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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>
<blockquote type=cite class=cite cite>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 href="http://www.berkeleysolar.com/" eudora="autourl">www.berkeleysolar.com</a>
<br><br>
-----Original Message-----<br>
From: re-wrenches-bounces@lists.re-wrenches.org<br>
[<a href="mailto:re-wrenches-bounces@lists.re-wrenches.org" eudora="autourl">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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