4 x 2.5mm = 10mm (Just Thinking Out Loud)

[OnlQQker]

So there I am trying to get to sleep at 2am this morning when a thought popped into my head. (I'm one exciting person to be with ) Not that I am thinking of doing it, ever done it or seen it done, I will still ask the question as otherwise I will be up again all night trying to work it out.

So 4 x 2.5mm cable = 10mm, now sparks I have been around swear about the 10mm stuff, especially when it comes to connecting it up in pull cords and wall sockets. The only reason I could think of why people don't use 4 x 2.5mm cables is...

1/ You've got to run 4 cables
2/ Somebody in the future might connect a socket into it
3/ 4 x 2.5mm cable is more expensive

But, though this is only a hypothetical question, would it still perform in the exact same way as a 10mm cable, or am I missing something?

 

[pc1966]

Off the top of my head I see the main problems (aside from the extra space all of the quadruple insulation is taking) are:

• Harder to be sure all 4 are properly gripped by the terminals
• Maintaining an equal share of current (especially if one core has a poor or non-existent connection, how do you check)

Also I am not sure about it being identical to 10mm even though it has the same CSA as the thermal resistance from conductor to ambient is probably different.

 

[westward10]

Same csa but 3×2.5 will have greater ccc than a single 10.0 so four are not necessary. 40A shower circuit with a suitable ref method you could use 2×2.5

 

[static zap]

If a joint fails and all the current finds its way into 1 of 3/4 cables ..will whole cable run let out the smoke ?

 

[Lucien Nunes]

Parallel-connected cables are used in the 100s of mm² range. As mentioned above, a key factor is making sure that the paralleling connections are reliable enough to absolutely minimise the chance of the full current being carried by less than the full number of conductors. On the large circuits there is also the question of short-circuit currents, which can distribute unevenly due to differences in inductance caused by the physical layout of the conductors in situ, all of which have to be planned for.

The reason for the apparent reduction in total conductor CSA needed, e.g. 3x 2.5mm² instead of 10mm² is that the bottlenecks for heat dissipation, and hence CCC, are the interface between conductor and insulation (where the insulation is hottest, and from insulation / sheath to ambient. For a constant radial thickness of insulation, the smaller the individual conductors are, the greater the circumference per unit CSA, so the greater heat-dissipating capability and the higher the current-density at which the conductor can operate. Grouping factors apply to paralleled cables just as for separate circuits. In extreme cases, because of the thermal resistance of all the extra insulation and air pockets bundled up amongst the conductors, the CCC might be lower than for the single cable despite the apparent increase in overall circumference (I haven't checked the numbers.)

 

[Ray Pooley]

So there I am trying to get to sleep at 2am this morning when a thought popped into my head. (I'm one exciting person to be with ) Not that I am thinking of doing it, ever done it or seen it done, I will still ask the question as otherwise I will be up again all night trying to work it out.

So 4 x 2.5mm cable = 10mm, now sparks I have been around swear about the 10mm stuff, especially when it comes to connecting it up in pull cords and wall sockets. The only reason I could think of why people don't use 4 x 2.5mm cables is...

1/ You've got to run 4 cables
2/ Somebody in the future might connect a socket into it
3/ 4 x 2.5mm cable is more expensive

But, though this is only a hypothetical question, would it still perform in the exact same way as a 10mm cable, or am I missing something?

Don't see why not. It's the cross sectional area that matters. Voltage stays the same and current is shared evenly. Whether it is all in one or split into quarters is irrelevant. It's basically the difference between single core and multi strand cable. The latter tends to be more flexible.

 

[7029 dave]

I have never used this method of bunching cables together to maximise csa, and never will.

 

[telectrix]

it would work.... but so did the titanic.

but back to the 4 cables.... you might have all the right cables, but not necessarily in the right order.

or a bit like four candles fork handles.

 

[Ray Pooley]

I have never used this method of bunching cables together to maximise csa, and never will.

To be fair it was just a hypothetical question. You would have to be desperate to do it but if you were desperate it wouldn't be a problem technically speaking.

 

[Simon47]

It is allowed by the regs, and if you think about it, is used to some extent in nearly all homes by way of an RFC. 533.7, 433.4, 434.4
That last one is the killer though - calculate the requirements for fault protection if one of (say) 3 cables gets damaged along it's length. That damaged cable will be fed from the supply end from an OPD that may be rated for the total overall CCC of the 3 cables - but now a single cable is carrying the fault current. The damaged cable is also fed from it's load end via the other two cables. That's not to say you can't just use a single device - but you'd need to work out what the fault impedance would be under worst conditions, what OPD would be appropriate for that, and do the adiabatic equation to check that the let through energy would be within the cable's capacity to handle it.
I did set out to try and work it out, but my head hurts now
And I guess that's why it's not common to do it ?
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Ah, sat down at the computer and worked part of it out. I reckon the worst case would be when the fault impedance is maximum - so minimum margin for the OPD tripping. For 3 cables in parallel, that's when the fault is 3/4 of the way along from the supply end. If the fault is at the supply end then obviously the fault impedance is 0. If the fault is at the remote end, then the fault impedance is 1/3 of that for a single cable - equal current sharing, so again no problem.
As you move from the supply end, the impedance increases and reaches 1/3 at half way along - so a fault in the first half of the run is no worse (for disconnection times) than one at the far end. But in terms of adiabatic cable heating, the faulty cable will take far more than it's fair share of the fault current.
After that, the fault impedance carries on rising till it reaches a paek at 3/4 of the run, and then reduces again to 1/3 of a single cable at the far end.

For 4 cables, the maximum fault impedance is at 2/3 of the run.

I guess the next bit is to work out the current share and find the worst case. But then it's not "simple" maths problem, as it involves the tripping characteristics and supply impedance to work out the effect on the faulty cable. It's fairly obvious that the highest fault current occurs when the fault is close to the supply - but then the OPD is going to trip/blow fastest under those conditions.