13A 1362 fuses and flex

I would be much more worried about installation conditions than exact fuse ratings

I recently saw an extension reel in a very sorry condition because someone had run it fairly near full load without unreeling it first. A couple of hours and the whole thing had sagged with the cables in the middle fused into each other

Have a look at Section 10.14 and Table 10.6 of the IET Code of Practice for In-Service Inspection and Testing of Electrical Equipment.

A 13 A or 10 A fuse may be fitted for appliances rated at 10 A, with 0.75 sq mm flex up to 2 m in length. Someone else has already done the sums to demonstrate this is probably going to be OK, including for fault current protection. BS 7671 permits the non-adiabatic approach in BS 7454 for fault current protection as well (but don't forget there may well be a requirement for protection against overload current).

And do we ignore conduction heat losses to the outside world via the supply cables?

Well the electricity has to join or leave at the ends of the test section of cable, depending how resistively that is done heat may flow in or out. But let us concern ourselves with the cable in the middle, and note that in these Gedankenversuchen  (*) we can simply imagine making  the experiment longer & longer until the end effects are far from the section of interest and so become safe to ignore.

In a real experiment, necessarily short, we would instead externally heat or cool the wire ends near the terminations to set a temperature at the ends close to the measured centre temperature, so there was no great lengthways thermal gradient to drive heat either way in that direction. In the language of the maths dept. we would impose a boundary condition of no significant longitudinal  heat flow. Its a lot cheaper than asking site facilities to provide an infinitely long thin lab bench and vacuum vessel ?

Mike

 * something like an experiment in the mind only, literally "thought-experiement"

Thanks for all your contributions chaps. I'm going to have to re-read some of these a few times. I'm particularly loving the disparity between what I would consider a fag packet computation and yours though! ? But then I'm one of those who, at least historically, has followed the numbers because I do err in favour of an easy life. Food for though though and clearly some new sums to learn… 

Very well said Mike and particularly Chris, I think that you are about right. This reinforces the long piece I wrote above, the time component of overload is very important, and usually not even realised to be there. 

mapj1: 
 

I assume your vacuum vessel is mirror lined so no radiation loss ?.

 

 

And do we ignore conduction heat losses to the outside world via the supply cables?

Z.

A real world example. 

A few months ago I found myself with mapj1 junior in the car at a remote place with a defective battery. The jump leads were in Mrs mapj1s car at the time. Then it began to rain.  That sort of day.

I was able to rustle up a short length of 2.5mm cable off the end of a reel, and the use of another car battery from an alarm engineer.

Thought process ran like this..

The car draws 250 to 300A to spin the starter.  How long can I crank the engine while my son and the nervous alarm man hold the 2.5mm onto the 2 batteries without burning either of them ?

Now I did not quite use Chris's figure of 3 mins at 20 A, but if I had it would have scaled to 3mins/100 at 200A, or about 2 seconds for a 40 degree rise.  I had a more optimistic figure, allowing a bit for the PVC, and as I did not have the regs with me, so used a pidooma figure from memory.

But I know the cable starts cold, I'm a bit more gung ho than the book ratings, as I do not care if the cable is ‘lifed’ afterwards  and so I can allow myself  a larger change in temp, perhaps 50 to 70C. 

That “sum” suggested 

“a-one and a-two and a-three.”  for cranking time, And relax and cool for 15 -20 seconds before retry  so average current is perhaps 20-30A.

As it happened, the engine started midway through the  2nd attempt, and the cable was blood warm.

When I described the thought process, the wide-eyed alarm installer said 

“ I'd never have dared to do anything like that -  2.5mm is for 16 amps innit ?” well maybe under some condition, not sure what alarm chaps get taught about installation methods and de-rating..

The only place that came slightly unstuck was one of the ends where the contact area to the battery was much less than the copper cross-section, and it spot welded. Well, sometimes wire cutters are a man's best friend, just keep the revs up a bit mapj1 junior while I find them.. 

I now have a shiny new car battery without a tag of 2,5mm wire stuck to the terminal. 

Mike.

PS and the jump leads are back in the boot. all is well with the world.

edited for a couple of nasty typos and readability. 

mapj1: 
I assume your vacuum vessel is mirror lined so no radiation loss ?.

Then I agree with the several mins figure for a 40 degree rise.

I just knew that one of you would mention radiation. ? But I am greatly reassured that my fag packet and pencil are not awry.

It all goes to show that the 10 min shower, 10 min drying cycle can be repeated several times without worrying.

I assume your vacuum vessel is mirror lined so no radiation loss ?.

Then I agree with the several mins figure for a 40 degree rise. The conductor only version is the basis of our old friend the adiabatic calculation, where it is so fast that the heat does not have time to stabilise to  a uniform temp throughout the PVC, so we pretend that no heat is lost to the insulation at all, as the temperature rise calculated that way is always a safe over estimate, so we safely under estimate the let-through energy the cable can stand.

In reality only AWE or CERN will install a cable in vacuum like that, and of course any real cable will heat quite a lot more slowly.

The heat up rates for the different mounting methods are variations on the problem of filling a leaky bath, with the complication that the rate of leak varies with the depth of water. Some things, like heating a 4 inch thick brick wall on one  side of the cable, have very long time constants, and others, like a cable sandwiched in between layers of something like Kingspan insulation are much closer to the ‘vacuum’ case, as the stuff has almost no heat capacity.

Mike

If I have my decimal point in the wrong place, please forgive me.

Take 1 m of 2.5 sqmm T&E and pass 20 A in a vacuum - so no heat losses.

Live conductors 2.5 x 1000 cumm = 2.5 cc

1 cc water weighs 1 g

Relative density of copper = 8.93

Mass of copper in one live conductor = 2.5 x 8.93 = 22 g

Total mass of copper = ((2.5+2.5+1.5) / 2.5) x 22 = 58 g

Mass of cable = 120 g/m (Eland Cables)

Mass of PVC = 62 g

Specific heat capacity of copper = 0.380 J/g/K

Heat up cable from 20 - 70 deg => 1103 J

Power loss in one live conductor = I2R = 20 x 20 x 7.41/1000 (OSG) = 2.96 W

Power loss in both conductors = 5.93 W

1 W = 1 J/s

Time taken to heat conductors = 1103/5.93 = 186 s = 3 minutes

Specific heat capacity of PVC = 0.9 J/g/K i.e. about 2.5 times copper

Time taken to heat whole cable = 10 minutes

Assume steady state at 70 deg C RM A (Table 4D5) so heat loss = 5.93 W

At the beginning, 100% of the heat generated in the live conductors is available to heat the cable. At 21 deg C, only 98% of the heat is available because the loss is 1/50 of the amount at the steady state. At 22 deg C, only 96% is available and so on. So looking at 1 deg C increments, it takes in fact 45 min to get to 69 deg C and you never quite get to 70 deg C. Please forgive me for having forgotten the calculus which I learned 45 years ago.

So there we are, a 2.5 sqmm cable loaded at 20 A will take over 45 minutes to get up to 70 deg C from 20 deg C enclosed in conduit in an insulated wall. Any smaller loading or less lagging and it will never get there because the heat generated in the conductors will always be less than the available heat loss.

That's my back-of-a-fag packet take on the situation.