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Thread: Sizing MCA per UL 1995 - Fan Powered HVAC units

  1. #11
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    Quote Originally Posted by david luchini View Post
    In his example, it's a #10 vs #10 conductor...same size required in both calculations.
    Yep, you are correct. Don't know where I got #12.

  2. #12
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    Quote Originally Posted by david luchini View Post
    #10 should be sufficient for an MCA of 31.6A.
    Yes, but the breaker will need to be 35A. Wouldn't NEC 240.4(D)(7) require the conductors be #8?

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    Quote Originally Posted by Ohno Raccoon View Post
    Yes, but the breaker will need to be 35A. Wouldn't NEC 240.4(D)(7) require the conductors be #8?
    See 240.4(G) - Specific conductor applications.

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    Quote Originally Posted by david luchini View Post
    See 240.4(G) - Specific conductor applications.
    Got it, thank you.

    Does anyone have any input regarding the calculation of MCA to cover the "worst case line", L1? Based on the explanation from the manufacturer's rep and the provided diagram, L1 serves both the heaters and the fan; L2 serves just the heaters. I admit that my initial calculation inaccurately divides the fan load evenly across both lines.

    But I figured the load for L1 would be:
    (1523VA + 4750VA) = 6273VA
    L1-N = 22.6A at 277V
    MCA for L1-N = 1.25*22.6A = 28.3A at 277V

    And the load for L2 would be just 4750VA:
    L2-N = 17.1A at 277V
    MCA for L2-N = 1.25*17.1A = 21.4A at 277V

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    Quote Originally Posted by Ohno Raccoon View Post

    And the load for L2 would be just 4750VA:
    L2-N = 17.1A at 277V
    MCA for L2-N = 1.25*17.1A = 21.4A at 277V
    The heater current is 19.8A, not 17.1A.

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    Quote Originally Posted by david luchini View Post
    The heater current is 19.8A, not 17.1A.
    Right, the heater current is 19.8A at 480V, 1PH (L1-L2). I was suggesting the heater current between L2-N is 17.1A.

    What about the current on L1? I'm still hung up with their explanation that the current on L1 is simply 19.8A at 480V, 1PH plus 5.5A at 277V, 1PH. Is that really an accurate calculation for getting the line current on L1?

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    Quote Originally Posted by Ohno Raccoon View Post
    Right, the heater current is 19.8A at 480V, 1PH (L1-L2). I was suggesting the heater current between L2-N is 17.1A.
    It's not 17.1A. There is no heater current between L2-N (or L1-N).


    Quote Originally Posted by Ohno Raccoon View Post
    What about the current on L1? I'm still hung up with their explanation that the current on L1 is simply 19.8A at 480V, 1PH plus 5.5A at 277V, 1PH. Is that really an accurate calculation for getting the line current on L1?
    If you wanted a precise number, you would need to know the power factor of the motor. But for simplicity's sake, the current on L1 is the heater current plus the motor current.

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    You have a single phase multiwire circuit - doesn't matter if it is 277/480 or 120/240, one of the ungrounded conductors has more load connected to it than the other, at the very least that one conductor needs higher ampacity than the other. Figuring out the net KVA and then dividing by line to line volts does not reflect actual load on each conductor here.
    I live for today, I'm just a day behind.

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    Quote Originally Posted by david luchini View Post
    It's not 17.1A. There is no heater current between L2-N (or L1-N).




    If you wanted a precise number, you would need to know the power factor of the motor. But for simplicity's sake, the current on L1 is the heater current plus the motor current.
    Thanks for the reminder. Yes, when you put it that way it makes sense. I was having a mental block adding current ratings across different operating voltages.

    Quote Originally Posted by kwired View Post
    You have a single phase multiwire circuit - doesn't matter if it is 277/480 or 120/240, one of the ungrounded conductors has more load connected to it than the other, at the very least that one conductor needs higher ampacity than the other. Figuring out the net KVA and then dividing by line to line volts does not reflect actual load on each conductor here.
    Yes, I understand and admitted that taking the net KVA and dividing by line to line volts was not accurate.

    Thanks for the input everyone.

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