Question 3 of 8: Impressed-Current Cathodic Protection — Anode-Bed Resistance
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).
Given. An impressed-current cathodic-protection installation. The rectifier delivers $V=1\ \text{V}$ at $I=5\ \text{A}$; the soil (electrolyte) resistance is 80% of the total external circuit resistance. The single cylindrical anode is $L=4\ \text{m}$ long and $d=0.10\ \text{m}$ in diameter.
Quantity
Value
Rectifier output voltage, $V$
$1\ \text{V}$
Rectifier output current, $I$
$5\ \text{A}$
Soil fraction of external resistance
$0.80$
Anode length, $L$
$4\ \text{m}$
Anode diameter, $d$
$0.10\ \text{m}$
Find. (a) The soil resistance; (b) the current density at the anode surface.
Figure 3.1 — Impressed-current CP circuit. The rectifier drives protective current from the anode bed, through the soil (the dominant resistance), onto the buried steel which becomes the cathode; the metallic return completes the loop. By Ohm's law the total external resistance is $V/I$, of which the soil accounts for 80%.
Approach
Ohm's law on the rectifier output gives the total external circuit resistance; 80% of that is the soil resistance. The anode current density is the delivered current divided by the anode's exposed (lateral) surface area.
Total external resistance (Ohm's law). The rectifier drives 5 A at 1 V, so the whole external circuit presents
$$R_{ext} = \frac{V}{I} = \frac{1}{5} = 0.20\ \Omega$$
(a) Soil resistance. The soil (anode-to-earth electrolytic path) is 80% of the external resistance:
$$R_{soil} = 0.80\,R_{ext} = 0.80\times0.20 = 0.16\ \Omega$$
==**$R_{soil} = 0.16\ \Omega$**== The remaining 20% (0.04 Ω) is the metallic path — cables, connections and anode/structure interface.
Anode surface area. The exposed area of the cylindrical anode is its lateral surface (end effects neglected):
$$A = \pi\,d\,L = \pi\,(0.10)(4) = 1.257\ \text{m}^2$$
(b) Current density. The protective current spread over that area gives
$$i = \frac{I}{A} = \frac{5}{1.257} = 3.98\ \text{A/m}^2$$
==**$i \approx 3.98\ \text{A/m}^2$**== ($\approx 0.40\ \text{mA/cm}^2$), a modest discharge density well within the range for graphite or high-silicon-iron groundbed anodes.
