Csa

mpc4000 · May 21, 2009

hi all i'm having the blondest of blonde days to day can anyone point me in the direction of how to work out the following :

"two conductors of 6mm and 4mm are connected in parallel, thier effective CSA will be" ?

I'm strugging to define my A**S from a hole in the ground so any pointers will be much appreciated

Thanks in advance

Mark.W · May 21, 2009

csa = r^2*pi

In other words r*r*3.14

so on 6mm its 6/2 therefore r = 3 so it's 3*3*3.14 = 28.27mm squared
and on 4mm it 4/2 therefore r = 2 so it's 2*2*3.14 = 12.56mm squared

I think their effective csa therefore would be the 2 together, not sure but I think it's 28.27+12.56=40.82
Hope that makes sense

scotsparky · May 21, 2009

sounds about right notabright

Mark.W · May 21, 2009

scotsparky said:
sounds about right notabright

LOL a man of many talents, you're a poet too Scotsparky

stretchyboy · May 21, 2009

surely its just add them together if you are looking for equivalent CSA.

If you had add one square mile plus two square miles you would get three square miles whichever way you look at it.

On a completely other topic, you could answer that you aren't supposed to install parallel feeders of differing sizes ;-)

Mac · May 21, 2009

Cross-sectional area to diameter conversion circle electric cable wire diameter and wiring and calculation cross section AGW American Wire Gauge - sengpielaudio Sengpiel Berlin

Spot on notabrightspark.Here is a calculator that will do it all

stretchyboy · May 21, 2009

you are adding the cables together not adding the two radiuses together to make a bigger single cable though are you!

think about it.
You cant gain cross sectional area it would be like adding one football pitch to another football pitch and then saying that it was twice as wide and twice as long, therefore being the size of four football pitches

Mac · May 21, 2009

stretchyboy said:
you are adding the cables together not adding the two radiuses together to make a bigger single cable though are you!

think about it.
You cant gain cross sectional area it would be like adding one football pitch to another football pitch and then saying that it was twice as wide and twice as long, therefore being the size of four football pitches

Did you have a look at the page i posted

Mark.W · May 21, 2009

I disagree!.

Mac · May 21, 2009

notabrightspark said:
I disagree, I think if you had 2 conductors of the same size then you could add them together, but seeing as you have 2 cables of different sizes, the csa's need to be calculated seperately.

Spot on mate,the figures you put out equated to the post i posted

ajelec10 · May 21, 2009

Hold on a miniute i agree with strechey boy as you said if the coductors are the same size you add them togther, for eg. 6mm2 + 6mm2 = 12mm2 so 6+4 can not equal more than twelve, remember the guy will be talking in mm2 not diameters

Mark.W · May 21, 2009

If you add the cables together the csa of a 10mm cable is on its own is 78.54mm.

The csa of a 6mm cable is 28.27mm and the csa of a 4mm cable is 12.57mm, the 2 together = 40.84mm

ajelec10 · May 21, 2009

Right you are getting confused with resistances in parell and cable diameters in parrell, a 6mm2 cable has a diemeter of 2.763 and a 4mm2 has a diemeter of 2.256 its simple circles A=pi x r2
its like having a two squares one with an area of 6 and one with an area of 4 put them together an you will get 10
you are right if the origaly post was talking about the diemeter of the cable but i really dout that
hopr this helps

Mark.W · May 21, 2009

Ok so you have given me the diameter of a 6mm cable = 2.763
and the diameter of a 4mm cable which = 2.256

so csa = r^2 x pi

d of 6mm cable = 2.763 therefore r = 1.38 therefore 1.38*1.38*3.142 = 5.99mm so I agree
d of 4mm cable = 2.256 therefore r = 1.13 therefore 1.30*1.30*3.142 = 3.99mm so I agree

The OP asked for the csa of a 6mm conductor and a 4mm conductor. He didnt give the csa and ask for the diameter!

Bit of a grey area on the OP but looking at the black and white of it....