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

[Andrew Truitt]

August / Eric - My understanding of this requirement is that all equipment
in the circuit must be rated to handle the maximum fault current that could
flow at a given point.  So, if your power source / overcurrent protection
scheme is: 50A (inverter 1) + 50A (inverter 2) + 100A (utility grid via
fused disconnect) then 200A is the max fault current at any point in that
circuit (including conductors and switches) and should be used in your 120%
rule calculation (as J.W. does).  This can definitely present issues,
especially when attempting to interconnect at existing subpanels with
feeders that were not sized with the future addition of PV in mind, but I
think one intent of this article is to ensure that if there were a fault in
those feeders, and the PV inverters continued to operate (unlikely), that
the conductor could handle the sum of the fault currents (PV + utility).


For a brighter energy future,


Andrew Truitt
NABCEP Certified PV Installer™ (ID# 032407-66)****

Principal
Truitt Renewable Energy Consulting****

(202) 486-7507****

http://www.linkedin.com/pub/andrew-truitt/8/622/713


"Don't get me wrong: I love nuclear energy! It's just that I prefer fusion
to fission. And it just so happens that there's an enormous fusion reactor
safely banked a few million miles from us. It delivers more than we could
ever use in just about 8 minutes. And it's wireless!"

~William McDonough



On Mon, May 7, 2012 at 10:55 AM, August Goers <august at luminalt.com> wrote:

> 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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