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<TITLE>Re: [RE-wrenches] voltag drop</TITLE>
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<FONT SIZE="5"><FONT FACE="Calibri, Verdana, Helvetica, Arial"><SPAN STYLE='font-size:12pt'>Marco,<BR>
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The formulas I learned in school are:<BR>
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Single-Phase Vd = (2 * d * k * I) / cm<BR>
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where: d = distance; the constant k = 12.9 Ohms for Cu & 21.2 Ohms for AL; I = current; cm = circular mils <BR>
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3-Phase Vd = (1.73 * d * k * I) / cm<BR>
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My note are failing me, but I believe d is actually the one-way distance. This is multiplied by 2, in the single-phase calculation, to account for the round trip distance. In the 3-phase calculation 1.73 is used instead. Table 8 in the NEC has the conductor properties. This is where I always look up circular mils.<BR>
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In the past I’ve run these calculations against my ElectriCalc & had things check out. <BR>
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</SPAN></FONT></FONT><BLOCKQUOTE><FONT SIZE="5"><FONT FACE="Calibri, Verdana, Helvetica, Arial"><SPAN STYLE='font-size:12pt'><BR>
</SPAN></FONT></FONT><FONT FACE="Calibri, Verdana, Helvetica, Arial"><FONT SIZE="4"><SPAN STYLE='font-size:11pt'>I feel kinda foolish in asking this.....but when calculating voltage drop in a conductor over distance, do I use the one-way or round-trip distance? </SPAN></FONT><FONT SIZE="5"><SPAN STYLE='font-size:12pt'> </SPAN></FONT><FONT SIZE="4"><SPAN STYLE='font-size:11pt'>thanks, marco </SPAN></FONT><FONT SIZE="5"><SPAN STYLE='font-size:12pt'> </SPAN></FONT><FONT SIZE="4"><SPAN STYLE='font-size:11pt'>Marco Mangelsdorf, Ph.D.</SPAN></FONT><FONT SIZE="5"><SPAN STYLE='font-size:12pt'>, </SPAN></FONT><FONT SIZE="4"><SPAN STYLE='font-size:11pt'>President</SPAN></FONT><FONT SIZE="5"><SPAN STYLE='font-size:12pt'> <BR>
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