Page 380 - Handbook of Electrical Engineering

368 HANDBOOK OF ELECTRICAL ENGINEERING

below that has the resistivity (ρ). This term is:-

R 2Fs = 6(ρ s )C s (h s ,K) for use in the step voltage case, and
R 2Fp = 1.5(ρ s )C s (h s ,K) for use in the touch voltage case

Where : − h s = the thickness of the crushed rock layer
K = (ρ − ρ s )/(ρ + ρ s ).

Which is negative when the upper layer is more resistive then the lower layer. If no crushed
rock is used then C s (h s ,K) = 1. The resistances R 2Fs and R 2Fp are added to the 1000 ohms in
equation (13.2) and the resulting threshold voltages are then denoted as E step50 and as E touch50 (using
the same notation as in IEEE80). The function C s (h s ,K) is derived from a convergent infinite series
and can be expressed as:-

n=∞ n
1 K
C s (h s ,K) = 1 + 2 (13.3)
0.96 u s
n=1
Where

2
2nh s
u s = 1 +
0.08
13.3.5.2 Soil resistivity

Soil resistivity varies greatly with the material, e.g. rocks, sand, clay, and its moisture content, as
in coastal areas, high annual rainfall, dry deserts. Table 1 of BS7430 gives comprehensive values
for these variations. For dry desert conditions a value of 1000 ohm-metres is generally considered
acceptable for design calculations, unless site measurement data are available. Table 3 of IEEE80
gives typical values of crushed rock that would be used as a surface layer, and recommends in its
sub-section 10.5 a value of 3000 ohm-metres for a wetted layer. Hence a dry layer would be very
7
6
much higher e.g. 10 to 10 ohm-metres.
13.3.5.3 Resistance to earth
The resistance to earth R e as measured or calculated for a conductor buried in the ground depends
upon its shape, volume and orientation in the ground. In favourable conditions the resistance should
be less than one ohm. With unfavourable conditions and small sites such as the bases of pylons a
value between 1 and 5 ohms should be considered. For simple shapes such as uniform rods, strips
and plates, there are formulae available for calculating the resistance. For example a vertical round
rod or hollow pipe the resistance is:-
ρ 8L
R e = log e − 1 ohms (13.4)
2πL d

Where ρ is the soil resistivity in ohm-metres
L is the buried length of the rod or pipe in metres
d is the diameter of the rod or pipe in metres.