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Filtration and removal: California T3 water treatment operator math

Filtration and loading rates, filter production, backwash, UFRV, percent and log removal.

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 plant filters 13.4 MGD through 3 identical filters, each 29 ft × 26 ft, all in service. What is the loading rate on each filter in GPM/ft²?

  1. A) 4.11 GPM/ft²
  2. B) 2.06 GPM/ft²
  3. C) 2.66 GPM/ft²
  4. D) 0.550 GPM/ft²
Show the answer and worked solution

Answer: A) 4.11 GPM/ft²

1. Plant flow = 13.4 × 694.4 = 9,304.96 GPM [Sheet p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"]; per filter = 9,304.96 ÷ 3 = 3,102 GPM. 2. Filter area = 29 × 26 = 754 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 3. Loading Rate = Flow Rate (GPM) ÷ Filter Area [Sheet p.2, Filtration: "Loading Rate (GPM/Ft2) = Flow Rate (GPM) / Filter Area (Ft2)"] = 3,102 ÷ 754 = 4.114 GPM/ft². 4. Answer: 4.11 GPM/ft² (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"; p.1, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Loading Rate (GPM/Ft2) = Flow Rate (GPM) / Filter Area (Ft2)"

Example 2. A 16 ft × 21 ft filter ran 59 hours at an average of 1,110 GPM before backwashing. What was the unit filter run volume in gal/ft²?

  1. A) 11,700 gal/ft²
  2. B) 281,000 gal/ft²
  3. C) 195 gal/ft²
  4. D) 1,560 gal/ft²
Show the answer and worked solution

Answer: A) 11,700 gal/ft²

1. Volume produced = 1,110 GPM × 59 hrs × 60 min/hr = 3,929,400 gal [Sheet p.1, Units and Conversion Factors, rows 8-9: 1 day = 24 hrs = 1440 min, so 1440 / 24 = 60 min/hr]. 2. Filter area = 16 × 21 = 336 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 3. UFRV = Volume Produced ÷ Filter Area [Sheet p.2, Filtration: "Unit Filter Run Volume (UFRV) = Volume Produced in a Filter Run (Gal) / Filter Area (Ft2)"] = 3,929,400 ÷ 336 = 11,695 gal/ft². 4. Answer: 11,700 gal/ft² (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, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Unit Filter Run Volume (UFRV) = Volume Produced in a Filter Run (Gal) / Filter Area (Ft2)"

Example 3. A treatment process removes 95.31% of Giardia cysts. What log removal is that?

  1. A) 1.33 log
  2. B) 0.665 log
  3. C) 0.443 log
  4. D) 0.998 log
Show the answer and worked solution

Answer: A) 1.33 log

1. Log Removal = log(1.0 − %Removal/100) × (−1) [Sheet p.2, C x T Calculations: "Log Removal = 1.0 - (% Removal/100) x Log Key x (-1)" (i.e. -log10(1 - %Removal/100))]. 2. 1.0 − 0.9531 = 0.0469; log(0.0469) = -1.329; × (−1) = 1.329 log. 3. Answer: 1.33 log (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Log Removal = 1.0 - (% Removal/100) x Log Key x (-1)" (i.e. -log10(1 - %Removal/100))

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