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<p class=MsoNormal><span style='font-size:11.0pt;font-family:"Calibri","sans-serif";
color:#1F497D'>Erika,<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'>The most correct answer (which is a really funny thing to say)
is to use John Wiles’ “5-step program”—that’s
what I call it. It is published in one of the appendices of his latest “Suggested
Practices” document. If you can actually follow what he leads you through
(that’s why I call it the 5-Step program), it will lead you to the
correct answer.<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'>My short answer, that can result in a larger than necessary
conductor, is to determine the required overcurrent protection device (OCPD) rating
and then size the conductor accordingly. The reason it may be conservative is
that the NEC allows you to round to the next larger standard OCPD in 240.4(B).
This is the way all conductors are sized in the electrical industry. I’m
all about simplicity and being a little conservative is always better than
being overly liberal (I’m not making a political statement here, but if
the shoe fits…).<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'>Bill.<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>

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<p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span
style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> re-wrenches-bounces@lists.re-wrenches.org
[mailto:re-wrenches-bounces@lists.re-wrenches.org] <b>On Behalf Of </b>Erika M.
Weliczko<br>
<b>Sent:</b> Friday, April 02, 2010 6:27 AM<br>
<b>To:</b> 'RE-wrenches'<br>
<b>Subject:</b> [RE-wrenches] DC wire sizing<o:p></o:p></span></p>

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<p class=MsoNormal><o:p> </o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>To
my understanding the 156% on PV source and output circuits is related to the
ability of PV to deliver more than rated and be continuous.</span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>Therefore,
the wire has to be able to carry this current, so now the temperature and fill
corrections are applied to find the wire capable of the 156%.</span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'> </span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>I
am in a debate where the question is why correct for temp and fill on 156% of
ISC and spend all that extra money when the normal operating is at Imp. Or why
correct the 156% but why not correct the Isc or Imp.</span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'> </span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>I
am going to stick to the fact that the circuit has to carry the 156% under all
conditions…</span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>Thoughts?
</span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'> </span><o:p></o:p></p>

<p class=MsoNormal><span style='font-size:10.0pt;font-family:"Century Gothic","sans-serif"'>Erika</span><o:p></o:p></p>

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