appendixb.calculationoflineandcableconstants

7/29/2019 AppendixB.CalculationofLineandCableConstants

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Appendix BCalculation of Line and Cable

Constants

This appendix presents an overview of calculations of line and cable constants with

an emphasis on three-phase models and transformation matrices. Practically, the

transmission or cable system parameters will be calculated using computer-based

subroutine programs. For simple systems the data are available in tabulated form for

various conductor types, sizes, and construction [1-4]. Nevertheless, the basis of these

calculations and required transformations are of interest to a power system engineer.

B.1 AC RESISTANCE

As we have seen, the conductor ac resistance is dependent upon frequency and

proximity effects, temperature, spiraling and bundle conductor effects, which

increase the length of wound conductor in spiral shape with a certain pitch. The

resistance increases linearly with temperature and is given by the following equation:

R2 R1T

t

2T t!

B:1where R2 is the resistance at temperature t2, R1 is the resistance at temperature t1, T

is the temperature coefficient, which depends on the conductor material. It is 234.5

for annealed copper, 241.5 for hard drawn copper, and 228.1 for aluminum. The

resistance is read from manufacturers data, databases in computer programs, or

generalized tables.

B.2 INDUCTANCE

The internalinductance of a solid, smooth, round metallic cylinder of infinite length is

due to its internal magnetic field when carrying an alternating current and is given by

Lint 0

8H=m Henry per meter B:2

Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.


2/20

where 0 is the permeability 4 107(H/m). Its external inductance is due to theflux outside the conductor and is given by

Lext 0

2ln

D

r

H=m B:3

where D is any point at a distance D from the surface of the conductor, and r is the

conductor radius. In most inductance tables, D is equal to 1 ft and adjustment

factors are tabulated for higher conductor spacings. The total reactance is

L 02

1

4 ln D

r

! 0

2ln

D

e1=4r

! 0

2ln

D

GMR

!H=m B:4

where GMR is called the geometric mean radius and is 0.7788r. It can be defined as

the radius of a tubular conductor with an infinitesimally thin wall that has the same

external flux out to a radius of 1 ft as the external and internal flux of a solid

conductor to the same distance.

B.2.1 Inductance of a Three-Phase Line

We can write the inductance matrix of a three-phase line in terms of flux linkages a,

b, and c:

abc

Laa Lab LacLba Lbb LbcLca Lcb Lcc

IaIbIc

B:5

The flux linkage a .b, and c are given by

a 0

2Ia ln

1

GMRa

Ib ln

1

Dab

Ic ln

1

Dac

!

b 0

2Ia ln

1

Dba

Ib ln

1

GMRb

Ic ln

1

Dbc

!

c 0

2Ia ln

1

Dca

Ib ln

1

Dcb

Ic ln

1

GMRc

!B:6

where Dab, Dac, . . ., are the distances between conductor of a phase with respect to

conductors of b and c phases; Laa, Lbb, and Lcc are the self-inductances of the

conductors, and Lab, Lac, . . ., are the mutual inductances. If we assume a symme-

trical line, i.e., the GMR of all three conductors is equal and also the spacing

between the conductors is equal. The equivalent inductance per phase is

L 02

lnD

GMR

H=m B:7

The phase-to-neutral inductance of a three-phase symmetrical line is the same as the

inductance per conductor of a two-phase line.

B.2.2 Transposed Line

A transposed line is shown in Fig. B-1. Each phase conductor occupies the position

of two other phase conductors for one-third of the length. The purpose is to equalize

the phase inductances and reduce unbalance. The inductance derived for a symme-



3/20

trical line is still valid and the distance D in Eq. (B.7) is substituted by GMD

(geometric mean distance). It is given by

GMD DabDbcDca1=3

B:8A detailed treatment of transposed lines with rotation matrices is given in Ref. 5.

B.2.3 Composite Conductors

A transmission line with composite conductors is shown in Fig. B-2. Consider that

group X is composed ofn conductors in parallel and each conductor carries 1/n of

the line current. The group Y is composed ofm parallel conductors, each of which

carries 1=m of the return current. Then Lx, the inductance of conductor group Xis

Lx 2 107 lnnm ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Daa 0 ; Dab 0 ; Dac 0 ;. . .

;Dam; . . . ; Dna 0 ; Dnb 0 ; Dnc 0 ; . . . ; Dnmpn2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiDaaDabDac; . . . ; Dan; . . . ; DnaDnbDnc; . . . ; Dnn

pB:9

Henry per meter.

Figure B-1 A transposed transmission line.

Figure B-2 Inductance of composite conductors.



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We write Eq. (B.9) as

Lx 2 107 lnDm

Dsx

H=m B:10

Similarly,

Ly 2 107 lnDm

Dsy

H=m B:11

The total inductance is

L Lx LyH=m B:12

B.3 IMPEDANCE MATRIX

In Chap. 1 we decoupled a symmetrical three-phase line 3 3 matrix having equalself-impedances and mutual impedances [see Eq. (1.37)].We showed that the off-

diagonal elements of the sequence impedance matrix are zero. In high-voltage trans-

mission lines which are transposed, this is generally true and the mutual couplings

between phases are almost equal. However, the same cannot be said of distribution

lines and these may have unequal off-diagonal terms. In many cases the off-diagonal

terms are smaller than the diagonal terms and the errors introduced in ignoring these

will be small. Sometimes an equivalence can be drawn by the equations:

Zs Zaa Zbb Zcc3

Zm Zab Zbc Zca

3

B:13

i.e., an average of the self- and mutual impedances can be taken. The sequence

impedance matrix then has only diagonal terms. (See Example B.1.)

B.4 THREE-PHASE LINE WITH GROUND CONDUCTORS

A three-phase transmission line has couplings between phase-to-phase conductors

and also between phase-to-ground conductors. Consider a three-phase line with two

ground conductors, as shown in Fig. B-3. The voltage Va can be written as

Va RaIa j!LaIa j!LabIb j!LacIc j!LawIw j!LavIvj!LanIn V0a RnIn j!LnIn j!LanIa j!LbnIbj!LcnIc j!Lwniw j!lvnIv

B:14

where:

Ra; Rb; . . . ; Rn are resistances of phases a; b; . . . ; n

La; Lb; . . . ; Ln are the self inductances

Lab; Lac; . . . ; Lan are the mutual inductances

http://dke506_ch1.pdf/http://dke506_ch1.pdf/


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This can be written as

Va Ra RnIa RnIb RnIc j!La Ln 2LanIaj!Lab Ln Lan LbnIb j!Lac Ln Lan LcnIc RnIw

j!Law Ln Lan LwnIw RnIv j!Lav Ln Lan LvnIv V0a ZaagIa ZabgIb ZacgIc ZawgIw ZavgIv V0aB:15

where Zaa-g and Zbb-g are the self-impedances of a conductor with ground return,

and Zab-g and Zac-g are the mutual impedances between two conductors with com-

mon earth return. Similar equations apply to the voltages of other phases and

ground wires. The following matrix then holds for the voltage differentials between

terminals marked w; ; a; b; and c, and w 0; 0; a 0; b 0; and c 0:

VaVbVcVwVw

Zaag Zabg Zacg Zawg ZagZbag Zbbg Zbcg Zbwg ZbgZcag Zcbg Zccg Zcwg ZcgZwag Zwbg Zwcg Zwwg ZwgZag Zbg Zcg Zwg Zg

IaIbIcIwIv

B:16

In the partitioned form this matrix can be written as

"VVabc "VVw

"ZZA "ZZB

"ZZC "ZZD

"IIabc"IIw

B:17

Considering that the ground wire voltages are zero:

"VVabc "ZZA "IIabc "ZZB "IIw0 "ZZc "IIabc "ZZD "IIw

B:18

Figure B-3 Transmission line section with two ground conductors.



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Thus:

"IIw "ZZ1D "ZZC "IIabc B:19 "VVabc "ZZA "ZZB "ZZ1D "ZZC "IIabc B:20

This can be written as:

"VVabc

"ZZabc

"IIabc

B:21

"ZZabc "ZZA "ZZB "ZZ1D "ZZC Zaa 0g Zab 0g Zac 0gZba 0g Zbb 0g Zbc 0gZca 0g Zcb 0g Zcc 0g

B:22

The five-conductor circuit is reduced to an equivalent three-conductor circuit. The

technique is applicable to circuits with any number of ground wires provided that the

voltages are zero in the lower portion of the voltage vector.

B.5 BUNDLE CONDUCTORS

Consider bundle conductors, consisting of two conductors per phase (Fig. B-4). The

original circuit of conductors a, b, c and a 0, b 0, c 0 can be transformed into anequivalent conductor system of a 00, b 00, and c 00.

Each conductor in the bundle carries a different current and has a different

self- and mutual impedance because of its specific location. Let the currents in the

conductors be Ia, Ib, and Ic, and I0

a , I0

b , and I0

c , respectively. The following primitive

matrix equation can be written:

Va

VbVc

V0aV0bV0c

Zaa Zab Zac Zaa 0 Zab 0 Zac 0

Zba Zbb Zbc Zba 0 Zbb 0 Zbc 0

Zca Zcb Zcc Zca 0 Zcb 0 Zcc 0

Za 0a Za 0b Za 0c Za 0a 0 Za 0b 0 Za 0c 0

Zb 0a Zb 0b Zb 0c Zb 0a 0 Zb 0b 0 Zb 0c 0

Zc 0a Zc 0b Zc 0c Zc 0a 0 Zc 0b 0 Zc 0c 0

Ia

IbIc

Ia 0

Ib 0

Ic 0

B:23

This can be partitioned so that

"VVabc"VVa 0b 0c 0

"ZZ1 "ZZ2

"ZZ2 "ZZ4

Iabc"IIa 0b 0c 0

B:24

for symmetrical arrangement of bundle conductors"

ZZ1 "ZZ4.

Figure B-4 Transformation of bundle conductors to equivalent single conductors.



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Modify so that the lower portion of the vector goes to zero. Assume that

Va V0a V00aVb V0b V00bVc V0c V00c

B:25

The upper part of the matrix can then be subtracted from the lower part:

VaVbVc

0

0

0

Zaa Zab Zac Zaa 0 Zab 0 Zac 0

Zba Zbb Zbc Zba 0 Zbb 0 Zbc 0

Zca Zcb Zcc Zca 0 Zcb 0 Zcc 0

Za 0a Zaa Za 0b Zab Za 0c Zac Za 0a 0 Zaa 0 Za 0b 0 Zab 0 Za 0c 0 Zac 0Zb 0a Zba Zb 0b Zbb Zb 0c Zbc Zb 0a 0 Zba 0 Zb 0b 0 Zbb 0 Zb 0c 0 Zbc 0Zc 0a Zca Zc 0b Zcb Zc 0c Zcc Zc 0a 0 Zca 0 Zc 0b 0 Zcb 0 Zc 0c 0 Zcc 0

IaIbIc

Ia 0

Ib 0

Ic 0

B:26

We can write it in the partitioned form as

"VVabc0

"ZZ1 "ZZ2 "ZZt2 "ZZ1 Z4 Z2

"IIabc"IIa 0b 0c 0

B:27

I00a Ia I0aI00b Ib I0bI00c Ic I0c

B:28

The matrix is modified as shown below:

VaVb

Vc

0

0

0

Zaa Zab Zac Zaa0

Zaa Zab0

Zab Zac0

ZacZba Zbb Zbc Zba 0 Zba Zbb 0 Zbb Zbc 0 ZbcZca Zcb Zcc Zca 0 Zca Zcb 0 Zcb Zcc 0 Zcc

Za 0 a Zaa Za 0 b Zab Za 0 c Zac Za 0 a 0 Zaa 0 Za 0 a Zaa Za 0 b 0 Zab 0 Za 0 b Zab Za 0 c 0 Zac 0 Za 0c ZacZb 0 a Zba Zb 0 b Zbb Zb 0 c Zbc Zb 0 a 0 Zba 0 Zb 0 a Zba Zb 0 b 0 Zbb 0 Zb 0 b Zbb Zb 0 c 0 Zbc 0 Zb 0c ZbcZc 0 a Zca Zc 0 b Zcb Zc 0 c Zcc Zc 0 a 0 Zca 0 Zc 0 a Zca Zc 0b 0 Zcb 0 Zc 0 b Zcb Zc 0 c 0 Zcc 0 Zc 0 c Zcc

Ia

I!

aIb I!bIc I!c

Ia 0Ib 0

Ic 0

B:29

or in partitioned form:

"VVabc0

Z1 Z2 Z1

Zt2 Z1 Z4 Z2 Zt2 Z1

I00abcI0abc

B:30

This can now be reduced to following 3 3 matrix as before:V00aV00bV00c

Z00aa Z00

ab Z00

ac

Z00ba Z00

bb Z00

bc

Z00ca Z00

cb Z00

cc

I00aI00bIc

B:31

B.6 CARSONS FORMULA

The theoretical value ofZabc-g can be calculated by Carsons formula (c. 1926). Thisis of importance even today in calculations of line constants. For an n-conductor

configuration, the earth is assumed as an infinite uniform solid with a constant

resistivity. Figure B-5 shows image conductors in the ground at a distance equal



8/20

to the height of the conductors above ground and exactly in the same formation,with the same spacings between the conductors. A flat conductor formation is shown

in Fig. B-5.

Zii Ri 4!PiiG j Xi 2!G lnSii

ri 4!QiiG

!=mile B:32

Zij

4!PiiG

j 2!G ln

Sij

Dij 4!QijG !=mile

B:33

where:

Zii = the self-impedance of conductor iwith earth return (ohms/mile)

Zij = mutual impedance between conductors i and j (ohms/mile)

Ri = resistance of conductor in ohms/mile

Sii = conductor to image distance of the ith conductor to its own image

Dij = distance between conductors iand j

ri = radius of conductor (in ft)

! = angular frequencyG = 0:1609347 107 ohm-cmGMRi = geometric mean radius of conductor i

= soil resistivity

ij = angle as shown in Fig. B-5

Figure B-5 Conductors and their images; Carsons formula.



9/20

Expressions for P and Q are

P 8

13ffiffiffi

2p k cos k

2

16cos2 0:6728 ln 2

k

k

2

16sin k

3 cos3

45ffiffiffi

2p k

4 cos4

1536

B:34

Q 0:0386 12

ln2

k 1

3ffiffiffi

2p cos k

2 cos2

64 K

3 cos3

45ffiffiffi

2p

k4 sin4

384 k

4 cos4

384ln

2

k 1:0895

B:35

where

k 8:565x104Sij ffiffiffiffiffiffiffiffif=

pB:36

Sij is in feet and is soil resistivity in ohms-meter, and f is the system frequency. This

shows dependence on frequency as well as on soil resistivity.

B.6.1 Approximations to Carsons Equations

These approximations involve P and Q and the expressions are given by:

Pij

8B:37

Qij 0:03860 1

2ln

2

kijB:38

Using these assumptions, f 60 Hz and soil resistivity 100-m, the equationsreduce to:

Zii Ri 0:0953 j0:12134 ln1

GMRi 7:93402

=mile B:39

Zij 0:0953 j0:12134 ln 1Dij

7:93402

=mile B:40

Equations (B.39) and (B.40) are of practical significane for calculations of line impe-

dances.

Example B.1

Consider an unsymmetrical overhead line configuration, as shown in Fig. B-6. The

phase conductors consist of 556.5 KCMIL (556,500 circular mils) of ACSR con-ductor consisting of 26 strands of aluminum, two layers and seven strands of steel.

From the properties of ACSR conductor tables, the conductor has a resistance of

0.1807 ohms at 60 Hz and its GMR is 0.0313 ft at 60 Hz; conductor diameter

= 0.927 in. The neutral consists of 336.4 KCMIL, ACSR conductor, resistance



10/20

0.259 ohms per mile at 60 Hz and 508C and GMR 0.0278 ft, and conductor diameter

0.806 in. It is required to form a primitive Z matrix, convert it into a 3 3 Zabcmatrix, and then to sequence impedance matrix Z012.

Using Eq. (B.39) and (B.40):

Zaa Zbb Zcc 0:1859 j1:3831Znn 0:3543 j1:3974Zab Zba 0:0953 j0:8515Zbc Zcb 0:0953 j0:7674Zca Zac 0:0953 j0:7182Zan Zna 0:0953 j0:7539Zbn Znb 0:0953 j0:7674Zcn Znc 0:0953 j0:7237

Therefore, the primitive impedance matrix is

"ZZprim 0:1859 j1:3831 0:0953 j0:08515 0:0953 j0:7182 0:0953 j0:75390:0953 j0:8515 0:1859 j1:3831 0:0953 j0:7624 0:0953 j0:76740:0953 j0:7182 0:0953 j0:7624 0:1859 j1:3831 0:0953 j0:72370:0953 j0:7539 0:0953 j0:7674 0:0953 j0:7237 0:3543 j1:3974

Figure B-6 Distribution line configuration for calculation of line parameters (Examples B.1and B.3).



11/20

Eliminate the last row and column using Eq. (B.22):

"ZZabc 0:1846 j0:9825 0:0949 j0:4439 0:0921 j0:33340:0949 j0:4439 0:1864 j0:9683 0:0929 j0:37090:0921 j0:3334 0:0929 j0:3709 0:1809 j1:0135

=mile

Convert to Z012 by using the transformation equation (1.29):

"ZZ012 0:3705 j1:7536 0:0194 j0:0007 0:0183 j0:0055

0:0183 j0:0055 0:0907 j0:6054 0:0769 j0:01460:0194 j0:0007 0:0767 j0:0147 0:0907 j0:6054

This shows the mutual coupling between sequence impedances. We could average

out the self- and mutual impedances according to Eq. (B.13):

Zs Zaa Zbb Zcc

3 0:184 j0:9973

Zm Zab

Zbc

Zca

3 0:0933 0:38271The matrix Zabc then becomes:

"ZZabc 0:184 j0:9973 0:933 j0:3827 0:933 j0:38270:933 j0:3827 0:184 j0:9973 0:933 j0:38270:933 j0:3827 0:933 j0:3827 0:184 j0:9973

=mile

and this gives

"ZZ012

0:3706 j1:7627 0 00 0:0907

j0:6146 0

0 0 0:0907 j0:6146 =mile

Example B.2

Figure B-7 shows a high-voltage line with two 636,000 mils ACSR bundle conduc-

tors per phase. Conductor GMR = 0.0329 ft, resistance = 0.1688 ohms per mile,

diameter = 0.977 in., and spacings are as shown in Fig. B-7. Calculate the primitive

impedance matrix and reduce it to a 3 3 matrix, then convert it into a sequencecomponent matrix.

Figure B-7 Configuration of bundle conductors (Example B.2).



12/20

From Eqs. (B.39) and (B.40) and the specified spacings in Fig. B7, matrix Z1 is

"ZZ1 0:164 j1:3770 0:0953 j0:5500 0:0953 j0:46590:953 j0:5500 0:164 j1:3770 0:0953 j0:5500

0:0953 j0:4659 0:0953 j0:5500 0:164 j1:3770

This is also equal to Z4, as the bundle conductors are identical and symmetricallyspaced. Matrix Z2 of Eq. (B.27) is

"ZZ2 0:0953 j0:8786 0:0953 j0:5348 0:0953 j0:45810:0953 j0:5674 0:0953 j0:8786 0:0953 j0:53480:0953 j0:4743 0:0953 j0:08786 0:0953 j0:8786

The primitive matrix is 6 6 given by Eq. (B.23) formed by partitioned matricesaccording to Eq. (B.24).Thus, from "ZZ1 and "ZZ2 the primitive matrix can be written.

From these two matrices, we will calculate matrix equation (B.30):

"ZZ1 "ZZ2 0:069 j0:498 j0:0150 j0:0079

j0:0171 0:069 j0:498 j0:0150j0:00841 j0:0170 0:069 j0:498

and

"ZZk "ZZ1 "ZZ2 "ZZt2 "ZZ1 0:138 j0:997 j0:0022 j0:0005

j0:0022 0:138 j0:997 j0:0022j0:0005 j0:0022 0:138 j0:997

The inverse is

"ZZ1k 0:136 j0:984 0:000589 j0:002092 0:0001357 j0:0004797

0:0005891 j0:002092 0:136 j0:981 0:0005891 j0:0020920:0001357 j0:0004797 0:0005891 j0:002092 0:136 j0:981

then, the matrix "ZZ2 "ZZ1 "ZZ1k "ZZt2 "ZZ1 is0:034 j0:2500 0:000018 j0:000419 0:0000363 j0:0003871

0:000018

j0:000419 0:034

j0:2500

0:000018

j0:000419

0:0000363 j0:000387 0:000018 j0:000419 0:034 j0:2500

Note that the off-diagonal elements are relatively small as compared to the

diagonal elements. The required 3 3 transformed matrix is then Z1 minus theabove matrix:

"ZZtransformed 0:13 j1:127 0:095 j0:55 0:095 j0:4660:095 j0:55 0:13 j1:127 0:095 j0:55

0:095 j0:466 0:095 j0:55 0:13 j1:127

=mile

Using Eq. (1.29), the sequence impedance matrix is

"ZZ0:12 0:32 j2:171 0:024 j0:014 0:024 j0:014

0:024 j0:014 0:035 j0:605 0:048 j0:0280:024 j0:014 0:048 j0:028 0:035 j0:605

=mile



13/20

B.7 CAPACITANCE OF LINES

The shunt capacitance per unit length of a two-wire, single-phase transmission

line is

C "0lnD=r F=m Farads per meter B:41

where "0 is the permittivity of free space 8:854 1012F/m, and other symbols areas defined before. For a three-phase line with equilaterally spaced conductors, the

line-to-neutral capacitance is

C 2"0lnD=r F=m B:42

For unequal spacings, D is replaced with GMD from Eq. (B.7). The capaci-

tance is affected by the ground and the effect is simulated by a mirror image of the

conductors exactly at the same depth as the height above the ground. These mirror-

image conductors carry charges which are of opposite polarity to conductors above

the ground (Fig. B-8). From this figure, the capacitance to ground is

Cn 2"0

lnGMD=r ln3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSab 0; Sbc 0 ;Sca 0 ;p =3 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiSaa 0 ; Sbb 0 ; Scc 0 ; p B:43Using the notations in Eq. (B.10), this can be written as

Cn 2"0

lnDn=Ds 10

9

18lnDm=DsF=m B:44

B.7.1 Capacitance Matrix

The capacitance matrix of a three-phase line is

"CCabc Caa Cab Cac

Cba Cbb CbcCca Ccb Ccc

B:45

This is diagrammatically shown in Fig. B-9(a). The capacitance between the phase

conductor a and b is Cab and the capacitance between conductor a and ground is:

Caa Cab Cac. If the line is perfectly symmetrical, all the diagonal elements are thesame and all off-diagonal elements of the capacitance matrix are identical:

"CCabc C C0 C0

C0 C C0C0 C0 C

B:46

Symmetrical component transformation is used to diagionalize the matrix:

"CC012 "TT1s "CCabc "TTs C 2C0 0 0

0 C C0 00 0 C C0

B:47

The zero, positive, and negative sequence networks of capacitance of a symmetricaltransmission line are shown in Fig. B-9(b). The eigenvalues are C 2C0, C C0, andC C0. The capacitance C C0 can be written as 3C0 C 2C0, i.e., it is equiva-lent to the line capacitance of a three-conductor system plus the line-to-ground

capacitance of a three-conductor system.



14/20

In a capacitor, V Q=C. The capacitance matrix can be written as"VV

abc "PP

abc

"QQabc

"CC1abc

"QQabc

B:48

where "PP is called the potential coefficient matrix, i.e,

VaVbVc

Paa Pab PacPba Pbb PbcPca Pcb Pcc

QaQbQc

B:49

where

Pii 1

2"0ln

Sii

ri 11:17689 ln Sii

riB:50

Pij 12"0

ln Sij

Dij 11:17689 ln Sij

DijB:51

where:

Sij = conductor-to-image distance below ground (in ft)

Figure B-8 Calculation of capacitances, conductors, mirror images, spacings, and charges.



15/20

Dij = conductor-to-conductor distance (in ft)

ri = radius of the conductor (in feet)

"0 = permitivity of the medium surrounding the conductor 1:424 108F/mile for air.

For sine-wave voltage and charge, the equation can be expressed as

IaIbIc

j!

Caa Cab CacCba Cbb CbcCca Ccb Ccc

VaVbVc

B:52

Figure B-9 (a) Capacitances of a three-phase line; (b) equivalent positive, negative, and zerosequence networks of capacitances.



16/20

The capacitance of three-phase lines with ground wires and with bundle conductors

can be addressed as in the calculations of inductances. The primitive P matrix can be

partitioned and reduces to a 3 3 matrix.

Example B.3

Calculate the matrices P and C for Example B.1. The neutral is 30 ft above groundand the configuration of Fig. B-6 is applicable.

The mirror images of the conductors are drawn in Fig. B-6. This facilitates

calculation of the spacings required in Eqs. (B.50) and (B.51) for the P matrix. Based

on the geometric distances and conductor diameter the primitive P matrix is

"PP

Paa Pab Pac Pan

Pba Pbb Pbc Pbn

Pca Pcb Pcc Pcn

Pna

Pnb

Pnc

Pnn

80:0922 33:5387 21:4230 23:3288

33:5387 80:0922 25:7913 24:5581

21:4230 25:7913 80:0922 20:7547

23:3288 24:5581 20:7547 79:1615

This is reduced to a 3 3 matrix

P 73:2172 26:3015 15:3066

26:3015 72:4736 19:3526

15:3066 19:3526 74:6507

Therefore, the required "CC matrix is inverse of "PP, and "YYabc is

"YYabc j! "PP1 j6:0141 j1:9911 j0:7170j1:9911 j6:2479 j1:2114j0:7170 j1:2114 j5:5111

siemens=mile

B.8 CABLE CONSTANTS

The construction of cables varies widely; it is mainly a function of insulation type,method of laying, and voltage of application. For high-voltage applications above

230 kV, oil-filled paper insulated cables are used, though recent trends see the devel-

opment of solid dielectic cables up to 345 kV. A three-phase solid dielectic cable has

three conductors enclosed within a sheath and because the conductors are much

closer to each other than those in an overhead line and the permittivity of insulating

medium is much higher than that of air, the shunt capacitive reactance is much lower

as compared to an overhead line. Thus, use of a T or model is required even for

shorter cable lengths.

The inductance per unit length of a single conductor cable is given by

L 02

lnr1

r2H=m B:53

where r1 is the radius of the conductor and r2 is the radius of the sheath, i.e., the

cable outside diameter divided by 2.



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When single conductor cables are installed in magnetic conduits the reactance

may increase by a factor of 1.5. Reactance is also dependent on conductor shape, i.e.,

circular or sector, and on the magnetic binders in three-conductor cables.

B.8.1 Concentric Neutral Underground Cable

We will consider a concentric neutral construction as shown in Fig. B-10(a). Theneutral is concentric to the conductor and consists of a number of copper strands

that are wound helically over the insulation. Such cables are used for underground

distribution, directly buried or installed in ducts. Referring to Fig. B-10(a), d is the

diameter of the conductor, d0 is the outside diameter of the cable over the concentric

neutral strands, and ds is the diameter of an individual neutral strand. Three cables in

flat formation are shown in Fig. B-10(b). The GMR of a phase conductor and a

neutral strand are given by the expression:

GMRcn ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiGMRsnRn

1nq B:54

where GMRcn is the equivalent GMR of the concentric neutral, GMRs is the GMR

of a single neutral strand, n is the number of concentric neutral strands, and R is the

Figure B-10 (a): Construction of a concentric neutral cable; (b) configuration for calcula-tion of cable series reactance (Example B.4).



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radius of a circle passing through the concentric neutral strands (see Fig. B-10(a)) d0 ds=2 (in ft).

The resistance of the concentric neutral is equal to the resistance of a single

strand divided by the number of strands.

The geometric mean distance between concentric neutral and adjacent phase

conductors is

Dij ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Dnmn Rnnp

B:55

where Dij is the equivalent center-to-center distance of the cable spacing. Note that it

is less than Dmn, the center-to-center spacing of the adjacent conductors, Fig. B-

10(b). Carsons formula can be applied and the calculations are similar to those in

Example B.1.

Example B.4

A concentric neutral cable system for 13.8 kV has a center-to-center spacing of 8 in.The cables are 500 KCMIL, with 16 strands of #12 copper wires. The following data

are supplied by the manufacturer:

GMR phase conductor = 0.00195 ft

GMR of neutral strand = 0.0030 ft

Resistance of phase conductor = 0.20 ohm/mile

Resistance of neutral strand = 10.76 ohms/mile. Therefore, the resistance of

the concentric neutral = 10.76/16= 0.6725 ohm/mile.

Diameter of neutral strand = 0.092 in.

Overall diameter of cable = 1.490 in.Therefore, R 1:490 0:092=24 0:0708 ft.

The effective conductor phase-to-phase spacing is approximately 8 in., from Eq.

(B.55).

The primitive matrix is a 6 6 matrix, similar to Eq. (B.16). In the partitionedform, Eq. (B.17), the matrices are

"ZZa 0:2953 j1:7199 0:0953 j1:0119 0:0953 j0:92780:0953 j1:0119 0:2953 j1:7199 0:0953 j1:01190:0953 j0:9278 0:0953 j1:0119 0:2953 j1:7199

The spacing between the concentric neutral and the phase conductors is approxi-

mately equal to the phase-to-phase spacing of the conductors. Therefore,

"ZZB 0:0953 j1:284 0:0953 j1:0119 0:0953 j0:9278

0:0953 j1:0119 0:0953 j1:284 0:0953 j1:01190:0953 j0:9278 0:0953 j1:0119 0:0953 j1:284

Matrix "ZZc "ZZB and matrix "ZZD is given by

"ZZD 0:7678 j1:2870 0:0953 j1:0119 0:0953 j0:9278

0:0953 j1:0119 0:7678 j1:2870 0:0953 j1:01190:0953 j0:9278 0:0953 j1:0119 0:7678 j1:2870

This primitive matrix can be reduced to a 3 3 matrix, as in other examples.



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B.8.2 Capacitance of Cables

In a single conductor cable, the capacitance per unit length is given by

C 2"0"lnr1=r2

F=m B:56

Note that " is the permittivity of the dilectric medium relative to air. The capaci-tances in a three-conductor cable are shown in Fig. B-11. This assumes a symmetrical

construction, and the capacitances between conductors and from conductors to the

sheath are equal. The circuit of Fig. B-11(a) is successively transformed and Fig.

B-11(d) shows that the net capacitance per phase C1 3C2.

Figure B-11 (a) Capacitances in a three-conductor cable; (b) and ( c) equivalent circuits; (d)final capacitance circuit.



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By change of units, Eq. (B.56) can be expressed as

C 7:35"logr1=r2

pF=ft B:57

This gives the capacitance of a single-conductor shielded cable. Table B-1 gives

values of " for various cable insulation types.

REFERENCES

1. DG Fink (Ed.). Standard Handbook for Electrical Engineers. 10th ed. New York:

McGraw-Hill, 1969.

2. Croft, Carr, Watt. American Electricians Handbook. 9th ed. New York: McGraw-Hill,

1970.

3. Central Station Engineers. Electrical Transmission and Distribution Reference Book. 4th

ed. East Pittsburgh, PA: Westinghouse Corp., 1964.

4. The Aluminum Association. Aluminum Conductor Handbook. 2nd ed. Washington, DC,

1982.

5. PM Anderson. Analysis of Faulted Systems. Ames, IA: Iowa State University Press, 1973.

Table B-1 Typical Values for Dielectric Constants of Cable Insulation

Type of insulation Permittivity (")

Polyvinyl chloride (PVC) 3.58.0

Ethyelene-propylene insulation (EP) 2.83.5

Polyethylene insulation 2.3

Cross-linked polyethylene 2.36.0

Impregnated paper 3.33.7

appendixb.calculationoflineandcableconstants

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