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    donf's Avatar
    donf Posts: 5,679, Reputation: 582
    Printers & Electronics Expert
     
    #21

    May 26, 2008, 12:41 PM
    Ron,

    A grounded conductor is still a conductor as was pointed out by Stan. You have to count the three conductors? Wouldn't you considered the 120 to be a derivative of the 240?

    The 120 VAC represents a line to Neutral, the 240 is the Line to Line connection.

    What is your formula for Ed?
    Washington1's Avatar
    Washington1 Posts: 798, Reputation: 36
    Senior Member
     
    #22

    May 26, 2008, 05:49 PM
    Don,

    Formula method:
    2KId/CM
    2=Two lenghts of wire to the load
    K=12.9 for copper
    I=Amperage
    D=Distance
    CM=Circular Mil

    You can also use Ohm's law to get VD.
    Yet, in this case you would need the conductors resistance.

    The formula would be:
    I(R)=VD

    Voltage drop is only recommended by the NEC, but is not required. What's recommended:
    -Max combined volatge drop for both feeder and branch circuit should not exceed 5%
    -Max on the feeder or branch circuit should not exceed 3%

    So this means. The conductor volatge (feeder only) drop on a 120V source would be: 120(.03)=3.6V then 120-3.6=116v recommended for the feeder.
    If 240, then: 240(.03)=7.2V then 240-7.2=233V recommended
    stanfortyman's Avatar
    stanfortyman Posts: 5,598, Reputation: 279
    Electrical & Lighting Expert
     
    #23

    May 26, 2008, 06:16 PM
    I use Lectri-Calc for voltage drop.

    Much as I like math, the math to figure VD it too tedious for me. :D
    KISS's Avatar
    KISS Posts: 12,510, Reputation: 839
    Uber Member
     
    #24

    May 26, 2008, 07:58 PM
    A lot of simplifications come from 2 basic equations:

    R = pL/A; Resistance = Resistivity*Length*(Cross sectional area) and
    V= IR

    p=16.8 nΩ-m; yep nano-Ohms-meter
    KISS's Avatar
    KISS Posts: 12,510, Reputation: 839
    Uber Member
     
    #25

    May 26, 2008, 10:26 PM
    Don:

    This should give you an intuitive idea as to why you would not consider the conductors as 3. I made the math EASY by saying that each conductor has 1 ohm of resistance.
    Note that when a 240 volt load is connected you are dissipating 800 W and when a single 120 V load is connected your also dissipating 800 W. And when you allocate between all three conductors only 350 watts are dissipated.

    So, you can hopefully see that you must use 3% of 120 V because of case 2 which is the worst 120 V load case. In case 1, you would take 3% of 240, Case 2 and 3, 3% of 120.
    In case #1, assume the neutral wire isn't run.

    .... Amp...0hms......Power (I^2*R)
    L1... 20... 1... 400
    L2... 20... 1... 400
    N... 0... 1... 0
    Ptotal......................800 W Case 1

    L1... 20... 1... 400
    L2... 0... 1... 0
    N... 20... 1... 400
    Ptotal......................800 W Case 2

    L1... 15... 1... 225
    L2... 5... 1... 25
    N... 10... 1... 100
    Ptotal .......................350 W Case 3

    What this says is that with a 20 A protection device on a dual breaker, you cannot exceed 800 W.

    Remember used 1 ohm to make the math very EASY.

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