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Thread: Can someone please explain this to me?

  1. #11
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    Quote Originally Posted by Carultch View Post
    Bear in mind that when you get questions like this, the voltage drop information could be a red herring. You could go through your voltage drop calculation and determine a size of conductor, that is insufficient for the circuit amps. Sometimes the voltage drop might be the factor that governs the answer, other times ampacity may govern it instead. In this case, all sizes indicated are sufficient for 30A noncontinuous, so voltage drop is the only calculation that will matter.

    For lack of additional information, we assume it is a single phase circuit, and we assume copper wires. Which means the total round trip length of the circuit, is twice the distance given. Call the given length L. The round trip length is 2*L.

    The wire has a resistance per unit length. Call it r. The total resistance (R) can be calculated with R = 2*L*r. This is a value in the lookup tables in units of thousand feet (kft) and kilometers, so you will often divide by 1000 to take care of this.

    Call the operating current, in units of amps, I.

    Construct Ohm's Law for voltage drop (dV):
    dV = I*R
    dV = I*2*r*L

    Solve for r:
    r = dV/(2*I*L)

    The NEC recommends (not requires, but recommends) not exceeding 3% voltage drop. For dV, plug in 3% of 230V. Plug in 30A for I, and plug in 100 ft for L. Our result is 0.0015 Ohms/ft, which translates to 1.15 Ohms/kft.

    We need a wire resistance per unit length that is does not exceed 1.15 Ohms/kft, which is #8 at 0.778 Ohms/kft.
    0.0015 Ohms/ft = 1.5 Ohms per kft, not 1.15. #10 has .999 Ohms/1kft
    Electricians do it until it Hertz!

  2. #12
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    Quote Originally Posted by JFletcher View Post
    I get #10 as well. There are a few ways I can think of that they got #8 tho:

    1) Total recommended feeder and branch circuit drop is 5%. If you split that equally amongst the 2, you would need #8 to be below 2.5% VD for the branch circuit; #10 gets you to 2.63% per the Southwire calculator:

    1 conductors per phase utilizing a #10 Copper conductor will limit the voltage drop to 2.63% or less when supplying 30.0 amps for 100 feet on a 240 volt system.

    2) The load is a continuous load, so you would be limited to 80% of the breaker's capacity. You cant put #10 on a 40A breaker per 240.4(D) unless it's a motor, thus the need for #8.

    3) The answer is wrong.

    Changing the input variables from 230 or 240V, buried or conduit, even 1 or 3 phase, dont appreciable change the numbers you get. Making the run length 200' jumps you to #6 conductor. The only way you get #8 with the values given is if you use Al wire.

    eta: given the typo in the OPs question ("be" vs "by"), it isnt directly copy/pasted, thus there may also be a key piece of information missing.
    I agree with the 5%. They must assume a voltage drop already existing on the feeder.

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  3. #13
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    In my opinion, it is about aluminum conductor single_phase 230 V
    From NEC 2014 Tb 9 for10# Z=1.8 and for #8 Z=1.1
    If DV=2*Z*30*100/1000 for #10 will be 10.4 V [4.5%] and for #8 will be 6.8 V 2.95%
    As per art.210.19 (1) Informative note 4 3% it is recommended.

  4. #14
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    Funny, I dont buy the Al wire or VD % arguments I made. Nobody these days is pulling small ga Al wire and 2.xx% VD is still under 3% (which is NOT a Code req). I'm more inclined to go with continuous load, even tho that is not stated either. That or #3, wrong answer.

    If the question were:

    "A 30 amp 230 volt continuous 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?

    or

    "A 30 amp 230 volt lighting 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?

    or

    "A 30 amp 230 volt heating 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?

    the answer would be #8 no problem. One word difference. As I mentioned before, could be a copy/paste or memory error from the OP.
    Electricians do it until it Hertz!

  5. #15
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    Quote Originally Posted by JFletcher View Post
    Funny, I dont buy the Al wire or VD % arguments I made. Nobody these days is pulling small ga Al wire and 2.xx% VD is still under 3% (which is NOT a Code req). I'm more inclined to go with continuous load, even tho that is not stated either. That or #3, wrong answer.

    If the question were:

    "A 30 amp 230 volt continuous 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?

    or

    "A 30 amp 230 volt lighting 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?

    or

    "A 30 amp 230 volt heating 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?

    the answer would be #8 no problem. One word difference. As I mentioned before, could be a copy/paste or memory error from the OP.
    I agree the question isn't complete. Need a little bit more information to make an accurate decision. Once a person told me that Voltage drop wasn't in forcible because it was a fine print note! That was about 8 years ago VA was probably under the 2002 code at the time.

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  6. #16
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    Quote Originally Posted by JFletcher View Post
    0.0015 Ohms/ft = 1.5 Ohms per kft, not 1.15. #10 has .999 Ohms/1kft
    #10 Cu wire has 1.24 Ohms/kft at 75C in the Chapter 9 table 8. I see my mistake, and now I notice that #10 is sufficient for voltage drop.


    The only explanation as to why you might need #8 Cu in this example, is if it is a continuous load (or other load requiring a safety factor) which would thus be connected to a 40A circuit. Something that should be specified.
    Last edited by Carultch; 08-07-17 at 10:26 AM.

  7. #17
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    Quote Originally Posted by Carultch View Post
    #10 Cu wire has 1.24 Ohms/kft at 75C in the Chapter 9 table 8. I see my mistake, and now I notice that #10 is sufficient for voltage drop.


    The only explanation as to why you might need #8 Cu in this example, is if it is a continuous load (or other load requiring a safety factor) which would thus be connected to a 40A circuit. Something that should be specified.
    Or if you had an existing 3% drop on service feeder.than you could only alow 2% on this ckt.

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  8. #18
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    Quote Originally Posted by Carultch View Post
    #10 Cu wire has 1.24 Ohms/kft at 75C in the Chapter 9 table 8. I see my mistake, and now I notice that #10 is sufficient for voltage drop.
    Humm, ....

    VD=I*R; I=30; R=1.24*.2=0.248

    VD= 30*0.248 = 7.44

    230*3% = 6.9

    So, according to the above, #10 is too small.

    Where did someone find that #10 has a smaller resistance than 1.24 ohms/1,000'?

    Where was your mistake? I figure I just made the same one.

  9. #19
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    Quote Originally Posted by oldsparky52 View Post
    Humm, ....

    VD=I*R; I=30; R=1.24*.2=0.248

    VD= 30*0.248 = 7.44

    230*3% = 6.9

    So, according to the above, #10 is too small.

    Where did someone find that #10 has a smaller resistance than 1.24 ohms/1,000'?

    Where was your mistake? I figure I just made the same one.
    My earlier post should've said "Our result is 0.00115 Ohms/ft, which translates to 1.15 Ohms/kft.". This would make #10 too small to maintain 3% voltage drop, thus requiring #8.

  10. #20
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    Quote Originally Posted by JFletcher View Post
    0.0015 Ohms/ft = 1.5 Ohms per kft, not 1.15. #10 has .999 Ohms/1kft
    Where did the .999 Ohms/1,000' come from?

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