Electric Power Generation, Transmission, and Distribution

½Š¼B ½ŠþU

1

2

½ŠZabc½ŠYabc

 1

½ŠZabc (21:63)

½Š¼U

100

010

001

2

4

3

(^5) (21:64)
In many cases the shunt admittance is so small that it can be neglected. However, for all underground
cables and for overhead lines longer than 15 miles, it is recommended that the shunt admittance be
included. When the shunt admittance is neglected, the [a], [b], [c], [d], [A], and [B] matrices become
½Š¼a ½ŠU (21:65)
½Š¼b ½ŠZabc (21:66)
½Š¼c ½Š 0 (21:67)
½Šd¼½ŠU (21:68)
½Š¼A ½ŠU (21:69)
½Š¼B ½ŠZabc (21:70)
When the shunt admittance is neglected, Eqs. (21.55), (21.56), and (21.61) become
½ŠVLGabcn¼½ŠVLGabcmþ½ŠZabc½ŠIabcm (21:71)
½ŠIabcn¼½ŠIabcm (21:72)
½ŠVLGabcm¼½ŠVLGabcn½ŠZabc½ŠIabcm (21:73)
If an accurate determination of the voltage drops down a line segment is to be made, it is essential that
the phase impedance matrix [Zabc] be computed based upon the actual configuration and phasing of the
overhead or underground lines. No assumptions should be made, such as transposition. The reason for
this is best demonstrated by an example.
Example 21.7
The phase impedance matrix for the line configuration in Example 21.1 was computed to be
½Š¼zabc
0 : 4576 þj 1 :0780 0: 1560 þj 0 :5017 0: 1535 þj 0 : 3849
0 : 1560 þj 0 :5017 0: 4666 þj 1 :0482 0: 1580 þj 0 : 4236
0 : 1535 þj 0 :3849 0: 1580 þj 0 :4236 0: 4615 þj 1 : 0651
2
4
3
(^5) V=mile
Assume that a 12.47 kV substation serves a load 1.5 miles from the substation. The metered output at the
substation is balanced 10,000 kVA at 12.47 kV and 0.9 lagging power factor. Compute the three-phase
line-to-ground voltages at the load end of the line and the voltage unbalance at the load.
Solution
The line-to-ground voltages and line currents at the substation are
½Š¼VLGabc
7200 ff 0
7200 ff 120
7200 ff^120
2
4
3
(^5) ½ŠIabcn¼
463 ff 25 : 84
463 ff 145 : 84
463 ff 94 : 16
2
(^64)
3
(^75)