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690.12 Rapid Shutdown question

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  • 690.12 Rapid Shutdown question

    I couldnt find an exact answer through search.

    As I understand it, the rapid shutdown device can be eliminated by placing the central inverter within 10 feet of an array. To me this means 10 feet radial. This makes sense to me (since firefighters are not going to sit there with a tape to get exact numbers) but I seem to find another interpretation on a few websites which think that the 10 feet is the length of conduit or DC wire from the array (thus not a radial distance). Anyone have any experience dealing with this? Which one is right?

    Another question, what happens when there are two arrays on my roof? One on one side of a roof and one on a the little roof above my garage? To me, the 10 foot rule still means 10 feet from AN (either or) array. But one site suggests that it is 10 feet from BOTH arrays which is harder to comply with even with rapid shutdown devices.

    Thanks in advance.

  • #2
    For the first question, whatever your AHJ says is right, is right. The text of 2014 NEC 690.12(1) is

    (1) Requirements for controlled conductors shall apply only to PV system conductors of more than 1.5 m (5 ft) in length inside a building, or more than 3 m (10ft) from a PV array.
    For the 2nd question, if you have the rapid shutdown box located within 10 ft of any edge of the array, all of the necessary wires get shut down. It isn't that the box itself needs to be located within 10 ft of both arrays, you just need to make sure all of the conductors outside the 10 ft envelope are shut down.

    Note that 2017 NEC drops the 10 ft requirement to just 1 ft.
    Last edited by sensij; 09-13-2017, 08:04 PM.
    CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

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    • #3
      Originally posted by sensij View Post
      For the first question, whatever your AHJ says is right, is right. The text of 2014 NEC 690.12(1) is

      For the 2nd question, if you have the rapid shutdown box located within 10 ft of any edge of the array, all of the necessary wires get shut down. It isn't that the box itself needs to be located within 10 ft of both arrays, you just need to make sure all of the conductors in the 10 ft envelope are shut down.

      Note that 2017 NEC drops the 10 ft requirement to just 1 ft.
      Okay so its conductors. But I am still a bit confused about the 2nd part. So the way I am drawing the systems right now, the inverter is going to be about 8 feet of conduit from the 1st array (on top of garage) but about 15 feet from the edge of the 2nd array (on second floor). Does this mean I am ok or do I need a device between them?

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      • #4
        Originally posted by thejumpingsheep View Post

        Okay so its conductors. But I am still a bit confused about the 2nd part. So the way I am drawing the systems right now, the inverter is going to be about 8 feet of conduit from the 1st array (on top of garage) but about 15 feet from the edge of the 2nd array (on second floor). Does this mean I am ok or do I need a device between them?
        All conductors outside a 10 ft envelope around the array need to be shutdown. If there are conductors running from the 2nd array directly to the inverter by a path that does not contain them within 10 ft envelope of either array (or if they run more than 5 ft in the interior), they would need to be controlled by a rapid shutdown box.
        Last edited by sensij; 09-13-2017, 08:03 PM.
        CS6P-260P/SE3000 - http://tiny.cc/ed5ozx

        Comment


        • #5
          Originally posted by sensij View Post

          All conductors outside a 10 ft envelope around the array need to be shutdown. If there are conductors running from the 2nd array directly to the inverter by a path that does not contain them within 10 ft envelope of either array (or if they run more than 5 ft in the interior), they would need to be controlled by a rapid shutdown box.
          Got it. Thanks again for the help.

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