Calculate Numbers of Plate/Pipe/Strip Earthings (Part1)
March 8, 2013 25 Comments
Introduction:
 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 no of Earth pits required.
 There is no general rule to calculate the exact no 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 equipments like Transformer, DG set etc., the General concept is to have 4 no 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.
 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.
 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 Breaker.
 For example if main incomer to breaker is 400A, then Body earthing for panel can have a strip size of 25×6 mm Which can easily carry 100A.
 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 & No of Rod 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)
 Where ρ=Resistivity of Soil (Ω Meter),
 L=Length of Electrode (Meter),
 D=Diameter of Electrode (Meter)
 Example:
 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:

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 ΩMeters.
 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 & No of Rod 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)).
 Example:
 Calculate GI Strip having width of 12mm , length of 2200 Meter buried in ground at depth of 200mm,Soil Resistivity is 72.44 ΩMeter
 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 Pipe (Rp) = 0.133 Ω. And it connected to bury Earthing Strip. Here Net Earthing Resistance =(RpxRe)/(Rp+Re)
 Net Earthing Resistance= =(0.133×0.05)/(0.133+0.05)= 0.036 Ω
(C) Total Earthing Resistance & No of Electrode for Group of Electrode (Parallel):
 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 Electrode in different configurations as per BS 7430:

Ra=R (1+λa/n) Where a= ρ/2X3.14XRXS
 Where S= Distance between adjustment Rod (Meter),
 λ =Factor Given in Table,
 n= Number of Electrode,
 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) 
2 
1.0 
3 
1.66 
4 
2.15 
5 
2.54 
6 
2.87 
7 
3.15 
8 
3.39 
9 
3.61 
10 
3.8 
 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) 
2 
2.71 
3 
4.51 
4 
5.48 
5 
6.13 
6 
6.63 
7 
7.03 
8 
7.36 
9 
7.65 
10 
7.9 
12 
8.3 
14 
8.6 
16 
8.9 
18 
9.2 
20 
9.4 
 For Hollow Square Total Number of Electrode (N) = (4n1).
 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.
 In practice, the effective earth resistance will usually be higher than Calculation. 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
 Example:
 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=(4n1) so n=49 No.
 Ra (In Hollow Square) =136.23x (1+9.4×0.146/200) =1.61 Ω.
HI Sir,
Thank you for the valuable info.. can u please forward documents required for IES exam.
Rgds,
Sravanthi.D
Dear Sir,
My question is not related to the topic.
It is related to cable sizing
How will we choose the cable size in sqmm if only the ampere is given ,with out the help of any charts
i mean like a thumb rule.
i did’nt see any provision to ask questions in the website that is why i put my query here.
I hope you will give me the answer.
Thanks
Rahul
=For selection of cable is totally depends on Distance,Load,Type of media.
You can not choose cable without considering above factors.
Ex. If you have load of 10KW at 10 meter Distance and Same Load at 200 meter Distance in both case Size of Cable is different.
= without any chart you can not decide size of cable however as a thumb rule we can say that
(1) for Copper wire (Up to 30 sq.mm) Amp capacity of Wire = 6Xsq.mm.
Ex. for 1.5 Sq.mm Wire, Amp Capacity =6X1.5 =App 9 Amp
(2) For Cable, Amp Capacity of Cable = 4X Size of Cable (Sq.mm).
Ex. 2.5 Sq.mm Cable = 4X2.5=10 Amp
Mr. Jignesh Kumar,
Thank you so much for keeping such a helpful blog active.
Hi Sir
How to calculate the grounding rods which is copper clad material. And I many doubts in Earthing. How can I convey that to you sir!
Hi sir,
I am very much confusing about battery calculation required for solar power plant, please suggest ideal battery calculation. Also is it possible to keep system voltage and battery voltage different
The calculation seems no applicable to MEN earthing system.
The most critical earth fault path is on the MEN link between the earth bar with the main earthing conductor and the neutral bar of the Main Switchboard.
Not getting your Point.
This is for TT Type Power Distribution System
You are right the calculation is limited to TT type but suitable for MEN earthing system.
dear sir, earth pit different size using flow chart 100kva , 200kva , 500kva ,,,,
Thanks a lot sir, for your information.
sir can u tell me plzz
how many earth pits are required for HT panel body and neutral?
It is already given in Blog for Panel Body earthing 2 Nos GI and for Neutral (TC) 2 Nos Copper..Please use “search” option of this Blog
Dear sir ,
My question is how to calculate earthing strip size for DG and transformer…..
Already given Excel Sheet and Calculation method with example in this Blog..
Dear Jignesh, I was just browing for some data for earthing pits and fortunately I got the reference of this blog. I am very much impressed about the the way you are sharing knowledge and keeping time to help others.
Sir, I have one query at this moment..I was reading “Calculate Numbers of Plate/Pipe/Strip Earthings (Part1)” which are unequivocally explicit. but please tell me how to calculate ‘FAULT Current’…Regards, Rahul Goel (Faridabad)
Can I have common earth pits two nos. for body of DG set and body of transformer . For neutrals I will have seperate earth pits .
You can make grid of Body earthing of any Equipments.
Sir, I want Know about fundametal requirments Tan / Delta Testing, limitation Value, Before to value, on meachine, is there any aspected Calculation regarding the mechine (Tansformer, Motor, Generator) or any IS, IEC, IEA.referance.
Dear Sir, i just want to know is there any formula for calculating specific resistance of soil or it is taken from the table.
Thank you, Sir i like to know testing methods of Current transformer. In my past experience i used primary injection method with loading transformer setup and Omicron. how omicron used to determine Vk, Imag, ISF, ALF etc., Provide any document available.
Mr Jignesh, Why are we considering the short circuit current rather than the Earth fault current for the sizing of earthing strip ?
Secondly, is it not necessary to verify the step and touch potential? why?
Sir,
Please let me the best brand ; quality wise available. A typical Dia is 88 mm and the length is 3 meter.
Regards
Kane
Dear sir,
i have a query kindly help me, i have installed a mechanical equipment with motor of 15kw and have given connection through a panel, i have given earthing strip of gi to the panel from the mechanical equipment through cable tray, won’t the leakage of current pass through the cable tray where i have laid the gi strip? I am a student of different stream so kindly help me with this
Max. Allowable Current Density I = 7.57×1000/(√ρxt) A/m2 . In this formula what is “7.57” factor? what will be the formula for Copper & G.I earthing plate? How do you calculate Max. allowable current density of Cu or G.I strip or plate??