COLORADO WW OPERATOR D EXAM ACTUAL EXAM QUESTIONS WITH
COMPLETE SOLUTION GUIDE (A+ GRADED 100% VERIFIED) LATEST VERSION
2025/2026!!
Question 1
What is your first course of action if you observe a slow, constant drip of water coming
from a pump shaft packing seal?
A) Immediately tighten the packing gland to stop the drip.
B) Shut down the pump and replace the packing material.
C) Switch the pump to a mechanical seal system.
D) Make no adjustment; this is a normal condition.
E) Increase the speed of the pump to utilize the leaking water.
Correct Answer: D) Make no adjustment; this is a normal condition.
Rationale: Packing seals are designed to leak a small amount of water (typically 40 to 60
drops per minute). This water is essential for cooling the shaft and lubricating the packing
material to prevent friction burns and equipment failure. Stopping the drip completely by
tightening the gland would cause the packing to overheat and score the pump shaft.
Question 2
In a hydraulic system, which of the following changes would cause the greatest increase in
pipe friction loss?
A) Decreasing the temperature of the wastewater.
B) Increasing the diameter of the pipe.
C) Increasing the velocity of the wastewater.
D) Decreasing the viscosity of the wastewater.
E) Replacing the pipe with a smoother material.
Correct Answer: C) Increasing the velocity of the wastewater.
Rationale: Friction loss in a pipe increases significantly as the velocity of the flow increases.
According to the Hazen-Williams or Darcy-Weisbach equations, friction loss is
proportional to the square of the velocity. Therefore, even a small increase in speed results
in a large increase in friction/head loss.
Question 3
A pressure gauge at the bottom of a water holding tank reads 15 psi. If the tank is 15 ft in
, 2
diameter and 40 ft high, approximately how many feet of water are currently in the tank?
A) 6.5 ft
B) 15.0 ft
C) 34.6 ft
D) 40.0 ft
E) 92.4 ft
Correct Answer: C) 34.6 ft
Rationale: To convert pressure (psi) to head (feet of water), use the conversion factor: 1 psi
= 2.31 ft. Calculation: 15 psi × 2.31 ft/psi = 34.65 ft. The diameter and total height of the
tank are extraneous information for this specific calculation.
Question 4
Calculate the detention time in a stabilization pond given the following data: Influent flow
rate = 0.785 MGD, Pond depth = 4.5 feet, Pond area = 17 acres.
A) 15 days
B) 24 days
C) 32 days
D) 45 days
E) 60 days
Correct Answer: C) 32 days
Rationale: First, calculate the volume of the pond in gallons.
Volume = Area (acres) × 43,560 ft²/acre × Depth (ft) × 7.48 gal/ft³.
Volume = 17 × 43,560 × 4.5 × 7.48 = 24,924,992 gallons.
Next, divide volume by flow rate (0.785 MGD = 785,000 gal/day).
Detention Time = 24,924,992 gal / 785,000 gal/day = 31.75 days (rounds to 32).
Question 5
How many pounds per day of suspended solids (TSS) are removed by a primary clarifier
given the following: Flow rate = 2.7 MGD, Influent TSS = 230 mg/L, Primary Effluent TSS
= 110 mg/L?
A) 1,200 lbs/day
, 3
B) 2,702 lbs/day
C) 4,050 lbs/day
D) 5,175 lbs/day
E) 6,300 lbs/day
Correct Answer: B) 2,702 lbs/day
Rationale: First, determine the concentration of solids removed: 230 mg/L (Influent) - 110
mg/L (Effluent) = 120 mg/L removed.
Next, use the "Pounds Formula": lbs/day = Concentration (mg/L) × Flow (MGD) × 8.34
lbs/gal.
Calculation: 120 mg/L × 2.7 MGD × 8.34 = 2,702.16 lbs/day.
Question 6
A 42-inch diameter pipe is flowing at a velocity of 6.5 ft/sec. What is the approximate flow
rate in cubic feet per second (cfs)?
A) 45.2 cfs
B) 55.8 cfs
C) 62.5 cfs
D) 78.4 cfs
E) 85.1 cfs
Correct Answer: C) 62.5 cfs
Rationale: The formula is Q = A × V. First, calculate the Area (A) of the pipe in square feet.
Diameter = 42 inches = 3.5 ft.
Area = 0.785 × (Diameter)² = 0.785 × (3.5)² = 0.785 × 12.25 = 9.616 ft².
Flow (Q) = 9.616 ft² × 6.5 ft/sec = 62.5 cfs.
Question 7
Regardless of the shape of the container, 1 acre-foot of media in a trickling filter is
mathematically equal to:
A) 7.48 ft³
B) 325,851 ft³
C) 43,560 ft³
COMPLETE SOLUTION GUIDE (A+ GRADED 100% VERIFIED) LATEST VERSION
2025/2026!!
Question 1
What is your first course of action if you observe a slow, constant drip of water coming
from a pump shaft packing seal?
A) Immediately tighten the packing gland to stop the drip.
B) Shut down the pump and replace the packing material.
C) Switch the pump to a mechanical seal system.
D) Make no adjustment; this is a normal condition.
E) Increase the speed of the pump to utilize the leaking water.
Correct Answer: D) Make no adjustment; this is a normal condition.
Rationale: Packing seals are designed to leak a small amount of water (typically 40 to 60
drops per minute). This water is essential for cooling the shaft and lubricating the packing
material to prevent friction burns and equipment failure. Stopping the drip completely by
tightening the gland would cause the packing to overheat and score the pump shaft.
Question 2
In a hydraulic system, which of the following changes would cause the greatest increase in
pipe friction loss?
A) Decreasing the temperature of the wastewater.
B) Increasing the diameter of the pipe.
C) Increasing the velocity of the wastewater.
D) Decreasing the viscosity of the wastewater.
E) Replacing the pipe with a smoother material.
Correct Answer: C) Increasing the velocity of the wastewater.
Rationale: Friction loss in a pipe increases significantly as the velocity of the flow increases.
According to the Hazen-Williams or Darcy-Weisbach equations, friction loss is
proportional to the square of the velocity. Therefore, even a small increase in speed results
in a large increase in friction/head loss.
Question 3
A pressure gauge at the bottom of a water holding tank reads 15 psi. If the tank is 15 ft in
, 2
diameter and 40 ft high, approximately how many feet of water are currently in the tank?
A) 6.5 ft
B) 15.0 ft
C) 34.6 ft
D) 40.0 ft
E) 92.4 ft
Correct Answer: C) 34.6 ft
Rationale: To convert pressure (psi) to head (feet of water), use the conversion factor: 1 psi
= 2.31 ft. Calculation: 15 psi × 2.31 ft/psi = 34.65 ft. The diameter and total height of the
tank are extraneous information for this specific calculation.
Question 4
Calculate the detention time in a stabilization pond given the following data: Influent flow
rate = 0.785 MGD, Pond depth = 4.5 feet, Pond area = 17 acres.
A) 15 days
B) 24 days
C) 32 days
D) 45 days
E) 60 days
Correct Answer: C) 32 days
Rationale: First, calculate the volume of the pond in gallons.
Volume = Area (acres) × 43,560 ft²/acre × Depth (ft) × 7.48 gal/ft³.
Volume = 17 × 43,560 × 4.5 × 7.48 = 24,924,992 gallons.
Next, divide volume by flow rate (0.785 MGD = 785,000 gal/day).
Detention Time = 24,924,992 gal / 785,000 gal/day = 31.75 days (rounds to 32).
Question 5
How many pounds per day of suspended solids (TSS) are removed by a primary clarifier
given the following: Flow rate = 2.7 MGD, Influent TSS = 230 mg/L, Primary Effluent TSS
= 110 mg/L?
A) 1,200 lbs/day
, 3
B) 2,702 lbs/day
C) 4,050 lbs/day
D) 5,175 lbs/day
E) 6,300 lbs/day
Correct Answer: B) 2,702 lbs/day
Rationale: First, determine the concentration of solids removed: 230 mg/L (Influent) - 110
mg/L (Effluent) = 120 mg/L removed.
Next, use the "Pounds Formula": lbs/day = Concentration (mg/L) × Flow (MGD) × 8.34
lbs/gal.
Calculation: 120 mg/L × 2.7 MGD × 8.34 = 2,702.16 lbs/day.
Question 6
A 42-inch diameter pipe is flowing at a velocity of 6.5 ft/sec. What is the approximate flow
rate in cubic feet per second (cfs)?
A) 45.2 cfs
B) 55.8 cfs
C) 62.5 cfs
D) 78.4 cfs
E) 85.1 cfs
Correct Answer: C) 62.5 cfs
Rationale: The formula is Q = A × V. First, calculate the Area (A) of the pipe in square feet.
Diameter = 42 inches = 3.5 ft.
Area = 0.785 × (Diameter)² = 0.785 × (3.5)² = 0.785 × 12.25 = 9.616 ft².
Flow (Q) = 9.616 ft² × 6.5 ft/sec = 62.5 cfs.
Question 7
Regardless of the shape of the container, 1 acre-foot of media in a trickling filter is
mathematically equal to:
A) 7.48 ft³
B) 325,851 ft³
C) 43,560 ft³
