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</o:shapelayout></xml><![endif]--></head><body bgcolor=white lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>It all boils down to how ampacity is determined in the NEC.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Ampacity is really related to temperature as far as the NEC is concerned. The ampacity of a 1/0 wire at 90c is the constant current it can carry in free air (30 C air) and not achieve an internal temperature of more than 90 C. the 75 c ampacity is the same, the amperage it can carry while not going over 75 C.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Now, if you double the diameter of a circle, the circumference also doubles, but the cross sectional Area actually goes up much more (4x more in fact) because cross sectional area is based on the square of the diameter and circumference is simply based on the diameter ^1 power.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Wire dissipates heat from its surface only, so the dimension critical for the amount of heat a wire can dissipate is circumference, not cross sectional area. So even though the wire is larger and has a much lower resistance, the heat dissipating area does not increase by as much, so in the end the larger wire has a lower current carrying capacity per unit cross sectional area, than a smaller wire.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Here is an example that lets me keep the math simple: (the #’s are all round #’s and not based on real ampacities/resistances, just to keep the math simple.)<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Wire 1:<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>A diameter of 10 units and can carry 100A through it and stay at 90C. It has a resistance of 1 ohm/1000ft.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Wire 2:<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>A diameter of 20 units. It has 4times the cross sectional area, and double the circumference (which means 2x the outer surface area) to dissipate heat. It has a resistance of .25 ohm/1000ft (1/4 that of wire 1 since it has 4x the amount of copper to carry current).<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>For wire 1 to stay at 90 C, it has to dissipate (P=I^2*R), P=(100amps^2)*1 omh=10,000 watts per unit of outer surface area.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>So if Wire 2 has double the surface area to dissipate heat, it can dissipate about 2x the energy, or 20,000watts. So if we work it backwards (P=I^2*R is the same as I=sqrt[P/R]) I=sqrt[20,000/.25]=sqrt[80000]=282amps<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>So wire 2 can handle (282amps/100amps)=2.8 times the amperage, even though it has 4times the cross sectional area, all because it only has 2 times the surface are to get rid of heat. There are other factors with heat transfer that make the larger wire have even lower ampacity, but this demonstrates the main contributing factor.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>[In the NEC table 310.15(B) we see that a 250kcmill copper wire handles 255A @75C, and a 100kcmill (4x the area) can only take 545A, or 2.2 times the current, so not too far from my example].<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I hope the above helps more than it hurts.<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><div><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>With Regards,<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>Daniel Young, <o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>NABCEP Certified PV Installation Professional<sup>TM</sup>: Cert #031508-90<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>NABCEP Certified Solar Heating Installer<sup>TM</sup>: Cert #SH031409-13<o:p></o:p></span></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> <o:p></o:p></span></p></div><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:windowtext'>From:</span></b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:windowtext'> RE-wrenches [mailto:re-wrenches-bounces@lists.re-wrenches.org] <b>On Behalf Of </b>Larry<br><b>Sent:</b> Tuesday, December 02, 2014 4:49 PM<br><b>To:</b> RE-wrenches<br><b>Subject:</b> Re: [RE-wrenches] Parallel Wire combining<o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><div><p class=MsoNormal>That's the part that is throwing me off. If I have 3 times the circular mil and compare that to a single conductor of similar circular mil, how do I have 3 times the ampacity? These are very different numbers.<br><br>Example: 1/0 @ 90c is 170 amps x 3 = 510 amps. 510A is what a conductor just over 750 AWG will carry. The circular mils of 3 1/0 cables is only 325,050 or a 325 AWG cable which would be rated at about 330 amps. <br><br>So, is it 510 amps (3x the ampacity) or 330 amps (3x circular mil)? And, more importantly, why?<br><br>Larry<br><br><br><o:p></o:p></p><pre><o:p> </o:p></pre><p class=MsoNormal>On 12/2/14 11:14 AM, Bill Turberville wrote:<o:p></o:p></p></div><blockquote style='margin-top:5.0pt;margin-bottom:5.0pt'><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'>I am sorry. Bad fingers. Three times the ampacity under the same conditions.</span><o:p></o:p></p><p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D'> </span><o:p></o:p></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:windowtext'>From:</span></b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif";color:windowtext'> RE-wrenches [mailto:<a href="mailto:re-wrenches-bounces@lists.re-wrenches.org">re-wrenches-bounces@lists.re-wrenches.org</a>] <b>On Behalf Of </b>Larry<br><b>Sent:</b> Tuesday, December 02, 2014 12:09 PM<br><b>To:</b> RE-wrenches<br><b>Subject:</b> Re: [RE-wrenches] Parallel Wire combining</span><o:p></o:p></p></div></div><p class=MsoNormal> <o:p></o:p></p><div><p class=MsoNormal style='margin-bottom:12.0pt'>OK, let's use 1/0 for the example. 108,350 x 3 = 325,050. Do I now have a cable between 300 and 350 AWG?<o:p></o:p></p><pre>Thank you,<o:p></o:p></pre><pre> <o:p></o:p></pre><pre>Larry<o:p></o:p></pre><p class=MsoNormal><o:p> </o:p></p></div></blockquote><div class=MsoNormal align=center style='text-align:center'><hr size=1 width="100%" noshade style='color:#A0A0A0' align=center></div><p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'>No virus found in this message.<br>Checked by AVG - <a href="http://www.avg.com">www.avg.com</a><br>Version: 2012.0.2249 / Virus Database: 4189/8166 - Release Date: 12/02/14<o:p></o:p></p></div></body></html>