Cooling Tower Water Loss
A small textile factory uses a cooling tower to remove waste heat from its dyeing machines. The tower circulates water at a rate of 12,000 liters per hour. The water enters the tower at 45°C and leaves at 30°C. The ambient air is 25°C dry-bulb and 18°C wet-bulb. The factory operates 16 hours a day, 300 days a year. The management wants to reduce water consumption because the local water supply is limited and expensive. They propose two options: (1) install a conductivity controller to automate blowdown, or (2) add a side-stream filtration system to remove suspended solids. Currently, blowdown is done manually once per shift (three times per day) by opening a valve for 10 minutes, during which water flows at 200 liters per minute. Evaporation loss is estimated at 1.5% of the circulation rate per hour. Drift loss is negligible. The cost of water is 0.5 Taka per liter, and the factory pays a fixed sewer charge of 0.2 Taka per liter for all water used, including blowdown. The conductivity controller costs 150,000 Taka and reduces blowdown water to the minimum required to maintain 3 cycles of concentration. The filtration system costs 250,000 Taka and does not affect blowdown volume but reduces the need for chemical treatment by 20,000 Taka per year. Assume the plant operates as described and that the cycles of concentration are defined as the ratio of dissolved solids in the circulating water to that in the makeup water.
1. Calculate the current annual water consumption (in liters) and the annual cost (in Taka) for water and sewer combined. Show your steps.
Read the worked answer
Step 1: Calculate evaporation loss per hour.
Evaporation = 1.5% of 12,000 L/h = 0.015 × 12,000 = 180 L/h.
Step 2: Calculate blowdown volume per day.
Blowdown occurs 3 times per day, each for 10 minutes at 200 L/min.
Volume per blowdown = 200 L/min × 10 min = 2,000 L.
Daily blowdown = 3 × 2,000 = 6,000 L/day.
Step 3: Calculate evaporation per day.
Daily evaporation = 180 L/h × 16 h = 2,880 L/day.
Step 4: Total water loss per day.
Drift is negligible, so total = evaporation + blowdown = 2,880 + 6,000 = 8,880 L/day.
Step 5: Annual water consumption.
Operating days = 300.
Annual water = 8,880 L/day × 300 days = 2,664,000 L/year.
Step 6: Annual cost.
Cost per liter = water 0.5 + sewer 0.2 = 0.7 Taka/L.
Total cost = 2,664,000 L × 0.7 Taka/L = 1,864,800 Taka/year.
Answer: Current annual water consumption is 2,664,000 liters, and the annual cost is 1,864,800 Taka.
2. Determine the annual water savings and net annual savings (in Taka) if the conductivity controller is installed. Assume the controller reduces blowdown to the minimum required for 3 cycles of concentration, and that evaporation and drift remain the same. Also, account for the fact that the sewer charge applies to all water used, including blowdown. Show your calculations.
Read the worked answer
Step 1: Understand cycles of concentration (CoC).
CoC = (dissolved solids in circulating water) / (dissolved solids in makeup water).
At steady state, the mass balance for dissolved solids gives:
Blowdown = Evaporation / (CoC – 1).
This assumes drift is negligible.
Step 2: Calculate minimum blowdown with CoC = 3.
Evaporation per hour = 180 L/h.
Blowdown per hour = 180 / (3 – 1) = 180 / 2 = 90 L/h.
Daily blowdown = 90 L/h × 16 h = 1,440 L/day.
Step 3: Compare with current blowdown.
Current daily blowdown = 6,000 L/day.
Savings in blowdown = 6,000 – 1,440 = 4,560 L/day.
Step 4: Annual water savings.
Annual savings = 4,560 L/day × 300 days = 1,368,000 L/year.
Step 5: Annual cost savings.
Cost per liter = 0.7 Taka/L (water + sewer).
Annual savings = 1,368,000 × 0.7 = 957,600 Taka/year.
Step 6: Net annual savings after controller cost.
Controller cost = 150,000 Taka (one-time).
Net first-year savings = 957,600 – 150,000 = 807,600 Taka.
In subsequent years, savings = 957,600 Taka/year.
Answer: The conductivity controller saves 1,368,000 liters per year, worth 957,600 Taka annually. After subtracting the controller cost, the first-year net savings is 807,600 Taka, and subsequent years save 957,600 Taka annually.
3. Evaluate the filtration system option. It costs 250,000 Taka and does not change blowdown volume but reduces chemical treatment costs by 20,000 Taka per year. Compare its net annual savings with the conductivity controller over a 5-year period. Which option is more cost-effective? Justify your answer with calculations.
Read the worked answer
Step 1: Filtration system savings.
Water savings = 0 (blowdown unchanged).
Chemical savings = 20,000 Taka/year.
Over 5 years, total savings = 20,000 × 5 = 100,000 Taka.
Net savings over 5 years = 100,000 – 250,000 = –150,000 Taka (a net cost).
Step 2: Conductivity controller savings over 5 years.
Annual water+sewer savings = 957,600 Taka.
Over 5 years, total savings = 957,600 × 5 = 4,788,000 Taka.
Net savings over 5 years = 4,788,000 – 150,000 = 4,638,000 Taka.
Step 3: Comparison.
The conductivity controller yields a net positive savings of 4,638,000 Taka over 5 years, while the filtration system results in a net cost of 150,000 Taka over the same period. Therefore, the conductivity controller is far more cost-effective.
Step 4: Additional considerations.
The filtration system might have other benefits (e.g., improved heat transfer, reduced maintenance), but based solely on the given numbers, it is not economically justified. The conductivity controller directly reduces water and sewer costs, which dominate the operating expenses.
Answer: The conductivity controller is more cost-effective, with a 5-year net savings of 4,638,000 Taka compared to a net cost of 150,000 Taka for the filtration system.