to the 16th Edition IEE Regulations

chapter 5

chapter 6

Installation control and protection
  3.1- Introduction 3.5 - High temperature protection
  3.2 - Switching 3.6 - Overload currents
  3.3 - Isolation 3.7 - Protection from faults
3.4 - Electric shock protection

3.8 - Short circuit and overload
------- protection

3.7.3- Operating time

Not only must the short-circuit protection system open the circuit to cut off a fault, but it must do so quickly enough to prevent both a damaging rise in the conductor insulation temperature and mechanical damage due to cable movement under the influence of electro-mechanical force. The time taken for the operation of fuses and circuit breakers of various types and ratings is shown in {Figs 3.13 to 3.19}. When the prospective short circuit current (PSC) for the point at which the protection is installed is less than its breaking capacity there will be no problem.

When a short circuit occurs there will be a high current which must be interrupted quickly to prevent a rapid rise in conductor temperature.

The position is complicated because the rise in conductor temperature results in an increase in resistance which leads to an increased loss of energy and increased heating (W = lRt), where W is the energy (J) and R is the resistance W. The Regulations make use of the adiabatic equation which assumes that all the energy dissipated in the conductor remains within it in the form of heat, because the faulty circuit is opened 50 quickly. The equation is:

t =

where t = the time for fault current to raise conductor temperature to the highest permissible level

k = a factor which varies with the type of cable

S = the cross-sectional area of the conductor (mm)

I = the fault current value (A) - this will be the PSC

Some cable temperatures and values of k for common cables are given in {Table 3.7}.

Table 3.7 Cable temperatures and k values (copper cable)

Insulation material
Assumed initial
temperature (C)
Limiting final
85C p.v.c
90C thermosetting
Mineral, exposed to touch
or p.v.c. covered
Mineral not exposed to touch

As an example, consider a 10 mm cable with p.v.c. insulation protected by a 40 A fuse to BS 88 Part 2 in an installation where the loop impedance between lines at the point where the fuse is installed is 0.12 W. If the supply is 415 V three phase, the prospective short circuit current (PSC) will be:

= UL A =
415 A
= 3.46 kA

from {Table 3.7}, k = 115
= kS=
115 x 10
= 0.110s

{Figure 3.15} shows that a 40 A fuse to BS 88 Part 2 will operate in 0.1 5 when carrying a current of 400 A. Since the calculated PSC at 3460 A is much greater than 400 A. the fuse will almost certainly clear the fault in a good deal less than 0.1 s. As this time is less than that calculated by using the adiabatic equation (0.11 s) the cable will be unharmed in the event of a short circuit fault.

It is important to appreciate that the adiabatic equation applies to all cables, regardless of size. Provided that a protective device on the load side of a circuit has a breaking capacity equal to or larger than the PSC of the circuit then that circuit complies with the PSC requirements of the Regulations (see {Fig 3.22} and see also the note in {7.15.1} concerning the use of dual rated fuses for motor protection).



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Extracted from The Electricians Guide Fifth Edition
by John Whitfield
Published by EPA Press Click Here to order your Copy

Click here for list of abbreviations