NCEES PE Civil: Water Resources and Environmental MCQs NCEES PE Civil: Water Resources and Environmental TestPrep NCEES PE Civil: Water Resources and Environmental Study Guide NCEES PE Civil: Water Resources and Environmental Practice Test
NCEES PE Civil: Water Resources and Environmental Exam Questions
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NCEES PE Civil: Water Resources and Environmental
NCEES - PE Civil Engineering - Water Resources and Environmental 2024
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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 m³/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
BOD Load = Concentration × Flow Rate × Time
Convert mg/L to kg/m³:
Concentration = 200 mg/L = 0.2 kg/m3
Thus,
BOD Load = 0.2 kg/m3 × 3 m3/s × 86, 400 s = 51, 840 kg/day
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 µg/L to 100 µg/L over a distance of 50 m, what is the attenuation factor?
0.1
0.5
0.7
10.0
C1
Attenuation Factor =
C2
1000 μg/L
=
100 μg/L
= 10
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?
20%
25%
37%
35%
Initial DO − Final DO PercentageChange=
Initial DO
× 100
Thus,
Percentage Change =
8 − 5
8
× 100 = 37.5%
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 m³ and the total phosphorus concentration is 0.2 mg/L, what is the total phosphorus load in kilograms?
0.2 kg
2 kg
20 kg
200 kg
Load = Concentration × Volume
Convert concentration to kg/m³:
Concentration = 0.2 mg/L = 0.0002 kg/m3
Thus,
Load = 0.0002 kg/m3 × 1, 000, 000 m3 = 200 kg
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?
0.01
0.05
0.1
0.15
Specific Yield =
Drawdown
×
Time
1
Hydraulic Conductivity
Thus,
Specific Yield =
5 m 12 × 3600 s
1
×
10 m/day
= 0.05
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 | m³/yr |
B. | 1500 | m³/yr |
C. | 1600 | m³/yr |
D. 1700 m³/yr
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 m2
Thus,
Sustainable Yield = 0.1 m/yr × 20, 000 m2 = 2, 000 m³/yr
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?
0.36 m/day
0.48 m/day
0.56 m/day
0.72 m/day
v = K ⋅ i
Where K is hydraulic conductivity and i is hydraulic gradient. Thus,
v = 12 m/day × 0.03 = 0.36 m/day
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?
2 m
3 m
4 m
5 m
Drawdown = Initial Water Level − Final Water Level
Thus,
Drawdown = 15 m − 10 m = 5 m
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 m³/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?
1288 kg
1576 kg
1036 kg
1296 kg
Mass Loss = (Upstream DO − Downstream DO) × Flow Rate × Tim
Where:
Mass Loss = (8 mg/L − 5 mg/L) × 4 m3/s × 86, 400 s
Convert mg/L to kg/m³:
Mass Loss = 3 mg/L × 4 × 86, 400 = 1036.8 kg
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?
25%
32%
40%
50%
Current Load − TMDL
Reduction =
Current Load
× 100
Thus,
Reduction =
2200 − 1500
2200
× 100 ≈ 31.82%
A lake has a total phosphorus concentration of 0.15 mg/L. If the lake has a
volume of 500,000 m³, what is the total phosphorus load in kilograms?
A. | 10.75 | kg |
B. | 11.25 | kg |
C. | 15.00 | kg |
D. 75.0 kg
Load = Concentration × Volume
Convert concentration to kg/m³:
Load = 0.15 mg/L × 500, 000 m3 = 75 kg
In a groundwater contamination study, a monitoring well shows a concentration of benzene at 5 µg/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?
0.15 mg
0.25 mg
0.50 mg
1.50 mg
Mass = Concentration × Flow Rate × Time
Convert flow rate to L/h:
Mass = 5 μg/L × 10 L/min × 30 min = 1, 500 μg = 1.5 mg
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?
4 mg/L
6 mg/L
8 mg/L
10 mg/L
Increase in BOD = Post-Pollution BOD − Natural BOD
Thus,
Increase in BOD = 12 mg/L − 4 mg/L = 8 mg/L
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?
15.33 L/s/m
26.67 L/s/m
10.00 L/s/m
12.00 L/s/m
Discharge Rate
Specific Capacity =
Drawdown
Thus,
Specific Capacity =
80 L/s
3 m
≈ 26.67 L/s/m
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?
0.15 mg/L
0.75 mg/L
91.00 mg/L
750 mg/L
Area = 1 hectare = 10, 000 m2
The volume of the pond is:
Volume = Area × Depth = 10, 000 m2 × 2 m = 20, 000 m3
Convert kg to mg:
Concentration =
15 kg × 1, 000, 000 mg/kg
20, 000 m3
= 750 mg/L
A stream has a flow rate of 1.5 m³/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?
518.4 kg
288 kg
864 kg
1,728 kg
Mass Loss = (Upstream DO − Downstream DO) × Flow Rate × Time
Thus,
Mass Loss = (9 mg/L − 5 mg/L) × 1.5 m3/s × 86, 400 s
Convert mg/L to kg/m³:
Mass Loss = 4 mg/L × 1.5 m3/s × 86, 400s = 518, 400 mg = 518.4kg
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 m³/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
Load = Concentration × Flow Rate × Time
Thus,
Load = 12 mg/L × 2.5 m3/s × 86, 400 s = 1, 036, 800 mg = 1, 036.8 kg
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