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Flow, velocity and detention: California T1 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. Water flows through a 10-inch main at 2.4 ft/sec. What is the flow in GPM?

  1. A) 9.79 GPM
  2. B) 2,350 GPM
  3. C) 705 GPM
  4. D) 587 GPM
Show the answer and worked solution

Answer: D) 587 GPM

1. A = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (10 ÷ 12)² = 0.5451 ft² [12 in = 1 ft]. 2. Q = A × V [Sheet p.1, Flow & Velocity: "(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)"] = 0.5451 × 2.4 = 1.30824 ft³/sec. 3. 1 CFS = 448.8 GPM [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]: 1.30824 × 448.8 = 587.1 GPM. 4. Answer: 587 GPM (rounded to 3 significant figures). 5. If you used 1.55 CFS per MGD (Units row 28) and 694.4 GPM per MGD (row 29) instead of 448.8 GPM per CFS (row 7), you get 586 GPM. 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, Area and Volume: "Area of Circle = .785 X D2"; p.1, Flow & Velocity: "(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)"; p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"

Example 2. A 20-inch main carries 4,695 GPM. What is the velocity in ft/sec?

  1. A) 3.10 ft/sec
  2. B) 1.20 ft/sec
  3. C) 4.80 ft/sec
  4. D) 2.40 ft/sec
Show the answer and worked solution

Answer: C) 4.80 ft/sec

1. Q = 4,695 GPM ÷ 448.8 GPM/CFS = 10.46 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (20 ÷ 12)² = 2.181 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)"] = 10.46 ÷ 2.181 = 4.796 ft/sec. 4. Answer: 4.80 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 4.81 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 round clearwell is 90 ft in diameter with water 14 ft deep. Flow through it is 5,350 GPM. What is the detention time in minutes?

  1. A) 124 min
  2. B) 498 min
  3. C) 186 min
  4. D) 248 min
Show the answer and worked solution

Answer: A) 124 min

1. Volume = .785 × 90² × 14 × 7.48 = 665,862 gal [Sheet p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"]. 2. DT (Min) = Volume (Gal) ÷ Flow (GPM) [Sheet p.1, Sedimentation: "DT (Min) = Volume (Gal) / Flow (GPM)"] = 665,862 ÷ 5,350 = 124.46 min. 3. Answer: 124 min (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"; p.1, Sedimentation: "DT (Min) = Volume (Gal) / Flow (GPM)"

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