On 17/08/2018 14:21, Dave Plowman (News) wrote:
In article ,
Andy Burns wrote:
Dave Plowman wrote:

Clive Arthur wrote:

Looking at the cable, it does say, among other things, 'KEMA-KEUR
300/300V 2X0.75mm2'

Think the regs have changed since such leads became common. All the
ones I've seen these days are capable of the full rated current of the
connector, regardless of how little the appliance uses.

With 0.7mm^2 cable, it's probably (hopefully?) only a 5A fuse in the
plug.

Aren't all flexes fitted to a 13 amp plug meant to be able to blow even a
13 amp fuse in event of a short? But not necessarily handle the full 13
amps long term?

Could be instructive to do some sums...

If one takes the max ELI for a circuit protected by a B32 MCB as 1.44
ohms (that's the 17th edition figure before it was slightly reduced by
the Cmin reductions in amendment 3).

Also lets assume the resistance of a 0.7mm^2 cable is in the order of 26
mOhms/m (1.0mm^2 CSA has a tabulated value of 18.10 mOhm/m). We can add
a further 46 mOhms to the circuit total.

That gets us a ELI with a fault at the end of the lead of 1.49 Ohms.

Our prospective fault current will therefore be 230 / 1.49 = 154A

That is slightly shy of the 160A needed for a "instant" trip on the MCB,

http://wiki.diyfaq.org.uk/index.php/...e-MCBTypeB.png

so let's look at the fuse:

http://wiki.diyfaq.org.uk/images/d/d...FusingTime.png

That looks well below the 0.1 sec time on that. So lets take a
pessimistic worst case of 0.1 secs.

So the remaining question is what size conductor is required to carry
154A for 0.1 sec? If we assume PVC insulation hence a k factor of 115 we
get:

sqrt( 154^2 x 0.1 ) / 115 = 0.42 mm^2

So we can conclude that the flex will withstand that current for long
enough to blow the plug fuse.


--
Cheers,

John.

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