Page 596 - Handbook of Electrical Engineering

EARTHING CURRENT AND ELECTRIC SHOCK HAZARD POTENTIAL DIFFERENCE 589

• Use the method described in IEEE80 sub-section 14.4 even though it is more applicable to much
larger sites, but include the earthing rods.
• If possible limit the maximum resistance to earth at the site to 5 ohms.

Step 3. Find the resistance R ep at the pole.

The following calculations are based on the methods given in IEEE80 Appendix C. The same
symbols and notation are generally used to avoid confusion with the reference. The design data and
constraints are:-

• Fault duration t s = 0.5 sec
• Resistivity of lower layer ρ = 100 to 1000 ohm-m
• Resistivity of upper layer ρ s = 1000 to 5000 ohm-m
• Thickness of upper layer h s = 0.2 and 1.0 m
• Depth of burial of grid h = 0.5 and 1.0 m
• Site area A = 36 to 256 m 2
• Diameter of rods d r = 0.02 and 0.2 m
• Depth of each rod l r = 10 and50m
• Number of meshes in each side of the grid N mesh = 3to 8
• Spacing between the mesh nodes d sp = 2.0 m

The results of the calculations are shown in Table H.1c; Case C.3 is used for the worked
example. In this case the following additional information was used:-

• Number of outer peripheral rods N rod1 = 4
• Number of inner rods N rod2 = 0
• Number of rods on each side of grid N = 2
• Diameter of grid conductors d m = 0.01 m

√
Calculate the resistivity derating factor C s {h s ,K} from (13.3), in which u s = (1 +
2
(2mh s /0.08) ). The number of terms m is taken to be 25 in order to obtain good convergence
of the factor. The reflection factor K is found from ρ and ρ s to be −0.6667 per-unit. C s {h s ,K} is
found to be 0.8338 for this example.
The approximate grid resistance R epo without the earthing rods can be found from Figure H.1a,
which was derived from Figure B.1 of IEEE80 but applicable to small sizes of the mesh R epo ,is
approximately:-
ρ51.94
R epo = = 51.94 ohms,
1000
which is too high and indicates the need for rods.

At this stage the 50 kg step and touch voltages can be calculated from C s {h s ,K},since ρ s
and t s are constants. The step voltage E step50 is: -

(1000 + 6C s ρ s ) 0.116
E step50 = √ = 4267 volts.
t s