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Chlorination and CT: California T3 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 chlorinator feeds 621 lbs/day into 13.8 MGD. The residual measured after the contact time is 1.1 mg/L. What is the chlorine demand?

  1. A) 4.30 mg/L
  2. B) 6.50 mg/L
  3. C) 2.15 mg/L
  4. D) 5.40 mg/L
Show the answer and worked solution

Answer: A) 4.30 mg/L

1. Dose (mg/L) = lbs ÷ (MG × 8.34) [Sheet p.1, Solutions: "mg/L = lbs / (MG x 8.34)"] = 621 ÷ (13.8 × 8.34) = 5.396 mg/L. 2. Dosage = Demand + Residual [Sheet p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"], so Demand = 5.396 − 1.1 = 4.296 mg/L. 3. Answer: 4.30 mg/L (rounded to 3 significant figures).

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

Example 2. A required CT is 56 mg·min/L and the free chlorine residual is 2.15 mg/L. What contact time is needed?

  1. A) 39.0 min
  2. B) 120 min
  3. C) 26.0 min
  4. D) 53.9 min
Show the answer and worked solution

Answer: C) 26.0 min

1. Time (min) = C × T ÷ Chlorine Residual [Sheet p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"] = 56 ÷ 2.15 = 26.047 min. 2. Answer: 26.0 min (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"

Example 3. A well pumps 1,350 GPM around the clock. How many lbs/day of chlorine are needed for a dose of 2 mg/L?

  1. A) 32.4 lbs/day
  2. B) 1.35 lbs/day
  3. C) 50.2 lbs/day
  4. D) 3.89 lbs/day
Show the answer and worked solution

Answer: A) 32.4 lbs/day

1. Flow = 1,350 ÷ 694.4 = 1.944 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"] = 1.944 × 2 × 8.34 = 32.43 lbs/day. 3. Answer: 32.4 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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