# Thread: Article: Cathodic Protection for Bare Steel Pipe

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## Article: Cathodic Protection for Bare Steel Pipe

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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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Hi,

How to calculate the current density or is there a table to follow?

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Cy

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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.