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Chapter 8 · 1 hour

Environmental Effects

Practice questions

Practice questions and answers

2 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 6 marks

Explain galvanic corrosion and stress corrosion cracking. State the conditions under which each occurs and give one example of each.

Answer

Corrosion is the deterioration of a metal by chemical or electrochemical reaction with its environment.

Galvanic corrosion

When two dissimilar metals are in electrical contact in the presence of an electrolyte (water, moisture), a galvanic cell forms. The more active metal (anode) corrodes and the more noble metal (cathode) is protected.

  • Anode: M→Mn++ne−M \rightarrow M^{n+} + n e^-
  • Cathode (in neutral water): O2+2H2O+4e−→4OH−O_2 + 2H_2O + 4e^- \rightarrow 4OH^-

The metal's position in the galvanic series decides which is the anode (e.g., Mg, Zn, Al are anodic to steel; steel is anodic to copper, stainless steel).

Severity increases with larger potential difference, small anode area compared with cathode area, and high conductivity of the electrolyte.

Example: steel bolts used on a copper plate or an aluminium sheet riveted with steel rivets in sea water.

Prevention: select metals close in the galvanic series, insulate them, keep the anode area large, apply coatings, or use sacrificial anodes.

Stress corrosion cracking (SCC)

SCC is brittle cracking of a normally ductile metal under the combined action of tensile stress (applied or residual) and a specific corrosive environment.

ConditionExplanation
Tensile stressResidual stress from welding or cold work is enough
Specific environmente.g., chlorides for austenitic stainless steel; ammonia for brass (season cracking); caustic for carbon steel (caustic embrittlement)
Susceptible alloyPure metals are generally immune

Cracks form at the crack tip where the protective film is broken by stress, and the exposed metal dissolves, so the crack advances (intergranular or transgranular) with little overall metal loss.

Example: season cracking of brass cartridge cases, and chloride SCC of 304 stainless steel pipes.

Prevention: stress-relief annealing, shot peening, changing the alloy, controlling the environment.

  • Practice · 4+4 marks

(a) Describe four methods of protecting metals from corrosion. (b) A mild steel coupon (density 7.87 g/cm3^3) with exposed area 12 cm2^2 lost 0.45 g after being immersed in an acid for 240 hours. Calculate the corrosion penetration rate in mm/year.

Answer

(a) Corrosion protection methods

  1. Protective coatings: paint, enamel, plastic, or metallic coatings. Galvanising (zinc on steel) protects both as a barrier and sacrificially, since zinc is anodic to iron even if scratched. Tin plating protects only as a barrier.
  2. Cathodic protection: make the structure the cathode.
    • Sacrificial anode: connect Mg, Zn or Al blocks to buried pipes or ship hulls; they corrode instead of the steel.
    • Impressed current: a DC source drives current from an inert anode to the structure.
  3. Inhibitors: small additions (chromates, phosphates, amines) to the environment that form a protective film or slow the electrode reaction.
  4. Material selection and alloying: use stainless steel, aluminium, or Cu alloys; design to avoid crevices, galvanic couples, and stagnant water.
    • Other: anodising of aluminium, controlling temperature/humidity.

(b) Corrosion rate

Volume of metal lost:

V=Wρ=0.457.87=0.05718 cm3V = \frac{W}{\rho} = \frac{0.45}{7.87} = 0.05718\ \text{cm}^3

Thickness lost over 240 h:

x=VA=0.0571812=0.004765 cm=0.04765 mmx = \frac{V}{A} = \frac{0.05718}{12} = 0.004765\ \text{cm} = 0.04765\ \text{mm}

Hours in a year =8760= 8760, so scaling to a year:

CPR=0.04765×8760240=0.04765×36.5=1.74 mm/yearCPR = 0.04765 \times \frac{8760}{240} = 0.04765 \times 36.5 = 1.74\ \text{mm/year}

This is the same as the standard formula CPR=87.6 WρAtCPR = \dfrac{87.6\,W}{\rho A t} with WW in mg, ρ\rho in g/cm3^3, AA in cm2^2, tt in hours: 87.6×4507.87×12×240=1.74\dfrac{87.6 \times 450}{7.87 \times 12 \times 240} = 1.74 mm/yr.

Answer: corrosion penetration rate = 1.74 mm/year (about 68 mpy).

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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