In this article, we will discuss how to determine the number of earth pits, the size of the earth grid conductor, and the overall resistance of the earthing system.

## Electrical Earthing Calculations

Sizing of earth grid conductor shall be done by considering fault current 1 sec. The uniform soil model assumption is based on a 3-meter long pipe electrode used at this installation which cannot penetrate to the second layer of lower soil resistivity.

The material of the earth grid conductor shall be galvanized iron (GI). GI strip shall be considered as the main earth conductor and earth electrodes shall be 80 mm dia pipe meeting the requirements of IS 3043.

The maximum value of a single line to earth fault current is considered as 80 kA. Based on soil resistivity testing at various locations, average soil resistivity **12.46 ohm-m **shall be considered for earthing system calculation which is done by wenner’s method.

Electric Resistivity in ohm-m = 2πRS

Where,

S: Distance between two electrodes in meter (m)

R: V/I: Resistance in Ohm Ω

## Earth Conductor Sizing

The cross-sectional area of the earth conductor shall be such that the current density value determined by the following formula is not exceeded.

**(I/S) = k (1/√ t)**

Where;

I = Value of fault current which can flow through the protective device in amperes

S = Cross sectional area in sq.mm.

k = Factor dependent on the material of the protective conductor, the Insulation and other parts, and the initial and final temperatures.

t = Operating time of the disconnecting device in seconds

The factor k is determined from the formula:

**k = sqrt [{Qc (B+20) / δ****20)} x ln {1+ (θ****f – θ****1) / (B + θ****1)}]**

Qc = Volumetric heat capacity of conductor material (J/ ºC mm3)

B = Reciprocal of temperature coefficient of Resistivity at 0ºC for the conductor (0ºC)

δ20 = Electrical Resistivity of conductor material at 20ºC (Ω-mm)

θ1 = Initial temperature of conductor (ºC)

θf = Final temperature of conductor (ºC)

For steel conductor;

Qc = 3.8 x 10-3 (J/ ºC mm3)

B = 202 ºC

δ20 = 138 x 10-6 (Ω-mm)

θ1 = 40 ºC

θf = 500 ºC

Put all values as given in equation, so we get,

**k = 80.68 ≈ 80**

Now, let’s find the cross-sectional area of the earth strip

S = (I x √ t) / k

(80000 x 0.6 x√ 1) /80 [As a total 60% of fault current will flow through the Earth conductor and the remaining 40% will flow through the soil. This is as per BS 7354 ]

**S = 600 mm2**

Hence S = 600 X 1.15 = 690 sq.mm. (Considering 15 % corrosion margin)

Hence GI Bar of size 36 X 36 sq. mm. size is considered adequate for the main buried earth grid conductor for the plant area.

## Surface Area of Solid Rod Electrode

Ae = (π x D x l) m2

Where,

D = Diameter of electrode (m)

l = Length of electrode (m)

Ae = (π x 0.08 x 3.00) m2 (for 80 mm diameter electrode)

**Ae = 0.8294 m2**

## Number of Earth Electrodes

Number of earth electrodes required to dissipate the fault current of 80kA are

Area of electrodes required to dissipate fault current** / **Area of each electrode

Number of earth electrodes = 48.49 /0.8294

Number of earth electrodes = 58.46 Nos. (**≈ ** 60 Nos.)

## Resistance of Electrode

R = (1/2 πl) x [(ρ – ρc) (ln (8L / D) – 1) + ρc (ln (8L / d) – 1)]

Where,

ρ = Resistivity of soil in Ω-m

ρc = Resistivity of infill material in Ω-m

d = Diameter of Electrode in meters

D = Diameter of infill material in meters

L = Driven length of electrode in meters

R = (1/ 2 π x 3) x [ (12.46 – 3) (ln (8×3/0.45)-1) + 3 (ln ((8×3/0.08)-1)]

R = 0.59 Ω

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Great

To calculate no of rods u use area required to dissipate fault current by 48 though it is calculated by 600mm2,how

I WANT SEE THE REPLY