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1. Homework Statement
Iron sample with dimensions 10mm x 20mm x 2mm is exposed to a corrosive environment (neglect thickness for surface area). Both sides of the sample are exposed to corrosive environment. Given http://i.imgur.com/tGRw4.jpg"
(a) Calculate corrosion rate mm/y
(b) Calculate corrosion rate if one of the sample surfaces is painted (not in contact with corrosive environment)
2. Homework Equations
M=56g,
t=3.1536e7 sec/yr
number of electrons taking part n=2
F=96500C
density=7.87g/cm3
A=10*20=200mm2
From Faraday's Law: W=MtI/nF (W=weight loss, I=current in this case corrosion current)
3. The Attempt at a Solution
(a) From Tafel Plot log(Icorr)=0.05mA/cm2 so Icorr=1.12e-3 A/cm2
I want to convert Icorr to just units of A, but since both surfaces are exposed should I multiply it by double the surface area?
i.e.: Icorr=1.12e-3*200mm2*2*(1cm/10mm)^2 = 4.48e-3 A
substituting in W=Mt*Icorr/nF = 40.99 g/yr
Corrosion Rate CR=W/density*Surface Area
Again I multiplied the Surface Area by 2 because both sides of the Iron sample is exposed
CR=(40.99g/yr)/[(7.87g/cm3)*(200mm2*2)*(1cm/10mm)^3)]=13.02mm/yr
(b) For only one side exposed:
I'm going to use half of Icorr since only 1 side is exposed so Icorr=2.24e-3 A
then W=20.49g/yr
Using only 200mm2 as the surface area
CR=(20.49g/yr)/[(7.87g/cm3)*(200mm2)*(1cm/10mm)^3)]=13.02mm/yr
I get the same corrosion rate...is this correct?!
Iron sample with dimensions 10mm x 20mm x 2mm is exposed to a corrosive environment (neglect thickness for surface area). Both sides of the sample are exposed to corrosive environment. Given http://i.imgur.com/tGRw4.jpg"
(a) Calculate corrosion rate mm/y
(b) Calculate corrosion rate if one of the sample surfaces is painted (not in contact with corrosive environment)
2. Homework Equations
M=56g,
t=3.1536e7 sec/yr
number of electrons taking part n=2
F=96500C
density=7.87g/cm3
A=10*20=200mm2
From Faraday's Law: W=MtI/nF (W=weight loss, I=current in this case corrosion current)
3. The Attempt at a Solution
(a) From Tafel Plot log(Icorr)=0.05mA/cm2 so Icorr=1.12e-3 A/cm2
I want to convert Icorr to just units of A, but since both surfaces are exposed should I multiply it by double the surface area?
i.e.: Icorr=1.12e-3*200mm2*2*(1cm/10mm)^2 = 4.48e-3 A
substituting in W=Mt*Icorr/nF = 40.99 g/yr
Corrosion Rate CR=W/density*Surface Area
Again I multiplied the Surface Area by 2 because both sides of the Iron sample is exposed
CR=(40.99g/yr)/[(7.87g/cm3)*(200mm2*2)*(1cm/10mm)^3)]=13.02mm/yr
(b) For only one side exposed:
I'm going to use half of Icorr since only 1 side is exposed so Icorr=2.24e-3 A
then W=20.49g/yr
Using only 200mm2 as the surface area
CR=(20.49g/yr)/[(7.87g/cm3)*(200mm2)*(1cm/10mm)^3)]=13.02mm/yr
I get the same corrosion rate...is this correct?!
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