Conductor Sizing for Single Phase Inverters on 3PH System

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ptrombley

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I have 3 - 6KW inverters that output 208V single phase each (L1 & L2 + N & Gnd). Per the spec sheet the maximum output current is 29A per inverter.. at 208V.
The electrical service that I will be connecting them to is 120/208V 3PH 4W (wye). If I connect the 1st inverter to A-B, 2nd to B-C and 3rd to A-C (at an AC combiner panel).. what will the load be on each phase?
My thought is 58A each, but I'm not so sure that is correct with a 3PH system.. and I'm not an engineer. :)
 

GoldDigger

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I have 3 - 6KW inverters that output 208V single phase each (L1 & L2 + N & Gnd). Per the spec sheet the maximum output current is 29A per inverter.. at 208V.
The electrical service that I will be connecting them to is 120/208V 3PH 4W (wye). If I connect the 1st inverter to A-B, 2nd to B-C and 3rd to A-C (at an AC combiner panel).. what will the load be on each phase?
My thought is 58A each, but I'm not so sure that is correct with a 3PH system.. and I'm not an engineer. :)
First, for clarity, you are asking what the load will be on each phase conductor (A,B,C), not on each of the phases, which is normally considered to be A-B,B-C and A-C.

The two 29A output currents are not in phase with each other but instead 120 degrees out of phase, and part of the current from each inverter connected to a single wire will cancel the corresponding current from the other inverter attached to that wire.
End result: Current in each phase wire will be 29 times 2 times sqrt(3)/2. Or 29 times sqrt(3). A little over 50A.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
First, for clarity, you are asking what the load will be on each phase conductor (A,B,C), not on each of the phases, which is normally considered to be A-B,B-C and A-C.

The two 29A output currents are not in phase with each other but instead 120 degrees out of phase, and part of the current from each inverter connected to a single wire will cancel the corresponding current from the other inverter attached to that wire.
End result: Current in each phase wire will be 29 times 2 times sqrt(3)/2. Or 29 times sqrt(3). A little over 50A.
Alternatively, you can consider the inverter block to be a single three phase 18kW inverter. 18000W/(208V)(sqrt3) = 49.96A. Same answer, more or less.
 

Carultch

Senior Member
Location
Massachusetts
I have 3 - 6KW inverters that output 208V single phase each (L1 & L2 + N & Gnd). Per the spec sheet the maximum output current is 29A per inverter.. at 208V.
The electrical service that I will be connecting them to is 120/208V 3PH 4W (wye). If I connect the 1st inverter to A-B, 2nd to B-C and 3rd to A-C (at an AC combiner panel).. what will the load be on each phase?
My thought is 58A each, but I'm not so sure that is correct with a 3PH system.. and I'm not an engineer. :)

Once you have a balanced configuration of phase-to-phase inverters on a three-phase system, the calculation of per conductor current is total power/voltage/sqrt(3).

6 kW * 3 = 18 kW

18000W / 208V / sqrt(3) = 50A

If for some reason, the inverter datasheets have a larger current than power rating alone would perscribe, increase this by the above nominal current ratio. One example is a hypothetical 6 kW inverter having 32A at 208V. Because 6 kW at exactly 240V is really 29A. This hypothetical datasheet current is 10% higher than power rating alone would predict, and one strategic reason for this, is to avoid clipping during undervoltage operation. If you triplet three of these hypothetical inverters, the current becomes 3*6000W/sqrt(3)/208V * (32/29) = 55A.


If you DO NOT have a balanced current situation, the formal procedure for adding up phase-to-phase currents is as follows:
Add up all A-B currents. Call it Iab.
Add up all B-C currents. Call it Ibc.
Add up all C-A currents. Call it Ica.

Calculate current on conductor A, by using all terms pertaining to phase A, which are Iab and Ica.
Ia = sqrt(Iab^2 + Ica^2 + Iab*Ica)

Calculate current on conductor B, by using all terms pertaining to phase B, which are Iab and Ibc.
Ib = sqrt(Iab^2 + Ibc^2 + Iab*Ibc)

Calculate current on conductor A, by using all terms pertaining to phase A, which are Iab and Ica.
Ic = sqrt(Ibc^2 + Ica^2 + Ibc*Ica)

Add in any applicable phase-to-neutral currents to each of the above values.
Add in any nominal per phase currents from any applicable three phase circuits, to each of the above values.

Take the largest current, and that becomes the operating current for the entire mix of loads. All OCPDs, panels, disconnects, and wire corresponding to the aggregate feeder, are to be sized to that value. On all phases.
 
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Smart $

Esteemed Member
Location
Ohio
The down side is, Code does not currently permit taking the 2 to 1.732 reduction for sizing wire, ocpd, bus, etc.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Once you have a balanced configuration of phase-to-phase inverters on a three-phase system, the calculation of per conductor current is total power/voltage/sqrt(3).

6 kW * 3 = 18 kW

18000W / 208V / sqrt(3) = 50A

If for some reason, the inverter datasheets have a larger current than power rating alone would perscribe, increase this by the above nominal current ratio. One example is a hypothetical 6 kW inverter having 32A at 208V. Because 6 kW at exactly 240V is really 29A. This hypothetical datasheet current is 10% higher than power rating alone would predict, and one strategic reason for this, is to avoid clipping during undervoltage operation. If you triplet three of these hypothetical inverters, the current becomes 3*6000W/sqrt(3)/208V * (32/29) = 55A.


If you DO NOT have a balanced current situation, the formal procedure for adding up phase-to-phase currents is as follows:
Add up all A-B currents. Call it Iab.
Add up all B-C currents. Call it Ibc.
Add up all C-A currents. Call it Ica.

Calculate current on conductor A, by using all terms pertaining to phase A, which are Iab and Ica.
Ia = sqrt(Iab^2 + Ica^2 + Iab*Ica)

Calculate current on conductor B, by using all terms pertaining to phase B, which are Iab and Ibc.
Ib = sqrt(Iab^2 + Ibc^2 + Iab*Ibc)

Calculate current on conductor A, by using all terms pertaining to phase A, which are Iab and Ica.
Ic = sqrt(Ibc^2 + Ica^2 + Ibc*Ica)

Add in any applicable phase-to-neutral currents to each of the above values.
Add in any nominal per phase currents from any applicable three phase circuits, to each of the above values.

Take the largest current, and that becomes the operating current for the entire mix of loads. All OCPDs, panels, disconnects, and wire corresponding to the aggregate feeder, are to be sized to that value. On all phases.
That of course is the accurate way to calculate those currents, but if the imbalance is not too great you can do it more simply and get close without undersizing anything. If you have, for example, two 7kW inverters and a single 6kW inverter, for figuring OCPD you can consider it to be a single 21kW three phase inverter. 21000W/208sqrt3 = 58.3A and 58.3A X 1.25 = 72.9A, so use 80A OCPD and make sure the minimum derated ampacity of your conductors is greater than 70A. I'll leave doing it "right" as an exercise for someone less lazy than I on a Sunday morning, but it will be close to this result and won't yield OCPD that is too small.
 

Smart $

Esteemed Member
Location
Ohio
That of course is the accurate way to calculate those currents, but if the imbalance is not too great you can do it more simply and get close without undersizing anything. If you have, for example, two 7kW inverters and a single 6kW inverter, for figuring OCPD you can consider it to be a single 21kW three phase inverter. 21000W/208sqrt3 = 58.3A and 58.3A X 1.25 = 72.9A, so use 80A OCPD and make sure the minimum derated ampacity of your conductors is greater than 70A. I'll leave doing it "right" as an exercise for someone less lazy than I on a Sunday morning, but it will be close to this result and won't yield OCPD that is too small.
But where does Code allow you to combine outputs before the ocpd? And then isn't any ocpd after that on the utility side based either on the sum of the output ratings @125% or the sum of the ocpd ratings?
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
But where does Code allow you to combine outputs before the ocpd?
Perhaps I am misunderstanding your question. AC combiner panels where inverter outputs are combined are commonplace in PV systems, and the conductors from the AC combiner panel must be protected by OCPD at the point of interconnection. Each inverter also has its own OCPD at the combiner panel, of course.
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
But where does Code allow you to combine outputs before the ocpd?

Where does it prohibit it?

And then isn't any ocpd after that on the utility side based either on the sum of the output ratings @125% or the sum of the ocpd ratings?

There is no code section in any code that I'm aware of that stipulates how to calculate the size of a utility side OCPD. That OCPD can be much larger than the inverter(s) output if the inverter(s) OCPD is in a panel with loads, which is typical. Making it smaller than the inverters output might not be functional (i.e. nuisance trip at full output) but I still can't figure a code section that prohibits it.

If an AHJ wants to be strict then they can require the use of sums when figuring how to apply the 120% rule. But I can't see any part of the code that requires the utility side OCPD to be figured on sums, in any code cycle.
 

Smart $

Esteemed Member
Location
Ohio
... Each inverter also has its own OCPD at the combiner panel, of course.
Where does it prohibit it?
That was my point. Each inverter must have its own ocpd. You cannot combine outputs before the first ocpd.


There is no code section in any code that I'm aware of that stipulates how to calculate the size of a utility side OCPD. That OCPD can be much larger than the inverter(s) output if the inverter(s) OCPD is in a panel with loads, which is typical. Making it smaller than the inverters output might not be functional (i.e. nuisance trip at full output) but I still can't figure a code section that prohibits it.

If an AHJ wants to be strict then they can require the use of sums when figuring how to apply the 120% rule. But I can't see any part of the code that requires the utility side OCPD to be figured on sums, in any code cycle.
Bad choice of word(s) on my part. I meant bus and feeder sizing (aka 120% rule).
 

Carultch

Senior Member
Location
Massachusetts
But where does Code allow you to combine outputs before the ocpd? And then isn't any ocpd after that on the utility side based either on the sum of the output ratings @125% or the sum of the ocpd ratings?


There is no reason to accumulate rounding errors, unless you do not have a master OCPD.
13 inverters at 18A each with a 25A breaker each = 234A protected by a 300A OCPD.
Using a 350A OCPD (because 13*25 = 325, which rounds up to 350) is overkill.

The overcurrent device is sized per 1.25*max continuous current. Sum up the max continuous currents of each inverter, with appropriate phase adjustments and grouping if applicable, and then apply the 1.25 factor to the largest per conductor current.


One place where you might not have a master OCPD, is if you are interconnecting two PV systems, both by a separate supply side tap. The sum of the OCPDs protecting each tap, shall not exceed the service. In this case, you would need to accumulate the rounding errors, of just those two taps. Not of all the inverters connected in to each of them.
 
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Smart $

Esteemed Member
Location
Ohio
There is no reason to accumulate rounding errors...
Not quite sure where you are coming up with this "rounding errors" term in this regard (yes, I know what rounding errors are). Nevertheless, your example is indicative of 1Ø aggregation. I'm talking about 3Ø aggregation and application of the 2 to 1.732 (when balanced) reduction in current (i.e. arithmetic vs. vector summed output currents).
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
JMHO the pre-2014 code requires you (for 120% rule and some other purposes) to sum the numeric values of the nearest OCPDs to the inverters, and as mentioned earlier, allows no leeway to do a vector addition. The sizes of the OCPDs are not vectors.
 

Carultch

Senior Member
Location
Massachusetts
Not quite sure where you are coming up with this "rounding errors" term in this regard (yes, I know what rounding errors are). Nevertheless, your example is indicative of 1Ø aggregation. I'm talking about 3Ø aggregation and application of the 2 to 1.732 (when balanced) reduction in current (i.e. arithmetic vs. vector summed output currents).

I'm talking about the fact that there is no need to accumulate the rounding up of the branch breakers, when determining the main breaker.

My example could apply to single phase, or three phase with three phase inverters.

Three phase generated from single phase inverters is a bit more complicated, where you would need to use a vector sum of currents. Rounding errors shouldn't exist in this calculation, if you carry enough digits (4 is plenty).
 

Smart $

Esteemed Member
Location
Ohio
...
Three phase generated from single phase inverters is a bit more complicated, where you would need to use a vector sum of currents. Rounding errors shouldn't exist in this calculation, if you carry enough digits (4 is plenty).
This is what we are discussing. So where does the 120% rule permit one to use the vector sum?
 
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