120% rule clarification

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I'm having some disagreement with some co-workers about the interpretation of 705.12 (D)(2). We had an installer connect a PV system to a 100 amp meter/main, which is protected by a 100 amp main breaker. Using the 120% rule, my call was that the max. size breaker he could use for the PV would be a 20 amp breaker. He has installed a 60 amp breaker. The code states "The sum of the ampere ratings of overcurrent devices in circuits supplying power to a busbar or conductor shall not exceed 120 percent of the rating of the busbar or conductor". My call is that the panel, which is rated at 100 amps, is now being supplied through a 100 amp ocd (from the utility) and a 60 amp ocd (PV), exceeding the 120% rule. I'm being told that I am misinterpreting the requirement.
 

GoldDigger

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You are correct.
But if you are lucky enough to have a 100 amp panel that uses 125 amp busbars (and you can prove that to the AHJ) then you could go as high as a 50 amp PV breaker. The 120% is applied to the busbar rating, which is not necessarily the same as the panel nameplate rating.
 

jaggedben

Senior Member
Location
Northern California
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Solar and Energy Storage Installer
The only way you could be misinterpreting the requirement is if the 60A breaker is installed as a supply side connection (nothing between it and the utility). The way you describe it sounds more like a load side connection (busbar protected by the 100A breaker, meaning the 100A breaker is between the utility and the PV breaker), in which case you are correct.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Can we set aside the NEC just for a moment and play "What's wrong with this picture"?

Let's say I have to design a system which combines the output of a couple of SatCon 100kW inverters at 480VAC whose maximum output current is 121A, and the interconnect is a tap between the utility meter and the MDP for the building.

(121A)(1.25) = 151.25A
(2)(151.25A) = 302.5A

Let's say I combine their output through 175A breakers in a 400A panel with no main breaker, and I run conductors rated at 400A from the panel to a 400A rated fused disconnect with 350A fuses next to the tap box.

Code compliance notwithstanding, is there a situation which could arise whereby that system could cause a fire or otherwise become unsafe?
 

Smart $

Esteemed Member
Location
Ohio
Can we set aside the NEC just for a moment and play "What's wrong with this picture"?

Let's say I have to design a system which combines the output of a couple of SatCon 100kW inverters at 480VAC whose maximum output current is 121A, and the interconnect is a tap between the utility meter and the MDP for the building.

(121A)(1.25) = 151.25A
(2)(151.25A) = 302.5A

Let's say I combine their output through 175A breakers in a 400A panel with no main breaker, and I run conductors rated at 400A from the panel to a 400A rated fused disconnect with 350A fuses next to the tap box.

Code compliance notwithstanding, is there a situation which could arise whereby that system could cause a fire or otherwise become unsafe?
Let me answer with what I hope to be a clarifying question. If a fault occurred between the 350A fuses of the disconnect and the 175A breakers in the aggregation panel, how much current could be fed to the fault without blowing the fuses and tripping the breakers?
 

SolarPro

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Location
Austin, TX
Fault currents from the utility grid will cause the 350 A fuse to open. In a bolted fault condition, this would be high current/short duration event.

The interactive inverters in the interconnected electric power system will not operate in the absence of the utility.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Let me answer with what I hope to be a clarifying question. If a fault occurred between the 350A fuses of the disconnect and the 175A breakers in the aggregation panel, how much current could be fed to the fault without blowing the fuses and tripping the breakers?
350A from the service. In the event of a fault in that location, the inverters would shut down.
 

Smart $

Esteemed Member
Location
Ohio
350A from the service. In the event of a fault in that location, the inverters would shut down.
I'm talking a resistive fault, not an open or bolted fault. Inverters do not sense faults. They sense a loss of voltage, which can be a result of an open or bolted fault. A resistive fault may still supply enough voltage to the inverter that it does not sense a fault condition. For the question at hand, a high enough resistive fault that the inverters do not sense a fault and no OCPD's trip.
 

jaggedben

Senior Member
Location
Northern California
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Solar and Energy Storage Installer
ggunn, I gather you intend your question to preclude any answers that begin with 'What if at some point another electrician comes along and makes changes such as...'. In that case, I believe the answer is no. There is no way that the system can be unsafe or cause a fire simply because the panelboard is rated at 400A instead of 800A.

Smart$, resistive faults might be an interesting problem, but the rating of the panelboard busbar in this example will have no effect on whether a resistive fault would start a fire.
 

Smart $

Esteemed Member
Location
Ohio
...

Smart$, resistive faults might be an interesting problem, but the rating of the panelboard busbar in this example will have no effect on whether a resistive fault would start a fire.
No? I'd say it depends on where the resistive fault is located. Aren't busbars typically isolated from the frame nowadays with thermoplastic barriers. Enough heat from a sustained resistive fault could melt or otherwise degrade the insulator, resulting in a full-on bolted fault (i.e. thinking arc flash here).

Also, the 120% rule includes conductors. The proposed scenario included conductors...

ggunn said:
Let's say I combine their output through 175A breakers in a 400A panel with no main breaker, and I run conductors rated at 400A from the panel to a 400A rated fused disconnect with 350A fuses next to the tap box.
 

SolarPro

Senior Member
Location
Austin, TX
For the question at hand, a high enough resistive fault that the inverters do not sense a fault and no OCPD's trip.

So a fault that's not a fault? Like a high resistance connection in the circuit? Or impending insulation breakdown?

These could lead to a fault. A short-circuit or ground-fault will open the 350 A fuse, according to the time-current curves. An open-circuit...well that's self-explanatory.

Do think that an inverter should sense a fault on the ac side of the system? A fuse doesn't sense a fault per se. It responds to current vs. time. It doesn't care why, where or how the trip event occurs. Similarly, an inverter needs stable voltage and frequency within set limits. It doesn't care how, where or why things go out of range. If they do, it shuts down. Response times vary by inverter capacity and by how far different parameters are outside of normal (magnitude vs time).

Interactive inverters detect ground-faults on the dc side of the system. Inverters with series arc-fault detection are being rolled out for PV systems on buildings. Parallel arc-fault detection may be added in NEC 2014. But I've not heard anyone suggest that protection on the ac side needs improvement?that ac arc-fault detection is required, for example.
 

Smart $

Esteemed Member
Location
Ohio
So a fault that's not a fault? Like a high resistance connection in the circuit? Or impending insulation breakdown?
A resistive fault that mimicks a heavy load between the OCPD's. In your language, it doesn't matter why, where or how. ;) As you establish in the rest of your post, the utility-side OCPD responds to current vs. time, but for sake of discussion let's say at or slightly below their rating of 350A. Because the conductors are rated 400A, there will not be enough voltage drop (or change in frequency) for the inverters to shut down. The inverters continue to operate and their current adds to the fault from the opposite direction. The current from either side is not enough to exceed the ampacity of the conductors rating, but it does exceed the rating at the location of the fault (let's say the resistive fault is drawing 650A from all sources). The insulation will continue to degrade, increasing the current magnitude until one or more OCPD feeding the fault opens, or the inverters shut down. The sum of OCPD's feeding this fault is 700A.

These could lead to a fault. A short-circuit or ground-fault will open the 350 A fuse, according to the time-current curves. An open-circuit...well that's self-explanatory.

Do think that an inverter should sense a fault on the ac side of the system? A fuse doesn't sense a fault per se. It responds to current vs. time. It doesn't care why, where or how the trip event occurs. Similarly, an inverter needs stable voltage and frequency within set limits. It doesn't care how, where or why things go out of range. If they do, it shuts down. Response times vary by inverter capacity and by how far different parameters are outside of normal (magnitude vs time).

Interactive inverters detect ground-faults on the dc side of the system. Inverters with series arc-fault detection are being rolled out for PV systems on buildings. Parallel arc-fault detection may be added in NEC 2014. But I've not heard anyone suggest that protection on the ac side needs improvement—that ac arc-fault detection is required, for example.
Now compare the proposed scenario to one where the only difference is the fused disconnect is instead a backfed 350A breaker in a 400A MCB service panel. Would there be a vioation of the 120% rule on either side of the breaker or both? Keep in mind, everything is the same on the PV-side of this backfed breaker.
 

SolarPro

Senior Member
Location
Austin, TX
let's say the resistive fault is drawing 650A from all sources

Sure, and for the sake of argument this happens during the day when the PV system can contribute to the fault current. Now the sun starts to set, meaning that the inverter output throttles back. The fault current will increasingly come from the utility until the 350 A OCPD trips. (It shouldn't mater whether the OCPD is a fuse or a CB, as far as I can tell.)

The proposed changes to Section 705.12(D)(2) in the NEC 2014 draft are revealing:

http://www.nfpa.org/Assets/files/AboutTheCodes/70/70-A2013-ROPDraft.pdf

While some alternatives to the basic 120% rule are proposed, the oversized busbar and feeder requirements remain very much in place.

Supersize it, baby. :thumbsup:
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
the rating of the panelboard busbar in this example will have no effect on whether a resistive fault would start a fire.
No? I'd say it depends on where the resistive fault is located.

I will admit I'm no expert on fault conditions, and in fact the notion of a resistive fault is new to me (although self explanatory, I think). But it seems to me, intuitively, that the chances are infinitesimal that the rating of the panelboard or conductor would make any practical difference in this example.

It seems to me that you are positing the following:

-That a resistive fault would occur that would just happen to conduct between 400A and 800A. (Is this even remotely likely by itself?)
-That such a large resistive fault would not cause a fire in and of itself (rather than because it overloads conductors or panelboards).

It seems to me that whatever material was conducting the resistive fault would be almost certain to ignite something in the vicinity well before the panelboard or conductors were overloaded. I mean, you're positing more than 192kW of pure heat passing through some unengineered path, right? Seems to me that's going to set something on fire no matter what (and no matter the location). If anything, having higher rated conductors and busbars might even make the situation worse by prolonging the amount of time that the resistive fault could 'operate' on flammable materials that might not even be electrical equipment.

This is all intuitive mind you, lacking any documentation concerning the likelihood and behavior of a resistive fault in an installation of this size and voltage.
 

Smart $

Esteemed Member
Location
Ohio
Sure, and for the sake of argument this happens during the day when the PV system can contribute to the fault current. Now the sun starts to set, meaning that the inverter output throttles back. The fault current will increasingly come from the utility until the 350 A OCPD trips. (It shouldn't mater whether the OCPD is a fuse or a CB, as far as I can tell.)...
That's a plausible scenario. However, that's only if during the day the fault didn't cause a fire that burnt the structure and electrical equipment to a cinder. :slaphead:
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
In this resistive fault model, would it make much difference what the current rating of the conductors and busbars are to the likelihood of a fire?
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
That's a plausible scenario. However, that's only if during the day the fault didn't cause a fire that burnt the structure and electrical equipment to a cinder. :slaphead:

Which it probably would do anyway, not because the panel or conductor rating was too small.

Also...

the 120% rule includes conductors. The proposed scenario included conductors...

In the proposed scenario the conductors could not physically be overloaded in any case. The only possible point of overloading is on the busbars, opposite the point where the utility connects, if not covered up by the solar breakers.
 
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ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Which it probably would do anyway, not because the panel or conductor rating was too small.

Also...



In the proposed scenario the conductors could not physically be overloaded in any case. The only possible point of overloading is on the busbars, opposite the point where the utility connects, if not covered up by the solar breakers.
I think he was suggesting that the resistive fault could be in the conductors, but still, (as you say) although the fault itself could be fed by both sources, the conductors on either side of it couldn't be. Hence my question of what difference it would make what the rating of the conductors would be. A resistive fault capable of dissipating 750A is going to generate a lot of heat no matter what the conductors feeding it are rated. My opinion is that it would very quickly develop into a dead short and shut everything down.
 
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