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Thread: Effective Current calculation

  1. #1
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    Effective Current calculation

    I am reviewing plans for a floating marina project and need some clarification. The drawings provide a chart for the power pedestal branch circuit calculations that seems to use "effective current" to size overcurrent protection. To be brief, an example single phase 120/240v circuit has a load of 270 amps (9 x 30a receptacles). After the adjustments from NEC 555.12 (80% load factor & 90% meter factor) I arrive at 194.4 amps. The chart lists an effective current of 108 amps and a 125 amp circuit breaker. I am questioning the effective current. Is the value correct? And should that be used here?

    Thank you,
    JB78

  2. #2
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    Effective current concerns RMS, DC, and conversion formulas.

    It really does not apply here.

    I played with the numbers a bit and can get 108 from 270, none of which make sense NEC values correctly.

    I looked at table in 555 and see the .8 adjustment and I think that is correct.

    Where did you come up with this .9 for meters? Never heard of it before.

    The 108 value seems to me that some fat fingered the calculator or doubled the adjustment factor. 108 is 40% of 270.
    "Electricity is really just organized lightning." George Carlin


    Derek

  3. #3
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    looks like 80% x 30 x 9 x 50%(duty cycle) = 108 A

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    Quote Originally Posted by jumper View Post
    Where did you come up with this .9 for meters? Never heard of it before.
    Note 2 following Table 555.12.

    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

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    Quote Originally Posted by Ingenieur View Post
    looks like 80% x 30 x 9 x 50%(duty cycle) = 108 A
    The arithmetic works. But where did the 50% duty cycle come from? It's not in 555.

    Charles E. Beck, P.E., Seattle
    Comments based on 2017 NEC unless otherwise noted.

  6. #6
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    Quote Originally Posted by Ingenieur View Post
    looks like 80% x 30 x 9 x 50%(duty cycle) = 108 A
    Okay, did not see a duty cycle adjustment, but I skimmed quickly through the article.

    Quote Originally Posted by charlie b View Post
    Note 2 following Table 555.12.

    Are you implying I should pay attention to the fine print stuff?
    Durn details.....
    "Electricity is really just organized lightning." George Carlin


    Derek

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    Quote Originally Posted by charlie b View Post
    The arithmetic works. But where did the 50% duty cycle come from? It's not in 555.

    not saying it is correct but I've seen it down

  8. #8
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    Quote Originally Posted by Ingenieur View Post
    not saying it is correct but I've seen it down
    Sorry, just getting back to this now. Thanks for all the input. I was trying to attach part of the chart that I was referring to but was having difficulty. Let me elaborate a bit. I stopped just short of calculating the "effective current" (at 194.4a) in my original question. My understanding is that the calculation for "effective current" is I/sqrt(2). If that's correct, eff. current would be 137.5 amps. The person creating this chart would be way off with the 108a. Also, proving the chart further, I find that the "meter factor" of 90% was only used sometimes. Actually, if you drop the meter factor in the example I provided and just use 270a x 0.8 = 216a, it looks like he divided the current by (2), arriving at 108a, and not the square root of (2). I don't know for sure but there are (14) circuits and the pattern is the same. I am just trying to understand this chart before pointing-out any discrepancies. Does this help at all?

    Thanks,
    JB78

  9. #9
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    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.

    .
    Last edited by gar; 01-15-18 at 09:49 AM.

  10. #10
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    effective ~ root mean square or rms
    rms = peak/sqrt2 (for a sinusoidal waveform)

    but the values we are discussing are already the rms or effective values
    the sqrt2 should not factor in

    fyi https://www.raeng.org.uk/publications/other/8-rms
    Last edited by Ingenieur; 01-15-18 at 09:55 AM.

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