Armoured cable size

David, this is has cowboy written all over it.

BASIC ELECTRICS
Protective device, lets say its a 20amp MCB. The rule is this must not exceed the current-carrying capacity of the attached cables.
Can 1.5mm cable hack 20amp? NO.
Can you 100% guarantee that some/of the load will go down the 2.5mm cable, even under fault conditions? NO
So can you run the two cables off the same CPD and sleep at night? NO

End of. Get back on your horse and gallop off.

 

I see your point, thanks. I'd be better off just hooking up the 2.5mm then and having an appropriately rated MCB (16 amp or whatever).

 

It can become imbalanced UNDER FAULT CONDITIONS.

You always design electrical circuits bearing in mind what can happen. You know, maybe one conductor of the 2.5mm becomes disconnected(its not unknown!) or Mick chops through the cable with his mini digger, alien spacecraft landing leg crushes the SWA, santa's reindeer catches his antlers in the 2.5mm as it enters the shed and rips it out.

 

You need to look very carefully at the volt drop though

5KW
2.5mm SWA
30metres?

I've not done the calcs, but my gut tells me that doesn't fly.

 

Extend a ring final 30mts up/down the garden. .!

Not how a sparks would design and install it!

 

Running that install in, I would be kicked out the club !

Love to see page 6 of the EIC on sizing of Line conductors (2.5 and a 1.5) not a chance with my name on it!

 

4mm - needs to be done by an electrician. So you can dig trench etc and lay cable without filling in, and the sparks will hook up and verify/certificate after seeing back filling is carried out correctly.

 

I may sound a bit pedantic, but some of the ideas that DIY;ers (if your not a sparks then thats what everyone else is) is sometimes hard to believe on their self design principles. .

We have a set of regulations that must be adhered to. .not just us, you lot as well!

 

4mm is good up to 4.9kw, so 16A MCB should be fine (3.7kw)

 

But you do not have 4mm2 - unless you run s new cable.

 

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I wasn't making any comment on the design, just about the application of Part P. The way the OP wants to do it may contravene principles of good circuit design but it does not require notification under Part P.

 

Sorry I meant 2.5mm is good up to 4.9kw.

 

I was thinking about this a little more while in bed (dreams have an odd way of crystallising thoughts sometimes), and I thought - how does this apply to a final ring circuit where the MCB is rated at 32A but there are 2 x 2.5T&E. If, under fault conditions as you say, one of these cables were to become disconnected then it would be a similar situation: a 32A MCB protecting a cable that is rated to around 20A. Is this not the same situation?

 

Yes, but we do not design and install circuits on what is basically an unsafe premise.
Also, a continuity test on the ring final would reveal a broken "leg". But a similar test on the OP's bodge would not reveal any issues.

 

I think you may have identified the reason that radials are just better design.

 

Did a quick google and found this on tlc, consistent with other places:

"Cables may otherwise be connected in parallel provided that they are of exactly the same type, run together throughout their length, have no branches and are expected to share the total circuit current in proportion to their cross-sectional area. It is not recommended that cables are connected in parallel at all except in ring final circuits.."

So I would conclude - it can be done if done "properly", but its not really a good idea.

 

Question 1

From the diagram below calculate the current that will flow in cable 1 and cable 2 to a resistive load of 2Kw, given the following conditions will .
a) Cable 1 is 30m 2.5mm (assume the characteristics of T&E)
Cable 2 is 30m 1.5mm (assume the characteristics of T&E)
b) Cable 1 is 30m 2.5mm (assume the characteristics of T&E)
Cable 2 is 30m 2.5mm (assume the characteristics of T&E)

*for the purposes of calculation the resistance of Cable 3 can be taken as 0

 

Ignoring the length and diameter of each cable, since I don't think that'll make much difference for the purpose of this exercise (the resistance/impendence of each cable will be the same give or take in each case)....each cable 4.35A in each case I would say.

Current = power/volts = 2000/230 = 8.7A. divide by 2 as there are two cables = 4.35A

So irrespective of the diameter of each cable, once has to assume that both cables are the diameter of the smallest.

 

But the Question asks you not to do that. You have only answered part b

 

I have answered part (a) but made some assumptions in doing so. The question then is whether my assumptions are valid or at least "reasonable" i.e. whether the answer would otherwise be materially different.

The impendence of the 1.5mm cable will be a bit higher, so slightly less current will flow through it compared to the 2.5mm cable. Lengths can be ignored since it is the same in each case.

So I'm guessing the current will be 4.4/4.3 rather than 4.35/4.35...not materially different. The current in both is substantially the same.