Chapter A General rules of electrical installation design

G32
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G - Sizing and protection of conductors
In practice this means that the length of circuit
downstream of the protective device must not
exceed a calculated maximum length:
0.8 U Sph
Lmax
=
2ρI m
(1) For larger c.s.a.’s, the resistance calculated for the
conductors must be increased to account for the non-uniform
current density in the conductor (due to “skin” and “proximity”
effects)
Suitable values are as follows:
150 mm 2 : R + 15%
185 mm 2 : R + 20%
240 mm 2 : R + 25%
300 mm 2 : R + 30%
(2) Or for aluminium according to conductor material
(3) The high value for resistivity is due to the elevated
temperature of the conductor when passing short-circuit
current
5 Particular cases of short-circuit
current
Practical method of calculating Lmax
The limiting effect of the impedance of long circuit conductors on the value of
short-circuit currents must be checked and the length of a circuit must be restricted
accordingly.
The method of calculating the maximum permitted length has already been
demonstrated in TN- and IT- earthed schemes for single and double earth faults,
respectively (see Chapter F Sub-clauses 6.2 and 7.2). Two cases are considered
below:
1 - Calculation of L max for a 3-phase 3-wire circuit
The minimum short-circuit current will occur when two phase wires are shortcircuited
at the remote end of the circuit (see Fig. G46).
Schneider Electric - Electrical installation guide 2008
P
0.8 U
Fig G46 : Definition of L for a 3-phase 3-wire circuit
L
Load
Using the “conventional method”, the voltage at the point of protection P is assumed
to be 80% of the nominal voltage during a short-circuit fault, so that 0.8 U = Isc Zd,
where:
Zd = impedance of the fault loop
Isc = short-circuit current (ph/ph)
U = phase-to-phase nominal voltage
For cables y 120 mm2 , reactance may be neglected, so that
2L
Zd = ρ (1)
Sph
where:
ρ = resistivity of copper (2) at the average temperature during a short-circuit,
Sph = c.s.a. of a phase conductor in mm 2
L = length in metres
The condition for the cable protection is Im y Isc with Im = magnetic trip current
setting of the CB.
This leads to Im y
0.8 U
which gives L y 0.8 U Sph
Zd
Lmax
=
2ρI m
with U = 400 V
ρ = 1.25 x 0.018 = 0.023 Ω.mm 2 /m (3)
Lmax = maximum circuit length in metres
k Sph
L
m
max = I
2 - Calculation of Lmax for a 3-phase 4-wire 230/400 V circuit
The minimum Isc will occur when the short-circuit is between a phase conductor and
the neutral.
A calculation similar to that of example 1 above is required, but using the following
formulae (for cable y 120 mm2 (1) ).
b Where Sn for the neutral conductor = Sph for the phase conductor
3,333 Sph
Lmax
=
Im
b If Sn for the neutral conductor < Sph, then
L 6,666 Sph 1
max =
where m =
Im
1+ m
Sph
Sn
For larger c.s.a.’s than those listed, reactance values must be combined with those of
resistance to give an impedance. Reactance may be taken as 0.08 mΩ/m for cables
(at 50 Hz). At 60 Hz the value is 0.096 mΩ/m.