re cable correction factors

[electropete]

hi guys, this is my first question so please be patient,


as part of a b-tech study i have to design a circuit and state the size of cable used, iam going wrong somewhere but where so here is an example.

Ib = 9000/230 = 40 amps (rounded off)
In = 45 amp RCD.

now this is where i sort of loose the plot

Iz = In/correctionn factors.
say for grouping (3 circuits) Cg = 0.7
say for thermal insulation Ci = 0.5 (thermal isulation the whole route) so

Iz =In/Cg x Ci = 45/.7 x .5 = 45/.35 = 128. and thats got to be wrong.

please help

peter.

 

[i=p/u]

means ur cable has to carry at least 128 amps, could be right, straight away i see your ci is fully surrounded so is doubling ALREADY

 

[ringer]

The problem your design is causing is trying to run 9kW (a lot of power) through thermal insulation.

Regulation 523.7 "A cable should preferably not be installed in a location where it is liable to be covered by thermal insulation...."

It looks like your calculations are right, and this is why the above regulation exists - if we ran all cables through thermal insulation for their entire length, we would be using an awful lot of copper!

 

[i=p/u]

1 the design current Ib must be found first
2 the overcurrent device rating In is then selected so that In is greater than or equal to Ib

In > Ib

then the tabulated current carrying capacity of the selected cable cable It is given by

It > In/ca,cg,ci,cc

 

[electropete]

thanks guys for the speedy replies

 

[Sparky Ninja]

Your calculations are on the right track. Many a time Ci will create huge change in the cable size.

What is the load? Is it a fixed load?

If you can determine that the load is not subject to possible overloading then you can stick with Ib instead of In.