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Thread: Can someone please explain this to me?

  1. #21
    Join Date
    Sep 2007
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    Omaha, NE
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    Quote Originally Posted by kefox81 View Post
    I am still studying for my WV Master Electricians Exam. I have been going along pretty good but I just hit a brick wall. Here is the question from my practice test.
    A 30 amp 230 volt load is located 100 feet from the source. What is the minimum size branch circuit conductor required to operate within the limits for voltage drop that is recommended be the NEC?
    a. # 10 conductor
    b. # 8 conductor
    c. # 6 conductor
    d. #4 conductor

    The book says the correct answer is B. I have no idea how to come to this conclusion. Can anyone help me?
    I suspect they are expecting you to use the method that uses DC-constant resistance for determining voltage drop to calculate the minimum wire area, in cmils, required.

    For single phase circuits (assumed because that information is missing in the test in the question), the equation is:

    VD = 2 x K x I x D / CM

    Rearranging the equation to solve for CM results in the equation:

    CM = 2 x K x I x D / VD

    Making assumptions for copper conductors and VD% cannot exceed 3 percent because the information is missing in the test question:

    K = 12.9 ohms
    I = 30 amps
    D = 100 feet
    VD = 230 volts x 0.3 = 6.9 volts

    Therefore:

    CM = 2 x 12.9 x 30 x 100 / 6.9 = 11217.4 cmils.

    Looking at Chapter 9, Table 8, the smallest conductor with this area is #8 AWG conductor at 16510 cmils. A #10 conductor has an area of 10380 cmils, which is too small, at least according to this calculation method.
    Jason Rohe, P.E.

  2. #22
    Join Date
    Jun 2003
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    6,601
    I get "B" or #8 if the PF = 1.

    #10 gives a 3.1% drop. #8 gives 2%.

    This is using X and R from NEC table #9 to calculate %Z. I get 1.2 ohms/1000 feet for Z for #10 wire, and 0.69 for #8 wire. (The conduit type only has a minor effect on the numbers).

    Then I use the formula:

    Vdrop Line to Line = 2x distance x amps x Z/1000.

    Then %drop = 100 x VdropL-L / 230.

    Actually, I have this all in an excel sheet.

    If you use a PF of .85 (maybe by using the values of Z in table #9) you get #10 wire.

  3. #23
    Join Date
    Jul 2017
    Location
    Western Grove, AR Newton County
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    3

    Lots to remember. Break the Question down into parts.

    kefox81,

    I read your question and the replies and moved on. You had received good replies and how to get them. After a while I realized that I might be able to offer something. You have lots to remember and 'The Test' often adds anxiety and what if you forget something?

    My suggestion is to first realize that the question is a Two Part Question. Part one is the Amp Capacity of the Conductor. The second part is the 100 Ft... or every 100 Ft. (What if you face a 200 Ft distance question?)

    If you break the question down that way, it will be easier to remember:
    30 Amps = #10 Conductor.
    Each 100 Ft, next size larger Conductor, #8.
    Answer, B would be correct for this question.

    That might be easier to remember.

    JimO

  4. #24
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    Quote Originally Posted by Little Bill View Post
    Well since the question said 100', that's what I put in.
    You've got 200 feet of conductor...... two times one hundred.

  5. #25
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    Quote Originally Posted by Little Bill View Post
    Well since the question said 100', that's what I put in.
    Once the current goes out 100 feet, it's gotta come back that 100 feet.

  6. #26
    Join Date
    May 2012
    Location
    Amissville, Virginia USA
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    Never heard that before. It's drop at the appliance not baking at source.

    Sent from my SM-T350 using Tapatalk

  7. #27
    Join Date
    May 2012
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    Amissville, Virginia USA
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    Quote Originally Posted by Daniel malack View Post
    Never heard that before. It's drop at the appliance not baking at source.

    Sent from my SM-T350 using Tapatalk
    Not back at source. Spell check got me.

    Sent from my SM-T350 using Tapatalk

  8. #28
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    Quote Originally Posted by Daniel malack View Post
    Not back at source. Spell check got me.

    Sent from my SM-T350 using Tapatalk
    There is voltage drop along both conductors. But it really just depends what the equation or calculator you are using is set up for.

    The equation I posted has a factor of 2x included. So it uses the one way length, and calculates twice the voltage drop.

    I haven't used the Mike Holt calculator, so I can't say which it expects.

    Like equation or calculator, you just have to make sure you enter the right numbers.

  9. #29
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    May 2012
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    Amissville, Virginia USA
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    What would be the concern for voltage drop on way back? It is irellavent at that point.

    Sent from my SM-T350 using Tapatalk

  10. #30
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    Quote Originally Posted by Daniel malack View Post
    What would be the concern for voltage drop on way back? It is irellavent at that point.

    Sent from my SM-T350 using Tapatalk
    No, it's not. Picture it as three resistors in series with the first and third as the conductors and the middle one as the load. The sum of the voltage drops across all three loads equals the supply voltage, therefore the voltage drop across the load is the supply voltage minus the other two.

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