Cable size help

[morrisndlovu]

Is there a formula that one can use to calculate the size of a cable if one knows the circumference of the cable? The formula for area of a circle does not seem to work, I’ve tried it on number of cable(mostly single cores).

 

[VEBill]

Are you taking into account the thickness of the insulation?

 

[morrisndlovu]

Yes, I do, and I get a much bigger value than the actual size of the cable. I use this formula: A = ?r², which is the formula for a circle.

 

[DRWeig]

Hi Morris,

Are you trying to compute size as in AWG (American Wire Gauge)? Give us an example of what you've calculated versus what you think the answer should be.

Good on ya,

Goober Dave

 

[desertfox]

Hi morrisndovu

If you know the circumference of the cable then you can workout the diameter from:-

circumference = pi*Diameter

therefore Diameter = circumference/pi


desertfox

 

[7anoter4]

IEC 60502 "Power cables with extruded insulation and their accessories for rated voltages from 1 kV (Um = 1,2 kV) up to 30 kV (Um = 36 kV)"

http://webstore.iec.ch/preview/info_iec60502-1{ed2.0}en_d.pdf

may assist you to calculated the cable overall diameter.
The conductor has to be extract from IEC 60228[Conductors of insulated cables]:

http://webstore.iec.ch/preview/info_iec60228{ed3.0}en_d.pdf

 

[7anoter4]

First of all you have to pick up the conductor diameter.
The total cross sectional conductor area is not 100% nominal but 95.5-93% due technological tolerances.
The conductor may be: solid [one single strand] or stranded [7, 19, 37, 61 strands].
Stranded in turn may be compressed, compacted or circular class II stranded.
Let's say the nominal cross sectional area of conductor is 50 sqr.mm [or 1/0 AWG approx.53.5 sqr.mm].
According to IEC60228 the conductor diameter of solid [one single strand] =7.8 mm [only 47.78 sqr.mm]
According to IEC60228 the class II conductor diameter will be 9.1 mm [19 strands].
The conductor diameter for class II could be approximated using formula: dia=sqrt[scu*4/pi()/1.2] mm
scu =nominal cross sectional area [sqr.mm].
compressed will be 0.97*class II dia=0.97*9.1=8.83 compacted 0.9*9.1=8.19 mm
According to ASTM B8-86 class B stranded the conductor diameter will be 0.372 [9.45 mm] compressed =0.361 inches[9.17 mm] compacted =0.336 inches[8.5 mm].

 

[MacGyverS2000]

The total cross sectional conductor area is not 100% nominal but 95.5-93% due technological tolerances.

Uhm, why? If manufacturing tolerances always downplay a Gauge thickness by 5-7%, wouldn't the manufacturer compensate by using a 5-8% larger draw die?

Dan - Owner

http://www.Hi-TecDesigns.com

 

[7anoter4]

I have to recognize I'm not sure that technological tolerance is the only cause of this.
If we'll take again the 50 sqr.mm class II stranded, the IEC 60228 standard states the maximum diameter as 9.1.
The minimum diameter results from the maximum permissible resistance at 20oC 0.387 ohm/km and the conductor diameter will be 8.8 mm.
So, the die, for the first time is suitable for 8.8 mm [19 strands of 1.76 each one.]
With the time the die hole-never the less is diamond provided-will increase up to 9.1[19 strands of 1.82 each]. Then one has to change the die.
Why the maximum diameter is 9.1 mm? I think it is because the inner diameter of the lug it is 9.5 mm and we need a little clearance in order to fit the conductor in the lug.
But this is only a speculation. The standard is the standard and the nominal cross section of the cable cannot be the actual one.

 

[zappedagain]

On a side note, cables sometimes get defined with the area in units of circular mils:

Diameter [mils] = sqrt( area [circular mils])

[units are in brackets]; mil = mils = 0.001"

So a #30 wire with an area of 100.5 circular mils has a 10 mil diameter.

No PI for lunch here!

John D