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Topic Title: Buried supply to a shed/garage
Topic Summary: List of requirements
Created On: 11 May 2009 02:57 PM
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 11 May 2009 07:27 PM
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OMS

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Are you now stating that using the SWA as a CPC does not comply with BS7671?


Not at all - the armouring usually makes a very fine protective conductor - but we are discussing bonding as far as I'm aware.

Table 54.8 is quite clear, for a TN-C-S system the minimum bonding conductor size is 10mm2 or equivalent conductance. The dominant case here if the armour is going to be both CPC and bonding conductor is that of bonding conductor

Why should we assume a ratio of 9:1?


Because that is about the ratio of conductance between steel and copper - you appear to making some equivalence based on thermal capacity not conductance

Sorry, should have been 115, which gives a csa for the SWA of Aprox 22mm²


My pleasure

Regards

OMS

-------------------------
Failure is always an option

Edited: 11 May 2009 at 07:29 PM by OMS
 11 May 2009 07:44 PM
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spinlondon

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Originally posted by: OMS
Are you now stating that using the SWA as a CPC does not comply with BS7671?


Not at all - the armouring usually makes a very fine protective conductor - but we are discussing bonding as far as I'm aware.

Are we? I thought we were discusing the exportation of the Earth to the Garage?

Table 54.8 is quite clear, for a TN-C-S system the minimum bonding conductor size is 10mm2 or equivalent conductance. The dominant case here if the armour is going to be both CPC and bonding conductor is that of bonding conductor

Sorry, SWA is acceptable for use when buried. Are you now stating that the only acceptable size would be 95mm²?
Considering that the minimum CSA for a buried earthing conductor protected against corrosin is 16mm², then by your calculations the only CSA of SWA taht can be used is 150mm².


Why should we assume a ratio of 9:1?


Because that is about the ratio of conductance between steel and copper - you appear to making some equivalence based on thermal capacity not conductance

Sorry, should have been 115, which gives a csa for the SWA of Aprox 22mm²


My pleasure

Regards

OMS




Edited: 12 May 2009 at 12:08 AM by spinlondon
 12 May 2009 06:15 PM
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OMS

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Are we? I thought we were discusing the exportation of the Earth to the Garage?


We are - we are exploring the scenario where the TN-C-S earth is to be used and that the garage contains extraneous conductive parts that require equipotential bonding.

Sorry, SWA is acceptable for use when buried. Are you now stating that the only acceptable size would be 95mm²?


The use of buried SWA is perfectly acceptable - if you wish to use the armouring as both the means of earthing and as a bonding conductor the dominant case is that of bonding.

That leads us to Table 54.8 which requires a minimum conductor size of 10mm2 if made of copper or of equivalent conductance if another metal.

As the armouring is steel we need the equivalent of 10mm2 of copper which using a conductivity ratio of 9:1 leads to the use of a minimum of 90mm2 of steel. Using XLPE/SWA we find the nearest cable would be a 95mm2 2 core.

Considering that the minimum CSA for a buried earthing conductor protected against corrosin is 16mm², then by your calculations the only CSA of SWA taht can be used is 150mm².


Sorry, I don't follow that - Table 54.1 tells me that the protective conductor (in this case the cable armour) which is protected against corrosion by a sheath needs to be a minimum of 16mm2 of coated steel - in our case we have around 110mm2 of steel so clearly we comply.

Regards


OMS

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Failure is always an option
 12 May 2009 06:59 PM
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spinlondon

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Originally posted by: OMS
Are we? I thought we were discusing the exportation of the Earth to the Garage?


We are - we are exploring the scenario where the TN-C-S earth is to be used and that the garage contains extraneous conductive parts that require equipotential bonding.

Sorry, SWA is acceptable for use when buried. Are you now stating that the only acceptable size would be 95mm²?


The use of buried SWA is perfectly acceptable - if you wish to use the armouring as both the means of earthing and as a bonding conductor the dominant case is that of bonding.

That leads us to Table 54.8 which requires a minimum conductor size of 10mm2 if made of copper or of equivalent conductance if another metal.

As the armouring is steel we need the equivalent of 10mm2 of copper which using a conductivity ratio of 9:1 leads to the use of a minimum of 90mm2 of steel. Using XLPE/SWA we find the nearest cable would be a 95mm2 2 core.

Considering that the minimum CSA for a buried earthing conductor protected against corrosin is 16mm², then by your calculations the only CSA of SWA taht can be used is 150mm².


Sorry, I don't follow that - Table 54.1 tells me that the protective conductor (in this case the cable armour) which is protected against corrosion by a sheath needs to be a minimum of 16mm2 of coated steel - in our case we have around 110mm2 of steel so clearly we comply.

Regards

OMS


I suggest you go and re-read the part of the Regs. pertaining to Protective Conductors.
 12 May 2009 07:04 PM
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OMS

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I suggest you go and re-read the part of the Regs. pertaining to Protective Conductors.


OK, done that - which reg in particular were you thinking of and which part of my assessment above don't you agree with

OMS

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 12 May 2009 07:15 PM
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spinlondon

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By exporting the Earth are we changing the TNCS into PME?
Would the SWA be classed as a protective conductor or an equipotential bonding conductor?
How would you calculate the CSA of the SWA as a protective conductor?
Why would you use a diferent calculation when calculating the CSA of the SWA as an equipotential bonding conductor?
 12 May 2009 07:24 PM
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OMS

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By exporting the Earth are we changing the TNCS into PME?


Take a look at the little drawing in Fig 2.4 (Page 33) and read the third sentence below that drawing

Would the SWA be classed as a protective conductor or an equipotential bonding conductor?


It would serve the function of both a protective conductor and a main equipotential conductor.

How would you calculate the CSA of the SWA as a protective conductor?


By means of the adiabatic expression or by reference to Table 54.7 depending on the particular circumstances.

Why would you use a diferent calculation when calculating the CSA of the SWA as an equipotential bonding conductor?


Because the armouring is acting as both a Circuit Protective conductor and as a bonding conductor.

Therefore the dominant case is the requirements for a bonding conductor in a PME installation. I would refer to Table 54.8 which tells me that a 10mm2 copper equivalent is needed, as this requirement is more onerous than the requirement for a CPC it clearly needs to meet 54.8 and would also meet 54.7 by default

Regards

OMS

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Failure is always an option
 12 May 2009 07:28 PM
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spinlondon

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Originally posted by: OMS

Because the armouring is acting as both a Circuit Protective conductor and as a bonding conductor.

Therefore the dominant case is the requirements for a bonding conductor in a PME installation. I would refer to Table 54.8 which tells me that a 10mm2 copper equivalent is needed, as this requirement is more onerous than the requirement for a CPC it clearly needs to meet 54.8 and would also meet 54.7 by default

And where does table 54.8 tell you that the calculation for determining equivelant conductance is diferent from the one at 54.7?

Regards



OMS




Edited: 12 May 2009 at 07:29 PM by spinlondon
 12 May 2009 07:33 PM
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OMS

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And where does table 54.8 tell you that the calculation for determining equivelant conductance is diferent from the one at 54.7?


You are are kidding I hope - since when did K factor translate to conductance.

K factor is a dimensionless number that attempts to bring together resistivity, temperature coefficient and volumetric heat capacity.

OMS

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Failure is always an option
 12 May 2009 07:41 PM
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spinlondon

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Originally posted by: OMS
And where does table 54.8 tell you that the calculation for determining equivelant conductance is diferent from the one at 54.7?


You are are kidding I hope - since when did K factor translate to conductance.

K factor is a dimensionless number that attempts to bring together resistivity, temperature coefficient and volumetric heat capacity.

OMS


It is what is used to determine the CSA equivalance for a protective conductor. Why should it not also be used for an equipotential bonding conductor?
 12 May 2009 07:48 PM
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OMS

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Read the note to table 54.8 then google "conductivity of metals" - ignore any hits that relate to "thermal" and search within results for "electrical".

Regards

OMS

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 12 May 2009 08:02 PM
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spinlondon

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Originally posted by: OMS
Read the note to table 54.8 then google "conductivity of metals" - ignore any hits that relate to "thermal" and search within results for "electrical".

Regards

OMS


Sorry, which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?
 13 May 2009 08:22 AM
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OMS

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Sorry, which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?


If a conductor is to act as a bonding conductor (as well as a CPC in this case) in a TN-C-S (or PME) system then you would need to refer to regulation 544.1.1. Read the second paragraph as PME conditions apply, that will direct you to Table 54.8. You will note the "*" after the text"Minimumm Copper equivalent........".

When you refer to "*" at the foot of the page you should see that the figures used in the table are for copper but it is allowable to use another metal affording equivalent conductance.

If you had googled conductance as I suggested you would have discovered that the relationship between the conductivity of steel and the conductivity of copper is around 9:1. From this you should recognise that for every 10mm2 of copper you would need to use around 90mm2 of steel to comply with the requirements of the regulation.

To answer fully, you should also recognise that main equipotential bonding conductor sizes are usually not calculated - they are selected based on knowledge of the distributors supply neutral size in a deemed to comply fashion in accordance with 544.1.1

Regards

OMS

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Failure is always an option
 13 May 2009 10:22 AM
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spinlondon

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Originally posted by: OMS
Sorry, which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?


If a conductor is to act as a bonding conductor (as well as a CPC in this case) in a TN-C-S (or PME) system then you would need to refer to regulation 544.1.1. Read the second paragraph as PME conditions apply, that will direct you to Table 54.8. You will note the "*" after the text"Minimumm Copper equivalent........".

Yes......

When you refer to "*" at the foot of the page you should see that the figures used in the table are for copper but it is allowable to use another metal affording equivalent conductance.

Agin Yes......

If you had googled conductance as I suggested you would have discovered that the relationship between the conductivity of steel and the conductivity of copper is around 9:1. From this you should recognise that for every 10mm2 of copper you would need to use around 90mm2 of steel to comply with the requirements of the regulation.

As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?

To answer fully, you should also recognise that main equipotential bonding conductor sizes are usually not calculated - they are selected based on knowledge of the distributors supply neutral size in a deemed to comply fashion in accordance with 544.1.1

Is there actually a point to this post?
As far as I can see, you have re-hashed previous posts, made a long winded and unecessary attempt at explaining how to use the Regs. Yet still failed to answer my question.
The figures you have quoted of 9:1, are they for anneald or hard drawn copper? Do they relate to flat steel bar or to multi-cored circular wire?
The Regs. contain a formula to calculate the CSA equivelance of Steel and Copper. Why should we not use that formula, why should we do as you say and Google?

Regards

OMS


 13 May 2009 02:09 PM
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AJJewsbury

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As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?

543.1.1 "The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be ...."

- Andy.
 13 May 2009 04:33 PM
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spinlondon

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Originally posted by: AJJewsbury
As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?


543.1.1 "The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be ...."

- Andy.


I've tried telling OMS that, but he states that we shouldn't use that. He states that we should use the formula 9:1.
 13 May 2009 04:44 PM
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OMS

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As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?


and as I said,

If a conductor is to act as a bonding conductor (as well as a CPC in this case) in a TN-C-S (or PME) system then you would need to refer to regulation 544.1.1. Read the second paragraph as PME conditions apply, that will direct you to Table 54.8. You will note the "*" after the text"Minimumm Copper equivalent........".


and as Andy kindly pointed out if you approach the problem the other way:

543.1.1 "The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be ...."


Is there actually a point to this post?


Well the point to this post was to point out that your comment:

If you are going to export the Earth, then you will need 10mm² SWA. However, if you TT the Garage, you could probably get away with 2.5mm². Do the cable calcs. for volt drop etc.


was a little erroneous if the garage contained extraneous conductive parts.

As for long winded, that's one opinion, although if you believe I haven't answered your question then I can only conclude you don't understand what the question is.

Perhaps if you take a look at figure 5.16 in Guidance Note 8 and examine building B3 then you may get the point.

The figures you have quoted of 9:1, are they for anneald or hard drawn copper? Do they relate to flat steel bar or to multi-cored circular wire?


I actually said, around 9:1 - I couldn't tell you the exact figure without knowledge of the specific material although it won't vary much.

The Regs. contain a formula to calculate the CSA equivelance of Steel and Copper. Why should we not use that formula, why should we do as you say and Google?


Because we are talking about the bonding conductor sizing being the dominant case - and as I tried to point out, we don't calculate that - we select it.

OMS

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Failure is always an option
 13 May 2009 04:48 PM
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OMS

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Originally posted by: spinlondon

Originally posted by: AJJewsbury

As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?




543.1.1 "The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be ...."



- Andy.




I've tried telling OMS that, but he states that we shouldn't use that. He states that we should use the formula 9:1.


Have you read it - it refers to the adiabatic expression or selection in accordance with 543.1.4 (ie Table 54.7) and it specifically tells you that the adaibatic expression or selection in accordance with 54.7 is not applicable to bonding conductors - the point Andy was subtly trying to make I suggest

OMS

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 13 May 2009 05:04 PM
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AJJewsbury

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point Andy was subtly trying to make I suggest

It's not often I'm accused of being subtle!

Yes, I was trying to point out the reg which specifically says you can't use the calculation you'd usually use for CPCs to size a bonding conductor.

I'm with OMS on this one - 544.1 is what's needed for main protective bonding conductors, for PME it says to refer to 54.8. Table 54.8 says 10mm2 Cu or equivalent conductance if other metals. AFAIK BS 7671 doesn't give the relative conductance of copper and steel, so we have to look elsewhere. The Boy's own Almanac of Metallurgical Data might be one source, but if you don't have a copy to hand and happen to be in front of a internet connected PC, then google might be quite convenient.

- Andy.
 13 May 2009 05:12 PM
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spinlondon

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Originally posted by: OMS
As I said. "Which Reg. is it that tells me I shouldn't calculate the CSA as per the formulae and tables in the Regs., but should Google?


and as I said,

If a conductor is to act as a bonding conductor (as well as a CPC in this case) in a TN-C-S (or PME) system then you would need to refer to regulation 544.1.1. Read the second paragraph as PME conditions apply, that will direct you to Table 54.8. You will note the "*" after the text"Minimumm Copper equivalent........".


We've done this? You are not answering my question, just re-hashing your previous posts

and as Andy kindly pointed out if you approach the problem the other way:

543.1.1 "The cross-sectional area of every protective conductor, other than a protective bonding conductor, shall be ...."


I don't consider that mine and Andy's way are "the otherway".

Is there actually a point to this post?


Well the point to this post was to point out that your comment:

If you are going to export the Earth, then you will need 10mm² SWA. However, if you TT the Garage, you could probably get away with 2.5mm². Do the cable calcs. for volt drop etc.


was a little erroneous if the garage contained extraneous conductive parts.

As far as I'm concerned, the Garage contains extraneous conductive parts.

As for long winded, that's one opinion, although if you believe I haven't answered your question then I can only conclude you don't understand what the question is.

I understand the question. However it appears that you don't.
It is why should we ignore the tables and formulae in the Regs. and instead Google?


Perhaps if you take a look at figure 5.16 in Guidance Note 8 and examine building B3 then you may get the point.

I don't have guidance note 8. I prefer to use the Regs. themselves, rather than someones presis based on their opinion.

The figures you have quoted of 9:1, are they for anneald or hard drawn copper? Do they relate to flat steel bar or to multi-cored circular wire?


I actually said, around 9:1 - I couldn't tell you the exact figure without knowledge of the specific material although it won't vary much.

I would prefer to use the tables and formulae contained in the Regs. Rather than some ratio that you have determined from insuficient knowledge that you state won't "vary much".

The Regs. contain a formula to calculate the CSA equivelance of Steel and Copper. Why should we not use that formula, why should we do as you say and Google?


Because we are talking about the bonding conductor sizing being the dominant case - and as I tried to point out, we don't calculate that - we select it.

If we are to "select it". not calculate it. Why are you advocating that we calculate it using a ratio of 9:1?
Please try to be consitant.


OMS


IET » Wiring and the regulations » Buried supply to a shed/garage

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