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Thread: Effective Current calculation

  1. #11
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    180115-1043 EST

    I have to assume each 30 A outlet has its own breaker or fuse.

    .

  2. #12
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    If the marina is going to be busy, you really need to keep the Information Note in mind.
    555.12 ... Informational Note: These demand factors may be inadequate in areas of extreme hot or cold temperatures with loaded circuits for heating, air-conditioning, or refrigerating equipment.
    Don, Illinois
    Ego is the anesthesia that deadens the pain of stupidity. Dr. Rick Rigsby
    (All code citations are 2017 unless otherwise noted)

  3. #13
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    Quote Originally Posted by gar View Post
    180115-0920 EST

    JB78:

    Nothing makes particular sense from what you have said.

    The wording "effective" current is usually used in a classroom setting to try to have a student understand that for a particular AC waveform that a simple multiplier can be used to relate that AC waveform to a DC current based on having the same heating effect in a resistor. As the AC waveform changes that multiplier changes. For a sine wave a ratio is Ipeak * 0.707 = Irms. Theoretically one can calculate this ratio with calculus. The 0.707 value is an approximation of the exact value of sq-root of 2 divided by 2.

    In general I would use the words "effective" and RMS as being identical.

    If you have a split phase system based on a center tapped 240 V secondary, your loads are 120 V, and you try to approximately balance the loads on you source, then one phase is loaded to 4*30 A and the other to 5*30 A. So you would use an RMS load current of 150 A. The 150 is the greater of the two split loads and will determine breaker and wire sizes.

    .
    I agree. It doesn't make sense. I have never seen this done before. My main reason for questioning it, along with the calculated load being reduced so far. You mentioned splitting the loads for 120v. Maybe that has something to do with it?? One other thing about this pedestal feeder: as the chart indicates, it is (9) 120/240v 30a receptacles for the load calculation. But each of the (5) pedestals has a 20a/120v GFCI receptacle that is dropped from the calculation, as allowed by NEC table 555.12, note-1. Any thoughts on this additional info?
    Again, I want to understand why he chose this method, as I may have to split-up or add some circuits and want to stay consistent.

    Thank you,
    JB78

  4. #14
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    how many pedestals?
    how many and what type of recept on each ?

    where does the 9 come from if there are 5 pedestals?
    The difference between genius and stupidity is that genius has its limits.

  5. #15
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    Are these 240 V receptacles? If so then 9*30 would be the number to use. But, still if the source is 270 A, and the receptacles are 30 A, then protection is needed at each receptacle.

    .

  6. #16
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    Quote Originally Posted by gar View Post
    180115-1229 EST

    Are these 240 V receptacles? If so then 9*30 would be the number to use. But, still if the source is 270 A, and the receptacles are 30 A, then protection is needed at each receptacle.

    .
    Right. Five pedestals on this circuit. Four of them have (2) 30a, 120/240v recept. and (1) 20a/120v. The fifth pedestal has (1) 30a, 120/240v and (1) 20a/120v. That's where we get (9) 30a, 120/240v for calculations - the 20a receptacles are not counted (table 555.12, note-2). All receptacles have individual OC protection in the pedestal.

    JB78

  7. #17
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    he is only using 1 30 from each pedestal
    not 2 if it has 2
    5 x 30 x 0.8 x 0.9 = 108 A

    the note is not clear
    it says use the largest
    it does not address if you have 2 'largest'
    The difference between genius and stupidity is that genius has its limits.

  8. #18
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    Quote Originally Posted by Ingenieur View Post
    he is only using 1 30 from each pedestal
    not 2 if it has 2
    5 x 30 x 0.8 x 0.9 = 108 A

    the note is not clear
    it says use the largest
    it does not address if you have 2 'largest'
    Although it does not say it explicitly, my impression is that the reason for only counting the largest circuit is the assumption that if there are two circuits of different size per pedestal that only one of them will be used at a time and the two circuits are to accommodate different connection needs for different boats, not that two boats will be connected to each pedestal.

    If the OP's design is for 35 pedestals for a 70 boat marina, then he would have to count both before applying any diversity factor.

  9. #19
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    Quote Originally Posted by GoldDigger View Post
    Although it does not say it explicitly, my impression is that the reason for only counting the largest circuit is the assumption that if there are two circuits of different size per pedestal that only one of them will be used at a time and the two circuits are to accommodate different connection needs for different boats, not that two boats will be connected to each pedestal.

    If the OP's design is for 35 pedestals for a 70 boat marina, then he would have to count both before applying any diversity factor.
    that is logical
    The difference between genius and stupidity is that genius has its limits.

  10. #20
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    Quote Originally Posted by Ingenieur View Post
    that is logical
    And therefore probably not relevant for an analysis of the NEC.


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