Changes to <http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact =8&ved=0CEoQFjAC&url=http%3A%2F%2Fwww.slideshare.net%2Fsolpowerpeople%2Finte rconnection-rule&ei=FZcYU5e_LsSi0QH5yICQCQ&usg=AFQjCNGypW0SbS6-POHt4d82Sxc-- kZ0ig&sig2=Kz2661LzDghXtx9GwuhOfw&bvm=bv.62577051,d.dmQ> 2014 NEC Interconnection Rule 705.12
Here's a link to a powerpoint (slideshare.net) by Richard Stovall which clarifies 705.12 using some excellent illustrations. Maybe some have seen it already. Works for me. Contact me off-list if you'd like a pdf version. Kirk Herander VT Solar, LLC dba Vermont Solar Engineering NABCEPTM Certified Inaugural Certificant NYSERDA-eligible Installer VT RE Incentive Program Partner 802.863.1202 From: [email protected] [mailto:[email protected]] On Behalf Of Brian Mehalic Sent: Wednesday, March 05, 2014 10:02 AM To: RE-wrenches Subject: Re: [RE-wrenches] conductors and the 120% rule A very clarifying change is what I'd call it! First off, rather than being based on the actual breaker size on the inverter output circuit, calcs are instead based on 125% of the inverter rated output current. 705.12(D)(2)(1) addresses "Feeders" - but only applies when the inverter output circuit connection is made somewhere other than the opposite end of the feeder from the utility supply. This addresses concerns about whether the feeder conductor needs to be larger due to the presence of the additional source of supply, and so long as the inverter isn't connected to the feeder in the middle of it then the existing conductor size should be okay (because if it is at the opposite end of the feeder than there is nowhere where the utility and inverter current will be additive). 705.12(D)(2)(3) addresses "Busbars" and allows several options, including the familiar "120% rule" as you stated in your original post. Also check out 705.12(D)(2)(3)(c) - depending on the load breakers in the subpanel, the 120% rule may not even need to be used (if the sum of the inverter and load breakers is less than or equal to the busbar rating). And remember, even if your AHJ hasn't adopted 2014 yet it is worth having a conversation with them to see if they'll allow you to design the system based on the new Code - after all, in a certain sense, the 2014 NEC is what "they" meant the 2011 NEC to say! Cheers, Brian Mehalic NABCEP Certified Solar PV Installation ProfessionalT R031508-59 IREC ISPQ Certified Affiliated Instructor/PV US-0132 PV Curriculum Developer and Instructor Solar Energy International http://www.solarenergy.org On Tue, Mar 4, 2014 at 5:21 PM, Kirk <[email protected]> wrote: Is this a code change in 2014 vs 2011 or merely a clarification? Vt has not adopted 2014 yet. What was the original rationale for the 120% rule to apply to conductors in addition to a panel bus? Kirk Herander VSE On Mar 4, 2014, at 6:20 PM, Brian Mehalic <[email protected]> wrote: If the subpanel is at the end if the feeder, and there are no taps in between the main and the sub then I don't see any reason that the conductors need to be any larger than 200 A as there is no where on the feeder conductors where grid and PV current will be additive. The changes in 705.12 in 2014 address this in large part. Brian On Mar 4, 2014, at 2:41 PM, "Kirk Herander" <[email protected]> wrote: Solaredge 20 kw, 480 3-phase. Good point, but that may be irrelevant. The feed-in subpanel is also powering unrelated loads, which use the neutral as a conductor from the main panel. So 4 conductors from the main. Kirk Herander VT Solar, LLC dba Vermont Solar Engineering NABCEPTM Certified Inaugural Certificant NYSERDA-eligible Installer VT RE Incentive Program Partner 802.863.1202 From: [email protected] [mailto:[email protected]] On Behalf Of Allen Frishman Sent: Tuesday, March 04, 2014 4:32 PM To: RE-wrenches Subject: Re: [RE-wrenches] conductors and the 120% rule what inverter(s) are you using? In many cases the Neutral is not considered a Current Carying Conductor by the Manufacturer and therefore you only have 3 CCC. Al Frishman AeonSolar (917) 699-6641 <tel:%28917%29%20699-6641> - cell (888) 460-2867 <tel:%28888%29%20460-2867> www.aeonsolar.com <http://www.aeonsolar.com/> On Mar 4, 2014, at 4:20 PM, Kirk Herander wrote: Approx.. 50 - 60ft. Kirk Herander VT Solar, LLC dba Vermont Solar Engineering NABCEPTM Certified Inaugural Certificant NYSERDA-eligible Installer VT RE Incentive Program Partner 802.863.1202 From: [email protected] [mailto:[email protected]] On Behalf Of Ray Walters Sent: Tuesday, March 04, 2014 4:05 PM To: RE-wrenches Subject: Re: [RE-wrenches] conductors and the 120% rule What is the length of the conduit to the subpanel? That will determine whether to apply the derates. R.Ray Walters CTO, Solarray, Inc Nabcep Certified PV Installer, Licensed Master Electrician Solar Design Engineer 303 505-8760 <tel:303%20505-8760> On 3/4/2014 1:34 PM, Kirk Herander wrote: Whether or not a further derate has to be applied is the killer here, as I am working with existing panels and conductors. In an old Code Corner(HP140) J. Wiles goes through a similar scenario and calls out the allowable current rating and conductor in 310.15, but makes no mention of applying additional derate factors. The .8 derate for 4-6 conductors(l1,l2,l3, & n) will put the existing 4/0 cable between feed-in and main panel at 208 amps, less than the allowable 217. I'd hate to need to upsize the wire to 250 mcm. Kirk Herander VT Solar, LLC dba Vermont Solar Engineering NABCEPTM Certified Inaugural Certificant NYSERDA-eligible Installer VT RE Incentive Program Partner 802.863.1202 From: [email protected] [mailto:[email protected]] On Behalf Of Jason Szumlanski Sent: Tuesday, March 04, 2014 2:57 PM To: RE-wrenches Subject: Re: [RE-wrenches] conductors and the 120% rule Both the bus and conductors need to be rated for 217 amps minimum. As you mentioned, the bus is not a problem. The way I interpret it, the conductor size required would be after derate factors are applied. The rating of the conductor is ultimately dependent on the derate factors. If you can locate your subpanel adjacent to the main distribution panel, you may be able to use Exception #3 to 310.15(B)(2) by connecting the panels with a short nipple. I assume you are just looking at a number of conductor derate and not an ambient temperature derate. Jason Szumlanski Fafco Solar <image001.jpg> On Tue, Mar 4, 2014 at 12:05 PM, Kirk Herander <[email protected]> wrote: Hello, I have a 225 amp 3-phase main lug sub-panel protected by a 200 amp breaker. My inverter breaker feeding the sub panel is 60 amps. So 225 a bus x 1.2 = 270 amps. That's less than the sum of the two breakers of 260 amps, so no issue there. The conductors between sub and main panel have to be rated for at least 260/1.2 = 217 amps, correct? Is this 217 amps before or after derating the conductor? 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