Hi there,
Need your your help in determining the maximum size of inverter (in AC amp rating will do) if the connection is done on a feeder.
I know there there is a 120% rule and a 100% rule if connections are doen on a buss bar or on the electrical panel butI am not sure how to determine is the connection is on the feeder.
I have attached an illustration to show my point.
Your help will be appreciated
For this particular setup, you also have the 120% rule that applies on the 1200A panelboard. You apply the 120% rule on all panels from the point of interconnection, all the way up to the service point, which is why you usually want to interconnect on the main panel instead of subpanels. In NEC2014 and later, this would limit you to 80% of 240A, i.e. 192A. That would imply up to a 250A OCPD for the interconnection tap circuit, but the total operating current of the source would be limited to 80% of 240A instead of 80% of 250A.
If upstream equipment doesn't have its own rules that otherwise limit you, the rule for interconnecting on feeders in general is as follows. There are 3 cases, depending on whether an OCPD is at both ends of the feeder to protect from potential overload, and depending on whether 240.4(B) was applied when sizing the feeder. Typical of all 3 cases, the short section of conductor between the feeder and the first PV-specific OCPD would be constructed to comply with one of the 240.21(B) tap rules. Maintain symmetry when parallel conductors are involved.
Case 1: There is an OCPD at both ends of the feeder, and feeder conductors meet or exceed the OCPD protecting them (i.e. it sizing is not taking credit for 240.4(B)):
You can connect anything you can fit behind an OCPD, up to the full rating of the OCPD feeder. So a 600A circuit, feeding a panel protected by a 600A main, could have anything up to a full 600A of interconnected PV, which is 480A total current.
Case #2: There is an OCPD at both ends, but the feeder conductors aren't quite equal to it. Whomever sized the feeder, took credit for 240.4(B). In your 600A example, you would see this if the 600A feeder were built with 2 sets of 300 kcmil Cu (which add up to 570A).
Either A: With a load calculation to justify, decrease the load-side OCPD of the feeder, so it is less than the ampacity of the feeder conductors.
Or B: Replace the conductors from the tap point to the load-side OCPD, so they meet or exceed the downstream OCPD.
Or C: Replace the feeder conductors entirely, so they meet or exceed both OCPD's, making your situation be the same as Case #1.
For solution A and B to Case #2, you could then connect up to 500A worth of PV OCPD (i.e. 400A operating current). I.e. the standard OCPD rating, that is doesn't exceed the feeder's ampacity.
Case #3: There is NO OCPD at the load side, the subpanel is main lug only (MLO):
The feeder from the tap point up until the panel, must have an ampacity equaling PV OCPD + main supply OCPD. Same with the busbar of the MLO subpanel. So in your 600A example, if you desired to connect 200A of PV OCPD, you'd need and 800A rating of the MLO subpanel and 800A worth of conductors (such as 2x 600 kcmil Cu). The reason why this is an issue, is that the main supply + PV source could load that panel up to 800A, in the unlikely event that the user turns on more loads than the load calculation considers. One way to work around this, is to put a set of 600A fuses or separate 600A breaker, just before the MLO subpanel, and set it up to be like Case #1 or Case #2 instead.