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RIch;<br>
<br>
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 <span class="moz-smiley-s3"><span>
;-) </span></span><br>
Now you have to convince the AHJ, so you might want to show them
Mike's book on the subject.<br>
<br>
Ray Walters<br>
<br>
On 1/22/2012 2:13 PM, Rich Nicol wrote:
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<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">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.<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Thanks<o:p></o:p></span></p>
<p class="MsoNormal"><span
style="font-size:11.0pt;font-family:"Calibri","sans-serif";color:#1F497D">Rich<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"">
<a moz-do-not-send="true"
href="mailto:re-wrenches-bounces@lists.re-wrenches.org">re-wrenches-bounces@lists.re-wrenches.org</a>
<a moz-do-not-send="true"
href="mailto:[mailto:re-wrenches-bounces@lists.re-wrenches.org]">[mailto:re-wrenches-bounces@lists.re-wrenches.org]</a>
<b>On Behalf Of </b>William Miller<br>
<b>Sent:</b> Sunday, January 22, 2012 2:28 PM<br>
<b>To:</b> RE-wrenches<br>
<b>Subject:</b> Re: [RE-wrenches] Paralleling Multiple
Inverter<o:p></o:p></span></p>
</div>
</div>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Mark:<br>
<br>
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.<br>
<br>
Pending hearing otherwise from you, here is what I know, based
on 2008 code, and assuming load side connection (2008
citations in italics):<br>
<br>
1. <i>690.64(B)(1) Dedicated Overcurrent and Disconnect. Each
source<br>
interconnection shall be made at a dedicated circuit breaker<br>
or fusible disconnecting means.<br>
<br>
</i>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.<br>
<br>
<i>2. 690.64(B)(2) Bus or Conductor Rating. The sum of the
ampere<br>
ratings of overcurrent devices in circuits supplying power<br>
to a busbar or conductor shall not exceed 120 percent of the<br>
rating of the busbar or conductor.<br>
<br>
</i>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. <br>
<br>
3. <i>In systems with panelboards<br>
connected in series, the rating of the first overcurrent<br>
device directly connected to the output of a utility
interactive<br>
inverter(s) shall be used in the calculations for<br>
all busbars and conductors.<br>
<br>
</i>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.<br>
<br>
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.<br>
<br>
Sincerely,<br>
<br>
William Miller<br>
<br>
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.<br>
<br>
This is why I am very interested in any knowledge that might
prove otherwise in this scenario.<br>
<br>
Wm<br>
<br>
<br>
At 10:36 AM 1/22/2012, you wrote:<br>
<br>
<o:p></o:p></p>
<p class="MsoNormal">William,<br>
<br>
The SMA6000 @ 240V has an AC output rating of 25A. 4 times 25
times 1.25<br>
equals 125 A. If the solar accumulation panel is a dedicated
load center,<br>
then the load center, feeder and feeder breaker need only be
rated to 125 A.<br>
If the dedicated load center is located in the same building
as the service,<br>
then the dedicated load center can be main lug without main
breaker. If the<br>
dedicated load center is located on a buidling remote from the
the service,<br>
then it will need a main breaker rated at 125A.<br>
<br>
<br>
Mark Frye<br>
Berkeley Solar Electric Systems<br>
303 Redbud Way<br>
Nevada City, CA 95959<br>
(530) 401-8024<br>
<a moz-do-not-send="true" href="http://www.berkeleysolar.com/">www.berkeleysolar.com</a>
<br>
<br>
-----Original Message-----<br>
From: <a moz-do-not-send="true"
href="mailto:re-wrenches-bounces@lists.re-wrenches.org">re-wrenches-bounces@lists.re-wrenches.org</a><br>
[<a moz-do-not-send="true"
href="mailto:re-wrenches-bounces@lists.re-wrenches.org">mailto:re-wrenches-bounces@lists.re-wrenches.org</a>]
On Behalf Of William<br>
Miller<br>
Sent: Sunday, January 22, 2012 10:20 AM<br>
To: RE-wrenches<br>
Subject: Re: [RE-wrenches] Paralleling Multiple Inverter<br>
<br>
Rich:<br>
<br>
Here is the first consideration you might want to make: You
state a value<br>
of 100 Amps for this system, which I assume is the maximum AC
output at 240<br>
VAC (25 Amps) times four inverters. The value of 25 amps per
inverter is<br>
irrelevant to your design, however. You must take the 25 amps
times 1.25<br>
(for continuous duty) which means your required breaker is 40
amps (unless<br>
you can find 35 amps breakers). 40 Amps is the relevant
value. 40 times 4<br>
equals 160 amps. Therefore, for purposes of 690.64(B)
calculations, you<br>
have a 160 amp system.<br>
<br>
Others on this list will no doubt provide advise on how to
deal with that<br>
160 amps.<br>
<br>
William Miller<br>
<br>
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