Earthing Resistance calculation Analysis Based on British Standard BS 7430:1991
A. Selection of earthing conductor and connection to an electrode
I t
S ----------------------BS 7430, Page 28
k
Where
I = Fault current, in amperes
t = Fault current duration, in seconds.
S = Conductor cross sectional area, in mm2.
2500
I= = 60140 Amperes.
3x0.4 x0.06
Where
2500 = Transformer KVA Rating
0.06 = Transformer impedance
0.4 = Transformer Secondary KV
K = 176 From table 10 page 29 based on maximum temperature of 250°C and one (1) second
fault duration.
So, S=
60140 1
= 341mm2
176
Use 2 x 300 mm2 copper conductor to each ground bus.
B. Resistance of one earth rod
p
R= ---------------BS 7430, Page 13
2l
Where
L = Length of rod, in meter = 3m
d = Diameter of rod, in meter = 0.02m
p = Resistivity of the soil, in ohm. m
R= Resistance of one rod in ohm.
Based on the nature of the soil of this project, 80.67 ohm.m soil resistivity may be assumed
and used.
Refer to table 1 page 11.
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So, R= )-1]
= 4.28 [
= 26.1 ohm
C. Combined resistance of number of rods in parallel
Rn = R ---------------- BS 7430, Page 15
In which a = ------------- BS 7430, Page 15
Where
Rn = Resistance of n rods in parallel, in ohm.
R = Resistance of one rod, in ohm.
S = Distance between adjacent rods, in m= 20 m.
p = Resistivity of soil, in ohm. m = 80.67 ohm. m.
λ = Factor given in table 3, Page 15.
n in table 3 is given by (Total number of electrode on the rectangle / 4 +1)
n= +1) = 7
λ = 7.03 for 7 rods
a= = 0.024
So, Rn = 26.1 x(
D. Resistance of horizontal Underground earth two conductors at 90o
R= ---------------BS 7430, Page 19
pL
Where
L = Length of conductor, in meter = 80m.
H = Depth of conductor, in meter = 0.8m.
w = Diameter of conductor, in meter = 0.0223m, for 300 conductor selected under (A).
p = Resistivity of soil, in ohm. m = 80.67 ohm. m
P and Q = Coefficients given in table 5 Page 19.
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P=4
Q = 0.9 for two conductors at 90o
Where
L = Length of one Conductors in meter = 80m.
So, R= = 1.15
There are two of horizontal underground earth two conductor at 90 o are connected in parallel each
one has earthing resistance value of 1.15 ohm.
So, the final total earthing resistance shall be less than the value of 1.15 ohm.
E. Conclusion
1. Earthing resistance value estimated by formulas under items (C) above is 1.2 ohm as a result
of use number of earth rods connected in parallel regardless the size of conductor used for
interconnection.
2. Earthing resistance value estimated by formulas under (D) above is 1.15 ohm as a result of use
horizontal conductors only without the effect of earth rods in the calculations
3. Combination of the two values estimated under items (C) and item (D) shall be less than any
individual value of (C) or (D) because they are naturally connected in parallel.
4. So, the final total earthing resistance shall be less than the smallest value of 1.2 and 1.15 ohms
i.e. less than 1.15 ohms.
F. The maximum permissible current density
The maximum permissible current density J, in amperes per square meter (A/m2), is given by the following
equation:
J=
where:
t is the duration of the earth fault, in seconds (s);
p is the resistivity of the soil, in ohm meters (Ωm).
J= 845.57 A/m2
G. The maximum fault current
The maximum fault current which the earthing network can carry
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= J x Area
= Jx DL = 845.57x x 0.0223 x 1400 = 82.9 KA
The maximum fault current which the Earthing network can carry > fault current of the transformer.
The design of Earthing network is satisfied.
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