Earthing BS 7430 - Vertical Rods in a Hollow Square

This thread is surprising, because the answers to the questions are best obtained by applying the content of the standard.

thank's 67model

you said "Given the practical variations caused by earth resistivity, tabling greater than 20 is a pointless illusion of accuracy."

when you have the value of resistivy of soil is 3000Ω.m and I want get 5 Ω as resistance and you have rectangle 30m length and 20m width how you can do it without use number of electrode greater than 20 ?

BS7621:2018 "Requirements for Electrical Installations" Table 54.1 gives minimum 25 sq.mm copper [50mm2 steel] for buried earthing conductors <not protected against corrosion>. Your national electrical code may have a different minimum.

I think you mean BS 7671 not BS7621 I didn't find BS 7621 ?

Cordialy

Absolutely right, it was BS7671 [which does show that giving the title as well as the number reduces errors - you found the correct one].

And I read 2 into the formula not lambda (λ) [I should have realised that λ must be in the formula]. Unfortunately, CR4 cuts the resolution when you give it a complex picture.

If I may go through a thought experiment on R_TOT formula.....

R_{TOT = {R/N}*[1 + λα] - but λ has declining effect with no. of rods - say it has maximum 10.}

_{Using → to mean <converges to> [symbol is right-pointing arrow? but not visually distinct]}

_So R_{TOT → {R/N}*[1 + 10α] - but α ≅ 0.53/s for 50mm diameter 3m deep galvanised iron rod which is 30 ohm in 100 ohm-m soil (ρ cancels out in α, because R is proportional to ρ).}

_So R_{TOT → {R/N}*[1 + (5.3/s)] - but if s is small, the 1 in [brackets] can be discounted.}

_So R_{TOT → {R/N}*[5.3/s] - but if B is the boundary length of the rectangle of rods, N is nearly B/s.}

_So R_{TOT → {R/B}*5.3 --- {a}}

_{So, even if you bang in lots of rods, the only options to get lower resistance are reducing R or increasing B.}

_{Maybe your boundary cannot be increased, but maybe R can be reduced by longer rods which penetrate to water table or different kind of soil.}

_{Your extreme 3000 ohm-m corresponds, in my limited delvings, to rock. It is worth looking at the basic resistance of a spherical metal electrode (a half-buried sphere). Taking your site & supposing a 15m radius r [the neighbours will not mind the overlap, "bonding" improves the Earth].}

_{R = ρ/2πr - for r = 15m, ρ =3000 Ω-m, π = 3.14;} R = 3000/30π = 32 ohm.

It seems that getting a resistance of 5 ohm is practically impossible.

That hemisphere has a surface of 1413 sq.m - even if you can apply the whole site being a concrete pad of relatively high conductivity filled with steel reinforcing bars (mentioned by BS7430), its area is only 600 sq.m, so greater than 32 ohm above.

I think your only chance for a miracle is porous rock filled with salty water not far under the surface rock.

Sympathetically,

67model

P.S. Have you measured 3000 ohm-m yourself? Is it only at small depth in a very dry surface layer? There may be hope yet.

Hello 67model

I have found a solution, I will use rod electrode 7 meter with the marconite , the resistivy of marconite is 0.001Ω.m and I will install the rod electrode between 1.14m and 10.33m depth.

BS7621:2018 "Requirements for Electrical Installations" Table 54.1 gives minimum 25 sq.mm copper [50mm² steel] for buried earthing conductors <not protected against corrosion>. Your national electrical code may have a different minimum.

unfortunately, I do not have this standard (BS7671) is there another technical document that has spoken on this subject ?

thnaks

below the table shows the value of the resistive according to the depth

Hello again Aminelm,

Congratulations on finding good earth at depth. Marconite looks very useful.

This is the section/table of BS7671 [which applies for most installations below 1000VAC in UK].

As I wrote, the national code for country of your installation is likely to have requirements for earthing conductors - most regulations are based on IEC now.

Noting the language of your table, I think Earthing conductors are covered by 542.3 of French Standard NFC_15-100 which is equivalent to BS7671 - note similarity of section number. Correspondence with IEC60364-5-54 is indicated. Both agree with same European harmonisation document, which has same no. as IEC but there may be national differences.

The ABB-FURSE document I referenced in earlier post has a list of standards in section 16 page 22 approx.

This is also a comprehensive guide which includes references to standards....

https://www.dehn-international.com/en/downloads#lightning-protection-guide

Regards,

67model

<...do not have this standard (BS7671)...>

Then buy a copy. Sheesh!

thank's 7anoter4

If you are interested in resistance calculation beyond this standard limits you may try IEEE 80/2013 –for instance.

but this standard is for AC substation only ,

Cordialy.

According to IEEE-80/2000 [ or 80/2013] the resistance of one rod does not depends on the depth but only on rod length. See Fagan-Lee formula for concrete embedded vertical rod:

RCE-rod=1/2/π/Lr*{ρ*[ln(8*Lr/Dc)-1]+ρc*[ln(8*Lr/d)-1]-ρc*[ln(8*Lr/Dc)-1]} [ formula 58/2000 or 65/2013]. Where:

ρc=concrete resistivity [Ωm]

ρ=resistivity of the soil [Ωm]

Lr=length of the rod [m]

d=rod diameter[m]

Dc=concrete shell diameter [m]

For example if ρc=0.001 ;ρ=100;Lr=7;d=0.019;Dc=0.1 then one rod resistance will be 12.11 Ω

If the rod is vertical of length L, surely resistance does depend on depth L?

Have you missed something in copying formula? Should everything including Lr before the first * be in brackets? Else you are dividing by ρ and the higher ρ, the lower RCE?

The formula seems over complex, because...

ρc*[ln(8*Lr/d)-1]-ρc*[ln(8*Lr/Dc)-1]} reduces to ρc*ln[Dc/d] since ln(a/b) = ln(a) - ln(b) and (ln(f) - 1) = ln(f/e) where e=2.7183... N.B. ln(e) =1

This is the original. It seems to me it is the same_ just modified a bit to be close to

Excel

it is better now, I think.

It seems that increasing the diameter of the rod does not reduce earthing resistance. So, if you use a conductive filler around the metal rod, not too much resistance change is expected.

According to 67model link [ABB Furse page 7/3] up to 1" the resistance decrease to 80% and further remains constant.

Calculated according to Schwarz equation from IEEE 80[ if the horizontal cable will not be buried at all !!!] the minimum resistance of 50 embedded rods [without concrete filling] will be 48.4 ohm [in an area of 30*20 m 3000 Ωm soil resistivity]. Occupying the entire area with electrodes [16 in length and 11 in width=325] and filling 20 cm around each rod marconite the total resistance will decrease up to 47.9 ohm only.

Why do you need to achieve 5 Ω only. No standard [for substation design] as far as I know will require such a thing. NEC recommends to add a supplementary electrode if the measured earth [grounding] resistance is more than 25 Ω. BS7671 requires RA*IA<=50 V for IT and TT system and Zs*Ia<=Uo for TN [however the loop does not include Earth resistance].

For higher than 1 kV installation one has to follow EN 50522 standard [BS EN 50522/2012]

7anoter4,

All sources seem to agree that thicker rod has little effect on resistance and the resistance of plates is related to length of side not area

I reached same "impractical" conclusion for 5 ohm with 3000 ohm-m by the ideal application of 15m radius half-buried metal sphere, bigger than site, in my post #6.

However, in post #7, the OP reported rod to increasing depth, and got low resistance readings.

I tried to calculate theoretical resistance of single rod of depth 7 m & got 6 ohm for ρ = 42 Ω-m, 0.1 m diameter - compared to 0.07 ohm measured & ρ = 42 Ω-m (calculated? how?) in table of post #7, so I am sceptical of the table figures now.

67model

Hello 67model and 7anoter ( sorry for the delay, I was very busy )

firstly thank you very much 67model for your table in the your answer#8

in my case, to reduce the value of resistance I used marconite. with the formula below, I got a good result.

let me get back to the subject, the cable section that connects the electrod rods between them

In BS 7430:2011+A1:2015

6.7 Earthing conductors (page 16)

Earthing conductors are protective conductors and should be sized in the same

way as other protective conductors (see 6.6), but with specified minimum

cross-sectional areas when buried in the ground specified in BS 7671:2008+A3, Table 54.1.

and in paragraph 6.6 there is :

I would like to protect human against static electricity, for a storage tank (oil and gas)

the BS 7671:2008+A3 give minimum value of Earthing conductors , and if I want to calculate the exact value ,according to BS 7430 paragraph 6.7 I can calculate the cable section with the formula given in paragraph 6.6

Earthing conductors are protective conductors and should be sized in the same way as other protective conductor

but in my case there is not a circuit breaker, time of the disconnecting device, I only have a storage tank,

in this case the formula given by BS 7430 (paragraph 6.6) is usable or I must use the values given by the standard 7671:2008+A3 ?

Cordialy

According to NEC 2017 Art.250.66 (A) (A) Connections to a Rod, Pipe, or Plate Electrode(s).

not be required to be larger than 6 AWG copper wire or 4 AWG aluminum.

6 AWG [copper wire] 13.3 mm^2

4 AWG [aluminum wire] 21.5 mm^2

BS 7671/2008 Reg. 542.3.2. states:

The connection of an earthing conductor to an earth electrode or other means shall be:

soundly made and be electrical and mechanically satisfactory.

In my opinion, if you don't intend to connect to earth equipment which the supply conductor will be more than 16 sqr.mm copper you don't have to use more than 16 mm^2 copper [Table 54.7]

If no mechanical protection will be provided and no corrosion means then according to Table 54.1 a 25 mm^2 copper it is required.

The table I copied in my post #8 was from BS7671:2018.

You have now revealed that your installation is oil & gas tanks, but apparently has no electricity supply [no known supply current] so 6.6 would reduce to s = 0. However, the risk of explosion/fire due to lightning looks considerable. There is also the requirement for earthing of trucks filling or being filled & bonding to the tank, since it is known that delivery nozzles can generate static charges able to ignite fuel-air.

You have to look at possible lightning currents & durations [unless you have a guarantee from God it will never be struck]. The following come from...

https://en.wikipedia.org/wiki/Lightning#Transient_currents_during_flash

.....click on 5.4.3 of contents.

1. The electric current within a typical negative CG Cloud-Ground lightning discharge rises very quickly to its peak value in 1–10 microseconds, then decays more slowly over 50–200 microseconds.
2. Positive lightning is less common than negative lightning, and on average makes up less than 5% of all lightning strikes.
3. Positive lightning strikes tend to be much more intense than their negative counterparts.
4. The average positive ground flash has roughly double the peak current of a typical negative flash, and can produce peak currents up to 400 kA and charges of several hundred coulombs.^[45][46] Furthermore, positive ground flashes with high peak currents are commonly followed by long continuing currents, a correlation not seen in negative ground flashes.^[47]

The typical decay from peak is exponential & test voltages for electrical equipment have figures like 1.2 microsecond rise & decay 50 microseconds for 10% to 90% of peak.

Taking 4. above....

• Assume a positive strike of 400 kA & 400 coulombs. N.B. from memory, aircraft engines are type-tested at 200kA peak.
• Charge is current x time. Assume a square current pulse, calculation gives t = 1.0 millisecond for pulse length N.B. 400.10³ amps x 10^-3 seconds = 400 Coulombs.
• Apply adiabatic equation, s ≥ { I√t }/k with k = 115 from Table 43.1 of BS7671:2018 (that is for temperature rise of < 300 mm² PVC insulated copper from 70 to 160 Celsius, will be over-estimate for lower starting temperatures). Note Table 54.6 gives k=228 for bare copper with final temperature 500'C
• s >= 4.10⁵ x √(10^-3)/115 = 110 mm².
• 110 mm² is an overestimate, because the current decays exponentially, however, I can see why Furse have a minimum cross section of 50 mm² copper[ 8mm ∅] for connections, which with k = 228 for bare copper seems adequatefor the worst strikes, when there is no risk of 500'C causing a fire [Lightning protection system kept away from any flammable materials].

Please note that 25 mm² copper in Table 54.1 is a minimum for a buried bare earth conductor not protected against corrosion. It is apparent that not all parts of an earthing system carry the full current when it is an electricity supply earth, but with lightning it is not known which part will be struck when there are multiple air spikes and ground rods.

The equation for s you quote [and used above] is in current standard BS7671:2018 clause 543.1.3 , no-change.

According to BS EN IEC 62305 if we take Imax=200 kA [LPL I] and the standard wave of 10/350 μs then for a copper wire of 30oC ambient and max.250 oC according to BS7430 the required cross section area will be: S=Imax.√t/k k=176 S=21.56 mm^2.

By the way, IEEE 80 takes the maximum copper wire temperature 1084 then S=13.3 mm^2

thank's very much 7anoter and 67model.

just I find something important

the standard IEC 60364 make difference between protective conductor and earthing conductror.

they give directly the value of earthing conductor like that : (BS7430 give the formula or use table BS 7671:2008+A3, Table 54.1. )

IEC 60364

The conductors may be:

• Copper: Bare cable ( ≥ 25 mm²) or multiple-strip (≥ 25 mm²) and (≥ 2 mm thick)
• Aluminium with lead jacket: Cable ( ≥ 35 mm²)
• Galvanised-steel cable: Bare cable ( ≥ 95 mm²) or multiple-strip ( ≥ 100 mm² and ≥ 3 mm thick)

and for protective conductor they use the formula like BS 7430 or it chosen like that

This method is based on PE conductor sizes being related to those of the corresponding circuit phase conductors, assuming that the same conductor material is used in each case.Thus, in for:Sph ≤ 16 mm² : S_PE = S_ph16 < Sph ≤ 35 mm² : S_PE = 16 mm²Sph > 35 mm² : S_PE = S_ph / 2

what do you think about this difference between stardand ?

Earthing conducteur

http://www.electrical-installation.org/enwiki/Installation_and_measurements_of_earth_electrodes

Proctective Conductor

http://www.electrical-installation.org/enwiki/Sizing_of_protective_earthing_conductor

The conductors are for different risks, one a known supply short-circuit current, for up to 5 seconds, lightning is very high current for an instant. In both cases the adiabatic equation & resulting temperature rise are needed to size a conductor but most cases can be covered by a minimum requirement. Inside most installations, bare earth conductors accessible to touch are exceptional, but for lightning bare, outdoor is usual - so a difference in requirement.

Protective conductor

The guiding principle of a protective conductor [PC or PE] is that exposed metalwork of electrical equipment is connected back to the place at which the supply source is earthed & from which the neutral of the supply is taken (commonly neutral of 3 phase star winding of supply transformer in Europe).

Modern UK town supplies combine neutral & earth outside consumer's building with supplier installing multiple earths along run [TN-C-S (PME) earthing] , consumer's "earth" terminal is connected to neutral at entry to installation N.B. the maximum current from supply is known from its capacity & commonly a fuse or circuit breaker will cut-off current within half-cycle before it gets to theoretical short-circuit peak.

This is TT - where supply does not provide earth (I believe this is typical to French national standard). These pictures are from the "on-Site Guide" to BS7671:2018 & for majority of home supplies where capacity is 100 amp [25 mm2 consumer tails] or less.

Except under fault, this conductor does not carry current* - note this means it will be at lower temperature in some cases, although difference will be insignificant if it is part of a 3 core L-N-E cable. Traditionally, this has been called earth or ground wire/connection, but regulations call it protective conductor now.

* Other than cable capacitance current from live, including (increasingly) Radio interference filter capacitors, less than 10mA in most installations.

The essential point is that in most installations, particularly wooden floors/wooden buildings the structure is non-conducting or not connected to earth, it would not matter if it were floating in air (It does not matter in a wood or metal aircraft and, if struck by lightning, safety lies in everything metallic being bonded together).

The prime safety feature is that under short circuit fault in the equipment the voltage on exposed metalwork to metalwork of un-faulted equipment is reduced and of short duration to minimise risk someone is touching both e.g. less than 0.4 seconds for socket circuits, 5 sec for fixed equipment. The protective conductor is usually >= 0.6 times mm² line conductor , so metal voltage is less than [(1/0.6)/(1 + 1/0.6)] = 1.66/2.66 = 0.65 x 230V = 150V in single phase installations. Cords on plug-in equipment have "earth" of same cross section as live so volts to unfaulted plug-in equipment near enough to touch, typically same circuit, is 115V (often less because local voltage difference is due to cord volt drop only e.g. a 230V/16 amp fused circuit blows in 0.4 sec at 100 amp, 1mm² x 3m cord earth conductor drops 3 x 0.022 ohm x 100 amp = 6.6 volts - most circuits are short enough for fusing time << 0.4 sec) N.B. 10 metre run of 1 mm² has S/C current of 500 amp - 6.6 x 5 = 33 volts in cord, but fuse blows in <0.01 sec.

Earthing Conductor

This is an actual connection to an earth electrode from the installation (or part of the earth electrode system). In the case of an electricity supply, the current is known & the size is e.g. 16mm² for 25mm² line conductors, plus allowance for mechanical strength or corrosion where necessary (outdoors/buried).

However, where lightning protection is involved, currents can be far higher & one must allow for the possible current & duration & if conductors are close to flammable building materials. Lightning frequency increases from poles to equator of earth & other risk factors are quantified in standards, such as value of installation & cost of repair, danger of injury.

Hello

Thank you veeery much for details 67model , it's very useful form me

I have question about the best value of earthing resistance

the rule in french standard that the resistance must be R < 50.IΔn for dry place

and for wet place R < 25.IΔn

IΔn : the operating current in amperes for RCDs

but about static electricity for a storage tank (oil and gas ) some people said R should be 10 ohms another said R should be 5 ohms

In which standard I can find a good answer ?

thank's

There is something strange about your formula - the higher the trip current of the protection, the higher the resistance!!!

BS7671:2018 (UK Wiring Regulations) clause 411.4.4 requires earth loop impedance Zs for a protective device of U₀ x 0.95/Ia where Uo is supply voltage & Ia is operate current of protective device e.g. for 230V & 0.1 amp RCD, Zs < 230 x 0.95/0.1 = 2185 ohms. But Table 41.5 requires <500 ohm for a TT system & its note 2 indicates that any earth electrode resistance over 200 ohm is not reliable. The formula given in 411.5.3 is R x Ia < 50 volts which matches Table 41.5 values.

This code of practice from our local gods of Health & Safety.....

The storage of flammable liquids in tanks

[free download] requires 10 ohms...

http://www.hse.gov.uk/pubns/books/hsg176.htm

It has a lot of information about requirements for design, installation & use of fuel tanks.

This may be of interest....

https://www.lightningprotection.com/wp-content/uploads/2017/04/lightning-protection-article.pdf

Two points it makes I note ...

a) A tank has so much contact area with ground that it makes its own earth electrode - OK if the ground is not very high resistivity!

b) Avoiding a lightning strike on or near tank is better than elaborate nearby towers or wires, which may attract lightning which diverts to tank - Multiple sharp points to dissipate local charge are effective to avoid strike (DAS Dissipation Array Systems).

N.B. Aircraft have dissipators to avoid a charged aircraft having a flashover as it lands, the Van de Graaff static generator machine works by moving a charge away from its original position.

Hello

thank you very much for your documents.

about the formula here is the true one R x IΔn ≤50V (or25V). (Sorry it's my mistake )

So I read clause 411.4.4 and 411.5 from BS7671:2018 it's the same with NFC 15-100

NOTE 2: • The resistance of the installation earth electrode should be as low as practicable. A value exceeding 200 ohms

may not be stable. Refer to Regulation 542.2.4.

542.2.4 The type and embedded depth of an earth electrode shall be such that soil drying and freezing will not increase its resistance above the required value.

What is a good ground resistance value?

https://www.ecmweb.com/sites/ecmweb.com/files/uploads/2014/03/4346628a_en_EarthGround-tutorial-e.pdf

" There is not one standard ground resistance threshold that is recognized by all agencies. However, the NFPA and IEEE have recommended a ground resistance value of 5.0 ohms or less. The NEC has stated to “Make sure that system impedance to ground is less than 25 ohms specified in NEC 250.56. In facilities with sensitive equipment it should be 5.0 ohms or less.” The Telecommunications industry has often used 5.0 ohms or less as their value for grounding and bonding. The goal in ground resistance is to achieve the lowest ground resistance value possible that makes sense economically and physically "

why NFPA and IEEE they don't take in consideration the formule given by

BS7671:2018 , R x IΔn ≤50V and they prefer to chose 5 ohms ? I think this formule

is more practical then 5 ohms, I know there no contradiction between them but

sometimes when we can't get 5ohms we can change IΔn of RCD and respect this

condition R x IΔn ≤50V

what do you think ?