Problem 7.6: A cooling tower runs at 6 cycles of

From Chapter 7 Problem Set. Level 2: Applied Thinking.

Question

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.

Answer

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(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.