X
 
  • Time
  • Show
Clear All
new posts

  • HX_Guy
    replied
    Originally posted by sensij
    You've almost got it. It looks like these documents are a newer revision than the code compliance document I listed earlier, and eliminate the alternatives for calculating the DC output current. 15 A makes sense to use as the rated output.

    You only need the 1.25 * 1.25 correction when looking at PV Source and PV Output circuits [690.8(A) and 690.8(B)(1)(a)]. Look at the document SolarEdge just provided you. The PV Source circuit is between the panels and the optimizers. The output of the optimizers is a DC output circuit, and only needs a single 1.25 correction, even when there are multiple strings combined. Because this wasn't clear in 2011 NEC, some additional definitions were added in 2014 NEC that help. However, the interpretation should not have changed.
    Yeah, but my city is using 2011 NEC and apparently, at least according to the designer, is using the 1.25 twice for DC. I guess they could be educated on the matter but still the numbers are pretty high.

    If you have 4 strings, that would need to be calculated as 60A x 1.25 = 75A. A #2 wire on the roof would be rated for up to 84.5A after temp correction (based on 130A @ 90º C). Makes sense why the permit says 2AWG then I guess.

    If I were to redo my permits, a #6 would still work apparently though, but I would have to show the new string configurations and everything, and hope they understand to use 1.25 only once, not twice. I don't have a lot of faith in that happening. Though with the conduit now in the attic, the derate is much lower (is it the standard .82 or something else?) and it may still work even with doing 1.25 twice.

    Leave a comment:


  • sensij
    replied
    Originally posted by HX_Guy
    Ok, so even SolarEdge says to use the maximum output of the optimizers for sizing wires, regardless of what the inverter input is.
    Well the maximum output of the inverters is 15A per string. With the original 4 strings, that would be 60A that need to be accounted for, and since it's DC, the city wants to see x1.25 and 1.25 again...so you've got 60A x 1.25 x 1.25 = 93.75A, and that's before any temperature correction!! Am I understanding that right?

    Now for my system of 3 strings, it would be 45A x 1.25 x 12.5 = 70.31A which is still astronomical. Seems the only way to satisfy a normal size conductor would be with 2 strings, or 30A total which would give you 46.88A before temp corrections.
    You've almost got it. It looks like these documents are a newer revision than the code compliance document I listed earlier, and eliminate the alternatives for calculating the DC output current. 15 A makes sense to use as the rated output.

    You only need the 1.25 * 1.25 correction when looking at PV Source and PV Output circuits [690.8(A) and 690.8(B)(1)(a)]. Look at the document SolarEdge just provided you. The PV Source circuit is between the panels and the optimizers. The output of the optimizers is a DC output circuit, and only needs a single 1.25 correction, even when there are multiple strings combined. Because this wasn't clear in 2011 NEC, some additional definitions were added in 2014 NEC that help. However, the interpretation should not have changed.

    Leave a comment:


  • HX_Guy
    replied
    Alright guys I have some fresh new content here this morning from SolarEdge.
    SolarEdge send me a few PDFs on how their system is supposed to be calculated and items that I should be able to send to the city if I redo the permit.


    Calculating Vmp, Imp, Voc and Isc

    Voc and Isc Explained

    Concept of Operation

    So, it seems what the City really cares about here is Isc, the other ones really don't matter as far as I can tell in figuring out what conductor size to use.
    Looking at this though, it seems ever more confusing however than before because in the Voc/Isc Explained document able, it specifically says:

    "For cabling calculations related to maximum current (i.e. wire gauges), use the maximum output current as appear in the power optimizer data sheet"

    Additionally, in the Calculation PDF, it says:

    "The maximum current value of 15 amps should be used to determine DC output circuit conductor size and overcorrect protection requirements. The value of this labeling requirement should be 15 amps per string of power optimizers. Since SolarEdge inverters can be fully loaded with either 1 or 2 strings, the value would be either 15 or 30 amps."

    How is that in any way clear for the City?

    So even SolarEdge says to use the maximum output of the optimizers for sizing wires, regardless of what the inverter input is.
    Well the maximum output of the inverters is 15A per string. With the original 4 strings, that would be 60A that need to be accounted for, and since it's DC, the city wants to see x1.25 and 1.25 again...so you've got 60A x 1.25 x 1.25 = 93.75A, and that's before any temperature correction!! Am I understanding that right?

    Now for my system of 3 strings, it would be 45A x 1.25 x 12.5 = 70.31A which is still astronomical. Seems the only way to satisfy a normal size conductor would be with 2 strings, or 30A total which would give you 46.88A before temp corrections.

    Leave a comment:


  • HX_Guy
    replied
    Originally posted by Ben25
    How many strings do you have? (I'm guessing 3?)The Solaredge inverter will accept 2 strings, so you could just combine 2 of them on the roof and run one down separately... That would take another revision, but get you away from 2awg.

    Ben
    Hey Ben, that's correct, I have 3 strings (16, 13 & 11 panels per). I cannot go to 2 strings because the maximum power allowed per string is 5250W and my system is over 10,500W.

    Leave a comment:


  • Ben25
    replied
    How many strings do you have? (I'm guessing 3?)The Solaredge inverter will accept 2 strings, so you could just combine 2 of them on the roof and run one down separately... That would take another revision, but get you away from 2awg.

    Ben

    Leave a comment:


  • HX_Guy
    replied
    Originally posted by Living Large
    I suggested scraping off more paint in one of the pics you posted a few days ago. Don't know if this is the same one.
    Yep, same one.

    Leave a comment:


  • Living Large
    replied
    Originally posted by HX_Guy
    she also mentioned needs to have more paint scraped off (but she was ok with the green ground screw).
    I suggested scraping off more paint in one of the pics you posted a few days ago. Don't know if this is the same one.

    Leave a comment:


  • HX_Guy
    replied
    Yeah, I caught it too and edited my last post.

    So looks like what's already in place is fine, now the question is what is the easiest route to get it resolved...try to teach the city or just comply with what they want?

    By the way, the inspector also said we need a bond bushing for the DC disconnect. Is that accurate? (See arrow)

    She also didn't like the the double lugged ground wire I ran in the AC disconnect, I think I remember you guys warning me about that. She said she needs paperwork that that's allowed or we need to loop the ground wire that's already there around that ground screw, which she also mentioned needs to have more paint scraped off (but she was ok with the green ground screw).

    Leave a comment:


  • sensij
    replied
    Originally posted by HX_Guy
    Either I'm too tired here or not sure why I'm not grasping it.

    So you take the inverter max output, 47.5A in this case, and multiply by 1.25 which give you 59.38A.
    Then separately calculate the derated wire capacity, in this case 65A x .82 = 53.3A.

    What am I missing?

    Base ampacity for 6 AWG THWN-2 is 75 A, not 65 Amp. 75 A * 0.82 = 61.5 A

    When you are look at the terminations at 75 deg, you do not temp correct, you just make sure it is higher than the conductor. Termination ampacity is 65 A, which is higher than 61.5 A and is OK.

    Leave a comment:


  • HX_Guy
    replied
    Originally posted by sensij
    Seriously? The example is exactly the calculation I just told you to do.

    the base ampacity of 8 AWG THWN-2 is 55 A. The temp correction they used is 82%, which was looked up from the table cited. 55 * 0.82 = 45.1 A

    The load *1.25 must be less than that 45.1 A.

    With respect to the termination, 8 AWG @ 75 deg is 50 A. Because the termination ampacity is higher than the corrected conductor ampacity, it is OK.
    Either I'm too tired here or not sure why I'm not grasping it.

    So you take the inverter max output, 47.5A in this case, and multiply by 1.25 which give you 59.38A.
    Then separately calculate the derated wire capacity, in this case 65A x .82 = 53.3A.

    What am I missing?

    EDIT: Wait, I think I caught my own mistake. When calculating the conductor size, you go by the maximum amp rating of the wire? Not by what the terminations are (which would be 75º)?

    If that's the case, then it would be...

    47.5A x 1.25 = 59.38A
    75A x .82 = 61.5A

    61.5A > 59.38A so all is good?

    Leave a comment:


  • sensij
    replied
    Originally posted by HX_Guy
    I guess I can see how the city could be struggling to accept this then. I mean the current that the source is rated to produce is what the inverter tell is to. If the inverter isn't working properly, then the optimizers will put out 1V. They won't work unless the inverter is working properly. And if they speculate a what if scenario where they think the optimizers will put out power without the inverter..then the calculation would be insane (they will put out up to 85V! and 15A). And if they further think what if the inverter and the optimizers fail, and somehow power direct from the panels makes its way through the conductors to the inverter, then I guess you would use the panels specs for all the number...but that is impossible to ever happen.
    The SolarEdge code compliance document is very clear about how to calculate the inverter input current. Presumably, they are not making up wild numbers, but are giving realistic failure scenarios.

    Leave a comment:


  • sensij
    replied
    Originally posted by HX_Guy
    Well, that may be the case but I have a feeling my city doesn't know or won't accept that. Apparently my designer did have to account for temp corrections and as I posted in the other thread, I have a copy of a permit a friend of mine submitted for his solar and it was also accounted for on there too.

    Seriously? The example is exactly the calculation I just told you to do.

    the base ampacity of 8 AWG THWN-2 is 55 A. The temp correction they used is 82%, which was looked up from the table cited. 55 * 0.82 = 45.1 A

    The load *1.25 must be less than that 45.1 A.

    With respect to the termination, 8 AWG @ 75 deg is 50 A. Because the termination ampacity is higher than the corrected conductor ampacity, it is OK.

    Leave a comment:


  • inetdog
    replied
    Originally posted by HX_Guy
    I guess I can see how the city could be struggling to accept this then. I mean the current that the source is rated to produce is what the inverter tell is to. If the inverter isn't working properly, then the optimizers will put out 1V. They won't work unless the inverter is working properly.
    Yes and no....

    They will not work unless the inverter keeps giving them command updates. But it does not have to be working properly.
    Also, if there is a medium resistance short circuit along the wire, the inverter will command the optimizers based on the current it is receiving, not the optimizer output. Whether that scenario is actually possible depends on the details of the interaction between inverter and optimizers.

    Leave a comment:


  • HX_Guy
    replied
    Originally posted by sensij
    Yes, correct.

    It could be higher if the inverter fails to limit the current properly. That is why code looks at the current that the source is rated to produce.
    I guess I can see how the city could be struggling to accept this then. I mean the current that the source is rated to produce is what the inverter tell is to. If the inverter isn't working properly, then the optimizers will put out 1V. They won't work unless the inverter is working properly. And if they speculate a what if scenario where they think the optimizers will put out power without the inverter..then the calculation would be insane (they will put out up to 85V! and 15A). And if they further think what if the inverter and the optimizers fail, and somehow power direct from the panels makes its way through the conductors to the inverter, then I guess you would use the panels specs for all the number...but that is impossible to ever happen.

    Leave a comment:


  • HX_Guy
    replied
    Originally posted by sensij
    You do not need #4 wire when you do the calculation correctly. Temperature corrections do not apply to the load, they apply to the base ampacity of the conductor. Every time you do something like load * 1.25 / (temp correction) you will get the wrong answer. Load gets multiplied by 1.25. Temp correction gets applied to the base ampacity of the conductor. Compare the two.

    The termination ampacity must be greater than (1.25 * the load), no temp corrections are required.
    Well, that may be the case but I have a feeling my city doesn't know or won't accept that. Apparently my designer did have to account for temp corrections and as I posted in the other thread, I have a copy of a permit a friend of mine submitted for his solar and it was also accounted for on there too.

    Leave a comment:

Working...