gauge (AWG sizes

conductors, ranging from a minimum of no.

!" to a ma#imum of no. !$" (which is the

same as %""""& for solid (single wire type

conductors. The smaller the gauge number,the larger the conductor diameter.

'or conductor sizes above !$", sizes are

given in ) (thousands of circular mil or

*ust cmils.

+


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What is a cmil A cmil is a unit of measure

for area and corresponds to the area of a

circle having a diameter of + mil, where+ mil-+"/inches.

The area of such a circle is 0r1- 0(d$11, or

0(+"/$11-2.34!#+"2inches1

+ 5mil-+ inch.

+ cmil-(+ mil1 and so corresponds to a

conductor having diameter of + mil-+"/in.

+"""5cmil-(+""" mils1and so corresponds

to a conductor having diameter of +"""

mils-+ in.

To determine diameter of conductor in

inches, ta5e s6uare root of cmils and then

divide by +"/78iameter in inches- 3

10

cmils.

1


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Stranded conductors (multiple strands of

wire for distribution purposes usually range

from a minimum of no. 9 to a ma#imum of+,""",""" cmil.

:t is 6uite common to specify conductors in

terms of the size (or gauge, the strands, and

the layers. 'or e#ample, !$" 9$+ indicates

AWG size """", with 9 strands and + layer.

The table below shows conductor sizes

commonly used in distribution systems.

/


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10.1 Example

!


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8etermine the phase impedance matri# ;


6/12

frjfRZ iiii

ln

2

16786.7

1ln1213.0001!8836.0"

fD

jfZij

ij

ln

2

16786.7

1ln1213.0001!8836.0"

With f-9" and B-+"", these e6uations are7

#301.7

1ln1213.00#!3016.0"

i

iiir

jRZ

#301.71ln1213.00#!3016.0"ij

ijD

jZ

So we need to compute the self terms for the

four conductors (which will all be the same

since they all have the same G?. We alsoneed to compute the mutuals, which will be

the harder part because we need to obtain a

mutual for each pairwise combination. This

is ! things ta5en 1 at a time, which is

!C$1C(!1C-9.So we have 9 calculations to ma5e.

'irst, lets do the self term, since it is easiest.

!6.1!0!3.0

#301.70171.0

1ln1213.00#!3016.01.0

#301.71

ln1213.00#!3016.0"

j

j

rjRZ

i

iii

To get the mutuals, we need to determine

the distance between conductors.

a to b7 s6rt(!1D11-!.!21+

9


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7810.00#!3.0

#301.7721.

1ln1213.00#!3016.0"

j

jZab

a to c7 1D1-!

7#!.00#!3.0

#301.71ln1213.00#!3016.0"

j

jZac

a to n7 s6rt(!1D91-2.1+++

7230.00#!3.0

#301.72111.7

1ln1213.00#!3016.0"

j

jZan

b to c7 s6rt(!1D11-!.!21+. Same as a to b.

b to n7 1D1-!. Same as a to c.c to n7 s6rt(!1D11-!.!21+. Same as a to b.

Therefore, the primitive impedance matri# is

=

!6.1!0!3.07810.00#!3.07#!.00#!3.07230.00#!3.0

7810.00#!3.0!6.1!0!3.07810.00#!3.07#!.00#!3.0

7#!.00#!3.07810.00#!3.0!6.1!0!3.07810.00#!3.0

7230.00#!3.07#!.00#!3.07810.00#!3.0!6.1!0!3.0

"

jjjj

jjjj

jjjj

jjjj

Zprim

The abc matri# is then found by Eron

reduction based on the formula we

developed7 ] ] ]

abcabcnpnnpnppabc VIZZZZV """" 1

where we define7 npnnpnppabc ZZZZZ """" 1:n our problem, the necessary matrices are7

2


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=!6.1!0!3.07810.00#!3.07#!.00#!3.0

7810.00#!3.0!6.1!0!3.07810.00#!3.0

7#!.00#!3.07810.00#!3.0!6.1!0!3.0

"

jjj

jjj

jjj

Zpp

=7810.00#!3.0

7#!.00#!3.0

7230.00#!3.0

"

j

j

j

Zpn

j0.6129-0.2126jZnn =11 !6.1!0!3.0" 7810.00#!3.07#!.00#!3.07230.00#!3.0" jjjZnp

Flugging in to get


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the a phase sees loo5ing at bphase is the

same that bphase sees loo5ing at aphase.

ow letHs compute the se6uence matri#


10/12

:f one is given a single phase or two phase

configuration, with or without neutrals, then

one may apply the same procedure as for the/phase configuration, i.e.,

+. 8etermine resistance per mile and G?

of each conductor.

1. 8etermine distance between conductors.

/. )ompute primitive impedance matri#

using e6s. (/@,!" of notes @ withappropriate value of resistivity B.

!. Ferform Eron reduction to eliminate the

presence of the neutral and thus obtain the

phase impedance matri#.

otes7 The primitive impedance matri# of step /

will be s6uare with dimension e6ual to the

+"


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number of conductors that you have in the

configuration.

The phase impedance matri# of step ! willbe s6uare with dimension e6ual to the

number of phase conductors that you have

in the configuration.

So a single phase configuration, with

neutral, will result in a 1#1 primitiveimpedance matri# and a +#+ phase

impedance matri#.

:f by chance a single or twophase

configuration e#tends from a three phaseconfiguration without changing the positions

of the remaining conductors, then one may

*ust pull out the appropriate elements from

the /#/ phase impedance matri#

corresponding to the / phase configuration.

'or e#ample, if

cccbca

bbbbba

acabaa

abc

ZZZ

ZZZ

ZZZ

Z

++


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is the phase impedance matri# for the /

phase configuration, if we run a 1phase

lateral using phases a I c without changingthe a I c phase I the neutral position, then

=

ccca

acaa

abc

ZZ

ZZ

Z

0

000

0

Ji5ewise, if we run a single phase lateral

using phase b with without changing the

phase b position relative to the neutral, then

=000

00

000

bbabc ZZ

:f positions do change, then non" elements

of above matrices will need to be

recomputed. owever, we will be able to

use /#/ matrices for any line configuration.

+1

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