Enphase IQ series +7 inverter

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hhsting said:
You mean the main breaker does not stop the fed from utility? I thought main breaker would allow one way fed from solar to grid.

Also more of the 705.12(D) quote....

"...where this distribution equipment is capable of supplying mulitple branch circuits or feeders or both......"

So where are the branch circuits or feeders in the 200A panelboard? I have none. So how can it apply?

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No you're going down a wrong road here.

First, it says 'capable of', not actually supplying.

Second, the circuits to the aggregators are feeders, and you have a branch circuit for the Envoy.
These definitions don't stop applying because they're connected to sources instead of loads.
 
hhsting said:
You mean the main breaker does not stop the fed from utility? I thought main breaker would allow one way fed from solar to grid.
Click to expand...

No. A breaker does not control the direction of current flow. We backfeed them all the time.
 
Post #1 what if conductor connecting the main bldg service disconnect ground bar conductor connecting the AC solar panel is removed.

Instead neutral from neutral bar is brought to the AC solar panel as shown on attached sketch. Ground bus both disconnects is grounded.

Understand that bonding of neutral and ground is depending on AHJ in AC solar panel.

Would the attached sketch removal of conductor connecting the main bldg service disconnect ground bar conductor connecting the AC solar panel and bring in neutral be per code?
175e9c294ec9e4308036824531c0e1bb.jpg


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So, on sizing this "Solar Subpanel" that has a line side connection to the service:

1) Suppose for the moment there's no 20A circuit for an Envoy. Then in terms of 125% of the inverter output current, the panel could be 200A as I understand it: The Enphase IQ7+ datasheet lists a "maximum continuous output power" of 290 VA. Each 208V/20A circuit can support 11 of the IQ7+ microinverters (20*208/290/1.25, rounded down). That means the panel would have at most 198 connected microinverters, balanced on the 3 phases, which is the maximum a 200A bus can support (3*120*200/290/1.25, rounded down to a multiple of 3).

However, if the panel is subject to 705.12(D), none of the options in 705.12(D)(2)(3) seem to work for the panel, as 705.12(D)(2)(3)(c) is written in terms of breaker rating, not inverter output current rating. Does the presence or absence of a main breaker in this panel really affect whether or not it is subject to 705.12(D)? Seems like the panel should be subject to 705.12(D) regardless.

But wouldn't a 225A panel be sufficient under 705.12(D)(2)(3)(c)? When adding the current on the A bus, for example, the current from the A-B breakers will not add fully with the current from the A-C breakers, but there will be a factor of sqrt(3)/2. So with (6) 60A 2-pole breakers, the current on each bus could be at most 2*60*sqrt(3) = 208A.

2) Now the 20A circuit for the Envoy would push sum of the breakers on that bus over 225A. A 15A breaker for the Envoy would work, or if one of the 60A breakers could be downsized to 50A because there are 193 or fewer IQ7+ microinverters in the system, and the Envoy is connected to the phase(s) with the 50A breaker, that would work. Is that correct?

[BTW, this cycle I wrote a late proposed change to 705.12(D)(2)(3)(c) that would permit ignoring the smallest breaker in the panel, on the basis of a counting argument, Kirchoff's current law, and that a single breaker can't simultaneously be both a load and a source. I don't think the committee took the time to read and understand it, as the committee response was a non-sequitur. But it would be useful precisely for this 20A Envoy circuit breaker.]

Cheers, Wayne
 
Engineer has shown following calculation:
Each inverter output current is 1.39A and output power is 325 watts.

Aggregator max output power is 290 watts.

1. I have 11 inverters in one string. Each aggregator has 3 inputs and takes 3 strings. So 11x3 = 33 inverters each branch circuits.

33x1.39A= 45.9A each feeder. 45.9×1.25= 57A so next size is 60 A.

2. I have one string with 8 inverters and two inputs to aggregators. So 8x2=16 inverters.

16x1.39=22.24A

22.24ax1.25= 27.6A so next size is 30A.

3. Panel sizing. Total inverters I have is

a. Phase A (L1 and L2) from two aggregators: 66x290W = 19.14 kW.

b. Phase B (L1 and L2) from two aggregators is 66x290W = 19.14kw.

c. Phase C (L1 and L2): 49x290W = 14.21kw.

Total is 14.21kw + 19.14kw +19.14kw = 52.94 kw.

For three phase 208V panel 52.94kw not counting Envoy: 145A so 200A panel is sufficient. Not sure why 705.12(D) not concur.

Also appreciate if code expert answers my post # 23 question.

Thank you in advance
 
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Thanks for reminding us about the three phase factor. Can't believe I forgot :slaphead:

wwhitney said:
So, on sizing this "Solar Subpanel" that has a line side connection to the service:

1) Suppose for the moment there's no 20A circuit for an Envoy. Then in terms of 125% of the inverter output current, the panel could be 200A as I understand it: The Enphase IQ7+ datasheet lists a "maximum continuous output power" of 290 VA. Each 208V/20A circuit can support 11 of the IQ7+ microinverters (20*208/290/1.25, rounded down). That means the panel would have at most 198 connected microinverters, balanced on the 3 phases, which is the maximum a 200A bus can support (3*120*200/290/1.25, rounded down to a multiple of 3).

...

But wouldn't a 225A panel be sufficient under 705.12(D)(2)(3)(c)? When adding the current on the A bus, for example, the current from the A-B breakers will not add fully with the current from the A-C breakers, but there will be a factor of sqrt(3)/2. So with (6) 60A 2-pole breakers, the current on each bus could be at most 2*60*sqrt(3) = 208A.
Click to expand...

Your calcs are correct as far as I see, but there's no code rule that allows you to calculate it this way and qualify the panel.

Does the presence or absence of a main breaker in this panel really affect whether or not it is subject to 705.12(D)? Seems like the panel should be subject to 705.12(D) regardless.
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His original sketch didn't show an upstream OCPD.
Thus if we remove that main breaker it would be supply side, with six breakers allowed. NOTE the supply side rule 705.12A goes by overcurrent device and not 125percent of inverter output! So your above calc is irrelevant to that, too.
 
jaggedben said:
Your calcs are correct as far as I see, but there's no code rule that allows you to calculate it this way and qualify the panel.
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Well, all you have to do is read "The sum of the ampere ratings of all overcurrent devices" to be (a) per busbar and (b) a vectorial sum for 3-phase. Both seem obvious to an informed reader. :)

Cheers, Wayne
 
wwhitney said:
Well, all you have to do is read "The sum of the ampere ratings of all overcurrent devices" to be (a) per busbar and (b) a vectorial sum for 3-phase. Both seem obvious to an informed reader. :)

Cheers, Wayne
Click to expand...

Where is vectorial sum of three phase written?

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hhsting said:
Post #1 what if conductor connecting the main bldg service disconnect ground bar conductor connecting the AC solar panel is removed.

Instead neutral from neutral bar is brought to the AC solar panel as shown on attached sketch. Ground bus both disconnects is grounded.

Understand that bonding of neutral and ground is depending on AHJ in AC solar panel.

Would the attached sketch removal of conductor connecting the main bldg service disconnect ground bar conductor connecting the AC solar panel and bring in neutral be per code?
175e9c294ec9e4308036824531c0e1bb.jpg


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Click to expand...

Post #23 question please see above got lost in this thread to answer.
 
hhsting said:
Where is vectorial sum of three phase written?
Click to expand...

It isn't.

Per code, providing a three phase calculation for "125% of inverter output" seems a lot more reasonable than applying a vector calc to "the sum of overcurrent devices". Really depends on the attitude of the AHJ.
 
jaggedben said:
Please rephrase, I think you mistyped since it is not grammatical.
Click to expand...
Post #1 attachment the conductor connecting the main bldg service disconnect ground bar with the AC solar panel is removed.

Instead neutral from neutral bar is brought to the AC solar panel as shown on attached sketch this post. Both panels ground bus are grounded with grounding electrode system.

Understand that bonding of neutral and ground is depending on AHJ in AC solar panel.

Would the attached sketch removal of conductor connecting the main bldg service disconnect ground bar conductor connecting the AC solar panel and bring in neutral be per code?
b9919f30aca82d828db4fc9ae7940cad.jpg


Sent from my SM-G935U using Tapatalk
 
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hhsting said:
Engineer has shown following calculation:
Each inverter output current is 1.39A and output power is 325 watts.

Aggregator max output power is 290 watts.

1. I have 11 inverters in one string. Each aggregator has 3 inputs and takes 3 strings. So 11x3 = 33 inverters each branch circuits.

33x1.39A= 45.9A each feeder. 45.9×1.25= 57A so next size is 60 A.

2. I have one string with 8 inverters and two inputs to aggregators. So 8x2=16 inverters.

16x1.39=22.24A

22.24ax1.25= 27.6A so next size is 30A.

3. Panel sizing. Total inverters I have is

a. Phase A (L1 and L2) from two aggregators: 66x290W = 19.14 kW.

b. Phase B (L1 and L2) from two aggregators is 66x290W = 19.14kw.

c. Phase C (L1 and L2): 49x290W = 14.21kw.

Total is 14.21kw + 19.14kw +19.14kw = 52.94 kw.

For three phase 208V panel 52.94kw not counting Envoy: 145A so 200A panel is sufficient. Not sure why 705.12(D) not concur.

Also appreciate if code expert answers my post # 23 question.
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One problem is that his/her outputs aren't balanced so s/he cannot use a uniform 1.73 divisor on the total KW. The L1 current is actually going to be higher than 145A at full output.

Also, s/h e left out the 125% factor on the 145A calc. When the proper vector calc is applied to L1 current and multiplied by 125% does it still come out under 200A?

Problem #2 is still that no part of 705.12(D) concurs. That is, unless you accept Wayne's proposition of allowing a vector sum to be applied to the overcurrent devices. For an unbalanced system this is even less straightforward and more of a stretch of the code language. Unless you are the ultimate authority on allowing this, you probably can't count on doing it.
 
jaggedben said:
One problem is that his/her outputs aren't balanced so s/he cannot use a uniform 1.73 divisor on the total KW. The L1 current is actually going to be higher than 145A at full output.

Also, s/h e left out the 125% factor on the 145A calc. When the proper vector calc is applied to L1 current and multiplied by 125% does it still come out under 200A?

Problem #2 is still that no part of 705.12(D) concurs. That is, unless you accept Wayne's proposition of allowing a vector sum to be applied to the overcurrent devices. For an unbalanced system this is even less straightforward and more of a stretch of the code language. Unless you are the ultimate authority on allowing this, you probably can't count on doing it.
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Waynes prop of vector sum exactly what is that?

L1 has 19.14 kw or L1 and L2 has 19.14kw?

Anyways L1 19.14kwx1.25 = 23.95kw. This is 120/208V system so I am assuming L1 voltage to be 120V to neutral so 23.95kw/0.12kv = 199.35A still under 200A.

Also adding and with 1.732 get 145A. 145Ax1.25 = 181A.

No?
 
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hhsting said:
Waynes prop of vector sum exactly what is that?

L1 has 19.14 kw or L1 and L2 has 19.14kw?

Anyways L1 19.14kwx1.25 = 23.95kw. This is 120/208V system so I am assuming L1 voltage to be 120V to neutral so 23.95kw/0.12kv = 199.35A still under 200A.

Also adding and with 1.732 get 145A. 145Ax1.25 = 181A.

No?
Click to expand...

No. It's more complicated.

There was a definitive article in the Dec/Jan 2010 issue of Solar Pro magazine by Marvin Harmon on the the math for calculating unbalanced line currents . It's no longer readily available online as a webpage but I believe you can register for free at solarpro.com and find the issue to download.
 
jaggedben said:
No. It's more complicated.

There was a definitive article in the Dec/Jan 2010 issue of Solar Pro magazine by Marvin Harmon on the the math for calculating unbalanced line currents . It's no longer readily available online as a webpage but I believe you can register for free at solarpro.com and find the issue to download.
Click to expand...
I dont have delta I have 208/120V which is wye and will be bringing neutral from utility side no?
 
jaggedben said:
No. It's more complicated.

There was a definitive article in the Dec/Jan 2010 issue of Solar Pro magazine by Marvin Harmon on the the math for calculating unbalanced line currents . It's no longer readily available online as a webpage but I believe you can register for free at solarpro.com and find the issue to download.
Click to expand...
I read solarpro article calculated worst case with 1.25 multiply but I got worst case 199.18A line current excluding Envoy which is still less than 200A. Am I missing something?

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jaggedben said:
The removal of which conductor? You need to fix the sentence structure.
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Post #1 attachment the conductor connecting the main bldg service disconnect ground bar with the AC solar panel is removed.

Instead neutral from neutral bar is brought to the AC solar panel. Both panels ground bus are grounded with grounding electrode system.

The result of all tha above is attached sketch this post.

Understand that bonding of neutral and ground is depending on AHJ in AC solar panel.

Would the attached sketch on this post be code compliant?
d554bff6f9340a2d4ef988a9e7ccc799.jpg


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