Understanding how current flows

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Carultch

Senior Member
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Massachusetts
Therefore the service panel feeding this sub-panel is being backfed from the sub-panel and that sub-panel feeder breaker needs to also be on the opposite end of the bus as the breaker feeding it?

Indeed. Assume worst-case scenario that all local load might diminish to zero, and the interconnected source at full capacity propagates to the higher ranking panels. "Rank" is my term, for where a panel is located in the network of panelboards, with the main panel being the highest rank. Therefore, all panelboards upstream, all the way from the point of interconnection to the service point, need to comply with a busbar protection rule in 705.12. It isn't the breaker in the main panel that feeds the subpanel, that needs to be used in the 120% rule, but rather the factor that sizes the original interconnection breaker. I.e. 125% of total inverter current. This eliminates rounding errors from being a show-stopper, and allows subpanel interconnections to comply, when the rule propagates to the main panel.

This is why there is an advantage to connect on as high-ranking of a panelboard as practical. You are more likely to be able to comply with the 120% rule, on a higher ampacity panelboard, and you eliminate the need to analyze additional panels.

An example of where you might see this. Suppose there is a main house with a 200A main panelboard (200A MCB & 200A bus). There is also a guest house, with a 100A subpanel (100A bus MLO, fed from a 100A branch breaker on the main panel). It is of interest to build an array on the guest house, and avoid the need to build a new feeder back to the main house. Because we need to apply the 120% rules, we have to relocate the guest house branch breaker to the opposite end of the busbar, if it isn't there already. A guest house is intermittently occupied, so it is a very likely situation that all local load diminishes to zero, when the source is at full capacity.

Apply the 120% rule to the guest house. 100 + x = 120%*100. Solution = 20A.
Apply the 120% rule to the main house. 200 + x = 120%*200. Solution = 40A.

20A limits us to a 3.8kW system. Suppose we wanted to build a 7.6kW system on the guest house, which would be the limit of the 120% rule capacity of interconnecting to the main house. Our options are as follows:
A: Using a load calculation or 15-minute interval load history to justify it, reduce the guest house feeder's branch breaker to 80A.
B: Interconnect on a feeder tap onto the existing 100A guest house feeder. Implement a breaker or disconnect, between the tap point and the subpanel, so it is protected from exceeding 100A.
C: Swap the guest house subpanel, with a minimum 125A subpanel.

An obvious loophole to part A, is to take your data for 30 days when the guest house is vacant. It would seemingly meet the wording of the rules, but defeat the intent of the rule entirely, and not something I would advise. A load calculation would be much more practical for this example.
 

Ravenvalor

Senior Member
Wow, that reads almost like “who’s on first”...

I would set it up that way to make sure the bus never”sees” more amps than it’s rated for regardless of which source the energy comes from.

Because if the pv system is (for instance) net metering (selling power to the utility) and the rating of the pv system is much higher than the rating of any bus on the property in which it's current flows through then it could burn up the bus feeding power to the unlimited load of the grid?
 
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