(2017) 705.12(B)(2)(2) Example Computations

[wwhitney]

Hello,

In trying to understand (2017) 705.12(B)(2)(2), taps to feeders with power output connections, I made up the example one-line diagram below. It has a feeder with 3 splices along its length, dividing the feeder into 7 segments. Assume each segment is between 10' and 25' long. I then tried to compute the minimum conductor ampacity allowed for each segment, using the tap rules where permissible. For two segments I wasn't sure which way to take the computation. Thoughts on those two or on any errors would be appreciated.

Thanks,
Wayne


Utility
|
Meter Main 150A OCPD
|
A
|
|--B-- 25A Breaker for Photovoltaic Power, output current is 18A
|
C
|
|--D-- MLO Load Panel with 225A Bus
|
E
|
|--F-- 60A Breaker for Energy Storage, output current is 42A
|
G
|
Load Panel with 60A Main Breaker


Minimum Ampacities for Feeder Segments:

A) 150
B) 50 (or 67.5?)
C) 172.5
D) 225
E) 202.5 (or 225?)
F) 60
G) 75

 

[pv_n00b]

Hello,

In trying to understand (2017) 705.12(B)(2)(2), taps to feeders with power output connections, I made up the example one-line diagram below. It has a feeder with 3 splices along its length, dividing the feeder into 7 segments. Assume each segment is between 10' and 25' long. I then tried to compute the minimum conductor ampacity allowed for each segment, using the tap rules where permissible. For two segments I wasn't sure which way to take the computation. Thoughts on those two or on any errors would be appreciated.

Thanks,
Wayne


Utility
|
Meter Main 150A OCPD
|
A
|
|--B-- 25A Breaker for Photovoltaic Power, output current is 18A
|
C
|
|--D-- MLO Load Panel with 225A Bus
|
E
|
|--F-- 60A Breaker for Energy Storage, output current is 42A
|
G
|
Load Panel with 60A Main Breaker


Minimum Ampacities for Feeder Segments:

A) 150
B) 50 (or 67.5?)
C) 172.5
D) 225
E) 202.5 (or 225?)
F) 60
G) 75

If I understand you correctly then A-C-E-G is one feeder conductor feeding the 60A load panel from the 150A main. Then B, D, and F are taps from that conductor that are between 10' and 25' long. Assuming that is correct then I'll break it down as follows:

A is exposed to the main or the PV plus the energy storage current depending on if there is export or not. (150 or 75)
C is exposed to the main and PV or the energy storage current depending on if there is export or not. (172.5 or 52.5)
E is exposed to the main and PV or the energy storage current depending on if there is export or not. (172.5 or 52.5)
G is exposed to the main, PV and energy storage current (225)
B, D, and F are exposed to the main, PV and energy storage current (225)

If the tap conductors are between 10' and 25' then they all fall under 240.21(B)(2) and D would not be allowed. The tap conductor has to terminate at a single OCPD.
B and F can be sized for 125A (1/3*150+75).

 

[jaggedben]

Good luck getting certainty on this. I attempted to get 705.12(B)(2)(2) 'fixed' in the 2017 code. To be blunt, I don't think the CMP understands how bad the language is, or had the attention span to think about it.

My opinion, for what it's worth:

A: 150
B: 75*
C: 172.5 (if the panel fed by 'D' remains MLO)
D: 235, to address pv_noobs objection about a tap terminating at an OCPD. At 235 it's definitely not a tap since that is the sum of all OCPDs that can feed it. Or put an MCB in the panel and match that rating, minimum 75*. I notice that the 225A panel is exactly rated to meet 705.12(D)(2)(1).
E: 75* This segement only sees a 60A load or a 60A backfeed, so I think it just needs to meet the tap minimum.
F: 75*
G: 75*

The asterisks are wherever you have a 'tap' conductor that is calculated in some way that's highly debatable. My proposed language would have (in my intent) called for taking all relevant inverter outputs*1.25, then adding that to the feeder OCPD, then using that number in 240.21(B) calcs. That was based on my understanding of what 705.12(B)(2)(2) was supposed to mean. That gets you a minimum of 75A for anything that's just a tap not subject to other rules. (((18+42)*1.25)+150). But this is just my opinion and interpretation. So that 75A could be totally different if the AHJ reads 705.12(B)(2)(2) differently. At worst, I think they could plausibly read it as being one third of 150A plus 125% of inverters, i.e. 50+75=125.

 

[wwhitney]

Well, the range of answers so far definitely shows the meaning of this section isn't clear.

Here are my thoughts on the segments that we disagree on:

B) This qualifies to be a feeder tap, so we have to determine the current rating to apply the 1/3 minimum to. 705.12(B)(2)(2) tells us to include 125% of the power source output current, but which power sources? One could argue for (1) none, as there are no loads at the terminal end of this segment (so 50A minimum tap ampacity); (b) just the ESS inverter output current of 42A, as the current from the PV inverter can't contribute additively in this segment to current from the 150A main OCPD (so 67.5A minimum tap ampacity); or (c) both power source output currents totalling 60A, because the language doesn't recognize the physics (so 75A minimum tap ampacity)

I lean towards (1) on the idea that 705.12(B)(2)(2) implicitly only applies to segments that meet the "that portion of the feeder" clause of 705.12(B)(2)(1)

D) This seems clear cut to me at 225A, because of the lack of a main breaker, 705.12(B)(2)(1)(a) is the only option. It instructs us to use 125% of the power source output currents, the values of the breakers on those sources are immaterial.

E) I rule out making this a tap, as I'd like to make F and G taps, and you can't "tap a tap". 705.12(B)(2)(1)(a) controls. Segment E is on the load side of the ESS power output and on the load side of the PV power output, so that would suggest including either the greater of 125% of the power output currents (giving 202.5A minimum ampacity) or their sum (225A minimum ampacity). Of course, physics-wise in normal operation it can't carry more than 60A continuous, so I think this case is just too complicated for the code language to handle reasonably.

F) I see this segment as not meeting the "that portion of the feeder" clause of 705.12(B)(2)(1) for either power source output current, so as a tap it could 50A. But it also needs to be at least 125% of the ESS power source output current, or 52.5A. I initially said 60A because of the breaker, but that doesn't matter, does it?

G) My intention is that this could be a tap based on a length of between 10' and 25'. So here 705.12(B)(2)(2) controls, and because this segment meets the "that portion of the feeder" clause of 705.12(B)(1)(a) for both power source output currents, the computation is (150 + 1.25*(42+18))/3 = 75A.

Cheers, Wayne

 

[jaggedben]

Is this related to an actual project, or just a brain teaser exercise? :lol:

...

B) This qualifies to be a feeder tap, so we have to determine the current rating to apply the 1/3 minimum to. 705.12(B)(2)(2) tells us to include 125% of the power source output current, but which power sources? One could argue for (1) none, as there are no loads at the terminal end of this segment (so 50A minimum tap ampacity); (b) just the ESS inverter output current of 42A, as the current from the PV inverter can't contribute additively in this segment to current from the 150A main OCPD (so 67.5A minimum tap ampacity); or (c) both power source output currents totalling 60A, because the language doesn't recognize the physics (so 75A minimum tap ampacity)

I lean towards (1) on the idea that 705.12(B)(2)(2) implicitly only applies to segments that meet the "that portion of the feeder" clause of 705.12(B)(2)(1)

Er, I disagree. You are conflating fault considerations with operating considerations. Ask yourself, what is the logic behind 1/10th, 1/3rd, etc in 240.21(B). Now, I'll be the first to admit I don't really know the answer to that. But presumably it has something to do with assuring that fault current will trip the feeder breaker, in this case the 150A upstream. If you have a fault anywhere on any of the segments you've lettered, all three sources could conceivably contribute to fault current. So it has nothing to do with which operating currents can add together towards loads in a non-fault scenario. To my mind the options are:
a) Add up the maximum possible current from all sources and then apply math per 240.21.
b) Declare that the fault current contributions of interactive inverters are insignificant to whether the breaker will trip, and use 150A.
c) Do some ridiculous amount of research and develop some complicated formula to establish a scientifically supported compromise in between (a) and (b), and submit that as a revision to the code.

To me, (a) seems like the approach that is currently practical without risking some unknown reduction in safety.

D) This seems clear cut to me at 225A, because of the lack of a main breaker, 705.12(B)(2)(1)(a) is the only option. It instructs us to use 125% of the power source output currents, the values of the breakers on those sources are immaterial.

My position is as before: I agree as long as you can convince the AHJ that it isn't a tap and that you are not violating the tap rule.

E) I rule out making this a tap, as I'd like to make F and G taps, and you can't "tap a tap". 705.12(B)(2)(1)(a) controls. Segment E is on the load side of the ESS power output and on the load side of the PV power output, so that would suggest including either the greater of 125% of the power output currents (giving 202.5A minimum ampacity) or their sum (225A minimum ampacity). Of course, physics-wise in normal operation it can't carry more than 60A continuous, so I think this case is just too complicated for the code language to handle reasonably.

Okay, this is a bit fraught. After reviewing 240.21(B), I agree that you can't tap a 25ft tap, but not that you can't tap a 10ft tap! But you said assume everything is 10-25ft. Okay, so let's look at 705.12(B)(2)(1). You comply with 705.12(B)(2)(1)(b) by way of your downstream 60A load breaker. So all you need to do to is make it protected by the feeder breaker, which is 150A. You don't need to do 705.12(B)(2)(1)(a) calc.

F) I see this segment as not meeting the "that portion of the feeder" clause of 705.12(B)(2)(1) for either power source output current, so as a tap it could 50A. But it also needs to be at least 125% of the ESS power source output current, or 52.5A. I initially said 60A because of the breaker, but that doesn't matter, does it?

See (B).

G) My intention is that this could be a tap based on a length of between 10' and 25'. So here 705.12(B)(2)(2) controls, and because this segment meets the "that portion of the feeder" clause of 705.12(B)(1)(a) for both power source output currents, the computation is (150 + 1.25*(42+18))/3 = 75A.

See (B). Your logic here actually applies to all segments, except those that are governed by some other rule requiring a higher ampacity.

 

[pv_n00b]

I think the OP needs to be more clear about the design. Based on the answers everyone is interpreting the OPs design differently. If it's a spaghetti of taps and feeders then the OP is on their own on that.

 

[jaggedben]

I think the OP needs to be more clear about the design. Based on the answers everyone is interpreting the OPs design differently. If the OP is saying there are taps on taps on feeders then they are on their own on that.

Er, I disagree. He gave us a clear diagram and said each segment is 10-25ft long. I think there's genuine questions about how to apply the rules.

 

[pv_n00b]

Er, I disagree. He gave us a clear diagram and said each segment is 10-25ft long. I think there's genuine questions about how to apply the rules.

Have fun then. I can see this being nothing but pages of arguing about what is what.

 

[jaggedben]

Oh, that I agree with. :lol:

 

[wwhitney]

Is this related to an actual project, or just a brain teaser exercise?

It's a thought exercise that occurred to me after reading another thread on this board and reading about the Tesla Powerwall.

You are conflating fault considerations with operating considerations.

Ah, thank you, that is very helpful. I agree now with your answers for (A)-(C), (F) and (G).

I agree as long as you can convince the AHJ that it isn't a tap and that you are not violating the tap rule.

I'm confused: whether (D) is a tap or not depends on deciding what minimum ampacity is required for the segment, ignoring the tap rules. Doesn't 705.12(B)(2)(1)(a) exactly tell us the answer to that? Is there any verbiage in the NEC that would direct us to add up the breakers instead?

On segment (E), it is on the load side of the point of interconnection of the ESS, and the panel fed by (D) does not have a main breaker, so I don't see how to qualify this segment other than via 705.12(B)(2)(1)(a).

Cheers, Wayne

 

[jaggedben]

I'm confused: whether (D) is a tap or not depends on deciding what minimum ampacity is required for the segment, ignoring the tap rules. Doesn't 705.12(B)(2)(1)(a) exactly tell us the answer to that? Is there any verbiage in the NEC that would direct us to add up the breakers instead?

The verbiage would be the definition of a tap. Mind you, I agree with you that 705.12(B)(2)(1)(a) should govern. I could just see it being a point of controversy.

On segment (E), it is on the load side of the point of interconnection of the ESS, and the panel fed by (D) does not have a main breaker, so I don't see how to qualify this segment other than via 705.12(B)(2)(1)(a).

How is (E) on the load side of the ESS? The ESS is at the opposite end of (E) as the other two sources. With respect to what 705.12(B)(2)(1)(a) is trying to do, there is no concern about additive current from the ESS. At most, 705.12(B)(2)(1)(a) requires it to be rated 172.5A using the same math as (C).

I do think (E) is where you've really stretched the ability of the code to apply sensibly. From an operating standpoint, it only needs to be rated 60A. From a fault current standpoint, it's whatever is needed to trip the breaker, I guess 75A in one agrees with with way we're applying 240.21(B). It is only because this segment violates the tap rules than you're forced to treat it as a main trunk of the 'feeder' and rate it ... something else. At that point, I just don't know what it needs to be. 705.12(B)(2)(1)(a) and 705.12(B)(2)(1)(b) could both apply. In my opinion there is just no rule in the code that was written with that segment in mind.

 

[wwhitney]

How is (E) on the load side of the ESS? The ESS is at the opposite end of (E) as the other two sources.

For (E), the panel (D) with loads is on one end, and the ESS power source is at the other end. So it is a "portion of the feeder on the load side of the power source output connection" of the ESS. Physics-wise the current from the ESS can't increase the current in E above 150A, but the language of 705.12(B)(2)(1) isn't comprehensive enough to reflect that. As you say it just doesn't contemplate feeder segments where there are loads attached at both ends and there are power sources attached at both ends. That was what I was trying to consider with segment (E) in my example.

Cheers, Wayne

 

[jaggedben]

For (E), the panel (D) with loads is on one end, and the ESS power source is at the other end. So it is a "portion of the feeder on the load side of the power source output connection" of the ESS. Physics-wise the current from the ESS can't increase the current in E above 150A, but the language of 705.12(B)(2)(1) isn't comprehensive enough to reflect that. As you say it just doesn't contemplate feeder segments where there are loads attached at both ends and there are power sources attached at both ends. That was what I was trying to consider with segment (E) in my example.

Cheers, Wayne

Right. Except that with the 60A breaker downstream of G, the ESS can't even increase the operating current in E beyond 60A.

 

[pv_n00b]

Er, I disagree. You are conflating fault considerations with operating considerations. Ask yourself, what is the logic behind 1/10th, 1/3rd, etc in 240.21(B). Now, I'll be the first to admit I don't really know the answer to that. But presumably it has something to do with assuring that fault current will trip the feeder breaker, in this case the 150A upstream.

The purpose of LV protection is two-fold, protection against overloading a conductor under normal operation and protection against fault current. Normally this is done with a single OCPD at the source end of the conductor. The tap rules allow these two protections to be pulled apart under some circumstances and with additional physical protection of the conductor. So the OCPD on the source end protects the conductor during a fault but not from overload, and can be omitted completely if a fault in the conductor is not a large risk such as short conductor lengths in an enclosure or underground outside. The OCPD at the load end has to protect the conductor from overload under normal operation so it will match the conductor rating.

The reason for the tap rules is so that a smaller conductor can be used after an OCPD that would not normally protect the conductor. If a fully rated conductor were used all the time then the tap rules would not be needed. So it's a bit of a cheat, but an authorized one.