Chapter 13 · Heating Things Up · Ten playable problems

Ch 13 · P1 — Half the oxygen, twice the trouble

PROBLEM 121 · HEATING THINGS UP — THE 140°F PARADOX

At 140°F the feedwater holds about 4 mg/L of oxygen — half its room-temperature load. Why is corrosion risk higher, not lower?

Live model

the paradox resolves only at the boil140°F water holds half the oxygen — why is the risk higher, not lower?

Ch 13 · P2 — A credit card of scale

PROBLEM 122 · HEATING THINGS UP — WHAT 1/32" COSTS

The boiler burns 100 billion BTU a year, and 1/32-inch of CaCO₃ steals 2% of its efficiency. Dial the annual cost at $10/MMBTU.

Live model

insulation where you never wanted it100 billion BTU/yr, 1/32-inch of scale costing 2% efficiency, $10/MMBTU. Dial the annual cost in $k.

Ch 13 · P3 — The 7 AM water hammer

PROBLEM 123 · HEATING THINGS UP — PRIMING OR FOAMING

Water hammer every weekday at 7 AM, never on weekends, chemistry inside ASME limits. Click the culprit.

Live model

chemistry in range · schedule tells the storyWater hammer at 7 AM weekdays, never weekends, chemistry within limits. Click the culprit.

Ch 13 · P4 — Four hits on the circuit

PROBLEM 124 · HEATING THINGS UP — LOSING HALF THE CONDENSATE

Half the condensate return is gone. Match each consequence to the link of the boiler circuit it damages.

Live model

one lost return line, every link paysHalf the condensate return is gone. Match each consequence to the link it damages.

Ch 13 · P5 — Sulfite for demand and residual

PROBLEM 125 · HEATING THINGS UP — THE SCAVENGER TARGET

3 ppm O₂ in the feedwater, 7.88 ppm sulfite per ppm O₂, a 30 ppm boiler residual wanted at 10 cycles. Dial the total feedwater sulfite target.

Live model

demand + residual ÷ cycles3 ppm O₂, ×7.88 stoichiometry, 30 ppm boiler residual wanted at 10 cycles. Dial the feedwater sulfite target.

Ch 13 · P6 — What the return line earns

PROBLEM 126 · HEATING THINGS UP — CONDENSATE IS FUEL

8,000 lb/hr of condensate returns at 170°F instead of 60°F makeup. 82% efficiency, $12/MMBTU, 8,000 hours. Dial the annual savings.

Live model

110°F of head start, 8,000 times an hour8,000 lb/hr of 170°F condensate vs 60°F makeup, 82% efficiency, $12/MMBTU, 8,000 hr/yr. Dial the annual savings in $k.

Ch 13 · P7 — Same number, different boiler

PROBLEM 127 · HEATING THINGS UP — 2,000 µS TWO WAYS

50 µS feedwater at 40 cycles, or 200 µS feedwater at 10 cycles — both read 2,000 µS in the drum. Same situation?

Live model

conductivity counts ions — it cannot name themTwo scenarios, same 2,000 µS boiler water. Same risk?

Ch 13 · P8 — The controller that lied

PROBLEM 128 · HEATING THINGS UP — 2,000 VS 3,800

The blowdown controller reads 2,000 µS; your handheld reads 3,800 on a cooled sample; the valve has been shut for hours. Feedwater is 80 µS. What happened?

Live model

a controller is only as honest as its probeController says 2,000 µS; your handheld says 3,800; the blowdown valve has been shut for hours. What happened?

Ch 13 · P9 — Do not touch the pumps

PROBLEM 129 · HEATING THINGS UP — THREE SYMPTOMS, ONE CAUSE

Sulfite at zero, scale inhibitor depleted, condensate pH 5.5. Click the diagnostic walk in order — the root cause is waiting at the end.

Live model

diagnose the circuit, not the chemical pumpsSulfite zero, inhibitor gone, condensate pH 5.5. Click the diagnostic walk in order — do NOT touch the pumps.

Ch 13 · P10 — The Perfect Storm, priced

PROBLEM 130 · HEATING THINGS UP — 90% RETURN FALLS TO 37%

25,000 lb/hr of steam; condensate at 180°F, makeup at 55°F; 80% efficiency, $10/MMBTU, 8,760 hours. Work the weighted feedwater temperatures, then dial the annual fuel cost of the failure.

Live model

the last of 130 — dimensional analysis to the endThe Perfect Storm: 90% return falls to 37% at 25,000 lb/hr. Work the weighted temperatures, then dial the annual fuel cost in $k.

Problem set

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Problem 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.
Authored answer
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.
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Problem 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.
Authored answer
2% of 100 billion BTU = 2 billion BTU/yr. At $10/MMBTU: 2,000 MMBTU × $10 = $20,000/yr from a credit-card’s thickness of scale.
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Problem 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.
Authored answer
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.
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Problem 4
List four consequences of losing 50% of condensate return, and for each, identify which link in the boiler circuit is affected.
Authored answer
(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.
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Problem 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?
Authored answer
(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.
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Problem 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.
Authored answer
(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 × $12 = $103,024/yr.
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Problem 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.
Authored answer
(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.
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Problem 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?
Authored answer
(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.
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Problem 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?
Authored answer
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.
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Problem 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.
Authored answer
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 × $10 = $181,500/yr. This exceeds the $95,000 estimate in the Perfect Storm because the original used actual boiler steaming rates (which were lower), a shorter operating season, and lower fuel cost. The point is that the magnitude is enormous relative to the $10,000 pump replacement.
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