⟵ 12 Cooling Things Down (Problem Set) · Contents · 13 Heating Things Up ⟶

Cooling Things Down: Answer Key

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 – it 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 = 7,524,420 gal/yr ≈ 7.52 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 = 6,022,968 gal/yr ≈ 6.02 Mgal. Savings = 7.52 − 6.02 = 1.50 Mgal/yr. Nearly 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 × $0.11=\ast \ast$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 × $0.11=\ast \ast$256,608**. Annual penalty from fouling = $256,608-$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 × $6.00/kgal=$38,400. Sewer = 1,600 kgal × $8.00/kgal=$12,800. Chemical = 6,400 kgal × $1.80/kgal=$11,520. Total = $62,720\ast \ast .At6cycles:Water=5,760kgal\times$6.00/kgal = $34,560.Sewer=960kgal\times$8.00/kgal = $7,680.Chemical=5,760kgal\times$2.50/kgal = $14,400.Total=\ast \ast$56,640.

(c) Net annual savings = $62,720-$56,640 = **$6,080\ast \ast .Waterandsewersavingstotal$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. Three possible explanations: (1) System leaks – piping leaks, valve weeping, overflow events, or basin cracks allow concentrated tower water to leave the system without being metered as blowdown. This requires additional makeup to maintain basin level. Investigate by performing a basin-level hold test with all known outflows closed and checking for unexplained level drop. (2) Drift losses higher than assumed – drift eliminators may be damaged, missing, or improperly seated, allowing entrained water droplets to escape with the exhaust air. Investigate by visually inspecting drift eliminators during operation and checking for excessive visible plume or wet surfaces near the tower discharge. (3) Unmetered water draw – another system or process (basin washdown, makeup to a secondary system, landscape irrigation) may be connected to the same makeup supply upstream of or bypassing the meter. Investigate by tracing the makeup piping and confirming that no branch connections exist between the meter and the tower. Additional possibility: the makeup meter itself may be reading high due to calibration drift or mechanical wear – verify by comparing meter reading against a timed bucket test or a secondary flow measurement.

---

⟵ 12 Cooling Things Down (Problem Set) · Contents · 13 Heating Things Up ⟶