Array wiring, conduit, and grounding questions

[foo1bar]

In either case, you should be fine with 1/2" EMT, if you can pull the wires.

I didn't check the numbers/calcs - but IMO going with 3/4 just makes life easier for 4x10AWG or 8x12AWG.
And isn't much more $.
BTW - that include a ground wire in the fill calc's? Or planning on running that outside the conduit? If it's inside the conduit it has to be counted for conduit fill (but wouldn't count for current de-rating purposes)
I think 9x12AWG still fit's in theory in 1/2" EMT - but it'll be a PITA I think.

[sensij]

I didn't check the numbers/calcs - but IMO going with 3/4 just makes life easier for 4x10AWG or 8x12AWG.
And isn't much more $.
BTW - that include a ground wire in the fill calc's? Or planning on running that outside the conduit? If it's inside the conduit it has to be counted for conduit fill (but wouldn't count for current de-rating purposes)
I think 9x12AWG still fit's in theory in 1/2" EMT - but it'll be a PITA I think.

Yeah... I would also recommend 3/4", just for the PITA factor.

[sunnyguy]

Yeah... I think you are right, the extra 1.25 and the conditions of use corrections do not get stacked.

OK, I think we are agreed that 14 AWG PV Wire is OK for the open air runs from the panels to the pass-though into conduit, and you would prefer 12 AWG to limit the voltage drop. Your question is whether it is better to combine and transition to THWN-2, running 4 conductors (+ ground) in the home run, or whether you should not combine and run 8 conductors (+ground).

Let's look at the conduit fill:
1/2" EMT can be used with up to 9 x 12 AWG THWN-2 conductors and meet the 40% fill requirement. No difference there.

Now look at conductor ampacity.
For the 4 conductor (combined) version:
Isc of each string = 8.62 A
690.8(A)(2) = 2*8.62*1.25 = 21.55 A
690.8(B)(2)(a) = 1.25 * 21.55 A = 26.9 A. Uncorrected, you are limited by the 75 deg rating of the terminals (assuming the terminals are not 60 deg), which requires 14 AWG. OK
690.8(B)(2)(b) = Carry 21.55 A after conditions of use applied. 10 AWG @ 90 deg = 55 A * (0.65 * 0.8) = 28.6 A. Terminals at 75 deg = 50 A. OK. Maximum of 6 x 10 AWG THWN-2 in 1/2" EMT. OK.

So, for this version, 10 AWG is required.

For the 8 conductor (uncombined) version:
Isc = 8.62 A
690.8(A)(2) = 8.62*1.25 = 10.8 A
690.8(B)(2)(a) = 1.25 * 10.8 = 13.5 A. Uncorrected 14 AWG OK.
690.8(B)(2)(b) = Carry 10.8 after conditions of use applied. Conduit fill correction is 0.7. 14 AWG @ 90 deg = 25 A * (0.65 *0.7) = 13.6 A. Terminal at 75 deg = 20 A. OK.

So, for this version, 14 AWG required, although as you said, 12 AWG would be better for voltage drop.

Based on this, it looks like your choices are 8 x 12 AWG without a combiner, or 4 x 10 AWG with a combiner. In either case, you should be fine with 1/2" EMT, if you can pull the wires.

Really, before you buy anything, you should submit the permit application and make sure what you propose is accepted by your AHJ.

Thanks again for working through the calcs with me.

I think I would prefer using the 8 x 12 AWG.

For sizing the EGC per 250.122:

(C) Multiple Circuits. Where a single equipment grounding conductor is run with multiple circuits in the same raceway, cable, or cable tray, it shall be sized for the largest overcurrent device protecting conductors in the raceway, cable, or cable tray.

So since the Isc of each parallel circuit is 8.62 does that mean the EGC only needs to meet that single rating and not 4x that rating? Is it typical to use the EMT as the EGC or to run an EGC conductor inside the EMT?

Now for the AC side wiring and combined GEC connection:

Inverter max continuous output current = 34.2A

240V Conductors:
690.8(B)(2)(a): 34.2A * 1.25 = 42.75A ( > 40A 10AWG so can't use, < 55A 8AWG so OK)
690.8(B)(2)(b): 8AWG 90C copper: 55A * 0.87 (temp adj. @ 45C) = 47.85A ( > 34.2A so OK)

So looks like I have to use 8AWG here, right?

Neutral (Sense) Conductor: 14 AWG (minimum supported by terminal) ?? This is not a current carrying conductor, right? EDIT: Manual says to size same as other AC conductors.

GEC 690.47(C.3): 8 AWG per 250.166(B) ?

Upsizing to 3/4" EMT recommendation noted and agreed.

I definitely am not going to buy anything until the permit is approved.

Thanks to all contributors.

[foo1bar]

Neutral (Sense) Conductor: 14 AWG (minimum supported by terminal) ?? This is not a current carrying conductor, right?

If-and-only-if the installation instructions from the manufacturer say it's OK would I do smaller for the neutral.

Also - are you going to have 4 current carrying conductors in the conduit for the AC side? I thought this was a single inverter, which would likely mean 2 current carrying conductors.

And have you checked the voltage drop (voltage rise in this case if you look at it from perspective of voltage at your panel)

Also realize that less power lost in cables == more power delivered to your panel, so you get some economic benefit from upsizing (probably not as much as it costs to upsize, depending on prices of wires)

I don't know if you need to size wire for 40A (breaker) or 34.2A (inverter).

[sunnyguy]

Now look at conductor ampacity.
For the 4 conductor (combined) version:
Isc of each string = 8.62 A
690.8(A)(2) = 2*8.62*1.25 = 21.55 A
690.8(B)(2)(a) = 1.25 * 21.55 A = 26.9 A. Uncorrected, you are limited by the 75 deg rating of the terminals (assuming the terminals are not 60 deg), which requires 14 AWG. OK
690.8(B)(2)(b) = Carry 21.55 A after conditions of use applied. 10 AWG @ 90 deg = 55 A * (0.65 * 0.8) = 28.6 A. Terminals at 75 deg = 50 A. OK. Maximum of 6 x 10 AWG THWN-2 in 1/2" EMT. OK.

So, for this version, 10 AWG is required.

Also just want to point out that per table 310.15(B)(16), 10AWG @ 90C = 40A so 10AWG does not pass the derating test at 40*0.65*0.8 = 20.8A.

[sunnyguy]

If-and-only-if the installation instructions from the manufacturer say it's OK would I do smaller for the neutral.

Also - are you going to have 4 current carrying conductors in the conduit for the AC side? I thought this was a single inverter, which would likely mean 2 current carrying conductors.

And have you checked the voltage drop (voltage rise in this case if you look at it from perspective of voltage at your panel)

Also realize that less power lost in cables == more power delivered to your panel, so you get some economic benefit from upsizing (probably not as much as it costs to upsize, depending on prices of wires)

I don't know if you need to size wire for 40A (breaker) or 34.2A (inverter).

Yes there are only 2 current carrying conductors. Unfortunately the installation manual does not say anything about the neutral connection. EDIT: Yes it does: If the neutral conductor is too small it can adversely affect the inverter's feed-in of power to the mains. The neutral conductor must therefore be the same size as the other live conductors.

I think I need to use 8AWG from the inverter to the panel anyway which when adjusted rates at 47.85A so should be ok for the 45A breaker too.

[sensij]

Thanks again for working through the calcs with me.

I think I would prefer using the 8 x 12 AWG.

For sizing the EGC per 250.122:

(C) Multiple Circuits. Where a single equipment grounding conductor is run with multiple circuits in the same raceway, cable, or cable tray, it shall be sized for the largest overcurrent device protecting conductors in the raceway, cable, or cable tray.

So since the Isc of each parallel circuit is 8.62 does that mean the EGC only needs to meet that single rating and not 4x that rating? Is it typical to use the EMT as the EGC or to run an EGC conductor inside the EMT?

Read 690.45(A) again. You are not sizing EGC per section 250.122, it is per *Table* 250.122. Specifically, this is the sentence that matters to you:

Where no overcurrent protective device is used in the circuit, an assumed overcurrent device rated at the photovoltaic rated short-circuit current shall be used in Table 250.122.

Now for the AC side wiring and combined GEC connection:

Inverter max continuous output current = 34.2A

240V Conductors:
690.8(B)(2)(a): 34.2A * 1.25 = 42.75A ( > 40A 10AWG so can't use, < 55A 8AWG so OK)
690.8(B)(2)(b): 8AWG 90C copper: 55A * 0.87 (temp adj. @ 45C) = 47.85A ( > 34.2A so OK)

So looks like I have to use 8AWG here, right?

Yes, 8 AWG looks right, subject to comment about the breaker below.

GEC 690.47(C.3): 8 AWG per 250.166(B) ?

Yes, looks right.

Also just want to point out that per table 310.15(B)(16), 10AWG @ 90C = 40A so 10AWG does not pass the derating test at 40*0.65*0.8 = 20.8A.

Thanks, that was sloppy of me.

One other thing to consider... your breaker will need to be greater than 42.75 A. You'll have to see what you can find to fit your main service panel. If it is too big, you will run into 690.8(B)(2)(c). For example, if you end up with a 50 A breaker, 8 AWG is too small now after the temp correction and your AC conductors would have to go to 6 AWG.

The other thing you'll need to look at is whether your main service panel will need an upgrade, or to have the main breaker de-rated. With a top fed panel, you would put the PV breaker on the bottom, and the total source OCPD rating can not exceed 120% of the busbar rating. For example, a 200 A bus with a 200 main circuit breaker could only have 40 A max PV breaker. You would have to drop down to a 175 A main breaker. On the other hand, with a 225 A bus, you would be ok with a 200 A main breaker and 60 A PV breaker.

[sunnyguy]

Read 690.45(A) again. You are not sizing EGC per section 250.122, it is per *Table* 250.122.
One other thing to consider... your breaker will need to be greater than 42.75 A. You'll have to see what you can find to fit your main service panel. If it is too big, you will run into 690.8(B)(2)(c). For example, if you end up with a 50 A breaker, 8 AWG is too small now after the temp correction and your AC conductors would have to go to 6 AWG.

The other thing you'll need to look at is whether your main service panel will need an upgrade, or to have the main breaker de-rated. With a top fed panel, you would put the PV breaker on the bottom, and the total source OCPD rating can not exceed 120% of the busbar rating. For example, a 200 A bus with a 200 main circuit breaker could only have 40 A max PV breaker. You would have to drop down to a 175 A main breaker. On the other hand, with a 225 A bus, you would be ok with a 200 A main breaker and 60 A PV breaker.

I am still not clear per 690.45(A) whether that is the combined Isc of all circuits sharing the ground or the maximum Isc of all circuits sharing the ground. So 8.62A or 8.62A * 4?

The recommended breaker size per the inverter manual is 45A. My panel does have a 225A bus bar and a 200A main breaker so I think I am ok there.

Thanks

[sensij]

8.62 a

[sunnyguy]

Thanks. Good sanity check. So the reasoning is similar to 250.122(C). I guess that means I can use 14AWG EGC. What is the best way to mate the 6AWG solid bare exposed and 14AWG THWN-2 in the JBs?

Is it better to buy some pre-terminated MC4 extension cables and cut them in half or to buy the connectors, tool, and wire and make custom lengths for the array to JB connections?

Finally, for the AC disconnect, does Siemens LNF222R 3R 60A 240V 2P seem appropriate?

Thanks

[foo1bar]

Thanks. Good sanity check. So the reasoning is similar to 250.122(C). I guess that means I can use 14AWG EGC. What is the best way to mate the 6AWG solid bare exposed and 14AWG THWN-2 in the JBs?

I used split bolts for connecting solid #6 (connecting rails/modules to the box) to THWN #10 (running to each box)

[sunnyguy]

I'd appreciate any feedback on my one-line and three-line diagrams.

Thanks