[RE-wrenches] Paralleling Multiple Inverter

Ray Walters ray at solarray.com
Sun Jan 22 17:04:13 PST 2012


RIch;

I think you're exactly right on your interpretation.  Mike Holt's  
"Guide to 2011 NEC requirements for PV..." has an 'author's comment' on 
page 305, that the 120% buss bar rule of article 705 only applies if 
there are loads in the panel.  If all slots are filled with PV breakers, 
you're off the hook, at least according to Mike Holt, and my self for 
the little that matters ;-)
Now you have to convince the AHJ, so you might want to show them Mike's 
book on the subject.

Ray Walters

On 1/22/2012 2:13 PM, Rich Nicol wrote:
>
> My take on it in that respect is a 200 amp panel, with no additional 
> slots other than those for the 4 inverters with clear signage that no 
> load circuits are permitted in this panel. A second small panel would 
> then be employed if any load circuits were required for lighting or 
> monitoring devices.
>
> Thanks
>
> Rich
>
> *From:*re-wrenches-bounces at lists.re-wrenches.org 
> <mailto:re-wrenches-bounces at lists.re-wrenches.org> 
> [mailto:re-wrenches-bounces at lists.re-wrenches.org] 
> <mailto:[mailto:re-wrenches-bounces at lists.re-wrenches.org]> *On Behalf 
> Of *William Miller
> *Sent:* Sunday, January 22, 2012 2:28 PM
> *To:* RE-wrenches
> *Subject:* Re: [RE-wrenches] Paralleling Multiple Inverter
>
> Mark:
>
> I would respectfully disagree, based on my reading.  If you have code 
> citations that inform me otherwise, I would be very grateful to expand 
> my knowledge.
>
> Pending hearing otherwise from you, here is what I know, based on 2008 
> code, and assuming load side connection  (2008 citations in italics):
>
> 1. /690.64(B)(1) Dedicated Overcurrent and Disconnect. Each source
> interconnection shall be made at a dedicated circuit breaker
> or fusible disconnecting means.
>
> /Each inverter will require a 40 amp circuit breaker.  The value of 
> that breaker is calculated by adding a 25% continuous duty rating to 
> the maximum AC output, which is 25 amps: 25 * 1.25 = 31.25.  The next 
> breaker size up is 40 amps, so you need 40 amp breakers.
>
> /2. 690.64(B)(2) Bus or Conductor Rating. The sum of the ampere
> ratings of overcurrent devices in circuits supplying power
> to a busbar or conductor shall not exceed 120 percent of the
> rating of the busbar or conductor.
>
> /The designer must add the values of the circuit breakers to determine 
> the back-feed value.  We just calculated the circuit breaker size in 
> step 1, above.  Four inverters means four 40 amp circuit breakers, 
> therefore: 4 * 40 = 160.
>
> 3. /In systems with panelboards
> connected in series, the rating of the first overcurrent
> device directly connected to the output of a utility interactive
> inverter(s) shall be used in the calculations for
> all busbars and conductors.
>
> /Therefore the designer must use 160 amps as the total back-feed value 
> for all panels and feeders in series all the way back to the service.
>
> Mark, I would really appreciate it if you could reply today with any 
> information I am missing.  I could use my new-found knowledge to 
> modify the permit application I am submitting tomorrow morning.
>
> Sincerely,
>
> William Miller
>
> PS:  I used to think, erroneously, that I need only consider the 
> actual maximum AC amperage from a given inverter.  Some time ago I bid 
> on and started a job based on that fallacy.  Mid-way into the job the 
> AHJ informed me that my calculations were incorrect, that I needed to 
> use the breaker value.  This job used SB6000 inverters and the value 
> required was 40 amps.  I researched this thoroughly and discovered 
> they were right.  In order to comply with 690.640(B), I had to 
> downgrade the main breaker at my own expense.  The breaker was not 
> inexpensive, so this is a lesson I learned the hard way.
>
> This is why I am very interested in any knowledge that might prove 
> otherwise in this scenario.
>
> Wm
>
>
> At 10:36 AM 1/22/2012, you wrote:
>
> William,
>
> The SMA6000 @ 240V has an AC output rating of 25A. 4 times 25 times 1.25
> equals 125 A. If the solar accumulation panel is a dedicated load center,
> then the load center, feeder and feeder breaker need only be rated to 
> 125 A.
> If the dedicated load center is located in the same building as the 
> service,
> then the dedicated load center can be main lug without main breaker. 
> If the
> dedicated load center is located on a buidling remote from the the 
> service,
> then it will need a main breaker rated at 125A.
>
>
> Mark Frye
> Berkeley Solar Electric Systems
> 303 Redbud Way
> Nevada City,  CA 95959
> (530) 401-8024
> www.berkeleysolar.com <http://www.berkeleysolar.com/>
>
> -----Original Message-----
> From: re-wrenches-bounces at lists.re-wrenches.org 
> <mailto:re-wrenches-bounces at lists.re-wrenches.org>
> [mailto:re-wrenches-bounces at lists.re-wrenches.org] On Behalf Of William
> Miller
> Sent: Sunday, January 22, 2012 10:20 AM
> To: RE-wrenches
> Subject: Re: [RE-wrenches] Paralleling Multiple Inverter
>
> Rich:
>
> Here is the first consideration you might want to make:  You state a value
> of 100 Amps for this system, which I assume is the maximum AC output 
> at 240
> VAC (25 Amps) times four inverters.  The value of 25 amps per inverter is
> irrelevant to your design, however.  You must take the 25 amps times 1.25
> (for continuous duty) which means your required breaker is 40 amps (unless
> you can find 35 amps breakers).  40 Amps is the relevant value.  40 
> times 4
> equals 160 amps.  Therefore, for purposes of 690.64(B) calculations, you
> have a 160 amp system.
>
> Others on this list will no doubt provide advise on how to deal with that
> 160 amps.
>
> William Miller
>
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