Building Services Engineering 5th Edition Handbook

Electrical installations 299
For 400 kV:
current = 10 × 106 W
400 × 10 3 V × 1 √
3
= 14.4 A
For 33 kV:
current = 10 × 106 W
33 × 10 3 V × 1 √
3
= 175 A
This demonstrates the advantages of high-voltage transmission of electrical power as smaller
cable sizes can be used for the long distances involved.
Cable capacity and voltage drop
The maximum current-carrying capacities and actual voltage drops according to the LEE
Regulations for Electrical Installations (16th edn, 1991) for unenclosed copper cables which are
twin-sheathed in PVC, clipped to the surface of the building, are given in Table 13.1. Flexible
connections to appliances may use 0.5 mm 2 conductors for 3 A and 0.75 mm 2 conductors for
6 A loads. The maximum voltage drop allowed is 4% of the 240 V nominal supply (Jenkins,
1991).
EXAMPLE 13.6
Find the maximum lengths of 1, 1.5 and 2.5 mm 2 copper cable which can be used on a
240 V circuit to a3kWimmersion heater.
current = 3000 W
240 V = 12.5 A
allowed voltage drop = 4 × 240 = 9.6 V
100
maximum voltage drop allowed mV
maximum length or run =
load current A × voltage drop mV/A m
Table 13.1 Electrical cable capacities.
Nominal cross-sectional
area of conductor (mm 2 )
Maximum current
rating (A)
Voltage drop
in cable (mV/Am)
1 15 44
1.5 19.5 29
2.5 27 18
4 36 11
6 46 7.3
10 63 4.4
16 85 2.8