Electric Power Generation, Transmission, and Distribution

Example 21.2

Three concentric neutral cables are buried in a trench with spacings as shown in Fig. 21.8. The cables are
15 kV, 250,000 CM stranded all aluminum with 13 strands of #14 annealed coated copper wires (1= 3
neutral). The data for the phase conductor and neutral strands from a conductor data table are

250,000 AA phase conductor: GMRp¼0.0171 ft, resistance¼0.4100V=mile

#14 copper neutral strands: GMRs¼0.00208 ft, resistance¼14.87V=mile

Diameter (ds)¼0.0641 in.

The equivalent GMR of the concentric neutral [Eq. (21.37)]¼0.04864 ft
The radius of the circle passing through strands [Eq. (21.38)]¼0.0511 ft
The equivalent resistance of the concentric neutral [Eq. (21.39)]¼1.1440V=mile
SinceR(0.0511 ft) is much less thanD 12 (0.5 ft) andD 13 (1.0 ft), then the distances between concentric
neutrals and adjacent phase conductors are the center-to-center distances of the cables.
Applying Carson’s equations results in a 66 primitive impedance matrix. This matrix in partitioned
form [Eq. (21.12)] is:

zij



¼

0 : 5053 þj 1 :4564 0: 0953 þj 1 :0468 0: 0953 þj 0 : 9627

0 : 0953 þj 1 :0468 0: 5053 þj 1 :4564 0: 0953 þj 1 : 0468

0 : 0953 þj 0 :9627 0: 0953 þj 1 :0468 0: 5053 þj 1 : 4564

2

(^64)
3
(^75)
½Š¼zin
0 : 0953 þj 1 :3236 0: 0953 þj 1 :0468 0: 0953 þj 0 : 9627
0 : 0953 þj 1 :0468 0: 0953 þj 1 :3236 0: 0953 þj 1 : 0468
0 : 0953 þj 0 :9627 0: 0953 þj 1 :0468 0: 0953 þj 1 : 3236
2
6
4
3
7
5
znj

¼½Šzin
½Š¼znn
1 : 2393 þj 1 :3296 0: 0953 þj 1 :0468 0: 0953 þj 0 : 9627
0 : 0953 þj 1 :0468 1: 2393 þj 1 :3296 0: 0953 þj 1 : 0468
0 : 0953 þj 0 :9627 0: 0953 þj 1 :0468 1: 2393 þj 1 : 3296
2
6
4
3
7
5
Using the Kron reduction [Eq. (21.13)] results in the phase impedance matrix
½Šzabc¼
0 : 7982 þj 0 :4463 0: 3192 þj 0 :0328 0: 2849 j 0 : 0143
0 : 3192 þj 0 :0328 0: 7891 þj 0 :4041 0: 3192 þj 0 : 0328
0 : 2849 j 0 :0143 0: 3192 þj 0 :0328 0: 7982 þj 0 : 4463
2
4
3
(^5) V=mile
The sequence impedance matrix for the concentric neutral three-phase line is determined usingEq.
(21.3). The resulting sequence impedance matrix is
½Š¼z 012
1 : 4106 þj 0 : 4665  0 : 0028 j 0 : 0081  0 : 0056 þj 0 : 0065
 0 : 0056 þj 0 :0065 0: 4874 þj 0 : 4151  0 : 0264 þj 0 : 0451
 0 : 0028 j 0 :0081 0: 0523 þj 0 :0003 0: 4867 þj 0 : 4151
2
4
3
(^5) V=mile
6 0 6 0
FIGURE 21.8 Three-phase concentric neutral cable spacing.