Lollipop circuit on a 40A RCBO

[HappyHippyDad]

2x 2.5 in parallel is not the same as 1x 5mm cable, this is why parallel cables are used for larger supplies.

That was sort of my point, well, the fact that we cannot just 'double up'. However, could you explain why a 2 x 2.5mm could take more current than 1 x 5mm as you've stated?

The current carrying capacity of the 6.0 mm², 'stick' of the lollipop is well protected by the 40 A OCPD.

How then do you explain the 32 A OCPD of a 2.5 mm² RFC when a single piece of cable may be limited to a CCC of 13.5A? I suspect that in this case, the OCPD should be 'downrated' to 20 A.

GB, you've gone from saying it can take 54A to 20A? I'm confused :confused5:

 

[GB.Kayak]

That was sort of my point, well, the fact that we cannot just 'double up'. However, could you explain why a 2 x 2.5mm could take more current than 1 x 5mm as you've stated?



GB, you've gone from saying it can take 54A to 20A? I'm confused :confused5:

I am sorry for causing confusion. M's, concern ... I think was for the 'ring' in 2.5 mm² at the 'head' of the 'lollipop'. The 40 A OCPD provides adequate protection for 'stick' in 6.0 mm² cable if it has an allowable CCC of 54 A. However, the cable at the 'head' will have a much lower CCC, less than the 40 A of the circuit's OCPD. When this cable is arranged as a RFC, the circuit's CCC is higher in non-fault conditions than that of a radial circuit using the same cable. The question is, how much higher? In some cases, depending on the installation method, a 32 A OCPD does not provide adequate protection to a RFC cabled in 2.5 mm², particularly if that circuit is enclosed in insulation which reduces the allowable CCC of the cable; hence the requirement for a lower rated OCPD ... and the next lower is 20 A in my memory serves me correctly.

 

[davesparks]

That was sort of my point, well, the fact that we cannot just 'double up'. However, could you explain why a 2 x 2.5mm could take more current than 1 x 5mm as you've stated?
:

The thing is you can just double up for a parallel supply, a ring is slightly different though.

I can explain why, but probably not very well after consuming so much Guinness!
If you look through tables in the regs you will see that the ccc of a cable will increase roughly 1.5x for a 2x increase in csa, this is to do with the thermal performance of the cable amongst other things.

 

[HappyHippyDad]

The thing is you can just double up for a parallel supply, a ring is slightly different though.

I can explain why, but probably not very well after consuming so much Guinness!
If you look through tables in the regs you will see that the ccc of a cable will increase roughly 1.5x for a 2x increase in csa, this is to do with the thermal performance of the cable amongst other things.

Yes I can see that ccc increases roughly 1.5x for a 2x increase in csa.. see pretty (non-linear) graph just drawn:



But I cant see how 2x2.5mm gives more CCC than 1x5mm, they both have the same CSA?

Bit late actually Dave, so I'll see if you've written anything tomorrow as I'm off to bed!

Ps.. I think I get it! A ring would give more CCC than a 5mm due to the thermal performance (and other) that you've mentioned, as it is 2 smaller CSA's compared to 1 larger CSA (even though the total CSA would be the same). The 1.5/2 ratio of CCC to CSA (as in the graph) shows this.

That could be absolute nonsense and I really have had enough now!

 

[davesparks]

Yes I can see that ccc increases roughly 1.5x for a 2x increase in csa.. see pretty (non-linear) graph just drawn:

View attachment 28684

But I cant see how 2x2.5mm gives more CCC than 1x5mm, they both have the same CSA?

Bit late actually Dave, so I'll see if you've written anything tomorrow as I'm off to bed!

It's got a lot to do with the ability of the cable to dissipate the heat generated by the flow of current. Heat dissipation is proportional to the surface area of a cable, and the surface area does not increase in a linear manner when compared to the increase in csa, all sorts of funky numbers like pi come in to play.
Current ratings in the book are the maximum current which can flow without taking the conductor above the rated temperature for the insulation.

 

[Engineer54]

From first principles I would interpret that a 40 A OCPD might be acceptable on a 2.5 mm² RFC or lollipop circuit, if the CCC of the cable allows. I am open to correction but it seems to me that BS 7671, TABLE 4D5, would allow 2.5 mm² T+CPC cable and a 40 A OCPD to be used 'clipped direct' for example for the circuit in question, the CCC of the cable being 27 A for this installation method and therefore in a ring capable of carrying up to 54 A. However, if the cable were wrapped in insulation at the centre of an insulated stud wall then a 32 A OCPD would not give adequate protection as the CCC is reduced to 13.5 A requiring at most a 20 A OCPD in a RFC.

A RFC is not a true parallel circuit and shouldn't really be considered as such. A RFC has multiple points of load along it's length and therefore cannot meet the requirements of a parallel circuit, unlike a true parallel circuit that has a single point load. There is a formula that that is appropriate to sizing RFC cable sizes but i'm buggered if i can remember it!! lol!!

Staggered ring final circuits are far better at dispersing loads over the whole ring circuit than standard rings that that are wired CU - socket to socket - CU, as there are no long runs back to the CU..... Depending on the overall length of the 2.5mm section and other circuit based criteria, i doubt if a 40A OCPD would cause any problems Apart from anything else, it is extremely rare for normal domestic ring to exceed it's 32 A rating. I'd still change out the existing 40A for a 32A OCPD mind, as i wouldn't have a clue how the circuit has been installed...

 

[GB.Kayak]

Surely it's all down to the assumptions made about where on the ring the load is.

If all the load is at the middle of the ring, then the current will be split equally between the two routes between origin and load. If you have multiple loads spread evenly along the ring, then again you'll have an equal current split.

However, if you have all the load at one location near to the origin, then the current will split in the ratio of the resistance of the two routes which, assuming the same cable used throughout, will be in proportion to the two lengths involved.

For example, if you put a 32A load 15% along the ring, you'll get about 27A in the short leg and about 5A in the long one. A load of 40A at the same point on the ring will give 34A and 6A.

... that said, I think that I am correct in saying that the CCC of a cable is related to the heating effect of the power dissipated due to its resistance exceeding the material property limits of the insulation.

That power is given by:

P = I² x R

However, the effective resistance of a RFC is 1/4 that of a radial circuit of the same length and therefore the current required to reach the same temperature in a RFC is ~ twice that in a radial circuit.

 

[GB.Kayak]

... but i'm buggered if i can remember it!! lol!!

Reputation quotient is much reduced in my estimation! ;-P

 

[GB.Kayak]

It's got a lot to do with the ability of the cable to dissipate the heat generated by the flow of current. Heat dissipation is proportional to the surface area of a cable, and the surface area does not increase in a linear manner when compared to the increase in csa, all sorts of funky numbers like pi come in to play.
Current ratings in the book are the maximum current which can flow without taking the conductor above the rated temperature for the insulation.

d, I hadn't read this when I wrote my post ... honest!

 

[Silly Sausage]

funky numbers like pi

Transcendental Man, transcendental.

e is much cooler!

 

[Engineer54]

Reputation quotient is much reduced in my estimation! ;-P

It unfortunately all comes down with age!! It may very well spring back into the light. ...lol!!

 

[HandySparks]

I've always found it odd that we have such specific tables for all the different CSA of cable but nothing for a ring, apart from appendix 15 which states a 30A or 32A OPD.

For the usual 32A ring circuit, this is covered by BYB 433.1.204, which states that the circuit meets the requirements of 433.1.1 (protection against overload) if the CCC of the cable, as installed, (Iz) is not less than 20A and if the load current in any part of the circuit is unlikely to exceed Iz for long periods.

ie, don't design in a heavy load immediately adjacent to the origin, as the current won't be shared adequately.

 

[HandySparks]

It's got a lot to do with the ability of the cable to dissipate the heat generated by the flow of current. Heat dissipation is proportional to the surface area of a cable, and the surface area does not increase in a linear manner when compared to the increase in csa, all sorts of funky numbers like pi come in to play.
Current ratings in the book are the maximum current which can flow without taking the conductor above the rated temperature for the insulation.

You're right that the CCC is down to the ability to dissipate heat. The relevant 'funky numbers' here are that, for solid core, the csa of the core is pi r², whereas the circumference is 2 pi r, where r is the radius of the core.

So, the circumference (and therefore the surface area) is proportional to the square root of the csa.

Stranded cores will be slightly different, but not much.


Nice graph, HHD.

 

[HandySparks]

... that said, I think that I am correct in saying that the CCC of a cable is related to the heating effect of the power dissipated due to its resistance exceeding the material property limits of the insulation.

That power is given by:

P = I² x R

However, the effective resistance of a RFC is 1/4 that of a radial circuit of the same length and therefore the current required to reach the same temperature in a RFC is ~ twice that in a radial circuit.

You need to determine the capacity of a ring by looking at the actual current in the various sections of the ring and working out where the most heavily loaded part will be. For this you need to determine how the load is distributed around the ring, or make reasonable assumptions. The IET clearly allow for a certain amount of 'unbalance' in the loading of the ring, although not to an extreme extent.

If you could simply double the CCC of the cable, the regs would be showing a 40A (or 50A?) breaker in Appendix 15 for a 2.5mm² ring, not a 32A.

 

[HappyHippyDad]

So from the graph we can see that 1x5mm cable could take 42.5A. If 2x2.5mm in parallel can take more than that due to being able to dissipate heat more efficiently than 1x5mm then its CCC is more, so in answer to Murdochs question (regulation's aside as stated by Mr DS) it should be a C3 as its not dangerous??

Although, its a ring with many points coming off of it (which could be very unbalanced) rather than just one end load (so not strickly speaking in parrallel as E54 showed), so perhaps it is a C2 as we cant assume the CCC>42.5A??

Glad we've cleared that up :smile5: