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Discuss Adibatic equation for 6mm earth in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

S

Scott F

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As a matter of intrest how is the calc worked out for working out whether a 6mm main bond on a Tns complies
 
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A

amberleaf

  • Thread Starter Thread Starter
  • #2
try this
4.0mm2 / 1.5mm2 cable
6.omm2 / 2.5mm2 cable < this is what your looking for CPC

k2 - 54.2 143 div 66 x 95 div 2 = 147.6mm2
143 div 46 = 3.10 copper / 90 oC thermoplastic
95 div 2 = 47.5
3.10 x 47.5 = 147.25mm2
---------
S = V 144 X2 x 2
---------- = 1.77mm2
115 -------------
144 x2 x 2 = 41472 V = 203.646753 div 115 = 1.77mm2

ps ive got more some where ill keep looking
i hope this put you in the right path

-----------------
S = V 114 X2 X 2 ------------
---------------- = 1.77mm2 ( the v pyp s
115
 
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G

Guest123

  • Thread Starter Thread Starter
  • #3
As a matter of intrest how is the calc worked out for working out whether a 6mm main bond on a Tns complies
Mate, calcs are not needed to determine the CSA of bonding conductors...do you mean a bonding conductor or main earthing conductor??

If it is a main earth and not a bond you want to work out then you need to measure your Ze and find out the type of cut-out fuse you have.

As a rule on a 230V supply with a Ze of 0.40 and a 60A cut-out fuse to BS 1361 the result is 9.017mm so a 10mm is needed.

I would say that a 6mm is too small on a TN-S anyway (main earth and bond) without the calc as it should be sized in relation to the supply conductors.

Hope this helps.
 
S

Scott F

  • Thread Starter Thread Starter
  • #4
Hi Lenny .In some older properties the main bond to the Gas/water is sometimes 6mm .If its TNCS then it needs to be 10mm .If it is a TNS system.Have been told it may comply but first you would have to do the adibiatic equation for it ?

Have i got this wrong/or my wires crossed?
 
S

Spudnik

  • Thread Starter Thread Starter
  • #5
Not sure calculations are valid anymore as the minimum sizes are in table 4.1 page 28 OSG.
 
G

Guest123

  • Thread Starter Thread Starter
  • #6
Hi mate.

Think you might have been mis-informed there. No calculation is needed for main equipotential bonding conductors just select from the relaevant table in ol' red. 10mm earth is the smallest size to use upto 35mm tails, then 16, 25, 35 and 50mm. The largest size you would ever need to go upto for a bonding conductor is 50mm for tails of 150mm or over.

The adiabatic equation is only used to initially size/confirm the size of the main/circuit protective conductors.
 
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S

Scott F

  • Thread Starter Thread Starter
  • #7
Yes but do you know the calc for working it out . Have been reading some old threads One school of thought is if it complied you would not neccassary have to replace with 10mm
 

Des 56

-
Arms
Esteemed
Rather than trying to explain mysef ,here is a working example

Why bother?

It is about making sure the CPC of a circuit will be able to carry the fault current produced during a short circuit or earth fault, without deterioration, or heat/ fire damage. Remember we are talking about fault currents possibly in 100’s if not 1000’s of amps, and even though the CPC will only carry the fault current for a short while, it must not cause fires or heat damage and remain intact so as to enable the protective device to clear the fault.



Isn’t there an easier way of checking this out?

Well actually, Yes! Look at table 54G (page 116 of BS7671: 2001 (2004). Here we have clear guidance; in a nutshell the table shows that for:
·[FONT=&quot] [/FONT]all cable up to 16mm2 the CPC must be the same size as the phase conductor
·[FONT=&quot] [/FONT]phase conductors between 16mm2 and 35mm2 require a CPC of 16mm2
·[FONT=&quot] [/FONT]a phase conductor above 16mm2 requires a CPC of at least ½ it’s size



So where’s the problem and why all the mathematical jargon?

Think about it, every PVC/PVC sheathed cable your likely to install in a domestic situation (other than 1mm2) has a CPC between a third and two thirds the size of the phase conductor (not meeting the requirements of section 543 of BS7671).
So what are we to do?
Well, for certain circuits at least, if we select all our cables and all our circuit arrangements in accordance with the IEE On-Site Guide, then we are told no further calculations will be necessary. Remember compliance with the On-Site Guide does not ensure compliance with BS7671 (particularly section 513) and it only covers installations with a supply cut-out of up to 100Amps. When using the standard circuit arrangements described in the On-Site Guide (Appendix 8) it is still the responsibility of designers and installers to satisfy themselves of amongst other things the suitability of the size of protective conductors (Chapter 54).





What about other circuit arrangements? This is where we might have to apply the adiabatic equation, so let’s have a go with a standard radial socket circuit arrangement to see, if in fact pushed to the limit, the 1.5mm2 CPC is sufficient in size.

Circuit design
An existing double socket outlet wired in 25 metres of PVC/PVC 2.5/1.5mm2 cable and protected by a 20A BS 1361 cartridge fuse. The Zs measured at the socket is 1.40 Ω.


Adiabatic equation :
[FONT=&quot]S =[/FONT]
Ö [FONT=&quot](I² t) [/FONT][FONT=&quot][/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot] k[/FONT]
[FONT=&quot] [/FONT]
S is the minimum protective conductor size (mm2)

I is the fault current (Amp)

t is the opening time of the protective device (sec)

k is a factor depending on the conductor material and insulation,
and the initial and maximum insulation temperatures
(typically 115 for copper conductors)


The problem is that when you are using the adiabatic equation you will normally be able to obtain a value for the fault current I, but then you have to obtain a protective device operating time (at that fault current) from the graphs in Appendix 3 of BS7671.

However, to continue:

Fault current I = Uo = 230 = 164 Amp
Zs 1.40

From the graph depicted in fig 3.1 (BS7671 Appendix 3), a fault current of 164 Amps would cause a 20 Amp BS1361 type fuse to operate in just less than 0.2 sec.

So the minimum size : [FONT=&quot]S =[/FONT]
Ö [FONT=&quot](164² x 0.2)[/FONT][FONT=&quot] = [/FONT][FONT=&quot]0.64[/FONT] mm2[FONT=&quot][/FONT]
[FONT=&quot] [/FONT]
[FONT=&quot] 115[/FONT]
Therefore our CPC of 1.5mm2 meets the requirements of Section 543.
 
G

Guest123

  • Thread Starter Thread Starter
  • #9
Thats a good example Des.

The op however is regarding a bonding conductor not a CPC.

As far as I'm aware there is no calc for bonding cinductor size only comparison tables.
 

Des 56

-
Arms
Esteemed
Agreed
Thought he wanted to verify a 6mm earth conductor for a TNS
Not applicable to bondinc conductor sorry
 
S

Scott F

  • Thread Starter Thread Starter
  • #11
You should go into teaching Des.
A very through and thought out explanation .
Thank you for your time and input
Would of taken me a week to type that out (but seriously thanks)
Think i have got some wires crossed misread something (late at night blah,blah,blah)
 
G

Guest123

  • Thread Starter Thread Starter
  • #12
Think he may have copied and pasted it from somewhere as it refers to BS 7671 2004.
 
M

maddfridge

  • Thread Starter Thread Starter
  • #13
hi there

the use of the adibatic equation is to ensure you have an adeqaute main earthing conductor and is usually used at the design stage to try and cut down the size of conductor on large sizes cables. this is where such as the sheath of an swa cannot carry the fualt current etc.

cost is also an isssue like always

chers
 
E

ezzzekiel

  • Thread Starter Thread Starter
  • #15
since when did the adiabitic account for main bond sizes......
 
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