Andy Wade wrote:
Eeeek - if you had 0.3 ohm contact resistance in a couple of connections
you'd have over 300 W being dissipated in inappropriate places (at 32
A). The short-circuit then would be the one that occurred as a result
of the fire...
You're right, of course - I was trying to handwave an "even if there's
notable contact resistance along the way" without thinking at all
seriously about probable values for same.
You've forgotten the source impedance at the point of supply. The
worst
case condition here is 0.8 ohm external earth fault loop impedance (Ze)
for TN-S earthing, so the total earth fault loop impedance (Zs) becomes
0.95 ohm, making the fault current only about 240 A.
Again "d'oh" applies - thanks for picking it up.
It doesn't take you into risk at all if you do the sums, provided that
the MCB is Type B. For Type B 'instant tripping' (i.e. 0.1 s) will
take place at = 5 * In, which is 200 A for the 40 A device. Since our
fault current is more than that, even with the worst case Ze value, we
know that the fault will clear within 0.1 s and thus that the
disconnection time requirement of BS 7671 is met.
It now just remains to check that the 1.5mm^2 CPC in the bit of 2.5 T&E
won't fry. The relevant limiting temperature for a PVC cable, by the
way, is 160 deg. 140 deg. applies for cables over 300 mm^2, which would
normally be thought of as beyond DIY territory :-). The simplest way to
tackle this is to use the adiabatic equation backwards to determine for
how long a 1.5 mm^2 conductor will stand 240 A. The equation is given
in BS 7671 as S = sqrt(I^2 * t)/k. Turning it round to find t gives t =
(k * S / I )^2. (k is the constant from Table 54C, S is the CSA of the
conductor and I is the fault current.) Hence t = (115 * 1.5 / 240)
which works out at a tad over half a second. With the B 40 A MCB we've
already determined that the fault will clear in 0.1 s, so all is well.
If you work through the same process using the same cable lengths, but
for different devices you'll find as follows:
- 30 A BS 1361 (cartridge) or BS 3036 (rewireable) fuse - OK
- 45 A fuse of either kind - not OK
- Type C MCB - not OK.
If the supply is PME we can take the max. Ze value to be 0.35 ohm rather
than 0.8. Zs now becomes 0.5 ohm and the fault current is 460 A. The
1.5 mm^2 CPC will stand this for 0.14 s. This doesn't change much
though. The 45 A fuses would still fail to protect the CPC, but a Type
C MCB would be OK (which is a bit academic because there's no reason
whatever to need to use a Type C device in this application).
There is now a clear answer to Tony's question (as modified):
- provided that the original cooker circuit is wired in 6 (or 10) mm^2
T&E, and
- the circuit length from the CU is not vastly in excess of 10 m, and
- the circuit is protected by a 30 A fuse or 32 A or 40 A Type B MCB,
- then the proposal to use short 4 mm^2 T&E cable 'tails' to the
appliances is perfectly OK. (I've said 4 mm^2 here because of the
concerns over the current rating of 2.5 when ambient temperature and
grouping factors are applied.)
QEF
Paul Cook couldn't've put it better ;-) Now, whether it's got too many
Numbers, Equazhuns, and Square Roots to be circulated as-is in NICEIC
helpful-hints is another question.
Stefek
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