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Csa

Discuss Csa in the Periodic Inspection Reporting & Certification area at ElectriciansForums.net

M

mpc4000

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
 
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
 
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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 ;-)
 
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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 :eek:
 
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 :eek:
Click to expand...
Did you have a look at the page i posted
 
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
 
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
 
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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
 
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....:D
 
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"two conductors of 6mm and 4mm are connected in parallel, thier effective CSA will be" ?

So asuming the guy ment to say the cables are 6mm2 and 4mm2 the effictive CSA of the two conductors in parrell would be 10mm2

Would you agree with that?
 
Please are you winding me up. all cable sizes are given in mm2 this is the cross sectional area of the cable. you dont need to square them all you do is add them together
i cant make it any simpler.
Take two squares one with a base of 3 and a height of 2 the area of that square is 6
take anoter square base 2 height 2 area is 4
add the two square together you get a square with an area of 10, its the same with circles
could some one help me explain?:)
 
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Sorry mate this is doing my head in.
6mm2 cable.CSA=pi 6 squared divided by 4=3.142x36/4=28.278
4mm2 cable.CSA=pi 4 squared divided by 4=3.142x16/4=12.568
I have had enough now.Think i'll dig my books out from 25 years ago:eek:These are my final answers now
 
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I did look at the weblink you posted and it is for getting the wire csa from the diameter ..............

Where does it calculate two csa's added together. Do you really not understand that you just simply add csa together. Can you explain why you are adding the radiuses together? It doesn't make sense, it's not as if the two cables are melting together

[ElectriciansForums.net] Csa

You are adding diameter A to Diameter B and then working the cross section area which will give you the area of the big circle round the outside of the other two doesn't it?
 
ajelec10 said:
So asuming the guy ment to say the cables are 6mm2 and 4mm2
Click to expand...

But you can't just assume that is what he meant!. The black and white is that he wanted the csa of a 6mm conductor and a 4mm conductor.


Ok so after some investigation, I can see where stretchyboy is coming from in that you have to add the 2 csa's together. But then we have to agree that we have to calculate the csa of the condutors seperately and then add them together. Not add the conductors together then calculate the csa.
 
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So if you go in to the whole saler and ask for some 6mm T+E what would you expect to get?
The csa of the phase and nuturals would be 6mm2 not 113.112,
We always refer to the cables as CSA not diemeter
Take some 25mm2 tails the diameter is not 25mm across is it!
Can we now agree please that the guy was talking about a 6mm2 and a 4mm2 and the effictive CSA of these two conductors are 10mm2

Regards

AJ
 

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