# Effect of earthing electrodes configuration on Earth Resistance Continued from How to Determine correct number of earthing electrodes

Contents

## Effect of Different configuration of Earthing Pipes on Value of Earth Resistance

In case single Electrode does not provide desired earth resistance value, then more than one electrodes are used. The separation of electrodes should be 4m from each other. These electrodes are arranged in different configurations. The total resistance of group of electrodes in different configuration as per BS7430 are \\\;\\R_{a}=\frac{R (1+\lambda a)}{n}\\\;\\Where \\\;\\a =\frac{\rho}{2\pi RS}

S = Distance between adjacent rods (meter)\\\lambda= factor given in a table\\n = number of electrodes\\\rho=resistivity of soil (ohm meter)\\R = Resistance of single rod (ohm)

## Value of λ as per BS7430

Factors for parallel electrodes in Line (BS 7430)

For Electrodes equally Space around a hollow space e.g. around the perimeter of building, the value of λ will be taken from the table given below.

For three rods place in equilateral or in L formation value of λ = 1.66

For Hollow Square total number of electrodes (N) = 4n – 1

where n = number of electrodes on each side

Example: 200 number of earthing rods are arranged in parallel around a hollow square with separation of 4 meter to each other. The length of each earthing rod is 4 meters and having diameter of 12.2mm. the resistivity of soil is 500Ω. Calculate total earth resistance.

Earth Resistance of single Rod\\\;\\R=\frac{\rho}{2\pi L}\times \left [log_e \left(\frac{8\times L}{d} \right)-1 \right ]

Earth Resistance of single Rod\\\;\\R=\frac{500}{8\pi }\times \left [log_e \left(\frac{8\times 4}{0.0125} \right)-1 \right ]=136.30 \Omega

Total Resistance of 200 earthing rods in parallel conditions:

a = \frac{\rho}{2 \pi R S} = \frac {500}{2 \pi \times 136.30 \times 4} = 0.146 \\\;\\R_{a} = \frac{R \times \left(1+ \lambda a \right)}{n}\\\;\\R_{a}= \frac{136.30 \times \left(1+(10 \times 0.146) \right)}{200}=1.67 \Omega

If the Rods are arranged around a hollow square than rod on each side is N = 4n-1 \\200 = 4n-1\\n=49 No.

Ra (In hollow square ) = \frac{136.30 \times \left(1+(9.4 \times .146) \right)}{200}=1.62\Omega

## Cross Section Area of Earthing Conductor

As per the IS 3043 the area of earth conductor is \\A\;=\;\frac{I_{f}\; \times \sqrt{t}}{K}sq.mm

where\\ If = fault current (Amperes)\\ t = time of fault (seconds)\\K = Material Constant (For GI = 80, Copper = 205, Aluminium = 126)

Example: Calculate cross section area of GI earthing conductor for system has 50KA fault current for 1 second. Corrosion will be 1.0% per year and number of year for replacement is 20 years.

Cross section Area of Conductor, \\A\;=\;\frac{I_{f}\; \times \sqrt{t}}{K}\\Here\\If = 50000 A, t = 1 second, K= 80

1. A = \frac{50000\times \sqrt{1}}{80} = 625sq.mm
2. Allowance for Corrosion = 1.0% per year and Number of year before replacement, say = 20 years
3. Total allowance = 20 \times1% =20%
4. Safety factor = 1.5
5. Required earthing conductor size = Cross sectional Area \times\left(1+Total\;Allowance\right)\timesSafety factor
6. Required Earthing conductor size = 625 (1 .20) x 1.5 = 1125 Approximately 1200sq.mm
7. Hence consider 1No x 12 x 100mm GI strip or 2 No x 6 x 100mm GI Strips