copper tube

[kenputer]

I have lots of 3/8" soft copper tube that has a plastic coating on it to protect it.

 I want to use it to tie a bank of 12 volt batterys together in parrell and what I would like to know is there a way to determine a size wire that it would be equal to?

[twidget]

I think a close approximation would be done by calculating the copper cross sectional area of the copper tubing, outside dia. area - inside dia. area = cross sectional area. Then compare this area to the cross sectional area of wire. This should give a good idea of the equivelent wire size. Don't expect an electrical inspector to go for it though.

Just my 2 cents

Twidget

[wpowokal]

Yes by area, A=3.147 X radius squared. So take the radius of the outside diamiter (ie half diamiter) work out its area.

 Then take the area of the inside diamiter away from that, the resultant square mm (if you are working in mm) is equal to the equivelent cable.

Use this link to evaluate

http://www.powerstream.com/Wire_Size.htm

allan

[BeenzMeenzWind]

I've used copper tube to carry audio signals around. We wanted a speaker in the bathroom at my friend's house, and as the heating pipes running into there were in plastic fitting at each end we just used them to carry the audio. They should have been grounded, now I think about it.

The other posts are correct about the effective cross sectional area. I suspect you may get less EM radiation from an AC line if it's made of tubes, too. Could be wrong there, though. Wave guide theory suggests the E-field should be mainly contained within the tube, but that may not be the case as the frequency is so low.

[John II]

Hello kenputer:

I recommend doing it by weight. Using a postage scale or something, I'd weigh a foot or two of your copper tubing. then peal the insulation off of a copper wire and weight it. You might have to add more 'equal lengths' of wire to match you copper pipe. But if you can find a copper wire that weights the same amount as your tubing would give you a good idea right there.

Now if you were running high frequency A.C. the copper tubing having a larger diamenter and "skin effect" probably would carry more current then an equivelent weight of wire... but sense you will be using D.C. this is not a concern.

John II

[kell]

3.147 is off.  Pi = 3.14159... round off where you think you need the accuracy, like 3.14 or 3.142

[kenputer]

Thanks for the help I will size it that way. I will be using the battery bank at the cottage so don't think I have to worry about the inspectors there. thanks again everybody :-)

[John II]

I rounded some pi off yesterday. It left more room in the frig and I felt good about that ; )

John II

[Ungrounded Lightning Rod]

I suspect you may get less EM radiation from an AC line if it's made of tubes, too. Could be wrong there, though. Wave guide theory suggests the E-field should be mainly contained within the tube, but that may not be the case as the frequency is so low.

Nope.

The e-field you injected into the pipe stays in there.  But when you're using it as a conductor the e-field is all from the pipe to the other conductor and the surrounds.  The e-field OUTSIDE the pipe stays OUTSIDE, too.

[BeenzMeenzWind]

Oh well, wrong again. What about if you had one conductor inside the other, just out of interest? Would that work like co-ax? I just buy wire, never thought about how it works. lol

[Ungrounded Lightning Rod]

What about if you had one conductor inside the other, just out of interest? Would that work like co-ax?

Yep.

In fact, if you kept the inner one at the center with some kind of spacer it would BE co-ax.  B-)  Term means coaxial, i.e. they share an axis.

The E and B fields would stay confined between the inner and outer conductors if the outer conductor was a superconductor.  But only a miniscule bit leaks due to voltage drop along the shield because it's built with regular, slightly resistive, metal.