Number of Earthing Electrode and Earthing Resistance depends on the resistivity of soil and time for fault current to pass through (1 sec or 3 sec). If we divide the area for earthing required by the area of one earth plate gives the number of earth pits required.
There is no general rule to calculate the exact number of earth pits and size of earthing strip, but discharging of leakage current is certainly dependent on the cross section area of the material so for any equipment the earth strip size is calculated on the current to be carried by that strip.
First the leakage current to be carried is calculated and then size of the strip is determined.
For most of the electrical equipment like transformer, diesel generator set etc., the general concept is to have 4 number of earth pits. 2 no’s for body earthing with 2 separate strips with the pits shorted and 2 nos for Neutral with 2 separate strips with the pits shorted.
[pullquote_left]The Size of Neutral Earthing Strip should be capable to carry neutral current of that equipment.
The Size of Body Earthing should be capable to carry half of neutral Current.[/pullquote_left]
For example for 100kVA transformer, the full load current is around 140A.
The strip connected should be capable to carry at least 70A (neutral current) which means a strip of GI 25x3mm should be enough to carry the current and for body a strip of 25×3 will do the needful. Normally we consider the strip size that is generally used as standards.
However a strip with lesser size which can carry a current of 35A can be used for body earthing. The reason for using 2 earth pits for each body and neutral and then shorting them is to serve as back up. If one strip gets corroded and cuts the continuity is broken and the other leakage current flows through the other run thery by completing the circuit.
Similarly for panels the no of pits should be 2 nos. The size can be decided on the main incomer circuit breaker.
Number of earth pits is decided by considering the total fault current to be dissipated to the ground in case of fault and the current that can be dissipated by each earth pit. Normally the density of current for GI strip can be roughly 200 amps per square cam. Based on the length and dia of the pipe used the number of earthing pits can be finalized.
1. Calculate numbers of pipe earthing
A. Earthing resistance and number of rods for isolated earth pit
(without buried earthing strip)
The earth resistance of single rod or pipe electrode is calculated as per BS 7430:
R=ρ/2×3.14xL (loge (8xL/d)-1)
ρ = Resistivity of soil (Ω meter),
L = Length of electrode (meter),
D = Diameter of electrode (meter)
Calculate isolated earthing rod resistance. The earthing rod is 4 meter long and having 12.2mm diameter, soil resistivity 500 Ω meter.
R=500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =156.19 Ω.
The earth resistance of single rod or pipe electrode is calculated as per IS 3040:
ρ = Resistivity of soil (Ω meter),
L = Length of electrode (cm),
D = Diameter of electrode (cm)
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.
B. Earthing resistance and number of rods for isolated earth pit
(with buried earthing strip)
Resistance of earth strip (R) As per IS 3043:
R=ρ/2×3.14xLx (loge (2xLxL/wt))
Calculate GI strip having width of 12mm , length of 2200 meter buried in ground at depth of 200mm, soil resistivity is 72.44 Ω-meter.
- Resistance of earth strip(Re) = 72.44/2×3.14x2200x(loge (2x2200x2200/.2x.012)) = 0.050 Ω
- From above calculation overall resistance of 60 no of earthing pipes (Rp) = 0.133 Ω.
And it connected to bury earthing strip. Here net earthing resistance = (RpxRe)/(Rp+Re)
- Net eatrthing resistance = (0.133×0.05)/(0.133+0.05) = 0.036 Ω
C. Total earthing resistance and number of electrode for group
In cases where a single electrode is not sufficient to provide the desired earth resistance, more than one electrode shall be used. The separation of the electrodes shall be about 4 m. The combined resistance of parallel electrodes is a complex function of several factors, such as the number and configuration of electrode the array.
The total resistance of group of electrodes in different configurations as per BS 7430:
Ra=R (1+λa/n) where a=ρ/2X3.14xRxS
S = Distance between adjustment rod (meter),
λ = Factor given in table below,
n = Number of electrodes,
ρ = Resistivity of soil (Ω meter),
R = Resistance of single rod in isolation (Ω)
|Factors for parallel electrodes in line (BS 7430)|
|Number of electrodes (n)||Factor (λ)|
For electrodes equally spaced around a hollow square, e.g. around the perimeter of a building, the equations given above are used with a value of λ taken from following table.
For three rods placed in an equilateral triangle, or in an L formation, a value of λ = 1.66 may be assumed.
|Factors for electrodes in a hollow square (BS 7430)|
|Number of electrodes (n)||Factor (λ)|
For Hollow square total number of electrodes (N) = (4n-1).
The rule of thumb is that rods in parallel should be spaced at least twice their length to utilize the full benefit of the additional rods. If the separation of the electrodes is much larger than their lengths and only a few electrodes are in parallel, then the resultant earth resistance can be calculated using the ordinary equation for resistances in parallel.
Typically, a 4 spike array may provide an improvement 2.5 to 3 times. An 8 spike array will typically give an improvement of maybe 5 to 6 times.
The Resistance of Original Earthing Rod will be lowered by Total of 40% for Second Rod, 60% for third Rod,66% for forth rod.
Calculate Total Earthing Rod Resistance of 200 Number arranges in Parallel having 4 Meter Space of each and if it connects in Hollow Square arrangement. The Earthing Rod is 4 Meter Long and having 12.2mm Diameter, Soil Resistivity 500 Ω.
First Calculate Single Earthing Rod Resistance:
- R = 500/ (2×3.14×4) x (Loge (8×4/0.0125)-1) =136.23 Ω.
Now calculate total resistance of earthing rod of 200 number in parallel condition:
- a = 500/(2×3.14x136x4) =0.146
- Ra (Parallel in Line) =136.23x (1+10×0.146/200) = 1.67 Ω.
If earthing rod is connected in Hollow square than rod in each side of square is 200 = (4n-1) so n = 49 No.
Ra (in hollow square) =136.23x (1+9.4×0.146/200) = 1.61 Ω.
Originally published at – Electrical Notes – Calculate Numbers of Plate/Pipe/Strip Earthings (Part-1)
There are a lot of mistakes in either the given Formulas or the calculations. So please, get them corrected.
I think it is improper to calculate the size of earthing strip on the basis of rating of main incomer.The short ckt rating of the system should be taken into consideration.The earthing strip should be able to carry fault current for 1 sec.
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Our Company is construction company, and we construct building by latest Technic by using puff i.e metallic wall, so please tell me how much earth pit required as per area.
why same unit is not used for length and diameter in below formula?
R = 100×72.44 /2×3.14x300x(loge (4X300/10))
R = 100×72.44 /2×3.14x300x(loge (4X300/10)) can’t find the same answer as yours. How do you get the answer?
Dear Sir , In above problems , in formulae there is natural logarithm i.e. ln , what in calculation you are using log base 10 ???????
Sir i have a doubt regarding the earthing.
One of my work site, the earth resistance requirement is less than 3 ohm. The site is a low tension project. I don’t know how to calculate the number of earth pit for that site. Please tell me the equation to find out the no. of earth pit required. The power is available from electricity board transformer having 160 kVA. Please help me.
You wrote that for 100kVA transformer, 70A would be neutral current which means a strip of GI 25x3mm should be sufficient. But, as i know Strip should be sized as per neutral fault current because I is fualt current in strip sizing formula as per IS 3043.
Formula : S=I x root t / K
The Formula you have mentioned is for Short circuit current. I believe the size of the copper tape was selected based on its ‘Ampacity’ (Capacity to carry current). This information can be found in the catalog.
Hi. Very useful info, thank you. I think there may be an error in how you interpreted the loge formula for the calculation of electrode resistance. Your calculations work out if you use log10 but not natural log (log e). Please re-check
Dear Mr. Raj,
Pls advise how you got Lambda λ equals to 10 from the equation of the Ra:
Ra (Parallel in Line) =136.23 x (1+10×0.146/200) = 1.67 Ω.
Usefull information every time getting from your articles.
I need information on Cu. bonded earth electrode and GI earth electrode fault current carrying capacity.
We are facing issues of GI earth electrode corrosion so we need to change with Cu. bonded earth electrode with carbon base back fill compound.
Nowadays chemical type Earth Electrodes are popular.
I want to design an Earthing system based on chemical type Earth Electrodes, for which no standard is available..
Can I get guidance for the design?
DILIPKUMAR D DESAI
Thanks for your question.
Chemical Maintenance free earthing is based on the IEC standrad 62561-1,2 & 6. The seletion of earth electrode is as per the standrad current rating as well as fault current. May I know for what rating of equipment you are going to use. regards
hello, please tell me how did you get the lamda factor 10 for 200 electrodes in the last example. The table has no.s up to 20 electrode for hollow square shape. thanks
V Nice work done !! congratulations to share the knowledge. Quick note the values I got for the calc is 136 not 156.