Cable ampacity table or calculation - Page 1
Forum section: electronics / projects

author: mcinque
author_rank: Unknown
timestamp: December 29, 2013, 02:42:38 pm
content: Until now I've always used some photocopied tables that a friend of mine found in a book many years ago to determine wich cable diameter use in my projects, and when in doubt I've always used a bigger size cable.Now I'd like to learn how totechnicallychoose the wire size.I've searched a lot on the Internet looking for a table or a calculator to determine easily cable ampacity.I've found many, many tables and calculators regarding this.The matter is that almost every table and every calculator provide different results from the other and I don't know who trust.Since is better to understand the principle instead merely apply some written numbers, maybe it's time I start to learn more about this.I've searched some formulas to understand more about this argument, and I've learn that the calculation must consider:- the diameter of the wire conductor- the type of the wire conductor (copper, silver etc)- the thermal resistance in DC of the wire conductor- the thermal resistance of the ambient where the cable should work (of course air for me)- ambient temperature- max cable temperatureOf course I understand is all a matter of heat dissipation, so the last point is very important.Can you please suggest me some literature (or some advices) to learn how to make this kind of calculation?Thank you in advance for your help.

author: Dave
author_rank: Super Contributor
timestamp: December 29, 2013, 06:16:02 pm
content: Quote from: mcinque on December 29, 2013, 02:42:38 pmI'm afraid it's not that simple.You have a conductor inside that is being heated by the current passing through it, we express this as power per unit of length (W/m).You then need to take into account the thermal resistance of the insulation, you know the inner and outer diameter of the insulation, the specific thermal resistance is probably expressed in the cable's datasheet, one simple integral and you have the thermal resistance per unit lenght (K/(W*m)). Everything is clear so far.This is where it gets tricky - the cable dissipates the heat in two ways:-convection (heating the air around it)-radiation (by radiating in the infrared spectrum)Convection (Newton's law of cooling):Let's simplify this.The differential of heat over time is power. Let's assume a constant temperature of the surface of the cable, so we can drop the function of time.What we have now is:P = h * A * (Tcable- Tambient) (power = heat transfer coefficient * surface area * temperature difference)Let's divide it by lenght.P/l = h * C * (Tcable- Tambient) (power per unit length = heat transfer coefficient * circumference * temperature difference)Let's say we know both temperatures and the outer diameter of the insulation. We are lacking one crucial piece of information: heat transfer coefficient. It depends on many, many variables. You could probably write a dissertation on that topic.Radiation (Stefan's law)(in case you didn't know, Jožef Stefan was also Slovenian; just wanted to point it out):There is an asterisk next to the j, because we also need to take into account theemissivityof the surface. A black body (completely absorbant) would have the emissivity of 1, while a white body (completely reflective) would have an emissivity of 0. Our cable's insulation would have an emissivity somewhere between 0 and 1.Let's rewrite this a bit:J = sigma * Tcable4* epsilonP/A = sigma * Tcable4* epsilonP/l = sigma * Tcable4* epsilon * C (power per unit length = Stefan's constant * (Temperature of cable)4* emissivity * outer circumference)So here we have the equation to calculate the power dissipated through infrared radiation. The problem is we don't know the emissivity of the insulation. You certainly won't find it in the datasheet.And there is also one more problem: The same way the insulation radiates the heat, it also absorbs the radiated heat from the surrounding components. Try to write an equation for that!If we managed to find all of the unknown coefficients, we would need to cram in all into a single bulky equation and calculate the temperatures of the different parts (conductor, insulation surface, etc.) and make sure that they doesn't exceed the maximum rating (with a safety margin, of course). There's an idea for your second dissertation.Oh, and let's not forget that when the cable would heat up, the conductor's resistance would change, and all of your calculations would go to hell.As you probably figured out already, calculating that in theory would be pretty damn difficult. You would be much better off just using the tables and picking slightly thicker wire gauges, just to be sure.

author: qno
author_rank: Frequent Contributor
timestamp: December 29, 2013, 07:33:37 pm
content: Its not so much the heat the wires can handle but also the voltage drop you allow.the resistance of a wire is R= (rho x Length)/ wire cross area (R= rho * l / a)Google for "wire resistance" and various calculaters will popup.With ohms law you can determine the voltage drop you allow during max current.This will work from DC to a couple of 100 kHz. Above that you have to account for the skin effect.

author: jmaja
author_rank: Frequent Contributor
timestamp: December 29, 2013, 08:19:13 pm
content: Quote from: Dave on December 29, 2013, 06:16:02 pmA few months ago I wrote a software, which does all that and a bit more. Yes it wasn't very easy, but I wouldn't call it "pretty damn difficult". It took me about a week (~40 hours) to write the software from scratch. It was for a power plant conductors with 10 000 A currents and the results were very close to the measured temperatures.The most problematic part is to define the operating conditions and the worst case scenario. Even a very short insulated part (say some tape or a plastic clamp) may lead to problems. That's why you need a lot of safety margin.

author: Dave
author_rank: Super Contributor
timestamp: December 30, 2013, 02:34:17 am
content: Interesting. Could you upload your program here (including the source code), if that wouldn't be too much trouble?

author: mrkev
author_rank: Regular Contributor
timestamp: December 30, 2013, 02:51:07 am
content: Well I think that those tables you have shows different results because they all use rounding and assume different conditions.The poppular rule of thumb is to have about 8A/1mm2with copper. Ofc, it could take much more, deppending on the situation, lenght, airflow, etc.Do you need it because you have really high current, or because you wanna save money

author: peter.mitchell
author_rank: Super Contributor
timestamp: December 30, 2013, 04:54:49 am
content: I usually base my sizing off "test it, does it seem ok? yeah, she'll be right mate"

author: jmaja
author_rank: Frequent Contributor
timestamp: December 30, 2013, 07:00:31 am
content: Quote from: Dave on December 30, 2013, 02:34:17 amSorry, it's confidential.

author: mcinque
author_rank: Unknown
timestamp: December 30, 2013, 12:21:03 pm
content: Dave, thank you very much indeed for your detailed reply, I appreciate the time you've spent writing the post.Peter.mitchell: that's probably the better methodqno: you're right, voltage drop is another important point.I will test and apply your suggestions, thank you again.

author: tszaboo
author_rank: Super Contributor
timestamp: December 30, 2013, 12:54:35 pm
content: I usually "google" for AWG to get to this page:http://en.wikipedia.org/wiki/American_wire_gaugeThen I take 50% safety margin, and then another 50% when I look for the cable in the warehouse. That being said, I am not an electrician, I usually make decisions for cables, if they are produced in really small quantities, and if I under-size the cable, that mistake costs way more money than the money spent on the cable. Hell, even me thinking about it costs more money.

author: TerraHertz
author_rank: Super Contributor
timestamp: December 31, 2013, 01:04:16 am
content: There are yet more factors, related to the thermal behavior of the cable's surroundings. Is it in a bundle of other cables? They may be dissipating heat too. Is it in a conduit? What is the thermal conductivity of the conduit? Is the conduit horizontal or vertical? If vertical, heated air will rise in it, and may result in very uneven temperatures.Also, the resistance of metals generally increases with temperature. So dissipation for a fixed current isn't a constant.Then there's the dropout capacity of the protection device. Under overload, will the protection device cut out before the cable becomes a fire hazard? What sort of insulation does the wire have, and how hot can it get before it is damaged (or catch fire)?Here are tables from the Australian Standard SAA Wiring Rules, AS 3000-1991