Page 248 - Handbook of Electrical Engineering

232 HANDBOOK OF ELECTRICAL ENGINEERING

Table 9.28. Cable data for the worked example
◦
Cable conductor Resistance at 90 C Reactance at 60 Hz 0.033r + 0.944x
2
area (mm ) (ohm/km) (ohm/km) (ohm/km)
95 0.247 0.0872 0.16383
120 0.197 0.0868 0.14694
150 0.160 0.0874 0.13531
185 0.128 0.0876 0.12493
240 0.0989 0.0866 0.11439
300 0.0802 0.0860 0.10765

2
Hence from Tables 9.16 and 9.17 the small size of cable could be 95 mm . Table 9.28 can be prepared
2
for cables of 95 mm and above, routed in air.
2
Hence a 300 mm would be necessary for the starting duty. Check the actual volt-drop for
both starting and running currents.
√
31408(0.0802 × 0.33 + 0.086 × 0.944)0.25 × 100
V start =
440
= 14.92 = 14.9%
√
3245.4(0.0802 × 0.92 + 0.086 × 0.3919)0.25 × 100
V run =
440
= 2.596 = 2.6%

2
The choice of a 300 mm may just be acceptable for a running volt-drop of 2.6%, but satisfies
the required starting volt-drop.
Calculate the cut-off capability of the 400 A fuses. The same approach is used as in the
previous Example 9.4.3.5.1. The fault impedance Z f in (9.8) is higher due to operation of only one
transformer,
Z f = Z bs + Z t = 0.001054 + j0.006821 ohms/phase

The magnitude of this is 0.006902 ohms/phase, which has an X/R ratio of 6.4715. The dou-
bling factor is,

√ −π
D = 2 1.0 + e 6.4715 = 2.2846

The prospective RMS fault current at or near to the fuses is I f ,

460
V os
I f = √ = √ = 38478.9 amps
3Z f 3 × 0.006902
Hence the peak value of the symmetrical fault current

I fpka = 2.2846 × 38478.9 = 87908.8 amps