Question 2 of 8: Corrosion Rate of Iron from i corr
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Open-book exam, 3 hours; any non-communicating calculator permitted. Eight questions of equal value (10 marks each) constitute a complete paper; full solutions to all eight are given here. Questions 1–3 are quantitative (a galvanic-cell Nernst calculation, a Faraday's-law corrosion-rate conversion, and an impressed-current cathodic-protection circuit); Questions 4–8 are short "corrosion-consultant" case studies answered as reasoned engineering judgements.
Reference texts: M. G. Fontana, Corrosion Engineering (3rd ed., McGraw-Hill) — the classic text behind this syllabus (electrode potentials and the EMF series Ch. 9; corrosion-rate expressions and Faraday's law Ch. 9–10; the eight forms of corrosion Ch. 3; materials selection and the sulfuric-acid/HCl case problems Ch. 12; cathodic protection and inhibitors Ch. 6–11); D. A. Jones, Principles and Prevention of Corrosion (2nd ed., Prentice Hall) — mixed-potential theory, Tafel extrapolation and CP design; A. W. Peabody, Control of Pipeline Corrosion (2nd ed., NACE) — anode-bed resistance and current density; ASM Handbook Vol. 13, Corrosion for materials-selection charts. Canadian practice: potable-water corrosion control follows the CCME/Health Canada guidelines and the AWWA carbonate-saturation approach (Question 8).
Question 2: Corrosion Rate of Iron from icorr(10 marks)
Given. Iron corroding uniformly, with a corrosion current density measured by Tafel extrapolation of $i_{corr}=3.74\times10^{-4}\ \text{A/m}^2$. Iron dissolves as Fe → Fe2+ + 2e−.
Quantity
Value
Corrosion current density $i_{corr}$
$3.74\times10^{-4}\ \text{A/m}^2$
Atomic weight of iron, $M$
$55.85\ \text{g/mol}$
Electrons per atom, $n$
$2$
Faraday constant, $F$
$96\,485\ \text{C/mol}$
Density of iron, $\rho$
$7.87\ \text{g/cm}^3 = 7870\ \text{kg/m}^3$
Find. The uniform corrosion rate expressed as (a) penetration in mm/year and (b) mass loss in mdd (mg dm−2 day−1).
Approach
Faraday's law converts current density to a molar (then mass) dissolution flux; dividing by density gives the penetration rate, and expressing the mass flux per dm2 per day gives the mdd figure.
Mass-loss flux from Faraday's law. The metal removed per unit area per unit time is
$$w = \frac{i_{corr}\,M}{n\,F} = \frac{(3.74\times10^{-4})(55.85)}{(2)(96\,485)} = 1.082\times10^{-7}\ \text{g}\,\text{m}^{-2}\text{s}^{-1}$$
using $M/n = 27.93$ g per equivalent (the equivalent weight of iron).
(b) Convert the mass flux to mdd. Multiply by 86 400 s/day, convert m2→dm2 (divide by 100) and g→mg (×1000):
$$r_{mdd} = 1.082\times10^{-7}\times\frac{86\,400}{100}\times1000 = 0.0935\ \text{mdd}$$
==**Mass-loss rate $\approx 0.094$ mdd**== (mg dm−2 day−1).
(a) Convert to penetration rate. Dividing the mass flux by the density gives a recession velocity; in consistent SI units,
$$r_{pen} = \frac{i_{corr}\,M}{n\,F\,\rho} = \frac{(3.74\times10^{-4})(55.85\times10^{-3})}{(2)(96\,485)(7870)} = 1.375\times10^{-14}\ \text{m/s}$$
Multiplying by $10^{3}\ \text{mm/m}$ and $3.154\times10^{7}\ \text{s/yr}$:
$$r_{pen} = 4.34\times10^{-4}\ \text{mm/yr}$$
==**Penetration rate $\approx 4.3\times10^{-4}$ mm/yr**== ($\approx 0.43\ \mu\text{m/yr}$, i.e. effectively negligible attack).
Check: the current density is taken exactly as printed, $3.74\times10^{-4}\ \text{A/m}^2$, which is an unusually low value — it corresponds to near-immunity (well under 0.01 mm/yr, "outstanding" on the Fontana rating scale). Most published forms of this problem quote the Tafel current in $\text{A/cm}^2$; had the intended units been $3.74\times10^{-4}\ \text{A/cm}^2$, both results scale by $10^{4}$ to $r_{pen}\approx4.3$ mm/yr and $r_{mdd}\approx935$ mdd — the more typical magnitude for actively corroding iron. Both readings; the printed A/m2 value is reported as the primary answer.
