i did my apprenticeship with a domestic sparky so have no idea unfortuneatly
Minimum tabulated current required:a single phase load of a 13a is supplied via a 15 a bs 1361 fuse using single core 70 degree c theromplastic copper cables installed to method 4 (ref method B) the rating factor for grouping is 0.7 and for ambient temperature 0.87 and overload protection is to be provided. the minimum acceptable size cable would be
a)1mm sq
b)1.5mm sq
c)2.5mm sq
d0 4 mm sq
ive used the brb and its told me to go to app 4
I know In is going to be 15
so 15 / ca ci cc
ca 0.87
ci no idea
cc 0.9
this gives me this for an answer but its obviously wrong
15/0.87x0.9 = 19.16
can anyone show me where im going wrong please
Minimum tabulated current required:
It=>In/CaCiCgCc
It=>15A/0.87x1x0.7x1
It=>24.63A
From Table 4D1A for Singles, 70°C, Copper.
It=24A for 2.5mm[SUP]2[/SUP] Does not comply with calculated It.
It=32A for 4mm[SUP]2 [/SUP]selected conductor size, complies with calculated It.
I've just successfully completed my 17th edition paper, and one of the parts which confused me slightly was the correction factors (amongst many other parts not helped by the legal lingo BS7671 is written in) Having read this thread it all seems to make sense now thanks. One thing, presumably being an accumulative calculation, the correction factors have to be applied in the order they are stated e.g CaxCixCgxCcxCt ?
A I understand it, If the circuits affected by the various self heating effects that would demand correction factors but do not affect the conductors concurrently then apply each correction factor separately and take the worst case (lowest correction factor; highest current value)
If the self heating effect of the environment ie. circuits run together in a group and surrounded by thermal insulation as well as an ambient temperature higher than 30 deg C. then apply all the factors together.
Imagine the example.
10 * 50W downlights on a circuit which is run with 12 other T&E circuits from a distribution board covered in 300mm of thermal insulation
You have 2.2A constant load running in trunking (Method B)
Cg = 0.45
Ci = 0.5
In = 6A
It= 6/(0.45*0.5) = 26.7A or
That would mean you would need to wire the first part of the circuit in 2.5mm2 and then possibly use a smaller conductor csa after that!
If you separated out the self heating conditions then you would only apply the worst case such as It=6/(0.45) = 13.3A.So you would be Ok with 1.0mm2 conductors.
How or where do you apply table 52.2 derating factors, does it replace ci when the length of cable in insulation does not exceed 0.5m? So take a cable clipped direct to timbers in a loft space, which travels 200mm through insulation, would you ignore the ci for clipped direct and use 0.63 in its place?
I would only apply table 52.2 when the cable passes through thermal insulation and not in thermal insulation.
If the cable now only run through 300mm and not in or surrounded by 300mm then you might apply a derating factor of 0.57, but since the grouping factor is lower and not concurrent then you would only apply the worst case which would be the grouping factor of 0.5.
Hope that makes sense
Cheers, Sam
Reply to the thread, titled "cg,ca,ci,cc" which is posted in Electrical Wiring, Theories and Regulations on Electricians Forums.