6mm Earth and Main protective bonding conductors

[Michaelwgroves]

So my equation looks like this;

I=230/0.22(Ze) = 1,045
I2=1,045*1,045 = 1,092,025
I2*0.1 (t) = 109,202
Square Rooted = 330
330/143 (70 thermoplastic not bunched) = 2.3

So I only need 2.3 mm earth, something must be wrong?

 

[davesparks]

Ian1981, sorry read and forgot your first post, got tied up with bedtime.......

I don't get how you get your 2300 amps, my Ze is 0.22, what calculation do I need to do?

Yes it's 60A, but how do I calculate fuse disconnect time?

I don't have the red version. Just checked latest version, it says replaced by BS 88-3
So I'm now on the right table, but don't know how to read it?

This is ridiculous!

You don't know how to calculate current, you don't know how to establish fuse disconnection times, you can't read the table once you've found it.
Seriously, stop attempting to do electrical work and hire a competent electrician.

 

[westward10]

Back to #1, I don't think you can ever prove the main protective bonding was adequate at the time and hence that it can be considered safe for continued use. The 15th Edition 1981 never gave a minimum size for main equipotential bonds if PME was employed just that you should consult the supply company, without digging out the book that is from memory so I stand to be corrected.

 

[davesparks]

So my equation looks like this;

I=230/0.22(Ze) = 1,045
I2=1,045*1,045 = 1,092,025
I2*0.1 (t) = 109,202
Square Rooted = 330
330/143 (70 thermoplastic not bunched) = 2.3

So I only need 2.3 mm earth, something must be wrong?

Why must something be wrong? I haven't checked the figures but that is within the ballpark of what you would normally expect for a small supply like this.

 

[Ian1981]

So my equation looks like this;

I=230/0.22(Ze) = 1,045
I2=1,045*1,045 = 1,092,025
I2*0.1 (t) = 109,202
Square Rooted = 330
330/143 (70 thermoplastic not bunched) = 2.3

So I only need 2.3 mm earth, something must be wrong?

No that is correct.

 

[Michaelwgroves]

Why must something be wrong? I haven't checked the figures but that is within the ballpark of what you would normally expect for a small supply like this.

Just seems so small when new installations require 16mm

 

[Ian1981]

Just seems so small when new installations require 16mm

Because everyone used the tables in bs 7671 rather than the calculation
Plus the DNO may stipulate the CSA

 

[Michaelwgroves]

Just so this sinks in, I calculated fault current 1,045A using measured Ze. But I could have equally used the fault current calculated by my meter of 1,080 A ?

 

[davesparks]

Just seems so small when new installations require 16mm

Instead of blindly following a book try actually thinking about what determines the size and what the conductor is required to do.

It is not required to carry the current for any appreciable length of time, it only has to carry it for maybe one tenth of a second.
To take a very simplistic view look at a piece of 60A fuse wire, as long as your earthing conductor is bigger than that it will do its job.

 

[Michaelwgroves]

I like you analogy.