Lollipop final circuit

[Paul_C]

At 50 Hz the difference is small enough to be almost insignifcant. But not completely insignificant.

As needs to be considered when talking about very large cables, and more so when designing power systems for higher frequencies such as the 400Hz used on board some ships etc.

But it's insignificant enough at 50Hz for the cable sizes being discussed here that it can be ignored.

 

[JohnW2]

Good points!

Something about current being a flow of electrons which are all negatively charged and try to repel away from eachother, meaning more of the electrons gather around the outside of the copper, and the core of the conductor has very few. (that's one of the reasons why multi strand wires can carry more current than single strand/solids.)

Indeed. It's called 'surface effect' or 'skin effect'.

As such the current carrying capability of a conductor is more related to its circumference than its cross sectional area.

That is certainly the tendency in terms of impedance of the conductor, hence voltage drop for a given current, but not very significant at mains frequency. Current-carrying capacity (CCC)is a different matter, since it's primarily about temperature rise in the cable. Even if it's the peiphery of the conductor that carries most of the current, the generated heat will distribute throughout the whole of the cross sectional area (CSA), hence that also becomes a factor in determining temperature rise, hence CCC..

The circumference of a 2.5mm2 conductor is 2.8mm, while the circumference of a 6mm2 conductor is 4.3mm. The combined circumferences of TWO 2.5mm2 conductors is therefore 5.6mm, which means that this will carry MORE current than a single 6mm2 conductor.

As above, the calculations are not anything like as simple as that. However, the officially tabulated figurse for current carrying capacity are certainly anything but linear in terms of CSA. (using 'clipped direct' T&E cable as the example throughout), in terms of your example, it does roughly work as you say - 2 x 2.5mm² cables have a total CCC of 54A, as compared with 46A for 6mm². Moving up, 4 x 2.5mm² cables would have a CCC of 108A, compared with only 63A for 10mm².

I think the bottom line is that determination of CC (which is probably empirical, rather than theoretical) is a complicated matter - but you are certainly correct that, in a discussion as that here, people should be looking at the combined current carryng capacities of two cables, not the combined CSA.

Kind Regards, John

 

[JohnW2]

But it's insignificant enough at 50Hz for the cable sizes being discussed here that it can be ignored.

I'm sure that's true - but, as I've just written, what is also true is that the BS7671 tabulated CCCs are (for other reasons) anything but linear in terms of CSA, so we should avoid thinking in terms of 'combined CSA' in discussions like thhe one in this thread - and I've been at least as big a culprit as anyone else!

Kind Regards, John.

 

[EFLImpudence]

The circumference of a 2.5mm2 conductor is 2.8mm, while the circumference of a 6mm2 conductor is 4.3mm.

You have made an error to a factor of 2.

2.5mm² solid core has a circumference of 5.6mm
6.0mm² solid core has a circumference of 8.7mm

However, a seven strand
6mm² has a combined circumference of 22.97mm.

 

[ban-all-sheds]

But not a CCC 4.1x that of 2.5mm²....

 

[bernardgreen]

The most important parameter is the ratio of surface area to volume when heat disipation is critical.

Heat disipation from current carrying metal is proportional to surface area while heat generation is proportional to volume.

Double the radius of the wire and the volume per unit length increases to times ( A = pi x R squared ) but the surface area of the same length only doubles ( C = 2 x pi x R )

 

[ban-all-sheds]