Dropping cable down cavity.

[Ian Stirling]

It also helps that the wire is very small, generally, in relation to the insulation - the apparent effectiveness of the insulator drops off as you go out.

Considering the simple case of a 1cm diameter cable in the middle of an infinite fiberglass insulator. Imagine the insulator split into millimeter thick cylinders. The first millimeter of the insulation has a perimeter of 32mm or so. The 50th mm has a perimeter of about 340mm.

So, the 50th millimeter adds 1/10th as much as the first one.

As an example, with fiberglass, 0.04W/m/K, that's 40w/mm/k for a 1mm thick 1m square sheet.

Or about 1.5w/m/k for 1m of flex insulated with 1mm of fiberglass.

10mm gets you about .21w/m/k, 50mm about .1, and 100mm 0.08.

A whole meter only about halves this.

1mm^2 cable has a resistance of 1.7*10^-2 ohms (.017 ohms). 2.5mm^2 cable, for both directions has about .013 ohms per meter.

Taking 30A through this dissipates 12W or so, or allows at most 10mm of insulation or so (temp rise of 60C ish).

Derating to half gives a power per meter of 3W. With 10mm, a rise of 14C, even 100mm is quite safe, with a rise of only

40C. You have to get pretty near a meter of insulation on each side of the wire, to get close to being dangerous.

A simple metal heat spreader, next to the wire, even if it's buried in insulation can help lots. Actually, it doesn't even need to be metal, it just needs to be a comparatively poor insulator. Timber helps lots.

(If you use this as the basis for running cables through insulation, and your house catches on fire, I reserve the right to point and laugh)

[ironer]

/snip/

/snip/

interesting calculations. I usually acept and rely on calculated results but in this case I regret I'm not convinced enough. The derating factors in the table in the IEE regs just don't level off in the way you normally get when a limiting value is near. On top of that I just don't trust FTE in close proximity to good insulation either.

One question I always consider before doing any work is 'If this job is done that way, what's the chance it will have to be done again while I'm still around?" In this case the answer is too near Yes for me to risk it.

[Ian Stirling]

Fair enough, I am pretty sure that they reflect the assumptions accurately, it doesn't get much simpler - I simply took the area of 1m long

1mm thick cylinders, from the surface of the cable out. Obtain U values for each shell, invert them to get R values, add all the R values, and then invert for an overall U value, and multiply to get the temp rise for a given power.

I'm pretty confident that this will give a fairly accurate figure.

(I needed to write this one-liner anyway, as I wanted to work out how much adding an extra 500mm of insulation on my hot water tank would do)

However. Under fault conditions, where higher than the design current can flow, even relatively small amounts, for relatively short times, this can lead to significant heating over the above calculated figures.

An interesting point about this is that thicker cables, with the same copper content have a significantly higher current carrying capacity, when insulated - as the plastic the cable is made of is a very poor thermal insulator, and acts to spread the heat.

[ironer]

/snip/

. On top of that

because I burnt my fingers on that one once!

Just like to make it clear I'm not casting doubt on your calculations & obvious expertise, but sometimes theory needs backing up with experiment. Maybe this is one such case where a practical demonstration would show all's well with the theory & nothing disastrous like a rewire will be needed n years down the line..