Electric Power Generation, Transmission, and Distribution

Example 21.4

Determine the shunt admittance matrix for the overhead line of Example 21.1. Assume that the neutral
conductor is 25 ft above ground.

Solution

For this configuration, the image spacing matrix is computed to be

½ŠS¼

58 58 :0539 58:4209 54: 1479

58 : 0539 58 58 :1743 54: 0208

58 :4209 58: 1743 58 54 : 0833

54 :1479 54:0208 54: 0835 50

2

6

(^64)
3
7
(^75) ft
The primitive potential coefficient matrix is computed to be
Pprimitive

¼
84 :56 35:1522 23:7147 25: 2469
35 :4522 84:56 28:6058 28: 359
23 :7147 28:6058 84: 56 26 : 6131
25 :2469 28:359 26:6131 85: 6659
2
6
(^64)
3
7
(^75)
Kron reduce to a 33 matrix
½Š¼P
77 :1194 26:7944 15: 8714
26 :7944 75:172 19: 7957
15 :8714 19:7957 76: 2923
2
4
3
5
Invert [P] to determine the shunt capacitance matrix
½Š¼Yabc j 376 : 9911 ½Š¼Cabc
j 5 : 6711 j 1 : 8362 j 0 : 7033
j 1 : 8362 j 5 : 9774 j 1 : 169
j 0 : 7033 j 1 : 169 j 5 : 391
2
4
3
(^5) mS=mile
Multiply [Cabc] by the radian frequency to determine the final three-phase shunt admittance matrix.
21.1.2.2 Underground Lines
Because the electric fields of underground cables are confined to the space between the phase conductor
and its concentric neutral to tape shield, the calculation of the shunt admittance matrix requires only the
determination of the ‘‘self’’ admittance terms.
21.1.2.3 Concentric Neutral
The self-admittance inmS=mile for a concentric neutral cable is given by
Ycn¼j
77 : 582
ln
Rb
Ra


1
k
ln
kRn
Rb
 (21:53)
whereRb¼radius of a circle to center of concentric neutral strands (ft)
Ra¼radius of phase conductor (ft)
Rn¼radius of concentric neutral strand (ft)
k ¼number of concentric neutral strands