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I have read Lennys sticky on old cables, but did red and black ever have the CPC less than 1.5 on 2.5 T&E,

Pict
 
Yes,, they used to be 1mm.. and with BS3036 fuses protecting such cable there can be issues due to the current carrying capacity under fault conditions... (gets ready to defend my new post)
 
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Taken from appendix c of Approved Document P.


2.5 mm2 twin-and-earth cables

incorporating circuit protective
conductor of only 1.0 mm2

For some years, 2.5 mm2 twin & earth pvc/pvc
cables to BS 6004 were manufactured with a
circuit protective conductor (cpc) of only 1 mm2,
rather than 1.5 mm2 as is incorporated today.
The size of the cpc was increased to 1.5 mm2
in BS 6004 because in certain circumstances
the 1 mm2 cpc may not always be properly
protected against thermal effects in the event
of an earth fault. This is where the cable is
used in a ring final circuit protected by a 30 A
semi-enclosed (rewirable) fuse. If this is the
case, a competent electrician should be
consulted about upgrading the cables and/or
the consumer unit.
 
Given the nominal current carrying capacity of a 1.0mm cable at 10 amps and a fusing factor of x1.5 on a rewirable fuse , ring final sub circuit = 2x1.0mm =20 amps approx, whereas the 30 am fuse actually blows at 30x1.5=45 amps, this makes a lot of sense. Thanks Lenny
 
Just thought you lot might like to know how we get from a fusing factor of approximately 2 to a correction factor for Cc of 0.725

If ff = I2/In and I2 = 1.45 Iz
Then
I2 = In * ff
so In * ff = I2 = 1.45 Iz
Therefore
In * ff = 1.45 Iz
and
In = 1.45 Iz / ff

substitute 2 for ff
and we have
In = 1.45/2 Iz
In = 0.725 Iz

So as long as the fuse rating is approximately 3/4 of the maximum current carrying capacity of the conductors the BS3036 fuse will provide overload protection

As far as earth fault protection is concerned

application of the adiabatic equation for a 30A protective device gives (√210² × 0.4) / 115 = 1.15mm²
Which means a 1.0mm2 cpc is insufficient to stop damage to the insulation of the circuit conductors

Now if a 20A (Yellow) fuse was substituted then (√130² × 0.4) / 115 = 0.72mm²

and the circuit would now comply!
 
very informative, ackbar. problem is in obtaining 20A fusewire. and before anyone else suggests it, a 15A and a 5A in parallel just ain't worth thinking about.LOL.
 
... ring final sub circuit = 2x1.0mm =20 amps approx, whereas the 30 am fuse actually blows at 30x1.5=45 amps, this makes a lot of sense. Thanks Lenny

Not forgetting that rings are rarely completely balanced in practice - most likely one leg of the 1mm would carry more than an exact 50% of a fault current, especially in a badly-designed installation with, for example, a kitchen, or a heavy fixed load at one end of the ring.

BTW, for £5 here's enough 20A fuse wire for a lifetime :)

TLC-direct .co .uk / products / TLFW20.html
 
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