First fix in a timber frame wall

[^woody^]

I need to install the first fix electrics in a timber frame wall. Its a bedroom and there will be 4 double sockets as an extension to the existing ring

The timber frame is 90mm with 90mm celotex between the studs. There will be a further 50mm celotex across the face of the frame, and then 12.5 plasterboard.

Should I run the cables in 20mm conduit up the face of the frame (ie within the 50mm insulation) or should the cable be on the surface of the insulation board. If on the surface how can it be fixed to the insulation?

Cheers

 

[ericmark]

See page 265 BS7671:2008 Table 4A2 the method 100 to 103 has a huge affect on permitted current see also page 282 Table 4D5 between 102 to 103 2.5mm drops from 21 amp to 13.5 amp so if not cooled by the timber or wall I would say the ring would run at 20 amp max rather than 32 amp rating. Can't really see how conduit helps still needs to cool and unless the conduit is in contact with a cooling surface it would not help.
To me it shows it clipped to wood look at diagram and see what you think?
Eric

 

[GaryMo]

In conduit, a 2.5mm cable will carry 20A compared to 21A for the cable not in conduit but touching the inner wall surface.
Both methods will afford compliance (just) for a 30/32A ring final circuit wired in 2.5mm t&e assuming no other correction factors need to be applied.

 

[ericmark]

I think you are looking at page 261 Table 4A2 number 2 giving ref method A but at 10 W/m2K but table 4A2 page 265 ref 103 is at 0.1 W/m2K and it is this U factor which will be used in selection the correct table. Although I could well have made a mistake unless people quote where they have got their information from it makes it very hard to cross check.
Here it will need an input from ^woody^ as to what the value of U is for that wall. And if unknown I would think only option is to play safe.
Eric

 

[Pensdown]

Table 4D5 ref 102# = 21amps. 102# in table 4D5 refers back to table 4A2 on page 265

 

[^woody^]

I've just logged on to our technical indices, and it appears that BS7671 2008 is not on there yet, for some reason. So I can't check the tables

I'm not sure how the wall u-value is used in the cable determination, as the cable will be about 1/3 into it.

The wall as a whole is going to be round about 0.25 W/m2/K ... with the 2/3 thickness to the external face being about 0.31 W/m2/K

 

[Pensdown]

If you follow GaryMo's post you will be OK with or without conduit.

But if not in conduit, the cable must be touching the inner face of the plasterboard so that will mean a small channel in the insulation.

 

[GaryMo]

Although I could well have made a mistake unless people quote where they have got their information from it makes it very hard to cross check.

The only table quoting current carrying capacity for twin & earth - table 4D5.
Considering it's the only table relating to twin & earth it shouldn't be that hard to cross check.

 

[^woody^]

So, I could just make a small groove in the surface of the insulation and then hold the cable in with some aluminum foil tape (to maintain the vapour check), and then just plasterboard over it?

That looks promising

 

[rssteve]

no the cable needs to be touching the plasterboard to dissipate the heat, also this will be going on an rcd if not already?

 

[GaryMo]

also this will be going on an rcd if not already?

There's no getting away from it as socket outlets are rated under 20A and will be used by ordinary persons so require additional protection via a 30mA RCD.

 

[Pensdown]

I've never tried it but I have seen plastic capping used the other way around. It was cut into the insulation with the cable way facing the plasterboard. It was held in place with foil tape. The cables weren't in place so I don't know how they were held but your foil tape would do it.

It looked like a good way of keeping the cables away from the insulation and in contact with the plasterboard.

 

[ericmark]

Looking at page 104 523.7 it seems one multiplies by 0.5 the current carrying capacity shown ref C so 27 x 0.5 = 13.5 Amp thats odd it's same as ref 103.

Nothing is easy in this world.

Eric

 

[GaryMo]

Looking at page 104 523.7 it seems one multiplies by 0.5 the current carrying capacity shown ref C so 27 x 0.5 = 13.5 Amp thats odd it's same as ref 103.

Nothing is easy in this world.

Eric

That's for a cable totally surrounded by insulation.

Method 103# is for cables in an insulated wall and not touching the inner surface so fully surrounded by insulation.

The OP is going to install his cables with them touching the inner wall surface so table 4D5, page 282, ref method 102#.