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Thread: Lay factor Medium Voltage Cable

  1. #1
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    Lay factor Medium Voltage Cable

    Hello,

    I am trying to calculate the resistance of a concentric neutral with a helically configuration for a medium voltage conductor with copper wires.

    However, I can not find a correct formula to calculate the "lay factor."

    I wonder if some of you have made this calculation before and could help me?, I do not understand the manufacturer's explanation.

    The cable is a #1/0 AWG, EPR, 34.kV, CU

    Attached is my spreadsheet, also is attached the explanation of the manufacturer for the lay factor calculation.

    Regards,
    Attached Files Attached Files

  2. #2
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    At first the article Helically Applied Cu Wire CN it is for shield only and not for conductors.
    From Lay Direction and Length:
    m=1.0619541
    UL 1072/1995 7 Resistance
    7.1 The direct-current resistance of any length of conductor in ohms per thousand conductor feet or in ohms per conductor kilometer shall not be higher than the maximum acceptable (nominal x 1.02) resistance indicated in Tables 7.1 - 7.10 at 20°C (68°F) or at 25°C (77°F). If, as provided for in 9.2, metal-coated wires are used in only the outer layer of an uncoated copper conductor, the direct-current resistance of the wires are used in only the outer layer of an uncoated copper conductor, the direct-current resistance of the construction.
    If you are a customer the maximum resistance at 20oC of the copper conductor has to be 0.102*1.02=0.10404 ohm/kft
    If you are a manufacturer the maximum resistance at 20oC of a copper strand before stranding has to be 0.102*1.02/1.0619541*19=1.861436 ohm/kft
    Then after stranding conductor resistance will be : 1.861436/19*1.0619541=0.10404 ohm/kft.

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    I am sorry I did not remark you intend to calculate the resistance of a concentric neutral with a helically configuration and not the conductors. Let's start from the beginning.

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    Hello,

    Thanks for your answer,

    Area you sure about this?': At first the article Helically Applied Cu Wire CN it is for shield only and not for conductors.?

    Quote Originally Posted by Julius Right View Post
    At first the article Helically Applied Cu Wire CN it is for shield only and not for conductors.
    From Lay Direction and Length:
    m=1.0619541
    UL 1072/1995 7 Resistance
    7.1 The direct-current resistance of any length of conductor in ohms per thousand conductor feet or in ohms per conductor kilometer shall not be higher than the maximum acceptable (nominal x 1.02) resistance indicated in Tables 7.1 - 7.10 at 20°C (68°F) or at 25°C (77°F). If, as provided for in 9.2, metal-coated wires are used in only the outer layer of an uncoated copper conductor, the direct-current resistance of the wires are used in only the outer layer of an uncoated copper conductor, the direct-current resistance of the construction.
    If you are a customer the maximum resistance at 20oC of the copper conductor has to be 0.102*1.02=0.10404 ohm/kft
    If you are a manufacturer the maximum resistance at 20oC of a copper strand before stranding has to be 0.102*1.02/1.0619541*19=1.861436 ohm/kft
    Then after stranding conductor resistance will be : 1.861436/19*1.0619541=0.10404 ohm/kft.

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    I think the problem is the lay factor. In order to calculate the concentric wire resistance you need to know the actual wire length for a foot length of the cable.
    The lay factor then will be m=wirelength/laylength
    wirelength=√[Laylength^2+(πDinscore)^2]
    m=√[1+(πDinscore)^2/Laylength^2]
    laylength=no*Dinscore
    where Dinscore it is the cable insulation core[here approx.1.13 inch including half of the nonmetallic insulation shield]
    no=8_16 ; I think no=12 is good.
    tan(a)=laylength/πDinscore ; tan(a)=no/π ; a=atan(no/ π)
    If no=12 a=75.3 degrees.
    m=sqrt(1+(pi()/12)^2)=1.0337
    Click image for larger version. 

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    Click image for larger version. 

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    Quote Originally Posted by Julius Right View Post
    Click image for larger version. 

Name:	Not scaled wire wound.jpg 
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ID:	22564
    Thanks a lot!

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