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Thread: Reducing AIC at a machine.

  1. #11
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    The AIC rating of the breaker has nothing to do with the available fault current, other than the breaker rating has to exceed the available fault current. You need a short circuit current study to find out how much current is actually available at the machine.
    Don, Illinois
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    (All code citations are 2017 unless otherwise noted)

  2. #12
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    you may be alright
    you may not
    you need a fault study before you do anything

    do you have any info on the service xfmr? kva, pu Z, etc?
    The difference between genius and stupidity is that genius has its limits.

  3. #13
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    Quote Originally Posted by Jraef View Post
    • AFC = Available Fault Current, also known as Short Circuit Amps (SCA); the term for how much fault current (typically in kA or thousands of amps) that are available at the line side terminals of something. It's derived from the transformer size, the capacity of the primary circuit of that transformer, the impedance of it and any other impedance values between that and the line terminals, typically the resistance of the length of cables (and anything like reactors added).

    I would add that the available fault current has to be obtained from the utility and be careful that you are not provided with the standard utility response of an infinite bus fault current based on transformer size.

    If there was an arc flash study done, the information you are seeking should be available.
    Last edited by charlie b; 01-09-18 at 05:49 PM. Reason: Repaired the quotation

  4. #14
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    I know I'll get guff over this
    op, please do not do this, no offense but you lack the fundemental understanding

    I do this as a sanity check
    start the machine, control panel, etc
    measure v at cb and at machine connection point
    messure all 3 ph and avg
    do the same for current, doesn't matter which point

    example:
    cb 480 vac
    load 470 vac
    i = 40 A
    v drop = 10
    line Z = 10/40 = 0.25 Ohm

    max avail fault i inf bus = 480/0.25 = 1920 A
    it will be less because under a fault due to the high i lowering the 480 v but Z will be the same
    depending on loads you may need to add a motor fault i contribution, I use 4 x sum(connected fla)
    The difference between genius and stupidity is that genius has its limits.

  5. #15
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    Quote Originally Posted by Ingenieur View Post
    I know I'll get guff over this
    op, please do not do this, no offense but you lack the fundemental understanding

    I do this as a sanity check
    start the machine, control panel, etc
    measure v at cb and at machine connection point
    messure all 3 ph and avg
    do the same for current, doesn't matter which point

    example:
    cb 480 vac
    load 470 vac
    i = 40 A
    v drop = 10
    line Z = 10/40 = 0.25 Ohm

    max avail fault i inf bus = 480/0.25 = 1920 A
    it will be less because under a fault due to the high i lowering the 480 v but Z will be the same
    depending on loads you may need to add a motor fault i contribution, I use 4 x sum(connected fla)
    Hmmm...
    Having the correct SCCR at the panel and having it meet or exceed the AFC is now a CODE REQUIREMENT.
    Your methodology, while probably good for a guesstimate (I'm not doing the math), is NOT going to suffice in lieu of a proper fault study and the required labeling.
    __________________________________________________ ____________________________
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  6. #16
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    Quote Originally Posted by wbdvt View Post
    [/LIST]
    I would add that the available fault current has to be obtained from the utility and be careful that you are not provided with the standard utility response of an infinite bus fault current based on transformer size.

    If there was an arc flash study done, the information you are seeking should be available.
    Good points.
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  7. #17
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    SCC is limited by impedance of the conductors between the source and the point of interest. Impedance of source factors in also.
    As mentioned it wont take all that much conductor length for 30-60 amp conductors to significantly reduce available SCC at the load end.

    Also note available current at panel in question could be more then 14 kA, but the 14kA breakers might be series rated with whatever is ahead of them.

  8. #18
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    Quote Originally Posted by Jraef View Post
    Hmmm...
    Having the correct SCCR at the panel and having it meet or exceed the AFC is now a CODE REQUIREMENT.
    Your methodology, while probably good for a guesstimate (I'm not doing the math), is NOT going to suffice in lieu of a proper fault study and the required labeling.
    depends what edition your juristiction adopted
    and I've used it to satisfy equipment supplier concerns
    the numbers are easy to validate

    assume a 500 kva, xfmr 480, Z 5% pu
    Z = 0.05 x sqrt3 480^2 / 500000 = 0.0399 ohm
    i fault = 500000/(sqrt 3 480 0.05) = 12028 A
    same as 480/0.0399 = 12028

    assume we add the 0.25 as determined above
    40 x 0.25 = 10 or 10/480 x 100 = 2.08% drop, within a common range

    ignoring the feeder to the pnl the cb is in and only using the xfmr and load ckt
    0.0399 + 0.25 = 0.2899
    i fault at load = 480/0.2899 = 1656 A < 1920 A
    Last edited by Ingenieur; 01-09-18 at 07:51 PM.
    The difference between genius and stupidity is that genius has its limits.

  9. #19
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    Quote Originally Posted by Ingenieur View Post
    depends what edition your juristiction adopted
    and I've used it to satisfy equipment supplier concerns
    the numbers are easy to validate
    Point taken. The OP is in Indiana, so they are still on the 2008 code and the latest changes wouldn't apply, unless the machine is considered "industrial" (Article 409 covers "Industrial" control panels and was introduced in the 2005 code).

    PS: Actually, the relevant wording hasn't really changed since the 2005 code. 409 requires it for "industrial controls", then defines that. But 430.8 requires it as well, for "motor controllers" in general.
    430.8 Marking on Controllers. A controller shall be marked
    with the manufacturer’s name or identification, the voltage,
    the current or horsepower rating, the short-circuit current rating,
    and such other necessary data to properly indicate the
    applications for which it is suitable.
    Last edited by Jraef; 01-09-18 at 08:02 PM.
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  10. #20
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    now to get really crazy lol
    if you know
    load 40 A
    and vdrop 10/480
    i fault (ignoring any upstream Z) = 40/(10/480) = 1920 A lol
    a pu calc using the 40 A as base

    pu Z % = prim v required to get sec rated current / rated prim v
    if pu 5% on a 12470:480...623.5 prim volt produces rated sec i..623.5/12470 = 0.05
    The difference between genius and stupidity is that genius has its limits.

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