Problem 12.8: A cooling tower evaporates 20 GPM and operates

From Chapter 12 Problem Set. Level 3: Optimization.

Question

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

Answer

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(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.