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13A 1362 fuses and flex

Good evening everybody. 

I've been cogitating on the fusing factor of 1362 fuses (specifically 13A fuses)  and how this correlates with the protection of a 1.5mm2 flexible cable. As ever I am hoping you can shine a light!

The code of practice for the in service inspection and testing of equipment  Table 15.6 states that for flexes to be protected by the fuse in a BS1363 plug there is no limit to their length providing that their csa's are as in table 15.6 which states a minimum flex size of 1.25mm when using a 13A fuse. I am minded that it is quite common site to see a multi-gang extension lead on sale using 1.5mm2 flex where there is obviously potential for overload given the unknown nature of what would be plugged into them (even though there will be a warning not to intentionally do so).

From the Beama guide:

2.5 The BS1362 Fuse
The UK uses a fused plug which must be fitted with a BS 1362 fuse. For domestic
installations the use of the BS 1363 plug and socket system and the fitting of a BS 1362 fuse
into a plug is a legal requirement under the UK Plug and Socket Safety Regulations, 1995.
With a correctly fused BS 1363 plug, the flexible cable connected to equipment is always fully
protected against the effects of overload or small overcurrents as follows:
3A fuse protects 0.5mm² cords
5A (6A) fuse protects 0.75mm² cords
13A fuse protects 1.25mm2 cords
Protection against excessive damage by a short circuit is still achieved even if the smaller
cord sizes are inadvertently protected by a 13A fuse. In addition, it has been accepted in the
UK that some marginal damage to small flexible cords is tolerable under short circuit
conditions, for example where a 0.22mm² cord is used with a 13A BS 1362 fuse.

As far as I understand it the fusing factor of a 1362 fuse BS 1362 fuse is 1.9 (0.763) although in fairness I have seen lower fusing factors quoted (1.66?? which removes the particular problem I am wrestling with.)

Reference 4F3A a 1.5mm2 single phase AC flexible cable has a tabulated current carrying capacity of 16amps. 16x0.763 = 12.208 amps which is obviously less than the 13 amp rating of the fuse. 

I find it quite common to see 13A 1362 fuses inline on 32A cooker circuits protecting 1.5mm2 flexes to ovens. Is this deemed acceptable even though the oven isn't strictly speaking a fixed load (fan motor etc.)?

Is there another factor at play here which I am missing? Or do I just have the wrong fusing factor!

Thanks for your help in advance.

Parents
  • 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.

Reply
  • 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.

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