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Problem 1
Name the four components of a corrosion cell. In an open cooling tower, which cathodic reactant is effectively impossible to eliminate, and why?
Authored answer
Anode (metal dissolves), cathode (site of electron consumption), electrolyte (ionic conduction through water), metallic path (electron conduction through metal). Oxygen is the most difficult cathodic reactant to remove because it is replenished from the atmosphere.
Read the connected sectionProblem 2
The chapter says “passivation is not immunity – it is a ceasefire.” Name four mechanisms (film saboteurs) that can break the ceasefire and briefly state how each works.
Authored answer
Mechanical disruption (flow/erosion): shear forces strip the oxide layer at elbows, tees, and pump impellers. Chemical dissolution (low pH): hydrogen ions protonate the oxide, converting it back to soluble species. Chloride penetration: small, mobile Cl⁻ ions pierce the film and form soluble metal–chloride complexes, preventing repassivation. Sulfate’s two paths: Sulfate can interfere with protective film stability by forming soluble metal-complexes. It also aids in microbiologically influenced corrosion (MIC) when SRB use sulfate as an electron acceptor to consume electrons associated with iron oxidation.
Read the connected sectionProblem 3
A contractor connects a new carbon steel header directly to existing copper piping with no dielectric union. Which metal will corrode preferentially, and why?
Authored answer
Carbon steel corrodes. It is less noble than copper on the galvanic series. When both metals share an electrolyte and metallic path, the less noble metal (steel) becomes the anode.
Read the connected sectionProblem 4
A gasket creates a crevice on a carbon steel flange in oxygenated water. Where does the anode develop? Where does the cathode develop? Why does the attack accelerate over time?
Authored answer
Anode: inside the crevice (low oxygen). Cathode: exposed surface outside the crevice (high oxygen). Oxygen differential creates voltage. Over time, chloride migrates into the pit to balance charge, acidity rises locally, and repassivation becomes impossible. The pit worsens.
Read the connected sectionProblem 5
A carbon steel coupon (density 7.87 g/cm³, surface area 3.5 in²) is exposed for 90 days. Initial weight: 12.4500 g. Final weight: 12.1500 g. Calculate the corrosion rate in MPY and classify it for a cooling tower.
Authored answer
W = (12.4500 − 12.1500) × 1,000 = 300 mg. T = 90 × 24 = 2,160 hr. MPY = (534 × 300) / (7.87 × 3.5 × 2,160) = 160,200 / 59,497.2 = 2.69 MPY. Classification: “Very Good”(1-3 MPY)
Problem 6
A cooling tower runs at 6 cycles of concentration. The makeup water contains Cl⁻ = 40 mg/L, SO₄²⁻ = 25 mg/L, and M-alkalinity = 50 ppm as CaCO₃. (a) Calculate the Larson-Skold ratio for the makeup water and for the recirculating water at 6 cycles. (b) The two values are identical. Explain why, despite the identical LS value, the recirculating water may present much higher corrosion stress than the makeup. (c) What does this reveal about using LS, or any bulk-water index, to predict corrosion in a cycled system? Assume chloride, sulfate, and alkalinity all concentrate by exactly six times.
Authored answer
(a) Cycling multiplies every ion by the cycle factor, so at 6 cycles Cl⁻ = 240, SO₄²⁻ = 150, alkalinity = 300 ppm as CaCO₃. Makeup: (40/35.5 + 25/48) / (50/50) = (1.13 + 0.52) / 1.00 = 1.65. 6 cycles: (240/35.5 + 150/48) / (300/50) = (6.76 + 3.13) / 6.00 = 1.65. The values are identical. (b) LS is a ratio; cycling scales numerator and denominator by the same factor, so the ratio is unchanged. But the metal does not respond to a ratio. It responds to concentration. At 6 cycles the chloride driving pitting is six times higher, crevices and deposits concentrate it further, and localized attack becomes far more likely. (c) A bulk-water ratio cannot predict corrosion in a cycled system. It is blind to the absolute concentrations and entirely blind to the local surface conditions where corrosion actually occurs. LS is useful only to demonstrate that chloride and sulfate matter, not to predict what they will do.
Read the connected sectionProblem 7
For each action, identify whether it primarily targets the anode, cathode, or electron path: (a) oxygen scavenger in boiler feedwater, (b) nitrite in a closed loop, (c) calcium/alkalinity adjustment in a cooling tower, (d) dielectric union between dissimilar metals.
Authored answer
(a) Cathode: removes the cathodic reactant (oxygen). (b) Anode: strengthens the passive film, locking the lattice. (c) Cathode: increases resistance via precipitating a protective film. (d) Electron path: breaks the metallic circuit between dissimilar metals.
Read the connected sectionProblem 8
A small carbon steel bolt is threaded into a large copper header. In a separate system, a large carbon steel pipe is connected to a small copper instrument line. Which configuration produces more severe corrosion of the steel? Explain using current density.
Authored answer
The small steel bolt in a large copper header is far worse. The large copper cathode drives a high total cathodic current (oxygen reduction over a large area). That entire current must be balanced by anodic dissolution concentrated on the tiny bolt surface, leading to rapid, localized metal loss. In the reverse case, the large steel anode spreads dissolution over a much larger area, reducing current density and slowing the rate of attack.
Read the connected sectionProblem 9
A cooling tower shows the following conditions at the same time: elevated chloride, intermittent low pH excursions, biofilm deposits, and 6 MPY carbon steel coupon loss. Using the corrosion-cell model, explain why these conditions do not merely add corrosion risk, they multiply it. In your answer, identify which factors are increasing voltage, which are decreasing resistance, and why the damage is likely to localize rather than remain uniform.
Authored answer
Chloride, low pH, and biofilm deposits weaken or remove passive films, which lowers resistance. Biofilms also create oxygen differential cells, which increase voltage by making the covered area anodic and the exposed area cathodic. Corrosion localizes because a small anode is driven by a large cathode, increasing current density and causing pitting rather than uniform metal loss. The 6 MPY is an average weight-loss rate, pitting may be much more severe.
Read the connected sectionProblem 10
Using Ohm’s Law applied to the corrosion cell (Current = Voltage / Resistance), explain why pitting accelerates once a passive film is breached. Address both the voltage and resistance terms.
Authored answer
When the passive film is breached, resistance drops because the protective barrier no longer impedes electron transfer and ion migration. Simultaneously, voltage increases because the exposed bare metal has a much lower reduction potential than the surrounding passivated surface, creating a large potential difference. By Ohm’s Law, lower resistance and higher voltage both drive higher current. That current concentrates on the small exposed anode, producing the intense localized dissolution that defines pitting.
Read the connected section