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8
In the previous section we
illustrated the requirements for
complying with voltage drop
and automatic disconnection.
In this section we will examine
the other two factors to be
considered, namely that the
circuit protective conductor
is large enough to carry the
earth fault currents and the
selection and erection of a
wiring system.
Q
Why do I need to protect
against the effects of fault
current?
Fault currents generate heat,
so you must ensure that the
circuit protective conductor
(CPC) can carry the earth
fault currents without thermal
damage until the overcurrent
device operates.
Q
How do I protect against
the effects of fault current?
If the cross sectional area of
the cpc has been worked out
by applying table 54.7 of BS
7671, and the overcurrent
protective device is providing
protection against overload
currents and fault currents,
no further checks are needed.
Table 54.7 details the
minimum crosssectional area
of the protective conductor in
relation to the crosssectional
area of the associated line
conductor.
Q
If the overcurrent
protective device is not
providing protection against
overload current and I have a
cpc that does not comply with
table 54.7, what do I do?
You need to apply the formula
in regulation
543.1.3
:
S = √(I
2
t) or I
2
t = k
2
S
2
–––––
k
S = nominal cross sectional
area of the cpc in mm
2
;
I = fault current in amperes;
t= operating time of
disconnecting device in
seconds;
k = factor taken from BS 7671.
Q
Is there a quick and
simple method of applying I
2
t
= k
2
S
2
?
Yes. Use the manufacturer’s
I
2
t characteristics for the
overcurrent protective
device. Calculate k
2
S
2
and
superimpose this value as a
horizontal line on the graph
showing the protective
device’s I
2
t characteristics.
This was illustrated earlier.
Hager has already completed
these calculations. They are
available upon request.
Earth fault current
protection