William,
You have 9 M190s each running at 800 mA according to spec sheet. That equals 7.2 A nominal output current. Plus you have 10 M250s (if you look at your drawing) each running at 1 A = 10 A. 7.2 A + 10 A = 17.2 A * 1.25 = 21.5 A min breaker size. So, you'll need to round your breaker to a 25 A or 30 A breaker in the sub. You don't have any problem in the sub panel but the meter main is going to be a problem if it is a 100 A meter/main. Your drawing and case study notes don't seem to correspond - you mention M215s in the case study and show M250s on the drawing. If you indeed have 10 M215s then you're at 10 * 0.9 A = 9 A. 7.2 A + 9 A = 16.2 A * 1.25 = 20.25 A min breaker size. This is where it gets tricky. My fellow engineer just pointed out NEC 220.5(B) which allows you to drop fractions of amperes less than 0.5. So, I think you'd be okay with a 20 A solar breaker in this case. Best, August *From:* [email protected] [mailto: [email protected]] *On Behalf Of *William Miller *Sent:* Wednesday, October 09, 2013 3:07 PM *To:* RE-wrenches *Subject:* [RE-wrenches] Point of connection question Friends: At the risk of asking a question that may have been asked and answered here in the past, I wish to pose a question about Point of Connection. Rather than outline the situation here I put it all on a web page. If any of you have a moment to review this, I would be most appreciative. http://www.millersolar.com/MillerSolar/case_studies/POC/_POC_Question.html Thanks in advance, William Miller
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