View the thread, titled "Bonding water services" which is posted in Electrical Wiring, Theories and Regulations on Electricians Forums.

Correct, so on a TN-S system to size a bonding conductor the main earthing conductor must be sized correctly. The OP said he just whacked in some 16mm but I would bet he gave this very little thought. The amount of times I've actually put in a 16mm MEC in a domestic situation because of requirement I could probably count on one hand.

Chances are, the original 10mm MEC was absolutely fine. If so, he could leave the 6mm main bond in place.
 
Correct, so on a TN-S system to size a bonding conductor the main earthing conductor must be sized correctly. The OP said he just whacked in some 16mm but I would bet he gave this very little thought. The amount of times I've actually put in a 16mm MEC in a domestic situation because of requirement I could probably count on one hand.

Chances are, the original 10mm MEC was absolutely fine. If so, he could leave the 6mm main bond in place.

You're absolutely correct (and you win the bet), I didn't think too much about installing 16mm² for the MEC because I thought I had to! Have I misunderstood something somewhere?
I'm going to read chapter 54 in the bgb again. I've lost count how many times I've read it, and unfortuneatly the misunderstood is sticking.
 
In short the MEC is a protective conductor, therefore it can be sized using an adiabatic equation. The bonding conductors can then be sized as not less than half the REQUIRED size of the MEC. TN-S system that is. Ergo sum, if the required size of the MEC is less than 10mm, then your main bonding conductors can be 6mm.
 
Isn't adiabatic for main earthing conductors and cpc? Didn't think you could use that for bonding?

bgb 543.1.1 ... , other than a protective bonding conductor, ...

supports keniff's post, unless I've misunderstood, in which case please clarify.
 
calculate the min. size required for main earth using adiabatic equation. then the bonding needs to be > or = to 50% of that
 
calculate the min. size required for main earth using adiabatic equation. then the bonding needs to be > or = to 50% of that

Z[SUB]s[/SUB] = 0.26 (at cu - included to verify voltage)

I[SUB]pf [/SUB]= 889 (measured)
t = 5 (main protective device > 32A)
k = 143 (from bgb table 54.2)

therefore:

S = (SQR (889[SUP]2[/SUP] x 5)) ÷ 143

= 13.9mm[SUP]2[/SUP]
Nearest larger conductor csa = 16mm[SUP]2[/SUP] and the bonding conductors 10mm[SUP]2[/SUP] (such that 10 >= 16 ÷ 2)

If my calculations are correct I've upgraded the cable correctly and now understand why! Please let me know if I've miscalculated.

I like Rampantchilli's solution below and bgb 544.1.2 applies if it's impracticable to connect within 600mm of the incoming water service. The property doesn't have a water meter.

Will this satisfy the niceic inspector?

quote_icon.png
Originally Posted by Rampantchilli
why not bond across all of your boiler pipes, including water, make a note on the cert



Read more: http://www.electriciansforums.net/newreply.php?do=newreply&p=696945#ixzz2Izwpi1Ap
 
Thanks for all your suggestions. I have several balls in the air at the moment and unable to sit in front of my computer.
It's a tn-s system.
I haven't checked the conductor resistance yet but I will a bit later when I've dropped some of the balls.
As far as I'm aware, the minimum CSA for bonding conductors is 6mm² for TN-S earthing systems.
TN-C-S systems require 10mm².

When you conduct the adiabatic equation, you either use measured values and the disconnection times for those measured values.
Or use those from the tables in appendix 3, not a mish mosh of the two.
As far as I can see, the current required for a 100A BS1361 fuse to operate within 5 secs is 630A, 889A would cause operation within about 1.5secs.
As you haven't supplied the information as to the rating or type of device, a BS1361 100A fuse is most likely the worst case scenario.
Either use the tabulated value of 630A along with the time 5secs, or use the measured value of 889A and the time of 1.5secs.

Using 889A and 1.5secs will result in an earth conductor CSA of 7.6mm².
Using 630A and 5secs will result in a CSA of 9.85mm².
Bonding conductors are required to be half of the CSA of that required for the earth conductor.
Either half of 7.6mm² = 3.8mm², or
half of 9.85mm² = 4.925mm².
In any event 6mm² is more than sufficient.
 
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Use the time from the time/current graphs in the appendix. This will make your calc more accurate.

This looks like the kiddie. I've got the csa of the mec to < 10mm[SUP]2[/SUP] so a 6mm[SUP]2[/SUP] bonding conductor should be fine.

Please sanity check my method. If it's right it's stuck and shouldn't give me any problems in the future. If it's not right, hopefully someone will correct me. Almost there:6:

Select Overcurrent Protective Device from bgb App3 e.g. Type B
Scan the 5 second line until it crosses the fuse rating curve e.g. 100A
Read from bottom e.g. 500A
Perform adiabatic equation using this value

(SQR(500[SUP]2[/SUP] x 5)) ÷ 143 = 7.8mm[SUP]2[/SUP]

This is hyperthetical as I don't have the relevant information at hand.
 
As far as I'm aware, the minimum CSA for bonding conductors is 6mm² for TN-S earthing systems.
TN-C-S systems require 10mm².

When you conduct the adiabatic equation, you either use measured values and the disconnection times for those measured values.
Or use those from the tables in appendix 3, not a mish mosh of the two.
As far as I can see, the current required for a 100A BS1361 fuse to operate within 5 secs is 630A, 889A would cause operation within about 1.5secs.
As you haven't supplied the information as to the rating or type of device, a BS1361 100A fuse is most likely the worst case scenario.
Either use the tabulated value of 630A along with the time 5secs, or use the measured value of 889A and the time of 1.5secs.

Using 889A and 1.5secs will result in an earth conductor CSA of 7.6mm².
Using 630A and 5secs will result in a CSA of 9.85mm².
Bonding conductors are required to be half of the CSA of that required for the earth conductor.
Either half of 7.6mm² = 3.8mm², or
half of 9.85mm² = 4.925mm².
In any event 6mm² is more than sufficient.

Thanks Spin, ships in the night spring to mind. Looking at your logic I understand and I believe my hyperthetical workings are similar to yours. Please let me know if you disagree.

If I've finally got it I'd like to thank all contributors to the thread and hope I can provide assistance to others in the future.
 

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