Page 223 - Handbook of Electrical Engineering

CABLES, WIRES AND CABLE INSTALLATION PRACTICES 207

The 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 K air = 0.84
b) For ground temperature K grd = 0.86
c) For grouping cables in air K ga = 1.00
d) For grouping cables in the ground K gg = 0.65
e) For ground thermal resistivity K gth = 0.75
f) For depth of burial K bury = 0.98
g) For using ducts in ground K duct = 0.875
h) Overall derating factor for air K a = K air × K ga = 0.84
i) Overall derating factor for ground K g = K grd × K gg × K gth × K bury × K duct
= 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.
5000000
Primary current I p = √ = 262.4amps
3 × 11000
Overall derating factor = K g = 0.36

I p 262.4
◦
Cable equivalent current at 25 C = I c25 = = = 728.9amps
K g 0.36
From Table 9.22 the nearest cable rated current equal to or greater than 728.9 amps for cables run
2
in air is 740 amps for a 400 mm 3-core cable. This choice would have a spare capacity in the
2
cable of only 1.5%, which is rather low for a practical design. A 400 mm 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.
◦ I c25
Cable equivalent current at 25 C = per cable = 364.4amps.
2
2
From Table 9.22 a suitable cable size to provide at least a 10% margin is 150 mm , 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
11000
I s = 262.4 × = 418.3amps
6900
Overall derating factor = K a = 0.84
I s 418.3
◦
Cable equivalent current at 25 C = I c25 = = = 498.0amps
K a 0.84