Can we use the armour of an SWA as the CPC??? | Page 3 | on ElectriciansForums

Discuss Can we use the armour of an SWA as the CPC??? in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

Show me the calcs Dave!

Seriously though lately i've been trying to gen up on using the adiabatic more lately as like you say the cost can differ massively on some jobs

Cheers
 
Hello everyone just been reading this thread and i find it very interesting......

I must admit i´m a little surprised that only one person has mentioned using the adiabitc equation for calculating the size of a cpc... It doesnt take long and i have found i have never needed to run a seperate cpc when making calculations for distribution cables. As a general rule though, if the cable is to be run outside or where other external influences are present then i always run a copper earth (not a seperate single core cable, an extra internal core) due to corrosion.

I undertook the 2391-20 design course where a lot of cable calculations were made and i always used the adiabatic, as by using the table at the start of this thread you will generally find that you run in an earth that frankly is oversized. If you want to price a job competitively (im talking larger commercial, industrial jobs) then using the adiabatic is alot more accurate, if were talking about large SWA cables from say 95mm upwards you will find a seperate earth costs a fair bit... 50mm BS6491x say 150m.... NOT CHEAP! a simple calc.. oh its not required so why bother....?

If anyone wants a bit more clarity on this like exactly how the calculation is made (can show examples) then let me know as i would be more than happy to help....!

cheers Dave


An SWA will still have it's armour though.....which is what the thread was started about.
 
hi Lenny! or Hola! (living in Barcelona! dont ask about the electrics over here, was just running my finger along my missus parents very shiny metal cooker hood today and i could feel a lovely 50hz... what do you mean earthing? ha ha.....)

I dont really understand what you mean by an SWA having its armour.... of course it will....? can you explain what you mean? bit lost as usual... ha ha!

cheers my friend

Dave
 
Show me the calcs Dave!

Seriously though lately i've been trying to gen up on using the adiabatic more lately as like you say the cost can differ massively on some jobs

Cheers

Yeah of course no probs mate... Ill dig out some calcs from my project on distribution circuits and put up some nice detailed examples..... Watch this space....
 
Hello again sparkies, here is a snippet of my project submitted for 2391-20 qualification regarding minimum size of CPC, proving the armouring is sufficient....

Minimum Size of CPC

Here I will calculate the minimum size of CPC in accordance with regulation 541.1.3 of BS7671.
Primarily I will need to calculate the minimum size of CPC (or confirm the armouring is adequate) of the distribution cables. I already know that the external loop impedance value is 0.08Ω, so I will now focus on the distribution cable to the distribution board, and secondly the supply cable between the suppliers cut-out and the CCU.
I have used data from AEI cables (this can be found in the index) to ascertain the values for conductor and armour resistance.
Distribution circuit to DB3
I will use the equation: Zs = Ze(R1+R2)
Ze = 0.08
R1 = 0.342 mΩ/m (at 90°C, i will not need to apply a correction factor of 1.28)
R2 = 1.2 mΩ/m (armour resistance)
Circuit length = 24m
Therefore: 0.342 + 1.2 = 1.542 mΩ/m × 24 = 37mΩ or 0.037Ω
0.037 + 0.08 = 0.117Ω
I can now calculate the fault current using the following equation:

If= Uo/Zs
Where If = fault current, Uo = line voltage to earth and Zs = total loop impedance
If= 230/0.117=1966A
Using Amtech software at work I have sourced the relevant time current graphs for SQUARE-D BSEN 60947-3 MCCB’s which I will use in the installation (I will go into more detail in part 7, time current graphs can be found in the index)
Using time current graph number 1: t = 0.1s @ 1966A
This is perfectly acceptable, maximum disconnection time = 5 seconds (over 32A)

I will now use the adiabatic equation below to confirm that the armouring of the cable can withstand the level of fault current:
S= √(I²×t)/k

Values for k can be found on p129 BS7671 (in this case table 54.4, the armour of a cable)

S= √(1966²×0.1)/46=13.5mm²

Using the table provided by AEI I can see that a 70mm² four core cable has a 131mm² armour CSA, therefore this is perfectly acceptable.


This shows that the armouring can handle 10x the amount of fault current that is required....!! no need for seperate CPC ;-)

Please note: values for armour and conductor resistance are from AEI cables, vvv useful!!! as you dont find them in the regs book, OSG etc. (OSG up to 50mm and doesnt show armour resistance.... there is a way of calculating conductor resistance but i cant remember how.....

Was going to post link to AEI site but you need to be a member to view the datasheets.... if anyone wants to see them and needs assistance then let me know ;)

hope this helps AMP DAVID!!!
 
I'm supprised that no one has mentioned that running a separate CPC contravenes BS7671.
Specifically Regulation 521.5.2.

Indeed it does however IEE Guidance Note 8 (9.3.4) describes the procedure for sizing 'a separate green-and-yellow covered copper conductor'

While the Guidance Notes are not BS7671, they are published by the same organisation and the whole series should be treated with the same respect afforded to Guidance Note 3.
 
These requirements do not preclude the use of an additional protective conductor in parallel with the steel wire armouring
of a cable where such is required to comply with the requirements of the appropriate regulations in Chapters 41 and 54. It

is permitted for such an additional protective conductor to enter the ferrous enclosure individually.

We live in hope!


 
I've always found it a bit odd, not allowed a separate CPC, but are allowed a supplementary bonding conductor.
As if the bonding conductor wouldn't carry any of the fault current.
 
Where does that reg state that you can't run a separate cpc.

It's referring specifically to single core armoured cable in an ac circuit.
It doesn't just refer specifically to single core armoured cable.
If you read more than the first sentance you would note that it also refers to conductors in ferromagnetic enclosures, and that with an a.c. circuit; all Line conductors, the Neutral and the appropriate protective conductor have to be contained in the same enclosure.
Now unless you are suggesting that SWA is not ferromagnetic, and the armour does not meet the definition in Part 2 of an enclosure, then the Regulation applies.
 
I take your point sinlondon but I would see the armour as mechanical protection and not the enclousure. Also the cpc does not carry current in normal use so there would not be a problem.

It's not that clear is it, but I would say the words 'conductors of an ac circuit installed in a ferromagnetic enclosure suggests that the regs are referring to cable in trunking or conduit etc.
 
Unfortunately whether the intention was to include SWA or not, the actual wording entails SWA is included.
SWA is ferromagnetic and meets the definition of an enclosure in Part 2, because it provides mechanical protection.
I believe IQ Electrical in post 40 has quoted the wording from the proposed amendment.
As you can see this is intended to remove this requirement from applying to SWA.
As yet the proposed amendment content and wording has not been finalised to my knowledge.
 

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