Some conversions useful for the corrosion calculations

Submitted by alexL on Wed, 08/27/2014 - 09:58

When we make calculations of the corrosion rates of Mg alloys, we often need to perform some simple conversions. The results are included in essentially any research  paper in the field, while the logics of the development often remains obscured. For that reason I found useful to show the development here.

1. Conversion of the corrosion current to the corrosion rate

Faraday's law: $m=\frac{M}{n}\cdot \frac{1}{F}\cdot it $

=> $ \frac{h, cm}{t,s}=\frac{M, g/mol}{n}\cdot\frac{1}{{A, cm^2}\cdot{\rho, g/cm^3}}\cdot\frac{1}{F}\cdot i,A$

$$Rate, mm/y=3.154\cdot {10^8} s \cdot\frac{24 g/mol}{2}\cdot \frac{1}{{1.1 cm^2}\cdot{1.8 g/cm^3}\cdot{96500 C/mol}}\cdot i (A) $$

Finally, Rate (mm/y)=1.981· 104·i(A)

This conversion is based on the Faraday's laws of Electrolysis and some basic conversions of units (centimeters to millimeters and seconds to years). The final formula was developed for the particular case when M = 24 g/mol, n = 2, ρ= 1.8 g/cm3 (Mg) and the area of a standard specimen A = 1.1 cm2. This values should, of course, be changed if necessary.

2. Calculation of the corrosion rate from the mass-loss (Pw)

$$ P_w (mm/y)=\frac{\Delta m (g)}{1.8 g/cm^3 \cdot A (cm^2) \cdot t (days)}\cdot 365 \cdot 10 $$

$$ P_w (mm/y)=2.03 \cdot {10^3} \cdot \frac{\Delta m (g)}{A (cm^2) \cdot t (days)}$$

This only includes the conversion of units: days to years, centimeters to millimeters. The density is taken  ρ= 1.8 g/cm3 (and should be adjusted if necessary) and the surface area is variable (A, cm2).

 

3. Calculation of the corrosion rate from the gas evolution data (PH)

m(Mg) = n(mol) ·24 g/mol

$ n(Mg)=n(H_2)=\frac{p}{RT}\cdot V =\frac{1.013\cdot 10^5 Pa}{8.314 J/mol-K \cdot 298K \cdot 10^6} \cdot {V(H_2) mL}  \\= 4.09 \cdot 10_{-5} \cdot V mL $

m(Mg), g = 9.8· 10-4 V(H2), mL

$ P_H = 2.03 \cdot 10^3 \cdot 9.8 \cdot 10^{-4} \cdot \frac {V(H_2), mL}{A, cm^2 \cdot t, days} $

$ P_H , mm/y= 1.99 \cdot \frac {V(H_2), mL}{A, cm^2 \cdot t, days} $

A bit more assumptions have been made here. First, the gas is assumed to be hydrogen formed in the simplified reaction:

1. Mg + H2O --> Mg(OH)2 + H2 (the important point here is the mole ratio Mg : H2 = 1 :1, the other things need not to be exactly as in the equation)

2. The other than Mg components of the alloy are considered inert (or at least their contribution to the gas evolution negligible).

3. The gas is considered ideal and SATP conditions are taken (T = 298K, p = 1 atm).