does this comply?

[ianmacd]

2. This is the most interesting thread that's been on the forum for months in my opinion.

You got a reg to confirm that

 

[ADS]

ive said before, it only complies if the cable can withstand a fault current which needs confirmation with adiabatic equation. You also need to know the pfc . the Zs and Ze.Where we take the pfc i am not sure, normally you would do it at the cable end, but i dont understand that cos who knows where the fault will occur.

Eg we may find that a 2.5 radial done in singles with the same size cpc in conduit complies, but 2.5 t&e doesnot. Who knows until the equations are done. Come on lets get this done. ray

These are irrelevant as the cables are protected by an RCBO.

The only figure that could matter is the PSCC, which, on short circuit, will certainly trip the 32A RCBO within the specified time.

I offer the same reasoning as I did for the overload protection -

The scenario is now a 4mm cable feeding a FCU and a 2.5mm cable feeding a double socket outlet, both from a 32A RCBO.

Basically, a 4mm radial with a 2.5mm unfused spur

Appendix 15

(4mm Radial Circuit)

Unfused spur

An unfused spur run in 2.5 mm2 cable should feed one single or one twin socket-outlet only.

An unfused spur may be connected to the origin of the circuit in the distribution board

How can the above circuit comply with the regs, yet the ops scenario doesn't?

How's it possible for the PSCC to effectively trip the RCBO, in the correct time (in the above circuit), yet in the ops scenario you question it?

And the biggy, to win the ultimate prize, What is providing 'overload' protection for the 2.5mm cable in the above scenario??

Answers on a post card ha ha:^O

 

[ianmacd]

I would afree the Zs of the circuit is not relevent here.

Also your logic does not follow. They are different circuits. Your one complies but the op's does not. There is no point asking us why, we didn't write the regs.

 

[Riggy]

You got a reg to confirm that

No, but no-one has quoted a reg saying it's not so I win! ROTFWL

 

[ADS]

You not having a go at the 'biggy', Ian:innocent

 

[ianmacd]

Well you ref appnx 15 but not a proper reg so actually you can not prove its complaint either.

 

[ADS]

Well you ref appnx 15 but not a proper reg so actually you can not prove its complaint either.

Are you a part-time politician? You swerve questions well

The Big Question:

In the scenario that I have given, and shown to comply in Appendix 15 of the Regs,

i.e. A 4mm cable feeding a FCU and a 2.5mm cable feeding a double socket outlet, both from a 32A RCBO.

Basically, a 4mm radial with a 2.5mm unfused spur

Appendix 15

(4mm Radial Circuit)

Unfused spur

An unfused spur run in 2.5 mm2 cable should feed one single or one twin socket-outlet only.

An unfused spur may be connected to the origin of the circuit in the distribution board

What is providing 'Overload' protection for the 2.5mm cable? bad day explode

 

[ADS]

C'mon Ian.

I'll give you a clue ...... It begins with BS

Do you want another?? ..... It contains 4 numbers BS _ _ _ _

Another?? ....... It's found in a plug-top. (Aw c**p, I've gone too far, you'll get it now )

C'mon Ian, I want to see you write it

 

[Nicky Tesla]

take this onboard please.

1...433.3.1...a device (mcb) for protection against overload need not be provided

(ii) for a conductor which, because of the charactristics of the load or supply, is not likely to carry overload current (which it is not) provided that the conductor is protected by "FAULT" current in accordance with reg 434 (so lets see wot that is)

2...434.1.....The PFC shall be determined at every relevant point.

.....434.2.2...The device protecting a conductor may be installed on the supply side of the point where a change in csa occurs provided that it possesses an operating characteristic that it protects the wiring on the load side against fault current in accordance with reg, 434.5.2 ( so lets see wot that is)

3....434.5.2...a fault occuring at any point in a cct shall be interupted in a time such that a fault does not cause the limiting temp of the cable conductors to be exceeded.

for a fault of very short duration, (less than 0.1 sec) k sqd x s sqd shall be greater than be value of let through energy( i sqd t) quoted for the protective device (MCB)

the time t, in which a given fault current ( the pfc) can raise the conductors to the highest permissible temp (befor they melt) can be calculated using the formula

t = k sqd x s sqd divided by i(pfc) sqd

so when we calculate it, the time must be over 0.1 seconds. This will show that the MCB will have tripped before the conductors have melted.

t = 115 sqd x 2.5 sqd / lets say on this occasion 300A sqd (0.8 Ze)

then t = 0.9.......therfore when Ze is 0.8 then 2.5 is ok

but when pfc is 3000A t = 0.009 so this time it may not comply

( im not 100% sure on these calcs but i would like to be corrected if need be)

So as explained previously, 2.5 will comply in this scenario as long as the pfc is below the limiting value:coffee

 

[ADS]

take this onboard please.

That's easy for you to say

I'm having trouble just reading it, so I'm going to take your word for it

 

[Nicky Tesla]

the pfc is the key

 

[Badger]

C'mon Ian.I'll give you a clue ...... It begins with BS

Do you want another?? ..... It contains 4 numbers BS 1234

Another?? ....... It's found in a plug-top. (Aw c**p, I've gone too far, you'll get it now )

C'mon Ian, I want to see you write it

] ROTFWL

 

[Lee321]

i like the fault calc nicky i just need it confirmed as that is how you calc fault protection only. wow 300 i said it would be high just never thought that high

 

[steptoe]

OK, enough rubbish,

now we are gonna have the LAW,

would any person complying with the LAW that is EAWR 16 design such a circuit.?

IMO if you did design such a circuit then you would be in breach of EAWR,

and therefore,

could NOT by default comply with BS7671,

as 7671 requires you to adhere to the law at all times.

 

[Badger]

OK, enough rubbish,now we are gonna have the LAW,

would any person complying with the LAW that is EAWR 16 design such a circuit.?

IMO if you did design such a circuit then you would be in breach of EAWR,

and therefore,

could NOT by default comply with BS7671,

as 7671 requires you to adhere to the law at all times.

well said that man!! finally some sense :Applaud :Applaud

 

[ADS]

OK, enough rubbish,now we are gonna have the LAW,

would any person complying with the LAW that is EAWR 16 design such a circuit.?

IMO if you did design such a circuit then you would be in breach of EAWR,

and therefore,

could NOT by default comply with BS7671,

as 7671 requires you to adhere to the law at all times.

And how, exactly, is it not complying with the law.

Firstly, everyone on here has stated they wouldn't design a circuit this way - the question was, 'Does it comply with BS7671'

You could come across it on a PIR.

Maybe you'ld like to answer the questions posed in 'Post #342', as Ian doesn't seem to want to. ; \

Then you can tell us how the ops example is breaking the electricity at work regs.

Persons to be competent to prevent danger and injury16. No person shall be engaged in any work activity where technical knowledge or experience is necessary to prevent danger or, where appropriate, injury, unless he possesses such knowledge or experience, or is under such degree of supervision as may be appropriate having regard to the nature of the work.

As we have proven that the circuit doesn't present 'danger or risk of injury' and complies with BS7671, then we have proven our competence and this doesn't really apply.

 

[ianmacd]

Maybe you'ld like to answer the questions posed in 'Post #342', as Ian doesn't seem to want to. ; \

I already did and am bored with repeating myself. Nothing suitable is but that seems to be allowed under the regs.

You still have not provided a reg that allows that either in your 4mm scenario.

 

[ADS]

Ian, if you are correct, then you, me and every other spark on here need to go around and remove every unfused spur we've ever installed on a ring-final or 4mm radial.

And if you come across an unfused spur on a PIR, the installation fails, is in breach of the EAWR and gains a code 1. ; \

You can't say that one scenario complies and is within the law, yet the other one doesn't - I've proven that much.

They both contain 2.5mm radials off a 32 A MCB - They either both comply, or they both don't

 

[ianmacd]

Not that simple is it and you know it. 2.5mm spur on a ring is allowed by reg 433.1.5 but as far as I can see and no one else has offered anything to contradict this, no other reg exists to support the use of a spur on a radial.

 

[Nicky Tesla]

I have, your reg is irrelevant. go back and check all my regs i put down a few posts ago.

Also i did it today, i used the reg. There was a 4mm radial fitting a number of sockets, but i needed to fit a spur for a doorbell Tx using the 32A mcb. I had no 4mm so i used 2.5 which is not a problem now that we all know it is perfectly ok. Great, saved me a trip back to the van.