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 = l²Rt), 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:
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
(mm²)
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
temperature(°C)
|
k
|
|
|
|
|
|
| p.v.c |
70
|
160
|
115
|
| 85°C p.v.c |
85
|
160
|
104
|
| 90°C thermosetting |
90
|
250
|
143
|
Mineral, exposed to touch
or p.v.c. covered |
70
|
160
|
115
|
| Mineral not exposed to
touch |
105
|
250
|
135
|
As an
example, consider a 10 mm² 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:
|
I
|
= UL A =
|
415 A
|
= 3.46 kA |
| |
Z
|
0.12
|
|
| from {Table 3.7}, k =
115 |
|
t
|
= k²S² =
|
115² x 10²
|
= 0.110s |
| |
-I²
|
34602
|
|
{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).