Page 235 - Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 219

And from Figure 9.1,

AB = 349.91 × 2.50 × 0.9500 = 831.04
BE = 349.91 × 2.50 × 0.3123 = 273.19
EF = 349.91 × 2.75 × 0.3123 = 300.51
DF = 349.91 × 2.75 × 0.9500 = 914.14
√
3(831.04 + 300.51) 100
V = 5.816%
33700
Alternatively the volt-drop can be calculated by solving the circuit conditions shown in Figure 9.4,
as follows:-
By simple proportions V r can be found from V s as follows,

V r Z L
=
V s R + X l + Z L
51.728 + j17.002
=
2.5 + j2.75 + 51.728 + j17.002
= 0.94299 + j0.02995

Therefore,
V r = (0.94299 + j0.02995)(19456.7)
= 18347.47 + j582.73

And
|V r |= 18356.73 volts/phase
The total actual volt-drop

|V s |− |V r |
V = × 100
|V s |
(19456.7 − 18356.73) × 100
= = 5.653%
19456.7

Figure 9.4 Equivalent simple circuit of a long cable. A long cable as a simple series circuit.