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Flow, velocity and detention: California T2 water treatment operator math

Q = A x V, fill rates, detention time, surface loading.

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 storage tank gains 420,000 gallons in 6 hours while nothing is drawn from it. What is the average inflow in GPM?

  1. A) 1,170 GPM
  2. B) 70,000 GPM
  3. C) 19.4 GPM
  4. D) 48.6 GPM
Show the answer and worked solution

Answer: A) 1,170 GPM

1. Time in minutes = 6 hrs × 60 min/hr = 360 min [Sheet p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr]. 2. GPM = Volume ÷ Time [Sheet p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"] = 420,000 ÷ 360 = 1,167 GPM. 3. Answer: 1,170 GPM (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr; p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"

Example 2. A 6-inch main carries 245 GPM. What is the velocity in ft/sec?

  1. A) 2.78 ft/sec
  2. B) 0.695 ft/sec
  3. C) 1.39 ft/sec
  4. D) 1.80 ft/sec
Show the answer and worked solution

Answer: A) 2.78 ft/sec

1. Q = 245 GPM ÷ 448.8 GPM/CFS = 0.5459 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (6 ÷ 12)² = 0.19625 ft² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"]. 3. V = Q ÷ A [Sheet p.1, Flow & Velocity: "V (Ft/Sec) = Q (Ft3/Sec) / A (Ft2)"] = 0.5459 ÷ 0.19625 = 2.782 ft/sec. 4. Answer: 2.78 ft/sec (rounded to 3 significant figures). 5. If you used 694.4 GPM per MGD (Units row 29) and 1.55 CFS per MGD (row 28) instead of 448.8 GPM per CFS (row 7), you get 2.79 ft/sec. The sheet prints both factors, so either answer is right; pick the option closest to yours.

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"; p.1, Area and Volume: "Area of Circle = .785 X D2"; p.1, Flow & Velocity: "V (Ft/Sec) = Q (Ft3/Sec) / A (Ft2)"

Example 3. A storage tank gains 425,000 gallons in 17.5 hours while nothing is drawn from it. What is the average inflow in GPM?

  1. A) 24,300 GPM
  2. B) 16.9 GPM
  3. C) 6.75 GPM
  4. D) 405 GPM
Show the answer and worked solution

Answer: D) 405 GPM

1. Time in minutes = 17.5 hrs × 60 min/hr = 1,050 min [Sheet p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr]. 2. GPM = Volume ÷ Time [Sheet p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"] = 425,000 ÷ 1,050 = 404.8 GPM. 3. Answer: 405 GPM (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr; p.1, Flow & Velocity: "GPM = Volume (Gal) / Time (Min)"

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