Practice › Conversion sheet

The CA exam conversion sheet, formula by formula

In short

The State Water Board publishes the Exam Conversion Sheet for the drinking water treatment and distribution operator exams. The current one is revised 4/30/2026, two pages. Download it and keep it next to you while you practice.
What changed in 2026: one number. Row 29 now reads 1 MGD = 694.4 GPM (the 6/2/2025 sheet said 694).
Watch out: in four places the sheet disagrees with itself (listed below). Use the factor printed on the line you are working from.
Below, each formula our problems use (48 of them) has a worked example, and the grades that use it link to their practice pages.

Where the sheet disagrees with itself

The units column says 1 Gal = 3.79 L, but the feed-pump formula (page 1, Solutions) uses 3.785 L/gal.
1 CFS = 448.8 GPM and 1 MGD = 1.55 CFS, yet 1 MGD = 694.4 GPM (1.55 × 448.8 = 695.6).
1 AF = 325,851 Gal, while 43,560 Ft² × 7.48 = 325,829.
“1 gm = 1000 mg/L” is printed; read it as 1 gm = 1,000 mg.

Every worked solution on this site uses the factor printed on the line it cites and never mixes two lines for the same step, so your answer matches the sheet you are given.

Area and Volume (sheet p.1)

Area of Circle = .785 X D2

Used in 544 problems: T1, T2, T3, D1, D2, D3.

Example 1. What is the cross-sectional area, in square feet, of a 16-inch diameter pipe?

A) 1.40 ft²
B) 0.700 ft²
C) 0.467 ft²
D) 1.05 ft²

Show the answer and worked solution

Answer: A) 1.40 ft²

1. Convert the diameter to feet: D = 16 in ÷ 12 in/ft = 1.333 ft [12 in = 1 ft]. 2. Area = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (16 ÷ 12)² = 1.396 ft². 3. Answer: 1.40 ft² (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Circle = .785 X D2"

Area of Rectangle = Length x Width

Used in 275 problems: T2, T3, D2, D3.

Example 2. A 32 ft × 18 ft filter is backwashed at 6,600 GPM. What is the backwash rate in GPM/ft²?

A) 132 GPM/ft²
B) 18.4 GPM/ft²
C) 1.53 GPM/ft²
D) 11.5 GPM/ft²

Show the answer and worked solution

Answer: D) 11.5 GPM/ft²

1. Filter area = 32 × 18 = 576 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 2. Backwash Rate = Flow Rate (GPM) ÷ Filter Area [Sheet p.2, Filtration: "Backwash Rate (GPM/Ft2) = Flow Rate (GPM) / Filter Area (Ft2)"] = 6,600 ÷ 576 = 11.46 GPM/ft². 3. Answer: 11.5 GPM/ft² (rounded to 3 significant figures).

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

More filtration and removal problems for T2

Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48

Used in 525 problems: T1, T2, T3, D1, D2, D3.

Example 3. A cylindrical storage tank is 25 ft in diameter with water 26 ft deep. How many gallons does it hold?

A) 106,000 gal
B) 382,000 gal
C) 12,800 gal
D) 95,400 gal

Show the answer and worked solution

Answer: D) 95,400 gal

1. Cylinder Volume (Gal) = .785 × D² × Height × 7.48 [Sheet p.1, Area and Volume: "Cylinder Volume (Gal) = .785 x D2 x Height or Length (Ft) x 7.48"]. 2. = .785 × 25 × 25 × 26 × 7.48 = 95,417 gal. 3. Answer: 95,400 gal (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"

Rectangular Basin (Gal) = Length (Ft) x Width (Ft) x Height (Ft) x 7.48

Used in 150 problems: T1, T2, T3, D1, D2, D3.

Example 4. A rectangular basin is 80 ft long and 25 ft wide, with water 9 ft deep. How many gallons does it hold?

A) 135,000 gal
B) 18,000 gal
C) 150,000 gal
D) 1,120,000 gal

Show the answer and worked solution

Answer: A) 135,000 gal

1. Rectangular Basin (Gal) = L × W × H × 7.48 [Sheet p.1, Area and Volume: "Rectangular Basin (Gal) = Length (Ft) x Width (Ft) x Height (Ft) x 7.48"]. 2. = 80 × 25 × 9 × 7.48 = 134,640 gal. 3. Answer: 135,000 gal (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Rectangular Basin (Gal) = Length (Ft) x Width (Ft) x Height (Ft) x 7.48"

Circumference of a Circle = 3.14 x D

Used in 75 problems: T3, D2, D3.

Example 5. A circular clarifier 40 ft in diameter has a weir around its full circumference and treats 2.2 MGD. What is the weir overflow rate in GPD per foot?

A) 70,100 GPD/ft
B) 8,750 GPD/ft
C) 17,500 GPD/ft
D) 55,000 GPD/ft

Show the answer and worked solution

Answer: C) 17,500 GPD/ft

1. Weir length = 3.14 × D [Sheet p.1, Area and Volume: "Circumference of a Circle = 3.14 x D"] = 3.14 × 40 = 125.6 ft. 2. Weir Overflow Rate = Flow (GPD) ÷ Weir Length [Sheet p.1, Sedimentation: "Weir Overflow Rate (GPD/Ft) = Flow (GPD) / Weir Length (Ft)"] = 2,200,000 ÷ 125.6 = 17,516 GPD/ft. 3. Answer: 17,500 GPD/ft (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Circumference of a Circle = 3.14 x D"; p.1, Sedimentation: "Weir Overflow Rate (GPD/Ft) = Flow (GPD) / Weir Length (Ft)"

More flow, velocity and detention problems for T3

Chlorination (sheet p.1)

Gas (lbs) = MG x mg/L x 8.34

Used in 500 problems: T1, T2, T3, D1, D2, D3.

Example 6. A system treats 6.05 MGD with a chlorine dose of 3.4 mg/L. How many pounds of chlorine gas are fed per day?

A) 264 lbs/day
B) 154 lbs/day
C) 20.6 lbs/day
D) 172 lbs/day

Show the answer and worked solution

Answer: D) 172 lbs/day

1. Gas (lbs) = MG × mg/L × 8.34 [Sheet p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"]; with MGD it gives lbs/day. 2. = 6.05 × 3.4 × 8.34 = 171.6 lbs/day. 3. Answer: 172 lbs/day (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Gas (lbs) = MG x mg/L x 8.34"

More chlorination and ct problems for T1

HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)

Used in 300 problems: T1, T2, T3, D1, D2, D3.

Example 7. A 1,280,000-gallon storage tank is to be disinfected at 25 mg/L using 65% HTH (calcium hypochlorite). How many pounds of HTH are needed?

A) 267 lbs
B) 206 lbs
C) 173 lbs
D) 411 lbs

Show the answer and worked solution

Answer: D) 411 lbs

1. MG = 1,280,000 ÷ 1,000,000 = 1.28 MG [Sheet p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal)]. 2. HTH Solid (lbs) = MG × mg/L × 8.34 ÷ (Strength/100) [Sheet p.1, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"]. 3. = 1.28 × 25 × 8.34 ÷ 0.65 = 266.88 ÷ 0.65 = 410.6 lbs. 4. Answer: 411 lbs (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, Chlorination: "HTH Solid (lbs) = MG x mg/L x 8.34 / (Strength (%)/100)"

More chlorination and ct problems for T1

Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))

Used in 225 problems: T1, T2, T3, D1, D2, D3.

Example 8. An 11.9 MGD system is dosed at 2.8 mg/L with 12.5% sodium hypochlorite that weighs 10.3 lbs/gal. How many gallons of hypochlorite are fed per day?

A) 216 GPD
B) 2,220 GPD
C) 267 GPD
D) 432 GPD

Show the answer and worked solution

Answer: A) 216 GPD

1. Liquid (Gal) = MG × mg/L × 8.34 ÷ (Strength/100 × Chemical Weight) [Sheet p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"]. 2. = 11.9 × 2.8 × 8.34 ÷ (0.125 × 10.3) = 277.9 ÷ 1.2875 = 215.8 GPD. 3. Answer: 216 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Chlorination: "Liquid (Gal) = MG x mg/L x 8.34 / (Strength (%)/100 x Chemical Weight (lbs/gal))"

More chlorination and ct problems for T1

Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)

Used in 300 problems: T1, T2, T3, D1, D2, D3.

Example 9. A chlorinator feeds 26 lbs/day into 1 MGD. The residual measured after the contact time is 0.7 mg/L. What is the chlorine demand?

A) 2.42 mg/L
B) 3.82 mg/L
C) 1.21 mg/L
D) 3.12 mg/L

Show the answer and worked solution

Answer: A) 2.42 mg/L

1. Dose (mg/L) = lbs ÷ (MG × 8.34) [Sheet p.1, Solutions: "mg/L = lbs / (MG x 8.34)"] = 26 ÷ (1 × 8.34) = 3.118 mg/L. 2. Dosage = Demand + Residual [Sheet p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"], so Demand = 3.118 − 0.7 = 2.418 mg/L. 3. Answer: 2.42 mg/L (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "mg/L = lbs / (MG x 8.34)"; p.1, Chlorination: "Dosage (mg/L) = Demand (mg/L) + Residual (mg/L)"

More chlorination and ct problems for T1

Flow & Velocity (sheet p.1)

(Q) Flow (Ft3/Sec) = (A) Area (Ft2) x (V) Velocity (Ft/Sec)

Used in 200 problems: T1, T2, T3, D1, D2, D3.

Example 10. Water flows through a 36-inch main at 1 ft/sec. What is the flow in GPM?

A) 1,060 GPM
B) 12,700 GPM
C) 3,170 GPM
D) 52.8 GPM

Show the answer and worked solution

Answer: C) 3,170 GPM

1. A = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × (36 ÷ 12)² = 7.065 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)"] = 7.065 × 1 = 7.065 ft³/sec. 3. 1 CFS = 448.8 GPM [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]: 7.065 × 448.8 = 3,171 GPM. 4. Answer: 3,170 GPM (rounded to 3 significant figures).

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"

More flow, velocity and detention problems for T1

V (Ft/Sec) = Q (Ft3/Sec) / A (Ft2)

Used in 125 problems: T1, T2, T3, D2, D3.

Example 11. A 10-inch main carries 955 GPM. What is the velocity in ft/sec?

A) 7.80 ft/sec
B) 3.90 ft/sec
C) 234 ft/sec
D) 5.85 ft/sec

Show the answer and worked solution

Answer: B) 3.90 ft/sec

1. Q = 955 GPM ÷ 448.8 GPM/CFS = 2.128 ft³/sec [Sheet p.1, Units and Conversion Factors, row 7: "1 CFS = 448.8 GPM"]. 2. A = .785 × (10 ÷ 12)² = 0.5451 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)"] = 2.128 ÷ 0.5451 = 3.904 ft/sec. 4. Answer: 3.90 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 3.91 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)"

More flow, velocity and detention problems for T1

GPM = Volume (Gal) / Time (Min)

Used in 150 problems: T1, T2, T3, D1, D2, D3.

Example 12. A storage tank gains 50,000 gallons in 17 hours while nothing is drawn from it. What is the average inflow in GPM?

A) 393 GPM
B) 2,940 GPM
C) 98.0 GPM
D) 49.0 GPM

Show the answer and worked solution

Answer: D) 49.0 GPM

1. Time in minutes = 17 hrs × 60 min/hr = 1,020 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)"] = 50,000 ÷ 1,020 = 49.02 GPM. 3. Answer: 49.0 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)"

More flow, velocity and detention problems for T1

Dilution/Blending (sheet p.1)

C1 x V1 = C2 x V2

Used in 50 problems: T2, T3.

Example 13. How many gallons of 12% sodium hypochlorite are needed to make 135 gallons of a 0.5% solution?

A) 1.88 gal
B) 4.22 gal
C) 2.82 gal
D) 5.63 gal

Show the answer and worked solution

Answer: D) 5.63 gal

1. C1 × V1 = C2 × V2 [Sheet p.1, Dilution/Blending: "C1 x V1 = C2 x V2"], so V1 = C2 × V2 ÷ C1. 2. V1 = 0.5 × 135 ÷ 12 = 5.625 gal. 3. Answer: 5.63 gal (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Dilution/Blending: "C1 x V1 = C2 x V2"

More chemical feed and solutions problems for T2

(C1 x V1) + (C2 x V2) = C3 x V3

Used in 25 problems: T3.

Example 14. Well 1 pumps 400 GPM with 16 mg/L nitrate (as N). Well 2 pumps 800 GPM with 3.5 mg/L nitrate (as N). The two flows are blended. What is the nitrate concentration of the blend?

A) 7.67 mg/L
B) 15.3 mg/L
C) 11.8 mg/L
D) 9.75 mg/L

Show the answer and worked solution

Answer: A) 7.67 mg/L

1. (C1 × V1) + (C2 × V2) = C3 × V3 [Sheet p.1, Dilution/Blending: "(C1 x V1) + (C2 x V2) = C3 x V3"], with flows in place of volumes. 2. C1 × V1 = 16 × 400 = 6,400; C2 × V2 = 3.5 × 800 = 2,800. 3. V3 = 400 + 800 = 1,200 GPM; C3 = 9,200 ÷ 1,200 = 7.667 mg/L. 4. That is below the 10 mg/L nitrate (as N) MCL, which is the point of blending. 5. Answer: 7.67 mg/L (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Dilution/Blending: "(C1 x V1) + (C2 x V2) = C3 x V3"

More chemical feed and solutions problems for T3

Electrical/SCADA (sheet p.1)

Process Units = (Live Signal (mA) - 4 mA) x Process Range / 16 mA

Used in 100 problems: T2, T3, D2, D3.

Example 15. A flow transmitter is scaled 4 mA = 0 GPM and 20 mA = 1,600 GPM. The SCADA input reads 14.9 mA. What flow does that represent?

A) 1,090 GPM
B) 545 GPM
C) 872 GPM
D) 363 GPM

Show the answer and worked solution

Answer: A) 1,090 GPM

1. Process Units = (Live Signal − 4 mA) × Process Range ÷ 16 mA [Sheet p.1, Electrical/SCADA: "Process Units = (Live Signal (mA) - 4 mA) x Process Range / 16 mA"]. 2. (14.9 − 4) = 10.9 mA; 10.9 × 1,600 ÷ 16 = 1,090 GPM. 3. Answer: 1,090 GPM.

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Electrical/SCADA: "Process Units = (Live Signal (mA) - 4 mA) x Process Range / 16 mA"

More unit conversions and pressure problems for T2

Other (sheet p.1)

Cost/Day ($) = Lbs/Day x Cost/lb ($)

Used in 50 problems: T3, D3.

Example 16. A 7.5 MGD system doses dry hydrated lime at 38.5 mg/L. It costs $0.43 per lb. What is the chemical cost per day?

A) $929 per day
B) $1,040 per day
C) $124 per day
D) $2,410 per day

Show the answer and worked solution

Answer: B) $1,040 per day

1. lbs/day = MGD × mg/L × 8.34 (the sheet's "ppm or mg/L" line solved for lbs) [Sheet p.2, Chemical Dosage Calculations: "ppm or mg/L = lbs / (MG x 8.34)"] = 7.5 × 38.5 × 8.34 = 2,408 lbs/day. 2. Cost/Day = Lbs/Day × Cost/lb [Sheet p.1, Other: "Cost/Day ($) = Lbs/Day x Cost/lb ($)"] = 2,408 × $0.43 = $1,035.44 per day. 3. Answer: $1,040 per day (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, Chemical Dosage Calculations: "ppm or mg/L = lbs / (MG x 8.34)"; p.1, Other: "Cost/Day ($) = Lbs/Day x Cost/lb ($)"

More chemical feed and solutions problems for T3

Removal (%) = (In - Out) / In x 100

Used in 25 problems: T3.

Example 17. Raw water turbidity is 59 NTU and settled water turbidity is 15.3 NTU. What is the percent removal across sedimentation?

A) 92.6%
B) 74.1%
C) 43.7%
D) 25.9%

Show the answer and worked solution

Answer: B) 74.1%

1. Removal (%) = (In − Out) ÷ In × 100 [Sheet p.1, Other: "Removal (%) = (In - Out) / In x 100"] = (59 − 15.3) ÷ 59 × 100 = 74.07%. 2. Answer: 74.1% (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Other: "Removal (%) = (In - Out) / In x 100"

More filtration and removal problems for T3

Solutions (sheet p.1)

(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)

Used in 75 problems: T1, T2, T3.

Example 18. A tank holds 970 gallons of a 50% solution with a specific gravity of 1.52. How many pounds of active chemical does it contain?

A) 4,040 lbs
B) 737 lbs
C) 5,510 lbs
D) 6,150 lbs

Show the answer and worked solution

Answer: D) 6,150 lbs

1. (lbs) Chemical = SG × 8.34 × Solution (Gal) × Solution (%/100) [Sheet p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"]. 2. = 1.52 × 8.34 × 970 × 0.5 = 6,148 lbs. 3. Answer: 6,150 lbs (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"

More chemical feed and solutions problems for T1

Specific Gravity (SG) = Chemical Weight (Lbs/Gal) / 8.34

Used in 75 problems: T1, T2, T3.

Example 19. A liquid chemical weighs 12.55 lbs per gallon. What is its specific gravity?

A) 1.50
B) 0.665
C) 0.201
D) 1.68

Show the answer and worked solution

Answer: A) 1.50

1. Specific Gravity = Chemical Weight (lbs/gal) ÷ 8.34 [Sheet p.1, Solutions: "Specific Gravity (SG) = Chemical Weight (Lbs/Gal) / 8.34"] = 12.55 ÷ 8.34 = 1.5048. 2. Answer: 1.50 (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "Specific Gravity (SG) = Chemical Weight (Lbs/Gal) / 8.34"

More chemical feed and solutions problems for T1

Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)

Used in 75 problems: T1, T2, T3.

Example 20. An operator dissolves 118 lbs of soda ash in 80 gallons of water. What is the percent strength of the solution by weight?

A) 59.6%
B) 15.0%
C) 17.7%
D) 7.50%

Show the answer and worked solution

Answer: B) 15.0%

1. Water weight = 80 gal × 8.34 lbs/gal = 667.2 lbs [Sheet p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"]. 2. Solution total weight = 118 + 667.2 = 785.2 lbs. 3. Chemical in Solution (%) = Dry Chemical × 100 ÷ Solution Total Weight [Sheet p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"] = 118 × 100 ÷ 785.2 = 15.03%. 4. Answer: 15.0% (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"; p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"

More chemical feed and solutions problems for T1

GPD = MGD x mg/L x 8.34 / ((% purity/100) x Chemical Wt. (lbs/gal))

Used in 75 problems: T1, T2, T3.

Example 21. A plant treats 13 MGD and feeds ferric chloride solution at 20.5 mg/L. The product is 40% pure and weighs 11.8 lbs/gal. How many gallons per day of product are needed?

A) 5,560 GPD
B) 666 GPD
C) 2,220 GPD
D) 471 GPD

Show the answer and worked solution

Answer: D) 471 GPD

1. GPD = MGD × mg/L × 8.34 ÷ ((% purity/100) × Chemical Wt.) [Sheet p.1, Solutions: "GPD = MGD x mg/L x 8.34 / ((% purity/100) x Chemical Wt. (lbs/gal))"]. 2. Numerator: 13 × 20.5 × 8.34 = 2,222.61 lbs/day. 3. Denominator: 0.4 × 11.8 = 4.72 lbs active per gal. 4. GPD = 2,222.61 ÷ 4.72 = 470.9 GPD. 5. Answer: 471 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "GPD = MGD x mg/L x 8.34 / ((% purity/100) x Chemical Wt. (lbs/gal))"

More chemical feed and solutions problems for T1

GPD = (Feed, ml/min. x 1,440 min/day) / (1,000 ml/L x 3.785 L/gal)

Used in 75 problems: T1, T2, T3.

Example 22. A chemical metering pump is calibrated at 390 mL/min. How many gallons per day does it feed?

A) 148 GPD
B) 6.18 GPD
C) 562 GPD
D) 2,130 GPD

Show the answer and worked solution

Answer: A) 148 GPD

1. GPD = (Feed, mL/min × 1,440 min/day) ÷ (1,000 mL/L × 3.785 L/gal) [Sheet p.1, Solutions: "GPD = (Feed, ml/min. x 1,440 min/day) / (1,000 ml/L x 3.785 L/gal)"]. 2. = 390 × 1,440 ÷ 3,785 = 561,600 ÷ 3,785 = 148.4 GPD. 3. Answer: 148 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "GPD = (Feed, ml/min. x 1,440 min/day) / (1,000 ml/L x 3.785 L/gal)"

More chemical feed and solutions problems for T1

mg/L = lbs / (MG x 8.34)

Used in 300 problems: T1, T2, T3, D1, D2, D3.

Example 23. A chlorinator fed 31 lbs of chlorine in one day to a flow of 1.1 MGD. What was the dose in mg/L?

A) 3.38 mg/L
B) 28.2 mg/L
C) 0.296 mg/L
D) 235 mg/L

Show the answer and worked solution

Answer: A) 3.38 mg/L

1. mg/L = lbs ÷ (MG × 8.34) [Sheet p.1, Solutions: "mg/L = lbs / (MG x 8.34)"] = 31 ÷ (1.1 × 8.34) = 31 ÷ 9.174 = 3.379 mg/L. 2. Answer: 3.38 mg/L (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "mg/L = lbs / (MG x 8.34)"

More chlorination and ct problems for T1

Pumping (sheet p.1)

Water HP = GPM x Head (Ft) / 3960

Used in 25 problems: D3.

Example 24. A pump delivers 2,080 GPM against a total head of 310 ft. What is the water horsepower?

A) 376 HP
B) 326 HP
C) 218 HP
D) 163 HP

Show the answer and worked solution

Answer: D) 163 HP

1. Water HP = GPM × Head (Ft) ÷ 3960 [Sheet p.1, Pumping: "Water HP = GPM x Head (Ft) / 3960"] = 2,080 × 310 ÷ 3960 = 162.8 HP. 2. Answer: 163 HP (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Water HP = GPM x Head (Ft) / 3960"

More wells, pumps and power problems for D3

Brake HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100))

Used in 25 problems: D3.

Example 25. A pump delivers 1,420 GPM against 90 ft of total head. The pump is 82% efficient. What is the brake horsepower?

A) 39.4 HP
B) 78.8 HP
C) 59.1 HP
D) 49.3 HP

Show the answer and worked solution

Answer: A) 39.4 HP

1. Brake HP = GPM × Head ÷ (3960 × Pump Eff/100) [Sheet p.1, Pumping: "Brake HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100))"]. 2. = 1,420 × 90 ÷ (3960 × 0.82) = 127,800 ÷ 3,247.2 = 39.36 HP. 3. Answer: 39.4 HP (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Brake HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100))"

More wells, pumps and power problems for D3

Motor HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100) x (Motor % Eff/100))

Used in 25 problems: D3.

Example 26. A pump delivers 580 GPM against 205 ft of total head. The pump is 66% efficient and the motor is 92% efficient. What is the motor horsepower?

A) 49.4 HP
B) 61.8 HP
C) 74.1 HP
D) 98.8 HP

Show the answer and worked solution

Answer: A) 49.4 HP

1. Motor HP = GPM × Head ÷ (3960 × Pump Eff/100 × Motor Eff/100) [Sheet p.1, Pumping: "Motor HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100) x (Motor % Eff/100))"]. 2. = 118,900 ÷ (3960 × 0.66 × 0.92) = 118,900 ÷ 2,405 = 49.44 HP. 3. Answer: 49.4 HP (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Motor HP = GPM x Head (Ft) / (3960 x (Pump % Eff/100) x (Motor % Eff/100))"

More wells, pumps and power problems for D3

WWE (%) = (Pump % Eff/100) x (Motor % Eff/100)

Used in 25 problems: D3.

Example 27. A pump is 84% efficient and its motor is 86% efficient. What is the wire-to-water efficiency, in percent?

A) 36.1%
B) 54.2%
C) 27.8%
D) 72.2%

Show the answer and worked solution

Answer: D) 72.2%

1. WWE = (Pump Eff/100) × (Motor Eff/100) [Sheet p.1, Pumping: "WWE (%) = (Pump % Eff/100) x (Motor % Eff/100)"] = 0.84 × 0.86 = 0.7224. 2. The sheet labels WWE "(%)", which means you multiply the decimal by 100 to get a percent: 0.7224 × 100 = 72.24%. 3. Answer: 72.2% (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "WWE (%) = (Pump % Eff/100) x (Motor % Eff/100)"

More wells, pumps and power problems for D3

Cost ($) = HP x .746 x Operating Hours (hrs) x $/kWh

Used in 25 problems: T3.

Example 28. A 60 HP motor runs 22 hours a day for 30 days. Power costs $0.15 per kWh. Ignoring motor efficiency, what is the power cost for the period?

A) $2,220
B) $3,320
C) $4,430
D) $148

Show the answer and worked solution

Answer: C) $4,430

1. Operating hours = 22 × 30 = 660 hrs. 2. Cost = HP × .746 × Operating Hours × $/kWh [Sheet p.1, Pumping: "Cost ($) = HP x .746 x Operating Hours (hrs) x $/kWh"] = 60 × .746 × 660 × 0.15 = $4,431.24. 3. Answer: $4,430 (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Cost ($) = HP x .746 x Operating Hours (hrs) x $/kWh"

More wells, pumps and power problems for T3

Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)

Used in 550 problems: T1, T2, T3, D1, D2, D3.

Example 29. Before pumping, the water level in a well is 44 ft below ground. After pumping for several hours it stabilizes at 128 ft below ground. What is the drawdown?

A) 44.0 ft
B) 172 ft
C) 128 ft
D) 84.0 ft

Show the answer and worked solution

Answer: D) 84.0 ft

1. Drawdown = pumping water level − static water level (the denominator of [Sheet p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"]). 2. = 128 − 44 = 84 ft. 3. Answer: 84.0 ft.

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Pumping: "Specific Capacity (GPM/Ft) = Well Yield (GPM) / Drawdown (Ft)"

More wells, pumps and power problems for T1

Sedimentation (sheet p.1)

DT (Min) = Volume (Gal) / Flow (GPM)

Used in 75 problems: T1, T2, T3.

Example 30. A round clearwell is 80 ft in diameter with water 28 ft deep. Flow through it is 14,050 GPM. What is the detention time in minutes?

A) 112 min
B) 150 min
C) 300 min
D) 74.9 min

Show the answer and worked solution

Answer: D) 74.9 min

1. Volume = .785 × 80² × 28 × 7.48 = 1,052,227 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)"] = 1,052,227 ÷ 14,050 = 74.89 min. 3. Answer: 74.9 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)"

More flow, velocity and detention problems for T1

DT (hrs) = Ft3 x 7.48 x 24 / Gal/Day

Used in 75 problems: T1, T2, T3.

Example 31. A sedimentation basin is 40 ft long, 18 ft wide and 9 ft deep. The plant flow is 0.2 MGD. What is the detention time in hours?

A) 5.82 hours
B) 0.778 hours
C) 6.49 hours
D) 0.242 hours

Show the answer and worked solution

Answer: A) 5.82 hours

1. Volume = 40 × 18 × 9 = 6,480 ft³. 2. Flow = 0.2 MGD × 1,000,000 = 200,000 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"] = 6,480 × 7.48 × 24 ÷ 200,000 = 5.816 hours. 4. Answer: 5.82 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"

More flow, velocity and detention problems for T1

SLR (GPD/Ft2) = Total Flow (GPD) / Surface Area (Ft2)

Used in 25 problems: T3.

Example 32. A circular clarifier 100 ft in diameter treats 4.6 MGD. What is the surface loading rate in GPD/ft²?

A) 586 GPD/ft²
B) 146 GPD/ft²
C) 0.407 GPD/ft²
D) 293 GPD/ft²

Show the answer and worked solution

Answer: A) 586 GPD/ft²

1. Surface area = .785 × 100² = 7,850 ft² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"]. 2. Flow = 4.6 × 1,000,000 = 4,600,000 GPD [Sheet p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal)]. 3. SLR = Total Flow (GPD) ÷ Surface Area [Sheet p.1, Sedimentation: "SLR (GPD/Ft2) = Total Flow (GPD) / Surface Area (Ft2)"] = 4,600,000 ÷ 7,850 = 586 GPD/ft². 4. Answer: 586 GPD/ft² (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Circle = .785 X D2"; p.1, Unit Definition: "MG Million Gallons" (1 MG = 1,000,000 gal); p.1, Sedimentation: "SLR (GPD/Ft2) = Total Flow (GPD) / Surface Area (Ft2)"

More flow, velocity and detention problems for T3

Weir Overflow Rate (GPD/Ft) = Flow (GPD) / Weir Length (Ft)

Used in 25 problems: T3.

Example 33. A circular clarifier 40 ft in diameter has a weir around its full circumference and treats 2.2 MGD. What is the weir overflow rate in GPD per foot?

A) 70,100 GPD/ft
B) 8,750 GPD/ft
C) 17,500 GPD/ft
D) 55,000 GPD/ft

Show the answer and worked solution

Answer: C) 17,500 GPD/ft

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Circumference of a Circle = 3.14 x D"; p.1, Sedimentation: "Weir Overflow Rate (GPD/Ft) = Flow (GPD) / Weir Length (Ft)"

More flow, velocity and detention problems for T3

Pounds Force (sheet p.1)

lbs Force = PSI x Area (in2)

Used in 50 problems: D2, D3.

Example 34. A closed 30-inch gate valve holds back water at 141 psi. What is the total force on the valve face, in pounds?

A) 692 lbs
B) 3,320 lbs
C) 398,000 lbs
D) 99,600 lbs

Show the answer and worked solution

Answer: D) 99,600 lbs

1. Area of the valve face = .785 × D² [Sheet p.1, Area and Volume: "Area of Circle = .785 X D2"] = .785 × 30 × 30 = 706.5 in². 2. lbs Force = PSI × Area (in²) [Sheet p.1, Pounds Force: "lbs Force = PSI x Area (in2)"] = 141 × 706.5 = 99,617 lbs. 3. Answer: 99,600 lbs (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Circle = .785 X D2"; p.1, Pounds Force: "lbs Force = PSI x Area (in2)"

Chemical Dosage Calculations (sheet p.2)

Liquid (Gal) = MG x mg/L x 8.34 / (Commercial Purity (%)/100 x Ion Purity (%)/100 x Chemical Weight (Lbs/gal))

Used in 25 problems: T3.

Example 35. A 0.9 MGD plant raises fluoride by 0.55 mg/L using fluorosilicic acid that is 25% commercial purity, 79.2% fluoride ion purity, and weighs 10.1 lbs/gal. How many gallons per day of acid are fed?

A) 0.516 GPD
B) 1.63 GPD
C) 2.06 GPD
D) 4.12 GPD

Show the answer and worked solution

Answer: C) 2.06 GPD

1. Liquid (Gal) = MG × mg/L × 8.34 ÷ (Commercial Purity/100 × Ion Purity/100 × Chemical Weight) [Sheet p.2, Chemical Dosage Calculations: "Liquid (Gal) = MG x mg/L x 8.34 / (Commercial Purity (%)/100 x Ion Purity (%)/100 x Chemical Weight (Lbs/gal))"]. 2. Numerator: 0.9 × 0.55 × 8.34 = 4.1283 lbs/day of fluoride ion. 3. Denominator: 0.25 × 0.792 × 10.1 = 1.9998 lbs of fluoride ion per gallon. 4. = 4.1283 ÷ 1.9998 = 2.064 GPD. 5. Answer: 2.06 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, Chemical Dosage Calculations: "Liquid (Gal) = MG x mg/L x 8.34 / (Commercial Purity (%)/100 x Ion Purity (%)/100 x Chemical Weight (Lbs/gal))"

More chemical feed and solutions problems for T3

ppm or mg/L = lbs / (MG x 8.34)

Used in 75 problems: T3, D3.

Example 36. A 7.5 MGD system doses dry hydrated lime at 38.5 mg/L. It costs $0.43 per lb. What is the chemical cost per day?

A) $929 per day
B) $1,040 per day
C) $124 per day
D) $2,410 per day

Show the answer and worked solution

Answer: B) $1,040 per day

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, Chemical Dosage Calculations: "ppm or mg/L = lbs / (MG x 8.34)"; p.1, Other: "Cost/Day ($) = Lbs/Day x Cost/lb ($)"

More chemical feed and solutions problems for T3

C x T Calculations (sheet p.2)

C x T = Chlorine Residual (mg/L) x Time (min)

Used in 75 problems: T2, T3, D3.

Example 37. Water leaving a contact tank has a free chlorine residual of 2.3 mg/L after a contact time (T10) of 87 minutes. What is the CT value?

A) 87.0 mg·min/L
B) 37.8 mg·min/L
C) 200 mg·min/L
D) 3.34 mg·min/L

Show the answer and worked solution

Answer: C) 200 mg·min/L

1. C × T = Chlorine Residual × Time [Sheet p.2, C x T Calculations: "C x T = Chlorine Residual (mg/L) x Time (min)"] = 2.3 × 87 = 200.1 mg·min/L. 2. Answer: 200 mg·min/L (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "C x T = Chlorine Residual (mg/L) x Time (min)"

More chlorination and ct problems for T2

Time (min) = C x T / Chlorine Residual (mg/L)

Used in 75 problems: T2, T3, D3.

Example 38. A required CT is 10 mg·min/L and the free chlorine residual is 0.75 mg/L. What contact time is needed?

A) 9.25 min
B) 13.3 min
C) 0.222 min
D) 7.50 min

Show the answer and worked solution

Answer: B) 13.3 min

1. Time (min) = C × T ÷ Chlorine Residual [Sheet p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"] = 10 ÷ 0.75 = 13.33 min. 2. Answer: 13.3 min (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "Time (min) = C x T / Chlorine Residual (mg/L)"

More chlorination and ct problems for T2

Inactivation Ratio = Actual C x T / Table

Read as E" C x T". Used in 75 problems: T2, T3, D3.

Example 39. A clearwell has a contact time (T10) of 86 minutes and a free chlorine residual of 0.8 mg/L. The required CT from the regulatory table is 44 mg·min/L. What is the inactivation ratio?

A) 1.56
B) 0.780
C) 0.512
D) 0.640

Show the answer and worked solution

Answer: A) 1.56

1. C × T Calculated = T10 × Chlorine Residual [Sheet p.2, C x T Calculations: "C x T Calculated = T10 Value, min x Chlorine Residual (mg/L)"] = 86 × 0.8 = 68.8 mg·min/L. 2. Inactivation Ratio = Actual C × T ÷ Table "E" C × T [Sheet p.2, C x T Calculations: "Inactivation Ratio = Actual C x T / Table "E" C x T"] = 68.8 ÷ 44 = 1.564. 3. A ratio of 1.0 or more means the required inactivation is being met. 4. Answer: 1.56 (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, C x T Calculations: "C x T Calculated = T10 Value, min x Chlorine Residual (mg/L)"; p.2, C x T Calculations: "Inactivation Ratio = Actual C x T / Table "E" C x T"

More chlorination and ct problems for T2

C x T Calculated = T10 Value, min x Chlorine Residual (mg/L)

Used in 75 problems: T2, T3, D3.

Example 40. A clearwell has a contact time (T10) of 86 minutes and a free chlorine residual of 0.8 mg/L. The required CT from the regulatory table is 44 mg·min/L. What is the inactivation ratio?

A) 1.56
B) 0.780
C) 0.512
D) 0.640

Show the answer and worked solution

Answer: A) 1.56

More chlorination and ct problems for T2

Log Removal = 1.0 - (% Removal/100) x Log Key x (-1)

Read as -log10(1 - %Removal/100). Used in 25 problems: T3.

Example 41. A treatment process removes 97.37% of Giardia cysts. What log removal is that?

A) 3.64 log
B) 1.58 log
C) 1.99 log
D) 2.63 log

Show the answer and worked solution

Answer: B) 1.58 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 − 0.9737 = 0.0263; log(0.0263) = -1.58; × (−1) = 1.58 log. 3. Answer: 1.58 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))

More filtration and removal problems for T3

Filtration (sheet p.2)

Filtration Rate (GPM/Ft2) = Filter Production (GPD) / (Filter Area (Ft2) x 1440)

Used in 25 problems: T3.

Example 42. A 34 ft × 30 ft filter produced 3,819,000 gallons in 24 hours. What was the filtration rate in GPM/ft²?

A) 0.348 GPM/ft²
B) 2.60 GPM/ft²
C) 1.30 GPM/ft²
D) 1.95 GPM/ft²

Show the answer and worked solution

Answer: B) 2.60 GPM/ft²

1. Filter area = 34 × 30 = 1020 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 2. Filtration Rate = Filter Production (GPD) ÷ (Filter Area × 1440) [Sheet p.2, Filtration: "Filtration Rate (GPM/Ft2) = Filter Production (GPD) / (Filter Area (Ft2) x 1440)"] = 3,819,000 ÷ (1020 × 1440) = 2.6 GPM/ft². 3. Answer: 2.60 GPM/ft² (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Filtration Rate (GPM/Ft2) = Filter Production (GPD) / (Filter Area (Ft2) x 1440)"

More filtration and removal problems for T3

Loading Rate (GPM/Ft2) = Flow Rate (GPM) / Filter Area (Ft2)

Used in 25 problems: T3.

Example 43. A plant filters 9.2 MGD through 6 identical filters, each 17 ft × 17 ft, all in service. What is the loading rate on each filter in GPM/ft²?

A) 2.38 GPM/ft²
B) 22.1 GPM/ft²
C) 3.68 GPM/ft²
D) 88.4 GPM/ft²

Show the answer and worked solution

Answer: C) 3.68 GPM/ft²

1. Plant flow = 9.2 × 694.4 = 6,388.48 GPM [Sheet p.1, Units and Conversion Factors, row 29: "1 MGD = 694.4 GPM"]; per filter = 6,388.48 ÷ 6 = 1,064.7 GPM. 2. Filter area = 17 × 17 = 289 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)"] = 1,064.7 ÷ 289 = 3.6841 GPM/ft². 4. Answer: 3.68 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)"

More filtration and removal problems for T3

Daily Filter Production (GPD) = Filter Area (Ft2) x (GPM/Ft2) x 1440

Used in 50 problems: T2, T3.

Example 44. A 36 ft × 25 ft filter runs all day at 2.9 GPM/ft². How many gallons does it produce per day?

A) 28,100,000 GPD
B) 157,000 GPD
C) 255,000 GPD
D) 3,760,000 GPD

Show the answer and worked solution

Answer: D) 3,760,000 GPD

1. Filter area = 36 × 25 = 900 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 2. Daily Filter Production = Filter Area × (GPM/ft²) × 1440 [Sheet p.2, Filtration: "Daily Filter Production (GPD) = Filter Area (Ft2) x (GPM/Ft2) x 1440"] = 900 × 2.9 × 1440 = 3,758,400 GPD. 3. Answer: 3,760,000 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Daily Filter Production (GPD) = Filter Area (Ft2) x (GPM/Ft2) x 1440"

More filtration and removal problems for T2

Backwash Volume (Gal) = Filter Area (Ft2) x Backwash Rate (GPM/Ft2) x Time (min)

Used in 50 problems: T2, T3.

Example 45. A 37 ft × 26 ft filter is backwashed at 22.5 GPM/ft² for 13 minutes. How many gallons of backwash water are used?

A) 4,690 gal
B) 281,000 gal
C) 37,600 gal
D) 21,600 gal

Show the answer and worked solution

Answer: B) 281,000 gal

1. Filter area = 37 × 26 = 962 ft² [Sheet p.1, Area and Volume: "Area of Rectangle = Length x Width"]. 2. Backwash Volume = Filter Area × Backwash Rate × Time [Sheet p.2, Filtration: "Backwash Volume (Gal) = Filter Area (Ft2) x Backwash Rate (GPM/Ft2) x Time (min)"] = 962 × 22.5 × 13 = 281,385 gal. 3. Answer: 281,000 gal (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Area and Volume: "Area of Rectangle = Length x Width"; p.2, Filtration: "Backwash Volume (Gal) = Filter Area (Ft2) x Backwash Rate (GPM/Ft2) x Time (min)"

More filtration and removal problems for T2

Backwash Rate (GPM/Ft2) = Flow Rate (GPM) / Filter Area (Ft2)

Used in 50 problems: T2, T3.

Example 46. A 32 ft × 18 ft filter is backwashed at 6,600 GPM. What is the backwash rate in GPM/ft²?

A) 132 GPM/ft²
B) 18.4 GPM/ft²
C) 1.53 GPM/ft²
D) 11.5 GPM/ft²

Show the answer and worked solution

Answer: D) 11.5 GPM/ft²

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

More filtration and removal problems for T2

Rate of Rise (Inches/min) = Backwash Rate (GPM/Ft2) x 12 (in/Ft) / 7.48 (gal/Ft3)

Used in 50 problems: T2, T3.

Example 47. A filter is backwashed at 23 GPM/ft². What is the rate of rise of the water in inches per minute?

A) 3.07 in/min
B) 276 in/min
C) 36.9 in/min
D) 14.3 in/min

Show the answer and worked solution

Answer: C) 36.9 in/min

1. Rate of Rise = Backwash Rate × 12 (in/ft) ÷ 7.48 (gal/ft³) [Sheet p.2, Filtration: "Rate of Rise (Inches/min) = Backwash Rate (GPM/Ft2) x 12 (in/Ft) / 7.48 (gal/Ft3)"]. 2. = 23 × 12 ÷ 7.48 = 276 ÷ 7.48 = 36.9 in/min. 3. Answer: 36.9 in/min (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.2, Filtration: "Rate of Rise (Inches/min) = Backwash Rate (GPM/Ft2) x 12 (in/Ft) / 7.48 (gal/Ft3)"

More filtration and removal problems for T2

Unit Filter Run Volume (UFRV) = Volume Produced in a Filter Run (Gal) / Filter Area (Ft2)

Used in 25 problems: T3.

Example 48. A 37 ft × 24 ft filter ran 52 hours at an average of 2,840 GPM before backwashing. What was the unit filter run volume in gal/ft²?

A) 20,000 gal/ft²
B) 239,000 gal/ft²
C) 15,000 gal/ft²
D) 9,980 gal/ft²

Show the answer and worked solution

Answer: D) 9,980 gal/ft²

1. Volume produced = 2,840 GPM × 52 hrs × 60 min/hr = 8,860,800 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 = 37 × 24 = 888 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)"] = 8,860,800 ÷ 888 = 9,978 gal/ft². 4. Answer: 9,980 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)"

More filtration and removal problems for T3

Conversion factors the problems use

From the Units and Conversion Factors column on page 1, by row number.

Row	Printed	Problems
1	1 % = 10,000 mg/L	150
4	1 AF = 325,851 Gal	150
6	1 CCF = 748 Gal	150
7	1 CFS = 448.8 GPM	425
12	1 Ft (H2O) = 0.433 PSI	150
16	1 Gal (H2O) = 8.34 lbs	75
17	1 gm = 1000 mg/L	25
18	1 gpg = 17.1 mg/L	150
22	1 L = 1000 mL	25
28	1 MGD = 1.55 CFS	200
29	1 MGD = 694.4 GPM	475
31	1 PSI = 2.31 Ft	275
34	1 yd3 = 27 Ft3	75

On the sheet, no problems here yet

These lines are printed on the sheet but none of our problems use them yet:

Slope (%) = Rise / Distance x 100
Area of Right Triangle = (Base x Height)/2
Gal/Capita/Day = GPD / Population
(lbs/Gal) = Solution (%)/100 x 8.34 x Specific Gravity (SG)
Flow = Volume / Time
Gas Feed Dry (Lbs) = MG x mg/L x 8.34
Solid (Lbs) = MG x mg/L x 8.34 / (Strength (%)/100)
Backwash Pumping Rate (GPM) = Filter Area (Ft2) x Backwash Rate (GPM/Ft2)
A = Q / V
Q1/V1 = Q2/V2
the temperature conversions
Chlorine Residual = C x T / Time

Conversion sheet questions

Where do I download the California water operator exam conversion sheet?

From the State Water Board's Drinking Water Operator Certification page, where it is listed as "New Format: Exam Formula Conversion Sheet". The current revision is dated 4/30/2026 and has two pages.

What changed on the 4/30/2026 conversion sheet?

One number. Row 29 of the units column now reads 1 MGD = 694.4 GPM; the 6/2/2025 sheet said 694 GPM. Everything else is the same apart from layout.

What formulas are on the conversion sheet?

A column of unit conversion factors plus formula boxes: Area and Volume, Chlorination, Flow & Velocity, Dilution/Blending, Electrical/SCADA, Other, Solutions, Pumping, Sedimentation, Pounds Force, Chemical Dosage Calculations, C x T Calculations, Filtration. This page shows 48 of the formulas with a worked example each.

Which number do I use when the sheet gives two different factors?

The one printed on the line you are using. For example, the feed-pump formula uses 3.785 L/gal, while the units column says 1 Gal = 3.79 L. Every worked solution here uses the factor printed on the line it cites and never mixes two lines in one step.

Official sheet: CA Exam Conversion Sheet (revised 4/30/2026), on the State Water Board operator certification page.

WaterExamMath is independent practice software. It is not affiliated with, endorsed by or connected to the California State Water Resources Control Board. It covers the math portion of the drinking water operator exams only, not the whole exam, and passing is not guaranteed. Worked examples by WaterExamMath.