5.3.6  Maximum circuit conductor length
The complete earthfault
loop path is made up of a large number of parts as shown
in {Fig
5.7}, many of which are external to the installation
and outside the control of the installer. These external
parts make up the external loop impedance (Ze). The rest
of the earthfault loop impedance of the installation consists
of the impedance of the phase and protective conductors
from the intake position to the point at which the loop
impedance is required.
Since an earth
fault may occur at the point farthest from the intake position,
where the impedance of the circuit conductors will be at
their highest value, this is the point which must be considered
when measuring or calculating the earthfault loop impedance
for the installation. Measurement of the impedance will
he considered in {8.6.2}.
Provided that the external fault loop impedance value for
the installation is known, total impedance can be calculated
by adding the external
Table 5.5 
Resistance per metre of copper conductors at 20°C for
 calculation
of R1 + R2 
Conductor crosssectional area
(mm²)

Resistance per metre run
(m ohms / m)

1.0

18.1

1.5

12.10

2.5

7.41

4.0

4.61

6.0

3.08

10.0

1.83

16.0

1.15

25.0

0.727

Note that to allow for the increase
in resistance with increased temperature under fault
conditions the values of {Table 5.5} must be multiplied
by 1.2 for p.v.c. insulated cables (see {Table 8.7})

impedance
to that of the installation conductors to the point concerned.
The combined resistance of the phase and protective conductors
is known as R1+ R2. The same term is sometimes used for
the combined resistance of neutral and protective conductors
(see
{8.4.1}). In the vast majority of cases phase and neutral
conductors have the same crosssectional area and hence
the same resistance.
For the majority
of installations, these conductors will be too small for
their reactance to have any effect (below 25 mm² crosssectional
area reactance is very small), so only their resistance's
will be of importance. This can be measured by the method
indicated in {Fig
5.9}, remembering that this time we are interested in
the combined resistance of phase and protective conductors,
or can be calculated if we measure the cable length and
can find data concerning the resistance of various standard
cables. These data are given here as {Table
5.5}.
The resistance
values given in {Table
5.5} are for conductors at 20°C. Under fault conditions
the high fault current will cause the temperature of the
conductors to rise and result in an increase in resistance.
To allow for this changed resistance, the values in {Table
5.5} must be multiplied by the appropriate correction
factor from Table
{8.7}. It should be mentioned that the practice which
has been adopted here of adding impedance and resistance
values arithmetically is not strictly correct. Phasor addition
is the only perfectly correct method since the phase angle
associated with resistance is likely to he different from
those associated with impedance, and in addition impedance
phase angles will differ from one another. However, if the
phase angles are similar, and this will be so in the vast
majority of cases where electrical installations are concerned,
the error will be acceptably small.
It is often assumed
that higher conductor temperatures are associated with the
higher levels of fault current. In most cases this is untrue.
A lower fault level will result in a longer period of time
before the protective device operates to clear it, and this
often results in higher conductor temperature.
Example 5.1
A 7 kW shower heater is to be installed in a house
fed with a 240 V TNS supply system with an external loop
impedance (Ze) of 0.8 Ohms. The heater is to be fed from
a 32 A BS 88 cartridge fuse. Calculate a suitable size for
the p.v.c. insulated and sheathed cable to be used and determine
the maximum possible length of run for this cable. It may
be assumed that the cable will not be subject to any correction
factors and is clipped direct to a heat conducting surface
throughout its run.
First calculate the circuit
current. 
I

= P =

7000 A

= 29.2 A 

U

240


Next,
select the cable size from {Table
4.7} (which is based on [Table 4D2A]) from which we
can see that 4 mm² cable of this kind clipped direct has
a rating of 36 A. The 2.5 mm² rating of 27 A is not large
enough. We shall assume that a 2.5 mm² protective conductor
is included within the sheath of the cable.
Now we must find
the maximum acceptable earthfault loop impedance for the
circuit. Since this is a shower, a maximum disconnection
time of 0.4 s will apply, so we need to consult {Table
5.1}, which gives a maximum loop impedance of 1.09 Ohms
for this situation. Since the external loop impedance is
0.8 Ohms, we can calculate the maximum resistance of the
cable.
R cable 
= max. loop impedance  external impedance 

=1.09  0.8 Ohms = 0.29 Ohms 
This
assumes that resistance and impedance phase angles are identical,
which is not strictly the case. However, the difference
is unimportant.
The phase conductor
(4 mm²) has a resistance of 4.6 m Ohms/m and the 2.5 mm²
protective conductor 7.4 mOhms/m, so the combined resistance
per metre is 4.6 + 7.4 = 12.0 m Ohms/m. The cable will get
hot under fault conditions, so we must apply the multiplier
of 1.2.
Effective resistance
of cable per metre 
= 12.0 x 1.2 m Ohms/m 

= 14.4 m Ohms/m 
The
maximum length of run in metres is thus the number of times
14.4 mOhms will divide into 0.29 Ohms.
Maximum length of run 
= 0.29 x 1000 m =

20.1 m 

14.4


This
is not quite the end of our calculation, because we must
check that this length of run will not result in an excessive
volt drop. From
{Table 4.7} based on [Table 4D2B] a 4 mm² cable of this
kind gives a volt drop of 11 mV/A/m.
Volt drop = 
11 x circuit current
(A) x length of run (m) divided by 1000 
Volt drop = 
11 x 29.2 x 20.1

V = 6.46 V 

1000


Since
the permissible volt drop is 4% of 240 V or 9.6 V, this
length of run is acceptable.