Problem Sets and Answer Keys
Every problem set and its answer key from the final manuscript, in one place. Narrative and Engineering Notes are excluded by design.
13 problem sets · 130 questions · 130 answers. Chapter 14 has no problem set.
These are only provided to reinforce understanding. The Engineering Notes contain some of the information that will be required. Make sure to employ dimensional analysis for all unit conversions.
— printed above every problem set in the book
Provenance
Extracted verbatim from
The Blueprint of Water - Final.docxon 2026-08-10. This file reflects the final text. The individual notes inProblems/predate it and no longer match on every problem, so treat this file as authoritative.
Contents
Chapter 1: Building Blocks
Chapter note: 01 Building Blocks
Chapter 1 Problem Set: Building Blocks
Level 1: Conceptual Mastery
1. Why do atoms form bonds? Your answer should reference lower total energy, not attraction alone.
2. What distinguishes an ionic bond from a covalent bond? Describe each in terms of what happens to the electrons.
3. Both water (H₂O) and carbon dioxide (CO₂) contain polar covalent bonds, yet only water is a polar molecule. Why?
4. What is the Hydrogen-Bond Network? Is it made of true chemical bonds? Why does it matter that each water molecule can participate in up to four hydrogen bonds?
Level 2: Applied Thinking
5. Using dimensional analysis, convert 10 gallons of water to (a) pounds and (b) grams. Show units cancelling at each step.
6. Calculate the molar mass of CaCO₃ using the periodic table. How many moles are in 50 grams?
7. A cooling tower basin holds 5,000 gallons of water. How many moles of water is that? Show the full conversion chain.
Level 3: System Integration
8. Calcium carbonate (CaCO₃) contains both ionic and covalent bonds within the same compound. Identify which bonds are ionic and which are covalent, and explain why this combination produces a crystal lattice that is difficult to pull apart.
9. Using the periodic table, explain why sodium (Group 1) and calcium (Group 2) both form positive ions, but calcium’s ion carries a +2 charge while sodium’s carries +1. Connect your answer to valence electrons.
10. The chapter describes the water molecule as “asymmetric, polar, and carrying stored electrical tension.” In 2–3 sentences, explain how each of these properties traces back to oxygen’s lone electron pairs and the resulting molecular geometry.
Chapter 1 Answer Key: Building Blocks
1. Atoms bond when the resulting arrangement places the combined system at a lower total energy than the separated atoms. Incomplete valence shells are high-energy states; bonding allows atoms to fill or empty their outer shells and settle into stability.
2. In ionic bonding, electrons are transferred from one atom to another, creating oppositely charged ions that attract. In covalent bonding, electrons are shared between atoms. The distinction is transfer vs. sharing.
3. CO₂ is linear: its two polar bonds point in opposite directions and cancel. Water is bent at 104.5°, so its polar bonds point in the same general direction and the charge imbalance survives. Geometry determines whether local polarity becomes molecular polarity.
4. The Hydrogen-Bond Network is the persistent web of intermolecular attractions between water molecules, formed by the partial positive charge on hydrogen attracting the partial negative charge on a neighboring oxygen. It is not a true chemical bond, it is roughly 1/10th the strength of a covalent bond. Because each molecule can form up to four hydrogen bonds, the network is continuous and interconnected, giving water its collective behavior.
5. (a) 10 gal × 8.34 lb/gal = 83.4 lb. (b) 83.4 lb × 453.6 g/lb = 37,830 g.
6. Ca: 40.08 g/mol, C: 12.01 g/mol, O: 16.00 g/mol. (40.08)+(12.01)+ (3×16.00) = 100.09 g/mol. 50 g ÷ 100.09 g/mol = 0.50 mol.
7. 5,000 gal × 8.34 lb/gal = 41,700 lb. 41,700 lb × 453.6 g/lb = 18,915,120 g. 18,915,120 g ÷ 18.015 g/mol = 1,050,000 mol (1.05 × 10⁶ mol).
8. Covalent bonds: the three C–O bonds within the carbonate ion (CO₃²⁻), forming a rigid, symmetric internal framework. Ionic bond: the electrostatic attraction between Ca²⁺ and CO₃²⁻. The rigid covalent unit does not deform, and the strong ionic attraction locks these units into a dense, interlocking crystal lattice, exceptionally stable and difficult to dissolve.
9. Sodium (Group 1) has one valence electron and donates it to achieve a full outer shell, producing Na⁺. Calcium (Group 2) has two valence electrons and must donate both to reach stability, producing Ca²⁺. Ion charge directly reflects the number of valence electrons an atom must lose.
10. Oxygen carries two lone electron pairs that repel the bonding pairs shared with hydrogen, forcing the molecule into a bent shape (104.5°) rather than a straight line. This asymmetry means the two polar O–H bonds do not cancel, making the molecule polar. The permanent separation of partial positive and partial negative charge is the “stored electrical tension,” it is locked into the geometry by the lone pairs. This is the basis of hydrogen bonding in water.
Chapter 2: The Hero of Heat Transfer
Chapter note: 02 The Hero of Heat Transfer
Chapter 2 Problem Set: The Hero of Heat Transfer
Level 1: Conceptual Mastery
1. At a heat exchanger, what roles do conduction, convection, and wetting each play in transferring heat into flowing water?
2. The chapter describes the Hydrogen-Bond Network as a “molecular shock absorber.” In your own words, explain what this means and why it gives water a high specific heat.
3. What is the difference between sensible heat and latent heat? Which one changes temperature, and which one changes state?
4. Why does raising pressure increase the temperature at which water boils? How does lowering pressure affect the boiling point?
Level 2: Applied Thinking
5. How many BTUs are required to heat 500 gallons of water from 60°F to 180°F? (Convert gallons to pounds first.)
6. How much energy is required to (a) heat 1 pound of water from 72°F to 212°F, and (b) completely vaporize that same pound of water at 212°F? How many times more energy does vaporization require than heating the water from 72°F to 212°F?
7. A cooling tower removes 6,000,000 BTU/hr. How many pounds of water must evaporate per hour? Assume that all heat is removed by evaporation and use a field latent-heat value of 1,000 BTU/lb. Convert to GPM.
Level 3: System Integration
8. Using the 1% evaporation rule of thumb, a tower recirculates 1,500 GPM with a 10°F temperature drop. Estimate the evaporation rate in GPM. Then verify your answer using an idealized latent-only energy balance and 1,000 BTU/lb. (Q = m × cₚ × ΔT)
9. A boiler produces steam at atmospheric pressure. The feedwater enters at 50°F. How many total BTUs per pound are required to produce steam? Break your answer into the sensible heat portion and the latent heat portion.
10. Water has both a high specific heat and a high latent heat of vaporization. In 2–3 sentences, explain why these two properties make water useful in hot-water loops, chilled-water loops, boilers, and cooling towers.
Chapter 2 Answer Key: The Hero of Heat Transfer
1. Conduction transfers energy through the heat-exchanger wall and into the thin layer of water touching it. Convection carries the warmed water away and replaces it with cooler bulk water. Wetting helps maintain contact between the water and the surface, increasing the area available for conduction.
2. Energy entering liquid water is distributed through molecular motion and continual rearrangement of the Hydrogen-Bond Network. Because the energy is spread across these modes rather than producing an immediate large increase in molecular speed, water’s temperature rises relatively slowly. That dampened response is the “shock absorber,” and it is the reason water’s specific heat is 1.0 BTU/lb·°F. Far higher than most common materials.
3. Sensible heat changes temperature without changing phase; it increases the energy contained within the existing state. Latent heat changes state without substantially changing temperature.
4. Water boils when its vapor pressure equals the pressure surrounding it. At higher pressure, water must reach a higher temperature before vapor bubbles can form and survive. Lowering the surrounding pressure allows boiling to occur at a lower temperature.
5. 500 gal × 8.34 lb/gal = 4,170 lb. Q = 4,170 lb × 1.0 BTU/lb·°F × (180 − 60)°F = 4,170 × 120 = 500,400 BTU.
6. (a) Sensible: 1 lb × 1.0 × (212 − 72) = 140 BTU. (b) Latent: 1 lb × 970 BTU/lb = 970 BTU. Vaporization requires 970/140 = 6.9 times more energy than heating to the boiling point.
7. 6,000,000 BTU/hr ÷ 1,000 BTU/lb = 6,000 lb/hr. 6,000 lb/hr ÷ 8.34 lb/gal = 719.4 gal/hr ÷ 60 min/hr = 12.0 GPM.
8. Rule of thumb: 1% of 1,500 GPM = 15 GPM evaporation for a 10°F drop. Verification: Q = 1,500 GPM × 8.34 lb/gal × 60 min/hr × 1.0 × 10°F = 7,506,000 BTU/hr. Evap = 7,506,000 ÷ 1,000 = 7,506 lb/hr ÷ 8.34 ÷ 60 = 15 GPM evaporation. The rule of thumb holds.
9. Sensible: 1 lb × 1.0 BTU/lb·°F × (212°F − 50°F) = 162 BTU. Latent: 970 BTU. Total: 162 + 970 = 1,132 BTU per pound of steam. Latent heat accounts for 86% of the total energy investment.
10. Water’s high specific heat allows it to absorb and transport large thermal loads with relatively modest temperature changes, which makes it effective in hot-water and chilled-water loops. Its high latent heat allows a small mass of water to carry enormous amounts of energy during vaporization and condensation, which makes it effective in boilers and cooling towers.
Chapter 3: The Solvent
Chapter note: 03 The Solvent
Chapter 3 Problem Set: The Solvent
Level 1: Conceptual Mastery
1. What is the difference between suspended solids and dissolved solids? Why can’t dissolved solids be removed by conventional filtration?
2. The chapter describes dissolution as a “transaction.” What are the three energetic terms involved, and which one represents the cost, the payoff, and the wildcard?
3. Sodium chloride dissolves readily in water while calcium carbonate dissolves reluctantly. Using the language of lattice energy and hydration energy, explain why.
4. Why does conductivity provide a useful estimate of TDS, and what is its most important limitation?
Level 2: Applied Thinking
5. Your conductivity meter reads 1,450 µS/cm. Estimate the TDS in ppm.
6. You add 5 pounds of NaCl to a 1,000-gallon cooling loop. Assuming complete dissolution, estimate the increase in concentration (ppm).
7. A 12,000-gallon cooling loop needs to be dosed to 150 ppm of a corrosion inhibitor that is 35% active by mass. How many pounds of product must be added? Show your dimensional analysis.
Level 3: System Integration
8. A deaerator heats boiler feedwater and vents a small amount of steam to atmosphere. Using Henry’s Law and the solubility behavior described in this chapter, explain why both the heating and the venting reduce dissolved oxygen. Then explain why the same equipment does nothing to remove dissolved calcium.
9. Two cooling towers both read 2,000 µS/cm conductivity. Tower A’s makeup is high in sodium chloride. Tower B’s makeup is high in calcium bicarbonate. Which tower faces a greater scaling risk, and why does identical conductivity not mean identical behavior?
10. A cooling tower’s makeup water reads 400 µS/cm. The circulating water reads 1,600 µS/cm. Estimate the cycles of concentration and the circulating TDS. If the makeup contains 80 mg/L calcium, what is the approximate circulating calcium, and why does the answer matter more at the heat exchanger surface than in the basin?
Chapter 3 Answer Key: The Solvent
1. Suspended solids are particles physically carried by water and defined as the material retained by a TSS filter. Dissolved solids pass through that filter and remain as residue after the water is evaporated. Conventional granular or cartridge filters cannot remove dissolved ions because the ions exist at molecular scale and travel with the water. Removing them requires processes such as ion exchange, reverse osmosis, electrodialysis, or distillation. Colloids occupy the gray zone and may require coagulation or membrane filtration.
2. Dissolution involves three energetic terms. Lattice energy is the cost of pulling a crystal apart. Hydration energy is the payoff released when water stabilizes the separated species. Entropy is the wildcard: dispersion tends to favor dissolution, while solvent ordering can oppose it. Gibbs free energy combines these effects and tells you whether dissolution is favored.
3. Sodium chloride has a manageable lattice cost, sufficient hydration payoff, and often favorable entropy. Dissolution is favored under normal conditions. Calcium carbonate has a much higher lattice cost because calcium and carbonate bind strongly in a stable crystal structure. Water can hydrate the separated ions, but the payoff is often conditional and not enough to overcome the strength of the lattice, especially as temperature, pH, calcium, alkalinity, and carbon dioxide shift the balance toward solid scale.
4. Dissolved ions are charge carriers. More ions means more current and higher conductivity, making it a fast, continuous proxy for TDS. Its limitation: conductivity measures total ionic load but not which ions are present. Two waters at the same conductivity can behave very differently. One may scale, the other may corrode.
5. TDS ≈ 0.7 × 1,450 = 1,015 ppm.
6. Water mass: 1,000 gal × 8.34 lb/gal = 8,340 lb. Weight of resulting solution = 8,340 lb + 5 lb = 8,345 lb. ppm increase: (5 lb / 8,345 lb) × (1,000,000) ≈ 599 ppm.
7. Water mass: 12,000 gal × 8.34 lb/gal = 100,080 lb. Active ingredient required: 100,080 lb × (150 / 1,000,000) = 15.0 lb active. Product required: 15.0 lb ÷ 0.35 = 42.9 lb of product. The percent-active conversion matters. Dosing 15 lb of product instead of 15 lb of active would leave the system at roughly 53 ppm, about a third of target.
8. Oxygen is held by physical equilibrium, not hydration. Heating lowers its solubility, so the water can hold less; venting steam lowers the partial pressure of oxygen above the water. Both actions push the equilibrium toward the gas phase. Calcium is not governed by equilibrium with the atmosphere. It was dismantled from a solid and is held in solution by hydration shells, so no amount of heating or venting will drive it out as a gas.
9. Identical conductivity means equal electrical conductance under the measurement conditions, not identical total ionic concentration or composition. Ion charge and mobility both affect the reading. Tower B faces the greater scaling risk because calcium and bicarbonate can concentrate and shift toward calcium carbonate precipitation, while sodium chloride generally remains soluble.
10. Cycles = 1,600 ÷ 400 = 4 cycles. Circulating TDS ≈ 0.7 × 1,600 = 1,120 ppm. Calcium ≈ 80 × 4 = 320 mg/L. Calcium carbonate has inverse solubility, so the hottest surface in the system is where ΔG turns over first. The basin may stay clear while the exchanger tubes scale.
Chapter 4: The Power of Hydrogen
Chapter note: 04 The Power of Hydrogen
Chapter 4 Problem Set: The Power of Hydrogen
Level 1: Conceptual Mastery
1. What is self-ionization? Why does pure water contain both H⁺ and OH⁻, even though no acid or base has been added?
2. The chapter says “We are measuring one side of a linked equilibrium.” Explain what this means in your own words.
3. Pure water at 100°C has a pH of approximately 6.14. Is it acidic? Why or why not?
4. The chapter distinguishes pH from alkalinity: “pH describes the current intensity of equilibrium. Alkalinity describes the future capacity to resist change.” Why is this distinction important?
Level 2: Applied Thinking
5. What is the concentration of [H⁺] and [OH⁻] at pH 4? At pH 10? Show how pH + pOH = 14 holds for each.
6. How many times more hydrogen ions are present at pH 4 than at pH 7?
7. A sample has pH 9 at 25°C. Calculate (a) the pOH, (b) [OH⁻], and (c) [H⁺].
Level 3: System Integration
8. During an acid cleaning, your solution sits at pH 2.5. You add caustic and the pH moves to 3.5. By what factor did the hydrogen concentration decrease? Explain why a technician might mistakenly think this change was trivial.
9. You must neutralize 1,000 gallons of acid cleaning solution from pH 1.0 to pH 7.0 using 100% KOH (56.1 g/mol). How many pounds of KOH are required? If the starting pH had been 2.0 instead, how many pounds would be needed? (Assume an ideal solution with no buffering).
Field Warning: Never calculate the neutralization requirement of a real acid-cleaning solution from pH alone. Determine demand through titration and add incrementally.
10. The chapter describes pH as reorganizing the “rules of chemistry” rather than simply adjusting a dial. Name two specific consequences of a pH shift mentioned in the chapter and explain why even a small change matters.
Chapter 4 Answer Key: The Power of Hydrogen
1. Self-ionization is the process by which water molecules react with one another to form hydronium and hydroxide. One water molecule effectively hands a proton to another, producing H₃O⁺ and OH⁻. The ions rapidly recombine, establishing a dynamic equilibrium. No acid or base is needed. The process arises from water’s own molecular behavior.
2. pH does not describe a fixed characteristic of water. It measures where the ongoing self-ionization equilibrium currently sits: the balance between H⁺/H₃O⁺ and OH⁻ at a given moment. Adding acid or base drags that balance to a new position, and pH reports where it settled. It is a measurement of equilibrium, not a simple tally of chemical added.
3. It is not acidic. Neutral means [H⁺] = [OH⁻], not “pH = 7.” At 100°C, Kw increases, so both [H⁺] and [OH⁻] are higher than 10⁻⁷, but they are equal. The neutral point shifts to ≈6.14 while the water remains perfectly neutral.
4. pH tells you the current hydrogen concentration: the intensity right now. Alkalinity tells you how much acid the water can absorb before the pH actually moves. A high-pH solution with little buffering capacity can shift sharply when acid is added, while a moderately alkaline solution with strong bicarbonate buffering can resist change. They are functionally different.
5. pH 4: [H⁺] = 10⁻⁴ mol/L, pOH = 10, [OH⁻] = 10⁻¹⁰ mol/L. Check: 4 + 10 = 14.
pH 10: [H⁺] = 10⁻¹⁰ mol/L, pOH = 4, [OH⁻] = 10⁻⁴ mol/L. Check: 10 + 4 = 14.
6. pH 4 vs pH 7 = 3 pH units difference. 10³ = 1,000 times more H⁺ at pH 4.
7. (a) pOH = 14 − 9 = 5. (b) [OH⁻] = 10⁻⁵ mol/L. (c) [H⁺] = 10⁻⁹ mol/L.
8. pH 2.5 → 3.5 = 1 pH unit = 10× decrease in [H⁺]. The technician sees “one point” and thinks the change is small because the pH scale is logarithmic. It compresses enormous changes into small numbers. In reality, 90% of the hydrogen has already been consumed.
9. At pH 1.0: [H⁺] = 0.1 mol/L. Volume: 1,000 gal × 3.785 L/gal = 3,785 L. Moles H⁺: 3,785 × 0.1 = 378.5 mol. KOH reacts 1:1, so 378.5 mol × 56.1 g/mol = 21,234 g ÷ 453.6 g/lb = 46.8 lb. At pH 2.0: [H⁺] = 0.01 mol/L. Moles: 3,785 × 0.01 = 37.85 mol. Mass: 37.85 × 56.1 = 2,123 g = 4.68 lb. One pH unit difference → 10× less chemical needed.
10. Two consequences from the chapter: (1) Protective oxide films on metals rely on hydroxide to form and persist. A pH drop collapses hydroxide concentration, dissolving the film faster than it rebuilds, accelerating corrosion. (2) Mineral scale precipitation depends on hydroxide availability. A pH rise shifts equilibrium toward carbonate formation, and scale deposits where it wasn’t forming before. Even a small pH change matters because each unit represents a 10× shift in both [H⁺] and [OH⁻], reorganizing which reactions are thermodynamically favored.
Chapter 5: Carbonated Water
Chapter note: 05 Carbonated Water
Chapter 5 Problem Set: Carbonated Water
Level 1: Conceptual Mastery
1. Henry’s Law governs gas solubility. How does it differ from the way ionic solids dissolve? Use the language of “pushed” versus “pulled.” (Refer to Chapter 3 if needed).
2. Write the full carbonate equilibrium expression. In one sentence, which species dominates at low, neutral, and high pH.
3. Why is bicarbonate considered a buffer? What dual role does it play?
4. A cooling tower operator stops feeding acid. No chemicals are added, yet the pH rises from 7.5 to 8.8. Explain why, using the concepts of CO₂ stripping and equilibrium shift.
Level 2: Applied Thinking
5. A cooling tower makeup has 250 ppm alkalinity as CaCO₃. Convert to (a) meq/L and (b) eq/L.
6. A water sample shows P-alkalinity = 80 ppm and M-alkalinity = 240 ppm (both as CaCO₃). Which alkalinity species are present? Calculate the bicarbonate and carbonate alkalinity.
7. A boiler’s feedwater M-alkalinity is 15 ppm as CaCO₃. Estimate the CO₂ concentration in the steam/condensate and the neutralizing amine demand at 2.5 ppm amine (active) per 1 ppm CO₂.
Level 3: System Integration
8. A boiler sample shows P-alkalinity = 400 ppm and M-alkalinity = 450 ppm (both as CaCO₃). Identify the species present, calculate the OH-alkalinity, and estimate the pH. Remember that for hydroxide, eq/L = mol/L.
9. The chapter describes two strategies for managing carbonate in boilers: neutralize alkalinity upstream or neutralize acidity downstream. In 2–3 sentences, explain what each approach targets and why alkalinity must be controlled either way.
10. The chapter opens with a story about 2,000 pounds of scale depositing on cooling tower fill after an acid feed failure. Using carbonate equilibrium, explain the chain of events: what left the water, what shifted, what formed, and why pH was the gatekeeper.
Chapter 5 Answer Key: Carbonated Water
1. Ionic solids are pulled into solution. Water’s polarity dismantles the crystal lattice and hydration shells stabilize the freed ions. Gases are pushed into solution by pressure. They occupy gaps in the Hydrogen-Bond Network but are not electrostatically stabilized. When pressure drops or temperature rises, gases leave because nothing is holding them in place.
2. CO₂(g) ⇌ CO₂(aq) ⇌ H₂CO₃ ⇌ HCO₃⁻ + H⁺ ⇌ CO₃²⁻ + 2H⁺. pH determines which species dominates: low pH favors CO₂/carbonic acid, neutral pH favors bicarbonate, and high pH favors carbonate.
3. Bicarbonate can absorb hydrogen (becoming carbonic acid) when acid is added, or donate hydrogen (becoming carbonate) when base is added. This dual role allows it to resist pH change in both directions, which is the definition of a buffer.
4. The tower continuously strips CO₂ from the water through warm temperature, massive airflow, and enormous surface area. As CO₂ leaves, equilibrium shifts to replace it: carbonic acid decomposes, bicarbonate consumes H⁺. The decrease in hydrogen increases the pH.
5. (a) 250 / 50 = 5.0 meq/L. (b) 250 / 50,000 = 0.0050 eq/L.
6. ½M = 120. Since P (80) < ½M (120), bicarbonate + carbonate are present. Carbonate alk = 2P = 160 ppm as CaCO₃. Bicarbonate alk = M − 2P = 240 − 160 = 80 ppm as CaCO₃.
7. CO₂ ≈ 15 × 0.79 = 11.85 ppm. Amine demand = 2.5 × 11.85 = 29.6 ppm active amine.
8. ½M = 225. Since P (400) > ½M (225), hydroxide + carbonate are present. OH– alk = 2P − M = 800 − 450 = 350 ppm as CaCO₃. [OH⁻] = 350 / 50,000 = 0.0070 eq/L = mol/L. pOH = −log10(0.0070) = 2.15. pH = 14 − 2.15 = 11.85.
9. Upstream treatment, such as reverse osmosis or dealkalization, removes bicarbonate before it enters the boiler. This greatly reduces alkalinity-derived CO₂ formation and helps control the condensate corrosion pathway. Downstream treatment uses neutralizing amines that volatilize with steam and condense with water, raising condensate pH and neutralizing acidity before carbonic acid can attack metal. Either way, alkalinity must be controlled because bicarbonate can thermally decompose into CO₂ inside the boiler, and that CO₂ can travel with steam, re-dissolve in condensate, lower pH, and create a corrosive environment.
10. When acid feed stopped, CO₂ stripping continued unchecked. As CO₂ left, equilibrium shifted: carbonic acid decomposed, bicarbonate consumed H⁺, and pH rose. As pH climbed through the 8s, the carbonate fraction increased sharply, making calcium carbonate precipitation increasingly favorable. With calcium already present and 60,000 sq ft of wetted surface area, CaCO₃ precipitated naturally and relentlessly. pH did not cause the scale, but it gave carbonate permission.
Chapter 6: Destructive Oxidation
Chapter note: 06 Destructive Oxidation
Chapter 6 Problem Set: Destructive Oxidation
Level 1: Conceptual Mastery
1. What is the difference between transactional oxidation and destructive oxidation? Why is the formation of NaCl considered transactional while the rusting of iron is destructive?
2. The chapter states that oxidants are not loyal to their intended target. Explain what this means and why it creates a dilemma for water treatment.
3. Four requirements must be present for electrochemical corrosion to occur in water. Name all four and explain why removing any single one stops the reaction.
4. In one sentence, explain the temperature paradox: why can corrosion get worse as water warms from 70°F to 120°F even though oxygen solubility decreases?
Level 2: Applied Thinking
5. Oxygen solubility is approximately 8.3 mg/L at 77°F and 4.4 mg/L at 140°F. Assuming linear behavior, estimate the solubility at 100°F. (Note: actual solubility follows a curve. This question uses linear approximation for simplicity.)
6. Using the Arrhenius rule of thumb (rate doubles per 10°C rise), calculate the approximate kinetic rate multiplier from 20°C to 50°C.
7. The simplified rust reaction is: 4Fe + 3O₂ → 2Fe₂O₃. To produce 10 pounds of rust (Fe₂O₃), how many pounds of iron must be destroyed? (Fe = 55.8, O = 16, Fe₂O₃ = 159.6 g/mol)
Level 3: System Integration
8. Two cooling towers operate at the same pH and conductivity. Tower A reads +650 mV and Tower B reads +200 mV. What can you reasonably conclude from these readings? What additional information would you need before comparing biological control or corrosion risk?
9. A deaerator normally heats boiler feedwater to 220°F, removing nearly all oxygen. The steam valve fails and feedwater drops to 140°F. If the system flows 50,000 lb/hr, how many pounds of dissolved oxygen per hour are now entering the boiler? Assume the feedwater is exposed to air and reaches approximately air-saturated dissolved oxygen at 140°F.
10. The chapter says oxygen is not the strongest oxidant but has two advantages no applied oxidant can match. Name those advantages, and explain why they make oxygen the most persistent corrosion threat in industrial water systems.
Chapter 6 Answer Key: Destructive Oxidation
1. Transactional oxidation transfers electrons without destroying a useful structure. Sodium and chlorine exchange electrons to form a stable salt, but no engineered material or living structure has lost its function. Destructive oxidation strips electrons from a working material, forcing it into a new and less useful form. Rusting converts iron metal into iron oxide; the pipe wall does not spontaneously rebuild itself under normal system conditions.
2. Oxidizers are added to attack vulnerable biological structures, but the same electron-accepting chemistry can also drive metal corrosion when system conditions permit. Their reactions are selective, but that selectivity is controlled by chemistry rather than by the treatment objective. This creates a fundamental tension: enough oxidizing power must be maintained for biological control without creating unacceptable damage to system metallurgy.
3. Anode (a material capable of being oxidized), cathode (the electrically connected surface where electrons are consumed), electrolyte (the water, which allows dissolved ions to move), and metal path (allows electrons to move). Remove the anode: no material to oxidize. Remove the cathode: no electron acceptor. Remove the electrolyte: no ionic pathway to complete circuit. Remove the metal path: electrons cannot flow between anode and cathode. Without all four, electrochemical corrosion in water comes to a halt.
4. Warming decreases oxygen solubility but increases reaction kinetics. In warm water, there may still be enough dissolved oxygen present for the faster kinetics to dominate, producing higher net oxidation rates.
5. Linear interpolation: 100°F is 23/63 of the way from 77°F to 140°F. Δ = 8.3 − 4.4 = 3.9 mg/L. Decrease: 3.9 × (23/63) = 1.4 mg/L. Estimated DO at 100°F ≈ 8.3 − 1.4 = 6.9 mg/L.
6. 20°C → 50°C = +30°C = three 10°C steps. Rate multiplier = 2³ = 8×.
7. Mass fraction of Fe in Fe₂O₃: (2 × 55.8) / 159.6 = 111.6 / 159.6 = 0.699 (≈70%). 10 lb rust × 0.699 = 6.99 lb of iron destroyed.
8. Tower A is exhibiting a more oxidizing ORP response at the probe. That supports the conclusion that its bulk redox environment is more oxidizing at the time of measurement. However, ORP alone cannot establish oxidizer concentration, biological kill, or corrosion rate. The oxidizer used, residual concentration, temperature, demand, contact time, metallurgy, inhibitor program, flow conditions, and corrosion measurements would also be needed.
9. At 140°F, dissolved O₂ ≈ 4.4 mg/L = 4.4 ppm. 1 ppm = 1 lb O₂ per million lb water. 50,000 lb/hr ÷ 1,000,000 × 4.4 = 0.22 lb O₂/hr entering the boiler.
10. Oxygen’s two unmatched advantages: (1) it is dissolved in almost every water system on Earth, and (2) it is constantly replenished by contact with air. Applied oxidants like chlorine or ozone must be dosed and maintained. Oxygen requires no deliberate addition. It is present by default in most waters and can be continuously replenished whenever the system contacts air.
Chapter 7: Corrosion
Chapter note: 07 Corrosion
Chapter 7 Problem Set: Corrosion
Level 1: Conceptual Mastery
1. Name the four components of a corrosion cell. In an open cooling tower, which cathodic reactant is effectively impossible to eliminate, and why?
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.
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?
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?
Level 2: Applied Thinking
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.
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.
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.
Level 3: System Integration
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.
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.
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.
Chapter 7 Answer Key: Corrosion
1. 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.
2. 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.
3. 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.
4. 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.
5. 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)
6. (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.
7. (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.
8. 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.
9. 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.
10. 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.
Chapter 8: Scale
Chapter note: 08 Scale
Chapter 8 Problem Set: Scale
Level 1: Conceptual Mastery
1. A tower is running at LSI = +1.1 but has no visible scaling. Give two operational reasons this can be true.
2. Explain why CaCO₃ often precipitates on heat-transfer surfaces even when the bulk water looks stable.
3. Define supersaturation in one sentence. Then explain why supersaturated water can remain clear for weeks.
4. Explain how a phosphate-based inhibitor can itself become a source of scale. Name the scaling salt involved.
Level 2: Applied Thinking
5. A water treater says, “Our LSI is negative, so we don’t have a scale problem.” The makeup water contains 80 mg/L silica and the tower is running at 5 cycles. Why might the treater be wrong?
6. Makeup conductivity = 850 µS/cm; tower conductivity = 3,400 µS/cm. Estimate cycles of concentration.
7. A tower uses sulfuric acid for pH control. Makeup sulfate = 50 mg/L. At 6 cycles, tower sulfate is measured at 480 mg/L, significantly higher than 50 × 6 = 300 mg/L. Why is the measured value higher than predicted?
Level 3: System Integration
8. Compute pHₛ and LSI for: pH = 8.6, T = 40°C, TDS = 2,200 mg/L, Ca hardness = 320 mg/L as CaCO₃, M-alkalinity = 220 mg/L as CaCO₃. Round logs to 2 decimals.
9. Using pHₛ from Question 8, compute RSI and interpret.
10. Using alkalinity from Question 8, compute pHeq and PSI. Interpret.
Chapter 8 Answer Key: Scale
1. Any two: effective scale inhibitor program extending induction time; good dispersion and low residence time; clean surfaces with few nucleation sites; high turbulence preventing particle adhesion; side-stream filtration removing seed crystals; effective control of hot-spot temperatures.
2. CaCO₃ exhibits inverse solubility: its solubility decreases as temperature rises. Heat-exchange surfaces are the hottest points in the system. Additionally, boundary layers at the surface concentrate ions, local pH can rise at cathodic sites, and the metal surface provides nucleation and anchoring points. Bulk water may be below saturation while the surface microenvironment is above it.
3. Supersaturation means dissolved ions exceed their equilibrium solubility but no stable solid has yet formed. Water can remain clear because nucleation requires an activation barrier to be overcome: ions must collide with compatible orientation and persist long enough to reach a critical cluster size. Until that occurs, thermodynamic permission exists but kinetic action has not started.
4. Polyphosphates hydrolyze into orthophosphate at elevated temperatures. Phosphonates can also revert under thermal or oxidative stress. The resulting orthophosphate reacts with calcium at elevated pH to precipitate calcium phosphate, Ca₃(PO₄)₂. The inhibitor chemistry itself becomes the source of a scaling ion.
5. LSI measures only CaCO₃ scaling pressure. At 5 cycles, silica concentration reaches 80 × 5 = 400 mg/L, far above the ~150 mg/L conservative operating limit for near-neutral water. Silica polymerization and deposition are likely, regardless of LSI. A negative LSI means the water is safe from calcium carbonate scale, not from all scale.
6. COC ≈ 3,400 / 850 = 4.0 cycles.
7. Sulfuric acid (H₂SO₄) adds sulfate to the system in addition to what enters with the makeup water. At each cycle, both the makeup sulfate and the acid-contributed sulfate concentrate. The measured value is higher because the acid feed is introducing sulfate ions not present in the makeup analysis. This is why CaSO₄ risk must be checked in any acid-fed system.
8. A = (log₁₀(2,200) − 1) / 10 = (3.34 − 1.00) / 10 = 0.23. B = −13.12 × log₁₀(313) + 34.55 = −13.12 × 2.50 + 34.55 = 1.75. C = log₁₀(320) − 0.4 = 2.51 − 0.40 = 2.11. D = log₁₀(220) = 2.34. pHₛ = (9.30 + 0.23 + 1.75) − (2.11 + 2.34) = 11.28 − 4.45 = 6.83. LSI = 8.60 − 6.83 = +1.77. Strongly scale-forming. (values are rounding-sensitive; full-precision yields LSI ≈ 1.70, same interpretation)
9. RSI = 2 × 6.83 − 8.60 = 13.66 − 8.60 = 5.06. RSI < 6 indicates scale-forming tendency.
10. pHeq = 1.465 × log₁₀(220) + 4.54 = 1.465 × 2.34 + 4.54 = 3.43 + 4.54 = 7.97. PSI = 2 × 6.83 − 7.97 = 13.66 − 7.97 = 5.69. PSI < 6 indicates scale-forming tendency, consistent with both LSI and RSI. PSI’s contribution here is nuance rather than contradiction. By substituting the equilibrium pH for the measured pH, it accounts for the buffering capacity of the alkalinity during precipitation. The result is still scale-forming, but the PSI value (5.69) sits closer to the borderline than RSI (5.06), reflecting that this water’s alkalinity provides some resistance to runaway precipitation compared to a low-alkalinity water at the same LSI.
Chapter 9: Microbiology
Chapter note: 09 Microbiology
Chapter 9 Problem Set: Microbiology
Level 1: Conceptual Mastery
1. In two sentences, explain why biological control requires fundamentally different thinking than scale or corrosion control.
2. A mature biofilm on a condenser tube is treated with a high dose of chlorine. Planktonic counts drop to near zero within 30 minutes. Two days later, counts are back to pre-treatment levels. Explain what happened using the concepts of EPS, diffusion limitation, and survivor repopulation.
3. Using the oxygen differential cell from the Corrosion chapter, explain how a biofilm creates the same corrosion mechanism as a deposit, but sustains it continuously.
Level 2: Applied Thinking
4. A tower runs at pH 8.4; a colleague says the chlorine is useless and wants to switch to bromine. Under continuous feed, is he right? What does the biofilm’s internal pH do to his argument?
5. A 4-inch branch off a cooling tower recirculation loop has been valved off for 6 months. The water inside is at 85°F. No biocide reaches it. Describe the biological progression you would expect from month 1 through month 6, and explain why this dead leg is a Legionella risk.
6. A facility has used the same non-oxidizing biocide (isothiazolone) on a weekly slug-dose schedule for 18 months. Biological counts have been rising steadily for the last 6 months despite consistent dosing. Explain what is likely happening and propose a modified program.
7. After adding a dispersant to a cooling tower, turbidity spikes from 15 to 50 NTU and the ATP reading in the bulk water triples. The operator wants to stop the dispersant because “it made things worse.” Explain what is actually happening and whether this is a problem or a success.
Level 3: System Integration
8. On Friday afternoon, a cooling tower’s biocide feed pump fails. It won’t be repaired until Monday morning. The water is warm. The system keeps recirculating. For 72 hours, nothing is fed. Using the biofilm lifecycle from this chapter, and what you already know about scale and corrosion, trace the likely sequence of events across the weekend. Where does biology go first? What does its growth do to the heat-transfer surfaces, and how does that pull scale and corrosion in behind it?
9. A hospital cooling tower has excellent bulk water chemistry: free chlorine at 0.5 mg/L, conductivity in range, coupons acceptable, no visible scaling. Legionella culture comes back at 1,000 CFU/mL. Explain how this is possible in a “well-treated” system. Identify at least three potential harboring locations.
10. You are treating a cooling tower with no sidestream filtration, inconsistent biocide feed, and visible biofilm on fill media. The customer asks you to “double the biocide.” Write a 4–6 sentence explanation of why this approach alone will not work, and what additional steps are required. Reference access, shelter, and at least one pillar interaction.
Chapter 9 Answer Key: Microbiology
1. Scale and corrosion respond predictably to changes in conditions: adjust pH, temperature, or concentration and the problem slows or stops. Biology adapts: it responds to pressure by reorganizing, building shelter (biofilm), and recovering from partial disruption, meaning that changing conditions alone is rarely sufficient for lasting control.
2. The high chlorine dose cleared the water column of planktonic organisms, which is what the bulk water test measures. However, the established biofilm on the condenser tube was protected by its EPS matrix, which consumed the oxidant through reaction before it could diffuse to the inner cells. Surviving organisms deep in the biofilm structure were never exposed to lethal concentrations. Once chlorine levels dropped, these survivors repopulated the EPS scaffold and shed new planktonic cells into the water column, returning counts to pre-treatment levels within 48 hours. The treatment killed the messengers but left the city intact.
3. A deposit creates an oxygen differential cell by physically separating aerated bulk water from the oxygen-depleted zone beneath it. The covered metal becomes anodic while the surrounding exposed metal becomes cathodic. A biofilm creates the same differential, but actively maintains it: microbial respiration continuously consumes oxygen at the metal interface, ensuring the anodic zone never re-passivates. Additionally, the biofilm generates metabolic acids that lower pH and, in the case of SRB, produces hydrogen sulfide that directly attacks the metal surface. Unlike a passive deposit, the biofilm regenerates the corrosive conditions around the clock.
4. He is not correct that “chlorine is useless.” At pH 8.4, the bulk water contains a much smaller fraction of chlorine as HOCl, so chlorine’s short-contact kill rate is reduced. That is where bromine has a real advantage: short contact time, high pH, and a need for rapid free-halogen activity. But a continuously fed cooling tower is not judged by short-contact speciation alone. If free chlorine residual is maintained for hours, contact time, residual persistence, oxidant demand, and surface access matter as much as the instantaneous HOCl percentage. The biofilm also weakens his argument because the pH inside and near the biofilm is often lower than the bulk water, which shifts more chlorine back toward HOCl where organisms are actually sheltered.
5. Month 1: Planktonic organisms already present in the stagnant water begin attaching to pipe surfaces. Initial EPS production begins. Month 2–3: Without biocide access or flow to disrupt attachment, biofilm matures. Oxygen is consumed at the metal surface, creating anaerobic zones. SRB may begin colonizing the base of the biofilm. Month 4–6: A mature, multi-species community develops with established gradients. The 85°F water temperature is within Legionella’s growth range, though below its optimal growth range (~95–105°F). Legionella replicates inside protozoa (amoebae) that thrive within the biofilm. No biocide reaches the branch. No flow disrupts the community. No thermal treatment exceeds growth temperature. This dead leg is a Legionella reservoir that, if reconnected to the system, could release organisms into the circulating water and potentially into aerosol-generating components.
6. Eighteen months of the same non-oxidizing biocide has created sustained selection pressure. Organisms tolerant to isothiazolone have been preferentially selected. The community has adapted, not disappeared. A modified program should: (a) introduce a different non-oxidizing chemistry (e.g., glutaraldehyde or DBNPA) to disrupt the selection pressure, (b) alternate between at least two non-oxidizing chemistries on a rotating schedule, (c) add a dispersant to expose biofilm-sheltered organisms to chemical treatment, and (d) consider mechanical cleaning of accessible surfaces to physically remove the adapted biofilm community before restarting chemical treatment.
7. The dispersant is working. The turbidity spike and increased ATP in bulk water indicate that the dispersant is lifting biofilm material from pipe and fill surfaces into the water column. The “worse” numbers reflect what was already present on the surfaces but invisible to bulk water testing. This is a success: the shelter has been disrupted and the previously protected organisms are now exposed to the biocide program. The correct next step is to follow the dispersant application with a biocide treatment to kill the newly exposed organisms, and to increase blowdown or filtration to remove the suspended biological material from the system.
8. The order is reliable; the timing is not. A biocide residual suppresses the population, but it never empties the system. Pull it, and the first thing that happens is chemical, not biological: the last oxidant is consumed and nothing replaces it. Only then does biology move, and it moves in a set order. Freed planktonic cells multiply in the warm water, find surfaces, and attach; EPS production ramps up and any suppressed biofilm resumes growth, thickening where it matters most, on the heat-transfer surfaces. That film is an insulator. The surface beneath it runs hotter than the bulk water, and you’ve met both consequences: calcium carbonate has inverse solubility, so a hotter surface scales more eagerly, while respiration beneath the film consumes the oxygen the metal can’t replenish, setting up the oxygen-differential cell from the Corrosion chapter. Under-deposit corrosion begins. How far a system travels in three days depends on temperature, nutrient load, and how much biofilm was already there when the pump failed. A warm, fouled system moves far in a weekend; a cold, clean one barely stirs. The lesson is the direction, not the calendar: biology reaches the surface first, and scale and corrosion follow it in. Killing the organisms Monday won’t undo that. The shelter is already built.
9. Excellent bulk water chemistry does not guarantee clean surfaces. The 0.5 mg/L free chlorine residual maintains the water column but may not penetrate established biofilms on pipe surfaces, fill media, or basin walls. Legionella replicates inside amoebae within biofilms, where it is doubly protected: by the host organism and by the EPS matrix. Three potential harboring locations: (1) Dead legs or low-use branches with stagnant, warm water and no biocide circulation. (2) Fill media surfaces where biofilm accumulates in low-flow zones between distribution nozzles. (3) The basin sump or areas beneath the tower where sediment and biological debris settle, providing shelter and nutrients. The system is “well-treated” by bulk water standards but has not been evaluated by surface standards. The Legionella result is a surface problem, not a water column problem.
10. Doubling the biocide will not solve this problem because the limiting factor is access, not potency. Visible biofilm on the fill media means established EPS infrastructure is present. Oxidant will react with the EPS matrix before reaching the organisms inside, and higher doses simply mean more chemistry consumed by the shelter rather than more organisms killed. Without sidestream filtration, suspended solids and biological debris recirculate continuously, providing nutrients and new attachment sites. The correct approach is: (1) mechanical cleaning of the fill and basin to physically remove the biofilm scaffold, (2) addition of a dispersant to lift remaining deposits from surfaces, (3) restoration of consistent biocide feed with rotation between oxidizing and non-oxidizing chemistries, and (4) installation or repair of sidestream filtration to remove suspended biological material and reduce the nutrient load that fuels regrowth. The biofilm is also creating conditions that accelerate corrosion through oxygen differential cells at the metal surface, something biocide alone cannot address.
Chapter 10: Suspended Solids
Chapter note: 10 Suspended Solids
Chapter 10 Problem Set: Suspended Solids
Level 1: Conceptual Mastery
1. Explain in one sentence why dissolved solids go wherever the water goes, but suspended solids go wherever the system lets them. What physical principle governs each?
2. A cooling tower has excellent water chemistry: inhibitor residuals on target, biocide fed on schedule, pH and conductivity within range. Corrosion coupons look acceptable, but during shutdown you find aggressive pitting beneath silt deposits on horizontal pipe runs. The basin has 3 inches of settled silt. Explain how these observations are consistent.
3. Describe the potential sequence of consequences through all three pillars (corrosion, biology, scale) for a windstorm that deposits a heavy layer of dust in a cooling tower basin if the solids are not removed.
4. A closed chilled water loop shows brown, turbid water at 45 NTU despite having been in operation for two years. Is this a suspended solids problem or something else? What would you investigate first?
Level 2: Applied Thinking
5. A cooling tower has a system volume of 3,000 gallons and a recirculation rate of 800 GPM. (a) What sidestream flow rate is needed to turn over the system volume every 4 hours? (b) What percentage of the recirculation rate is this? (c) If the tower is near an active construction site, would you recommend increasing this target?
6. Using the concept from Stokes’ Law, explain why a hydrocyclone separator is effective for removing rust particles (density ~5,200 kg/m³, diameter ~50 µm) but ineffective for biological debris (density ~1,050 kg/m³, diameter ~20 µm). You do not need to calculate, just explain the reasoning.
7. Your corrosion coupon program shows carbon steel rates of 1.2 MPY, within the “good” range. But during a shutdown, you find deep pitting beneath silt deposits on horizontal pipe runs. Explain why the coupons missed this, and propose a change to the monitoring program.
Level 3: System Integration
8. A 1,000-ton chiller system operates 4,000 hours per year at 8,000 installed. If it restores the condenser to clean conditions within one cooling season, what is the simple payback period?
9. A water treater increases the dispersant dose in a heavily fouled cooling tower. Within 48 hours, turbidity increases from 20 to 60 NTU and condenser approach temperature actually improves by 1°F. Explain what is happening and whether this is a problem or a success.
10. You are called to a cooling tower that has not had sidestream filtration, has been running without consistent biocide treatment, and has visible silt deposits in the basin. The customer wants you to “fix the chemistry.” Write a brief explanation (4–6 sentences) of why adjusting chemical feed rates alone will not solve this problem. Reference at least two amplification pathways.
Chapter 10 Answer Key: Suspended Solids
1. Dissolved solids have been dismantled at the molecular level and are governed by chemistry (equilibrium, solubility, reaction kinetics). They move wherever the water moves because they are part of the solution. Suspended solids are intact particles governed by physics (gravity, flow velocity, inertia). They settle wherever the system’s hydraulics allow them to stop moving.
2. The bulk water chemistry is excellent, but it is only measuring the water column. The 3 inches of settled silt have created deposits on horizontal pipe surfaces. Beneath those deposits, oxygen is consumed but cannot be replenished, creating oxygen differential cells that drive under-deposit corrosion. The corrosion inhibitor, pH, and biocide measured in the bulk water are not reaching the metal surface beneath the deposits. The coupons, which are typically placed in the flowing water column on clean holders, are measuring the chemistry the pipes are not actually receiving. Excellent bulk chemistry and aggressive pitting are entirely consistent when deposits are present.
3. Dust settles in low-flow zones and on pipe surfaces, creating deposits. The deposits create oxygen differential cells that drive under-deposit corrosion (Corrosion pillar). Corrosion products (iron oxide) generate additional suspended solids, accelerating the cycle. The deposits also provide shelter for biological attachment: prefabricated homes for biofilm formation that are shielded from biocides in the water column (Biology pillar). The biofilm produces EPS that traps additional particles, further stabilizing the deposit. Meanwhile, the deposit surfaces provide heterogeneous nucleation sites that lower the kinetic barrier for scale formation, particularly on heat-exchange surfaces where temperature is highest and hydration is most stressed (Scale pillar). A single windstorm, left unaddressed, cascades through all three pillars.
4. In a closed system that has been operating for two years, 45 NTU is a suspended-solids problem, but the source is probably corrosion rather than external dirt. Closed systems are not continuously exposed to atmospheric dust, but they can still accumulate solids from makeup water, construction debris, maintenance activities, process leaks, line breaks, and internal corrosion. In this case, the brown, turbid water most likely indicates iron oxide generated by active corrosion of carbon-steel piping. Investigate first: (a) check the corrosion inhibitor residual: is the treatment program being fed consistently? (b) filter a sample and examine the particulate: brown/red suggests iron oxide, black suggests magnetite or SRB activity, green suggests copper corrosion products. (c) check whether the system was properly flushed after construction: construction debris from two years ago can continue circulating indefinitely.
5. (a) Required throughput = 3,000 gal ÷ 4 hr = 750 gal/hr = 12.5 GPM. (b) 12.5 GPM ÷ 800 GPM = 1.56% of the recirculation rate. (c) Yes. Near an active construction site, solids loading will be significantly higher than normal. Increasing the sidestream flow rate to turn over the system volume every 2–3 hours (18.75–25 GPM, or roughly 2.3–3.1% of the recirc rate) would be prudent. Additionally, check the filter type: construction dust may include fine particles that require a sand or multimedia filter rather than a screen or disk filter.
6. Stokes’ Law shows that settling velocity depends on the square of the particle diameter and the density difference between the particle and water. Rust particles have a very high density difference (ρp − ρw ≈ 5,200 − 998 = 4,202 kg/m³) and a moderate diameter (50 µm). The hydrocyclone generates centrifugal force that accelerates settling. Heavy, dense particles are thrown to the outer wall and separated. Biological debris has a density barely above water (ρp − ρw ≈ 1,050 – 998 = 52 kg/m³), roughly 80× less density difference, and a smaller diameter (20 µm). The centrifugal force is not sufficient to separate particles with such a small density difference from the water. They simply recirculate with the flow.
7. Standard corrosion coupons are mounted in the flowing water column on clean, smooth holders. They measure uniform corrosion in the bulk water environment. Under-deposit corrosion is localized: it occurs beneath silt deposits on horizontal pipe surfaces where the metal is isolated from the treatment program. The coupon never sees this condition because it has no deposit on it. To detect under-deposit corrosion, the monitoring program should be supplemented with: (a) visual inspection of horizontal pipe runs during shutdowns, looking for deposits and localized pitting beneath them; (b) ultrasonic thickness testing on deposit-prone sections; (c) placing some coupons horizontally with intentional deposits or in spool pieces that mimic pipe conditions rather than in clean, vertical holders.
8. (a) Energy penalty = 3°F × 1.5%/°F = 4.5%. (b) Baseline energy: 1,000 tons × 0.65 kW/ton = 650 kW. Annual energy: 650 kW × 4,000 hr = 2,600,000 kWh. Annual cost: 2,600,000 kWh × 312,000**. Energy penalty cost: 14,040/year**. (c) Simple payback: 14,040/yr = 0.57 years, or approximately 7 months. The filter pays for itself well within the first cooling season. And this calculation does not include reduced chemical cost, lower maintenance labor, or extended equipment life.
9. The dispersant is working. The turbidity spike from 20 to 60 NTU indicates that the dispersant is lifting deposits from heat-exchange surfaces and piping into the water column. The 1°F improvement in condenser approach temperature confirms this. The insulating deposit layer on the condenser tubes has been partially removed, restoring heat transfer. The “worse” turbidity reading is actually showing material that was already in the system but hidden on surfaces, invisible to bulk water monitoring. This is a success. The correct next steps are to increase blowdown temporarily or ensure sidestream filtration is running to capture the suspended material, and to follow with a biocide treatment to kill organisms that were sheltered beneath the deposits and are now exposed.
10. Adjusting chemical feed rates alone will not solve this problem because the suspended solids have created deposits that isolate surfaces from whatever chemistry is in the water. Increasing corrosion inhibitor concentration does not help when the inhibitor cannot reach the metal beneath the silt deposits. Under-deposit corrosion will continue regardless of bulk inhibitor residual. Increasing biocide dose does not help when the biocide is consumed by EPS and organic debris in the deposits before it can reach the organisms sheltered within them. The deposits must be physically removed first: the basin needs to be cleaned, sidestream filtration must be installed to prevent reaccumulation, and a dispersant program should be used to lift existing deposits off surfaces. Only then will the chemistry operate in the environment it was designed for. The customer is asking to fix a mechanical problem with a chemical solution, and the amplifier effect guarantees that approach will fail.
Chapter 11: The Pillars of Water Treatment
Chapter note: 11 The Pillars of Water Treatment
Chapter 11 Problem Set: The Pillars of Water Treatment
Level 1: Conceptual Mastery
1. A cooling tower suddenly shows higher conductivity, higher pH, and weaker oxidant residual. Before adjusting any feed pump, list:
- What may have changed in what enters?
- What may have changed in what leaves?
- What may now be accumulating?
- What transformations may be occurring faster than before?
2. For each of the following components, write one sentence defining the system boundary and two likely inputs and outputs:
- Cooling tower basin
- Condenser tube
- Boiler feedwater tank
- Dead leg in a chilled water system
3. For each case, identify the primary pillar and the most likely amplifier:
- Deep pitting beneath brown tubercles on mild steel
- White mineral deposit on a warm tube with soft slime beneath it
- Rapid oxidant demand increase after weeks of poor filtration
- Stable bulk chemistry but increasing condenser approach temperature
Level 2: Applied Thinking
4. A feedwater tank normally receives 80% condensate return at 180°F and 20% makeup at 65°F. Suddenly the condensate return drops to 35%, while total flow remains unchanged. Explain qualitatively what happens to:
- Feedwater temperature
- Conductivity
- Dissolved oxygen load
- Sulfite demand
Then explain why increasing sulfite feed alone may not solve the problem.
5. A condenser tube receives tower water at 4 cycles, pH 8.3, calcium hardness 220 mg/L as CaCO₃ in the makeup, visible suspended solids, and intermittent oxidant feed. Write a free body diagram that includes the following, along with which pillar you would address first.
- System boundary
- Inputs
- Outputs
- Accumulations
- Transformations
- Likely surface conditions not captured by the bulk water tests
6. A valved-off 4-inch branch line has been stagnant for 2 years in a nitrite-treated chilled water loop. Using the four questions, predict:
- What no longer enters
- What no longer leaves
- What may have accumulated
- What chemical and biological transformations likely occurred
Then identify the dominant failure mode.
7. A facility reports:
- Rising condenser approach
- Higher coupon corrosion rates
- Higher biocide demand
- Visible silt in the tower basin
Use the free body diagram approach to identify the most likely root cause and explain how one problem can create all four symptoms.
Level 3: System Integration
8. A cooling tower has high calcium hardness and a phosphate-based corrosion inhibitor. Scale pressure is rising. Lowering pH would help with CaCO₃ control, but may destabilize other parts of the treatment strategy.
Give two alternative responses besides simply feeding more acid. For each, explain the consequences across all four pillars.
9. A condenser system shows:
- Inhibitor residual on target
- pH in range
- Conductivity in range
- Biocide schedule maintained
- No visible heavy scale
But corrosion coupons show 6.5 MPY. Explain at least two ways this can happen using the distinction between bulk-water chemistry and surface conditions.
10. Design a 6-step first-visit walkthrough for a cooling tower system using the logic of this chapter. For each stop, state:
- What component you are evaluating
- What you are looking for
- Which of the four questions you are answering
- Which pillar(s) you are primarily evaluating
Chapter 11 Answer Key: The Pillars of Water Treatment
1. Likely possibilities include:
- What enters: Makeup fraction may have increased, incoming solids may have increased, contamination may have entered, or inlet temperature may have changed.
- What leaves: Blowdown may have decreased, a valve may have failed, filtration may have lost effectiveness, or another removal pathway may have changed.
- What accumulates: Dissolved solids, deposits, biomass, and sludge may now be accumulating.
- What changes form: Bicarbonate may be shifting toward carbonate at higher pH, oxidant demand may be increasing through reaction with organics or biofilm, and dissolved species may be precipitating as solids.
2. A strong answer clearly defines each component as a bounded control volume and identifies plausible inflows and outflows appropriate to that component.
The key is not one exact answer, but correct diagnostic framing:
- a clear system boundary
- realistic inputs
- realistic outputs
- treats the component as part of a larger system
3. These cases allow for some interpretation, but strong answers will generally follow this logic:
- Primary: corrosion. Amplifier: solids/deposit formation. The deposit creates the oxygen differential and helps localize attack.
- Primary: scale. Amplifier: biology. The mineral deposit is the dominant feature, but biofilm likely contributed by insulating the surface and trapping ions.
- Primary: biology. Amplifier: solids. Poor solids control creates shelter and surface area for biological growth, increasing oxidant demand.
- Primary: scale or solids. Amplifier: biology, if biofilm is part of the insulating layer. The key insight is that the problem is likely at the surface rather than in the bulk water.
4. When condensate return collapses and makeup rises:
- Feedwater temperature decreases
- Conductivity increases
- Dissolved oxygen load increases
- Sulfite demand increases
Increasing sulfite feed addresses one symptom, but the deeper problem is the change in the system’s mass and heat balance. The correct diagnosis is not simply “the sulfite program is failing,” but “the feedwater system has shifted to a different input balance.”
5. A strong answer includes:
- System boundary: condenser tube or condenser tube bundle
- Inputs: warm recirculating water, dissolved calcium and alkalinity, suspended solids, intermittent oxidant, heat-transfer load
- Outputs: water leaving at changed temperature, consumed chemistry, any released corrosion products or biological fragments
- Accumulations: scale, solids, biofilm, corrosion products
- Transformations: precipitation, oxidant consumption, biological attachment and growth, corrosion reactions
- Surface conditions not captured by bulk tests: elevated skin temperature, sheltered deposits, localized concentration effects, reduced chemistry penetration beneath deposits
The most defensible first priority is often solids control if visible suspended matter is high, because solids amplify deposition, shelter biology, and promote under-deposit corrosion. A stronger oxidant strategy may also be necessary, but cleaning and solids control often come first.
6. Strong answers may include:
- What no longer enters: Fresh inhibitor, fresh water, oxidant, or meaningful flow
- What no longer leaves: Water, corrosion byproducts, or biological material
- What may have accumulated: Corrosion products, stagnant water, depleted chemistry, biomass
- What likely changed form: Dissolved oxygen is consumed, inhibitor residual is depleted, localized microbial communities may establish, and corrosion proceeds in chemically isolated conditions
Dominant failure mode: localized corrosion, often with significant MIC risk.
7. Most likely root cause: failure of solids management, often due to filtration bypass, poor basin cleanliness, or excessive suspended loading.
A strong cascade explanation would be:
- Solids accumulate in the basin and low-flow zones
- Deposits form on heat-transfer and metal surfaces
- Deposits create differential aeration and under-deposit corrosion
- Deposits shelter biology, which raises biocide demand
- Deposit and biofilm insulation increase condenser approach
This is exactly the kind of multi-symptom problem that the free body diagram is meant to untangle.
8. Strong answers may include two or more of the following:
- Increase blowdown to reduce concentration of calcium and alkalinity
- Improve solids removal and deposit control
- Optimize dispersant or polymer feed
- Shift to a different corrosion inhibitor strategy
- Improve control stability and monitoring rather than making a large acid adjustment
The best answers trace the consequences through all four pillars:
- reduced scale pressure
- possible effects on corrosion protection
- indirect effects on biology through cleaner surfaces
- changes in solids accumulation and transport
9. Two strong explanations are:
- Under-deposit corrosion: Bulk chemistry may be acceptable, but deposits isolate the metal beneath them from inhibitor protection and create local oxygen differentials.
- MIC or localized biofilm attack: The water column may test well while the surface microenvironment remains biologically active and corrosive.
Other acceptable explanations include:
- flow-related erosion-corrosion
- galvanic effects
- poor coupon location
- coupon exposure that reflects a particularly aggressive part of the system rather than the average system condition
The key idea is that bulk water data does not guarantee surface protection.
10. A strong answer should move physically and logically through the system. A common sequence would include:
- Basin or sump
- Fill or distribution deck
- Sample point and control station
- Condenser-side performance review
- Blowdown and conductivity control equipment
- Filtration or side-stream solids removal equipment
The best answers explain, at each stop:
- what the component is
- what evidence is being gathered
- which diagnostic question is being answered
- which pillar or pillars are being evaluated
Chapter 12: Cooling Things Down
Chapter note: 12 Cooling Things Down
Chapter 12 Problem Set: Cooling Things Down
Level 1: Conceptual Mastery
1. Why do large cooling systems move cooling with water rather than air? In your answer, reference specific heat, density, and the practical size of the distribution system.
2. In the mass balance, evaporation removes pure water while blowdown removes concentrated water. Explain why this distinction matters for how dissolved solids accumulate, and why the water treater controls blowdown (not evaporation) to manage chemistry.
3. A facility manager notices that the conductivity controller is holding its setpoint perfectly, and concludes that the cooling tower chemistry must be under control. Explain why this conclusion might be wrong. What diagnostic would you use to verify the mass balance, and what would the results tell you?
Level 2: Applied Thinking
4. A cooling tower serves a 750-ton chiller (750 refrigeration tons). The tower range is 10°F. Using an evaporation factor of 0.85:
- Calculate the cooling tower heat load in BTU/hr.
- Estimate the recirculation rate in GPM.
- Calculate the evaporation rate in GPM.
- Calculate blowdown and total makeup at 4 cycles of concentration.
5. Using the evaporation rate from Problem 4, calculate the total annual makeup water (in gallons) at 3 cycles and at 6 cycles, assuming 4,380 hours of operation per year. What is the annual water savings from increasing cycles?
6. A tower operates at 5 cycles based on conductivity. Lab results show:
MAKEUP: Chloride = 48 mg/L; Calcium = 160 mg/L as CaCO₃
TOWER: Chloride = 245 mg/L; Calcium = 640 mg/L as CaCO₃
- Calculate the chloride-based COC and the calcium-based COC.
- Is the mass balance intact? If not, what percentage of calcium is leaving the water as scale?
- What would you recommend as the next diagnostic step or treatment response?
7. An 800-ton chiller operates at 0.60 kW/ton under clean conditions for 4,500 hours per year at $0.11/kWh. Condenser tube fouling has increased the approach temperature by 4°F. Using the upper-end estimate (2% energy increase per °F of approach rise), calculate:
- Annual energy cost under clean conditions.
- Annual energy cost under fouled conditions.
- The annual cost penalty from fouling.
Level 3: Optimization
8. A cooling tower evaporates 20 GPM and operates 4,000 hours per year. The facility is evaluating whether to increase from 4 cycles to 6 cycles of concentration. The following costs apply:
Water: $6.00 per 1,000 gallons (Applies to Makeup)
Sewer: $8.00 per 1,000 gallons (Applies to Blowdown)
Chemical treatment at 4 cycles: $1.80 per 1,000 gallons of makeup
Chemical treatment at 6 cycles: $2.50 per 1,000 gallons of makeup
-
Calculate annual blowdown and makeup at each cycle count.
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Calculate annual water cost, sewer cost, and chemical cost at each cycle count.
-
What is the net annual savings from increasing to 6 cycles?
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In one or two sentences, explain why the chemical cost increases even though total makeup decreases.
9. Your makeup water contains 320 mg/L calcium hardness as CaCO₃, 210 mg/L alkalinity as CaCO₃, and 150 mg/L chloride. Your scale inhibitor is rated for a maximum of 1,400 mg/L calcium hardness and 1,000 mg/L alkalinity in the tower water.
- What is the maximum number of cycles allowed by the calcium hardness limit?
- What is the maximum number of cycles allowed by the alkalinity limit?
- Which constraint governs, and at that cycle count, what are the tower concentrations of calcium, alkalinity, and chloride?
- Explain why the actual maximum safe cycle count may be lower than the value you calculated, considering pH, temperature, and the interactions between scaling parameters.
10. You are called to evaluate a cooling tower that is consuming significantly more makeup water than expected. The conductivity controller is functioning and holding tower conductivity at 2,400 µS/cm. Makeup conductivity is 600 µS/cm. Based on the calculated cycles, the makeup meter shows water consumption approximately 30% higher than the mass balance predicts. Identify at least three possible explanations for the discrepancy and describe how you would investigate each one.
Chapter 12 Answer Key: Cooling Things Down
1. Air has a specific heat of approximately 0.24 BTU/lb·°F versus water’s 1.0, and is roughly 800 times less dense. To move the same BTU load, air-based systems require vastly greater volumetric flow through massive ductwork with high-powered fans. A chilled water system consolidates cooling into a central plant with compact piping – a six-inch chilled water pipe carries the same capacity as a four-foot air duct – with a fraction of the pumping energy and far simpler redundancy.
2. Evaporation removes only pure water – dissolved solids cannot evaporate and remain behind in the tower. This is why concentration increases even though no solids are being added beyond what arrives in the makeup. Blowdown removes concentrated water, carrying dissolved solids out of the system and relieving the accumulation. The water treater controls blowdown because evaporation is fixed by the heat load. Evaporation cannot be reduced without reducing the cooling demand itself. Blowdown is the only controllable mechanism for managing how concentrated the tower water becomes.
3. Conductivity tracks total dissolved solids but does not distinguish between species. If calcium carbonate is precipitating as scale, its contribution to conductivity disappears – but the controller does not know this. Makeup continues entering, the conductivity setpoint holds, and the true concentration of scaling species drifts from what conductivity reports. The diagnostic is to compare cycles of concentration calculated from a soluble tracer ion (chloride) against cycles calculated from a scaling-risk ion (calcium). If chloride-based COC is significantly higher than calcium-based COC, calcium is leaving the water as scale. The divergence quantifies how much.
4. (a) Q = 750 tons × 15,000 BTU/hr per CT ton = 11,250,000 BTU/hr. (b) Recirc = 750 × 3 GPM/ton = 2,250 GPM. (c) E = 2,250 × 10 × 0.85 ÷ 1,000 = 19.1 GPM. (d) B = 19.1 ÷ (4 − 1) = 6.37 GPM. M = 19.1 + 6.37 = 25.5 GPM. Check: M = 19.1 × 4 ÷ 3 = 25.5 GPM. ✓
5. At 3 cycles: B = 19.1 ÷ (3 − 1) = 9.55 GPM. M = 19.1 + 9.55 = 28.65 GPM. Annual = 28.65 × 60 × 4,380 = 7,529,220 gal/yr ≈ 7.53 Mgal. At 6 cycles: B = 19.1 ÷ (6 − 1) = 3.82 GPM. M = 19.1 + 3.82 = 22.92 GPM. Annual = 22.92 × 60 × 4,380 = 6,023,376 gal/yr ≈ 6.02 Mgal. Savings = 7.53 − 6.02 = 1.51 Mgal/yr. Over 1.5 million gallons saved on a single tower by increasing from 3 to 6 cycles.
6. Chloride COC = 245 ÷ 48 = 5.10. Calcium COC = 640 ÷ 160 = 4.00. The mass balance is not intact. Chloride-based COC exceeds calcium-based COC, meaning calcium is leaving the water as scale. Expected calcium at 5.10 cycles = 160 × 5.10 = 816 mg/L as CaCO₃. Actual = 640 mg/L. Calcium lost = (816 − 640) ÷ 816 = 21.6% precipitating as scale. Next steps: evaluate LSI/RSI at current conditions, check condenser approach temperature for fouling evidence, review scale inhibitor residual and dosing, and consider whether cycles need to be reduced or inhibitor chemistry adjusted.
7. Clean conditions: 800 tons × 0.60 kW/ton = 480 kW. Annual energy = 480 × 4,500 = 2,160,000 kWh. Annual cost = 2,160,000 × 237,600**. Fouled conditions: 4°F × 2% = 8% energy increase. Additional power = 480 × 0.08 = 38.4 kW. Total = 518.4 kW. Annual energy = 518.4 × 4,500 = 2,332,800 kWh. Annual cost = 2,332,800 × 256,608**. Annual penalty from fouling = 237,600 = $19,008.
8. (a) At 4 cycles: B = 20 ÷ (4 − 1) = 6.67 GPM. M = 20 + 6.67 = 26.67 GPM. Annual makeup = 26.67 × 60 × 4,000 = 6,400,000 gal. Annual blowdown = 6.67 × 60 × 4,000 = 1,600,000 gal. At 6 cycles: B = 20 ÷ (6 − 1) = 4.0 GPM. M = 20 + 4.0 = 24.0 GPM. Annual makeup = 24.0 × 60 × 4,000 = 5,760,000 gal. Annual blowdown = 4.0 × 60 × 4,000 = 960,000 gal.
(b) At 4 cycles: Water = 6,400 kgal × 38,400. Sewer = 1,600 kgal × 12,800. Chemical = 6,400 kgal × 11,520. Total = 6.00/kgal = 8.00/kgal = 2.50/kgal = 56,640.
(c) Net annual savings = 56,640 = **8,960. Chemical costs increase by $2,880. The net savings are positive.
(d) Higher cycles concentrate every dissolved species in the tower water – calcium, alkalinity, silica, chloride, sulfate – increasing supersaturation and scaling pressure. The treatment program must work harder to keep those species in solution, which requires more aggressive (and more expensive) inhibitor chemistry even though the total volume of water being treated is lower.
9. Calcium limit: 1,400 ÷ 320 = 4.38 cycles. Alkalinity limit: 1,000 ÷ 210 = 4.76 cycles. Calcium governs – round down to 4 cycles for safety margin. At 4 cycles: calcium = 320 × 4 = 1,280 mg/L as CaCO₃; alkalinity = 210 × 4 = 840 mg/L as CaCO₃; chloride = 150 × 4 = 600 mg/L. The actual safe maximum may be lower because CO₂ stripping in the tower drives pH upward. At 840 mg/L alkalinity, pH will likely exceed 8.3, shifting the carbonate equilibrium strongly toward CO₃²⁻ and increasing CaCO₃ supersaturation beyond what calcium hardness alone would suggest. The LSI will be significantly positive. Additionally, condenser tube surface temperatures are higher than bulk water temperature, and inverse solubility means calcium carbonate is least soluble exactly where heat transfer matters most. The true operating limit is determined by the interaction of hardness, alkalinity, pH, and temperature – not by any single parameter.
10. Tower cycles are 2,400 ÷ 600 = 4 cycles. If the tower is truly holding 4 cycles, then excessive drift or a tower-water leak would not normally explain makeup above the predicted value. Those losses remove concentrated tower water, so they act like uncontrolled blowdown. The conductivity controller should respond by reducing controlled blowdown, and total makeup should remain governed by evaporation and cycles. If the concentrated losses became larger than the required blowdown, tower conductivity would fall below setpoint.
More consistent explanations are: (1) Evaporation is higher than assumed. The heat load, condenser water flow, tower range, or operating hours may be higher than the values used in the calculation. Investigate by verifying condenser water flow, entering and leaving condenser water temperatures, tower range, and chiller load. (2) The makeup meter is over-reading. The meter may have an incorrect multiplier, scaling error, installation issue, or mechanical problem. Investigate by checking the meter setup and comparing it against a portable flow meter or controlled fill test. (3) Some metered makeup water is not actually reaching the tower. A branch line, hose bib, washdown connection, or other user may be connected downstream of the meter. Investigate by tracing the piping and isolating branch lines. (4) A makeup-line leak may exist downstream of the meter but upstream of the tower. This would increase metered water use without changing tower conductivity. Investigate by closing the tower makeup isolation valve and checking whether the meter continues to move.
Chapter 13: Heating Things Up
Chapter note: 13 Heating Things Up
Chapter 13 Problem Set: Heating Things Up
Level 1: Conceptual Mastery
1. At 140°F, feedwater contains roughly 4 mg/L dissolved oxygen, which is half of what it held at room temperature. Explain why corrosion risk is higher at 140°F, not lower.
2. A boiler burns 100 billion BTU per year. If 1/32” of calcium carbonate scale reduces efficiency by 2%, estimate the annual cost of that inefficiency at $10/MMBTU.
3. A facility reports water hammer every weekday morning at 7 AM but never on weekends. The boiler chemistry is within ASME limits. Is the problem more likely priming or foaming? Explain.
4. List four consequences of losing 50% of condensate return, and for each, identify which link in the boiler circuit is affected.
Level 2: Applied Thinking
5. Feedwater contains 3 ppm dissolved oxygen. (a) Calculate the stoichiometric sulfite requirement. (b) If the boiler is operating at 10 cycles of concentration and the target boiler residual is 30 ppm, what total sulfite concentration would you target in the boiler feedwater?
6. A plant returns 8,000 lb/hr of condensate at 170°F. Cold makeup is 60°F. Boiler efficiency is 82%. Natural gas costs $12/MMBTU. The plant operates 8,000 hours per year. (a) Calculate the hourly energy savings from condensate return. (b) Calculate the annual fuel savings in dollars.
7. A boiler produces 20,000 lb/hr of steam. Feedwater conductivity = 50 µS. (a) At 40 cycles, what is the boiler water conductivity? (b) If cycles drop to 10 due to lost condensate, and feedwater conductivity rises to 200 µS, what is the new boiler water conductivity? (c) Using the ASME table, is this still within guidelines for a 150 psig boiler? Explain why conductivity alone does not tell the full story.
Level 3: System Integration
8. Your blowdown conductivity controller reads 2,000 µS. Your handheld meter reads 3,800 µS on a cooled grab sample. The blowdown valve has been closed for hours. (a) What has happened? (b) Estimate the actual cycles if feedwater conductivity is 80 µS. (c) Using the ASME table, is this boiler (300 psig) at risk? For what?
9. A technician reports: sulfite residual is zero in the boiler, scale inhibitor is depleted, and condensate pH has dropped to 5.5. Instead of increasing all three chemical pumps, describe the diagnostic sequence you would follow to identify the root cause. What single failure could explain all three symptoms?
10. Using the Perfect Storm scenario (90% condensate return dropping to 37%), estimate the fuel cost of heating the additional cold makeup water. Assume: steam production = 25,000 lb/hr, condensate returns at 180°F, makeup temperature = 55°F, boiler efficiency = 80%, gas cost = $10/MMBTU, 8,760 hours/year operation. Show your work using dimensional analysis.
Chapter 13 Answer Key: Heating Things Up
1. Corrosion rates double every 18°F. At 140°F (~70°F above ambient) ≈ 3.9 doublings → ~23.9 ≈ 15×. Although half the oxygen is gone, the remaining 4 mg/L is reacting 15× faster. Net corrosion potential is approximately 7.5x compared to room temperature. The paradox does not resolve until the boiling point eliminates oxygen almost entirely.
2. 2% of 100 billion BTU = 2 billion BTU/yr. At 10 = $20,000/yr from a credit-card’s thickness of scale.
3. Priming. The pattern (weekday mornings only) correlates with a sudden load swing: occupants arriving, thermostats activating, steam demand surging from low fire. This is mechanical surging, not chemical foaming. Chemistry is within limits, and the problem does not occur during steady-state weekend operation.
4. (a) Softener overload / hardness breakthrough → pretreatment link. (b) Cold deaerator / elevated dissolved oxygen → feedwater/deaeration link. (c) Low cycles / increased blowdown → boiler vessel link. (d) Elevated CO₂ / low condensate pH → steam-condensate link. Also: increased makeup volume, increased fuel cost, increased chemical consumption.
5. (a) Stoichiometric sulfite requirement: 3 ppm O₂ × 7.88 = 23.6 ppm sodium sulfite in the feedwater to neutralize the dissolved oxygen. (b) The sulfite that reacts with oxygen is consumed – it does not concentrate in the boiler. Only the excess sulfite that survives the reaction concentrates with the boiler water. To achieve a 30 ppm residual in the boiler at 10 cycles, the excess sulfite in the feedwater must be: 30 ppm ÷ 10 = 3.0 ppm excess in feedwater. Total feedwater sulfite target = stoichiometric demand + excess for residual = 23.6 + 3.0 = 26.6 ppm sodium sulfite.
6. (a) Q = 8,000 × 1.0 × (170 − 60) = 880,000 BTU/hr. (b) Annual energy = 880,000 × 8,000 = 7.04 × 10⁹ BTU = 7,040 MMBTU. Adjusted for efficiency: 7,040 / 0.82 = 8,585 MMBTU. Cost: 8,585 × 103,024/yr**.
7. (a) 50 µS × 40 = 2,000 µS. (b) 200 µS × 10 = 2,000 µS. (c) The conductivity is the same numerically, but the underlying chemistry is radically different. At 40 cycles on clean feedwater, TDS is composed of benign sodium salts. At 10 cycles on dirty feedwater, TDS includes calcium, magnesium, and elevated alkalinity, the species the softener should have removed. Estimated TDS ≈ 0.7 × 2,000 µS/cm = 1,400 mg/L. This is below the stated 3,500 mg/L TDS limit for <300 psig, but the composition, not the conductivity, determines the risk. Hardness breakthrough means scaling risk is severe regardless of the conductivity reading.
8. (a) The conductivity sensor is fouled or out of calibration, reading artificially low. The controller believes the water is clean and has closed the blowdown valve, allowing cycles to climb unchecked. (b) Actual cycles = 3,800 / 80 = 47.5. (c) At 300 psig, the simplified reference table in this chapter gives a boiler-water TDS ceiling of 3,500 mg/L. Using the approximate conversion TDS ≈ 0.7 × conductivity, 3,800 µS/cm corresponds to approximately 2,660 mg/L TDS, which is below that simplified ceiling. However, the actual cycles have climbed to 47.5, the controller is operating from a false reading, and TDS alone does not establish safety. Silica, alkalinity, foaming tendency, steam-purity requirements, and the boiler manufacturer’s limits must also be evaluated. Immediate action remains appropriate: validate the manual result, restore controlled blowdown, clean and recalibrate the sensor, and confirm all boiler-water parameters before returning to normal control.
9. Do not increase chemical pumps. Diagnose the circuit. Step 1: Check deaerator temperature (explains sulfite consumption if DA is cold: elevated dissolved oxygen is overwhelming the scavenger). Step 2: Check softener output and brine tank (explains inhibitor depletion if hardness is breaking through: the scale inhibitor is being consumed by calcium instead of maintaining a residual). Step 3: Check condensate return rate (explains elevated CO₂ and low condensate pH: increased makeup alkalinity means more bicarbonate entering the boiler, more CO₂ in the steam, more carbonic acid in the condensate). A single root cause, loss of condensate return, explains all three symptoms simultaneously, exactly as in the Perfect Storm scenario. The circuit reveals the cause; the chemical tests only showed the symptoms.
10. At 90% return: condensate = 22,500 lb/hr at 180°F, makeup = 2,500 lb/hr at 55°F. Weighted feedwater temp = (22,500 × 180 + 2,500 × 55) / 25,000 = (4,050,000 + 137,500) / 25,000 = 167.5°F. At 37% return: condensate = 9,250 lb/hr at 180°F, makeup = 15,750 lb/hr at 55°F. Weighted feedwater temp = (9,250 × 180 + 15,750 × 55) / 25,000 = (1,665,000 + 866,250) / 25,000 = 101.2°F. Additional heating required per pound: 167.5 − 101.2 = 66.3°F. Additional fuel load = 25,000 lb/hr × 66.3 BTU/lb = 1,657,500 BTU/hr. Annual = 1,657,500 × 8,760 = 14.52 × 10⁹ BTU = 14,520 MMBTU. At 80% boiler efficiency: 14,520 / 0.80 = 18,150 MMBTU of fuel input. Cost: 18,150 × 181,500/yr**. This exceeds the 10,000 pump replacement.