Understanding 705.12

[kaveenkw123]

Hello All,
I have been trying to understand the reasoning for 705.12 interconnection rules but having a had time grasping the idea and explaining to clients why they need to upgrade their switchgear.

I get that its for protection of the bus bar but considering a scenario where the service disconnect is 600A and the busbar is also 600A along with a 200A PV System;

If the load draw is more than 600A; wouldnt the service disconnect trip saving the busbar and feeders regardless of the PV system?
I would really appreciate if someone can direct me towards any resources for understanding this or can dumb this down for me.

Thankyou

 

[don_resqcapt19]

Hello All,
I have been trying to understand the reasoning for 705.12 interconnection rules but having a had time grasping the idea and explaining to clients why they need to upgrade their switchgear.

I get that its for protection of the bus bar but considering a scenario where the service disconnect is 600A and the busbar is also 600A along with a 200A PV System;

If the load draw is more than 600A; wouldnt the service disconnect trip saving the busbar and feeders regardless of the PV system?
I would really appreciate if someone can direct me towards any resources for understanding this or can dumb this down for me.

Thankyou

The 600 amp main could supply 600 amps and the 200 amp solar could supply 200 amps, so with an 800 amp load, there would be no reason for either breaker to trip.

However, I don't see where you could have more than 600 amps of current at any point on the busbar.

 

[jaggedben]

These rules have always been very conservative. Essentially someone would have to deliberately schedule increased load when the sun was shining in order to have any chance of overloading a busbar. So that is effectively what the code makers have decided to protect against.

 

[wwhitney]

Say you have a 600A busbar (600A is arbitrary, we could just call it I_busbar). With two only connections to it, one is a source, one is a load, and you just need to limit each connection (or rather just one in this case) to 600A with a 600A OCPD.

With three connections, it still suffices to limit each connection with a 600A OCPD. If only one connection is a source, its OCPD will limit the inflow to the bus to 600A. Whereas if only one connection is a load, its OCPD will limit the outflow to 600A. Either way, there is at most 600A flowing through the bus.

With four connections, the story changes. You could have (2) 600A sources and (2) 600A loads. In which case, if you arrange them on the bus in the order source-source-load-load, you could actually have 1200A flowing through the bus between the adjacent source and load. That particular problem is solved by arranging the 4 connections as source - load - load - source. Now no section of the busbar will carry more than 600A.

However, suppose that each connection is carrying a full 600A. With 4 such connections to the busbar, you get 4 times the I_busbar² * R heating from the connection resistance that you get from a single connection. Whereas in the two or three connection case, or the one source and arbitrary number of loads case, the worst case heating is 2 times the worst case heating from a single connection. If the product was designed for only one source connection, then the extra connection heating could raise the temperature in the enclosure, affecting OCPD performance.

That brings us to the 120% rule, where the two sources are required to be at the opposite ends of the busbar, and the sum of the sources are limited to 120% of the busbar rating. Now the worst case is one source is 100% of the busbar rating, and the other source is 20% of the busbar rating. The worst case heating is (100%)² + (20%)² = 104% of the single source worst case heating. So the writers of 705.12's 120% rule decided that an arbitrary product can be pushed to a 4% heating overload without ill consequence, and allowed that for all products.

Cheers, Wayne

 

[wwhitney]

Now the worst case is one source is 100% of the busbar rating, and the other source is 20% of the busbar rating. The worst case heating is (100%)² + (20%)² = 104% of the single source worst case heating.

Math-wise it's not hard to prove that 100% + 20% is the worst case for the sum of the squares. For example, if you compare 90% + 30%, then the sum of the squares is 90%, rather than 104%.

So maybe I should submit a PI that changes the 120% to a 104% sum of squares rule: "the sum of the square of 125 percent of the
power source(s) output circuit current and the square of the rating of the overcurrent device protecting the busbar shall not exceed
120 104 percent of the square of the ampacity of the busbar."

Think that would go over well with the CMP? : - )

[Would presumably actually be helpful for real-world installations. If you have a 200A main breaker and 200A bus, then the maximum allowable power source output current when downsizing the main breaker to 175A would go from 52A to 84A. And if downsizing the main breaker to 150A, it would go from 72A to 111A.]

Cheers, Wayne

 

[ggunn]

Hello All,
I have been trying to understand the reasoning for 705.12 interconnection rules but having a had time grasping the idea and explaining to clients why they need to upgrade their switchgear.

I get that its for protection of the bus bar but considering a scenario where the service disconnect is 600A and the busbar is also 600A along with a 200A PV System;

If the load draw is more than 600A; wouldnt the service disconnect trip saving the busbar and feeders regardless of the PV system?
I would really appreciate if someone can direct me towards any resources for understanding this or can dumb this down for me.

If the solar feed is on the load side of the main breaker in the panel, then the panel would have to draw 600A plus the output of the PV in order to trip the main.

 

[don_resqcapt19]

....

That brings us to the 120% rule, where the two sources are required to be at the opposite ends of the busbar, and the sum of the sources are limited to 120% of the busbar rating. Now the worst case is one source is 100% of the busbar rating, and the other source is 20% of the busbar rating. The worst case heating is (100%)² + (20%)² = 104% of the single source worst case heating. So the writers of 705.12's 120% rule decided that an arbitrary product can be pushed to a 4% heating overload without ill consequence, and allowed that for all products.

Cheers, Wayne

But unless I have a single 120% between the two sources at the opposite ends of the bus, there would be no point where the bus is carrying more than 100%.

 

[ggunn]

But unless I have a single 120% between the two sources at the opposite ends of the bus, there would be no point where the bus is carrying more than 100%.

We get that and we have discussed it ad nauseum, but the code is the code.

 

[wwhitney]

But unless I have a single 120% between the two sources at the opposite ends of the bus, there would be no point where the bus is carrying more than 100%.

That is correct, and the post you quoted said that, two paragraphs up from your quote.

But as I pointed out, avoiding an overload of the bus is not the only physics behind the 120% rule. If it were, it could be a 200% rule and there would be much rejoicing.

The busbar itself is not the only source of I² * R heating, there are also the connections to the bus. Counting those up in the worst case and comparing to the worst case single source case (100% source at one end, 100% load at the other end) is a necessary analysis.

Cheers, Wayne

 

[jaggedben]

...

Think that would go over well with the CMP? : - )

Nope. But please try anyway? Haha.

[Would presumably actually be helpful for real-world installations. If you have a 200A main breaker and 200A bus, then the maximum allowable power source output current when downsizing the main breaker to 175A would go from 52A to 84A. And if downsizing the main breaker to 150A, it would go from 72A to 111A.]

This is most interesting when you add ESS.
For example with a 100/125 panel, we could have 60A of inverter output which would mean for example a total of say 25A of solar and 32A of ESS would be fine on a 100A service without PCS.

With a 200/225 'solar ready' panel, we could have 82 instead of 56.

 

[wwhitney]

For example with a 100/125 panel, we could have 60A of inverter output
. . .
With a 200/225 'solar ready' panel, we could have 82 instead of 56.

I'm getting 63A for the 100A MB / 125A bus case, and 90A for the 200A MB / 225A bus case.

Did you omit the 104% factor? That would give you heating equivalence to the 100% rule (but with breakers at opposite ends), not heating equivalence to the 120% rule.

Cheers, Wayne

 

[jaggedben]

Yeah I used 100%. Lol
See there already is a reason for the CMP to reject. If I can't understand it...

 

[wwhitney]

But please try anyway?

OK, I wrote up an argument here: https://forums.mikeholt.com/threads/2029-future-pi-705-12-b-2.2579081/ So if I don't remember to submit the PI 2.5 years from you, you're welcome to do so, and amend it as you see fit.

Yeah I used 100%.

A conservative mistake, heating wise, that still is more flexible than the linear 120% rule we have, when the main breaker is not full-sized.

Cheers, Wayne

 

[Carultch]

Hello All,
I have been trying to understand the reasoning for 705.12 interconnection rules but having a had time grasping the idea and explaining to clients why they need to upgrade their switchgear.

I get that its for protection of the bus bar but considering a scenario where the service disconnect is 600A and the busbar is also 600A along with a 200A PV System;

If the load draw is more than 600A; wouldnt the service disconnect trip saving the busbar and feeders regardless of the PV system?
I would really appreciate if someone can direct me towards any resources for understanding this or can dumb this down for me.

Thankyou

The most commonly used rule for interconnecting, is the famous 120% rule. It allows taking partial credit for Kirchhoff's current law, feeding from opposite sides. Given a busbar with a matching main, you have 20% headroom available for interconnecting, after accounting for the standard 125% continuous load factor. The idea is that busbar current can be subtractive, rather than additive, when properly placed. The 120% limit is an industry compromise, that has to do with mitigating the risk of mutual heating of the breakers. For a 600A panel with a matching main, this means you can have up to 96A of total inverter current, since 125% of 96A = 120A, and (600A + 120A) is 120% of the 600A bus.

Panelboards are routinely filled with a lot more branch breakers than their full rating. The underlying idea is that not all of them draw full current at once. Engineers have nuanced ways of adding up the loads for circuit & eqpt sizing, and the main breaker is expected to be the fail-safe in the unlikely event that somehow the loads do exceed what the engineers anticipated. By introducing a source on the busbar, you create a blindspot that the main breaker will not detect. Given a 200A PV system at full power (160A continuous) and a 600A busbar and matching main breaker, this means if you draw 700A among the loads, it will be business as usual for the main breaker, and the main breaker will not trip. Putting both these breakers on the same side of the bus, means you'll have the full 700A on the 600A bus.

Putting them on opposite ends, mitigates this problem, but the industry compromise of the 120% rule still limits you to 96A of inverter current in this example. This would be a 120A breaker if it existed, but as of NEC2014, they reworded it so that rounding to the standard size of 125A isn't a show stopper.

 

[kaveenkw123]

The most commonly used rule for interconnecting, is the famous 120% rule. It allows taking partial credit for Kirchhoff's current law, feeding from opposite sides. Given a busbar with a matching main, you have 20% headroom available for interconnecting, after accounting for the standard 125% continuous load factor. The idea is that busbar current can be subtractive, rather than additive, when properly placed. The 120% limit is an industry compromise, that has to do with mitigating the risk of mutual heating of the breakers. For a 600A panel with a matching main, this means you can have up to 96A of total inverter current, since 125% of 96A = 120A, and (600A + 120A) is 120% of the 600A bus.

Panelboards are routinely filled with a lot more branch breakers than their full rating. The underlying idea is that not all of them draw full current at once. Engineers have nuanced ways of adding up the loads for circuit & eqpt sizing, and the main breaker is expected to be the fail-safe in the unlikely event that somehow the loads do exceed what the engineers anticipated. By introducing a source on the busbar, you create a blindspot that the main breaker will not detect. Given a 200A PV system at full power (160A continuous) and a 600A busbar and matching main breaker, this means if you draw 700A among the loads, it will be business as usual for the main breaker, and the main breaker will not trip. Putting both these breakers on the same side of the bus, means you'll have the full 700A on the 600A bus.

Putting them on opposite ends, mitigates this problem, but the industry compromise of the 120% rule still limits you to 96A of inverter current in this example. This would be a 120A breaker if it existed, but as of NEC2014, they reworded it so that rounding to the standard size of 125A isn't a show stopper.

Thankyou. This was very helpful!