Load-side interconnection where there are no spaces in panel

[msheets]

We are installing a ~100kW PV system on the roof of a building. For interconnection we had planned to tie the inverter outputs to a new panel on the roof and bring the feeders down into the building at the 800A service panel. The only available breaker space will not accommodate the 110A breaker that we would need, so our backup plan is to mount a fused disconnect next to the service panel and feed it off the panel's load-side lugs. Is there something in the code that would restrict this plan? It seems to me that if we were to connect to the supply side, there would be no issue, but I can't seem to find anything for or against connecting to the load side lugs of the service panel.

Thanks in advance.

 

[GoldDigger]

If the wire size you use will not be protected by the main breaker you will have to decide which of the several tap rule scenarios you will take advantage of.
I assume that the load side lugs you mention are deliberately designed for feed through use?
If so you should be OK.

 

[msheets]

If the wire size you use will not be protected by the main breaker you will have to decide which of the several tap rule scenarios you will take advantage of.
I assume that the load side lugs you mention are deliberately designed for feed through use?
If so you should be OK.

Yes, the lugs are designed for feed through use.

I'm not sure which tap rule scenario to use. It looks like the busway tap fits best, but it may not quite be what I'm looking for.

The bus and main breaker are rated at 800A, and the conductors between the panel and the disconnect are #2 CU. The conductors are only about 6' or 7' in length. Does that count as a reduction in ampacity size of busway (368.17(B)), which would mean I need a OCPD between the bus and the conductors?

Unless I'm missing something, it seems like it would be easier to move the connection point to the supply side of the main breaker and call it a second service.

 

[GoldDigger]

You could use the 10 foot tap rule to good advantage, since you have only a short distance to go. See whether your wire sizes and ampacity work out.

You cannot apply the busway rules when what you are actually using is wires in a raceway.

 

[Smart $]

Yes, the lugs are designed for feed through use.

I'm not sure which tap rule scenario to use. It looks like the busway tap fits best, but it may not quite be what I'm looking for.

The bus and main breaker are rated at 800A, and the conductors between the panel and the disconnect are #2 CU. The conductors are only about 6' or 7' in length. Does that count as a reduction in ampacity size of busway (368.17(B)), which would mean I need a OCPD between the bus and the conductors?

Unless I'm missing something, it seems like it would be easier to move the connection point to the supply side of the main breaker and call it a second service.

To use as a tap, 240.4(E)(5) sends you to 368.17(C)... and Exception No.1 thereto sends you back to 240.21. So I believe as long as you follow the tap rules of 240.21(B) you are good to go.

 

[Zee]

We are installing a ~100kW PV system............110A breaker that we would need.

Three points briefly:
A 100 kw system would need about a 400 Amp breaker or fuses......why do you say 110 Amp?

Are the lugs located opposite feed? (at the opposite end of the panel busbar from the main breaker)

Finally, lucky you have lugs.

 

[ggunn]

Three points briefly:
A 100 kw system would need about a 400 Amp breaker or fuses......why do you say 110 Amp?

Are the lugs located opposite feed? (at the opposite end of the panel busbar from the main breaker)

Finally, lucky you have lugs.

If it's 100kW at 480VAC three phase, it looks to me like it could just barely fit into a 150A OCPD. (100,000)(1.25)/(480)(sqrt3) = 150.35A. About 73kW is the most that would fit into a 110A OCPD at 480V. Point, though.

I don't know how he would reconcile with the opposite end of the busbar requirement; my interpretation of the OP is that he was wanting to lug onto the output of the main breaker.

If it were I (and it has been), I would try to lug onto the supply side of the MDP main breaker/disco.

 

[GoldDigger]

I don't know how he would reconcile with the opposite end of the busbar requirement; my interpretation of the OP is that he was wanting to lug onto the output of the main breaker.

If you do not connect to the opposite end of the bus bar you would be subject to the 100% rule instead of the 120% rule. But you could still make the connection with that limitation as long as the rating of the main is low enough compared to the bus rating.

 

[ggunn]

If you do not connect to the opposite end of the bus bar you would be subject to the 100% rule instead of the 120% rule. But you could still make the connection with that limitation as long as the rating of the main is low enough compared to the bus rating.

That assumes facts not in evidence, your honor.

 

[jaggedben]

If you do not connect to the opposite end of the bus bar you would be subject to the 100% rule instead of the 120% rule. But you could still make the connection with that limitation as long as the rating of the main is low enough compared to the bus rating.

He did say that the main and bus were both rated 800A. In that case he can only connect to the lugs if they are on the opposite end.

If it's 100kW at 480VAC three phase, it looks to me like it could just barely fit into a 150A OCPD. (100,000)(1.25)/(480)(sqrt3) = 150.35A. About 73kW is the most that would fit into a 110A OCPD at 480V. Point, though.

It could easily be 100 kW DC and 73 kW AC. Anyway the fuse rating can be up to 160A if the lugs are on the opposite end, or up to 800A for a supply side connection, which appear to be the only two options. So as long as the output is 480V is does not appear that the OP has made a fatal error in this regard.

 

[ggunn]

He did say that the main and bus were both rated 800A. In that case he can only connect to the lugs if they are on the opposite end.



It could easily be 100 kW DC and 73 kW AC. Anyway the fuse rating can be up to 160A if the lugs are on the opposite end, or up to 800A for a supply side connection, which appear to be the only two options. So as long as the output is 480V is does not appear that the OP has made a fatal error in this regard.

Agreed on all points. I have run into this exact problem, and a supply side tap on the input busbars of the MDP main breaker was the only viable solution. That, or add another section to the MDP, but that would have been prohibitively expensive.

 

[msheets]

Sorry for the prolonged silence.

Yes, I listed the DC rating, not the AC (which is 69.6kW and 480V), and the lugs are on the opposite side of the bus bars from the main breaker.

So, if I am understanding everyone correctly, I should be okay with landing the #2 AWG CU feeders on the feed-through lugs, as long as they are no more than 10' long, with 110A fuses in the disconnect.

Thanks.

 

[Smart $]

Sorry for the prolonged silence.

Yes, I listed the DC rating, not the AC (which is 69.6kW and 480V), and the lugs are on the opposite side of the bus bars from the main breaker.

So, if I am understanding everyone correctly, I should be okay with landing the #2 AWG CU feeders on the feed-through lugs, as long as they are no more than 10' long, with 110A fuses in the disconnect.

Thanks.

Your AHJ may and likely will not agree... IMO. There is nothing in 690 or 705 of the 2011 which says you can use tap rules between two sources of power. You will likely be restricted to 800A conductors minimum between feed-through lugs and disconnect.

 

[jaggedben]

Your AHJ may and likely will not agree... IMO. There is nothing in 690 or 705 of the 2011 which says you can use tap rules between two sources of power. You will likely be restricted to 800A conductors minimum between feed-through lugs and disconnect.

Which is really unnecessary in my opinion. So... what's your reading of 705.12(D)(2)(2) in the 2014 code? I just read it and can't figure it out. I would like to think that it was meant to head-off your interpretation above. But I'm not sure what it does.

Taps. In systems where inverter output connections are made at feeders, any taps shall be sized based on the sum of 125 percent of the inverter(s) output circuit current and the rating of the overcurrent device protecting the feeder conductors as calculated in 240.21(B).

 

[Smart $]

Which is really unnecessary in my opinion.

I agree... to a degree. We are looking at conductors between two sources of power. Prior to 2014 edition, the Code makes essentially no concessions for these conductors to be sized only at one or the other's rating.

So... what's your reading of 705.12(D)(2)(2) in the 2014 code? I just read it and can't figure it out. I would like to think that it was meant to head-off your interpretation above. But I'm not sure what it does.

I'm not sure what to make of it either. Perhaps it means the larger of the feeder ocpd rating or 125% of inverter(s) output rating is to be considered as the supply ocpd rating, while the smaller of the two is to be considered the tap ocpd. More often than not, I'd say the feeder ocpd rating would be the larger of the two... but we do have to consider the possibility 125% of the inverter(s) output is the larger.

 

[GoldDigger]

... More often than not, I'd say the feeder ocpd rating would be the larger of the two... but we do have to consider the possibility 125% of the inverter(s) output is the larger.

Since the inverter output is a continuous current (although not a load), the NEC requires, I believe, that the breaker be sized at least to 125% of that rated output. Without even considering the 80% breaker issue, there is no way that the nominal breaker size could be less than 125% of the rated inverter output.

 

[ggunn]

Your AHJ may and likely will not agree... IMO. There is nothing in 690 or 705 of the 2011 which says you can use tap rules between two sources of power. You will likely be restricted to 800A conductors minimum between feed-through lugs and disconnect.

Why (logically) would the conductors have to be sized larger than if they were connected to a supply side tap?

 

[jaggedben]

... Perhaps it means the larger of the feeder ocpd rating or 125% of inverter(s) output rating is to be considered as the supply ocpd rating, while the smaller of the two is to be considered the tap ocpd. More often than not, I'd say the feeder ocpd rating would be the larger of the two... but we do have to consider the possibility 125% of the inverter(s) output is the larger.

First of all, the feeder OCPD shouldn't be smaller than the 125% of the output of the inverter, because then you have the possibility of tripping the feeder OCPD at full inverter output and no load. In my opinion the feeder is part of the inverter output circuit and is therefore required to be sized equal or greater than the tap.

As for what they meant, I think...or rather I should say hope ... that they meant this:

Taps. In systems where inverter output connections are made at feeders, any taps shall be sized according to 240.21(B) using an assumed rating for the overcurrent device protecting the feeder. The assumed rating shall be the sum of 125 percent of the inverter(s) output circuit current and the rating of the actual overcurrent device protecting the feeder.

...

Why (logically) would the conductors have to be sized larger than if they were connected to a supply side tap?

I know that you know better than to ask for a logical explanation of why the code might require something.

 

[Smart $]

Since the inverter output is a continuous current (although not a load), the NEC requires, I believe, that the breaker be sized at least to 125% of that rated output. Without even considering the 80% breaker issue, there is no way that the nominal breaker size could be less than 125% of the rated inverter output.

Just to clarify, feeder ocpd is at POCO end of feeder... typically.

And I realize the inverter output ocpd(s) must be at least 125% of inverter(s) output rating... but 705.12(D)(2)(2) does not reference the ocpd rating. I take that as meaning to substitute 125% of inverter(s) output rating for the tap ocpd (or feeder ocpd, if larger than POCO end ocpd) in any calculation for 240.21(B).

 

[Smart $]

First of all, the feeder OCPD shouldn't be smaller than the 125% of the output of the inverter, because then you have the possibility of tripping the feeder OCPD at full inverter output and no load. In my opinion the feeder is part of the inverter output circuit and is therefore required to be sized equal or greater than the tap.

Correct, utility-end ocpd shouldn't be equal or smaller than inverter(s) ocpd... but you have to consider some creative engineering may be attempted somewhere.

Also, according to the tap definition, the tap conductor has to be small enough to be considered unprotected by the feeder ocpd... so 'equal' and 'larger' is not in the scope of a tap rule.