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Flow, velocity and detention: California D3 water distribution 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. What is the capacity, in MGD, of a 4-inch main if velocity is limited to 8 ft/sec?

  1. A) 0.450 MGD
  2. B) 0.225 MGD
  3. C) 0.150 MGD
  4. D) 0.338 MGD
Show the answer and worked solution

Answer: A) 0.450 MGD

1. A = .785 × (4 ÷ 12)² = 0.08722 ft² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"]. 2. Q = A × V [Sheet p.1, Flow & Velocity: "(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)"] = 0.08722 × 8 = 0.69776 ft³/sec. 3. 1 MGD = 1.55 CFS [Sheet p.1, Units and Conversion Factors, row 28: "1 MGD = 1.55 CFS"]: 0.69776 ÷ 1.55 = 0.4502 MGD. 4. Answer: 0.450 MGD (rounded to 3 significant figures). 5. If you used 448.8 GPM per CFS (Units row 7) and 694.4 GPM per MGD (row 29) instead of 1.55 CFS per MGD (row 28), you get 0.451 MGD. 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 28: "1 MGD = 1.55 CFS"

Example 2. A storage tank gains 170,000 gallons in 6 hours while nothing is drawn from it. What is the average inflow in GPM?

  1. A) 19.7 GPM
  2. B) 7.87 GPM
  3. C) 472 GPM
  4. D) 236 GPM
Show the answer and worked solution

Answer: C) 472 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)"] = 170,000 ÷ 360 = 472.2 GPM. 3. Answer: 472 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 3. A 20-inch main carries 4,010 GPM. What is the velocity in ft/sec?

  1. A) 246 ft/sec
  2. B) 6.83 ft/sec
  3. C) 1.02 ft/sec
  4. D) 4.10 ft/sec
Show the answer and worked solution

Answer: D) 4.10 ft/sec

1. Q = 4,010 GPM ÷ 448.8 GPM/CFS = 8.935 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)"] = 8.935 ÷ 2.181 = 4.097 ft/sec. 4. Answer: 4.10 ft/sec (rounded to 3 significant figures).

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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