Solar panelboard

wwhitney said:
That is an interpretation of 705.12(B)(3), but vectorially is another interpretation. Just like the words of 705.12(B)(3) are silent on the question of "per pole" or not, they are silent on the question of arithmetically vs vectorially. In both cases, we can be guided by the physics in choosing the proper interpretation.


Cheers, Wayne
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I'm really out after this, but seeing as most AHJ inspectors only have a high school education in addition to whatever trade certifications they have I don't think you'd have much luck arguing a vector addition interpretation to 705.12(B)(3). I know that I would never try it.
 
jaggedben said:
Unfortunately no, because the particular rule under discussion references overcurrent devices. For all the other rules that go by 125% of inverter output current I agree with your approach.
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The rating in question of the overcurrent devices is a current rating, so I see no reason that such ratings shouldn't add like currents themselves add.

Cheers, Wayne
 
ggunn said:
I'm really out after this, but seeing as most AHJ inspectors only have a high school education in addition to whatever trade certifications they have I don't think you'd have much luck arguing a vector addition interpretation to 705.12(B)(3). I know that I would never try it.
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You certainly can do it via 2023 NEC 705.12(B)(6) "under engineering supervision" since you are an engineer. I would still argue that 705.12(B)(6) is not required, and that a non-engineer can do the same thing just using the wording of 705.12(B)(3) and the proper understanding of the word "addition."

An inspector needs to understand that a 125A panel containing (3) 70A double pole breakers for (3) 70A 208V loads is not overloaded, while the same panel and breaker arrangement used for (3) pairs of 70A 120V loads is overloaded. Once you have that understanding, the proper interpretation of 705.12(B)(3) follows.

Cheers, Wayne
 
wwhitney said:
The rating in question of the overcurrent devices is a current rating, so I see no reason that such ratings shouldn't add like currents themselves add.

Cheers, Wayne
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Case: Does a solitary single phase inverter connected to the A and B phases of a three phase panel contribute any current to the C phase busbar? I wouldn't think so, but show me how it would.
 
ggunn said:
Case: Does a solitary single phase inverter connected to the A and B phases of a three phase panel contribute any current to the C phase busbar? I wouldn't think so, but show me how it would.
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Obviously it would not, so I have no idea what you are driving at.

Cheers, Wayne
 
wwhitney said:
Obviously it would not, so I have no idea what you are driving at.

Cheers, Wayne
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Now add another inverter on the B and C phases. The current on A and C are the current from the individual inverters and the current on B is, as you say, added vectorially, but the rating of the CBs connected to B are the sum of the two for the purposes of 705.12(B)(3). Good luck convincing an inspector that 50A + 50A = 78A.
 
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ggunn said:
Now add another inverter on the B and C phases.
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Your currents are now in the ratio of (1 : sqrt(3) : 1) on legs (A, B, C), respectively. (*)

Cheers, Wayne

(*) This assumes both inverters are outputting at power factor 1, or at least at the same power factor. It is true that if one inverter is outputting at power factor 1, and the other at power factor 0.5 (power factor angle 60 degrees), the current on B could be twice the current on A or C. But I don't believe that's a plausible scenario.
 
wwhitney said:
As per the second paragraph of post #23, anyone inspecting 3 phase installations already needs to understand that sometimes 50A + 50A = 87A.

Cheers, Wayne
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And I do, of course, but the currents have nothing to do with the sum of all breakers rule; it's not a sum of all currents rule. This horse is officially a greasy spot on the ground and I apologize for my role in the overflogging. :D
 
This rule could use some reform. It seems like overkill to have phantom amps govern a significantly larger panelboard, whether due to rounding errors, and/or insignificant loads that rarely draw more than an individual ampere.

A recommendation I'd make, is to allow the 120% rule to work in reverse. A busbar be sized at 125% of the total inverter current (or vector sum as applicable), with a provision for the sum of auxiliary load breakers to up to 20% of the busbar rating, and loads arranged to subtract current. And with appropriate labelling that it's dedicated to the PV system with its restrictions for future use.
 
Carultch said:
This rule could use some reform. It seems like overkill to have phantom amps govern a significantly larger panelboard, whether due to rounding errors, and/or insignificant loads that rarely draw more than an individual ampere.

A recommendation I'd make, is to allow the 120% rule to work in reverse. A busbar be sized at 125% of the total inverter current (or vector sum as applicable), with a provision for the sum of auxiliary load breakers to up to 20% of the busbar rating, and loads arranged to subtract current. And with appropriate labelling that it's dedicated to the PV system with its restrictions for future use.
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That's nice for the PV combiner panel like the OP's but not for other uses of the rule. One thing we've sort of ignored so far is that the sum of all overcurrent devices also applies to loads, and there's no ironclad reason load currents should obey the same vector math as line-line inverter outputs.
 
jaggedben said:
One thing we've sort of ignored so far is that the sum of all overcurrent devices also applies to loads, and there's no ironclad reason load currents should obey the same vector math as line-line inverter outputs.
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Okay, I think you've got me there.

Take a 125A 3-phase panel, and put in it all double pole breakers in balanced sets of 3: one set of 10A and two sets of 30A, so 9 double pole breakers. If everything is power factor 1, load or source, this arrangement can't overload the bus without overloading a breaker, as 70 * sqrt(3) < 125A.

But suppose 1 10A breaker is a source and all the other breakers are loads of possibly varying power factor. Now we need to add the load breaker ratings arithmetically. So on two legs, we get 135A of supply and 130A of load. The bus can be overloaded without overloading any breaker.

Since the point of 705.12(B)(3) is to be agnostic as to whether breakers are loads or sources, vector addition of breaker ratings is not appropriate. Use 705.12(B)(6) to do that.

Cheers, Wayne
 
Im just a low level electrician (and I dont understand the 50 + 50 = 87:), but I read the sum of all breakers code rule as "..shall not exceed THE BUSBAR..". In my interpretation I'd agree with Ggunn- that "busbar" would be referring to an individual busbar, i.e. phase A busbar. So if I had a three phase panelboard with three 3 pole breakers each 60A. And I had a 2 pole 20A breaker for a DAS on phase A-B and a 1 pole 20A breaker for carport lighting on phase C then I would be okay with a 200A rated panelboard/busbar. The sum of the breakers that are connected to each phase busbar would be 200A at that point, instead of adding all the breakers together for all three busbars which would be 220A. Would I be wrong in my interpretation?
 
electro7 said:
Im just a low level electrician (and I dont understand the 50 + 50 = 87:), but I read the sum of all breakers code rule as "..shall not exceed THE BUSBAR..". In my interpretation I'd agree with Ggunn- that "busbar" would be referring to an individual busbar, i.e. phase A busbar. So if I had a three phase panelboard with three 3 pole breakers each 60A. And I had a 2 pole 20A breaker for a DAS on phase A-B and a 1 pole 20A breaker for carport lighting on phase C then I would be okay with a 200A rated panelboard/busbar. The sum of the breakers that are connected to each phase busbar would be 200A at that point, instead of adding all the breakers together for all three busbars which would be 220A. Would I be wrong in my interpretation?
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Nope, not wrong. We agree with you on this point.
 
electro7 said:
Im just a low level electrician (and I dont understand the 50 + 50 = 87:)
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Put a stake in the ground, and tie three ropes to the top of it. Pull on the ropes horizontally.

Ropes 1 and 2 are to be 60 degrees apart.
Rope 3 is to be opposite the two of them, and an equal angle of 150 degrees from each.

Pull on rope 1 with 50 pounds, pull on rope 2 with 50 pounds. How many pounds are needed to pull on rope 3, so that the stake isn't pulled over in any direction? Answer: 87 pounds.

Current adds up the same way, i.e. with vector math, when accounting for phase (direction in this analogy).

Here's the vector math of what's happening:
<50, 0> + <50*cos(60 deg), 50*sin(60 deg)> =

After adding up corresponding components:
<75, 25*sqrt(3)> = 25*<3, sqrt(3)>

Magnitude of <3, sqrt(3)> = sqrt(3^2 + sqrt(3)^2) = 2*sqrt(3)
Magnitude of 25*<3, sqrt(3)> = 50*sqrt(3)
 
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There is a good article on finding three phase line currents when using single phase inverters in a three phase panel in Solar Professional magazine. It is in issue 3.1 titled Three-Phase Line Currents. Still lots of good content in those old SolarPro mags. You can download them all for free at https://www.solarprofessional.com/
 
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