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Flow, velocity and detention: California D2 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. A 20-inch main carries 5,285 GPM. What is the velocity in ft/sec?

  1. A) 3.49 ft/sec
  2. B) 5.40 ft/sec
  3. C) 1.35 ft/sec
  4. D) 2.70 ft/sec
Show the answer and worked solution

Answer: B) 5.40 ft/sec

1. Q = 5,285 GPM ÷ 448.8 GPM/CFS = 11.78 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)"] = 11.78 ÷ 2.181 = 5.401 ft/sec. 4. Answer: 5.40 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 5.41 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 2. Water flows through an 8-inch main at 2.9 ft/sec. What is the flow in GPM?

  1. A) 1,820 GPM
  2. B) 7.57 GPM
  3. C) 454 GPM
  4. D) 681 GPM
Show the answer and worked solution

Answer: C) 454 GPM

1. A = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (8 ÷ 12)² = 0.3489 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)"] = 0.3489 × 2.9 = 1.01181 ft³/sec. 3. 1 CFS = 448.8 GPM [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]: 1.01181 × 448.8 = 454.1 GPM. 4. Answer: 454 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 453 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 3. What is the capacity, in MGD, of an 18-inch main if velocity is limited to 3.5 ft/sec?

  1. A) 9.58 MGD
  2. B) 16.0 MGD
  3. C) 6.18 MGD
  4. D) 3.99 MGD
Show the answer and worked solution

Answer: D) 3.99 MGD

1. A = .785 × (18 ÷ 12)² = 1.76625 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)"] = 1.76625 × 3.5 = 6.182 ft³/sec. 3. 1 MGD = 1.55 CFS [Sheet p.1, Units and Conversion Factors, row 28: "1 MGD = 1.55 CFS"]: 6.182 ÷ 1.55 = 3.988 MGD. 4. Answer: 3.99 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 4.00 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"

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