CABLES, WIRES AND CABLE INSTALLATION PRACTICES 207The secondary cable runs on its own cable rack to a switchboard. Both cable routes are short enough
to neglect volt-drop considerations. Find suitable Cu/XLPE/PVC/SWA/PVC cable conductor sizes.
Suitable derating factors:-a) For air ambient temperature Kair = 0.84
b) For ground temperature Kgrd = 0.86
c) For grouping cables in air Kga = 1.00
d) For grouping cables in the ground Kgg = 0.65
e) For ground thermal resistivity Kgth = 0.75
f) For depth of burial Kbury = 0.98
g) For using ducts in ground Kduct = 0.875
h) Overall derating factor for air Ka = Kair×Kga= 0. 84
i) Overall derating factor for ground Kg = Kgrd×Kgg×Kgth×Kbury×Kduct
= 0. 86 × 0. 65 × 0. 75 × 0. 98 × 0. 875 = 0. 36
Solution for primary cable:
Calculate the primary current for the ONAF loading of 5 MVA.
Primary currentIp=
5000000
√
3 × 11000
= 262 .4ampsOverall derating factor=Kg= 0. 36
Cable equivalent current at 25◦C=Ic 25 =
Ip
Kg=
262. 4
0. 36
= 728 .9ampsFrom Table 9.22 the nearest cable rated current equal to or greater than 728.9 amps for cables run
in air is 740 amps for a 400 mm^2 3-core cable. This choice would have a spare capacity in the
cable of only 1.5%, which is rather low for a practical design. A 400 mm^2 high voltage cable is
also difficult to manipulate during laying. A better choice would be two cables in parallel. The same
overall derating factor can be used if the two cables are spaced sufficiently far apart.
Cable equivalent current at 25◦C=Ic 25
2per cable= 364 .4amps.From Table 9.22 a suitable cable size to provide at least a 10% margin is 150 mm^2 , giving a
rated current in air of 430 amps. Hence the appropriate choice for the primary is 2×3c×150 mm^2
cables. The margin will allow for short duration overloading of the transformer.
Solution for the secondary cable:The corresponding secondary current
Is= 262. 4 ×11000
6900
= 418 .3ampsOverall derating factor=Ka= 0. 84
Cable equivalent current at 25◦C=Ic25=
Is
Ka=
418. 3
0. 84
= 498 .0amps