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NCEES PE Civil Engineering: Water Resources and Environmental Practice Test

NCEES-PE-Civil-WRE test Format | Course Contents | Course Outline | test Syllabus | test Objectives

- Quantity take-off methods
- Cost estimating
- Project schedules
- Activity identification and sequencing
- Economic and sustainability analysis
- present worth
- lifecycle costs
- comparison of alternatives
- Lateral earth pressure
- Soil consolidation and compaction
- Bearing capacity
- Settlement
- Slope stability
- Soil classification and boring log interpretation
- Soil properties
- strength
- permeability
- compressibility
- phase relationships
- Concrete
- nonreinforced
- reinforced
- Piping materials
- Material test methods and specification conformance
- Mass balance
- Hydraulic loading
- Solids loading
- sediment loading
- sludge
- Hydraulic flow measurement
- Energy and/or continuity equation
- Bernoulli
- grade line analyses
- momentum equation
- Pressure conduit
- single pipe
- force mains
- Hazen-Williams
- Darcy-Weisbach
- major and minor losses
- Pump application and analysis
- wet wells
- lift stations
- cavitation
- Pipe network analysis
- series
- parallel
- loop networks
- Open-channel flow
- Hydraulic grade lines and energy dissipation
- plunge pool
- drop structure
- culvert outlet
- Stormwater collection and drainage
- culvert
- stormwater inlets
- gutter flow
- street flow
- storm sewer pipes
- Sub- and supercritical flow
- Storm characteristics
- storm frequency
- rainfall measurement
- distribution
- Runoff analysis
- rational
- SCS/NRCS methods
- Hydrograph development and applications- including synthetic hydrographs
- Rainfall intensity- duration- frequency- and probability of exceedance
- Time of concentration
- Rainfall and stream gauging stations
- Depletions
- evaporation
- detention
- percolation
- diversions
- Stormwater management and treatment
- detention and retention ponds
- infiltration
- swales
- constructed wetlands
- Aquifers
- Groundwater flow
- Well and drawdown analysis
- Stream degradation and oxygen dynamics
- Total maximum daily load (TMDL)
- nutrient contamination
- DO
- load allocation
- Biological and chemical contaminants

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NCEES PE Civil: Water Resources and Environmental
NCEES - PE Civil Engineering - Water Resources and
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Question 386:
An environmental engineer is assessing the impact of a sewage treatment plant
on a nearby stream. If the plant discharges effluent with a biochemical oxygen
demand (BOD) of 200 mg/L and the stream's flow is 3 m�/s, what is the total BOD
load entering the stream from the plant in kilograms per day?
A. 51,840 kg/day
B. 21,880 kg/day
C. 45,320 kg/day
D. 65,000 kg/day
Answer: A
Explanation: The BOD load can be calculated as:
BOD Load = Concentration � Flow Rate � Time
Convert mg/L to kg/m�:
Concentration = 200 mg/L = 0.2 kg/m 3
Thus,
3
3
BOD Load = 0.2 kg/m � 3 m /s � 86, 400 s = 51, 840 kg/day
Question 387:
A groundwater engineer is evaluating the effects of a contaminant plume in a
confined aquifer. If the hydraulic conductivity is 20 m/day and the contaminant
concentration decreases from 1,000 �g/L to 100 �g/L over a distance of 50 m,
what is the attenuation factor?
A. 0.1
B. 0.5
C. 0.7
D. 10.0
Answer: D
Explanation: The attenuation factor is calculated as:
C 1 1000 �g/L
Attenuation Factor = = = 10
C 2 100 �g/L
Question 388:
A civil engineer is assessing the effect of urban runoff on a stream's DO levels. If
the stream's DO was 8 mg/L before the runoff event and dropped to 5 mg/L
after, what is the percentage change in DO?
A. 20%
B. 25%
C. 37%
D. 35%
Answer: C
Explanation: The percentage change in DO is calculated as:
Initial DO - Final DO
Percentage Change = � 100
Initial DO
Thus,
8 - 5
Percentage Change = � 100 = 37.5%
8
Question 389:
An environmental scientist is evaluating the impact of nutrients on a lake's water
quality. If the lake has a volume of 1,000,000 m� and the total phosphorus
concentration is 0.2 mg/L, what is the total phosphorus load in kilograms?
A. 0.2 kg
B. 2 kg
C. 20 kg
D. 200 kg
Answer: D
Explanation: The total phosphorus load can be calculated as:
Load = Concentration � Volume
Convert concentration to kg/m�:
Concentration = 0.2 mg/L = 0.0002 kg/m 3
Thus,
3
3
Load = 0.0002 kg/m � 1, 000, 000 m = 200 kg
Question 390:
A groundwater model indicates that a well is experiencing a drawdown of 5 m
after 12 hours of continuous pumping. If the well has a radius of 0.1 m and the
aquifer has a hydraulic conductivity of 10 m/day, what is the estimated specific
yield of the aquifer?
A. 0.01
B. 0.05
C. 0.1
D. 0.15
Answer: B
Explanation: The specific yield can be calculated using the relationship:
Drawdown 1
Specific Yield = �
Time Hydraulic Conductivity
Thus,
5 m 1
Specific Yield = � = 0.05
12 � 3600 s 10 m/day
Question 391:
A hydrogeologist is evaluating a confined aquifer that has a hydraulic
conductivity of 25 m/day and a thickness of 30 m. If the aquifer is being
recharged at a rate of 0.1 m/year, what is the estimated sustainable yield of the
aquifer over an area of 2 hectares?
A. 2000 m�/yr
B. 1500 m�/yr
C. 1600 m�/yr
D. 1700 m�/yr
Answer: A
Explanation: The sustainable yield can be estimated using:
Sustainable Yield = Recharge Rate � Area
Convert the recharge rate to meters:
Recharge Rate = 0.1 m/yr
Convert area to square meters:
Area = 2 hectares = 20, 000 m 2
Thus,
2
Sustainable Yield = 0.1 m/yr � 20, 000 m = 2, 000 m�/yr
Question 392:
An engineer is analyzing groundwater flow through a heterogeneous aquifer. The
hydraulic gradient in one section of the aquifer is measured at 0.03, and the
hydraulic conductivity is 12 m/day. What is the groundwater flow velocity in that
section?
A. 0.36 m/day
B. 0.48 m/day
C. 0.56 m/day
D. 0.72 m/day
Answer: A
Explanation: Groundwater flow velocity can be calculated using Darcy's law:
v = K � i
Where K is hydraulic conductivity and i is hydraulic gradient.
Thus,
v = 12 m/day � 0.03 = 0.36 m/day
Question 393:
A well in an unconfined aquifer is pumped at a rate of 100 L/s. After 48 hours of
continuous pumping, the water level in the well has dropped from 15 m to 10 m.
What is the total drawdown experienced by the well?
A. 2 m
B. 3 m
C. 4 m
D. 5 m
Answer: D
Explanation: The drawdown is calculated as:
Drawdown = Initial Water Level - Final Water Level
Thus,
Drawdown = 15 m - 10 m = 5 m
Question 394:
A civil engineer is studying the impact of a wastewater discharge on a river�s
dissolved oxygen (DO) levels. If the river has a flow rate of 4 m�/s and the DO
concentration downstream of the discharge is 5 mg/L, while the upstream
concentration is 8 mg/L, what is the total mass of oxygen depleted over a
24-hour period?
A. 1288 kg
B. 1576 kg
C. 1036 kg
D. 1296 kg
Answer: C
Explanation: The mass of oxygen lost can be calculated as:
Mass Loss = (Upstream DO - Downstream DO) � Flow Rate � Time
Where:
3
Mass Loss = (8 mg/L - 5 mg/L) � 4 m /s � 86, 400 s
Convert mg/L to kg/m�:
Mass Loss = 3 mg/L � 4 � 86, 400 = 1036.8 kg
Question 395:
An environmental scientist is calculating the Total Maximum Daily Load (TMDL)
for nitrogen in a river. The current nitrogen load is 2,200 kg/year, and the TMDL is
set at 1,500 kg/year. What is the percentage reduction needed to meet the
TMDL?
A. 25%
B. 32%
C. 40%
D. 50%
Answer: B
Explanation: The percentage reduction can be calculated as:
Current Load - TMDL
Reduction = � 100
Current Load
Thus,
2200 - 1500
Reduction = � 100 � 31.82%
2200
Question 396:
A lake has a total phosphorus concentration of 0.15 mg/L. If the lake has a
volume of 500,000 m�, what is the total phosphorus load in kilograms?
A. 10.75 kg
B. 11.25 kg
C. 15.00 kg
D. 75.0 kg
Answer: D
Explanation: The total phosphorus load can be calculated as:
Load = Concentration � Volume
Convert concentration to kg/m�:
3
Load = 0.15 mg/L � 500, 000 m = 75 kg
Question 397:
In a groundwater contamination study, a monitoring well shows a concentration
of benzene at 5 �g/L. If the well extracts water at a rate of 10 L/min, what is the
total mass of benzene extracted in a 30-minute sampling period?
A. 0.15 mg
B. 0.25 mg
C. 0.50 mg
D. 1.50 mg
Answer: D
Explanation: The total mass can be calculated as:
Mass = Concentration � Flow Rate � Time
Convert flow rate to L/h:
Mass = 5 �g/L � 10 L/min � 30 min = 1, 500 �g = 1.5 mg
Question 398:
A civil engineer is evaluating a stream's health by assessing its biological oxygen
demand (BOD). If the natural BOD of the stream is 4 mg/L and the BOD after a
pollutant influx is measured at 12 mg/L, what is the increase in BOD due to the
pollutants?
A. 4 mg/L
B. 6 mg/L
C. 8 mg/L
D. 10 mg/L
Answer: C
Explanation: The increase in BOD is calculated as:
Increase in BOD = Post-Pollution BOD - Natural BOD
Thus,
Increase in BOD = 12 mg/L - 4 mg/L = 8 mg/L
Question 399:
A groundwater model reveals that a well has a drawdown of 3 m after 24 hours of
pumping at a rate of 80 L/s. If the well has a radius of 0.15 m, what is the specific
capacity of the well in L/s/m?
A. 15.33 L/s/m
B. 26.67 L/s/m
C. 10.00 L/s/m
D. 12.00 L/s/m
Answer: B
Explanation: Specific capacity can be calculated using:
Discharge Rate
Specific Capacity =
Drawdown
Thus,
80 L/s
Specific Capacity = � 26.67 L/s/m
3 m
Question 400:
An environmental engineer is assessing the impact of nutrient runoff on a pond.
If the pond has a surface area of 1 hectare and receives 15 kg of phosphorus
from runoff annually, what is the concentration of phosphorus in mg/L, assuming
an average depth of 2 m?
A. 0.15 mg/L
B. 0.75 mg/L
C. 91.00 mg/L
D. 750 mg/L
Answer: D
Explanation: Convert area to square meters:
Area = 1 hectare = 10, 000 m 2
The volume of the pond is:
2 3
Volume = Area � Depth = 10, 000 m � 2 m = 20, 000 m
Convert kg to mg:
15 kg � 1, 000, 000 mg/kg
Concentration = = 750 mg/L
20, 000 m 3
Question 401:
A stream has a flow rate of 1.5 m�/s and a dissolved oxygen (DO) concentration
of 9 mg/L upstream. If the DO concentration drops to 5 mg/L downstream after
discharge from a wastewater treatment plant, what is the total mass of oxygen
lost in kilograms over 24 hours?
A. 518.4 kg
B. 288 kg
C. 864 kg
D. 1,728 kg
Answer: A
Explanation: The mass of oxygen lost can be calculated as:
Mass Loss = (Upstream DO - Downstream DO) � Flow Rate � Time
Thus,
3
Mass Loss = (9 mg/L - 5 mg/L) � 1.5 m /s � 86, 400 s
Convert mg/L to kg/m�:
3
Mass Loss = 4 mg/L � 1.5 m /s � 86, 400s = 518, 400 mg = 518.4kg
Question 402:
In a water quality assessment, a river's total nitrogen concentration is measured
at 12 mg/L. If the river has a flow rate of 2.5 m�/s, what is the total nitrogen load
in kilograms per day?
A. 1250 kg/day
B. 1300 kg/day
C. 1036 kg/day
D. 1600 kg/day
Answer: C
Explanation: The nitrogen load can be calculated as:
Load = Concentration � Flow Rate � Time
Thus,
3
Load = 12 mg/L � 2.5 m /s � 86, 400 s = 1, 036, 800 mg = 1, 036.8 kg /
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