Colorado WW Operator D Exam Actual
Exam QUESTIONS AND ANSWERS 2026 |
Complete Solution Guide A+ Graded | Pass
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Formula Reference Sheet
Table
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Formula Description
Volume (rectangular) V=L×W×D
Volume (circular) V = 0.785 × D² × H
Flow rate Q = A × V (Area × Velocity)
Detention time DT = Volume / Flow
Chemical dosage lbs/day = mg/L × MGD × 8.34
Chlorine dose Dose = Demand + Residual
Surface loading rate SLR = Flow / Surface Area
Weir loading rate WLR = Flow / Weir Length
, Formula Description
Pump horsepower HP = (Flow × Head) / (3960 × Efficiency)
Pressure to head Head (ft) = PSI × 2.31
mg/L to ppm 1 mg/L = 1 ppm (in water)
Temperature conversion °C = (°F - 32) × 5/9
Conversions:
• 1 cubic foot = 7.48 gallons
• 1 MGD = 1,000,000 gallons/day = 694.4 gpm
• 1 acre-foot = 325,851 gallons
• 1 psi = 2.31 feet of head
• 8.34 lbs = weight of 1 gallon of water
Section 1: Basic Mathematics and Calculations
Questions 1-15
Question 1 Convert 2,500 gallons to cubic feet. (1 ft³ = 7.48 gallons)
A. 18.7 ft³ B. 334.2 ft³ [CORRECT] C. 18,700 ft³ D. 334.2 MGD
Correct Answer: B
Rationale:
• Concept: Unit conversion between gallons and cubic feet.
• Step-by-Step Solution:
o Volume in ft³ = 2,500 gallons ÷ 7.48 gal/ft³
o 2,500 ÷ 7.48 = 334.224... ≈ 334.2 ft³
• Why B is correct: Proper division conversion factor.
, • Why A is incorrect: This is 2,500 ÷ 133.7 (using gallons per cubic foot reversed) or
calculation error.
• Why C is incorrect: This multiplies instead of divides (2,500 × 7.48).
• Why D is incorrect: MGD (million gallons per day) is a flow rate unit, not volume.
Practical Application: Tank volumes are often calculated in cubic feet but reported in gallons.
Always verify units on pump curves and tank specifications.
Question 2 A rectangular sedimentation basin is 40 feet long, 20 feet wide, and 10 feet deep.
What is the volume in gallons when filled to 8 feet depth?
A. 6,400 gallons B. 47,872 gallons [CORRECT] C. 59,840 gallons D. 478,720 gallons
Correct Answer: B
Rationale:
• Concept: Volume calculation with partial depth.
• Step-by-Step Solution:
o Volume in ft³ = Length × Width × Depth = 40 ft × 20 ft × 8 ft = 6,400 ft³
o Volume in gallons = 6,400 ft³ × 7.48 gal/ft³ = 47,872 gallons
• Why B is correct: Correct calculation using actual water depth (8 ft), not total basin
depth.
• Why A is incorrect: This is the volume in cubic feet, not converted to gallons.
• Why C is incorrect: This uses full 10 ft depth: 40 × 20 × 10 × 7.48 = 59,840 gallons.
• Why D is incorrect: Decimal error—likely multiplied by 74.8 instead of 7.48.
Practical Application: Always use actual operating depth, not total basin depth, unless
calculating maximum capacity.
Question 3 A pipeline has a diameter of 12 inches and water flows at 3 feet per second. What is
the flow rate in gallons per minute (gpm)?
A. 105.6 gpm B. 211.5 gpm C. 1,058 gpm [CORRECT] D. 1,270 gpm
Correct Answer: C
Rationale:
, • Concept: Flow rate calculation using Q = A × V.
• Step-by-Step Solution:
o Convert diameter to feet: 12 inches = 1 foot, radius = 0.5 ft
o Area = π × r² = 3.1416 × (0.5)² = 0.7854 ft²
o Or use formula: Area = 0.785 × D² = 0.785 × 1² = 0.785 ft²
o Flow in cfs = 0.785 ft² × 3 ft/sec = 2.355 ft³/sec
o Convert to gpm: 2.355 ft³/sec × 7.48 gal/ft³ × 60 sec/min = 1,057.3 gpm ≈ 1,058
gpm
• Why C is correct: Correct application of Q = A × V with proper unit conversions.
• Why A is incorrect: This omits the 60 seconds/minute conversion (gives cfs × 7.48
only).
• Why B is incorrect: This uses diameter instead of radius squared, or uses 6 inches as
radius.
• Why D is incorrect: This might use 12 inches as 1.5 feet or similar error.
Practical Application: Velocity measurements with flow meters require pipe diameter
knowledge. Always verify pipe ID, not nominal diameter.
Question 4 A water treatment plant treats 3.5 MGD. The alum dosage is 8 mg/L. How many
pounds of alum are needed per day?
A. 29.2 lbs B. 233.5 lbs [CORRECT] C. 2,335 lbs D. 29,190 lbs
Correct Answer: B
Rationale:
• Concept: Chemical dosage calculation using pounds formula.
• Step-by-Step Solution:
o Formula: lbs/day = mg/L × MGD × 8.34
o lbs/day = 8 mg/L × 3.5 MGD × 8.34
o lbs/day = 8 × 3.5 × 8.34 = 28 × 8.34 = 233.52 lbs ≈ 233.5 lbs
• Why B is correct: Proper application of dosage formula.
• Why A is incorrect: This might be 8 × 3.5 = 28 (missing 8.34 factor).
Exam QUESTIONS AND ANSWERS 2026 |
Complete Solution Guide A+ Graded | Pass
Guaranteed - A+ Graded
Formula Reference Sheet
Table
Copy
Formula Description
Volume (rectangular) V=L×W×D
Volume (circular) V = 0.785 × D² × H
Flow rate Q = A × V (Area × Velocity)
Detention time DT = Volume / Flow
Chemical dosage lbs/day = mg/L × MGD × 8.34
Chlorine dose Dose = Demand + Residual
Surface loading rate SLR = Flow / Surface Area
Weir loading rate WLR = Flow / Weir Length
, Formula Description
Pump horsepower HP = (Flow × Head) / (3960 × Efficiency)
Pressure to head Head (ft) = PSI × 2.31
mg/L to ppm 1 mg/L = 1 ppm (in water)
Temperature conversion °C = (°F - 32) × 5/9
Conversions:
• 1 cubic foot = 7.48 gallons
• 1 MGD = 1,000,000 gallons/day = 694.4 gpm
• 1 acre-foot = 325,851 gallons
• 1 psi = 2.31 feet of head
• 8.34 lbs = weight of 1 gallon of water
Section 1: Basic Mathematics and Calculations
Questions 1-15
Question 1 Convert 2,500 gallons to cubic feet. (1 ft³ = 7.48 gallons)
A. 18.7 ft³ B. 334.2 ft³ [CORRECT] C. 18,700 ft³ D. 334.2 MGD
Correct Answer: B
Rationale:
• Concept: Unit conversion between gallons and cubic feet.
• Step-by-Step Solution:
o Volume in ft³ = 2,500 gallons ÷ 7.48 gal/ft³
o 2,500 ÷ 7.48 = 334.224... ≈ 334.2 ft³
• Why B is correct: Proper division conversion factor.
, • Why A is incorrect: This is 2,500 ÷ 133.7 (using gallons per cubic foot reversed) or
calculation error.
• Why C is incorrect: This multiplies instead of divides (2,500 × 7.48).
• Why D is incorrect: MGD (million gallons per day) is a flow rate unit, not volume.
Practical Application: Tank volumes are often calculated in cubic feet but reported in gallons.
Always verify units on pump curves and tank specifications.
Question 2 A rectangular sedimentation basin is 40 feet long, 20 feet wide, and 10 feet deep.
What is the volume in gallons when filled to 8 feet depth?
A. 6,400 gallons B. 47,872 gallons [CORRECT] C. 59,840 gallons D. 478,720 gallons
Correct Answer: B
Rationale:
• Concept: Volume calculation with partial depth.
• Step-by-Step Solution:
o Volume in ft³ = Length × Width × Depth = 40 ft × 20 ft × 8 ft = 6,400 ft³
o Volume in gallons = 6,400 ft³ × 7.48 gal/ft³ = 47,872 gallons
• Why B is correct: Correct calculation using actual water depth (8 ft), not total basin
depth.
• Why A is incorrect: This is the volume in cubic feet, not converted to gallons.
• Why C is incorrect: This uses full 10 ft depth: 40 × 20 × 10 × 7.48 = 59,840 gallons.
• Why D is incorrect: Decimal error—likely multiplied by 74.8 instead of 7.48.
Practical Application: Always use actual operating depth, not total basin depth, unless
calculating maximum capacity.
Question 3 A pipeline has a diameter of 12 inches and water flows at 3 feet per second. What is
the flow rate in gallons per minute (gpm)?
A. 105.6 gpm B. 211.5 gpm C. 1,058 gpm [CORRECT] D. 1,270 gpm
Correct Answer: C
Rationale:
, • Concept: Flow rate calculation using Q = A × V.
• Step-by-Step Solution:
o Convert diameter to feet: 12 inches = 1 foot, radius = 0.5 ft
o Area = π × r² = 3.1416 × (0.5)² = 0.7854 ft²
o Or use formula: Area = 0.785 × D² = 0.785 × 1² = 0.785 ft²
o Flow in cfs = 0.785 ft² × 3 ft/sec = 2.355 ft³/sec
o Convert to gpm: 2.355 ft³/sec × 7.48 gal/ft³ × 60 sec/min = 1,057.3 gpm ≈ 1,058
gpm
• Why C is correct: Correct application of Q = A × V with proper unit conversions.
• Why A is incorrect: This omits the 60 seconds/minute conversion (gives cfs × 7.48
only).
• Why B is incorrect: This uses diameter instead of radius squared, or uses 6 inches as
radius.
• Why D is incorrect: This might use 12 inches as 1.5 feet or similar error.
Practical Application: Velocity measurements with flow meters require pipe diameter
knowledge. Always verify pipe ID, not nominal diameter.
Question 4 A water treatment plant treats 3.5 MGD. The alum dosage is 8 mg/L. How many
pounds of alum are needed per day?
A. 29.2 lbs B. 233.5 lbs [CORRECT] C. 2,335 lbs D. 29,190 lbs
Correct Answer: B
Rationale:
• Concept: Chemical dosage calculation using pounds formula.
• Step-by-Step Solution:
o Formula: lbs/day = mg/L × MGD × 8.34
o lbs/day = 8 mg/L × 3.5 MGD × 8.34
o lbs/day = 8 × 3.5 × 8.34 = 28 × 8.34 = 233.52 lbs ≈ 233.5 lbs
• Why B is correct: Proper application of dosage formula.
• Why A is incorrect: This might be 8 × 3.5 = 28 (missing 8.34 factor).
