If it's a TT supply, then PME conditions can not apply, so a 2.5mm² main earth is sufficient providing it is protected against mechanical damage.
No Earth from Supply on front of terraced house
- Thread starter scouseneil
- Start date
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see post above, but yes you can, for a supply with 2.5mm² tails..
not very likely..
again, see above but yes, as long as the supply is 2.5mm²
Table 4.2 in the OSG
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screw the OSG.. it's exactly that a guide..
the regs state quite clearly what I've stated above, if the guy writing the OSG can't interpret that then he shouldn't be writing it..
regs book..
fig 2.1 in definitions..
number 3 - earthing conductor
number E - means of earthing ( TN systems )
number T - earth electrode ( TT and IT systems )
number B - main earth terminal
so "earthing conductor" connects the MET to the source of earthing..
regulation 542.3.1 - Every "earthing conductor" shall comply with Section 543
543 - Protective conductors - the cross section of a protective conductor, other than a bonding conductor, shall be;
(i) calculated in accordance with Regulation 543.1.3 ( the adiabatic ), or
(ii) selected in accordance with Regulation 543.1.4 ( selected from table 54.7 the jist of which I posted above )
makes no discrimination between TT or TN systems..
so the OSG quite clearly has it wrong..
if you have a 2.5mm supply then you can have a 2.5mm burried earthing conductor, providing it's mechanically protected, 16mm if it's not protected..
the regs state quite clearly what I've stated above, if the guy writing the OSG can't interpret that then he shouldn't be writing it..
regs book..
fig 2.1 in definitions..
number 3 - earthing conductor
number E - means of earthing ( TN systems )
number T - earth electrode ( TT and IT systems )
number B - main earth terminal
so "earthing conductor" connects the MET to the source of earthing..
regulation 542.3.1 - Every "earthing conductor" shall comply with Section 543
543 - Protective conductors - the cross section of a protective conductor, other than a bonding conductor, shall be;
(i) calculated in accordance with Regulation 543.1.3 ( the adiabatic ), or
(ii) selected in accordance with Regulation 543.1.4 ( selected from table 54.7 the jist of which I posted above )
makes no discrimination between TT or TN systems..
so the OSG quite clearly has it wrong..
if you have a 2.5mm supply then you can have a 2.5mm burried earthing conductor, providing it's mechanically protected, 16mm if it's not protected..
S
scotsparky
You got is Coljack
ifyou follow the genaral rule of thumb which is
Main earth---16mm
Main Bond----10mm
Suplimetary---4
In a domestic propert
you cant go far wrong
The ONS does contain good information but it has to be used in conjunction with the big book
ifyou follow the genaral rule of thumb which is
Main earth---16mm
Main Bond----10mm
Suplimetary---4
In a domestic propert
you cant go far wrong
The ONS does contain good information but it has to be used in conjunction with the big book
But by using the adiabatic, on a TT supply a 2.5mm² is sufficient to comply with section 543
go on then, prove it..
where do you get the I value from anyway? i can't remember...
where do you get the I value from anyway? i can't remember...
543.1.1 contains a very important small word: OR
You CAN use table 54.7 - in which case the minimum size is generally half that of the line conductor
OR you can determine the minimum size by calculation. NOT both.
For a TT system, the worst possible scenario is a disconnection time of 1 second (411.3.2.4).
In reality it will be much less, since the protective device will be an RCD, and the disconnection time would have to be less than 0.2 seconds for most circuits anyway (Table 41.1)
This reduces the top half of the adiabatic equation to just I (since t is 1, and the square root of I squared is just I)
Table 54.2 gives the value of k as 143
Therefore you have S = I/143
Since the minumum value of S is 2.5mm, 2.5 = I/143
With simple rearrangement, 2.5 x 143 = I, so I = 357.5A
This is the maximum fault current which can be permitted using a 2.5mm protective conductor.
To get this, you would require an earth electrode with an impedance of about 0.64 ohms.
Clearly this is not going to happen, most earth electrodes will be many tens of ohms at best, and quite likely into the 100s of ohms.
With disconnection times of 0.2s, the required fault current is approximately 800A, or an electrode impedance of 0.28 ohms. Not going to happen.
You CAN use table 54.7 - in which case the minimum size is generally half that of the line conductor
OR you can determine the minimum size by calculation. NOT both.
For a TT system, the worst possible scenario is a disconnection time of 1 second (411.3.2.4).
In reality it will be much less, since the protective device will be an RCD, and the disconnection time would have to be less than 0.2 seconds for most circuits anyway (Table 41.1)
This reduces the top half of the adiabatic equation to just I (since t is 1, and the square root of I squared is just I)
Table 54.2 gives the value of k as 143
Therefore you have S = I/143
Since the minumum value of S is 2.5mm, 2.5 = I/143
With simple rearrangement, 2.5 x 143 = I, so I = 357.5A
This is the maximum fault current which can be permitted using a 2.5mm protective conductor.
To get this, you would require an earth electrode with an impedance of about 0.64 ohms.
Clearly this is not going to happen, most earth electrodes will be many tens of ohms at best, and quite likely into the 100s of ohms.
With disconnection times of 0.2s, the required fault current is approximately 800A, or an electrode impedance of 0.28 ohms. Not going to happen.
was that an agrument for or aginst 2.5? i can't tell
and where do you get the value for I from? I can't remember.. it's been a while since I had to use it.. generally rule of thumb it.. it's easier..
and where do you get the value for I from? I can't remember.. it's been a while since I had to use it.. generally rule of thumb it.. it's easier..
In favour of 2.5mm - the point is that to require a mnimum size greater than 2.5mm you need a fault current of at least 357.5A, and probably 800A for most circuits.
This will not be happening on a TT supply, so 2.5mm is acceptable.
The value of I is the fault current (defined on page 128).
This will not be happening on a TT supply, so 2.5mm is acceptable.
The value of I is the fault current (defined on page 128).
I'd just worked that out too
FYI, I=Uo/Ze
FYI, I=Uo/Ze
S
scotsparky
If you going worst case then the maximum Eli from electricity copanies are
TN_C_S 0.35OHMS
TN-S 0.8OHMS
TT 21OHMS
TN_C_S 0.35OHMS
TN-S 0.8OHMS
TT 21OHMS
The value of I is simply the supply voltage divided by the resistance of earth fault path via the electrode.
If you electrode is 100 ohms then 230/100 = 2.3 amps.
If you electrode is 50 ohms then 230/50 = 4.6 amps.
Stick it into the adiabatic and you'll get a rediculously low minimum size! (not suggesting you should use anything less than those in the tables tho!)
If you electrode is 100 ohms then 230/100 = 2.3 amps.
If you electrode is 50 ohms then 230/50 = 4.6 amps.
Stick it into the adiabatic and you'll get a rediculously low minimum size! (not suggesting you should use anything less than those in the tables tho!)
21 ohms is the maximum resistance to ground of their electrode, it doesn't include your electrode.
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