TCEQ CLASS C WATER LICENSE EXAM
PREP 150 MASTER Q AND AS WITH
EMBEDDED BOLD ANSWERS AND
ITALICIZED RATIONALES
TCEQ Class C Water Operator Practice Exam (1-150)
1. A rectangular sedimentation basin is 50 feet long, 20
feet wide, and 12 feet deep. If the flow into the basin is
500 gallons per minute (gpm), what is the detention
time in minutes?
• 179.5 minutes Correct Answer: Volume = 50 ft ×
20 ft × 12 ft = 12,000 cu ft. Gallons = 12,000 cu ft × 7.48
gal/cu ft = 89,760 gallons. Detention time = 89,760
gallons / 500 gpm = 179.52 minutes.
• 36 minutes Incorrect. The volume must be calculated
first, then converted to gallons, and finally divided by the
flow rate.
• 45 minutes Incorrect. Ensure you use the correct
conversion factor for cubic feet to gallons.
• 220 minutes Incorrect. The calculated detention time is
less than 200 minutes.
2. A water treatment plant is calculating its
Disinfection Contact Time (CT) value. The free
chlorine residual is 1.5 mg/L, the baffling factor is 0.5,
and the peak hourly flow detention time is 60 minutes.
What is the actual CT value?
• 30 mg-min/L Incorrect. Did you account for the baffling
factor when calculating the effective contact time (T10)?
, • 45 mg-min/L Correct Answer: T10 (Effective
Contact Time) = Detention Time × Baffling Factor = 60
minutes × 0.5 = 30 minutes. CT = Residual × T10 = 1.5
mg/L × 30 minutes = 45 mg-min/L.
• 60 mg-min/L Incorrect. The actual detention time must
be adjusted by the baffling factor before multiplying by
the chlorine residual.
• 90 mg-min/L Incorrect. This would be the CT value if
there was no baffling effect and the factor was 1.0.
3. What is the water horsepower (WHP) required to
pump 1,200 gallons per minute (gpm) against a total
dynamic head (TDH) of 150 feet?
• 45.5 WHP Correct Answer: Water Horsepower
(WHP) = (Flow in gpm × TDH in ft) / 3,960. WHP =
(1,200 × 150) / 3,960 = 45.45 WHP.
• 35.2 WHP Incorrect. The formula for WHP is (Flow ×
Head) / 3,960. Double-check your calculation.
• 55.8 WHP Incorrect. The result is lower than 55. WHP
represents the useful power actually delivered to the
water.
• 75.0 WHP Incorrect. This calculation overestimates the
power needed for this specific flow and head.
4. A well has a static water level of 45 feet below the
surface. When the well pump is operating at 250 gpm,
the pumping water level drops to 70 feet. What is the
well drawdown?
• 15 feet Incorrect. Drawdown is the distance the water
level drops from static, not a static measurement.
, • 35 feet Incorrect. Drawdown is simply the difference
between the pumping water level and the static water
level.
• 115 feet Incorrect. This would be if you added the
pumping level and static level, which is incorrect for
determining drawdown.
• 25 feet Correct Answer: Drawdown is the distance
the water level drops from the static level. Drawdown =
Pumping Water Level (70 ft) - Static Water Level (45 ft) =
25 feet.
5. A well is pumping at 400 gpm. The static water level
is 50 feet, and the pumping water level is 90 feet. What
is the specific capacity of the well in gpm per foot of
drawdown?
• 4.5 gpm/ft Incorrect. Recalculate the drawdown first,
then divide the pumping rate by that drawdown.
• 8.0 gpm/ft Incorrect. This calculation would mean the
drawdown was 50 feet, which is incorrect.
• 12.5 gpm/ft Incorrect. Check your drawdown calculation
and try again.
• 10.0 gpm/ft Correct Answer: Drawdown =
Pumping Level (90 ft) - Static Level (50 ft) = 40 ft.
Specific Capacity = Flow (gpm) / Drawdown (ft) = 400
gpm / 40 ft = 10.0 gpm/ft.
6. You need to dose a water pipeline with 5 mg/L of
chlorine. The flow through the pipeline is 2.0 million
gallons per day (MGD). How many pounds of chlorine
are required per day?
• 83.4 lbs/day Correct Answer: Pounds per day =
Flow (MGD) × Dose (mg/L) × 8.34 lbs/gal. Lbs/day = 2.0
MGD × 5 mg/L × 8.34 = 83.4 lbs/day.
, • 41.7 lbs/day Incorrect. Did you multiply by the conversion
factor of 8.34 lbs/gal?
• 104.5 lbs/day Incorrect. The math results in less than 100
lbs/day.
• 166.8 lbs/day Incorrect. This would be the requirement if
the flow was 4 MGD or the dose was 10 mg/L.
7. A hypochlorinator is pumping a 12.5% sodium
hypochlorite solution. To achieve a dosage of 3.0 mg/L
in a flow of 1.0 MGD, how many pounds of the 12.5%
solution are required per day?
• 12.5 lbs/day Incorrect. Remember to account for the fact
that the solution is only 12.5% pure chlorine.
• 200.2 lbs/day Correct Answer: Pure chlorine
needed = Flow (MGD) × Dose (mg/L) × 8.34 = 1.0 × 3.0 ×
8.34 = 25.02 lbs/day. Solution needed = Pure chlorine
needed / Purity (0.125) = 25..125 = 200.16 lbs/day.
• 25.0 lbs/day Incorrect. This is the amount of pure
chlorine needed, but the product used is only 12.5%
chlorine. You will need a larger amount of the
commercial solution.
• 312.5 lbs/day Incorrect. The calculation overestimates the
required solution weight.
8. A rapid sand filter is 20 feet wide and 30 feet long. If
the plant treats 2,500 gpm, what is the filtration rate
in gallons per minute per square foot (gpm/sq ft) of
filter area?
• 2.5 gpm/sq ft Incorrect. Check the calculation of your
total filter area.
PREP 150 MASTER Q AND AS WITH
EMBEDDED BOLD ANSWERS AND
ITALICIZED RATIONALES
TCEQ Class C Water Operator Practice Exam (1-150)
1. A rectangular sedimentation basin is 50 feet long, 20
feet wide, and 12 feet deep. If the flow into the basin is
500 gallons per minute (gpm), what is the detention
time in minutes?
• 179.5 minutes Correct Answer: Volume = 50 ft ×
20 ft × 12 ft = 12,000 cu ft. Gallons = 12,000 cu ft × 7.48
gal/cu ft = 89,760 gallons. Detention time = 89,760
gallons / 500 gpm = 179.52 minutes.
• 36 minutes Incorrect. The volume must be calculated
first, then converted to gallons, and finally divided by the
flow rate.
• 45 minutes Incorrect. Ensure you use the correct
conversion factor for cubic feet to gallons.
• 220 minutes Incorrect. The calculated detention time is
less than 200 minutes.
2. A water treatment plant is calculating its
Disinfection Contact Time (CT) value. The free
chlorine residual is 1.5 mg/L, the baffling factor is 0.5,
and the peak hourly flow detention time is 60 minutes.
What is the actual CT value?
• 30 mg-min/L Incorrect. Did you account for the baffling
factor when calculating the effective contact time (T10)?
, • 45 mg-min/L Correct Answer: T10 (Effective
Contact Time) = Detention Time × Baffling Factor = 60
minutes × 0.5 = 30 minutes. CT = Residual × T10 = 1.5
mg/L × 30 minutes = 45 mg-min/L.
• 60 mg-min/L Incorrect. The actual detention time must
be adjusted by the baffling factor before multiplying by
the chlorine residual.
• 90 mg-min/L Incorrect. This would be the CT value if
there was no baffling effect and the factor was 1.0.
3. What is the water horsepower (WHP) required to
pump 1,200 gallons per minute (gpm) against a total
dynamic head (TDH) of 150 feet?
• 45.5 WHP Correct Answer: Water Horsepower
(WHP) = (Flow in gpm × TDH in ft) / 3,960. WHP =
(1,200 × 150) / 3,960 = 45.45 WHP.
• 35.2 WHP Incorrect. The formula for WHP is (Flow ×
Head) / 3,960. Double-check your calculation.
• 55.8 WHP Incorrect. The result is lower than 55. WHP
represents the useful power actually delivered to the
water.
• 75.0 WHP Incorrect. This calculation overestimates the
power needed for this specific flow and head.
4. A well has a static water level of 45 feet below the
surface. When the well pump is operating at 250 gpm,
the pumping water level drops to 70 feet. What is the
well drawdown?
• 15 feet Incorrect. Drawdown is the distance the water
level drops from static, not a static measurement.
, • 35 feet Incorrect. Drawdown is simply the difference
between the pumping water level and the static water
level.
• 115 feet Incorrect. This would be if you added the
pumping level and static level, which is incorrect for
determining drawdown.
• 25 feet Correct Answer: Drawdown is the distance
the water level drops from the static level. Drawdown =
Pumping Water Level (70 ft) - Static Water Level (45 ft) =
25 feet.
5. A well is pumping at 400 gpm. The static water level
is 50 feet, and the pumping water level is 90 feet. What
is the specific capacity of the well in gpm per foot of
drawdown?
• 4.5 gpm/ft Incorrect. Recalculate the drawdown first,
then divide the pumping rate by that drawdown.
• 8.0 gpm/ft Incorrect. This calculation would mean the
drawdown was 50 feet, which is incorrect.
• 12.5 gpm/ft Incorrect. Check your drawdown calculation
and try again.
• 10.0 gpm/ft Correct Answer: Drawdown =
Pumping Level (90 ft) - Static Level (50 ft) = 40 ft.
Specific Capacity = Flow (gpm) / Drawdown (ft) = 400
gpm / 40 ft = 10.0 gpm/ft.
6. You need to dose a water pipeline with 5 mg/L of
chlorine. The flow through the pipeline is 2.0 million
gallons per day (MGD). How many pounds of chlorine
are required per day?
• 83.4 lbs/day Correct Answer: Pounds per day =
Flow (MGD) × Dose (mg/L) × 8.34 lbs/gal. Lbs/day = 2.0
MGD × 5 mg/L × 8.34 = 83.4 lbs/day.
, • 41.7 lbs/day Incorrect. Did you multiply by the conversion
factor of 8.34 lbs/gal?
• 104.5 lbs/day Incorrect. The math results in less than 100
lbs/day.
• 166.8 lbs/day Incorrect. This would be the requirement if
the flow was 4 MGD or the dose was 10 mg/L.
7. A hypochlorinator is pumping a 12.5% sodium
hypochlorite solution. To achieve a dosage of 3.0 mg/L
in a flow of 1.0 MGD, how many pounds of the 12.5%
solution are required per day?
• 12.5 lbs/day Incorrect. Remember to account for the fact
that the solution is only 12.5% pure chlorine.
• 200.2 lbs/day Correct Answer: Pure chlorine
needed = Flow (MGD) × Dose (mg/L) × 8.34 = 1.0 × 3.0 ×
8.34 = 25.02 lbs/day. Solution needed = Pure chlorine
needed / Purity (0.125) = 25..125 = 200.16 lbs/day.
• 25.0 lbs/day Incorrect. This is the amount of pure
chlorine needed, but the product used is only 12.5%
chlorine. You will need a larger amount of the
commercial solution.
• 312.5 lbs/day Incorrect. The calculation overestimates the
required solution weight.
8. A rapid sand filter is 20 feet wide and 30 feet long. If
the plant treats 2,500 gpm, what is the filtration rate
in gallons per minute per square foot (gpm/sq ft) of
filter area?
• 2.5 gpm/sq ft Incorrect. Check the calculation of your
total filter area.
