Problem 3.7: A 12000gallon cooling loop needs to be dosed

From Chapter 3 Problem Set. Level 2: Applied Thinking.

Question

A 12,000-gallon cooling loop needs to be dosed to 150 ppm of a corrosion inhibitor that is 35% active by mass. How many pounds of product must be added? Show your dimensional analysis.

Answer

Show answer

Water mass: 12,000 gal × 8.34 lb/gal = 100,080 lb. Active ingredient required: 100,080 lb × (150 / 1,000,000) = 15.0 lb active. Product required: 15.0 lb ÷ 0.35 = 42.9 lb of product. The percent-active conversion matters. Dosing 15 lb of product instead of 15 lb of active would leave the system at roughly 53 ppm, about a third of target.