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Thread: Article: Cathodic Protection for Bare Steel Pipe

  1. #1

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    Article: Cathodic Protection for Bare Steel Pipe

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  3. Hi
    first calculate requird current due to pipe surface area and cuurent density for bare pipe according to your std. -----> I
    Y: system life (year)
    C:consumption rate (Kg/A.year) see nace refernce book
    E: anode efficiency see nace reference book
    U: Utilization factor according to your std

    Mass of anodes = I Y C / (E U)

    for example 500m2 bare steel, 20mA/m2 current density, design life=2 years
    I=500x20=10000mA=10A

    *****galvanic method*****
    C=8 Kg/A.year
    E=50%=0.5
    U=0.8
    So Mass of anodes = 10x2x8/(0.5x0.8)=400Kg
    number of anodes = 400 /20(mass for each anode) = 20
    for protection of bare pipes or when design life is high, using galvani method is not economical.

    *****Impressed current method*****
    C=0.2 Kg/A.year for HSCI anode
    Y=2 year
    E=0.8, U=0.5

    mass of anodes=10x2x0.2/(0.4)=10kg

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  5. #3

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  6. #4

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    ghezeljeh

    I've been calculating by using the above formulas, however it will resulted tons of anodes. But I was considered to use coating.

    thx bro for attention.

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