Help with Review of contractors conductor calculations

[cf400]

Hello all,


I'm looking for a second opinion here regarding my contractors wire size calculations below. I have a 26 panel Solaredge system being installed with 2 strings of 13 panels each. The calculations below in the DC CONDUCTOR AMPACITY CALC'S ROOF CONDUIT shows "# OF STRINGS PER INVERTER: 1" and "# OF CURRENT CARRYING CONDUCTORS: 2"

I was thinking that should read 2 on the strings per inverter (since the plans say 2 strings of 13 panels) and 4 on the current carrying conductors (since it's 2 strings, not 1). If that is in fact what it should show doesn't that mean the calculations are incorrect on the REQUIRED CIRCUIT CONDUCTOR AMPACITY? Should the calc be (1.25)(1.25)(15)(2) = 46.9A instead of (1.25)(1.25)(15) = 23.4A since there are actually going to be 4 conductors in the conduit instead of 2 as shown or is not an issue with Solaredge?

If true then I'm thinking the wire sizing is incorrect since this calulation below (0.76)(1.0)(40) = 30.4A > 23.4A OK, would actually be (0.76)(.8)(40) = 24.4A > 46.9A and that's NOT OK using 10AWG THWN-2 wire. I'm coming up a need for using 4AWG THWN-2 size wire from the NEC Table 310.15(B)(16). (.76)(.8)(95) = 57.7A > 46.9A.

Am I doing this correctly?
I attached the calculations from the submitted and permitted plans below for reference. (the blue highlighting is mine)

Thoughts please,
Thank you

--------------------------

ELECTRICAL CALCULATIONS
SOLAREDGE INVERTERS ARE OPERATING IN A FIXED
STRING VOLTAGE POINT, THEREFORE THE STRING
VOLTAGE IS ALWAYS THE SAME, REGARDLESS OF THE
NUMBER OF MODULES CONNECTED IN THE STRING OR
THE IRRADIANCE POWER OR MODULES TEMPERATURE.
FROM SOLAREDGE INVERTER SPEC SHEET

MAX SYSTEM CURRENT: 15A
MAX SYSTEM VOLTAGE: 350V

DC CONDUCTOR AMPACITY CALC'S ROOF CONDUIT
# OF STRINGS PER INVERTER: 1
EXPECTED MAX AMBIENT TEMP. (°C): 53
TEMP. CORRECTION NEC, TABLE 310.15(B)(2)(a): 0.76
CIRCUIT CONDUCTOR SIZE/TYPE: 10AWG, THWN-2, 90°C
# OF CURRENT CARRYING CONDUCTORS: 2
CONDUIT FILL CORR., NEC TABLE 310.15(B)(3)(a): 1.0
CIRCUIT CONDUCTOR AMPACITY, TABLE 310.15(B)(16): 40A

REQUIRED CIRCUIT CONDUCTOR AMPACITY, NEC 690.8(B):
(1.25)(1.25)(MAX CONTINUOUS DC CONDUCTOR CURRENT)
(1.25)(1.25)(15) = 23.4A

ADJUSTED AMPACITIES FOR CONDUCTORS, NEC 310.15:
TEMP. CORRECTION, NEC TABLE 310.15(B)(2)(a) x
CONDUIT FILL CORRECTION, NEC 310.15(B)(3)(a) x
CONDUCTOR AMPACITY, NEC TABLE 10.15(B)(16)
(0.76)(1.0)(40) = 30.4A > 23.4A OK
10AWG, THWN-2, 90°C WIRE IS OK

[HX_Guy]

This sounds very similar to the confusion on my SolarEdge permit and how the permit installer came up with 2AWG wire instead of 6AWG.

Someone else with more knowledge than me will have to chime in, but with SolarEdge, the inverter is considered ungrounded and therefore both DC+ and DC- are "hot" is true, I don't think you're supposed to count the amps twice. If each circuit has 15A, then it's 15A max...it's not 15A per conductor therefore 2 conductors = 30A. I believe your installers calculations are correct.

BTW though, the max system voltage is 500V, not 350V.

[sensij]

You might want to familiarize yourself with the way the SolarEdge system works, in the context of the NEC. They have published a white paper that explains it in detail.

If the two strings are not combined at any point, then 15 A is correct for the max rated current on the DC circuit. That circuit requires only a single 1.25 correction, since it is not technically a photovoltaic circuit. The only photovoltaic circuits in a SolarEdge system are between each solar panel and its optimizer. The optimizer is DC utilization equipment, so the circuit including the optimizers and inverters can be treated like a regular DC circuit, only subject to whatever constraints might exist in section 690 on inverter input circuit design.

If all 4 conductors run all the way to the inverter, then the calculation should be:
(Temp correction) * (Fill correction) * (base ampacity of 10 AWG THWN-2) = corrected ampacity.
(0.76) * (0.8) * (40) = 24.32

The corrected ampacity must be greater than or equal to 1.25 * 15 = 18.75, which it is, so you are OK.

If the two strings are actually combined somewhere ahead of the inverter, then you need a calculation based on two conductors carrying 30 A.

(0.76) * (1) * (40) = 30.4
1.25 * 30 = 37.5
10 AWG would be too small for the combined strings, you would have to increase to 8 AWG.

It seems from permits that have been getting approved around here that not many installers or AHJ's really understand the SolarEdge system, and choose to include the extra 1.25 factor, among other misunderstandings. Since the AHJ is always right, what I've posted here might just be academic, there is a decent chance your permit will get approved as-is.

[sensij]

This sounds very similar to the confusion on my SolarEdge permit and how the permit installer came up with 2AWG wire instead of 6AWG.

Someone else with more knowledge than me will have to chime in, but with SolarEdge, the inverter is considered ungrounded and therefore both DC+ and DC- are "hot" is true, I don't think you're supposed to count the amps twice. If each circuit has 15A, then it's 15A max...it's not 15A per conductor therefore 2 conductors = 30A. I believe your installers calculations are correct.

BTW though, the max system voltage is 500V, not 350V.

The max system voltage doesn't matter much, 350 or 500 V it will not affect the design in any way. Ungrounded or not has no bearing on this calculation. You never "count the amps twice" for a supply and return conductor in a single phase circuit, regardless of whether either of the conductors is ground referenced.

The one thing that isn't in the installer's calculation that might be a gotcha is that rooftop conduit needs an ambient temperature correction from table 310.15(B)(3)(c). The 53 deg C design temp may already reflect this, but it seems a bit low if it does. The adder is 14 deg C if the conduit is 1 ft off the roof surface, or 17 deg C if between 3.5" and 1 ft, which suggests that 97 - 102 deg F was the maximum ambient being considered.

[HX_Guy]

One thing I find peculiar with SolarEdge inverters is that they list that the inverter inputs for 2 strings (any more than 2 and they recommend using a combiner box) but inside their DC disconnect, there seems to only be 1 input.



The DIN terminal block does have two inputs on it as you can see in the pictures, but doesn't that in essence combine the strings?

[sensij]

One thing I find peculiar with SolarEdge inverters is that they list that the inverter inputs for 2 strings (any more than 2 and they recommend using a combiner box) but inside their DC disconnect, there seems to only be 1 input.

The DIN terminal block does have two inputs on it as you can see in the pictures, but doesn't that in essence combine the strings?

Yes, it combines the strings by connecting them in the terminal block. However, the internal wiring of the inverter is built to handle that. It does not affect the outside wiring calculations at all.

[cf400]

You might want to familiarize yourself with the way the SolarEdge system works, in the context of the NEC. They have published a white paper that explains it in detail.

If the two strings are not combined at any point, then 15 A is correct for the max rated current on the DC circuit. That circuit requires only a single 1.25 correction, since it is not technically a photovoltaic circuit. The only photovoltaic circuits in a SolarEdge system are between each solar panel and its optimizer. The optimizer is DC utilization equipment, so the circuit including the optimizers and inverters can be treated like a regular DC circuit, only subject to whatever constraints might exist in section 690 on inverter input circuit design.

If all 4 conductors run all the way to the inverter, then the calculation should be:
(Temp correction) * (Fill correction) * (base ampacity of 10 AWG THWN-2) = corrected ampacity.
(0.76) * (0.8) * (40) = 24.32

The corrected ampacity must be greater than or equal to 1.25 * 15 = 18.75, which it is, so you are OK.

If the two strings are actually combined somewhere ahead of the inverter, then you need a calculation based on two conductors carrying 30 A.

(0.76) * (1) * (40) = 30.4
1.25 * 30 = 37.5
10 AWG would be too small for the combined strings, you would have to increase to 8 AWG.

It seems from permits that have been getting approved around here that not many installers or AHJ's really understand the SolarEdge system, and choose to include the extra 1.25 factor, among other misunderstandings. Since the AHJ is always right, what I've posted here might just be academic, there is a decent chance your permit will get approved as-is.

Thank you for the response. I see where I went wrong in the calculation by doubling the amp calculation for sizing. My concern was the contractor called for only one string on the plans, but on the phone now says it'll be two to the inverter. I see that one string won't work at 10AWG

The permit has been approved with the above calculations, so we'll see if the inspector catches that...

Thanks again for the white paper link, I read it a couple months ago but had forgotten about it!