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Thread: Protection of Conductors under 705.12.(D) (2)

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    Protection of Conductors under 705.12.(D) (2)

    I have a plan the has a 100A disconnect wired with #1 in between of a 200A protected Tesla gateway (on the utility side of the disconnect) and a 125A breaker feeding out of a panel recieving the input from 2-30A Tesla Powerwalls and a 35A Inverter breaker (solar input side of the disconnect). 200A potential on one side and 125A on the other. If any conductor grounded out (unlikely in the 3' segment of the system) the potential would be 300A +/- to ground over a #1 THWN and thru a 100A rated disconnect. Shouldn't these components be protected to a 325 ++Amp level (125 + 200)x125%=? )
    Thanks

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    Your description of the proposed arrangement is a bit unclear, perhaps you could post a one-line diagram? ASCII art can work for that.

    If the only protection on the utility side of the #1 copper feeder is a 200A breaker, then the feeder would have to comply with the feeder tap rules in 240.21(B). Is the 100A disconnect fused or unfused?

    As for the utility interactive inverters, the sum of the breakers is only 95A, so the #1 copper feeder should have adequate ampacity for them (barring unusual adjustment factors).

    How are the loads connected to this system? If via breakers in the panel with the 125A main and the utility interactive inverter connections, that panel needs to have a 200A bus with proper breaker location, or a 225A bus.

    Cheers, Wayne

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    To clarify.

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    Quote Originally Posted by Dcoon@cityofsparks.us View Post
    I have a plan the has a 100A disconnect wired with #1 in between of a 200A protected Tesla gateway (on the utility side of the disconnect) and a 125A breaker feeding out of a panel recieving the input from 2-30A Tesla Powerwalls and a 35A Inverter breaker (solar input side of the disconnect). 200A potential on one side and 125A on the other. If any conductor grounded out (unlikely in the 3' segment of the system) the potential would be 300A +/- to ground over a #1 THWN and thru a 100A rated disconnect. Shouldn't these components be protected to a 325 ++Amp level (125 + 200)x125%=? )
    Thanks
    I attached an image but it appears to small to identify the features. The disconnect is not fused. I will investgate the tap rule and watch for additional post.
    Thanks for your help.

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    Quote Originally Posted by Dcoon@cityofsparks.us View Post
    I have a plan the has a 100A disconnect wired with #1 in between of a 200A protected Tesla gateway (on the utility side of the disconnect) and a 125A breaker feeding out of a panel recieving the input from 2-30A Tesla Powerwalls and a 35A Inverter breaker (solar input side of the disconnect). 200A potential on one side and 125A on the other. If any conductor grounded out (unlikely in the 3' segment of the system) the potential would be 300A +/- to ground over a #1 THWN and thru a 100A rated disconnect. Shouldn't these components be protected to a 325 ++Amp level (125 + 200)x125%=? )
    Thanks
    No. A grid tied inverter cannot feed a short. In the event of a fault the inverter will shut down and the conductors will only see fault current from the service.

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    I agree with that the wiring doesn't need to be sized for 325A, and that the tap rule may apply. But the plan does seem a bit weird to me. I'm not sure why the feed through the disconnect to the PV/Battery panel is not simply connected at the opposite end of the 200A distribution panel with a 100A breaker. And I'm not sure why it's main is 125A instead of 100. That's how I'd probably have done it. Maybe there are physical location constraints.

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    Quote Originally Posted by ggunn View Post
    No. A grid tied inverter cannot feed a short. In the event of a fault the inverter will shut down and the conductors will only see fault current from the service.
    The NEC has always required that the fault current from grid tied inverters be used in all calculations. True, the inverter should shut down, but the NEC assumes it's going to supply its maximum current to a fault and size accordingly. But the utility fault current is usually so much larger than the inverter current that the rating based on the utility fault current will give a larger conductor so the inverter contribution can be ignored, in most cases.
    Last edited by pv_n00b; 01-17-18 at 10:04 PM.

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    Quote Originally Posted by Dcoon@cityofsparks.us View Post
    I have a plan the has a 100A disconnect wired with #1 in between of a 200A protected Tesla gateway (on the utility side of the disconnect) and a 125A breaker feeding out of a panel recieving the input from 2-30A Tesla Powerwalls and a 35A Inverter breaker (solar input side of the disconnect). 200A potential on one side and 125A on the other. If any conductor grounded out (unlikely in the 3' segment of the system) the potential would be 300A +/- to ground over a #1 THWN and thru a 100A rated disconnect. Shouldn't these components be protected to a 325 ++Amp level (125 + 200)x125%=? )
    Thanks
    The 100A disconnect is undersized. It's protected from the utility by 200A OCPDs so it needs to be rated for 200A.

    When rating conductors like those on either side of the 100A disconnect for fault current you do the calculations twice, once for fault current from the utility and once for fault current from the PV+Storage side. Choose the larger of the two conductors you get. You do not need to add the fault currents together and size for that since at any point in the conductor under fault it will only have to carry fault current from one side. Only the point of fault will carry the sum of the currents.

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    Quote Originally Posted by pv_n00b View Post
    The NEC has always required that the fault current from grid tied inverters be used in all calculations. True, the inverter should shut down, but the NEC assumes it's going to supply its maximum current to a fault and size accordingly. But the utility fault current is usually so much larger than the inverter current that the rating based on the utility fault current will give a larger conductor so the inverter contribution can be ignored, in most cases.
    No. The NEC requires the wiring be sized to handle 125% of the maximum inverter output under normal operating conditions and sized to be protected from fault current from the service by the OCPD under fault conditions. Nowhere does it require the wiring to handle both at the same time. It's a moot point, anyway, if you consider that Kirchoff tells us that in the case of a point fault no part of the conductor could see the sum of the two currents.

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    Quote Originally Posted by ggunn View Post
    No. The NEC requires the wiring be sized to handle 125% of the maximum inverter output under normal operating conditions and sized to be protected from fault current from the service by the OCPD under fault conditions. Nowhere does it require the wiring to handle both at the same time. It's a moot point, anyway, if you consider that Kirchoff tells us that in the case of a point fault no part of the conductor could see the sum of the two currents.
    I've had to do too many C&I systems where the PV system fault contribution has to be included with the utility fault contribution when calculating the available fault current to equipment to believe this. You are talking about designing to normal operating current requirements not fault calculations. Completely different part of the design. When you choose OCPD they have to be rated to break the available fault current from all sources feeding through them. Since the utility fault current is on the order of tens of kA and the inverter fault current is only around 1.2* Imax of the inverter it's so small that in most systems it can be safely ignored. Most, but not large commercial or utility scale systems.
    Last edited by pv_n00b; 01-23-18 at 03:50 PM.

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    Quote Originally Posted by pv_n00b View Post
    I've had to do too many C&I systems where the PV system fault contribution has to be included with the utility fault contribution when calculating the available fault current to equipment to believe this. You are talking about designing to normal operating current requirements not fault calculations. Completely different part of the design. When you choose OCPD they have to be rated to break the available fault current from all sources feeding through them. Since the utility fault current is on the order of tens of kA and the inverter fault current is only around 1.2* Imax of the inverter it's so small that in most systems it can be safely ignored. Most, but not large commercial or utility scale systems.
    But in the OP's example, the disconnect wouldn't have to interrupt both fault currents at the same time, would it? When would that happen?

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