[RE-wrenches] 120% Rule - applying to multiple load centers etc

[August Goers]

Hi Wrenches,



Please see question below forwarded from one of my fellow engineers.
Thanks, August



*From:* Eric Schoonbaert [mailto:eric at luminalt.com]
*Sent:* Monday, May 07, 2012 9:38 AM
*To:* August Goers
*Subject:* 120% Rule



The 120% rule, and how it is applied to panel board bus size has been
widely discussed. There is one part of the rule (2008 and 2011 NEC quoted
below) that gets much less attention and is the subject of this email. That
is, how and when is the rule applied to a conductor? The heading and text
both clearly say bus or *conductor *rating [emphasis added].

* *

*2011 NEC 705.12 Point of Connection* *(D)* *(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.



*2008 690.64 Point of Connection (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. 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



In Homepower 140 (December 2010 & January 2011), John Wiles has several
examples of the application of the 120% rule in his code corner article. I
have included one below.



For example: Two inverters each require a 50 A backfed breaker in a main
lug inverter combining load center to meet 690.8 requirements. A
supply-side connection is going to be made with a 100 A fused disconnect.
The rating of the combining load center and the ampacity of the
conductor to the 100 A fused disconnect must follow the
690.64(B)(2) requirements. As noted, even with a supply-side connection, as
soon as the circuit passes through the service
entrance disconnect/overcurrent device, all load-side requirements apply,
because the PC circuit is now on the load side of the service disconnect.

(50 A + 50 A +100 A) ÷ 1.2 = 200 A ÷ 1.2 = 166.7 A



The numbers indicate that a 200 A inverter load center/panel would be
needed because there is no 175 A option available. Assuming a 75°C rated
conductor, a 2/0 AWG conductor should be used between that panel and the
100 A fused disconnect



My particular question concerns the application of this rule to the
ampacity of the conductor. This is the only example I have come across
where the rule is applied to a conductor in this way. Is it the intention
of the code language to size the conductor between a main lug sub panel
used to combine solar circuits and the OCPD for the panel using the 120%
rule? Or is the inclusion of the word *conductor* to ensure that load
centers that have conductors integral to them are covered?



What about other equipment between along the conductor such as a utility of
fire department disconnect? In the John Wiles example above, if an utility
disconnect was to be located along the conductor, would 200 A rated
disconnect be necessary?



Applying this rule to the conductor does not seem to follow the same logic
(at least as I understand it) as a panel board bus bar which has spots for
additional circuits between the OCPD's supplying power to it while a
conductor is typically does not.
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