Exercise set

Corrosion and Surface Protection Engineering Exercises

Worked materials engineering exercises for corrosion and surface protection covering wall loss, corrosion allowance, galvanic area ratio, coating thickness, zinc consumption, cathodic protection current, oxidation growth, pit severity, inspection coverage, lifecycle cost, and validation evidence.

These exercises practise corrosion and surface protection as engineering control problems. They cover wall loss, corrosion allowance, galvanic area ratio, coating thickness, zinc consumption, cathodic protection current, oxidation growth, pit severity, inspection coverage, lifecycle cost, and validation evidence.

The goal is not only to compute a corrosion rate. The goal is to decide whether the material, surface condition, geometry, coating system, environment, and inspection plan keep degradation slow, detectable, repairable, and compatible with the failure consequence.

Assume simplified screening models unless an exercise states otherwise. Real corrosion assessments should also check chemistry, pH, chloride level, dissolved oxygen, temperature, flow, deposits, crevices, surface preparation, coating damage, welding, stress, fatigue, inspection probability of detection, and field maintenance practice.

How to Use These Exercises

For each calculation, define:

  1. the exposed material, surface condition, and product form;
  2. the electrolyte, atmosphere, chemical exposure, temperature, and wet-dry cycle;
  3. the corrosion mechanism assumed by the simplified model;
  4. the geometric detail where degradation matters most;
  5. the inspection or maintenance action triggered by the result.

The common mistake is treating corrosion resistance as a fixed material property. Corrosion behavior is a system property of material, environment, geometry, protection, manufacturing, inspection, and time.

Use the exercises as decision gates, not isolated calculations: accept or reject a wall-loss margin, shorten an inspection interval, redesign an unfavorable galvanic detail, hold a coating system for repair, size cathodic-protection current, investigate a localized pit, compare lifecycle protection options, or refuse release when corrosion-control evidence is incomplete.

Exercise 1: Uniform Wall Loss Over Service Life

A carbon-steel component has initial wall thickness:

t_0=10.0\ \text{mm}

The minimum allowable wall thickness from stress analysis is:

t_{min}=7.5\ \text{mm}

The measured uniform corrosion rate is:

r=0.08\ \text{mm/year}

Estimate the wall thickness after 20 years and decide whether the uniform-corrosion screen passes.

Solution

Uniform loss is:

t_{loss}=rt

Substitute:

t_{loss}=(0.08)(20)=1.6\ \text{mm}

Remaining wall thickness is:

t_{rem}=t_0-t_{loss}
t_{rem}=10.0-1.6=8.4\ \text{mm}

Compare with the minimum:

8.4\ \text{mm}>7.5\ \text{mm}

The uniform-corrosion screen passes with margin:

M=t_{rem}-t_{min}=8.4-7.5=0.9\ \text{mm}

Engineering Comment

The margin is modest. This result does not prove acceptability for pitting, crevice corrosion, erosion-corrosion, coating holidays, weld attack, or corrosion fatigue. The inspection plan should still define where thickness is measured, what grid resolution is used, and what wall-thickness trigger starts repair or replacement.

Exercise 2: Corrosion Allowance and Inspection Interval

A pressure-boundary component has corrosion allowance:

CA=3.0\ \text{mm}

The expected corrosion rate is:

r=0.12\ \text{mm/year}

The inspection program requires at least 1.0 mm of unused allowance at the time of inspection so corrective action can be planned before the limit is reached.

Find the maximum inspection interval for this screening rule.

Solution

The allowable loss before inspection is:

t_{usable}=CA-1.0
t_{usable}=3.0-1.0=2.0\ \text{mm}

The maximum interval is:

\displaystyle T_{max}=\frac{t_{usable}}{r}
\displaystyle T_{max}=\frac{2.0}{0.12}=16.67\ \text{years}

A practical screening interval would be no longer than about 16 years, and usually shorter after applying uncertainty, consequence, and inspection-access factors.

Engineering Comment

Inspection interval should not be selected from average corrosion rate alone. If the environment can change, if rates are based on few measurements, or if localized attack is credible, the interval should be reduced and supplemented with focused inspection at high-risk locations.

Exercise 3: Galvanic Area Ratio Screen

A small aluminum bracket is electrically connected to a stainless-steel panel in a wet chloride environment. The exposed anodic aluminum area is:

A_a=40\ \text{cm}^2

The exposed cathodic stainless-steel area is:

A_c=800\ \text{cm}^2

Compute the cathode-to-anode area ratio and interpret the risk qualitatively.

Solution

The area ratio is:

\displaystyle R_A=\frac{A_c}{A_a}
\displaystyle R_A=\frac{800}{40}=20

The cathode area is 20 times the anodic area.

Engineering Comment

This is an unfavorable galvanic geometry because a small anodic component may have to support current associated with a large cathodic surface. The design should consider isolation, compatible fasteners, sealants, drainage, coating continuity, sacrificial protection, or material substitution. A coating defect on the aluminum side would be especially important because it can concentrate attack into a small exposed region.

Exercise 4: Coating Thickness Loss

A coated steel structure has specified dry film thickness:

DFT_0=240\ \mu\text{m}

Field measurements after service show average thickness:

DFT_{meas}=165\ \mu\text{m}

The maintenance specification requires recoating when average remaining thickness falls below:

DFT_{min}=140\ \mu\text{m}

Calculate the thickness loss and the remaining margin to the maintenance trigger.

Solution

Coating loss is:

\Delta DFT=DFT_0-DFT_{meas}
\Delta DFT=240-165=75\ \mu\text{m}

Remaining margin to the trigger is:

M=DFT_{meas}-DFT_{min}
M=165-140=25\ \mu\text{m}

The coating has lost 75 micrometres and has 25 micrometres of average-thickness margin before the trigger.

Engineering Comment

Average thickness can hide weak details. Edges, weld toes, bolt heads, drain holes, field cuts, and repaired areas may be below the trigger even when the average passes. Maintenance decisions should combine dry-film thickness with visual condition, adhesion, holiday detection, edge coverage, and active corrosion evidence.

Exercise 5: Zinc Coating Consumption Life

A galvanized steel component has zinc coating thickness:

h_0=85\ \mu\text{m}

The environment consumes zinc at an estimated rate:

r_z=4.0\ \mu\text{m/year}

Estimate the time until the zinc layer is consumed in a simplified uniform-consumption model.

Solution

The coating life is:

\displaystyle T=\frac{h_0}{r_z}
\displaystyle T=\frac{85}{4.0}=21.25\ \text{years}

The simplified zinc-consumption life is about 21 years.

Engineering Comment

This model assumes uniform zinc loss and does not include abrasion, trapped moisture, crevices, chloride concentration, acidic exposure, cut edges, fastener damage, or local coating defects. A real galvanizing assessment should also define inspection frequency and repair method before red rust or section loss becomes structurally relevant.

Exercise 6: Cathodic Protection Current Requirement

A buried steel surface requiring cathodic protection has exposed area:

A=55\ \text{m}^2

The design current density is:

i_d=12\ \text{mA/m}^2

Estimate the total protection current.

Solution

Total current is:

I=i_d A
I=(12)(55)=660\ \text{mA}

Convert to amperes:

I=0.660\ \text{A}

The estimated protection current is 0.66 A.

Engineering Comment

This is only a sizing screen. A cathodic protection design also needs soil or electrolyte resistivity, coating condition, anode output, current distribution, electrical continuity, interference risks, test-point layout, reference electrode readings, and verification after installation.

Exercise 7: Oxide Scale Growth

A high-temperature alloy develops oxide scale according to a simplified parabolic law:

x^2=k_p t

where oxide thickness (x) is in micrometres, time (t) is in hours, and:

k_p=0.18\ \mu\text{m}^2/\text{hour}

Estimate the oxide thickness after:

t=500\ \text{hours}

Solution

Use:

x=\sqrt{k_p t}

Substitute:

x=\sqrt{(0.18)(500)}
x=\sqrt{90}=9.49\ \mu\text{m}

The predicted oxide thickness is about 9.5 micrometres.

Engineering Comment

Parabolic growth suggests diffusion-controlled oxidation, but the simple law may fail after thermal cycling, scale cracking, spallation, contaminant attack, or phase changes. For hot components, oxide thickness should be connected to metal loss, heat transfer, fatigue, creep, and inspection evidence.

Exercise 8: Pit Depth and Remaining Ligament

A plate has nominal thickness:

t_0=6.0\ \text{mm}

Inspection finds a corrosion pit with maximum depth:

d_p=1.4\ \text{mm}

The local minimum allowable remaining ligament is:

t_{lig,min}=5.0\ \text{mm}

Calculate the remaining ligament and decide whether the pit passes this local screen.

Solution

The remaining ligament is:

t_{lig}=t_0-d_p
t_{lig}=6.0-1.4=4.6\ \text{mm}

Compare with the minimum:

4.6\ \text{mm}<5.0\ \text{mm}

The pit fails the local remaining-ligament screen.

Engineering Comment

Pit depth can be more important than average wall loss because it creates a local section reduction and a stress concentration. The engineering response may include local repair, replacement, derating, focused crack inspection, fatigue review, or root-cause correction such as drainage improvement or coating repair.

Exercise 9: Coating Holiday Density

A coated tank floor has inspected area:

A=120\ \text{m}^2

Holiday testing finds:

N=18\ \text{holidays}

The project acceptance criterion limits holiday density to:

D_{max}=0.10\ \text{holidays/m}^2

Calculate the holiday density and decide whether the inspected area passes.

Solution

Holiday density is:

\displaystyle D=\frac{N}{A}
\displaystyle D=\frac{18}{120}=0.15\ \text{holidays/m}^2

Compare:

0.15>0.10

The inspected area fails the acceptance criterion.

Engineering Comment

The engineering action should not stop at counting holidays. The team should locate the defects, repair them using the specified procedure, retest the repaired area, and investigate whether surface preparation, cure, handling damage, edge geometry, or applicator technique caused the defect density.

Exercise 10: Lifecycle Cost of Two Protection Systems

Two coating systems are proposed for the same steel structure.

System A has initial application cost:

C_{A,0}=42{,}000\ \text{USD}

and recoating cost every 8 years:

C_{A,r}=26{,}000\ \text{USD}

System B has initial application cost:

C_{B,0}=62{,}000\ \text{USD}

and recoating cost every 15 years:

C_{B,r}=31{,}000\ \text{USD}

Ignoring discounting, compare total coating cost over 30 years. Assume recoating occurs at the end of each full interval before year 30.

Solution

System A requires recoating at years 8, 16, and 24:

C_A=C_{A,0}+3C_{A,r}
C_A=42{,}000+3(26{,}000)=120{,}000\ \text{USD}

System B requires recoating at year 15:

C_B=C_{B,0}+1C_{B,r}
C_B=62{,}000+31{,}000=93{,}000\ \text{USD}

System B is lower by:

\Delta C=C_A-C_B=120{,}000-93{,}000=27{,}000\ \text{USD}

Engineering Comment

The higher initial-cost system is cheaper in this simplified 30-year screen. A real lifecycle decision should also include downtime, access cost, containment, surface preparation, environmental controls, inspection cost, repair quality, failure consequence, risk of schedule disruption, and whether performance evidence is valid for the actual exposure.

Exercise 11: Validation Evidence Gap

A corrosion-control plan lists six required evidence items before release:

  1. exposure definition;
  2. material compatibility review;
  3. coating specification;
  4. surface-preparation record;
  5. inspection trigger limits;
  6. repair procedure.

Only four items are complete at design freeze. Calculate the completion fraction and decide whether the plan is ready for release if the project requires at least 90 percent completion before production.

Solution

Completion fraction is:

\displaystyle f=\frac{4}{6}=0.667

Convert to percentage:

f=66.7\%

Compare with the requirement:

66.7\%<90\%

The corrosion-control evidence is not ready for production release.

Engineering Comment

Missing corrosion evidence is not administrative noise. If inspection triggers or repair procedures are undefined, the system may enter service without a clear way to detect and control degradation. For safety-critical, pressure-boundary, marine, infrastructure, or implantable systems, evidence gaps should be treated as engineering risk.

Review Checklist

When reviewing corrosion and surface-protection evidence, ask:

  • Is the exposure basis specific enough to reproduce the corrosion assumption?
  • Does the assessment distinguish uniform corrosion, pitting, crevice corrosion, galvanic corrosion, oxidation, erosion-corrosion, and corrosion fatigue?
  • Are coatings, zinc systems, isolation details, cathodic protection, drainage, and inspection access treated as part of one control system?
  • Are welds, edges, fasteners, cut surfaces, field repairs, and hidden interfaces checked separately from ideal flat surfaces?
  • Are measured corrosion rates connected to wall-thickness limits, local pit limits, crack detection, and maintenance triggers?
  • Is the inspection method capable of detecting the failure mode that matters?
  • Are uncertainty, access limits, probability of detection, and measurement repeatability included before extending service life?
  • Are repair records and exposure changes fed back into the next corrosion review?
  • Is closeout based on repaired defects, retested protection, updated limits, and assigned maintenance ownership rather than on the original calculation alone?

Good corrosion engineering makes degradation measurable and actionable. A material or coating is not protected by its datasheet; it is protected by a design basis, an inspection plan, and disciplined maintenance evidence.

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See also