View the thread, titled "Cross sectional area current capacity" which is posted in UK Electrical Forum on Electricians Forums.

G

Gardner

How would you go about figuring the current carry capacity of 2.08mm2 copper conductor? In so far no reg tables have none standard size.
 
You wouldn't happen to know what equation is used in other regs like the IEC? My understanding is, and I could be wrong, that table 4D5 and others were derived via thermodynamic equations made by the IEC.
I would agree with the above posts about cable sizing too many variable in standards compliance and cable design.

However this webpage seems to discuss the methods used in IEC60287. I feel it is not worth working it out when the standard cables have been calculated (probably well on the side of caution!) for you.

Edit just found this
IEC METHOD
IEC 60287 provides method for calculation of current carrying capacity of cable for different design parameters of cable (cross section, thermal resistivity, thickness of layer … etc.) as well as for different environment (ambient temperature, soil condition, depth of cable, spacing of phases … etc.).The current rating (ampere) using in the IEC standard is:

CCC calculation IEC60287.jpg
where Δθ [[SUP]o[/SUP] C] is temperature rise between ambient temperature and cable conductor temperature, W[SUB]d[/SUB] [W/m] is dielectric loss of cable insulation, T1, T2, T3 [K.m/W] are equivalent thermal resistances calculated from the cable material’s thermal properties, T4 [K.m/W] is the cable external thermal resistance, R[SUB]ac[/SUB] [Ω/m] is the AC electrical resistance of the cable conductor at maximum temperature and λ1 and λ2 are the ratio of losses in the metal sheath to total losses in all conductors and the ratio of losses in the armouring to total losses in all conductors, respectively.
 
Last edited:
How would you go about figuring the current carry capacity of 2.08mm2 copper conductor? In so far no reg tables have none standard size.

I would start by ripping this 2.08 mm or 14 AWG whatever it is straight out. Then replace with 1.5 mm or 2.5 mm depends what the load is.
 
.........The current rating (ampere) using in the IEC standard is:

View attachment 29359
where Δθ [[SUP]o[/SUP] C] is temperature rise between ambient temperature and cable conductor temperature, W[SUB]d[/SUB] [W/m] is dielectric loss of cable insulation, T1, T2, T3 [K.m/W] are equivalent thermal resistances calculated from the cable material’s thermal properties, T4 [K.m/W] is the cable external thermal resistance, R[SUB]ac[/SUB] [Ω/m] is the AC electrical resistance of the cable conductor at maximum temperature and λ1 and λ2 are the ratio of losses in the metal sheath to total losses in all conductors and the ratio of losses in the armouring to total losses in all conductors, respectively.
Not sure I can see that formula catching on TBH :)
 
Is this for the cabins you were making which were intended for UK use? If so, wouldn't it be better to use standard UK wiring types? Daz
 
I would agree with the above posts about cable sizing too many variable in standards compliance and cable design.

However this webpage seems to discuss the methods used in IEC60287. I feel it is not worth working it out when the standard cables have been calculated (probably well on the side of caution!) for you.

Edit just found this
IEC METHOD
IEC 60287 provides method for calculation of current carrying capacity of cable for different design parameters of cable (cross section, thermal resistivity, thickness of layer … etc.) as well as for different environment (ambient temperature, soil condition, depth of cable, spacing of phases … etc.).The current rating (ampere) using in the IEC standard is:

View attachment 29359
where Δθ [[SUP]o[/SUP] C] is temperature rise between ambient temperature and cable conductor temperature, W[SUB]d[/SUB] [W/m] is dielectric loss of cable insulation, T1, T2, T3 [K.m/W] are equivalent thermal resistances calculated from the cable material’s thermal properties, T4 [K.m/W] is the cable external thermal resistance, R[SUB]ac[/SUB] [Ω/m] is the AC electrical resistance of the cable conductor at maximum temperature and λ1 and λ2 are the ratio of losses in the metal sheath to total losses in all conductors and the ratio of losses in the armouring to total losses in all conductors, respectively.

Or you could just just my super scientific graph :) (please dont!)

current.cableGraph.jpg
 
This is because its under different regulations and the cable standards are also manufactured to differing regulation too, we have our own BS standards we follow and the cable we buy is manufactured to either EU or BS requirements.


I think the insulation might have something to do with it... My understanding is that typical PVC insulated wires used in the UK are made for 70*C operation, while the 2.08mm2 PVC seems to be made for 90*C operation hence the higher current allowance in other regs.
 
Also to note here is that the cables may be required a derating factor if connected to equipment rated at 70c, through experience that is most switchgear etc out there.
 

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