how to calculate the earthing conductor & no. of earth pits required

how to calculate the earthing conductor & no. of earth pits
required

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how to calculate the earthing conductor & no. of earth pits required ..

Answer / mrutyunjay mohanty

Dear Sir,

Greetings of the day !

Recently I have been through one answer shared by you and got your email id at AllInterview.com. Please kindly help me in my below concern :

Permissible current density equation i = (7.57*1000) / sqrt(resistivity * duration of earth fault)

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how to calculate the earthing conductor & no. of earth pits required ..

Answer / mrutyunjay

Dear Sudhakar Sir,

Greetings of the day !

Recently I have been through one answer shared by you and got your email id at AllInterview.com. Please kindly help me in my below concern :

Permissible current density equation i = (7.57*1000) / sqrt(resistivity * duration of earth fault) as per IS 3043

Is this equation to be taken for all types of electrodes ( Plate, Rod or Pipe & Strip ) ?

Because , as per IS 3043, same equation is appropriate for plate electrode

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how to calculate the earthing conductor & no. of earth pits required ..

Answer / sudhakar.g

The earth resistance of single rod or pipe electrode is calculated as per IS 3040:
R=100xρ/2×3.14xL (loge(4xL/d))
Where:
ρ = Resistivity of soil (Ω meter),
L = Length of electrode (cm),
D = Diameter of electrode (cm)

Example:
Calculate number of CI earthing pipe of 100mm diameter, 3 meter length. System has fault current 50KA for 1 sec and soil resistivity is 72.44 Ω-Meters.
Current Density At The Surface of Earth Electrode (As per IS 3043):
• Max. allowable current density I = 7.57×1000/(√ρxt) A/m2
• Max. allowable current density = 7.57×1000/(√72.44X1) = 889.419 A/m2
• Surface area of one 100mm dia. 3 meter Pipe = 2 x 3.14 x r x L = 2 x 3.14 x 0.05 x3 = 0.942 m2
• Max. current dissipated by one Earthing Pipe = Current Density x Surface area of electrode
• Max. current dissipated by one earthing pipe = 889.419x 0.942 = 837.83 A say 838 Amps
• Number of earthing pipe required = Fault Current / Max.current dissipated by one earthing pipe.
• Number of earthing pipe required = 50000/838 = 59.66 Say 60 No’s.
• Total number of earthing pipe required = 60 No’s.
• Resistance of earthing pipe (isolated) R = 100xρ/2×3.14xLx(loge (4XL/d))
• Resistance of earthing pipe (isolated) R = 100×72.44 /2×3.14x300x(loge (4X300/10)) = 7.99 Ω/Pipe
• Overall resistance of 60 no of earthing pipe = 7.99/60 = 0.133 Ω.

If you have any doubt regarding this ...pls mail me sudhakar28485@gmail.com