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Chlorination and CT: California T1 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. The chlorine demand of a 12.7 MGD supply is 2.5 mg/L and the target free residual is 1 mg/L. How many lbs/day of chlorine must be fed?

  1. A) 159 lbs/day
  2. B) 265 lbs/day
  3. C) 332 lbs/day
  4. D) 371 lbs/day
Show the answer and worked solution

Answer: D) 371 lbs/day

1. Dosage = Demand + Residual [Sheet p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"] = 2.5 + 1 = 3.5 mg/L. 2. lbs/day = MGD × mg/L × 8.34 [Sheet p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"] = 12.7 × 3.5 × 8.34 = 370.713 lbs/day. 3. Answer: 371 lbs/day (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"; p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"

Example 2. A 1,280,000-gallon storage tank is to be disinfected at 25 mg/L using 65% HTH (calcium hypochlorite). How many pounds of HTH are needed?

  1. A) 267 lbs
  2. B) 206 lbs
  3. C) 173 lbs
  4. D) 411 lbs
Show the answer and worked solution

Answer: D) 411 lbs

1. MG = 1,280,000 ÷ 1,000,000 = 1.28 MG [Sheet p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal)]. 2. HTH Solid (lbs) = MG × mg/L × 8.34 ÷ (Strength/100) [Sheet p.1, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"]. 3. = 1.28 × 25 × 8.34 ÷ 0.65 = 266.88 ÷ 0.65 = 410.6 lbs. 4. Answer: 411 lbs (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal); p.1, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"

Example 3. A well pumps 3,000 GPM around the clock. How many lbs/day of chlorine are needed for a dose of 2.5 mg/L?

  1. A) 139 lbs/day
  2. B) 90.1 lbs/day
  3. C) 10.8 lbs/day
  4. D) 3.75 lbs/day
Show the answer and worked solution

Answer: B) 90.1 lbs/day

1. Flow = 3,000 ÷ 694.4 = 4.32 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"] = 4.32 × 2.5 × 8.34 = 90.072 lbs/day. 3. Answer: 90.1 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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