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Chlorination and CT: California T2 water treatment operator math

Chlorine feed, demand and residual, hypochlorite, well and main disinfection, CT, chloramines.

Below are 3 worked examples. Each answer is computed with the formulas and constants on the CA Exam Conversion Sheet (revised 4/30/2026), and the solution names the line it uses.

Formulas from the sheet for this topic

Example 1. A 14.7 MGD system is dosed at 2.2 mg/L with 15% sodium hypochlorite that weighs 9.6 lbs/gal. How many gallons of hypochlorite are fed per day?

  1. A) 374 GPD
  2. B) 1,800 GPD
  3. C) 216 GPD
  4. D) 187 GPD
Show the answer and worked solution

Answer: D) 187 GPD

1. Liquid (Gal) = MG × mg/L × 8.34 ÷ (Strength/100 × Chemical Weight) [Sheet p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"]. 2. = 14.7 × 2.2 × 8.34 ÷ (0.15 × 9.6) = 269.7 ÷ 1.44 = 187.3 GPD. 3. Answer: 187 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"

Example 2. An 11.1 MGD system is dosed at 2.3 mg/L with 12.5% sodium hypochlorite that weighs 9.9 lbs/gal. How many gallons of hypochlorite are fed per day?

  1. A) 172 GPD
  2. B) 344 GPD
  3. C) 21.5 GPD
  4. D) 204 GPD
Show the answer and worked solution

Answer: A) 172 GPD

1. Liquid (Gal) = MG × mg/L × 8.34 ÷ (Strength/100 × Chemical Weight) [Sheet p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"]. 2. = 11.1 × 2.3 × 8.34 ÷ (0.125 × 9.9) = 212.9 ÷ 1.2375 = 172 GPD. 3. Answer: 172 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"

Example 3. A well pumps 1,600 GPM around the clock. How many lbs/day of chlorine are needed for a dose of 2.5 mg/L?

  1. A) 5.76 lbs/day
  2. B) 74.3 lbs/day
  3. C) 48.0 lbs/day
  4. D) 2.00 lbs/day
Show the answer and worked solution

Answer: C) 48.0 lbs/day

1. Flow = 1,600 ÷ 694.4 = 2.304 MGD [Sheet p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"]. 2. Gas (lbs) = MG × mg/L × 8.34 [Sheet p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"] = 2.304 × 2.5 × 8.34 = 48.0384 lbs/day. 3. Answer: 48.0 lbs/day (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"; p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"

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