Electric Power Generation, Transmission, and Distribution

In applying Carson’s equations for both concentric neutral and tape shielded cables, the numbering
of conductors and neutrals is important. For example, a three-phase underground circuit with an
additional neutral conductor must be numbered as

1 ¼phase conductor #1

2 ¼phase conductor #2

3 ¼phase conductor #3

4 ¼neutral of conductor #1

5 ¼neutral of conductor #2

6 ¼neutral of conductor #3

7 ¼additional neutral conductor (if present)

Example 21.3

A single-phase circuit consists of a 1=0 AA tape shielded cable and a 1=0 CU neutral conductor as shown
in Fig. 21.10.
Cable Data: 1=0AA

Inside diameter of tape shield¼ds¼1.084 in.

Resistance¼0.97V=mile

GMRp¼0.0111 ft

Tape shield thickness¼T¼8 mils

Neutral Data: 1=0 Copper, 7 strand

Resistance¼0.607V=mile

GMRn¼0.01113 ft

Distance between cable and neutral¼Dnm¼3 in.
The resistance of the tape shield is computed according to Eq. (21.41):

rshield¼

18 : 826

dsT

¼

18 : 826

1 : 084  8

¼ 2 : 1705 V=mile

The GMR of the tape shield is computed according toEq. (21.42):

GMRshield¼

ds

2



T

2000

12

¼

1 : 084

2



8

2000

12

¼ 0 :0455 ft

Using the relations defined in Eqs. (21.43) through (21.45) and Carson’s equations results in a 3 3
primitive impedance matrix:

zprimitive¼

1 : 0653 þj 1 :5088 0: 0953 þj 1 :3377 0: 0953 þ 1 : 1309

0 : 0953 þj 1 :3377 2: 2658 þj 1 :3377 0: 0953 þj 1 : 1309

0 : 0953 þj 1 :1309 0: 0953 þj 1 :1309 0: 7023 þj 1 : 5085

2

4

3

(^5) V=mile
Applying Kron’s reduction method will result in a single impedance that
represents the equivalent single-phase impedance of the tape shield cable and
the neutral conductor.
zlp¼ 1 : 3368 þj 0 : 6028 V=mile
3 0
FIGURE 21.10 Single-
phase tape shield with
neutral.