Electric Power Generation, Transmission, and Distribution

Example 21.5

Determine the three-phase shunt admittance matrix for the concentric neutral line of Example 21.2.

Solution

Rb¼R¼ 0 :0511 ft

Diameter of the 250,000 AA phase conductor¼0.567 in.

Ra¼

0 : 567

24

¼ 0 :0236 ft

Diameter of the #14 CU concentric neutral strand¼0.0641 in.

Rn¼

0 : 0641

24

¼ 0 :0027 ft

Substitute into Eq. (21.53):

Ycn¼j

77 : 582

ln

Rb

Ra





1

k

ln

kRn

Rb

¼j^77 :^582

ln

0 : 0511

0 : 0236





1

13

ln

13  0 : 0027

0 : 0511

¼j 96 : 8847

The three-phase shunt admittance matrix is:

½Š¼Yabc

j 96 : 8847 0 0

0 j 96 : 8847 0

00 j 96 : 8847

2

4

3

(^5) mS=mile
21.1.2.4 Tape Shield Cable
The shunt admittance inmS=mile for tape shielded cables is given by
Yts¼j
77 : 586
ln
Rb
Ra
mS=mile (21:54)
whereRb¼inside radius of the tape shield
Ra¼radius of phase conductor
Example 21.6
Determine the shunt admittance of the single-phase tape shielded cable of Example 21.3 in
Section 21.1.1.
Solution
Rb¼
ds
24
¼
1 : 084
24
¼ 0 : 0452
The diameter of the 1=0 AA phase conductor¼0.368 in.
Ra¼
dp
24
¼
0 : 368
24
¼ 0 : 0153