Page 600 - Handbook of Electrical Engineering

EARTHING CURRENT AND ELECTRIC SHOCK HAZARD POTENTIAL DIFFERENCE 593

Calculate the constants K h , K ii and K m for use in equation 68 from IEEE80.

0.5
K h = (1 + h) = 1.2247

Use the following auxiliary equations to simplify the work:-
2
d sp
U 1 = = 50.0
16hd m
(d sp + 2h) 2
U 2 = = 56.25
8d sp d m
h
U 3 = = 12.5
4d m
8
U 4 = = 0.2315
π(2n − 1)
K ii
U 5 = = 0.8165
K h
Where K ii in this example is 1.

log (U 1 + U 2 − U 3 ) + U 5 log (U 4 )
e
e
K m = = 0.5325
2π
Also from the explanation in sub-section 14.5.1 in IEEE80 the correction factor K i is required,
which is:-
K i = 0.656 + 0.172N n = 1.688
Where, N n = 6 – number of parallel conductors in each direction of the grid, which equals the
number of nodes on each side of the grid.
K h = 1.2247 – correction factor for the depth of the grid.
K ii = 1.0 – correction factor if the rods are placed inside the grid area.
K m = 0.5325 – spacing factor for the mesh voltage.
K i = 1.688 – correction factor for the grid geometry as a function of the number of nodes
on each side of the grid.

Having found R ep , K m and K i it is now possible to find the mesh voltage E m as follows.
The resistance R ep is substituted into the fault current equations (H.1.1) and (H.1.2), to give the total
fault current I f . The earth return circuit between the pole at point A in Figure 13.12 and the earthing
connection at point B at the source is a parallel circuit of the resistances to earth R en and R ep and
the overhead earth return line impedance Z eoh . The parallel combination is:-
Z eoh (R en + R ep )
Z e = = 11.019 + j5.5597 ohms
Z eoh + R en + R ep
Where
Z eoh = route length × Z 0pkm = 10.5 + j22.5 ohms.