Current capacity calculation

[phuct78]

Hello All

I was wondering if anyone can help with calculating the current carrying capacity of a copper tube.

I have found info that states it is that a tube of Ø130mm x 12mm wall, will have a DC current capacity of 3080A. The thing that throws me is that it doesn't seem to take into account the length of the tube and seems to only go on the X-section area. Unfortunately the software I used doesn't show the calculation.

I also know that it is dependant on the surrounding ambient temperature and this has been set at 58°C and I am allowed a 20°C temperature rise.

Is there anything I can use?

[omulki]

I wonder, why would the length of the pipe be relevant. The length should reduce itself in the equation (hope I put that correctly... I mean reduce in terms of eliminate).
Regarding the temperature dependency: alpha is only around 0.4 % / K in the proximity of 50 ° C. Eight percent delta R in your stated 20 degree range should be easy to take into account, plus easily fit in the huge safety margin you would throw in anyways.

Is liquid cooling like shown here an option? I belive, internal cooling is the primary reason to use a hollow formfactor for a high current conductor in the first place. (Probably apart from some skin effect thingy).

There are some high current specialists on board here - they'll know.

Regards, OM.

[Rerouter]

One way you can approach it is to go to equivalent cross sectional area, the current value it gives you will be the minimum its rated for,

Some math says its CSA is 4448.5 mm2
This is equivilent to a wire with a radius of 37.63 mm

For this, online calculators say its good for 3900A. and will be radiating 60W per meter.
http://circuitcalculator.com/wordpress/2007/09/20/wire-parameter-calculator/

http://www.lapptannehill.com/wp-content/uploads/2014/02/CU_AL_Ampacity_Chart_Final.pdf
Moving along, 11 parrellel conductors of 750 KCMIL (weird american unit), equivilent to 380mm2 CSA, is rated for 4000A, 380 * 11 = 4180 < 4485mm2, though you do end up with less outer surface area, 11 * 69.12 mm2 / mm = 760mm2 of surface area to atmosphere, vs, 408mm2 / mm for your pipe, meaning it will be harder for it to radiate the heat, but it will be somewhat cooler. 0.00000395 ohms vs (0.0000462 / 11) = 0.0000042 = 63.8W / m

The calcs i was using assumed a maximum temperature of 70C, if you can allow hotter you will be well within margin, if there is some airflow through the tube, then the surface area exceeds the bundle and your more than fine.

[phuct78]

Thank you both for your input. Let me say that I am heavily mechanical bias with a basic understanding of electrics/electronics, so I may ask stupid things about what you reply.

Maybe I should try and explain the . My company manufactures electrical rotary joints/slip rings and we generally have a sliding contact on a copper ring in 2 positions. We generally use an equation that uses the rectangular CSA for the point of contact but we have a new enquiry that will require sliding contacts in parallel in 10-12 positions around the ring meaning our normal equation is not suitable.

@omulki The reason I query the length is that I would say the a 1mm thick section can't take as much current a say a 100mm length. The cooling of the tube isn't a factor that I want to use in the equation.

@Rerouter I understand the first part of your reply but the parallel conductors part has gone over my head.

[Rerouter]

The last link i gave listed 4000A in the bottom left corner when you have 11 seperate wires of 380 mm2 CSA,

your cross sectional area does exceed this, but because you only have 1 conductor, your surface area is smaller than the 11 smaller ones, meaning it will be harder for it to radiate away the 60W / m of heat. again bugger all in the scheme of things.

[ajb]

The capacity depends on what your limiting parameters are.  The current capacity given by various standards will depend on the conditions that those standards assume.  Electrical codes, for instance, will give different capacities depending on free air or in conduit with different conduit fill factors.  Standards usually base current capacities on allowable power loss per unit length, but your application may require a different max power loss, or you may have a very long conductor and be more concerned about voltage drop.  It's important to understand the conditions that the standard assumes and the limiting parameters they use, because the conditions in your application may differ substantially.

If you have a maximum voltage drop specification, then you need to determine the resistance which would, in fact, take into account the length of the conductor.  However, I would suspect that the contact resistance would be a much larger factor than conductor resistance, unless your contacts have to move across a very long conductor.

If you need to determine the total resistance of a set of parallel resistances (IE, your 10-12 contact resistances in parallel), there's a simple equation for that: R_total = 1/( 1/R₁ + 1/R₂ + ... + 1/R_n ).  If all of the resistances are identical, then for n resistances this simplifies to R_total = R₁/n.

[ajb]

One more thing:

We generally use an equation that uses the rectangular CSA for the point of contact but we have a new enquiry that will require sliding contacts in parallel in 10-12 positions around the ring meaning our normal equation is not suitable.

It sounds like you might be basing the current capacity of the contact on the contact patch area and the material resistance, but this is a very incomplete picture of the matter.  The actual effective contact resistance (and therefore current capacity) will depend on the quality of the mating surfaces, surface contamination/oxidation, contact pressure, etc.  In short, it's the kind of thing that you probably want to measure under a variety of conditions rather than simply calculate.  Calculate, but verify.

[IanB]

Thank you both for your input. Let me say that I am heavily mechanical bias with a basic understanding of electrics/electronics, so I may ask stupid things about what you reply.

Maybe I should try and explain the . My company manufactures electrical rotary joints/slip rings and we generally have a sliding contact on a copper ring in 2 positions. We generally use an equation that uses the rectangular CSA for the point of contact but we have a new enquiry that will require sliding contacts in parallel in 10-12 positions around the ring meaning our normal equation is not suitable.

@omulki The reason I query the length is that I would say the a 1mm thick section can't take as much current a say a 100mm length. The cooling of the tube isn't a factor that I want to use in the equation.

@Rerouter I understand the first part of your reply but the parallel conductors part has gone over my head.

When asking "What is the current carrying capacity of a conductor?", then with no other information given the answer to this question is all about heating. Current going through the conductor heats it up, and this heat has to be carried away to the surroundings. Therefore the answer to the question is entirely mechanical: what is the overall heat transfer capability from the conductor to the surroundings, by conduction and/or by radiation. That tells you how hot the conductor will get, and how hot it gets is the limiting factor. Conductors in enclosed ducts or surrounded by insulation can carry less current than free conductors in the open air.

Now the length of the conductor doesn't matter in this calculation because everything can be normalized to unit length. The heat dissipated by the current is proportional to the length, and the heat removed to the surroundings is also proportional to the length.

If there is some other factor at play, such as a sliding shoe taking off current, then the question becomes about the contact resistance between the shoe and the conductor. So it is necessary to define the parameters of the problem before being able to answer it.

[Zero999]

Length is often important because that gives the resistance and can be used to work out the voltage drop, which is often an issue before overheating of the material surrounding the conductor is.

[IanB]

Length is often important because that gives the resistance and can be used to work out the voltage drop, which is often an issue before overheating of the material surrounding the conductor is.

Very true. I wasn't thinking about that dimension.

[mc172]

Completely off-topic but your username is brilliant.