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Flow, velocity and detention: California T3 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 sedimentation basin is 90 ft long, 33 ft wide and 16 ft deep. The plant flow is 1.8 MGD. What is the detention time in hours?

  1. A) 4.74 hours
  2. B) 0.197 hours
  3. C) 5.28 hours
  4. D) 0.634 hours
Show the answer and worked solution

Answer: A) 4.74 hours

1. Volume = 90 × 33 × 16 = 47,520 ft³. 2. Flow = 1.8 MGD × 1,000,000 = 1,800,000 gal/day [Sheet p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal)]. 3. DT (hrs) = Ft³ × 7.48 × 24 ÷ Gal/Day [Sheet p.1, Sedimentation: "DT (hrs) = Ft3 x 7.48 x 24 / Gal/Day"] = 47,520 × 7.48 × 24 ÷ 1,800,000 = 4.739 hours. 4. Answer: 4.74 hours (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, Sedimentation: "DT (hrs) = Ft3 x 7.48 x 24 / Gal/Day"

Example 2. A 4-inch main carries 125 GPM. What is the velocity in ft/sec?

  1. A) 4.79 ft/sec
  2. B) 192 ft/sec
  3. C) 3.19 ft/sec
  4. D) 6.38 ft/sec
Show the answer and worked solution

Answer: C) 3.19 ft/sec

1. Q = 125 GPM ÷ 448.8 GPM/CFS = 0.2785 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (4 ÷ 12)² = 0.08722 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)"] = 0.2785 ÷ 0.08722 = 3.193 ft/sec. 4. Answer: 3.19 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 3.20 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 16-inch main carries 5,010 GPM. What is the velocity in ft/sec?

  1. A) 8.00 ft/sec
  2. B) 10.7 ft/sec
  3. C) 2.00 ft/sec
  4. D) 480 ft/sec
Show the answer and worked solution

Answer: A) 8.00 ft/sec

1. Q = 5,010 GPM ÷ 448.8 GPM/CFS = 11.163 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (16 ÷ 12)² = 1.3956 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)"] = 11.163 ÷ 1.3956 = 7.9987 ft/sec. 4. Answer: 8.00 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 8.01 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)"

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