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What is the way of thinking to chose the proper wire gauge for the following system?:

enter image description here

There are calculators in web but I don't know the reasoning behind clearly. One example here: http://www.powerstream.com/Wire_Size.htm

As far as I know, the current is the only parameter which causes the heating and voltage drop. But voltage also taken into account when sizing the wires. I think they want to size the wire that it would cause %3 voltage drop.

Here is my example. Is my way of calculation correct?:

Lets say the intrinsic resistance of copper is rho. So for my system for the voltage drop ΔV to be 3%:

ΔV = 48 * 0.03

The total resistance is: [2*L*rho/(wire area)] so

ΔV = 48 * 0.03 = [2*L*rho/(wire area)] * I

(wire area) = [2*L*rho/(48 * 0.03)] * I

So in my case:

I = 3A

rho = 1.68 * 10^-8 (copper)

lets say L = 10 meters

wire area = [2*10*(1.68 * 10^-8)/(48 * 0.03)] * (3)

wire area = 7 * 10^-9 meter^2

Is that correct?

EDIT

The following two calculators calculate totally different results:

http://www.energymatters.com.au/climate-data/cable-sizing-calculator.php?main_page=wire_calc&calc_action=cable_size

http://www.solar-wind.co.uk/cable-sizing-DC-cables.html

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  • \$\begingroup\$ You got it wrong, U=R*I, not R*(I^2). The latter is P. Never skip the units, you can easily see such errors then. \$\endgroup\$ – Janka Oct 24 '16 at 23:02
  • \$\begingroup\$ thnks i edited it. how about now? and please see this link: solar-wind.co.uk/cable-sizing-DC-cables.html \$\endgroup\$ – HelpMee Oct 24 '16 at 23:04
  • \$\begingroup\$ according to this link:solar-wind.co.uk/cable-sizing-DC-cables.html i find 1.5mm^2 but my calculation yields something very different. \$\endgroup\$ – HelpMee Oct 24 '16 at 23:05
  • \$\begingroup\$ i insert 48V 3A 10m in that calculator. \$\endgroup\$ – HelpMee Oct 24 '16 at 23:07
  • \$\begingroup\$ @jbord39 this is DC current skin effect is for AC isnt it? \$\endgroup\$ – HelpMee Oct 24 '16 at 23:09
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The reasoning behind wire sizing is to provide less than allowable voltage drop over the distance between power source and destination.

You are doing all your calculations without understanding this goal, jumping between clearly defined terms as voltage drop, and undefined terms as conductance of unspecified copper alloy.

Start with a clear goal: voltage drop must be less than 3%. At 48 V this gives 1.44V, which you skipped to calculate and post.

Now, you have 2 wires, hot wire, and return wire. To get less than 1.44V at the load, you need to divide this value by two, since both wires will have the voltage drop. Which gives 0.72V per wire, assuming the wires are identical.

This gives you the limit on how much wire resistance you can afford at 3A load: 0.72V/3A = 0.24 Ohms. If your distance is 10m, then the wire must have 0.024 Ohm/m, a clearly identifiable parameter. This concludes the first part of the task.

Now which wire to select for this job, is up to unspecified condition in your exam task - it could be a copper-based alloy wire, it could be aluminum wire, or could be copper-coated steel wire. The rated carrying capacity will also depend on wire insulation and requirements for overheating. So it is a plus-minus guardbands set by manufacturer.

The first calculator is goofy because it does not count for allowable voltage drop, and its output is nonsense. The second calculator gives 16AWG copper for 2% loss (they don't have 1.5% entry), which looks reasonable.

So, what is the question again?

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  • \$\begingroup\$ i know my question wast put clear enough beacuse of panic. in my question i write ΔV = 48 * 0.03 isnt it %3 voltage drop? you explained great how i wanted, you wrote "0.024 Ohm/m" can you cpme to a conclusion as finding AWG after 0.24 Ohms calculation. do we need a table to find AWG after finding resistance here? could u elaborate on that? lets say we dont have calculator and we wanna find the awg roughly.. thanks \$\endgroup\$ – HelpMee Oct 25 '16 at 0:44
  • \$\begingroup\$ You go Google and ask for, say, "wire AWG resistance per foot". You can get something like this, daycounter.com/Calculators/AWG.phtml \$\endgroup\$ – Ale..chenski Oct 25 '16 at 1:01
  • \$\begingroup\$ it seems like there is a tabl from ohm/meter to awg for copper wires. i dont need to use conductance stuff as you said. so after finding Ohm/m the rest is following the AWG tables for copper right?i hope i got it. thanks for great answer \$\endgroup\$ – HelpMee Oct 25 '16 at 1:07
  • \$\begingroup\$ Yes, just the wires could be not necessary copper wires, so you will need to dig deeper for other wires. But the main limitation comes with your conditions, 0.024 Ohms per meter. The rest follows. \$\endgroup\$ – Ale..chenski Oct 25 '16 at 1:12
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There are three factors considered in wire sizing.

  1. Maximum wire temperature is affected by current, wire resistivity, type of insulation and number of conductors in a cable. Ambient temperature must also be considered. Allowable temperature is determined by type of insulation. Selection is generally made by consulting a table for the specific type of wire to be used etc.

  2. Voltage drop is affected by current, wire length and resistivity. The maximum voltage drop is often limited to 3%. The requirements of the load and expected source voltage variation are a considered in determining the maximum allowable voltage drop.

  3. Mechanical strength of the wire is also considered.

The selected wire is the largest wire indicated by each of the above factors. Depending on location and application, electrical codes may provide all of the required information and procedures.

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  • \$\begingroup\$ is "Cable run length" total length including return or the average length of plus and minus wires? \$\endgroup\$ – HelpMee Oct 24 '16 at 23:38
  • \$\begingroup\$ Cable run length would be the the length of the cable run between the source and load, not the length of the individual wires. There would be separate tables for single-phase and three-phase AC and a selection in a calculation program. For voltage drop, there is also a power factor selection. For DC you would probably use the single-phase, unity power-factor table or selection. \$\endgroup\$ – Charles Cowie Oct 24 '16 at 23:45
  • \$\begingroup\$ there is a calculator here; 12voltplanet.co.uk/cable-sizing-selection.html all the calculators give differnt results, and when i calculate myself i got another result. i will go crazy.. \$\endgroup\$ – HelpMee Oct 24 '16 at 23:46
  • \$\begingroup\$ Various tables may have different assumptions regarding insulation rating, ambient temperature etc. There can even different resistivity of copper wire because of different alloys. Pure copper is too soft for wire, so all "copper" wire is alloyed for a required strength and flexibility. You have to use your bets estimate to match the table you use with your wire and other factors. \$\endgroup\$ – Charles Cowie Oct 24 '16 at 23:53
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AWG choices can be many. Thermal rise, %Voltage loss, %power Loss, Ohm's Law, cable cost, availability, skin effect etc. Beware which criteria you NEED before making a selection. Your LINK was based on %V loss and not "Ampacity" std. which is defined by regional standards" and determined by a greater temp rise of an insulated conductor. eg. 40'C rise or 60'C etc., which depends on thermal insulation too. e.g. household wiring stds.

Your question was due to a lack of understanding on how AWG is selected.

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