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ted_yosem
Sound technical content, curated with aloha by
Ted Mooney, P.E. RET
Pine Beach, NJ
finishing.com -- The Home Page of the Finishing Industry

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Calculating required galvanizing thickness with Faraday's Law





Q. Under particular corrosive conditions it is estimated that the current density of galvanised steel sheet is 6X10^-3 A/m^2. Estimate the thickness of the galvanised layer to give rust protection for a minimum of 5 years?

This question seems unsolvable because there is no mass or area given
Can anyone help me out?

Answer: thickness> 45 micrometer

Edwin Pang Jian Xiang
STUDENT - Petaling Jaya, Selangor , Malaysia
2005


A. I have never seen a student question quite like that one, Edwin; thanks for sharing it with us!

We rarely do people's homework for them, but this one is interesting so let me try it.

6 x 10^-3 A of current, running for 5 years (1.58 x 10^8 seconds) would be 948,000 coulombs or about 9.82 Faradays. That should consume 9.82 gram equivalent weights of zinc or 4.91 molecular weights or 321 grams or 45 cubic centimeters.

So, if each square meter of surface area must be coated with 45 cubic centimeters of zinc, it is coated with 45 cm/m^2 or 45 cm/(100x100)cm^2 or .0045 cm thickness or 45 micrometers. Note that galvanizing is invariably two-sided, so the thickness on each side is 22.5 micrometers.

Now my question is, how can we measure that current density -- I don't think there is any practical way.

Ted Mooney, finishing.com
Ted Mooney, P.E.
Striving to live Aloha
finishing.com - Pine Beach, New Jersey
2005




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