Handbook of Electrical Engineering

232 HANDBOOK OF ELECTRICAL ENGINEERING

Table 9.28. Cable data for the worked example

Cable conductor

area (mm^2 )

Resistance at 90◦C

(ohm/km)

Reactance at 60 Hz

(ohm/km)

0. 033 r+ 0. 944 x

(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

Hence from Tables 9.16 and 9.17 the small size of cable could be 95 mm^2. Table 9.28 can be prepared
for cables of 95 mm^2 and above, routed in air.

Hence a 300 mm^2 would be necessary for the starting duty. Check the actual volt-drop for
both starting and running currents.

Vstart=

√

31408 ( 0. 0802 × 0. 33 + 0. 086 × 0. 944 ) 0. 25 × 100

440

= 14. 92 = 14 .9%

Vrun=

√

3245. 4 ( 0. 0802 × 0. 92 + 0. 086 × 0. 3919 ) 0. 25 × 100

440

= 2. 596 = 2 .6%

The choice of a 300 mm^2 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 impedanceZfin (9.8) is higher due to operation of only one
transformer,

Zf=Zbs+Zt= 0. 001054 +j 0 .006821 ohms/phase

The magnitude of this is 0.006902 ohms/phase, which has anX/Rratio of 6.4715. The dou-
bling factor is,

D=

√

2

[

1. 0 +e

−π

6. 4715

]

= 2. 2846

The prospective RMS fault current at or near to the fuses isIf,

If=

Vos

√

3 Zf

=

460

√

3 × 0. 006902

=38478.9 amps

Hence the peak value of the symmetrical fault current

Ifpka= 2. 2846 × 38478. 9 =87908.8 amps